\n\nI want to know if the right side is a new identity, before starting to write a chapter about this in my article
\n
This identity is known to Mathematica:
\n![\"enter](\"https://i.sstatic.net/oJ6SxgFA.png\")
For positive integer $n$, how many integer points can an $n$-by-$n$ square in the plane possibly cover?
\nLet $E$ be the set of all numbers that an $n$-by-$n$ square can possibly cover. I guess\n\\begin{equation*}\nE=[n^2,n(n+1)]\\cup\\{(n+1)^2\\}.\n\\end{equation*}\nThe square $[0.5,n+0.5]\\times[0.5,n+0.5]$ covers $n^2$ integer points; the square $[0,n]\\times[0,n]$ covers $(n+1)^2$ integer points. If the latter is slightly tilted, it seems to miss at least $(n+1)$ integer points. It is not difficult to prove the number can \"continuously\" change from $n^2$ to $n(n+1)$, but whether $n^2$ is the minimum and whether $n(n+1)$ is maximal other than $(n+1)^2$ are not so clear to me.
\nCase 1: all 4 sides cover integer points: By Pick's theorem this number is given by $n^2+\\frac{4n}{2}+1=(n+1)^2$. Case 2: no side cover integer points: We look at the smaller included $(n-1)\\times(n-1)$ square with vertices with integer coordinates and obtain, by the same argument as in Case 1, $n^2$ covered integer points. Case 3: exactly two opposite sides cover integers: This is equivalent to Case 1, where by tilting $(n+1)$ integer points miss (side shifted away) and so the result is $(n+1)^2-(n+1)=n(n+1)$.
\nLet $Y$ be an oriented $3$-dimesional manifold, given by a link of a surface quotient singularity. This means that $Y=S^3/G$ for some finite subgroup $G$ of $U(2)$, and the cone $CY=D^4/G$ of $Y$ naturally inherits a complex orbifold structure, with a unique singular point (the quotient singularity).
\nSuppose we have an almost complex structure on the cylinder $Y\\times [0,1]$, which is a smooth, oriented $4$-manifold with boundary $Y\\times \\{0\\} \\amalg Y\\times \\{1\\}$. The orientation convention is that the obvious map $Y\\times \\{0\\} \\amalg Y\\times \\{1\\}\\to -Y\\amalg Y$ is orientation-preserving, where $-Y$ is the orientation reversal of $Y$. Can we extend the almost complex structure on $Y\\times [0,1]$ to an almost complex structure over the orbifold $Y\\times [0,1]\\cup CY$, where $CY$ is attached to the cylinder along $Y\\times \\{0\\}$ (the oriented boundary of $CY$ is $Y$)?
\nThe almost complex structure on $CY$ need not be the one coming from the complex structure, but I would also like to know when we can assume that the almost complex structure on $CY$ is the canonical one.
\nI have a basic question about the nonlinear wave equation. I am looking to prove the existence of a self similar solution to a wave equation; say $u=u(x,t)$\n$$ u_{tt}-\\Delta_x u = |u|^{p-2} u$$ for $ x \\in \\mathbb{R}^N$ and $ t \\in \\mathbb{R}$ or a subset of the time interval (I am looking for backward of forward self similar). In the end I get an elliptic equation for $w(y)$. I have searched a bit and occasionally see some stuff that hints at solving the equation for $|y|<1$ is enough because $|y|=1$ is a characteristic. So my question is: if I solve the elliptic problem up to the boundary and show the solution is continuous on $B_1$ with $ w=0$ on $ \\partial B_1$ is this enough to claim I have found some sort of weak solution to the original problem or do I need $w$ to satisfy some specific BC's on $|y|=1$. Maybe this question doesn't belong here and rather on Mathstack exchange (I really don't know anything about the wave equation). Is there any basic references which talk about this? thanks
\nEDIT. Let me be more specific. I have a pde similar to the above wave equation and when I look for self similar solutions I get some degenerate elliptic problem that degenerate on $|y|=1$ ($y$ is the self similar variable). I can solve this elliptic problem on $|y|<1$ with with $ w=0$ on $ |y|=1$ and maybe also a $ \\partial_\\nu w =0$ on $|y|=1$ and the solution on the elliptic problem has some nice properties. So I am interested if this gives new results on the wave equation and hence I want to know if solving on $B_1$ is sufficient. ChatGPT seems to say for the backward self similar it might be; for the forward it seems to say its not.
\nI am reading about the Simon-Smith min-max method for constructing minimal surfaces in 3-manifolds, following this survey by Colding and De Lellis.
\nHere is what I understand:
\nIn light of this, I am looking for an example of a min-max sequence of surfaces in a 3-manifold, whose limiting varifold is stationary but not a smooth minimal surface. If you can also explain intuitively how the almost minimizing property helps avoiding this situation, that would help me a lot.
\nTake $S^3(\\sqrt{2})\\#S^3(1)$ namely join these two 3-spheres with a tiny neck. Consider a sweepout that starts in the $S^3(1)$ region and at time $t_0$ looks like two crossing equatorial $S^2$'s. Then push one of the $S^2$'s back down and continue sweeping the other one upwards. If you do this correctly it will cross the equator in $S^3(\\sqrt{2})$ at $t_1$. Thus, this sweepout will have two max times, $t_0,t_1$ (both have area $8\\pi$). Both are stationary varifolds. However, the $t_0$ stationary varifold is not smooth (it's crossing of two smooth surfaces).
\nWorking out a picture of how this fails to be almost minimizing will be quite instructive.
\nIf we look at Wikipedia's list of hypergeometric identities, we find classic results such as Saalschütz's theorem—\n$${}_3F_2 (a,b, -n;c, 1+a+b-c-n;1)= \\frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n}$$\n—and Clausen's formula—\n$${}_3F_2(2c-2s-1, 2s, c-1/2; 2c-1, c; x)=\\, {}_2F_1(c-s-1/2,s; c; x)^2.$$\nThe vague question I have, which I will try to make more precise below, is whether all such identities have been discovered, or whether it's possible that a new identity might be discovered in the future.
\nThe first step toward making the question more precise is to specify the precise form of the identities under consideration. Let's say we fix the underlying field of constants to be $\\mathbb{Q}$. The simplest thing I can think of is to fix a finite number of indeterminates $a,b,c,\\ldots$ and allow a linear combination of ${}_mF_n$ expressions where the parameters are affine functions of the indeterminates, and the coefficients are rational functions of the indeterminates and of $x$. The \"constant term\" is trickier to specify; I guess we have to allow an integer parameter $n$ and the constant term will have the property that the ratio between the value of the term at $n+1$ and the value of the term at $n$ is a rational function of the indeterminates. I'm not sure exactly how to get all the details right here, but the point is to come up with the simplest possible family of expressions that includes all the classic identities.
\nHaving specified the class of allowable expressions, we can define some algebra of formal symbols, and hypothesize something like, the known identities generate all possible identities. That is, if we map the formal symbols to actual hypergeometric functions, then the kernel is generated by known identities.
\nAnyway, I'm sure I haven't formulated the question completely correctly, but I'm wondering if anything like this could possibly be true, or even better, whether there's a known theorem along these lines.
\nI did what I should have done before posting to MO, which is to ask Doron Zeilberger directly. Most of what follows is due to him or to references that he pointed to.
\nThere is a distinction between identities such as Saalschütz's theorem, which concern values of a hypergeometric function at a special point (usually a simple rational number), and transformation formulas such as Clausen's formula, which is valid for all arguments.
\nA classification theorem for identities is probably hopeless, since it seems to be possible to generate an endless stream of sporadic examples if you put in enough effort. An early reference, which predates Wilf–Zeilberger, is Strange evaluations of hypergeometric series, by Ira Gessel and Dennis Stanton, SIAM J. Math. Anal. 13 (1982), 295–308. Other references include \nForty \"Strange\" Computer-Discovered [and Computer-Proved (of course!)]\nHypergeometric Series Evaluations by Shalosh B. Ekhad, Special values of the hypergeometric series by Akihito Ebisu (Mem. Amer. Math. Soc. 248 (2017)), and (Strange) gamma evaluations by Wadim Zudilin.
\nBy contrast, transformation formulas are much rarer. A good reference is Beyond the Beta Integral Method: Transformation Formulas for Hypergeometric Functions via Meijer's $G$ Function, by Dmitrii Karp and Elena Prilepkina, Symmetry 14.8 (2022), 1541. Referring to Bailey's classic book, Generalized Hypergeometric Series, Karp and Prilepkina write:
\n\n\nHis student, Lucy Joan Slater, attributes to L. J. Rogers the statement that after Bailey's work, \"nothing remains to be done in the field of hypergeometric series.\"
\n
However, Bailey did not prove any kind of classification theorem, and the point of Karp and Prilepkina's paper is to present a method that enables them to derive a large number of transformation formulas. They write:
\n\n\nIt is typically rather difficult to claim that a hypergeometric\ntransformation is new, as the literature is vast and there could always be a hidden trick\nas to how a “new” transformation can be derived from a known one. Hence, we simply\npresent all the formulas that we identified as interesting with the hope that some of them\nare indeed new.
\n
From this we can infer that even if we restrict to transformation formulas, there is no known theorem of the kind I was asking for. Perhaps, though, it is not hopeless for some such theorem to eventually be proved—though of course one would first need to state a precise conjecture, and even that does not seem to have been done.
\nThis may well be what you are looking for:
\nH. S. Wilf & D. Zeilberger, An algorithmic proof theory for hypergeometric\n(ordinary and \"q\") multisum/integral identities (1992).\nTheir finding is that \"a large class of hypergeometric identities can be embedded in a class of holonomic function identities, the elements of which are specifiable by a finite amount of data.\"
\nMy understanding is that these identities are generated by a finite basis of differential/recurrence relations, at least for fixed parameters; if you allow arbitrary shifts in parameters and arguments the set of hypergeometric identities is not finitely generated.
At the risk of sinking out of my depth, let me try and follow up this question whether \"the set of all hypergeometric identities is finitely generated\". The classic example is the set of contiguous relations of Gauss hypergeometric functions, which are generated by a finite set of 15 primitive contiguous relations. My understanding of the WZ paper is that this \"finite generation\" is generalized to a much broader class of hypergeometric identities, of the form\n$$S(\\mathbf{\\alpha})=\\sum_k F(k,\\mathbf{\\alpha}),\\tag{1}$$\nwhere $F(k,\\alpha)$ is a general hypergeometric function in $k$, depending on parameters $\\mathbf{\\alpha}=(\\alpha_1,\\alpha_2,\\ldots\\alpha_m)$. \"Hypergeometric\" in this context means that $F(k+1,\\mathbf{\\alpha})/F(k,\\mathbf{\\alpha})$ is a rational function of $k$. The sum over $k$ is equated to an explicit closed form $S(\\mathbf{\\alpha})$. The WZ‑algorithm determines all linear recurrence relations in $\\mathbf{\\alpha}$ satisfied by $S(\\mathbf{\\alpha})$ and finds that these recurrences are generated by a finite set of primitive recurrences. In this algorithmic sense the infinite set of identities of the form (1) is \"finitely generated\".
\nAs far as I have been able to pick up from the literature, the requirement of a rational term ratio is essential for the finite generation. Identities for q-hypergeometric functions, with a term ratio that is rational in $q^k$ rather than in $k$, have infinitely many primitive identities.
\nBasic question: Is it correct that the $p$-adic Langlands correspondence is known for $GL_2$ only over $Q_p$ but not other $p$-adic fields? If so, I would like to request some light to be shed on this restriction, i.e., why only $Q_p$ but not its extensions. Any reference for this (and prospects) would be much appreciated.
\nThanks!
\nYes, this is correct.
\nThe problem is that when you replace $Q_p$ by an extension, the dimension of $GL_2(F)$ as a $p$-adic analytic group increases. This also means that the cohomological dimension of its open subgroups increases. This leads to representation theory of $GL_2(F)$\nof being much more complicated than $GL_2(Q_p)$. For example, smooth irreducible $\\overline{\\mathbb F}_p$-representations have not been classified if $F\\neq Q_p$.
\nPrototypical examle: Let $\\mathbb G=\\mathbb G_a$, and let $K=\\mathbb{G}(\\mathcal O_F)$.\nSo that $K$ is $(\\mathcal O_F, +)$. Then the completed group agebra $\\mathcal O[[K]]$\nis isomorphic to $\\mathcal O[[x_1, ..., x_d]]$, where $d=[F:Q_p]$, where $\\mathcal O$ is a ring of inegers in a finite extension of $Q_p$. The theory of modules of $\\mathcal O[[K]]$ is much easier, when $d=1$. If you want to see this in action have a look at\nEmerton's\n\"On a class of coherent rings, with applications to the smooth representation theory of GL_2(Q_p) in characteristic p\", available on his\n website .
\nSince $GL_2(F)$ is locally pro-$p$ this problem does not arise if you are working over $\\mathbb C$ or $\\mathbb F_l$, $l\\neq p$.
\nRegarding prospects for extending the correspondence to $GL_2(F)$ for other $F$,\none could look at Paškūnas's papers \"Coefficient systems and supersingular representations\", \"Towards a modulo $p$ Langlands correspondence for $GL_2(F)$\" (joint with C. Breuil),\nand \"Admissible unitary completions of locally $\\mathbb Q_p$-rational representations of\n$GL_2(F)$\", available on his website and/or the arXiv.
\nThere is also Breuil's ICM talk from last summer, \"The emerging $p$-adic Langlands program\", available at his website.\nThis gives a very nice survey of the whole state of the theory (which has remained relatively stable since then).
\nSome commentary on Paškūnas's papers: In the $GL_2(\\mathbb Q_p)$ case, Breuil found that the numbers of irred. supersingular reps. of $GL_2(\\mathbb Q_p)$ mod $p$ matches with the numbers of $2$-dim'l irred. mod $p$ reps. of $G_{\\mathbb Q_p}$, and that there is even a natural way to match them (which is e.g. compatible with Serre's conjecture on weights of modular forms giving rise to mod $p$ global Galois reps.).
\nIt was then natural to conjecture that the same was true for $GL_2(F)$.\nThe first of these papers has the goal of verifying this conjecture. Indeed,\nit succeeds in constructing the right number\nof supersingular reps. mod $p$ of $GL_2(F)$. However, it was later realized that there was\nno way to match these with irred. Galois reps. in any way that is compatible with the Buzzard--Diamond--Jarvis (BDJ) conjecture (the generalization of Serre's conjecture to Hilbert modular forms).
\nThe second paper extends the techniques of the first, and shows in fact that when\n$F \\neq \\mathbb Q_p$ there are many, many more supersingulars than there are $2$-dim'l.\nirreps of $G_F$. It attempts to find order among this chaos by identifying certain classes of supersingulars which seem to have something to do with the Galois side (in the sense\nthat they match with the predictions of the BDJ conjecture).
\nThe third paper shows how to lift mod $p$ representations to $p$-adic Banach spaces\nrepresentations in interesting ways, and so can be thought of as (i) giving some evidence\nthat there will be a $p$-adic local Langlands for $GL_2(F)$, but also (ii) showing that\nunderstanding it will be at least as difficult as understanding the mod $p$ situation.
\nI am working with a binomial sum that arises in some combinatorial arguments (and also appears in certain generating‐function manipulations). Specifically, I have this identity
\n$$\n\\sum_{j=0}^{n}\n\\binom{k}{j}\\,\\binom{k}{n-j}\n\\;=\\;\n\\sum_{j=0}^{n}\n\\,(-1)^{\\,n-j}\n\\,\\frac{k-n}\n{\\displaystyle k - j}\\displaystyle \\binom{n}{j}\\,\\binom{k-1}{n}\\,\\binom{j+k}{n}\n\\,\\;=\\;\n\\binom{2k}{n}\\\n$$
\nI’m trying to understand why these two sums are equal from a combinatorial perspective, and also which binomial-coefficient identities or transformations are used to rewrite one side into the other.
\nIs there a known binomial identity that takes us directly from one to the other, or do I need a more intricate counting argument? Any insights or references to a standard identity would be really helpful.
\nIf it's a new one please propose a name for it
\nThank you everyone
\nThe identity is a special case of Saalschütz's theorem.\nThe second sum may be written as\n$$(-1)^n \\binom{k-1}{n}^2{}_3F_2\\left({-n,-k,k+1\\atop -k+1,k+1-n} \\biggm|1\\right)$$\nand the $_3F_2$ can be evaluated by Saalschütz's theorem.
\nAdded later: I missed Hjalmar Rosengren's comment, in which he said the same thing earlier.
\nI arrived late to this game. Let's prove the identity with the help of the so-called Wilf-Zeilberger method of automatic proof.
\nTo this end, divide through by the factor $\\binom{2k}n$ so that we show\nthe two identities\n\\begin{align*}\n\\sum_{j=0}^{n}\n\\frac{\\binom{k}{j}\\binom{k}{n-j}}{\\binom{2k}n}=1 \\qquad \\text{and} \\qquad\n\\sum_{j=0}^{n}\n(-1)^{n-j}\n\\frac{(k-n)\\binom{n}{j}\\binom{k}{n}\\binom{j+k}{n}}{(k-j)\\binom{2k}n}=1.\n\\end{align*}\nDefine the functions $F_1(n,j):=\\frac{\\binom{k}{j}\\binom{k}{n-j}}{\\binom{2k}n}$\nand $G_1(n,j):= - \\frac12 \\frac{\\binom{k-1}{j-1}\\binom{k}{n+1-j}}{\\binom{2k-1}n}$. One can routinely check that $F_1(n+1,j)-F_1(n,j)=G_1(n,j+1)-G_1(n,j)$. Then, sum both sides over all integers $j$ (actually these are finite sums). The immediate impact is: the right-hand side vanishes. So, we obtain that $f_1(n+1)=f_1(n)$ where $f_1(n):=\\sum_{j=0}^n F_1(n,j)$. Compute the value at, say $n=0$ (and get $f_1(0)=1$) to confirm the first promised identity.
\nI'll leave the 2nd identity because the procedure is very similar.
\n\n\nI want to know if the right side is a new identity, before starting to write a chapter about this in my article
\n
This identity is known to Mathematica:
\n
Let $\\Omega = \\mathbb B_n,$ the unit ball in $\\mathbb C^n$ and $L^2_a(\\Omega)$ be the Bergman space endowed with the normalized volume measure on $\\Omega.$ Let $k_{\\lambda}$ be the associated Bergman reproducing kernel. Let $\\varphi_{\\lambda}$ be an analytic automorphism of $\\Omega$ having the properties $:$
\n\n\n$(1)$ $\\varphi_{\\lambda} (\\lambda) = 0$
\n$(2)$ $\\varphi_{\\lambda} \\circ \\varphi_{\\lambda} = \\text{Id}_{\\Omega}.$
\n
It is proved in Section $2.2$ in Rudin's Function Theory in the Unit Ball of $\\mathbb C^n$ that the real Jacobian of $\\varphi_{\\lambda}$ has the following form $:$\n$$J_{\\mathbb R} (\\varphi_{\\lambda}) (z) = \\frac {\\left \\lvert k_{\\lambda} (z) \\right \\rvert^2} {k_{\\lambda} (\\lambda)}.$$
\nFor a given $f \\in L^{\\infty} (\\Omega)$ we define the Berezin transform $\\widetilde f$ of $f$ in the following way $:$
\n$$\\begin{align*} \\widetilde {f} (\\lambda) & = \\left \\langle f \\frac {k_{\\lambda}} {\\|k_{\\lambda}\\|_2}, \\frac {k_{\\lambda}} {\\|k_{\\lambda}\\|_2} \\right \\rangle \\\\ & = \\frac {1} {k_{\\lambda} (\\lambda)} \\int_{\\Omega} f(z)\\ \\left \\lvert k_{\\lambda} (z) \\right \\rvert^2\\ dV(z) \\end{align*}$$
\nBy change of variable formula we have $$\\widetilde {f} (\\lambda) = \\int_{\\Omega} f \\circ \\varphi_{\\lambda}\\ dV.$$
\nIn Corollary $10$ of the paper Toeplitz and Hankel Operators on Bergman Spaces authored by Karel Stroethoff and Dechao Zheng it is claimed that if $P$ denotes the Bergman projection then
\n$$\\left\\|P\\left(f\\circ\\varphi_{\\lambda}\\right)-\\widetilde{f}(\\lambda)\\right\\|_p=\\left\\|P\\left(f\\circ\\varphi_{\\lambda}-\\overline{P\\left(\\overline{f}\\circ\\varphi_{\\lambda}\\right)}\\right)\\right\\|_p.$$
\nBut I am having hard time following this claim. Any help in this regard would be warmly appreciated.
\nThanks for your time.
\nI think it goes like this.
\n\\begin{align*}\nP \\big(\\overline{P(\\overline{f} \\circ \\varphi_\\lambda}) \\big) (w) & = \\int_\\Omega \\overline{P(\\overline{f} \\circ \\varphi_\\lambda}) (z) \\overline{k_w(z)} dV(z) \\\\ \n& = \\int_\\Omega \\int_\\Omega f \\circ\\varphi_\\lambda(\\eta) k_z(\\eta) dV(\\eta) \\overline{k_w(z)} dV(z) \\\\\n& = \\int_\\Omega f \\circ \\varphi_\\lambda(\\eta) \\overline{\\int_\\Omega k_\\eta(z)k_w(z) dV(z)} dV(\\eta) \\\\ \n& = \\int_\\Omega f \\circ \\varphi_\\lambda (\\eta) dV(\\eta) = \\tilde{f}(\\lambda).\n\\end{align*}
\nThe second to last inequality comes from the fact that for the unit ball and the polydisc if $F$ is a holomorphic function and $dV$ is the normalized area measure then $\\int_\\Omega F(z) dV(z) = F(0) $.
\nIn Grothendieck's Sketch of a Programme he spends a few pages discussing polyhedra over arbitrary rings and concludes with some intriguing remarks on specializing polyhedra over their \"most singular characteristics\". I am having trouble understanding what he means or finding other references.
\nHere is a link to the version of Sketch I was reading.\nSketch of a Programme
\nMore specifically, Grothendieck begins (p. 252):
\n\n\nIn 1977 and 1978, in parallel with two C4 courses on the geometry of the cube and that of the icosahedron, I began to become interested in regular polyhedra, which then appeared to me as particularly concrete “geometric realizations” of combinatorial maps, the vertices, edges and faces being realised as points, lines and plans respectively in a suitable 3-dimensional affine space, and respecting incidence relations. This notion of a geometric realisation of a combinatorial map keeps its meaning over an arbitrary base field, and even over an arbitrary base ring. It also keeps its meaning …
\n
My understanding is that he is viewing the polyhedron as a configuration of affine subspaces—what does he mean by \"combinatorial maps\"? This seems to be the key to understanding the last line, on the concept making sense over an arbitrary ring.
\nThe most curious remark comes at the end of this section where he writes (p. 255):
\n\n\n… on the already known cases. Thus, examining the Pythagorean polyhedra one after the other, I saw that the same small miracle was repeated each time, which I called the combinatorial paradigm [underlined in original] of the polyhedra under consideration. Roughly speaking, it can be described by saying that when we consider the specialisation of the polyhedra in the or one of the most singular characteristic(s) (namely characteristics 2 and 5 for the icosahedron, characteristic 2 for the octahedron), we read off from the geometric regular polyhedron over the finite field ($\\mathbb F_4$ and $\\mathbb F_5$ for the icosahedron, $\\mathbb F_2$ for the octahedron) a particularly elegant (and unexpected) description of the combinatorics of the polyhedron. It seems to me that I perceived there a principle of great generality, which I believed I found again for example in a later reflection on the combinatorics of the system of 27 lines on a cubic surface, and its relations with the root system $E_7$. Whether it happens that such a principle really exists, and even that we succeed in uncovering it from its cloak of fog, or that it recedes as we pursue it and ends up vanishing like a Fata Morgana, I find in it for my part a force of motivation, a rare fascination, perhaps similar to that of dreams. No doubt that following such an unformulated call, the unformulated seeking form, from an elusive glimpse which seems to take pleasure in simultaneously hiding and revealing itself — can only lead far, although no one could predict where…
\n
How is he determining the \"most singular characteristics\" of these polyhedra? If I can understand the combinatorial maps comment above, it may make more sense how he is considering the specializations of icosahedra over finite fields. What is the elegant description of their combinatorics Grothendieck refers to?
\nIf anyone can explain Grothendieck's comments or point to other references, I would be appreciative.
\nNot an exact answer to your question but maybe it can be helpful to know:
\nLa théorie combinatoire de l'icosaèdre by V Diekert
\nThis was a \"Rapport pour un DES d'université de l'année 1977/78\" under Grothendieck.
\nI have been exploring a combinatorial approach to express Chebyshev polynomials and generalizing them through a recurrence relation. I would like to know whether this recurrence relation can be proven using induction or a combinatorial approach.
\n\\begin{eqnarray}\n\\Re[e^{in\\theta}] &=& \\cos(n\\theta) \\\\\n&=& \\Re\\left[\\sum_{k=0}^{n} \\binom{n}{k} \\cos^{n-k}(\\theta) \\cdot (i \\sin \\theta)^k\\right] \\\\\n&=& \\Re\\left[\\sum_{0 \\leq 2k \\leq n} \\binom{n}{2k} \\cos^{n-2k}(\\theta) \\cdot (-\\sin^2 \\theta)^k\\right] \\\\\n&=& \\sum_{0 \\leq 2k \\leq n} \\binom{n}{2k} \\cos^{n-2k}(\\theta) \\cdot (\\cos^2 \\theta - 1)^k \\\\\n&=& \\sum_{0 \\leq 2j \\leq n} (-1)^j \\cos^{n-2j}(\\theta) \\sum_{2j \\leq 2k \\leq n} \\binom{n}{2k} \\binom{k}{j}.\n\\end{eqnarray}
\nThus, we can express $\\cos(n\\theta)$ in terms of powers of $\\cos(\\theta)$. Substituting $\\cos(\\theta) = x$, we obtain a polynomial, proving the existence of $T_n(x)$:
\n[Symbolic Representation of Chebyshev Polynomials]\n\\begin{eqnarray}\nT_n(x) = \\sum_{0 \\leq 2j \\leq n} (-1)^j x^{n-2j} \\sum_{2j \\leq 2k \\leq n} \\binom{n}{2k} \\binom{k}{j}.\n\\end{eqnarray}
\nThe recursive relation for Chebyshev polynomials $T_n(x)$ is given by:\n\\begin{eqnarray}\nT_n(x) = 2xT_{n-1}(x) - T_{n-2}(x)\n\\end{eqnarray}\nwith initial conditions:\n$$T_0(x) = 1, \\quad T_1(x) = x.$$
\nUsing \"constructive\" approaches, we can generalize the Chebyshev polynomials by starting with the combinatorial representation of $T_n(x)$.
\n[Generalized Symbolic Chebyshev Polynomials]\n\\begin{eqnarray}\nG_{n,m,d}(h,t) = \\sum_{0 \\leq mj \\leq n} \\sum_{mj \\leq mp \\leq n-dj} \\binom{n-dj}{mp} \\binom{p}{j} h^{n-j(d+2)} t^j.\n\\end{eqnarray}
\nIf $m = 2p$, where $p \\in \\mathbb{N}$, then:\n\\begin{eqnarray}\nG_{n,m,d}(h,t) = \\left(\\sum_{k=1}^{m-1} \\binom{m}{k} h^k (-1)^{k+1} G_{n-k,m,d}(h,t)\\right) + h^{m-2} t G_{n-m-d,m,d}(h,t).\n\\end{eqnarray}
\nIf $m = 2p + 1$, where $p \\in \\mathbb{N}$, then:\n\\begin{eqnarray}\nG_{n,m,d}(h,t) = \\left(\\sum_{k=1}^{m-1} \\binom{m}{k} h^k (-1)^{k+1} G_{n-k,m,d}(h,t)\\right) + 2h^m G_{n-m,m,d}(h,t) + h^{m-2} t G_{n-m-d,m,d}(h,t).\n\\end{eqnarray}
\nTo define the first $m+d$ terms, we use the explicit formula. Additionally, when evaluating the function $G$ for certain parameters, we obtain the following equality:\n$$G_{n,2,0}(x,-1) = T_n(x).$$
\nQuestions:
\nMore generally, is this line of research valuable? I am not a mathematician and I am working alone without support (I don't have formal guidance from peers), so I wonder whether this is a direction worth exploring further.
\nAny insights or references would be greatly appreciated!
\nI want to share all my work with the community (this only one chapter), but it's not ready and I’m struggling with writing my article because I don’t know which parts are important for other mathematicians. I find it hard to structure everything clearly because, to me, every aspect feels important. Additionally, since a lot of my work is constructive, there are actual proofs, but I struggle to write them down properly. Initially, I was working on computational irreducibility in cellular automata, like Rule 30.
\nExample of other identity involved in the article Binomial coefficient C(2k,n-1) alternative formula equivalent to the Vandermonde identity?
\nExample of cellular automaton generated when evaluating G with random parameters:
\n![\"Example](\"https://i.sstatic.net/rWyK8kZ1.png\"/)
Update 3/12/2025:
\nWe can derive the general term of the Fibonacci sequence (or K-bonacci sequence) defined by the recurrence relation:
\n\\begin{align}\nf_{n,k+1} &= \\sum_{j=1}^{k+1} f_{n-j,k+1}\\\\\n f_{n,k+1} &= \\sum _ {j=0}^{\\left\\lfloor\\frac{n}{k+1}\\right \\rfloor } (-1)^{j+1} 2^{n-j (k+2)} \\left(\\binom{n-(k+1) j}{j}-2 \\binom{-((k+1) j)+n+1}{j} \\right) \\\\\n&= \\sum _ {j=0}^{\\left\\lfloor \\frac{n+1}{2}\\right \\rfloor } \\left(\\sum _ {p=j}^{\\left\\lfloor \\frac{1}{2} (-j k+n+1)\\right \\rfloor } (-1)^j \\binom{p}{j} \\binom{-j k+n+1}{2 p} \\right) \\\\\n&= \\sum _ {j=0}^{\\left\\lfloor \\frac{n+1}{2}\\right \\rfloor } (-1)^j \\bigg(-\\binom{-j k+n+1}{2 \\left\\lfloor \\frac{1}{2} (-j k+n+1)\\right \\rfloor } \\binom{\\left\\lfloor \\frac{1}{2} (-j k+n+1)\\right \\rfloor }{j} \\, \\\\\n&\\quad \\,_3F_2\\left(1,\\frac{j k}{2}+\\left\\lfloor \\frac{1}{2} (-j k+n+1)\\right \\rfloor -\\frac{n}{2}-\\frac{1}{2},\\frac{j k}{2}+\\left\\lfloor \\frac{1}{2} (-j k+n+1)\\right \\rfloor -\\frac{n}{2};\\right. \\\\\n&\\quad \\left. \\left\\lfloor \\frac{1}{2} (-j k+n+1)\\right \\rfloor +\\frac{1}{2},-j+\\left\\lfloor \\frac{1}{2} (-j k+n+1)\\right \\rfloor +1;1\\right) \\\\\n&\\quad +\\binom{-j k+n+1}{2 \\left\\lfloor \\frac{1}{2} (-j k+n+1)\\right \\rfloor } \\binom{\\left\\lfloor \\frac{1}{2} (-j k+n+1)\\right \\rfloor }{j} \\\\\n&\\quad +\\binom{-j k+n+1}{2 j} \\, \\,_2F_1\\left(\\frac{1}{2} ((k+2) j-n-1),\\frac{k j}{2}+j-\\frac{n}{2};j+\\frac{1}{2};1\\right) \\bigg).\n\\end{align}
\nProof. Given the generalized Chebyshev polynomial:\n$$G_{n,m,d}(h,t) = \\sum_{0 \\leq mj \\leq n} \\sum_{mj \\leq ml \\leq n-dj} \\binom{n-dj}{ml} \\binom{l}{j} h^{n-j(d+2)} t^j$$\nThe recurrence relation is defined by the parity of $m$:\n
\nIf $m = 2p$ ($p \\in \\mathbb{N}$):\n$$G_{n,m,d}(h,t) = \\left(\\sum_{k=1}^{m-1} \\binom{m}{k} h^k (-1)^{k+1} G_{n-k,m,d}(h,t)\\right) + h^{m-2} t G_{n-m-d,m,d}(h,t)$$\n
\nIf $m = 2p + 1$ ($p \\in \\mathbb{N}$):\n$$G_{n,m,d}(h,t) = \\left(\\sum_{k=1}^{m-1} \\binom{m}{k} h^k (-1)^{k+1} G_{n-k,m,d}(h,t)\\right) + 2h^m G_{n-m,m,d}(h,t) + h^{m-2} t G_{n-m-d,m,d}(h,t)$$
To establish this recurrence, we employ a finite difference approach combined with a classic binomial identity. Let us rename the inner summation index from $p$ to $l$ to avoid confusion with the parity parameter $p$ where $m = 2p$ or $m = 2p+1$.
\nStep 1: Set Up the m-th Difference Sum
\nWe evaluate the following sum, which represents the m-th order finite difference, which we will call $S$:
\n$$S = \\sum_{k=0}^m \\binom{m}{k} (-h)^k G_{n-k,m,d}(h,t)$$\nSubstitute the definition of $G_{n-k,m,d}(h,t)$ into $S$:
\n$$S = \\sum_{k=0}^m \\binom{m}{k} (-h)^k \\sum_{j} \\sum_{l} \\binom{n-k-dj}{ml} \\binom{l}{j} h^{n-k-j(d+2)} t^j$$\nThe powers of $h$ combine such that $(-h)^k \\cdot h^{n-k-j(d+2)} = (-1)^k h^{n-j(d+2)}$. Swapping the order of summation and pulling the $h$ and $t$ terms out:
\n$$S = \\sum_{j} \\sum_{l} \\binom{l}{j} h^{n-j(d+2)} t^j \\left[ \\sum_{k=0}^m (-1)^k \\binom{m}{k} \\binom{n-dj-k}{ml} \\right]$$\nStep 2: Apply the Binomial Identity
\nThe inner bracket is a known binomial identity related to the m-th difference of a polynomial. For any integers $A$ and $B$:
\n$$\\sum_{k=0}^m (-1)^k \\binom{m}{k} \\binom{A-k}{B} = \\binom{A-m}{B-m}$$\nBy setting $A = n-dj$ and $B = ml$, the bracket simplifies to:
\n$$\\binom{n-dj-m}{ml-m} = \\binom{n-dj-m}{m(l-1)}$$\nSubstituting this back into $S$:
\n$$S = \\sum_{j} \\sum_{l} \\binom{n-dj-m}{m(l-1)} \\binom{l}{j} h^{n-j(d+2)} t^j$$\nStep 3: Index Shifting and Pascal's Rule
\nShift the index $l$ down by 1. Let $l' = l - 1$, so $l = l' + 1$:
\n$$S = \\sum_{j} \\sum_{l'} \\binom{n-m-dj}{ml'} \\binom{l'+1}{j} h^{n-j(d+2)} t^j$$\nUsing Pascal's Identity, $\\binom{l'+1}{j} = \\binom{l'}{j} + \\binom{l'}{j-1}$. This splits $S$ into two separate sums, $S_1$ and $S_2$:
\n$$S = S_1 + S_2$$\nEvaluating $S_1$:
\n$$S_1 = \\sum_{j} \\sum_{l'} \\binom{n-m-dj}{ml'} \\binom{l'}{j} h^{n-j(d+2)} t^j$$\nBy factoring $h^m$ out of the exponent where $n-j(d+2) = (n-m)-j(d+2)+m$, we get the definition of $G_{n-m}$:
\n$$S_1 = h^m G_{n-m, m, d}(h,t)$$\nEvaluating $S_2$:
\n$$S_2 = \\sum_{j} \\sum_{l'} \\binom{n-m-dj}{ml'} \\binom{l'}{j-1} h^{n-j(d+2)} t^j$$\nShift the index $j$ down by 1. Let $j' = j - 1$, so $j = j' + 1$:
\n$$S_2 = \\sum_{j'} \\sum_{l'} \\binom{n-m-d(j'+1)}{ml'} \\binom{l'}{j'} h^{n-(j'+1)(d+2)} t^{j'+1}$$\nSimplifying the exponents and binomial top, and factoring out $h^{m-2}t$, we are left with the definition of $G_{n-m-d}$:
\n$$S_2 = h^{m-2} t G_{n-m-d, m, d}(h,t)$$\nStep 4: Isolate $G_n$ and Evaluate Parity
\nWe have proven that:
\n$$\\sum_{k=0}^m \\binom{m}{k} (-h)^k G_{n-k} = h^m G_{n-m} + h^{m-2} t G_{n-m-d}$$\nPull the $k=0$ term out of the sum, and move the rest to the right side:
\n$$G_n = -\\sum_{k=1}^{m} \\binom{m}{k} (-h)^k G_{n-k} + h^m G_{n-m} + h^{m-2} t G_{n-m-d}$$\nExtract the $k=m$ term from the sum so it matches the upper bound $m-1$:
\n$$G_n = \\sum_{k=1}^{m-1} \\binom{m}{k} (-1)^{k+1} h^k G_{n-k} - (-1)^m h^m G_{n-m} + h^m G_{n-m} + h^{m-2} t G_{n-m-d}$$\nCombine the $G_{n-m}$ terms:
\n$$G_n = \\sum_{k=1}^{m-1} \\binom{m}{k} (-1)^{k+1} h^k G_{n-k} + h^m(1 - (-1)^m) G_{n-m} + h^{m-2} t G_{n-m-d}$$\nThe recurrence depends on the parity of $m$:
\nIn the diagram, circles of the same color are congruent.
\n![\"Sequence](\"https://i.sstatic.net/4XXO2ILj.png\")
I have a nonintuitive proof that the radii converge. Is there an intuitive explanation?
\nAs noted in the comment, apparently I misunderstood the question. I did not notice the ellipsis in the picture in the OP and therefore thought that the process begins from the red circles and goes right-to-left.
\nSo, I will retain the right-to-left-interpretation solution but will now add -- based on the same idea -- a simple and complete left-to-right-interpretation proof.
\nFirst here is the original, right-to-left solution, given while apparently misunderstanding the question:
\nThe radii will be decreasing (and hence converging) iff the angle between the line through the centers of the red circles with the negative horizontal axis will be $<\\pi/3$.
\nSo, it is enough to show the angle between the line through the centers of the green circles with the negative horizontal axis is less than the angle between the line through the centers of the red circles with the negative horizontal axis.
\nBut clearly $a<\\pi/3<b$ in the picture below, so that $a<b$. $\\quad\\Box$
\n![\"enter](\"https://i.sstatic.net/Y1S9Q0x7.png\")
Now, with the ellipsis noticed, it is easy to adapt the idea used above, to the opposite, left-to-right direction, to a get a simple complete proof of the apparently desired result. Indeed, now let $a_n:=a<\\pi/3<b=:b_n$, and also let $r_n$ and $r_{n+1}$ denote the respective radii of the green and red circles. Then the angle between the line through the centers of the green circles and the line through the centers of the red circles is $b_n-a_n$, so that $\\sum_n(b_n-a_n)<\\infty$ and hence $a_n,b_n\\to\\pi/3$. On the other hand, considering the two isosceles triangles in the picture above and using the first-order Taylor expansion of $\\cos$ at $\\pi/3$, we see that\n$$\\frac{r_{n+1}}{r_n}=\\frac{\\cos a_n}{\\cos b_n}\n=\\exp\\big((\\sqrt3+o(1))\\,(b_n-a_n)\\big).$$\nIt remains to recall that $\\sum_n(b_n-a_n)<\\infty$. $\\quad\\Box$
\nThe intuitive explanation is that the sequence approaches the hexagonal close packing, two staggered rows of identical circles:
\n![](\"https://i.sstatic.net/yrwWWty0.png\")
\nBecause the sequence of radii is monotonically increasing and bounded by this \"perfect fit\" geometry, it must converge to a finite limit.
\nHere is a proof that the radii converge without using any big formulas.
\nLet $r>0$, and consider the following setup in $\\mathbb{R}^2$, where the two smaller circles have radius $r$:
\n![](\"https://i.sstatic.net/pBFu0tkf.png\")
\nThe angles $\\alpha(r),\\beta(r)$ are decreasing with $r$, in particular $\\alpha'(r),\\beta'(r)<0$ for all $r\\in(0,\\infty)$. Let $\\gamma(r)=\\alpha(r)+\\beta(r)-\\pi/2$.
\nNow, in your figure, let $A_n,B_n$ be the centers of the $n^{\\text{th}}$ circle, $r_n$ the $n^{\\text{th}}$ radius and $\\alpha_n$ the angle between the the $x$-axis and the ray $B_nA_n$ (so in the picture $\\alpha_0=\\pi$ and the sequence $(\\alpha_n)$ looks decreasing in the first few steps). Note the function $f:(0,\\pi)\\to(0,\\pi)$ that gives $\\alpha_{n+1}$ in terms of $\\alpha_n$ has fixed point $f(2\\pi/3)=2\\pi/3$ (see Carlo Beenaker's answer), and $f(\\pi)<\\pi$. Moreover, $f$ is increasing in $[2\\pi/3,\\pi]$; this is equivalent to saying that the angle $\\theta(r)$ from the picture is increasing whenever it is defined, which is true because when $r$ increases, the point $F$ moves clockwise.
\nThus, $\\alpha_n$ is decreasing and converges when $n\\to\\infty$, and $r_{n+1}/r_n\\to1$. But on the other hand, we have that $\\alpha_{n+1}=\\alpha_n+\\gamma(r_{n+1}/r_n)$. As $\\gamma'(1)<0$, there is $\\varepsilon>0$ such that, for big enough $n$, we have $|\\alpha_{n+1}-\\alpha_n|>\\varepsilon\\left|1-\\frac{r_{n+1}}{r_n}\\right|$.
\nThus, as $\\sum_{n}|\\alpha_{n+1}-\\alpha_n|<\\infty$, we also have $\\sum_n\\left|1-\\frac{r_{n+1}}{r_n}\\right|<\\infty$, which implies $\\prod_n\\frac{r_{n+1}}{r_n}<\\infty$, that is, the radii converge.
\nThis proof is closely related to Saul RM's one, but I think it's a bit simpler to see what's going on.
\nFor $0\\leq \\alpha<\\pi/2$, let $f(\\alpha)$ be the ratio of radii of circles when they are placed in the following configuration:
\n![\"enter](\"https://i.sstatic.net/TQGFkqJj.png\")
One can compute $f(\\alpha)$ explicitly in terms of trigonometric functions, but all we need to know is that it is differentiable at $0$, which hopefully is believable. This in particular implies that for $\\alpha$ small and some constant $C$ (which we can take to be $2f'(\\alpha)$) we have\n$$f(\\alpha)\\leq f(0)+C\\alpha=1+C\\alpha.$$
\nConsidering now the diagram in the OP, let $r_n$ be the radius of the $n$-th pair of circles, and $\\alpha_n$ to be the angle between the lines connecting the centers of the $n$-th and $n+1$-st pair. By the definition of $f$, this means $r_n=r_{n-1}\\cdot f(\\alpha_{n-1})$, and so we find\n$$r_n=r_1\\cdot\\prod_{i=1}^{n-1}f(\\alpha_i)\\leq r_1\\cdot\\prod_{i=1}^{n-1}(1+C\\alpha_i).$$\nBy a well-known connection between convergence of sums and products (easily shown with help of the inequality $1+x\\leq e^x$) this product converges as $n\\to\\infty$ iff the sum $\\sum_{i=1}^\\infty\\alpha_i$ converges. But in our case, the sum $\\sum_{i=1}^{n-1}\\alpha_i$ represents the angle between the line connecting the centers of the $n$-th pair of circles with the horizontal line, and it's not hard to convince yourself that this angle tends to $\\pi/3$, giving the desired convergence.
\nGiven functors $F : \\mathbf A \\to \\mathbf C$ and $G : \\mathbf B \\to \\mathbf C$, it is well known that the comma category $F \\downarrow G$ is cocomplete assuming that $\\mathbf A$ and $\\mathbf B$ are cocomplete and $F$ is cocontinuous, in which case the projection functors are also cocontinuous.
\nI would like a reference for the existence of colimits in comma categories in which $F$ is not assumed cocontinuous (in which case the projections will no longer be necessarily cocontinuous).
\nIn the case where $F$ is an endofunctor and $G$ is the identity, one reference is section 14.1 of Kelly's A unified treatment of transfinite constructions [...] (and Kelly's construction works more generally for $F$ an arbitrary functor) but I would like a reference in which $G$ may not be the identity.
\nI'm looking for a source to help me better understand Shelah-Harrington-Makkai's proof of Vaught's conjecture for $\\omega$-stable theories (DOI, link at Shelah's page). An obvious candidate is Makkai's survey paper DOI link, but I felt it was pretty dense (and maybe I need to see a little bit more of motivation behind all the definitions).
\nSo are there other sources that would enable me to understand the proof and maybe unpack things a little more than Makkai (or use a more modern exposition/viewpoint)? I feel that Baldwin's Fundamentals of Stability Theory could be good, but the more the better.
\nCross-posted from Math.SE.
\nWe try to go a step beyond Cutting squares and rectangles into mutually similar but non-congruent rectangles and record a couple of more questions.
\n![\"enter](\"https://i.sstatic.net/JpvRxwU2.png\")
This construction easily generalizes into cutting the same big rectangle into 5 pieces (simply cut the smallest rectangle in the above layout into 3 pieces in exactly the same manner) and 7 pieces and so on... Question: Are there rectangles that can be cut into n mutually similar and non-congruent rectangles that are all similar to the big rectangle where n is some even number? If so, is there any rectangle that can be so cut for all even values with n>2? And for even and odd values of n?
\nFurther question: Is there any rectangle that can be cut into n mutually self-similar rectangles that are not similar to the big rectangle for any positive integer value of n greater than 2?
\nAs noted in a comment, but for completeness: if we partition into two rectangles which are similar to each other, they cannot also be similar to the original rectangle.
\nIf we extend the same construction of a slicing floorplan from three to four rooms:
\n c a^2 1\n+-------------------------------------+-----+-+\n| | | | a\n| +-----+-+\n| | |\n| | |\n| | |\n| | | b\n| | |\n| | |\n| | |\n+-------------------------------------+-------+\nwe get the constraints \\begin{eqnarray*}\\frac{b}{a^2+1} &=& a^{\\pm 1} \\\\\n\\frac{c}{a+b} &=& a^{\\pm 1} \\\\\n\\frac{1+a^2+c}{a+b} &=& a^{\\pm 1}\n\\end{eqnarray*}
\nClearly the third ratio is larger than the second so if we take $a > 1$ we have \\begin{eqnarray*}\\frac{c}{a+b} &=& a^{-1} \\\\\n\\frac{1+a^2+c}{a+b} &=& a\n\\end{eqnarray*} and we eliminate $c$ to get \\begin{eqnarray*}c &=& 1 + a^{-1}b \\\\\n2a+b &=& a^2b\n\\end{eqnarray*}
\nTaking the case $\\frac{b}{a^2+1} = a^{-1}$ gives solution $a = \\sqrt{1 + \\sqrt{2}}$, and taking the case $\\frac{b}{a^2+1} = a$ gives solution $a = \\sqrt[4]{3}$.
\nIf we instead take $a < 1$ then we get the reciprocals $a = \\sqrt[4]{\\frac13}$ or $a = \\sqrt{\\sqrt{2}-1}$.
\nA similar recursive construction to the one described in the question extends this to dissections into $4 + 3k$ parts.
\nA similar process to this answer but with fewer constraints on the layouts tested allows us to enumerate ratios which have suitable dissections. Many layouts gave parameterised families of solutions; I believe that these correspond to degenerate cases where one or more row or column dimension is zero, but I haven't tested and proven this systematically. On the assumption that it is the case, the following lists of possible ratios are exhaustive for dissection size ($n$) up to six:
\n\\begin{array}{ccc}n & \\textrm{Approx. ratio} & \\textrm{minpoly} \\\\\n3 & 1.272019649514069 & x^4 - x^2 - 1 \\\\\n\\hline\n4 & 1.316074012952493 & x^4 - 3 \\\\\n & 1.553773974030038 & x^4 - 2x^2 - 1 \\\\\n\\hline\n5 & 1.116682409513765 & x^6 + x^4 - 2x^2 - 1 \\\\\n & 1.217043263032180 & x^6 + x^4 - 3x^2 - 1 \\\\\n & 1.272019649514069 & x^4 - x^2 - 1 \\\\\n & 1.303697874614866 & x^6 + x^4 - 4x^2 - 1 \\\\\n & 1.370906722418348 & x^6 - 3x^2 - 1 \\\\\n & 1.517489913551980 & x^4 - x^2 - 3 \\\\\n\\hline\n6 & 1.099010977042511 & x^4 + x^2 - \\tfrac83 \\\\\n & 1.100083692217721 & x^6 + 4x^4 - 3x^2 - 4 \\\\\n & 1.123299713015174 & x^6 + 3x^4 - 3x^2 - 3 \\\\\n & 1.124171968973597 & x^4 - x^2 - \\tfrac13 \\\\\n & 1.141391973746090 & x^4 + x^2 - 3 \\\\\n & 1.145905333901386 & x^6 + \\tfrac12x^4 - 2x^2 - \\tfrac12 \\\\\n & 1.177890439184729 & x^4 - \\tfrac23x^2 - 1 \\\\\n & 1.179568726539875 & x^6 + 2x^4 - 4x^2 - 1 \\\\\n & 1.217043263032180 & x^6 + x^4 - 3x^2 - 1 \\\\\n & 1.224744871391589 & x^2 - \\tfrac32 \\\\\n & 1.277886208492545 & x^4 - \\tfrac83 \\\\\n & 1.292154727825668 & x^6 + 3x^4 - 6x^2 - 3 \\\\\n & 1.307392580458113 & x^6 + 2x^4 - 4x^2 - 4 \\\\\n & 1.317874861923462 & x^6 - \\tfrac23x^4 - \\tfrac53x^2 - \\tfrac13 \\\\\n & 1.331685928306913 & x^6 + 2x^4 - 5x^2 - 3 \\\\\n & 1.370906722418348 & x^6 - 3x^2 - 1 \\\\\n & 1.382833749401712 & x^6 + x^4 - 4x^2 - 3 \\\\\n & 1.441583361214096 & x^6 - x^4 - 2x^2 - \\tfrac12 \\\\\n & 1.444344210239199 & x^6 + x^4 - 5x^2 - 3 \\\\\n & 1.485875205951003 & x^4 - x^2 - \\tfrac83 \\\\\n & 1.489154360470618 & x^6 + 2x^4 - 8x^2 - 3 \\\\\n & 1.526452992911339 & x^6 - 5x^2 - 1 \\\\\n & 1.578247007081285 & x^6 - 5x^2 - 3 \\\\\n & 1.606592303630324 & x^4 - 2x^2 - \\tfrac32 \\\\\n & 1.628217862860243 & x^6 - x^4 - 4x^2 - 1 \\\\\n & 1.692985015441372 & x^6 - x^4 - 5x^2 - 1 \\\\\n & 1.922600525160414 & x^6 - 2x^4 - 6x^2 - 1 \\\\\n & 1.971294228756678 & x^4 - \\tfrac72x^2 - \\tfrac32 \\\\\n & 2.012192172612324 & x^6 - 3x^4 - 4x^2 - 1 \\end{array}
\nNote that none of the ratios for $n=4$ coincide with the ratios for $n=5$ or $n=6$. However, there are two non-trivial coincidences: ratio $\\sim 1.217043263032180$ with minpoly $x^6 + x^4 - 3x^2 - 1$ and ratio $\\sim 1.370906722418348$ with minpoly $x^6 - 3x^2 - 1$ give partitions into $5$ or $6$ similar rectangles, whence by recursive construction they cover $1+4k+5m$, and in particular they cover every integer in $\\mathbb{N} \\setminus \\{ 0, 2, 3, 4, 7, 8, 12 \\}$.
\n![\"Dissections](\"https://cheddarmonk.org/maths/self-similar-dissections.png\"/)
There are $181$ ratios for partitions into $7$ noncongruent similar rectangles (uploaded elsewhere because that's too long for this post). The only non-trivial coincidences are with partitions into $6$, and these leave more gaps than the $5,6$ coincidences. Fewer gaps would be possible if there's a partition into $8$ or $10$ with ratio $\\sqrt{\\frac{1 + \\sqrt 5}2}$ (i.e. the same ratio as $n=3$); otherwise ties for fewest gaps are possible with $3,12$ or $4,8$. However, the search for $n=7$ took 29 hours so I'm not going to run a search for $n=8$ unless I can optimise it a lot.
\nAs Wikipedia says:
\n\n\nIn Grothendieck's retrospective Récoltes et Semailles, he identified twelve of his contributions which he believed qualified as \"great ideas\". In chronological order, they are:
\n\n
\n- Topological tensor products and nuclear spaces.
\n- \"Continuous\" and \"discrete\" duality (derived categories, \"six operations\").
\n- Yoga of the Grothendieck–Riemann–Roch theorem (K-theory, relation with intersection theory).
\n- Schemes.
\n- Topoi.
\n- Étale cohomology and l-adic cohomology.
\n- Motives and the motivic Galois group (Grothendieck ⊗-categories).
\n- Crystals and crystalline cohomology, yoga of \"de Rham coefficients\", \"Hodge coefficients\", ...
\n- \"Topological algebra\": ∞-stacks, derivators; cohomological formalism of topoi as inspiration for a new homotopical algebra.
\n- Tame topology.
\n- Yoga of anabelian algebraic geometry, Galois–Teichmüller theory.
\n- \"Schematic\" or \"arithmetic\" point of view for regular polyhedra and regular configurations of all kinds.
\n
What are some modern, concise expository texts on these topics that students can use to learn the great ideas of Grothendieck? EGA, SGA, Les Dérivateurs, La Longue Marche, and so on don't count, because they are French, overwhelming, and maybe a bit outdated (I'm not sure) - anyway, it's not realistic for a student to go through them in their free time in addition to the courses they take for their degree. If you want to give a textbook, make sure it's as short as possible.
\nPlease only one topic with expository text per answer.
\nLet me start by giving two examples of the kind of answers I am expecting:
\nLeinster's An informal introduction to topos theory is an amazing introduction to the basic ideas surrounding topoi (and their connections to logic and geometry). And it is concise! This is idea 5.
\nMilne's lecture notes on étale cohomology. Not as short as Leinster's paper, but better (for a student) than 1000 pages of SGA 4 and 4 1⁄2. This is idea 6.
\nA good roadmap for FGA (topic 4, with glimpses on topic 6) is
\nFantechi, Barbara; Göttsche, Lothar; Illusie, Luc; Kleiman, Steven L.; Nitsure, Nitin; Vistoli, Angelo, Fundamental algebraic geometry: Grothendieck’s FGA explained, Mathematical Surveys and Monographs 123. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3541-6/hbk). x, 339 p. (2005). ZBL1085.14001.
\nFor topic 4 (schemes), I'd suggest Schemes: The Language of Algebraic Geometry, by Eisenbud and Harris. This book is out of print, and is generally regarded as having been \"superseded\" by its successor, The Geometry of Schemes. But for your purposes, I think the earlier book is actually better, since it's only 160 pages, and if you just want the main ideas, the first two chapters should suffice. The authors specifically aimed at making the topic accessible to students encountering schemes for the first time.
\nLet me suggest having looking at Grothendieck's own article The cohomology theory of abstract algebraic varieties in the proceedings of the 1958 ICM. This is both short (15 pages) and in English. Although written quite early on, it explains the motivations and intuitions for many of the developments that lay in the future in a lucid way. For instance, he compares and contrast his notion of scheme (referred to here as a pre-schema) with the more classical notions. He hints at \"a definition of the Weil cohomology (involving both 'spatial' and Galois cohomology)...\", which would eventually become étale cohomology.
\nFor topic 3, I would say Hartshorne's Appendix A is pretty good as a VERY brief introduction. After that, as also referred to there, I would recommend Manin's Lectures on the K-functor in algebraic geometry.
\nTopic 10 got developed further via the concept of o-minimality. See e.g. \"Tame topology and o-minimal structures\" by\nLou van den Dries, 1998 CUP, http://dx.doi.org/10.1017/CBO9780511525919
\nA shorter text is by Michel Coste: http://perso.univ-rennes1.fr/michel.coste/polyens/OMIN.pdf
\nSee also nLab article on tame topology: https://ncatlab.org/nlab/show/tame+topology
\nFor topic 5, try Leinster's nice expository article https://arxiv.org/abs/1012.5647.
\nFor topic 7, Milne's article https://www.jmilne.org/math/xnotes/MOT.pdf could be useful.
\nFor topic 6, there are Milne's notes: https://www.jmilne.org/math/CourseNotes/LEC.pdf
\nIn the front matter Milne writes:
\n\n\nThese are the notes for a course taught at the University of Michigan in 1989 and 1998.\nIn comparison with my book, the emphasis is on heuristic arguments rather than formal\nproofs and on varieties rather than schemes.
\n
For learning about topoi, I'd strongly recommend Goldblatt's Topoi: The Categorical Analysis of Logic, which goes over the subject and a few interesting applications in a very approachable and pleasant manner. It's not particularly concise at 556 pages, though you needn't read the entire text to get a solid foundational concept of topoi.
\nFor 5. Topoi. There is a very readable notice of the AMS: What is... a topos?, Luc Illusie, and with M Raynaud about Schemes at Grothendieck and Algebraic Geometry. In fact, the webpage of Professor L. Illusie contains very valuable material regarding the question.
\nYou can read (online) about various of the listed topics in Lectures grothendieckiennes Edited by Frédéric Jaëck
\nAnd find a lot of original material at Thèmes pour une Harmonie!
\nOn topic 1, I'd like to recommend Ray Ryan's Introduction to tensor products of Banach spaces, Springer 2002. Zbl 1090.46001
\nBefore learning about infinity stacks, one ought to learn about stacks. For this, look at section 4.1.1 of Vistoli's notes, Notes on Grothendieck topologies, fibred categories and descent theory, which is an exposition of one concrete case. This section is around a page and a half. He states:
\n\n\nthat the fact we can glue continuous maps and topological spaces says $(Cont)$ is a stack over $Top$.
\n
This in fact a generalisation of the clutching construction of bundles. And you can learn about bundles in the very readable first chapter of Grothendiecks Kansas notes on fibre bundles.
\nThere is also an expository note by Barbara Fantechi, Stacks for Everybody.
\nFor topic 1, the book by Diestel, Fourie, Swart: https://books.google.de/books?id=VBg5cGSngRoC&printsec=copyright&redir_esc=y#v=onepage&q&f=false provides a good introduction.
\nThe same question was asked on Mathematics Stack exchange but has not received any response or comments.
\nLet $M$ be a smooth $n$-dimensional manifold and $X$ be a smooth vector field on it. Consider a chart $(U, x^1, \\ldots, x^n)$ and another set of coordinates $(y^1,\\ldots,y^n)$ on the same chart (or on an open subset of $U$ in which case we will just shrink $U$). Since $X$ is smooth, there exists smooth functions $a^i : U \\to \\mathbb{R}$ and $b^i: U \\to \\mathbb{R}$, with $i \\in \\{1, \\ldots, n\\}$, such that, for every $p \\in U$:
\n\\begin{equation*}\nX_p = \\sum_{i=1}^n a^i(p) \\frac{\\partial }{\\partial x^i} \\mid_{p} = \\sum_{i=1}^n b^i(p) \\frac{\\partial }{\\partial y^i} \\mid_{p}.\n\\end{equation*}
\nApply to both sides the function $x^i$'s to get, for all $p \\in U$:
\n\\begin{equation*}\n\\begin{bmatrix}\n a^1(p) \\\\\n \\vdots \\\\\n a^n(p) \n\\end{bmatrix} = \n\\begin{bmatrix}\n \\frac{\\partial x^1}{\\partial y^1} & \\cdots & \\frac{\\partial x^1}{\\partial y^n}\\\\\n \\vdots & \\ddots & \\vdots \\\\\n \\frac{\\partial x^n}{\\partial y^1} & \\cdots & \\frac{\\partial x^n}{\\partial y^n} \n\\end{bmatrix}_p\n\\begin{bmatrix}\n b^1(p) \\\\\n \\vdots \\\\\n b^n(p) \n\\end{bmatrix}.\n\\end{equation*}
\nUsually, one is given $(x^1, \\ldots, x^n)$, $(y^1, \\ldots, y^n)$, and $a^1, \\ldots, a^n$.\nOne is then required to compute $b^1, \\ldots, b^n$ in order to represent the vector field $X$ in the other set of coordinates.
\nMy question is about the reverse. Suppose that we know $(x^1, \\ldots, x^n)$ and $a^1, \\ldots, a^n$. With a specific set of smooth functions $b^1, \\ldots, b^n$ I chose, can we always guarantee the existence of coordinates $(y^1, \\ldots, y^n)$ such that my vector field can be represented in the way I specified?
\nThis is definitely not always doable. For example, if $X$ is the zero vector field and $b^i$'s are nonzero, then it is clearly impossible. One must impose some conditions on $(x^1, \\ldots, x^n)$ or on $a^1, \\ldots, a^n$. After searching, I found a theorem regarding the representation of a vector field around a regular point: Assume $p \\in M$ is such that $X_p \\neq 0$, the theorem says that we can always find a chart $(U, y^1, \\ldots, y^n)$ about $p$ such that $X_q = \\frac{\\partial}{\\partial y_1} \\mid_q$ for all $q \\in U$. In our language, this means $b^1 \\equiv 1$, $b^i \\equiv 0$ for $i \\neq 1$, and $a^i$'s are not always $0$. So this should be a solid starting point of a more general result.
\nSo let me rephrase the question :
\nAs my own attempt, it seems question 1 has to do with whether one can show the existence of solutions in a set of partial differential equation. But I am not familiar with the relevant theory. As for question 2, I have no clue on how to start.
\nFor question 2., even for analytic (or polynomial) vector fields no universal family of models is known. And even if such a family where known, nothing would assert that it is computable given the data you provide.
\nIn most cases your vector field is linearizable by an analytic change of coordinates (e.g. by Poincaré theorem), meaning that your $a_i$ are linear functions. But as soon as (quasi-)resonances arise between eigenvalues of the linearized vector field this is no longer the case. It is known that the set of different analytic conjugacy classes (up to local change of variables) is uncountable (for instance, in the case of resonances in dimension 2, it is in natural bijection with functional spaces like the space of germs of an analytic function $\\mathbb{R}\\{h\\}$, see the work of Martinet and Ramis).
\nIn dimension 2, there are a few results that allow to \"desingularize\" à la Hironaka by successive ponctual blowups (see Seidenberg's algortihm), but this is not guaranteed to be explicit. Even then, you end up with local models, called \"reduced\" vector fields (roughly speaking, the linear part is non-nilpotent), for which no simple local form exist in general. In specific cases there does exist universal local analytic models (look for \"analytic normal forms\" in planar vector fields). Moreover, I remember that two decades ago an alogrithm was devised by Dumortier and al. to visualize the flow of polynomial vector fields in $\\mathbb{R}^2$, based on Seidenberg's reduction, but I don't recollect more than that.
\nIn greater dimension the theory is not done, save for specific cases, and even then the results are much more complex than in the planar setting. And that is only for isolated singularities. To the best of my knowledge not much is known in the vicinity of a singular point belonging to a non-discrete manifold of singularities.
\nIf you allow less regularity for your local charts (e.g. smooth instead of analytic), then there exist partial results of conjugacy between $X$ and one of its finite jet in dimension 2 (Dumortier, Roussarie, Rousseau, Takens etc.), but some of the results only work for $C^k$ local charts with $k<\\infty$. Not sure how these generalize in higher dimension. It is known that some $X$ are not smoothly conjugate to one of their finite jets.
\nAs a conclusion, I'd say that what you ask is beyond our present knowledge as soon as singularities are involved.
\nFor question 1, yes. Let’s turn the notation around. Suppose that $X$ is $\\frac{\\partial}{\\partial y^1}$ for some coordinate system $y^1,\\dots ,y^n$, and let $a^1,\\dots ,a^n$ be any functions (of $y$), not all $0$. For each $i$ make a function $x^i$ such that $\\frac{\\partial x^i}{\\partial y^1}=a^i$. If we can arrange for the $x^i$ to be a coordinate system, then in $x$ coordinates $X$ will be $\\sum_i a^i\\frac{\\partial}{\\partial x^i}$. To do so, we try to alter $x^i$ without changing $\\frac{\\partial x^i}{\\partial x^1}$, in such a way that the Jacobian matrix $\\frac{\\partial x^i}{\\partial y^j}$ at the origin becomes invertible. We can do this by adding a suitable linear function of $L(y^2,\\dots ,y^n)$ to each $x^i$.
\nInspired by Question 508663, I discovered the following identity
\n$$\\sum_{k=1}^\\infty\\frac{H_{2k}-H_{k}}{k^2\\binom{2k}k}=2 \\zeta(3)-\\frac{\\pi \\sqrt{3}\\, \\Psi^{\\left(1\\right)}\\! \\left(\\frac{1}{3}\\right)}{24}+\\frac{\\pi \\sqrt{3}\\, \\Psi^{\\left(1\\right)}\\! \\left(\\frac{5}{6}\\right)}{72}\\tag{1},$$\nwhere $H_n$ denotes the harmonic numbers and $\\Psi^{(n)}(x)$ is the $n$th polygamma function.
\nMy questions:
\nThe formula is not new. It follows from (3.2) and (3.3) in Conjecture 3.1 of this 2015 paper of mine. The conjecture has been confirmed by several authors including J. Ablinger, Wenchang Chu, and Kam Cheong Au. One may search their papers via MathSciNet or arXiv. For example,\nJakob Ablinger's related paper appreared as a preprint in arXiv\n(see (73) and (62) of this paper) and its published form can be found here.
\n$\\newcommand{\\mfrac}[2]{{\\textstyle\\frac{#1}{#2}}}$\nThis is an edited version of an answer supplied to me by ChatGPT 5.2 'Extended Pro'. I have checked\nevery line and believe it is correct.\nLet $S$ be the sum and let\n$$\nF(z)=\\sum_{k=1}^\\infty \\frac{z^k}{k^2\\binom{2k}{k}} $$\nfor $0 \\le z \\le 1$. From the standard Taylor series\n$$\\arcsin^2 x = \\frac{1}{2} \\sum_{k=1}^\\infty \\frac{1}{k^2 \\binom{2k}{k}} (2x)^{2k},$$\nsee for instance this MathStackexchange answer (although ChatGPT was keen to share its own proof with me), we have $F(z)=2\\arcsin^2\\bigl(\\frac{\\sqrt z}{2}\\bigr)$.\nUsing the integral formula for the Harmonic numbers we have\n$$ H_{2k}-H_k\n= \\int_0^1 \\frac{x^k-x^{2k}}{1-x}\\,\\mathrm{d}x $$\nand so\n$$ S =\\int_0^1 \\frac{F(x)-F(x^2)}{1-x}\\,\\mathrm{d}x. $$\nNow\n$F(x)-F(x^2)=\\int_{x^2}^x F'(t)\\,\\mathrm{d}t $\nso we have, swapping the order of integration using\n$x^2 \\le t \\le x \\iff t \\le x \\le \\sqrt{t}$ to get the second line,\n$$ \\begin{aligned} S &= \\int_0^1 \\int_{x^2}^x F'(t)\\, \\mathrm{d}t\\, \\frac{\\mathrm{d}x}{1-x} \\\\ &=\n\\int_0^1 F'(t)\\bigl(\\int_t^{\\sqrt t}\\frac{dx}{1-x}\\bigr)\\mathrm{d}t \\\\ &=\n\\int_0^1 F'(t)\\log \\bigl(\\frac{1-t}{1-\\sqrt t}\\bigr)dt \\\\ \n&= \\int_0^1 F'(t)\\log(1+\\sqrt t)\\,\\mathrm{d}t \\\\\n&= \\int_0^1 F'(u^2) \\log (1+u)\\, 2u\\, \\mathrm{d}u \\\\\n&= \\int_0^{\\pi/6} F'(4\\sin^2\\theta) \\log(1 + 2\\sin \\theta) 8\\sin\\theta \\cos \\theta \\,\\mathrm{d}\\theta \\\\\n&= \\int_0^{\\pi/6} \\frac{\\theta}{\\sin 2 \\theta} \\log (1 + 2 \\sin \\theta) 4 \\sin 2\\theta \\,\\mathrm{d}\\theta \\\\ \n&= 4\\int_0^{\\pi/6} \\theta \\log (1 + 2 \\sin \\theta) \\,\\mathrm{d}\\theta.\n \\end{aligned}$$\nThis used the substitution $u^2 = t$, then $u = 2\\sin \\theta$ and\nthen in the penultimate step\n$$ F'(z) = \\frac{\\arcsin \\frac{\\sqrt{z}}{2} }{\\sqrt{1-z/4} \\sqrt{z}} $$\nto get $F'(4\\sin^2 \\theta) = \\frac{\\theta}{\\sin 2\\theta}$.\nNow factor\n$$\n1+2\\sin\\theta =\n4\\sin\\!\\left(\\mfrac{\\theta}{2}+\\mfrac{\\pi}{12}\\right)\n \\sin\\!\\left(\\mfrac{\\theta}{2}+\\mfrac{5\\pi}{12}\\right) $$\nto get\n$$\n\\log(1+2\\sin\\theta) =\n\\log \\Bigr(2\\sin\\bigl(\\mfrac{\\theta}{2}+\\mfrac{\\pi}{12}\\bigr)\\Bigr) +\n\\log \\Bigr(2\\sin\\bigl(\\mfrac{\\theta}{2}+\\mfrac{5\\pi}{12}\\bigr)\\Bigr). $$
\nThe standard Fourier series\n$$\n\\log(2\\sin \\phi) =-\\sum_{n=1}^\\infty \\frac{\\cos(2n\\phi )}{n}\n$$\nis uniformly convergent for $\\frac{\\theta}{2} + \\frac{\\pi}{12}$\nand $\\frac{\\theta}{2} + \\frac{5\\pi}{12}$ for $\\theta \\in [0, \\frac{\\pi}{6}]$.\nSubstituting for each summand in the right-hand side for $\\log (1 + 2 \\sin \\theta)$ above we get\n$$\n\\log(1+2\\sin\\theta)\n=\n-\\sum_{n=1}^\\infty\n\\frac{\\cos\\bigl(n(\\theta+\\frac{\\pi}{6})\\bigr)+\\cos\\bigl((\\theta+\\frac{5\\pi}{6})n\\bigr)}{n}.\n$$\nUsing\n$\n\\cos\\bigl( n (\\theta+\\frac{\\pi}{6}) \\bigr) + \\cos\\bigl(n(\\theta+\\frac{5\\pi}{6})\\bigr)\n= 2\\cos\\bigl(\\frac{\\pi}{3}n\\bigr)\\cos (\\theta+\\frac{\\pi}{2})n\n$\nwe obtain\n$$\nS\n=\n-8\\sum_{n=1}^\\infty \\frac{\\cos(\\frac{\\pi}{3}n)}{n}\n\\int_0^{\\pi/6}\\theta\\cos \\bigl( n (\\theta+\\frac{\\pi}{2}) \\bigr) \\,\\mathrm{d}\\theta.\n$$\nA direct integration, which I verified in Mathematica, gives\n$$\n\\int_0^{\\pi/6}\\theta\\cos\\bigl (n (\\theta+\\mfrac{\\pi}{2}) \\bigr) \\,\\mathrm{d}\\theta\n=\n\\frac{\\pi}{6n}\\sin\\!\\left(\\frac{2n\\pi}{3}\\right)\n+\n\\frac{\\cos(\\frac{2\\pi}{3} n)-\\cos( \\frac{\\pi}{2}n ) }{n^2}.\n$$\nTherefore using the uniform convergence mentioned above,\n$$\nS = -\\frac{8\\pi}{6}\\sum_{n=1}^\\infty \\frac{\\cos(\\frac{\\pi}{3}n)\\sin(\\frac{2\\pi}{3} n)}{n^2} -8\\sum_{n=1}^\\infty \\frac{a_n}{n^3},\n$$\nwhere\n$$ a_n = \\cos \\bigl( \\mfrac{n\\pi}{3} \\bigr) \\Bigl( \n\\cos\\bigl(\\mfrac{2n\\pi}{3}\\bigr)-\\cos\\bigl(\\mfrac{n\\pi}{2}\\bigr) \\Bigr). $$\nSince\n$ \\cos\\bigl(\\mfrac{n\\pi}{3}\\bigr)\\sin \\bigl(\\mfrac{2n\\pi}{3}\\bigr)\n= \\frac{1}{2}\\sin\\bigl( \\mfrac{n\\pi}{3}\\bigr) $\nthis becomes\n$$ S =\n-\\frac{2\\pi}{3}\\sum_{n=1}^\\infty \\frac{\\sin( \\frac{\\pi}{3} n)}{n^2}\n-8\\sum_{n=1}^\\infty \\frac{a_n}{n^3}.\n$$
\nNow the sequence $(a_n)$ is periodic with period $12$.\nThe values $a_0,\\ldots, a_{11}$ are\n$0,-\\frac{1}{4},-\\frac{1}{4},-1, \\frac{3}{4}, -\\frac{1}{4}, 2, -\\frac{1}{4}, \\frac{3}{4}, -1, -\\frac{1}{4},\n-\\frac{1}{4}$. (Again I checked this with Mathematica.)\nHence\n$$\na_n=\n-\\frac{1}{4}-\\frac{3}{4} [3 \\mid n] + \n[ 4 \\mid n] + 3 [6 \\mid n] - 3 [12 \\mid n]\n$$\nin Iverson bracket notation.\nIt follows that\n$$\n\\sum_{n=1}^\\infty \\frac{a_n}{n^3} =\n\\Bigl( -\\frac{1}{4} -\\frac{3}{4\\cdot 3^3} +\\frac{1}{4^3} +\\frac{3}{6^3} -\\frac{3}{12^3}\n\\Bigr)\\zeta(3)\n= -\\frac{1}{4}\\zeta(3).\n$$\nWe have shown that\n$$ S = 2\\zeta(3)-\\frac{2\\pi}{3}\\sum_{n=1}^\\infty \\frac{\\sin(\\frac{\\pi}{3} n)}{n^2}. \\qquad(\\star) $$\nIt remains to evaluate\n$ C=\\sum_{n=1}^\\infty \\frac{\\sin(\\frac{\\pi}{3}n)}{n^2}$.\nUsing the values of $\\sin \\frac{\\pi}{3}n$, which have period $6$, we get\n$$\nC = \\frac{\\sqrt{3}}{2}\\sum_{m=0}^\\infty\n\\bigl( \\frac{1}{(6m+1)^2} +\\frac{1}{(6m+2)^2} -\\frac{1}{(6m+4)^2} -\\frac{1}{(6m+5)^2} \\bigr). $$\nBy one definition of the trigamma function $\\psi_1$ we have\n$\\psi_1(z)=\\sum_{m=0}^\\infty \\frac1{(m+z)^2}$,\nso we may rewrite the previous equation for $C$ as\n$$\nC = \\frac{\\sqrt{3}}{72}\n\\Bigl( \\psi_1\\bigl(\\mfrac{1}{6} \\bigr) + \\psi_1\\bigl(\\mfrac{1}{3} \\bigr) - \\psi_1\\bigl(\\mfrac{2}{3} \\bigr) \n-\\psi_1\\bigl(\\mfrac{5}{6} \\bigr) \\Bigr) $$\nThe trigamma duplication formula states that\n$$\n\\psi_1(z)+\\psi_1\\!\\left(z+ \\mfrac{1}{2} \\right)=4\\psi_1(2z). $$\nApply this with $z=\\mfrac{1}{6}$ and $z = \\mfrac{1}{3}$ to get\n$ \\psi_1\\bigl( \\mfrac{1}{6} \\bigr) + \\psi_1 \\bigl( \\mfrac{2}{3} \\bigr) = 4 \\psi_1 \\bigl( \\mfrac{1}{3} \\bigr)$\nand similarly\n$ \\psi_1\\bigl( \\mfrac{1}{3} \\bigr) + \\psi_1 \\bigl( \\mfrac{5}{6} \\bigr) = 4 \\psi_1 \\bigl( \\mfrac{2}{3} \\bigr)$.\nHence\n$$ \\begin{aligned} C &=\n\\frac{\\sqrt{3}}{72}\n\\Bigl( 4 \\psi_1\\bigl( \\mfrac{1}{3} \\bigr) - \\psi_1 \\bigl( \\mfrac{2}{3} \\bigr) \n+ \\psi_1\\bigl(\\mfrac{1}{3} \\bigr) - \\psi_1\\bigl(\\mfrac{2}{3} \\bigr) \n-\\psi_1\\bigl(\\mfrac{5}{6} \\bigr) \\Bigr) \\\\\n&=\n\\frac{\\sqrt{3}}{72}\n\\Bigl( \n5\\psi_1\\bigl(\\mfrac{1}{3} \\bigr) -2\\psi_1\\bigl(\\mfrac{2}{3} \\bigr) -\\psi_1\\bigl( \\mfrac{5}{6} \\bigr) \\Bigr) \n\\\\ &=\n\\frac{\\sqrt{3}}{72}\n\\Bigl( \n5\\psi_1\\bigl( \\mfrac{1}{3} \\bigr) - \\mfrac{1}{2} \\psi_1 \\bigl( \\mfrac{1}{3} \\bigr) \n- \\mfrac{1}{2} \\psi_1 \\bigl( \\mfrac{5}{6} \\bigr) - \\psi_1 \\bigl( \\mfrac{5}{6} \\bigr) \n\\Bigr) \\\\\n&=\n\\frac{\\sqrt3}{48}\n\\bigl( \n3\\psi_1\\bigl( \\mfrac{1}{3} \\bigr) - \\psi_1 \\bigl( \\mfrac{5}{6} \\bigr) \\bigr)\n\\end{aligned}\n$$\nSubstituting this into the formula for $S$ marked ($\\star$) above we arrive at\n$$ S =\n2\\zeta(3) -\\frac{2\\pi}{3}C = 2\\zeta(3)\n-\\frac{\\pi\\sqrt{3}}{24}\\psi_1\\bigl(\\mfrac{1}{3} \\bigr) \n+\\frac{\\pi\\sqrt{3}}{72}\\psi_1\\bigl( \\mfrac{5}{6} \\bigr)\n$$\nas required.
\nI am trying to find a copy of a picture \"Mathematische Gesellschaft: \nGroup Portrait, Faculty, University of Göttingen (1899).\"
\nThis picture was published by Springer-Verlag as a poster in 1985,\nbut Springer has been unable to find a copy for me. I also sent an email\nquery to the Mathematics library at Göttingen, but received no response.
\nSo, as a last resort, I would like to ask whether any MO members have\naccess to a copy of the picture. I want to use it in a talk about Max \nDehn I am giving in Frankfurt in June, because Max Dehn appears in the \npicture.
\nAs a matter of fact, the poster is right on the other side of the hall in front of my office. Here is a picture of Dehn that I've made with my phone:
\n![\"Max](\"https://regularize.files.wordpress.com/2012/06/foto0138.jpg\"/)
I could make a better picture on Monday if that would be helpful...
\nEdit: Ok, my next better camera produced this picture:
\n![\"Max](\"https://regularize.files.wordpress.com/2012/06/dscn0044.jpg\"/)
Probably that's already near the best you can get because you can already see the halftoning effects pretty clear...
\nThere is also the same poster in my classroom. However, I can't find it in good quality on internet.
\n\nA soft copy of the photograph is available in Max Dehn Polyphonic Portrait by Jemma Lorenat, John McCleary, Volker R. Remmert, David E. Rowe, and Marjorie Senechal. It was published by the American Mathematical Society. Check page 8 of the preface. Unlike the Mathematical Intelligencer poster, Hilbert does not a hold copy of the magazine as a reader noted.
\nAlso, if you wish, you can contact the Josef and Anni Albers Foundation who provided this figure to the book authors.
\nEbook is available at https://bookstore.ams.org/HMATH/46
\nLet $(K, |\\cdot|)$ be a complete ultrametric field of characteristic $0$ with residue field $k$ of characteristic $p > 0$. Assume $K$ contains a primitive $p$-th root of unity $\\zeta$. Consider a Kummer extension $L = K(a^{1/p})$ where $a \\in O_K^\\times$. Suppose the extension of residue fields $l/k$ is of degree $p$ and separable. It is well-known that $l/k$ is an Artin-Schreier extension of the form $z^p - z = \\bar{\\gamma}$. How can we find $\\gamma$ from $a$?
\nMore precisely, for the case where $|.|$ is not discret, how can we found $\\gamma\\in O_K$ such that $a=(1+X)^p(1+p(\\zeta-1)\\gamma)$?
\nFor the case where $|.|$ is discret and $\\pi$ is the uniformizer, we can construct an algorithm such that if $|a_i-1|>|p(\\zeta-1)|$ we can construct $u_i$ with $a_i=u_i^p a_{i+1}$ and $|a_{i+1}-1|\\leq |\\pi(a_i-1)|$. Therefore at some stage we obtain $|a_i-1|\\leq |p(\\zeta-1)|$.\nIn the general case, however, we have\n$|a_{i+1}| \\leq q_i \\, |a_i - 1|$ for some $q_i \\in (0,1)$.\nThere is no concrete information about the constants $q_i$.\nIf we assume that\n$|a_i - 1| > |p(\\zeta - 1)|$ for all $i$,\nwe can only prove that\n$|a_i - 1| \\to |p(\\zeta - 1)|$,\nand so far I have not been able to derive a contradiction.
\nIf $a$ has the special form $a = 1 + p(\\zeta-1) b$ for $b \\in \\mathcal O_K^\\times$ then we can expand $(1+(\\zeta-1) x)^p= a$ using the binomial theorem as
\n$$ p (\\zeta-1) x + \\dots + (\\zeta-1)^p x^p = p (\\zeta-1) b $$
\nwhere the omitted terms are divisible by $p(\\zeta-1)^2$. From the minimal polynomial\n$$\\frac{ (y+1)^p -1}{y} = y^{p-1} + \\binom{p}{1} y^{p-2} + \\dots + \\binom{p}{2} y+ \\binom{p}{1} $$ of $\\zeta-1$, we see that $$(\\zeta-1)^{p-1} \\equiv - p \\pmod {p (\\zeta-1)},$$ so\n$$(\\zeta-1)^p \\equiv - p (\\zeta-1) \\pmod{ p (\\zeta-1)^2}.$$
\nThus, dividing by $p(\\zeta-1)$ and modding out by $(\\zeta-1)$ gives
\n$$x - x^p \\equiv b \\pmod {(\\zeta-1)} $$\nwhich is the desired Artin–Schreier form.
\nFor the general case, one can observe that every $a$ for which the Kummer extension is unramified can be expressed as a $p$th power times a number of this special form. To do this, one just has to check that $a_0 = 1 + p(\\zeta-1) b_0$ for $b_0$ chosen to be congruent mod the maximal ideal to an element that generates a nontrivial Artin-Schreier extension does in fact generate a nontrivial degree $p$ unramified extension, and there is only one such extension, so every such $a$ must be a power of $a_0$ times a $p$th power by a general uniqueness statement in Kummer theory.
\nFrom A Mathematician’s Apology, G. H. Hardy, 1940:\n\"I had better say something here about this question of age, since it is particularly important for mathematicians. No mathematician should ever allow himself to forget that mathematics, more than any other art or science, is a young man's game. ... I do not know an instance of a major mathematical advance initiated by a man past fifty. If a man of mature age loses interest in and abandons mathematics, the loss is not likely to be very serious either for mathematics or for himself.\"
\nHave matters improved for the elderly mathematician? Please answer with major discoveries made by mathematicians past 50.
\nRoger Apery was 62 when he proved the irrationality of $\\zeta(3)$.
\nAn answer of particular contemporary relevance would be Yitang Zhang, who established earlier this year (2013) that there are infinitely many pairs of primes which differ by less than 70 million (this constant has subsequently been improved to 246 by the team of a Polymath project). He was born in 1955 and had only two previous journal publications.
\nWeierstrass approximation theorem was proved by Karl Weierstrass when he was 70 years old
\nSince no one has mentioned A.N. Kolmogorov (born 1903), I hope I may be \nforgiven for a second answer. The following is from Kolmogorov's\nWikipedia biography.
\nIn classical mechanics, he is best known for the Kolmogorov–Arnold–Moser \ntheorem (first presented in 1954 at the International Congress of \nMathematicians). In 1957 he solved Hilbert's thirteenth problem (a joint \nwork with his student V. I. Arnold). He was a founder of algorithmic \ncomplexity theory, often referred to as Kolmogorov complexity theory, \nwhich he began to develop around this time.
\nKurt Heegner published his only, extremely influential paper, in 1952 when he was 59. However it took nearly 20 years for the mathematical community to realize what a gem it was.
\nLeonhard Euler. According to the Wikipedia page, he still managed to produce one paper per week in the year 1775 (at age 68), despite deteriorating eyesight. As a concrete example, at age 65 he proved that $2^{31} − 1$ is a Mersenne prime, which may have remained the largest known prime for the next 95 years.
\nMarina Ratner (b. 1938) proved Ratner's Theorems around 1990. They are some of the biggest advances in ergodic theory for quite a long time.
\nP. S. Novikov was 54 when he gave the first proof (143 pages!) of the unsolvability of the word problem for groups in 1955, and 58 when he co-solved the Burnside problem with S. I. Adian.
\nThere are many examples of people doing significant work into their 60s and 70s, but fewer great discoveries. Here are a couple of my favorites:
\nAugust Ferdinand Möbius discovered the Möbius band in 1858 at age 68 (the date referenced in Wikipedia). Other sources place the discovery even later: in 1861 he submitted to the French Academy prize competition a paper on it that passed unnoticed. As John Stillwell pointed out, in 1863 (age 73), Möbius published the classification of surfaces by genus (and in 1865 he finally described the Möbius band and the notion of orientability in print). Johann Benedict Listing turned 54 in 1862, the year in which he published a memoir discussing a 4-dimensional generalization of Euler's formula and described the Möbius band which he discovered independently.
\nJulius Plücker was 64 in 1865, when he \"returned to the field of geometry\" after a hiatus of nearly 20 years (Wikipedia, McTutor, Cajori) and discovered the \"line geometry\" (it is possible that the roots of this discovery go back to his 1846 monograph). The first volume of his book Neue Geometrie des Raumes describing it was published in 1868 and the second volume was completed and published posthumously by Felix Klein in 1869. The idea of using higher-dimensional objects as points in new \"geometry\" made profound impact on Klein and Sophus Lie and led to the Erlangen program and, by route of Lie sphere geometry, to Lie's general theory of transformation groups. This also marked one of the first appearances of higer-dimensional spaces in geometry.
\nAccording to wiki, Mihailescu got his PhD at the age of 42; and then proved Catalan's conjecture in 2002, age 47, so almost 50.
\nZariski proved what might be arguably his greatest result, the theorem on formal functions, just after turning fifty. He also initiated a whole field of enquiry, the theory of equisingularity, in his late 60's.
\nLouis de Branges solved the Bieberbach conjecture in 1985 when he was 53.
\nAlthough I concede that there is some truth to the belief that the greatest conceptual breakthroughs in mathematics are made by younger mathematicians, I think it has led to the mistaken idea that older mathematicians rarely do anything significant.
\nI just don't think it's that uncommon for top mathematicians today to be productive after they're 50. Atiyah and Bott did great work after they were 50. It seems to me that so did Singer. Although most mathematicians slow down after they are 50, so do most non-mathematicians. But there are not a few exceptions to this.
\nAnd is any of this that different from other fields?
\nTheorema Egregium was published by Gauss in 1828. Since Gauss was born in 1777, he ought to have been a little over 50 then.
\nRef: Disquisitiones generales circa superficies curva (1828)
\nFurtwängler proved the principal ideal theorem when he was almost 60. No small feat given that Artin and Schreier simultaneously were working on it.
\nPaul Erdős continued to do work in many fields including combinatorics after his 50th birthday. Some of his papers are here
\nSomething fitting this description that I haven't seen mentioned here is Norman Levinson's proof that asymptotically 1/3 of the zeroes of the Riemann zeta function lie on the critical line, which was the best result of its kind at the time. He was a little over 60 when he proved this, shortly before his death. What I find most remarkable about this is that he didn't really do much number theory until his last few years.
\nChristos Papadimitriou is in his late 50's now (I can't find his exact age, which is a little strange), and in just the past few years he's done major work in algorithmic game theory, a field at least somewhat removed from the one he made his career in. Technically, he's a theoretical computer scientist - I say this is close enough though.
\nThis is not really an answer but an objection to most of the answers at this pages and in particular to not so well formed question (it does not do justice to Hardy's book in my opinion).
\nIf you read the whole chapter of Hardy's book where the excerpt is from, Hardy explains somewhere that he does not know a highest class mathematician whose best discoveries came after 50. I recall after reading the whole chapter that I was convinced with the bulk of text that Hardy meant that there are no major advances by a mathematician after 50, unless they had major discoveries also before 50. So Euler and Poincare are not counterexamples to Hardy's experience, and some other answers in this column are not as well! Of course some people completed earlier work after 50, or continued with major advances while they already became major mathematicians before, but do you really a know a mathematician who done no major research before 50 and done such world class research after 50 ?? Also do not look the publication dates but the creation dates.
\nPhilip Hall published his paper with Higman, as well as his \"Theorems like Sylow's\", after he was 50. These are arguably his two biggest papers (and the Hall-Higman paper is arguably one of the most important papers in group theory).
\nSteve
\nKarl Dickman (born 1862) published the only math paper in 1930 (age 68) about distribution of prime factors. \nHe discovered the asymptotic distribution of the largest prime divisor of n, where n is chosen uniformly from $1,...,N$ and $N\\to\\infty$ (this is Dickman distribution).\nMuch later the distribution of other prime divisors was described. This is related to the famous Poisson-Dirichlet distribution.\n(see also \"The Poisson–Dirichlet Distribution and its Relatives Revisited\" by Lars Holst).
\nTibor Rado introduced the busy beaver function and proved its noncomputablity at the age of 67.
\nAnd of course, Dennis Sullivan and James Stasheff, both well into their 60's and 70's, are still both major contributors to topology and categorical algebra.
\nWhen Khare and Wintenberger proved Serre's conjecture, Wintenberger was older than fifty.
\nConnes has initiated whole new areas of mathematics since turning 50: spectral triples, and his novel approach to the Riemann hypothesis, for example.
\nA recent example (you may or may not think it's a major advance - but it is certainly big news in fundamental game theory): William Press (64) and the legendary Freeman Dyson (89) have shown that iterated Prisoner’s Dilemma contains strategies that dominate any evolutionary opponent (in the paper bearing the same title).
\nThe Fermat number $F_6$ was shown to have nontrivial factorization, by Landry at the age of 82. And apparently it was Landry's only mathematical publication.
\n(Source: Ribenboim, Prime number records(the smaller book).)
\nThis is perhaps not a \"major mathematical advance\" in the sense of Hardy; but is inspiring nonetheless. I have seen a good number of elderly retired people with dreams of solving Fermat's Last Theorem or other such theorems in a simple way, and doggedly keep on trying and without getting disheartened by the lack of recognition for their efforts.
\nAndre Weil lay the modern foundation of \"theta series\" in Acta math. (1964/65) when he was almost 60 years old!
\nThis \"almost\" answers Zoran Škoda's question: Otto Grün (his theorems in group theory are still well known) published his first paper at the age of 46.
\nBurnside proved the $p^aq^b$ theorem at age 53.
\nRoughly 1990, Lusztig wrote a series of papers on quantum groups. Perhaps the result that the braid groups acts on $U_q(\\mathfrak{g})$ is the proof which is least conceptual.
\nThe original paper contains a case by case check for $m=2,3,4,6$ as far as I understand.
\nIf it is not a coincidence, then it needs an explanation. My question is
\n\n\nAny conceptual proof up to now is known?
\n
At least not one by one.
\nFor example, can we define quantum groups for the type $I_n$? (The naive version is to work in the field of Puiseux series.)
\nPerhaps we have very limited 'conceptual' understanding on quantum groups up to now. There is a lot on the negative part $\\mathfrak{f}$ (geomatric realization, categorification, etc.). But the braid group action is on the complete quantum group. The classic one is the Hall algebra. For this are there any interpolation of the braid group actions?
\nThe only other result I know is the work of Nakajima which showed that the whole algebra can be realized by the equivariant K-theory of Nakajima varieties. (any result about how this action reflecting the geometry side?)
\nFor example, the simplest case, in the springer realization of $U_q(\\mathfrak{sl}_3)$, does the braid group action have any geometric meaning?
\nFor the braid group actions on Yangians and on quantum untwisted loop algebras (so effectively, on quantum Kac-Moody algebras of untwisted affine types), we can now refer to the paper https://arxiv.org/abs/2401.06402.
\nGiven a topological group $G$ and a closed subgroup $H \\le G$, one can ask when the projection map $G \\to G/H$ is a fiber bundle. In Steenrod's book on fiber bundles he gives a necessary and sufficient condition, being when there is a locally defined continuous section in the neighborhood of the identity coset, called a local cross-section.
\nHe refers to an unpublished example of Hanner of a compact abelian group of infinite dimension $G$ and a zero-dimensional subgroup $H$ which fails this condition. Mostert similarly hints at this result in 1952, and Karube gives a similar but supposedly different example in 1958.
\nDoes anyone know the precise folklore construction of Hanner?
\nAssume $X:=X_d$ is an hypersurface of degree $d$ in $\\mathbb{P}^n$. Assume in addition that $d<n+1$, hence $X$ is a Fano variety.
\nThe hyperplane section theorem of Lefschetz ensures that the only interesting part of the cohomology is $H^{n-1}(X,\\mathbb{Q})$, and as $X$ is Fano, it is well known that $H^{n-1,0}(X)=0$ : the Hodge structure is of coniveau $1$.
\nRecall that an hodge structure $H$ is of niveau $\\alpha$ if $H^{p,q}=0$ for $\\mid p-q \\mid > \\alpha$.
\nNow for very low degree hypersurfaces, the niveau is bigger. I observed the following facts with direct computations :
\nFurthermore on all this examples this bound is optimal, except from the case of cubic surfaces, so maybe we should neglect cases with a very small $n$.
\nSo I was looking for a precise result about the niveau of this Hodge structure. As we have explicit formulas I tried the two following methods :
\nWith intersection theory
\nWe can easily compute the Chern classes of $X$ as it is an hypersurface, and so Riemann-Roch formula gives an explicit element $P\\in \\mathbb{Q}(x,y)$ such that in a formal power extension, the coefficient of $x^py^q$ is exactly the vanishing part of $h^{p,q}$.
\nHowever it seems really annoying to expend this fraction for a general couple $(d,n)$.
\nWith vanishing of the pole
\nLet $S=\\mathbb{C}[x_0,\\cdots,x_n]$, $f$ the homogenous polynomial defining $X$ and $J_f$ the homogenous ideal generated by the jacobian of $f$. Finally, let $R=S/J_f$. Then a vanishing of the pole argument gives an isomorphism
\n$$R^{kd-n-1}\\simeq H^{n-k,k-1}_{van}(X)$$
\nwhere $R^m$ is the part of degree $m$, and $van$ denotes the vanishing part of the cohomology.
\nNow if I choose $f= \\sum_{i=0}^n x_i^d$, then the jacobian is really easy to compute, so we can find the degree where $R^m$ vanishes. More precisely, we have $R^m=0$ exactly when
\n$$m>(n+1)(d-2).$$
\nSo $H^{n-k,k-1}_{van}(X)=0$ for $k>(n+1)(d-1)/d$. It is exactly the niveau bound I was looking for !
\nConclusion of the argument and question
\nI have computed the niveau bound for one example of hypersurface. However, from Riemann-Roch theorem and computation of the Chern numbers of hypersurfaces, we know that the Hodge numbers depend only on $(d,n)$, so my bound is true for any hypersurface of degree $d$.
\nSo I have found the result I was looking for, but I am annoyed because the proof used a vanishing of the pole argument, and so the Bott vanishing theorem. I would like to extend this computation to other kind of hypersurfaces (at least for the vanishing part), but in general this vanishing theorem is wrong.
\nAlso it is rather disappointing to prove the result only for a precise choice of $f$, and then use the independance of the choice. I would rather prove the result for any choice of $f$, or directly with the computation of Chern numbers.
\nDo you have any idea to find a better argument ?
\nSince $H^{p,q}(X) = H^q(X,\\Omega^p_X)$ and since for $p + q \\ne 1$ the only nontrivial cohomology of $\\Omega^p(X)$ is $H^p(X, \\Omega^p_X) = H^{p,p}(X)$, and it is 1-dimensional, one has\n$$\nH^{p,q}(X) = 0 \n\\iff\n\\chi(\\Omega^p_X) = (-1)^p.\n$$\nOn the other hand, $\\chi(\\Omega^p_X)$ can be computed by Riemann--Roch.
\nSo, I'm not sure to what extent this is a thing. John Baez mentions in this blog post that common large countable ordinals beyond the large Veblen ordinal can also \"be defined as fixed points\". He doesn't expand on that but I take it that means fixed points of a further Veblen-like / klammersymbol-like construction.
\nBut, I haven't been able to find any good account of this. Is this a known/standard thing? I want to know just how far the Veblen construction can be pushed -- like, beyond the large Veblen ordinal, is there an \"ultimate Veblen ordinal\" somewhere; and might it be equal to an already-named / well-known one, such as Bachmann-Howard? (I suppose a negative answer here would be if arbitrarily large computable ordinals could be obtained this way, making the \"ultimate Veblen ordinal\" just $\\omega_1^{\\mathrm{CK}}$.)
\nNow I can certainly think of ways of continuing beyond the large Veblen ordinal myself... but I don't really want to reinvent the wheel here when I expect others have likely already done it better. So, is there a good account of this somewhere?
\nThank you all!
\nHere is some relevant information by a grandmaster on the subject:\nhttp://www.mathematik.uni-muenchen.de/~buchholz/articles/jaegerfestschr_buchholz3.pdf
\nOK, these people define the \"dimensional Veblen\" which they show can approach the Bachmann-Howard ordinal: https://arxiv.org/abs/2310.12832
\nSo, that's pretty cool. And it does indeed seem like it's hard to go further than this, because the depth of indexing is always finite -- if it could be any ordinal, then you could take fixed points and go further, but as they've set it up, you can't do that. Moreover, I don't think there is any sensible way to do that; sure, you could potentially allow non-uniform indexing rank, which they don't, but because every relation allowed is finite, this doesn't actually let you get to \"indexing rank $\\omega$\"; there will always be a maximum, finite, indexing rank in each relation even if you allow non-uniformity. So this might be the answer!
\nHowever, there is a comment towards the end about going further by adding more types of brackets? I have to admit I don't understand what that means, though.
\nHello MathOverflow Community!
\nI have just created my account, and this is my very first post on here, so please excuse me if this is formatted incorrectly or not the right place for such a question.
\nAlso my highest level of education is high-school level mathematics, so please answer in a way that I may understand if possible. Although the question itself can be understood by much younger math enthusiasts than that! Here goes!
\nI’ve been pondering the Goldbach Conjecture for several years now
\n(which states that every even number greater than 2 is the sum of two primes)
\nI believe I have discovered/created a deceptively simple yet very interesting function which seemingly always produces two unique prime numbers which sum to any even number greater than 8.\nI have not proven this function definitively for all even numbers equal to or greater than 8, but have asked AI to check up to about 50,000 which has been exceedingly validating so far.
\nThis is the function
\nFor any even number ‘N’ equal to or greater than 8,
\nFirst subtract any arbitrary prime number that is both
\nIf this produces a prime number, congratulations the function halts/terminates and it has found two unique prime numbers that sum to N.
\nIf however this produces a composite number, then this is where it becomes more fun… Then subtract one of the prime factors of this new composite number from the original number N.
\nThis will either produce a prime number and halt the function, or yet another composite number in which case keep iterating… but be cautious to avoid subtracting a prime factor that has already been attempted at any previous step of the function; as this could create an obvious/trivial loop. However it seems as though there will always be at least one ‘as of yet untested’ unique prime factor of each new composite number to try each step until eventually halting at just a prime number.
\nI call this the subtract-factor-subtract method, and AI calls this a prime factorization feedback loop.\nDespite my best efforts so far I can’t seem to prove it halts for all even numbers, nor can I see how it would be mathematically possible to not halt, such as a theoretical counterexample of a loop in which a composite number generated at a later step in the function is comprised only of previously-tested prime factors , but I’ve not yet encountered any counterexamples of this happening.
\nThere are quite a bit more interesting properties of this function I’d love to discuss (one such property is that even for numbers where the function takes many steps to halt, each composite number along the way contains at least one unique prime factor not found within any of the other composite numbers of all of the other steps), but I hope this post so far covers the highlights.
\nI don’t have a specific question about this function, but here’s a few general questions that come to mind:
\nI understand that proving this function would also prove the Goldbach conjecture, however I believe disproving this function (or finding a counterexample) doesn’t necessarily disprove the Goldbach conjecture.
\nAny and all feedback on this idea is welcome! Math is a big hobby of mine, and I hope to pursue it someday at a higher academic level. Thank you so much for reading!
\n\n\nThe purpose of this question is to collect examples where large language models (LLMs) like ChatGPT have led to notable mathematical developments.
\n
The emphasis in this question is on LLMs, but answers about other machine-learning tools are also welcome.
\nThis question complements two questions that I asked before: Experimental mathematics leading to major advances (January 2010) and The use of computers leading to major mathematical advances II (June 2021). I think it will be useful to keep track of mathematical achievements based on LLMs or assisted by LLMs since it is considered a serious possibility that LLM's have the potential to change (and automatize) or at least assist research in mathematics.
\nI relaxed the threshold from \"major\" (in the previous two questions) to \"notable\" to allow more answers.
\nA related question specifically about Deep Mind is this: What mathematical problems can be attacked using DeepMind's recent mathematical breakthroughs? ; Another related question referring to deep learning is What are possible applications of deep learning to research mathematics? See also What mathematical problems can be attacked using DeepMind's recent mathematical breakthroughs? .
\nBoris Alexeev and Dustin Mixon posted last week their paper Forbidden Sidon subsets of perfect difference sets, featuring a human-assisted proof, where they had an LLM generate the Lean formalization of their proof. In my view this is one of the promising uses of LLMs, because the verifier naturally guards against hallucinations.
\nThe problem is notable: they give a counterexample to a $1000 Erdös problem (as well as noting that Marshall Hall had published a counterexample before Erdös made the conjecture).
\nMy caveat: a human must still verify that the definitions and the statement of the main theorem are correct, lest the LLM generate a correct proof, but of a different theorem.
\nHere is an example Counterexample to majority optimality in NICD with erasures
\nFrom the abstract:
\nWe asked GPT-5 Pro to look for counterexamples among a public list of open problems (the Simons ``Real Analysis in Computer Science'' collection). After several numerical experiments, it suggested a counterexample for the Non-Interactive Correlation Distillation (NICD) with erasures question: namely, a Boolean function on 5 bits that achieves a strictly larger value of E|f(z)| than the 5-bit majority function when the erasure parameter is p=0.40. In this very short note we record the finding, state the problem precisely, give the explicit function, and verify the computation step by step by hand so that it can be checked without a computer. In addition, we show that for each fixed odd n the majority is optimal (among unbiased Boolean functions) in a neighborhood of p=0. We view this as a little spark of an AI contribution in Theoretical Computer Science: while modern Large Language Models (LLMs) often assist with literature and numerics, here a concrete finite counterexample emerged.
\nThis paper
\nSergey Avvakumov, Roman Karasev,\nTensor rank of the determinant and periodic triangulations of $\\mathbb{R}^n$
\nhttps://arxiv.org/abs/2509.22333
\nincludes in the Acknowledgments \"We also thank ChatGPT 5 for pointing out that the lower bound in the proof of Theorem 1.5 can be stated in tensor language and is thus equal to the determinant’s tensor rank.\"
\nAt the request of Gil Kalai, I'm converting a comment to an answer.
\nThe paper \"Mathematical exploration and discovery at scale\" by Bogdan Georgiev, Javier Gómez-Serrano, Terence Tao, and Adam Zsolt Wagner was just posted to the arXiv: https://arxiv.org/abs/2511.02864.
\nBelow is the abstract of the paper.
\n\n\nAlphaEvolve is a generic evolutionary coding agent that combines the generative capabilities of LLMs with automated evaluation in an iterative evolutionary framework that proposes, tests, and refines algorithmic solutions to challenging scientific and practical problems. In this paper we showcase AlphaEvolve as a tool for autonomously discovering novel mathematical constructions and advancing our understanding of long-standing open problems.\nTo demonstrate its breadth, we considered a list of 67 problems spanning mathematical analysis, combinatorics, geometry, and number theory. The system rediscovered the best known solutions in most of the cases and discovered improved solutions in several. In some instances, AlphaEvolve is also able to generalize results for a finite number of input values into a formula valid for all input values. Furthermore, we are able to combine this methodology with Deep Think and AlphaProof in a broader framework where the additional proof-assistants and reasoning systems provide automated proof generation and further mathematical insights.\nThese results demonstrate that large language model-guided evolutionary search can autonomously discover mathematical constructions that complement human intuition, at times matching or even improving the best known results, highlighting the potential for significant new ways of interaction between mathematicians and AI systems. We present AlphaEvolve as a powerful new tool for mathematical discovery, capable of exploring vast search spaces to solve complex optimization problems at scale, often with significantly reduced requirements on preparation and computation time.
\n
EDIT:
\nSome further developments are in the paper \"New Nikodym set constructions over finite fields\" by Terence Tao (https://arxiv.org/abs/2511.07721) whose abstract reads
\n\n\nFor any fixed dimension $d \\geq 3$ we construct a Nikodym set in $\\mathbb{F}_q^d$ of cardinality $q^d - \n(\\frac{d-2}{\\log 2} +1+o(1)) q^{d-1} \\log q$ in the limit $q \\to \\infty$, when $q$ is an odd prime power. This improves upon the naive random construction, which gives a set of cardinality\n$q^d - (d-1+o(1)) q^{d-1} \\log q$, and is new in the regime where $\\mathbb{F}_q$ has unbounded characteristic and $q$ not a perfect square. While the final proofs are completely human generated, the initial ideas of the construction were inspired by output from the tools AlphaEvolve and DeepThink. We also give a new construction of Nikodym sets in $\\mathbb{F}_q^2$ for $q$ a perfect square that match the existing bounds of $q^2 - q^{3/2} + O(q \\log q)$, assuming that $q$ is not the square of a prime $p \\equiv 3 \\pmod{4}$.
\n
And also in the paper \"Sum-difference exponents for boundedly many slopes, and rational complexity\" again by Terence Tao (https://arxiv.org/abs/2511.15135) whose abstract reads
\n\n\nThe dimension of Kakeya sets can be bounded using sum-difference exponents $\\mathrm{SD}(R;s)$ for various sets of rational slopes $R$ and output slope $s$; the arithmetic Kakeya conjecture, which implies the Kakeya conjecture in all dimensions, asserts that the infimum of such exponents is $1$. The best upper bound on this infimum currently is $1.67513\\dots$. In this note, inspired by numerical explorations from the tool AlphaEvolve, we study the regime where the cardinality of the set of slopes $R$ is bounded. In this regime, we establish that these exponents converge to $2$ at a rate controlled by the rational complexity of $s$ relative to $R$, which measures how efficiently $s$ can be expressed as a rational combination of slopes in $R$.
\n
The paper “Point Convergence of Nesterov's Accelerated Gradient Method: An AI-Assisted Proof” by Uijeong Jang and Ernest Ryu, posted to Arxiv October 27, 2025, states in the abstract:
\n\n\nThe Nesterov accelerated gradient method, introduced in 1983, has been\na cornerstone of optimization theory and practice. Yet the question of\nits point convergence had remained open. In this work, we resolve this\nlongstanding open problem in the affirmative. The discovery of the\nproof was heavily assisted by ChatGPT, a proprietary large language\nmodel, and we describe the process through which its assistance was\nelicited.
\n
https://arxiv.org/abs/2510.23513
\nSee also this discussion by Damek Davis that helps put the result in perspective: https://x.com/damekdavis/status/1982529760505782510?s=46
\nScott Aaronson Phillip Harris, Freek Witteveen have a recent paper on the bounds of amplification of QMA (quantum Merlin-Arthur). A critical part of the paper involved a linear algebra trick suggested by GPT5. See Aaronson's blog entry here.
\nThere have been some notable recent examples of LLMs playing an important role in solving certain Erdős problems, e.g., Problem 124 and Problem 481 (although I think the latter turned out to be implied by a result of Klarner in 1982), which were purportedly solved entirely by Aristotle, and Problem 367, which was reduced by Wouter van Doorn to a lemma that was solved by Gemini Deepthink. While these are not famous Erdős problems and turned out to have relatively short and simple solutions, they differ from Olympiad problems in that the computer solved problems without known solutions.
\nNot exactly a notable result, but in my recent preprint Evaluation of GPT-5 on an Advanced Extension of Kashihara's Problem I describe how GPT-5 has been able to improve the general version of an extended combinatorial problem I originally solved in 2010.
\nThe abstract of\nEarly science acceleration experiments with GPT-5 by Sébastien Bubeck, Christian Coester, Ronen Eldan, Timothy Gowers, Yin Tat Lee,\nAlexandru Lupsasca, Mehtaab Sawhney, Robert Scherrer, Mark Sellke,\nBrian K. Spears, Derya Unutmaz, Kevin Weil, Steven Yin, and Nikita Zhivotovskiy states in part,\n\"Of note, this paper includes four new\nresults in mathematics (carefully verified by the human authors), underscoring how GPT-5\ncan help human mathematicians settle previously unsolved problems.\"
\nI have hesitated to post this example because I don't think it's really a \"notable mathematical development\" as such, but after seeing the other answers, I think this one is worth mentioning.
\nAs reported in Scientific American, Epoch AI invited several mathematicians, including Ken Ono, to a meeting designed to generate challenge problems for \"FrontierMath\". Among other things, Ono came up with what he thought was a Ph.D.-thesis-level problem: \"What is the 5th power moment of Tamagawa numbers of elliptic curves over $\\mathbb{Q}$?\" To Ono's amazement, the AI autonomously solved the problem. You can read Ono's account on his Facebook page (also reproduced below), or listen to him talk about it here.
\nEven if this is a cherry-picked example—the best one from the whole meeting—this strikes me as a very impressive achievement. But see also this tweet by Daniel Litt, who was also one of the invited mathematicians but was not too impressed when he read over the chat log.\n![\"enter](\"https://i.sstatic.net/LR9D59gd.jpg\")
This is a modest development and a modest use of AI, but nonetheless, since it may seem that only discretish mathematics is mentioned in most answers, the proof of Lemma 6 in https://arxiv.org/pdf/2511.06849 is due to ChatGPT 5. The techniques are standard but their use is elegant. Honestly, we were stressed out about the proof (when we discovered that an earlier one we had was flawed), and ChatGPT came to our rescue.
\nhere is a paper on bottleneck duality in flow networks with lattice coefficients from fall 2024.
\nhttps://arxiv.org/abs/2410.00315
\nthe appendix to this paper details how the main result and the proof were generated by GPT-o1-mini in september 2024. it was very difficult to get a correct proof at the time; current models nail a correct proof immediately.
\nThe \"notability\" here may be less because of the problem itself, and more because Donald Knuth was involved. Knuth wrote:
\n\n\nShock! Shock! I learned yesterday that an open problem I’d been working on for several weeks had just\nbeen solved by Claude Opus 4.6— Anthropic’s hybrid reasoning model that had been released three weeks\nearlier! It seems that I’ll have to revise my opinions about “generative AI” one of these days. What a joy\nit is to learn not only that my conjecture has a nice solution but also to celebrate this dramatic advance in\nautomatic deduction and creative problem solving. I’ll try to tell the story briefly in this note.
\n
\nHere’s the problem, which came up while I was writing about directed Hamiltonian cycles for a future\nvolume of The Art of Computer Programming:
\nConsider the digraph with $m^3$ vertices $ijk$ for $0 \\le i, j, k < m$, and three arcs from\neach vertex, namely to $i^+jk$, $ij^+k$, and $ijk^+$, where $i^+ = (i+1) \\bmod m$. Try to find\na general decomposition of the arcs into three directed $m^3$-cycles, for all $m > 2$.
Knuth then went on to to describe how Filip Stappers used Claude interactively to discover a conjectural solution, which was then rigorously proved.
\nUsing deep neural networks, Deepmind & collaborators numerically found a class of unstable singularities of the porous media and 3D Euler (with boundary) equations. Notable here is the fact that the level of precision of their solutions \"meets the stringent requirements for rigorous mathematical validation via computer-assisted proofs\" (quote from the paper).
\nPaper here: https://arxiv.org/abs/2509.14185\nArticle: https://deepmind.google/discover/blog/discovering-new-solutions-to-century-old-problems-in-fluid-dynamics/
\nA recent paper by Nagda, Raghavan, and Thakurta: \"Reinforced generation of combinatorial structures: Applications to complexity theory\"\nThey received help from AlphaEvolve to improve the best-known bound for Max-3CUT and Max-4CUT. Their idea seems quite general, so I would not be surprised if more complexity theory results would be improved.
\nThe paper Solving a Research Problem in Mathematical Statistics with AI Assistance by Edgar Dobriban documents how GPT-5 Pro helped close a gap between upper and lower bounds in robust density estimation under Wasserstein contaminations.
\nThe mathematical results appear in version 3 of Minimax Statistical Estimation under Wasserstein Contamination (Chao–Dobriban). Notably, the math paper itself contains no mention of AI assistance—the role of GPT-5 is documented only in the separate meta-paper.
\nThe key AI contributions were:
\nThe meta-paper provides detailed transcripts and a balanced discussion of limitations (hallucinated references, glossed-over details requiring days to verify).
\nSee also the author's Twitter/X thread summarizing the experience.
\nI initiated a meta.MO discussion on whether it's worthwhile to continue posting answers here. Denis T suggested that a minimum threshold would be \"a defendable case that the same result could not be achieved by using Google a few times.\" Ravi Vakil commented that he has definitely seen such examples, so I think it is worth posting a link to the following paper, which was co-authored by Vakil.
\nThe abstract of The motivic class of the space of genus 0 maps to the flag variety by Jim Bryan, Balázs Elek, Freddie Manners, George Salafatinos, and Ravi Vakil says in part, \"The proof of this result was obtained in conjunction with Google Gemini and related tools. We briefly discuss this research interaction, which may be of independent interest. However, the treatment in this paper is entirely human-authored (aside from excerpts in an appendix which are clearly marked as such).\"
\nThe computational complexity paper \"Search versus Decision for $S_2^P$\" by Lance Fortnow writes in the acknowledgements:
\n\n\nWhile the results are fully due to the author, this paper was mostly generated using the large\nlanguage model Gemini 3 Pro with prompting from the author. The author takes full responsibility\nfor its contents.
\n
EDIT (additional context):\nThe author further elaborated on Twitter/X: when asked \"It looks like you only told it the theorem statement and didn't give it the sketch.\" the author replied \"Yes, it came up with the proof on its own. Surprised me as well.\".
\n(I know that there was some discussion of whether to keep answering this question with new examples, but this one seemed cool enough to me to include here.)
\nIn \"Bruhat intervals that are large hypercubes\" (https://arxiv.org/abs/2601.01235), Jordan Ellenberg, Nicolas Libedinsky, David Plaza, José Simental, and Geordie Williamson find large hypercubes (of order $n \\log n$, in particular, superlinear) inside of Bruhat graphs of the symmetric group $S_n$, a problem of interest for various connections including to cluster algebras of Richardson varieties and the combinatorial invariance conjecture for Kazhdan-Lusztig polynomials. They used the AlphaEvolve AI system to find these intervals - more specifically, they studied examples of algorithms based on data for small $n$, and then were able to give a natural \"interpretation\" to what the machine was doing (at least in the case of $n=2^m$ a power of two) to find a natural class of permutations (the \"dyadically well-distributed\" ones) which form a big hypercube interval in the Bruhat graph.
\nThe abstract of their paper is
\n\n\nWe study the question of finding big Bruhat intervals that are poset hypercubes in the symmetric group $S_n$. Using permutations suggested by AlphaEvolve (an evolutionary coding agent developed by Google DeepMind), we were led to an unusual situation in which the agent produced a pattern which performed well for the $n$ tested, and which we show works well for general $n$. When $n$ is a power of 2 we exhibit a hypercube of dimension $O(n\\log n)$, matching the largest possible dimension up to a constant multiple.\nFurthermore, we give an exact characterization of the vertices of this hypercube: they are precisely the dyadically well-distributed permutations, a simple digitwise property that already appeared in connection with Monte Carlo integration and mathematical finance. The maximal dimension of a Bruhat interval that is an hypercube in $S_n$ gives a lower bound (and possibly is equal to) the maximal possible coefficient of the second-highest degree term in the Kazhdan--Lusztig $R$-polynomial in $S_n$. As a surprising consequence, we obtain a new lower bound of order $n\\log n$ for the maximal number of frozen variables appearing in the cluster algebras attached to the open Richardson varieties in $S_n$, and a similar result for moduli spaces of embeddings of Bruhat graphs.
\n
This does not seem to be widely circulating (not by one of the big tech companies), that the lower bounds of the kissing numbers in dimensions 25 through 31 have been broken, using a two-player game-theoretic/reinforcement-learning approach.
\nhttps://arxiv.org/abs/2511.13391
\nThe table on wikipedia and the one maintained by Henry Cohn have been updated.
\nLet $\\text{I}(\\omega)$ denote the collection of strictly increasing functions $f:\\omega\\to\\omega$ and consider the poset $\\big(\\text{I}(\\omega),\\leq_\\omega\\big)$ where $f \\leq_\\omega g$ for $f, g \\in \\text{I}(\\omega)$ if and only if $f(n) \\leq g(n)$ for all $n\\in \\omega$.
\nWhat is the smallest ordinal not embeddable into $\\text{I}(\\omega)$?
\nThe answer is $\\omega_1$, the least uncountable ordinal.
\nIndeed, suppose $f_\\alpha,\\alpha<\\omega_1$ is an increasing sequence in $I(\\omega)$. I claim that for each $n$, the value $f_\\alpha(n)$ eventually stabilizes. Indeed, if not, let $\\alpha_k$ be the least such that $f_{\\alpha_k}(n)>k$. Since $\\omega_1$ is regular, there is a countable ordinal $\\alpha>\\alpha_k$ for all $k$. But then $f_\\alpha(n)>k$ for all $k$, which is impossible.
\nTherefore for each $n$ there is an ordinal $\\beta_n$ such that $f_\\alpha(n)=f_{\\beta_n}(n)$ for all $\\alpha>\\beta_n$. But using regularity once again and taking $\\beta>\\beta_n$ for all $n$, we find $f_\\alpha=f_\\beta$ for all $\\alpha>\\beta$, which is a contradiction.
\nConversely, for any countable ordinal $\\eta$ there is an order-preserving embedding $c:\\eta\\to\\Bbb Q$, and we can assume $c(\\alpha)>1$ for all $\\alpha\\in\\eta$. Setting $f_\\alpha(n)=\\lfloor c(\\alpha)n\\rfloor$ we get an order-embedding $\\eta\\to I(\\omega),\\alpha\\mapsto f_\\alpha$.
\nTheorem. A linearly ordered set $A$ is embeddable in $I(\\omega)$ if and only if $A$ is embeddable in $\\mathbb R$.
\nProof. As in Wojowu's answer we can embed $[1,\\infty)$ into $I(\\omega)$ by setting $f_t(n)=\\lfloor nt\\rfloor$ for $1\\le t\\lt\\infty$ and $n\\lt\\omega$. For the converse, choose an injection $\\pi:\\omega\\times\\omega\\to\\omega$ and define a map $\\varphi:I(\\omega)\\to\\mathbb R$ by setting\n$$\\varphi(f)=\\sum_{n\\lt f(m)}2^{-\\pi(m,n)}.$$\nSince $f\\lt_\\omega g\\implies\\varphi(f)\\lt\\varphi(g)$, every chain in $I(\\omega)$ is embeddable in $\\mathbb R$.
\nCorollary 1. Every countable linear order is embeddable in $I(\\omega)$.
\nProof. By a theorem of Cantor every countable linear order is embeddable in $\\mathbb Q$ and therefore in $\\mathbb R$. Alternatively, if $A$ is a linearly ordered set and $i:A\\to\\omega$ is injective, then the map\n$$a\\mapsto\\sum_{x\\lt a}2^{-i(x)}$$\nis an order-embedding of $A$ into $\\mathbb R$.
\nCorollary 2. $\\omega_1$ is not embeddable in $I(\\omega)$.
\nProof. $\\omega_1$ is not embeddable in $\\mathbb R$.
\nRemark. For posets $P$ and $Q$ let us write $P\\le Q$ if there is a map $\\varphi:P\\to Q$ such that $x\\lt y\\implies\\varphi(x)\\lt\\varphi(y)$. The theorem was proved by showing that $\\mathbb R\\le I(\\omega)$ and $I(\\omega)\\le\\mathbb R$. Hence, for any poset $P$, we have $P\\le I(\\omega)$ if and only if $P\\le\\mathbb R$.
\nMotivated by Zeilberger's series\n$$\\sum_{k=1}^\\infty\\frac{21k-8}{k^3\\binom{2k}k^3}=\\zeta(2)$$\nand Questions 508743 and 508768\nI have discovered several identities involving higher-order derivatives of the function\n$$Z(k):=\\frac{(21k-8)\\Gamma(k+1)^6}{k^3\\Gamma(2k+1)^3}.$$\nNamely,\n$$\\sum_{k=1}^\\infty Z'(k)=-6\\zeta(3),\\tag{1}$$\n$$\\sum_{k=1}^\\infty Z''(k)=\\frac{57}2\\zeta(4),\\tag{2}$$\n$$\\sum_{k=1}^\\infty Z^{(3)}(k)=18\\pi^2\\zeta(3)-324\\zeta(4),\\tag{3}$$\n$$\\sum_{k=1}^\\infty Z^{(4)}(k)=\\frac{1959}2\\zeta(6)-432\\zeta(3)^2,\\tag{4}$$\n$$\\sum_{k=1}^\\infty Z^{(5)}(k)=19440\\zeta(2)\\zeta(5)-540\\zeta(3)\\zeta(4)-31860\\zeta(7),\\tag{5}$$\nand\n$$\\sum_{k=1}^\\infty Z^{(6)}(k)=31104\\zeta(5,3)+38880\\zeta(2)\\zeta(3)^2-77760\\zeta(3)\\zeta(5)-\\frac{84807}4\\zeta(8),\\tag{6}$$\nwhere $$\\zeta(5,3):=\\sum_{k_1>k_2>0}\\frac1{k_1^5k_2^3}=0.0377\\ldots.$$
\nNote that $(1)$ is equivalent to a formula obtained by K. C. Au in 2022.
\nQUESTION. Are the identities $(2)-(6)$ new? If some of them are new, how to prove them?
\nYour comments are welcome!
\nMotivated by Question 508743 and the known identities
\n$$\\sum_{k=1}^\\infty\\frac{(-1)^k}{k^3\\binom{2k}k}=-\\frac25\\zeta(3)\\ \\ \\text{and}\\ \\ \\sum_{k=1}^\\infty\\frac1{k^4\\binom{2k}k}=\\frac{17}{36}\\zeta(4),$$\nI found the following identities involving higher-order derivatives of the functions\n$$f(k)=\\frac{e^{\\pi ik}}{k^3\\binom{2k}k}=\\frac{e^{\\pi ik}\\Gamma(k+1)^2}{\\Gamma(2k+1)}\n\\ \\ \\text{and}\\ \\ g(k)=\\frac1{k^4\\binom{2k}k}=\\frac{\\Gamma(k+1)^2}{k^4\\Gamma(2k+1)}.$$
\nConjecture 1. We have\n$$\\sum_{k=1}^\\infty f'(k)=\\frac{9}{5}\\zeta(4)-\\frac{2}5\\pi\\zeta(3)i.\\tag{I1}$$\n$$\\sum_{k=1}^\\infty f''(k)=\\frac{\\pi^5}{25}i+\\frac{8}{15}\\pi^2\\zeta(3)-\\frac{52}5\\zeta(5).\\tag{I2}$$
\nConjecture II. We have\n$$\\sum_{k=1}^\\infty g'(k)=-\\frac{22}9\\zeta(5),\\tag{II1}$$\n$$\\sum_{k=1}^\\infty g''(k)=\\frac{3073}{216}\\zeta(6),\\tag{II2}$$\n$$\\sum_{k=1}^\\infty g^{(3)}(k)=\\frac{176\\zeta(2)\\zeta(5)-1423\\zeta(7)}{12},\\tag{II3}$$\n$$\\sum_{k=1}^\\infty g^{(4)}(k)=\\frac{752537\\zeta(8)+25344\\zeta(5,3)}{1080},\\tag{II4}$$\n$$\\sum_{k=1}^\\infty g^{(5)}(k)=660\\zeta(4)\\zeta(5)+\\frac{7115}3\\zeta(2)\\zeta(7)-\\frac{283010}{27}\\zeta(9),\\tag{II5}$$\nwhere\n$$\\zeta(5,3):=\\sum_{k_1>k_2>0}\\frac1{k_1^5k_2^3}=0.0377\\ldots.$$
\nQUESTION. Are the conjectural identities known? If some of them are new, how to prove them?
\nLet $\\mathbb M^n = \\mathbb R^{1,n-1}$ be $n$-dimensional Minkowski space and $\\eta\\colon \\mathbb M^n \\times \\mathbb M^n \\to \\mathbb R $ the corresponding indefinite inner product.
\nHow do I see for a given $n$-simplex $\\Delta \\subseteq \\mathbb R^{1,n-1}$ whether it is space like. More precisely, let $x_1, \\dots, x_n \\in \\mathbb R^{1,n-1}$ and let $\\Delta :=\\operatorname{Conv}(\\{x_1,\\dots,x_n\\})$ be the convex hull. How do I know $\\Delta$ is spacelike, i.e., $\\forall x,y \\in \\Delta\\colon \\eta(x-y, x-y) \\geq 0$ (using signature $(-+\\dots+)$).
\nWhat I am looking for, is an easy way to see whether this is the case given the points $x_1,\\dots, x_n$.
\nFor $n = 2$ it is sufficient that $\\eta(x_1-x_2, x_1-x_2)\\geq 0$. But for higher dimensions it is not sufficient to show, that the points $\\{x_1,\\dots, x_n\\}$ are pairwise spacelike. E.g., the points\n$$\\begin{pmatrix}\n0\\\\2\\\\0\n\\end{pmatrix},\\begin{pmatrix}\n0\\\\-2\\\\0\n\\end{pmatrix},\n\\begin{pmatrix}\n1\\\\0\\\\0\n\\end{pmatrix}\n\\in \\mathbb R^{1,3}\n$$\nare clearly pairwise spacelike but there convex hull contains a timelike direction.
\nIn my previous question, I asked about the quadratic forms of class number 8. While it answered my question about how to get the correct splitting polynomial, I'm still curious about conditions on the representable primes.
\nSpecificially, I would like to know a proof (or counterexample) to the conjectures:
\nConjecture 1: If $q$ and $r$ are $5$(mod $8$), with $q=a^2+b^2$ and $r=c^2+d^2$, with $a+bi$ and $c+di$ primary ($b,d>0$ are even and $a+b$ and $c+d$ are congruent to $1$ mod $4$) and $h(-qr)=8$, then $p=x^2+qry^2$ if and only if $q, r,$ and $(a+bi)(c+di)$ are quadratic residues mod $p$.
\nFor an example: $q=5$ and $r=13$. The corresponding primary Gaussian primes are $-1+2i$ and $3+2i$. Multiplying them you get $-7+4i$. This is what must be a quadratic residue mod $p$ (along with $q$ and $r$) so that $p=x^2+qry^2$
\nEDIT: The main thing that concerns me is dealing with the $C_4$ part of the form class group. The remaining cases can be split with congruence conditions. For example, for $n=65$, as Will Jagy pointed out, the main cases to distinguish are $x^2+65y^2$ and $9x^2+8xy+9y^2$. I conjecture that the quadratic residuacity of $(a+bi)(c+di)$ is enough to distinguish them. This is what I'm asking for a proof of.
\nLet\n$$f(k)=\\frac{1}{k^2\\binom{2 k}{k}},$$\nand\n$$f^{(n)}(k)=\\frac{d^n}{dk^n} f(k).$$\nInspired by Question 508718, I discovered the following identities
\n$$\\sum_{k=1}^\\infty f^{(1)}(k)=-\\frac{4}{3}\\zeta \\! \\left(3\\right),\\tag{1}$$\n$$\\sum_{k=1}^\\infty f^{(2)}(k)=\\frac{31}{6}\\zeta(4),\\tag{2}$$\n$$\\sum_{k=1}^\\infty f^{(3)}(k)=-38 \\zeta \\! \\left(5\\right)+8 \\zeta \\! \\left(2\\right) \\zeta \\! \\left(3\\right),\\tag{3}$$\n$$\\sum_{k=1}^\\infty f^{(4)}(k)=-32 \\zeta \\! \\left(3\\right)^{2}+\\frac{979}{6}\\zeta \\! \\left(6\\right),\\tag{4}$$\n$$\\sum_{k=1}^\\infty f^{(5)}(k)=760 \\zeta \\! \\left(2\\right) \\zeta \\! \\left(5\\right)+360 \\zeta \\! \\left(3\\right) \\zeta \\! \\left(4\\right)-2465 \\zeta \\! \\left(7\\right)\n.\\tag{5}$$
\nSince $f^{(1)}(k)=-\\frac{2}{k^{3}{\\binom{2 k}{k}}}-2\\frac{H_{2k}-H_{k}}{k^2\\binom{2k}k}$ , we can deduce that $(1)$ is a direct corollary of I. J. Zucker's series (2.9) $\\sum_{k=1}^\\infty \\frac{1}{k^3\\binom{2 k}{k}}=\\sqrt{3}\\frac{\\pi}{2}L_{-3}(2)-\\frac{4}{3}L_{1}(3)\n $ and Question 508718.
\nMy questions:
\nThis is a partial answer to question 3., that does not fit as a comment. Given the form of the relations for $1\\leq n \\leq 5$, it is natural to write the following ansatz for $n=6$\n$$\n\\sum_{k=1}^\\infty f^{(6)}(k) = c_0 \\zeta(8) + c_1 \\zeta(5)\\zeta(3) + c_2 \\zeta(3)^2 \\zeta(2)\n$$\nand find the appropriate rational $c_i$. This problem can be solved by an en.wikipedia.org/wiki/Integer_relation_algorithm, as was suggested to me in this comment to this question (409365).
\nI tried this using Mathematica, which implements an integer relation algorithm via LatticeReduce[]. To my surprise, I did not find a solution for the case $n=6$, even when using 300 digit precision. However, for $n=7$, I found the following relation (for which 100 digit precision suffices):\n$$\n\\sum_{k=1}^\\infty f^{(7)}(k) = -\\frac{974470}{3} \\zeta(9) + 103530 \\zeta(7) \\zeta(2) + 33180 \\zeta(6) \\zeta(3) + 71820 \\zeta(5) \\zeta(4) - 4480 \\zeta(3)^3 \\ .\n$$
\nThese results suggest that at least for $n=6$, the structure for the sum $\\sum_{k=1}^\\infty f^{(n)}(k)$ as suggested by the results of the OP is not general enough. I did not (yet) try cases $n\\geq 8$, so the general picture is not clear at the moment.
\nEdit As pointed out by Henri Cohen in this comment, in general one needs the en.wikipedia.org/wiki/Multiple_zeta_function.
\nI am interested in exploring polynomial matings via the Medusa algorithm, as described by Christian Henriksen, David Farris, and Kuan Ju Liu at Cornell University (1998) Cornell dynamics webpage.
\nThe original programs were written in C++ for Unix systems, with a minimal interface, and the source code is available as Medusa.tar.gz.
\nAre there any known ready-to-run implementations of the Medusa algorithm for Windows, or any maintained ports of these programs to non-Unix systems? Alternatively, are there modern open-source equivalents for experimenting with polynomial matings that are compatible with typical desktop environments?
\nThanks
\ncurl call will get you most of the way there. The only thing you might be missing at that point is to install the C compiler and make.",
"created_date": "2026-02-27 08:31:39Z"
},
{
"comment_id": 1325059,
"author_name": "Carlo Beenakker",
"author_id": 11260,
"author_url": "https://mathoverflow.net/users/11260/carlo-beenakker",
"score": 2,
"body_html": "I tried to compile the source code but failed; it's using obsolete C++ headers and memory‑management functions, requiring significant updates.",
"created_date": "2026-02-27 10:51:49Z"
}
],
"answers": [
{
"answer_id": 508773,
"question_id": 508468,
"score": 6,
"is_accepted": false,
"author_name": "LegionMammal978",
"author_id": 101028,
"author_url": "https://mathoverflow.net/users/101028/legionmammal978",
"author_reputation": 741,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-05 06:30:00Z",
"last_edited_date": "2026-03-05 06:56:33Z",
"body_html": "I have uploaded a set of patches at https://github.com/LegionMammal978/medusa-matings to allow the original source code to be built on modern systems. I have tested this workflow on Windows 11:
\npacman -S git make mingw-w64-ucrt-x86_64-gcc patch.git clone https://github.com/LegionMammal978/medusa-matings.git\ncd medusa-matings\n./download.sh\ncd Medusa\nmake all\n./mate_interact, ./draw_pb_sphere, ./drawpullback, or ./find_matings according to their documentation..eps files generated by draw_pb_sphere and drawpullback, you can install Ghostscript with pacman -S mingw-w64-ucrt-x86_64-ghostscript and convert to PDF with ps2pdf image.eps image.pdf. Alternatively, you can use any other PostScript tool available for Windows.Using this, I have been able to reproduce the five example pictures available from https://pi.math.cornell.edu/~dynamics/Matings/Pics/index.html, the first one nearly exactly and the other four exactly. Here are the parameters I deduced:
\nmate_interact; 144 dpi with draw_pb_sphere.mate_interact; 144 dpi with draw_pb_sphere.mate_interact; 144 dpi with draw_pb_sphere.mate_interact; 144 dpi with draw_pb_sphere.mate_interact; 144 dpi with draw_pb_sphere.Does a homeomorphism between pairs of spaces induce an isomorphism of the fundamental quandle?
\nIs it consistent that there exists a $\\subseteq^*$-decreasing chain $(X_\\alpha)_{\\alpha<\\omega_1}$ of infinite subsets of $\\omega$ such that for all ordinals $\\alpha_0 < \\alpha_1 < \\cdots < \\alpha_\\omega$ with $\\alpha_\\omega = \\sup_{n<\\omega} \\alpha_n$, $\\bigcap_{\\gamma \\leq \\omega} X_{\\alpha_\\gamma}$ is finite?
\nNote that if $(X_\\alpha)_{\\alpha<\\omega_1}$ has a pseudo-intersection (i.e. some infinite $Y \\subseteq \\omega$ such that $Y \\subseteq^* X_\\alpha$ for all $\\alpha$), then this chain doesn't have such a sequence of ordinals. Therefore, $\\omega_1 < \\mathfrak{t}$ implies the non-existence of such a chain.
\nNo.
\nIf given $(X_\\alpha)_{\\alpha < \\omega_1}$, then there is a ccc forcing $\\mathbb{P}$ that adds a pseudointersection $Y$. In the forcing extension, the sets $A_n = \\{ \\alpha < \\omega_1 : Y \\setminus X_\\alpha \\subseteq n \\}$ form a countable cover of $\\omega_1$ (which is still the real $\\omega_1$), so one of them is stationary. Say $A_m$ is stationary. This means that there is $\\alpha \\in A_m \\cap \\lim A_m$ and in particular there is an increasing sequence $\\langle \\alpha_n : n \\leq \\omega \\rangle$ completely contained within $A_m$ with $\\alpha = \\alpha_\\omega = \\sup \\alpha_n$ and $\\bigcap_{n \\leq \\omega} X_{\\alpha_n}$ is infinite, as it contains $Y \\setminus m$.
\nNow this all happens in the forcing extension. But you have names in the ground model for these objects and all that it suffices to do is to define a decreasing sequence $(p_n)$ in $\\mathbb{P}$ deciding more and more of the $\\alpha_n$'s, ensuring these become cofinal in what $\\alpha_\\omega < \\omega_1$ has been decided to be, and deciding bigger and bigger naturals that should be in the intersection of these $X_{\\alpha_n}$, $n \\leq \\omega$.
\nI search to prove ($n\\in \\mathbb{N}_0$)
\n$$ |f_n(x)| \\le 1 \\text{ for } x\\in[0,1]$$
\nwhere
\n$$\nf_n(x) = 1 + \\sum_{j=1}^n k_{n,j}x^{1/j}\n\\quad\n\\text{ with }\n\\quad\nk_{n,j}=(-1)^{n+j+1} \\binom{n}{j} \\binom{n+j}{j}.$$
\nI would be happy with any $n$-independent bound. There is a strong numerical evidence the bounding is true. Two remarks:
\nThe coefficients become large and a large global (non-telescopic) cancellation occurs.\nI would prefer a computational/algebraic/technical proof rather than a highly theoretical (functional analysis, Chebyshev/Muntz Systems), although I doubt such a proof (i.e. algrebraic) is feasible (I cannot find such even for Legendre polynomials).
\nUpdate:
\nFirst, I want to lower my expectations, I would by happy with any bound that rises slower than a power, e.g. logarithmic $|f_m (x)|<K\\log(m)$, but valid for all $x$ from the closed interval $[0,1]$. Quickly one realizes the problem comes from the neighborhood of $x=0$ ,where oscillations accumulate.
\nHere are some possibly useful equations (established by AI, checked by me):
\n$$\nf_{n}(x)-f_{n-1}(x)=nx\\left[f'_{n}(x)+f'_{n-1}(x)\\right]\\text{ for }n\\ge2. \n$$\n$$\nxf'_{n}(x)=(-1)^{n}+\\frac{1}{n}f_{n}(x)+\\sum_{k=1}^{n-1}(-1)^{n-k}\\frac{2k+1}{k(k+1)}f_{k}(x)\n$$\n$$\nf_{n}(x)=-f_{n-1}(x)+\\frac{2}{n}\\,x^{1/n}\\int_{x}^{1}t^{-1-1/n}\\,f_{n-1}(t)\\,dt\n$$\n$$\n\\int_{0}^{1}f_{n}^{2}(x)dx+\\frac{n}{2}\\int_{0}^{1}\\bigl(f_{n}(x)+f_{n-1}(x)\\bigr)^{2}dx=\\int_{0}^{1}f_{n-1}^{2}(x)dx\n$$\n$$\n\\int_{0}^{1}f_{n}^{2}(x)dx\\le\\int_{0}^{1}f_{1}^{2}(x)dx=\\frac{1}{3}.\n$$\n$$\n\\left|f_{n}(x)\\right|\\le C_{\\varepsilon}(1+\\ln n) \\text{ for } x\\in[\\varepsilon,1]\n$$
\nUpdate 2:
\nThe functions $(-1)^nf_n(x^\\alpha)$ with some appropriate $\\alpha$ (denoted as $R_m$ and $r_m$ in the following code) have very similar behavior to the shifted Legendre polynomials. Yet slightly differ in value and maxima/minima positions. Match is astonishing but not perfect. Here is the WxMaxima code:
\nkill(all);\nL(n,x) := legendre_p(n,2*x-1);\nk[n,j] := (-1)^(j+n+1)*binomial(n+j,j)*binomial(n,j)$\nR(n,x) := 1 + sum(k[n,j]*x^(1/j), j, 1, n)$\nr(n,x) := (-1)^n*R(n,x);\nM : 5;\nwxplot2d([L(M,x),r(M,x^(6.820))],[x,0,1]);\nM : 6;\nwxplot2d([L(M,x),r(M,x^(9.231))],[x,0,1]);\nAnd here are pictures
\n\n\nConsider the ring $R$ of Laurent polynomials in $n$ variables over the finite field of two elements, $R= \\mathbb F_2[x_0,x_0^{-1},x_1,x_1^{-1},\\ldots]$, with the standard involution by inverting monomials. Concretely, I'm interested in a small number of variables, like $n\\in\\{1,2,3,4\\}$. Consider \"hermitian\" $R$-valued matrices $M$, i.e., $M=M^\\dagger$ where $\\dagger$ denotes both transposition and ring involution. I'm interested in the equivalence classes of such matrices under both of the following two equivalences:
\n(1) Congruence, that is, $M\\sim A^\\dagger MA$ for some invertible $R$-valued matrix $A$.
\n(2) Schur complement, that is, if $M$ has block form,\n$$M=\\begin{pmatrix}X&Y\\\\Y^\\dagger&Z\\end{pmatrix}\\;,$$\nand $Z$ has an inverse $Z^{-1}$, then $M\\sim X-YZ^{-1}Y^\\dagger$.\nI believe it may be sufficient to consider this equivalence for $Y=1$ and $Y=\\begin{pmatrix}0&1\\\\1&0\\end{pmatrix}$.
\nQuestion: Do you believe that there could be a reasonable classification of hermitian matrices under the equivalence relations above? Do you know of similar classification problems (such as for other rings $R$) where classification is known to be feasible or unfeasible? Is the Schur complement of bilinear forms something that has been considered anywhere in the literature as an equivalence relation?
\nI have reasons to believe that the classification should be possible as it corresponds to some well-known problem in theoretical physics, namely the classification of topological phases restricted to \"free\" qubit models, which is known to some extent (at least for $n\\leq 3$). On the other hand, this also suggests that the classification is not too simple.
\n$\\newcommand\\Ksymb{\\genfrac(){}{}}$For $s=1,2,3,\\dotsc$, let us define\n$$L_{-3}(s):=L\\left(s,\\Ksymb{-3}\\cdot\\right)=\\sum_{k=1}^\\infty\\frac{\\Ksymb{-3}k}{k^s}=\\sum_{n=0}^\\infty\\left(\\frac1{(3n+1)^s}-\\frac1{(3n+2)^s}\\right),$$\nwhere $\\Ksymb{-3}k$ is the Kronecker symbol which coincides with the Legendre symbol $\\genfrac(){}{}k3$.
\nIt is well known that $$\\sum_{k=1}^\\infty\\frac1{k\\binom{2k}k}=\\frac{\\pi}{3\\sqrt3}=L_{-3}(1).\\tag{0}\\label{508769_0}$$\nMotivated by this and Questions 508743 and 508768, I make the following conjecture involving higher-order derivatives of the function\n$$h(k)=\\frac1{k\\binom{2k}k}=\\frac{\\Gamma(k+1)^2}{k\\Gamma(2k+1)}.$$
\nConjecture. We have\n$$\\sum_{k=1}^\\infty h'(k)=-\\frac32L_{-3}(2),\\tag{1}\\label{508769_1}$$\n$$\\sum_{k=1}^\\infty h''(k)=3L_{-3}(3)=\\frac{4\\pi^3}{27\\sqrt3},\\tag2\\label{508769_2}$$\nand\n$$\\sum_{k=1}^\\infty h'''(k)=\\frac34\\left(2\\pi^2L_{-3}(2)-27L_{-3}(4)\\right).\\tag{3}\\label{508769_3}$$
\nQUESTION. Are the three conjectural identities known? If some of them are new, how to prove them?
\nYour comments are welcome!
\nI am turning 72 this year, an abundant number between a pair of twin primes. We do not know yet whether or not there are infinitely many twin primes, but can I be certain that at least most numbers, more than half inside the pairs of twin primes (.e. $\\ n\\ $ such that $\\ n\\pm1\\ $ are both prime), are abundant?
\nEvery multiple of $6$ greater than $6$ is abundant. Every twin prime pair greater than $(3, 5)$ is of the form $(6n-1, 6n+1)$. So every number between a prime pair except $4$ and $6$ is abundant.
\n$\\DeclareMathOperator\\Ker{Ker}$Given two Banach spaces $X$ and $Y$ with a continuous inclusion $X\\subset Y$, and another couple $X’ \\subset Y’$ with the same properties. Take $f : Y \\longrightarrow Y’$ linear continuous, such that $f_{\\mid X}$ induces a linear continuous map from $X$ to $X'$. My question is for $0<s<1$ and $p>1$, is the following true:
\n$$(X\\cap \\Ker(f),\\Ker(f))_{s,p}=\\Ker(f) \\cap (X,Y)_{s,p}\\;\\;?$$
\nHere I'm considering the K-method for the interpolation.
\nThe inclusion $(X\\cap \\Ker(f),\\Ker(f))_{s,p}\\subset \\Ker(f) \\cap (X,Y)_{s,p}$ follows directly from the definition. My problem is the other inclusion.
\nIt's clear that if we have $Z\\subset Y$ then in general the following is not true:
\n$$(X\\cap Z,Z)_{s,p}=Z \\cap (X,Y)_{s,p},$$
\none can take $X=H^{2}(U)$, $Z=H^{1}(U)$, $Y=L^{2}(U)$, $s=\\frac{1}{2}$, and $p=2$. But this does not contradict our case (because $Z$ here is not even closed in $Y$).
\nIf what I'm asking is not true in general, is it true under the following assumptions:
\n$f(X)$ is closed in $X'$, and $f(Y)$ is closed in $Y'$, and
\n$f$ is open onto $f(Y)$, and $f_{\\mid X}$ is open onto $f(X)$.
\n(Making CW as this is an extended comment.)
\nLet's generalize slightly1: let $X\\hookrightarrow Y$ be an embedding of Banach spaces, and let $S\\subset Y$ be a closed subspace with the induced norm.
\nYour question essentially boils down to comparing the $K$ functionals $K$ and $K^{(S)}$, where\n$$ K(t,z) = \\inf_{z = x + y} \\|x\\|_X + \\|y\\|_Y $$\nand $K^{(S)}$ is analogously defined, except that instead of allowing arbitrary decompositions $z = x+y$ with $x\\in X$ and $y\\in Y$, you are looking at decompositions with $x \\in X\\cap S$ and $y\\in S$. (Obviously here $z\\in S$ as well.)
\nClearly $K(t,z) \\leq K^{(S)}(t,z)$ for $z\\in S$.
\nYour question will have a positive answer if $K(t,z) \\gtrsim K^{(S)}(t,z)$ for some implicit constant independent of $t$ and $z$.
\n(This is why I objected to your \"$H^1$ counterexample; the $K$-functional for the $(X\\cap Z,Z)$ interpolation has almost nothing to do with the functional for the $(X,Y)$ interpolation in that case. Compared to this case where it is essentially a geometry question about how $S$ is situated in $X$ and in $Y$.)
\nA sufficient condition is therefore that there exists a projection $\\Pi:Y\\to S$ with the property that $\\Pi(X) \\subseteq X\\cap S$ and the operator norms $\\|\\Pi\\|_{Y\\to S}$ and $\\|\\Pi\\|_{X\\to X\\cap S}$ are both bounded.
\n(The existence of such projection operators is in general a non-trivial assumption, as for Banach spaces there are generally closed subspaces which are not images of continuous projections, and for Hilbert spaces non-orthogonal projections may fail to be bounded.2)
\nThis would be the case if, for example, $X, Y$ were Hilbert, and $S$ is such that whenever $x\\in X$ is orthogonal to $S$ w.r.t. the $X$ inner product, it is also orthogonal w.r.t. the $Y$ inner product.
\nAs Hannes noted in a comment below: the result alluded to above is given as Theorem 1 in Section 1.17.1 of Interpolation Theory, Function Spaces, Differential Operators (ed. Hans Triebel, North-Holland (1978)).
\n1: When $Y$ is separable, every closed subspace is a kernel, so in this case it is not more general. https://arxiv.org/abs/1811.02399
\n2: I don't see how to relate the existence of these projections to the additional conditions you are willing to assume, as stated in the question. But chances are this is because I don't know very much about complemented subspaces.
\nLet $Z(g_1)$ and $Z(g_2)$ be the centralizers of two permutations $g_1$ and $g_2$ in the symmetric group $S_n$. Is there an algorithm which calculates the intersection $Z(g_1) \\cap Z(g_2)$ as a subgroup of $S_n$? Any references will be highly appreciated.
\nThe intersection of the centralizers of $g_1$ and $g_2$ is just the centralizer of the subgroup $\\langle g_1,g_2 \\rangle$ of $S_n$. Yes there is an efficient algorithm for that problem. It can be done in polynomial time.
\nIt is described in detail in Section 6.1.2 of the book \"Permutation Group Algorithms\" by Akos Seress.
\n\n\nConjecture. Let $D_n$ denote the number of derangements on $n$ and $S$ denote the set of all $k$ such that $k \\nmid D_{2n-1}$ for all $1 < n \\leq k$. Then we have the following: (a) if $a \\in S$ and $p \\mid a$ is prime then $p \\in S$, (b) if $a,b \\in S$ and $9 \\nmid ab$ then $ab \\in S$.
\n
For reference, this is the set $S$ up to $n = 100$: $$\\{1,3,5,7,15,17,19,21,23,25,29,35,43,49,51,57,59,61,69,71,73,75,85,87,95\\}.$$
\nNow I can prove (b) under the stronger assumption that $\\gcd(a,b) = 1$ and $a,b \\in S$ implies that $ab \\in S$.
\nIt can be shown that $$D_{n+k} = (-1)^kD_n \\pmod{k} \\implies D_{n+2k} = D_n \\pmod{k}.$$ This means that $$D_{2(n+k)-1} = D_{(2n-1)+2k} = D_{2n-1} \\pmod{k}$$ and so the sequence $x_n = D_{2n-1} \\pmod{k}$ has period $k$.
\nNow we show that if $k \\in S, k \\mid D_{2n-1} \\implies n = 1 \\pmod{k}$.
\nThis is since $$D_{2n-1} = D_{2\\alpha-1} \\pmod{k}$$ where $\\alpha \\equiv n \\mod{k}$ and $\\alpha \\in \\{1,\\ldots,k\\}$ and so $k \\mid D_{2\\alpha-1}$. But this must mean that $n = 1 \\pmod{k}$ since $k \\nmid D_{2\\alpha-1}$ for each $1 < \\alpha \\leq k$.
\nLet $a,b \\in S$ with $\\gcd(a,b) = 1$. To show $ab \\in S$ suppose for sake of contradiction that $ab \\not \\in S$. Then there exists $n$ such that $1 < n \\leq ab$ and $ab \\mid D_{2n-1}$ and so $a \\mid D_{2n-1}$ and $b \\mid D_{2n-1}$. Now by the last fact this means that $n = 1 \\pmod{a}$ and $n = 1 \\pmod{b}$ and since $\\gcd(a,b) = 1$ it means that $n = 1 \\pmod{ab}$, which is a contradiction if $1 < n \\leq ab$.
\nLet $p: B \\to M$ be a smooth fiber bundle. Denote $\\Gamma(B)$ the set of smooth sections. What is the natural topology on this set?
\nIt is often viewed as a Frechet manifold. For example, in a paper by Richard Hamilton \"Inverse function theorem by Nash and Moser\" its charts are given (example 4.1.2). Yet, what bothers me, is that its topology is not stated explicitly in advance. So, formally, in this example it is just a charted set with smooth transition functions. Is its topology inherited from the maximal atlas somehow?
\nEdit: if possible, give citations, please.
\nPeople usually consider one of the two Whitney topologies.
\nThe weak one is the natural topology that makes sequences converge if and only if all derivatives converge locally uniformly (it is the only separable, locally convex, Hausdorff topology satisfying this property), and it is Fréchet. One way of defining it is to say it is induced by the family of seminorms given by the supremum of $|\\partial^\\alpha f_\\phi|$ over $K$, where $K$ is compact inside an open set $U$ for a trivializing chart $\\phi:\\pi_B^{-1}U\\to V\\times\\mathbb R^r$, $f_\\phi$ is the section expressed in those coordinates, and $\\alpha\\in\\mathbb R^d$ is a multi-index.
\nThe strong one is defined to make sense of “globally uniform convergence”, so they coincide when the base space is compact. In this case the space is not Fréchet when the base space is not compact, and one choice of a defining family of seminorms is to take $\\sup_{\\phi,\\alpha,K}|\\partial^\\alpha f_\\phi|$ with the argument like before, over some collection of $(\\phi,\\alpha,K)$ such that the domains $U$ of the trivializations $\\phi$ are locally finite, and $\\alpha$ takes only finitely many values (equivalently, the partial derivatives have bounded order).
\nI have the following conjecture.
\nConjecture. Let $A$ be a finite set in some abelian group $G$, and let $A = B\\cup B'$ be a partition of $A$ into two disjoint nonempty sets. We define a graph on the vertex set $B'$ by placing an edge between $b_1'$ and $b_2'$ if and only if $b_1' + b_2' \\notin A+B$. My conjecture is that there always exists some $b'\\in B'$ such that\n$$\\deg(b') \\le {|A+B|\\over |B|}.$$
\nHere's some vague reasoning why the inequality might be true. If $B$ is very small, then the right-hand side is close to $|A|$. If $B$ is very large, then $B'$ is very small and the degree on the left-hand side, which is bounded above by $|B|-1$, is small.
\nThe trouble arises when $A+B$ is small. In this case the inequality might seem more far-fetched, and indeed there are cases where some of the $b'$ have degree $> |A+B|/|B|$, but still I've always been able to find some $b'$ with small degree. (And I've never actually found any cases where equality holds for the minimiser $b'$, it's always a strict inequality.)
\nI thought that maybe some basic averaging argument would do the trick, but that doesn't seem to be the case. Then I thought maybe the conjecture is false, but at least in the case $G={\\bf Z}$ that interests me most, I have been unable to find any counterexample using Sage. (Of course I haven't done an exhaustive search, I've just generated some random examples, and then also tried certain \"structured\" choices of $A$, like arithmetic progressions and geometric rogressions, combinations of the two, etc.)
\nCounterexamples over other groups would be interesting to me as well, as they would rule out many of the basic \"Ruzsa calculus\" techniques.
\nHere is a family of counterexamples using elementary abelian $2$-groups, i.e. vector spaces over $\\mathbb{F}_2$.
\nTake $G = \\mathbb{F}_2^n$ and let $G = B' \\oplus \\langle v \\rangle$, so $B'$ is a codimension $1$ hyperplane in $G$ with a complement spanned by $v$. Let $C$ be a codimension $1$ hyperplane in $B'$ and let $B = C + v$. Thus $|B'| = 2^{n-1}$, $|B| = |C| = 2^{n-2}$ and since $B \\cap B' = \\varnothing$, we have $|A| = |B \\cup B'| = |B| + |B'| = 3 \\times 2^{n-2}$.
\nObserve that
\n$$A+ B = (B \\cup B') + B = (B + B) \\cup (B' + B) = C \\cup (B' + v),$$
\nwhere we used $B + B = (C+ v) + (C+ v) = C$ and, since $C \\subseteq B'$,
\n$$B' + B = B' + C + v = B' + v.$$
\nThus $|A+B| = 3 \\times 2^{n-2}$ and the ratio in the question is $|A+B|/|B| = 3$.
\nFor the graph we have an edge $\\{b_1', b_2'\\}$ for $b_1', b_2' \\in B'$ if and only if
\n$$b_1' + b_2' \\not\\in A + B = C \\cup (B' + v).$$
\nSince $B'$ is a hyperplane, we never have $b_1' + b_2' \\in B'+ v$. Hence there is an edge $\\{b_1', b_2'\\}$ if and only if $b_1'+b_2' \\not\\in C$, the codimension $1$ hyperplane inside $B'$. This is the case if and only if $b_1', b_2'$ come from different cosets of $C$ in the union $B' = C \\cup (B' \\backslash C)$. Therefore the graph is is the complete bipartite graph with bipartition $\\{C, B' \\backslash C\\}$. In particular, every vertex has degree $|C| = |B' \\backslash C| = 2^{n-2}$. Therefore we get a counterexample if and only if $2^{n-2} \\not\\le 3$. The smallest such $n$ is $n= 4$.
\nI've seen it written that the existence of the Joyal model structure on simplicial sets is one of the foundational results on which the theory of $\\infty$-categories (or, rather, the theory of quasi-categories as a model of $\\infty$-categories) rests. On the other hand, I've worked through (what seems to me like) most of the foundational results which set up the theory of quasi-categories, and I have yet to encounter any serious/non-trivial use of the Joyal model structure. This leads me to ask:
\n\n\nWhat is the Joyal model structure actually useful for?
\n
In fact, I don't even have a good understanding of what the existence of the Joyal model structure really means, so a related question (which may share the same answer) would be:
\n\n\nWhat is the principal mathematical content in the existence of the Joyal model structure?
\n
Or to be really concrete:
\n\n\nName a foundational result about quasi-categories whose proof uses the existence of the Joyal model structure (or something equivalent to it)? Or, convince me that this is the \"wrong\" question to be asking.
\n
I guess that's three questions, but they're basically all asking the same thing.
\nIf you’re reading HTT say, you might not find the existence of the Joyal model structure explicitly invoked very often. But for almost any definition you find in the book, you can ask “why is this the right definition?”. Lurie’s answer is generally “well, just watch— I’m about to show you how this definition leads to a successful generalization of 1-category theory”. Then you can sit there and say “wow, it works because it works, and Joyal and Lurie managed to steal a copy of the Book from the SF!”. But lurking in the background, there’s often some meta-considerations suggesting why such a definition. should work, and these considerations are typically of a model categorical flavor.
\nFor example, consider the definition of limits and colimits in terms of the join. A priori, it’s hard to imagine that such a concrete, “ point-set “ definition should be correct! it even specializes in the 1-categorical case to the usual definition (up to isomorphism, not equivalence). But it becomes less mysterious when you realize that join with a fixed object is left Quillen, so adjointly, the cone construction is homotopy-invariant on fibrant objects (and hence doesn’t need to be “more derived” at least on fibrant objects). Since the cofibrations are easy to understand (and every object is cofibrant), mysterious coherence questions about a right-adjoint construction are translated into simple combinatorial questions on the left adjoint. Maybe Lurie doesn’t often explicitly invoke this in his arguments, but these kinds of considerations allow the reader to believe that what he’s doing should be possible (and before that presumably allowed Joyal and Lurie some hope that it might be possible).
\nSome of this sort of thinking is made precise in Riehl and Verity’s infinity cosmos work. They work with less than a full model structure— to be sure, you don’t need a model structure per se. But the model categorical philosophy still feels evident in motivating why you write one set of axioms and not another.
\nI think I’ve been addressing primarily your first question above.
\nFor your second question, I think I will refer back to my comment above — for pretty much any model category, the “principal mathematical content” of its existence is a matter of perspective, and you probably need to clarify what sort of answer you’re looking for there.
\nFor your third question, I think the examples given by Tyler Lawson and Phil Tosteson in the comments are great. Phil’s is particularly punchy — you should not take for granted that a functor category $Fun(A,B)$ between quasicategories is literally just the internal hom of simplicial sets! The canonical proof of that fact is “the Joyal model structure is cartesian and every object is cofibrant”. As Phil says, in simplicial categories, for example, you can’t just write down a functor category without the migraines associated with cofibrantly replacing $A$…
\n(Tyler’s “(b)” gives a great example of how things are even better since in fact $A$ doesn’t need to be fibrant. Another example that comes to mind here is computing a sequential colimit as a coequalizer of two maps between coproducts, which comes from the fact that the nerve of the poset $\\mathbb N$ is Joyal-equivalent to its Hasse diagram (viewed as a 1-skeletal simplicial set).)
\n![\"<span](\"https://i.sstatic.net/65SI6RQB.png\")
$r_5$ and ![\"<span](\"https://i.sstatic.net/Ix7RiuEW.png\")
$r_6$ and ![https://i.sstatic.net/CUpGsNtr.jpg](\"https://i.sstatic.net/CUpGsNtr.jpg\"){kind=link}