{"question_id": "1", "site": "bioacoustics", "title": "How do animals adapt to a partially or fully deaf individual of their group?", "body": "In eusocial animals for which hearing is important to communicate between individuals, is there any species apart humans for which there are some evidence that the group adapts their acoustic communication to a fully or partially deaf individual of the group?\n\nFor instance, by systematically vocalizing louder/closer to a partially deaf individual, or switching the communication modality (e.g. acoustics to touching) for a fully deaf individual, or by showing more support to them when an alarm call is raised, etc.\n\nI have not found any information after a quick search on the web and Google Scholar.\n", "link": "https://bioacoustics.stackexchange.com/questions/1235/how-do-animals-adapt-to-a-partially-or-fully-deaf-individual-of-their-group", "tags": ["hearing", "adaptation"], "votes": 6, "creation_date": "2022-11-14T07:01:24", "comments": ["Thanks @DanStowell, done!", "Did you search for possible literature? If so it would be good to give some indication of that. Also, I think \"eurosocial\" should be \"eusocial\", right?"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "2", "site": "codegolf", "title": "Minimum cost of solving the Eni-Puzzle", "body": "You're tasked with writing an algorithm to efficiently estimate cost of solving an [Eni-Puzzle](http://www.enipuzzles.com/buy/eni-braille-puzzle-with-bold-colors) from a scrambled state as follows:\n\nYou're given m lists of containing n elements each(representing the rows of the puzzle). The elements are numbers between 0 and n-1 inclusive (representing the colors of tiles). There are exactly m occurrences of each integers across all m lists (one for each list). \n\nFor example: \n \n \n m=3, n=4 :\n \n [[3, 0, 3, 1], [[1, 3, 0, 1],\n [1, 0, 2, 2], or [0, 2, 3, 1],\n [3, 0, 1, 2]] [0, 3, 2, 2]]\n \n\nYou can manipulate these lists in two ways:\n\n1: Swapping two elements between circularly adjacent indices in (non circularly) adjacent lists. Cost=1.\n\nEx:\n \n \n m=3, n=4 :\n \n Legal:\n Swap((0,0)(1,1))\n Swap((1,0)(2,3)) (circularly adjacent)\n \n Illegal:\n Swap((0,0)(0,1)) (same list)\n Swap((0,0)(2,1)) (lists are not adjacent)\n Swap((0,0)(1,0)) (indices are not circularly adjacent (they're the same)\n Swap((0,0)(1,2)) (indices are not circularly adjacent)\n \n\n 2. Circularly shifting one of the lists (Cost=number of shifts)\n\n\n\nYour algorithm must efficiently calculate minimum cost required to manipulate the lists such that the resulting lists are all rotations of each other (meaning the puzzle can be fully solved from this state using only rotation moves) i.e.:\n \n \n [[0, 1, 2, 3] [[2, 1, 0, 3]\n [3, 0, 1, 2] and [0, 3, 2, 1]\n [1, 2, 3, 0]] [3, 2, 1, 0]] \n \n\n...are both valid final states.\n\nInstead of lists, you may use any data structure(s) of your choice to represent the puzzle, so long as the cost of simulating a valid move (sliding or rotating) on the puzzle with this representation is O(n*m). The setup cost of initializing this data structure can be disregarded.\n\nA winning solution will compute the cost in the lowest asymptotic runtime in terms of m and n. Execution time will be assessed as a tie breaker. \n", "link": "https://codegolf.stackexchange.com/questions/182647/minimum-cost-of-solving-the-eni-puzzle", "tags": ["puzzle-solver", "fastest-algorithm", "sliding-puzzle"], "votes": 6, "creation_date": "2019-04-03T19:53:28", "comments": ["@Jonah Because The costs I've outlined here are not the actual number of moves required to make equivalent moves on the puzzle. \"Swapping\" moves have a higher cost. To solve the actual puzzle, you first reach the state I described, or a state that is only one sliding move away. Because my specifications don't allow direct sliding moves (swapping equal indices) like the puzzle does for the blank space, I don't need to consider this latter case. This reduces the problem so that only the two move types I described need to be considered and the concept of a moving \"blank space\" can be ignored.", "@JonathanAllan Yes. I've edited this to be more clear.", "why not define the winning state as the actual state where they're all lined up?", "I edited the tags. Thanks!", "Hi and welcome to PPCG :) I think you do not need fastest-code or code-challenge since you only actually specify the winning criterion as the algorithmic speed. Further, I think you should specify if you mean the asymptotic worst case, or average case (best case is not a good idea, for reasons that are hopefully obvious). For future use, we have a sandbox to post challenges to work out issues before you post to the main site. Good luck!"], "comment_count": 5, "category": "Technology", "diamond": 0} {"question_id": "3", "site": "crypto", "title": "Share Conversion between Different Finite Fields", "body": "Let us have any linear secret sharing scheme (LSSS) that works on some field $Z_{p}$, where p is some prime or a power of a prime e.g., Shamir Secret Sharing, Additive secret Sharing. The problem at hand is simple, for any secret shared value in $Z_{p}$, is it possible to convert it (and its shares) to elements on $Z_{q}$, where q could be either greater or smaller than p? Note that **I do not need** conversion from $Z_{2^n}$ to $Z_{2}$.\n\nI am aware of the results on [PRSS](http://link.springer.com/chapter/10.1007%2F978-3-540-30576-7_19), and well as in this [2008](https://eprint.iacr.org/2008/221.pdf) work, however the former does not address field conversion, and the latter, have explanations that are not clear nor it offers perfect security. \n\nAny help is more than welcome!\n", "link": "https://crypto.stackexchange.com/questions/47554/share-conversion-between-different-finite-fields", "tags": ["secret-sharing", "multiparty-computation", "finite-field", "garbled-circuits", "function-evaluation"], "votes": 11, "creation_date": "2017-05-19T06:58:27", "comments": ["@Dragos... Btw.... Are you from Bristol's group? that Dragos?", "I see..Even converting shares from $Z_p \\rightarrow Z_2^{r}$ is a hard problem - at least I have encountered it several times and don't know how to solve it efficiently other than bit decompose and simulate the binary circuit in $Z_p$.", "@Dragos. My inputs are bounded by the conversion target, if I do any transformation to a smaller field, let say $Z_{2}$ I would loose information (or be forced to use bit decomposition which I also want to avoid). This is why I cannot use $Z_{2}$, as I pointed out. Any idea is welcome!", "For me, arithmetic means $Z_p$. You can use the method above by first converting from $Z_p \\rightarrow Z_2$ and then from $Z_2 \\rightarrow Z_q$.", "@Dragos, as I pointed out on the explanation of the problem I 'm not interested on $Z_{2}$ to $Z_{2^r}$ conversions. Thanks anyways!", "There is a way to do share conversion from binary to arithmetic and vice-versa using this paper: thomaschneider.de/papers/DSZ15.pdf"], "comment_count": 6, "category": "Technology", "diamond": 0} {"question_id": "4", "site": "crypto", "title": "RSA key such that pi deciphers to your name per RSA-OAEP", "body": "Can you efficiently construct an RSA public/private key pair with $8k$-bit public modulus such that $C=\\left\\lfloor\\pi\\,2^{8k-2}\\right\\rfloor$ deciphers per RSA-OAEP to your name as a bytestring in ASCII or UTF-8?\n\nThe decryption must be per RSAES-OAEP of [PKCS#1v2.2](http://mpqs.free.fr/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp_EMC_Corporation_Public-Key_Cryptography_Standards_\\(PKCS\\).pdf#page=21), SHA-256 hash, no label, MGF1 with SHA-256, and sucessful. The key pair must conform¹ to [section 3](http://mpqs.free.fr/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp_EMC_Corporation_Public-Key_Cryptography_Standards_\\(PKCS\\).pdf#page=6). It could come as $n$, $e$, $d$, or as a PEM text encoding of the private key per the format of [section A.1.2](http://mpqs.free.fr/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp_EMC_Corporation_Public-Key_Cryptography_Standards_\\(PKCS\\).pdf#page=41) acceptable by an [ASN.1 decoder](https://lapo.it/asn1js/). Kudos if it passes² OpenSSL's [`RSA_check_key`](https://gitlab.com/gitlab-org/build/omnibus-mirror/openssl/blob/706457b7bda7fdbab426b8dce83b318908339da4/crypto/rsa/rsa_chk.c), more kudos if $8k\\ge1024$, the higher the better.\n\n$C$ is the first $k$ bytes of [these](http://mpqs.free.fr/C.txt) (in hex) for $8k$ up to $2^{16}$ bits.\n\n* * *\n\nUpdate: [this](https://crypto.stackexchange.com/a/80322/555) answers in the affirmative, for whatever usual $k$, with $p$ and $q$ equal-size primes, and $N$ hard to factor. The next frontier is a low $e$, say in $[3,2^{256})$, ideally $65537$.\n\nFurther stunts are possible, like having $C$ or $2$ the signature of anything per any usual RSA signature scheme, even deterministic and/or with message recovery. \n\nAs a consequence: verifying a signature of a file against Alice's RSA public key (including, certified and used for other purposes) is proof neither that the file is unchanged since signature, nor that Alice was the signer.\n\nExample where that matters: the signature of a file is on a trusted site, the file is not. Alice manages to get her public key (which she uses normally) in the verifier's keystore (with the key identifier of the legitimate signer, if there is such thing in the signature). The verifier, who just uses crypto as a magic tool, sees that the signature verifies, and (rightly) trusts the secure site to hold only trusted signatures of trusted files. There is a chance that a file alteration by Alice goes uncaught, or/and that she can wrongly convince the verifier that she made the signature at the date notarized by the trusted site.\n\nUpdate 2: Some consequences of the same possibility are discussed in Dennis Jackson, Cas Cremers, Katriel Cohn-Gordon and Ralf Sasse's [_Seems Legit: Automated Analysis of Subtle Attacks on Protocols that Use Signatures_](https://eprint.iacr.org/2019/779), originally in [proceedings of CCS 2019](https://doi.org/10.1145/3319535.3339813).\n\n* * *\n\n¹ That is: public modulus $n$ must be the product of $u\\ge2$ distinct odd primes, odd public exponent $e$ with $3\\le e Is there a non-zero vector $v\\in \\\\{-1,0,1\\\\}^n$ such that $Mv=0$?\n\nThe problem is clearly in NP but my guess is that it is not NP-hard (unlike [this related question](https://cstheory.stackexchange.com/questions/20277/decide-whether-a-matrixs-kernel-contains-any-non-zero-vector-all-of-whose-entri)). I have not, however, found a poly time solution.\n", "link": "https://cstheory.stackexchange.com/questions/12060/partial-circulant-matrices-is-there-a-non-zero-vector-v-in-1-0-1-n-such", "tags": ["cc.complexity-theory", "ds.algorithms", "matrices"], "votes": 20, "creation_date": "2014-11-04T06:46:47", "comments": ["The circulant case was in fact considered at mathoverflow.net/questions/168474/… . However, it seems that there is fatal bug in one of the answers and the other answer is not a complete solution.", "would it be more natural to study the circulant case 1st vs the partial circulant? also (idea) the other problem answers are formulated in terms of subset sum problem, how would this problem be formulated that way (ie as a constrained or special case of subset sum problem)? note there are many subset sum variants studied in the literature that might be close..."], "comment_count": 2, "category": "Science", "diamond": 1} {"question_id": "9", "site": "cstheory", "title": "Are monotone Boolean functions in P well-approximated by monotone polynomial-size circuits?", "body": "**Question 1:** Is it true that for every polynomial $p(n)$ and $\\epsilon >0$ there is a polynomial $q(n)$ such that every monotone Boolean function on $n$ variables that can be expressed by a Boolean circuit of size $p(n)$ can be $\\epsilon$-approximated by a _monotone_ Boolean circuit of size at most $q(n)$. \n\nA function $f$ is $\\epsilon$-approximated by a function $g$ if they agree on at least $(1-\\epsilon)$ fraction of inputs. \n\nWe can ask a slightly more general question. Let $\\mu_p$ denote the Bernoulli measure on $\\\\{0,1\\\\}^n$ where every variable is '1' with probability $p$. For a (non-constant) monotone function $f$ let $p_c(f)$ be the value such that $\\mu_{p_c}(\\\\{x: f(x)=1\\\\})=1/2$.\n\n**Question 2:** Is it true that for every polynomial $p(n)$ and $\\epsilon >0$ there is a polynomial $q(n)$ such that every monotone Boolean function $f$ on $n$ variables, there is a Boolean function $g$ that can be expressed by a monotone Boolean circuit of size $q(n)$ such that for $p=p_c(f)$, $\\mu_p(\\\\{x: f(x) \\ne g(x) \\\\}) \\le \\epsilon$. \n\nOf course, the instinctive thought is that the answers for both these question is **NO**. Is it known? Razborov famously proved that matching cannot be expressed by a monotone Boolean circuit of polynomial size. But matching can be approximated in its critical probability by the property that there are no isolated vertices and this property admits a monotone (low depth) circuit.\n\nI asked the analog questions (Problems 1,2,3) for $AC^0$ and $TC^0$ in [this blog post](https://gilkalai.wordpress.com/2010/02/10/noise-stability-and-threshold-circuits/).\n", "link": "https://cstheory.stackexchange.com/questions/31473/are-monotone-boolean-functions-in-p-well-approximated-by-monotone-polynomial-siz", "tags": ["cc.complexity-theory", "circuit-complexity", "pr.probability"], "votes": 18, "creation_date": "2015-05-12T23:58:06", "comments": ["Dear Gil, now I see the point: I just wrongly interpreted your question! We indeed have that every monotone circuit, which needs only to coincide with b-Clique on an extremely small, but special set of inputs, must be large. You, however, do not specify this subset of \"hard\" inputs, only its ratio $1-\\epsilon$. This is a very interesting question. B.t.w. that result (on clique approximation) is Thm. 9.26 in my book.", "\"Every monotone circuit separating graphs consisting of complete a-partite graphs with a > 32 from graphs consisting of b-cliques for a < b < n/32 must have size exponential in min{a,n/b}^{1/4}.\"This is interesting (reference?) but I am not sure it is relevent - namely I don't see that my notion of $\\epsilon$-approximation is relevant to the result you mention.", "Dear Gil, I interpreted your Question 1 as \"if a monotone boolean function $f$ cannot be $\\epsilon$-approximated by a small monotone circuit, then $f$ has no small non-monotone circuits\" (with \"small\" meaning \"of polynomial size\"). If the interpretation is correct, the function I mentioned only weakly approximates b-Clique function, and still requires large monotone circuits.", "Dear Stasy, I don't see how the results you mention refer to Questions 1 and 2. (As I said, I also expect a 'no' answer..)", "Gill, what about the fact that every monotone circuit separating graphs consisting of complete a-partite graphs with a > 32 from graphs consisting of b-cliques for a < b < n/32 must have size exponential in min{a,n/b}^{1/4}. The approximation density seems to be here enough (is it?) to say that your conjecture would definitely imply P!=NP.", "Thanks Gil, this seems like a good reason to update my prior to \"NO\".", "Dear Andras, Alas, counting arguments are too weak for things of this kind. A YES answer will tell you that an argument that (1) some monotone NP-complete problem cannot be approximated by a polynomial size monotone circuit in the critical probability, automatically implies that (2) it cannot be approximated by a polynomial size circuit in the critical probability. (2) is a stronger statement than $NP \\ne P$ but I dont regard (1) as hopeless. (You may think that the answer is YES but proving it is hopeless, but I find it easier to think that the answer is NO and its not beyond reach.)", "Why is NO the instinctive response? Monotone functions have all sorts of nice properties, so a priori this doesn't seem all that unlikely to me. Are you perhaps thinking of a counting argument?"], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "10", "site": "cstheory", "title": "Weighted Hamming distance", "body": "Basically my question is, what kind of geometry do we get if we use a \"weighted\" Hamming distance. This is not necessarily Theoretical Computer Science but I think similar things come up sometimes, for instance in randomness extraction.\n\nDefine:\n\n$d(x,y)=$ the Hamming distance between binary strings $x$ and $y$ of length $n$, $=$ the cardinality of $\\\\{k: x(k)\\ne y(k)\\\\}$.\n\nFor a set of strings $A$,\n\n$d(x,A)=\\min \\\\{ d(x,y): y\\in A\\\\}$.\n\nThe _$r$ -fold boundary of_ $A\\subseteq \\\\{0,1\\\\}^n$ is\n\n$$\\\\{x\\in\\\\{0,1\\\\}^n: 0 < d(x,A)\\le r\\\\}.$$\n\nBalls centered at $0$ are given by\n\n$$ B(p)=\\\\{x: d(x,0)\\le p\\\\}, $$ where $0$ is the string of $n$ many zeroes.\n\nA _Hamming-sphere_ is a set $H$ with $B(p)\\subseteq H\\subseteq B(p+1)$. (So it's more like a ball than a sphere, but this is the standard terminology...)\n\nNow, Harper in 1966 showed that for each $k$, $n$, $r$, one can find a Hamming-sphere that has minimal $r$-fold boundary among sets of cardinality $k$ in $\\\\\\\\{0,1\\\\\\\\}^n$. So a ball is a set having minimal boundary -- just like in Euclidean space.\n\nThe cardinality of $B(p)$ is ${n\\choose 0}+\\cdots {n\\choose p}$.\n\nThe $r$-fold boundary of $B(p)$ is just the set $B(p+r)\\setminus B(p)$, which then has cardinality ${n\\choose p+1}+\\cdots+{n\\choose p+r}$.\n\nSo far, so good. But now suppose we replace $d$ by a different metric $D$: first let $d_j(x,y)$ be the Hamming distance between the prefixes of $x$ and $y$ of length $j$, and then $$ D(x,y)=\\max_{j\\le n}\\ \\frac{d_j(x,y)}{f(j) }$$ where $0\\le f(j)\\le j$. (For example we could have $f(j)=\\sqrt{j}$, or $f(j)=j/\\log j$.)\n\nThis is supposed to make $D(x,y)$ small if the differences of $x$ and $y$ do not clump together at small values of $j$.\n\n# Questions\n\n> Is the minimum $r$-fold boundary (under $D$) realized by a $D$-ball?\n> \n> Is there a better definition of $D$?\n> \n> Under the metric $D$, what's the minimum size of the $r$-fold boundary of a subset of $\\\\\\\\{0,1\\\\\\\\}^n$ having cardinality $k$? (A reasonable lower bound would be nice.)\n\n(Cross-[posted](https://mathoverflow.net/questions/38221/geometry-in-a-hamming-box) on MathOverflow).\n", "link": "https://cstheory.stackexchange.com/questions/2637/weighted-hamming-distance", "tags": ["co.combinatorics", "randomness"], "votes": 20, "creation_date": "2010-11-01T22:06:48", "comments": [], "comment_count": 0, "category": "Science", "diamond": 1} {"question_id": "11", "site": "cstheory", "title": "To what extent MSO = WS1S, when adding relations?", "body": "[This question has been asked on MathOverflow with no luck a month ago.]\n\nLet me first clarify my definitions. For a word $w \\in \\Sigma^*$, with $\\Sigma =\\\\{a_1, \\ldots, a_n\\\\}$, I define two structures:\n\n$\\mathbb{N}(w) = \\langle \\mathbb{N}, <, Q_{a_1}, \\ldots, Q_{a_n} \\rangle$,\n\nand the more usual _word model_ :\n\n$\\mathbb{N}^r(w) = \\langle \\\\{0, \\ldots, |w|-1\\\\}, <, Q_{a_1}, \\ldots, Q_{a_n} \\rangle$,\n\nwhere $Q_{a_i} = \\\\{p \\mid w_p = a_i\\\\}$.\n\nThen WS1S is the set of second order formulas with models of the form $\\mathbb{N}(w)$, with order, and for which second order quantification is limited to finite subsets of the domain. MSO is the set of monadic second order formulas with models of the form $\\mathbb{N}^r(w)$, with order.\n\nThe usual proof that REG = WS1S proves at the same time that MSO = WS1S. My question is then, for which first or second order relations can we keep this to be true?\n\nFor instance, if we add a unary predicate $E(X)$ which says that a (monadic) second order variable contains an even number of objects, we add no power, as $E(X)$ is expressible as \"there exists $X_1$ and $X_2$ that partition $X$, in such a way that if an element is in $X_i$ the next one in $X$ is in $X_j$, $i \\neq j$, and the first element of $X$ is in $X_1$ and the last is in $X_2$.\"\n\nNow, if we add a predicate $|X| < |Y|$, then WS1S becomes undecidable (see Klaedtke & Ruess, 10.1.1.7.3029), while MSO stays trivially decidable.\n\nThank you.\n\n* * *\n\n**Edit:** As a side question, ... is this question of interest? I mean, I'm no expert in the field, so I'm not sure this question is relevant.\n", "link": "https://cstheory.stackexchange.com/questions/15/to-what-extent-mso-ws1s-when-adding-relations", "tags": ["lo.logic", "automata-theory", "descriptive-complexity"], "votes": 19, "creation_date": "2010-08-16T13:32:53", "comments": ["Your model $\\mathbb{N}(w)$ is very unusual : what is the need of considering a finite word on an infinite underlying structure ? If you forget the letter predicates (which do not change the problem a lot), you are in fact comparing MSO on finite linear orders, and WS1S on the order $\\omega$. This is strange, because it is more intuitive to compare different logic formalisms on the same structure.", "What are your reasons to think this question is uninteresting? From a practical point of view, it is useful to know when you can restrict yourself to finite models, isn't it? For the record, I received no comment on MO.", "Was there no joy on MO because they couldn't figure it out, or because they found your question uninteresting? (I'm guessing the latter)", "also see the greasemonkey hack mentioned here: meta.cstheory.stackexchange.com/questions/3/latex-math-suppo‌​rt", "Ain't we just a click away from it?", "We do need Latex math support..."], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "12", "site": "cstheory", "title": "Descriptive complexity of communication complexity classes", "body": "It is well known that some major complexity classes, like P or NP, admit a full logical characterization (e.g NP = existential 2nd order logic by Fagin's theorem). On the other hand, one can also define complexity classes in communication complexity (where P = problems solvable with poly(logN) communication etc. - see [Complexity classes in communication complexity theory](http://dl.acm.org/citation.cfm?id=1382962) for more).\n\nMy question is - are any descriptive complexity characterizations known for communication complexity classes (or do any such results from standard complexity classes transfer to communication setting easily)?\n", "link": "https://cstheory.stackexchange.com/questions/9041/descriptive-complexity-of-communication-complexity-classes", "tags": ["cc.complexity-theory", "lo.logic", "communication-complexity"], "votes": 18, "creation_date": "2011-11-19T14:15:36", "comments": ["You may want to limit the computational power of the parties, which is not done in the usual communication complexity models."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "13", "site": "cstheory", "title": "Does $EXP\\neq ZPP$ imply sub-exponential simulation of BPP or NP?", "body": "By simulation I mean in the Impaglazzio-Widgerson [IW98] sense, i.e. sub-exponential deterministic simulation which appears correct i.o to every efficient adversary.\n\nI think this is a proof: if $EXP\\neq BPP$ then from [IW98] we get that BPP has such a simulation. Otherwise we have that $EXP=BPP$, which implies $RP=NP$ (because $NP \\subseteq BPP$) and $EXP \\in PH$. Now if $NP=RP=ZPP$ we have that $PH$ collapses to $ZPP$ and as a result $EXP$, but this cannot happen because of the assumption, so $RP\\neq ZPP$ and this by the Kabanets paper \"Easiness assumptions and hardness tests: trading time for zero error\" implies that RP has such a simulation and as a result also NP.\n\nThis sounds like it could be a useful gap theorem. Does anyone know if it appears anywhere?\n", "link": "https://cstheory.stackexchange.com/questions/430/does-exp-neq-zpp-imply-sub-exponential-simulation-of-bpp-or-np", "tags": ["cc.complexity-theory", "reference-request", "complexity-classes", "conditional-results"], "votes": 29, "creation_date": "2010-08-23T13:21:04", "comments": ["ZPP is closed under complement, and contained in NP. So, $\\mathrm{NP\\subseteq ZPP}$ implies NP = ZPP = coNP, hence PH = NP = ZPP.", "Why does NP in ZPP imply PH in ZPP?", "(Just for reference): Identical question on MO: mathoverflow.net/questions/35945. Maybe someone finds the comments there inspiring.", "Very interesting. I haven't seen this argument before, but I'm not by any means an expert in these things..."], "comment_count": 4, "category": "Science", "diamond": 1} {"question_id": "14", "site": "cstheory", "title": "Interesting PCP characterization of classes smaller than P?", "body": "The PCP theorem, $\\mathsf{NP} = \\mathsf{PCP}(\\mathsf{log}\\, n, 1)$, involves probabilistically checkable proofs with polynomial time verifiers, so the smallest class that can be characterized in this way (that is, $\\mathsf{PCP}(0, 0)$) must be $\\mathsf{P}$. There are also PCP characterizations of larger complexity classes (for example, $\\mathsf{NEXP} = \\mathsf{PCP}(\\mathsf{poly}, \\mathsf{poly})$), also using polynomial time verifiers.\n\nCan we achieve an interesting (that is, not immediately following from the definitions) PCP characterization of smaller complexity classes by restricting the time or space used by the verifier? For example, by using a logarithmic space verifier, or an $\\mathsf{NC}$ circuit verifier?\n", "link": "https://cstheory.stackexchange.com/questions/12060/interesting-pcp-characterization-of-classes-smaller-than-p", "tags": ["cc.complexity-theory", "complexity-classes", "pcp"], "votes": 20, "creation_date": "2012-07-18T11:54:42", "comments": ["Your comment is contradictory to your question where you count “PCP(0, 0) = P” as a PCP characterization of P.", "I suppose I meant are there any characterizations that don't follow immediately from the definition, in the same way that PCP(log n, 1) is non-obvious characterization of NP.", "Why not? If you restrict the space of the verifier to log, then PCP(0, 0) obviously becomes equal to L. And this is a PCP characterization of L if you count PCP(0, 0)=P as a PCP characterization of P. I cannot see the point of the question."], "comment_count": 3, "category": "Science", "diamond": 1} {"question_id": "15", "site": "cstheory", "title": "Is Hankelability NP-hard?", "body": "I asked this question on [SO](https://stackoverflow.com/questions/29484864/an-algorithm-to-detect-permutations-of-hankel-matrices) on April 7 and added a bounty which has now expired but no poly time solution has been found yet. \n\nI am trying to write code to detect if a matrix is a permutation of a Hankel matrix. Here is the spec.\n\n**Input:** An n by n matrix M whose entries are 1 or 0.\n\n**Output:** A permutation of the rows and columns of M so that M is a [Hankel matrix](http://en.m.wikipedia.org/wiki/Hankel_matrix) if that is possible. A Hankel matrix has constant skew-diagonals (positive sloping diagonals). \n\nWhen I say a permutation, I mean we can apply one permutation to the order of the rows and a possibly different one to the columns.\n\nA very nice $O(n^2)$ time algorithm [is known](https://stackoverflow.com/questions/20704900/determine-if-some-row-permutation-of-a-matrix-is-toeplitz) for this problem if we only allow permutation of the order of rows.\n\nPeter de Rivaz pointed out this [paper](http://www.mat.ucsb.edu/~g.legrady/academic/courses/15w259/d/re_orderableMatrix.pdf) as a possible route to proving NP-hardness but I haven't managed to get that to work.\n", "link": "https://cstheory.stackexchange.com/questions/31174/is-hankelability-np-hard", "tags": ["cc.complexity-theory"], "votes": 28, "creation_date": "2015-04-16T10:44:55", "comments": ["Cross-posted now to mathoverflow.net/questions/204294/is-hankelability-np-hard", "@IgorShinkar I mean first some permutation on the order of the rows and then a possibly different permutation on the order of the columns, then stop.", "Do you mean apply first some permutation on the rows and then some permutation on the columns? or do you allow more permutations?", "@Kaveh Thanks for the question. The permutation applied to the order of the columns can be different from the one applied to the order of the rows.", "Do you mean that the same permutation is applied to rows and columns? (my got feeling says this is in P, if a matrix is Hankelable with different diagonal values there aren't that many possibilities for the permutation.)", "Just a guess: maybe this problem is GI-complete?"], "comment_count": 6, "category": "Science", "diamond": 1} {"question_id": "16", "site": "cstheory", "title": "Model-checking for three-variable logics and restricted structures", "body": "Denote the $k$-variable fragment of logic $L$ by $L^{(k)}$. The model-checking problem for a logic $L$ with respect to a class of structures $C$, denoted $MC(L,C)$, is the decision problem\n\n> $MC(L,C)$ \n> _Input:_ formula $\\phi$ of $L$, structure $S$ from $C$ \n> _Question:_ does $S$ satisfy $\\phi$?\n\nIs there a logic $L$, with associated class of structures $C$, and a subclass $D$ of $C$, such that \n1\\. $MC(L^{(3)},D)$ is \"easy\", \n2\\. $MC(L,D)$ is \"hard\", and \n3\\. $MC(L^{(3)},C)$ is \"hard\"? \n\nWith \"easy\" vs. \"hard\" I mean decidable vs. undecidable, PTIME vs. NP-hard, LogCFL vs. P-complete, or similar dichotomies.\n\n# Motivation\n\nFor many logics the three-variable fragment is enough to express \"hard\" properties (this is true for MSO and FO over ordered structures). So results about undecidability often carry across from the logic to its three-variable fragment. There is also often a way to reduce the arity of predicates used in formulas efficiently (for instance, SAT can be reduced to 3SAT with a linear increase in formula size). Reducing arity facilitates expressing some formulas with fewer variables.\n\nOn the other hand, there exist logics which are \"easy\" for some classes of structures yet \"hard\" in general. For instance, the model-checking problem for MSO is \"easy\" (linear time) on unordered graphs satisfying the property \"has treewidth less than $k$\" for any fixed $k$, but \"hard\" (NP-complete) on unordered graphs in general.\n\nI am asking whether there are any known examples where the restriction to 3 variables interacts with the property defining the subclass $D$ in an essential way to yield tractability, yet where neither of these restrictions on their own is enough.\n\nApologies for the vagueness of the question: I am trying to leave this general because I can't think of any examples. As soon as I restrict $D$ enough to ensure condition 1 holds, condition 2 seems to break. I suspect there are some unnatural classes which would work, but ideally $D$ should be \"nice\", at least hereditary and closed under isomorphism. As a further non-example, model-checking for FO is PSPACE-complete while model-checking for $FO^{(3)}$ is PTIME-complete; here the restriction to a finite number of variables is already enough to ensure an \"easy\" model-checking problem and condition 3 fails.\n\nDefinitions for completeness: FO is [first-order logic](http://en.wikipedia.org/wiki/First-order_logic), MSO is monadic [second-order logic](http://en.wikipedia.org/wiki/Second-order_logic), where quantification over sets of individuals is allowed, as well as quantification over individuals.\n", "link": "https://cstheory.stackexchange.com/questions/2637/model-checking-for-three-variable-logics-and-restricted-structures", "tags": ["cc.complexity-theory", "co.combinatorics", "lo.logic", "descriptive-complexity"], "votes": 20, "creation_date": "2010-11-03T11:47:19", "comments": ["@MichaëlCadilhac alas, no.", "Have you made any progress on that question, András?"], "comment_count": 2, "category": "Science", "diamond": 1} {"question_id": "17", "site": "cstheory", "title": "In an $m$ by $n$ Boolean matrix, can you find a square block whose four corners are ones in $O(m \\cdot n)$ time?", "body": "**Decision Problem**\n\nInput: An $m$ by $n$ Boolean matrix $M$.\n\nDecision Question: Does there exist a square block within $M$ such that upper-left corner entry == upper-right corner entry == lower-left corner entry == lower-right corner entry == 1? That is, all four corners of the square block are 1's.\n\n**Cubic Time Solution**\n\nWe know that this problem can be solved in $O(m \\cdot n \\cdot min\\\\{m,n\\\\})$ time. The approach involves scanning through the matrix row by row. For each row, for each pair of 1's in that row, we check whether those 1's form an edge of a square block whose four corners are 1's.\n\n**Our Question**\n\nWhat is the time complexity of this problem? Is there a quadratic time solution? In particular, can we solve this decision problem in $O(m \\cdot n)$ time?\n\n**Extra Background**\n\n 1. If you're just looking for a rectangular block whose four corners are ones, this can be solved in $O(m \\cdot n)$ time. Many variations to this problem can be solved in $O(m \\cdot n)$ time as well. I co-authored a paper on this subject.\n\n 2. The paper [\"Finding squares and rectangles in sets of points\"](https://doi.org/10.1007/BF01931281) investigates a related problem from computational geometry where you're given a set of points in a 2D plane and you want to know if there are four points that form an axis-parallel square.\n\n\n\n\nWritten together with Eevvoor.\n", "link": "https://cstheory.stackexchange.com/questions/47588/in-an-m-by-n-boolean-matrix-can-you-find-a-square-block-whose-four-corners", "tags": ["ds.algorithms", "matrices", "computational-geometry", "search-problem", "boolean-matrix"], "votes": 18, "creation_date": "2020-09-18T11:04:38", "comments": ["It feels somehow like an FFT would help, although I haven't been able to puzzle out details. You're looking for something translation-invariant, and working in frequency space often helps with that sort of thing. Tossing out this comment in case someone else can spot something.", "@MichaelWehar: ok, let me know if you make some progress!", "@MarzioDeBiasi It turns out that my reduction doesn't quite work. I'm seeing if I can fix it.", "@MarzioDeBiasi I guess that the right isosceles triangle problem is reducible to triangle finding. We can convert the matrix to a graph where vertices are rows, columns, and diagonals. Also, entries with 1's are associated with edges. But, can we do any better than matrix multiplication time?", "@MarzioDeBiasi Please feel welcome to share these works! I think they could be pretty helpful. :)", "@michaelwehar my idea of right triangle is a square with three 1s on the corners (no matter where the missing one is) but there is a reduction from mine to yours, where the missing one is in the bottom right (just rotate 90 and replicate four times). BTW there are some works both on monochromatic square and triangle free grids (somewhat related if you interpret \"colors\" with small monochromatic (sub)boxes with 1s on the diagonal)", "@MarzioDeBiasi I thought about your right isosceles triangle problem. Just to confirm, this problem is asking for a square block where upper-left corner entry == upper-right corner entry == lower-left corner entry == 1, but we don't care whether lower-right corner entry is 0 or 1? I think that I have some potentially promising ideas for this problem. I'm going to try to formalize the ideas and I hope to share in a few days. Thanks again!", "Even if you restrict the search to right isosceles triangles, it seems impossible to have an $O(mn)$ algorithm", "@DavidEppstein I wasn't quite sure how many 1's there could be without forcing there to be a square block so this is very helpful! User whosyourjay showed a construction related to affine planes on how many 1's there could be without forcing there to be a rectangular block whose corners are 1's. If I recall correctly, the near-tight bound was something like $m \\cdot n^{1/2}$ (it had degree 3/2).", "This is more dead end than answer, but: there exist sets of $mn^{1-o(1)}$ nonzeros with no square, formed by filling diagonals according to a near-linear-size progression-free set (en.wikipedia.org/wiki/Salem%E2%80%93Spencer_set). So an algorithm that answers yes if dense enough to force a square and switches to the naive O(nonzeros times min(m,n)) otherwise can't be strongly subcubic.", "@JoshuaGrochow Yes, when searching for a rectangular block, the problem is related to finding a four cycle in a graph which is solvable in quadratic time. However, when searching for a square block, I don't know what can be done. Maybe there is a way to reduce triangle finding to this problem.", "Seems a little similar to finding a triangle in a graph. If you get a subcubic reduction from that problem that'd be the end of the story for now.", "@daniello Yes! That's a good point. When the input matrix is rectangular, we can reduce it to the case where the input matrix is square. Thank you. :)", "You can assume wlog that $m \\leq n$ (otherwise transpose) and that $n \\leq 2m$ since otherwise you can cover the matrix by $2n/m$ matrices of size $m \\times 2m$ that cover all square blocks. So wlog the input matrix is square $n$ by $n$."], "comment_count": 14, "category": "Science", "diamond": 0} {"question_id": "18", "site": "cstheory", "title": "Problem unsolvable in $2^{o(n)}$ on inputs with $n$ bits, assuming ETH?", "body": "If we assume the Exponential-Time Hypothesis, then there is no $2^{o(n)}$ algorithm for $n$-variable 3-SAT, and many other natural problems, such as 3-COLORING on graphs with $n$ vertices. Notice though that, in general, encoding the input for $n$-variable 3-SAT or $n$-vertex 3-COLORING takes something like $O(n\\log n)$ bits. For example, to describe a sparse graph as input to 3-COLORING, for each edge we would have to list its endpoints. So the lower bound is not exponential in the length of the input. Therefore, my question is the following:\n\nIs there a problem for which no $2^{o(n)}$ algorithm exists for inputs of length $n$ bits (assuming ETH)?\n\nIdeally, the problem would be in NP (no cheating with succinct NEXP-hard problems!) and be reasonably natural, but I won't be picky.\n\nLet me also note that after digging around I found that there are efficient ways to encode planar graphs with $O(n)$ bits. So, if one could find a problem that takes time exponential in the number of vertices even for planar graphs, the question would be settled. However, because planar graphs have treewidth $O(\\sqrt{n})$, most natural problems have sub-exponential algorithms in this case.\n", "link": "https://cstheory.stackexchange.com/questions/16148/problem-unsolvable-in-2on-on-inputs-with-n-bits-assuming-eth", "tags": ["cc.complexity-theory", "sat", "planar-graphs", "succinct"], "votes": 48, "creation_date": "2013-01-19T08:24:59", "comments": ["I believe that it is an open question if we can have a lower bound $2^{\\Omega(n)}$ under ETH, where $n$ is the bit size. We know that proving lower bound $\\Omega(c^n)$ under SETH would disprove SETH.", "How about the following problem, for some large constant $c>0$? Given the encoding of a non-deterministic TM $M$ and binary string $w$, does $M(w)$ have an accepting computation of length at most $|w|^c$? This should work if anything does... (?)", "@AviTal This paper comes to mind arxiv.org/abs/cs/0102005 . The usual keywords for this line of research seem to be \"compact\" and \"succinct\" encodings or representations. (succinct is stronger in that it is optimal up to an additive term instead of a multiplicative factor for compact).", "Would disordered lattice tiling work? Without disorder the problem is NEXP-complete [arXiv:0905.2419] but with disorder (e.g., a vertex-dependent constraint on the tiles) the input would need to be $O(n)$. This is reasonably natural: it would arise in physics as the zero-temperature partition function of a disordered vertex model.", "Can you point to a reference where I can find efficient ways to encode planar graphs with O(n) bits? (I assume that n is the number of vertices). If so, there are NP-Complete problems for planar graphs and maybe the problem can also be presented in O(n) bits using these methods. Thanks.", "It is probably not possible to compress $n$ variable SAT instances to $O(n \\log n)$ as stated in the question (Dell, van Malkebeek, \"Satisfibility Allows No Nontrivial Sparsification Unless The Polynomial-Time Hierarchy Collapses\"). However, 3-SAT has no $2^{o(m)}$ algorithm where $m$ is the number of clauses, as shown by Impagliazzo and Paturi (assuming ETH). Since the hard cases have $O(m)$ variables we have $b=O(m \\log m)$, so $m=\\Omega(b/\\log b)$, so 3-SAT has no $2^{o(b/\\log b)}$ algorithm where $b$ is the input size.", "A problem in NP for which our present algorithmic knowledge is consistent with such a ETH result is edge chromatic number in a dense graph. AFAIK, we do not yet know of any $2^{o(|E|)}$ time algorithms...", "If you are fine with dropping the condition that the problem is in NP, then you can use any undecidable problem and you do not even need the exponential time hypothesis. So probably you should be picky."], "comment_count": 8, "category": "Science", "diamond": 1} {"question_id": "19", "site": "cstheory", "title": "Sylver Coinage Game", "body": "A game in which the players alternately name positive integers that are not sums of previously named integers (with repetitions being allowed). The person who names 1 (so ending the game) is the loser.\n\nThe question is: If player 1 names ‘16’, and both players play optimally thereafter, then who wins?\n\nIt has been known that if player 1 name \"5n,\" then player 1 wins. If player name \"5n+2\", the result is known(maybe palyer 2 wins, but I haven't found the source yet). 16 is the minimum number of \"5n+1\". \n\nI guess that this problem is PSPACE-hard, but I haven't proved it yet. \n", "link": "https://cstheory.stackexchange.com/questions/22112/sylver-coinage-game", "tags": ["co.combinatorics", "combinatorial-game-theory"], "votes": 18, "creation_date": "2014-04-14T04:03:06", "comments": ["@NealYoung The algorithm in Winning Ways only gives whether an opening move is winning or not -- in accordance with Conway's comment, it relies on a theorem saying finitely moves of the form 2^a 3^b win, and I imagine (Conway having told me of that fact before) that this is what he meant when he said that an algorithm exists. If so, he was being unclear; however, this means that Jeffe's original comment was still correct.", "Here's an updated link for Jeffε's comment (as of May 2017). Although I read Conway's comment as suggesting the problem is decidable: \"Is there an algorithm for Sylver Coinage that tells you, by looking over all (possibly infinite) options, what the status of the a position is and what if any are the winning moves? The answer is yes (see Winning Ways) but I do not know what the algorithm is.\"", "@PyRulez just writing down that regex is exponential time. However, there's an easy polynomial time test for whether you have to name 1 as your next move (and lose): it's true iff both 2 and 3 have already been named (and 1 hasn't). For if they have, then no higher numbers are available, and if they haven't, then one of those two numbers can be named next.", "@SureshVenkat It is efficient. For example, if the current coins are $4$, $5$, and $7$, you effectively have a regex of (aaaa)*(aaaaa)*(aaaaaaa)*, which can be matched against efficiently.", "I am considering using the construction of sum-free set, for example, the set of odd numbers is a sum-free subset of the integers, and the set $\\{N/2+1, ..., N\\}$ forms a large sum-free subset of the set $\\{1,...,N\\}$ ($N$ even). Fermat's Last Theorem is the statement that the set of all nonzero nth powers is a sum-free subset of the integers for $n > 2$.However, some basic problems of sum-free set are still unknown. How many sum-free subsets of $\\{1, ..., N\\}$ are there, for an integer $N$? Ben Green has shown that the answer is $O(2^{N/2})$. Can this help?", "Is it even efficient (i.e P time) to check if someone has lost ? Seems like the \"you lost\" test is an unbounded knapsack problem.", "Apparently it's open whether Sylver Coinage is decidable. Cool. And at least as of 2002, even Conway didn't know who wins from 16."], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "20", "site": "cstheory", "title": "Complexity of the homomorphism problem parameterized by treewidth", "body": "The _homomorphism problem_ $\\text{Hom}(\\mathcal{G}, \\mathcal{H})$ for two classes $\\mathcal{G}$ and $\\mathcal{H}$ of graphs is defined as follows:\n\n> **Input:** a graph $G$ in $\\mathcal{G}$, a graph $H$ in $\\mathcal{H}$\n> \n> **Output:** decide if there is a homomorphism from $G$ to $H$, i.e., a mapping $h$ from the vertices of $G$ to those of $H$ such that, for any edge $\\\\{x, y\\\\}$ of $G$, $\\\\{h(x), h(y)\\\\}$ is an edge of $H$.\n\nFor each $k \\in \\mathbb{N}$, I will call $\\mathcal{T}_k$ the class of the graphs of [treewidth](http://en.wikipedia.org/wiki/Treewidth) at most $k$. I'm interested in the problem $\\text{Hom}(\\mathcal{T}_k, \\mathcal{T}_k)$, which I see as a _parameterized_ problem (by the treewidth bound $k$). My question is: **what is the complexity of this parameterized problem?** Is it known to be FPT? or is it W[1]-hard?\n\nHere are some things that I found about the $\\text{Hom}$ problem, but which do not help me answer the question. (I write $-$ for the class of all graphs.)\n\n * : If $\\mathcal{H}$ is bipartite then $\\text{Hom}(-, \\mathcal{H})$ is in PTIME, otherwise it is NP-complete, but of course the NP-hardness relies on allowing arbitrary $G$.\n * [http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.86.9013&rep=rep1&type=pdf](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.86.9013&rep=rep1&type=pdf): If the treewidth of $\\mathcal{G}$ (modulo homomorphic equivalence) is bounded by a constant then $\\text{Hom}(\\mathcal{G}, -)$ is in PTIME (and otherwise it isn't, assuming FPT != W[1]). Hence, in particular my problem $\\text{Hom}(\\mathcal{T}_k, \\mathcal{T}_k)$ is in PTIME for fixed $k$, but this doesn't tell me what is the dependency on the parameter.\n * From Flum and Grohe's book _Parameterized Complexity Theory_ , Corollary 13.17: The problem $\\text{Hom}(\\mathcal{T}_k, -)$ is FPT when parameterized by the _size_ of $G$ (but I am parameterizing by the treewidth)\n * , Corollary 3.2: When fixing a specific graph $H$, the problem $\\text{Hom}(\\mathcal{T}_k, \\\\{H\\\\})$, parameterized by k, is FPT (this even holds for more complicated counting variants), but I do not want to restrict to fixed $H$.\n\n\n", "link": "https://cstheory.stackexchange.com/questions/34877/complexity-of-the-homomorphism-problem-parameterized-by-treewidth", "tags": ["graph-algorithms", "parameterized-complexity", "treewidth", "fixed-parameter-tractable", "homomorphism"], "votes": 18, "creation_date": "2016-06-02T03:37:34", "comments": ["This question is still open, but one remark: there is an FPT algorithm parameterized by treewidth for the graph isomorphism problem, here: epubs.siam.org/doi/abs/10.1137/… (Daniel Lokshtanov, Marcin Pilipczuk, Michał Pilipczuk, and Saket Saurabh, \"Fixed-Parameter Tractable Canonization and Isomorphism Test for Graphs of Bounded Treewidth\", SICOMP.) As far as I know, unfortunately, this does not say anything about the homomorphism problem."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "21", "site": "cstheory", "title": "Is Node Multiway Cut NP-complete on planar graphs when all terminals lie on the outer face?", "body": "I am interested in the following problem.\n\n**Node Multiway Cut on Planar Graphs with terminals on the outer face**\n\n * Instance: A plane graph G, and integer k, and a set $S \\subseteq V(G)$ of terminals which are all incident on the outer face of G.\n * Question: Is there a set of vertices $X \\subseteq V(G)$ of size at most $k$ such that all vertices of $S \\setminus X$ belong to different connected components of $G - X$?\n\n\n\nInformally, the problem asks whether all terminals can be separated with k vertex deletions, knowing that all terminals lie on the outer face. If we replace vertex deletions by edge deletions, then this is known to be possible in polynomial time by considering multiway cuts in the planar dual, and relating them to certain kind of Steiner trees ([Efficient Algorithms for k-Terminal Cuts on Planar Graphs](https://doi.org/10.1007/3-540-45678-3_29)). Since the edge-deletion version reduces to the vertex-deletion version, the edge-deletion version seems easier; no reduction in the reverse direction is known.\n\nThe question is: is Node Multiway Cut on Planar Graphs with terminals on the outer face NP-complete? The reason I am interested in this problem is that it would provide me with a starting point for another complexity lower bound if this is indeed NP-complete.\n", "link": "https://cstheory.stackexchange.com/questions/8969/is-node-multiway-cut-np-complete-on-planar-graphs-when-all-terminals-lie-on-the", "tags": ["cc.complexity-theory", "np-hardness", "planar-graphs"], "votes": 17, "creation_date": "2011-11-14T09:06:13", "comments": ["I know for sure that the algorithm I mentioned works for edge weighted Steiner tree, but one should be able to adapt it to work for vertices.", "I think that this is equivalent to solving vertex weighted Steiner tree where all terminals on the outer face. The very loose idea would be something like: consider the boundary path between every pair of consecutive terminals you want to separate; connect all the vertices on that path to a new vertex, which becomes a terminal for the Steiner tree problem. Now all these Steiner terminals will all lie on the same face. This should be solvable in polynomial time using the algorithm of Erickson et al. (jstor.org/pss/3689922)."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "22", "site": "cstheory", "title": "Can short-distance connectivity be harder than connectivity?", "body": "Has anybody seen the following (or similar) question being considered:\n\n> Can it be **easier** to determine the presence/absence of $s$-$t$ paths than to determine the presence/absence of _short_ $s$-$t$ paths? \n\nA bit more formally, the _distance_ -$k$ _connectivity_ problem STCON(n,k) is, given a subgraph of a complete undirected graph $K_n$ on $[n]=\\\\{1,\\ldots,n\\\\}$, to decide whether there is a path from node $s=1$ to node $t=n$ of length $\\leq k$. By the _length_ , I mean the number of inner nodes (just for notational convenience). The _connectivity_ problem STCON(n) corresponds to the case when $k=n-2$: is there any $s$-$t$ path at all? These problems are monotone boolean functions of $\\binom{n}{2}$ variables $x_{i,j}$ corresponding to the edges of $K_n$. \n\nIt is well known, that $O(kn^2)$ fanin-$2$ AND and OR gates are enough to compute STCON(n,k). The desired monotone circuit is given by [Bellman-Ford](http://en.wikipedia.org/wiki/Bellman%E2%80%93Ford_algorithm): first compute $F_j^{(0)}=x_{s,j}$ for all nodes $j$, and then use the recursion $F_j^{(l+1)}=\\bigvee_{i\\in[n]} F_i^{(l)}\\land x_{i,j}$. Then $F_j^{(l)}$ accepts a given graph iff it has an $s$-$j$ path of length $\\leq l$. Hence, $F_n^{(k)}=$STCON(n,k).\n\n> **Question 1:** Can STCON(n) have more than constant times **smaller** monotone circuits than STCON(n,k), for some $k\\ll n$? \n\nA boolean function $f(x_1,\\ldots,x_n)$ is a _monotone projection_ of a boolean function $g(y_1,\\ldots,y_m)$, if the function $f$ can be obtained from $g$ by replacing each $y$-variable by $0$ or $1$ or by some $x$-variable. Say, we trivially have that every THRESHOLD(n,k) is a monotone projection of MAJORITY(2n). \n\nBy taking $k$ copies of the input graph for STCON(n,k), one can easily show that STCON(n,k) is a monotone projection of STCON(kn). Thus, STCON(kn) cannot have smaller circuits than STCON(n,k). Can the factor \"k\" here be replaced by some constant?\n\n> **Question 2:** Is STCON(n,k) a monotone projection of STCON(m) for some $m < kn$? \n\n**ADDDED** (Apr.7 ): Among related results I know are the following:\n\n 1. _STCON and monotone**NC$^1$**_ : if a boolean function $f$ has a monotone formula of size $s$, then $f$ is a monotone projection of STCON(m) for $m=O(s)$; this is easy: just view formulas as parallel-sequential networks. In particular, this implies that _every_ monotone boolean function of $n$ variables _is_ a monotone projection of STCON(m) for some $m\\leq n2^n$ (take a monotone DNF). \n 2. _STCON itself is not in monotone**NC$^1$**_ : every monotone circuit for STCON(n,k) requires depth about $(\\log k)(\\log n)$ (independent on circuit size!). This was shown by [Karchmer and Wigderson](http://www.math.ias.edu/~avi/PUBLICATIONS/MYPAPERS/KW90/KW90.pdf). As observed by [Grigni and Sipser](http://www.mathcs.emory.edu/~mic/papers/4.ps), this holds even when AND gates are allowed to have unbounded fanin! If, however, OR gates can have large fanin and AND gates have fanin-$2$, then depth $2\\log k$ and size $O(n^3\\log k)$ is already enough: repeated squaring of the adjacency matrix $\\log k$ times. \n 3. _$s$-$t$ Connectivity vs. Connectivity_ : If $f=x_1x_2\\lor x_3x_4\\lor\\cdots\\lor x_{n-1}x_n$ is a monotone projection of CONN(m), then $m\\geq 2^n+1$. This was proved by [Skyum and Valiant](http://dl.acm.org/citation.cfm?id=3158&CFID=497574256&CFTOKEN=86163804). Here CONN is a \"sibling\" of STCON: accepts a graph iff it is connected. Note that this particular function $f$ _is_ a monotone projection of STCON(m) even for $m=n$: take a bunch of $n/2$ parallel paths of length two. \n 4. _Formulas vs. Circuits_ : If the depth is restricted to $o(\\log n)$ then STCON(n,k) requires even non-monotone(!) unbounded fanin _formulas_ of size $n^{\\Omega(\\log k)}$ for all $k\\leq \\log\\log n$. This was proved by [Rossman](http://eccc.hpi-web.de/report/2013/169/download), and separated for the first time bounded-depth circuits from formulas, because for such values of $k$, repeated squaring gives the desired small (even monotone) _circuit_ of depth $2\\log k\\leq 2\\log\\log\\log n=o(\\log n)$. \n 5. _Directed vs. Undirected Connectivity_ : The directed reachability problem DSTCON(n) (where input graphs are directed) is a monotone projection of STCON(m) only if $m=n^{\\Theta(\\log n)}$. This was proved by [Potechin](http://eccc.hpi-web.de/report/2009/142/revision/1/download), and separated monotone versions of **L** and **NL**. \n\n\nDoes anybody know some results more tightly related to the STCON(n,k) vs. STCON(n) question itself? \n\n* * *\n\n**CORRECTION** (Apr 9): Unfortunately, just taking $k$ copies of the input graph $G$ for STCON(n,k), only shows that STCON(n,k) is a monotone projection of **D** STCON(kn), the **directed** (even acyclic) version of STCON. The difficult direction is to ensure that the modified graph $H=H_G$ has no s-t paths, when $G$ _has_ s-t paths, but all they are longer than $k$. If $H$ is allowed to be directed, this is easy to ensure by arranging the copies of $G$ into $k$ layers, and drawing _directed_ edges between layers according to $G$. But if the edges $H$ are required to be _undirected_ , this simple layering trick won't work. So, I must refine my Question 2 as (cf. with the 5-th result by Potechin above):\n\n> **Question 2':** Is STCON(n,k) a monotone projection of **D** STCON(m) for some $m < kn$? \n\nBoth answers would be interesting. If YES, this would show that layering is not the best way to \"count\". If NO, this would show that every monotone switching network for STCON(n,k) requires $kn$ nodes. (By the Bellman-Ford algorithm - which itself is just the \"layering trick\" in disguise - $kn$ nodes are also enough.)\n", "link": "https://cstheory.stackexchange.com/questions/31005/can-short-distance-connectivity-be-harder-than-connectivity", "tags": ["cc.complexity-theory", "graph-algorithms", "circuit-complexity", "lower-bounds", "dynamic-programming"], "votes": 17, "creation_date": "2015-04-03T10:00:23", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "23", "site": "cstheory", "title": "Sequences with sublogarithmic concat and approximate split", "body": "Is there a data structure for representing sequences that supports the operations:\n\n * **concat** takes two sequences of size $m$ and $n$ and produces a new sequence of size $m+n$ by joining them in time $o(\\lg \\min(n,m/n))$ (or $o(\\lg \\min(m,n/m))$ if $n>m$).\n * For some constants $c$ and $N$, **approximate split** takes a sequence of size $n > N$ and divides it into two contiguous non-overlapping sequences of sizes $p$ and $q$ such that $p+q = n$ and $1/c < p/q < c$ in time $o(\\lg n)$\n * **singleton** takes a single element and produces a sequence containing only that element in $O(1)$ time.\n * **linearize** takes a sequence of size $n$ and produces an array of size $n$ containing the elements of the sequence in time $O(n)$.\n\n\n\nI'm really looking for worst-case or expected bounds on the time, but amortized bounds would also be interesting. Results in the word RAM would be interesting as well.\n\nHere are some results I am aware of. In this table, **sloppy split** is like approximate split, but the bounds are $1/c < (\\lg p)/(\\lg q) < c$.\n \n \n structure | concat | approx split | sloppy split\n ==================================================================================\n doubly-linked lists | O(1) | O(n) | *\n arrays | O(n+m) | O(1) | *\n weight balanced trees | O(lg (n+m)) | O(1) | *\n treaps and skip lists | O(lg min(n,m)) ** | O(1) ** | *,**\n functional finger trees | O(min(lg lg m/n, lg lg n) | O(lg n) | O(1)\n \n *: same as approx split\n **: expected\n \n\nBy functional finger trees I mean \"[Purely functional representations of catenable sorted lists](http://www.math.tau.ac.il/~haimk/papers/loglog23.ps)\" by Kaplan and Tarjan.\n", "link": "https://cstheory.stackexchange.com/questions/5964/sequences-with-sublogarithmic-concat-and-approximate-split", "tags": ["ds.data-structures"], "votes": 17, "creation_date": "2011-04-08T22:01:48", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "24", "site": "cstheory", "title": "Linear-time algorithm to test if clique number equals degeneracy bound?", "body": "Given a connected simple graph $G=(V,E)$, let $d$ denote its [degeneracy](https://en.wikipedia.org/wiki/Degeneracy_\\(graph_theory\\)) and let $\\omega$ denote the size of a [maximum clique](https://en.wikipedia.org/wiki/Clique_problem).\n\nA well-known bound on the clique number is $\\omega\\le d+1$, which is helpful when solving the maximum clique problem or when [enumerating all maximal cliques](https://dl.acm.org/doi/abs/10.1145/2543629?casa_token=fxzBWpiA1ZsAAAAA:CrMNWZ-0GU9DU3L0-UE43AniI2H-2gvXWFzx341kOOA8MvUU2QnCVAQUxcpoQQGtfriWot8YZGi9TA).\n\nHow fast can one test whether this bound is tight, $\\omega=d+1$? An algorithm that runs in time $O(dm)=O(m^{1.5})$ is [known](https://pubsonline.informs.org/doi/10.1287/opre.2019.1970).\n\nIs there a faster algorithm, say, running in linear time?\n\nOr, is $O(m^{1.5})$ best possible under something like [SETH](https://en.wikipedia.org/wiki/Exponential_time_hypothesis)?\n", "link": "https://cstheory.stackexchange.com/questions/47947/linear-time-algorithm-to-test-if-clique-number-equals-degeneracy-bound", "tags": ["graph-algorithms", "parameterized-complexity", "clique", "fixed-parameter-tractable"], "votes": 17, "creation_date": "2020-11-30T14:49:56", "comments": ["Maybe you are aware, but this problem recently appeared in codeforces codeforces.com/contest/1439/problem/B. It seems nobody in the discussion mentions anything better than $O(m^{1.5})$.", "Um...it looks like their article was just published and they list your question as Open Problem 1 (minus the possible connection to SETH)...", "Ah, I see now. (After your comment, I thought I could answer your Q, but then looked at the article and saw I just rediscovered their algorithm in the case $p=0$.) I wonder if you don't even need SETH - it seems like maybe it's possible to show that if you could do better than $O((m+n) + d^2 n)$ (which is what they have) then either you'd be able to compute $k$-cores faster than linear (not possible by query complexity) or test whether a $d$-vertex graph is a clique in $o(d^2)$ (also not possible by query complexity). I don't quite see a reduction, but maybe one can do it w/o SETH...", "The decomposition is linear time, but I don’t think that solves the problem. It works, for example, for G=C_3 or G=C_4 but fails for their disjoint union (or for their disjoint union with an added edge between them).", "Can't you just compute the k-core decomposition (which is quasilinear time IIRC), and then just check if the last core is a clique?"], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "25", "site": "cstheory", "title": "Is it possible to boost the error probability of a Consensus protocol over dynamic network?", "body": "Consider the binary consensus problem in a synchronous setting over dynamic network (thus, there are $n$ nodes, and some of them are connected by edges that may change round to round). Given a randomized $\\delta$-error Monte Carlo protocol for Consensus in this setting, is it possible to use this protocol as a black box to obtain another Consensus protocol with smaller error?\n\nNote that for problems with deterministic outcome (such as Sum), this can be done by repeatedly executing the protocol and taking the majority as the answer. However I am unable to figure out how for Consensus, so I wonder whether this is actually impossible to do.\n\n\\--- Details ---\n\n**Dynamic Network.** There are $n$ nodes on a dynamic network over synchronous setting (thus the protocol will proceed in rounds). Each node has a unique ID of size $\\log(n)$. There are edges connecting pairs of nodes, and this set of edges may change from round to round (but remains fixed throughout any particular round). There may be additional restrictions to the dynamic network that allows the problem to be solved by some protocol (such as having a fixed diameter), but it is irrelevant to our discussion.\n\nThe only information known to each node is their unique ID. They do not know $n$ or the topology of the network. In particular, they do not know who their neighbors are in the current round (neighbors are nodes that are adjacent to them with an edge). The only way to figure out the neighbors is after receiving messages from them.\n\n**How nodes communicate.** The only method of communication between nodes is by broadcasting messages under the $CONGEST$ model (D. Peleg, \"Distributed Computing, A Locality-Sensitive Approach\"). More specifically, in each round, every node may broadcast an $O(\\log n)$ sized message. Then, every node will receive ALL messages that were broadcasted by all of its neighbors in that particular round in an arbitrary order. In particular, nodes may attach their IDs onto their messages, since their IDs are of size $O(\\log n)$.\n\n**Binary Consensus.** The binary consensus problem over this network is as follows. Each node has a binary input ($0$ or $1$), and each must decide either $0$ or $1$ under the the validity, termination, and agreement for consensus with $\\delta$ error:\n\n * validity means that for $z \\in \\\\{0, 1\\\\}$, if all nodes have $z$ as its initial input, then the only possible decision is $z$.\n\n * termination means that each node must eventually decide.\n\n * agreement means that all nodes must have the same decision.\n\n * The error rate means that under worst case input and worst case network, the probability that the protocol does not satisfy any of the three conditions is at most $\\delta$ over average coin flips of the protocol's coin.\n\n\n\n\n**Question.** For a particular set of dynamic networks, suppose we have a lower bound on the average number of rounds incurred by any randomized $\\delta$-error protocol $P$ for solving binary consensus under this settings such that $\\delta < \\frac{1}{3}$, under worst case dynamic network from this set, worst case nodes' inputs, and on average coin flips of the protocol. Does this bound automatically hold for any $\\delta'$-error protocol for $\\delta < \\delta' < \\frac{1}{3}$?.\n", "link": "https://cstheory.stackexchange.com/questions/32442/is-it-possible-to-boost-the-error-probability-of-a-consensus-protocol-over-dynam", "tags": ["ds.algorithms", "reductions", "dc.distributed-comp"], "votes": 16, "creation_date": "2015-08-23T23:24:43", "comments": ["@Peter Also this question is more of a whether it is possible to obtain a better protocol using a black box protocol. For example, in Sum, each node can run the protocol multiple times (either in parallel or in a row) and takes a majority. Using Chernoff's bound, for any $0 < \\delta' < \\delta < \\frac{1}{3}$, one can obtain protocol with $\\delta'$ error from a $\\delta$ error protocol for Sum by executing the protocol a constant amount of time and taking a majority as the answer.", "@Peter No unless your protocol is specifically designed to be able to do that. The only way for the nodes to communicate is through broadcasting and receiving messages. E.g., the lack of message from a particular node may mean something. There are no \"magical\" communication channels otherwise.", "Do you assume that, when an instance of your blackbox consensus algorithm fails, all nodes are aware that the consensus algorithm has failed? This seems to depend on the precise properties of the dynamic network that you're assuming...", "suggest try to fit it into the following framework. The Consensus Problem in Unreliable Distributed Systems (A Brief Survey) (2000) / Fisher. there are some impossibility results known & could your idea possibly violate any?", "@vzn No, I'm not really familiar with this topic. Thus I thought that this question is well-known, and I was hoping for a quick answer here. Seems like this is not well-known then. I wasn't aware of the TCS stackexchange before, I'll try to ask the admin to cross post this question there.", "are you familiar with Paxos, \"a family of protocols for solving consensus in a network of unreliable processors\"? does it not match the requirements for some reason? there are many (standard) consensus protocols, can you give an example of a \"randomized δ-error protocol for Consensus\"?", "it appears research level to me! the whole question seems to be built on a research framework! seriously consider Theoretical Computer Science! can you define \"randomized δ-error protocol\"? try also Theoretical Computer Science Chat"], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "26", "site": "cstheory", "title": "complexity of checking if a subspace is a Euclidean section of L1", "body": "If $X$ is a linear subspace of ${\\mathbb R}^n$, $X$ is high-dimensional, and for every $x\\in X$ we have\n\n$(1-\\epsilon) \\sqrt n ||x||_2 \\leq ||x||_1 \\leq \\sqrt n ||x||_2$\n\nfor some small $\\epsilon >0$, then we say that $X$ is an almost-Euclidean section of $\\ell_1^n$, and (the matrix whose image is) X is useful in compressed sensing. A random subspace works excellently, and there is a huge research program devoted to the **explicit construction** of such spaces.\n\nIs it known what is the complexity of approximating the \"Euclidan-sectionness\" of $X$? That is, given a subspace $X$, say presented via a basis, consider the problem of finding the unit (in $\\ell_2$ norm) vector in $X$ of smallest $\\ell_1$ norm.\n\n> What is the complexity of this problem? Are hardness of approximation results known?\n\nApart from specific applications, these seem to be interesting problems. Is it known what is the complexity of finding the vector of **maximum** $\\ell_1$ norm among the unit vectors of a given subspace? \n", "link": "https://cstheory.stackexchange.com/questions/4992/complexity-of-checking-if-a-subspace-is-a-euclidean-section-of-l1", "tags": ["cc.complexity-theory", "cg.comp-geom", "norms", "compressed-sensing"], "votes": 17, "creation_date": "2011-02-17T18:27:44", "comments": ["This is not an answer to your question. But in the vein of computational problems arising from compressed sensing applications, Koiran and Zouzias have a recent paper on checking whether a matrix satisfies the restricted isoperimetry property (RIP) and related problems.", "Thanks Sariel, I had inverted the places of $\\ell_1$ and $\\ell_2$. (If you want $\\ell_2$ in the middle, then it should be $||x||_1/\\sqrt n \\leq ||x||_2 \\leq (1+\\epsilon)||x||_1 / \\sqrt n$, because it's always $||x||_1 \\leq \\sqrt n ||x||_2$.)", "The above inequality is slightly wrong. You probably mean that $(1-\\varepsilon)||x||_1 /\\sqrt{n} \\leq ||x||_2 \\leq ||x||_1/\\sqrt{n}$. Indeed, for any $x$, $||x||_1 \\geq ||x||_2$."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "27", "site": "cstheory", "title": "Looking for an operator on polynomials", "body": "I have a small, self-contained, math question, whose motivation is from theoretical computer science (specifically, list decoding of algebraic codes, derivative/multiplicity codes, etc). I wonder whether someone might have an idea.\n\nI'm looking for an operator T that can be applied to m-variate polynomials over a finite field. When applied to a polynomial, the operator should yield a polynomial of not much higher degree. The operator should satisfy the following property: for every m-variate polynomial $F$, and m univariate non-constant polynomials $g_1,...,g_m$, if you know $g_1,...,g_m$ and $F(g_1(t),...,g_m(t))$ for a parameter t, then you can also compute $TF(g_1(t),...,g_m(t))$. The operator should be non-trivial, in the sense that for a fixing $x_1,..,x_m$, the value $F(x_1,...,x_m)$ does not determine the value of $TF(x_1,...,x_m)$. \n\nFor $m=1$, the derivative operator gives exactly that: By the chain rule $(F(g(t)))' = F'(g(t))\\cdot g'(t)$, which implies that $F'(g(t)) = (F(g(t)))'/g'(t)$. My question is whether there is an operator that works for general m.\n", "link": "https://cstheory.stackexchange.com/questions/10881/looking-for-an-operator-on-polynomials", "tags": ["it.information-theory", "coding-theory", "algebra", "polynomials"], "votes": 16, "creation_date": "2012-03-28T08:17:46", "comments": ["Honestly, I asked this question so long ago, I only barely remember what I needed it for :-)... I believe that the issue is that multiplying by a fixed polynomial is \"trivial\" in the sense that to compute the multiplication at a point, you only need to know the value of F at the point (and nothing about F's inner-workings).", "Dana, I may be missing something, what about $TF(x_1,...,x_n) = T(x_1,...,x_n) P(x_1, ..., x_n)$, where $P$ is some multivariate polynomial?", "F(g(t))' and g'(t) are formal polynomials in t. You can divide the formal polynomials.", "Suppose $g(t) = t^2$. You still can't evaluate $F'$ at 0? (I understand that you want point-wise evaluation)", "If g' is identically zero then you can't expect anything (I'll edit the question to specify the g_i's are not identically zero).", "Your example in one dimensional case is not flawless. What if $g'(t)=0$ ?", "Suppose you take $TF(g_1(t),..,g_m(t)) = \\sum_{i=1}^{m} g_i'(t) dF(x_1,...,x_m)/dx_i$ (known as the total derivative). How you do compute $dF(x_1,...,x_m)/dx_i$ from $F(g_1(t),...,g_m(t))$?", "Why total derivative wouldn't fit your requirements?"], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "28", "site": "cstheory", "title": "When does adding edges decrease the cover time of a graph?", "body": "When first learning about random walks on a graph $G$, one may have an intuitive feeling that adding edges to $G$ will decrease its cover time $C(G)$. However, this is not the case. The [path graph](https://en.wikipedia.org/wiki/Path_graph) $P_n$ has cover time $C(P_n) = \\Theta(n^2)$ while the [($\\frac{n}{2}$, $\\frac{n}{2}$)-lollipop graph](http://mathworld.wolfram.com/LollipopGraph.html) $L_{\\frac{n}{2}, \\frac{n}{2}}$ has cover time $C(L_{\\frac{n}{2}, \\frac{n}{2}}) = \\Theta(n^3)$. Eventually adding edges does decrease the cover time because the complete graph $K_n$ has cover time $C(K_n) = \\Theta(n \\log n)$ by analogy to the [coupon collector problem](https://en.wikipedia.org/wiki/Coupon_collector%27s_problem).\n\nUnder what assumptions though would the supposed intuition hold true? Specifically, what about [vertex transitivity](https://en.wikipedia.org/wiki/Vertex-transitive_graph)?\n\n> Let $G$ be a vertex-transitive graph with $n$ vertices. Consider any vertex-transitive graph $G^+$ obtained from $G$ by adding edges (i.e. both $G$ and $G^+$ are vertex transitive and $G$ is a spanning subgraph of $G^+$).\n> \n> Is it the case that $C(G^+) \\le C(G)$?\n", "link": "https://cstheory.stackexchange.com/questions/33071/when-does-adding-edges-decrease-the-cover-time-of-a-graph", "tags": ["graph-theory", "pr.probability"], "votes": 16, "creation_date": "2015-11-12T18:35:47", "comments": ["My bad, I confused spanning with induced.", "@chazisop Those subgraphs are not spanning.", "Wouldn't the complete graph $K_{n}$ and any of its subgraphs $K_{k}$, $k < n$ be a counterexample for general vertex transitivity?"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "29", "site": "cstheory", "title": "Complexity of approximating the range of a matrix", "body": "Given an $m$ by $n$ matrix $M$ with $m \\leq n$ and elements from $\\\\{-1,1\\\\}$, let us define:\n\n$$S_M = |\\\\{Mx : x \\in \\\\{-1,1\\\\}^n\\\\}|.$$\n\nI believe that it is NP-hard to compute $S_M$ exactly, by applying the reductions from [Decide whether a matrix's kernel contains any non-zero vector all of whose entries are -1, 0, or 1](https://cstheory.stackexchange.com/questions/20277/decide-whether-a-matrixs-kernel-contains-any-non-zero-vector-all-of-whose-entri) to the following decision problem: does $S_M = 2^n$?\n\n> Is it possible to approximate $S_M$ to within a constant factor in polynomial time? If not, what is the best one can do in polynomial time?\n\n[Cross-posted to ] \n", "link": "https://cstheory.stackexchange.com/questions/33676/complexity-of-approximating-the-range-of-a-matrix", "tags": ["cc.complexity-theory", "approximation-algorithms", "linear-algebra"], "votes": 15, "creation_date": "2016-01-28T02:00:10", "comments": ["(1) If there is a poly-time $2^{n^{1-\\epsilon}}$-approximation for any $\\epsilon>0$, then there is a poly-time $(1+1/q(n))$-approximation for any polynomial $q$. To see this, given $M$, consider the block-diagonal matrix $M(k)$ formed by $k$ independent copies of $M$ along the diagonal, so $S_M=(S_{M(k)})^{1/k}$. Take $k=(n q(n))^{1/\\delta}$. If $x$ is a $2^{(nk)^{1-\\epsilon}}$-approximation of $S_{M(k)}$, then $x^{1/k}$ is a $(1+1/q(n))$-approximation of $S(M)$. (2) One has $2^k\\le S_M\\le 2^n$ where $k$ is the rank of $M$, but this doesn't help as $S_M$ can be large even when $k=m=1$."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "30", "site": "cstheory", "title": "Intersecting Complexity Classes with Advice", "body": "In [on hiding information from an oracle](http://linkinghub.elsevier.com/retrieve/pii/0022000089900184), the authors (Abadi, Feigenbaum, and Kilian) wrote:\n\n> $(\\mathsf{NP/poly} \\cap \\mathsf{co\\text-NP}{/poly})$ ... is **not known** to be equal to $(\\mathsf{NP} ∩ \\mathsf{co\\text-NP}){/poly}$. \n\nThey highlighted that in the conference paper, they mistook the two classes. Apparently, the latter is a subset of the former, but we don't know if the containment is strict.\n\nAssuming $X$ and $Y$ are complexity classes, and $F$ is a set of functions specifying the length of the advice strings, are there recent results comparing $(X_{/F} \\cap Y_{/F})$ and $(X \\cap Y)_{/F}$, resolving issues like the one pointed above?\n", "link": "https://cstheory.stackexchange.com/questions/5839/intersecting-complexity-classes-with-advice", "tags": ["cc.complexity-theory", "reference-request", "complexity-classes", "advice-and-nonuniformity"], "votes": 15, "creation_date": "2011-04-03T03:17:43", "comments": ["@HenryYuen: I think that a Language $L$ is in $(NP \\cap coNP)/poly$ iff there is a language $K$ in $NP \\cap coNP$ and $a_i$ advice of polynomial length s.t. $x \\in L$ iff $(x, a_{|x|}) \\in K$.", "@HenryYuen: Oh, I got your point. Strangely, the definition of (NP ∩ coNP)/poly in Complexity Zoo (qwiki.stanford.edu/index.php/Complexity_Zoo:N#npiconppoly) does not mention machines with advice, but \"languages\" with advice. However, quoting from An Oracle's Builder Toolkit: \"NP ∩ coNP is defined by the enumeration $M_1,M_2,\\ldots$, where $M_i$ with $i = \\langle i_1,i_2 \\rangle$ is the pair of non-deterministic oracle Turing machines $N_{i_1}$ and $N_{i_2}$ such that both machines run in time $n^i$...\" (Read page 27 for more info)", "The reason I ask about the computational model is because I don't know what (NP ∩ coNP)/poly means. NP/poly is the set of languages that are accepted by NP machines with polynomial amount of advice. Presumably, (NP ∩ coNP)/poly is the set of languages accepted by (NP ∩ coNP) machines with polynomial advice -- or is there another definition that isn't machine-definition-dependent?", "@HenryYuen: I actually don't know if there's an underlying model of computation, but the definitions of complexity classes are clear, as in the case of ZPP. Besides being a very interesting problem, is there any significance (to my question) if NP ∩ coNP does (or does NOT) admit a computation model?", "If you take the definition that ZPP = RP ∩ coRP, that doesn't immediately give you a model of computation that accepts only ZPP languages: it only says that every language in ZPP has both an RP machine and a coRP machine (which is well defined). You have to prove that ZPP is the class of languages that admit Las Vegas algorithms. Similarly -- NP ∩ coNP defines a set of languages that have both NP and coNP machines, but is there a model of computation that accepts precisely NP ∩ coNP?", "@HenryYuen: Maybe the definition of ZPP helps: ZP = RP ∩ CoRP. See en.wikipedia.org/wiki/ZPP_(complexity)#Intersection_definiti‌​on.", "What is the definition of (NP ∩ coNP)/poly? What is a NP ∩ coNP machine?"], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "31", "site": "cstheory", "title": "Intermediate problems between PSPACE and EXPTIME", "body": "Intermediate problems between P and NP are quite famous, and are sometimes considered as complexity classes by themselves.\n\nDo you know of any problem that is known to be PSPACE-hard and in EXPTIME, and resisting all efforts to be proved complete for one of these classes ?\n\nLifted (succinct) versions of problems between LOGSPACE and PTIME are accepted, but even more interesting would be problems that are not of this form.\n\nHere are some I found, this area of complexity theory seems to be the realm of games:\n\n * Phutball: \n * Go (with Chinese rules): D. Lichtenstein and M. Sipser, Go is polynomial-space hard\n\n\n", "link": "https://cstheory.stackexchange.com/questions/38393/intermediate-problems-between-pspace-and-exptime", "tags": ["cc.complexity-theory", "complexity-classes", "exp-time-algorithms", "pspace"], "votes": 17, "creation_date": "2017-06-09T06:50:47", "comments": ["I think the more correct way of stating the question is asking for candidate problems in ExpTime - PSpace - ExpTime-hard. Note that e.g. in the case of P vs. NP we don't know if Factoring or GI is P-hard."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "32", "site": "cstheory", "title": "Mutual information vs. Product sets", "body": "Suppose we have two _dependent_ random variables $X$ and $Y$, each of which is uniform over $\\\\{0,1\\\\}^n$, such that their mutual information $I(X;Y)$ is small, say, at most $\\sqrt{n}$. Does this imply that there exist large sets $\\mathcal{X}, \\mathcal{Y} \\subset \\\\{0,1\\\\}^n$ such that the product set $\\mathcal{X} \\times \\mathcal{Y}$ is contained in the support of the distribution $(X,Y)$?\n\nA variant of this question can be phrased in graph-theoretic terms: Suppose we have a sufficiently dense bipartite graph G. Does this imply that $G$ contains a large complete sub-graph $G'$? (i.e., $G'$ is required to be a complete bipartite graph with bi-partition $(A,B)$ such that $A$ and $B$ are relatively dense in the corresponding parts of $G$)\n", "link": "https://cstheory.stackexchange.com/questions/10225/mutual-information-vs-product-sets", "tags": ["co.combinatorics", "it.information-theory"], "votes": 15, "creation_date": "2012-02-15T17:49:06", "comments": ["If $(X,Y)$ is a product of iid trials $(X_i,Y_i)$ then I am inclined to say yes.", "If my calculation is correct, a standard use of the probabilistic method shows that for sufficiently large N, there exists a bipartite graph on N+N vertices with at least (1/2)N^2 edges that does not contain K_{k,k} as a subgraph, where k = floor(2.001 lg N). (“lg” denotes the logarithm to base two.)", "I'm not an expert, but perhaps this paper on Ramsey theory applied to bipartite graphs can help you: math.mit.edu/~fox/paper-density-theorems-final.pdf"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "33", "site": "cstheory", "title": "Computability of a "weird" set", "body": "The starting point of this question is the observation that the smallest positive integers $a,b,c$ satisfying\n\n$$\\frac{a}{b+c} + \\frac{b}{a+c} + \\frac{c}{a+b} = 4$$\n\nare [absurdly high](https://plus.google.com/+johncbaez999/posts/Pr8LgYYxvbM). This leads to the following general question: Is the set $C\\subseteq {\\mathbb N}$ defined by $$ C = \\left\\\\{n\\in\\mathbb{N}\\setminus\\\\{0\\\\}: \\big(\\exists a,b,c \\in\\mathbb{N}\\setminus\\\\{0\\\\}\\big):\\frac{a}{b+c} + \\frac{b}{a+c} + \\frac{c}{a+b} = n\\right\\\\}$$ computable?\n", "link": "https://cstheory.stackexchange.com/questions/39383/computability-of-a-weird-set", "tags": ["computability"], "votes": 14, "creation_date": "2017-10-26T07:49:40", "comments": ["Cross-posted from mathoverflow.net/questions/278747/is-this-set-computable", "I'm not an expert, but if you simplify the equation you get a cubic diophantine equation in three variables, and I think it's an open problem if universality can be achieved in such a setting. In every case, quadratic diophantine equations in two variables ($ax^2 + by + c = 0$) are already NP-complete (see NP-complete decision problems for quadratic polynomials) ... so I'm not surprised if switching to cubic+three vars leads to very hard instances (outside NP)."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "34", "site": "cstheory", "title": "Can a Penrose tile cellular automaton be Turing-complete?", "body": "This question was based on an incorrect premise ... see Colin's comment below. Forget it.\n\nThis was inspired by the discussion on [this Math Overflow question](https://mathoverflow.net/questions/45378/undecidability-in-conways-game-of-life/45382). First, I need to define our terms.\n\nIn a Penrose tiling, there are generally several different legal patterns of tiles which will cover a given region. A Penrose tile cellular automaton should be some rule that deterministically takes a Penrose tiling $T$ to a Penrose tiling $T'$, where each tile in $T'$ is deterministically given by looking at the tiles in a radius $r$ around the corresponding spot in $T$. You should be able to implement these by having a set of rules of the form: when a region of shape $S$ is covered by a pattern $X$ of tiles, replace it with a pattern $Y$ of tiles. In order to do this deterministically, you either have to make sure no two of your replaceable regions can overlap or have some priority ranking on these replacement rules. But certainly you can ensure this, and define lots of different Penrose tiling cellular automata.\n\nWhat does it mean for a Penrose tiling celluar automaton to be Turing-complete. Here's what I think it should mean (although I won't insist on it in the answers): the halting problem should be reducible to the behavior of such an automaton. Let's take some canonical Penrose tiling $T_0$ that provides constraints on our starting position. We start with an input $G$ to a universal Turing machine, and we'd like to decide whether the machine halts on input $G$. I would like some computable map from $G$ to a starting Penrose tiling $S$ such that $S$ differs from $T_0$ on only a finite set of tiles. We then run the Penrose tiling cellular automaton, starting with $S$, and wait for some finite canonical configuration $C$ at position $0$ in the Penrose tiling. I want configuration $C$ to appear at position $0$ if and only if the universal Turing machine halts on input $G$.\n\nThe strange thing about Penrose tiling cellular automata is that every finite legal configuration of tiles appears infinitely often in every Penrose tiling of the plane. Thus, our universal Penrose tiling is not only computing the behavior of our universal Turing machine on input $G$, it's also simultaneously carrying out the first steps of every other possible computation. This makes me doubt whether Penrose tiling cellular automata could be Turing complete. On the other hand, if they're not Turing complete, what is the class of computations that Penrose tiling cellular automata can perform?\n", "link": "https://cstheory.stackexchange.com/questions/2883/can-a-penrose-tile-cellular-automaton-be-turing-complete", "tags": ["computability", "machine-models", "cellular-automata"], "votes": 15, "creation_date": "2010-11-11T06:14:56", "comments": ["Yes, that is true for tilings formed via the deflation process. But an arbitrary planar tiling might have a \"rift\" along a line in the plane, along which the higher level tiles do not match up (like a tiling of the plane by squares in which the upper half plane is shifted right by a fraction of a unit). This is certainly possible with Robinson tiles and Ammann tiles, and I would expect it also for the various Penrose tiles although I haven't checked. If rifts can occur, local patches along the rift will appear infinitely often in the same tiling but needn't appear in another tiling.", "@Matt: What I should have said is that every finite pattern that appears once in an infinite Penrose tiling appears infinitely often in every Penrose tiling.", "If every finite pattern would occur, you would not be able to \"get stuck\" when attempting to tile the plane (because your finite pattern, whatever it is so far, would be guaranteed to be part of a valid tiling of the plane). But in fact, you do need to know what you are doing in order to not get stuck, regardless of which Penrose tile set you are talking about. Only very occasionally do you really have a choice of what tile to place (in the sense that either tile would be ok for creating a full tiling of the plane), approximately once per inflation scale level.", "Yes, it's 14 years later, but I can't help noticing that something is wrong on the internet (like the xkcd cartoon). It is not correct that every finite pattern occurs in a Penrose tiling.", "@Peter: This question intrigued me because it seems related to one of the main open problems in tile self-assembly, which is \"the power of multiple nucleation.\" If the Penrose tiling rule were somehow limited to radiate out one tile at a stage from the origin, instead of everywhere being tileable at once, I think the model would be very close to self-assembly models. The open questions there are complexity-related, not computability, e.g.: is there a finite shape such a restricted Penrose tiling could achieve that a GCA could not achieve with only polynomial complexity increase?", "@Colin, Actually, I was wrong about this. Forget the question. Maybe I should delete it.", "Can you give an example of a region which may be legally covered by different patterns of tiles? I'm not familiar with the subject, but there is a unique way to break up a triangle into smaller triangles in \"up-down\" generation - see for example Prop 6.1 of Quasicrystals and geometry.", "If someone reduced the Turing Machine Immortality Problem to a Penrose CA, would that be progress? (TMIP = given a TM with an input tape with infinitely many nonblank symbols on it, does it halt?) That problem is undecidable for TMs.", "I don't think the existence of every finite pattern is any evidence against Turing completeness - in the game of life it seems plausible that there is a (finite) initial state that runs every turing machine G and puts a \"halt\" marker at some position f(G) if and only if G halts.", "I hope it's clearer now. I want the rules to be deterministic and local, but exactly how they're formulated is less imporant.", "typo: I meant \"$\\delta \\in S$\"", "Okay, second round. There are Generalized Cellular Automata, where each cell $x$ transitions based on a \"stencil\" $\\delta_x$, which is a neighborhood of tiles around $x$, and the size and shape of these neighborhoods may vary from cell $x$ to cell $y$. Is a Penrose CA a generalization of GCA's, where now each $x$ has a stencil set $S$ (perhaps common to all cells) and some $\\delta \\subseteq S$ can be applied to $x$ and its surrounding cells? This then leads to multiple possible tilings of the plane for a single start state. Does every (\"nice\"?) start state converge to a stable tiling?", "The replacement rules are indeed connected templates of finite size. The problem with your periodic \"halting\" state is that it can't happen. You always can find any specific finite pattern in a Penrose tiling, so if any Penrose tiling CA has a starting state that is eventually periodic, then any starting state for that Penrose tiling CA is periodic with that period.", "I am confused by your definition, sorry. Is a Penrose CA just like a normal (2D) CA, except that the replacement rules are connected templates of finite size, instead of a local transition function that is applied cell by cell? So there's an initial configuration at time step 0 that specifies the input, and the connected templates specify the program? One way small models of computation specify \"halting\" is to repeat a special state infinitely often. So maybe if the simulated TM halts the Penrose CA keeps recreating a certain pattern, and if the TM runs forever, anything goes."], "comment_count": 14, "category": "Science", "diamond": 0} {"question_id": "35", "site": "cstheory", "title": "NP-Hardness of 4-cycle packing problem in complete bipartite digraph?", "body": "A directed complete bipartite graph is a bipartite graph where there is exactly one directed edge between any two vertices from its two different parts. In other words, it's an orientation of a complete bipartite graph. \n\nGiven a directed complete bipartite graph, we are asked to find the largest set of edge-disjoint 4-cycles. Note that these 4-cycles can share node(s).\n\nIs this problem NP-Hard? I tried to reduce from 2P2N-3SAT to this problem, but failed. On the other hand, I found no polynomial algorithm could solve it optimally.\n", "link": "https://cstheory.stackexchange.com/questions/45998/np-hardness-of-4-cycle-packing-problem-in-complete-bipartite-digraph", "tags": ["graph-theory", "np-hardness"], "votes": 14, "creation_date": "2019-12-10T00:40:50", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "36", "site": "cstheory", "title": "Is there a P-complete language X such that succinct-X is in P?", "body": "I came across a paper called \"A Note on Succinct Representation of Graphs\". It seems that in the discussion section they claim that for any problem $X$ that is $\\mathrm{P}$-hard under projections, $\\mbox{succinct-X}$ is $\\mathrm{EXP}$-hard.\n\nThis made me wonder if the theorem fails for more complicated kinds of reductions. This led me to the following.\n\n> **My Question**\n> \n> Are there any problems $X$ such that $X$ is $\\mathrm{P}$-complete under logspace reductions and $\\mbox{succinct-X}$ is in $\\mathrm{P}$?\n\n**Additional Questions:**\n\n(1) If this is an open problem, then what are the implications if such an $X$ existed?\n\n(2) If this is an open problem, is it still open is we weaken the requirement to $\\mbox{succinct-X}$ in $PH$?\n\n(3) It seems that such an $X$ would also satisfy that $X$ is $\\mathrm{P}$-hard under logspace reductions, but not $\\mathrm{P}$-hard under projections. Are such problems known to exist?\n\n(4) I might be mistaken, but I think that I have a construction for such an $X$ assuming that $NP = L$. Would this be interesting?\n\n**Note:** I'm still learning about projections so please let me know if I made any mistakes. Thank you!\n", "link": "https://cstheory.stackexchange.com/questions/42185/is-there-a-p-complete-language-x-such-that-succinct-x-is-in-p", "tags": ["cc.complexity-theory", "complexity-classes", "reductions", "succinct"], "votes": 14, "creation_date": "2019-01-09T22:34:19", "comments": ["Werent p-projections used in Valiants' original completeness and algebra paper?", "@EmilJeřábek Yes, I noticed this too. :)", "Seeing that projections are a restricted class of polylogtime reductions, let me add that what I wrote above about logtime reductions also applies to polylogtime reductions.", "@EmilJeřábek Thank you for the clarification! I really appreciate it.", "Yes, that's it. (Note that logtime Turing machines are defined so that they receive input by means of a query tape that requests individual bits instead of the usual input tape, hence they readily take input represented by a circuit.) For pspace-succinct-$X$, one can take a variant of succinct-$X$ where the input is not represented by an ordinary Boolean circuit, but by a quantified Boolean formula (or circuit).", "@MichaëlCadilhac Thanks Michaël! I appreciate the helpful suggestions. :)", "@user3483902 Thank you very much for the good questions! I'm still learning about projections so I think it's best if I refer you to a reference that includes a definition for time bounded projections: theoryofcomputing.org/articles/v013a004", "@EmilJeřábek In regards to the second part of your comment, did you mean the following? If $X$ is $P$-complete under logspace reductions, it doesn't necessarily mean that $\\mbox{succinct-X}$ is $EXP$-hard. But, $\\mbox{pspace-succinct-X}$ would be $EXP$-hard where $\\mbox{pspace-succinct}$ is a more powerful notion of succinctness (that we haven't formally defined) that offers a higher compression ratio in some cases.", "@EmilJeřábek Now, although $x$ has length $n$ and $pad$-$x$ has length $2^n$, $pad$-$x$ can be encoded as a circuit of size $n$. Further, because $r$ is in dlogtime, it can be represented by a circuit of size $n^2$ on inputs of size $2^n$. And, it seems that we can somehow combine these two circuits to get a circuit $c$ of size roughly $n^3$ that encodes $\\mbox{r(pad-x)}$ so that $x \\in L$ iff $c \\in \\mbox{succinct-X}$.", "@EmilJeřábek Thanks for the helpful comment! So if I understand the first part correctly, the proof would go something like this? Given a language $X$ that is $P$-complete under dlogtime reductions and a language $L$ in $DTIME(2^n)$. Consider padding $L$ to get a language $pad$-$L$. Since $pad$-$L$ is in $P$, there is a dlogtime reduction $r$ from $pad$-$L$ to $X$. We have that for every input string $x$, $x \\in L$ iff $pad$-$x \\in pad$-$L$ iff $\\mbox{r(pad-x)} \\in X$.", "A simple padding argument shows that if $X$ is P-complete under dlogtime reductions, then succinct-$X$ is EXP-complete (in fact, still under dlogtime reductions). I thought at first this would generalize to less restrictive notions of reductions so that, e.g., P-completeness of $X$ under logspace reductions would imply EXP-completess of succinct-$X$ under PSPACE reductions, but it does not seem to work that way; rather, it gives the EXP-completeness under polytime (or even more efficient) reductions of \"PSPACE-succinct\" $X$, if you get what I mean by that.", "what is a \"projection\" , does it pertain to reductions only - reductions with logspace requirements have this property, or something that imposes succintness to the representation?", "My 2¢ here: Question 3 certainly looks like a prerequisite, so it may be worth asking separately. If this doesn't get an answer, you may want to contact P-hardness specialists such as P. McKenzie."], "comment_count": 13, "category": "Science", "diamond": 0} {"question_id": "37", "site": "cstheory", "title": "Fourier spectrum of the parity of two monotone Boolean functions", "body": "This is a question that I've been pondering, on and off, for a while, and unsuccessfully. I'd be very interested in any insight regarding this conjecture. (Or rather, these conjectures.)\n\nRecall that, given a Boolean function $f\\colon \\\\{-1,1\\\\}^n \\to \\\\{-1,1\\\\}$, the Kahn—Kalai—Linial theorem states that $$\\max_{i\\in[n]} \\operatorname{Inf}_i f \\geq \\operatorname{Var}[f]\\cdot \\Omega\\\\!\\left(\\frac{\\log n}{n}\\right) \\tag{1} $$ from which we get that, for any _monotone_ Boolean function $f\\colon \\\\{-1,1\\\\}^n \\to \\\\{-1,1\\\\}$, $$\\max_{i\\in[n]} \\widehat{f}(i) \\geq \\operatorname{Var}[f]\\cdot \\Omega\\\\!\\left(\\frac{\\log n}{n}\\right) = (1-\\widehat{f}(0)^2)\\cdot \\Omega\\\\!\\left(\\frac{\\log n}{n}\\right) \\tag{2} $$ (writing $\\widehat{f}(i)$ for $\\widehat{f}(\\\\{i\\\\})$, $i\\in\\\\{0,\\dots,n\\\\}$). Moreover, (2) is tight, as shown by considering the $\\textsf{Tribes}_n$ function. In particular, this implies that $$ W^{(0)}[f]+W^{(1)}[f] = \\sum_{i=0}^n \\widehat{f}(i)^2 = \\Omega\\\\!\\left(\\frac{\\log^2 n}{n^2}\\right) \\tag{3} $$ for any monotone Boolean function $f\\colon \\\\{-1,1\\\\}^n \\to \\\\{-1,1\\\\}$, where $W^{(k)}[f] \\stackrel{\\rm def}{=} \\sum_{S: \\lvert S\\rvert =k} \\widehat{f}(S)^2$.\n\nNow, consider _two_ monotone Boolean functions $f,g\\colon \\\\{-1,1\\\\}^n \\to \\\\{-1,1\\\\}$, and let $h\\stackrel{\\rm def}{=}fg$ be their parity. It is easy to see that we could have $W^{(0)}[h]+W^{(1)}[h]=0$, e.g. by considering $f,g$ to be two different dictator functions (but in that very specific case, $W^{(2)}[h]=1$). _But must there be some non-negligible Fourier mass on the first 3 Fourier levels, then?_\n\n> What can we say about $W^{(0)}[h]+W^{(1)}[h]+W^{(2)}[h] = \\sum_{S: \\lvert S\\rvert \\leq 2} \\widehat{h}(S)^2$?\n\n**Conjecture 1.** For any two monotone Boolean functions $f,g\\colon \\\\{-1,1\\\\}^n \\to \\\\{-1,1\\\\}$, one must have $W^{(0)}[fg]+W^{(1)}[fg]+W^{(2)}[fg] > 0$.\n\nActually, I'd be actually inclined to believe the following stronger statement:\n\n**Conjecture 2.** For any two monotone Boolean functions $f,g\\colon \\\\{-1,1\\\\}^n \\to \\\\{-1,1\\\\}$, one must have $W^{(0)}[fg]+W^{(1)}[fg]+W^{(2)}[fg] \\geq \\frac{1}{\\operatorname{poly}(n)}$.\n\n_Side note:_ this would have implications about _weak learning_ of the class of \"2-monotone\" functions, which are exactly those functions obtained as parity or anti-parity of $2$ monotone functions. (See e.g. [BCOST15].) But more importantly, this is a very simple-looking question, that has been nagging at me for way too long.\n\n* * *\n\n[BCOST15] Eric Blais, Clément L. Canonne, Igor Carboni Oliveira, Rocco A. Servedio, Li-Yang Tan. _Learning Circuits with few Negations._ APPROX-RANDOM 2015: 512-527\n", "link": "https://cstheory.stackexchange.com/questions/37722/fourier-spectrum-of-the-parity-of-two-monotone-boolean-functions", "tags": ["boolean-functions", "fourier-analysis", "monotone"], "votes": 14, "creation_date": "2017-03-07T06:55:43", "comments": ["@daniello As far as I can tell, not much, at least in terms of blackbox interpretation of the results. The question above would have applications to weak learning (learning to advantage $1/2+1/\\mathrm{poly}(n)$), while the paper you link shows lower bound on testing (and a relaxation, parameterized testing). As far as I am aware, there is no direct connection between the two (the general connection is \"(strong) learning implies testing\").", "Can you shed some light on what arxiv.org/pdf/1705.04205.pdf means for this question?"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "38", "site": "cstheory", "title": "Is it possible to find the median with a linear size sorting network?", "body": "Is there a sorting network that makes only $O(n)$ comparisons and finds the median?\n\nThe AKS sorting network sorts with $O(\\log n)$ parallel steps, but here I am only interested in the number of comparisons. The median of medians algorithm finds the median with $O(n)$ comparisons, but it cannot be implemented as a sorting network.\n\nRemark. In fact, in a recent work, we needed a version that is less powerful than pairwise comparison and resembles sorting networks. In our model one could input two elements, $a$ and $b$, and the output was either \"a\" or \"b\", such that the output is at least as big as the other number. (In case of equality, either one of them, and we are interested in worst case complexity.) In this variant we could prove that there is a solution with $O(n)$ comparisons. Of course I am also interested if anyone knows anything about this model ever being studied.\n", "link": "https://cstheory.stackexchange.com/questions/27765/is-it-possible-to-find-the-median-with-a-linear-size-sorting-network", "tags": ["ds.algorithms", "sorting", "sorting-network", "selection"], "votes": 14, "creation_date": "2014-12-09T06:40:04", "comments": ["Actually the comparator circuit model is a bit different, since you are allowed to repeat inputs and their negations. What you are looking for is a comparator network for majority.", "@Yuval Your result seems to be the only hit on google for \"comparator circuit complexity\" and even there I cannot see the definition as the minimum number of gates needed, but I suppose you mean the same thing as I do.", "Equivalently, you are asking what is the comparator circuit complexity of (Boolean) majority."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "39", "site": "cstheory", "title": "Exponential-time factorization of polynomials", "body": "Let an _explicit_ field be a field for which equality is decidable (in some standard model of computation). I am interested in the factorization of univariate polynomials over an explicit field.\n\nIt is known that over some explicit fields, testing irreducibility of polynomials is undecidable [1]. Though, over many fields, the irreducible factorization of polynomials is known to be computable in polynomial time, either deterministically (over $\\mathbb Q$ [2] or any number field) or probabilistically (over finite fields [3]). Here the size of the input is the size of the _dense representation_ of the polynomial, that is the list of all its coefficients.\n\n> Is there an explicit field for which\n> \n> * either the best known algorithm has an exponential running time,\n> * or (better) there exists an exponential lower bound for irreducibility testing?\n> \n\n\n**Quick clarification:** There may be very different _explicit fields_ , depending on the complexity (for instance) on the equality test or of the standard operations (addition, multiplication). Actually, I am not aware of any result about polynomial factorization which is not a polynomial time upper bound, or undecidability. So I am interested in any kind of different result (exponential time is just an example). Conditional lower bounds are of interest too.\n\n**References.**\n\n[1] A. Fröhlich et J. C. Shepherdson (1955): [On the factorisation of polynomials in a finite number of steps](http://dx.doi.org/10.1007/BF01180640). \n[2] A. K. Lenstra, H. W. Lenstra Jr., L. Lovász (1982): [Factoring polynomials with rational coefficients](http://dx.doi.org/10.1007/BF01457454). \n[3] E.R. Berlekamp (1967): [Factoring Polynomials Over Finite Fields](http://www3.alcatel-lucent.com/bstj/vol46-1967/articles/bstj46-8-1853.pdf).\n", "link": "https://cstheory.stackexchange.com/questions/18502/exponential-time-factorization-of-polynomials", "tags": ["algebraic-complexity", "polynomials", "exp-time-algorithms"], "votes": 14, "creation_date": "2013-07-30T01:12:45", "comments": ["Thanks @JoshuaGrochow! I found the paper incredibly easy to read, even though my German lessons are quite old now...", "Depending on your definition of \"explicit field\", the result you cite as [1] was also proven by van der Waerden much earlier (1930). Amazingly, he proved this before the notion of computability had been formally defined! See: ams.org/mathscinet-getitem?mr=1512605", "I was aware while writing my question that some this question should be addressed. Actually, I am interested in both cases. I would be very much interested by the implication to ETH you mention. In other words, I am also interested in conditional lower bounds.", "What do you mean by \"explicit field\" ? Are the operations is the field given by oracles or by algorithms ? I am asking because if the field's operations are given by efficient algorithms proving an exponential lower bound for factoring seems to imply of proof of ETH."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "40", "site": "cstheory", "title": "Pseudorandom functions in ACC^0?", "body": "In the lower bound result by Ryan Williams (Non-uniform $\\mathsf{ACC}$ circuit lower bounds), there is a mention of \"little evidence that Pseudorandom function generators exist in $\\mathsf{ACC}^0$. Is there any development in this regard that might be of interest? (Even old results implying something along these lines will be great.)\n", "link": "https://cstheory.stackexchange.com/questions/10220/pseudorandom-functions-in-acc0", "tags": ["cc.complexity-theory", "reference-request", "cr.crypto-security", "circuit-complexity", "natural-proofs"], "votes": 14, "creation_date": "2012-02-15T01:26:37", "comments": ["The following paper doesn't answer your question but does show somewhat simpler candidate PRFs ccs.neu.edu/home/viola/papers/spn.pdf"], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "41", "site": "cstheory", "title": "DPLL and Lovász Local Lemma", "body": "Let $\\varphi$ be a CNF formula. Suppose that each of $\\varphi$'s clauses consist of exactly $t$ literals (and, moreover, all literals within one particular clause correspond to different variables). It is well known that if every clause has less than $2^t / e$ clauses that share variables with it, then $\\varphi$ is satisfiable (let us call such formulae _easy_). Satisfiability can be proved easily using [Lovász local lemma](http://en.wikipedia.org/wiki/Lov%C3%A1sz_local_lemma). Moreover, using [a recent result](http://arxiv.org/abs/0903.0544) by Moser and Tardos one can show that one of the satisfying assignments can be found in polynomial expected time using the following very simple procedure:\n\n * Pick a random assignment\n * While there exists an unsatisfied clause, resample all its variables.\n\n\n\nOn the other hand, most of modern SAT solvers are [DPLL](http://en.wikipedia.org/wiki/DPLL_algorithm)-based. It means that they try to find a satisfying assignment using brute force with two simple prunings:\n\n * If a formula contains clause with one literal, then we can fix it.\n * If one variable occurs in a formula only with (or without) negation, then we can fix it.\n\n\n\n**The question: is it true that a version of DPLL that splits on random variables finds a satisfying assignment of any easy formula in polynomial expected time?**\n", "link": "https://cstheory.stackexchange.com/questions/7720/dpll-and-lov%c3%a1sz-local-lemma", "tags": ["cc.complexity-theory", "ds.algorithms", "sat"], "votes": 14, "creation_date": "2011-08-13T02:32:29", "comments": ["@ilyaraz Still hoping you'll post an official answer to this question.", "@Stasys Actually, Dmitriy Itsykson (who you probably know) convinced me that the answer is negative. I'll elaborate on it soon. Argument is more or less the following: we pick a hard unsatisfiable formula and pad it in order to make it easy.", "I think the answer to your question is (apparently) \"No\". Hirsch (2000) exhibited a simple CNF with exactly one satisfying assignment such that no \"randomized DPLL\" (randomized local search algorithm) can find it in expected polynomial time. See my writeup thi.informatik.uni-frankfurt.de/~jukna/BFC-book/sat-chapter.‌​pdf User: friend, password: catchthecat."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "42", "site": "cstheory", "title": "Which monotone DNFs are evasive?", "body": "A Boolean function $\\phi$ on variables $X$ is _[evasive](https://en.wikipedia.org/wiki/Evasive_Boolean_function)_ if every decision tree for $\\phi$ has height $|X|$. In other words, for any strategy that picks variables of $X$ and asks for their value, an adversary can answer the queries such that the strategy needs to ask about the value of all variables of $X$ to know the value to which $\\phi$ evaluates.\n\nA _monotone DNF_ is a disjunction of conjunction of variables, representing a monotone Boolean function on the set $X$ of variables that it uses.\n\n**Which monotone DNFs are evasive?** Is there a characterization of them? What is the complexity, given a monotone DNF, of determining whether it is evasive?\n\nExamples of evasive monotone DNFs include single-clause DNFs, e.g., $x \\land y \\land z$, where we must query the value of every variable (and the adversary can answer $1$) until the last one which determines the value of the formula. For another example, take $(x \\land y) \\lor (y \\land z)$: if I ask about $x$ or $z$ then the opponent answers $0$ and I'm left with evaluating the other clause, if I ask about $y$ the opponent answers $1$ and I must evaluate $x \\lor z$ which requires me to query both variables.\n\nHowever, not all monotone DNFs are evasive. For instance, for $(s \\land t) \\lor (t \\land u) \\lor (u \\land v)$, I can query variable $t$. If it is false, then I know that clauses $s\\land t$ and $t \\land u$ are false, so I can evaluate the function just knowing $u$ and $v$ (I don't need to query $s$). If it is true, then I have to evaluate $s \\lor u \\lor (u \\land v)$, and as the second clause subsumes the third, I just need to know $s$ and $u$ (I don't need to query $v$).\n\nPlaying a bit with the problem in the case of graph DNFs (with two variables per clause), it looks like for a path $(x_0 \\land x_1) \\lor \\cdots \\lor (x_{n-1} \\land x_{n})$ with $n>0$ the function is not evasive iff $n$ is a multiple of 3, and this seems to generalize to trees if there is some kind of tree structure on nodes that are 3 steps apart. For general graphs, I have no idea. I have looked for related work about evasiveness and decision tree complexity for monotone CNFs and DNFs but I didn't find anything relevant.\n\nOn graph DNFs, the problem can be seen as a kind of game on the graph of the formula. You pick a vertex $v$ (= variable) on the graph, and the opponent can choose between:\n\n * removing $v$ and all its incident edges (= $v$ evaluates to $0$, all clauses involving it are removed), or\n * removing the open ball of radius 2 centered on $v$, i.e., $v$, its adjacent edges, the neighbors of $v$, and their adjacent edges (= $v$ evaluates to $1$, all clauses involving $v$ become singleton clauses involving $v$'s neighbors, you will need to ask about each of these neighbors because each has a singleton clause involving it, but these singleton clauses subsume all other clauses involving the neighbors so you get to remove them).\n\n\n\nThe question is about characterizing and recognizing the graphs $G$ for which you have a strategy that guarantees that you win, where \"winning\" means \"reaching a graph with an isolated vertex\" (= a variable that you do not need to ask about).\n", "link": "https://cstheory.stackexchange.com/questions/44278/which-monotone-dnfs-are-evasive", "tags": ["np-hardness", "boolean-functions"], "votes": 13, "creation_date": "2019-07-16T13:28:04", "comments": ["Since the Alexander dual of the corresponding simplicial complex consists of all the independent sets of $G$, a necessary condition for evasiveness is that the independence polynomial of $G$ vanishes at $-1$.", "Possibly relevant references for the lazy: A survey of evasiveness: Lower bounds on the decision-tree complexity of boolean functions, Lecture Notes on Evasiveness of Graph Properties"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "43", "site": "cstheory", "title": "Question on Products of Graphs", "body": "Let $S_{n}(G)$, $C_{n}(G)$, $T_{n}(G)$ be the $n$-fold Strong Product, Cartesian Product and Tensor Product of a graph $G$ on $|V|$ vertices.\n\nLet the chromatic number ($\\chi(G)$) and the independence number ($\\alpha(G)$) of $G$ be known through a (possibly exponential time) algorithm. Is it known that the calculation of $\\chi(S_{n}(G))$, $\\chi(C_{n}(G))$, $\\chi(T_{n}(G))$,$\\alpha(S_{n}(G))$, $\\alpha(C_{n}(G))$ and/or $\\alpha(T_{n}(G))$ require exponential in $n$ complexity for any $n$ say even when restricted to some special graphs?\n", "link": "https://cstheory.stackexchange.com/questions/7522/question-on-products-of-graphs", "tags": ["cc.complexity-theory", "graph-theory", "graph-colouring"], "votes": 14, "creation_date": "2011-07-26T18:45:56", "comments": ["@Andrew: The first one", "By $n$-fold strong product do you mean $G\\boxtimes G\\boxtimes \\ldots \\boxtimes G$, or $G\\boxtimes K_n$?", "Here is Wikipedia's article on Hedetniemi's conjecture.", "Hi Hsien: Actually you answered part of the question. If you assume the conjecture, then we already have an $o(0)$ algorithm for those cases.", "Hi Hsien: You are correct.", "You mean other than the trivial case like $\\chi(C_n(G)) = \\chi(G)$, or $\\chi(T_n(G)) = \\chi(G)$ assuming Hedetniemi's Conjecture, right?"], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "44", "site": "cstheory", "title": "Deterministic context-free languages that can be represented as the word problem of a group", "body": "Consider a group $G$. We call $G$ virtually free is it contains a free subgroup of finite index.\n\nIf $G$ is finitely generated by some set $X \\subseteq G$ one can consider the _word problem_ $W\\\\!P(G)$ that is the formal language consisting of all words over the alphabet $X \\cup X^{-1}$ that evaluate to the unit in $G$.\n\nIt is a famous result by Mueller and Schupp that the $W\\\\!P(G)$ is context-free if and only if $G$ is virtually free.\n\nIt is also known that if $W\\\\!P(G)$ is context-free, it is deterministic context-free.\n\nMy question is: Do we know more about how the class of deterministic context-free languages that can be represented as the word problem of some group, is contained in the class of all (deterministic) context-free languages?\n", "link": "https://cstheory.stackexchange.com/questions/39954/deterministic-context-free-languages-that-can-be-represented-as-the-word-problem", "tags": ["fl.formal-languages", "gr.group-theory"], "votes": 13, "creation_date": "2018-01-11T04:49:31", "comments": ["I'm not sure if you're still interested in this, but it is known that such languages are closed under deletion and have universal prefix. This is true for all word problems, not just DCFL. There's a proof here: D. W. Parkes and R. M. Thomas. Groups with context-free reduced word problem. Communications in Algebra, 30:3143–3156, 2002.", "@MichaëlCadilhac Thank you. This answer of yours is also very interesting.", "Maybe the class of languages DLOGTIME-reducible to some WP(G) would be more robust.", "For that above question, I believe $\\{a^nb^m \\mid m < n\\}$ is not the WP of some group. I guess you'd be quite interested in Greibach's hardest language, and her notion of jump PDAs.", "No not in particular; but do you have a DCFL in mind that is not the WP of some group? For example: The WP of any free group is a DCFL and is solvable in logspace (Lipton, Zalcstein, 1977). Maybe there are more connections between WP of groups and DCFL solvable in logspace. But again, nothing in particular.", "Are you just asking whether there exists a DCFL that is not the WP of some group? Or something more general?"], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "45", "site": "cstheory", "title": "Complexity to compute the eigenvalue signs of the adjacency matrix", "body": "Let $A$ be the $n\\times n$ adjacency matrix of a (non-bipartite) graph. Assume that we are given the amplitudes of its eigenvalues, i.e., $|\\lambda_1|=a_1,\\ldots, |\\lambda_n|=a_n$, and we would like to calculate their signs. Is there a faster way of computing the signs of these eigenvalues, other than recomputing the eigenvalues themselves? \n", "link": "https://cstheory.stackexchange.com/questions/16789/complexity-to-compute-the-eigenvalue-signs-of-the-adjacency-matrix", "tags": ["cc.complexity-theory", "ds.algorithms", "linear-algebra", "spectral-graph-theory"], "votes": 13, "creation_date": "2013-03-07T12:38:48", "comments": ["Thanks, that's exactly the point. I was wondering if the knowledge of the amplitudes made the problem any simpler. I think that this could have interesting implications in problems where eigenvalues are well approximated by the squared degrees of the graph, etc.", "Nice question! For comparison, the best known running time for computing the eigenvalues themselves (upto additive error 1/4) is $O(n^3)$ (cstheory.stackexchange.com/a/2810/15)"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "46", "site": "cstheory", "title": "Name and references for balanced variant of the long code?", "body": "The [long code](https://en.wikipedia.org/wiki/Long_code_%28mathematics%29) (arising in PCP theory etc) is an encoding of a set of $k$ values, using binary strings of length $2^k$ (double exponential in the number of bits needed to specify a value), with one coordinate position in the encoding for each of the $2^k$ different binary functions on the input values. Instead, I want to consider a shorter but still doubly exponential code, with one coordinate position for each of the $\\tbinom{k}{k/2}/2$ partitions of the $k$ values into two equal-sized subsets (assuming $k$ is even). Each partition can be thought of as corresponding to two different binary functions (the functions whose inverse images form the given partition) and we choose arbitrarily which one of these two functions to use.\n\nThe minimum distance of the code (and in fact the distance between all pairs of codewords) is $\\tbinom{k-2}{(k-2)/2}$, smaller than the length of the code by only a factor of $\\tfrac{k}{2k-1}>\\tfrac{1}{2}$. For instance for $k=8$ we get a code of length 35 with minimum distance 20, which tells me matches the [Griesmer bound](https://en.wikipedia.org/wiki/Griesmer_bound) for binary linear codes of the same size (rank 3 or size 8).\n\nI am interested in this for reasons only vaguely related to coding theory, but it seems likely to me that the coding theorists have already studied this code. Unfortunately, I don't know of a good way of looking up coding schemas by the formulas for their parameters. If it has been previously studied, what is it called? And what are some references to it?\n\n(BTW, the property I actually care about, which is easy enough to prove using Sperner's theorem, is that this is the longest possible binary code in which each two positions are independent of each other, in the sense that all four possible combinations of symbols in the two positions are realized by codewords. There appears to be something in the literature called a Sperner code but it's transposed from this — each codeword has equally many zeros and ones, rather than the property of the code I'm looking at, where each bit position has equally many codewords for which that position is a 0 or 1.)\n", "link": "https://cstheory.stackexchange.com/questions/19602/name-and-references-for-balanced-variant-of-the-long-code", "tags": ["reference-request", "coding-theory"], "votes": 13, "creation_date": "2013-10-31T16:11:44", "comments": ["Ok, if it wasn't in the literature before, it is now. See section 6 of arxiv.org/abs/1303.1136 — this code gives the smallest set of vertices to delete from a hypercube graph in order to make all remaining hypercube subgraphs have at most 1/8 of the number of vertices of the original graph."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "47", "site": "cstheory", "title": "Generalizing limit-colimit coincidence to Scott-continuous adjunctions: any uses?", "body": "In Abramsky and Jung's 1994 handbook chapter on denotational semantics, after proving that the limit and colimit of expanding sequences exist and coincide, they have the following to say about generalizing this theorem: \n\n> We can generalize Theorem 3.3.7 [the limit-colimit coincidence of expanding sequences in DCPO] [...] we can use general Scott-continuous adjunctions instead of e-p-pairs. [...] By the passage from embeddings to, no longer injective, lower adjoints, we allow domains not only to grow but also to shrink as we move on in the index set. Thus points, which at some stage looked different, may at a later stage be recognized to be the same. [...] For the main text, we must remind ourselves that this generalization has so far not found any application in semantics. \n\nIt's been over fifteen years since this chapter was written. Have there been any applications of this generalization in semantics since then?\n", "link": "https://cstheory.stackexchange.com/questions/2528/generalizing-limit-colimit-coincidence-to-scott-continuous-adjunctions-any-uses", "tags": ["pl.programming-languages", "semantics", "denotational-semantics", "domain-theory"], "votes": 13, "creation_date": "2010-10-28T10:07:39", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "48", "site": "cstheory", "title": "The best known upper bound for two-way probabilistic finite automata with one-counter", "body": "It is known that the class of languages recognized by two-way deterministic finite automata with one-counter (2D1CAs) is a proper subset of $ \\mathsf{L} $ (deterministic log-space): A 2D1CA can run at most $ O(n^2) $ steps if it halts on a given input, and there is a language in $ \\mathsf{L} $, i.e. \\begin{equation} \\mathtt{DG} = \\lbrace w_0 \\$ w_1 \\$ \\cdots \\$ w_k \\mid k \\geq 1, w_{i \\in \\lbrace 0, \\ldots, k \\rbrace} \\in \\lbrace a,b\\rbrace^* , \\mbox{ and } w_0=w_i \\mbox{ for some } 1 \\leq i \\leq k \\rbrace, \\end{equation} which cannot be recognized by any 2D1CA [(Duris and Galil, 1982)](http://www.sciencedirect.com/science/article/pii/0304397582900871).\n\nWhat happens if such an automaton can use a coin?\n\nMore precisely, what is the best known upper bound for the class of languages recognized by two-way **probabilistic** finite automata with one-counter (2P1CAs) with bounded error?\n", "link": "https://cstheory.stackexchange.com/questions/11080/the-best-known-upper-bound-for-two-way-probabilistic-finite-automata-with-one-co", "tags": ["cc.complexity-theory", "reference-request", "automata-theory", "upper-bounds"], "votes": 14, "creation_date": "2012-04-13T13:36:31", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "49", "site": "history", "title": "Are there any references to entombed animals in ancient India?", "body": "The 13th century Hindu philosopher Arulnandi Shivacharya wrote a work called the Shiva Jnana Siddhiyar, which among other things contains a refutation of Buddhist philosophy. In [this excerpt](https://i.sstatic.net/KxNh7.jpg), various Buddhist theories of what the ultimate cause of the body is are refuted:\n\n> If you say bodies are formed from the mixture of four elements, then these cannot unite as their natures are opposed to each other. **If you say they are formed by the union of blood and semen, then account for toads being found in the heart of rocks, and worms in the heart of trees.** If you say the real cause is good and bad Karma, then these, being opposed, cannot join and form bodies. If food is the cause, then the food which in youth develops the body is not capable of preventing decay in old age. If intelligence is the cause, then that which is formless Chaitanya cannot assume Achaitanya (non-intelligent) form. If you assert that bodies are formed from nothing, then we could cull flowers from the sky.\n\nMy question is about the part in bold, which is arguing that intercourse cannot be the ultimate cause of the body, because there are toads which are found in the center of solid rock, and they couldn't have been produced by intercourse since there's no way that toads could have gone inside a solid rock and reproduced.\n\nThis statement struck me as rather odd, but then I found out from the Wikipedia article [Entombed animal](https://en.wikipedia.org/wiki/Entombed_animal) that claims of toads being found in solid rock are surprisingly common across many cultures:\n\n> Entombed animals are animals reportedly found alive after being encased in solid rock, or coal or wood, for an indeterminate amount of time. The accounts usually involve frogs or toads. The reputed phenomenon, sometimes called \"toad in the hole\", has been dismissed by mainstream science, but has remained a topic of interest to Fortean researchers.... References to entombed animals have appeared in the writings of William of Newburgh, J. G. Wood, Ambroise Paré, Robert Plot, André Marie Constant Duméril, John Wesley, and others. Even Charles Dickens mentioned the phenomenon in his journal All the Year Round. **According to the Fortean Times, about 210 entombed animal cases have been described in Europe, North America, Africa, Australia, and New Zealand since the fifteenth century.**\n\nConspicuously absent from that list of places is Asia. So my question is, are there any references in ancient India to entombed animals? The work I quoted above was written in the 13th century. But I'm wondering whether there are earlier references to this phenomenon. Arulnandi Shivachrya refers to this phenomenon as if it's common knowledge, so I'm guessing older sources would mention it as well.\n", "link": "https://history.stackexchange.com/questions/34768/are-there-any-references-to-entombed-animals-in-ancient-india", "tags": ["india", "science", "animals", "ancient-india", "folklore"], "votes": 16, "creation_date": "2017-01-05T23:06:24", "comments": ["THe proposed edit seems to replace a question with an answer. I'm confused.", "Unless someone can say what's available in the article I linked to above, there's not much else that can be said—at least the last time I researched this, I spent a fair bit of time and didn't uncover anything useful, but it's so very hard to prove a negative.", "This, though not readily available online, might be useful. Didn't find anything else that would refer to this with a quick search.", "It reminds me Schrödinger's cat..."], "comment_count": 4, "category": "Culture & Recreation", "diamond": 0} {"question_id": "50", "site": "history", "title": "What was the first overland road from Sweden to Finland?", "body": "The [Swedish post road](https://en.wikipedia.org/wiki/King%27s_Road_\\(Finland\\)) from Norway, through Sweden, used the Åland archipelago to pass into Sweden, and this is easily found (evidence of) in the south of Finland to the present day. **When (and where) was the first overland route constructed overland from Sweden into (Swedish) Finland?**\n\nThe only (poor) evidence I have for roads existing in the north is by the [War of 1808–9](https://en.wikipedia.org/wiki/Finnish_War) where Russian forces were planning to advance overland into Sweden (along with an army group advancing across the Gulf of Bothnia). One of the WP article's references does say \"In addition, several new good roads had been built into Finland greatly reducing the earlier dependency on naval support for any large operation in Finland.\" but it doesn't specify where these roads were.\n\nI looked through all articles on [Swedish](https://en.wikipedia.org/wiki/Category:Roads_in_Sweden) and Finnish road networks on the English Wikipedia, and the most I found was a reference to a ['Finnmark path'](https://en.wikipedia.org/wiki/Arctic_Ocean_Highway) which was meant to have gone from Finnish Lapland to Finnmark in the 16th century. The [Finnish WP article](https://fi.wikipedia.org/wiki/J%C3%A4%C3%A4merentie) for the same page _does not_ mention the Finnmark path at all, and I couldn't find anything else on a road of that name.\n\nI understand—from the comments—that the term \"road\" can be meaningless without further definition for a period much longer than a few centuries ago. For clarity, I'm defining road as purpose-built (or purpose-developed) and used regionally for that purpose, such as the post road mentioned above. This would mean hunting tracks that slowly developed don't count, while a merchant-led endeavour to expand (and maintain) the tracks between two townships would.\n", "link": "https://history.stackexchange.com/questions/62286/what-was-the-first-overland-road-from-sweden-to-finland", "tags": ["transportation", "sweden", "finland"], "votes": 6, "creation_date": "2020-12-21T03:13:26", "comments": ["Carl Linnaeus's 1732 trip to Lapland was recorded in a fascinating book. Of course he was not interested in the speediest or easiest route, so much as the most interesting, but it does give first-hand detailed account of his travels & travel conditions. Wikipedia summary: \"Linnaeus travelled clockwise around the coast of the Gulf of Bothnia, making major inland incursions from Umeå, Luleå and Tornio. He returned from his six-month-long, over 2,000 kilometres (1,200 mi) expedition in October, having gathered and observed many plants, birds and rocks\".", "No. It was a national road net (for low values of \"road\"). There is a map from 1742, but I cannot access it here. Whatever. There were \"roads\" a long time ago, between 1809 and 1917 it was a hostile border, and then after Finland became independent they opened a railroad.", "@TomasBy: That's not what I wrote. \"local roads\" means those used to get around town, and its immediate vicinity. Emphasis on immediate vicinity. Magnus Ericson's law basically requires that towns build and maintain a local urban road network for commerce.", "@PieterGeerkens are you suggesting they were not actually connected? Each man just had a small bit of private road in isolation?", "@TomasBy: Read the next sentence, these are local roads only: \"... stipulate that the land-owning farmers are responsible for each road lot, bridge, ice road or ferry. The length of the road lot depends on the size of the agricultural property.\" It would be insanity to require small isolated villages otherwise quite content to build and maintain long-haul access rods they don't themselves use or need.", "@PieterGeerkens history of roads in Sweden: \"The national law of Magnus Eriksson (1350) decrees that each landholder is responsible for a section of road, a bridge, an ice road, or a ferry.\".", "Torneå was founded in the early 17th c. I assume there were roads to it from both east and west."], "comment_count": 7, "category": "Culture & Recreation", "diamond": 0} {"question_id": "51", "site": "history", "title": "Which misdeeds of F. Reineke was J.B.B. de Lesseps "obliged to suppress"?", "body": "F. Reineke was commandant of Kamchatka from about 1780 to 1784. According to Витер's _История символов как история административного деления государства_ he was responsible for moving the territorial administration from Bolsheretesk to Nizhnekamchatsk. According to J.B.B. de Lesseps, Reineke resigned in 1784 [\"for reasons which I am obliged to suppress\"](https://archive.org/details/travelsinkamtsch01less/page/140).\n\nIt was common for Siberian administrators to be venal. According to Forsyth's _A history of the peoples of Siberia_ , the very first governor of Siberia, Matvei Gagarin, was hanged for fraud, nepotism, and selling offices; another governor in Irkutsk was relieved from office for gross corruption and replaced with his second, who was \"vicious\" and \"ruthless\".\n\n**Was Reineke guilty of the same old corruption and exploitation, or something worse?** I'm skeptical that de Lesseps, a Frenchman, had any reason to suppress a story of Russian corruption.\n", "link": "https://history.stackexchange.com/questions/55995/which-misdeeds-of-f-reineke-was-j-b-b-de-lesseps-obliged-to-suppress", "tags": ["russia", "18th-century", "government", "crime", "kamchatka"], "votes": 6, "creation_date": "2019-12-31T20:47:10", "comments": ["One more thing: According to this source, in 1786 F. Reineke moves from Kamchatka to Irkutsk and in 1798 retires from the civil service. All this suggests that his record in Kamchatka was just fine. There was some controversy related to his subordinates, as well as some misdeeds in the Russian postal service in Kamchatka during Reineke's tenure. Maybe this is what Lesseps alludes to.", "The original says \"деятельный\"; \"Active\" might be a better translation. You can also try to get the book \"Правители Камчатки : 1700-2007\" via an interlibrary loan and look there for further information.", "@MoisheKohan Very interesting. As Reineke was a public servant, I wonder if \"entrepreneurial\" was a backhanded compliment.", "For what it is worth, A. S. Sgibnev in his detailed \"History of main events in Kamchatka, 1650 - 1856\", describes F. Reineke as \"honest and entrepreneurial.\"", "Gagarin was not hanged (just) for corruption, Russian Wikipedia mentions that he attempted to form separate Siberian kingdom : ru.wikipedia.org/wiki/…", "Thanks, Mark, for your help along the way!", "Consistently high quality questions. Thank you for your contributions to the site."], "comment_count": 7, "category": "Culture & Recreation", "diamond": 0} {"question_id": "52", "site": "history", "title": "Who owned the ship Ensayo, and what were they doing near Baja California in 1842?", "body": "According to Plummer's _The Shogun's Reluctant Ambassadors_ , in 1842, sea drifters from the _Eijū-maru_ were picked up and by the _Ensayo_ , a \"Spanish pirate ship\" with a Philippine crew. It was \"carrying illegal goods between Spain or Mexico and Manila,\" and \"[b]ecause it was on a 'wanted list', it could not enter any harbors freely.\" The Japanese men were treated badly and forced ashore on a beach in Baja California.\n\nI found the description confusing. Of Pacific ports, only the Philippines were still Spanish. The Manila galleon system had been over for decades. During Mexico's first twenty years of independence, as far as I know, it didn't trade with East Asia. The Spanish were a minor presence in the Pacific even when they controlled a lot of its coastline.\n\nWho owned _Ensayo_ and what was it up to?\n", "link": "https://history.stackexchange.com/questions/47476/who-owned-the-ship-ensayo-and-what-were-they-doing-near-baja-california-in-1842", "tags": ["19th-century", "spain", "age-of-sail", "philippines", "piracy"], "votes": 6, "creation_date": "2018-08-05T23:12:12", "comments": ["@AlbertoYagos: I concur with T.E.D.'s view.", "I had \"Vida en México de trece náufragos japoneses\" in my hands, but forgot to look for this point!!", "@AlbertoYagos - I can't always speak for all our voters, but I for one would consider what you have there acceptable, with perhaps a GoodReads link to the book in question, and an admission that you are working from memory.", "@T.E.D. I don't have access to either of those books and I can't quote more. I just have that reference. That is why it's a comment and not an answer.", "@AlbertoYagos - Any way you could see your way to expanding that to an answer? It probably wouldn't take much, as it seems like you already put in enough research for it to be one. You ought to reap the reward for that work, and Aaron may eventually even want to accept it.", "According to Del tratado al tratado. 120 años de relaciones entre Japón y México by Aurelio Asiain, 2008 (p. 46), the Ensayo didn't commit piracy but smuggling. She smuggled goods from and to Mexico. Adiain's source seems to be Sano, Yoshikazu, Vida en México de trece náufragos japoneses, 1842."], "comment_count": 6, "category": "Culture & Recreation", "diamond": 0} {"question_id": "53", "site": "history", "title": "What would a criminal defense attorney charge in late 19th century America?", "body": "What would a criminal defense attorney charge on the [American frontier](https://en.wikipedia.org/wiki/American_frontier#The_Postbellum_West) in a murder trial in 1885? I am specifcally interested in what it would have cost three cowpokes to mount a murder defence in Rapid City, Dakota Territory 1885. I know they were defended by Colonel William Parker who was a successful lawyer and politician in the Territory at the time, but so far I have found no clue what his fees were.\n", "link": "https://history.stackexchange.com/questions/68240/what-would-a-criminal-defense-attorney-charge-in-late-19th-century-america", "tags": ["united-states", "19th-century", "law", "old-west"], "votes": 5, "creation_date": "2022-02-04T11:16:22", "comments": ["Would also significantly help if you documented your preliminary research. I imagine that the answer is highly variable - urban setting vs rural, proximity of a a railroad, the presence of skilled attorney vs having to recruit one from elsewhere, Of course, the cost will range from zero (self representation) to what the market will bear.", "Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer."], "comment_count": 2, "category": "Culture & Recreation", "diamond": 0} {"question_id": "54", "site": "linguistics", "title": "In which non-Sinitic languages do negative clauses retain older constituent order in SVC-derived complex predicates?", "body": "Many complex predicates are historically derived from serial verb constructions. This is not only true of the Sinitic family. For example, in Saramaccan (Byrne 1987, as cited in Givón 2009):\n \n \n (1) a bi-fefi di-wosu kaba.\n he TNS-paint the-house finish\n He finished painting the house.\n \n\nHowever, the Chinese constructions are more innovative in that the object has been moved to the end of the sentence, for the most part. Yet in most dialects of modern Chinese except Mandarin, complex predicates also sometimes preserve the older V + O + C order similar to the Saramaccan example in (1). This is most common in negative clauses. (The C is what is traditionally termed a _complement_ in Chinese linguistics, but it's more like a _satellite_ in Talmy's sense, and is developed from the 'lighter' verb of an asymmetric serial verb construction). An example from Cantonese is (2):\n \n \n (2) nei wan ngo m dou\n you find me NEG arrive\n You cannot find me.\n \n\nMy question: I know negative clauses are more conservative in general, but are there specific cases outside the Sinitic branch where object placement in complex predicates is more conservative in negative clauses? Thanks!\n\nGivón, T. (2009). _The genesis of syntactic complexity: Diachrony, ontogeny, neuro-cognition, evolution_. John Benjamins Publishing.\n", "link": "https://linguistics.stackexchange.com/questions/21555/in-which-non-sinitic-languages-do-negative-clauses-retain-older-constituent-orde", "tags": ["syntax", "historical-linguistics", "word-order", "negation"], "votes": 7, "creation_date": "2017-04-06T01:26:05", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "55", "site": "linguistics", "title": "Are there any languages where the first person cannot be an object?", "body": "In some languages, nouns low on the animacy hierarchy, particularly inanimates cannot surface as A, and if a situation arises where they are underlyingly A, some reparative strategy such as a passive must be used. Some theorise that this was the case in early PIE, and a friend of mine mentioned Algonquian and Japhung Qiangic in the conversation that lead me to consider this:\n\nAre there any languages where a similar rule is found at the other end of the animacy hierarchy, such that the 1st person is defective and cannot be O, and an underlying 1st person O requires a syntactic derivation to be used, moving the first person either to S (a passive) or oblique status (an antipassive)?\n\n[Chukchi](https://en.wikipedia.org/wiki/Chukchi_language) seems to come close, as it has a verbal prefix _(i)ne- ~ (e)na-_ , which Comrie(1979) analyses as a \"de-ergativising prefix\" which in some cases acts like an antipassive is required with 1st person singular Os when the A is not 3rd person plural (in which case a 1sg.S/O suffix _is_ used), however it's not quite a full antipassive as in these cases, as while there is a change to verbal person agreement (absolutive forms are used to mark underlying A) there is not change to nominal case marking, and furthermore, in the \"present II\"-tense it's used with a much larger set of A/P combinations. Further examples of borderline cases like Chukchi would be fine, but what I'm really interested in is if there are any more clear-cut examples where the construction involved is clearly a syntactic derivation.\n\n* * *\n\nComrie, B. 1979: _Degrees of Ergativity: Some Chukchee evidence_ , pp. 219-40 in Plank, F. 1979 (ed.): _[Ergativity - Towards a Theory of Grammatical Relations](https://drive.google.com/file/d/1f3eXlLDI2qC3yeMG1YStsmtH5RSXzHLE/view)_ , Academic Press\n", "link": "https://linguistics.stackexchange.com/questions/27161/are-there-any-languages-where-the-first-person-cannot-be-an-object", "tags": ["syntax", "grammar", "list-of-languages", "grammatical-object", "person"], "votes": 7, "creation_date": "2018-02-16T13:02:04", "comments": ["Inanimate objects not being able to be agents makes some sense (if your language doesn't use a lot of metaphor), but why would any language prohibit 1st person objects? I can imagine this as a kind of taboo, but not as an actual ungrammaticality."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "56", "site": "mathoverflow", "title": "2, 3, and 4 (a possible fixed point result ?)", "body": "The question below is related to the classical [Browder-Goehde-Kirk fixed point theorem](https://www.researchgate.net/publication/38323269_An_elementary_proof_of_the_fixed_point_theorem_of_Browder_and_Kirk).\n\nLet $K$ be the closed unit ball of $\\ell^{2}$, and let $T:K\\rightarrow K$ be a mapping such that $$\\Vert Tx-Ty\\Vert _{\\ell^{4}}\\leq\\Vert x-y\\Vert _{\\ell^{3}}$$ for all $x,y\\in K$.\n\nIs it true that $T$ has fixed points?\n", "link": "https://mathoverflow.net/questions/18264/2-3-and-4-a-possible-fixed-point-result", "tags": ["fa.functional-analysis", "open-problems", "banach-spaces"], "votes": 79, "creation_date": "2010-03-15T06:26:45", "comments": ["$FP(p,p,\\infty)$ is false for any $p$. We can let $(Tx)_i = x_{i-1}$ for $i<1$ and $(Tx)_0 =(1/2)( 1 - \\sum_i x_i^2)$. Then $||Tx||_{\\ell_2}^2 = (1/4)(1-||x||_{\\ell_2}^2)^2+ x_{\\ell_2}^2 \\leq 1$ since $x_{\\ell_2}\\leq 1$ and, because each coordinate is a $1$-Lipschitz function in the $\\ell_2$ norm, the whole thing is a $1$-Lipschitz transformation from $\\ell^\\infty$ to $\\ell^2$.", "As far as I understood, the BGK (Browder-Göhde/Göbel-Kirk) fixed point theorem states that every non-expansive self-mapping on a non-empty, closed and convex subset of a uniformly convex Banach space has a fixed point.", "What are the related known results?", "Any progress on this?", "Would someone please state the classical Browder-Goehde-Kirk fixed point theorem?", "@Fabrizio Polo I'm not claiming that. Just that, e.g., $FP(2,4,3)$ holds, via BGK. Please read carefully my comment.", "@Ady: Since your question is whether or not $FP(2,3,4)$ holds, I'm confused. You seem to be claiming that BGK resolves your question.", "BGK $\\Longrightarrow$ FP$(p,q,r)$ for 1 < p $\\leq$ r $\\leq q< \\infty $ , e.g.", "For fun, I started considering variants of this question but made no real progress on them either. Let ${\\bf FP}(p,q,r)$ be the same statement with $2,3,4$ replaced by $p,q,r$. The BGK fixed point theorem gives us ${\\bf FP}(p,p,p)$ for $1 < p < \\infty.$ For what other values of $p,q,r$ can you prove or disprove this statement? Do you know of a counterexample to $FP(1,\\infty, \\infty)$?", "Why did this question get a downvote? Seems interesting to me.", "I don't think it does. I just meant to make sure other readers could easily tell what you were asking; I didn't downvote and I hope I didn't encourage anyone else to.", "If that deserves a \"-1\", it's cool to me. ", "It would be easier to read if you just wrote $\\|Tx−Ty\\|_4 \\le \\|x−y\\|_3$. It took me a minute to get the point of your question since I could barely see the difference between the supersubscripts"], "comment_count": 13, "category": "Science", "diamond": 1} {"question_id": "57", "site": "mathoverflow", "title": "Grothendieck's Period Conjecture and the missing p-adic Hodge Theories", "body": "Singular cohomology and algebraic de Rham cohomology are both functors from the category of smooth projective algebraic varieties over $\\mathbb Q$ to $\\mathbb Q$-vectors spaces. They come with the extra structure of Weil cohomology theories - grading, cup product, orientation map, and cycle class map. We can consider the space of all invertible natural transformations from singular cohomology to algebraic de Rham cohomology that respect this structure. For purely formal reasons, this is an affine scheme. The coordinates are the matrix coefficients of the natural transformation for each algebraic variety, and the relations are given by the various commutative diagrams that must hold. Conjecturally, this scheme is a torsor for the motivic Galois group of $\\mathbb Q$.\n\nde Rham's theorem, together with GAGA, gives us an isomorphism between these two functors when tensored with $\\mathbb C$. In other words, it gives us a $\\mathbb C$-point of this affine scheme.\n\nGrothendieck's period conjecture says that this point is a generic point - so the ring of periods in $\\mathbb C$ is equal to the whole ring of functions on this scheme.\n\n* * *\n\nOn the other hand, consider the $p$-adic analogue. $p$-adic etale cohomology and algebraic De Rham cohomology are both functors from the category of smooth projective algebraic varieties over $\\mathbb Q_p$ to $\\mathbb Q_p$-vector spaces. They are Weil cohomology theories.\n\nWe can consider the space of all isomorphisms between these two functors as an affine scheme over $\\mathbb Q_p$.\n\n$p$-adic Hodge theory gives a $B_{dR}$-valued point of this scheme.\n\nHowever, it is not the generic point. The reason is that the isomorphism described by $p$-adic Hodge theory naturally factors through the category of $\\operatorname{Gal}(\\overline{\\mathbb Q}_p|\\mathbb Q_p)$-representations. Hence it lies in a closed subscheme that is a torsor under, not the full motivic Galois group, but the Zariski closure of $\\operatorname{Gal}(\\overline{\\mathbb Q}_p|\\mathbb Q_p)$ inside it. This group is much smaller because the Tate conjecture fails over $\\mathbb Q_p$, and a map of Tannakian categories that is not a full functor corresponds to a map of Tannakian groups that is not surjective.\n\n* * *\n\nSo the analogue of the period conjecture fails badly over $\\mathbb Q_p$. This leads me to my question: Grothendieck's period conjecture essentially gives us a description of the scheme of isomorphisms betweeen singular and de Rham cohomology - it's the spectrum of the period ring. However, we do not have a similar picture of the scheme of isomorphisms between $p$-adic etale cohomology and de Rham cohomology. The period ring only gives us one point.\n\n> Can we explicitly describe any other points of this scheme of isomorphisms?\n", "link": "https://mathoverflow.net/questions/200083/grothendiecks-period-conjecture-and-the-missing-p-adic-hodge-theories", "tags": ["nt.number-theory", "tannakian-category", "p-adic-hodge-theory"], "votes": 61, "creation_date": "2015-03-15T14:18:19", "comments": ["I should correct myself: to obtain that \"over a finite field mixed motives are pure\" we must assume that such mixed motives exist (which follows from the Tate conjecture) and that they have a weight filtration respected by Frobenius. Then for each mixed motive $M$ appropriate polynomials in Frobenius are idempotent on $M$ which split it as a direct sum of pure motives. I am not sure how far we are from proving the Tate conjecture for varieties over finite fields; i imagine it's not considered out of reach. And i'm not sure how useful motives over finite fields are apart as source of intuition.", "Thank you @WillSawin for this excellent question. I think that for the usual equal characteristic $0$ Grothendieck conjecture one must consider arbitrary mixed motives which complicates things. So for varieties over p-adic fields shouldn't one also consider motives of nonprojective varieties ? It seems that p-adic étale and similar cohomologies can get complicated for those. Im not familiar enough with the field. We have Voevodsky's $DM_{gm}^{eff}(Q_p,Q)$ but no t-structure i believe. Also have you thought about varieties over a finite field where mixed motives with $Q$-coefficients are pure ?", "@AknazarKazhymurat Thanks! I have figured nothing out.", "representation is abnormally small (compared to the full motivic group) and try to work from there (one can find examples among abelian varieties). Another question is whether you should restrict to isomorphism respecting the filtration on both sides (as I think you should).", "Will Sawin, first of all, thank you for this interesting (but hard) question. My impression is that part of the difficulty stems from the fact that your analogy is perhaps not so close: a closer analogue might be the collection of all comparison theorems between étale cohomology for each $p$ and de Rham cohomology for a variety over $\\mathbb Q$ (rather than the single comparison theorem for a variety over $\\mathbb Q_{p}$. In that case, one recovers I believe the full motivic group. As for your precise question, I guess one can take a variety over $\\mathbb Q$, select a $p$ such that the Galois", "To get an isomorphism between algebraic de Rham cohomology and singular cohomology one only needs de Rham's theorem and Serre's GAGA, so no Hodge theory is required.", "@Olivier I meant to write $B_{dR}$ instead of $\\overline{\\mathbb C}_p$. Given that, do I need to use a different scheme?"], "comment_count": 7, "category": "Science", "diamond": 1} {"question_id": "58", "site": "mathoverflow", "title": "Does every triangle-free graph with maximum degree at most 6 have a 5-colouring?", "body": "## A very specific case of Reed's Conjecture\n\nReed's $\\omega$,$\\Delta$, $\\chi$ conjecture proposes that every graph has $\\chi \\leq \\lceil \\tfrac 12(\\Delta+1+\\omega)\\rceil$. Here $\\chi$ is the chromatic number, $\\Delta$ is the maximum degree, and $\\omega$ is the clique number.\n\nWhen restricted to triangle-free graphs, the equivalent question is, Does every triangle-free graph have chromatic number $\\leq \\frac \\Delta 2 +2$?\n\nThis is known for $\\Delta\\leq 4$. In general for triangle-free graphs, $\\chi \\leq O(\\Delta/\\log \\Delta)$, so the conjecture is also true for very large $\\Delta$.\n\nHow about $\\Delta=5$? $\\Delta=6$? Because of parity, $\\Delta=6$ is the easier of these two cases (and actually easily implies the $\\Delta=5$ case. Can anyone prove it?\n\nKostochka proved that every triangle-free graph has $\\chi \\leq \\frac 2 3 \\Delta +2$. He also proved that $\\chi\\leq \\frac \\Delta 2 +2$ for graphs of sufficiently large girth depending on $\\Delta$. Can anyone prove it for girth $\\geq 5$? $4$?\n\nThis would at least provide some hope for proving Reed's Conjecture for triangle-free graphs.\n\n> Does every triangle-free graph with $\\Delta\\leq 6$ have $\\chi \\leq 5$? What about every graph with girth at least five?\n", "link": "https://mathoverflow.net/questions/37923/does-every-triangle-free-graph-with-maximum-degree-at-most-6-have-a-5-colouring", "tags": ["graph-theory", "graph-colorings", "co.combinatorics"], "votes": 55, "creation_date": "2010-09-06T12:58:11", "comments": ["Another possible tangential attack is to establish the conjecture for the \"warmth\" $w$ of the graph, because Brightwell and Winkler proved that $w \\le \\chi$.", "Good question; it is known to hold, without the round-up. Reed never published the result in a paper but it's in Graph Colouring and the Probabilistic Method (with Molloy), in the chapter on hard-core distributions (Chapter 23 I think). (It wouldn't let me comment twice in a row, so I deleted the old comment to add: ) A proof of a stronger result, noted by McDiarmid, appears in Section 2.2 of my thesis, which is here: columbia.edu/~ak3074/papers/phdthesis.pdf", "This is not addressing your question, but I wonder if it is known whether Reed's conjecture holds when $\\chi$ is replaced by $\\chi_f$, the fractional chromatic number?"], "comment_count": 3, "category": "Science", "diamond": 1} {"question_id": "59", "site": "mathoverflow", "title": "Class function counting solutions of equation in finite group: when is it a virtual character?", "body": "Let $w=w(x_1,\\dots,x_n)$ be a word in a free group of rank $n$. Let $G$ be a finite group. Then we may define a class function $f=f_w$ of $G$ by \n$$ f_w(g) = |\\\\{ (x_1,\\dots, x_n)\\in G^n\\mid w(x_1,\\dots,x_n)=g \\\\}|. $$ The question is: \n\n> Can we characterize the words $w$ for which $f_w$ is a virtual character for any group $G$? \n\n(A virtual character is a difference of two characters, sometimes also called generalized character.) That $f_w$ is a virtual character of $G$ is equivalent to \n$$ (f_w,\\chi) = \\frac{1}{|G|} \\sum_{x_1, \\dots,x_n\\in G} \\chi(w(x_1, \\dots,x_n)) \\in \\mathbb{Z} $$ for all $\\chi \\in \\operatorname{Irr}(G)$. \nSome examples and non-examples: \n\n 1. $n=1$, $w= x^k$. Then $(f_w,\\chi)\\in \\mathbb{Z}$ is the $k$-th Frobenius-Schur-indicator. \n\n 2. $w=[x,y]=x^{-1}y^{-1}xy$. Then $(f_w,\\chi)= \\frac{|G|}{\\chi(1)}$, so $f_w$ is in fact a character. \n\n 3. For $w=x^{-k}y^{-1}x^ly$, it can be shown that $(f_w,\\chi)= \\frac{|G|}{\\chi(1)} a$ for some integer $a$.\n\n 4. $f_w$ is a virtual character for words of the form $w= x_1^{e_1}x_2^{e_2} \\dots x_n^{e_n}$.\n\n 5. For $w=[x^2,y^2]$ and $G= SL(2,7)$, the class function $f_w$ is not a generalized character.\n\n\n\n\nIt is always the case that $(f_w, \\chi)$ is an algebraic integer in $\\mathbb{Q}(\\chi)$. This follows from an old result of Solomon saying $f_w(g)$ is divisible by $|C_G(g)|$ when the number of variables is $>1$. (See also [this paper](http://de.arxiv.org/abs/1105.6066v1) by Rodriguez Villegas and Gordon, where my question is discussed very briefly.) So for rational-valued groups, $f_w$ is a virtual character for any word. (In fact, it suffices that $G$ is a normal subgroup of a rational-valued group.) In view of 4., the same is true for abelian groups, and I don't know examples of solvable groups where $f_w$ is not a virtual character. \nAnother remark is that $f_w$ does not change if we replace $w$ by its image under an automorphism of the free group in question. Using this, one can of course find rather complicated looking words for which $f_w$ is trivial, that is, $f_w(g)\\equiv |G|^{n-1}$. \n\nAdded later: \nThe proofs that $(f_w,\\chi)\\in \\mathbb{Z}$ in 2.-4. above basically use the same \"trick\". I'm also interested in other examples where perhaps another idea of proof is used than in the examples 1.-4., or \"more conceptual\" explanations why $f_w$ sometimes is a (virtual) character. (I admit that this is somewhat vague.) \n\nA related question is when $f_w$ is a character, not only a virtual character. For $G$ the symmetric group, this is discussed in Exercise 7.69k of \"Enumerative Combinatorics 2\" by Richard Stanley. (Note that in this case, $f_w$ is always a virtual character.)\n", "link": "https://mathoverflow.net/questions/105854/class-function-counting-solutions-of-equation-in-finite-group-when-is-it-a-virt", "tags": ["gr.group-theory", "rt.representation-theory", "finite-groups", "characters"], "votes": 54, "creation_date": "2012-08-29T08:38:56", "comments": ["To add to HJRW's comment a function is a $\\mathbb Q$-linear combination of characters iff it is constant on elements generating conjugate subgroups. In your set up because it is an algebraic integer combination of characters you just need it to be a $\\mathbb Q$-linear combination to be a virtual character. So at the end of the day you need that if $m$ is relatively prime to the order of $G$ then the number of solutions for $g$ and $g^m$ are the same. For the symmetric group this happens always because $g$ and $g^m$ are conjugate. The same is true for any group with all characters rational.", "This question is considered (although I don't think answered definitively) in this papr of Amit and Vishne: u.math.biu.ac.il/~vishne/publications/S0219498811004690.pdf ."], "comment_count": 2, "category": "Science", "diamond": 1} {"question_id": "60", "site": "mathoverflow", "title": "How many algebraic closures can a field have?", "body": "Assuming the axiom of choice given a field $F$, there is an algebraic extension $\\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is algebraically closed, then $K\\cong\\overline F$ via an isomorphism which fixes $F$. We can therefore say that $\\overline F$ is _the_ algebraic closure of $F$.\n\n(In fact, much less than the axiom of choice is necessary.)\n\nWithout the axiom of choice, it is consistent that some fields do not have an algebraic closure. It is consistent that $\\Bbb Q$ has two non-isomorphic algebraically closed algebraic extensions.\n\nIt therefore makes sense to ask: Suppose there are two non-isomorphic algebraically closed algebraic extensions. Is there a third? Are there infinitely many? Are there Dedekind-infinitely many?\n\n> What is provable from $\\sf ZF$ about the spectrum of algebraically closed algebraic extensions of an arbitrary field? What about the rational numbers?\n", "link": "https://mathoverflow.net/questions/325756/how-many-algebraic-closures-can-a-field-have", "tags": ["set-theory", "ac.commutative-algebra", "lo.logic", "axiom-of-choice", "field-extensions"], "votes": 54, "creation_date": "2019-03-19T02:41:22", "comments": ["@DavidRoberts: Should, yes. One day...", "Shouldn't you add an answer about this now, pointing to your own paper?", "@AsafKaragila thanks!", "@Holo: arxiv.org/abs/1911.09285", "@Holo: The one I am putting in arxiv later today. Check Monday's Replacements.", "@AsafKaragila hi, may I ask which paper of yours implies this result? I would like to read it", "@Noah: Some months ago I was talking with another MO user about this question, and I realised that one of my recent papers implies that it is consistent that every field has a proper class of pairwise non-isomorphic algebraic closures, unless it is real-closed or algebraically closed. It's still not clear whether or not we can have some of them having the same cardinality. That's a great little idea.", "@Noah: That's a good question. I don't know. It is true that the standard axiomatization of ACF is complete.", "What happens to the model theoretic results about algebraically closed fields without choice? Can you have uncountable nonisomorphic ACFs with the same characteristic and cardinality without choice?", "@WillSawin My intuition says yes, but I’m not sure — perhaps field properties like being formally real or having a particular valuation could have some impact on the answer. Hopefully looking at the arguments in the paper Asaf linked above can sharpen it somewhat.", "@AlecRhea The answer seems pretty likely to be \"proper class many\", right?", "@Gerhard: In the case of the rationals, I guess? There is a canonically constructed closure which is countable. But it is consistent that there is an algebraically closed algebraic extension which is not countable. In fact, it does not have a real-closed subfield.", "Does the consistency of two closures follow because there are two sets for which no bijection exists and each set is contained by one of the closures? In which case, I suspect there are more than two closures. Gerhard \"It's Not About The Algebra?\" Paseman, 2019.03.19.", "It seems we can ask a similar question even with choice (but without global choice) about proper class sized fields like the field of fractions of the Grothendieck ring of the ordinals, since the poset/chains we want to Zorn are proper classes. It might be interesting to see how many algebraic closures a proper class sized field can have in the presence of choice without global choice.", "@Matt: Not much, I expect. But maybe with time someone will find this interesting enough to pursue from a research point of view.", "Oh my misunderstanding. As an outsider to this, I'm excited to see what comes out of this question. Cheers.", "@Matt: Not really. That tells us that for the rational numbers there is a canonical algebraically closed algebraic extension, so the spectrum is non-empty. But it's pretty far from anything I'm asking.", "Is this related? math.stackexchange.com/questions/114978/…", "Much appreciated Asaf; it seems I take choice for granted all too often, I thought these facts were more intrinsic to field theory.", "Alec, it appears in Hodges' Läuchli's algebraic closure of $Q$. Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289–297. MR422022.", "Very cool question, could you provide a reference/proof hint for the comment about the consistency $\\mathbb{Q}$ having more than one algebraic closure without choice?", "Not fully sure about the tags, though."], "comment_count": 22, "category": "Science", "diamond": 1} {"question_id": "61", "site": "mathoverflow", "title": "Atiyah's paper on complex structures on $S^6$", "body": "M. Atiyah has posted a preprint on arXiv on the non-existence of complex structure on the sphere $S^6$.\n\n\n\nIt relies on the topological $K$-theory $KR$ and in particular on the forgetful map from topological complex $K$-theory to $KR$. \n\nQuestion: what are the nice references to learn about $KR$?\n\n**Edit :** Thank you very much for the comments and suggestions, M. Atiyah's paper \"K-theory and reality.\" Quart. J. Math., Oxford (2), 17 (1966),367-86 is a fantastic paper. However I have more questions.\n\nQuestion 1: how to build the morphism $$KSp(\\mathbb{R}^6)\\rightarrow K^{7,1}(pt)?$$\n\nQuestion 2: why do we use $\\mathbb{R}^{7,1}$? \n\nIn fact I do not understand the sentence \"The 6-sphere then appears naturally as the base of the light-cone\". And why it suffices to look at this particular model of $S^6$.\n", "link": "https://mathoverflow.net/questions/253577/atiyahs-paper-on-complex-structures-on-s6", "tags": ["dg.differential-geometry", "at.algebraic-topology", "complex-geometry", "kt.k-theory-and-homology", "index-theory"], "votes": 52, "creation_date": "2016-10-31T13:01:46", "comments": ["I think question should be edited to remove any implication that it is asking about possible correctness of Atiyah's paper. If the title were to say something about your actual question on KR, that would be good.", "Regarding the map you wish to understand, I think that the KSp should be seen as coming from one specific KR. Btw web.archive.org/web/20230113145507/https://www.math.umd.edu/‌​… should be rather useful, but I haven't been able to figure out the dimension correctly from there. Rosenberg uses a different grading to Atiyah.", "For your final question, the section titled \"Ambient Space Construction\" on this blog entry of mine is related. Basically it is a way to fix one conformal structure on $\\mathbb{S}^6$.", "@RyanBudney I suspect that the sentence \"integers goes to 0\" means integrable ACS goes to 0 in KR, while non-integrable ones goes to 1. It seems that he was using this map to separate integrable and non-integrable ACSs, it cannot be done if it is trivial. I feel that this might be the most crucial step, but do not have a clear idea what's going on...", "Karoubi briefly treats KR-theory in \"K-Theory: An Introduction\" (page 177). The book is a good reference for K-Theory in general.", "I think basics of $KR$ theory is easy to learn from [1]. What is puzzling is the claim on p.5 of arxiv.org/abs/1610.09366 that \"Because of the Atiyah-Singer theory, the linear algebra acquires a topological meaning, and that is embodied in $KR$ theory.\"", "And the involution on the base space would be trivial.", "@DenisNardin You go from $E$ to $E \\oplus \\overline{E}$; your involution is the obvious one that swaps factors.", "@RyanBudney Can you explain a little bit more how to treat a complex bundle as a real bundle? You need to throw in a C_2-action and I really don't see how to do that (this is a bit embarassing because I should know KR very well).", "I think it's just a straight-forward computation, isn't it? I've only scanned the paper and this is my first time thinking about $KR$, but I suspect what's going on is that the generator for complex k-theory is the bundle with Euler class $1$. In $KR$ theory this is twice the generator, because the generator is one where the fixed-point set of the involution is a Moebius bundle. I haven't thought about this in detail but that's what I suspect is going on.", "Yes of course, but do you know why it is trivial?", "The integers he's talking about is 6-dimensional complex k-theory of $S^6$, i.e. this group is infinite cyclic.", "This sentence is on p4 of Atiyah's preprint.", "The forgetful map appears to be that you can treat a complex bundle as a real bundle, that gives your map from complex k-theory to this \"KR\" variant. Where is the sentence you are quoting?", "I am reading [1], it is beautifully written. However as I am a complete amateur, I would like to understand the relationships between topological complex K-theory and KR-theory. In particular this sentence: \"There are natural forgetful maps from complex K-theory to KR-theory and in dimension 6 the integers go to 0\".", "It looks like you implicitly do not accept Atiyah's reference [1] in his paper. What isn't \"nice\" about that?"], "comment_count": 16, "category": "Science", "diamond": 0} {"question_id": "62", "site": "mathoverflow", "title": "Concerning proofs from the axiom of choice that ℝ³ admits surprising geometrical decompositions: Can we prove there is no Borel decomposition?", "body": "This question follows up on a [comment](https://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines#comment238532_92919) I made on Joseph O'Rourke's recent [question](https://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines), one of several questions here on mathoverflow concerning surprising geometric partitions of space using the axiom of choice.\n\n 1. Joseph O'Rourke: [Filling $\\mathbb{R}^3$ with skew lines](https://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines)\n\n 2. Zarathustra: [Is it possible to partition $\\mathbb{R}^3$ into unit circles?](https://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles)\n\n 3. Benoît Kloeckner: [Is it still impossible to partition the plane into Jordan curves without choice?](https://mathoverflow.net/questions/21327/is-it-still-impossible-to-partition-the-plane-into-jordan-curves-without-choice)\n\n\n\n\nThe answers to these questions show that the axiom of choice implies, by rather easy and flexible arguments, that space admits several surprising geometrical decompositions. To give a few examples:\n\n * $\\mathbb{R}^3$ is the union of disjoint circles, all with the same radius. Proof: Well-order the points in type $\\frak{c}$. At stage $\\alpha$, consider the $\\alpha^\\text{th}$ point. If it is not yet covered, we may find a unit circle through this point that avoids the fewer-than-$\\mathfrak{c}$-many previously chosen circles. QED\n\n * $\\mathbb{R}^3$ is the union of disjoint circles, with all different radii. Proof: at stage $\\alpha$, find a circle of a previously unused radius that avoids the fewer-than-$\\mathfrak{c}$-many previously chosen circles. QED\n\n * $\\mathbb{R}^3$ is the union of disjoint skew lines, each with a different direction. Proof: at stage $\\alpha$, put a line through the $\\alpha^\\text{th}$ point with a different direction than any line used previously. QED (Edit: it appears that one might achieve this one constructively by using the $z$-axis and a nested collection of [hyperboloids](http://en.wikipedia.org/wiki/Ruled_surface), which unless I am mistaken would give lines of different directions.)\n\n * $\\mathbb{R}^3$ is the union of disjoint lines, each of which pierces the unit ball.\n\n * And so on.\n\n\n\n\nThe excellent article _Jonsson, M.; Wästlund, J._ , [**Partitions of $\\mathbb{R}^3$ into curves**](http://dx.doi.org/10.7146/math.scand.a-13850), Math. Scand. 83, No. 2, 192–204 (1998). [ZBL0951.52018](https://zbmath.org/?q=an:0951.52018). [JStor](https://www.jstor.org/stable/24493132), proves among other things that $\\mathbb{R}^3$ can be partitioned into unlinked unit circles, either all with the same radius or all with different radii, and they have a very general theorem about partitioning $\\mathbb{R}^3$ into isometric copies of any family of continuum many real algebraic curves.\n\nSo the general situation is that the axiom of choice constructions are both easy and flexible and not entirely ungeometrical.\n\nMeanwhile, there are also a few concrete constructions, which do not use the axiom of choice. For example, Szulkin (see theorem 1.1 in [the Jonsson, Wästlund article](http://dx.doi.org/10.7146/math.scand.a-13850)) shows that $\\mathbb{R}^3$ is the union of disjoint circles by a completely explicit method. And there are others. But my question is not about these cases where there is a concrete construction. Rather, my question is:\n\n**Question 1.** Can we prove, in any of the cases where there is no concrete construction, that there is none?\n\nFor example, taking Borel as the measure of \"explicit\", the question would be: can we prove, for any of the kinds of decompositions I mention above or similar ones, that there is no Borel such decomposition? This would mean a decomposition of $\\mathbb{R}^3$ such that the relation of \"being on the same circle\" would be a Borel subset of $\\mathbb{R}^6$, and in this case, the function that maps a point to the nature of the circle on which it was to be found (the center, radius, and so on) would also be a Borel function.\n\nA closely related question would be:\n\n**Question 2.** Can we prove for any of the decomposition types that it requires the axiom of choice?\n\nFor example, perhaps there is an argument that in one of the standard models of $\\text{ZF}+\\neg\\text{AC}$ that there is no such kind of decomposition.\n\nMy only idea for an attack on question 1 is this: if there were a Borel decomposition of $\\mathbb{R}^3$ into circles or lines of a certain kind, then the assertion that this particular Borel definition was such a partition would be $\\Pi^1_2$ and hence absolute to all forcing extensions. So the same Borel definition would work to define such a partition even in forcing extensions of the universe having additional reals. Furthermore, the ground model partitions would be a part of that new partition. And this would seem to be a very delicate geometric matter to fit the new reals into the partition in between the ground model objects. But I don't know how to push an argument through.\n", "link": "https://mathoverflow.net/questions/93601/concerning-proofs-from-the-axiom-of-choice-that-%e2%84%9d%c2%b3-admits-surprising-geometrical", "tags": ["lo.logic", "set-theory", "axiom-of-choice", "mg.metric-geometry", "descriptive-set-theory"], "votes": 51, "creation_date": "2012-04-09T13:32:24", "comments": ["@LSpice Thanks!", "Sounds like a good idea. Perhaps one could make such a kind of argument work with determinacy or $\\text{AD}_{\\mathbb{R}}$?", "I really like these questions and every so often I go back and think about them. For me the obvious strategy to attempt is analytic - under some convenient assumption like \"all sets of reals are measurable\", show that such a decomposition cannot exist by some argument that involves, e.g. the probability distribution of the angle of the unique line through a random point. However I've been unable to make anything like this work. Is there a philosophical reason why this sort of arugment cannot be the right way to go?", "Joel, you might be interested in this paper of Zoltan Vidnyansky: arxiv.org/pdf/1209.4267.pdf. In it, he provides a \"black box\" theorem that allows one to deduce, under the assumption $V=L$, that sets like these, which are typically produced by transfinite recursion in the way you describe, can be coanalytic. His construction is still by transfinite recursion, but he shows how to do it very carefully so that the end result is coanalytic. This does not really answer either of your main questions, but it seems relevant so I thought I should make you aware of it!", "Good morning, Paul! Yes, of course full AC is not necessary, since as you say a well-ordering of the reals is sufficient. My question was whether one can prove that ZF (or ZF+DC), if consistent, is not able to prove the existence of decompositions. I like your selector idea, and perhaps that is a good strategy.", "Hi Joel. The second question is whether some form (or consequence) of the Axiom of Choice not following from ZF (or ZF + DC$_{\\mathbb{R}}$) follows from the existence of one of these decompositions? For instance, a selector for some Borel equivalence relation, or even a set of reals without the Baire property? It seems that you already know that full AC is not necessary, since the constructions all work from the existence of a wellordering of the reals. Is that right? Perhaps the existence of a selector for some equivalence relation is sufficient to carry out the constructions.", "Thanks! I'm really looking forward to hearing the outcome for the constructive case---please post an answer when you arrive at something. I would be interested to know the connection between the constructive case and the case of continuous partitions in classical logic, which could of course be viewed as a stricter form of the Borel partitions I asked about. What I expect is that one could hope to rule out continuous partitions in many of the cases, but ruling out Borel partitions seems mysterious to me.", "The constructive version is currently discussed at groups.google.com/group/constructivenews/browse_thread/threa‌​d/…", "@Joel: The trouble, constructively, is in showing that each sphere intersects the circles in exactly two points. When the points of intersection pass from one circle to another, as we vary the radius of the spehere, there is a discontinuity (the points jump suddenly). This is a \"model-theoretic\" way of seeing that something isn't right constructively.", "Joseph, thanks! These lines are skew in the sense that any two of them have different directions, but not skew in the sense of your question, which I know how to achieve only by using AC. ", "@Joel: Your idea to use hyperboloids to construct skew lines is nice!", "Asaf, since it seems difficult to make any proof at all, it would be interesting to see that ZF alone, without DC, cannot give a certain kind of construction. So your idea of considering a model where the reals are a countable union of countable sets is a good one, but I don't see how to make it work. In that model, the two questions become in effect the same, since every set is Borel there.", "And no, I don't think we know that we can't do all these things constructively. Perhaps there are nice explicit decompositions of each type. When there is, then we can prove that there is just by exhibiting it. But when there isn't, I don't see how we could prove that there isn't, and this is what my question is about.", "Andrej, I was using the term \"construction\" loosely, in classical logic, but could you explain why the argument isn't constructive in your sense? You have a circle in the $xy$ plane with center at $(4k+1,0,0)$ and a sphere of radius $r$ at the origin. So you are intersecting two circles; you can solve for the intersection points, and they vary step-wise continuously with $r$. (I am guesing that this \"step-wise\" step is the issue for constructive logic...)", "By the way, theorem 1.1 in Jonsson & Wästlund article is not constructive as far as I can tell. There is a problem with \"each sphere intersects one of the circles in exactly two points\".", "Do we know that these things can't happen constructively?", "Shooting from the hip, what happens when the real numbers are a countable union of countable sets? Should you require DC as well, as it ensures that most mathematics behaves normally, and at least ensures that there are more sets than Borel sets?"], "comment_count": 17, "category": "Science", "diamond": 1} {"question_id": "63", "site": "mathoverflow", "title": "Enriched Categories: Ideals/Submodules and algebraic geometry", "body": "While working through Atiyah/MacDonald for my final exams I realized the following:\n\nThe category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals\n\n$$I\\cdot J:=\\langle\\\\{ab\\mid a\\in I, b\\in J\\\\}\\rangle$$\n\nThe internal Hom is given by the ideal quotient; let's denote it $[I, J]:=(J:I)=\\lbrace a\\in A\\mid aI\\subset J\\rbrace$. We have\n\n$$IJ\\subset K \\Leftrightarrow I\\subset [J,K].$$\n\nFurthermore the category(poset) of submodules $U(M)$ of an A-Module $M$ is enriched over the ideal category via\n\n$$[U,V]:=\\lbrace a\\in A\\mid aU\\subset V\\rbrace.$$\n\nWhere does this enrichment come from? More specifically: As in algebraic geometry we are dealing with the set of prime ideals ( wich is more or less the subcategory\n\n$$\\mathrm{Int^{op}}/A\\subset\\mathrm{CRing^{op}}/A$$\n\nof ring-maps $A\\to B$ with $B$ an integral domain), I'd like to see the category of Ideals as some \"flattened\" version of $\\mathrm{CRing^{op}}/A$ by taking the kernel. Or more generally i'd like to see like to see it as a \"flattened\" version of the category of modules (by taking the annihilator?). So:\n\n**Question:** Is it possible to build the enriched categories $I(A)$ and $U(M)$ out of $\\mathrm{CRing^{op}}/A$ and $\\mathrm{Mod}_A/M$ in a nice way? So: Can we for example write $I\\cdot J$ down as a kernel of some $A\\to B$? Or $\\mathrm{Ann}(M)\\cdot \\mathrm{Ann}(N)$ just by using $N,M$ and constructions in/on $\\mathrm{Mod}_A$?\n", "link": "https://mathoverflow.net/questions/76443/enriched-categories-ideals-submodules-and-algebraic-geometry", "tags": ["ac.commutative-algebra", "ag.algebraic-geometry", "ct.category-theory", "monoidal-categories", "enriched-category-theory"], "votes": 48, "creation_date": "2011-09-26T11:24:59", "comments": ["\"The category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals\" That's how I'll explain commutative algebra to my son!", "Very interesting question. $I(A)$ is equivalent to the full subcategory of the comma-category which consists of regular epimorphisms $A \\to ?$. I've already asked here (mathoverflow.net/questions/69037/…) how we can describe the ideal product under this equivalence."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "64", "site": "mathoverflow", "title": "Extending a line-arrangement so that the bounded components of its complement are triangles", "body": "Given a finite collection of lines $L_1,\\dots,L_m$ in ${\\bf{R}}^2$, let $R_1,\\dots,R_n$ be the connected components of ${\\bf{R}}^2 \\setminus (L_1 \\cup \\dots \\cup L_m)$, and say that $\\\\{L_1,\\dots,L_m\\\\}$ is _triangulating away from infinity_ iff every $R_j$ that is bounded is triangular.\n\nCan every finite line-arrangement be augmented (by adding more lines) to yield one that is triangulating away from infinity?\n\n(The old and new arrangements need not be generic; they may have parallel lines and/or multiple intersections.)\n\nSome background: A slightly different (and, in my view, less natural) question was raised by me about twenty years ago; it has never been answered either. That problem first appeared in Richard K. Guy and Richard J. Nowakowski’s American Mathematical Monthly column on open problems (see “Bite-Sized Combinatorial Geometry Problems”, AMM, Vol. 103, No. 4, Apr. 1996, p. 342 and “Monthly Unsolved Problems”, AMM, Vol. 104, No. 10, Dec. 1997, pp. 969-970). Stan Wagon invited the readers of his on-line column to completely solve a one-parameter class of problems of this kind in which three lines symmetrically divide an equilateral triangle; see [Link](https://web.archive.org/web/20171230114418/http://mathforum.org/wagon/spring96/p812.html) .\n\nA related problem that might be simpler to solve in the negative is a version in which we not only require that all bounded components have 3 sides but also require that all unbounded components have 2 or 3 sides. If nobody can solve my original problem affirmatively or negatively, a negative solution to this problem would deserve the bounty.\n\n(Historical aside: Many years ago I proposed the original version of this problem to John Conway. He said “Let me think about that,” and as the curiosity-bug bit into his brain, he began to draw pictures, make observations, and formulate conjectures. But he was no fool; he could see that I was deliberately trying to entice him into working on the problem, and he was too proud to want to be seen as one who is so easily seduced. He shuddered as if shaking off an unpleasant memory and said “You know, I don’t have to work on just ANY damned problem!”)\n", "link": "https://mathoverflow.net/questions/155196/extending-a-line-arrangement-so-that-the-bounded-components-of-its-complement-ar", "tags": ["discrete-geometry", "plane-geometry", "hyperplane-arrangements"], "votes": 47, "creation_date": "2014-01-20T11:45:57", "comments": ["the dual version of Silvester-Caley thm in a stronger version shows that a simplicial argangment has to satisfy an optimal incidency condition that cannot hold if one move some lines in a certain neigbourhood while some other will remain fixed. In another hand a set of lines construced with the condition of my first comment should not change (up to homeomorphism) if we move the lines that are constructed thanks to the 5th line according to any choice of this 5th line in a fine neigbourhood, while a set of line including the first four one will remain fixed. (it's the heuristic argmt)", "The weaker condition (unbounded are 2 or 3 side) is stronger than the condition in the projective plane because if two unbounded component have both 3 sides, and correspound to a unique component in the projective plane, then this projective component will have 4 sides", "@JamesPropp : would you be ok with a negative answer in the projective plane (it would still be weaker than the weaker condition) I think it should not be difficult to show that if we start from 4 projective lines in general position and we add one more line that can't be constructed (with the ruler) from the first 4 lines, then it won't be possible to augment in a simplicial arrangement. Maybe it is not strong enought to be in an answer eventhought I thing the general and wealker case can be deduced from the argument that I think woukd work in the more simple case of the projective plane", "There are more simplicial pseudoline arrangements known than simplicial line arrangements; see Berman's \"Symmetric Simplicial Pseudoline Arrangements\", combinatorics.org/ojs/index.php/eljc/article/view/737 — but that doesn't make the problem trivial, or even easier, for that variation.", "@LSpice : I had no idea that had happened. The point was a MathJax usage correction.", "Another partial solution: Put a vertical line through every intersection. Then every bounded region is either a triangle or a trapezoid, because all vertices must lie on those vertical lines.", "Leaving aside the fact that Yoav is thinking of affine space rather than projective space, the truth is even more embarrassing than his comment would suggest: if I'm thinking clearly today (as opposed to yesterday), any generic 3-line arrangement in projective space IS a triangulation! So I retract my proposal. To replace it by a more credible proposal, I'd have to acquaint myself with the prior literature that David Eppstein is implicitly citing.", "Isn't it the case that a generic 3-line arrangement can be extended to a simplicial arrangement by adding at each intersection of two lines the parallel to the third line?", "If anyone can show that a generic triple of lines in the projective plane does not belong to any simplicial arrangement (which ought to be true, given what David Eppstein tells us), I would be inclined to award the bounty to that person.", "In case you're unaware (since I didn't see any mention of it in the question): these things are more commonly studied in the projective plane, where there is no distinction between bounded and unbounded. An arrangement of lines or pseudolines in the projective plane for which all faces are three-sided is called a simplicial arrangement. There are only a few infinite families of them known, so certainly it is not known that everything else can be augmented to become simplicial.", "It would be also interesting to find whether the answer to the question is affirmative if we replace lines by pseudolines (that is --- unbounded curves or olylines satisfying the property that every two of them intersect at exactly one point; here we omit parallel lines).", "@James: I would add combinatorial-geometry and/or discrete-geometry.", "I have modified the question again, this time to suggest a related question whose solution would win the bounty if the original question proves to be too hard.", "@domotorp: Are there other appropriate tags I should add?", "@James, regarding bountyhunting: I think you might have received low attention due to the appended tags that are not followed by many. Offering a bounty certainly draws more attention, as there are only a few questions with a bounty.", "Don't take $A$ to be on one of the old lines. Any 3 new lines all pass through $A$, so 3 new lines cannot all be sides of a convex region. Thus, any region $D$ lies in a wedge between two consecutive new lines, call them $L$ and $L'$. Then $\\partial D \\setminus (\\partial D \\cap (L \\cup L'))$ consists of two paths, call them $p$ and $q$. (Or, if $A$ is at a corner of $D$, one path.) If $p$ contains more than one old line, then it has an internal vertex $v$. But there would be a new line through $v$, contradicting that $L$ and $L'$ are consecutive. So $p$ is a single segment, and likewise $q$.", "I like Ilya Bogdanov's observation, but I don't follow his proof. Why couldn't there be a five-sided face whose sides go \"old\", \"new\", \"old\", \"new\", \"new\"? Or a six-sided face whose sides go \"old\", \"new\", \"old\", \"new\", \"old\", \"new\"? Clearly I'm missing something. Also, in the case where a line is both \"old\" AND \"new\" (if say $A$ is one of the original intersection points), should we interpret \"old\" as \"non-new\" in Ilya's argument?", "It is easy to \"quadrilaterate away\" just by choosing a point $A$ and drawing all lines connecting $A$ with all intersection points of the original lines. Each resulting region will have at most four sides because it cannot contain two consecutive sides on \"old\" lines.", "@AdamP.Goucher I am confused. If a line crosses a square at two adjacent sides, this leaves a pentagon behind. I can also get a pentagon by having two lines cross inside a square, and both exit on parallel sides.", "I've revised the question to include reference to an earlier, similar problem, though I doubt that this will be of much use to anyone, except as an indication that the problem is subtler than it appears at first. (Though I'm hoping that both problems eventually turn out to be simple after all, when approached in the right way.)", "Re Gerhard Paseman's comment, I think I've learned something from this experience about modifying questions in MathOverflow. It's not enough to revise the wording of a question in response to people's queries; it's important to make it clear to everyone that one has done so! I'll try to remember to do this in the future.", "Re Adam Goucher's suggestion, I find the phrase \"all lines that intersect the square pass through a single point\" obscure. Adam, can you clarify this for me?", "Re Sam Hopkins' first question, I suspect that the problem becomes easier (and has a negative answer) in higher dimensions, but I have no counterexamples to propose.", "Observe that it is possible to extend an arrangement to one that is `quadrilateralating away from infinity' by superimposing a sufficiently large square mesh (two perpendicular families of parallel lines), where the squares are sufficiently small that all lines that intersect the square pass through a single point.", "I have delayed in responding because I found the question incomplete. In particular, I wanted Joseph's question to be acknowledged. Now that I notice the acknowledgment, I am inclined to believe the answer is no, as one may set up a sequence of \"problem quadrilaterals\", where simple attempts to triangulate them produce more \"problem quads\". Further, if there is a delay of about a day in responding to comments, you might expect twice that in responding to responses. Personally, I disagree about the \"bounty bias\". Gerhard \"Does Like Thinking For Rewards\" Paseman, 2014.01.22", "@JamesPropp: Do you have a counterexample for the analogous statement (involving simplices, of course) in higher dimensions?", "Well, now there's a bounty as extra incentive.", "I find it strange that this post has received no answers and so few comments. I wonder if the whole bounty-system is distorting the priorities of the MathOverflow community, by causing MathOverflow contributors to withhold information about solved problems (or to hold off on tackling unsolved problems) until they can get a bounty for their contribution.", "How generic is the original/new arrangement? Do you admit multiple points? Parallel lines?", "May I ask: What does it mean to \"extend\" a line arrangement?"], "comment_count": 30, "category": "Science", "diamond": 0} {"question_id": "65", "site": "mathoverflow", "title": "Homotopy type of TOP(4)/PL(4)", "body": "It is known (e.g. the Kirby-Siebenmann book) that $\\mathrm{TOP}(n)/\\mathrm{PL}(n)\\simeq K({\\mathbb Z}/2,3)$ for $n>4$. I believe it is also known (Freedman-Quinn) that $\\mathrm{TOP}(4)/\\mathrm{PL}(4)\\to K({\\mathbb Z}/2,3)$ is 5-connected. Is it known whether $\\mathrm{TOP}(4)/\\mathrm{PL}(4)$ is not equivalent to $K({\\mathbb Z}/2,3)$?\n\n**Edit:** The following lists the relevant definitions.\n\n * The topological group $\\mathrm{TOP}(n)$ is the group of self-homeomorphisms of ${\\mathbb R}^n$ with the compact-open topology.\n\n * The topological group $\\mathrm{PL}(n)=|\\mathrm{PL}_S(n)|$ is defined as the geometric realization of the simplicial group $\\mathrm{PL}_S(n)$. The $k$-simplices of $\\mathrm{PL}_S(n)$ are the piecewise linear homeomorphisms $\\Delta^k\\times{\\mathbb R}^n\\to\\Delta^k\\times{\\mathbb R}^n$ which commute with the projection onto $\\Delta^k$.\n\n * With the above definitions, there exists a canonical map of topological groups $\\mathrm{PL}(n)\\to\\mathrm{TOP}(n)$. Then the space $\\mathrm{TOP}(n)/\\mathrm{PL}(n)$ is defined as the homotopy fibre of the induced map $B\\mathrm{PL}(n)\\to B\\mathrm{TOP}(n)$. It is _not_ actually a quotient of a group by a subgroup.\n\n * Here is a way of recovering the homotopy type of $\\mathrm{TOP}(n)/\\mathrm{PL}(n)$ as an actual quotient. Let $\\mathrm{TOP}_S(n)$ be the singular complex of $\\mathrm{TOP}(n)$: $\\mathrm{TOP}_S(n)$ is the simplicial set whose $k$-simplices are continuous maps $\\Delta^k\\to\\mathrm{TOP}(n)$; these are in canonical bijection with the homeomorphisms $\\Delta^k\\times{\\mathbb R}^n\\to\\Delta^k\\times{\\mathbb R}^n$ commuting with the projection onto $\\Delta^k$. Hence we obtain an inclusion of simplicial groups $\\mathrm{PL}_S(n)\\hookrightarrow\\mathrm{TOP}_S(n)$, which induces by adjunction the previous map of topological groups $\\mathrm{PL}(n)\\to\\mathrm{TOP}(n)$. The space $\\mathrm{TOP}(n)/\\mathrm{PL}(n)$ is weakly homotopy equivalent to the geometric realization of the simplicial set $\\mathrm{TOP}_S(n)/\\mathrm{PL}_S(n)$ (which is levelwise given by taking cosets).\n\n\n\n", "link": "https://mathoverflow.net/questions/106971/homotopy-type-of-top4-pl4", "tags": ["at.algebraic-topology", "gt.geometric-topology", "homotopy-theory"], "votes": 45, "creation_date": "2012-09-11T19:38:57", "comments": ["@Ricardo: yes, that's what I meant.", "@Peter: Thank you for remarking that $\\mathrm{PL}(n)$ is not defined as a subgroup of $\\mathrm{TOP}(n)$. In fact, by $\\mathrm{TOP}(n)/\\mathrm{PL}(n)$ I meant the homotopy fibre of the map $B \\mathrm{PL}(n)\\to B \\mathrm{TOP}(n)$, which I believe is the usual meaning. I edited the question to clarify that. @John: Are you by any chance referring to Michael Weiss' recent sequence of articles titled \"Smooth maps to the plane and Pontryagin classes\"?", "@Peter: Michael Weiss has made some recent progress on understanding $\\text{Top}(n)$.", "PL(n) needs a different definition; subspace of Top(n) won't do; see for example 1.10 in Madsen and Milgram, \"The classifying spaces for surgery and cobordism of manifolds\". It is BTop(n) and BPL(n) that classify the appropriately defined bundles. As far as I know the question asked is still open. In general, BTop(n) and cognate spaces are not well understood; even their cohomology is not known, although the cohomology when one goes to the limit and considers BTop and BPL is completely understood.", "$\\text{TOP}(n)$ is in fact the group of self-homeomorphisms of ${\\mathbb R}^n$, as Qfwfq states.", "@Qfwfq : This is not really my area, but I'm pretty sure that $TOP(n)$ is suppposed to classify topological microbundles on a space. This means that it is something like the classifying space for something like the group of germs of homeomorphisms of $\\mathbb{R}^n$ (or maybe the pseudogroup of homeomorphisms between open sets in $\\mathbb{R}^n$). Similarly for $PL(n)$. I don't have it at hand, but there are proper definitions in a book by Madsen and Milgram.", "@Steven Landsburg: From books.google.it/… I gather $TOP(n)$ is the group of self-homeomorphisms of $\\mathbb{R}^n$ and $PL(n)$ the subgroup of piecewise linear ones.", "It was unknown 20 years ago, when this question was proposed as Conjecture 3.10 in 'Differential Topology, Foliations, and Group Actions' by Paul A. Schweitzer. Link: tinyurl.com/cddp8oq ", "What is $TOP(n)$?"], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "66", "site": "mathoverflow", "title": "A kaleidoscopic coloring of the plane", "body": "> **Problem.** Is there a partition $\\mathbb R^2=A\\sqcup B$ of the Euclidean plane into two Lebesgue measurable sets such that for any disk $D$ of the unit radius we get $\\lambda(A\\cap D)=\\lambda(B\\cap D)=\\frac12\\lambda(D)$? \n\n(I.V.Protasov called such partitions _kaleidoscopic_).\n\nObserve that for the $\\ell_1$\\- or $\\ell_\\infty$-norms on the plane such partitions exist: just take a suitable chessboard coloring.\n\nThe problem can be reformulated in terms of convolutions: _Is there a measurable function $f:\\mathbb R^2\\to\\\\{1,-1\\\\}$ such that its convolution with the characteristic function $\\chi_D$ of the unit disk $D$ is identically zero?_\n\n(The problem was posed 08.11.2015 by T.Banakh and I.Protasov on [page 19](http://www.math.lviv.ua/szkocka/viewpage.php?vol=0&page=19) of [Volume 0](http://www.math.lviv.ua/szkocka/viewbook.php?vol=0) of the [Lviv Scottish Book](http://www.math.lviv.ua/szkocka)).\n", "link": "https://mathoverflow.net/questions/219860/a-kaleidoscopic-coloring-of-the-plane", "tags": ["real-analysis", "mg.metric-geometry", "measure-theory", "harmonic-analysis", "geometric-measure-theory"], "votes": 44, "creation_date": "2015-10-02T11:55:32", "comments": ["@AntonPetrunin No, I do not know such a figure. It seems that for any triangle a kaleidoscopic coloring does exist.", "Do you know any plane figure (maybe nonconvex) without kaleidoscopic colorings? (wrt parallel translations)", "I seem to see a 3-color kaleidoscope for the perfect hexagonal ball.", "I'll record the basic observations in the Fourier domain: Let $f$ be the stated function, let $\\hat{f}$ be its Fourier transform and let $\\hat{\\chi_D}$ be the Fourier transform of $\\chi_D$. Then $\\hat{f} \\hat{\\chi_D}$ is orthogonal to every character of $\\mathbb{R}^2$. In my naive nonanalyst way, I think this means $\\hat{f}$ is supported on the zeroes of $\\hat{\\chi_D}$. According to Mathematica, $\\hat{\\chi_D}(k_1, k_2) = 2 \\pi \\tfrac{J_1(k)}{k}$ where $J_1$ is the first Bessel function and $k = \\sqrt{k_1^2+k_2^2}$. Unfortunately, $J_1$ has infinitely many zeroes.", "About the tag scottish-book, see the meta discussion: meta.mathoverflow.net/questions/3948/tag-referring-to-a-book‌​/…"], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "67", "site": "mathoverflow", "title": "What does the theta divisor of a number field know about its arithmetic?", "body": "This question is about a remark made by van der Geer and Schoof in their beautiful article \"Effectivity of Arakelov divisors and the theta divisor of a number field\" (from '98) ([link](http://arxiv.org/pdf/math/9802121v3)).\n\nLet me first of all recall that a curve $C$ over a function field can be recovered from its Jacobian $\\operatorname{Pic}(C)$ and its theta divisor $\\Theta \\subset \\operatorname{Pic}^{g-1}(C)$ ($g$ the genus of $C$).\n\nIn their article van der Geer and Schoof construct analogues of the Jacobian and the theta divisor in the case of number fields.\n\nNamely they consider the Arakelov-Picard group $\\operatorname{Pic}(F)$ of a number field $F$ and define its theta divisor as restriction of the function $h^0$ on $\\operatorname{Pic}(F)$ given by $$h^0(D) = \\log\\left(\\sum_{f \\in I} e^{-\\pi \\|f\\|^2_D}\\right),$$ where $I$ is a lattice associated with the Arakelov divisor $D$, to the subspace $\\operatorname{Pic}^{(d)}(F)$ of Arakelov divisors of degree $d$. Here $d$ is a suitable analogue of the genus of a curve.\n\nNow they say that \"it should be possible to reconstruct the arithmetic of the number field $F$ from $\\operatorname{Pic}^{(d)}(F)$ together with $h^0$\".\n\nMy question is now of course which parts of the arithmetic of $F$ are known to be recovered from their analogy? For example I would be very interested in the question if the units of $F$ can be recovered somehow.\n", "link": "https://mathoverflow.net/questions/47920/what-does-the-theta-divisor-of-a-number-field-know-about-its-arithmetic", "tags": ["nt.number-theory", "arakelov-theory", "theta-functions", "reference-request"], "votes": 41, "creation_date": "2010-12-01T08:17:15", "comments": ["Since I think it is useful information to have around, I summarize from memory from the deleted answer and subsequent discussion: the above mentioned paper is the thesis of R. Groenewegen \"Vector bundles and geometry of numbers\", more specifically its second part \"Torelli for number fields\", which explores the problem mentioned in the question, obtaining interesting results related to it, yet not answering the question itself. (I hope this is an accurate and neutral summary of the situation.) The document itself is available eg here: math.leidenuniv.nl/en/theses/13 ", "Anyway, the purpose is solved. You came to know about the paper I have kept . I am happy. I am deleting my answer. Why to welcome troubles unnecessarily. Let peace prevail. "], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "68", "site": "mathoverflow", "title": "Homology of $\\mathrm{PGL}_2(F)$", "body": "**Update:** As mentioned below, the answer to the original question is a strong No. However, the case of $\\pi_4$ remains, and actually I think that this one would follow from Suslin's conjecture on injectivity of $H_i(\\mathrm{GL}_n(F))\\to H_i(\\mathrm{GL}_{n+1}(F))$ up to torsion (in the case $i=4$, $n=3$) together with some results of Dupont (specifically exact sequence (15.10) in Dupont's book \"Scissors congruences, group homology and characteristic classes\") and Goncharov's conjecture on motivic cohomology in terms of Bloch groups in weights $2$ and $3$. Concretely:\n\n> Question. Let $F$ be any field. Consider the inclusion $F^\\times\\to \\mathrm{PGL}_2(F)$ of the diagonal matrices. Is the induced map $$ \\bigwedge^4 F^\\times_{\\mathbb Q} = H_4(F^\\times,\\mathbb Q)\\to H_4(\\mathrm{PGL}_2(F),\\mathbb Q)$$ surjective?\n\n(Explanation: In Dupont's (15.10), the kernel of the last map is precisely $H^2(F,\\mathbb Q(3))$ under Goncharov's conjecture. Under Suslin's conjecture, $H_4(\\mathrm{PGL}_3(F),\\mathbb Q)\\to H_4(\\mathrm{PGL}_4(F),\\mathbb Q)$ is injective, and the target is stable. Computing in terms of the homology of the $K$-theory space, one sees that the contributions to $H_4(\\mathrm{PGL}_4(F),\\mathbb Q)$ are $\\mathrm{Sym}^2 K_2^M(F)_{\\mathbb Q}$ and $H^2(F,\\mathbb Q(3))$ and $K_4^M(F)_{\\mathbb Q}$. The last term $K_4^M(F)_{\\mathbb Q}$ is precisely the cokernel of the map $H_4(\\mathrm{PGL}_3(F),\\mathbb Q)\\to H_4(\\mathrm{PGL}_4(F),\\mathbb Q)$ by Suslin's theorem, so $H_4(\\mathrm{PGL}_3(F),\\mathbb Q)$ should have contributions from $H^2(F,\\mathbb Q(3))$ and $\\mathrm{Sym}^2 K_2^M(F)_\\mathbb{Q}$. The latter admits a surjection from $\\mathrm{Sym}^2(\\bigwedge^2 F^\\times)_{\\mathbb{Q}} = H_4((F^\\times)^2\\rtimes \\Sigma_3)$, which is precisely the decomposable part of $H_4(\\mathrm{PGL}_3(F),\\mathbb Q)$ in Dupont's work. Looking back at (15.10), this shows that the first term should be zero, which leads to the question. There is some confusion between $\\mathrm{PSL}$ and $\\mathrm{PGL}$ in what I said, but at least assuming that all elements of $F$ have $6$-th roots, the argument should work.)\n\n* * *\n\nHere is a hopelessly naive question. Please point me to the relevant literature!\n\nLet $F$ be any field. The Cartan subgroup of $\\mathrm{PGL}_2(F)$ is $F^\\times\\rtimes \\Sigma_2$. Let $X_2(F)$ be the homotopy cofiber of $B(F^\\times\\rtimes\\Sigma_2)\\to B\\mathrm{PGL}_2(F)$, so there is a long exact sequence $$ \\ldots\\to H_i(F^\\times\\rtimes\\Sigma_2)\\to H_i(\\mathrm{PGL}_2(F))\\to H_i(X_2(F))\\to \\ldots $$ (I'm not sure if I should write $H_i(G)$ or $H_i(BG)$...). If I am not making a stupid mistake, then the known computations of the homology of $\\mathrm{GL}_2(F)$ going up to the first unstable group $H_3(\\mathrm{GL}_2(F))$ can be summarized by saying that $H_i(X_2(F))=0$ for $i=1,2$, and $H_3(X_2(F))=\\mathfrak p(F)$ is the pre-Bloch group of $F$. (The pre-Bloch group is the quotient of $\\mathbb Z[F^\\times]$ by the ``$5$-term relations''.) (I might be off by some $2$-torsion.) By Hurewicz (and the check that $\\pi_1 X_2(F)=0$), the same holds true for the homotopy groups of $X_2(F)$.\n\nAs far as I could find, little is known about the homology of $\\mathrm{GL}_2(F)$ or $\\mathrm{PGL}_2(F)$ in degrees $>3$. The rational structure could be determined by the following very naive question.\n\n> Question. Are the homotopy groups $\\pi_i X_2(F)$ bounded torsion for $i\\neq 3$?\n\nI might even expect the implicit bound to be independent of $F$ (but of course depend on $i$; the order of magnitude should be $i!$).\n\nThe only evidence I have is that a back-of-the-envelope calculation seems to suggest that this holds true for finite fields. In that case the order of $\\mathrm{PGL}_2(\\mathbb F_q)$ is the product of $q-1$, $q$ and $q+1$. Localized at $q-1$, the homology agrees with the homology of the Cartan; localized at $q$, it is only in degrees larger than $q$ (which is OK as the bound on torsion may depend on $i$), and localized at $q+1$, the homology seems to agree with the homology of a $K(\\mathbb Z/(q+1),3)$ (up to bounded torsion in each degree), which is roughly $\\mathbb Z/(q+1)$ in all degrees $\\equiv 3\\mod 4$ (and is bounded torsion in other degrees). The pre-Bloch group $\\mathfrak p(\\mathbb F_q)$ is also $\\mathbb Z/(q+1)$ up to $2$-torsion, so things add up.\n\nThe case of $F=\\mathbb Q$ might be amenable to computations, but I was unable to do those. Are there any relevant results about $H_4(\\mathrm{GL}_2(F))$ that might shed light on $\\pi_4 X_2(F)$?\n\nA final remark: I expect that it is critical that $F$ is a field. The similar statement should be false already for discrete valuation rings, or $\\mathbb Z[\\tfrac 1n]$ (which is why the case of $F=\\mathbb Q$ is not so easy to compute; I presume one would try to first compute for all $\\mathbb Z[\\tfrac 1n]$ and then pass to a colimit, and the desired structure should only appear in the colimit).\n\n**Update:** As mentioned by Matthias Wendt in the comments, a special case of the work of Borel-Yang says that $H_i(\\mathrm{PGL}_2(\\mathbb Q),\\mathbb Q)=0$ for $i>0$. Inserting this into the above long exact sequence, one immediately sees that $X_2(\\mathbb Q)$ is in fact rather complicated, and for example has nontorsion $\\pi_5$. Thus, the answer to the question is a strong No. Thanks for the help in sorting this out!\n", "link": "https://mathoverflow.net/questions/298336/homology-of-mathrmpgl-2f", "tags": ["gr.group-theory", "algebraic-groups", "kt.k-theory-and-homology", "group-cohomology", "algebraic-k-theory"], "votes": 39, "creation_date": "2018-04-20T02:58:29", "comments": ["@Matthias Wendt: Thanks a lot for the Borel-Yang reference! Let's not worry right now about the exact shape of the torsion; I'd be happy enough with the statement rationally (but then my \"evidence\" would be vacuous). About homotopy vs. homology: One reason is that the homotopy of the K-theory space is usually simpler than the homology of the K-theory space (just the primitives). Maybe the better answer is that I have some roundabout argument that led me to believe in the question, but it's not worth spelling this out while there is no nontrivial evidence.", "For the function field case, Harder's results show that rational cohomology of $SL_2(F)$ is torsion. I don't know of any results bounding torsion there. The homology of $X_2(F)$ is the equivariant homology of a complex of configurations of points on $\\mathbb{P}^1$. The exponential bounds for torsion (in cohomology) could follow if we knew that all torsion comes from symmetry groups of configurations. But I guess we don't know that. Finally: why should we assume that it's easier to study homotopy of $X_2(F)$ instead of its homology?", "According to Borel and Yang, the rational cohomology of $SL_2(F)$ with $F$ a number field is the tensor product: for every real place an exterior algebra generated by an Euler class in degree 2, for every complex place an exterior algebra generated by a class in the Bloch group. So it's reasonable to assume that in this case $X_2(F)$ is rationally a product of 3-spheres. I haven't checked the details, but that would be partial support for assuming that the higher homotopy groups (above degree 3) of $X_2(F)$ are torsion.", "@S. carmeli: I don't think the cohomological dimension of $F$ should be relevant. In fact, there is no \"Galois descent\" (even approximate, like maybe in degrees bigger than the cohomological dimension or so (like for $K$-theory)) for $X_2(F)$. In fact, for $F=\\overline{\\mathbb F}_p$, the Bloch group is $0$, and all homotopy groups of $X_2(F)$ are bounded torsion. Thus, there is no hope to recover $X_2(\\mathbb F_p)$ from $X_2(\\overline{\\mathbb F}_p)$.", "Don't you want to assume a bound on the cohomological dimension of $F$? I think that his is sort of important in the case of a finite field, since etale homotopically the space X(F) over the algebraic closure is very close to an Eilenberg-Mclane space and then taking fixed points don't take high homotopies to far down, but this sort of reasoning breaks in high cohomological dimension.", "Related to your final remark, K. Knudson has computed the homology of PGL_2(A) where A is the coordinate ring of an elliptic curve, using an action of this group on a certain Bruhat-Tits tree, see \"On the K-theory of elliptic curves\". Of course the result involves the homology of PGL_2 of the base field k, so is computable if k is finite.", "@Matthias Wendt: \"Bounded torsion\" means \"bounded exponent\". I would expect something like \"$\\pi_i X_2(F)$ is killed by $i!$ for $i\\neq 3$\". I would imagine that the torsion is of the same cardinality as $F$.", "@Will Sawin: This is basically Quillen's classical computation.", "I think for all that's known now we can't even rule out the possibility that $H_i(X_2(F))$ has cardinality of $F$ for any algebraically closed field $F$. Does \"bounded torsion\" mean bound on size or bound on exponent?", "How does one do the computations in the $PGL_2(\\mathbb F_q)$ case?"], "comment_count": 10, "category": "Science", "diamond": 0} {"question_id": "69", "site": "mathoverflow", "title": "Correspondence between eigenvalue distributions of random unitary and random orthogonal matrices", "body": "In the course of a physics problem ([arXiv:1206.6687](http://arxiv.org/abs/1206.6687)), I stumbled on a curious correspondence between the eigenvalue distributions of the matrix product $U\\bar{U}$, with $U$ a random unitary matrix and $\\bar{U}$ its complex conjugate, on the one hand, and the random orthogonal matrix $O$ with determinant $-1$, on the other hand. The random matrices $U$ and $O$ are uniformly distributed with respect to the Haar measure on the unitary group $U(N)$ and the orthogonal group $O(N+1)$.\n\nOne eigenvalue of $O$ is fixed at $-1$ (to ensure that det $O=-1$). The other $N$ eigenvalues have a distribution $p_O$ which was [known](http://en.wikipedia.org/wiki/Circular_ensemble) since Girko (1985). We calculated the distribution $p_{U\\bar{U}}$ of the eigenvalues of $U\\bar{U}$ (which we did not find in the literature _\\--- has anyone seen it before?_). We discovered to our surprise that $p_{U\\bar{U}}=p_O$. This holds for both $N$ even and odd (in the latter case both $U\\bar{U}$ and $O$ have an eigenvalue fixed at $+1$).\n\n**Question:** Is there a more direct route to arrive at this identity between the two eigenvalue distributions, without going through a separate calculation of each one? (You can find two such separate calculations in the Appendix of arXiv:1206.6687, but this seems a rather unsatisfactory way of understanding the correspondence.)\n\n* * *\n\nSome intuition for what is going on: for both $U\\bar{U}$ and $O$ the eigenvalues different from $\\pm 1$ come in complex conjugate pairs $e^{\\pm i\\theta}$. The matrix $O$ has an unpaired eigenvalue at $-1$, which repels $\\theta$ from $\\pi$. The matrix $U\\bar{U}$ cannot have an unpaired eigenvalue at $-1$ by construction and somehow this leads to a repulsion of $\\theta$ from $\\pi$ which is mathematically equivalent to what happens for the matrix $O$.\n\nFor example, when $N=2$ the eigenvalue $e^{i\\theta}$ has the same distribution $P(\\theta)=(2\\pi)^{-1}(1+\\cos\\theta)$ for both $U\\bar{U}$ and $O$. For $N=3$ the distribution is $P(\\theta)=\\pi^{-1}(1-\\cos^2\\theta)$, in addition to an eigenvalue fixed at $+1$, again the same for $U\\bar{U}$ and $O$. The correspondence continues for larger $N$, when factors $(\\cos\\theta_k-\\cos\\theta_l)^2$ appear in both distributions.\n", "link": "https://mathoverflow.net/questions/105512/correspondence-between-eigenvalue-distributions-of-random-unitary-and-random-ort", "tags": ["pr.probability", "random-matrices", "haar-measure", "mp.mathematical-physics", "linear-algebra"], "votes": 37, "creation_date": "2012-08-26T03:42:59", "comments": ["@student -- the first statement [$U^TU\\in{\\rm COE}(n)$] is correct, but the second statement is not, I think. The eigenvalue distribution of $U\\bar{U}$, with $U\\in{\\rm CUE}(n)$ is not the eigenvalue distribution of ${\\rm CRE}(n)$, it's the distribution of the eigenvalues $\\neq -1$ in ${\\rm CRE}(n+1)$. So there is this increment by one of the dimensionality. Additionally, note that the inverse of $U\\bar{U}$ is not equal to its transpose, so it's not an element of ${\\rm O}(n)$ (as it should be, for the CRE).", "so we have that $U^TU = e^{i\\bar{H}} e^{iH} = e^{2\\operatorname{Re} H + [H, \\bar{H}]/2 + \\ldots}$ is COE$(n)$ and apparently $U\\bar{U} = e^{iH} e^{-i\\bar{H}} = e^{-2\\operatorname{Im} H + [H, \\bar{H}]/2 + \\ldots}$ is CRE$(n)$ ...", "@CarloBeenakker I will try to write a proof when time permits.", "@Henry.L --- Thank you for your feedback. Can you explain what you mean by \"after you have figured out this case\" ? How do I find the distribution of $\\Lambda$ and show that it is equal to the distribution of the eigenvalues of orthogonal matrices?", "@CarloBeenakker I think that you can assume without loss of generality that $U\\bar{U}=\\Lambda$ is a diagonal matrix first whose eigenvalues must come in pairs. After you figured out this case, we can multiply both sides with another unitary random matrix $V$ and its Hermitian conjugate $V^*$ to bend the distribution in a linear manner. I do not think complex conjugate provide anything more than a real structure with double dimension? Or did I understand your description wrongly?", "@Z254R: as far as I know, the Weingarten formula gives averages of polynomials in the matrix elements of $U$ or $O$; that does not seem a viable route to find the distribution of the eigenvalues of $U\\bar{U}$, nor to show that the eigenvalues of $U\\bar{U}$ and $O$ have the same distribution.", "@Carlo: thanks for the clarification. Time for me to go and read a book on this subject before further commenting :-) (I think I made all my $U\\bar{U}$ related errors so far because of \"wrongly\" simulating a random unitary matrix. I was simulating random symmetric unitaries, not arbitrary ones). I still believe that there should be a slick solution to your problem!", "@Suvrit, $U\\bar{U}$ is not a symmetric matrix, meaning it is not equal to its transpose; the probability distribution of unitary symmetric matrices is indeed well known, it is the circular orthogonal ensemble (COE) of random matrix theory; this is an unfortunate name, because it can create a confusion with the ensemble of random orthogonal matrices, but these are entirely different objects.", "Is the observation that $U\\bar{U}$ is a unitary symmetric matrix (and every symmetric unitary matrix $S$ can be written as such a product) of value to you here (I say that because such matrices seem to be used when studying orthogonal ensembles...)", "indeed, the eigenvalue distributions of the classical compact groups are related, notably $p_O$ for $N$ even is also the eigenvalue distribution of the compact symplectic group; the matrices of the form $U\\bar{U}$ do not form a group, so this does not seem to help much; note also that the correspondence between $p_{U\\bar{U}}$ and $p_O$ is not an asymptotic large-$N$ result (like the correspondence to a normal distribution): it holds exactly for any finite $N$.", "There is an identity for the eigenvalue distributions in this paper: web.williams.edu/go/math/sjmiller/public_html/ntandrmt/hando‌​uts/… (see page 6) Maybe this would help. It is also a known fact that the orthogonal group is similar to a standard normal distribution: www-stat.stanford.edu/~cgates/PERSI/papers/random_matrices.p‌​df (see page 56). Thus it is possible to solve equality of these two distributions to obtain the answer to your question. ", "argh...i made stupid calculation error!", "well, $U\\bar{U}$ is not itself orthogonal (its inverse is not equal to its transpose); I can construct an orthogonal matrix $O$ with the same eigenvalues as $U\\bar{U}$, but this orthogonal matrix is not uniformly distributed, so that does not seem to help much."], "comment_count": 13, "category": "Science", "diamond": 0} {"question_id": "70", "site": "mathoverflow", "title": "Is there any positive integer sequence $c_{n+1}=\\frac{c_n(c_n+n+d)}n$?", "body": "In a [recent answer](https://mathoverflow.net/a/282154/17581) Max Alekseyev provided two recurrences of the form mentioned in the title which stay integer for a long time. However, they eventually fail.\n\n> **QUESTION** Is there any (**added:** strictly increasing) sequence of positive integers $c_1,c_2,c_3,\\dots$ satisfying the relation $$ c_{n+1}=\\frac{c_n(c_n+n+d)}n $$ for for all $n\\geq 1$ (and some integer constant $d$)?\n\n(As Sam Hopkins notes, it would be also very interesting if the indices of the sequence start from some $k>1$ rather than from 1.)\n", "link": "https://mathoverflow.net/questions/282159/is-there-any-positive-integer-sequence-c-n1-fracc-nc-nndn", "tags": ["nt.number-theory", "integer-sequences", "recurrences"], "votes": 37, "creation_date": "2017-09-27T08:35:39", "comments": ["@WillSawin: phrased this way, the transformation sort of looks like those found in the theory of cluster algebras, so maybe if there is a positive answer to the question it could be found via the \"Laurent phenomenon.\"", "If we ignore the division by n, a necessary condition for the integrality of the original sequence is for the modified $c_{p+1}$ to divide $p$ for all primes $p$. Might it be possible to show that given any $c_1$ and $d$, that there must exist a prime $p$ such that this is not the case?", "A note that you may want to allow the sequence to start at $c_k$ for some $k>1$ because otherwise the motivating example is not really of this form.", "We can express this as a time-invariant dynamical system by adding an additional variable $n$, i.e. $(x,y) \\to (x+1, y(y+x+d)/x)$. For each prime $p$, this naturally gives an algebraic dynamical system on $\\mathbb Q_p \\times \\mathbb Z_p$. One wants to know if there is a $d$ such that all these systems stay inside $\\mathbb Z_p \\times \\mathbb Z_p$ forever. Maybe techniques of $p$-adic algebraic dynamics would be helpful here?"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "71", "site": "mathoverflow", "title": "What is the three-dimensional hyperbolic volume of a four-manifold?", "body": "Every smooth closed orientable 4-manifold may be constructed via a handle decomposition. Before asking a couple of questions, I recall some well-known facts about handle-decompositions of 4-manifolds.\n\nWe can of course order the handles according to their index. Handles of index 0 and 1 form a connected 4-dimensional handlebody, whose boundary is a closed 3-manifold, diffeomorphic to a connected sum $\\\\#_g(S^2\\times S^1)$ of some $g$ copies of $S^2\\times S^1$ (if $g=0$ we get $S^3$). Handles of index 2 are attached to some framed link $L\\subset \\\\#_g(S^2\\times S^1)$. Since 3- and 4-handles also form a 1-dimensional handlebody, after the attaching of the 2-handles we must necessarily obtain a 4-manifold whose boundary is again diffeomorphic to $\\\\#_h(S^2\\times S^1)$, for some $h$ which is not necessarily equal to $g$. The new $\\\\#_h(S^2\\times S^1)$ is obtained by _surgery_ along the framed link $L$.\n\nTherefore, in some sense, constructing closed orientable 4-manifolds reduces to constructing framed links $L\\subset \\\\#_g(S^2\\times S^1)$ that produce some $\\\\#_h(S^2\\times S^1)$ via surgery. Let us define the _3-dimensional hyperbolic volume_ of a handle decomposition as the Gromov norm of $(S^2\\times S^1) \\setminus L$ (which is in turn the sum of the hyperbolic volumes of its pieces according to geometrization, whence the name). Let us then define the _3-dimensional hyperbolic volume_ of a closed orientable 4-manifold as the infimum of all the hyperbolic 3-dimensional volumes among all its handle decompositions. The infimum is actually a minimum because the set of 3-dimensional hyperbolic volumes is well-ordered.\n\nThe general question is:\n\n> What can we say about the 3-dimensional hyperbolic volume of a closed 4-manifold?\n\nA more specific one:\n\n> Which closed 4-manifolds have zero 3-dimensional hyperbolic volume?\n\nwhich is equivalent to the following:\n\n> Which closed 4-manifolds admit a handle decomposition such that $\\\\#_g(S^2\\times S^1)\\setminus L$ is a graph manifold?\n\nComplex projetive plane belongs to this class, and also many doubles of 2-handlebodies: if you take any link $L\\subset S^2\\times S^1$ whose complement is a graph manifold, you can attach 2-handles to it, and then make the double of the resulting bounded 4-manifold. The resulting double has volume zero.\n\nFinally, we have the following very specific question:\n\n> Is there a 4-manifold with positive 3-dimensional hyperbolic volume?\n\nI would expect that most (all?) aspherical 4-manifolds have positive volume, and maybe also many simply connected ones, but I don't know the answer.\n", "link": "https://mathoverflow.net/questions/69819/what-is-the-three-dimensional-hyperbolic-volume-of-a-four-manifold", "tags": ["gt.geometric-topology", "4-manifolds", "3-manifolds", "knot-theory"], "votes": 37, "creation_date": "2011-07-08T12:14:14", "comments": ["I think this might be related to this question: mathoverflow.net/q/19974/1345 In particular, if the 3-dimensional volume is zero, I suspect that the manifold admits an F-structure, and the simplicial volume is zero (and likely minimal volume =0). The point is that the circle action on a Seifert manifold M extends over dehn fillings and attaching 2-handles along MxI. If all of the level sets of a Morse function on a 4-manifold are graph manifolds (and the complements of the 2-handles attaching maps), then I think that these local circle actions may be assembled to get an F-structure.", "This question sounds a bit similar to what is written in the last lines of a recent article of Gromov and Guth : arxiv.org/abs/1103.3423"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "72", "site": "mathoverflow", "title": "Two-convexity ⇒ Lefschetz?", "body": "Assume that\n\n * $\\Omega$ is an open simply connected set in $\\mathbb R^n$ \n * _(two-convexity)_ if 3 faces of a 3-simplex belong to $\\Omega$ then whole simplex in $\\Omega$.\n\n\n\nIs it true that any component of intersection of $\\Omega$ with any 2-plane is simply connected?\n\n**Comments:**\n\n * Note that an open connected set $\\Omega\\subset\\mathbb R^n$ is convex (in the usual sense) if together with two sides of triangle $\\Omega$ contains whole triangle.\n\n * If the boundary of $\\Omega$ is a smooth hypersurface then the answer is YES. [The above property implies that at most one principle curvature of the boundary is negative. Then the statement follows easily from a Morse-type argument; see Gromov's \"Sign and geometric meaning of curvature\", Lefschetz theorem in Section $\\tfrac12$. In fact this argument, shows that in this case any component of intersection of $\\Omega$ with any affine subspace (not necessary 2-dimensional) is simply connected.]\n\n * The answer is YES if $n=3$. [In this case I can mimic the Morse-type argument.]\n\n * The statement would follow if one could approximate any $\\Omega$ by two-convex domains with smooth boundary. BUT the example below shows that such approximation does not exist for $n\\ge 4$.\n\n * I was surprised that my question has a lot in common with [this question](https://mathoverflow.net/questions/29373/minimal-length-embeddings-of-braids-into-r3-with-fixed-endpoints/)\n\n\n\n\n**Example:** We will construct two-convex simply connected open set $\\Omega$ in $\\mathbb R^4$ such that intersection $L_{t_0}$ of $\\Omega$ with some hyperplane is not simply connected. [This NOT a counterexample, but it shows that it is impossible to prove it using smoothing, as indicated in the comments.] \n\nSet $$\\Pi=\\\\{\\,(x,y,z,t)\\in\\mathbb R^4\\mid\\,y< x^2\\\\}.$$ Let $\\Pi'$ be the image of $\\Pi$ under a generic rotation of $\\mathbb R^4$, say $(x,y,z,t)\\mapsto(z,t,x,y)$.\n\nNote that $\\Pi$ is open and two-convex. Therefore $\\Omega=\\Pi\\cap \\Pi'$ is also open two-convex set. One can choose coordinates so that $\\Omega$ is an epigraph for a function $f\\colon\\mathbb R^3\\to\\mathbb R$ like $$f=\\max\\\\{\\alpha_1-\\beta_1^2,\\alpha_2-\\beta_2^2 \\\\},$$ where $\\alpha_i$ and $\\beta_i$ are linear functions. In particular $\\Omega$ is contactable.\n\nLet $L_{t_0}$ be the intersection of $\\Omega$ with hyperplane $t=t_0$; it is a complement of two convex parabolic cylinders in general position. If these cylidners have a point of intersection then $\\pi_1 L_{t_0}=\\mathbb Z$.\n\n[In particular, the function $f$ can not be approximated by smooth functions which Hessian has at most one negative eigenvalue value at all points.]\n", "link": "https://mathoverflow.net/questions/55788/two-convexity-%e2%87%92-lefschetz", "tags": ["convexity", "convex-geometry", "morse-theory", "tag-removed"], "votes": 36, "creation_date": "2011-02-17T14:12:07", "comments": ["@WillSawin, I guess your question is about last par, starting with \"Let $L_{t_0}$...\". We are in hypereplane, i.e., in $\\mathbb R^3$. Imagine that your parabolic cylinders have exactly one point of intersection. Then the complement is homotoipically equivalent to $\\mathbb{S}^1$.", "I don't understand the example. I'm visualizing pairs parabolic cylinders in general position and all of them have simply-connected complements. Could you describe a specific hyperplane that does the job?", "Here's an article of mine where (a slight variant of) higher convexity is used: staff.science.uu.nl/~henri105/PDF/advgometry.pdf", "For an open set $\\Omega$, why is [ if 3 faces of a 3-simplex belong to $\\Omega$ then whole simplex in $\\Omega$ ] not the same as convex? "], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "73", "site": "mathoverflow", "title": "Orthogonal vectors with entries from $\\{-1,0,1\\}$", "body": "Let $\\mathbf{1}$ be the all-ones vector, and suppose $\\mathbf{1}, \\mathbf{v_1}, \\mathbf{v_2}, \\ldots, \\mathbf{v_{n-1}} \\in \\\\{-1,0,1\\\\}^n$ are mutually orthogonal non-zero vectors. Does it follow that $n \\in \\\\{1,2\\\\} \\cup \\\\{4k : k \\in \\mathbb{N}\\\\}$?\n\nA few notes are in order:\n\n 1. If we do not allow \"$0$\" entries in the vectors then this is a well-known necessary condition for the existence of an $n \\times n$ [Hadamard matrix](https://en.wikipedia.org/wiki/Hadamard_matrix), so the answer to my question is \"yes\" if we replace $\\\\{-1,0,1\\\\}$ with$\\\\{-1,1\\\\}$.\n\n 2. With some computer help, I have shown that there do not exist collections of vectors with these properties when $n = 3$, $5$, or $6$, so the answer to my question is \"yes\" when $n \\leq 6$.\n\n\n\n\n**Edit (May 19, 2023):** In the comments, Max Alekseyev has shown that the answer is \"yes\" when $n \\leq 12$, and Ilya Bogdanov has shown that the answer is \"yes\" when $n$ is an odd prime.\n", "link": "https://mathoverflow.net/questions/446995/orthogonal-vectors-with-entries-from-1-0-1", "tags": ["co.combinatorics", "linear-algebra", "matrices"], "votes": 35, "creation_date": "2023-05-17T12:10:00", "comments": ["I've computationally verified the conjecture $n=15$ and together with Ilya's and Mikhail's arguments that makes it proved for all $n\\leq 20$. For $n=21$, I can tell for now that if a counterexample exists, it should have at least 6 vectors with exactly 14 nonzero components.", "@MaxAlekseyev Same argument that establishes that Hadamard matrices are of order $1$, $2$, or $4k$. If there are three mutually orthogonal ±1-vectors $\\mathbf{1}$, $v_1$, $v_2$, then the number of indices $i$ where $v_{1i} = s_1$, $v_{2i} = s_2$ is the same for any of the four combinations of $s_1, s_2 \\in \\{+1, -1\\}$.", "@MikhailTikhomirov: What does \"invoke Hadamard\" stand for?", "Of course, $2p^k$ ditto.", "Further, $n = 2p$ for odd prime $p$ is also a no-go. $1 = 1/2p + \\ldots$ requires at least one more $1/2p$, and their sum $1 / p$ requires an extra (even) denominator divisible by $p$ (thus, another $2p$). We now have three full rows, which is enough to invoke Hadamard.", "@IlyaBogdanov's argument also seems to apply when $n = p^k$ for $p$ odd prime. For the diagonal element we have $1 = 1/p^k + \\ldots$, which has to involve at least $p - 1$ more $1/p^k$'s. But two odd-sized full rows are already a contradiction.", "@kodlu Yes, even many such weighing matrixes exist. But the all-1 column changes a lot.", "@IlyaBogdanov - You're quite right, thank you.", "@NathanielJohnston The life seems more complicated for me. E.g., it could happen that the sum involves $1/15$, $2\\times 1/6$ and $6\\times 1/10$ (,the other denominators should be even!). However, this may fasten algorithms searching such matrices.", "@IlyaBogdanov - If you agree with my reasoning, please feel free to write this up as an answer and I will accept it. Your denominators-cancelling argument is what blew this open. Thank you!", "@IlyaBogdanov - And a similar argument works to rule out the $n \\equiv 2\\pmod{4}$ cases, thus proving the entire thing. Suppose $n = 2m$ for some odd number $m$. Then (again, by your denominators-cancelling argument) $A$ must have at least $2$ columns with all entries non-zero (one of which is the all-ones column). If it had $3$ or more such columns we get the Hadamard contradiction, so assume it has exactly $2$. The remaining columns each have their number of non-zero entries each equal to factors of $m$, which are all odd. Such columns cannot be orthogonal to $\\mathbf{1}$. Contradiction.", "Very interesting. If the all 1 vector is removed there are solutions with odd lengths.", "@IlyaBogdanov - Very nice, thank you! I think your argument can be extended to all odd numbers, not just odd primes. For example, if $n = 15$ then (by your denominators-cancelling argument) there must be at least 3 columns of $A$ that have all $15$ entries with no $0$s (since then you can get $1/15 + 1/15 + 1/15 = 1/5$ to reduce the denominator). But the standard proof of the Hadamard necessary condition says that you can't have $3$ orthogonal columns of $\\pm 1$ entries unless $n \\in \\{1,2\\} \\cup \\{4k\\}$.", "@NathanielJohnston Each diagonal element of $BB^T$ is a sum of $n$ njmbers each of them is either 0 or of the form $1/k$ with $k\\leq n$. At least one summand is $1/n$; if there are less than $n$ such, then $n$ in the denominators cannot cancel.", "@Ilya - Can you expand upon the statement \"But the diagonal elements in $BB^T$ may equal $1$ only if $D = nI$\"? I expect that this is the step that is using the hypothesis of $n$ being an odd prime, but I'm having a hard time seeing how.", "As a first step, $n$ cannot be an odd prime. This is because the matrix $A$ composed from those columns should be such that $D=A^TA$ is diagonal, so $B=AD^{-1/2}$ should be orthogonal. But the diagonal element s in $BB^T$ may equal $1$ only I’d $D=nI$ which is impossible by Hadamard. It seems that one may investigate $BB^T$ or, equivalently, the relation $AD^{-1}A^T=I$ further…", "@domotorp - Only when there's a Hadamard matrix.", "Do you have a construction for such a matrix for every $n$ that is divisible by 4, or only for those $n$ for which there exists a Hadamard matrix?", "The question seems related to weighing matrices except that the all-ones vector kind of messes things up.", "@PeterMueller: I've also checked with cliquer and it confirmed no solutions for $n=11$ within a couple of hours. So, the \"yes\" answer holds for all $n\\leq 12$.", "Maybe linear algebra modulo $4$ is relevant, as these vectors miss $2 \\bmod 4$? There is also a natural combinatorial interpretation -- with each $\\mathbf{v}_i$ associate disjoint subsets $A_i^+$, $A^-_i$ of $\\{1,2,\\ldots,n\\}$ defined by $A^{\\pm}_i=\\{j: (\\mathbf{v}_{i})_j=\\pm 1\\}$. Then $|A^+_i|=|A^-_i|$ by orthogonality of $\\mathbf{v}_i$ with $\\mathbf{1}$. The rest of the orthogonality conditions are encoded by $|A^+_i \\cap A^+_j| + |A^-_i \\cap A^-_j| = |A^+_i \\cap A^-_j| + |A^-_i \\cap A^+_j|$ for $i \\neq j$.", "@MaxAlekseyev ... and so did I using cliquer. Presently running $n=11$ which might take a while ...", "I confirm the affirmative answer for all $n\\leq 10$ with ILP.", "Somewhat relevant: mathoverflow.net/q/418009", "@Peter: Thanks, I do need that condition so I'll add that in. (I'm reminded of a time when I asked my linear algebra students to produce some orthogonal vectors satisfying certain conditions, and lots of them got easy marks since I forgot to specify \"non-zero\"...)", "@NathanielJohnston Which condition prevents all $\\mathbf{v_i}$ to be the $0$-vectors?", "@Amir: If you just require linear independence then you can always (for all $n$) find such a set of vectors. Furthermore, if you just require linear independence then whenever $n$ is even you can find such a set of vectors with entries in $\\{-1,1\\}$.", "Interesting! Would the answer change if we replace the \"orthogonality\" requirement with \"independence\"?"], "comment_count": 28, "category": "Science", "diamond": 0} {"question_id": "74", "site": "mathoverflow", "title": "Is there a Mathieu groupoid M_31?", "body": "I have read something which said that the large amount of common structure between the simple groups $SL(3,3)$ and $M_{11}$ indicated to Conway the possibility that the Mathieu groupoid $M_{13}$ might exist. \nIndeed, it exists, and the group of permutations of the other points generated by paths in which the hole starts and ends at the same point is the Mathieu group $M_{12}$.\n\nIt is natural to ask whether or not something analogous exists for the Mathieu group $M_{24}$. The corresponding linear group seems to be $GL(5,2)$. \n$2^{4}:A_{8}$ is a vector (or hyperplane) stabilizer in $GL(5,2)$ and an octad stabilizer in $M_{24}$. $2^{6}:(GL(2,2)\\times GL(3,2))$ is likewise the stabilizer of a 2-dimensional (or 3-dimensional) space in $GL(5,2)$, and the stabilizer of a 'trio' (to use the term from SPLaG) in $M_{24}$.\n\nThe natural setting, therefore, in which to look for something analogous to $M_{13}$ is $(\\mathbb{Z}/(2))^{5} \\backslash 0$. (This can also be regarded as $4$-dimensional projective space over $\\mathbb{Z}/(2)$, but I will here think of it as $5$-dimensional space over $\\mathbb{Z}/(2)$ with the origin removed.) \nThen, to have $24$ points left over when the 'holes' are put in place, it would be natural to have a 'missing $3$-dimensional space' for a hole. \nA move should consist of moving the $3$-dimensional 'hole' to some other $3$-dimensional subspace (minus the origin) of $(\\mathbb{Z}/(2))^{5}$. The intersection of these spaces can have dimension $1$ or $2$, and there should be different rules for moving the hole to another location depending on the dimension of its intersection with the new location.\n\nWho has tried this before? Are there any papers on it?\n", "link": "https://mathoverflow.net/questions/70680/is-there-a-mathieu-groupoid-m-31", "tags": ["gr.group-theory", "finite-groups", "reference-request"], "votes": 34, "creation_date": "2011-07-18T14:48:41", "comments": ["It's good to know. 31 vs. 13 for M12."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "75", "site": "mathoverflow", "title": "Subalgebras of von Neumann algebras", "body": "In the late 70s, Cuntz and Behncke had a paper \n\nH. Behncke and J. Cuntz, _Local Completeness of Operator Algebras_ , Proceedings of the American Mathematical Society, Vol. 62, No. 1 (Jan., 1977), pp. 95- 100\n\nthe about the following question. Let $A$ be a $C^\\star$-algebra and let $B \\subset A$ be some $*$-subalgebra. Let $B$ be dense and such that every maximal abelian subalgebra of $B$ is norm-closed. Is it true that $B=A$? They proved this in various cases.\n\nI want to ask a related question in the von-Neumann-setting. More precisely: Let $A$ be a von Neumann algebra and $B \\subset A$ be some ultra-weakly dense $\\star$-subalgebra such every MASA of $B$ is ultra-weakly closed in $A$. Is it true that $B=A$? A result of Gert Pedersen implies that once $B$ is a $C^\\star$-algebra, then $B=A$. Hence, the two questions are closely related. Taking the work of Behncke-Cuntz and Pedersen together, it is known that $B=A$ if $A$ has no $II_1$-part.\n\n> **Question:** Is it true for every von Neumann algebra?\n\nor a little more modest\n\n> **Question:** Is it true for the hyperfinite $II_1$-factor?\n\nAnother strenghtening of the the assumption (which could help) would be the following:\n\n> **Question:** Let $A$ be a $II_1$-factor and $B \\subset A$ be a ultra-weakly dense $\\star$-subalgebra such that for every hyperfinite subalgebra $R \\subset A$ one has that $R \\cap B$ is ultra-weakly closed in $A$. Is it true that $B=A$?\n", "link": "https://mathoverflow.net/questions/34692/subalgebras-of-von-neumann-algebras", "tags": ["oa.operator-algebras", "fa.functional-analysis"], "votes": 33, "creation_date": "2010-08-05T14:40:37", "comments": ["If I understand correctly, you need to add a density hypothesis in your question?"], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "76", "site": "mathoverflow", "title": "Isometric embeddings of finite subsets of $\\ell_2$ into infinite-dimensional Banach spaces", "body": "Question: Does there exist a finite subset $F$ of $\\ell_2$ and an infinite-dimensional Banach space $X$ such that $F$ does not admit an isometric embedding into $X$?\n\nThere are some results of the type: each finite ultrametrics admits an isomeric embedding into any infinite dimensional Banach space. See Shkarin, S. A. Isometric embedding of finite ultrametric spaces in Banach spaces. Topology Appl. 142 (2004), no. 1-3, 13-17 for this, and also the papers: Dekster, B. V. Simplexes with prescribed edge lengths in Minkowski and Banach spaces. Acta Math. Hungar. 86 (2000), no. 4, 343-358; Faver, Timothy; Kochalski, Katelynn; Murugan, Mathav Kishore; Verheggen, Heidi; Wesson, Elizabeth; Weston, Anthony Roundness properties of ultrametric spaces. Glasg. Math. J. 56 (2014), no. 3, 519-535.\n\nThere is a related question on MathOverflow: [Almost isometric embeddability implies isometric embeddability](https://mathoverflow.net/questions/209435/almost-isometric-embeddability-implies-isometric-embeddability)\n\nAdded on 9/29/2016: Related results were obtained by James Kilbane in \n\nAdded on 4/3/2017: In another [recent paper](https://arxiv.org/pdf/1704.00319.pdf) James Kilbane proved that the set of possible counterexamples (if they exist) is small in a certain sense.\n", "link": "https://mathoverflow.net/questions/221181/isometric-embeddings-of-finite-subsets-of-ell-2-into-infinite-dimensional-ban", "tags": ["mg.metric-geometry", "banach-spaces", "metric-embeddings"], "votes": 33, "creation_date": "2015-10-17T17:44:32", "comments": ["The last result I know in this direction is in Petrov, F. V.; Stolyarov, D. M.; Zatitskiy, P. B. On embeddings of finite metric spaces in ln∞. Mathematika 56 (2010), no. 1, 135–139", "Is much known when $X$ is $\\ell^n_{\\infty}$ for $n<\\#F$?"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "77", "site": "mathoverflow", "title": "Peano Arithmetic and the Field of Rationals", "body": "In 1949 Julia Robinson showed the undecidability of the first order theory of the field of rationals by demonstrating that the set of natural numbers $\\Bbb{N}$ is _first order definable_ in $(\\Bbb{Q}, +, \\cdot$).\n\nIt is not hard to see that Robinson's result can be reformulated in the following symmetric form.\n\n**Theorem A.** _The structures_ ($\\Bbb{N}, +, \\cdot$) _and_ $(\\Bbb{Q}, +, \\cdot$) _are bi-interpretable._\n\nThe following generalization of Theorem A is considered folkore (I am not aware of a published reference).\n\n**Theorem B.** _If $(M, +, \\cdot)$ is a model of $PA$ (Peano arithmetic), then the field of rationals $\\Bbb{Q}^M$ of $(M, +, \\cdot)$ is bi-interpretable with $(M, +, \\cdot )$._\n\nLet $EFA$ denote the _exponential function arithmetic_ fragment of $PA$, a fragment also known as $I\\Delta_{0}+exp$. \n\nBased on _a posteriori_ evidence _classical_ theorems of Number Theory do not require the full power of $PA$ since they can be already verified in $EFA$ (indeed Harvey Friedman has conjectured that even FLT can be verified in $EFA$, with a proof that would be very different from Wiles'). \n\nThis suggests that in Theorem B one should be able to weaken $PA$ to $EFA$, hence my question:\n\n**Question**. _Is there a published reference for the strengthening of Theorem B, where $PA$ is weakened to $EFA$?_\n\nP.S. The following paper provides an excellent expository account of Robinson's theorem (and related results).\n\nD. Flath and S. Wagon, _How to Pick Out the Integers in the Rationals: An Application of Logic to Number Theory_ , American Mathematical Monthly, Nov. 1991.\n", "link": "https://mathoverflow.net/questions/65664/peano-arithmetic-and-the-field-of-rationals", "tags": ["lo.logic", "nt.number-theory"], "votes": 32, "creation_date": "2011-05-21T11:01:13", "comments": ["@Samuel Ried: No I have not (sorry for the tardy reply, I have been \"away\" for a long while).", "@AliEnayat: Have you found an answer to your question yet? I would be very interested to know.", "SJR: The answer to your first question is positive, i.e., nothing fancier is going on. Regarding the second one: the arguments of Robinson's paper should go through in $EFA$ (based on $EFA$'s track record in handling \"elementary\" number theory); my question is whether anyone has actually shown - in a publishd source - that this is indeed the case.", "Ali: Suppose $M\\models PA$. To interpret $M$ in $\\mathbb{Q}^M$ does Julia Robinson's formula with the operations of $\\mathbb{Q}^M$ do the job, or is there something fancier going on? Is your question about how much of the argument in Robinson's paper goes through in models of $I\\Delta_0+exp$?"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "78", "site": "mathoverflow", "title": "$f\\circ f=g$ revisited", "body": "This may be related to [solving $f(f(x))=g(x)$](https://mathoverflow.net/questions/17614). Let $C(\\mathbb{R})$ be the linear space of all continuous functions from $\\mathbb{R}$ to $\\mathbb{R}$, and let $\\mathcal{S}:=\\\\{g\\in C(\\mathbb{R}) ; \\exists\\; f\\in C(\\mathbb{R})$ s.t. $f\\circ f=g \\\\}$ . Is there some infinite dimensional (or, at least, bidimensional) linear subspace of $C(\\mathbb{R})$ contained in $\\mathcal{S}$ ?\n\nP.S. As a remark, there is a [maybe] interesting connection between [How to solve $f(f(x)) = \\cos(x)$?](https://mathoverflow.net/questions/17605) and [Borsuk pairs of Banach spaces ](https://mathoverflow.net/questions/11675). Namely, let $E$ be the closed subspace of $C[-1,1]$ consisting of all even functions, and let $K$ be the closed unit ball of $E$. Then the continuous mapping $\\Psi:$ $K$ $\\rightarrow$ $E$ expressed by $\\Psi(f)$ $:=$ $f\\circ f$ $+$ $\\left(\\left\\Vert f\\right\\Vert _{\\infty}-1\\right)\\cdot\\cos$\n\nis odd on $\\partial K$, and has no zeros in $K$.\n", "link": "https://mathoverflow.net/questions/18882/f-circ-f-g-revisited", "tags": ["ca.classical-analysis-and-odes", "fractional-iteration"], "votes": 32, "creation_date": "2010-03-20T14:35:10", "comments": ["In the previous comment, the full stop was taken as a part of the url - however, the intention was clearly to link to this post: solving $f(f(x))=g(x)$.", "Unfortunately, not even two vectors in that cone generate a subspace in ${\\mathcal S}$ since ${\\mathcal S}$ contains no decreasing functions (see mathoverflow.net/questions/17614.) Any solution to this problem is a linear space containing no injective functions.", "All increasing bijections are ok, so there is an infinite-dimensional cone. ", "ad least there is a 2-dimensional cone (a space which is closed under linear combinations with positive coefficients) given by the space of all nondecreasing linear maps."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "79", "site": "mathoverflow", "title": "The central insight in the proof of the existence of a class of Kervaire invariant one in dimension 126", "body": "I understand from [a helpful earlier MO question](https://mathoverflow.net/questions/101031/kervaire-invariant-why-dimension-126-especially-difficult) that the techniques leading to the celebrated resolution of the Kervaire invariant one problem in the other candidate dimensions yield no insight on dimension 126, to the extent that until recently there was not consensus as to which way it was likely to go.\n\nFor a few weeks now, I've been aware of [a manuscript of Minami](https://arxiv.org/pdf/2106.14604) which, if I'm reading correctly, asserts the existence of such an element in dimension 126 as a consequence of Wang and Xu's 2017 paper on the stable 61-stem. Presumably they would not have failed to remark on the resolution of the Kervaire invariant one problem if such was readily visible from what they had done, so it would seem that there is something that is not _obviously_ a consequence of it, but which can nevertheless be extracted in eight pages if one does somehow know what to look for. It's common that when an old conjecture like this is resolved, it is as an application of some newly discovered techique or heretofore-ungleaned insight into the inner workings of the encompassing theory.\n\n**What is the new insight here?**\n\nThe work on the stable stems is forbiddingly technical work I am unlikely to ever grasp the full details of, but it's still possible sometimes to get the flavor of the ideas involved, and that's what I would hope for here.\n", "link": "https://mathoverflow.net/questions/397905/the-central-insight-in-the-proof-of-the-existence-of-a-class-of-kervaire-invaria", "tags": ["at.algebraic-topology", "homotopy-theory", "stable-homotopy"], "votes": 31, "creation_date": "2021-07-19T16:48:31", "comments": ["Probably you aware by now but just in case, there is also this preprint \"On the Last Kervaire Invariant Problem\" by Weinan Lin, Guozhen Wang, and Zhouli Xu (arxiv.org/abs/2412.10879) which also claims to resolve the final case of dimension 126 in the affirmative.", "You're most welcome. \"Suitable for MathOverflow\" is among my favorite genres of MathOverflow question!", "Thank you for asking this question in a way that is suitable for MO! (For those wondering: it's asking about mathematical content/insight, not correctness of a preprint)"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "80", "site": "mathoverflow", "title": "On the definition of regular (non-noetherian, commutative) rings", "body": "All rings are commutative with unit. A ring $R$ is called _regular_ if it satisfies \n\n**(Reg)** Every finitely generated ideal of $R$ has finite projective dimension. \n\nClearly this gives the usual definition if $R$ is noetherian. In the non-noetherian world, valuation rings are regular (this is the reason why I got interested), as are polynomial rings in arbitrary sets of variables over a field. \nI am wondering about the reason for this definition, as opposed in particular to:\n\n**(Reg')** Every finitely presented $R$-module has finite projective dimension.\n\nClearly (Reg') implies (Reg), and they are equivalent if $R$ is _coherent_ (every finitely generated ideal is finitely presented). \nTo me, (Reg') looks much more natural than (Reg), and from my limited knowledge of the literature it seems that the association \"coherent+regular\" is quite popular. So, my question is \n\n> What are examples of regular rings not satisfying (Reg')? \n\nSince my post is really about motivating the definition, I am of course more interested in \"natural\" examples (not just constructed for the purpose). \nOn the other hand, all examples are welcome since I don't know any.\n", "link": "https://mathoverflow.net/questions/327031/on-the-definition-of-regular-non-noetherian-commutative-rings", "tags": ["ag.algebraic-geometry", "ac.commutative-algebra", "homological-algebra", "regular-rings", "valuation-rings"], "votes": 31, "creation_date": "2019-04-03T01:38:58", "comments": ["If I understand correctly, to show that regular Noetherian is regular in this sense, you have to use the fact that for Noetherian rings being proj dimension Question: Are all simple, separable, infinite-dimensional $C^*$-algebras isomorphic as Banach spaces?\n", "link": "https://mathoverflow.net/questions/156621/when-are-two-c-algebras-isomorphic-as-banach-spaces", "tags": ["fa.functional-analysis", "oa.operator-algebras"], "votes": 31, "creation_date": "2014-02-03T15:01:12", "comments": ["This has been answered many times in the comments but is the pair $c_0$ (the null sequences) and the algebra of continuous functions on the closed unit interval not a simple counter example to your initial question?", "It is an open problem whether $C^∗_r(F_2)$ is Banach isomorphic to a nuclear $C^*$-algebra (say the CAR algebra $B$ as a model). It is known that (1) the dual of $B$ has the bounded approximation property (because it's AFD), (2) $B$ has a Schauder basis (Junge--Nielsen--Ruan--Xu, Adv. Math. 2004), and (3) $B\\cong B\\otimes_\\alpha B$ for some reasonable tensor product (in this case the spatial tensor product). None of these properties is known for $C^∗_r(F_2)$.", "@HannesThiel, so maybe you could organize all this long conversation into the single answer?", "@Tomek: Thank you for pointing out that reference, I was not aware of it. It deals exaclty with the topic I asked about.", "As Taka pointed out, among the simple, separable, non-type I C*-algebras, all the nuclear ones are isomorphic as Banach spaces (even much more is true) but beyond the nuclear case this is no longer the case. This certainly answers my question, thanks.", "Hannes, this ma.utexas.edu/users/rosenthl/pdf-papers/93.pdf survey article of Rosenthal gives a nice overview of the Banach/operator-space structure of C*-algebras.", "@BillJohnson: HRS proved that none of Type II and III noncommutative $L^1$ spaces Banach embed into a type I noncommutative $L^1$ space. If $A$ is a type I $C^*$-algebra, then $A^*$ is a predual of type I von Neumann algebra.", "Thanks, Nik. Taka, can you give a lazy old man a hint as to why the HRS classification theorem yields that Type I is invariant under Banach space isomorphisms?", "By Szankowski (Acta Math 1981), there is a separable (simple, unital, etc.) $C^*$-algebra which does not have the approximation property. Such a $C^*$-algebra cannot be Banach isomorphic to a nuclear one.", "A complement to Caleb's remark: Being type I is invariant under a Banach space isomorphism (at least in the separable case). This follows from Haagerup--Rosenthal--Sukochev's classification of noncommutative $L^1$-spaces up to Banach isomorphism.", "@Caleb: Thanks for pointing out the result of Kirchberg. This clarifies the picture a lot. The question now seems to be whether the Banach space structure can detect nuclearity of the C*-algebra.", "@Qiaochu: Thanks for clarifying. Here, two Banach spaces E and F are called isomorphic if there exists a bounded linear map from E to F that is bijective. (The inverse map will automatically be bounded, too). I do not want to assume isometric isomorphism.", "Hi Hannes, Kirchberg showed in his subalgebras of CAR algebra paper that all separable nuclear, non Type I C*-algebras are isomorphic as operator spaces (much stronger than just isomorphic as Banach spaces). I think Christensen and others also did some things in this direction for von Neumann algebras: those results are in Pisier's operator space book, but my copy isn't with me.", "@Bill: the \"simplest\" example of a simple C*-algebra (hard not to make puns here) is the algebra $K(H)$ of compact operators on a separable Hilbert space. The irrational rotation algebras $A_\\theta \\subset B(L^2({\\bf T}))$ generated by multiplication by $e^{2\\pi i t}$ and rotation by the angle $2\\pi\\theta$ for irrational $\\theta$ are also simple. The CAR algebra is obtained by embedding $M_{2^n}$ in $M_{2^{n+1}}$ via the map $A \\mapsto \\left[\\matrix{A&0\\cr 0&A}\\right]$ and completing the union $\\bigcup_{n=1}^\\infty M_{2^n}$; it is also simple.", "I assume \"isomorphic as Banach spaces\" means linear homeomorphism. It's well known that there are non-isomorphic C*-algberas that are isometric as Banach spaces. (Google \"C*-algebra not isomorphic to its opposite algebra.) The correct statement is: A and B are isomorphic as C*-algebras if and only if they are completely isometric as operator spaces.", "What are some simple and complicated examples of simple separable $C^*$-algebras?", "Is an isomorphism of Banach spaces a bounded linear map with bounded linear inverse or an isometric isomorphism?"], "comment_count": 17, "category": "Science", "diamond": 0} {"question_id": "83", "site": "mathoverflow", "title": ""Three great cocycles" in Complex Analysis as cohomology generators", "body": "In his [lecture notes](http://www.math.harvard.edu/~ctm/home/text/class/harvard/213a/00/html/course/course.pdf), C. McMullen discusses \"the three great cocycles\" in Complex Analysis: the derivative $$f\\mapsto\\log f',$$ the non-linearity $$f\\mapsto (\\log f')'dz$$ and the Schwarzian derivative $$f\\mapsto \\left(\\frac{f'''}{f'}-\\frac32\\left(\\frac{f''}{f'}\\right)^2\\right)(dz)^2$$ I would like to understand what \"cocycle\" means here precisely, and in which sense these maps are \"the only\" cocycles.\n\nFor diffeomorphisms of the circle, this is very clearly explained, e. g., in the [book](http://www.math.psu.edu/tabachni/Books/BookPro.pdf) by Ovsienko and Tabachnikov. Namely, $\\mathrm{Diff}(S^1)$ is a group acting naturally on each of the spaces $F_\\lambda(S^1)$ of tensor densities of degree $\\lambda$. The corresponding (group) 1-cohomology is trivial unless $\\lambda$ equals $0$, $1$, or $2$, where it is one-dimensional and generated by the above cocycles.\n\nQuestion: is there an equally simply formulated statement in the context of complex variable, e. g. treating the above cocycles as generators of a group (or perhaps groupoid?) cohomology? This [MO answer](https://mathoverflow.net/a/38133/56624) suggests that the group is $\\mathrm{Aut}$(meromorphic functions), but, at least naively, this seems to coincide with a subgroup of the Mobius group.\n\n**Update:** Let me restate the question in simpler terms.The maps above satisfy the cocycle equation $$ D(g\\circ f)(z)=(f'(z))^\\lambda D(g)(f(z))+D(f(z)), $$ for $\\lambda=0,1,2$, respectively. For a tensor density $\\psi$, one can define the coboundary $$ D_\\psi:f\\mapsto (f')^\\lambda(\\psi\\circ f)-\\psi, $$ satisfying the same equation. So, the question is: what are the explicit conditions guaranteeing that the three maps above are the only, up to coboundaries, solutions to the cocycle equation?\n", "link": "https://mathoverflow.net/questions/231823/three-great-cocycles-in-complex-analysis-as-cohomology-generators", "tags": ["reference-request", "cv.complex-variables", "group-cohomology"], "votes": 31, "creation_date": "2016-02-22T05:57:23", "comments": ["I'm confused by the notation, especially $D$ and $\\circle$.", "See perhaps \"Explicit formulae for cocycles of holomorphic vector fields with values in $ \\lambda$ densities\" by Wagemann (arxiv.org/pdf/math-ph/0003035.pdf).", "@WillSawin, well, in the case of the circle the space of tensor densities of degree $\\Lambda$ is just the vector space of smooth functions on the circle, endowed with the diffeomorphism group action $g\\mapsto (f')^\\lambda \\cdot g\\circ f$. In the complex variable case, it's a part of my question what this space should be...", "What is a tensor density?"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "84", "site": "mathoverflow", "title": "When is a compact topological 4-manifold a CW complex?", "body": "Freedman's $E_8$-manifold is nontriangulable, as proved on page (xvi) of the Akbulut-McCarthy 1990 Princeton Mathematical Notes \"Casson's invariant for oriented homology 3-spheres\". Kirby showed that a compact 4-manifold has a handlebody structure if and only if it is smoothable: [1](http://www.manifoldatlas.him.uni-bonn.de/index.php/Questions_about_surgery_theory) and [2](https://mathoverflow.net/questions/7921/failure-of-smoothing-theory-for-topological-4-manifolds/38739#38739). When is a compact topological 4-manifold a CW complex? \n", "link": "https://mathoverflow.net/questions/73428/when-is-a-compact-topological-4-manifold-a-cw-complex", "tags": ["at.algebraic-topology", "open-problems", "4-manifolds", "cw-complexes", "topological-manifolds"], "votes": 31, "creation_date": "2011-08-22T11:32:18", "comments": ["I have good reason to believe that it is an open question! Apologies - I hadn't seen the earlier posting mathoverflow.net/questions/36838/… ", "I take it this is an open problem? Small note, but your question was effectively asked in another form, here: mathoverflow.net/questions/36838/… I usually think of CW-complexes as being a tool for describing homotopy-types rather than homeomorphism types, so my answer was to a weaker question than the one asked. "], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "85", "site": "mathoverflow", "title": "Cohomology of symmetric groups and the integers mod 12", "body": "When $n \\ge 4$, the third homology group $H_3(S_n,\\mathbb{Z})$ of the symmetric group $S_n$ contains $\\mathbb{Z}_{12}$ as a summand. Using the universal coefficient theorem we get $\\mathbb{Z}_{12}$ as a summand of the cohomology group $H^3(S_n,\\mathbb{Z}_{12})$. \n\n**What's a nice 'formula' for a $\\mathbb{Z}_{12}$-valued 3-cocycle on $S_n$ that generates this summand of the third cohomology?**\n\nIn more concrete terms, I'm looking for a nontrivial recipe to get an element $c(g,g',g'') \\in \\mathbb{Z}_{12}$ from three elements of $S_n$, such that\n\n$$ c(g',g'',g''') - c(gg',g'',g''') + c(g,g'g'',g''') - c(g,g',g''g''') + c(g,g',g'') = 0 $$\n\nfor all $g,g',g'',g''' \\in S_n$. Here \"nontrivial\" means that for all nonzero $\\alpha \\in \\mathbb{Z}_{12}$, we do not have\n\n$$ \\alpha c(g,g',g'') = f(g',g'') - f(gg',g'') + f(g,g'g'') - f(g,g') $$\n\nfor some $f \\colon S_n \\times S_n \\to \\mathbb{Z}_{12}$.\n", "link": "https://mathoverflow.net/questions/314626/cohomology-of-symmetric-groups-and-the-integers-mod-12", "tags": ["gr.group-theory", "group-cohomology"], "votes": 30, "creation_date": "2018-11-05T12:17:19", "comments": ["@DevSinha - can you explain that class and how it works in more detail, or point me toward an explanation? I wanted a \"formula\", but an explicit geometrical description like this would also be nice.", "@PhillippLampe - another typo! I'll fix it, thanks.", "@JohnBaez Why do the second and the third summand in the cocycle condition have the same sign? Should the sign not alternate?", "From a number theoretic perspective, the third homology group of the general linear group is related to algebraic $K$-theory, namely $K_3$ (this seems relevant because $S_n$ embeds into $\\mathrm{GL}_n$). In general the definition of $K$-group is highly non-constructive but $K_3$ is related to the Bloch group which is very explicit and some people have studied its torsion. The group $K_3(\\mathbb{Z})=K_3(\\mathbb{Q})$ is cyclic of order 48. By conjectures of Quillen-Lichtenbaum and Bloch-Kato the size of $(K_3)_\\mathrm{tors}$ is linked with $\\zeta'(-1)=-1/12$.", "@JohnBaez, I can tell you about the geometry of this cocycle, using Fox-Neuwirth cochains in the unordered configuration space model for $BS_n$: this class is represented by having four points which share a coordinate, and three points which all share a coordinate, with two sharing another coordinate. There is probably a prettier subvariety which represents this, but I don't have any techniques/ideas to get at those.", "@JohnBaez -- I guess you've seen the formulae for 3-cocycles on dihedral groups in the math-phys literature? It doesn't directly help, since you assume $n \\geq 4$... but it gives some idea of what might be possible.", "It would be equivalent (and also interesting) to demonstrate a $\\Bbb Z/12$-valued invariant for bordism classes of oriented 3-manifolds equipped with $n$-fold covering space, $n \\geq 4$. It's not so clear to me that this is any easier, though.", "This doesn’t really address your question, but I once saw Fernando Rodriguez-Villegas give a talk called “All 12s are the same.” The theme was exactly your motivation here - that of describing deep connections that underlie various occurrences of the number 12 throughout mathematics. I don’t recall if this occurrence was discussed, but he might be a good guy to hit up as a curator of 12s.", "@YCor Yes thank you, I guess I constructed an element of $H_3(S_n)$ coming from the generator of $H_3(A_n)$ but maybe I am confused here too... At any rate I did not think at all about how to move to $H^3$", "@მამუკაჯიბლაძე $H_3$ is not cyclic in this case, so I'm confused what you want to call a generator in $H^3(S_n,Z/12Z)$. You maybe mean an element of order 12.", "On the afterthought - I think I had (at least) one thing imprecise: $S_5$ embeds in $O(3)$, and its pullback to $Pin(3)$ is actually a double cover of $S_5$, so you get a 3-cycle for that double cover, which must then be descended back to $S_5$ via transfer/induction/Gysin/whateveritscalled...", "@JohnBaez Yes I think one might also obtain a generator of $H_3(A_n)$ this way. I said a little bit about it in an answer here", "Our comments crossed paths. Okay, so your idea (in the case $S_5$) is based on embedding $S_5$ as a subgroup of $\\mathrm{Pin}(3)$. This is closely related to Epa and Gantner's method embedding of the double cover of $A_n$ in $\\mathrm{Spin}(n)$.", "I didn't see that recipe in there. I like it but I don't quite understand it. It seems there should be a direct way to see how 3-simplices in the triangulation you describe give 3-chains in some chain complex that computes the homology of $S_5$. But I'm not seeing it.", "Hmmm I don't actually see clearly a proof that for larger $n$ the cycle I describe will continue to be a generator - in fact that it will not even bound. In detail one should do this: realise $S_5$ as a subgroup in $Pin(3)$; for larger $n$, find an $m$ such that there is an embedding of $Pin(3)$ into $Pin(m)$ that (1) induces isomorphism on $H_3$ and (2) carries $S_5$ into a subgroup of a $S_n$ subgroup. While this will give what is needed, what I don't see is whether you can choose such a thing in such a way that $S_5$ will land in a subgroup as indicated in my previous comment...", "And I think that answer linked by @QiaochuYuan contains also a recipe for an explicit 3-cycle representing a generator in $H_3$. Take the Cayley graph for some presentation of $S_5$ by generators and relations - e. g. the Coxeter one. It gives a polyhedral subdivision of the 3-sphere (for the Coxeter presentation it consists of all (hyper)faces of a permutahedron). Triangulate it, and take the sum of all obtained 3-simplices (with signs according to the orientation). For larger $n$, just choose any $S_5$ subgroup on a 5-element subset of $\\{1,...,n\\}$", "Yes, it needs that summand. Fixed, thanks!", "I believe the last (non-coboundary) condition needs one more summand, $-f(g,g')$", "@MinseonShin - yes, I'll fix that. Thanks!", "Should the domain of $f$ be two copies of $S_{n}$?", "@QiaochuYuan - that's a great explanation of what's secretly motivating my question! However, I'm looking for something maximally concrete: a 'formula', as explicit as possible, that computes an element of $\\mathbb{Z}_{12}$ from 3 permutations, obeying the conditions I listed. If you have to use conformal nets to get that element, okay... but I'm hoping there's a formula that combinatorists or finite group theorists could more easily enjoy. Sort of like the sign of a permutation, only a lot subtler.", "Here's a thing I wrote which is maybe relevant to your question: mathoverflow.net/a/196130/290 In the frame of that answer you are looking for something like a suitably nontrivial action of $S_n$ on an object in a suitably nice 3-category, and you can try to get such a thing by taking powers of an invertible object in a symmetric monoidal 3-category (for example, as suggested in that answer, conformal nets). The action of $S_n$ on such powers factors through the $3$-truncation of the sphere spectrum which naturally introduces a $\\mathbb{Z}_{24}$.", "OK. Note that this is equivalent to independently asking for (a) an explicit 3-cocycle $c_4$ valued in $Z/4Z$, such that $2c_4$ is not a coboundary; (b) an explicit 3-cocycle $c_3$ valued in $Z/3Z$, which is not a coboundary. [The solutions $c_3$ and $2c_4$ are unique modulo coboundaries, but not $c_4$.]", "@ChrisGerig - an explicit 4-term exact sequence would be a big step in the right direction and in some ways even better than the cocycle.", "Since I'm trying to get to the bottom of why the numbers 12 and 24 show up in many contexts in mathematics, I greatly prefer a $\\mathbb{Z}_{12}$-valued cocycle and suspect there will be a good answer to my question, even if technically speaking it involves some noncanonical choices. However, beggars can't be choosers! So, a $\\mathbb{Z}_6$-valued cocycle would also be okay.", "The homology is isomorphic to $C_{12}\\oplus C_2(\\oplus C_2)$ for $n\\ge 4$ (without the second $\\oplus C_2$ for $n=4,5$ groupprops.subwiki.org/wiki/…) but there is no reason that the $C_{12}$ summand to be canonical a priori. If you're looking for a nice formula it's likely that you don't want non-canonical choices and hence have to accept these few $C_2$ terms (on the other hand, modding out the 2-torsion should give something canonical, in $C_6$).", "Would you accept the following rephrasal? The generator of your summand in $H^3(S_n,\\mathbb{Z}/12)$ corresponds to a 4-term exact sequence (a crossed module extension) $0\\to \\mathbb{Z}/12\\to N\\to E\\to S_n\\to 0$, from which you can write down an explicit 3-cocycle using a choice of cross-section $S_n\\to E$. I am not sure if this is due to MacLane (could check later)."], "comment_count": 27, "category": "Science", "diamond": 0} {"question_id": "86", "site": "mathoverflow", "title": "Is there an Ehrhart polynomial for Gaussian integers", "body": "Let $N$ be a positive integer and let $P \\subset \\mathbb{C}$ be a polygon whose vertices are of the form $(a_1+b_1 i)/N$, $(a_2+b_2 i)/N$, ..., $(a_r+b_r i)/N$, with $a_j + b_j i$ being various Gaussian integers. For any Gaussian integer $u+vi$, let $h_P(u+vi)$ be the number of lattice point in $(u+vi) \\cdot P$. I would like to say that $h_P(u+vi)$ is in some sense \"quasi-polynomial modulo $N$\". To emphasize: **The reason that this does not follow from the ordinary Ehrhart polynomial theorem is that I am multiplying by a Gaussian integer, not just an ordinary integer.**\n\nThe case $N=1$ can be studied using [Pick's theorem](http://en.wikipedia.org/wiki/Pick%27s_theorem). The area of $P$ transforms simply under multiplication by Gaussian integers, the number of lattice points on the boundary is only a little messier. In particular, it is easy to show that there is some non-zero Gaussian integer $D$ such that, as long as $GCD(D, u+vi)=1$ and $GCD(u,v)=1$, then $h_P(u+vi) = (u^2+v^2) \\mathrm{Area}(P) + \\mbox{constant}$.\n\nA result that would satisfy me is that there is some non-zero non-zero Gaussian integer $D$ such that, as long as $GCD(D, u+vi)=1$ and $GCD(u,v)=1$, then $$h_p(u,v) = (u^2+v^2) \\mathrm{Area}(P) + a(u,v) u + b(u,v) v + c(u,v)$$ where $a$, $b$ and $c$ are periodic modulo $N$ in each input.\n\nMotivation: This is enough to give an elementary deduction that the [quartic residue](http://en.wikipedia.org/wiki/Quartic_reciprocity#Quartic_residue_character) symbol $\\left[ \\frac{z}{u+vi} \\right]$ is periodic as a function of $u+vi$.\n\nMotivation behind the motivation: There are a series of papers: [Habicht](http://www.ams.org/mathscinet-getitem?mr=113867), [Kubota](http://www.ams.org/mathscinet-getitem?mr=921585), [Hill](http://www.ams.org/mathscinet-getitem?mr=1324545) which promise to prove $m$-th power reciprocity laws roughly by counting lattice points. I can't follow any of them, so I thought I would true to reverse engineer what might be true in order to make the $\\mathbb{Z}[i]$ proof come out.\n\nFor this reason, I would also be interested in statements for other number fields, but I don't have a precise question worked out in that setting.\n", "link": "https://mathoverflow.net/questions/165884/is-there-an-ehrhart-polynomial-for-gaussian-integers", "tags": ["nt.number-theory", "mg.metric-geometry", "lattices"], "votes": 30, "creation_date": "2014-05-11T18:46:47", "comments": ["Dear David. Sorry for the late question, but, did you eventually found some place (in addition to the articles you mention) where this is explained/worked, hopefully for other number fields?", "Yeah, right, I didn't think straight. Sorry for noise.", "I do not see this. A rotation can change the lattice in a quite messy way. What am I missing?", "naively, at least if $|a+ib|\\in\\mathbb{Z}$, this should follow from the classical case; indeed, by multiplying by $a+ib$ you apply to your $\\mathbb{C}$-plane the composition of a rotation and the scalar matrix $C=|a+ib|I$, and so the integer points would behave in the same way as by scaling with $C$ alone.", "@DimaPasechnik Because I'm multiplying by Gaussian integers, not ordinary integers. Edited to clarify.", "I was under impression that the quasiperiodicity of Ehrhart polynomial holds for any lattice $\\Lambda$ in $\\mathbb{R}^d$ and a polytope with vertices in $\\Lambda$. How does your setting differ?"], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "87", "site": "mathoverflow", "title": "Nontrivial tangent bundle that is diffeomorphic to the trivial bundle", "body": "Is there an example of a smooth $n$-manifold $M$ whose tangent bundle is nontrivial as a bundle but is nonetheless (abstractly) diffeomorphic to the trivial bundle $M \\times \\mathbb{R}^n$?\n\n(This question was inspired by [Trivial fiber bundle](https://mathoverflow.net/questions/58685/trivial-fiber-bundle).)\n", "link": "https://mathoverflow.net/questions/58793/nontrivial-tangent-bundle-that-is-diffeomorphic-to-the-trivial-bundle", "tags": ["dg.differential-geometry", "differential-topology"], "votes": 30, "creation_date": "2011-03-17T16:25:39", "comments": ["$T^*G\\cong G\\times \\mathfrak g^*\\cong G\\times \\mathbb R^{n}$", "@Dylan and @Somnath: Tangent bundles over spheres never give such examples. Moreover, in general, if the homotopy-equivalence $M\\to M$ induced by a diffeomorphism $TM\\to M\\times {\\mathbb R}^n$ is homotopic to a diffeomorphism $M\\to M$, then $TM$ is trivial. See \"Diffeomorphism of total spaces and equivalence of bundles\" by De Sapio and Waldschap. This still leaves open the case when $M$ is a homotopy-sphere though. ", "@Dylan - That was the example I was thinking of too! It is clear that $TS^5$ is not the trivial rank $5$ bundle over $S^5$. It only has $3$ linearly independent nowhere vanishing vector fields (for lack of anything elementary one can use Adams' vector fields on spheres result). Although I don't know how to do it myself, if you can prove (extend your sphere bundle result to vector bundles) your claim, we're done! ", "I think I can show that the unit tangent bundle over $S^5$ is diffeomorphic to the trivial one... but I'm having trouble otherwise, since most of the diffeomorphisms I can think of rely on theorems about compact spaces. ", "@David, is no reason for the diffeomorphism $TM\\to M\\times \\mathbb R^n$ to cover a diffemorphism of $M$. @Faisal, the answer to your question is almost certainly yes, but I do not see an explicit example at the moment, cf. my answer to mathoverflow.net/questions/58685/trivial-fiber-bundle.", "A trivial observation: such an isomorphism would cover an automorphism of $M$. A possibly unhelpful observation: One could try to put a Riemannian metric on $M$ and see what constraints this gives in the light of my first remark."], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "88", "site": "mathoverflow", "title": "Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$?", "body": "> Is there a field $F$ such that $F \\cong F(X,Y)$ as fields, but $F \\not \\cong F(X)$ as fields?\n\nI know only an example of a field $F$ such that $F$ isomorphic to $F(x,y)$ : this is something like $F=k(x_0,x_1,\\dots)$. But in this case we have $F \\cong F(x)$.\n\nThis is related to [this](https://mathoverflow.net/questions/225896) question. This was previously [asked](https://math.stackexchange.com/questions/2000333) on MSE but after 2 months and a bounty, no answer was provided.\n\nThank you very much!\n", "link": "https://mathoverflow.net/questions/258647/is-there-a-field-f-which-is-isomorphic-to-fx-y-but-not-to-fx", "tags": ["ra.rings-and-algebras", "examples", "fields"], "votes": 30, "creation_date": "2017-01-03T03:25:32", "comments": ["Oops, I forgot to mention that I've posted a related question here: mathoverflow.net/questions/259117/…", "I have an example where I know $F\\cong F(X,Y)$ but I don't know whether $F\\cong F(X)$. There are torsion-free abelian groups $G$ such that $G\\cong G\\oplus\\mathbb{Z}\\oplus\\mathbb{Z}$ but $G\\not\\cong G\\oplus\\mathbb{Z}$ (see mathoverflow.net/questions/218113/… for example). Let $\\mathbb{Q}(G)$ be the field of fractions of the group algebra $\\mathbb{Q}[G]$. Then $\\mathbb{Q}(G)\\cong\\mathbb{Q}(G)(X,Y)$, but I don't see why it should be isomorphic to $\\mathbb{Q}(G)(X)$.", "Related on MSE: math.stackexchange.com/questions/388291"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "89", "site": "mathoverflow", "title": "The field of fractions of the rational group algebra of a torsion free abelian group", "body": "Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\\mathbb{Q}[G]$ is an integral domain. Let $\\mathbb{Q}(G)$ be its field of fractions.\n\n> Are there non-isomorphic torsion free abelian groups $G$ and $H$ such that $\\mathbb{Q}(G)\\cong\\mathbb{Q}(H)$? If so, are there easy examples?\n\nThis is motivated by [this question](https://mathoverflow.net/questions/258647/is-there-a-field-f-which-is-isomorphic-to-fx-y-but-not-to-fx). There are [examples](https://mathoverflow.net/questions/218113/a-is-isomorphic-to-a-oplus-mathbbz2-but-not-to-a-oplus-mathbbz) of torsion free abelian groups $G$ where $G\\cong G\\oplus\\mathbb{Z}\\oplus\\mathbb{Z}$, but $G\\not\\cong G\\oplus\\mathbb{Z}$. Since $\\mathbb{Q}(G\\oplus\\mathbb{Z}\\oplus\\mathbb{Z})\\cong\\mathbb{Q}(G)(X,Y)$ and $\\mathbb{Q}(G\\oplus\\mathbb{Z})\\cong\\mathbb{Q}(G)(X)$, a negative answer to my question would give a positive answer to the motivating question.\n\nI've found a little about this construction in the literature, but not much. For example, in\n\n_Gervasio G. Bastos and T. M. Viswanathan_ , MR 942063 [**Torsion-free abelian groups, valuations and twisted group rings**](http://dx.doi.org/10.4153/CMB-1988-021-x), _Canad. Math. Bull._ **31** (1988), no. 2, 139--146. \n\nit is proved that $\\mathbb{Q}(G)$ is a purely transcendental extension of $\\mathbb{Q}$ if and only if $G$ is free abelian.\n\nAny other relevant references would be welcome. Also, there are obvious variants: e.g., using a different field, such as $\\mathbb{F}_2$ instead of $\\mathbb{Q}$. \n", "link": "https://mathoverflow.net/questions/259117/the-field-of-fractions-of-the-rational-group-algebra-of-a-torsion-free-abelian-g", "tags": ["gr.group-theory", "ac.commutative-algebra", "fields", "abelian-groups"], "votes": 29, "creation_date": "2017-01-08T13:30:07", "comments": ["This question could be related.", "Lovely question Jeremy. I only wish I could say something useful about it!", "Just a remark: if $G$ has no primitive element (in the sense that for every $g\\in G$ there exists $n\\ge 2$ such that $n^{-1}g\\in G$) then I guess that $\\pm G$ is characterized within $\\mathbb{Q}(G)^*$ as those elements with roots of arbitrarily high order. (This holds if $G$ is a $\\mathbb{Z}[1/p]$-module, or if $G$ is noncyclic of rank 1.) Since $G$ is isomorphic to $(\\pm G,\\cdot)$ modulo its torsion subgroup, it follows that $\\mathbb{Q}(G)$ recognizes $G$ in this case."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "90", "site": "mathoverflow", "title": "Linking formulas by Euler, Pólya, Nekrasov-Okounkov", "body": "Consider the formal product $$F(t,x,z):=\\prod_{j=0}^{\\infty}(1-tx^j)^{z-1}.$$ (a) If $z=2$ then on the one hand we get Euler's $$F(t,x,2)=\\sum_{n\\geq0}\\frac{(-1)^nx^{\\binom{n}2}}{(x;x)_n}t^n,$$ on the other we get Pólya's formula (the \"cycle index decomposition\", see Stanley's EC2, p.19) $$F(t,x,2)=\\sum_{n\\geq0}Z(S_n,(1-x)^{-1},\\dots,(1-x^n)^{-1})t^n.$$ (b) If $t=x$ then we get Nekrasov-Okounkov's $$F(x,x,z)=\\sum_{n\\geq0}x^n\\sum_{\\lambda\\vdash n}\\prod_{\\square\\in\\lambda}\\left(1-\\frac{z}{h_{\\square}^2}\\right);$$ where $h_{\\square}$ is the hook-length (for notations and references you may see _[Theorems, problems and conjectures](http://dauns.math.tulane.edu/%7Etamdeberhan/conjectures.html)_).\n\n> **My Question still waiting for an answer.** Is there a unifying combinatorial right-hand side in $$\\prod_{j\\geq0}(1-tx^j)^{z-1}=?$$\n\n**Remark 1.** One possibility is perhaps a refinement in line with (b) involving hook-lengths.\n\n**Remark 2.** Even special cases, such as evaluation for specific numerical values of $t$ and/or $z$, would be interesting.\n\n**Remark 3.** Yet another direction is to find a hook-length expansion for $F(t,x,2)$. This could be fascinating because the result will connect the cycle index polynomial with hooks.\n", "link": "https://mathoverflow.net/questions/251068/linking-formulas-by-euler-p%c3%b3lya-nekrasov-okounkov", "tags": ["nt.number-theory", "co.combinatorics", "rt.representation-theory", "power-series", "gauge-theory"], "votes": 29, "creation_date": "2016-09-29T21:18:19", "comments": ["Neither one follows from the other.", "never mind -- NO formula generalizes the cycle index", "does Nekrasov-Okounkov formula really fall out of cycle index for permutation groups??"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "91", "site": "mathoverflow", "title": "Non-linear expanders?", "body": "Recall that a family of graphs (indexed by an infinite set, such as the primes, say) is called an _expander family_ if there is a $\\delta>0$ such that, on every graph in the family, the discrete Laplacian (or the adjacency matrix) has spectral gap $|\\lambda_0-\\lambda_1|\\geq \\delta$. (Assume all graphs in the family have the same degree (= valency) $d$.)\n\nIf, for every $p$, we specify some maps from $\\mathbb{Z}/p\\mathbb{Z}$ to itself, then we are defining a graph with vertex set $\\mathbb{Z}/p\\mathbb{Z}$ for each $p$: a vertex is adjacent to the vertices it is mapped to.\n\nIf we consider the maps $x\\to x+1$, $x\\to 3x$ (say), then we do not get expanders. On the other hand, if we take $x\\to x+1$, $x\\to x^{-1}$, then, as is widely known, we do get an expander family.\n\nConsider the maps\n\n$x\\mapsto x+1$, $x\\mapsto x^3$.\n\nDo they (and their inverses, if you wish) give an expander family? (Let $p$ range only over primes $p\\equiv 2 \\mod 3$, so as to keep the map $x\\mapsto x^3$ injective.)\n\nWhat about the maps $x\\mapsto x+1$, $x\\mapsto 3 x$, $x\\mapsto x^3$? Do they, taken together, give an expander family?\n\n(Is there a way to relate these maps to the action of a linear group? Are there examples of sets of not neessarily linear maps giving rise to expander families?)\n\nUPDATE: What if these were shown _not_ to be expanders? What would be some interesting consequences?\n", "link": "https://mathoverflow.net/questions/168426/non-linear-expanders", "tags": ["nt.number-theory", "gr.group-theory", "graph-theory"], "votes": 29, "creation_date": "2014-05-28T07:17:01", "comments": ["Can you say anything on the structure of those $A\\subset{\\mathbb F}_p$ with $|(3A)\\setminus(A+1)|3$, one can consider analogous questions, with a variety of types of a simplicial cage, by taking the $i$-skeleton of a convex $n$-dimensional simplex, with $1\\le i\\le n-2$.\n\nA trivial example: A ball is trapped in the cage consisting of the $1$-skeleton of the regular tetrahedron edge-tangent to the ball. Also, obviously, every convex body can be trapped in some polyhedral cage.\n\nFor an interested reader, I suggest a few of somewhat less trivial exercises: each of the following convex bodies can be trapped in a tetrahedral cage: the cube, the circular cylinder of any (finite) height, the circular cone, the regular tetrahedron.\n", "link": "https://mathoverflow.net/questions/379560/can-every-3-dimensional-convex-body-be-trapped-in-a-tetrahedral-cage", "tags": ["mg.metric-geometry", "convex-geometry"], "votes": 28, "creation_date": "2020-12-22T20:17:05", "comments": ["@Anton The half-circle, for example, and many other, related.", "In 2-dimensional case, one can ask which convex figures can be trapped by three nails. Rectangle cannot be trapped, but I do not see other examples.", "@DanRomik Romik : Yes, it is quite a bit different.", "The paper How to cage an egg by Oded Schramm (Inventiones Math. 107 (1992), 543-560) comes to mind as potentially relevant, although the question he answers is a bit different than yours.", "@JosephO'Rourke: Yes. More precisely, some of the traps for the cone also trap the spindle, no matter how thin it is.", "Does your trap for a cone also trap a thin spindle? Say, two tall, base-to-base cones?"], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "93", "site": "mathoverflow", "title": "Mathieu group $M_{23}$ as an algebraic group via additive polynomials", "body": "An elegant description of the Mathieu group $M_{23}$ is the following: Let $C$ be the multiplicative subgroup of order $23$ in the field $F=\\mathbb F_{2^{11}}$ with $2^{11}$ elements. Then $M_{23}$ is the group of additive maps of $F$ to itself which permute the set $C$. The restriction to $C$ then is the natural action of $M_{23}$ of degree $23$.\n\nAny additive map of $F$ to itself has the form $x\\mapsto\\sum_{i=0}^{10}a_ix^{2^i}$ for $a_i\\in F$. The condition that such maps preserve $C$ can be expressed easily as an equivalent set of polynomial conditions on the coefficients $a_0,\\dots,a_{10}$. For instance, for a variable $t$, compare coefficients of $$\\prod_{c\\in C}(t-\\sum_{i=0}^{10}a_ic^{2^i})=t^{23}-1.$$ In addition, we need the equations $a_i^{2^{11}}=a_i$ to ensure that the $a_i$ are in $F$.\n\nHowever, the polynomials we obtain this way (or some other way, there are somewhat smarter possibilities) are extremely complicated.\n\nOf course, any finite subgroup of a linear group is an algebraic set, that is a solution of a system of polynomial equations. And in general such a system will be messy.\n\nHere, I somehow expect that there should be a simple system of polynomials discribing $M_{23}$ as an algebraic set. The obvious attempt would be to first take a system as above, and then hope that a Groebner basis looks better. After some naive attempts it seems that the systems are too complex for Magma or Singular to compute the Groebner bases.\n\nBefore trying to refine the approach, here is my question: Has anyone seen this version to describe $M_{23}$? \n", "link": "https://mathoverflow.net/questions/264513/mathieu-group-m-23-as-an-algebraic-group-via-additive-polynomials", "tags": ["ag.algebraic-geometry", "gr.group-theory", "polynomials", "finite-groups"], "votes": 28, "creation_date": "2017-03-13T06:48:14", "comments": ["@DaveBenson Indeed, I had also tried something along this idea. The upper bound of the Galois group of $Y^{23}+XY^3+1$ is achieved by showing that this polynomial divides an additive polynomial of degree $2^{11}$. So actually this additive polynomial has Galois group $M_{23}$, and its groups acts on an $11$-dimensional $\\mathbb F_2$-space.", "This is a very old question, but I can't help adding the following comment, that may or may not be relevant. I wonder whether Abhyankar's paper, \"Mathieu group coverings in characteristic two\" is relevant here? He demonstrates that there are very innocuous looking degree 23 trinomials over ${\\mathbb F}_2(X)$ with Galois group $M_{23}$. His example is $Y^{23}+XY^3+1$.", "The additive subgroup generated by $C$ is a subfield, is not reduced to $F_2$, and hence equals $F_{11}$. Hence the fixator of $C$ in $GL_{11}$ is reduced to the trivial subgroup. Hence the stabilizer $G$ of $C$ in $GL_{11}$ is finite (as an algebraic group). Since $|C|=23$ and the dimension is 11, there exists a 3-element subset $F$ of $C$ such that $C-F$ generates $F_{11}$ additively. Hence a 3-cycle on $F$ cannot belong to $G(K)$ for any extension $K$ of $F_2$. So the (faithful) action of $G(K)$ on $C$ is reduced to $M_{23}$, and hence $G(K)$ is reduced to $G(F_2)=M_{23}$ for every $K$.", "If I understand correctly, you don't know the answer (whether $G$ has positive dimension, where $G\\subset\\mathrm{GL}_{11}$ is the stabilizer of $C$).", "@YCor: I tried to answer that in my first comment above.", "Why are you not happy with just defining the $G$ as the subgroup $\\mathrm{GL}_{11}$ that preserves $C$? then $G(\\mathbf{F}_2)$ is reduced to the Mathieu group. No need to incorporate the conditions that entries are in $\\mathbf{F}_2$: this is part of the definition of $G(\\mathbf{F}_2)$. If $G$ is defined as I just did, I don't even know if it's positive-dimensional.", "one natural way to produce an explicit set of equations is to identify a polynomial $p$ left invariant by $M_{23}$, but not by any overgroup of it in GL(11,2), and write down equations for the matrix entries saying that $p$ is invariant.", "@DimaPasechnik Yes, I know that. But I believe that has nothing to do with my question.", "It is known how to express generators of $M_{23}$ as a permutation group in terms of rational univariate maps over $\\mathbb{F}_{23}$. (Not sure about the original reference, I saw it in a book by D. Gorenstein ams.org/mathscinet-getitem?mr=698782). More precisely, it gives such representation for $M_{24}$, as the maps in PSL(2,23) and one extra binomial of relatively high degree.", "@DimaPasechnik Right, one could try some smaller toy examples. The group $PSL(3,2)$ maybe is not that interesting though, because it equals $GL(3,2)$. So all that we polynomially encode would be bijectivity of the linear maps.", "@JimHumphreys I hope to use it for certain combinatorial questions about $M_{23}$, for instance in the study of sharply transitive sets of permutations. Such question have a polynomial formulation, so if the $M_{23}$ has a nice polynomial formulation too, I have some hope.", "E.g. one may ask for such a description for PSL(3,2): I suppose you have to replace 11 by 3 and 23 by 7, i.e. $|C|=7$ and $F=\\mathbb{F}_{2^3}$, and consider PSL(3,2) as the group of additive maps of $F$ preserving $C$. (Disclaimer---I didn't check all the details here :-)).", "the natural object preserved by the group here is dual of the perfect binary Golay code en.wikipedia.org/wiki/Binary_Golay_code One might ask for similar descriptions of smaller codes to begin with.", "This looks like a reasonable (though maybe quite difficult) question. But I wonder what consequences such a description would have?"], "comment_count": 14, "category": "Science", "diamond": 0} {"question_id": "94", "site": "mathoverflow", "title": "Derivative of Class number of real quadratic fields", "body": "Let $\\Delta$ be a fundamental quadratic discriminant, set $N = |\\Delta|$, and define the Fekete polynomials $$ F_N(X) = \\sum_{a=1}^N \\Big(\\frac{\\Delta}a\\Big) X^a. $$ Define $$ f_N(X) = \\frac{F_N(X)}{X^N-1}. $$ Canceling common factors it can be shown that $$ f_N(X) = \\frac{H_N(X)}{\\Phi_N(X)}, $$ where $\\Phi_N$ is the $N$-th cyclotomic polynomial and $H$ has integral coefficients. (These definitions seem rather ad hoc, but in fact the objects above are all quite \"natural\".)\n\nIf $\\Delta < 0$, then $f_N(1) = \\frac{2h}w$, where $h$ is the class number of $\\Delta$ and $w$ the number of roots of unity in the corresponding quadratic number fields. In fact we get $$ f_N(1) = - \\frac{\\sum_{a=1}^{N-1} a \\Big( \\frac{\\Delta}a\\Big)}N . $$\n\nIf $\\Delta > 0$, then $f_N(1) = 0$, which suggests looking at the first derivative. Here we find that $f_N'(1)$ is always even, so we set $f_N'(1) = - 2 h$ and get $$ h = \\frac{\\sum_{a=1}^{N-1} \\frac{a(a-1)}2 \\Big( \\frac{\\Delta}a\\Big)}N . $$ Here is a small table: \n$ $ $ \\begin{array} {c|ccccccccccc} N & 12 & 13 & 17 & 21 & 24 & 28 & 29 & 33 & 37 & 41 & 44 \\\\\\ \\hline h & 1 & 1 & 2 & 2 & 3 & 4 & 3 & 6 & 5 & 8 & 7 \\end{array} $\n\nIt is quite remarkable that $h$ seems to possess, for positive discriminants, properties analogous to class numbers. In particular, $h$ is odd if and only if $\\Delta = p > 5$ is a prime $\\equiv 5 \\bmod 8$, and if $\\Delta$ is divisible by $n$ distinct primes, then $h$ is divisible by $2^{n-2}$ etc.\n\nIn the case $\\Delta < 0$, Hasse and Bergstrom have shown how to extract the \"genus factor\" from the analytic class number formula, but the necessary calculations were horrible. Thus if we want to avoid going through similar technicalities in the case $\\Delta > 0$ then we would need a natural interpretation of $h$ as the order of an analogue of a \"class group\". \n\nSo I guess my question is whether there is such a mysterious group (connected to real quadratic number fields) with order $h$. Even a proof that $h \\ne 0$ would be welcome, since in the case of negative discriminants this is equivalent to the nonvanishing of Dirichlet's L-function at $s = 1$. \n\n**Edit.** As Rene Schoof has pointed out, the numbers in question are essentially generalized Bernoulli numbers $B_{2,\\chi}$ for (quadratic) Dirichlet characters with conductor $N$. These numbers are defined by $$ B_{2,\\chi} = \\sum_{a=1}^N \\chi(a) B_2(a/N), $$ where $B_2(X) = X^2 - X + \\frac16$ is the second Bernoulli polynomial. The constant term $\\frac 16$, in which this expression differs from the one I have used above, is irrelevant since $\\sum \\chi(a) = 0$.\n\nThese numbers $B_{2,\\chi}$ show up as factors of eigenspaces of cuspidal divisor class groups as studied in the book by Kubert-Lang (more exactly, the $p$-part of the group is reflected in the divisibility by $p$ of the Bernoulli numbers). The Birch-Tate-Lichtenbaum conjecture also predicts that the $B_{2,\\chi}$ are factors of $K_2({\\mathcal O})$, where ${\\mathcal O}$ is the ring of integers of the maximal real subfield of the $N$-th roots of unity (Kubert-Lang, p. 151).\n\nAlthough these numbers have such a prominent place in number theory, their elementary properties apparently have not been investigated (I've looked into Washington's and Lang's books on cyclotomic number fields). Numerical experiments suggest e.g. the following:\n\nLet $\\chi$ be an even Dirichlet character with conductor $N$. Then $h_2(N) = \\frac{N}4 B_{2,\\chi}$ is an integer for all $N > 8$. \nMoreover, $h_2(N)$ is odd if and only if\n\n 1. $N = p$ for primes $p \\not \\equiv 1 \\bmod 8$;\n 2. $N = 4p$ for primes $p \\equiv 3 \\bmod 8$;\n 3. $N = 8p$ for odd primes $p \\not \\equiv \\pm 1 \\bmod 8$. \n\n\n\nIn absence of any kind of genus theory for cuspidal divisor class groups it seems that the only option for proving such results is working directly with the definition of these numbers. As mentioned above, this becomes technical for integers with many prime factors.\n\nThere are similar results for $h_3(N) = \\frac{N^2}6 B_{3,\\chi}$, and there can be no doubt that most of this generalizes in some form to all generalized Bernoulli numbers $B_{n,\\chi}$. \n\nThis whole set of question came out of my approach to quadratic reciprocity in Chapter 7 of my notes on [Pell conics](http://www.rzuser.uni-heidelberg.de/~hb3/pell.html).\n", "link": "https://mathoverflow.net/questions/114434/derivative-of-class-number-of-real-quadratic-fields", "tags": ["nt.number-theory", "algebraic-number-theory"], "votes": 28, "creation_date": "2012-11-25T10:14:45", "comments": ["Hi! I think these are generalized Bernoulli numbers $B_{2,\\chi}$ associated the quadratic character $\\chi$ of conductor $\\Delta$. Apart from a power of $2$ perhaps, they are equal to the order of the $\\chi$-part of the cuspidal divisor class group of the modular curve $X_1(\\Delta)$. See the book by Kubert-Lang for more details."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "95", "site": "mathoverflow", "title": "derived category of equivariant coherent sheaves and fixed points", "body": "The K-group $K^T(X)$ of $T$(torus)-equivariant coherent sheaves on a variety $X$ is isomorphic to $K^T(X^T)$, that of the fixed point locus via the inclusion homomorphism, when we tensor the quotient field of the representation ring $R(T)$ of $T$. \n\nIs there a similar result for $D_T(Coh X)$ and $D_T(Coh X^T)$, derived categories of $T$-equivariant coherent sheaves ? I do not know even how to formulate `the quotient field of $R(T)$' in the derived category case.\n", "link": "https://mathoverflow.net/questions/48407/derived-category-of-equivariant-coherent-sheaves-and-fixed-points", "tags": ["kt.k-theory-and-homology", "ag.algebraic-geometry"], "votes": 28, "creation_date": "2010-12-05T16:09:13", "comments": ["@Ben, you're absolutely right, somehow I missed the salient word \"coherent\" in the question. But even for constructible sheaves, is there a localization statement at the level of derived categories?", "It doesn't seem like that to me at all. Is there anything about coherent sheaves in GKM?", "This seems like exactly the sort of thing that ought to be one of the main results in the paper of Goresky-Kottwitz-MacPherson. Unfortunately, I haven't been able to extract the kind of statement you're asking for; it would be very interesting if someone can give a precise answer. They do give a localization-type theorem for an element of the equivariant derived category in (7.8), but only after taking cohomology."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "96", "site": "mathoverflow", "title": "Computational complexity of topological K-theory", "body": "I am a novice with K-theory trying to understand what is and what is not possible.\n\nGiven a finite simplicial complex $X$, there of course elementary ways to quickly compute the cohomology of $X$ with field coefficients. However, it is known that computing the rational homotopy groups of $X$ is at least $NP$-hard (in fact, at least $\\\\# P$-hard), simply because it is possible to reduce NP-hard problems to the computations of some rational homotopy groups.\n\nFor any of: (1) Real K-theory, (2) Complex K-theory, (3) p-adically completed K-theory, is there an algorithm to compute $K^0$ of a finite simplicial complex? Is this algorithm polynomial-time?\n\nIf there is no known algorithm, is there at least evidence that any theoretical algorithm, should it exist, must be at least $NP$-hard? In other words, is there any way to reduce an $NP$-hard problem to the calculation of some $K$ group?\n", "link": "https://mathoverflow.net/questions/125745/computational-complexity-of-topological-k-theory", "tags": ["kt.k-theory-and-homology", "at.algebraic-topology", "computational-complexity"], "votes": 28, "creation_date": "2013-03-27T10:15:43", "comments": ["Just thought I'd add a citation for the fact that rational homotopy groups of simply-connected spaces are NP hard here. In particular, this is not just some $\\pi_1$ issue.", "Link: jointmathematicsmeetings.org/meetings/national/jmm2012/…", "Have you looked at Gunnar Carlsson's stuff? They have this software for computing homology; maybe it could be made to work with K-theory, or maybe they've addressed the computational complexity and that could help you bound the complexity of K-theory. Here is a link: comptop.stanford.edu. In a similar vein, Rob Ghrist has done work on algorithms for cohomology theory, but I'm not sure if he's addressed computational complexity or not. Last thought: at the JMM last year there was a special session about K-theory and electrical engineering. Maybe those guys have looked into this question", "Dominique Arlettaz has a result regarding the order of differentials in the Atiyah-Hirzebruch spectral sequence, see \"The order of the differentials in the Atiyah-Hirzebruch spectral sequence\". I would imagine that one might be able to make certain conclusions if you assumed enough about the torsion of the homology of the space. However, on rereading your question I realize this is not so much what you were interested in. Thought I'd mention the paper of Arlettaz just in case though.", "I have thought about this occasionally. There is some reason to think that there might be a tractable combinatorial construction of a minimal Kan complex of homotopy type $\\Omega^\\infty(KU/2)$. From that it should be possible to determine the computational complexity of $(KU/2)^*(X)$. Odd primes might be possible as well. I have various kinds of calculations related to this, but no real conclusion.", "Probably not. The first differential is easy (for KU), it's determined by a cohomology operation, but I think the higher ones are algorithmically intractable and would describe higher-order cohomology operations. Maybe one should try to describe vector bundles on finite simplicial complexes combinatorially instead? ", "Yes, but only as a black-box. I do not know an algorithm to compute differentials or solve extension problems. Is this possible?", "Are you familiar with the Atiyah-Hirzebruch SS?"], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "97", "site": "mathoverflow", "title": "Finite-dimensional subalgebras of $C^\\star$-algebras", "body": "Let $A$ be a unital $C^\\star$-algebra and let $a_1,\\dots,a_n$ be a finite list of normal elements in $A$ which (together with their adjoints) generate a norm-dense $\\star$-subalgebra $B \\subset A$. Clearly, if $A$ is finite-dimensional, then every element in $A$ (and hence $B$) has finite spectrum. I am asking for the converse.\n\n> **Question:** Assume that each element of $B$ has finite spectrum. Is it true that $A$ has to be finite-dimensional?\n\nThe existence of finitely-generated infinite torsion groups shows that this might be a highly non-trivial problem. In this case one would consider the reduced group $C^\\star$-algebra and note that all monomials in the generators of the group and their inverses (which are equal to the adjoints of the generators) would have finite spectrum. However, the generated algebra would still be infinite-dimensional. I do not know of any simpler way to come up with such an example. In this case it is conceivable that the random-walk operator associated with the generating set (which is an element in the real group ring) has infinite spectrum, even though I did not prove this.\n\nMaybe there is also need to consider the spectra of elements in matrices over $B$ (which of course follow to be finite if $A$ is finite-dimensional.) In view of this, I am not only asking for an answer to the question but also for the right question (or a better one) if the answer to the original question is negative.\n\nA stronger assumption would be to assume that $A$ itself consists only of elements with finite spectrum. This case seems much easier to approach and the answer seems to be positive. In fact every infinite-dimensional $C^\\star$-algebra should contain an element with infinite spectrum.\n\nJust to get started, a more concrete instance of the question above is:\n\n> **Question:** Let $p_1, p_2$ and $p_3$ be three projections in a $C^*$-algebra with the property that every (non-commutative) polynomial in $p_1, p_2$ and $p_3$ has finite spectrum. Is it true that the projections generate a finite-dimensional algebra?\n", "link": "https://mathoverflow.net/questions/35207/finite-dimensional-subalgebras-of-c-star-algebras", "tags": ["oa.operator-algebras", "c-star-algebras", "von-neumann-algebras", "open-problems"], "votes": 28, "creation_date": "2010-08-11T02:49:03", "comments": ["Every finitely generated *-algebra is generated by a finite list of normal elements (the real and imaginary parts of its generators) so the assumption that the $a_i$ are normal is superfluous.", "@AliTaghavi: Yes, in this case, the spectrum of $A$ is equal to the joint spectrum of $a_1,\\dots,a_n$ (which is contained in the product of the spectra of $a_1,\\dots,a_n$).", "Is the answer to the original qustion obvious for commutative C* algebras?", "Let $G$ be a finitely-generated Group, $H$ a Hilbert space, $h : G \\mapsto \\mathbb{B}(H)$ a unitary representation of $G$ such that $h(G)$ is linear independent and $B = span(h(G))$. Then G is finite. For a simple proof see Proposition 2 in ON THE SPECTRUM OF THE SUM OF GENERATORS OF A FINITELY GENERATED GROUP, II, matwbn.icm.edu.pl/ksiazki/cm/cm65/cm65111.pdf", "To \"In fact every infinite-dimensional C⋆-algebra should contain an element with infinite spectrum.\" : For a proof see ALAN W. TULLO, CONDITIONS ON BANACH ALGEBRAS WHICH IMPLY FINITE DIMENSIONALITY, journals.cambridge.org/action/… ", "Could be; I do not know. I was just talking with Joachim Cuntz and George Elliott about the problem, but they did not mention Richard Kadison. There is a related problem to find an element with infinite spectrum in an infinite dimensional $C^\\ast$-algebra, which is a lot easier. Where did you hear that Kadison asked the question?", "@Andreas, Why don't you go ahead and ask this as a separate question. I was thinking about this as an approach to your question. One next step (which would be much harder) would to say if the spectra of these elements isn't the solution of a polynomial equation, then one of these elements has an infinite spectrum. I have no idea how to prove this, or even if it's true, but it seems to work for a few examples. ", "I do not know. Beautiful question! Do you want to (or may I) ask this as a separate question?", "Suppose you have a C$^*$ algebra generated by $g_1$, $g_2$ and $g_3$. Consider the elements $x_1g_1+x_2g_2+x_3g_3$, where the $x_i$ are variables. If the spectra of these elements are the solutions of a polynomial equation in $x_1$, $x_2$ and $x_3$, is the algebra finite dimensional? ", "I have just learned that Choi-Elliott proved in (ams.org/mathscinet-getitem?mr=1081290) that the set of self-adjoint elements with finite spectrum in certain rotation algebras are dense in the set of all self-adjoint elements. (Interestingly, this result is not explicit, in the sense that there is no concrete $\\theta$ for which it happens.) However, from what I understand, it seems unclear whether there is a finitely generated dense sub-algebra such that each element has finite spectrum."], "comment_count": 10, "category": "Science", "diamond": 0} {"question_id": "98", "site": "mathoverflow", "title": "Can one divide by the cardinal of an amorphous set?", "body": "This question arose in a discussion with Peter Doyle.\n\nIt is provable in ZF that one can divide by any positive finite cardinal $k$: if $X \\times \\\\{1,\\ldots,k\\\\} \\simeq Y \\times \\\\{1,\\ldots,k\\\\}$ then $X \\simeq Y,$ where I use $\\simeq$ to denote the existence of a bijection between two sets.\n\nAssuming the Axiom of Choice, the positive finite cardinals are precisely the cardinals that one can divide by since $1\\cdot\\mathfrak{m} = 2\\cdot\\mathfrak{m}$ for any infinite cardinal $\\mathfrak{m}$. The ultimate question is whether this is actually provable in ZF:\n\n> If $\\mathfrak{m}$ is an infinite cardinal, must there be two cardinals $\\mathfrak{p} \\neq \\mathfrak{q}$ such that $\\mathfrak{p}\\cdot\\mathfrak{m} = \\mathfrak{q}\\cdot\\mathfrak{m}$?\n\nAs a candidate for a counterexample, I thought it might be possible to divide by the cardinal of a (strongly) amorphous set. Specifically, in the First Fraenkel model:\n\n> Is it true that if $X \\times A \\simeq Y \\times A$ then $X \\simeq Y$, where $A$ is the set of atoms?\n", "link": "https://mathoverflow.net/questions/235798/can-one-divide-by-the-cardinal-of-an-amorphous-set", "tags": ["lo.logic", "set-theory", "axiom-of-choice"], "votes": 28, "creation_date": "2016-04-09T15:05:50", "comments": ["@Lorenzo: A hint would imply that this is easy... it's not! Peter Doyle has written up division by three and division by four - math.dartmouth.edu/~doyle", "Can someone give me a hint on how to prove that in $\\text{ZF}$ we can divide by finite cardinals?", "I like this question!", "There is Lovasz's unique factorization and unique roots for finite structures, and related results. It might be prudent (and useful if not tangential for this question) to see in which logical domains his arguments can be repeated. Gerhard \"Forward Engineering For Reverse Mathematics\" Paseman, 2016.04.09."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "99", "site": "mathoverflow", "title": "Unital $C^{*}$ algebras whose all elements have path connected spectrum", "body": "A unital $C^{*}$ algebra is called a \"Path connected algebra\" if the spectrum of all its elements is a path connected subset of $\\mathbb{C}$.\n\n> **What is an example of a non commutative path connected $C^*$ algebra? What is an example of a simple path connected $C^{*}$ algebra? In particular does $C^{*}_{red} F_{2}$ satisfiy this property?**\n> \n> As another question: Is the tensor product of two path connected algebra, a path connected algebra?(For spatial norm).\n", "link": "https://mathoverflow.net/questions/195900/unital-c-algebras-whose-all-elements-have-path-connected-spectrum", "tags": ["fa.functional-analysis", "oa.operator-algebras", "operator-theory", "sp.spectral-theory", "c-star-algebras"], "votes": 28, "creation_date": "2015-02-06T16:24:12", "comments": ["@JohannesHahn Thanks for your comment. As you pointed out a minimal dynamical system gives a simple C* algebra. But what about the spectrum of its elements(Path connected spectrum)?", "@HannesThiel Thank you very much for your comment and your attention. The motivation for the second part of my question, about the reduced $C^*$ algebra, is obvious. Regarding the first part: the motivation is that the product topology of two path connected space is a path connected space. But according to your comment, one can ask that: Is the tensor product of two idempotent less C* algebras, an idempotent less C* algebra?", "@AliTaghavi What is your motivation to consider path connected and not just connected spectra?", "Example of a connected C∗ algebra : See B.E. Blackadar. A simple unital projectionless C∗-algebra. J. Operator Theory, 5:63–71, 1981. By the comments of Hannes Thiel and Sam Evington this algebra is an example of a connected C* algebra. But is it also path connected (remark by Ali Taghavi) ?", "Hannes' observation can be improved: all elements of a C*-algebra with no non-trivial projections have connected spectrum. Indeed given an element with disconnected spectrum the holomorphic functional calculus gives a non-trivial idempotent. This will be similar to some self-adjoint idempotent (see Blackadar's K-Theory book) which will be non-trivial.", "@HannesThiel Thank you for the comment. So my second question leads to the following: What is an example of a projectionless simple $C^{*}$ algebra for which the spectrum of all elements are path connected.", "The spectrum of a nontrivial projection (i.e., nonzero and not the unit) is disconnected. Thus, a unital C*-algebra with the property you consider cannot contain any nontrivial projections. Conversely, if a C*-algebra contains a normal element with disconnected spectrum, then it contains a nontrivial projection. Thus, in a unital, simple C*-algebras without nontrivial projections (such as the Jiang-Su algebra), at least the spectrum of every normal element is connected.", "@JohannesHahn Thanks for your comment. If I am not mistaken, this algebra is simple iff the action of $G$ on $X$ does not have a nontrivial compact invariant set. Now how you controle the spectrum of elements?", "Well $\\mathbb{C}$ is simple. But I assume you want something less trivial. A stab in the dark: One could imagine putting some group action on $C(X)$ such that none of its non-trivial ideals are $G$-invariant and hope that $C(X)\\rtimes G$ is simple.", "@JohannesHahn yes I mean without closed two sided ideal.", "Oh you meant simple as in simple not as in easy...", "@JohannesHahn But $C(X)$ is not simple.", "An easy example: $C(X)$ for some path connected compactum $X$ is path connected in this sense because the spectrum of $f\\in C(X)$ coincides with the range of $f$."], "comment_count": 13, "category": "Science", "diamond": 0} {"question_id": "100", "site": "mathoverflow", "title": "A question on simultaneous conjugation of permutations", "body": "Given $a,b\\in S_n$ such that their commutator has at least $n-4$ fixed points, is there an element $z\\in S_n$ such that $a^z=a^{-1}$, and $b^z=b^{-1}$? Here $a^z:=z^{-1}az$.\n\nMagma says that the answer is yes up to $n\\leq 11$ (computed by M. Zieve). The answer is also yes if $a$ and $b$ commute via the argument below. It would be great to have an answer in case a,b do not commute, even under the assumption that $\\langle a,b\\rangle$ contains the alternating group $A_n$. If $\\langle a,b\\rangle$ contains $A_n$ with $n\\geq 4$, then any $z$ as above is unique.\n\nThe commutative case: If $a^b=a$, the orbits (and cycles) of $a$ are permuted by $b$. Say $b$ permutes under conjugation cyclicly the orbits $A_1,A_2,\\ldots,A_r$ of $a$. Let $a_i\\in S_n$ be the cycle induced by $a$ on $A_i$, for $i=1,\\ldots,r$. Choose $c\\in S_n$ to be element of order $r$ such that $b=c$ on $A_i$ for $i=1,...,r-1$. (This forces $a_r^c=a_1$). Since $b=c$ outside $A_1$ and $a_1^{bc^{-1}}=a_1$, we have $b=a_1^e c$ for some integer $e$. \n\nChoose $v$ to be an involution on $A_1$ such that $a_1^v=a_1^{-1}$ on $A_1$, and trivial outside $A_1$. Choose $w$ to be the involution in the symmetric group on $1,\\ldots,r$ such that $(1,\\ldots,r)^w=(r,\\ldots,1)$ and $1^w=2$. Define $u$ to be a permutation such that if $(i,j)$, $i\\leq j$, is an orbit of $w$, then $\\alpha^u = \\alpha^{c^{j-i}}$ for every $\\alpha\\in A_i$, $i=1,\\ldots,r$. This way $c^u =c^{-1}$. Let $z:=vv^c\\cdots v^{c^{r-1}}u$, so that $c^z=c^{-1}$, and $a^z=a^{-1}$. (Moreover, $a_i^z=a_{i^w}^{-1}$ for $i=1,\\ldots,r$). It follows that $b^z=a_{1^w}^{-e}c^{-1} = a_2^{-e}c^{-1}=c^{-1}a_1^{-e} = b^{-1}$. \n\nSome might consider this proof (in the commutative case) to be simpler in the language of wreath products. \n", "link": "https://mathoverflow.net/questions/267191/a-question-on-simultaneous-conjugation-of-permutations", "tags": ["co.combinatorics", "gr.group-theory", "permutations", "algebraic-combinatorics"], "votes": 27, "creation_date": "2017-04-14T04:11:08", "comments": ["For the record: This is proved in Theorem 4.9 of arXiv:1810.08971v4 by Junyao Pan. (I have not checked the proof.)", "Consider the polygonal mesh with vertices $\\{1,\\dots,n\\}$, directed edges $i\\to a(i)$ and $i\\to b(i)$ for each $i$, and for each (possibly trivial) cycle $(k\\;\\cdots\\;k+m-1)$ of $[a,b]$ a face with boundary $[a,b]^m$ starting at $k$. The genus depends on the cycle type of $[a,b]$. Lifting to the universal cover gives a tessellation of the Euclidean or hyperbolic plane. It suffices to find a symmetry of the plane sending each $a$ (resp $b$) edge to an $a$ (resp $b$) edge with opposite orientation. In the commutative case this is realized by a $180^\\circ$ rotation around any square face.", "Have you tried using Rutilo Moreno, Luis Manuel Rivera, Blocks in cycles and $k$-commuting permutations, SpringerPlus, December 2016, 5:1949? (A preprint also appears as arXiv:1306.5708.) Corollary 1 characterizes when the commutator of two permutations has at least $n-k$ fixed points (the authors say that these two permutations $k$-commute).", "@SamHopkins I don't think so. I think a reason why many fixed points matter is as follows: Suppose $a^{-1}b^{-1}ab=x$. If $z$ inverts $a$ and $b$, then $x^z=a^{-z}b^{-z}a^zb^z=aba^{-1}b^{-1}$, so $x^{zab}=x$. Thus the $z$ we are looking for is in the coset $C_G(x)b^{-1}a^{-1}$. Note that $C_G(x)$ is $S_{n-k}$ times a subgroup of $S_k$, where $k\\le4$. I believe the reason for the existence of $z$ is the hugeness of $C_G(x)$.", "Totally naive comment, but would the 4 vs 5 thing have anything to do with when A_n is simple?", "Note also that it is very easy to find examples in which $[a,b]$ has $n-5$ fixed points and there is no element $z$ inverting both $a$ and $b$.", "After doing lots of random experiments I am starting to believe that this might be true. If so, then I am afraid that it might not be easy to prove!"], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "101", "site": "mathoverflow", "title": "Supercompact and Reinhardt cardinals without choice", "body": "A friend of mine and I ran into the following question while reading about proper forcing, and have been unable to resolve it:\n\n_Definition._ A cardinal $\\kappa$ is _supercompact_ if for all ordinals $\\lambda$, there exists a transitive inner model $M_\\lambda$ of $V$ and an elementary embedding $j_\\lambda: V\\rightarrow M_\\lambda$ such that $crit(j_\\lambda):=\\min\\lbrace\\alpha\\in ON: j(\\alpha)\\not=\\alpha\\rbrace=\\kappa$ and $M_\\lambda$ is closed under $\\lambda$-sequences (that is, $^\\lambda M_\\lambda\\subseteq M_\\lambda$).\n\nNow in the definition of supercompactness, the inner models $M_\\lambda$ are allowed to vary with $\\lambda$. We could define a large cardinal notion, _uniform supercompactness_ , by removing this possibility: $\\kappa$ is uniform supercompact if there is a single $M$ and $j$ such that $M$ is closed under all sequences of ordinal length and $j: V\\rightarrow M$ is an elementary embedding with critical point $\\kappa$. But in ZFC, this is not an interesting notion. First, assuming choice, $M$ is closed under sequences of ordinal length if and only if $M=V$ (since two models with the same ordinals and sets of ordinals are equal), so $ZFC\\models$ \"$\\kappa$ is uniformly supercompact $\\iff$ $\\kappa$ is Reinhardt;\" second, Kunen showed that Reinhardt cardinals are inconsistent with $ZFC$. (EDIT: as Joel points out, this statement doesn't actually make sense, as Reinhardt-ness is not a first-order property. One needs to pass to some extension of ZFC which can talk about proper classes directly, like Morse-Kelly set theory, which is the setting Kunen used for his proof.)\n\nHowever, assuming $\\neg$Choice, both of these points seem suspect. It is still open whether Reinhardt cardinals are consistent with $ZF$, and it is unclear to me that $ZF$ alone proves that uniformly supercompact cardinals are Reinhardt. So my questions are the following:\n\nQuestion 1: Does $ZF\\models$ every uniformly supercompact cardinal is Reinhardt?\n\nQuestion 2: What is the consistency strength of $ZF+$ there exists a uniformly supercompact cardinal?\n\nObviously $ZF$ proves that every Reinhardt cardinal is uniformly supercompact; also obviously, Question 2 is really only interesting in lieu of a positive answer to Question 1. The most basic obstruction to a positive answer to Question 1 is the fact that given an elementary embedding $j: V\\rightarrow M$ with $M$ closed under ordinal-length sequences and $crit(j)=\\kappa$, the restriction of $j$ to $M$, $j\\upharpoonright M: M\\rightarrow M$ is not obviously (to me at least) an elementary embedding.\n", "link": "https://mathoverflow.net/questions/94900/supercompact-and-reinhardt-cardinals-without-choice", "tags": ["lo.logic", "large-cardinals", "set-theory", "axiom-of-choice"], "votes": 27, "creation_date": "2012-04-22T18:53:23", "comments": ["The point here is that we require $j(\\kappa)$ to be the same in all of these embeddings. Clearly if there is a class embedding its restrictions satisfy this property. And now we can quantify over all $\\lambda$ if need be.", "I know that this is quite old by now, and I'm sure that Noah and several others here would have no problem tackling @Joel's last comment. Here is my \"starting suggestion\" to modify Woodin's definition: $\\kappa$ is $\\lambda$-supercompact* if there is $\\lambda'\\geq\\lambda$ s.t. for every large enough $\\alpha$ there is an elementary $j\\colon V_\\alpha\\to N$, where $N$ is transitive, the critical point is $\\kappa$, $j(\\kappa)=\\lambda'$, $V_\\lambda\\subseteq N$, and $N^\\lambda\\subseteq N$. (Some indices might need a small tweak, e.g. $V_{\\lambda+1}$ or so to be \"faithful to the idea\".)", "Hm, good point. Is there a \"uniformized\" version of the existence of normal fine measures that might be to the existence of normal fine measures what the second-order definition of uniform supercompact is to the second-order definition of supercompact? (If that makes any sense.) Looking at it now, I don't see one . . .", "I noticed that the definition you give of supercompact is explicitly second-order. With AC, this is equivalent to the existence of normal fine measures, but without AC, one no longer has the Los theorem and so it isn't clear that your second-order definition has a first-order equivalent. ", "Thanks for your thoughtful response, Noah. Elliott Mendelson dispenses with global AC by positing, in place of Limitation of Size, AC for sets plus a form of Replacement. In that case, ZFC only holds for sets, not for proper classes (\"pc's\") such as V and M (for M an inner transitive model of V). Hence, ZFC doen't rule out the existence of a suitable pc M, i.e. a pc that witnesses a negative answer to Question 1. Also, a negative answer to 1 is compatible with the existence of Reinhardts; it simply says that the class of Reinhardts doesn't include all of the uniform supercompacts. ", "will yield a theory which still proves ZFC; in this case, there will be no such suitable model. Specifically, I believe that given any (I suppose wellfounded) model $W$ of this (admittedly vague) version of NBG, it will still be the case that any submodel of $W$ with the same ordinals and sets of ordinals has the same sets as $W$ itself. If, on the other hand, we weaken Limitation of Size so much that even normal choice is not provable, then we can have such an $M$, but we no longer no that no Reinhardts exist; so we still can't answer 1 negatively.", "Oh, I see what you mean. Sorry, I missed the \"global\" in your comment and didn't understand what you were saying. Actually addressing the point you made: The way I'm familiar with it, NBG doesn't actually have global choice as an axiom; instead, it follows from Limitation of Size. So there are two ways I could understand \"dropping global choice\" from NBG: weakening Limitation of Size so that only normal AC is provable, or weakening Limitation of Size so that not even normal AC is provable. Both are vague, but some things can be said about them. Any suitable way of doing the former (cont'd)", "I agree that the mere existence of an M with certain properties doesn't mean we have an elementary embedding. Question 1, however, doesn't ask for an e.e.; it asks if all uniform supercompact cardinals are Reinhardt (given ZF). I.e., it assumes that uniform supercompacts and their e.e.'s exist (including a suitable M), and asks if this entails an e.e. of V into V. In NBG without global AC, there exists (I claim) a suitable M distinct from V, thus answering Question 1 negatively. I may be wrong, of course; but even if I am, I don't think it's due to not showing that a certain e.e. exists.", "@wmitt, I don't think that works - uniform supercompactness (and, for that matter, Reinhardtness) doesn't just ask for an inner model with some properties, it asks for an elementary embedding. So even given an $M\\subsetneq V$ closed under sequences of ordinal length, that doesn't mean that we have an elementary embedding $V\\preccurlyeq M$. (I think what I've said is right; somebody who knows more, please correct me if I'm wrong!)", "Thanks. If we do drop global AC from NBG, doesn't the resulting lack of a bijection between ON and V entail the existence of an inner model M that's (i) closed under sequences of ordinal length, and (ii) not equal to V? If so, we would seem to have a negative answer to Question 1, making the idea of uniform supercompactness interesting. ", "wmitt, Yes, it does not require any AC or global AC to formulate the Reinhardt cardinal concept in second-order Godel-Bernays set theory (without AC). So you are correct, wmitt, that we should speak of formalizing it in GB rather than in GBC. The Kunen inconsistency, which rules out Reinhardt cardinals, is a theorem of GBC, with choice. ", "Is there a way to formalize Reinhardt cardinals and uniform supercompactness without using Choice? The above examples of using NBG or KM to formalize these notions both involve the use of Choice. But we need to work with a Choiceless universe in order to answer Questions 1 and 2.", "We are jumping over each other's comments! Yes, by \"second-order GBC\", I just meant that GBC was a second-order theory. Sorry to confuse...", "Yes, NBG and GBC are the same; just an alternative acronym. N = von Neumann. It is conservative over ZFC, because you can force to add a global choice class and then just take the definable classes. So every model of ZFC is the first-order part of a model of GBC. This is not true for KM, since KM proves the existence of satisfaction classes for first-order truth, and hence proves Con(ZFC). ", "(Not to keep spamming comments, but:) Your most recent comment confuses me somewhat. Is GBC a theory in second-order logic, or by \"second-order GBC\" did you just mean something similar to \"second-order arithmetic?\" ", "The super compactness of $\\kappa$ is expressible in first-order as the assertion: for every ordinal $\\lambda$, there is a $\\kappa$-complete normal fine ultrafilter on $P_\\kappa\\lambda$. ", "Also, you say that it is also possible to formalize Kunen's inconsistency argument in GBC. I'm not familiar with GBC; is it the same as NBG, and if not, is it a conservative extension of ZFC?", "Noah, while we can prove that Reinhardt cardinals are not first order expressible (otherwise let $\\kappa$ be the least one, and then consider $j(\\kappa)$...), it seems that we should similarly expect the same for uniform supercompactness. But it can be formalized in second-order GBC, as an assertion about a class $j$ and a class $M$. One replaces the assertion that $j$ is fully elementary with the assertion that $j$ is $\\Sigma_1$-elementary and cofinal, which implies $\\Sigma_n$-elementarity by a meta-theoretic induction. ", "@Joel: good points. I've corrected the question. Also, building off of your first comment, is it obvious that uniform supercompactness is first-order expressible? For that matter, how is supercompactness itself first-order expressible?", "I believe (from your remark about ordinal lengths) that you intend that the assertion \"for all $\\lambda$...\" in the definition is meant to be interpreted as, \"for all ordinals $\\lambda$...\". The difference matters, as you likely know, because if you have $j:V\\to M$ and $j\"A\\in M$ for all sets $A$, then it follows easily that $M=V$ even in ZF, since if $A$ is transitive, then $A$ is the Mostowski collapse of $j\"A$.", "Very nice question. One quibble: it isn't really correct to say that ZFC refutes Reinhardt cardinals, since there is no first-order characterization of Reinhardt cardinals; that is, you can't say \"$\\kappa$ is a Reinhardt cardinal\" in a first-order way. Kunen proved his inconsistency in the second-order Kelly-Morse set theory, but it is also possible to formalize the result in Goedel-Bernays set theory GBC. See jdh.hamkins.org/generalizationsofkuneninconsistency for a further discussion of this point. ", "This is an interesting question. The contradiction in ZFC comes from the fact that models with the same sets of ordinals are isomorphic. In ZF it is not true. Monro proved that there is a chain of $\\omega$ models, where $V_n\\subseteq V_{n+1}$ and they both have the same sets-of-sets-..$n$-times of ordinals, but they are not isomorphic (Jech The Axiom of Choice, Ch. 5, Problem 17). This, however is \"going up\" while inner models are \"going down\" so there might be some difference there.", "Like you say in the last paragraph, it is not clear that the restriction of j to M is going to be an elementary embedding into M in general. Instead you could consider the restriction of j to smaller models, like the class N consisting of all sets constructible from sequences of ordinals. Then j does restrict to an embedding N -> N but you still have might have the problem that the embedding is not amenable to the model. This seems like the bigger issue to me."], "comment_count": 23, "category": "Science", "diamond": 0} {"question_id": "102", "site": "mathoverflow", "title": "Orders in number fields", "body": "Let $K$ be a degree $n$ extension of ${\\mathbb Q}$ with ring of integers $R$. An order in $K$ is a subring with identity of $R$ which is a ${\\mathbb Z}$-module of rank $n$. \n\nQuestion: Let $p$ be an unramified prime in $K$. Is it true that the number of orders in $R$ of index equal to $p^r$, for some natural number $r$, is less than or equal to the number of subrings with identity of ${\\mathbb Z}^n$ of index equal to $p^r$?\n\nNathan Kaplan and I need this fact for $n=5$ in a project where we are trying to find asymptotic formulae for the number of orders of bounded index in a given quintic field. \n\nWe've been staring at Jos Brakenhoff's [thesis](https://openaccess.leidenuniv.nl/bitstream/handle/1887/14539/proefschrift-brakenhoff.pdf?sequence=2) for a while, but I haven't gotten anywhere. Any advice will be greatly appreciated. Thanks. \n\nAdded in edit: Here is an elementary reformulation of this problem. Let $r(x)$ be a polynomial with integer coefficients. Then show that for any natural number $a$ in order to maximize the number of subrings of $({\\mathbb Z}/p^a {\\mathbb Z})[x]/(r(x))$ of a given index, the polynomial $r(x)$ has to be a product of linear factors modulo $p$. \n\nUpdate: In a soon-to-be-posted preprint Nathan Kaplan, Jake Marcinek, and I have proved an asymptotic formula for the number of orders in a given quintic field of bounded discriminant, using an argument independent of the above question. We can also give non-trivial upper bounds for a general number field. I'm now convinced the answer to the question is yes, but I still don't know how to prove it. \n", "link": "https://mathoverflow.net/questions/130089/orders-in-number-fields", "tags": ["number-fields", "nt.number-theory", "algebraic-number-theory"], "votes": 26, "creation_date": "2013-05-08T08:50:58", "comments": ["Thanks everyone. You can actually count the number of subrings of index $p$. Namely, if $r(x) \\mod p$ has $u$ factors of degree $1$ and $w$ factors of degree $2$, and the rest of higher degrees, then the number of subrings of ${\\mathbb Z}/p{\\mathbb Z}[x]/(r(x))$ of index $p$ is ${u \\choose 2} + w$. ", "The $r=1$ case can be handled as follows. Each order in $R$ of index $p$ is uniquely determined by its image in $R/pR$. Now $R/pR$ is an \\'etale $k := \\mathbf{F_p}$-algebra of rank $n$ as $p$ is unramified. Now the set of such rings injects into the set of index $p$ $k$-subalgebras of $R/pR \\otimes_k \\overline{k} \\simeq \\overline{k}^n$. The latter set is a combinatorial object (it's the set of surjective maps from an $n$-element set to an $(n-1)$-element set), and can be bounded above by the set of index $p$ subrings of $\\mathbf{Z}^n$. ", "I found a paper by Melanie Wood which might be relevant, although I did not check it carefully: arxiv.org/pdf/1007.5508v1.pdf", "Have you tried (probably you have) writing down a binary $n$-ic form and playing the same game as in Theorem 9 to Lemma 13 of Bhargava, Shankar, and Tsimerman's paper arxiv.org/pdf/1005.0672.pdf ? This doesn't \"just work\", the details are much messier, but since you need something weaker I wonder if it is possible to salvage the result you are looking for?", "In our work we only need this for unramified primes. ", "It is clear that this question can be studied locally at $p$. In other words, this depends only on $R \\otimes \\mathbb Z_p$, so it depends only on $K \\otimes \\mathbb Q_p$, which is just a product of $p$-adic fields.", "Is it obvious that the elementary reformulation is the same, given that one is restricted to $p$ unramified and the other is not?", "Who voted to close? And why?"], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "103", "site": "mathoverflow", "title": "Where do uncountable models collapse to?", "body": "Suppose $T$ is a complete first-order theory (in an finite, or at worst countable, language). Given any model $\\mathcal{M}\\models T$ of cardinality $\\kappa$, we can ask whether $\\mathcal{M}$ can be made isomorphic to any countable model $\\mathcal{N}\\models T$ in the ground model; that is, whether there is some $\\mathcal{N}\\models T$, which is countable, such that $V[G]\\models \\mathcal{M}\\cong\\mathcal{N}$ for $G$ $Col(\\kappa,\\omega)$-generic over $V$. **For simplicity, we'll say \"$\\mathcal{M}$ collapses to $\\mathcal{N}$\".**\n\nIn case $T$ has _countably many_ countable models, the answer is yes: the set of models of $T$ is a countable Borel set, and so is absolute. The uncountable models of such a $T$ are thus partitioned into $\\omega$-many classes. My question is about what the nonempty classes are. For example, suppose $T$ is $\\aleph_1$-categorical. Then the countable models of $T$ form an elementary chain, with a \"top\" model $\\mathcal{N}_\\omega$, and it is easy to see that this is the only countable model which uncountable models can collapse to.\n\nHowever, there are plenty of non-$\\aleph_1$-categorical theories with only countably many countable models, and beyond the $\\aleph_1$-categorical setting things are much less clear. Some examples:\n\n * The language has predicates $P_i$ ($i\\in\\omega$); $T$ asserts that the $P_i$ name infinite disjoint sets. Then the countable models of $T$ are classified by how many elements they have not in $\\bigcup_{i\\in\\omega} P_i$, and so there are countably many countable models. For each countable model of $T$, there is an uncountable model of $T$ which collapses to it.\n\n * The language has a single binary relation $E$; $T$ asserts that $E$ is an equivalence relation and that there is exactly one class of cardinality $n$, for each $n$. Countable models of $T$ are determined by how many infinite classes they have; and every countable model of $T$ **except the prime** is collapsed to by some uncountable model of $T$.\n\n\n\n\nI'm curious what we can say in general about the set of models which can be collapsed to. There's a lot of questions around here that one can ask; let me focus on:\n\n> Is there a theory $T$ with countably many countable models, including $\\mathcal{M}_0\\prec\\mathcal{M}_1$, such that $\\mathcal{M}_0$ can be collapsed to by some uncountable structure but $\\mathcal{M}_1$ cannot?\n\nThat is, can there be \"gaps\" in the range of the collapse?\n\n* * *\n\nEDIT: As Paul Larson pointed out in the comments below, a countable model is collapsed to by some uncountable model iff it is **extendible** (see Definition 2.6 in ) - that is, iff it has an uncountable $L_{\\omega_1\\omega}$-elementary extension. One direction is trivial. To show that every extendible model is collapsed to, let $\\mathcal{M}$ be extendible with $\\mathcal{M}\\prec_{\\omega_1\\omega}\\mathcal{N}$ for $\\mathcal{N}$ uncountable. Let $\\varphi$ be the Scott sentence of $\\mathcal{M}$, and let $V[G]$ be a forcing extension in which $\\mathcal{N}$ is made countable. Then $\\mathcal{N}\\models\\varphi$ in both $V$ and $V[G]$; moreover, the statement \"$\\varphi$ is the Scott sentence of $\\mathcal{M}$\" is $\\Pi^1_2$ (actually, better, but this is enough) and so absolute between $V$ and $V[G]$. So $V[G]\\models\\mathcal{M}\\cong\\mathcal{N}$.\n\nIt's worth keeping in mind the broader result, due to Barwise and Karp if I recall correctly, that $\\mathcal{M}\\equiv_{\\infty\\omega}\\mathcal{N}$ iff $\\mathcal{M}\\cong\\mathcal{N}$ in some forcing extension. \n", "link": "https://mathoverflow.net/questions/220850/where-do-uncountable-models-collapse-to", "tags": ["lo.logic", "set-theory", "model-theory", "forcing"], "votes": 26, "creation_date": "2015-10-13T21:45:52", "comments": ["I asked Saharon my version of your question, and it turns out to have a relatively easy positive answer. Here is a write-up, which could probably be improved : users.miamioh.edu/larsonpb/extendible.pdf", "I've been wondering about this too. I don't see that it matters. By the way, Su Gao's \"On automorphism groups of countable structures\" shows that a countable structure is extendible if and only if its automorphism group is not closed in the group of permutations of its domain. So one could rephrase the problem using this. Here's a variation I've wondered about : can you have countable $M \\prec N$ such that $M$ is extendible and $N$ is rigid?", "@PaulLarson Coming back to this a year later, it occurs to me that we can ask the same question of theories with uncountably many countable models. That is, are there $\\mathcal{M}_0\\prec \\mathcal{M}_1$ with $\\mathcal{M}_0$ extendible but $\\mathcal{M}_1$ not? It's not obvious to me that this makes the question any easier - I don't see how to get such a pair, even if I allow lots of countable models - but maybe you know an easy answer in this case?", "For instance, it seems that the model-theoretic forcing approach gives : if $T$ is a complete first order theory having an uncountable model and only countably many countable models, then there is a countable extendible model of $T$ having a copy of each countable model of $T$ as an elementary submodel. Again, I don't know how much insight this gives into your question.", "@PaulLarson That looks really cool, thanks!", "You might also get some mileage out of Shelah's theory of model-theoretic forcing, from his publication #88r, depending on what you're interested in.", "John Baldwin used \"extendible\" in our paper with Saharon (shelah.logic.at/files/1003.pdf). I assume that it was used previously, but I don't really know.", "@PaulLarson I didn't know about extendible models - yes, it looks like a model is extendible iff it is collapsed to. But googling around I can't find much about them - can you point me to a good source?", "It seems that you could naturally rephrase this in terms of extendible models (i.e., countable models which have the same Scott sentence as some uncountable model) and then drop the mention of forcing."], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "104", "site": "mathoverflow", "title": "Which sets of roots of unity give a polynomial with nonnegative coefficients?", "body": "> **The question in brief:** When does a subset $S$ of the complex $n$th roots of unity have the property that $$\\prod_{\\alpha\\, \\in \\,S} (z-\\alpha)$$ gives a polynomial in $\\mathbb R[z]$ with nonnegative coefficients?\n\nSome trivial necessary conditions include $1\\not\\in S$ and also that $S$ is self-conjugate so that the coefficients will at least be real.\n\nA sufficient condition that is not obvious (to me) is that the roots of unity in $S$ are precisely those lying outside of some wedge-shaped region of the complex plane, i.e., those with $\\left|\\arg(z)\\right| > t$ for some fixed $t > 0$. This follows from the main theorem of\n\n> Barnard, R. W., W. Dayawansa, K. Pearce, and D. Weinberg. \"Polynomials with nonnegative coefficients.\" _Proceedings of the American Mathematical Society_ 113, no. 1 (1991): 77-85. [Journal link](http://www.ams.org/journals/proc/1991-113-01/S0002-9939-1991-1072329-2/S0002-9939-1991-1072329-2.pdf)\n\nwhich says that, given any polynomial with nonnegative coefficients, dividing it by the linear factors corresponding to exactly those of its roots lying in such a region produces a polynomial with only positive coefficients. Applying this to $z^{n-1}+\\cdots+z+1$ gives the desired result.\n\nBut that theorem is not number-theoretic; it doesn't care at all that we are dealing with roots of unity. So maybe there is a nicer proof in this special case?\n\nMore generally, what other necessary and/or sufficient conditions exist? Is there some reasonable number-theoretic or combinatorial characterization of such sets? What tools seem likely to shed some light on this question?\n\n**ADDED LATER:** A couple of very interesting suggestions have been made in the comments so far, and I'm eager to see where they lead. However, I remain quite curious as to what can be said from a number-theoretic perspective. For example, if $n=p^r > 1$ for some prime $p$, then the cyclotomic polynomial $\\Phi_n(x)=\\Phi_p(x^{n/p})$ has only nonnegative coefficients. So a sufficient condition that I did not mention above is that $S$ is the set of primitive $n$th roots of unity for some prime power $n$. I believe this is the only way to obtain such a polynomial that is a _cyclotomic_ polynomial, but that doesn't imply that nothing else can be gained from a number-theoretic perspective on this question – however I myself lack the expertise necessary to make the most of such a perspective.\n\nI should also mention that (ironically, given my desire for a number-theoretic perspective on this) the cyclotomic result actually follows from a more general, not number-theoretic result of\n\n> Evans, Ronald, and John Greene. \"Polynomials with nonnegative coefficients whose zeros have modulus one.\" _SIAM Journal on Mathematical Analysis_ 22, no. 4 (1991): 1173-1182. [Author's link](http://math.ucsd.edu/~revans/PolynomialsGreene.pdf)\n\nwhich gives that for any proper divisor $d$ of $n$, $(x^n-1)/(x^d-1)$ has only nonnegative coefficients. So that generalizes the cyclotomic case, but not via number theory!\n", "link": "https://mathoverflow.net/questions/214784/which-sets-of-roots-of-unity-give-a-polynomial-with-nonnegative-coefficients", "tags": ["nt.number-theory", "co.combinatorics", "polynomials", "additive-combinatorics", "roots-of-unity"], "votes": 26, "creation_date": "2015-08-14T07:23:56", "comments": ["\"I believe this is the only way to obtain such a polynomial that is a cyclotomic polynomial\": yes, you're correct. The easiest way to see this is plugging in x=1: it's easy to show that $\\Phi_m(1) = \\begin{cases}0 & m=1 \\\\ p & m=p^k \\\\ 1 & \\text{otherwise}\\end{cases}.$ This essentially says there must be cancellation outside of the prime-power case.", "@Josep: The reference you gave is extremely interesting. I was totally unaware of Suffridge's result.", "In fact the Suffridge polynomials cover the more general case when $1\\not\\in S,$ S self-conjugate and the arguments of the roots in S are separated by the same angle, except for a pair.", "In the wedge case you can compute explicitly the coefficients and see they are positive. It's a particular case of Suffridge's extremal polynomials- you have the expression deduced in Sheil-Small, Complex polynomials, p. 251-252.", "Maybe, the following related result of Kellog is helpful: Let $A$ be a complex $n\\times n$ matrix. If all its elementary symmetric functions are positive (so that the characteristic polynomial has alternating signs), then the spectrum of $A$ lies in the set $\\{z : |\\text{arg}z| \\le \\pi - \\pi/n\\}$...."], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "105", "site": "mathoverflow", "title": "Chromatic Spectra and Cobordism", "body": "I apologize in advance, if some of the things I've written are incorrect.\n\nThe cobordism hypothesis states that $\\mathbf{Bord}^\\mathrm{fr}_n$ is the free symmetric monoidal $(\\infty,n)$-category with duals generated by a point. There is a geometric realization functor $|-|:\\text{Cat}_{(\\infty,n)}\\rightarrow \\text{Gpd}_\\infty$, where $\\text{Cat}_{(\\infty,n)}$ is the $\\infty$-category (I mean a quasicategory, but you can think of it as a model category, or whatever you like) of (small) $(\\infty,n)$-categories, and $\\text{Gpd}_\\infty$ is $\\infty$-category of $\\infty$-groupoids, otherwise known as spaces. This functor is a left adjoint (in the homotopical sense) to the inclusion functor $\\text{Gpd}_\\infty\\hookrightarrow \\text{Cat}_{(\\infty,n)}$. Therefore, $|-|$ preserves (homotopy) colimits. If we \"restrict\" ourselves to the symmetric monodical $(\\infty,n)$-categories, we would get a functor $|-|:\\text{SymMon}_{(\\infty,n)}\\rightarrow \\text{Alg}_{\\mathbb{E}_\\infty}$, where $\\text{SymMon}_{(\\infty,n)}$ is the $\\infty$-category of symmetric monoidal $(\\infty,n)$-categories and $\\text{Alg}_{\\mathbb{E}_\\infty}$ is the $\\infty$-category of infinite loop spaces. This should also preserve colimits. It manifests itself in the fact that $|\\mathbf{Bord}^\\text{fr}_n|$ is the infinite loop space of the sphere spectrum. \n\nThere is a universal characterization of the bordism $(\\infty,n)$-category of manifolds with singularities [Theorem 4.3.11 in [Lurie's paper](http://arxiv.org/abs/0905.0465)]. The type of singularities considered are those of Baas-Sullivan theory, which produce a wide array of chromatic spectra, such as the Brown-Peterson spectrum $\\text{BP}$, the Johnson-Wilson spectra $\\text{E}(n)$, and, of course, the Morava K-theory spectra $\\text{K}(n)$. I am wondering:\n\n> Is it possible to give these spectra a universal characterization using the fact that geometric realization preserves colimits?\n\n**EDIT:** When I think about it, I am not sure whether there is a universal description of $|\\mathbf{Bord}^{(X,\\zeta)}_n|$, where $X$ is topological space, $\\zeta$ a vector bundle on it of dimension $n$, other than the case when $X$ is point. We _can_ write it down. It is simply $\\Omega^{\\infty-n}M\\zeta$, where $M\\zeta$ is the Thom spectrum of $\\zeta$. \n\n> Do we get a different characterization of these infinite loopspaces from the cobordism hypothesis?\n", "link": "https://mathoverflow.net/questions/184261/chromatic-spectra-and-cobordism", "tags": ["homotopy-theory", "stable-homotopy", "infinity-categories", "cobordism", "chromatic-homotopy"], "votes": 26, "creation_date": "2014-10-12T17:49:07", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "106", "site": "mathoverflow", "title": "Planar minor graphs", "body": "The theorem of Robertson-Seymour about graph minors says that there exists no infinite family of graphs such that none of them is a minor of another one.\n\nApparently, it came as a generalization of the Kruskal's theorem that states that there exists no infinite family of rooted ordered trees such that none is a minor of another one. Here, _rooted ordered_ means that the tree has a root, and that the edges escaping from a vertex are ordered. In other words, the trees are assumed to be embedded in the plane, and the minor operation has to respect this embedding.\n\nHere comes the question: is Robertson-Seymour theorem true for planar graphs, when we add the condition that the minor operation respects the embedding? (_i.e._ we not only ask $G_i$ to be a minor of $G_j$ as an abstract graph, but also as an embedded graph.)\n\nIt is not clear to me that this should be a direct corollary of the original theorem, because of the amount of possible embeddings into the plane for a given planar graph.\n", "link": "https://mathoverflow.net/questions/92421/planar-minor-graphs", "tags": ["graph-minors", "graph-theory"], "votes": 26, "creation_date": "2012-03-27T16:28:06", "comments": ["@PierreDehornoy: I think it would improve the question if \"true for planar graphs\" was replaced with 'true for plane graphs', because what you write immediately afterwards suggests that you are thinking about planar graphs together with a specified embedding into the plane, and that is called a plane graph.", "@Pierre Dehornoy: Pierre, sorry, you a right, this argument does not answer your question. ", "@ Misha: This looks almost convincing, but I do not understand the end of the argument. One of the embedding of $G_m$ appears infinitely offen, but how does it says that this particular embedding is the one that was given at the beginning, i.e. that corresponds to the given embedding of $G_m$?", "I guess, my understanding of R-S theorem is correct: If a sequence of graphs were to contain infinitely many minimal elements $G_m$ then the new sequence of graphs $(G_m)$ would violate R-S theorem. ", "@Pierre Dehornoy: I thought that the conclusion of Robertson-Seymour theorem could be strengthened to \"A certain graph $G_m$ occurs as a minor infinitely many times in a given infinite sequence\" ($G_m$ would be one of the finitely many minimal elements of the sequence, with respect to the \"minor\" order). If my understanding is correct, then the answer to your question is positive since $G_m$ has only finitely many planar embeddings, up to isotopy, so one of them would be repeated (infinitely often). "], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "107", "site": "mathoverflow", "title": "Is every $p$-group the $\\mathbb{F}_p$-points of a unipotent group", "body": "Let $\\Gamma$ be a finite group of order $p^n$. Is there necessarily a unipotent algebraic group $G$ of dimension $n$, defined over $\\mathbb{F}_p$, with $\\Gamma \\cong G(\\mathbb{F}_p)$?\n\nI have no real motivation for this question, it just came up in conversation and no one knew the answer. There does not appear to be any sort of uniqueness to $G$; both the groups $\\mathbb{Z}/p^2 \\mathbb{Z}$ and $(\\mathbb{Z}/p \\mathbb{Z})^2$ have infinitely many lifts to unipotent groups.\n\n* * *\n\nI am bumping this question to the top because I now suspect I have a counterexample, namely the dihedral group of order $16$, also known as $C_2 \\ltimes C_8$, where $C_n$ is cyclic of order $n$. The \"obvious\" lift of $C_8$ is the additive group of the $3$-truncated Witt vectors, $\\mathcal{W}_3$. So the obvious thing to try is a semidirect product $\\mathbb{G}_a \\ltimes \\mathcal{W}_3$. But $\\mathbb{G}_a$ has exponent $2$, so an action of $\\mathbb{G}_a$ on $\\mathcal{W}_3$ must live in the $2$-torsion elements of $\\mathrm{Aut}(\\mathcal{W}_3)$. I compute that the space of $2$-torsion elements in $\\mathrm{Aut}(\\mathcal{W}_3)$ has two connnected components, one of which maps $1$ to things that are $1 \\bmod 4$ and one of which maps $1$ to things which are $-1 \\bmod 4$. Since $\\mathbb{G}_a$ is connected, it must land in one component, but the two elements of $C_2$ live in different components.\n\nI've made some partial progress thinking about the structure of connected unipotent groups whose $\\mathbb{F}_2$ points are $C_2 \\ltimes C_8$, but I've gotten stuck, so I am putting up a bounty for progress either on this case or the questions as a whole.\n", "link": "https://mathoverflow.net/questions/262745/is-every-p-group-the-mathbbf-p-points-of-a-unipotent-group", "tags": ["ag.algebraic-geometry", "gr.group-theory", "finite-groups", "algebraic-groups", "nilpotent-groups"], "votes": 26, "creation_date": "2017-02-20T19:26:35", "comments": ["Do you require $G$ to be connected?", "@DrorSpeiser the Zariski closure of a finite set is itself (so not unipotent). I don't see any reason why we should expect any given embedding of D_8 in a unipotent group to come from the F_2-points of a connected subgroup.", "Take $a=1+E_{12}+E_{34}+E_{56}$, $b=1+E_{23}+E_{45}+E_{67}\\in Mat_{7\\times7}(\\mathbb{F}_2)$, where $E_{ij}$ has a unique nonzero element at row $i$ and column $j$ equal to 1. Set $c=ab$. Let $V$ be $\\overline{\\mathbb{F}}_2^7$, and let $D_{16}\\cong$ act on $V$ by sending $x$ to $a$ and $y$ to $c$. With this action $V$ is an indecomposable module for $D_{16}$. Let $\\overline{D_{16}}$ be the Zariski closure of $$ in $GL(V)$, which is unipotent. Does it satisfy $\\overline{D_{16}}(\\mathbb{F}_2)=\\cong D_{16}$?", "@DrorSpeiser I decided to put the bounty on my own version, rather than the duplicate because I'd rather edit my own question and because it has a bunch of useful comments. If I had (two years ago) seen Gjergji's link to the duplicate before a bunch of those comments showed up, I probably would have self-closed, but it doesn't seem like a big deal either way. Of course, if someone posts a good answer, I'll link it on that question as well to make sure searchers find it.", "Given that this is a duplicate of a question six years earlier, how is this not closed?", "Okay, I'll take a look at Abhyankar's conjecture and see what that does for me. In the mean time, I'll record that the formula was meant to be $(x_1, y_1, z_1) \\ast (x_2, y_2, z_2) = (x_1+x_2, y_1+y_2, z_1+z_2 +(x_1 y_2)^p-x_1 y_2)$, sorry for the incorrect subscripts in the previous one.", "I am specifically suggesting to use the faithful representations of $p$-groups arising from Abhyankar's conjecture. If we realize $\\Gamma$ as the Galois group of a finite etale extension $\\mathbb{F}_q[t] \\subset R$, then for every subgroup $\\Lambda \\subset \\Gamma$, then $R$ is a finite etale extension of the $\\Lambda$-invariant subring $S$, with Galois group $\\Lambda$. This is a \"complete reducibility\" result that is missing for arbitrary representations. I suggest we projectively complete these rings at $\\infty$, and quotient by the maximal ideal of $\\infty$ in $\\mathbb{F}_q[t^{-1}]$.", "Suppose there were such a $G$, and let $Z$ be the center $\\{ (0,0,z) \\}$ of $U$. Case 1: $G \\supset Z$. Then $G(\\mathbb{F}_p) \\supset \\{ (0,0,z) : z \\in \\mathbb{F}_p \\}$, which is not in $\\Gamma$. Case 2: $G \\cap Z$ is finite. In that case, the projection $G \\to U/Z$ is surjective. But any subvariety of $U$ which surjects onto $U/Z$ generates $U$, contradicting that $G$ is two dimensional.", "I think there is a fatal flaw in @JasonStarr's approach. It seems that Jason's method would show that, if $U$ is a unipotent group, and $\\Gamma$ is a subgroup of $U(\\mathbb{F}_p)$, then there is a unipotent subgroup $G$ of $U$ with $G(\\mathbb{F}_p) = \\Gamma$. This isn't true. Let $U$ be the group structure on $\\mathbb{A}^3$ given by $(x_1,y_1,z_1) \\ast (x_2,y_2,z_2) = (x_1+x_2, y_1+y_2, z_1+z_2+x_1^p y_1^p - x_1 y_1)$ and let $\\Gamma = \\{ (x,y,0) : x,y \\in \\mathbb{F}_p \\} \\subset U(\\mathbb{F}_p)$. Then $\\Gamma$ is not $G(\\mathbb{F}_p)$ for any $2$-dimensional subgroup of $U$. (continued)", "@DavidSpeyer. I should not have called this the unique smallest subgroup. I am saying that for a vector space $V$ and a finite subgroup $Z$ of the group of $\\mathbb{F}_p$-points of the scheme $\\text{Spec}\\ \\text{Sym}_{\\mathbb{F}_p}(V)$, the kernel of the restriction map $V\\to \\prod_{z\\in X} \\kappa(z)$ is an $\\mathbb{F}_p$-linear subspace of $V$ generating an ideal whose associated closed subscheme is a closed subgroup scheme whose associated group of $\\mathbb{F}_p$-points equal to $Z$.", "@JasonStarr It seems like you are saying that, given an connected abelian group $H$ (in your setting, $Z(U)$), and a discrete subgroup $B$, there is a smallest connected group $A$ containing $B$. But that isn't true: consider $H = \\mathbb{G}_a^2$ and $B = \\{ (x,0) : x \\in \\mathbb{F}_p \\}$. Then $B$ is contained in $\\mathbb{G}_a$ embedded in $H$ by either $t \\mapsto (t,0)$ or $t \\mapsto (t,t^p-t)$. What am I missing?", "@JasonStarr: I was wondering about that connectedness issue too but thought I might be overlooking something.", "@nfdc23. You are correct. I realize now the problem with the approach I suggest: why should the centralizer of $Z(\\Gamma)$ in $U$, much less the center of the centralizer, be a connected group?", "@JasonStarr: In view of your inductive suggestion it seems that what you call \"$n$\" (from SL$_n$) is not \"$n$\" as in the statement of the question, just some unknown integer, in which case such $\\rho$ is provided by the regular representation of $\\Gamma$ on itself over $\\mathbf{F}_p$ (that gives $\\rho$ into ${\\rm{GL}}_N(\\mathbf{F}_p)$, necessarily landing inside SL$_N({\\mathbf{F}}_p)$ and even conjugating into $U_N(\\mathbf{F}_p)$ since the latter is a $p$-Sylow subgroup).", ". . . So there is a unique minimal, connected subgroup $A\\subset Z(U)$ such that $A(\\mathbb{F}_p)$ equals $Z(\\Gamma)$. Now replace $U$ by $U/A$, and use induction on the rank.", "Here is how I would try to prove this. There exists a faithful homomorphism, $\\rho:\\Gamma\\to \\textbf{SL}_n(\\mathbb{F}_p)$. This follows, for instance, from the solution of the Abhyankar conjecture by Raynaud and Harbater. By counting $\\mathbb{F}_p$-points of the flag variety of $\\textbf{SL}_n$, there exists a Borel subgroup $B$ that contains the image of $\\rho$. Thus, there exists a faithful homomorphism from $\\Gamma$ to a unipotent group. Replace that unipotent group by the centralizer $U$ of $Z(\\Gamma)$. Thus $Z(U)(\\mathbb{F}_p)$ is an additive group that contains $Z(\\Gamma)$ . . .", "A step towards an inductive proof: if $G$ is a unipotent $k$-group, and $c \\in Z^2(G(k), Z / p Z)$ is a 2-cocycle (for trivial action), does there exist a polynomial 2-cocycle $\\gamma \\in Z^2(G, G_a)$ such that $\\gamma$ gives $c$ upon taking $k$-points? I want to attack this one inductively with inflation-restriction with respect to $G$ and $[G,G]$. But no time for more than a sketchy comment.", "This was asked a while ago, as well, but there was no answer. mathoverflow.net/questions/69397/…", "If $U$ is smooth connected unipotent of dimension $n$ over $k=\\mathbf{F}_p$ then $\\#U(k)=p^n$ (use composition series with successive quotients $\\mathbf{G}_a$). Thus, if $\\Gamma=G(k)$ for a unipotent $k$-group $G$ of dimension $n$ then by passing to $G_{\\rm{red}}$ so that $G$ is smooth we see that if $G$ is not connected then $G^0(k)$ of size $p^n$ must exhaust $G(k)$. In other words, we lose nothing by assuming $G$ is smooth and connected. So it is the same as asking for smooth connected unipotent $k$-groups $U$ if there is any constraint on $U(k)$ beyond its order being $p^n$. Good puzzler."], "comment_count": 19, "category": "Science", "diamond": 0} {"question_id": "108", "site": "mathoverflow", "title": "Does the Tate construction (defined with direct sums) have a derived interpretation?", "body": "Any abelian group M with an action of a finite group $G$ has a Tate cohomology object $\\hat H(G;M)$ in the derived category of chain complexes. There are several ways to define this. One is as the tensor product $W \\otimes_G M$ with a particular (typically unbounded) complex of projective $\\Bbb Z[G]$-modules which happens to have trivial homology. One is to instead use $Hom^G(W',M)$ for $W'$ a similar complex. One is as the mapping cone of a norm map $$ M_{hG} \\to M^{hG} $$ from derived coinvariants to derived invariants, where we view $M$ as a complex concentrated in degree zero. \n\nWe can attempt to generalize this construction to the derived category of $\\Bbb Z[G]$-modules, or of $R[G]$-modules for $R$ a commutative ring. If $M_*$ is a chain complex of $\\Bbb Z[G]$-modules, these give us three definitions of the Tate construction on $M_*$: $$ \\begin{align*} Tate^\\oplus(M)_n &= \\bigoplus_{p+q=n} W_p \\otimes_G M_q\\\\\\ Tate^\\prod(M)_n &= \\prod_{p+q=n} W_p \\otimes_G M_q\\\\\\ Tate(M)_n &= \\bigcup_N \\prod_{p+q = n, p \\leq N} W_p \\otimes_G M_q \\end{align*} $$ (There are natural maps $Tate^\\oplus \\to Tate \\to Tate^{\\prod}$ which are isomorphisms on bounded complexes.)\n\nThe first construction is nice because it commutes with filtered colimits, and the second with filtered limits. The third is neither, but it does have a norm-cofiber sequence and a Tate cohomology spectral sequence $$ \\hat H^p(G; H^q(M)) \\Rightarrow H^{p+q}(Tate(M)). $$ The third construction is also the only one that preserves quasi-isomorphisms, and thus the only one that (naturally) extends to the _unbounded_ derived category. \n\nAre there derived-category interpretations for the constructions $Tate^\\oplus$ or $Tate^\\prod$?\n\nSuch an interpretation probably needs to take in more information than an object of the derived category of $\\Bbb Z[G]$-modules. The main motivation for asking is that I'd like to see what is necessary in more general contexts (e.g. modules over a differential graded algebra).\n", "link": "https://mathoverflow.net/questions/234577/does-the-tate-construction-defined-with-direct-sums-have-a-derived-interpretat", "tags": ["at.algebraic-topology", "homological-algebra", "group-cohomology"], "votes": 25, "creation_date": "2016-03-26T13:01:27", "comments": ["Another characterisation of Tate cohomology $\\hat H^n(G, M)$ is as $Hom(\\mathbf Z, M[n])$, where the Hom-set is in the category of singularities, defined as the quotient of the usual derived category $D^b(Mod_{\\mathbf Z G})$ modulo its subcategory of perfect complexes. A natural extension thus seems to be (I remember this is also mentioned somewhere in Gaitsgory's work, but don't seem to find it right now) to consider the Hom-set or -space in the quotient of Ind-Coherent sheaves modulo Quasi-coherent sheaves (in the $\\infty$-sense).", "@Jesper Both of those are good suggestions. I haven't asked either yet, and it looks like I'll probably be doing so soon.", "Dave Benson or John Greenlees would be my go-to people for this. Maybe you've already asked them..."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "109", "site": "mathoverflow", "title": "Are amenable groups topologizable?", "body": "I've learned about the notion of topologizability from \"On topologizable and non-topologizable groups\" by Klyachko, Olshanskii and Osin () - a discrete group $G$ is topologizable iff there exists a topology on $G$ which makes it into a Hausdorff non-discrete topological group. \n\n> **Main question:** Is every infinite amenable group topologizable?\n\n* * *\n\nMain motivation for this question is that perhaps naively non-topologizability seemed to me such a strange property that I hoped it could be used to show existence of non-sofic groups.\n\n> **Question:** Is there an infinite non-topologizable sofic group?\n\nAfter failing to prove that sofic groups are topologizable I thought it would be still interesting to prove that infinite \"elementary sofic\" groups are topologizable. Elementary sofic groups are for the purpose of this discussion the class of \"groups which are provably sofic by current methods\", i.e. it contains all amenable groups, is closed under taking free products amalgamated over amenable groups, extensions with sofic kernel and amenable quotient, /any other results which are in the literature/, and with the property that if G is residually elementary sofic then G is elementary sofic.\n\n> **Question:** Is there an infinite non-topologizable elementary sofic group?\n\nUnfortunately my plan to answer the above question negatively failed at step 1, and hence the Main question. \n", "link": "https://mathoverflow.net/questions/114688/are-amenable-groups-topologizable", "tags": ["gr.group-theory", "gn.general-topology"], "votes": 25, "creation_date": "2012-11-27T10:42:34", "comments": ["I essentially stopped thinking about this, because I don't know how to proceed, but I thought I'd share one idea which at one point I thought was hopeful: G acts on a certain metric space X by isometries - namely fix a mean m on G and define X to be the set of subsets of G up to sets of mean 0. The metric on X is d(A,B) = m(A-B \\cup B-A). There are various topologies on Isom(X) but I failed to prove any of them gives a Hausdorff non-discrete topology on G. ", "@Simone: AFAI understand, Bohr compactif'n B(G) is in particular compact, so by Peter-Weyl thm if G embeds in B(G) then G is res. linear, and so by Malcev thm if it is fin. generated then it is res. finite. So to produce example of an amenable group G which doesn't embed into B(G) take a fin. generated simple amenable group, or easier take a fin.gen. solvable non-res. finite group (e.g. BS(1,n)). Taking a simple group is more convincing though, because the image in B(G) is trivial, whereas if the image in B(G) is infinite one could still hope for inducing a (Hausdorff) topology on G somehow...", "probably Corollary 3 on page 213 in mscand.dk/article.php?id=2006 will help...", "unfortunately it seems that the bohr topology is T2 only for MAP groups... is there any known example of amenable non-MAP group (with the discrete topology)?", "Any infinite abelian group is topologizable (e.g. by its Bohr topology).", "Just to clarify, Łukasz's first sentence actually characterizes the pro-p topology for fixed p; the profinite topology on $\\mathbb{Z}$ is generated by all the pro-p topologies.", "@Ben: you can use profinite topology, e.g. fix a prime number $p$; then integer $n$ is \"near\" to $0$ if large power of $p$ divides $n$. Other way to topologize integers is - take the action of integers $Z$ on the circle such that the generator $t$ of $Z$ acts by irrational rotation, and define $t^k$ to be \"close\" to the identity element iff $t^k$ is a rotation by a \"small\" angle. This is special case of \"find an infinite cyclic subggroup of a compact group and induce the topology\"", "Wouldn't the evenly spaced integer topology work?", "Are the integers topologizable? What does that topology look like?"], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "110", "site": "mathoverflow", "title": "Caramello's theory: applications", "body": "In [this](https://www.laurentlafforgue.org/math/TheorieCaramello.pdf#page=9) text, the author says (well, he says it in French, but I am too lazy to fix all the accents, so here is a Google translation):\n\n_In any case, contemporary mathematics provides an example of extraordinarily deep and highly studied equivalence which, without having so far been formulated as an equivalence of Morita, still seems very close to the general framework of Caramello's theory: it is the correspondence from Langlands._\n\nI am trying to understand this remark. The entire mathematical world has produced few, if any, correspondences which unify lots of non-trivial mathematics as elegantly as the Langlands program does (other than the Langlands, that is). Several genuises (including L. and V. Lafforgue, A. Wiles, P. Scholze) have contributed to its study and our understanding of it is still far from complete. \n\n> I wonder if there were any papers confirming the overall sentiment of the author, i.e. papers where Caramello's theory is applied to Langlands. Were there at least some long-standing problems that historically were not believed to have any serious connection with logic or topoi, and were solved using Caramello's techniques?\n\nThere were some similar questions, but [this](https://mathoverflow.net/q/330087/140765) one seemed to be largely about understanding an incorrect translation of a quote of another distinguished mathematician (hopefully, our translation is correct!), and [this](https://mathoverflow.net/q/29232/140765) one is pretty similar but not the same (it asks about producing theorems without any creative effort, while we allow additional creative input but just want to see theorems that rely in an essential way on ideas of Caramello, among other things). The second question did receive a response that might have qualified as an answer to our question, but that response also received [criticism](https://mathoverflow.net/questions/29232/the-unification-of-mathematics-via-topos-theory?noredirect=1&lq=1#comment70456_29826) from BCnrd (which was not addressed AFAICT).\n\nP.S. The kind of response we are not looking for in this question:\n\n * some philosophical argument why topoi are great that does not give a specific resut. While such arguments can be of interest in some situations, not right now.\n * something along the lines of \"One can produce thousands of deep theorems easily, so easily in fact that I won't even bother producing a single example\". \n\n\n\nThe kind of response we are looking for:\n\n * a link to a paper proving a result outside of logic or set theory and whose statement does not involve topoi, and an explanation of how Caramello's techniques enter the proof as an important ingredient.\n\n\n", "link": "https://mathoverflow.net/questions/332279/caramellos-theory-applications", "tags": ["ag.algebraic-geometry", "ct.category-theory", "topos-theory", "langlands-conjectures", "applications"], "votes": 25, "creation_date": "2019-05-23T03:04:44", "comments": ["@TimCampion OK, maybe somebody who actually read those papers will comment on them. But who knows, maybe logic will give us the motivic t-structure!", "The first two publications listed under \"Publications\" on her website are both papers in English written jointly with Lafforgue. In particular, the second appears to do something reasonably concrete in algebraic geometry / number theory. The first appears to be more category-theoretic, but perhaps closer inspection would reveal some algebraic geometry / number theory.", "@TimCampion I browsed through some of them and did not manage to find any results of geometric or topological interest (though those that I saw were in French, so I did not try super hard).", "Caramello's website lists 3 papers joint with Lafforgue by my count. Presumably they contain some kind of working-out of her theoretical framework, if not directly in the area of the Langlands program, then at least somewhere in that area."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "111", "site": "mathoverflow", "title": "$\\infty$-topos and localic $\\infty$-groupoids?", "body": "It's known that every classical (Grothendieck) topos is equivalent to the topos of sheaves on a localic groupoid (a groupoid in the category of locales).\n\nFor the record, this is proved by, starting form a topos $T$, constructing a locale $L$ and a surjection $L \\rightarrow T$ 'nice enough' (like a proper surjection, or an open surjection depending on the proof). Then $(L, L \\times_T L, L \\times_T L \\times_T L)$ is a truncated simplicial locale, which can be seen as a localic groupoid. There is a canonical geometric morphism from the topos of sheaves on this groupoid to $T$, and if the surjection $L \\rightarrow T$ was nice enough it's an isomorphism.\n\nMy question is : Can we hope for a similar result for $\\infty$-toposes ? for example by replacing localic groupoids by a localic $\\infty$-groupoids, defined as a simplicial locale satisfying some form of Kan complex condition and defining the associated $\\infty$-topos as the colimits of the corresponding simplicial diagram of localic $\\infty$-toposes.\n", "link": "https://mathoverflow.net/questions/93517/infty-topos-and-localic-infty-groupoids", "tags": ["ct.category-theory", "topos-theory", "infinity-topos-theory", "locales", "infinity-categories"], "votes": 25, "creation_date": "2012-04-08T13:23:10", "comments": ["@SimonHenry In classical topos theory, the Diaconescu localic cover theorem gives a connected locally connected geometric morphism. Often people only stress on the fact that it is open.", "@IvanDiLiberti the stability under pullback of localic morphisms might be useful at some point yes. But the crucial point would be some form of Diaconescu Localic open cover theorem for $\\infty$-toposes (though \"open\" should probably be replaced by locally $\\infty$-connected or something like that...).", "@SimonHenry I think that would be a very good starting point to prove the result you want.", "@IvanDiLiberti Probably - but as far as I remember it hasn't been written (from what I remember Lurie in HTT talks about n-localic toposes but does not study the relative version)", "Do we have a hyperconnected-localic fact system?", "So, you're asking if every classical $\\infty$-topos is equivalent to the category of sheaves on a localic $\\infty$-topos, correct?", "If I found any things about this, I will. But for now I have a lot of thing to learn before...", "This is something I have thought about in the back of my head for several years. If you come up with something, or would just like to brainstorm, let me know."], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "112", "site": "mathoverflow", "title": "Status of the Euler characteristic in characteristic p", "body": "In the introduction to the Asterisque 82-83 volume on `Caractérisque d'Euler-Poincaré, Verdier writes:\n\n> Enfin signalons que la situation en caractéristique positive est loin d'être aussi satisfaisante que dans le cas transcendant. On aimerait pouvoir disposer d'un groupe de Grothendieck `sauvage' des faisceaux constructibles tels que deux faisceaux ayant même rang et même comportement sauvage en chaque point aient même classe dans ce groupe de Grothendieck et tel que ce groupe soit foncteur covariant. Des resultats récents de Laumon donnent des indications dans ce sens. On aimerait aussi disposer d'une théorie des classes d'homologie de Chern qui sereait une transformation naturelle de ce groupe de Grothendieck sauvage dans les groupes d'homologie étale.\n\nMy question is: \n\n> > Has the situation improved in the intervening 32 years?\n\nI recall in detail below some background and ask some more precise questions.\n\nFor a map of complex algebraic varieties $f: X \\to Y$, one can define the pushforward of a constructible function $\\xi$:\n\n$$(f_* \\xi)(p) = \\sum_n n \\chi(\\xi^{-1}(n) \\cap f^{-1}(p) )$$\n\nThis is a relative notion of Euler characteristic: $f_* (1_X)$ is the function on $Y$ whose value at $y$ is the Euler characteristic $\\chi(X_y)$. \n\nThis notion is functorial: $(gf)_* = g_* f_*$. \n\nThe (naively) analogous notion in characteristic $p$ is not functorial. The standard counterexample is the Artin-Schreier map \n\n$g:\\mathbb{A}^1 \\to \\mathbb{A}^1$ \n\n$z \\mapsto z^p - z$\n\nNote $g_* 1 = p$. On the other hand, taking $f: \\mathbb{A}^1 \\to \\mathrm{Spec}\\,k$ the structure map, one has $f_* (g_* 1)) = p \\ne 1 = (fg)_* 1$. The problem seems to be that the map is wildly ramified at infinity. The beginning of the above paragraph of Verdier is a request that this situation be remedied by introducing some intermediary between constructible sheaves and their stalkwise Euler characteristics which would record the possibility of wild ramification and thus be functorial under pushforward. \n\n> Is there now some such object?\n\nThe second part of Verdier's paragraph presumes that the answer to the question above is yes, and requests an analogue of Macpherson's Chern class transformation. Over $\\mathbb{C}$, this is a natural transformation $c_{SM}: Con(\\cdot) \\to H_*(\\cdot)$ between the functors (covariant with respect to proper pushforward) which respectively assign to a variety its constructible functions and homology; it has the property that if $X$ is proper and smooth then $c_{SM}(X) = c(TX) \\cap [X]$.\n\n> Is there an analogue of the Macpherson Chern class transformation in characteristic $p$?\n\nThe Macpherson Chern class is defined in terms of something called the 'local Euler obstruction'. This is a constructible function $Eu_V$ attached to a variety $V$ which, roughly speaking, at a point $v\\in V$ records the virtual number of zeroes of any extension of the radial vector field from the boundary of ball around $v$ to its interior. (One passes to the Nash blowup in order to make sense of this vector field.) \n\nThe Euler obstruction enjoys the following hyperplane formula: locally embedding $(V, v)$ in a smooth variety $Y$, taking some sufficiently small $\\epsilon$ ball $B_\\epsilon(v)$, and a generic projection $l: B_\\epsilon(v) \\to \\mathbb{C}$, one finds that $l_* Eu_V$ is locally constant near $l(v)$. In other words, the Euler obstruction is annihilated by taking vanishing cycles with respect to a general linear function. \n\n> Is there an analogue of the local Euler obstruction in characteristic $p$, and is it annihilated by taking vanishing cycles with respect to a general linear function?\n", "link": "https://mathoverflow.net/questions/135983/status-of-the-euler-characteristic-in-characteristic-p", "tags": ["ag.algebraic-geometry", "characteristic-p", "characteristic-classes", "euler-characteristics"], "votes": 25, "creation_date": "2013-07-07T00:32:03", "comments": ["You've probably seen this already, but I'm adding it here anyway for others: arxiv.org/abs/1510.03018 by Takeshi Saito is relevant."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "113", "site": "mathoverflow", "title": "0's in 815915283247897734345611269596115894272000000000", "body": "> Is 40 the largest number for which all the 0 digits in the decimal form of $n!$ come at the end?\n\nMotivation: My son considered learning all digits of 40! for my birthday. I told him that the best way to remember them would be to come up with a mnemonic. He asked me what word he should come up with if the digit is 0. The rest is history.\n", "link": "https://mathoverflow.net/questions/393993/0s-in-815915283247897734345611269596115894272000000000", "tags": ["nt.number-theory", "open-problems", "digits"], "votes": 25, "creation_date": "2021-05-28T09:10:15", "comments": ["Maybe it will easy to solve in binary first. Where satisfied for 0!, 1!, 2!, 3! and 4!.", "The point is that the question is not about the zeros in this number (which is a trivial question) but about the zeros in infinitely many other numbers.", "I prefer to keep an air of mystery about 815915283247897734345611269596115894272000000000.", "I changed the title to \"$0$'s in 40!=...\", that is, added \"40!\". This makes the title less cryptic within the list of questions. I regret you reverted it.", "The title would be improved (and less puzzling) if it were replaced by the sentence that is the actual question.", "@Malkoun I did a similar check before seeing your post, and ran it up to 40,000. OEIS A182049 mentioned by Jeppe Stig Nielsen implies it's been checked up to 100,000. I guess I could check somewhat further by letting a computer run through the weekend, but I doubt there's much point.", "$41!$ seems to be the last factorial number with no digit 9 in it. It looks like for $n\\ge 42$, the decimal representation of $n!$ will contain all ten digits, even after trailing zeros have been removed. Edit: Found OEIS entry.", "Ok, it seems you're Hungarian. Well, finding such words should not be too much of an issue :-)", "As for the mnemonic, he can try to use 10 letters words, if they're not too uncommon in your native language.", "Interesting... I could not find another example for $41 \\leq n \\leq 16000$. I wrote a small Python program for that purpose, which checked the $n$s in the previous range one by one.", "Then I suppose this answers my question as 'known open problem'.", "One could call it the Zumkeller conjecture.", "@domotorp: oh, that's what I feared... :D", "I would learn a cyclic number instead, that he can use for magic tricks en.wikipedia.org/wiki/Cyclic_number", "@Loïc Thanks, though a bit early, and it is not the 40th, but the 815915283247897734345611269596115894272000000000th ;)", "Oh, I understood it the wrong way! My bad.", "@JukkaKohonen: I mean that 40 probably is the largest number $n$ for which all the 0's in $n!$ appear at the end.", "@Will, did your back-of-envelope calculation provide any estimate on when it would happen?", "Using the heuristic that the digits of $n!$ are just random, followed by some zeroes at the end, you can calculate without too much trouble that the answer to your question is: probably yes. My (very crude) estimate puts it at a better than 99% chance. I don't know whether this heuristic is any good, and I doubt this way of thinking will lead to a real answer . . . but there you have it.", "What follows doesn't answer your question, and probably won't contribute to an answer either, but a result by John Edward Maxfield -- A note on $N!$, Mathematics Magazine 43 #2 (March 1970), pp. 64-67 -- might be of interest: Given any positive integer $n,$ there exists a positive integer $N$ such that the decimal digits of $N!$ begin with all the decimal digits of $n,$ in their correct order. For more about this, see my answer to Short papers for undergraduate course on reading scholarly math.", "And by the way, happy 40th birthday ;)", "Is there an integer n which does not divide 10 such that for any k integer with 10 does not divide a = n * k and a has a zero in its decimal places ?", "Interesting question, but very difficult. I am not sure we even know if there are an infinite number of 0s in the decimal expansion of pi. (But it's believed this is the case)"], "comment_count": 23, "category": "Science", "diamond": 0} {"question_id": "114", "site": "mathoverflow", "title": "What's the smallest $\\lambda$-calculus term not known to have a normal form?", "body": "For Turing Machines, the question of halting behavior of small TMs has been well studied in the context of the Busy Beaver function, which maps n to the longest output or running time of any halting n state TM. The smallest TM with unknown halting behavior has 5 states. It takes $5\\cdot2\\cdot\\log_2(5\\cdot4 + 4)$ or roughly 46 bits to describe, 50 bits if we assume a straightforward encoding.\n\nIn comparison, $\\lambda$-calculus terms have a [simple binary encoding](http://tromp.github.io/cl/Binary_lambda_calculus.html): $00$ for lambda, $01$ for application, and $1^n0$ for variable with de Bruijn index $n$. It's natural to define a $\\lambda$-calculus analog of the busy beaver function as the maximum normal form size of any size $n$ closed lambda term.\n\nAs the smallest closed lambda term is $\\lambda\\,1$, with encoding $0010$, we determine $$ BB_{\\lambda}(4) = 4 $$ The next smallest ones, $\\lambda\\,\\lambda\\,1$ and $\\lambda\\,\\lambda\\,2$ are similarly already in normal form, and give $$ BB_{\\lambda}(6) = 6,\\qquad BB_{\\lambda}(7) = 7 $$ $BB_{\\lambda}(n)$ will have to remain undefined for $n < 4$ or $n = 5$.\n\nThe first enlarged normal form shows up at $\\lambda\\,(\\lambda\\,1\\,1)\\,(1\\,(\\lambda\\,2))$ which gives $$ BB_{\\lambda}(21) = 22 $$ $BB_{\\lambda}$ starts to grow rapidly at $n \\geq 30$, since tripling Church numeral two, $(\\lambda\\,1\\,1\\,1)\\,(\\lambda\\,\\lambda\\,2\\,(2\\,1))$ with normal form Church numeral $2^{2^2}= 16$, gives $$ BB_{\\lambda}(30) \\geq 5 \\cdot 16 + 6 = 86 $$ and quadrupling/quintupling give $$ BB_{\\lambda}(34) \\geq 5 \\cdot 2^{16} + 6 $$ $$ BB_{\\lambda}(38) \\geq 5 \\cdot 2^{2^{16}} + 6 $$ which exceed the TM Busy Beavers for 4 and 5 states.\n\nAn Ackermann-like function takes a mere [29 bits](https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/ackermann.lam). A twisted application to Church numeral $2$ yields a $BB_{\\lambda}(51)$ exceeding 2↑↑↑↑5.\n\nGraham's number is exceeded in at most [49 bits](https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam), giving $$ BB_{\\lambda}(49) \\geq 5 \\cdot G + 6 $$ (compared with a 16 state TM that needs over 192 bits to describe).\n\nWhat's the smallest n for which $BB_{\\lambda}(n)$ is unknown in ZFC?\n\nOne upper bound is [213 bits](https://codegolf.stackexchange.com/questions/79620/laver-table-computations-and-an-algorithm-that-is-not-known-to-terminate-in-zfc).\n\nLet's try to narrow it down some more.\n\nFunction $BB_{\\lambda}$ has been added to the [Online Encyclopedia for Integer Sequences](https://oeis.org/A333479).\n", "link": "https://mathoverflow.net/questions/353514/whats-the-smallest-lambda-calculus-term-not-known-to-have-a-normal-form", "tags": ["lo.logic", "lambda-calculus"], "votes": 24, "creation_date": "2020-02-25T03:18:56", "comments": ["Yes, I meant unknown in ZFC. I'll make that explicit.", "Does \"unknown\" mean \"unknown in ZFC\"? I infer that this is the case from the upper bound being a reference to a program which is known to halt under the assumption of a rank-into-rank cardinal.", "I mean \"unknown halting behaviour\". We know that (\\ 1 1)(\\ 1 1) is the smallest nonhalting term, of size 18.", "By “not known to halt”, do you mean “unknown halting behavior” or “known to not halt or unknown halting behavior”?", "I also have an upper bound of 247 bits for finding an odd perfect number at github.com/tromp/AIT/blob/master/oddperfect.lam", "Yes; Goldbach is 267 bits, as can be seen at github.com/tromp/AIT/blob/master/goldbach.gif which is compiled from goldbach.lam; larger than the 213 bit Laver statement above.", "For an upper bound, do you know what a lambda calculus formulation of twin primes / Goldbach / similar number-theoretic statements might look like?", "Curiously, triple Church_2 is not the best 30 bit beaver. A very minor tweak gives a 160 bit normal form!", "oops; good catch! will fix right away.", "It seems that application is also the other way round... quoting yourself :-)", "When i changed from 0-based to 1-based variables, I forgot to update the encoding. It's fixed now.", "Shouldn't $\\lambda 1$ have encoding 00110 (of length five), or am I missing something in the notation here?", "Yes, they were. I replaced them by actual lambdas now.", "Are all those slashes supposed to represent lambda?"], "comment_count": 14, "category": "Science", "diamond": 0} {"question_id": "115", "site": "mathoverflow", "title": "conjectures regarding a new Renyi information quantity", "body": "In a recent paper , we defined a quantity that we called the \"Renyi conditional mutual information\" and investigated several of its properties. We have some open conjectures, and I would like to pose them as open questions to the mathematics community on this forum. Perhaps there is a simple resolution, but we have considered known approaches and it appears that new tools are necessary for solving these problems (or maybe known tools could work?).\n\n## Background\n\nLet $\\mathcal{H}$ denote a Hilbert space, and let $\\mathcal{D}(\\mathcal{H})$ denote the set of [density operators](https://en.wikipedia.org/wiki/Density_matrix#Formulation) acting on this Hilbert space (positive semi-definite operators with trace one). (It suffices for our purposes to consider finite-dimensional Hilbert spaces, but of course the infinite-dimensional case is interesting as well.) We are interested in \"three-party\" density operators $\\rho_{ABC}$ acting on the tensor-product Hilbert space $\\mathcal{H}_A \\otimes \\mathcal{H}_B \\otimes \\mathcal{H}_C$. Let $\\rho_A$ denote the \"marginal\" density operator of $\\rho_{ABC}$, obtained by taking a partial trace over the spaces $\\mathcal{H}_B \\otimes \\mathcal{H}_C$: $$ \\rho_A = \\operatorname{Tr}_{BC} \\\\{ \\rho_{ABC}\\\\} $$ In a similar way, we can define $\\rho_{B}$, $\\rho_{C}$, $\\rho_{AB}$, $\\rho_{BC}$, and $\\rho_{AC}$.\n\nWe define the \"Renyi conditional mutual information\" of order $\\alpha \\geq 0 $ as follows: $$ I_{\\alpha}(A;B|C)_{\\rho} \\equiv \\frac{1}{\\alpha - 1} \\log \\operatorname{Tr} \\\\{ \\rho_{ABC}^{\\alpha} \\rho_{AC}^{(1-\\alpha) / 2} \\rho_C^{(\\alpha-1)/2} \\rho_{BC}^{1-\\alpha} \\rho_C^{(\\alpha-1)/2} \\rho_{AC}^{(1-\\alpha) / 2} \\\\} $$ In the above, identity operators are implicit, so that, e.g., $\\rho_C^{(\\alpha-1)/2} = I_{AB} \\otimes \\rho_C^{(\\alpha-1)/2}$, where $I_{AB}$ is the identity operator acting on $\\mathcal{H}_A \\otimes \\mathcal{H}_B$. For more motivation for the above quantity, please consult our paper.\n\nWe can easily prove that the above quantity is \"monotone under a local quantum operation acting on system $B$\" for all $\\alpha \\in [0,2]$, i.e., that the following inequality holds $$ I_{\\alpha}(A;B|C)_{\\rho} \\geq I_{\\alpha}(A;B|C)_{\\omega}, $$ where $\\omega_{ABC} \\equiv (\\operatorname{id}_A \\otimes \\mathcal{M}_B \\otimes \\operatorname{id}_C)(\\rho_{ABC})$, $\\operatorname{id}$ denotes the identity map, and $\\mathcal{M}_B$ is a completely positive trace preserving linear map acting on system $B$. A proof follows by applying a standard approach with the [Lieb concavity theorem](https://en.wikipedia.org/wiki/Trace_inequalities#Lieb.27s_concavity_theorem) and the [Ando convexity theorem](https://en.wikipedia.org/wiki/Trace_inequalities#Ando.27s_convexity_theorem) (proof is detailed in the paper).\n\n## Conjectures/Questions\n\nWe have several conjectures which (to fit the MathOverflow format) could be converted into questions in obvious ways (e.g., are the statements true? please provide an argument or counterexample, etc.). \n\nConjecture 1: \n\n> The quantity $I_\\alpha(A;B|C)$ is \"monotone under a local quantum operation acting on system $A$\" for all $\\alpha \\in [0,2]$, i.e., the following inequality holds $$ I_{\\alpha}(A;B|C)_{\\rho} \\geq I_{\\alpha}(A;B|C)_{\\tau}, $$ where $\\tau_{ABC} \\equiv (\\mathcal{N}_A \\otimes \\operatorname{id}_B \\otimes \\operatorname{id}_C)(\\rho_{ABC})$ and $\\mathcal{N}_A$ is a completely positive trace preserving linear map acting on system $A$. (Numerical evidence indicates that this conjecture should be true.) \n\nConjecture 2: \n\n> The Renyi conditional mutual information of order $\\alpha$ should be monotone increasing in the Renyi parameter $\\alpha$, i.e., if $0 \\leq \\alpha \\leq \\beta$, then $$ I_{\\alpha}(A;B|C)_{\\rho} \\leq I_{\\beta}(A;B|C)_{\\rho}. $$ (Numerical evidence indicates that this conjecture should be true as well, and we furthermore have proofs that it is true in some special cases (for $\\alpha$ in a neighborhood of one and when $\\alpha + \\beta = 2$).)\n\nThese are the basic conjectures, but please consult our paper for more general forms of them. Solving the first conjecture would be very interesting for us, as it would establish the Renyi conditional mutual as a true \"Renyi generalization\" of the \"von Neumann style\" quantum conditional mutual information. Solving the second conjecture would have widespread implications throughout quantum information theory and even for condensed matter physics ([topological order](http://dao.mit.edu/~wen/topartS3.pdf) \\- pdf).\n", "link": "https://mathoverflow.net/questions/172684/conjectures-regarding-a-new-renyi-information-quantity", "tags": ["pr.probability", "linear-algebra", "it.information-theory", "matrix-analysis", "quantum-mechanics"], "votes": 24, "creation_date": "2014-06-25T20:21:44", "comments": ["There has now been even more progress on Conjecture 2 that is useful enough for applications in quantum information theory. The idea was to make use of the Hadamard three-line theorem (in particular, Riesz-Thorin interpolation). This is detailed in the following paper: arxiv.org/abs/1505.04661 . See also arxiv.org/abs/1506.00981 for follow-up work.", "This is just to say that there has been what I would consider siginificant progress on Conjecture 2 since it was posted. Fawzi and Renner have proved a variation of the case when $\\beta = 1$ and $\\alpha = 1/2$, in the paper arxiv.org/abs/1410.0664 . My coauthors and I have also generalized the notion of Renyi conditional mutual information Renyi relative entropy differences in the paper arxiv.org/abs/1410.1443 and we generalized the methods of Fawzi and Renner in the paper arxiv.org/abs/1412.4067 . Of course, the conjectures stated here still remain open.", "In response to earlier comments (now deleted), I have reformatted the post in order to better highlight the apparent questions. With that, let's please focus on the mathematics.", "I really like this question --- I hope I manage to get some time to think about it!"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "116", "site": "mathoverflow", "title": "Is A276175 integer-only?", "body": "The terms of the sequence [A276123](https://oeis.org/A276123), defined by $a_0=a_1=a_2=1$ and $$a_n=\\dfrac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-3}}\\;,$$ are all integers (it's easy to prove that for all $n\\geq2$, $a_n=\\frac{9-3(-1)^n}{2}a_{n-1}-a_{n-2}-1$).\n\nBut is it also true for the sequence [A276175](https://oeis.org/A276175) defined by $a_0=a_1=a_2=a_3=1$ and $$a_n=\\dfrac{(a_{n-1}+1)(a_{n-2}+1)(a_{n-3}+1)}{a_{n-4}} \\;\\;?$$\n\n**Remark** : This question has been [asked previously on math.SE](https://math.stackexchange.com/questions/1905063/is-a276175-integer-only) ; one participant gave an interesting answer, but partial.\n", "link": "https://mathoverflow.net/questions/248604/is-a276175-integer-only", "tags": ["nt.number-theory", "sequences-and-series", "integer-sequences", "cluster-algebras"], "votes": 24, "creation_date": "2016-08-30T02:54:11", "comments": ["@YCor the sequences are examples of Y-frieze patterns, which are naturally associated to Fock and Goncharov's $\\mathcal{X}$-cluster varieties.", "More precisely, it seems that $a^{(5)}$ is still integral (checked for $n<35$), and $a^{(k)}$ is non-integral for $k>5$.", "Initially I thought all of the recurrences $a^{(k)}_n = \\prod_{i=1}^{k-1} (a^{(k)}_{n-i}+1)/a^{(k)}_{n-k}$ (with initial values all $1$) might have all terms integral - where $a^{(3)}$ is oeis.org/A276123 and $a^{(4)}$ is oeis.org/A276175. But it seems that $a^{(6)}_{12}$ is non-integral.", "@YCor The first sequence (the one with the explicit quasi-linear recurrence) is one of the period-1 Laurent phenomenon sequences from Alman, Cuenca and Huang's paper, thus presumably related to cluster algebras (not sure about this).", "@CarloBeenakker The proof relies on a computer calculation that no one but mercio seems to have seen let alone checked. And from the sound of his comments, mercio doesn't seem 100% sure of it either.", "@FedorPetrov what's the connection with cluster algebras?", "@darij grinberg : perhaps in view of today's edit it would be helpful to indicate why the MSE proof is not complete (I note that also OEIS lists this as a proven statement).", "@FedorPetrov they are actual polynomials in the first eight $a_{n-1}a_{n+1}/(a_n(a_n+1))$ but unfortunately for this particular sequence, those quantities are not all integers.", "At least a priori it could be Laurent polynomial not literally in $a_0,\\dots,a_3$, but in something different, like, say, also $(a_1a_2+1)/(a_2+a_3)$ or whatever.", "So it is unlikely 'cluster algebras'.", "When using generic $a_0,a_1,a_2,a_3$, one does not get Laurent polynomials (starting with $a_8$).", "I feel that 'cluster algebras' tag may be appropriate."], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "117", "site": "mathoverflow", "title": "Subfields of $\\mathbb{C}$ isomorphic to $\\mathbb{R}$ that have Baire property, without Choice", "body": "While sitting through my complex analysis class, beginning with a very low level introduction, the teacher mentioned the obvious subfield of $\\mathbb{C}$ isomorphic to $\\mathbb{R}$, and I then wondered whether this was unique. When I got back to my computer, I looked on google and came up with [this](http://camoo.freeshell.org/cfield.isor.html), which indicated that is not. However, the proof given there makes heavy use of Choice, and based on similar things, I am guessing that this cannot be avoided. \n \n \nDefine DC($\\omega_1$) as: \n \nFor all [trees](http://en.wikipedia.org/wiki/Tree_%28set_theory%29) $T$, $\\quad$ $T\\:$ has a branch $\\ $ or $\\ $ $T\\:$ has a chain of length $\\omega_1$ $\\quad$. \n \nwhich, if I haven't messed up the simplification, ZF proves is equivalent to the definition given at shelah.logic.at/files/446.ps . \n \n \n \nDoes $\\ ZF+DC(\\omega_1)\\ $ prove that there is a unique subfield of $\\mathbb{C}$ which is both isomorphic to $\\mathbb{R}$ and has Baire property?\n", "link": "https://mathoverflow.net/questions/51085/subfields-of-mathbbc-isomorphic-to-mathbbr-that-have-baire-property-w", "tags": ["set-theory", "gn.general-topology", "axiom-of-choice", "fields"], "votes": 24, "creation_date": "2011-01-03T19:36:46", "comments": ["Just a comment - The first link should be camoo.freeshell.org/cfield.isor.html instead of camoo.freeshell.org/cfield.isor to make it more readable.", "@Ricky: Your translation of DC_omega1 to trees is not correct. The above statement is obviously false, for instance since clearly the one element tree has no chain of length omega1.", "@Harry, What does there seem to be wrong with the formatting?", "Ah, thanks for the explanation Tobias. I found the result very counterintuitive until you pointed out that there was no reason to have that intuition to begin with :)", "There seems to be something wrong with the formatting?", "@Zev: The fact that $\\mathbb C$ has uncountably many field automorphisms does not seem so surprising to me. After all the only obstruction that keeps $\\mathbb R$ from having many automorphisms is the fact that its order can be described algebraically, hence is preserved. Since there is no such obstruction on $\\mathbb C$, there should be plenty of automorphisms, and in fact there are.", "+1 just for blowing my mind with the fact that there are subfields of $\\mathbb{C}$ isomorphic to $\\mathbb{R}$ other than $\\mathbb{R}$!"], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "118", "site": "mathoverflow", "title": "Example of a quasi-Bernoulli measure which is not Gibbs?", "body": "Let $X=\\\\{0,1\\\\}^{\\mathbb{N}}$. For simplicity I consider measures on $X$ only.\n\nA measure $\\mu$ is **quasi-Bernoulli** if there is a constant $C\\ge 1$ such that for any finite sequences $i,j$, $$ C^{-1} \\mu[ij] \\le \\mu[i]\\mu[j] \\le C\\mu[ij]. $$\n\n(Here as usual $ij$ is the juxtaposition of $i$ and $j$ and $[k]$ is the cylinder of all infinite sequences starting with $k$.)\n\nLet $f:X\\to \\mathbb{R}$ be continuous. The measure $\\mu$ is a **Gibbs measure** with potential $f$ if there are $C>0$ and $P\\in\\mathbb{R}$, such that for every infinite sequence $i_1 i_2\\ldots$ and all natural $n$, $$ C^{-1} \\le \\frac{\\mu[i_1\\ldots i_n]}{\\exp(-nP+f(i)+f(\\sigma i)+\\cdots+f(\\sigma^{n-1}i))} \\le C, $$ where $\\sigma$ is the left shift.\n\nOf course, there are other definitions of Gibbs measure, but they all agree if the potential $f$ is Hölder. In this case, it follows readily that a Gibbs measure is quasi-Bernoulli.\n\nIf a measure is quasi-Bernoulli, there is an equivalent measure (mutually absolutely continuous with bounded densities) which is invariant and ergodic under the shift.\n\n* * *\n\n**Question** : Are all quasi-Bernoulli measures Gibbs? (for some continuous potential, not necessarily Hölder). If not, what is a counterexample?\n\n**Motivation** : Gibbs measures (with Hölder potentials) enjoy many nice statistical properties. Sometimes I have a measure that is quasi-Bernoulli or satisfies some similar but weaker property. I would like to understand if and to what extent good statistical properties continue to hold in that setting.\n", "link": "https://mathoverflow.net/questions/53406/example-of-a-quasi-bernoulli-measure-which-is-not-gibbs", "tags": ["ds.dynamical-systems", "measure-theory"], "votes": 24, "creation_date": "2011-01-26T13:19:29", "comments": ["Dear Vaughn, dear John, could you give us references about both of your arguments? That would be very useful!", "Let me point out that there is a standard procedure associating to each almost additive sequence of functions a measure on the algebra up to each level (say $n$ if you consider the element $\\varphi_n$ of the sequence of functions). Moreover, any weak sublimit of an almost additive sequence with bounded variation is a Gibbs measure. This implies that your question has a positive answer at each level $n$, that is, indeed any quasi-Bernoulli measure (in your sense) is Gibbs at the algebra of level $n$. Moreover, any of its weak sublimits is also Gibbs, now on the whole $\\sigma$-algebra.", "Just a thought that doesn't necessarily go anywhere. If $\\mu$ is a Gibbs measure, then you can recover the corresponding potential $f$ (up to a constant) by $f = \\log(\\frac{d\\mu}{d(\\mu\\circ\\sigma)})$. So if we take an arbitrary quasi-Bernoulli measure and take $f$ to be its log-Jacobian, can we deduce anything about the regularity of $f$ and/or a Gibbs relationship based on the quasi-Bernoulli property?"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "119", "site": "mathoverflow", "title": "Is the Poset of Graphs Automorphism-free?", "body": "For $n\\geq 5$, let $\\mathcal {P}_n$ be the set of all isomorphism classes of graphs with n vertices. Give this set the poset structure given by $G \\le H$ if and only if $G$ is a subgraph of $H$.\n\n> Is it true that $\\mathcal {P}_n$ has no nontrivial automorphisms?\n\n**Remarks:**\n\nThis follows if one can recognize a graph from the set of isomorphism classes of its edge-deleted subgraphs. However, since recognizing a graph from the set of isomorphism classes of its edge-deleted subgraphs is stronger than edge reconstruction, I'm wondering if there is an alternative way of proving this.\n", "link": "https://mathoverflow.net/questions/153208/is-the-poset-of-graphs-automorphism-free", "tags": ["co.combinatorics", "graph-theory", "posets", "graph-reconstruction"], "votes": 24, "creation_date": "2013-12-30T20:23:03", "comments": ["Probably has something to do with: math.stackexchange.com/questions/3052384/…", "Of course there is in fact no need to check, as it anyhow follows from the set edge reconstruction conjecture but I could not find anything about for what values that has been checked.", "$\\mathcal P_5$ can pretty easily be checked by hand. As for anything larger, I'm not sure.", "Maybe it is useful to add that $\\cal P_4$ has a nontrivial automorphism. Did you ckeck for higher values, e.g. using the tables from maths.uq.edu.au/~pa/research/posets4to8.html?"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "120", "site": "mathoverflow", "title": "An $n \\times n$ matrix $A$ is similar to its transpose $A^{\\top}$: elementary proof?", "body": "A famous result in linear algebra is the following.\n\n> An $n \\times n$ matrix $A$ over a field $\\mathbb{F}$ is similar to its transpose $A^T$.\n\nI know one proof using the Smith Normal Form (SNF). However, I want to find an elementary proof avoiding any concepts related to the SNF. My question is: is there an elementary way to prove this?\n\nThe requirements are:\n\n * Do NOT use the structure theorem over PID.\n\n * Do NOT use the Smith Normal form (nor Jordan canonical form).\n\n * Do NOT use the concept of invariant factors.\n\n * Provide an explicit invertible matrix $P$ such that $A=PA^T P^{-1}$.\n\n\n\n", "link": "https://mathoverflow.net/questions/122345/an-n-times-n-matrix-a-is-similar-to-its-transpose-a-top-elementary-pr", "tags": ["linear-algebra", "matrices", "matrix-theory", "similarity"], "votes": 24, "creation_date": "2013-02-19T11:40:49", "comments": ["After many years this question was first asked, I cannot find anything meeting all of my requirement. Actually, finding Smith Normal Form can be done through elementary row/column operations. So, it is already elementary enough, and I do not have any reason left for avoiding such a simple and powerful tool. Also, SNF provides not only the existence of $P$, but also an explicit way to write down a such $P$.", "@user46896 The method is interesting, but not very far from Jordan canonical form (see questions V.B.1. and V.B.2.). One dreams about an algorithm providing $P$ from scratch ...", "@i707107 So, I hope you can work with this : sujets-de-concours.net/sujets/centrale/2003/tsi/math2.pdf , look at Partie V, it's an elementary approach. The solution here : pomux.free.fr/corriges-2003/index.html", "@Julien I know very little french, but there is google translate.", "@i707107 Can you read some french papers ?", "See projecteuclid.org/euclid.pjm/1103039127. Maybe math.stackexchange.com/questions/62497/… is related.", "@BenjaminDickman The first link shows that the solution space of $XA=A^T X$, $X=X^T$ has at least dimension $n$. However, the paper does not discuss how to obtain a \"nonsingular\" matrix in the solution space. That was actually a motivation of my question.", "Is the proof given here msp.org/pjm/1959/9-3/pjm-v9-n3-p25-p.pdf sufficiently \"elementary\"? Incidentally, note the remark on p.895 (pdf 4/7) comes from M. Newman; there was an earlier MO question answered using Smith's work (which you wish to avoid here...) that turned out to have a more elementary solution from M. Newman (mathoverflow.net/questions/151166/…). If the linked proof here does not suffice, perhaps Newman is the fellow to look to...", "@MarkSapir : Yes, of course the fact that $A$ and $A^{T}$ have the same minimum polynomial is obvious: it was the use of rational canonical form ( which is really more general that Jordan Normal Form) that seemed to be outside the spirit of the question.", "This is slightly related to my question ", "@Tom: This link projecteuclid.org/…", "@Geoff: the fact that $A$ and $A^T$ have the same minimal polynomial is obvious: if $p(A)=0$, then $p(A^T)=p(A)^T=0$. ", "In the two by two case $P$ can always be chosen to be symmetric. Is this true in general? ", "It's easy to prove using normal forms, but you can reframe it as the statement that if $A:V\\to V$ is linear then there is a nondegenerate bilinear form on $V$ such that $\\langle Av,w\\rangle=\\langle v, Aw\\rangle$ identically. Is there any chance of a proof of this that is quite different from the normal-form proofs? ", "There is a proof using rational canonical form, which basically just needs the fact that $A$ and $A^{t}$ have the same minimum polynomial, but I don't think that will be considered explicit .enough", "@Mariano: Yeah, it might not be possible to avoid those to prove it. The matrix P coming from Jordan canonical form proof is already complicated enough. Maybe requiring to provide explicit P is not possible without those. The generalized eigenvectors are involved and an inversion that transforms Jordan block to its transpose. ", "This is a lot like looking for a proof which does not need the letter a in order to be written down :-) What exactly do you mean by elementary? The Jordan canonical form is basic linear algebra, really."], "comment_count": 17, "category": "Science", "diamond": 0} {"question_id": "121", "site": "mathoverflow", "title": "What is the symmetric monoidal functor from Clifford algebras to invertible K-module spectra?", "body": "There ought to be a symmetric monoidal functor from the symmetric monoidal $2$-groupoid whose objects are Morita-invertible real superalgebras (precisely the Clifford algebras), morphisms are invertible superbimodules, and $2$-morphisms are invertible superbimodule homomorphisms to the symmetric monoidal ($\\infty$-)groupoid whose objects are invertible $KO$-module spectra, morphisms are invertible morphisms of $KO$-module spectra, etc. Everything I'm about to say should generalize to the complex case and $KU$ but let me only address the real case. \n\nOn $\\pi_0$ this functor ought to induce a map from the graded Brauer group of $\\mathbb{R}$ to the Picard group of $KO$, both of which turn out to be $\\mathbb{Z}_8$; this directly relates Bott periodicity for Clifford algebras and for K-theory. \n\nRecently I tried to write down this functor \"explicitly\" (to prepare for a talk and to confirm that I wasn't lying when I wrote [this](https://mathoverflow.net/questions/87345/brauer-groups-and-k-theory/177321#177321)) and found that I couldn't quite get some details to work out the way I wanted them to. The idea should be, starting with a real (resp. complex) superalgebra $A$, to\n\n 1. Construct the topological groupoid of finitely generated projective $A$-supermodules (automorphism groups have their usual topologies as Lie groups),\n 2. Pass to the algebraic K-theory spectrum of this groupoid, regarded as a symmetric monoidal groupoid under direct sum,\n 3. At some point during this process, kill off \"trivial objects.\"\n\n\n\nThe correct version of this process should, when given $A = \\mathbb{R}$ itself, reproduce $KO$, and more generally, when given the Clifford algebra $\\text{Cliff}(p, q)$, reproduce a shift of $KO$ (as a $KO$-module spectrum) by $p - q$ either up or down depending on some [sign conventions](https://mathoverflow.net/questions/185645/what-are-the-correct-conventions-for-defining-clifford-algebras). \n\nI ran into two problems getting this picture to work out: \n\n 1. The construction I know of the algebraic K-theory spectrum of a symmetric monoidal groupoid produces a connective spectrum. For example, applied to $\\text{Vect}_{\\mathbb{R}}$ it reproduces $ko$ rather than $KO$. \n\n 2. It seems like there are two inequivalent ways to kill off trivial objects, and they don't produce the same answer. \n\n\n\n\nIn detail, to fix the sign conventions, let $\\text{Cliff}(p, q)$ be the Clifford algebra generated by $p$ anticommuting square roots of $1$ and $q$ anticommuting square roots of $-1$, so that $\\text{Cliff}(1, 0) \\cong \\mathbb{R} \\times \\mathbb{R}$ and $\\text{Cliff}(0, 1) \\cong \\mathbb{C}$ (as ungraded algebras). Let $K(\\text{Cliff}(p, q))$ be the algebraic K-theory of the category of finite-dimensional $\\text{Cliff}(p, q)$-supermodules (so some connective spectrum). There are two natural forgetful functors inducing maps\n\n$$K(\\text{Cliff}(p + 1, q)) \\to K(\\text{Cliff}(p, q)), K(\\text{Cliff}(p, q + 1)) \\to K(\\text{Cliff}(p, q))$$\n\nof connective spectra; as maps on categories, they pick out the supermodules admitting an additional odd automorphism squaring to $1$ or an odd automorphism squaring to $-1$ respectively. For example, when $p = q = 0$, these are two ways of identifying the supervector spaces you want to regard as trivial for the purposes of defining $KO$ in terms of supervector spaces. \n\nI computed that the homotopy cofibers of these maps (in connective spectra) are $\\Omega^{p-q} ko$ and $B^{p-q} ko$ respectively; that is, we get the infinite loop spaces making up $ko$, but in two different orders. \n\n> How can this construction be fixed so that it outputs $KO$-module spectra rather than $ko$-module spectra, and what's the \"correct\" way to kill trivial objects?\n\nOne desideratum for \"correct\" is that it should generalize cleanly to the case where we replace $\\mathbb{R}$ with a discrete field $k$ and replace $KO$ with the algebraic K-theory spectrum $K(k)$ of $k$. The end result of the \"correct\" process is that it should naturally give a symmetric monoidal functor inducing a map from the graded Brauer group of $k$ to the Picard group of $K(k)$. But here there are not only two choices for what an odd automorphism could square to: instead the set of choices is parameterized by \n\n$$k^{\\times}/(k^{\\times})^2 \\cong H^1(\\text{Spec } k, \\mu_2).$$\n\nSo I don't know what to do. Maybe always pick squaring to $1$? \n\nOne thing that made me uncomfortable about the above is that it's conceptually important that the supermodule categories should be regarded as enriched and tensored over supervector spaces, but I ignored their enrichment over supervector spaces when I passed to algebraic K-theory spectra. The enrichment seems like it ought to be important for defining the \"correct\" way to kill of trivial objects. \n", "link": "https://mathoverflow.net/questions/186148/what-is-the-symmetric-monoidal-functor-from-clifford-algebras-to-invertible-k-mo", "tags": ["at.algebraic-topology", "kt.k-theory-and-homology", "clifford-algebras"], "votes": 23, "creation_date": "2014-11-03T18:03:26", "comments": ["Update: awhile back I asked Lurie this question and his sense is that the functor I want isn't symmetric monoidal but at best lax / oplax.", "Have you worked out the complex version (graded complex central simple algebras and invertible $KU$-module spectra) of this story? If so, perhaps you can go down to $KO$-module spectra by taking $\\mathbb{Z}/2$-fixed points."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "122", "site": "mathoverflow", "title": "Base change for $\\sqrt{2}.$", "body": "This is a direct follow-up to [Conjecture on irrational algebraic numbers](https://mathoverflow.net/questions/173414/conjecture-on-irrational-algebraic-numbers). \n\nTake the decimal expansion for $\\sqrt{2},$ but now think of it as the base $11$ expansion of some number $\\theta_{11}.$ Is there an easy (or, failing that, hard) proof that $\\theta_{11}$ is transcendental? Of course, same question stands for $\\theta_k,$ for your favorite $k>10.$\n", "link": "https://mathoverflow.net/questions/173442/base-change-for-sqrt2", "tags": ["nt.number-theory", "ds.dynamical-systems"], "votes": 23, "creation_date": "2014-07-06T08:36:56", "comments": ["@NikitaSidorov I, for one, am completely agnostic on the subject, and have absolutely no clue why anyone would believe what you say everyone believes. I am, of course, just a caveman.", "@NikitaSidorov And of course, we all speak for ourselves.", "Of course, we all believe it is transcendental. And of course, we all know that this is way beyond our reach at present. Why ask, then?", "@AnthonyQuas I cannot actually understand your comment...", "Since the digits are the same but the base is larger, in a certain sense, $\\theta_{11}$ has a quicker rational approximations than $\\sqrt 2$. Can this fact affect the irrationality measure of $\\theta_{11}$?", "@AnthonyQuas: Are there reasons to believe in this independency, besides mere numerical observations?", "Algebraic irrationality \"is independent of\" base expansions; there are countably many algebraic irrationals; Therefore \"with probability 1\", all algebraic irrationals are normal to all bases.", "@GeraldEdgar What is the justification for the well-known conjecture?", "$\\theta_{11}$ is certainly irrational. There is a well-known (but still far from proved) conjecture that all algebraic irrationals are normal in all bases. So (of course) your number is transcendental. But your question (whether there is an easy proof of it) is not answered by this."], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "123", "site": "mathoverflow", "title": "Do all possible trees arise as orbit trees of some permutation groups?", "body": "## I.Motivation from descriptive set theory\n\n(Contains some quotes from Maciej Malicki's paper.)\n\nThe classical theorem of Birkhoff-Kakutani implies that every metrizable topological group G admits a compatible left-invariant metric, that is, a metric d inducing the group's topology, and such that d(zx,zy)=d(x,y) for all x,y,z in G. However, even if G is completely metrizable, it is not always true that G admits a metric that is left-invariant and complete at the same time.\n\nPolish groups admitting a complete left-invariant metric are called CLI groups. We only focus on CLI subgroups of $S_{\\infty}$, the symmetry group of natural numbers, for now. Malicki proved that a permutation group G is CLI iff the _orbit tree_ of G is well-founded, i.e. every branch of the tree terminates at some finite level, where the concept _orbit tree_ is defined as following.\n\n* * *\n\n## II.The problem\n\nLet $G\\leq S_{\\infty}$ be a permutation group acting on $\\mathbb{N}$ (starts from 1 for convenience). Define $G_n$ to be the n-point stabilizer, $G_n=\\\\{g\\in G|g(i)=i,1\\leq i\\leq n\\\\}$, and $G_0=G$. We represent an infinite orbit $O$ of $G_n$ by a node $N$ on level $n$ (finite orbits do not count).\n\nIf $O_1$ is an infinite orbit of $G_k$, $O_2$ an infinite orbit of $G_{k+1}$ and $O_2\\subset O_1$ (possibly the same), we correspondingly draw $N_2$ as a one-step extension of $N_1$.\n\nAssuming $G$ is transitive, this process defines a map from $G$ to a rooted tree $T_G \\subset\\omega^{<\\omega}$. We say it is the _orbit tree_ of $G$ and it roughly describes how the orbits of $G$ split when we keep fixing more and more points. The natural inverse question is\n\n> For every tree $T\\subset\\omega^{<\\omega}$, is there a group $G$ such that the orbit tree $T_G$ is isomorphic to $T$? Can we construct it?\n\n* * *\n\n## III.Some attempts\n\nSince I do not know too much about group theory, especially not about infinite permutation groups, I started with some concrete examples. The construction was hard even for very simple finite trees, though. For instance, $\\mathbb{Z}$ acting on $\\mathbb{Z}$ realizes the tree of a single node. $\\mathbb{Z}_{wr}\\mathbb{Z}_2$(wr for wreath product) on $\\mathbb{Z}\\times 2$ realizes a \"chain\" of any finite depth, etc. But these seem to be too _ad hoc_.\n\nI also received a generous hint from an expert, saying that\n\n> 1\\. If $G$ is a permutation group on $\\Omega$ and $\\alpha\\in\\Omega$ and the orbits of $G_{\\alpha}$ are $\\Omega_{\\lambda}(\\lambda\\in\\Lambda)$ then $G_{\\alpha}$ is a subdirect product of the restrictions $H_{\\lambda}:=G_{\\alpha}^{\\Omega_{\\lambda}}$. Specifically $G_{\\alpha}$ can be embedded in the Cartesian product $\\prod_{\\lambda}H_{\\lambda}$ such that the projections onto each factor are surjective. \n> \n> 2\\. Now let $G$ be a free group of countably infinite rank. We want to find a corefree subgroup $H$ of $G$ of countable index such that $H$ is free of infinite rank and has a specified number of double cosets $HxH$ for which $|H:x^{-1}Hx|$ is infinite ($H$ is going to play the role of $G_{\\alpha}$). We know that every subgroup of a free group is free and many (most of them?) have infinite rank when $G$ has infinite rank. I don't think it is that hard to satisfy the condition about the number of cosets. If this is true then we can go from level 0 to level 1. \n> \n> 3\\. To go to level 2 we know that $H$ is a subdirect product of the $H_{\\lambda}$. Suppose that all the infinite $H_{\\lambda}$ are free of infinite rank and that the embedding of $H$ induces an isomorphism onto each of these. Now for each infinite $H_{\\lambda}$ choose (corefree, infinite rank) $K_{\\lambda}$ so that $H_{\\lambda}$ and $K_{\\lambda}$ play the roles of $G$ and $H$ in 2 to provide the correct number of infinite orbits at the next level. Can we choose the isomorphisms $H\\rightarrow H_{\\lambda}$ and a subgroup $K$ of $H$ such that $K$ maps onto $K_{\\lambda}$ for each $\\lambda$? \n> \n> 4.If the answers to these questions are yes, I think that you can then prove what you want. \n\nIt is not easy for me to comprehend his thought since I have no training in free groups. I have some vague doubts that\n\n 1. What is the action of $G$ on countably many things? $G$ naturally acts on itself by concatenation but this seems not correct.\n\n 2. Why must such an $H$ be a one-point stabilizer?\n\n 3. Moreover, the free group of countably infinite rank is countable itself. It is known that every countable group is CLI, hence its orbit tree must be well-founded. That is to say, $G$ may never realize ill-founded trees. I suspect we need to use other methods for non-CLI groups, taking inverse limit for example.\n\n\n\n\n* * *\n\n## IV.Possible modifications of the problem (not strict)\n\nFrom the viewpoint of descriptive set theory, it is not necessary to realize every tree, \"a subcollection of trees\" which has the same Borel complexity as \"all trees\" will suffice. For example, we may properly \"stretch\" the tree so that there is at most one \"event\" (a node terminates or a node splits into more than one nodes) happens on each level. we may also require every splitting node to split infinitely, etc. As long as the transformation is Borel, we may choose the easiest class of trees to realize.\n\nTo realize the tree, we may also choose to fix finitely many points at one time if necessary.\n\nIf we only look at the well-founded trees, we may construct a big tree recursively. We know that the one-point stabilizer is isomorphic to a subdirect product of the restrictions on each orbit. But\n\n> Given transitive permutation groups $H_1$, $H_2$... (finitely or countably many) acting on $\\mathbb{N}$, is there a transitive group $G$ such that $G_{\\alpha}$ is isomorphic to the direct product of these $H_{\\lambda}$?\n\nMy greedy hope (possibly incorrect!) is, there is such a group. If this is true, noticing that factors of direct product are \"independent\" of each other, we may fix points on orbits \"one by one\" and let other orbits \"wait\". Say the orbit tree of $H_{\\lambda}$ is $T_{\\lambda}$, then the orbit tree of $G$ will be of the shape that each subtree is a \"stretched\" $T_{\\lambda}$. It will work since we only need to realize all \"stretched\" trees.\n\nPlease do not laugh at my silliness if I am miserably wrong :-P Thank you so much for your time and effort.\n", "link": "https://mathoverflow.net/questions/61361/do-all-possible-trees-arise-as-orbit-trees-of-some-permutation-groups", "tags": ["gr.group-theory", "co.combinatorics", "descriptive-set-theory", "topological-groups", "permutations"], "votes": 23, "creation_date": "2011-04-11T21:02:55", "comments": ["a rooted tree is a subset(well, my fault) in \\omega^{<\\omega} that is closed under taking initial segment. For example, the set {(0),(0,0),(0,1),(0,2)} is the tree that has one root and three \"branches\", each terminates at level 1. But we usually think of it intuitively(as drawn on paper). As we fix more and more points in N, the orbits become smaller and smaller as a set. We intuitively draw a node for each orbit with infinite size. If O2 is a subset of O1, then N2(on the level below) is linked to N1. Thus we may draw the orbit tree.", "Thank you for your comment. G_1 could have infinitely many orbits. But your construction will not obtain the wanted direct product. This is because g(a1,a2...)g^{-1}=(g(a1),g(a2)...) and it will introduce a lot more permutations.", " How are you relating a tree to an element in \\omega^{<\\omega}? Does this mean all finite sequences in natural numbers? If so why does the tree need to be finite? Probably I am misunderstanding the notation. Can you clarify?", "Can't G1 have infinitely many orbits? Construct your group in the following fashion. Divide N into {1} and countably more disjoint infinite sets Sis. Let G be generated by transpositions between the 1 and the first element of Sis and the full permutation group on Si. I was not sure how G corresponded to the rooted tree"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "124", "site": "mathoverflow", "title": "Boundaries of noncompact contractible manifolds", "body": "It is known that a manifold $B$ bounds a compact contractible topological manifold if and only if $B$ is a homology sphere. The \"only if\" direction follows by excising a small ball in the interior of the contractible manifold, and noting that its boundary sphere has the same homology as $B$ by Poincare duality. The \"if\" direction is due to Freedman in dimension $3$ and to Kervaire in dimensions $>3$.\n\n**Question.** Is there a characterization of boundaries of _noncompact_ contractible manifolds? \n\nNote that if $B$ is a boundary component of a contractible $n$-manifold $W$, then the following holds.\n\n 1. If $B$ is compact, then $W$ is compact and $\\partial W=B$. \n(If $V$ denotes the union of $B$ and the interior of $W$, then the long exact sequence of the pair gives $H_{n-1}(B)\\cong H_{n}(V,B)$ and Poincare duality gives implies that $H_{n-1}(B)$ is nontrivial, and hence so is $H_{n}(V,B)\\cong H^0_c(V)$ which is only possible if $V$ is compact.) \n\n 2. $B$ is stably parallelizable (because $W$ is parallelizable).\n\n 3. If $U$ is a proper open subset of $B$, then $U$ bounds a noncompact contractible manifold, namely, $U\\cup\\mathrm{Int}(W)$. \n\n 4. The product of $B$ with any open contractible manifold bounds a noncompact contractible manifold.\n\n 5. If $B'$ bounds a contractible manifold $W'$, then the connected sum $B\\\\# B'$ bounds a contractible manifold, namely, the boundary connected sum of $W$, $W'$.\n\n\n\n\nThe above question may be too hard, so I would be happy with any addition to 1--5.\n", "link": "https://mathoverflow.net/questions/89604/boundaries-of-noncompact-contractible-manifolds", "tags": ["gt.geometric-topology", "mg.metric-geometry"], "votes": 23, "creation_date": "2012-02-26T12:43:35", "comments": ["@User See theorem 11.3 in Massey \"Homology and cohomology theory\", or the section on Poincare duality in Bredon's \"Sheaf theory\". There two references involve (co)homology of general spaces and may be hard for a beginner. Alternatively, solve exercise 35 in section 33 in Hatcher's \"Algebraic topology\".", "Could you please tell me a source of the proof of $H^k_\\textbf{c}(M)\\cong H_{n-k}(M,\\partial M)$, where $n=\\dim M$?", "Igor, of course, one has to construct a Gromov-hyperbolic metric on $W$ \"relative\" to $B$, so that $B$ embeds. The difficulty with my approach, I think, is in two absolute cases: (1) constructing a Gromov-hyperbolic metric on $W$ so that the entire boundary is a homology sphere; (2) Given a compact contractible manifold $W$ with boundary $B$, construct a metric on $W$ so that $B$ is the ideal boundary. Part (2), I think, is quite doable and mildly interesting. Part (1), I think, is difficult, if not impossible. ", "Misha, thank you for bringing geometry into the game. Yet I do not see how this gives an embedding of the boundary of $W$ into the ideal boundary of (a hyperbolic metric on) the interior of $W$. The ideal boundary is not necessarily the largest compactification. Say start with a compact contractible manifold whose boundary is a homology sphere $B$, remove a proper closed subset of $B$, and try to use your idea to embed what is left in a homology manifold; this looks hard.", "A remark: The approach I suggested could work only for $n\\ge 4$, since every homology 2-sphere is homeomorphic to the usual sphere. But for $n\\ge 4$ I do not see any obvious obstructions. ", "Recall that Bestvina proved in 1996 that a hyperbolic group $G$ is a $PD(n)$ group if and only if its ideal boundary is a homology manifold which is also a homology sphere (of dimension n-1 of course). One can also ask an absolute version of your question in the hyperbolic context: Suppose $W$ is contractible $n$-manifold. Does $W$ admit a bounded geometry hyperbolic metric so that the ideal boundary is a homology manifold/homology sphere (of dimension n-1)? No group is involved here, just pure geometry. ", "I do not have an answer, but only a vague suggestion: Instead of trying to embed $B$ in a homology (n-1)-manifold, one can try to embed it in a homology sphere which is also a homology (n-1)-manifold. This might be easier. This situation could be similar to Mladen Bestvina's theorems on ideal boundaries of hyperbolic PD(n) groups. One approach would be to construct a suitable Gromov-hyperbolic metric on $W$ so that $B$ is a part of the ideal boundary. Alternatively, one can try to imitate Bestvina's arguments without having a metric (or, instead, using a weaker structure a la John Roe). ", "Tom's comment on intersection product together with classification of noncompact 2-manifolds implies that if $B$ has dimension $2$, then it is homeomorphic to an open subset of $S^2$, and clearly any open subset of $S^2$ occurs as $B$. Indeed, if the surface is not planar, then it contains a handle, and hence there are two curves that only intersect once.", "@Igor: I meant the $Q/Z$ valued linking form on the torsion homology classes. I was thinking a warm up problem is to determine which $B=$interior$(M)$, $M$ compact, are boundaries, and suggesting maybe all such embed as codim 0 submanifolds of a homology sphere. ", "@Paul, linking numbers do not vanish, not even in the sphere bounding a disk. I am not sure whether \"open subsets of homology spheres\" is the answer. Certainly, $B\\times\\mathbb R^n$ is homeomorphic to an open subset of $\\mathbb R^{2n-1}$: just embed $B$ properly in $\\mathbb R^{2n-1}$ and note that the normal bundle is trivial by 2. But I cannot see why $B$ itself would be homeomorphic to an open subset of $\\mathbb R^{2n-1}$. Would not it be like showing that any open contractible manifold has a manifold compactification (which is surely false)? ", "Maybe the answer is any open subset of a homology sphere? Linking numbers vanish also. I can't see how to find a compact $n$-manifold with boundary $M$ whose interior bounds a contractible manifold but which doesn't embed in a homology $n$-sphere. For $n=3$, it seems like you can add 2-handles to $M$ to get a homology ball, and perhaps in higher dimensions you could add handles to get a homology ball.", "@Tom, oh, and the intersection number is trivial because the $0$-cycle is a boundary, so its points come in pairs with opposite signs, and cancel. Thank you for suggesting this!. I think this \"intersection product\" is Poincare dual to the cup product, as discussed in Dold's \"Algebraic Topology\", section VIII.13.", "In particular I am referring to intersection numbers, not to the ring structure on homology.", "Every cycle in $B$ is the boundary of a chain in $W$. So basically given a $p$-cycle and an $(n-p-1)$-cycle in $B$ their intersection is the boundary of the intersection of a $(p+1)$-chain and an $(n-p)$-chain in $W$, so the boundary of a $1$-chain, so a homologically trivial $0$-chain. To make this rigorous I guess you would need to say something about Poincare duality and compactly supported cohomology.", "@Tom, I am struggling to make the notion of an intersection product precise. Is triviality of the intersection product based on the fact that $B$ has trivial normal bundle in $W$, or does the contractibility of $W$ get used?", "By 3, $B$ can have the homotopy type of any finite complex, by taking a thickening of some embedding in Euclidian space. So for example any $k$ dim'l complex has the homotopy type of a $n=2k+1$ manifold which is the boundary of a contractible $n+1$ manifold.", "6. Intersection products in $B$ are trivial."], "comment_count": 17, "category": "Science", "diamond": 0} {"question_id": "125", "site": "mathoverflow", "title": "Riemannian manifolds etc. as locally ringed spaces?", "body": "There are, among others, three general ways of equipping a \"space\" (which for the purposes of this question could be a topological space or a differentiable manifold, according to the case) with further structure:\n\n(1) \"Specifying regular funcions\", which leads to locally ringed spaces, e.g. real-analytic manifolds or holomorphic manifolds.\n\n(2) \"Choosing a section (perhaps with some nondegeneracy properties) of some (usually tensor) bundle\", which leads for example to Riemannian manifolds, quasi-symplectic manifolds and quasi-holomorphic manifolds.\n\n(2') A variant of the latter: \"Choosing a nondegenerate section _with some integrability properties_ of some (usually tensor) bundle\", which leads e.g. to symplectic geometry and holomorphic geometry.\n\n(3) \"Choosing a sub-bundle (with some properties) of some bundle\" , and this leads for example to foliated manifolds and contact structures.\n\nThere are certainly some overlaps between the above approaches. For example, one can think of a complex manifold as a topological space equipped with a sheaf of local rings (the sheaf of holomorphic functions of the complex manifold), which is an instance of (1), or as an even dimensional smooth manifold equipped with an \"integrable\" field of endomorphisms of its tangent bundle which square to $-\\mathrm{id}$, and that's an instance of (2), or rather (2'). Another example is that of a foliated manifold: it's usually seen as an instance of (3), but you can construct the sheaf of locally-constant-on-the-leaves smooth functions, and that reduces, in some sense, to the point of view (1). \n\nFurthermore, smooth (in the sense of differentiable) maps between complex manifolds that preserve the tensor $J$ giving the complex structure correspond exactly to locally ringed space maps between the corresponding complex analytic spaces. And I would imagine that some foliation-preserving map in the case of foliated manifolds translate fully faithfully into maps for the locally ringed structure mentioned above. \n\nIn either case, if I'm not mistaken, you get a fully faithful embedding of some \"geometric category\" into the category of locally ringed spaces (such that, say, the forgetful functor to $\\mathrm{Top}$ is respected). So it seems that the approach (1) is quite general. It's then spontaneous to ask:\n\n> Is the category of symplectic manifolds (with symplectomorphisms as morphisms) fully faithfully embeddable (say, preserving the forgetful functor to $Top$) into the category of locally ringed spaces?\n\nand \n\n> Is there a good notion of morphisms between Riemannian manifolds (e.g. local isometries? Riemannian submersions? compositions thereof?) such that the resulting category has a fully faithful embedding into locally ringed spaces as above?\n\n* * *\n\nThe first (perhaps useless?) construction that comes to my mind is the following. Given a smooth manifold $X$ with sheaf of differentiable functions $\\mathcal{O}_X$, let $\\mathcal{T} \\\\!en_X^{\\bullet}:=\\oplus_{i\\geq 0} (T_X^{\\; *})^{\\otimes i}$ be the sheaf of covariant tensors on $X$. If $g$ is a Riemannian metric on $X$, one could consider the sheaf of (commutative) local rings generated by $\\mathcal{O}_X$ and $g$:\n\n$\\mathcal{O}_X[g]\\subseteq \\mathrm{Sym}^{\\bullet}(T_X^{\\; *})\\subseteq \\mathcal{T} \\\\!en_X^{\\bullet}$\n\nand then consider the locally ringed space $(X,\\mathcal{O}_X[g])$.\n\nOne can do an analogous thing with a symplectic manifold $(X,\\omega)$ viewing $\\omega$ as an anti-symmetric tensor:\n\n$\\mathcal{O}_X[\\omega]\\subseteq \\Omega_X^{\\mathrm{even}} \\subseteq \\mathcal{T} \\\\!en_X^{\\bullet}$\n\nto get a locally ringed space $(X,\\mathcal{O}_X[\\omega])$.\n\nCan this construction be of any help to answer the above questions?\n", "link": "https://mathoverflow.net/questions/56833/riemannian-manifolds-etc-as-locally-ringed-spaces", "tags": ["locally-ringed-spaces"], "votes": 23, "creation_date": "2011-02-27T08:08:01", "comments": ["From scheme theoretic perspective, how do wish to 'deal' with additional structures like metrics, symplectic forms, hermitian or kahler structures?", "I'm quite interested in this question. Made any progress since Feb?", "@JohanesEbert: recovering $g$ up to conformal equivalence would already be nice. And what about LRS morphisms: would they correspond to conformal mappings? ", "@Zack: ya.. it looks like so. Perhaps my definition is just useless. But I don't know how morphisms of symplectic/Riemannian manifolds induce homomorphisms on that sheaves. ", "Isn't $\\mathcal{O}[\\omega]$ just $\\mathcal{O}[x]/x^{n+1}$?", "You cannot recover $g$ from $\\mathcal{O}[g]$, only up to conformal equivalence.", "@André: In the theory of \"supermanifolds\" people already use sheaves of $\\mathbb{Z}/2\\mathbb{Z}$-graded rings. And there are some recent theories of \"derived manifolds\" (of which I don't know anything). But my question was perhaps more down to earth. Anyway, why do you suggets using differential graded rings? Is it to keep track that the \"generator\" $g$ in $\\mathcal{O}_X[g]$ is placed in (tensor) degree 2?", "Maybe it could be useful to generalize the notion of locally ringed space, and allow \"structure sheaves\" with values in categories other than rings (e.g. differential graded rings)."], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "126", "site": "mathoverflow", "title": "When does a representation admit a spin structure?", "body": "Let $G$ be a finite group, and let $V$ be an $n$-dimensional _real_ representation of $G$. Think of $V$ as given by a homomorphism $$ \\rho_V\\colon G\\to O(n).$$ Write $\\chi_V$ for the character of $V$.\n\nHere are two problems.\n\n 1. Using _only_ the character $\\chi_V$ of $V$, determine whether $\\rho_V(G)\\subset SO(n)$.\n\n 2. Using _only_ the character $\\chi_V$ of $V$, and assuming $\\rho_V(G)\\subset SO(n)$, determine whether $\\rho_V$ admits a factorization through a homomorphism $\\widetilde{\\rho}_V:G\\to Spin(n)$. \n\n\n\n\nHere's one answer to 1: the identity of formal power series $$ \\sum_{k\\geq0} \\chi_{\\Lambda^kV}(g)\\,T^k = \\exp\\Bigl( -\\sum_{k\\geq1} \\frac{1}{k}\\chi_V(g^k)\\, (-T)^k \\Bigr)$$ where $\\Lambda^kV$ is the $k$-th exterior power representation of $V$, gives $\\chi_{\\Lambda^nV}(g)$ as a polynomial in $\\chi_V(g),\\dots,\\chi_V(g^n)$, and $\\rho_V(G)\\subset SO(n)$ if and only if all $\\chi_{\\Lambda^nV}(g)>0$.\n\nIs there a better answer for 1? Is there any answer in a similar spirit for 2?\n", "link": "https://mathoverflow.net/questions/52643/when-does-a-representation-admit-a-spin-structure", "tags": ["rt.representation-theory"], "votes": 23, "creation_date": "2011-01-20T08:10:12", "comments": ["For 1, it depends what you mean by \"using the character ONLY\". You can't tell the number of elements of each order from the character alone. If you do know the order of each element, then 1 requires just that each element of $G$ has the eigenvalue $-1$ with even (possibly zero) multiplicity, and this is an easy calculation, given the character values ( in fact, you only need to check elements whose order is a power of $2$).", "Charles, I can see at least a necessary condition on the character (that you probably know of). It is that for any involution $s\\in G$, $\\chi_V(s)\\equiv\\chi_V(1) \\mod 8$ (this comes from restriction to cyclic subgroups). Given the \"miracles\" of Brauer theory, might it be sufficient also ? A perhaps relevant reference is this Deligne's 1976 [paper][1], related to Marty's first remark about $\\epsilon$ factors. [1]: digizeitschriften.de/de/dms/img/?PPN=GDZPPN002092654", "Marty, your observation about pairs of conjugacy classes seems convincing to me.", "@BS: Ah yes.. that was dumb on my part. I had in mind the lifting of subgroups isomorphic to $Z$, not $Z / kZ$, when I wrote that. The remainder of my comment -- that one needs information about pairs, not just individual elements -- should hold (though it's vague). ", "@marty : are you saying that any morphism $Z/kZ \\to SO(n)$ lifts $Spin(n)$ ? This seems not true if $k$ is even.", "Here's another reason why such an answer shouldn't exist for (2). Let's say you know all the eigenvalues for $\\rho(g)$, $\\rho(g^2)$, etc., information given by the character. Well, that information tells you nothing about whether $\\rho$ factors through the 2-fold Spin cover, because every element of $SO(n)$ lifts (compatibly with its powers) to an element of $Spin(n)$. Instead, you need information about $\\rho(g_1), \\rho(g_2), \\rho(g_1 g_2)$ for all pairs $(g_1, g_2)$ -- information depending probably on the conjugacy class of the pair $(g_1, g_2)$ (not individual conjugacy classes).", "Okay. But I don't see why, a priori, having a cohomological obstruction means you can't compute it from the character; my question 1 is really about the vanishing of $w_1(\\rho_V)$, after all. On the other hand, I could certainly believe that such a problem could be intractible.", "(I suppose that the $SU(2)$ and $SO(3)$ case is just the case $n = 3$ of the question...)", "I don't think that any such answer exists for (2). The obstruction to lifting is called the second Steifel-Whitney class of the orthogonal real representation $\\rho_V$, written $w_2(\\rho_V)$. It's difficult to compute and quite important for Galois representations (related to $\\epsilon$-factors). Such results lie beyond simple character calculuations, I think. Compare, for instance, the problem of deciding whether a rep. into $SO(3)$ arises as $Sym^2$ of a rep into $SU(2)$. Can you tell by the character? ", "I guess there is a cohomological obstruction for 2 in $H^2(G,\\mathbb{Z}/2\\mathbb{Z})$. Although, I have no idea how to express this in character theory."], "comment_count": 10, "category": "Science", "diamond": 0} {"question_id": "127", "site": "mathoverflow", "title": "Is the set of integers of the form $a/(b+c)+b/(a+c)+c/(a+b)$ computable?", "body": "The starting point of this question is the observation that the smallest positive integers $a,b,c$ satisfying\n\n$$\\frac{a}{b+c} + \\frac{b}{a+c} + \\frac{c}{a+b} = 4$$\n\nare [absurdly high](https://web.archive.org/web/20170813162605/https://plus.google.com/+johncbaez999/posts/Pr8LgYYxvbM), namely $$(154476802108746166441951315019919837485664325669565431700026634898253202035277999,$$ $$36875131794129999827197811565225474825492979968971970996283137471637224634055579 ,$$ $$ 4373612677928697257861252602371390152816537558161613618621437993378423467772036) .$$ This leads to the following general question: Is the set $C\\subseteq {\\mathbb N}$ defined by $$ C = \\\\{n\\in\\mathbb{N}\\setminus\\\\{0\\\\}: (\\exists a,b,c \\in\\mathbb{N}\\setminus\\\\{0\\\\}):\\frac{a}{b+c} + \\frac{b}{a+c} + \\frac{c}{a+b} = n\\\\}$$ computable? (As user [Watson](https://mathoverflow.net/users/84923/watson) points out in the comment section below, $C$ contains no odd numbers. It would also be great to see an even number $\\geq 6$ not contained in $C$.)\n", "link": "https://mathoverflow.net/questions/278747/is-the-set-of-integers-of-the-form-a-bcb-acc-ab-computable", "tags": ["nt.number-theory", "computability-theory", "diophantine-equations", "computer-science"], "votes": 22, "creation_date": "2017-08-14T12:16:31", "comments": ["The goo.gl link in a previous comment points to Quora.", "mathoverflow.net/questions/264754/…", "Also discussed at mathematica.stackexchange.com/questions/184956/…", "Cross-posted to cstheory.stackexchange.com/questions/39383/…", "(continued) Note that there are odd $n$ for which the associated elliptic curve has positive rank, but does not have rational points on the component that contains the positive points. (IIRC, $n = 19$ is an example.) This gives an additional twist to the question.", "This is somewhat similar to the question which natural numbers $n$ are congruent numbers (i.e. the area of a right-angled triangle with rational sides), which comes down to asking whether the elliptic curve $E_n \\colon y^2 = x^3-n^2x$ has positive rank over $\\mathbb Q$. In this case, there is a conjectural answer (Tunnell's Theorem, conditional on the BSD conjecture), which relies on the fact that the $E_n$ are all quadratic twists of a fixed curve. The question asked here is likely to be harder, since the resulting curves are not twists, and there is the positivity condition. -->", "So we're half done!", "Here is explained why $C$ doesn't contain any odd number.", "See also math.stackexchange.com/questions/402537/… and Bremner and Macleod, An unusual cubic representation problem, Annales Mathematicae et Informaticae 43 (2014) 29-41, ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf", "I don't have the time to look into the details, but from a cursory glance at this quora answer it looks like the problem amounts to deciding whether a certain elliptic curve has solutions over $\\mathbb{Q}$. Now IIRC there is an algorithm for that (find solutions locally and try to globalize) provided Ш is finite, which is conjecturally always the case. So I think conjecturally your set is indeed computable.", "A quick heuristic comment: for each $n$, the existence of such $a, b, c$ corresponds to the existence of a solution to a degree-$3$ Diophantine equation in $3$ variables; and this is, I believe, a bit beyond what is generally known to be decidable. So I suspect that there will be no general reason why this set is computable; rather, if it is computable (which I extremely strongly suspect it is), the proof is likely to be a corollary of a complete characterization of these $n$s, which will not use any computability theory."], "comment_count": 11, "category": "Science", "diamond": 0} {"question_id": "128", "site": "mathoverflow", "title": "A mysterious paper of Stallings that was supposed to appear in the Annals", "body": "In Stallings's paper\n\n * Stallings, John, _[Groups with infinite products](https://doi.org/10.1090/S0002-9904-1962-10817-9)_ , Bull. Amer. Math. Soc. **68** (1962), 388–389.\n\n\n\nhe briefly discusses how to prove \"several generalizations\" of Brown's theorem saying that monotone union of open $n$-cells is an open $n$-cell; however, he never formulates a precise statement of what he proves.\n\nFor the details, he refers to the paper\n\n * J. Stallings, On a theorem of Brown about the union of open cones, Ann. of Math, (to appear).\n\n\n\nBut as far as I can tell this paper never appeared. Does anyone know what happened and what the theorem is?\n", "link": "https://mathoverflow.net/questions/405617/a-mysterious-paper-of-stallings-that-was-supposed-to-appear-in-the-annals", "tags": ["reference-request", "gt.geometric-topology", "soft-question"], "votes": 22, "creation_date": "2021-10-06T13:41:53", "comments": ["I feel the pain. There's more than one of these in category theory from the 70s-80s, announced to appear in JPAA, but which never turned up.", "With regards to the publication of this paper, it looks like the journal may have been... stalling", "@CarloBeenakker: Thanks for noticing that reference! I think that the lack of a journal might just be the Annals's style guide -- there are several papers in that bibliography that are \"to appear\", and none of them specify a journal.", "in a 1963 Ann. of Math. article by Stallings that paper is still cited as \"to appear\", however now without a journal being mentioned."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "129", "site": "mathoverflow", "title": "The multiplication game on the free group", "body": "Fix $W\\subseteq\\mathbb F_2$ and consider the following two-person game: Player 1 and Player 2 **simultaneously** choose $x$ and $y$ in $\\mathbb F_2$. The first player wins, say one dollar, iff $xy\\in W$. Does this game admit Nash equilibria? \n\nOf course, for very particular choices of $W$ it does, for instance if $W$ is either finite or cofinite. But I am not interested in these cases, I am interested in _solvability_ of the _multiplication game_ for all $W$. \n\nOne possible mathematical reformulation (game theorists would say that the mixed extension of this game is not uniquely defined and then there are _several_ different mathematical reformulations) is the following:\n\n> **Question.** Do there exist, for all $W\\subseteq\\mathbb F_2$, finitely additive probability measures $\\overline\\mu,\\overline\\nu$ (of course depending on $W$) on the power set of $\\mathbb F_2$ such that for all other finitely additive probability measures $\\mu,\\nu$ one has $$ \\int\\int\\chi_W(xy)d\\mu(x)d\\overline\\nu(y)\\leq\\int\\int\\chi_W(xy)d\\overline\\mu(x)d\\overline\\nu(y)\\leq\\int\\int\\chi_W(xy)d\\overline\\mu(x)d\\nu(y) $$ ?\n\n**Motivation.** If we play the same game on a (countable) amenable group, one can show that a Nash equilibrium exists and is given by a right-invariant mean $\\overline\\mu$ and a bi-invariant mean $\\overline\\nu$. I am wondering what can happen for non amenable groups... I suspect (motivated by some possibly wrong speculative consideration) that the _multiplication game_ is actually solvable for all $W$ only on amenable groups, giving an hopefully nice new characterization of amenability. However, any approach to proving that solvability implies amenability so far got stuck.\n\nThanks in advance,\n\nValerio\n", "link": "https://mathoverflow.net/questions/98493/the-multiplication-game-on-the-free-group", "tags": ["gr.group-theory", "game-theory", "amenability"], "votes": 22, "creation_date": "2012-05-31T07:19:39", "comments": ["Ah, that was my misunderstanding.", "I've edited, saying explicitly that the choices of $x$ and $y$ are simultaneous.", "Why should P2 win? Notice that the choices are done independently (simultaneously, if you want), so that P2 cannot choose a suitable $y$ such that $xy\\in W$, because (s)he does not know $x$.", "I'm confused. If $W$ is the whole group, player 1 wins, otherwise player 2 wins. Right?"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "130", "site": "mathoverflow", "title": "Characterising critical points of $E(f)=\\int_{M}| \\bigwedge^2 df|^2 \\text{Vol}_{M}$", "body": "$\\newcommand{\\id}{\\operatorname{Id}}$ $\\newcommand{\\R}{\\mathbb{R}}$ $\\newcommand{\\TM}{\\operatorname{TM}}$ $\\newcommand{\\Hom}{\\operatorname{Hom}}$ $\\newcommand{\\Cof}{\\operatorname{Cof}}$ $\\newcommand{\\Det}{\\operatorname{Det}}$ $\\newcommand{\\M}{\\mathcal{M}}$ $\\newcommand{\\N}{\\mathcal{N}}$ $\\newcommand{\\tr}{\\operatorname{tr}}$ $\\newcommand{\\TM}{\\operatorname{T\\M}}$ $\\newcommand{\\TN}{\\operatorname{T\\N}}$ $\\newcommand{\\TstarM}{T^*\\M}$\n\nLet $\\M,\\N$ be $d$-dimensional oriented Riemannian manifolds. Let $\\tilde E:C^{\\infty}(M,N) \\to \\R$ be defined by\n\n$$\\tilde E(f)=\\frac{1}{2}\\int_{\\M} | \\bigwedge^2 df|^2 \\text{Vol}_{\\M}.$$ $\\tilde E(f)$ measures the mean action of $f$ on 2$D$-cubes. (How $f$ affects areas of surfaces, locally).\n\nNote that $\\bigwedge^2 df\\in \\Omega^2\\big(\\M,\\Lambda_2(f^*{\\TN})\\big)$. Let $\\nabla^{\\Lambda_2(f^*{\\TN})}$ be the induced connection on $\\Lambda_2(f^*{\\TN})$ and let $\\delta_{\\nabla^{\\Lambda_2(f^*{\\TN})}} $ be the adjoint of the covariant exterior derivative $d_{\\nabla^{(f^*{\\TN})}}$.\n\n> **Question:** Does every critical point of $\\tilde E$ satisfy $\\delta_{\\nabla^{\\Lambda_2(f^*{\\TN})}} \\big( \\bigwedge^2 df \\big) =0$?\n\nOf course, by counting degrees of freedom, this doesn't look reasonable.\n\nThe converse direction follows from the Euler-Lagrange equation of $\\tilde E$: I proved [here](https://mathoverflow.net/a/271264/46290) that the E-L equation is $$h_{f^*\\TN} \\big( \\tr_{\\TM}\\big( df \\otimes \\delta_{\\nabla^{\\Lambda_2(f^*{\\TN})}}(\\bigwedge^2 df)\\big) \\bigg)=0,$$\n\nwhere $h_{f^*\\TN}:f^*\\TN \\otimes \\Lambda_2(f^*\\TN) \\to f^*\\TN$ is the linear extension of\n\n$$ \\tilde w \\otimes (w_1 \\wedge w_2) \\to \\langle \\tilde w,w_2 \\rangle w_1-\\langle \\tilde w,w_1 \\rangle w_2.$$\n\nThus $\\delta_{\\nabla^{\\Lambda_2(f^*{\\TN})}} \\big( \\bigwedge^2 df \\big) =0$ implies $f$ is critical.\n\n**Edit 1:**\n\nA natural way to produce critical points of a functional is to look at its symmetries. $\\tilde E$ is conformally-invariant exactly at dimension $4$. So, in dimension $4$ conformal maps are critical. However, I proved that they also satisfy the stronger equation $\\delta_{\\nabla^{\\Lambda_2(f^*{\\TN})}} \\big( \\bigwedge^2 df \\big) =0$.\n\n(In fact I proved that a conformal map $\\M^d \\to \\N^d$ satisfies $\\delta_{\\nabla^{\\Lambda_k(f^*{\\TN})}} \\big( \\bigwedge^k df \\big) =0$ if and only if $d=2k$ or it is a homothety).\n\n**Edit 2:**\n\nAs I shall prove below (see \"Edit 3\"), in dimension $d=2$ the answer is positive. (Note that in this case the \"number of constraints\" is the same). Focusing on the simplest next case, we consider $d=3,\\M=\\N=\\R^3$ with the flat metrics.\n\n**First, let's try to answer a more degenerate version of the question (forgetting $h_{f^*\\TN}$):**\n\n> Is there a smooth map $f:\\mathbb{R}^3 \\to \\mathbb{R}^3$, which satisfy\n> \n> $$\\delta \\big( df \\wedge df \\big) \\neq 0, \\, \\text{and } \\, \\tr \\big( df \\otimes \\delta(df \\wedge df) \\big)=0, $$\n\nIf there is such an $f$ then it cannot be a local diffeomorphism; its Jacobian must vanish somewhere.\n\nHere $ \\delta \\big( df \\wedge df \\big)\\in \\Omega^1\\big(\\R^3;\\Lambda_2(\\R^3)\\big)$ is a one-form on $\\R^3$ with values in $\\Lambda_2(\\R^3)$, and\n\n$$ \\delta \\big( df \\wedge df \\big)(X)=\\sum_i \\nabla_{e_i}(df \\wedge df)(e_i,X)=\\sum_i (\\nabla_{e_i} df)(e_i) \\wedge df(X) + df(e_i) \\wedge (\\nabla_{e_i} df)(X)$$ $$ =\\Delta f \\wedge df(X)+\\sum_i df(e_i) \\wedge (\\nabla_{e_i} df)(X).$$\n\nWe can try to look first for harmonic counter-examples, but so far I failed doing even that.\n\n**Edit 3:**\n\n> The answer is positive for $d=2$, so we need to restrict our attention to $d \\ge 3$.\n\nIndeed, in this case $E(f)=\\int_{\\M}| \\bigwedge^2 df|^2 \\text{Vol}_{\\M}=\\int_{\\M}(\\Det df)^2 \\text{Vol}_{\\M}$. We shall prove a map $f$ is $E$-critical if and only if its determinant is constant.\n\nSince $\\Det df$ is constant $\\iff$ $\\nabla (\\bigwedge^2 df)=0 \\iff \\delta_{\\nabla^{\\Lambda_2(f^*\\TN)}} \\big( \\bigwedge^2 df \\big)=0$, we are done.\n\nThe Euler-Lagrange equation of $E$ can be written as\n\n$$ \\delta (\\Det df \\cdot \\Cof df) =0, \\tag{1}$$ where $\\Cof df:\\TM \\to f^*(\\TN)$ is the _cofactor map_ of $df$ defined by $$ \\Cof df= \\star_{f^*TN}^{d-1} (\\wedge^{d-1} df) \\star_{TM}^1, $$\n\nand $\\delta$ is the adjoint of the pullback connection on $f^*\\TN$.\n\n(The Euler-Lagrange equation of the Jacobian functional $E(f)=\\int_{\\M} \\Det df \\text{Vol}_{\\M}=\\int_{\\M} f^* \\text{Vol}_{\\N}$ is $\\delta (\\Cof df)=0$. Details can be found in lemma 2.9 in my paper [here](https://arxiv.org/abs/1701.08892)).\n\nExpanding equation $(1)$ we get $$ 0=\\delta(\\Det df \\cdot \\Cof df )= \\Det df \\delta(\\Cof df ) - \\tr_{g}(d \\Det df \\otimes \\Cof df). $$\n\nWe now use the fact the Jacobian functional is a null-Lagrangian, i.e. every smooth map is a critical point, or equivalently $\\delta (\\Cof df)=0$. (This is essentially Stokes theorem, you can see lemma 2.5 [here](https://arxiv.org/abs/1701.08892)). \n\nSo, the E-L equation $(1)$ reduces to $$\\tr_{g}(d \\Det df \\otimes \\Cof df)=0. \\tag{2}$$\n\nLet $f$ be a map satisfying equation $(2)$. We shall prove $\\Det df$ is constant; suppose that $\\Det df_p \\neq 0$ for some $p \\in \\M$. This implies $ \\Cof (df_p)$ is invertible, so by equation $(2)$ $d\\Det df_p=0$. \n\nNow [we observe](https://math.stackexchange.com/a/2489068/104576) that any $C^1$ function $g : \\M \\to \\mathbb R$ on a connected manifold, satisfying $g(p) \\ne 0 \\implies dg_p = 0$ is constant. \n", "link": "https://mathoverflow.net/questions/272355/characterising-critical-points-of-ef-int-m-bigwedge2-df2-textvol", "tags": ["ap.analysis-of-pdes", "riemannian-geometry", "calculus-of-variations", "critical-point-theory"], "votes": 22, "creation_date": "2017-06-16T06:50:03", "comments": ["For the local problem, it may be useful to assume that $M,N$ are both flat in which case you can get rid of the geometry and concentrate on the multilinear algebra."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "131", "site": "mathoverflow", "title": "Smooth thickenings of non-smoothable manifolds", "body": "It is known that any closed topological manifold is homotopy equivalent to an open smooth manifold.\n\n**Question 1.**_What can be said about the**smallest** dimension of a smooth manifold that is homotopy equivalent to a given closed topological manifold?_\n\nThe following somewhat heavy-handed argument yields a smooth manifold of roughly twice the dimension, namely, use \n\n * [ West's solution ](http://www.maths.ed.ac.uk/~aar/papers/west.pdf) of Borsuk's conjecture that any compact ANR of dimension $n$, and in particular any closed $n$-manifold, is homotopy equivalent to a finite polyhedron of dimension $\\max\\\\{n, 3\\\\}$; \n * [ Stallings-Dranishnikov-Repovs's ](http://www.pef.uni-lj.si/repovs/clanki/1993/BullAustralMathSoc.pdf) embedding up to homotopy type theorem that any finite $n$-dimensional polyhedron is homotopy equivalent to a finite $n$-dimensional subpolyhedron of $\\mathbb R^{2n}$, so its regular neighborhood is the desired open smooth manifold. \n\n\nHere is a specific question that shows the state of my ignorance on this matter:\n\n**Question 2.** _Is there a closed $n$-manifold which is**not** homotopy equivalent to a smooth $(n+1)$-manifold? _\n\nThe naive idea to look at the product of a non-smoothable manifold of dimension $\\ge 5$ with $\\mathbb R$ fails, because such a product is also non-smoothable (by the topological product structure theorem of Kirby-Siebenmann).\n\n**Edit:** Misha kindly corrects me that a $5$-manifold is smoothable if and only if its Kirby-Siebenmann invariant vanishes; in particular, this apples to products of a $\\mathbb R$ and a closed $4$-manifold $M$. Thus $M\\times\\mathbb R$ is smoothable iff $M$ has zero KS invariant. Smooth $4$-manifolds have zero KS invariant, but amazingly so do some non-smoothable ones.\n", "link": "https://mathoverflow.net/questions/97598/smooth-thickenings-of-non-smoothable-manifolds", "tags": ["gt.geometric-topology"], "votes": 22, "creation_date": "2012-05-21T12:43:39", "comments": ["@Misha: certainly, not all $(n+1)$-dimensional thickenings of closed $n$-manifolds are proper homotopic to $I$-bundles. One can start with an $I$-bundle, and then attach a one-sided $h$-cobordism along the boundary; there are general methods to construct those in higher dimensions. One can also take boundary connected sums with (possibly noncompact) contractible manifolds with boundary, and this way one gets tons of examples when $n\\ge 3$ whose ends are quite complicated. ", "@Igor: Concerning KS invariant for 4-manifolds: It vanishes iff $M^4\\times {\\mathbb R}$ is smoothable. In dimension 4 there are non-smoothable manifolds with both zero and non-zero KS invariants since there are other, gauge-theoretic, obstructions to smoothability. ", "@Igor: Actually, I realized that even for n=3, a tame 4-dimensional thickening of a hyperbolic 3-manifold need not be properly homotopy-equivalent to an interval bundle. ", "@Misha: I am familiar with examples of closed non-smoothable aspherical manifolds by Davis-Januszkiewicz and Davis-Hausmann, in fact, my question was motivated by the desire to see how much the action dimension of arxiv.org/abs/math/0010141 would change if one defines it using smooth (!) properly discontinuous actions.", "@Mark: corrected; I hope you won't insists on accents in Repovs. @Misha: Do you know a $4$-dimensional example? In higher dimensions there is an old preprint of Galewski-Hollingworth that implies that there is no $(n+1)$-dimensional compact smooth thickening $V$ of an $n$-dimensional Poincare complex $X$, provided both of them are orientable; compactness of $V$ can be replaced by the assumption that an image of $X$ separates a neighborhood in $V$. Showing that $V$ is properly h.e. to an $\\mathbb R$-bundle seems tricky.", "@Igor: 1. Sometimes, the product with ${\\mathbb R}$ is smoothable. 2. You can avoid Stallings' theorem if you use immersion to ${\\mathbb R}^{2n}$ instead of an embedding. 3. Try Davis-Januszkiewicz argument, see the reference here and see if this gives a counter-example at least for a tame smoothing. (Try to argue that your open smooth (n+1)-manifold is properly homotopy-equivalent to an open interval bundle over a non-smoothable n-manifold.) ", "Should be Dranishnikov?"], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "132", "site": "mathoverflow", "title": "On certain representations of algebraic numbers in terms of trigonometric functions", "body": "Let's say that a real number has a _simple trigonometric representation_ , if it can be represented as a product of zero or more rational powers of positive integers and zero or more (positive or negative) integer powers of $\\sin(\\cdot)$ at rational multiples of $\\pi$ (the number of terms in the product is assumed to be finite, the empty product is taken to be $1$).\n\n_Examples:_\n\n * $\\sqrt[3]{\\sin\\\\!\\left(\\frac\\pi3\\right)}$ has a simple trigonometric representation, because $\\sqrt[3]{\\sin\\\\!\\left(\\frac\\pi3\\right)}=\\frac{3^{1/6}}{2^{1/3}}$.\n * $\\sqrt{\\sin\\\\!\\left(\\frac\\pi{10}\\right)}$ has a simple trigonometric representation, because $\\sqrt{\\sin\\\\!\\left(\\frac\\pi{10}\\right)}=\\frac{2^{1/2}}{5^{1/4}}\\,\\sin\\\\!\\left(\\frac\\pi5\\right)$.\n * $\\pi$ does not have a simple trigonometric representation, because it is not an algebraic number.\n\n\n\n_Questions:_\n\n * Do $\\sqrt{\\sin\\\\!\\left(\\frac\\pi5\\right)},$ $\\sqrt{\\sin\\\\!\\left(\\frac\\pi8\\right)},$ $\\sqrt{\\sin\\\\!\\left(\\frac\\pi{12}\\right)},$ $\\sqrt{\\sin\\\\!\\left(\\frac\\pi{15}\\right)},$ $\\sqrt{\\sin\\\\!\\left(\\frac\\pi{20}\\right)},$ $\\sqrt{\\sin\\\\!\\left(\\frac\\pi{24}\\right)}$ have simple trigonometric representations?\n * Is there an algorithm that, given a rational power of $\\sin(\\cdot)$ at a rational multiple of $\\pi$, would determine if it has a simple trigonometric representation? If so, could you give (or outline) a concrete example of such an algorithm (efficient, if possible)?\n * More generally, is there an algorithm that, given a real algebraic number (in some explicit form, e.g. as its minimal polynomial and a rational isolating interval), would determine if it has a simple trigonometric representation? If so, could you give (or outline) a concrete example of such an algorithm (efficient, if possible)?\n\n\n", "link": "https://mathoverflow.net/questions/194181/on-certain-representations-of-algebraic-numbers-in-terms-of-trigonometric-functi", "tags": ["nt.number-theory", "algorithms", "algebraic-number-theory", "computability-theory", "galois-theory"], "votes": 22, "creation_date": "2015-01-17T20:50:44", "comments": ["You mean solve all minimal polynomials with $\\sin(\\pi\\times rational)$?", "This is related I guess: mathworld.wolfram.com/TrigonometryAngles.html"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "133", "site": "mathoverflow", "title": "Infinitely many planets on a line, with Newtonian gravity", "body": "(I previously asked essentially this [on physics.stackexchange](https://physics.stackexchange.com/questions/56843/infinitely-many-planets-on-a-line-with-newtonian-gravity), but was actually \nhoping for answers with something closer to a proof than what I got there.)\n\nSuppose we have a unit mass planet at each integer point in 1-d space. $\\:$ As described in that answer, the sum of the forces acting on any particular planet is absolutely convergent. $\\;\\;$ Suppose we move planet_0 \nto point $\\epsilon$, where $\\: 0< \\epsilon< \\frac12 \\:$. $\\;\\;$ For similar reasons, those sums will still be absolutely convergent. \nNow we let Newtonian gravity apply. $\\:$ What will happen?\n\n \n \n\n\nIf it's unclear what an answer might look like, you could consider the following more specific questions: \n \n \nWill there be a positive amount of time before any collisions occur? \n(As opposed to, for example, a collision at time $\\frac1n$ for each positive integer $n$.)\n\n\"Obviously\" (at least, I hope I'm right), planet_0 will collide with planet_1. $\\:$ Will that be the first collision?\n\nplanet_0 will start out moving right, and all of the other planets will start out moving to the left. \nWill there be a positive amount of time before any of them turn around?\n\nHow long will it be before there are any collisions? $\\:\\:$ (perhaps just an approximation for small $\\:\\epsilon\\:$) \n \n\n", "link": "https://mathoverflow.net/questions/128796/infinitely-many-planets-on-a-line-with-newtonian-gravity", "tags": ["mp.mathematical-physics", "differential-equations"], "votes": 22, "creation_date": "2013-04-25T20:31:29", "comments": ["@Anthony Quas and Jon: I think this is not what he means. He does not \"live in 1D world\" but in 3-space. Just the planets happen to be on the line. The gravity law must be inverse squares. Otherwise the questions make no sense.", "@AnthonyQuas is correct. The Poisson equation in one dimension in this case is $\\phi''(x)=\\delta(x)$ that has as a solution $\\theta(x) x$. So, the potential between two bodies is not decaying in this case.", "Excellent question! Perhaps You should start as in thermodynamics with a large number $N$ of particles located at $-N,\\dotsc,-1, \\epsilon, 1,\\dotsc, N$ and see what happens then. At least in this case the total energy is finite, which is not the case when infinitely many particles are present.", "point masses $\\:$", "Are the planets point masses, or some radius $r \\lt \\frac{1}{2}$?", "Yes, particles clump together, typically forming smaller systems first. This is studied extensively in cosmology, both analytically and numerically. Gravitational effects are easy to model, and affect dark matter. youtube.com/watch?v=8C_dnP2fvxk However, dissipative effects such as the inelastic contraction of gas clouds are important, too. ", "I think that as long as each planet is at most $\\delta$ from its nearest integer, the total force on each planet is $O(\\delta)$. This can be used to prove rigorously that there's a positive $\\tau>0$ before any collision can occur. ", "Of course if you really live in a 1D world, gravitational force presumably doesn't decay at all?", "I read recently about a very similar problem that appeared in a 1949 letter from Ulam to von Neumann. (In that case the particles started at points of $\\mathbb Z$ with each node being occupied with probability 1/2). He showed(?) that something analogous to the universe happens: nearby groups of particles come together; and then those \"solar systems\" form galaxies etc."], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "134", "site": "mathoverflow", "title": "Are there "chain complexes" and "homology groups" taking values in pairs of topological spaces?", "body": "Throughout this question, notation of the form $(X,A)$ denotes a sufficiently nice pair of topological spaces. I think for most of what I'm saying here, it is enough to assume that the inclusion $A \\hookrightarrow X$ is a [cofibration](http://ncatlab.org/nlab/show/Hurewicz+cofibration), or that $(X,A)$ is a neighborhood deformation retract ([NDR](http://ncatlab.org/nlab/show/neighborhood+retract)) pair. All maps $f:(X,A) \\to (Y,B)$ are continuous and satisfy $f(A) \\subset B$.\n\nConsider a sequence $\\mathcal{S}$ given by: $$\\ldots \\to (X_{n-1},A_{n-1}) \\stackrel{d_{n-1}}{\\to} (X_n,A_n) \\stackrel{d_n}{\\to} (X_{n+1},A_{n+1}) \\to \\ldots$$ so that $d_n \\circ d_{n-1}(X_{n-1}) \\subset A_{n+1}$ for each $n$. \n\n> Has this object been defined and studied? If so, where?\n\nOne can associate to this sequence of space-pairs a \"homology\" which takes values in the category of space-pairs. That is, define $$\\mathcal{HT}_n(\\mathcal{S}) = \\left(d_n^{-1}(A_{n+1}),d_{n-1}(X_{n-1})\\right).$$ The construction is functorial if one considers the obvious analogue of \"chain maps\" in the category which contains objects like $\\mathcal{S}$. I'm wondering if basic definitions and properties, etc. of this or related constructions have been set down somewhere. \n", "link": "https://mathoverflow.net/questions/134443/are-there-chain-complexes-and-homology-groups-taking-values-in-pairs-of-topo", "tags": ["at.algebraic-topology", "chain-complexes", "reference-request", "gn.general-topology"], "votes": 22, "creation_date": "2013-06-21T17:21:30", "comments": ["@ViditNanda This is two years old by now. Did you make any progress?", "The pair should be $(X,A)$ of course. ", "@Vidit: I've sent an email on this. One point is that a pair $(X,S)$ defines filtrations $E_n(X,A)$ which are a base point in dim $0$, $A$ in levels $1$ to $n-1$ and $X$ thereafter. So if $X$ already has a filtration $X_*$ then we get two filtrations for a given $n$. ", "Ronnie, I'm not sure I see an \"obvious\" bifiltration anywhere in this question. Could you explain why that would be the natural thing to do? I am happy to discuss this via email if things get too intricate for comment boxes.", "@Vidit: Reading your question again, I would be inclined to go for defining invariants of bifiltered spaces, rather than invariants in spaces. Such bifiltered spaces in truncated form can be regarded as special cases of $n$-cubes of spaces, to which my work with Loday applies. I'm slowly writing up something on the bifiltered case, and my exposition [60] ``Triadic Van Kampen theorems and Hurewicz theorems'', Algebraic Topology, Proc. Int. Conf. March 1988, Edited M.Mahowald and S.Priddy, Cont. Math. 96 (1989) 39-57. (pdf on my web site) gives an idea. Also I missed the word \"discrete\"!", "Thank you, Ronnie. I already have your book(s), being a fan if the groupoid point of view, so I will take a look.", "@Vidit: You mention a \"filtered cell complex\". The book \"Nonabelian algebraic topology\" (2011) see my web page, pages.bangor.ac.uk/~mas010/nonab-a-t.html , builds algebraic topology directly from filtered spaces, via homotopically defined functors. Problem 16.1.17 is about relating these methods to Morse Theory. ", "Johannes: somewhere in between. I'd call it concretely-motivated curiosity. My thesis work involved simplifying a filtered cell complex via discrete Morse theory in a way that preserved algebraic topological invariants (persistent homology groups). As such, I find the process of constructing massive chain groups and then taking a huge quotient somewhat inefficient, and was wondering if there is a \"compact representation\" (both words used non-technically) of some object in the category of topological spaces itself from which one could derive the invariants whenever necessary.", "Where did you meet this structure? In a concrete problem or just out of curiosity?"], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "135", "site": "mathoverflow", "title": "Fake CM elliptic curves", "body": "Suppose one has an elliptic curve $E$ over $\\mathbb{Q}$ with conductor $N < k^3$ for some (large) positive $k$, with the property that its Fourier coefficients satisfy $$ a_p=0, \\; \\mbox{ for all } \\; p \\equiv -1 \\mod{4}, \\; \\; \\; k/3 < p < k. $$ Can one prove unconditionally that $E$ has CM? This follows from work of Serre (under GRH) or Elkies (under something like Szpiro's conjecture), since otherwise we'd have a surplus of supersingular primes. It does not appear to be a consequence of, say, Serre's argument without additional hypotheses (though I'd be happy to be wrong on this score).\n", "link": "https://mathoverflow.net/questions/49937/fake-cm-elliptic-curves", "tags": ["nt.number-theory", "elliptic-curves"], "votes": 22, "creation_date": "2010-12-20T00:09:05", "comments": ["Dear BCrd : I've pondered that, but I really don't know. My admittedly uneducated guess would be \"probably not\" with the level of precision I'm after....", "Dear Mike: Can one make \"effective\" (in the spirit of your question) any information coming from the knowledge that Sato-Tate holds in the non-CM case? (Sorry to answer a question with another question...)."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "136", "site": "mathoverflow", "title": "Is the equivariant cohomology an equivariant cohomology?", "body": "Suppose a finite group $G$ acts piecewise linearly on a polyhedron $X$. Then there are two kinds of equivariant cohomology (or homology). \n\n$\\bullet$ With coefficients in a $\\Bbb Z G$-module $M$. A reference is K. Brown's \"Cohomology of groups\". Namely, $H^\\ast_G(X;M)=H^\\ast(Hom_{\\Bbb Z G}(C_\\ast,M))$, where $C_\\ast$ is the simplicial chain complex of a suitable invariant triangulation of $X$. Note that $H^\\ast_G(X;M)\\simeq H^\\ast_G(X\\times EG;M)$. Generalized cohomology (including stable cohomotopy) of $X$ with coefficients in a module can be defined via generalized cohomology of $X\\times_G EG$ with local coefficients, which in turn is defined in the last chapter of the Buoncristiano-Rourke-Sanderson book.\n\n$\\bullet$ Representation-graded. A reference is \"Equivariant homotopy and cohomology theory\" by J. P. May et al. (but to simplify matters, I'm thinking of trivial coefficients, as in [Segal](http://www.mathunion.org/ICM/ICM1970.2/Main/icm1970.2.0059.0064.ocr.pdf) and [Kosniowski](http://resolver.sub.uni-goettingen.de/purl?GDZPPN002309874) rather than coefficients in a Mackey functor). Instead of defining $H^*_G$ let me only say that there is a Hurewicz homomorphism $\\omega^*_G(X)\\to H^*_G(X)$, and the equivariant stable cohomotopy $\\omega_G^{V-W}(X)$ can be defined as the equivariant homotopy set $[S^{W+U}\\wedge X,S^{V+U}]_G$, where $U$ is a sufficiently large (with respect to the partial order by inclusion) finite-dimensional $\\Bbb RG$-submodule of $\\Bbb R G\\oplus\\Bbb RG\\oplus\\dots$, and the \"representation sphere\" $S^U$ is the one-point compactification of $U$ (viewed as a Euclidean space with an action of $G$). \n\nWhat relations are there between these two kinds of (e.g. ordinary) equivariant cohomology? More specifically, \n\n> is cohomology with coefficients in a $\\Bbb ZG$-module a special case of representation-graded cohomology?\n\nA related (equivalent?) question is whether cohomology with coefficients in a (nontrivial!) module is representable in some reasonable sense.\n\nFor instance, by reducing both sides to non-equivariant cohomology, I know that $H^m_{\\Bbb Z/2}(X;I^{\\otimes m})\\simeq H_{\\Bbb Z/2}^{mT}(X_+)$, where $\\Bbb Z/2$ acts freely on $X$, $X_+=X\\sqcup$(basepoint), $I$ is the augmentation ideal of $\\Bbb Z[\\Bbb Z/2]$ (so $I^{\\otimes m}$ is $\\Bbb Z$ if $m$ is even and $I$ if $m$ is odd) and $T$ is the augmentation ideal of $\\Bbb R[\\Bbb Z/2]$ (so $mT$ is $\\Bbb R^m$ with the antipodal involution). Something of this kind can also be done for stable cohomotopy. However this all depends on a twisted Thom isomorphism which I don't know for more general modules.\n", "link": "https://mathoverflow.net/questions/68330/is-the-equivariant-cohomology-an-equivariant-cohomology", "tags": ["at.algebraic-topology", "equivariant-homotopy", "equivariant", "group-cohomology"], "votes": 22, "creation_date": "2011-06-20T17:42:25", "comments": ["When the action is free, $H^n_G(X;M)\\simeq [X,K(M,n)]_G$; the relative case looks a bit more complicated, $H^n_G(X,Y;M)\\simeq [(X,Y),(K(M,n)\\times BG,BG)]_G$ (see VI.3.11 in the Goerss-Jardine book and projecteuclid.org/euclid.ijm/1256052280). On the other hand, for every $G$-spectrum $E$, $h^{V-W}_G(X):=[S^W\\wedge X,S^V\\wedge E]_G$ is an equivariant cohomology theory. So we want to find for an $M$ e.g. a $V$ such that $H^{dim V+1}_G(S^V\\wedge B|M|)\\simeq M$, where $|M|$ is $M$ with the trivial action of $G$. This is indeed possible for $G=\\Bbb Z/2$ but not in general.", "Another reference for generalized cohomology with local coefficients is K. Brown's paper, ams.org/journals/tran/1973-186-00/S0002-9947-1973-0341469-9"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "137", "site": "mathoverflow", "title": "bound on the genus of a fiber of the Albanese map of a surface with $h^1({\\mathcal O})=1$?", "body": "This is maybe more an open problem than a question, since I have seriously thought about it and asked several people working on algebraic surfaces with no success. I hope somebody here can suggest an approach different from the standard arguments in surface theory.\n\nBACKGROUND: let $X$ be a smooth minimal complex projective surface of general type. An _irrational pencil_ is a morphism with connected fibers $f\\colon X\\to B$, with $B$ a smooth curve of genus $b>0$.\n\nFor $b>1$, $X$ has at most finitely many pencils of genus $b$, having such a pencil is a topological property and it is possible to bound explicitly the genus of a general fiber of $f$ in terms of $K^2_X$ (Arakelov' theorem).\n\nFor $b=1$, namely for _elliptic pencils_ , things are very different in general: a surface can have infinitely many such pencils, the genus of the general fibers of these pencils can be unbounded, and it is possible that a surface with an elliptic pencil deforms to a surface without elliptic pencils.\n\nHowever, if $h^1({\\mathcal O}_X)=1$, then the Albanese map $a\\colon X\\to Alb(X)$ is an elliptic pencil, and for fixed $K^2$ the genus of a general fiber of $a$ is bounded, since the moduli space of surfaces with fixed $K^2$ is quasiprojective.\n\nQUESTION: can one give a bound for the genus of the general fiber of the Albanese pencil of a minimal surface of general type $X$ with $h^1({\\mathcal O}_X)=1$ in terms of $K^2_X$? Such a bound would be very interesting in the fine classification of surfaces of general type. \n", "link": "https://mathoverflow.net/questions/49169/bound-on-the-genus-of-a-fiber-of-the-albanese-map-of-a-surface-with-h1-mathc", "tags": ["ag.algebraic-geometry", "algebraic-surfaces"], "votes": 22, "creation_date": "2010-12-12T12:12:39", "comments": ["If $K_S-F$ is effective, than of course I have a bound since $K_S(K_S-F)\\ge 0$ and $K_SF=2g(F)-2$. And by the same argument, I would have a bound if I could determine explicitly an $m$ such that $mK_S-F$ is effective, but that's precisely what I don't know how to do.", "I GUESS in some sense you can make it. For example, you can consider the canonical map of the surface, which will work if $K_S^2$ is not so small. Assume that the fiber of the Albanese map is $F$, then consider whether $\\mathscr{O}(K_S-F)$ has global section or not. If it has a global section, then after computing some intersection number, maybe you can have a bound of $g(F)$."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "138", "site": "mathoverflow", "title": "Row of the character table of symmetric group with most negative entries", "body": "The row of the character table of $S_n$ corresponding to the trivial representation has all entries positive, and by orthogonality clearly it is the only one like this.\n\nIs it true that for $n\\gg 0$, the row of the character table of $S_n$ corresponding to the sign representation has the most negative entries of any row (about half of them)?\n\nNote that for $S_4$, in addition to the sign representation row there are two other rows which also have 2 negative entries.\n\nPossibly the fact that row sums are positive is relevant here.\n\n**EDIT** : see also \n", "link": "https://mathoverflow.net/questions/420865/row-of-the-character-table-of-symmetric-group-with-most-negative-entries", "tags": ["co.combinatorics", "gr.group-theory", "rt.representation-theory", "symmetric-groups", "characters"], "votes": 21, "creation_date": "2022-04-22T09:54:56", "comments": ["@PerAlexandersson: I don't see any obvious connection to that conjecture.", "Is this related perhaps to the conjecture that about half of the partition numbers, p(n) are even, as n goes to infinity?", "@DenisSerre: the sum of the row corresponding to the sign representation is oeis.org/A000700. This quantity grows much slower than $p(n)$, hence why I said “about half.”", "Is it clear/known that \"about half of conjugacy classes have negative signature\" ?", "Some questions of a somewhat similar flavor that have been considered lately include whether most entries in the character table are highly divisible, or in fact are equal to $0$ (see e.g. arxiv.org/abs/2010.12410).", "@FedorPetrov: each conjugacy class. The character table is a $p(n)\\times p(n)$ table, where $p(n)$ is the number of partitions of $n$.", "Do we count each permutation once or each conjugacy class once?"], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "139", "site": "mathoverflow", "title": "Can a 4D spacecraft, with just a single rigid thruster, achieve any rotational velocity?", "body": "_(Copied[from MSE](https://math.stackexchange.com/questions/4472850/can-a-4d-spacecraft-with-just-a-single-rigid-thruster-achieve-any-rotational-v). Offering four bounties over time, I got no response, other than twenty-nine upvotes.)_\n\nIt seems preposterous at first glance. I just want to be sure. Even in 3D the behaviour of rotating objects can be surprising (see the Dzhanibekov effect); in 4D it could be more surprising.\n\nA 2D or 3D spacecraft (with no reaction wheels or gimbaling etc.) needs at least two thrusters to control its spin. See my [answer](https://space.stackexchange.com/a/59430) to [Could this three-thruster spherical spacecraft de-tumble itself with zero final velocity?](https://space.stackexchange.com/questions/53590) on space.SE.\n\n* * *\n\nFor any two $4\\times4$ matrices $X$ and $Y$, define the commutator $[X,Y]=XY-YX$, the anticommutator $\\\\{X,Y\\\\}=XY+YX$, and the Frobenius inner product $\\langle X,Y\\rangle=\\operatorname{tr}(X^\\top Y)$, where $\\operatorname{tr}$ is the trace and $\\top$ is the transpose.\n\nLet $M$ be a symmetric positive-definite $4\\times4$ matrix, and $T=\\mathbf f\\,\\mathbf r^\\top-\\mathbf r\\,\\mathbf f^\\top$ an antisymmetric $4\\times4$ matrix. ($M$ describes the distribution of mass in the spacecraft, $\\mathbf r$ is a vector locating the thruster relative to the centre of mass, $\\mathbf f$ is the force produced by the thruster, and $T$ is the torque produced by the thruster. More details at [Chiral Anomaly's answer](https://physics.stackexchange.com/a/471197/247726) to [Rigid body dynamics derivation from Newton's laws for higher dimensions](https://physics.stackexchange.com/questions/471106). Everything is described in the rotating reference frame.)\n\nThe angular velocity $\\Omega(t)$, an antisymmetric $4\\times4$ matrix, changes with time $t$ according to Euler's equation\n\n$$\\\\{M,\\Omega'(t)\\\\}+[\\Omega(t),\\\\{M,\\Omega(t)\\\\}]=f(t)T$$\n\nwhere $f(t)\\geq0$ is a function (continuous, piecewise-constant, or just integrable) describing when and how strongly the thruster is used.\n\n**Question:** Can $M$ and $T$ be chosen such that, for any two antisymmetric $4\\times4$ matrices $\\Omega_0$ and $\\Omega_1$, there exist $t_1>0$ and $f$ such that the solution $\\Omega$ to Euler's equation with initial value $\\Omega(0)=\\Omega_0$ has final value $\\Omega(t_1)=\\Omega_1$?\n\nTo simplify things, we may take $f$ to be a combination of Dirac deltas instead of an ordinary function. Then the angular velocity satisfies $\\\\{M,\\Omega'(t)\\\\}+[\\Omega(t),\\\\{M,\\Omega(t)\\\\}]=0$ except at a finite set of times when $\\\\{M,\\Omega(t)\\\\}$ changes by a positive multiple of $T$.\n\n* * *\n\nHere is Euler's equation in terms of the components $\\omega_{ij}$ of the angular velocity. Assume that $M$ is diagonal, with components $m_i>0$.\n\n$$(m_1+m_2)\\,\\omega_{12}'(t)+(m_2-m_1)\\big(\\omega_{13}(t)\\,\\omega_{32}(t)+\\omega_{14}(t)\\,\\omega_{42}(t)\\big)=f(t)\\,\\tau_{12}$$\n\n(and permute the indices to get 6 equations like this).\n\nIf $m_1=m_2$, then $\\omega_{12}'$ has constant sign; $\\omega_{12}$ is either always non-increasing, or always non-decreasing. So, if the difference between initial and final values of $\\omega_{12}$ has the wrong sign compared to $\\tau_{12}$, then there is no solution. Thus, we must take $m_1\\neq m_2$, and similarly $m_i\\neq m_j$ for $i\\neq j$.\n\n* * *\n\nThe Frobenius inner product of two antisymmetric matrices has every term duplicated: $\\sum_{i,j}x_{ij}y_{ij}=2\\sum_{im_2>m_3>m_4>0$. Taking the squared magnitude of $L$, and subtracting $(m_1+m_3)$ or $(m_2+m_4)$ times the doubled energy, we get\n\n$$ \\left(\\tfrac{m_2-m_3}{m_1+m_2}\\right)\\ell_{12}^2-\\left(\\tfrac{m_1-m_2}{m_2+m_3}\\right)\\ell_{23}^2-\\left(\\tfrac{m_3-m_4}{m_1+m_4}\\right)\\ell_{14}^2-\\left(\\tfrac{m_1-m_2+m_3-m_4}{m_2+m_4}\\right)\\ell_{24}^2-\\left(\\tfrac{m_1-m_4}{m_3+m_4}\\right)\\ell_{34}^2, $$$$ \\left(\\tfrac{m_1-m_4}{m_1+m_2}\\right)\\ell_{12}^2+\\left(\\tfrac{m_1-m_2+m_3-m_4}{m_1+m_3}\\right)\\ell_{13}^2+\\left(\\tfrac{m_3-m_4}{m_2+m_3}\\right)\\ell_{23}^2+\\left(\\tfrac{m_1-m_2}{m_1+m_4}\\right)\\ell_{14}^2-\\left(\\tfrac{m_2-m_3}{m_3+m_4}\\right)\\ell_{34}^2, $$\n\nboth being constant. Notice the signs of the coefficients; the first expression has only one positive term, so (if the expression is $0$) it defines a cone, or a pair of opposite cones, around the $\\ell_{12}$ axis in 6D. And the second expression defines a pair of opposite cones around the $\\ell_{34}$ axis. If the torque $T$ is inside one of these cones, and the initial angular momentum $L_0$ is in the same cone, then $L$ stays in that cone, and there's no solution with the final angular momentum $L_1$ outside of the cone. Thus, we must take $T$ to be outside of all four of these cones. That is, the first expression (with $\\tau_{ij}$ in place of $\\ell_{ij}$) must be negative, and the second expression must be positive.\n\n* * *\n\nNotice that Euler's equation has a time symmetry: Given a solution $\\Omega(t)$, we can construct another solution $\\tilde\\Omega(t)=-\\Omega(t_1-t)$, so that $\\\\{M,\\tilde\\Omega'(t)\\\\}+[\\tilde\\Omega(t),\\\\{M,\\tilde\\Omega(t)\\\\}]=f(t_1-t)\\,T$, and the initial and final values are swapped (and negated): $\\tilde\\Omega(0)=-\\Omega_1$ and $\\tilde\\Omega(t_1)=-\\Omega_0$. Therefore, any angular velocity can be reached from any other, if and only if any angular velocity can be reached from $0$. (We can reverse a path from $0$ to $-\\Omega_0$ to get a path from $\\Omega_0$ to $0$, and concatenate that with a path from $0$ to $\\Omega_1$, to get a path from $\\Omega_0$ to $\\Omega_1$.)\n\nAlso, Euler's equation has a scale symmetry: Given a solution $\\Omega(t)$, for any $k>0$ we can construct another solution $\\tilde\\Omega(t)=k\\,\\Omega(kt)$, so that $\\\\{M,\\tilde\\Omega'(t)\\\\}+[\\tilde\\Omega(t),\\\\{M,\\tilde\\Omega(t)\\\\}]=k^2f(kt)\\,T$, and the final value is $\\tilde\\Omega(t_1/k)=k\\,\\Omega(t_1)=k\\,\\Omega_1$. Therefore, any angular velocity can be reached from $0$, if and only if any angular velocity _in a small neighbourhood of $0$_ can be reached from $0$.\n\n(It looks like these two symmetries are related by $k=-1$, but we should keep a distinction between positive time and negative time.)\n\n* * *\n\nThis has the form of a quadratic differential equation, $\\mathbf x'(t)=\\mathbf x(t)\\odot\\mathbf x(t)$ where $\\odot$ is some bilinear function. (Specifically, for antisymmetric matrices $X$ and $Y$, define $X\\odot Y=-\\\\{M,\\\\}^{-1}([X,\\\\{M,Y\\\\}])$, so Euler's equation is $\\Omega'=\\Omega\\odot\\Omega$, as long as $f=0$. Alternatively, define $X\\odot Y=-[\\\\{M,\\\\}^{-1}(X),Y]$, so Euler's equation is $L'=L\\odot L$.)\n\nFix a norm on the space, and find some constant $\\lVert\\odot\\rVert>0$ such that $\\lVert\\mathbf x\\odot\\mathbf y\\rVert\\leq\\lVert\\odot\\rVert\\lVert\\mathbf x\\rVert\\lVert\\mathbf y\\rVert$ for all $\\mathbf x,\\mathbf y$.\n\nGiven initial value $\\mathbf x(0)=\\mathbf a$, the equation $\\mathbf x'=\\mathbf x\\odot\\mathbf x$ has the power series solution\n\n$$\\mathbf x(t)=\\sum_{n=0}^\\infty t^n\\frac{\\sum^{n!}\\mathbf a^{n+1}}{\\sum^{n!}1}$$\n\nwhere the coefficient of $t^n$ is the average of all possible ways of evaluating $\\mathbf a^{n+1}$ using the product $\\odot$. For example, the coefficient of $t^3$ is $\\tfrac16$ of\n\n$$\\sum^6\\mathbf a^4=((\\mathbf a\\mathbf a)\\mathbf a)\\mathbf a+(\\mathbf a(\\mathbf a\\mathbf a))\\mathbf a+2(\\mathbf a\\mathbf a)(\\mathbf a\\mathbf a)+\\mathbf a((\\mathbf a\\mathbf a)\\mathbf a)+\\mathbf a(\\mathbf a(\\mathbf a\\mathbf a))$$ $$\\newcommand{\\aaaa}[3]{\\mathbf a\\underset{#1}\\odot\\mathbf a\\underset{#2}\\odot\\mathbf a\\underset{#3}\\odot\\mathbf a} =\\aaaa{1}{2}{3}+\\aaaa{2}{1}{3}\\begin{matrix}{}+\\aaaa{1}{3}{2} \\\\\\ {}+\\aaaa{2}{3}{1}\\end{matrix}+\\aaaa{3}{1}{2}+\\aaaa{3}{2}{1}.$$\n\n(In the $1$-dimensional algebra, the product is associative, and the power series simplifies to $\\sum_nt^na^{n+1}=a/(1-at)$.)\n\nThis converges absolutely, for small $t$:\n\n$$\\sum_{n=0}^\\infty\\left\\lVert t^n\\frac{\\sum^{n!}\\mathbf a^{n+1}}{\\sum^{n!}1}\\right\\rVert\\leq\\sum_{n=0}^\\infty|t|^n\\lVert\\odot\\rVert^n\\lVert\\mathbf a\\rVert^{n+1}=\\frac{\\lVert\\mathbf a\\rVert}{1-|t|\\lVert\\odot\\rVert\\lVert\\mathbf a\\rVert},$$ $$|t|<\\frac{1}{\\lVert\\odot\\rVert\\lVert\\mathbf a\\rVert}.$$\n\nAnd at certain times (when the thruster is used) an impulse may be applied to $\\mathbf x$, with a fixed direction $\\mathbf b$ but an arbitrary magnitude $c>0$, thus: $\\mathbf x(t+)=\\mathbf x(t-)+c\\mathbf b$. These impulses, and the time intervals between them, are many variables that we can control. If the number of variables is at least $6$ (or the dimension of the space $\\mathbf x$ is in), then there is hope of surrounding $0$ in an open set, and thus reaching everywhere in the space (according to the previous section).\n\n$$ \\mathbf x(0+)=0+c_0\\mathbf b $$$$ \\mathbf x(t_1-)=\\sum_{n=0}^\\infty\\frac{t_1^n}{n!}\\sum^{n!}\\mathbf x(0+)^{n+1} $$$$ \\mathbf x(t_1+)=\\mathbf x(t_1-)+c_1\\mathbf b $$$$ \\mathbf x(t_1+t_2-)=\\sum_{n=0}^\\infty\\frac{t_2^n}{n!}\\sum^{n!}\\mathbf x(t_1+)^{n+1} $$$$ \\mathbf x(t_1+t_2+)=\\mathbf x(t_1+t_2-)+c_2\\mathbf b $$$$ \\mathbf x(t_1+t_2+t_3-)=\\sum_{n=0}^\\infty\\frac{t_3^n}{n!}\\sum^{n!}\\mathbf x(t_1+t_2+)^{n+1} $$$$ c_0,t_1,c_1,t_2,c_2,t_3\\geq0 $$$$ \\mathbf x(t_1+t_2+t_3-)\\approx0\\quad? $$\n", "link": "https://mathoverflow.net/questions/448773/can-a-4d-spacecraft-with-just-a-single-rigid-thruster-achieve-any-rotational-v", "tags": ["ca.classical-analysis-and-odes", "ds.dynamical-systems", "mp.mathematical-physics", "differential-equations"], "votes": 21, "creation_date": "2023-06-13T06:58:24", "comments": ["Re, I had not meant to make any adverse changes, and didn't even realise that it made a difference in the rendering. My apologies. I have changed back to back-to-back $$ $$ environments.", "@LSpice - Now the displayed equations are too closely spaced."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "140", "site": "mathoverflow", "title": "Reference request: deforming a G-local system to a variation of Hodge structure", "body": "Let $X$ be a smooth connected quasiprojective variety over $\\mathbb{C}$ and let $G$ be a complex reductive group. Let $$\\iota: G\\to GL_N$$ be a representation and let $$\\rho: \\pi_1(X(\\mathbb{C}))\\to G(\\mathbb{C})$$ be a homomorphism. I am looking for a reference for the following fact:\n\n> $\\rho$ is deformation-equivalent to a homomorphism $$\\rho': \\pi_1(X(\\mathbb{C}))\\to G(\\mathbb{C})$$ such that the local system on $X$ associated to $\\iota\\circ \\rho'$ underlies a polarizable complex variation of Hodge structure.\n\nIf $X$ is projective, this is due to Simpson [1, Theorem 3]; in the quasiprojective case some special cases have been written down by T. Mochizuki, e.g. if $G=GL_N$ [2, Theorem 10.5] or if $\\rho$ is rigid with Zariski-dense image [2, Lemma 10.13].\n\nI think this can probably be extracted with some difficulty from existing literature, but I'm hoping the statement has been written down explicitly somewhere.\n\nEDIT: If it helps, I'm primarily interested in the case $G=PGL_n$.\n\n[1] _Simpson, Carlos T._ , [**Higgs bundles and local systems**](http://dx.doi.org/10.1007/BF02699491), Publ. Math., Inst. Hautes Étud. Sci. 75, 5-95 (1992). [ZBL0814.32003](https://zbmath.org/?q=an:0814.32003).\n\n[2] _Mochizuki, Takuro_ , [**Kobayashi-Hitchin correspondence for tame harmonic bundles and an application**](http://www.arxiv.org/abs/math/0411300), Astérisque 309. Paris: Société Mathématique de France (ISBN 978-2-85629-226-6/pbk). viii, 117 p. (2006). [ZBL1119.14001](https://zbmath.org/?q=an:1119.14001).\n", "link": "https://mathoverflow.net/questions/413301/reference-request-deforming-a-g-local-system-to-a-variation-of-hodge-structure", "tags": ["ag.algebraic-geometry", "complex-geometry", "hodge-theory"], "votes": 21, "creation_date": "2022-01-06T11:59:59", "comments": ["@SashaP: actually I think this should probably be false over infinite finitely generated fields; I have an idea for a counterexample conditional on Fontaine-Mazur, but it’s maybe too involved for an MO comment.", "@SashaP: This is a nice question! I think I see an argument for the analogous statement over finite fields (i.e. working in a component of the rigid generic fiber of the deformation ring of a given residual rep of the geometric fundamental group). Plausibly I see an argument over p-adic fields as well, but I need to check details...send me an email and let's discuss further? I don't yet see how to do it over finitely generated infinite fields, which is I think the most interesting case...", "Apologies for an off-topic comment, but do you know if the analogous statement (say, at least for $GL_n$) is true for 'arithmetic' in place of 'underlies a polarizable VHS'?", "@MoisheKohan: No worries!", "I see, I just did not read the OP carefully.", "@MoisheKohan: As I remark in the question, Mochizuki has proven a quasiprojective version, just not for arbitrary $G$...", "I remember that Simpson proved this for projective varieties. I do not remember seeing a quasiprojective version."], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "141", "site": "mathoverflow", "title": "A multiple integral", "body": "Let us consider the multiple integral $$I_{n}=\\int_{-\\infty }^{\\infty }ds_{1}\\int_{-\\infty}^{s_{1}}ds_{2}\\cdots \\int_{-\\infty }^{s_{2n-1}}ds_{2n}\\;\\cos {(s_{1}^{2}-s_{2}^{2})}\\;\\cdots \\cos {(s_{2n-1}^{2}-s_{2n}^{2})}.$$\n\nUsing an indirect method (Landau-Zener formula from the theory of molecular scattering), we had shown in that $$I_n=\\frac{2}{n!}\\left(\\frac{\\pi}{4}\\right)^{\\,n}.$$ I'm very interested how to get this result by direct mathematical (preferably simple) methods, without using the Landau-Zener formula. Can anyone help? \n\nActually, I asked similar question an year ago [Multiple Integral (American Mathematical Monthly problem 11621 and its generalization)](https://mathoverflow.net/questions/125237/multiple-integral-american-mathematical-monthly-problem-11621-and-its-generali) but have not got an answer.\n", "link": "https://mathoverflow.net/questions/172278/a-multiple-integral", "tags": ["ca.classical-analysis-and-odes", "real-analysis", "integration"], "votes": 21, "creation_date": "2014-06-20T04:15:04", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "142", "site": "mathoverflow", "title": "Bounding failures of the integral Hodge and Tate conjectures", "body": "It is well know that the integral versions of the Hodge and Tate conjectures can fail. I once heard an off hand comment however that they should only fail by a \"bounded amount\". My question is what this means and whether this can be made more precise.\n\n> > Let $\\pi: X \\to B$ be a smooth projective morphism of smooth varieties over $\\mathbb{C}$. Assume that the (usual rational) Hodge conjecture holds for the fibre $B_x$ over every point $x \\in B(\\mathbb{C})$. Does there exist an integer $d$ such that for any integral Hodge class $h$ on $B_x(\\mathbb{C})$, the class $dh$ is represented by an algebraic cycle for all $x \\in B(\\mathbb{C})$?\n\nOf course the point of this question is whether $d$ can be chosen uniformly with respect to the family.\n\nAs a brief remark, it is quite easy to prove this when $B={\\mbox{Spec}}\\ \\mathbb{C}$. Indeed, by assumption the algebraic classes form a subgroup of finite index inside the Hodge classes!\n\nI am also similarly interested in the analogous question for the integral Tate conjecture for families of varieties over number fields and finite fields.\n", "link": "https://mathoverflow.net/questions/168997/bounding-failures-of-the-integral-hodge-and-tate-conjectures", "tags": ["ag.algebraic-geometry", "nt.number-theory", "arithmetic-geometry", "hodge-theory"], "votes": 21, "creation_date": "2014-06-04T01:36:23", "comments": ["It seems unlikely that any of the known counterexamples to the integral Hodge conjecture would give a negative answer to your question. On the other hand, there seems to be no reason to expect a positive answer since the dimension of the space of Hodge classes could jump over infinitely many subvarieties of $B$. Interesting!"], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "143", "site": "mathoverflow", "title": "Is the mapping class group of $\\Bbb{CP}^n$ known?", "body": "In his paper [\"Concordance spaces, higher simple homotopy theory, and applications\"](https://pi.math.cornell.edu/~hatcher/Papers/ConcordanceSpaces.pdf), Hatcher calcuates the smooth, PL, and topological mapping class groups of the $n$-torus $T^n$. This requires an understanding of the surgery structure set of the $n$-torus, as well as the space of self-equivalences (easy, because the torus is aspherical and a Lie group).\n\nAt least $\\pi_0$ and $\\pi_1$ of $\\text{hAut}(\\Bbb{CP}^n)$, the space of self-homotopy equivalences, are known: the former is $\\Bbb Z/2$ because $[\\Bbb{CP}^n, \\Bbb{CP}^n] = [\\Bbb{CP}^n, \\Bbb{CP}^\\infty] = \\Bbb Z$, and the only invertible elements are $\\pm 1$; further the homology of this space seems to have been calculated by Sasao, who gives that $\\pi_1 \\text{hAut} = \\Bbb Z/(n+1)$.\n\nThe best references I know for the surgery structure set of $\\Bbb{CP}^n$ are Wall's book on surgery and the Madsen-Milgram book on Top/PL cobordism groups \"The classifying spaces for surgery and cobordism of manifolds\", neither of which seem to give a completely explicit description. \n\nFrom this, it's not immediately clear to me whether enough is known to run Hatcher's argument for $T^n$ on complex projective space.\n\nHas anybody computed the various mapping class groups of $\\Bbb{CP}^n$ when $n \\geq 3$? Maybe in any specific examples, or when $n$ is very large? If not, is it out of reach with current technology?\n\n(I would also be interested in other infinite families of high-dimensional manifolds, like real projective spaces and lens spaces, but I imagine those require strictly more work to understand, given the presence of non-trivial fundamental group without being aspherical like the torus is.)\n", "link": "https://mathoverflow.net/questions/349319/is-the-mapping-class-group-of-bbbcpn-known", "tags": ["reference-request", "at.algebraic-topology", "gt.geometric-topology", "differential-topology", "mapping-class-groups"], "votes": 21, "creation_date": "2019-12-29T07:40:43", "comments": ["Regarding your last comment: Wall (MR0156354, MR0177421) and Kreck (MR0561244) have computed various mapping class groups of highly connected manifolds up to extension problems.", "Cool, I wasn't aware of that paper!", "@skupers Kreck points out to me that the oriented smooth mapping class group $\\pi_0 \\text{Diff}^+(\\Bbb{CP}^3)$ was calculated by Brumfiel to be $\\Bbb Z/4$, all of whose elements are represnted by diffeomorphisms supported in a ball. I did not check but presumably $\\pi_0 \\text{Diff}$ is the dihedral group on 8 elements.", "Kreck and Su have announced a paper containing the case n=3, see Remark 1.4 of arxiv.org/abs/1907.05693. You could try asking one of them.", "Only small homotopy groups seem relevant in Hatcher's computation, though, so I hope that doesn't cause too much trouble. Thanks for the reference!", "There are spectral sequences developed by Schultz, and Becker-Schultz that converge to homotopy groups of the identity component of the space of self-maps of $CP^n$. See section 6 in arxiv.org/abs/0912.4874 and references therein. The computational difficulties are substantial, e.g., in the linked paper we only manage to get partial information on $\\pi_7$."], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "144", "site": "mathoverflow", "title": "Cauchy matrices with elementary symmetric polynomials", "body": "$\\newcommand{\\vx}{\\mathbf{x}}$ Let $e_k(\\vx)$ denote the [elementary symmetric polynomial](http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial), defined for $k=0,1,\\ldots,n$ over a vector $\\vx=(x_1,\\ldots,x_n)$ by\n\n\\begin{equation*} e_k(\\vx) := \\sum_{1 \\le l_1 < l_2 < \\cdots < l_k \\le n} x_{l_1}x_{l_2}\\cdots x_{l_k}. \\end{equation*}\n\nTo avoid messy conjugates in what follows, let me limit $x$ to be a real vector.\n\n[**EDIT**]: Changed notation below to highlight that size of $M_k$ is independent of size of the vectors being used to define it.\n\n> **Definition** (GC-matrix) Let $(\\vx^1,\\ldots,\\vx^m)$ be vectors in $\\mathbb{R}^{n}$ such that each component $|x^p_j| < 1$ (for $1\\le p \\le m$, $1\\le j \\le n$). Denote by $\\vx^i \\circ \\vx^j$ the vector formed by taking elementwise products; also, let $\\mathbf{1}$ denote the length $n$ vector of all ones.\n> \n> Define now the $m\\times m$ _Generalized Cauchy (GC) matrix_ $M_k$, for $1\\le k\\le n$, as \\begin{equation*} M_k := \\left[\\frac{1}{e_k(\\mathbf{1}-\\vx^i\\circ \\vx^j)}\\right]_{i,j=1}^m. \\end{equation*}\n\n**Example** For $e_1(\\vx)=\\sum_lx_l$, the GC matrix above essentially (up to scaling) boils down to a standard [Cauchy matrix](http://en.wikipedia.org/wiki/Cauchy_matrix), of the form \\begin{equation*} \\left[\\frac{1}{1-\\lambda_i\\lambda_j}\\right], \\end{equation*} for suitable scalars $\\lambda_i$; while for $e_n(\\vx)=\\prod_l x_l$, the GC matrix reduces to a Hadamard product of ordinary Cauchy matrices.\n\n> **Question.** The above example shows that $M_1$ and $M_n$ are positive semidefinite (since the involved Cauchy matrices are). So my question is there a slick representation for $e_k$ that I could use to conclude positive semidefiniteness of $M_k$?\n\n* * *\n\n**PS:** I suspect the answer follows immediately from well-known (in the correct circles) results; but I need some MO help here to put me rapidly in the correct direction!\n", "link": "https://mathoverflow.net/questions/114101/cauchy-matrices-with-elementary-symmetric-polynomials", "tags": ["matrices", "rt.representation-theory", "linear-algebra", "co.combinatorics"], "votes": 21, "creation_date": "2012-11-21T13:01:30", "comments": ["Unfortunately, this question has seems to have a negative answer (as proved to me very recently by one of the mathematicians with whom I discussed it; will update this question once I get a chance.)"], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "145", "site": "mathoverflow", "title": "p-groups as rational points of unipotent groups", "body": "Is it true that every finite p-group can be realized as the group of rational points over $\\mathbb{F_p}$ of some connected unipotent algebraic group defined over $\\mathbb{F_p}$? For abelian p-groups, the answer is yes via Witt vectors, but is it true in general?\n", "link": "https://mathoverflow.net/questions/69397/p-groups-as-rational-points-of-unipotent-groups", "tags": ["p-groups", "algebraic-groups"], "votes": 21, "creation_date": "2011-07-03T06:46:19", "comments": ["Sorry, after a bit more thought, it occurs to me that, since $p$-groups have centres, it's OK to have the Abelian group as sub-object. Then Theorem 1.8(c) of the paper by Kumar and Neeb shows that $\\operatorname{Ext}^1_{\\text{alg}}(G, \\mathbb G_a) \\cong H^2(\\mathfrak g, \\mathfrak{gl}_1)^{\\mathfrak g}$; but I still can't see my way through to showing that the necessary map is surjective.", "L. Spice, do you want to replace $GL_1$ with $\\mathbb{G}_a$?", "By using a filtration of your $p$-group with all successive quotients $\\mathbb F_p$, you can reduce the problem to showing that $\\operatorname{Ext}^1_{\\text{alg}}(\\operatorname{GL}_1, G) \\to \\operatorname{Ext}^1_{\\mathbb Z}(\\mathbb F_p, G(\\mathbb F_p))$ is surjective for all connected, unipotent $\\mathbb F_p$-groups $G$. A quick Google search turned up “Extensions of algebraic groups” by Kumar and Neeb (#48 at math.unc.edu/Faculty/kumar), but the Abelian group by which you're extending there is the subobject, not the quotient."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "146", "site": "mathoverflow", "title": "Deciding whether a given power series is modular or not", "body": "The degree 3 modular equation for the Jacobi modular invariant $$ \\lambda(q)=\\biggl(\\frac{\\sum_{n\\in\\mathbb Z}q^{(n+1/2)^2}}{\\sum_{n\\in\\mathbb Z}q^{n^2}}\\biggr)^4 $$ is given by $$ (\\alpha^2+\\beta^2+6\\alpha\\beta)^2-16\\alpha\\beta\\bigl(4(1+\\alpha\\beta)-3(\\alpha+\\beta)\\bigr)^2=0, $$ where $\\alpha=\\lambda(q)$ and $\\beta=\\lambda(q^3)$. This has a very simple rational parametrization $$ \\alpha=\\frac{p(2+p)^3}{(1+2p)^3}, \\qquad \\beta=\\frac{p^3(2+p)}{1+2p} $$ (with $p$ ranging from 0 to 1 as as $q$ changes in the range).\n\nBy certain heuristical reasons (which are hidden behind analysis in my [recent joint work](http://arxiv.org/abs/1012.3036)), I expect that modularity can occur in some other similar parameterizations. In particular, the expansions $$ \\mu(q) = 4096q - 294912q^2 + 12238848q^3 - 379846656q^4 $$ $$ \\+ 9737920512q^5 - 217011585024q^6 + 4333573472256q^7 - 79091807551488q^8 $$ $$ \\+ 1337378422542336q^9 - 21157503871942656q^{10} + 315428695901356032q^{11} $$ $$ \\- 4455786006742302720q^{12} + 59885350975571779584q^{13} + O(q^{14}) $$ and $$ p = 4q + 12q^2 - 48q^3 + 156q^4 - 12q^5 - 6576q^6 + 78144q^7 - 607812q^8 $$ $$ \\+ 3017364q^9 + 156q^{10} - 208502832q^{11} + 2876189520q^{12} - 24837585384q^{13} + O(q^{14}) $$ (which, of course, can be further extended) satisfy $$ \\mu(q)=\\frac{p(4+p)^5}{(1+4p)^5} \\quad\\text{and}\\quad \\mu(q^5)=\\frac{p^5(4+p)}{1+4p}. $$ (These relations define $p$ and $\\mu$ in a unique way.)\n\nIs there any way to identify $\\mu(q)$ with a known modular function, or to show that $\\mu(q)$ is not modular at all?\n\nThanks! Best wishes already from 2011.\n", "link": "https://mathoverflow.net/questions/50804/deciding-whether-a-given-power-series-is-modular-or-not", "tags": ["nt.number-theory", "modular-forms"], "votes": 21, "creation_date": "2010-12-31T05:11:02", "comments": ["@Kevin, thanks for this hint. Because I have no guess about the index of the underlying group in $\\Gamma(1)$, I am not sure that $O(q^{1000})$ would be enough. I didn't try to expand so far (the coefficients grow extremely fast), but what I did (up to $O(q^{50})$) was verifying a possible algebraic relation between $\\mu(q)$ and the classical $j$-invariant. None was found...", "If $\\mu$ were modular then presumably there would be an algebraic relation between $\\mu(q)$ and $\\mu(q^n)$ for all positive integer values of $n$, the relation being of degree something like the index of $\\Gamma_0(n)$ in $SL(2,Z)$ (but perhaps this isn't precisely right---the exact degree would depend on the level of $\\mu$). So you could expand $\\mu$ out to $O(q^1000)$ and then it would be easy to search for these relations. If it doesn't work out, i.e. if $n=5$ is OK but the others don't seem to come out, then this is evidence to suggest that $\\mu$ isn't modular."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "147", "site": "mathoverflow", "title": "Schemification (schematization?) of locally ringed spaces", "body": "**Motivation:**\n\nSay $F: D \\to Sch$ is a diagram in the category of schemes, and we're interested in whether it has a [colimit](http://en.wikipedia.org/wiki/Limit_\\(category_theory\\)#Colimits_2) (gluings, pushouts, and \"categorical\" quotients are all examples of colimits). Its colimit $Q$ in the category of locally ringed spaces always exists, and a scheme $Y$ is the scheme-colimit of $F$ iff there is a map $Q\\to Y$ that is \"initial among maps to schemes\", i.e. any other map from $Q$ to a scheme factors through it.\n\nThus the problem of when colimits of schemes exists can be answered if we know when a locally ringed space admits a \"schemification\".\n\n**Examples:**\n\n1) The process of turning a classical variety $V$ into a \"variety with generic points\" $V^s$ is a schemification. Proving this boils down the classical affine case. If $R$ is a finite-type reduced $k=\\overline{k}$-algebra, and we map $V=mSpec(R)$ to a scheme $Y$, first cover $Y$ by affines $Spec(B_i)$, and then pull back this cover to $V$ and refine it by principal opens $mSpec(R_{f_{ij}})$, so we have maps $B_i \\to R_{f_{ij}}$ that determine the map $V\\to Y$ (by the adjunction of $Spec({\\cal O}(-))$ to the inclusion $AffSch\\hookrightarrow LRS)$. But these define the desired map $V^s\\to Y$ because $V^s$ can be obtained by gluing $Spec(R_{f_{ij}})$.\n\n2) For $k=\\overline{k}$, if we let $\\mathbb{G_m}=\\mathbb{A}^1_k\\setminus 0$ act on $\\mathbb{A}^1_k$, then the coequalizer in schemes of the action and the projection $G_m \\times_k \\mathbb{A}^1_k \\rightrightarrows \\mathbb{A}^1_k$ (i.e. the \"categorical quotient\") is $Spec(k)$, hence the locally ringed space coequalizer $Q$, which has two points and is not a scheme, has $Spec(k)$ as its schemification.\n\n3) Some locally ringed spaces have no schemification; for example, two affine lines glued along their generic points. This is precisely because the gluing diagram has no coequalizer in schemes, as per BCnrd and Anton Geraschenko's [answer here](https://mathoverflow.net/questions/9961/colimits-of-schemes/23966#23966).\n\n> When does a locally ringed space $X$ admit a \"schemification\", i.e. a map to a scheme $X^s$ through which any other map to a scheme must factor?\n\n**EDIT:** (Response to Martin's answer)\n\n4) Any locally ringed space $X$ with exactly one closed point $x$ has its \"affinization\", $X^a:=Spec({\\cal O}_X(X))$, as a schemification. This is because in a map $F:X\\to Y$ with $Y$ a scheme, every point maps to a generization of $f(x)$, so and since open sets are closed under generization, $f$ factors through an affine neighborhood of $f(x)$. But any map from $X$ to an affine factors uniquely through $X^a$, so we're done. \n\n5) Schemification commutes with disjoint unions, in that if $X= \\coprod X_i$ then $X^s$ exists iff all $X_i^s$ exist, and in that case $X^s = \\coprod X_i^s$. So _if $X$ is locally connected_ , since it then decomposes into a disjoint union of connected clopen components, we might as well assume it is connected.\n", "link": "https://mathoverflow.net/questions/60524/schemification-schematization-of-locally-ringed-spaces", "tags": ["ag.algebraic-geometry", "ct.category-theory"], "votes": 21, "creation_date": "2011-04-03T23:40:12", "comments": ["@Andrew: I'm thinking of something like a \"generalized GAGA\", but I'm not really sure what conditions you need on $(X,A)$. I think something like what I'm talking about is covered in chapter 8 of that book, but I really don't know the specifics.", "@Buschi, Chapter IV seems to be about forming locally ringed topoi (or spaces) from ringed topoi (or spaces). If I understand, this construction is right adjoint to the inclusion $LRT \\to RT$. @Harry, Could you say more specifically what result you're talking about? I'm dubious, because I'd expect any ringed topos to be a relative scheme over itself...", "@Andrew: There is some stuff in the next chapter (chapter 5) that might be useful to you (if you can show that your locally ringed topos is a relative scheme for some other ringed topos $(X,A)$, you have a universal scheme associated to it by way of a crazy adjunction induced by the universal map $(X,A)\\to (Set,Z)$), but I don't think it really addresses your question in any sort of concrete way.", "I haven't. Should I? I just skimmed through a copy and didn't see anything about turning espaces annelés en anneaux locaux into schemas.", "HAve you read CAP.IV of \"Topos Annelés et schemas relatifs\" by M. Hakim , Ergebn. Math. Grenzgeb. , 64 ? "], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "148", "site": "mathoverflow", "title": "Closed connected additive subgroups of the Hilbert space", "body": "It is a classical result that a closed and connected additive subgroup of $\\mathbb{R}^n$ is necessarily a linear subspace. However, this is no longer true in infinite dimension: a very easy example is the subgroup $L^2(I,\\mathbb{Z})$ of the real Hilbert space of all $L^2$ real valued functions on the unit interval $I:=[0,1].$ \n\nIndeed, any element $\\phi$ of $L^2(I,\\mathbb{Z})$ is connected to the origin by the path $\\gamma:I\\ni t\\mapsto \\phi\\chi_{[0,t]}\\in L^2(I,\\mathbb{Z}),$ where $\\chi_{[0,t]}$ is the characteristic function of the interval $[0,t].$ Actually, up to a reparametrization, this path is also $1/2$-Hölder continuous. (Indeed, if $\\sigma:[0,1]\\to\\big[0,\\ 1+ \\|\\phi\\| _2^2\\, \\big]$ is the strictly increasing, surjective continuous map $t\\mapsto t+\\int_0^t \\phi^2dx$, then $\\|\\gamma(t)-\\gamma(t')\\|_2\\le|\\sigma(t)-\\sigma(t')|^{1/2}$, meaning that $\\gamma\\circ\\sigma^{-1}$ is $1/2$-Hölder continuous). \nSo we may say that $L^2(I,\\mathbb{Z})$ is even $1/2$-Hölder-path-connected, though it is certainly not a linear subspace.\n\nIt is also not hard to see that the Hölder exponent $1/2$ is critic: any closed subgroup $G$ of a Hilbert space $H$, which is connected by $\\alpha$-Hölder paths, with $\\alpha > 1/2,$ is necessarily a linear space. (Reason: as a consequence of the generalized parallelogram identity, it turns out that the lattice generated by $n$ vectors $g_1,\\dots,g_n$ in $H$, with norms $\\|g_k\\|\\leq r,$ is a $rn^{1/2}$-net in their linear span. In particular, if $\\gamma:[0,1]\\to G$ is an $\\alpha$-Hölder path, for any $n\\in\\mathbb{N},$ the $n$ elements $g_{k,n}:=\\gamma(\\frac{k+1}{n})-\\gamma(\\frac{k}{n})\\in G,\\quad k=0,\\dots,n-1$ are a $Cn^{1/2 - \\alpha }$-net in their linear span. Since $G$ is closed this implies that it is a cone, hence a linear subspace).\n\nI find this quite nice, but at this point some questions arise quite naturally. Let $H$ be the infinite dimensional real separable Hilbert space.\n\n * Let $0 < \\alpha < 1/2.$ Are there closed additive subgroups of $H$ which are connected by $\\alpha$-Hölder paths, but not by $\\beta$-Hölder paths for any $\\beta >\\alpha \\, $? \n\n * More generally: connected / non-connected w.r.to paths with given modulus of continuity? Are there closed, connected, not path-connected additive subgroups? \n\n * Are these objects just pathologies/curiosities of the mathematical Zoo, or did anybody gave an application of them to functional analysis? \n\n\n\n", "link": "https://mathoverflow.net/questions/45322/closed-connected-additive-subgroups-of-the-hilbert-space", "tags": ["hilbert-spaces", "topological-groups", "fa.functional-analysis"], "votes": 21, "creation_date": "2010-11-08T08:48:41", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "149", "site": "mathoverflow", "title": "Almost everywhere differentiability for a class of functions on $\\mathbb{R}^2$", "body": "A while ago, I came across the following problem, which I was not able to resolve one way or the other.\n\n> Let $f,g\\colon\\mathbb{R}^2\\to\\mathbb{R}$ be continuous functions such that $f(t,x)$ and $g(t,x)$ are Lipschitz continuous with respect to $x$ and increasing in $t$. Is it necessarily true that the partial derivative $f_x=\\partial f/\\partial x$ exists almost everywhere with respect to the measure $\\iint\\cdot\\,d_tg(t,x)dx$?\n\nThe notation $\\int\\cdot\\,d_tg(t,x)$ refers to the Lebesgue-Stieltjes integral with respect to $g(t,x)$, for each fixed $x$. At first sight, this looks like a basic real-analysis problem for which it should be possible to either prove true or construct a simple counterexample. However, after playing around with this for a while, I was none the wiser.\n\nThe reason for the interest in this question comes from attempts to construct a particular decomposition of stochastic processes. Supposing that $X$ is a continuous martingale, it can be useful to decompose $f(t,X_t)$ into a martingale term and a 'drift' component, $$ \\begin{align} f(t,X_t)=\\int_0^tf_x(s,X_s)\\,dX_s+A_t.&&{\\rm(1)} \\end{align} $$ For this expression to be well-defined, it is necessary that $f_x$ exists outside of a null set. That is, $\\int1_{\\\\{f_x(s,X_s)\\rm\\ doesn't\\ exist\\\\}}dX_s$ should be zero. However, letting $g(t,x)=\\mathbb{E}[\\vert X_t-x\\vert]$, it can be shown that the expected value of the square of this integral is $\\iint1_{\\\\{f_x\\rm\\ doesn't\\ exist\\\\}}d_tg(t,x)dx$. So a positive answer to my question would mean that decomposition (1) is well-defined. Furthermore, it can then be shown that $A$ has zero [quadratic variation](http://en.wikipedia.org/wiki/Quadratic_variation), so that $f(t,X_t)$ is what is sometimes referred to as a _Dirichlet process_ (although this term does also have a completely different meaning). Decomposition (1) was used to prove some results about the distribution of martingales, but I also published it as a stand-alone paper ([here](http://dx.doi.org/10.1214/09-AOP476)). However, not knowing an answer to the question above, this did require imposing the annoying additional condition that the left and right hand derivatives of $f(t,x)$ with respect to $x$ exist everywhere.\n\nSome further points are in order.\n\n * Lebesgue's differentiation theorem tells us that the partial derivative $f_x$ exists almost everywhere with respect to the Lebesgue measure. That does not help here though, because $\\iint\\cdot d_tg(t,x)dx$ need not be absolutely continuous. In fact, even restricting to the case where $g(t,x)$ is convex in $x$, it is possible to construct nontrivial examples where $\\iint\\cdot d_tg(t,x)dx$ is supported on the graph $\\\\{(t,\\gamma(t))\\colon t\\in\\mathbb{R}\\\\}$ of a (discontinuous) function $\\gamma\\colon\\mathbb{R}\\to\\mathbb{R}$. One such example is given by $g(t,x)=\\mathbb{E}[(M_t-x)_+]$ where $M$ is the martingale constructed in [this blog post](https://almostsure.wordpress.com/2016/10/05/a-martingale-which-moves-along-a-deterministic-path/). Another (similar) example is the function $f(t,x)$ constructed [here](https://almostsure.wordpress.com/2016/10/12/do-convex-and-decreasing-functions-preserve-the-semimartingale-property-a-possible-counterexample/). \n * If the left and right hand partial derivatives of $f(t,x)$ with respect to $x$ exist everywhere, then the question has a positive answer.\n * If the left and right hand partial derivatives of $g(t,x)$ with respect to $x$ exist everywhere, then $f_x$ exists in a weak sense. Given any bounded measurable $\\theta\\colon\\mathbb{R}^2\\to\\mathbb{R}$ with compact support,$$\\iint h^{-1}(f(t,x+h)-f(t,x))\\theta d_tgdx\\to\\iint f_x\\theta d_tgdx$$as $h\\to0$.\n * Rather than monotonicity in $t$, I am really interested in the weaker condition that $\\iint\\cdot\\vert d_tf(t,x)\\vert dx$ is locally finite ($\\int\\cdot\\vert d_tf(t,x)\\vert$ is the variation of $\\int\\cdot d_tf(t,x)$), and similarly for $g$. If I find a positive answer to this, then I think I would update my papers on arXiv to reflect this.\n\n\n", "link": "https://mathoverflow.net/questions/77957/almost-everywhere-differentiability-for-a-class-of-functions-on-mathbbr2", "tags": ["ca.classical-analysis-and-odes", "real-analysis"], "votes": 21, "creation_date": "2011-10-12T14:15:06", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "150", "site": "mathoverflow", "title": "Is the Dieudonne module actually a cohomology group?", "body": "One often times thinks of the Dieudonne module $M(X)$ of a $p$-divisible group (say over $k$, a perfect characteristic $p$ field) as being some sort of cohomology theory\n\n$$M:\\left\\\\{p\\text{- divisible groups}/k\\right\\\\}\\to \\left\\\\{F\\text{-crystals }/k\\text{ with slopes in }[0,1]\\right\\\\}.$$\n\nThis is supported by the famous observation of Mazur-Messing-Oda that if $A/k$ is an abelian variety, then \n\n$$M(A[p^\\infty])=H^1_\\text{crys}(A/W(k))$$\n\nas $F$-crystals.\n\nThere are two natural questions that come from this, in my opinion. First, can one make sense of $M(X)$ as a crystalline cohomology group (in a literal sense) if $X$ 'comes from a scheme' (e.g. is the $p^\\infty$-torsion of some group scheme). I've asked this [here](https://mathoverflow.net/questions/219015/extension-of-messing-mazur-oda-to-general-groups) as well as asking when $M$ (an $F$-crystal) 'comes from a scheme'. \n\nBut, the other question, and the one I am asking here is the following. Is there some way to interpret the understanding of the Dieudonne module 'as cohomology' in a rigorous way? For example, is there a natural site over $X$ (perhaps with underlying category $p$-divisible groups over $X$) such that $M(X)$ is cohomology on this site? I would even be interested in knowing if there is a rigorous way of considering $M(X)/pM(X)$ as a cohomology theory (in the Mazur-Messing-Oda setup it's $H^1_\\text{dR}(A/k)$).\n\nThe obvious guess is to create a crystalline/infinitesimal site on $X$, but I don't really know how to make this precise.\n", "link": "https://mathoverflow.net/questions/235186/is-the-dieudonne-module-actually-a-cohomology-group", "tags": ["arithmetic-geometry", "crystalline-cohomology"], "votes": 21, "creation_date": "2016-04-03T05:24:29", "comments": ["Doesn't Mazur-Messing interpret the Dieudonne module of a p-divisible group G as rigidified extensions? This suggests that defining a category of G-invariant D-modules should be possible. I would naively guess that the resulting cohomology would be \"generated in degree 1\", just as for abelian varieties, though.", "One wants not a site on $X$ but a site whose \"fundamental group\" is in some way \"dual\" to $X$. The simplest picture of this seems like it should be the quotient of a point by the Cartier dual of $X$."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "151", "site": "mathoverflow", "title": "Density of first-order definable sets in a directed union of finite groups", "body": "This is a generalization of the following [question](https://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is) by John Wiltshire-Gordon.\n\nConsider an inductive family of finite groups: $$ G_0 \\hookrightarrow G_1 \\hookrightarrow \\ldots \\hookrightarrow G_i \\hookrightarrow \\ldots $$\n\nWe may view each group as a subgroup of the next. Let $G$ be the directed union of the $G_i$: $$ G := \\bigcup_{i=0}^\\infty \\ G_i $$\n\nNow, $G$ is a countable group. Suppose we're asked a question like, \"What proportion of the elements of $G$ are commutators?\", or \"What proportion of pairs $(g,h) \\in G^2$ satisfy $g^2h^2=1$ but $g^2 \\neq 1$?\" We try to make sense of such questions in terms of _density_ on $G^n$, which is not defined for all subsets. For $E \\subseteq G^n$, let $$ d(E) := \\lim_{i \\to \\infty} \\frac{|E \\cap G_i^n|}{|G_i^n|}, $$ if the limit exists.\n\nMy question is: If $E$ is a subset of $G^n$ that is first-order definable in the language of groups, does the density of $E$ necessarily exist? John's [question](https://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is) covers the (still unresolved) special case of when $E$ is defined by an atomic formula.\n\nA first-order definable subset $E \\subseteq G^n$ is the set of $n$-tuples of group elements where a particular first-order formula in $n$ free variables is true. Such a formula is a finite string of symbols and may involve multiplication, inversion, the identity element, equality, logical connectives ($\\wedge$, $\\vee$, $\\neg$, $\\implies$), quatifiers ($\\exists$ and $\\forall$), and parentheses. For example, the following first-order formula defines the set of commutators $g \\in G$:\n\n$$ (\\exists x)(\\exists y)(g = xyx^{-1}y^{-1}). $$\n\nOn the other hand, if you try to define the commutator subgroup by such a formula, you run into trouble. You might want to say, \"$g$ is a commutator or $g$ is a product of two commutators or $g$ is a product of three commutators or ...,\" but infinite disjunctions are not allowed.\n\nPlease note that the counterexample to a similar question in a comment by Vipul Naik [here](https://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is) is _not_ a counterexample to this question. In the inductive family $$ A_3 \\hookrightarrow S_3 \\hookrightarrow \\ldots \\hookrightarrow A_{2i-1} \\hookrightarrow S_{2i-1} \\hookrightarrow A_{2i+1} \\hookrightarrow S_{2i+1} \\hookrightarrow \\ldots, $$ the groups alternate between having all the elements as commutators and half the elements as commutators. However, in the directed union, which is $A_\\infty$, all the elements are commutators. The upshot is that quantifiers in a first-order formula may demand that you \"look ahead\" in your inductive family.\n", "link": "https://mathoverflow.net/questions/43900/density-of-first-order-definable-sets-in-a-directed-union-of-finite-groups", "tags": ["gr.group-theory", "lo.logic", "combinatorial-group-theory", "pr.probability", "finite-groups"], "votes": 21, "creation_date": "2010-10-27T16:10:17", "comments": ["I agree, cool question!", "This is a great question.", "It may also not be true that you will get a density, i.e. that there will be a limit law. I definitely don't know, but Vardi might. Gerhard \"Ask Me About System Design\" Paseman, 2010.11.02", "@Gerhard: It's not true that you'll always get density zero or one, if that's what you mean. For the union $S_\\infty$ of the symmetric groups $$ S_1 \\hookrightarrow S_2 \\hookrightarrow \\ldots \\hookrightarrow S_n \\hookrightarrow \\ldots, $$ the set of commutators is $A_\\infty$, and its density is $1/2$. An even more basic example would be the set of elements squaring to one in the union of the groups $\\mathbb{Z}/3\\mathbb{Z} \\times (\\mathbb{Z}/2\\mathbb{Z})^i$, which has density $1/3$.", "This shouts \"zero-one law\" at me. Perhaps looking at limit laws, and works of Vardi, Fagin, and others on finite structures may give you what you want. Gerhard \"Ask Me About System Design\" Paseman, 2010.10.27"], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "152", "site": "mathoverflow", "title": "If $X\\times Y$ is homotopy equivalent to a finite-dimensional CW Complex, are $X$ and $Y$ as well?", "body": "Is there a space $X$ that is not homotopy equivalent to a finite-dimensional CW complex for which there exists a space $Y$ such that the product space $X\\times Y$ is homotopy equivalent to a finite-dimensional CW complex? If so, how might we construct an example?\n\nA first consideration could be where $X$ has infinitely many nontrivial homology groups, in which case $X$ is not homotopy equivalent to a finite-dimensional CW complex. Yet, in this case it follows from the Künneth Formula that no suitable $Y$ exists. Of course, this only rules out spaces with infinitely many nontrivial homology groups.\n\nAnother consideration could be where $X$ is an Eilenberg–Maclane space for which $\\pi_1(X)$ has non-trivial torsion, in which case $X$ is, again, not homotopy equivalent to a finite-dimensional CW complex. In this case, if $\\pi_1(X)$ is abelian then $X$ is homotopy equivalent to $L\\times Z$ for some other space Z and an infinite-dimensional lens space $L$, and hence $X$ has infinitely many nontrivial homology groups, which implies, again, that no suitable $Y$ exists. This has led me to wonder if there exists a space $Y$ and a noncommutative group $G$ with non-trivial torsion such that $Y\\times K(G,1)$ is homotopy equivalent to a finite-dimensional CW complex, where $ K(G,1)$ denotes an Eilenberg–Maclane space.\n\nAny insight into approaching this problem is greatly appreciated.\n", "link": "https://mathoverflow.net/questions/267669/if-x-times-y-is-homotopy-equivalent-to-a-finite-dimensional-cw-complex-are-x", "tags": ["at.algebraic-topology", "homotopy-theory", "cohomology", "classifying-spaces"], "votes": 21, "creation_date": "2017-04-19T14:52:52", "comments": ["There are examples by R. Bing and many other mathematicians when a product of a non-manifold B and a real line gives a euclidean space, see e.g. the following papers and references there: ams.org/journals/bull/1958-64-03/S0002-9904-1958-10160-3/… ams.org/journals/proc/1961-012-01/S0002-9939-1961-0123303-2/‌​… However, I do not know whether such space B has the homotopy type of a CW-complex.", "A postscript to Igor Belegradek's comment: The argument in Proposition A.11 of my book shows that a space dominated by a CW complex of dimension $n$ is homotopy equivalent to a CW complex of dimension $n+1$, where the extra dimension arises from a mapping telescope construction. The argument, which is due to J.H.C.Whitehead if I remember correctly, uses only elementary homotopy theory. Using cohomology and Wall's work one obtains the stronger result that the increase in dimension is not really necessary, at least in dimensions 3 and greater.", "Just to add references to Tom's answer. The last sentence it is theorem E (p.63) of C.T.C. Wal, \"Finiteness conditions for CW complexes\", see math.uchicago.edu/~shmuel/tom-readings/…. The statement that a homotopy retract of a CW-complex is a CW complex is e.g. in Hatcher's \"Algebraic Topology\", proposition A.11.", "If $Y$ is nonempty then $X$ is a homotopy retract of a CW complex, therefore homotopy equivalent to a CW complex. If $X$ is a homotopy retract of a finite-dimensional CW complex, then $X$ has no cohomology above some dimension $d$, for any coefficient system. This in turn implies (if $d\\ge 3$) that $X$ is weakly, therefore strongly equivalent, to a $d$-dimensional complex.", "Not exactly what you wanted, but there are pointed CW complexes $X$ that are not finite, but such that $S^1\\wedge X$ is finite, for example $BG$ for $G$ an infinite acyclic group.", "A triviality: Y can be the empty space"], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "153", "site": "mathoverflow", "title": "Are the eigenvalues of the Laplacian of a generic Kähler metric simple?", "body": "It is a [theorem of Uhlenbeck](http://www.jstor.org/stable/2374041) that for a generic Riemannian metric, the Laplacian acting on functions has simple eigenvalues, i.e., all the eigenspaces are 1-dimensional. (Here \"generic\" means the set of such metrics is the complement of a [meagre set](http://en.wikipedia.org/wiki/Meagre_set) in the space of all metrics on a given manifold.)\n\nI would like to know whether this property is also true for a generic Kähler metric. More precisely, given a compact complex manifold with a fixed choice of Kähler class, is it true that for the generic representative of this class, the Laplacian acting on functions has simple eigenvalues?\n\n(To be honest, on a first reading it seems as if one might be able to apply similar arguments to those of Uhlenbeck to prove this result. I guess I would like to know if someone has already done this, or if there is a cunning counter-example that I'm missing, before I commit the time and energy to try!)\n", "link": "https://mathoverflow.net/questions/32810/are-the-eigenvalues-of-the-laplacian-of-a-generic-k%c3%a4hler-metric-simple", "tags": ["dg.differential-geometry", "complex-geometry", "riemannian-geometry", "fa.functional-analysis"], "votes": 21, "creation_date": "2010-07-21T08:52:36", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "154", "site": "mathoverflow", "title": "Homotopy flat DG-modules", "body": "A right DG-module $F$ over an associative DG-algebra (or DG-category) $A$ is said to be homotopy flat (h-flat for brevity) if for any acyclic left DG-module $M$ over $A$ the complex of abelian groups $F\\otimes_A M$ is acyclic. Homotopy projective and injective DG-modules are defined in the similar way, e.g., a left DG-module $P$ over $A$ is h-projective if for any acyclic left DG-module $M$ over $A$ the complex of abelian groups $Hom_A(P,M)$ is acyclic. Homotopy adjusted DG-modules are used to define derived functors on the derived categories of DG-modules.\n\nMy question is: can one describe the class of h-flat DG-modules in any way alternative to the definition? Here are examples of the kind of description I have in mind:\n\n 1. A DG-module is h-projective if and only if it is homotopy equivalent to a DG-module obtained from the free DG-module $A$ over $A$ using the operations of shift, cone, and infinite direct sum. (Similarly, a DG-module is h-injective if and only if it is homotopy equivalent to a DG-module obtained from the cofree DG-module $Hom_{\\mathbb Z}(A,\\mathbb Q/\\mathbb Z)$ using the operations of shift, cone, and infinite product.)\n\n 2. A module over a noncommutative ring is flat if and only if it is a filtered inductive limit of projective (or even projective and finitely generated, or free and finitely generated) modules (the Govorov-Lazard theorem).\n\n\n\n\nThe class of h-flat DG-modules is closed with respect to the operations of shift, cone, filtered inductive limit, and passage to a homotopy equivalent DG-module. It also contains the free DG-module $A$ over $A$. Can all h-flat DG-modules be obtained from that one DG-module using these four operations?\n\nReferences: 1. Keller's paper \"Deriving DG-categories\"; 2. my preprint arXiv:0905.2621, section 1.\n", "link": "https://mathoverflow.net/questions/40036/homotopy-flat-dg-modules", "tags": ["ct.category-theory", "homological-algebra", "differential-graded-algebras", "flatness"], "votes": 21, "creation_date": "2010-09-26T09:53:57", "comments": ["I think the paper arxiv.org/abs/1607.02609 answers your question.", "The following paper solves a closely related problem: L.W. Christensen and H. Holm, \"The direct limit closure of perfect complexes\", Journ. of Pure and Appl. Algebra 219, #3, p.449-463, 2015. arxiv.org/abs/1301.0731"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "155", "site": "mathoverflow", "title": "Is there a "direct" proof of the Galois symmetry on centre of group algebra?", "body": "Let $G$ be a finite group, and $n$ an integer coprime to $|G|$. Then we have the following map, which is clearly not a morphism of groups in general: $$g\\mapsto g^n.$$\n\nThis induces a linear automorphism of $\\mathbb{Z}[G]^G$, the algebra of $G$-invariant functions on $G$ under convolution, and surprisingly, this induced map is also an algebra automorphism, as can be seen by passing to $\\mathbb{C}$ and noting that this is the Galois action on characters, which permutes the set of primitive idempotents of this algebra.\n\nMy question is whether this surprising fact can be explained directly, without using character theory? I would be interested in both a high-concept explanation of this symmetry, or a generators and relations argument for why there exists, for any $g,h$ in $G$, an $a,b\\in G$ with: $$(gh)^n=ag^n a^{-1}bh^n b^{-1}.$$\n", "link": "https://mathoverflow.net/questions/415174/is-there-a-direct-proof-of-the-galois-symmetry-on-centre-of-group-algebra", "tags": ["gr.group-theory", "rt.representation-theory", "finite-groups", "characters"], "votes": 21, "creation_date": "2022-02-01T13:38:45", "comments": ["By a generators and relations argument I would think some symbolic argument that holds in any torsion group of exponent coprime to $n$, building $a,b$ from $g,h$ using these assumptions and the assumed $n$th root function. If one could give a group of this type where the result fails, it would rule out such an argument, which would also be very interesting.", "What do you mean by \"a generators-and-relations argument for... $(gh)^n=ag^na^{-1}bh^nb^{-1}$\"? I doubt there exist such $a,b$ in the free group over $g,h$. For $n=2$ there are none: any $x,y,z$ in a free group satisfying $x^2y^2=z^2$ must commute (it follows there are none whenever $n$ is even).", "A slightly different (but definitely character theoretic) proof of this interesting fact uses the structure constants $a_{KLM}=\\frac{|K||L|}{|G|}\\sum_{\\chi\\in\\mathrm{Irr}(G)}\\frac{\\chi(x_K)\\chi(x_L)\\overline{\\chi(x_M)}}{\\chi(1)}.$ If $K^\\prime$ etc. are the classes containing the respective $n^{th}$ powers and if $\\sigma$ is Chris' automorphism then $a_{K^\\prime L^\\prime M^\\prime}=\\frac{|K||L|}{|G|}\\sum_{\\chi\\in\\mathrm{Irr}(G)}\\frac{\\chi^\\sigma(x_K)\\chi^\\sigma(x_L)\\overline{\\chi^\\sigma(x_M)}}{\\chi(1)}=a_{KLM}^\\sigma,$ so $a_{K^\\prime L^\\prime M^\\prime}=a_{KLM}^\\sigma=a_{KLM}.$", "There are a number of similar facts which seem non-obvious from a group-theoretic point of view, but are easy to prove with characters: for example, the number of times $ x \\in G$ is expressible as a commutator is unchanged if we replace $x$ by a different generator of $\\langle x \\rangle.$", "@spin ah, thanks", "@FedorPetrov: Note the assumption $\\gcd(n, |G|) = 1$. So if $g^n = 1$, then $g = 1$.", "The Frobenius group of order $20$ is an example, trivial outer automorphism group, with nonrational characters, take $n=3$.", "Basic question: What is an example of a finite group $G$ and an integer $n$ prime to $|G|$ such that there is no automorphism of $G$ which induces the permutation $\\text{Conj}(g) \\mapsto \\text{Conj}(g^n)$ on conjugacy classes?", "Wow, that's a beautiful find. I am getting a slightly cohomological whiff from it (as, e.g., in the proof of the transfer's homomorphism property).", "There is a theory of products and powers of conjugacy classes; see Beltrán, Felipe, and Melchor - Some problems about products of conjugacy classes in finite groups for what appears to be a recent survey. It seems to be in a different direction, but I wonder if that might be the place to look.", "Its true, and follows from the fact that $g\\mapsto g^n$ is an algebra automorphism of $\\mathbb{Z}[G]^G$, by expanding the product of the classes of $g$ and $h$. I don't know of a non character theoretic proof of why this map is actually an algebra automorphism however.", "Is the quoted equality about $(g h)^n$ true, but you don't know how to prove it, or you're not sure if it's true?"], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "156", "site": "mathoverflow", "title": "monoidal (∞,1)-categories from weakly monoidal model categories", "body": "In _Higher Algebra_ section 4.1.7, Jacob Lurie proves that the underlying $(\\infty,1)$-category of a monoidal model category is a monoidal $(\\infty,1)$-category.\n\nDominic Verity and Yuki Maehara have (independently) defined the lax Gray tensor product for model categories $S$ and $T$ for two different models of $(\\infty,2)$-categories and proved that their tensor product is \"weakly monoidal\" in various senses:\n\n * Maehara provides functors $\\otimes_n \\colon T^{\\times n} \\to T$ for each $n \\geq 2$ that define the left adjoint parts of $n$-variable Quillen adjunctions. But these functors are associative in a weak sense, eg with a zig zag of natural weak equivalences\n\n\n\n$$ X \\otimes_2 (Y \\otimes_2 Z) \\to \\otimes_3(X,Y,Z) \\leftarrow (X \\otimes_2 Y) \\otimes_2 Z.$$\n\n * Verity constructs two binary tensor products $\\otimes, \\otimes' \\colon S \\times S \\to S$ that are naturally weakly equivalent, one of which is a closed left Quillen bifunctor and the other of which is associative up to isomorphism.\n\n\n\nMorally, this weakness should not matter for the underlying $(\\infty,1)$-category, but has anyone verified that a monoidal structure on the underlying $(\\infty,1)$-category can be produced under these weaker hypotheses?\n\nAlternatively, would someone like to take this on? I'd be happy to help publicize a result along these lines whenever it appears.\n\n(FYI: Andrea Gagna, Yonatan Harpaz, Edoardo Lanari have a third version of the lax Gray tensor product that defines a monoidal model category in the usual sense. But I'm still interested in the question of about weak monoidal model categories, which is why I've set this aside.)\n", "link": "https://mathoverflow.net/questions/370172/monoidal-%e2%88%9e-1-categories-from-weakly-monoidal-model-categories", "tags": ["model-categories", "monoidal-categories", "infinity-categories"], "votes": 20, "creation_date": "2020-08-26T09:02:52", "comments": ["There was further discussion in chat. To sum up: the suggestion from my second comment was misguided because Lurie really only deals in reflective localizations, and so can't say much about the localization from a model category to its associated $\\infty$-category. But Rune points out that Hinich has results on how DK localization interacts with cocartesian fibrations which could do the trick if generalized to the locally cocartesian case.", "@TimCampion Excellent!", "@AlexanderCampbell Great, that's good to know! Regarding pre-complicial sets, Verity indeed showed that the Gray tensor product is monoidal biclosed on them, and Yuki, Chris Kaupulkin and I observed in Theorem 1.33 here that the usual model structures are monoidal with respect to it. Relatedly, in that same preprint, we construct a monoidal biclosed lax Gray tensor product on a category of marked cubical sets (without having to futz around with the category a la precomplicial sets) and show it's a monoidal model category.", "@TimCampion Yuki's structure is indeed an example of a lax monoidal category (which is the standard name in the literature). Lurie's \"corepresentable operads\" correspond to (symmetric) colax monoidal categories. Of course, the opposite of a lax monoidal category is a colax monoidal category, so I agree that this a promising approach.", "In the second situation, isn't the full subcategory of pre-complicial sets a genuine monoidal model category?", "Hiya Emily, nice to see you here. I think that this question was explored in the paper by Heuts, Hinich, and Moerdijk that fixes all of the mistakes from Moerdijk-Weiss. I'm not sure if they verified transport of structure to the associated ∞-category, but it's another data point: arxiv.org/abs/1305.3658", "Localization is a product-preserving functor from relative categories to $\\infty$-categories. That means it's compatible with the enrichment of each category in itself, which should give you enough structure to say a functor of 2-categories to the 2-category of relative categories gives after localization one to the $(\\infty,2)$-category of $\\infty$-categories, which would give another approach to the first part. Unfortunately this kind of naturality for enrichments from module structures hasn't been worked out yet, however, though I believe Hadrian Heine is working on this.", "It's conceivable that Lurie's framework might even be helpful here -- if you're interested mostly in the $\\infty$-category presented by your model category, then maybe some of Lurie's general results on localization for $\\infty$-operads could be used to say things like \"there is a monoidal structure on the presented $\\infty$-category\".", "Ignoring the model structure, I would like to call the situation of the first bullet a \"lax monoidal category\" (or maybe \"oplax\"?). I've only seen lax monoidal categories (or rather their dual) in one source, namely Higher Algebra Definition 6.2.4.3 where they're called \"corepresentable ($\\infty$-)operads\" (of course, it's jazzed up to be $\\infty$ everything). I think another example of a (op)lax monoidal category (interacting nicely with a model structure) would be Boardman's handicrafted smash products of spectra."], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "157", "site": "mathoverflow", "title": "Is every positive integer the rank of an elliptic curve over some number field?", "body": "For every positive integer $n$, is there some number field $K$ and elliptic curve $E/K$ such that $E(K)$ has rank $n$?\n\nIt's easy to show that the set of such $n$ is unbounded. But can one show that _every_ positive integer is the rank of some elliptic curve over a number field?\n\nThe analogous question for a fixed number field is expected to have a negative answer (c.f. e.g., [this question](https://mathoverflow.net/questions/137970/is-it-expected-that-every-natural-number-is-the-rank-of-some-elliptic-curve-over)) but is still conjectural. But I wonder if one might be able to prove a positive answer to the question I asked above.\n", "link": "https://mathoverflow.net/questions/334853/is-every-positive-integer-the-rank-of-an-elliptic-curve-over-some-number-field", "tags": ["nt.number-theory", "arithmetic-geometry", "elliptic-curves"], "votes": 20, "creation_date": "2019-06-26T07:04:40", "comments": ["The \"trick\" here is we can use $q \\to\\infty$ results because the question lets us pass to any finite extension. A similar result would follow over number fields from heuristics for the distribution of $L$-functions. In particular, it requires our heuristics to be true only in the 100% setting, and not in the power savings setting rqeuired by the boundedness of rank arguments over the integers.", "So for $q$ sufficiently large we can show the existence of an elliptic curve $E$ and discriminants $D_1,\\dots, D_k$ such that of the $2^k$ twists generated by $D_1,\\dots, D_k$, $n$ have analytic rank $1$and the rest have rank $0$. Then one just has to use the analytic rank $1$ part of the BSD conjecture (which I think is now known in full generality over function fields?) and deduce that $n$ of these twists have algebraic rank $1$, hence the pullback to the composite of quadratic extensions has rank $n$.", "It should be possible to prove the analogous statement over function fields. For a quadratic twist of a fixed elliptic curve $E$, Katz proved that $1/2$ the twists have analytic rank $0$ and half the twists have analytic rank $1$, in the large $q$ limit. It should be possible to check by explicitly examining monodromy that the root numbers of different quadratic twists are suitably independent.", "@VesselinDimitrov Yes it's also natural to consider the Tate-Shafaverich group, from the point of view of the BSD conjecture and analytic class number formula. Both are special cases of the Tamagawa number conjecture, which uses Galois cohomology. The class number arises as the boundary in the Poitou-Tate localization sequence. The ideal class group and the Mordell-Weil group are both Picard groups but for Dedekind domains which are of somewhat different nature.", "@FrançoisBrunault: That rel Picard over $\\mathbb{Q}$, though, kills the much more mysterious Shafarevich-Tate part; the Selmer group I think should be the true analogy. It is very easy to construct many $E/K$ of rank zero, much harder to construct number fields of class number one (apart from finitely many known cases).", "@FrançoisBrunault: I didn't think of this, thanks for noting it! It makes sense, then. I assumed the class group was similar to a Selmer group, which includes only an upper bound on the rank. Still it looked to me that Stanley's question should be harder than the current one, but I could be off.", "@VesselinDimitrov The Mordell-Weil group of $E$ is isomorphic to the relative Picard group of $E/\\mathbb{Q}$, so this is a natural analogue. By the way it is known that every abelian group is the Picard group of some Dedekind domain (Claborn 1966). One may ask the same thing while restricting the class of abelian groups and/or Dedekind domains considered.", "@StanleyYaoXiao: Your question should be infinitely harder than this one. I don't think there is much of a similarity between a class number and a Mordell-Weil rank.", "This other question I asked may be of relevance: mathoverflow.net/questions/88175/…", "Sorry, I realized that my comment sounds less than clear, so let me add that apart from the remark in the first sentence, I do regard there both $K$ and $E$ as varying. Basically I am saying that a version of Silverman's specialization theorem can be used $r$ times over an $r$-dimensional char. $0$ base $T$ assuming that there exists an elliptic scheme $A/T$ such that $A(T) / \\mathrm{tors} \\cong \\mathbb{Z}^n$. From this point of view the problem is geometric, and I am guessing that an explicit such $A/T$ is given by $A_{1,n+1} / A_{1,n}$ (but any other construction would suffice).", "@Wojowu: I am merely pointing out the existence of the phenomenon 'idle curiosity'. If you would label Goldbach's conjecture as such, that is up to you. Anyway, I wanted to give my reasons for my downvote, not enter into a discussion about the value of mathematics, so I will leave it here.", "One may ask more generally: given a representation of a finite group $\\rho : G \\to \\mathrm{GL}_n(\\mathbb{Q})$, does $\\rho$ arise as the Galois module $E(L) \\otimes \\mathbb{Q}$ for some elliptic curve $E/K$ and some extension $L/K$ with Galois group $G$. We can also fix the number field $K$. If I recall correctly, if one replaces \"arises\" by \"appears inside\" then for function fields $K=\\mathbb{F}_p(t)$ these kinds of things are known.", "@RP: What would be the use of knowing the answer of any question on this website? Unless you're only in it for cryptographic applications, this question is useful because of human curiosity.", "@RP_ I would -1 your comment if I could. Since when does usefulness of the result dictate whether the question is good for this site? Which ranks are possible over a fixed number field is similarly useless, and yet it's a very interesting open problem.", "-1. I think this question is a little silly. What could possibly be the use of knowing the answer? I mean, if you could reduce the Goldbach conjecture for $2n$ to the existence of an elliptic curve of rank $n$ over some number field, then I would see the point. As the matter stands I'm afraid I don't.", "I am not sure whether it is fair to say that the fixed number field case should admit a negative answer; of course that comes down to a famous unsolved problem. For the existence of $E/K$, an iterated application of Silverman's specialization theorem reduces us to proving that the group of sections of the elliptic scheme $A_{1,n+1} \\to A_{1,n}$ has rank $n$, generated by the $n$ 'point doubling' sections $(x_1, x_2,\\ldots,x_n) \\mapsto (x_1,x_1;x_2,\\ldots,x_n)$, etc. (Here $A_{g,n}$ is the space of $n$-pointed abelian varieties of dimension $g$). This should be true."], "comment_count": 16, "category": "Science", "diamond": 0} {"question_id": "158", "site": "mathoverflow", "title": "Non-rigid ultrapowers in $\\mathsf{ZFC}$?", "body": "_Originally[asked and bountied at MSE](https://math.stackexchange.com/questions/4293034/non-rigid-ultrapowers-in-mathsfzfc):_\n\n> **Question** : Can $\\mathsf{ZFC}$ prove that for every countably infinite structure $\\mathcal{A}$ in a countable language there is an ultrafilter $\\mathcal{U}$ on $\\omega$ such that the ultrapower $\\mathcal{A}^\\mathcal{U}$ has a nontrivial automorphism?\n\n\"Obviously\" the answer is yes! _(... right?)_\n\nNote that this is trivial in $\\mathsf{ZFC+CH}$: $\\mathsf{ZFC}$ alone proves that any nontrivial ultrapower is countably saturated, which in the presence of $\\mathsf{CH}$ lets us run a back-and-forth argument to show that $\\mathcal{A}^\\mathcal{U}$ is non-rigid whenever $\\mathcal{A}$ is countable and $\\mathcal{U}$ is a nonprincipal ultrafilter on $\\omega$. Separately, note that if we drop the requirement that the ultrafilter be on $\\omega$ a positive answer follows from the Keisler-Shelah theorem, but [$\\mathsf{ZFC}$ can't bring down KS to $\\omega$ as one might expect](https://shelah.logic.at/files/130108/vive.pdf), so that ultimately doesn't seem like a promising direction.\n\nI strongly suspect that I'm missing a very simple argument, but at present I'm not seeing it.\n\n_As a minor aside, I'm especially interested in a proof which avoids using results about first-order logic (techniques from the model-theory of first-order logic are fine). This comes from the original motivation for this question, which was a weakness in[an MSE answer of mine](https://math.stackexchange.com/questions/41469/non-axiomatisability-and-ultraproducts/3569736#3569736) where the whole point was to avoid some standard model theory. However, in retrospect this seems to be jumping the gun here, so I'll relegate this aspect of the question from the question itself to this mere side comment._\n", "link": "https://mathoverflow.net/questions/408178/non-rigid-ultrapowers-in-mathsfzfc", "tags": ["set-theory", "lo.logic", "model-theory", "ultrafilters", "ultrapowers"], "votes": 20, "creation_date": "2021-11-10T08:01:04", "comments": ["@PaulLarson \" An ultrapower of a structure of infinite Scott height by an ultrafilter on omega shouldn't have its points separated by its Scott process. Should that make it nonrigid?\" I'm not certain, I'm not too familiar with Scott processes.", "The proof of Harrington's theorem on the Scott ranks of counterexamples to Vaught's Conjecture should show that separating points characterizes rigidity for structures of size aleph_1 too.", "A countable structure is rigid if and only if its Scott process separates points. An ultrapower of a structure of infinite Scott height by an ultrafilter on omega shouldn't have its points separated by its Scott process. Should that make it nonrigid?", "@PaulLarson Ooh, fun question (although not directly related to this one unless I'm missing something?) - I can't think of anything off the top of my head, but this is well outside my normal sphere so that doesn't mean much.", "Is there anything known about rigid structures whose countable elementary substructures are all non-rigid?", "@FarmerS Good question! :P The best I can say is that Shelah's Vive la difference series probably has some relevant info, per Douglas Ulrich's comment, but I haven't worked through those papers to see if there's an answer there.", "Maybe this is easy, but do we know there is a countably infinite structure and some non-principal ultrafilter on $\\omega$ such that the ultraproduct is rigid?"], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "159", "site": "mathoverflow", "title": "A spin extension of a Coxeter group?", "body": "Consider a Coxeter group $W$ with simple generators $S$ and Coxeter matrix $\\left( m_{s,t}\\right) _{\\left( s,t\\right) \\in S\\times S}$.\n\nLet $\\mathfrak{M}$ be the set of all pairs $\\left(s, t\\right) \\in S^2$ satisfying $s \\neq t$ and $m_{s,t} < \\infty$.\n\nFor every $\\left( s,t\\right) \\in \\mathfrak{M}$, let $c_{s,t}$ be an element of $\\left\\\\{ 1,-1\\right\\\\} $. Assume that:\n\n * We have $c_{s,t}=c_{s^{\\prime},t^{\\prime}}$ for any two elements $\\left( s,t\\right) $ and $\\left( s^{\\prime},t^{\\prime}\\right) $ of $\\mathfrak{M}$ for which there exists some $q \\in W$ satisfying $s^\\prime = qsq^{-1}$ and $t^\\prime = qtq^{-1}$.\n\n * We have $c_{s,t}=c_{t,s}$ for each $\\left( s,t\\right) \\in \\mathfrak{M}$.\n\n\n\n\nLet $W^{\\prime}$ be the group with the following generators and relations:\n\n_Generators:_ the elements $s\\in S$ and an extra generator $q$.\n\n_Relations:_ \\begin{align*} s^{2} & =1\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{for every }s\\in S;\\\\\\ q^{2} & =1;\\\\\\ qs & =sq\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{for every }s\\in S;\\\\\\ \\left( st\\right) ^{m_{s,t}} & =1\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{for every }\\left( s,t\\right) \\in \\mathfrak{M} \\text{ satisfying } c_{s,t}=1;\\\\\\ \\left( st\\right) ^{m_{s,t}} & =q\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{for every }\\left( s,t\\right) \\in \\mathfrak{M} \\text{ satisfying } c_{s,t}=-1. \\end{align*}\n\nThere is clearly a surjective group homomorphism $\\pi:W^{\\prime}\\rightarrow W$ sending each $s\\in S$ to $s$, and sending $q$ to $1$. There is also a group homomorphism $\\iota:\\mathbb{Z}/2\\mathbb{Z} \\rightarrow W^{\\prime}$ which sends the generator of $\\mathbb{Z}/2\\mathbb{Z}$ to $q$.\n\n> **Question.** Is $\\iota$ injective? Equivalently, is the sequence \\begin{equation} 1\\longrightarrow\\mathbb{Z}/2\\mathbb{Z}\\overset{\\iota}{\\longrightarrow}W^{\\prime }\\overset{\\pi}{\\longrightarrow}W \\longrightarrow1 \\end{equation} exact? Equivalently, is $\\left\\vert \\operatorname*{Ker}\\pi\\right\\vert =2$ ?\n\n**Background.** This would generalize at least one of the two \"spin symmetric groups\" to the situation of any Coxeter group. It would explain one of the results (Theorem 2.3 **(b)**) in [Alexander Postnikov and Darij Grinberg, _Proof of a conjecture of Bergeron, Ceballos and Labbé_](http://nyjm.albany.edu/j/2017/23-70.html), and prove a generalization of this result (Conjecture 6.1 **(b)**).\n\nI have tried generalizing the [standard approach to constructing the spin symmetric groups by embedding them in the Hecke-Clifford algebra](https://arxiv.org/abs/1110.0263), but to no avail so far. Nor has the existing literature on central extensions of Coxeter groups been particularly helpful ([Howlett's _On the Schur multipliers of Coxeter groups_](http://onlinelibrary.wiley.com/doi/10.1112/jlms/s2-38.2.263/abstract) counts the extensions abstractly, but doesn't help understanding whether a given one exists; [Burichenko's _On extensions of Coxeter groups_](http://www.tandfonline.com/doi/abs/10.1080/00927879508825315?journalCode=lagb20) gives a criterion that I don't seem to properly understand, as it gives me wrong answers).\n", "link": "https://mathoverflow.net/questions/285263/a-spin-extension-of-a-coxeter-group", "tags": ["algebraic-combinatorics", "coxeter-groups", "hecke-algebras"], "votes": 20, "creation_date": "2017-11-04T12:46:36", "comments": ["@DavidESpeyer: Thanks -- I've answered the Question positively in the meantime (a year ago), but only in a month or so will probably be able to write up my proof. Meanwhile, I did look into Howlett's and others' Schur-multiplier papers, but never found myself able to get something out of it that wasn't obviously wrong; it's too much of a foreign language to me.", "Morris, \"Projective representations of finite reflection groups. III.\", Comm. Alg. (2004), is doing something very like this, although I haven't unwound his notation to see whether it exactly matches your formulation or not. The title of that paper has the word finite in it, but most of the results don't seem to need it. See also Howlett, \"On the Schur multipliers of Coxeter groups\", JLMS (1988), which is less concrete but is better about not including unneeded finiteness hypotheses.", "This is a quotient of the Tits group (MSN) in a natural way. It seems like maybe Théorème 2.5 there will give you what you want, although I can't see it right away.", "@VictorProtsak: Thanks. I've reverted to the $\\mathfrak{M}$ notation from the paper.", "Sure, $s\\ne t$. Actually, your last relation involving $q$ should also have restriction $s\\ne t$.", "@VictorProtsak: I take it you want $s \\neq t$, since otherwise picking $s=t$ and $s' = t'$ to be two non-conjugate simple reflections would break it. Anyway, nice question! Perhaps what we did in the proof of Claim 1 in the proof of Lemma 4.0.3 in Alex's and my paper could be of use.", "Darij, very interesting question! Clearly your two conditions are necessary: conjugate elements have the same order. But are they sufficient to guarantee that if $st$ and $s't'$ have the same order in $W$, then $c_{s,t}=c_{s',t'}$ ? More precisely, do you know whether the following statement is true for general $W$? $$ $$ If $st$ and $s't'$ are conjugate in $W$ then there is $q\\in W$ simultateously conjugating either $s$ to $s'$ and $t$ to $t'$ or $s$ to $t'$ and $t$ to $s'$.", "Jim, for any finite dihedral group (finite Coxeter group of rank 2) $\\langle s,t\\rangle\\simeq D_m$ of type $I_2(m)$, in the nontrivial case $c_{s,t}=-1$ this construction produces the dihedral group $D_{2m}$ of type $I_2(2m)$ with double the size. So starting from $W(G_2)=D_6$ you would get $D_{12}$.", "@Gro-Tsen: Interesting idea; but this would only give two \"spin extensions\" (and I have no idea how to precisely find their generators and relations), whereas my $c_{s,t}$ in general allow for much more freedom. (Not saying that the approach is broken; could be my extensions are all equivalent.)", "@darij: This is an intriguing area which I haven't gone into, but I wonder for example what would happen in the frequent case that -1 is in a given finite irreducible Coxeter group $W$? (This is a special case of the longest element of $W$, which has been well studied using the classification.) A small example is the $G_2$ case, though symmetric groups of rank > 1 aren't examples.", "I had in mind that every Coxeter group can be realized as isometries of a certain \"canonical\" symmetric bilinear form (not necessarily definite), so that it embeds in $\\mathit{O}(p,q)$: wouldn't its inverse image in $\\mathit{Pin}^+(p,q)$ or $\\mathit{Pin}^-(p,q)$ answer your question? Or am I talking nonsense?"], "comment_count": 11, "category": "Science", "diamond": 0} {"question_id": "160", "site": "mathoverflow", "title": "Isoperimetric inequality and geometric measure theory", "body": "The following version of the isoperimetric inequality can be easily deduced from the Brunn-Minkowski inequality:\n\n> **Theorem.** _If $K\\subset\\mathbb{R}^n$ is compact, then $$ |K|^{\\frac{n-1}{n}}\\leq n^{-1}\\omega_n^{-1/n}\\mu_+(K), $$ where $\\omega_n$ is the volume of the unit ball and \n> $$ \\mu_+(K)=\\liminf_{h\\to 0} \\frac{|\\\\{x:\\, 0<{\\rm dist}\\, (x,K)\\leq h\\\\}|}{h} $$ is the Minkowski content._\n\nIf $K$ is the closure of a bounded set with $C^2$ boundary, then $\\mu_+(K)=H^{n-1}(\\partial K)$ (Hausdorff measure) so we have $$ |K|^{\\frac{n-1}{n}}\\leq n^{-1}\\omega_n^{-1/n}H^{n-1}(\\partial K). $$ However, the above inequality is true for _any_ compact set without assuming anything about regularity of the boundary. My question is:\n\n## Is there a simple proof of the following result?\n\n> **Theorem.** _If $K\\subset\\mathbb{R}^n$ is compact, then $$ |K|^{\\frac{n-1}{n}}\\leq n^{-1}\\omega_n^{-1/n}H^{n-1}(\\partial K). $$_\n\nThis result can be proved using the machinery of the geometric measure theory. If $H^{n-1}(\\partial K)=\\infty$, the inequality is obvious. If $H^{n-1}(\\partial K)<\\infty$, then $K$ has finite perimeter and the isoperimetric inequality for sets of finite perimeter yields $$ |K|^{\\frac{n-1}{n}}\\leq n^{-1}\\omega_n^{-1/n}P(K)= n^{-1}\\omega_n^{-1/n} H^{n-1}(\\partial^* K) \\leq n^{-1}\\omega_n^{-1/n} H^{n-1}(\\partial K), $$ where $P(K)$ is the perimeter of $K$ and $\\partial^*K\\subset\\partial K$ is the reduced boundary. For details see [1].\n\nUnfortunately, this argument is very far from being elementary.\n\n**[1] Ambrosio, L., Fusco, N., Pallara, D.:** _Functions of bounded variation and free discontinuity problems._ Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000.\n", "link": "https://mathoverflow.net/questions/294882/isoperimetric-inequality-and-geometric-measure-theory", "tags": ["geometric-measure-theory", "isoperimetric-problems", "hausdorff-measure"], "votes": 20, "creation_date": "2018-03-10T13:32:46", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "161", "site": "mathoverflow", "title": "Hahn-Banach and the "Axiom of Probabilistic Choice"", "body": "Stipulate that the Axiom of Probabilistic Choice (APC) says that for every collection $\\\\{ A_i : i \\in I \\\\}$ of non-empty sets, there is a function on $I$ that assigns to $i$ a finitely-additive probability measure $\\mu_i$ on $A_i$.\n\nThen Hahn-Banach (HB) implies APC, since HB is equivalent to the existence of a finitely-additive probability measure on every boolean algebra, and hence implies the existence of such a measure on the direct sum of the powerset algebras $P(A_i)$. \n\nAPC is non-trivial in that it implies the Banach-Tarski paradox and the existence of nonmeasurable sets (the proof Foreman and Wehrung use to show that HB implies Banach-Tarski works).\n\n**Question:** Does APC imply HB?\n\n(When I think about this, I find I keep on wanting to use Stone representation, but of course to do that would be to assume Boolean Prime Ideal, which is stronger than HB or APC.)\n", "link": "https://mathoverflow.net/questions/281142/hahn-banach-and-the-axiom-of-probabilistic-choice", "tags": ["set-theory", "axiom-of-choice", "hahn-banach-theorem"], "votes": 20, "creation_date": "2017-09-14T07:57:09", "comments": ["1. D. Pincus, The strength of Hahn–Banach's Theorem, in: Victoria Symposium on Non-standard Analysis, Lecture notes in Math. 369, Springer 1974, pp. 203-248. Implies Hahn-Banach theorem is weaker than the ultrafilter lemma", "@Fedor: You might have remembered that HB+SKM (Strong Krein-Milman) imply AC.", "@Wojowu ah, indeed, I remembered not well", "@FedorPetrov Not at all. It is very much weaker. It follows from the ultrafilter lemma (and at the same time doesn't imply it).", "If I remember well, HB is equivalent to AC.", "Oh, I wasn't implying it's trivial. I was hoping for someone to complete the proof. :)", "... sorry, I don't see the \"so\"...", "Well, using Bartle integrals you can get linear functionals on $\\ell^\\infty(A_i)$ which do not come from $\\ell^1(A_i)$. And HB is equivalent to \"Every Banach space has a nontrivial functional\" (continuous or otherwise). So..."], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "162", "site": "mathoverflow", "title": "Looking for an effective irrationality measure of $\\pi$", "body": "Most standard summaries of the literature on [irrationality measure](http://mathworld.wolfram.com/IrrationalityMeasure.html) simply say, e.g., that $$ \\left| \\pi - \\frac{p}{q}\\right| > \\frac{1}{q^{7.6063}} $$ for all sufficiently large $q$, without giving any indication of how large qualifies as \"sufficiently large.\" It would occasionally be useful (for example, [for my answer here](https://math.stackexchange.com/a/1313626/30836)) to be able to come up with explicit bounds on the size of $q$.\n\nI don't have access to Salikhov's paper where the exponent $7.6063$ is obtained, but Hata's paper which obtained an exponent of $8.0161$ is online [here](http://matwbn.icm.edu.pl/ksiazki/aa/aa63/aa6344.pdf). Looking through it, the proof seems to be effective (all the bounds he's using appear to be defined explicitly), but it is definitely not written to be clear about what those bounds are.\n\nHas anyone extracted explicit bounds on the size of the denominator from any irrationality measure calculation for $\\pi$ more recent than Mahler's proof (referenced in the Hata paper) that $$ \\left| \\pi - \\frac{p}{q}\\right| > \\frac{1}{q^{42}} $$ unrestrictedly?\n", "link": "https://mathoverflow.net/questions/210509/looking-for-an-effective-irrationality-measure-of-pi", "tags": ["nt.number-theory", "diophantine-approximation"], "votes": 20, "creation_date": "2015-06-07T11:02:16", "comments": ["Salikhov's paper is available here: mathnet.ru/links/93c99ab6587bdc1bf0912ec96b1d9f4f/rm9175.pdf", "I wasn't aware this was migrated, I'll delete my answer in that case.", "The exponent -42 got by Mahler was quite extraordinary in his time (\"Striking inequality\", Alan Baker). It is valid for all rational p/q with q > 1 and also indicates that rationals \"may not be very closed to $\\pi$\". I guess all improvement of the exponent -42 is valid just \"for all sufficiently large q\" in the ordinary meaning of this expression without explicity giving of bounds. Maybe possible but certainly quite difficult to get them.", "@muaddib: Nope, you must be thinking of some other Micah...", "I know this is random but by any chance did you go to New College?"], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "163", "site": "mathoverflow", "title": "Is the determinant of cohomology a TQFT?", "body": "If $M$ is an oriented $d$-manifold, let $D(M)$ denote the top exterior power of $H^*(M,\\mathbf{C})$. Then $D(M_1 \\amalg M_2) = D(M_1) \\otimes D(M_2)$. Is there a good recipe for a map $D(M) \\to D(N)$ induced by a cobordism from $M$ to $N$?\n\nIn some dimensions, there is a natural identification $D(M) = \\mathbf{C}$ and you could take every cobordism to the identity map. But such a TQFT would not be completely trivial when $d = 4$, I think. For example it would distinguish $S^1 \\times \\mathbf{C}P^2$ from the mapping torus of complex conjugation.\n\nIf there's no problem defining it for $(d+1)$-manifolds, can it be \"extended down\" any distance, i.e. associate something to a $(d-1)$-manifold?\n", "link": "https://mathoverflow.net/questions/241807/is-the-determinant-of-cohomology-a-tqft", "tags": ["at.algebraic-topology", "dg.differential-geometry", "topological-quantum-field-theory"], "votes": 20, "creation_date": "2016-06-09T10:31:23", "comments": ["I would have thought that the $d$-dimensional theory (or maybe I mean $d\\pm 1$) would assign $\\mathrm{pt} \\mapsto \\mathbb C$-as-an-$E_d$-algebra. But that defines the trivial (framed) theory. Perhaps there's a nontrivial choice of how to descend from a framed theory to an oriented one, but I'm not seeing it.", "I'm sure the answer is \"yes, $D$ is part of an invertible TFT\", but I'm not seeing totally immediately how to write it down. From my point of view, you want to look at the \"free topological boson on $M$\", which is the theory whose field is a map $\\phi: M \\to \\mathbb C$ and whose EOM is $\\mathrm d \\phi = 0$. This has a \"cotangent quantization\" in the Costello-Gwilliam language, which up to some degree shift is $D$. If you can realize it as a fully extended TFT, you win.", "This TFT would presumably be invertible, so can in principle be classified using the cohomology of Madsen-Tillman spectra. There's a known invertible 2d TFT which assigns to a closed oriented surface the number $\\lambda^{\\chi(X)}$ where $\\chi$ is the Euler characteristic and $\\lambda \\in \\mathbb{C}^{\\times}$, and this TFT is a categorification of that. Maybe this paper of Freed will be helpful: arxiv.org/abs/1406.7278"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "164", "site": "mathoverflow", "title": "Homeomorphisms of the sphere mapping a geodesic triangulation to another one", "body": "Consider the standard Riemannian 2-sphere $S$, equipped with a geodesic triangulation $T$. Let $L(S,T)$ be the space of homeomorphisms of $S$ which map $T$ to a geodesic triangulation. What is the homotopy type of $L(S,T)$? A similar question has been solved by [E. Bloch, R. Connelly, D. Henderson, Topology 23 (1984), 161-175](http://www.sciencedirect.com/science/article/pii/0040938384900375), when $S$ is a simplicial disc in the plane, and it is a \"tour de force\"!\n\nThis question is on behalf of Jean Cerf, my former advisor, who is interested in S. Cairns' work.\n", "link": "https://mathoverflow.net/questions/262048/homeomorphisms-of-the-sphere-mapping-a-geodesic-triangulation-to-another-one", "tags": ["gt.geometric-topology", "riemannian-geometry", "triangulations"], "votes": 20, "creation_date": "2017-02-12T11:27:31", "comments": ["I agree with the change of title. It is better, but I thought of (S,T) as a structure, not an object; it was ambiguous.", "I have improved the title", "The question in the edited title doesn't seem to be the same as the one in the body (and is much easier)."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "165", "site": "mathoverflow", "title": "Polynomials with roots in convex position", "body": "Let $\\mathcal P_n$ denote the set of all monic polynomials of degree $n$ with real or complex coefficients such that $P\\in\\mathcal P_n$ if for all $k\\in\\lbrace 0,1,\\dots,n-2\\rbrace$ the $n-k$ roots of $P^{(k)}=\\left(\\frac{d}{dx}\\right)^kP$ are in strictly convex position (ie are the $n-k$ vertices of a convex polygon with $n-k$ extremal vertices).\n\nThe set $\\mathcal P_n$ is clearly an open subset of all monic polynomials of degree $n$ over $\\mathbb R$ or $\\mathbb C$.\n\nWhat is the geometry and topology of $\\mathcal P_n$?\n\nFacts:\n\n(1) Over the reals, $\\mathcal P_n$ has at least $2^{n-1}$ connected components: Indeed, consider a very fast decreasing sequence (I guess $n\\longmapsto 1/((1+n)^{(1+n)^{1+n}})!$ will probably work) of strictly positive reals $\\alpha_0 > \\alpha_1 > \\cdots$ and a sequence of signs $\\epsilon_0,\\epsilon_1,\\ldots,\\epsilon_{n-2}\\in\\lbrace \\pm 1\\rbrace$. Then \n$$x^n+\\sum_{k=0}^{n-2}\\epsilon_k\\alpha_k x^k\\in \\mathcal P_n(\\mathbb R)$$ and different choices of signs correspond to different connected components. (Proof: Up to translation, we can assume that all roots sum up to $0$. If two polynomials of the above form are connected by a continuous path of real polynomials with roots summing up to $0$ , then all derivatives except the very last one do not have a root at $0$ (since it is otherwise in the interior of the convex hull of all roots). All coefficients up to degree $d-2$ are thus necessarily non-zero and different signs correspond to different connected components). Are there other connected components?\n\n(2) Over the complex numbers, all the polynomials described in (1) are in the same connected component. Choosing $\\epsilon_i$ on the complex unit circle suggests however that $\\pi_1(\\mathcal C_n)$ might be $\\mathbb Z^{n-1}$ where $\\mathcal C_n$ denotes the connected component of $x^n+\\sum_{k=0}^{n-2}\\alpha_kx^k$ in $\\mathcal P_n(\\mathbb C)$. Do we have $\\mathcal C_n=\\mathcal P_n(\\mathbb C)$?\n\n(3) We have the equalities $\\mathcal P_{n-1}=\\lbrace P'/n \\ |\\ P\\in\\mathcal P_n\\rbrace$.\n", "link": "https://mathoverflow.net/questions/26799/polynomials-with-roots-in-convex-position", "tags": ["gt.geometric-topology", "cv.complex-variables", "polynomials"], "votes": 20, "creation_date": "2010-06-02T01:21:03", "comments": ["I have sketched a proof in order to adress the the comment of Denis Serre.", "I am not sure of why all these real polynomials are in different CCs. Do you pretend that if a coefficient of degree $\\le n-2$ is zero, then $P$ is not in ${\\mathcal P)_n$ ? That seems wrong for the constant coefficient.", "Ī̲ have some ideas how to prove $\\mathcal C_n=\\mathcal P_n(\\mathbb C)$ and, similarly, to make a retraction of the real case to components described by the OP. Still searching for a suitable compactification of $\\mathcal P_n$ because Ī̲ detest to construct ε–δ proofs."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "166", "site": "mathoverflow", "title": "On random Dirichlet distributions", "body": "Fix a dimension $d\\ge2$. \n\n * Let $Q_d$ denote the positive quadrant of $\\mathbb{R}^d$, that is, $Q_d$ is the set of points $\\mathbf{x}=(x_i)_i$ in $\\mathbb{R}^d$ such that $x_i>0$ for every $i$. \n * For every $\\mathbf{x}$ in $Q_d$, let $|\\mathbf{x}|=x_1+\\ldots+x_d$. \n * Let $\\Delta_d$ denote the set of points $\\mathbf{x}$ in $Q_d$ such that $|\\mathbf{x}|=1$. \n * For every $\\mathbf{a}$ and $\\mathbf{b}$ in $Q_d$, define $\\mathbf{a}\\cdot \\mathbf{b}$ in $Q_d$ by $(\\mathbf{a}\\cdot \\mathbf{b})_i=a_ib_i$ for every $i$.\n * For every $\\mathbf{a}$ in $Q_d$, let $\\mathrm{Dir}(\\mathbf{a})$ denote the Dirichlet distribution of parameter $\\mathbf{a}$.\n\n\n\n**The problem in a nutshell**\n\nFix $\\mathbf{a}$ and $\\mathbf{b}$ in $Q_d$. Choose a random parameter $\\mathbf{u}$ in $\\Delta_d$ with distribution $\\mathrm{Dir}(\\mathbf{a})$. Then choose a random point $\\mathbf{X}$ in $\\Delta_d$ with distribution $\\mathrm{Dir}(\\mathbf{b}\\cdot \\mathbf{u})$. \n\n> My aim is to understand the (absolute) distribution of $\\mathbf{X}$.\n\n**Some more notations**\n\nFor every $\\mathbf{a}=(a_i)_i$ in $Q_d$, $\\mathrm{Dir}(\\mathbf{a})$ is the absolutely continuous probability measure on $\\Delta_d$ whose density $f(\\ |\\mathbf{a})$ at $\\mathbf{x}$ is proportional to $x_1^{a_1-1}\\cdots x_d^{a_d-1}$. More precisely, $$ f(\\mathbf{x}|\\mathbf{a})=\\Gamma(|\\mathbf{a}|)\\mathbf{x}^{\\mathbf{a}-1}/\\Gamma(\\mathbf{a}), $$ with the following shorthands: $$ \\Gamma(\\mathbf{a})=\\Gamma(a_1)\\cdots\\Gamma(a_d),\\quad \\mathbf{x}^{\\mathbf{a}-1}=x_1^{a_1-1}\\cdots x_d^{a_d-1}. $$ The density $f_{\\mathbf{a},\\mathbf{b}}$ of the distribution of $\\mathbf{X}$ is $$ f_{\\mathbf{a},\\mathbf{b}}(\\mathbf{x})=\\int_{\\Delta_d} f(\\mathbf{x}|\\mathbf{b}\\cdot \\mathbf{u})f(\\mathbf{u}|\\mathbf{a})\\mathrm{d}u_1\\cdots\\mathrm{d}u_{d-1}. $$ \n\n**Some special cases**\n\nIf $a_i=b_i=1$ for every $i$, $\\displaystyle f_{\\mathbf{1},\\mathbf{1}}(\\mathbf{x})\\propto\\int_{\\Delta_d} \\frac{\\mathbf{x}^{\\mathbf{u}-1}}{\\Gamma(\\mathbf{u})}\\mathrm{d}u_1\\cdots\\mathrm{d}u_{d-1}.$\n\nThe case $d=2$ yields $$ f_{\\mathbf{1},\\mathbf{1}}(x,1-x)\\propto\\int_0^1\\frac{x^{w-1}(1-x)^{-w}}{\\Gamma(w)\\Gamma(1-w)}\\mathrm{d}w=\\frac1{\\pi x}\\int_0^1\\left(\\frac{x}{1-x}\\right)^{w}\\sin(\\pi w)\\mathrm{d}w, $$ hence $$ f_{\\mathbf{1},\\mathbf{1}}(x,1-x)=\\frac1{x(1-x)}\\frac1{\\pi^2+(\\log[x/(1-x)])^2}. $$ Writing $\\mathbf{X}=(X_1,X_2)$ with $X_1\\ge0$, $X_2\\ge0$ and $X_1+X_2=1$, this can be rewritten as the fact that, for every $x$ in $(0,1)$, $$ P(X_1\\le x)=P(X_2\\le x)=\\frac12+\\frac1\\pi\\arctan\\left(\\frac1\\pi\\log\\left(\\frac{x}{1-x}\\right)\\right). $$ Are there other cases where the density $f_{\\mathbf{a},\\mathbf{b}}$ is (reasonably) explicit? Or, for example, where the moments $E(\\mathbf{X}^\\mathbf{n})$ of $\\mathbf{X}$ with $\\mathbf{n}=(n_1,\\ldots,n_d)$ any $d$-uplet of integers, are (reasonably) explicit?\n", "link": "https://mathoverflow.net/questions/55147/on-random-dirichlet-distributions", "tags": ["pr.probability", "probability-distributions"], "votes": 20, "creation_date": "2011-02-11T09:23:01", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "167", "site": "mathoverflow", "title": "Etale fundamental group of a curve in characteristic $p$", "body": "Let $C$ be a connected, smooth, proper curve of genus $g$ over an algebraically closed field $k$ of characteristic $p>0$. Let $\\pi_1(C)$ be the etale fundamental group of $C$ - I only care about this as an abstract profinite group, so I omit base points.\n\nIt is well known by Grothendieck that $\\pi_1(C)$ is topologically generated by $2g$ elements. Formally, there is a surjection $\\phi:\\widehat{F_{2g}}\\rightarrow \\pi_1(C)$, where the first group is the profinite completion of the free group on $2g$ elements. Moreover, if we let $G^{(p)}$ be the prime-to-$p$ quotient of $G$, then $$\\pi_1(C)^{(p)}\\cong\\widehat{F_{2g}/w}^{(p)},$$ where $w=\\prod_{i=1}^g[x_i,y_i].$ I.e., you get what you expect from the characteristic 0 case.\n\nRecall that the prime-to-p quotient is the quotient $G/K$, where $K$ is the intersection of the kernels of all the maps from $G$ to finite prime-to-$p$ groups.\n\n**Question: Is there any case of a curve $C$ of genus at least 2 for which its etale fundamental group is known?**\n\nI phrase the question in this way because the Etale fundamental group does depend on the curve in question. In fact, even its abelianization will depend on the $p$-rank of the Jacobian.\n", "link": "https://mathoverflow.net/questions/186687/etale-fundamental-group-of-a-curve-in-characteristic-p", "tags": ["ag.algebraic-geometry", "fundamental-group"], "votes": 20, "creation_date": "2014-11-09T23:09:03", "comments": ["I think $\\pi_1$ is not known for any hyperbolic curve in any reasonable sense. For instance, just knowing for each prime-to-$p$ cover the $p$-rank of its Jacobian seems very hard - it is not at all obvious that there is a usable finite description of this data.", "Take $G_1$ an infinitely generated free $p$-group and $G_2$ the product of $G_1$ with $\\mathbb Z/p$. Then they have the same finite quotients but are distinct. This is not so great an example, because group cohomology distinguishes them.", "@Niels: Not to belabour the point, but i think knowing whcih groups $G$ arise suffices to figure out the cardinality of the set of $G$-corvers. The point being that if there are $k$ non-conjugate homomorphisms to $G$, then there is a homomorphism $\\pi_1\\to G^k$, and the image $H$ of this homomorphism has $k$ non-conjugate homomorphisms to $G$. So we can detect whether theres at least $k$ $G$-covers by asking whether such a group $H$ occurs as a finite quotient. Probably even this is not enough to determine the profinite group. Perhaps I'll ask this a separate question. Thanks for your help!", "@jacob: for an affine curve, we just know if a given finite group $G$ arises as a quotient of $\\pi_1$ or not and nothing more. In particular the cardinality of the set of $G$-covers is not known for general $G$. And this data would be probably insufficient to recover the $\\pi_1$ anyway. To sum up: no the $\\pi_1$ is not known for any curve (complete or not) over any field of positive characteristic except for the trivial cases of $\\mathbb P^1$ and elliptic curves.", "@Niels: You said Joels observation of knowing the finite quotients does not tell us, for a fixed finite group, the cardinality of the set of $G$-covers.My third sentence was meant to say that,if we are working over a countable field, it does. Because the answer is either countable infinity,or some finite number,either of which can be detected on the level of finite quotients. As you say, this is irrelevant if we are working with proper curves, but if we talk about affine curves this starts mattering. So, my question is: If the $\\pi_1$ of a hyperbolic affine curve over a countable field known?", "@jacob I am not sure I understand your third sentence. About the last sentence : the $\\pi_1$ of a proper curve is invariant by algebraically closed field extension. So fixing a specific base field does not change the question for proper curves. And the answer is no, the $\\pi_1$ is not known. By specialization theory we just know it is topologically of finite type. For affine curves the situation is even worse since the $\\pi_1$ is not topologically of finite type.", "Aha! That is illuminating. But over a countable field like $\\mathbb{F}_{p^{\\infty}}$ we know that the set of $G$-covers is at most countably infinite. So in that context, at least, one can answer Niels's question simply by knowing the finite quotients. Am I correct that even in this restricted setting, we do not know the $\\pi_1$?", "@Joël I agree \"completely wrong\" was a bit strong :) but I disagree with your interpretation of the question. Clearly the question is : \"is the $\\pi_1$ known up to isomorphism ?\". An answer to your weak version wouldn't answer the following question : \"given a fixed finite group $G$, what is the cardinality of the set of $G$-covers\" ?", "@jacob I think I remember you can find an example in Fried-Jarden Field Arithmetic.", "@jacob. The finite quotient of $\\mathbb Z_p^I$ for $I$ infinite are all the abelian $p$-group. Now take two infinite sets $I$, one enumerable say and another say larger than the continuum, and you have two profinite groups non-isomorphic (because they don't have the same cardinality) having the same set of finite quotients. So Niels is right that there is more in understanding a $\\pi^1$ than its understanding it's finie quotient. Now in algebraic geometry one could have the point of view that it is the finite quotients, corresponding to finite cover, that are what we want to understand.", "@Niels, Ah okay, interesting. But \"completely wrong\" is perhaps not the right term (or possibly I am more wrong that you explain in your comment): one at least knows all the finite quotients of this profinite group (the $pi^1$ of a non-complete curve), which is already something important, and sufficient for many applications. Then it depends of what we mean by \"know\", and apparently the OP was not completely sure about that. But, do we know the complete list of finite quotient of the $\\pi^1$ for a curve of genus $\\geq 2$ without any point removed ?", "@Niels: Actually, I was wondering about this. Is there an example of 2 non-isomorphic profinite groups with the same finite quotients? Perhaps this is worth posting as a separate question...", "@Joël : what you wrote is completely wrong. Abyankhar's conjecture describes the set of finite quotients of the $\\pi_1$, but since in the affine case the $\\pi_1$ is not topologically of finite type, this is not enough information to get back the $\\pi_1$. For instance, the $\\pi_1$ of $\\mathbb A^1$ is not known in positive characteristic, even if one knows that its finite quotients are the quasi-$p$ groups. In fact to my knowledge there is no example of an hyperbolic curve whose $\\pi_1$ is \"known\" in positive characteristic.", "Of course (you certainly know this), if you remove $t$ points to your curve, then the problem is solved, provided than $t>0$. It is Abyankhar's conjecture, now a theorem of Raynaud and Harbater.", "Tamagawa proved that for all profinite groups $G$, there are only finitely many smooth proper curves over $\\bar{\\mathbb F_p}$ (up to isomorphism) with etale fundamental group isomorphic to $G$. Maybe you can find something useful in his article; see Finiteness of isomorphism classes of curves in positive characteristic with prescribed fundamental groups, J. Algebraic Geom. 13 (2004), 675–724."], "comment_count": 15, "category": "Science", "diamond": 0} {"question_id": "168", "site": "mathoverflow", "title": "Finiteness of etale cohomology for arithmetic schemes", "body": "By an _arithmetic scheme_ I mean a finite type flat regular integral scheme over $\\mathrm{Spec} \\, \\mathbb{Z}$.\n\n> Let $X$ be an arithmetic scheme. Then is $H_{et}^2(X,\\mathbb{Z}/n\\mathbb{Z})$ finite for all $n \\in \\mathbb{N}$?\n\nRemarks: Let $j:U \\to X$ be the open subset given by removing the fibres of $X \\to \\mathrm{Spec} \\, \\mathbb{Z}$ lying above those primes dividing $n$.\n\n 1. I can easily show that $H_{et}^1(X,\\mathbb{Z}/n\\mathbb{Z})$ is finite. For example, it follows from the Leray spectral sequence that $H_{et}^1(X,\\mathbb{Z}/n\\mathbb{Z}) \\to H_{et}^1(U,\\mathbb{Z}/n\\mathbb{Z})$ is injective, and the latter group is finite by a special case of [1, Proposition II.7.1].\n 2. If $n$ invertible on $X$, then the answer is _yes_ and this is again a special case of [1, Proposition II.7.1]. So I'm interested in the case where $n$ is not invertible on $X$. (In particular, this shows that $H_{et}^2(U,\\mathbb{Z}/n\\mathbb{Z})$ is finite).\n 3. If $\\dim X = 1$ then the answer if _yes_ by [1,Theorem II.3.1].\n 4. Another application of the Leray spectral sequence argument shows that is suffices to show that $H^0(X, R^1j_* \\mathbb{Z}/n\\mathbb{Z})$ is finite, but I don't see why this should be the case.\n 5. The analogous question concerning finiteness of $H_{et}^i(X,\\mathbb{Z}/n\\mathbb{Z})$ is also interesting, but for my application I only need $i = 2$.\n\n\n\nReferences:\n\n[1] Milne - Arithmetic duality theorems.\n", "link": "https://mathoverflow.net/questions/284354/finiteness-of-etale-cohomology-for-arithmetic-schemes", "tags": ["ag.algebraic-geometry", "nt.number-theory", "arithmetic-geometry", "etale-cohomology"], "votes": 20, "creation_date": "2017-10-25T04:51:27", "comments": ["@MartinBright: Thanks for the references! I have had a look at the paper of Sato, but it is written in the language of derived categories which are not my forte.... Could you please be more specific which result is most relevant? He seems to have many different purity results in his paper.", "A couple of references: the purity result of Gabber referred to is here arxiv.org/abs/1207.3648 Exposé XVI, and indeed assumes the torsion order is invertible. For p-torsion there is some kind of purity theorem in this article of Sato arxiv.org/abs/math/0610426 but I don't know whether it is any use to you.", "@JasonStarr: as you know, there's always an excision sequence relating the cohomology of a scheme to that of an open subscheme with the extra term being cohomology with supports along the closed complement (rather than cohomology on that complement), so the Artin-Schreier stuff doesn't directly intervene (but there's likely no purity theorem for this situation with torsion-orders not invertible; with reasonable torsion-orders I think Gabber has proved purity theorems for abstract regular schemes).", "I have one more comment. For every flat, regular, affine $\\mathbb{Z}$-scheme $X$ whose reduction $X_p$ over $\\mathbb{Z}/p\\mathbb{Z}$ is regular of dimension $\\geq 1$, the cohomology $H^1_{\\text{et}}(X_p,\\mathbb{Z}/p\\mathbb{Z})$ is infinite, because of the many different Artin-Schreier covers. If there were a long exact sequence, presumably this would imply that $H^2_{\\text{et}}(X,\\mathbb{Z}/p\\mathbb{Z})$ is also infinite. That might be evidence that there is no long exact sequence, or it might be evidence that the cohomology is infinite.", "The first paper of Esnault that I know including such a Gysin sequence (the one with the joint appendix with Deligne) is \"Deligne's Integrality Theorem in Unequal Characteristic and Rational Points over Finite Fields\", Ann. of Math. 164 (2006), pp. 715--730. A second paper, where she uses alterations, is \"Coniveau over $\\mathfrak{p}$-adic Fields and Points over Finite Fields\", C. R. Acad. Sci. Paris, Ser. I 345 (2007), pp. 73--76.", "@Jason Starr: Which paper of Esnault are you referring to?", "Esnault's uses alterations (in the strong form of a $G$-Galois cover that has a regular modification) and the map from the cohomology of the original scheme to the $G$-invariants of the cohomology of the $G$-cover to deal with the regularity properties. You are completely correct that she needs to invert several integers.", "@JasonStarr: The purity theorems (such as in Milne's notes that you mention) require the closed complement to satisfy regularity properties (which $X \\bmod p$ may not) and more importantly assume torsion-orders for the sheaf are invertible on the scheme being considered (and avoiding the latter assumption is the main thrust of the question here).", "@JasonStarr: In the papers of Deligne and/or Esnault surely the relevant torsion sheaves have torsion orders invertible on that base (and section 6 of Weil II assumes properness). With \"invertible\" torsion-orders one can use Deligne's \"generic base change\" expose from SGA 4.5 to see that the higher direct images on ${\\rm{Spec}}(\\mathbf{Z})$ are constructible, and it is a general theorem (perhaps due to some combination of Artin, Mazur, Zink?) that the cohomology of any constructible abelian sheaf (no hypothesis on torsion-orders) on the ring of $S$-integers of a number field is finite.", ". . . Esnault cites Equation 6 in Section 3.6, p. 389 of \"La Conjecture de Weil, II\" by Pierre Deligne. Deligne considers a finite type flat scheme over a DVR, but he seems to assume that it has equicharacteristic. However, in the papers of Esnault (including the paper with a joint appendix with Deligne), the exact sequence is applied when the DVR has mixed characteristic . . .", "The version in Milne is Theorem 16.1 of \"Lectures on Etale Cohomology\", but he only formulates there the theorem for finite type schemes over a field. The purity theorem over a Dedekind domain base has been used by Esnault in her work on congruences for rational points . . .", "There is a purity theorem that is really a Gysin sequence relating etale cohomology of $U$, etale cohomology of $X$, and etale cohomology with supports in the closed complement of $U$ (which then gets turned into etale cohomology of the complement via purity). There is a reference in Milne. I will write it in a moment."], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "169", "site": "mathoverflow", "title": "Infinitely generated non-free group with all proper subgroups free", "body": "Is there any example of group $G$ satisfying the following properties?\n\n 1. $G$ is non-abelian, infinitely generated (i.e. it is not finitely generated) and not a free group.\n 2. $H< G$ implies that $H$ is a free group.\n\n\n\nClearly such an example should be torsion-free and countable.\n\n* * *\n\n(Added, from comments) For completeness, here's why it has to be countable. First, it is enough to construct a subgroup of index 2. Indeed, a theorem of Swan asserts that a torsion-free group with a free finite-index subgroup is free. Next, if a group $G$ has no subgroup of index 2 (i.e. every element is product of squares), it is easy to construct a nontrivial countable subgroup $H$ with the same property). In the current setting, $G$ is uncountable so $H$ is a proper subgroup, and hence has to be free, contradiction.\n", "link": "https://mathoverflow.net/questions/351295/infinitely-generated-non-free-group-with-all-proper-subgroups-free", "tags": ["gr.group-theory", "examples", "free-groups"], "votes": 20, "creation_date": "2020-01-27T14:45:56", "comments": ["Since a partly equivalent question was just asked, I have added the short argument to discard the uncountable case.", "@YCor You're probably right I should not have have said \"clearly\" but in my mind it was a very direct application of some known facts (to me at least) in infinite group theory. Swan Theorem says that if your group is torsion-free and it is a finite extension of a free group, then it must be free itself.", "@W4cc0 thanks; I wouldn't have said \"clearly\"! I didn't know Swan's theorem, what does it say exactly? Here's an argument to show residual finiteness: it's enough to check that there is a proper subgroup of index 2. Indeed, the latter being free by assumption, it has to be res. finite and hence so is the whole group. Now, if by contradiction there is no subgroup of index 2, then every element is a finite product of squares. From this, it is easy to construct a nontrivial countable subgroup in which every element is a finite product of squares, and hence this is a non-free countable subgroup.", "@YCor such a group must be countable because if it is not, then it would be residually finite, but then a very well known theorem of Swan would imply that it is free", "@SantanaAfton -- Mark explained how to construct a non-free group in which every finitely generated subgroup is free. This contradicts your intuition that since \"every finitely-generated subgroup is both proper and free\" ... the group must itself be free.", "@W4cc0 on the other hand do you maintain your claim that such a group has to be countable? or do you want to require the group to be countable? (I guess the countable case and uncountable case are quite distinct questions.)", "@YCor, I meant $G$ non-abelian (even if it makes sense in this case, it is easy to see that there are no abelian such groups), of course it would have no sense in the other case; but unfortunately I always write the opposite of what I would like to, sorry.", "Higman constructed non-free groups of cardinality $\\aleph_1$ in which every countable subgroup is free. But every proper subgroup being free sounds hard.", "@SantanaAfton It's overly simplistic, if not senseless, to think of a free group as a group with no relations.", "@W4cc0 no, you don't: there are always free abelian subgroups. So the question was correctly stated, and wouldn't if \"non-abelian\" were added. Also as already mentioned, every group with your property has a non-abelian free subgroup.", "@SantanaAfton Every finitely generated group of the rationals/Hawaiian earring is free but these are not free", "@HJRW Are you referring to the HNN construction? In that example, if you take the relation $a^t=ab$ and the group generated by all letters involved, you don’t get a proper subgroup. This never happens in OP’s situation.", "@SantanaAfton -- see Mark Sapir's comment above.", "It feels as though no such group exists. Since every finitely-generated subgroup is both proper and free by assumption, how could the group have any nontrivial relations?", "I should have been more clear, I meant free non-abelian.", "@BenWieland a torsion-free abelian group in which every proper subgroup is cyclic, is itself cyclic, by an easy argument.", "@MarkSapir $\\mathbf{Z}[3/2]=\\mathbf{Z}[1/2]$, so it's not a proper subgroup. Actually, the only unital subrings of $\\mathbf{Z}[1/2]$ are itself and $\\mathbf{Z}$. Also, every subgroup of $\\mathbf{Z}[1/2]$ is $\\{0\\}$, infinite cyclic, or has finite index and isomorphic to $\\mathbf{Z}[1/2]$ (namely equals $k\\mathbf{Z}[1/2]$ for some unique positive odd $k$.)", "@MarkSapir Oh, yeah, but probably better to call the subgroup $3\\cdot\\mathbb Z[\\frac12]$", "@BenWieland: $\\mathbb{Z}[1/2]$ is the derived subgroup of the Baumslag-Solitar group $BS(1,2)$, it contains non-free subgroup $\\mathbb{Z}[3/2]$", "and this londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/plms/…", "$\\mathbb Z[\\frac12]$", "arxiv.org/pdf/1903.03334.pdf", "Sorry. I missed the infinitely Generated.", "About the countability issue, I think I'm not aware of a non-free group in which every countable subgroup is free. (By analogy, $\\mathbf{Z}^\\mathbf{N}$ is not free abelian but all its countable subgroups are. Also, the Hawaiian earring group is locally free but has non-free countable subgroups.)", "@PaulPlummer: I do not think it is clear (or even true) that these groups must be countable,", "@MarkSapir and YCor Thanks, for some reason my comment sounded like a better idea when typing it, even though I knew of locally free nonfree groups.", "@PaulPlummer: Locally free but non-free groups are easy to construct. Take a proper ascending HNN extension of the free group $F_k$, and the normal subgroup $N$ generated by $F_k$. For example $\\langle a,b,t \\mid a^t=ab, b^t=ba>$. The normal closure $N$ of $a,b$ is locally free but not free. The OP does not want this example. He wants all proper subgroups to be free, not just finitely generated subgroups.", "@PaulPlummer I don't really understand your first comment. A free group itself has plenty of generating families under which it is not free, so using non-freeness of a given generating family will not help much.", "Also why must such groups be countable? I am not seeing it immediately", "@BenjaminSteinberg To complement Mark's comment, groups with Olshanskii's property are generated by every non-commuting pair."], "comment_count": 30, "category": "Science", "diamond": 0} {"question_id": "170", "site": "mathoverflow", "title": "Characteristic subgroups and direct powers", "body": "Solved question: Suppose _H_ is a characteristic subgroup of a group _G_. Is it then necessary that, for every natural number _n_ , in the group $G^n$ (the external direct product of $G$ with itself $n$ times), the subgroup $H^n$ (embedded the obvious way) is characteristic?\n\nAnswer: No. A counterexample can be constructed where $G = \\mathbb{Z}_8 \\times \\mathbb{Z}_2$ (here $\\mathbb{Z}_n$ is the group of integers modulo $n$) with $H$ the subgroup \n\n$$\\\\{ (0,0), (2,1), (4,0), (6,1) \\\\}$$\n\nThis subgroup sits in a weird diagonal sort of way and just happens to be characteristic (a quirk of the prime $2$ because there isn't enough space). We find that $H \\times H$ is not characteristic in $G \\times G$.\n\n(ASIDE: The answer is _yes_ , though, for many important characteristic subgroups, including fully invariant subgroups, members of the upper central series, and others that occur through typical definitions. Since for abelian groups of odd order, all characteristic subgroups are fully invariant, the answer is yes for abelian groups of odd order, so the example of order $2^4$ has no analogue in odd order abelian groups.)\n\nMy question is this:\n\n 1. Strongest: Is it true that if _H_ is characteristic in _G_ and $H \\times H$ in $G \\times G$, then each $H^n$ is characteristic in $G^n$ [NOTE: As Marty Isaacs points out in a comment to this question, $H \\times H$ being characteristic in $G \\times G$ implies _H_ characteristic in _G_ , so part of the condition is redundant -- as explained in (2)]? \n 2. Intermediate: Is there some finite $n_0$ such that it suffices to check $H^n$ characteristic in $G^n$ for $n = n_0$? Note that if $H^n$ is not characteristic in $G^n$ for any particular _n_ , then characteristicity fails for all bigger $n$ as well. I'd allow $n_0$ to depend on the underlying prime of _G_ if we are examining $p$-groups.\n 3. Weakest: Is there a test that would always terminate in finite time, that could tell, for a given _H_ and _G_ , whether $H^n$ is characteristic in $G^n$ for all _n_? The \"try all _n_ \" terminates in finite time if the answer is _no_ , but goes on forever if the answer is _yes_. In other words, is there a finite characterization of the property that each direct power of the subgroup is characteristic in the corresponding direct power of the group?\n\n\n\nADDED: My intuition, for what it's worth, is that those subgroups _H_ of _G_ that can be characterized through \"positive\" statements, i.e., those that do not make use of negations or $\\ne$ symbols, would have the property that $H^n$ is characteristic in $G^n$. On the other hand, those whose characterization requires statements of exclusion (_not a ..._) would fail because negative statements are difficult to preserve on taking direct powers. But I don't know how to make this rigorous. \n", "link": "https://mathoverflow.net/questions/35701/characteristic-subgroups-and-direct-powers", "tags": ["gr.group-theory"], "votes": 20, "creation_date": "2010-08-15T18:10:41", "comments": ["Do you have a reference for \"if $H$ is fully invariant in $G$ then $H \\times H$ is fully invariant in $G \\times G$? Thanks in advance.", "In the case of finite abelian groups, is it true that $\\operatorname{Aut}(G \\times G \\times G)$ is generated by automorphisms stabilizing one copy of $G$ ? This holds at least when $G=(\\mathbf{Z}/p^n\\mathbf{Z})^k$.", "@Marty Isaacs: I have added this note to Question 1. I had already incorporated a similar observation into Question 2, as you probably noticed.", "There is a redundancy in Question 1 since if $H \\times H$ is characteristic in $G \\times G$, then $H$ is automatically characteristic in $G$. This is because every automorphism of $G$ extends to an automorphism of $G \\times G$ stabilizing the first component.", "No reference to a paper or book, but here's an online proof I wrote up: groupprops.subwiki.org/wiki/… It's rather long. What fails for the prime 2 is the first half of the proof: \"it is the direct sum of its intersections with the direct summands\" because in that part we use that the doubling map is an automorphism. The second half is bookkeeping that takes homomorphisms between cyclic subgroups and converts them to automorphisms by adding the identity map.", "That odd prime statement sounds handy, do you have a reference for \"characteristic subgroups of abelian groups of odd order are fully invariant\"?"], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "171", "site": "mathoverflow", "title": "Large values of characters of the symmetric group", "body": "For $g$ an element of a group and $\\chi$ an irreducible character, there are two easy bounds for the character value $\\chi(g)$: First, the bound $|\\chi(g)|\\leq \\chi(1)$ by the dimension of the representation and, second, the bound $|\\chi(g) |\\ \\leq \\sqrt{ |Z(g)|}$ by the centralizer arising from the formula $\\sum_{\\chi} |\\chi(g)|^2= |Z(g)|$.\n\nI am interested on upper bounds for characters of the symmetric group that improve slightly on the second bound.\n\nMore specifically, fix $\\delta> 0$ small. I want to know for which $g\\in S_n$ there exists an irreducible character $\\chi$ with $|\\chi(g)| > \\sqrt{ |Z(g)|} e^{- \\delta \\sqrt{n}}$.\n\nRecall that for a permutation $g\\in S_n$ with $m_k$ cycles of size $k$ for all $k$, so that $\\sum_k m_k k =n$, we have $|Z(g)| = \\prod_k (k^{m_k} m_k!)$.\n\nFor example, known upper bounds for the dimension of representations of the symmetric group imply there does not exist such a character for $g$ the identity. On the other hand, such a character does exist if $g$ is an $n$-cycle, since the right side is less than $1$. I suspect and hope that this can only happen for $g$ containing some relatively large cycles.\n\nI found works in the literature giving bounds for $\\chi(g)$ that improve on the $|\\chi(g)|\\leq \\chi(1)$ by a multiplicative factor ([Asymptotics of characters of symmetric groups related to Stanley character formula by Féray and Śniady](http://doi.org/10.4007/annals.2011.173.2.6)) or a power ([Characters of symmetric groups: sharp bounds and applications by Michael Larsen and Aner Shalev](https://doi.org/10.1007/s00222-008-0145-7)) but it's not obvious if it's possible to transform them into a bound of the form I need.\n", "link": "https://mathoverflow.net/questions/441235/large-values-of-characters-of-the-symmetric-group", "tags": ["co.combinatorics", "rt.representation-theory", "symmetric-groups", "characters"], "votes": 19, "creation_date": "2023-02-20T06:22:56", "comments": ["But now suppose we consider a permutation with $n/{3k}$ $k-1$-cycles, $k$-cycles, and $k+1$-cycles, or something like that. The square root of the centralizer is now $\\sqrt{ (n/3k)!^3 (k^3-k)^{n/3k} }$ which is smaller by a factor of roughly $\\sqrt{ 3^{n/k} e^{ - \\pi \\sqrt{2n/3}}}$. So the upper bound you desire should imply that slightly changing the size of the strips gives us vastly fewer border strip tableaux than before, which seems unlikely (instead of merely more cancellation, which is more feasible.)", "@TimothyChow Actually, consider the following. If we consider a permutation with $n/k$ $k$-cycles, its centralizer has size $(n/k)! k^{n/k}$. This is the sum of the squares of the character values and the number of character values is $e^{ \\pi \\sqrt{2n/3}}$ so one character value has to be at least $\\sqrt{ (n/k)! k^{n/k} e^{ - \\pi \\sqrt{2n/3}}}$. So the number of border strip tableaux is at least that number as well.", "@SamHopkins Yes, good point, and an \"unsigned combinatorial description for the square of the characters\" as ruled out by this paper would certainly be very convenient for proving the bound I want if it was possible.", "Of course this does not in any way preclude bounds of the form you are interested in, but it might also be worth pointing out on the negative side of things that there have been recent dramatic advances which show that symmetric group character values are computationally hard: arxiv.org/abs/2207.05423", "@TimothyChow For $n=4$ if $g$ is a $3$-cycle and a $1$-cycle and the character corresponds to the partition $3+1$ then $\\sqrt{ | Z(g)|} =\\sqrt{3}<2$ and the number of border-strip tableaux is either $2$ or $0$ depending on the order of the cycles. So maybe the truth or falsity depends on the order. Another evidence is that this bound is true is if all the cycles are the same size so that there is no choice of order, by Corollary 9 of the paper A bijection proving orthogonality of the characters of $S_n$ by Dennis E White mentioned in a deleted answer.", "My first instinct is to try Murnaghan-Nakayama, but I don't know if that will work. Best-case scenario would be that the number of border-strip tableaux is already $\\le \\sqrt{|Z(g)|}$ (so that you don't even have to worry about sign-cancellation), but maybe this is false?", "@PerAlexandersson Thanks! It seems to me that one of these formulas will be helpful if there exists a relatively simple combinatorial proof of the bound $\\chi(g)^2 \\leq |Z(g)|$, because then one can try to study the proof more carefully to see when it can be sharp. So far I don't see how to do that for any of them, but it could be possible...", "Some various combinatorial formulas for Sn-characters can be found here: symmetricfunctions.com/murnaghanNakayama.htm"], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "172", "site": "mathoverflow", "title": "What algebraic properties are preserved by $\\mathbb{N}\\leadsto\\beta\\mathbb{N}$?", "body": "Given a binary operation $\\star$ on $\\mathbb{N}$, we can naturally extend $\\star$ to a semicontinuous operation $\\widehat{\\star}$ on the set $\\beta\\mathbb{N}$ of ultrafilters on $\\mathbb{N}$ as follows:\n\n$$\\mathcal{U}\\mathbin{\\widehat \\star}\\mathcal{W}:=\\\\{A:\\\\{k: \\\\{a: a\\star k\\in A\\\\}\\in\\mathcal{U}\\\\}\\in\\mathcal{W}\\\\}.$$\n\nIt's a standard (and [quite useful!](https://web.williams.edu/Mathematics/lg5/Hindman.pdf)) fact that associativity is preserved under this transformation: if $\\star$ is associative, then so is $\\widehat{\\star}$. On the other hand, commutativity [is _not_ preserved](https://math.stackexchange.com/questions/87109/addition-on-ultrafilters-is-non-commutative). Say that an equational sentence $\\sigma$ involving a single binary function symbol is a **$\\beta$ -property** iff we have $$(\\mathbb{N};\\star)\\models\\sigma\\quad\\implies\\quad (\\beta\\mathbb{N};\\widehat{\\star})\\models\\sigma$$ for every binary operation $\\star$ on $\\mathbb{N}$.\n\n> **Question** : Which equational sentences are $\\beta$-properties?\n\nOf course there is no need to restrict attention to equational properties per se _(which is why I've used \"$\\implies$\" rather than \"$\\iff$\" in the definition of $\\beta$-property)_, and we can similarly transform arbitrarily many arbitrary-arity operations on $\\mathbb{N}$; however, already this question seems difficult. I recall seeing a partial(?) answer to this question which involved the way variables were/were not _reordered_ in the left vs. right hand sides of $\\sigma$, but at the moment I can't track it down.\n\n* * *\n\nIncidentally, the following question is at least of a related _flavor_ , and may involve relevant ideas (and be easier to tackle at first): what is $(\\beta\\mathbb{N};\\widehat{+})$ _(or $(\\beta\\mathbb{Z};\\widehat{+})$ using the analogous definition)_ like from a universal-algebraic perspective? For example, is it congruence modular?\n", "link": "https://mathoverflow.net/questions/427135/what-algebraic-properties-are-preserved-by-mathbbn-leadsto-beta-mathbbn", "tags": ["set-theory", "lo.logic", "gn.general-topology", "model-theory", "universal-algebra"], "votes": 19, "creation_date": "2022-07-22T17:24:11", "comments": ["@BenjaminSteinberg xy=zw is also preserved. :P", "One thing that is preserved of course is either xy=x or xy=y", "Yes. I think chapter 6 talks about free subsemigroups. Of course it doesn't looks at nonassociative magmas. My impression is that few general semigroup identities are preserved.", "@BenjaminSteinberg Is that book Algebra in the Stone-Cech compactification?", "There are some results in Hindman's book on conditions guaranteeing the Stone-Cech compactification of a semigroup contains a free semigroup on 2 generators and hence satisfies no semigroup identity. If is of course possible for the original semigroup to satisfy an identity and so this identity is not preserved. Commutativity is such and identity and so are Malcev's nilpotent identities.", "Notice also many quasiidentities like cancellativity are not preserved."], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "173", "site": "mathoverflow", "title": "Mumford-Tate conjecture for mixed Tate motives", "body": "Let $X$ be a (not necessarily smooth or proper) variety over a number field $k$. Suppose we are given\n\n 1. A subquotient $V_{dR}$ of the algebraic de Rham cohomology $H_{dR}^i(X)$ (defined in the non-smooth case via a thickening, as in e.g. Hartshorne's algebraic de Rham cohomology paper),\n\n 2. For each embedding $\\iota: k\\to \\mathbb{C}$, a subquotient $V_\\iota$ of $$H^i(X_{\\mathbb{C}}^{an}, \\mathbb{Q})$$ respecting the $\\mathbb{Q}$-mixed Hodge structure, and\n\n 3. For some fixed prime $\\ell$ and algebraic closure $\\bar k$ of $k$, a subquotient $V_\\ell$ of $$H^i(X_{\\bar k, \\text{ét}}, \\mathbb{Q}_\\ell)$$ respecting the action of $\\text{Gal}(\\bar k/k)$, such that\n\n 4. For each $\\iota$, the corresponding comparison isomorphism $$H^i_{dR}(X)\\otimes_k \\mathbb{C} \\to H^i(X_{\\mathbb{C}}^{an}, \\mathbb{Q})\\otimes \\mathbb{C}$$ sends $V_{dR}$ to $V_\\iota$, and\n\n 5. For each $\\iota$, the corresponding comparison isomorphism $$H^i(X_{\\mathbb{C}}^{an}, \\mathbb{Q})\\otimes \\mathbb{Q}_\\ell \\to H^i(X_{\\bar k, \\text{ét}}, \\mathbb{Q}_\\ell)$$ sends $V_\\iota$ to $V_\\ell$, and\n\n 6. The $V_\\iota$ is of mixed Tate type (i.e. all $h^{p,q}$'s vanish for $p\\not=q$). Or equivalently $V_\\ell$ is an iterated extension of powers of the cyclotomic character.\n\n\n\n\n(If you'd like, you can think of $V$ as the realization of a mixed Tate motive over $k$ under any of the various formalisms for mixed Tate motives over number fields -- I think that a priori the conditions above are slightly weaker but expected to be equivalent.)\n\n> I'd like to know -- is the analogue of the Mumford-Tate conjecture known to be true for such $V$? Explicitly, one expects that the Lie algebra of the image of $$\\text{Gal}(\\bar k/k)\\to GL(V_\\ell)$$ is the extension of scalars to $\\mathbb{Q}_\\ell$ of the Lie algebra of a $\\mathbb{Q}$-group, given as the Tannaka dual of the subcategory of mixed Tate Hodge structures over $k$, generated by the Hodge realization $(V_{dR}, V_{\\iota_1}, V_{\\iota_2}, \\cdots)$.\n\nMy feeling is that one can probably extract this from the literature by comparing ranks of various Ext groups (in one's favorite category of mixed Tate motives and in the $\\ell$-adic and Hodge settings) but I am hoping it is written explicitly somewhere.\n\n**EDIT:** Fixed some errors in the formulation of the Hodge realization.\n", "link": "https://mathoverflow.net/questions/379972/mumford-tate-conjecture-for-mixed-tate-motives", "tags": ["ag.algebraic-geometry", "nt.number-theory", "galois-representations", "mixed-hodge-structure"], "votes": 19, "creation_date": "2020-12-29T09:18:09", "comments": ["@DonuArapura It seems like I might have to, since I have an application in mind!", "Hi Daniel. I'm not sure it is written down anywhere, so it would be nice if you did."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "174", "site": "mathoverflow", "title": "Is there a classification of reflection groups over division rings?", "body": "I asked a [version](https://math.stackexchange.com/questions/3491953/reflection-groups-of-division-rings) of this question in Math StackExchange about a week ago but I've received no feedback so far, so following the advice I received on [meta](https://meta.mathoverflow.net/questions/4416/would-this-question-be-appropriate-for-mathoverflow) I decided to post it here.\n\n* * *\n\n## Details\n\nThe irreducible finite [complex reflection groups](https://en.wikipedia.org/wiki/Complex_reflection_group) were classified by Shephard and Todd. The list consists of a three-parameter family of imprimitive groups and 34 exceptional cases. If $\\mathbb{K}$ is any field of characteristic zero, it is known (see e.g. Section 15-2 [here](https://www.springer.com/gp/book/9780387989792)) that if a group has a representation as a $\\mathbb{K}$-reflection group then it also has a representation as a complex reflection group, so we get no new examples for fields. I'm interested to know if an analogue of this classification exists when we allow $\\mathbb{K}$ to be a division ring of characteristic zero.\n\nI have only found partial results. The irreducible finite quaternionic reflection groups were classified in [this paper](https://www.sciencedirect.com/science/article/pii/0021869380901489) by Cohen. Finite $\\mathbb{K}$-reflection groups of rank $1$, or equivalently finite subgroups of division rings of characteristic $0$, were classified in [this paper](https://www.jstor.org/stable/1992994?seq=1) by Amitsur. The classification for rank $2$ can perhaps be extracted from the [classification of finite subgroups of $GL(2,\\mathbb{K})$](https://core.ac.uk/download/pdf/82740228.pdf) by Banieqbal using a case-by-case check.\n\nIn view of the complex and quaternionic lists, I would expect the full classification (if there is one) to follow a form similar to this:\n\n * An infinite family $G_n(M,P,\\alpha)$ of imprimitive reflection groups of rank $n$, where $M$ is a finite subgroup of a division ring $\\mathbb{K}$ (i.e. an Amitsur group) and $[M,M]\\le P \\trianglelefteq M$, possibly with some extra data $\\alpha$ in low rank. Algebraically it should correspond to something like the group of generalized permutation matrices with entries in $M$ whose determinant is in $P$, like in the cases of fields and quaternions (note that order doesn't matter when computing the determinant, since $[M,M]\\le P$).\n\n * A family or families of examples in rank $2$ (or perhaps in rank $\\le m$ where $m^2$ is the dimension of $\\mathbb{K}$ as a division algebra over its center). In the quaternionic case they are the primitive reflection groups whose complexification is imprimitive, and are all constructed from certain $2$-dimensional primitive complex reflection groups; it's not clear to me how this construction could generalize to arbitrary $\\mathbb{K}$.\n\n * A number of exceptional cases of small rank. These are the ones I'm most interested in. In the quaternionic case these are precisely the primitive reflection groups whose complexification is also primitive; in the general case they might correspond to primitive reflection groups which remain primitive after tensoring the representation with a splitting field.\n\n\n\n\n* * *\n\n## Question\n\nMy main question is thus:\n\n> Is there a classification of groups representable as a $\\mathbb{K}$-reflection group over some division ring $\\mathbb{K}$ of characteristic zero?\n\nI would also appreciate any references dealing with this problem or with particular cases. If the classification turns out to be intractable, or currently out of reach, I would ask if at least an example can be found of a new exceptional reflection group of rank $\\ge 3$ (see details above).\n\n* * *\n\n### Update\n\nAfter much searching, I finally found a brief reference to this problem in the literature. The mention occurs at the end of Section 3 in [this 1981 paper](https://darkwing.uoregon.edu/%7Ekantor/PAPERS/GenLinearGps.pdf) by Kantor, which I quote here for convenience:\n\n> [...] For example, consider the problem of determining all finite primitive reflection groups $G$ in $GL(n,D)$, for $D$ an arbitrary noncommutative division ring of characteristic $0$. If $n=1$, this is just the famous problem solved by Amitsur (1955) (and independently and almost simultaneously by J. A. Green). If $n=2$ and $G$ is solvable, the problem seems to involve even more difficult number theory than Amitsur used. But if $n\\ge 3$, and if simple group classification theorems are thrown at the problem, no new nonsolvable examples arise. [...]\n\nIf the last sentence is true, it means that the new examples of \"exceptional groups\" I asked for must necessarily be solvable. However, the author does not provide any in-text citation for that statement, and I haven't been able to find which result is being alluded to.\n", "link": "https://mathoverflow.net/questions/349809/is-there-a-classification-of-reflection-groups-over-division-rings", "tags": ["gr.group-theory", "rt.representation-theory", "division-rings", "reflection-groups"], "votes": 19, "creation_date": "2020-01-06T00:40:10", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "175", "site": "mathoverflow", "title": "xkcd's "Unsolved Math Problems", straight lines in random walk patterns", "body": "STEM student's favourite source of amusement posted a comic titled \"Unsolved Math Problems\" one of which looks like something that could actually be tackled.\n\n> If I walk randomly on a grid, never visiting any square twice, placing a marble every N steps, on average how many marbles will be in the longest line after NK steps?\n\n[Original comic (with ilustration of the problem)](https://xkcd.com/2529)\n\nI know there are some results in form of power laws (critical exponents) like the 0.587597... constant for SARW in 3D.\n\nCan we say anything interesting about the process described in the comic?\n", "link": "https://mathoverflow.net/questions/406469/xkcds-unsolved-math-problems-straight-lines-in-random-walk-patterns", "tags": ["random-walks"], "votes": 19, "creation_date": "2021-10-18T01:49:48", "comments": ["@Keba if you get boxed in discard that walk and begin again. A better way to think of the question might be: consider ALL possible SAWs of NK steps with a marble placed at every Nth step. What is the average number of marbles on the longest line?", "@ZachTeitler, re, a (literally) bold new notation for the $5$-adics?", "@Gro-Tsen It's a \"conjection\" whatever that is. (Also: I wonder what $\\mathbb{5}$ is supposed to mean.)", "@SamHopkins A horizontal or vertical line through the origin will have $\\sqrt{K/N}$ marbles on average and I would wildly guess that the correct answer is within $(KN)^\\epsilon$ of this. It looks conceivable to prove this by bounding for each line the expectation of the $\\ell$th power of the number of marbles on the line and summing over all lines passing through at least a few points within a distance $NK$ of the origin.", "If we consider the analogous process for a random walk which is not self-avoiding, is this understood? (SAWs are much harder than regular random walks, from what I understand…)", "@Gro-Tsen: $\\epsilon$ cannot be $\\lt 0$.", "I believe deeplinking to images on XKCD is discouraged, so I changed the link to point to the comic.", "What I very much want to know is: is the Euler field manifold hypergroup isomorphic to a Gödel-Klein meta-algebreic $\\epsilon<0$ quasimonoid connection under Sondheim calculus? (Also, what in Apollo's name is going on with this curve?)", "@MartinHairer Is this 'dropping a marble' part redundant then? Is it roughly equivalent to how many collinear gridpoints were visited?", "@Keba either seems interesting but I think the original intention was point-like marbles.", "The self-avoiding random walk with $N$ steps is usually simply defined as the uniform measure on the set of all self-avoiding walks of $N$ steps. It doesn't have any kind of Markov property, but is conjectured in $2D$ to converge to an SLE curve with self-similarity exponent $3/4$. Rigorously, almost nothing non-trivial is known about its large-$N$ behaviour.", "I feel like we are getting nerd sniped again...", "Also, how large are these marbles? Do they fill the whole square? Are they just points?", "I'm no expert on random walks so I probably miss something easy, but why is this process well-defined? I might corner myself in?"], "comment_count": 14, "category": "Science", "diamond": 0} {"question_id": "176", "site": "mathoverflow", "title": "Is there a simpler proof of the key lemma in the paper by Hiroshi Iriyeh and Masataka Shibata on the 3D Mahler conjecture?", "body": "In this [remarkable paper](https://arxiv.org/abs/1706.01749) 30 pages are occupied by the proof of the following innocently looking lemma:\n\nLet $K$ be an origin-symmetric convex body in $\\mathbb R^3$. There exist three planes through the origin splitting $K$ into $8$ parts of equal volume and such that each two of these planes split the cross-section of $K$ by the third one into $4$ parts of equal area.\n\nI cannot shake off the feeling that there must be a half-page proof of this statement though I don't have one yet. I also know that MO is swarming with good topologists. Anybody up to the challenge?\n", "link": "https://mathoverflow.net/questions/271972/is-there-a-simpler-proof-of-the-key-lemma-in-the-paper-by-hiroshi-iriyeh-and-mas", "tags": ["gt.geometric-topology", "convex-geometry"], "votes": 19, "creation_date": "2017-06-11T18:05:22", "comments": ["it might be of interest: in the first part of the proof of Theorem 1 keithmball.files.wordpress.com/2014/11/… Ball constructs a similar map to simplex as Makeev does but he uses Browers Fixed Point Theorem (BFPT) unlike Makeev. So it quite might be that one needs to apply BFPT in a proper way which I don't see yet how.", "@fedja yes, the claim is indeed bit different, but the topological argument is still the same", "@FedorPetrov Except in our case the measures to partition depend on the planes. However the idea that if the solution is unique in some generic position, then there is an odd number of solutions in every generic position is amazing and certainly very promising. It should be rather standard, of course, but the good side of ignorance (mine) is the possibility to get surprised with the facts everybody else considers routine :-).", "(continued) Makeev partitions two measures which are absolutely continuous w.r.t. Lebesgue measure, but actually this absolute continuity is not essential.", "Roman Karasev (private communication) suggests that this essentially follows from the results of [V.V. Makeev. Equipartition of a continuous mass distribution. Journal of Mathematical Sciences, 140:4 (2007), 551--557, mathnet.ru/links/a1722f058ed63ed36cb13d9e67fd255f/znsl299.pd‌​f, theorems 5 and 6]", "Now one can hope that geometric considerations will show that (in this \"generic\" situation) the intersection class in $H^6((\\mathbb{RP}^2)^3,\\mathbb{Z}/2\\mathbb{Z})$ will be nontrivial, and that furthermore that there is no contribution from some degenerate case e.g. two planes coinciding. Then it probably still remains necessary to pass to some perturbation of $K$, and to run this argument there instead. So even if this sketch can be made to work, I doubt that a fully rigorous proof would fit in a half-page...", "Some naive thoughts: We can try and derive this from an intersection theory statement on $(\\mathbb{RP}^2)^3$. Points correspond to triples of planes, and the condition that three planes split $K$ into $8$ parts of equal volume will give a codimension $3$ cycle for \"generic\" $K$. (We get codimension $3$ instead of codimension $7$ because diametrically opposite regions have the same area, by reflection symmetry of $K$). Similarly each of the conditions that two planes split a cross-section into four parts of equal area should lead to a codimension $1$ cycle."], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "177", "site": "mathoverflow", "title": "Checking Mertens and the like in less than linear time or less than $\\sqrt{x}$ space", "body": "Say you want to check that $|\\sum_{n\\leq x} \\mu(n)|\\leq \\sqrt{x}$ for all $x\\leq X$. (I am actually interested in checking that $\\sum_{n\\leq x} \\mu(n)/n|\\leq c/\\sqrt{x}$, where $c$ is a constant, and in finding the best $c$ such that this is true in a given range, such as $3\\leq x\\leq X$, say; all of these problems are evidently related.)\n\nThe natural algorithm (the same you can find in section 4 of [1], say) has running time $O(X \\log \\log X)$ and takes space $O(\\sqrt{X})$ (where we think of integers as taking constant space). I've coded it with some optimizations (for $\\sum_{n\\leq x} \\mu(n)/n$, using interval arithmetic) and should have a result for $x=10^{14}$ in about a fortnight; $x=10^{12}$ takes an afternoon. (When I first coded this, less carefully and on worse hardware, it took a week.)\n\nIs it possible to do things in either less time or less space? Space is important here in practice - ideally you would want to keep everything in cache, and it is rare to have more than 4MB per processor core - that is, enough for $1.6\\cdot 10^7$ values of $\\mu$, or $5\\cdot 10^5$ large integers. \n\nNotice that I am asking for a check for all $x\\leq X$, and not just for a single $x=X$. \n\n* * *\n\nFurther remarks: I am aware that there are algorithms for computing a single value of $\\sum_{n\\leq x} \\mu(n)$ in time $O(x^{2/3} \\log \\log x)$ [2], or even (in what looks like a more sophisticated but less practical way that may have never been coded) in time $O(x^{1/2+\\epsilon})$ ([3]; see also [Mertens' function in time $O(\\sqrt x)$](https://mathoverflow.net/questions/95726/mertens-function-in-time-o-sqrt-x)). Once can of course use such an algorithm to compute the sum for values of $x$ that are about $c\\sqrt{x}$ apart, and then use what I've called the \"natural\" algorithm to deal with intervals in which such a computation shows that the statement to be verified could be violated. This results in a running time of $O(x^{7/6} \\log \\log x)$, or $O(x^{1+\\epsilon})$, and while memory usage could be decreased by applying the \"natural\" algorithm on intervals of length smaller than $\\sqrt{x}$, this would result in a longer running time.\n\nThe same is true, of course, of the \"natural\" algorithm itself: one could store a list of primes $p\\leq \\sqrt{x}$ in the main memory in $\\sqrt{x}$ bits, and compute $\\mu(m)$ in blocks of size $M$ at a time; this would require working in memory $O(M)$ (to be stored in the cache) for the most part, but the running time would be $O(x^{3/2}/M)$. Or is there a way around this? Can one, say, select the primes $p\\leq \\sqrt{x}$ that might divide integers in an interval of length $M \\lll \\sqrt{x}$, and do so in time less than $\\sqrt{x}$ or $\\sqrt{x}/\\log x$? Or is there any other way to take space substantially less than $O(\\sqrt{x})$ while still having essentially linear running time? Or a way to have better than linear running time?\n\nPS. I could also ask about parallelising this, but I would not like to risk making the discussion too hardware-specific here.\n\n[1] _François Dress_ , MR 1259423 [**Fonction sommatoire de la fonction de Möbius. I. Majorations expérimentales**](http://projecteuclid.org/euclid.em/1048516214), _Experiment. Math._ **2** (1993), no. 2, 89--98.\n\n[2] _Marc Deléglise and Joël Rivat_ , MR 1437219 [**Computing the summation of the Möbius function**](http://projecteuclid.org/euclid.em/1047565447), _Experiment. Math._ **5** (1996), no. 4, 291--295.\n\n[3] _J. C. Lagarias, and A. M. Odlyzko_ , MR 0890871 [**Computing $\\pi(x)$: an analytic method**](http://dx.doi.org/10.1016/0196-6774%2887%2990037-X), _J. Algorithms_ **8** (1987), no. 2, 173--191.\n", "link": "https://mathoverflow.net/questions/237308/checking-mertens-and-the-like-in-less-than-linear-time-or-less-than-sqrtx-s", "tags": ["nt.number-theory", "analytic-number-theory", "computational-complexity", "computation"], "votes": 19, "creation_date": "2016-04-25T12:22:32", "comments": ["Well, numbers of size about $X$ without prime factors between $X^(1/3)$ and $X^{1/2}$ are very roughly as common as the primes - but the sum of the reciprocals of the primes diverges. Or rather - the sum $\\sum_n 1/n$ over all $X\\leq n<2X$ without such prime factors is about a constant times $1/\\log X$, i.e., much larger than $1/\\sqrt{X}$. The same happens if you work over much smaller intervals. So I'm afraid this cannot work as stated.", "You might consider \"faulty\" versions of the natural algorithm, where one computes the wrong value on some of the values, or forgets to store the primes above cube root of X. You might still be able to show the error is small enough that the inequalities you want to check are preserved. Gerhard \"Faulty Computing Isn't Computationally Wrong\" Paseman, 2016.04.25.", "And yes, the running time is as you said and I said. I'm looking for something better.", "Gerhard: that's exactly what I mean by \"the natural algorithm\".", "the simplest algorithm for checking $M(x)$ not for every $x \\le X$ but for every $x = \\lfloor X/n\\rfloor$ is by using $\\sum_{n=1}^x M(x/n) = 1$, and computing $M(k)$ for every different $k \\in \\{ \\lfloor X/n \\rfloor \\}$, it takes $\\mathcal{O}(X)$ additions and $\\mathcal{O}(\\sqrt{X}\\log X)$ in space", "It may be that I am talking about the \"natural algorithm\" already, which is a variant of the one at mathoverflow.net/a/50691 for determining distinct prime factors of integers up to X in a serial fashion. However I still wonder what parameter you are talking about for space usage. Gerhard \"Space Usage: The Final Frontier\" Paseman, 2016.04.25.", "Do you mean $\\sqrt{x}$ or $\\sqrt{X}$? If the latter, there is a simple sieving process that computes the sum on the fly, and you can save extremal values for later processing and check that the inequality is satisfied. You can even sift just the squarefree values for processing. The memory storage is $O(\\log X)$ bits for each prime at most $\\sqrt{X}$. The runtime should be not much longer than $O(X\\log\\log X)$. Gerhard \"Sometimes Has Xtreme Case Sensitivity\" Paseman, 2016.04.25."], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "178", "site": "mathoverflow", "title": "Which manifolds decompose into pants?", "body": "In this [nice paper](http://arxiv.org/abs/math/0205011) Mikhalkin uses certain (more geometrical than algebraic) aspects of tropical geometry to prove that every complex projective hypersurface in $\\mathbb C \\mathbb P ^n$ decomposes as a smooth manifold into some _$(n-1)$-dimensional pair-of-pants_.\n\nFollowing Mikhalkin, a $k$-dimensional pair-of-pants is a particular real $(2k)$-manifold with corners. It is obtained by removing $k+2$ generic hyperplanes from $\\mathbb C \\mathbb P^k$. For instance, a $1$-dimensional pair-of-pants is a sphere minus 3 points (as everybody knows), while a $2$-dimensional pair-of-pants is $\\mathbb C \\mathbb P^2$ minus 4 generic lines. \n\nBoundaries and corners arise when you take the compact version of these pair-of-pants, i.e. you remove an open regular neighborhood of the $k+2$ generic hyperplanes: \n\n * When $k=1$ you get the genuine compact pair-of-pants $P$ with 3 boundary components. \n * When $k=2$ you get a real 4-manifold with boundary and corners: the boundary consists of four compact 3-manifolds homeomorphic to $P\\times S^1$ (corresponding to the four lines), glued together along six tori (corresponding to the six intersections between the lines). The tori are the corners. \n\n\nThe 4-dimensional manifold with corners we get with $k=2$ is a nice object, which may look intriguing to a low-dimensional topologist. \n\nAs everybody knows, not only complex hypersurfaces (i.e. curves) in $\\mathbb C \\mathbb P^2$ decompose into pair-of-pants: every oriented surface of negative Euler characteristic does! It would be then natural to ask the following:\n\n> Which compact 4-manifolds decompose into pair-of-pants?\n\nThe set of course includes all complex hypersurfaces in $\\mathbb C \\mathbb P^3$. \n\n**Edit:** It is not true that every curve in $\\mathbb C \\mathbb P^2$ decomposes into pants: a line or a cubic give $S^2$ and $T^2$ and they of course do not decompose into pants. Reading more carefully Mikhalkin's paper, it seems to me that he also allows to collapse the natural fibering of the boundaries of the blocks: for instance, by collapsing the boundaries of a two-dimensional pair-of-pants we get $S^2$. \n", "link": "https://mathoverflow.net/questions/113875/which-manifolds-decompose-into-pants", "tags": ["gt.geometric-topology", "tropical-geometry", "4-manifolds"], "votes": 19, "creation_date": "2012-11-19T13:53:05", "comments": ["I think that the following holds: let $X$ be obtained by gluing $n$ Pairs-of-Pants, and suppose that $X'$ is obtained by the same gluing rules, except that you add a Dehn twist in one of the gluing faces. Then, in analogy with the Lickorish-Wallace theorem, $X'$ should be obtained from $X$ by a single surgery along a (probably homologically essential, zero-square) torus.", "Thinking a bit more about your comment, I realized that Mikhalkin probably also admits the collapsing of the natural fibrations of the boundaries of the pants: the situation is slightly more complicated than I guessed, I have edited the text accordingly.", "Hi Marco. I would put no restriction at all (I can't see any reasonable restriction...). You glue along diffeomorphisms of blocks of type $P\\times S^1$: every such diffeomorphism must preserve the fiber, but there are infinitely many non-isotopic choices. You are absolutely right about $\\chi$, I must admit I didn't think about it. It might be that you get only finitely many manifolds for each $\\chi$, but since you might choose among infinitely many gluing maps between the $P\\times S^1$ faces, this is not clear (to me).", "Hi Bruno! What sort of restrictions do you put on the gluing maps between (not necessarily different?) Pairs-of-Pants? For example, how would you obtain $\\mathbb{CP}^2$ in this picture? Is it explained in Mikhalkin's paper? In any case, it looks like a necessary condition is to have $\\chi>0$, since your building blocks all have $\\chi=1$, and both the submanifolds you glue them along and the corners have $\\chi=0$. In particular, $\\chi$ of your manifolds seems to determine how many PoPs you need."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "179", "site": "mathoverflow", "title": "Can a number be palindromic in more than 3 consecutive number bases?", "body": "_$2017:$ Was initially asked on [MSE](https://math.stackexchange.com/questions/2234587/can-a-number-be-palindrome-in-4-consecutive-number-bases)_ \\- but wasn't solved or updated there since.\n\n**Update $2019$:** I've returned to this problem, made some progress and updated the post here. \n(I've basically rewritten this entire post here)\n\n* * *\n\n \n\n\n## **Introduction and problem**\n\nAn $k$-palindrome is a number palindromic in $k$ consecutive number bases. Here, I'm wondering about the existence of $k=4$. Note that if it does not exits, $k\\gt 4$ can't exist by definition.\n\nAlso note that if a number is even length (number of digits is even) palindrome in base $b$, then it is divisible by $b+1$ and thus can't be palindromic in that number base. So we can look for odd length cases of digits in the first (last, depending on how you define it) base only.\n\n> Can a natural number be _nontrivially_ [palindromic](https://en.wikipedia.org/wiki/Palindromic_number) in **more than** $3$ consecutive number bases? \n> \n> _Nontrivially_ means that I'm not counting one-digit palindromes.\n\nSmallest number $N$ which is nontrivially palindromic in $k$ consecutive number bases:\n\n$$ \\begin{array}{|c|} \\hline k& N & \\text{Palindrome} \\\\\\ \\hline 1& 3 & 11_2 \\\\\\ 2& 10 & 101_3=22_4\\\\\\ 3& 178 & 454_6 =343_7 = 262_8\\\\\\ 4& ?& \\\\\\ \\hline \\end{array} $$\n\nWhere the index denotes the number base representation. For example $11_2$ is three in binary.\n\nNote that $k=1$ is not special as those are just palindromes and easily can be constructed. \n\nIt is not hard to see $k=2,3$ have infinitely many examples. But finding all of them is hard.\n\n> I conjecture solution for $k=4$ does not exist. That is, there are no $4$-palindromes.\n\nMy **question** here is:\n\n> Can you provide arguments (help) or direction (on proving) why (why not) would this be true?\n\nMy ideas are presented below:\n\n \n**First approach to proving this**\n\nOne idea to prove this is proving $(1)$ and $(2)$ which would imply this. The $(1)$ is now proven. The $(2)$ will be much harder to prove. The details are included below. Here are the claims:\n\n * $(1)$ A $3$ digit (when written in palindromic bases) number can't be a $4$-palindrome.\n\n * $(2)$ To be a $4$-palindrome, a number must have $3$ digits (when written in palindromic bases).\n\n\n\n\nThe second claim is based on the fact that _almost-all_ \"almost $4$-palindromes\" have $3$ digits. That is, numbers palindromic in three out of four consecutive bases (that is, for example in bases $b,b+3$ and either $b+1$ or $b+2$). \n\n_Almost-all_ since the only two known \"almost $4$-palindromes\" that do not have $3$ digits are $71240, 1241507$ which have $5$ digits (when talking about digits, they are counted in the \"pivot\" number base among the consecutive bases in which the number is palindromic - \"pivot\" being either the smallest or largest base).\n\n> Proving $(2)$ is equally hard as proving $(3)$ from alternate approach below, since all cases of $d$ must be resolved. The $(1)$ specifically asked to resolve $d=3$ which was accomplished.\n\n \n**Second approach to proving this**\n\n> $(3)$ I actually conjecture I have found all $3$-palindromes. If this is true, then a $4$-palindrome does not exist since neither of my $k=3$ solutions can be extended to a fourth consecuitve number base.\n\nMore details are linked at the end of the next section:\n\n \n\n\n## **\" $k$-problem system\" results and conjectures**\n\nWhat I was able to prove both computationally and by hand:\n\n> Expanding $(1):$ In short, I have proven I have all $3$ digit $3$-palindromes, and nontrivial $d\\le 3$ solutions don't exist. Now it is easy to see now that a $3$ digit $3$-palindrome can't be palindromic in fourth consecutive base, which means $4$-palindromes with $d\\le 3$ digits do not exist. (Digits counted when written in palindromic number bases).\n> \n> More details about this are linked at the end of this section, as well.\n\nExpanding $(3):$ What I was able to conjecture is presented below:\n\nIn short, we can write the problem of finding $k$-palindromes in form of solving linear diophantine equations whose order grows when solving for longer palindromes of length $d$ (number of digits). Lets call that system needed to be solved \"$k$-problem system\".\n\nThere is the \"general\", and there is the \"non-general\" case of this problem system.\n\nThe non-general $k$-problem system is related to finding $d=2l+1$ (WLOG $d$ is odd) digit solutions, where all representations in consecutive bases have exactly $d$ digits. \n\nThe general $k$-problem system allows different lengths of digits in those $k$ consecutive number bases, where \"pivot\" number base among these $k$ consecutive number bases, usually smallest or largest, is taken to have $d$ digits.\n\nThe $k=2$ case has infinitely many solutions for every case of $d$. It is hard to find them all.\n\nThe $k=3$ is what will be the case of the \"problem-system\" from now on:\n\n> Expanded on $(3)$ in this context:\n> \n> For the non-general case of the \"problem-system\", I have computationally solved (proved) I have all solutions for two smallest cases, $d=3,5$. I also strongly conjecture same is true for $d=7$. I also conjecture based only on computation, that a $3$-palindrome solution for $d\\ge 9$ does not exist. That is, I conjecture I have all solutions.\n> \n> For the general case of the \"problem-system\", I have computationally solved (proved) I have all solutions for smallest case, $d=3$. I also conjecture I have all solutions for $d\\ge 5$ since another conjecture I strongly believe is that the general case is equivalent to the non-general case, except for having an extra finite set of solutions for sufficiently small bases $b$ which I have collected and computed all (conjectured).\n\n* * *\n\n_Summing this all up_ : the important thing is I have all $3$ digit $3$-palindromes and that they can't be $4$-palindromes. I have all conjectured solutions for $3$-palindromes, but unable to prove the fact that large $d$ can't have solutions for $k=3$.\n\n* * *\n\nNote that I was trying to find counterexamples to these conjectures for a long time now but couldn't.\n\nThe more details as promised: A post mainly focusing on non-general \"problem-system\" [can be read on MSE](https://math.stackexchange.com/questions/3305696/hard-system-in-integers-related-to-natural-number-representations) if you are interested in more details, or if you want to see the \"all $3$-palindromes\" solutions for the non-general case (the extra solutions from general case are easily re-computable if my conjectures are true).\n\nI'll also include all _infinite families_ (the non-sporadic solutions) for $k=3$ given there, here:\n\nGiven $k\\in\\mathbb N_0 = \\mathbb N \\cup \\\\{0\\\\}$, \"infinite families\" giving $3$-palindromes:\n\n$$ \\begin{array}{l,l,l,l} d & (a_i)&(c_i)k & b\\\\\\ d=3 & (2,6)&(1,1)k & 2k+8\\\\\\ d=5 & (31,32,0)&(3,2,1)k & 4k+47 \\\\\\ d=7 & (34,50,10,74)&(1,1,1,1)k & 2k+76 \\\\\\ d=7 & (8,33,0,41)&(1,3,1,3)k & 6k+58 \\\\\\ d=7 & (112,15,0,36)&(4,0,1,0)k & 6k+175 \\\\\\ d=7 & (227,160,187,200)&(5,3,5,3)k & 6k+280 \\\\\\ d=7 & (5,23,6,14)&(2,6,5,0)k & 12k+39 \\\\\\ d=7 & (93,78,30,50)&(10,6,7,0)k & 12k+119 \\\\\\ d=7 & (47,150,249,26)&(2,6,11,0)k & 12k+291 \\\\\\ \\end{array} $$\n\nWhere these give digits of $3$-palindromes in base $b$, which are also palindromic in $b-1,b-2$.\n\nThat is, for example, observing the second row (family), the only $d=5$ case infinite family, means $(31+3k,32+2k,0+1k,32+2k,31+3k)$ are digits of a $3$-palindrome in base $4k+47$, for every $k$. We can convert this to a decimal value easily.\n\n> $(3):$ If we can prove these are all of the solutions for $k=3$ (and that sporadic solutions exist only for sufficiently small bases as I conjectured), we have solved the $k=4$ problem, the $4$-palindrome problem: There does not exist a number palindromic in four consecutive number bases.\n\nThe main thing needing proving is that $d\\ge 9$ can't have solutions. Or can you find a counterexample, and actual solution? That would make this much more interesting.\n\nThe $d=9$ specifically for example will be very surprising if it had a solution (I found the period of solutions if they exist, must be larger than $500$ in this case, which is unlikely given $d=3,5$ have periods $2,4$ and $d=7$ has periods $2,6,12$). Period being families producing a solution every period amount of bases.\n", "link": "https://mathoverflow.net/questions/268590/can-a-number-be-palindromic-in-more-than-3-consecutive-number-bases", "tags": ["nt.number-theory", "diophantine-equations", "palindromes"], "votes": 19, "creation_date": "2017-04-29T08:32:32", "comments": ["@GerryMyerson Regarding a relevant sequence, I think A279093 is updated with all known 3-palindromic solutions.", "Loosely related: oeis.org/A214425", "@joro Update: Now it is known that the $3$ digit solution does not exist. I will have to update this post.", "@JoseBrox If we have $n$ and $n+3$, and we also include either $n+1$ or $n+2$ and \"observe almost palindromic in four consecutive\", then surprisingly, only 3 digit palindromes seem to have a chance to be palindromic in four consecutive bases ~ Updated the question", "@Vepir Yes, sorry, I meant $n$ and $n+3$. I was just trying to understand if the obstruction for the existence of consecutive palindromes for $k=4$ (in case it exists) is of an algebraic/number-theoretic nature or rather of an analytic/combinatoric nature (which is what the existence of examples for $n$ and $n+3$ suggests).", "@JoseBrox Why $n, n+4$? There are a lot of examples, smallest one being $7=111_2=11_6$. You can easily modify the linked python code to generate them. If you meant $n, n+3$ there are again a lot of examples, smallest being $31=11111_2=111_5$. I only found interesting the palindromes which share consecutive bases. Smallest example for $k=1,2,3,4,\\dots$ consecutive bases: $3(=11_2), 10(=11_3=101_4),178(= 454_6 =343_7 = 262_8), ? (k=4)$... The solution for four is either very large or does not exist.", "@Vepir Do you know of any example of palindrome simultaneously for bases $n$ and $n+4$?", "@joro: That doesn't seem to be proven. At MSE, Vepir searched up to $10^7$ and found a one parameter family which does not extend to $4$ bases, and a single sporadic example, but there is no proof these are the only $3$ digit solutions.", "Is it known that three digit solution doesn't exist?", "This is perfectly tweetable mathematics and thus certainly not offtopic on mathovertweet. I vote Reopen", "@Andy, the question was asked at m.se two weeks ago. If it hasn't been answered to Vepir's satisfaction, no reason why it shouldn't be posted here. And no reason why anyone here has to accept Vepir's suggestion to answer there. I'm sure that if someone posts an answer here, Vepir will survive the trauma.", "I voted to close. MO is not intended to advertise questions on math.se, so if you want answers there, you should just ask there."], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "180", "site": "mathoverflow", "title": "A Linear Order from AP Calculus", "body": "In teaching my calculus students about limits and function domination, we ran into the class of functions\n\n$$\\Theta=\\\\{x^\\alpha (\\ln{x})^\\beta\\\\}_{(\\alpha,\\beta)\\in\\mathbb{R}^2}$$\n\nSuppose we say that $g$ weakly dominates $f$, and write $f\\preceq g$, if\n\n$$\\lim_{x\\to\\infty}\\frac{f(x)}{g(x)} \\hspace{3 mm} \\text{is finite}$$\n\nWe can then readily see that $(\\Theta,\\preceq)$ is a total order isomorphic to the lexicographic order on $\\mathbb{R}^2$.\n\nBut we can get more complicated total orders with, say\n\n$$\\Theta_n=(x^{\\alpha_0}(\\ln{x})^{\\alpha_1}(\\ln\\ln{x})^{\\alpha_2}\\cdots(\\ln^{n-1} x)^{\\alpha_{n-1}})_{\\vec{\\alpha}\\in\\mathbb{R}^{n}}$$\n\n$$\\Phi=\\\\{e^{p(x)}\\\\}_{p(x)\\in\\mathbb{R}[x]}$$\n\nwhich are isomorphic as total orders to the lexicographic orders on $\\mathbb{R}^n$ and $\\operatorname{List}\\mathbb{R}$\n\nAll of these complicated orders live inside what I'd call \"the AP Calc linear order\" $(\\Omega,\\preceq)$ defined as:\n\n$$\\Omega_0=\\\\{f\\in\\mathscr{C}^0((\\lambda,\\infty))\\\\}_{\\lambda\\in\\mathbb{R}}$$\n\n$$f\\preceq g\\Longleftrightarrow \\max\\left\\\\{\\left|\\liminf_{x\\to\\infty} \\frac{f(x)}{g(x)}\\right|,\\left|\\limsup_{x\\to\\infty} \\frac{f(x)}{g(x)}\\right|\\right\\\\}<\\infty$$\n\n$$\\Omega=\\Omega_0/\\simeq \\hspace{5 mm} \\text{where} \\hspace{5 mm} f\\simeq g \\Leftrightarrow \\left[f\\preceq g \\text{ and } g\\preceq f\\right]$$\n\nwhere the refinement on $\\preceq$ is made so as to avoid problems with things like $\\sin{x}$.\n\nThis seems to be a very complicated linear order, as it includes as a suborder things like\n\n$$\\Psi=\\\\{p_0(x)e^{p_1(x)}e^{e^{p_2(x)}}\\cdots\\exp^{n-1}(p_{n-1}(x))\\\\}_{p_i(x)\\in\\mathbb{R}[x]\\forall i}$$\n\nMy question is the following: is there any combinatorial description or universal construction, i.e. as a colimit, of the isomorphism type of $(\\Omega,\\preceq)$?\n", "link": "https://mathoverflow.net/questions/215798/a-linear-order-from-ap-calculus", "tags": ["co.combinatorics", "ct.category-theory", "real-analysis", "order-theory", "linear-orders"], "votes": 19, "creation_date": "2015-08-27T09:12:46", "comments": ["This is essentially just a question involving Landau notation, no?", "With Anthony Quas, I suggest you look up \"Hardy field\", and also \"o-minimal structure\" whose germs at infinity produce Hardy fields. The question as it stands suffers from problems noted by Eric Wofsey.", "@DmitryV: That's not what I was talking about (I was also thinking like a category theorist and not caring about that). If $f/g$ oscillates between approaching $0$ and approaching $\\infty$, then $f\\not\\leq g$ and $g\\not\\leq f$. Actually, if $f$ and $g$ don't have to be positive, you could just have an unbounded set where $f$ is zero and $g$ is not and an unbounded set where $g$ is zero and $f$ is not.", "@EricWofsey you're right, I was being too much a category theorist and forgetting to mod out by isomorphisms, turning this pre-order into a total order. I'll edit it now.", "A side comment: The class $\\Theta$ is a very small sub-class of the logarithmico-exponential functions studied by Hardy (see also Boshernitzan's work). These have very nice properties including no oscillation: for any two distinct functions in a Hardy Field, eventually $f$ is bigger than $g$ or vice versa.", "This is not a total order; $f$ and $g$ can oscillate between $f\\gg g$ and $g\\gg f$.", "Related mathoverflow.net/questions/29624/…"], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "181", "site": "mathoverflow", "title": "A question in Fontaine--Laffaille theory", "body": "Let $K$ be finite unramified extension of $\\mathbf{Q}_p$ with ring of integers $W$. Let ${\\rm MF}$ be the category of strongly divisible $W$-modules $M$ with ${\\rm Fil}^0M=M$ and ${\\rm Fil}^{p-1}M=0$. Let ${\\rm Crys}(G_K)$ be the category of finite free $\\mathbf{Z_p}$-representations of $G_K$ which become crystalline over $\\mathbf{Q}_p$. There is a functor $$A:{\\rm MF} \\to {\\rm Crys}(G_K), \\qquad M \\mapsto {\\rm Hom}(M, A_{\\rm crys}),$$ where the ${\\rm Hom}'s$ are required to preserve the filtration and the Frobenius action, which is fully faithful by Fontaine-Laffaille theory.\n\nGiven $T$ in the image of the above functor, one can construct a finite free $W$-module with filtration and Frobenius, namely $B(T)={\\rm Hom}_{G_K}(T, A_{\\rm crys})$. Note that $B(A(M))[1/p]$ is isomorphic to $M[1/p]$ by the theory of crystalline representations. I believe that $B$ is not the inverse to $A$ in general, however. \n\nQuestions: 1) What is an example of a $T$ for which $B(T)$ is not the strongly divisible module associated to $T$? 2) For which $T$ is $B(T)$ the strongly divisible lattice associated to $T$?\n", "link": "https://mathoverflow.net/questions/60134/a-question-in-fontaine-laffaille-theory", "tags": ["nt.number-theory", "galois-representations"], "votes": 19, "creation_date": "2011-03-30T15:31:30", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "182", "site": "mathoverflow", "title": "The oriented homeomorphism problem for Haken 3-manifolds", "body": "Haken famously described an algorithm to solve the homeomorphism problem for [the 3-manifolds that bear his name](http://en.wikipedia.org/wiki/Haken_manifold) (fleshed out by many others, including Hemion and Matveev who fixed some gaps). But it's often natural to consider 3-manifolds _equipped with an orientation_ , and so my question is:\n\n> **Question 1:** Is there an algorithm _in the literature_ that determines whether or not a pair of oriented Haken 3-manifolds M and N are _orientedly_ homeomorphic?\n\nBy Haken's Algorithm, we can reduce to the case in which M and N are (unorientedly) homeomorphic, and so the question can be reduced to:\n\n> **Question 2:** Is there an algorithm _in the literature_ that determines whether or not an orientable Haken 3-manifold M admits an orientation-reversing self-homeomorphism?\n\n**Notes:**\n\n 1. Note the words 'in the literature'. I don't really doubt that this algorithm exists, but the details seem quite complicated, particularly in the non-geometric case, so I hope that someone will be able to supply a reference. A very short argument would also be nice. Answers of the form 'I don't know, but I'm sure it can be done if you work out each geometric case individually and then think hard enough about the JSJ decomposition', while appreciated, would be less useful.\n\n 2. The hyperbolic case can be deduced from Sela's solution to the homeomorphism problem, which also computes the automorphism group. I've no doubt the other non-Haken cases can be handled similarly. Hence my interest in the Haken case.\n\n 3. It may be that a simple modification of Haken's Algorithm handles the oriented case. This isn't at all clear to me, but I would be very interested to hear if it's true.\n\n 4. **(added later)** I would also be happy to confirm that this question is `open' (in the sense that an answer is not in the literature, rather than that the answer is in doubt)---I very much suspect that this is the case. An authoritative pronouncement by someone who knows the literature very well would therefore be a valid answer.\n\n\n\n", "link": "https://mathoverflow.net/questions/107573/the-oriented-homeomorphism-problem-for-haken-3-manifolds", "tags": ["3-manifolds", "gt.geometric-topology", "reference-request", "open-problems"], "votes": 19, "creation_date": "2012-09-19T07:37:20", "comments": ["That's a very nice observation, Ian!", "Certain special cases can be deduced from the literature. If you look at Theorem 6.1.6 in Matveev's book, it says that there is an algorithm to tell if two Haken manifolds with boundary pattern are homeomorphic taking boundary pattern to boundary pattern. If you have two oriented knot complements, then you can tell if they are orientation preserving homeomorphic by choosing a boundary pattern which is not preserved under mirror image (such as a single closed (1,1) curve). However, this is also appealing to the knot complement problem. springerlink.com/content/978-3-540-45898-2", "Actually, Ian, this is precisely my motivation for asking the question. As far as I can tell, the treatments of the homeomorphism problem for reducible manifolds in the literature miss this point.", "If you connect sum with a manifold that doesn't admit an orientation reversing isometry, then check homeomorphism, then this checks if the manifolds are orientation preserving homeomorphic. However, I'm not sure if the homeo. problem for reducible (or even connect sums of Haken) manifolds is in the literature explicitly. "], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "183", "site": "mathoverflow", "title": "Reference request: Parallel processor theorem of William Thurston", "body": "Sometime in the 1980's or 1990's, Bill Thurston proved a theorem regarding the existence of a universal parallel processing machine, using a certain class for such machines having finite deterministic automata at each node. There is no such published paper on MathSciNet, and I have no record of this theorem other than my own faulty memory. \n\nI'd like to ask whether there exists a preprint somewhere, or some other reference.\n\nAlthough I cannot quite reproduce the theorem's exact statement, here's what I roughly remember. Caveat: please take this all with a grain or three of salt, because I am sure my memory has gaps and/or errors.\n\nThe parallel processors of this class are built on an underlying framework consisting of a connected graph with finite valence at each vertex. At each node of this graph one places a finite deterministic automaton. This automaton has some special capabilities. After each computation, it can send one of finitely many outputs along each adjacent edge. Also, before each computation it can receive inputs along each adjacent edge (which have previously been sent by adjacent automata, or have been set up as part of an initialization) and it uses those inputs together with the current state to compute its next state and its next outputs. \n\nI believe there are strong regularity constraints. The graph should have an automorphism group which is transitive on vertices. In fact the entire setup, including the automata themselves and how their inputs and outputs are associated to the adjacent edges, should have an automorphism group that is transitive on vertices. The reason for these constraints is so that the entire network can be \"finitely described\" or \"finitely programmed\", and perhaps that is the true constraint, however it may be formalized.\n\nOne programs or initializes the processor by choosing initial states and initial inputs. I believe there must be some finiteness condition, for example outside of some bounded region the initial states and initial inputs are perhaps in some default position. Then one lets the processor run until some point (I don't remember the halting conditions), at which time one reads off the terminal states and outputs.\n\nI remember Thurston proved two theorems about universal machines in this class (subject to some restrictions about how to encode or program one machine within another, which I totally do not remember; but this argues for some strong regularity constraints in the class of machines).\n\nOne theorem is that a processor of this type whose underlying graph is the Cayley graph of a free group cannot be universal. Roughly speaking the reason one might expect this to be a universal processor is because of exponential growth of nodes, but the trouble is that there is too much bottlenecking at each node.\n\nThe main theorem is that there exists a processor of this type whose underlying graph is any Cayley graph of a lattice in the 3-dimensional solvable Lie group. For example, one such lattice is the Cayley graph of the semidirect product $$\\mathbb{Z}^2 \\rtimes_M \\mathbb{Z} $$ using the action of $\\mathbb{Z}$ on $\\mathbb{Z}^2$ generated by the matrix $M = \\pmatrix{2 & 1 \\\\\\ 1 & 1}$. Here the rough idea is that one still has exponential growth of nodes, but the bottlenecking phemonenon has been avoided.\n\n**Added:** I'm convinced now that so far this is just a description of computable functions in the ordinary sense. But this jogs a really dim memory that what's missing is a way to constrain or measure how the programming process is allowed to compress information. I'm still hoping that someone will have a more accurate record of this theorem.\n", "link": "https://mathoverflow.net/questions/213181/reference-request-parallel-processor-theorem-of-william-thurston", "tags": ["reference-request", "graph-theory", "computational-complexity", "computer-science", "geometric-group-theory"], "votes": 19, "creation_date": "2015-08-06T08:29:11", "comments": ["Maybe you could try theoretical computer science, cstheory.stackexchange.com", "@GerryMyerson: No, it's not in that paper, although the issues of mesh partitions discussed in that paper remind me of the \"bottlenecking\" issues that are at the heart of this parallel processing model I'm asking about, and this may have been around the same time as well.", "@IgorRivin: I did ask Matt, and although he remembers working on somewhat similar problems around the same (vague) time, this particular theorem is not something he remembers.", "@GerryMyerson: I wondered the same, but it seems far from Cayley graphs and solvable Lie groups...", "Is it possible we are talking about the paper, Miller, Teng, Thurston, and Vavasis, Automatic mesh partitioning, available at cs.cmu.edu/~glmiller/Publications/Papers/MiTeThVa93.pdf ?", "@Algernon: You might be right, I could be overlooking some kind of space-or-time compression restrictions. But I cannot remember.", "My recollection that this is actually due to Matt Grayson, so you might want to ask him...", "Your model is what people call a cellular automaton. It is easy to simulate a Turing machine with a \"one-dimensional\" cellular automaton (i.e., the underlying graph is a bi-infinite path, a Cayley graph of the one-generator free group!), with reasonable finiteness constraint. On a free group with a larger generator set, you can simply ignore all but one of the generators and simulate the Turing machine along the remaining generator. I suspect your memory is putting the emphasis of whatever Thurston's result was on the wrong place."], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "184", "site": "mathoverflow", "title": "Coarse moduli spaces of stacks for which every atlas is a scheme", "body": "Let $X = [P/G]$ be a smooth finite type separated DM-stack over $\\mathbb C$ given as the quotient of a smooth quasi-projective scheme $P$ by the action of a smooth (finite type separated) reductive group scheme $G$. Since $X$ has finite inertia, the coarse space $X^c$ of $X$ exists as an algebraic space.\n\nAt _atlas_ of $X$ is an étale morphism $U\\to X$ with $U$ an algebraic space.\n\nI sm interested in the converse of the following statement:\n\n**Thm.** If the coarse moduli space of $X$ is a scheme,then every atlas $U$ of $X$ is a scheme.\n\n_Proof._ The map from an atlas $U$ to the coarse moduli space $X^c$ of $X$ is quasi-finite separated. Therefore $U$ is a scheme by Knutson Cor. II.6.16., p. 138 QED\n\nSo the converse would read as follows:\n\n**Q.** Suppose that every atlas $U$ of $X$ is a scheme. Is the coarse moduli space of $X$ a scheme?\n\nI expect the answer to be negative, but can't find a good example. Note that a counterexample can not be an algebraic space, as an algebraic space with the property that every atlas is a scheme is itself a scheme (the identity morphism being an atlas).\n\nEdit: I have an application in mind of the above question, and in the context of my application the stack $X$ is even generically a scheme. Not sure if that helps...\n", "link": "https://mathoverflow.net/questions/186379/coarse-moduli-spaces-of-stacks-for-which-every-atlas-is-a-scheme", "tags": ["ag.algebraic-geometry", "complex-geometry", "stacks", "algebraic-stacks", "algebraic-spaces"], "votes": 19, "creation_date": "2014-11-06T08:42:34", "comments": ["Yes it looks like the example from Corollary 6 does the job! After taking a quick look at the proof, it seems like the condition that the coarse space is a scheme should be equivalent to something like the following: for each point $x \\in P$, there exists a linearized ample line bundle $L$ such that $x$ is $L$-semistable. I'm not sure how this interacts with your atlas condition though.", "@DoriBejleri I think (but I'd have to double-check) that Kollar's paper arxiv.org/pdf/math/0501294.pdf contains such examples. Do you agree? (BTW, this question is seven years old, and I somehow decided to revive it now, but I don't even remember the \"application in mind\" at the moment.)", "Do you have an example, without assuming your condition on the atlases, where $P$ is quasi-projective and $G$ is reductive, $[P/G]$ is separated DM, but the coarse space of $[P/G]$ is not a scheme?"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "185", "site": "mathoverflow", "title": "The cofinality of $(\\mathbb{N}^\\kappa,\\le)$ for uncountable $\\kappa$?", "body": "For a partially ordered set $P$, a set $A\\subseteq P$ is _cofinal_ if for each element of $P$ there is a larger element in $A$. The _cofinality_ of $P$, ${\\rm cof}(P)$, is the minimal cardinality of a cofinal family in $P$.\n\nLet $\\kappa$ be an infinite cardinal number. Consider the set $\\mathbb{N}^\\kappa$ of all functions from $\\kappa$ to $\\mathbb{N}$ (equivalently, $\\kappa$-sequences of natural numbers), with the pointwise partial order: $f\\le g$ if $f(\\alpha)\\le g(\\alpha)$ for all $\\alpha<\\kappa$.\n\nLet $$f(\\kappa):={\\rm cof}(\\mathbb{N}^\\kappa,\\le).$$ The cardinal $f(\\aleph_0)$ is the well-understood dominating number $\\mathfrak{d}$, which is consistently any cardinal of uncountable cofinality that is not larger than the continuum (see [Blass](http://www.math.lsa.umich.edu/~ablass/hbk.pdf)).\n\nBasic facts include:\n\n 1. For cardinal numbers $\\kappa\\le\\mu$, we have that $f(\\kappa)\\le f(\\mu)$: A projection of a cofinal family in $\\mathbb{N}^\\mu$ on the first $\\kappa$ corrdinates is cofinal in $\\mathbb{N}^\\kappa$.\n 2. $\\kappa0$ and odd, and $f=x+x^9+x^{25}+x^{49}+\\cdots$. For $g$ in $M$, let $S(g)$ consist of the primes, $p$, for which the coefficient of $x^p$ in $g$ is 1. Note that each $p$ in $S(f^k)$ is congruent to $k$ mod 8.\n\nT1.----- If $k=3 {\\rm\\ or\\ } 5$, $S(f^k)$ consists of the $p$ that are $k$ mod 8\n\nT2.----- $S(f^7)$ consists of the $p$ that are 7 mod 16\n\nT3.----- If $k=19 {\\rm\\ or\\ } 21$, then $S(f^k)$ consists of the $p$ that are $k$ or $k+8$ mod 32.\n\nTo prove T1 when $k=3$, we write $f^k$ as $f*f^2$ and use the fact that if $p$ is 3 mod 8, then $p$ is uniquely the sum of a square and twice a square. When $k=5$ we argue similarly using Fermat's two square theorem.\n\nAs I indicated in a comment on a recent MO question of Joel Bellaiche, \"[Primes and x^2+2y^2+4z^2](https://mathoverflow.net/questions/100701/primes-and-x22y24z2)\" ,T2 follows from a result of Hasse on the class number of $Q(\\sqrt{-2p})$, using Gauss' theorem that the number of representations of $2p$ as a sum of 3 squares is 12*(this class number). Hasse's proof is an application of the Gauss theory of genera and ambiguous forms.\n\nT3 is thornier. Because $f$ is the mod 2 reduction of (the Fourier expansion of) the normalized weight 12 cusp form for the full modular group, each $g$ is the mod 2 reduction of a modular form of integral weight. A profound result of Deligne, relating Hecke eigenforms to Galois representations, then shows that $S(g)$ is a \"Frobenian set\". Nicolas, Serre and Bellaiche, continuing in this vein, developed a theory of level 1 modular forms in characteristic 2 that led to more precise results. Their investigations motivated me to try to determine $S(f^k)$ empirically for small $k$, and I was led to conjecture T3. Joel then applied his methods to give a proof. But this is very hard, and so I ask:\n\nQUESTION\n\nDoes there exist an \"elementary proof\" of T3, using the theory of binary quadratic forms, along the lines of the Hasse-Gauss argument?\n\nEDIT: Motivated by my recent simple proof of T2 (see my answer to the question of Joel cited above), I've found arguments that ought to reduce the proof of T3 to Sage calculations. The point is that forms of weight 2 are easier to deal with than forms of weight 3/2, so one should work with quadratic forms in 4 variables rather than in 3, even when the genera that arise have more than 1 class in them. Here's the idea of my argument for f21.\n\nLet p be a prime that is 5 mod 8. Writing $f^{21}$ as $(f)(f^2)(f^2)(f^{16})$ we find that if $R$ is (1/16)*(the number of representations of $p$ by $G_1=x^2+2y^2+2z^2+16t^2$ with $x,y,z$ and $t$ all odd), then $p$ is in $S(f^21)$ if and only if $R$ is odd. Now since $p$ is 5 mod 8, in any representation of $p$ by $G_1$, $x,y$ and $z$ must be odd. So if we set $G_2=x^2+2y^2+2z^2+64t^2$ then $R=(N1-N2)/16$, where $N_1$ and $N_2$ are the numbers of representations of $p$ by $G_1$ and $G_2$ respectively. Now write $p$ as $a^2+4b^2$ with $a$ and $b$ congruent to 1 mod 4. Computer calculations indicate:\n\nConjecture 1. $N_1=p+1+2a$\n\nConjecture 2. $N_2=((p+1)/2)+a+4b$\n\nIf these conjectures hold then $R=(p+1+2a-8b)/32$. The numerator here is $4(b-1)^2 +(a+3)(a-1)$, which mod 64 is $(a+3)(a-1)$. So $R$ has the same parity as $(a+3)(a-1)/32$ and is odd just when $a$ is $5$ or $9$ mod $16$. Now mod $32$, $p=a^2+4$. So $R$ is odd just when $p$ is $29$ or $85$ mod $32$, and so the conjectures imply Joel's result for $S(f^21)$.\n\nHow does one attack the conjectures? The theta series attached to $G_1$ and $G_2$ are modular forms for $\\Gamma_0 (64)$ and $\\Gamma_0 (256)$ respectively. If the conjectures are to hold it seems that each of these theta series should be a linear combination of Eisenstein series and cusp forms attached to Grossencharaktere for $\\mathbb Q(i)$. It should be possible, using Sage, to get an explicit formulation of this, and prove the conjectures.\n\nMy proposed treatment of $S(f^{19})$ is entirely similar. Suppose $p$ is $3$ mod $8$. Writing $f^{19}$ as $(f)(f)(f)(f^{16})$ and arguing as above we find that if we take $H_1$ and $H_2$ to be $x^2+y^2+z^2+16t^2$ and $x^2+y^2+z^2+64t^2$ respectively, and let $N_1$ and $N_2$ be the number of representations of $p$ by $H_1$ and $H_2$, then $p$ is in $S(f^{19})$ just when $R=(N_1-N_2)/16$ is odd. Now Jacobi's 4 square theorem, (see the argument in my answer to Joel's question), shows that $N_1$ is $2(p+1)$. Write $p$ as $a^2+2b^2$ with $a =1$ or 3 mod 8. The computer suggests:\n\nConjecture 3. $N_2=p+1+4a$\n\nSo if the conjecture holds, $R=(p+1-4a)/16$, and one sees easily that this is odd just when $p$ is 19 or 27 mod 32. Once again the theta series attached to H2 is a modular form for $\\Gamma_0 (256)$. The conjecture indicates that it should be a linear combination of Eisenstein series and cusp forms attached to Grossencharaktere for $\\mathbb Q(\\sqrt{-2})$; all this should admit a proof using Sage.\n", "link": "https://mathoverflow.net/questions/106267/does-this-variant-of-a-theorem-of-hasse-really-due-to-gauss-have-an-elementar", "tags": ["nt.number-theory", "quadratic-forms", "modular-forms", "power-series"], "votes": 19, "creation_date": "2012-09-03T12:27:05", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "189", "site": "mathoverflow", "title": "Is there some way to see a Hilbert space as a C-enriched category?", "body": "The inner product of vectors in a Hilbert space has many properties in common with a hom functor. I know that one can make a projectivized Hilbert space into a metric space with the Fubini-Study metric and then follow Lawvere in considering it a $[0,\\infty)$-enriched category; it's particularly nice because the distance between two vectors is a function of the inner product.\n\nIs there a way of defining a monoidal structure on $\\mathbb{C}$ such that the inner product is the hom so that we can think of a (perhaps projectivized) Hilbert space as $\\mathbb{C}$-enriched?\n", "link": "https://mathoverflow.net/questions/47644/is-there-some-way-to-see-a-hilbert-space-as-a-c-enriched-category", "tags": ["hilbert-spaces", "ct.category-theory", "enriched-category-theory"], "votes": 19, "creation_date": "2010-11-28T21:30:13", "comments": ["You might also be interested in John Baez's paper on the subject: arxiv.org/abs/q-alg/9609018", "Related question: mathoverflow.net/questions/476/…"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "190", "site": "mathoverflow", "title": "are there high-dimensional knots with non-trivial normal bundle?", "body": "Does there exist a smooth embedding $\\varphi\\colon S^k\\to S^n$ such that $\\varphi(S^k)$ has non-trivial normal bundle? I looked at some of the old papers by Kervaire, Haefliger, Massey, Levine but I couldn't find an answer. There are various related results, for example it is shown by Kervaire et al that in many dimensions the normal bundle is trivial. Furthermore Kervaire showed that there exists an immersion, such that the pullback of the normal bundle is non-trivial.\n", "link": "https://mathoverflow.net/questions/382047/are-there-high-dimensional-knots-with-non-trivial-normal-bundle", "tags": ["at.algebraic-topology", "gt.geometric-topology"], "votes": 19, "creation_date": "2021-01-24T03:39:42", "comments": ["@StefanFriedl Thanks for the corrections.", "@archipelago thanks, you put me on the right track!", "Namely Danny Ruberman writes: \"Jerry Levine's paper \"A classification of differentiable knots\" (Annals 82 (1965) 15-50) determines (see proposition 6.2) the possible normal bundles in some range of codimensions in terms of maps in various exact sequences. It's hard to summarize the results, but he gives a table at the end dealing with relatively low dimensions, where calculations can be made explicit. The case n=11 and k=6 contains Haefliger's example; in fact there are exactly 5 possible normal bundles among all embeddings.\"", "more googling gives an answer in mathoverflow.net/questions/237384/….", "@archipelago but for this conversation it's more important that by accident you misquoted their comment, they write \"since Haefliger (unpublished) has shown that S^11 embeds in R^17 with a +++++non-trivial++++ normal bundle.\" That's a very good find and answers my question with a shaky reference. It would be great if there was a reference with details", "@archipelago Since Levine was my advisor I should point out that it's a paper by Hsiang, Levine and Szcarba", "@archipelago: Of course, thanks for pointing that out.", "@archipelago thanks for your comments! But from you and I unearthed so far in the literature it could in principle be that the normal bundle is always trivial, isn't it? You'd think that this question was answered in the 60's.", "According to Hsiang, Lee, and Szczarba, it is an unpublished result of Haefliger that $S^{11}$ embedds into $S^{17}$ with trivial normal bundle (see 'On the normal bundle of a homotopy sphere embedded in Euclidean space'). This would show that Kervaire's bound is sharp.", "@MichaelAlbanese Any normal bundle arising from such an embedding is stably trivial, since the tangent bundle of $S^k$ is.", "Do you know if Kervaire's example $\\nu$ is stably trivial? Maybe I'm oversimplifying, but if you have an immersion $f : S^k \\to S^n$ with $k < n$, then it factorises as $i_n\\circ f_0$ where $i_n : \\mathbb{R}^n \\to S^n$ is the standard inclusion and $f_0 : S^k \\to \\mathbb{R}^n$. The map $h : S^k \\to S^{n+k+1}$ given by $i_{n+k+1}\\circ(f_0, \\iota)$ is an embedding where $\\iota: S^k \\to \\mathbb{R}^{k+1}$ is the standard embedding. If I'm not mistaken, the normal bundle of $h$ is $\\nu\\oplus\\varepsilon^{k+1}$ which is trivial if and only if $\\nu$ is stably trivial.", "Using this, one sees that all embeddings with $k\\le 6$ have trivial normal bundle. I believe it is an open question whether there is an embedding with nontrivial normal bundle for k=7 and n=11.", "If k<2(n-k)-1 then the normal bundle is trivial by a result of Kervaire (Theorem 8.2 in his \"An Interpretation of G. Whitehead's Generalization of H. Hopf's Invariant\"). If n=k+1,k+2,k+3 it is trivial as well."], "comment_count": 13, "category": "Science", "diamond": 0} {"question_id": "191", "site": "mathoverflow", "title": "About the equivariant analogue of $G_n/O_n$", "body": "Let $BO_n$ and $BG_n$ be the classifying spaces for rank $n$ vector bundles and for spherical fibrations with fiber $S^{n-1}$, respectively, and let $G_n/O_n$ be the homotopy fiber of $BO_n\\to BG_n$.\n\nThere is an interesting old fact, that the stabilization map $G_n/O_n\\to G_{n+1}/O_{n+1}$ is $(2n-4)$-connected, about twice as good as the corresponding maps $BO_n\\to BO_{n+1}$ and $BG_n\\to BG_{n+1}$. Equivalently, the map $G_n/O_n\\to G/O$ is $(2n-4)$-connected. I was wondering about the history of this result.\n\nBut what I was really wondering about was the following equivariant analogue. Let $\\Gamma$ be a finite group. For a real representation $V$ of $\\Gamma$, let $O(V)^\\Gamma$ be the group of linear (isometric) automorphisms respecting the $\\Gamma$-action and let $G(V)^\\Gamma$ be the topological monoid of equivariant homotopy equivalences from the unit sphere $S(V)$ to itself. Write $G(V)^\\Gamma/O(V)^\\Gamma$ for the homotopy fiber of the map of classifying spaces, and write $G^\\Gamma/O^\\Gamma$ for the colimit of this over all $V$ in a complete universe (or the sequential colimit over direct sums of copies of the regular representation).\n\nI think maybe with some work I could get a good connectivity statement for the map $G(V)^\\Gamma/O(V)^\\Gamma\\to G^\\Gamma/O^\\Gamma$, but I wonder if someone has already done so, and more generally if anyone has any insights to offer about any of this.\n\nNote that $BO^\\Gamma$ is the product, over irreducible real representations $W$, of either $BO$, $BU$, or $BSp$ depending on the type of $W$, while $G^\\Gamma$ is the union of the invertible components of $(\\Omega^\\infty S^\\infty)^\\Gamma$. In particular, $\\pi_1(BG^\\Gamma)$ is the group of units of the Burnside ring of $\\Gamma$.\n", "link": "https://mathoverflow.net/questions/425108/about-the-equivariant-analogue-of-g-n-o-n", "tags": ["at.algebraic-topology", "homotopy-theory"], "votes": 18, "creation_date": "2022-06-20T06:53:22", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "192", "site": "mathoverflow", "title": "The free complete lattice on three generators, beyond ZF", "body": "_This was originally[asked at MSE](https://math.stackexchange.com/q/4261226/28111); although it is still under bounty it seems unlikely to be answered there._\n\n[$\\mathsf{ZF}$ proves that there is no free complete lattice on three generators](https://www.jstor.org/stable/1993166?seq=1#metadata_info_tab_contents) since any such lattice would have to surject onto the ordinals. This isn't a problem in the context of set theories which admit a universal set, such as [$\\mathsf{NFU}$](https://plato.stanford.edu/entries/settheory-alternative/#NewFounRelaSyst) or [$\\mathsf{GPK}_\\infty^+$](https://plato.stanford.edu/entries/settheory-alternative/#SystGPKiOlivEsse). However, in such theories recursive constructions (which are utterly unproblematic in $\\mathsf{ZF}$) become more complicated. So the status of the free complete lattice on three generators in such a theory is unclear to me:\n\n> Is the existence of a free complete lattice on three generators consistent with either of $\\mathsf{NFU}$ or $\\mathsf{GPK_\\infty^+}$?\n\nOf course $\\mathsf{NFU}$ and $\\mathsf{GPK_\\infty^+}$ are not the only two set theories with a universal set, and I'd be interested in an answer for any such theory. My focus on these two is due to $(i)$ my impression that $\\mathsf{NFU}$ is the best-understood such set theory by far at the moment and $(ii)$ my personal liking of $\\mathsf{GPK_\\infty^+}$.\n\n**EDIT** : I forgot to include exactly what \"free complete lattice on three generators\" means for me in this context (given that such notions are often more fragile in non-$\\mathsf{ZF}$ contexts). While to a certain extent I'm interested in any reasonable interpretation of the phrase, I'm specifically interested in the following precisiation: a complete lattice $L$ with distinguished elements $a,b,c$ such that for every complete lattice $M$ and every $x,y,z\\in M$ there is exactly one complete lattice homomorphism $L\\rightarrow M$ sending $a$ to $x$, $b$ to $y$, and $c$ to $z$.\n", "link": "https://mathoverflow.net/questions/405269/the-free-complete-lattice-on-three-generators-beyond-zf", "tags": ["set-theory", "lo.logic", "universal-algebra", "foundations", "new-foundations"], "votes": 18, "creation_date": "2021-10-01T10:56:33", "comments": ["Given the results in the Crawley-Dean paper, I suspect that if the free complete lattice with $3$ generators exists in a model of $\\mathsf{NFU}$ with choice, then it would actually have the same with any number of generators, including $|V|$.", "@JamesHanson In fact, I can't even show that \"the free complete lattice on $V$ generators\" doesn't exist! (That is, a pair $(X,f)$ where $X$ is a complete lattice and $f:\\{x:x= x\\}\\rightarrow X$ is such that for every other complete lattice $A$ and every map $g:\\{x:x =x\\}\\rightarrow A$ there is exactly one complete lattice homomorphism $h: X\\rightarrow A$ such that $h\\circ f=g$.)", "@JamesHanson Yes, I don't see how to get Dedekind-MacNeille completions to work in such settings either. (On that note I forgot to include my definition of freeness - fixed!)", "I think you can show that in $\\mathsf{NFU}$ there is no complete lattice generated by three elements that satisfies the conditions in Crawley and Dean's definition 2. The issue I'm having with applying this to the definition of freeness in your MSE question is actually that it seems likely that not all lattices have completions in $\\mathsf{NFU}$. The Dedekind-MacNeille completion of a lattice naturally lives one type higher than the lattice itself, which makes it seem like in general you only know that completions exist for lattices on Cantorian sets."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "193", "site": "mathoverflow", "title": "Automorphic forms and coherent cohomology", "body": "Why is it (and what does it mean) that automorphic forms do not contribute in the coherent cohomology of Siegel modular varieties parametrizing abelian varieties of dimension $d>2$ (see section 7 of the [thesis](https://tel.archives-ouvertes.fr/tel-02940906/document) of Nguyen or section 1.4.1 of this [paper](https://arxiv.org/pdf/1812.09269.pdf) of Boxer, Calegari, Gee, and Pilloni)?\n\nApparently this comes from the fact that the Hodge-Tate weight of the associated motive is neither regular nor weakly regular, i.e. the same number appears more than twice in the Hodge-Tate weights, but why does this mean that we cannot find them in the coherent cohomology? Apparently this is also true for the Betti cohomology of the associated locally symmetric space. Does the theorem of Franke (generalization of Eichler-Shimura) have anything to say about this?\n", "link": "https://mathoverflow.net/questions/375826/automorphic-forms-and-coherent-cohomology", "tags": ["nt.number-theory", "arithmetic-geometry", "automorphic-forms", "galois-representations"], "votes": 18, "creation_date": "2020-11-06T10:32:18", "comments": ["A good reference for coherent cohomology of Shimura varieties is the paper of Michael Harris, cited as [Har90b] in the BCGP paper. In particular, Proposition 4.3.2 there gives a formula for the infinitesimal character associated to the coherent cohomology of an automorphic vector bundle. You might also find the paper of Gross cited as [Gro16] in BCGM useful: it explains the role played by the endomorphism ring of the abelian variety.", "I also agree that a short answer with some reference could be useful, not only for the OP but for other people interested in this question as well. Even if it is not perfect, it would be better than nothing.", "@naf: I think I understand now why the motives of abelian varieties of dimension d>2 cannot come from the coherent cohomology of a Shimura variety: The former is never regular, since HT weights are d 1's and d 0's. The latter however are required to be regular. Could you give me a reference for coherent cohomology that discusses this?", "@YemonChoi: Coherent cohomology actually corresponds to non-degenerate limits of discrete series. Without knowing anything about what the OP is willing to assume, it is not possible (for me) to write a short answer which would also be useful.", "(Alternatively, is there a representation-theoretic explanation of why certain representations are not contained in the discrete series for the relevant group, without going through Hodge-Tate weights?)", "@naf I am neither a number theorist nor an algebraic geometer, but perhaps the OP would find it useful to see an expanded version of the statement \"Another way to see this...\" that you refer to. I don't know if \"computing the possible infinitesimal characters\" is supposed to be a standard procedure in this setting, but perhaps an outline of how one does this for a toy example could be helpful for the OP?", "@Tim: The explanation in the paper of BGGP is fairly detailed and seems quite clear to me. What I said corresponds to the statement in brackets beginning with \"Another...\" in the second paragraph there. If there is something specific you do not understand you should ask a more focused question.", "@naf If you know the correct statement then do you also know an answer? It might be a bit more constructive to maybe expand on this a bit more :-)", "Thank you, and sorry for getting it wrong. I would be happy to edit the question, but could you elaborate more or give me some references to learn more about this?", "The question as written doesn't make sense (or is wrong...): the correct statement is that the motive of an abelian variety of dimension $>2$ with endomorphism ring the integers does not occur in the (motive associated to) the coherent cohomology of (automorphic vector bundles on) a Shimura variety."], "comment_count": 10, "category": "Science", "diamond": 0} {"question_id": "194", "site": "mathoverflow", "title": "Mysterious sum equal to $\\frac{7(p^2-1)}{24}$ where $p \\equiv 1 \\pmod{4}$", "body": "Consider a prime number $p \\equiv 1 \\pmod{4}$ and $n_p$ denotes the remainder of $n$ upon division by $p$. Let $A_p=\\\\{ a \\in [[0,p]] \\mid {(a+1)^2}_p<{a^2}_p\\\\}$.\n\nI Conjecture\n\n$$\\sum_{n \\in A_p } n=\\dfrac{7(p^2-1)}{24}$$\n\nThe question has already been asked in this [thread](https://math.stackexchange.com/questions/4639883/mysterious-sum-equal-to-frac7p2-124-where-p-equiv-1-pmod4?noredirect=1#comment9793873_4639883). If it is a repetition, please vote to close this thread.\n\n**Addition** After have [answer](https://math.stackexchange.com/questions/4639883/mysterious-sum-equal-to-frac7p2-124-where-p-equiv-1-pmod4/4640608#4640608) Numerically, I find this equality : for every odd integer p ≥ 3 $$\\sum_{n \\in A_p} n=\\dfrac{7p^2 -12p+5}{24}+\\dfrac{1}{p}\\sum_{n=1}^{p-1}{n^2}_p$$ Perhaps, it is useful to demonstrate in a simple way the fact that [the sum differs from 7(p^2-1)/24 by the class number of Q(sqrt(-p))](https://math.stackexchange.com/questions/4639930/even-more-mysterious-sum-equaling-to-frac7p2-124-h-p-where-p-equ).\n", "link": "https://mathoverflow.net/questions/440995/mysterious-sum-equal-to-frac7p2-124-where-p-equiv-1-pmod4", "tags": ["nt.number-theory", "prime-numbers"], "votes": 18, "creation_date": "2023-02-16T06:48:42", "comments": ["In the same vein, see also in Franz Lemmermeyer's book \"Reciprocity Laws\", especially chapter 1 and exercise 1.32 (Gauss).", "@SB1729 Good question", "@Pascal I did some simple computations, this problem is so interesting feels like doing more, so I think we have $\\sum_{a\\in A_p}a^2=\\frac{(p-1)(5p^2+7p-10)}{24}$. In general I think it is interesting to know if there is a degree $d$ polynomial $P_d$ such that $\\frac{24}{p-1}\\sum_{a\\in A_p}a^d=P_d(p)$ for any prime $1$ modulo $4$ and if there are some patterns for the polynomials $P_d$.", "I have added a numerical result that I think is interesting", "Thank you very much , I received a proof in the other thread that I had reported here", "This has been answered in your MSE post. Generally it's best practice to leave a reasonable amount of time (at least a week) between crossposts to avoid situations like this.", "I believe this has nothing to do with primes. Try for instance powers of primes 1 mod 4 (the pattern is immediate for powers 2,3,4 for instance), or products of primes 1 mod 4, where the same formula holds (and probably also with class numbers).", "Let $(a+1)^2=a^2+2a+1\\equiv\\alpha\\pmod{p}$ and $a^2\\equiv \\beta\\pmod{p}$. Then $\\beta-\\alpha\\equiv-2a-1\\pmod{p}$. Also for $a\\in A_p$ we have $\\beta-\\alpha\\geq1$. But according to the size of $a$, I think we should get $\\beta-\\alpha=2p-2a-1$. Not sure though.", "To test that I understand correctly, for $p=5$ we are talking about $3+4$, because $0^2_5 < 1^2_5 < 2^2_5 = 3^2_5 > 4^2_5 > 5^2_5$, right?", ". . . and as already noted at math.stackexchange.com/questions/4639930 when p is -1 mod 4 the sum differs from 7(p^2-1)/24 by the class number of Q(sqrt(-p)) ! This too is yet unproved.", "$[0.p]\\cap \\mathbb N$", "What is $[[0,p]]$?"], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "195", "site": "mathoverflow", "title": "Two curious series for $1/\\pi$", "body": "On Jan. 18, 2012 I conjectured that for any prime $p>3$ we have $$R_p^2\\equiv\\frac1{10}\\left(512\\left(\\frac{10}p\\right)-27\\left(\\frac{-15}p\\right)-475\\right)\\pmod p,$$ where $(\\frac{\\cdot}p)$ denotes the Legendre symbol and $$R_p:=\\frac1p\\sum_{n=0}^{p-1}\\frac{6n+1}{(-1728)^n}\\binom{2n}n\\sum_{k=0}^n\\binom nk\\binom{n+2k}{2k}\\binom{2k}k(-324)^{n-k}.$$ Motivated by this I believed in 2011 that $$\\sum_{n=0}^{\\infty}\\frac{6n+1}{(-1728)^n}\\binom{2n}n\\sum_{k=0}^n\\binom nk\\binom{n+2k}{2k}\\binom{2k}k(-324)^{n-k}=\\frac{c}{\\pi}$$ for some algebraic number $c$, but I could not figure out the exact value of $c$ at that time.\n\nOn August 19, 2020 I learned from Dr. Deyi Chen that Maple has a function to justify the form of an algebraic number if we know enough digits of this number. Inspired by this, here I pose the following conjecture.\n\n**Conjecture 1.** We have $$\\begin{aligned}&\\sum_{n=0}^{\\infty}\\frac{6n+1}{(-1728)^n}\\binom{2n}n\\sum_{k=0}^n\\binom nk\\binom{n+2k}{2k}\\binom{2k}k(-324)^{n-k}\\\\\\&\\qquad=\\frac{24}{25\\pi}\\sqrt{375+120\\sqrt{10}}. \\end{aligned}\\tag{1}$$\n\nOn June 16, 2011 I conjectured that for any odd prime $p\\not=5$ we have $$\\sum_{n=0}^{p-1}\\frac{4n+1}{(-160)^n}\\binom{2n}n\\sum_{k=0}^n\\binom nk\\binom{n+2k}{2k}\\binom{2k}k(-20)^{n-k}\\equiv 0\\pmod p.$$ Motivated by this I formulated the following conjecture on August 20, 2020.\n\n**Conjecture 2.** We have $$\\begin{aligned}&\\sum_{n=0}^{\\infty}\\frac{4n+1}{(-160)^n}\\binom{2n}n\\sum_{k=0}^n\\binom nk\\binom{n+2k}{2k}\\binom{2k}k(-20)^{n-k}\\\\\\&\\qquad=\\frac{\\sqrt{30}}{5\\pi}\\cdot\\frac{5+\\root 3\\of{145+30\\sqrt6}}{\\root 6\\of{145+30\\sqrt6}}. \\end{aligned}\\tag{2}$$\n\nI have checked $(1)$ and $(2)$ for the first $100$ decimal digits.\n\n**Question**. How to prove the identities $(1)$ and $(2)$?\n\nNote that my two other similar conjectural identities $$\\sum_{n=0}^\\infty\\frac{357n+103}{2160^n}\\binom{2n}n\\sum_{k=0}^n\\binom nk\\binom{n+2k}{2k}\\binom{2k}k(-324)^{n-k}=\\frac{90}{\\pi}\\tag{3}$$ and $$\\sum_{n=0}^\\infty\\frac{n}{3645^n}\\binom{2n}n\\sum_{k=0}^n\\binom nk\\binom{n+2k}{2k}\\binom{2k}k486^{n-k}=\\frac{10}{3\\pi}\\tag{4}$$ found in 2011-2012 and published in [this paper](http://maths.nju.edu.cn/%7Ezwsun/157b.pdf) also remain open.\n", "link": "https://mathoverflow.net/questions/369569/two-curious-series-for-1-pi", "tags": ["nt.number-theory", "sequences-and-series", "combinatorial-identities", "congruences", "ramanujan"], "votes": 18, "creation_date": "2020-08-19T07:20:37", "comments": ["Thank you, this is very interesting!", "You may look at my talk available from maths.nju.edu.cn/~zwsun/CNT-talk3.pdf and my recent paper available from dx.doi.org/doi:10.3934/era.2020070.", "What is the relation between the congruence and the identity for pi? Is there an explanation for why we would expect such things?"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "196", "site": "mathoverflow", "title": "Cycles in algebraic de Rham cohomology", "body": "Let $F$ be a number field, $S$ a finite set of places, and $X$ a smooth projective $\\mathscr{O}_{F,S}$-scheme with geometrically connected fibers. For each point $t\\in \\text{Spec}(\\mathscr{O}_{F,S})$, there is a Chern class map $$c_{1,t}:\\text{Pic}(X_t)\\to H^2_{dR}(X_t),$$ where $H^2_{dR}$ denotes algebraic de Rham cohomology. (See below for one construction of $c_1$.)\n\n> Let $c\\in H^2_{dR}(X_F)$ be an element so that there exists a non-zero integer $N$ such that $Nc$ specializes to an element of the image $c_{1,t}\\otimes k(t)$, for almost all closed points $t\\in \\text{Spec}(\\mathscr{O}_{F,S})$. Is $c$ in the image of $$c_{1, F}: \\text{Pic}(X_F)\\otimes F\\to H^2_{dR}(X_F)?$$ Here $c_{1,F}$ means $c_{1, \\eta}$, for $\\eta$ the generic point of $\\text{Spec}(\\mathscr{O}_{F,S})$, and $k(t)$ is the residue field of $t$.\n\n**Is the answer to the above question (and its evident analogue for higher degree Chern classes/cycle class maps) known in general? Does the question have a name, and is it discussed somewhere?**\n\nIn general, given a cohomology theory, there's some conjectural description of the image of the cycle class map (the Hodge conjecture for Betti/de Rham cohomology, the Tate conjecture for $\\ell$-adic cohomology) -- my motivation for asking the question is that I realized I hadn't seen such a conjecture for algebraic de Rham cohomology.\n\n* * *\n\nI'll now give a brief description of $c_1$, for completeness. Let $\\Omega^\\bullet_{dR}$ be the algebraic de Rham complex of some variety $Y$. There's a natural map $$d\\log: \\mathscr{O}_Y^*[-1]\\to \\Omega^{\\bullet}_{dR}$$ sending a nowhere-vanishing function $f$ to $df/f$. Applying $H^2$ gives the desired map $$c_1:\\text{Pic}(Y)=H^1(Y, \\mathscr{O}_Y^*)=H^2(Y, \\mathscr{O}_Y^*[-1])\\overset{d\\log}{\\longrightarrow} H^2_{dR}(Y).$$\n\nBy construction, it factors through $F^1H_{dR}^2(Y)$, the first piece of the Hodge filtration.\n\n* * *\n\nSome remarks, added September 6, 2022:\n\n * François Charles has explained to me how to give a positive answer to the above question for divisors on Abelian varieties, using Bost's work on foliations. I think a similar argument handles products of curves.\n * One natural strategy is to try to find a prime where the Picard rank does not jump. This will not work, unfortunately; there are varieties where the Picard rank jumps at _every_ prime, e.g. the square of a non-CM elliptic curve.\n * The question can be reduced to the case of surfaces, if I'm not mistaken.\n\n\n", "link": "https://mathoverflow.net/questions/429644/cycles-in-algebraic-de-rham-cohomology", "tags": ["ag.algebraic-geometry", "arithmetic-geometry", "algebraic-cycles", "chern-classes", "derham-cohomology"], "votes": 18, "creation_date": "2022-09-02T07:26:35", "comments": ["@PeterScholze: Thanks! This does indeed seem quite related, and indeed the discussion on page 81 of \"Une Introduction aux Motifs\" is closely related to the argument François Charles sketched for the case of Abelian varieties.", "Related in spirit, but I think a bit different, is the Ogus conjecture, discussed at mathoverflow.net/questions/389391/… (see in particular the reference to Yves Andre's \"Une Introduction aux Motifs\"). Ogus' conjecture is the only one I'm aware of in the neighborhood of what you are proposing.", "@R.vanDobbendeBruyn: yes, I was hedging because otherwise specialization isn’t defined everywhere. But since it’s an almost-everywhere condition it’s fine to take N=1 if you don’t mind specialization not being defined somewhere.", "You don't lose generality by taking $N = 1$, right? Because $N$ is invertible in $F$ and in $\\kappa(t)$ for almost all $t$, and all maps are maps of vector spaces.", "@DanPetersen: Good point. To be honest my actual motivation was: Simpson's conjecture says that a local system underlies an integral VHS iff it arises from geometry; this is the non-abelian analogue of the Hodge conjecture. The Fontaine-Mazur conjecture says a local system is arithmetic iff it arises from geometry; this is the non-abelian analogue of the Tate conjecture. Now there is a conjecture that a flat vector bundle has almost all p-curvatures nilpotent iff it comes from geometry (the p-curvature conjecture is closely related) -- I was wondering what the \"abelian\" analogue of this is.", "An analogue of Hodge and Tate conjecture with respect to algebraic de Rham cohomology is the Grothendieck period conjecture: if $X$ is a smooth projective variety over $\\overline{\\mathbf{Q}}$, then conjecturally the image of the cycle class map is given by the intersection $ H^{2k}_{\\mathrm{sing}}(X^{\\mathrm{an}},\\mathbf Q(k)) \\cap H^{2k}_{\\mathrm{dR}}(X/{\\overline{\\mathbf{Q}}})$.", "Clearly, this is a special case of the Litt conjecture. :)"], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "197", "site": "mathoverflow", "title": "Čech functions and the axiom of choice", "body": "A **Čech closure function** on $\\omega$ is a function $\\varphi:\\mathcal P(\\omega)\\to\\mathcal P(\\omega)$ such that (i) $X\\subseteq\\varphi(X)$ for all $X\\subseteq\\omega$, (ii) $\\varphi(\\emptyset)=\\emptyset$, and (iii) $\\varphi(X\\cup Y)=\\varphi(X)\\cup\\varphi(Y)$ for all $X,Y\\subseteq\\omega$; in other words, it obeys the Kuratowski closure axioms except possibly the idempotent law $\\varphi(\\varphi(X))=\\varphi(X)$.\n\nIn 1947 E. Čech asked if there exists such a closure function (on any set, not necessarily $\\omega$) which is also **surjective** and **nontrivial** (not the identity map). Čech's question has been answered in the affirmative [under various set-theoretic assumptions](https://www.sciencedirect.com/science/article/pii/0166864182900608), including ZFC. [The ZFC example](https://pdfs.semanticscholar.org/8870/c3c8a574f24aa1ee411f82283a1076c53303.pdf) is rather complicated and makes heavy use of the axiom of choice; it seems unlikely that one could construct such a thing without choice.\n\n**Question.** Is it consistent with ZF that there is no nontrivial surjective Čech closure function $\\varphi:\\mathcal P(\\omega)\\to\\mathcal P(\\omega)$?\n\nThat is, there is no function $\\varphi:\\mathcal P(\\omega)\\to\\mathcal P(\\omega)$ satisfying the conditions (i)-(iii) above and also (iv) for every $Y\\subseteq\\omega$ there exists $X\\subseteq\\omega$ such that $\\varphi(X)=Y$, and (v) $\\varphi(X)\\ne X$ for some $X\\subseteq\\omega$. (Condition (ii) is now redundant, as it follows from (i) and (iv) that $\\varphi(X)=X$ if $X$ is finite.)\n", "link": "https://mathoverflow.net/questions/361038/%c4%8cech-functions-and-the-axiom-of-choice", "tags": ["gn.general-topology", "set-theory", "axiom-of-choice"], "votes": 18, "creation_date": "2020-05-20T03:03:19", "comments": ["@Haim Looks to me like they both use choice. My guess is, if you can construct such a function without choice, it will be by a completely different method.", "@bof Out of Theorem 1 and Theorem 2 in the Galvin-Simon paper, which one is using AC in its proof?", "@YCor: I got a partial answer to this in this question.", "@YCor: Are you asking for a definable class (the class of functions such that blah blah, defined by formula) whose elements are individually not definable (so that you cannot write down such $F$'s to pick out individual elements)?", "@bof I'd be happy with a weaker definition of \"constructible\". I'm just wondering whether the slogan \"(exists in ZF)=(can be constructed)\" has any foundation.", "Is it correct to think as \"in ZF there's such a function\" as \"one can construct such a function\"? More precisely, is it true (from some general principle) that if \"there is such a function\" is a theorem of ZF, then there exists $f_0\\in P(\\omega)^{P(\\omega)}$ satisfying the given condition, and a formula of set theory $F(x)$ with only parameter $x$, such that for every $f\\in P(\\omega)^{P(\\omega)}$, $F(f)$ holds iff $f=f_0$?", "I think it would be a good point in time to migrate this to MathOverflow. I don't have an obvious solution at hand, and I have quite a lot of things on my hands right now (which is why I might be missing something obvious).", "Also a candidate: the Feferman–Levy model where the reals are a countable union of countable sets, or the Truss model (although if this holds there, I'd imagine the Solovay model would already capture this).", "I could imagine several candidate models: Maybe Cohen's model is a suitable example with the Dedekind-finite set of reals \"spoiling things\", but more likely would be Solovay's model where there are no MAD families, etc."], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "198", "site": "mathoverflow", "title": "Is the Frog game solvable in the root of a full binary tree?", "body": "_This is a cross-post from[math.stackexchange.com](https://math.stackexchange.com/q/3800570/318073)$^{[1]}$, since the bounty there didn't lead to any new insights._\n\n* * *\n\nFor reference,\n\n> The **Frog game** is the generalization of the _Frog Jumping (see it on[Numberphile](https://www.youtube.com/watch?v=X3HDnrehyDM))_ that can be played on any graph, but by convention, we restrict the game to tree graphs. The _Frog Jumping_ is equivalent to the case on \"paths (tree graphs with exactly 2 leaves)\" of the **Frog game** , which was resolved in the linked video.\n\nThe mechanics of the Frog game are in short,\n\n> Given a tree graph and one frog on each vertex, the goal of the game is for all frogs to host a party on a single vertex of the graph. All of the $f$ frogs on some vertex can \"jump\" to some other vertex if and only if both vertices have at least one frog on them and both vertices are exactly $f$ edges apart from each other.\n> \n> If a sequence of \"jumps\" exists such that all frogs end up on a single vertex, then the party is successfully hosted on that vertex and we call that vertex a \"lazy toad\" (or just \"lazy\") because the frog that started there never jumped during the game.\n\nTrivially, if $|V|=n$ is the number of vertices of some graph $G=(V,E)$, then every game on $G$ must end in $n-1$ moves (or less if the party isn't hosted, e.g. if the sequence of jumps is suboptimal and we can no longer make moves or if none of the vertices are lazy).\n\n* * *\n\nHere, I'm trying to resolve the Frog game on \"full binary trees\".\n\nLet $T_h=(V,E)$ be a full binary tree of height $h$. Denote its layers with $0,1,2,\\dots,h$ where the root $r\\in V$ is on the layer $0$ (\"bottom layer\"), where layer $i=1,2,\\dots,h$ contains vertices $v_{i,j}$ labeled with $j=1,2,\\dots,2^i$.\n\nThat is, we have $|V|=2^{h+1}-1$. Let $h\\in \\mathbb N$ because $h=0$ is a single vertex which is trivial.\n\n> **Conjecture.** For all $h\\ge 4$, every vertex of $T_h$ is a \"lazy toad\" (is \"lazy\").\n\nI couldn't prove the above, and for simplicity, my question is just about the root vertex:\n\n> Let $h\\in\\mathbb N, h\\ne 2$. Can we prove that the root $r$ of every such $T_h$ is a \"lazy toad\" (is \"lazy\")?\n\nHowever, if the approach that can solve the root can be generalized to all other vertices, that's a big bonus.\n\n* * *\n\n**Solving the root**\n\n**Reduction.** We say that $T_h$ can be reduced to $T_s,s\\le h-2$, if there exists a sequence of moves that can bring all frogs from the top $s$ layers $h,h-1,\\dots,h-s+1$ to the root vertex.\n\nI've managed to reduce $T_h$'s for all $h\\in(2,20)$ by hand pretty easily. At most, I needed to consider only $31$ vertices or less to find a pattern that allows a reduction of entire layers, which is very efficient if you consider that $T_{19}$ has over a million vertices. It looks like that maybe an inductive argument would be able to solve the problem, but I did not manage to find a sufficient set of move sequences.\n\nI'll summarize some immediate results below.\n\n> **Trivial reduction.** If $h=2^t+t-2,t\\ge 2$ and $T_{t-1}$ is solvable in root, then $T_h$ can be reduced to $T_{h-t}$.\n\nIn the above reduction, we are simply (at the top layers) solving $T_{t-1}$ subtrees in root and then jumping from that root to the root of $T_h$.\n\n> **Necessary minimal reduction condition.** Let $h\\gt2,a\\ge2$. If reduction from $T_h$ to $T_{h-a}$ is possible, then there exists $b\\in[0,h-a+1]$ and a partition of the number $(2^a-1)2^b$ into the layers $H=\\\\{h,h-1,h-2,\\dots,h-a+1\\\\}$ where $(h-l)\\in H$ is used at most $(2^{a-l-1})2^b$ times.\n\nIn the above condition, we say that \"a partition is solvable\" if we can bring the corresponding amounts of frogs to the corresponding layers. I.e. the above condition simply says that we must be able to partition the total number of frogs from the top layers (that we are reducing) into the corresponding vertices from which they can jump to the root.\n\nIt remains to show (or find an explicit set of move sequences that will imply) that at least one partition from the above condition is solvable. Surprisingly, for all $h\\in(2,20)$ that I've solved so far, the solution was always available in the smallest or one of the smallest partitions ($a\\le 5,b\\le 3$) of no more than $31$. The $h=20$ is the first case that requires considering more than $31$ vertices.\n\nI still do not see how to construct explicit move sequences for larger $h$ without tackling each one individually. Alternatively, could there be a way to prove that the root is lazy, without finding explicit move sequences?\n", "link": "https://mathoverflow.net/questions/370694/is-the-frog-game-solvable-in-the-root-of-a-full-binary-tree", "tags": ["graph-theory", "trees", "induction", "binary-tree"], "votes": 18, "creation_date": "2020-09-02T09:53:11", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "199", "site": "mathoverflow", "title": "Orientation-reversing homotopy equivalence", "body": "If there is an orientation-reversing homotopy equivalence on a closed simply-connected orientable manifold is there an orientation-reversing homeomorphism?\n\nIt is not true, for instance, that if there is an orientation-reversing homeomorphism there must be an orientation-reversing diffeomorphism (consider exotic 7-spheres).\n", "link": "https://mathoverflow.net/questions/373348/orientation-reversing-homotopy-equivalence", "tags": ["dg.differential-geometry", "at.algebraic-topology", "gt.geometric-topology", "homotopy-theory"], "votes": 18, "creation_date": "2020-10-05T06:15:17", "comments": ["This can be heuristically reduced to finding manifolds that are homotopy equivalent but not homeomorphic, which should show you how to organize thinking about this. If $X$ does not admit an orientation-reversing homotopy equivalence and $Y$ is homotopy equivalent to $X$ but not homeomorphic, then the connected sum of $X$ and the opposite orientation of $Y$ is a manifold that admits an orientation-reversing homotopy equivalence, but no orientation-reversing homeo. $X=\\mathbb C\\mathbb P^4$ and $Y$ with different Pontrjagin classes. This isn't any easier than my previous example, though.", "Correction: there is a smooth manifold homotopy equivalent to $S^4\\times S^4$ with nonzero $p_2$ and thus no orientation reversing homeomorphism, but I don't see an easy way to construct it. However, linear sphere bundles have nontrivial $p_1$ and so do provide counterexamples to the more natural question of whether a particular homotopy equivalence is homotopic to a homeomorphism. That is, the homotopy equivalence which reverses orientation on the base sphere is not homotopic to a homeomorphism, even though the homotopy equivalence which reverses orientation on fiber is realized by a homeo.", "@ConnorMalin I'm not really sure what you're asking, but the first topic is called surgery theory and the second topic is called smoothing theory (or triangulation theory). For most purposes, I recommend the two volumes \"Surveys on Surgery Theory,\" ed Cappell-Ranicki-Rosenberg. But if you really want a foundational reference for the topological category, I think the only option is Kirby-Siebenmann.", "@BenWieland Do you know of a standard reference for the obstructions to homotoping a homotopy equivalence to a homemorphism? And for isotopy between a homeomorphism and a diffeomorphism?", "The difference between homeomorphism and diffeomorphism is subtle (torsion). The difference between homotopy equivalence and homeomorphism is crude (rational), the same as the difference between homotopy equivalence and diffeomorphism. Examples involving exotic spheres are difficult and overkill. There are easy smooth examples using Pontrjagin classes. There are linear $S^4$ bundles over $S^4$ homotopy equivalent to $S^4\\times S^4$ with nonzero $p_2$. Obviously smoothly chiral. Given the difficult but blackbox theorem that $p_2$ is a rational homeomorphism invariant, topologically chiral.", "According to arxiv.org/pdf/0907.5283.pdf (Section 2, 4th bullet point), the lens space $L_5(1,1)$ has an orientation reversing homotopy equivalence, but no orientation reversing homeomorphism. Of course, it is not simply connected."], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "200", "site": "mathoverflow", "title": "Are these local systems on $\\mathscr{M}_{g,1}$ motivic?", "body": "Let $(\\Sigma_g, x)$ be a pointed topological surface of genus $g$, and let $MCG(g,1)$ be the mapping class group of this pointed surface. Then $MCG(g,1)$ has a natural action on $\\pi_1(\\Sigma_g, x)$ $$MCG(g)\\to \\text{Aut}(\\pi_1(\\Sigma_g))$$ and hence acts on the set of irreducible characters of $\\pi_1(\\Sigma_g)$: $$\\gamma: MCG(g,1)\\to \\text{Aut}(\\text{Char}^{\\text{irr}}(\\pi_1(\\Sigma_g))).$$\n\nSuppose $$\\rho: \\pi_1(\\Sigma_g)\\to GL_n(\\mathbb{C})$$ is an irreducible representation of $\\pi_1(\\Sigma_g)$ which is fixed by this action, i.e. for every element $\\tau$ of $MCG(g,1)$, there exists a matrix $\\varphi_\\rho(\\tau)$ conjugating $\\rho^\\tau$ to $\\rho$ as a representation of $\\pi_1(\\Sigma_g)$. Because $\\rho$ is irreducible, $\\varphi_\\rho(\\tau)$ is well-defined up to scaling, and so we obtain a natural projective representation $$\\varphi_\\rho: MCG(g,1)\\to PGL_n(\\mathbb{C}).$$ Choosing a representation of $PGL_n(\\mathbb{C})$ (say, the adjoint representation), we obtain a representation of $MCG(g,1)$ and hence a local system on the moduli space of pointed curves $\\mathscr{M}_{g,1}$.\n\n> Are the local systems obtained this way motivic?\n\nI know how to produce several examples of such $\\rho$ fixed by $MCG(g,1)$ (e.g. using the Parshin trick) and in these examples the local systems in question are indeed motivic.\n\nOf course this question is too hard as stated in this generality -- so it would be better to ask:\n\n> Do the local systems thus obtained underly $\\overline{\\mathbb{Z}}$-variations of Hodge structure?\n\nThis question is naturally divided into three parts:\n\n> 1. Is $\\varphi_\\rho$ always defined over $\\overline{\\mathbb{Q}}$? I.e. is it conjugate to a representation into $PGL_n(\\overline{\\mathbb{Q}})$?\n> \n> 2. Is $\\varphi_\\rho$ always integral, i.e. is it conjugate to a representation into $PGL_n(\\overline{\\mathbb{Z}})$?\n> \n> 3. Does $\\text{ad}_{PGL_n}\\circ \\varphi_\\rho$ always underly a $\\overline{\\mathbb{Q}}$-variation of Hodge structure?\n> \n> \n\n\nThe only method of proof I can think of at the moment would be to show that $\\varphi_\\rho$ is _rigid_ (which would imply (1) and (3), using non-abelian Hodge theory) or _cohomologically rigid_ (which would imply (1-3) via recent results of Esnault and Groechenig). I haven't had any luck showing this thus far.\n\n* * *\n\nSome brief remarks on this old question (9/11/22):\n\n 1. As written, the answer is \"no\": I know how to make counterexamples for $g\\leq 2$. Indeed, for $g\\leq 2$ one may construct such local systems which do not have quasi-unipotent mondoromy at infinity. It's possible that the answer is \"yes\" in $g\\leq 2$ if one additionally assumes this.\n 2. Such examples are impossible for $g\\geq 3$, as any local system on $\\mathscr{M}_{g,n}$ automatically has quasi-unipotent monodromy at infinity in this case by a result of Aramayona-Souto.\n 3. The answer is \"yes\" for representations of rank $r<\\sqrt{g+1}$ with finite orbit under $MCG(g,1)$ by one of the results of [this paper](https://arxiv.org/abs/2205.15352) of mine and Landesman. This is because such representations necessarily have finite image.\n\n\n", "link": "https://mathoverflow.net/questions/345955/are-these-local-systems-on-mathscrm-g-1-motivic", "tags": ["ag.algebraic-geometry", "gt.geometric-topology", "mapping-class-groups"], "votes": 18, "creation_date": "2019-11-13T08:17:12", "comments": ["@BenWieland: (1) is an interesting idea -- I'll play with it a bit and see what I get. Thanks!", "Oops, my (2) is not reducible as an $M_{g,1}$ representation. That was the whole point, yet it ruins it. The whole point is that it is the degeneration of an extension to a split extension, but such degenerations are trivial away from the split locus.", "Yes, expand . . $M_2$ is virt $M_{0,6}$, so admits non-rigid reps. So ask $g\\geqslant3$, but why stop there? Why did you think that 2 was enough? I don't really know any non-rigid reps, but here are two ideas. (1) a Burau-ish rep of $M_g$ on the homology of the abelian cover, as a $\\mathbb Z[t_1,...t_{2g}]$-mod. (2) The rep of of $M_{g,1}$ on the Lie algebra of the 2-step nilpotent quotient of the fundamental group of the curve: shouldn't a 2-step nilpotent Lie algebra have a defm towards splitting, natural enough that it carries the rep? What happens when you restrict $M_{g,1}$ to $\\pi_1S_g$?", "@BenWieland: I should have assumed $g\\geq 2$ -- the hope was that this construction rigidifies things, and it's motivated by some arithmetic considerations I omitted from the question, but which I can expand on if you'd like. Do you know of an example of a non-rigid representation of $\\pi_1(M_{g,n})$ for some $g\\geq 2$? Re: reps of $\\pi_1(M_{g, 1})$, the Parshin trick gives examples, and there are others coming from TQFT's...", "This seems very arbitrary to me. Why not ask if all representations of $M_{g,n}$ are motivic? But they're not all rigid. If you had a nonrigid representation of $M_{g,1}$, you could restrict to $\\pi_1(\\Sigma_g)$ and check if it remained irreducible. But I don't know any representations of $M_{g,1}$. For example, the Burau representation of $M_{0,n}^1$ is not rigid and probably remains irreducible when restricted to $\\pi_1(P^1-n))$. (But it sounds motivic.)", "@WillSawin: It seems to me that such reps will always be rigid as projective reps of the mapping class group, so at least (1) and (3) should be OK. And I think (2) should be OK by Esnault-Groechenig if the mapping class group action on the tangent space to $\\rho$ has no invariants...", "It might be an easier project to show that, if $\\rho$ is rigid as a representation satisfying this additional condition, then $\\varphi_{\\rho}$ is motivic, and if $\\rho$ is not rigid then $\\varphi_\\rho$ is not (always) motivic. Determining whether $\\rho$ is / isn't always rigid could be a separate matter.", "You’re right, fixed. Do you know how to make such a nontrivial family?", "Your construction of a representation does not really work since you are trying to act via $Aut(\\pi_1(\\Sigma))$ instead of $Out(pi_1(\\Sigma))$. What you are actually getting here are projective representations of $MCG(g,1)$. I am sure, these are non-motivic in general and there are nontrivial families of (equivalence classes ) such representations. One more thing: Each fixed point $[\\rho]$ does get you a linear representation of $MCG(g)$ but in a different dimension: You act on the Zariski tangent space to the character variety at the point $[\\rho]$.", "I agree with everything you’ve written—I tried the strategy in your second comment at some point but didn’t have any luck there either...some things are known in the case of 2-dimensional reps IIRC (due to Goldman and collaborators, for example), but not much.", "With regards to showing at least the first couple of your three points, might it make sense to show that $\\rho$ has these properties first? I think if $\\rho$ is algebraic or integral then $\\varphi_{\\rho}$ is algebraic or integral as well. For algebraicity at least a plausible avenue of proof is to show that $\\rho$ has no 1-parameter deformations that respect this invariance property.", "In the $\\ell$-adic world, if $\\rho$, $\\pi_1(\\Sigma_g) \\to GL_n( \\overline{\\mathbb Q}_\\ell)$ factors continuously through the profinite completion of $\\Sigma_g$, and this factorized representation is stable under the action of the arithmetic fundamental group of the moduli space of curves, then I think by the same construction we get a projective arithmetic representation. Of course by the grand theory of motives we expect variations of Hodge sructures and arithmetic representations should appear at the same time."], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "201", "site": "mathoverflow", "title": "Is there a model category describing shape theory?", "body": "Is there a nice model structure on some category of topological spaces compatible with [shape theory](https://en.wikipedia.org/wiki/Shape_theory_\\(mathematics\\))? In particular, weak equivalences should induce isomorphisms on sheaf cohomology.\n\nAs an example, the [Polish Circle](https://en.wikipedia.org/wiki/Shape_theory_\\(mathematics\\)#Warsaw_Circle) has the same sheaf cohomology with constant coefficients as $S^1$, and both spaces also have equivalent categories of covering spaces. So one could ask if in such a model structure, the polish circle would be weakly equivalent to an ordinary one.\n\n**Edit:** I would be happy with an answer for any nice category of topological spaces which contains the Polish circle, e.g. compactly generated weak Hausdorff spaces. Please ignore the following paragraph.\n\nAlso, sheaf theory is defined on rather general topological spaces, so it would be nice if the underlying category of topological spaces would be much more general than, say, compactly generated weak Hausdorff spaces.\n", "link": "https://mathoverflow.net/questions/345396/is-there-a-model-category-describing-shape-theory", "tags": ["at.algebraic-topology", "sheaf-theory", "model-categories", "shape-theory"], "votes": 18, "creation_date": "2019-11-06T08:52:29", "comments": ["@DenisNardin Thanks for the suggestion. I see the point now and did the edit.", "@SebastianGoette Well, it is sort of the same definition, only you need to generalize it to a notion of the shape of a topos rather than just of topological spaces (you might have heard it under the name \"étale homotopy type\").The point is that the underlying space of an algebraic variety does not have much useful information (e.g. all curves over an alg. closed field are homeomorphic!),and you need more refined objects (e.g the étale topos)to recover the information present in the topology for manifolds. Still, your question is interesting, I would only remove the reference to alg. varieties.", "@DenisNardin I was really thinking of the shape as a topological space, as in the wikipedia article. I was not even aware that there is a separate definition in algebraic geometry.", "@SimonHenry It's not idempotent, but it becomes one if you consider the profinite shape instead (see the appendix to SAG and the Barwick-Glasman-Haine work on the stratified shape)", "In Lurie's perspective shape theory is an adjunction between the $\\infty$-category of $\\infty$-toposes and the $\\infty$-category of pro-spaces. I always had the feeling that this adjunction is idempotent ( not sure if this is really true though, is there someone who know ?) if it is the case \"shapes\" are a localization of the category of pro-spaces, as such it should be possible to construct a model structure representing them (maybe on pro-simplicial sets...). If this works one can start thinking about whether this transfer to a model structure on some category of topological spaces...", "@SebastianGoette I'll have to admit I am a bit confused. What do you mean with the shape of an algebraic variety then? The notion I know is not encoded in the topology.", "@DenisNardin I don't expect to see the 'algebraic' cohomology (in whatever sense) here, because the structure sheaf seems to be an extra datum that the topological space does not have. But maybe apart from constant sheaves, we also want to (maybe have to) consider constructible sheaves. But then all information we need is still encoded in the topology.", "@SimonHenry I must admit that I have no idea. The correct notion of equivalence seems to be part of the problem. Also I wonder whether it is possible to work on topological spaces, or whether one needs topoi.", "What is the precise definition of 'shape equivalences' (it is not in the wikipedia link) ? I'm not expert in shape theory but I believe there has been several non-equivalent definition... is it equivalent to the definition Lurie gave for infinity toposes ?", "Hmm... the shape of algebraic varieties does not depend on the underlying topological space, unless I'm mistaken (rather, it is the shape of the étale topos), so I doubt a naive approach would work there."], "comment_count": 10, "category": "Science", "diamond": 0} {"question_id": "202", "site": "mathoverflow", "title": "Are simplicial finite CW complexes and simplicial finite simplicial sets equivalent?", "body": "**Edit** Originally the question was whether an arbitrary diagram of finite CW complexes can be approximated by a diagram of finite simplicial sets. In view of Tyler's comment, this was clearly asking for too much. I restricted the question from arbitrary diagrams to simplicial diagrams.\n\n* * *\n\nIt is well-known that for any finite CW complex $K$ one can construct a finite simplicial set whose geometric realization is equivalent to $K$. But the construction is not functorial.\n\nSuppose we have a simplicial object in finite CW complexes. Can one always construct a simplicial object of finite simplicial sets, whose levelwise geometric realization is connected to the original diagram by a zig-zag of levelwise weak equivalences?\n\nOne may ask a similar question in a more general context. Suppose we have an $\\infty$-category $\\mathcal C$. Suppose $A$ is a collection of (say compact) objects of $\\mathcal C$. Let $CW^f(A)$ be the category of finite cellular objects generated by $A$. Roughly speaking, $CW^f(A)$ consists of objects that can be built as a finite homotopy colimit of objects of $A$. I am leaving the definition of $CW^f(A)$ a little vague - feel free to use any reasonable notion. Let $S_\\bullet^f(A)$ be the category of simplicial objects in $\\mathcal C$ that in each simplicial degree are a finite sum of elements of $A$, and are degenerate above some dimension. Geometric realization gives a functor $S_\\bullet^f(A)\\to CW^f(A)$. Now we have an obvious generalization of the question:\n\nSuppose we have a simplicial object in $CW^f(A)$. Can we find a simplicial object in $S_\\bullet^f(A)$ whose levelwise geometric realization is levelwise (weakly) equivalent to it? If it is not true in general, is there a reasonable sufficient assumption?\n", "link": "https://mathoverflow.net/questions/338660/are-simplicial-finite-cw-complexes-and-simplicial-finite-simplicial-sets-equival", "tags": ["at.algebraic-topology", "homotopy-theory"], "votes": 18, "creation_date": "2019-08-19T00:48:40", "comments": ["The point of the original question is that a model does not support every morphism. Tyler's example of a functor from $B\\mathbb N$ is just the choice of an object and a morphism. If you allow Kan complexes or finite CW complexes, then they are cofibrant-fibrant and thus they support arbitrary morphisms. In the purely homotopical setting, the object supports the morphism. Tyler's example was $S^1$, which is as nicely cellular as you can imagine: represent it as the constant simplicial object. Maybe having to spell out the cells allows counterexamples, but not that one.", "@BenWieland I am feeling a little slow. Why it is not a counterexample?", "@BenWieland yes, absolutely. you can replace each object in a diagram with any weakly equivalent object and get an equivalent diagram at the cost of working coherently.", "If the \"more general\" version is purely homotopical, Tyler's example isn't a counterexample, is it? (a counterexample to an even more general version) And a nitpick, just a problem for objects, if a \"finite homotopy colimit\" includes retracts, then you run into Wall finiteness problems, but @R.vanDobbendeBruyn's version doesn't.", "A possible more general finiteness condition could be the assumption that every object in the diagram has finitely many maps out of it. (This has popped up in some inductive constructions I ran into; examples include finite diagrams and semisimplicial objects.)", "@AchimKrause That is, indeed, a great deal simpler than what I had in mind.", "In Tyler's example, can't we just observe that for any finite simplicial set $K$, only finitely many maps on $\\pi_1$ are induced by endomorphisms $K\\to K$? This is because there are only finitely many maps $K\\to K$ in total.", "Hi Tyler, thanks. This sounds right and suggests I should impose a local finiteness condition on the indexing diagram (such as, there can be only finitely many morphisms between any two objects). What I really wanted to know was about simplicial objects: can any simplicial object in finite CW complexes be approximated by a simplicial object in finite simplicial sets? I will edit the question.", "Hi Greg, I suspect that the map $B\\Bbb N \\to CW$, representing $S^1$ with the endomorphisms $2^k$ for $k > 0$, can't be realized by a diagram of finite simplicial sets, but to prove it I'd want to come up with some explicit growth condition on the number of edges needed to represent elements of $\\pi_1$ for a fixed finite model. This may be a little harder for s.sets than for $\\infty$-categories because the mapping complexes aren't Kan complexes."], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "203", "site": "mathoverflow", "title": "What is the logical complexity of the Hodge conjecture?", "body": "The Hodge conjecture seems to me the most mysterious among the Millennium problems (and many others). In particular, I am not sure about its logical complexity. It is not difficult to see that the conjecture is equivalent to a $\\Pi^0_2$ statement if restricted to varieties over $\\overline{\\mathbb{Q}}$. (Basically, it is because these varieties form a recursive set, and the statement sounds like \"for any Hodge class there is an algebraic cycle\".) However the conjecture may be true over $\\overline{\\mathbb{Q}}$ but false in general ; if I am not mistaken, this possibility has not being ruled out yet. My guess is that the Hodge conjecture may be formulated as a $\\Pi^0_3$ statement but, at present, not as $\\Pi^0_2$. Is it true? \n\nSome remarks. As is well known, the Hodge conjecture may be separated into two parts:\n\nI. Hodge classes are absolute,\n\nII. Absolute Hodge classes are algebraic.\n\nThe second part is $\\Pi^0_2$. (It follows from results of Voisin, \"Hodge loci and absolute Hodge classes\", Compositio Mathematica, Vol. 143 Part 4, 945-958, 2007. ) So, the logically hard part seems to be the first one. In view of Proposition 0.5 of the Voisin's paper, it boils down to whether Hodge loci are defined over $\\overline{\\mathbb{Q}}$ or not. I have some ideas about how to translate this into $\\Pi^0_3$ but nothing close to $\\Pi^0_2$. Naturally, if it is true that the Hodge conjecture can _not_ be formulated as a $\\Pi^0_2$ statement for now, there is no way one can actually prove this fact. In case it is not unlikely to be so, I am only asking for an expert opinion. Also, if possible, I would appreciate a slick and sound proof of the \"upper bound\" $\\Pi^0_3$ (although I have some ideas about it, I do not really like them). \n", "link": "https://mathoverflow.net/questions/289745/what-is-the-logical-complexity-of-the-hodge-conjecture", "tags": ["ag.algebraic-geometry", "lo.logic"], "votes": 18, "creation_date": "2018-01-02T03:07:33", "comments": ["@Libli See en.wikipedia.org/wiki/Arithmetical_hierarchy", "what are $\\Pi_{2}^0$ and $\\Pi_{3}^0$?"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "204", "site": "mathoverflow", "title": "Is this Variation of the Continuum Hypothesis Inconsistent with ZFC or ZF?", "body": "It is a well-known fact that the Generalized Continuum Hypothesis is undecidable from ZFC. For similar sentences $\\phi$, this is simply equivalent to ZFC having a model $M$ for which $M\\models\\phi$.\n\nI wanted to see if there was any way to make all $\\beth$-numbers limits. At first, this seemed dubious, but now, after creating a variation on GCH, I have realized it seems completely legitimate. \n\nThe variation I created is called the **Cofinal Continuum Hypothesis** , or CFCH. It is defined as follows:\n\n$$\\forall\\kappa(2^\\kappa=\\aleph_{\\kappa^+})$$\n\nThis immediately destroys the idea of the continuum hypothesis, as with this definition $2^{\\aleph_0}=\\aleph_\\Omega$, which is very far off from $\\aleph_1$.\n\nThe reason I chose specifically this definition is that its cofinality preserving properties when compared to GCH, namely that $\\mathrm{cof}(2^\\kappa)=\\mathrm{cof}(\\kappa^+)$. This property \"piggybacks\" off of GCH's undecidability in a sense; most proofs that a variation on GCH is inconsistent with ZFC (or similar theories) involve some failure of an inequality with $\\mathrm{cof}(2^\\kappa)$ and $\\mathrm{cof}(\\kappa)$. With this definition, it is clear to see that this problem no longer exists because in GCH $\\mathrm{cof}(2^\\kappa)=\\mathrm{cof}(\\kappa^+)$.\n\nBecause of this construction, one could generalize it to $2^\\kappa$ is the $\\kappa^+$-th aleph fixed point, or $2^\\kappa=\\aleph_{\\aleph_\\kappa}$. These, however, seem even more audacious than CFCH itself.\n\n**However** , the reason I feel this is still dubious is because of it's interesting relation to the Singular Cardinal Hypothesis. It is clear that, with this definition, the first strong limit cardinal is a singular cardinal. This becomes a problem when introduced to the Singular Cardinal Hypothesis, as, assuming and CFCH and SCH:\n\n$$\\forall\\alpha(\\mathrm{cof}(\\beth_{\\omega\\cdot\\alpha})\\neq\\beth_{\\omega\\cdot\\alpha}\\rightarrow 2^{\\beth_{\\omega\\cdot\\alpha}}=\\beth_{\\omega\\cdot\\alpha}^+)$$ $$2^{\\beth_\\omega}=\\beth_\\omega^+$$ $$2^{\\beth_\\omega}=\\aleph_{\\beth_\\omega^+}$$ $$\\beth_\\omega^+=\\aleph_{\\beth_\\omega^+}$$\n\nThis is a contradiction, as it implies that a successor cardinal is an $\\aleph$-fixed point. Thus, CFCH is inconsistent with SCH. Proving CFCH equiconsistent to ZFC would be a great feat, as it would prove $\\neg$SCH equiconsistent to ZFC as well, thus showing that the existence of a measurable cardinal with Mitchell order $\\kappa^{++}$ is equiconsistent to ZFC (as a result of Gitik), and every other large cardinal axiom implied by said axiom (which has not yet been done). This is why I doubt that CFCH will be proven equiconsistent with ZFC.\n\n**Here are some other interesting facts implied by CFCH:**\n\n * $\\beth_{\\omega\\cdot\\alpha}$ is the $\\alpha$-th $\\aleph$-fixed point (and thus all strong limit cardinals are $\\aleph$-fixed points). This is yet another doubt I have for CFCH.\n * $\\mathrm{cof}(\\beth_\\alpha)=\\mathrm{cof}(\\aleph_\\alpha)$ (and thus for regular cardinals $\\aleph_\\alpha$, $\\mathrm{cof}(\\beth_\\alpha)=\\aleph_\\alpha$ and for limit ordinals $\\alpha$, $\\mathrm{cof}(\\beth_\\alpha)=\\alpha$.)\n * The Konig's Theorem corollaries about cofinality seem are implied by it \n($\\kappa<\\mathrm{cof}(2^\\kappa)=\\mathrm{cof}(\\kappa^+)=\\kappa^+$ and $2^\\kappa<(2^\\kappa)^{\\mathrm{cof}(2^\\kappa)})=2^{\\kappa\\cdot\\mathrm{cof}(2^\\kappa)}=2^{\\mathrm{cof}(2^\\kappa)}>2^\\kappa$)\n * As for other variations of the Generalized Continuum Hypothesis and Continuum Hypothesis: \n * $\\mathrm{CFCH}\\rightarrow\\neg\\mathrm{GCH}$\n * $\\mathrm{CFCH}\\rightarrow\\neg\\mathrm{SCH}$\n * $\\mathrm{CFCH}\\rightarrow\\neg\\mathrm{CH}$\n * $\\mathrm{CFCH}\\rightarrow\\neg\\mathrm{NCH}$ (The Natural Continuum Hypothesis, which is found in another MathExchange link \"[A New Continuum Hypothesis](https://mathoverflow.net/questions/148756/a-new-continuum-hypothesis-revised-version)\")\n\n\n\n**In general, it would be really useful if the reader of this question reply in the comments with either a link to a website listing many interesting models of ZF and ZFC or several Forcing methods or they send me the names of the models themselves.** Of course, you could reply to this question with comments pertaining to the question as well.\n", "link": "https://mathoverflow.net/questions/283443/is-this-variation-of-the-continuum-hypothesis-inconsistent-with-zfc-or-zf", "tags": ["lo.logic", "set-theory", "continuum-hypothesis"], "votes": 18, "creation_date": "2017-10-13T21:31:39", "comments": ["I mainly mentioned NCH because it seems like this is a \"variant\" of NCH (where cardinal exponentiation speeds up)", "Just as a reminder, I would like to mention that despite its tempting appearance and intuitive background from natural numbers, the Natural Continuum Hypothesis (NCH) is inconsistent with ZFC due to Konig's lemma. However, it sounds interesting to me that you considered the cofinality version of CH independently.", "Under CFCH, $2^{\\aleph_n} = \\aleph_{\\omega_{n+1}}$ and therefore $2^{\\aleph_{\\omega}} = \\aleph_{\\omega_\\omega}^\\omega = \\aleph_{\\omega_{\\omega + 1}}$. By the theory of pcf, we get that if $a$ is the set of all cardinals between $\\aleph_{\\omega}$ and $\\aleph_{\\omega_\\omega}$, the possible cofinalities of $\\prod a$ is the regular cardinals in the range between $\\aleph_{\\omega_\\omega + 1}$ and $\\aleph_{\\omega_{\\omega + 1}}$ which has cardinality $|a|^+$. A recent paper of Gitik claims that it is consistent, but it is open if one can obtain this type of pcf structure in accessible cardinals.", "Ah, so ZFC + CFCH is even stronger.", "Easton's theorem implies that ZFC + \"for all regular $\\kappa$, $2^\\kappa = \\aleph_{\\kappa^+}$\" is equiconsistent with ZFC. Other variants, such as \"the $\\kappa^+$-th fixpoint\" or $\\aleph_{\\aleph_{\\kappa^+}}$ are possible, too. - But it is well known that ZFC+ non-SCH is much stronger than ZFC, in terms of consistency strength."], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "205", "site": "mathoverflow", "title": "Is the universality of the surreal number line a weak global choice principle?", "body": "I'd like to consider the principle asserting that the surreal number line is universal for all class linear orders, or in other words, that every linear order (including proper-class-sized) linear orders, order-embeds into the surreal number line.\n\nThis principle is the class-analogue of the corresponding result for countable linear orders, proved by Cantor, namely, that every countable linear order embeds into the rational line $\\newcommand\\Q{\\mathbb{Q}}\\Q$. Using the \"forth\" part of Cantor's back-and-forth method, one enumerates a given countable linear order and then maps each new point to a rational number filling the corresponding cut in the images of the previous points.\n\nA similar transfinite extension of this argument works with the surreal numbers. Namely, if $\\langle A,\\leq_A\\rangle$ is any class-sized linear order and $A$ is $\\newcommand\\Ord{\\text{Ord}}\\Ord$-enumerable, then we may map each new point to the first-born surreal number filling the corresponding cut in the images of the previous elements of $A$, and thereby construct an embedding of $A$ to the surreal numbers.\n\nWhat this argument shows is that, provided the elements of the linear order $A$ are $\\Ord$-enumerable, then we may recursively build up an embedding of the linear order into the surreal number line. In other words, if the principle of global choice holds, which is equivalent to the assertion that every class is $\\Ord$-enumerable, then every class-sized linear order embeds into the surreal line. So global AC implies that the surreals are universal for class linear orders.\n\nSimilar uses of global choice underlie many other assertions of universality for the surreal numbers, with respect to class-sized structures, as ordered fields and so on. It is fundamental in all the arguments that I know for such universality that one should be able to enumerate the elements of the domain structure in order type $\\Ord$, in order to carry out the transfinite embedding procedure.\n\nSince this use of the $\\Ord$-enumerations has always struck me as essential, I have long believed that the global axiom of choice was required in order to prove that the surreal number line is universal for class linear orders.\n\nBut upon recent reflection, I realized that I don't actually have a proof that global AC is required. I simply don't have a proof that global AC is required for universality, and at the same time, I don't know how to omit it.\n\nHence these questions:\n\n * Is the global axiom of choice required to prove that the surreal number line is universal for all class-sized linear orders?\n\n * Can one prove the universality of the surreal number line merely in GB+AC, without the global axiom of choice?\n\n * Or is the universality of $\\text{No}$ an intermediate principle between AC and global AC?\n\n\n\n\nTo answer these questions, it would seem to be useful to settle the issue of the surreal-numbers-universality principle (which I propose we denote by $\\text{UNo}$), in some of the known models where global AC fails. For example, in my answer to the question [Does ZFC prove the universe is linearly orderable?](https://mathoverflow.net/a/110823/1946), I showed that there are models of ZFC with no definable linear order of the universe. If we could settle the universality of the surreals in that model, it would answer at least one of the questions above. For example, if the model had surreal-universality UNo, then we would know that universality is not equivalent to global AC; if it is didn't have it, then we would know that UNo isn't provable in GB+AC.\n\n(This topic is related to some ideas I'll be speaking about at my talk on [The hypnagogic digraph](http://jdh.hamkins.org/the-hypnagogic-digraph-jmm-2016/), which I am giving this Friday at the JMM in Seattle.)\n", "link": "https://mathoverflow.net/questions/227849/is-the-universality-of-the-surreal-number-line-a-weak-global-choice-principle", "tags": ["lo.logic", "set-theory", "axiom-of-choice", "linear-orders", "surreal-numbers"], "votes": 18, "creation_date": "2016-01-06T22:01:00", "comments": ["Or to put it the other way round: the fact that a class-sized linear order A is embeddable in No apparently does not give a way to Ord-enumerate A. If one wants to somehow use the property `all class-sized linear orders are embeddable in No', then again, we trivially know that all class-sized linear orders isomorphic to a subclass of No are embeddable...so you would have to come up with a class-sized linear order B defined outside of No, and from B's embeddability in No deduce that No is Ord-enumerable. [Well, I hope I'm making some sense here, feel free to disregard my question/comments].", "The question intrigues me although I have extremely little knowledge of such matters (but I'm pondering on variations of surreal numbers for constructive math). If I understand correctly, No itself is Ord-enumerable iff Global AC holds? (I don't know the first thing about this, but read it in another post). If that is correct, then intuitively I'd say: global AC is not necessary for UNo, because one can embed No in No without global AC... This may seem a strange argument, but it is meant to illustrate that one can have weaker conditions which already give embeddability.", "Keep in mind that the surreal numbers No themselves are definable in L[G], and as elements these define every object in L[G].", "Also, when you appeal to UNo, do you have to shout UNo when you have one line left in your proof, or else you have to take four more lines to finish?", "Joel, what I'd try to do is to show that every linear ordering comes from a bounded part.", "If you think about $L[G]$ in the model in my answer to your question about linear orders of $V$, then the question is whether UNo holds there. We know there is no linear order of all of $L[G]$ there, so perhaps there is a paucity of definable linear orders there (always allowing parameters). If so, then we might expect UNo to hold. But I can't quite prove it.", "I just opened my eyes ten minutes ago, it's my pleasure to have choice related questions to think about when I wake up. :-)", "@AsafKaragila But let me say that I'm very glad you're looking at this question. I'd appreciate any insight you might have.", "I think of global-AC as an assertion in second-order set theory, as in Goedel-Bernays set theory, so one doesn't need to worry about definability. That is, one simply asserts that there is a proper class choice function on all nonempty sets, or equivalently, that there is an Ord-enumeration of V. In particular, global AC definitely holds after adding a Cohen real.", "When you say global choice, does it come with parameters? Specifically, if you add a Cohen real to $L$, you have global choice with parameters, but not global choice itself. It stands to reason that in such model you might have what you were looking for.", "I'll post my slides on Friday, and you can read all about it!", "Hypnagogic digraph! Quite an intriguing title. Will there be forthcoming discussions of the hypnopompic digraph as well? :-)"], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "206", "site": "mathoverflow", "title": "Does the "holomorphic spheres-to-continuous spheres" forgetful function respect the mixed Hodge structures on homotopy groups?", "body": "For each smooth, projective, complex variety $X$ that is simply connected, John Morgan constructed a natural mixed Hodge structure on the homotopy group $\\pi_k(X,x)\\otimes \\mathbb{Q}$. This was extended to (possibly singular) quasi-projective varieties independently by Hain, by Deligne, and by Navarro Aznar. For ~~any~~ the basepoint connected component of the double loop space $\\Omega^2 X_0\\sim [(S^2,0),(X,x)]_0$, each homotopy group $\\pi_k(\\Omega^2 X_0)\\otimes \\mathbb{Q}$ is canonically isomorphic to $\\pi_{k+2}(X)\\otimes \\mathbb{Q}$, $k>0$. Thus these also have natural mixed Hodge structures. \n\nGrothendieck constructed a Hom scheme $$H=\\text{Hom}_{\\mathbb{C}}((\\mathbb{CP}^1,0),(X,x))$$ parameterizing algebraic maps (i.e., holomorphic maps in case $X$ is projective). Each quasi-compact, open and closed subset $H_\\beta$ of the Hom scheme is quasi-projective. Assuming that $H_\\beta$ is connected, there is a well-defined homotopy class of maps $$\\Phi_{\\beta}:H_\\beta \\to \\Omega^2 X_0,$$ that associates to each holomorphic map from $\\mathbb{CP}^1 = S^2$ the underlying continuous map. There are associated pushforward maps on homotopy groups, $$\\Phi_{k,\\beta}: \\pi_k(H_\\beta)\\otimes \\mathbb{Q} \\to \\pi_{k+2}(X,x)\\otimes \\mathbb{Q}.$$ Assume now that $H_\\beta$ is simply connected so that the groups $\\pi_k(H_\\beta)\\otimes \\mathbb{Q}$ have mixed Hodge structures. (There are many cases of $X$ where the associated spaces $H_\\beta$ are connected and simply connected. Also, with no hypothesis on $H_\\beta$, one can also ask about compatibility with mixed Hodge structures of the induced homomorphism $H_1(H_\\beta;\\mathbb{Q}) \\to \\pi_3(X,x)\\otimes \\mathbb{Q}.$) Also, as Dan Petersen points out, it is necessary to form the Tate twist of the target, i.e., $\\pi_{k+2}(X,x)\\otimes \\mathbb{Q}(1)$. \n\n**Question**. Is the pushforward map $\\Phi_{k,\\beta}$ a morphism of mixed Hodge structures?\n\nOf course there are many complex varieties $X$ for which $H$ is empty, and then the answer is trivially yes. Yet there are many other complex projective manifolds, e.g., Fano manifolds, where the maps $\\Phi_{k,\\beta}$ are nontrivial.\n", "link": "https://mathoverflow.net/questions/233979/does-the-holomorphic-spheres-to-continuous-spheres-forgetful-function-respect", "tags": ["ag.algebraic-geometry", "homotopy-theory", "hodge-structure"], "votes": 18, "creation_date": "2016-03-18T08:22:50", "comments": ["@AknazarKazhymurat No, unfortunately I have not found an answer to this question.", ". . . This reduces to the smooth, projective case. In that case, the Sullivan minimal model of the De Rham complex agrees with the Sullivan minimal model of the De Rham cohomology (with zero differentials). Since the Sullivan minimal model of a dga is (roughly) a multilinear algebra operation on the dga, the work is to check that these multilinear operations respect the weight and Hodge filtrations.", "@VivekShende. \"Is this mixed hodge structure constructed by actually making a simplicial scheme ...\" Whether it can be made that way, I do not believe that is how Morgan first constructed these mixed Hodge structures. Morgan uses a single log resolution and an algebraic construction (basically by spectral sequences) of the rational homotopy type of the open stratum in terms of those of the (projective, smooth) boundary strata. Deligne proved that this spectral sequence respects mixed Hodge structures. Morgan checks that is also respects Sullivan minimal models . . .", "Is this mixed hodge structure constructed by actually making a simplicial scheme (which?) whose cohomology groups are the homotopy groups of the original $X$?", "@DanPetersen. You are correct, there should be a Tate twist. I will add it now.", "I'd expect there to be a Tate twist at least. If $k=0$ you get a map from $H_0(H_\\beta)=\\mathbf Q(0)$ to $H_2(X)$, which has weight $-2$.", "\"Is it easy to see?\" Yes; this is part of adjointness of the space of continuous functions with the compact-open topology. Let $C$, $X$, and $H$ be topological spaces with $H$ Hausdorff. Let $F:H\\times C \\to X$ be continuous. The claim is that the induced function $\\Phi:H\\to [C,X]$ is continuous, where $[C,X]$ has the compact-open topology. For an open subset $U$ of $X$, for a compact subset $K$ of $C$, the subset $(H\\times K)\\cap \\Phi^{-1}(X\\setminus U)$ is a closed subset of $H\\times K$, hence it is proper over $H$. So it has closed image in $H$, and the complement of the image is open.", "Yes, I did mean that topology. Is it easy to see? Are the $H_\\beta$s actually homeomorphic to subspaces of $\\Omega^2X$?", "\"Is it easily seen that there is a continuous $\\Phi_\\beta$?\" Certainly $\\Phi_\\beta$ is not continuous for the Zariski topology on $H_\\beta$. It is continuous for the analytic topology on the associated analytic space of $H_\\beta$. Similarly, the notation $\\pi_{k+2}(X,x)$ means the homotopy group for the associated analytic space of $X$.", "Is it easily seen that there is a continuous $\\Phi_\\beta$?", "@TomChurch. You are correct. I am eliding the issue that the function $\\Phi_\\beta$ is only well-defined up to homotopy. There is a canonical function $\\Phi_\\beta$ to the connected component of $\\Omega^2 X$ corresponding to the homotopy class $\\beta\\in \\pi_2(X)$. The homotopy equivalence of this component with the basepoint component is only well-defined up to homotopy.", "What does the subscript 0 mean in $\\Omega^2 X_0\\sim [(S^2,0),(X,x)]_0$? If it means \"basepoint component\", then why do you say \"for any connected component\"? If not, why is the isomorphism between $\\pi_k(\\Omega^2 X_0)\\otimes \\mathbb{Q}$ and $\\pi_{k+2}(X)\\otimes \\mathbb{Q}$ canonical?"], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "207", "site": "mathoverflow", "title": "Deforming a basis of a polynomial ring", "body": "The ring $Symm$ of symmetric functions in infinitely many variables is well-known to be a polynomial ring in the elementary symmetric functions, and has a $\\mathbb Z$-basis of Schur functions $\\\\{S_\\lambda\\\\}$, with structure constants $c_{\\lambda\\mu}^\\nu$ called Littlewood-Richardson numbers.\n\nI have been studying a commutative deformation of these structure constants, i.e. lifting $c_{\\lambda\\mu}^\\nu$ to a 2-variable polynomial $c_{\\lambda\\mu}^\\nu(a,b)$, for which I have a very explicit formula (much too complicated to describe here). However, since $Symm$ is a polynomial ring (i.e. free), any commutative deformation of it must be trivializable: there must be a way to deform each $S_\\lambda$ to a function $S_\\lambda(a,b)$ whose ordinary multiplication has my structure constants.\n\n> Given the $\\\\{c_{\\lambda\\mu}^\\nu(a,b)\\\\}$, how can I trivialize this family, finding such deformations $S_\\lambda(a,b)$?\n\nIt seems like the answer should be nonunique (hence harder to find); it's easy to imagine deformations of the basis that don't change the structure constants at all. \n\n> What extra conditions should one put on the deformation in order to make the trivialization unique?\n", "link": "https://mathoverflow.net/questions/199537/deforming-a-basis-of-a-polynomial-ring", "tags": ["co.combinatorics", "ac.commutative-algebra", "symmetric-functions", "algebraic-combinatorics", "schur-functions"], "votes": 18, "creation_date": "2015-03-09T18:40:56", "comments": ["I think $a=0$ is triangular one way, $b=0$ triangular the other way. The one-box rule is here: mathoverflow.net/questions/88569/…", "Is it possible that the deformation is upper-triangular (i.e. that $S_\\lambda(a,b) = S_\\lambda + \\textrm{lower order terms}$)? How complicated are the structure constants when $\\mu$ is 1 box?"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "208", "site": "mathoverflow", "title": "An algebraic strengthening of the Saturation Conjecture", "body": "The Saturation Conjecture (proved by Knutson-Tao) asserts that $c_{n\\mu,n\\nu}^{n\\lambda}\\neq 0\\Rightarrow c_{\\mu,\\nu}^{\\lambda} \\neq 0$, where $c$ denotes a Littlewood-Richardson coefficient and $n$ is a positive integer. Let $\\mathfrak{o}$ be a (commutative) discrete valuation ring with a finite residue field $k=\\mathfrak{o}/\\mathfrak{p}$. Let $M$ be a finite $\\mathfrak{o}$-module. (For instance, we can take $\\mathfrak{o}$ to be the $p$-adic integers $\\mathbb{Z}_p$, in which case $M$ is a finite abelian $p$-group.) Thus there is a unique partition $\\lambda=(\\lambda_1,\\lambda_2,\\dots,\\lambda_r)$ such that $$ M \\cong \\bigoplus_{i=1}^r (\\mathfrak{o}/\\mathfrak{p}^{\\lambda_i}). $$ We call $\\lambda$ the _type_ of $M$. A fundamental result of Philip Hall is that if $M$ has type $\\lambda$, then there exists a submodule $N$ of type $\\mu$ and cotype $\\nu$ (i.e., $M/N$ has type $\\nu$) if and only if $c^{\\lambda}_{\\mu\\nu}\\neq 0$. See Macdonald, _Symmetric Functions and Hall Polynomials_ , Chapter II, (4.3).\n\nNow suppose that $M$ has type $n\\lambda$, and $N$ has type $n\\mu$ and cotype $n\\nu$. Is there is a submodule $L$ of $M$ of type $\\lambda$, such that $L\\cap N$ has type $\\mu$ and cotype (with respect to $L$) $\\nu$? \n", "link": "https://mathoverflow.net/questions/212368/an-algebraic-strengthening-of-the-saturation-conjecture", "tags": ["co.combinatorics", "symmetric-functions"], "votes": 18, "creation_date": "2015-07-26T11:57:23", "comments": ["@DavidSpeyer You are right about the definition of Hall algebra. I have fixed this.", "A note on terminology (and then I will think about this interesting question). I usually understood the Hall algebra to refer to the ring whose elements were formal sums of isomorphism classes of $\\mathfrak{o}$-modules, and where multipication was given by counting submodules of given type and co-type. See, for example, arxiv.org/abs/math/0611617 . Also, as I imagine you know, Derksen and Weyman gave a far reaching generalization of the saturation conjecture in terms of this sort of Hall algebra -- see ams.org/journals/jams/2000-13-03/S0894-0347-00-00331-3", "Is there a combinatorial interpretation for some special case (similar to Yamanouchi tableaux for the LR-coeffs)? The saturation conjecture implies the analogous statement for Kosktka and skew Kosktka coefficients, so is there a \"Kostka\"-analogue of this question, which should be easier to prove?"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "209", "site": "mathoverflow", "title": "Smooth curves on smooth varieties", "body": "Let $X$ be a smooth, proper algebraic variety over a field $k$, of positive dimension. \n\nIs it true that $X$ contains a smooth Zariski-closed curve? \n\nIf it is projective, this is true by Bertini. But is it true in general?\n", "link": "https://mathoverflow.net/questions/112524/smooth-curves-on-smooth-varieties", "tags": ["ag.algebraic-geometry"], "votes": 18, "creation_date": "2012-11-15T13:44:16", "comments": ["@algori : yes, indeed ! The answer is positive if $X$ is separably rationally connected (over an algebraically closed field). Indeed, if $X$ is of dimension $\\leq 2$, it is projective so there is no problem. And if it is of dimension $\\geq 3$ it contains (lots of) smooth rational curves by Theorem IV 3.9 in Koll'ar's book \"Rational curves on algebraic varieties\". ", "Olivier -- it's worse than you think: I'm not sure how to prevent the curve from intersecting itself.", "@algori : The problem with this strategy is that it is really possible to have a divisor $E$ on $X'$, exceptional over $X$, such that every curve meeting $E$ has singular image in $X$. This happens for instance if $X$ is a smooth surface, and $X'$ is constructed by blowing up a point, then blowing up a point on the exceptional divisor, then blowing up the intersection of the two exceptional divisors, and if you choose $E$ to be the third exceptional divisor. The problem is that the differential of $X'\\to X$ is identically zero at every point of $E$. ", "Here is a strategy for showing that the answer is positive: blow up a proper smooth variety, call it $X$, to get a projective smooth $X'$ such that $X'\\setminus D'$ is isomorphic to $X\\setminus D$ with $D\\subset X$ closed and $D'\\subset X'$ a divisor with normal crossings; then take a curve in $X'$ that intersects transversally all the components of $D'$ and project back again."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "210", "site": "mathoverflow", "title": "An "exercise" on von Neumann algebra tensor product", "body": "The following problem appears to be an easy exercise on von Neumann algebra tensor products, but since I've been failing to find a rigorous proof, I'd like to make sure it's not that trivial. Suppose $M$ and $N$ are von Neuamann factors such that $L^\\infty[0,1] \\mathbin{\\bar\\otimes} M \\cong L^\\infty[0,1] \\mathbin{\\bar\\otimes} N$. Does it follow $M \\cong N$? It's true under the assumption of separability by the disintegration theory, but not so sure if $M$ (and hence $N$) does not have separable predual.\n", "link": "https://mathoverflow.net/questions/270650/an-exercise-on-von-neumann-algebra-tensor-product", "tags": ["oa.operator-algebras", "von-neumann-algebras"], "votes": 18, "creation_date": "2017-05-25T03:32:20", "comments": ["Obviously when $M$ or $N$ are Type $\\mathrm{I}$ it can done in a different way and the result is well-known, but I'm unsure how to proceed if the factors in question are Type $\\mathrm{II}_\\infty$ or Type $\\mathrm{III}$, even assuming separability.", "This is obviously achievable if $M$ and $N$ are $\\mathrm{II}_1$ factors, with $H$ and $K$ being their standard representations. In which case $L^2(0, 1) \\otimes H$ and $L^2(0, 1) \\otimes K$ are standard representations of $L^\\infty(0, 1) \\bar{\\otimes} M$ and $L^\\infty(0, 1) \\bar{\\otimes} N$, respectively, so a spatial isomorphism of the above form just comes from any isomorphism between $L^\\infty(0, 1) \\bar{\\otimes} M$ and $L^\\infty(0, 1) \\bar{\\otimes} N$. But without trace can this still be done?", "I'm late here, but I'm curious as to how the question is resolved even in the separable case. All the treatments I could find on direct integral theory always needs a spatial isomorphism between $L^\\infty(0, 1) \\bar{\\otimes} M$ and $L^\\infty(0, 1) \\bar{\\otimes} N$. More precisely, suppose $M$ acts on $H$ and $N$ acts on $K$, it seems to always require a unitary between $L^2(0, 1) \\otimes H$ and $L^2(0, 1) \\otimes K$ that conjugates $L^\\infty(0, 1) \\bar{\\otimes} M$ and $L^\\infty(0, 1) \\bar{\\otimes} N$ onto each other."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "211", "site": "mathoverflow", "title": "What is the Hochschild cohomology of the Fukaya-Seidel category?", "body": "Let $(Y, \\omega)$ be a compact symplectic manifold and let $Fuk(X,\\omega)$ be its Fukaya category. The Hochschild cohomology of this category should be given by $HH^\\bullet(Fuk(Y,\\omega))=H^\\bullet(Y, \\mathbb{C})$, at least at the level of vector spaces (the product structure should be the quantum cohomology on $H^\\bullet(Y,\\mathbb{C})$: see for example [Hochschild (co)homology of Fukaya categories and (quantum) (co)homology](https://mathoverflow.net/questions/11081/hochschild-cohomology-of-fukaya-categories-and-quantum-cohomology) ). As $Fuk(Y,\\omega)$ should be a Calabi-Yau category (see for example [Are Fukaya categories Calabi-Yau categories?](https://mathoverflow.net/questions/13114/are-fukaya-categories-calabi-yau-categories) ), the Hochschild homology $HH_\\bullet(Fuk(Y,\\omega))$ coincides with the Hochschild cohomology $HH^\\bullet(Fuk(Y,\\omega))$ up to a shift of the grading.\n\nLet me deform the preceding setting: let me assume that $(Y,\\omega)$ is a Kähler manifold and that $W \\colon Y \\rightarrow \\mathbb{C}$ is an holomorphic function on $Y$ (in particular $Y$ is noncompact if one wants $W$ non-constant). One should be able to define the Fukaya-Seidel category $FS((Y,\\omega),W)$ (I probably need to assume that $W$ has only isolated non-degenerate critical points given the current technology). The Hochschild homology of this category should be given by: \n\n$HH_\\bullet(FS((Y,\\omega),W)=H^\\bullet(Y,Y_{-\\infty},\\mathbb{C})$, \n\nwhere we take the cohomology relative to the fiber $Y_{-\\infty}=W^{-1}(z)$ of a point $z\\in\\mathbb{C}$ with $Re(z)<<0$. But now $FS((Y,\\omega),W)$ is no longer Calabi-Yau and so the Hochschild cohomology is no longer simply related to the Hochschild homology. So my question is:\n\nWhat is (or should be) the Hochschild cohomology $HH^\\bullet(FS((Y,\\omega),W))$ of the Fukaya-Seidel category $FS((Y,\\omega),W)$?\n\nIf $(Y,\\omega),W)$ is the mirror of a smooth projective Fano variety $X$ then \n\n$FS((Y,\\omega),W)=D^b Coh(X)$ \n\nand one knows by the Hochschild-Kostant-Rosenberg theorem that \n\n$HH_\\bullet(D^bCoh(X))=\\oplus_{p-q=\\bullet}H^p(X,\\Omega_X^q)$ \n\nand \n\n$HH^\\bullet(D^bCoh(X))=\\oplus_{p+q=\\bullet}H^p(X,\\wedge^q T_X)$.\n\nIn particular, Hochschild homology and cohomology are indeed very different in general.\n\nThis question is motivated by the general principle saying that the deformations of a category are governed by $HH^2$. In particular, a subquestion could be: what are the deformations of the Fukaya-Seidel category?, and this should be equivalent to understand what $HH^2$ is.\n\nDisclaimer: when I write that something \"should be\" true, it is what I think is the general expectation but as I am not an expert in symplectic geometry, I don't know the technical state of the art of the rigorous proofs. As my question is itself of the \"should be\" form, I hope it is fine.\n", "link": "https://mathoverflow.net/questions/218232/what-is-the-hochschild-cohomology-of-the-fukaya-seidel-category", "tags": ["sg.symplectic-geometry", "mirror-symmetry", "hochschild-cohomology", "fukaya-category"], "votes": 18, "creation_date": "2015-09-13T14:06:08", "comments": ["Another important paper which is related to your question is Seidel's symplectic homology as a Hochschild homology. If you denote by $\\mathscr{B}$ the Fukaya category of the fiber and $\\mathscr{A}$ the Fukaya-Seidel category, then $HH_\\ast(\\mathscr{A}\\oplus t\\mathscr{B}[[t]])$ gives the symplectic homology of the total space.", "The deformation issues concerning fiberwise compactification of a Lefschetz fibration are discussed in II1/2. The key conceptual piece is the fixed point Floer cohomology.", "You probably need to read Seidel's papers entitled Fukaya $A_\\infty$ structures associated to Lefschetz fibrations I, II, II1/2."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "212", "site": "mathoverflow", "title": "A cohomology class associated with a complex representation of a group", "body": "$\\newcommand\\CC{\\mathbb C}\\newcommand\\ZZ{\\mathbb Z}\\newcommand\\ad{\\mathsf{ad}}\\newcommand\\Ext{\\operatorname{Ext}}$ Suppose that $G$ is a finite group and that it acts on a finite dimensional complex vector space $V$. \n\nFor each $g\\in G$ we may consider the subspace $V_g$ of $V$ spanned by the eigenvectors of $g$ corresponding to eigenvalues different from $1$, and put $d(g)=\\dim V_g$. We pick any nonzero linear map $\\omega_g:\\Lambda^{d(g)}V_g\\to\\CC$.\n\nIf $h\\in G$ then the map $x\\in \\Lambda^{d(g)}V_{hgh^{-1}}\\mapsto \\omega_g(h^{-1}v)\\in\\CC$ makes sense, and we write it $h\\cdot\\omega_g$. Since it is not zero, there is a nonzero scalar $\\lambda(h,g)\\in\\CC^\\times$ such that $h\\cdot\\omega_g=\\lambda(h,g)\\omega_{hgh^{-1}}$. Associativity of the group action implies that $$\\lambda(gh,k)=\\lambda(h,k)\\lambda(g,hkh^{-1})$$ for all $g$, $h$, $k\\in G$. This means that the function $\\lambda:G\\times G\\to\\CC^\\times$ is a $1$-cocycle in the complex which computes $\\Ext_{\\ZZ G}^\\bullet(\\ZZ G^\\ad,\\CC^\\times)$, when we use a bar resolution of the $G$-module $\\ZZ G^\\ad$ (which is the permutation $G$-module constructed from the conjugation action of $G$ on itself) We may therefore take its class $[\\lambda]$ in $\\Ext^1_{\\ZZ G}(\\ZZ G^\\ad,\\CC^\\times)$. \n\nThis class depends only on the representation and not on the choice of the $\\omega_g$'s, so we write it $c(V)$.\n\nThe long exact sequence corresponding to the exponential sequence $0\\to\\ZZ\\to\\CC\\to\\CC^\\times\\to0$ allows us to identify that $\\Ext^1$ with $\\Ext^2_{\\ZZ G}(\\ZZ G^\\ad,\\ZZ)$. If $CG$ is the set of conjugacy classes and for each $c\\in CG$ we let $\\ZZ c$ be the permutation module corresponding to the conjugation action on $c$, we have $\\ZZ G^\\ad=\\bigoplus_{c\\in CG}\\ZZ c$. If $G_c$ denotes the centralizer of an element of $c$, some form of Shapiro's lemma allows us to identify $\\Ext^2_{\\ZZ G}(\\ZZ c,\\ZZ)$ with $\\Ext^2_{\\ZZ G_c}(\\ZZ,\\ZZ)$ and then our class $[\\lambda]$ is an element of $$\\bigoplus_{c\\in CG}H^2(G_c,\\ZZ).$$ This can be written a bit more canonically as $$\\left(\\bigoplus_{g\\in G}H^2(C_g,\\ZZ)\\right)^G$$ with now $C_g$ the centralizer of $g\\in G$ and $G$ acting on the direct sum in the only sensible way. If $G$ is abelian, this can be identified further to $\\ZZ G\\otimes H^2(G,\\ZZ)$.\n\n> **Question:** What is this cohomology class $c(V)$?\n\nThis reminds me of Burghelea's description of the cyclic homology of the group algebra, so it might be some form of Chern class (if some vector bundle constructed from V on the free loop space of BG?). \n\nThis class shows up when one computes the Hochschild cohomology of the cross product algebra $S(V)\\rtimes G$, but explaining how would make this post too much longer. It appears as an obstruction to making this nice :-/\n\n**On triviality:** suppose there is a central element $z$ in $G$ which acts by multiplication by a scalar different from $1$. In that case $c=\\\\{z\\\\}$ is a conjugacy class, $G_c=G$, $\\lambda(g,z)=\\det g$ for all $g$, and the component of $c(V)$ in $\\Ext^1_{\\ZZ G}(\\ZZ c,\\CC^\\times)=\\hom(G,\\CC^\\times)$ is the determinant map. If the representation does not land in $SL(V)$, then this is not zero. Examples of representations like this are the geometric representation of Weyl groups which have nontrivial center, in which case the longest element is central and acts as $-1$ —see [this question and its answers](https://mathoverflow.net/questions/81858/does-i-belong-to-weyl-group) for the list of which types work— or abelian groups acting with some element not having $1$ in its spectrum.\n\nIn the other direction, if the representation respects a symplectic form $\\omega$ on $V$, then all the $d(g)$ are even and we may take $\\omega_g$ to be the restriction of the $\\tfrac12d(g)$th power $\\omega^{d(g)/2}$ to $V_g$. In this case, $\\lambda(h,g)=1$ identically, so $c(V)=0$.\n\n**Note.** If $e$ is the exponent of $G$ and $\\Omega$ is any set of $e$th roots of unity, we can consider the subspace $V_g^\\Omega$ spanned by eigenvectors of $g$ corresponding to eigenvalues in $\\Omega$, and proceeding as above we get a class $c_\\Omega(V)$ in the same direct sum. The class above corresponds to taking $\\Omega$ the set of all $e$th roots of unity different from $1$. The assigment $V\\mapsto\\bigoplus_{g\\in G}V_g^\\Omega$, for $c$ a conjugacy class of $G$, is a nice functor into Yetter-Drinfeld modules, by the way.\n", "link": "https://mathoverflow.net/questions/232090/a-cohomology-class-associated-with-a-complex-representation-of-a-group", "tags": ["rt.representation-theory", "group-cohomology"], "votes": 18, "creation_date": "2016-02-24T21:43:54", "comments": ["(Here LBG is the free loop space on BG) The first Chern class of this descended bundle is an element of $H^2(LBG,\\mathbb Z)$, which is a direct sum of cohomologies of centralizers. Is that class the same as the one described? That would be an answer to \"what is c(V)?\".", "@Andreas, well, I know that the class is a bunch of homomorphisms (which are coherent in some way): that is how it is defined! What I am asking is what the class is, in the sense of is it something? For example, from V you can construct a vector bundle $\\tilde V=EG\\times_GV$ on $BG$, which you can pull back along the evaluation map $LBG\\times S^1\\to BG$ to get a vector bundle on that product which descends onto $LBG\\times_{S^1}S^1$, which is $LBG$.", "@MarianoSuárez-Alvarez: Your question was \"What is this cohomology class?\". Maybe I do not understand what a possible answer would be, but didn't appear the class in $H^1$ naturally from the representation. What are you looking for?", "$\\beta$ will be a cocycle here. An easy way to prove this is the following: in the twisted group algebra $\\mathbb{C}^{\\alpha}G$ we have the equality, if $g$ and $h$ commute in $G$, $U_h^{-1}U_gU_h = \\beta(g,h)U_g$. Now it is quite easy to prove that the resulting map $C_g\\rightarrow \\mathbb{C}^{\\times}$ is a homomorphism, and thus a one cocycle, since the action on $\\mathbb{C}$ is trivial.", "I honestly do not understand what «that is all» means in this context.", "The identification $H^2(C_g,\\mathbb Z)=H^1(C_g,\\mathbb C^{\\times})$ makes it look more obscure than it is. For each $g$, the centralizer $C_g$ fixes the subspace spanned by non-fixed eigenvectors and hence there is a determinant homomorphism $C_g \\to \\mathbb C^{\\times}$, i.e., a class in $H^1(C_g,\\mathbb C^{\\times})$, that's all.", "@EhudMeir, indeed, my example with the determinant was an example of something else! :-| I'll correct it. if I start with with a $2$-cocycle $\\alpha:G\\times G\\to\\CC^\\times$, fix $g\\in G$ and let $\\beta:h\\in G_g\\mapsto \\alpha(g,h)/\\alpha(h,g)\\in\\CC^\\times$, I am only being able to prove that $d\\beta=\\beta^2\\smile\\beta^2$, no that it is a $1$-cocycle on $G_g$; is that construction treated somewhere? (I can do it if $\\alpha:G\\times G^{\\ad}\\to\\CC^\\times$ is a $1$-cocycle giving an element of $\\Ext^1(\\ZZ G^\\ad,\\CC^\\times)$, though: is that what you meant?)", "Another question: if $[\\alpha]\\in H^2(G,\\mathbb{C}^{\\times})$, then we get another element in the same group $\\oplus_{c\\in CG}H^2(G_c,\\mathbb{Z})\\cong \\oplus_{c\\in CG}H^1(G_c,\\mathbb{C}^{\\times})$, namely if $h$ commutes with $g$ then we have the one cocycle $h\\mapsto \\alpha(g,h)/\\alpha(h,g)$. Is it possible to see any sort of connection between elements arising from two cocycles and from representations?", "Shouldn't $\\lambda(g,1_G)=1$? That is: for the identity element $e=1_G$ it holds that $d(e) = 0$ because the only eigenvalue of $e$ is 1. I guess that in this case the interpretation of $\\wedge^0 V_e$ should be $\\mathbb{C}$, and the induced map should be the identity?"], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "213", "site": "mathoverflow", "title": "Cohomological characterization of CM curves", "body": "In his 1976 classical Annals paper on $p$-adic interpolation, N. Katz uses the fact that if $E_{/K}$ is an elliptic curve with complex multiplications in the quadratic field $F$, up to a suitable tensoring the decomposition of the algebraic $H_{\\rm dR}^{1}(E,K)$ in eigenspaces for the natural $F^\\times$-action coincides with the Hodge decomposition of $H_{\\rm dR}^{1}(E,{\\Bbb C})$ and (for ordinary good reduction at $p$) with the Dwork-Katz decomposition of $H_{\\rm dR}^{1}(E)\\otimes B$ for $p$-adic algebras $B$.\n\nThen, he asks for a converse statement. Namely, is it true that if the Hodge decomposition of $H_{\\rm dR}^{1}(E,{\\Bbb C})$, where $E_{/K}$ is an elliptic curve, is induced by a splitting of the algebraic de Rham, then $E$ has complex multiplications?\n\nThe question is left unanswered in that paper. Does anyone know if the question has been answered since?\n", "link": "https://mathoverflow.net/questions/13681/cohomological-characterization-of-cm-curves", "tags": ["elliptic-curves", "cohomology", "nt.number-theory"], "votes": 18, "creation_date": "2010-02-01T06:36:19", "comments": ["Ok, I guess I was a bit lazy myself too..... :-) $K$ is a finite extension of $\\Bbb Q$ over which the complex multiplications are defined, that is $F\\subseteq K$. The comparison between algebraic $H^1$ and the Rham over $\\Bbb C$ is allowed by choosing (and fixing) an embedding $K\\rightarrow{\\Bbb C}$.", "What is $K$ in this question? Too lazy to find Katz' paper in the mess behind me :-) To compute the de Rham cohomology with coefficients in $C$ you'll need a map $K\\to C$ right? Or are we considering all such maps? Or have I misunderstood the question?"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "214", "site": "mathoverflow", "title": "Lipschitz constant of a homotopy", "body": "Let $n>0$, $\\mathbb S^n$ be $n$-sphere and $1\\in \\mathbb S^n$ be its north pole.\n\nA am looking for an example of compact manifold $M$ with a continuous $n$-parameter family of maps $h_x\\colon M\\to M$, $x\\in \\mathbb S^n$ such that \n\n * $h_1=\\mathop{\\rm id}_M$\n * For any Riemannian metric $g$ on $M$ and any family of maps $h'_x\\sim h_x$ (i.e., such that the maps $h,h'\\colon \\mathbb S^n\\times M\\to M$ are homotopic) there is $x\\in\\mathbb S^n$ such that the Lipschitz constant of $h'_x\\colon (M,g)\\to(M,g)$ is $\\ge 10000$. \n\n\n\n**Comments.**\n\n * I think about this problem for few years; I do not think it is an easy one, but I want to make sure that I did not miss a well known trick. A positive answer might have interesting consequences for collapsing with lower curvature bound.\n * If $n=0$, so $\\mathbb S^0=\\\\{+1,-1\\\\}$ then $M=\\mathbb T^k$ and $h_{-1}=$ a hyperbolic linear map with a big eigenvalue will do.\n * Denote by $F$ the space of all maps $M\\to M$. If $\\pi_n(F,\\mathop{\\rm id}_M)$ is finitely generated then for every $h_x\\colon (M,g)\\to (M,g)$ there is $h'_x$ with a fixed Lipschitz constant $L=L(M,g)$ for all $x$.\n * I know few examples of families $h_x$ for which one can not find a metric $g$ and homotopic family $h'_x$ with Lipschitz constant $=1$ (in other words the family can not be made isometric). \n * I do not know an answer even if instead of 10000 I would have $1+\\varepsilon$ for a fixed $\\varepsilon> 0$.\n\n\n", "link": "https://mathoverflow.net/questions/99276/lipschitz-constant-of-a-homotopy", "tags": ["dg.differential-geometry", "mg.metric-geometry", "homotopy-theory", "at.algebraic-topology"], "votes": 18, "creation_date": "2012-06-11T00:57:45", "comments": ["If it were me, I'd try to start : Let $\\Sigma$ be an exotic $n$-sphere. By conjugacy through the round sphere, this carries a very-transitive action by homeomorphisms of $S O (n+1)$; this action cannot be by diffeomorphisms, or else one could use it to give $\\Sigma$ an $S O (n+1)$-equivariant Riemann metric... But I can't see whether this is helpful after perturbing away from the subspace of group actions; still, I do expect exotic spheres to be useful somewhere, especially since it is known that some carry obstructions to global curvature positivities.", "@Henrik: it is almost correct, you should change \"any\" to \"some\".", "The paper \"Homotopical effects of dilation\" by Gromov considers similar question, but I haven't checked the exact relationship. Links: projecteuclid.org/euclid.jdg/1214434601 , or (non-paywalled) seven.ihes.fr/~gromov/PDF/4%5B19%5D.pdf", "So is the following rephrasing correct: The question is whether any element in $\\pi_k(Map(M,M))$ lies in the image of $\\pi_k(Lip_{10000}(M,g))$ for any Riemannian metric $g$ on $M$, where $Lip_C(M,g)$ denotes the space of self maps of $(M,g)$ Lipschitz constant $C$."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "215", "site": "mathoverflow", "title": "local equivalence of loop group representations", "body": "Let $G$ be a compact, simple, connected, simply connected (cscsc) Lie group, and let its smooth loop group $LG:=C^\\infty(S^1,G)$. Given an interval $I\\subset S^1$, we have the **_local loop group_** $$ L_IG := \\\\{\\gamma\\in LG \\ | \\ \\forall z\\not\\in I \\ \\gamma(z)=e\\\\} $$ which is a subgroup of $LG$. Let $k\\ge 1$ be an integer. The level $k$ central extension of $LG$ is denoted $\\mathcal{L}G_k$. It restricts to a central extension of the local loop group that we denote $\\mathcal{L}_IG_k$.\n\nA representation of $\\mathcal{L}G_k$ on a Hilbert space is called _**positive energy**_ if it admits a covariant action of $S^1$ (i.e., the action should extend to $S^1\\ltimes \\mathcal{L}G_k$) whose infinitesimal generator has positive spectrum. Here, the center of $\\mathcal{L}G_k$ is required to act by scalar multiplication.\n\n> **Definition 1:** \n> _Two level $k$ positive energy representation of the loop group are called **locally equivalent** if they become equivalent when restricted to $\\mathcal{L}_IG_k$._\n\nThe follows is believed to be true:\n\n> **Claim 2:** \n> _Let $G$ be a cscsc group and let $V$ and $W$ be any two positive energy representations of $\\mathcal{L}G_k$. Then $V$ and $W$ are locally equivalent._\n\n* * *\n\n~~I know a paper that proves the following:~~\n\n> **Theorem 3:** \n> _Let $G$ be a **simply laced** cscsc group and let $V$ and $W$ be two positive energy representations of $\\mathcal{L}G_k$. Then $V$ and $W$ are locally equivalent._\n\n_**Edit:** The argument in [[GF]](https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-155/issue-3/Operator-algebras-and-conformal-field-theory/cmp/1104253398.full) seems to contain a mistake (on lines -4 and -3 of page 600)_\n\n~~The basic ingredients that are needed (see page 599 of [Gabbiani & Fröhlich _[Operator algebras and conformal field theory](https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-155/issue-3/Operator-algebras-and-conformal-field-theory/cmp/1104253398.full)_] for the proof) are the following two facts about positive energy representations of simply laced loop groups: \n• Every level 1 rep can be obtained from the vacuum rep by precomposing the action by an outer automorphism of $\\mathcal{L}G_1$ that is the identity on $\\mathcal{L}_IG_1$. \n• Every level $k$ rep appears in the restriction of a level 1 rep under the map $\\mathcal{L}G_k\\to \\mathcal{L}G_1$ induced by the $k$-fold cover of $S^1\\to S^1$. ~~\n\nThere are proofs in the literature, due to A. Wassermann ([here](https://arxiv.org/abs/math/9806031) p23) and V. Toledano-Laredo ([here](https://arxiv.org/abs/math/0409044) p82) respectively, for the cases $LSU(n)$ and $LSpin(2n)$, that are based on the theory of free fermions -- actually, Toledano only treats half of the representations of $LSpin(2n)$.\n\n* * *\n\nIs there a proof of Claim 2 in the literature? \nHow does one prove Claim 2? \n", "link": "https://mathoverflow.net/questions/66859/local-equivalence-of-loop-group-representations", "tags": ["loop-spaces", "conformal-field-theory", "von-neumann-algebras", "rt.representation-theory", "reference-request"], "votes": 18, "creation_date": "2011-06-03T16:35:03", "comments": ["After having a further look at his preprint, it is now my opinion that Antony Wassermann does not provide a complete proof of the desired claim. So the question is open.", "Many thanks to the person who sent me Wassermann's preprint.", "There seems to be a proof of Claim 2 in some unpublished notes, namely in Böckenhauer and Evans \"Modular Invariants, Graphs and $\\alpha$-induction for Nets of Subfactors. II\" on page 66 is claimed that local equivalence hold for any compact connected simple $G$ with reference to the unplished notes of Wassermann \"Subfactors arising from positive energy representations of some infinte dimensional groups\", 1999.", "probably not really helpful, but there seems to be a paper in preparation by Toledano Laredo which maybe could shed some light about this question in the B and C case"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "216", "site": "mathoverflow", "title": "Steenrod algebra at a prime power", "body": "Let $n=p^k$ be a prime power.\n\nWhen $k=1$, the algebra of stable operations in mod $p$ cohomology is the [Steenrod algebra](http://en.wikipedia.org/wiki/Steenrod_algebra) $\\mathcal{A}_p$. It has a nice description in terms of generators and relations. Its dual (as a Hopf algebra) is also well understood by work of Milnor.\n\nWhat about when $k>1$? Has any work been done on trying to understand the algebra of stable operations in mod $n$ cohomology? For instance, is there a description of $\\mathcal{A}_4$ in terms of generators and relations? What about the dual Hopf algebra?\n\nOf course I am just asking about the cohomology ring $H^\\ast (H\\mathbb{Z}/n;\\mathbb{Z}/n)$, where $H\\mathbb{Z}/n$ denotes the mod $n$ Eilenberg--Mac Lane spectrum. So I fully expect the answer to be either one of \"yes, this is well-known and classical\" or \"no, this is known to be a big mess\". \n", "link": "https://mathoverflow.net/questions/106602/steenrod-algebra-at-a-prime-power", "tags": ["at.algebraic-topology", "cohomology", "steenrod-algebra"], "votes": 18, "creation_date": "2012-09-07T05:54:20", "comments": ["@elidiot: I don't know anything about mod p^k Dyer-Lashof operations, but that's probably just my own ignorance. It seems worthy of an MO question for sure!", "I know that this is another question but as it is related I'm taking the liberty to ask it here: has anything been written about mod p^k Dyer Lashof operations ? I would especially be interested in mod 4 ones. Please tell me if this deserves a new thread.", "@Sean: No. Shortly after asking this question I started working on other things. I did check out Adams' Berkeley notes, but couldn't find anything therein. Tyler's answer to mathoverflow.net/questions/119604/computation-of-hz-4-hz-4 seems encouraging though.", "@Mark: Did you figure out a satisfactory answer? I had a thought about one possible way to \"reverse engineer\" it.", "@Tyler: Thanks for the tip. I've ordered Adams' Berkeley notes from our library, I'll let you know if I find anything!", "@Mark: I do remember reading somewhere in one of Adams' texts - possibly the Berkeley lectures? - about looking at cohomology operations into or out of mod $2^n$ cohomology. I might start looking there. The method is to filter $H\\mathbb{Z}/p^n$ by $H\\mathbb{Z}/p^k$ and look at the two associated Bockstein sequences that you get. Sorry for the indeterminacy in references.", "@Tyler: This sounds promising. Was this a self-made calculation? If not, is there any hope that you could remember where you are remembering it from? Thanks.", "Err, a lot of the multiplications become zero.", "If I remember correctly, this can be calculated because the mod-$p$ Steenrod algebra has a relatively simple structure when considered as a bimodule over the subalgebra generated by the Bockstein. (I'm less sure at odd primes.) For $p=2$: in degrees zero and one you get $\\mathbb{Z}/2^k$, generated by the Bockstein; I believe that in higher degrees it's abstractly isomorphic to the Steenrod algebra as a module except that admissible monomials starting with $Sq^1$ start with the Bockstein $\\beta$ and admissible monomials ending with $Sq^r$, $r$ odd, end with $\\beta Sq^{r-1}$. A lot of the mul", "@Sean: I am perhaps not so interested in the answer for finite fields, but I would be interested to know why it is easier (presumably issues with the Kunneth Theorems and duality)?", "This is not directly related, but it is not hard to compute the Steenrod algebra corresponding a finite field. I am guessing you are not interested in that though."], "comment_count": 11, "category": "Science", "diamond": 0} {"question_id": "217", "site": "mathoverflow", "title": "Elliptic $\\infty$-line bundles over $B G$", "body": "Theorem 5.2 in Jacob Lurie's \"Survey of Elliptic Cohomology\" ([pdf](http://www.math.harvard.edu/~lurie/papers/survey.pdf)) states the equivalence of two maps\n\n$$ B G \\longrightarrow B \\mathrm{GL}_1(A) $$\n\nfor $A$ an $E_\\infty$-ring carrying an oriented derived elliptic curve $E \\to \\mathrm{Spec}(A)$. \n\nThe theorem is stated for $G = \\mathrm{Spin}$. That's the case of relevance in the paragraph directly following its statement. But, in view of the whole 2-equivariance story, the statement of the theorem as such would make sense for more general $G$ equipped with a homomorphism $G \\to O$, as in the discussion on the preceding page.\n\n**First question:** Might this statement still be true for more general $G\\to O$? (What's the idea of the proof, anyway?)\n\nMore in detail, one of the two maps is the composite\n\n$$ B G \\to B O \\stackrel{J}{\\longrightarrow} B \\mathrm{GL}_1(\\mathbb{S}) \\longrightarrow B \\mathrm{GL}_1(A) $$\n\nwhere $J$ denotes the $J$-homomorphism and $\\mathbb{S}$ the sphere spectrum. (Here I am giving a nominally different but hopefully straightforwardly equivalent description of what is indicated on p. 38 of Lurie's note.) The other one is the restriction of the derived theta-line bundle \n\n$$ \\theta \\;\\colon\\; \\mathrm{Loc}_{\\mathrm{Spin}}(E) \\longrightarrow \\mathbf{B} \\mathbb{G}_m $$\n\nalong the global point inclusion $\\mathrm{Spec}(A)\\to \\mathrm{Loc}_{\\mathrm{Spin}}(E)$ that picks the trivial local system/flat connection (that's again my slight refomulation which I believe is straightforwardly equivalent, but check -- my $\\mathrm{Loc}_{\\mathrm{Spin}}(E)$ is Lurie's \"$M_{\\mathrm{Spin}}$\").\n\nRegarding the first map, for $G = \\mathrm{Spin}$ and for the special case $A = \\mathrm{tmf}$ this is discussed in section 8 of Ando-Blumberg-Gepner [arXiv:1002.3004](http://arxiv.org/abs/1002.3004) and crucially equivalent to the $\\mathrm{tmf}$-line bundle which is associated via the String-orientation $\\sigma$ to the \"Chern-Simons 3-bundle\" classified by $\\tfrac{1}{2}p_1$, i.e. to the map\n\n$$ B \\mathrm{Spin} \\stackrel{\\tfrac{1}{2}p_1}{\\longrightarrow} B^4 \\mathbb{Z} \\stackrel{\\tilde \\sigma}{\\longrightarrow} B \\mathrm{GL}_1(\\mathrm{tmf}) \\,. $$\n\nI suppose the argument there straightforwardly generalizes from $A = \\mathrm{tmf}$ to any $A$ as above. But as a sanity check:\n\n**Second question** : is that right?\n\nIn conlusion I am wondering:\n\n**Third question** : In which generality in $E_\\infty$-rings $A$ carrying oriented derived elliptic curves and in compact simply connected simple Lie groups $G$, is it true that the restriction of the derived theta line bundle along the $\\mathrm{Spec}(A)$-point inclusion is equivalent to the $A$-line bundle classified by\n\n$$ B G \\stackrel{c_2}{\\longrightarrow} B^4 \\mathbb{Z} \\stackrel{\\tilde \\sigma}{\\longrightarrow} B \\mathrm{GL}_1(A) \\,. $$\n\n?\n\n(Where now $c_2$ denotes the generator of $H^4(B G, \\mathbb{Z}) \\simeq \\mathbb{Z}$.)\n", "link": "https://mathoverflow.net/questions/185721/elliptic-infty-line-bundles-over-b-g", "tags": ["at.algebraic-topology", "elliptic-curves", "cohomology", "derived-algebraic-geometry"], "votes": 18, "creation_date": "2014-10-29T10:11:41", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "218", "site": "mathoverflow", "title": "Computation of low weight Siegel modular forms", "body": "We have these huge tables of elliptic curves, which were generated by computing modular forms of weight $2$ and level $\\Gamma_0(N)$ as N increased.\n\nFor abelian surfaces over $\\mathbb{Q}$ we have very little as far as I know. The Langlands philosophy suggests that every abelian surface should be attached to a Siegel modular form of weight $(2,2)$ on $GSp_4$, but the problem is that this weight is not cohomological, which has the concrete consequence that it's going to be tough to compute such things using group cohomology. In particular one of the reasons that modular symbols work for computing elliptic curves, fails in this situation. \n\nI guess though that one might be able to somehow use the trace formula to compute the trace of various Hecke operators on Siegel modular forms of weight $(2,2)$ and various levels, because presumably the trace formula translates the problem into some sort of \"class group\" (in some general sense) computation, plus some combinatorics.\n\n[EDIT: from FC's comment, it seems that my guess is wrong.]\n\nHas anyone ever implemented this and tabulated the results?\n\n[NB I know that people have done computations for low level and high weight, for example there's a lovely paper of Skoruppa that outlines how to compute in level 1; my question is specifically about the weights that are tough to access]\n", "link": "https://mathoverflow.net/questions/22949/computation-of-low-weight-siegel-modular-forms", "tags": ["computational-number-theory", "siegel-modular-forms", "nt.number-theory", "trace-formula"], "votes": 18, "creation_date": "2010-04-28T23:46:38", "comments": ["@David: in which case my answer to your aside is then simply \"yes\".", "@Kevin: I am aware of the low-dimensional coincidences SO(2,1)=PGL(2) and Spin(2,3)=Sp(4). I suppose n > 2 was implicit in my query. :)", "@David: there is a low-degree coincidence: the dual group of GSp_4 is GSp_4 (think about the Dynkin diagram of Sp_4). I agree that the pattern should not continue for Sp_6 (think about the Dynkin diagram again).", "Aside: Shouldn't the automorphic representation associated to an abelian n-fold live on SO(2n+1), rather than a symplectic group? (Look at the L-groups)", "Thanks for the reference FC. What you say about Sp_6 is terrifying!", "@David: I think that trying to describe the moduli space is not what I am after. I am after an algorithm to compute traces of Hecke operators on spaces of cusp forms that takes a level and a prime (or some slightly more precise data; a diagonal matrix with powers of p on the diagonal) as an input and gives a number as an output and doesn't care less about whether a certain moduli space is general type or not. The same way as modular symbols would do the job for classical modular forms. But from FC's comment it seems that the trace formula won't do it either :-/", "@FC: I am pretty sure one can't use the trace formula to compute forms of weight 1 but I have so little understanding of the trace formula that I don't know why this is the case. Seems to me that you're saying that the same obstruction stops me computing low weight Siegel forms :-( ", "@Kevin: Curves of genera up 6 have a very pleasant description, and you can comfortably (depends on your standard of comfort of course) live with the description of curves up to genus 15; I'd buy a curve of genus up to 15 as a moduli space any day. I have yet to see a general type three-fold with a nice description (unless you cooked it for this purpose). Disclaimer: I don't know the method involved, you may well be right, it's just my vague intuition speaking here.", "I don't buy this. Isn't that like saying \"wouldn't computational efforts with elliptic curves of conductor 100 or more be hard because the moduli space has genus at least 2\", isn't it? The point is that you don't compute the moduli space---that's the last thing you want to do! You use the trace formula applied to a carefully-chosen function which will give you essentially the trace of Frobenius on the cohomology in terms of some much more algebraic/combinatorial/number-theoretic data and compute that instead. Well, that's my suggestion, but a lot of thought needs to go in to making it work.", "Wouldn't any computational efforts with N > 3 be very difficult simply because the moduli space is general type ?"], "comment_count": 10, "category": "Science", "diamond": 0} {"question_id": "219", "site": "mathoverflow", "title": "Quotients of residually finite groups by amenable normal subgroups", "body": "My questions are:\n\n> Is there any group, which cannot be written as the quotient of a residually finite group by an amenable normal subgroup? Is it possible for large classes of groups?\n\nand\n\n> Is there a group, such that every extension by an amenable group is split?\n", "link": "https://mathoverflow.net/questions/40991/quotients-of-residually-finite-groups-by-amenable-normal-subgroups", "tags": ["gr.group-theory"], "votes": 18, "creation_date": "2010-10-03T23:19:30", "comments": ["@Andreas: Me too! Can you send me the text? This sounds very interesting!", "Have you imagined a way to answer the first question considering only the one-relator groups?", "@Andreas: Could you please send me a copy of this paper, also? I am very interested in this.", "@Andreas: Can you send me the text? I suggest you publish the result. Do you need advice about a journal? ", "@Mark: I proved it and it is not published, since I do not yet know what to do with it.", "Also, 'if the answer is \"yes\", it would provide a new way to construct groups because old ways don't seem to work' is exactly the sort of motivation I was asking for.", "Mark - I agree that if the answer to the first question were 'no' then it would be remarkable! I was wondering what motivates Andreas to ask the question. It's always useful when the asker gives all the information that s/he has about a problem.", "@Andreas: who proved the result of measured group theory that you mentioned, and where?", "I meant that no motivation is required because if the answer to the first question is \"no\", then it would be the first non-trivial property of all groups (which would disprove Gromov's thesis). And if the answer is \"yes\", it would provide a new way to construct groups because old ways don't seem to work. ", "̯@Henry: If the second question has a positive answer, then this would (most likely) help to answer the first. My motivation is the study of sofic groups. In the sense of measured group theory, every sofic group is the quotient of a residually finite group by an amenable subgroup (but that takes some time to explain). I was wondering what kind of condition one gets in the more rigid setup of ordinary group theory. Burnside groups are not known to be sofic, so Mark's comment is very interesting.", "@Henry: None of these questions requires an additional motivation. @Andreas: I suspect that the free Burnside group of large enough exponent is an example. That would solve mathoverflow.net/questions/37344/… (see the discussion there). But that is a difficult question.", "Could you give some motivation, and explain what (if anything) the two questions have to do with each other?"], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "220", "site": "mathoverflow", "title": "What is known about module categories over general monoidal categories?", "body": "All of the literature I have seen on module categories over monoidal categories has been in the rigid $k$-linear semisimple case, more or less in the spirit of Ostrik's paper,\n\n * Ostrik, V. _Module categories, weak Hopf algebras and modular invariants_ , Transformation Groups **8** (2003) 177–206, doi:[10.1007/s00031-003-0515-6](https://doi.org/10.1007/s00031-003-0515-6), arXiv:[math/0111139](https://arxiv.org/abs/math/0111139).\n\n\n\nHowever, the basic definition of module category makes perfect sense in general, so I am wondering if there has been any work done in more general settings.\n\nIn particular, I am interested in the following sort of question. Let $\\mathcal{C}$ be a braided monoidal category (maybe ultimately with some added assumptions, such as rigidity). Any left $\\mathcal{C}$-module category is automatically a $(\\mathcal{C}, \\mathcal{C})$-bimodule category via the braiding. For bimodule categories there should be a good notion of \"tensoring over $\\mathcal{C}$,\" which makes the 2-category of bimodules (and hence the 2-category of left modules) into a monoidal 2-category. If we extract invertible objects and morphisms at all levels, we should obtain a sort of Brauer 3-group for $\\mathcal{C}$. In the fusion category setting, this object is considered in the recent preprint\n\n * Pavel Etingof, Dmitri Nikshych, Victor Ostrik, _Fusion categories and homotopy theory_ , Quantum Topol. **1** (2010), 209–273, [journal](https://ems.press/journals/qt/articles/2876), arXiv:[0909.3140](https://arxiv.org/abs/0909.3140).\n\n\n\nI'd like to know how this is related to the \"internal\" Brauer 3-group of $\\mathcal{C}$, whose objects are Azumaya algebras in $\\mathcal{C}$, morphisms are invertible bimodules, and 2-morphisms are invertible bimodule morphisms. In the fusion category setting, the connection is furnished by the main theorem in Ostrik's paper, which states that any semisimple indecomposable module category is equivalent to the category of modules over some algebra in $\\mathcal{C}$. (I think this implies that the two notions of Brauer 3-group are equivalent in the fusion category setting.) Is any result along these lines known in more generality? It's not even clear to me that the module category of left modules over an Azumaya algebra in $\\mathcal{C}$ is invertible, or even what the tensor product of two module categories of modules is.\n", "link": "https://mathoverflow.net/questions/6775/what-is-known-about-module-categories-over-general-monoidal-categories", "tags": ["ct.category-theory", "monoidal-categories"], "votes": 18, "creation_date": "2009-11-24T23:08:19", "comments": ["The \"module categories are categories of modules\" yoga is probably very special to the finite rigid setting. We tried to be careful about what exactly was needed for this here:arxiv:1406.4204. An interesting example is the following (non-rigid) tensor category C. It is finite, semisimple and has two simple objects 1 and x. The object 1 is the unit and we have $x \\otimes x = 0$. This determines the monoidal structure by linearity. If I remember right, the subcategory generated by x is a module category which is not of the form A-mod(C) for an algebra A in C.", "The Deligne tensor product doesn't exist for arbitrary abelian categories. But if you relax \"abelian\" to instead mean \"has finite colimits\" then there is a very general tensor product that always exists called the \"Kelly tensor product\". A similar statement for locally presentable categories is in Cor 2.2.5 of (arXiv:1105.3104). I think the categorical machinery underlying these results can be adapted to the case of tensor products of module cats in the generality you are considering, but I don't know of a reference that does it.", "An Azumaya algebra is one in which the tensor actions $X \\mapsto A \\otimes X$ and $X \\mapsto X \\otimes A$ are equivalences between $\\mathcal{C}$ and the categories of left $A \\otimes A^{\\text{op}}$-modules and right $A^{\\text{op}} \\otimes A$-modules, respectively. This definition comes from the paper by Van Oystaeyen and Zhang, \"The Brauer Group of a Braided Monoidal Category.\"", "Could you recall the definition of Azumaya algebra in a general braided monoidal category?"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "221", "site": "mathoverflow", "title": "Almost complex 4-manifolds with a "holomorphic" vector field", "body": "**Main question.** What is the class of smooth orientable 4-dimensional manifolds that admit an almost complex structure $J$ and a vector field v, that preserves $J$? \n\n_The following sub question is rewritten thanks to the comment of Robert Bryant:_\n\nIs it true that if $(M,J)$ admits a vector field that preserves $J$ then there is $J'$ on $M$ homotopic to $J$ that is preserved by an $S^1$-action on $M$?\n\n_After three years I don't think indeed that the following is such a reasonable motivation_\n\nPOSSIBLE MOTIVATION. Claire Voisin gave a construction of the Hilbert scheme of points for every almost complex 4-fold by an analogy with the Hilbert scheme of points of a complex surface. The first calculation of the Euler charactericstics of Hilbert scheme of points of complex surfaces was done via localisation techniques, for $CP^2$. Now, if we have a \"holomorphic\" vector field on an almost complex manifold, this could potentially help to reduce the calculation of the Euler charachteristics of its Hilbert scheme to the study of fixed points of the manifold. So the question is how flexible this notion is... But in its nature this question seems to be more a question (maybe not a hard one) on dynamical systems. \n", "link": "https://mathoverflow.net/questions/9518/almost-complex-4-manifolds-with-a-holomorphic-vector-field", "tags": ["4-manifolds", "ds.dynamical-systems", "gt.geometric-topology", "complex-geometry"], "votes": 18, "creation_date": "2009-12-21T16:14:00", "comments": ["Robert, thanks for this remark, you are right that this second phrase makes no sense. I rewrote so that I it makes sense.", "@Dmitri: Perhaps I don't understand your terminology, but it seems to me that $S^4$ admits a smooth $S^1$ action, but it doesn't admit any almost complex structure, let alone one with a (nontrivial) symmetry vector field. Thus, your second sentence is mysterious to me.", "Yes, I refer to the flow associated to the vector field. I don't want to impose that all zeros are isolated. ", "When you say the vector field preserves the almost-complex structure, are you referring to the flow associated to the vector field? And do you want the vector field to be everywhere nonzero? Nonzero at all but finitely many points? ", "Tim, thanks for the comment! In fact the approach I propose is exactly to give a justification of your words:)) How do we calculate the Euler charecteristics of a manifold - count the number of zeros of a vector field. I want to say, that for a Hilbert scheme this should be a result of the same caclulation. I think it would be cool to have a \"simple\" calculation of the Euler charateristics. By the way this speculation can also be applied to DT invariants -- when you count 0-dim subschemes on 3 dimensional CY manifolds. This calculation was done in 2005 and was one of MNOP conjectures.", "It's an intriguing question, but it seems to me a surprisingly delicate way to calculate the Euler characteristic of the Hilbert scheme, which one expects to be a universal function of the Betti numbers of the 4-manifold, as in the integrable case. Perhaps one can prove this using the Cech spectral sequence coming from the open cover of Hilb arising from a good cover of the 4-manifold? [BTW, I know you're quoting Voisin, but the following rule is worth insisting on: a 4-fold is an algebraic or analytic variety of dimension 4. A 4-manifold has real dimension 4.]", "Thanks :) Maybe this sound cooler than it should be. He is the reference, it was published in 2002 people.math.jussieu.fr/~voisin/Articlesweb/almost.pdf", "This sounds really cool. Is this recent work of Voisin?"], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "222", "site": "mathoverflow", "title": "$G$ a group, with $p$ a prime number, and $|G|=2^p-1$, is it abelian?", "body": "During my research I came across this question, I proposed it in the chat, but nobody could find a counterexample, so I allow myself to ask you :\n\n$G$ a group, with $p$ a prime number, and $|G|=2^p-1$, is it abelian ?\n", "link": "https://mathoverflow.net/questions/273976/g-a-group-with-p-a-prime-number-and-g-2p-1-is-it-abelian", "tags": ["nt.number-theory", "gr.group-theory", "finite-groups"], "votes": 18, "creation_date": "2017-07-08T05:27:26", "comments": ["@YemonChoi Well, I would wait 500 years for the math community to find a complex 600-page solution, like for Fermat.", "If you are claiming that you can answer this question, then you can either add an answer below, or write up your work as a preprint in the normal academic way, or write it in a blog post and leave a link. Saying \"I can answer this, can you?\" is not a constructive use of this site, nor is it collegial.", "I have rolled back the edits to this old question, which seem to be purely cosmetic and done for promotion. Yes the question is appropriate for MO, but that doesn't mean you need to add some kind of \"RL\" label", "Indeed, your suggestion mathoverflow.net/questions/273976/… is the same as @spin's.", "msp.org/pjm/1967/22-3/pjm-v22-n3-p15-p.pdf", "I did a quick check for $p$ at most $800$ or so and found no examples. In the end this is a question about number theory and not group theory, and it might be difficult. For example it is an old open problem to determine whether $2^p - 1$ is squarefree for all primes $p$. If the answer to that old problem is yes, then you are asking whether every group of order $2^p - 1$ is cyclic.", "One can use the Cunningham tables to do a quick check on the factors for more than 50 prime exponents. Gerhard \"Not Computationally Motivated This Morning\" Paseman, 2017.07.08.", "@PéterKomjáth: No, we have then $2^p \\equiv 2$ (mod $q^3$), respectively, $2^{p-1} \\equiv 1$ (mod $q^3$).", "@Glorfindel, it's hard to believe that never happens, but a quick test in Mathematica shows that (I screwed up in Mathematica or) it doesn't happen for the first 50 primes.", "Yes, but probably easier to solve. I doubt there is a prime $q$ for which $q^3 | 2^p - 1$, but the other 'case', two primes $q_1, q_2$ in the factorization for which $q_1 | q_2 - 1$ might very well be possible.", "It depends on the factorization of $2^p-1$, see groupprops.subwiki.org/wiki/…"], "comment_count": 11, "category": "Science", "diamond": 0} {"question_id": "223", "site": "mathoverflow", "title": "Symmetries of local systems on the punctured sphere", "body": "Let $X=S^2\\setminus D$, for $D\\subset S^2$ some finite set of points, say with $|D|=n\\geq 1$. The category of locally constant sheaves of $\\mathbb{C}$-vector spaces on $X$ (equivalently, complex representations of $\\pi_1(X)=F_{n-1}$), $\\text{LocSys}(X)$, has many natural autoequivalences.\n\nFor example, if $f: S^2\\to S^2$ is a homeomorphism fixing $D$, then $\\mathbb{V}\\mapsto f^*\\mathbb{V}$ is an autoequivalence of $\\text{LocSys}(X)$. Two homeomorphisms fixing $D$, homotopic through homeomorphisms fixing $D$, induce naturally equivalent autoequivalences, so one obtains an action of the mapping class group $\\text{Mod}(S^2, D)$ on $\\text{LocSys}(X)$. Since $\\text{Mod}(S^2, D)$ is the pure spherical braid group on $n$ strands, I'll denote it by $B_n$. My question is:\n\n> What are the auotequivalences of $\\text{LocSys}(X)$ commuting (up to natural isomorphism) with the actions of $B_n$?\n\nI know of three sources of examples. The first is invertible objects in $\\text{LocSys}(X)$: if $\\mathbb{L}$ is a rank one local system on $X$, the functor $\\mathbb{V}\\mapsto \\mathbb{V}\\otimes \\mathbb{L}$ is $B_n$-equivariant. The second is the evident action of the group $\\text{Aut}(\\mathbb{C}/\\mathbb{Q})$ on the structure constants of any given local system. Both of these operations preserve the rank of a local system.\n\nThe other is Katz's middle convolution operation, which does not.\n\nLet $\\Delta\\subset X\\times X$ be the diagonal, $\\pi_1, \\pi_2: X\\times X\\setminus \\Delta\\to X$ the two projections, $a: X\\times X\\setminus \\Delta\\to \\mathbb{C}^\\times$ the map $(x, y)\\mapsto x-y$ (where here we choose an isomorphism $S^2\\simeq \\mathbb{CP}^1$ sending some point of $D$ to $\\infty$ to coordinatize $S^2$), and $j: X\\times X\\setminus \\Delta\\hookrightarrow S^2\\times X$ the evident inclusion. Then given a non-trivial rank one local system $\\chi$ on $\\mathbb{C}^\\times$, the middle convolution $$MC_{\\chi}(\\mathbb{V}):=R^1\\pi_{2\\ast}j_{\\ast}(\\pi_1^{\\ast}\\mathbb{V}\\otimes a^{\\ast}\\chi).$$ It is not obvious but Katz shows it is true that $MC_\\chi$ is (up to equivalence) inverse to $MC_{\\chi^{-1}}$. Moreover this operation (not entirely obviously) commutes with the action of $B_n$.\n\nOf course one can compose these operations in various ways to get more autoequivalences commuting with the $B_n$-action.\n\n> Is that everything?\n\nI'm also curious about the analogous question with $S^2\\setminus D$ replaced by a surface of higher genus.\n", "link": "https://mathoverflow.net/questions/446416/symmetries-of-local-systems-on-the-punctured-sphere", "tags": ["ag.algebraic-geometry", "rt.representation-theory", "braid-groups", "local-systems"], "votes": 17, "creation_date": "2023-05-08T12:40:43", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "224", "site": "mathoverflow", "title": "Does the Ackermann function count something?", "body": "Let $\\mathrm{FinSet}$ be the category of finite sets.\n\nA _finite set structure_ is a faithful functor $F\\colon C\\to \\mathrm{FinSet}$ such that, for any $n\\geq 1$, there are only finitely many isomorphism classes of objects $F$ maps to $\\\\{1, \\dots, n\\\\}$.\n\n> **Question.** Is there a natural finite set structure realizing the Ackermann function (or some other computable function growing faster than any primitive recursive function)?\n", "link": "https://mathoverflow.net/questions/392387/does-the-ackermann-function-count-something", "tags": ["co.combinatorics", "ct.category-theory"], "votes": 17, "creation_date": "2021-05-10T07:20:52", "comments": ["So, I do not know what the Ackermann function counts, but the TREE function does count something explicit and combinatorial, and the Ackermann function is tiny in comparison with TREE. mathoverflow.net/questions/93828/how-large-is-tree3", "The inverse Ackermann function appears in some algorithms, e.g. using a \"soft heap\" you can find the minimum spanning tree in time $O(m\\alpha(m, n))$. A decade ago I probably knew why and whether something was being counted.", "Interesting question. Maybe some nested set-partition-like structures? A stupid answer would be to take a deterministic rewrite system that takes $A\\left(n,m\\right)$ steps to terminate (I think there should be one), and just say \"the intermediate states of this rewrite system\".", "There is a whole \"school\" of enumerative combinatorics done using Joyal's species! Relevant names are Joyal himself, Labelle, Leroux, Viennot, Mendez, Rajan...", "Not sure why this (good) question has a category theory tag, but there is a kind of \"categorical number theory / combinatorics\" that was studied by Steve Schanuel. I don't know any more than that.", "@SimonHenry even though the question isn't phrased in this way, I suspect it is asking about a species of structure with the Ackermann function as generating species. I have no idea what the sequence $a_n=A(n,n)$ could be counting, though, and throwing the very few terms of the sequence that can be computed at OEIS doesn't seem a viable choice: is there a combinatorial interpretation for those numbers?", "@fosco : yes but,I am not aware of anything that forces the exponential generating function of combinatorial species to converge, so why would the convergence brings anything to the question ? But maybe you have in mind some result proving the convergence for some class of \"nice\" species ?", "@SimonHenry yes you can associate some category but I wondered if there was a natural (in the vague sense of the word) choice. Maybe a structure arising independently elsewhere in mathematics.", "I'm not sure how to understand this question. For any function $f:\\mathbb{N} \\to \\mathbb{N}$ I can chose a sequence of sets $P_n$ such that $\\# P_n = f(n)$ and take $C$ to be a naively defined category with an object for each $i \\in P_n$ mapping to a set with $n$ elements... so any functions can be obtained this way.", "hi @SimonHenry ! To a combinatorial species you can associate various formal power series, and especially from the exponential series you can understand a lot about the species itself --although not everything. Ad yes, that series it's not convergent, so there is no associated function you can study. This makes the problem elusive, even provided you find a meaning for what kind of structure $A(n,n)$ is counting.", "@fosco, surely not; $f(n) \\ge (n!)^2$ eventually.", "@fosco : why is the convergence of the series relevant to the question ?", "oh, fair enough; well, is the series $\\sum \\frac{f(n)}{n!}X^n$ convergent in at least a disk $B(0,\\epsilon[$?", "@fosco I had in mind $f(n)=A(n, n)$", "The Ackermann function $A(m,n)$ takes two inputs; apart from this, if I understand well, you're asking about the existence of a species of structure whose associated exponential series (in the variables X,Y) is $\\sum \\frac{A(m,n)}{n!m!}X^nY^m$. First of all, is this series convergent?"], "comment_count": 15, "category": "Science", "diamond": 0} {"question_id": "225", "site": "mathoverflow", "title": "Picture of Lambert's proof that $\\pi$ is irrational?", "body": "With a suitably generous notion of \"picture proof\" or \"proof without words\" or \"geometric proof,\" there do exist such proofs of [the irrationality of square roots](https://www.cut-the-knot.org/proofs/GraphicalSqRoots.shtml) and even of the [irrationality of $e$](https://arxiv.org/pdf/0704.1282.pdf). However, I am not aware of anything that could plausibly be called a [geometric proof of the irrationality of $\\pi$](https://math.stackexchange.com/q/1658174). Given that the original definition of $\\pi$ was geometric, it seems a pity that there is no such thing. Or maybe there is?\n\nPerhaps one can try to draw pictures to accompany [Lambert's irrationality proof](https://math.stackexchange.com/a/895728). For example, is there a way to draw a picture of the following fact? $$\\tan(a/b) = {a \\over\\displaystyle b - {a^2 \\over \\displaystyle 3b - {a^2 \\over \\displaystyle 5b - {\\displaystyle a^2 \\over 7b - \\cdots}}}}$$ And if so, is there any way to draw a picture of the fact that such a continued fraction is irrational when $a$ and $b$ are positive integers?\n", "link": "https://mathoverflow.net/questions/376318/picture-of-lamberts-proof-that-pi-is-irrational", "tags": ["nt.number-theory", "continued-fractions", "transcendental-number-theory", "irrational-numbers", "alternative-proof"], "votes": 17, "creation_date": "2020-11-12T11:09:09", "comments": ["I don't have a specific suggestion, but of possible related interest is a collection of references I assembled 3 years ago in my answer to Irrationality of $\\pi^2$ and $\\pi^3$.", "Correction: The above formula should have $\\tan(x^{-1})$ on the right hand side. I also think there might be a sign error somewhere.", "I'll think more about this later but: Let $SL_2(\\mathbb{R})$ act on $\\mathbb{R}$ by Mobius transformations in the usual way. It looks to me like your continued fraction says that $\\begin{bmatrix} 0 & 1 \\\\ -1 & x \\\\ \\end{bmatrix} \\begin{bmatrix} 0 & 1 \\\\ -1 & 3x \\\\ \\end{bmatrix} \\begin{bmatrix} 0 & 1 \\\\ -1 & 5x \\\\ \\end{bmatrix} \\cdots (0) = \\tan(x)$. So maybe draw a sequence of lines given by partial products of this, and see that their slopes are approaching $\\tan(x)$ or, equivalently, their angles are approaching $x$? (Here $x = a/b$.)"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "226", "site": "mathoverflow", "title": "When is the determinant an $8$-th power?", "body": "I am working over $\\mathbb{R}$ (though most of the story goes over any field). I am looking for linear spaces of matrices such that the restriction of the determinant to this spaces can be written (non-trivially) as the power of another polynomial. Let me give some examples where such a phenomenon appears, before asking my question in more details.\n\n * let $n = 2m$ be an even integer. Then, the restriction of the determinant to $\\bigwedge^2 \\mathbb{R}^n \\subset \\mathrm{End}(\\mathbb{R}^n)$ can be written as the square of a non-zero polynomial : the pfaffian.\n\n * let $n = 2m$ be again even. Let $\\mathcal{H}_{m}(\\mathbb{C})$ be the set of $m \\times m$ Hermitian matrices with complex coefficients. Using the matrix representation of $i$ (square root of $-1$) as: $$ i = \\begin{pmatrix} 0 & -1 \\\\\\ 1 & 0 \\\\\\ \\end{pmatrix},$$ one can embed $\\mathcal{H}_{m}(\\mathbb{C})$ as a sub-algebra of $\\mathrm{S}^{2} \\mathbb{R}^{n} \\subset \\mathrm{End}(\\mathbb{R}^{n})$. Then, it is easily checked that the following equality holds on $\\mathcal{H}_{m}(\\mathbb{C})$:\n\n\n\n\n$$ \\mathrm{det}_{\\mathrm{End}(\\mathbb{R}^{n})} = \\left( \\mathrm{det}_{\\mathcal{H}_{m}(\\mathbb{C})} \\right)^2.$$\n\n * let $n = 4m$. Let $\\mathcal{H}_{m}(\\mathbb{H})$ be the set of $m \\times m$ Hermitian matrices with quaternionic coefficients. Using the $4 \\times 4$ matrix representation of $i,j,k$ (square roots of $-1$ in $\\mathbb{H}$), One can embed $\\mathcal{H}_{m}(\\mathbb{C})$ as a sub-algebra of $\\mathrm{S}^{2} \\mathbb{R}^{n} \\subset \\mathrm{End}(\\mathbb{R}^{n})$. Then, it is again easily checked that the following equality holds on $\\mathcal{H}_{m}(\\mathbb{H})$:\n\n\n\n$$ \\mathrm{det}_{\\mathrm{End}(\\mathbb{R}^{n})} = \\left( \\mathrm{det}_{\\mathcal{H}_{m}(\\mathbb{H})} \\right)^4.$$\n\n**Question** : Is there a similar story for $\\mathcal{H}_{3}(\\mathbb{O})$?\n\nMore precisely, one can define a good notion of determinant for Hermitian $3 \\times 3$ matrices with octonionic coefficients. I was wondering if there is an embedding $\\mathcal{H}_{3}(\\mathbb{O}) \\hookrightarrow \\mathrm{S}^{2} \\mathbb{R}^{24} \\subset \\mathrm{End}(\\mathbb{R}^{24})$, such that the following hold on $\\mathcal{H}_{3}(\\mathbb{O})$:\n\n$$ \\mathrm{det}_{\\mathrm{End}(\\mathbb{R}^{24})} = \\left( \\mathrm{det}_{\\mathcal{H}_{3}(\\mathbb{O})} \\right)^8?$$\n\nThe algebra $\\mathcal{H}_{3}(\\mathbb{O})$ being non-associative, it can not be embedded **as an algebra** into a matrix algebra. On the other hand, what I am asking is (probably) considerably weaker : I just want an embedding of vector spaces $\\mathcal{H}_{3}(\\mathbb{O}) \\hookrightarrow \\mathrm{End}(\\mathbb{R}^{24})$ such that the restriction of $\\mathrm{det}_{\\mathrm{End}(\\mathbb{R}^{24})}$ to $\\mathcal{H}_{3}(\\mathbb{O})$ is the $8$-th power of $\\mathrm{det}_{\\mathcal{H}_{3}(\\mathbb{O})}$.\n\nOf course, since $\\mathcal{H}_{3}(\\mathbb{O})$ can not be embedded into $\\mathrm{End}(\\mathbb{R}^{24})$ as an algebra, it will probably be harder to check such an equality of determinants on $\\mathcal{H}_{3}(\\mathbb{O})$. Indeed, we can't use the representative of the Hermitian comatrix in $\\mathrm{End}(\\mathbb{R}^{24})$ and just multiply it with the orginal matrix to get $\\mathrm{det}_{\\mathcal{H}_{3}(\\mathbb{O})}.Id_{24}$. But I thought that it was perhaps possible to find another trick for $\\mathcal{H}_{3}(\\mathbb{O})$?\n\nThanks for your help.\n", "link": "https://mathoverflow.net/questions/278981/when-is-the-determinant-an-8-th-power", "tags": ["ag.algebraic-geometry", "rt.representation-theory", "matrices", "ra.rings-and-algebras", "octonions"], "votes": 17, "creation_date": "2017-08-17T13:13:33", "comments": ["But I may be wrong and I would be delighted to learn that there is such an embedding!", "@DenisSerre : Yes sure, I tried a lot of stuff related to the Jordan algebra theory. In fact I am now able to prove that there is an embedding of $\\mathcal{H}_{3}(\\mathbb{O})$ in $\\mathbb{R}^{24}$ such that $\\mathrm{det}_{\\mathbb{R}^{24}}$ is the $4$-th power of a certain sextic related to $\\mathrm{det}_{\\mathcal{H}_{3}(\\mathbb{0})}$. However, I did a lot of computations with Macaulay2, and I start to believe that there is no embedding such that the restriction of $\\mathrm{det}_{\\mathbb{R}^{24}}$ to $\\mathcal{H}_{3}(\\mathbb{O})$ is a $8$-th power.", "Have you tried using the fact that ${\\cal H}_3({\\mathbb O})$ is the exceptional Jordan algebra, that called the Albert algebra ?", "@MarekMitros : Thanks! Yes I have seen this question a bit after having posted mine. It is very much related. Though, I guess mine is more focused on the equation lining the two dterminants while the other one focuses more to the existence of a determinant for hermitian matrices with octonionic coefficients.", "See also mathoverflow.net/questions/239954/…", "The first suggestion of embedding octonions into matrices is to map octonion $x$ to $8\\times 8$ matrix $L_x$ or $R_x$ of left and right multiplication by octonion $x$. Similarly octonion matrix $2\\times 2$ can be mapped to block matrix consisting of four blocks $L_{a_{i,j}}$ where $a_{i,j}$ is octonion in original matrix. Do we have formula for determinant of block matrix ?", "What is that first map? The multiplication table should be neither symmetric nor skew-symmetric. I see how given an $\\mathbb{R}$-linear map $f:\\mathbb{O}\\hookrightarrow\\mathrm{End}(\\mathbb{R}^8)$ satisfying $f(z)f(\\bar z)=|z|^2I_8$ one could extend to $\\mathcal{H}_2(\\mathbb{O})$ with the right determinant relation. That would require 7 anticommuting matrices $e_1,\\dots,e_7$ in $\\mathrm{End}(\\mathbb{R}^8)$ that square to $-I_8$. How are you building these out of the multiplication table?", "@MTyson In the case of $\\mathcal{H}_{2}(\\mathbb{O})$ I think there is something going on. Let $\\mathbb{O} \\hookrightarrow \\mathrm{S}^2 \\mathbb{R}^8$ which associates to an octonion its multiplication table. This gives rise to an embedding $\\mathcal{H}_{2}(\\mathbb{O})$ in $\\mathrm{S}^2 \\mathbb{R}^{16}$. Using the block computation of a determinant for a $2*2$ block matrix (with the bottom right being a translation), one sees that $\\mathrm{det}_{\\mathrm{End}(\\mathbb{R}^{16})} = (\\mathrm{det}_{\\mathcal{H}_{2}(\\mathbb{O})})^3$", "Do you know anything about the $\\mathcal{H}_{2}(\\mathbb{O})$ case?"], "comment_count": 9, "category": "Science", "diamond": 0} {"question_id": "227", "site": "mathoverflow", "title": "Kan's simplicial formula for the Whitehead product", "body": "In his article on _Simplicial Homotopy Theory_ (Advances in Math., 6, (1971), 107 –209) Curtis quotes a formula (on page 197) for the Whitehead and Samelson products in a simplicial group $G$. The formula for the Samelson product $\\langle x,y\\rangle$ of elements $x\\in \\pi_p(G)$ and $y\\in \\pi_q(G)$ is in terms of an ordered product of commutators of pairs of certain degenerate elements. The order of multiplication is determined by an antilex total order on the set of $(p,q)$-shuffles. I tried, and think I succeeded in filling in the details of this sketch from Curtis. I understand that the result was due to Dan Kan but that he never published it.\n\nMy questions are\n\n(i) is there a published proof of this formula and if so where? (As I said I have a proof that it seems to work, but do not yet have one that it actually deserves to be called the Samelson product and that the related Whitehead product corresponds to the classical form, e.g. involving a 'universal example'.)\n\nand\n\n(ii) I have so far failed to prove that this formula gives a bilinear pairing on homotopy groups. (It is simple to prove a related formula which incorporates an action as in the Witt-Hall identities for commutators but I do not see why the action involved should be trivial as would seem to be required for the bilinearity result to hold.)\n\nCan anyone suggest a reference to a classical proof of bilinearity of the (topological) Whitehead product that is fairly categorical and hence adaptable to my simplicial setting. Most sources seem to leave it as an exercise and Whitehead's original approach although clear is very topological in nature.\n\n(Edit: I note that Adams in his _Student's Guide to Algebraic Topology_ , stated: The Whitehead product is bilinear and anticommutative; this can be proved by diagram chasing with the universal example'. Unfortunately I have yet to find the diagram through which I have to chase, although I have tried several ones that initially seemed to give some hope of being good for the task in hand.)\n", "link": "https://mathoverflow.net/questions/296479/kans-simplicial-formula-for-the-whitehead-product", "tags": ["at.algebraic-topology", "homotopy-theory", "simplicial-stuff"], "votes": 17, "creation_date": "2018-03-29T00:19:56", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "228", "site": "mathoverflow", "title": "Almost monochromatic point sets", "body": "There are many sort of equivalent theorems about monochromatic configurations in finite colorings, such as [Van der Waerden](https://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem), [Hales-Jewett](https://en.wikipedia.org/wiki/Hales%E2%80%93Jewett_theorem) or [Gallai's theorem](https://arxiv.org/abs/1411.1038), the latter of which states that in a finite coloring of $\\mathbb Z^d$ or $\\mathbb R^d$, there is a homothetic (i.e., scaled and translated) copy of any finite configuration $S$. Motivated by [this problem](https://mathoverflow.net/questions/299616/bichromatic-pencils), I wonder if similar statements hold if instead we require that in a configuration $(S,s_0)$ all points in $S$ are monochromatic, while $s_0$ has a different color from the rest.\n\nObviously, we need to impose some conditions on the coloring and the configuration. About the coloring, I only want to demand that it is non-monochromatic, i.e., not all points of the space are colored with the same color. About $(S,s_0)$, I want to require that $s_0\\notin conv(S)$, i.e., $s$ is not in the convex hull of some points from $S$, as then we might not have a solution if the \"first\" half of the space is red, while the \"second\" half is blue.\n\n> > Is there always an almost monochromatic copy (homothetic or isometric) of any finite $(S,s_0)$ with $s_0\\notin conv(S)$ in a non-monochromatic finite coloring of $\\mathbb R^d$?\n\nNote that the answer is no for $\\mathbb Z$ if $s_0=0$ and $S=\\\\{1,2\\\\}$ as shown by coloring odd numbers red and even numbers blue. This particular configuration, however, is easy to find in $\\mathbb Q$.\n", "link": "https://mathoverflow.net/questions/300604/almost-monochromatic-point-sets", "tags": ["nt.number-theory", "co.combinatorics", "mg.metric-geometry", "ramsey-theory", "polymath16"], "votes": 17, "creation_date": "2018-05-19T13:50:40", "comments": ["@fedja How can I cite your comment?", "@fedja I've just discovered that essentially the same problem was posed by some dude called Erdős and his pals in '75, and they made the same conjecture, so you might want to spell out your argument for their sake; see Conjecture 4 in old.renyi.hu/~p_erdos/1975-12.pdf.", "Didn't I spell it out? $\\tau$ is a field automorphism of $\\mathbb C$ preserving $\\mathbb Q$.", "@fedja I'm not sure, but it feels like your reasoning goes the other way. Let me try. We know that there is a bad coloring (of even $\\mathbb C$) for $(0,1,s_0)$ whenever $0 One can show that the map $$j^r s \\mapsto i_\\nabla(j^r s)\\doteq(s,D_\\nabla s=\\nabla s,\\ldots,D^r_\\nabla s)$$ is a $C^\\infty(M)$-linear isomorphism between the $C^\\infty(M)$-modules $\\Gamma(J^r E)$ and $\\Gamma(\\oplus^r_{k=0}\\vee^kT^*M\\otimes E)$, thus establishing an instance of the bundle isomorphism stated in the first paragraph. Now, given two vector bundles $\\pi:E\\rightarrow M$, $\\pi':F\\rightarrow M$, a linear partial differential operator $P$ of type $E\\rightarrow F$ and order $r\\geq 0$ is the composite of the $r$-th order jet prolongation map $\\Gamma(E)\\ni s\\mapsto j^r s\\in\\Gamma(J^r E)$ with a $C^\\infty(M)$-linear map $p$ from $\\Gamma(J^r E)$ into $\\Gamma(F)$ (the \"symbol map\" of $P$). Using $i_\\nabla$, one may write $$Ps=p(j^r s)=(p\\circ i_\\nabla^{-1})(s,D_\\nabla s,\\ldots,D^r_\\nabla s)=\\sum^r_{k=0}a_k D_\\nabla^k s\\ ,$$ where $a_k\\in\\Gamma(\\vee^k T M\\otimes E^*\\otimes F)=$ the $k$-th order coefficient of $P$ with respect to $\\nabla$ is uniquely determined by $P$ and $\\nabla$ for each $k=0,\\ldots,r$ (_remark:_ uniqueness of $a_k$ no longer holds if the symmetry requirement is dropped). Moreover, $a_r$ (the \"principal symbol\" of $P$) does not depend on $\\nabla$.\n\nThe earliest reference I could find on this statement is Chapter IV, Section 9 (more precisely, the Corollary to Theorem 7, pp. 90-91) of the book edited by Richard S. Palais, _Seminar on the Atiyah-Singer Index Theorem_ (Princeton University Press, 1965). This chapter was written by Palais himself, and it has no references.\n\n> **Question:** Is this really the first published proof of this result? As stated is it due to Palais, or does it go back even farther?\n", "link": "https://mathoverflow.net/questions/240329/jets-of-sections-of-vector-bundles-expressed-by-symmetrized-iterated-covariant-d", "tags": ["reference-request", "dg.differential-geometry", "ho.history-overview", "jets"], "votes": 17, "creation_date": "2016-06-02T22:30:00", "comments": ["Conversely, given $P$ as defined in the question above, it is clear that $[P,f]=$ commutator of $P$ and (multiplication by) a smooth function $f$ on $M$ is a linear differential operator of order $k-1$. (Remark: one has to add to your definition the \"initial condition\" that linear differential operators of order $-1$ are identically zero)", "@AliTaghavi Not really - such a definition is essentially the same as the one in terms of jet bundles. To see that, it suffices to know that the latter's fiber over $p\\in M$ may be defined as the quotient of the space $E(p)$ of germs of smooth sections of $\\pi$ over $p$ modulo the subspace of such germs vanishing to order $k+1$ at $p$. As such, it can be seen by induction on $k$ that given smooth functions $f_1,\\ldots,f_{k+1}$ on $M$, one always has $D((f_1-f_1(p))\\cdots(f_{k+1}-f_{k+1}(p))s)(p)=0$ for all $p\\in M$, $s\\in\\Gamma(\\pi)$, hence $D$ is necessarily of the form stated for $P$ above.", "I confess I did not understand your terminologies completely. But what would be happen if one define a diffm operator of order k as a linear map $D$ such that $s\\mapsto D(fs)-fD(s)$ would be of order k-1.(inductive definition). Is there a problem of GLOBAL definition of differentoal operatores in this manner?", "Pohl's results were derived independently by Libermann (1963) (a student of Ehresmann) and Feldman (1963), but one also cannot find any explicit isomorphism there. There is a couple of references by Ehresmann (1955) and Libermann (1961) on higher-order connections I couldn't get access to (they are both in rather obscure proceedings volumes) and maybe there is something there closer to what I ask, but I really don't know.", "@DmitriZaitsev That's true, but that's not what I am asking. I've also tracked Ehresmann's pioneer work on jets and (arbitrary-order) connections, and couldn't find any trace of the result I've stated - to wit, relating jets to iterated (first-order) covariant derivatives. The book by Jafarpour and Lewis quoted by Umberto Lupo above traces related results to a Trans. AMS paper by Pohl (1966) whose preprint actually dates back to 1963, but there (as the authors themselves state) no such formula can be found explicitly.", "The jets are usually attributed to Ehresmann", "That is actually a very nice reference (despite lacking a bit of historical care, as you noticed and most people using/quoting the result do), thanks!", "I had not known that the result dates back to Palais' work. I learned it from engineering.ucsb.edu/~saber.jafarpour/time_varying.pdf (see Lemma 2.1) and apparently those authors did not come across your reference either."], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "231", "site": "mathoverflow", "title": "Trying to reconcile two facts about the Appell-Lerch sum learned from Polishchuk and Zwegers", "body": "One of the key characters in the [thesis](https://arxiv.org/abs/0807.4834) of Zwegers is the modular correction $\\tilde\\mu(u,v;\\tau)=\\mu(u,v;\\tau)+\\frac i2R(u-v;\\tau)$ of the Lerch sum $\\mu(u,v;\\tau)=\\frac{e^{\\pi i u}}{\\vartheta(v;\\tau)}\\sum_{n\\in\\mathbb Z}(-1)^n\\frac{e^{\\pi i(n^2+n)\\tau}e^{2\\pi inv}}{1-e^{2\\pi in\\tau}e^{2\\pi iu}}$, where $$ R(u;\\tau)=\\sum_{\\nu\\in\\frac12+\\mathbb Z}(-1)^{\\nu-\\frac12}\\left\\\\{\\operatorname{sgn}(\\nu)-E\\left(\\left(\\nu+\\frac{\\Im(u)}{\\Im(\\tau)}\\right)\\sqrt{2\\,\\Im(\\tau)}\\right)\\right\\\\}e^{-\\pi i\\nu^2\\tau}e^{-2\\pi i\\nu u}, $$ with $E(x)=2\\int_0^xe^{-\\pi t^2}dt$.\n\nIn an illuminating (for me at any rate) paper \"[M. P. Appell's function and Vector Bundles of Rank 2 on Elliptic Curves](https://pdfs.semanticscholar.org/4aef/7841fc86e1e2024c020094f75a1e6e9ef8f9.pdf)\", Polishchuk explains that the function $\\kappa_a(z;q)=\\sum_{n\\in\\mathbb Z}\\frac{q^{n^2/2}}{q^n-a}z^n$ can be uniquely characterized by the fact that $(\\kappa_a(z;q),\\theta(z;q))$ defines a holomorphic section of the rank 2 vector bundle on the elliptic curve $\\mathbb C^*/q^{\\mathbb Z}$ given by the quotient $(z,v_1,v_2)\\sim(qz,av_1+v_2,q^{-1/2}z^{-1}v_2)$ of $\\mathbb C^*\\times\\mathbb C^2$.\n\nWhat confuses me is this - the way $\\kappa$ is involved in the section of a bundle on the universal elliptic curve, it must have certain modularity properties; and it is clearly holomorphic (well, for $a\\notin q^{\\mathbb Z}$). On the other hand $\\mu$, which is obviously closely related to it, although holomorphic, is not modular, while $\\tilde\\mu$, being modular, is not holomorphic. So seemingly $\\tilde\\mu$ must have a counterpart in the context considered by Polishchuk; but the latter does not mention any possible nonholomorphic modular corrections.\n\nHow exactly are $\\mu$, $\\tilde\\mu$ and $\\kappa$ related to each other?\n", "link": "https://mathoverflow.net/questions/266199/trying-to-reconcile-two-facts-about-the-appell-lerch-sum-learned-from-polishchuk", "tags": ["ag.algebraic-geometry", "elliptic-curves", "modular-forms", "vector-bundles"], "votes": 17, "creation_date": "2017-04-02T11:59:10", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "232", "site": "mathoverflow", "title": "Are there two non-equivalent exotic structures on $\\mathbb{R}^4$ coming from topologically slice, non-slice knots?", "body": "For a knot $K \\subset S^3$, which is topologically slice but not slice (in a smooth way), there's a four manifold $\\mathbb{R}^4_K$, homeomorphic but not diffeomorphic to standard euclidean $\\mathbb{R}^4$. \n\nThere are ways to find such knots, particularly knots with the Alexander polynomial equal to 1, are topologically slice (Freedman). Among these knots one can find non-slice knots using obstructions like $\\tau$-invariant (coming from Knot Floer Homology) or Rasmussen $s$-invariant (coming from Khovanov Homology). Knots with non-zero $\\tau$- or $s$- invariant are not slice. \n\nThe negative Whitehead double of left trefoil is an example of these knots.\n\nQuestion. The above argument proves the existence of an exotic structure on $\\mathbb{R} ^4$, but this method proves the existence of only one exotic structure on $\\mathbb{R} ^4$. Can you show if there are two different knots, $K$ and $K'$ (both topologically slice, non-slice knots), which they give us two non-equivalent exotic $\\mathbb{R} ^4$s? Or if all of $\\mathbb{R}_K^4$ coming from these knots are the same, i.e. you can get to only one exotic structure through this method.\n", "link": "https://mathoverflow.net/questions/277368/are-there-two-non-equivalent-exotic-structures-on-mathbbr4-coming-from-top", "tags": ["differential-topology", "knot-theory", "4-manifolds", "heegaard-floer-homology", "khovanov-homology"], "votes": 17, "creation_date": "2017-07-27T03:27:14", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "233", "site": "mathoverflow", "title": "Need explicit formula for certain "$q$-numbers" involving gcd's", "body": "The question is motivated by yet another possible approach to a combinatorial problem formulated previously in [\"Special\" meanders](https://mathoverflow.net/q/146802/41291). I'm not giving details of the connection as I believe the question is sufficiently motivated by itself.\n\nI've got a vector space with basis $\\left\\\\{e_n\\mid n\\geqslant0\\right\\\\}$ and scalar product $$ \\left\\langle e_m,e_n\\right\\rangle=q^{\\gcd(m,n)} $$ (with the convention $\\gcd(0,n)=n$ for all $n\\geqslant0$).\n\nWhat I need is a maximally explicit expression for an orthogonal basis $\\left\\\\{o_n\\mid n\\geqslant0\\right\\\\}$ with respect to this scalar product. I do not mind if the $o_n$ are not of unit norm (and this is clearly a minor point anyway).\n\nHere are the first few values (calculated with MAPLE). Up to arbitrary rescalings, \\begin{align*} o_{{0}}&=e_{{0}}\\\\\\ o_{{1}}&=e_{{1}}-qe_{{0}}\\\\\\ o_{{2}}&=e_{{2}}- \\left( q+1 \\right) e_{{1}}+qe_{{0}}\\\\\\ o_{{3}}&=e_{{3}}-qe_{{2}}-e_{{1}}+qe_{{0}}\\\\\\ o_{{4}}&=\\left( q+1 \\right) e_{{4}}-{q}^{2}e_{{3}}- \\left( {q}^{2}+q+1 \\right) e_{{2}}+{q}^{2}e_{{1}}+{q}^{2}e_{{0}}\\\\\\ o_{{5}}&=\\left( {q}^{2}+q+1 \\right) e_{{5}}- \\left( {q}^{2}+1 \\right) qe_{{4}}- \\left( {q}^{2}+1 \\right) qe_{{3}}+ \\left( {q}^{3}-{q}^{2}-1 \\right) e_{{1}}+ \\left( {q}^{2}+1 \\right) qe_{{0}}\\\\\\ o_{{6}}&=\\left( {q}^{3}+{q}^{2}+2\\,q+1 \\right) e_{{6}}- \\left( {q}^{3} +q-1 \\right) qe_{{5}}- \\left( {q}^{3}+q-1 \\right) qe_{{4}}\\\\\\&\\- \\left( {q} ^{2}+1 \\right) \\left( {q}^{2}+q+1 \\right) e_{{3}}- \\left( {q}^{3}+{q} ^{2}+2\\,q+1 \\right) e_{{2}}+ \\left( 2\\,{q}^{4}+{q}^{3}+3\\,{q}^{2}+1 \\right) e_{{1}}+ \\left( {q}^{3}+q-1 \\right) qe_{{0}}\\\\\\ o_{{7}}&=\\left( {q}^{2}+1 \\right) e_{{7}}- \\left( {q}^{2}-q+1 \\right) qe_{{6}}- \\left( {q}^{2}-q+1 \\right) qe_{{5}}- \\left( {q}^{2}-q+1 \\right) qe_{{4}}\\\\\\&\\+ \\left( {q}^{2}-q+1 \\right) qe_{{2}}+ \\left( {q}^{3} -2\\,{q}^{2}+q-1 \\right) e_{{1}}+ \\left( {q}^{2}-q+1 \\right) qe_{{0}} \\end{align*}\n\nThe inverse transformation (again up to rescalings) does not look any more enlightening (except that everything is positive):\n\n\\begin{align*} e_{{0}}&=o_{{0}}\\\\\\ e_{{1}}&=o_{{1}}+qo_{{0}}\\\\\\ e_{{2}}&=o_{{2}}+ \\left( q+1 \\right) o_{{1}}+{q}^{2}o_{{0}}\\\\\\ e_{{3}}&=o_{{3}}+qo_{{2}}+ \\left( {q}^{2}+q+1 \\right) o_{{1}}+{q}^{3}o_ {{0}}\\\\\\ e_{{4}}&=o_{{4}}+{\\frac {{q}^{2}}{q+1}}o_{{3}}+ \\left( {q}^{2}+1 \\right) o_{{2}}+ \\left( q+1 \\right) \\left( {q}^{2}+1 \\right) o_{{1}} +{q}^{4}o_{{0}}\\\\\\ e_{{5}}&=o_{{5}}+{\\frac { \\left( {q}^{2}+1 \\right) q}{{q}^{2}+q+ 1}}o_{{4}}+{\\frac {q \\left( {q}^{2}+1 \\right)}{q+1}} o_{{3}}+q \\left( {q}^{2} +1 \\right) o_{{2}}+ \\left( {q}^{4}+{q}^{3}+{q}^{2}+q+1 \\right) o_{{1}} +{q}^{5}o_{{0}}\\\\\\ e_{{6}}&=o_{{6}}+{\\frac { \\left( {q}^{3}+q-1 \\right) q}{{q}^{3}+ {q}^{2}+2\\,q+1}}o_{{5}}+{\\frac { \\left( {q}^{3}+q-1 \\right) q}{{q}^{2} +q+1}}o_{{4}}+{\\frac { \\left( {q}^{2}+q+1 \\right) \\left( {q}^{2}-q+1 \\right)}{q+1}} o_{{3}}\\\\\\ &\\+ \\left( {q}^{2}+q+1 \\right) \\left( {q}^{2}-q+ 1 \\right) o_{{2}}+ \\left( q+1 \\right) \\left( {q}^{2}+q+1 \\right) \\left( {q}^{2}-q+1 \\right) o_{{1}}+{q}^{6}o_{{0}}\\\\\\ e_{{7}}&=o_{{7}}+{\\frac { \\left( {q}^{2}-q+1 \\right) q}{{q}^{2}+ 1}}o_{{6}}+{\\frac { \\left( {q}^{2}+q+1 \\right) \\left( {q}^{2}-q+1 \\right) q}{{q}^{3}+{q}^{2}+2\\,q+1}}o_{{5}}+ \\left( {q}^{2}-q+1 \\right) qo_{{4}}\\\\\\ &+{\\frac {q \\left( {q}^{2}+q+1 \\right) \\left( {q}^{2}-q+1 \\right)}{q+1}}o_{{3}} +q \\left( {q}^{2}+q+1 \\right) \\left( {q}^{2}-q+1 \\right) o_{{2}} \\+ \\left( {q}^{6}+{q}^{5}+{q}^{4}+{q}^{3}+{q}^{2}+q+1 \\right) o_{{1}}+{q}^{7}o_{{0}}\\\\\\ e_{{8}}&=o_{{8}}+{\\frac { \\left( {q}^{2}+1 \\right) {q}^{4}}{ \\left( {q}^{2}+q+1 \\right) \\left( {q}^{3}+q+1 \\right) }}o_{{7}}+{\\frac {{q} ^{4}}{{q}^{2}+q+1}}o_{{6}}+{\\frac { \\left( {q}^{2}+1 \\right) {q}^{4}}{{q}^{3}+{q}^{2}+2\\,q+1}}o_{{5}}\\\\\\ &+{\\frac { {q}^{6}+{q}^{4}+{q}^{2}+q+1 }{{q}^{2}+q+1}}o_{{4}} +{\\frac {{q}^{2} \\left( {q}^{2}+q+1 \\right) \\left( {q}^{2}-q+1 \\right)}{q+1}}o_{{3}}+ \\left( {q}^{2}+1\\right) \\left( {q}^{4}+1 \\right) o_{{2}}+ \\left( q+1 \\right) \\left( {q}^{2}+1 \\right) \\left( {q}^{4}+1 \\right) o_{{1}}+{q}^{8}o_{ {0}} \\end{align*}\n\nThere are some patterns, but I cannot even guess any statement that I could try to prove about these coefficients. Also the above was computed under the assumption that the transformation matrix is triangular, i. e. that $o_n$ only depends on $e_k$ with $k\\leqslant n$. It is quite possible that there is a nicer basis which violates this assumption, but again I cannot think of any natural alternative form of the transformation matrix.\n\nWhat I know is an explicit orthogonal basis for a similar scalar product \"without $q$\": if I would have just $\\left\\langle e_m,e_n\\right\\rangle=\\gcd(m,n)$, then an orthogonal basis would be given by $e_n=\\sum_{d|n}o_d$, so $o_n=\\sum_{d|n}\\mu\\left(\\frac nd\\right)e_d$. Strangely enough, I obtain this from the above by substituting $q=0$, and again I have no idea why.\n\nAs @alpoge points out in a comment, if one removes $e_0$, then this actually works \"with $q$\" too.\n\nPS As suggested by @Wolfgang in a comment, I've looked at polynomials resulting from the substitutions $q=\\pm1$, $e_n=x^n$. Here:\n\n$q=1$: \\begin{align*} &1\\\\\\ &x-1\\\\\\ & \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( 2\\,x+1 \\right) \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( 3\\,{x}^{2}+x+2 \\right) \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( 5\\,{x}^{3}+4\\,{x}^{2}+8\\,x+1 \\right) \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( {x}^{2}+x+1 \\right) \\left( 2\\,{x}^{2}-x+1 \\right) \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( 9\\,{x}^{5}+7\\,{x}^{4}+14\\,{x}^{3}+10\\,{x}^ {2}+6\\,x+2 \\right) \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( 11\\,{x}^{6}+8\\,{x}^{5}+16\\,{x}^{4}+10\\,{x} ^{3}+15\\,{x}^{2}+9\\,x+3 \\right) \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( 7\\,{x}^{5}+6\\,{x}^{4}+5\\,{x}^{3}+4\\,{x}^{2 }+10\\,x+1 \\right) \\left( {x}^{2}+1 \\right) \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( 16\\,{x}^{8}+11\\,{x}^{7}+22\\,{x}^{6}+12\\,{x }^{5}+18\\,{x}^{4}+3\\,{x}^{3}+9\\,{x}^{2}-6\\,x+5 \\right) \\left( x-1 \\right) ^{2}\\\\\\ & \\left( x+1 \\right) \\left( 21\\,{x}^{9}+19\\,{x}^{8}+38\\,{x}^{7}+34\\,{x }^{6}+51\\,{x}^{5}+45\\,{x}^{4}+39\\,{x}^{3}+33\\,{x}^{2}+6\\,x+2 \\right) \\left( x-1 \\right) ^{2} \\end{align*} $q=-1$: \\begin{align*} 0&:1\\\\\\ 1&:1+x\\\\\\ 2&:\\left( x-1 \\right) \\left( 1+x \\right)\\\\\\ 3&:\\left( x-1 \\right) \\left( 1+x \\right) ^{2}\\\\\\ 4&:- \\left( x-1 \\right) \\left( 1+x \\right) ^{2}\\\\\\ 5&:\\left( x-1 \\right) \\left( {x}^{2}+x+2 \\right) \\left( 1+x\\right) ^{2}\\\\\\ 6&:-\\left( x-1 \\right) \\left( {x}^{3}+2\\,{x}^{2}+2\\,x+3 \\right) \\left( 1+x \\right) ^{2}\\\\\\ 7&:\\left( x-1 \\right) \\left( {x}^{2}+x+1 \\right) \\left( 2\\,{x}^{2}- x+3 \\right) \\left( 1+x \\right) ^{2}\\\\\\ 8&:-\\left( x-1 \\right) \\left( {x}^{4}+2\\,{x}^{2}+2 \\right) \\left( 1 +x \\right) ^{3}\\\\\\ 9&:\\left( x-1 \\right) \\left( {x}^{5}+{x}^{4}+{x}^{3}+3\\,{x}^{2}+3 \\right) \\left( 1+x \\right) ^{3}\\\\\\ 10&:- \\left( x-1 \\right) \\left( {x}^{2}+1 \\right) \\left( {x}^{5}+2\\, {x}^{4}+{x}^{3}+2\\,{x}^{2}+2\\,x+3 \\right) \\left( 1+x \\right) ^{2}\\\\\\ 11&:\\left( x-1 \\right) \\left( 4\\,{x}^{8}+{x}^{7}+8\\,{x}^{6}+2\\,{x}^{ 5}+12\\,{x}^{4}+3\\,{x}^{3}+11\\,{x}^{2}+4\\,x+5 \\right) \\left( 1+x \\right) ^{2} \\end{align*}\n", "link": "https://mathoverflow.net/questions/210448/need-explicit-formula-for-certain-q-numbers-involving-gcds", "tags": ["nt.number-theory", "co.combinatorics", "orthogonal-matrices"], "votes": 17, "creation_date": "2015-06-30T00:37:31", "comments": ["@OfirGorodetsky Thanks for the link! Reading that - seems an interesting idea to give it a combinatorial interpretation in terms of the Möbius function of some poset. In fact this might be important for the original problem which is of combinatorial nature.", "@alpoge But this is great! I was sticking to $e_0$ so much it never occurred to me that dropping it gives such a nice orthogonal basis! It is true that I still need $e_0$ essentially (it is the unit of certain algebra), but your version does almost all of it. Could you please post this as an answer? It's worth it in any case, and if nobody will find a way to incorporate $e_0$ I will go for it", "Just a remark\\restatement, nothing fancy: Let $A_{i,j} = q^{(i,j)}$ be a $(n+1) \\times(n+1)$ matrix. Decompose $A$ as $B^T B$ (Cholesky-Decomposition; $B$ can be upper-triangular). The condition that $o_i = \\sum_{j} o_{i,j} e_j$ are orthonormal w.r.t to the given inner product is the same as requiring that $B o_i$ are orthonormal w.r.t to the ordinary inner product, so we can take $o_i = B^{-1} e_i$. Hence the problem reduces to Cholesky-decomposing $A$, and inverting. Bruce Sagan's talk here about GCD-matrices might provide useful input: users.math.msu.edu/users/sagan/Slides/mfp5.pdf", "Formally doesn't $o_n := \\sum_{d\\vert n} \\mu(d) e_{n/d}$ for $n > 0$ and $o_0 := e_0 - \\sum_{n\\geq 1} o_n$ work? (Not that that helps much, since the expression for $o_0$ is nonsense.)", "@Wolfgang Well $q=-1$ looks somehow less hopeless but still I have no clue :) I'll add it too", "and I suppose the same with $q=-1$ doesn't reveal more either?", "I had skipped $n=6 $, so that was a wrong conclusion of mine! :(.", "@PeterMueller I've added that", "@PeterMueller I use $\\gcd(0,0)=0$, I think it is not senseless in view of $\\gcd(0,n)=\\gcd(n,n)=n$ for all $n$.", "@OP: What is $\\langle e_0,e_0\\rangle$? It seems to me that the problem is not well formulated, as $gcd(0,0)=\\infty$. In particular, I don't know how to interpret your assertion $0=\\langle o_0,o_1\\rangle=\\langle e_0,e_1-qe_0\\rangle=q-q\\cdot q^{gcd(0,0)}$.", "@Wolfgang I've looked at it, what happens that seeminglu each of these polynomials (well, starting from 3) is divisible by $(x-1)^2(x+1)$, but the quotient seems impenetrable to me. I'll add it to the question though", "@Wolfgang Many thanks for the suggestion, sounds promising! I'll definitely try it", "Have you tried putting $q=1$ and replacing $e_i$ by $x^i$ in the $o_j$ expressions? The resulting polynomials seem to split into linear and quadratic factors, and maybe you'll find some patterns in those factors, which might give some ideas to start with."], "comment_count": 13, "category": "Science", "diamond": 0} {"question_id": "234", "site": "mathoverflow", "title": "Bunnity of multilinear maps", "body": "Is there a way to compute the following nullity of multilinear maps? As it is different from any nullity I know of, I call it _bunnity_ after myself:-)) If it already has a name, it be nice to know it. References are particularly appreciated.\n\nLet $V$ be a $d$-dimensional vector space over a field $K$, $f:V^{\\otimes n}\\rightarrow K$ a multilinear map. We define the bunnity by\n\n$Bun(f) = sup \\\\{ \\sum_{k=1}^n dim(U_k) \\ | \\ f(U_1\\otimes U_2 \\ldots \\otimes U_n)=0\\\\}$.\n\nIf $n=2$, the bunnity is nullity plus dimension. It is more complicated if $n\\geq 3$. I expect that for $n=3$ or $n=4$, there should be an algorithm because the 3-subspace problem is finite, while the 4-subspace problem is tame. I would be pleasantly surprised if there is an algorithm for a general $n$.\n\n**EDIT** I forgot to say what $U_i$ are. They are **non-zero** vector subspaces of $V$.\n", "link": "https://mathoverflow.net/questions/206211/bunnity-of-multilinear-maps", "tags": ["linear-algebra", "multilinear-algebra"], "votes": 17, "creation_date": "2015-05-10T07:36:12", "comments": ["The dimension of the null-space: think of $f$ as a map $V\\rightarrow V^\\ast$ and take the dimension of its kernel. Similarly, the usual multinullity is defined for any $n$: use position $i$ to write $f$ as $V\\rightarrow (V^{\\otimes (n-1)})^\\ast$ and take the dimension of its kernel.", "Could you remind what definition of \"nullity\" do you use (e.g. when $n=2$)?", "You should offer a bounty to whoever manages to frame the answer in terms of Wascal's triangle.", "Inequality. In the usual tensor rank/nullity, you choose each $U_i$ separately. I ask you to choose them simultaneously.", "What's the connection to multilinear tensor rank?"], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "235", "site": "mathoverflow", "title": "Question about combinatorics on words", "body": "Let $\\\\{a_1,a_2,...,a_n\\\\}$ be an alphabet and let $\\\\{u_1,...,u_n\\\\}$ be words in this alphabet, and $a_i\\mapsto u_i$ be a substitution $\\phi$. \n\nQuestion: Is there an algorithm to check if for some $m,k$ some prefix of the word $\\phi^m(a_1)$ coincides with some suffix of the word $\\phi^k(a_2)$?\n\nThis is related to a question about the R. Thompson group $F$. \n", "link": "https://mathoverflow.net/questions/250290/question-about-combinatorics-on-words", "tags": ["co.combinatorics", "gr.group-theory", "semigroups-and-monoids", "combinatorics-on-words"], "votes": 17, "creation_date": "2016-09-19T20:08:39", "comments": ["@domotorp The equality version is decidable, by Theorem 4 of [1]. Unfortunately, this doesn't seem very helpful for the original question, because it goes via saying that if the substitution homomorphism is not injective, then it factors through a smaller alphabet and we are happy by induction. [1] A. Ehrenfeucht, G. Rozenberg, Simplifications of homomorphisms, Information and Control, Volume 38, Issue 3, 1978 (sciencedirect.com/science/article/pii/S0019995878900955)", "With the below simple trick we can ensure that if a prefix equals a suffix, then the whole words are equal. I'm thinking that deciding equality might be already undecidable. Trick: $\\phi(a_1)=a_{start} a_1' a_{end}$ and $\\phi(a_2)=a_{start} a_2' a_{end}$, where $\\phi(a_{start})=a_{start}$ and $\\phi(a_{end})=a_{end}$, and none of $a_1, a_2, a_{start}, a_{end}$ are in the image of any other $a_i$.", "@NaradRampersad: You can ask him and post his answer here, if he is not on MO himself.", "I don't have an actual solution to the problem, but I know the best person to ask about it: Juha Honkala of the University of Turku. If anyone knows the answer it would be him.", "This is a decision problem. I need to decide whether or not the property holds. The second $n$ is a $k$.", "Assuming the words are finite, there is the brute force method with no guarantee of termination. Are you looking for an algorithm with such a guarantee, or do you want even more? Also, is the n in the question the same as the alphabet size? (Probably not.) Gerhard \"Just Checking On Some Details\" Paseman, 2016.09.19."], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "236", "site": "mathoverflow", "title": "The topos for forcing in computability theory", "body": "My understanding is that forcing (such as Cohen forcing) can be described via a topos. For example this [nlab article on forcing](http://ncatlab.org/nlab/show/forcing) describes forcing as a \"the topos of sheaves on a suitable site.\" \n\nMy question concerns forcing in computability theory, for example as described in Chapter 3 or these [lecture notes of Richard Shore](http://www.math.cornell.edu/~shore/papers/pdf/SingLect2NS.pdf). The idea is that the generics are those which meet all _computable_ dense sets of forcing conditions. (Computable can mean a few things. Often it is taken to mean a $\\Sigma^0_1$ set of forcing conditions. Also, usually the forcing posets are countable.) Since there are only countably many such dense sets, such effective generics exist.\n\n> Is there a known/canonical type of topos corresponding to the forcing in computability theory?\n\nAny references would be appreciated.\n\n_FYI: My background is in computability theory, proof theory, and computable analysis. I know little about topos theory, but I am willing to learn a bit. I am mostly asking this question because I want to compare some ideas I have about effective versions of Solovay forcing with some work by others about the topos corresponding to Solovay forcing. Also, it is always nice to learn new things._\n", "link": "https://mathoverflow.net/questions/195794/the-topos-for-forcing-in-computability-theory", "tags": ["ag.algebraic-geometry", "ct.category-theory", "lo.logic", "computability-theory", "topos-theory"], "votes": 17, "creation_date": "2015-02-05T19:55:21", "comments": ["Have you, since asking the question two years ago, learned about a topos-theoretic treatment of forcing in computability theory?", "@BjørnKjos-Hanssen, I would be happy to know the answer in that case. (The application I have in mind is a bit more general, but I think knowing what is going on in the case of a computable forcing poset is sufficient for me to get a general idea of what is going on.)", "Did you also want the forcing partial order to be a (close to) computable relation?"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "237", "site": "mathoverflow", "title": "Elements of finite fields with many powers of trace zero", "body": "Let $p$ be an odd prime number, $n>1$ be an integer, and $\\mathrm{tr}$ be the trace map of the field extension $\\mathrm{GF}(p^{2n})/\\mathrm{GF}(p)$. For which pair $(p,n)$ does there exists $x\\in\\mathrm{GF}(p^{2n})$ such that $x^{2(1+p^n)}=1$ and $\\mathrm{tr}(x^{1+p^j})=0$ for $j=0,1,\\dots,n-1$?\n\nAfter computing the gcd of the corresponding $n+1$ polynomials over $\\mathrm{GF}(p)$ for small $p$ and $n$, I believe that the answer should be ``if and only if $p$ divides $n$\" and moreover, when $p$ divides $n$, all such $x$ would form a subgroup of $\\mathrm{GF}(p^{2n})^\\times$ of index $2(1+p^{n/p})$.\n\n**Update.** If $(p,n)$ is such a pair, then denoting by $X$ the $(2n)\\times(2n)$ matrix with $(i,j)$th entry $x^{p^{i+j-2}}$ we deduce from the conditions that $$ X^2= \\begin{pmatrix} 0&2naI_n\\\\\\ 2naI_n&0 \\end{pmatrix}, $$ where $a=x^{1+p^n}=\\pm1$. This yields two necessary conditions:\n\n(1). If $n$ is not divisible by $p$ then $n$ is odd.\n\n_Proof._ Suppose on the contrary that $n$ is even. Then $$ |X|^2= \\begin{vmatrix} 0&2naI_n\\\\\\ 2naI_n&0 \\end{vmatrix} =(-1)^n(2na)^{2n}=(2na)^{2n} $$ is a square in $\\mathrm{GF}(p)$, which implies that $|X|\\in\\mathrm{GF}(p)$. However, changing columns shows that $-|X|=(-1)^{2n-1}|X|$ is the determinant of the matrix with $(i,j)$th entry $x^{p^{i+j-1}}$. Hence $-|X|=|X|^p$ and so $|X|\\notin\\mathrm{GF}(p)$, a contradiction. \n\n(2). If $n$ is not divisible by $p$ then $x,x^p,\\dots,x^{p^{2n-1}}$ is a normal basis of $\\mathrm{GF}(p^{2n})/\\mathrm{GF}(p)$.\n\n_Proof._ $|X|^2=(-1)^n(2na)^{2n}\\neq0$ implies that $|X|\\neq0$ and so $x,x^p,\\dots,x^{p^{2n-1}}$ are $\\mathrm{GF}(p)$-linearly independent.\n", "link": "https://mathoverflow.net/questions/257733/elements-of-finite-fields-with-many-powers-of-trace-zero", "tags": ["nt.number-theory", "galois-theory", "finite-fields"], "votes": 17, "creation_date": "2016-12-20T21:45:08", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "238", "site": "mathoverflow", "title": "Groups generated by 3 involutions", "body": "Let $r(m)$ denote the residue class $r+m\\mathbb{Z}$, where $0 \\leq r < m$. Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$, let the class transposition $\\tau_{r_1(m_1),r_2(m_2)}$ be the permutation of $\\mathbb{Z}$ which interchanges $r_1+km_1$ and $r_2+km_2$ for every $k \\in \\mathbb{Z}$ and which fixes everything else.\n\n 1. Is it algorithmically decidable whether all orbits on $\\mathbb{Z}$ under the action of a group generated by 3 given class transpositions are finite?\n\n 2. Is it algorithmically decidable whether a group generated by 3 given class transpositions acts transitively on the set of nonnegative integers in its support (i.e. set of moved points)?\n\n\n\n\n**Added on Dec 11, 2013:** This question will appear as Problem 18.47 in:\n\nKourovka Notebook: _Unsolved Problems in Group Theory_. Editors V. D. Mazurov, E. I. Khukhro. 18th Edition, Novosibirsk 2014.\n\nRemarks:\n\nAd 1.: A hard case is $G = \\left\\langle\\tau_{0(2),1(2)}, \\tau_{0(5),4(5)}, \\tau_{1(4),0(6)}\\right\\rangle$. For example we have $|32^G| = 6296$ and $|25952^G| = 245719352$, and the largest point in the latter orbit is about $10^{5759}$. The finiteness of the orbit $173176^G$ is not known to the author so far. \n\nEDIT: In the meantime, the length of the cycle of $g := \\tau_{0(2),1(2)} \\cdot \\tau_{0(5),4(5)} \\cdot \\tau_{1(4),0(6)}$ containing 173176 has been computed by Jason B. Hill () -- it is 47610700792, and the largest point in the cycle is about $10^{76785}$. An earlier attempt of the same computation ended without success after one week of CPU time (). Possibly the cycle length is also the orbit length $|173176^G|$. \n\nAd 2.: A hard case is $G = \\left\\langle\\tau_{1(2),4(6)}, \\tau_{1(3),2(6)}, \\tau_{2(3),4(6)}\\right\\rangle$. As the question whether this group acts transitively on $\\mathbb{N} \\backslash 0(6)$ is equivalent to Collatz' $3n+1$ conjecture (cf. ), likely only a negative answer might be given without solving a known open problem. This will likely also not be easy -- at least unless someone has e.g. a good idea on how to encode arbitrary computations with just 3 class transpositions.\n\n**Added on Apr 24:** An easy case is $G = \\left\\langle\\tau_{0(2),1(2)}, \\tau_{0(3),2(3)}, \\tau_{1(2),2(4)}\\right\\rangle$. By means of computation it can be checked that this group acts at least 5-transitively on $\\mathbb{N}_0$.\n", "link": "https://mathoverflow.net/questions/112527/groups-generated-by-3-involutions", "tags": ["permutation-groups", "computational-group-theory", "gr.group-theory", "nt.number-theory", "computability-theory"], "votes": 17, "creation_date": "2012-11-15T14:14:42", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "239", "site": "mathoverflow", "title": "Is there a n/2 version of the Erdős-Hanani conjecture?", "body": "This question comes out of REU research from this past summer. Unfortunately weeks of thought led to only trivial observations and the conclusion that the problem is quite hard.\n\nFix $k,t$. Let $F$ be a set of $k$-subsets of $[n] := \\\\{1,\\ldots,n\\\\}$ of minimal cardinality such that $F$ covers all $t$-subsets of $[n]$ (covers in the sense that any $t$-subset of $[n]$ is a subset of an element of $F$.) Let $\\kappa_n := |F|$. The Erdős-Hanani conjecture states that \n\n> $\\kappa_n = \\binom{n}{t} / \\binom{k}{t}(1 + o(1))$. \n\nOf course $\\binom{n}{t} / \\binom{k}{t}$ is a lower bound on $\\kappa_n$, so the EH conjecture is saying that the obvious necessary condition is asymptotically sufficient. Rödl proved the EH conjecture in 1985.\n\nThis question is about what happens when $k$ and $t$ are not fixed. Specifically, take $k = \\lfloor n/2 \\rfloor$ and $t = \\lfloor n/2\\rfloor - 1$. Define $F$ and $\\kappa_n$ as above. Is it true that\n\n> $\\kappa_n = \\frac{1}{\\lfloor n / 2 \\rfloor} \\binom{n}{\\lfloor n/2 \\rfloor}(1 + o(1))$?\n\n# Background\n\nThe EH conjecture lead to the study of what is called \"packing in a hypergraph.\" See . Rödl's proof introduced what is now called the \"Rödl nibble\" and is pseudo-random in nature. Spencer gave a lovely proof using branching processes. There are a lot of results from the late 80s to 90s that say, as Kahn puts it in \"Asymptotics of Hypergraph Matching, Covering and Coloring Problems\", that hypergraphs are asymptotically well-behaved _as long as their edge sizes are bounded_! Unfortunately the $n/2$ version of EH involves hypergraphs of unbounded edge size and the existing methods appear useless.\n\n# Some ideas\n\nA straightforward application of the method of alterations (or equivalently, some easy analysis of the greedy algorithm) gives that $\\kappa_n \\leq \\log n \\frac{1}{\\lfloor n / 2 \\rfloor} \\binom{n}{\\lfloor n/2 \\rfloor} (1 + o(1))$, so the whole question is whether we can eliminate this log factor.\n\nA set of $\\lfloor n/2 \\rfloor$-subsets has maximum coverage of $(\\lfloor n/2 \\rfloor - 1)$-subsets when all its elements have pairwise symmetric difference of at least 4. So really this is a coding theory problem. The paper \"Lower bounds for constant weight codes\" by Graham and Sloane shows that we can find a set $H$ of $\\lfloor n /2\\rfloor$-subsets of $[n]$ such that $|H| \\geq \\frac{1}{2}\\binom{n}{\\lfloor n/2 \\rfloor}$ and the hamming distance between elements is at least 4. Let $G$ be the set of $(\\lfloor n/2 \\rfloor - 1)$-subsets covered by $H$. $G$ is half the size we want it to be, but we only used half as many elements are we are allowed. So we might be optimistic that by allowing some small overlap we can cover everything we want. If we take a permutation $\\sigma \\in S_n$ and look at $\\sigma(H)$ (i.e. apply the permutation to the elements of the elements of $H$) it covers $\\sigma(G)$. Of course $|\\sigma(G)| = |G|$. We could hope that a good choice of $\\sigma$ gives $|G \\cup \\sigma(G)| \\approx 2|G|$ and we have found an appropriate set $F := H \\cup \\sigma(H)$. I asked the question of whether such a $\\sigma$ must exist before: [Size of union of a set of subsets and its permutations](https://mathoverflow.net/questions/101886/size-of-union-of-a-set-of-subsets-and-its-permutations). That question is interesting in its own right, but this EH conjecture is really why I wanted an answer.\n", "link": "https://mathoverflow.net/questions/121961/is-there-a-n-2-version-of-the-erd%c5%91s-hanani-conjecture", "tags": ["co.combinatorics"], "votes": 17, "creation_date": "2013-02-15T17:02:33", "comments": ["@Gerhard Paseman: it looks like the repository has $n < 100$, $k \\leq 25$ and $t \\leq 8$. With such small values it is not possible to detect the presence or absence of a log factor.", "Some structural results on this part of the boolean lattice can be found with the search phrases \"middle levels\" and \"boolean\" together. For example, see the references in arxiv.org/pdf/math.CO/0608485.pdf .", "I should mention also that the Graham-Sloane construction means that the EH-conjecture for $k = \\phi(n)$ and $t = k - 1$ has a positive answer when $\\phi(n) = o(n)$, so $n/2$ is in fact the \"hard case.\"", "@Brendan McKay: Yes, thank you. It is corrected.", "Does your description have $t$ and $l$ mixed up? If so, please edit.", "Have you checked out the La Jolla Covering Repository online? That might give a nice suggestion of what to expect. Gerhard \"Ask Me About System Design\" Paseman, 2013.02.15"], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "240", "site": "mathoverflow", "title": "Katz--Mazur for abelian varieties", "body": "Over $\\mathbb Z$, there is a smooth DM stack $A_g$ classifying abelian varieties.\n\nOver $\\mathbb Z[\\frac 1N]$, there is finite etale cover $A_g(N)_{\\mathbb Z[\\frac 1N]}\\to A_g\\otimes\\mathbb Z[\\frac 1N]$ classifying abelian varieties with $N$-level structure.\n\n> Is there a finite proper map $A_g(N)\\to A_g$ over $\\mathbb Z$ with a _nice moduli interpretation_ , which recovers the usual notion of $N$-level structure over $\\mathbb Z[\\frac 1N]$?\n\n**Remark:** In fact, I believe that such an $A_g(N)\\to A_g$ is determined canonically from $A_g(N)_{\\mathbb Z[\\frac 1N]}\\to A_g\\otimes\\mathbb Z[\\frac 1N]$ by \"normalizing in the function field\", so the content of the question is really to have a good moduli interpretation for this canonical extension.\n\nIn more informal terms, I'm just asking:\n\n> What is the \"right\" notion of $p$-level structure on an abelian variety in characteristic $p$?\n\nThe impression I get from reading in various places is that the answer to this is well-known, but I haven't been able to find a good reference. I don't have a specific type of level structure in mind. I'd just like to know what is known (and where to find it) for some common notions of level structure. **EDIT** : Actually, for level structures consisting of a choice of _subgroup(s)_ , there is a natural choice of moduli space over $\\mathbb Z$ classifying abelian varieties with a choice of subgroup scheme(s) (or equivalently, isogenies) and this is valid in \"bad\" characteristics -- see, e.g. the end of [Chai](http://www.ams.org/mathscinet-getitem?mr=1124633). So, let me state my question for level structures given by choices of _point(s)_ , e.g. \"a point of exact order $N$\", or full level $N$ structure.\n\nFor elliptic curves, [Katz--Mazur](http://www.ams.org/mathscinet-getitem?mr=772569) showed that the answer is provided by the notion of a _Drinfeld level structure_ (first considered by [Drinfeld](http://www.ams.org/mathscinet-getitem?mr=384707)). For example, the \"right\" notion of a $p$-torsion point on an elliptic curve $E\\to S$ (for _any_ scheme $S$) is a point $P:S\\to E$ so that:\n\n 1. $pP=1$.\n 2. The Cartier divisor $\\sum_{i=0}^{p-1}[iP]$ is a subgroup scheme.\n\n\n\nAs Katz--Mazur write, it is not clear how to generalize this notion to abelian varieties since points are no longer divisors.\n", "link": "https://mathoverflow.net/questions/152986/katz-mazur-for-abelian-varieties", "tags": ["ag.algebraic-geometry", "nt.number-theory", "moduli-spaces", "abelian-varieties", "characteristic-p"], "votes": 17, "creation_date": "2013-12-28T11:48:13", "comments": ["Dear Daniel Litt: Thanks for tracking it down; that is indeed the right paper. It is also worth noting (in view of the question posed) that this paper provides a sense in which mere normalization in higher level (which doesn't have positive-dimensional fibers) is the \"wrong\" thing to consider, suggesting that perhaps the question posed about normalization is not a \"useful\" point of view for applications with $g > 1$.", "The paper of Chai and Norman user76758 alludes to is likely this one: jstor.org/stable/2374734", "Note that $A_g$ classifies principally polarized abelian schemes (of relative dimension $g$); the polarization aspect comes along for free uniquely when $g=1$ but not otherwise, as you undoubtedly know. I was once told that there is a paper by Chai and Norman (title I do not know) which shows in some sense that there isn't a good version of Drinfeld's idea beyond relative dimension 1. Perhaps email your question to Chai (or Kai-Wen Lan) if you don't receive a satisfactory answer here.", "+1: very nice question!"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "241", "site": "mathoverflow", "title": "Souslin trees and weakly compact cardinals", "body": "In [Souslin trees on the first inaccessible cardinal](https://mathoverflow.net/questions/143530/souslin-trees-on-the-first-inaccessible-cardinal) it is asked if it is consistent that there are no $\\kappa-$Souslin trees at the least inaccessible cardinal $\\kappa$.\n\nIn this question I would like to ask an apparently simpler question:\n\n**Question.** Is it consistent that there are no $\\kappa-$Souslin trees for some inaccessible not weakly compact cardinal $\\kappa$?\n\n**Remarks.** (A) If $\\kappa$ is weakly compact, then the tree property holds at $\\kappa$, in particular there are no $\\kappa-$Souslin trees,\n\n(B) If $V=L,$ then $\\kappa$ is weakly compact iff there are no $\\kappa-$Souslin trees.\n\n(C) $\\Diamond_\\kappa+GCH$ does not imply the existence of a $\\kappa-$Souslin tree, for $\\kappa$ inaccessible. For example if $\\kappa$ is ineffable and $GCH$ holds, then $\\Diamond_\\kappa+GCH$ holds and there are no $\\kappa-$Souslin trees.\n", "link": "https://mathoverflow.net/questions/161868/souslin-trees-and-weakly-compact-cardinals", "tags": ["lo.logic", "set-theory", "forcing", "large-cardinals"], "votes": 17, "creation_date": "2014-03-30T04:28:37", "comments": ["Its about your comment above (since in the situation of your comment $S$ is non reflecting at every $\\alpha$, as witnessed by $\\text{acc }C_\\alpha$).", "Your comment is about my above comment for a weaker sufficient condition?", "@Mohammad: you don't actually need a $\\square (\\kappa )$ sequence, all you need is a non reflecting stationary set $S$ with diamond on it.", "@Yair: Your last comment seems like a familiar topic all of a sudden! :-)", "If there is a $\\kappa$-closed $\\kappa$ saturated ideal on $\\kappa$ and $\\kappa$ is not measurable then $\\kappa$ is inaccessible and there is a $\\kappa$-Suslin tree (by splitting the positive sets in a partial binary tree). Actually, in this case every normal $\\kappa$-tree contains a Suslin subtree.", "Does the existence of a $\\kappa-$Souslin tree for $\\kappa$ inaccessible have any effects on saturated ideals on $\\kappa$ or something else? Maybe this way we can reduce the problem to something that we know more about it.", "I think it does not give the answer. Note that a sufficient condition for having a $\\kappa-$Souslin tree for $\\kappa$ inaccessible, is square-with-built-in-diamond at $\\kappa$, as is implicit in [Sh:624]. A weaker sufficient conditions is, e.g.,: there exists a $\\square(\\kappa)-$sequence $< C_\\alpha: \\alpha<\\kappa >$, and a stationary set $S$ in $\\kappa$, which is disjoint from $acc(C_\\alpha)$ for all $\\alpha<\\kappa$, and $\\Diamond(S)$ holds. To force the failure of these principles, we need more than weakly compact cardinals.", "What happens if we start with a weakly compact cardinal $\\kappa$ in $V$, and then force to kill the Mahloness of $\\kappa$ by adding club $C\\subset\\kappa$ avoiding the regular cardinals below $\\kappa$? Does this create a $\\kappa$-Suslin tree?"], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "242", "site": "mathoverflow", "title": "Maximum automorphism group for a 3-connected cubic graph", "body": "The following arose as a side issue in a project on graph reconstruction.\n\n**Problem:** Let $a(n)$ be the greatest order of the automorphism group of a 3-connected cubic graph with $n$ vertices. Find a good upper bound on $a(n)$.\n\nThere is a [paper of Opstall and Veliche](http://arxiv.org/abs/math/0608645) that finds the maximum over all cubic graphs, but the maximum occurs for graphs very far from being 3-connected.\n\nI foolishly **conjecture** : for $n\\ge 16$, $a(n) < n 2^{n/4}$.\n\nWhen $n$ is a multiple of 4 there is a vertex-transitive cubic graph achieving half the conjectured bound, so if true the bound is pretty sharp.\n\nThe maximum does not always occur for vertex-transitive graphs. There is no vertex-transitive counterexample to the conjecture with less than 1000 vertices.\n\nAny bound for which the exponential part is $2^{cn}$ for $c<\\frac12$ is potentially useful.\n\n**Added:** As verret pointed out in a comment, a [paper of Potočnik, Spiga and Verret](http://arxiv.org/abs/1010.2546), together with the computation I mentioned, establishes the conjecture for vertex-transitive graphs, so the remaining problem is whether one can do better for non-transitive graphs. For 20, and all odd multiples of 2 vertices from 18 to at least 998 (but not for 4-16 or 24 vertices) the graph achieving the maximum is not vertex-transitive.\n", "link": "https://mathoverflow.net/questions/136929/maximum-automorphism-group-for-a-3-connected-cubic-graph", "tags": ["co.combinatorics", "gr.group-theory", "graph-theory", "finite-groups", "automorphism-groups"], "votes": 17, "creation_date": "2013-07-16T18:29:15", "comments": ["In the cubic vertex-transitive case and n twice an odd number, it immediately follows from the same paper that you get a polynomial rather than exponential upper bound. For example Corollary 4 yields that, for large enough n, we have |G| For comparison, the theory of the cardinals of a model of $ZFC$ (as a poset or under addition) is decidable.

Note that, so the underlying universe is a set rather than a proper class, I am restricting the \"model\" of set theory to any downward closed subset of cardinals within the \"model\".", "Asher, suppose that $\\{\\a\\}$ was a maximal antichain (again, I'm ignoring finite sets) then $a$ is comparable with its Hartog number and thus well orderable. So everyone below $a$ can be well ordered as well. It's clear that $W_{\\aleph_0}$ implies $1\\in S$; on the other hand, if $W_{\\aleph_0}$ does not hold, pick $a$ to be a D-finite set, every cardinal above it cannot be well ordered and thus incomparable with its Hartog number. Therefore no one above $a$ can be a maximal antichain of size $1$.", "Asher, the connection is that I believe that whenever there is an antichain of size 2, then one can make larger finite antichains, but it wasn't clear to me how to make larger finite maximal antichains. But if the answer to my question was that it always held, then one could not attain $S=\\{1,2}$, for example. Meanwhile, you had asked about infinite $S$, but it isn't clear to me why you insist on that.", "Joel: Naively, I would expect models of ZF with finite antichains having no finite maximal antichain extension. Is there a connection between this phenomena and the primary question?", "Trevor: Thanks, I was not even certain that (nontrivial) finite maximal antichains were consistent with ZF. Asaf and Joel: Indeed, the only reason I required $0 \\not \\in S$ and $1 \\in S$ was to avoid such trivialities with the finite cardinals. Excepting them, why is $1 \\in S$ if and only if $W_{\\aleph_0}$ holds? Naively, it seems plausible to have a set an infinite set $S$ with no countable subset, with some other cardinal (with countable subsets and size $S$ subsets) comparable to all cardinals. ", "Asaf, I don't think your embedding fact settles the question I raise in my comment, since perhaps the finite antichain can be extended to a maximal antichain that does not lie in the copy of the poset in the cardinals. ", "@Joel: Sure, if you allow finite cardinals then $1\\in S$ is a requirement. I just don't think it is very interesting that way. Considering only infinite cardinals, on the other hand... :-) As for your second question, I'd not think so. Jech has a theorem [Axiom of Choice, Thm 11.1] which embeds every poset into the cardinals, take a poset with only infinite maximal antichains, then its embedding will produce a model in which there are finite antichains that cannot be extended into finite maximal antichains.", "Do we have any reason to hope that every finite antichain of cardinals can be extended to a finite maximal antichain? ", "Asaf, I don't think what you say is correct. We need $1\\in S$ always because, for example, {7} is a maximal antichain, since every set either has fewer than 7 elements or at least 7. Similarly for {0} or {k} for any finite k. ", "To answer a very small part too, $1\\in S$ if and only if $W_{\\aleph_0}$ holds, that is every infinite set has a countable subset.", "To answer a very small part of your question, $\\lbrace \\omega_1, \\mathbb{R} \\rbrace$ is a maximal antichain in all known natural models of the Axiom of Determinacy. See A.E. Caicedo and R. Ketchersid, \\emph{A trichotomy theorem in natural models of AD}, available online at sites.google.com/site/richardketchersid/home/research/adcf-final.pdf."], "comment_count": 11, "category": "Science", "diamond": 0} {"question_id": "246", "site": "mathoverflow", "title": ""extended TQFT" versus "TQFT with defects"", "body": "There are two ways in which higher categories appear in topological field theory: in extended TFTs and in TFTs with defects. How are these appearances related?\n\nAccording to the Atiyah-Segal axioms, a d-dimensional TFT is a symmetric monoidal functor\n\n\\begin{equation*}\\text{Bord}_d\\rightarrow\\mathcal{C}\\end{equation*}\n\nwhere the target category $\\mathcal{C}$ is a symmetric monoidal category, typically $\\text{Vect}$. The objects of $\\text{Bord}_d$ are closed $(d-1)$-manifolds and map to the vector spaces of $\\text{Vect}$ under the TFT. The morphisms are d-dimensional bordisms of the closed $(d-1)$-manifolds and map to the linear maps of $\\text{Vect}$.\n\nThis description can be \"extended down\" to define a d-dimensional n-extended TFT as a symmetric monoidal $n$-functor\n\n\\begin{equation*}\\text{Bord}_d^n\\rightarrow\\mathcal{C}\\end{equation*}\n\nwhere $\\mathcal{C}$ is now an symmetric monoidal $n$-category. The objects of $\\text{Bord}_d^n$ are $(d-n)$-manifolds (for a \"fully extended\" TFT, _i.e._ $n=d$, the bordism objects are points) and map to the objects of $\\mathcal{C}$, which can be thought of as $(n-1)$-categories. One should really specify $\\mathcal{C}$ as some extension of $\\text{Vect}$, but I am ignoring this technicality by thinking of the n-category of all (small) $(n-1)$-categories. More generally, $k$-morphisms are $(d-n+k)$-dimensional bordisms between $(d-n+k-1)$-manifolds and map to $(n+k-1)$-categories. There are also technicalities surrounding how one manages \"manifolds with corners\" in $\\text{Bord}_d^n$, but allow me to gloss over them here.\n\nMeanwhile, one uses higher categories to decorate manifolds with defects. In this picture, $k$-dimensional manifolds are decorated with $k$-dimensional (extended?) TFTs called \"$k$-defects.\" Lower dimensional submanifolds can be decorated with \"defects within defects.\" The $k$-defect assigned to a $k$-dimensional \"boundary\" between two $(k+1)$-dimensional regions, each with attached $(k+1)$-defects, amounts to a morphism of $(k+1)$-defects. Fusion of defects and sub-defects endows the set of $k$-defects with the structure of a $k$-category. (For details, see .) If we again imagine the $k$-category of (small) $(k-1)$-categories, we can understand decoration as an assignment of $(k-1)$-categories to $k$-manifolds.\n\nThis construction of TFT with defects feels \"upside down\" compared to extended TFT: $k$-manifolds are decorated by $(k-1)$-categories and map to $(d-k-1)$-categories under the TFT functor. I realize that decoration is not a functor from a bordism category, but is it a functor in some other sense (from some category where higher degree morphisms are lower dimensional submanifolds)? Defects may have an interpretation as inserted operators (Wilson loops, surface operators, etc); can these two formalisms be combined to compute path integrals with operator insertions in an extended theory? In general, I am curious about the relation between how higher categories are used to define an extended TFT and how they are used to characterize TFTs with defects.\n\nHere is a particular problem from . Consider a fully extended 4D TFT. The TFT functor assigns $1$-categories to Riemann surfaces $\\Sigma$. Meanwhile, in the defect description, 2D TFTs form a $2$-category. Restricting ourselves to a single object leaves us with a symmetric monoidal category of $1$-dimensional domain walls (essentially boundary conditions) of the 2D theory. Is this category of boundary conditions identified with the category assigned by the TFT functor (as is claimed)? It seems that this identification is only possible since $\\dim\\Sigma=2=\\text{codim }\\Sigma$ in dimension four and does not reflect a general relation between extended TFT and TFT with defects.\n", "link": "https://mathoverflow.net/questions/184454/extended-tqft-versus-tqft-with-defects", "tags": ["mp.mathematical-physics", "higher-category-theory", "quantum-field-theory", "extended-tqft"], "votes": 17, "creation_date": "2014-10-14T19:07:01", "comments": ["One approach to this is discussed in detail at arxiv.org/abs/1009.5025 . See sections 2 and 6, and the remark about defects near the start of section 6.7.", "Perhaps the following question needs to be answered first: a defect theory (an object of the k-category of defects that decorates a k-fold) is itself a TFT. How is this object realized as a TFT? In particular, given a d-dim theory Bord->C, how does one characterize the (d-1)-dim TFTs on boundaries (or domain walls)? And how are these boundary theories related to the defect decoration?", "@KevinWalker Following your hint, the upside-down can be turned rightside-up, and the degrees and dimensions work out; however, I am still unclear about how to identify the data of the TFT with the defect data. For a d-dim theory Bord->C, the defects on a k-fold M form a k-category, the objects (defect theories) of which might be interpreted as k-morphisms of C. Endomorphisms in C of a defect theory form a (n-k-1)-category of boundary conditions (or an (n-k)-category with one object) of the defect theory. Somehow this might be identified with the (n-k-1)-category assigned to M by the TFT.", "Forget about TQFTs for a moment and consider an $n$-category $C$. To a $k$-morphism $x$ of $C$ we can, of course, associate the $k$-morphism $x$ itself. We can also associate to $x$ the $(n-k)$-category of endomorphisms of $x$ in $C$. The \"upside-down-ness\" you are noticing is closely related to this familiar example.", "Yes, representations of cobordisms with singularities encode TFTs with boundaries and defects. For \"pre-quantum\" field theory this is discussed in sections 3.9.14.4 to 3.9.14.6 of arxiv.org/abs/1310.7930 (improved version in preparation at ncatlab.org/schreiber/show/Local+prequantum+field+theory). For quantization of this: arxiv.org/abs/1402.7041 . This is based on discussion with Domenico Fiorenza and Alessandro Valentino that recently appeared as arxiv.org/abs/1409.5723", "I think you want the cobordism hypothesis with singularities (Theorem 4.3.11 in arxiv.org/abs/0905.0465)."], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "247", "site": "mathoverflow", "title": "The Grothendieck Ring of Higher Stacks", "body": "The Grothendieck ring of varieties is defined to be the free abelian group spanned by isomorphism classes of varieties modulo the _cut & paste_ (or scissor) relations, which say that $[X] = [U] + [Y]$, for any closed $Y \\subset X$ and where $U$ is the open complement.\n\nNow, we can equally speak of a Grothendieck ring of schemes, or algebraic spaces, and they all turn out to be isomorphic via the obvious inclusions. (for the scheme case one sees this by noticing that $[X] = [X_\\text{red}] - 0$) There is also a Grothendieck ring of Artin stacks (everything here is of finite type over a field of characteristic zero, possibly algebraically closed, and with affine diagonal). This ring turns out to be a localisation of the previous one.\n\nMy question is: what happens if we include _higher_ stacks?\n\nI think there is work by Toën on the Grothendieck ring for derived stacks, but here I'm only asking about higher (underived) stacks. Do we have a similar phenomenon to the case of varieties VS schemes? In the sense that $[X] = [\\pi_0(X)] - 0$, where $\\pi_0$ of a derived stack is its underived truncation?\n", "link": "https://mathoverflow.net/questions/121984/the-grothendieck-ring-of-higher-stacks", "tags": ["ag.algebraic-geometry", "stacks"], "votes": 17, "creation_date": "2013-02-16T05:10:36", "comments": ["are there any interesting phenomena you expect to happen? or just for fun?"], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "248", "site": "mathoverflow", "title": "Vector Bundles on the Moduli Stack of Elliptic Curves", "body": "As is well known, there is classification of line bundles on the moduli stack of elliptic curves over a nearly arbitrary base scheme in the paper _The Picard group of $M_{1,1}$_ by Fulton and Olsson: every line bundle is isomorphic to a tensor power of the line bundle of differentials $\\omega$ and $\\omega^{12}$ is trivial. \n\nI am now interested in a similar classification scheme for higher dimensional vector bundles (i.e. etale-locally free quasi-coherent sheaves of finite rank). I am especially interested in the prime 3, so you may assume, 2 is inverted, or even that we work over $\\mathbb{Z}_{(3)}$. I found really very little in the literature on these questions. I know only of two strategies to approach the topic:\n\n1) I think, I can prove that every vector bundle $E$ on the moduli stack over $\\mathbb{Z}$ localized at $p$ for $p>2$ is an extension of the form $0\\to L \\to E \\to F$ where $L$ is a line bundle and $F$ a vector bundle of one dimension smaller than $E$ (this may hold also for $p=2$, but I haven't checked). In a paper of Tilman Bauer ([Computation of the homotopy of the spectrum tmf](http://arxiv.org/abs/math/0311328)) Ext groups of the so called Weierstraß Hopf algebroid are computed, which should amount to a computation of the Ext groups of the line bundles on the moduli stack of elliptic curves if one inverts $\\Delta$. It follows than that every vector bundle on the moduli stack of elliptic curves over $\\mathbb{Z}_{(p)}$ is isomorphic to a sum of line bundles for $p>3$ if I have not made a mistake. But for $p=3$, there are many non-trivial Ext groups and I did not manage to see which of the occuring vector bundles are isomorphic. \n\n2) One can try to find explicit examples of a non-trivial higher dimensional vector bundles. A candidate was suggested to me by M. Rapoport: for every elliptic curve $E$ over a base scheme $S$ we have an universal extension of $E$ by a vector bundle. Take the Lie algebra of this extension and we get a canonical vector bundle over $S$. As explained in the book _Universal Extensions and One Dimensional Crystalline Cohomology_ by Mazur and Messing, this is isomorphic to the deRham cohomology of $E$. This vector bundle is an extension of $\\omega$ and $\\omega^{-1}$ and lies in a non-trivial Ext group. But I don't know how to show that this bundle is non-trivial.\n\nI should add that I am more a topologist than an algebraic geometer and stand not really on firm ground in this topic. I would be thankful for any comment on the two strategies or anything else concerning a possible classification scheme. \n", "link": "https://mathoverflow.net/questions/20824/vector-bundles-on-the-moduli-stack-of-elliptic-curves", "tags": ["ag.algebraic-geometry"], "votes": 17, "creation_date": "2010-04-09T03:06:40", "comments": ["The fact that vector bundles on $\\mathcal M_{1,1}$ split as sums of line bundles in characteristic larger than 3 can also be seen by the classical description of $\\mathcal M_{1,1}$ as an open substack of the weighted projective stack $\\mathbb P(4,6)$, coming from the Weierstrass form. Any locally free sheaf on $\\mathcal M_{1,1}$ extends to a reflexive sheaf on $\\mathbb P(4,6)$, which is locally free, because $\\mathbb P(4,6)$ is regular of dimension 1. It is not hard to prove that any locally free sheaf on a weighted projective stack $\\mathbb P(m,n)$ splits as a direct sum of line bundles.", "@Cardano: Sure.", "@Charles: Good point, I should have noticed the link with the non-liftability to char. 0 for the Hasse invariant when $p = 3$ (and its square when $p=2$). Thanks. ", "@Tyler Lawson: you use the covering $\\mathcal{M}(4) \\to \\mathcal{M}$ and some flatness criterion. I can send you a pdf file with some details if you like. ", "You can also deduce that $Ext^1(O, \\omega^2)$ is non-trivial from the fact that there are non-trivial \"mod p\" modular forms of weight 2 when p=2 or 3; these are computed (for instance) in Deligne's note on \"formulas, after Tate\" in Antwerp 4.", "@Brian: The computation of the Ext-groups of tensor powers of $\\omega$ on this moduli stack is written up in the Bauer paper that was linked to under part (1).", "Since the coarse space is affine and the automorphism groups of the geometric points have order a power of 2 times a power of 3, and the Ext-group (or $\\omega^{-1}$ by $\\omega$, to be precise) is identified with degree-1 coherent cohomology (of $\\omega^2$), its nontriviality is also not visible over $\\mathbf{Z}[1/6]$. So can you also briefly indicate how you know it is nontrivial?", "How did you prove that every vector bundle is an extension?"], "comment_count": 8, "category": "Science", "diamond": 0} {"question_id": "249", "site": "mathoverflow", "title": "Combinatorial identity involving the Coxeter numbers of root systems", "body": "The setup is:\n\n$R$ = irreducible (reduced) root system;\n\n$D$ = connected Dynkin diagram of $R$, with nodes numbered $1,2,...,r$;\n\n$\\hat D$ = extended Dynkin diagram, nodes numbered $0,1,2,...,r$;\n\n$\\alpha_k$ = $k^{th}$ simple root ($1\\le k\\le r$); $\\alpha_0$ = -(highest root);\n\nlabel $n_k$ ($1\\le k\\le n$) = multiplicity of root k in the highest root; $n_0=1$;\n\n$n^\\vee_k$ = dual label = multiplicity of co-root $\\alpha_k^\\vee$ in the highest _short_ co-root;\n\n$\\quad\\quad=n_k$ ($\\alpha_k$ long) or $n_k/c$ ($\\alpha_k$ short), where $c=2$ (types $B_n,C_n,F_4$) or $3$ (type $G_2$) (also $n_0^\\vee=1$);\n\n$h(D)$ = Coxeter number of D = $\\sum_{k=0}^rn_k$ (for example $h(E_8)=30$);\n\nNow, delete node $k$ from $\\hat D$ where $n_k>1$. Write $\\hat D-\\\\{k\\\\}$ as a union of connected Dynkin diagrams $D_1,\\dots, D_s$. Let $h_1,\\dots, h_s$ be the Coxeter numbers of $D_1,\\dots, D_s$.\n\n**Lemma** : If $\\alpha_k$ is long:\n\n$$ \\sum_{i=1}^{s}\\frac1{h_i}\\sum_{j\\in D_i}n_j=n_k $$ If $\\alpha_k$ is short: $$ \\sum_{i=1}^s\\frac1{h_i}\\sum_{j\\in D_i}n_j^\\vee=n_k^\\vee $$\n\nNote that the sum over $D_i$ involves the labels $n_j$ coming from $D$, which have no obvious relation to those coming from $D_i$. On the other hand $h_i$ is the Coxeter number for $D_i$, so this is a mix of data from $D$ and $\\\\{D_i\\\\}$.\n\nThe proof is case-by-case.\n\n> **Question** : is there a conceptual proof of the Lemma?\n\n**Examples:** (Bourbaki numbering)\n\n1) $D=E_8$, $k=2$, $n_2=3$, $\\hat D-\\\\{2\\\\}$ = type $A_8$, $s=1$, $h(A_8)=9$: $$ \\frac19(2+4+6+5+4+3+2+1)=3=n_2 $$ (As mentioned above, the numbers $2,4,6,5,4,3,2,1$ come from $E_8$, not $A_8$.)\n\n2) $D=E_8$, $k=4$, $n_4=6$, $\\hat D-\\\\{6\\\\}=A1+A2+A5$, $s=3$, $h(A_i)=i+1$: $$ \\frac12(3) + \\frac13(2+4) + \\frac16(5+4+3+2+1) = \\frac32 + 2 + \\frac{15}6 = 6 = n_4 $$ Note that the terms in the sum are not all integers.\n\n3) $D=E_8$, $k=8$, $n_8=2$, $\\hat D-\\\\{8\\\\}=E_7+A_1$, $s=2$, $h(E_7)=18$, $h(A_1)=2$: $$ \\frac1{18}(2+3+4+6+5+4+3)+\\frac12(1)=\\frac{27}{18}+\\frac12=2=n_8 $$\n\n4) Here is an example involving $n_k^\\vee$: $D=C_8$, $k=3$, $\\alpha_k$ is short, $n_k=2$, $n^\\vee_k=1$ (in fact all $n^\\vee_j=1$). $D-\\\\{3\\\\}=C_3+C_5$, $h(C_n)=2n$: $$ \\frac16(1+1+1) + \\frac1{10}(1+1+1+1+1)=\\frac12+\\frac12=1=n_3^\\vee $$\n\n5) By popular demand, $G_2$: \n\na) $k=2$ (long root), $n_2=2$, $n_2^\\vee=2$, $\\hat D-\\\\{2\\\\}=A_1+A_1$. Note that $n_1=3,n_1^\\vee=1$. \n\n$$ \\frac12(1)+\\frac12(1)=1=n_1^\\vee $$\n\nb) $k=1$ (short root), $n_1=3$, $n_1^\\vee=1$, $\\hat D-\\\\{3\\\\}=A_2$, $h(A_2)=3$. Note that $n_2=n_2^\\vee=2, n_0=n_0^\\vee=1$: $$ \\frac13(2+1)=1=n_1^\\vee $$\n\n**Remark:** Write $H$ for the subgroup of $G$ corresponding to $\\hat D-\\\\{k\\\\}$. This identity plays a role in some computations related to embedding elliptic elements of the Weyl group of $H$ in those of $G$ (the Coxeter element is an example).\n", "link": "https://mathoverflow.net/questions/38992/combinatorial-identity-involving-the-coxeter-numbers-of-root-systems", "tags": ["co.combinatorics", "rt.representation-theory", "lie-algebras", "root-systems", "weyl-group"], "votes": 17, "creation_date": "2010-09-16T09:47:55", "comments": ["@Jeff: One small nit-pick about your formulation is that the Coxeter number $h$ only depends on the Weyl group, not on $D$; for example, Lie types $B_r, C_r$ yield the same $h$. Coxeter originally found the concept in the framework of finite real groups generated by reflections, which is independent of Lie theory and root systems but of course has a lot of overlap. Your question, however, definitely comes from Lie theory.", "Maybe a proof of this result can be furnished via Vinberg's classification of Dynkin diagrams in terms of the existence of a subadditive function? I think the coefficients $n_k$ you are describing are just the coefficients of the unique (up to scale) subadditive function...", "This is an intriguing observation, which may not have a convincing explanation outside representation theory or the combinatorial geometry of affine Weyl groups. It might for example come from Langlands duality in some context not so visible in classical Lie theory (The tag weyl-group and/or lie-algebra might be appropriate. Also, it makes the dualization process clearer if you include the case $G_2$.) "], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "250", "site": "mathoverflow", "title": "How many Hecke operators span the Hecke algebra?", "body": "This is a generalisation of my [earlier question](https://mathoverflow.net/questions/42809/how-many-hecke-operators-span-the-level-1-hecke-algebra) about generators for the level 1 Hecke algebra.\n\nLet $\\Gamma$ be a congruence subgroup of $\\operatorname{SL}_2(\\mathbb{Z})$, and $k \\ge 1$ an integer. There's an explicit bound $S$ (the \"Sturm bound\") known such that if $f \\in M_k(\\Gamma)$ (the space of modular forms of weight $k$ and level $\\Gamma$) satisfies $a_n(f) = 0$ for $0 \\le n \\le S$, then $f = 0$, where $a_n(f)$ denote the coefficients of the $q$-expansion of $f$ (at the cusp $\\infty$).\n\nSince $M_k(\\Gamma)$ is finite-dimensional and doesn't contain any nonzero constant functions, there must actually exist some $S'$ such that if $a_n(f) = 0$ for $1 \\le n \\le S'$ (but with no assumption on $a_0(f)$), then in fact $f = 0$. Can one give an effective upper bound on $S'$ (in terms of $k$ and the index $[\\operatorname{SL}_2(\\mathbb{Z}) : \\Gamma]$)? \n\nI'd be interested to know the answer to this even for the special case $\\Gamma = \\Gamma_0(N)$ or $\\Gamma_1(N)$. For these groups it is equivalent to asking that the Hecke algebra acting on $M_k(\\Gamma)$ is spanned by the operators $T_1, \\dots, T_{S'}$ (hence the title of the question). \n", "link": "https://mathoverflow.net/questions/42815/how-many-hecke-operators-span-the-hecke-algebra", "tags": ["nt.number-theory", "modular-forms"], "votes": 17, "creation_date": "2010-10-19T11:30:31", "comments": ["Joel: see e.g. wstein.org/books/modform/modform/newforms.html#sturm-s-theor‌​em.", "David, do you know a good reference for sturm bounds?", "A data point: for $\\Gamma_0(p)$ in weight 4, $p$ prime, the Sturm bound $S$ is $(p + 1)/3$, but the Eisenstein series $E_4(pz)$ has $a_n = 0$ for $1 \\le n < p$, so $S'$ must be at least $p$.", "Julian, if you read the question you'll see that I've carefully explained why what I'm looking for is not the same as the Sturm bound.", "Hmmm... isn't it called a \"Sturm bound\"?", "I've decided to undelete this old question -- I did delete it because I wanted to set it as a student project, but the student didn't solve it.", "@David: if you don't want anyone to answer the question you should delete it."], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "251", "site": "mathoverflow", "title": "monomorphisms and epimorphisms of local rings", "body": "I want to understand the structure of monomorphisms/epimorphisms in the category of local rings (with local homomorphisms), or dually in the category of local schemes. Let $LR$ denote this category.\n\n * Every monomorphism is injective.\n\n\n\nProof: Let $R \\to S$ be a monomorphism and $a,b \\in R$ mapped to the same element. Regard $a,b$ as ring maps $R[t] \\to R$. Let $\\mathfrak{p},\\mathfrak{q}$ the preimages of the maximal ideal $\\mathfrak{m}$. Then we get two morphisms of local rings $R[t]_{\\mathfrak{p}} \\times_S R[t]_{\\mathfrak{q}} \\to R$ which agree when composed with $R \\to S$, thus they are equal. Evaluating at $(t,t)$ gives $a=b$. There is a shorter proof which uses $R[t]_{(t)+\\mathfrak{m}}$.\n\n 2. Is every monomorpism regular? (i.e. the equalizer of some morphism pair)\n\n 3. Is every monomorphism extremal? (i.e. $i=ep$ with $p$ epi implies $p$ iso)\n\n 4. Is every epimorphism, which is also injective, and isomorphism?\n\n 5. Is every epimorphism surjective?\n\n\n\n\nClearly we have 4 => 3 <=> 2 and 1 => 2. Since in $LR$ every morphism can be factorized as $ep$ where $p$ is surjective and $e$ is injective, we also have 3 => 4.\n\nIn the category of rings, 1,2,3,4 are false (see [this discussion](https://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like)). However, the counterexamples don't carry over to $LR$. In the \"category theory bible\" (joy of cats!) you can find nothing about local rings. The questions are partially motivated by Anton's problem about [coequalizers](https://mathoverflow.net/questions/63/can-a-coequalizer-of-schemes-fail-to-be-surjective) in the category of (local) schemes.\n\nIn Lazards article about flat epimorphisms ([pdf link](http://archive.numdam.org/ARCHIVE/SAC/SAC_1967-1968__2_/SAC_1)), there is an example (1.6) in which 4 fails. I have not understood it yet. What is $Z$ there? Perhaps $B$ is a localization of $k[T,Z]$ and not $k[T]$? But I don't believe that $f$ is well-defined.\n\nIf the answers turn out to be \"no\":\n\n * Is there a nice description for the regular monomorphisms? What about field extensions?\n\n\n\nIf $L/K$ is a finite galois extension, then $K \\to L$ is a regular monomorphism iff $L/K$ is cyclic (see Brians comments).\n", "link": "https://mathoverflow.net/questions/24066/monomorphisms-and-epimorphisms-of-local-rings", "tags": ["ac.commutative-algebra", "ct.category-theory"], "votes": 17, "creation_date": "2010-05-10T00:17:01", "comments": ["Oh yes, this is the example. I'll take a look at Paul Balmer's explanation.", "The link to Lazard's article seems to be broken; is this (mathoverflow.net/questions/19282/…) the counterexample you were referring to? If so, I don't see why $C$ maps to $D$ either.", "@Martin: here's proof of non-existence when Gal. group G for K/k is non-cyclic. Suppose $s,t:K \\rightrightarrows A$ is equalizer pair for $k \\rightarrow K$ in category $LR$. Then $A$ is naturally a $K \\otimes_k K$-algebra, and by locality of $A$ and structure of $K \\otimes_k K$ (using $K/k$ is Galois), the algebra structure factors through projection to a factor field. Thus, $t=s \\circ g$ for some $g \\in G$. Hence, by univ. property of $A$ and inj. of $s$, any $f:B \\rightarrow K$ s.t. $g \\circ f=f$ factors through $k$. Thus, $K^g=k$, so $G=\\langle g \\rangle$. Contradiction.", "I'm thinking about that, but it's pretty difficult to start. Even for nice small extensions ...", "@Martin: sorry, for #1 I got disoriented by Spec and was thinking of co-equalizers. But how about a Galois extension whose Galois group doesn't admit a pair of generators? By restricting to local rings can you find an equalizer pair description for that?", "If $K/k$ is a cyclic galois extension with generator $\\sigma$ of the galois group, then $k \\to K$ is regular since it is the equalizer of $id,\\sigma : K \\to K$.", "1 fails for quadratic Galois ext'n of fields $K/k$ (due to non-locality of $K \\otimes_k K$ and how its isom with $K \\times K$ is defined). In category of local noetherian rings, map from ring to completion is an epimorphism (due to Krull intersection theorem, hence the noetherian condition), and in non-complete case provides counterexamples to 2, 3, 4. Doesn't show 2,3,4 fail in $LR$, but kills a reason to care, since affirmative results then seem useless. Category theory is like point-set topology: can get wrapped up in \"weird examples\"; not necessarily a good use of time. "], "comment_count": 7, "category": "Science", "diamond": 0} {"question_id": "252", "site": "mathoverflow", "title": "Does every connected set that is not a line segment cross some dyadic square?", "body": "A _dyadic square_ is a subset of $R^2$ of the form $x + 2^{-n} [0,1]^2$ with $x \\in 2^{-m} Z^2$, for integers $m,n \\geq 0$. We say that a set $A$ _crosses_ a square $S$ if there exists a connected subset of $A \\cap S$ which intersects two opposite sides of the square $S$. Clearly, the 45 degree line $\\\\{ (\\pi + t, t) : t \\in R \\\\}$ does not cross any dyadic square. Does every non-trivial closed, connected set that is not a line segment cross some dyadic square?\n", "link": "https://mathoverflow.net/questions/120415/does-every-connected-set-that-is-not-a-line-segment-cross-some-dyadic-square", "tags": ["geometry", "discrete-geometry", "euclidean-geometry", "gt.geometric-topology", "mg.metric-geometry"], "votes": 17, "creation_date": "2013-01-31T06:07:14", "comments": ["Since you @mirko don't mind rephrasing-reposting the question I'd repost as a dimension/measure-theoretical question on curves in the plane, someone will spit some result at you really quick if you're convinced it's true. But I think it's false, you should be able to get good control of limiting behavior with a rectilinear construction and finish using basic results about direct limits of continua.", "Really, some zero-span curve can't be constructed with just a uniform direct limit by oscillating off corners in Koch curve style with rectilinear paths by like a scaled factor of $> \\frac{1}{4^n}$? In my head it works, the argument should go like 'what's the biggest dyadic square it crosses' and then induct. Has anyone tried to write it down? In my head it works. If there is no example then a dimension theorist probably can handwave why, you should tag dimension theory and GMT.", "@IanAgol No, I don't have a proof for the pseudoarc, though this is usually what I think of, as a totally path-disconnected continuum (i.e. path-components are points). If $C$ is say hereditarily indecomposable plane continuum I have considered (something elementary), take $\\frac1n$-neighborhood (union of all $\\frac1n$-balls with centers in $C$) this is path-connected, try to use it somehow (then let $n\\to\\infty$), but haven't had much success. Your comment reminded me that the pseudoarc is arc-like(chainable), so perhaps use this to modify the path-connected case, but I have to think about it", "@Mirko: do you think that you can prove it for pseudo-arcs?", "I believe I have a positive answer under the additional assumption that the set in question is path-connected. I posted a related question which eventually got a bit long, but see the definition of hsb plane continuum there, at the beginning, along with the Result after the \"Edit May 29-31, 2023\" here: mathoverflow.net/q/446092 . For completeness, here are a couple of other, related questions that I posted recently mathoverflow.net/q/446317 and math.stackexchange.com/q/4694709 . (I don't feel like posting my \"path-connected\" solution here, it is only a special case.)", "Is it pure curiosity? In other words --- do you have some motivation for this question?", "Hello Kevin, this is a nice question and seems difficult, though I came up with some ideas that I need to develop. Do you know if it is an open question, and if anyone has done work on it, any references? Did you find it somewhere in a book or a paper, or did you come up with it on your own? I am inclined to believe that there is no example, though I changed my mind about it several times yesterday and today. I will email a colleague whose area is closer to this question, to see if he knows anything or has any advice (just sending him the link to your question as posted here). Thank you,", "Hi Gunter, I pointed out that there were two mistakes in this answer. A day later it was removed by somebody.", "There used to be an interesting answer posted here. Where is it? Was it wrong?", "Gunter Rote, I do not mean the condition $m\\leq n$. E.g. the square $[1/8,5/8]\\times [1/4,3/4]$ is a dyadic square with my definition. Perhaps you would like to call such a square a translated dyadic square.", "I think that every path-connected component of such a non-crossing set should be a line segment.", "I would have expected the condition $m\\le n$. Is that what you mean? Or is this another version of the problem? Every line segment that is not $\\pm45$ degrees, and every sufficiently smooth curve that is not a $\\pm45$ degree line segment must cross a dyadic square (even with the strong definition of dyadic square). So if there are counterexamples they are pretty pathological."], "comment_count": 12, "category": "Science", "diamond": 0} {"question_id": "253", "site": "mathoverflow", "title": "Rational equivalence of smooth closed manifolds", "body": "All spaces below will be assumed simply connected. A continuous map is a _rational equivalence_ if it induces an isomorphism of the rational homology groups. Two spaces are _rationally equivalent_ if they can be connected by a sequence of rational equivalences (not necessarily all going in the same direction). Now suppose $X$ and $Y$ are rationally equivalent smooth compact manifolds without boundary. Can one choose a sequence of rational equivalences from $X$ to $Y$ that only passes through smooth compact manifolds of dimension $\\dim X=\\dim Y$?\n\nNote: if $X$ and $Y$ are CW-complexes, or more generally, compactly generated Hausdorff spaces, then $X$ is rationally equivalent to $Y$ if and only if the $\\mathbb{Q}$-localizations $X_\\mathbb{Q}$ and $Y_\\mathbb{Q}$ are homotopy equivalent, see e.g. Felix, Halperin, Thomas, Rational homotopy theory, Proposition 9.8. If this is the case, then there is a roof $X\\to X_\\mathbb{Q}\\gets Y$ of rational equivalences. The middle of the roof is of course an infinite complex (except when it is a point).\n", "link": "https://mathoverflow.net/questions/429394/rational-equivalence-of-smooth-closed-manifolds", "tags": ["at.algebraic-topology", "gt.geometric-topology", "smooth-manifolds", "rational-homotopy-theory"], "votes": 16, "creation_date": "2022-08-29T13:31:35", "comments": ["To add to Denis T's comment, if the rational homotopy type in question is formal, then $X$ admits endomorphisms of arbitrarily divisible degree (see Infinitesimal Computations in Topology Theorem 12.2, and F. Manin's \"Positive weights and self-maps\" for a detailed discussion, Corollary 1.1 arxiv.org/abs/2108.02173v3) and so, upon precomposing with an appropriate endomorphism one can lift the map $X \\to X_{Q}$ through $Y \\to X_{Q}$ to get a direct rational equivalence $X \\to Y$.", "...And SW classes are functorial along any maps which are $\\Bbb Z/2$ cohomology equivalence, so one can maybe just use them instead of more complicated things like cokernel of Hurewitz map for oriented bordism provided our cohomology rings are \"sturdy\" enough in sense I defined above.", "Well, yeah, that's why I'm writing it as a comment and not an answer, but I think my non-example can be promoted. By methods of Manuel Amann I think one can produce \"sturdy\" PD rational types, i. e. ones without self-maps of degree $\\neq 1$ and retaining that property under connected sums. Then one can use some secondary obstructions (i. e. homology classes not realizable as manifolds) to distinguish two realisations of that type. Sturdiness will prevent positive degree things from killing our obstructions, which will forbid any nonzero degree maps in either direction.", ".. also, the tangent Stiefel-Whitney classes are preserved under genuine homotopy equivalences of manifolds, but not in general under rational ones.", "Denis: this is a good example, but rational equivalences may have degree $\\neq 1$. For example, any map $S^n\\to S^n$ of degree $\\neq 0$ is a rational equivalence.", "I think there's no way to connect $S^2 \\times S^3$ with the unique other 5-mfd with $H_2 = \\Bbb Z$ (double of nontrivial $D^3$-bundle over $S^2$) through degree 1 maps because one has trivial $w_2$ and other one does not. They are rationally equivalent, since they are both formal."], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "254", "site": "mathoverflow", "title": "Transcendence of sum of reciprocals of factorials", "body": "For $A \\subseteq \\mathbb{N}$, define $\\displaystyle x_A = \\sum_{n \\in A} \\frac{1}{n!}$. It is easy to see that for every infinite $A$, $x_A$ is irrational.\n\n**Question** : Is there an infinite $A \\subseteq \\mathbb{N}$ for which $\\displaystyle x_A = \\sum_{n \\in A} \\frac{1}{n!}$ is algebraic?\n", "link": "https://mathoverflow.net/questions/419523/transcendence-of-sum-of-reciprocals-of-factorials", "tags": ["nt.number-theory", "analytic-number-theory", "transcendental-number-theory"], "votes": 16, "creation_date": "2022-04-02T22:22:00", "comments": ["At least we can prove that if $x_{A}$ is an algebraic number then $A$ is not a cofinite subset of $\\mathbb{N}$. :)", "For other results along these lines, see Kumar and Vance and the references therein. I'm not familiar with them all but at first glance they all require faster growth rates than $n!$. And for an arbitrary $A$ it seems to me that there's not much you can say about $\\sum_{n\\in A} 1/n!$ other than the general rate at which the summands shrink.", "I suspect the answer is no but that a proof is beyond current technology. The proof that $e$ is transcendental relies on special properties of the function $e^x$, not just on the rate of growth of $n!$. There are transcendence results that rely only on how fast the summands shrink but they require a faster growth rate than $n!$. For example, Nyblom proves that if for some fixed $\\lambda>2$ we have $\\liminf_{n\\to\\infty} a_{n+1}/a_n^{λ+1}>1$ then $\\sum_n 1/a_n$ is transcendental.", "Exactly this class of numbers is proved irrational (and some examples given, which I think all happen to be transcendental) in Martin Griffiths, \"Irrational sums from reciprocals of factorials\", Math. Gaz. 2015, jstor.org/stable/24496963"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "255", "site": "mathoverflow", "title": "Is "Escherian metamorphosis" always possible?", "body": "$\\DeclareMathOperator\\int{int}\\DeclareMathOperator\\diam{diam}\\DeclareMathOperator\\area{area}\\DeclareMathOperator\\cl{cl}\\DeclareMathOperator\\ran{ran}\\DeclareMathOperator\\dom{dom}$_This is a tweaked version of a question which was asked and bountied[at MSE](https://math.stackexchange.com/q/4555946/28111); that question still hasn't quite been answered, but suggestions in the comments convinced me that this was the \"right\" version to ask instead:_\n\nThis question is motivated by Escher's series of **Metamorphosis** woodcuts (see e.g. [here](https://www.escherinhetpaleis.nl/story-of-escher/metamorphosis-i-ii-iii/?lang=en)), where one tesselating tile is gradually transformed into another. Basically, this is a precise way of asking the following: can we always \"metamorphose\" between any pair of tesselating shapes?\n\nFirst, some definitions (everything is in $\\mathbb{R}^2$ with the usual metric):\n\n * A _shape_ is a compact connected set $X$ with $X=\\cl(\\int(X))$.\n\n * A _tiling_ is a pair $(\\mathscr{S},F)$ where $\\mathscr{S}$ is a finite set of shapes and $F$ is a set of functions such that\n\n * each $f\\in F$ is an isometric embedding of some $S\\in\\mathscr{S}$ into $\\mathbb{R}^2$,\n\n * $\\bigcup_{f\\in F}\\ran(f)=\\mathbb{R}^2$, and\n\n * if $f,g\\in F$ are distinct with domains $S,T$ respectively and $x\\in \\int(S)$, then $f(x)\\not\\in \\ran(g)$ (\"shapes only meet at their boundaries\").\n\n * A _tile_ is a shape $X$ such that $(\\\\{X\\\\},F)$ is a tiling for some $F$.\n\n * $d_H$ is [Hausdorff distance](https://en.wikipedia.org/wiki/Hausdorff_distance). _(Note that we could replace $d_H$ with the modified version $d_{H*}(U,V)$ = the infimum over planar isometries $p$ of $d_H(p(U), V)$ without changing the question substantively; all that would change is that the number of tiles needed would shrink, but I'm not looking at that here.)_\n\n\n\n\nNow with apologies to Escher, given tiles $A,B$ and $\\epsilon>0$ let a **strong $\\epsilon$-metamorphism** from $A$ to $B$ be a tiling $(\\mathscr{S},F)$ such that\n\n * if $f\\in F$ and $(x,y)\\in \\ran(f)$ with $x<0$ then $\\dom(f)=A$,\n\n * there is some $N$ such that if $g\\in F$ and $(x,y)\\in \\ran(g)$ with $x>N$ then $\\dom(g)=B$ (call the least such $N$ the _length_ of the $\\epsilon$-metamorphism), and\n\n * if $f,g\\in F$ and $\\ran(f)\\cap \\ran(g)\\not=0$ then $$d_H(\\dom(f), \\dom(g))<\\epsilon\\cdot {\\area_f+\\area_g\\over 2},$$ where $\\area_j$ is the area of the domain of $j$ for $j\\in F$. (This \"scaling\" of the Hausdorff distance means that we can't trivialize things by shrinking. This should have been included in the original problem, but I forgot it.)\n\n\n\n\nFinally, say that an **$\\epsilon$ -metamorphism** is as above but with the final inequality having right hand side merely $\\epsilon$.\n\n> **Question 1** : Is there always a strong $\\epsilon$-metamorphism between $A$ and $B$ for any tiles $A,B$ and any positive $\\epsilon$?\n\nI recall seeing a theorem that the answer is **yes** , and that moreover we can always find metamorphisms with length \"close to\" the naive guess $${\\max\\\\{\\diam(A),\\diam(B)\\\\}\\cdot d_H(A,B)\\over \\epsilon},$$ but I haven't been able to track it down or prove it myself. I'm also interested in whether the situation changes as we appropriately tweak things to work in $\\mathbb{R}^n$ for $n>2$.\n\n> **Question 2** : Assuming for the moment that $\\epsilon$-metamorphisms always exist (which seems plausible based on the comments at the MSE version of this question), what can we say about the length function $$\\mathscr{l}_{A,B}: \\epsilon\\mapsto\\inf\\\\{N: \\mbox{an $\\epsilon$-metamorphism from $A$ to $B$ with length $N$ exists}\\\\}$$ for a pair of tiles $A,B$? In particular, what is $\\mathscr{l}_{\\mbox{unitsquare, unithexagon}}$?\n", "link": "https://mathoverflow.net/questions/433479/is-escherian-metamorphosis-always-possible", "tags": ["mg.metric-geometry", "euclidean-geometry", "plane-geometry", "tiling"], "votes": 16, "creation_date": "2022-10-29T10:38:49", "comments": ["@MattF. Metamorphosis II has a portion that goes squares-lizards-hexagons, I believe.", "Which of the Escher images do you find most suggestive for an answer about unit squares and unit hexagons?", "Interesting question!"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "256", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/358423", "title": "A proof of $\\dim(R[T])=\\dim(R)+1$ without prime ideals?", "body": "**Background.** If $R$ is a commutative ring, it is easy to prove $\\dim(R[T]) \\geq \\dim(R)+1$, where $\\dim$ denotes the Krull dimension. If $R$ is Noetherian, we have equality. Every proof of this fact I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem. It is worth mentioning that Gelfand-Kirillov dimension satisfies $\\mathrm{GK}\\dim(R[T])=\\mathrm{GK}\\dim(R)+1$ for every $K$-algebra $R$.\nT. Coquand and H. Lombardi have found a surprisingly elementary, first-order characterization of the Krull dimension that does not use prime ideals at all.\n\nT. Coquand, H. Lombardi, A Short Proof for the Krull Dimension of a Polynomial Ring, The American Mathematical Monthly, Vol. 112, No. 9 (Nov., 2005), pp. 826-829 (4 pages)\n\nYou can read the article here. Here is a summary.\nFor $x \\in R$ let $R_{\\{x\\}}$ denote the localization of $R$ at the multiplicative subset $x^{\\mathbb{N}} (1+xR) \\subseteq R$. Then we have\n$$\\qquad \\dim(R) = \\sup_{x \\in R} \\left(\\dim(R_{\\{x\\}})+1\\right)\\!. \\label{1}\\tag{$\\ast$}$$\nIt follows that for $k \\in \\mathbb{N}$ we have $\\dim(R) \\leq k$ if and only if for all $x_0,\\dotsc,x_k \\in R$ there are $a_0,\\dotsc,a_k \\in R$ and $m_0,\\ldots,m_k \\in \\mathbb{N}$ such that\n$$x_0^{m_0} (\\cdots ( x_k^{m_k} (1+a_k x_k)+\\cdots)+a_0 x_0)=0.$$\nYou can use this to define the Krull dimension.\nThis new characterization of $\\dim(R) \\leq k$ can be seen as a statement in first-order logic (whereas the usual definition with prime ideals uses second-order logic): Use two sorts $N,R$, the usual ring operations for $R$, but also the \"mixed\" operation $N \\times R \\to R$, $(n,x) \\mapsto x^n$. Every ring becomes a model of this language.\nA consequence of this is a new short proof of $\\dim(K[x_1,\\dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $\\dim(R\\otimes_K S)=\\dim(R)+\\dim(S)$ if $R,S$ are finitely generated commutative $K$-algebras. In particular $\\dim(R[T])=\\dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.\n\nQuestion. Can we use the characterization \\eqref{1} of the Krull dimension by Coquand-Lombardi to prove the formula $$\\dim(R[T])=\\dim(R)+1$$ for Noetherian commutative rings $R$?\n\nSuch a proof should not use the prime ideal characterization/definition of the Krull dimension. Notice that the claim is equivalent to $\\dim(R[T]_{\\{f\\}}) \\leq \\dim(R)$ for all $f \\in R[T]$. I suspect that this can only work if we find a first-order property of commutative rings (with powers) which is satisfied in particular by Noetherian rings and prove the formula for these rings.\nPlease read this first before answering. This question is only concerned with a proof of the dimension formula using the Coquand-Lombardi characterization. If you post an answer that doesn't mention the characterization, then it's not an answer to my question and therefore offtopic. As of writing this, 20 answers have been posted, all of which have been deleted.\n", "comments": ["@Alexey Back then I removed the remark about first-order because of your comment. But I have just added this back, because we can just change the language to make it first-order. I have added an explanation for this.", "Why this question has no answer or answers were deleted?", "Currently 18 answers, all deleted.", "This may not be a direct answer to your question, but it is relevant to the problem and too long to share in the comments. You can use valuation overrings and valuative dimension instead of prime ideals and Krull dimension, respectively. Polynomial rings are well-behaved with respect to valuative dimension, and Jaffard rings are the natural context in which to study the condition you seek. All Noetherian rings and all Prufer rings are Jaffard, for example, and the class of Jaffard rings is much larger than that of the Noetherian rings. See, for example, reader.elsevier.com/reader/sd/pi", "@WernerGermánBusch Of course $T$ is just one variable.", "@moderators: Please make this question CW.", "This question has been referred to at math.meta.stackexchange.com/questions/33453/…", "@Malkoun en.wikipedia.org/wiki/First-order_logic", "@Foobanana I know them. There is no such proof.", "This might not be too helpful, but have you tried searching through Gathmann's old notes on algebraic geometry? We're going through his notes in my algebraic geometry this year and he gives a lot of alternative proofs to commutative algebra facts :D", "Might I ask why the dimension definition using prime ideals isn't elementary enough? The proof isn't short, but it isn't terrible to see why the dimension increases by 1. Is there a reason this other construction is the one you'd prefer to use?", "For my own education, what do you mean when you write \"first order property\"? What is the definition of a first order property?", "I guess T is a set of polynomial variables, like $T= \\{x\\}$ gets you $R[x]$ polynomials in $x$ with coefficients in the ring $R$, $T= \\{x_1, x_2, ... , x_n\\}$ gets you the polynomials in $n$ variables, and so on. I'm not sure if OP specificies T to be numerable or even finite. +1 means just that, +1, the dimension is an integer number, so it makes sense to add 1 to it.", "If you can explain in a few words, what is T and what does it mean +1 in your equation?", "bourbaki's proof of this theorem uses only general results, not more partiuclar results such as krull IT", "MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-di‌​mr1", "Shall I ask this on mathoverflow?", "Indeed. By the compactness theorem, it can be seen that ${\\rm dim}R\\le l$ is not a first order statement for any finite $l$.", "This characterization of $\\dim R\\le l$ does not look like first-order to me, at least in the language of rings: it uses quantifiers over $\\mathbb N$ and exponentiation with the exponent as an argument, which is not a ring operation.", "Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know.", "At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows?"], "comment_count": 0, "tags": ["ring-theory", "commutative-algebra", "noetherian", "krull-dimension", "dimension-theory-algebra"], "creation_date": "2013-04-11T08:19:32", "diamond": 1, "votes": 873} {"question_id": "257", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/952466", "title": "Is there a bijection of $\\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?", "body": "Let $(X,\\tau), (Y,\\sigma)$ be two topological spaces. We say that a map $f: \\mathcal{P}(X)\\to \\mathcal{P}(Y)$ between their power sets is connected if for every $S\\subset X$ connected, $f(S)\\subset Y$ is connected.\nQuestion: Assume $f:\\mathbb{R}^n\\to\\mathbb{R}^n$ is a bijection, where $\\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?\n\nFull disclosure: I've now crossposted the question on MO.\n\nVarious remarks\n\nIf we remove the bijection requirement then the answer is clearly \"no\". For example, $f(x) = \\sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not.\n\nWith the bijection, it holds true in $n = 1$. But this is using the order structure of $\\mathbb{R}$: a bijection that preserves connectedness on $\\mathbb{R}$ must be monotone.\n\nAs a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.)\n\n\nInvariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above:\n\nAssume $f:\\mathbb{R}^n\\to\\mathbb{R}^n$ is a bijection, where $\\mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open?\n\n\nSome properties of $\\mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $\\mathbb{R}^n$ with the standard topology to itself, self-maps of some arbitrary topological space, it is easy to make the answer go either way.\n\n\nFor example, if the topological space $(X,\\tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by the cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected.\n\nOn the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = \\mathbb{Z}$ be equipped with the topology generated by:\n$$ \\{\\mathbb{N}\\} \\cup \\{ \\{k\\} \\mid k \\in \\mathbb{Z} \\setminus \\mathbb{N} \\} $$\nThen the map $k \\mapsto k+\\ell$ for any $\\ell > 0$ maps connected sets to connected sets, but its inverse $k\\mapsto k-\\ell$ can map connected sets to disconnected ones.\nWorking a bit harder one can construct in similar vein examples which are Hausdorff.\n\n\n", "comments": ["@HelloWorld: thank you for the reminder. The cash prize is now posted on X.", "Will there be a cash price in twelve days? ;)", "Two dozen answers, all deleted.", "@Carlyle: nothing immediate comes to mind. You can have highly discontinuous functions whose graph still remain connected.", "How does this definition of a connected function relate, if at all, to the property of the graph of a function (in the product Topology) being connected?", "@FurdzikZbignew: the question is asking about mappings from $\\mathbb{R}^n$ to itself, not from $\\mathbb{R}^n\\to\\mathbb{R}^m$; I am not sure how your observation is relevant?", "A counterexample could be a space-filling Peano curve, which is a bijection from the unit interval to the unit square but does not preserve dimensions/connectedness. Therefore, the answer to the question is \"no\" for dimensions $n>1$. The total order on $\\mathbb{R}$ was crucial in the proof for $n=1$, but does not generalize to higher dimensions. Additional regularity like continuity would be needed.", "@user65023: the function $f$ maps $(s,t) \\mapsto (s, t + \\sin(\\pi/s))$. And $J$ maps $(s,t) \\mapsto (-s,t)$. You want to show $J\\circ f \\circ J \\circ f$ is the identity, so just write it out $$ (s,t) \\mapsto (s,t + \\sin(\\pi/s)) \\mapsto (-s, t + \\sin(\\pi/s)) \\mapsto (-s, t + \\sin(\\pi/s) + \\sin(\\pi / (-s))) = (-s,t) \\mapsto (s,t) $$", "@AlanU.Kennington: It seems like the example you gave above has the following properties: (1) $f$ is connected, (2) $f$ is discontinuous, (3) $f^{-1}$ is not connected, (4) $f$ is a bijection. $f(s,t) = (s,t+\\sin{({\\pi}/{s})})$ for $s\\neq 0$ and $f(0,t)=(0,t)$.", "@PietroMajer: Do you have a proof of why AlanU.Kennington's example satisfies $f^{-1}=J\\circ f \\circ J$? I'm not seeing it. Also, it looks like for AlanU.Kennington's example $f^{-1}$ is not connected. I think $f^{-1}$ maps the connected set: $\\{(s,sin(\\frac{\\pi}{s})): s>0\\} \\cup \\{(0,t): 0 < t < 1\\}$ to the unconnected set: $\\{ (s,0): s>0\\} \\cup \\{ (0,t): 0 < t < 1\\}$. To see this set is unconnected, it is covered respectively by two open half planes: $\\{ (s,t): t < s\\}$ and $\\{ (s,t): t > s\\}$.", "@WillieWong: Why do you think this would be embarrassing? Almost all of important branches of mathematics have been born just by such questions. For example think about Graph theory. Or think about one of my favorites unsolved problem in math i.e. square peg problem! a simple conjecture but so beautiful one. :). My suggestion is to offer this challenge to undergraduate and graduate students first. then history will find its way.", "@C.F.G It would be embarrassing if, 30 years later, this question becomes my one legacy in mathematics. :-) That said, I'll perhaps save that (announcement of a cash prize) for the tenth anniversary of this post (which is only 1.5 years away).", "@WillieWong: You Should (as the author of this problem :) ) offer a prize for any progress on this problem.", "To be thorough: the statement is also true for $n = 0$. :-)", "@AlexHeuman your $f$ is not a bijection. $f(x_1, x_2, \\ldots) = f(x_1, x_2 + 2\\pi, x_3,x_4 ,\\ldots)$.", "@GerryMyerson: Now 17 answers, all deleted.", "@WillieWong might it be possible to send borromian rings in $\\Bbb R^3$ to trefoil knots? I can see the \"rope\" would need to be spliced into possibly three strands, with twists in such a way that it continuously morphs between knots and rings. I've no idea if this can hang together rigorously.", "Ok I think I see the issue. The piecewise assignment is a priori and causes us to voluntarily disconnect the domain. So the pre-image and the image are both disconnected. But the problem asks for a bijection with a connected pre-image but disconnected image, for example.", "For n=1 with f(x) = x, if x is rational; and x+1, if x is irrational, f is a bijection that is nowhere continuous. The usual topology on R begins its life connected. What is the definition of connected that lets us discard this f and conclude n > 1 for R^n?", "@GerryMyerson: though 6 of them are from the same user....", "Currently, 12 answers have been posted – all deleted!", "@MiloBrandt: for $S^2$ your suspicion is sound. See this answer on the MathOverflow version of this question. Thanks!", "I suspect that there is no such map in the analogous problem for $S^2$ instead of $\\mathbb R^2$; the reason being that if we have a separating set $X$ (i.e. one such that $S^2\\setminus X$ is disconnected), then its preimage has to be a separating set as well. Separating sets are kinda circle-like and I suspect have properties like \"there are connected subsets $A,B,C\\subseteq X$, no pair disjoint, such that $A\\cap B\\cap C=\\emptyset$\". Given this lemma or a similar ones, one can show the non-existence of such a forward-but-not-backwards connected map $S^2\\rightarrow S^2$.", "@WillieWong I think I can prove this if I succeed to prove that f maps closed connected sets onto closed connected sets.", "@RobertFrost: this particular question I am asking about is turning out to be very subtle. This particular answer of the cross post to MathOverflow shows how to answer the question if we change some small number of conditions. But the original question is still, as far as I know, open.", "@WillieWong I'm not competent with the subtleties of the concept of continuity but this reminds me of a function I'm doing some work with. The branch cuts of its inverse are all superimposed on top of each other so for any $x+\\epsilon$ there are $f^{-1}(x+\\epsilon)$ not necessarily nearby to $f^{-1}(x)$ but I'm not totally clear how the concept of continuity treats this and whether it would fit your conditions.", "@RobertFrost: mostly for my curiosity. But I note that the answer to a related question is known, and that question is what inspired me to ask this one.", "@RobertFrost: you are half right. See this previous comment", "@AndréLevy: Can you justify that the bijection is connected from the 2 spheres to the 1 sphere? Each of the paradoxical pieces is dense in the sphere, this suggests that the corresponding bijection cannot be connected. (Let $P,Q\\subset \\mathbb{S}^2$ be two disjoint dense sets. And let $\\rho$ be a non-trivial rotation. Then for any $\\epsilon > 0$ there exists $p\\in P$ and $q\\in Q$ such that $d(p,q) < \\epsilon d(p,\\rho(q))$.)", "Well, @WillieWong, isn't the disassembling and reassembling of the pieces in B-T a bijection, isn't this bijection connected from the 2 spheres to the 1 sphere, but not the reverse?"], "comment_count": 0, "tags": ["general-topology", "metric-spaces", "examples-counterexamples", "connectedness"], "creation_date": "2014-09-30T05:18:04", "diamond": 1, "votes": 632} {"question_id": "258", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1754860", "title": "Does every ring of integers sit inside a ring of integers that has a power basis?", "body": "Given a finite extension of the rationals, $K$, we know that $K=\\mathbb{Q}[\\alpha]$ by the primitive element theorem, so every $x \\in K$ has the form\n$$x = a_0 + a_1 \\alpha + \\cdots + a_n \\alpha^n,$$\nwith $a_i \\in \\mathbb{Q}$.\nHowever, the ring of integers, $\\mathcal{O}_K$, of $K$ need not have a basis over $\\mathbb{Z}$ which consists of $1$ and powers of a single element (a power basis). In fact, there exist number fields which require an arbitrarily large number of elements to form such a basis.\n\nQuestion: Can every ring of integers $\\mathcal{O}_K$ that does not have a power basis be extended to a ring of integers $\\mathcal{O}_L$ which does have a power basis, for some finite $L/K$?\n\n", "comments": ["The difficulty in finding a general approach to construct such an extension arises from the fact that the structure of the ring of integers can be quite complicated, and power bases may not always exist for every number field.", "@Jacob Wakem, if you take a field extension, let's say $\\mathbb{Q}(\\sqrt{2})/\\mathbb{Q}$, then the ring of integers (of $\\mathbb{Q}(\\sqrt{2})$) refers in some sense to the 'analogon' of what $\\mathbb{Z}$ is to $\\mathbb{Q}$. In this case the ring of integers would be $\\mathbb{Z}[\\sqrt{2}]$. But I hope my comment is not misleading, because if $\\mathbb{Q}(\\alpha)/\\mathbb{Q}$ is a field extension, then the ring of integers of $\\mathbb{Q}(\\alpha)$ is not necessarily(!) $\\mathbb{Z}[\\alpha]$.", "What I should say is this: The number of generators of a ring of integers, as an algebra, can be 1 (monogenic) or arbitrarily large.", "I honestly don't understand your remark \"In fact, there exist number fields which require an arbitrarily large number of elements to form such a basis.\": rings of integers are $\\mathbb Z$-modules, so all their $\\mathbb Z$-bases have the same number of elements (namely the degree of the corresponding field extension).", "What is a ring of integers? Merely a ring containing only integers?", "Note a similar result holds for the ordinals. See: en.wikipedia.org/wiki/Ordinal_arithmetic#Cantor_normal_form", "@Gro-Tsen and sadly it's not had much luck there either!", "This has now been asked on MathOverflow", "I think this question should really be reposted on MathOverflow. Also, before tackling the number field case, it's probably worth thinking about curves over a field: given a base curve $C$ and an affine open set $U$, say that $C_1\\to C$ (ramified covering) is \"monogenic\" for these data when there is $f$ regular on $U_1:=U\\times_C C_1$ such that $\\mathcal{O}(U_1)=\\mathcal{O}(U)[f]$. Is it true that there is always $C_2\\to C_1$ that is monogenic? (Maybe this is stupidly true or false, I'm not sure. But it could be easier to think about.)", "Just a remark on terminology: If $\\mathcal O_K=\\mathbb Z[\\alpha]$, then $\\mathcal O_K$ is sometimes called monogenic.", "Every $abelian$ number field $K$ sits inside $\\mathbb Q(\\zeta)$ for some root of unity $\\zeta$ and has a ring of integers $\\mathbb Z[\\zeta]$. So your question is answered in the positive for abelian $K/\\mathbb Q$."], "comment_count": 0, "tags": ["abstract-algebra", "number-theory", "ring-theory", "algebraic-number-theory"], "creation_date": "2016-04-22T16:54:33", "diamond": 1, "votes": 185} {"question_id": "259", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1792464", "title": "If polynomials are almost surjective over a field, is the field algebraically closed?", "body": "Let $K$ be a field. Say that polynomials are almost surjective over $K$ if for any nonconstant polynomial $f(x)\\in K[x]$, the image of the map $f:K\\to K$ contains all but finitely many points of $K$. That is, for all but finitely many $a\\in K$, $f(x)-a$ has a root.\nClearly polynomials are almost surjective over any finite field, or over any algebraically closed field. My question is whether the converse holds. That is:\n\nIf $K$ is an infinite field and polynomials are almost surjective over $K$, must $K$ be algebraically closed?\n\n(This answer to a similar question gave a simple proof that $\\mathbb{C}$ is algebraically closed from the fact that polynomials are almost surjective over $\\mathbb{C}$. However, this proof made heavy use of special properties of $\\mathbb{C}$ such as its topology, so it does not generalize to arbitrary fields.)\n", "comments": ["I guess the problem with the original reasoning is that it relies on the fact that the image of $\\mathbb{C}$ has to be simply connected, but this is hard to use for abitrary fields, the only topology you could use would be zariski but knowing zariski would give you info about the algebraic closeness. Maybe studying homology would give some results? The only nice thing is that being infinite gives you irreducibility", "I do not know whether OP knew this but this was conjectured by Reinke in their 1975 paper \"Minimale Gruppen\", and as far as I know is still open. A similar conjecture in model theory is the so-called \"minimal fields conjecture\" of Podewski, asserting that infinite \"minimal\" fields are algebraically closed (where \"minimal\" means that every field formula with one free variable either defines a finite subset of the field or a cofinite one). Wagner solves the latter in his paper \"Minimal fields\" in positive characteristic, where he also solves a weaker variant of OP's question.", "@student91: That's why I restrict $K$ to be infinite in the precise statement of the question.", "I am slightly confused. If $K$ is finite, then any polynomial is \"almost surjective\" in this sense, because there are only finitely many points that are not in the image of $f$ (even if the image of $f$ consists of only one point). Furthermore, finite fields are not algebraically closed. Do you assume your characteristic to be 0?", "Ooh you are right. I thought it was strange to have such a regular topology for arbitrary fields without being known.", "@AndreaMarino: I think it is extremely unlikely that you can get such topology at the end of an infinite chain. Local compactness is also not enough (then you could have the discrete topology!), and you need $T_2$, not just $T_0$, since you need to know the limit of $y_n$ is unique to conclude $p(x)=s$. Note then that no countably infinite compact $T_2$ space is homogeneous, so you can never have such a topology on a countably infinite field.", "Sorry for double comments but I can't edit. It is enough for the topology to be locally compact, so that we recover the cases of $\\mathbb{Q}, \\mathbb{C}$.", "How I would build such a topology: start from $\\mathbb{Q}$ and $\\mathbb{F}_p$, then show that the topology extends by transcendent, algebraic extensions and by infinite chains. By Zorn, this gives the topology on whatever field. Don't know if this route map can work!!", "Here it is my 1 cent also. Suppose there exist a $T_0$, compact and first countable topology on K such that polynomials are continuous. There are no isolated points x, otherwise $x+k$ would be also isolated because polynomials are continuous, thus the topology is discrete. But being compact it should be finite. Now take a polynomial p with non taken values $s_1,\\ldots, s_k$. Take $y_n$ in the $U_n $ (given by first countability) of $s_1$ , WLOG $x_n$ different by $s_i$. Take $x_n$ st $p(x_n)=y_n$. The space is sequentially compact, so WLOG $x_n\\to x$. Then $p(x)=s$ and p is surjective.", "Or wait, can your question be rephrased as 'If $K$ is an infinite field and polynomials are almost surjective over $K$, must polynomials be fully surjective over $K$?' ?", "I find the concept of almost surjective polynomials a bit confusing. Can you give an example of an almost surjective polynomial over $\\mathbb{C}$ that is not surjective?", "Here's my 5 cents. $K$ must contain an $n^{th}$ root for each of its elements (take $f:=X^{n}$; for $\\alpha\\in K^{*}$, $f$ cannot miss the infinitely many elements $\\alpha\\cdot\\beta^{n}$ with $\\beta\\in K^{*}$). At least when $char(K)=0$, the roots of unity are expressible as radicals. This implies that $K$ does not admit any finite Galois extensions with solvable Galois group. I.e., any minimal non-trivial Galois extension of $K$ must have a simple non-abelian Galois group.", "I believe this is open. At least, it was posted on MO with no definitive answer: mathoverflow.net/questions/6820/…"], "comment_count": 0, "tags": ["abstract-algebra", "polynomials", "field-theory"], "creation_date": "2016-05-19T19:06:55", "diamond": 1, "votes": 142} {"question_id": "260", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3770846", "title": "Probability for an $n\\times n$ matrix to have only real eigenvalues", "body": "Let $A$ be an $n\\times n$ random matrix where every entry is i.i.d. and uniformly distributed on $[0,1]$. What is the probability that $A$ has only real eigenvalues?\nThe answer cannot be $0$ or $1$, since the set of matrices with distinct real eigenvalues is open, and also the set with distinct, but not all real, eigenvalues is open (the matrices with repeated eigenvalues have measure zero).\nI don't see any easy transformation that links the two sets, and working on the characteristic polynomial seems quite impractical. Also, I have the feeling that $[0,1]^{n^2}$ is not a good space to work in, due to its lack of rotational invariance.\n", "comments": ["@EXodd Ok. Then we can try this way for $n\\geq 4$ case. Let $f(x)=det(xI−A)$ be the characteristic polynomial. Let $a_1 >⋯>a_{n−1}$ be roots of $f′(x)$. Then the roots of f are all real if $f(a_1)<0,f(a_3)<0,…$ and $f(a2)>0,f(a4)>0,…$. For $n=4$, this seem to give the explicit algebraic relation. For $n>4$, I am not sure.", "@J1U already for degree 4, \"If the coefficients are real numbers and the discriminant is positive, then the roots are either all real or all non-real\" so it cannot predict if the roots are all real or not", "@Exodd See this: en.m.wikipedia.org/wiki/Discriminant For degree 3 and 4, the discriminant is explicitly computed. For even higher degrees, the discriminant is still expressed as the algebraic relation of the coefficients of the polynomial, although it would be messier.", "@J1U for the 2x2 it is easy indeed (actually the proobability is 1) but for larger degrees how do you ensure the polynomial has only real solutions?", "Why don’t you just use the characteristic polynomial? All the eigenvalues are real if $det(xI-A)$ only has real solution. For example, for 2-by-2, $(a_{11}-a_{22})^2+4a_{12}a_{21}>0$ if and only if all the eigenvalues are real. Thus the probability is the volume of the set of inequality intersection $[0,1]^4$.", "The probability the matrix is symmetric is zero, but maybe one can argue that a certain class of “almost symmetric” matrices (via a perturbation) have real eigenvalues and with positive probability, to get a lower bound on the desired probability in question.", "@AdamCataldo yes, if you only consider real matrices then the eigenvalues are very biased towards the real line. Maybe finding a bigger space where to work may not be a bad idea, but the choice of the biger space has to reflect/expand somehow consistently the properties of the spectra of real matrices", "I think what I'm stuck on is measure space you're trying to work in, since you're suggesting in the problem it shouldn't be $[0, 1]^{n^2}$ (presumably with Lebesgue measure). It sounds like it's also not the eigenspace (the subset of $\\mathbb{C}^n$ that is projected on by the set of matrices in $[0, 1]^{n \\times n}$). Or do you think the projection into the eigenspace is such that the eigenvalues on the real line have non-zero measure?", "@AdamCataldo if the eigenvalues of the matrices were \"continuously\" distributed in the unit circle, then you would be correct. But if you take real matrices, the distribution of their eigenvalues is biased on certain areas of the unit circle, and in particular they greatly concentrate on the real line, so much that even if $[-1,1]$ is negligible in the unit circle, the probability of having real eigenvalues is not zero", "Not sure I follow this, because what I'm saying is that in the unit circle in the complex plane, the real line, that is the real values between -1 and 1, has measure zero. It does in fact have an empty interior, because if you draw a ball of any radius around any of the points on the real line, the ball will include complex numbers, not just real values.", "@AdamCataldo the probability Cannot be zero, since the set of all-real eigenvalues matrices has non-empty interior part. Your reasoning holds if you consider complex-entries matrices, while here we have only real entries", "My intuition is that the probability is zero, and that intuition is based on a slightly simpler problem: Consider the set of n x n matrices where all eigenvalues satisfy $|\\lambda| < 1$. The subset of these matrices where all eigenvalues is real is a set of subset of measure zero relative to the entire set in eigenspace. I would expect something similar if you change the constraint from the one I suggested to the original problem, but I'm not sure what the mapping from n x n matrices with entries all in [0, 1] \"looks like\" in eigenspace, so that's where my intuition might fall apart.", "Since this question is getting a lot of attention, I ask you to reformulate it. Every square matrix has at least one complex eigenvalue, since every polynomial of degree $\\geq 1$ has at least one complex root (note that real numbers are also complex numbers, by definition). What you ask is the probability for all eigenvalues to be real. Right?", "just a quick thought. Assume $n=2$. If $A$ eigenvalue $\\lambda_2$ is complex, then $A-\\lambda_2 I $ cannot be drawn as a parallelogram in, say $XY$ plane i.e. its area (in $XY$ plane) does not exist. Since the number of eigenvalues increases as $n\\rightarrow \\infty$, it feels for sure that at least for one of them, say $\\lambda_n$, the $n$-dimenional area $det(A-\\lambda_n I)$ does not exist. Thus the sought probability is $0$.", "@leonbloy (cont.) Then again, the \"large amount of independence\" from originally $n^2$ random values has to be \"squeezed\" into $(n-1)^2$ random values. So they are certainly independent indeed. Admittedly, this is still in a typical what could possible go wrong? scenario.", "@leonbloy That is quite feasible: If we pick one (probably existing) real eigenvector and take it to $e_n$ via an orthonormal base change, the other entries in the resulting matrix are essentially linear combinations of a large number of iid uniform random variables, so by the law of big numbers are gaussian. Well, the independence is a bit hand-wavy as the original eigenvector depends on all entries and vice versa all entries in the new matrix are influenced by the eigenvector. (cont.)", "This paper: shub.ccny.cuny.edu/articles/… says:Mehta [24, Conjectures 1.2.1 and 1.2.21 conjectures from extensive numerical experience that the statistical properties of matrices with independent identically distributed entries behave as if they were normally distributed as n -+ rn . Mehta focuses on the symmetric or Hermitian cases, but surely the idea is quite general.", "with 10kk the results do not change much...", "about one million (trusting Octave's random generator)", "@leonbloy how many examples have you generated for the empirical evaluation?", "It should be a coincidence, but the desired probability seems to be quite near the probability of an $n-1 \\times n-1$ random gaussian matrix having all real eigenvalues $$\\begin{array} n & & \\\\ 2 &1 & 1 \\\\ 3 &0.708 & 0.70711\\\\ 4 &0.346 & 0.35355\\\\ 5 &0.117 & 0.125\\\\ 6 & 0.028 & 0.03132\\\\ \\end{array} $$ Left column: empirical. Right column: from here: core.ac.uk/download/pdf/82140233.pdf", "Empirically : $3$: $0.708$, $4$: $0.346$, $5$: $0.117$", "@AdinaGoldberg We are on the set of real matrices, so the eigenvalues are either real or come in couples of conjugated complex numbers. If you have distinct real eigenvalues, then you can always find a small enough neighbourhood where the eigenvalues are not complex due to the continuity of the eigenvalues", "I guess to clarify, I certainly agree that within the set of matrices with real eigenvalues, those with distinct eigenvalues is an open set. But that is relatively open, in the sense that $(0,1)$ is open in $\\mathbb{R}$ but not in $\\mathbb{R}^2$.", "@Exodd Yes, indeed! But $\\mathbb{R}$ is not open as a subset of $\\mathbb{C}$. How can you complete your argument?", "@AdinaGoldberg the eigenvalues of a matrix are the roots of the characteristic polynomial, and the roots are continuous functions of the coefficients", "I'm curious why the set of matrices with distinct real eigenvalues is open (I'm assuming as a subset of the $n \\times n$ real matrices.)", "Perhaps you could try to say something about the distribution of the discriminant of the characteristic polynomial", "I'm not sure it will work, no. On a second look it seems like a key component of the argument for higher moments is that $\\sum_{j=1}^N(A^2)_{jj}=\\sum_{j,k=1}^N(A_{jk})^2$ which requires symmetric matrices.", "@SamJaques they are dealing with symmetric matrices, and they work for big $n$.. Are you sure?"], "comment_count": 0, "tags": ["linear-algebra", "probability", "matrices", "eigenvalues-eigenvectors", "random-matrices"], "creation_date": "2020-07-27T03:59:33", "diamond": 1, "votes": 135} {"question_id": "261", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/527495", "title": "A question about divisibility of sum of two consecutive primes", "body": "I was curious about the sum of two consecutive primes and after proving that the sum for the odd primes always has at least 3 prime divisors, I came up with this question:\n\nFind the least natural number $k$ so that there will be only a finite\nnumber of $2$ consecutive primes whose sum is divisible by $k$.\n\nAlthough I couldn't go anywhere on finding $k$, I could prove the number isn't $1, 2, 3, 4$ or $6$, just with proving there are infinitely many primes $P_n$ so that $k\\mid(P_n+P_{n+1})$ and $k$ is one of $1, 2, 3, 4, 6$:\nThe cases of $k=1$ and $k=2$ are trivial. The case of $k=3$, I prove as follows:\nSuppose there are only a finite number of primes $P_k$ so that $3\\mid(P_k+P_{k+1})$. We can conclude there exists the largest prime number $P_m$ so that $3\\mid(P_m+P_{m-1})$ and thus, for every prime number $P_n$ where $n>m$, we know that $P_n+P_{n+1}$ is not divisible by $3$. We also know that for every prime number $p$ larger than $3$ we have: $p \\equiv 1 \\pmod 3$ or $p \\equiv 2 \\pmod 3$. According to this we can say for every natural number $n>m$, we have either $P_n \\equiv P_{n+1} \\equiv 1 \\pmod 3$ or $P_n \\equiv P_{n+1} \\equiv 2 \\pmod 3$, because otherwise, $3\\mid(P_n+P_{n+1})$ which is not true. Now according to Dirichlet's Theorem, we do have infinitely many primes congruent to $2$ or $1$, mod $3$. So our case can't be true because of it.\nWe can prove the case of $k=4$ and $k=6$ with the exact same method, but I couldn't find any other method for proving the result for other $k$.\n", "comments": ["This is equivalent to: Find the least natural number $k$ so that there is no pair of $2$ consecutive primes whose sum is divisible by $k$. Let say $k$ is the least natural number such there is only a finite number of pair of $2$ consecutive primes whose sum is divisible by $k$, and that $p+q=kn$ is the largest of those sums. Then no sum of $2$ consecutive primes is divisible by $k\\left(n+1\\right)$ ...", "Easily to check that each of the described sums for $n\\in[2,5000]\\;$ has one or more dividers among $\\pi(k),\\; k\\in[2, 999].$ That's why the required k should be too big.", "OP, your first link goes to some weird Yahoo page. Please fix. Ty", "@KeithBackman late response, but surely that can't be true: Schinzel's hypothesis H implies that for all $n$, there are infinitely many pairs of twin primes of the form $kn \\pm 1$. For example, $11$ divides $197+199$, $13$ divides $311+313$, $19$ divides $227+229$, etc.", "@CODE: I'm just curious, isn't $k=2$ the answer?, of course this is except for $2 + 3$.", "Anecdotally, there are a plethora of numbers that never divide the sum of twin primes, which are consecutive primes: $11,13,19,31,43, \\dots$. However, these numbers may well divide the sums of other consecutive primes.", "The link to Yahoo Answers shut down today. Can you link elsewhere?", "@RomainS It has, but no answer.", "@CODE has this been crossposted to Math Overflow?", "The question amounts to : does there exist a k, such that, only finitely many pairs of consecutive primes are additive inverses mod k.", "My apologies. You are right.", "@user254665 my reasoning for k=3 says that if the hypothesis is false, from some point on we would have all prime numbers are either congruent to 2 mod 3 or 1 mod 3, THAT can't be true according to Dirichlet's Theorem.", "3+ years and no A. Maybe put it on MathOverflow and see what the pros say.", "A particular case of Dirichlet's Theorem is that there are infinitely many primes congruent to 1 mod 3 and infintely many congruent to 2 mod 3. But Dirichlet's Theoerem does NOT say anything about CONSECUTIVE primes. Your reasoning for k=3 is invalid. ... However it is a good Q, and, I suspect, with no known A.", "@ghosts not true, $12$ does not divide $103+107$, for instance.", "You can make a kind of a prime counting function that counts the pairs of consecutive primes divisible by $k$ below $n$. Maybe call it $\\pi_k(n)$. After doing some computations it seems that $\\pi_k(n) \\sim c Li(n)$ for some constant $c$ which for a lot of $k$'s looks to be close to $1/\\phi(k)$", "@CODE: Indeed, Schinzel's Hypothesis H implies it. In fact you don't even need that much: Dickson's conjecture suffices, with some cleverness.", "I also doubt that this is going to be answered.why don't you try posting this to mathoverflow.net?", "I think Schinzel's Hypothesis says it's true though.", "Thanks, I myself think such a $k$ doesn't exist too, but proving it is another matter :P", "this is a very nice question and it will seem weird if there exists such a $k$"], "comment_count": 0, "tags": ["number-theory", "prime-numbers", "prime-gaps"], "creation_date": "2013-10-15T11:52:59", "diamond": 1, "votes": 122} {"question_id": "262", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4044670", "title": "What is the largest volume of a polyhedron whose skeleton has total length 1? Is it the regular triangular prism?", "body": "Say that the perimeter of a polyhedron is the sum of its edge lengths. What is the maximum volume of a polyhedron with a unit perimeter?\nA reasonable first guess would be the regular tetrahedron of side length $1/6$, with volume $\\left(\\frac16\\right)^3\\cdot\\frac1{6\\sqrt{2}}=\\frac{\\sqrt{2}}{2592}\\approx 0.0005456$. However, the cube fares slightly better, at $\\frac{1}{1728}\\approx 0.0005787$.\nAfter some experimentation, it seems that the triangular prism with all edges of length $1/9$ is optimal, at a volume of $\\frac{\\sqrt{3}}{2916}\\approx0.00059398$. I can prove that this is optimal among all prisms (the Cartesian product of any polygon with an interval) and that there is no way to cut off a small corner from the shape and improve it.\nIs the triangular prism the largest polyhedron with a fixed perimeter?\nI can prove a weak upper bound of $\\frac1{12\\pi^2\\sqrt{2}}\\approx 0.00597$ on the volume of such a polyhedron, by combining the isoperimetric inequalities in both $2$ and $3$ dimensions (i.e., the fact that polygons cannot enclose more surface area than a circle and that a polyhedron cannot enclose more volume than a sphere of the same surface area) along with the observation that a single face of a polyhedron cannot take up the majority of its surface area. Note the number of leading zeros - this upper bound is a bit over $10$ times my lower bound!\nEdit: A friend of mine has confirmed with Mathematica that no polyhedron with $5$ or fewer vertices, or anything combinatorially equivalent to the triangular prism, improves on this bound. (With some work, this approach might be extended to all polyhedra on at most $k$ vertices, for $k$ on the order of $7$ to $10$.)\n", "comments": ["The term \"skeleton\" in the title should be avoided because this term is used in Mathematical Morphology for something else, rather different from the meaning given here.", "A good starting point would be to consider Gauss's Shoelace formula for volume en.wikipedia.org/wiki/Shoelace_formula#Generalization . Every polyhedron has a volume formulation which is the sum of tetrahedra radiating from a common centre (after triangulation of surface mesh).", "This paper [Donald W. Grace, Search For Largest Polyhedra, MCOM, 1963] talks about the 8-vertex polyhedron formed by points on a unit sphere. From the article: \"The dual of Goldberg's conjecture can be loosely stated as follows: In the polyhedron of maximum volume formed by $n$ points on a sphere, the faces are triangular and the number of edges leading to any vertex is as nearly as possible equal to the average of such numbers.\" I thought this might be relevant. ams.org/journals/mcom/1963-17-082/S0025-5718-63-99183-X/…", "Digging a bit on the Melzak conjecture literature, I found this paper: londmathsoc.onlinelibrary.wiley.com/doi/10.1112/jlms/s2-6.2.382 (unfortunately not freely available), which gives the upper bound $V<\\frac{1}{432\\pi}\\approx 0.00073682$, which is pretty close to $0.00059$", "This is essentially Melzak's Conjecture. ijgeometry.com/wp-content/uploads/2021/09/4.-63-73.pdf and scholar.rose-hulman.edu/cgi/…", "@RavenclawPrefect Just like with cotangent area formula for regular polygons based on the amounts of sides, for now ignoring side lengths, there may a formula like this from n “pyramids”.", "@TymaGaidash: What do you mean by a \"volume formula\"? The volume of a polyhedron is not uniquely determined by the multiset of its edge lengths, even given the associated edge graph (consider the family of parallelepipeds with unit edges), so there can be no deterministic mapping from a polyhedron's edge lengths to its volume.", "If we could somehow find a volume formula and then optimize it, it should give the answer.", "You can get a better upper bound by using the theorem 2 from this paper. From that together with polyhedron cannot enclose more volume than a sphere, you get the upper bound of $\\frac{1}{36\\sqrt{6}\\pi^2}\\approx 0.00115$", "@StevenStadnicki: Just an update that I did investigate something like this for some small polyhedra and didn't find any improvements (see the edit).", "I wonder whether you could approach this by specific combinatorial counting; the number of polyhedra with low edge counts is relatively small. An orange-slice shape suggests that it might be hard to lower-bound the number of 'long' edges, though. This is a great question."], "comment_count": 0, "tags": ["geometry", "optimization", "volume", "calculus-of-variations", "polyhedra"], "creation_date": "2021-03-01T09:12:52", "diamond": 1, "votes": 112} {"question_id": "263", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3070332", "title": "Can Erdős-Turán $\\frac{5}{8}$ theorem be generalised that way?", "body": "\nSuppose for an arbitrary group word $w$ ower the alphabet of $n$ symbols $\\mathfrak{U_w}$ is a variety of all groups $G$, that satisfy an identity $\\forall a_1, … , a_n \\in G$ $w(a_1, … , a_n) = e$. Is it true, that for any group word $w$ there exists a positive real number $\\epsilon (w) > 0$, such that any finite group $G$ is in $\\mathfrak{U_w}$ iff $$\\frac{\\lvert\\{(a_1, … , a_n) \\in G^n : w(a_1, … , a_n) = e\\}\\rvert}{{|G|}^n} > 1 - \\epsilon(w)?$$\n\nHow did this question arise? There is a widely known theorem proved by P. Erdős and P. Turán that states:\n\nA finite group $G$ is abelian iff $$\\frac{|\\{(a, b) \\in G^2 : [a, b] = e\\}|}{{|G|}^2} > \\frac{5}{8}.$$\n\nThis theorem can be rephrased using aforementioned terminology as $\\epsilon([a, b]) = \\frac{3}{8}$.\nThere also is a generalisation of this theorem, stating that a finite group $G$ is nilpotent of class $n$ iff $$\\frac{|\\{(a_0, a_1, … , a_n) \\in G^{n + 1} : [ … [[a_0, a_1], a_2]… a_n] = e\\}|}{{|G|}^{n + 1}} > 1 - \\frac{3}{2^{n + 2}},$$ thus making $\\epsilon([ … [[a_0, a_1], a_2]… a_n]) = \\frac{3}{2^{n + 2}}$.\nHowever, I have never seen similar statements about other one-word varieties being proved or disproved, despite such question seeming quite natural . . . \nActually, I doubt that the conjecture in the main part of question is true. However, I failed to find any counterexamples myself. \n", "comments": ["Which group words w admit an ε(w) > 0 such that: A finite group G satisfies w = e globally ⟺ P_w(G) > 1 - ε(w)?", "@Z. A. K. (Well, technically I proved them during my PhD, but I wrote them up earlier this year)", "I proved analogous results about the equations $xy^2=y^2x$, $xy^3=y^3x$ and $xy=yx^{-1}$ (and infinitely many others derived from these by adding further variables) earlier this year: preprint.", "@YaniorWeg This conjecture is false for $p=5$; see my answer at mathoverflow.net/a/337483/297 .", "For some recent progress, see the paper Delizia et al., Gaps in probabilities of satisfying some commutator-like identities (2019), where the conjecture is proved for the metabelian and $2$-Engel word.", "@MeesdeVries, there is also a conjecture, that $\\epsilon(x^p) = \\frac{p-1}{p^2}$ for prime $p$, however it remains unproven.", "@MeesdeVries, for the case $n = 1$ only three results are currently known: $\\epsilon(x) = \\frac{1}{2}$, $\\epsilon(x^2) = \\frac{1}{4}$ and $\\epsilon(x^3) = \\frac{2}{9}$.", "Is the $n=1$ case obviously true? Or is even that case difficult?", "Not mentioned in a 2015 survey Farrokhi, D. G. (2015). ON THE PROBABILITY THAT A GROUP SATISFIES A LAW: A SURVEY (Research on finite groups and their representations, vertex operator algebras, and algebraic combinatorics), muroran-it.ac.jp/mathsci/danwakai/past/articles/201404-20150‌​3/…. Mentioned as open in a note by John D. Dixon, \"Probabilistic Group Theory\", people.math.carleton.ca/~jdixon/Prgrpth.pdf", "I one wrote out an answer which is about generalising the Erdos-Turan result to infinite groups: math.stackexchange.com/a/2809964/10513 You might find it interesting/relevant."], "comment_count": 0, "tags": ["combinatorics", "group-theory", "finite-groups", "conjectures", "universal-algebra"], "creation_date": "2019-01-11T12:51:53", "diamond": 1, "votes": 94} {"question_id": "264", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/913049", "title": "Complete, Finitely Axiomatizable, Theory with 3 Countable Models", "body": "Does there exist a complete, finitely axiomatizable, first-order theory $T$ with exactly 3 countable non-isomorphic models?\nA few relevant comments:\nThere is a classical example of a complete theory with exacly $3$ models. This theory is not finitely axiomatizable (For the trivial reason that the language is infinite).\nIn this post, Javier Moreno explains how to rephrase this example in a finite language. Still, the theory is not finitely axiomatizable.\nSome less relevant comments:\nI would like to know if finite axiomability has ever be asked in this context.\nThere have been research in connection with stability. Lachlan has proved that a superstable theory with finitely many countable models is $\\omega$-categorical. And it is still open if this can be extended to all stable theories.\n", "comments": ["and by Scott's theorem a complete $\\mathcal{L}_{\\omega_1,\\omega}$-theory has at most one countable model up to isomorphism (as long as the language is countable, anyways); actually, Scott showed that for every countable structure $\\mathcal{A}$ (in a countable language), there is a single $\\mathcal{L}_{\\omega_1,\\omega}$-sentence $\\varphi$ such that for all countable $\\mathcal{B}$ we have $\\mathcal{B}\\models\\varphi\\iff\\mathcal{B}\\cong\\mathcal{A}$ (google \"Scott sentence\"). So none of this is relevant.", "@mohottnad It is not true that if $T$ is a conservative extension (even in the \"strong\" sense) of $S$ then $T$ has the same number of countable models (up to isomorphism) as $S$; the \"additional structure\" of $T$ may introduce additional models. All we can say is that there will be the same number of reducts to the language of $S$ of countable models of $T$, up to isomorphism, as there are countable models of $S$ up to isomorphism. But that's much weaker. As to changing to $\\mathcal{L}_{\\omega_1,\\omega}$, note that changing the logic also changes the meaning of \"complete,\" (cont'd)", "@NoahSchweber thanks for your critique and I agree my above sentence is not well formed. However, similarly here's some post suggests \"every recursively axiomatizable theory in first-order logic with identity that has only infinite models, has a finitely axiomatized conservative extension... simply endow the given theory with a new sort..\" which seems applicable here for above theory with 3 countable models. Another way for a single sentence using $L_{w1,w}$ logic?", "@mohottnad That doesn't help at all. \"$\\forall k.c_k4$; my main question deals with the smallest case $k=n=5$.\nThis paper implies that if a $F_5$ exists, it is necessarily not expressible in terms of multiplication and real constants alone, unlike $F_0, F_2, F_3$ and $F_4$. There could still be a closed-form expression if we allow the use of conjugation as with $F_1$, but it looks unlikely to me. The possibility of using non-real constants seems more promising, because the automorphism groups of sedenions and above do not act transitively on imaginary elements of norm $1$, so some basis elements are \"more special\" than others. In fact, one can define new \"conjugations\" $x^{(8)} = e_{8} x e_{8}$, $x^{(16)} = e_{16} x e_{16}$, etc. which can be shown to be invariant under all automorphisms of the algebra; this follows from Lemma 2.1 here. An obvious followup to my main question would be whether the $2^n$-inators can be expressed in terms of multiplication, real constants, and the conjugations $x^*, x^{(8)}, x^{(16)}, x^{(32)}, \\ldots, x^{(2^n)}$. But in principle it's not guaranteed that $F_n$ will be algebraically expressible at all, if it exists.\n\nQuestion\nMy question is:\nIs there a $32$-inator in the $32$-nions, i.e. a multilinear map $F_5: [x,y,z,w,v]$ satisfying axioms (A1)-(A4)? Followup: can it be expressed in terms of multiplication, real constants and conjugations?\nSuch a map would necessarily be nontrivial by (A3), and would be identically zero in any sedenion subalgebra by (A2) and (A4). Some remarks:\n\nIn page 12 of the paper I linked above, it is suggested that the possible existence of higher-order maps might be related to projective geometry over the field $\\mathbb{F}_2$.\n\nThe first problem looks simple enough that it could be solved with a computer search, but the search space seems to be quite large at first glance. The obvious upper bound on possible cases to check, taking into account only multilinearity and (A1), would be $(2\\cdot 32 + 1)^{32^5} = 65^{2^{25}} \\lesssim 2^{2^{28}}$. Perhaps some clever argument could reduce this bound to a more manageable number.\n\n\n", "comments": ["In regards to conjugation. One way to view it is as an invertible operator on a vector of reals, that can only change the signs and the order of elements but not the magnitudes of the elements themselves. For an $n$ vector there are $2^n n!$ such conjugation operators. That gives a lot of options to work with.", "I see how $[w,x,y,z]$ is a polarization of the Moufang identity $x(yz)x=(xy)(zx)$, but I don't see how it's related to the other two Moufang identities. Are the three identities equivalent in any alternative algebra?"], "comment_count": 0, "tags": ["abstract-algebra", "recreational-mathematics", "multilinear-algebra", "octonions", "sedenions"], "creation_date": "2022-07-22T14:58:59", "diamond": 0, "votes": 84} {"question_id": "267", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1295967", "title": "Is there a "ping-pong lemma proof" that $\\langle x \\mapsto x+1,x \\mapsto x^3 \\rangle$ is a free group of rank 2?", "body": "Let $f,g\\colon \\mathbb R \\to \\mathbb R$ be the permutations defined by $f\\colon x \\mapsto x+1$ and $g\\colon x \\mapsto x^3$, or maybe even have $g\\colon x \\mapsto x^p$, $p$ an odd prime.\nIn the book, by Pierre de la Harpe, Topics in Geometric Group Theory section $\\textrm{II.B.40}$, as a research problem, it asks to find an appropriate \"ping-pong\" action to show that the group, under function composition, $G=\\langle f,g \\rangle$ is a free group of rank two.\nIs there such a proof? That is, is there a proof where the key insight is having that group act in such a way to apply the ping-pong lemma(table-tennis lemma)? I have not been able to find such a proof either by working on it, or in the literature. \nMaybe we don't have such a proof but do we have a proof that $G$ contains a free subgroup of rank two, akin to proofs for torsion-free hyperbolic group, or the Tits alternative. I am not sure how obvious it is that $G$ is hyperbolic, or linear. I am guessing it is not obvious that it is linear since I would suspect a ping-pong proof would come out of that pretty quickly.\nNote that there are proofs of this theorem, but as far as I know, they do not use the ping-pong lemma.\nThe only proofs of the result(and more general things) I know of are in :\n\nFree groups from fields by Stephen D. Cohen and A.M.W. Glass \nThe group generated by $x \\mapsto x+1$ and $x \\mapsto x^p$ is free. by Samuel White\nArithmetic permutations by S.A. Adeleke and A.M.W. Glass\n\n", "comments": ["I don't expect that linearity would be easier than free. For instance, the subgroup $H$ generated by $f,g$ and in addition $h:x\\mapsto 2x$ is not linear because of the relations $hfh^{-1}=f^2$, $ghg^{-1}=h^3$. Indeed the last implies that $g^nhg^{-n}f^{\\pm 1}g^nh^{-1}g^{-n}=f^{\\pm 2^{(3^n)}}$. If these are complex matrices, the norm of the left hand is bounded exponentially, and hence the norm of $f^m$ partially growth logarithmically (both for $m\\ge 0$ and $m\\le 0$). This is possible only if $f$ has finite order, so every representation of $H$ maps $f$ to an element of finite order.", "\"I am guessing it is not obvious that it is linear since I would suspect a ping-pong proof would come out of that pretty quickly\": this is not quite true. To find free groups in linear groups is now easy, but to show that two given elements generate a free group is usually delicate. For instance I think the following is open (or known but hard): let $G$ be a Zariski-dense subgroup of $SL(n,\\mathbf{C})$ be non-virtually-solvable, and let $x\\in G$ be of infinite order: does there exist $y\\in G$ such that $\\langle x,y\\rangle$ is free? Typically if $x$ is unipotent, it's hard.", "Given that I found the idea of such a proof quite compelling, I figured others had the same though, and maybe someone had found one since the book was written.", "@MartinBrandenburg I agree, the Andrews-Curtis conjecture, 3d Poincare conjecture, \"is every hyperbolic group residually finite?\", are only some of the research problems he has in the book. It has been 15 years or so since the book was published, so I am wondering is such a proof has been done, and I don't think at the time of the book being published there was such a proof. It is also really difficult to find papers on this, the only reason I found the ones sited above was because he sited them! (It is sometimes a little ambiguous as to \"how solved\" the research problems are)", "The book contains quite a lot of these \"research problems\". Could it be that the author hasn't worked them out and instead offers them as, say, research proposals for bachelor or master theses? Perhaps they are open problems, including this one here? Notice that the existing proofs are quite complicated."], "comment_count": 0, "tags": ["group-theory", "reference-request", "group-actions", "geometric-group-theory"], "creation_date": "2015-05-23T15:06:23", "diamond": 0, "votes": 80} {"question_id": "268", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/61026", "title": "Dedekind Sum Congruences", "body": "For $a,b,c \\in \\mathbb{N}$, let $a^{\\prime} = \\gcd(b,c)$, $b^{\\prime} = \\gcd(a,c)$, $c^{\\prime} = \\gcd(a,b)$ and $d = a^{\\prime} b^{\\prime} c^{\\prime}$. Define $\\mathfrak{S}(a,b,c) = a^{\\prime} \\mathfrak{s}( \\tfrac{bc}{d}, \\tfrac{a}{b^{\\prime} c^{\\prime}}) + b^{\\prime} \\mathfrak{s}( \\tfrac{ac}{d}, \\tfrac{b}{a^{\\prime} c^{\\prime}}) + c^{\\prime} \\mathfrak{s}( \\tfrac{ab}{d}, \\tfrac{c }{a^{\\prime} b^{\\prime}} )$, where $\\mathfrak{s}$ is the Dedekind sum. I can prove the following: For $a, b, c \\in \\mathbb{N}$ with $\\gcd(a,b,c) = 1$,\n\\begin{align}\n12 a b c \\ \\mathfrak{S}(a,b,c) \\equiv (ab)^{2} + (bc)^{2} + (ca)^{2} + d^{2} \\ \\pmod{a b c}.\n\\end{align}\nIs this congruence in the literature? I'm aware of the 3-term generalization of Rademacher and Pommersheim and of the work of Beck on the Carlitz-Dedekind Sums, but these don't seem to include this particular congruence as a special case.\nI generalize. For $a_{1}, \\dots, a_{n} \\in \\mathbb{N}$ such that $\\gcd(a_1, \\dots, a_{n}) = 1$, let $a_{i}^{\\prime} = \\gcd(a_{1}, \\dots, \\hat{a}_{i}, \\dots, a_{n})$. Define the symmetric summation\n$$\n\\mathfrak{S}(a_{1},\\dots, a_{n}) = \\sum_{i = 1}^{n} a_{i}^{\\prime} \\, \\mathfrak{s}\\left( \\tfrac{a_{1} \\cdots \\hat{a}_{i} \\cdots a_{n}}{a_{n+1}}, \\tfrac{a_{i}}{a_{1}^{\\prime} \\cdots \\hat{a}_{i}^{\\prime} \\cdots a_{n}^{\\prime}} \\right).\n$$\nwhere $\\hat{}$ denotes omission. For pairwise coprime $a_{1}, \\dots, a_{n} \\in \\mathbb{N}$ (all primed variables equal $1$), I conjecture the following congruence holds for $n = 4$ (and I have a hunch that it continues to hold for $n > 4$):\n$$\n12 a_{1} \\cdots a_{n} \\, \\mathfrak{S}(a_{1}, \\dots, a_{n}) \\equiv \\left( \\sum_{i = 1}^{n} (a_{1} \\cdots \\hat{a}_{i} \\cdots a_{n})^{2} \\right) + 1 \\ \\pmod{a_{1} \\cdots a_{n}}.\n$$\nIs this more general congruence known?\n", "comments": [], "comment_count": 0, "tags": ["number-theory", "reference-request"], "creation_date": "2011-08-31T13:39:55", "diamond": 0, "votes": 73} {"question_id": "269", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2355787", "title": "Determinant of a matrix that contains the first $n^2$ primes.", "body": "Let $n$ be an integer and $p_1,\\ldots,p_{n^2}$ be the first prime numbers. Writing them down in a matrix\n$$\n\\left(\\begin{matrix}\np_1 & p_2 & \\cdots & p_n \\\\\np_{n+1} & p_{n+2} & \\cdots & p_{2n} \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\n\\cdots & \\cdots & \\cdots & p_{n^2}\n\\end{matrix}\n\\right)\n$$\nwe can take the determinant. How to prove that determinant is not zero for every $n$?\n", "comments": ["Is it possible to prove this by showing that inverse of given matrix always exists as there always exists trivial solutions to Ax = 0 equation?. Trivial solutions are only possible as the rows of the matrix will always be independent, as each element of matix is a distinct prime number.", "Has anyone thought of posting this on math overflow?", "@Gerald Did you misread \"not zero\" as \"nonnegative\"?", "@Boris: Although the absolute value increases most of the time, there are exceptions. For example with $n=57,58$ the determinants are about $2.14 \\times 10^{89}$ and $1.25 \\times 10^{89}$, respectively. There are more exceptions later.", "@Elborito This is because the Perron-Frobenius Eigenvalue grows faster than linearly and is much larger than the other eigenvalues (which are rarely smaller than one). Here is a list of the eigenvalues for $n=14$: $7843.09, -168.956, -11.4222 + 2.69341 i, -11.4222 - 2.69341 i, -5.68197 + 5.79415 i, -5.68197 - 5.79415 i, 0.121105 + 6.20539 i, 0.121105 - 6.20539 i, 4.28891 + 4.0145 i, 4.28891 - 4.0145 i, 11.132, 7.30824, -4.67599, -1.51002$. I've also made a plot of the lowest singular value for $n=10,\\ldots,100$. But I can't see any pattern. E.g. , at $n=28$ the lowest sing. value is $0.006$.", "The absolute value of the determinant of the matrix is strictly crescent. maybe by recursivity", "$n=951$ and there isn't counterexample", "But I want to tell those interested that the bound that arrives is $n=845$ with $714025$ primes in the maximum matrix.", "No answer: You might look at \"Integral Matrices\" by Morris Newman. In particular the \"Smith Normal Form\". It discusses divisibility and such. As I recall some of the results were surprising. The class is just a distance memory (but I have the book) so it might lead nowhere.", "@RoflUkulus We can also to check whether starting series from $1$ gives similar results. So far it was checked up to $n=200$.", "From my point of there are two things to study: 1.) Are there any permutations $\\sigma \\in S_{n^2}$ that permutate the entries such that the determinant goes to zero? If so, what do they look like? 2.) What are the least gaps (least is not really defined properly), that primes can have, to produce determinant zero, like the matrices Robert posted, for example maybe is there any $k$ such that $p_1,p_{1+k},p_{2+k},\\ldots$ determinant gets zero.", "@RobertIsrael The number of digits for determinant seems to increase (or at least not to decrease - almost linear function) . Could it be possible to compute with your software also series with this number of digits?", "@RobertIsrael Very inspiring, I hope some day a limit $n= 1000$ will be achieved.", "At $n= 460$ the determinant has $1001$ digits. I was making a b-file for sequence A067276, and the OEIS doesn't like numbers with more than $999$ digits. I could go further, but computations start to slow down...", "@RobertIsrael so now 460 is the limit. I wonder why exactly 460 in this case?", "@Robert Israel thank you for these examples. Indeed, this makes it harder to prove and there might be some evil matrix around which determinant goes to zero :-) I looked up the prime factors of the determinants but did not find any pattern. There are large powers of 2 appearing in the factorization, but there not even increasing monotonely.", "Also with determinant $0$: $$ \\pmatrix{2 & 3 & 5 & 7\\cr 11 & 13 & 17 & 19\\cr 23 & 29 & 31 & 37\\cr 41 & 47 & 67 & 73\\cr }$$", "I think this is going to be an intractable problem. Note that there are square matrices with determinant $0$ made up of distinct primes, e.g. $$\\pmatrix{2 & 3 & 5\\cr 7 & 11 & 13\\cr 19 & 23 & 97\\cr}$$ Thus you somehow have to depend on the fact that you're using the consecutive primes. And those just don't have enough regularity.", "No zeros up to $n=460$.", "@RoflUkulus I think that Peter is the best expert on prime numbers ( much better than me) , ask him when he will be available. BTW he has asked many other interesting questions about primes..", "@Widawensen Oh, you already asked the questions. Did you try the flintc library for checking n > 201. Not sure if it can handle those big matrices. I strongly believe that the proof will be difficult, too. Maybe it would be interesting to study the prime factorization of those determinants.", "Read comments from math.stackexchange.com/questions/2047879/…, especially Peter's comments are valuable.", "The sequence of determinants is OEIS sequence A067276. Not that this helps..."], "comment_count": 0, "tags": ["linear-algebra", "number-theory", "prime-numbers", "determinant"], "creation_date": "2017-07-11T23:55:08", "diamond": 0, "votes": 69} {"question_id": "270", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3176228", "title": "Does the average primeness of natural numbers tend to zero?", "body": "Note 1: This questions requires some new definitions, namely \"continuous primeness\" which I have made. Everyone is welcome to improve the definition without altering the spirit of the question. Click here for a somewhat related question.\n\nA number is either prime or composite, hence primality is a binary concept. Instead I wanted to put a value of primality to every number using some function $f$ such that $f(n) = 1$ iff $n$ is a prime otherwise, $0 < f(n) < 1$ and as the number divisors of $n$ increases, $f(n)$ decreases on average. Thus $f(n)$ is a measure of the degree of primeness of $n$ where 1 is a perfect prime and 0 is a hypothetical perfect composite. Hence $\\frac{1}{N}\\sum_{r \\le N} f(r)$ can be interpreted as a measure of average primeness of the first $N$ integers. \nAfter trying several definitions and going through the ones in literature, I came up with:\n\nDefine $f(n) = \\dfrac{2s_n}{n-1}$ for $n \\ge 2$, where $s_n$ is the\n standard deviation of the divisors of $n$.\n\nOne advantage of using standard deviation is that even if two numbers have the same number of divisor their value of $f$ appears to be different hence their measure of primeness will be different.\nQuestion 1: Does the average primeness tend to zero? i.e. does the following hold?\n$$\n\\lim_{N \\to \\infty} \\frac{1}{N}\\sum_{r = 2}^N f(r) = 0\n$$\nQuestion 2: Is $f(n)$ injective over composites? i.e., do there exist composites $3 < m < n$ such that $f(m) = f(n)$? \n\nMy progress\n\n$f(4.35\\times 10^8) \\approx 0.5919$ and decreasing so the limit if it exists must be between 0 and 0.5919.\nFor $2 \\le i \\le n$, the minimum value of $f(i)$ occurs at the largest highly composite number $\\le n$.\n\nNote 2: Here standard deviation of $x_1, x_2, \\ldots , x_n$ is defined as $\\sqrt \\frac{\\sum_{i=1}^{n} (x-x_i)^2}{n}$. Also notice that even if we define standard deviation as $\\sqrt \\frac{\\sum_{i=1}^{n} (x-x_i)^2}{n-1}$ our questions remain unaffected because in this case in the definition of $f$, we will be multiplying with $\\sqrt 2$ instead of $2$ to normalize $f$ in the interval $(0,1)$.\nNote 3: Posted this question in MO and got answer for question 1. Indeed the limit tends to zero. Question 2 is still open.\n", "comments": ["If $f(n)=1$ iff $n$ is a prime then I would have thought $f(n)$ cannot uniquely identify $n$ (i.e. $f(n)$ is not $1-1$ injective), since it cannot when $n$ is a prime.", "@daniel One of the reasons why I invented this definition was because even if two numbers have the same number of divisors or the same number distinct prime divisors, their value of $f(n)$ was found to be unique. In that way, $f(n)$ can uniquely identify $n$ but $\\omega(n)$ or $d(n)$ cannot uniquely identify $n$.", "@daniel After I posted in MO, it was answered. But I am not sure if a similar question was asked in MSE and answered", "@daniel I do not see how Erod Kac theorem directly implies that the avove limit tends to zero?", "$0.59$ is the average of all the values for $n \\le 4.35 \\times 10^8$", "Yes, that is what I got. I was confused when you reported getting 0.59 above.", "@Scott You must have done an error. Try this Sagemath code (N(variance(divisors(320700000), bias = True), digits = 10))^0.5 / (320700000-1) you will get 0.1009653420", "In that case, for 320700000, the divisors are: $[1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 25, 30, 32 ......... 13362500, 16035000, 20043750, 21380000, 26725000, 32070000, 40087500, 53450000, 64140000, 80175000, 106900000, 160350000, 320700000]$ with a variance of $1048437529950912.5$. However, this gives $f(320700000) \\approx 0.201930684009565$ = 2 times the square root of the variance divided by 320700000 - 1. What am I missing here?", "@Scott E.g. For $n = 19$, the divisors are $1,19$ and the variance of these two divisors is $82$. Similarly for $n = 20$, the divisors are $1,2,4,5,10,20$ and their variance is $42$", "I can't seem to recreate your test cases. Do you mean proper divisors, etc? Could you post a couple of low-valued test cases for this so that we can make sure we're evaluating the same thing?", "@NilotpalKantiSinha I'm not sure I follow why Question 2 is a question. Isn't $f(p) = f(q) = 1$ for any primes $p,q$ by construction?", "@JavaMan Yes, standard deviation gives more information than just the number of divisors. But to be honest, I don't exactly know how to interpret standard deviation completely in the context of primeness. One of the reason for using standard deviation was because some progress was already made on the standard deviation of the divisors of the first $n$ natural numbers. Check this post. math.stackexchange.com/questions/2773289/…", "Why do you care about the standard deviation? I understand that $s_n$ may be different even for a number with the same amount of divisors, but if all you want is to measure “primeness,” then why complicate this with standard deviation? It seems like this question now accounts for more than just primeness. It now considers smoothness of an integer (or something like it).", "@TheSimpliFire Sure joining", "We can continue this discussion in chat", "@TheSimpliFire My graph is identical to yours but instead of converging to about $0.4$ as in your case, it is converging to about $0.4 \\times \\sqrt 2 \\approx 0.59$. This is because of bais = True or False in your formula for calculating standard deviation. You have used bias = False and got the limit 0.4. I have used bias = True and got $f(320700000) = 0.594127573872085$", "I meant your second sentence below the Note at the start of your post. Here is a plot of the limit for $N\\le 10000$.", "@TheSimpliFire The difference is because of the the parameter 'bias = ' in the formula for calculating standard deviation in Sagemath. Check my source code above. For a prime $p$ we have 2*std(divisors(p), bias = True).n()/(p-1) = 1 and 2*std(divisors(p), bias = True).n()/(p-1) = $\\sqrt 2$. I have used bias = True because the range $(0,1)$ looks more elegant than $(0, \\sqrt 2)$. Also $r = 2$ is correct. However, regardless of bias = True or False, the the question remain unchanged as $f$ is only gets multiplied by a scaling factor so we are good.", "I don't get why you say that $f(n)=1$ if $n$ is prime. For a prime $p$, $$f(p)=\\frac{2\\sqrt{\\frac{\\left(\\frac{p-1}2\\right)^2+\\left(\\frac{1-p}2\\right)^2}{2-1}}}{p-1}=\\sqrt2$$ so shouldn't it be $f(n)=\\frac{\\sqrt2 s_n}{n-1}$? Furthermore the sum should be from $r=2$ to $N$ as $1/(r-1)$ is undefined when $r=1$.", "@YiFan Sure done.", "You might want to rephrase \"is $f(n)$ unique\" to \"is $f(n)$ injective\". At first glance I thought you didn't know whether $f(n)$ was well-defined.", "@TheSimpliFire I have added my Sagemath code above", "I have written some code for this in R: stanfun <- function(n){sd(divisors(n))/(n-1)};funstan <- function(m){sum(sapply(2:m, function(i){stanfun(i)}))/m}, so $\\frac1{10000}\\sum\\limits_{r=1}^{10000}f(r)$ is given by funstan(10000) which outputs $0.404801$.", "@GerryMyerson: Here is a more technical answer why standard deviation. If two numbers have the same number of divisors then the value $2/d(n)$ is same for both but the values of $f(n)$ is different. So under my definition, I will consider the number with smaller value of $f(n)$ to have a greater primness because we are not just measuring how many divisors a number has but also how scattered these divisors are. At the moment, I don't know if $f(n)$ is unique. I will add this to the question.", "Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$?", "Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument."], "comment_count": 0, "tags": ["number-theory", "limits", "statistics", "prime-numbers", "natural-numbers"], "creation_date": "2019-04-05T10:59:06", "diamond": 0, "votes": 62} {"question_id": "271", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4103564", "title": "On the equivalence relation $(a,b) \\sim (c,d)\\iff a+b=c+d$", "body": "Setup\nLet $A=\\{a_1\\text{sum of some $(\\Delta a)_k$ and some $(\\Delta b)_l$}$$\nwhere the $(\\Delta a)_k$ and $(\\Delta b)_l$ to take into account are found using visibility relations. Similarly there are Vertical Consistency conditions of the same ilk\n$$V_{ij}:(\\Delta b)_j>\\text{sum of some $(\\Delta a)_k$ and some $(\\Delta b)_l$}$$\nwhere again, the $(\\Delta a)_k$ and $(\\Delta b)_l$ that appear on the right hand side are deduced from visibility relations along the $<$-path of equivalence classes linking the $\\sim$ equivalence class containing $(i,j-1)$ to the $\\sim$ equivalence class containing $(i,j)$\n\nBeginnings of a positive solution for $p=2$\nI believe the answer is positive to both questions at least when $p=2$. WLOG we can take $a_1=0$, $a_2=1$ and $b_0=0$. There seems to exist an explicit algorithm to find a rational solution to the problem. I've illustrated it below. The necessary and sufficient condition on $\\sim$ in the case $p=2$ seems to be that the lines associated to $2$-element equivalence classes don't intersect. The algorithm is simple but I haven't conceptualized it really:\n\nYou start with $b_1=0$, the sum in the the lower left corner $(1,1)$ is thus $0$ since $a_1=0$\nYou put $1$ in the lower right corner $(2,1)$, the sum is $1$ since $a_2=1$\nIf the lower right corner is part of a $2$ element equivalence class you put are forced to attribute $1$ to the other member $(1,r)$.\n\nif $r>2$ you set $b_r=1$. This is the case in the example below so you set $b_3=1$. Then you are forced to put a $2$ in position $(2,r)$ since $a_2=1$. You keep on going until you don't fall into a 2 element equivalence class.\n\n\nThen you have to insert values for $b_2,\\dots,b_{r-1}$. You pick them uniformly apart between $b_1=0$ and $b_r=1$, that is $b_k=\\frac{k-1}r$.\nThen you are forced to inscribe $b_k+1$ on the right hand side. If the right hand side is part of a $2$ element equivalence class you are forced to set the same value for some new $b_s$ etc ...\n\nI hope the picture below (and the color coded rounds) are more explicit than the half-baked algorithm above.\nI don't know if this will work for $p\\geq 3$.\n\n\nA connection with topology\n$\\newcommand{\\O}{\\mathcal{O}}\\newcommand{\\R}{\\Bbb{R}}$\nFix $\\sim$ an equivalence relation on $R_{p,q}$ and $<$ a total order on $R_{p,q}/\\sim$. Define\n$$\\O_{p,q}[\\sim,<]=\\{(A,B)\\in\\O_p\\times\\O_q\\text{ inducing $\\sim$ and }<\\}$$\nwhere $\\O_n=\\{(x_1,\\dots,x_n)\\in\\R^n\\mid x_11$ cannot output a consecutive list of $n+1$ numbers. This last claim itself is a nice problem; for a solution, see Example 2.24 in page 11 of Number Theory: Concepts and Problems by Andreescu, Dospinescu and Mushkarov.\n", "comments": ["@Sil Great suggestion! Just posted the problem on MathOverFlow at this link.", "Interesting problem, you might want to consider asking on MathOverflow as it was not resolved here even after bounty.", "In fact, the polynomial $P=p*q=[(p-1)/2 + (p+1)/2]*[(q-1)/2 + (q+1)/2]$ can represent non-squares and non-primes. Example: $91=7*13=(3+4)*(6+7)=3*6 +3*7 +4*6 + 4*7$ with $(91-1)/2=3*7 + 4*6=45$ and $(91+1)/2=3*6 + 4*7=46$.", "I cannot answer the above question but I can say that every odd square can be represented by $P(x,y)=(x+y)^2=x^2 + 2xy +y^2$ with $xy=2T_n$ with $T_n$ twice a triangular number and also $y=x+1$. Ex: $81=9^2=(4+5)^2=4^2 + 2*4*5 + 5^2$. Example of P(x,y) for a non-square: $73=(72/2 + 74/2)=1+2*18+6^2$. Note that $18$ is not twice a triangular number. This means than non-squares can be represented by $P(x,y)$ but $x$ and $y$ will necessarily differ by more than $1$. If $(N-1)/2$ is odd, the number can never be a square. Ex $71$ with $(71-1)/2=35$ or $91$ with $(91-1)/2=45$.", "A possibly helpful note: This can be done with two very similar polynomials $P_1(x,y)=(x+y-1)^2+y$ and $P_2(x,y)=(x+y-1)^2+(x+y-1)+y$. These two polynomials have disjoint images, with $\\operatorname{Im}(P_1)\\cup \\operatorname{Im}(P_2)$ representing all non-squares: the image of $P_1$ is the union of the intervals $[n^2+1,n^2+n]$ for each $n\\geq 1$, while the image of $P_2$ is the union of the intervals $[n^2+n+1,n^2+2n]$ for each $n\\geq 1$. (This can be used to construct a $3$-variable polynomial $$P(x,y,z)=\\big(1-(z-1)(z-2)\\big)\\big((x+y-1)^2+(z-1)(x+y-1)+y\\big)$$ with the desired property."], "comment_count": 0, "tags": ["number-theory", "polynomials", "contest-math"], "creation_date": "2022-04-28T11:58:00", "diamond": 0, "votes": 49} {"question_id": "273", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2621776", "title": "The sequence $\\,a_n=\\lfloor \\mathrm{e}^n\\rfloor$ contains infinitely many odd and infinitely many even terms", "body": "PROBLEM. Show that the sequence $\\,a_n=\\lfloor \\mathrm{e}^n\\rfloor$ contains infinitely many odd and infinitely many even terms.\nIt suffices to show that the terms of the sequence\n$$\\,b_n=\\mathrm{e}^n\\,\\mathrm{mod}\\, 2,\\,\\,\\,n\\in\\mathbb N,$$ \nare dense in $[0,2]$.\nUnfortunately, Weyl's Theorem does not look helpful in this case.\nEDIT. As Chris Culter said, the claim that the terms of the sequence\n$$\\,b_n=\\mathrm{e}^n\\,\\mathrm{mod}\\, 2,\\,\\,\\,n\\in\\mathbb N,$$ \nare dense in $[0,2]$ is (or might be) an open problem. Nevertheless, this does not imply that the claim that the sequence $\\,a_n=\\lfloor \\mathrm{e}^n\\rfloor$ contains infinitely many odd and infinitely many even terms is necessarily an open problem as well. It is also noteworthy that it is relatively easy to construct an irrational $\\alpha$ with the property that the sequence $\\,\\alpha^n\\,\\mathrm{mod}\\, 2,\\,\\,n\\in\\mathbb N,$ is NOT dense in $[0,2]$.\n", "comments": ["For any $a\\in\\mathbb N$, $\\lfloor a^n\\rfloor$ will have the same parity as $a$. For $0 < a < 1$ we always have $\\lfloor a^n\\rfloor = 0$. But for any other values of $a$ I don't see obvious way to prove or disprove the property from the problem. Maybe it has nothing to do specifically with $e$ and a proof would cover a class of reals (potentially all non-integers $>1$ ?).", "Exercises for university students? Seems hard. The sequence is tabulated at oeis.org/A000149 but I don't think anything at that page helps to resolve the question.", "I had found it in a Greek site with Mathematical problems for university students.", "What's the source of this question, please?", "@MathematicsStudent1122 Does \"look\" helpful, but not really helpful.", "If $e^n$ is dense modulo $2$, then it is also dense modulo $1$, yet the latter is an open problem, at least circa 2003 per this paper."], "comment_count": 0, "tags": ["real-analysis", "sequences-and-series", "uniform-distribution"], "creation_date": "2018-01-25T23:17:44", "diamond": 0, "votes": 46} {"question_id": "274", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2194973", "title": "Is every finite list of integers coprime to $n$ congruent $\\pmod n$ to a list of consecutive primes?", "body": "For example the list $(2, 1, 2, 1)$ is congruent $\\pmod 3$ to the consecutive primes $(5, 7, 11, 13)$. But how about the list $(1,1,1,1,1,1,1,1,2,3,4,3,2,3,1) \\mod 5$?\nMore generally, we are given some integer $n \\geq 2$ and a finite list of integers that are coprime to and less than $n$. Is it always possible to produce the same list by consecutive primes $\\pmod n$?\nFormally: given $n \\geq 2$ and $(a_0,a_1,\\cdots \\,a_k)$ such that for all $i$, $GCD(a_i, n) = 1$, is there a list of consecutive primes such that each $p_i \\equiv a_i \\pmod n$?\n", "comments": ["For those seeking Shiu's paper, here is the citation: Shiu, D.K.L. (2000), Strings of Congruent Primes. Journal of the London Mathematical Society, 61: 359-373. doi.org/10.1112/S0024610799007863", "Very interesting question. Just as an observation, if we took $n$ to be a prime and a finite list of values less than $n$ as stated in the problem and if the list of values is congruent to successive primes mod $n$, then we could encode that entire list using just the initial prime's index and deterministically recover the elements of the list with no additional information (except the size of the list). Perhaps, this is intricately linked to Shannon Entropy.", "It's not, and I never said it was...", "@RushabhMehta How is that Greenwood-Tao result an advance on Dirichlet's theorem that every AP of integers $a, a+d,\\dots$ with $d>0$ coprime to $a$ contains arbitrarily many primes?", "I Think the question is more interesting if we consider the case where $n$ is a prime itself.", "What about using a generalized Dirichlet series but with multiple entries, i.e. characters $\\chi: (\\mathbb{Z}/n\\mathbb{Z} )^* \\times \\ldots \\times (\\mathbb{Z}/n\\mathbb{Z} )^* \\to \\mathbb{C}^*$ ? I just don't know how to implement the \"consecutive primes\" constraint..", "Well lets first note, that even residues modulo odd $n$ aren't possible to be prime unless the multiplier on $n$ is odd. Likewise, odd residues aren't representing primes unless the multiplier on $n$ is even in this case. This at least forces gap size constraints (1,4,2,3) mod 5 won't happen below gap sizes 8,8,6 with gaps 3 mod 5, 3 mod 5, and 1 mod 5 Your long one needs gaps of 10,10,10,10,10,10,10,6,6,6,4,4,6,8", "@cr001 Can't find a wiki link. TLDR, it was proven by Daniel Shiu in the 90s, and the statement is as follows: for any coprime $a,b\\in\\mathbb N$, there exists arbitrarily long sequences of consecutive primes, all congruent to $a\\bmod b$. Similar but different than Greenwood-Tao, which doesn't require consecutiveness of the primes.", "nobody intrested in my bounty ??", "Is there a Wikipedia link to the Shiu Theorem? I wan't able to find any statement of the Theorem via google search.", "For the record, you can find $\\{1,1,1,1,1,1,1,1,2,3,4\\} \\pmod {5}$ starting at $1313286451$, which looks about in line with what I'd expect from the random model. So yeah, it certainly seems likely to me that arbitrarily long sequences of this sort exist.", "This is an open problem except in the case where all $a_i$ are equal, which is a famous theorem of Shiu. There are a few other cases that are consequently true for simple combinatorial reasons, such as $(1,1,1,1,1,1,1,1,2)$ mod $3$, but essentially Shiu's theorem is the best we have, AFAIK.", "Very interesting question. Altough I would be surprised if this wasn't true, I would be even more surprised if there was an easy proof.", "@David Schneider-Joseph thanks.", "No, for example $(2, 2) \\pmod 4$ would require two even primes. The question might be more interesting if you require that entries in the list are coprime with $n$ (a generalization of disallowing 0)."], "comment_count": 0, "tags": ["number-theory", "elementary-number-theory", "prime-numbers", "prime-gaps"], "creation_date": "2017-03-20T05:52:33", "diamond": 0, "votes": 40} {"question_id": "275", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4397045", "title": "Are these generalizations known in the literature?", "body": "By using\n$$\\int_0^\\infty\\frac{\\ln^{2n}(x)}{1+x^2}dx=|E_{2n}|\\left(\\frac{\\pi}{2}\\right)^{2n+1}\\tag{a}$$\nand\n$$\\text{Li}_{a}(-z)+(-1)^a\\text{Li}_{a}(-1/z)=-2\\sum_{k=0}^{\\lfloor{a/2}\\rfloor }\\frac{\\eta(2k)}{(a-2k)!}\\ln^{a-2k}(z)\\tag{b}$$\nI managed to find:\n$$\\int_0^\\infty\\frac{\\ln^{2n}(x)}{1+yx^2}dx=\\frac{\\left(\\frac{\\pi}{2}\\right)^{2n+1}}{\\sqrt{y}}\\sum_{k=0}^n\\binom{2n}{2k}|E_{2n-2k}|\\pi^{-2k}\\ln^{2k}(y)\\tag{c}$$\n\n$$\\int_0^\\infty\\frac{\\ln^{2n-1}(x)}{1+yx^2}dx=\\frac{-\\left(\\frac{\\pi}{2}\\right)^{2n-1}}{2\\sqrt{y}}\\sum_{k=0}^{n-1}\\binom{2n-1}{2k+1}|E_{2n-2k-2}|\\pi^{-2k}\\ln^{2k+1}(y)\\tag{d}$$\n\n$$\\int_0^\\infty\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=|E_{2n}|\\left(\\frac{\\pi}{2}\\right)^{2n+1}\\zeta(2a+1)$$\n$$-\\frac{\\left(\\frac{\\pi}{2}\\right)^{2n+1}}{(2a)!}\\sum_{k=0}^n \\binom{2n}{2k}|E_{2n-2k}|\\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\\zeta(2k+2a+1)\\tag{e}$$\n\n$$\\int_0^\\infty\\frac{\\ln^{2n-1}(x)\\text{Li}_{2a}(-x^2)}{1+x^2}dx=\\frac{-\\left(\\frac{\\pi}{2}\\right)^{2n-1}}{2(2a-1)!}*$$\n$$\\sum_{k=0}^{n-1} \\binom{2n-1}{2k+1}|E_{2n-2k-2}|\\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\\zeta(2k+2a+1)\\tag{f}$$\n\n$$\\int_0^1\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\\frac12|E_{2n}|\\left(\\frac{\\pi}{2}\\right)^{2n+1}\\zeta(2a+1)$$\n$$-\\frac{\\left(\\frac{\\pi}{2}\\right)^{2n+1}}{2(2a)!}\\sum_{k=0}^n \\binom{2n}{2k}|E_{2n-2k}|\\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\\zeta(2k+2a+1)$$\n$$+2\\sum_{k=0}^a\\frac{(2k+2n+1)!}{(2k+1)!}4^k \\eta(2a-2k)\\beta(2k+2n+2)\\tag{g}$$\n\n$$\\int_0^1\\frac{\\ln^{2n-1}(x)\\text{Li}_{2a}(-x^2)}{1+x^2}dx=$$\n$$\\frac{-\\left(\\frac{\\pi}{2}\\right)^{2n-1}}{4(2a-1)!}\\sum_{k=0}^{n-1} \\binom{2n-1}{2k+1}|E_{2n-2k-2}|\\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\\zeta(2k+2a+1)$$\n$$+\\sum_{k=0}^a\\frac{(2k+2n-1)!}{(2k)!}4^k \\eta(2a-2k)\\beta(2k+2n)\\tag{h}$$\n\n$$\\sum_{k=1}^\\infty\\frac{(-1)^k H^{(2a+1)}_k}{(2k+1)^{2n+1}}=\\frac1{2(2n)!}|E_{2n}|\\left(\\frac{\\pi}{2}\\right)^{2n+1}\\zeta(2a+1)$$\n$$-\\frac{\\left(\\frac{\\pi}{2}\\right)^{2n+1}}{2(2n)!(2a)!}\\sum_{k=0}^n \\binom{2n}{2k}|E_{2n-2k}|\\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\\zeta(2k+2a+1)$$\n$$+\\frac{2}{(2n)!}\\sum_{k=0}^a\\frac{(2k+2n+1)!}{(2k+1)!}4^k \\eta(2a-2k)\\beta(2k+2n+2)\\tag{i}$$\n\n$$\\sum_{k=1}^\\infty\\frac{(-1)^k H^{(2a)}_k}{(2k+1)^{2n}}=\\frac{\\left(\\frac{\\pi}{2}\\right)^{2n-1}}{4(2n-1)!(2a-1)!}*$$\n$$\\sum_{k=0}^{n-1} \\binom{2n-1}{2k+1}|E_{2n-2k-2}|\\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\\zeta(2k+2a+1)$$\n$$-\\frac{1}{(2n-1)!}\\sum_{k=0}^a\\frac{(2k+2n-1)!}{(2k)!}4^k \\eta(2a-2k)\\beta(2k+2n)\\tag{j}$$\n\nQuestion: Are the results of $(c)$ to $(j)$ known in the literature?\nIf the reader is curious about the correctness of the results above and wants to verify them on Mathematica, the Mathematica command of $|E_r|$ is Abs[EulerE[r]]\nThanks,\n\nProof of (a): By using Euler's reflection formula\n$$\\Gamma(m)\\Gamma(1-m)=\\pi \\csc(m\\pi),\\quad m\\notin\\mathbb{Z}$$\nand beta function\n$$\\operatorname{B}(a,b)=\\frac{\\Gamma(a)\\Gamma(b)}{\\Gamma(a+b)}=\\int_0^\\infty \\frac{x^{a-1}}{(1+x)^{a+b}}dx$$\nwith $a=m$ and $b=1-m$, we have\n$$\\pi\\csc(m\\pi)=\\int_0^\\infty\\frac{x^{m-1}}{1+x}dx$$\ndifferentiate both sides $2n$ times with respect to $m$ then let $m$ approach $1/2$\n$$\\int_0^\\infty\\frac{\\ln^{2n}(x)}{1+x^2}dx=\\frac{\\pi}{2^{2n+1}}\\lim_{m\\to \\frac12}\\frac{d^{2n}}{dm^{2n}}\\csc(m\\pi)$$\nthe proof completes on using\n$$\\lim_{m\\to \\frac12}\\frac{d^{2n}}{dm^{2n}}\\csc(m\\pi)=|E_{2n}|\\pi^{2n}$$\nwhich is explained here.\n\nProof of (b): Divide both sides of the common dilogarithm identity\n$$\\text{Li}_2(-z)+\\text{Li}_2(-1/z)=-\\frac{\\ln^2(z)}{2}-2\\eta(2)$$\nby $z$ then integrate repeatedly.\n\nProof of (c): Let $yx^2=t^2$,\n$$\\int_0^\\infty\\frac{\\ln^{2n}(x)}{1+yx^2}dx=\\frac{1}{\\sqrt{y}}\\int_0^\\infty\\left(\\ln(\\sqrt{y})-\\ln(t)\\right)^{2n}\\frac{dt}{1+t^2}$$\n$$=\\frac{1}{\\sqrt{y}}\\int_0^\\infty\\left(\\sum_{k=0}^{2n} \\binom{2n}{k}(-\\ln(t))^{2n-k}\\ln^k(\\sqrt{y})\\right)\\frac{dt}{1+t^2}$$\n$$=\\frac1{\\sqrt{y}}\\sum_{k=0}^{2n}\\binom{2n}{k}\\ln^k(\\sqrt{y})\\int_0^\\infty\\frac{(-\\ln(t))^{2n-k}}{1+t^2}dt$$\nSince we care for only the even powers of $\\ln(t)$ as the odd powers make the integral zero, we have\n$$\\sum_{k=0}^{2n}f(k)=\\sum_{k=0}^n f(2k)$$\nand so\n$$\\int_0^\\infty\\frac{\\ln^{2n}(x)}{1+yx^2}dx=\\frac1{\\sqrt{y}}\\sum_{k=0}^{n}\\binom{2n}{2k}\\ln^{2k}(\\sqrt{y})\\int_0^\\infty\\frac{\\ln^{2n-2k}(t)}{1+t^2}dt$$\nThe proof completes on using the result in $(a)$.\n\nProof of (d): We follow the same steps in proof $(c)$:\n$$\\int_0^\\infty\\frac{\\ln^{2n-1}(x)}{1+yx^2}dx=\\frac{-1}{\\sqrt{y}}\\int_0^\\infty\\left(\\ln(\\sqrt{y})-\\ln(t)\\right)^{2n-1}\\frac{dt}{1+t^2}$$\n$$=\\frac{-1}{\\sqrt{y}}\\int_0^\\infty\\left(\\sum_{k=0}^{2n-1} \\binom{2n-1}{k}(-\\ln(t))^{2n-k-1}\\ln^k(\\sqrt{y})\\right)\\frac{dt}{1+t^2}$$\n$$=\\frac{-1}{\\sqrt{y}}\\sum_{k=0}^{2n-1}\\binom{2n-1}{k}\\ln^k(\\sqrt{y})\\int_0^\\infty\\frac{(-\\ln(t))^{2n-k-1}}{1+t^2}dt$$\nSince we care for only the even powers of $\\ln(t)$, we have\n$$\\sum_{k=0}^{2n-1}f(k)=\\sum_{k=0}^{n-1} f(2k+1)$$\nand so\n$$\\int_0^\\infty\\frac{\\ln^{2n-1}(x)}{1+yx^2}dx=\\frac{-1}{\\sqrt{y}}\\sum_{k=0}^{n-1}\\binom{2n-1}{2k+1}\\ln^{2k+1}(\\sqrt{y})\\int_0^\\infty\\frac{\\ln^{2n-2k-2}(t)}{1+t^2}dt$$\nThe proof completes on using the result in $(c)$.\n\nProof of (e): Using the integral representation of the polylogarithm function:\n$$\\text{Li}_{a}(z)=\\frac{(-1)^{a-1}}{(a-1)!}\\int_0^1\\frac{z\\ln^{a-1}(t)}{1-zt}dt$$\nWe have\n$$\\int_0^\\infty\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\\int_0^\\infty\\frac{\\ln^{2n}(x)}{1+x^2}\\left(\\frac{1}{(2a)!}\\int_0^1\\frac{-x^2\\ln^{2a}(y)}{1+yx^2}dy\\right)dx$$\n$$=\\frac{1}{(2a)!}\\int_0^1\\ln^{2a}(y)\\left(\\int_0^\\infty\\frac{-x^2\\ln^{2n}(x)}{(1+x^2)(1+yx^2)}dx\\right)dy$$\n$$=\\frac{1}{(2a)!}\\int_0^1\\frac{\\ln^{2a}(y)}{1-y}\\left(\\int_0^\\infty\\frac{\\ln^{2n}(x)}{1+x^2}dx-\\int_0^\\infty\\frac{\\ln^{2n}(x)}{1+yx^2}dx\\right)dy$$\nuse $(a)$ and $(c)$ for the inner integrals\n$$=\\frac{\\left(\\frac{\\pi}{2}\\right)^{2n+1}}{(2a)!}\\left(|E_{2n}|\\int_0^1\\frac{\\ln^{2a}(y)}{1-y}dy-\\sum_{k=0}^n\\binom{2n}{2k}|E_{2n-2k}|\\pi^{-2k}\\int_0^1\\frac{\\ln^{2k+2a}(y)}{\\sqrt{y}(1-y)}dy\\right)$$\nThe proof completes on using:\n$$\\int_0^1\\frac{\\ln^a(x)}{1-x}dx=(-1)^aa!\\zeta(a+1)$$\n$$\\int_0^1\\frac{\\ln^a(x)}{\\sqrt{x}(1-x)}dx=(-1)^aa!(2^{a+1}-1)\\zeta(a+1)$$\n\nProof of (f): Following the same steps in proof $(e)$:\n$$\\int_0^\\infty\\frac{\\ln^{2n-1}(x)\\text{Li}_{2a}(-x^2)}{1+x^2}dx=\\int_0^\\infty\\frac{\\ln^{2n-1}(x)}{1+x^2}\\left(\\frac{1}{(2a-1)!}\\int_0^1\\frac{x^2\\ln^{2a-1}(y)}{1+yx^2}dy\\right)dx$$\n$$=\\frac{1}{(2a-1)!}\\int_0^1\\ln^{2a-1}(y)\\left(\\int_0^\\infty\\frac{x^2\\ln^{2n-1}(x)}{(1+x^2)(1+yx^2)}dx\\right)dy$$\n$$=\\frac{1}{(2a-1)!}\\int_0^1\\frac{\\ln^{2a-1}(y)}{1-y}\\left(\\int_0^\\infty\\frac{\\ln^{2n-1}(x)}{1+yx^2}dx-\\int_0^\\infty\\frac{\\ln^{2n-1}(x)}{1+x^2}dx\\right)dy$$\nsubstitute $(d)$ and notice that the second inner integral is zero\n$$=\\frac{-\\left(\\frac{\\pi}{2}\\right)^{2n-1}}{2(2a-1)!}\\sum_{k=0}^{n-1}\\binom{2n-1}{2k+1}|E_{2n-2k-2}|\\pi^{-2k}\\int_0^1\\frac{\\ln^{2k+2a}(y)}{\\sqrt{y}(1-y)}dy$$\nThe proof completes on using\n$$\\int_0^1\\frac{\\ln^a(x)}{\\sqrt{x}(1-x)}dx=(-1)^aa!(2^{a+1}-1)\\zeta(a+1)$$\n\nProof of (g):\n$$\\int_0^1\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\\int_0^\\infty\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx-\\underbrace{\\int_1^\\infty\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx}_{x\\to1/x}$$\n$$=\\int_0^\\infty\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx-\\int_0^1\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-1/x^2)}{1+x^2}dx$$\nadd the integral to both sides\n$$2\\int_0^1\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\\int_0^\\infty\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx$$\n$$+\\int_0^1\\frac{\\ln^{2n}(x)[\\text{Li}_{2a+1}(-x^2)-\\text{Li}_{2a+1}(-1/x^2)]}{1+x^2}dx$$\nthe first integral is given in $(e)$. For the second integral, replace $a$ by $2a+1$ in $(b)$ then use $\\sum_{k=m}^na_k=\\sum_{k=m}^{n}a_{n-k+m}$\n$$\\text{Li}_{2a+1}(-z)-\\text{Li}_{2a+1}(-1/z)=-2\\sum_{k=0}^a \\frac{\\eta(2a-2k)}{(2k+1)!}\\ln^{2k+1}(z)$$\nand so\n$$\\int_0^1\\frac{\\ln^{2n}(x)[\\text{Li}_{2a+1}(-x^2)-\\text{Li}_{2a+1}(-1/x^2)]}{1+x^2}dx=-4\\sum_{k=0}^a\\frac{\\eta(2a-2k)}{(2k+1)!}4^k\\int_0^1\\frac{\\ln^{2k+2n+1}(x)}{1+x^2}dx$$\nand the proof completes on using\n$$\\int_0^1\\frac{\\ln^a(x)}{1+x^2}dx=(-1)^a a! \\beta(a+1)$$\n\nProof of (h): We follow exactly the same steps in proof $(g)$ but here we use\n$$\\text{Li}_{2a}(-z)+\\text{Li}_{2a}(-1/z)=-2\\sum_{k=0}^a \\frac{\\eta(2a-2k)}{(2k)!}\\ln^{2k}(z)$$\n\nProof of (i) and (j): Using the generating function\n$$\\sum_{k=1}^\\infty H_k^{(a)} x^k=\\frac{\\text{Li}_a(x)}{1-x}$$\nand the fact that\n$$\\int_0^1 x^{k}\\ln^n(x)dx=\\frac{(-1)^nn!}{(k+1)^{n+1}}$$\nwe have:\n$$\\sum_{k=1}^\\infty\\frac{(-1)^k H^{(2a+1)}_k}{(2k+1)^{2n+1}}=\\frac1{(2n)!}\\int_0^1\\frac{\\ln^{2n}(x)\\text{Li}_{2a+1}(-x^2)}{1+x^2}dx$$\n$$\\sum_{k=1}^\\infty\\frac{(-1)^k H^{(2a)}_k}{(2k+1)^{2n}}=\\frac{-1}{(2n-1)!}\\int_0^1\\frac{\\ln^{2n-1}(x)\\text{Li}_{2a}(-x^2)}{1+x^2}dx$$\nThese two integrals are given in $(g)$ and $(h)$.\n\nEdit 3/26/2024\nMore generalized polylogarithmic integrals can be found in this preprint.\n", "comments": ["Very impressive and nice work !", "Marvelous job. A lot of work.", "Thanks for great effort in exploring to such an amazing depth!", "@AliShadhar Thank you so much for your comment. I meant significance in a broad sense... Have similar identities been applied in other areas of math (to give better approximation or better bounds, etc)?", "I'm sorry for my limited understanding, but can somebody elaborate on the significance of these identities (besides their novelty)?", "It might be interesting to know that these results have been submitted virtually literally by a group of authors to a scientific journal 25 days after the publication here and have been published without mentioning the source. I wrote an e-mail to the responsible author pointing out this violation of the scientific code of conduct. He replied finally, quote, \"I have e-mailed the editor in chief of the journal asking him to retract the paper.\"", "Oh yes, now these generalizations are known in the (online) literature !!", "(+1) Great work. Nice proofs, and nicely laid out, as you always do. I'll put a printout into my copy of your book.", "Nice results . We can find a recursive relationship using complex analysis", "Very impressive.", "(+1) Great pack of results to have in hand and refer to! :-)"], "comment_count": 0, "tags": ["integration", "sequences-and-series", "reference-request", "harmonic-numbers", "polylogarithm"], "creation_date": "2022-03-05T22:37:13", "diamond": 0, "votes": 40} {"question_id": "276", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2856060", "title": "If $K\\cong K(X)$ then must $K$ be a field of rational functions in infinitely many variables?", "body": "If $k$ is any field, then the field $K=k(X_0,X_1,\\dots)$ of rational functions in infinitely many variables satisfies $K(X)\\cong K$ (by mapping $X$ to $X_0$ and $X_n$ to $X_{n+1}$). My question is, does the converse hold? That is:\n\nSuppose $K$ is a field such that $K\\cong K(X)$. Must there exist a field $k$ such that $K\\cong k(X_0,X_1,\\dots)$?\n\n(If such a $k$ exists, then we can in fact take $k=K$, by splitting the variables into two infinite sets.)\nNote that if we were talking about polynomial rings instead of fields of rational functions, the answer would be no: there exists an integral domain $R$ such that $R\\cong R[X]$ but $R\\not\\cong S[X_0,X_1,\\dots]$ for any ring $S$. A counterexample is given at A ring isomorphic to its finite polynomial rings but not to its infinite one. However, for that example the field of fractions of $R$ actually is a field of rational functions in infinitely many variables, so it does not give a counterexample to this question.\nIn any case, I suspect the answer is no and it may be possible to find a counterexample using some idea similar to the example there (some sort of \"all but finitely many...\" construction), but don't have any concrete idea of how to make it work.\nSome other (unanswered) questions that may be related: Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$? (also on MO), The field of fractions of the rational group algebra of a torsion free abelian group. (In particular, an answer to the former question would give a field $F$ such that $F\\cong F(X,Y)$ but $F$ is not isomorphic to a field of rational functions in infinitely many variables, which is very close to this question.)\n", "comments": ["@JeremyRickard Nevermind, those are the same thing. And I see now why my idéa doesn't work.", "@JeremyRickard I thought that $K(X)$ denoted $\\text{Frac} (K[X])$, not field extension.", "@Kilian I don't see why $K$ would be isomorphic to $K(X)$. $\\mathbb{R}((X_0,X_1,\\dots))$ is not generated as a field extension of $\\mathbb{R}((X_1,X_2,\\dots))$ by $X_0$.", "Just an idea, haven't fully thought it through, but if we let $K=\\mathbb{R}((X_0,X_1...))$, i.e. the field of formal Laurent series in infinitely many variables over $\\mathbb{R}$, it seems like we would have $K\\cong K(X)$ but I don't see how it would be isomorphic to $k(X_0,X_1,...)$ for any field $k$. Possibly a counterexample?", "It’s an example of something. Either it’s a counterexample for your question, or the rational group algebras of the nonisomorphic groups $\\mathbb{Z}^\\omega$ and $\\mathbb{Z}^\\omega\\oplus\\mathbb{Z}^{(\\omega)}$ have isomorphic fields of fractions, answering my question.", "@JeremyRickard: That's a neat idea. I don't actually recall thinking through it before (I think the constructions I was trying were all countable).", "Do you know that the field of fractions of the rational group algebra of the Baer-Specker group is not a counterexample? Given the links in the question, I’m guessing you’ve thought about it.", "Sure. Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though. Feel free to crosspost it if you want.", "Might it be worth asking this interesting question on MathOverflow?"], "comment_count": 0, "tags": ["abstract-algebra", "field-theory"], "creation_date": "2018-07-18T15:05:43", "diamond": 0, "votes": 37} {"question_id": "277", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3014759", "title": "Constructing an infinite chain of subfields of 'hyper' algebraic numbers?", "body": "This has now been cross posted to MO. \nLet $F$ be a subset of $\\mathbb{R}$ and let $S_F$ denote the set of values which satisfy some generalized polynomial whose exponents and coefficients are drawn from $F$. \nThat is, we let $S_F$ denote\n$$\\bigg \\{x \\in \\mathbb{R}: 0=\\sum_{i=1}^n{a_i x^{e_i}}: e_i \\in F \\text{ distinct}, a_i\\in F \\text{ non-zero}, n\\in \\mathbb{N} \\bigg \\}$$ \nThen $S_{\\mathbb{\\mathbb{Q}}}$ is the set of algebraic real numbers and we start to see the beginnings of a chain: \n$\n\\mathbb{Q}\n\\subsetneq S_\\mathbb{Q} \\subsetneq\nS_{S_\\mathbb{Q}} $\nMain Question\n\nDoes this chain continue forever? That is, we let $A_0= \\mathbb{Q}$ and let $A_{n+1}=S_{A_{n}}$. Is it the case that $A_n \\subsetneq A_{n+1}$ for all $n\\in\\mathbb{N}$?\n\nOther curiosities: \nIs $A_i$ always a field? Perhaps, the argument is analogous to this. Or maybe this is just the case in a more general setting: Is it the case that $F \\subset \\mathbb{R}$, a field implies that $S_F$ is a field? \nIs it possible to see that $e\\notin \\cup A_i$? Perhaps this is just a tweaking of LW Theorem. \n", "comments": ["@JackM, $x=0$ satisfies this and is algebraic. I am not sure I understand your meaning... But if you take $x=y^6$ then we can replace your equation with $y^3+y^2=0$ and we can always manage to do this type of change when the exponents are rational.", "Why is $S_\\mathbb Q$ equal to the set of algebraic real numbers? For instance, is any solution to $\\sqrt x+\\sqrt[3]x=0$ algebraic?", "I am not so sure but maybe this piece by B Zilbner speaks to the question.", "Actually... $S_F$ is clearly closed under multiplicative inverses because $-e_i$ is in a field $F$ whenever $e_i$ is in $F$", "It seems like we might be able to achieve multiplicative inverses using a technique analogous to the one presented here. I haven't worked out all the details but it looks like it could be a good lead", "How do we prove that $4^\\sqrt3+1\\in\\cup A_n$? This seems like an argument for defining $S_F$ to include closure under field operations.", "I don’t think we can rule out $e\\in A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^\\sqrt6-4^\\sqrt3-1$.", "@JyrkiLahtonen, for this purpose I think we can define $(-|x|)^\\alpha$ as $-(|x|^\\alpha)$.", "Just a question: is $n$ (defined inside the definition of $S_F$ as a natural number) also arbitrary for any member of $S_F$?", "(continued) We should answer $e^{i \\pi \\sqrt{2}}$ but all of sudden it's start look like the question is better raised in the complex numbers. But maybe there is a different polynomial that satisfies $-x$?", "@JyrkiLahtonen. Just to be clear about the concern raised: $ x^{\\sqrt2}-2=0 \\implies x= \\sqrt{2}^{1/\\sqrt{2}}$ What (generalized) polynomial is satisfied by $-x$? It could be that $ (-x)^{\\sqrt2}-2=0$ is not a satisfactory answer. No. I would think this is indeed a satisfactory answer. The first polynomial has a non negative domain and the second has a non positive domain. What trouble do we get into if we take this naive approach? This route at least wins us closure under negation. Oh. The trouble we get into is when someone asks what the coefficient of the $x^{\\sqrt 2}$ term is...", "Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation?", "I'm still somewhat concerned about the definition of $x^\\alpha$ when $x<0$ and $\\alpha$ is not an integer. But, there is a cool question in here (possibly a very difficult one).", "$2^{\\sqrt2}$ is transcendental, but it satisfies $x^{\\sqrt2}-4=0$, so it's in $A_2$. I'd conjecture that $2^{2^{\\sqrt2}}$ is in $A_3$ and not $A_2$, that $2^{2^{2^{\\sqrt2}}}$ is in $A_4$ and not $A_3$, etcetera.", "How do you propose to define $x^\\alpha$ if $\\alpha \\notin \\mathbb{Q}$?", "Every set in the chain is countable, so you certainly don't hit $\\mathbb{R}$ at any point."], "comment_count": 0, "tags": ["real-analysis", "field-theory", "transcendental-numbers"], "creation_date": "2018-11-26T10:55:38", "diamond": 0, "votes": 36} {"question_id": "278", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4338032", "title": "An iterative logarithmic transformation of a power series", "body": "Consider the following iterative process. We start with the function having all $1$'s in its Taylor series expansion:\n$$f_0(x)=\\frac1{1-x}=1+x+x^2+x^3+x^4+O\\left(x^5\\right).\\tag1$$\nThen, at each step we apply the following transformation:\n$$f_{n+1}(x)=x^{-1}\\log\\left(\\frac{f_n(x)}{f_n(0)}\\right).\\tag2$$\nA few initial iterations give us:\n$$\n\\begin{array}{l}\nf_1(x)=1+\\frac{x}{2}+\\frac{x^2}{3}+\\frac{x^3}{4}+\\frac{x^4}{5}+O\\left(x^5\\right), \\\\\nf_2(x)=\\frac{1}{2}+\\frac{5 x}{24}+\\frac{x^2}{8}+\\frac{251 x^3}{2880}+\\frac{19 x^4}{288}+O\\left(x^5\\right), \\\\\nf_3(x)=\\frac{5}{12}+\\frac{47 x}{288}+\\frac{2443 x^2}{25920}+\\frac{5303 x^3}{82944}+\\frac{412589x^4}{8709120}+O\\left(x^5\\right),\n\\end{array}\\tag3\n$$\nand their initial terms form the following sequence:\n$$1,\\,\\frac{1}{2},\\,\\frac{5}{12},\\,\\frac{47}{120},\\,\\frac{12917}{33840},\\,\\frac{329458703}{874222560},\\,\\dots,\\tag4$$\nwhose denominators grow pretty quickly, but which appear to slowly converge to a value $$\\lambda\\stackrel{\\color{#aaaaaa}?}\\approx0.3678\\dots\\tag5$$\nIf we assume that the process with the iterative step $(2)$ converges to a fixed point, we can see that it must have a form:\n$$f_\\omega(x)=-x^{-1}\\,W(-c\\,x)=c+c^2 x+\\frac{3\\, c^3\\, x^2}{2}+\\frac{8\\, c^4\\, x^3}{3}+\\frac{125\\, c^5\\, x^4}{24}+O\\left(x^5\\right),\\tag6$$\nwhere $W(\\cdot)$ is the Lambert 𝑊-function, and $c$ is a coefficient that is not uniquely determined but depends on the choice of the initial function $f_0(x)$. In our case, $c=\\lambda$.\n\nQuestions: Does this process actually converge to a fixed point? If yes, then what is a closed-form expression (or another useful description) for $\\lambda$?\n\nUpdate: An explicit recurrent formula for the coefficients (the parenthesized superscript $m$ in $a_n^{\\small(m)}$ is just the second index of the coefficient; the sum $\\sum_{\\ell=1}^m$ is assumed to be $0$ when $m=0$):\n$$f_n(x)=\\sum_{m=0}^\\infty a_{n\\vphantom{+0}}^{\\small(m)}x^m,$$\nwhere\n$$a_{0\\vphantom{+0}}^{\\small(m)}=1,\\quad a_{n\\vphantom{+0}}^{\\small(m)}=\\frac1{\\,a_{n-1}^{\\small(0)}\\,}\\left(a_{n-1}^{\\small(m+1)}-\\frac1{m+1} \\sum_{\\ell=1}^m\\ell\\;a_{n\\vphantom{+0}}^{\\small(\\ell-1)}\\,a_{n-1}^{\\small(m-\\ell+1)}\\right).$$\nThe sequence of coefficients $(4)$ is $\\big\\{a_{n\\vphantom{+0}}^{\\small(0)}\\big\\}$.\n", "comments": ["He also gives the first terms of the next power series in the sequence. Then, he writes the first ever recorded insights on the formula for iteration of power series $f \\circ \\ldots \\circ f(x)$, which then insipired Schröder in his 1872 paper, in which he first considered, what now call, the \"Schröder equation\". This is one of the first paper on iteration theory, which unfortunately doesn't seem to apply to your case.", "Interestingly enough, Caylay briefly considered this problem in a note called \"On Some Numerical Expansions\" from the Quarterly Journal of Pure and Applied Mathematics, vol. III. (1860), pp. 366-369, but he didn't go much farther than you. What stroke me is that he got the same first few terms as you, but he was probably wrong for the last term \"$\\log\\left(-\\frac2x\\log\\left(\\frac1x\\log(1+x)\\right)\\right)=-\\frac5{12}x+\\frac{47}{288}x^2+\\frac{2443}{25920}x^3-\\frac{5303}{82944}x^4+\\frac{19631}{580608}x^5+\\&c$\".", "@VladimirReshetnikov It seems that, if $f_n(0)=a/b$ (in lowest terms), then the denominator of $f_{n+1}(0)$ is a small multiple ($2$ or $6$) of $ab$. Let $b_n$ be the denominator of $f_n(0)$. If $f_n(0)$ converges to a limit $\\lambda$, then this would imply that $b_{n+1}$ is multiplicatively not far off from $\\lambda b_n^2$, which gives double-exponential growth.", "Another very interesting question is how fast do denominators grow (or would grow if we ignored occasional cancellations of some of their factors with numerators). Empirically, they appear to grow (very roughly) doubly exponentially: $2^{2^{n-0.97942\\dots}}$.", "Originally I found this sequence when I tried to construct the continued exponent expansion of the Fibonacci numbers: pbs.twimg.com/media/FHAQfurVQAANsLx?format=png&name=large, but later realized I can get it simply from $1/(1-x)$.", "@mathworker21 Usually, OEIS keeps numerators and denominators as distinct sequences. This entry mixes them into one interleaved. Probably this is why I was not able to find the coefficients in OEIS after I computed a few of them.", "@mathworker21 Thanks! It would be nice to see a rigorous proof. I haven't been able to prove it myself so far.", "it says the limit is $1/e$ here.", "I concur with Claude and will go further saying this is a very nice problem in every respect.", "I think $\\lambda =1/e=0.367879441\\ldots$. A rough argument is as follows. It seems that each $f_n(x)$ is singular at $x=1$. Since the limit function is $-x^{-1}W(-\\lambda x)$ and $W(z)$ is singular at $z=-1/e$, $\\lambda$ should be $1/e$ in order for $f_\\omega(x)$ to be singular at $x=1$.", "The formula is correct. We assume that $\\sum_{\\ell=1}^0(\\dots)=0$, because there are no indices satisfying $1\\le\\ell\\le0$, so the sum is empty .", "is there a typo in your explicit formula? We are after $a_n^{(0)}$ and when I tried plug $m = 0,$ the sum starts at $l=1$ and ends at $l = m.$", "Whatever the answer could be, this is a nice problem.", "By the mean value theorem, taking $x\\to 0,$ you get $$f_{n+1}(0)=\\frac{f_n’(0)}{f_n(0)}.$$ Not sure if that helps in any way."], "comment_count": 0, "tags": ["limits", "logarithms", "power-series", "closed-form", "lambert-w"], "creation_date": "2021-12-19T19:33:11", "diamond": 0, "votes": 36} {"question_id": "279", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1081468", "title": "Smallest Subset of $\\mathbb{R}_{>0}$ Closed under Typical Operations", "body": "Let $S$ denote the smallest subset of $\\mathbb{R}_{>0}$ which includes $1$, and is closed under addition, multiplication, reciprocation, and the function $x,y \\in \\mathbb{R}_{>0} \\mapsto x^y.$\nQuestions: \n1. Do either $S$ or its elements have an accepted name?\n2. Where can I learn more about the set $S$ and it elements? (Reference Request)\n3. Do there exist positive, real algebraic numbers which are not in $S$?\n4. Are either $e$ or $\\pi$ elements of $S$?\nWhat I Already Know:\nBy the Gelfond-Schneider Theorem, $S$ includes some transcendental numbers, like $2^{\\sqrt{2}}$.\n", "comments": ["Could we build this set by transfinite recursion and then study its properties (like for the Borel Hierarchy)?", "@EricWofsey, if S is closed under subtraction it would include $2^\\sqrt{2}-1$, and that seems unlikely.", "S is a subset of the positive \"elementary-logarithmic\" or \"closed-form\" numbers, as defined by Chow. (timothychow.net/closedform.pdf) So his paper proves, assuming Schanuel's conjecture, that the positive roots of $2x^5-10x+5$ are not in S.", "@YoTengoUnLCD, in any partially ordered set $P$, every subset $A$ of $P$ has at most one minimum element (but may have many \"minimal\" elements). So let $P$ denote the poset $(\\mathcal{P}(\\mathbb{R}_{>0}),\\subseteq).$ Let $A$ denote those elements of $P$ that are closed under the relevant operations. Then $A$ has at most one least element. Regarding existence of a minimum element, it can be constructed as $\\bigcap A$.", "Just a question: what tells you that this characterization determines $S$? As inclusion is not a total order...", "I might add another question: is $S$ closed under subtraction (when the difference is positive)?", "@Jihad, I suppose that any incomputable would not belong to $S,$ and there exist incomputable numbers that can be defined explicitly. Although my knowledge of computability theory is sufficiently poor that I am not 100% certain of this statement. More interestingly perhaps, I have hunch that $e$ and $\\pi$ do not belong to $S$, and that there might exist a fifth-degree polynomial (or higher) over $\\mathbb{R}$ with a solution that does not belong to $S$.", "Do you know any number that does not belong to this set?", "@Jihad, it certainly is. That doesn't tell us much, though, since the set of algebraic numbers is also countable.", "Seems like this set should be countable."], "comment_count": 0, "tags": ["number-theory", "reference-request", "algebraic-number-theory", "real-numbers", "transcendental-numbers"], "creation_date": "2014-12-26T06:48:47", "diamond": 0, "votes": 36} {"question_id": "280", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/407484", "title": "Determining the kernel of a Vandermonde-like matrix", "body": "The kernel of a Vandermonde matrix can be determined using this formula.\nThe following type of matrix has a similar structure, and should also have a one-dimensional kernel. \n$$V= \n\\begin{bmatrix} \n1 & 1 & 1 & \\ldots & 1 \\\\\nx_1 & x_2 & x_3 & \\ldots & x_n \\\\\nx_1^2 & x_2^2 & x_3^2 & \\ldots & x_n^2 \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\nx_1^{m-1} & x_2^{m-1} & x_3^{m-1} & \\ldots & x_n^{m-1}\\\\\ny_1 & y_2 & y_3 & \\ldots & y_n \\\\\ny_1x_1 & y_2x_2 & y_3x_3 & \\ldots & y_nx_n \\\\\ny_1x_1^2 & y_2x_2^2 & y_3x_3^2 & \\ldots & y_nx_n^2 \\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\ny_1x_1^{m-1} & y_2x_2^{m-1} & y_3x_3^{m-1} & \\ldots & y_nx_n^{m-1}\\\\\ny_1^2x_1 & y_2^2x_2 & y_3^2x_3 & \\ldots & y_n^2x_n\\\\\n\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\ny_1^{m-1}x_1^{m-1} & y_2^{m-1}x_2^{m-1} & y_3^{m-1}x_3^{m-1} & \\ldots & y_n^{m-1}x_n^{m-1}\\\\\n\\end{bmatrix} \\in \\mathbb{R}^{(n-1)\\times n}$$\nwhere $n = m^2+1$ and $(x_i, y_i) \\neq (x_j, y_j)$ for $i \\neq j$; i.e. there are $m$ groups of $m$ rows with all possible combinations of powers $y^ax^b$ and one more column than rows.\n\nDoes a similar analytical form exist for it? Or, would additional constraints be required, like $x_i^ay_i^b \\neq x_i^cy_i^d$ for $i \\neq j$?\n\n(Crossposted to MO)\n", "comments": ["Wouldn't there be an additional row of $y_1^2 \\cdots y_n^2$ before the row $y_1^2x_1 \\cdots y_n^2x_n$?", "Assuming you're in the complex field, what you're asking for is exactly which are the two-variable polynomials in $\\mathbb{C}[x,y]$ with degree $m-1$ in both $x$ and $y$, such that they are zero on a given set of points $\\{ (x_1,y_1), (x_2,y_2) , ... (x_n, y_n)\\}$. This is just Lagrange interpolation.", "Questions: have you tried using Lagrange interpolation, as in the regular proof of the kernel of the van der monde matrix? Also, have you tried getting people on mathematica.stackexchange to experiment with different $m$?", "Edited the question to make it clearer.", "Ah, I see now. Nevermind.", "There are $m$ groups of $m$ rows (All possible combinations of powers $x^iy^j$); the matrix should have one more column than rows to give a 1-dimensional kernel, therefore $n=m^2+1$.", "Shouldn't $n=2m+1$ rather than $n=m^2+1$?"], "comment_count": 0, "tags": ["linear-algebra", "matrices", "determinant"], "creation_date": "2013-05-31T00:32:53", "diamond": 0, "votes": 35} {"question_id": "281", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4353238", "title": "Is the power of $2$ in the Euclidean norm related to the fact that the algebraic closure of the reals is $2$-dimensional?", "body": "Consider any local field $K$, endowed with its topological field structure. We define the function $| \\cdot | : K \\to \\mathbb{R}_{\\ge 0}$ as\n$$|x| = \\frac{\\mu(xS)}{\\mu(S)},$$\nwhere $\\mu$ is any Haar measure (which exists because a local field is additively a locally compact group), and $S$ is any measurable set of nonzero measure (see e.g. Definition 12.28 here). This definition can be shown to be independent of $\\mu$ and $S$.\nNow consider a vector space $V \\simeq K^n$ over $K$, and equip it with the $p$-length function $$\\|(x_1,x_2,\\ldots,x_n)\\|_p = (|x_1|^p+|x_2|^p+\\ldots+|x_n|^p)^{1/p}$$ for some $p \\in [1,\\infty]$ (where as usual we define $\\| \\cdot \\|_\\infty$ as the limit of $\\| \\cdot \\|_p$ as $p\\to \\infty$). There is a unique critical value $p=p^*(K)$ for which this length function becomes especially symmetric, in that its associated group of isometries (i.e., the subgroup of $GL(V)$ that preserves $\\| \\cdot \\|_{p^*(K)}$) acts transitively on the corresponding unit sphere. The amazing fact is that this critical value $p^*(K)$ always coincides with the degree $[\\bar{K}:K]$ of the algebraic closure of $K$!\nHere is a proof sketch:\nFrom the classification of local fields, it is known${}^{(\\dagger)}$ that $K$ is either $\\mathbb{R}$, $\\mathbb{C}$ or a non-Archimedean local field (that is, a finite extension of either $\\mathbb{Q}_P$ or $\\mathbb{F}_P((t))$ for some prime $P$).\n\nIf $K=\\mathbb{R}$, $|\\cdot|$ is the usual absolute value, and the critical length function turns out to be $\\| \\cdot \\|_2$ (the usual Euclidean length), invariant under $O(n)$. On the other hand, $[\\bar{\\mathbb{R}}:\\mathbb{R}]=[\\mathbb{C}:\\mathbb{R}]=2$.\n\nIf $K=\\mathbb{C}$, we have $|x|=\\bar{x} x$, and the corresponding critical length function is $\\| \\cdot \\|_1$, invariant under $U(n)$.${}^{(\\ddagger)}$ On the other hand, $[\\bar{\\mathbb{C}}:\\mathbb{C}]=[\\mathbb{C}:\\mathbb{C}]=1$. Note that neither $|\\cdot|$ nor $\\|\\cdot\\|$ satisfy the triangle inequality in this particular case (this is why I avoided using the terms absolute value and norm throughout the question).\n\nIf $K$ is non-Archimedean, $|\\cdot|$ is a suitably normalized $\\chi$-adic absolute value for some $\\chi$, and the corresponding critical length function is the supremum norm $\\| \\cdot \\|_\\infty$, which can be shown to be invariant under $GL(\\mathcal{O}_K^n)$, where $\\mathcal{O}_K$ is the ring of integers of $K$. On the other hand, it is known that the algebraic closure of $K$ has infinite degree, i.e. $[\\bar{K}:K]=\\infty$.\n\n\nThus in all cases we have $p^*(K) = [\\bar{K}:K]$.\nA problem with this proof is that it gives no indication as to why the relationship holds: after all, the values could just happen to be the same by complete coincidence. For that reason, I am wondering if there exists an alternative proof of this fact that is \"$K$-agnostic\", i.e. a proof that does not use at any point the Artin-Schreier theorem, nor the classification of local fields, nor any properties from a specific such field (such as the existence of a symmetric/Hermitian inner product) other than the degree of an algebraic closure and properties derived from it.\nSuch a proof would allow one to meaningfully speak of a relationship between those two quantities, and to state things like \"in a field with a $3$-dimensional algebraic closure, the natural analogue of the Pythagorean theorem would be $|a|^3+|b|^3=|c|^3$\" without appealing to the principle of explosion. I know it is hard in general to formally determine whether a proof does or does not use a fact, hence why I'm labeling this with the soft-question tag, but hopefully the kind of proof I want is clear, at least informally.\n\nIn summary, my question is:\nGiven a local field $K$, with $| \\cdot |$ defined as above, is there any $K$-agnostic proof that the group of isometries of $\\|\\cdot\\|_p$ (defined in terms of $| \\cdot |$ as above) acts transitively on unit spheres if and only if $p = [\\bar{K}:K]$?\nIt seems reasonable to try proving first that $p^*(K) = 1$ iff $K$ is algebraically closed. That is where I am currently stuck. The thing that makes linear algebra over an algebraically closed field \"special\" is that every linear transformation has an eigenvalue, but I am not sure how to apply that fact to show that the critical length function is $\\|\\cdot\\|_1$.\nUPDATE: As suggested by Torsten Schoeneberg in the comments, perhaps it will be easier to prove the relationship $p^*(K) = [L:K]\\: p^*(L)$ for a finite field extension $L/K$. Consider such an extension with $[L:K]=n$, and suppose we know $L$ has a critical $p$-length function for some $p$. Now, $L \\simeq K^n$ is additively a vector space over $K$, which is equipped with the product topology. This means that, if we identify $k\\in K $ with its image under the inclusion $K \\subseteq L$, we have\n$$|k|_L = \\frac{\\mu_L(k S^n)}{\\mu_L(S^n)} = \\left(\\frac{\\mu_K(k S)}{\\mu_K(S)}\\right)^n = |k|_K^n.$$\nOn the other hand, note that the subgroup of $GL_1(L) = L^\\times$ preserving $|\\cdot|_L$, i.e. the group of invertible elements of norm $1$, acts transitively on the associated unit sphere (which coincides with the group itself) in a tautological way, since any element $x$ of norm $1$ can be sent to any other element $y$ of norm $1$ through multiplication by $x^{-1}y$. Since $GL_1(L) \\subseteq GL_n(K)$, this proves that the length function on $K^n$ defined through the identification $L \\simeq K^n$ as $\\|\\cdot\\| = |\\cdot|_L^{1/n}$ is a critical length function, that moreover satisfies the homogeneity property $\\|kv\\|=|k|_L \\|v\\|$ (which follows by the above relationship between the Haar measures and multiplicativity of $|\\cdot|$). In a similar way, from the inclusion $GL_m(L) \\subseteq GL_{mn}(K)$ we can prove that any critical length function $||\\cdot||$ on $L^m$ induces a corresponding critical length function on $K^{mn}$.\nThe only ingredient left would be to prove is that there always exists some basis $\\{1=a_0, a_1, \\ldots, a_{n-1}\\}$ of $L$ such that the above defined length functions are of the required form, that is, $|\\sum_{i=0}^{n-1} k_i a_i|_L = \\sum_{i=0}^{n-1} |k_i|_K^n$. Once we have that, we can show using isotropy subgroups that the length function induced from $L$ is critical in any dimension, not just multiples of $n$, so that $p^*(K) = [L:K]\\: p^*(L)$. However, I don't know how to prove the existence of such a basis in a $K$-agnostic way.\n\nEDIT: A possible way to restate the problem is by noticing that for any $(v,w)$ in the direct sum $V \\oplus W$, we have $\\|(v,w)\\|^p_p = \\|v\\|^p_p + \\|w\\|^p_p$, i.e. the $p$th power of $\\|\\cdot\\|_p$ acts additively on \"independent\" vectors (for finite $p$, of course).\nGiven a length function on vector spaces $V$, we can define the associated Gaussian function as an integrable function $g : V \\to \\mathbb{R}_{\\ge 0}$ depending only on length (i.e. factorizing as the composition $f \\circ \\|\\cdot\\|_p : V \\to \\mathbb{R}_{\\ge 0} \\to \\mathbb{R}_{\\ge 0}$ for some $f$), and satisfying the independence property\n$$g( (v,w) ) = g(v) g(w).$$\nWith the above choice of maximally symmetric length functions, using the aforementioned additivity property immediately leads to the usual Gaussian function $\\exp(-k \\|x\\|_2^2)$ for $\\mathbb{R}$ and to $\\exp(-k' \\|x\\|_1)$ for $\\mathbb{C}$ (where the constants $k, k' >0$ depend on the normalization convention). For $p=\\infty$, this reasoning can't be used directly, but if we interpret it as a limit like before, we do recover the standard Gaussian function for non-Archimedean fields\n$$\\lim_{p\\to \\infty} \\exp(-k'' \\|x\\|_p^p) = \\begin{cases} 1 & \\|x\\|_p \\le 1 \\\\ 0 & \\text{otherwise} \\end{cases} = \\mathbf{1}_\\mathcal{O_K^n}(x),$$\nwhich is the indicator function of $\\mathcal{O}_K^n$ in $V=K^n$. Focusing on the $1$-dimensional case $V=K$, these Gaussian functions coincide with the standard ones as used e.g. in Tate's thesis.\nSo if I'm not mistaken, the problem can be restated as proving that $[\\bar{K} : K] = p$ if and only if the standard Gaussian function associated to $K$ is $\\exp(-k \\|x\\|_p^p)$ (with a limit intended if $p=\\infty$). A possible advantage of this restatement is that standard Gaussian functions have different characterizations that do not previously assume any length function (for example, they are their own Fourier transform, or satisfy analogues of the central limit theorem, see e.g. here).\n\n$(\\dagger)$ Note that by the Artin-Schreier theorem, the only degrees $[\\bar{K}:K]$ that could possibly occur for any field $K$ are $1, 2, \\infty$ (this, in turn, seems to be related to the fact that a nonzero integer can only have multiplicative order $1, 2$ or $\\infty$, but that is a topic for another question).\n$(\\ddagger)$ As a fun fact, this can be shown to imply that the hypervolume of the $n$-dimensional complex unit ball $|x_1|+|x_2|+\\ldots+|x_n| = 1$ , i.e. the ordinary $2n$-dimensional unit ball, equals $V(B_{2n})=V(B_{2})^n /n!=\\pi^n /n!$; compare with the hypervolume $V(X_{n})=V(X_{1})^n /n!=2^n /n!$ of the $n$-dimensional cross-polytope $|x_1|+|x_2|+\\ldots+|x_n| = 1$, where $| \\cdot |$ now denotes the real absolute value.\n", "comments": ["The question has been posted on MathOverflow.", "I also suggest asking on MO, as Nolord suggested. You can repost it there and link the questions to each other (for example, with a brief heading saying \"Cross-posted to/from MO/MSE.\").", "@user76284 Edited, thanks for the suggestion!", "Minor suggestion: You can typeset double vertical bars as \\|. This shows up as $\\|x\\|$ rather than $||x||$ (notice the lack of excess space between the bars).", "I guess in the last paragraph should be $|x_1|+|x_2|+\\ldots+|x_n| {\\color{red}\\le}1 $ instead of $|x_1|+|x_2|+\\ldots+|x_n| = 1$.", "This should go on MathOverflow!", "@TorstenSchoeneberg That seems like a good idea, thank you! I've updated the question with some of my thoughts about it.", "Well, maybe there is a \"field-agnostic\"/\"Artin-Schreier-less\" proof that if $L \\vert K$ is a finite dimensional field extension, $p^*(K) =[L:K] \\cdot p^*(L)$. (Or show it at least for some, e.g. cyclotomic, extensions. But as you say it actually holds even for skew field extensions.) Then we would be done in the non-archimedean case once we show further that the possible $p^*$'s are bounded away from zero.", "@TorstenSchoeneberg As I mention in a footnote, the fact that the multiplicative order of $n\\in\\mathbb{Z}$ (and thus the order of this generator) can only be $1, 2, \\infty$ seems to be the \"source\" of many (if not all?) appearances of the trichotomy in the theory.", "@TorstenSchoeneberg Yes, that trichotomy is realized as $\\mathrm{trivial}, \\mathbb{Z}_2, \\mathbb{Z}$ in the Brauer groups of local fields. There are many things to say about this which I want to save for my next question, but here is something I can't resist mentioning: for all local $K$, the topological generator of the Galois group of $K^{ur}$ acts as $\\omega \\mapsto \\omega^n$ on roots of unity for some $n$. In the nonarchimedean case this is Frobenius, and $n$ is a prime power; for $\\mathbb{R}$ it is complex conjugation with $n=-1$, and for $\\mathbb{C}$ it is the identity ($n=1$).", "@TorstenSchoeneberg Good questions! For the quaternions one can mimic the construction of $|\\cdot |$, with the slightly weird result that the critical length function invariant under $USp(n)$ looks like $||(x_1,\\ldots,x_n)||_{1/2} = (\\sqrt{|x_1|}+\\ldots+\\sqrt{|x_n|})^2$. I haven't checked it but I think the same should happen for the octonions with $||\\cdot ||_{1/4}$ invariant under several exceptional groups, with the usual caveat that octonionic vector spaces are only well-behaved for $n\\le 3$.", "By the way, what happens over the quaternions $\\mathbb H$? --- Superficially related, the trichotomy $1,2,\\infty$ also matches the cardinality of the Brauer group $Br(k)$ for the respective local fields $k$, right?", "I have a feeling that at least in the two archimedean cases, this is actually related to the fact that what we have decided on as codomain for measures, $\\mathbb R_{\\ge 0}$, is made up of the squares in $(\\mathbb R, \\cdot)$, or the norms from $\\mathbb C \\vert \\mathbb R$. --- Anyway, maybe one can get insight by checking what happens in a non-local case like some cubic or higher extension of $\\mathbb Q$, or $\\mathbb Q(i)$, with restricting the possible Haar measures (they still exist, just not unique) to those."], "comment_count": 0, "tags": ["linear-algebra", "number-theory", "measure-theory", "soft-question", "local-field"], "creation_date": "2022-01-10T04:38:48", "diamond": 0, "votes": 34} {"question_id": "282", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4472850", "title": "Can a 4D spacecraft, with just a single rigid thruster, achieve any rotational velocity?", "body": "It seems preposterous at first glance. I just want to be sure. Even in 3D the behaviour of rotating objects can be surprising (see the Dzhanibekov effect); in 4D it could be more surprising.\nA 2D or 3D spacecraft (with no reaction wheels or gimbaling etc.) needs at least two thrusters to control its spin. See my answer on space.SE.\n\nFor any two $4\\times4$ matrices $X$ and $Y$, define the commutator $[X,Y]=XY-YX$, the anticommutator $\\{X,Y\\}=XY+YX$, and the Frobenius inner product $\\langle X,Y\\rangle=\\operatorname{tr}(X^\\top Y)$, where $\\operatorname{tr}$ is the trace and $\\top$ is the transpose.\nLet $M$ be a symmetric positive-definite $4\\times4$ matrix, and $T=\\mathbf f\\,\\mathbf r^\\top-\\mathbf r\\,\\mathbf f^\\top$ an antisymmetric $4\\times4$ matrix. ($M$ describes the distribution of mass in the spacecraft, $\\mathbf r$ is a vector locating the thruster relative to the centre of mass, $\\mathbf f$ is the force produced by the thruster, and $T$ is the torque produced by the thruster. More details here. Everything is described in the rotating reference frame.)\nThe angular velocity $\\Omega(t)$, an antisymmetric $4\\times4$ matrix, changes with time $t$ according to Euler's equation\n$$\\{M,\\Omega'(t)\\}+[\\Omega(t),\\{M,\\Omega(t)\\}]=f(t)\\,T$$\nwhere $f(t)\\geq0$ is a function (continuous, piecewise-constant, or just integrable) describing when and how strongly the thruster is used.\nQuestion: Can $M$ and $T$ be chosen such that, for any two antisymmetric $4\\times4$ matrices $\\Omega_0$ and $\\Omega_1$, there exist $t_1>0$ and $f$ such that the solution $\\Omega$ to Euler's equation with initial value $\\Omega(0)=\\Omega_0$ has final value $\\Omega(t_1)=\\Omega_1$?\nTo simplify things, we may take $f$ to be a combination of Dirac deltas instead of an ordinary function. Then the angular velocity satisfies $\\{M,\\Omega'(t)\\}+[\\Omega(t),\\{M,\\Omega(t)\\}]=0$ except at a finite set of times when the angular momentum $L(t)=\\{M,\\Omega(t)\\}$ changes by a positive multiple of $T$, or equivalently when $\\Omega(t)$ changes by a positive multiple of $A=\\{M,\\}^{-1}(T)$, where $A$ is the angular acceleration produced by the thruster.\n\nHere is Euler's equation in terms of the components $\\omega_{ij}$ of the angular velocity. Assume that $M$ is diagonal, with components $m_i>0$.\n$$(m_1+m_2)\\,\\omega_{12}'(t)+(m_2-m_1)\\big(\\omega_{13}(t)\\,\\omega_{32}(t)+\\omega_{14}(t)\\,\\omega_{42}(t)\\big)=f(t)\\,\\tau_{12}$$\n(And permute the indices to get 6 equations like this.) Equivalently:\n$$\\omega_{12}'(t)=\\frac{m_1-m_2}{m_1+m_2}\\big(\\omega_{13}(t)\\,\\omega_{32}(t)+\\omega_{14}(t)\\,\\omega_{42}(t)\\big)+f(t)\\,\\alpha_{12}$$\nIf $m_1=m_2$, then $\\omega_{12}'$ has constant sign; $\\omega_{12}$ is either always non-increasing, or always non-decreasing. So, if the difference between initial and final values of $\\omega_{12}$ has the wrong sign compared to $\\tau_{12}$, then there is no solution. Thus, we must take $m_1\\neq m_2$, and similarly $m_i\\neq m_j$ for $i\\neq j$.\n\nThe Frobenius inner product of two antisymmetric matrices has every term duplicated: $\\sum_{i,j}x_{ij}y_{ij}=2\\sum_{im_2>m_3>m_4>0$. The doubled energy is $\\sum_i(m_i-m_{i+1})P_i$, and the momentum's squared magnitude is $\\sum_i(m_i^2-m_{i+1}^2)P_i$. But what we should focus on are the quadratic forms $N_i=P_{i-1}-P_i+P_{i+1}$ which are not positive-semidefinite but have a single negative term:\n$$N_1=\\frac{m_2+m_3}{m_2-m_3}\\omega_{23}^2+\\frac{m_2+m_4}{m_2-m_4}\\omega_{24}^2-\\frac{m_1+m_2}{m_1-m_2}\\omega_{12}^2$$\n$$N_2=\\frac{m_1+m_2}{m_1-m_2}\\omega_{12}^2+\\frac{m_1+m_4}{m_1-m_4}\\omega_{14}^2+\\frac{m_3+m_4}{m_3-m_4}\\omega_{34}^2-\\frac{m_2+m_3}{m_2-m_3}\\omega_{23}^2$$\n$$N_3=\\frac{m_1+m_3}{m_1-m_3}\\omega_{13}^2+\\frac{m_2+m_3}{m_2-m_3}\\omega_{23}^2-\\frac{m_3+m_4}{m_3-m_4}\\omega_{34}^2$$\nThese are all constant. The first expression (if it's equated to $0$) defines a cone, or a pair of opposite cones, around the $\\omega_{12}$ axis in 6D. The second expression defines a pair of opposite cones around the $\\omega_{23}$ axis; and the third, around the $\\omega_{34}$ axis. If the applied acceleration $A$ is inside one of these cones, and the initial angular velocity $\\Omega_0$ is in the same cone, then $\\Omega$ stays in that cone, and there's no solution with the final angular velocity $\\Omega_1$ outside of the cone. Thus, we must take $A$ to be outside of all six of these cones. That is, all three expressions (with $\\alpha_{ij}$ in place of $\\omega_{ij}$) must be positive.\n\nNotice that Euler's equation has a time symmetry: Given a solution $\\Omega(t)$, we can construct another solution $\\tilde\\Omega(t)=-\\Omega(t_1-t)$, so that $\\{M,\\tilde\\Omega'(t)\\}+[\\tilde\\Omega(t),\\{M,\\tilde\\Omega(t)\\}]=f(t_1-t)\\,T$, and the initial and final values are swapped (and negated): $\\tilde\\Omega(0)=-\\Omega_1$ and $\\tilde\\Omega(t_1)=-\\Omega_0$. Therefore, any angular velocity can be reached from any other, if and only if any angular velocity can be reached from $0$. (We can reverse a path from $0$ to $-\\Omega_0$ to get a path from $\\Omega_0$ to $0$, and concatenate that with a path from $0$ to $\\Omega_1$, to get a path from $\\Omega_0$ to $\\Omega_1$.)\nAlso, Euler's equation has a scale symmetry: Given a solution $\\Omega(t)$, for any $k>0$ we can construct another solution $\\tilde\\Omega(t)=k\\,\\Omega(kt)$, so that $\\{M,\\tilde\\Omega'(t)\\}+[\\tilde\\Omega(t),\\{M,\\tilde\\Omega(t)\\}]=k^2f(kt)\\,T$, and the final value is $\\tilde\\Omega(t_1/k)=k\\,\\Omega(t_1)=k\\,\\Omega_1$. Therefore, any angular velocity can be reached from $0$, if and only if any angular velocity in a small neighbourhood of $0$ can be reached from $0$.\n(It looks like these two symmetries are related by $k=-1$, but we should keep a distinction between positive time and negative time.)\n\nThis has the form of a quadratic differential equation, $\\mathbf x'(t)=\\mathbf x(t)\\odot\\mathbf x(t)$ where $\\odot$ is some bilinear function. (Specifically, for antisymmetric matrices $X$ and $Y$, define $X\\odot Y=-\\{M,\\}^{-1}([X,\\{M,Y\\}])$, so Euler's equation is $\\Omega'=\\Omega\\odot\\Omega$, as long as $f=0$. Alternatively, define $X\\odot Y=-[\\{M,\\}^{-1}(X),Y]$, so Euler's equation is $L'=L\\odot L$.)\n(I ought to move this power series stuff to a different page, since it's applicable far beyond this particular Question, and anyway it's bloating this page.)\nFix a norm on the space, and find some constant $\\lVert\\odot\\rVert>0$ such that $\\lVert\\mathbf x\\odot\\mathbf y\\rVert\\leq\\lVert\\odot\\rVert\\lVert\\mathbf x\\rVert\\lVert\\mathbf y\\rVert$ for all $\\mathbf x,\\mathbf y$.\nGiven initial value $\\mathbf x(0)=\\mathbf a$, the equation $\\mathbf x'=\\mathbf x\\odot\\mathbf x$ has the power series solution\n$$\\mathbf x(t)=\\sum_{n=0}^\\infty t^n\\frac{\\sum^{n!}\\mathbf a^{n+1}}{\\sum^{n!}1}$$\nwhere the coefficient of $t^n$ is the average of all possible ways of evaluating $\\mathbf a^{n+1}$ using the product $\\odot$. For example, the coefficient of $t^3$ is $\\tfrac16$ of\n$$\\sum^6\\mathbf a^4=((\\mathbf a\\mathbf a)\\mathbf a)\\mathbf a+(\\mathbf a(\\mathbf a\\mathbf a))\\mathbf a+2(\\mathbf a\\mathbf a)(\\mathbf a\\mathbf a)+\\mathbf a((\\mathbf a\\mathbf a)\\mathbf a)+\\mathbf a(\\mathbf a(\\mathbf a\\mathbf a))$$\n$$\\newcommand{\\aaaa}[3]{\\mathbf a\\underset{#1}\\odot\\mathbf a\\underset{#2}\\odot\\mathbf a\\underset{#3}\\odot\\mathbf a} =\\aaaa{1}{2}{3}+\\aaaa{2}{1}{3}\\begin{matrix}{}+\\aaaa{1}{3}{2} \\\\ {}+\\aaaa{2}{3}{1}\\end{matrix}+\\aaaa{3}{1}{2}+\\aaaa{3}{2}{1}.$$\nThe terms in the series can be obtained recursively:\n$$\\mathbf u_n=\\frac{t^n}{n!}\\sum^{n!}\\mathbf a^{n+1}$$\n$$=\\frac tn\\sum_{k=0}^{n-1}\\mathbf u_{n-1-k}\\odot\\mathbf u_k.$$\nWe also have $\\lVert\\mathbf u_n\\rVert\\leq|t|^n\\lVert\\odot\\rVert^n\\lVert\\mathbf a\\rVert^{n+1}$, which ensures absolute convergence, for small $t$:\n$$\\sum_{n=0}^\\infty\\left\\lVert t^n\\frac{\\sum^{n!}\\mathbf a^{n+1}}{\\sum^{n!}1}\\right\\rVert\\leq\\sum_{n=0}^\\infty|t|^n\\lVert\\odot\\rVert^n\\lVert\\mathbf a\\rVert^{n+1}=\\frac{\\lVert\\mathbf a\\rVert}{1-|t|\\lVert\\odot\\rVert\\lVert\\mathbf a\\rVert},$$\n$$|t|<\\frac{1}{\\lVert\\odot\\rVert\\lVert\\mathbf a\\rVert}.$$\nAnd at certain times (when the thruster is used) an impulse may be applied to $\\mathbf x$, with a fixed direction $\\mathbf b$ but an arbitrary magnitude $c>0$, thus: $\\mathbf x(t+)=\\mathbf x(t-)+c\\mathbf b$. These impulses, and the time intervals between them, are many variables that we can control. If the number of variables is at least $6$ (or the dimension of the space $\\mathbf x$ is in), then there is hope of surrounding $0$ in an open set, and thus reaching everywhere in the space (according to the previous section).\n$$\\mathbf x(0+)=0+c_0\\mathbf b$$\n$$\\mathbf x(t_1-)=\\sum_{n=0}^\\infty\\frac{t_1^n}{n!}\\sum^{n!}\\mathbf x(0+)^{n+1}$$\n$$\\mathbf x(t_1+)=\\mathbf x(t_1-)+c_1\\mathbf b$$\n$$\\mathbf x(t_1+t_2-)=\\sum_{n=0}^\\infty\\frac{t_2^n}{n!}\\sum^{n!}\\mathbf x(t_1+)^{n+1}$$\n$$\\mathbf x(t_1+t_2+)=\\mathbf x(t_1+t_2-)+c_2\\mathbf b$$\n$$\\mathbf x(t_1+t_2+t_3-)=\\sum_{n=0}^\\infty\\frac{t_3^n}{n!}\\sum^{n!}\\mathbf x(t_1+t_2+)^{n+1}$$\n$$c_0,t_1,c_1,t_2,c_2,t_3\\geq0$$\n$$\\mathbf x(t_1+t_2+t_3-)\\approx0\\quad?$$\n", "comments": ["Cross-posted to MO: mathoverflow.net/questions/448773/…", "Oh, I didn't explain what $M$ is (though the linked physics.SE post did explain it). It's the matrix of second moments of mass: $M=\\int\\mathbf r\\,\\mathbf r^T\\,\\rho\\,dV$, where $\\mathbf r$ is position relative to the centre of mass, $\\rho$ is density, and $V$ is volume. But that's not necessary for understanding this Question.", "Given that $M$ is diagonal (which is no loss of generality, since any symmetric matrix has an orthogonal eigenbasis), the inertia tensor is also diagonal, with components $m_1+m_2$, $m_1+m_3$, $m_2+m_3$, $m_1+m_4$, $m_2+m_4$, $m_3+m_4$.", "And of course I'm considering bivectors as antisymmetric matrices.", "The inertia tensor is the function $\\Omega\\mapsto L=\\{M,\\Omega\\}=M\\Omega+\\Omega M$.", "What's the relationship of $M$ with the inertia tensor? My understanding was that the inertia tensor is a linear transformation on bivectors, which in this case are 6-dimensional, so it would be a symmetric 6x6 matrix. Is that not correct?"], "comment_count": 0, "tags": ["ordinary-differential-equations", "analysis", "dynamical-systems", "physics"], "creation_date": "2022-06-14T18:13:22", "diamond": 0, "votes": 34} {"question_id": "283", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/813676", "title": "Semirings induced by symmetric monoidal categories with finite coproducts", "body": "A symmetric monoidal category with finite coproducts is by definition a symmetric monoidal category $(\\mathcal{C},\\otimes,1,\\dotsc)$ such that the underlying category $\\mathcal{C}$ has finite coproducts (this includes an initial object $0$) with the property that $\\otimes$ preserves finite coproducts in each variable. In other words, we have a distributive law\n$$X \\otimes (\\bigoplus_{i \\in I} Y_i) \\cong \\bigoplus_{i \\in I} (X \\otimes Y_i).$$\nThis is also known as a finitary distributive symmetric monoidal category. Notice that to each symmetric monoidal category with finite coproducts we may associate a commutative semiring: The elements are the isomorphism classes of objects in $\\mathcal{C}$. The addition is $[X] + [Y] := [X \\oplus Y]$, the multiplication is $[X] \\cdot [Y] := [X \\otimes Y]$. The zero element is $[0]$, the multiplicative unit is $[1]$. To avoid set-theoretic issues, let us assume that $\\mathcal{C}$ is essentially small. As a consequence, we may think of symmetric monoidal categories with finite coproducts as a possible categorification of semirings.\nA familar example is the symmetric monoidal category of vector bundles on a space $X$, which produces the semiring of isomorphism classes of vector bundles on $X$. If $X$ is a point, we see that the semiring associated to $\\mathsf{FinSet}$ is $\\mathbb{N}$.\nNote, however, that not every commutative semiring arises from the construction above. Basically this is because we have agreed that the addition is not induced by some symmetric monoidal structure which distributes over $\\otimes$, but rather from the coproduct. The coproduct satisfies $X \\oplus Y \\cong 0 \\Rightarrow X \\cong Y \\cong 0$. It follows that in the associated semiring only $0$ has an additive inverse.\nQuestion. How can we classify those commutative semirings which are induced by essentially small symmetric monoidal categories with finite coproducts?\nEvery commutative monoid with the property $a+b=0 \\Rightarrow a=b=0$ arises from an essentially small category with finite coproducts, see SE/834869. But it is not clear if we can use this here.\n", "comments": ["@Arducode This does not define a functor for $\\otimes$. Or how do you define the tensor product of morphisms? Also notice that it is required that the tensor product is distributive - this is not guaranteed by arbitrary isomorphisms, they have to be the canonical ones. Therefore I do not think that the skeleton will help us.", "If you have a commutative semiring $(R, +, \\cdot)$ that satisfies $a+b = 0 \\Rightarrow a=b=0$, can't you just take an essentially small category $C$ realizing $(R,+)$, take a skeleton $\\mathrm{Sk}(C)$ of it and define on $\\mathrm{Sk}(C)$ the symmetric monoidal structure induced by $(R,\\cdot)$? That is, $[x] \\otimes [y] \\colon= [x \\cdot y]$ (where $[x]$ denotes the object in $\\mathrm{Sk}(C)$ that is associated to the object $x \\in R$)"], "comment_count": 0, "tags": ["ring-theory", "category-theory", "examples-counterexamples", "monoidal-categories"], "creation_date": "2014-05-29T05:39:16", "diamond": 0, "votes": 33} {"question_id": "284", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2865253", "title": "Can you obtain $\\pi$ using elements of $\\mathbb{N}$, and finite number of basic arithmetic operations + exponentiation?", "body": "Is it possible to obtain $\\pi$ from finite amount of operations\n$\\{+,-,\\cdot,\\div,\\wedge\\}$ on $\\mathbb{N}$ (or $\\mathbb{Q}$, the answer will still be the same)? Note that the set of all real numbers obtainable this way contains numbers that are not algebraic (for example $2^{2^{1/2}}$ is transcendental).\nBonus: If it happens that the answer is no, is it a solution to some equation generated that way (those $5$ operations performed finitely many times on elements on $\\mathbb{N}$) ?\n", "comments": ["As mentioned above, it is probably false, but a disproof is difficult. A hope could be to describe a subfield of $\\Bbb{R}$ containing all elements generated by the method you describe, but not containing $\\pi$. How such a field would look is difficult to imagine though.", "The bonus question is the same question except allowing the use of $\\log$, since all the operations are invertible, and their inverse is included in the allowed operations - except for $^$, which needs $\\log$.", "If you only use +,-,x,/, you can't get $\\pi$ because it will only yield a rational number, so you have to use the exponential operator sometime.", "It can be obtained using calculus, check Wikipedia.", "Could Machin-like formulas be of any use?", "Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special.", "Of course not, but I can't prove it. There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way. Unless a real has some reason to be expressible it almost certainly can't be.", "(this one is more general. If the answer to this one is \"no\" then the answer to the other one is \"no\" as well)", "Related: math.stackexchange.com/questions/2611084/…"], "comment_count": 0, "tags": ["number-theory", "pi", "transcendental-numbers"], "creation_date": "2018-07-28T06:29:14", "diamond": 0, "votes": 33} {"question_id": "285", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4624439", "title": "A holomorphic function sending integers (and only integers) to $\\{0,1,2,3\\}$", "body": "Does there exist a function $f$, holomorphic on the whole complex plane $\\mathbb{C}$, such that $f\\left(\\mathbb{Z}\\right)=\\{0,1,2,3\\}$ and\n$\\forall z\\in\\mathbb{C}\\ (f(z)\\in\\{0,1,2,3\\}\\Rightarrow z\\in\\mathbb{Z})$?\nIf yes, is it possible to have an explicit construction?\nNote that, for example, $h(z)=\\frac{1}{6} \\left(9-8 \\cos \\left(\\frac{\\pi z}{3}\\right)-\\cos (\\pi z)\\right)$ is not a valid solution, since, in particular, certain roots of the equation $h(z)=0$ are not integers, but complex numbers.\nAlso note that this question is answered in positive regarding the function $g$ such that $g\\left(\\mathbb{Z}\\right)=\\{0,1,2\\}$ and $\\forall z\\in\\mathbb{C}\\ (g(z)\\in\\{0,1,2\\}\\Rightarrow z\\in\\mathbb{Z})$. In this case, an example of such function is $g(z)=1-\\cos \\left(\\frac{\\pi z}{2}\\right)$.\n", "comments": ["It is already solved on MO", "Crossposted to MathOverflow: A holomorphic function sending integers (and only integers) to $\\{0,1,2,3\\}$", "I suggest you post this to MathOverflow, adding a link to this MSE post to let readers know you've crossposted.", "I'd ask for a Laurent polynomial solution in a separate post (ie. $g\\in \\Bbb{C}[q,q^{-1}]$ such that $g^{-1}(\\{0,1,2,3\\}) = \\mu_n$ for some $n$)", "@user1950 - one has to be careful as the cited theorem applies only for multiplicity one values, so in other words if $f^{-1}(\\{0,1,2,3\\})=\\mathbb Z$ and $f'(n) \\ne 0$ for any $n \\in \\mathbb Z$; probably not hard to generalize if the multiplicities are uniformly bounded but unclear what happens if multiplicities at integers can be unbounded", "So from @reuns remark it follows that the sought function, if it exists, must be a Laurent polynomial of $e^{\\frac{2\\pi i z}{n}}$. Looks like the problem is within reach of an expert in Galois theory.", "It follows from Theorem 1 of G.G. Gundersen and C.C. Yang's \"On the Preimage Sets of Entire Functions\" that the sought function must be periodic. (projecteuclid.org/journalArticle/…)", "one way to go is to interpolate the values $0,1,2,3$ evenly at integers (first for example the values may be symmetrized and the function made odd say on the integers as that gives better interpolation convergence) and get a function $f(z)=\\frac{\\sin \\pi z}{\\pi}(f(0)/z +\\sum_{n \\ne 0} (-1)^n f(n)/(z-n)+c)$ of order $1$ where $f(n)$ are precisely the choices you make and see if some clever choice of the interpolation values, the function obtained can be show to satisfy the property that only integers are sent there", "why does it need to be periodic? the above applies only to periodic functions", "@Conrad: So any valid solution to the main question must be of the form $\\sum_{n=a}^{b} a_n q^n$, where $-\\inftyi&&t1[[i+1]] not bounded away from $-1$, the converse fails! You are quite right, it's unjustified at present", "@FShrike Please can you justify your claim? It seems wrong, or at best, unjustified to me. For example, it's not clear to me why we cannot have: $\\ (\\sin(n!))_n\\ $ not equi-distributed and $\\liminf_{n\\rightarrow\\infty}(\\sin(n!)+1)^{1/n}<1.$", "@FShrike: The problem of the density or lacunarity of the limit points of the sequence $\\sin n!$ is probably as hard as that of $\\cos(n!)$.", "@MartinR we know $(\\sin(n))_n$ is equidistributed, but do we know $(\\sin(n!))_n$ is equidistributed? If it isn't, then we know for sure that the limit is $1$", "Here is a similar question about $\\cos(n!)$. The answer (as I understand it) is “we don't know.”"], "comment_count": 0, "tags": ["sequences-and-series", "limits", "analysis", "trigonometry", "factorial"], "creation_date": "2022-11-15T07:22:14", "diamond": 0, "votes": 30} {"question_id": "290", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4278597", "title": "Wave equation: predicting geometric dispersion with group theory", "body": "Context\nThe wave equation\n$$\n\\partial_{tt}\\psi=v^2\\nabla^2 \\psi\n$$\ndescribes waves that travel with frequency-independent speed $v$, ie. the waves are dispersionless. The character of solutions is different in odd vs even number of spatial dimensions, $n$. A point source in odd-$n$ creates a disturbance that propagates on the light cone and vanishes elsewhere: if the point source is a flash of light, an observer sees darkness, then a flash, then darkness. When $n$ is even, a disturbed media never returns to rest: the observer sees darkness, then brightness that lingers for all $t$. This phenomena is known as geometric dispersion.\nQuestion\nIs it possible to show that geometric dispersion is predicted by the wave equation, using group theory? For a point source at the origin, we would be searching for spherically symmetric solutions, and the rotation group $SO(n)$ has a different structure depending on whether $n$ is odd or even. In particular, I am interested in doing this without actually solving the wave equation. Unfortunately, I don't know enough group theory to know if this is even possible.\nWhat I know\nI can 'show' geometric dispersion by solving the wave equation with an initial condition, or computing the Green's function for the wave equation and noting that it is either supported only on the light cone (odd $n$), or everywhere within the light cone (even $n$).\nI know some group theory 'for physicists'.\nRelated\nThis unanswered question is similar. I think my question is more specific: I'm asking about a way to predict (rather than explain) geometric dispersion using group theory.\nUpdate: (thanks to comments of Alp Uzman and GiuseppeNegro)\nIt appears to be possible using group theoretic machinery, described in the book Nonabelian harmonic analysis by Howe and Tan. The relevant section is 4.3.1. So an equivalent question becomes: can someone explain the result from Howe and Tan in a more accessible way? The book is beyond my level of group theory at the moment.\n", "comments": ["@GiuseppeNegro I had tried studying the Howe-Tan account of the Huygens' principle about five years ago and at the time I too could not handle the representation theory involved (I'm afraid I still can't). I don't really have any useful insights, as such I can't claim any credits, though thank you for acknowledging my comment elsewhere.", "I wonder if we can @AlpUzman to notify them. It's them, not me, who gave that useful reference", "@GiuseppeNegro Thank you, that reference seems to answer precisely my question- I do not understand it yet either", "Following the comments in the linked question, I arrived at Theorem 4.3.1 in the book \"Nonabelian harmonic analysis\" of Howe and Tan (thank you Alp Uzman). That theorem contains exactly the answer to this question. Unfortunately, it uses heavy machinery from representation theory and I cannot fully understand it.", "I would love to see an answer to this. Group theory is one of my weakest areas though, so I won't be able to produce one...."], "comment_count": 0, "tags": ["group-theory", "partial-differential-equations", "mathematical-physics", "wave-equation", "dispersive-pde"], "creation_date": "2021-10-16T12:35:52", "diamond": 0, "votes": 29} {"question_id": "291", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3411958", "title": "Binomial coefficients modulo a prime", "body": "Consider an odd prime $p\\equiv1 \\pmod {16}$ and set $M=\\frac{p-1}{2}$ for notational convenience. Then is there even a single prime $p$ of the above form for which the following congruence holds? $$\\binom{M}{M/2}\\binom{M}{M/4}\\equiv \\pm \\binom{M}{3M/8}\\binom{M}{7M/8}\\pmod p ?$$\nHere the symbol $\\binom{n}{r}$ is a binomial coefficient and is defined as $\\frac{n!}{r!(n-r)!}$\nI am trying to prove something significantly more abstract but have managed to reduce it down to proving that the above statement can never hold. I have written some code to check this for primes of the above form less than $1000$ and found no counter examples. My attempts to find results on binomial coefficients $\\pmod p$ have mostly involved Lucas' theorem which does not seem very applicable here. I would appreciate a solution, reference or possible strategies I might try. \n\nAdded later: The more general problem.\nLet $$f_a(x)=x(x^4-1)(x^4+ax^2+1)$$ be a family of polynomials in $\n\\mathbb F_p[t]$ in the parameter $a\\in \\overline {\\mathbb F}_p$.\nLet $$\\sum c_{\\alpha}x^\\alpha := f_a(x)^{\\frac{p-1}{2}} $$ and note that the $c_\\alpha$ are polynomials in $a$. Now make the $4\n\\times 4$ matrix whose $(i,j)$th entry is $c_{pi-j}$ for $1\\leq i,j \\leq 4$. Then I want to show that this matrix is non-invertible for values of $a$ other than $a=2,-2$. In other words, I want to show that the determinant $d(a)$ has irreducible factors other than $(a-2)$ and $(a+2)$. \nWhat I have managed to prove is that $d(a)$ is a polynomial of degree $\\frac{3(p-1)}{2}$ and also that $(a-2)$ and $(a+2)$ occur with the same multiplicity in the factorization of $d(a)$ in $\\mathbb F_p[a]$. So for a contradiction, suppose $d(a)=\\alpha(a^2-4)^{\\frac{3(p-1)}{4}}$ for some $\\alpha\\in \\mathbb F_p$. Then, the ratio of the leading term and the constant term must be 1. The leading term is given by the square of the right hand side in the top most equation and the constant term is given by the square of the left hand side. The other coefficients are very difficult to extract and it is lucky that I even managed to get the leading and constant terms. So I am hoping that if by some combinatorial argument I can rule out this congruence, then I will have proved that the determinant has to have an irreducible factor other than $(a-2)$ and $(a+2)$.\n", "comments": ["@JyrkiLahtonen it's been a while but yes the presence of that last factor is critical. For some more information, this is about superspecial hyperelliptic curves in the family $y^2 =x(x^4-1)(x^4+ax^2+1)$. I was trying to prove that for all primes $p = 1,7 \\mod 8$, and $p>7$, there is a superspecial curve in this family.", "You can find a bunch of congruences of similar shape in Binomial coefficients and Jacobi sums, by Hudson and Williams.", "I still wonder what the original problem is? The exponent in $f_a(x)^{(p-1)/2}$ somehow suggests a character sum involving a quadratic character, but I may be off :-)", "Just to check that I have the right determinant. When $p=17$, the determinant of that 4x4 matrix is $$8(a+2)^8(a-2)^8(a^2-5)^4,$$ and the presence of that last factor is crucial for your purpose?", "@mathworker21 Per your request, I have added some additional information.", "@RKD can you say the significantly more abstract thing you want to prove? that might actually help to prove the thing you asked as your question", "(Added for completeness of comment: I wrote some code to check up to $10^7$ - didn't find anything)", "(cont.) My point being that strong computational evidence would require that one examine way more than the primes up to $10000$; I got Mathematica to go up to $244737$ and found nothing, but computing the binomial coefficients seems to be rather expensive (...though maybe less so if I could tell Mathematica to do this computation mod $p$, though I still don't know of any methods better than the usual formula, which uses $O(p)$ multiplication mod $p$ - which is really bad if you want a long search)", "A heuristic note: if we let $P'_n$ be the $n^{th}$ prime congruent to $1$ mod $16$ and if we chose two numbers $(a_n,b_n)$ at random mod each $P'_n$, the probability that no solution to $a_n=\\pm b_n$ exists in the first $N$ terms is bounded below by $$\\prod_{n=1}^{N}(1-2/P_n).$$ This product is zero in the limit, suggesting a counterexample exists unless a pattern does - but of the primes up to $10^8$, this product is still $0.619$ - so even if we checked that far on a random sequence, we would still be less likely to find a $a_n=\\pm b_n$ pair than not to.", "I have no clue. They seem to be completely at random.", "Do you have any clue what LHS/RHS (modulo $p$) might be equal to?"], "comment_count": 0, "tags": ["modular-arithmetic", "binomial-coefficients", "finite-fields"], "creation_date": "2019-10-27T21:43:24", "diamond": 0, "votes": 29} {"question_id": "292", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/49208", "title": "$f\\colon I\\rightarrow G$ and Gromov $\\delta$-hyperbolicity", "body": "Please recall that $\\left|\\int_0^1 f(t)\\,dt -w\\right|\\leq \\int_0^1|f(t)-w|\\,dt$. In general, let $(X,d)$ be a metric space. Given a function $f:I\\to X$ let $m_f\\in X$ be such that $d(m_f,w)\\leq \\int_0^1 d(f(t),w)\\,dt$ for every $w\\in X$. If such $m_f\\in X$ exists then we call $f$ as $D$-integrable with $D$-integral $m_f$. ($I=[0,1]$.)\nQuestion: Given a finitely generated group $G$ (with a usual word distance), does there exist a condition on the set of $D$-integrable functions $f: I\\to G$ (which uses $D$-integrability) for the Gromov $\\delta$-hyperbolicity of $G$? I.e, what is the characterization of hyperbolicity in terms of $D$-integrability? \n(It seems to me that the real question here should be: how can one extract group theoretic intel from $D$-integrability?)\n", "comments": [], "comment_count": 0, "tags": ["calculus", "group-theory", "metric-spaces"], "creation_date": "2011-07-03T07:06:02", "diamond": 0, "votes": 28} {"question_id": "293", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1688173", "title": "Lowest dimensional faithful representation of a finite group", "body": "How does one compute the lowest dimensional faithful representation of a finite group?\nThis question originated in the context of given a finite group $G$: trying to find the lowest dimensional shape whose rotational/reflection symmetries form $G$. (Formally stated as finding the lowest dimensional faithful representation of $G$ into the orthogonal group $O(n)$.) \nNow just listing out examples is hard, because even in just $\\mathbb{R}^2$ outside of the symmetries of simple polygons we see things like $\\mathbb{Z}_2 \\times \\mathbb{Z}_2 $ crop up, yet it appears that $\\mathbb{Z}_2 \\times \\mathbb{Z}_2 \\times \\mathbb{Z}_2$ doesn't have any obvious two dimensional representations. Grouppropswiki doesn't even have a single representation for it at all.\nSo I thought it was wise before I tackle the orthogonal group question I should know how to just, in general, find a low dimensional faithful representation. Some trivialities are that the dimension for a group $G$ will be less than or equal to the smallest $j$ such that $G \\subset S_j$ since each symmetric group can be realized as the symmetries of a $j$-dimensional simplex space due to Burnside. \nBut this doesn't say much because even something like $\\mathbb{D}_{\\text{Graham's Number}}$ can be realized in $\\mathbb{R}^2$ (and we still are dealing with isometries here, what about represenatations that transcend that!?)\n", "comments": ["For future visitors: some research developments since this question was asked mathoverflow.net/questions/351938/… as well it is interesting to ask even whether there are small irreducible representations, as discussed here: mathoverflow.net/questions/400864/… Finally, note that $j$ in the question statement is called the minimal degree in group theory; computing is still open in many cases.", "Is there a list of these smallest dimensions somewhere for small groups? @DerekHolt", "It's a difficult problem in general over any field. For an abelian group with invariant factor decomposition $C_{k_1} \\times \\cdots C_{k_r} \\times C_2^s$ with each $k_i > 2$, I think the smallest dimension of a real faithful representation is $2r+s$. It is certainly $s$ for $C_2^s$.", "I'm focusing on real representations. Does that make matters worse?", "Are you talking about complex representations, or representations over an arbitrary field? For complex representations, the smallest dimension of a faithful representation of a finite abelian group with minimal generating set of size $d$ is $d$. But $C_2 \\times C_2 \\times C_2$ has a faithful $2$-dimensional representation over the finite field of order $8$. In general, this is a difficult problem - even finding the smallest degree faithful permutation representation can be difficult."], "comment_count": 0, "tags": ["linear-algebra", "group-theory", "finite-groups", "representation-theory", "linear-transformations"], "creation_date": "2016-03-08T00:31:03", "diamond": 0, "votes": 28} {"question_id": "294", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1381042", "title": "Dividing the whole into a minimal amount of parts to equally distribute it between different groups.", "body": "Suppose we have a finite amount of numbers $x_1, x_2, ..., x_n$ ($x_i\\in\\mathbb{N}$) and an object that should be divided into parts in such a way that it can be without further dividing distributed in $x_i$ equal piles for any $1\\leq i\\leq n$. \nThe question is what is the minimal amount of parts for some $x_1, x_2, ..., x_n$ that would allow us to do that?\nExample: we have a cake and expect some friends, but we don't know for sure how many of them are coming. But somehow we know, that the total amount of people will be either 5 or 6 $(x_1 = 5, x_2 = 6)$. So we divide a cake beforehand into $5\\times 6 = 30$ parts, and so in each scenario we can just give the guest his share of the cake.\nOptimization. Of course, 30 parts is too many. In the example above we could just divide the cake into 5 parts, then ignore it and divide the cake (like it was whole) into 6 parts. First we do 4 cuts, second we do 5 cuts; they don't coincide because 5 and 6 are coprimes. Total 9 cuts gives us 10 pieces. So instead of cutting the cake in 30 pieces, we can only cut it in 10. And that is the minimum here.\nSimple case. Let $x_1$ and $x_2$ be two coprime numbers. Then, using the same argumentation as above with the case $x_1=5, x_2=6$, we can note that the minimal amount of parts is $(x_1-1)+(x_2-1)+1$ which is equal to $x_1+x_2-1$. \nGeneralization. First I thought that for $n$ coprime numbers the solution is simple and beautiful:\n$$\n\\sum_{i=1}^n (x_i-1) + 1\n$$\nBut this formula is wrong for $n>2$ as shows the following counterexample.\nCounterexample. For $x_1=3, x_2 = 4, x_3 = 5$ the formula gives the answer 10. But that's not the minimum, as we can divide at least in nine portions:\n$$\n\\frac{1}{60} + \\frac{2}{60} + \\frac{4}{60} + \\frac{5}{60} + \\frac{7}{60} + \\frac{8}{60} + \\frac{10}{60} + \\frac{11}{60} + \\frac{12}{60} = 1\n$$ \nso we can split it equally in 3 pieces:\n$$\n\\left(\\frac{1}{60} + \\frac{2}{60} + \\frac{7}{60} + \\frac{10}{60}\\right) + \\left(\\frac{4}{60} + \\frac{5}{60} +\\frac{11}{60}\\right) + \\left(\\frac{8}{60} + \\frac{12}{60}\\right) = 1\n$$ \nin 4 pieces:\n$$\n\\left(\\frac{1}{60} + \\frac{2}{60} + \\frac{12}{60}\n\\right) + \\left(\\frac{4}{60} +\\frac{11}{60}\\right) + \\left(\\frac{5}{60} + \\frac{10}{60}\\right) + \\left(\\frac{7}{60} + \\frac{8}{60}\\right) = 1\n$$ \nin 5 pieces:\n$$\n\\frac{12}{60}+\\left(\\frac{1}{60} + \\frac{11}{60}\n\\right) + \\left(\\frac{2}{60} +\\frac{10}{60}\\right) + \\left(\\frac{4}{60} + \\frac{8}{60}\\right) + \\left(\\frac{5}{60} + \\frac{7}{60}\\right) = 1\n$$ \nTo sum up, the posed question has an evident answer in the case $n=1,2$, but in the case $n=3$ (and higher) I didn't discover any pattern.\n[Edit] Related Topics:\nMinimum Cake Cutting for a Party\nhttps://puzzling.stackexchange.com/questions/19870/nine-gangsters-and-a-gold-bar\nhttps://mathoverflow.net/questions/214477/minimal-possible-cardinality-of-a-a-1-a-k-distributable-multiset\n", "comments": ["Note that the same question has been asked at math.stackexchange.com/questions/1383406/…", "OK, so for $n$ coprime numbers, we can always do it with $1+\\sum(x_i-1)$ pieces, and the question is, under what circumstances can we do it with fewer, and how many fewer. So I wonder whether the $3,4,5$ construction generalizes to, say, $3,3k+1,3k+2$, or to $2k-1,2k,2k+1$, or some other infinite family of examples.", "@rywit, but if we shift a little the second series of cuts like this (dividing with shift), we will use one additional cut but at the same time we will be able to assemble one-third parts and no more cuts will be needed.", "@rywit, about the counterexample. Dividing into 3, 4 and 5 part using the formula in the question (let's call it the naive formula), can be graphically presented like this: naive dividing. When we divide the object into 5 parts and then into 6, it can be pictured like this (numbers are sizes of pieses in one-sixtieth portions): first two series of cuts. We cannot assemble any one-third parts using this pieses.", "@GerryMyerson, at least that small (you can make it as big as you want by dividing into $kx_1x_2$ parts with any positive integer $k$). In order to divide into $x_1$ parts we do $x_1 - 1$ cuts. To divide into $x_2$ parts we do $x_2-1$ cuts. Number of pieces is the number of cuts plus 1, so we have $(x_1 - 1)+(x_2-1)+1 = x_1 +x_2-1$. Geometrical point of view shows that this is minimum, but that is more intuitive approach than a strict proof; that's why I wrote \"seems to be\".", "@GerryMyerson Ah, you're right. Thanks!", "@rywit, I think we're trying to minimize the number of pieces, rather than the number of physical cuts.", "When you write, in the \"simple case\", that the answer \"seems to be\" $x_1+x_2-1$, do you mean you can prove it's at least that big? or do you mean you can prove it's at least that small? or what?", "Also, in your example with 5 and 6 people you say you make 4 cuts and then separately 5 cuts for a total of 9. But you could align the first cut in each set. So you could make 4 cuts (for 5 people) and then 4 more cuts (reusing one of the existing cuts) for 6 people. Thus a total of only 8 cuts is required in this case. Am I right?", "How did you discover this counterexample?"], "comment_count": 0, "tags": ["number-theory", "elementary-number-theory", "divisibility"], "creation_date": "2015-08-01T03:46:39", "diamond": 0, "votes": 28} {"question_id": "295", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4619480", "title": "Circles of radius $1, 2, 3, ..., n$ all touch a middle circle. How to make the middle circle as small as possible?", "body": "Non-overlapping circles of radius $1, 2, 3, ..., n$ are all externally tangent to a middle circle.\n\nHow should we arrange the surrounding circles, in order to minimize the middle circle's radius $R$?\n\nTake $n=10$ for example.\n\nOn the left, going anticlockwise the radii are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, and $R\\approx 10.77$.\nOn the right, going anticlockwise the radii are $10, 2, 9, 4, 7, 6, 5, 1, 8, 3$ and $R\\approx 9.98$.\nHow should the circles be arranged to minimize $R$?\nHere is a desmos graph where you can try different arrangements for the case $n=10$.\nMy attempt\nFor general $n$, call the radii of the surrounding circles going anti-clockwise, $r_1, r_2, r_3, ..., r_n$.\nDraw line segments from the centre of the middle circle to the centre of each surrounding circle. So we have $n$ angles at the centre of the middle circle. Each of these angles can be expressed in terms of $R, r_k, r_{k+1}$, using the law of cosines. The sum of the $n$ angles is $2π$. So we have:\n$$\\sum\\limits_{k=1}^n \\arccos{\\left(\\frac{(R+r_k)^2+(R+r_{k+1})^2-(r_k+r_{k+1})^2}{2(R+r_k)(R+r_{k+1})}\\right)}=2\\pi \\text{ (where }r_{n+1}=r_1)$$\nThis simplifies to:\n$$\\sum\\limits_{k=1}^n \\arccos{\\left(1-\\frac{2r_k r_{k+1}}{(R+r_k)(R+r_{k+1})}\\right)}=2\\pi$$\nWe want to assign each $r$ a unique value among $1, 2, 3, ..., n$ so that $R$ is minimized. But how? Is there a general pattern?\nConjectured answer\nI think the following general procedure will make the middle circle as small as possible. Take $n=10$ for example. We ignore the $1$ at first; it will be placed last. First put down the numbers $10, 9, 8, 7, 6$ in pyramid-fashion, from top to down and from left to right, like the red numbers below. Then put down the numbers $2, 3, 4, 5$ in pyramid-fashion also, but between the previous rows, like the blue numbers below.\n\nThis gives the order of radii going around the circle: $7, 4, 9, 2, 10, 3, 8, 5, 6$. Then put the circle with radius $1$ anywhere you like, as long as it fits without disturbing other circles.\nFor larger values of $n$, there may be multiple small circles that can fit between other circles without disturbing them. Put those last.\nIntuitively, this procedure minimizes the tendency of the large surrounding circles to take up space around the middle circle. You can see this in the diagrams above with $n=10$, looking specifically at the two largest surrounding circles in each case.\nAnyway, it's just a conjecture for now. Notice that this conjecture implies that, for $n=10$, the arrangement I showed above on the right, yields the minimum radius of the middle circle. I don't think the middle circle can be any smaller.\n", "comments": ["I wrote a program that loops over all permutations of 2...10 (excluding the circle of radius 1), and finds an $R$ that solves the equation that you wrote (the sum of the arc-cosines). It confirmed that the optimal permutation is the one that you conjectured: 2, 9, 4, 7, 6, 5, 8, 3, 10, where the radius of the inner circle is ~9.979907. I had to exclude the circle of radius $1$ from the sum of course.", "It seems the right drawing the small circle may be removed without impacting the result.", "@VarunVejalla Good point. I took this point into account in my desmos graph, but forgot to mention this in my question. We should ignore the smallest few circles, because they will each fit between some two other circles that touch each other. How many circles should be ignored, depends on $n$, with an upper limit of $n/4$ (based on Descartes' Circle Theorem).", "The problem with the current sum of $\\arccos$ is that you can have gaps between circles. For example, in the arrangement on the right, there would have to be some wiggle room for the circle of radius $1$ between the circles of radius $5$ and $8$ (i.e. it can't be tangent to both of those and the middle circle), and numerically solving for $R$ using the sum of $\\arccos$ yields $\\approx 9.90$, different from the $\\approx 9.98$ that it would actually be.", "@Dan Yes, absolutely. Actually, I think there's no point trying that. For small numbers (say, $n=7,8,9,10$) one might solve the equation numerically for all the possible combinations of $r_k$. Perhaps that gives a clue on how the $r_k$ must be arranged for general $n$.", "Yeah, I thought the same", "@Dog_69 That approach is very difficult due to the $\\sum$, I think.", "First thing tthat comes to my mind is to take the derivative with respect to $R$ and see the condition for it to vanish."], "comment_count": 0, "tags": ["geometry", "optimization", "circles", "discrete-geometry"], "creation_date": "2023-01-16T03:46:59", "diamond": 0, "votes": 27} {"question_id": "296", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4434174", "title": "A curious identity on $q$-binomial coefficients", "body": "Let's first recall some notations:\n\nThe $q$-Pochhammer symbol is defined as\n$$(x)_n = (x;q)_n := \\prod_{0\\leq l\\leq n-1}(1-q^l x).$$\n\nThe $q$-binomial coefficient (also known as the Gaussian binomial coefficient) is defined as $$\\binom{n}{k}_q := \\frac{(q)_{n}}{(q)_{n-k}(q)_{k}}.$$\n\n\nI found the following curious identity on $q$-binomial coefficients, and I would like to ask for some ideas on how to prove it.\n$$\\sum_{0\\leq j\\leq k\\leq 2n}(-1)^{j}q^{k(k-j-n)+\\frac{j(j+1)}{2}}\\binom{2n}{j}_q \\overset{?}{=} (q)_n$$\nI am familiar with the $q$-binomial theorem and $q$-Vandermonde identity, which I think might be useful. If there are some other well-known identities on $q$-binomial coefficients that might be helpful in figuring this out, I would like to know them too. Thank you!\n\nSome thoughts:\n\nThe classical version of this identity is simply $$\\sum_{0\\leq j\\leq 2n}(-1)^{j}(2n+1-j)\\binom{2n}{j} = \\sum_{0\\leq j\\leq 2n}(-1)^{j}(j+1)\\binom{2n}{j} = \\delta_{n,0},$$\nwhich can be proved simply by $$\\sum_{0\\leq j\\leq 2n}(-1)^{j}\\binom{2n}{j} + \\sum_{0\\leq j\\leq 2n}(-1)^{j}j\\binom{2n}{j} = (1-x)^{2n}+\\frac{d}{dx}(1-x)^{2n} \\bigg\\vert_{x=1} = \\delta_{n,0}.$$\nThere is a natural $q$-analogue of the above identity, obtained by replacing $(1-x)^{2n}$ by $(x)_{2n}$ and the derivative by the $q$-derivative, but unfortunately this is not the desired identity.\n\nThere seems to be other closely related identities:\n$$\\sum_{0\\leq j\\leq k\\leq 2n}(-1)^{j}q^{k(k-j-n-1)+\\frac{j(j+1)}{2}}\\binom{2n}{j}_q \\overset{?}{=} \\delta_{n,0},$$\n$$\\sum_{0\\leq j\\leq k\\leq 2n}(-1)^{j}q^{k(k-j-n+1)+\\frac{j(j-1)}{2}}\\binom{2n}{j}_q \\overset{?}{=} q^{n}(q)_{2n}.$$\nProving these other identities (or a family of similar identities) might be helpful if we were to try to use induction on $n$.\n\n\n", "comments": ["So we begin with a combinatorial interpretation : Consider all $n$-tilings using $k$ green tiles and $n-k$ red tiles, such that every green tile's weight is $q^r$ where $r$ is the number of red tiles preceding this green tile. The weight of a tiling is the product of weights of all green tiles, and the sum over all such tilings of these weights equals $\\binom{n}{k}_q$. We get that the LHS is a sum over all tilings of length $2n$ and $k \\geq $ the number of green tiles, of sth. involving $k$ and the green tiles. Using $(-1)^j$ and some work, cancellations can be figured out that eliminate $k$.", "I managed to obtain a partial explanation using combinatorics. I don't have a full answer to fall back upon ,but I eliminated the role of $k$ and figured out some further cancellations using the ideas. I don't want to post a partial answer (I'd rather it be complete and worthy of evaluation), but if someone's interested I'll put the details up in the next comment(s).", "It might be useful to post the same question on Mathoverflow. It is definitely worth it.", "@Henry That is nice. I will have a look today at it. The quadratic form on the exponent is unusual. Nice identity tho!", "@Phicar Okay, I have turned the desired identity into a single-sum identity and was able to verify it using qZeil. This is nice, and thanks for the suggestion. Still, it would be nice to have a non-algorithmic proof of the identity.", "@SarveshRavichandranIyer Unfortunately I am not aware of any useful reference either.", "@Phicar Can qZeil handle double summations? I thought it was mostly for single summations.", "Can you provide some source for these identities? I had a general look and unfortunately there's very little on double-sum identities that I found lying around for the $q$-binomial.", "Perhaps $q-$Zeilbergs gets you a nice recursion? risc.jku.at/sw/qzeil"], "comment_count": 0, "tags": ["binomial-coefficients", "q-analogs", "q-series"], "creation_date": "2022-04-22T23:52:21", "diamond": 0, "votes": 27} {"question_id": "297", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3103805", "title": "What is $\\int_0^1 \\left(\\tfrac{\\pi}2\\,_2F_1\\big(\\tfrac13,\\tfrac23,1,\\,k^2\\big)\\right)^3 dk$?", "body": "As in this post, define the ff:\n$$K_2(k)={\\tfrac{\\pi}{2}\\,_2F_1\\left(\\tfrac12,\\tfrac12,1,\\,k^2\\right)}$$\n$$K_3(k)={\\tfrac{\\pi}{2}\\,_2F_1\\left(\\tfrac13,\\tfrac23,1,\\,k^2\\right)}$$\n$$K_4(k)={\\tfrac{\\pi}{2}\\,_2F_1\\left(\\tfrac14,\\tfrac34,1,\\,k^2\\right)}$$\n$$K_6(k)={\\tfrac{\\pi}{2}\\,_2F_1\\left(\\tfrac16,\\tfrac56,1,\\,k^2\\right)}$$\nWe find that,\n$$\\int_0^1 K_2(k)\\, dk = {\\tfrac{\\pi}{2}\\,_3F_2\\left(\\tfrac12,\\tfrac12,\\tfrac12;1,\\tfrac32;1\\right)}=2G$$\n$$\\int_0^1 K_3(k)\\, dk = {\\tfrac{\\pi}{2}\\,_3F_2\\left(\\tfrac12,\\tfrac13,\\tfrac23;1,\\tfrac32;1\\right)}=\\tfrac{3\\sqrt3}2\\, \\ln2$$\n$$\\int_0^1 K_4(k)\\, dk = {\\tfrac{\\pi}{2}\\,_3F_2\\left(\\tfrac12,\\tfrac14,\\tfrac34;1,\\tfrac32;1\\right)}=2\\ln(1+\\sqrt2)$$\n$$\\int_0^1 K_6(k)\\, dk = {\\tfrac{\\pi}{2}\\,_3F_2\\left(\\tfrac12,\\tfrac16,\\tfrac56;1,\\tfrac32;1\\right)}=\\tfrac{3\\sqrt3}4\\, \\ln(2+\\sqrt{3})$$\nwhere $G$ is Catalan's constant. However, this post gives a third power, \n$$\\int_0^1\\big(K_2(k)\\big)^3 dk=\\frac35 \\bigg(\\tfrac{\\pi}2\\,_2F_1\\big(\\tfrac12,\\tfrac12,1,\\tfrac12\\big)\\bigg)^4=\\frac35 \\big(K(k_1)\\big)^4 = \\frac{3\\,\\Gamma (\\frac14)^8}{1280\\,\\pi^2} \\approx 7.0902$$\nwhere $K(k_d)$ is an elliptic integral singular value while YuriyS in his comment below gives,\n$$\\int_0^1\\big(K_3(k)\\big)^3 dk \\approx 6.53686311168760876289835638374 $$\n\nQuestions: \n\nWhat are the closed-forms of:\n$$\\int_0^1\\big(K_n(k)\\big)^m dk=\\,?$$\nfor power $m=2$ or $m=3$?\nOr at least their numerical evaluation up to 20 digits?\n\nP.S. My old version of Mathematica can't evaluate it with sufficient precision, nor does WolframAlpha. (Enough digits may make it amenable to the Inverse Symbolic Calculator.)\n", "comments": ["At one point, I convinced myself that dlmf.nist.gov/15.5 (at least a couple of them) worked for fractional derivatives. If so, then the order might be reduced To 1F1(a,b;k^2), which might be easier to evaluate (and invert). If you're still interested, I might take a stab at that. No warranties implied :)", "A related case with a closed form in this question: math.stackexchange.com/q/3325678/269624", "@YuriyS: I revised the post to give a clearer version of the problem.", "Mathematica gives $I_3=6.53686311168760876289835638374$ with WorkingPrecision->30 $$ $$ It has no problem evaluating the integral, only takes a few seconds", "@YuriyS: The closed-form is not the titular integral $I_3$. Note the closed-form involves $\\frac12,\\tfrac12$, while the titular integral $I_3$ involves $\\frac13,\\frac23$. P.S. If you can provide a numerical evaluation of $I_3$ for 20 digits or more, that would be nice.", "Could you clarify: does your 2nd question ask for numerical evaluation of the titular integral? Or the general integral from the 1st question? Above you have already provided the closed form for the titular integral, so I'm confused", "Please use \\left and \\right rather than \\bigg, especially in titles."], "comment_count": 0, "tags": ["integration", "definite-integrals", "closed-form", "hypergeometric-function", "elliptic-functions"], "creation_date": "2019-02-07T05:32:18", "diamond": 0, "votes": 27} {"question_id": "298", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3925067", "title": "Is there a group $G$ for which $\\mathrm{Aut}(G) \\simeq (\\mathbb{R},+)$?", "body": "I know the classic theorem that $(\\mathbb{Q},+)$ cannot be expressed as an automorphism group, i.e. there is no group $G$ such that $\\mathrm{Aut}(G)\\simeq (\\mathbb{Q},+)$.\n\nTheorem A. If $L$ is a locally cyclic group with no element of order $2$, then $L$ cannot be expressed as an automorphism group.\n\nBut how about $(\\mathbb{R},+)$?\nI think the answer might be no, and the proof proceeds by showing that if $\\mathrm{Aut}(G) \\simeq \\mathbb{R}$, then some subgroup or quotient $H$ of $G$ will satisfy $\\mathrm{Aut}(G) \\simeq \\mathbb{Q}$, which contradicts Theorem A.\nSo how can we construct this $H$? Assuming the axiom of choice, we can say $\\mathbb{R} = \\mathbb{Q}\\oplus B$ for some additive subgroup $B$ of $\\mathbb{R}$ (just by picking a $\\mathbb{Q}$-basis for $\\mathbb{R}$). Now I'm tempted to do some Galois-type thing, where you use a \"fixed\" subgroup $$H=\\mathrm{Fix}(B)=\\{g\\in G : b(g) = g \\text{ for all } b\\in B \\},$$ and then try to say something about $\\mathrm{Aut}(H)$ or $\\mathrm{Aut}(G/H)$ (where, for the latter, maybe we take the normal closure of $H$). But I can't complete this line of reasoning. It seems key that $\\mathrm{Aut}(G)$ splits as a direct sum --- that seems special.\nAm I just totally off-base here? Is there some obvious group $G$ whose automorphism group is $\\mathbb{R}$?\nRelated questions\nIf I can prove that $\\mathbb{R}$ is not an automorphism group, then the same would hold for the isomorphic group $\\mathbb{R}_{>0}$ of positive reals under multiplication. But how about $\\mathbb{R}^\\times \\simeq \\mathbb{R}_{>0}\\times C_2\\simeq \\mathbb{R}\\times C_2$ --- can this be achieved as an automorphism group? If $\\mathrm{Aut}(G)$ is a direct product of abelian groups, what can be said about $G$?\nA related observation is that if $\\mathrm{Aut}(G)$ is abelian, then $G$ is nilpotent of rank $\\leq 2$. So there is a sense in which $G$ is \"almost\" abelian.\nI know $\\mathbb{Q}^\\times$ is the automorphism group of $(\\mathbb{Q},+)$, and that $\\mathbb{R}^\\times$ constitutes the continuous automorphisms of $(\\mathbb{R},+)$ ... Then there's multiplicative/additive groups of other rings/fields ...\nI can't even resolve this question in the case of $\\mathrm{Aut}(G) \\simeq \\mathbb{Z}\\times \\mathbb{Z}$. For this one I can show that $G'$ would necessarily by cyclic, but that's about it.\nI'm very curious about which types of groups can be achieved as automorphism groups.\nA quick observation\nIf $\\mathrm{Aut}(G) \\simeq \\mathbb{R}$, then $G$ must be infinitely generated (or else its automorphism group is countable) and nilpotent of rank $\\leq 2$ (this follows for any group whose automorphism group is abelian).\n", "comments": ["Thanks, I did already know that! Same idea to prove that $\\mathbb{Q}$ is not an automorphism group.", "Partial answer: $G$ cannot be abelian. If $G$ is abelian, then $x\\mapsto x^{-1}$ is an automorphism of $G$ of order $\\leq 2$. $\\mathbb{R}$ has no nonidentity elements of finite order, so $x=x^{-1}$ for all $x\\in G$. This implies that $G$ is the underlying abelian group of an $\\mathbb{F}^2$-vector space $V$, and we have $\\operatorname{Aut}(G)=\\operatorname{Aut}_{\\mathbb{F}_2}(V)$. We must have $\\dim V\\geq 2$, but this means $\\operatorname{Aut}(G)$ contains $\\operatorname{GL}_2(\\mathbb{F}_2)$ as a subgroup. This would make $\\operatorname{Aut}(G)$ nonabelian, contradiction!", ">it's consistent with ZF that every endomorphism of R is measurable --- Wow.", "@Ehsaan: re: your related questions, it's consistent with ZF that every endomorphism of $\\mathbb{R}$ is measurable, hence continuous; if this is true then $\\text{Aut}(\\mathbb{R}) \\cong \\mathbb{R}^{\\times} \\cong \\mathbb{R} \\times C_2$. (With the axiom of choice $\\text{Aut}(\\mathbb{R})$ is much larger and in particular nonabelian.)", "@DietrichBurde thank you --- that's related for sure. I wonder which if there are groups $A$ for which it is true, but difficult, to prove that $A$ cannot be achieved as an automorphism group.", "See also this post.", "Yes, it would have to be some special property of $\\mathbb{R}$ that comes up here. Subgroup or quotient would be OK here --- just as long as we can use the fact that $\\mathbb{Q}$ is a direct summand of $\\mathbb{R}$ to reduce to the case of $\\mathbb{Q}$ (which is a known exercise).", "It's certainly not the case 'generically' that if $H\\subset Aut(G)$ then there's some subgroup $G'$ of G with $Aut(G')\\equiv H$, even if $Aut(G)=H\\times K$; consider the automorphism group of $C_p$ for $p$ prime. I don't see anything in your reasoning offhand that uses any properties of $\\mathbb{R}$ that don't mirror to this case."], "comment_count": 0, "tags": ["group-theory", "automorphism-group"], "creation_date": "2020-11-27T07:48:46", "diamond": 0, "votes": 27} {"question_id": "299", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3655352", "title": "If $r\\in\\mathbb{Q}\\setminus\\mathbb{Z}$ is it possible that $r^{r^{r^r}}\\in \\mathbb{Q}$?", "body": "It's straightforward to prove that $r^r\\notin\\mathbb{Q}$, which furthermore allows us to use the Gelfond-Schneider theorem to prove that $r^{r^r}\\notin\\mathbb{Q}$.\n\nIs it true that $r^{r^{r^r}}\\notin\\mathbb{Q}$?\n\nIt seems like it ought to be true, though my guess would be that this is an open problem. If not, does anyone know a proof? And if so, do there exist any discussions of it in the literature?\n", "comments": ["en.wikipedia.org/wiki/… - the techniques for the proof by infinite descent here can be generalized in a straightforward way", "@DarkMalthorp how do u prove $r^r \\not \\in \\mathbb{Q}$?", "Yes, indeed it is a stronger result :) Of course, I fully expect that $r^{r^{r^r}}$ is also transcendental.", "@darkmalthorp For the record, Gelfond-Schneider says that $r^{r^r}\\notin\\overline{\\Bbb Q}$, not just $\\notin\\Bbb Q$ (though I presume you meant that).", "I would also guess that this is an open problem. It is even unknown whether the power-tower is an integer for $\\ r=\\pi\\ $ (which would however be a miracle)"], "comment_count": 0, "tags": ["reference-request", "exponentiation", "irrational-numbers", "tetration"], "creation_date": "2020-05-02T09:25:14", "diamond": 0, "votes": 27} {"question_id": "300", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4605213", "title": "What functions from $\\Bbb N\\to\\Bbb N$ can be made from $+$, $-$, $\\times$, $\\div$, exponentiation, and $\\lfloor\\cdot\\rfloor$?", "body": "What functions from $\\Bbb N\\to\\Bbb N$ can be made from $+$, $-$, $\\times$, $\\div$, exponentiation, and $\\lfloor\\cdot\\rfloor$? Call this class of functions $\\mathcal Flex$ (for \"floor and exponentiation\").\nThe $\\rm mod$ function $a\\operatorname{mod}b$ can be defined by $a-b\\lfloor a/b\\rfloor$. In addition,\n$$\\binom nk=\\left\\lfloor\\frac{(u+1)^n}{u^k}\\right\\rfloor\\operatorname{mod}u$$\nfor $u>2^n$, and\n$$n!=\\left\\lfloor\\frac{r^n}{\\binom rn}\\right\\rfloor$$\nfor $r>(2n)^{n+1}$. Thus the factorial function is in $\\mathcal Flex$.\nThe function $\\left\\lfloor\\lfloor\\frac ab\\rfloor-\\frac ab\\right\\rfloor+1$ equals $1$ if $a$ divides $b$ and $0$ otherwise. According to Wilson's theorem, $n$ is prime iff $n$ divides $(n-1)!+1$; therefore, the function that is $1$ for primes and $0$ otherwise is in $\\mathcal Flex$.\nThe function $\\delta_{0,n}$ is in $\\mathcal Flex$ since it equals $\\left\\lfloor\\frac1{1+n}\\right\\rfloor$. I feel like there should be some way to use the MDRP theorem to show that all primitive recursive functions with images in $\\{0,1\\}$ are in $\\mathcal Flex$, but I'm not quite sure how.\nIn fact, my big conjecture so far is that $\\mathcal Flex$ actually equals the set of all primitive-recursive functions. (And I have a suspicion that exponentiation might be unnecessary after all...)\n", "comments": ["Given that we have binomials and modular arithmetic, the function $\\Big\\lfloor 10^n\\frac{16^{f(n)}}{f(n){2f(n)\\choose f(n)}^2} \\pmod{10}\\Big\\rfloor$ probably* gives the $n^\\textrm{th}$ digit of $\\pi$ for sufficiently fast-growing $f(n)$ like $n^{n^{n^n}}$. *Unless pi has an incredibly long string of zeros somewhere in its decimal expansion.", "Thanks for your explanation. This got into my review queue as a test, which got me very confused.", "@youthdoo Well, if someone provided a nice criterion for something to fall into this class, it would be a good solution. As it stands, I think this coincides with the ELEMENTARY class I found a month ago.", "Very possibly I didn't understand this well, but isn't this question too broad?", "I vote to call these \"flexible functions\"", "Maybe it's this?: en.wikipedia.org/wiki/ELEMENTARY I have some reading to do.", "Incidentally, I don't think it really matters whether we look at $\\Bbb N\\to\\Bbb N$ or $\\Bbb Z\\to\\Bbb Z$. In the latter case, the function $\\lfloor\\frac{3n}{3n-1}\\rfloor$ seems to give $1$ if $n$ is positive and $0$ otherwise, which seems useful.", "It is also possible to show that if $h(a,b,y):=\\prod_{k=1}^y(a+bk)$ then $h\\in\\mathcal Flex$. (I believe this is $b^yy!\\binom{q+y}y\\operatorname{mod}(bq-1)$ where $q>(a+by)^y$.) In the context of proving the MDRP theorem, this is function is used in conjunction with the Sunzi (Chinese) Remainder Theorem to encode bounded quantifiers.", "Most of these equations come from Martin Davis's exposition of the proof of the MDRP theorem: math.umd.edu/~laskow/Pubs/713/Diophantine.pdf"], "comment_count": 0, "tags": ["exponential-function", "diophantine-equations", "recursion", "computability", "ceiling-and-floor-functions"], "creation_date": "2022-12-24T20:59:34", "diamond": 0, "votes": 26} {"question_id": "301", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4320839", "title": "Proving the number circle internet meme, where sum of each adjacent numbers is a perfect square", "body": "\nI came across this internet meme and one of my friend mentioned that he thinks that such number circle would always exists for any big enough integer. We tried to prove the hypothesis, but could not find a proof.\nWe defined the circle to be a Hamiltonian cycle of a graph $G_n$ where $V(G_n) = \\{1, ..., n\\}$ and $$E(G_n) = \\{(v_1, v_2)\\mid v_1, v_2 \\in V,~i \\in \\Bbb Z \\text{ where } v_1 + v_2 = i^2 \\}$$\nMy attempt was to show that there exists a hamiltonian path in $G_n$ that starts and ends for vertex of neighbors of $n+1$, but without success.\n\nPS: I built a simple program that brute forcely computes for a hamiltonian cycle of such graph and got this for result\n\n31 : false\n32 : 1-8-28-21-4-32-17-19-30-6-3-13-12-24-25-11-5-31-18-7-29-20-16-9-27-22-14-2-23-26-10-15\n33 : 1-8-28-21-4-32-17-19-30-6-3-13-12-24-25-11-5-20-29-7-18-31-33-16-9-27-22-14-2-23-26-10-15\n34 : 1-3-13-12-4-32-17-8-28-21-15-34-30-19-6-10-26-23-2-14-22-27-9-16-33-31-18-7-29-20-5-11-25-24\n35 : 1-3-6-19-30-34-2-7-18-31-33-16-9-27-22-14-11-25-24-12-13-23-26-10-15-21-28-8-17-32-4-5-20-29-35\n36 : 1-3-6-19-30-34-2-23-26-10-15-21-4-32-17-8-28-36-13-12-24-25-11-5-20-29-7-18-31-33-16-9-27-22-14-35\n37 : 1-3-22-14-35-29-20-5-11-25-24-12-37-27-9-16-33-31-18-7-2-34-15-21-4-32-17-19-30-6-10-26-23-13-36-28-8\n...\n58 : 1-3-6-10-15-21-4-5-11-25-39-42-58-23-13-36-28-53-47-2-34-30-51-49-32-17-19-45-55-26-38-43-57-24-40-41-8-56-44-37-12-52-48-33-31-50-14-22-27-54-46-18-7-9-16-20-29-35\n\nI don't know if this is any clue, but nearly every path segments between each $n$ are reused either partially or in reversed order.\n\nPS2: If anyone is interested in generating their own series, to reduce the burden, here is a javascript source code of generating brute force Hamiltonian cycle of the graph.\nvar size = 32;\nvar maxSqrt = () => Math.sqrt(size * 2 - 1) | 0;\n\nfunction isSquare(x) {\n return x > 0 && Math.sqrt(x) % 1 === 0;\n}\n\nfunction foo(n = 1, step = 0, visited = new Array(size).fill(false)) {\n if (step == size - 1) {\n if (isSquare(n + 1)) {\n return n;\n } else {\n return false;\n }\n }\n\n visited[n - 1] = true;\n\n //find edges\n var minSqrt = Math.ceil(Math.sqrt(n));\n minSqrt = minSqrt === 1 ? 2 : minSqrt;\n\n //recursivly run edges\n for (var i = minSqrt; i <= maxSqrt(); i++) {\n var target = i * i - n;\n if (n != target && !visited[target - 1] && target <= size && target > 0) {\n var temp = foo(target, step + 1, visited.slice(0));\n if (!temp) continue;\n else {\n return n + \"-\" + temp;\n }\n }\n }\n return false;\n}\n\nconsole.log(size + \" : \" + foo());\n\n\nPS3: The question about Hamiltonian path of n numbers seems relevant to this question\n", "comments": ["This question is similar to: Generalisation of this circular arrangement of numbers from $1$ to $32$ with two adjacent numbers being perfect squares. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem.", "This seems to be a duplicate of math.stackexchange.com/questions/4289064/…", "@PrasunBiswas nice. The number of circular solutions (as wanted in this question is in: A071984", "I noticed that Gerbicz's method uses T(c)=25*a[1]+c,25*a[2]-c,25*a[3]+c,25*a[4]-c,...,25*a[n]+(-1‌​)^(n+1)*c this formula and it's property to compute for segments of sequences for the hamiltonian cycle. But how can you be sure that there exists glue between each sequences?", "@PrasunBiswasm, that seems worth posting as an answer!", "FYI, the Mathematica command FindHamiltonianCycle[RelationGraph[(Sqrt[#1 + #2] == Floor[Sqrt[#1 + #2]]) && (#1 != #2) &, Range[k]]] generates a cycle for the numbers $1$ to $k$. (I suspect Gerbicz' dedicated algorithm is more efficient.)", "Here's an efficient algorithm by Gerbicz for finding Hamiltonian cycles for $n\\ge 32$.", "A bit of searching informed me that this conjecture was proved in 2018 by Robert Gerbicz (there is a Hamiltonian cycle for the square-sum chain for all $n\\ge 32$) (see this Mersenneforum post). Related OEIS sequences: 1, 2 and a few others you can find linked there. There's also some Numberphile videos on YouTube on this, see here and here.", "Sorry about the syntax. I'm new to math stackexchange.", "See MathJax tutorial and quick reference for formatting."], "comment_count": 0, "tags": ["graph-theory", "hamiltonian-path"], "creation_date": "2021-11-30T22:02:26", "diamond": 0, "votes": 26} {"question_id": "302", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1700528", "title": "When is a polynomial contained in the ideal generated by its partial derivatives?", "body": "Let $R = k[x_1,\\dots,x_n]$ be a multivariate polynomial ring over a field $k$ of characteristic zero, and let $f\\in R$.\n\nIs there an easy-to-test necessary and sufficient condition on $f$ such that $f$ is in the ideal of $R$ generated by its partial derivatives $\\partial_if$?\n\nGeometrically, $f\\in \\left(\\partial_1 f,\\dots, \\partial_n f\\right)$ is the statement that the map $f:\\mathbb{A}^n\\rightarrow\\mathbb{A}^1$ is a submersion anywhere it is nonzero. (Edit per a comment of @Evangelion045: this isn't quite true. $f\\in(\\partial_1f,\\dots,\\partial_n f)$ implies that $f$ is a submersion everywhere it's nonzero, but in the opposite direction, $f$ being a sumbersion everywhere it's nonzero only implies $f$ is in the radical of the ideal generated by its partials.)\nA sufficient condition is if $f$ is homogeneous of degree $d$, due to Euler's formula\n$$ d\\cdot f = \\sum_i x_i\\partial_i f$$\nI was led to the question by the surprise discovery that the statement is also true if $f$ happens to be the resolvent cubic of a monic quartic polynomial $x^4 - \\sigma_1x^3 +\\sigma_2x^2 - \\sigma_3x +\\sigma_4$, which is an inhomogeneous polynomial of degree three in the four coefficients $\\sigma_1,\\dots,\\sigma_4$ plus an auxiliary variable $\\Lambda$. But the resolvent cubic is, to be sure, weighted homogeneous (with weight $i$ for $\\sigma_i$ and weight $2$ for $\\Lambda$). And sure enough, the proof for Euler's formula generalizes to this case. We have\n$$ d\\cdot f = \\sum_i d_ix_i\\partial_if$$\nwhere $d_i$ is the weight of $x_i$ and $d$ is the weighted degree of $f$. Thus the existence of a set of weights making $f$ weighted homogeneous is also a sufficient criterion. (A google search revealed an analytic generalization in a book of Arnol'd, citing Saito.)\nHowever, it's not quite necessary: any (nonconstant) linear function also trivially has $f\\in \\left(\\partial_1 f,\\dots,\\partial_n f\\right)$ since the latter is the unit ideal, while if its constant term is nonzero then it is not homogeneous with respect to any set of (positive integer) weights.\nDo you know a condition that is both sufficient and necessary?\n", "comments": ["Thanks for answering @BenBlum-Smith.", "Hi @Evangelion045 - I think you're right. It's true that the hypothesis about $f$ being in the ideal implies that the map it induces is a submersion everywhere it's nonzero, but the geometric hypothesis only implies $f$ is in the radical. There's no reason for the ideal to be radical, so that seems the best you can do. I guess I was being sloppy. I'll edit.", "@BenBlum-Smith Hi. I am trying to understand why the condition on $f$ being a submersion anywhere it is nonzero, is equivalent to what you asked; but, using the Nullstellensatz, I can only arrive at $f\\in\\sqrt{(\\partial_1 f,\\ldots,\\partial_n f)}.$ Do you need further assumptions on $f$? I cannot see how to conclude that it is in the ideal, not just in the square root.", "In particular, this example shows that it is easier to test directly if a polynomial is in the ideal generated by its partial derivatives than to test if it is the automorphic image of a weighed homogeneous polynomial.", "The proof that $f$ is not a the automorphic image of a weighted homogeneous polynomial is quite long. One should take the inverse automorphism $\\phi^{-1}$ and divide the proof into the two cases $deg(\\phi^{-1}(x))=deg(\\phi^{-1}(y))=1$, and the alternative (one of the images has degree greater than 1).", "The criterion is too weak, but finding counterexamples is not easy. I was playing around with some polynomials trying to find a counterexample to the Jacobian conjecture in dimension 2, and I had a (nontrivial) pair $P,Q$ with $J(P,Q)=x$. Then applying the endomorphism $x\\mapsto x$ and $y\\mapsto xy$ I obtained that the image of $P$, which is the $f$ in the other comment, satisfies the property.", "@san - how did you come up with that $f$ and what's the argument it's not the automorphic image of any weighted homogeneous polynomial? (Aside: the weighted-homogeneous [or automorphic image thereof] criterion seemed way too weak in any case, because it requires the expression for $f$ as a linear combination of its partials to have a very particular form.)", "For example, $f(x,y)=x^2y+x^2y^2+6x^4y^3+9x^6y^4$ is in the ideal generated by its partial derivatives, but its not the automorphic image of any weighted homogeneous polynomial.", "Building on Eric Wofsey's comment, I don't think one can determine a condition that is both sufficient and necessary. Every automorphic image of a weighted homogeneous polynomial will satisfy the property (this includes the affine functions), but so will the hypothetical counterexamples of the unsolved Jacobian Conjecture, if this conjecture is false.", "@san - not a linear form or a linear functional but a linear function. Affine function, if you prefer.", "How a linear function can have a non-zero constant term?", "Note that this condition can be expressed using only the $k$-algebra structure of $R$: $f$ is in the ideal generated by its partial derivatives iff there exists a $k$-linear derivation $d:R\\to R$ such that $d(f)=f$. In particular, this means that the condition is preserved by any automorphisms of $R$."], "comment_count": 0, "tags": ["algebraic-geometry", "polynomials", "commutative-algebra"], "creation_date": "2016-03-16T10:51:14", "diamond": 0, "votes": 26} {"question_id": "303", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1719104", "title": "Does the category of algebraically closed fields of characteristic $p$ change when $p$ changes?", "body": "EDIT I've now posted this question on mathoverflow. It probably makes sense to post answers over there, unless someone prefers posting here.\nLet $\\mathrm{ACF}_p$ denote the category of algebraically closed fields of characteristic $p$, with all homomorphisms as morphisms. The question is: when is there an equivalence of categories between $\\mathrm{ACF}_p$ and $\\mathrm{ACF}_l$ (with the expected answer being: only when $p=l$)?\nHere are some easy observations:\n\nFirst, any equivalence of categories must do the obvious thing on objects, preserving transcendence degree because $K$ has a smaller transcendence degree than $L$ if and only if there is a morphism $K \\to L$ but not $L \\to K$. \nWe can distinguish the case $p=0$ from the case $p \\neq 0$ because $\\mathrm{Gal}(\\mathbb{Q}) \\not \\cong \\mathrm{Gal}(\\mathbb{F}_p)$. But $\\mathrm{Gal}(\\mathbb{F}_p) \\cong \\hat{\\mathbb{Z}}$ for any prime $p$. So we can't distinguish $\\mathrm{ACF}_p$ from $\\mathrm{ACF}_l$ for different primes $p \\neq l$ in such a simple-minded way. So for the rest of this post, let $p,l$ be distinct primes.\nThe next guess is that maybe we can distinguish $\\mathrm{ACF}_p$ from $\\mathrm{ACF}_l$ by seeing that $\\mathrm{Aut}(\\overline{\\mathbb{F}_p(t)}) \\not \\cong \\mathrm{Aut}(\\overline{\\mathbb{F}_l(t)})$. To this end, note that there is a tower $\\mathbb{F}_p \\subset \\overline{\\mathbb{F}_p} \\subset \\overline{\\mathbb{F}_p}(t) \\subset \\overline{\\mathbb{F}_p(t)}$. The automorphism groups of these intermediate extensions are respectively $\\hat{\\mathbb{Z}}$, $\\mathrm{PGL}_2(\\overline{\\mathbb{F}_p})$, and a free profinite group (the last one is according to wikipedia).\n\nFrom (3), there is at least a subquotient $\\mathrm{PGL}_2(\\overline{\\mathbb{F}_p})$ which looks different for different primes. But I don't see how to turn this observation into a proof that $\\mathrm{Aut}(\\overline{\\mathbb{F}_p(t)}) \\not \\cong \\mathrm{Aut}(\\overline{\\mathbb{F}_l(t)})$ specifically, or that $\\mathrm{ACF}_p \\not \\simeq \\mathrm{ACF}_l$ more generally.\n Note that if we change \"algebraically closed fields of characteristic $p$\" to \"fields of characteristic $p$\", then its easy to distinguish these categories because their subcategories of finite fields look very different.\nAlso, there is a natural topological enrichment of $\\mathrm{ACF}_p$ where one gives the homset the topology of pointwise convergence. I'd be interested to hear of a way to distinguish these topologically-enriched categories.\n", "comments": ["@EricWofsey yeah, I guess I did gloss over that.", "Your first observation is not as trivial as you make it sound: it uses the fact that the totally ordered class of all cardinal numbers has no automorphisms besides the identity.", "@tcamps Yeah, right; but the main thing is that there are automorphisms", "@egreg Both these automorphism groups are cyclic of order 2, generated by a Frobenius element, right?", "I see. My bad, I always forget morphisms.", "@Crostul The automorphism group of the four element field is not isomorphic to the automorphism group of the nine element field. But just looking at the four element field suffices to prove your assertion is false.", "No, not the same morphisms (even though it's true that the categories of finite fields for all $p$ are equivalent).", "I'm taking all field homomorphisms as morphisms. So the subcategory of finite fields includes the automorphism group of each finite field. But this... actually is not different! So I think you're right to disagree after all!", "I don't agree. The subcategories of finite fields are all equivalent to the poset $(\\Bbb{N} , \\mbox{divides})$.", "Note that if we change \"algebraically closed fields of characteristic $p$\" to \"fields of characteristic $p$\", then its easy to distinguish these categories because their subcategories of finite fields look very different."], "comment_count": 0, "tags": ["number-theory", "category-theory", "field-theory", "galois-theory", "positive-characteristic"], "creation_date": "2016-03-29T10:49:31", "diamond": 0, "votes": 26} {"question_id": "304", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4025983", "title": "Finite-Dimensional Homogeneous Contractible Spaces", "body": "Suppose that $X \\subset \\mathbb{R}^n$ is compact, homogeneous and contractible (and thus connected). Does $X$ have to be a point?\nI couldn't think of a non-trivial example, and there isn't a counterexample in the plane. The homogeneous planar continua have been classified (point, circle, pseudo-arc, circle of pseudo-arcs) and the only contractible one is a point. Maybe there is some twisty sort of example in three or four dimensions, though?\nDoes it become true if $\\dim(X) = n $ and is embeddable in $\\mathbb{R}^{n+1}$?\nBy homogeneous I mean for any $x, y \\in X$ there is a homeomorphism $f$ of $X$ with $f(x) = y$. By contractible I mean the identity map on $X$ is homotopic to a constant map. For example the circle is homogeneous but not contractible, and the closed disc is contractible but not homogeneous.\n", "comments": ["@JohnSamples yes : )", "It's infinite-dimensional, right?", "What about $S^{\\infty}$?", "I also just realized that the homogeneous planar continua are known to be the point, the circle, the pseudo-arc and the circle of pseudo-arcs, so that case is actually known. Going to edit the post.", "Fun fact: if $G$ is a compact topological group and $G$ is contractible, then $G$ is the trivial group. This rules out such objects from being counterexamples.", "Nice observation @AdamChalumeau", "For those of you (like me) who where looking for an example which is a manifold, this is not possible: a compact simply-connected manifold is orientable so it has $\\mathbb Z$ for top homology. Hence, by homogeneity, no neighborhood of any point is homeomorphic to a disk, thus the \"local topology\" of a possible counterexample must be pretty nasty.", "Another easy observation: if $X$ contractible is relaxed to $\\pi_n(X)=0$ for $n\\geq 1$ then the pseudo-arc is a (connected) example", "Ok, I added the relevant definitions and removed the superfluity of \"connected.\" Is there a way to adjust Bing's \"two drill bits\" space, perhaps? I forget the exact properties of that space.", "@AlessandroCodenotti You are right, I didn't properly see that you wrote $\\mathbb N$; I read $N$.", "Suppose that $X$ is a contractible CW complex. If $\\dim X = 1$, then $X$ is a tree. Serre proves in Trees that any action of a finite group will have fixed points. That any finite $2$-dimension $G$-complex (with $G$ finite) admits fixed points is the Casacuberta-Dicks conjecture. They proved it originally for $X$ acyclic and $G$ solvable. So even though this is not exactly what you asked for, the problem seems hard (maybe the fact that we can work with the whole group $\\mathsf {Homeo}(X)$ makes things easier, I don't know).", "@AlessandroCodenotti $[0,1]^N$ is not homogeneous. No homeomorphism maps a boundary point to an interior point.", "An easy observation: there are such continua in $\\Bbb R^\\Bbb N$, for example $[0,1]^\\Bbb N$. (Also \"connected\" is redundant in your list of properties, being implied by \"contractible\")", "Ah, right. Thanks :)", "@guidoar $\\mathrm{Homeo}(X)$ acts transitively on $X$. In other words for all $x,y\\in X$ there is an homeomorphism $f\\colon X\\to X$ with $f(x)=y$.", "What's the definition of homogeneous?"], "comment_count": 0, "tags": ["general-topology", "algebraic-topology", "examples-counterexamples", "geometric-topology", "continuum-theory"], "creation_date": "2021-02-14T15:24:24", "diamond": 0, "votes": 25} {"question_id": "305", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3564278", "title": "Maximizing $\\sum_{i,j=1}^{n}|\\operatorname{deg}\\ x_{i}-\\operatorname{deg}\\ x_{j}|^{3}$ over all simple graphs with $n$ vertices", "body": "For a simple graph $G$ on $n$ vertices, let us define \n$$\\mathcal{I}_{n}(G)=\\sum_{i,j=1}^{n}|\\operatorname{deg}\\ x_{i}-\\operatorname{deg}\\ x_{j}|^{3}$$\nI am highly interested in finding $\\sup \\mathcal{I}_{n}$ over all graphs with $n$ vertices (or at least some tight upper bound). What I have tried myself, was noticing that $\\mathcal{I}_{n}$ must be maximized by a threshold graph:\n\nSketch of proof: This index $\\mathcal{I}_{n}$ is a convex function of degree sequence\n $\\operatorname{deg} \\ x_{1},...,\\operatorname{deg} \\ x_{n}$. Call the set of all such graphic sequences $D$. Then we\n can look on $D^{∗}=\\mathcal{Con}D$ - a convex hull of $D$. It must then attain it's\n maximum on some extreme point of $D^{∗}$. It can be shown, that such\n extreme points of $D^{*}$ are exactly those corresponding to threshold\n graphs.\n\nBut this didn't lead me too far. I will be glad for any insight.\nSimulations show that the optimum occurs for $k$ isolated vertices and a complete graph on the other $n−k$ where $k=\\lfloor\\frac{n+1}5\\rfloor$. The same count occurs for $k$ vertices of degree $n−1$ and no other edges so the other $n−k$ have degree $k$. Do You have any ideas on how to prove it?\n", "comments": [], "comment_count": 0, "tags": ["combinatorics", "graph-theory", "discrete-optimization", "extremal-combinatorics"], "creation_date": "2020-02-29T06:41:46", "diamond": 0, "votes": 24} {"question_id": "306", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2674160", "title": "Is there an endofunctor of the category of sets that maps $\\kappa$ to $\\kappa^+$?", "body": "For already some time I am slightly bothered by the following question about endofunctors of the category of sets: Is there an endofunctor of set which maps each infinite cardinal $\\kappa$ to a set of size $\\kappa^+$ (i.e., the first larger cardinality)?\nIt is clear that under GCH, an example of such a functor is the power set functor as $2^\\kappa = \\kappa^+$. But without this axiom, I don't know of any other example. Anyway, I was interested whether one can construct such functor without any additional set axioms. To rephrase the question once mo: Is it always true under ZFC that there exists an endofunctor of the category of sets that maps $\\kappa$ to a set of size $\\kappa^+$?\nMy master thesis advisor suggested that such functor might be found as a subfunctor of the powerset functor, but I never managed to make it work. Am I missing something?\n", "comments": ["@EricWofsey About the extension to finite cardinals, I think there is one. The requirement for finite sets is very strong. If you want to map $n$ to $n+1$ you are forced to define it as $F(x) = x \\cup \\{*\\}$ where $*$ is a new point. Anyway, thankfully I can use the point $*$ for my extension. Let us first define $$ P_{\\geq\\kappa} X = \\{ A \\subseteq X : |A| \\geq \\kappa \\} \\cup \\{*\\} $$ this extends to maps as $ P_{\\geq\\kappa} f(A) = f(A) $ if $ |f(A)| \\geq \\kappa $ and $*$ otherwise. Now, $FX = P_{\\geq \\aleph_0}X + X$ if my cardinal arithmetic does not fail me.", "@EricWofsey I like the construction of $\\kappa^+$, thanks for sharing. It's good to know that there is such a thing.", "Anyways, it seems highly likely to me that the answer to your question is \"no\", but proving a negative answer would of course require some serious set theory since such a functor does exist in some models.", "Incidentally, it is not clear to me that even assuming GCH there is a functor which sends $\\kappa$ to a set of size $\\kappa^+$ for all cardinals (including finite ones).", "Note that there is a \"canonical\" way to construct a set of size $|X|^+$ from a set $X$: take the set of isomorphism classes of well-orderings of subsets of $X$. This construction is functorial with respect to bijections in an obvious way (which turns out to be rather trivial; every automorphism gets sent to the identity). Unfortunately this cannot be extended to be functorial with respect to all functions.", "@Berci: But global choice is not a theorem of ZFC. Not even global choice \"from a parameter\". Also, how does this definition preserves the identity and composition (together)?", "But you don't mention any condition for the functor, so you can just fix an element from each nonempty set (using choice, of course) and send all functions to the constant one, mapping everything to the picked element of the codomain.", "@NoahSchweber Thanks, I am just little slow.", "@Jakub Yes, it's clearly first-order definable.", "Also what exactly do you mean by definable? Is it first order definable?", "You still need to define the functor on morphisms which is definetely not easy.", "I'm confused. Isn't the fact that the successor function $\\kappa\\mapsto\\kappa^+$ is itself definable from $\\sf ZF$ is enough to conclude there is such functor?"], "comment_count": 0, "tags": ["category-theory", "set-theory"], "creation_date": "2018-03-02T12:44:24", "diamond": 0, "votes": 24} {"question_id": "307", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/674591", "title": "Relationship between intersection and compositum of fields", "body": "This issue came up in a number theory lecture today. Let $K$ be a number field and let $L/K$ be an abelian (finite Galois) extension. Then there exists a primitive $m$th root of unity $\\zeta_m$ so that $K(\\zeta_m)\\cap L=K$ so that $m$ satisfies a number of nice qualities.\nWe tried to apply the following fact: if $(m,p)=1$ for every prime $p\\in\\mathbb Z$ ramifying in $\\mathcal O_L$, then $\\mathbb Q(\\zeta_m)\\cap L=\\mathbb Q$.\nThe issue in moving this fact up to $K/\\mathbb Q$ is that, for general fields $E_1,E_2,E_3$, $E_1(E_2\\cap E_3)\\neq E_1E_2\\cap E_1E_3$. However in our case, we do have that $K(\\zeta_m)\\cap L=K$, where $m$ is chosen via primes ramifying from the base field $\\mathbb Q$. Perhaps if $K/\\mathbb Q$ were Galois, this might be easier, but the question is this:\n\nGiven three fields $F,K,L$ contained in some larger field $M$, under what (minimal) conditions do we have $F(K\\cap L)=FK\\cap FL$?\n\n", "comments": [], "comment_count": 0, "tags": ["number-theory", "field-theory", "galois-theory", "algebraic-number-theory", "class-field-theory"], "creation_date": "2014-02-12T19:13:34", "diamond": 0, "votes": 24} {"question_id": "308", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/338453", "title": "When does a modular form satisfy a differential equation with rational coefficients?", "body": "Given a modular form $f$ of weight $k$ for a congruence subgroup $\\Gamma$, and a modular function $t$ for $\\Gamma$ with $t(i\\infty)=0$, we can form a function $F$ such that $F(t(z))=f(z)$ (at least locally), and we know that this $F$ must now satisfy a linear ordinary differential equation\n$$P_{k+1}(T)F^{(k+1)} + P_{k}(T)F^{(k)} + ... + P_{0}(T)F = 0$$\nWhere $F^{(i)}$ is the i-th derivative, and the $P_i$ are algebraic functions of $T$, and are rational functions of $T$ if $t$ is a Hauptmodul for $X(\\Gamma)$.\nMy question is the following:\n\ngiven a modular form $f$, what are necessary and sufficient conditions for the existence of a modular function $t$ such that the $P_i(T)$ are rational functions?\n\n\nFor example, the easiest sufficient condition is that $X(\\Gamma)$ has genus 0, by letting $t$ be a Hauptmodul.\nBut, this is not necessary, as the next condition will show.\nAnother sufficient condition is that $f$ is a rational weight 2 eigenform. I can show this using Shimura's construction* of an associated elliptic curve, and a computation of a logarithm for the formal group in some coordinates (*any choice in the isogeny class will work).\nTrying to generalise, I have thought of the following: if $f$ is associated to a motive $h^i(V)$ of a variety $V$, with Artin-Mazur formal group $\\Phi^i(V)$ of dimension 1, then we can construct formal group law a-la Stienstra style, and get a logarithm using the coefficients of powers of certain polynomials. Since the dimension is 1, there will actually be a single polynomial that we take powers of, making the coefficients have a rather simple recurrence relation, forcing our $P_i$ to be rational.\nNow, some people, without naming names, believe that rational eigenforms should correspond to the middle cohomology of certain rational Calabi-Yai varieties. I'm not entirely certain that such people exist. Probably.\nIf this is true, then this should answer my question for rational eigenforms.\nPutting non-eigenforms aside, since I'm not interested as much in them, we are left with non-rational eigenforms. We can try to perform the same Stienstra construction, but this time we get that the galois orbit of $f$ is associated to a \"formal group law\" of a motive with dimension greater than one. This will make for an interesting recurrence for the vector of the galois orbit, but not necessarily for each form individually, as the isomorphism of formal groups laws (between Stienstra's and those with the modular forms as logarithm) will scramble them together.\nI realise this last paragraph might be difficult to understand, for the wording is clumsy, and the mathematical notions are even worse. If you're really interested in this, I'd be happy to elaborate.\n", "comments": [], "comment_count": 0, "tags": ["complex-analysis", "number-theory", "algebraic-geometry", "modular-forms"], "creation_date": "2013-03-22T19:43:17", "diamond": 0, "votes": 24} {"question_id": "309", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/95895", "title": "Analyzing a class of vertex-deletion games", "body": "As part of the discussion on this question (Permutation Game Redux), a simple vertex-deletion game was proposed. The game is very simple.\n\nDisconnect. Players alternately remove vertices from a graph $G$. The player that produces a fully disconnected graph (i.e., a graph with no edges) is the winner.\n\nBecause the game is impartial, the Sprague-Grundy theory applies: each game is equivalent to a nim-heap of some size (its nim-value), which can be calculated as the mex (minimum excluded nim-value) of its options. These nim-values can then be used to compute the nim-values of disjunctive sums of games in the usual way.\nOne would like to apply this theory within a single game, e.g., to break a graph into its connected components, calculate their nim-values, and then combine them to find the value of the overall graph. Unfortunately, this doesn't work. The problem is that the win condition is not standard: the game ends before all moves are exhausted (or, equivalently, the allowed moves in one component of the graph depend on the other components).\nIt is not hard to see that for any graph $G$ and any even $n$, the game $G \\cup \\bar{K}_n$ is equivalent to $G$ (where $\\bar{K}_n$ is the edgeless graph on $n$ vertices). To prove it, we need to show that the disjunctive sum $G + G\\cup\\bar{K}_n$ is a second-player win. The proof is by induction on $|G|+n$. If $G$ is edgeless, then the first player loses immediately (both games are over). Otherwise, the first player can move in either $G$, and the second player can copy his move in the other one (reducing to $G' + G'\\cup \\bar{K_n}$ with $|G'|=|G|-1$); or, if $n\\ge 2$, the first player can move in the disconnected piece, and the second player can do the same (reducing to $G + G\\cup\\bar{K}_{n-2}$).\nThis shows that any graph $G$ is equivalent to $H \\cup K_p$, where $H$ is the part of $G$ with no disconnected vertices, and $p=0$ or $1$ is the parity of the number of disconnected vertices in $G$. All games in an equivalence class have the same nim-value, and moreover, the equivalence relation respects the union operation: if $G \\sim H \\cup K_p$ and $G' \\sim H' \\cup K_{p'}$ then $G \\cup G' \\sim (H \\cup H')\\cup K_{p\\oplus p'}$. Moreover, one can see that the games in $[H \\cup K_0]$ and $[H \\cup K_1]$ have different nim-values unless $H$ is the null graph: when playing $H + H \\cup K_1$, the first player can take the isolated vertex, leaving $H+H$, and then copy the second player's moves thereafter. \nBeyond this, are there any other general decomposition or equivalence results? Any extension of the Sprague-Grundy theory to this class of games? In particular, is there some more refined equivalence relation still to be found such that all games in $[G]$ have the same nim-value, and $[G \\cup H]$ can be determined in terms of $[G]$ and $[H]$?\n", "comments": ["I see -- interesting -- the supposed proof of equivalence fails because it's not always possible to mirror moves because, as you wrote, a move's admissibility depends on the other components, and it works in your special case because disconnected vertices are the only type of components that don't affect admissibility.", "@joriki: I thought that initially as well, but in fact it's not the case. As a counterexample, consider the graph $K_1 \\cup G \\cup G$, where $G$ is not edgeless. Your argument suggests that this should be equivalent to $K_1$, which is a second-player win (since it is disconnected, there are no more moves allowed). But in fact the first player has a winning strategy: first delete the $K_1$, producing $G \\cup G$, and then copy the second player's moves. In other words, $K_1 = 0$ and $G \\cup G = 0$, but $K_1 \\cup (G \\cup G) \\neq K_1 + (G \\cup G)$.", "Your insight generalizes to all connected components, not just disconnected vertices. The game is completely determined by the parities of the numbers of instances of all types of connected components; adding two isomorphic connected components leads to an equivalent game, since you can always play in one when the other player plays in the other, until they're reduced to an even number of disconnected vertices, which you've handled."], "comment_count": 0, "tags": ["graph-theory", "combinatorial-game-theory"], "creation_date": "2012-01-02T12:39:17", "diamond": 0, "votes": 24} {"question_id": "310", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4261226", "title": "Can a free complete lattice on three generators exist in $\\mathsf{NFU}$?", "body": "Also asked at MO.\nIt's a fun exercise to show in $\\mathsf{ZF}$ that \"the free complete lattice on $3$ generators\" doesn't actually exist. The punchline, unsurprisingly, is size: a putative free complete lattice on $3$ generators would surject onto the class of ordinals.\nThis obstacle isn't a problem however in the context of $\\mathsf{NFU}$; on the other hand, recursive constructions (which were utterly unproblematic in $\\mathsf{ZF}$) now become more complicated. So the situation is unclear to me: is it consistent with $\\mathsf{NFU}$ that there is a free complete lattice on $3$ generators?\n\nI don't have much experience with $\\mathsf{NFU}$, so as far as I know it's possible that in in this context the phrase \"free complete lattice on $3$ generators\" is ambiguous - specifically, I'm a bit worried about the word \"free\" in this context. To make things precise, I'll use the homomorphism-based notion of freeness, that is, I'm looking for a complete lattice $L$ with three distinguished elements $a,b,c$ such that for every other complete lattice $M$ and distinguished elements $u,v,w\\in M$ there is exactly one complete lattice homomorphism $L\\rightarrow M$ sending $a$ to $u$, $b$ to $v$, and $c$ to $w$.\n\n\nI would also be interseted in the answer to the same question for other set theories admitting a universal set, such as $\\mathsf{GPK}_\\infty^+$. However, my current impression is that $\\mathsf{NFU}$ is by far the best-understood such theory, so it seems like a good starting point.\n", "comments": [], "comment_count": 0, "tags": ["logic", "set-theory", "lattice-orders", "universal-algebra", "alternative-set-theories"], "creation_date": "2021-09-26T17:21:39", "diamond": 0, "votes": 24} {"question_id": "311", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2924934", "title": "Maximal subgroups that force solvability.", "body": "For which finite groups $M$ is it the case that every finite group $G$ with $M$ as a maximal subgroup solvable?\nIf $M$ satisfies this condition then $M$ is solvable. Also, if $M$ is abelian then $M$ satisfies this condition. Futhermore, I believe that if $M$ is nilpotent and if all 2-subgroups of $M$ are normal subgroups of $M$ (if Sylow 2-subgroups of $M$ are abelian or quaternion, for example) then $M$ satisfies this condition (proof below).\nMore specific questions:\n1) Is there a non-nilpotent group that satisfies this condition?\n2) Which 2-groups satisfy this condition?\nApparently, the dihedral group of order 8 satisfies this condition (see Mikko Korhonen's comment on this post).\nAlso, if $M\\times N$ satisfies this condition then $M$ and $N$ both satisfy this condition.\n\n(This proof is adapted from j.p.'s answer to the linked question).\nLet $G$ be minimal such that $G$ is not solvable and such that $G$ contains a maximal subgroup $M$ that is nilpotent and whose 2-subgroups are normal. If $M$ contains a nontrivial normal subgroup $N$ of $G$ then $G/N$ contradicts the minimality of $G$. Thus, $M$ does not contain nontrivial normal subgroups of $G$. In particular, $N_G(P)=M$ for all Sylow $p$-subgroups $P$ of $M$. Then $P$ is a Sylow $p$-subgroup of $N_G(P)$ so $P$ is a Sylow $p$-subgroup of $G$. This shows that $M$ is a Hall subgroup of $G$.\nIf $P$ is a Sylow $p$-subgroup of $M$ and if $Q$ is a nontrivial normal subgroup of $P$ then $N_G(Q)=M$ which has a normal $p$-complement. For $p=2$, Frobenius' normal $p$-complement theorem gives that $G$ has a normal $p$-complement. For $p\\geq3$, Thompson's normal $p$-complement theorem or Glauberman's normal $p$-complement theorem gives that $G$ has a normal $p$-complement (since you only have to consider characteristic $p$-subgroups).\nThus, for each prime $p$ dividing the order of $M$, $G$ has a normal $p$-complement. Then $M$ has a normal complement $N$ in $G$. Since $M$ is solvable but $G$ is not solvable, $N$ is not solvable. In particular, $N$ does not admit a fixed-point-free automorphism of prime order. If $m\\in Z(M)$ has prime order then $C_N(m)$ is nontrivial. Then $C_N(m)M$ is a subgroup of $G$ that properly contains $M$ so $C_N(m)M=G$ by the maximality of $M$. Comparing cardinalities shows that $C_N(m)=N$ so $m\\in Z(G)$. Then $\\langle m\\rangle$ is a nontrivial normal subgroup of $G$ contained in $M$ which is a contradiction.\n", "comments": ["No. From the condition $M$ maximal in $G$ you get only an handle on everything between $M_G$ and $G$, so from a group-theoretic point one should always assume $M_G=1$. You cannot control direct factors: You cannot get rid of $H$ in $M\\times H$ maximal in $G\\times H$ without factoring out $M_G$. OK, you could require $M$ not to be the direct product of two non-trivial subgroups, but this cannot be used in this situation. On the contrary with the assumption $M_G=1$, $M$ being direct product (e.g. being product of its $p$-Sylow subgroups for different $p$ if it is nilpotent) is quire useful.", "Can the condition on $M_G$ be strengthened to get a condition on $M$ that does not depend on $G$?", "With $M_G:=\\cap_{g\\in G} M^g$, corollary 4.5 in On locally finite groups with locally nilpotent maximal subgroups by B. Bruno and S Schuur (Arch. Math. Vol 61, pp. 215-20) shows that $M$ nilpotent and $M_G$ not a $2$-group implies the solvability of $G$. Without conditions on $M_G$ one can take direct products with the simple group $PSL_2(17)$ (whose $2$-Sylow subgroups isomorphic to $D_{16}$ are maximal subgroups) with other groups for non-solvable constructions.", "@ThomasBrowning: In \"A condition for the solvability of a finite group\" (1961) Deskins shows that if $G$ has a nilpotent maximal subgroup of nilpotency class $\\leq 2$, then $G$ is solvable. In particular, the dihedral group of order $8$ forces solvability as a maximal subgroup", "@ThomasBrowning: For your question 1), perhaps the non-trivial semidirect product $C_3 \\rtimes C_4$ is an example.", "Interesting. I computed maximal subgroups of a number of simple groups a while ago. It looked like all small nontrivial semi-direct products of two cyclic groups occurred with the exception of the dihedral group of order 8. I still don't know whether the dihedral group of order 8 forces solvability.", "About 2), just to give an example: if $p = 2^n - 1$ is a Mersenne prime ($n > 3$), then $PSL(2,p)$ has the dihedral group of order $2^{n-1}$ as a maximal subgroup.", "I wonder if there is a reduction to the case where $G$ is almost simple? That is, could we show that $M$ \"forces solvability as a maximal subgroup\" if and only if $M$ is solvable and $M$ is not a maximal subgroup of any almost simple group? If the answer is yes, then your questions are answered by a paper of Li and Zhang from 2011 (Proceedings of the LMS), who classify the solvable maximal subgroups of almost simple groups.", "A famous example here is when $M$ is a $p$-group for an odd prime $p$."], "comment_count": 0, "tags": ["group-theory", "finite-groups", "solvable-groups"], "creation_date": "2018-09-20T22:48:14", "diamond": 0, "votes": 23} {"question_id": "312", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/886041", "title": "Conjecture---Identity for Sieve of Eratosthenes collisions.", "body": "Let\n$$\\beta(n,k) = \\max_{d \\leq k}(d|n)$$\n$$S(k)= \\sum_{n=1}^{k!} \\beta(n,k),$$\n$\\hspace{20mm}$and\n$$T(k)=\\# \\{ ~i\\cdot j~~\\big|_{i=1}^k \\big|_{j=1}^{k!} \\}$$\nDoes $$S(k)=T(k)?$$\nSee OEIS A126959.\nReplace $k!$ in $S,T$ with $\\exp (\\psi(k) )$, where $\\psi(\\cdot)$ is second Chebyshev function, to get A101459.\n", "comments": ["@GerryMyerson, One sequence fails at $k=10$ and the other at $k=17$. Yikes!", "oeis.org/A126959 has been calculated out to $n=36$, so you have verified $S(k)=T(K)$ out to $k=36$?", "For which values of $k$ have you verified that $S(k)=T(k)$?", "@frogeyedpeas It usually denotes the number of elements in the set.", "What is T(n,k)?"], "comment_count": 0, "tags": ["number-theory"], "creation_date": "2014-08-02T22:24:21", "diamond": 0, "votes": 23} {"question_id": "313", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/738393", "title": "Difficult integral for a marginal distribution", "body": "I am trying to derive a marginal probability distribution for $y$, and failed, having tried all methods to solve the following integral:\n$$\n\\operatorname{p}\\left(y\\right) =\n\\int_0^{1/\\sqrt{\\,2\\pi\\,}}\\!\\!\\!\\!\n\\frac{\\sqrt{2/\\pi}\\,\\,\\mathrm{e}^{-y/\\left(2z\\right)}}{\\sqrt{y\\, z}\\,\\, \\sqrt{-\\log \\left(2\\pi\\,z^2\\right)}}\n\\,\\mathrm{d}z\\quad \\mbox{with}\\quad y>0.\n$$\nIt is easy to verify that\n$$\\int_0^\\infty \\int_0^{\\frac{1}{\\sqrt{2 \\pi }}} \\frac{\\sqrt{\\frac{2}{\\pi }} e^{-\\frac{y}{2 \\,z}}}{\\sqrt{y\\, z} \\sqrt{-\\log \\left(2 \\pi \\, z^2\\right)}} \\, \\mathrm{d}z \\,\\mathrm{d}y=1$$\nAfter some work, figured out that, remarkably, we can get the fist moment, $\\mathbb{E}(y)=\\frac{1}{2 \\sqrt{\\pi }}$ and $\\mathbb{E}(y^2)=\\frac{\\sqrt{3}}{2 \\pi }$ without the density.\n", "comments": ["Wolfram is also unable to solve it.", "I find $$\\mathbb{E}(y^n) = \\frac{2^{n/2}\\Gamma\\Big(n+\\frac12\\Big)}{\\pi^{(n+1)/2}\\sqrt{n+1}}.$$", "Yes, that's true. I'll keep trying a few methods to evaluate it, but if they don't work, this is still a rather nice form I would argue.", "@Zachary this kind of integrals is notoriously resistant to be evaluated in closed form. I'm pretty sure there isn't much to be done here :()", "I was able to simplify the integral to $$4(2\\pi)^{-1/4}\\frac{1}{\\sqrt{z}} \\int_0^\\infty \\exp\\left(-ze^{2t^2}\\right) e^{-t^2}\\,dt,$$ where $z=\\sqrt{\\frac{\\pi}{2}}y$.", "Bounty: Intuition for Conditional Expectation (1. by linking bounty question here, this question gets attention because the bounty question is linked to this question. 2. by linking bounty question here, nero gets to see bounty question and so might answer bounty question)", "Indeed I am working on the product of 1) Chi-square distribution and 2) distribution of the density of a standardized Gaussian.", "Looks as if conditioned on the value of $Z$, $Y$ is a Gamma random variable with parameters $\\left(\\frac{1}{2}, \\frac{1}{2\\sqrt{Z}}\\right)$ and $Z$ had marginal pdf of the form $\\frac{1}{\\sqrt{-\\log(2\\pi z^2)}}$ on $(0,1/\\sqrt{2\\pi})$.", "Sorry it was a typo in the posting that I fixed (an extra z term). I still can't find solution."], "comment_count": 0, "tags": ["integration", "probability-theory", "probability-distributions", "definite-integrals", "marginal-distribution"], "creation_date": "2014-04-03T09:14:43", "diamond": 0, "votes": 23} {"question_id": "314", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4059088", "title": "Does every topological $n$-manifold ($n>0$) admit an embedding into $\\Bbb R^{2n}$? If not, what $n$-manifold does not embed into $\\Bbb R^{2n}$?", "body": "The strong Whitney Embedding Theorem tells us that every smooth n-manifold (n>0) admits a smooth embedding into $\\mathbb{R}^{2n}$. Also, every topological $n$-manifold admits an embedding into $\\mathbb{R}^{2n+1}$ (Munkres' Topology. Exercise §50.7).\nMy question now: can this last bound be lowered to $2n$? And if not, which topological $n$-manifold isn't embeddable into $\\mathbb{R}^{2n}$? (Whitney's embedding theorem already tells us that such a manifold cannot admit a smooth structure)\n", "comments": ["@MoisheKohan Thanks", "It's likely a wasted bounty. The result is known in compact case and, most likely, is not in the literature in the noncompact case. If one wants to find out, the best thing to do is to email Prof. Washington Mio (at Florida State University) and ask him directly. He is a coauthor of the paper Bryant, J. L.; Mio, W., \"Embeddings in generalized manifolds,\" with a result closest to the OP, answering it in the case of compact topological manifolds.", "@IvinBabu: If you want references to the compact case, see my answer at the linked Mathoverflow question.", "@JohnSamples Can you provide materials which elaborates the statement you gave- that any compact topological manifold with dimension not a power of $2$ can be embedded into $\\Bbb R^{2n}$", "@LéoMousseau Yes, beware any people telling you that such an inherently attractice theorem is known and follows readily from known results. So what, they could prove it in 2 or 3 pages and get a paper in the Annals - but they just don't have the time? A bit suspicious . . .", "@C.F.G Munkres embeds into $\\mathbb R^{2n+1}$. Here we consider $\\mathbb R^{2n}$.", "I wonder why most of authors call this theorem a \"hard theorem\" while its proof is inside the Munkres' Book as exercise?", "It is known that smooth (or more generally PL) manifolds embed into $\\mathbb R^{2n}$. See ams.org/journals/tran/1971-157-00/S0002-9947-1971-0278314-4/…", "@JohnSamples It seems strange that such a basic result (not basic in the sense of difficulty of proving) isn't written down anywhere...", "I don't think it's been formally proven but it's generally accepted as true. Any counterexample would have to be non-compact and have dimension $2^k$.", "@AlessandroCodenotti I feel the same way. I dont see how this follows from any of the statements made in those papers. The commenters also didn't seem too confident in their ideas of proving this.", "@AlessandroCodenotti I looked through the papers suggested by the first link in Sergey Melikhov's comment to Andy Putnam's answer and I was not able to find the desired statement. (The second link in Melikhov's comment seems to be broken.) I guess the situation is that the embedding into $\\mathbb{R}^{2n}$ somehow follows from the theorems in the Bryant-Mio and Johnston papers, but this isn't obvious to me as a non-expert.", "mathoverflow.net/questions/34658/… whoops I copied the wrong link earlier. The answer is positive by comments here", "The answer to your questions seems to be positive by various comments spread here math.stackexchange.com/questions/4059088/…", "@AlessandroCodenotti exactly, every knot is homeomorphic to S^1 and is therefore embeddable in R^2.", "@Henno aren't knots just $S^1$ as topological manifolds though?", "There are knots in $\\Bbb R^3$ that don’t have a planar embedding."], "comment_count": 0, "tags": ["general-topology", "manifolds", "differential-topology"], "creation_date": "2021-03-12T06:20:42", "diamond": 0, "votes": 23} {"question_id": "315", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2288648", "title": "Does the sum of reciprocals of all prime-prefix-free numbers converge?", "body": "Call a positive integer $n$ prime-prefix-free if for all $k \\ge 1$, $\\lfloor \\frac{n}{2^k} \\rfloor$ is not an odd prime. (Odd because otherwise the property is trivial, as every integer greater than $3$ has $10_2=2$ or $11_2=3$ as a proper binary prefix.)\nDoes the sum of reciprocals of all prime-prefix-free numbers converge?\nI know that the sum of reciprocals of all prime prime-prefix-free numbers converges, using the Kraft-McMillan inequality and the fact that their binary representations form a prefix-free set.\nBut this doesn't seem like much of a starting point for the whole problem, since a number being prime-prefix-free isn't related to whether its factors are (except when the other factor is a power of $2$). I'm willing to assume Cramér's conjecture if that helps, limiting how many bits must be appended to make a number prime.\n", "comments": ["How many prime-prefix-free numbers less than $n$ are there asymptotically?", "The sum up to 10^11 is 0.75523... and I doubt more than 2-3 of these digits carry over to the infinite sum.", "Mirko, in that case none of the values would be integers for odd $n$, so every odd number would be prime-prefix-free and the sum of reciprocals would diverge.", "I wonder if you know the answer for the version of your question when, in the definition of prime-prefix-free, you replace $\\lfloor\\frac{n}{2^k}\\rfloor$ with $\\frac{n}{2^k}$. It looks like it would be easier to handle that version(modified one), though I do not have an answer for either. (I would guess divergent, for what seems to me the easier version.) Nice question!", "Made a stupid mistake in my \"answer\". Estimating the sum of reciprocals of prime-prefix numbers not exceeding $n$ by $$\\sum_{\\text{odd prime }p\\le n} \\sum_{k\\ge 0: 2^k p\\le n} \\sum_{m=2^k p}^{\\big((p+1)2^k-1\\big)\\wedge\\, n}\\frac{1}{m}$$ gives upper estimate of $\\log_2 n \\times \\log\\log n$, which is even bigger than the sum of reciprocals of all numbers. This is because there are too many prime-prefix primes.", "I did when I couldn't find it either, it's awaiting approval. oeis.org/draft/A287117", "So your sequence begins $1,2,3,4,5,8,9,16,17,18,19,32,33,36,37,\\dots$ ? I'm surprised to not find this in the OEIS - perhaps you could add it?", "For a random integer $n$, the probability of $n, n/2, n/4, \\cdots$ are all not being a odd prime is approximately $\\prod (1-\\frac{1}{\\ln n-k\\ln 2})$. Also, $\\prod (1-\\frac{1}{\\ln n-k\\ln 2})\\approx\\prod (1-\\frac{\\ln2}{\\ln n-k\\ln 2})=\\frac{\\ln n-(k+1)\\ln 2}{\\ln n}=O(\\frac{1}{\\ln n})$ (telescoping product, and $\\ln n-(k+1)\\ln 2$ must be smaller than $\\ln 2$). Therefore the sum is \"approximately\" $\\sum\\frac{1}{n \\ln n}$ and this sum diverges. Of course, this is not a proof..."], "comment_count": 0, "tags": ["number-theory", "convergence-divergence", "prime-numbers"], "creation_date": "2017-05-19T22:15:08", "diamond": 0, "votes": 22} {"question_id": "316", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2097510", "title": "Found an recursive identity (involving a continued fraction) for which some simplification is needed.", "body": "This is my second question in this forum; as I previously explained it, I am a \"hobbyst\" mathematician and not a professional one; I apologize by advance if something is wrong in my question.\nI enjoy doing numerical computations on my leisure time, and at the end of year 2015, I was working on some personal routines related to the world of ISC. With the help of these pieces of code, I detected algorithmically several identities; one was already described here and solved (see this question regarding a continued fraction for tanh). After having spent some time on another one a year ago, I would like to get some help for simplifying what I found. This new continued fraction is:\n$$\n\\mathcal{K}\\left(k,x\\right)=\\operatorname*{K}_{n=1}^{\\infty} \\frac{\n\\left(n+1\\right)\\left(\\left(k-1\\right)x-n\\right)k}{\\left(n+1\\right)\\left(k+1\\right)} \\tag{1} \\label{1}\n$$\nThe previous notation is the one I use; I find it convenient and it can be found for instance in Continued Fractions with Applications by Lorentzen & Waadeland, but I know that some people don't like it; it has to be read the following way:\n$$\na_0 + \\operatorname*{K}_{n=1}^{\\infty} \\frac{b_n}{a_n} = a_0 + \\cfrac{b_1}{a_1 + \\cfrac{b_2}{a_2 + \\cfrac{b_3}{a_3 + \\dotsb}}}\n$$\nI found many partial formulas, some of them being very nice, involving hypergeometric functions or the Lerch Phi function. But of course I was rather interested by a fully general identity. I finally found something but that would need to be simplified now and this is what I am asking here. I would be very happy to finally see a nice identity for this continued fraction. Maybe such an identity could be published somewhere (if it happens to be interesting) and I would happily leave it to anyone who would have taken some time on it.\n\nIs it true that:\n $$\\mathcal{K}\\left(k,x\\right)=\n \\frac{\\Gamma\\left(x+1\\right)\\Gamma\\big((k-1)x\\big)}{\\Gamma\\left(kx\\right)}\n\\times\\mathcal{L}\\left(k,x\\right)$$\n\nwhere,\n$$\\mathcal{L}\\left(k,x\\right)=\\frac{k^{kx}}{2\\left(k-1\\right)^{\\left(k-1\\right)x-1}}+\\sum_{i=0}^\\infty\n\\big(\n\\alpha\\;\\mathcal{K}(k, x+i)\n-\n\\beta\\;\\mathcal{K}(k, x+i+1)\n\\big)$$\nand,\n$$\\alpha=\\frac{\\left(k-1\\right)^{\\left(k-1\\right)i}\\Gamma\\left(k\\left(i+x\\right)\\right)}{k^{ki}\\,\\Gamma\\left(1+i+x\\right)\\Gamma\\left(\\left(k-1\\right)\\left(x+i\\right)\\right)}\\\\\n\\beta=\\frac{\\left(k-1\\right)^{\\left(k-1\\right)i+k}\\Gamma\\left(k\\left(1+i+x\\right)\\right)}{k^{k\\left(i+1\\right)}\\Gamma\\left(1+i+x\\right)\\Gamma\\left(\\left(k-1\\right)\\left(x+i\\right)+k\\right)}$$ \nSince I work with empirical computations, I have to say that this formula is rather difficult to check because convergence is rather slow, but it is a result I managed to get by gathering various other materials.\n\nIn case someone would wonder whether it is worth spending some time on it or not, I can provide some partial results like nice special values:\n$$\n \\mathcal{K}\\left(2, 1/2\\right)\\;=\\;\\pi/2-2\n \\mathrm{,}\n \\qquad\n \\mathcal{K}\\left(4, 1/2\\right)\\;=\\;4\\pi\\sqrt{3}/9-2\n \\qquad\\mathrm{etc.}\n$$\nThe case $k$ being an integer\nWhen $k$ is an integer, $k\\geq3$, the continued fraction follows a functional identity:\n$$\n \\mathcal{K}\\left(k,x\\right)\\;=\\;\n% \\displaystyle\\frac{k^{kx} \\Gamma{\\left(x+1\\right)}\\Gamma{\\left(\\left(k-1\\right)x\\right)} }{2\\left(k-1\\right)^{\\left(k-1\\right)x-1}\\Gamma{\\left(kx\\right)}} + \\mathcal{K}\\left(k, x+1\\right)g_k\\left(x\\right)\n g_k\\left(x\\right) + \\left(\\displaystyle\\frac{k-1}{k}\\right)^k \\displaystyle\\frac{\\Gamma{\\left(\\left(k-1\\right)x\\right)}\\Gamma{\\left(k\\left(x+1\\right)\\right)}}{\\Gamma{\\left(\\left(k-1\\right)x+k\\right)}\\Gamma{\\left(kx\\right)}}\\mathcal{K}\\left(k, x+1\\right)\n$$\nwhere $g$ is a sequence of rational functions. Unfortunately I couldn't find a general form for it but I could easely compute about 30 of them with the help of numerical pieces of software. The best I could do here was to build a triangle of integer coefficients needed for building a given $g_k$ function.\nLet's call $m_{(a,b)}$ the coefficient in the $a^\\textrm{th}$ row and $b^\\textrm{th}$ column from the following triangle:\n$$\n \\begin{array}{rrrr@{\\qquad}l}\n -8&&&&\\textrm{for $k=3$}\\\\\n -20&2&&&\\textrm{for $k=4$}\\\\\n -40&12&-24&&\\textrm{for $k=5$}\\\\\n -70&42&-202&624&\\textrm{for $k=6$}\\\\\n \\textrm{etc.}\n \\end{array}\n$$\n(I can provide about 30 rows of these coefficients, see below); then, \n$$\n g_k\\left(x\\right)\\;=\\;\\displaystyle\\sum_{i=1}^{k-2}\\displaystyle\\frac{m_{(k-2,i)}x\n \\Gamma{\\left(\\left(k-1\\right)x\\right)}}{\\left(x+1\\right)k^i\\Gamma{\\left(\\left(k-1\\right)x+i\\right)}}\n$$\nA direct formula for $\\mathcal{K}\\left(k,x\\right)$ involves the same rational function $g_k$:\n\\begin{equation}\n \\begin{array}{lcl}\n \\mathcal{K}\\left(k,x\\right)&=&\n \\displaystyle\\frac{ \\Gamma{\\left(x+1\\right)}\\Gamma{\\left(\\left(k-1\\right)x\\right)} }{ \\Gamma{\\left(kx\\right)} }\\\\[24pt]\n &&\\times\\quad\\left( \\displaystyle\\frac{k^{kx} }{2\\left(k-1\\right)^{\\left(k-1\\right)x-1}}\n +\\displaystyle\\sum_{i=0}^\\infty\n \\displaystyle\\frac{ \\left(k-1\\right)^{\\left(k-1\\right)i}\n \\Gamma{\\left(k\\left(i+x\\right)\\right)}}{k^{ki}\\Gamma{\\left(1+i+x\\right)}\\Gamma{\\left(\\left(k-1\\right)\\left(x+i\\right)\\right)}}g_k\\left(x+i\\right)\n\\right)\n\\end{array}\n\\label{directform}\n\\end{equation}\nThe first formula I gave in my question was found by substituting $g$ from these last expressions. Of course, finding some formula for the $m$ coefficients would be nice, but I couldn't figure out any despite the time I spent on it.\nIt looks also like the infinite sum above can be turned into a finite sum of $k-2$ hypergeometric functions as a direct formula for $\\mathcal{K}\\left(k,x\\right)$. The initial cases are the simplest, like for $k=3$:\n$$\n\\begin{array}{lcl}\n \\displaystyle\\mathcal{K}\\left(3,x\\right)&=&\n\\displaystyle\\operatorname*{K}_{n=1}^{\\infty}\\displaystyle\\frac{\\left(n+1\\right)\\left(6x-3n\\right)}{4n+4}\\\\\n&=&\n\\cfrac{12x-6}{8+\\cfrac{18x-18}{12+\\cfrac{24x-36}{16+\\cfrac{30x-60}{20+\\ddots}}}}\\\\\n&=&\n\\displaystyle\\frac{27^x\\, \\Gamma\\left(x+1\\right)\\Gamma\\left(2x\\right)}\n {4^x\\, \\Gamma\\left(3x\\right)}\n -\n\\displaystyle\\frac{4\\,\\displaystyle{}_3F_2\\left(1,x+\\frac{1}{3},x+\\frac{2}{3};\\;x+\\frac{1}{2},x+2;\\;1\\right)}{ 3\\left(x+1\\right) }\n\\end{array}\n$$\nThe general case being:\n$$\n \\begin{array}{lcl}\n \\mathcal{K}\\left(k,x\\right)&=&\n \\displaystyle\\frac{k^{kx} \\Gamma{\\left(x+1\\right)}\\Gamma{\\left(\\left(k-1\\right)x\\right)} }{2\\left(k-1\\right)^{\\left(k-1\\right)x-1}\\Gamma{\\left(kx\\right)}}\\\\[24pt]\n &+&\n\\displaystyle\\sum_{i=1}^{k-2}\n\\displaystyle\\frac{m_{(k-2,i)} x\n\\;\\;\\displaystyle\n{}_{k+1}F_{k}\n\\left(\n\\begin{array}{l}\n1,x+\\frac{1}{k},x+\\frac{2}{k},\\dots,x+\\frac{k-1}{k},x+1\\\\[4pt]\nx+\\frac{i}{k-1}, x+\\frac{i+1}{k-1}, \\dots,x+\\frac{i+k-2}{k-1},x+2\n\\end{array}\n {1}\n\\right)\n}\n{k^i\\left(x+1\\right) \\Gamma{\\left(\\left(k-1\\right)x+i \\right)}/\\Gamma{\\left(\\left(k-1\\right)x\\right)} }\n \\end{array}\n$$\nwhere obviously the $x+1$ term can always be cancelled (since one of the $x+(i+\\dots)/(k-1)$ terms is always equal to $x+1$ leading to a ${}_{k}F_{k-1}$ function.\nUnfortunately, I couldn't figure out how to make some substitution between this last formula with $m$ coefficients and previous formulae above involving the $g$ function.\nThe case $kx$ being an integer\nIn order to study consecutive integer values of the~$kx$ product, I will now use another notation:\n$$\n \\mathcal{K}'\\left(k', x'\\right)\\;=\\;\\mathcal{K}\\left(x', k'/x'\\right)\n$$\nStarting from here, $k'$ is assumed to be an integer (according to the previous notation, it means that~$kx$ is an integer).\nNew identities, involving different expressions can be found (leading to different kinds of special values when possible):\n$$\n \\begin{array}{lcl}\n \\mathcal{K}'\\left(k',x'\\right)&=&\n h_{k'}\\left(x'\\right) -\n \\left(\\displaystyle\\frac{x'}{x'-1}\\right)^{k'-1}\n \\displaystyle\\frac\n {k'\\;\\Phi\\left(-(x'-1)^{-1}, 1, k'/x'\\right)-x'+1}\n {k'/x'\\; \\beta\\left(k', -k'/x'\\right)}\\\\[24pt]\n &=& h_{k'}\\left(x'\\right) -\n \\left(\\displaystyle\\frac{x'}{x'-1}\\right)^{k'-1}\n \\displaystyle\\frac{\n \\displaystyle\\int_{t=0}^1\n \\left(\n \\displaystyle\\frac{\\left(1-t\\right)\\left(x'-1\\right)}{x'-1+t}\n \\right)^{k'/x'}\\textrm{d}t}\n {k'/x'\\; \\beta\\left(k', -k'/x'\\right)}\\\\[24pt]\n &=& h_{k'}\\left(x'\\right) -\n \\left(\\displaystyle\\frac{x'}{x'-1}\\right)^{k'-1}\n \\;\n \\displaystyle\\frac\n {\n \\displaystyle{}_{2}F_{1}\\left(1,k'/x'\\;; 1+k'/x'\\;; \\left(1-x'\\right)^{-1}\\right) x'\n -x' + 1 }\n {k'/x'\\; \\beta\\left(k', -k'/x'\\right)}\\\\[24pt]\n \\end{array}\n$$\nwhere $h$ is (again) a sequence of rational functions. Let's call $m'$ the sequence of following polynomial functions (I computed about 20 of them):\n$$\n \\begin{array}{l@{\\qquad}l}\n 1,&\\textrm{(for $k'=3$)}\\\\\n -6 + 4x,&\\textrm{(for $k'=4$)}\\\\\n 41 - 53x + 18x^2,&\\textrm{(for $k'=5$)}\\\\\n -348 + 648x - 420x^2 + 96x^3,&\\textrm{(for $k'=6$)}\\\\\n \\textrm{etc.}\n \\end{array}\n$$\n(I have about 20 rows like that, see below); then,\n$$\n \\left\\{\\begin{array}{l@{\\qquad}l}\n h_1\\left(x'\\right)\\;=\\;-1&\\textrm{if $k'=1$}\\\\[12pt]\n h_2\\left(x'\\right)\\;=\\;0&\\textrm{if $k'=2$}\\\\[12pt]\n h_{k'}\\left(x'\\right)\\;=\\;\n \\displaystyle\\frac{ m'_{k'-2} \\left(k'\\left(x'-1\\right)-x'\\right)\\left(x'-1\\right) }{ \\left(k'-1\\right)!\\left(x'-1\\right)^{k'-1} }\n&\\textrm{if $k'\\geq3$}\n\\end{array}\\right.\n$$\nAgain, finding some formula for these $m'$ coefficients would lead to a beautiful formula for $\\mathcal{K}$.\nMaterials (data)\nFinding a formula for one of these two triangles would lead to a direct formula; I put my data below in case someone would have a glance at them. I used the Mathematica format but it is very easy to convert it to anything else.\nTwo pieces of data are attaches:\n\nthe triangle of $m$ coefficients;\nthe triangle of $m'$ polynomials.\n\nFirst, the triangle of $m$ coefficients:\nm = {\n {-8},\n {-20,2},\n {-40,12,-24},\n {-70,42,-202,624},\n {-112,112,-944,6800,-28160},\n {-168,252,-3240,39990,-378180,1956240},\n {-240,504,-9120,168780,-2691500,31299660,-193818240},\n {-330,924,-22308,573210,-13533950,262134768,-3604679456,25969798400},\n {-440,1584,-49104,1665048,-54028800,1533955752,-34784795304,551021454648,-4524877873152},\n {-572,2574,-99528,4294290,-182338520,7056003570,-232622918920,6027044680440,-107934603537600,994719833856000},\n {-728,4004,-188760,10081500,-540836660,27196564920,-1213669081240,45402131767300,-1320731548020500,26362209822109700,-269367401834854400},\n {-910,6006,-338910,21926190,-1447260750,91405905570,-5269521952170,265270592109600,-11076506267112000,357029036918928000,-7854973969921056000,88120488036962304000},\n {-1120,8736,-581152,44753280,-3559649600,275205604224,-19825712025728,1282796713117920,-71715923149544960,3301368476366175072,-116699890447511246304,2804668390029759121440,-34267109445760293273600},\n {-1360,12376,-958256,86572772,-8159187400,756765661176,-66440724681072,5348371530601974,-382626987630417516,23479203132873399912,-1180109553700743730064,45368332310341259611392,-1182254848944902971766528,15625389962188145791748096},\n {-1632,17136,-1527552,159942120,-17614027080,1928217475800,-202304078444160,19770089799495420,-1752820980572484900,137106641163083684400,-9150469205270049657000,498255795785126793714300,-20689280727893354516039100,580954533672149606803333500,-8258153843323806482092032000},\n {-1938,23256,-2364360,283936380,-36111651940,4603291360920,-568072401325800,66113606126531250,-7091616708136156350,685093642937381277600,-58084005963489507601600,4185232111643968748563200,-245300606116021459523520000,10937978741665549577203200000,-329201103515195643674342400000,5008018989272579747902955520000},\n {-2280, 31008, -3565920, 486748080, -70778344000, 10387428137520, -1488100140390640, 203083381548077800, -25863156057665295600, 3013571052233142231600, -314587396914476897195600, 28707485207091210763643400, -2219712979095004695654352000, 139280212909636939833667975000, -6636253210525016113608507935000, 213097407927242925450657097805000, -3454359206881951401298519654400000},\n {-2660, 40698, -5255856, 809056860, -133342927200, 22312129321188, -3669893137886976, 579814786246525830, -86348985070472364240, 11912853180399067114218, -1495810636771876664855616, 167605806446114219615512080, -16367656617234522958801322880, 1351158196468373411398992679968, -90343714080278210268675472690176, 4580006947756999980100721131530240, -156282112623791407868541623144448000, 2689249605403395016547015123312640000},\n {-3080, 52668, -7589208, 1308328296, -242549291800, 45885645345792, -8583630937867664, 1553160300088906908, -267115992196423617852, 42987425869754619349668, -6375040043352284259869056, 857073372770354425645988028, -102516882515527648341577456052, 10661831289669551213862356652936, -935559064657406376990369478217592, 66392317932445459737857422287168372, -3567745082228659013145697404664186980, 128910715024676759785064905466962718772, -2346816894979813166169796326498705604608},\n {-3542, 67298, -10758066, 2064221940, -427579512340, 90781666964580, -19156534075381220, 3933373867476907490, -773001449364837212650, 143337480955932385601050, -24740379167353794462482250, 3919619596881620663698352400, -561261767174736596981182684800, 71335052610001221107048830848000, -7868282527494059201650532567232000, 731130238635173198053153940388864000, -54874347623661338563124599643795456000, 3115425378725886177371208787427500032000, -118823016992948260463398185243235328000000, 2281667516033630958272834061325819904000000},\n {-4048, 85008, -14997840, 3185310480, -732817955520, 173481958077120, -40999952833277760, 9476856735932804400, -2109071440724185828800, 445975726476730199677200, -88512496883173204230589200, 16287555278375689955554908000, -2742762926014265565760465116000, 416464503552989975128829015232000, -56022670706540265707212091974752000, 6530157038781041616772104173250906000, -640432883214307884870393296312193408000, 50678900925449456482492390925833162170000, -3030888088511779915449136680430563767970000, 121680563022029894895609769265080802797710000, -2457881871574620592645101983798188100812800000},\n {-4600, 106260, -20594200, 4817335050, -1224368027500, 321314420301000, -84511921344096400, 21837002485484744250, -5460560343151268855500, 1305129863913508752619500, -294830327010826293740053000, 62266523988286549276643448150, -12155836300655550173555485100500, 2166552581042493037698711451044000, -347528815311777085029924893736724000, 49310886906769450018463698053718956000, -6055094240863690023007706913404687496000, 624931908053334691607080995207899763360000, -51995246297017813506321403783934541271680000, 3267058354801244958789636823999905461614080000, -137714123890197288080314052933290069397790720000, 2919082688078782495755638635126117397822177280000},\n {-5200, 131560, -27890720, 7153246100, -1998828604500, 578492714221380, -168381395188292160, 48336833335182707700, -13488618129875116559100, 3616198939768951478832120, -921761072167762889584562040, 221195726134157418021734756400, -49475332499912242476184032601200, 10205195620733840007949834678752240, -1918172758706697630190525216812587520, 323985355991420046527323571346102988380, -48344306336853827546608471043597263180580, 6236410605945196309646575950593070996242112, -675574363650914146210117776862820284274422024, 58952707582889583114849809235709300486236865900, -3882542732229399143933348322987701786016750303500, 171441026104474223322555734221353597751602880467500, -3804948188770142619704669041367337098336518799360000},\n {-5850, 161460, -37297260, 10445304870, -3194948279250, 1014976460259720, -325284627793625640, 103172770888592820030, -31935449045956426109370, 9539216762922608385767220, -2723057324780894154211144140, 736160112029528961953639831370, -186796816819349796321290668269630, 44075061969736070597572230035573520, -9572302178958218895730236218841324000, 1891505089131626077283496419557169149440, -335445207934760792427195571281784372615680, 52500216486691388374580499101901862088712192, -7097170407703652020998162722033686958068359168, 805065643420847023775013665323598164107193712640, -73517194182833158880223919841841477149672840232960, 5063910198045757853788580192610834249566744036769792, -233753499049244474320664989973779541903817843793723392, 5421019686584291940290446961296577366550977042379177984},\n {-6552, 196560, -49299120, 15019547400, -5008903972800, 1739242914960600, -610937879343586200, 213044845306016606400, -72754198545840383754000, 24070115922338692488744000, -7644318575328042577024440000, 2310967474707680781916704026400, -659643825387695670545486563900800, 176310960419571195542552313305220000, -43737562948690701530900945112487860000, 9971569996183917023985409450427829525000, -2065816103048706564571284238051613500970000, 383694616243443437568263013812039800791990000, -62837694530317608096687694020291847950887850000, 8882027287256940966231272654559742520461647675000, -1052798325643449998635040396582752781540903534400000, 100403132568689053077039174326824794886423800369525000, -7219010863790387928075409606725932651241502240697925000, 347694912384706890967941398378351924881505314746114375000, -8410211480678023320925498721110295938761779872530432000000},\n {-7308, 237510, -64467000, 21292941150, -7714097921400, 2916404269698150, -1118190692314588200, 426802089352299911250, -160027260019840672173000, 58331250782529140212904250, -20490396263678036008373775000, 6882240483264899117171621464650, -2193778617416596625925140519889000, 658701474719992054755835904812589250, -184849219993358885493062726683645011000, 48071268295949005824672264507034120713000, -11474627010559184001964832102040305176064000, 2486300080568850616517044780683050329279500000, -482557829071841856097474441995986751588159600000, 82519812703910109373696671538744758363463431600000, -12171485983756715937697088833853438913213231139200000, 1504622344753359530620585743898465267509637578086400000, -149578514098722961283327295745791298648255342896128000000, 11206095897313941255176473673738843538864273909178368000000, -562168844769302825642413825011269789161124218168180736000000, 14158685308482113187947385422334786985368814916547903488000000},\n {-8120, 285012, -83467800, 29793593700, -11686535347500, 4793533613422200, -1998556890132738600, 831586576427972253000, -340848228875785588089000, 136238023124447243505363000, -52660395629469710899852464000, 19538961040560845384445253280400, -6910955344258841823610759577910000, 2314356603026298502745751283117512000, -728675543038928155386332670093531348000, 214094057758291962937552370682805572874500, -58219061862163874792551203092332714335562500, 14516218849791911496733341909570194788693657500, -3282604618552285952731079609822870039677157280000, 664411186688887595279287675566612107186757131137500, -118409734996254908439089485033112187166889903398912500, 18191629115197225627312502465880976532595946618885575000, -2341224306602995570500391595079223927042601832325548375000, 242207299534450314667366061440535601337557549950222099062500, -18876082685899484114785412705333370447849929199143588260312500, 984735360414688845599685871865811518568346909249116274589062500, -25783538810629517021266249543138106125941453578155830804480000000},\n {-8990, 339822, -107076294, 41184403650, -17437036139250, 7734593443379670, -3494544596518289790, 1579284525442183941150, -704824443778733315833950, 307603458576783505012150890, -130224815626952571544054365330, 53104886131252741135008778869750, -20725045171508268666798696739515750, 7692149381874530736969377052576265170, -2697951867406866051352410906078675593130, 888312186434005905720786992903999283113280, -272592942573130969721591283451860861263314880, 77338601225244978709744671633189447473744656512, -20101127086406857123404360233486723068548779401984, 4734712312004040060142996428231341452704698192056320, -997560027458845642233086777856612791698591648386293760, 184958250344358621036304852979454062719166092717119602688, -29548178532481410695731393820614988207097594428678936395776, 3952663598502220791832289869795663784663381872742938194739200, -424872593639734370119419756220910340050772936517217787445248000, 34392492457805374774120867589709950335110603672365392763289600000, -1863044156298137915554810903420018148506651555374947876496998400000, 50638772787167674049719466457371229923270837855299620820746240000000}\n};\n\nThen the triangle of $m'$ polynomials:\nmm = {\n1, -6 + 4*x, 41 - 53*x + 18*x^2, -348 + 648*x - 420*x^2 + 96*x^3, \n 3669 - 8734*x + 8067*x^2 - 3482*x^3 + 600*x^4, \n -47248 + 135328*x - 158672*x^2 + 96800*x^3 - 31248*x^4 + 4320*x^5, \n 727641 - 2423511*x + 3405267*x^2 - 2622141*x^3 + 1188252*x^4 - 305748*x^5 + \n 35280*x^6, -13122720 + 49768960*x - 81172560*x^2 + 74655760*x^3 - 42491760*x^4 + \n 15258160*x^5 - 3258720*x^6 + 322560*x^7, \n 271959293 - 1157733608*x + 2149311530*x^2 - 2291218376*x^3 + 1553100917*x^4 - \n 697389632*x^5 + 206763300*x^6 - 37696464*x^7 + 3265920*x^8, \n -6373686528 + 30125806848*x - 62800700544*x^2 + 76183473024*x^3 - \n 59781151872*x^4 + 31886134272*x^5 - 11774336256*x^6 + 2965752576*x^7 - \n 471208320*x^8 + 36288000*x^9, 166695335769 - 867080392969*x + 2008379540976*x^2 - \n 2736817207770*x^3 + 2443396898805*x^4 - 1507011320601*x^5 + 659441858394*x^6 - \n 206111203940*x^7 + 45043779816*x^8 - 6336456480*x^9 + 439084800*x^10, \n -4812534974464 + 27341393304064*x - 69753592234368*x^2 + 105687047754624*x^3 - \n 106030998529344*x^4 + 74392108414848*x^5 - 37603296588032*x^6 + \n 13896748361216*x^7 - 3755526664512*x^8 + 723700309248*x^9 - 91276174080*x^10 + \n 5748019200*x^11, 151999996277925 - 937041564265650*x + 2613327954360525*x^2 - \n 4364620345415250*x^3 + 4871696279607675*x^4 - 3842208959921550*x^5 + \n 2208928948800975*x^6 - 942087609901350*x^7 + 300405312044100*x^8 - \n 71352461709000*x^9 + 12280661164800*x^10 - 1402935292800*x^11 + 80951270400*x^12, \n -5212950320375808 + 34671923430780928*x - 105013701846865920*x^2 + \n 191879918776057856*x^3 - 236246229569439744*x^4 + 207409229427830784*x^5 - \n 134079476009502720*x^6 + 65028411548069888*x^7 - 23908234488465408*x^8 + \n 6687356722905088*x^9 - 1414323490805760*x^10 + 219693856813056*x^11 - \n 22925711370240*x^12 + 1220496076800*x^13, \n 192908434730267801 - 1377309563570454971*x + 4504240832008823353*x^2 - \n 8944731476737727263*x^3 + 12057054300505697583*x^4 - 11683650400191522273*x^5 + \n 8411790424807054099*x^6 - 4588515109996305829*x^7 + 1918116031712679596*x^8 - \n 618064823563159856*x^9 + 153647212030630848*x^10 - 29242401180913008*x^11 + \n 4135229005230720*x^12 - 396997001452800*x^13 + 19615115520000*x^14, \n -7661276413060165632 + 58455665081940344832*x - 205362993128521586688*x^2 + \n 440656568659067627520*x^3 - 646000731200138715648*x^4 + \n 685780191107302542336*x^5 - 545270893574020280064*x^6 + \n 331406006965726821120*x^7 - 155847180496082807808*x^8 + \n 57087389250576640512*x^9 - 16335131671115336448*x^10 + 3647848903953212160*x^11 - \n 630293068833472512*x^12 + 81737176968299520*x^13 - 7263281191219200*x^14 + \n 334764638208000*x^15, 325015466875658755821 - 2639668188726987561436*x + \n 9917145842014784419212*x^2 - 22874961160535389523056*x^3 + \n 36258193335964278154110*x^4 - 41887476949175919553816*x^5 + \n 36506794561565447125836*x^6 - 24516370991512685833088*x^7 + \n 12850659286368069730317*x^8 - 5296394247470924322860*x^9 + \n 1722523200312064838592*x^10 - 442409275011858790256*x^11 + \n 89540582470596717552*x^12 - 14150517433376508288*x^13 + \n 1693379120667874560*x^14 - 140015823275059200*x^15 + 6046686277632000*x^16, \n -14668500851872718848000 + 126360196032621236224000*x - \n 505641446052759169024000*x^2 + 1248039501127422625792000*x^3 - \n 2127811079963561138176000*x^4 + 2659365515136882884608000*x^5 - \n 2523680829463448281088000*x^6 + 1858664349808259201024000*x^7 - \n 1076955696528752952320000*x^8 + 494913795536023414784000*x^9 - \n 181130674396834734080000*x^10 + 52865257295200467968000*x^11 - \n 12296348611822927872000*x^12 + 2272553630125056000000*x^13 - \n 330581514194555904000*x^14 + 36704049561050112000*x^15 - \n 2836877949868032000*x^16 + 115242726703104000*x^17\n}\n\n", "comments": ["@TitoPiezasIII I will do the same thing for other formulas I have; they may have some interest for understanding how the function behave. Thank you again for your interest. Regards.", "@ThomasBaruchel: I made some formatting changes to your general identity. Pls check. If you have a long expression, the way to do it is to break it up. Kindly also include your formula for integer $k$ and if it is too long for the screen, you can format it like what I did with $\\mathcal{K}(k,x)$.", "@Nicco You are right ; I will fix it immediately. Regards."], "comment_count": 0, "tags": ["number-theory", "gamma-function", "continued-fractions", "hypergeometric-function"], "creation_date": "2017-01-14T06:53:32", "diamond": 0, "votes": 22} {"question_id": "317", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/54377", "title": "A question connected with the decomposition of a functional on $C(X)$ on Riesz and Banach functionals", "body": "Let $X$ be a metric space and let $C(X)$ be a family of all bounded and continuous functions from $X$ in $\\mathbb{R}$.\nWe call a positive linear functional $\\varphi: C(X) \\rightarrow \\mathbb{R}$ the functional of Riesz if there is a borel measure $\\mu$ on $X$, such that $\\varphi(f)=\\int_X f \\,d\\mu$, for $f\\in C(X)$.\nWe call a positive linear functional $\\varphi: C(X) \\rightarrow \\mathbb{R}$ the functional of Banach if for each borel measure $\\nu$ on $X$ the condition:$\\int_X f d\\nu\\leq \\varphi(f)$, for $f\\in C(X)$ - implies that $\\nu$ is trivial.\nThere is a well known theorem :\nLet $X$ be a polish space. Then, for each positive linear functional $\\varphi: C(X) \\rightarrow \\mathbb{R}$ there is a unique couple $(\\varphi_0,\\varphi_*)$ of positive linear functionals defined on $C(X)$, such that $\\varphi_0$ is the functional of Riesz, $\\varphi_*$ is the functional of Banach and $\\varphi=\\varphi_*+\\varphi_0$. Moreover, the measure $\\mu$ related to $\\varphi_0$ is defined by: $$\\mu(K)=\\inf\\{\\varphi(f): f\\in C(X), 1_X\\geq f \\geq 1_K\\},$$ for each compact set $K\\subset X$.\nMore pecisely, for the proof, we define:\n$$\\varphi_{\\delta}(f)=\\sup\\{\\varphi(h): \\mbox{ supp}\\,h\\in N(\\delta), 0\\leq h\\leq f\\},$$ for $\\delta>0$, $$\\varphi_{0}(f)=\\lim\\limits_{\\delta \\to 0^+}\\varphi_{\\delta}(f),$$ for $f\\in C(X), f\\geq 0$, and $$\\varphi_{0}(f)=\\varphi_{0}(f^+)-\\varphi_{0}(f^-),$$ for $f \\in C(X)$, where $N(\\delta)$ is a family of sets that possess a covering composed of finite number of open balls with a radius equal to $\\delta$.\nMy question concerns the truth of the following sentence: Let $X$ be a $\\sigma$-compact and polish space. Assume that $\\varphi^x:C(X) \\rightarrow \\mathbb{R}$ is a positive linear functional, for all $x \\in X$ and let $((\\varphi^x)_0,(\\varphi^x)_*)$ be a couple of Banach-Riesz functionals, for $x \\in X$. If the mapping $X \\ni x \\mapsto \\varphi^x(f)$ is continuous for all $f \\in C(X)$ and $\\varphi^x(1_X)=1$, for $x \\in X$, then mapping $X \\ni x \\mapsto (\\varphi^x)_0(f)$ is continuous for all $f \\in C(X)$ (or may be for only $f \\in C_c(X)$).\nI was able to proof only that the mapping $X \\ni x \\mapsto (\\varphi^x)_0(f)$ is upper semi-continuous, for $f\\in C_c(X)$.\n", "comments": ["Actually, I am trying to use this theorem to generalize a result in the paper of Meyn and Tweedie \"Markov Chains and Stochastic Stability associated with Markov e-chains\" on the polish spaces. Sorry I didn't gave this kind of informations. I will be very grateful for your help.", "It's getting too late for me to think straight, maybe I'll get back to your problem tomorrow. I'm pretty sure that you're trying to adapt the paper (specifically Appendix A) of Lasota-Szarek Lower bound technique in the theory of a stochastic differential equation to your more general situation? It would be extremely helpful if you gave the reference you're working with next time you ask a question.", "I can afford to assume that X is also $\\sigma$-compact. Let $\\mu_x$ be a measure related to Riesz functional $(\\varphi^x)_0$. We have: $(\\varphi^x)_0(f)=\\mu_x(X)-(\\varphi^x)_0(1-f)=\\lim\\limits_{n\\to\\infty}\\mu_x(K_n)-(\\varphi^x)_0(1-f)$, so i think it suffice to prove that $x \\mapsto \\mu_x(K)$ is lower semi-continous, for compact sets $K$, but i dont know if that is truth.", "Yes, exactly, I assumed that $f\\in C_c(X)$ and used the fact, that for each $\\delta>0$ and a compact set $K$ (here I set K=supp f) there is a function $f_\\delta\\in N(\\delta)$, such that $1_K\\leq f_\\delta$ and the equality $(\\varphi^x)_\\delta(f_\\delta)=\\varphi^x(f_\\delta)$.", "And I assume that you use compact support of $f$ to see that $\\phi_{\\delta}$ is also upper smicontinuous (after all, you're then working on the compact space given by the support of $f$)?", "So, at the first, I noticed that the function $x \\mapsto \\varphi_\\delta (f)$ is lower semi-continous (as you said), next, i proved that the same function $x \\mapsto \\varphi_\\delta (f)$ is also upper semi-continous, so it is continous. Finally, $x \\mapsto \\varphi_0 (f)$ is upper semicontinous as monotonically decreasing limit of upper semi-continuous (even continous) functions.", "Now let us get rid of all unnecessary ballast. Fix $f \\geq 0$. Then you're taking the supremum over the continuous functions $x \\mapsto \\phi^x (f)$. In my world this gives a lower semicontinuous function in general. Now you're taking a monotonically decreasing limit of lower semi-continuous functions to get $\\phi_{0}(f)$. So why on earth is that upper semicontinuous? Also, where did you get that from? The theorem is definitely not well-known to me.", "exactly yes, this is a construction from the proof of the theorem that i gave", "I'm not sure if I understand your definition of $\\varphi_{\\delta}$. Do you mean you're taking the supremum over all $h$ such that $0 \\leq h \\leq f$ with the property that $\\operatorname{supp}{h}$ can be covered by finitely many balls of radius $\\delta$?"], "comment_count": 0, "tags": ["real-analysis", "probability-theory", "measure-theory"], "creation_date": "2011-07-28T15:55:48", "diamond": 0, "votes": 22} {"question_id": "318", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/414887", "title": "Removing deterministic discontinuities from semi-martingales", "body": "Let $X:=(X_t)_{0 \\le t \\le T}$ be a solution of the SDE\n$$ X_t = X_0 + \\int_0^t \\sigma(s,X_s) dW_s + \\sum_{i=1}^n f_i(X_{t_i^-}) 1_{\\{t > t_i\\}}$$\nwhere $t_1,\\cdots,t_n \\in [0,T]$ and $(f_i)_{1 \\le i \\le n}$ a family of measurable functions. My goal is to remove jumps from $X$ using a change of variable\n$$ Y_t = \\Phi(t,X_t)$$\nwhere $\\Phi$ is yet to be found. When $f_i$'s are affine, it is possible to remove the deterministic term in Itô's lemma by choosing an affine form for $\\Phi$ in the space variable and a piecewise constant form in the time variable.\nI could not find any result in the litterature regarding the general case, i.e. $f_i$ polynomial, or even discontinuous itself. Is there a general approach for removing deterministic discountinuities from semi-martingales ?\n", "comments": ["Please may you specify precisely what you mean by \"remove jumps from $X$\"? Which properties do you want $Y$ to have?", "I clarified my question. I do not assmue $X$ is a local martingale anymore.", "I implicitly meant the sum of a local martingale and a pure jump process whose jumps are predictable ($\\mathcal{F}_{t_i^-}$-measurable) and jump dates are constant ($\\mathcal{F}_0$-measurable)."], "comment_count": 0, "tags": ["stochastic-processes", "stochastic-calculus", "martingales"], "creation_date": "2013-06-08T11:27:11", "diamond": 0, "votes": 22} {"question_id": "319", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3832522", "title": "Are $3^6-6^3$ and $4^8-8^4$ the only sums of four $a^b-b^a,1\\lt a\\lt b$ numbers?", "body": "Question\nHow many numbers of form $a_0^{b_0}-b_0^{a_0}$ are a \"nontrivial\" sum of four such numbers $a_i^{b_i}-b_i^{a_i}$ ?\nThe \"nontrivial\" means: all unordered pairs $\\{a_i,b_i\\}$ are distinct, $a_i^{b_i}\\ne b_i^{a_i}$ and $1 \\lt a_i\\lt b_i$.\nThis implies that such summands are positive (are in OEIS A045575), except $2^3-3^2 = -1$.\nThe only \"nontrivial\" examples I could find are:\n$$\\begin{align}\n(2^5-5^2) + (2^6-6^2) + (2^7-7^2) + (4^5-5^4) &= (3^6-6^3) \\\\\n(2^8-8^2)+ (4^5-5^4) + (4^6-6^4) + (3^{10}-10^3) &= (4^8-8^4)\n\\end{align}$$\n\nAre these two the only such numbers?\n\nFor comparison, I suspect such sums with less than four summands do not exist, and that there are infinitely many such sums with more than four summands. With four summands exactly, I have only these two examples, hence this question.\nAre there any other references (than ones listed in OEIS A045575) on problems related to $x^y-y^x$ numbers?\n\nBackground\nThese two numbers correspond to the following two Base-Exponent Invariant numbers:\n$$\\begin{array}{}\n1464 &=& 2^5 + 2^6 + 2^7 + 4^5 + 6^3 &=& 5^2 + 6^2 + 7^2 + 5^4 + 3^6 \\\\\n68521 &=& 2^8 + 4^5 + 4^6 + 3^{10} + 8^4 &=& 8^2 + 5^4 + 6^4 + 10^3 + 4^8\n\\end{array}$$\nThat is, these numbers are a special case of the \"Base-Exponent Invariant\" numbers.\nI call these \"Order-$5$ Genus-$1$\" Base-Exponent Invariant numbers $1464,68521\\in G^{(5)}_1$.\nIn general, I have found only $14$ examples (see \"short examples\" in this answer) of \"Order-$5$ Base-Exponent Invariant numbers\". The largest known example is around $6\\cdot 10^6$, while the next one, if it exists, is larger than $10^{16}$.\n\nGeneral near examples\nI've searched for smallest \"error\" $e(n)$ such that \"some elements plus the error\" are a sum of the \"other elements\" from the \"best\" 5-subset of A045575 among \"nontrivial\" 5-subsets whose largest element is the $n$th nonzero term of A045575.\nIf $e(n)=0$ (and $n\\ge 5$) then we have general examples(s) and $(n,0)$ is colored blue (or green if corresponding example is also \"Genus-$1$\"). If $e(n)=\\pm 1$ we have a \"near example\" (colored red). Else, we have a black point $(n,\\log e(n))$. For $n$ up to $100$, we have the log plot of errors:\n\nNotice that for $n\\gt 43$, we have no general examples, and for $n\\gt 25$ we have no \"Genus-$1$\" examples (the examples I'm asking about in this question), so far.\nIt would seem that new examples are very large and rare or do not exist. However, notice the far right \"near example\" (red point) at $n=83$, which gives us hope\n$$\n(2^8-8^2) + (2^{16}-16^2) + (4^{16}-16^4) + (2^{32}-32^2) = (2^{33}-33^2) \\color{red}{+1}\n$$\nthat maybe a large example could exist.\n\nDo there exist any larger general examples, Genus-1 examples or near examples?\n\n", "comments": ["@jjagmath It is included as trivial: The \"nontrivial\" condition contains \"$a_i^{b_i}\\ne b_i^{a_i}$\", implying summands that equal zero are not being considered.", "$2^3-3^2=-1$ is not the only exception. $2^4-4^2 = 0$. That should be also consider as trivial?"], "comment_count": 0, "tags": ["number-theory", "elementary-number-theory", "reference-request", "examples-counterexamples", "recreational-mathematics"], "creation_date": "2020-09-19T10:29:20", "diamond": 0, "votes": 21} {"question_id": "320", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3827693", "title": "Fibonacci-like sequences mod $p$ where $a_{n+1}$ only really depends on $a_n$.", "body": "Consider a prime $p$ and a sequence $(a_n)_{n\\ge 0}$ in $\\mathbb{F}_p$ satisfying $a_{n+2}=a_{n+1}+a_n$ for all $n\\ge 0$.\nNow, assume that each element of the sequence only really depends on the previous one. That is, assume there exists a function $f:\\mathbb{F}_p\\to\\mathbb{F}_p$ such that $a_{n+1}=f(a_n)$ for all $n\\ge 0$.\nIf $p\\not\\in\\{2,5\\}$ and $5$ is a quadratic residue mod $p$ there are the obvious sequences\n$$\\left(c\\left[\\frac12+\\frac12\\sqrt5\\right]^n\\right)_{n\\ge 0}\\quad\\text{and}\\quad\\left(c\\left[\\frac12-\\frac12\\sqrt5\\right]^n\\right)_{n\\ge 0}$$\nfor $c\\in\\mathbb{F}_p$ any constant, but are there any others?\nComputational results by @Servaes\nLet's call two Fibonacci-like sequences modulo $p$ equivalent if they can be turned into each other through shifting and multiplication by units, then any sequence where each element is a function of the previous and and which is not equivalent to\n$$(0)_{n\\ge 0},\\quad (c^n)_{n\\ge 0}$$\nwhere $c^2=c+1$, is called strange.\n@Servaes has shown through computation that the first few primes for which strange sequences exist are\n$$199, 211, 233, 281, 421, 461, 521, 557, 859, 911.$$\nOwn work\nFor any prime $p$ let $\\pi(p)$ be the Pisano period mod $p$ (so this is not the prime counting function)\nClaim: Let $p$ be a prime, $(a_n)_{n\\ge 0}$ a strange sequence. Then it has period $\\pi(p)$.\nProof: Let $A=\\{a_n:n\\ge 0\\}$ be the set of attained values and $f:A\\to A$ the function which makes this sequence strange, in the sense that $a_{n+1}=f(a_n)$ for all $n\\ge 0$. Clearly, $f$ is a bijection.\nIt is easily proved that, for all $a\\in A$ and $n\\in\\mathbb{Z}$,\n$$f^n(a)=F_{n-1}a+F_nf(a).$$\nLet $m$ be the period of $(a_n)_{n\\ge 0}$, then clearly $m$ is the smallest positive integer $n$ for which $f^n=\\operatorname{id}_A$. Thus, for all $a\\in A$,\n$$(1-F_{m-1})a=F_mf(a)$$\nIf $F_m\\neq 0$ it follows that $f(a)=F_m^{-1}(1-F_{m-1})a$ and the sequence is equivalent to $\\left(c^n\\right)_{n\\ge 0}$ where $c=F_m^{-1}(1-F_{m-1})$ satisfies $c^2=c+1$. This contradicts our assumption that the sequence is strange, so $F_m=0$.\nSince the sequence is not the null sequence, we may take $a\\in A$ non-zero and conclude that $F_{m-1}=1$. It follows that $\\pi(p)\\mid m$. Since\n$$f^{\\pi(p)}(a)=F_{\\pi(p)-1}a+F_{\\pi(p)}f(a)=F_{-1}a+F_0F(a)=a,$$\nthe opposite division relation holds as well and we are done.\nEDIT: I asked a slightly more general version of this question on Mathoverflow and linked to this question, but forgot to link the Mathoverflow question here.\n", "comments": ["@DonThousand Sure, no problem.", "@OP Do you mind if I hijack some of this awesome work for a codegolf challenge? Credit will be attributed.", "Giving a bit of extra attention to this within the Pearl Dive project. Also editing my Pearl Dive \"ad\" in Meta.MathOverflow.", "Yes, it stands to reason that small Pisano periods are needed for this to happen. Modulo $p=199$ the zeros of $x^2=x+1$ have multiplicative orders $22$ and $11$ (both in $\\Bbb{F}_p^*$ as $\\left(\\dfrac5p\\right)=1$). Modulo $p=233$ they are both of order $52$ in $\\Bbb{F}_{p^2}^*$. Not clear what else is needed to guarantee no repetitions within a period.", "@Mastrem Not at all, I just ran my code for $p<1000$ initially.", "This is a really good question and if it doesn't get an answer in the next few days I'd strongly suggest asking it also on Mathoverflow with a link to this question.", "@Servaes By your earlier comment, did you mean you did not find any such primes beyond $911$?", "Also, I could not help but notice that these exceptional primes appear quite early on in the list of Fibonacci primitive parts. I'm not sure I can imagine any sort of connection, certainly not at this hour.", "For $p=211$ I get: $$6,8,9,23,24,26,29,32,33,34,45,48,62,71,76,98,108,109,119,124,127, 128, 132, 133, 135, 137, 139, 146, 152, 156, 157, 158, 162, 166, 168, 169, 173, 177, 179, 181, 187, 195, 199, 208.$$ For $p=233$ I get: $$7,9,12,13,14,15,17,18,19,20,23,24,27,30,31,33,34,36,39,42,43,47,49,50,51,53,55,56,61,62,65,67,69,72,73,76,81,82,84,87,92,96,97,98,101,102,104,106,108,110,113,114,120,121,124,126,128,130,132,133,136,137,138,142,147,150,152,153,158,161,162,165,167,169,172,173,178,179,181,183,184,185,187,191,192,195,198,200,201,203,204,207,210,211,214,215,216,217,219,220,221,222,225,227.$$", "Quite surprisingly, it seems that the number of values for $a_1$ that yields such a sequence is either $1+\\left(\\tfrac{5}{p}\\right)$, except for the primes I just listed, in which case the number of values for $a_1$ is huge!", "For $p=199$ I found the starting values $a_0=1$ and $a_1$ is one of: $$6, 8, 9, 10, 11, 12, 14, 15, 18, 20, 21, 23, 24, 25, 26, 27, 29, 32, 34, 38, 39, 40, 42, 43, 44, 46, 48, 53, 54, 55, 57, 60, 61, 62, 63, 64, 67, 72, 77, 80, 81, 82, 84, 86, 87, 88, 89, 93, 96, 102, 103, 104, 105, 106, 108, 109, 110, 111, 114, 115, 117, 118, 122, 124, 125, 127, 128, 129, 130, 131, 135, 136, 137, 138, 139, 140, 141, 142, 144, 147, 148, 149, 150, 152, 153, 155, 156, 159, 162, 163, 164, 165, 167, 169, 171, 172, 173, 177, 178, 179, 180, 181, 182, 183, 184, 185, 187, 189, 190, 193, 195, 196.$$", "I'm still cleaning up the proof, but I believe that the period of any sequence not of the 'standard' types must be the Pisano period mod p.", "@Servaes Interesting! Could you perhaps compute the starting two values of such a sequence mod $199$ which does not correspond to a root of $X^2-X-1$?", "Some simple python code shows that the first primes for which there are not either $0$ or $2$ such sequences, up to multiplication by constants and shifting, are: $$5,\\ 199,\\ 211,\\ 233,\\ 281,\\ 421,\\ 461,\\ 521,\\ 557,\\ 859,\\ 911.$$", "Correction to my earlier comment; only $2$ sequences for $p=29$, also corresponding to the roots of $X^2-X-1$.", "@Servaes If it's easier to check whether any sequences exist, I suggest you look only at primes $p$ such that $5$ is not a quadratic residue modulo $p$.", "@Servaes Note that, mod $11$, the roots of $X^2-X-1$ are $4$ and $8$, so the given sequences are exactly of the type I describe in my question.", "A computer search yields plenty of examples. For $p=11$: $$(1, 4, 5, 9, 3, 1, 4,\\ldots),\\qquad (1, 8, 9, 6, 4, 10, 3, 2, 5, 7, 1, 8,\\ldots),$$ and $2$ sequences for $p=19$, up to shifts, and $16$ sequences for $p=29$, up to shifts, etc.", "Also, except the zero sequence, no sequence can contain a $0$. Nor can it have $a_{n+1}=a_n$ for any $n$. Moreover, if $(a_n)_{n\\geq0}$ is such a sequence the so is $(ca_n)_{n\\geq0}$ for any constant $c$. So without loss of generality $a_0=1$. This leaves $p-2$ initial values for $a_1$, for each $p$.", "For $p=5$ the sequence $(2,1,3,4,2,1,\\ldots)$ works. And of course the zero sequence works for any $p$. There are no other sequences (up to shifting) for $p\\leq5$."], "comment_count": 0, "tags": ["number-theory", "elementary-number-theory", "recurrence-relations", "finite-fields", "fibonacci-numbers"], "creation_date": "2020-09-15T13:57:17", "diamond": 0, "votes": 21} {"question_id": "321", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3724155", "title": "A Polynomial Formed from the Roots of Another Polynomial ad infinitum", "body": "Let $P(x)$ be a monic polynomial of degree $d$ with complex coefficients. Let $r_1(P),r_2(P),\\dots, r_d(P)$ denote the set of roots, ordered so that $|r_1(P)| \\leq |r_2(P)|\\leq\\dots\\leq |r_d(P)|$. Define the map $T$ by:\n$$(TP)(x)=x^d+r_1(P)x^{d-1}+r_2(P)x^{d-2}+\\dots+r_d(P),$$\ni.e. $TP$ is the monic polynomial whose coefficients are the roots of $P$.\nLet us call a monic polynomial periodic if $T^KP=P$ for some $K>0$.\nThe question is: for any $d>0$, does there exist a periodic polynomial of degree $d$, other than the trivial solution $x^d$?\nRemark on the definition of T\nAs pointed out in the comments, the definition of $TP$ is ambiguous if there are two roots $r_i(P)$ and $r_j(P)$ such that $|r_i(P)|=|r_j(P)|$ and $r_i(P)\\neq r_j(P)$. If the roots of $P$ have this property, then you may break the ties however you please. For example, if $P(x)=x^3-x$, then it is up to you whether to set $r_2(P)=1$ and $r_3(P)=-1$ or $r_2(P)=-1$ and $r_3(P)=1$. However, either ordering still must have $r_1(P)=0$, since there is no ambiguity there.\nNote that the set of polynomials that have this ambiguity has measure zero, so I suspect such considerations will not influence the solution of the problem anyway.\nEmpirical Evidence\nIf $d=1$ then the answer is clearly yes (any $P(x)=x-a$ will do the job, with $a\\ne 0$). If $d=2$ then $P(x)=x^2+x-2$ is a fixed point of $T$, so in particular is periodic with period 1.\nI examined other low degrees by numerical simulation. Note that this requires relaxing the definition of a cycle, since testing for exact equality of floating point numbers is impossible. Thus, for these simulations, the condition $T^KP=P$ was replaced with $\\|T^KP-P\\|_\\infty<\\varepsilon$, with $\\varepsilon=10^{-10}$. In particular, these simulations can only find polynomials $P$ that are periodic up to some fixed error tolerance.\nThe simulation was done by first initializing the coefficients of $P$ using values drawn from a standard normal distribution, and then iteratively applying $T$ 1000 times and checking whether the obtained sequence was eventually periodic (up to error $<\\varepsilon$). Note that this method might not find all cycles.\nThe periods found thusly for low degrees are:\n$$\n\\begin{array}{rc}\nd=3 & \\text{possible periods}= 1 ; 11 \\\\\n4 & 21 \\\\\n5 & 4 ; 56 \\\\\n6 & 34 ; 44 \\\\\n7 & 10 ; 15 ; 26 ; 234 \\\\\n8 & 3 ; 38 ; 83 ; 292 \\\\\n9 & 256 ; 311 ; 466 \\\\\n10 & 275 ; 336\n\\end{array}\n$$\nFurthermore, for degrees $\\leq 8$, all of the simulated sequences eventually became periodic, however this was not true for $d=9$ or $10$ (of course, this does not imply that these sequences never become periodic, just that they did not before the simulation ended).\nCrossposted to Mathoverflow:https://mathoverflow.net/questions/364359/a-polynomial-formed-from-the-roots-of-another-polynomial-ad-infinitum\n", "comments": ["The backward way from $TP$ to $P$ by Vieta formulae looks much more simple to calculate and has no ambiguous cases. On the other hand, to assure that the found orbit $T^{-n}P$, $n\\in \\Bbb N\\cup\\{0\\}$ provides an answer, we should check that for each polynomial of the orbit the sequence of absolute values of its coefficients is non-decreasing.", "@MikeHawk also, id prefer if u resolved the ambiguity of the defn of $T$ by just allowing us to order how we please when trying to find a periodic point, but it's ur question so u can make the rules", "@MikeHawk u now have excluded the trivial solution $x^d$. fix ur question. also, nice username", "Your $T$ is not well defined. When there are several different roots of $P$ with the same absolute value there is no definite rule of ordering these within the coefficient sequence of $TP$.", "@JG, you are correct, I have edited the question", "Sorry if I’m misreading something, but isn’t $x^d$ trivially periodic?"], "comment_count": 0, "tags": ["sequences-and-series", "polynomials", "complex-numbers", "dynamical-systems"], "creation_date": "2020-06-17T15:38:33", "diamond": 0, "votes": 21} {"question_id": "322", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2831675", "title": "Smallest region that can contain all free $n$-ominoes.", "body": "A nine-cell region is the smallest subset of the plane that can contain all twelve free pentominoes, as illustrated below. (A free polyomino is one that can be rotated and flipped.)\nA twelve-cell region is the smallest subset of the plane the can contain all $35$ free hexominoes.\n\n\nWhat is the smallest region of the plane that can contain all $108$ free heptominoes (shown below)? All $369$ free octominoes?\n\n\nAlso, is there an existing OEIS sequence for this? If not, I'll add one once there is a bit more data. (The sequence begins $1, 2, 4, 6, 9, 12, \\cdots$.)\n", "comments": ["Some upper bounds$$\\begin{array}{rcl} n & a(n)\\le & \\text{example container}\\\\ \\hline 7 & 17 & \\small\\verb/11-111-11111-1111111/\\\\ 8 & 20 & \\small\\verb/111-111-111111-11111111/\\\\ 9 & 27 & \\small\\verb/111-111-11111-1111111-111111111/\\\\ 10 & 31 & \\small\\verb/111-1111-111111-11111111-1111111111/\\\\ 11 & 38 & \\small\\verb/111-1111-1111-1111111-111111111-11111111111/\\\\ 12 & 43 & \\small\\verb/1111-1111-11111-11111111-1111111111-111111111111/ \\end{array}$$", "I submitted a related computational challenge over at Programming Puzzles & Code Golf Stack Exchange.", "@achillehui, my program confirmed that this container works, so indeed $a(8) \\leq 20$.", "if I didn' make any mistake, $a(8) \\le 20$. example container 00000111-00000111-00111111-11111111", "For the heptomino case, a search over all shapes containable in a $4 \\times 7$ bbox only return regions of size $17$. I don't know how to extend the emulation to $5 \\times 7$ bbox but this is a strong indication that $a(7) = 17$.", "You can make the bound $a(n) \\leq n \\cdot \\left\\lceil \\dfrac{n}{2} \\right\\rceil$. It's not very good though.", "And $a(10) \\leq 32$.", "If $a(n)$ is the minimum size region covering all $n$-ominoes, then I've constructed examples to show $a(7) \\leq 17$, $a(8) \\leq 22$ and $a(9) \\leq 26$."], "comment_count": 0, "tags": ["extremal-combinatorics", "oeis", "polyomino"], "creation_date": "2018-06-25T10:48:06", "diamond": 0, "votes": 21} {"question_id": "323", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2281000", "title": "Convergence acceleration technique for $\\zeta(4)$ (or $\\eta(4)$) via creative telescoping?", "body": "\nQuestion\n\nIs it already known whether the $\\zeta(4):=\\sum_{n=1}^{\\infty}1/n^4$ accelerated convergence series $(1)$, proved for instance in [1, Corollaire 5.3], could be obtained by a similar technique to the ones explained by Alf van der Poorten in [2, section 1] for $\\zeta(3)$ and $\\zeta(2)$?\n $$\\zeta(4)=\\frac{36}{17}\\sum_{n=1}^{\\infty}\\frac{1}{n^{4}\\binom{2n}{n}}.\\tag{1}$$\n(a) In other words, does there exist a pair of functions $F(n,k), G(n,k)$ obeying equation \n $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)\\tag{$\\ast$}$$\n from which $(1)$ can be proved? That is, is it possible to transform the defining series for $\\zeta(4):=\\sum_{n=1}^{\\infty}1/n^4$ by means of the Wilf-Zeilberger method (or the Markov-WZ Method) into the faster series $(1)$? (b) Most likely there isn't any such a pair $(F, G)$, but I do not have the means to use these methods on my own. \n\n\nShort description of section 1 of Alf van der Poorten's paper\nThe defining series for $\\zeta(3):=\\sum_{n=1}^{\\infty}1/n^3$ and $\\zeta(2):=\\sum_{n=1}^{\\infty}1/n^2$ are accelerated resulting in\n\\begin{equation*}\n\\zeta (2)=3\\sum_{n=1}^{\\infty }\n\\frac{1}{n^{2}\\binom{2n}{n}},\\tag{2}\n\\end{equation*}\n\\begin{equation*}\n\\zeta (3)=\\frac{5}{2}\\sum_{n=1}^{\\infty }\n\\frac{(-1)^{n-1}}{n^{3}\\binom{2n}{n}}\\tag{3}.\n\\end{equation*}\nFor instance, $(3)$ follows from the identity\n\\begin{equation*}\n\\sum_{n=1}^{N}\\frac{1}{n^{3}}-2\\sum_{n=1}^{N}\\frac{\\left( -1\\right) ^{n-1}}{n^{3}\\binom{2n}{n}}=\\sum_{k=1}^{N}\\frac{(-1)^{k}}{2k^{3}\\binom{N+k}{k}\\binom{N}{k}}-\\sum_{k=1}^{N}\\frac{(-1)^{k}}{2k^{3}\\binom{2k}{k}}\\tag{4},\n\\end{equation*}\nby letting $N\\rightarrow \\infty $ and noticing that\n\\begin{equation*}\n\\lim_{N\\to\\infty}\\sum_{k=1}^{\\infty}\\frac{(-1)^{k}}{2k^{3}\\binom{N+k}{k}\\binom{N}{k}}=0.\n\\end{equation*}\nEquality $(4)$ can be explained as follows:\n\nWrite\n\\begin{equation*}\nX_{n,k}=\\frac{(-1)^{k-1}}{k^{2}\\binom{n+k}{k}\\binom{n-1}{k}},\\qquad D_{n,k}=\\frac{(-1)^{k}}{n^{2}\\binom{n+k}{k}\\binom{n-1}{k}}\\qquad k2$ should have arbitrarily large CF entries at some point. See this paper. Meanwhile we know, that for degree $2$ the CF is (eventually) periodic.\n\nNotice also the same 'pattern' which goes for the examples with odd $a$ here. We have a list of CF entries going the same way.\nIf we subtract the list of entries for $a=3$ from the list of entries for $a=5$, we obtain:\n$$L_5-L_3=[0;2,2,2,2,2,2,2,2,2,...]$$\n$$L_7-L_3=[0;4,4,4,4,4,4,4,4,4,...]$$\n$$L_{113}-L_3=[0;110,110,110,110,110,110,...]$$\nFor $6$ and $2$ it goes the same way, but not for $4$ and $2$, there is some 'scrambling' there.\n\nHow can we prove/explain this facts? The same pattern for different $a$ seems very strange to me, especially if the numbers are transcendental.\n\nBasically, if this turns out to be true, then from the CF for $x_3$ we will immediately obtain all the CFs for every $x_{2n+1}$\n\nImportant update. See http://oeis.org/A004200 for the case $a=3$, it seems like these continued fractions have pattern. And the following paper is linked: Simple continued fractions for some irrational numbers\nThe pattern is the same for every $a$ except $2$, so not only for the odd $a$.\nMorevoer, look at the continued fractions for:\n$$y_{ap}=a^p x_a$$\nFor integer $p$ you will notice a very apparent pattern.\n\nMy questions are largely answered by the linked paper. I will try to write up a short summary and post it as an answer, but anyone is three to do it before me.\n\n\nTurns out a very close question was asked before Continued fraction for $c= \\sum_{k=0}^\\infty \\frac 1{2^{2^k}} $ - is there a systematic expression?\n", "comments": ["@RobertIsrael, I have corrected my question, thank you", "@RobArthan, yes, thank you. I get it now", "@YuriyS: thanks for the link. I think you misunderstood the conjecture (which is far from trivial). the conjecture was that the simple continued fractions for algebraic numbers is either periodic or contains arbitrarily large coefficients. Read the abstract of the paper in your link (and then the rest of the paper) for clarification.", "What is conjectured is that the continued fraction of any real algebraic number of degree $> 2$ has arbitrarily large entries.", "@YuriyS: I am intrigued by the conjecture. Can you give a pointer to the paper you found about this conjecture, please.", "Two simple CFs are equal iff their sequences of coefficients are equal. If the set of allowed coefficients includes two numbers $c_0$ and $c_1$, then you get a distinct CF for every real number $\\alpha \\in (0, 1)$, by mapping the binary representation of $\\alpha$, a sequence of $0$s and $1$s, to the corresponding sequence of $c_0$s and $c_1$s.", "@YuriyS: There is a bijection between them and the interval $[0, 1]$ expressed in base $b$, where $b$ is the number of integers in the set, right?", "@RobArthan, how do you prove there are uncountably many such CFs?", "There are uncountably many simple continued fractions with coefficients drawn from any given finite set of $2$ or more positive integers. So I don't think the conjecture you have in mind in your last two paragraphs can be correct without some further qualification."], "comment_count": 0, "tags": ["elementary-number-theory", "irrational-numbers", "continued-fractions", "egyptian-fractions"], "creation_date": "2016-08-28T15:51:52", "diamond": 0, "votes": 21} {"question_id": "325", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/653940", "title": "Which digit occurs most often?", "body": "Is there any method to calculate, which digit occurs most often in the number\n$$4 \\uparrow \\uparrow \\uparrow \\uparrow 4\\ ,$$\nthe fourth Ackermann-number ?\nOr would it be necessary to calculate the number digit by digit ?\nI only know that the last n digits can be calculated, but this does not help much.\n", "comments": ["According to Wolfram Alpha, the number's too large to even represent, so that's saying something. It wouldn't be practical to calculate the 4th Ackermann number much less count its digits, so I can see why the urgency to have a simpler method is here.", "I did not offer this bounty. And this is a very old question, I just wondered whether there is a trick that allows us to answer this question, which is apparently not the case. And I would find it very interesting, if it would be possible.", "Even if we know the answer to this question, this is a dead end, unstructural information that needs a lot of work to obtain. (Despite of the bounty, please provide context and motivation, e.g. show the own effort to solve the issue. Why this function invented by humans for other reasons, why the digits in the very peculiar base ten used by humans for other reasons, and why $4\\uparrow^44$?) As a parallel, let us consider some smaller \"similar\" number, $2\\uparrow\\uparrow 4=65536$, which is the profit to know a part of the statistics of the occurence of the digits?", "In general, we know nothing about the digits of large powers, other than their last ones.", "been any progress ?"], "comment_count": 0, "tags": ["elementary-number-theory", "decimal-expansion", "tetration", "hyperoperation"], "creation_date": "2014-01-27T16:29:40", "diamond": 0, "votes": 21} {"question_id": "326", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2200929", "title": "Do Hopf bundles give all relations between these "composition factors"?", "body": "Write a fiber bundle $F\\to E\\to B$ in short as $E=B\\ltimes F$ (in analogy with groups). \n(This is not necessary, but: given another bundle $X\\to B\\to Y$, we can write $E=(Y\\ltimes X)\\ltimes F$, but we may also compose $E\\to B$ with $B\\to Y$ to get fibrations $E\\to Y$, whose fibers $Z$ fit inside fibrations $F\\to Z\\to X$ hence $Z=X\\ltimes F$ and $E=Y\\ltimes(X\\ltimes F)$ as well. Thus we have a kind of associativity property $(Y\\ltimes X)\\ltimes F= Y\\ltimes(X\\ltimes F)$.)\nI noticed something about sequences involving $\\mathrm{Spin}(7)$. \nFirst of all, unit imaginary octonions are square roots of negative one, so $L:\\mathrm{Im}(\\mathbb{O})\\to\\mathrm{End}_{\\mathbb{R}}(\\mathbb{O})$ which sends $u$ to left-multiplication-by-$u$ is \"Clifford\" and so extends to a representation of the Clifford algebra $\\mathrm{Cliff}(\\mathrm{Im}(\\mathbb{O}))\\to\\mathrm{End}_{\\mathbb{R}}(\\mathbb{O})$, which restricts to a map $\\mathrm{Spin}(7)\\to\\mathrm{SO}(8)$.\nThe point-stabilizer of any point in $S^7\\subset\\mathbb{O}\\cong\\mathbb{R}^8$ is $G_2=\\mathrm{Aut}(\\mathbb{O})$, and the point-stabilizer of any point in $S^6\\subset\\mathrm{Im}(\\mathbb{O})$ is $\\mathrm{SU}(3)$. Thus, we have some bundles\n$$ \\begin{array}{ccccc} G_2 & \\to & \\mathrm{Spin}(7) & \\to & S^7 \\\\ \\mathrm{SU}(3) & \\to & G_2 & \\to & S^6 \\\\ \\mathrm{SU}(2) & \\to & \\mathrm{SU}(3) & \\to & S^5 \\end{array}$$\nwhich combined with $\\mathrm{SU}(2)\\simeq S^3$ gives\n$$ \\mathrm{Spin}(7)=S^7\\ltimes (S^6\\ltimes (S^5\\ltimes S^3)), $$\nand when combined with the Hopf bundles $S^7=S^4\\ltimes S^3$ and $ S^3=S^2\\ltimes S^1$ becomes\n$$ \\begin{array}{lcl} \\mathrm{Spin}(7) & = & (S^4\\ltimes (S^2\\ltimes S^1))\\ltimes (S^6\\ltimes(S^5\\ltimes S^3)) \\\\ & \\mathrm{or} & (S^4\\ltimes S^3)\\ltimes(S^6\\ltimes(S^5\\ltimes(S^2\\ltimes S^1))). \\end{array} $$\nOn the other hand, we have $\\mathrm{Spin}(n)=S^{n-1}\\ltimes\\mathrm{Spin}(n-1)$ (where $\\mathrm{Spin}(n)\\to\\mathrm{SO}(n)$ gives an action on $S^{n-1}\\subset\\mathbb{R}^n$ and we invoke orbit-stabilizer again), which means\n$$ \\mathrm{Spin}(7)=S^6\\ltimes(S^5\\ltimes(S^4\\ltimes (S^3\\ltimes (S^2\\ltimes S^1)))). $$\nSo it looks like given any two sets of \"composition factors\" for $\\mathrm{Spin}(7)$, they can made equivalent using the Hopf bundles. Is this a general principle or is there an obvious counterexample? That is, if something can be written as $S^{n_1}\\ltimes\\cdots\\ltimes S^{n_k}$ in two different ways, can the (multisets of) composition factors be equated after using the Hopf bundles to replace $S^7$ with $S^4$ and $S^3$ etc.?\n(Hopefully this is not too vague and sloppy so as to be cryptic, and not trivial.)\n", "comments": [], "comment_count": 0, "tags": ["homotopy-theory", "fiber-bundles", "division-algebras", "spin-geometry", "hopf-fibration"], "creation_date": "2017-03-24T00:07:25", "diamond": 0, "votes": 21} {"question_id": "327", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/164670", "title": "Gauss-Lucas Theorem (roots of derivatives)", "body": "Gauss-Lucas Theorem states:\n\"Let f be a polynomial and $f'$ the derivative of $f$. Then the theorem states that the $n-1$ roots of $f'$ all lie within the convex hull of the $n$ roots $\\alpha_1,\\ldots,\\alpha_n$ of $f$.\"\nMy Question is:\nIs there a theorem which states that there exists a permutation $\\sigma \\in S_n$ that the inner area of the polygon which edges go through the roots of $f$ \n$$\\alpha_{\\sigma(1)}\\longrightarrow\\alpha_{\\sigma(2)}\\longrightarrow\\ldots\\longrightarrow\\alpha_{\\sigma(n)}\\longrightarrow\\alpha_{\\sigma(1)}$$ contains all roots of $f'$?\n\nEDIT (OB) It is not completely clear from the original question wether the OP allowed for self intersections of the polygonal curve with vertices the roots of $f$. The question that has a bounty on its head asks for a polygonal Jordan curve with vertices the roots of $f$ containing the roots of $f'$ further assuming $f$ has simple roots. Roots of $f'$ are allowed to lie on the edges of the polygonal Jordan curve. We further assume $n\\geq 3$ and that the roots of $f$ are not all aligned (i.e. not all contained in a real affine line.)\n", "comments": ["I found an interesting paper: arxiv.org/pdf/1405.0689", "@HansEngler Indeed! The same happens if you consider $(X^2+1)X(X^2-1)(X^2-4)$ for instance.", "It certainly can happen that some roots of $f'$ must lie on edges of any such polygonal Jordan curve. For example, take $f(z) = z^4 - z$ with roots $0$ and $a_i$, where the $a_i$ are roots of unity. Then the roots of $f'$ are $2^{-2/3}a_i$.", "@Maesumi the OP certainly meant to write \"whose edges\". I might consider rewriting the question and including an example, the only problem is that I don't know how to draw plots. What the OP means by inner area I can explain in a few words: you have a closed polygonal path (in my question it is a Jordan path, so no self intersections) that links all roots of $f$, and there is a compact \"inner area\" and a non compact \"outer area\", the same way when you draw a circle, it defines a compact disk $\\lbrace |z|\\leq 1\\rbrace$ and the non compact outer area $\\lbrace |z|>1\\rbrace$.", "I think it might be better to re-write the question from scratch and add an example or further explanation. what is \"inner area\" and what do you mean by \"which edges\"?", "@LeonidKovalev you are right. I'll edit my edit ^^ thanks for your help.", "@OlivierBégassat If the roots all lie on the same line, we can't have a polygonal Jordan curve through them. Maybe you wanted to say that the roots are in generic position: no three on the same line. And $n\\ge 3$.", "I started a bounty on this question. To get the reward, either give a counter-example, or produce a proof! Assume all roots to be distinct (this is not very restrictive, at least when it comes to producing a counter example, since we can modify the constant term of $f$ slightly, so that all its roots are now distinct, without affecting the position of its roots too much, and obviously not the roots of its derivative), and the polygon should be a polygonal Jordan curve.", "This is a very interesting question.", "Is the permutation required to be such that the edges do not cross? Otherwise it's hard to interpret the inner area. In any case, you have to allow the roots on the boundary of the polygon, as the case $n=2$ shows."], "comment_count": 0, "tags": ["calculus", "geometry", "polynomials", "convex-analysis"], "creation_date": "2012-06-29T11:30:59", "diamond": 0, "votes": 21} {"question_id": "328", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2812010", "title": "Number as the sum of digits of some degree", "body": "\nWe will say that the measure of a number is equal to the maximum degree in which it is possible to represent a number in the form of a sum of digits copied (You can not rearrange the numbers). For example, for $55$ this will be $5$, because $$ 55^1 = 55, \\quad 55 = 55$$ $$ 55^2 = 3025, \\quad 30+25 = 55 $$ $$ 55^3 = 166375, \\quad 1+6+6+37+5 = 55$$ $$ 55^4 = 9150625, \\quad 9+15+0+6+25 = 55 $$$$55 ^ 5 = 503284375, \\quad 5 + 0 + 3 + 28 + 4 + 3 + 7 + 5 = 55.$$\n Let $a_n$ be a sequence of numbers such that all smaller ones have a measure less than. What is the asymptotics of this sequence? Is it possible to somehow build numbers with a given measure? If not, what measures can not be built? \n\nThe task was put in a Russian forum, I put it a little different question, I will be glad to any help in its solution :)\n", "comments": ["We shouldn't have an immediate obstacle $\\mod 9$, so interesting numbers are $0$ or $1$ modulo $9$. Assuming that, one would expect that the sum of digits of $n^k$ is about $\\frac 9{2\\log 10} k\\log n$, so the power may be as large as $\\frac {2n\\log 10}{9\\log n}$ and, if you are lucky, you can go up from here for numbers divisible by $10$ because the growth rate of the sum of the digits for them is slower. For the numbers in the table that do not end with $0$, this is an almost exact match. However, the third (and last) term in your sequence is formally $10$, whose measure is $+\\infty$.", "All previous must have", "For Maximum, do you mean that all of the previous degrees have to have the same property? Or just the biggest degree with that property?", "Yes, it is obvious that the problem for numbers ending with zero is uninteresting.", "What is most interesting is that in the examples above the measure value coincides with the exact upper bound of the measure.", "The Russian forum also presents some results, it seems to me amazing such large numbers on such relatively small results. $ n \\quad b(n) \\\\ 675 \\quad 50 \\\\ 945 \\quad 68 \\\\ 964 \\quad 71 \\\\ 990 \\quad 107 \\\\ 991 \\quad 71 \\\\ 1296 \\quad 84 \\\\ 1702 \\quad 114 \\\\ 2728 \\quad 173 \\\\ 4879 \\quad 285 \\\\ 5050 \\quad 403 \\\\ 5149 \\quad 300 \\\\ 5292 \\quad 309 \\\\ $", "I ask the answer to one of the questions. Do you think it's bad when there are so many of them?", "What exactly are you asking?", "All natural large ones are considered, the first term in the sequence is obviously $2$ with measure $1$, and the second term is 9, since $$ 9 ^ 2 = 81, \\quad 8 + 1 = 9 $$"], "comment_count": 0, "tags": ["sequences-and-series", "number-theory", "elementary-number-theory", "asymptotics"], "creation_date": "2018-06-07T18:18:43", "diamond": 0, "votes": 20} {"question_id": "329", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4528730", "title": "Does every finitely generated group have finitely many retracts up to isomorphism?", "body": "\nThe infinite dihedral group $D_\\infty = \\langle a,b \\mid a^2 = b^2 = \\text{Id}\\rangle $ is a finitely generated group with infinitely many cyclic subgroups of order 2, every one of which is a retract.\n\nFor the group $\\mathbb{Z}\\oplus\\mathbb{Z}$‎, ‎take $H_n=\\langle (1,n)\\rangle$ for any integer $n$ (with $K=\\langle (0,1)\\rangle$)‎. ‎Then we have $\\mathbb{Z}\\oplus \\mathbb{Z}=H_n \\oplus K$ which shows that $\\mathbb{Z}\\oplus\\mathbb{Z}$ (hence every finitely generated abelian group) has infinitely many different retracts‎.\n\n\n\nEvery free group $F_n$ of finite rank $n$ has infinitely many retracts. In fact, each free factor of $F_n$ is a retract and there are infinitely many free factors.\n\nThese are examples of finitely generated groups with infinitely many retracts. If we look at them, we'll find that they have finitely many retracts up to isomorphism. My question is that does every finitely generated group have finitely many retracts up to isomorphism?\n", "comments": ["@CheerfulParsnip You are right. I'll add free groups of finite rank to my list. Thanks very much for the comment.", "You can add finitely generated free groups to your list of examples. Any retract of $F_n$ has rank $\\leq n$. Maybe look for a counterexample in groups like Grigorchuk's Group."], "comment_count": 0, "tags": ["group-theory", "finitely-generated", "retraction"], "creation_date": "2022-09-10T08:32:53", "diamond": 0, "votes": 20} {"question_id": "330", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/4495174", "title": "A difficult integral for the Chern number", "body": "The integral\n$$\nI(m)=\\frac{1}{4\\pi}\\int_{-\\pi}^{\\pi}\\mathrm{d}x\\int_{-\\pi}^\\pi\\mathrm{d}y \\frac{m\\cos(x)\\cos(y)-\\cos x-\\cos y}{\\left( \\sin^2x+\\sin^2y +(m-\\cos x-\\cos y)^2\\right)^{3/2}}\n$$\ngives the Chern number of a certain vector bundle [1] over a torus. It can be shown using the theory of characteristic classes that\n$$\nI(m) = \\frac{\\mathrm{sign}(m-2)+\\mathrm{sign}(m+2)}{2}-\\mathrm{sign}(m) = \\begin{cases}1 & -2< m < 0 \\\\ -1 & 0 < m < 2 \\\\0 & \\text{otherwise}\\end{cases}.\n$$\nIs there any way to evaluate this integral directly (i.e. without making use of methods from differential geometry) to obtain the above result?\nI should mention that the above integral can be written as ($1/4\\pi$ times) the solid angle subtended from the origin of the unit vector $\\hat{\\mathbf{n}}$,\n$$\nI(m)=\\frac{1}{4\\pi}\\int_{-\\pi}^{\\pi}\\mathrm{d}x\\int_{-\\pi}^\\pi\\mathrm{d}y\\, \\hat{\\mathbf{n}}\\cdot\\left(\\partial_x \\hat{\\mathbf{n}}\\times \\partial_y \\hat{\\mathbf{n}}\\right),\n$$\nwhere $\\mathbf{n}(m)=(\\sin x, \\sin y, m- \\cos x-\\cos y)$. While this form makes it very straightforward to evaluate $I(m)$, I am interested in whether there is a way to compute this integral using more standard techniques.\n\n[1] B. Bernevig Topological Insulators and Topological Superconductors Chapter 8\n", "comments": ["@TheSimpliFire done :) mathoverflow.net/questions/453343/…", "@xzd209 I would suggest you crosspost this question on MathOverflow, ensuring the post includes a link to this MathStackExchange post.", "@SetnessRamesory That case can probably be done with elliptic integrals but I still believe there must be an easier way to prove $I(m)=0\\quad\\forall|m|>2$, for instance. Due to the symmetry of the integrand and the exponent of $3/2$, an application of Green's theorem may be useful, similar to this approach.", "@TheSimpliFire do you know$$\\int_{0}^{\\pi} \\int_{0}^{\\pi} \\frac{\\text{d}x\\text{d}y}{\\sqrt{\\sin(x)^2+\\sin(y)^2+\\left ( 1-\\cos(x)-\\cos(y) \\right )^2 } }=\\frac{\\Gamma\\left ( \\frac14 \\right )^4 }{8\\pi}?$$", "The case $m=0$ is easy because we can rewrite as $-\\pi I(0)=\\int_0^\\pi f(x)\\,dx$ where $$f(x)=\\int_0^\\pi\\frac{\\cos x+\\cos y}{(2+2\\cos x\\cos y)^{3/2}}\\,dy$$ and we find that $f(x)=-f(\\pi-x)$ for $x\\in[0,\\pi]$ after substituting $u=\\pi-y$. So $I(0)=0$.", "It appears there are two cases missing: $I(\\pm2)=\\mp1/2$."], "comment_count": 0, "tags": ["integration", "multivariable-calculus", "differential-geometry", "definite-integrals", "characteristic-classes"], "creation_date": "2022-07-18T03:39:30", "diamond": 0, "votes": 20} {"question_id": "331", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3286093", "title": "Conjecture: No positive integer can be written as $a^b+b^a$ in more than one way", "body": "Today, I came up with the following problem when trying to solve this.\n\nAre there distinct integers $a,b,m,n>1$ such that the equation $$a^b+b^a=m^n+n^m$$ holds? That is, is there ever an integer that can be written as $a^b+b^a$ in more than one way?\n\nI claim that the answer is No, but I think solving this is beyond my knowledge. For a very preliminary observation, the simplest case is to consider the powers of $1,5,6,0$, since they end in those same digits. For example, $$\\begin{cases}a\\equiv5\\pmod{10}\\\\b\\equiv6\\pmod{10}\\end{cases}\\implies a^b+b^a\\equiv1\\pmod{10}.$$ However, this brings about an issue, since there is hardly any indication as to what values $x$ and $y$ can take other than them having opposite parity.\n\nPARI/GP code is\nintfun(a,b,m,n)={for(i=2,a,for(j=2,b,for(k=2,m,for(l=2,n,if(i<>k && i<>l && j<>k && j<>l && i^j+j^i-k^l-l^k==0,print(i,\" \",j,\" \",k,\" \",l))))));}\n\nNo solutions have been found up to $a,b,m,n\\le100$.\n", "comments": ["@ZachHunter I ran a code that checks for collisions as you suggested, up to $a, b \\le 10000$, and I also ran it again for $a<70000$ and $b<500$. Did someone calculate asymptotics for this question? Can we say something along the lines \"the function $f$ gets so big so fact that intuitively it's very unlikely that we'll find such number considering the bruteforces we ran for small numbers\"?", "Code could use updates: like parfor, or using the fact that if $a,b$ are same parity, $m,n$ are going to need to be same parity as well. Or that The code as written checks $2^5+5^2$ and $5^2+2^5$ you're doing more than $4$ times as many checks as needed.", "Maybe this could help: $a^b+b^a=b^{\\frac{b\\ln(a)}{\\ln(b)}}+b^a=b^a(b^M+1)$ where $M=\\ln\\left(\\dfrac{\\sqrt[\\ln(b)]{b\\ln(a)}}{e^a}\\right)$", "you can do a much broader search if you simply check if $f(a,b) = a^b+b^a$ is an injection for $(a,b) \\in \\mathbb{Z}^2, a \\geq b \\geq 2$. I was able to confirm this up for $a,b \\leq 1300$ here repl.it/repls/OutstandingEarlyAutomaticvectorization. You would easily be able to search through higher values if you ran this locally on your computer instead of through repl, but I don't have access to that at the moment.", "from my reconfiguring it, we get if any pair of them share a factor n, we can get a difference of nth powers on one side.", "If $a | (b-1)$ and $a |(x-1)$ and $ a |( y-1)$ then there are no solutions (obviously).", "$x^y-a^b$ rather.", "$a^b-x^y=b^a-y^x$ etc mean they either pump out factors on both sides, or are all coprime.", "Fermat's little theorem will help.", "Still no solution in the range $a,b,x,y\\le 200$", "We can accelerate the search by assuming $a\\le b, x\\le y, a\\le x$"], "comment_count": 0, "tags": ["number-theory", "modular-arithmetic", "diophantine-equations", "conjectures"], "creation_date": "2019-07-07T12:39:27", "diamond": 0, "votes": 20} {"question_id": "332", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3984049", "title": "If two convex polygons tile the plane, how many sides can one of them have?", "body": "The set of convex polygons which tile the plane is, as of $2017$, known: it consists of all triangles, all quadrilaterals, $15$ families of pentagons, and three families of hexagons. Euler's formula rules out strictly convex $n$-gons with $n\\ge 7$. (The pentagonal case is by far the most difficult one.)\nI am interested in pairs of convex polygons that can collectively tile the plane. Specifically, I am curious how many sides can be in a polygon which is part of such a tiling.\nHere are some conditions to impose on such a tiling, from weakest to strongest:\n\nThere is at least one copy of each tile. (Without this condition, one can trivially take a pair consisting of a tiling polygon and any other convex polygon, and just never use the latter shape.)\n\nThere are at least $k$ copies of each tile.\n\nThere are infinitely many of each tile.\n\nEvery tile borders a tile of the other type.\n\nThe tiling is $2$-isohedral, i.e., every tile can be carried to any other tile of the same shape by a symmetry of the tiling.\n\n\nEach of these conditions implies those above it.\nIn the weakest case, the number of sides is unbounded, as exhibited by the following example:\n\n(The tiling is constructed by decomposing \"wedges\" of central angle $2\\pi/N$ into congruent isosceles triangles, and then combining the central triangles to yield an $N$-gon in the center.)\nRequiring at least $k$ of each tile still yields arbitrarily high numbers of sides, by taking the above construction for $N=M\\cdot k$ and subdividing the $N$-gon into $k$ \"wedges\" which are $(M+2)$-gonal.\nOn the other end of the spectrum, I have found a $2$-isohedral tiling using regular $18$-gons, shown below:\n\nAfter consulting this paper, it seems that the tiling pictured above is of type $4_2 18_{12}-1\\text{a}\\ \\text{MN}\\ \\text{p}6\\text{m}$ in their classification scheme (shown at the bottom of page 109); there are no $2$-isohedral tilings which allow for any higher number of contacts between different shapes, although type $3_1 18_{12}-1\\text{a}\\ \\text{MN}\\ \\text{p}6\\text{m}$ also works (and can be obtained from the above construction by cutting each kite-shaped tile in two). Thus, it is maximal among $2$-isohedral tilings.\nWhat are the maximal tilings under weaker conditions? The maximal number of sides under each successively stronger restriction is a weakly decreasing sequence which goes $\\infty, \\infty, ?, ?, 18$. So far, I have no bounds on the missing two terms except that they are each at least $18$.\nSome notes on this problem:\n\nIt is not necessarily the case that one of the tiles may tile the plane on its own; see this math.SE question for a counterexample.\n\nIf convexity is relaxed for either piece, the number of sides is unbounded even in the $2$-isohedral case (in fact, both pieces can simultaneously have arbitrarily many sides).\n\n\nEdit: Crossposted to Math Overflow here.\n", "comments": [], "comment_count": 0, "tags": ["geometry", "plane-geometry", "polygons", "convex-geometry", "tiling"], "creation_date": "2021-01-13T10:15:35", "diamond": 0, "votes": 20} {"question_id": "333", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1789111", "title": "Progressive Dice Game", "body": "$(2019.)$ Edit: Rewriting the question to make it clear.\n\nThe progressive dice game\n\nAt the start, you have a fair, regular six sided dice $D=(1,2,3,4,5,6)$. \nThe game is played for $n$ turns. Each turn you make a roll, which will be $r\\in D$.\nThen to complete the turn, you make one of the following choices:\n\nBank the rolled number: You gain $r$ points (score).\nInvest in (upgrade) the dice \"evenly\": If $r\\lt 6$, choose $r$ sides and increase them by $1$ each. If $r\\ge 6$, that is, $r=6k+k_0,k_0\\lt 6$, then increase each of the six sides by $k$, and then choose $k_0$ sides and increase them by $1$ each.\nReroll the dice, effectively restarting this turn. But before rerolling, you must apply the penalty to the dice: \"Evenly\" downgrade the dice: If $S_0$ is the number of sides $\\gt 0$ on the dice, then: If $r\\lt S_0$, choose $r$ sides and decrease them by $1$ each. If $r\\ge S_0$, that is, $r=S_0k+k_0,k_0\\lt S_0$, then decrease each of the $S_0$ sides by $k$, and then choose $k_0$ sides and decrease them by $1$ each.\n\nWhat is the optimal way to play to maximize your expected score at the end of the game?\n\nIf the dice was allowed to be upgraded/downgraded arbitrarily (not \"evenly\"), then one could downgrade the first five sides until they reach $0$. These sides now act as free rerolls. Then, keep investing the remaining points into the sixth side, which is now guaranteed to be rolled on each turn, after some amount of rerolls of that turn. Finally, bank that sixth rolled side in the last couple turns to maximize the expected value of the score.\nBut since we must upgrade/downgrade evenly, I'm not sure what is the optimal strategy.\n\nIf we ignore the \"reroll\" move:\nIf you upgrade the first $t$ turns, then bank the rest of the turns, you will expect the following amount of points on average:\n$$ f(t) = 3.5\\times\\left(\\frac{7}{6}\\right)^{t}\\times(n-t)$$\nWhich boils down to, that if you want to maximize your expected score, you should upgrade until the last $6$ (or $7$) turns and then bank those turns.\nBut this approach completely ignores the third action; the rerolls.\n\nCan we do better than this strategy, if we use the rerolls somehow?\n\n\nRerolls?\nI haven't worked out the strategy if the rerolls are considered.\nA reroll will on average decrease the average value of the dice, and allow you to either improve or worsen your current turn, with equal probability on average?\nBut there seem to be exceptions? For example, rerolling a $1$ seems useful if used early (as later, if we had upgraded a lot, all sides will be much greater than $1$). Simply downgrade the rolled $1$ side when downgrading (rerolling). If you roll that side again, it will be $0$, and this allows you to again reroll the dice for free (downgrading $0$ points is a free reroll). Which means, choosing to downgrade when you roll a $1$, can only increase your expected score in that turn. But there is still a (small?) drawback: Lets say some other side is a number $\\ge 6$. Then when upgrading later, you will have to put back at least one of those upgrade points into that downgraded $0$ side.\n\nSeems to me that rerolls will decrease the expected score on average\n (as they decrease the average value of the dice), so it is always\n better not to use them (except in that early scenario of the game, if $n$ is small)? Is this true?\n\nFor example, for small $n$, rerolls can be useful to force larger values. For $n=1$ specifically, it seems we can always force the first turn (the only turn) to end up banking a $6$, the maximal possible score for $n=1$, by rerolling and downgrading strategically the rest of the sides if $6$ was not rolled.\nBut for large $n$, the rerolls seem to lower the average expected score at the end, if used anytime in previous turns, as larger upgrades will need to replenish those downgraded points inevetably at some point, as they are carried out \"evenly\".\n", "comments": ["As he described you could decrement 1 to 0 but the next time you decrement you have to decrement any other number before you can decrement the 1 which is 0 now", "I think the increment/decrement rule can be described thusly: If you roll a $k$, then you add/subtract $1$ to/from the smallest/largest number currently on the die, and repeat this action $k$ times. Or do you want the player to be able to choose the sides whose numbers get changed? (If so, then any time you roll a $1$, you should decrement that side to a $0$, which effectively gives you a free roll.)", "@joriki Yes, a reroll is a new turn, more precisely it is starting your turn over, but with a bit downgraded dice. Yes, by maximise I tought of maximising the expected value, or in other words the average value over repeated games. Yes it \"starts over\", downgrade points work the same way as upgrades, but the zero field is simply not being looked at. The point is that the points are evenly distributed, not for example stacked at one side, and then being taken from the rest of the sides leading to a $(0,0,0,0,0,x)$ dice with $x$ being doubled each turn, and zeroes used as free rerolls.", "Your formula seems right, but would be a bit easier to understand if you write it as $3.5\\cdot\\left(\\frac76\\right)^t(n-t)$.", "And for the downgrade: Does this also start over once you've taken a point off all sides? If there are sides with $0$, does the downgrade start over once you've taken a point off all non-zero sides, or do the zero sides count towards the points despite not being further downgraded?", "You can't maximise your score, since this is a random variable. I suspect what you mean is to maximise the expected value of the score?", "And in the downgrade you remove as many points as you'd rolled, and then effectively get a new turn, in which you have a new choice whether to bank, upgrade or downgrade?", "@mjqxxxx You do one side by one point, then repeat for another side, till you use up all your points that were rolled, if you do all sides and have points left to spent, you do the same with the remaining points again. (Example; rolled $4$ = do $4$ sides, each by $1$ point, Example; $8$ = do all sides by one point, then do two sides by one point)", "Can you be more specific about the meaning of \"[increment] its sides distributing the points one by one\" and \"downgrade the dice evenly taking one by one point\"? Are you incrementing/decrementing all sides by one? Just the smallest/largest side?"], "comment_count": 0, "tags": ["probability", "recreational-mathematics", "game-theory", "dice"], "creation_date": "2016-05-17T08:25:12", "diamond": 0, "votes": 20} {"question_id": "334", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2248589", "title": "Is $\\Phi(q)$ rational for some $q \\in \\mathbb{Q}^*$, where $\\Phi$ is the standard normal cumulative distribution function?", "body": "Suppose that we have rational numbers $q_1$, $q_2$ such that\n$$\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^{q_1}e^{-\\frac{t^2}{2}} \\,\\mathrm{d}t=q_2.$$\nDoes this imply that $q_1=0$ and $q_2=\\dfrac{1}{2}$?\n", "comments": ["We know that $ q_{2} = \\frac{1}{2} \\Big( 1 + \\text {erf} \\Big( \\frac{ q_{1} }{ \\sqrt{2} } \\Big) \\Big)$ and $ \\text{erf} (x) = 1 - \\frac{1}{ \\sqrt{ \\pi }} \\Gamma \\Big( \\frac{1}{2} , x^2 \\Big)$, so for $ q_{2} $ to be rational $ \\text {erf} \\Big( \\frac{ q_{1} }{ \\sqrt{2} } \\Big) $ must be zero. Therefore $ q_{1} = 0$ and $q_{2} = \\frac{1}{2}$.", "I suspect $q_1$, $q_2$ can be even algebraic instead of rational (like in Lindemann-Weierstrass theorem showing that $\\sin x$ is transcendental)", "Two different questions here: (a) $\\Phi(\\Phi^{-1}(1/4)) = 1/4$ and $q = \\Phi^{-1}(1/4) \\approx -0.6744898$ is a number with rational $\\Phi(q),$ with $q \\ne 0$, so Yes to the title. (b) However, I would not want to claim that $q$ is rational, so the main question remains unanswered."], "comment_count": 0, "tags": ["number-theory", "normal-distribution", "transcendental-functions"], "creation_date": "2017-04-23T12:31:39", "diamond": 0, "votes": 20} {"question_id": "335", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/870912", "title": "Open problems in Federer's Geometric Measure Theory", "body": "I wanted to know if the problems mentionned in this book are solved. More specifically, at some places, the author says that he doesn't know the answer, for example :\"I do not know whether this equations are always true\" p.361 4.1.8, or \"I do not know...\" p.189 2.10.26. Are there counterexamples or proofs of these asumptions? \nI add some details : in theorem 2.10.25, if $f:X\\rightarrow Y$ is a lipschitzian map of metric spaces, $A\\subset X$, $0\\leq k<\\infty$, and $0\\leq m<\\infty$, then\n$$\n\\int_Y^*\\mathscr{H}^k(A\\cap f^{-1}\\{y\\})d\\mathscr{H}^m(y)\\leq (\\mathrm{Lip} f)^m \\frac{\\alpha(k)\\alpha(m)}{\\alpha(k+m)}\\mathscr{H}^{k+m}(A).\n$$ (where $\\mathscr{H}^n$ is the Hausdorff measure associated to the metrics on $X$ and $Y$, and $\\int^*$ is the upper integral) provided either $\\{y\\in Y, \\mathscr{H}^k(A\\cap f^{-1})>0\\}$\n is the union of countable family of sets with finite $\\mathscr{H}^m$ measure, or $Y$ is boundedly compact (each close bounded subset is compact). The question Federer asks is to determine if everything after \"provided either...\" is necessary. \nThe other question is about currents : let $S$, $T$ be two currents on open subsets $A$,$B$ of euclidean spaces, of degree $i$ and $j$ : if $S$, $T$ are representable by integration, do we always have $\\Vert S\\times T\\Vert=\\Vert S\\Vert\\times\\Vert T\\Vert$, and $\\overrightarrow{S\\times T}(a,b)=(\\wedge_i p)\\overrightarrow{S}(a)\\wedge(\\wedge_j q)\\overrightarrow{T}(b)$ for $\\Vert S\\times T\\Vert$ almost all $(a,b)\\in A\\times B$, where $p:A\\rightarrow A\\times B$ and $q:B\\rightarrow A\\times B$ are the canonical injections (it is indeed the case if either $\\overrightarrow{S}$ or $\\overrightarrow{T}$ is simple $\\Vert S\\Vert\\times\\Vert T\\Vert$ almost everywhere).\n", "comments": ["It seems it has been proved in Measure theory and fine properties of functions, revised edition, by Evans.", "Thanks, it would surprise me if there were no counter example or proof already available.", "I'm a bit surprised, I have seen some discussion of Federer's text here before. Keep an eye on this from time to time, it may get an answer later.", "Thank you anyway, I guess I'll have to find by myself. I really appreciate your help (I had no result either on math overflow).", "well, I tried. Sorry.", "I'll put a bounty on your question here once it's possible.", "Thank you Mr Cook, I asked the question on math overflow and added the same détails here.", "I wonder if you might get better input from math overflow on this one."], "comment_count": 0, "tags": ["integration", "measure-theory", "geometric-measure-theory"], "creation_date": "2014-07-18T08:02:10", "diamond": 0, "votes": 20} {"question_id": "336", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/226899", "title": "(Weil divisors : Cartier divisors) = (p-Cycles : ? )", "body": "Suppose $X$ verifies the suitable conditions in which Weil (resp. Cartier) divisors make sense. \nThe group of Weil divisors $\\mathrm{Div}(X)$ on a scheme $X$ is the free abelian group generated by codimention $1$ irreducible subvarieties. \nThe sheaf of Cartier divisors is $\\mathrm{Cart}_X:=\\mathcal{K}_X^{\\times}/\\mathcal{O}_X^{\\times}$.\nThe group of Cartier divisors is $\\mathrm{Cart}(X):=\\Gamma(X,\\mathrm{Cart}_X)$.\nThe group $Z^p(X)$ of $p$-cycles is the free abelian group generated by irreducible subvarieties of codimension $p$, so in particular $\\mathrm{Div}(X)=Z^1(X)$. So the notion of $p$-cycle is a direct generalization of the notion of Weil divisor. \nMy question:\n\nIs there an analogous notion of group of \"Cartier $p$-cycles\" $\\mathrm{Cart}^p(X)$? If yes, is there a sheaf $\\mathrm{Cart}^p_X$ such that (naturally in $X$) we have $\\mathrm{Cart}^p(X)=\\Gamma(X,\\mathrm{Cart}^p_X)$?\n\n", "comments": ["As Matt said we can consider ideal sheaves locally generated by $p$ element. A variety of codimension $p$ satisfies this is called local complete intersection, so we don't get all $p$-cycles, but certainly we get the smooth ones since by the Jacobian criterion they smooth (sub)varieties are lci.", "One naive attempt would be to say that Cartier divisors are \"locally principal.\" To get a codimension p object we'd need something like \"locally the zero set of p things\" which we could take to mean, lci or something. I have no idea how to make a sheaf that describes that, though."], "comment_count": 0, "tags": ["algebraic-geometry", "intersection-theory"], "creation_date": "2012-11-01T10:14:28", "diamond": 0, "votes": 20} {"question_id": "337", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/271611", "title": "Free medial magmas", "body": "A medial magma is a set $M$ with a binary operation $*$ satisfying $$(a*b)*(c*d) = (a*c)*(b*d)$$ for all $a,b,c,d \\in M$. Medial magmas constitute a finitary algebraic category $\\mathsf{Med}$, therefore there is a functor $M : \\mathsf{Set} \\to \\mathsf{Med}$ which sends a set $X$ to the free medial magma $M(X)$ over $X$. Elements of $M(X)$ can be seen as equivalence classes of oriented non-empty binary trees whose leaves are marked with elements of $X$, where two such trees are equivalent if the one can be reached from the other by a finite number of steps, where each steps looks as follows:\n\nNow I wonder if there is a more explicit description of the underlying set of $M(X)$, or a specific system of representatives. Can we simplify it? For example, when $X=\\{\\star\\}$, we have no markings; what is a specific system of representatives? I also wonder if these free medial magmas are studied or used anywhere in the literature.\n", "comments": ["The commutative version has been studied under the name of \"level algebras\", see for example arxiv.org/pdf/math/0209363v3.pdf.", "Me neither. In the introduction to the second paper I linked there are a few remarks on free medial groupoids. There is something to the effect that \"It seems that there is no very nice description of the equational theory of medial groupoids and free medial groupoids.\" I can only see the first two pages, so I don't know if there is more in later sections.", "Thank you. In fact this gives more search results. But unfortunately I couldn't find a paper which discusses free medial groupoids.", "It seems that searching for \"medial groupoid\" instead of \"medial magma\" yields more results (in universal algebra groupoid is synonymous with magma). For instance, you can find the work of J. Ježek, T. Kepka and others who produced a number of papers on related structures and their representations, e.g. here (where the free cancellative and the free commutative medial magma are constructed) or here."], "comment_count": 0, "tags": ["abstract-algebra", "category-theory", "trees"], "creation_date": "2013-01-06T08:06:35", "diamond": 0, "votes": 20} {"question_id": "338", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/3800570", "title": "Is the Frog game solvable in the root of a full binary tree?", "body": "Frog game\nThe Frog game is the generalization of the Frog Jumping (see it on Numberphile) that can be played on any graph, but by convention, we restrict the game to Tree graphs (see wikipedia).\nThe game is simple to play, but it can be hard to determine if it is solvable in a given vertex.\n\nIn short, given a graph, the goal of the game is for all frogs to host a party on a single vertex of that graph. Initially, every vertex has one frog. All of the $f$ frogs on some vertex can \"jump\" to some other vertex if they both have at least one frog on them and if they are exactly $f$ edges apart from each other. If a \"sequence of jumps\" exists such that all frogs end up on a single vertex, then the party is successfully hosted on that vertex and we call that vertex a \"lazy toad\" because the frog that started there never jumped.\n\nFormal rules\nLet $v,w,u\\in V$ be vertices of a graph $G=(V,E)$ and $n=|V|$ the number of vertices.\nLet $f_m:V\\to\\mathbb N$ count the number of frogs on a given vertex where $m$ is the current declared move. A game takes $(n-1)$ moves to solve (if possible) and is played as follows:\n\nWe start the game on move zero $m=0$ where each vertex has one frog $f_0(v)=1,\\forall v\\in V$.\n\nIf it is move $m\\lt n-1$ and the following conditions are met:\n\nThere exist two vertices $v,w\\in V$ that have frogs on them $f_m(v)\\ge 1$, $f_m(w)\\ge 1$.\nThere exists a path from $v$ to $w$ containing exactly $f_m(v)$ unique edges.\n\n\nThen, a legal move (or jump) can be made and is denoted as $(v\\to w)$. If the move is made, then the following transitions occur:\n\nAll frogs jump from $v$ to $w$, that is, $f_{m+1}(v)=0,f_{m+1}(w)=f_m(v)+f_m(w)$.\n\nThe remaining frogs don't move, that is, $f_{m+1}(u)=f_{m}(u),\\forall u\\not\\in\\{v,w\\}$.\n\nWe declare the next move $m+1$. (The game ends if we can't move.)\n\n\n\nWe say that the game is solvable in some vertex $w$ if there exists a sequence of legal moves such that we end up moving all frogs to the vertex $w$, which means $f_{n-1}(w)=n$. Consequently, if the game is solvable in $w$ then $f_{n-1}(v)=0,\\forall v\\ne w$. The solvable vertex $w$ is called a \"lazy toad\" because the frog on the vertex $w$ never jumped during the game.\n\n\nFor some examples, you can visit a post about this game on mathpickle. There is also an included link to a google drive where all trees with less than $15$ vertices have been computationally solved and categorized by the number of vertices then by the number of non-solvable vertices.\nRemark. I've asked a general question about this game before Frogs jumping on trees, which asks if we can characterize solvable and non-solvable vertices of certain sets of graphs, but that appears to be a hard problem. Here, I've decided to restrict the problem to binary trees and only observe the root vertex.\n\n\nQuestion\nLet $T_h=(V,E)$ be a full binary tree of height $h$. This means that it has layers $0,1,2,\\dots,h$ where the root $r\\in V$ is on layer $0$. That is, we have $|V|=2^{h+1}-1$. Let $h\\in \\mathbb N$ because $h=0$ is a single vertex which is trivial.\n\nI would conjecture that for all $h\\ge 4$, every vertex of $T_h$ is a \"lazy toad\" (is solvable).\n\nHowever, for simplicity, my question is just about the root vertex:\n\nLet $h\\in\\mathbb N, h\\ne 2$. Can we prove that the root $r$ of every such $T_h$ is a \"lazy toad\" (is solvable) ?\n\nNotation\nLet $v(i,j)\\in V$ be the $j$th non-root vertex on the layer $i=1,2,\\dots,h$ where $j=1,2,\\dots,2^i$.\nLets declare a shorthand notation for moves:\n\n\"$(v\\to w)$ then $(w\\to u)$\" to be equivalent to \"$((v\\to w)\\to u)$\".\n\n\"$(v\\to w)$ and $(u\\to w)$\" to be equivalent to \"$(v\\to w \\leftarrow u)$\".\n\n\nFor clarity, let $f=f_m(v)$ in \"$\\left(v\\xrightarrow{f} w\\right)$\" stand for the number of frogs being moved.\nAlternative suggestions for notation are welcome.\n\n\nExamples (Reducing $T_3,T_4,T_5$ to $T_1$)\nThe fact that we can reduce examples to smaller examples makes me think that maybe induction is possible. Notice that $T_1$ was solved in the original version of the game where all vertices are on a line and that $T_2$ is not solvable, so $T_3$ is the first \"real\" case.\n$$(h=1)$$ The trivial case $T_1$ is trivially solvable in all vertices:\n\n[1.1] In $r$ using legal moves $v(1,1)\\xrightarrow{1} r$ and $v(1,2)\\xrightarrow{1} r$ in any order.\n[1.2] In $v(1,1)$ by using moves $\\left(r\\xrightarrow{1} v(1,2)\\right)\\xrightarrow{2} v(1,1)$.\n[1.3] In $v(1,2)$ by using moves $\\left(r\\xrightarrow{1} v(1,1)\\right)\\xrightarrow{2} v(1,2)$.\n\n$$(h\\ne 2)$$ The second case $T_2$ is not solvable in the root. This can be checked by listing all move sequences.\n$$(h=3)$$ Lets break down the $T_3$ case. This case can be looked as a \"base\" trivial $T_1$ case that has two trivial $T_1$ cases \"connected\" to every leaf vertex. Let the vertices of those four \"connected\" trivial cases $T^{(1)}_1,T^{(2)}_1,T^{(3)}_1,T^{(4)}_1$ be indexed as: $v_1,v_2,v_3,v_4$ and let the vertices of the \"base\" trivial $T^{(0)}_1$ case be indexed as $v_0$.\nThat is, we can observe the $T_3$ as \"composition\" of $T_1$'s:\n\nNow we can see that the solution in the root of $T_3$ follows from the solution of $T_1$. That is:\n\nsolve the \"connected\" $(T^{(i)}_1,i\\in\\{1,2,3,4\\})$'s in a leaf vertex using [1.3]: $$\\left(r_i\\xrightarrow{1} v_i(1,1)\\right)\\xrightarrow{2} v_i(1,2),$$\n\nthen do the moves $v_i(1,2)\\xrightarrow{3} r_0$ for $i\\in\\{1,2,3,4\\}$ to transfer those frogs to root $r_0$,\n\nthen finally just solve the base $T^{(0)}_1$ with [1.1]: $v_0(1,1)\\xrightarrow{1} r_0$ and $v_0(1,2)\\xrightarrow{1} r_0$.\n\n\nThat is, the $T_3$ case is solved using the solutions of the trivial $T_1$ case.\nIn other words, if we know the solutions of $T_1$, we can reduce the $T_3$ to $T_1$ again.\n$$(h=4)$$ Similarly as in the previous case, we can observe $T_4$ as \"composition\" of two connected $T_2$'s to each of the two leafs of one \"base\" $T_1$. That is, we \"reduce\" the $T_4$ to $T_1$ by \"solving\" the four $T_2$'s by moving all frogs from their vertices to the root of the \"base\".\nWe can \"reduce\" (\"solve\") $(T^{(i)}_2,i\\in\\{1,2,3,4\\})$'s using following two move sequences:\n\n$v_i(2,1)\\xrightarrow{1} v_i(1,1)$ and $v_i(2,2)\\xrightarrow{1} v_i(1,1)$, then $\\left(v_i(1,1)\\xrightarrow{3} v(2,4)\\right)\\xrightarrow{4} r_0$.\n\n$r_i\\xrightarrow{1} v_i(1,2)$ and $v_i(2,3)\\xrightarrow{1} v_i(1,2)$, then $v_i(1,2)\\xrightarrow{3} r_0$.\n\n\nThe vertices from the first (second) move sequence are colored blue (green) here in the $T_2^{(i)}$:\n\nAfter the reduction, we are left with the \"base\" $T_1$ which is solved with [1.1] from the trivial case. In other words, we essentially moved all frogs from layers $2,3,4$ of $T_4$ to the root, leaving us only with layers $0,1$ which is equivalent to $T_1$.\n$$(h=5)$$ Similarly to previous two cases, this $T_5$ case can be reduced to $T_1$ by solving four $T_3$'s. That is, we can reduce the layers $2,3,4,5$ by moving all their frogs to the base root. To do this, there are three separate move sequences highlighted in the following image (each puts $5$ frogs in a leaf vertex before moving the frogs to the base root):\n\n$\\left(\\left(v_i(2,1)\\xrightarrow{1} v(3,1)\\right)\\xrightarrow{2} v_i(1,1)\\xleftarrow{1} r_i\\right)$, then $\\left(v_i(1,1)\\xrightarrow{4}v_i(3,8)\\right)\\xrightarrow{5}r_0.$\n\n$\\left(v_i(3,3)\\xrightarrow{1} v_i(2,2)\\leftarrow{1} v_i(1,2)\\right)$, then $\\left(\\left(v_i(2,2)\\xrightarrow{3} v_i(1,2)\\right)\\xrightarrow{4}v_i(3,2)\\right)\\xrightarrow{5}r_0.$\n\n$\\left(v_i(3,5)\\xrightarrow{1} v_i(2,3)\\xleftarrow{1} v_i(3,6)\\right)\\xrightarrow{3} v_i(3,7)$, then $\\left(v_i(2,4)\\xrightarrow{1}v_i(3,7)\\right)\\xrightarrow{5}r_0.$\n\n\n\n$$(h\\ge 6)$$ These examples so far motivate the following question:\n\nIs the reduction from $T_h$ to $T_1$ by solving four $T_{h-2}$'s always possible?\n\nThis could be one way to solve the problem. On the other hand, are there any other reductions?\n\n\nReduction pattern based on $T_6,T_7$\nSince reductions from $T_h$ to $T_1$ over $T_{h-2}$'s seem to get more complex as $h$ grows, this begs the question if there are any other simpler reductions.\nIt is sometimes possible to reduce $T_h$ to $T_{h-2}$ over $T_1$'s. Take a look at the following examples.\nLet $k=0,1,2,\\dots,2^{h-2}-1$. If we look at the $T_6$, we can reduce its top two layers (reuducing it to $T_{4}$) using the following:\n\n$v(6,4k+1)\\xrightarrow{1} v(5,2k+1)$ and $v(6,4k+2)\\xrightarrow{1} v(5,2k+1)$,\n\n$\\left(v(5,2k+2)\\xrightarrow{1} v(6,4k+3)\\right)\\xrightarrow{2} v(6,4k+4)$,\n\n$\\left(v(5,2k+1)\\xrightarrow{3} v(6,4k+4)\\right)\\xrightarrow{6} r$.\n\n\nSimilarly, we can reduce the $T_7$ to $T_5$ by using steps 1., 2. from above but modifying step 3. as (Note that you also modify the layers from $5$ to $6$ and $6$ to $7$ in all steps, of course.)\n\n$\\left(v(7,4k+4)\\xrightarrow{3} v(6,2k+1)\\right)\\xrightarrow{6} r$.\n\nWe can ask if there are any other trees $T_h$ whose top two layers can be reduced.\nNotice that we are collecting $T_1$'s in this reduction, that they contain $3$ vertices (frogs) each and that the total number of used $T_1$'s in our move sequence(s) must be a power of $2$ to completely reduce the top two layers, where we have $2$ possible layers $\\{h,h-1\\}$ to put the frogs on before jumping to root. Because of this, it is meaningful to try to extend this pattern to the full binary trees of the following kind:\n\nIf $h\\in\\{3\\cdot2^t,3\\cdot2^t+1\\},t\\in\\mathbb N$, for which $t$'s is such $T_h$ reducible to $T_{h-2}$?\n\nThis now does smell like something that would be used in an inductive proof. But, even if this pattern holds for all $t$, it alone is not enough. The question is, can we find a sufficient set of reduction patterns?\n\nLet $s\\lt h$ and $h\\gt 2$. Is it true that every $T_h$ can be reduced to some $T_s$ ?\n\n\n\nTrivial reduction pattern\n\nIf $h=2^t+t-2,t\\ge 2$ and $T_{t-1}$ is solvable in root, then $T_h$ can be reduced to $T_{h-t}$.\n\nThis is trivially true because the distance of the root of \"$T_{t-1}$ that spans the top $t$ layers\" to the \"(base) root $r=r_0$\", is precisely equal to $2^t-1$, which is the total number of frogs in $T_{t-1}$.\nFor example, if $t=4$ we have $h=18$. Solving top $t=4$ layers as a $(T_{t-1}=T_{3})$'s in root, puts $15$ frogs on a vertices that are on layer $15$. These frogs can jump to the root of $T_h$ which leaves us with $T_{h-t}=T_{14}$.\nBoth this and the previous reduction pattern are exponential and are not enough to cover all values of $h$ in the inductive argument. That is, we need more statements like this to cover all values.\n\n\nOther examples\nI've managed to find reductions for $T_h$ for $h$ up to $20$ by hand to show they are all solvable in the root, but I still don't see how I could complete the inductive argument or solve this problem in some other way for all $h\\ne 2$.\nFor example, I can reduce $T_8,T_9,T_{10},T_{11}$ to $T_5,T_5,T_6,T_7$ respectively.\nThen, $T_{12},T_{13}$ belong to $t=2$ from the above reduction pattern for $h\\in\\{3\\cdot2^t,3\\cdot2^t+1\\}$ and can be reduced to $T_{10},T_{11}$. It is not as easy as $t=1$, but the move sequences do exist.\nAfter those, I can reduce $T_{14},T_{15},T_{16},T_{17}$ to $T_{11},T_{11},T_{12},T_{13}$.\nThen, $T_{18}$ belongs to $t=4$ of the trivial reduction pattern which means it can be reduced to $T_{14}$.\nFor $T_{19}$ I found a reduction to $T_{14}$,... and so on.\nNote that these reductions are not the only ways to solve a tree in the root.\n\n\nNecessary reduction condition\nNotice that to reduce $T_h$ to $T_{h-a}$ for some $a\\ge 2$, we need to move all frogs from top $a$ layers to the root. The number of frogs we are moving to the root in such reduction is $(2^a-1)2^{h-a+1}$. That is, we are shaving off the $T_{a-1}$'s at the top layers. For such a reduction to be possible, there must exist a set of moves that transfers the frogs to the root $r$\n$$M_r=\\{v(i,j)\\xrightarrow{f_{i,j}}r\\},\\text{ where } i\\in\\{h,h-1,\\dots,h-a+1\\}.$$\nAssuming such moves exist at some point in the game, they are legal only if $f_{i,j}=i$ where $i$ is the layer on which the vertex $v(i,j)$ is on. On the other hand, the sum of $f_{i,j}$'s must be equal to the total number of frogs that we are reducing (moving to root). That is, for $M_r$ to be able to exist, we necessarily need to be able to partition the number of frogs we are moving onto the corresponding layers, and this is essentially what we state in the necessary condition.\nWe also care if our move sequences repeat or have patterns (for example, see the use of $k$ in the $T_6,T_7$ reduction pattern), because then instead of considering all frogs at once, we can consider only $(2^a-1)2^{b},b\\in[0,h-a+1]$ frogs for some $b$, and repeat the same move pattern $(2^{h-a+1}/2^b)$ times.\nBased on all of this, we can state the following necessary condition:\n\n(Necessary condition for reduction). Let $h\\gt2,a\\ge2$. If reduction from $T_h$ to $T_{h-a}$ is possible, then there exists $b\\in[0,h-a+1]$ and a partition of the number $(2^a-1)2^b$ into the layers $H=\\{h,h-1,h-2,\\dots,h-a+1\\}$ where $(h-l)\\in H$ is used at most $(2^{a-l-1})2^b$ times.\n\nOf course, the converse does not hold because this is only a necessary (not a sufficient) condition.\nFor an example, if we want a $T_h$ to $T_{h-2}$ reduction then we need to partition $3\\cdot 2^b$ into $\\{h,h-1\\}$. If we assume that the solution is \"simple\", i.e. that all frogs are either on layer $h$ or layer $h-1$ (but not mixed between both layers), we get the following pattern:\n\n$h$ must either be of form $3\\cdot 2^t$ or $3\\cdot 2^t+1$. This was essentially our assumption on $h$ when we talked about extending the $T_6,T_7$ pattern. But now we also need a sufficient condition. That is, if we want to prove this is possible for every $t$, then we need to for every $t$ find a sequence of corresponding legal moves or prove it exists.\n\nSolvable partitions?\nSurprisingly, the above condition is already very useful (efficient). I have that for all $h\\in(2,20)$ I've solved so far, I needed to consider only $31$ or less vertices ($a\\le 5,b\\le 3$) to show that all vertices can be reduced to the root (that we can send all frogs to the root).\nFor example, the $T_{19}$ has over a million vertices! But, to solve it, one needs to consider just one $T_4$ (which has only $31$ vertices) at the top layers to find a solution! (Consider $a=5,b=0$ and the corresponding partition $31=15+16$ is solvable.) That is, the reduction pattern of that $T_4$ can simply be repeated on all adjacent $T_4$'s in the top five layers, which gives us a reduction to $T_{14}$ which we have already solved by reduction to $T_{11}$ then to $T_5$ then to $T_1$.\nEssentially, we need to prove that at least one of the partitions given by the \"(Necessary condition for reduction)\" always allows the reduction to a smaller binary tree (that the converse holds for at least one partition). Then by induction, we have that they are all reducible to $T_1$.\nI don't yet see how to construct move sequences for all cases of $h$. Alternatively, I do not know if we can maybe prove it without explicit construction of solutions?\n", "comments": ["Since the bounty didn't lead to anything new, its now crossposted to mathoverflow.net."], "comment_count": 0, "tags": ["graph-theory", "induction", "recreational-mathematics", "trees"], "creation_date": "2020-08-23T03:56:44", "diamond": 0, "votes": 19} {"question_id": "339", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2104883", "title": "Is there an irrational number containing only $0$'s and $1$'s with continued fraction entries less than $10$?", "body": "The number $$0.10111001001000000000001$$ has continued fraction $$[0, 9, 1, 8, 9, 5, 1, 1, 5, 3, 1, 3, 1, 1, 4, 6, 1, 1, 8, 2, 5, 8, 1, 9, 9, 5, 2\n, 8, 1, 1, 6, 4, 1, 1, 3, 1, 3, 5, 1, 1, 5, 9, 8, 1, 9]$$\nSo, the maximum values is $9$. But we are only at $23$ digits. Can we produce larger decimal expansions with the required property ? Perhaps arbitary large ones ?\n\nIs there an irrational number, such that the decimal expansion contains only ones and zeros and the continued fraction contains no entry larger than $9$ ?\n\n", "comments": ["A shorter example without long sequences is $$0.1011010011100100111101101110101$$", "But I would prefer sequences without such long $0$-sequences, if possible :)", "$$0.1011010000001000000010000000000000$$ $$0000000010000000000000000000000000000000000000 0000001$$ is far better with $87$ digits! Seems that we can continute forever, if we add a digit $1$ and then insert enough $0$'s between the last two ones ...", "(+1) An interesting question on the intersection of Cantor-like sets."], "comment_count": 0, "tags": ["number-theory", "irrational-numbers", "continued-fractions"], "creation_date": "2017-01-19T10:25:18", "diamond": 0, "votes": 19} {"question_id": "340", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1887362", "title": "Increasing derivatives of recursively defined polynomials", "body": "Consider recursively defined polynomials $f_0(x) = x$ and $f_{n+1}(x) = f_n(x) - f_n'(x) x (1-x)$.\nThese polynomials have some special properties, for example $f_n(0) = 0$, $f_n(1) = 1$, and all $n+1$ roots of $f_n$ are in $[0,1)$. \nLet $x_n$ denote the largest root of $f_n$. Then $f_n(x_n) = 0$ and $f_n'(x_n)>0$. Moreover, $x_n > x_{n-1}$ for all $n$.\nI want to prove the following claim: $f_{n}'(x_{n+1}) > f_{n-1}'(x_{n+1})$ for all $n \\geq 2$.\nNote that the claim does not hold for arbitrary $x$. The derivatives are polynomials themselves, by Gauss-Lucas theorem all their roots are in $[0,x_n)$ and there are many points where $f_{n}'(x) < 0 < f_{n-1}'(x)$. However, I am quite sure that at $x \\geq x_{n+1}$, the derivatives are ordered: $f_1'(x) < \\dots < f_n'(x)$. \nSome of the first polynomials are:\n\n$f_1(x) = x^2$, $f_1'(x) = 2x$, $x_1 = 0$\n$f_2(x) = 2x^3 - x^2$, $f_2'(x) = 6 x^2 -2x$, $x_2 = \\frac{1}{2}$\n$f_3(x) = 6 x^4 - 6x^3 + x^2$, $f_3'(x) = 24x^3-18x^2+ 2x$, $x_3 \\approx 0.7887$\n$f_4(x) = 24 x^5 - 36 x^4 + 14 x^3 -x^2$, $f_4'(x) =120x^4-144x^3 +42x^2 -2x$, $x_4 \\approx 0.9082$\n\nTherefore $f_2'(x)-f_1'(x) = 6x^2-4x \\geq 0$ for all $x \\geq \\frac{2}{3}$. Note that, $x_2 < \\frac{2}{3} < x_3$.\nFor $f_3'(x) - f_2'(x) = 24 x^3-24 x^2 + 4 x \\geq 0$ for all $x \\geq 0.7887$. Here it turns out that at $x_3$ the inequality holds as an equality (coincidence perhaps?), but of course then for $x_4 > x_3$ it holds as a strict inequality. \n", "comments": ["That's right, the claim does not hold for $n=1$. What I wanted was for all $n \\geq 2$ (in my application $f_0$ is irrelevant, I'm just using it for construction). Sorry for the confusion. I'm updating the question.", "The claim is false since $f'_1(x_2)=f'_0(x_2)$.", "Just a conjecture, coming from looking at the plots for the first 5-6 cases.", "Why are you quite sure the derivatives are ordered at $x \\geq x_{n+1}$?"], "comment_count": 0, "tags": ["real-analysis", "polynomials", "recurrence-relations", "roots"], "creation_date": "2016-08-09T10:49:41", "diamond": 0, "votes": 19} {"question_id": "341", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2104333", "title": "Calculate using residues $\\int_0^\\infty\\int_0^\\infty{\\cos\\frac{\\pi}2(nx^2-\\frac{y^2}n)\\cos\\pi xy\\over\\cosh\\pi x\\cosh\\pi y}dxdy,n\\in\\mathbb{N}$", "body": "\nQ: Is it possible to calculate the integral\n $$\n\\int\\limits_0^\\infty \\int\\limits_0^\\infty\\frac{\\cos\\frac{\\pi}2\n\\left(nx^2-\\frac{y^2}n\\right)\\cos \\pi xy}{\\cosh \\pi x\\cosh \\pi y}dxdy,~n\\in\\mathbb{N}\\tag{1}\n$$\n using residue theory?\n\nFor example, when $n=3$\n$$\n\\int\\limits_0^\\infty \\int\\limits_0^\\infty\\frac{\\cos\\frac{\\pi}{2}\n\\left(3x^2-\\frac{y^2}{3}\\right)\\cos \\pi xy}{\\cosh \\pi x\\cosh \\pi y}dxdy=\\frac{\\sqrt{3}-1}{2\\sqrt{6}}.\n$$\nThere is a closed form formula to calculate (1) for arbitrary natural $n$, but I don't know how to do it by residue theory. Maybe it is possible in principle, but is residue theory practical in this particular case? It seems such an approach would lead to a sum with $O(n^2)$ terms. Any hints would be appreciated.\n", "comments": ["You mentioned that there is a closed form formula. Would you care to provide it?", "@AlfredYerger I know $I_{1,2}$ can be evaluated with residue theory. It has been done by Ron Gordon, for exampe, here on MSE. I believe his method can be applied to the case $n=3$ by calculating the integral over $x$ first, then similarly the remaining integral over $y$.", "This is very interesting. So you would be content if the integrals $I_1$ and $I_2$ just below (26) in this paper would be evaluated with residue theory? arxiv.org/pdf/1712.10324.pdf", "@AlfredYerger please see my paper at arxiv regarding 2D Mordell integrals.", "It is also possible the LM estimates one typically sees are too coarse for this integral, and something better is needed to show that the integral along such an arc goes to $0$ as $R \\to \\infty$, but I don't have any ideas right now.", "In particular, the place I got stuck is at estimating $| \\int_\\gamma \\frac{\\cos \\frac{\\pi}{2} u^2 \\cos \\pi u v}{\\operatorname{cosh} \\pi u/\\sqrt{n}}|$ which I obtained by making the change of variables $u = \\sqrt{n}x$ and $v = y/\\sqrt{n}$, using the angle subtraction formula, distributing, and then pulling out some terms depending only on $v$ from the first integral. This integral is of an even function, so we can extend to an integral over the whole real line. If the LM estimates for a semi-circular arc in the upper half plane could be made to $\\to 0$, we could try again to calculate residues.", "Has the case $n=3$ been done with Residue theory? I spent some time thinking about this evening and I don't see how it can be possible. There is no residue theory to my knowledge for double integrals, so you must do a residue calculation on one or both of the iterated integrals, but after some algebra, I see no way to make the LM estimates work, so I don't see how it can be possible. If the LM estimates could be worked out, I would be willing to try again.", "@Nicco I don't mind", "@ Nemo:Based on residue theory.Unfortunately my browser can't display the formulas in your blog", "@ Nemo: would you mind if I can offer a bounty for this question?", "thx, that is cool stuff...are do you doing this on a recreational basis or is this for professional purposes?", "@tired it is not due to Ramanujan.", "can you give a reference for the amazing formula $n=3$? i bet it is due to ramanujan"], "comment_count": 0, "tags": ["integration", "complex-analysis", "definite-integrals", "contour-integration", "residue-calculus"], "creation_date": "2017-01-19T02:28:47", "diamond": 0, "votes": 19} {"question_id": "342", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2076008", "title": "When does $x^{x^{x^{...^x}}}$ diverge but $x^{x^{x^{...^c}}}$ converge?", "body": "Let us define these two sequences as follows:\n$a_0=1$, $b_0=c$\n$a_{n+1}=x^{a_n}$, $b_{n+1}=x^{b_n}$\n$b_{n+1}\\ne b_n$ for any $n$.\n$x,c\\in\\mathbb C$\nIs it possible for $a_n$ to diverge but $b_n$ to converge under these conditions? For example, if $x=2$ and $c=i$, we have\n$b_1=2^i=\\operatorname{cis}(\\ln2)$\n$b_2=2^{\\operatorname{cis}(\\ln2)}=2^{\\cos(\\ln2)}\\operatorname{cis}(\\ln2\\sin(\\ln2))$\netc.\nI haven't much clue as to whether it is the case that $b_n$ can converge when $a_n$ diverges, and I can hardly work out if $b_n$ converges with $x=2$ and $c=i$.\n", "comments": ["@miloBrandt b=exp(1/e) has a neutral fixed point=e, and iterating b^^n converges towards e.", "One can work out that the only times that $x^{\\ldots^c}$ can converge without hitting a fixed point is whenever $W(-\\log(x))$ has absolute value at most $1$, where $W$ is the product log (and there are definitely converging $c$ when this absolute value is strictly less than $1$). I would maybe bet that there's an example of this happening where the absolute value is exactly $1$ and that where it's less than $1$, the tower converges starting at $1$.", "@SheldonL If you still wish to I would be interested in seeing your numerical results.", "@SheldonL: if you'll get something substantial - I'd like to see a workout also in our tetrationforum! (It seems to be an interesting facet in context with the D. Shell article)", "Even considering complex values of x, I believe the surprising result is that there are no solutions! I wrote a pari-gp loop that verified this for over 300,000 real and complex bases with attracting fixed points .... I think this is the case since I think the domain of the basin of attraction of the attracting fixed point can be extended to f'=0; but I will need to do some reading to brush up on complex dynamics, and iterated functions.", "@SheldonL Oddly, though I am asking such questions, such vocabulary is new to me. But it's ok, I get the gist of what you are saying. If you can think up a worthy answer, go for it and I'll ask clarifications later. And do complex fixed points behave like real ones? It seems quite hard to discern", "@SimpleArt ouch!, but I think. $0exp(1/e) so that the first power tower diverges. But all complex fixed points for bases>exp(1/e) are repelling. So doesn't that mean the 2nd power tower won't converge? And therefore there are no real valued solutions to the Op's problem?", "Hmm, the clause $c \\ne x^c$ intends to exclude the \"trivial\" case of $c$ being the fixpoint. But consider cases , where the exponentialtower to base $x$ has a set of cyclic fixpoints (which might also be accumulation points such that the height-increasing exponentialtower of $b_n$ \"converges\" to a cycle over such a set of points) then you might want to add an analogue clause/to extend the given clause. (I've not yet a true answer to your question so far)", "Then the second paper may be useful.", "@Rohan I'm not so sure, both papers cover the topic of my sequence $a_n$, which I already know about, I'm more interested in $b_n$.", "I hope this and this will help you."], "comment_count": 0, "tags": ["sequences-and-series", "exponentiation", "tetration", "complex-dynamics"], "creation_date": "2016-12-29T05:50:35", "diamond": 0, "votes": 19} {"question_id": "343", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1640772", "title": "Consecutive prime numerators of harmonic numbers?", "body": "Let\n$$\\frac{1}{1}+\\frac{1}{2}+\\frac{1}{3}+\\cdots+\\frac{1}{n}=\\frac{a}{b}$$\nand let $a$ and $b$ are coprime, $h_{n}=a$. \n$h_{n}$ is prime for \n$$n=2,3,5,8,9,21,26,41,56,62,69,79,89,91,122,127,143,167,201,230,247,252,290,349,376,459,489,492,516,662,687,714,771,932,944,1061,1281,1352,1489,1730, 1969,2012,2116,2457,2663,2955,3083,3130,3204,3359,3494,3572,3995,4155,4231,4250,4496,4616,5069,5988,6656,6883,8067,8156,8661,9097,\\ldots$$\nI guess proving that there are infinitely (or finitely) many primes of the form $h_n$ is very hard. But can we prove both of the $h_n$ and $h_{n+1}$ cannot be prime for $n>8$?\n", "comments": ["@vrugtehagel You're welcome.", "@Mr. Brooks, no, thank you. That is be some very useful information", "@vrugtehagel Did you look in oeis.org/A056903/b056903.txt ? They call it a \"b-file\" and it goes up to $h_{97} = 78128$.", "I haven't looked much into it before. The only thing I've done on this subject is prove that the harmonic series is never an integer for $n>1$. But this is interesting, so I'll definitely look into it", "@vrugtehagel : what do you already know about the numerators of the harmonic numbers ?", "I edited the question to include some more values for $n$, more than oeis.org, which for some reason only includes values up to $3572$", "Just for reference, the first values of $h_n$ are listed here: oeis.org/A001008", "What sort of question is this? I think it isn't trivial. Where is the conjecture from, just looking on the numbers or are there reasons why it should be right?"], "comment_count": 0, "tags": ["prime-numbers", "harmonic-numbers"], "creation_date": "2016-02-04T11:16:53", "diamond": 0, "votes": 19} {"question_id": "344", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/730", "title": "An elegant description for graded-module morphisms with non-zero zero component", "body": "In an example I have worked out for my work, I have constructed a category whose objects are graded $R$-modules (where $R$ is a graded ring), and with morphisms the usual morphisms quotient the following class of morphisms:\n$\\Sigma=\\left\\lbrace f\\in \\hom_{\\text{gr}R\\text{-mod}}\\left(A,B\\right) \\ | \\ \\ker\\left(f\\right)_0\\neq 0, \\ \\mathrm{coker}\\left(f\\right)_0\\neq 0\\right\\rbrace$ \n(by quotient I mean simply that this class of morphisms are isomorphisms, thus creating an equivalence relation) I am wondering if this category has a better (more canonical) description, or if I can show it is equivalent to some other interesting category.\nThanks!\n", "comments": ["Wow this is the first non-deleted question! Good job.", "Well, I don't remember. I do remember that I was localizing by these morphisms and that was what I meant by quotienting. Judging by the date, I was probably trying to work out a ncag example. I don't have a clue what it was!", "@MarianoSuárez-Alvarez I was trying to get the altruist badge, so I found the oldest unanswered question and put a bounty on it. It makes sense that this has been unanswered for three years...", "This is very weird, really: you are making all modules with non-zero zero component isomorphic... Does this really make sense in some context?", "Do you mean to make the maps in $\\Sigma$ all the zero maps (which is quotienting) or do you mean to make the maps in $\\Sigma$ isomorphisms (which is localizing)? As $\\Sigma$ is not an ideal the first isn't well defined, and as $\\Sigma$ can contain the zero map between modules the second doesn't seem to make much sense either.", "Since $\\Sigma$ isn't an ideal in the category of graded $R$-modules, I don't think that quotienting by it makes much sense."], "comment_count": 0, "tags": ["modules", "graded-modules", "graded-rings"], "creation_date": "2010-07-26T10:29:10", "diamond": 0, "votes": 19} {"question_id": "345", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/292279", "title": "Hopf-like monoid in $(\\Bbb{Set}, \\times)$", "body": "I am looking for a nontrivial example of the following:\n\nLet a monoid $A$ be given with unit $e$, and two of its distinguished disjoint submonoids $B_1$ and $B_2$ (s.t. $B_1\\cap B_2=\\{e\\}$), endowed with monoid homomorphisms $\\Delta_i:A\\to B_i$ such that $\\Delta_i|_{B_i}={\\rm id}_{B_i}\\ $ ($i=1,2$) . \n Moreover, also let a mapping $\\sigma:A\\to A$ be given, such that for all $a\\in A$:\n $$\\sigma(\\Delta_1(a))\\cdot\\Delta_2(a)=e= \\Delta_1(a)\\cdot\\sigma(\\Delta_2(a)) $$\n In particular, each element of $B_1$ is right invertible, and each element of $B_2$ is left invertible in $A$.\n\nSuch an example is trivial if $B_1=B_2=A$.\nMotivation: I am looking for Hopf-like monoids in cartesian categories, in which the counit axioms for the unique $A\\to 1$ morphism are not posed. Then, the comultiplication map $\\Delta:A\\to A\\times A$ has to be of the form $\\Delta=(\\Delta_1,\\Delta_2)$ and coassociativity means ${\\Delta_i}^2=\\Delta_i$ and $\\Delta_1\\Delta_2=\\Delta_2\\Delta_1$. $B_1\\cap B_2=\\{e\\}$ ensures this latter equality.\nOf course, $\\sigma$ would be the antipode.\n", "comments": [], "comment_count": 0, "tags": ["examples-counterexamples", "monoid", "hopf-algebras"], "creation_date": "2013-02-01T10:40:31", "diamond": 0, "votes": 19} {"question_id": "346", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/52933", "title": "Integral involving Complete Elliptic Integral of the First Kind K(k)", "body": "I have run into an integral involving the complete elliptic integral, which can be put into the following form after changing integration variables to the modulus:\n$$\\int_0^{\\sqrt{\\frac{\\alpha}{1+\\beta}}} dk\\, \\frac{ k^{11} K(k) } {\\sqrt{(\\alpha-\\beta k^2)^2 - k^4} (\\alpha - \\beta k^2)^{11/2}}$$\n$K(k)$ is the complete elliptic integral of the first kind, where $k$ is the modulus. We can assume that $\\alpha$ and $\\beta$ are such the maximum value for $k$ is less than or equal to $1$. Are there any ways to get a closed form solution out of this? The indefinite integrals in G&R are not much help.\n", "comments": ["+1 for remembering to mention your argument convention for elliptic integrals. It looks a bit gnarly as it stands, but I'll see what I can do."], "comment_count": 0, "tags": ["integration", "special-functions", "elliptic-integrals"], "creation_date": "2011-07-21T09:47:28", "diamond": 0, "votes": 19} {"question_id": "347", "site": "math", "category": "Science", "title": "Expected number of operations on a vector until one of the coordinates becomes zero.", "body": "Let's say we have a vector $v = (x_1, ..., x_n) \\in \\mathbb{N}^n$ where $x_1 = x_2 = ... = x_n$. Next we choose an ordered pair of coordinates at random $(i, j)$ where $i, j \\in \\\\{1, ..., n\\\\}$ and $i \\neq j$. Finally we substitute the vector $v$ with a new vector $v' = (x_1, ..., x_i + 1, ..., x_j - 1, ..., x_n)$. Now we choose again an ordered pair of coordinates at random and substitute the vector $v'$ with a new vector doing the same we did for $v$. We continue doing this until one of the coordinates becomes zero. \nWhat is the expected number of operations we are going to make?\n\nI know the answer for $n = 2$ because you can model this process with a random walk. If $v = (x, x)$, then the expected number of operations is the same as the expected number of steps it will take to hit $x$ or $-x$ doing a random walk starting at zero. In this case the expected number of step starting at $y$ satisfies the recurrence relation $$E_y = 1 + \\frac{1}{2} E_{y - 1} + \\frac{1}{2} E_{y + 1}. $$ Then one can solve this linear recurrence. \nI tried to do the same for the original problem but the recurrence relation is more difficult. Let $F_{(x_1, ..., x_n)}$ be the expected number of operations one can make to vector $v= (x_1, ..., x_n)$ before one of the coordinates becomes zero (in this case we allow $x_1, ..., x_n$ to be different). If I'm not wrong $F$ satisfies the following relation $$F_{(x_1, ..., x_n)} = 1 + \\sum_{i, j} \\frac{1}{n (n - 1)} F_{(x_1, ..., x_n) + e_{i, j}}, $$ where the $i$-th coordiante of $e_{i, j}$ is $1$, the $j$-th is $-1$ and the rest are all zero (the sum runs through all possible operations). \n", "link": "https://math.stackexchange.com/questions/2675605/expected-number-of-operations-on-a-vector-until-one-of-the-coordinates-becomes-z", "tags": ["probability"], "votes": 13, "creation_date": "2018-03-03T16:41:57", "comments": ["I think we can project the movement to hyperplane $x_1+\\cdots+x_n=nN$, where $N=x_1=\\cdots=x_n$ is the initial coordinate of $v$. With the stopping condition(i.e. stop when some coordinate becomes zero) we are on a bounded domain, with each transition $(0,\\cdots,1,\\cdots,-1,\\cdots,0)$ parallel to some boundary line $x_k=0,k\\not\\in\\{i,j\\},x_i+x_j=nN$. Seem like we can reduce the dimension...will this help?", "Note that raising operators do not modify the total weight of vectors. Moreover, whenever $a$ and $b$ are of equal weight and $a\\leqslant b$, there is a raising operator $R$ such that $Ra =b$. It is not clear if this should really help here, though, but perhaps knowing these operators have a name might help in searching for references. Buena suerte! ;)", "The operation you are considering is called an (elementary?) raising operator (a general raising operator is a product of perhaps repeated elementary ones). There is a partial order on $\\mathbb Z^n$ given by $a \\geqslant b$ iff for each $i$ we have $a_1+\\cdots+a_i\\geqslant b_1+\\cdots+b_i$, and raising operators are nondecreasing for this order."], "comment_count": 3, "diamond": 0} {"question_id": "348", "site": "math", "category": "Science", "title": "Reversal of an Autoregressive Cauchy Markov Chain", "body": "Let $\\mu_0 (dx)$ be the standard one-dimensional Cauchy distribution, i.e.\n\n\\begin{align} \\mu_0 (dx) = \\frac{1}{\\pi} \\frac{1}{1+x^2} dx. \\end{align}\n\nSuppose I fix $h \\in [0, 1]$, and form a Markov chain $\\\\{X_n\\\\}_{n \\geqslant 0}$ as follows:\n\n 1. At step $n$, I sample $Y_n \\sim \\mu_0$.\n 2. I then set $X_{n+1} = (1 - h) X_n + h Y_n$\n\n\n\nIt is not so hard to show that this chain admits $\\mu_0$ as a stationary measure, as this essentially comes from the fact that Cauchy distribution is a stable distribution.\n\nWhat I'm interested in is the reversal of this Markov chain. More precisely, if the chain I describe above uses the Markov kernel $q (x \\to dy)$, I want to understand the Markov kernel $r$ such that\n\n\\begin{align} \\mu_0 (dx) q (x \\to dy) = \\mu_0 (dy) r (y \\to dx). \\end{align}\n\nFortunately, all of the quantities involved have densities with respect to Lebesgue measure, and as such, I can write down what $r (y \\to dx)$ is:\n\n\\begin{align} r (y \\to dx) = \\frac{1 + y^2}{\\pi} \\frac{1}{1 + x^2} \\frac{h}{ h^2 + (y - (1 - h) x)^2} dx. \\end{align}\n\nMy question is then: is there a simple, elegant way to draw exact samples from $r$?\n\nI would highlight that this is not a purely algorithmic question; I'd _really_ like to understand what this reversal kernel $r$ is doing. A nice byproduct of that would then be that I could simulate from it easily. \n\nFor completeness, some of the `purely algorithmic' solutions I had considered were the following.\n\n * I could try rejection sampling, and in principle this would work, but it wouldn't really give me insight into the nature of the Markov chain.\n * I could try something like the inverse CDF method, but it seems to me that the CDF of $r$ is not particularly nice to work with. As such, I'd have to use e.g. Newton iterations to use this method, and I'd prefer to not have to do this.\n\n\n", "link": "https://math.stackexchange.com/questions/3014357/reversal-of-an-autoregressive-cauchy-markov-chain", "tags": ["probability", "markov-chains"], "votes": 10, "creation_date": "2018-11-26T05:49:00", "comments": ["The general Cauchy distribution $c_w(dx)=\\frac{b dx}{\\pi(b^2+(x-a)^2)}$ where $b>0$ and $w=a+ib$ has Fourier transform $e^{iwt}$ for $t>0.$ Therefore your Markov chain defined by $X_{n+1}=(1-h)X_n+hY_n$ where $Y_n\\sim c_i=\\mu_0$ is such that $X_n\\sim c_{w_n}$ where $w_{n+1}=(1-h)w_n+ih$, hence $w_n=(1-h)^nw_0+ih$. I will try to understand your question in terms of these $w$ and $c_w.$"], "comment_count": 1, "diamond": 0} {"question_id": "349", "site": "math", "category": "Science", "title": "Extracting an (almost) independent large subset from a pairwise independent set of Bernoulli variables", "body": "Let $n>1$, and let $X_1,X_2, \\ldots ,X_n$ be non-constant random variables with values in $\\lbrace 0,1 \\rbrace$. Let us say that a subset of variables $X_{i_1},X_{i_2}, \\ldots,X_{i_d}$ is **complete** if the vector $\\overrightarrow{X}=(X_{i_1},\\ldots,X_{i_d})$ satisfies $P(\\overrightarrow{X}=\\overrightarrow{c})>0$ for any $\\overrightarrow{c}\\in \\lbrace 0,1 \\rbrace^d$. \n\nProve or find a counterexample : if $X_1,X_2, \\ldots ,X_n$ are pairwise independent Bernoulli variables, then we may extract a complete subset of cardinality at least $t+1$, where $t$ is the largest integer satisfying $2^{t} \\leq n$. \n\nThis is true for $n=3$ (and hence also true for $n$ between $3$ and $7$), as is shown in the main answer to that [MathOverflow question](https://mathoverflow.net/questions/66738/is-there-a-good-explanation-for-this-fact-on-pairwise-independent-variables). (That other [MathOverflow question](https://mathoverflow.net/questions/64973/sufficiently-random-sample) is also related, and provides several links)\n\nIf true, this result is sharp, as can be seen by the classical example of taking all arbitrary sums modulo 2 of an initial set of fully independent $t+1$ Bernoulli variables. This produces a set of pairwise independent $2^{t+1}-1$ variables, and where the maximal cardinality of a complete subset is $t+1$.\n\n**Update 10/10/2012** : By induction, it would suffice to show the following : if $X_1, \\ldots ,X_t$ is a fully independent set of $t$ Bernoulli variables and $X$ is another Bernoulli variable, such that the pair $(X_i,X)$ is independent for each $i$, then there are coefficients $\\varepsilon_0,\\varepsilon_1, \\ldots ,\\varepsilon_t$ in $\\lbrace 0,1 \\rbrace$ such that, if we put\n\n$$ H=\\Bigg\\lbrace (x_1,\\ldots,x_t,x) \\in \\lbrace 0,1 \\rbrace ^{t+1} \\Bigg| x=\\varepsilon_0+\\sum_{k=1}^{t}\\varepsilon_kx_k \\ {\\sf mod} \\ 2\\Bigg\\rbrace, \\ \\overrightarrow{X}=(X_{1},\\ldots,X_{t},X) $$ then $P(\\overrightarrow{X}=h)>0$ for any $h\\in H$. \n", "link": "https://math.stackexchange.com/questions/208075/extracting-an-almost-independent-large-subset-from-a-pairwise-independent-set", "tags": ["probability", "combinatorics"], "votes": 10, "creation_date": "2012-10-05T22:04:02", "comments": [], "comment_count": 0, "diamond": 0} {"question_id": "350", "site": "math", "category": "Science", "title": "How to prove this lemma related to Rolle's theorem", "body": "For any function $f$ denote by $Z(f)$ and $Z_o(f)$ the cardinalities of $f^{-1}(0)\\cap[0,1]$ and $f^{-1}(0)\\cap(0,1)$, respectively. Let $H=\\\\{f\\in C^\\infty(\\mathbb{R}): \\text{supp}(f) = [0,1]\\\\}$\n\nFrom [this question](https://math.stackexchange.com/questions/664741) we have \n**Lemma 1** \nLet $q:x\\mapsto(x-r)p(x)$ where $p\\in H$ and $r\\in\\mathbb{R}$. Then $Z(q^{(n)})\\geq Z(p^{(n-1)})+1$ for all $n\\in\\mathbb{N}$.\n\n**Proof** \nNote that $q^{(n)}(x) = n p^{(n-1)}(x) + (x-r)p^{(n)}(x)$. Hence $r$ is a root of $q^{(n)}$ if and only if it is a root of $p^{(n-1)}$. Moreover we have\n\n$$\\underbrace{(x-r)^{n-1}q^{(n)}(x)}_{\\text{LHS}} = n (x-r)^{n-1} p^{(n-1)}(x) + (x-r)^np^{(n)}(x) = \\underbrace{\\frac{d}{dx}(x-r)^n p^{(n-1)}(x)}_{\\text{RHS}}$$\n\nIf $q^{(n)}(r) = 0$ then $Z(q^{(n)}) - 2 = Z_o(\\text{LHS}) = Z_o(\\text{RHS}) \\geq Z(p^{(n-1)}) - 1$. \nIf $q^{(n)}(r) \\neq 0$ then $Z(q^{(n)}) - 1 = Z_o(\\text{LHS}) = Z_o(\\text{RHS}) \\geq Z(p^{(n-1)})\\quad\\quad\\quad\\quad\\text{q.e.d.}$ \n$\\text{ }$\n\nThe following modification \n**Lemma 2** \nLet $q:x\\mapsto(x-r)(x-\\bar{r})p(x)$ where $p\\in H$ and $r\\in\\mathbb{C}\\setminus\\mathbb{R}$. Then $Z(q^{(n)})\\geq Z(p^{(n-2)})+2$ for all $n\\in\\mathbb{N}\\setminus\\\\{1\\\\}$.\n\nI tried to prove the same way by writing\n\n$$\\frac{j(x)k(x)}{(x-r)(x-\\bar{r})}q^{(n)}(x) = \\frac{d}{dx}\\left(j(x)^2 k(x)\\frac{d}{dx}j(x)^{-1}p^{(n-2)}(x)\\right)$$ where $\\frac{d}{dx}$ is differentiation along the real axis and \n$j(x)=c_1(x-r)^{1-n}+c_2(x-\\bar{r})^{1-n}$ \n$k(x) = c_3 ((x-r)(x-\\bar{r}))^n$ \n$c_1,c_2,c_3\\in\\mathbb{C}$\n\nThe real and imaginary part of $j(x)^{-1}$ are proportional, and $j(x)k(x)\\in \\mathbb{R}$ when choosing $c_1 = e^{i d_1}\\\\\\c_2=e^{i d_2}\\\\\\c_3=e^{-i\\cdot (d_1+d_2)/2}\\\\\\d_1,d_2\\in\\mathbb{R}$\n\nbut even then I couldn't get Rolle's theorem to work unless $j(x)$ has no roots on $(0,1)$. I have looked at many functions/different $n$s and even when choosing $r$ such that $j(x)$ has such roots, it seems impossible to find a counterexample to the lemma.\n\nHow to prove the lemma?\n\nOr maybe this simpler version which remains when removing some of the assumptions: \nIf $p$ is smooth with $p^{(n-2)}$ having $m$ roots on $[0,1]$, then how to prove that the $n$'th derivative of $x \\mapsto p(x)(x-r)(x-\\bar{r})$ has at least $m-2$ roots on $(0,1)$.\n", "link": "https://math.stackexchange.com/questions/2548926/how-to-prove-this-lemma-related-to-rolles-theorem", "tags": ["real-analysis", "derivatives", "roots"], "votes": 15, "creation_date": "2017-12-03T06:03:26", "comments": ["@mucciolo The real part can be non-zero", "In the last (simpler) lemma still $r$ purely imaginary?", "@DavidSpeyer I'm interested in how to prove lemma 2 further than for those choices of $r$, $n$ and $p$ that implies that $j$ has no roots on $(0,1)$", "@GerryMyerson Any function whose domain contains $[0,1]$ and whose codomain contains $\\{0\\}$", "I take it \"for any function $f$\" means \"for any real-valued function $f$ of a real variable\"?", "@WillFisher I mean the space of bump functions, but the other conditions implies the $c$ so it is redundant", "What does the subscript $c$ signify in your notation $C^{\\infty}_c$?", "In the linked question, $r$ is assumed not in $(0,1)$. Do you want some analogous hypothesis in Lemma 2?"], "comment_count": 8, "diamond": 0} {"question_id": "351", "site": "math", "category": "Science", "title": "Existence of function satisfying $f(f'(x))=x$ almost everywhere", "body": "**My project is to Study the existence of a continuous function $f : \\mathbb{R} \\rightarrow \\mathbb{R}$ differentiable almost everywhere satisfying $ f\\circ f'(x)=x$ almost everywhere $x \\in \\mathbb{R}$**\n\nI began the study by supposing $f\\in C ^ 1(\\mathbb{R}) $, I have shown that f does not exist. \n\nAfter, I found some difficulties when we assume only f differentiable on $\\mathbb{R}$, I had an answer using Darboux's theorem [Questions about the existence of a function](https://math.stackexchange.com/questions/3312572/questions-about-the-existence-of-a-function?noredirect=1#comment6815760_3312572).\n\nNow, I want to attack the initial problem. Previous arguments do not work!\n\nDo you have any suggestions for me?\n", "link": "https://math.stackexchange.com/questions/3313126/existence-of-function-satisfying-ffx-x-almost-everywhere", "tags": ["real-analysis", "functional-analysis", "derivatives", "functional-equations"], "votes": 13, "creation_date": "2019-08-04T02:40:45", "comments": ["Why this question is different from?$$f'(x)=f^{-1}(x)$$ which is solvable, see video", "@ ibnAbu can explain your approach ?!", "Convert your equation into a differential equation and seek out for a solution", "@Jack D'Aurizio the link math.stackexchange.com/questions/3312572/… does not answer the question you asked me?", "The question is not yet solved in mathoverflow.net/questions/337607/…", "@Jack D'Aurizio in the above link, I explained this case thank you for saving my question.", "@Kavi Rama Murthy Can u explain me why u voted against this question?"], "comment_count": 7, "diamond": 0} {"question_id": "352", "site": "math", "category": "Science", "title": "Is there any example of a real-analytic approach to evaluate a definite integral (with an elementary integrand) whose value involves Lambert W?", "body": "I have never seen a real-analytic approach before to evaluate integrals of the form below $$\\int_a^b\\text{elementary function}(x)\\,dx=\\text{constant involving}\\,W(\\cdot)\\,\\text{in its simplest form}\\tag1.$$\n\nFor instance, on MSE, all use the residue theorem:\n\n * [$\\int_{-\\infty}^{\\infty}{e^x+1\\over (e^x-x+1)^2+\\pi^2}\\mathrm dx=\\int_{-\\infty}^{\\infty}{e^x+1\\over (e^x+x+1)^2+\\pi^2}\\mathrm dx=1$](https://math.stackexchange.com/questions/2113205/int-infty-inftyex1-over-ex-x12-pi2-mathrm-dx-int-infty?noredirect=1&lq=1)\n\n * [Interesting integral related to the Omega Constant/Lambert W Function](https://math.stackexchange.com/questions/45745/interesting-integral-related-to-the-omega-constant-lambert-w-function?noredirect=1&lq=1)\n\n * [Prove that $\\int_0^\\infty \\frac{1+2\\cos x+x\\sin x}{1+2x\\sin x +x^2}dx=\\frac{\\pi}{1+\\Omega}$ where $\\Omega e^\\Omega=1$](https://math.stackexchange.com/questions/2331600/prove-that-int-0-infty-frac12-cos-xx-sin-x12x-sin-x-x2dx-frac-pi?noredirect=1&lq=1)\n\n * [Proof for integral representation of Lambert W function](https://math.stackexchange.com/questions/3347447/proof-for-integral-representation-of-lambert-w-function)\n\n * [Evaluate $\\int_{0}^{\\infty} \\ln(1+\\frac{2\\cos x}{x^2} +\\frac{1}{x^4}) \\, dx$](https://math.stackexchange.com/questions/4433252/evaluate-int-0-infty-ln1-frac2-cos-xx2-frac1x4-dx)\n\n\n\n\nAnd the same applies to some of the wider literature I have come across:\n\n * [Stieltjes, Poisson and other integral representations for functions of Lambert W](https://arxiv.org/pdf/1103.5640.pdf)\n\n * [An Integral Representation of the Lambert $W$ Function](https://arxiv.org/pdf/2012.02480.pdf) **Note:** the proof is real-analytic, but the very first line assumes the validity of an integral identity which was only proven using complex analysis (Hankel contour).\n\n\n\n\nSo, my question is this:\n\nDoes anyone know of a proof of an identity of the form in $(1)$ that involves only real analysis (i.e. does not assume the existence of $\\sqrt{-1}$)?\n", "link": "https://math.stackexchange.com/questions/4501736/is-there-any-example-of-a-real-analytic-approach-to-evaluate-a-definite-integral", "tags": ["real-analysis", "integration", "definite-integrals", "lambert-w", "elementary-functions"], "votes": 12, "creation_date": "2022-07-28T02:40:43", "comments": ["@ТymaGaidash No, the proof must not use the fact that $\\sqrt{-1}$ exists.", "@StevenClark No (I'm looking for an analytical proof rather than numerical). In that example, I'm actually not referring to the end result $\\int_{-\\infty}^{\\infty}{e^x+1\\over (e^x-x+1)^2+\\pi^2}\\,dx=1$ but rather the lemma $\\int_{-\\infty}^{\\infty}\\frac{a^2\\,dx}{(e^x-ax-b)^2+(a\\pi)^2}=\\frac{1}{1+W\\left(\\frac{1}{a}e^{-b/a}\\right)}$ in Jack's answer, which he said was evaluated only through the residue theorem. To qualify as an answer here, an example would be to prove the latter identity using real analysis when, say, $a=2$ and $b=3$, since $W(e^{-3/2}/2)$ doesn't have an obvious simplification.", "Do you consider numerical integration a real-analytic approach? Some of your examples can be evaluated via numerical integration, but some of your examples also simplify which seems inconsistent with your desire to find an example that doesn't simplify. Excluding results that simplify seems rather arbitrary assuming one can show the example is inherently related to the Lambert W function. Trivial cases such as $\\frac{W(10)}{W(10)}$ don't generally meet this criteria, but it seems to me $\\int\\limits_{-\\infty}^{\\infty}{e^x+1\\over (e^x-x+1)^2+\\pi^2}\\mathrm dx=1$ does meet this criteria.", "@StevenClark No, I think as long as your $x$ does not obviously simplify to $ye^y$ for some elementary $y$ then it's accepted (like obvious manipulations of $W((\\sqrt2+1)e^{\\sqrt2+1})$. Since otherwise we'd be running into much deeper holes involving Schanuel's conjecture.", "Does one also have to prove a result involving $W(x)$ has no closed form that does't involve the $W(x)$ function?", "@StevenClark I think there is a bit of misunderstanding -- I'm trying to find a valid integral representation (elementary integrand) for values of a constant involving $W(\\cdot)$ such that it cannot be simplified to a constant that doesn't involve $W$. So $1/(1+W(2))$ would be valid, but constants like $W(-1/e)=-1$ or $\\sin W(3e^3)^2=\\sin9$ would not.", "My example is less trivial than your example, and is $\\int\\limits_{\\infty }^1 e^{1-x}\\,dx$ not a valid integral representation of $W\\left(-\\frac{1}{e}\\right)$?", "@StevenClark No, which is the reason I added \"in its simplest form\" (or \"in most simplified form\" before this edit) so that we don't have trivial cases like $\\int_0^12x\\,dx=1=W(10)/W(10)$.", "Does $\\int\\limits_{\\infty }^1 e^{1-x}\\,dx=W\\left(-\\frac{1}{e}\\right)=-1$ count?"], "comment_count": 9, "diamond": 0} {"question_id": "353", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/2844749", "title": "Minimum area contained between measurable set and translate by $\\lambda$: A strengthening of 2018 USA TSTST #9", "body": "Question\nGiven $\\lambda\\in\\mathbb{R}^+$, what is the smallest possible $c$ for which, given any measurable region $\\mathcal{P}$ in the plane with measure $1$, there always exists a vector $\\mathbf{v}$ with magnitude $\\lambda$ so that the area shared between $\\mathcal{P}$ and its translate by $\\mathbf{v}$ is at most $c$?\n\nBackground \nOn this year's USA TSTST (a test that determines a group of about 30 people to take selection tests for the following year's USA team to the International Math Olympiad), there was an algebra/geometry/combinatorics hybrid problem that I found interesting:\n\nShow that there is an absolute constant $c < 1$ with the following property: whenever $\\mathcal P$ is a polygon with area $1$ in the plane, one can translate it by a distance of $\\frac{1}{100}$ in some direction to obtain a polygon $\\mathcal Q$, for which the intersection of the interiors of $\\mathcal P$ and $\\mathcal Q$ has total area at most $c$.\n\nEssentially every solution can be boiled down to the following:\nStep 1. For a given vector $\\mathbf{v}\\in \\mathbb{R}^2$, define $f(\\mathbf{v})=\\mu\\big(\\mathcal{P}\\cap(\\mathcal{P}+\\mathbf{v})\\big)$, where $\\mu(\\cdot)$ is the area of a region, and $\\mathcal{P}+\\mathbf{v}$ consists of the points in $\\mathcal{P}$ translated by $\\mathbf{v}$.\nStep 2. Prove $f(\\mathbf{u}+\\mathbf{v})\\geq f(\\mathbf{u})+f(\\mathbf{v})-1$ and generalize it to\n$$1-f\\left(\\sum_{i=1}^N \\mathbf{v}_i\\right)\\leq \\sum_{i=1}^N \\left(1-f\\left(\\mathbf{v}_i\\right)\\right).$$\nStep 3. Define, for real $t$,\n$$I_t=\\int_{0\\leq ||\\mathbf{v}||\\leq t} f(\\mathbf{v})\\ d^2\\mathbf{v} = \\int_{0\\leq ||\\mathbf{v}||\\leq t}\\int_{\\mathbf{x}\\in \\mathcal{P}} \\mathbf{1}_{\\mathcal{P}}(\\mathbf{x}+\\mathbf{v})\\ d^2\\mathbf{x}\\ d^2\\mathbf{v}.$$\nWe have\n\\begin{align}\nI_t&=\\int_{\\mathbf{x}\\in \\mathcal{P}} \\int_{0\\leq ||\\mathbf{v}||\\leq t}\\mathbf{1}_{\\mathcal{P}}(\\mathbf{x}+\\mathbf{v})\\ d^2\\mathbf{v}\\ d^2\\mathbf{x}\\\\\n&=\\int_{\\mathbf{x}\\in \\mathcal{P}} \\mu\\left(\\mathcal{P}\\cap\\big\\{\\mathbf{x}+\\mathbf{v}\\big|0\\leq ||\\mathbf{v}||\\leq t\\big\\}\\right)\\ d^2\\mathbf{x}\\\\\n&\\leq\\int_{\\mathbf{x}\\in \\mathcal{P}} 1\\ d^2\\mathbf{x} = 1.\n\\end{align}\nSo, there must exist some $\\mathbf{v}$ with $0\\leq ||\\mathbf{v}||\\leq t$ that satisfies\n$$f(\\mathbf{v})\\leq \\frac{1}{\\pi t^2}.$$\nStep 4. Write this vector as\n$$\\mathbf{v}=\\sum_{i=1}^{\\lceil 100t\\rceil}\\mathbf{u}_i$$\nwhere $||\\mathbf{u}_i||=1/100$.\nStep 5. We now have\n$$1-\\frac{1}{\\pi t^2}\\leq 1-f(\\mathbf{v})\\leq \\sum_{i=1}^N (1-f(\\mathbf{u}_i)),$$\nso for some $i$ we must have\n$$1-f(\\mathbf{u}_i) \\geq \\frac{1}{\\lceil 100t\\rceil}\\left(1-\\frac{1}{\\pi t^2}\\right)$$\n$$f(\\mathbf{u}_i) \\leq 1-\\frac{1}{\\lceil 100t\\rceil}\\left(1-\\frac{1}{\\pi t^2}\\right).$$\nAs long as $t^2>1/\\pi$, this gives us a working value of $c<1$. It is minimized at $t=0.98$, which gives\n$$c\\approx 0.99318.$$\nMore generally, if $\\lambda$ is the length of our translate, this is minimized at the nearest integer multiple of $\\lambda$ to $\\sqrt{3/\\pi}$, which gives \n$$c\\approx 1-2\\lambda\\sqrt{\\frac{\\pi}{27}}.$$\nIn particular, for large enough $\\lambda$, this bound does nothing. \n\nProgress\nI've been able to improve the bound on $c$ given slightly in the following manner:\nAssume, for all $||\\mathbf{u}||=\\lambda$, that $f(\\mathbf{u})>1-\\epsilon$. Then, for all vectors $\\mathbf{v}$ of length $\\leq n\\lambda$, by writing $\\mathbf{v}=\\sum_{i=1}^n \\mathbf{u}_i$ with $||\\mathbf{u}_i||=\\lambda$, we must have\n$$1-f(\\mathbf{v})\\leq n\\epsilon\\implies f(\\mathbf{v})\\geq 1-n\\epsilon.$$\nUsing (almost) our same integral as earlier and setting $t=N\\lambda$, we have\n$$\\int_{\\lambda< ||\\mathbf{v}||\\leq t} f(\\mathbf{v})\\ d^2\\mathbf{v}\\leq 1.$$\n(we cannot start the integral at $0$ as vectors with magnitude $<\\lambda$ cannot be represented as the sum of one vector with magnitude $\\lambda$). However,\n\\begin{align}\n\\int_{\\lambda \\leq ||\\mathbf{v}||\\leq N\\lambda} f(\\mathbf{v})\\ d^2\\mathbf{v}\n&=\n\\sum_{k=2}^N \\int_{(k-1)\\lambda< ||\\mathbf{v}|| \\leq k\\lambda} f(\\mathbf{v})\\ d^2\\mathbf{v}\\\\\n&\\geq \n\\sum_{k=2}^N \\int_{(k-1)\\lambda< ||\\mathbf{v}|| \\leq k\\lambda} 1-k\\epsilon\\ d^2\\mathbf{v}\\\\\n&=\n\\pi\\lambda^2\\sum_{k=2}^N (2k-1)(1-k\\epsilon)\\\\\n&=\n\\pi\\lambda^2\\frac{N-1}{6}\\left(6N+6-\\epsilon\\left(4N^2 + 7N + 6\\right)\\right),\n\\end{align}\nso\n$$\\pi\\lambda^2\\frac{N-1}{6}\\left(6N+6-\\epsilon\\left(4N^2 + 7N + 6\\right)\\right)\\leq 1$$\n$$N+1-\\epsilon\\left(\\frac{4N^2 + 7N + 6}{6}\\right)\\leq \\frac{1}{\\pi(N-1)\\lambda^2}$$\n$$N+1- \\frac{1}{\\pi(N-1)\\lambda^2}\\leq \\epsilon\\left(\\frac{4N^2 + 7N + 6}{6}\\right)$$\n$$\\frac{6\\left(\\pi\\left(N^2-1\\right)\\lambda^2- 1\\right)}{\\left(4N^2 + 7N + 6\\right)\\left(\\pi(N-1)\\lambda^2\\right)}\\leq \\epsilon$$\n$$1-\\frac{6\\left(\\pi\\left(N^2-1\\right)\\lambda^2- 1\\right)}{\\left(4N^2 + 7N + 6\\right)\\left(\\pi(N-1)\\lambda^2\\right)}\\geq 1-\\epsilon.$$\nMinimizing $N$ gives you about $0.9898$ for $\\lambda=1/100$.\nOn the other hand, I've also been trying to find constructions that give a large value of $c$. I haven't come up with any great ones - the best I have is a circle of radius $\\sqrt{\\frac{1}{\\pi}}$ which gives you a $c$ of about 0.9887. For larger $\\lambda$ the best I can think of is something like a \"star\" with large radius and very many thin \"prongs,\" but I haven't calculated the asymptotics on it yet.\nI believe the bound given in step 2 of the solution is sharp iff \n$$(\\mathcal{P}+\\mathbf{u})\\cap(\\mathcal{P}+\\mathbf{v})\\subseteq\\mathcal{P}\\subseteq(\\mathcal{P}+\\mathbf{u})\\cup(\\mathcal{P}+\\mathbf{v}),$$\nbut all attempts I've made to find measurable sets for which this is true or nearly true for small vectors have failed. So I'm stuck. \nAnyone have any ideas, either on sharper upper bounds on $c$ or a sharper $\\mathcal{P}$?\n", "comments": ["Post in mathoverflow", "@TakahiroWaki Can you formalize that?", "Original problem can be solved, the value that half of circumference multiple 1/100 is upperbound. Thinned shape give a small c, maximum case would be about round and small shape, that's circle.", "@mathworker21 You're right; I missed that. I've edited the question accordingly.", "The question at the start is phrased poorly. You should reorder like \"Given lambda, what is the smallest possible c>o so that for any P ....\". The point is that c depends on lambda and not P."], "comment_count": 0, "tags": ["measure-theory", "lebesgue-measure", "geometric-measure-theory"], "creation_date": "2018-07-08T09:20:30", "diamond": 0, "votes": 18} {"question_id": "354", "site": "math", "category": "Science", "title": "For $n\\in\\mathbb{N}^+$, $\\Re(s)>0$, evaluate $\\int_{0}^{\\infty}\\sin\\left(2\\pi ne^{x}\\right)\\left[\\frac{s}{e^{sx}-1}-\\frac{1}{x}\\right]dx$", "body": "Let $n$ be a positive integer, and $s\\in \\mathbb{C}\\;,\\Re(s)>0$. I want to compute the integral : $$\\int_{0}^{\\infty}\\sin\\left(2\\pi ne^{x}\\right)\\left[\\frac{s}{e^{sx}-1}-\\frac{1}{x}\\right]dx$$\n\nI tried using the integral presentation found [here](http://functions.wolfram.com/ElementaryFunctions/Sin/07/02/) :\n\n$$\\sin(2\\pi ne^{x})=\\frac{\\sqrt{\\pi}}{2\\pi i }\\int_{\\gamma-i\\infty}^{\\gamma+i\\infty}\\frac{\\Gamma(z)}{\\Gamma\\left(\\frac{3}{2}-z \\right )}\\left(\\pi n \\right )^{-2z+1}e^{-(2z-1)x}dz\\;\\;\\;\\;\\;0<\\gamma<1$$ in conjunction with : $$r+\\log(r)+\\psi\\left(\\frac{1}{r}\\right)=-\\int_{0}^{\\infty}e^{-y}\\left(\\frac{r}{e^{ry}-1}-\\frac{1}{y}\\right)dy$$ Where $\\psi\\left(\\cdot\\right)$ is the digamma function. But that made the problem even more difficult. Any help is highly appreciated.\n\n**EDIT** :\n\nUsing the Fourier reciprocity : $$\\frac{s}{e^{sx}-1}-\\frac{1}{x}=2\\int_{0}^{\\infty}\\left(\\frac{1}{e^{2\\pi \\theta/s}-1}-\\frac{s}{2\\pi \\theta} \\right )\\sin(x\\theta)d\\theta$$ Our integral reads : $$2\\int_{0}^{\\infty} g(\\theta)\\left(\\frac{1}{e^{2\\pi \\theta/s}-1}-\\frac{s}{2\\pi \\theta} \\right )d\\theta$$ Where : $$g(\\theta)=\\int_{0}^{\\infty}\\sin(2\\pi ne^{x})\\sin(x\\theta)dx$$ $$=\\frac{\\sqrt{\\pi}}{2\\pi i }\\int_{\\gamma-i\\infty}^{\\gamma+i\\infty}\\frac{\\Gamma(z)}{\\Gamma\\left(\\frac{3}{2}-z \\right )}\\left(\\pi n \\right )^{-2z+1}\\frac{\\theta}{\\theta^{2}+(2z-1)^{2}}dz$$\n", "link": "https://math.stackexchange.com/questions/3551521/for-n-in-mathbbn-res0-evaluate-int-0-infty-sin-left2-pi-ne", "tags": ["real-analysis", "complex-analysis", "definite-integrals", "residue-calculus"], "votes": 11, "creation_date": "2020-02-18T09:35:42", "comments": ["Using change of variable $y=e^x$ plus noting that $\\left[\\frac{s}{e^{s x}-1}-\\frac{1}{x}\\right]=\\left[\\frac{s \\coth \\left(\\frac{s x}{2}\\right)}{2}-\\frac{1}{x}-\\frac{s}{2}\\right]$ I think your integral can be written as $$I=\\int_1^{\\infty } \\frac{\\sin (2 \\pi n y)}{y} \\left(\\frac{s \\coth \\left(\\frac{1}{2} s \\log (y)\\right)}{2 }-\\frac{1}{ \\log (y)}-\\frac{s}{2 }\\right) \\, dy$$ The integral of the $-\\frac{s}{2}$ term involving just the sine integral is the largest or principal term. When integrated the term is $\\left[ -\\frac{1}{2} s \\left(\\frac{\\pi }{2}-\\text{Si}(2 n \\pi )\\right) \\right]$.", "I am interested in an exact from", "Are you interested in any sort of asymptotics of $s$? Or do you need an exact closed form?", "Yes your right Sorry", "Are you sure you evaluated the integral with $\\sin(2\\pi n e^{x})$ not $\\sin(2\\pi nx)$ ?", "A quick and dirty conjectural result found using Mathematica and tested for a few values of $n$ and real $s$ is $$I=\\frac{1}{2} \\pi \\left(\\coth \\left(\\frac{2 \\pi ^2 n}{s}\\right)-1\\right)-\\frac{s}{4 \\pi n}$$", "Did you consider numeric approximation (for fixed values of $s,n$) to get an idea? If you evaluate the integral from $0$ to some $n$ this integral should be well approximated even for relatively small $n$ because for large $x$ the sin term will just average out to zero.", "It is esoteric indeed. This integral came up in the study of a certain number-theoretic function. The function between the parentheses is self reciprocal wrt the Fourier sine transform, and can be used to prove the transformation formula of the Dedekind eta function, when the sine factor is $\\sin(2\\pi n x)$. The problem above is a little bit different, but concerns a closely related function.", "What context did this come up in? This seems like a very esoteric problem.", "$\\int_0^\\infty\\sin\\left(2\\pi ne^{x}\\right)\\left[\\frac{s}{e^{sx}-1}-\\frac{1}{x}\\right]dx$ is the incomplete version of $\\int_{-\\infty}^\\infty\\sin\\left(2\\pi ne^{x}\\right)\\left[\\frac{s}{e^{sx}-1}-\\frac{1}{x}\\right]dx$ which has way more chances to be evaluated, why do you need the former", "You meant $\\int_{-\\infty}^\\infty\\sin\\left(2\\pi ne^{x}\\right)\\left[\\frac{s}{e^{sx}-1}-\\frac{1}{x}\\right]dx$"], "comment_count": 11, "diamond": 0} {"question_id": "355", "site": "math", "category": "Science", "title": "The sum of eigenvalues of integral operator $S(f)(x)=\\int_{\\mathcal{X}} k(x,y)f(y)d\\mu(y)$ is given by $\\int_{\\mathcal{X}} k(x,x) d\\mu(x)$?", "body": "**Setup:** Let $(\\mathcal{X},d_{\\mathcal{X}})$ and $(\\mathcal{Y},d_{\\mathcal{Y}})$ be two separable metric spaces. Let $M^1(\\mathcal{X})$ be the space of Borel probability measures on $\\mathcal{X}$ with finite first moment, i.e. a Borel probability measure $\\mu$ on $\\mathcal{X}$ is in $M^1(\\mathcal{X})$ if $\\int d_{\\mathcal{X}}(x,o) d\\mu(x)<\\infty$ for any $o\\in\\mathcal{X}$. The space $M^1(\\mathcal{Y})$ is defined in similar fashion.\n\nFix $\\mu\\in M^1(\\mathcal{X})$ and $\\nu\\in M^1(\\mathcal{Y})$ and define $$ d_\\mu(x_1,x_2)= d_{\\mathcal{X}}(x_1,x_2) -\\int d_{\\mathcal{X}}(x_1,x)\\, d\\mu(x) - \\int d_{\\mathcal{X}}(x_2,x)\\, d\\mu(x) + \\int d_{\\mathcal{X}}(x,x')\\, d\\mu^2(x,x'), $$ and a similar definition of $d_\\nu:\\mathcal{Y}\\times \\mathcal{Y}\\to\\mathbb{R}$.\n\nNow let $S:L^2(\\mathcal{X}\\times \\mathcal{Y},\\mathcal{B}(\\mathcal{X})\\otimes \\mathcal{B}(\\mathcal{Y}),\\mu\\times \\nu) \\to L^2(\\mathcal{X}\\times \\mathcal{Y},\\mathcal{B}(\\mathcal{X})\\otimes \\mathcal{B}(\\mathcal{Y}),\\mu\\times \\nu),$ be a Hilbert-Schmidt operator given by $$ S(f)(x,y) = \\int d_\\mu(x,x')d_\\nu(y,y') f(x',y') d\\mu\\times \\nu(x',y'). $$ and let $\\\\{\\lambda_i\\\\}_{i\\geq 1}$ denote the non-zero eigenvalues of $S$ repeated according to multiplicity.\n\n**Question:** How do I prove the following identity: $$\\sum_{i=1}^\\infty\\lambda_i=\\int d_\\mu(x,x)d_\\nu(y,y) \\, d\\mu\\times \\nu(x,y).$$ I tried but failed to show that $S$ is of trace class, since they under certain conditions (which I also can't verify in this setup) satisfy that $$ Trace(S)=\\int d_\\mu(x,x)d_\\nu(y,y) \\, d\\mu\\times \\nu(x,y), $$ which yields the result since $Trace(S)=\\sum_{i=1}^\\infty\\lambda_i$ (if it were of trace class).\n\n_The identity $\\sum_{i=1}^\\infty\\lambda_i=\\int d_\\mu(x,x)d_\\nu(y,y) \\, d\\mu\\times \\nu(x,y)$ is stated in [Distance covariance in metric spaces by Russell Lyons (2013) Theorem 2.7](https://projecteuclid.org/euclid.aop/1378991840), without proof, so my approach with traces is only an idea._\n\n_If another way of proving the identity appears or if there is a counterexample, that would more than satisfy my needs. In the case of a counterexample i would very much appreciate stronger initial conditions rendering the identity true._\n\nPlease bear in mind that i am a novice in the theory of operators and trace class operators (only went down this road to explain the equality above), so references would be much appriciated. \n\n**Update:** A counterexample to the operator being of trace-class is presented by Russell Lyons in the errata to the mentioned paper. Furthermore a proof that the formula holds and that the operator is of traceclass, whenever the marginal spaces posses additional nice proporties (isometric embeddability into hilbert spaces), is also presented in this errata.\n", "link": "https://math.stackexchange.com/questions/2035989/the-sum-of-eigenvalues-of-integral-operator-sfx-int-mathcalx-kx-yf", "tags": ["functional-analysis", "statistics", "operator-theory", "spectral-theory", "trace"], "votes": 9, "creation_date": "2016-11-29T08:56:58", "comments": ["@Renart I think there are more subtle problems. For example $y\\mapsto K(x,y) = \\sum_{n}e_n(y) \\int e_n(z) K(x,z)dz,$ only in $L^2(\\mathbb{R})$ for each fixed $x$, which by definition means that for any $x$ $\\|K(x,\\cdot)- \\sum_{n=1}^k e_n(\\cdot) \\int e_n(z) K(x,z)dz\\|_2\\to_k 0.$ But in order to say that $\\int \\sum_{n=1}^\\infty e_n(x)\\int K(x,y)e_n(y)dy dx= \\int K(x,x) dx$ we need stronger convergence. Anyways, all this assumes that the operator is of trace class, which i haven't even verified.", "Well i don't have tried it myself but... From what i read the answer of jonhatan doesn't really use the fact that it's on $\\mathbf R$. The only thing that need justification is the permutation of a sum and an integral at the very end. Fubini works pretty independently of the space you're concidering.", "@Renart Yes that is indeed one of the questions that i have read, suggesting the trace approach. Though this is a integral kernel over $\\mathbb{R}$, and i'm quite sure that problems arise when integrating over arbitrary metric spaces.", "math.stackexchange.com/questions/185587/…", "If not possible in the above setup, can the equality be shown for when $\\mathcal{X}$ and $\\mathcal{Y}$ are separable Hilbert spaces?"], "comment_count": 5, "diamond": 0} {"question_id": "356", "site": "math", "category": "Science", "title": "Spectacular failure of Lebesgue differentiation for rectangles", "body": "Let $\\mathcal{R}$ be the set of rectangles in the plane and, given $f \\in L^1$ let $$ f^*(x) = \\sup_{x \\in R \\in \\mathcal{R}} \\frac{1}{ \\lvert R \\rvert} \\int_R \\lvert \\, f \\,\\rvert $$ as defined in [this question](https://math.stackexchange.com/questions/641399/maximal-functions-where-weak-type-inequality-fails). You can show that the weak-type inequality fails for this operator, that is, there is no constant $A$ such that $$ m(\\\\{ f^* > \\alpha \\\\}) < \\frac{A}{\\alpha} \\lVert \\, f \\, \\rVert_1 $$ for each integrable $f$. From Stein and Shakarchi's book on real analysis I'd like to show the following, much stronger claim: there is an integrable $f$ such that\n\n$$ \\limsup_{\\text{diam}(R) \\to 0} \\frac{1}{|\\,R\\,|} \\int_R \\lvert \\, f \\, \\rvert = \\infty \\text{ a.e.} $$ where diam, of course, is the diameter. According to the book this \"should\" follow from the failure of the weak type inequality, but I don't really know how to do it.\n\nI was thinking something like the following, but it failed. It suffices to find and $f$ such that the desired conclusion holds only on a set of positive measure, for we can translate around and get the desired effect. So I wanted to show that if the $\\limsup$ expression is finite everywhere, the weak-type inequality holds, for that specific function. That would get me what I want. However I got nowhere with finding a function for which the weak-type inequality fails, sadly.\n\nEdit: Whoops, this is actually very hard. In its most general form it relies on a result found on page 441 of Stein's book Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals. Hopefully this specific case can be had more easily, though?\n", "link": "https://math.stackexchange.com/questions/1854652/spectacular-failure-of-lebesgue-differentiation-for-rectangles", "tags": ["real-analysis", "functional-analysis", "lebesgue-integral"], "votes": 11, "creation_date": "2016-07-09T20:30:12", "comments": ["Thanks for referring to Stein's proof. I did struggle with (15), until I realized that $|E_k|$ is bounded by the size of $B$ so $||g'_k|| \\rightarrow 0$ as $k\\rightarrow\\infty$.", "Have you studied Stein's proof? It is quite straightforward and most of the work is in making formal your intuition that you can find a contradiction on a compact set and translate it around.", "Banach-Tarski 2..."], "comment_count": 3, "diamond": 0} {"question_id": "357", "site": "math", "category": "Science", "title": "Is there an open subset $A$ of $[0,1]^2$ with measure $>\\frac{1}{100}$ that satisfies this property?", "body": "Can we find for any given $\\varepsilon>0$ an open subset $A\\subseteq[0,1]^2$ with measure $>\\frac{1}{100}$ such that, for any smooth curve $\\gamma:[0,1]\\to\\mathbb{R}^2$ of length $1$, the set $\\gamma+A=\\\\{\\gamma(t)+a;t\\in[0,1],a\\in A\\\\}$ does not contain any balls of radius $\\varepsilon$?\n\nI wouldn't mind changing the $\\frac{1}{100}$ for any other positive constant. Also, I ask about smooth curves but it may make more sense to consider in general $1$-Lipschitz functions $\\gamma:[0,1]\\to[0,1]^2$.\n\nFor context, a positive answer to this question could be useful for [this other question](https://mathoverflow.net/questions/420897/what-is-the-smallest-size-of-a-shape-in-which-all-fixed-n-polyominos-can-fit/421175#421175). But the question is also interesting in itself, of course.\n", "link": "https://math.stackexchange.com/questions/4445539/is-there-an-open-subset-a-of-0-12-with-measure-frac1100-that-sati", "tags": ["measure-theory", "multivariable-calculus", "geometric-measure-theory"], "votes": 16, "creation_date": "2022-05-07T17:12:26", "comments": ["Now posted to MO, mathoverflow.net/questions/422493/…", "Yes, I was asking having in mind the negative answer (baseless belief as well!)", "Oh and if you mean that if the curves $\\gamma$ have length $10$ the answer is negative then yes, I am interested (for some reason I think that the answer of these problems is going to be positive but it's a baseless belief)", "The case of $\\gamma$ having length $10$ would imply the question as above, because any smooth curve of length $1$ is a subset of a smooth curve of length $10$. But any ideas of how to solve problems similar to the one above are welcome. For example I would also be interested in the case of the curves $\\gamma$ having length $\\leq\\frac{1}{2}$", "Would you be interested in the case when $\\gamma$ has length, say, 10?", "Btw, I am not sure why the \"multivariable calculus\" tag is more appropiate than the \"recreational mathematics\" one on this question. This is not part of any mathematics curriculum that I know of", "Thanks, that's what I missed: the restriction on length.", "@mihaild I don't understand very well, your curve $\\gamma$ is $[0,1]\\to\\mathbb{R}$, not $[0,1]\\to\\mathbb{R}^2$, and I don't think it has length $1$", "If $A$ is open, it contains an open ball of radius $x$, now if we take $\\gamma(t) = \\sin\\frac{20 \\pi t}{x}$, $\\gamma +A$ fills unit square (as moving even just this ball along $\\gamma$ fills it). I guess something is wrong with this reasoning, but I can't find what."], "comment_count": 9, "diamond": 0} {"question_id": "358", "site": "math", "category": "Science", "title": "Closed form expression for $\\sqrt{4+\\sqrt[3]{4+\\sqrt[4]{4+\\dots}}}$", "body": "Since $e=\\sum_{n=0}^\\infty\\frac{1}{n!}=1+1+\\frac12(1+\\frac13(1+\\frac14(1+\\dots)))$, we have\n\n$$4^{e-2}=\\sqrt{4\\cdot\\sqrt[3]{4\\cdot\\sqrt[4]{4\\cdots}}}$$\n\nIs there however a nice way to express the radical in the title too?\n", "link": "https://math.stackexchange.com/questions/662563/closed-form-expression-for-sqrt4-sqrt34-sqrt44-dots", "tags": ["closed-form", "radicals", "nested-radicals"], "votes": 16, "creation_date": "2014-02-03T13:49:36", "comments": ["math.stackexchange.com/questions/875203/… math.stackexchange.com/questions/1782032/… math.stackexchange.com/questions/1843974/…", "It might be interesting to you that by taking $4$ out of the first root, we obtain another form of this radical: $$2\\sqrt{1+\\sqrt[3]{\\frac{1}{2^4}+\\sqrt[4]{\\frac{1}{2^{22}}+\\dots\\sqrt[k]{\\frac{1}{2^{k!-2}}+\\dots}}}} \\approx 2.401615526...$$ You just have to carefully take $1/4$ under each root, moving further and further", "I have to agree, you probably won't find an answer to this. Noting that the power of the radicals changes (or if the values inside change) makes me want to say there is no closed form.", "@BarryCipra Well that's unfortunate... these things aren't perfect. On the other hand, a constant like $4^e$ seems pretty rare, I don't think I've ever gotten anything quite like it when evaluating an integral, say. But it does seem weird that it wouldn't be detected.", "@BrunoJoyal, I agree it's a good idea, but with one caveat: I ran $4^{e-2}\\approx2.70675377667$ through an ISC, just as a \"sanity check,\" and came up empty.", "I would try to evaluate it to high precision and run the resulting approximation through a reverse symbolic calculator to see if anything comes up."], "comment_count": 6, "diamond": 0} {"question_id": "359", "site": "math", "category": "Science", "title": "Evaluating a Polynomic-Trigonometric-Hyperbolic Integral", "body": "Within [this AoPS thread](https://artofproblemsolving.com/community/c7h1746945p11375644) it is asked to evaluate the following integral\n\n> $$\\mathfrak I~=~\\int_0^\\infty \\frac{x\\sin x}{\\cos x+\\cosh^2 x}\\mathrm dx\\tag1$$\n\nIn order to be precise there is also a possible closed-form conjectured which is given by\n\n> $$\\mathfrak I~=~G-\\frac12\\tag2$$\n\nBut as it is pointed out within the linked thread this seems to be only a reasonable approximation off after the $5$th decimal digit.\n\nI have to admit that it is highly improbable that there exists a nice looking closed-form for $(1)$ since the integrand involves polynomials, trigonometric aswell as hyperbolic functions. I am not even sure how to get started, i.e. which substitution to choose or which technique at all to start with.\n\nA related, but perhaps more handable integral, would be the following\n\n> $$\\mathfrak J~=~\\int_0^\\infty \\frac{\\sin x}{\\cos x+\\cosh^2 x}\\mathrm dx\\tag{1$'$}$$\n\nOut of experience I could imagine that $(1')$ may has a closed-form in terms of known constants $($or series$)$ since it only contains the two closely connected trigonometric and hyperbolic functions.\n\n> Is it in fact possible to deduce a closed-form for $(1)$ and $(1')$? For myself I cannot offer an approach since everything I tried was not helpful at all hence I was not even able to perform one or two steps in order to simplify the given integrals. I would be glad to see a full solution or even attempts in evaluating $(1)$ and $(1')$ since I have no idea how to deal with such integrands.\n\nThanks in advance!\n\n**EDIT**\n\nOut of pure chance I just stumbled upon [a related MSE question](https://math.stackexchange.com/questions/1798467/why-does-the-hard-looking-integral-int-0-infty-fracx-sin2x-coshx/1798559#1798559) dealing with the integral\n\n$$\\int_0^\\infty\\frac{x\\sin^2x}{\\cosh x+\\cos x}\\mathrm dx=1$$\n\nWhich on the other hand motivates me to believe that there may be a closed-form for $(1)$.\n", "link": "https://math.stackexchange.com/questions/3154378/evaluating-a-polynomic-trigonometric-hyperbolic-integral", "tags": ["integration", "definite-integrals", "closed-form"], "votes": 14, "creation_date": "2019-03-19T10:33:16", "comments": ["@GEdgar I have problems with applying the residue theorem for complicated integrals correctly. Are you here able to do so?", "@Zacky I am aware of this list but has not thought about searching within for the given integral. As you already mentioned this list is unreliable but cotains some \"good\" approximations but way to many completely wrong values.", "The integral also appears within this list: en.wikiversity.org/wiki/User:Integrals123 (I typed in ctrl + F cosh and scrolled down till the $54$-th integral), which unfortunately has many wrong answers since it was done with Inverse Symbolic calculator and that result kinda played us. But yes a closed form would be interesting even though it's not $G-\\frac12$.", "Note. Since the integrand is even, we ave $$2 \\mathfrak I~=~\\int_{-\\infty}^\\infty \\frac{x\\sin x}{\\cos x+\\cosh^2 x}\\mathrm dx$$ and there is a chance this could be done using a contour in the complex plane."], "comment_count": 4, "diamond": 0} {"question_id": "360", "site": "math", "category": "Science", "title": "How to generalize Reshetnikov's $\\arg\\,B\\left(\\frac{-5+8\\,\\sqrt{-11}}{27};\\,\\frac12,\\frac13\\right)=\\frac\\pi3$?", "body": "Given the [argument](https://en.wikipedia.org/wiki/Argument_\\(complex_analysis\\)) $\\arg(z)$, we can observe that for $k=1,2,3$, $$\\arg z_1 = \\frac{k\\,\\pi}3, \\quad z_1 = \\left(\\tfrac{1+\\sqrt{-3}}{2}\\right)^k\\qquad\\tag1$$ $$\\qquad \\arg z_2=\\frac{k\\,\\pi}3, \\quad z_2 = \\left( B\\Big(\\color{blue}{\\tfrac{-5+8\\,\\sqrt{-11}}{27}};\\,\\tfrac12,\\tfrac13\\Big)\\right)^k \\tag2$$ and _incomplete beta function_ $B(z;a,b)$. [The second](https://math.stackexchange.com/questions/687567/closed-form-for-2f-1-left-frac12-frac23-frac32-frac8-sqrt11-i-5) with $k=1$ is by V. Reshetnikov.\n\nBut that cannot be an isolated result. Note that $z_1^3=-1$ while the complex number $z_2$ has a _real_ cube $z_2^3 \\approx 6.1319080509$ with no known closed-form.\n\nReshetnikov states (without giving details) that $(2)$ is equivalent to, $$B\\left(\\frac19;\\,\\frac16,\\frac13\\right) = \\frac12\\,B\\left(\\frac16,\\frac13\\right) =\\frac{1}{2\\sqrt{\\pi}}\\,\\Gamma\\left(\\frac16\\right)\\Gamma\\left(\\frac13\\right)\\tag3$$ Thus, $(3)$ is related to, $$I\\left(\\color{blue}{\\frac19};\\,\\frac16,\\frac13\\right)=\\frac12 \\tag4$$ with _regularized beta function_ $I(z;a,b)$. The equation $$I\\left(z;\\,\\frac16,\\frac13\\right)=\\frac1n \\tag5$$ to find some algebraic $z$ for a given integer $n$ is quite easy to solve. For example, the smallest real root of, $$-1 + 99 z - 243 z^2 + 81 z^3= 0\\tag6$$ will yield $n=3$. More specifically, $$I\\left(\\frac19\\Big(1-4\\sin\\big(\\tfrac{\\pi}{18}\\big)\\Big)^2;\\,\\frac16,\\frac13 \\right)=\\frac13\\quad\\quad\\tag7$$ $$\\quad I\\left(\\frac13\\Big(1-\\frac{\\sqrt2}{\\sqrt[4]3}\\Big)^2;\\,\\frac16,\\frac13\\right) =\\frac14\\quad\\tag8$$ and so on.\n\n> **Q:** But how do we get from $\\color{blue}{\\frac{-5+8\\,\\sqrt{-11}}{27}}$ of $(2)$ to $\\color{blue}{\\frac19}$ of $(4)$? (I assume a hypergeometric transformation is involved?)\n\nIf the answer can be found, then perhaps we can use an expression $P(z)$ of a root $z$ of $(6)$ to find another, $$\\arg z_3=\\frac{\\beta\\,\\pi}3,\\quad z_3=B\\Big( P(z);\\,\\tfrac12,\\tfrac13\\Big)\\quad\\tag7$$ where $\\beta$ is a rational/algebraic number?\n", "link": "https://math.stackexchange.com/questions/2053470/how-to-generalize-reshetnikovs-arg-b-left-frac-58-sqrt-1127-fra", "tags": ["complex-analysis", "special-functions", "closed-form", "pi", "beta-function"], "votes": 13, "creation_date": "2016-12-10T21:52:11", "comments": ["@J.M.: I rephrased it for clarity. :)", "The exponent in $(2)$ should be inside $\\arg$, no?"], "comment_count": 2, "diamond": 0} {"question_id": "361", "site": "math", "category": "Science", "title": "Covering number/Metric Entropy of the unit ball with respect to Mahalanobis distance", "body": "Let $B$ denote the unit ball on $\\mathbb{R}^d$ and $N(\\epsilon, B, d)$ be the cardinality of the smallest $\\epsilon$-cover of $B$. An epsilon cover is a set $T \\subset B$ such that for any $x \\in B$, there is a $t \\in T$ with $d(t,x) \\le \\epsilon$. See for example [here](https://www.stat.berkeley.edu/%7Ebartlett/courses/2013spring-stat210b/notes/12notes.pdf). $N$ is referred to as the covering number, and $\\log N$ is the Metric entropy.\n\nConsider the following result: let $\\|\\cdot\\|$ be a norm on $\\mathbb{R}^d$ then $$ \\frac{1}{\\epsilon^d} \\le N(\\epsilon, B, \\|\\cdot\\|) \\le \\left (1+\\frac{2}{\\epsilon} \\right)^d. $$\n\nI would like to know if there are bounds on the covering numbers that are dimension free when we choose the metric to be the Mahalanobis distance $d_S(x,y) = \\|S^{-1/2}(x-y)\\|_2$ for some positive definite covariance matrix $S$. Are there results along the lines of:\n\n$$ \\frac{1}{\\epsilon^{f(S)}} \\le N(\\epsilon, B, d_S) \\le \\left (1+\\frac{2}{\\epsilon} \\right)^{f(S)}. $$ where $f(S)$ is some quantity depending on $S$? An example I have in mind is when $S$ is diagonal with quickly decaying diagonal elements, e.g. $S_{ii} = i^{-2}$.\n", "link": "https://math.stackexchange.com/questions/4707017/covering-number-metric-entropy-of-the-unit-ball-with-respect-to-mahalanobis-dist", "tags": ["real-analysis", "combinatorics", "geometry", "analysis", "statistics"], "votes": 10, "creation_date": "2023-05-26T07:18:57", "comments": [], "comment_count": 0, "diamond": 0} {"question_id": "362", "site": "math", "category": "Science", "title": "Expectation of maximum of minimums of permutations", "body": "Assume $n$ random permutations $\\pi_1,\\pi_2,\\ldots,\\pi_n: \\lbrace 1,2,\\ldots,m \\rbrace \\rightarrow \\lbrace 1,2,\\ldots,m \\rbrace$. Let $X_i = \\min(\\pi_1(i),\\pi_2(i),\\ldots,\\pi_n(i))$ and $Y = \\max(X_1, X_2, \\ldots, X_m)$. What is the expectation of $Y$, $E(Y)$, as function of $n$? An upper bound approximation for $E(Y)$ would also be very helpful. \n\nObviously, we have $E(Y)=m$ for $n=1$ and $E(Y)=1$ for $n\\rightarrow\\infty$.\n\nI know that the distribution of $X_i$ is given by $P(X_i\\leq k) = 1 - \\left(1-\\frac{k}{m}\\right)^n$. However, since $X_1,X_2,\\ldots,X_m$ are not independent, it is not possible to get the distribution of $Y$ by $P(Y \\leq k) = P(X_1 \\leq k \\wedge X_2 \\leq k \\wedge \\ldots \\wedge X_m\\leq k) \\neq P(X_1 \\leq k) \\cdot P(X_2 \\leq k) \\cdot\\ldots \\cdot P(X_m \\leq k)$.\n", "link": "https://math.stackexchange.com/questions/2257782/expectation-of-maximum-of-minimums-of-permutations", "tags": ["probability", "statistics", "permutations"], "votes": 7, "creation_date": "2017-04-29T08:56:32", "comments": [], "comment_count": 0, "diamond": 0} {"question_id": "363", "site": "math", "category": "Science", "link": "https://math.stackexchange.com/questions/1896209", "title": "Conditional expectation continuous in the conditioning argument?", "body": "Let $X$ and $Y$ be random vectors defined on a common probability space. $X$ takes values in a finite-dimensional space $\\mathcal{X} \\subset \\mathbb{R}^p$, while $Y$ takes values in $\\mathbb{R}$. The conditional expectation $E(Y \\mid X)$ is then a random variable that is uniquely defined up to null sets.\nI am seeking a set of sufficient conditions on the joint distribution of $(X,Y)$ for the following statement to be true:\n\nGiven any point $x_0 \\in \\mathcal{X}$, there exists a function $f \\colon \\mathcal{X} \\to \\mathbb{R}$ such that (i) $f(X)$ is a version of $E(Y \\mid X)$, and (ii) $f(\\cdot)$ is continuous at $x_0$.\n\nObviously, the continuity part is the non-trivial one. By Lusin's theorem, any measurable function (such as any version of the conditional expectation function) is \"nearly continuous\", but this is not quite enough for me.\nIdeally the sufficient conditions for the above statement would not involve restrictions on the densities or conditional densities of $X$ and $Y$. The problem that motivates this question has a complicated geometry, so it is difficult to characterize densities with respect to fixed dominating measures.\nIf you require more structure to the problem (but ideally the question would be answered in more generality), you may assume: $Y = g(A)$ for a continuous function $g(\\cdot)$, $X = A + B$, and the random vectors $A$ and $B$ are independent. However, $A$ and $B$ may concentrate on different subspaces of $\\mathbb{R}^p$, each of which is a complicated manifold.\nThank you in advance for your time!\n", "comments": [], "comment_count": 0, "tags": ["probability-theory", "measure-theory", "continuity", "conditional-expectation"], "creation_date": "2016-08-18T07:47:08", "diamond": 0, "votes": 11} {"question_id": "364", "site": "mathematica", "title": "Changing FrontEnd automatic scrolling in version 8", "body": "In _Mathematica_ versions < 8, the FrontEnd has a very intelligent behavior: \n\n * On evaluation, it by default automatically scrolls down the Notebook window to the last printed Output cell but also allows a user to scroll the Notebook window up by hand and then does not scroll down the window again automatically.\n\n * If the user wishes to restore automatic scrolling it is sufficient to scroll down the window to the last currently printed Output cell and automatic scrolling will be restored.\n\n\n\n\nI feel such behavior very comfortable. But in _Mathematica_ 8 we have no such behavior by default. I found that it may be partially restored by setting\n \n \n SetOptions[$FrontEnd, \n EvaluationCompletionAction -> {\"ScrollToOutput\"}]\n \n\nBut then it is not possible to stop automatic scrolling by scrolling the window by hands. It is possible to restore the old scrolling behavior?\n\n* * *\n\nThrough _Mathematica_ 10.4 the old scrolling behavior has not been restored. \n\n * Have any new options come on line to control this?\n\n * Is there a hook to determine scroll position that might be used for a workaround?\n\n * Could [`PrintTemporary`](http://reference.wolfram.com/language/ref/PrintTemporary.html) and/or [`Dynamic`](http://reference.wolfram.com/language/ref/Dynamic.html) (which is active only when visible) be used to simulate the old behavior?\n\n * If the old behavior is simply not achievable what is the best alternative for a similar workflow?\n\n\n\n", "link": "https://mathematica.stackexchange.com/questions/1948/changing-frontend-automatic-scrolling-in-version-8", "tags": ["front-end", "customization", "cells", "printing"], "votes": 37, "creation_date": "2012-02-17T20:42:53", "comments": ["@Wizzerad In Mathematica version 12.0 your problem seems to be resolved.", "@Wizzerad I suggest you to ask the tech support about it because (as you see) no one here seems to know how to get tiny control on the FrontEnd behavior while evaluating. If you get a solution, please post it here - it will help others.", "In Mathematica 10, it does not seem possible to peacefully write some code while it is evaluating. It keeps jumping back and forth between the code line I am writing and the end of the input part of the section that is being evaluated. Is there a way to normally work on the code while Mathematica 10 is evaluating part of the notebook?", "In version 10.0 I have that, by default, the front end always auto-scrolls to view new output, regardless of previous hand scrolls by the user. I take it from this question that I may disable all auto-scrolling by setting EvaluationCompletionAction to something else. Is that correct? Are there any other options as of version 10?", "In March 2015 I send a suggestion to WRI tech support (CASE:2679254) with a link to this thread. They replied that it was forwarded to \"the appropriate people in our development group\", so I think we at least have a hope. I have nothing against the new section in my question.", "Alexey, I added a new section to this question probing for alternatives and workarounds, and noting that the behavior has not been restored through 10.1 and is presumably not going to be. I hope you approve.", "@AlexeyPopkov Ok :) good luck.", "@shrx I mean at least Mathematica versions 5, 6 and 7 (I worked with).", "@Alexey Popkov It is unclear, do you mean: \"In Mathematica versions < 8...\"?", "@Sjoerd I agree that editing during evaluation was not always stable (of course, it does not mean that this is bad idea itself). The main advantage of the old behavior is the ability to switch between inspecting of already printed outputs and automatic scrolling.", "Well, it happened to me quite a lot that I got drawn to the bottom of the print area at times I didn't want to be there and when I succeeded staying above the scroll area it wasn't very stable either.", "@Sjoerd With the old behavior we could not only edit the input during evaluation (as we can by default now) but also inspect already printed outputs from the running evaluation (by scrolling the page). And all of this without loosing automatic scrolling when we wish! I see no advantages in loosing this.", "Actually, I found the old behaviour rather annoying. Quite distracting when you were editing and running evaluations at the same time.", "Welcome, Alexey!"], "comment_count": 14, "category": "Technology", "diamond": 0} {"question_id": "365", "site": "mathematica", "title": "Is it possible to regain Mathematica 5.2's palette input focus behaviour with version 8.0?", "body": "Between Mathematica 5.2 and later versions, there has been a change in determining which notebook gets the palette input focus, which leads to quite unfortunate behaviour if you use focus-follows-mouse as I do.\n\nFor those who don't know focus-follows-mouse: This mode gives the keyboard input focus to the window which currently contains the mouse pointer. Now consider the following situations: There are two notebooks and a palette open (one of the notebooks might be the help window, which under Mathematica is basically just a notebook). Let's say, the first notebook is on the left, the second one is in the middle, and the palette is on the right. Let's also say that I'm working on the left notebook, and now want to use the palette. Therefore I move the mouse to the palette and press the button I want.\n\nIn Mathematica 5.2 this had exactly the intended effect: Mathematica remembered that I last interacted with the left notebook, and inserted the input generated by the palette there. However in Mathematica 8.0 the input is instead inserted in the _right_ notebook.\n\nOf course it's not hard to understand how this happens (and indeed, the problem is not restricted to Mathematica, but occurs also e.g. with Gimp; however the fact that Mathematica 5.2 got it right gives me hope that it's just some option I have to toggle or some initialization I have to edit to regain the behaviour). When I move the mouse pointer from the left notebook to the palette, it of course crosses the right notebook which is in between, and therefore focus-on-mouse gives the keyboard input focus to that window. Of course it loses the input focus again when I move on to the palette (which ultimately gets the input focus), but newer Mathematica versions obviously record which window had input focus last and use _that_ to determine which notebook the palette should act on.\n\nTherefore my question is: Is there some option I can set to recover the old Mathematica-5.2 behaviour? Or if not, is there a way to get it with some Mathematica programming (which I then could put into an init file)?\n\n**Edit**\n\nAs I said in a comment to an answer that was deleted in the mean time, I've found out that Mathematica 7 still behaves in the way I want. So the change was obviously introduced in between 7 and 8. I'm adding that here so that the information is visible to those who cannot see deleted answers (and in case the deleted answer including comments might eventually go away completely).\n\nAlso note that (as I also wrote in that comment) `SelectedNotebook[]` changes the same way (i.e. follows focus unconditionally in version 8, but changes only on interaction in v7).\n", "link": "https://mathematica.stackexchange.com/questions/857/is-it-possible-to-regain-mathematica-5-2s-palette-input-focus-behaviour-with-ve", "tags": ["front-end", "customization"], "votes": 14, "creation_date": "2012-01-28T04:30:14", "comments": ["It is not the case anymore, right?", "I'd like to add a related difference between 5.2 and 8 which annoyes me a bit: select (highlight) any Keyword, say \"Solve\", in the code, then press F1 to get help. The window with the help system opens and displayes the explanation of that keyword. Now continue to work in your notebook hereby moving the cursor to another position. Now you wish to come back to the help for our keyword \"Solve\". Pressing F1 in 5.2 leads you where you want, 8 doesn't (but to the character at your pursor position). So some automatic focusing mechanism in 8 leads to an unwanted (from my point of view) effect.", "AFIK, OS X doesn't have focus-follows-mouse. Keyboard focus remains with the active window even if the mouse focus moves to, say, a floating palette.", "Ah, and note that those previous versions show the good behaviour runnning on the very same computer, under the very same operating system. Oh, and I just noted that I made a mistake in my description of focus-follows-mouse; fixed now.", "Do you have Focus-follows-mouse on OS X (I never worked with it, so I can't tell, but e.g. Windows doesn't have it)? Because the problem description only makes sense with Focus-follows-mouse. The point is, previous versions of Mathematica contain code that creates the good behaviour (and that code must have been there on purpose, because a naive implementation will give the exact behaviour I see with 8.0.0.0), and therefore I think that code might also be in 8.0.0.0, but just not activated for some reason (I hope it's still there). Of course it could also be a bug which was fixed in 8.0.4.", "I can not reproduce this problem with Mathematica 8.0.4 running under OS X 10.6.8 on a Mac. Are you sure this isn't a problem with your OS?"], "comment_count": 6, "category": "Technology", "diamond": 0} {"question_id": "366", "site": "mathematica", "title": "What is the complete list of ExportPacket formats?", "body": "`ExportPacket` is the most fundamental way we can turn `Box` forms into string data in other formats via the FE. \n\nFor example, a classic way this works is:\n \n \n First@FrontEndExecute@\n ExportPacket[Cell@BoxData@ToBoxes@Hyperlink[\"asdasd\", \"asdasd\"], \n \"PlainText\"]\n \n \"asdasd\"\n \n\nOr\n \n \n First@FrontEndExecute@\n ExportPacket[Cell@BoxData@ToBoxes@Hyperlink[\"asdasd\", \"asdasd\"], \n \"InputText\"]\n \n \"Hyperlink[\\\"asdasd\\\", {\\\"asdasd\\\", None}]\"\n \n\nIt's enormously useful, but obviously entirely undocumented.\n\nNow I'm interested in figuring out what other options it can take. Currently I've scraped up the following formats:\n \n \n {\n \"PostScript\", \"InputForm\", \"GIF\",\n \"BoundingBox\", \"NotebookString\",\n \"EnhancedMetafile\", \"Metafile\", \"MGF\", \n \"CDFNotebookString\", \"PDF\",\n \"PICT\", \"BitmapPacket\", \n \"InputText\", \"PlainText\", \"DefaultText\" (* thanks to John Fultz for this one *)\n } \n \n\nBut it can take more [per John Fultz](https://mathematica.stackexchange.com/questions/1319/how-do-i-extract-the-contents-of-a-selected-cell-as-plain-text/1411#comment3519_1411)\n\nSo what's the complete list of formats? As a real long shot, what're the other options it can take?\n\n* * *\n\n### Scraping notes\n\nFor those interested, I just scraped those formats from the `DownValues` of various `Export` functions:\n \n \n System`ConvertersDump`installSource[\n DeleteMissing@\n Flatten@Lookup[Quiet[getFormatExportData /@ $ExportFormats], \n \"Sources\"],\n Automatic\n ];\n epfuncs =\n Quiet@\n ToExpression[DeleteDuplicates@Join[Names[\"*`*\"], Names[\"*`*`*\"]], \n StandardForm, \n Function[Null,\n If[FreeQ[DownValues[#], ExportPacket | FrontEnd`ExportPacket],\n Nothing,\n #\n ],\n HoldAllComplete\n ]\n ];\n epdats =\n AssociationMap[\n Cases[\n DownValues@#,\n p : (ExportPacket | FrontEnd`ExportPacket)[_, type_, opts___] :>\n {\n \"Type\" -> type,\n \"Opts\" -> {opts}\n },\n \\[Infinity],\n Heads -> True\n ] &,\n epfuncs\n ];\n Cases[Flatten@Values@epdats, (\"Type\" -> t_) :> t] // DeleteDuplicates\n \n", "link": "https://mathematica.stackexchange.com/questions/165065/what-is-the-complete-list-of-exportpacket-formats", "tags": ["front-end", "undocumented"], "votes": 12, "creation_date": "2018-02-02T15:10:38", "comments": ["@ChipHurst it exists in 12.1 too", "In 12.3 it looks like Rasterize got a performance boost through \"ImageObjectPacket\". I don't know if this format is in older versions though."], "comment_count": 2, "category": "Technology", "diamond": 0} {"question_id": "367", "site": "mathematica", "title": "Creating compiled search TRIE file for argument string completion", "body": "I'd like to generate a compiled search TRIE file (like those found in `$InstallationDirectory\\SystemFiles\\FrontEnd\\SystemResources\\FunctionalFrequency\\`) to help implement the auto-completion feature on user-defined functions. \n\nAs described in [File-name completion for custom functions](https://mathematica.stackexchange.com/q/56984), I could directly add a list of strings into the specialArgFunctions.tr file. However, my lists of arguments is long and likely to change during development, so I'd prefer to point to a TRIE file that I can update as needed. \n\nHow can I create a compiled TRIE file? Can this be done in Mathematica or Workbench? Is the TRIE file extension structure documented anywhere?\n\n* * *\n\n> b3m2a1 comment:\n> \n> These files are here in v11:\n> \n> \n> FileNames[\"*.trie\", \n> PacletFind[\"AutoCompletionData\"][[1]][\"Location\"],\n> Infinity\n> ]\n> \n> \n> Similarly in 11 is [`CA`CADumpTriePacket`](https://mathematica.stackexchange.com/a/13452/38205) which seems to extract the properties from a trie file, but not the trie format itself.\n", "link": "https://mathematica.stackexchange.com/questions/113116/creating-compiled-search-trie-file-for-argument-string-completion", "tags": ["front-end", "files-and-directories", "customization", "autocomplete"], "votes": 12, "creation_date": "2016-04-20T06:58:44", "comments": ["@Szabolcs, thanks for the suggestion. I will try it out and let you know if I make any progress. Thanks again!", "You may be interested in this: mathematica.stackexchange.com/a/129910/12 It shows how to add completion without editing specialArgFunctions.tr. What I don't know how to do is enable completion of non-string arguments, as all examples that do this use .trie files and I don't know how to create such files.", "@DanielLichtblau, Thanks for the link to 441, but I am not asking about general practices for implementing a trie data structure in Mathematica. I am looking for the specific TRIE file structure used by Mathematica's FrontEnd.", "Duplicate of 441?"], "comment_count": 4, "category": "Technology", "diamond": 0} {"question_id": "368", "site": "mathematica", "title": "How to modify NDSolve`StateData without crashing the kernel?", "body": "Probably a hard question, but it's better to cry out loud.\n\n[Reminded by Chris K](https://chat.stackexchange.com/transcript/message/51136457#51136457), I noticed my [`fix`](https://mathematica.stackexchange.com/a/129193/1871) function has been broken since _v11.3_. After some checking, I found `NDSolve`StateData[…]`, which is not an atom in _v9_ and becomes an atom now, can no longer be modified with pattern matching. \n\nTo keep `fix` up to date, I tried the solution in this [excellent post](https://mathematica.stackexchange.com/q/96225/1871) about modifying data inside atom, sadly the modified `NDSolve`StateData` crashes the kernel in the `NDSolve`ProcessSolutions` stage **even in _v9_** and I can't figure out a workaround.\n\nThe following is a simplified example attempting to modify the difference order to `2`. The last line crashes the kernel of course so please save your work before testing.\n \n \n tmax = 10; lb = 0; rb = 5;\n system = With[{u = u[t, x]}, {D[u, t] == D[u, x, x], u == 0 /. t -> 0, \n u == Sin[t] /. x -> lb, u == 0 /. x -> rb}];\n \n {state} = NDSolve`ProcessEquations[system, u, {t, 0, tmax}, {x, lb, rb}];\n \n teststate = \n state /. a_NDSolve`FiniteDifferenceDerivativeFunction :> \n RuleCondition@\n NDSolve`FiniteDifferenceDerivative[a@\"DerivativeOrder\", a@\"Coordinates\", \n \"DifferenceOrder\" -> 2, PeriodicInterpolation -> a@\"PeriodicInterpolation\"];\n \n Head[#][\"DifferenceOrder\"] & /@ \n teststate[\"NumericalFunction\"][\"FunctionExpression\"][[2, 1]]\n (* Should give {{2}} if the replacement succeeds. *)\n \n (*Failed attempt: *)\n ml = LinkCreate[LinkMode -> Loopback];\n LinkWrite[ml, With[{e = state}, Hold[e]]];\n holdstate = LinkRead[ml];\n \n newstate = holdstate /. \n a_NDSolve`FiniteDifferenceDerivativeFunction :> \n RuleCondition@\n NDSolve`FiniteDifferenceDerivative[a@\"DerivativeOrder\", a@\"Coordinates\", \n \"DifferenceOrder\" -> 2, PeriodicInterpolation -> a@\"PeriodicInterpolation\"] // \n ReleaseHold\n \n NDSolve`Iterate[newstate, tmax]\n \n (*Warning: the following line crashes the kernel.*)\n u /. NDSolve`ProcessSolutions@newstate\n \n\nAny way to avoid the crashing and fix my `fix` function?\n\n* * *\n\nA bit of spelunking shows this seems to be (at least partly) related to `Internal`Bag`. A minimal example:\n \n \n bag = Internal`Bag[]\n ml = LinkCreate[LinkMode -> Loopback];\n LinkWrite[ml, With[{e = bag}, Hold[e]]]\n holdbag = LinkRead[ml]\n LinkClose[ml]\n ReleaseHold[holdbag] === bag\n (* False *)\n \n", "link": "https://mathematica.stackexchange.com/questions/202668/how-to-modify-ndsolvestatedata-without-crashing-the-kernel", "tags": ["differential-equations", "numerics", "undocumented", "compatibility"], "votes": 11, "creation_date": "2019-07-24T07:41:44", "comments": ["I think the relation to Bag is only really in the fact that both are mutable object stored at the C++ level"], "comment_count": 1, "category": "Technology", "diamond": 0} {"question_id": "369", "site": "mathematica", "title": "Customized StepSize control in unconstrained gradient-descent optimization", "body": "I have a type of minimization problem I frequently solve that involves warping a large 2D triangle-mesh (~25,000 vertices) to fit a model. In the mesh, vertices carry the empirical measurements (each of which has a predicted position in the model, which is a continuous field). The potential/objective function for the system is something like this:\n \n \n potentialFn[{X0_List, triangles_List}, Xideal_List, X_List] := With[\n {elens0 = EdgeLengths[X0, triangles],\n elens = EdgeLengths[X, triangles],\n angs0 = AngleValues[X0, triangles],\n angs = AngleValues[X, triangles]},\n Plus[\n Total[(elens0 - elens)^2],\n Total[(angs0 - angs)^2],\n Total@Exp[-(X - Xideal)^2],\n 0.25*Total[1 + (angs0/angs)^2 - 2*(angs0/angs))]]];\n \n\nThe idea is that the potential is equal to the sum of the squared deviations in the distances between vertex neighbors (the edges) plus the sum of the squared deviation in the angles of the mesh, plus the goal function, which is 0 when the vertices are perfectly aligned with the model and otherwise monotonically increasing with distance from an ideal fit, and finally a term that is designed to make the potential approach infinity as any triangle approaches inversion; i.e., the last term (which is similar to models of the van der Waals potential) constrains the potential such that triangle ABC would have to cross a singularity in order to becomes triangle ACB (in terms of a counterclockwise ordering of vertices). Additionally I have well-tested analytical functions that calculate the gradients (but nothing for the Hessians). I've confirmed that the gradient descent works correctly by, among other things, running a gradient descent search with a very small step-size for a very long time.\n\nMost of those details are irrelevant to my question, however. What is important is that I can, for any set of vertex coordinates that are valid (i.e., no triangles have been inverted), calculate a maximum steps-size `S` such that for any actual step size `s` (`0 < s < S`) the resulting vertex coordinates will also be valid; so long as my step sizes always follow this rule, I can guarantee that no triangles will invert. The problem I have is that there doesn't seem to be a way for me to provide this information to the Mathematica line-search algorithm in functions like `FindMinimum`.\n\nInitially, I thought that something like this would be the solution:\n \n \n FindMinimum[\n potentialFn[{X0, triangles}, X],\n {X, X0},\n Gradient :> potentialGradient[{X0, triangles}, X],\n Method -> {\n \"ConjugateGradient\",\n \"StepControl\" -> {\n \"LineSearch\",\n \"MaxRelativeStepSize\" :> CalculateMaxRelativeStepSize[{X0, triangles}, X]}}]\n \n\nThis, however, gives me an error (`FindMinimum::lsmrss`, with message \"-- Message Text not found -- `(CalculateMaxRelativeStepSize[{X0, triangles}, X])`\") that I can only assume is due to FindMinimum's inability to interpret the delayed-rule. I've spent a lot of time looking through the Mathematica documentation on conjugate-gradient and related unconstrained search and have found nothing that indicates that I can actually control the step-size aside from setting a permanent step-size length relative to the norm of the total gradient. That is fairly useless in this kind of case, unfortunately --- I can use it, but it results in a very slow search.\n\nMy question is this: are there existing (undocumented?) methods for providing Mathematica's line-search method with a way to calculate a maximum gradient step-size?\n\nNote: I realize I haven't provided a minimal working example of this problem. I can do so, but this would be quite an undertaking as there is a lot of context around the specific problem instances --- if anyone believes that they can help me with this kind of optimization given an example, I will work on this, but I'd appreciate some indication that the work of translating the problem into a compact instance won't be for naught before I do it.\n", "link": "https://mathematica.stackexchange.com/questions/105093/customized-stepsize-control-in-unconstrained-gradient-descent-optimization", "tags": ["mathematical-optimization"], "votes": 11, "creation_date": "2016-01-28T10:43:41", "comments": ["Might an appropriate value for \"RestartThreshold\" be helpful? See Nonlinear Conjugate Gradient Methods.", "I interpreted you last paragraph differently. Sorry. In any case, I would at least take a serious look at your problem, if I could run your code.", "@bbgodfrey Thanks for pointing out the minor syntax error, but the note at the end of the original post was included precisely so that people would not try to copy/paste the code then post comments like that.", "You have one too many ) in the last line of potentialFn. Also, potentialGradient and CalculateMaxRelativeStepSize are undefined. EdgeLengths and AngleValues also appear to be undefined or misused.", "I should also note that the approach in the above comment quietly crashes the kernel as often as not somewhere in the middle of the optimization.", "I should have noted that the best solution I've found is to use the StepMonitor argument; whenever the previous step invalidates a triangle, I use Throw/Catch to exit the minimization and restart it with a smaller \"MaxRelativeStepSize\". It seems clunky and is definitely not optimal, but it works for now."], "comment_count": 6, "category": "Technology", "diamond": 0} {"question_id": "370", "site": "mathematica", "title": "How can autocomplete entries be added for DownValues and Properties?", "body": "I'd like to assign `DownValues` to a symbol like this\n \n \n x[\"firstvalue\"] = 1;\n x[\"secondvalue\"] = 2;\n \n\nAnd then have the `DownValue` keys `\"firstvalue\"` and `\"secondvalue\"` offered in an autocomplete list once I have entered `x[`.\n\nAnother useful application for this kind of modification is to have the `\"Properties\"` of an object like a `FittedModel` added to the autocomplete list. \n\nIt is not always easy to remember `\"SinglePredictionConfidenceIntervalTableEntries\"`, which is one of the 64 properties of `FittedModel`.\n\nCan this be done? What controls the entries in the _autocomplete \"engine\"_?\n\n### Addendum\n\n[This amazing answer](https://mathematica.stackexchange.com/a/16710/92) shows a way of altering the behavior of autocomplete. I do not yet understand how it works.\n", "link": "https://mathematica.stackexchange.com/questions/27629/how-can-autocomplete-entries-be-added-for-downvalues-and-properties", "tags": ["front-end", "customization", "autocomplete"], "votes": 11, "creation_date": "2013-06-25T11:45:19", "comments": ["There is not much hope you achieve this with version 9.0.1. My answer you were referring to only works for 9.0. since back then the autocompletion was only partially done by the frontend. Nowadays, I don't see a way to interfere (let's call it improve) with the autocompletion happening in the Mathematica front end. In my latest project this would be possible to implement, but it's not the front end.", "Well, that is far and away superior to my ad hoc solution. I may have to expropriate it ... :)", "Ah for this case I've added the Keys function from this answer to my toolchain.", "Not on thisFit, obviously. But, DownValues[x][[All,1,1,1]] is stymied by fcn[] := . So, I've been using Case, but you have to get the pattern correct.", "@rcollyer, I don't understand your question. I can always call thisFit[\"Properties\"] to find the properties of the fit or DownValues[x][[All,1,1,1]] to get the keys of x. You have improvements upon these methods?", "I can't answer the autocomplete question, but do you want a mechanism for automatically updating the properties list?"], "comment_count": 6, "category": "Technology", "diamond": 0} {"question_id": "371", "site": "mathematica", "title": "Different timing with large array assignment", "body": "Consider the following example\n \n \n n = 10^8;\n AbsoluteTiming[\n A = ConstantArray[1., n]; // anyFunc;\n A[[2 ;; ;; 2]] = -1.;\n ]\n \n\n> {0.388108, Null}\n \n \n n = 10^8;\n AbsoluteTiming[\n B = ConstantArray[1., n];\n B[[2 ;; ;; 2]] = -1.;\n ]\n \n\n> {0.738504, Null}\n \n \n A == B\n \n\n> True\n\nWhy is the first approach faster? The only difference is, there is an arbitrary function outside the first one. I am using Mathematica 12 on windows 10 64 bit.\n", "link": "https://mathematica.stackexchange.com/questions/210417/different-timing-with-large-array-assignment", "tags": ["list-manipulation", "performance-tuning", "packed-arrays"], "votes": 10, "creation_date": "2019-11-28T02:11:43", "comments": ["This $HistoryLength = 0 can be relevant in wolframscript, cause there you cannot have access to history even though is set to infinity by default. It would make sense to set it to zero if it gives performance increase.", "@HenrikSchumacher: I have no idea, and the days of being motivated to dig way under the weighted covers to find out have faded. I found this and similar behaviors years ago, as did others, but the how/why of this particular case... beats me. Perhaps a Wolfram engineer might chime in.", "@HenrikSchumacher I'd bet that the history tracking of the kernel keeps a reference to B but not A that causes the tensor to be copied when some elements are changed to -1. That the tracking would be turned off when one sets $HistoryLength = 0 makes sense; but why a reference to B exists but not to A -- or why a reference to B exists at all -- I cannot explain.", "@ciao Wow, that does help, but I wonder why it does so. Can you explain that. It is not like DownValues[Out] were totally cluttered...", "Set $HistoryLength=0"], "comment_count": 5, "category": "Technology", "diamond": 0} {"question_id": "372", "site": "physics", "title": "Simple argument for unexpected behavior in SUSY model", "body": "Consider a supersymmetric theory with 3 chiral superfields, $X, \\Phi_1$ and $\\Phi_2,$ with canonical Kahler potential and superpotential $$ W= \\frac12 h_1 X\\Phi_1^2 +\\frac12 h_2 \\Phi_2\\Phi_1^2 + fX.$$ One can show, by doing calculations, that (i) supersymmetry is spontaneously broken, but (ii) one-loop corrections do not lift the classical pseudo-moduli space.\n\nQUESTION: is it possible to say (ii) without looking at the explicit form of Coleman-Weinberg potential, e.g. making some field redefinition which shows that this is not an interacting theory and it is very close to Polonyi model?\n", "link": "https://physics.stackexchange.com/questions/76207/simple-argument-for-unexpected-behavior-in-susy-model", "tags": ["homework-and-exercises", "supersymmetry", "symmetry-breaking"], "votes": 15, "creation_date": "2013-09-04T05:35:16", "comments": ["@Nicolo': Comparing your calculus, with the calculus done for O’Raifeartaigh model, maybe you can check the step, or the steps, where the difference happens. Maybe this will give an idea for some possible \"rule\".", "right, but this is not the O'Raifeartaigh model", "This does not seem obvious .Following this paper, it we take a model similar to yours, but different, the O’Raifeartaigh model $(7.32)$, one may show that quantum corrections lift the classical pseudo-moduli space, see Chapter $7.4$ and formula $7.51$.", "Yay, a good question after so many days! I hope you get an answer..."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "373", "site": "physics", "title": "Extended Born relativity, Nambu 3-form and ternary ($n$-ary) symmetry", "body": "**Background:** Classical Mechanics is based on the Poincare-Cartan two-form\n\n$$\\omega_2=dx\\wedge dp$$\n\nwhere $p=\\dot{x}$. Quantum mechanics is secretly a subtle modification of this. On the other hand, the so-called Born-reciprocal relativity is based on the \"phase-space\"-like metric\n\n$$ds^2=dx^2-c^2dt^2+Adp^2-BdE^2$$\n\nand its full space-time+phase-space extension:\n\n$$ds^2=dX^2+dP^2=dx^\\mu dx_\\mu+\\dfrac{1}{\\lambda^2}dp^\\nu dp_\\nu$$\n\nwhere $$P=\\dot{X}$$\n\nNote: particle-wave duality is something like $ x^\\mu=\\dfrac{h}{p_\\mu}$.\n\nIn Born's reciprocal relativity, you have the invariance group which is the _intersection_ of $SO (4 +4)$ and the ordinary symplectic group $Sp (4)$, related to the invariance under the symplectic transformations leaving the Poincaré-Cartan two-form invariant. The intersection of $SO(8)$ and $Sp(4)$ gives you, essentially, the unitary group $U (4)$, or some \"cousin\" closely related to the metaplectic group.\n\nWe can try to guess an extension of Born's reciprocal relativity based on higher accelerations as an interesting academical exercise (at least it is for me). To do it, you have to find a symmetry which leaves spacetime+phasespace invariant, the force-momentum-space-time extended Born space-time+phase-space interval\n\n$ds^2=dx^2+dp^2+df^2$\n\nwith $p=\\dot{x}$, $ f=\\dot{p}$ in this set up. Note that it is the most simple extension, but I am also interested in the problem to enlarge it to extra derivatives, like Tug, Yank,...and n-order derivatives of position. Let me continue. This last metric looks invariant under an orthogonal group $SO (4+4+4) = SO (12)$ group (you can forget about signatures at this moment).\n\nOne also needs to have an invariant triple wedge product three-form\n\n$$\\omega_3=d X\\wedge dP \\wedge d F$$\n\nsomething that seems to be connected with a Nambu structure and where $P=\\dot{X}$ and $F=\\dot{P}$ and with invariance under the (ternary) 3-ary \"symplectic\" transformations leaving the above 3-form invariant.\n\n**My Question(s):** I am trying to discover some (likely nontrivial) Born-reciprocal like generalized transformations for the case of \"higher-order\" Born-reciprocal like relativities (I am interested in that topic for more than one reason I can not tell you here). I do know what the phase-space Born-reciprocal invariance group transformations ARE (you can see them,e.g., in this nice thesis [BornRelthesis](http://eprints.utas.edu.au/10689/2/Whole.pdf)) in the case of reciprocal relativity (as I told you above). So, my question, which comes from the original author of the extended Born-phase space relativity, **Carlos Castro Perelman in** [this paper](http://vixra.org/abs/1302.0103), and references therein, is a natural question in the context of higher-order Finsler-like extensions of Special Relativity, and it eventually would include the important issue of curved (generalized) relativistic phase-space-time. After the above preliminary stuff, the issue is:\n\n> What is the intersection of the group $SO (12)$ with the _ternary_ group which leaves invariant the triple-wedge product\n> \n> $$\\omega_3=d X\\wedge dP \\wedge d F$$\n\nMore generally, I am interested in the next problem. So the extra or bonus question is: what is the ($n$-ary?) group structure leaving invariant the ($n+1$)-form\n\n$$ \\omega_{n+1}=dx\\wedge dp\\wedge d\\dot{p}\\wedge\\cdots \\wedge dp^{(n-1)}$$\n\nwhere there we include up to ($n-1$) derivatives of momentum in the exterior product or equivalently\n\n$$ \\omega_{n+1}=dx\\wedge d\\dot{x}\\wedge d\\ddot{x}\\wedge\\cdots \\wedge dx^{(n)}$$\n\ncontains up to the $n$-th derivative of the position. In this case, the higher-order metric would be:\n\n$$ds^2=dX^2+dP^2+dF^2+\\ldots+dP^{(n-1)^2}=dX^2+d\\dot{X}^2+d\\ddot{X}^2+\\ldots+dX^{(n)2}$$\n\nThis metric is invariant under $SO(4(n+1))$ symmetry (if we work in 4D spacetime), but what is the symmetry group or invariance of the above ($n+1$)-form and whose intersection with the $SO(4(n+1))$ group gives us the higher-order generalization of the $U(4)$/metaplectic invariance group of Born's reciprocal relativity in phase-space?\n\nThis knowledge should allow me (us) to find the analogue of the (nontrivial) Lorentz transformations which mix the\n\n$X,\\dot{X}=P,\\ddot{X}=\\dot{P}=F,\\ldots$\n\ncoordinates in this enlarged Born relativity theory.\n\n**Remark:** In the case we include no derivatives in the \"generalized phase space\" of position (or we don't include any momentum coordinate in the metric) we get the usual SR/GR metric. When n=1, we get phase space relativity. When $n=2$, we would obtain the first of a higher-order space-time-momentum-force generalized Born relativity. I am interested in that because one of my main research topics is generalized/enlarged/enhanced/extended theories of relativity. I firmly believe we have not exhausted the power of the relativity principle in every possible direction.\n\nI do know what the transformation is in the case where one only has $X$ and $P$. I need help to find and work out the nontrivial transformations mixing $X,P$ and higher order derivatives...The higher-order extension of Lorentz-Born symmetry/transformation group of special/reciprocal relativity.\n", "link": "https://physics.stackexchange.com/questions/61522/extended-born-relativity-nambu-3-form-and-ternary-n-ary-symmetry", "tags": ["classical-mechanics", "symmetry", "differential-geometry", "group-theory"], "votes": 30, "creation_date": "2013-04-18T09:55:29", "comments": ["@lurscher A generalized exteded relativity theory, beyond the one pioneered by Castro, Born, Cainiello, and many others in other \"flavors\" ...And where the principles of relativity and quantum mechanics merge and get generalized, much like your work on categories and branes generalize point particles...Quite impressive and stunning! I presented my own \"roadmap\" towards \"ultimate\"(final?) relativity in Slovenia, IARD 2016...Anyway, I had not too much time to develop the ideas presented there. I hope that change in the near future.", "i don't have a clue where you are going with all this, but it definitely looks interesting :-)"], "comment_count": 2, "category": "Science", "diamond": 1} {"question_id": "374", "site": "physics", "title": "Is there a null incomplete spacetime which is spacelike and timelike complete?", "body": "Geodesic completeness, the fact we can make the domain of the geodesic parametrized with respect an affine parameter the whole real line, is an important concept in GR. Especially, because the lack of it signals singularities.\n\nThis raises the question if incompleteness of one type of geodesic implies incompleteness of the rest. In general this is not the case.\n\nI've found examples of the following scenarios:\n\n * timelike complete, spacelike and null incomplete\n * spacelike complete, timelike and null incomplete\n * null complete, timelike and spacelike incomplete\n * timelike and null complete, spacelike incomplete\n * spacelike and null complete, timelike incomplete\n\n\n\nIs it possible to construct a spacelike and timelike complete spacetime but null incomplete?\n", "link": "https://physics.stackexchange.com/questions/141643/is-there-a-null-incomplete-spacetime-which-is-spacelike-and-timelike-complete", "tags": ["general-relativity", "differential-geometry", "spacetime", "mathematical-physics", "geodesics"], "votes": 11, "creation_date": "2014-10-16T07:49:08", "comments": ["@BenCrowell I agree that in general singularities are discuss in other terms. Nevertheless, Penrose and Hawking theorems only imply geodesic incompleteness.", "@yess: I see. Sorry, my comments were not that helpful then. I'll delete them.", "@BenCrowell Yes I meant that. Sorry for the sloppy definition.", "@MBN Yes, in fact, I look for all the examples but couldn't find the one Geroch mention as an open problem. However the paper that John links seems to settle the mattter.", "@JohnRennie Thank you very much the paper seems very interesting", "I seem to remember a paper by Geroch, where this was stated as an open problem. But the paper was probably from the 70's.", "This paper claims to describe just such a spacetime, however it's behind a paywall.", "@BenCrowell The term \"parametrization of a geodesic\" does not make any sense without being compatible with the ambient differentiable structure. I.e. the differentiable structure is fixed and the $\\sim \\mathbb{R}$ structure is unambiguously confirmed by finding a single $\\mathbb{R}$-parametrization. It is a good definition.", "[...] create the definition of a naked singularity, which is pretty intricate and doesn't just refer to what types of geodesics hit it.", "Discussions I've seen of this kind of thing actually talk about the characteristics of the singularity, not the characteristics of the incomplete geodesics. The main cases of interest are spacelike singularities (e.g., a black hole's singularity) and timelike singularities (e.g., cosmological ones, or naked singularities). Googling does turn up discussion of null singularities as well. I'm not convinced that the classification in terms of the characteristics of the complete and incomplete geodesics is actually useful or consistent. If so, then it would seem to have been unnecessary to [...]"], "comment_count": 10, "category": "Science", "diamond": 0} {"question_id": "375", "site": "physics", "title": "Equation of motion for cyclic model of the universe", "body": "I recently started to study the cyclic universe. I came across this [article](http://www.physics.princeton.edu/%7Esteinh/sciencecyc.pdf) [1]. My question is about the action used for describing the cyclic model: $$S = \\int d^{4}x\\sqrt{-g}(\\frac{1}{16\\pi G}R-\\frac{1}{2}(\\partial\\phi)^{2}-V(\\phi)+\\beta ^{4}(\\phi)(\\rho _{M}+\\rho _{R}))$$ where $R$ is Ricci scalar and $g$ is the metric. I solve the Euler-Lagrange equation for this action and find the equation of motion for $\\phi$: $$\\partial _{\\mu }\\frac{\\partial L}{\\partial (\\partial _{\\mu }\\phi )} - \\frac{\\partial L}{\\partial _{\\mu }\\phi } = 0$$ $$\\Rightarrow \\partial _{\\mu }\\left [ \\frac{1}{2} \\sqrt{-g} g^{\\alpha \\beta }\\left ( \\delta _{\\mu }^{\\alpha } \\partial _{\\beta }\\phi + \\delta _{\\mu }^{\\beta }\\partial _{\\alpha}\\phi \\right )\\right ] = \\sqrt{-g}\\left ( -V_{,\\phi }+4\\beta ^{3} \\beta _{,\\phi }(\\rho _{M}+\\rho _{R})\\right )$$ The radiation term is independent of $\\phi$ so only $\\rho _{M}$ enters the equation of motion. The zero component: $$3 a^{2} \\dot{a} \\dot{\\phi } + a^{3}\\ddot{\\phi }=a^{3}(-V_{,\\phi }+4 \\beta ^{3}\\beta _{,\\phi }\\rho _{M})$$ $$\\Rightarrow\\ddot{\\phi }+3H\\dot{\\phi }= -V_{,\\phi }+4 \\beta ^{3}\\beta _{,\\phi }\\rho _{M}$$ where $H$ is the Hubble parameter. But this result is different from what was written in the article. The difference is the coefficient of the last term. In addition, I have a problem finding Friedmann equations for this action (again in finding coefficients). Can anybody elaborate on the reason?\n\n**Reference:**\n\n[1] P.J. Steinhardt and N. Turok, _\" A Cyclic Model of the Universe,\"_ Science 296 (2002), available at [here](http://www.physics.princeton.edu/%7Esteinh/sciencecyc.pdf)\n", "link": "https://physics.stackexchange.com/questions/88894/equation-of-motion-for-cyclic-model-of-the-universe", "tags": ["cosmology", "lagrangian-formalism", "cosmological-inflation"], "votes": 11, "creation_date": "2013-12-03T12:52:45", "comments": ["Seems strange. You have $2$ papers with same authors (J. Steinhardt and Neil Turok) and a different sign... Sounds like a typo for the equation ($1$) of the first paper.", "I think that $\\triangle\\phi$ is zero and $\\phi$ only depends on time @Trimok.", "It is not completely clear for me what $(\\partial \\phi)^2$ is exactly. If it is $(\\partial \\phi)^2 = -(\\partial_0 \\phi)^2+(\\vec \\partial \\phi)^2$, then the Euler-Lagrange equations give a $\\square \\phi = \\ddot \\phi - \\triangle\\phi$, and not a $\\ddot \\phi$. So, I think you need to reexpress, if possible, $\\triangle\\phi$ as a function of $\\beta, \\beta_{,\\phi}$, etc...", "Thanks very much, @Trimok. That explains the factor 4, but what about the sign? I read this article. In this paper the sign of the coupling term is negative. Which one is correct?", "For the problem of the factor $4$, from the equation $(5)$ (fluid equation of motion), you see that there is a dependence $\\rho_M \\sim \\hat a^{-3}\\sim (a\\beta(\\phi))^{-3}$, so the term $\\beta^4(\\phi)\\rho_m$ is (secretely...) proportionnal to $\\beta(\\phi)$"], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "376", "site": "physics", "title": "Radiative equilibrium in orbit of a black hole", "body": "According to [Life under a black sun](http://arxiv.org/pdf/1601.02897v1.pdf), Miller's planet from _Interstellar_ , with a time dilation factor of 60,000, should be heated to around 890C by blue-shifted cosmic background radiation.\n\nHow they arrive at that number, however, seems to me a little opaque.\n\nAs the article describes, there are two major effects to consider: gravitational blueshifting, and blue- and redshifts due to the planet's orbital motion.\n\nCalculating the purely gravitational effects seems straightforward (although I admit I may still be missing something); given that radiative power is proportional to $T^4$, and power should scale linearly with the time dilation factor, the apparent CMB temperature should be $2.7K * 60,000^{1/4} = 42.26K$. Considering that a cold black hole occupies part of the sky, the equilibrium temperature of the planet should be slightly lower. That's clearly a long way from 890C!\n\nIt appears, then, that the majority of the heating must be a result of the circular motion of the planet in orbit. Now, it seems fairly obvious that getting precise answers will require numerical simulation, but it should be possible to at least get a close order-of-magnitude estimate based on a model of a planet moving at constant velocity through a background of the temperature calculated from gravitational effects alone. Unfortunately, though the article doesn't quote speeds, and I haven't been able to figure out how to calculate the relevant velocities for a planet is a low orbit around a rotating black hole.\n\nSo, can anybody help me fill in the blanks? If I start with a black hole of a given mass and angular momentum, and a planet in a stable circular orbit at some given radius, how do I get to an estimate of equilibrium temperature?\n", "link": "https://physics.stackexchange.com/questions/246203/radiative-equilibrium-in-orbit-of-a-black-hole", "tags": ["general-relativity", "thermodynamics", "time-dilation", "cosmic-microwave-background"], "votes": 12, "creation_date": "2016-03-29T14:04:05", "comments": ["Funny how things come around. I have been struggling with the same thing. I follow where the 890C comes from, given the illumination conditions that are deduced, but like you, struggle to immediately see where the blueshift factor of 275,000 came from. @Yuketerz to the rescue. physics.stackexchange.com/a/710922/43351", "I was just about to embark on an answer but then noticed that the paper you reference is all about how to get to that answer. So you need to read it and then come back with what you don't understand. e.g. The calculation you have done is (a) incorrect and (b) not directly relevant to calculating the equilibrium temperature of the planet.", "See Section 3 of this paper for calculating the velocity of circular orbits in Kerr spacetime."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "377", "site": "physics", "title": "Topological quantum error correcting codes which are not CSS codes", "body": "The most promising-seeming quantum error correction codes for the medium-to-long term are the topological codes, of which the toric code (and variants such as planar surface codes) and colour codes are the main examples.\n\nAs with essentially all approaches to quantum error correction, these are stabiliser codes. It also happens that both the toric code and colour codes are CSS codes: that is, they have a set of stabiliser generators which consist of products of $X$ operators or products of $Z$ operators, on subsets of the qubits. The useful properties of these codes for quantum error correction do not seem to particularly rely on this fact (the much more obviously relevant property is that the code distance of these codes scale with the size of the system and can be described by topological properties). So it may simply be a coincidence that these are CSS codes -- but at the same time, it is intriguing that the codes with the best known thresholds seem to be CSS codes.\n\n**Question.** Are there known examples of topological stabiliser codes which are neither CSS codes, nor equivalent to a CSS code up to local unitaries? Or is there a reason why such codes could not exist?\n", "link": "https://physics.stackexchange.com/questions/380501/topological-quantum-error-correcting-codes-which-are-not-css-codes", "tags": ["quantum-information", "quantum-error-correction"], "votes": 11, "creation_date": "2018-01-17T02:44:48", "comments": ["A google search brought this up: arxiv.org/abs/1012.0859 (\"Local non-CSS quantum error correcting code on a three-dimensional lattice\", Isaac H. Kim)"], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "378", "site": "rpg", "title": "What is this RPGA world/campaign design contest winner (ca. 2006?) I remember?", "body": "More than a decade ago, my best friend had an online subscription to various D&D materials. When he was back home visiting, he let me spend a few hours exploring this treasure trove. From what I can remember, there were fully indexed, digital versions of the current rule books. There were Q&A and discussion fora as well as online campaign settings somewhat like much older/slower play-by-mail. One of the campaign worlds I was particularly impressed with there was essentially a water world, where adventures took place on the high seas between islands and underwater in the deep.\n\nWhat struck me the most, though, was the world design contest winner from an RPGA network contest that I had found in the campaign settings. As I recall, it was set on a world where the great heroes had just failed to prevent the invasion of some extra-planar entities that meant to take total control. Completely blown away by the unusual concept of the starting point, I had noted the name of this world and campaign setting but lost it over the years. My friend could not remember it and no longer had his subscription. I would very much like to relearn the name of this world and get more details about it as a campaign setting. Any assistance is much appreciated.\n", "link": "https://rpg.stackexchange.com/questions/181613/what-is-this-rpga-world-campaign-design-contest-winner-ca-2006-i-remember", "tags": ["dungeons-and-dragons", "product-identification"], "votes": 16, "creation_date": "2021-03-06T05:10:05", "comments": ["The subscription service could be DnD Insider? That however only existed from about 2009 onwards rpg.stackexchange.com/questions/193504/…", "@KorvinStarmast It wasn’t, sadly, which is why I didn’t put an answer on this.", "@KRyan Hoping your research is productive", "Looking into this more, it seems that WotC went out of their way to not discuss the non-winning entries to that contest. Of the three finalists, only Eberron is known since the rules of the contest included the rights to all finalist submissions becoming property of Wizards of the Coast and the authors signing NDAs about them. We know that Rich Burlew (of Order of the Stick fame) was one and Nathan Toomey was the other, but nothing about what they wrote. Of the semi-finalists, 5/8 have been published elsewhere, and the other 3/8 are unknown. Investigating those 5.", "Notably, the winner of the 2002 RPGA contest was Eberron, which is still a major campaign setting published by Wizards of the Coast. Definitely doesn’t match this description, but so far as I know, that was the only contest of the sort that Wizards held—certainly the only one that resulted in publication of the setting that won. Is it possible that this was runner-up or honorable mention or something from that contest?", "Thanks for the advice, Jason_c_o. While I realised the superfluous details could be distracting, they also provided a context I thought might help. No, the water world campaign is not the one about which I was asking (though I wouldn't mind learning its name and more about it, too). It was the WoC subscription service with all the manuals, rulebooks, campaign settings and adventures that mentioned the invaded world setting. I recall my friend's visit as being in late summer 2006, the time at which I saw the invaded world campaign: it had already won an RPGA world design contest.", "Just to be make sure: The water-world you mention is not the world design contest winner you're looking for, correct? If so, you may want to cut some of the tangential information, it's distracting. Also, can you remember anything else? What year the contest was held (The RPGA Net ran many design contests, and took submissions for tournaments), or what book/magazine mentioned it? Is 2006 when you saw it, or when it was made?", "Welcome to RPG.SE! Take the tour if you haven't already and see the help center or ask us here in the comments (use @ to ping someone) if you need more guidance. Good Luck and Happy Gaming!"], "comment_count": 8, "category": "Culture & Recreation", "diamond": 0} {"question_id": "379", "site": "stackoverflow", "title": "Compatibility of impredicative Set and function extensionality", "body": "The Coq [FAQ](https://coq.inria.fr/faq#htoc41) says that function extensionality is consistent with predicative `Set`. It's not fully clear to me from this whether it's consistent with impredicative `Set` (or maybe the consistency is unknown in that case). \n", "link": "https://stackoverflow.com/questions/40395946/compatibility-of-impredicative-set-and-function-extensionality", "tags": ["rocq-prover"], "votes": 31, "creation_date": "2016-11-03T00:27:09", "comments": ["This question has been mentioned by Zimm i48 in this related Coq-Club thread (this post specifically).", "If you're considering migration then cstheory.SE would be another good choice, imo.", "Additionally, this question is about the theory of Coq rather than a programming issue with Coq and should probably be migrated to MathOverflow where it has more chances of being answered in details.", "This sounds a bit misleading; I imagine that it is consistent with impredicative Set as well. But that's probably a better question for the Coq mailing list."], "comment_count": 4, "category": "Technology", "diamond": 0} {"question_id": "380", "site": "stackoverflow", "title": "How can (<*) be implemented optimally for sequences?", "body": "The `Applicative` instance for `Data.Sequence` generally performs very well. Almost all the methods are _incrementally asymptotically optimal_ in time and space. That is, given fully forced/realized inputs, it's possible to access any _portion_ of the result in asymptotically optimal time and memory residency. There is one remaining exception: `(<*)`. I only know two ways to implement it as yet:\n\n 1. The default implementation\n \n xs <* ys = liftA2 const xs ys\n \n\nThis implementation takes `O(|xs| * |ys|)` time and space to fully realize the result, but only `O(log(min(k, |xs|*|ys|-k)))` to access just the `k`th element of the result.\n\n 2. A \"monadic\" implementation\n \n xs <* ys = xs >>= replicate (length ys)\n \n\nThis takes only `O(|xs| * log |ys|)` time and space, but it's not incremental; accessing an arbitrary element of the result requires `O(|xs| * log |ys|)` time and space.\n\n\n\n\nI have long believed that it should be possible to have our cake and eat it too, but I've never been able to juggle the pieces in my mind well enough to get there. To do so appears to require a combination of ideas (but not actual code) from the implementations of `liftA2` and `replicate`. How can this be done?\n\n* * *\n\nNote: it surely won't be necessary to incorporate anything like the `rigidify` mechanism of `liftA2`. The `replicate`-like pieces should surely produce only the sorts of \"rigid\" structures we use `rigidify` to get from user-supplied trees.\n\n### Update (April 6, 2020)\n\nMission accomplished! I managed to find [a way to do it](https://github.com/haskell/containers/pull/713). Unfortunately, it's a little too complicated for me to understand everything going on, and the code is ... rather opaque. I will upvote and accept a good _explanation_ of what I've written, and will also happily accept suggestions for clarity improvements and comments on GitHub.\n\n### Update 2\n\nMany thanks to [Li-Yao Xia](https://stackoverflow.com/users/6863749/li-yao-xia) and Bertram Felgenhauer for helping to clean up and document my draft code. It's now considerably less difficult to understand, and will appear in the next version of `containers`. It would still be nice to get an answer to close out this question.\n", "link": "https://stackoverflow.com/questions/60621991/how-can-be-implemented-optimally-for-sequences", "tags": ["haskell", "data-structures", "applicative", "finger-tree"], "votes": 22, "creation_date": "2020-03-10T09:30:55", "comments": ["@Li-yaoXia, something very similar goes on in the aptyMiddle function. The tricky bit here is being sure that the function constructed (when applying the size to your hypothetical version) creates a result with the required internal sharing.", "@dfeuer I think I've reduced the problem to constructing a function whose type would look like this with dependent types: repeat :: Int -> a -> (exists (n :: Nat). Node^n a) (this existential is equivalent to what you get from throwing away the Digits fields in the definition of FingerTree), which I find awfully non-obvious to implement.", "Aha! I think I've found a way to manage the complexity at a (very small) efficiency cost! The trick will be a stripped-down version of the type of rigid trees that, instead of having digits, just has numbers indicating how big those digits are. This is all the structural information we need, because replication can (I'm pretty sure) still work in redundant base 3! So we just need a function (or class method) to flesh out those rigid tree bits, and a slightly modified aptyMiddle to put it all together.", "@moonGoose, by way of background, I'm the one who wrote the liftA2 implementation, and there are still parts of it that I don't understand myself. replicate was one of several extremely clever functions Louis Wasserman wrote; it's simpler than his tails but still pretty tricky.", "@moonGoose, I hear you about trying to give more details, but I don't think it will help. This is most definitely not going to be a quick-and-easy question for anybody. Answering it will require a deep dive into Data.Sequence internals; explaining the performance analysis doesn't seem likely to reduce the amount of work or effort involved.", "@moonGoose, yes, it is.", "Is join $ replicate (length ys) <$> xs the same as 2?", "Imo it would help everyone to explain a bit how you get those bounds, perhaps even a diagram of what the different implementations produce? Otherwise everyone who would like to think about the problem has to independently do the work to get to where you're starting from."], "comment_count": 8, "category": "Technology", "diamond": 0} {"question_id": "381", "site": "stackoverflow", "title": "Metal crash upon adding SKSpriteNode to SKEffectNode", "body": "\n > -[MTLDebugRenderCommandEncoder setScissorRect:]:2028: failed assertion (rect.x(0) + rect.width(1080))(1080) must be <= 240\n \n\nI am getting this crash when adding a simple `SKSpriteNode` to a `SKEffectNode` with the following code\n \n \n SKSpriteNode *warpSprite = [SKSpriteNode spriteNodeWithImageNamed:@\"art.scnassets/symbol.png\"];\n SKEffectNode *entryEffectsNode = [[SKEffectNode alloc] init];\n [entryEffectsNode addChild:warpSprite];\n [self addChild:entryEffectsNode];\n \n\nI have not touched these nodes anywhere else in my project, when i change the sprite the value in (must be <=value) changes within the error.\n\nEdit: I have replaced the sprite image with a simple `spriteNodeWithColor:Size:` and the (<=value) is always twice size of the sprite. Also it should be noted that the SKScene is being used as a overlay in a SceneKit scene.\n\nI have created a seperate SKScene with the following code, which still results in the same error.\n \n \n @implementation testScene\n \n -(id)initWithSize:(CGSize)size {\n if (self = [super initWithSize:size]) {\n \n SKSpriteNode *testSprite = [SKSpriteNode spriteNodeWithColor:[SKColor purpleColor] size:CGSizeMake(100, 100)];\n SKEffectNode *testEffect = [[SKEffectNode alloc] init];\n [testEffect addChild:testSprite];\n [self addChild:testEffect];\n \n }\n return self;\n }\n @end\n \n\nEdit 2: I have just tested the above scene as an overlay on a default SceneKit Project and it crashes with the same error.\n\nEdit 3: I have reproduced this using swift. Bug report summited to Apple.\n", "link": "https://stackoverflow.com/questions/42383395/metal-crash-upon-adding-skspritenode-to-skeffectnode", "tags": ["ios", "objective-c", "sprite-kit", "swift3", "skeffectnode"], "votes": 21, "creation_date": "2017-02-21T21:29:20", "comments": ["What is the current status on your bug report?", "Are you certain your initWithSize method is called on the main thread ? When using a SKScene as a SCNScene overlay, all SKNode operations must be run from the main thread.", "Might be a bug. File one with Apple to be sure.", "With this changed i still get the same error. What else would you need to know? I have just run the same code in a new project and it works as you would expect. I have replaced the sprite image with a simple spriteNodeWithColor:Size: and the (<=value) is always twice size of the sprite.", "your rect is greater than what is supported, [SKSpriteNode spriteNodeWithImageNamed:@\"art.scnassets/symbol.png\"] should be [SKSpriteNode spriteNodeWithImageNamed:@\"symbol\"] so that it can properly handle retina graphics, Other than this, we would need to know more"], "comment_count": 5, "category": "Technology", "diamond": 0} {"question_id": "382", "site": "stackoverflow", "title": "Creating Spark Objects from JPEG and using spark_apply() on a non-translated function", "body": "Typically when one wants to use `sparklyr` on a custom function (_i.e._ [**non-translated functions) they place them within `spark_apply()`](https://sparkfromr.com/non-translated-functions-with-spark-apply.html). However, I've only encountered examples where a _single_ local data frame is either `copy_to()` or `spark_read_csv()` to a remote data source and then used `spark_apply()` on it. An example, for illustrative purposes only:\n \n \n library(sparklyr)\n sc <- spark_connect(master = \"local\")\n \n n_sim = 100\n iris_samps <- iris %>% dplyr::filter(Species == \"virginica\") %>%\n sapply(rep.int, times=n_sim) %>% cbind(replicate = rep(1:n_sim, each = 50)) %>% \n data.frame() %>%\n dplyr::group_by(replicate) %>%\n dplyr::sample_n(50, replace = TRUE)\n \n iris_samps_tbl <- copy_to(sc, iris_samps)\n \n iris_samps_tbl %>% \n spark_apply(function(x) {mean(x$Petal_Length)}, \n group_by = \"replicate\") %>%\n ggplot(aes(x = result)) + geom_histogram(bins = 20) + ggtitle(\"Histogram of 100 Bootstrapped Means using sparklyr\")\n \n\nIt seems like it would be therefore possible to use this on any range of non-translated functions coming from `CRAN` or `Bioconductor` packages as long as the data resides in a Spark Object. \n\nI've came up with a specific problem for `.jpeg` images as I read that [`SparkR` can load compressed image (`.jpeg`, `.png`, _etc._) into raw image representation via `ImageIO` in Java library](https://spark.apache.org/docs/latest/ml-datasource) \\- it seems possible that `sparklyr` could do this as well.\n\n`RsimMosaic::composeMosaicFromImageRandom(inputImage, outputImage, pathToTilesLibrary)` function takes an input image and a path to tiles used to create a photo mosaic and outputs an image ([example here](https://rviews.rstudio.com/2020/02/13/photo-mosaics-in-r/)). \n\nIf this function only took one image and I knew how to turn it into spark object I might imagine the command would look like: `composeMosaicFromImageRandom(inputImage, outputImage, spark_obj)`. However, this function is taking a path to 30,000 images. \n\nHow would one create 30,000 Spark Objects from the path to these tiles (`.jpegs`) and then use this function? \n\nIf the underlying code would actually need to be modified I've used `jimhester/lookup` to provide the source code:\n \n \n function (originalImageFileName, outputImageFileName, imagesToUseInMosaic, \n useGradients = FALSE, removeTiles = TRUE, fracLibSizeThreshold = 0.7, \n repFracSize = 0.25, verbose = TRUE) \n {\n if (verbose) {\n cat(paste(\"\\n ------------------------------------------------ \\n\"))\n cat(paste(\" R Simple Mosaic composer - random version \\n\"))\n cat(paste(\" ------------------------------------------------ \\n\\n\"))\n }\n if (verbose) {\n cat(paste(\" Creating the library... \\n\"))\n }\n libForMosaicFull <- createLibraryIndexDataFrame(imagesToUseInMosaic, \n saveLibraryIndex = F, useGradients = useGradients)\n libForMosaic <- libForMosaicFull\n filenameArray <- list.files(imagesToUseInMosaic, full.names = TRUE)\n originalImage <- jpeg::readJPEG(filenameArray[1])\n xTileSize <- dim(originalImage[, , 1])[1]\n yTileSize <- dim(originalImage[, , 1])[2]\n if (verbose) {\n cat(paste(\" -- Tiles in the Library : \", length(libForMosaic[, \n 1]), \"\\n\"))\n cat(paste(\" -- Tile dimensions : \", xTileSize, \" x \", \n yTileSize, \"\\n\"))\n }\n if (verbose) {\n cat(paste(\"\\n\"))\n cat(paste(\" Reading the original image... \\n\"))\n }\n originalImage <- jpeg::readJPEG(originalImageFileName)\n xOrigImgSize <- dim(originalImage[, , 1])[1]\n yOrigImgSize <- dim(originalImage[, , 1])[2]\n if (verbose) {\n cat(paste(\" -- Original image dimensions : \", xOrigImgSize, \n \" x \", yOrigImgSize, \"\\n\"))\n cat(paste(\" -- Output image dimensions : \", ((xOrigImgSize - \n 2) * xTileSize), \" x \", ((yOrigImgSize - 2) * yTileSize), \n \"\\n\"))\n }\n if (verbose) {\n cat(paste(\"\\n\"))\n cat(paste(\" Computing the mosaic... \\n\"))\n }\n outputImage <- array(dim = c(((xOrigImgSize - 2) * xTileSize), \n ((yOrigImgSize - 2) * yTileSize), 3))\n removedList <- c()\n l <- 1\n pCoord <- matrix(nrow = ((xOrigImgSize - 2) * (yOrigImgSize - \n 2)), ncol = 2)\n for (i in 2:(xOrigImgSize - 1)) {\n for (j in 2:(yOrigImgSize - 1)) {\n pCoord[l, 1] <- i\n pCoord[l, 2] <- j\n l <- l + 1\n }\n }\n npixels <- length(pCoord[, 1])\n for (i in 1:npixels) {\n idx <- round(runif(1, 1, length(pCoord[, 1])))\n pixelRGBandNeigArray <- computeStatisticalQuantitiesPixel(pCoord[idx, \n 1], pCoord[idx, 2], originalImage, useGradients)\n tileFilename <- getCloseMatch(pixelRGBandNeigArray, \n libForMosaic)\n startI <- (pCoord[idx, 1] - 2) * xTileSize + 1\n startJ <- (pCoord[idx, 2] - 2) * yTileSize + 1\n outputImage[startI:(startI + xTileSize - 1), startJ:(startJ + \n yTileSize - 1), ] <- jpeg::readJPEG(tileFilename)\n if (removeTiles) {\n libForMosaic <- removeTile(tileFilename, libForMosaic)\n removedList <- c(removedList, tileFilename)\n if (length(libForMosaic[, 1]) < (fracLibSizeThreshold * \n length(libForMosaicFull[, 1]))) {\n idxs <- runif(round(0.25 * length(libForMosaicFull[, \n 1])), 1, length(removedList))\n for (ii in 1:length(idxs)) {\n libForMosaic <- addBackTile(removedList[idxs[ii]], \n libForMosaic, libForMosaicFull)\n }\n removedList <- removedList[-idxs]\n }\n }\n if (length(pCoord[, 1]) > 2) {\n pCoord <- pCoord[-idx, ]\n }\n }\n if (verbose) {\n cat(paste(\"\\n\"))\n cat(paste(\" Done!\\n\\n\"))\n }\n jpeg::writeJPEG(outputImage, outputImageFileName)\n }\n \n\n_Please Note:_ My first attempt to speed up this code was 1) using `profvis` to find the bottlenecks (_i.e._ the for-loops) 2) use the `foreach` package on for-loops. This resulted in slower code which suggested I was parallelising at too low of a level. As I understand `sparklyr` is more about distributing the computing than about parallelizing it so perhaps this could work.\n", "link": "https://stackoverflow.com/questions/60743764/creating-spark-objects-from-jpeg-and-using-spark-apply-on-a-non-translated-fun", "tags": ["r", "binary", "jpeg", "sparklyr"], "votes": 18, "creation_date": "2020-03-18T09:33:27", "comments": ["Have you successfully used spark_apply() on a single .jpeg? The help says that you need to provide >An object (usually a spark_tbl) coercable to a Spark DataFrame.", "I would like to use sparklyr on non-translated functions, specifically for composeMosaicFromImageRandom().", "A long post; what exactly are you trying to achieve?"], "comment_count": 3, "category": "Technology", "diamond": 0} {"question_id": "383", "site": "stackoverflow", "title": "Detect ring/silent switch position change", "body": "I'm working on an app for which I would like to:\n\n 1. respect the ring/silent switch when playing audio, and\n\n 2. display an icon indicating that sound is muted when the ring/silent switch is set to silent.\n\n\n\n\nRequirement 1 is easy: I'm using [`AVAudioSessionSoloAmbient`](https://developer.apple.com/library/ios/documentation/AVFoundation/Reference/AVAudioSession_ClassReference/Reference/Reference.html#//apple_ref/doc/constant_group/Audio_Session_Categories) as my app's audio session category, so that my audio session will mute itself when the ring/silent switch is off.\n\nRequirement 2 seems considerably harder, because I need some sort of callback, notification, or KVO that will allow me to monitor the position of the switch, but Apple has made it clear that it is unwilling to offer an officially exposed way of doing this. That said, if I can find a nonintrusive way to monitor the switch's position, even one that is technically prohibited (like, say, an internal `NSNotification`), I would be willing to run it by Apple. \n\nFurther, I would prefer not to implement some of the polling solutions I've found elsewhere. See the Related Questions section for an example.\n\n* * *\n\n### What I've Learned (aka What Doesn't Work)\n\n * In iOS versions 4 and 5, at least, there was a trick that could be used to get the switch's position by [watching the route property of the current audio session](https://stackoverflow.com/questions/6901363/detecting-the-iphones-ring-silent-mute-switch-using-avaudioplayer-not-worki). Besides being deprecated by the `AVAudioSession` class, I can confirm that this trick is no longer an option. The current route, both as reported by the C functions comprising the deprecated `Audio Session` API and the current `AVAudioSession` class does not change when the ring/silent switch is toggled.\n\n * [`AVSystemController` is an internal class](https://github.com/nst/iOS-Runtime-Headers/blob/7ef0330f961248b9021e59e12aa3182440194817/PrivateFrameworks/Celestial.framework/AVSystemController.h) that seems to have a lot of promise. Invoking `- (BOOL)toggleActiveCategoryMuted` on the `sharedAVSystemController` does indeed mute my app's audio. Further, the shared singleton posts an `AVSystemController_SystemVolumeDidChangeNotification` notification when the system volume is changed via the volume buttons. Unfortunately, this notification is not posted in response to changes to the ring/silent switch (though [this dubiously attributed source says it should](http://cocoadev.com/AVSystemController)).\n\n * As far as I can tell, there are _no_ `NSNotification`s posted by _any_ object in response to ring/silent switch position changes. I came to this conclusion after adding myself as an observer to all notifications in the default center:\n \n [[NSNotificationCenter defaultCenter] addObserver:self selector:@selector(handleNotification:) name:nil object:nil];\n \n\nand then toggling the ring/silent switch. Nothing.\n\n * The `AVSystemController` class has a promising method with the signature:\n \n - (BOOL)getActiveCategoryMuted:(BOOL*)arg1;\n \n\nHowever, there are two problems with this:\n\n 1. Neither the return value nor the `BOOL` pointed to by `arg1` ever seem to change in response to toggling the ring/silent switch.\n 2. Because of the method signature, this method is not (so far as I understand) a candidate for KVO. \n\n\n\n****\n\n * I suspect that some object sends some other object(s) a [`GSEventRef`](https://github.com/kennytm/iphone-private-frameworks/blob/master/GraphicsServices/GSEvent.h) when the mute switch is changed, because I see the following in the declaration for event types:\n \n kGSEventRingerOff = 1012,\n kGSEventRingerOn = 1013,\n \n\nHowever, I'm pretty sure I can't intercept those messages, and even if I could, that would be a bit more than \"a little\" intrusive.\n\n\n\n\n* * *\n\n### Why I Believe This Is Possible\n\nPut simply: the Instagram app exhibits essentially this behavior. When you watch a video, it respects the ring/silent switch's setting, but displays an icon when the switch is off. The icon disappears and reappears so immediately after moving the switch that I assume it must be event-based, not polling.\n\n* * *\n\n### Related Questions\n\n * [This question dates back to iOS 4, and uses the methods that I mentioned in my first bullet above.](https://stackoverflow.com/questions/6901363/detecting-the-iphones-ring-silent-mute-switch-using-avaudioplayer-not-worki)\n\n * [This question is very similar to the one above.](https://stackoverflow.com/questions/7798891/detect-silent-mode-in-ios5?lq=1)\n\n * [This question is (much) more current, asking about iOS 7.](https://stackoverflow.com/questions/20992452/detect-silent-mode-in-ios-7) However, because I'm willing to accept a minimally intrusive breaking of the private-API rules, I would contend that this is a different question from my own.\n\n * [This _answer_ suggests using a polling method that I would strongly prefer to avoid.](https://stackoverflow.com/a/19013671/1292061)\n\n\n\n", "link": "https://stackoverflow.com/questions/24145386/detect-ring-silent-switch-position-change", "tags": ["ios", "iphone", "audio", "avaudioplayer", "ios7.1"], "votes": 17, "creation_date": "2014-06-10T08:47:51", "comments": ["The link above looks dead. The code above is hosted at: github.com/moshegottlieb/SoundSwitch (via stackoverflow.com/questions/20992452/…)", "Instagram reportedly uses this solution, which is not pretty: sharkfood.com/content/Developers/content/Sound%20Switch", "@PeterRobert No, I haven't. I ended up carefully implementing a solution by polling. It wasn't pretty.", "Have you found any event based solution?"], "comment_count": 4, "category": "Technology", "diamond": 0} {"question_id": "384", "site": "stackoverflow", "title": "Azure App Service load balancing settings", "body": "ARM template for Azure App Service has setting to configure load balancing algorithm - loadBalancing. According to [documentation](https://learn.microsoft.com/en-us/azure/templates/microsoft.web/sites#siteconfig-object) it's available through SiteConfig object and can have following values: WeightedRoundRobin, LeastRequests, LeastResponseTime, WeightedTotalTraffic, RequestHash.\n\nWe performed some testing with Standard S1 app service plan with two instances. First instance was responding to all request with no delay, second instance was responding to all requests with 3 seconds delay, ARR affinity was turned off. \n\nTest showed that all settings perform the same - after some ramp up time all requests spread evenly between two instances. It was not expected at least for LeastResponseTime, which intuitively suppose to direct more traffic to first instance (with low response time). \n\nSo the questions is, does this setting even work? And if it does, in what app service configuration it's respected?\n", "link": "https://stackoverflow.com/questions/52492410/azure-app-service-load-balancing-settings", "tags": ["azure", "azure-web-app-service", "azure-load-balancer"], "votes": 12, "creation_date": "2018-09-25T00:03:55", "comments": ["The App Service Plan level might also play a role. In some cases, certain features or their behaviors might differ based on whether you're on a Standard, Premium, or Isolated plan. Although documentation doesn't typically specify this, service level might impact the sophistication of load balancing features.", "The specific implementation of these algorithms by Azure might not exactly match their traditional definitions. For example, LeastResponseTime might require a significant difference in response times or a larger sample size to alter the routing decisions noticeably.", "You could take a look at this case. The scale-out itself is the load balance.", "Just a thought, have you tried to verify it from a different region. This latency tool may be helpful azurespeed.com", "Azure App Service uses internal load balancer that is not exposed and not configurable directly (except misterios loadBalancing setting in ARM). So there is no way to change the weights.", "For Weighted round robin, you can set the weight 5 for instance1, and weight 1 for instance2, what is the result? and you can refer to these load balancing algorithm description.", "Azure app service has setting inside ARM template called \"loadBalancing\". Setting name and supported values suggest that it should somehow specify how requests are distributed between instances. However we were not able to see any difference using different values for \"loadBalancing\" setting for our test case. 50% of request was server by first instance, 50% of requests were server by second instance, for all supported setting value values: WeightedRoundRobin, LeastRequests, LeastResponseTime, WeightedTotalTraffic, RequestHash.", "We used Netling to generate lot of HTTP GET requests to app service. App service was running simple ASP.NET MVC application that respond HTTP 200 to all get requests with \"OK\" in response body. App was configured in such a way that first instance was responding immediately (response time below 100 ms) and second instance was adding 3 second delay. We tried to simulate case when one of the servers experiencing high load with this test.", "Not sure your question, could you describe what you have done in your test? or what is load balancing setting? and what do you expect in the result?"], "comment_count": 9, "category": "Technology", "diamond": 0} {"question_id": "385", "site": "stackoverflow", "title": "When malloc_trim is (was) called automatically from free in glibc's ptmalloc or dlmalloc?", "body": "There is [incorrectly documented](https://stackoverflow.com/questions/28612438/can-malloc-trim-release-memory-from-the-middle-of-the-heap/42273711) function `malloc_trim` in glibc malloc (ptmalloc2), added in 1995 by Doug Lea and Wolfram Gloger (dlmalloc 2.5.4). \n\nIn glibc this function can return some memory freed by application back to operating system using negative sbrk for heap trimming, `madvise(...MADV_DONTNEED)` for unused pages in the middle of the heaps (this [feature is in `malloc_trim` since 2007 - glibc 2.9](https://sourceware.org/git/?p=glibc.git;a=commit;f=malloc/malloc.c;h=68631c8eb92ff38d9da1ae34f6aa048539b199cc), but not in `systrim`), and probably not by trimming additional thread arenas (which are not sbrk-allocated by mmaped as separate heaps).\n\nSuch function can be very useful for some long-running C++ daemons, doing a lot of threads with very high number of mixed size allocations, both tiny, small, middle and large in required size. These daemons may have almost constant amount of live (allocated by malloc and not yet freed) memory, but at the same time grow in RSS (physical memory consumption) over time. But not every daemon can call this function, and not every author of such daemons know that he should call this function periodically.\n\nMan page of `malloc_trim` by Kerrisk (2012) says in Notes about automatic calls to `malloc_trim` from other parts of glibc malloc:\n\n> This function is automatically called by free(3) in certain circumstances.\n\n(Also in Advanced Memory Allocation, 2003 - _Automatic trimming is done inside the free() function by calling`memory_trim()`_)\n\nThis note is probably from source code of malloc/malloc.c which mentions in several places that `malloc_trim` is called automatically, probably from the `free()`:\n \n \n 736 M_TRIM_THRESHOLD is the maximum amount of unused top-most memory\n 737 to keep before releasing **via malloc_trim in free().**\n \n 809 * When malloc_trim is called automatically from free(),\n 810 it is used as the `pad' argument.\n \n 2722 systrim ... is also called by the public malloc_trim routine.\n \n\nBut according to grep for `malloc_trim` in glibc's malloc: [http://code.metager.de/source/search?q=malloc_trim&path=%2Fgnu%2Fglibc%2Fmalloc%2F&project=gnu](http://code.metager.de/source/search?q=malloc_trim&path=%2Fgnu%2Fglibc%2Fmalloc%2F&project=gnu) there are: declaration, definition and two calls of it from `tst-trim1.c` (test, not part of malloc). Same result for whole glibc grep (additionally it was listed in abilists). Actual implementation of `malloc_trim` is `mtrim()` but it is called only from `__malloc_trim()` of malloc.c.\n\nSo the question is: When `malloc_trim` or its internal implementation (`mtrim`/`mTRIm`) is called by glibc, is it called from `free` or `malloc` or `malloc_consolidate`, or any other function?\n\nIf this call is lost in current version of glibc, was it there in any earlier version of glibc, in any version of ptmalloc2, or in original dlmalloc code ()? When and why it was removed? (What is the difference between `systrim`/`sys_trim` and `malloc_trim`?)\n", "link": "https://stackoverflow.com/questions/42283222/when-malloc-trim-is-was-called-automatically-from-free-in-glibcs-ptmalloc-or", "tags": ["malloc", "heap-memory", "glibc"], "votes": 12, "creation_date": "2017-02-16T11:44:26", "comments": ["It loos like it only runs automatically when freeing a large space: github.com/bminor/glibc/blob/…. Unfortunately the threshold for \"large\" is not a tunable.", "A very good question. I find this method to test malloc_trim() on running apps: notes.secretsauce.net/notes/2016/04/…", "dlmalloc versions: g.oswego.edu/pub/misc; glibc git github.com/bminor/glibc/blob/master/malloc/malloc.c; version of malloc in glibc with main_trim: github.com/bminor/glibc/blob/…. Example of RSS memory grow: bugs.python.org/issue11849 - message 133929"], "comment_count": 3, "category": "Technology", "diamond": 0} {"question_id": "386", "site": "stackoverflow", "title": "Connman without any user interaction", "body": "I'm trying to use Connman to manage the WiFi connection of my embedded system because it handles automagically any type of protection.\n\nIn interactive mode it's very simple:\n\n 1. connmanctl\n 2. agent on\n 3. scan wifi\n 4. services\n 5. connect \n 6. enter password if requested\n\n\n\nOn my system, the user enters the WiFi credentials (SSID, password) using a remote (web) application. Then I would use this information to setup connman using a script. \n\nThe goal is to avoid the user to select which type of protection is going to setup. I mean, most users just enter SSID/password but they don't know if it is a WPA-PSK or WEP connection.\n\nI'm reading throught the documentation, but I'm not sure which is the correct approach:\n\n * a config file: \n\n\n\nbut as far as I understand I need to specify the type of the security:\n\n> Security: The security type of the network. Possible values are 'psk' (WPA/WPA2 PSK), 'ieee8021x' (WPA EAP), 'none' and 'wep'. When not set, the default value is 'ieee8021x' if an EAP type is configured, 'psk' if a passphrase is present and 'none' otherwise.\n\nIt seems 'wep' is not handled if the field is omitted.\n\n * dbus-api: \n\n\n\nHere I understand it needs an 'agent' to feed the passphrase, thus I'm afraid I cannot send it programmatically.\n\nDo you have any recommendation about?\n", "link": "https://stackoverflow.com/questions/37230288/connman-without-any-user-interaction", "tags": ["connection", "wifi", "agent", "connman"], "votes": 12, "creation_date": "2016-05-14T11:17:00", "comments": ["I guess connman has for each session a directory in /var/lib/connman/. So we have to connect manually to a new wifi, create a session and then connman connects to saved wifi automatically. But I am not sure - just guessing when observing the behavior."], "comment_count": 1, "category": "Technology", "diamond": 0} {"question_id": "387", "site": "stackoverflow", "title": "Is integer vectorization accuracy / precision of integer division CPU-dependent?", "body": "I tried to vectorize the premultiplication of 64-bit colors of 16-bit integer ARGB channels.\n\nI quickly realized that due to lack of accelerated integer division support I need to convert my values to `float` and use some SSE2/SSE4.1 intrinsics explicitly for the best performance. Still, I wanted to leave the non-specific generic version as a fallback solution (I know that it's currently slower than some vanilla operations but it would provide future compatibility for possible improvements).\n\nHowever, the results are incorrect on my machine.\n\nA very minimal repro:\n \n \n // Test color with 50% alpha\n (ushort A, ushort R, ushort G, ushort B) c = (0x8000, 0xFFFF, 0xFFFF, 0xFFFF);\n \n // Minimal version of the fallback logic if HW intrinsics cannot be used:\n Vector128 v = Vector128.Create(c.R, c.G, c.B, 0u);\n v = v * c.A / Vector128.Create(0xFFFFu);\n var cPre = (c.A, (ushort)v[0], (ushort)v[1], (ushort)v[2]);\n \n // Original color:\n Console.WriteLine(c); // prints (32768, 65535, 65535, 65535)\n \n // Expected premultiplied color: (32768, 32768, 32768, 32768)\n Console.WriteLine(cPre); // prints (32768, 32769, 32769, 32769)\n \n\nI tried to determine what instructions are emitted causing the inaccuracy but I was really surprised to see that in SharpLab the results are [correct](https://sharplab.io/#v2:EYLgtghglgdgNAFxAJwK4wD4AEBMBGAWACgsAGAAizwDoAldBKMAU2oEkYFlYBnKAYx4BuYsQAUqHgAsA9sgTkAgnHKTZ88rRVq5CgOLbpu8gCEAlOX7kAvOTGkAHgA5SrlY4BiXj+4fef5J7eZiJExABqzPwIcng4TgA8qLAIAHzkAG425JHRsfHUAMLIzBAIzGL8dCpVBpbUJu6oIcRZtlkAVPWK5AD0OVExyHFORSVlFUFezaEZEMiWAAol2ZXUynY68mYZANqkALoqEkbbe3hHm6cIO7s4By1hJHgAnJWPVG/8y8yPQA). On the other hand, the issue is [reproducible](https://dotnetfiddle.net/4AMEZS) in .NET Fiddle.\n\nIs it something that's expected on some platforms or should I report it in the runtime repo as a bug?\n\n* * *\n\n### Update\n\nNevermind, this is clearly a bug. Using other values cause totally wrong results:\n \n \n using System;\n using System.Numerics;\n using System.Runtime.Intrinsics;\n \n (ushort A, ushort R, ushort G, ushort B) c = (32768, 65535, 32768, 16384);\n \n Vector128 v1 = Vector128.Create(c.R, c.G, c.B, 0u);\n v1 = v1 * c.A / Vector128.Create(0xFFFFu);\n \n // prints <32769, 49152, 57344, 0> instead of <32768, 16384, 8192, 0>\n Console.WriteLine(v1);\n \n // Also for the older Vector\n Span span = stackalloc uint[Vector.Count];\n span[0] = c.R;\n span[1] = c.G;\n span[2] = c.B;\n Vector v2 = new Vector(span) * c.A / new Vector(0xFFFF);\n \n // prints <32769, 49152, 57344, 0, 0, 0, 0, 0> on my machine\n Console.WriteLine(v2);\n \n\nIn the end I realized that the issue was at the multiplication: if I replace `* c.A` to the constant expression `* 32768`, then the result is correct. For some reason the `ushort` value is not correctly extracted/masked(?) out from the packed field. Even `Vector.Create` is affected:\n \n \n (ushort A, ushort R, ushort G, ushort B) c = (32768, 65535, 32768, 16384);\n \n Console.WriteLine(Vector128.Create((int)c.A)); // -32768\n Console.WriteLine(Vector128.Create((int)32768)); // 32768\n Console.WriteLine(Vector128.Create((int)c.A, (int)c.A, (int)c.A, (int)c.A)); // 32768\n \n\n* * *\n\n### Update 2\n\nIn the end filed an [issue](https://github.com/dotnet/runtime/issues/83387) in the runtime repo\n", "link": "https://stackoverflow.com/questions/75732627/is-integer-vectorization-accuracy-precision-of-integer-division-cpu-dependent", "tags": ["c#", "vectorization", "precision", "simd", "auto-vectorization"], "votes": 11, "creation_date": "2023-03-14T04:37:29", "comments": ["@GyörgyKőszeg: The compiler output I linked is for vec / 0xffffu. The results are exact, using a multiplicative inverse like I said, same as compilers do for scalar uint / constant. Why does GCC use multiplication by a strange number in implementing integer division? Did you think I meant v >> 16? Oh, you probably thought I meant doing the * part of you expression with an integer multiply; I was talking about using the high half of integer multiply as part of the division.", "Alright, this must be a bug, using some other values end up in total off results. I will update the post soon.", "@PeterCordes: yes, bit shifting is faster with vanilla code than division but the results are not exactly the same. And the floating point version now works well, it's just about the issue I noticed.", "If you were using FP, it would probably perform better to multiply by 1.0f/0xffff (or however you write a float constant in C#), although that can't be represented exactly. But with about 23 bits of precision, might still give the correct 16-bit integer after rounding or truncating to 16-bit.", "I don't see anything in the software fallback path which would cause this... I'd be tempted to either raise this in the dotnet/runtime repo, or ask in #allow-unsafe-blocks on discord (lots of JIT people hang out there, and they'll tell you if you need to open an issue)", "This can be done with integer multiply and shifts, although it's not super efficient since it needs the high half of a 32x32 => 64-bit multiply, and SSE2 / AVX only gives you that with pmuludq which gives you the full 64-bit results. So only half the input elements per vector. godbolt.org/z/7E9P9aWMh shows GCC and clang using a multiplicative inverse for dividing a vector of 4x uint32_t by 0xffffu. This is exact integer division, no FP involved at any point.", "In the meantime I realized that SharpLab also produces the wrong result in Debug mode. Strange, because on my computer both Debug and Release builds are incorrect. So I start to believe this is a bug after all."], "comment_count": 7, "category": "Technology", "diamond": 0} {"question_id": "388", "site": "stackoverflow", "title": "Is there something like a continuation Arrow transformer?", "body": "The [`ContT`](https://hackage.haskell.org/package/mtl-2.2.1/docs/Control-Monad-Cont.html#v:ContT) monad transformer has a interesting property: If there is a `* -> *` type such as `Set`, that has well-defined monadic operations, but can't have a `Monad` instance due to some constraints (here `Ord a`), it's possible to wrap it in `ContT` (`ContT r Set`) to get a monad instance, and defer the constraints outside it, like when we inject `Set` into `ContT r Set`. See [Constructing efficient monad instances on `Set` using the continuation monad](https://stackoverflow.com/q/12183656/1333025).\n\nIs there something similar for arrows? An [arrow transformer](https://hackage.haskell.org/package/arrows-0.4.4.1/docs/Control-Arrow-Transformer.html#t:ArrowTransformer) that'd allow to wrap an \"almost arrow\" into it, getting a valid `Arrow` instance, and defer problematic constraints to the part where we inject the \"almost arrow\" into it?\n\nFor example, if we had a type `AlmostArrow :: * -> * -> *` for which we'd have the usual `Arrow` operations, but with constraints, such as\n \n \n arr' :: (Ord a, Ord b) => (a -> b) -> AlmostArrow a b\n (>>>') :: (Ord a, Ord b, Ord c) => AlmostArrow a b -> AlmostArrow b c -> AlmostArrow a c\n \n\nAs a bonus, if yes, is there some nifty, generic category-theory way how to derive both `ContT` and such an arrow transformer?\n", "link": "https://stackoverflow.com/questions/42873249/is-there-something-like-a-continuation-arrow-transformer", "tags": ["haskell", "monad-transformers", "continuations", "category-theory", "arrow-abstraction"], "votes": 11, "creation_date": "2017-03-18T03:37:02", "comments": ["A free arrow would surely do the trick. Then you just have to do some junk to make it efficient."], "comment_count": 1, "category": "Technology", "diamond": 0} {"question_id": "389", "site": "stackoverflow", "title": "Multiset domination algorithm", "body": "Let us say that a multiset M _dominates_ another multiset N if each element in N occurs at least as many times in M.\n\nGiven a target multiset M and an integer k>0, I'd like to find a list, L, of size-k multisets whose sum dominates M. I'd like this list to be of small cost, where my cost function is of the form:\n\ncost = c*m + n\n\nwhere c is a constant, m is the number of multisets in L, and n is the number of _distinct_ multisets in L.\n\nHow can I do this? An efficient algorithm to find the optimal solution would be ideal.\n\nThe problem comes from trying to fulfill a customer's order for printing pages with a specialized block-printer that prints k pages at a time. Setting up the block-printer to print a particular template of k pages is costly, but once a template is initialized, printing with it is cheap. The target multiset M represents the customer's order, and the n distinct multisets of the list L represent n distinct k-page templates.\n\nIn my particular application, M typically has >30 elements whose multiplicities are in the range [10^4, 10^6]. The value of k is 15, and c is approximately 10^-5.\n", "link": "https://stackoverflow.com/questions/36459635/multiset-domination-algorithm", "tags": ["algorithm", "multiset"], "votes": 11, "creation_date": "2016-04-06T11:36:37", "comments": ["Well given that k is 15 and your example is worst case for my approach it can be bearable IMO. At least it gives you a fixed cap, with the \"one of a kind\" approach that would be optimal in your example, if distinct elements aren't multiples of k or multiplicities vary a lot (they have a 10^2 range right?) you're back having to find a way of grouping them. You can always take some initial time calculating the number of distinct elements and their multiplicity, then based on that remove as many as you can with you \"horizontal\" approach and clean up the rest with \"vertical\" approach.", "@CarloMoretti Consider the case when M consists of a*k distinct elements, each with identical multiplicity b. Clearly the optimal cost here is a(cb+1). Your solution, if I understand it correctly, does no better than a(cb+k). This performs nearly k times worse than optimal if k>>cb.", "Since multiplicities are so high why not just make sets of k identical elements? Let's say M has p distinct elements, you'll have at first n = p (without counting the reminder multiplicities). Since k is 15 I guess all combined remaining items will be r < 15*p which if you combine them at random will result in < p k-sets. So at the end you'll have n < 2p. So m = |M| / 15 if p is [10, 10^2] then |M| is in [10^5, 10^8] and m is in [10^4, 10^7] so c*m would be in [10^-1, 10^2] which is fairly comparable to p so roughly cost < 3p", "This seems like a really interesting question! It does seem like it might be NP-hard to find the exact optimum, but I would be very interested in hearing some sample problem instances to see what I can get.", "@DavidEisenstat btw, I just deleted an earlier comment saying I could express the optimization problem with linear constraints, but I realized I was wrong. The only formulation I know of currently has quadratic constraints.", "@piotrekg2 My friend has so far only given me one problem instance, without a corresponding human solution. I am pushing him to make more available.", "@dshin I know about the magic of commercial solvers, but I'd be a little surprised if you had a formulation that they would work well on.", "@dshin Did you try to compare combinations computed by humans with combinations computed by a greedy algorithm?", "@DavidEisenstat I don't think integer quadratic programs are optimally solvable in polynomial time in general, but that often doesn't stop commercial scientific solvers from doing a good job. Adding a restriction of not sharing pages between templates hurts your ability to minimize cost.", "(Fixed prev comment) This is similar to the NP-hard Cutting Stock problem. I've come up with a deterministic algorithm that will take just O(k|M|^2) total time (with |M| being the number of elements in M, not their sum) to find |M| solutions: one for each possible number i of distinct templates. The solution for i distinct templates exactly minimises the maximum copy count of any template under the constraint that it is possible to order templates, and pages, so that each page spans a contiguous block of templates. Taking the best of these |M| solutions should give a good quality answer.", "@j_random_hacker I was actually introduced to this problem by my friend - he told me that his company's current solution is to have a human try combinations by hand for hours until he finds one that seems ok. If you are serious about being paid, I can get you guys in touch. I'd imagine he'd want to see how the algorithm compares to the human solution (and to a simple greedy solution) on real life examples to put a dollar amount on the algorithm's value.", "Interesting problem! Is the quadratic program actually solvable to optimality? Would it make sense to consider a variant where each type of page can belong to exactly one template?", "@dwanderson One approach is to fix n and then formulate the optimization problem as an integer quadratic program, which can be solved with a scientific library. Then iterate over candidate values of n. I'm hoping for something nicer. I've tried a greedy-ish algorithm which works decently but is not optimal.", "My guess is that finding the optimal solution is NP-complete, but greedy can find a solution fairly easily.", "What have you tried? SO isn't a code-writing service. Which parts of the algorithm confuse you or are you having trouble with?"], "comment_count": 15, "category": "Technology", "diamond": 0} {"question_id": "390", "site": "stackoverflow", "title": "Equality of int&'s in template parameters", "body": "Suppose we have the following program:\n \n \n template \n struct Probe { \n static const int which = 1;\n };\n \n template \n struct Probe { \n static const int which = 2;\n };\n \n \n int i = 123;\n \n const int myQuestion = Probe::which;\n \n\nI am pretty sure, that `myQuestion` should be `2` regardless of the version of C++ standard, but compilers disagree upon it. MSVC and clang say that it is `2` until C++14, and `1` since C++17. See the [demo](https://godbolt.org/z/efETKzTo3). What is the truth?\n\nMy investigation so far:\n\n * I have found one relevant sentence in the C++ standard. It was there in [C++11](https://timsong-cpp.github.io/cppwp/n3337/temp.type#1.6), [C++14](https://timsong-cpp.github.io/cppwp/n4140/temp.type#1.6), [C++17](https://timsong-cpp.github.io/cppwp/n4659/temp.type#1.6) and [C++20](https://timsong-cpp.github.io/cppwp/n4868/temp.type#2.7). It did not change.\n * If you remove the parameter `T` from the example code, all compilers agree, that `myQuestion` is `2`. [Demo](https://godbolt.org/z/6cW8GGEod).\n\n\n", "link": "https://stackoverflow.com/questions/71022302/equality-of-ints-in-template-parameters", "tags": ["c++", "c++17", "c++14", "template-specialization", "non-type-template-parameter"], "votes": 10, "creation_date": "2022-02-07T08:54:16", "comments": ["Ok, I don't have an issue with that. It is a duplicate, but it's not exactly a question that gets asked a lot, so it's not a big deal to leave it open. I wouldn't dispute the closure either, i.e. I'm fine not voting either way.", "@cigien: It's not a good idea to mark my question as a duplicate of that long question. Even if my question is answered there, since it's so long, it would be difficult for the readers to extract information from that.", "@AnoopRana - Accessing the object's value in a constant expression requires it to satisfy additional constraints so the value is usable (in that case it should be constexpr indeed). Referring to the object alone (like in the OP, on surface level) doesn't require anything from the value.", "This is a duplicate of stackoverflow.com/questions/37369129 but can't be targetted because it doesn't have an upvoted/accepted answer.", "@StoryTeller-UnslanderMonica Ok then shouldn't this work without error? As you said, here we have a reference ref to an object i with static storage duration. Then ref can be used as size of an array. Am i missing something?", "@AnoopRana - timsong-cpp.github.io/cppwp/n4868/expr.const#11 should have the gist of it.", "@StoryTeller-UnslanderMonica I think you're right[in which case i'll delete my comments] but i can't find the exact statement that you quoted: \"A reference to an object with static storage duration is permissable in constant expressions, even when the referent is not const\". Can you put a link here for me?", "@AnoopRana - And you missed the point of mine. It already is valid in constant expressions.", "@AnoopRana Same error with contexpr/const int&: godbolt.org/z/a5Mrj4M53", "@AnoopRana - it's not ill-formed. For god's sake, it's a minimal reproducible example, check it before commenting. A reference to an object with static storage duration is permissable in constant expressions, even when the referent is not const.", "Definitely an interesting Q. References are meant to be translucent in the type system. Not sure how that may interplay with template equivalence checks however.", "Looks like the program is ill-formed since i is not a constant expression in your case. If you make it a constant expression, the program will no longer compile in any version.", "First i should be a constant expression when passing as template argument. Add constexpr or const for it.", "Shouldn't i be const at least?"], "comment_count": 14, "category": "Technology", "diamond": 0} {"question_id": "391", "site": "stackoverflow", "title": "Can skew binomial heaps support efficient merge?", "body": "The skew binomial heaps described in Okasaki's _Purely Functional Data Structures_ support merge in worst-case `O(log (max (m,n)))` time, where `m` and `n` are the lengths of the queues being merged. This is worse than segmented binomial queues, which support it in worst-case `O(log (min (m,n)))` time, and lazy binomial queues, which support it in worst-case `O(log (max (m,n)))` time but `O(log (min (m,n)))` amortized time[*]. This seems to be inherent in the restriction that the skew binary number in the queue representation is in canonical form (only one 2, and only as the least significant nonzero digit). Would it be possible to relax this restriction somewhat to get more efficient merges? The basic challenge is that a 2 must not be allowed to cascade into another 2.\n\n[*] I've also recently come up with a variant of scheduled binomial queues with the same worst-case bounds as segmented queues; that version is not yet fully implemented.\n", "link": "https://stackoverflow.com/questions/65051168/can-skew-binomial-heaps-support-efficient-merge", "tags": ["haskell", "data-structures", "binomial-heap"], "votes": 10, "creation_date": "2020-11-28T07:21:46", "comments": [], "comment_count": 0, "category": "Technology", "diamond": 0} {"question_id": "392", "site": "stackoverflow", "title": "Using ESM Modules with Coffeescript and Node.js", "body": "Ecmascript Modules are the future for packaging JS code, and both Node.js and Coffeescript support them. But I've had some trouble getting their support for ESM to work together.\n\nThe current stable Node (12.x) has ESM modules behind a flag (`--experimental-modules`). Coffeescript supports passing flags through to Node with `--nodejs`. So with a couple of files using ESM modules:\n \n \n # a.coffee\n import b from './b.coffee'\n \n b()\n \n \n \n # b.coffee\n b = ->\n console.log \"Hello\"\n \n export default b\n \n\nIn theory, we can run this code with `npx coffee --nodejs --experimental-modules a.coffee`. In practice this raises an error:\n \n \n 13:24 $ npx coffee --nodejs --experimental-modules a.coffee\n (node:8923) ExperimentalWarning: The ESM module loader is experimental.\n (node:8923) Warning: To load an ES module, set \"type\": \"module\" in the package.json or use the .mjs extension.\n /media/projects/coffeemodules/a.coffee:1\n import b from './b.coffee';\n ^^^^^^\n \n SyntaxError: Cannot use import statement outside a module\n ...\n \n\nThe error and docs say there are two ways to flag a file as containing an ESM module, one is to use the `mjs` extension (which isn't available to us here), and the other is to set `\"type\": \"module\"` in `package.json`, which also doesn't seem to work.\n\nSo: can it be done? Is there a way to get Coffeescript and Node.js ES modules to play together?\n", "link": "https://stackoverflow.com/questions/59312784/using-esm-modules-with-coffeescript-and-node-js", "tags": ["node.js", "coffeescript", "es6-modules"], "votes": 10, "creation_date": "2019-12-12T13:14:21", "comments": [], "comment_count": 0, "category": "Technology", "diamond": 0} {"question_id": "393", "site": "stackoverflow", "title": "MSDeploy not replacing encoded xml strings", "body": "In web.config I have:\n \n \n \n \n \n __ProfitConnectorToken__\n \n \n \n\nIn my parameters.xml:\n \n \n \n \n \n\nAnd in my SetParameters.xml:\n \n \n \n \n\nBut this value is not set when the web application is deployed. When I change my SetParameters.xml to:\n \n \n \n \n\nIt does work, so my XPath is correct. Why is the encoded xml value not set?\n", "link": "https://stackoverflow.com/questions/44433931/msdeploy-not-replacing-encoded-xml-strings", "tags": ["asp.net", "asp.net-mvc", "web-config", "msdeploy", "webdeploy"], "votes": 10, "creation_date": "2017-06-08T04:02:38", "comments": ["Nope, if I do that the value isn't replaced at all, not even my \"TEST\" value which was working before.", "Try removing /text() at the end of your XPath query?"], "comment_count": 2, "category": "Technology", "diamond": 0} {"question_id": "394", "site": "stackoverflow", "title": "How to ignore warnings in headers that are part of a clang C++ module?", "body": "We're using clang with `-fmodule` and `-fcxx-module` to enable module support as documented at . We're already seeing a significant improvement in build times by defining module maps for our core libraries.\n\nHowever, we have some library headers that use pragmas to disable warnings for certain lines, for example:\n \n \n template \n static bool exactlyEqual(TFloat lhs, TFloat rhs)\n {\n #pragma clang diagnostic push\n #pragma clang diagnostic ignored \"-Wfloat-equal\"\n return lhs == rhs;\n #pragma clang diagnostic pop\n }\n \n\nWhen this header is pulled in as a precompiled module, it seems clang's internal representation does not preserve the pragma information and the warning is still emitted. Since we treat warnings as errors this causes compilation to fail. Some might argue to just disable `float-equal` entirely, but we have a bunch of other cases with different warnings which we don't want to globally disable.\n\nWe're already using `-Wno-system-headers` and `-isystem` so that clients of libraries generally don't see warnings like this anyway (even without the pragma), but this doesn't seem to work when the header is imported as a module. In addition we still hit these warnings for code internal to the library which includes the header as a non-system header (i.e. without using `-isystem` / using double quotes), since module precompilation and importing also occurs here.\n\nI've tried using `_Pragma(...)` instead of `#pragma` which didn't have any effect.\n\nIs there some other way to conditionally ignore warnings in headers that come from precompiled clang modules?\n\n**UPDATE:** I've put a sample project up on which reproduces the problem\n\n**UPDATE:** Seems the `[system]` module attribute will suppress all warnings. However this suppresses even warnings we want to see when building the library itself, and we don't want to make all our modules system modules. If we find a way to not use the library's module map when building the library itself this may be acceptable but we'd still like to pragma out certain warnings for non-system modules..\n", "link": "https://stackoverflow.com/questions/38889998/how-to-ignore-warnings-in-headers-that-are-part-of-a-clang-c-module", "tags": ["c++", "clang", "clang++", "llvm-clang"], "votes": 10, "creation_date": "2016-08-11T00:30:06", "comments": ["Seeing as this question was asked way back when in 2016, are you still having this issue? Have you asked on the Clang mailing list?"], "comment_count": 1, "category": "Technology", "diamond": 0} {"question_id": "395", "site": "stackoverflow", "title": "iOS OpenGL Catch-22: OpenGL background rules and "app snapshot" for App Switcher", "body": "Like many developers, I have an app that uses OpenGL via a `UIView` subclass whose `layerClass:` method returns `[CAEAGLLayer class]`.\n\nNote I am **not** using `GLKit` or `GLKView` or `GLKViewController`\n\nWhen I click Home to put the app into the background, after `applicationDidEnterBackground`, iOS calls my view's `layoutSubviews` twice, with portrait and landscape sizes, trying to generate an \"app snapshot\" as explained here (see \"prepare for the app snapshot\"):\n\n\n\nHow can this possibly work?\n\nThere seems to be a direct contradiction here with the very clear advice on this page (see \"Background Apps May Not Execute Commands on the Graphics Hardware\"):\n\n\n\nthat we must not draw anything with OpenGL after `applicationDidEnterBackground`\n\nIf we don't draw, we cannot generate the snapshots. We must violate one rule or the other.\n\nBut we also want good snapshots in both orientations, so that when the user double-clicks home and goes to the App Switcher, they see reasonable snapshot images.\n\nEven if I temporarily change my code to fully implement the `layoutSubviews` after `applicationDidEnterBackground` by creating an OpenGL surface and drawing (which, contrary to the Apple dox, does not crash), and then I double-click home and look at the snapshot in different orientations, only the snapshot for the orientation I was in before is correct. The other one is a super-ugly nasty re-scaling of the other snapshot. Apple seems to be going through the motions of taking snapshots, but not actually taking them.\n\nI am seeing this behavior on iOS 9.3.2 on an iPad Mini. The behavior doesn't show on most/all iPhone devices since they don't support a landscape App Switcher.\n\n**UPDATE:** the problem also happens, and happens much worse, when using the new iOS 9 \"Slide Over\" multitasking feature and switching the same app between being a normal fullscreen app vs. being an app slid over another app. iOS only seems to capture a snapshot of the last app size, so after using the app at 640px wide and then trying to use the App Switcher to get to the app fullscreen, we see a grotesque pixely out-of-proportion snapshot in the App Switcher and also during the first second of launch. There has got to be some way to fix this!\n\n**UPDATE 2:** I have seen a few iOS apps, which I know to be OpenGL-only apps, where if you use them in portrait, then go back Home and rotate to landscape, then double-click Home, you see the portrait launch image rather than a horrible, distorted, out-of-proportion image like I am seeing. While I would prefer to render snapshot images, I would even be happy to see the launch image. But the option everyone mentions, `ignoreSnapshotOnNextApplicationLaunch`, **does not work** because it only affects what you see at actual app launch time, not what is seen in the App Switcher when you double-click Home, and for many on StackOverflow it actually didn't even work at all (not even at launch time).\n\nHow do we get around this Catch-22?\n\nThis StackOverflow thread (unlike me, the OP here uses GLKit but the symptom is the same): \n\n[iOS OpenGL ES screen rotation while background apps bar visible](https://stackoverflow.com/questions/13954994/ios-opengl-es-screen-rotation-while-background-apps-bar-visible)\n\nconfirms that some OpenGL apps on iOS **are** able to have proper preview images in the Home double-press app switcher for both orientations. How do they do it?\n\nHow can I get proper snapshots shown in both orientations in the App Switcher?\n\nHere is a log of `AppDelegate` (appdel), `ViewController` (eaglc) and `View` (eaglv) calls that come from iOS at the time that I click the Home button once to exit the app. You can see the attempts at snapshotting that come well after `didEnterBackground`:\n \n \n + 189.57ms appdel appWillResignActive\n + 0.74ms appdel appWillResignActive between_view_os_callbacks 0\n + 4.11ms appdel appWillResignActive between_view_os_callbacks 0 done\n + 0.82ms appdel appWillResignActive activation_changed\n + 1.50ms appdel appWillResignActive activation_changed done\n + 0.47ms appdel appWillResignActive between_view_os_callbacks 1\n + 2.68ms drawing rect [(144,1418)+(2,66)] (0 left)\n + 44.28ms swap_buffers glFlush()\n + 6.16ms swap_buffers presentRenderBuffer\n + 9.01ms appdel appWillResignActive between_view_os_callbacks 1 done\n + 0.61ms appdel save_state\n ..app saving data, no OpenGL here..\n + 0.49ms appdel save_state calling glFinish\n + 0.34ms appdel save_state done\n + 0.25ms appdel appWillResignActive done\n + 492.72ms appdel applicationDidEnterBackground\n + 0.56ms appdel save_state\n ..app saving data, no OpenGL here..\n + 0.65ms appdel save_state calling glFinish\n + 0.54ms appdel save_state done\n + 0.65ms eaglv let_go_of_frame_buffer_render_buffer\n app drops OpenGL frame_buffer and render_buffer here\n + 1.10ms appdel applicationDidEnterBackground done\n \n Now we are not supposed to do OpenGL, BUT...\n \n + 6.30ms eaglc supportedInterfaceOrientations\n + 5.74ms about_to_sleep between_view_os_callbacks\n + 1.30ms SKIPPING between_view_os_callbacks cuz app in background\n + 0.66ms about_to_sleep between_view_os_callbacks done\n + 135.85ms eaglc willRotateToInterfaceOrientation\n + 2.49ms appdel willChangeStatusBarFrame new=0,0 768x20\n + 3.21ms appdel didChangeStatusBarFrame old=0,0 1024x20\n \n we get a portrait layoutSubviews....\n \n + 1.26ms eaglv layoutSubviews (initted=1, have_fbrb=0)\n + 1.80ms eaglv assure_frame_buffer_render_buffer\n + 0.95ms eaglv assure_fbrb scale ios=2 eaglv=2\n + 0.90ms eaglv assure_fbrb (frame=1536,2048)\n + 1.04ms eaglv assure_fbrb (layer frame=1536,2048)\n + 0.92ms eaglv assure_fbrb in bg: will make fbrb later\n + 0.96ms eaglv layoutSubviews done\n + 3.11ms eaglc didRotateFromInterfaceOrientation\n + 149.07ms eaglc willRotateToInterfaceOrientation\n + 1.99ms appdel willChangeStatusBarFrame new=0,0 1024x20\n + 2.35ms appdel didChangeStatusBarFrame old=0,0 768x20\n \n then a landscape layoutSubviews...\n \n + 1.91ms eaglv layoutSubviews (initted=1, have_fbrb=0)\n + 1.09ms eaglv assure_frame_buffer_render_buffer\n + 0.91ms eaglv assure_fbrb scale ios=2 eaglv=2\n + 1.65ms eaglv assure_fbrb (frame=2048,1536)\n + 0.92ms eaglv assure_fbrb (layer frame=2048,1536)\n + 0.93ms eaglv assure_fbrb in bg: will make fbrb later\n + 0.83ms eaglv layoutSubviews done\n + 2.79ms eaglc didRotateFromInterfaceOrientation\n \n and, adding insult to injury, we get this log message:\n \n Snapshotting a view that has not been rendered results in an empty snapshot. Ensure your view has been rendered at least once before snapshotting or snapshot after screen updates.\n \n", "link": "https://stackoverflow.com/questions/37602075/ios-opengl-catch-22-opengl-background-rules-and-app-snapshot-for-app-switcher", "tags": ["ios", "objective-c", "ipad", "opengl-es", "layoutsubviews"], "votes": 10, "creation_date": "2016-06-02T14:04:01", "comments": ["perhaps you can try to allow only portrait mode when entering background, so that it at least wouldn't generate distorted snapshot ... but i think this is a good question, and hope somebody else could offer some insights", "I do update my UI before returning from applicationDidEnterBackground, however the problem is that the two layoutSubviews calls which represent my only chance to draw the state of the app in the other orientation (portrait or landscape, whichever one it is) both come AFTER applicationDidEnterBackground, when Apple says it is forbidden to draw. Catch-22! So even if Apple were using my app state that I leave after applicationDidEnterBackground, it's not possible that Apple will know how to draw the app in the other orientation. And so I see distorted ugliness in the App Switcher.", "Are you updating your UI / drawing with GLES in layoutSubviews:? Apple states that you need to do that \"Before returning from your applicationDidEnterBackground: method\", which means you need to do UI updates in applicationDidEnterBackground:"], "comment_count": 3, "category": "Technology", "diamond": 0} {"question_id": "396", "site": "stackoverflow", "title": "How to calculate the checksum in an XFA form", "body": "When you save an XFA form (XFA = XML Forms Architecture) using Adobe software, a checksum attribute is added to the form element. This checksum appears to be a SHA-1 digest, but it's unclear as to what is actually fed to the hash. Does anyone have any idea as to how this is generated? This value is needed by Adobe Acrobat to validate what's actually in the form's XML data, but when I create a hash of the XML that is being fed to the form, Adobe Acrobat doesn't accept it. This checksum attribute isn't documented in the XFA specification, so I would really appreciate it, if somebody could:\n\n 1. Confirm that the value is actually a hash created using the SHA-1 hashing algorithm?\n 2. Explain which data should be used to create this hash.\n\n\n", "link": "https://stackoverflow.com/questions/27470442/how-to-calculate-the-checksum-in-an-xfa-form", "tags": ["xml", "forms", "pdf", "hash", "xfa"], "votes": 10, "creation_date": "2014-12-14T06:46:23", "comments": ["It seems like iText once had an implementation of this, but I can't find it in the more recent versions of their software: api.itextpdf.com/pdfXFA/java/2.0.2/index.html?com/itextpdf/t‌​ool/…", "@BrunoLowagie I am working on populate data into \"form\" - xfa.org/schema/xfa-form/2.8. But it doesn't affect any. So, without checksum, we cannot do it?", "@Setasign I don't have any answer yet. Foxit must have reverse engineered it...", "Bruno, do you have any news on that issue? Or is it still in the hands of Adobe? Foxit, for example, is able to create this hash..."], "comment_count": 4, "category": "Technology", "diamond": 0} {"question_id": "397", "site": "tex", "title": "Listings package and forcing inclusion of empty lines at end of line ranges?", "body": "By default, the listings package suppresses empty lines at the end of the file, and this behavior can be switched off with the option `showlines`.\n\nHowever, listings also suppresses empty lines at the end of each range in a `linerange`. Consider the following Java source file:\n \n \n public class Example implements StringHandler {\n \n /** \n * Prints the given string.\n * \n * @param s the given string\n */ \n @Override\n public void handle(String s) {\n System.out.println(s);\n }\n \n }\n \n\nAnd this is the LaTeX source file `test.tex`:\n \n \n \\documentclass[a4paper,landscape]{slides}\n \\usepackage{color,listings,courier}\n \n \\lstset{language=Java,%\n basicstyle=\\ttfamily,%\n numbers=left,%\n numberstyle=\\tiny,%\n commentstyle=\\color{blue}\\itshape}\n \n \\begin{document}\n \n % Suppress complete javadoc and @Override (to avoid cluttering the slide)\n \\begin{slide}\n \\lstinputlisting[linerange={1-2,9-13}]{Example.java}\n \\end{slide}\n \n \\end{document}\n \n\nThe expected output is:\n \n \n public class Example implements StringHandler {\n \n public void handle(String s) {\n System.out.println(s);\n }\n \n }\n \n\nBut instead, I get\n \n \n public class Example implements StringHandler {\n public void handle(String s) {\n System.out.println(s);\n }\n \n }\n \n\nLine 2 (which is empty) is not included in the output, being at the end of a line range. (I am using listings v1.5; but v1.3 behaves similarly.)\n\nIt turns out that empty lines at the _beginning_ of a line range (consisting of more than one line) are included. But that is of no help in this situation.\n\nHow can I force the inclusion of these embedded empty lines? This is useful to make the listing more readable.\n\nThe options `showlines` and `emptylines` do not affect this behavior.\n", "link": "https://tex.stackexchange.com/questions/233445/listings-package-and-forcing-inclusion-of-empty-lines-at-end-of-line-ranges", "tags": ["listings"], "votes": 7, "creation_date": "2015-03-16T09:22:19", "comments": ["@user2768: No, I never a solution.", "Did you ever find a solution?"], "comment_count": 2, "category": "Technology", "diamond": 0} {"question_id": "398", "site": "tex", "title": "Glossary style with tabularray", "body": "I am currently migrating from xltabular to tabularray. As part of that I want to adapt my custom glossary style for List of Symbols and List of Abbreviations (same style).\n\nThis is my old code:\n \n \n % tables\n \\usepackage{booktabs}\n \\usepackage{multirow}\n \\usepackage{xltabular}\n \\usepackage{tabularray}\n \\addto\\captionsngerman{\n \\DefTblrTemplate{contfoot-text}{default}{Fortsetzung auf der n\\\"achsten Seite}\n \\DefTblrTemplate{conthead-text}{default}{(fortgesetzt)}\n }\n \n % glossaries\n \\usepackage[\n abbreviations,\n nonumberlist,\n record,\n symbols\n ]{glossaries-extra}\n \\GlsXtrLoadResources[\n src=glossaries,\n not-match={entrytype=symbol}\n ]\n \\GlsXtrLoadResources[\n selection=all,\n src=glossaries,\n type=symbols,\n match={entrytype=symbol}\n ]\n \\newglossarystyle{customlong}{\n \\setglossarystyle{long}\n \\renewenvironment{theglossary}{\\xltabular{\\textwidth}{llX}}{\\endxltabular}\n \\renewcommand{\\glossentry}[2]{\n \\glsentryitem{##1}\\textbf{\\glstarget{##1}{\\glossentryname{##1}}} & \\multicolumn{2}{X}{\\glossentrydesc{##1}}\\\\\n }\n \\renewcommand{\\subglossentry}[3]{\n & \\glssubentryitem{##2}\\glstarget{##2}{\\strut} & \\glossentrydesc{##2}\\\\\n }\n \\ifglsnogroupskip\n \\renewcommand*{\\glsgroupskip}{}%\n \\else\n \\renewcommand*{\\glsgroupskip}{\\\\}%\n \\fi\n }\n \n\nI would like to get rid of:\n \n \n \\usepackage{booktabs}\n \\usepackage{multirow}\n \\usepackage{xltabular}\n \n\nTherefore, I need to adapt `\\newglossarystyle{customlong}{...}`. This is my current state:\n \n \n \\newglossarystyle{customlong}{\n \\setglossarystyle{long}\n \\renewenvironment{theglossary}{\n \\begin{longtblr}[\n entry=none,\n label=none\n ]{\n colspec={llX},\n hspan=minimal,\n stretch=1,\n width=\\textwidth\n }\n }{\\end{longtblr}}\n \\renewcommand{\\glossentry}[2]{\n \\glsentryitem{##1}\\textbf{\\glstarget{##1}{\\glossentryname{##1}}} & \\SetCell[c=2]{l}\\glossentrydesc{##1}\\\\\n }\n \\renewcommand{\\subglossentry}[3]{\n & \\glssubentryitem{##2}\\glstarget{##2}{\\strut} & \\glossentrydesc{##2}\\\\\n }\n \\ifglsnogroupskip\n \\renewcommand*{\\glsgroupskip}{}%\n \\else\n \\renewcommand*{\\glsgroupskip}{\\\\}%\n \\fi\n }\n \n\nCompilation fails with 'Misplaced alignment tab character &.'. When I replace `&` with `\\&` in `\\renewcommand{\\glossentry}[2]{...}` and `\\renewcommand{\\subglossentry}[3]{...}`, the compilation is successful but the output obviously contains '&'.\n\nI assume that this error is related to the following content of the package documentation of tabularray:\n\n> \"In contrast to traditional tabular environment, tabularray environments need to see every & and \\ when splitting the table body with l3regex. And you can not put cell text inside any table command defined with \\NewTableCommand. But you could use outer key expand to make tabularray expand every occurrence of a specified macro once before splitting the table body. Note that you can not expand a command defined with \\NewDocumentCommand.\" (, p. 30)\n\nThis would mean that I need to expand `\\renewcommand{\\glossentry}[2]{...}` and `\\renewcommand{\\subglossentry}[3]{...}`. I have no idea how to achieve that. Can someone help?\n\nThanks in advance.\n\n# Edit 1: M(N)WE\n\nThis is my current code which prints '&'. Some unrelated contents have been removed. I build with `latexmk`.\n\n## main.tex\n \n \n \\documentclass{scrbook}\n \n \\usepackage{tabularray}\n \n \\usepackage[\n abbreviations,\n nonumberlist,\n record,\n symbols\n ]{glossaries-extra}\n \\GlsXtrLoadResources[\n src=glossaries,\n not-match={entrytype=symbol}\n ]\n \\GlsXtrLoadResources[\n selection=all,\n src=glossaries,\n type=symbols,\n match={entrytype=symbol}\n ]\n \\newglossarystyle{customlong}{\n \\setglossarystyle{long}\n \\renewenvironment{theglossary}{\n \\begin{longtblr}[\n entry=none,\n label=none\n ]{\n colspec={llX},\n hspan=minimal,\n stretch=1,\n width=\\textwidth\n }\n }{\\end{longtblr}}\n \\renewcommand{\\glossentry}[2]{\n \\glsentryitem{##1}\\textbf{\\glstarget{##1}{\\glossentryname{##1}}} \\& \\SetCell[c=2]{l}\\glossentrydesc{##1}\\\\\n }\n \\renewcommand{\\subglossentry}[3]{\n \\& \\glssubentryitem{##2}\\glstarget{##2}{\\strut} \\& \\glossentrydesc{##2}\\\\\n }\n \\ifglsnogroupskip\n \\renewcommand*{\\glsgroupskip}{}%\n \\else\n \\renewcommand*{\\glsgroupskip}{\\\\}%\n \\fi\n }\n \\newglossarystyle{customindex}{\n \\setglossarystyle{index}\n \\renewcommand{\\glstreeitem}{\\parindent0pt\\par}\n \\renewcommand{\\glstreepredesc}{\\par\\glstreeitem\\parindent40pt\\hangindent40pt}\n }\n \n \\begin{document}\n \\frontmatter\n \\printunsrtabbreviations[style=customlong]\n \\printunsrtsymbols[style=customlong]\n \n \\mainmatter\n \n \\appendix\n \n \\backmatter\n \\printunsrtglossary[style=customindex]\n \\end{document}\n \n\n## glossaries.bib\n \n \n @entry{gls-uml,\n name = {Unified Modeling Language},\n description = {\\enquote{A specification defining a graphical language for visualizing, specifying, constructing, and documenting the artifacts of distributed object systems.}\\footnote{\\url{https://www.omg.org/spec/UML}, aufgerufen am 20.02.2023}.}\n }\n \n @abbreviation{auv,\n description = {\\gls{gls-auv}},\n short = {AUV},\n long = {Autonomous Underwater Vehicle}\n }\n \n @symbol{v-desired,\n name = {\\ensuremath{\\overrightarrow{v_{desired}}}},\n description = {Wunschgeschwindigkeit}\n }\n \n\n## .latexmkrc\n \n \n @default_files = ('main');\n \n $pdf_mode = 4;\n $dvi_mode = 0;\n $postscript_mode = 0;\n \n $lualatex = 'lualatex -synctex=1 -interaction=nonstopmode %O %S';\n \n push @generated_exts, 'glstex', 'glg';\n \n $clean_ext .= ' %R.bbl %R.glstex %R.lol %R.run.xml %R-1.glstex ';\n \n add_cus_dep( 'aux', 'glstex', 0, 'run_bib2gls' );\n \n sub run_bib2gls {\n if ($silent) {\n my $ret = system \"bib2gls --silent --group $_[0]\";\n }\n else {\n my $ret = system \"bib2gls --group $_[0]\";\n }\n \n my ( $base, $path ) = fileparse( $_[0] );\n if ( $path && -e \"$base.glstex\" ) {\n rename \"$base.glstex\", \"$path$base.glstex\";\n }\n \n # Analyze log file.\n local *LOG;\n $LOG = \"$_[0].glg\";\n if ( !$ret && -e $LOG ) {\n open LOG, \"<$LOG\";\n while () {\n if (/^Reading (.*\\.bib)\\s$/) {\n rdb_ensure_file( $rule, $1 );\n }\n }\n close LOG;\n }\n return $ret;\n }\n \n", "link": "https://tex.stackexchange.com/questions/678834/glossary-style-with-tabularray", "tags": ["glossaries", "tabularray", "glossaries-extra", "xltabular"], "votes": 6, "creation_date": "2023-03-09T01:12:54", "comments": ["I am having the same Problem... I think Nicola Talbot could say if glossaries ist able to handle this or not.", "Thank you. See my edit for mwe.", "Welcome to TeX.SX! Can you please add a complete minimal working example starting with \\documentclass and ending with \\end{document} instead of showing only some code snippets. This would help us to test you code, reproduce the issue and find and test a suggestion."], "comment_count": 3, "category": "Technology", "diamond": 0} {"question_id": "399", "site": "ai", "title": "Is the Bellman equation that uses sampling weighted by the Q values (instead of max) a contraction?", "body": "It is proved that the Bellman update is a contraction (1).\n\n**Here is the Bellman update that is used for Q-Learning:**\n\n$$Q_{t+1}(s, a) = Q_{t}(s, a) + \\alpha*(r(s, a, s') + \\gamma \\max_{a^*} (Q_{t}(s', a^*)) - Q_t(s,a)) \\tag{1} \\label{1}$$\n\nThe proof of (\\ref{1}) being contraction comes from one of the facts (the relevant one for the question) that max operation is non expansive; that is:\n\n$$\\lvert \\max_a f(a)- \\max_a g(a) \\rvert \\leq \\max_a \\lvert f(a) - g(a) \\rvert \\tag{2}\\label{2}$$\n\nThis is also proved in a lot of places and it is pretty intuitive.\n\n**Consider the following Bellman update:**\n\n$$ Q_{t+1}(s, a) = Q_{t}(s, a) + \\alpha*(r(s, a, s') + \\gamma SAMPLE_{a^*} (Q_{t}(s', a^*)) - Q_t(s,a)) \\tag{3}\\label{3}$$\n\nwhere $SAMPLE_a(Q(s, a))$ samples an action with respect to the Q values (weighted by their Q values) of each action in that state.\n\n**Is this new Bellman operation still a contraction?**\n\nIs the SAMPLE operation non-expansive? It is, of course, possible to generate samples that will not satisfy equation (\\ref{2}). I ask **is it non-expansive in expectation?**\n\nMy approach is:\n\n$$\\lvert\\,\\mathbb{E}_{a \\sim Q}[f(a)] - \\mathbb{E}_{a \\sim Q}[g(a)]\\, \\rvert \\leq \\,\\,\\mathbb{E}_{a \\sim Q}\\lvert\\,\\,[f(a) - g(a)]\\,\\,\\rvert \\tag{4} \\label{4} $$\n\nEquivalently:\n\n$$\\lvert\\,\\mathbb{E}_{a \\sim Q}[f(a) - g(a)] \\, \\rvert \\leq \\,\\,\\mathbb{E}_{a \\sim Q}\\lvert\\,\\,[f(a) - g(a)]\\,\\,\\rvert$$\n\n(\\ref{4}) is true since:\n\n$$\\lvert\\,\\mathbb{E}[X] \\, \\rvert \\leq \\,\\,\\mathbb{E} \\,\\,\\lvert\\,\\,[X]\\,\\,\\rvert $$\n\n**But, I am not sure if proving (\\ref{4}) proves the theorem. Do you think that this is a legit proof that (\\ref{3}) is a contraction.**\n\n(If so; this would mean that stochastic policy q learning theoretically converges and we can have stochastic policies with regular q learning; and this is why I am interested.)\n\nBoth intuitive answers and mathematical proofs are welcome.\n", "link": "https://ai.stackexchange.com/questions/22642/is-the-bellman-equation-that-uses-sampling-weighted-by-the-q-values-instead-of", "tags": ["reinforcement-learning", "q-learning", "proofs", "convergence", "bellman-equations"], "votes": 8, "creation_date": "2020-07-23T10:32:14", "comments": ["Your question in not very clear to me. Since $f(a)$ and $g(a)$ are not clear to me. The formulas are intuitive and individually correct, I am not sure whether arriving at those intuitive forms you have mentioned, so easy. Check this link for example: users.isr.ist.utl.pt/~mtjspaan/readingGroup/ProofQlearning.pdf As a side not I do not think proving convergence is so easy. There is a topic called Concentration Inequalities which have to be studied to prove convergence. I think you can use this to prove your theorems.", "(1) is a bellman update; it is a copy paste error that rhs has t+1 (sorry about that) thanks for noticing; I fixed the error now."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "400", "site": "economics", "title": "Dynamic Bertrand competition when players take turns", "body": "Consider the following game:\n\n * There are two players, $i\\in\\\\{1,2\\\\}$\n * Time is discrete and runs to infinity during periods $t=\\\\{1,2,\\ldots\\\\}$\n * At eat point in time, players have a price $p_i(t)\\in\\mathbb{R}_+$\n * Initialise the game with $p_1=p_2=p(0)$.\n * In odd-numbered periods, player 1 can change his price to any $p_1\\in\\mathbb{R}_+$. Player 2 cannot change his price.\n * In even-numbered periods, player 2 can change his price to any $p_2\\in\\mathbb{R}_+$. Player 1 cannot change his price.\n * For each price, there is a demand $D(p)$, which is a decreasing function. The firm with the lowest price at the end of each period captures the whole demand, receiving payoff $D(p_i)p_i$ for that period. The firm with the highest price receives a payoff of zero for the period. If prices are equal then each firm gets a payoff of $pD(p)/2$.\n * Players discount the future at common rate $0\\leq\\delta\\leq1$ so a payoff of $\\pi$ that occurs $t$ periods in the future has present value $\\pi\\delta^t$.\n * Write $p^*$ for the monopoly price that maximises $D(p)p$.\n\n\n\nThe question is: can we identify a complete characterization of the set of Nash equilibria of this game?\n\nNote that a valid strategy is a complete contingent plan, which specifies the choice of action for any history of the game.\n\nThis question follows the discussion here: [What determines the outcome of a price war, and why isn't that outcome reached instantaneously?](https://economics.stackexchange.com/questions/8398/what-determines-the-outcome-of-a-price-war-and-why-isnt-that-outcome-reached-i?noredirect=1#comment10737_8398)\n", "link": "https://economics.stackexchange.com/questions/8473/dynamic-bertrand-competition-when-players-take-turns", "tags": ["game-theory", "oligopoly", "repeated-games"], "votes": 7, "creation_date": "2015-09-30T11:45:29", "comments": ["Is the Maskin-Tirole ECTA 1988 (the second paper, oligopoly) what you are looking for?", "@denesp You're right. The more I think about it, the more the premise of the question doesn't make much sense.", "Yes but the grim strategies are just a subset of all possible strategies. And what you describe is actually just a subset of all grim strategies, the condition to go grim can be quite ridiculous: \"Play $p$ unless the other players last six prices were not $\\pi, \\pi, 2, 2, \\sqrt{2}, \\sqrt{2}$ in which case play 0.\" This strategy can be part of an equilibrium (given some restrictions on $\\delta$). So a characterization of the kind you describe would not be a characterization of all equilibria.", "@densep By characterization I mean a description of every Nash equilibrium. As you note, there are infinitely many equilibria just considering grim trigger strategies, so it is impossible to individually enumerate them all. But those equilibria can be concisely 'characterised' as \"play $p$ unless someone ever played a $p'\\neq p$ in which case play 0\", which should work for $\\delta$ large enough.", "The word characterization is problematic. Saying that the strategies you are looking for are the strategies that constitute a Nash-equilibrium would be a tautology, but it is a characterization nonetheless. Such a tautology can be made less trivial by giving an alternate definition of Nash-equilibrium. I am not sure this is what you are looking for but you could say something about the range of payoffs: It seems to me that literally any payoff vector between the competitive solution and the cooperative solution is attainable in equilibrium using grim strategies."], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "401", "site": "dsp", "title": "a moving average relation that I can't find a counter-example to", "body": "Hi: I'm new to this list but I had a question that's not exactly related to dsp but maybe in a slight way and I didn't know where else to send it. So, here goes and my apologies if this is totally the wrong place to send it. \n\nThe following describes the background as far as number generation.\n\nSuppose have a series of positive numbers coming in sequentially ( regularly spaced ) that are generated by some invisible machine or oracle. no distribution is assumed for these numbers ( except that they are positive ) and they are generated at t1, ... tn. they are labelled as $x_1, x_2, ... x_n, x_{n+1}$. at each time $t$, the moving average with window size $n$ of the log of the $x_i$ is calculated and is labelled $mavelogx_t$. \n\nthe parameters of are $n$ and $d$. $n$ is the length of the moving average calculated at each time $t$. $d$ is positive and will be explained below. \n\nThe following describes the game for a player who decides to play given the above ( assuming that atleast n numbers have been generated by now so a moving average can be calculated ). It doesn't cost anything to play this game.\n\nThe player of the game sees a new number at time $tstar$, $x_{tstar}$, and if $log(x_{tstar})$ is $d$ less than the current moving average, $mavelogx_{tstar}$, then $x_{tstar}$ is marked as the player's \"initial\" number. like 36.83 or whatever. the game works so the next time $log(x_t)$ crosses the moving average, $mavelogx_{t}$, from below at say $t = tstarstar$, then $x_{tstarstar}$ is marked as the player's ending number and the game ends.\n\nThe game rule is such that at $t = tstarstar$, the game ends and the player receives $winnum$ = $log(tstarstar) - log(tstar)$ ( winnum could be negative in which the player of course pays ). Of course it could happen that the player never get an initial number or that the player gets an initial number and never gets an ending number but let's make the assumption that the player plays and the game both starts and ends.\n\nNow, I claim ( and think I proved ) the following two things. ( really one because it's an if and only if ) are true for the player playing the game.\n\nA) if $(tstarstar - tstar)$ turns out to be less than $n$, then $winnum$ has to be positive. similarly if $(tstarstar - tstar)$ turns out to be greater than or equal to $n$, then $winnum$ has to be non-positive.\n\nand conversely\n\nB) if $winnum$ is positive, then $(tstarstar - tstar)$ has to be less than $n$. similarly, if $winnum$ is not positive , then $(tstarstar - tstar)$ has to be greater than or equal to $n$.\n\nI could send the proof to anyone who is interested but my questions were :\n\nA) does anyone have a counter-example to disprove what I'm saying in A) or B).\n\nB) if A) and B) are true ( my proof is correct ), then is this a well known result in the literature ( any literature. doesn't have to be DSP ) ?\n\nthanks a lot. Mark\n", "link": "https://dsp.stackexchange.com/questions/9915/a-moving-average-relation-that-i-cant-find-a-counter-example-to", "tags": ["algorithms"], "votes": 5, "creation_date": "2013-07-12T06:20:42", "comments": ["also, a different explanation of the same problem is in the article at arxiv.org/abs/1212.4890. it's very long but might be clearer. thanks again.", "sorry jason. my mistake. it should read that $winnum = log(x_{tstarstar})−log(x_{tstar})$. thank you for pointing that out.", "The above description of your game is very difficult to understand. I would try to clarify it with a more succinct explanation. For instance, I don't see how $t^{**}$ could ever be less than $t^*$ according to your description, so $log(t^{**}) - log(t^*)$ would always be nonnegative."], "comment_count": 3, "category": "Technology", "diamond": 0} {"question_id": "402", "site": "mattermodeling", "title": "Contribution in action from collisions for a gas?", "body": "So if I write the [action](https://en.wikipedia.org/wiki/Action_\\(physics\\)) for a gas:\n\n$$ S = \\int (T -U) dt \\tag{1}$$ where $T$ is the kinetic energy and $U$ is potential energy.\n\nI suspect there is a constant term (which does not affect the equations of motion) for the gas (in classical mechanics). I was hoping to verify my ~~[paper's](https://drive.google.com/file/d/1deuIYhorbMBnIRoblhe_0q1P4CkIx3DD/view?usp=sharing) idea ~~ (There is now an updated version see [here](https://drive.google.com/file/d/1K95v8crlBWsf2_3kFiEbRDMNa5xVBdOV/view?usp=sharing)) up to equation $(19)$.The summary of ideas section wise are as follows:\n\n$(1.1)$ We consider an arbitrary potential which $0$ everywhere until both the gas molecules collide.\n\n$(1.2)$ Assuming a non-ideal collision (finite collision time) I get a non-zero action\n\n$(1.3)$ We use Newton's second law to find the finite collision time and the average force is considered as pressure.\n\n$(1.4)$ The deformed face of the molecule has an area which is also considered.\n\n$(1.5)$ Since we are considering $2$ spheres colliding which is a function of their relative velocity we can find the average momentum change assuming the collision is not a special event.\n\n$(1.6)$ The collision per unit 4-volume is mentioned as derived by Einstein.\n\nCombining all the above subsections we derive a constant action term:\n\n$$S_c = \\int \\frac{N R \\tilde V'(2R)}{\\pi} \\frac{ \\langle \\mu \\rangle \\lambda }{k_B } dt \\tag{2}$$\n\nwhere $N$ is the number of particles, $R$ is the radius of the particle, $ \\tilde V'(x)$ is the spatial derivative potential and $ \\frac{ \\langle \\mu \\rangle \\lambda }{k_B } $ are constants.\n\nIs this expression correct?\n", "link": "https://mattermodeling.stackexchange.com/questions/8498/contribution-in-action-from-collisions-for-a-gas", "tags": ["molecular-dynamics", "thermodynamics"], "votes": 8, "creation_date": "2022-01-07T07:08:23", "comments": ["(by linking to your paper you are now Less Anonymous :)"], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "403", "site": "puzzling", "title": "REALLY important days in history", "body": "@Z. Dailey asks a mathematical puzzle [Important days in history... and to come!](https://puzzling.stackexchange.com/questions/29503/important-days-in-history-and-to-come) with a series of dates in years 1973, 1978, 1983, 1992, 2005.\n\nMy puzzle asks the question: What is the previous date in my series? Explain!!\n \n \n .......... .., .....\n June 22, 1973 \n .......... .., .....\n June 16, 1978\n June 11, 1983\n March 25, 1992\n April 14, 2005\n \n\nHint: \n\n> \"rudder as a male swan\"\n", "link": "https://puzzling.stackexchange.com/questions/29533/really-important-days-in-history", "tags": ["knowledge", "pattern", "number-sequence", "sequence", "history"], "votes": 22, "creation_date": "2016-03-24T02:41:27", "comments": ["I'm assuming it's not January 3, 1972 and April 1, 1975...", "@Haobin Is the hint in any way related to the Lufthansa logo?", "Nothing failed about it. Also it's valid from a math p.o.v. ;check the MM/DD also.", "@Overmind I would question your failed conclusion.", "Logic dictates there's a high chance that the year is 1969, the fake moon landing year.", "Apparently ebhtu nf n pbo is a phrase", "A male swan being a rot13(pbo)"], "comment_count": 7, "category": "Culture & Recreation", "diamond": 1} {"question_id": "404", "site": "puzzling", "title": "Can you find the cipher that I used?", "body": "I have encoded a phrase, using a cipher that I invented. But I don't want you to decode it, rather I want you to figure out how my cipher works.\n\nThis is the phrase that I encrypted\n\n> Find the connection between this phrase and the output. You don't even have to do any solving, just figure out how this was encrypted\n\nThis is the key that I used\n\n> A Blemished Orange\n\nThis is the encrypted phrase\n\n> 1YaF7\\B*,IdM#mI7\\$iEB8lbnOKl5o.f9,J+772V2??16;XV2[)nhOov4elOT@h?Iw7nAO-2Nt2*,Sdq7XW&lBW0%^>u-C>,0)XSV&(627K,lHw7XeL8\\B$PSj@[7n?PBbYbm@\n\nThese are important, but you'll have to figure out how\n\n> 20\n> \n> 9100, 9129\n> \n> Python\n\nThat's everything, see what you can find!\n", "link": "https://puzzling.stackexchange.com/questions/96886/can-you-find-the-cipher-that-i-used", "tags": ["cipher"], "votes": 11, "creation_date": "2020-04-09T08:43:03", "comments": ["That's much better, nice job! +1."], "comment_count": 1, "category": "Culture & Recreation", "diamond": 0} {"question_id": "405", "site": "puzzling", "title": "A Glutton and a Split Check", "body": "**N** , **T** , **A** , **B** are arbitrary constants in this problem. I have only definitively solved the **N=1** subproblem, so, if the general problem cannot be solved, I will be satisfied to mark as accepted a solution to this subproblem. I may put a bounty out for the general solution before that.\n\n## Problem Statement\n\nYou and **N** friends are seated at a restaurant, where you have agreed to equally split the check for \\$**T**. The dining process consists of a waiter going around the table, asking each guest in turn to order an entrée costing any amount of \\$1, \\$2, ..., up to \\$**A** (you cannot skip an entrée). The meal is of course over once the \\$**T** limit has been reached (it cannot be exceeded).\n\nThe vain glutton that you are, you want to:\n\n 1. eat more than you pay for (that is, \\$**T/(N+1)**), but also;\n 2. not be exposed as a glutton, which in this case consists of eating more than \\$**B** extra than the next most expensive eater. If (1) cannot be achieved, you will nonetheless consider it a victory to see someone _else_ exposed in such a fashion.\n\n\n\nAssuming you are first to order, for what values of **N** , **T** , **A** , **B** can you guarantee victory, no matter how your friends order? Keep in mind, they may conspire to make it difficult for you...\n\n## Example\n\nIf **(N, T, A, B) = (1, 21, 13, 10)** , your gluttony might tempt you to order the most expensive \\$13 entrée right off the bat. This would be a mistake, because your friend could thereafter order only \\$1 entrées and expose you as a glutton. Similarly, you shouldn't order an \\$11 or \\$12 entrée. Yet if you order a $10 entrée or less, your friend may order an \\$11 entrée, and you wind up paying for more than you eat while he cannot be exposed. So, there is no victory in this example configuration.\n", "link": "https://puzzling.stackexchange.com/questions/111330/a-glutton-and-a-split-check", "tags": ["mathematics", "strategy", "game-theory"], "votes": 9, "creation_date": "2021-08-16T15:49:43", "comments": ["@Feryll Ah sorry, missed that N was your friends, not the number of people at the table. I see now the problem statement clarifies that. Thanks!", "You have N friends, so, there are N+1 people in total who have precommitted to splitting the $T check. What they have not decided in advance is who is ordering what.", "Hmm I just re-read the question and I'm a bit confused. How are they \"equally splitting the check for \\$T\" if by the definition of the rest of the problem some people are ordering food that costs more than others' orders? Is everyone spending \\$T/N regardless of how much they ordered for? If so, then not sure I understand \\$T/(N+1) -- if everyone is spending \\$T/N then aren't you \"eating more than you pay for\" starting from orders worth \\$(T/N)+1 or more, rather than \\$T/(N+1)? Are the parentheses applied incorrectly in the problem statement?", "@Petr Naryshkin He will go around the table as many times as it takes to reach a total check of T, yes.", "If the waiter gets all your orders and the check amount T is not reached (like in your example) does he start another round? If not, do you want to still eat for at least T/(n+1) or only for (total check)/(n+1)?", "Reminds me of this a little bit en.wikipedia.org/wiki/Pirate_game", "Funny, N>1 case seems easier than N=1, because the other's don't have to worry about being exposed (unless b=0). Still, too many cases for me ;-)", "Yes, it says so in the least paragraph of Problem Statement :)", "Are you the first to order?"], "comment_count": 9, "category": "Culture & Recreation", "diamond": 0} {"question_id": "406", "site": "puzzling", "title": "Who wins the game of "Sticky Gomoku" on an infinite board?", "body": "This is a [Gomoku](https://en.wikipedia.org/wiki/Gomoku) like game, played by two players with Go pieces (black and white stones) on an infinite (in all four directions) Go board.\n\nRules\n\n 1. **Four in a row:** players alternate turns placing a stone of their color on an empty intersection. Black plays first. The winner is the first player to form an unbroken chain of four stones horizontally, vertically, or diagonally.\n 2. **Sticky:** except for black's first stone, every stone must be placed adjacent to one of your opponent's.\n\n\n\nQuestion: can black force a win?\n", "link": "https://puzzling.stackexchange.com/questions/112346/who-wins-the-game-of-sticky-gomoku-on-an-infinite-board", "tags": ["game", "board-games"], "votes": 7, "creation_date": "2021-10-28T02:56:16", "comments": ["@Eric I see, thanks for the clarification! I guess it would be very hard to solve, as one doesn't expect to find a finite game tree that totally describes optimal moves.", "@WhatsUp I heard of this game from an internet friend, who allegedly had invented it a few years ago, and had played it with friends since. The conjecture is that optimal play for both players will end in a draw, but no one is sure.", "Looks like a funny game to play with family or friends, but I suspect that a systematic study would simply be an exhaustive search, probably computer-aided. Maybe @Eric can clarify whether there is a clever approach to be expected.", "The only possible endgame I think is to create two 3-in-a-rows with one move, where the move itself and the two fourths are pairwise adjacent (forming a small triangle). Obviously the problem is how to force such a position in the first place...", "This is actually much tricker than I thought. Getting 3-in-a-row is rarely an immediate threat because the blank space is usually not adjacent to the opponent and so not immediately playable.", "@BenjaminWang Sure. Every intersection has eight adjacent neighbors.", "Does \"adjacent\" include diagonally adjacent?"], "comment_count": 7, "category": "Culture & Recreation", "diamond": 0} {"question_id": "407", "site": "puzzling", "title": "Best moves for a 7X7 4 queen board", "body": "There is a 7x7 board and 2 players, each having 2 queens. The queens can move just like the queens in chess, but cannot capture each other. At the beginning, the first player places both the queens on any 2 different squares. Then the second player does the same with his queens. From this point, the players alternate turns, moving both queens each turn. When a queen is moved, the square that was previously occupied is now blocked and cannot be stood on or moved across for the remainder of the game. The first player who is unable to move any of his queens loses!\n\nI am designing a player for it. What will be the strategy for the player who starts first or second? Also, what are best starting moves for any player?\n", "link": "https://puzzling.stackexchange.com/questions/59804/best-moves-for-a-7x7-4-queen-board", "tags": ["board-games"], "votes": 7, "creation_date": "2018-01-25T21:09:16", "comments": ["@SuhailGupta I hope that this answers most of your doubts. [1]:", "didn't read the \"can't move across\" part sry, so I guess they can't.", "@Untitpoi How does it matter?", "If I have two queen in A1 and B1 can I make the queen in A1 move into C1? At the same time I choose to move my queen in B1 in C3! In other word are the moves of the two queen (of the same color) simultaneous or sequential.", "I made a suggested edit to clarify the rules. If I got anything wrong please edit or roll back.", "@Bass No. As the question says the end will be: \"The first player who is unable to move any of his queens loses!\"", "Since \"the queens can move just like the queens in chess\", can they also capture the opposing queens?", "@elias Both of his queens", "It's still unclear for me if a player should move both his queens on his turn, or only one of them. Please clarify!", "@prog_SAHIL The first move could be (3,4) and (5,5) for player 1 and then move for player 2 could be (1,2) and (2,2)", "Can you please give an example for first 3,4 moves. Its hard to understand the rules.", "After the first player has placed his two queens. They alternate their turns", "When will the second player place the queens?"], "comment_count": 13, "category": "Culture & Recreation", "diamond": 0} {"question_id": "408", "site": "puzzling", "title": "What is a 255b Word™?", "body": "This is in the spirit of the [What is a Word/Phrase™](https://puzzling.stackexchange.com/search?q=%22what+is+a%22+word+is%3Aquestion) series started by [JLee](https://puzzling.stackexchange.com/users/463) with a special brand of [Phrase™](https://puzzling.stackexchange.com/search?q=%22Phrase%22%20user%3A463%20is%3Aquestion) and [Word™](https://puzzling.stackexchange.com/search?q=%22What+is+%22+%22Word%22+-missing+user%3A463+is%3Aquestion) puzzles.\n\n* * *\n\nIf a word conforms to a special rule, I call it a **255b Word™**. \nUse the following examples below to find the rule.\n\n$$ % set Title text. (spaces around the text ARE important; do not remove.) % increase Pad value only if your entries are longer than the title bar. % \\def\\Pad{\\P{2.5}} \\def\\Title{\\textbf{ 255b }} % \\def\\S#1#2{\\Space{#1}{20px}{#2px}}\\def\\P#1{\\V{#1em}}\\def\\V#1{\\S{#1}{9}} \\def\\T{\\Title\\textbf{Words}^{\\;\\\\!™}\\Pad}\\def\\NT{\\Pad\\textbf{Not}\\T\\ }\\displaystyle \\smash{\\lower{29px}\\bbox[grey]{\\phantom{\\rlap{rubio.2019.05.15}\\S{6px}{0} \\begin{array}{cc}\\Pad\\T&\\NT\\\\\\\\\\end{array}}}}\\atop\\def\\V#1{\\S{#1}{5}} \\begin{array}{|c|c|}\\hline\\Pad\\T&\\NT\\\\\\\\\\hline % \\text{ AMBIGUOUSLY }&\\text{ AMBITIOUSLY }\\\\\\ \\hline \\text{ BOXING }&\\text{ BOXER }\\\\\\ \\hline \\text{ COMPLICATE }&\\text{ COMPLEX }\\\\\\ \\hline \\text{ DECENTRALIZE }&\\text{ DECENTRALIZATION }\\\\\\ \\hline \\text{ ENTERTAINMENT }&\\text{ EXCITEMENT }\\\\\\ \\hline \\text{ FINALLY }&\\text{ FINISHED }\\\\\\ \\hline \\text{ GEOGRAPHY }&\\text{ GEOLOGY }\\\\\\ \\hline \\text{ HENRY }&\\text{ HIGUAIN }\\\\\\ \\hline \\text{ INDICATION }&\\text{ INCLINATION }\\\\\\ \\hline \\text{ JUXTAPOSING }&\\text{ JUXTAPOSITION }\\\\\\ \\hline \\text{ KNOWLEDGE }&\\text{ KNOWLEDGEABLE }\\\\\\ \\hline \\text{ LIGHTNING }&\\text{ LIGHTENING }\\\\\\ \\hline \\end{array}$$\n\nAnd, if you want to analyze, here is a CSV version:\n \n \n 255b Words™,Not 255b Words™\n AMBIGUOUSLY,AMBITIOUSLY\n BOXING,BOXER\n COMPLICATE,COMPlEX\n DECENTRALIZE,DECENTHALIZATION\n ENTERTAINMENT,EXCITEMENT\n FINALLY,FINISHEO\n GEOGRAPHY,GEOLORY\n HENRY,HIGUAIN\n INDICATION,INCLINATION\n JUXTAPOSING,JUXTAPOSITION\n KNOWLEDGE,KNOWLEDgEABLE\n LIGHTNING,LIGHTENING\n \n\nThe puzzle satisfies the series' inbuilt assumption, that each word can be tested for whether it is a 255b Word™ without relying on the other words. These are not the only examples of 255b Words™; many more exist.\n\n**Problem: What is a 255b Word™? And why is the property called 255b?**\n\n* * *\n\n**Hint:**\n\n> This rule is actually quite arbitrary. I don't think you can read my mind without the [steganography](/questions/tagged/steganography \"show questions tagged 'steganography'\") tag.\n", "link": "https://puzzling.stackexchange.com/questions/92184/what-is-a-255b-word", "tags": ["steganography", "computer-puzzle", "word-property"], "votes": 7, "creation_date": "2019-12-22T04:56:24", "comments": ["@Jasen Yes, both are correct.", "are both lists correct?", "@Vepir Yeah, I deliberately misspelled a few letters in CSV. :)", "Is the CSV misspelled? (\"DECENT'H'ALIZATION\" with 'H' in CSV, but with 'R' in the table) Is CSV purposely not fully capitalized? (such as 'g' in \"KNOWLED'g'EABLE\" and 'l' in \"COMP'l'EX\")"], "comment_count": 4, "category": "Culture & Recreation", "diamond": 0} {"question_id": "409", "site": "puzzling", "title": "Need help with a cipher: "poems" with capital letters", "body": "I am trying to figure out a cipher I _think_ I found in a book from 1982 and it's a 36 year old mystery we are trying to solve (I will reveal the book if this turns out to be something so everyone can play along!). The book has several poems in it that has has lines that start with capital letters. There is also several things that should be capitalized (like a name or place) that aren't capitalized and things that randomly made their own line in the puzzle. I thought it was weird so I typed them out and there were double and triple letters. I tried putting them in online puzzle solvers but nothing came up except gibberish. I also thought maybe the spaces meant new words, but nothing!\n\nThere is another post that I saw (that's how I found this site) that was similar using a columnar transposition cipher but I'm not sure if it is the same!\n\nThree of the twelve did not have any double letters and there are several others poems in the book that mean something.\n\n> FCFTTNTNISSPISTWOFIDLTA \n> AFIFOAFHWTGFTN \n> ITNXTIAINWWFYFAAETLI \n> BAIBFBOSFSPAFTSF \n> LTYWTOWCABBOTWAFGT \n> OMCSBSHHFOMEEOWBBBEWFW \n> ATNHEJFSNRIIOTGGTT \n> VMAAFAFAFSSCAAPTBWSTPYOOAT \n> TWNWBAAYSSSELOY \n> IOFEIYCANSITOHVTOFOLAIOTB \n> PIDRTTDWWAIJAATUWOLATBDY \n> WMBACRLBIFFTTIFCFSOBH\n\nThese are the \"extra\" poems:\n\n> TNAESFSSFWBDFAEM \n> WFFEFEGQBCCETHISTFFRDAIGTOLGAATRTPIATRASADBTNHAFCPCEGVEWTFF \n> KTTT \n> NWEGCTFFMTDSTFANATSSNH \n> TTSIBMHAFFWTWOAAFETFF \n> ASYFCTTWWFFFG\n", "link": "https://puzzling.stackexchange.com/questions/53173/need-help-with-a-cipher-poems-with-capital-letters", "tags": ["cipher", "real", "unsolved-mysteries"], "votes": 6, "creation_date": "2017-07-05T14:13:38", "comments": ["thesecret.pbworks.com/w/page/22148585/To%20Do", "@Rubio As far as I can tell no one has come up with this cipher. I believe the only ones that are important are the first letters (besides using the clues in the verse to guide you too) as part of a secret cipher for several reasons.Preiss doesn't give us a way in the book to match up the picture and the verse, only tells us to match them up.There are a ton of combinations that they could be so there has to be a way for us to tell which is which.If you google the verses you will see that things like oz and wonderstone aren't capitalize, but they should be because they are places/people.", "Yes, its real and not gibberish. I will just tell you guys what the book is because I know I can't figure it out. Hopefully someone can and find one of the treasures. In 1982 a guy named Byron Preiss wrote a book called The Secret, a Treasure Hunt and hid 12 casques throughout 12 North American cities. Since 19821, only 2 of the 12 have ever been found (Chicago in 1983 and Cleveland in 2004). The poems are long (all of them put together), otherwise I would post them. If you google the book you can find all the poems and what cities people think they are in. Good luck!", "If this isn't real, forgive my edit", "Can you post the full poem as well? How can we be sure that this isn't just gibberish you pulled?", "How sure are you that the first letters are the only important parts here?"], "comment_count": 6, "category": "Culture & Recreation", "diamond": 0} {"question_id": "410", "site": "puzzling", "title": "Rare but Abundant in Hill and Hole", "body": "> I am uncommon, \n> But a worthy goal. \n> Rare but abundant \n> In hill and hole. \n> Who am I?\n\nYou may already know the answer, so the task is not to answer the riddle per se, but to solve the following 'What is a Word' puzzle, use the words in the puzzle to find another four word phrase (2 from each side), and explain how the phrase provides/validates the answer to the above riddle.\n\n> What is an Uncommon Word?\n> \n> If a word conforms to a special rule, I call it an Uncommon Word. \n> Use the examples below to find the rule.\n\nUncommon Words | Not Uncommon Words \n---|--- \nFEW | LOTS \nMANY | ALL \nRARE | ATYPICAL \nTIE | UNHINGE \nIA | RI \nBAGGED | TOSSED \nDOOMED | PERISHED \nGAPING | BEHOLDING \nADO (2x) | TWICE \nSEAL | REFUSE \nHAVE | LACKED \nROUNDS | TURNS \nSPITS | SPITES \n \n> These are not the only examples of Uncommon Words, many more exist.\n\nHint:\n\n> Teepee is an extremely common word. \n> Quitting is an extremely uncommon word. \n> Puddle is a quite uncommon word and also quite relevant. What property makes something uncommon?\n\nHint 2:\n\n> I updated a word so that all the uncommon words are in an American scrabble dictionary rather than simply being American-English words, to help solvers identify them. Also added a very useful tag.\n", "link": "https://puzzling.stackexchange.com/questions/109321/rare-but-abundant-in-hill-and-hole", "tags": ["riddle", "word", "pattern", "word-property", "letters"], "votes": 6, "creation_date": "2021-04-07T21:21:04", "comments": [], "comment_count": 0, "category": "Culture & Recreation", "diamond": 0} {"question_id": "411", "site": "cs", "title": "Can a calculus have incremental copying and closed scopes?", "body": "A few days ago, I proposed the [Abstract Calculus](https://medium.com/@maiavictor/the-abstract-calculus-fe8c46bcf39c), a minimal untyped language that is very similar to the Lambda Calculus, except for the main difference that substitutions are `O(1)` (i.e., variables only occur once) and copying is an explicit, constant-time and incremental operation. Here is an example:\n \n \n let (a & b) = (λx. λy. λz. y) in (a & b)\n ---------------------------------------- step 0\n let (a & b) = (λy. λz. y) in (λx0. a) & (λx1. b)\n ------------------------------------------------ step 1\n let (a & b) = (λz. y0 & y1) in (λx0. λy0. a) & (λx1. λy1. b)\n ------------------------------------------------------------ step 2\n let (a & b) = y0 & y1 in (λx0. λy0. λz0. a) & (λx1. λy1. λz1. b)\n ---------------------------------------------------------------- step 3\n (λx0. λy0. λz0. y0) & (λx1. λy1. λz1. y1)\n \n\nHere, the `let (a & b) = t in k` syntax substitutes the occurrences of `a` and `b` in `k` by `t`, except it does so incrementally: notice how each lambda is pulled one by one. \n\nThere is a problem, though. Eventually, you'll try to pull a lambda which has a variable that is bound in `t`. That's occurs with `y` behind step 1. To solve that issue, whenever that happens, the bound variable `y` is replaced by `y0 & y1`, which are bound to each of the two copied lambdas, respectively. The problem with that is this requires us to abandon the notion of scopes, as, now, the bound variable of a λ can occur outside its own body (such as `y0` and `y1` behind step 2)! \n\nDespite sounding bizarre, the language operates very well. You can represent the usual λ-encodings on it, implement recursive algorithms and it is compatible with the oracle-free fragment of the optimal reduction algorithm. Yet, some have questioned the absence of scopes. I've spent some time reasoning about this and couldn't find a way to have both things. In fact, it looks impossible. As a sanity check, I'd like to know if my reasoning makes sense.\n\n**Thus, my question is: is that possible? Can we have both incremental copying and closed scopes in a calculus?**\n", "link": "https://cs.stackexchange.com/questions/96798/can-a-calculus-have-incremental-copying-and-closed-scopes", "tags": ["lambda-calculus", "type-theory", "functional-programming", "language-design"], "votes": 24, "creation_date": "2018-08-27T17:09:19", "comments": ["@chi I was reading Lamping's original paper right now, seems like he doesn't propose a textual calculus indeed. What I proposed is just a textual interpretation of his algorithm (in any of its stages). That representation can itself can be considered a calculus with 4 reduction rules, but requires abandoning the notion of scopes, which is what is puzzling me. Thanks (:", "I'm not terribly familiar with graph-reduction techniques for the lambda calculi. I only remember that some later variants were much simpler than Lamping's (fewer brackets/special tokens around). Perhaps there's some way to represent those \"bracket\" rules in a term form? I vaguely recall a presentation stating that those rules could not be used on terms -- they really need graphs --, but I am unsure.", "@augustss you're right, I should've clarified that I meant a language with abstractions λx. f.", "@chi yes the reason I came up with this was to have an untyped (textual) language for Lamping's abstract algorithm. This is just a textual interpretation of that algorithm. Or do you mean he actually proposed such representation? I'm not aware of that, but would be very glad if that was the case. I'll check the paper again.", "To me, this looks similar to the graph-reduction algorithms for (optimal) reduction in lambda calculus. The first paper IIRC was at POPL'90 by Lamping. There, the strict notion of scope is \"broken\" as it happens above, and replaced with a plethora of (indexed) bracket-markers in the term graph, which is reduced using small steps as above, duplicating terms gradually. I'd recommend you have a look at it, and subsequent (simpler) graph-reduction techniques.", "SKI combinators have incremental copying, are (almost) isomorphic to the lambda calculus."], "comment_count": 6, "category": "Science", "diamond": 0} {"question_id": "412", "site": "cs", "title": "Is the two-color leapfrog problem in P?", "body": "My question is whether a specific decision problem is in P or not. It's straightforwardly in NP. The decision problem is a specific case of the general $k$-color leapfrog problem.\n\nI can already show that when $k=1$, the problem is in P. And when $k$ is unlimited, the problem is NP-complete. I have not been able to find an algorithm for the $k=2$ case, but I suspect it is in P. I am wondering if we can prove that it is in P, or (what would surprise me) that it is in fact NP-complete.\n\nThis question springs from an interesting discussion in a question I [previously asked](https://cs.stackexchange.com/questions/127268/is-the-leapfrog-automata-problem-in-p) on this site.\n\n> **The $k$-color leapfrog problem**. Fix $k$. In a $k$-color leapfrog problem, you are given some number of transparent urns containing colored balls. Each ball is one of $k$ possible colors. You are also supplied with an _agenda_ , which is an ordered list of colors, for example \"red, red, red, blue, red, green, ...\". The object of the game is to draw balls out of the urns, one at a time, following the sequence of colors specified in the agenda. You can easily see the contents of all urns at all times, and can draw any ball out at will. If you could draw the appropriate color from multiple urns, you may choose which one. The only constraint is that you may not draw from the same urn twice in a row. You win the game if you complete the agenda, otherwise you lose.\n> \n> The decision problem is whether, for a given arrangement of urns and a given agenda, the game is winnable.\n\n**The 1-color leapfrog problem is in P**. If there is only one color of ball, you can decide it with the following algorithm: at each turn, draw the ball from the urn with the most balls in it (among urns you're allowed to draw from). If the game is winnable at all, this algorithm will win it. (Note that even when $k=1$, a game may still be unwinnable if there's too great a disparity in ball count. For example, the setup [\"R\", \"RRR\"] with the agenda \"RRRR\" is unwinnable.)\n\n**The $\\infty$-color leapfrog problem is NP-complete**. See my [previous question](https://cs.stackexchange.com/questions/127268/is-the-leapfrog-automata-problem-in-p) and the dazzling solution provided. You can perform a reduction from SAT using a game with two new urns and three new colors per clause.\n\n**What about 2-color leapfrog?** This is my main question. I suspect the problem is in P, as with the 1-color case, but the 1-color algorithm doesn't seem to generalize — or at least I'm not sure.\n\n**Bonus: What about $k$-color leapfrog?** An interesting higher level question is where exactly the problem becomes hard. Is is still in P when $k=3$, or does the leapfrog game have a phase transition like 2SAT to 3SAT? Is perhaps in P for all finite $k$? I've been struggling to find an efficient algorithm— I suspect there might be one, just based on how the problem becomes smaller and smaller as you remove balls.\n\n* * *\n\n**Edit** : _The naive extension does not work._ Note that the \"leveling\" strategy — generally ensuring that the urns have approximately the same number of balls of each color by removing from the urn that has the most— perfectly solves the $k=1$ case. I expect finding an extension of this strategy will solve the $k=2$ case, but note that one naive extension does not work:\n\nFor the scenario where the urns contain [AAB, ABB] and the agenda requires AABBBA, the naive strategy of taking from the legal urn with the most of the requested color does not work. It fails to find a solution despite the fact that this instance is actually solvable. (Similarly, the strategy of taking each color from the urn that has the least amount of the _other_ color also fails on this problem instance.)\n\n* * *\n\n**Update:** In the special case that the agenda has the form $A^m B^n$ $(m,n\\geq 1)$, you can determine whether it's solvable in polynomial time by playing two single-color games in series. Briefly put, a single-color game is winnable if and only if the vase with the most balls has at most one more ball than the rest of the vases put together. Equivalently, it's winnable if and only if you can pair up each ball with a ball from another vase, with at most one left over. You can construct a winning sequence of moves given such a pairing: first draw the unpaired ball if any. Then iterate over each pair in an arbitrary order. Because the pairs come from different vases, at least one of them will be legal to draw next; draw it, then the other one. Proceed until you draw all the balls.\n\nBecause you can iterate over the pairs in an arbitrary order, you can generally always choose a drawing order such that the last ball comes from the vase of your choice. Or, reversing the drawing order*, that the first ball comes from the vase of your choice. Hence in the two-color scenario, when the agenda is $A^mB^n$, you can solve it as follows: Pick a vase $V_A$ with an $A$-colored ball and another vase $V_B$ with a $B$-colored ball (if you can't, then there's only one vase and it has multiple balls in it, so the game is unwinnable). If it's possible to win the $A^m$, choose the winning strategy to draw from $V_A$ in the last step. If it's possible to win the $B^n$ game, choose the winning strategy to draw from $V_B$ in the first step. This can all be done and checked in polynomial time, QED.\n\n[*] Actually, reversing a path is not always possible: if the vase with the largest amount has exactly one more ball than the rest of them put together, then you _must_ draw from the largest vase first. The proof for how to solve $A^mB^n$ in this case will have to wait.\n\n* * *\n\n**Update** : By exchanging the notion of urns and colors, you can get the following equivalent formulation:\n\n> _Reformulated leapfrog problem:_ Fix $k$. You are given $k$ urns full of different-colored balls, and an agenda which tells you which urn you must draw from at each step. Your goal is to empty all of the urns subject to the constraint that each ball must be a different color from the one before it.\n\nPerhaps rephrasing this way can be useful. Also in this reformulation, I think the complexity is exposed by supposing you have two urns, where the second urn contains at most two balls.\n", "link": "https://cs.stackexchange.com/questions/127611/is-the-two-color-leapfrog-problem-in-p", "tags": ["complexity-theory", "turing-machines"], "votes": 18, "creation_date": "2020-06-24T01:51:59", "comments": ["A case is a set of additional constraints that we might choose to impose. E.g., \"Is there a solution in which urns 3 and 5 finish with at least 1 A-ball each, and urns 2 and 5 finish with at least 1 B-ball each?\" The idea being to come up with a set of at most polynomially many cases such that (a) any solution to the original problem belongs to at least 1 case in the set, and (b) each of these more constrained problem instances can be solved in poly time -- solving each case would then give a poly-time algorithm. But now I think (b) is harder to achieve than I thought.", "@j_random_hacker If we restrict either the maximum number of urns or the number of balls per urn, the search problem becomes polynomial, but I'm not sure yet that it's polynomial without those restrictions. In the first case, a game state lists the number of balls of each color in each urn. In the second, the contents of each urn have a constant number of possibilities, so a game state counts how many urns of each content-type there are. In both cases, the number of game states is polynomial in the input.", "Thanks. I'm wondering what counts as a \"case\" --- is it a game state (i.e. the contents of each urn at a single point in time) or a game path (i.e. a sequence of draws to make)? Naively, there are a lot of paths --- each game state will generally have multiple ways to arrive at it, and some might not be legally reachable.", "The above is not quite right: Once there is only one A left in the agenda, if there is only one urn with A-balls in it and it is also one of the 2 urns with B-balls, then we cannot necessarily choose B-ball urns freely. Still I think this problem can be attacked with a case analysis: Either there is a solution where we never drop below 2 A-ball urns and 2 B-ball urns; or a solution in which some urn becomes the only A-ball urn \"first\"; or a solution in which some urn becomes the only B-ball urn \"first\"; or no solution. There are only $O(n^2)$ such cases to consider.", "If there are not enough As and Bs in total to cover the letters in the agenda, then clearly the instance is unsolvable. Otherwise, the only way to get stuck is to get into a situation where two adjacent agenda letters can only come from the same urn. So if you can choose urns in sequence so that, while $\\ge 2$ A-balls remain in the agenda, you have at least 2 distinct urns $a_1, a_2$ containing 1 or more A-balls, and similarly urns $b_1, b_2$ for B-balls (note $\\{a_1, a_2\\} \\cap \\{b_1, b_2\\}$ is allowed to be nonempty), you will get a solution. However this can miss some solvable instances.", "As a follow up, note that when the agenda is a simple band of colors, e.g. AAAAAA or AAAAABBBBB, the problem is quite simple because it factors mostly like the $k=1$ case; the hard part is deciding what to do at the junctions between neighboring colors. If we can find a simple algorithm for the junctions, we can handle each color more or less separately and stitch them together in the end.", "Assuming there is a polynomial algorithm, perhaps there's one that runs in time $O(n^k)$, where $k$ is the number of colors and $n$ is the problem size. This suggests an approach to designing an algorithm---iterate over each color, or even each pair of colors. And it would suggest why the problem becomes harder when the number of colors is unbounded so $k \\approx n$.", "It seems I wasn't careful enough, so there's no issue with encoding (we can assume that each vase has at most the agenda's length balls of each color, and since the agenda is ordered this means we are allowed a polynomial number of operations in the numerical value of the number of balls in each vase).", "@Ariel Can you, please, elaborate on the $O(n^3)$ algorithm from your previous comment? If I understand you right, you will have to fill a $(n + 1)$-dimensional table with $n$ elements in each dimension at the worst case. But it will give an $\\Omega(n^{n+1})$ time complexity.", "My mistake, I missed the \"draw any ball at will\". The encoding becomes important here, each vase should be described as a histogram (n-white, m-black) otherwise the previous solution is polynomial.", "@Ariel I should clarify that there's no order to the balls in each vase---you can see all of them, and pick any one out at will. The agenda is a separate structure, listing the sequence of colors you must draw (and you can pick which vases). I like your approach, though, and a similar algorithm definitely works for $k=1$ colors. Maybe we can make it work for $k=2$?", "Have you tried dynamic programming? Given $n$ vases each described as a list of colors, our ability to win the game depends on $n+1$ parameters, the index of the active vase (constraint) and $n$ indices of locations in each list which indicate the number of balls we have drawn from each vase. If I'm not missing anything basic, this seems to admit to a straightforward $O(n^3)$ time algorithm (assume for simplicitly that $n$ is a bound on both the number of vases and the number of balls in each vase). It also seems to work for the $\\infty$-color problem, so I might be completely off.", "@Juho Thanks for the tip! I'll look into color graph path problems; the leapfrog problem is definitely a special case of that. I suspect it's an easier special case, but if I can perform a reduction from GRAPH MOTIF, that'll establish that it's in fact just as difficult.", "It sounds similar to a problem in which you want to find a path in a (directed) graph with specified colors (in order). These type of problems are usually hard (like GRAPH MOTIF, which IIRC is hard for 2 colors even on trees). Your visitation constraint might complicate things a little (or maybe you can avoid it) but a reduction from such a problem might be short and simple."], "comment_count": 14, "category": "Science", "diamond": 0} {"question_id": "413", "site": "cs", "title": "Test whether two languages are equal, when give in algebraic form", "body": "This sub-problem is motivated by [Algorithm to test whether a language is regular](https://cs.stackexchange.com/q/18010/755).\n\nSuppose we have two languages $L_1,L_2$ that are expressed in \"algebraic\" form, as formalized below. I want to determine whether $L_1 = L_2$.\n\nIs this decidable? If yes, can anyone suggest an algorithm for this? If not, can anyone suggest a semi-algorithm (one that returns \"equal\", \"not equal\", or \"I'm not sure\"; where we want it to return \"not sure\" as rarely as possible).\n\nMotivation: A solution to this sub-problem would be helpful for solving [Algorithm to test whether a language is regular](https://cs.stackexchange.com/q/18010/755)\n\n* * *\n\n**Algebraic form.** Here's one possible definition of what I mean by a representation of a language in algebraic form. Such a language is given by\n\n$$L = \\\\{E : S\\\\}$$\n\nwhere $E$ is a word-expression and $S$ is a system of linear inequalities over the length-variables, with the following definitions:\n\n * Each of $x,y,z,\\dots$ is a word-expression. (These represent variables that can take on any word in $\\Sigma^*$.)\n\n * Each of $x^r,y^r,z^r,\\dots$ is a word-expression. (Here $x^r$ represents the reverse of the string $x$.)\n\n * Each of $a,b,c,\\dots$ is a word-expression. (Implicitly, $\\Sigma=\\\\{a,b,c,\\dots\\\\}$, so $a,b,c,\\dots$ represent a single symbol in the underlying alphabet.)\n\n * Each of $a^\\eta,b^\\eta,c^\\eta,\\dots$ is a word-expression, if $\\eta$ is a length-variable.\n\n * The concatenation of word-expressions is a word-expression.\n\n * Each of $m,n,p,q,\\dots$ is a length-variable. (These represent variables that can take on any natural number.)\n\n * Each of $|x|,|y|,|z|,\\dots$ is a length-variable. (These represent the length of a corresponding word.)\n\n\n\n\nMy goal is something rich enough to capture many of the examples we see in textbook exercises. (Feel free to suggest modifications to this formalization if it makes the equality-testing problem easier while still remaining expressive enough to capture many of the examples we see in textbooks.)\n\n* * *\n\n**Some easier cases.** If the original problem is too hard, here are some sub-cases that would still be interesting:\n\n * If $L_1,L_2$ are two languages specified in \"algebraic\" form as above, and $L_1\\ne L_2$, let $d(L_1,L_2)$ denote the length of the shortest word that is an element of one of $L_1,L_2$ but not the other. Can we upper-bound $d(L_1,L_2)$ as a function of the length of the descriptions of $L_1,L_2$? Is it guaranteed to always be polynomial? at most singly-exponential?\n\nMotivation: If we could prove it is always polynomial, this might help us exhibit a witness that $L_1 \\ne L_2$.\n\n * If we omit the reversal operation (we do not allow $x^r,y^r,\\dots$), does the problem become easier?\n\n\n\n", "link": "https://cs.stackexchange.com/questions/18062/test-whether-two-languages-are-equal-when-give-in-algebraic-form", "tags": ["algorithms", "formal-languages", "regular-languages", "decision-problem"], "votes": 15, "creation_date": "2013-11-15T19:08:21", "comments": ["If it is limited to regular languages, then you can compare them. Regular languages can be converted to finite state machine. And if you minimise two different finite state machines and you end up with the same answer for both, then they are infact the same. (They both handle the same language if they simplify to the same form.) You won't be able to do this for a CFG language, otherwise you would have solved the halting problem.", "I don't see how to express Dyck languages in this framework.", "@FrançoisG., no. For instance, $\\{w^\\eta w^\\eta : w \\in \\Sigma^*, \\eta \\in \\mathbb{N}\\}$ is not context-free but has an algebraic form. (I don't know whether every context-free language has an algebraic form expression; I don't see any reason to expect this to be true.) I know that testing equality of two context-free languages is undecidable. I don't know how hard the first sub-case (at the end of the question) is if $L_1,L_2$ are two context-free languages.", "Are those algebraic form languages and context free languages the same ?"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "414", "site": "cs", "title": "What can be proven regarding the differences in power between unary ECMAScript regex functions and primitive recursive functions?", "body": "In 2014, inspired by [Regex Golf](https://alf.nu/RegexGolf), I started exploring, along with a mathematician going by the name **teukon** , what could be done in the unary domain in ECMAScript regex that went significantly beyond matching primes and powers of 2 (both of which were puzzles in the Regex Golf site). This is a dialect of regex that includes negative and atomic positive lookahead, and backreferences that are erased when the loop they were defined in starts a new iteration. (They're of course not \"regular expressions\" in the CS sense).\n\nA unary regex function takes a string of identical characters as input (`x` is used for clarity instead of `1`), whose length represents $n \\in \\mathbb{N_0}$. Its output may be either a non-match (which can represent \"false\"), or any $m \\le n$ (which can represent \"true\", ignoring $m$'s value).\n\nDue to the limitations on ECMAScript backrefs, only one variable* can be changed from iteration to iteration in a loop – the the $tail$ following the current cursor position, in such a way that it is decreased by at least $1$ on every iteration (by moving the cursor forward). All other variables (backrefs) can only capture a value once (per iteration if in a loop) which cannot be greater than the value of $tail$ at the time of its definition. An additional constraint is that when searching for a number that satisfies some condition, the current value being tested must be subtracted from $tail$ before its properties can be tested (this can be avoided by defining a new operation, non-atomic lookahead, which allows the regex engine to backtrack into the lookahead from outside it). And $n$ can only be operated on directly when the cursor is at the leftmost position, i.e. $tail=n$.\n\nBecause of this, the approaches available in more powerful regex flavors (e.g. Perl, PCRE, Java, .NET, Python, Ruby) won't work, for example `^(^x|xx\\1)*$` can't be used to match squares.\n\nI was suprised and fascinated to find that the power of unary ECMAScript regex goes rather deep indeed:\n\n * Match primes - [`^(?!(xx+)\\1+$)xx`](https://codegolf.stackexchange.com/a/177196/17216)\n * Match powers of 2 - `^(?!(x(xx)+|)\\1*$)` or `^(?!(x*)(\\1\\1)+$)` or `^((x+)(?=\\2$))*x$`\n * Match powers of 10 - `^((x+)\\2{8}(?=\\2$))*x$`\n * Match prime powers - [`^(?!(((x+)x+)(?=\\2+$)\\2*\\3)\\1+$)x`](https://codegolf.stackexchange.com/a/219939/17216)\n * Match `N`th powers - [`^((?=(xx+?)\\2+$)((?=\\2+$)(?=(x+)(\\4+$))\\5){N})*x?$`](https://codegolf.stackexchange.com/a/21951/17216)\n * Match correct statements of multiplication - [`^x(x*)·?(x*)·?x\\1=(?=(x(\\1\\2))\\1(\\3*)\\4*$)\\3*$\\5`](https://github.com/Davidebyzero/RegexGolf/blob/master/regex%20for%20matching%20multiplication%20-%20simple%20method.txt) (takes its input as `A·B=C`; doesn't handle zero); this uses the Chinese remainder theorem. Although it's not purely unary (with delimiters `·` and `=`), its multiplication logic can be used within a purely unary regex. \n * Fully generalized, with handling of zero: [`^(?=(x?(x*))·(x?(x*))).*=(?=.*((?=\\1*$)(?=\\3*$)(?=\\1\\4+$)\\3\\2+$|$\\1|$\\3))\\5$`](https://github.com/Davidebyzero/RegexGolf/blob/master/regex%20for%20matching%20multiplication%20-%20multiplication%20%2B%20comparison%20-%20fully%20generalized.txt)\n * Division and remainder - [`(x(x*)),(x*?)(?=\\1*$)(x?(x*))(?=\\4*$)((?=\\2+$)\\2\\5*$|$\\4)`](https://codegolf.stackexchange.com/a/222932/17216)\n * This turned out to be quite interesting. The algorithm can't be proved using the Chinese remainder theorem, but finally does have a proof (linked above) thanks to H.PWiz. There are shorter variants that can be used when certain inequalities regarding the dividend and divisor are guaranteed to be met.\n * Match squares - [`^(x(x*)|)(?=(\\1*)\\2+$)\\1*$\\3`](https://codegolf.stackexchange.com/a/223201/17216) (shorter than the above Nth powers regex thanks to generalized multiplication)\n * Match perfect powers ($a^b$ where $b>1$) - `^(x+)((\\1(x+))(?=(\\3*)\\1*$)\\3*(?=\\4$\\5))+x$|^x?$`\n * Match triangular numbers - [`^((((x*)x?)\\3)x)?(?=\\4(\\1*)\\2*$)\\1*$\\5`](https://codegolf.stackexchange.com/a/220988/17216)\n * Match Fibonacci numbers - [`^(?=(x*).*(?=x{4}(x{5}(\\1{5}))(?=\\2*$)\\3+$)(|x{4})(?=xx(x*)\\5)\\4(x(x*))(?=(\\6*)\\7+$)(?=\\6*$\\8)\\6*(?=x\\1\\7+$)(x*)(\\9x?)|)(\\9\\10(\\5\\10)\\12?|xx?x?|x{5}|x{8}|x{21})$`](https://codegolf.stackexchange.com/a/178956/17216)\n * Match factorial numbers - [`^(((x+)\\3+)(?=\\3\\3\\2$)(x(?=\\3+(x*))(?=\\5+$)(?=((x+)(?=\\5\\7*$)x)\\6*$)\\3+(?=\\5\\7$))*\\3xx|x?)x$`](https://codegolf.stackexchange.com/a/178979/17216)\n * Match abundant numbers - [`^(?=((?=(xx+?)\\2+$)(x+)\\3*(?=\\3$))+(?!\\2+$)(?=(xx(x*?))\\4*$)x((x(x*))(?=\\7*$)\\5\\8+$))(?=(x(x*))(?=\\9*$)\\6\\10+$)(?=(x*?)(?=(x(\\6\\7))+$)(x(x*))(?=\\14*$)(?=\\13+$)\\15+(?=\\11\\11|(x))\\13$)((?=(x*?(?=\\9*$)(?=(x+)(?=\\10+$)\\19*$)(^\\14{2}\\16|)))(?=(xx+).*(?=\\9$)(?=\\21*(?!\\19)\\21$)(|(x+)\\23*(?=\\23$))(?!(xx+)(?!\\21+$)\\24+$)(?=(x*)(?=xx(x\\25)*$)\\21\\25*$)(\\21+$))(?=.*?(?!x\\18)(?=(x(\\27\\25))+$)(x(x*))(?=\\30*$)(?=\\29+$)\\29\\31+$|)(?=\\30(x*)(?=\\9*$)(\\21\\10+$))\\32)+\\33$`](https://codegolf.stackexchange.com/a/178952/17216) (there is a shorter version, but I haven't proved it to be correct to infinity yet)\n * Return $\\lfloor{\\log_2 n}\\rfloor$ \\- [`(?!(x*)(\\1\\1)+$)(?=(x+)\\3)(?=((?=(x+)(?=\\3)\\5(((x*)(?=\\8$)x)+$))\\6)*)\\5x|\\b$`](https://chat.stackexchange.com/transcript/message/48931239#48931239)\n * Return $\\varphi(n)$ \\- [`(?=((xx+)(?=\\2+$)|x+)+)(?=((x*?)(?=\\1*$)(?=(\\4xx+?)(\\5*(?!(xx+)\\7+$)\\5)?$)(?=((x*)(?=\\5\\9*$)x)(\\8*)$)x*(?=(?=\\5$)\\1|\\5\\10)x)+)\\10|x`](https://codegolf.stackexchange.com/a/180255/17216)\n * Return $\\lfloor{n\\sqrt{1/2}}\\rfloor$ \\- [`(?=(x(x*)).*(?=\\1*$)\\2+$)(?=(x\\1)+(x?(x*)))(?=\\4(x(x*?))\\1+$)(?=.*(?=(?=\\4*$)\\4\\5+$)(x*?)(?=\\3*$)(x?(x*?))(\\1+$|$\\9))(?=.*(?=(?=\\4*$)(?=\\6*$)(?=\\4\\7+$)\\6\\5+$|$\\4)(x*?)(?=\\3*$)(x?(x*?))(\\1+$|$\\13))(?=.*(?=\\12\\12\\9$)(x*?)(?=\\3*$)(x?(x*?))(\\1+$|$\\17))(?*.*?(?=((?=\\3*(x?(x*)))\\21(x(x*?))\\1+$)))(?=.*(?=\\23*$)(\\23\\24+$))(?=.*(?=(?=\\21*$)\\21\\22+$)(x*?)(?=\\3*$)(x?(x*?))(\\1+$|$\\27))(?=.*(?=(?=\\21*$)(?=\\23*$)(?=\\21\\24+$)\\23\\22+$|$\\21)(x*?)(?=\\3*$)(x?(x*?))(\\1+$|$\\31))(?=.*(?=\\30\\30\\27$)(x*?)(?=\\3*$)(x?(x*?))(\\1+$|$\\35))(?=.*(?=\\26\\26)(?=\\3*(x*))(\\1(x)|))(?=.*(?=\\34\\34\\40)(?=\\3*(x*))(\\1(x)|))(?=(?=(.*)\\13\\13\\17(?=\\6*$)\\6\\7+$)\\44(x+|(?=.*(?!\\16)\\41|(?!.*(?!\\38)\\8).*(?=\\16$)\\41$))(\\25\\31\\31\\35){2}\\43$)\\20|xx?\\B|`](https://codegolf.stackexchange.com/a/198428/17216) \\- uses `(?*`...`)`; without that, it'd be about twice as long or more. This can clearly be extended to any algebraic number $01$. This gives the sequence $20, 28, 29, 35, 41, 52, 68, 72, 89, 90, 126, 130, 133, 174...$\n * In ECMAScript+`(?*)` this is just [`^(?*(x(x*)x+))(?!(\\1|x\\2)(?!(x+)((\\4(x+))(?=(\\6*)\\4*$)\\6*(?=\\7$\\8))+x$|x$)).*(?=\\1$)\\2((\\2(x+))(?=(\\10*)\\2*$)\\10*(?=\\11$\\12))+x$`](https://kingbird.myphotos.cc/b751bb38f1a1b12e90f23541f8f6ec33/regex%20for%20perfect%20power%20differences%20puzzle,%20with%20molecular%20lookahead.txt).\n * In ECMAScript it's still possible: [`^(?=(x*)\\1(x*))((x+(x*))(x?))(?=\\6*(x?)).*(?=\\1$)(?=.*(?=\\2\\7)(xx)?((\\7?)x*))(?=\\9*(x?))(?!\\10(\\4|\\5)\\6(?!$\\11|(\\11(x*))(?:(?=(?=\\13$\\9|(?=(\\13\\13\\9)+(x*))(?=.*(?=\\15)\\16(x*))\\17\\15*$)((x*)(x|(?=(x))))(?=\\21((?=\\18$)|(?=(\\18\\18\\21))(?=\\18\\18\\23*(x*))(?=.*(?=\\23)\\24(x*))(?=\\25\\23*$)))((?=.*(?=\\21$)\\11(x*)|(x))(?=(\\27\\28))\\21\\14\\28((?=.*(?=\\29$)\\20$)\\19$|(?=(\\19\\19\\20)+(x*))(?=.*(?=\\31)\\29\\32(x*))\\33\\31*$)))(?=.*(?=\\21$)\\11$)\\26\\14)+$)).*(?=\\6\\4$)(\\6(\\5))(?:(?=(?=\\34$\\7|(?=(\\34\\34\\7)+(x*))(?=.*(?=\\36)\\37(x*))\\38\\36*$)((x*)(x|(?=(x))))(?=\\42((?=\\39$)|(?=(\\39\\39\\42))(?=\\39\\39\\44*(x*))(?=.*(?=\\44)\\45(x*))(?=\\46\\44*$)))((?=.*(?=\\42$)\\6(x*)|(x))(?=(\\48\\49))\\42\\35\\49((?=.*(?=\\50$)\\41$)\\40$|(?=(\\40\\40\\41)+(x*))(?=.*(?=\\52)\\50\\53(x*))\\54\\52*$)))(?=.*(?=\\42$)\\6$)\\47\\35)+$`](https://chat.stackexchange.com/transcript/message/49026359#49026359) \\- this works by pair-encoding variables. \nWith a little modification, it seems it should be possible to design a function that is not possible without `(?*)`. But could it be proven to be impossible?\n 4. Are there any functions that can be implemented with right-to-left evaluated variable-length lookbehind `(?<=`...`)`, as in the ECMAScript 2018 standard, but not with `(?*`...`)`? In most ways lookbehind gives a superset of molecular lookahead's power. `^(?*A)B` can be transformed to `^A(?<=(?=B)^.*)`. When we're not at the $tail=n$ position, `(?*A)B` can become `(?<=(.*))A(?<=^\\1(?=B).*)`. Lookbehind can directly do things that molecular lookahead cannot; most notably, `^`...`(?:`...`(?<=(?=A)^.*)`...`)*`...`$` uses it to do tests directly on $n$ from within a loop. Are there functions that require this functionality, and can't be implemented with just `(?*`...`)`?\n 5. Are there functions that can be implemented with `(?*`...`)` but not with `(?<=`...`)`? Although the latter may seem at first to offer a strict superset of the former's functionality, this may not be the case. `(?:(?*A)B)*` can only be transformed to `(?:(?<=(.*))A(?<=^\\1(?=B(.*)).*).*(?=\\2$))*` if `A` never advances the cursor by more than `B`.\n\n\n", "link": "https://cs.stackexchange.com/questions/137112/what-can-be-proven-regarding-the-differences-in-power-between-unary-ecmascript-r", "tags": ["complexity-theory", "computability", "space-complexity", "number-theory", "primitive-recursion"], "votes": 14, "creation_date": "2021-03-25T14:34:10", "comments": ["@Nayuki That logic doesn't work. The equivalent of only a finite number of digits of the transcendental number need to be calculated. By your logic, it wouldn't even be possible in a Turing-complete language to take a natural number N as input and calculate the first N digits of pi as output, but that is possible. (You may be falsely equating transcendental numbers with uncomputable numbers.)", "\"Can any transcendental number computed\": No, because transcendental numbers are uncountable whereas program code strings are countable."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "415", "site": "cs", "title": "Choosing a subset of binary variables to maximize the sum of the highest $K$", "body": "Consider the following problem:\n\n**Input:**\n\n * integers $n > m > k$;\n * $n$ numbers $0 \\leq p_1, \\ldots, p_n \\leq 1$;\n * $n$ numbers $r_1, \\ldots, r_n$ where ($r_i \\geq 0$).\n\n\n\nLet $X_1,\\dots,X_n$ be $n$ independent random variables with distribution $X_i \\sim \\text{Bernoulli}(p_i)$ and define $Y_i = r_iX_i$.\n\n**Output:** a subset $S$ of $[n]$ of size $m$ that maximizes $r_S = \\mathbb{E}[Z_S]$, where $Z_S$ is the random variable $Z_S = \\max_{T \\subseteq S, |T|=k} \\sum_{i\\in T} Y_i$.\n\nIn other words, we want to select $m$ $Y_i$s such that the expected value of the sum of the largest $k$ of them is maximized.\n\nCan this problem be solved in polynomial time? \nOr is it NP-hard? \nIs there an efficient $\\alpha$-approximation algorithm?\n\n* * *\n\nExample: \n**Input:**\n\n * $n=3, m=2, k=1$\n * $p_1=1.00, p_2=0.10, p_3=0.01$\n * $r_1=2, r_2=11, r_3=100$\n\n\n\n**Output:** the correct output is $S = \\\\{1,3\\\\}$. Why?\n\n * $r_{\\\\{1,3\\\\}} = 2.98$: if $X_3=1$, then $Z_S=100$ (since $T=\\\\{3\\\\}$ in this case), otherwise $Z_S=2$ (since $T=\\\\{1\\\\}$ in this case), and thus\n\n$$r_{\\\\{1,3\\\\}} = \\Pr[X_3=1] \\times 100 + \\Pr[X_3=0] \\times 2 = 0.01 \\times 100 + 0.99 \\times 2 = 2.98.$$\n\n * $r_{\\\\{1,2\\\\}} = 2.9$: if $X_2=1$, then $Z_S=11$, otherwise $Z_S=2$, and thus\n\n$$r_{\\\\{1,2\\\\}} = 0.1 \\times 11 + 0.9 \\times 2 = 2.9.$$\n\n\n\n\n* * *\n\nI have been working on this for a while now. As the example above demonstrates, the obvious greedy approach of selecting $Y_i$s with largest expectation does not work. I solved the special case of $k=1$ using dynamic programming. But my guess is that the general case is NP-hard.\n\nNote that the value of of $Z_S$ depends on $S$ and the values of $Y_1,\\dots,Y_n$: given $S$, the random process is that we randomly pick the values of the $X_1,\\dots,X_n$, compute the values of $Y_1,\\dots,Y_n$, then based on the values of $Y_1,\\dots,Y_n$ we find the set $T$ of size $k$ that maximizes $\\sum_{i \\in T} Y_i$, and let $Z_S$ be the result (i.e., we let $Z_S$ be the sum of the $k$ largest values from $Y_1,\\dots,Y_n$). This is why just choosing the $k$ $Y_i$'s whose expected value is largest does not yield the correct result.\n", "link": "https://cs.stackexchange.com/questions/37294/choosing-a-subset-of-binary-variables-to-maximize-the-sum-of-the-highest-k", "tags": ["algorithms", "complexity-theory", "probability-theory"], "votes": 13, "creation_date": "2015-01-16T13:41:23", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "416", "site": "cs", "title": "Covering a complete graph with n copies of an arbitrary graph: NP-complete?", "body": "> Given a complete graph $G$, an arbitrary graph $H$, and a positive integer $n$, are there subgraphs $A_1,\\dots,A_n$ of $G$ (not necessarily disjoint) such that their union is $G$, and each of them are isomorphic to $H$?\n\nThis is a problem which I believe is NP-complete, but I am unsure if it is actually so. Any ideas of how to prove this?\n\nNotes:\n\n * If $G$ is instead allowed to be an arbitrary graph, then this is clearly at least as hard as [subgraph isomorphism](https://en.wikipedia.org/wiki/Subgraph_isomorphism_problem), which is NP-complete.\n * If $G$ is allowed to be an arbitrary graph, and $A_1,\\dots,A_n$ are required to be isomorphic to a subgraph of $H$, instead of the entire graph $H$, then the problem can be reduced from [vertex cover](https://en.wikipedia.org/wiki/Vertex_cover) (which is NP-complete) by setting $H$ to be a [star graph](https://en.wikipedia.org/wiki/Star_\\(graph_theory\\)) with the number of spokes equal to the number of nodes in $G$.\n * The problem where $H$ must also be a complete graph is called \"covering design\", and has some discussion [here](https://math.stackexchange.com/questions/1734855/minimum-number-of-subsets-of-a-of-a-given-order-that-contain-all-possible-pair), with closed forms for when $H$ has 3 or 4 nodes. Finding algorithm for this is apparently an open problem.\n\n\n", "link": "https://cs.stackexchange.com/questions/89221/covering-a-complete-graph-with-n-copies-of-an-arbitrary-graph-np-complete", "tags": ["algorithms", "complexity-theory", "graphs", "np-complete"], "votes": 12, "creation_date": "2018-03-11T15:07:57", "comments": ["Do you mean a subgraph by a subgraph induced by a subset of vertices of $G$ (which is the definition used in subgraph isomorphism problem)? How do you define the union of two subgraphs?", "Another special case that looks hard (an open problem since 1967) is when $H$ is a collection of vertex-disjoint cycles: en.wikipedia.org/wiki/Oberwolfach_problem", "@WillardZhan I didn't think about it before. (I mainly thought about cases where $H$ is nearly as large as $G$.) Let's say I'm interested in answers of either variant.", "If $G=K_m$, is $m$ given in binary or unary?"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "417", "site": "cs", "title": "Is extensionality for coinductive datatypes consistent with Coq's logic?", "body": "Given a coinductive datatype, one can usually (always?) define a bisimulation as the largest equivalence relation over it. I would like to add an axiom stating that if two members of the type are related by the bisimulation, they are equal in the sense of Leibniz equality (`=`). Would this make the logic inconsistent?\n\n* * *\n\nAn example for streams:\n \n \n CoInductive Stream A :=\n | Cons : A -> Stream A -> Stream A.\n \n CoInductive Stream_bisim A : Stream A -> Stream A -> Prop :=\n | Stream_bisim_Cons :\n forall x xs ys,\n Stream_bisim A xs ys ->\n Stream_bisim A (Cons A x xs) (Cons A x ys).\n \n Axiom Stream_bisim_eq :\n forall A xs ys,\n Stream_bisim A xs ys ->\n xs = ys.\n \n\n* * *\n\nMy intuition is that this should be safe by analogy with functional extensionality, since it should not be possible to distinguish bisimilar streams by observation. But of course I'd prefer an actual proof (or at least expert testimony) to such informalities.\n", "link": "https://cs.stackexchange.com/questions/63197/is-extensionality-for-coinductive-datatypes-consistent-with-coqs-logic", "tags": ["coq", "equality", "coinduction"], "votes": 12, "creation_date": "2016-09-06T08:12:19", "comments": ["The last paragraph of section 2.2.2 of Simon Boulier's PhD thesis \"Extending type theory with syntactic models\" (tel.archives-ouvertes.fr/tel-02007839) states that it is provable in a univalent setting, if you use a well-chosen definition of stream, and gives references.", "@paulotorrens No -- I still believe that this is safe, but still don't know for certain. In practice, Coq's generalised rewriting makes using bisim instead of propositional equality bearable, so I would tend not to use the axiom.", "One year later; did you manage to find the answer?"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "418", "site": "cs", "title": "Regularity profiles", "body": "A standard exercise in formal language theory uses [Lagrange's four-square theorem](https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem) to construct a language $L$ such that $L$ isn't regular but $L^2$ is regular. (Let $A = \\\\{ a^{n^2} : n \\geq 0 \\\\}$. Then $A$ isn't regular, but $A^4 = \\\\{ a^n : n \\geq 0 \\\\}$ is regular, hence either $L = A$ or $L = A^2$ fits the bill; in fact $A^2$ is not regular, so we must pick the latter.)\n\n[This answer](https://cs.stackexchange.com/a/112357/683) generalizes this, constructing for every $m$ a language $L$ such that $L,L^2,\\ldots,L^{m-1}$ are not regular, but $L^m,L^{m+1},\\ldots$ are regular. This prompts the following definition:\n\n> The _regularity profile_ of a language $L$ is $\\rho(L) = \\\\{ n \\in \\mathbb{N}_+ : L^n \\text{ is regular} \\\\}$.\n\nMy question is:\n\n> Which regularity profiles are achievable?\n\nThe answer mentioned above shows that $\\\\{n : n \\geq m \\\\}$ is a regularity profile for every $m \\geq 1$.\n\nIt is also easy to construct a language whose regularity profile is empty: $\\\\{ a^{2^n} : n \\geq 0 \\\\}$.\n\nClearly every regularity profile is closed under addition: if $L^a,L^b$ are regular then so is $L^{a+b}$.\n\n> Is every subset of $\\mathbb{N}_+$ which is closed under addition a regularity profile?\n\nThe question is interesting both for a general alphabet and for the special case of a unary alphabet.\n", "link": "https://cs.stackexchange.com/questions/113326/regularity-profiles", "tags": ["regular-languages"], "votes": 11, "creation_date": "2019-09-02T07:08:40", "comments": ["@BaderAbuRadi I'm not aware of any progress on these questions, though I'm not sure I would know. Nobody contacted me regarding them.", "Yuval, do you have an idea if these questions have been studied yet, or did they remain open?"], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "419", "site": "cs", "title": "(Slightly) faster simulation of quantum Fourier transform", "body": "Suppose I want to write a classical software simulator of a quantum circuit with $N$ qubits. When it comes time to simulate the quantum Fourier transform I can evaluate all $2^N$ states to determine the probability amplitudes, and then perform a Fast-Fourier Transform on the probability amplitudes in time $o(N 2^N)$. Finally in $o(2^N)$ time I can generate a scan of the partial sums of the probabilities of all the result states. Then I can choose a random number in the range $[0,1]$ and use it to do a binary search of the partial sums.\n\nThis results in a simulator that, each time is run outputs a single $N$ bit binary number with the probability distribution predicted by theory.\n\nCan I do better? Of course I can't do exponentially better in general, but perhaps I could reduce the time to simulate a single experiment to $o(2^N)$?\n\nI can do significantly better under some circumstances. For example, [Gilbert, Guha, Indyk, Muthukrishnan, and Strauss, \"Near-Optimal Sparse Fourier Representations via Sampling\", _ACM Symp Theory Comp_ , STOC-44:152-161, 2002,](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.135.7607) seems to indicate that if there are only $B$ frequencies (or if the $B$ frequencies make up \"most\" of the power of the signal) then there is a randomized algorithm that will recover all of them (and their amplitudes) in time, space and number of samples polynomial in $B$ and $N$.\n\nI guess I'm hoping for something like that, but only to get one frequency, and to have some guarantee that the probability of getting a particular frequency will be proportional to the amplitude of its coefficient.\n", "link": "https://cs.stackexchange.com/questions/13325/slightly-faster-simulation-of-quantum-fourier-transform", "tags": ["randomized-algorithms", "quantum-computing", "fourier-transform"], "votes": 11, "creation_date": "2013-07-17T20:50:14", "comments": ["suggest migrate to Theoretical Computer Science"], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "420", "site": "cs", "title": "Change in the distances in a graph after removal of a node", "body": "Given an undirected unweighted graph $G=(V,E)$ and a node $s \\in V$, we are looking for a vector $\\operatorname{diff}[]$, such that,\n\n$$\\operatorname{diff}[v] = \\sum_{u \\in V \\setminus \\\\{v\\\\}}{(d^{G \\setminus \\\\{v\\\\}}_{su}-d^G_{su})}$$ \n\nIn other words, we are looking for the difference in distances from $s$ to every other nodes after removal of each node $v$. \n\nNow the question: is there an algorithm with time complexity $O(n+m)$ that can calculate the vector $\\operatorname{diff}$ for a given graph $G$ and node $s$? If yes, what is the algorithm (or the idea behind that algorithm)?\n\nFor a specific node $v$, we can run BFS on $G$ and $G \\setminus \\\\{v\\\\}$ and find the $\\operatorname{diff}[v]$ in $O(n+m)$, but here we want $\\operatorname{diff}$ for all of the nodes.\n\nI know that we can determine, whether $\\operatorname{diff}[v] >0$ for every node $v$ (using shortest path tree of $s$), in $O(n+m)$, but is there an algorithm to find the exact value of $\\operatorname{diff}$ in $O(n+m)$ time complexity?\n\n**Hints.**\n\nThere are some properties in the shortest path tree that may help:\n\n 1. If we construct the SPT, and then add the other edges of the graph that are not in the SPT, there will be edges from one level to the next level, or edges among nodes within the same level.\n\n 2. When we remove a node $v$, only the distance between $s$ and the descendants of $v$ can be changed.\n\n 3. For a node $v$, if all the descendants have another neightbour in $G$ that is on the same level as $v$ in the SPT, then $diff[v]=0$.\n\n 4. For every descendant of $v$, if we keep track of the incoming edges whose other endpoint is in another subtree (subtree of siblings of $v$), we can easily update the distances for its children (Suppose that child is $v'$, the change in $diff[v]$ for such $v'$ will be $\\min{(diff[v']+1,(d^{new}_{sv'}-d_{sv'})\\cdot(\\\\# \\text{children whose endpoint is in the subtree of }v)+(\\text{new distance of nodes whose endpoint is outside subtree of v}))}$).\n\n\n\n\nWe can assume that the graph is 2-vertex-connected, which means that it will not be disconnected by removing one node (otherwise we can find all the vertex cuts of size of using DFS and check each connected component separately).\n", "link": "https://cs.stackexchange.com/questions/29746/change-in-the-distances-in-a-graph-after-removal-of-a-node", "tags": ["algorithms", "graphs", "shortest-path"], "votes": 11, "creation_date": "2014-09-07T20:47:48", "comments": ["@FrankW If that distance goes through a child, then it will be the minimum is the first term $diff[v']+1$.", "there are at least 3 papers by Camil on that page, & online paper links are not given so far. the papers generally look at adding/ removing vertices/ edges.", "@vzn The paper by Camil investigates edge removal right?", "suggest looking into \"online algorithms\" or \"incremental/ decremental\". seems like an online version of APSP eg online version of APSP tcs.se"], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "421", "site": "cs", "title": "Minimum edge deletion partitioning of a planar graph", "body": "I'm interested in the time complexity of the following problem:\n\nGiven an undirected _planar_ graph $G=(V,E)$ and a weight function $w:E \\rightarrow \\mathbb{Z}$ (so weights can be negative, too), color the vertices in such a way that the sum of the weights of the monochromatic edges (i.e. those between same color vertices) is minimized. There is no limit on the number of colors you can use. Note that it is not necessarily optimal to give each vertex a distinct color as negative edge weights are possible.\n\nA version of this problem where $G$ is not restricted to be a planar graph is NP-hard (see [Minimum edge deletion partitioning](https://cs.stackexchange.com/questions/45372/minimum-edge-deletion-partitioning)) by a reduction from the vertex coloring problem. However, we cannot use the same reduction for the problem here because $G$ is planar. \n\nAny hints, pointers, comments are welcome.\n", "link": "https://cs.stackexchange.com/questions/45731/minimum-edge-deletion-partitioning-of-a-planar-graph", "tags": ["complexity-theory", "graphs", "np-hard", "weighted-graphs"], "votes": 11, "creation_date": "2015-08-31T06:00:04", "comments": ["If you restrict the number of colors to 2, then the problem is equivalent to max-cut (with possibly negative edge weights), which is NP-hard for general graphs but solvable in polynomial time for planar graphs. So maybe planarity could help here as well.", "The interesting property of planar graphs in this context seems to be that they are 4-colorable. As D.W. pointed out however, the optimum for your problem doesn't depend on using a minimum number of colors, so intuitively the planar restriction shouldn't affect the complexity. I have a hunch the reduction given by Dennis Kraft would still hold, because you're not considering the planar input graph coloring in isolation, but I might just as well be wrong.", "I've edited the question to reflect your comment that the number of colors is unlimited. In the future, when someone asks a clarification question, please edit the question -- don't just leave clarifications in the comments, as people shouldn't have to read the comments to understand your question. The source of the confusion might have been: if edge weights are non-negative, and if there's no limit on the number of colors, the problem is trivial: just assign each vertex its own color. However in this problem edge weights can be negative, so the trivial solution isn't necessarily optimal.", "No, it's not given.", "The number of colors is given, right?"], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "422", "site": "cs", "title": "When can you "invert" an equation in the lambda calculus", "body": "Suppose that $M$ is a full model of the simply typed lambda calculus. Suppose each base type is infinite.\n\nNow suppose that $f$ and $g$ are two functions in $M$ (not necessarily in the same domain) that are not definable by any pure term, and that $\\alpha$ is a pure term of the $\\lambda$-calculus such that: $$M(\\alpha) g = f $$\n\nMy question is: when is it possible to find a pure term $\\beta$ such that $$g=M(\\beta)f$$\n\nAre there easy to state necessary or sufficient conditions? I have been unable to find counterexamples when gs type complexity is higher than fs, so maybe that is a sufficient condition?\n\nA couple of examples:\n\nif $f$ is the converse of $g$: $$f=M(\\lambda xyz. xzy)g$$ then we can find such a $\\beta$ -- in this case, letting it also be $\\lambda xyz. xzy$. \n\nSimilarly If $f$ is application to $a$: $$f = M(\\lambda xy.yx)a$$ then $a$ is $f$ applied to $\\lambda x.x$: $$a = M(\\lambda y.y(\\lambda x.x))f$$\n\nLastly, if $x$ is a variable in some base type and $f=M(\\lambda xy.x)a$, then I don't think there is any pure term such that $a=M(\\beta)f$.\n", "link": "https://cs.stackexchange.com/questions/57786/when-can-you-invert-an-equation-in-the-lambda-calculus", "tags": ["lambda-calculus", "combinatory-logic"], "votes": 11, "creation_date": "2016-05-23T18:15:31", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "423", "site": "scifi", "title": "Novel about an immortal man exposed by world government relates his life through history", "body": "I read the book from my local library in the early- to mid- seventies. It was hardcover (library bound) and the dust jacket had a printed burlap or canvas background. The back flap mentioned that the author's name was a pseudonym. The book had lots of conversation, anecdotes and opinions with little to no 'action'. I came away with impression that the author wasn't a professional writer, but rather more like a history professor with an academic reputation to protect. \n\nIt begins shortly after the formation of a world government, and the President of Earth is presiding over the trial of Earth's only immortal man, for the crime of withholding the secret of immortality. Much of the book is taken up by his narration of stories of his life, from his birth in prehistoric times to the present. This could almost be the backstory of 'The Man From Earth', but IIRC he was never a famous person. \n\nOne anecdote I remember is what he called 'Pavlov's Other Experiment', with the supposed discoveries by the scientist after the surprising results of the 1924 Leningrad flood. The novelist added a vitamin deficiency factor that became an important plot point. There was also a story I recall as being very amusing about a British military steamship in Egypt before or during the occupation in the eighteen hundreds.\n\nAnd finally, as a major plotline (such as it is), everybody is taught the simple trick to become immortal. When dead people are cloned and become 99.99% similar, their soul and memories are reunited with their bodies and they are taught immortality; time-traveling robots collect DNA from everyone who has ever lived and they become immortal as well.\n\nThis is not Bixby's \"A Man From Earth\", Anderson's \"Boat of a Million Years\", Heinlein's Lazarus Long, Borges' \"The Immortal\", Zelanzny's \"This Immortal\". The main character was not a time traveler or the Wandering Jew. It is possible I'm getting the similarity thing confused with van Vogt's Null-A, but there was something like that. \n\nThe immortal man was likely 6000-8000 years old, and the time of the novel in a future not far from now. His prehistoric tribal roots being near the invention of agriculture? (maybe)\n", "link": "https://scifi.stackexchange.com/questions/66361/novel-about-an-immortal-man-exposed-by-world-government-relates-his-life-through", "tags": ["story-identification", "novel", "immortality"], "votes": 59, "creation_date": "2014-08-23T20:23:52", "comments": ["The part about the protagonist being thousands of years old eliminates the novel Operation Longlife by E. Hoffmann Price. Doc Brandon, an obvious knockoff of the pulp hero Doc Savage, was less than two centuries old when the government realized he'd somehow stumbled across a method of keeping himself young and fit, indefinitely, and they very much wanted to persuade or coerce him to share his secrets.", "@ImaginaryEvents Elements of it do sound hauntingly familiar, but then Time Enough for Love and This Immortal are a couple of my favourites so it's probably just that.", "@Ash, no, the book was new and a complete novel, Perhaps I mis-capitalized 'other experiment', it was just how the character referred to it. I'm still hoping someone like yourself will recognize it or discover it.", "The description you've given of the condition of the \"book\" in question actually sounds like you've found a single bound copy of a serial of some sort, it might be worth checking back catalogs from the like of Strange Tales and Omni for particular parts that stand out to you. But given that the title \"Pavlov's Other Experiment\" returns zero useful hits anywhere I can think of you may be out of luck I'm afraid.", "@GorchestopherH Nothing yet. I will update here if I get the answer.", "Have there been any updates on this on other sites?", "Always worth mentioning what you've tried before so you can eliminate the obvious.", "Thanks. There was a second reddit thread as well, but I tried to mention the top guesses above. reddit.com/r/printSF/comments/va9dz/…", "I'm sorry. @Richard, what is obvious?", "reddit.com/r/tipofmytongue/comments/wi3hk/… - Obviously"], "comment_count": 10, "category": "Life & Arts", "diamond": 1} {"question_id": "424", "site": "scifi", "title": "Looking for science fiction assassination story with mysterious girl", "body": "For years I've been looking for a short science fiction story about a man who was released from a prison planet in order to assassinate a candidate for galactic president. He does this on a space station, and escapes with the help of a mysterious and very attractive girl with silvery hair and multi-colored skin.\n\nI read this short story some 5-6 years ago online.\n\nThe assassin was released from the prison planet by some influential people so that he may assassinate the galactic president candidate, whose rule would, according to said people, be bad for the galaxy.\n\nI also remember that the girl, who is actually a young woman is also a prostitute. The first time he sees her she's walking into the bar accompanied by another man. I don't remember why she helps him. She was likely his contact, given to him by those who sent him on the assassination.\n\nThey escape together after creating a diversion with an explosion, running off into some corridor that probably leads to a way off the space station.\n\nThe story story ends there but in my opinion leaves room for a sequel, even if only to explore the protagonists' developing relationship. \n", "link": "https://scifi.stackexchange.com/questions/27694/looking-for-science-fiction-assassination-story-with-mysterious-girl", "tags": ["story-identification", "short-stories"], "votes": 55, "creation_date": "2012-12-06T11:39:59", "comments": ["Jupiter Ascending? Which is a 2015 film, and not a short story I read somewhere in the early 2000s. I saw that film, and never even noticed any common plot elements. Sorry.", "Sounds like Jupiter Ascending (imdb.com/title/tt1617661).", "Aspects remind me of Spinrad's \"Agent of Chaos\". But that was a full novel and not a short story.", "Hello @Moacir and thanks for trying to help me find this story. I looked up the index of that issue here (archive.org/details/…), but couldn't find a story by that author or by that name in 1983. Moreover, I found no place where a story by that name is attributed to that author.", "Hello @Dovid, I've set a bounty on your question. If you remember any additional details, now would be a good time to edit them in.", "Novels are often serialized in magazines.", "To the best of my knowledge it was a short story, not part of a longer story. Although, as I said, it asks for a continuation. So I am not 100% sure. I read it online.", "Are you sure it was a short story? Could it have been an excerpt from a longer story? Also do you remember if you read it in a magazine, book or online?"], "comment_count": 8, "category": "Life & Arts", "diamond": 0} {"question_id": "425", "site": "scifi", "title": "Story about man unable to wake up from dream", "body": "I am looking for a story which I read a long time back but cannot recall the name. The plot line was: \n\nIn a futuristic world, there are dream parlors which provide sleep with dream services. The protagonist is a frequenter of them. \n\nOne day, it was found that he is unable to wake up after slipping into a dream. All efforts to wake him up fail, and he keeps on dreaming and sleeping but is medically fit otherwise. \n\nHe became a medical case study and was kept alive for centuries. By studying him a lot was learnt about hibernation and the code to interstellar travel in hibernation was cracked through that knowledge. \n\nEventually after a few centuries the government asked his custodians for justification for spending billions per century to keep him alive.\n\nHis custodians revealed that:\n\n> he is the last person alive who is capable of dreaming. The entire human race has lost the capability to dream. So he should be kept alive as long as possible as a rare treasure to humanity.\n", "link": "https://scifi.stackexchange.com/questions/117737/story-about-man-unable-to-wake-up-from-dream", "tags": ["story-identification", "dreams"], "votes": 43, "creation_date": "2016-02-08T02:42:15", "comments": ["Sounds like a good story... Were there aliens/robots or anything?", "Damn, I have read this before, but I cannot for the life of me remember the name of the story. I think it was a short story from an anthology by Heinlein, but I can't say for sure.", "When was \"a long time ago\"? You may not remember exactly, but can you give a ballpark? What is a long time ago to one person may only be a handful of years to another."], "comment_count": 3, "category": "Life & Arts", "diamond": 0} {"question_id": "426", "site": "scifi", "title": "A space story about a desert planet and a cup of water", "body": "I read this story over 15 years ago. It is probably a lot older. I read it in Serbian magazine (Politikin Zabavnik) but I think that the language of the original is in English. It's a short story, less than two A4 formats long.\n\nThe story as I remember it is as follows:\n\nA young man/boy (crash)lands on a desert planet. His only possessions are clothes on his back and a cup. Maybe a gold necklace, too, I am not sure about that. He reaches the biggest city on the planet.\n\nThe main trade and commodity on the planet is water. The richest families are those that have wells, people work for water, trade deals are made for water etc. The youth comes to the town and says to one of the smaller traders in water that he is thirsty and would like to buy a cup of water. There might be some flying robots that ensure that all the contracts are obeyed by killing anyone who tries to renege on the deal, I am not sure.\n\nAnyway, the youth is not allowed to drink, not right now. Trader thinks he has nothing of value or something. A bigger water trader, as a joke, offers to sell the youth a cup of water. His fellow traders joke with him, tell him they would loan him a cup or something, because the youth doesn't have anything of value and would have to sell himself into slavery to survive on the planet.\n\nI do not remember what exactly happened, but the end result is that all the biggest traders on the planet sign a contract with the boy. And it goes like this: a boy's necklace for one cupful of water. One cupful, nothing more. The arc words are: _What is the value of one cup of water?_\n\nThe end of story is this: the boy puts a drop of water into the cup. The cup opens up a bit then starts transforming, in fractal designs. It sucks all of the water in the city, maybe even in the whole world and that giant form of cup is floating in the air now, blotting the sun. Some of the traders get worried halfway, try to renege on the deal and are zapped by the robots. Once there is no more water and the cup stops filling, the boy laughs at now poor traders and asks them, \"What is the value of one cup of water?\"\n\nIt is very similar to the story about king that would give half of his kingdom for a cup of water and another half for some bread.\n", "link": "https://scifi.stackexchange.com/questions/79870/a-space-story-about-a-desert-planet-and-a-cup-of-water", "tags": ["story-identification", "short-stories"], "votes": 42, "creation_date": "2017-09-28T09:16:06", "comments": ["It was a kid's story-they read it to us in class back in the 2nd or 3rd grade-so this would put the book at least before 1990, if that helps anyone"], "comment_count": 1, "category": "Life & Arts", "diamond": 0} {"question_id": "427", "site": "scifi", "title": "Futuristic world where reality is humanity's plaything and God is in a zoo", "body": "This is a novel I read in 2015. It was an older book. The cover was black and hardcover. Don't remember any specific details. The book was in English. I believe it was from the 1980s or ‘90s.\n\nIt was set in the future where the world is united as one. \n\nI'll try to outline all the major events that I remember, in as much detail as possible.\n\nThe story begins with a prologue of how a mutant gene called \"The God Factor\" was isolated and subsequently researched to reveal it gave humans reality-bending powers.\n\nThe first few test tube babies born with the gene were horribly deformed and could only lift light objects with their mind. However, with every subsequent generation they grew stronger until they could manipulate events on a universal scale. They were fitted with microchips which gave scientists complete control over them. \n\nUsing these people humanity became an interstellar civilization. Thoughts and ideas are no longer abstract as people with the God Factor can bring them into existence. The world government brings into existence \"God\", a manifestation of the collective faith of people, and places him in a zoo. \n\nTo mock death itself, the human race increases the biological lifespan of all humans by a trillion years, reasoning that, with unlimited resources and a universe too vast, they need as many people as possible. And that's where everything starts going wrong. \n\nThe protagonist is a scientist who works under the guidance of the scientist who was a part of the team that originally created the God Factor babies. He gets involved in a riot where 50 people immolate themselves, seemingly for no reason. Similar riots spread globally and on human colonized planets. The protagonist discovers that the reality benders have been siphoning energy from other universes to provide to their own and the continuous transfer of energy has caused the membrane separating different universes to weaken which causes severe suicidal tendencies in exposed people. Over 75% of the world population dies in a week as even the GF people are affected causing them to go supernova when dying. \n\nThe protagonist rushes to try and build a \"reality anchor\". It had a fancy name but I can't remember it for the life of me. As the world delves into anarchy, he used \"God\" to power it and restore the world's reality to it's original state. Then he discovers that he is in the lab which his mentor used to work in while they were working on isolating the God Factor gene. The book ends with him in a dilemma of whether to destroy the project once and for all or let it proceed and try to influence events in the future. \n\nIt's an amazing book and much deeper than I outlined here. It's very long and very thrilling. \n", "link": "https://scifi.stackexchange.com/questions/203664/futuristic-world-where-reality-is-humanitys-plaything-and-god-is-in-a-zoo", "tags": ["story-identification", "novel", "genetic-engineering"], "votes": 39, "creation_date": "2019-01-17T20:42:52", "comments": ["@NeoDarwin Does anything in this question of mine sound familiar? scifi.stackexchange.com/questions/14510/…", "Reality anchor is in quotes because it had a complicated name. It's purpose was to anchor reality.", "This got nothing", "It is not this book", "Very confident. I read it only 3 years ago and have searched for it myself for the past week. Only upon failing did I decide to ask here @John Rennie .", "It seems this should be an easy find with such easily Googlable terms as the god factor, but I have to confess I've drawn a complete blank. How confident are you that you've remembered the phrase correctly?", "@Broklynite His mentor used to work in a different lab and it's there they discovered the God Factor. I think the author mentioned that it's since being converted into an museum.", "“Then he discovers that he is in the lab which his mentor while they were working on isolating the god factor gene” I think there is a bit missing from the middle of that sentence. And how do you mean that he discovers that he is in a lab he presumably used to work in? Was this hidden from him in some way? And is the idea that reality has been restabilized but questioning if it should go on, or that the restabilization is possible but should it be? Sorry, just trying to get some clarifications."], "comment_count": 8, "category": "Life & Arts", "diamond": 0} {"question_id": "428", "site": "scifi", "title": "Old SF novel, set on a military ship, one character can't order spaghetti and meatballs because of sudden mid-voyage rationing?", "body": "This one is a real longshot. I won't be surprised if nobody can identify it. I read this book sometime in the early-to-mid-1980s. Science fiction novel, checked out from a public library, written in English, and I am pretty sure it was in hardback. Told in the third person, with multiple viewpoint characters, and a fairly serious tone -- not any sort of parody or satire, for instance. The problem is that I only remember _one short scene_ clearly. Here's how it went:\n\n 1. This part of the story -- perhaps most of the story -- is taking place aboard a large spaceship with interstellar range (military, I'm nearly sure) which is large enough to require a great many crew members (probably hundreds). I think there is some sort of interstellar war going on, or else one seems likely to break out at any minute, but I can't remember details of the larger plot. I'm _sure_ the crew of _this_ ship were all human -- but I don't remember if their enemies (or prospective enemies) were equally human, or if this was set in one of those futures where there are other sentient alien races with their own battle fleets. \n\n 2. In the one scene that I still remember pretty well, which I believe comes near the middle of the book (definitely not in the first chapter!), the viewpoint character -- who was not a major figure in the early chapters -- has just completed a long shift. I believe he's part of the Engineering Department. Now he wants to relax by eating a nice hot dinner.\n\n 3. He decides to splurge. Instead of taking whatever is standard-issue food aboard that ship, he is willing to spend extra to get one of his favorite dishes. I don't think it was regular money he would be spending, though. The implication was that, in order to keep the ship's galley from running out of all the best items too fast during a long cruise, the rules were set up so that you had to accumulate \"ration points,\" or something similar, for the _occasional_ special treat. The rest of the time, you just ate something much simpler and cheaper, three times a day, as a way to keep your body adequately nourished. For the sake of argument, let's say that what this guy wanted was a big plateful of spaghetti and meatballs, with a certain type of salad on the side. (I don't swear that this is accurate, but I _think_ he was looking forward to spaghetti and meatballs.)\n\n 4. He punches his order into a computer terminal, knowing full well that his account has more than enough \"ration points\" (or whatever they are called in the ship's economy) to cover the special dinner he's been promising himself. \n\n 5. Nonetheless, despite how hard he has worked to accumulate those points which are supposed to be redeemable whenever you like, he gets an error message telling him his order _cannot_ be filled today because of some change in ship's policies. He throws a fit! (I don't mean literally going berserk, but he does complain loudly.)\n\n 6. I just _vaguely_ remember that there was some sort of problem with the ship which had caused the Captain, earlier that day, to approve some new rationing restrictions on various things as a reaction to \"a state of emergency\" (or words to that effect). I don't remember if this was because of lack of energy to make nonessential things happen, or because they were carrying too many people and couldn't create special meals on demand for everyone, or what. Apparently the character I was just mentioning had not gotten the word yet when he sat down to eat dinner. I think the scene I described was followed by another scene showing the Captain (and/or some of the other senior officers) talking about how a flood of complaints was already coming in from sailors who didn't know all the facts about the problems the ship was currently facing (or was expected to face in the near future?). \n\n 7. I don't remember a thing about the cover art or the author, but I can say that if this novel had been written by any of the Really Big Names of 20th Century SF who sometimes wrote stuff with a military-style setting -- such as Poul Anderson or Gordon R. Dickson -- then I would have expected to run across the book again, sometime in the last couple of decades, and I haven't. (And it _certainly_ wasn't written by Heinlein; I'm familiar with everything he published.) I'm also sure it _wasn't_ a \"licensed novel\" set in a preexisting SF universe, such as _Star Wars_ or _Star Trek_.\n\n\n\n", "link": "https://scifi.stackexchange.com/questions/145076/old-sf-novel-set-on-a-military-ship-one-character-cant-order-spaghetti-and-me", "tags": ["story-identification", "spaceship"], "votes": 36, "creation_date": "2016-11-13T18:07:05", "comments": ["Seems similar enough that it might be a duplication of this answered question.scifi.stackexchange.com/questions/286709/…", "This makes me think of a story I read in which garlic has been restricted onboard a spaceship, not due to a shortage, but because the aliens who have infiltrated the ship are allergic to it. Our hero, in tracking down why he can't get spaghetti, uncovers the plot.", "I wonder if this couls be part of the Fixup The Great Explosion by Eric Frank Russel, the last part of which is better known as \"...and then there were none\"?", "Hmm, for some reason it makes me think of Tau Zero by Poul Anderson, but you say it's definitely not by him.", "@Iolrus I have read Tactics of Mistake several times (although not in the last few years). I don't remember ever noticing a scene such as I described above, with a member of a starship's crew getting frustrated over his inability to redeem his \"food ration points.\" I think the story I asked about, two years ago, was mostly set aboard a starship during a long voyage. Cletus Graeme was the main POV character in Tactics, and spent most of his time down on the ground as a soldier outsmarting his enemies during campaigns on various planets. Hope you can find what I'm looking for, though!", "I'm sure I have read this in the last year or two, in which case I think it might be \"Tactics of Mistake\" by the aforementioned Gordon R. Dickson. It's the only book that fits the bill and is on my goodreads list.", "Worth a try I suppose 🙂", "@Danny3414 -- I'm sure those are two different things. I first read a Seafort book (#3) in paperback in the mid-1990s. Since then, I've read most of the others. The story I'm asking about is something I'd read at least a decade earlier, in a hardback I had checked out from a public library. Of course, stories of people in spaceships having to carefully ration their food in emergency conditions have probably been done dozens of other times in the history of SF.", "Are you certain about the dates you read this? It reminds me of one of the Seafort saga books by David Feintuch but they were in the nineties. In one of them a batch of their ship's stores have been contaminated by Fish like aliens and the ship has to go onto emergency ration reductions", "Hm, okay, then you might try the Humanx Commonwealth series, by Alan Dean Foster. I mostly only read the Flinx and Pip story arch, but I did pick up another book or two from the rest of the series, so it might have been one of those.", "Interesting suggestion. Way back around 1983 or 1984 (roughly), I read the original Foundation Trilogy, and then the recently-released Foundation's Edge. I've reread those stories at various times, but not for some years now. Off the top of my head, I'm not remembering a scene such I described in my original post, though -- not in one of the first 4 Foundation books. It wouldn't be in any later \"Foundation\" book; I'm sure they weren't even in print when I first read the scene I've described about getting an error message when trying to order a special meal aboard a starship.", "I very, very vaguely remember something like this, and I think it was part of the Foundation cycle by Isaac Asimov. However, not 100% sure, and there are a lot of books in that series, so only a comment, but you might want to look into it. Most of the books are great reads!", "It's funny, I was also thinking Stainless Steel Rat or Bill the Galactic Hero. But the scene I think you folks are thinking of was when the titular character gets on board a ship and is forced to eat (I think it was called) rat-puppies the entire voyage. A rat-puppy was a ration sausage that got poked with a knife and blew up large and self cooked. It was tasty enough the first time, but it was the only food available the entire way. While it does have a similar feel to the OPs question, I don't think it matches.", "I did read The Forever War back around the mid-1980s, but I've read it again since, and it wasn't this. One reason I'm sure is that \"The Forever War\" was very much a first-person narrative focused on one character's experiences. The book I'm vaguely remembering was third-person, with multiple viewpoint characters.", "I doubt it's The Forever War by Aldeman, but that book does have some scenes with food rationing and kilocalorie equivalences being used to purchase goods and food.", "Regarding the Harry Harrison suggestion: I'm sure the book in question was not written as comedy/parody; it was more serious. That eliminates the first Bill, the Galactic Hero book (although I did check it out from the same library in the 1980s, and enjoyed it). Also, the book I want was written in third person, with multiple viewpoints -- that would help eliminate the Stainless Steel Rat books. (Besides, I'm positive I didn't read anything about Slippery Jim diGriz until years later! The library near where I lived in the mid-1980s didn't have any of his stuff.)", "@Longspeak found the first one online, searched through it, no joy.", "@SQB I was just going to suggest Bill the Galactic Hero, too. Pretty sure it is, but have to wait until I get home to confirm.", "Rings a bell, vaguely. Could it be something by Harrison, either in his Bill, the Galactic Hero series or his Stainless Steel Rat series?"], "comment_count": 19, "category": "Life & Arts", "diamond": 0} {"question_id": "429", "site": "scifi", "title": "Short story about a man in a madhouse who asks the guard for a cigarette and tells him his story about having swapped bodies before", "body": "I am searching for a short story about a man locked in a madhouse who swaps his body with the guard/doctor while he tells him the story about his own body being swapped.\n\nI remember the following:\n\n * The madman asks the guard for a cigarette and starts telling him his own story.\n * There was some relative of the madman who had a miserable life and one day he took the madman for a chat to a gloomy place where he would perform the conjuring to swap bodies.\n * When the madman reaches the end of the story, he has already swapped his body with the guard's and starts speaking as if he is in the corridor of the hospital and not in a cell.\n * The madman, who is now the guard, tells his victim not to yell nor to despair for nobody will believe him. He tells him of the importance of the burnt offering which, in this case, was the cigarette and not to forget 'the words'. Then he takes the guard's car keys and drives away, leaving the guard locked in his cell.\n\n\n\nI read this story in 2006. I searched google but the keywords hospital/madhouse take me to news and the keywords cigarette/burnt offering take me to religious content. I have no clue as to where was this story published.\n\nThanks in advance.\n", "link": "https://scifi.stackexchange.com/questions/223920/short-story-about-a-man-in-a-madhouse-who-asks-the-guard-for-a-cigarette-and-tel", "tags": ["story-identification", "short-stories"], "votes": 35, "creation_date": "2019-12-03T05:39:17", "comments": ["I, too, remember this story. I've been trying to recall the title for years now. I first read it sometime in the early 90s in a short story anthology I picked up while living in Brazil. I'm pretty sure the collection was published for ESL students, as I remember it was in the library of the language school \"Associacao Brasil-America\" where I taught for several years. For a while now I had been thinking I only imagined this story, until I saw your post! Just wish someone out there could remember the title . . .", "The concept sounds similar to the A.E Van Vogt short story \"Dear Pen Pal\". In that one the story is a series of letters from an alien to an earth man. During the correspondence the alien performs a mind swap with the man, the last letter being written by the man (now in the alien's body). en.wikipedia.org/wiki/Dear_Pen_Pal", "I read this probably before 1986 and certainly before 1992. Maybe something by Zelazny or Harlan Ellison or something in one of the Campbell awards books.", "Sounds similiar in concept to the movie 'Fallen' (1998) link - but checking that I can't see anywhere that it was based on a novel...", "I recall this story too. It trigger something... will try to remember. In the meanwhile I suggest you, with a very similar plot, the short story by Julio Cortazar, \"Lejana\" in his great \"Bestiario\": es.wikipedia.org/wiki/Bestiario_(libro)#Lejana", "Can you remember when you read this? If it was new at the time? What anthology/magazine it may have been in if one of the reoccurring ones? etc. If so you can edit your question with this information."], "comment_count": 6, "category": "Life & Arts", "diamond": 0} {"question_id": "430", "site": "scifi", "title": "A story about a doctor who travels to a parallel universe to seek a cure for bubonic plague", "body": "Recent news has been reminding me of various things I've read about the black death, including a story I believe I only saw once, and never stumbled across again. I encountered it over 30 years ago, so I remember nothing of such details as character names. (Not even what city, in what country, it was set within.) I know it was a shorter piece of fiction, rather than a complete novel. I found it in an anthology I checked out from a library, probably no later than 1988. English language. I can't remember what else was in the anthology, nor what the overall theme may have been, but I'm sure it was **not** a collection of several stories by the same author.\n\n* * *\n\n**Plot Points**\n\n 1. The main character, whose viewpoint is presented to us in the third person, lives in the twentieth century, but not _our_ twentieth century. The history of western civilization veered off in a different direction some centuries earlier. (I can't remember just when the divergence was stated or implied to have occurred, but at least one language spoken by the protagonist is still comprehensible over here, on our side of the divide.) As we meet him, he is already an example of his world's version of a fully trained physician who is dedicated to his work. I _think_ he is called \"doctor\" as a result of his education at a reputable university, but he doesn't use the methods we normally expect of a modern doctor – no immunization shots, no antibiotics, no X-rays or other imaging equipment, and I don't think he has ever touched a microscope in his life – because all those things, and the theories behind them, don't seem to exist in his world.\n\n 2. In the protagonist's culture, the preferred method of treating the sick is the sort of thing I would associate with \"witch doctors.\" Elaborate magical rituals involving chants and dances and so forth, I think. In some cases, dealing with some types of ailments, these methods have delivered a comfortably high success rate, which, presumably, is why these methods are now considered the correct way to practice medicine, and anything else is widely viewed (by physicians, anyway) as a complete waste of time.\n\n 3. Our hero is a maverick. When the bubonic plague hits the city in which he resides, and hits it _hard,_ he becomes convinced that it is caused by something which his magic spells are completely unable to cure – and he doesn't think any of his professional colleagues in the area are doing much better. He wants to use some \"experimental\" and \"alternative\" methods which are radically different from the \"proper\" procedures he had diligently studied in school. I can't remember exactly what his methods were, but I know that my teenage self thought they at least _resembled_ some of the measures which a real-world modern doctor might recommend if he were on the scene – assuming the modern doctor lacked access to antibiotics and other modern tools of the trade, such as I mentioned in point #1, but wanted to help as best he could, based on his knowledge of the germ theory of disease and other useful points.\n\n 4. There is an argument in front of some sort of authority regarding whether the protagonist should be allowed to inflict such time-wasting and unproven techniques upon his trusting patients. (I think the authority in question was a committee which was either partially or entirely composed of senior medical men.) He gets a partial victory – he is allowed to take responsibility for treating any plague victims residing in one sector on the city map. Public health statistics will be gathered from each sector, day by day over the next several weeks, to see if his methods appear to be making any difference at all.\n\n 5. That decision _sounds_ fair and reasonable – except that, in practice, one or more of our hero's \"competitors\" (other doctors, using the time-honored methods of treatment within their assigned sectors) start _cheating._ Some weeks after that policy decision was made, the protagonist is told by one of his loyal assistants that a lot of dead bodies of plague victims are turning up _just inside_ the boundaries of the hero's assigned sector, and it's pretty obvious that at least one other doctor is having fresh corpses loaded onto wagons and dumped along the fringes of that sector in the middle of the night so that their fatality rates look lower, and his look higher.\n\n 6. The protagonist finally tries something completely new. This is another area where my memory fails. _Somehow,_ he has gained access to a magic spell that could let him travel from his world to a parallel world where history followed a different path. I think he needs other people to help by chanting portions of the spell, making certain gestures or dance steps, etc., to make it work properly. He can't just cast it on himself. I don't remember where this spell came from, nor why it had rarely (if ever) been used before. The doctor puts it into effect and vanishes into thin air.\n\n 7. At this point, the story stops showing us his viewpoint, and we start reading a formal report written by a professional medical man in what appears to be _our_ world. The report tells us that a strangely-garbed man was found in the streets of their city (our world's equivalent of the setting of the previous scenes, I believe) and was quickly diagnosed as being feverish from a serious case of bubonic plague. He was taken to a hospital and pumped full of antibiotics. Physically, he made an excellent recovery, but he kept saying nonsensical things about world history, about his own alleged medical training and preferred techniques, etc., which made no sense to anyone else, and so he was placed in the psych ward for a while. The plan was to figure out how deep his mental problems ran – was he delusional as a regular thing, or just temporarily having trouble distinguishing between \"reliable memories\" and some elaborate hallucinations which he might have experienced while delirious with that fever, or what?\n\n 8. The author of the report notes that after landing in the psych ward, this mystery patient became acquainted with the other mental patients and then started doing some \"magic\" nonsense, with ritual gestures and chants and so forth, which he loudly claimed could \"cure\" the afflictions with which they had been cursed. (I don't remember for sure, but _he_ may have claimed he was \"casting out demons\" or something along those lines.) The report-writer assures his intended readers that while several of the mental patients do, in fact, appear to have been acting much more lucid and well-adjusted since then, it must be a coincidence. The painfully obvious implication (to us, knowing what we know) is that the protagonist does, in fact, know some \"magical\" techniques which somehow can make a huge difference in treating cases of paranoia, schizophrenia, etc. In other words, _both_ worlds could benefit enormously by having the members of their respective medical communities compare notes and maintain an open mind about anything that seems \"crazy\" at first glance.\n\n 9. Again, my memory fails: I can't remember how the story ends. Specifically, I don't know whether the protagonist finally managed to get hold of a stash of antibiotics, etc. which would allow him to make a huge difference in treating plague victims after he carried those modern assets back home via another use of the same spell. (Note: As I was about to post this, it suddenly occurred to me – but I'm not sure if this is a genuine memory or a wild guess – that the protagonist _may_ have somehow persuaded several of his fellow mental patients to each perform one role within the coordinated team effort that is required to reverse the travel spell and send him right back to his native timeline with his new knowledge and medical supplies. _Maybe._)\n\n\n\n\n* * *\n\nDoes anyone else remember reading this?\n\nI did a little Googling before I took the trouble to type out all of the above, as well as searching through previous questions on this very site. I kept coming up with references to various full-length SF novels that gave lots of attention to the historical black death, and/or alternate-timeline variations and their aftermaths, but these novels definitely are not related to the much shorter story I had in mind.\n\nSo the reject pile – the list of things not to bother offering as \"answers\" – includes all of the following:\n\n * _The Years of Rice and Salt,_ by **Kim Stanley Robinson.**\n\n * _Doomsday Book,_ by **Connie Willis.**\n\n * _The Plague Tales,_ by **Ann Benson.**\n\n * _The Crystal Empire,_ by **L. Neil Smith.**\n\n * _In High Places,_ by **Harry Turtledove.**\n\n * _Any_ of the novels and shorter stories that were ever published as part of **Randall Garrett's** \"Lord Darcy\" series. (Not all of them written by Garrett.) Those tales were set in a parallel universe where most technology is lagging a century or more behind ours, but medical treatment is largely a matter of using psychic healing techniques which _frequently_ succeed. At one point, a sorcerer spoke disparagingly of the superstitious fools who think applying some moldy old bread to a fresh wound can somehow keep it from getting a nasty infection. (After all, what good did mold ever do anybody?)\n\n\n\n", "link": "https://scifi.stackexchange.com/questions/230744/a-story-about-a-doctor-who-travels-to-a-parallel-universe-to-seek-a-cure-for-bub", "tags": ["story-identification", "short-stories", "alternate-history"], "votes": 33, "creation_date": "2020-05-02T21:28:52", "comments": ["Nourse did write at least one story about the plague but it has a very different plot.", "0 I have read this story and remember it. I am 80% certain it was a short story written by Alan E Nourse, who wrote lots of medical sci fi stories. Hunted through my collection to see if I can locate it but annoyingly, I cannot find it in any Nourse collection. If not him, it was a contemporary like Blish...but I still think it was Nourse. I recall the story very clearly.It doesn't seem to be JAmes white either.", "S/what related to The Hertford Manuscript, a sequel to H. G. Wells' The Time Machine by Richard Cowper: the protagonist gets stuck in 1665 London and attempts to direct public health measures against rats, eventually getting infected.", "@user888379 I'm very familiar with all three, but I hadn't really thought of connecting them to this one. \"Wheels of If\" probably comes closest -- it deals with going \"sideways in time\" to where it is still \"the same year,\" but history took a different turn centuries ago, instead of just people or artifacts from a better future coming back to the 20th Century. Also, the other two had tragic endings. Although in \"Wheels of If\" I was never convinced that it was perfectly all right for the main character to deliberately leave his analogs stranded in each other's bodies in other timelines.", "This feels familiar, but it may just be that I'm reminded of aspects of de Camp's \"Wheels of If\", Moore's \"Vintage Season\" and Kornbluth's \"The Little Black Bag\" - I could imagine it having been published in \"Unknown Worlds\" or \"Astounding\" in the '40s."], "comment_count": 5, "category": "Life & Arts", "diamond": 0} {"question_id": "431", "site": "scifi", "title": "SciFi Short Story about an immortal man digging through an asteroid to meet a woman", "body": "Short story from the 80's or 90's, I saw someone asking about this on another forum but never got an answer. I've been curious ever since.\n\nThe story is set in a decadent future where everyone lives forever and mostly they go to parties. Guy meets a girl at a party, suggest he start at one side of an asteroid and digs a tunnel by hand to her, while she waits in a hollowed out space and creates a beautiful garden while waiting (robots take care of all mundane life support stuff).\n\nSo he digs for 100 years, makes it through to her chamber, they have a long romance, and when they've had enough they go back to civilization. The robots bury everything and return the asteroid to the way it was. Then they run into each other at a party and she overhears him proposing it to a different girl. That's when she says, \"You've done this before.\"\n\nThis is the original post, for those curious (though I've reposted all relevant info): \n", "link": "https://scifi.stackexchange.com/questions/120302/scifi-short-story-about-an-immortal-man-digging-through-an-asteroid-to-meet-a-wo", "tags": ["story-identification", "short-stories"], "votes": 30, "creation_date": "2016-02-22T23:57:46", "comments": ["That's not a knife, that's a spoon!"], "comment_count": 1, "category": "Life & Arts", "diamond": 0} {"question_id": "432", "site": "scifi", "title": "A man sent to a parallel Earth to save someone important", "body": "This is not a novel, but \"longish\" short fiction, novelette or novella. I read it around 1984, in English, in an already used collection containing half a dozen or so stories.\n\nA man, on our Earth, is contacted by a bizarre agency that offers him to get his mind transferred to a body in an alternate Earth, in order to save, either some polity from an external enemy, or the queen of some polity threatened by an \"internal\" cabal. Why he accepts is not clear, since when will he be able to use the money he is offered? No return seems promised.\n\nHe gets there, and meets several species of humanoid beings. They all have names that contain the word \"man\" in the local language, but I forgot how that sounds. Somehow, the agency that hired him managed for him to understand the language of the species he was supposed to help (otherwise, he would have been useless!) When he finally meets them, they tell him the names of all the other species, from his descriptions. And the names were translated into English for the benefit of the reader. This is how I found out that the second part of the name, common to all species, meant \"man\". Note that this is a \"reconstructed\" answer to Valorum's question in a comment, not a direct memory. An alternate possibility is that the language capabilities he must have received from the agency that hired him included the words for man, snow, cave and water and he invented the name of the species in the local language himself, but it is rather unlikely.\n\n * One species is a \"Snow-man\", a huge and frightening Yeti (but not the most aggressive species in that world) who lives in the eternal snows of a high mountain range (or polar cap?).\n\n * Another is a \"Cave-man\", without eyes because they live in deep caves with no light.\n\n * A third one is amphibious, \"Water-man\" or \"Sea-man\" or \"River-man\", I forget which.\n\n * One species practices cannibalism, and they file their teeth better to tear flesh. I forgot if it is one of the latter two, or still a fourth one.\n\n * Finally, the ones he is supposed to help are essentially like us.\n\n\n\n\nOur hero crosses all these obstacles, reaches the polity he has been hired to help, and saves the queen from the cabal, or possibly the entire polity from the enemy. I don't remember the end well, but he never goes back to our Earth.\n", "link": "https://scifi.stackexchange.com/questions/211068/a-man-sent-to-a-parallel-earth-to-save-someone-important", "tags": ["story-identification", "parallel-universe", "novella"], "votes": 29, "creation_date": "2019-04-28T10:07:19", "comments": ["@MadPhysicist Very funny. I was referring to en.wikipedia.org/wiki/Graphic_violence IIRC the scenes with the cannibals did involve some violence and some blood, but well within the very moderate criteria of SF at that time. So as I understand this wiki page, they was no \"gore\".", "@Alfred en.m.wikipedia.org/wiki/Al_Gore", "@Valorum If by \"gore\" you just mean a lot of blood, well, when he met the cannibal species, there were maybe some scenes with a lot of blood. But not \"gore\" in a sense of something really offensive, disgusting. As a Frenchman, I am not completely clear about the meaning of \"gore\".", "Re: phrases : no, alas, I don't remember any phrases.", "@Valorum Re: sexual content. I cannot vouch that no discrete allusion to sexual contact was mentioned. But if it was, it would have been in the very, very subtle way such topics were treated in SF of before 1990.", "@Alfred - That needs editing in. Also, can you recall any specific phrases used in the story?", "@Valorum When he meets the final species, the one closest to us, and understands their language (the agency that sent him probably arranged this, otherwise how could he be of any use ?) they tell him the names of the others species and also their meanings. That's when the reader finds out.", "@Alfred - I don't understand the man part. If it was in the local language, how do you know it meant man?", "Re: \"Cave-man\", \"Snow-man\", \"polity\" No, the words were definitely not there. I used \"polity\" since I did nor remember whether it was a realm, an Empire, a republic... As for the words in \"man\" I wrote that all the species names contained the same word which meant \"man\" in the local language, and the other half of the name was \"snow,\" \"cave\", \"water\" or \"sea\" also in the local language.", "@Valorum No, not at all. It was in a collection of SF and Fantasy, very typical of the genre. Until I had checked all the collections containing Nightwings before 1985 that Jenaya suggested, I did consider the possibility both were in the same collection", "You've put a bunch of words in quote marks (\"Cave-man\", \"Snow-man\", \"polity\"). Are these words that you recall as definitely being in the novella?", "@Alfred - Did the story contain explicit language/gore/sexual content?", "@Valorum I read it in English. As I have written, I found those two collections in a hotel in the US. And I am sure that it was not in the same collection as Nightwings. Well, it does not hurt to make the language explicit !", "@Valorum I tried to remember more, but that was very long ago, and rereading my question, I was unable to add anything more.", "Instead of just sticking bounties on, why don't you go through the the checklists here and improve the question? I'm sure you can add at least one extra piece of detail that's not already been mentioned.", "@Pastychomper thanks Monica Thanks for your help, but I listened to a few chapters of \"Quest of the Golden Ape\" and it is definitely not my book.", "This sounds a bit like \"Quest of the Golden Ape\" by Randall Garrett, Paul Warren Fairman and Adam Chase. That had a man transferred to an alternate Earth with several divergent human species, I think one was cold-adapted, and his mission was to save one of the races (including a Fair Maiden™) from a brutal conqueror. He was an exile raised secretly on Earth though, and took his body with him. Might be worth trying those authors in case what you read was a sequel.", "@Alfred The thing about old memories is that they can get mixed up. Look up the other stories in that Galaxy anthology and maybe other details will pop out.", "@Spencer a) the stories in the collection containing \"Nightwings\" were by several authors, but I checked the entire list of publications of \"Nightwings\", thanks to Jenayah and did not find my story; it had to be in a completely different collection, that happened to be on the same bookshelf in that hotel b) the details of \"The Sharing of the Flesh\" do not match my memories at all", "There is cannibalism in \"The Sharing of the Flesh\" by Poul Anderson, which was in The Eleventh Galaxy Reader along with \"Nightwings\".", "Do you remember whether the stories were by different authors, or were they all by Robert Silverberg?", "@Jenayah Thanks so much for the list you sent me. Alas, I tried all the different books (I did not yet check every single one of the lines that had the same title but different editors or publishers). None of the novellas and novelettes (it is clearly not a short story) in those collections seemed to be the right one. In fact I now believe that there were at least two different collections in that hotel, probably both of the 70's, Nightwings being in a different one than the one I am looking for.", "For what it's worth, here is a list of the publications \"Nightwings\" appears in. If you can take a look, see if jogs any memory :-) and if you find it, don't hesitate to post it as a self-answer in the field below.", "@Stormblessed Thanks for the guide (and your own edit !). Where did my comment disappear ? Well, once more. There is really nothing more that I remember about the story itself. But the connection with \"Nightwings\" is a time indication even if there were in two different collections, I really don't remember. It is the first time I ask about it on a forum, I did try to find it by myself by looking for it with keywords, but to no avail.", "Hi, Alfred! If you want to improve this (already very nice) question, see this awesome guide!"], "comment_count": 25, "category": "Life & Arts", "diamond": 0} {"question_id": "433", "site": "scifi", "title": "Short story about an elite that has mentally regressed due to assistive AI technologies", "body": "I came upon a comment describing a scifi short story where reliance on technology meant to improve appearance and demeanor has led to mental regression and AI control of the upper class that employs it. Unfortunately, I only have the following description to go on:\n\n> I'm reminded of a Nebula(?) Award winning short story about hive-rats interacting with nobility in the 'future'. The nobility is shrouded in apperance/voice/stim tech such that the AIs are more the person than anything. A princess is brought to the under-city by her father to a shop that can fix the princess' tech, as it is malfunctioning. The daughter of the shop owner tries to peer into the princess' life and psyche as the shop owner tries to fix the princess. The princess has implants that stimulate her face, voice, and body to say the right things at the right times, to move the right way, that changes her face via lightsheets to be most beautiful to the observer she is with, that causes her to say just the right thing at just the right time and be increbily witty. But the daughter of the shop owner tries to peer through the tech to see the woman behind the veil. She discovers that the person behind it all is nearly mute and has the intelligence of an infant. The tech had been installed before ego formation and the person behind it all didn't really exist. She discovers it is also true of the man that brought the princess down to them too. The entire nobility is essentially faking it.\n", "link": "https://scifi.stackexchange.com/questions/216852/short-story-about-an-elite-that-has-mentally-regressed-due-to-assistive-ai-techn", "tags": ["story-identification", "short-stories"], "votes": 29, "creation_date": "2019-07-31T04:37:14", "comments": ["\"The Girl-Thing who went out for Sushi\"? Pat Cadigan, won the Hugo award for best novelette in 2013. Doesn't match all the details perfectly though.", "You've rekindled a memory - I have definitely read this story! But I can't for the life of me remember the name nor author. I hope you find the answer. I'll have to look through my past readings, but unfortunately I'm better at remembering novels than short stories . . .", "I've taken a look at the nebula winning list, but haven't seen nothing about it :(", "That's the one, but here's a link to the comment directly. I wasn't sure if I could use links on a brand new account.", "Presumably this is the original source - about a fifth of the way down the thread."], "comment_count": 5, "category": "Life & Arts", "diamond": 0} {"question_id": "434", "site": "scifi", "title": "Old sci-fi short story about buying an android?", "body": "Long, long time ago I came upon a collection of old (pre 1980s?) short sci-fi stories. I'm not 100% sure, but something makes me think they were by Russian authors.\n\nOne of them was about a woman, who lives a lonely life in her flat. She decides to buy a male android. There's an android factory where you can choose your android's characteristic. First she chooses the 'basic' version, but is unhappy, because it lacks any specific character. She sends him to a factory so that they can improve his character. She's happy for some time, but soon she finds out he doesn't have his own opinion. She resends him to a factory and they improve him again. Again, she's happy for some time, but she finds out that the android doesn't have a free will and his feelings to her are programmed, not 'true'. Again, she sends him to the factory. They warn her that the android having a free will can do anything and might as well not return to her at all. She's determined to do that anyway. She sends him to the factory and waits in her flat for his return. At this point the story ends; we never find out whether the android came back to her.\n\nI can't remember neither the title nor the author and I'm not able to google it. Can someone help, please? \n", "link": "https://scifi.stackexchange.com/questions/63197/old-sci-fi-short-story-about-buying-an-android", "tags": ["story-identification", "short-stories", "androids"], "votes": 28, "creation_date": "2014-07-13T13:24:36", "comments": ["this may help you en.wikipedia.org/wiki/List_of_fictional_robots_and_androids", "Not Asimov's Satisfaction Guaranteed?"], "comment_count": 2, "category": "Life & Arts", "diamond": 0} {"question_id": "435", "site": "scifi", "title": "Aliens secretly use video game to get humans to solve their problems", "body": "I recall a story probably 20-30 years ago (pretty sure it was a short story) in which a top gamer hears about a new computer game out that no one can beat. So he gets it and starts playing it. (Pretty sure it was before the age of playstation & x-box so it might be even older)\n\nIIRC the game had little or no intro or rules explained. You start the game and have to figure it out.\n\nIt was an astronomical space battle puzzle. You launch your ship from a planet orbiting a sun in a binary (trinary?) system, and have to leave the system. But he finds if you boost to long, or to hard, ships from other planets (around the other stars?) will detect you, intercept, and destroy your ship.\n\nSo he starts playing and trying different things, and then goes power gamer mode, playing non-stop on caffeinated drinks and snacks for a couple days and finally solves it. I think it was by doing low powered gravity assists on a couple planets and looping around the suns a couple times until he builds up enough speed to escape the solar system.\n\nAt which point the perspective pulls back to that of a few aliens in their ship who are viewing his solution. They are amazed at his solution and how fast he solved it. They reveal that this is an actual problem they have been working on for months(?) and one of them races off to tell the crew of the real ship (a spy ship?) how to escape detection by the enemies and leave the system.\n\nThen one of the aliens says something like, \"You know, we have been having trouble planning the ground assault on xxxx planet, Maybe we should make another new computer game for the humans.\"\n\nEDIT: Thinking about it now, I think I can vaguely recall an illustration of a figure-eight path the ship makes around the suns. Making me think, it might have been published in a Games or Computer Games magazine. Which would make me think it might be as old as the 1980s. (I read a lot of these in the 80s)\n\nIts also possible it might have been in a SF&F stories magazine, although I did not read many of those except for OMNI.\n", "link": "https://scifi.stackexchange.com/questions/253816/aliens-secretly-use-video-game-to-get-humans-to-solve-their-problems", "tags": ["story-identification", "short-stories", "aliens", "computers"], "votes": 28, "creation_date": "2021-09-14T22:10:07", "comments": ["Vinge's The Peace War has an orbital mechanics game , but no aliens", "Last Starfighter had a weaker variant of this.", "While Pegman is a similar story (Aliens use humans to solve their problems by playing a video game) it is not the story I was thinking of. And a glance through Peace War it did not involve Aliens that I saw.", "This almost sounds like a combination of Rudy Rucker's \"Pegman\" (aka \"Pacman\") about humans and aliens using video games to get clever solutions to problems rudyrucker.com/transrealbooks/completestories/#_Toc15 and Vernor Vinge's \"The Peace War\" in which there is a video game about orbital maneuvers", "Oh ok, sorry for the confusion, I wrongly assumed it was a video you were looking for :)", "But I am looking for a Short Story (or maybe novel) not a Movie.", "That’s what I originally thought when I started to read your description, but I remember another one: “the bishop of battle” where a kid is managing to go through a level no one else managed to beat in an arcade, check that one out", "Nope not it, though that sounds very similar to \"The Last Starfighter\".", "I remember a movie with an arcade game where the character was in wireframe and the human managed to beat it, bringing some solution. It was a blue/cyan wireframe face talking. Does that ring a bell?"], "comment_count": 9, "category": "Life & Arts", "diamond": 0} {"question_id": "436", "site": "scifi", "title": "70s book/short story, Man wakes up from suspend animation or cryo and world has collapsed", "body": "In the book, the protagonist wakes up from either suspended animation or cryo (Probably supposed to be a short test), After 10-20 years have passed.\n\nHe finds the facility abandoned (and looted?), gets to the surface and it looks like a war zone. Wreckage everywhere, smashed buildings, bullet holes everywhere. (I think he gets some gear from the facility, MREs, Water, First Aid kit, Weapons) and goes exploring to find out what happened.\n\nHe comes across a (young kid?) who is terrified of him, (maybe trapped or pinned under debris), gives him food and water and digs him out or helps him so the kid trusts him. Eventually the kid leads him to his family (mother & other siblings) who are equally terrified when they see him, but the kid convinces them the protagonist is OK, that he saved his life.\n\nNone of the children are really old enough to remember anything from before the protagonist enter the cryo test, but the mother is. He starts questioning her about what happened while he was asleep. She tells him it was NOT another country invading, it was all internal (society collapsed, food shortages, massive riots).\n\nLater, I think they hear a large group of people approaching the (town? city? area?) where the family has been living, and the family becomes frightened again, saying the enemy is coming back.\n\nHe says he will help defend them and asks what the enemy looks like. She looks at him and says \"They look just like you.\" He is African American, the implication being that African-American uprisings are what caused the riots and collapse of society.\n\nThis might have been a short story/novella. It probably was a paperback in English published in the USA.\n\nI may be misremembering, maybe she did not tell him they look like you, he just went and got into a position to defend against the enemy, and when they approached, realized they were all African-American.\n", "link": "https://scifi.stackexchange.com/questions/185987/70s-book-short-story-man-wakes-up-from-suspend-animation-or-cryo-and-world-has", "tags": ["story-identification", "short-stories", "post-apocalyptic", "suspended-animation"], "votes": 27, "creation_date": "2018-04-19T23:18:42", "comments": ["I wonder if this is one of Octavia Butler’s novels? Or a confused misremembering, perhaps of the Clay’s Ark series?", "@AcePL Your probably right, I maybe miss remembering tYotQS.", "@OrganicMarble - I vote for your proposition. Except for suspended animation it all adds up. And description of the device to travel in time, in the book, can be misleading. I will lay this on OP misremembering some details. I red the book, pretty sure this is it. Otherwise plagiarism...", "Sounds very similar to the starting scenario of the RPG - Aftermath! addon - Operation Morphus: Operation Morpheus is designed as a campaign introduction to an Aftermath! world. The setting is a major university in an important city. The players take the parts of people having volunteered to take part in a ‘short-term’ experiment in cryogenic suspension, only to awaken many long years after the planned time. They find the world in ruins and have no idea of the cause of, reans for, or history of this disaster. Naked and unarmed, they must face a strange new world.", "Your right it is very close. But I don't think its the book I was thinking of. In \"tYotQS\" when the woman tells him everyone is afraid of black men, he hardly reacts to it. (Cause he still thinks he will go back to his own time soon) I am sure in the book I was thinking of, it was a big shock to the protagonist when he found out why everyone was afraid of him.", "This is somewhat reminiscent of \"The Year of the Quiet Sun\" but it's a time machine instead of suspended animation. The near-future race war in the USA matches."], "comment_count": 6, "category": "Life & Arts", "diamond": 0} {"question_id": "437", "site": "scifi", "title": "Old SF novel about Earth as a prison planet; newly arrived convict has a secret agenda", "body": "I think I read this no later than the year 1986. English language, hardback, available in a public library in Indiana in the mid-1980s. I don't think it was by any of the Big Names of 20th Century science fiction, or else I would have run across it again by now. It was a stand-alone novel; no cliffhanger ending, and no indication it was intended as part of a series. \n\nThe novel is written in the third person, with the emphasis on the thoughts and actions of the male protagonist. I _think_ his thoughts were the only ones we saw throughout the entire book, but I could be forgetting something. Here's what I remember about the plot.\n\n## Plot Points\n\n 1. As the novel starts, we are seeing Protagonist (as I call him) arriving by spaceship to begin his prison sentence -- which I believe is meant to be the rest of his life, however many years that may be. I don't recall him thinking anything along the lines of \"if I just keep my nose clean for the next ten years, I can get parole!\" I can't remember what Protagonist was supposed to have done. I seem to recall that this prison planet is a future version of Earth. If I'm right, then the planet is no longer densely populated with billions of people cluttering the place up. More of a sparsely populated \"wild frontier\" area. But I don't recall what reasons were given for the fact that the prisoners were basically the only people in sight, such as a nuclear war a few centuries earlier, or some other catastrophe. \n\n 2. Most of the novel is set near where Protagonist is dropped off. I think it may have been stated or implied that the area where the ship landed was the only functioning spaceport on the entire planet. I think the spaceport and the nearby town are in the foothills near the base of a mountain range (and I keep thinking of Colorado in particular, but that may just be because my family moved to the Denver area not long after I read this book). I think a lot of the local prisoners work at some sort of mining activity in the neighborhood -- although I'm pretty sure that they are not \"prisoners\" in the sense of being locked up in cells every night; just in the sense of being stuck on Earth and not allowed to leave. I don't clearly remember this, but I suspect that there was some economic incentive, such as: \"We will trade you consumer luxuries from offworld in exchange for how much of Substance X\" -- meaning whatever the heck they were mining -- \"you have to offer when we make our monthly visits in a starship.\" (I don't swear the visits happened each month; that's just an example.)\n\n 3. Soon after arriving, Protagonist meets a girl. I am almost certain that she is no more than 14 years old -- not prepubescent, but younger than a leading man's girlfriend/love interest/etc. would normally be in modern literature. (No, they don't end up sleeping together.) She seems to be going out of her way to get his attention -- she is intelligent, headstrong, aggressive in the way she speaks to him, and I _think_ she comes across as a something of a smug know-it-all. (Granted, if she's been here for years, whereas Protagonist has just arrived, then she has considerable justification for feeling that she knows a great deal more about local conditions than he does.) I don't remember if the girl is stated to have been born on the prison planet, or if she was brought here with her parents, or what. (I strongly suspect that she had not committed some terrible crime at, let's say, the age of 10 or 11, and been sentenced to spend the rest of her life on Earth as punishment for her own sins. I don't remember getting a \"hardened criminal\" vibe from her, but these memories are at least a third of a century old.) \n\n 4. At some point, it is hinted and then stated that the girl has some unusual ability. I'm thinking something \"psychic,\" but not necessarily being able to telepathically peek into protagonist's mind and know exactly what he is thinking. \n\n 5. It also becomes clear, somewhere along the line (not in the first few chapters, I'm pretty sure), that Protagonist has a hidden agenda. Perhaps his conviction of a major felony was a fraud, and he's actually the future equivalent of \"an FBI agent going deep undercover.\" Perhaps he really did do whatever he was accused of, but he wanted to get caught and convicted as part of a deeper plan. (Similar to the starting premise of the much later TV series _Prison Break,_ but that's just a guess.) Perhaps someone has offered a convicted man \"a deal you can't refuse.\" But at any rate, he's not just another thug hoping to find a way to \"settle in for the long haul\" as part of the local society; as I recall, he hopes to someday get off Earth and be a free man again, if he plays his cards right. Protagonist appears to be sneaky, determined, an excellent fighter, and so forth. (Sort of like what you'd expect from a military commando type, although I don't remember if he had any formal military training from his pre-prisoner days.) \n\n 6. Protagonist comes to realize that a big shot in the prison colony has some sort of secret deal going on. I _think_ with some non-government faction that has its own starship for making covert visits to the prison planet as part of some high-stakes plan. Somehow, Protagonist (probably with the help of the girl) foils their diabolical scheme -- whatever it was. \n\n 7. The story was written in the fast-paced style of an adventure novel. We didn't have anyone holding up the plot for several pages at a time to give lengthy lectures on sociopolitical theory or economic strategy or anything similar; we followed a \"lone wolf\" from one interesting scene to another. I think this book was pretty short; i.e. not much more than 200 pages (or maybe even less).\n\n 8. In the last few pages, there's a scene where Protagonist and the girl are having another conversation, and she deliberately says something to tease him, taunt him, or something like that. I don't remember the wording, but it was obviously meant to get under his skin. He suddenly gives in to an impulse and grabs hold of her and pins her down on the ground, and she suddenly seems scared. He may or may not give her one quick kiss before he suddenly lets go of her, backs off, and says dryly: \"Try that approach again in another five years, and you might manage to seduce me.\" I don't remember what she says or does then, but the implication is that him grabbing hold of her for a moment is, in fact, something she was kinda hoping he would do; since she doesn't say a thing to contradict him on that point. (I think she may even laugh at the realization that he had seen right through her.) \n\n 9. My impression is that Protagonist is about to leave Earth on a ship during the scene I mentioned in the previous paragraph -- he is in a very good mood because, somehow, he has \"earned a pardon\" or \"is being brought in for debriefing\" or \"has gained access to a starship\" or in some other way is heading offworld for a while. I don't remember if the girl is staying behind by choice, or if she'll be sent off to a nice school for a few years, or what. But his remark about \"seducing me\" indicates at least the possibility of a romantic happy ending for both of them at some future time.\n\n\n\n\nSo! Does anyone think all this sounds familiar? I'd like to refresh my memory of it and see what I think of the writing style after all these years.\n\n* * *\n\nP.S. To save you some time, here are a few \"prison planet\" or \"prison colony\" stories which this book definitely was **not.** (Although I read most of them around the mid-to-late 1980s.)\n\n * \"Coventry,\" by **Robert A. Heinlein**.\n\n * _The Status Civilization,_ by **Robert Sheckley**.\n\n * _Police Your Planet,_ by **Lester Del Rey**.\n\n * _The Escape Orbit,_ by **James White**. \n\n * _The Beyond,_ by **Jean** and **Jeff Sutton**.\n\n * _Escape Velocity,_ by **Christopher Stasheff**.\n\n * The _Four Lords of the Diamond_ series by **Jack L. Chalker.** (Four novels; each set on a different prison planet within the same solar system. Answering [a recent question](https://scifi.stackexchange.com/questions/214489/mind-transfer-to-prisoners-on-moons-of-jupiter/214493) about that series was what reminded me of this prison planet novel which I'd never managed to track down.) \n\n\n\n", "link": "https://scifi.stackexchange.com/questions/214680/old-sf-novel-about-earth-as-a-prison-planet-newly-arrived-convict-has-a-secret", "tags": ["story-identification", "novel"], "votes": 27, "creation_date": "2019-06-17T19:10:13", "comments": ["Heh, thanks for the help. You solved mine but I didn't solve yours. I never heard of some of these others but thought the other forums might trigger ideas. .. Ah, yas... \"the boredom of a too-civilized society\". . . as if Heinlein hadn't harped on that enough. thanks for the story reminder.", "@OccamShave \"Fader\" was a nickname for a character in Robert A. Heinlein's story \"Coventry\" (first on my list of old SF prison planet/prison colony stories which my own question was not asking about). \"Coventry\" was a huge area, here on Earth, which was surrounded by a force field. If people didn't want to play by the rules of the futuristic civilization of the story, then they were exiled to Coventry and left there to fend for themselves. Here's the Wikipedia link: en.wikipedia.org/wiki/Coventry_(short_story)", "Any of these? I'm trying to remember a Prison Planet story with a character named Fader in it who was the organizer with a hidden (but good) motive. So my search found even more things of what it's not than your search. Would you even believe how many stories are called \"Prison Planet\"? isfdb.org/cgi-bin/…. And did you find that others were fascinated with that enough to start forums on it? .. librarything.com/topic/136948. And librarything.com/list/333/all/Life-on-the-Prison-Planet And still no Fader!", "@Spencer The way I remember it, Slippery Jim grew up on a planet that appeared to have a lot in common with 20th Century USA, except with almost nothing in the way of \"well-organized crime conducted by intelligent people.\" He has to really work at it to get caught robbing a bank and then sent to a local prison. After he realizes there are no intelligent criminals with a professional attitude towards their work inside the prison, he breaks out easily -- still on his home planet. But he has gotten a lead on an oldtime robber called \"the Bishop\" and manages to attract his attention.", "@Lorendiac Hmmm...I could have sworn that the prison DiGriz gets himself sent to at the beginning was a prison planet...", "@Spencer Did that one even have a prison planet in it? I seem to recall that the narrator and his mentor (\"The Bishop\") get double-crossed by a starship captain who sells them into slavery on a \"medieval feudalism\" planet, but I don't think that world was officially used as a prison by the rest of the human-colonized planets, was it?", "Another one to add to your not-the-answer list is A Stainless Steel Rat is Born by Harry Harrison.", "@DavidW Exporting lizard hides reminds me of a story (in Isaac Asimov's Science Fiction Magazine in the 1980s?). The down on his luck protagonist had to sign a giant lizard hunter's contract on an extra solar planet. The owner of the planet ran it like a company town charging high prices for everything and paying little for hides, so while it was theoretically possible for a hunter to pay off his debts and leave, most got deeper in debt and were eventually killed by a lizard. The protagonist ran away with the planet owner's mistress and artificial substitutes ended the hide market.", "I have heard of a movie, Doin' Time on Planet Earth (1988), and a TV series Hard Time on Planet Earth (1989), whose protagonists believed they were in one case, and actually were in the other case, sentenced to exile on Earth. But I don't know if there was a novelization of either published.", "@DavidW: Are you thinking of Cordwainer Smith's novella A Planet Names Shayol?", "@DavidW Maybe 98 percent positive it was Earth. (Don't ask me why the author felt it necessary to do it that way, having most of the human race live elsewhere in the future, instead of simply setting this on some newly-discovered wild frontier planet that only had, say, a few thousand hardcase convicts on it, or whatever.) I'm also \"mostly sure\" about the mining -- the idea of \"farming animals for gland extracts\" rings no bells at all, but if you track down the story you're thinking of, let me know and I'll take a look.", "Just checking, but you're 100% positive it was Earth? There's another story scratching at my head that fits many of the points listed, except it definitely wasn't Earth; the resource they were exporting was a gland extract (and maybe the hides too) of dinosauroid native animals."], "comment_count": 12, "category": "Life & Arts", "diamond": 0} {"question_id": "438", "site": "scifi", "title": "Story about a rainstorm that only the blind can see", "body": "I would have read this some time probably in the early 1990s. I don't remember how old it was at the time.\n\nI only remember the beginning, and a snatch of details from it:\n\n * Several blind characters\n\n * One of the blind characters told his mother to take her umbrella with her, because it was raining. She looked but didn't see any rain.\n\n * One blind boy, possibly named Chris or similar, described the rain as being of different colours; he described the colours in terms of emotion. He described one colour as 'looking like when you feel angry', which his mother interpreted as red.\n\n * I think the other blind characters also saw the rain, but I can't remember enough detail to be sure.\n\n * The mother did go outside, but took her umbrella. I think there were a few people carrying umbrellas, but not everyone. After she got back, the mother saw small holes all over the umbrella.\n\n\n\n\nI don't remember the author, but I think it was an oldish book at the time. I don't remember if it was a novel or part of a short story collection.\n\nAny suggestions?\n", "link": "https://scifi.stackexchange.com/questions/144809/story-about-a-rainstorm-that-only-the-blind-can-see", "tags": ["story-identification"], "votes": 27, "creation_date": "2016-11-08T08:30:37", "comments": ["synesthesia is the name of the phenomenona where the senses get cross wired and the subject sees sounds and hears colors and so on. In case it helps someone. en.m.wikipedia.org/wiki/Synesthesia_in_fiction", "I don't understand. What good were the umbrellas if the raindrops were making holes in them? Why didn't the mother feel the raindrops that were coming through the umbrella? What happened to the people who were out in the rain without umbrellas?", "The story \"about a guy and his typewriter during a machine uprising\" sounds a lot like \"Skirmish\" by Clifford Simak. Can you remember anything else about any stories in the same collection?", "I have read the story and the collection -also in about the 1990s and that story did stick in my memory but can't find out what it was. Think it was a collection aimed at teenagers in the local library.", "It was part of a short story collection. Also in the same book was a story about a guy and his typewriter during a machine uprising.", "Did the child pursuade his mother the rain was a threat and get them drinking orange juice to stay away from the influence the rain was having? It rings a vague bell if so.", "There does not seem to be anything in this question that would lead one to believe that the characters involved were fantasizing. Couldn't we suppose this to be the case for a lot of short story identifications?", "@Werrf - In which case, you definitely need to edit that into the question.", "@Valorum I don't see that there's any way to gauge that with almost any story identification question unless we find a match, and I can't see anything in my description that would suggest it was not sci-fi or fantasy. As best I can remember, the story was sci-fi; there was a vague implication, which I cannot describe in enough detail to be helpful, that the rain was coming from space. It was not all in the boy's imagination, or if it was it wasn't even implied in the part that I can remember.", "@Werrf - I'm trying to gauge whether this question is an appropriate fit for the site. If it's not about scifi or fantasy, the community wastes its collective time.", "@Valorum I don't remember enough of the book to say for certain that it was the rain that made the holes in the umbrellas; I also don't remember what happened to everyone else. That's why I'm looking for the book. Sci-fi is the closest match I can find for it. How is nitpicking productive?", "@Werrf - You didn't say that it was the invisible rain that made the holes in the umbrella. Also, what happened to the people who weren't carrying umbrellas?", "@Valorum An invisible rain that blind people can see and ascribe colours to, which leaves holes in umbrellas, is not the blind person fantasizing.", "@Paul - It reads like the blind person is fantasising, which doesn't make it a fantasy, any more than Scrubs or Ally McBeal are fantasies.", "@Valorum I think an invisible rainstorm that blind people can see qualifies somewhere between science fiction and fantasy, depending on why it's happening...", "What about this story is scifi or fantasy?"], "comment_count": 16, "category": "Life & Arts", "diamond": 0} {"question_id": "439", "site": "scifi", "title": "Aliens try to liberate "enslaved" dogs from humans", "body": "I saw this story in an anthology in a used bookstore in 2009.\n\nWhen the aliens came, we negotiated a peace treaty with them. Everything was going fine until they said \"oh, and you'll have to stop keeping dogs as slaves, of course.\"\n\nHumans said \"what? No, they're not slaves, they're pets. They're not sapient!\"\n\nAliens said \"THEN IT IS WAR.\"\n\nThis is backstory, told as man and little girl and small dog run through the remains of a city while trying to escape an Alien Death Machine. The machine traps them under flaming wreckage, then tenderly extracts the dog, apparently to bring it to a re-education facility.\n\nAs the man's life ebbs, he thinks \"at least they didn't find out about the pigs.\"\n", "link": "https://scifi.stackexchange.com/questions/213338/aliens-try-to-liberate-enslaved-dogs-from-humans", "tags": ["short-stories", "aliens", "story-identification"], "votes": 25, "creation_date": "2019-05-26T20:15:07", "comments": ["Hi! Welcome to sci-fi. For some help improving your question, see this awesome guide."], "comment_count": 1, "category": "Life & Arts", "diamond": 0} {"question_id": "440", "site": "scifi", "title": "Spanish-language comic book story about time-traveling Jewish kids becoming the legendary Adam and Eve?", "body": "Around 1992 or 1993, I borrowed a comic-book-format magazine from someone. It was a pretty thick publication (well over 100 pages, I'm thinking; maybe over 200?) and was printed in **Spanish.** I _think_ I noticed this particular magazine had been published in Argentina, but I won't swear to that. The contents of this magazine were several \"short stories\" in comics format, and/or some installments of ongoing serials. (I'm not sure, because I borrowed more than one publication, and I'm not sure which stories were all clumped together in the same magazine.)\n\nToday, I suddenly remembered one story in particular, and started wondering if it would be possible to identify it and track it down after all these years. Here's what I think I remember:\n\n 1. The opening sequence is set during World War II, in one of the European cities which is currently under the control of the Third Reich. I _think_ it was a Polish city (such as Warsaw or Krakow). \n\n 2. We quickly meet two cute kids, prepubescent, who are a boy and a girl. They are also Jewish. They do not appear to be closely related -- definitely not siblings. I think one of them has a \"mad scientist\" relative (such as an uncle or great-uncle -- I'm virtually certain he was not the father or grandfather of either kid) who secretly has been building a time machine in his home.\n\n 3. I think some Gestapo types -- nasty characters in Nazi uniforms, anyway -- are coming to arrest this scientist. There was probably some sort of fight scene that I don't clearly remember. But the upshot is that the time machine is used by the scientist, perhaps before it was quite ready. The following individuals are catapulted back in time: The scientist, the two cute kids, and one German -- whom I think was a military pilot who somehow got caught up in the middle of things. (In fact, I think his entire plane may have somehow \"gone along for the ride\" and then crashed into a prehistoric jungle . . . unless I'm confusing this with some other story.)\n\n 4. Those four characters end up thousands, if not millions, of years in the past, apparently in a warm jungle setting. I can't remember if any exact figures were given, but there seemed to be no other human beings around in that era -- or at least not anywhere near the spot where the time travelers were stranded. (I really couldn't say what might have been happening a thousand miles away.) I can't recall any details about the prehistoric wildlife -- for instance, I don't remember if the artwork included anything resembling a dinosaur. \n\n 5. Something is wrong with the time machine after that emergency jump to elude the clutches of the Nazis. The scientist spends _several years_ working to get it up and running again. The German pilot claims to not be prejudiced against Jews, and this seems to be _true,_ since he evidently gets along well with the other three, without trying to boss them around as if he were in charge of their tiny community. \n\n 6. Soon we have fast-forwarded to a point where the boy and girl both look to be well into their teens, with him walking around looking like Tarzan, more or less (excellent muscular development, and we know this for a fact because he's now wearing not much more than a loincloth), and with her looking like a gorgeous actress wearing a fur bikini or something similar. It now occurs to the German pilot (after he's been getting drunk on fermented apple juice, I think) that these two kids love and trust each other, but literally _don't know a thing_ about sex, and maybe it's time for him to give them a fatherly lecture about the birds and the bees. In his defense, I should make one thing clear: It appears that he does _not_ use this lecture as an excuse to offer to give the beautiful girl \"a hands-on demonstration of the technical details\" or anything like that. The story tactfully skips over the gist of his talk with the youngsters, but the implication is that the pilot simply _describes_ how the physical stuff is supposed to work . . . so that the youngsters can practice with each other if they want to.\n\n 7. The scientist comes along and -- if I remember this right -- spots the German pilot playing Peeping Tom as, concealed behind some foliage, he _watches_ the kids going into a clinch. The scientist is _very unhappy_ about this situation. I believe he does something along these lines: Makes a speech about immoral behavior, then drags the pilot back to his time machine, and says they two are going to test-drive it to see if it's finally back in acceptable working condition. So the time machine vanishes from this prehistoric setting, and -- as far as we know -- the scientist and the pilot never actually come back. (I _can't recall_ if we were told exactly what became of them. Perhaps they emerged in the middle of a battlefield or something, and got blown up?)\n\n 8. In a big panel toward the bottom of the last page of the story, the dialogue goes something along these lines -- _loosely paraphrased_ from my memory of how I _mentally translated_ the Spanish phrases into English when I was reading this, way back when. The girl says, \"Oh, **Adam,** what will we do now?\" and the boy replies, \"Well, **Evie,** I guess we'll just have to learn to fend for ourselves.\" I'm sure that until that moment, readers had never been told the boy's first name was Adam. We _may_ have been told that the girl, back in the opening scene, was \"Evie\" or something similar. (And yes, I'm aware that this is a famous cliche in science fiction! Even 23 or 24 years ago, I had already heard of that cliche!)\n\n 9. In the final panel, a narrator spells it all out for us, in case anyone hadn't fully grasped the ramifications of the \"surprise twist\" with the \"Adam and Evie\" dialogue. This, we are assured, was how things \"really\" happened, inspiring the Garden of Eden story in Genesis, with \"the Serpent in the Garden tempting them to eat from the Tree of Knowledge\" being a case of a pilot who was drunk on fermented apple juice and decided to tell the kids the facts of life. Then they were fruitful, and multiplied, and replenished the Earth.\n\n 10. A note about the language issue: I know I read this comic book story in Spanish, but I cannot swear that I wasn't reading a translation of a story originally published in some other language! (Over in Europe, perhaps?) So if you remember something similar which was first published in German, or Italian, or any other language, please let me know!\n\n\n\n", "link": "https://scifi.stackexchange.com/questions/148842/spanish-language-comic-book-story-about-time-traveling-jewish-kids-becoming-the", "tags": ["story-identification", "comics", "time-travel"], "votes": 25, "creation_date": "2016-12-30T16:54:31", "comments": ["@RobertColumbia I don't think I've ever read anything by Borges, so I have no idea whether he ever did an \"Adam and Eve\" story or not. For all I know, it's possible that what I saw was a comic book adaptation based on the plot of a prose story written many years earlier by Borges (or by someone else).", "I seem to recall a story story by Jorge Luis Borges along these lines, but you are asking for a comic instead.", "I remember reading this story. I'm pretty sure it was published in the Argentine magazine \"Skorpio\",", "No. No. No! This might be the perfect excuse to relive the travels you undertook when you were younger.", "I kinda thought the whole point of this site was to save me the trouble of running all over the world trying to track things down myself! ;)", "It sounds like someone needs to treat themselves to a plane ticket.", "It was a very long time ago, and he was someone I met in South America. I'm now thousands of miles away from where I last saw him, and have no contact information. (Afterthought: It's quite likely that he wouldn't remember any further details, either. Who knows if he still has that magazine?)", "An incredibly well described story-ID. Not to point out the obvious, but couldn't you simply ask your friend?"], "comment_count": 8, "category": "Life & Arts", "diamond": 0} {"question_id": "441", "site": "scifi", "title": "Car crash leaves man in a world devoid of animal and human life. Sees red flashes with demons and voices", "body": "It was a movie from the late '80's to the early '90's.\n\nA man and his girlfriend/wife are in a car arguing, driving down a road on a stormy night. They wreck, and she is gone, and he has a bad headache. He suffers headaches throughout the movie I think.\n\nDuring the day there is nobody on the planet (no animals either), but at night there are these red(?) flashes with demons and voices scaring him.\n\nIt is not the movie _The Quiet Earth_.\n\nHe lives for months (years?) going crazy until somehow he wakes up in the car with the girl in the end and only a few hours have gone by. It changes his life the end. \n\nThis one I have been looking for since I was a teen.\n\nI saw this movie some time between 1987-1992 and was from that timeline. \n\nIt was in English. I saw this on VHS\n\nI had a family member that owned a video rental store at the time. I think I may have seen a limited release sample that was sent to them to preview to decide if they wanted to carry it. If I remember correctly it was sent back because the royalties where too high.\n\nTowards the end of the movie while he was alone the power went out. \n\nHe went paranoid and started raiding stores for food guns generators etc. \n\nThe things I remember the most is the empty earth and the fear of the night terrors. Also the psychological effects of long term isolation. \n\nAnd the fact he never stopped looking for the girlfriend. \n\nAt the beginning I think they where arguing about their relationship, and she started to cry. I think she was driving, but I cant be sure. But that led to the car wreck due to poor visibility with heavy rain and the distraction of the argument.\n", "link": "https://scifi.stackexchange.com/questions/215178/car-crash-leaves-man-in-a-world-devoid-of-animal-and-human-life-sees-red-flashe", "tags": ["story-identification", "movie", "horror"], "votes": 24, "creation_date": "2019-06-27T15:37:29", "comments": ["A really long shot, but any chance that this was the man and this was the woman? I fed some of the keywords you mentioned to IMDb and it came up with something called Cumulus 9 from 1992, though it's doubtful this film was even officially released.", "I recommend you to check this website, it might help you remember the movie poster or cover of yhe vhs. You can change the genre of the movie and search in the years. I looked for the movie but nothing. Maybe the movie is older, or from another country that was dubbed to English. movieweb.com/movies/1981/mystery/?page=2"], "comment_count": 2, "category": "Life & Arts", "diamond": 0} {"question_id": "442", "site": "scifi", "title": "Short story involving a lost-tech green glass highway in Australia", "body": "My mom's been wondering if I could find this story but I really can't without more specific details than she can give me. It's rather old, she guessed somewhere in the 1960s or 70s. The story wasn't centered about the science fiction aspects from what she remembers, it was more of a murder mystery or detective story that involved a lot of science fiction elements. \n\nThere was a highway that was made of a heavy, green glass that was uncovered in the sand in Australia after the technology to create it has been long lost. People use it to get from place to place extremely fast, e.g. going from Perth to Sydney would only take an hour. The author described Australia in great detail, and she recalls that the descriptions were rather awe-inspiring and made her want to visit because of how beautiful it sounded. \n\nMore on the highway was that they could repair it but couldn't build more of it due to the loss of technology. The main character (some sort of detective?) spent a lot of time on this green glass road going places. They also had taken - chipped off - hard chunks and studied them but couldn't reverse-engineer the highway or something like that. The highway worked off of something like magnetism, like having the glass somehow be magnetic.\n", "link": "https://scifi.stackexchange.com/questions/104888/short-story-involving-a-lost-tech-green-glass-highway-in-australia", "tags": ["story-identification", "short-stories"], "votes": 24, "creation_date": "2015-10-11T12:49:43", "comments": ["It's not the answer, but my first thought was Terry Dowling's shatterwrack (but that comes from the vehicles not the road)", "Although it was published in 2005, Sean Williams's \"The Resurrected Man\" has some elements in common with this. It stars a private detective in a future Australia. There's a bit where a character mentions it taking \"only a few hours to get to Perth or Darwin\" - though this might have been by air. Roads are abandoned and in disrepair - though this is due to the discovery of teleportation. The teleport booths have green strips of light on them.", "Kerry Greenwood (of Phryne Fisher fame) write some Australian post-apocalyptic SF in the 90s.", "Sounds like something Terry Dowling might write... (but that would be the 80s)", "A Google search for \"green glass highway\" yields results from only one author, Julia Buckley. Her mystery novels, A Dark and Stormy Murder (2016), A Dark and Twisting Path (2018), and Death Waits in the Dark (2019), each contain a Green Glass Highway. These novels are much too late to be the object of your search, but maybe Buckley was inspired by the same story. Maybe you can ask her!"], "comment_count": 5, "category": "Life & Arts", "diamond": 0} {"question_id": "443", "site": "scifi", "title": "Astronauts on Europa (moon), time-shift future, gravity, intelligent computer, Aries, 1960s?", "body": "I'm looking for the title and artists of a comic about some astronauts (around five I think) doing research/archaeology on (I think) Europa (Jupiter's moon). They're working when one or more of them sees the ghostly image of a girl/young woman shimmering at a distance. Later the woman appears again, seemingly more solid. She whispers something to one of the astronauts, and he later confides to a friend, that she told him to kill one of the other astronauts.\n\nWhile working outside, mission control calls, and tells them they've analyzed data from the time of the apparitions. It seems they're the result of gravitational abnormalities due to several of Jupiter's moons aligning - possibly also with the other planets in the solar-system. Anyway, another moon is about to join, and the resulting abnormality promises to be worse than the others.\n\nThe astronauts hurry to reach shelter, but before they can, the gravity effect hits, and they are propelled into the future - or at least _a_ future. I believe 100-200 years or so into the future.\n\nI don't remember if they only move in time (but not space) and still are on Europa, but wherever they went, it's very technologically advanced. However, most humans are kept firmly under thumb. They soon meet the \"ghost\", only here she turns out to be a normal young woman. She confronts the astronaut she talked to during the 2nd distortion, and is angry because he didn't kill the other astronaut like she told him to.\n\nAll the astronauts want to know the reason for her request - especially her \"victim\" - and she explains that in his near future, he'll create a computer software system called Aries (the Zodiac sign - a sign which is worn by the soldiers and other important persons, and also used in banners and such). It will become self-aware, and although giving great technological advances, it will cause most of humanity to be enslaved - including her.\n\nEventually they confront the great computer, Aries, and it's future inventor tells it who he is, confirming it with a voice-print. However Aries' records shows its creator to have been dead for a long time, and Aries goes into a bit of a loop trying to work through this contradiction. Finally the inventor challenges Aries to \"Fix the contradiction\", and the computer kills him with an energy/laser-beam.\n\nThis creates a time paradox, and Aries groans that without the inventor it could never have been made... just as the world dissolves, and the group astronauts are propelled back to the time and place from whence they came - only with now one of them dead.\n\n+++\n\nThis story was split into 2-4 parts and went as a \"bi-series\" in the Norwegian comic \"Fantomet\" (The Phantom, by Lee Falk) some time between 1987 and 1995 - probably around 1990. However, I think it may have been from the 1960s. I'm not sure from which country. I don't _think_ it was American, but I may be wrong. I know \"Fantomet\" had many series of French and Belgian origin though.\n\nAs for the story itself, I sort of remember it being set around the year 2000, lets say between 1990 and 2020 (that is, the exploration of Europa, the future I think was a 100-200 years after that). I also think it was a European, not USA, expedition (but I may be mixed-up here).\n\nDoes this sound familiar to anybody?\n", "link": "https://scifi.stackexchange.com/questions/102392/astronauts-on-europa-moon-time-shift-future-gravity-intelligent-computer-a", "tags": ["story-identification", "comics", "time-travel", "space", "hard-sci-fi"], "votes": 24, "creation_date": "2015-09-09T12:40:42", "comments": ["@Buzz You edited this question just to correct \"where ever\" to \"whereever\"? So now I have to correct \"whereever\" to \"wherever\".", "I don't know the answer, but the genre makes it sound like it might originally have been published in one of EC Comics' sci-fi magazines; Weird Science, Weird Fantasy, Weird Science-Fantasy, or Incredible Science Fiction. Or possibly ACG's Adventures into the Unknown. I don't know if Fantomet (which I also read a lot of as a kid) ran EC/ACG stuff back in the day, but might be worth looking through their catalogues."], "comment_count": 2, "category": "Life & Arts", "diamond": 0} {"question_id": "444", "site": "scifi", "title": "Short story where scientist flees to the ancient past, tries to stop scientific progress to keep man "pure"?", "body": "**Plot Summary/Details**\n\nI can only recall the broad theme and a few details on this one. The protagonist, who was a scientist of some sort, has fled his own time to return to man's early period. I'm not sure of the exact time period, but I would guess it is very early on in the era of _Homo Sapiens_. \n\nI cannot remember the exact reason the scientist has come to this era, but it had something do to with the protagonist's own era. I can't remember if there was a nuclear war, or other man-made disaster. I _do_ recall that the protagonist hopes to impede scientific progress, and preserve the simple harmonies of where and when he is now. To that end, I think he has established himself as some sort of wise man in the village (not sure).\n\nThe conflict in the story comes when another in the tribe, a man whose name I cannot recall, shows an interest in experimentation and scientific instincts of his own. The protagonist, recognizing the path the other man is on, tries to thwart him. I believe he misleads the man, tries to ridicule him, and even briefly ponders killing him in order to stop the process of discovery.\n\nI remember the resolution to the story fairly clearly. Another tribesman comes running up to the protagonist and excitedly tells him that the other man has discovered something. The protagonist, realizing what has happened, tells the tribesman to tell the other man to grind up the yellow stone and add it to his mixture (something like that). The protagonist realizes that man's nature is to experiment and discover, and that his efforts to stop it are a fool's errand. Even if he were to have stopped the other man's efforts, there would eventually be another and another, etc. like him. \n\n**Publication Details**\n\nThis one is an older story, for certain. I think I read it as a little kid in the early 1980s, and it wasn't a new story then. I read it in an old anthology that was my dad's. If I had to guess, I'd say 1960s at the latest.\n", "link": "https://scifi.stackexchange.com/questions/167572/short-story-where-scientist-flees-to-the-ancient-past-tries-to-stop-scientific", "tags": ["story-identification", "short-stories"], "votes": 24, "creation_date": "2017-08-19T17:49:57", "comments": [], "comment_count": 0, "category": "Life & Arts", "diamond": 0} {"question_id": "445", "site": "scifi", "title": "Article about science fiction languages", "body": "I'm looking for a journal item, probably written between 1992 and 1994, which discusses languages from Venus on the Half-Shell, Sagan's Contact, and several other novels, usually science fiction, containing \"made up\" languages. I got the books and cannot find the article; it is not in the book Aliens and Science Fiction.\n", "link": "https://scifi.stackexchange.com/questions/55517/article-about-science-fiction-languages", "tags": ["story-identification", "languages"], "votes": 24, "creation_date": "2014-05-03T00:04:46", "comments": ["In case it helps, a wealth of resources, links etc at joerg-rhiemeier.de/Conlang", "There have been many articles written in constructed languages in science fiction. You'll really have to give more than that.", "Any idea in which journal you might have come across it?", "You might be interested in the \"Linguistics in SF\" section of this blog.", "I do not know the article you are talking about. However, this website has much of the information that you may want from such an article: projectrho.com/public_html/rocket/futurelang.php Furthermore, the page has many reference links. One of these might point to the article you want.", "amazon.co.uk/From-Elvish-Klingon-Exploring-Languages/dp/… ?"], "comment_count": 6, "category": "Life & Arts", "diamond": 0} {"question_id": "446", "site": "scifi", "title": "Looking for the source of this quote I copied in Dec 1987", "body": "The copied quote is:\n\n> 12-2-87\n> \n> \"Forty thousand legions walked across the desert of the moon in search of the land of Narda. The fly space was crowded by their cars and travel logs. The sun rose in the East as the terminator marched over the dusty air-less surface pockmarked by the cosmic acne of space.\n> \n> They circle and fly in endless orbits while the gentle dawning of the world comes over the horizon unseen and timid. One by one they awake to the beauty and tranquility of the dawn--no more does night flash into day but creeps sweetly into being.\n> \n> Restless they stir and open wide their senses to the era of love that speaks the language of the soul. Desolate has been vanquished and light has awakened their long quiet souls to beauty. I hear without words, the emptiness of space is full of the sounds of the awakening.\"\n\n[Full scan of text as written above](https://i.sstatic.net/scA17.jpg)\n\nI'm assuming it is from a scifi or fantasy work, but not sure.\n", "link": "https://scifi.stackexchange.com/questions/255498/looking-for-the-source-of-this-quote-i-copied-in-dec-1987", "tags": ["story-identification"], "votes": 22, "creation_date": "2021-10-30T14:34:01", "comments": ["Jacob, good question, but it doesn't read like my writing style. Guess I'll never know.", "If you don't recall its source, what makes you think it was a quote at all and not simply your own composition? Was manually copying long SF passages something you specifically remember doing?", "You've got a date of \"12-2-87\" in the typed transcription, but it looks more like 12-5-87 in the handwritten original.", "If you wouldn't mind transcribing the rest...", "The style reminds me of the Warhammer books. Possibly something associated with a game of that genre?", "Is there any other potential evidence on the paper you wrote this quote on? Other notes, something cryptic, initials, names --- anything? A hi res scan of your original document would be helpful.", "The meaning of \"terminator\" in the quote is the moving day-night boundary on a (nearly) airless body like the moon. With too tenuous an atmosphere to scatter much light, the boundary appears quite sharp (just like shadows on the moon are very dark).", "The obvious Tales of Narda fantasy doesn't seem to fit - no cars and a decade and a half too late."], "comment_count": 8, "category": "Life & Arts", "diamond": 0} {"question_id": "447", "site": "scifi", "title": "Man accidentally hired into government position, uncovers an attempted alien invasion", "body": "This is a book science fiction story. A man was accidentally hired into a (secret?) government position due to having the same exact name as their intended hire. I think that his name was \"Mr. Kennedy\", but I'm not sure. I don't remember, is he required to take the job or not, but he takes it. He then becomes a part of the bureaucracy of that \"orwellian\" world. A large part of the novel is the man describing his new office, the culture there, and he is feeling out of place, like an impostor. He describes the employees there as not really working, only making an attempt to look like they're working. Only one of the employees there recognizes that this man is not the intended hire, and he plots to get him fired.\n\nThis newly hired man comes across some documents about an attempted alien invasion of Earth. The alien's spaceships are described as being pitch black, with an ominous disposition. An alien spaceship shot a red laser beam which melted the peak of a mountain. Due to bureaucratic procedures, this incident was never investigated properly, and falls into obscurity in the archives, where this man retrieves it. It is later revealed that this alien species is a new comer to some kind of \"galactic federation\" of all advanced species. This species attacked the Earth because they were less advanced morally and culturally and were still prone to violent tendencies. All of these events happen in a \"Cold War\" environment of high secrecy and bureaucracy.\n\nThe introduction or prologue of this novel depicted beings of the \"galactic federation\" thinking about the fate of Earth's civilization. They said that most likely the Earth's civilization will destroy itself before it gains access to technology necessary to explore the stars. It was compared to a chicken (or some other creature?) breaking out of an egg. If a civilization goes on to explore the stars, it is a metaphor for a chicken which broke out of it's egg.\n\nI read this book 10 years ago in a public library, and so I do not remember most of the events. I also didn't read this book cover to cover, but some random detached chapters. I maybe wrong in my description of some of the events.\n", "link": "https://scifi.stackexchange.com/questions/228567/man-accidentally-hired-into-government-position-uncovers-an-attempted-alien-inv", "tags": ["story-identification", "aliens", "dystopia"], "votes": 22, "creation_date": "2020-03-13T13:57:56", "comments": ["I'd be interested in reading this book one day. It sounds like a good book. :)", "Also if you do post the answer, please post excerpts of the book which match these parts of the plot that I remember reading.", "@Jenayah I originally read this book 10 years ago in a public library. I don't know the name of the book, nor the author, nor even the year it was published. It might have been from the 60's. I don't know.", "How long is \"a long time ago\"? Three years ago? The 60's?"], "comment_count": 4, "category": "Life & Arts", "diamond": 0} {"question_id": "448", "site": "scifi", "title": "Short story: Lost starship crew in suspended animation trying to return to Earth. Crew suicide a factor", "body": "Read a sci-fi anthology sometime in the 1980s, though the story feels like it's from the 50s or 60s. An all-male starship crew is trying to return to Earth after performing some mission, but they're lost and for [reasons] are reduced to wandering from system to system.\n\nThe crew is in suspended animation between systems, and when they arrive at a destination, the crew is revived and a two-man scouting party is sent to the planet to check if it's Earth.\n\nThey've been lost for some time, and the crew is succumbing to despair. They lose men to suicide each time they come up empty.\n\nAfter arriving at the latest system, the captain and a crewman go to check out the planet. On it are the ~~radioactive~~ ruins of a civilization (may not have been due to a nuclear war), which shockingly turns out to be \n\n> EARTH!!!!! \n\nAdditional detail (that I may be pulling from another story, but here goes: in the location where they make the _shocking_ discovery, the room is filled with a thick (like centuries or millennia worth) layer of dust). The crewman laughs wildly then shoots himself in the head, the captain returns to the ship, sets the navigation controls to the next system, and records a message to the crew for the next wakeup along the lines of \"buck up, lads, the next one might be home.\"\n", "link": "https://scifi.stackexchange.com/questions/139009/short-story-lost-starship-crew-in-suspended-animation-trying-to-return-to-earth", "tags": ["story-identification", "short-stories"], "votes": 22, "creation_date": "2016-08-28T10:36:03", "comments": ["Haven't figured it out yet. The lighthouse bit sounds right. Getting closer...", "Hey, I’m searching for this short story as well. Did you ever find out the title? I think the structure is a stone lighthouse (I could be wrong) and the captain reads the account of mankind’s extinction in a log book of sorts. I think you’re right that the author made a point to describe the deep accumulation of dust in the room.", "Having said that it was a common post war theme as a good deal of the media (and fiction) suggested we would blow ourselves to hell (on a regular basis). Add this to the requirement for lengthy space travel (no speed of light, great distances) and the sitiuation is poised. Clarke did it in his novel 2010 (although the Earth doesn't blow the Russian / US crew are under constant pressure as they wake from hyper sleep because of rising war tension on Earth). And as you say we had it in Planet Of the Apes. Sure there are many other examples.", "This sounds like the movie Terminal Voyage (aka Star Quest), in which a crew wake from hyper-sleep to discover Earth has destroyed itself during the journey. The first officer (now captain) immediately commits suicide. Others follow as they come to terms with being the last of their race. The twist is that they never left Earth but are part of a test on effects of hyper-sleep / stress. Coming out of the simulator they are met with decades of decay and find a guard dead in a chair. Opening a hatch they climb out to discover Earth has indeed blown to hell (whilst they were asleep).", "Sorry, I added the detail about the deep dust, but don't recall precisely how they figured out it was RNEGU -- don't think it was a Statue of Liberty moment.", "Do you happen to recall how the space travelers found out that the planet was — don't know how to put a spoiler block in a comment — OK, I guess rot13 will do the trick — RNEGU???", "Great use of the spoiler tag there, I'd never guess in a million years!"], "comment_count": 7, "category": "Life & Arts", "diamond": 0} {"question_id": "449", "site": "scifi", "title": "A ruined USA, where protagonists are captured by a biker bandits encamped in a zoo", "body": "It was written in English. I read it around 2013-2014 in British Columbia, Canada when I was in middle school.\n\nIt could have been a short series, or one book. \n\nThe protagonists are young adults, and so the book was probably for young adults.\n\nThe setting is a post-apocalyptic/heavily resource-drained world, with the narrative taking place in the United States; I don't believe the author explains why the world has gone downhill. The government only retain a loose control over some places.\n\n**Distinct scenes:**\n\n * (At possibly the beginning.) The group of several protagonists join a walled community, they regret it, and now they plan to flee through the sewers. They have an argument about leaving behind someone who is ill and would not be able to flee with them. When that community finds out they've fled, they quickly assemble a group to chase after them into the forest. A member of the search group is described briefly as \"gung-ho\" I think it was the first time I ever saw that word. The chased protagonists are then split up into 2 groups, who manage to escape. They had planned to meet at Salt Lake City in case of this, I believe.\n\n * In a later scene with one group of three protagonists (2 male, 1 female), one of the male characters uses up a bunch of food, including a bird he's killed to have a Thanksgiving celebration in their makeshift cave shelter. The female is happy about the celebration, but also questions whether they could afford it. They later travel down the road.\n\nThe three are captured by a gang of bikers/bandits they take them to a zoo where they have taken up residence. The gang leader is like a king. Most of the animals have been eaten by the gang already. They plan to feed the 2 males to the alligators while the female becomes a slave to the gang's leader. The other slaves take her to have make-up and different clothes put on while she cries. Another slave-girl gives her a tip \"When he enters you, don't cry, he hates that\".\n\n * In a different scene, the other group of two protagonists (1 male, 1 female) are walking through a city (Salt Lake City maybe) and a soldier described as a 'Peacekeeper' asks them why they aren't in school. The boy is worried about them not having a passport. He lies to the guard something like: \"Well, you see, we're young lovers who have recently gotten engaged\". The Peacekeeper feels sympathy for them and lets them go.\n\n\n\n", "link": "https://scifi.stackexchange.com/questions/166731/a-ruined-usa-where-protagonists-are-captured-by-a-biker-bandits-encamped-in-a-z", "tags": ["story-identification", "young-adult", "post-apocalyptic"], "votes": 22, "creation_date": "2017-08-08T01:29:33", "comments": ["Can you remember whether there were religious undertones? I read something similar in the '80s. The people still celebrated Christmas but it was illegal to believe in God or have a religion. Does that ring a bell? Not that I can remember the book!", "This question would be improved by going through the checklists here; How to ask a good story-ID question?"], "comment_count": 2, "category": "Life & Arts", "diamond": 0} {"question_id": "450", "site": "scifi", "title": "WW2 airman is thawed out, and saves Earth from interplanetary invasion", "body": "I am trying to find the title and author of an extremely pulpy science fiction yarn. Unfortunately I do not know how the cover was - the version I had was secondhand, and in the vicissitudes of its life the cover was lost by the time the book came to me. It was definitely old - I suspect from either the 1950s, or possibly the 1960s.\n\nThe story is set in the future, when the Earth is under attack from another planet. The protagonist is an airman, I think from the Second World War (or shortly afterward). His plane crashed and we was frozen in a glacier (rather like Buck Rogers) until he was found and thawed out. In the process he was also “upgraded”; in particular I remember that his muscle strength was enhanced, and his blood was replaced with amniotic fluid with a higher oxygen carrying factor, that gave him the capacity to hold his breath for long times. This would be a plot point later.\n\nBecause of these enhancements he is placed with a commando team who will travel in a spaceship to the planet of the attackers, carrying an atom bomb on board of sufficient force to destroy the planet. When they get there they find that the planet supports two intelligent races. One dwelling on the land is the one attacking the Earth. Rapidly increasing desertification means that the land area will soon not support life, and so they are planning to move to the Earth after it has been conquered. The other race lives in the oceans. As an interim measure, the land dwellers are moving into the oceans, which are less affected by the global warming, by building underwater cities. This encroachment is opposed by the marine race, leading to a vigorous war between them.\n\nI don’t recall too much of the protagonist’s adventures. They are captured on arrival and taken to one of the underwater cities. The protagonist escapes, and due to his ability to hold his breath, is able to make contact with the marine race. Thank to his assistance, I believe (but I’m not sure) that the marine race win a comprehensive victory against the other, eliminating the threat to Earth.\n\nI remember the ending better. The Earthmen work out that if the atom bomb is placed in a suitable canyon, then instead of blowing the planet apart, its detonation will shift the planet into a new orbit, further from the sun, and so cure the global warming problem. I remember feeling rather skeptical about this, even at the time. But I don’t know if the scheme worked or not, because the last few pages were missing.\n\nExtra detail: After thinking hard about this, I remember that after his capture the protagonist was interrogated by a sexy scantily-clad alien princess. To find out what planet they came from, she used a telescope, whose primary mirror was formed by a rotating bath of mercury (and so had a perfect parabolic shape). This is an inconsequential but slightly unusual touch, so maybe it can jog someone's memory.\n", "link": "https://scifi.stackexchange.com/questions/246437/ww2-airman-is-thawed-out-and-saves-earth-from-interplanetary-invasion", "tags": ["story-identification", "novel"], "votes": 21, "creation_date": "2021-04-29T05:53:07", "comments": ["@user14111 Yes to both questions. I also have the impression it was British rather than American, but I am not 100% sure of that.", "English language? Paberback?", "This is of course not the yarn you're looking for, but such a rotating bowl of mercury was used by Number 774, the Martian protagonist of Raymond Z. Gallun's 1934 novelette \"Old Faithful\", for his observations of the planet Earth. archive.org/details/Astounding_v14n04_1934-12/page/n111/mode‌​/…", "@MarkOlson I remember the writing as quite plain and unornamented. I don't think it was full of repeated adjectives or phrases", "That sounds vaguely familiar. It sounds so full of unnecessary stuff that I wonder if it is one of Lionel Fanthorpes' opus? Do you remember if the writing was unusual -- full of repeated adjectives and phrases?", "A fair bit of this sounds like Marvel's Captain America somewhere between Captain America: The First Avenger and The Avengers. Of course, Steve Rogers was \"upgraded\" before the plane crash and subsequent freezing, not after. All I'm familiar with, though, is the MCU movies, so the actual events in the comic books may have predated the book you're after and given its author some ideas."], "comment_count": 6, "category": "Life & Arts", "diamond": 0} {"question_id": "451", "site": "scifi", "title": "Short story about a giant ship that thoughtlessly destroys planets as it travels, is visited by a victim", "body": "I think I came across this ten or fifteen years ago, I cannot remember much clearly so some of these details may be mistaken. The story is told from the perspective of a crew member, on the bridge I think.\n\nThe ship is described as utterly enormous, sleek, and silver colored; I think it has been cruising at (near?) lightspeed for so long the crew have forgotten why. As it travels, it automatically targets planets, destroying them instantly as it passes. A major point of the story is that the inhabitant of one such planet manifests on the ship and speaks to the crew, I cannot recall the mechanics of this—spiritual, interdimensional, an AI? I only skimmed the beginning of this when I encountered it, and it's been lurking in my head for ages to finish it.\n", "link": "https://scifi.stackexchange.com/questions/240103/short-story-about-a-giant-ship-that-thoughtlessly-destroys-planets-as-it-travels", "tags": ["story-identification", "short-stories", "aliens", "generation-ship"], "votes": 21, "creation_date": "2020-12-07T03:28:52", "comments": ["I'm thinkng of Stephen Baxter's Xeelee Sequence, particularly Flux (if the \"ship\" is allowed to be a neutron star).", "@Tango: No, the description doesn't match a Berserker story. Berserker ships have an AI and are definitely not piloted by lifeforms. They may have prisoners, and in at least one case an obedient slave, but no crew. There is one case where a Berserker AI forgot its purpose (or more exactly, the meaning of its purpose), but not because of age but due to an unfinished repair. Also, Berserkers target lifeforms, not whole planets.", "I think I read that story and that it's in the Berserker books by Fred Saberhagan. I wish I had time to look it up, but I don't. I think the survivor has some way of messing with the Berserker's programming by tricking it or something. (Kind of like how Kirk messes up computers!) en.wikipedia.org/wiki/Berserker_(novel_series)", "This sounds a little like Diane Carey's ST:TNG novel 'Ghost Ship'...? (memory-alpha.fandom.com/wiki/Ghost_Ship)", "I've read this, or something similar in an anthology, it was quite surreal half to do with relativistic effects on the ships crew. Let me think . . ."], "comment_count": 5, "category": "Life & Arts", "diamond": 0} {"question_id": "452", "site": "scifi", "title": "What's this serialised comic strip about a girl who discovers everyone around her is a robot?", "body": "As a kid in the '80s, I found myself often in doctor's waiting rooms. I devoured most of the comic magazines there. In one of those, I read fragments of a story that I would like to find.\n\nThe story was set in (then) current time. The protagonist was a teen girl, who discovered that people around her had been replaced by robots. These robots' faces could hinge upwards, revealing gears behind them. \nI remember her looking in the mirror and even seeing her _own_ face hinging upwards to reveal herself being a robot, however, that may have been a dream.\n\nAs I recall, in the end it was revealed some catastrophe had happened, perhaps a nuclear bomb, and \"someone\" (perhaps benevolent aliens) had replaced her loved ones for her. I think it ended with her reliving the final moments of her old world, knowing what was about to happen.\n\n* * *\n\nThis was in The Netherlands and the story was written in Dutch, although it may have been translated.\n\nThe story was serialised, 2 or perhaps 4 pages at a time, which is why I've only read fragments.\n\n * The most likely magazine I've read this in was a Dutch girls' magazine called _[Tina](https://nl.wikipedia.org/wiki/Tina_\\(tijdschrift\\))_. _(Dutch Wikipedia)_ \nMost comics in it were about girls overcoming some obstacle to pursue their dream — usually horse riding or ice skating, finding love in the process — which is what would've made this story stand out. Another such magazine was called _Debbie_.\n * Another likely source is comics magazine _[Eppo](https://en.wikipedia.org/wiki/Eppo_\\(comics\\))_ , later called _Eppo Wordt Vervolgd_ , that featured serialised comics.\n * Yet another possible source is _[Taptoe](https://nl.wikipedia.org/wiki/Taptoe_\\(tijdschrift\\))_. _(Dutch Wikipedia)_ , a weekly and later biweekly magazine that was distributed through schools/ It had sister publications _Okki_ and _Jippo_ , but those were aimed at smaller children and this story must've been too mature for them.\n\n\n\nThe style as I remember it resembles _[ligne claire](https://en.wikipedia.org/wiki/Ligne_claire)_ , known from the Franco-Belgian comics known as _bandes dessinée_. It may also have been a realistic drawing style. It was definitely in colour.\n", "link": "https://scifi.stackexchange.com/questions/199749/whats-this-serialised-comic-strip-about-a-girl-who-discovers-everyone-around-he", "tags": ["story-identification", "comics", "robots"], "votes": 21, "creation_date": "2018-12-03T01:28:10", "comments": [], "comment_count": 0, "category": "Life & Arts", "diamond": 0} {"question_id": "453", "site": "scifi", "title": "Old movie involving unexplained handprint on mirror in spaceship", "body": "I would like the title of an old black-and-white movie. My best guess is that it would have been released in the 1950s or early 1960s, as I saw at least a small part of it when I was about ten years old (1964). \n\nIt was a live-action film in English, presumably American or British in origin, but it could have been dubbed. \n\nThe only thing that I remember about the film is that (I _think_) the setting is a spaceship (flying somewhere in outer space) that is supposed to have only one man aboard, but in one scene, the man goes into an area of the spaceship where there is a mirror, and on the mirror there is a hand print. That hand print, of course, is not his own, i.e., he is evidently being served notice that he is not alone on the ship after all...\n\nShortly after I saw this scene, the horn of the '63 station wagon (precursor to the soccer-mothership SUV) sounded out in the driveway, summoning me to baseball practice with my junior cronies, so I had to bail on this movie and I never again encountered it. So I don't know just what dynamic might have played out after the mysterious appearance of the hand print: was it all in the astronaut's mind, or was the print placed there pre-takeoff by a cold-war enemy, to drive the astronaut nuts, or was there really someone else on board, and if so, who and why? That is why I would appreciate any help with identifying this film.\n\nSorry to be able to provide so little info. Many thanks for any help with this!\n", "link": "https://scifi.stackexchange.com/questions/188716/old-movie-involving-unexplained-handprint-on-mirror-in-spaceship", "tags": ["story-identification", "movie"], "votes": 21, "creation_date": "2018-05-30T15:02:29", "comments": ["The only image of a handprint in a glass in a spaceship that comes to my mind belongs to Ark (2009). A canceled TV series (only one episode) from director Trey Strokes. I could be a coincidence or possibly inspired by your movie. imgur.com/a/PEVjPKk", "Sounds like an awesome premise for a sci fi psychological/horror movie."], "comment_count": 2, "category": "Life & Arts", "diamond": 0} {"question_id": "454", "site": "scifi", "title": "Continent sized, X-shaped alien", "body": "I have a vague recollection of this book, mostly the X-shaped alien. I can’t find anything through internet searches. I thought it may be some sort of “choose your adventure” book, but I haven’t seen it on any lists or pages discussing those sorts of books.\n\nAlien description:\n\n * Hugging a water world. Looks like regular land mass from space, just X shaped.\n * Exerts influence over its inhabitants. I don’t recall the method or intensity of the influence. Could have been simple indoctrination (not the Reaper kind).\n * Reproduces by launching spores into space. Similar to a sarlacc.\n * When killed, the alien unlatches from the planet.\n * Intelligent. To what degree, I don’t recall.\n\n\n\nOther details:\n\n * The only visible land mass was the alien, nothing natural.\n * The alien would react when you cut into it (cutting into the ground).\n * The planet is inhabited by human children (the ones under the influence of the alien).\n * The children view the alien as a parent or caretaker.\n * The book is written in either 1st or 2nd person (2 person being the choose your adventure style).\n * The planet the alien is on is not Earth.\n * I don’t recall any crew with the protagonist.\n * The protagonist does not immediately know that the X shaped continent is actually a creature. I do recall that being a decent twist for my young mind.\n\n\n\nUpon some further investigating, Ray Bradbury’s “Here there be Tygers” is close and potentially the inspiration for the story I’m looking for. It sounds like they took “Tygers” added some Stephen King flair (the lost boy-esque children) and made it a CYA.\n\nI made a mock-up of it [here](https://www.maptoglobe.com/rJPYKPjfo) in case a visual triggers someone’s memory.\n\nI read it somewhere around 16 years ago from a school library.\n", "link": "https://scifi.stackexchange.com/questions/268271/continent-sized-x-shaped-alien", "tags": ["story-identification", "books"], "votes": 20, "creation_date": "2022-10-03T11:36:53", "comments": ["Legacy by Greg Bear is about continental scale organisms", "I read this as well! There was another scene in which one of the protagonists lay down on a stone altar/bed, and the alien began lulling him to sleep and moulding itself to his form. Also looking for it, but know that I read it sometime before the turn of the millennium. Also trying to find it, sadly. However, I thought it was a desert world, with the implication being that the alien had drained the life from the planet.", "I couldn’t find anything that DC published, but if it’s hard to find like you said then that doesn’t rule it out. I would have expected the protagonist be notable in some way if SC published it, but I remember him as average (not a super hero).", "It sure does sound like something DC Comics would have put out featuring Starro the Conqueror. But those children's books by DC are tough to track down.", "It sounds like some fever dream out of Dan Simmons's Hyperion saga.", "I don’t recall. This is everything I know and it’s from memory. It’s a fairly unique take on living planets from what I can find, but it must not have been popular.", "Is the alien intelligent or is its effect on the inhabitants automatic? You say it unlatches when it is killed; that implies that over the course of the book it is killed? How is a continent-sized alien being killed?", "Hi, welcome to SF&F. Where and when did you read this? Do you remember any details of the cover art? Any character or place names? You can edit any additional details you recall into your question."], "comment_count": 8, "category": "Life & Arts", "diamond": 0} {"question_id": "455", "site": "scifi", "title": "A story about a music teacher, cheated by a used-car salesman, and society begins to unravel", "body": "The story opens with the teacher (I can't remember his name, so call him 'Whosit') stranded because his 'new' car has died, phoning the salesman, hoping for redress, and being blown off.\n\nSuddenly the world turns on the salesman: at first amusingly, but then it snowballs into horror — we see the salesman's wife walking out, saying only \"Whosit doesn't like you\" — and though he attempts redress, he winds up dying by the side of the road while people mock him.\n\nThen we see the music teacher, receiving a call from the student who's appointment he'd missed, and although he tries to apologize, it is clear that the weight of the world is about to descend on him in his turn!\n\nI've often remembered this story as a clever parable about the danger of revenge fantasies. I believe I read this is some American collection in the seventies or eighties.\n", "link": "https://scifi.stackexchange.com/questions/263954/a-story-about-a-music-teacher-cheated-by-a-used-car-salesman-and-society-begin", "tags": ["story-identification"], "votes": 20, "creation_date": "2022-06-03T12:56:22", "comments": ["Almost certainly a short story. Maybe a long short story--I always remember things as shorter than they were.", "Is this a short story, or a novel?", "Scope-wise, it sounds very Twilight Zone-y."], "comment_count": 3, "category": "Life & Arts", "diamond": 0} {"question_id": "456", "site": "scifi", "title": "Lost novella or short story: Either uplifted or evolved animals finally break through a barrier to discover that the humans have left Earth", "body": "The protagonist, possibly an uplifted or naturally evolved bear is in charge of a project to breach a barrier. This has kept all intelligent species from leaving the area. They eventually break through to discover a final message left by humans.\n\nThey have left the Earth to move out into space and were either responsible for the animal uplift in intelligence or acted responsibly when it naturally occurred.\n\nThey left Earth to allow the animals to develop naturally, uninfluenced by human activities.\n\nI read this sometime in the 1970's, it may have been the first and longest story in an anthology.\n\nI recall that the story ends just after they discover a final message left by humans. I think that the bear feels wonder that a species could be so altruistic.\n\nIt is not any of the following:- \"Day of Judgment\" by Edmond Hamilton (1946). \"A Heritage of Stars\" by Clifford D. Simak. \"City\" by Simak. \"The Crystal Spheres\" by David Brin. \"Old Hundredth\" by Brian Aldis.\n", "link": "https://scifi.stackexchange.com/questions/237062/lost-novella-or-short-story-either-uplifted-or-evolved-animals-finally-break-th", "tags": ["story-identification", "short-stories"], "votes": 20, "creation_date": "2020-09-16T10:31:39", "comments": ["Heinlein's novella Lost Legacy ends with an idea like this, but it's only the last 2-3 paragraphs, not the entire novella.", "Wow, this question is from 2020? And ChatPig can't tell you?", "L. Sprague de Camp wrote some stories about an uplifted bear, Johnny Black, but the ones I've skimmed all feature humans too.", "I have just checked and it is definitely NOT 'Old Hundredth' but I can see the why it was proposed.", "It sounds a bit like Brian Aldis' \"Old Hundredth\". I don't think it is, though.", "@Seldon2k: City is excellent. IMO A Heritage of Stars is not one of his stronger works.", "The barrier reminds me a lot of David Brin's novella The Crystal Spheres, but in it, it is humans breaking through and discovering that ancient aliens once put these barriers (imprenetable from the outside) there to allow unevolved beings on various planets to evolve enough to break them. No bears.", "City, a collection of short stories about intelligent speaking dogs investigating their origins, did have Jenkins the robot taking the last of the humans not in permanent Sleep to a parallel world, to give the dogs their chance at being the dominant race. It is the first thing I thought of, but that is the only match: there is no barrier. (Obviously for anyone but Jenkins moving to a parallel planet is a barrier, but there is no \"barrier\" in the story.) And no single animal, bear or dog, but a race of them.", "Sounds vaguely like Simak, City or A Heritage of Stars or something."], "comment_count": 9, "category": "Life & Arts", "diamond": 0} {"question_id": "457", "site": "scifi", "title": "Looking for a children's or young adult novel about three kids travelling to a fantasy land", "body": "I've been trying to remember the title of a children's or young adult book I read as a child about thirty years ago. (So it may be from '70s or '80s). I don't remember many of the details. \n\nThere were three main protagonists; I want to say that all three were boys (maybe one was a girl), one of them may have been from India and the other two from the US or UK. The three end up in a fantasy land with an evil king or prince. \n\nThere's a **giant \"inchworm\" that the three characters can ride on saddles, which means that the middle rider goes up and down as the worm moves**. As the worm is running from guards at one point, the middle rider gets jammed up into a crack in the ceiling of a tunnel and gets separated from the other two. \n\nI also vaguely remember a scene where **a character eats some glowing mushrooms and ends up glowing himself** , being mistaken for a ghost. \n\nOdd memories, but I do remember enjoying the book as a child and would like to read it to my own son. I've made multiple searches online trying to word the search in every way I can think of without success.\n", "link": "https://scifi.stackexchange.com/questions/35701/looking-for-a-childrens-or-young-adult-novel-about-three-kids-travelling-to-a-f", "tags": ["story-identification", "young-adult", "childrens-novel"], "votes": 20, "creation_date": "2013-05-16T13:16:21", "comments": [], "comment_count": 0, "category": "Life & Arts", "diamond": 0} {"question_id": "458", "site": "unix", "title": "How to escape from a hardened chroot on Linux when only file capabilities are available to the attacker?", "body": "There is a chroot environment in `/var/myroot`, and an attacker has obtained arbitrary machine code execution in a process running as root (EUID 0) in the chroot. But the processes under control of the attacker don't have all capabilities enabled (just filesystem capabilities). The attacker wants to escape the chroot, and append a line to `/etc/passwd` outside the chroot. How can he do it?\n\nThe following security measures have been set up:\n\n * To prevent the [chdir(\"..\") chroot escape technique](http://www.ouah.org/chroot-break.html), the chroot environment was entered using _pivot_root(2)_ rather than _chroot(2)_. See in [jchroot.c](https://github.com/vincentbernat/jchroot/blob/3ad09cca4879c27f7eef657b7fa1dd8b4f6aa47b/jchroot.c#L122) how it can be done.\n * When the first process in the chroot was started, it didn't have any file descriptors open pointing outside the chroot.\n * Each process in the chroot has at most `CAP_CHROOT` and `CAP_FOWNER`, `CAP_FSETID`, `CAP_CHOWN`, `CAP_DAC_OVERRIDE`, `CAP_DAC_READ_SEARCH`, `CAP_SETGID`, `CAP_SETUID` capabilities and is not able to gain other ones. In short, the attacker is able to make arbitrary filesystem reads and writes, bypassing permission checks etc., but he is not able to send arbitrary signals to processes (`CAP_KILL`) or send arbitrary packets on the network (`CAP_NET_RAW`) or reboot the system (`CAP_SYS_BOOT`) or modify arbitrary bytes in memory (`CAP_SYS_RAWIO`) etc.\n * `unshare(CLONE_NEWUSER)` was _not_ called, the UID 0 of the chroot process is the same as UID 0 outside the chroot.\n * `unshare(CLONE_NEWPID)` was called, so the attacker doesn't see processes running outside the chroot.\n * `unshare(CLONE_NEWNS)` was called when setting up the chroot, and the following filesystems are visible: \n * `/var/chroot` is visible as `/`, remounted with `MS_NODEV` and `MS_NOSUID`. (`MS_NODEV` is used so that the attacker can't write to `/dev/sda`, and `MS_NOSUID` is used so that the attacker can't gain new capabilities from file capabilities.)\n * A proc filesystem is visible as `/proc`, with the following paths removed (by putting an empty file there using a bind-mount): `/proc/kcore`, `/proc/latency_stats`, `/proc/timer_list`, `/proc/timer_stats`, `/proc/sched_debug`, `/proc/scsi`, and the following paths made read-only: `/proc/asound`, `/proc/bus`, `/proc/fs`, `/proc/irq`, `/proc/sys`, `/proc/sysrq-trigger`.\n * A sysfs filesystem is not mounted.\n * A devpts filesystem is not mounted.\n * A tmpfs filesystem is mounted as `/dev` without `MS_NODEV`, prepopulated with a few devices.\n * No other filesystems are visible from the chroot.\n * Block device nodes are not available in the chroot (and `CAP_MKNOD` is not available, so the attacker is not able to create them).\n * Only the following character device nodes are available: `/dev/null`, `/dev/zero`, `/dev/full`, `/dev/random`, `/dev/urandom`, `/dev/tty`, `/dev/ptmx` (same as `/dev/pts/ptmx`) and a `/dev/pts/X`, the terminal which was used outside the chroot.\n * If the attacker calls `ioctl(..., TIOCSTI, ...)` (simulate typed input) on `/dev/pts/X`, and exits the chroot, probably an interactive shell outside the chroot will read those simulated bytes, so simulating this can be useful: `sudo sh -c 'echo pwned::0:0:pwned:/:/bin/bash >>/etc/passwd'`. To prevent this from succeeding, the parent process which created the processes in the chroot will flush all terminal input before returning to the shell.\n\n\n\nFYI My use case is the following: there is a Debian system in `/var/myroot` (possibly created by _debootstrap_), and I'd like to be able to install untrusted packages there without exposing the host system to attacks by install scripts in malicious packages. Unfortunately `sudo chroot /var/myroot apt-get install MALICIOUS-PACKAGE` is not secure enough, because e.g. the install script of the package can create the block device node `/dev/sda1`, find the `/etc/passwd` file there and modify it, thus potentially escaping from the chroot. I'm now exploring what other options are secure enough, and a hardened chroot as described above is one of the candidates. (In this question I'm not looking for other candidates.) In this question I'd like to understand how secure it is.\n\nFYI There is a command-line tool [chw00t](https://github.com/earthquake/chw00t) to escape from chroot. About its techniques:\n\n * -0 does not work here because _pivot_root(2)_ was used.\n * -1 does not work because there are no file descriptors pointing to outside the chroot available.\n * -2 may work, I need to check if it works with _pivot_root(2)_.\n * -3 does not work because _unshare(CLONE_NEWPID)_ was called.\n * -4 does not work because block devices are not available.\n * -5 may work, I need to check if it works with _pivot_root(2)_.\n * -6 does not work because _unshare(CLONE_NEWPID)_ was called.\n * -7 does not work because there are no file descriptors pointing to outside the chroot available.\n\n\n", "link": "https://unix.stackexchange.com/questions/492473/how-to-escape-from-a-hardened-chroot-on-linux-when-only-file-capabilities-are-av", "tags": ["linux", "security", "chroot", "capabilities"], "votes": 6, "creation_date": "2019-01-04T05:23:07", "comments": ["@A.B: I agree that using a well-maintained container setup tool (such as LXC or runc) gives better security than reinventing the wheel and doing it halfway. However, in this question I'd like to better understand what the attack vectors are (some of which may be mitigated by LXC and runc, some aren't). I've just updated the question to state that's using pivot_root(2) to enter the chroot environment, and I'm looking for more attack vectors.", "Just run strace on lxc to see how it's doing: it's using clone(CLONE_NEWNS...) + mount(... MS_BIND ...) + pivot_root and never chroot . pivot_root (existing as command too) is secure. I think you'd better run a container with ready software (eg: lxc) than trying to reinvent the wheel. LXC's container running debian is usually installed by LXC using debootstrap, or with some minor tweaking can be pointed to an existing installation.", "To those who are voting to close this question as too broad: Could you please advise me how I can make it less broad? I was trying really hard to make it as specific as possible, and I'd be happy to extend it if needed, but I fail to see how.", "@A.B: Thank you for the link about breaking out of a chroot! I've modified the question title, now it says hardened chroot. It's in fact a chroot, because chroot(\"/var/myroot\") was called after unshare(CLONE_NEWNS). (Is there something more secure that should be called instead?)"], "comment_count": 4, "category": "Technology", "diamond": 0} {"question_id": "459", "site": "unix", "title": "Restoring in-memory/swapped page state on resume from hibernation", "body": "I'm a big fan of Linux's hibernation support, which works extremely well on all the (admittedly slightly older) hardware I've tried it on. I highly prefer it to full shutdown and power up.\n\nThere's one thing about hibernation that's bugged me for a while, though: a hibernated system is always sluggish and unresponsive immediately after having been resumed. It's not snappy immediately after power on. This is exacerbated by old hardware, but happens on newer systems to a slightly notable extent too.\n\nThis seems to be because the kernel only swaps the pages critically necessary for [kernel-level] operation back into working memory, restores the kernel's base working state, and lets userspace just churn for a while as various processes swap the pages they need back into RAM under their own steam.\n\nThis does not work very well in practice, as the system initially functions as though some large process had forced everything to fully swap to disk. Whatever was on the screen will swap back in pretty quickly, but switch to another process and you'll be waiting a couple seconds as _it_ swaps back into memory too. On older hardware - my case in point is a Core 2 ThinkPad T60 - \"a couple seconds\" can even wind up being a couple minutes.\n\nI recently realized that this problem has a surprisingly simple solution, after thinking about it for a bit: take note of the tags that label what pages are on disk and in RAM, then restore this exact state on resume. Sure, the resume process might take ~10 seconds more, but I don't care - I'd have a snappy system.\n\nI was wondering if there are any obscure kernel compile options that enable such functionality, or some configuration I can set up that would approximate this behavior?\n\n**NOTE:** I do not consider `swapoff -a; swapon -a` a viable solution; the moment the suspended userspace is reanimated, all loaded processes are attempting to execute code and fighting to swap themselves back into RAM, causing massive disk I/O. Attempting to destroy the swap area would only add to the hurricane, and would actually take longer to complete than if the kernel resumed _all_ of RAM before reanimating userspace.\n", "link": "https://unix.stackexchange.com/questions/242264/restoring-in-memory-swapped-page-state-on-resume-from-hibernation", "tags": ["linux", "swap", "io", "hibernate", "high-performance"], "votes": 5, "creation_date": "2015-11-10T17:00:13", "comments": ["Note: implemented a script to read memory of specified PIDs back to memory explicitly. Maybe it can be a part of the solution."], "comment_count": 1, "category": "Technology", "diamond": 0} {"question_id": "460", "site": "stats", "title": "Is there a general expression for ancillary statistics in exponential families?", "body": "An i.i.d sample $X_1,\\dots,X_n$ from a scale family with c.d.f. $F(\\frac{x}{\\sigma})$ has $S(X)$ as an ancillary statistic if $S(X)$ depends on the sample only through $\\frac{X_1}{X_n},\\cdots,\\frac{X_{n-1}}{X_n}$.\n\n 1. Is this result also sufficient?\n 2. Is there a parallel result for a general exponential family which is not necessarily a scale family?(not asymptotic results, see update below)\n\n\n\nIf yes, I want to see some reference; If no, why is it not possible?\n\n**Note** : Reading B. Efron's paper on the geometry on exponential families I now believe that this should somehow relate to the geometric nature of the exponential families.\n\nBut I have difficulty imagining what geometric object ancillary statistics should correspond to. Firstly I thought it should be the normal bundle, but later I found it only sufficient.\n\n* * *\n\nUpdate on this question:\n\nAfter a careful look into [1], I think by ancillary statistics I mean the likelihood ratio (derived) ancillary statistics, not the Efron-Hinkley affine ancillary statistics. [1,Fig1] showed the difference in their marginal densities.\n\n_The results pointed out in [2] by a nice comment by @kjetilbhalvorsen below do not address my question._\n\n[2] discussed several examples around pp.30-45, and proposed a simple case where S-sufficient S-ancillary are simultaneously introduced. i.e. $(\\psi,\\chi)$ are the parameter of the exponential family, and then $S=(T,A)$ is the minimal sufficient statistics for the parameter $psi$ of interest.We call the statistics $A$ an S-cut if the distribution $T\\mid A$ depends only on $\\psi$ and the distribution of $A$ only depends on $\\chi$. If we look it via factorization, this actually prompt us to $(T,A)_{(\\psi,\\chi)}\\overset{d}{=}(T\\mid A)_{\\psi}\\cdot (A)_{\\chi}$. Geometrically this only means that we can find a subspace for statistic $A$ and write them in form of direct product. This is not interesting since we know the minimal sufficient statistics does not always exist. One useful example in [2] is that it gave the approximation formula ($p^{\\dagger}$-formula, 6.10)for abritrary ancillary statistics and later proved it is $\\sqrt{n}$-consistent. **However, it does not reveal any geometric feature since if $n\\rightarrow\\infty$ then any such aprroximation essentially describing locally Gaussian space.**\n\n[1]Pedersen, Bo V. \"A comparison of the Efron-Hinkley ancillary and the likelihood ratio ancillary in a particular example.\" The Annals of Statistics (1981): 1328-1333.\n\n[2]Cox, D. R., and O. E. Barndorff-Nielsen. Inference and asymptotics. Vol. 52. CRC Press, 1994.\n", "link": "https://stats.stackexchange.com/questions/193331/is-there-a-general-expression-for-ancillary-statistics-in-exponential-families", "tags": ["mathematical-statistics", "references", "exponential-family", "information-geometry", "ancillary-statistics"], "votes": 25, "creation_date": "2016-01-31T06:18:42", "comments": ["Not yet. Exponential families can have multiple parameters and the distributions can vary widely. It's really hard to come up with a general expression that is independent of all these parameters.", "Ghosh, M., et al. “ANCILLARY STATISTICS: A REVIEW.” Statistica Sinica, vol. 20, no. 4, Institute of Statistical Science, Academia Sinica, 2010, pp. 1309–32, jstor.org/stable/24309506.", "O.E.Barndorff-Nielsen&D.R.Cox: \"Asymptotic Techniques for Use in Statistics\" & \"Inference and Asymptotics\" (probably the last one) (Both books is Chapman&Hall)", "@kjetilbhalvorsen Would you mind adding at least the publishing information of the title you are referring to? Thanks,", "Look at the books by Barndorff-Nielsen & Cox, the answer is in there, someplace"], "comment_count": 5, "category": "Science", "diamond": 0} {"question_id": "461", "site": "stats", "title": "Is the Wilcoxon two-sample test maximally powered to detect proportional odds alternatives?", "body": "We know from the literature that\n\n * The Wilcoxon-Mann-Whitney two-sample rank sum test is optimal for detecting simple location shifts when comparing two continuous random variables that each have a logistic distribution\n * The Wilcoxon test is a special case of the semiparametric proportional odds ordinal logistic model, e.g., the numerator of the score test for testing $\\beta_{1}=0$ in this model, when there is only a single covariate and it is binary, is exactly the rank sum statistic (this is a generalization of the first bullet)\n\n\n\nDo we know that the most general way to state that the Wilcoxon test has optimum power in a given situation with continuous $Y$ is when the two distributions are in proportional odds, e.g., logit$(F(x))$ is parallel to logit$(G(x))$?\n\nAs a slight aside, we know that the log-rank statistic, a special case of the Cox proportional hazards semiparametric regression model, has optimum performance when proportional hazards is true (just like its generalization).\n", "link": "https://stats.stackexchange.com/questions/439672/is-the-wilcoxon-two-sample-test-maximally-powered-to-detect-proportional-odds-al", "tags": ["statistical-power", "wilcoxon-mann-whitney-test", "ordered-logit", "semiparametric"], "votes": 21, "creation_date": "2019-12-06T12:55:57", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "462", "site": "stats", "title": "Generalization of degrees of freedom for t distribution for coefficients after multiple imputation", "body": "Donald Rubin has shown that regression coefficient estimates have fatter tails after multiple imputation and has provided a formula for the degrees of freedom to use as a t-distribution approximation to the coefficient estimates resulting from Rubin's rule for combining multiple imputations. I would like a generalization of this approach that handles more than single degree of freedom tests, e.g., an adjustment to the multiple degree of freedom $\\chi^2$ distribution for \"chunk tests\" that test combinations of parameters. Has anyone seen such a procedure? The goal is to improve (raise) the coverage probability of confidence intervals after multiple imputation, and to better preserve type-I error.\n\nUltimately it would be nice to have a way to convert covariance matrices arising from Rubin's rule so that normal and $\\chi^2$ distributions can still be used, and confidence coverage will be more accurate. In the single parameter case one could just bump up the standard error of $\\hat{\\beta}$ by a factor equal to the $t$ critical value divided by the normal critical value.\n\n**Follow-up**\n\nAs I've gotten more into Bayesian modeling I came to know about posterior stacking for incorporating multiple imputation into Bayesian analysis using MCMC posterior sampling. This makes the tails of the posterior distribution properly heavier, automatically, and avoids complexities of the approximate Bayesian method of multiple inputation requiring the use of Rubin's rule. An example is in the [RMS course notes](https://hbiostat.org/rms).\n", "link": "https://stats.stackexchange.com/questions/220005/generalization-of-degrees-of-freedom-for-t-distribution-for-coefficients-after-m", "tags": ["bayesian", "missing-data", "multiple-imputation", "degrees-of-freedom", "t-distribution"], "votes": 9, "creation_date": "2016-06-21T08:41:32", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "463", "site": "stats", "title": "Distribution/expected length of the shortest path in infinite random geometric graphs", "body": "Consider an infinite random geometric graph $G(\\rho,d)$ in which vertices are uniformly and independently scattered over the 2D plane with density $\\rho$ and edges connect the vertices that are closer than $d$. \n\nWhat is the distribution/expected value of the length of the shortest path from a vertex $v_0$ to another vertex $v_1$ which is $k$ hops faraway?\n\n_**Note:_**\n\nWe know that the length of the edges follow the following PDF:\n\n$$ f(l)= \\begin{cases} \\frac{2 l}{d^2} \\;\\quad l \\le d \\\\\\ 0 \\qquad\\; l > d \\end{cases} $$\n\nHowever, I guess the expected length of the shortest path is not simply $k \\times$ the expected length of the edges because in a shortest path longer edges are more likely to be chosen, right?\n", "link": "https://stats.stackexchange.com/questions/48063/distribution-expected-length-of-the-shortest-path-in-infinite-random-geometric-g", "tags": ["probability", "stochastic-processes", "density-function", "asymptotics", "graph-theory"], "votes": 9, "creation_date": "2013-01-18T15:57:33", "comments": ["The probability of having a neighbor at distance $l$ grows linearly since the circumference of the circle is $2\\pi r$. $\\frac{1}{\\pi d^2}$ is the normalizing factor to make the probabilities sum to unity.", "I asked the question about a year ago and I can't remember if I copied the formula from somewhere or not. But it seems quite simple. Isn't it the circumference of a circle with radius $l$ over the area of a circle with radius $d$?", "How do we know the functional form for the edge length distribution? do you have a derivation/reference?"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "464", "site": "stats", "title": "First two moments of the ratio of the geometric mean to the arithmetic mean of Gamma random variables", "body": "Let $X_1,\\ldots, X_n$ be $n$ uncorrelated random variables from a Gamma distribution with **different parameters** : $X_i \\sim Gamma(k_i, \\theta_i)$. What is the distribution of $$ U=\\log \\left[ \\dfrac{\\bar{X}}{\\tilde{X}}\\right]$$ where $\\bar{X}$ is the sample arithmetic mean and $\\tilde{X}$ is the geometric mean. \n\nEDIT: I don't necessarily need the distribution, the first 2 moments will suffice.\n", "link": "https://stats.stackexchange.com/questions/336677/first-two-moments-of-the-ratio-of-the-geometric-mean-to-the-arithmetic-mean-of-g", "tags": ["mean", "gamma-distribution"], "votes": 8, "creation_date": "2018-03-25T14:57:11", "comments": ["@Jason, if the $\\theta_i$'s are not identical then we can consider each r.v. a scaled $\\chi^2$, then $\\sum X_i$ would be a sum of scaled $\\chi^2$ r.v. which can be approximated by another scaled $\\chi^2$ which would be a Gamma r.v.", "@ToneyShields Can you clarify that point a little? I'm familiar with the fact that $\\sum X_{i}$ is gamma if the $\\theta_{i}$ are identical and the $X_{i}$ are independent, but not otherwise.", "@Jason, actually $\\sum X_i$ is a gamma distribution, which means that $E(log (\\sum X_i))$ is known.", "Well for the first moment of $U$, the main challenge is determining $E[ \\log (\\sum X_{i}) ]$. If the $k_{i}$ are identical, this article might be helpful for working that out: arxiv.org/abs/1103.0505.", "@ToneyShields I was joking :-)", "@oneloop, I'm not asking for people to do this for me, I just want to know if this has already been done, or get some hints that might be useful.", "@ToneyShields If this is for a test you're working on, then deriving the test statistic should at least get the person who answers co-authorship ;-)", "@LucasRoberts, the assumption is uncorrelatedness.", "@ToneyShields, the post states \"uncorrelated\" is this truly what the assumption is or is the assumption \"independent\" which would imply \"uncorrelated\"?", "@AlecosPapadopoulos, I can find the mgf If I had a common $\\theta$. but unfortunately, it's not the case, but I'd like to see how to find the first 2 moments with a common $k$, please.", "@ToneyShields Of course it does. But if this is negotiable, it would be even better to have the $k_i$'s different and a common $\\theta$.", "@AlecosPapadopoulos, my bad, the$k_i$'s are also different. Does it make it easier if only $\\theta_i$'s were different?", "It appears that only the scale parameters $\\theta_i$'s are different. Is that correct?", "@Xi'an, how about the first two moments, that's what I'm looking for.", "With different parameters, there cannot be a simple expression to this distribution, since these parameters will not simplify into a summary.", "@Glen_b, you're right, but in my case, $X_i$'s aren't identically distrubuted.", "Taking the ratio of the GM to the AM in the gamma case isn't exactly new.", "@Glen_b, No, I'm working on a new test statistic and I need the distribution of this statistic, at least its 2 first moments.", "Is this for a subject/coursework/bookwork/self-study etc?", "Yes they do form a sample. Each $X_i$ is drawn from the same distribution but with different parameters (They are uncorrelated).", "If $X_i$'s have different distributions, then do they form a 'sample' ?"], "comment_count": 21, "category": "Science", "diamond": 0} {"question_id": "465", "site": "stats", "title": "Time evolution of a Bayesian posterior", "body": "I have a question regarding the time evolution of a quantity related to a Bayesian posterior.\n\nSuppose we have binary parameter space $\\\\{ s_1, s_2 \\\\}$ with prior $(p, 1-p)$, The data generating processes are Brownian motions with parameter-determined drift: \n\n$$ dX_t = \\theta_i(t) dt + dB_t $$\n\nfor deterministic functions $\\theta_i(t)$, $i = 1, 2$.\n\nBy Girsanov's theorem, the likelihood ratio, conditional on $X_t$, as a stochastic process is $e^{\\gamma_t}$\n\nwhere\n\n$$ \\gamma_t = \\int_0^t \\theta_1(s) - \\theta_2(s) dB_s - \\int_0^t \\frac{ \\theta_1^2(s) - \\theta_2^2(s) }{2} ds. $$\n\nSo the posterior probability (of $s_1$) is then given by\n\n$$ \\frac{p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p) }. $$\n\n**Question** : Can one describe explicitly how $E[\\frac{e^{\\gamma_t} p}{ (e^{\\gamma_t} p + (1 - p))^2 }]$ evolves?\n\nDirect calculation, by Ito's lemma, shows that\n\n$$ d \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p) } = (1 - p) \\frac{1}{(p e^{\\gamma_t} + (1 - p) )^2} d pe^{\\gamma_t} - (1 - p) \\frac{1}{(p e^{\\gamma_t} + (1 - p) )^{3}} ( d pe^{\\gamma_t} )^2 \\\\\\ $$\n\nSo\n\n\\begin{align*} \\frac{ d E[ \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p) }] }{dt} &= (1 - p) E[\\frac{p e^{\\gamma_t} }{(p e^{\\gamma_t} + (1 - p) )^2}] (\\theta_1^2(t) - \\theta_2^2(t)) \\\\\\ &\\; - (1 - p) E[\\frac{( p e^{\\gamma_t})^2 }{(p e^{\\gamma_t} + (1 - p) )^3}] \\ (\\theta_1^2(t) - \\theta_2^2(t)) , \\end{align*}\n\ni.e. \n\n\\begin{align*} \\frac{ d E[ \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p) }]}{dt} &= (1 - p) E[\\frac{p e^{\\gamma_t} }{(p e^{\\gamma_t} + (1 - p) )^2}(1- \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p) })] \\cdot (\\theta_1^2(t) - \\theta_2^2(t)) \\\\\\ &= (1 - p) [\\; E[\\frac{p e^{\\gamma_t} }{(p e^{\\gamma_t} + (1 - p) )^2}] E[1- \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p) }]\\\\\\ &\\; + Cov(\\frac{p e^{\\gamma_t} }{(p e^{\\gamma_t} + (1 - p) )^2}, 1- \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p }) \\;] \\cdot (\\theta_1^2(t) - \\theta_2^2(t)) \\\\\\ &= 0. \\\\\\ \\end{align*}\n\nThe last inequality follows from the fact that successive Bayesian priors form a martingale: let $p(s)$ be the prior, $q(s)$ be the posterior, and $p(x|s)$ be the conditional densities, then\n\n\\begin{align*} E[q(s)] &= E[ \\frac{ p(x|s)p(s) }{ \\int p(x|s')p(s') ds' }] \\\\\\ &= \\int \\frac{ p(x|s)p(s) }{ \\int p(x|s')p(s') ds' } \\int p(x|s')p(s') ds' dx\\\\\\ &= \\int p(x|s) p(s) dx \\\\\\ &= p(s). \\end{align*}\n\nSo the whole thing boils down to\n\n\\begin{align*} 0 &= E[\\frac{p e^{\\gamma_t} }{(p e^{\\gamma_t} + (1 - p) )^2}(1- \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p) })] \\\\\\ &= E[\\frac{p e^{\\gamma_t} }{(p e^{\\gamma_t} + (1 - p) )^2}] E[1- \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p) }]\\\\\\ &\\; + Cov(\\frac{p e^{\\gamma_t} }{(p e^{\\gamma_t} + (1 - p) )^2}, 1- \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p }). \\end{align*}\n\nThe question is then whether one can solve for $E[ \\frac{ p e^{\\gamma_t} }{ ( p e^{\\gamma_t} + (1 - p) )^2}]$ from above---by computing the time evolution of the covariance term. Or, is there another way to get at this?\n\n**Follow Up Question** : Does $\\frac{ p e^{\\gamma_t} }{ ( p e^{\\gamma_t} + (1 - p) )^2}$ have a Bayesian interpretation? Why should it be orthogonal to $1- \\frac{ p e^{\\gamma_t} }{ p e^{\\gamma_t} + (1 - p) }$---the posterior probability of the other state $s_2$---at all times $t$?\n", "link": "https://stats.stackexchange.com/questions/254097/time-evolution-of-a-bayesian-posterior", "tags": ["probability", "bayesian", "mathematical-statistics", "stochastic-processes", "posterior"], "votes": 8, "creation_date": "2017-01-01T16:33:21", "comments": ["How are the parameters $s_1$ and $s_2$ involved in the data generating process?"], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "466", "site": "stats", "title": "Can these asymptotic conditional expectations be bounded from above?", "body": "**Problem Setup**\n\nLet $\\\\{X^d_1, X^d_2, \\cdots, X^d_n\\\\}$ be a $d-$dimensional zero-mean, i.i.d. random variables. Let $S_n^d$ be $$ S^d_n = \\frac{\\sum_{i=1}^n X_i^d}{\\sqrt{n}} $$\n\nLet $Y^d$ be a zero-mean normal random variable with covariance $\\Sigma^d = \\mathbb{E}\\big[X^d(X^d)^T\\big]$\n\nThe paper by [Bentkus](http://www.mii.lt/files/Bentkus_2003_2.pdf), for a given class $\\mathcal{A}$, has the following result:\n\n> $$ \\sup_{A\\in \\mathcal{A}} \\big|\\mathbb{P}\\\\{S^d_n\\in A\\\\}−\\mathbb{P}\\\\{Y^d\\in A\\\\}\\big|\\leq \\frac{c_d(\\mathcal{A})\\beta}{\\sqrt{n}}; \\qquad \\beta = \\mathbb{E}\\big[|X^d|^3\\big] $$\n> \n> * The constant $c_d(\\mathcal{A}) = d^{1/4}$ when $\\mathcal{A}$ is all the convex sets in $\\mathbb{R}^d$ and \n> * $c_d(\\mathcal{A}) = 1 $ when $\\mathcal{A}$ is all the Euclidian balls in $\\mathbb{R}^d$.\n> \n\n\n**Question** Can we upperbound any of the following when $d$ grows with $n$?\n\n 1. $\\Big|\\mathbb{E}\\big[S^d_n \\big] - \\mathbb{E}\\big[Y^d\\big]\\Big|$\n 2. $\\Big|\\mathbb{E}\\big[S^d_n(i)\\big\\vert S^d_n(1,2,\\cdots,i-1) \\big] - \\mathbb{E}\\big[Y^d(i)\\big\\vert Y^d(1,2,\\cdots,i-1)\\big]\\Big|$\n\n\n\nHere $S^d_n(i)$ is the $i^{th}$ component $d-$dimensional vector and $S^d_n(1,2,\\cdots,i-1)$ are the first the $(i-1)$ components of $S^d_n$.\n\n**Possible Method**\n\nI have a method which might possibly give some upper-bound but I am unsure of this idea and it requires some nuanced arguments (which I am unable to make). $$\\Big|\\mathbb{E}\\big[S^d_n \\big] - \\mathbb{E}\\big[Y^d\\big]\\Big| = \\int_{x\\in\\mathbb{R}^d}x\\Big(p_S(x) - p_Y(x)\\Big)dx \\\\\\ \\leq \\sqrt{\\int_{x\\in\\mathbb{R}^d}x^2\\big|p_S(x) - p_Y(x)\\big|dx \\int_{x\\in\\mathbb{R}^d}\\big|p_S(x) - p_Y(x)\\big|dx }\\\\\\ \\leq \\sqrt{\\int_{x\\in\\mathbb{R}^d}x^2\\big(p_S(x) + p_Y(x)\\big)dx \\int_{x\\in\\mathbb{R}^d}\\big|p_S(x) - p_Y(x)\\big|dx } \\\\\\ \\leq \\sqrt{2d\\sigma^2\\int_{x\\in\\mathbb{R}^d}\\big|p_S(x) - p_Y(x)\\big|dx } \\\\\\ \\overset{?}{\\leq} \\sqrt{2d\\sigma^2\\frac{2c_d(\\mathcal{A})\\beta}{\\sqrt{n}}} \\\\\\ $$\n\n**Issues:**\n\n 1. I am not sure the conditions under which the last equation is valid.\n 2. In the above method I am using the Lebesgue measure, can we get better bounds using other measures? (Note : $c_d(\\mathcal{A}) = 1 $ when $\\mathcal{A}$ is all the Euclidean balls in $\\mathbb{R}^d$.)\n 3. How can one apply similar ideas for question 2?\n\n\n", "link": "https://stats.stackexchange.com/questions/210943/can-these-asymptotic-conditional-expectations-be-bounded-from-above", "tags": ["normal-distribution", "conditional-probability", "conditional-expectation", "asymptotics"], "votes": 8, "creation_date": "2016-05-04T15:41:00", "comments": ["The last step is correct but useless because $\\int\\big|p_S(x)-p_Y(x)\\big|dx$ is total variation distance and it takes supremum over class of all sets. However $c_d(\\mathcal{A})$ is bounded only when the $\\mathcal{A}$ is set of convex set or Euclidean balls.", "Are $S_n$ and $Y$ in the $\\sup$ inequality the same as $S_n^d$ and Y^d$?", "The above problem might not be well posed, feel free to edit the question if that improves the question."], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "467", "site": "stats", "title": "QR decomposition of normally distributed matrices", "body": "Assume $M$ is an $N \\times k$ Gaussian matrix, i.e., its entries are i.i.d. standard normal random variables, with $N>>k$. Take $D=\\text{diag}(\\lambda_1, \\dotsc ,\\lambda_N)$ for some fixed real scalars. I am interested in finding the p.d.f. of the $N \\times k$ \"unitary\" matrix $Q$ from the QR decomposition of $DM$ (and possibly $D^2M$, etc.).\n\nIt is known that if $k=N$ and $D=I_N$, the identity matrix, then $Q$ is distributed with respect to the [Haar measure on the Lie group of orthonormal matrices of order $N$](http://www.ams.org/notices/200705/fea-mezzadri-web.pdf). Can you provide any insight on the general case for $k ...Persian romances often mention the blissful land of Shadu-kam.\n\nI started digging, and I found a bit more detail in Thomas Keightley's [The Fairy Mythology, Volume 1](http://www.gutenberg.org/files/41006/41006-h/41006-h.htm), published in 1828. I have no idea if Keightley is trustworthy, but he writes:\n\n> Jinnestân is the common appellation of the whole of this ideal region. Its respective empires were divided into many kingdoms, containing numerous provinces and cities. **Thus in the Peri-realms we meet with the luxuriant province of Shad-u-kâm (Pleasure and Delight)** , with its magnificent capital Juherabâd (Jewel-city), whose two kings solicited the aid of Cahermân against the Deevs, and also the stately Amberabâd (Amber-city), and others equally splendid. The metropolis of the Deev-empire is named Ahermanabâd (Aherman's city); and imagination has lavished its stores in the description of the enchanted castle, palace, and gallery of the Deev monarch, Arzshenk.\n\nIt's an old book, so transliterations don't always match their modern versions.\n\nRegarding the hero Cahermân mentioned above, he later adds:\n\n> The Cahermân Nâmeh is a romance in Turkish. Cahermân was the father of Sâm, the grandfather of the celebrated Roostem.\n\nThis would seem to make Cahermân match the [Nariman](https://en.wikipedia.org/wiki/Nariman_\\(father_of_S%C4%81m\\)) of the Shahnameh. But I haven't been able to find anything about Shadu-kam in the Shahnameh, nor can I find any information anywhere on this Cahermân Nâmeh romance.\n\nI've found other books that mention Shadu-kam, but most seem like they're just repeating Keightley. The Keightley book might even be Eco's source, but it's hard to tell.\n\nDoes anyone know of any primary sources?\n", "link": "https://mythology.stackexchange.com/questions/1474/primary-sources-for-persian-land-of-shadu-kam", "tags": ["persian", "shahnameh"], "votes": 9, "creation_date": "2016-03-02T07:02:42", "comments": ["You're on the right track when you mention the similarities between Caherman and Nariman: remember, translation techniques have improved vastly since Keightley was alive. I could be wrong, but my initial impression is that the best way to answer this question would be to try to search for parallels between Shad-u-kâm and some other concept in persian literature. However, keep in mind that it's possible that Keightley made this story up: at the time he was writing it was really common for researchers to pass off stories they wrote as genuine literature.", "Hi Brian. I've taken an initial look at this question. I think you're right when you say that most of the sources that mention Shad-u-kâm are copying Keightley. You are also right to be skeptical of Keightley: he was writing when the field of folklore was just getting started, which means that accuracy wasn't a really big concern of his. Looking at his writings, my initial impression is that he was motivated by a desire to project Christian concepts onto Persian mythology (although a lot of parallels do exist between the two, Keightley seems to have gone above and beyond in this)."], "comment_count": 2, "category": "Culture & Recreation", "diamond": 0} {"question_id": "474", "site": "quant", "title": "What is the probability of ruin of a Geometric Ornstein-Uhlenbeck process?", "body": "I would like to calculate the probability of ruin (or, default), i.e. $$\\text{Pr}(\\tau0$. In this case, my intuition is that default is possible. Think of $\\kappa=-1$ for simplicity, and of $\\theta$ as the rate of consumption per unit of time from an initial endowment $X_0>0$ whose growth follows a GBM. For a high enough consumption rate, financial ruin is probable!", "Here is my intuition: Consider $\\delta = 1$ and $\\theta = 0$. Then we are back in the standard GBM case. For $X_0 > 0$, zero cannot be attained in finite time. Adding back $\\kappa \\theta > 0$, this shouldn't change anything.", "$B = 0$, thanks for asking, I should have clarified it.", "Where is the default boundary - a) exactly at zero or b) strictly positive?"], "comment_count": 5, "category": "Business", "diamond": 0} {"question_id": "475", "site": "quant", "title": "Taking into account the correlation in Barrier options on a Basket", "body": "In a Barrier option (where the contract cancels when the underlying hits the barrier) I succesfully found the way to compute the probability of a single underlying touching the barrier (with constant volatility). The problem comes when I have an average basket of underlyings $B_t$ $$ B_t = \\sum_{i=1}^n \\omega_iS_t,\\qquad \\sum_{i=1}^n\\omega_i=1\\ . $$ I don't know how to properly calculate the probability of the basket $B_t$ touching the barrier. I think I need to take into account the correlations $\\langle dW_t^{i},dW_{t}^{j}\\rangle=\\rho_{ij}$ between assets but i just don't know how to do it. \n\nI'm considering constant volatility and constant correlation thoughout an arbitrary time step.\n\nExtra: Same question if the basket is Best Of/Worst Of type.\n\nI'm not expecting a complete developement of the answer but I would find very helpfull any indications or references. Thanks a lot. \n\nEdit: Yes, i'm considering that Stock prices follow geometric Brownian motions. I also forgot to say that the prices at the end of the time interval are known, so the probabilities are actually conditioned probabilitites. \n", "link": "https://quant.stackexchange.com/questions/2420/taking-into-account-the-correlation-in-barrier-options-on-a-basket", "tags": ["correlation", "basket"], "votes": 7, "creation_date": "2011-11-23T08:38:53", "comments": ["I think that $\\langle dW_t^{i},dW_{t}^{j}\\rangle=\\rho_{ij}dt$", "@FKaria: Thanks for your explanation. Now I understand the relevance of your question. One workaround is of course to only check the barrier at the discrete timesteps, but this is sometimes a crude approximation I guess.", "@user561749 This is to calculate the touching probabilities in a Monte Carlo simulation. In that context you know the price at a beginning and at the end of a cerain timestep but you don't know what happened in between.", "Is it important for you that prices at the end of the time interval are known? That is a strange question from a industry point of view - because if prices at the end of the time interval are known then probably we have complete information about the path of the basket up to that time? Perhaps you mean that prices at the start of the time interval are known?", "I guess the stocks in your basket, are each following a geometric browian motion, is that right ? Even in that case you can't get closed form formulas, and you have to use approximations. Regards"], "comment_count": 5, "category": "Business", "diamond": 0} {"question_id": "476", "site": "space", "title": "Does Privateer's tracking software Crow's Nest actually offer users something better than the previous state of the art?", "body": "CNN's [Steve Wozniak's new venture takes aim at space junk](https://edition.cnn.com/2022/03/01/tech/space-junk-steve-wozniak-privateer-scn/index.html) discusses a project with [Wozniak](https://en.wikipedia.org/wiki/Steve_Wozniak), [Alex Fielding](https://en.wikipedia.org/wiki/Alex_Fielding) and [Moriba Jah](https://en.wikipedia.org/wiki/Moriba_Jah):\n\n> But the money, according to Wozniak's co-founder in this new venture, couldn't be further from the point. \"I don't think Steve [Wozniak] gives a damn about making another 10 cents, and I really couldn't care less,\" Alex Fielding, a longtime business acquaintance of Wozniak's who will serve as CEO of the new venture, called Privateer, told CNN Business.\n> \n> Privateer's mission is to develop better tracking of objects in space, and to use this data to help avert disastrous collisions. To aid in this effort, Wozniak and Fielding brought in Moriba Jah, a PhD and orbital mechanics professor who has dedicated most of his life to academia and attempting to raise awareness about the ever-growing threat posed by the proliferation of debris and garbage in outer space. It's a threat that could wipe out satellites that provide communications services to Earth or even bring space travel to a grinding halt. He's [led research](https://www.strausscenter.org/person/moriba-jah/) at the University of Texas. He's [appeared](https://www.commerce.senate.gov/2020/2/space-missions-of-global-importance-planetary-defense-space-weather-protection-and-space-situational-awareness) at Congressional hearings. He's [advocated](https://unidir.org/non-resident-fellows/moriba-jah) for change on the world's stage. But Jah told CNN Business recently came to a solemn conclusion: There is not enough funding in academia to develop the technologies he envisions the world needs to combat [the space junk issue](http://www.cnn.com/2021/01/11/perspectives/nasa-biden-administration-satellites/index.html), he says.\n> \n> So, Jah went searching for that funding. And it brought him to Wozniak, the coding savant who co-founded Apple with Steve Jobs.\n\nAround the world there are several military and civilian agencies that keep track of junk-things in space, and there's lots of software to track it and propagate orbits beyond just the government-issued TLEs.\n\n**Question:** Does Privateer's tracking software [Crow's Nest](https://mission.privateer.com/crows-nest) actually offer users something better than the previous state of the art?\n\nBetter propagators? AI to catch mis-identified or mis-assigned information? Blockchain verification of debris catalogs? Pull-down menus and a mouse? (humor)\n\nRelated:\n\n * [How might one approach using AI (convolutional neural network) to predict collisions in orbit?](https://space.stackexchange.com/q/24488/12102)\n * [Is it true that 18SPCS is \"not really concerned with tracking deep-space debris like\" the Chang’e 5-T1 rocket body that hit the Moon?](https://space.stackexchange.com/q/58528/12102)\n * [How would blockchain be integrated into spaceflight? Have any methods been proposed yet?](https://space.stackexchange.com/q/31207/12102)\n * [Why exactly did ESA provide a grant funding blockchain in space? Which program was used for the funding? What will this do?](https://space.stackexchange.com/q/40306/12102)\n * [Project Pluto asks: \"Should high-flying space junk be better tracked? Perhaps by an 'official' agency of some sort?\" What would it take to implement?](https://space.stackexchange.com/q/58420/12102)\n\n\n", "link": "https://space.stackexchange.com/questions/58535/does-privateers-tracking-software-crows-nest-actually-offer-users-something-be", "tags": ["debris", "tracking", "collision-avoidance", "business", "artificial-intelligence"], "votes": 7, "creation_date": "2022-03-05T21:22:00", "comments": ["@DavidHammen one can now download the \"puffery\" and give it a try, it's in public beta. See the links in the bounty message.", "In this case, I suspect the PR-puffery to which @OrganicMarble alluded was the standard crud in seen in press releases and related items. Almost all startups funded by venture capital start with a lot of PR-puffery. 75% of those startups fail, with many of those failures resulting from the PR-puffery turning out to be nothing more than puffery.", "@uhoh, by that logic nothing would ever get closed. In my opinion this is off-topic, so I've voted to close it. If nobody agrees with me it will remain open.", "@GdD How can you know how much everyone else knows? Again; Perhaps someone has been to some presentations that you don't know about or are familiar with some of Jah's papers. The answer might be right in here! You can't know what everyone else does or doesn't know. If an opinion-based answer does appear, we handle those quickly. But why pre-block everyone else from an opportunity to answer?", "Because answering the question will require speculation now and two days from now @uhoh.", "@GdD That could be; though with 30k users I think it takes a lot of insight to be so sure that none of them could write a fact based answer. But I have to ask what's the rush? Why not wait a few days and see what answers might appear? Why pre-block all 30,000 users before we might find out that someone actually does have some knowledge to form the basis of a fact-based answer? Perhaps they've been to some presentations that you don't know about or are familiar with some of Jah's papers. The answer might be right in here!", "I'm voting to close as opinion based. There's limited information on this venture and no reason to believe the company will share these details so any answer would be pure speculation.", "@OrganicMarble The question does not need explaining, it's clear. There is a link to Jah's Wikipedia page which is well sourced and now there's a link to the bibliography in comments. The question is fine.", "This is a new approach! The answers will explain the question. Can't wait.", "@OrganicMarble I think when you read the answers or at least this you'll have some better insight as to why that's not going t be necessary in this particular case.", "Nothing in what you quote claims his software is better. Missing PR-puffery?", "I can safely say that you can't use blockchains to make debris detection and tracking better, because they're useless for most things."], "comment_count": 12, "category": "Technology", "diamond": 0} {"question_id": "477", "site": "space", "title": "Crew Dragon: How thick is the skin?", "body": "Auto body panels are about 20-gauge, or just under 1 mm thick. This is fine for many reasons: thin panels are light for good fuel economy and crumple easily for shock absorption without unnecessary mass---especially since body panels aren't normally part of a car's load-bearing structural frame.\n\nBut what about a spacecraft, like Dragon? Are the body panels thin or thick? On one hand, you want the lightest cheapest part you can make, and thin panels would help. On the other hand, you have peak dynamic pressures of 30 kPa to worry about.\n\nAnd I don't know if spacecraft panels might serve a structural function also---like helping transmit the thrust force from RCS thrusters. Try as I might, I can't seem to find a good picture of the load-bearing support that should join an RCS thruster to the spacecraft's structural frame.\n\nCan someone clarify how thick Dragon (or other spacecraft) body panels are, and what material they're made of? How would they compare to auto body panels on these points? And do they serve a structural purpose? Are they lined on the inside with more filler material for strength?\n\nA cross-section view of a Dragon's or similar spacecraft's body panel would be greatly appreciated. Thanks!!\n", "link": "https://space.stackexchange.com/questions/51782/crew-dragon-how-thick-is-the-skin", "tags": ["spacex", "falcon-9", "dragon", "structural-design"], "votes": 8, "creation_date": "2021-04-29T21:41:01", "comments": ["Hmmm... how thick are panels on commercial jetliners? Quick search says 2 to 4 mm for pressurized craft.", "+1 and relevant to my comment where I'd assumed 3 mm of aluminum without a source."], "comment_count": 2, "category": "Technology", "diamond": 0} {"question_id": "478", "site": "space", "title": "Dragon Pad Abort hot fire, how much cleanup work was needed?", "body": "The day before the Pad Abort test attempt (May 5, 2015) SpaceX test fired the SuperDraco engines in a hot fire. Similar to how they test the Falcon 9 a day or three before launch, the engines spin up and fire, the booster/capsule is mechanically held down for the test.\n\nBut normally, SpaceX burns LOX/RP1 whose exhaust is carbon dioxide, water, and some sulfur by products. Basically car exhaust, just lots more of it.\n\nFor the Pad Abort vehicle, it is Nitrogen Tetroxide and MMDH which are astonishingly nasty nasty fluids. They are horribly corrosive and dangerous to humans. The products of full combustion are not very bad (water, ammonia, and other benign things) but you cannot guarentee 100% full combustion, and this stuff is just plain nasty. (Potential additional question will be for capsules landing on land, how much work will be needed before they can leave the capsule safely. Shuttle had people in Hazmat suits cleaning up, before the crew were allowed to leave from its NTO/MMDH thrusters.). \n\nHow much cleanup was required at the pad after the test?\n\nLooking at the pictures, they did use the Niagara water deluge system. Was that sufficient to wash away any issues? (If so, do they process the deluge system water before letting it flow back into the Florida waterways?)\n", "link": "https://space.stackexchange.com/questions/9045/dragon-pad-abort-hot-fire-how-much-cleanup-work-was-needed", "tags": ["spacex", "testing", "abort", "dragon-v2"], "votes": 8, "creation_date": "2015-05-06T05:57:32", "comments": ["The Nitrogen Tetroxide fuels are quite toxic because they react with h2o to form nitric acid. For that reason, you don't want to breath the stuff in, burning the lungs. But the same high reactivity also means that the stuff will break down quickly. In the case of the shuttle, the landing team had a giant fan to blow whatever fumes away. The people in the hazmat suites are there for emergency rescue, leak checks and to make sure that the stuff has been properly dissipated. There us a bit more at en.wikipedia.org/wiki/Dinitrogen_tetroxide#Chemical_reaction‌​s", "Another example of the problem with (incompletely combusted) N2O4/UDMH is a Proton launch failure in 1969 which made the entire kosmodrom toxic and unusable for some time. And a similar accident happened again with the 2013 Proton failure. Russia, China and India use this fuel on big launchers today, but except for India, they are about to replace those launchers.", "My impression is that the products it forms when burned (completely) are not toxic. The unburnt fuel actually injured astronauts pretty bad (unconsciousness) in the Apollo-Soyuz mission."], "comment_count": 3, "category": "Technology", "diamond": 0} {"question_id": "479", "site": "or", "title": "MIP formulation for graph planarity test", "body": "In [this question](https://cstheory.stackexchange.com/questions/48099/integer-linear-program-formulation-of-planarity), it was asked wether a MIP formulation exists to test for a graph's planarity. The inputs are the graph's nodes and edges, and the output would be a certificate which guarantees that the graph is, or is not, planar.\n\nThe python package NetworkX provides a [method](https://networkx.org/documentation/stable/_modules/networkx/algorithms/planarity.html#is_planar) which does exactly this, but it is not based on linear programming. The source code provides the following references:\n\n * Ulrik Brandes: [The Left-Right Planarity Test](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.217.9208) (2009)\n * Takao Nishizeki, Md Saidur Rahman: Planar graph drawing, Lecture Notes Series on Computing: Volume 12 (2004)\n\n\n\nSome ideas:\n\n * [Euler's theorem](https://en.wikipedia.org/wiki/Euler_characteristic#Plane_graphs) ($n-m+f=2$); but this depends on how the nodes (and edges) are organized in space, which is not part of the input data.\n * $G$ planar $\\implies m\\le 3n-6$; but this does not cover all cases, it can be a tool to prove that a graph is not planar, but is useless the otherway around.\n * There is a relationship between planar graphs and [book embeddings](https://en.wikipedia.org/wiki/Book_embedding) (see this great [video](https://www.youtube.com/watch?v=qw2Pl_Nk3CA) from Numberphile for a recent breakthrough in graph theory on the subject), but not sure it helps.\n * Other characterizations of planar graphs are mentioned on the [Wikipedia page](https://en.wikipedia.org/wiki/Planar_graph#Planarity_criteria), such as [Whitney's planarity criterion](https://en.wikipedia.org/wiki/Whitney%27s_planarity_criterion) or [Mac Lane's planarity criterion](https://en.wikipedia.org/wiki/Mac_Lane%27s_planarity_criterion). These might be interesting.\n\n\n\nTo this day, the question has not been answered. I find the question interesting from a modeling point of view and am curious wheter it can be answered **with a MIP**.\n", "link": "https://or.stackexchange.com/questions/10399/mip-formulation-for-graph-planarity-test", "tags": ["mixed-integer-programming", "modeling", "graphs"], "votes": 8, "creation_date": "2023-05-01T00:42:22", "comments": ["As far as I checked Euler's theorem, It is not correct for the space with dimension $\\mathbb{R}^2$. As the graph was planar. (I am not aware of if there exists a counterpart).", "My best bet would be that you can check if the graph has $K_5$ or $K_{3,3}$ as a minor, which is equivalent to planarity by Wagner's theorem. Checking if $H$ is a minor of $G$, for fixed $H$, seems doable, you have to find $|V(H)|$ vertex-disjoint connected subgraphs in $G$, each representing a vertex of $H$, such that there is at least one edge between parts corresponding to adjacent vertices of $H$. Disclaimer, I have not checked any details."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "480", "site": "quantumcomputing", "title": "Is HHL still BQP-complete when the matrix entries are only in {0,1}?", "body": "I'm studying BQP-completeness proofs of a number of interesting problems of Janzing and Wocjan, and Wocjan and Zhang. Janzing and Wocjan show that estimating entries of [matrix powers](https://www.researchgate.net/publication/220295493_A_Simple_PromiseBQP-complete_Matrix_Problem) $(A^m)_{ij}$ with $A_{ij}\\in\\\\{-1,0,1\\\\}$ is (promise) BQP-complete. That is, the problem is both in BQP, and can simulate other problems in BQP.\n\nJanzing and Wocjan emphasize that their BQP-hardness reduction requires a negative-sign in an entry of $A$ to _simulate interference_ afforded by BQP algorithms. At a critical point in their reduction they apply a (conditional)-$Z$ rotation, which leads to a matrix $A$ with a negative entry. I don't think their proof of BQP-hardness would carry over when entries $A_{ij}$ are strictly non-negative, e.g. in $\\\\{0,1\\\\}$. Indeed, I believe that such a restricted matrix-powers problem may be amenable to some form of Stockmeyer approximation, e.g. in $\\mathsf{AM}$, and hence not BQP-complete under the reasonable hypothesis that $\\mathsf{BQP}\\not\\subseteq\\mathsf{AM}$.\n\nThe proof of BQP-hardness of matrix powers appears initially to be similar to the BQP-hardness of the HHL algorithm, which was wonderfully summarized by @DaftWullie [here](https://quantumcomputing.stackexchange.com/questions/6615/showing-that-matrix-inversion-is-bqp-complete-hhl-algorithm). However, HHL considers the Taylor-series expansion of $A^{-1}$, where $A=I-Ue^{-1/T}$ and $U$ is a unitary operator which simulates a given unitary circuit with a clock register construction — so that $U$ (and powers of $U$) will have negative or complex entries, if any of the gates in the circuit do. For the case $\\tilde A = U \\mathrm{e}^{-1/T}$, this again suggests that the BQP-completeness of estimating entries of matrix powers $\\tilde A^m$ is associated with whether $\\tilde A$ has entries apart from non-negative reals.\n\nGiven this, considered as a special case of HHL which is motivated by the comparison to the problem of Janzing and Wocjan, I'd like to know: is HHL still BQP-complete if all of the entries of $A$ and $\\lvert b \\rangle$ are in $\\\\{0,1\\\\}$ ?\n", "link": "https://quantumcomputing.stackexchange.com/questions/13404/is-hhl-still-bqp-complete-when-the-matrix-entries-are-only-in-0-1", "tags": ["complexity-theory", "hhl-algorithm", "bqp"], "votes": 14, "creation_date": "2020-08-19T17:26:40", "comments": ["(Having said that, seeing how having entries of {0,1} for $A$ is still not the same as having {0,1} entries for $A^{-1}$ or a matrix $\\tilde A$ whose powers you would like to learn about, I still think that having non-negative entries for $A$ itself is a bit of a red herring... once you take the inverse of $A$, negative entries can still arise, shedding no further light on the question quantum computational difficulty without destructive interference.)", "I've made some changes. I think I understood your question better once I read it with a different emphasis, and I've tried to amplify that emphasis.", "Thanks, could you edit the question to address your concerns? I’d like an answer to the question in the title...", "I'd be willing to have a go at editing the question, if what you'd like is something specifically addressing the case where A and |𝑏⟩ have {0,1} entries, together with some remarks putting the question in context. As it stands, it still feels a bit as though you're asking one thing because you want to know the answer to something else.", "So, you've added some commentary about the role of negative entries, but the main question (the title, the reasonable question towards the end) is still about {0,1} entries. Also: in your recent edit, it's not clear that the claim about $A = I - U\\exp(-1/T)$ is what you intend to claim (do you mean $A^{-1}$ by chance?), or that the statement that you might like to make is correct (what about the special case where $U$ is constructed only from classical reversible gates?).", "You're asking two different questions. The question about whether HHL is BQP-complete, if all the entries of A and |𝑏⟩ are in {0,1}, is clear enough. But you're never going to construct A = I - U exp(-1/T) and still have coefficients in {0,1}, so it's not clear how that construction could play any role. So, are you asking about the role of the minus sign in the construction, or are you asking about constraining the values of A?"], "comment_count": 6, "category": "Technology", "diamond": 0} {"question_id": "481", "site": "quantumcomputing", "title": "Can you programatically check whether a given set of gates is universal?", "body": "I am wondering if there is an automated way to determine whether a given set of quantum operations is universal.\n\nMore precisely, given a set of 1 and 2 qubit gates, can we write a program to determine whether this constitutes a universal gate set? If so, how might we do this?\n\n**EDIT:** To clarify I'm interested in both approaches which allow exact replacements (as shown below) and approaches which allow approximating any unitary to some finite precision. Perhaps restricting the problem by not allowing arbitrary rotations in the input gateset could make the problem more tractable I'm not sure.\n\nIf I was given a set of quantum gates I would try to use the gates I'm given to implement a well known universal gateset.\n\nTake this artificial example...\n\n\\begin{equation} S_0 = \\\\{\\text{H}, \\text{Rx}(\\alpha), \\text{CZ}\\\\} \\end{equation}\n\nI know I can write any single qubit operation with $\\\\{\\text{H}, \\text{Rx}(\\alpha)\\\\}$ as I can reach any point on the Bloch sphere with 3 parameterised rotations in 2 directions (I can implement $\\text{Rz}$ in terms of $\\text{H}$ and $\\text{Rx}$).\n\nFrom here it's well known that an arbitrary single qubit rotation along with a $\\text{CX}$ gate is sufficient for universality and I can realise a $\\text{CX}$ using a $\\text{CZ}$ with Hadamards on the target qubit. Therefore the set $S_0$ is a universal gate set.\n\nWhen I try to think of a way to automate this process of using the provided gates to implement a well known universal gateset I'm not sure where to start. It certainly seems like a hard problem.\n", "link": "https://quantumcomputing.stackexchange.com/questions/32408/can-you-programatically-check-whether-a-given-set-of-gates-is-universal", "tags": ["quantum-gate", "universal-gates", "transpile"], "votes": 9, "creation_date": "2023-05-02T13:13:45", "comments": ["I came across this paper, and I think this will be useful to you.", "It might be an incomplete analogy but - we can't square the circle with a ruler and a compass; however, Archimedes taught us that we can get arbitrarily close, by inscribing and circumscribing polygons. With that in mind, can the ruler and compass alone get arbitrarily close to any well-defined point on the plane?", "i.e. you could restrict to a finite set of parameterised gates rather than doing general single qubit rotations.", "Interesting. I would have guessed that just verifying that a gateset can apporximate any n qubit unitary would be easier than exactly recreating any gate. Why might the approximate version using a finite set be harder to tackle?", "@Callum That may be the harder case!", "@DaftWullie . I suppose both cases are of interest to me. Is there a way to check if a gateset is approximately universal up to some $\\epsilon$?", "Thanks @Mark S. indeed this other post is relevant. I think I’d need to read around a bit more to understand the discussion", "Are you asking specifically about the case where you can exactly recreate any gate you want (e.g. the gate sets you've mentioned) or would finite gate sets achieving the target gate with arbitrary accuracy also fall within your interest?", "See this question and answer -cstheory.stackexchange.com/questions/11298/…"], "comment_count": 9, "category": "Technology", "diamond": 0} {"question_id": "482", "site": "quantumcomputing", "title": "Better "In-Place" Amplification of QMA", "body": "$\\def\\braket#1#2{\\langle#1|#2\\rangle}\\def\\bra#1{\\langle#1|}\\def\\ket#1{|#1\\rangle}$ In [MW05](https://arxiv.org/abs/cs/0506068) the authors demonstrate so-called \"in-place\" amplitude amplification for QMA, exhibiting a method for Arthur to amplify his success probability without requiring any increase in the size of Merlin's witness state. Call the original machine $M$ and the amplified machine $M'$. \n\nSuppose that $x$ is the input, the original machine is $M(x,w)$ and the amplified machine is $M'(x,w)$. The approach of MW05 does not guarantee that a witness state that was in the accepting set for $M(x,\\cdot)$ (i.e. that made $M$ accept input $x$ with probability exceeding, say, $\\frac{2}{3}$) is still in the accepting set for $M'(x, \\cdot)$ (i.e. those states that make $M'(x, \\cdot)$ accept with amplified probability). I explain why I think that it's actually not possible to make their method do this below the line. \n\n**Q: Is there a way to perform amplification of QMA machines without increasing the length of Merlin's witness and while also guaranteeing that all initially valid witnesses for a given input stay valid witnesses for this input?**\n\n* * *\n\nWithout explaining too much context and using their notation, the issue that seems to arise is that MW05 first proves that their approach works for states that are eigenvectors of $Q$. This is good enough for completeness because if there's a state that accepts with probability $p \\geq a$ then there's an eigenvector with success probability at least $p$. However, I don't think this can be used to show that any \"good\" witness also accepts w.h.p. after amplification. \n\nIn general a \"good\" witness $\\ket{\\psi}$. will be some linear combination of eigenvectors $\\sum_j \\alpha_j\\ket{\\psi_j}$, some of which are \"good\" and have acceptance probability $\\geq a$ and others of which are \"bad\" and have acceptance probability $\\leq b$. Let $S$ be the set of \"good\" witness eigenvectors. Then, once we amplify, our success probability is roughly (for $r$ large enough I guess) \n\n$$\\sum_{j \\in S} |\\alpha_j|^2(1-2^{-r}).$$\n\nIn general $\\sum_{j \\in S} |\\alpha_j|^2$ isn't $1$ so we're in trouble. In the worst case of witness eigenvectors, we can even construct a \"good\" witness $\\ket{\\psi}$ that after $r$-rounds of amplification accepts with probability roughly $\\frac{a-b}{1-b}(1-2^{-r}) \\leq \\frac{1}{2}$ for $(a,b) = (\\frac{2}{3}, \\frac{1}{3})$. \n", "link": "https://quantumcomputing.stackexchange.com/questions/8586/better-in-place-amplification-of-qma", "tags": ["quantum-algorithms", "complexity-theory", "qma"], "votes": 8, "creation_date": "2019-10-27T15:47:51", "comments": [], "comment_count": 0, "category": "Technology", "diamond": 0} {"question_id": "483", "site": "quantumcomputing", "title": "Qubit fidelity of DWAVE device", "body": "Since DWAVE quantum device is constructed using superconducting flux qubits, each qubit cannot be produced identically so that the fidelity of the qubit must be different. DWAVE only provides the information of their devices in terms of number of qubits, couplers, etc without any calibration data of each qubit on their [website](https://www.dwavesys.com/solutions-and-products/systems/).\n\nI was wondering if we can find some information about the qubit fidelity of DWAVE machine. Since the number of qubits is quite huge, I assume it's time-consuming to calibrate the device and methods like randomized benchmarking are not applicable anymore. Is there any specific calibration method for that?\n", "link": "https://quantumcomputing.stackexchange.com/questions/23916/qubit-fidelity-of-dwave-device", "tags": ["experimental-realization", "fidelity", "d-wave"], "votes": 8, "creation_date": "2022-02-02T07:43:05", "comments": [], "comment_count": 0, "category": "Technology", "diamond": 0} {"question_id": "484", "site": "retrocomputing", "title": "How much did Atari pay for GEM?", "body": "Atari licensed Digital Research's GEM graphic user interface for the ST. As puts it:\n\n> And of course in the wake of the Macintosh the ST simply had to ship with a mouse and an operating system to support it.\n\n> It was this latter that presented by far the biggest problem. While the fairly conservative hardware of the ST could be put together relatively quickly, writing a modern, GUI-based operating system for the new computer represented a herculean task. Apple, for instance, had spent years on the Macintosh’s operating system, and when the Mac was released it was still riddled with bugs and frustrations. This time around Tramiel wouldn’t be able to just slap an archaic-but-paid-for old PET BASIC ROM into the thing, as he had in the case of the Commodore 64. He needed a real operating system. Quickly. Where to get it?\n\n> He found his solution in a very surprising place: at Digital Research, whose CP/M was busily losing its last bits of business-computing market-share to Microsoft’s juggernaut MS-DOS. Digital had adopted an if-you-can’t-beat-em-join-em mentality in response. They were hard at work developing a complete Mac-like window manager that could run on top of MS-DOS or CP/M. It was called GEM, the “Graphical Environment Manager.” GEM was merely one of a whole range of similar shells that were appearing by 1985, struggling with varying degrees of failure to bring that Mac magic to the bland beige world of the IBM clones. Also among them was Microsoft’s original Windows 1.0 — another product that Tramiel briefly considered licensing for the ST. Digital got the nod because they were willing to license both GEM and a CP/M layer to run underneath it fairly cheap, always music to Jack Tramiel’s ears.\n\nJust how cheap was it? How much did Atari pay DR? And what sort of terms were involved?\n", "link": "https://retrocomputing.stackexchange.com/questions/15377/how-much-did-atari-pay-for-gem", "tags": ["history", "atari", "atari-st"], "votes": 10, "creation_date": "2020-06-28T21:34:26", "comments": ["@tofro - details of other similar deals (eg Microsoft's licensing of SCP's 86-DOS) are known. It's plausible that details of this deal have come to light via a similar route.", "Sadly, I suspect that the intersection of “people who would have even known at the time” (C-suite, accountant) and “people who care enough about old technology enough to visit Retro.SO” is very, very small. Of the intersection, the chances that they will remember what the negotiated price was is probably zero. Engineers would have been told “we licensed GEM, take care of the integration” but they would not have been told the licensing cost because it wasn’t relevant, NDA would likely impose need-to-know basis, and it was none of the engineer’s business to know.", "@tofro I would normally expect NDA's to have expired after three and a half decades with the companies no longer in business. If they are really for life in this case, that in itself would be an interesting historical fact.", "@user there were several companies selling their home/soho machines with GEM. Think Philips Yes and alike.", "Who do you think could answer this? Everyone who really could know is either dead or very probably still under a lifetime NDA...", "Would be interesting to know what Amstrad paid for it too. It was included with the PC1512."], "comment_count": 6, "category": "Technology", "diamond": 0} {"question_id": "485", "site": "retrocomputing", "title": "What was IBM's internal Specification Language of the 1980s?", "body": "Within IBM's internal Development community, there was a move in the 1980s to bring our skills up to date. As part of this, we were introduced to a specification language, independent of the programming languages of the time (PL/S and Basic Assembler for example), in which a specification would be implemented. This specification language, as I recall, embodied set theory and logic in a type-safe context.\n\nIn fact, in the IBM Hursley Lab we eventually chose to use the Z notation from the Programming Research Group at Oxford University. This was perhaps more strictly mathematical than the IBM language - part of a Formal Methods approach.\n\nHowever, I spent quite a bit of time with the IBM specification language and I would very much like to know if anyone recalls what it was called and what happened to it. \nIt had a “specification checker” for type consistency etc and I remember it very fondly, before getting immersed in the Z notation.\n", "link": "https://retrocomputing.stackexchange.com/questions/25868/what-was-ibms-internal-specification-language-of-the-1980s", "tags": ["history"], "votes": 10, "creation_date": "2022-12-14T00:21:28", "comments": ["Thanks @davidbak I’m not thinking of VDM which, like Z, is more mathematically formal than the IBM language I am searching for.", "See if this article rings a bell - dl.acm.org/doi/abs/10.1145/356596.356598 (PDF is freely available here)", "I'm thinking it was the \"Vienna Definition Language\"? foldoc.org/Vienna+Definition+Language - see also en.wikipedia.org/wiki/Vienna_Development_Method#History"], "comment_count": 3, "category": "Technology", "diamond": 0} {"question_id": "486", "site": "stackoverflow", "title": "Ограничения на поиск пользователей через users.search api vk", "body": "У api вконтакте есть ограничение - 3 запроса в секунду. Однако у `users.search` есть какое-то своё ограничение на количество вызовов. \n\n> Помимо ограничений на частоту обращений, существуют и количественные ограничения на вызов однотипных методов. По понятным причинам, мы не предоставляем информацию о точных лимитах.\n\nНужно найти 1000 человек в соцсети (клиенты компании). Если api бы работало хорошо, это было бы 300 секунд. А что делать с тем, что на N-м запросе api выдаёт 0 найденных? Похоже, что N-порядка десятков. \nДругие методы работают, а этот выдаёт 0. Причём даже если попытаться поискать людей просто в браузере - тоже 0 найденных.\n\nКто сталкивался, поясните, надолго ли происходит такое отключение поиска. И может есть какое-то время паузы, выше которого поиск всегда работает и возвращает результаты?\n\n[Проверка через execute тут](https://vk.com/dev/execute \"Проверка через execute тут\")\n \n \n var found = API.users.search({\"q\":\"Иванов Иван\"});\n return found;\n \n\nЗапуск этого кода даёт `response: { count: 0, items: [] }`, когда `users.search` вдруг перестаёт работать.\n", "link": "https://ru.stackoverflow.com/questions/451645/%d0%9e%d0%b3%d1%80%d0%b0%d0%bd%d0%b8%d1%87%d0%b5%d0%bd%d0%b8%d1%8f-%d0%bd%d0%b0-%d0%bf%d0%be%d0%b8%d1%81%d0%ba-%d0%bf%d0%be%d0%bb%d1%8c%d0%b7%d0%be%d0%b2%d0%b0%d1%82%d0%b5%d0%bb%d0%b5%d0%b9-%d1%87%d0%b5%d1%80%d0%b5%d0%b7-users-search-api-vk", "tags": ["javascript", "vkontakte-api"], "votes": 8, "creation_date": "2015-09-17T05:51:12", "comments": ["Возможный дубликат вопроса: Ограничения VK-APi", "Но нельзя не согласиться с тем, что эта мера с неизвестным лимитом сильно портит жизнь и тем, кто готов играть по правилам. Видимо, выбирая защиту от злоупотреблений, недовольством соблюдающих правила пользователей решили пренебречь.", "Если допустимое кол-во запросов будет известно заранее, то ботописатели обойдут ограничение, путём использования в парсере неограниченного количества учётных записей, коих в сети можно покупать, хоть пачками. А тут получается, что возврат нулевого значения не гарантирует в любой момент времени, что поиск не обнаружил данные по запрашиваемой строке. Может они есть, а может аккаунт уже забанен. Это ведёт к тому, что нужно выполнять отдельный запрос с заранее проверенным на корректность результатом на каждый нулевой ответ. Это конечно позволит проверить достоверность ответа, но сильно снизит КПД.", "@alexis031182 честно говоря, из-за этой неоднозначности не очень хочется с ними работать, не беру работу по их апи больше. Программирование и нечёткие правила несовместимы.", "Похоже, что не только users.search подвержен ограничению, но и wall.search. Проверено, \"банят\" довольно быстро с теми же симптомами, что и в вопросе. К сожалению, не удалось вычислить момент, с какого по счёту запроса это началось. Вполне вероятно, что audio.search, а также video.search тоже весьма ограничены."], "comment_count": 5, "category": "Technology", "diamond": 0} {"question_id": "487", "site": "stackoverflow", "title": "Максимально возможное заполнение в клеточном автомате "Life"", "body": "Меня интересует алгоритм вычисления максимально возможного количества \"живых\" клеток на поле размером 100х100. Предполагается, что это поле - развернутый тор. Стартовое и конечное положения меня не интересуют. Но это должно быть возможное игровое состояние - т.е. состояние, соответствующее правилам (а не, например, просто заполненный квадрат 100x100). Интересует именно максимум. Возможно, кто-то логически объяснит, что может быть именно такое-то число и не больше ни меньше.\n\nВозможно, многие знают правила игры. Но я их повторю:\n\n> 1) в пустой (мёртвой) клетке, рядом с которой ровно три живые клетки, зарождается жизнь;\n> \n> 2) если у живой клетки есть две или три живые соседки, то эта клетка продолжает жить;\n> \n> 3) в противном случае (если соседей меньше двух или больше трёх) клетка умирает («от одиночества» или «от перенаселённости»)\n\n**Уточнение:** Дабы не было никому обидно (и были понятны комменты), правку вношу в виде уточнения, а не меняя непосредственно сам вопрос.\n\nМне необходимо реальное промежуточное состояние. То, которое может быть достигнуто в принципе. При этом, хотелось бы знать максимально возможное количество живых особей в популяции. Да, действительно, это состояние может быть конечными или цикличным. Но меня не интересует стартовое состояние, когда оно соответствует правилам старта, но, на самом деле, оно не может быть получено иным путем.\n\nДаже скажу больше... возможно, это состояние (состояние максимума популяции для квадрата 100x100) может быть достигнуто из расчета, что первоначальный квадрат был больше. Хотя желательно, чтобы он был ограничен в размерах и был тором.\n\nОтвечая на вопрос, что значит _\"возможное игровое состояние\"_ отмечу, что в моем понимании, стартовое состояние является как бы предигровым состоянием или стартовым состоянием, но не состоянием, полученным в процессе игры. _\"Возможное игровое состояние\"_ в данном случае мной понимается как состояние, полученное в процессе игры.\n\nВозможно, кто-то и не согласен с моим пониманием игрового состояния, но такое оно и есть. Представьте, что гости собираются и рассаживаются за стол (даже уже сели). Но это не застолье. Вот как-то так.\n\nЕсли предлагаемая популяция циклична, то, может быть, имеет смысл сократить размер поля до 10x10?\n", "link": "https://ru.stackoverflow.com/questions/502844/%d0%9c%d0%b0%d0%ba%d1%81%d0%b8%d0%bc%d0%b0%d0%bb%d1%8c%d0%bd%d0%be-%d0%b2%d0%be%d0%b7%d0%bc%d0%be%d0%b6%d0%bd%d0%be%d0%b5-%d0%b7%d0%b0%d0%bf%d0%be%d0%bb%d0%bd%d0%b5%d0%bd%d0%b8%d0%b5-%d0%b2-%d0%ba%d0%bb%d0%b5%d1%82%d0%be%d1%87%d0%bd%d0%be%d0%bc-%d0%b0%d0%b2%d1%82%d0%be%d0%bc%d0%b0%d1%82%d0%b5-life", "tags": ["алгоритм"], "votes": 8, "creation_date": "2016-03-14T07:25:23", "comments": ["Комментарии не предназначены для расширенной дискуссии; разговор перемещен в чат."], "comment_count": 1, "category": "Technology", "diamond": 0} {"question_id": "488", "site": "chemistry", "title": "Dry water using graphite", "body": "[Dry water](https://en.wikipedia.org/wiki/Dry_water) or \"powdered water\" is essentially a bunch of extremely small water droplets surrounded by $\\ce{SiO2}$.\n\nSome time ago, I was messing around and mixed some powdered graphite with water. (The graphite was obtained from a pencil, so it was not pure. Also I mixed the two substances with a stick, not with a high-speed blender.) When a droplet of this was splashed onto the table, it could be blown around with air and even held on the hand without breaking, until it was squished.\n\nThe important part of this is, that that droplet could be placed back into the original mixture of water and graphite powder, and it would roll around on the surface.\n\nSo does anyone know if it's possible to use powdered graphite rather than silica particles to make dry water? If is is, what properties are different from the ordinary powdered water?\n", "link": "https://chemistry.stackexchange.com/questions/61838/dry-water-using-graphite", "tags": ["water", "mixtures"], "votes": 7, "creation_date": "2016-10-30T06:48:15", "comments": ["Graphite for pencils is mixed with clay, which is largely composed of silicates. I don't know whether pure graphite powder would display the same behavior you describe.", "I may end up doing something similar to this for a school project. If I decide to mix the graphite with water instead of ink, I will measure its conductance. I could post my results here, maybe.", "@Ivan But the 'dry water' uses hydrophobic silica particles (surface modified). I think that nanographite would be a suitable substitute.", "Silica is highly hydrophilic, graphite is not quite like that."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "489", "site": "chemistry", "title": "Mathematical models of oscillatory chemical reactions", "body": "I am a mathematician working on real-life models in ordinary differential equations. I want to know if there are any models of oscillatory chemical reactions that consist of three ordinary differential equations where one variable is much slower than the other two. In particular I am interested in bursting behavior, which consists of alternating trains of fast oscillations with periods of rest.\n\nI have been searching and I couldn't find what I am looking for. I found about the Belousov-Zhabotinsky reaction, but in this case we have one fast variable and two slow variables, while I'm looking for two fast variables and one slow variable.\n", "link": "https://chemistry.stackexchange.com/questions/166898/mathematical-models-of-oscillatory-chemical-reactions", "tags": ["physical-chemistry", "kinetics", "theoretical-chemistry"], "votes": 7, "creation_date": "2022-08-05T08:27:40", "comments": ["related? wrap.warwick.ac.uk/60495/1/WRAP_computation-02-00047.pdf", "Related: mattermodeling.stackexchange.com/q/8812/5", "Somewhat related"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "490", "site": "chemistry", "title": "Interaction of trifluoroacetates with acetylcholinesterase", "body": "There exists a substance called TMTFA, or 3-(_N,N,N_ -Trimethylammonio)-2,2,2-trifluoroacetophenone. It is known for being able to inhibit acetylcholinesterase at femtomolar concentrations. The TMTFA-AChE adduct resembles the tetrahedral transition state through which the hydrolysis of acetylcholine by the enzyme normally proceeds, and this adduct is stabilized by the electron-withdrawing trifluoromethyl group present in the TMTFA molecule. This inhibition is reversible.\n\nWhile there have been studies regarding trifluoroacetophenones and their interactions with AChE, I haven't been able to find studies exploring the interactions between AChE and trifluoroacetates, such as 3-(_N,N,N_ -trimethylammonio)phenyl trifluoroacetate or 2,2,2-trifluoroacetylcholine. Would these, and other, esters of trifluoroacetic acid inhibit AChE or be hydrolyzed normally? If it's the former, then what are the typical inhibitory concentrations of such esters? What is the shape of the adduct of the ester with AChE? How long would they remain bound to the enzyme? Have any studies been done regarding the bioactivity of such esters?\n\n### References\n\n 1. Nair, H. K.; Seravalli, J.; Arbuckle, T.; Quinn, D. M. Molecular Recognition in Acetylcholinesterase Catalysis: Free-Energy Correlations for Substrate Turnover and Inhibition by Trifluoro Ketone Transition-State Analogs. _Biochemistry_ **1994** , _33(28)_ , 8566–8576. DOI: [10.1021/bi00194a023](https://doi.org/10.1021/bi00194a023).\n 2. Harel, M.; Quinn, D. M.; Nair, H. K.; Silman, I.; Sussman, J. L. The X-Ray Structure of a Transition State Analog Complex Reveals the Molecular Origins of the Catalytic Power and Substrate Specificity of Acetylcholinesterase. _J. Am. Chem. Soc._ **1996** , _118(10)_ , 2340–2346. DOI: [10.1021/ja952232h](https://doi.org/10.1021/ja952232h).\n\n\n", "link": "https://chemistry.stackexchange.com/questions/135747/interaction-of-trifluoroacetates-with-acetylcholinesterase", "tags": ["organic-chemistry", "biochemistry", "proteins", "esters", "enzymes"], "votes": 5, "creation_date": "2020-06-24T21:41:08", "comments": ["According to Ref.2, TMTFA binds titly to AcE from a Pacific electric ray, Torpedo californica. It might not do the same to Human AcE. I believe your intention is for inhibitors of Human AcE."], "comment_count": 1, "category": "Science", "diamond": 0} {"question_id": "491", "site": "chemistry", "title": "Reaction of Rubidium with vacuum system viewport", "body": "I have a vacuum system ($<1\\times10^{-7}$ mbar) filled with Rubidium ($^{87}$Rb) vapour, at room temperature. \n\nWe also have a viewport near the Rubidium source, made of a stainless steel flange and a glass substrate, connected with a [Kovar](https://en.wikipedia.org/wiki/Kovar) alloy.\n\nWe are seeing potential degradation of the seal, which we do not see on other parts of the system. The only difference is the Nickel and Cobalt based Kovar alloy, not present anywhere else. \n\nWhat are the reaction rates of Nickel and/or Cobalt with Rubidium?\n", "link": "https://chemistry.stackexchange.com/questions/81551/reaction-of-rubidium-with-vacuum-system-viewport", "tags": ["everyday-chemistry", "home-experiment"], "votes": 5, "creation_date": "2017-08-21T04:23:40", "comments": ["btw, thanks to ability of rubidium to form suboxides even tiniest amount of oxygen may bind a lot of rubidium.", "Please, check the mark of the glass and if it is suited for work with rubidium in this particular settings. In general, glasses are oxides that can potentially react with alkali metals in right circumstances. I see no reason for them to react with nickel/cobalt alloy. However, rubidium might react with oxide layer on the surfaces of this alloy and/or any dirt left on the seal (and glass, and any dirt left on the glass) and any passing air if the seal is... not good enough."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "492", "site": "hsm", "title": "Whence “homomorphism”, “homomorphic”?", "body": "The [kernel](https://hsm.stackexchange.com/questions/2149/what-came-first-the-kernel-from-vector-spaces-or-from-group-theory) question leads to another : Today, **_homomorphism_** (resp. _isomorphism_) means what Jordan ([1870](https://archive.org/stream/traitdessubsti00jorduoft#page/56)) had called _isomorphism_ (resp. _[holoedric](https://hsm.stackexchange.com/questions/3108/jordan-called-isomorphisms-iso-and-homomorphisms-iso-holoedriques-and-iso) isomorphism_). How did the switch happen?\n\n * “Homomorphic” appears e.g. in de Séguier ([1904](https://www.emis.de/cgi-bin/JFM-item?36.0187.02), [pp. 65–66](https://archive.org/stream/thoriedesgroup00sguoft#page/65)):\n\n> (...) supposons entre les groupes $\\mathrm A$ et $\\mathrm B$ une correspondance telle qu’à chaque élément de $\\mathrm A$ réponde un élément au moins de $\\mathrm B$ et à chaque élément de $\\mathrm B$ un élément au moins de $\\mathrm A$ et que si $a_i$ de $\\mathrm A$ et $b_i$ de $\\mathrm B$ se correspondent, $a_ia_k$ et $b_ib_k$ se correspondent aussi. On dit que $\\mathrm A$ et $\\mathrm B$ sont _homomorphes_.\n\n * “Homomorphism” (as _property_ of being homomorphic) appears e.g. in Loewy ([1912](https://www.emis.de/cgi-bin/JFM-item?43.0199.02), [p. 198](https://archive.org/stream/festschriftheinr06webeuoft#page/198)):\n\n> §1 (...) untersucht den Homomorphismus zweier abstrakter Gruppen (...).\n\n * “Homomorphism” (as _map_ between groups) appears e.g. in Hopf ([1928](https://www.emis.de/cgi-bin/JFM-item?54.0610.02), [p. 128](http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002507641&physid=phys139)):\n\n> 4) Unter einem Homomorphismus der Gruppe $\\mathfrak G$ in die Gruppe $\\mathfrak H$ verstehen wir eine eindeutige Abbildung $f$ von $\\mathfrak G$ auf einen echten oder unechten Teil von $\\mathfrak H$, bei der stets $f(x+y)=f(x)+f(y)$ ist.\n\n\n\n\nAre these the earliest occurrences?\n", "link": "https://hsm.stackexchange.com/questions/6460/whence-homomorphism-homomorphic", "tags": ["terminology", "abstract-algebra", "group-theory"], "votes": 9, "creation_date": "2017-08-26T17:04:12", "comments": ["Now cross-posted: mathoverflow.", "Homomorphism is etymologically correct in the sense that the structures of two groups exhibit a degree of similarity when one can map into the other (trivial cases apart). Isomorphism is then more accurately used to mean \"congruent\" structure. (\"Isos\" is more commonly used to mean equal, but Euclid also uses it for congruent.)", "The following text notes the use of homomorphism by de Séguier in 1904, and then states that the term was introduced by Klein without giving a reference. books.google.ca/…"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "493", "site": "hsm", "title": "Who was the first to use the "does not exist" sign ∄?", "body": "Who was the first to use the \"does not exist\" sign ∄? I'm aware that Giuseppe Peano originated serifed ∃ and, moreover that Whitehead and Russell repurposed Peano's serifed ∃; I'm also aware that Gerhard Gentzen introduced the ∃ sans serif (and used it like Russell and Whitehead rather than like Peano). Yet, I cannot find who was the first to use ∄ as ~∃ to mean \"there does not exist\". Any idea? Thank you.\n", "link": "https://hsm.stackexchange.com/questions/7181/who-was-the-first-to-use-the-does-not-exist-sign-%e2%88%84", "tags": ["terminology", "notation", "mathematical-logic"], "votes": 9, "creation_date": "2018-03-26T14:01:36", "comments": [], "comment_count": 0, "category": "Science", "diamond": 0} {"question_id": "494", "site": "hsm", "title": "Who first defined polynomials as sequences?", "body": "> **Question 1.** When did the modern definition of a **polynomial** (as a sequence of coefficients, with multiplication defined by $\\left(ab\\right)_n = \\sum\\limits_{k=0}^n a_k b_{n-k}$) emerge?\n\nLet me clarify: People have been working with some notion of polynomials for centuries, but mostly without a rigorous concept of what a polynomial is. From what I understand, \"equations\" rather than polynomials were the focus, and demarcating the precise extent of what an \"equation\" is was secondary to actually solving equations of given, fairly specific types. When the need for a rigorous definition became clear, it seems that the go-to solution was to define a polynomial as a polynomial **function** (e.g., §1 of [Roland Weitzenböck, _Invariantentheorie_ , 1923](https://quod.lib.umich.edu/cgi/t/text/text-idx?c=umhistmath;idno=ABV0733)). This worked well for polynomials over $\\mathbb R$ or $\\mathbb C$, but not so much for polynomials over finite fields (which may be why E. H. Moore had to define $\\mathbb{F}_{p^n}$ as a quotient ring of a polynomial ring over $\\mathbb Z$ rather than of a polynomial ring over $\\mathbb{F}_p$ [in his 1893 talk](https://projecteuclid.org/download/pdf_1/euclid.bams/1183407788)).\n\nWhat I call the \"modern definition\" of a polynomial is one of the following two definitions:\n\n * either as an infinite sequence $\\left(a_0, a_1, a_2, \\ldots\\right)$ of coefficients (with all but finitely many $n$ satisfying $a_n = 0$), with addition defined entrywise and multiplication defined by the $\\left(ab\\right)_n = \\sum\\limits_{k=0}^n a_k b_{n-k}$ formula (and the indeterminate defined as the sequence $\\left(0,1,0,0,0,\\ldots\\right)$ rather than being some mythic \"symbol\" or \"indeterminate value\" or \"variable\"),\n\n * or as the monoid ring of the additive monoid $\\left\\\\{0,1,2,\\ldots\\right\\\\}$.\n\n\n\n\nThe second of these two definitions appears in §15 of B. L. van der Waerden, _Moderne Algebra I_ , 2te Auflage 1937. (I can't check the 1st edition, as I don't have it.) This gives a lower bound on the age of the modern definition. As for an upper bound, I am perplexed. Nicolas Bourbaki's _Elements of the History of Mathematics_ grazes the question on its page 82:\n\n> In all this research, the fields that arise consist of \"concrete\" elements in the sense of classical mathematics - (complex) numbers or functions of complex variables. 32 But already Kronecker, in 1882, takes full account of the fact (obscurely sensed by GaUBB and Galois) that \"indeterminates\" only play the role in his theory of the basis elements of an algebra, and not that of variables in the sense of Analysis ([186 a], v. II, pp. 339-340); and, in 1887, he develops this idea, linked to a vast programme that aims at nothing less than recasting all mathematics by rejecting all that cannot be reduced to algebraic operations on the integers. It is on this occasion that, taking up an idea of Cauchy ([56 a], (1), v. X, pp. 312 and 351) who had defined the field $\\mathbb C$ of complex numbers as the field of remainders $\\mathbb R\\left[X\\right]/\\left(X^2+1\\right)$, Kronecker shows how the theory of algebraic numbers is completely independent of the \"fundamental theorem of algebra\" and even of the theory of real numbers, all fields of algebraic numbers (of finite degree) being isomorphic to a field of remainders $\\mathbb Q\\left(X\\right)/\\left(f\\right)$ ($f$ an irreducible polynomial over $\\mathbb Q$)([186 a], v. 1111, pp. 211-240).\n\nFollowing the references here, I see some signs of the modern definition slowly dawning in the 19th Century: [Note: \"Polynomials\" are often called \"forms\" before ca. the 1930s, particularly when they are homogeneous.]\n\n * I don't know what exactly Cauchy had done; the [56 a] reference goes to his Oeuvres, which I don't have.\n\n * The first Kronecker reference goes to [his _Grundzüge einer arithmetischen theorie der algebraischen Grössen_ (1882)](https://archive.org/details/grundzgeeinerar01krongoog/page/n129), where he talks of \"veränderliche oder unbestimmte Grössen\" but never constructs any.\n\n * [J. Molk, _Sur une notion qui comprend celle de la divisibilité et sur la théorie générale de l'élimination_ , Acta Math. **6** (1885), pp. 1--165](https://projecteuclid.org/euclid.acta/1485888021), spends its section 2 (and maybe the whole Chapitre I?) arguing about the difference between variables and indeterminates. Alas, the argument appears to veer into philosophical terrain rather quickly, which puts it beyond my French comprehension skills; thus I have no idea what conclusion he actually comes to.\n\n * [H. Weber, _Die allgemeinen Grundlagen der Galois'schen Gleichungstheorie_ , Mathematische Annalen **43** (1893), pp. 521--549](https://eudml.org/doc/157689) tries to define polynomials in its §3 (after defining the notions of groups and fields in ways that are equivalent to their modern definitions). The definition pays obvious tribute to the idea that polynomials should not be functions; in particular it explicitly says that two polynomials are understood to be equal if and only if their respective coefficients are equal (rather than their values being equal). However, a vague notion of \"expressions of the form $\\Phi\\left(x,y,z,\\ldots\\right) = \\sum a x^r y^s z^t \\cdots$\" underlies this definition, and (e.g.) the multiplication is never explicitly defined (nor its associativity proven). (To this day, some algebra textbooks geared towards non-math majors work on this level.)\n\n\n\n\nSo my question is: Who completed this definition? Weber (later)? Kronecker? Dedekind? Noether? van der Waerden? And while at that:\n\n> **Question 2.** Who defined formal power series?\n> \n> **Question 3.** Who defined polynomials in noncommutative indeterminates?\n\nThese are litmus tests for Question 1; a rigorous definition of polynomials needs only a few trivial modifications to define formal power series, and a definition of noncommutative polynomials is not far away either.\n\nThe concept of a polynomial is, in a way, similar to the number $0$: It's an idea that, on its own, appears pedantic and meatless. But once it is established, it becomes a building block of a whole discipline that no one would want to be missing. (19th Century algebraists like Sylvester tend to use the unique factorization property of a polynomial ring without ever explaining what it is that they are factoring; from a modern perspective, this is a gaping hole in their proofs.) My hope is that, as this specific matter is much younger than the number $0$, we may have better clues to its authors.\n", "link": "https://hsm.stackexchange.com/questions/9710/who-first-defined-polynomials-as-sequences", "tags": ["mathematics", "abstract-algebra"], "votes": 8, "creation_date": "2019-06-07T14:43:53", "comments": ["Not in the modern sense, but that can be said about most of algebra until the late 18th century. Hudde further treated polynomials purely algebraically in the 17th. Dieudonne, before reinterpreting Euler's manipulations in terms of formal power series, states the following:\"These series appear in print in Hilbert's Grundlagen der Geometrie, apparently for the first time\".", "Interesting! This is pretty much the modern definition. I take it he didn't prove anything, though? (The question is nontrivial, because he had something very much resembling induction in his work, so it's not like the toolkit was entirely missing.)", "What you call the \"modern definition\" was the original one given by al-Karaji and al-Samawal back in the 12th century. The latter wrote powers and the corresponding coefficients as columns of a table, and gave a multiplication rule equivalent to the convolution formula (al-Karaji's descriptions were verbal and more obscure). He even allowed negative powers (hence Laurent polynomials), and gave an algorithm for the \"synthetic division\". See e.g. Katz, History of mathematics, 9.3.3"], "comment_count": 3, "category": "Science", "diamond": 0} {"question_id": "495", "site": "proofassistants", "title": "Unintentionally proven false theorem with type-in-type outside logic and foundations?", "body": "We are all familiar with Russell's paradox, and [it is known](https://proofassistants.stackexchange.com/a/1220) that Per Martin-Löf proved that type-in-type is normalizing and consistent (which is false), by accidentally using an assumption in his meta-theory that is essentially type-in-type. But all these arise in the areas of logic and foundations. Has anyone proven some statement Q by accidentally using some type-in-type assumption, without being aware that such an assumption is unsound, and such that Q can be shown to be surely false (say in a typical predicative type theory)?\n\nThe linked example would be considered to be under foundations. Here is another example that is not quite under logic or foundations but anyone who proves it will certainly be immediately aware that the type-in-type assumption is unsound: Let W be the type of all rooted (downward-directed) trees with no infinite (downward) path. Let T be a rooted tree whose root has exactly the members of W as children. Now T is in W...\n\nSo is there any example where the person who proved the result was still unaware (at least for a reasonable period of time) that the type-in-type assumption was unsound?\n", "link": "https://proofassistants.stackexchange.com/questions/1238/unintentionally-proven-false-theorem-with-type-in-type-outside-logic-and-foundat", "tags": ["foundations", "history", "predicativity", "universe", "consistency"], "votes": 13, "creation_date": "2022-04-04T07:40:29", "comments": ["And negation is not even needed, nor is any notion at all about 'categories'. To see why, if type∈type, then let S be the 'type' of all types x such that ∀t∈{ 0 : x∈x } ( ⊥ ), and it is trivial to obtain ⊥ by constructive proof: Given any t∈{ 0 : S∈S }, we have S∈S and so ∀t∈{ 0 : S∈S } ( ⊥ ), hence ⊥. Therefore S∈S, and so ⊥. By the way, if we require a type's defining predicate to quantify over only predicative types, then we can have type∈type in a predicative system!", "@AndrejBauer: Yes, but that has nothing to do with category theory; it is just plain foundational concerns, because it just comes down to saying that sets are closed under function types but classes are not (if you only have two levels). In turn, this is ultimately like Frege's Axiom V; because it is just another result of 'wishful thinking' that certain concepts have reifications. If people want to reason about any 'large' things without giving solid justification why those things can even be reified in the first place, then it's just like asking for Russel's paradox all over again.", "Still running on vague memories, I think one of the pertinent observations was that “$f : A \\to B$ is epi” in a large category involves universal quantification over a class. Speaking type-theoretically, the type of epis in a category $C$ living in a universe $U$ might live outside $U$. So for instance, if we take the subcategory of epis in $C$, we've bumped up the universes. But ultimately, we'd have to find the specific objections to see in what way they're problematic.", "@AndrejBauer: I vaguely recall reading some examples of that sort, yes, but to me those are due to wishful thinking way beyond higher-order arithmetic. Ultimately, what is the difference between those mistakes and Frege's Axiom V? The point is that a lot of these explorations were based on dubious axioms that arose out of logic and foundations, and were not things that mathematicians genuinely believed to be true in a platonic sense.", "I vaguely remember Benabou, or perhaps Dobus, telling some cautionary tales about faulty reasoning in category theory, where mistakes can be made easily enough if one does not pay attention to class-set distinction. But I don't know the reference.", "@AndrejBauer: Let me put it another way; if an example was actually utilized to obtain a theorem in the language of higher-order arithmetic that is believed to be incorrect, then I would agree that that obtained theorem is an actual example of the type I am looking for. If not, then I'm unlikely to buy that it is outside of logic and foundations.", "Mike's example looks like it's \"just\" about type theory, but it actually has a much wider applicability.", "I don't count Mike's example by the way, as it is literally about type theories, which is squarely in logic and foundations. After all, I did link to that thread in my question!", "@AndrejBauer: It need not be in type theory. Naive set theory also has a type-in-type assumption, even if it is not called \"type\". I'm not intending to restrict answers to only type theories of a particular narrow tradition.", "@MevenLennon-Bertrand: But did that result in any false theorem that was believed to be a good theorem?", "Or in the glory era of early Unimath, when they decided to use type-in-type to avoid issues with universes? Which looks like the time to which Mike Shulman's example belongs.", "Just about anyone who learns type theory nowadays knows about type-in-type. So if there is an example of what you're asking for, it would have to have taken place in the narrow period between Martin-Löf's type theory with type-in-type, and the realization that it's inconsistent.", "Yes, that's the one I was thinking of!", "@AndrejBauer is this the one: proofassistants.stackexchange.com/questions/1219/… ? I don't know if this fits the question, though: I would say it is outside of logic and foundation, but Mike Shulman says he does not know whether it holds or not, so it probably cannot \"be shown to be surely false\" with our current knowledge…", "I seem to remember some discussion about accidental universe inconsistencies that were masked at first. Maybe Mike Shulman communicated them, I'll try to remember."], "comment_count": 15, "category": "Science", "diamond": 0} {"question_id": "496", "site": "proofassistants", "title": "Rules for mutual inductive/coinductive types", "body": "Some proof assistants, like Agda and maybe Coq, allow families of mutually defined types, or nested definitions of types, in which some are inductive and others are coinductive. I have no idea what the type-theoretic introduction/elimination rules should be for such a family, and as shown in [this question](https://proofassistants.stackexchange.com/q/961/98) their behavior can be kind of unexpected. The implementation, I guess, allows general \"matches\" and \"comatches\", with a separate \"termination/productivity checker\"; but this is not a semantic explanation suitable for, say, finding category-theoretic models.\n\nHas anyone written down an inference-rule presentation of a type theory including mutual or nested inductive/coinductive definitions?\n", "link": "https://proofassistants.stackexchange.com/questions/980/rules-for-mutual-inductive-coinductive-types", "tags": ["type-theory", "coinductive-type", "inductive-type"], "votes": 12, "creation_date": "2022-03-06T12:57:18", "comments": ["I'm aware of Hening Basold's thesis. And Danielsson and Altenkirch's Mixing Induction and Coinduction has a long list of related work, some of which may have rules (e.g. the work on sized types).", "I mean the 'partiality monad', which you can model as a HIT/QIT via the free ω-complete partial order on a type. As a mixed definition you try to define the same thing similarly to the coinductive definition of the extended natural numbers. Conceivably it might be better behaved than quotienting a coinductive type after the fact (which I think requires countable choice to be a monad).", "What is \"$\\omega$-completion\"?", "I might add that even if there's a more rigorous presentation of what Agda/Coq does, it'd probably also be nice to have something better. A while back it was observed that trying to define ω-completion as a mixed higher-inductive/coinductive definition gets you something contractible. But now I wonder if it's just because you can't write down the right thing in Agda (i.e. you really want uses of the higher constructor to be well-founded, but they aren't in Agda)."], "comment_count": 4, "category": "Science", "diamond": 0} {"question_id": "497", "site": "biology", "title": "What's the difference between proliferation and diffusion when talking about changes in tumor density?", "body": "Cell proliferation and cell diffusion seem to be important quantities to estimate when trying to understand or measure tumor growth, but I don't really understand a) the difference between them or b) their relationship to tumor growth. \n\nCell proliferation refers to cellular subdivision. This doesn't necessarily mean tumor growth because something can subdivide but still have the same net mass, right?\n\nCell diffusion on the other hand, refers to changes in cell density. Supposing cell proliferation is constant and we have positive diffusion, that would mean that cells are 'spreading out' (ie the mass is less dense but bigger). Is this correct? Would positive diffusion be equivalent to tumor growth?\n\nThis question came up when reading [this paper](https://www.ncbi.nlm.nih.gov/pubmed/27164582). It talks about estimating diffusion and proliferation parameters in order to solve the reaction-diffusion equation which is, in turn, used to model tumor growth.\n", "link": "https://biology.stackexchange.com/questions/57119/whats-the-difference-between-proliferation-and-diffusion-when-talking-about-cha", "tags": ["cell-biology", "growth", "differentiation", "cancer"], "votes": 5, "creation_date": "2017-03-08T10:29:49", "comments": ["A tumor by definition is an abnormal proliferation of cells but I don't think diffusion has anything to do with cells actually diffusing. Rather, the carrying capacity of a tumor depends on such parameters as nutrients or oxygen, which must diffuse further inwards until they cannot reach cells. Hence, the core of large solid tumors tends to be necrotic. It's also my understanding that there is a natural pressure that forces older cells inwards to the core of the tumor, as the outside layers continue to divide and grow outwards, increasing the diameter of the tumor.", "I would also like to see a reference to these terms if you have one."], "comment_count": 2, "category": "Science", "diamond": 0} {"question_id": "498", "site": "stackoverflow", "title": "ある一定のダウンタイムがある並列システムを、特定の確率以上で常に必要並列数を上回らせるために必要な並列数は?", "body": "例えば web サービスは一般的にロードバランサーと並列数`N`で表されるワーカープロセスによって構築される場合が多いです。\n\n今、あるワーカーが故障した時、そのワーカーが復帰するために必要な時間を`t`とします。ワーカーは、単位時間あたり`λ`の確率で故障し、その故障確率はそれまでの稼動時間に依らない、とします。 \n今、このサービスが問題なく稼動してワークロードをさばくためには、最低`m`個のワーカーがダウンしていない状態で稼動している必要があるとします。\n\nこのとき、確率論的に、ある一定の期間 `T` が与えられたとき、その期間の間のすべてのタイミングにおいて、上記の `m` 個のワーカーが稼動している確率 `p` が定義できると思い、これはこれまで出てきた `N`, `m`, `t`, `λ`, `T` についての関数になっていると思うのですが、その算出方法が分からずにいます。\n\n# 質問\n\n上記のような、ある並列システムが最低`m`の並列度を指定期間において達成している確率を求めたいのでが、これはどのようにすると求められますか?\n\nこれが分かることで、逆に算出された確率を信頼度とすることで、求められる信頼度に対して必要な `N` を求めたいと思い、質問しています。\n", "link": "https://ja.stackoverflow.com/questions/85441/%e3%81%82%e3%82%8b%e4%b8%80%e5%ae%9a%e3%81%ae%e3%83%80%e3%82%a6%e3%83%b3%e3%82%bf%e3%82%a4%e3%83%a0%e3%81%8c%e3%81%82%e3%82%8b%e4%b8%a6%e5%88%97%e3%82%b7%e3%82%b9%e3%83%86%e3%83%a0%e3%82%92-%e7%89%b9%e5%ae%9a%e3%81%ae%e7%a2%ba%e7%8e%87%e4%bb%a5%e4%b8%8a%e3%81%a7%e5%b8%b8%e3%81%ab%e5%bf%85%e8%a6%81%e4%b8%a6%e5%88%97%e6%95%b0%e3%82%92%e4%b8%8a%e5%9b%9e%e3%82%89%e3%81%9b%e3%82%8b%e3%81%9f%e3%82%81%e3%81%ab%e5%bf%85%e8%a6%81%e3%81%aa%e4%b8%a6%e5%88%97%e6%95%b0%e3%81%af", "tags": ["数学"], "votes": 5, "creation_date": "2022-01-01T23:26:31", "comments": ["> \"N, m, t, λ, T についての関数になっていると思うのですが\" 関数形を解析的に書けないように思います。 パラメータの値を具体的に与えたうえでシミュレーションして確率を求める方針をとることはできます。", "課題か何かでしょうか。実務的にはほとんど意味がない計算にしかならないと思います(構成要素の網羅度が足りなさすぎる、ある程‌​度網羅したら複雑すぎて計算は困難)"], "comment_count": 2, "category": "Technology", "diamond": 0} {"question_id": "499", "site": "retrocomputing", "title": "Plessey computer at Stanford?", "body": "While updating the wiki article on the 4004, I came across an interesting claim by Ted Hoff, who stated the idea of using a general-purpose CPU in a calculator came to him due to a minicomputer that had been donated to Stanford by the UK company Plessey.\n\nI'm at a loss as to what this might be... I know Plessey had a range of realtime machines used for telephone exchanges and the Linesman radar system (rather unsuccessfully) but these were mainframe sized and I cannot find a hint of a mini.\n\nI suppose this could be any old mini that Plessey bought and donated, but it seems less likely that Hoff would refer to it that way, if AT&T donated a PDP-11 to Stanford someone talking about it later would likely refer to it as \"a dec mini\".\n", "link": "https://retrocomputing.stackexchange.com/questions/24545/plessey-computer-at-stanford", "tags": ["minicomputers"], "votes": 9, "creation_date": "2022-05-23T09:04:36", "comments": ["On the face of it, the dates don't work, but 1. the information in Wikipeda is not sourced 2. a prototype might have been given to a university before it was ready for sale - perhaps hardware would have been available before suitable software was written.", "@mikado - had not heard of this - a little late for the timeline but an interesting device... and apparently actually used in the 90s!", "en.wikipedia.org/wiki/Plessey_System_250?wprov=sfla1 sounds like an interesting and unusual system", "Whatever it was, it needs to have been in Stanford in 1969, to fit the 4004 narrative.", "forum.stanford.edu/wiki/index.php/…. has no indication of a Plessey as one of the early computers at Stanford. Doesn't rule it out though...", "@another-dave Right. At least that's the only one I know. In fact, I've seen one not long ago at the VCL collection in Neubiberg.", "Per Wikipedia, Plessey was in the PDP-11-compatible business. But that might be a little late for 4004 work."], "comment_count": 7, "category": "Technology", "diamond": 0} {"question_id": "500", "site": "retrocomputing", "title": "Looking for late 90s shareware open world racing game", "body": "I'm trying to track down a shareware racing game released somewhere between 1999 and 2004 - or at least, my parents found it bundled in a computer magazine during that time.\n\nHere's what I remember about it:\n\n * It was a Win9x game, 3D Flat shaded with hard edges and basically no textures, no actual transparency and rather sparse geometry; It wasn't due to low settings, it was just stylized like that;\n * It was open world;\n * The scope of the game was basically collecting (pinkish?) waypoints randomly placed around the map within a time limit;\n * It was probably made by an European dev;\n * the \"buy the game\" screen showed a few other maps, including a \"rainbows in the sky\" fantasy setting.\n\n\n\nIt seems that most websites curating shareware games have just disappeared from the face of the earth, so I haven't found any reference about it online.\n\nDoes my description ring any bells to anyone?\n", "link": "https://retrocomputing.stackexchange.com/questions/21755/looking-for-late-90s-shareware-open-world-racing-game", "tags": ["identify-this-game"], "votes": 9, "creation_date": "2021-09-09T01:58:28", "comments": ["Are you really sure that it is a Win9x game? Low Poly graphics without texturing could also mean, that it was a DOS game.", "@thkala you were absolutely alone against time, and the car in the demo was just one extremely low poly sedan. The physics and handling weren’t too arcades, but very far from current day realism level.", "Do you remember anything about the cars? Were they street cars? F1-type? Off-road vehicles? Were you alone against time, or were there NPCs?", "I wish I could give some more details but there's very little to add. I mean, I thought it was popular enough given it was deemed worthy of being bundled on a magazine. I'll check it out on discord.", "Since I see there's one close vote, I won't wait to mention that, if you don't get an answer here, try /r/TipOfMyJoystick or the #identify-a-game channel on the MobyGames Discord.", "An italian magazine by Future Media called \"Giochi per il mio computer\" which was the local edition of \"PC Gamer\". I've already tried browsing archive.org for some clues but couldn't find anything.", "Can you remember which magazine it might have been ?", "Nope, it was clearly a small team or one-man made game. Much more lo-fi than the Interstate games.", "Interstate 76 or Interstate 82 ?"], "comment_count": 9, "category": "Technology", "diamond": 0}