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{
"question_id": 508744,
"title": "Possible numbers of integer points that an $n$-by-$n$ square can cover",
"link": "https://mathoverflow.net/questions/508744/possible-numbers-of-integer-points-that-an-n-by-n-square-can-cover",
"score": 2,
"view_count": 60,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "discrete-geometry",
"creation_date": "2026-03-05 10:24:18Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For positive integer <span class=\"math-container\">$n$</span>, how many integer points can an <span class=\"math-container\">$n$</span>-by-<span class=\"math-container\">$n$</span> square in the plane possibly cover?</p>\n<p>Let <span class=\"math-container\">$E$</span> be the set of all numbers that an <span class=\"math-container\">$n$</span>-by-<span class=\"math-container\">$n$</span> square can possibly cover. I guess\n<span class=\"math-container\">\\begin{equation*}\nE=[n^2,n(n+1)]\\cup\\{(n+1)^2\\}.\n\\end{equation*}</span>\nThe square <span class=\"math-container\">$[0.5,n+0.5]\\times[0.5,n+0.5]$</span> covers <span class=\"math-container\">$n^2$</span> integer points; the square <span class=\"math-container\">$[0,n]\\times[0,n]$</span> covers <span class=\"math-container\">$(n+1)^2$</span> integer points. If the latter is slightly tilted, it seems to miss at least <span class=\"math-container\">$(n+1)$</span> integer points. It is not difficult to prove the number can \"continuously\" change from <span class=\"math-container\">$n^2$</span> to <span class=\"math-container\">$n(n+1)$</span>, but whether <span class=\"math-container\">$n^2$</span> is the minimum and whether <span class=\"math-container\">$n(n+1)$</span> is maximal other than <span class=\"math-container\">$(n+1)^2$</span> are not so clear to me.</p>\n</div>",
"comments": [
{
"comment_id": 1325623,
"author_name": "Oleg Orlov",
"author_id": 552730,
"author_url": "https://mathoverflow.net/users/552730/oleg-orlov",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">A square can cover less than $n^2$ integer points, e. g. for $n=1$ the tilted square with center $(0.5,0.5)$ doesn't cover integer points.</span>",
"created_date": "2026-03-04 14:24:08Z"
},
{
"comment_id": 1325630,
"author_name": "Haoran Chen",
"author_id": 115637,
"author_url": "https://mathoverflow.net/users/115637/haoran-chen",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@OlegOrlov Great point!</span>",
"created_date": "2026-03-04 15:12:13Z"
},
{
"comment_id": 1325660,
"author_name": "Benjamin Baily",
"author_id": 490938,
"author_url": "https://mathoverflow.net/users/490938/benjamin-baily",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The $3\\times 3$ square $\\{(x,y): |x| + |y| \\leq 3/\\sqrt{2}\\}$ contains 13 integer points.</span>",
"created_date": "2026-03-04 22:22:34Z"
},
{
"comment_id": 1325661,
"author_name": "Benjamin Baily",
"author_id": 490938,
"author_url": "https://mathoverflow.net/users/490938/benjamin-baily",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">In general, $E$ can take the value $2z^2 - 2z + 1$, where $z = \\lceil n/\\sqrt{2}\\rceil$. When $n/\\sqrt{2}$ is close to an integer from above (e.g. $n = 3, 10$) this is outside your proposed range. See <a href=\"https://oeis.org/A235337\" rel=\"nofollow noreferrer\">oeis.org/A235337</a>.</span>",
"created_date": "2026-03-04 22:36:53Z"
}
],
"answers": [
{
"answer_id": 508778,
"question_id": 508744,
"score": 0,
"is_accepted": false,
"author_name": "Elias Panholzer",
"author_id": 586494,
"author_url": "https://mathoverflow.net/users/586494/elias-panholzer",
"author_reputation": 140,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-05 10:09:52Z",
"last_edited_date": "2026-03-05 10:24:18Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Case 1: all 4 sides cover integer points: By Pick's theorem this number is given by <span class=\"math-container\">$n^2+\\frac{4n}{2}+1=(n+1)^2$</span>. Case 2: no side cover integer points: We look at the smaller included <span class=\"math-container\">$(n-1)\\times(n-1)$</span> square with vertices with integer coordinates and obtain, by the same argument as in Case 1, <span class=\"math-container\">$n^2$</span> covered integer points. Case 3: exactly two opposite sides cover integers: This is equivalent to Case 1, where by tilting <span class=\"math-container\">$(n+1)$</span> integer points miss (side shifted away) and so the result is <span class=\"math-container\">$(n+1)^2-(n+1)=n(n+1)$</span>.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508767,
"title": "Capping off an almost complex structure over a cone on a link of a quotient surface singularity",
"link": "https://mathoverflow.net/questions/508767/capping-off-an-almost-complex-structure-over-a-cone-on-a-link-of-a-quotient-surf",
"score": 1,
"view_count": 48,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "dg.differential-geometry;4-manifolds;orbifolds",
"creation_date": "2026-03-05 10:05:53Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Let <span class=\"math-container\">$Y$</span> be an oriented <span class=\"math-container\">$3$</span>-dimesional manifold, given by a link of a surface quotient singularity. This means that <span class=\"math-container\">$Y=S^3/G$</span> for some finite subgroup <span class=\"math-container\">$G$</span> of <span class=\"math-container\">$U(2)$</span>, and the cone <span class=\"math-container\">$CY=D^4/G$</span> of <span class=\"math-container\">$Y$</span> naturally inherits a complex orbifold structure, with a unique singular point (the quotient singularity).</p>\n<p>Suppose we have an almost complex structure on the cylinder <span class=\"math-container\">$Y\\times [0,1]$</span>, which is a smooth, oriented <span class=\"math-container\">$4$</span>-manifold with boundary <span class=\"math-container\">$Y\\times \\{0\\} \\amalg Y\\times \\{1\\}$</span>. The orientation convention is that the obvious map <span class=\"math-container\">$Y\\times \\{0\\} \\amalg Y\\times \\{1\\}\\to -Y\\amalg Y$</span> is orientation-preserving, where <span class=\"math-container\">$-Y$</span> is the orientation reversal of <span class=\"math-container\">$Y$</span>. Can we extend the almost complex structure on <span class=\"math-container\">$Y\\times [0,1]$</span> to an almost complex structure over the orbifold <span class=\"math-container\">$Y\\times [0,1]\\cup CY$</span>, where <span class=\"math-container\">$CY$</span> is attached to the cylinder along <span class=\"math-container\">$Y\\times \\{0\\}$</span> (the oriented boundary of <span class=\"math-container\">$CY$</span> is <span class=\"math-container\">$Y$</span>)?</p>\n<p>The almost complex structure on <span class=\"math-container\">$CY$</span> 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 <span class=\"math-container\">$CY$</span> is the canonical one.</p>\n</div>",
"comments": [
{
"comment_id": 1325709,
"author_name": "Johannes Nordström",
"author_id": 13061,
"author_url": "https://mathoverflow.net/users/13061/johannes-nordstr%c3%b6m",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">You say \"orbifold\", but you seem initially to not impose any conditions on Y that ensure that CY is an orbifold, making it hard to see what conditions you are hoping to satisfy near the vertex of the cone.</span>",
"created_date": "2026-03-05 09:31:14Z"
},
{
"comment_id": 1325710,
"author_name": "blancket",
"author_id": 168675,
"author_url": "https://mathoverflow.net/users/168675/blancket",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@JohannesNordström You're right. Maybe I should focus on the special case that I mentioned, in which the orbifold structure is clear</span>",
"created_date": "2026-03-05 09:34:32Z"
}
],
"answers": []
},
{
"question_id": 508689,
"title": "Nonlinear wave equation and the light cone",
"link": "https://mathoverflow.net/questions/508689/nonlinear-wave-equation-and-the-light-cone",
"score": 0,
"view_count": 113,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "reference-request;ap.analysis-of-pdes;wave-equation",
"creation_date": "2026-03-04 18:53:20Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I 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 <span class=\"math-container\">$u=u(x,t)$</span>\n<span class=\"math-container\">$$ u_{tt}-\\Delta_x u = |u|^{p-2} u$$</span> for <span class=\"math-container\">$ x \\in \\mathbb{R}^N$</span> and <span class=\"math-container\">$ t \\in \\mathbb{R}$</span> or a subset of the time interval (I am looking for backward of forward self similar). In the end I get an elliptic equation for <span class=\"math-container\">$w(y)$</span>. I have searched a bit and occasionally see some stuff that hints at solving the equation for <span class=\"math-container\">$|y|<1$</span> is enough because <span class=\"math-container\">$|y|=1$</span> is a characteristic. So my question is: if I solve the elliptic problem up to the boundary and show the solution is continuous on <span class=\"math-container\">$B_1$</span> with <span class=\"math-container\">$ w=0$</span> on <span class=\"math-container\">$ \\partial B_1$</span> is this enough to claim I have found some sort of weak solution to the original problem or do I need <span class=\"math-container\">$w$</span> to satisfy some specific BC's on <span class=\"math-container\">$|y|=1$</span>. Maybe this question doesn't belong here and rather on Mathstack exchange (I really don't know anything about the wave equation). Is there any basic references which talk about this? thanks</p>\n<p>EDIT. Let me be more specific. I have a pde similar to the above wave equation and when I look for self similar solutions I get some degenerate elliptic problem that degenerate on <span class=\"math-container\">$|y|=1$</span> (<span class=\"math-container\">$y$</span> is the self similar variable). I can solve this elliptic problem on <span class=\"math-container\">$|y|<1$</span> with with <span class=\"math-container\">$ w=0$</span> on <span class=\"math-container\">$ |y|=1$</span> and maybe also a <span class=\"math-container\">$ \\partial_\\nu w =0$</span> on <span class=\"math-container\">$|y|=1$</span> and the solution on the elliptic problem has some nice properties. So I am interested if this gives new results on the wave equation and hence I want to know if solving on <span class=\"math-container\">$B_1$</span> is sufficient. ChatGPT seems to say for the backward self similar it might be; for the forward it seems to say its not.</p>\n</div>",
"comments": [
{
"comment_id": 1325473,
"author_name": "Elias Panholzer",
"author_id": 586494,
"author_url": "https://mathoverflow.net/users/586494/elias-panholzer",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Vladimir Georgiev and Grozdena Todorova. \"Existence of a Solution of the Wave Equation with Nonlinear Damping and Source Terms\". In: Journal of Differential Equations, vol 109, iss 2: 295-308, 1994.</span>",
"created_date": "2026-03-03 08:41:41Z"
},
{
"comment_id": 1325519,
"author_name": "Math604",
"author_id": 66623,
"author_url": "https://mathoverflow.net/users/66623/math604",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks for the reference. Let me clarify what i am asking. I don't so much care about the result of of global existence versus and i don't even really care about this specific pde that much. I am really only interested in the specific stuff about the light cone and the boundary condition on. thanks</span>",
"created_date": "2026-03-03 15:28:34Z"
},
{
"comment_id": 1325588,
"author_name": "Carlo Beenakker",
"author_id": 11260,
"author_url": "https://mathoverflow.net/users/11260/carlo-beenakker",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">there is the <a href=\"https://en.wikipedia.org/wiki/Goursat_problem\" rel=\"nofollow noreferrer\">Goursat problem</a>, where you specify boundary conditions on characteristic surfaces; for the typical Cauchy problem the boundary conditions are on a space-like surface (say, initial time), and then no boundary conditions on the characteristic surfaces are allowed, these would make the problem overdetermined.</span>",
"created_date": "2026-03-04 07:14:26Z"
},
{
"comment_id": 1325648,
"author_name": "Math604",
"author_id": 66623,
"author_url": "https://mathoverflow.net/users/66623/math604",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks Carlo for the comment. Not sure I really understand it though (again I don't really know anything about the wave equation).</span>",
"created_date": "2026-03-04 18:50:12Z"
}
],
"answers": []
},
{
"question_id": 508160,
"title": "Counterexamples in Simon-Smith min-max theory",
"link": "https://mathoverflow.net/questions/508160/counterexamples-in-simon-smith-min-max-theory",
"score": 1,
"view_count": 88,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "riemannian-geometry;geometric-measure-theory;minimal-surfaces",
"creation_date": "2026-03-04 18:48:46Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I am reading about the Simon-Smith min-max method for constructing minimal surfaces in 3-manifolds, following <a href=\"https://www.math.ias.edu/delellis/sites/math.ias.edu.delellis/files/MinMax92.pdf\" rel=\"nofollow noreferrer\">this survey</a> by Colding and De Lellis.</p>\n<p>Here is what I understand:</p>\n<ul>\n<li>Given a min-max sequence of surfaces, up to a subsequence it always converges in the sense of varifolds.</li>\n<li>However, the limiting varifold can be very far from a smooth minimal surface. Most of the work goes into showing that, for a well-chosen min-max sequence, the limiting varifold is indeed a smooth minimal surface.</li>\n<li>The first step, which I understand, is to use the pull-tight process to generate min-max sequences which converge to <strong>stationary</strong> varifolds (Section 4 in the survey).</li>\n<li>However, even if the limiting varifold is stationary, it may still be far from a smooth minimal surface. To handle that, the authors introduce the notion of <strong>almost minimizing sequences</strong> in Section 5.</li>\n</ul>\n<p>In 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 <strong>not</strong> 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.</p>\n</div>",
"comments": [],
"answers": [
{
"answer_id": 508753,
"question_id": 508160,
"score": 3,
"is_accepted": false,
"author_name": "Otis Chodosh",
"author_id": 1540,
"author_url": "https://mathoverflow.net/users/1540/otis-chodosh",
"author_reputation": 7412,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-04 18:48:46Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Take <span class=\"math-container\">$S^3(\\sqrt{2})\\#S^3(1)$</span> namely join these two 3-spheres with a tiny neck. Consider a sweepout that starts in the <span class=\"math-container\">$S^3(1)$</span> region and at time <span class=\"math-container\">$t_0$</span> looks like two crossing equatorial <span class=\"math-container\">$S^2$</span>'s. Then push one of the <span class=\"math-container\">$S^2$</span>'s back down and continue sweeping the other one upwards. If you do this correctly it will cross the equator in <span class=\"math-container\">$S^3(\\sqrt{2})$</span> at <span class=\"math-container\">$t_1$</span>. Thus, this sweepout will have two max times, <span class=\"math-container\">$t_0,t_1$</span> (both have area <span class=\"math-container\">$8\\pi$</span>). Both are stationary varifolds. However, the <span class=\"math-container\">$t_0$</span> stationary varifold is not smooth (it's crossing of two smooth surfaces).</p>\n<p>Working out a picture of how this fails to be almost minimizing will be quite instructive.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508721,
"title": "Is the set of all hypergeometric identities finitely generated?",
"link": "https://mathoverflow.net/questions/508721/is-the-set-of-all-hypergeometric-identities-finitely-generated",
"score": 8,
"view_count": 352,
"answer_count": 2,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "hypergeometric-functions",
"creation_date": "2026-03-04 18:33:07Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>If we look at Wikipedia's <a href=\"https://en.wikipedia.org/wiki/Generalized_hypergeometric_function#Identities\" rel=\"noreferrer\">list of hypergeometric identities</a>, we find classic results such as Saalschütz's theorem—\n<span class=\"math-container\">$${}_3F_2 (a,b, -n;c, 1+a+b-c-n;1)= \\frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n}$$</span>\n—and Clausen's formula—\n<span class=\"math-container\">$${}_3F_2(2c-2s-1, 2s, c-1/2; 2c-1, c; x)=\\, {}_2F_1(c-s-1/2,s; c; x)^2.$$</span>\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.</p>\n<p>The 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 <span class=\"math-container\">$\\mathbb{Q}$</span>. The simplest thing I can think of is to fix a finite number of indeterminates <span class=\"math-container\">$a,b,c,\\ldots$</span> and allow a linear combination of <span class=\"math-container\">${}_mF_n$</span> expressions where the parameters are affine functions of the indeterminates, and the coefficients are rational functions of the indeterminates and of <span class=\"math-container\">$x$</span>. The \"constant term\" is trickier to specify; I guess we have to allow an integer parameter <span class=\"math-container\">$n$</span> and the constant term will have the property that the ratio between the value of the term at <span class=\"math-container\">$n+1$</span> and the value of the term at <span class=\"math-container\">$n$</span> 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.</p>\n<p>Having 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.</p>\n<p>Anyway, 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.</p>\n</div>",
"comments": [
{
"comment_id": 1325553,
"author_name": "Steven Stadnicki",
"author_id": 7092,
"author_url": "https://mathoverflow.net/users/7092/steven-stadnicki",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Would all the sorts of things you're thinking about be covered by W-Z pairs? <a href=\"https://en.wikipedia.org/wiki/Wilf%E2%80%93Zeilberger_pair\" rel=\"nofollow noreferrer\">en.wikipedia.org/wiki/Wilf%E2%80%93Zeilberger_pair</a></span>",
"created_date": "2026-03-03 20:17:33Z"
},
{
"comment_id": 1325559,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@StevenStadnicki Possibly, but even if so, my question is really about the structure of the space of all W-Z pairs (of a certain form, I guess...), which I haven't seen discussed in the papers about the W-Z method that I'm aware of.</span>",
"created_date": "2026-03-03 21:37:01Z"
}
],
"answers": [
{
"answer_id": 508745,
"question_id": 508721,
"score": 4,
"is_accepted": false,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"author_reputation": 91241,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-04 14:02:28Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I 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.</p>\n<p>There is a distinction between <i>identities</i> such as Saalschütz's theorem, which concern values of a hypergeometric function at a special point (usually a simple rational number), and <i>transformation formulas</i> such as Clausen's formula, which is valid for all arguments.</p>\n<p>A 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 <a href=\"https://dl.acm.org/doi/abs/10.1137/0513021\" rel=\"nofollow noreferrer\">Strange evaluations of hypergeometric series</a>, by Ira Gessel and Dennis Stanton, <i>SIAM J. Math. Anal.</i> 13 (1982), 295–308. Other references include <a href=\"https://sites.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/strange.html\" rel=\"nofollow noreferrer\">\nForty \"Strange\" Computer-Discovered [and Computer-Proved (of course!)]\nHypergeometric Series Evaluations</a> by Shalosh B. Ekhad, <a href=\"https://arxiv.org/abs/1308.5588\" rel=\"nofollow noreferrer\">Special values of the hypergeometric series</a> by Akihito Ebisu (<i>Mem. Amer. Math. Soc.</i> 248 (2017)), and <a href=\"https://arxiv.org/abs/2506.20289\" rel=\"nofollow noreferrer\">(Strange) gamma evaluations</a> by Wadim Zudilin.</p>\n<p>By contrast, transformation formulas are much rarer. A good reference is <a href=\"https://www.mdpi.com/2073-8994/14/8/1541\" rel=\"nofollow noreferrer\">Beyond the Beta Integral Method: Transformation Formulas for Hypergeometric Functions via Meijer's <span class=\"math-container\">$G$</span> Function</a>, by Dmitrii Karp and Elena Prilepkina, <i>Symmetry</i> 14.8 (2022), 1541. Referring to Bailey's classic book, <i>Generalized Hypergeometric Series</i>, Karp and Prilepkina write:</p>\n<blockquote>\n<p>His 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.\"</p>\n</blockquote>\n<p>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:</p>\n<blockquote>\n<p>It 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.</p>\n</blockquote>\n<p>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.</p>\n</div>",
"comments": [
{
"comment_id": 1325632,
"author_name": "Alexandre Eremenko",
"author_id": 25510,
"author_url": "https://mathoverflow.net/users/25510/alexandre-eremenko",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The question was whether the set of these identities is finitely generated, not \"how hard it is to find new ones\", or \"how to classify them\".</span>",
"created_date": "2026-03-04 15:38:52Z"
},
{
"comment_id": 1325637,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@AlexandreEremenko Precisely. There is apparently no such theorem, and it seems that the empirical evidence is that they are not finitely generated. An explicit finite presentation would be a classification of sorts. Notice that in my attempt to make the question more precise, I said \"something like, the known identities generate all possible identities\" (formally). If sporadic examples seem to be findable at will then it's unlikely that the known identities formally generate all possible identities.</span>",
"created_date": "2026-03-04 16:56:49Z"
},
{
"comment_id": 1325674,
"author_name": "Alexandre Eremenko",
"author_id": 25510,
"author_url": "https://mathoverflow.net/users/25510/alexandre-eremenko",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I understand. But there is a possibility that there is some non-constructive \"finiteness theorem\" with a \"soft\" proof.</span>",
"created_date": "2026-03-05 01:10:13Z"
},
{
"comment_id": 1325675,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@AlexandreEremenko That would be extremely interesting to me!</span>",
"created_date": "2026-03-05 01:11:06Z"
}
]
},
{
"answer_id": 508723,
"question_id": 508721,
"score": 4,
"is_accepted": false,
"author_name": "Carlo Beenakker",
"author_id": 11260,
"author_url": "https://mathoverflow.net/users/11260/carlo-beenakker",
"author_reputation": 206919,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-03 20:19:40Z",
"last_edited_date": "2026-03-04 18:33:07Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>This may well be what you are looking for:<br/>\nH. S. Wilf & D. Zeilberger, <a href=\"https://sites.math.rutgers.edu/~zeilberg/mamarimY/Zeilberger_y1992_p575.pdf\" rel=\"nofollow noreferrer\">An algorithmic proof theory for hypergeometric\n(ordinary and \"q\") multisum/integral identities</a> (1992).\nTheir finding is that \"<em>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.</em>\"<br/>\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.</p>\n<hr/>\n<p>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<span class=\"math-container\">$$S(\\mathbf{\\alpha})=\\sum_k F(k,\\mathbf{\\alpha}),\\tag{1}$$</span>\nwhere <span class=\"math-container\">$F(k,\\alpha)$</span> is a general hypergeometric function in <span class=\"math-container\">$k$</span>, depending on parameters <span class=\"math-container\">$\\mathbf{\\alpha}=(\\alpha_1,\\alpha_2,\\ldots\\alpha_m)$</span>. \"Hypergeometric\" in this context means that <span class=\"math-container\">$F(k+1,\\mathbf{\\alpha})/F(k,\\mathbf{\\alpha})$</span> is a <em>rational</em> function of <span class=\"math-container\">$k$</span>. The sum over <span class=\"math-container\">$k$</span> is equated to an explicit closed form <span class=\"math-container\">$S(\\mathbf{\\alpha})$</span>. The WZ‑algorithm determines all linear recurrence relations in <span class=\"math-container\">$\\mathbf{\\alpha}$</span> satisfied by <span class=\"math-container\">$S(\\mathbf{\\alpha})$</span> and finds that these recurrences are generated by a <em>finite</em> set of primitive recurrences. In this algorithmic sense the infinite set of identities of the form (1) is \"finitely generated\".</p>\n<p>As 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 <a href=\"https://mathworld.wolfram.com/q-HypergeometricFunction.html\" rel=\"nofollow noreferrer\">q-hypergeometric functions</a>, with a term ratio that is rational in <span class=\"math-container\">$q^k$</span> rather than in <span class=\"math-container\">$k$</span>, have infinitely many primitive identities.</p>\n</div>",
"comments": [
{
"comment_id": 1325558,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I have some familiarity with WZ theory, but I didn't think it answered my question. Can you elaborate on your italicized sentence? At first I thought you were quoting a theorem in the paper but I don't know exactly what result you mean.</span>",
"created_date": "2026-03-03 21:33:25Z"
},
{
"comment_id": 1325562,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Also, to clarify, I'm not asking about the space of all identities that can be proven by the WZ method. Saalschütz's theorem and Clausen's formula have a very specific form, and there seems to be a short list of formulas of this type.</span>",
"created_date": "2026-03-03 21:40:54Z"
},
{
"comment_id": 1325564,
"author_name": "Carlo Beenakker",
"author_id": 11260,
"author_url": "https://mathoverflow.net/users/11260/carlo-beenakker",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I clarified what is a quote from the WZ paper and what my understanding of the approach.</span>",
"created_date": "2026-03-03 21:54:04Z"
}
]
}
]
},
{
"question_id": 61773,
"title": "$p$-adic Langlands correspondence",
"link": "https://mathoverflow.net/questions/61773/p-adic-langlands-correspondence",
"score": 23,
"view_count": 3000,
"answer_count": 2,
"is_answered": true,
"has_accepted_answer": true,
"accepted_answer_id": 61808,
"tags": "reference-request;nt.number-theory;galois-representations",
"creation_date": "2026-03-04 18:00:02Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Basic 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. </p>\n<p>Thanks!</p>\n</div>",
"comments": [
{
"comment_id": 155189,
"author_name": "Chandan Singh Dalawat",
"author_id": 2821,
"author_url": "https://mathoverflow.net/users/2821/chandan-singh-dalawat",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Start with Pierre Berger's recent Bourbaki exposé 1017 <a href=\"http://arxiv.org/abs/1002.4111\" rel=\"nofollow noreferrer\">arxiv.org/abs/1002.4111</a>.</span>",
"created_date": "2011-04-15 02:28:36Z"
},
{
"comment_id": 155213,
"author_name": "Olivier",
"author_id": 2284,
"author_url": "https://mathoverflow.net/users/2284/olivier",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">If you read the proofs of the construction of the $p$-adic Langlands correspondence, you will see that the restriction $F=\\mathbb Q_{p}$ is ubiquitous, starting from (but not restricted to) the fact that you want the residual field to be cyclic. @Chandan Pierre Berger is an entrepreneur and benefactor unlikely ever to contribute to Bourbaki, it is of course Laurent you want.</span>",
"created_date": "2011-04-15 03:46:34Z"
},
{
"comment_id": 155324,
"author_name": "Chandan Singh Dalawat",
"author_id": 2821,
"author_url": "https://mathoverflow.net/users/2821/chandan-singh-dalawat",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Olivier : quelle bêtise de ma part ! C'est inexcusable, d'autant plus que je connais Laurent Berger, et j'ai beaucoup entendu parler de l'ami Pierre Bergé du feu Yves Saint-Laurent... Je demande pardon à Laurent.</span>",
"created_date": "2011-04-15 13:44:42Z"
},
{
"comment_id": 155326,
"author_name": "Laurent Berger",
"author_id": 5743,
"author_url": "https://mathoverflow.net/users/5743/laurent-berger",
"score": 7,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Chandan : tu es tout excusé!</span>",
"created_date": "2011-04-15 13:52:31Z"
},
{
"comment_id": 155327,
"author_name": "Laurent Berger",
"author_id": 5743,
"author_url": "https://mathoverflow.net/users/5743/laurent-berger",
"score": 7,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Breuil has written an article for the ICM (\"The emerging p-adic Langlands programme\" see his webpage <a href=\"http://www.math.u-psud.fr/~breuil/publications.html\" rel=\"nofollow noreferrer\">math.u-psud.fr/~breuil/publications.html</a>), where he explains the difficulties in going from $Q_p$ to an extension.</span>",
"created_date": "2011-04-15 13:54:40Z"
}
],
"answers": [
{
"answer_id": 61808,
"question_id": 61773,
"score": 23,
"is_accepted": true,
"author_name": "vytas",
"author_id": 13024,
"author_url": "https://mathoverflow.net/users/13024/vytas",
"author_reputation": 433,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2011-04-15 12:10:51Z",
"last_edited_date": "2026-03-04 18:00:02Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Yes, this is correct.</p>\n<p>The problem is that when you replace <span class=\"math-container\">$Q_p$</span> by an extension, the dimension of <span class=\"math-container\">$GL_2(F)$</span> as a <span class=\"math-container\">$p$</span>-adic analytic group increases. This also means that the cohomological dimension of its open subgroups increases. This leads to representation theory of <span class=\"math-container\">$GL_2(F)$</span>\nof being much more complicated than <span class=\"math-container\">$GL_2(Q_p)$</span>. For example, smooth irreducible <span class=\"math-container\">$\\overline{\\mathbb F}_p$</span>-representations have not been classified if <span class=\"math-container\">$F\\neq Q_p$</span>.</p>\n<p>Prototypical examle: Let <span class=\"math-container\">$\\mathbb G=\\mathbb G_a$</span>, and let <span class=\"math-container\">$K=\\mathbb{G}(\\mathcal O_F)$</span>.\nSo that <span class=\"math-container\">$K$</span> is <span class=\"math-container\">$(\\mathcal O_F, +)$</span>. Then the completed group agebra <span class=\"math-container\">$\\mathcal O[[K]]$</span>\nis isomorphic to <span class=\"math-container\">$\\mathcal O[[x_1, ..., x_d]]$</span>, where <span class=\"math-container\">$d=[F:Q_p]$</span>, where <span class=\"math-container\">$\\mathcal O$</span> is a ring of inegers in a finite extension of <span class=\"math-container\">$Q_p$</span>. The theory of modules of <span class=\"math-container\">$\\mathcal O[[K]]$</span> is much easier, when <span class=\"math-container\">$d=1$</span>. 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<a href=\"https://math.uchicago.edu/~emerton/preprints.html\" rel=\"nofollow noreferrer\"> website </a>.</p>\n<p>Since <span class=\"math-container\">$GL_2(F)$</span> is locally pro-<span class=\"math-container\">$p$</span> this problem does not arise if you are working over <span class=\"math-container\">$\\mathbb C$</span> or <span class=\"math-container\">$\\mathbb F_l$</span>, <span class=\"math-container\">$l\\neq p$</span>.</p>\n</div>",
"comments": []
},
{
"answer_id": 61811,
"question_id": 61773,
"score": 22,
"is_accepted": false,
"author_name": "Emerton",
"author_id": 2874,
"author_url": "https://mathoverflow.net/users/2874/emerton",
"author_reputation": 58752,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2011-04-15 12:35:29Z",
"last_edited_date": "2026-03-04 17:20:11Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Regarding prospects for extending the correspondence to <span class=\"math-container\">$GL_2(F)$</span> for other <span class=\"math-container\">$F$</span>,\none could look at Paškūnas's papers \"Coefficient systems and supersingular representations\", \"Towards a modulo <span class=\"math-container\">$p$</span> Langlands correspondence for <span class=\"math-container\">$GL_2(F)$</span>\" (joint with C. Breuil),\nand \"Admissible unitary completions of locally <span class=\"math-container\">$\\mathbb Q_p$</span>-rational representations of\n<span class=\"math-container\">$GL_2(F)$</span>\", available on his <a href=\"https://www.esaga.uni-due.de/vytautas.paskunas/publications/\" rel=\"nofollow noreferrer\">website</a> and/or the <a href=\"https://arxiv.org/search/math?query=paskunas&searchtype=author\" rel=\"nofollow noreferrer\">arXiv</a>.</p>\n<p>There is also Breuil's ICM talk from last summer, \"The emerging <span class=\"math-container\">$p$</span>-adic Langlands program\", available at his <a href=\"https://www.imo.universite-paris-saclay.fr/%7Echristophe.breuil/publications.html\" rel=\"nofollow noreferrer\">website</a>.\nThis gives a very nice survey of the whole state of the theory (which has remained relatively stable since then).</p>\n<hr/>\n<p>Some commentary on Paškūnas's papers: In the <span class=\"math-container\">$GL_2(\\mathbb Q_p)$</span> case, Breuil found that the numbers of irred. supersingular reps. of <span class=\"math-container\">$GL_2(\\mathbb Q_p)$</span> mod <span class=\"math-container\">$p$</span> matches with the numbers of <span class=\"math-container\">$2$</span>-dim'l irred. mod <span class=\"math-container\">$p$</span> reps. of <span class=\"math-container\">$G_{\\mathbb Q_p}$</span>, 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 <span class=\"math-container\">$p$</span> global Galois reps.).</p>\n<p>It was then natural to conjecture that the same was true for <span class=\"math-container\">$GL_2(F)$</span>.\nThe first of these papers has the goal of verifying this conjecture. Indeed,\nit succeeds in constructing the right number\nof supersingular reps. mod <span class=\"math-container\">$p$</span> of <span class=\"math-container\">$GL_2(F)$</span>. 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).</p>\n<p>The second paper extends the techniques of the first, and shows in fact that when\n<span class=\"math-container\">$F \\neq \\mathbb Q_p$</span> there are <em>many, many</em> more supersingulars than there are <span class=\"math-container\">$2$</span>-dim'l.\nirreps of <span class=\"math-container\">$G_F$</span>. 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).</p>\n<p>The third paper shows how to lift mod <span class=\"math-container\">$p$</span> representations to <span class=\"math-container\">$p$</span>-adic Banach spaces\nrepresentations in interesting ways, and so can be thought of as (i) giving some evidence\nthat there will be a <span class=\"math-container\">$p$</span>-adic local Langlands for <span class=\"math-container\">$GL_2(F)$</span>, but also (ii) showing that\nunderstanding it will be at least as difficult as understanding the mod <span class=\"math-container\">$p$</span> situation.</p>\n</div>",
"comments": [
{
"comment_id": 155315,
"author_name": "",
"author_id": -1,
"author_url": "",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">A very minor comment: I assume 'Breuil' instead of 'Breiul' is meant. </span>",
"created_date": "2011-04-15 13:02:14Z"
},
{
"comment_id": 155325,
"author_name": "Chandan Singh Dalawat",
"author_id": 2821,
"author_url": "https://mathoverflow.net/users/2821/chandan-singh-dalawat",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I have taken the liberty of correcting the spelling.</span>",
"created_date": "2011-04-15 13:46:36Z"
},
{
"comment_id": 155415,
"author_name": "SGP",
"author_id": 11786,
"author_url": "https://mathoverflow.net/users/11786/sgp",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thank you for a very clear and detailed answer! Thanks also for providing clear answers to many many questions on MO!</span>",
"created_date": "2011-04-15 21:11:27Z"
},
{
"comment_id": 155416,
"author_name": "SGP",
"author_id": 11786,
"author_url": "https://mathoverflow.net/users/11786/sgp",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I wish I could select both yours and Vytas answers. Thanks again</span>",
"created_date": "2011-04-15 21:13:12Z"
},
{
"comment_id": 155439,
"author_name": "Emerton",
"author_id": 2874,
"author_url": "https://mathoverflow.net/users/2874/emerton",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Dear SGP, You're welcome. Regards, Matthew</span>",
"created_date": "2011-04-16 00:10:42Z"
}
]
}
]
},
{
"question_id": 488840,
"title": "Binomial coefficient C(2k,n-1) alternative formula equivalent to the Vandermonde identity?",
"link": "https://mathoverflow.net/questions/488840/binomial-coefficient-c2k-n-1-alternative-formula-equivalent-to-the-vandermonde",
"score": 1,
"view_count": 467,
"answer_count": 3,
"is_answered": true,
"has_accepted_answer": true,
"accepted_answer_id": 488880,
"tags": "reference-request;co.combinatorics;linear-algebra;binomial-coefficients;recurrences",
"creation_date": "2026-03-04 17:31:53Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I 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</p>\n<p><span class=\"math-container\">$$\n\\sum_{j=0}^{n}\n\\binom{k}{j}\\,\\binom{k}{n-j}\n\\;=\\;\n\\sum_{j=0}^{n}\n\\,(-1)^{\\,n-j}\n\\,\\frac{k-n}\n{\\displaystyle k - j}\\displaystyle \\binom{n}{j}\\,\\binom{k-1}{n}\\,\\binom{j+k}{n}\n\\,\\;=\\;\n\\binom{2k}{n}\\\n$$</span></p>\n<p>I’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.</p>\n<p>Is 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.</p>\n<p>If it's a new one please propose a name for it</p>\n<p>Thank you everyone</p>\n</div>",
"comments": [
{
"comment_id": 1274651,
"author_name": "Max Alekseyev",
"author_id": 7076,
"author_url": "https://mathoverflow.net/users/7076/max-alekseyev",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Why is it important to link the two sums rather than link the second sum directly to the binomial coefficient $\\binom{2k}{n-1}$ ?</span>",
"created_date": "2025-03-04 15:04:24Z"
},
{
"comment_id": 1274652,
"author_name": "TigranNersissian",
"author_id": 556642,
"author_url": "https://mathoverflow.net/users/556642/tigrannersissian",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The goal here is to understand the combinatorial reasoning behind the equality of the two sums rather than just relying on an algebraic identity. If there is a direct counting argument that transforms one sum into the other, it would provide deeper insight into the underlying structure of the identity. Linking the second sum directly to $ \\binom{2k}{n-1}\\ $ might give a shortcut but wouldn’t necessarily reveal the combinatorial intuition behind it.</span>",
"created_date": "2025-03-04 15:11:22Z"
},
{
"comment_id": 1274653,
"author_name": "TigranNersissian",
"author_id": 556642,
"author_url": "https://mathoverflow.net/users/556642/tigrannersissian",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I want to know if the right side is a new identity, before starting to write a chapter about this in my article</span>",
"created_date": "2025-03-04 15:19:30Z"
},
{
"comment_id": 1274665,
"author_name": "Peter Taylor",
"author_id": 46140,
"author_url": "https://mathoverflow.net/users/46140/peter-taylor",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Maybe more natural to rewrite slightly as $$\\sum_{j=0}^{n} \\binom{k}{j}\\binom{k}{n-j} = \\sum_{j=0}^{n} (-1)^{n-j} \\frac{k-n}{k-j}\\displaystyle \\binom{n}{j} \\binom{k}{n}\\binom{j+k}{n}$$</span>",
"created_date": "2025-03-04 16:57:05Z"
},
{
"comment_id": 1274673,
"author_name": "Hjalmar Rosengren",
"author_id": 10846,
"author_url": "https://mathoverflow.net/users/10846/hjalmar-rosengren",
"score": 5,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">About your question whether the right identity is new. You can check this by writing it in standard hypergeometric notation. The sum is a multiple of $${}_3F_2\\left(\\begin{matrix}1-n,k+1,-k\\\\2+k-n,1-k\\end{matrix};1\\right).$$ This is a special case of the Pfaff-Saalschutz summation, see Eq. 16.4.3 on <a href=\"https://dlmf.nist.gov/16.4#ii\" rel=\"nofollow noreferrer\">dlmf.nist.gov/16.4#ii</a>.</span>",
"created_date": "2025-03-04 18:12:11Z"
}
],
"answers": [
{
"answer_id": 488880,
"question_id": 488840,
"score": 7,
"is_accepted": true,
"author_name": "Ira Gessel",
"author_id": 10744,
"author_url": "https://mathoverflow.net/users/10744/ira-gessel",
"author_reputation": 17728,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2025-03-05 04:42:58Z",
"last_edited_date": "2026-03-04 17:31:53Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The identity is a special case of <a href=\"https://en.wikipedia.org/wiki/Generalized_hypergeometric_function#Saalsch%C3%BCtz%27s_theorem\" rel=\"nofollow noreferrer\">Saalschütz's theorem</a>.\nThe second sum may be written as\n<span class=\"math-container\">$$(-1)^n \\binom{k-1}{n}^2{}_3F_2\\left({-n,-k,k+1\\atop -k+1,k+1-n} \\biggm|1\\right)$$</span>\nand the <span class=\"math-container\">$_3F_2$</span> can be evaluated by Saalschütz's theorem.</p>\n<p>Added later: I missed <a href=\"https://mathoverflow.net/questions/488840/binomial-coefficient-c2k-n-1-alternative-formula-equivalent-to-the-vandermonde#comment1274673_488840\">Hjalmar Rosengren's comment</a>, in which he said the same thing earlier.</p>\n</div>",
"comments": []
},
{
"answer_id": 501222,
"question_id": 488840,
"score": 3,
"is_accepted": false,
"author_name": "T. Amdeberhan",
"author_id": 66131,
"author_url": "https://mathoverflow.net/users/66131/t-amdeberhan",
"author_reputation": 44033,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-10-05 18:17:43Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I arrived late to this game. Let's prove the identity with the help of the so-called Wilf-Zeilberger method of automatic proof.</p>\n<p>To this end, divide through by the factor <span class=\"math-container\">$\\binom{2k}n$</span> so that we show\nthe two identities\n<span class=\"math-container\">\\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*}</span>\nDefine the functions <span class=\"math-container\">$F_1(n,j):=\\frac{\\binom{k}{j}\\binom{k}{n-j}}{\\binom{2k}n}$</span>\nand <span class=\"math-container\">$G_1(n,j):= - \\frac12 \\frac{\\binom{k-1}{j-1}\\binom{k}{n+1-j}}{\\binom{2k-1}n}$</span>. One can routinely check that <span class=\"math-container\">$F_1(n+1,j)-F_1(n,j)=G_1(n,j+1)-G_1(n,j)$</span>. Then, sum both sides over all integers <span class=\"math-container\">$j$</span> (actually these are finite sums). The immediate impact is: the right-hand side vanishes. So, we obtain that <span class=\"math-container\">$f_1(n+1)=f_1(n)$</span> where <span class=\"math-container\">$f_1(n):=\\sum_{j=0}^n F_1(n,j)$</span>. Compute the value at, say <span class=\"math-container\">$n=0$</span> (and get <span class=\"math-container\">$f_1(0)=1$</span>) to confirm the first promised identity.</p>\n<p>I'll leave the 2nd identity because the procedure is very similar.</p>\n</div>",
"comments": []
},
{
"answer_id": 488845,
"question_id": 488840,
"score": 1,
"is_accepted": false,
"author_name": "Iosif Pinelis",
"author_id": 36721,
"author_url": "https://mathoverflow.net/users/36721/iosif-pinelis",
"author_reputation": 146947,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-03-04 16:14:31Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<blockquote>\n<p>I want to know if the right side is a new identity, before starting to write a chapter about this in my article</p>\n</blockquote>\n<p>This identity is known to Mathematica:</p>\n<p><a href=\"https://i.sstatic.net/oJ6SxgFA.png\" rel=\"nofollow noreferrer\"><img alt=\"enter image description here\" src=\"https://i.sstatic.net/oJ6SxgFA.png\"/></a></p>\n</div>",
"comments": [
{
"comment_id": 1274664,
"author_name": "TigranNersissian",
"author_id": 556642,
"author_url": "https://mathoverflow.net/users/556642/tigrannersissian",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Yes Mathematica can simplify it to their factorial form, so the identity is True Sum[Binomial[k, j]*Binomial[k, (n - 1) - j], {j, 0, n - 1}] == -Sum[(k*(-1)^(n - j)*Binomial[n - 1, j]* Binomial[k - 1, n - 1]*Binomial[j + k, n - 1])/(k - j), {j, 0, n - 1}] return true But it does not imply that the identity is new or not right ? Thank you very much for your comment</span>",
"created_date": "2025-03-04 16:32:32Z"
},
{
"comment_id": 1274669,
"author_name": "Iosif Pinelis",
"author_id": 36721,
"author_url": "https://mathoverflow.net/users/36721/iosif-pinelis",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Fibonacci : Right. However, I suspect that Mathematica derived this identity using some rather simple manipulations based on a vast collection of binomial identities that Mathematica has.</span>",
"created_date": "2025-03-04 17:32:36Z"
},
{
"comment_id": 1274711,
"author_name": "TigranNersissian",
"author_id": 556642,
"author_url": "https://mathoverflow.net/users/556642/tigrannersissian",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Yes totally agree with you, It can be nice if the method FullSimplify show the \"stacktrace\"</span>",
"created_date": "2025-03-04 22:34:50Z"
},
{
"comment_id": 1274793,
"author_name": "Peter Taylor",
"author_id": 46140,
"author_url": "https://mathoverflow.net/users/46140/peter-taylor",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@IosifPinelis, unlikely. Searching for applicable identities is hard: Gosper's algorithm is easier and more general.</span>",
"created_date": "2025-03-05 08:17:39Z"
},
{
"comment_id": 1274847,
"author_name": "Iosif Pinelis",
"author_id": 36721,
"author_url": "https://mathoverflow.net/users/36721/iosif-pinelis",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@PeterTaylor : Thank you for your comment. However, the accepted answer suggests that, at least in this case, the identity was derived simply based on a known family of identities.</span>",
"created_date": "2025-03-05 15:52:05Z"
}
]
}
]
},
{
"question_id": 472830,
"title": "Show that $\\|P(f\\circ\\varphi_{\\lambda})-\\widetilde{f}(\\lambda)\\|_p=\\|P(f\\circ\\varphi_{\\lambda}-\\overline{P(\\overline{f}\\circ\\varphi_{\\lambda}}))\\|_p.$",
"link": "https://mathoverflow.net/questions/472830/show-that-pf-circ-varphi-lambda-widetildef-lambda-p-pf-circ-va",
"score": 2,
"view_count": 316,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "fa.functional-analysis;measure-theory;operator-theory;function-spaces;toeplitz-operators",
"creation_date": "2026-03-04 17:11:15Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Let <span class=\"math-container\">$\\Omega = \\mathbb B_n,$</span> the unit ball in <span class=\"math-container\">$\\mathbb C^n$</span> and <span class=\"math-container\">$L^2_a(\\Omega)$</span> be the Bergman space endowed with the normalized volume measure on <span class=\"math-container\">$\\Omega.$</span> Let <span class=\"math-container\">$k_{\\lambda}$</span> be the associated Bergman reproducing kernel. Let <span class=\"math-container\">$\\varphi_{\\lambda}$</span> be an analytic automorphism of <span class=\"math-container\">$\\Omega$</span> having the properties <span class=\"math-container\">$:$</span></p>\n<blockquote>\n<p><span class=\"math-container\">$(1)$</span> <span class=\"math-container\">$\\varphi_{\\lambda} (\\lambda) = 0$</span></p>\n<p><span class=\"math-container\">$(2)$</span> <span class=\"math-container\">$\\varphi_{\\lambda} \\circ \\varphi_{\\lambda} = \\text{Id}_{\\Omega}.$</span></p>\n</blockquote>\n<p>It is proved in <strong>Section</strong> <span class=\"math-container\">$2.2$</span> in Rudin's <em>Function Theory in the Unit Ball of <span class=\"math-container\">$\\mathbb C^n$</span></em> that the real Jacobian of <span class=\"math-container\">$\\varphi_{\\lambda}$</span> has the following form <span class=\"math-container\">$:$</span>\n<span class=\"math-container\">$$J_{\\mathbb R} (\\varphi_{\\lambda}) (z) = \\frac {\\left \\lvert k_{\\lambda} (z) \\right \\rvert^2} {k_{\\lambda} (\\lambda)}.$$</span></p>\n<p>For a given <span class=\"math-container\">$f \\in L^{\\infty} (\\Omega)$</span> we define the <strong>Berezin transform</strong> <span class=\"math-container\">$\\widetilde f$</span> of <span class=\"math-container\">$f$</span> in the following way <span class=\"math-container\">$:$</span></p>\n<p><span class=\"math-container\">$$\\begin{align*} \\widetilde {f} (\\lambda) & = \\left \\langle f \\frac {k_{\\lambda}} {\\|k_{\\lambda}\\|_2}, \\frac {k_{\\lambda}} {\\|k_{\\lambda}\\|_2} \\right \\rangle \\\\ & = \\frac {1} {k_{\\lambda} (\\lambda)} \\int_{\\Omega} f(z)\\ \\left \\lvert k_{\\lambda} (z) \\right \\rvert^2\\ dV(z) \\end{align*}$$</span></p>\n<p>By change of variable formula we have <span class=\"math-container\">$$\\widetilde {f} (\\lambda) = \\int_{\\Omega} f \\circ \\varphi_{\\lambda}\\ dV.$$</span></p>\n<p>In <strong>Corollary</strong> <span class=\"math-container\">$10$</span> of the paper <a href=\"https://acrobat.adobe.com/id/urn:aaid:sc:AP:e1a6c9e1-a34d-4463-9597-04e9731ad3b0\" rel=\"nofollow noreferrer\"><em>Toeplitz and Hankel Operators on Bergman Spaces</em></a> authored by Karel Stroethoff and Dechao Zheng it is claimed that if <span class=\"math-container\">$P$</span> denotes the Bergman projection then</p>\n<p><span class=\"math-container\">$$\\left\\|P\\left(f\\circ\\varphi_{\\lambda}\\right)-\\widetilde{f}(\\lambda)\\right\\|_p=\\left\\|P\\left(f\\circ\\varphi_{\\lambda}-\\overline{P\\left(\\overline{f}\\circ\\varphi_{\\lambda}\\right)}\\right)\\right\\|_p.$$</span></p>\n<p>But I am having hard time following this claim. Any help in this regard would be warmly appreciated.</p>\n<p>Thanks for your time.</p>\n</div>",
"comments": [],
"answers": [
{
"answer_id": 473110,
"question_id": 472830,
"score": 0,
"is_accepted": false,
"author_name": "an_ordinary_mathematician",
"author_id": 153260,
"author_url": "https://mathoverflow.net/users/153260/an-ordinary-mathematician",
"author_reputation": 3226,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2024-06-12 13:08:52Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I think it goes like this.</p>\n<p><span class=\"math-container\">\\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*}</span></p>\n<p>The second to last inequality comes from the fact that for the unit ball and the polydisc if <span class=\"math-container\">$F$</span> is a holomorphic function and <span class=\"math-container\">$dV$</span> is the normalized area measure then <span class=\"math-container\">$\\int_\\Omega F(z) dV(z) = F(0) $</span>.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 195938,
"title": "Grothendieck on polyhedra over finite fields",
"link": "https://mathoverflow.net/questions/195938/grothendieck-on-polyhedra-over-finite-fields",
"score": 26,
"view_count": 1000,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "reference-request;co.combinatorics;ho.history-overview;discrete-geometry;finite-fields",
"creation_date": "2026-03-04 16:04:28Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>In Grothendieck's <em>Sketch of a Programme</em> 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.</p>\n<p>Here is a link to the version of <em>Sketch</em> I was reading.\n<a href=\"http://matematicas.unex.es/%7Enavarro/res/esquisseeng.pdf\" rel=\"noreferrer\">Sketch of a Programme</a></p>\n<p>More specifically, Grothendieck begins (p. 252):</p>\n<blockquote>\n<p>In 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 …</p>\n</blockquote>\n<p>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.</p>\n<p>The most curious remark comes at the end of this section where he writes (p. 255):</p>\n<blockquote>\n<p>… 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 <em>combinatorial paradigm</em> [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 (<span class=\"math-container\">$\\mathbb F_4$</span> and <span class=\"math-container\">$\\mathbb F_5$</span> for the icosahedron, <span class=\"math-container\">$\\mathbb F_2$</span> for the octahedron) a particularly elegant (and unexpected) description of the combinatorics of the polyhedron. It seems to me that I perceived there a principle of great generality, which I believed I found again for example in a later reflection on the combinatorics of the system of 27 lines on a cubic surface, and its relations with the root system <span class=\"math-container\">$E_7$</span>. 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…</p>\n</blockquote>\n<p>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?</p>\n<p>If anyone can explain Grothendieck's comments or point to other references, I would be appreciative.</p>\n</div>",
"comments": [
{
"comment_id": 488061,
"author_name": "Matthias Wendt",
"author_id": 50846,
"author_url": "https://mathoverflow.net/users/50846/matthias-wendt",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Could the singular characteristic have something to do with the modular representation theory of the automorphism groups? Also, isn't it more natural to relate the 27 lines to $E_6$ instead of $E_7$?</span>",
"created_date": "2015-02-07 15:23:02Z"
},
{
"comment_id": 488062,
"author_name": "Matthias Wendt",
"author_id": 50846,
"author_url": "https://mathoverflow.net/users/50846/matthias-wendt",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Maybe the following paper is relevant: B. Monson, E. Schulte. Reflection groups and polytopes over finite fields I. Advances Applied Mathematics 33 (2004), 290--317</span>",
"created_date": "2015-02-07 15:27:22Z"
},
{
"comment_id": 1018270,
"author_name": "user149000",
"author_id": 50638,
"author_url": "https://mathoverflow.net/users/50638/user149000",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">He defines maps in a previous section essentially as graphs on surfaces. As he explains in the same section, one can view a regular polyhedron as defined simply by its (oriented) automorphism group acting on a flag (vertex + center of side + center of face), and each generator is simply a linear polynomial in the flag. The polyhedron chosen determines the angles, which determines the polynomial. One can consider these same equations in various characteristics, which gives you the specializations. However I don't know what description he refers to (maybe you need a projective context).</span>",
"created_date": "2021-07-14 22:25:38Z"
},
{
"comment_id": 1257449,
"author_name": "mr_e_man",
"author_id": 134099,
"author_url": "https://mathoverflow.net/users/134099/mr-e-man",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@user149000 \"and each generator is simply a linear polynomial in the flag\" - Can you clarify?</span>",
"created_date": "2024-11-20 20:00:29Z"
},
{
"comment_id": 1257647,
"author_name": "user149000",
"author_id": 50638,
"author_url": "https://mathoverflow.net/users/50638/user149000",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@mr_e_man See <a href=\"https://arxiv.org/abs/2304.03345\" rel=\"nofollow noreferrer\">arxiv.org/abs/2304.03345</a></span>",
"created_date": "2024-11-22 04:48:41Z"
}
],
"answers": [
{
"answer_id": 423573,
"question_id": 195938,
"score": 4,
"is_accepted": false,
"author_name": "user234212323",
"author_id": 429204,
"author_url": "https://mathoverflow.net/users/429204/user234212323",
"author_reputation": 934,
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"created_date": "2022-05-29 18:18:42Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Not an exact answer to your question but maybe it can be helpful to know:</p>\n<p><a href=\"https://csg.igrothendieck.org/wp-content/uploads/2024/02/diekertscan.pdf\" rel=\"nofollow noreferrer\">La théorie combinatoire de l'icosaèdre</a> by V Diekert</p>\n<p>This was a \"Rapport pour un DES d'université de l'année 1977/78\" under Grothendieck.</p>\n</div>",
"comments": [
{
"comment_id": 1257800,
"author_name": "Bumblebee",
"author_id": 54507,
"author_url": "https://mathoverflow.net/users/54507/bumblebee",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Link is not working.</span>",
"created_date": "2024-11-23 03:40:22Z"
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{
"question_id": 488866,
"title": "Generalization of Chebyshev polynomials with connection to K-bonacci sequence number",
"link": "https://mathoverflow.net/questions/488866/generalization-of-chebyshev-polynomials-with-connection-to-k-bonacci-sequence-nu",
"score": 3,
"view_count": 183,
"answer_count": 1,
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"tags": "co.combinatorics;computability-theory;binomial-coefficients;recurrences;trigonometric-polynomials",
"creation_date": "2026-03-04 15:49:13Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I 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.</p>\n<p><span class=\"math-container\">\\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}</span></p>\n<p>Thus, we can express <span class=\"math-container\">$\\cos(n\\theta)$</span> in terms of powers of <span class=\"math-container\">$\\cos(\\theta)$</span>. Substituting <span class=\"math-container\">$\\cos(\\theta) = x$</span>, we obtain a polynomial, proving the existence of <span class=\"math-container\">$T_n(x)$</span>:</p>\n<p>[Symbolic Representation of Chebyshev Polynomials]\n<span class=\"math-container\">\\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}</span></p>\n<p>The recursive relation for Chebyshev polynomials <span class=\"math-container\">$T_n(x)$</span> is given by:\n<span class=\"math-container\">\\begin{eqnarray}\nT_n(x) = 2xT_{n-1}(x) - T_{n-2}(x)\n\\end{eqnarray}</span>\nwith initial conditions:\n<span class=\"math-container\">$$T_0(x) = 1, \\quad T_1(x) = x.$$</span></p>\n<p>Using \"constructive\" approaches, we can generalize the Chebyshev polynomials by starting with the combinatorial representation of <span class=\"math-container\">$T_n(x)$</span>.</p>\n<p>[Generalized Symbolic Chebyshev Polynomials]\n<span class=\"math-container\">\\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}</span></p>\n<p>If <span class=\"math-container\">$m = 2p$</span>, where <span class=\"math-container\">$p \\in \\mathbb{N}$</span>, then:\n<span class=\"math-container\">\\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}</span></p>\n<p>If <span class=\"math-container\">$m = 2p + 1$</span>, where <span class=\"math-container\">$p \\in \\mathbb{N}$</span>, then:\n<span class=\"math-container\">\\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}</span></p>\n<p>To define the first <span class=\"math-container\">$m+d$</span> terms, we use the explicit formula. Additionally, when evaluating the function <span class=\"math-container\">$G$</span> for certain parameters, we obtain the following equality:\n<span class=\"math-container\">$$G_{n,2,0}(x,-1) = T_n(x).$$</span></p>\n<p>Questions:</p>\n<ul>\n<li>Can this recurrence relation for G be proven using mathematical induction?</li>\n<li>Is there a combinatorial proof that directly explains this recurrence? partially replied by @darij-grinberg</li>\n<li>Does this generalization have any known connections to combinatorial structures (e.g., lattice paths, binomial sum transformations)?</li>\n</ul>\n<p>More 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.</p>\n<p>Any insights or references would be greatly appreciated!</p>\n<p>I 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.</p>\n<p>Example of other identity involved in the article <a href=\"https://mathoverflow.net/questions/488840/binomial-coefficient-c2k-n-1-alternative-formula-equivalent-to-the-vandermonde\">Binomial coefficient C(2k,n-1) alternative formula equivalent to the Vandermonde identity?</a></p>\n<p>Example of cellular automaton generated when evaluating G with random parameters:</p>\n<p><img alt=\"Example of automaton generated by G\" src=\"https://i.sstatic.net/rWyK8kZ1.png\"/></p>\n<p><strong>Update 3/12/2025:</strong></p>\n<p>We can derive the general term of the Fibonacci sequence (or K-bonacci sequence) defined by the recurrence relation:</p>\n<p><span class=\"math-container\">\\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}</span></p>\n</div>",
"comments": [
{
"comment_id": 1274717,
"author_name": "darij grinberg",
"author_id": 2530,
"author_url": "https://mathoverflow.net/users/2530/darij-grinberg",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Without getting into anything deep, a few comments: Have you seen the work of Arthur Benjamin and others, e.g., <a href=\"https://scholarship.claremont.edu/cgi/viewcontent.cgi?article=1106&context=hmc_fac_pub\" rel=\"nofollow noreferrer\">scholarship.claremont.edu/cgi/…</a> ? And while I don't know where your \"Generalized Symbolic Chebyshev Polynomials\" are coming from, they look to me like a combinatorially expanded solution to a linear recurrence. There is likely a lot of work on that. For example, you can write a linear recurrence as powers of a matrix, but then you can do various matrix manipulations that result in binomial sums.</span>",
"created_date": "2025-03-04 23:11:01Z"
},
{
"comment_id": 1274719,
"author_name": "TigranNersissian",
"author_id": 556642,
"author_url": "https://mathoverflow.net/users/556642/tigrannersissian",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thank you @darijgrinberg for your comment! I haven’t yet looked into Arthur Benjamin’s work in detail, but I’ll definitely check the reference you shared. Regarding the 'Generalized Symbolic Chebyshev Polynomials,' they emerged from a combinatorial construction I’m working on. Initially, I was modifying the binomial coefficient formulation of Chebyshev polynomials and observed a structural change in the recurrence relation. This led me to conjecture that the recurrence could be generalized through a different combinatorial framework, which eventually resulted in the expressions I posted.</span>",
"created_date": "2025-03-04 23:23:40Z"
},
{
"comment_id": 1274720,
"author_name": "TigranNersissian",
"author_id": 556642,
"author_url": "https://mathoverflow.net/users/556642/tigrannersissian",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Your observation that this resembles a combinatorially expanded solution to a linear recurrence is very interesting—especially the connection to matrix powers and binomial manipulations, it makes sense in the context of structured linear recurrences. Would you happen to know any specific references or examples of such matrix-based approaches to binomial sums? That could help me refine my formulation further. Thanks again for your insights!</span>",
"created_date": "2025-03-04 23:24:22Z"
}
],
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{
"answer_id": 508749,
"question_id": 488866,
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"author_name": "TigranNersissian",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><b>Proof.</b> <i>Given the generalized Chebyshev polynomial:</i>\n<span class=\"math-container\">$$G_{n,m,d}(h,t) = \\sum_{0 \\leq mj \\leq n} \\sum_{mj \\leq ml \\leq n-dj} \\binom{n-dj}{ml} \\binom{l}{j} h^{n-j(d+2)} t^j$$</span>\n<i>The recurrence relation is defined by the parity of <span class=\"math-container\">$m$</span>:</i>\n<br/><br/>\n<i>If <span class=\"math-container\">$m = 2p$</span> (<span class=\"math-container\">$p \\in \\mathbb{N}$</span>):</i>\n<span class=\"math-container\">$$G_{n,m,d}(h,t) = \\left(\\sum_{k=1}^{m-1} \\binom{m}{k} h^k (-1)^{k+1} G_{n-k,m,d}(h,t)\\right) + h^{m-2} t G_{n-m-d,m,d}(h,t)$$</span>\n<br/>\n<i>If <span class=\"math-container\">$m = 2p + 1$</span> (<span class=\"math-container\">$p \\in \\mathbb{N}$</span>):</i>\n<span class=\"math-container\">$$G_{n,m,d}(h,t) = \\left(\\sum_{k=1}^{m-1} \\binom{m}{k} h^k (-1)^{k+1} G_{n-k,m,d}(h,t)\\right) + 2h^m G_{n-m,m,d}(h,t) + h^{m-2} t G_{n-m-d,m,d}(h,t)$$</span></p>\n<hr/>\n<p>To establish this recurrence, we employ a finite difference approach combined with a classic binomial identity. Let us rename the inner summation index from <span class=\"math-container\">$p$</span> to <span class=\"math-container\">$l$</span> to avoid confusion with the parity parameter <span class=\"math-container\">$p$</span> where <span class=\"math-container\">$m = 2p$</span> or <span class=\"math-container\">$m = 2p+1$</span>.</p>\n<p><b>Step 1: Set Up the m-th Difference Sum</b></p>\n<p>We evaluate the following sum, which represents the m-th order finite difference, which we will call <span class=\"math-container\">$S$</span>:</p>\n<span class=\"math-container\">$$S = \\sum_{k=0}^m \\binom{m}{k} (-h)^k G_{n-k,m,d}(h,t)$$</span>\n<p>Substitute the definition of <span class=\"math-container\">$G_{n-k,m,d}(h,t)$</span> into <span class=\"math-container\">$S$</span>:</p>\n<span class=\"math-container\">$$S = \\sum_{k=0}^m \\binom{m}{k} (-h)^k \\sum_{j} \\sum_{l} \\binom{n-k-dj}{ml} \\binom{l}{j} h^{n-k-j(d+2)} t^j$$</span>\n<p>The powers of <span class=\"math-container\">$h$</span> combine such that <span class=\"math-container\">$(-h)^k \\cdot h^{n-k-j(d+2)} = (-1)^k h^{n-j(d+2)}$</span>. Swapping the order of summation and pulling the <span class=\"math-container\">$h$</span> and <span class=\"math-container\">$t$</span> terms out:</p>\n<span class=\"math-container\">$$S = \\sum_{j} \\sum_{l} \\binom{l}{j} h^{n-j(d+2)} t^j \\left[ \\sum_{k=0}^m (-1)^k \\binom{m}{k} \\binom{n-dj-k}{ml} \\right]$$</span>\n<p><b>Step 2: Apply the Binomial Identity</b></p>\n<p>The inner bracket is a known binomial identity related to the m-th difference of a polynomial. For any integers <span class=\"math-container\">$A$</span> and <span class=\"math-container\">$B$</span>:</p>\n<span class=\"math-container\">$$\\sum_{k=0}^m (-1)^k \\binom{m}{k} \\binom{A-k}{B} = \\binom{A-m}{B-m}$$</span>\n<p>By setting <span class=\"math-container\">$A = n-dj$</span> and <span class=\"math-container\">$B = ml$</span>, the bracket simplifies to:</p>\n<span class=\"math-container\">$$\\binom{n-dj-m}{ml-m} = \\binom{n-dj-m}{m(l-1)}$$</span>\n<p>Substituting this back into <span class=\"math-container\">$S$</span>:</p>\n<span class=\"math-container\">$$S = \\sum_{j} \\sum_{l} \\binom{n-dj-m}{m(l-1)} \\binom{l}{j} h^{n-j(d+2)} t^j$$</span>\n<p><b>Step 3: Index Shifting and Pascal's Rule</b></p>\n<p>Shift the index <span class=\"math-container\">$l$</span> down by 1. Let <span class=\"math-container\">$l' = l - 1$</span>, so <span class=\"math-container\">$l = l' + 1$</span>:</p>\n<span class=\"math-container\">$$S = \\sum_{j} \\sum_{l'} \\binom{n-m-dj}{ml'} \\binom{l'+1}{j} h^{n-j(d+2)} t^j$$</span>\n<p>Using Pascal's Identity, <span class=\"math-container\">$\\binom{l'+1}{j} = \\binom{l'}{j} + \\binom{l'}{j-1}$</span>. This splits <span class=\"math-container\">$S$</span> into two separate sums, <span class=\"math-container\">$S_1$</span> and <span class=\"math-container\">$S_2$</span>:</p>\n<span class=\"math-container\">$$S = S_1 + S_2$$</span>\n<p>Evaluating <span class=\"math-container\">$S_1$</span>:</p>\n<span class=\"math-container\">$$S_1 = \\sum_{j} \\sum_{l'} \\binom{n-m-dj}{ml'} \\binom{l'}{j} h^{n-j(d+2)} t^j$$</span>\n<p>By factoring <span class=\"math-container\">$h^m$</span> out of the exponent where <span class=\"math-container\">$n-j(d+2) = (n-m)-j(d+2)+m$</span>, we get the definition of <span class=\"math-container\">$G_{n-m}$</span>:</p>\n<span class=\"math-container\">$$S_1 = h^m G_{n-m, m, d}(h,t)$$</span>\n<p>Evaluating <span class=\"math-container\">$S_2$</span>:</p>\n<span class=\"math-container\">$$S_2 = \\sum_{j} \\sum_{l'} \\binom{n-m-dj}{ml'} \\binom{l'}{j-1} h^{n-j(d+2)} t^j$$</span>\n<p>Shift the index <span class=\"math-container\">$j$</span> down by 1. Let <span class=\"math-container\">$j' = j - 1$</span>, so <span class=\"math-container\">$j = j' + 1$</span>:</p>\n<span class=\"math-container\">$$S_2 = \\sum_{j'} \\sum_{l'} \\binom{n-m-d(j'+1)}{ml'} \\binom{l'}{j'} h^{n-(j'+1)(d+2)} t^{j'+1}$$</span>\n<p>Simplifying the exponents and binomial top, and factoring out <span class=\"math-container\">$h^{m-2}t$</span>, we are left with the definition of <span class=\"math-container\">$G_{n-m-d}$</span>:</p>\n<span class=\"math-container\">$$S_2 = h^{m-2} t G_{n-m-d, m, d}(h,t)$$</span>\n<p><b>Step 4: Isolate <span class=\"math-container\">$G_n$</span> and Evaluate Parity</b></p>\n<p>We have proven that:</p>\n<span class=\"math-container\">$$\\sum_{k=0}^m \\binom{m}{k} (-h)^k G_{n-k} = h^m G_{n-m} + h^{m-2} t G_{n-m-d}$$</span>\n<p>Pull the <span class=\"math-container\">$k=0$</span> term out of the sum, and move the rest to the right side:</p>\n<span class=\"math-container\">$$G_n = -\\sum_{k=1}^{m} \\binom{m}{k} (-h)^k G_{n-k} + h^m G_{n-m} + h^{m-2} t G_{n-m-d}$$</span>\n<p>Extract the <span class=\"math-container\">$k=m$</span> term from the sum so it matches the upper bound <span class=\"math-container\">$m-1$</span>:</p>\n<span class=\"math-container\">$$G_n = \\sum_{k=1}^{m-1} \\binom{m}{k} (-1)^{k+1} h^k G_{n-k} - (-1)^m h^m G_{n-m} + h^m G_{n-m} + h^{m-2} t G_{n-m-d}$$</span>\n<p>Combine the <span class=\"math-container\">$G_{n-m}$</span> terms:</p>\n<span class=\"math-container\">$$G_n = \\sum_{k=1}^{m-1} \\binom{m}{k} (-1)^{k+1} h^k G_{n-k} + h^m(1 - (-1)^m) G_{n-m} + h^{m-2} t G_{n-m-d}$$</span>\n<p>The recurrence depends on the parity of <span class=\"math-container\">$m$</span>:</p>\n<ul>\n<li>If <span class=\"math-container\">$m$</span> is even (<span class=\"math-container\">$m = 2p$</span>): <span class=\"math-container\">$(-1)^m = 1$</span>, so <span class=\"math-container\">$(1 - 1) = 0$</span>. The middle term vanishes.</li>\n<li>If <span class=\"math-container\">$m$</span> is odd (<span class=\"math-container\">$m = 2p+1$</span>): <span class=\"math-container\">$(-1)^m = -1$</span>, so <span class=\"math-container\">$(1 - (-1)) = 2$</span>. The middle term becomes <span class=\"math-container\">$2h^m G_{n-m}$</span>.</li>\n</ul>\n</div>",
"comments": []
}
]
},
{
"question_id": 508702,
"title": "Intuitive explanation why the radii converge in this sequence of eights?",
"link": "https://mathoverflow.net/questions/508702/intuitive-explanation-why-the-radii-converge-in-this-sequence-of-eights",
"score": 19,
"view_count": 1000,
"answer_count": 4,
"is_answered": true,
"has_accepted_answer": true,
"accepted_answer_id": 508712,
"tags": "mg.metric-geometry;sequences-and-series;limits-and-convergence;intuition",
"creation_date": "2026-03-04 15:46:01Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>In the diagram, circles of the same color are congruent.</p>\n<p><a href=\"https://i.sstatic.net/4XXO2ILj.png\" rel=\"noreferrer\"><img alt=\"Sequence of nested figure eights\" src=\"https://i.sstatic.net/4XXO2ILj.png\"/></a></p>\n<p>I have a <a href=\"https://math.stackexchange.com/a/5124763/398708\">nonintuitive proof</a> that the radii converge. Is there an intuitive explanation?</p>\n</div>",
"comments": [
{
"comment_id": 1325491,
"author_name": "Saúl RM",
"author_id": 172802,
"author_url": "https://mathoverflow.net/users/172802/sa%c3%bal-rm",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Why is the proof non-intuitive? If you mean all the big formulas, they can be avoided in a proof, i.e. one does not need an explicit expression relating the radii to complete the proof</span>",
"created_date": "2026-03-03 11:42:12Z"
},
{
"comment_id": 1325497,
"author_name": "Dan",
"author_id": 494920,
"author_url": "https://mathoverflow.net/users/494920/dan",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@SaúlRM Yes, I mean all the big formulas. How can it be proved without an expression relating the radii?</span>",
"created_date": "2026-03-03 11:49:41Z"
},
{
"comment_id": 1325507,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@SaúlRM I think that's a sound approach. If you choose to write it up properly, I think it'd be better to write it in terms of radius as a function $\\beta$ of the angle. The radius of the $n$-th pair of circles is then a product of values of $\\beta$ at angle differences, and if you know $\\beta$ is differentiable, you can relate this product to the sum of angles, which as you say is bounded.</span>",
"created_date": "2026-03-03 13:04:26Z"
},
{
"comment_id": 1325527,
"author_name": "Saúl RM",
"author_id": 172802,
"author_url": "https://mathoverflow.net/users/172802/sa%c3%bal-rm",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I added an answer, I will delete my previous comments to save comment space</span>",
"created_date": "2026-03-03 16:19:10Z"
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{
"answer_id": 508712,
"question_id": 508702,
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"author_name": "Iosif Pinelis",
"author_id": 36721,
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"created_date": "2026-03-03 16:06:44Z",
"last_edited_date": "2026-03-04 15:46:01Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>As <a href=\"https://mathoverflow.net/questions/508702/intuitive-explanation-why-the-radii-converge-in-this-sequence-of-eights/508712#comment1325529_508712\">noted in the comment</a>, 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.</p>\n<p>So, 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.</p>\n<hr/>\n<p>First here is the original, right-to-left solution, given while apparently misunderstanding the question:</p>\n<p>The 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 <span class=\"math-container\">$<\\pi/3$</span>.</p>\n<p>So, 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.</p>\n<p>But clearly <span class=\"math-container\">$a<\\pi/3<b$</span> in the picture below, so that <span class=\"math-container\">$a<b$</span>. <span class=\"math-container\">$\\quad\\Box$</span></p>\n<p><a href=\"https://i.sstatic.net/Y1S9Q0x7.png\" rel=\"nofollow noreferrer\"><img alt=\"enter image description here\" src=\"https://i.sstatic.net/Y1S9Q0x7.png\"/></a></p>\n<hr/>\n<p>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 <span class=\"math-container\">$a_n:=a<\\pi/3<b=:b_n$</span>, and also let <span class=\"math-container\">$r_n$</span> and <span class=\"math-container\">$r_{n+1}$</span> 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 <span class=\"math-container\">$b_n-a_n$</span>, so that <span class=\"math-container\">$\\sum_n(b_n-a_n)<\\infty$</span> and hence <span class=\"math-container\">$a_n,b_n\\to\\pi/3$</span>. On the other hand, considering the two isosceles triangles in the picture above and using the first-order Taylor expansion of <span class=\"math-container\">$\\cos$</span> at <span class=\"math-container\">$\\pi/3$</span>, we see that\n<span class=\"math-container\">$$\\frac{r_{n+1}}{r_n}=\\frac{\\cos a_n}{\\cos b_n}\n=\\exp\\big((\\sqrt3+o(1))\\,(b_n-a_n)\\big).$$</span>\nIt remains to recall that <span class=\"math-container\">$\\sum_n(b_n-a_n)<\\infty$</span>. <span class=\"math-container\">$\\quad\\Box$</span></p>\n</div>",
"comments": [
{
"comment_id": 1325526,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I'm very confused by this answer. The radii will be quite clearly increasing with every step, the question is whether they remain bounded or not.</span>",
"created_date": "2026-03-03 16:18:19Z"
},
{
"comment_id": 1325529,
"author_name": "Iosif Pinelis",
"author_id": 36721,
"author_url": "https://mathoverflow.net/users/36721/iosif-pinelis",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Wojowu : I guess I misunderstood the question. I thought the process begins from the red circles.</span>",
"created_date": "2026-03-03 16:21:54Z"
},
{
"comment_id": 1325581,
"author_name": "Iosif Pinelis",
"author_id": 36721,
"author_url": "https://mathoverflow.net/users/36721/iosif-pinelis",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Wojowu : This misunderstanding is now completely fixed.</span>",
"created_date": "2026-03-04 04:12:30Z"
},
{
"comment_id": 1325617,
"author_name": "Dan",
"author_id": 494920,
"author_url": "https://mathoverflow.net/users/494920/dan",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Very elegant! Thank you.</span>",
"created_date": "2026-03-04 12:58:02Z"
},
{
"comment_id": 1325618,
"author_name": "Iosif Pinelis",
"author_id": 36721,
"author_url": "https://mathoverflow.net/users/36721/iosif-pinelis",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Dan : Thank you for your appreciation.</span>",
"created_date": "2026-03-04 13:13:52Z"
}
]
},
{
"answer_id": 508705,
"question_id": 508702,
"score": 6,
"is_accepted": false,
"author_name": "Carlo Beenakker",
"author_id": 11260,
"author_url": "https://mathoverflow.net/users/11260/carlo-beenakker",
"author_reputation": 206919,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-03 11:30:42Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The intuitive explanation is that the sequence approaches the hexagonal close packing, two staggered rows of identical circles:</p>\n<img src=\"https://i.sstatic.net/yrwWWty0.png\" width=\"400\">\n<p>Because the sequence of radii is monotonically increasing and bounded by this \"perfect fit\" geometry, it must converge to a finite limit.</p>\n</img></div>",
"comments": [
{
"comment_id": 1325490,
"author_name": "Dan",
"author_id": 494920,
"author_url": "https://mathoverflow.net/users/494920/dan",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">But why can't the circles grow without bound as they approach hexagonal close packing?</span>",
"created_date": "2026-03-03 11:39:38Z"
},
{
"comment_id": 1325502,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">If you try deforming the perfect packing by moving one circle and adjusting the size of all the other ones, then this forces all the ones on one of the sides to become larger. It is not clear at all that they will eventually approach a larger perfect tiling, rather than say growing logarithmically or so.</span>",
"created_date": "2026-03-03 12:05:46Z"
},
{
"comment_id": 1325503,
"author_name": "Saúl RM",
"author_id": 172802,
"author_url": "https://mathoverflow.net/users/172802/sa%c3%bal-rm",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Wojowu I believe that possibility is made impossible by my comments to the question (which perhaps I should have written as an answer but I did not have much time now)</span>",
"created_date": "2026-03-03 12:10:21Z"
},
{
"comment_id": 1325504,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">One can try to compare it with some other sort of iteration, say $(a,b)\\to(a+\\frac{(a-b)}{b},b+\\frac{(a-b)}{b})$. Naively you may think that this iteration \"approaches\" a constant sequence $(a,a)$, and indeed upon iteration we will have $a/b\\to 1$, but nevertheless this sequence will diverge for $a\\neq b$. So there must be something more going on in OP's problem to make such \"convergence in shape, divergence in size\" not occur here.</span>",
"created_date": "2026-03-03 12:12:07Z"
}
]
},
{
"answer_id": 508714,
"question_id": 508702,
"score": 5,
"is_accepted": false,
"author_name": "Saúl RM",
"author_id": 172802,
"author_url": "https://mathoverflow.net/users/172802/sa%c3%bal-rm",
"author_reputation": 13476,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-03 16:18:10Z",
"last_edited_date": "2026-03-03 19:35:28Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Here is a proof that the radii converge without using any big formulas.</p>\n<p>Let <span class=\"math-container\">$r>0$</span>, and consider the following setup in <span class=\"math-container\">$\\mathbb{R}^2$</span>, where the two smaller circles have radius <span class=\"math-container\">$r$</span>:</p>\n<img src=\"https://i.sstatic.net/pBFu0tkf.png\" width=\"500\">\n<p>The angles <span class=\"math-container\">$\\alpha(r),\\beta(r)$</span> are decreasing with <span class=\"math-container\">$r$</span>, in particular <span class=\"math-container\">$\\alpha'(r),\\beta'(r)<0$</span> for all <span class=\"math-container\">$r\\in(0,\\infty)$</span>. Let <span class=\"math-container\">$\\gamma(r)=\\alpha(r)+\\beta(r)-\\pi/2$</span>.</p>\n<p>Now, in your figure, let <span class=\"math-container\">$A_n,B_n$</span> be the centers of the <span class=\"math-container\">$n^{\\text{th}}$</span> circle, <span class=\"math-container\">$r_n$</span> the <span class=\"math-container\">$n^{\\text{th}}$</span> radius and <span class=\"math-container\">$\\alpha_n$</span> the angle between the the <span class=\"math-container\">$x$</span>-axis and the ray <span class=\"math-container\">$B_nA_n$</span> (so in the picture <span class=\"math-container\">$\\alpha_0=\\pi$</span> and the sequence <span class=\"math-container\">$(\\alpha_n)$</span> looks decreasing in the first few steps). Note the function <span class=\"math-container\">$f:(0,\\pi)\\to(0,\\pi)$</span> that gives <span class=\"math-container\">$\\alpha_{n+1}$</span> in terms of <span class=\"math-container\">$\\alpha_n$</span> has fixed point <span class=\"math-container\">$f(2\\pi/3)=2\\pi/3$</span> (see Carlo Beenaker's answer), and <span class=\"math-container\">$f(\\pi)<\\pi$</span>. Moreover, <span class=\"math-container\">$f$</span> is increasing in <span class=\"math-container\">$[2\\pi/3,\\pi]$</span>; this is equivalent to saying that the angle <span class=\"math-container\">$\\theta(r)$</span> from the picture is increasing whenever it is defined, which is true because when <span class=\"math-container\">$r$</span> increases, the point <span class=\"math-container\">$F$</span> moves clockwise.</p>\n<p>Thus, <span class=\"math-container\">$\\alpha_n$</span> is decreasing and converges when <span class=\"math-container\">$n\\to\\infty$</span>, and <span class=\"math-container\">$r_{n+1}/r_n\\to1$</span>. But on the other hand, we have that <span class=\"math-container\">$\\alpha_{n+1}=\\alpha_n+\\gamma(r_{n+1}/r_n)$</span>. As <span class=\"math-container\">$\\gamma'(1)<0$</span>, there is <span class=\"math-container\">$\\varepsilon>0$</span> such that, for big enough <span class=\"math-container\">$n$</span>, we have <span class=\"math-container\">$|\\alpha_{n+1}-\\alpha_n|>\\varepsilon\\left|1-\\frac{r_{n+1}}{r_n}\\right|$</span>.</p>\n<p>Thus, as <span class=\"math-container\">$\\sum_{n}|\\alpha_{n+1}-\\alpha_n|<\\infty$</span>, we also have <span class=\"math-container\">$\\sum_n\\left|1-\\frac{r_{n+1}}{r_n}\\right|<\\infty$</span>, which <a href=\"https://math.stackexchange.com/a/4193095/807670\">implies</a> <span class=\"math-container\">$\\prod_n\\frac{r_{n+1}}{r_n}<\\infty$</span>, that is, the radii converge.</p>\n</img></div>",
"comments": []
},
{
"answer_id": 508717,
"question_id": 508702,
"score": 4,
"is_accepted": false,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"author_reputation": 34953,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-03 18:03:28Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>This proof is closely related to Saul RM's one, but I think it's a bit simpler to see what's going on.</p>\n<p>For <span class=\"math-container\">$0\\leq \\alpha<\\pi/2$</span>, let <span class=\"math-container\">$f(\\alpha)$</span> be the ratio of radii of circles when they are placed in the following configuration:</p>\n<p><a href=\"https://i.sstatic.net/TQGFkqJj.png\" rel=\"nofollow noreferrer\"><img alt=\"enter image description here\" src=\"https://i.sstatic.net/TQGFkqJj.png\"/></a></p>\n<p>One can compute <span class=\"math-container\">$f(\\alpha)$</span> explicitly in terms of trigonometric functions, but all we need to know is that it is differentiable at <span class=\"math-container\">$0$</span>, which hopefully is believable. This in particular implies that for <span class=\"math-container\">$\\alpha$</span> small and some constant <span class=\"math-container\">$C$</span> (which we can take to be <span class=\"math-container\">$2f'(\\alpha)$</span>) we have\n<span class=\"math-container\">$$f(\\alpha)\\leq f(0)+C\\alpha=1+C\\alpha.$$</span></p>\n<p>Considering now the diagram in the OP, let <span class=\"math-container\">$r_n$</span> be the radius of the <span class=\"math-container\">$n$</span>-th pair of circles, and <span class=\"math-container\">$\\alpha_n$</span> to be the angle between the lines connecting the centers of the <span class=\"math-container\">$n$</span>-th and <span class=\"math-container\">$n+1$</span>-st pair. By the definition of <span class=\"math-container\">$f$</span>, this means <span class=\"math-container\">$r_n=r_{n-1}\\cdot f(\\alpha_{n-1})$</span>, and so we find\n<span class=\"math-container\">$$r_n=r_1\\cdot\\prod_{i=1}^{n-1}f(\\alpha_i)\\leq r_1\\cdot\\prod_{i=1}^{n-1}(1+C\\alpha_i).$$</span>\nBy a well-known connection between convergence of sums and products (easily shown with help of the inequality <span class=\"math-container\">$1+x\\leq e^x$</span>) this product converges as <span class=\"math-container\">$n\\to\\infty$</span> iff the sum <span class=\"math-container\">$\\sum_{i=1}^\\infty\\alpha_i$</span> converges. But in our case, the sum <span class=\"math-container\">$\\sum_{i=1}^{n-1}\\alpha_i$</span> represents the angle between the line connecting the centers of the <span class=\"math-container\">$n$</span>-th pair of circles with the horizontal line, and it's not hard to convince yourself that this angle tends to <span class=\"math-container\">$\\pi/3$</span>, giving the desired convergence.</p>\n</div>",
"comments": [
{
"comment_id": 1325548,
"author_name": "Saúl RM",
"author_id": 172802,
"author_url": "https://mathoverflow.net/users/172802/sa%c3%bal-rm",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I see, that is what you meant by writing things in terms of the radii. I wrote the other way mostly because differentiability at $0$ was not obvious to me. In any case, the more pictures the better</span>",
"created_date": "2026-03-03 19:26:54Z"
}
]
}
]
},
{
"question_id": 508490,
"title": "Proving Kashiwara–Schapira's micro-local cutoff lemma $\\infty$-categorically",
"link": "https://mathoverflow.net/questions/508490/proving-kashiwara-schapiras-micro-local-cutoff-lemma-infty-categorically",
"score": 7,
"view_count": 271,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "sheaf-theory;micro-local-analysis",
"creation_date": "2026-03-04 15:14:22Z",
"last_activity_date": "",
"body_html": "",
"comments": [],
"answers": []
},
{
"question_id": 508585,
"title": "Improperness of regular conditional probabilities",
"link": "https://mathoverflow.net/questions/508585/improperness-of-regular-conditional-probabilities",
"score": 1,
"view_count": 86,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "pr.probability;measure-theory",
"creation_date": "2026-03-04 14:47:17Z",
"last_activity_date": "",
"body_html": "",
"comments": [],
"answers": []
},
{
"question_id": 391124,
"title": "Are hypergeometric series not taught often at universities nowadays, and if so, why?",
"link": "https://mathoverflow.net/questions/391124/are-hypergeometric-series-not-taught-often-at-universities-nowadays-and-if-so",
"score": 52,
"view_count": 8000,
"answer_count": 7,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "mathematics-education;hypergeometric-functions",
"creation_date": "2026-03-04 14:28:14Z",
"last_activity_date": "",
"body_html": "",
"comments": [],
"answers": []
},
{
"question_id": 508738,
"title": "Reference for colimits in comma categories F ↓ G in which F and G are not assumed cocontinuous",
"link": "https://mathoverflow.net/questions/508738/reference-for-colimits-in-comma-categories-f-%e2%86%93-g-in-which-f-and-g-are-not-assume",
"score": 3,
"view_count": 57,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "reference-request;ct.category-theory;limits-and-colimits",
"creation_date": "2026-03-04 11:08:40Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Given functors <span class=\"math-container\">$F : \\mathbf A \\to \\mathbf C$</span> and <span class=\"math-container\">$G : \\mathbf B \\to \\mathbf C$</span>, it is <a href=\"https://ncatlab.org/nlab/show/comma+category#completeness_and_cocompleteness\" rel=\"nofollow noreferrer\">well known</a> that the comma category <span class=\"math-container\">$F \\downarrow G$</span> is cocomplete assuming that <span class=\"math-container\">$\\mathbf A$</span> and <span class=\"math-container\">$\\mathbf B$</span> are cocomplete and <span class=\"math-container\">$F$</span> is cocontinuous, in which case the projection functors are also cocontinuous.</p>\n<p>I would like a reference for the existence of colimits in comma categories in which <span class=\"math-container\">$F$</span> is <em>not</em> assumed cocontinuous (in which case the projections will no longer be necessarily cocontinuous).</p>\n<p>In the case where <span class=\"math-container\">$F$</span> is an endofunctor and <span class=\"math-container\">$G$</span> is the identity, one reference is section 14.1 of Kelly's <em><a href=\"https://doi.org/10.1017/S0004972700006353\" rel=\"nofollow noreferrer\">A unified treatment of transfinite constructions [...]</a></em> (and Kelly's construction works more generally for <span class=\"math-container\">$F$</span> an arbitrary functor) but I would like a reference in which <span class=\"math-container\">$G$</span> may not be the identity.</p>\n</div>",
"comments": [],
"answers": []
},
{
"question_id": 456780,
"title": "Reference for understanding Shelah-Harrington-Makkai's proof of Vaught's conjecture for $\\omega$-stable theories",
"link": "https://mathoverflow.net/questions/456780/reference-for-understanding-shelah-harrington-makkais-proof-of-vaughts-conject",
"score": 6,
"view_count": 359,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "reference-request;lo.logic;model-theory",
"creation_date": "2026-03-04 09:41:17Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I'm looking for a source to help me better understand Shelah-Harrington-Makkai's proof of Vaught's conjecture for <span class=\"math-container\">$\\omega$</span>-stable theories (<a href=\"https://doi.org/10.1007/BF02760651\" rel=\"nofollow noreferrer\">DOI</a>, <a href=\"https://shelah.logic.at/files/95409/158.pdf\" rel=\"nofollow noreferrer\">link at Shelah's page</a>). An obvious candidate is Makkai's survey paper <a href=\"https://doi.org/10.1007/BF02760649\" rel=\"nofollow noreferrer\">DOI link</a>, but I felt it was pretty dense (and maybe I need to see a little bit more of motivation behind all the definitions).</p>\n<p>So 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 <em>Fundamentals of Stability Theory</em> could be good, but the more the better.</p>\n<p>Cross-posted from <a href=\"https://math.stackexchange.com/questions/4790296/reference-for-understanding-shelahs-proof-of-vaughts-conjecture-for-omega-s\">Math.SE</a>.</p>\n</div>",
"comments": [
{
"comment_id": 1183086,
"author_name": "James E Hanson",
"author_id": 83901,
"author_url": "https://mathoverflow.net/users/83901/james-e-hanson",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Hopefully someone will come in with a reference but my impression is that there isn't a lot of good exposition of some of the more technical stuff in stability theory.</span>",
"created_date": "2023-10-20 04:01:50Z"
},
{
"comment_id": 1325563,
"author_name": "Jonas Frey",
"author_id": 51432,
"author_url": "https://mathoverflow.net/users/51432/jonas-frey",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">is there a reason why you're crediting only Shelah with the proof and not the three authors of the paper?</span>",
"created_date": "2026-03-03 21:48:27Z"
},
{
"comment_id": 1325568,
"author_name": "Tesla Daybreak",
"author_id": 170461,
"author_url": "https://mathoverflow.net/users/170461/tesla-daybreak",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@JonasFrey Good point, edited.</span>",
"created_date": "2026-03-03 23:31:26Z"
}
],
"answers": []
},
{
"question_id": 508666,
"title": "Cutting squares and rectangles into mutually similar but non-congruent rectangles -2",
"link": "https://mathoverflow.net/questions/508666/cutting-squares-and-rectangles-into-mutually-similar-but-non-congruent-rectangle",
"score": 3,
"view_count": 163,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": true,
"accepted_answer_id": 508673,
"tags": "co.combinatorics;discrete-geometry;tiling",
"creation_date": "2026-03-04 08:39:15Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>We try to go a step beyond <a href=\"https://mathoverflow.net/questions/508628/cutting-squares-and-rectangles-into-mutually-similar-but-non-congruent-rectangle\">Cutting squares and rectangles into mutually similar but non-congruent rectangles</a> and record a couple of more questions.</p>\n<ol>\n<li>In the answer to question 2 of above post, Sayan Dutta shows the following dissection of a rectangle into 3 rectangles all mutually non-congruent, mutually similar and also similar to the big rectangle.\n<a href=\"https://i.sstatic.net/JpvRxwU2.png\" rel=\"nofollow noreferrer\"><img alt=\"enter image description here\" src=\"https://i.sstatic.net/JpvRxwU2.png\"/></a></li>\n</ol>\n<p>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... <strong>Question</strong>: 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 <em>some</em> 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?</p>\n<p><strong>Further question:</strong> Is there any rectangle that can be cut into n mutually self-similar rectangles that are <em>not</em> similar to the big rectangle for any positive integer value of n greater than 2?</p>\n<ol start=\"2\">\n<li>The answer to question 1 of above post points to figure 7 in this paper: <a href=\"https://www.heldermann-verlag.de/jgg/jgg13/j13h2spin.pdf\" rel=\"nofollow noreferrer\">https://www.heldermann-verlag.de/jgg/jgg13/j13h2spin.pdf</a>. It shows a square being cut into 3 mutually non-congruent rectangles all similar to one another. What can be said about the square so partitioned into n pieces where n >3?</li>\n</ol>\n</div>",
"comments": [
{
"comment_id": 1325367,
"author_name": "Peter Taylor",
"author_id": 46140,
"author_url": "https://mathoverflow.net/users/46140/peter-taylor",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Trivially, if a rectangle is dissected into two rectangles which are similar to each other and have side ratio $a$ then the original rectangle has side ratio $a+\\frac{1}{a}$ so no rectangle can be dissected into two rectangles both similar to itself.</span>",
"created_date": "2026-03-02 08:22:40Z"
},
{
"comment_id": 1325369,
"author_name": "Nandakumar R",
"author_id": 142600,
"author_url": "https://mathoverflow.net/users/142600/nandakumar-r",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Yes. n=2 is trivially ruled out. One would want to know of higher even n. Thanks</span>",
"created_date": "2026-03-02 08:58:05Z"
},
{
"comment_id": 1325669,
"author_name": "Richard Stanley",
"author_id": 2807,
"author_url": "https://mathoverflow.net/users/2807/richard-stanley",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">A related result of Laczkovich and Szekeres, <i>Discrete Comput. Geom.</i> <b>13</b> (1995), 569-572, is that that a square can be tiled with finitely many similar copies of the $1\\times u$ rectangle if and only if $u$ is an algebraic numbers all of whose conjugates have positive real part. Thus for instance $u=\\sqrt{2}+\\frac{17}{12}$ is possible, but not $\\sqrt{2}+\\frac{4}{3}$.</span>",
"created_date": "2026-03-05 00:17:05Z"
}
],
"answers": [
{
"answer_id": 508673,
"question_id": 508666,
"score": 6,
"is_accepted": true,
"author_name": "Peter Taylor",
"author_id": 46140,
"author_url": "https://mathoverflow.net/users/46140/peter-taylor",
"author_reputation": 8178,
"license": "CC BY-SA 4.0",
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"created_date": "2026-03-02 10:03:10Z",
"last_edited_date": "2026-03-04 08:39:15Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>As 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.</p>\n<hr/>\n<p>If we extend the same construction of a slicing floorplan from three to four rooms:</p>\n<pre><code> c a^2 1\n+-------------------------------------+-----+-+\n| | | | a\n| +-----+-+\n| | |\n| | |\n| | |\n| | | b\n| | |\n| | |\n| | |\n+-------------------------------------+-------+\n</code></pre>\n<p>we get the constraints <span class=\"math-container\">\\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*}</span></p>\n<p>Clearly the third ratio is larger than the second so if we take <span class=\"math-container\">$a > 1$</span> we have <span class=\"math-container\">\\begin{eqnarray*}\\frac{c}{a+b} &=& a^{-1} \\\\\n\\frac{1+a^2+c}{a+b} &=& a\n\\end{eqnarray*}</span> and we eliminate <span class=\"math-container\">$c$</span> to get <span class=\"math-container\">\\begin{eqnarray*}c &=& 1 + a^{-1}b \\\\\n2a+b &=& a^2b\n\\end{eqnarray*}</span></p>\n<p>Taking the case <span class=\"math-container\">$\\frac{b}{a^2+1} = a^{-1}$</span> gives solution <span class=\"math-container\">$a = \\sqrt{1 + \\sqrt{2}}$</span>, and taking the case <span class=\"math-container\">$\\frac{b}{a^2+1} = a$</span> gives solution <span class=\"math-container\">$a = \\sqrt[4]{3}$</span>.</p>\n<p>If we instead take <span class=\"math-container\">$a < 1$</span> then we get the reciprocals <span class=\"math-container\">$a = \\sqrt[4]{\\frac13}$</span> or <span class=\"math-container\">$a = \\sqrt{\\sqrt{2}-1}$</span>.</p>\n<p>A similar recursive construction to the one described in the question extends this to dissections into <span class=\"math-container\">$4 + 3k$</span> parts.</p>\n<hr/>\n<p>A similar process to <a href=\"https://mathoverflow.net/a/507946/46140\">this answer</a> 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 (<span class=\"math-container\">$n$</span>) up to six:</p>\n<p><span class=\"math-container\">\\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}</span></p>\n<p>Note that none of the ratios for <span class=\"math-container\">$n=4$</span> coincide with the ratios for <span class=\"math-container\">$n=5$</span> or <span class=\"math-container\">$n=6$</span>. However, there are two non-trivial coincidences: ratio <span class=\"math-container\">$\\sim 1.217043263032180$</span> with minpoly <span class=\"math-container\">$x^6 + x^4 - 3x^2 - 1$</span> and ratio <span class=\"math-container\">$\\sim 1.370906722418348$</span> with minpoly <span class=\"math-container\">$x^6 - 3x^2 - 1$</span> give partitions into <span class=\"math-container\">$5$</span> or <span class=\"math-container\">$6$</span> similar rectangles, whence by recursive construction they cover <span class=\"math-container\">$1+4k+5m$</span>, and in particular they cover every integer in <span class=\"math-container\">$\\mathbb{N} \\setminus \\{ 0, 2, 3, 4, 7, 8, 12 \\}$</span>.</p>\n<p><img alt=\"Dissections into 5 and 6 self-similar rectangles for the two given ratios\" src=\"https://cheddarmonk.org/maths/self-similar-dissections.png\"/></p>\n<p>There are <a href=\"https://cheddarmonk.org/maths/self-similar-dissection-ratios.txt\" rel=\"nofollow noreferrer\"><span class=\"math-container\">$181$</span> ratios</a> for partitions into <span class=\"math-container\">$7$</span> noncongruent similar rectangles (uploaded elsewhere because that's too long for this post). The only non-trivial coincidences are with partitions into <span class=\"math-container\">$6$</span>, and these leave more gaps than the <span class=\"math-container\">$5,6$</span> coincidences. Fewer gaps would be possible if there's a partition into <span class=\"math-container\">$8$</span> or <span class=\"math-container\">$10$</span> with ratio <span class=\"math-container\">$\\sqrt{\\frac{1 + \\sqrt 5}2}$</span> (i.e. the same ratio as <span class=\"math-container\">$n=3$</span>); otherwise ties for fewest gaps are possible with <span class=\"math-container\">$3,12$</span> or <span class=\"math-container\">$4,8$</span>. However, the search for <span class=\"math-container\">$n=7$</span> took 29 hours so I'm not going to run a search for <span class=\"math-container\">$n=8$</span> unless I can optimise it a <em>lot</em>.</p>\n</div>",
"comments": [
{
"comment_id": 1325404,
"author_name": "Nandakumar R",
"author_id": 142600,
"author_url": "https://mathoverflow.net/users/142600/nandakumar-r",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Could you clarify if there is <i>any</i> even number n for which there is no rectangle that can be cut into n self-similar and mutually non congruent rectangles?</span>",
"created_date": "2026-03-02 15:26:03Z"
},
{
"comment_id": 1325706,
"author_name": "Nandakumar R",
"author_id": 142600,
"author_url": "https://mathoverflow.net/users/142600/nandakumar-r",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I suspect there is no rectangle that allows partition into n rectangles that are all smililar to itself and mutually non-congruent for ALL even n.</span>",
"created_date": "2026-03-05 09:19:53Z"
},
{
"comment_id": 1325707,
"author_name": "Nandakumar R",
"author_id": 142600,
"author_url": "https://mathoverflow.net/users/142600/nandakumar-r",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">thanks. marking question as answered although question 2 remains.</span>",
"created_date": "2026-03-05 09:20:40Z"
}
]
}
]
},
{
"question_id": 415209,
"title": "Exposition of Grothendieck's mathematics",
"link": "https://mathoverflow.net/questions/415209/exposition-of-grothendiecks-mathematics",
"score": 40,
"view_count": 7000,
"answer_count": 13,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "ag.algebraic-geometry;soft-question;textbook-recommendation;books;exposition",
"creation_date": "2026-03-04 08:04:30Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>As <a href=\"https://en.wikipedia.org/wiki/Alexander_Grothendieck\" rel=\"noreferrer\">Wikipedia</a> says:</p>\n<blockquote>\n<p>In Grothendieck's retrospective <em>Récoltes et Semailles</em>, he identified twelve of his contributions which he believed qualified as \"great ideas\". In chronological order, they are:</p>\n<ol>\n<li>Topological tensor products and nuclear spaces.</li>\n<li>\"Continuous\" and \"discrete\" duality (derived categories, \"six operations\").</li>\n<li>Yoga of the Grothendieck–Riemann–Roch theorem (K-theory, relation with intersection theory).</li>\n<li>Schemes.</li>\n<li>Topoi.</li>\n<li>Étale cohomology and l-adic cohomology.</li>\n<li>Motives and the motivic Galois group (Grothendieck ⊗-categories).</li>\n<li>Crystals and crystalline cohomology, yoga of \"de Rham coefficients\", \"Hodge coefficients\", ...</li>\n<li>\"Topological algebra\": ∞-stacks, derivators; cohomological formalism of topoi as inspiration for a new homotopical algebra.</li>\n<li>Tame topology.</li>\n<li>Yoga of anabelian algebraic geometry, Galois–Teichmüller theory.</li>\n<li>\"Schematic\" or \"arithmetic\" point of view for regular polyhedra and regular configurations of all kinds.</li>\n</ol>\n</blockquote>\n<p>What are some <em>modern</em>, <em>concise</em> expository texts on these topics that students can use to learn the great ideas of Grothendieck? <em>EGA</em>, <em>SGA</em>, <em>Les Dérivateurs</em>, <em>La Longue Marche</em>, 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.</p>\n<p><em>Please only one topic with expository text per answer.</em></p>\n<p>Let me start by giving two examples of the kind of answers I am expecting:</p>\n<ul>\n<li><p>Leinster's <a href=\"https://arxiv.org/abs/1012.5647\" rel=\"noreferrer\"><em>An informal introduction to topos theory</em></a> 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.</p>\n</li>\n<li><p>Milne's <a href=\"https://www.jmilne.org/math/CourseNotes/LEC.pdf\" rel=\"noreferrer\">lecture notes on étale cohomology</a>. 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.</p>\n</li>\n</ul>\n</div>",
"comments": [
{
"comment_id": 1065064,
"author_name": "Aravindh Krishnamoorthy",
"author_id": 148448,
"author_url": "https://mathoverflow.net/users/148448/aravindh-krishnamoorthy",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">May I suggest to include the two examples into (separate) answers just to know (as someone else suggested) the relative interests of MathOverflow audience... thanks!</span>",
"created_date": "2022-02-03 02:04:27Z"
},
{
"comment_id": 1065077,
"author_name": "Kapil",
"author_id": 124862,
"author_url": "https://mathoverflow.net/users/124862/kapil",
"score": 6,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Apocryphally Gauss said, \"There is no royal road to mathematics.\" There is also <i>realistically</i> no way to learn <i>all</i> that someone has to say by reading other people's expositions! Finally, a lot of good mathematics is written in French. Learn to read it. Seriously! Do not be put off by the length. After some time you learn to read faster.</span>",
"created_date": "2022-02-03 03:20:13Z"
},
{
"comment_id": 1065159,
"author_name": "user476368",
"author_id": 476368,
"author_url": "https://mathoverflow.net/users/476368/user476368",
"score": 7,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Kapil I only agree if you give this piece of advice to somehow who is really sure he wants to do research in algebraic geometry. Then the EGA's and SGA's may be better then modern and concise expositions.</span>",
"created_date": "2022-02-03 13:19:04Z"
},
{
"comment_id": 1065178,
"author_name": "Kimball",
"author_id": 6518,
"author_url": "https://mathoverflow.net/users/6518/kimball",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\"><i>because they are French</i> - this does contradict your modern, concise, for students conditions. If you want to specify in English, just specify that. Otherwise, is Italian or Japanese okay?</span>",
"created_date": "2022-02-03 14:07:14Z"
},
{
"comment_id": 1065259,
"author_name": "user476368",
"author_id": 476368,
"author_url": "https://mathoverflow.net/users/476368/user476368",
"score": 5,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@AlexanderWoo Everybody decides for themselves what is a waste of time for them. If someone enjoys learning algebraic geometry without doing research in it, why not.</span>",
"created_date": "2022-02-03 21:58:35Z"
}
],
"answers": [
{
"answer_id": 415212,
"question_id": 415209,
"score": 13,
"is_accepted": false,
"author_name": "Francesco Polizzi",
"author_id": null,
"author_url": "",
"author_reputation": 68975,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-02-02 14:41:05Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>A good roadmap for FGA (topic 4, with glimpses on topic 6) is</p>\n<p><em>Fantechi, Barbara; Göttsche, Lothar; Illusie, Luc; Kleiman, Steven L.; Nitsure, Nitin; Vistoli, Angelo</em>, 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). <a href=\"https://zbmath.org/?q=an:1085.14001\" rel=\"noreferrer\">ZBL1085.14001</a>.</p>\n</div>",
"comments": [
{
"comment_id": 1064903,
"author_name": "user476368",
"author_id": 476368,
"author_url": "https://mathoverflow.net/users/476368/user476368",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks! Please give one of the topics 1-12 this gets associated to.</span>",
"created_date": "2022-02-02 14:34:50Z"
}
]
},
{
"answer_id": 415263,
"question_id": 415209,
"score": 12,
"is_accepted": false,
"author_name": "Timothy Chow",
"author_id": null,
"author_url": "",
"author_reputation": 91241,
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"created_date": "2022-02-03 00:01:26Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For topic 4 (schemes), I'd suggest <i>Schemes: The Language of Algebraic Geometry</i>, by Eisenbud and Harris. This book is out of print, and is generally regarded as having been \"superseded\" by its successor, <i>The Geometry of Schemes</i>. 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.</p>\n</div>",
"comments": [
{
"comment_id": 1065057,
"author_name": "Jason Starr",
"author_id": 13265,
"author_url": "https://mathoverflow.net/users/13265/jason-starr",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I believe the survey in Mumford's \"Lectures on curves in an algebraic surface\" is even shorter (and, of course, it is written by Mumford).</span>",
"created_date": "2022-02-03 00:50:17Z"
},
{
"comment_id": 1065058,
"author_name": "Paul Siegel",
"author_id": 4362,
"author_url": "https://mathoverflow.net/users/4362/paul-siegel",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I've always felt a bit self-conscious about schemes, because in spite of having taken a course or two back in the day I don't feel like I could direct a short infomercial on why schemes are important - this is in contrast to, say, manifolds or Hilbert spaces, where I am more confident I could put together a good sales pitch. Would Eisenbud-Harris' or Mumford's books help with this?</span>",
"created_date": "2022-02-03 00:56:49Z"
},
{
"comment_id": 1065074,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@PaulSiegel I do think that Eisenbud-Harris (Chapter 2 especially) would help you. But for an elevator pitch, maybe the short section on schemes by Danilov in <i>Algebraic Curves, Algebraic Manifolds and Schemes</i> is even more direct. See also <a href=\"https://math.stackexchange.com/questions/99605/why-study-schemes\">this math.SE question</a>.</span>",
"created_date": "2022-02-03 02:58:17Z"
},
{
"comment_id": 1065080,
"author_name": "Jason Starr",
"author_id": 13265,
"author_url": "https://mathoverflow.net/users/13265/jason-starr",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Mumford’s book is precisely about presenting Grothendieck’s proof, via schemes, of the “Completeness Theorem” that led to such bitter fight between Severi and Enriques.</span>",
"created_date": "2022-02-03 03:33:29Z"
},
{
"comment_id": 1299673,
"author_name": "Vincent Tran",
"author_id": 537043,
"author_url": "https://mathoverflow.net/users/537043/vincent-tran",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Also I think a simple elevator pitch can be seen here: <a href=\"https://math.stackexchange.com/a/2312107/1360259\">math.stackexchange.com/a/2312107/1360259</a> Basically, schemes allow for \"correct\" categorical things like kernels</span>",
"created_date": "2025-08-07 16:02:01Z"
}
]
},
{
"answer_id": 415300,
"question_id": 415209,
"score": 12,
"is_accepted": false,
"author_name": "Donu Arapura",
"author_id": null,
"author_url": "",
"author_reputation": 36407,
"license": "CC BY-SA 4.0",
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"created_date": "2022-02-03 13:04:31Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Let me suggest having looking at Grothendieck's own article <em>The cohomology theory of abstract algebraic varieties</em> 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.</p>\n</div>",
"comments": [
{
"comment_id": 1085232,
"author_name": "roy smith",
"author_id": 9449,
"author_url": "https://mathoverflow.net/users/9449/roy-smith",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">this is a wonderful suggestion. In addition, although I may be the only person with a copy, Dieudonne's 1962 Maryland notes, formerly available from Harvard math dept, are a summary of EGA at about 105 pages. I myself however strongly recommend reading at least selected pages of the longer versions, EGA, SGA, etc as absolutely unmatched treatments written by the original creators of these ideas. You will be very glad of every word you read by Grothendieck, or Mumford.</span>",
"created_date": "2022-05-12 02:37:09Z"
},
{
"comment_id": 1299671,
"author_name": "Vincent Tran",
"author_id": 537043,
"author_url": "https://mathoverflow.net/users/537043/vincent-tran",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@roysmith That sounds interesting. Would you mind sending a copy of Dieudonne's notes my way?</span>",
"created_date": "2025-08-07 15:52:59Z"
},
{
"comment_id": 1299894,
"author_name": "roy smith",
"author_id": 9449,
"author_url": "https://mathoverflow.net/users/9449/roy-smith",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Vincent Tran: I only have a physical copy. There is a copy for sale at abebooks: search on books titled Algebraic Geometry, by Jean Dieudonne.</span>",
"created_date": "2025-08-09 00:08:02Z"
},
{
"comment_id": 1299895,
"author_name": "roy smith",
"author_id": 9449,
"author_url": "https://mathoverflow.net/users/9449/roy-smith",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Or contact Grey Matter books, HADLEY, MASS.</span>",
"created_date": "2025-08-09 00:13:34Z"
},
{
"comment_id": 1300269,
"author_name": "Vincent Tran",
"author_id": 537043,
"author_url": "https://mathoverflow.net/users/537043/vincent-tran",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Ah I see, thank you!</span>",
"created_date": "2025-08-11 18:10:47Z"
}
]
},
{
"answer_id": 415224,
"question_id": 415209,
"score": 8,
"is_accepted": false,
"author_name": "Balazs",
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"created_date": "2022-02-02 16:08:09Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For topic 3, I would say Hartshorne's <em>Appendix A</em> is pretty good as a VERY brief introduction. After that, as also referred to there, I would recommend Manin's <em>Lectures on the K-functor in algebraic geometry</em>.</p>\n</div>",
"comments": []
},
{
"answer_id": 415307,
"question_id": 415209,
"score": 6,
"is_accepted": false,
"author_name": "Dima Pasechnik",
"author_id": null,
"author_url": "https://mathoverflow.net/users/11100",
"author_reputation": 15030,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-04 08:04:30Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Topic 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, <a href=\"http://dx.doi.org/10.1017/CBO9780511525919\" rel=\"nofollow noreferrer\">http://dx.doi.org/10.1017/CBO9780511525919</a></p>\n<p>A shorter text is by Michel Coste: <a href=\"http://perso.univ-rennes1.fr/michel.coste/polyens/OMIN.pdf\" rel=\"nofollow noreferrer\">http://perso.univ-rennes1.fr/michel.coste/polyens/OMIN.pdf</a></p>\n<p>See also nLab article on tame topology: <a href=\"https://ncatlab.org/nlab/show/tame+topology\" rel=\"nofollow noreferrer\">https://ncatlab.org/nlab/show/tame+topology</a></p>\n</div>",
"comments": [
{
"comment_id": 1065319,
"author_name": "Hvjurthuk",
"author_id": 158098,
"author_url": "https://mathoverflow.net/users/158098/hvjurthuk",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Note the intersection of Coste's mentioned text with his book Real Algebraic Geometry, by Jacek Bochnak, Michel Coste, and Marie-Francoise Roy.</span>",
"created_date": "2022-02-04 06:30:54Z"
}
]
},
{
"answer_id": 415303,
"question_id": 415209,
"score": 5,
"is_accepted": false,
"author_name": "user476368",
"author_id": null,
"author_url": "",
"author_reputation": 211,
"license": "CC BY-SA 4.0",
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"created_date": "2022-02-03 13:20:18Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For topic 5, try Leinster's nice expository article <a href=\"https://arxiv.org/abs/1012.5647\" rel=\"noreferrer\">https://arxiv.org/abs/1012.5647</a>.</p>\n</div>",
"comments": []
},
{
"answer_id": 415305,
"question_id": 415209,
"score": 2,
"is_accepted": false,
"author_name": "user476368",
"author_id": null,
"author_url": "",
"author_reputation": 211,
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"is_edited": false,
"created_date": "2022-02-03 13:23:37Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For topic 7, Milne's article <a href=\"https://www.jmilne.org/math/xnotes/MOT.pdf\" rel=\"nofollow noreferrer\">https://www.jmilne.org/math/xnotes/MOT.pdf</a> could be useful.</p>\n</div>",
"comments": []
},
{
"answer_id": 415304,
"question_id": 415209,
"score": 2,
"is_accepted": false,
"author_name": "user476368",
"author_id": null,
"author_url": "https://mathoverflow.net/users/476368",
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"is_edited": true,
"created_date": "2022-02-03 15:41:15Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For topic 6, there are Milne's notes: <a href=\"https://www.jmilne.org/math/CourseNotes/LEC.pdf\" rel=\"nofollow noreferrer\">https://www.jmilne.org/math/CourseNotes/LEC.pdf</a></p>\n<p>In the front matter Milne writes:</p>\n<blockquote>\n<p>These 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.</p>\n</blockquote>\n</div>",
"comments": []
},
{
"answer_id": 415332,
"question_id": 415209,
"score": 2,
"is_accepted": false,
"author_name": "albert",
"author_id": null,
"author_url": "https://mathoverflow.net/users/112219",
"author_reputation": 101,
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"is_edited": true,
"created_date": "2022-02-03 20:29:18Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For learning about topoi, I'd strongly recommend Goldblatt's <em>Topoi: The Categorical Analysis of Logic</em>, 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.</p>\n</div>",
"comments": []
},
{
"answer_id": 415402,
"question_id": 415209,
"score": 2,
"is_accepted": false,
"author_name": "user234212323",
"author_id": null,
"author_url": "",
"author_reputation": 934,
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"created_date": "2022-02-04 18:04:19Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For 5. Topoi. There is a very readable notice of the AMS: <a href=\"https://www.ams.org/notices/200409/what-is-illusie.pdf\" rel=\"nofollow noreferrer\">What is... a topos?</a>, Luc Illusie, and with M Raynaud about Schemes at <a href=\"http://www.asiapacific-mathnews.com/05/0501/0001_0005.pdf\" rel=\"nofollow noreferrer\">Grothendieck and Algebraic Geometry</a>. In fact, the webpage of Professor <a href=\"https://www.imo.universite-paris-saclay.fr/%7Eillusie/\" rel=\"nofollow noreferrer\">L. Illusie</a> contains very valuable material regarding the question.</p>\n<p>You can read (online) about various of the listed topics in <a href=\"https://spartacus-idh.com/094.html\" rel=\"nofollow noreferrer\">Lectures grothendieckiennes</a> Edited by Frédéric Jaëck</p>\n<p>And find a lot of original material at <a href=\"https://agrothendieck.github.io/\" rel=\"nofollow noreferrer\">Thèmes pour une Harmonie</a>!</p>\n</div>",
"comments": []
},
{
"answer_id": 422229,
"question_id": 415209,
"score": 1,
"is_accepted": false,
"author_name": "Dirk Werner",
"author_id": null,
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"created_date": "2022-05-10 19:53:32Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>On topic 1, I'd like to recommend Ray Ryan's Introduction to tensor products of Banach spaces, Springer 2002. <a href=\"https://zbmath.org/?q=an%3A1090.46001\" rel=\"nofollow noreferrer\">Zbl 1090.46001</a></p>\n</div>",
"comments": []
},
{
"answer_id": 415309,
"question_id": 415209,
"score": 0,
"is_accepted": false,
"author_name": "Mozibur Ullah",
"author_id": null,
"author_url": "",
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"created_date": "2022-02-03 15:06:22Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<ol start=\"9\">\n<li>Infinity-stacks</li>\n</ol>\n<p>Before learning about infinity stacks, one ought to learn about stacks. For this, look at section 4.1.1 of Vistoli's notes, <em>Notes on Grothendieck topologies, fibred categories and descent theory</em>, which is an exposition of one concrete case. This section is around a page and a half. He states:</p>\n<blockquote>\n<p>that the fact we can glue continuous maps and topological spaces says <span class=\"math-container\">$(Cont)$</span> is a stack over <span class=\"math-container\">$Top$</span>.</p>\n</blockquote>\n<p>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.</p>\n<p>There is also an expository note by Barbara Fantechi, <em>Stacks for Everybody</em>.</p>\n</div>",
"comments": [
{
"comment_id": 1065191,
"author_name": "user476368",
"author_id": 476368,
"author_url": "https://mathoverflow.net/users/476368/user476368",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks! Concerning topic 9, I wonder what he means by \"cohomological formalism of topoi as inspiration for a new homotopical algebra\" - what have topoi to do with homotopical algebra? Could he mean $\\infty$-topoi à la Lurie?</span>",
"created_date": "2022-02-03 15:08:36Z"
},
{
"comment_id": 1065193,
"author_name": "Mozibur Ullah",
"author_id": 35706,
"author_url": "https://mathoverflow.net/users/35706/mozibur-ullah",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@user476368: There is a duality between bundles and sheafs. Likewise with higher versions of these. A 2-sheaf is exactly a stack. According to Nlab, Quillen showed how to do abstract homotopy via model categories and this was called homotopical algebra. Model categories are a presentation of infinity topoi which suggests homotopical algebra can be generalised to this setting.</span>",
"created_date": "2022-02-03 15:25:22Z"
},
{
"comment_id": 1065198,
"author_name": "Denis Nardin",
"author_id": 43054,
"author_url": "https://mathoverflow.net/users/43054/denis-nardin",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@MoziburUllah Model categories do not (necessarily) present ∞-topoi, only those called model topoi do. In general model categories present complete and cocomplete ∞-categories (exactly which of those is not fully known, IIRC - certainly the presentable ones though)</span>",
"created_date": "2022-02-03 16:16:42Z"
}
]
},
{
"answer_id": 415313,
"question_id": 415209,
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"is_accepted": false,
"author_name": "M1011",
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"created_date": "2022-02-03 15:54:30Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For topic 1, the book by Diestel, Fourie, Swart: <a href=\"https://books.google.de/books?id=VBg5cGSngRoC&printsec=copyright&redir_esc=y#v=onepage&q&f=false\" rel=\"nofollow noreferrer\">https://books.google.de/books?id=VBg5cGSngRoC&printsec=copyright&redir_esc=y#v=onepage&q&f=false</a> provides a good introduction.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508715,
"title": "Representing a given vector field in any way I like around regular points",
"link": "https://mathoverflow.net/questions/508715/representing-a-given-vector-field-in-any-way-i-like-around-regular-points",
"score": 6,
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"tags": "dg.differential-geometry;ap.analysis-of-pdes;differential-topology;differential-equations;smooth-manifolds",
"creation_date": "2026-03-04 07:52:53Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The same question was asked <a href=\"https://math.stackexchange.com/questions/5126824/representing-vector-field-in-anyway-i-like-around-regular-points\">on Mathematics Stack exchange</a> but has not received any response or comments.</p>\n<p>Let <span class=\"math-container\">$M$</span> be a smooth <span class=\"math-container\">$n$</span>-dimensional manifold and <span class=\"math-container\">$X$</span> be a smooth vector field on it. Consider a chart <span class=\"math-container\">$(U, x^1, \\ldots, x^n)$</span> and another set of coordinates <span class=\"math-container\">$(y^1,\\ldots,y^n)$</span> on the same chart (or on an open subset of <span class=\"math-container\">$U$</span> in which case we will just shrink <span class=\"math-container\">$U$</span>). Since <span class=\"math-container\">$X$</span> is smooth, there exists smooth functions <span class=\"math-container\">$a^i : U \\to \\mathbb{R}$</span> and <span class=\"math-container\">$b^i: U \\to \\mathbb{R}$</span>, with <span class=\"math-container\">$i \\in \\{1, \\ldots, n\\}$</span>, such that, for every <span class=\"math-container\">$p \\in U$</span>:</p>\n<p><span class=\"math-container\">\\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*}</span></p>\n<p>Apply to both sides the function <span class=\"math-container\">$x^i$</span>'s to get, for all <span class=\"math-container\">$p \\in U$</span>:</p>\n<p><span class=\"math-container\">\\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*}</span></p>\n<p>Usually, one is given <span class=\"math-container\">$(x^1, \\ldots, x^n)$</span>, <span class=\"math-container\">$(y^1, \\ldots, y^n)$</span>, and <span class=\"math-container\">$a^1, \\ldots, a^n$</span>.\nOne is then required to compute <span class=\"math-container\">$b^1, \\ldots, b^n$</span> in order to represent the vector field <span class=\"math-container\">$X$</span> in the other set of coordinates.</p>\n<p>My question is about the reverse. Suppose that we know <span class=\"math-container\">$(x^1, \\ldots, x^n)$</span> and <span class=\"math-container\">$a^1, \\ldots, a^n$</span>. With a specific set of smooth functions <span class=\"math-container\">$b^1, \\ldots, b^n$</span> I chose, can we always guarantee the existence of coordinates <span class=\"math-container\">$(y^1, \\ldots, y^n)$</span> such that my vector field can be represented in the way I specified?</p>\n<p>This is definitely not always doable. For example, if <span class=\"math-container\">$X$</span> is the zero vector field and <span class=\"math-container\">$b^i$</span>'s are nonzero, then it is clearly impossible. One must impose some conditions on <span class=\"math-container\">$(x^1, \\ldots, x^n)$</span> or on <span class=\"math-container\">$a^1, \\ldots, a^n$</span>. After searching, I found a theorem regarding the representation of a vector field around a regular point: Assume <span class=\"math-container\">$p \\in M$</span> is such that <span class=\"math-container\">$X_p \\neq 0$</span>, the theorem says that we can always find a chart <span class=\"math-container\">$(U, y^1, \\ldots, y^n)$</span> about <span class=\"math-container\">$p$</span> such that <span class=\"math-container\">$X_q = \\frac{\\partial}{\\partial y_1} \\mid_q$</span> for all <span class=\"math-container\">$q \\in U$</span>. In our language, this means <span class=\"math-container\">$b^1 \\equiv 1$</span>, <span class=\"math-container\">$b^i \\equiv 0$</span> for <span class=\"math-container\">$i \\neq 1$</span>, and <span class=\"math-container\">$a^i$</span>'s are not always <span class=\"math-container\">$0$</span>. So this should be a solid starting point of a more general result.</p>\n<p>So let me rephrase the question :</p>\n<ol>\n<li>Suppose that we are in a neighborhood <span class=\"math-container\">$(U, x^1, \\ldots, x^n)$</span> of a regular point <span class=\"math-container\">$p$</span>, <span class=\"math-container\">$a^1 \\equiv 1$</span>, <span class=\"math-container\">$a^i \\equiv 0$</span> if <span class=\"math-container\">$i \\neq 1$</span>. Is it true that for every set of <span class=\"math-container\">$n$</span> smooth functions <span class=\"math-container\">$b^i:U \\to \\mathbb{R}$</span>, not all zero, we can always find a possibly smaller neighborhood of <span class=\"math-container\">$p$</span> along with local coordinates <span class=\"math-container\">$(y^1, \\ldots, y^n)$</span>, representing our vector field using functions <span class=\"math-container\">$b^i$</span>?</li>\n<li>What about other points? If <span class=\"math-container\">$X$</span>\nis zero in a neighborhood, then the only possible choice is <span class=\"math-container\">$b^i≡0$</span>.\nConsider a point <span class=\"math-container\">$p$</span> such that <span class=\"math-container\">$X_p=0$</span> but <span class=\"math-container\">$X$</span> is not zero in any neighborhood of <span class=\"math-container\">$p$</span>. What are conditions on <span class=\"math-container\">$(x_1, \\ldots,x_n)$</span>\nor on <span class=\"math-container\">$a_1, \\ldots,a_n$</span> such that <span class=\"math-container\">$X$</span> can be represented by any <span class=\"math-container\">$b_i$</span>'s in a neighborhood of <span class=\"math-container\">$p$</span>?</li>\n</ol>\n<p>As 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.</p>\n</div>",
"comments": [
{
"comment_id": 1325539,
"author_name": "Daniel Asimov",
"author_id": 5484,
"author_url": "https://mathoverflow.net/users/5484/daniel-asimov",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Shouldn't 1. read \"𝑎^1 ≡ 1, 𝑎^i ≡ 0 if i ≠ 1\" ?</span>",
"created_date": "2026-03-03 17:58:34Z"
},
{
"comment_id": 1325543,
"author_name": "温泽海",
"author_id": 368992,
"author_url": "https://mathoverflow.net/users/368992/%e6%b8%a9%e6%b3%bd%e6%b5%b7",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">That is right. I have corrected the typo.</span>",
"created_date": "2026-03-03 19:14:22Z"
}
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"answers": [
{
"answer_id": 508735,
"question_id": 508715,
"score": 5,
"is_accepted": false,
"author_name": "Loïc Teyssier",
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"created_date": "2026-03-04 07:52:53Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For 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.</p>\n<p>In most cases your vector field is linearizable by an analytic change of coordinates (<em>e.g.</em> by Poincaré theorem), meaning that your <span class=\"math-container\">$a_i$</span> 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 <span class=\"math-container\">$\\mathbb{R}\\{h\\}$</span>, see the work of Martinet and Ramis).</p>\n<p>In dimension 2, there are a few results that allow to \"desingularize\" <em>à la</em> 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 <em>al.</em> to visualize the flow of polynomial vector fields in <span class=\"math-container\">$\\mathbb{R}^2$</span>, based on Seidenberg's reduction, but I don't recollect more than that.</p>\n<p>In 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 <em>isolated</em> 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.</p>\n<p>If you allow less regularity for your local charts (<em>e.g.</em> smooth instead of analytic), then there exist partial results of conjugacy between <span class=\"math-container\">$X$</span> and one of its finite jet in dimension 2 (Dumortier, Roussarie, Rousseau, Takens <em>etc.</em>), but some of the results only work for <span class=\"math-container\">$C^k$</span> local charts with <span class=\"math-container\">$k<\\infty$</span>. Not sure how these generalize in higher dimension. It is known that some <span class=\"math-container\">$X$</span> are not smoothly conjugate to one of their finite jets.</p>\n<p>As a conclusion, I'd say that what you ask is beyond our present knowledge as soon as singularities are involved.</p>\n</div>",
"comments": []
},
{
"answer_id": 508734,
"question_id": 508715,
"score": 2,
"is_accepted": false,
"author_name": "Tom Goodwillie",
"author_id": 6666,
"author_url": "https://mathoverflow.net/users/6666/tom-goodwillie",
"author_reputation": 57946,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-04 03:47:06Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For question 1, yes. Let’s turn the notation around. Suppose that <span class=\"math-container\">$X$</span> is <span class=\"math-container\">$\\frac{\\partial}{\\partial y^1}$</span> for some coordinate system <span class=\"math-container\">$y^1,\\dots ,y^n$</span>, and let <span class=\"math-container\">$a^1,\\dots ,a^n$</span> be any functions (of <span class=\"math-container\">$y$</span>), not all <span class=\"math-container\">$0$</span>. For each <span class=\"math-container\">$i$</span> make a function <span class=\"math-container\">$x^i$</span> such that <span class=\"math-container\">$\\frac{\\partial x^i}{\\partial y^1}=a^i$</span>. If we can arrange for the <span class=\"math-container\">$x^i$</span> to be a coordinate system, then in <span class=\"math-container\">$x$</span> coordinates <span class=\"math-container\">$X$</span> will be <span class=\"math-container\">$\\sum_i a^i\\frac{\\partial}{\\partial x^i}$</span>. To do so, we try to alter <span class=\"math-container\">$x^i$</span> without changing <span class=\"math-container\">$\\frac{\\partial x^i}{\\partial x^1}$</span>, in such a way that the Jacobian matrix <span class=\"math-container\">$\\frac{\\partial x^i}{\\partial y^j}$</span> at the origin becomes invertible. We can do this by adding a suitable linear function of <span class=\"math-container\">$L(y^2,\\dots ,y^n)$</span> to each <span class=\"math-container\">$x^i$</span>.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508718,
"title": "A series related to $\\zeta(3)$",
"link": "https://mathoverflow.net/questions/508718/a-series-related-to-zeta3",
"score": 3,
"view_count": 379,
"answer_count": 2,
"is_answered": true,
"has_accepted_answer": true,
"accepted_answer_id": 508731,
"tags": "nt.number-theory;sequences-and-series;riemann-zeta-function;binomial-coefficients",
"creation_date": "2026-03-04 06:33:42Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Inspired by <a href=\"https://mathoverflow.net/questions/508663\">Question 508663</a>, I discovered the following identity</p>\n<p><span class=\"math-container\">$$\\sum_{k=1}^\\infty\\frac{H_{2k}-H_{k}}{k^2\\binom{2k}k}=2 \\zeta(3)-\\frac{\\pi \\sqrt{3}\\, \\Psi^{\\left(1\\right)}\\! \\left(\\frac{1}{3}\\right)}{24}+\\frac{\\pi \\sqrt{3}\\, \\Psi^{\\left(1\\right)}\\! \\left(\\frac{5}{6}\\right)}{72}\\tag{1},$$</span>\nwhere <span class=\"math-container\">$H_n$</span> denotes the <a href=\"https://mathworld.wolfram.com/HarmonicNumber.html\" rel=\"nofollow noreferrer\">harmonic numbers</a> and <span class=\"math-container\">$\\Psi^{(n)}(x)$</span> is the <span class=\"math-container\">$n$</span>th <a href=\"https://mathworld.wolfram.com/PolygammaFunction.html\" rel=\"nofollow noreferrer\">polygamma function</a>.</p>\n<p>My questions:</p>\n<ol>\n<li>Is the series (1) known?</li>\n<li>How can it be proven?</li>\n</ol>\n</div>",
"comments": [
{
"comment_id": 1325547,
"author_name": "Steven Clark",
"author_id": 110710,
"author_url": "https://mathoverflow.net/users/110710/steven-clark",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Mathematica gives the result $$\\sum _{k=1}^{\\infty } \\frac{H_{2 k}-H_k}{k^2 \\binom{2 k}{k}}=2 \\zeta (3)-\\frac{2 i \\pi ^3}{81}+\\frac{1}{216} \\pi \\left(2 (-1)^{5/6} \\psi ^{(1)}\\left(\\frac{1}{6}\\right)+\\left(\\sqrt{3}+i\\right) \\psi ^{(1)}\\left(\\frac{5}{6}\\right)+\\left(7 \\sqrt{3}+i\\right) \\psi ^{(1)}\\left(\\frac{2}{3}\\right)+\\left(-7 \\sqrt{3}+i\\right) \\psi ^{(1)}\\left(\\frac{1}{3}\\right)\\right)$$ which seems to evaluate consistent with your formula (1) numerically.</span>",
"created_date": "2026-03-03 19:26:28Z"
},
{
"comment_id": 1325549,
"author_name": "Henri Cohen",
"author_id": 81776,
"author_url": "https://mathoverflow.net/users/81776/henri-cohen",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The OP's formula checks out to hundreds of decimal places. It's probably a simple consequence of Mathematica's result (if correct) and standard properties of $\\psi'$. In fact the $(-1)^{5/6}$ is a giveaway that Mma did not simplify since the imaginary part should vanish.</span>",
"created_date": "2026-03-03 19:28:06Z"
},
{
"comment_id": 1325560,
"author_name": "Gerry Myerson",
"author_id": 3684,
"author_url": "https://mathoverflow.net/users/3684/gerry-myerson",
"score": 5,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">How does one discover an identity like (1)?</span>",
"created_date": "2026-03-03 21:37:26Z"
},
{
"comment_id": 1325561,
"author_name": "P. Grabowski",
"author_id": 127260,
"author_url": "https://mathoverflow.net/users/127260/p-grabowski",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I saw the previous post. I see the formula here. I am flabbergasted. How do you \"I discovered the following identity\" and you do not know how to prove it? Doesn't one come up with such things by being immersed in doing such things? However, I would then expect some discussion of, like, \"I did this and that; you know the standard stuff,\" or so. And then a person answering credits an LLM. I don't understand this dynamic... :(</span>",
"created_date": "2026-03-03 21:38:13Z"
},
{
"comment_id": 1325569,
"author_name": "Mark Wildon",
"author_id": 7709,
"author_url": "https://mathoverflow.net/users/7709/mark-wildon",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I can understand the dislike for LLMs: wading through their walls of text is frustrating, and often unrewarding. But they're going to become part of the mathematicians's toolkit, and I'd rather make the effort now to understand their strengths and limitations. I think the answer below is interesting for its mixture of routine steps (that are helped by an encyclopaedic grasp of the literature) and what may seem like genuine flashes of intuition, for example the factorization of $1 + 2 \\sin \\theta$ using $\\pi/12$ and $5\\pi/12$.</span>",
"created_date": "2026-03-03 23:37:08Z"
}
],
"answers": [
{
"answer_id": 508731,
"question_id": 508718,
"score": 2,
"is_accepted": true,
"author_name": "Zhi-Wei Sun",
"author_id": 124654,
"author_url": "https://mathoverflow.net/users/124654/zhi-wei-sun",
"author_reputation": 18678,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-04 01:44:15Z",
"last_edited_date": "2026-03-04 06:33:42Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The formula is not new. It follows from (3.2) and (3.3) in Conjecture 3.1 of <a href=\"http://maths.nju.edu.cn/%7Ezwsun/165s.pdf\" rel=\"nofollow noreferrer\">this 2015 paper of mine</a>. 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 href=\"https://arxiv.org/abs/1507.01703\" rel=\"nofollow noreferrer\">a preprint in arXiv</a>\n(see (73) and (62) of this paper) and its published form can be found <a href=\"https://www.tandfonline.com/doi/full/10.1080/10586458.2015.1116028\" rel=\"nofollow noreferrer\">here</a>.</p>\n</div>",
"comments": [
{
"comment_id": 1325579,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Can you give a link to at least one paper that proves your Conjecture 3.1?</span>",
"created_date": "2026-03-04 03:37:10Z"
},
{
"comment_id": 1325582,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Here's what zbMath knows about papers citing yours <a href=\"https://zbmath.org/?q=rf%3A6613107\" rel=\"nofollow noreferrer\">zbmath.org/?q=rf%3A6613107</a>. It's not clear without sitting down and reading a bunch of them which one should look at.</span>",
"created_date": "2026-03-04 04:34:03Z"
},
{
"comment_id": 1325587,
"author_name": "Zhi-Wei Sun",
"author_id": 124654,
"author_url": "https://mathoverflow.net/users/124654/zhi-wei-sun",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I have added links to Ablinger's paper.</span>",
"created_date": "2026-03-04 06:35:09Z"
},
{
"comment_id": 1325615,
"author_name": "Deyi Chen",
"author_id": 287674,
"author_url": "https://mathoverflow.net/users/287674/deyi-chen",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thank you to Professor Sun for the articles provided; your conjectures and the articles are both very interesting and inspiring!</span>",
"created_date": "2026-03-04 12:47:11Z"
}
]
},
{
"answer_id": 508725,
"question_id": 508718,
"score": 6,
"is_accepted": false,
"author_name": "Mark Wildon",
"author_id": null,
"author_url": "https://mathoverflow.net/users/7709",
"author_reputation": 12112,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-03 21:23:14Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><span class=\"math-container\">$\\newcommand{\\mfrac}[2]{{\\textstyle\\frac{#1}{#2}}}$</span>\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 <span class=\"math-container\">$S$</span> be the sum and let\n<span class=\"math-container\">$$\nF(z)=\\sum_{k=1}^\\infty \\frac{z^k}{k^2\\binom{2k}{k}} $$</span>\nfor <span class=\"math-container\">$0 \\le z \\le 1$</span>. From the standard Taylor series\n<span class=\"math-container\">$$\\arcsin^2 x = \\frac{1}{2} \\sum_{k=1}^\\infty \\frac{1}{k^2 \\binom{2k}{k}} (2x)^{2k},$$</span>\nsee for instance <a href=\"https://math.stackexchange.com/questions/1448822/taylor-expansion-for-arcsin2x\">this MathStackexchange answer</a> (although ChatGPT was keen to share its own proof with me), we have <span class=\"math-container\">$F(z)=2\\arcsin^2\\bigl(\\frac{\\sqrt z}{2}\\bigr)$</span>.\nUsing the integral formula for the Harmonic numbers we have\n<span class=\"math-container\">$$ H_{2k}-H_k\n= \\int_0^1 \\frac{x^k-x^{2k}}{1-x}\\,\\mathrm{d}x $$</span>\nand so\n<span class=\"math-container\">$$ S =\\int_0^1 \\frac{F(x)-F(x^2)}{1-x}\\,\\mathrm{d}x. $$</span>\nNow\n<span class=\"math-container\">$F(x)-F(x^2)=\\int_{x^2}^x F'(t)\\,\\mathrm{d}t $</span>\nso we have, swapping the order of integration using\n<span class=\"math-container\">$x^2 \\le t \\le x \\iff t \\le x \\le \\sqrt{t}$</span> to get the second line,\n<span class=\"math-container\">$$ \\begin{aligned} S &= \\int_0^1 \\int_{x^2}^x F'(t)\\, \\mathrm{d}t\\, \\frac{\\mathrm{d}x}{1-x} \\\\ &=\n\\int_0^1 F'(t)\\bigl(\\int_t^{\\sqrt t}\\frac{dx}{1-x}\\bigr)\\mathrm{d}t \\\\ &=\n\\int_0^1 F'(t)\\log \\bigl(\\frac{1-t}{1-\\sqrt t}\\bigr)dt \\\\ \n&= \\int_0^1 F'(t)\\log(1+\\sqrt t)\\,\\mathrm{d}t \\\\\n&= \\int_0^1 F'(u^2) \\log (1+u)\\, 2u\\, \\mathrm{d}u \\\\\n&= \\int_0^{\\pi/6} F'(4\\sin^2\\theta) \\log(1 + 2\\sin \\theta) 8\\sin\\theta \\cos \\theta \\,\\mathrm{d}\\theta \\\\\n&= \\int_0^{\\pi/6} \\frac{\\theta}{\\sin 2 \\theta} \\log (1 + 2 \\sin \\theta) 4 \\sin 2\\theta \\,\\mathrm{d}\\theta \\\\ \n&= 4\\int_0^{\\pi/6} \\theta \\log (1 + 2 \\sin \\theta) \\,\\mathrm{d}\\theta.\n \\end{aligned}$$</span>\nThis used the substitution <span class=\"math-container\">$u^2 = t$</span>, then <span class=\"math-container\">$u = 2\\sin \\theta$</span> and\nthen in the penultimate step\n<span class=\"math-container\">$$ F'(z) = \\frac{\\arcsin \\frac{\\sqrt{z}}{2} }{\\sqrt{1-z/4} \\sqrt{z}} $$</span>\nto get <span class=\"math-container\">$F'(4\\sin^2 \\theta) = \\frac{\\theta}{\\sin 2\\theta}$</span>.\nNow factor\n<span class=\"math-container\">$$\n1+2\\sin\\theta =\n4\\sin\\!\\left(\\mfrac{\\theta}{2}+\\mfrac{\\pi}{12}\\right)\n \\sin\\!\\left(\\mfrac{\\theta}{2}+\\mfrac{5\\pi}{12}\\right) $$</span>\nto get\n<span class=\"math-container\">$$\n\\log(1+2\\sin\\theta) =\n\\log \\Bigr(2\\sin\\bigl(\\mfrac{\\theta}{2}+\\mfrac{\\pi}{12}\\bigr)\\Bigr) +\n\\log \\Bigr(2\\sin\\bigl(\\mfrac{\\theta}{2}+\\mfrac{5\\pi}{12}\\bigr)\\Bigr). $$</span></p>\n<p>The standard Fourier series\n<span class=\"math-container\">$$\n\\log(2\\sin \\phi) =-\\sum_{n=1}^\\infty \\frac{\\cos(2n\\phi )}{n}\n$$</span>\nis uniformly convergent for <span class=\"math-container\">$\\frac{\\theta}{2} + \\frac{\\pi}{12}$</span>\nand <span class=\"math-container\">$\\frac{\\theta}{2} + \\frac{5\\pi}{12}$</span> for <span class=\"math-container\">$\\theta \\in [0, \\frac{\\pi}{6}]$</span>.\nSubstituting for each summand in the right-hand side for <span class=\"math-container\">$\\log (1 + 2 \\sin \\theta)$</span> above we get\n<span class=\"math-container\">$$\n\\log(1+2\\sin\\theta)\n=\n-\\sum_{n=1}^\\infty\n\\frac{\\cos\\bigl(n(\\theta+\\frac{\\pi}{6})\\bigr)+\\cos\\bigl((\\theta+\\frac{5\\pi}{6})n\\bigr)}{n}.\n$$</span>\nUsing\n<span class=\"math-container\">$\n\\cos\\bigl( n (\\theta+\\frac{\\pi}{6}) \\bigr) + \\cos\\bigl(n(\\theta+\\frac{5\\pi}{6})\\bigr)\n= 2\\cos\\bigl(\\frac{\\pi}{3}n\\bigr)\\cos (\\theta+\\frac{\\pi}{2})n\n$</span>\nwe obtain\n<span class=\"math-container\">$$\nS\n=\n-8\\sum_{n=1}^\\infty \\frac{\\cos(\\frac{\\pi}{3}n)}{n}\n\\int_0^{\\pi/6}\\theta\\cos \\bigl( n (\\theta+\\frac{\\pi}{2}) \\bigr) \\,\\mathrm{d}\\theta.\n$$</span>\nA direct integration, which I verified in Mathematica, gives\n<span class=\"math-container\">$$\n\\int_0^{\\pi/6}\\theta\\cos\\bigl (n (\\theta+\\mfrac{\\pi}{2}) \\bigr) \\,\\mathrm{d}\\theta\n=\n\\frac{\\pi}{6n}\\sin\\!\\left(\\frac{2n\\pi}{3}\\right)\n+\n\\frac{\\cos(\\frac{2\\pi}{3} n)-\\cos( \\frac{\\pi}{2}n ) }{n^2}.\n$$</span>\nTherefore using the uniform convergence mentioned above,\n<span class=\"math-container\">$$\nS = -\\frac{8\\pi}{6}\\sum_{n=1}^\\infty \\frac{\\cos(\\frac{\\pi}{3}n)\\sin(\\frac{2\\pi}{3} n)}{n^2} -8\\sum_{n=1}^\\infty \\frac{a_n}{n^3},\n$$</span>\nwhere\n<span class=\"math-container\">$$ a_n = \\cos \\bigl( \\mfrac{n\\pi}{3} \\bigr) \\Bigl( \n\\cos\\bigl(\\mfrac{2n\\pi}{3}\\bigr)-\\cos\\bigl(\\mfrac{n\\pi}{2}\\bigr) \\Bigr). $$</span>\nSince\n<span class=\"math-container\">$ \\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) $</span>\nthis becomes\n<span class=\"math-container\">$$ S =\n-\\frac{2\\pi}{3}\\sum_{n=1}^\\infty \\frac{\\sin( \\frac{\\pi}{3} n)}{n^2}\n-8\\sum_{n=1}^\\infty \\frac{a_n}{n^3}.\n$$</span></p>\n<p>Now the sequence <span class=\"math-container\">$(a_n)$</span> is periodic with period <span class=\"math-container\">$12$</span>.\nThe values <span class=\"math-container\">$a_0,\\ldots, a_{11}$</span> are\n<span class=\"math-container\">$0,-\\frac{1}{4},-\\frac{1}{4},-1, \\frac{3}{4}, -\\frac{1}{4}, 2, -\\frac{1}{4}, \\frac{3}{4}, -1, -\\frac{1}{4},\n-\\frac{1}{4}$</span>. (Again I checked this with Mathematica.)\nHence\n<span class=\"math-container\">$$\na_n=\n-\\frac{1}{4}-\\frac{3}{4} [3 \\mid n] + \n[ 4 \\mid n] + 3 [6 \\mid n] - 3 [12 \\mid n]\n$$</span>\nin Iverson bracket notation.\nIt follows that\n<span class=\"math-container\">$$\n\\sum_{n=1}^\\infty \\frac{a_n}{n^3} =\n\\Bigl( -\\frac{1}{4} -\\frac{3}{4\\cdot 3^3} +\\frac{1}{4^3} +\\frac{3}{6^3} -\\frac{3}{12^3}\n\\Bigr)\\zeta(3)\n= -\\frac{1}{4}\\zeta(3).\n$$</span>\nWe have shown that\n<span class=\"math-container\">$$ S = 2\\zeta(3)-\\frac{2\\pi}{3}\\sum_{n=1}^\\infty \\frac{\\sin(\\frac{\\pi}{3} n)}{n^2}. \\qquad(\\star) $$</span>\nIt remains to evaluate\n<span class=\"math-container\">$ C=\\sum_{n=1}^\\infty \\frac{\\sin(\\frac{\\pi}{3}n)}{n^2}$</span>.\nUsing the values of <span class=\"math-container\">$\\sin \\frac{\\pi}{3}n$</span>, which have period <span class=\"math-container\">$6$</span>, we get\n<span class=\"math-container\">$$\nC = \\frac{\\sqrt{3}}{2}\\sum_{m=0}^\\infty\n\\bigl( \\frac{1}{(6m+1)^2} +\\frac{1}{(6m+2)^2} -\\frac{1}{(6m+4)^2} -\\frac{1}{(6m+5)^2} \\bigr). $$</span>\nBy one definition of the trigamma function <span class=\"math-container\">$\\psi_1$</span> we have\n<span class=\"math-container\">$\\psi_1(z)=\\sum_{m=0}^\\infty \\frac1{(m+z)^2}$</span>,\nso we may rewrite the previous equation for <span class=\"math-container\">$C$</span> as\n<span class=\"math-container\">$$\nC = \\frac{\\sqrt{3}}{72}\n\\Bigl( \\psi_1\\bigl(\\mfrac{1}{6} \\bigr) + \\psi_1\\bigl(\\mfrac{1}{3} \\bigr) - \\psi_1\\bigl(\\mfrac{2}{3} \\bigr) \n-\\psi_1\\bigl(\\mfrac{5}{6} \\bigr) \\Bigr) $$</span>\nThe <a href=\"https://www.johndcook.com/blog/gamma_identities/\" rel=\"nofollow noreferrer\">trigamma duplication formula</a> states that\n<span class=\"math-container\">$$\n\\psi_1(z)+\\psi_1\\!\\left(z+ \\mfrac{1}{2} \\right)=4\\psi_1(2z). $$</span>\nApply this with <span class=\"math-container\">$z=\\mfrac{1}{6}$</span> and <span class=\"math-container\">$z = \\mfrac{1}{3}$</span> to get\n<span class=\"math-container\">$ \\psi_1\\bigl( \\mfrac{1}{6} \\bigr) + \\psi_1 \\bigl( \\mfrac{2}{3} \\bigr) = 4 \\psi_1 \\bigl( \\mfrac{1}{3} \\bigr)$</span>\nand similarly\n<span class=\"math-container\">$ \\psi_1\\bigl( \\mfrac{1}{3} \\bigr) + \\psi_1 \\bigl( \\mfrac{5}{6} \\bigr) = 4 \\psi_1 \\bigl( \\mfrac{2}{3} \\bigr)$</span>.\nHence\n<span class=\"math-container\">$$ \\begin{aligned} C &=\n\\frac{\\sqrt{3}}{72}\n\\Bigl( 4 \\psi_1\\bigl( \\mfrac{1}{3} \\bigr) - \\psi_1 \\bigl( \\mfrac{2}{3} \\bigr) \n+ \\psi_1\\bigl(\\mfrac{1}{3} \\bigr) - \\psi_1\\bigl(\\mfrac{2}{3} \\bigr) \n-\\psi_1\\bigl(\\mfrac{5}{6} \\bigr) \\Bigr) \\\\\n&=\n\\frac{\\sqrt{3}}{72}\n\\Bigl( \n5\\psi_1\\bigl(\\mfrac{1}{3} \\bigr) -2\\psi_1\\bigl(\\mfrac{2}{3} \\bigr) -\\psi_1\\bigl( \\mfrac{5}{6} \\bigr) \\Bigr) \n\\\\ &=\n\\frac{\\sqrt{3}}{72}\n\\Bigl( \n5\\psi_1\\bigl( \\mfrac{1}{3} \\bigr) - \\mfrac{1}{2} \\psi_1 \\bigl( \\mfrac{1}{3} \\bigr) \n- \\mfrac{1}{2} \\psi_1 \\bigl( \\mfrac{5}{6} \\bigr) - \\psi_1 \\bigl( \\mfrac{5}{6} \\bigr) \n\\Bigr) \\\\\n&=\n\\frac{\\sqrt3}{48}\n\\bigl( \n3\\psi_1\\bigl( \\mfrac{1}{3} \\bigr) - \\psi_1 \\bigl( \\mfrac{5}{6} \\bigr) \\bigr)\n\\end{aligned}\n$$</span>\nSubstituting this into the formula for <span class=\"math-container\">$S$</span> marked (<span class=\"math-container\">$\\star$</span>) above we arrive at\n<span class=\"math-container\">$$ S =\n2\\zeta(3) -\\frac{2\\pi}{3}C = 2\\zeta(3)\n-\\frac{\\pi\\sqrt{3}}{24}\\psi_1\\bigl(\\mfrac{1}{3} \\bigr) \n+\\frac{\\pi\\sqrt{3}}{72}\\psi_1\\bigl( \\mfrac{5}{6} \\bigr)\n$$</span>\nas required.</p>\n</div>",
"comments": [
{
"comment_id": 1325557,
"author_name": "Mark Wildon",
"author_id": 7709,
"author_url": "https://mathoverflow.net/users/7709/mark-wildon",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Again since any reputation should go to the LLM, I have made this answer community wiki. I think it is hard not to be at least a little impressed by what it can do, although I doubt this is the shortest proof.</span>",
"created_date": "2026-03-03 21:27:15Z"
},
{
"comment_id": 1325574,
"author_name": "mathworker21",
"author_id": 129185,
"author_url": "https://mathoverflow.net/users/129185/mathworker21",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Wow! Can you post a link to the GPT chat? If not, can you please share how long GPT thought for before providing the answer?</span>",
"created_date": "2026-03-04 01:29:13Z"
},
{
"comment_id": 1325575,
"author_name": "Mark Wildon",
"author_id": 7709,
"author_url": "https://mathoverflow.net/users/7709/mark-wildon",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Here's the link: <a href=\"https://chatgpt.com/share/69a78e23-d658-8008-8e3e-8173ef444b60\" rel=\"nofollow noreferrer\">chatgpt.com/share/69a78e23-d658-8008-8e3e-8173ef444b60</a>. Thinking time: 40m 47s. I typically see between 30m and 65m on problems like this.</span>",
"created_date": "2026-03-04 01:44:39Z"
},
{
"comment_id": 1325578,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@MarkWildon thank you for your care in this as well as transparency. This reminds me that there is still the unresolved question <a href=\"https://meta.mathoverflow.net/questions/6285/proposal-to-permit-verified-numerical-ai-output-as-a-component-to-a-mathoverflow\" title=\"proposal to permit verified numerical ai output as a component to a mathoverflow\">meta.mathoverflow.net/questions/6285/…</a> on meta. MathOverflow hasn't yet got a good, robust general policy about use of AI leading to valid answers. For readers of this answer: please be aware that an established mathematician who has carefully checked (and is willing to vouch under their real name the correctness of) the output is no license to post genAI answers.</span>",
"created_date": "2026-03-04 02:59:44Z"
},
{
"comment_id": 1325583,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Given Sun's answer (now with a reference to a paper that has enough to prove the claimed result) it might be that there was enough in the literature that ChatGPT was able to extract the techniques.</span>",
"created_date": "2026-03-04 04:34:16Z"
}
]
}
]
},
{
"question_id": 98548,
"title": "In search of an early picture of Max Dehn",
"link": "https://mathoverflow.net/questions/98548/in-search-of-an-early-picture-of-max-dehn",
"score": 20,
"view_count": 1000,
"answer_count": 3,
"is_answered": true,
"has_accepted_answer": true,
"accepted_answer_id": 98550,
"tags": "ho.history-overview",
"creation_date": "2026-03-04 06:16:24Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I am trying to find a copy of a picture \"Mathematische Gesellschaft: \nGroup Portrait, Faculty, University of Göttingen (1899).\"</p>\n<p>This 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.</p>\n<p>So, 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.</p>\n</div>",
"comments": [],
"answers": [
{
"answer_id": 98550,
"question_id": 98548,
"score": 30,
"is_accepted": true,
"author_name": "Dirk",
"author_id": 9652,
"author_url": "https://mathoverflow.net/users/9652/dirk",
"author_reputation": 13198,
"license": "CC BY-SA 3.0",
"is_edited": true,
"created_date": "2012-06-01 08:42:42Z",
"last_edited_date": "2020-06-15 07:27:00Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>As 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:</p>\n<p><img alt=\"Max Dehn\" src=\"https://regularize.files.wordpress.com/2012/06/foto0138.jpg\"/></p>\n<p>I could make a better picture on Monday if that would be helpful...</p>\n<p><strong>Edit:</strong> Ok, my next better camera produced this picture:</p>\n<p><img alt=\"Max Dehn\" src=\"https://regularize.files.wordpress.com/2012/06/dscn0044.jpg\"/></p>\n<p>Probably that's already near the best you can get because you can already see the halftoning effects pretty clear...</p>\n</div>",
"comments": [
{
"comment_id": 252893,
"author_name": "John Stillwell",
"author_id": 1587,
"author_url": "https://mathoverflow.net/users/1587/john-stillwell",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks, Dirk, this is wonderful! If it is possible to make a sharper picture I would appreciate it, but this is already good enough for me to use.</span>",
"created_date": "2012-06-01 11:45:07Z"
},
{
"comment_id": 253517,
"author_name": "John Stillwell",
"author_id": 1587,
"author_url": "https://mathoverflow.net/users/1587/john-stillwell",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks for the second attempt; I'm now convinced that the first one is close to best possible. Back in 1899, photographs were usually not very sharp, and Dehn was only a small part of a large group picture.</span>",
"created_date": "2012-06-04 10:12:41Z"
},
{
"comment_id": 1073209,
"author_name": "José Hdz. Stgo.",
"author_id": 1593,
"author_url": "https://mathoverflow.net/users/1593/jos%c3%a9-hdz-stgo",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Dirk: Would you be so kind as to share a photo of the whole poster with me? Best regards...</span>",
"created_date": "2022-03-14 00:15:13Z"
},
{
"comment_id": 1073264,
"author_name": "Dirk",
"author_id": 9652,
"author_url": "https://mathoverflow.net/users/9652/dirk",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Sorry, the building where this picture was has meanwhile been completely remodeled. While my office is in this building again, the poster is not. I don't know where it is now…</span>",
"created_date": "2022-03-14 10:36:31Z"
}
]
},
{
"answer_id": 503749,
"question_id": 98548,
"score": 6,
"is_accepted": false,
"author_name": "Astabricot",
"author_id": 578415,
"author_url": "https://mathoverflow.net/users/578415/astabricot",
"author_reputation": 61,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2025-11-11 20:16:09Z",
"last_edited_date": "2025-11-11 20:16:30Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>There is also the same poster in my classroom. However, I can't find it in good quality on internet.</p>\n<p><a href=\"https://i.sstatic.net/CUpGsNtr.jpg\" rel=\"noreferrer\">https://i.sstatic.net/CUpGsNtr.jpg</a></p>\n</div>",
"comments": [
{
"comment_id": 1312765,
"author_name": "Nate Eldredge",
"author_id": 4832,
"author_url": "https://mathoverflow.net/users/4832/nate-eldredge",
"score": 5,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I take it that the copy of \"The Mathematical Intelligencer\" in Hilbert's hand is fake? The current journal by that name only started in 1977, and I found no indication of any earlier ones. Perhaps the poster was created to promote the journal?</span>",
"created_date": "2025-11-11 21:47:08Z"
},
{
"comment_id": 1312788,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@NateEldredge it does look composited in, yes. The colouring is off for a B&W photo</span>",
"created_date": "2025-11-12 03:45:15Z"
}
]
},
{
"answer_id": 503760,
"question_id": 98548,
"score": 6,
"is_accepted": false,
"author_name": "ACR",
"author_id": 142414,
"author_url": "https://mathoverflow.net/users/142414/acr",
"author_reputation": 943,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2025-11-12 02:39:07Z",
"last_edited_date": "2026-03-04 06:16:24Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>A soft copy of the photograph is available in <em>Max Dehn Polyphonic Portrait</em> 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 <em>Mathematical Intelligencer</em> poster, Hilbert does not a hold copy of the magazine as a reader noted.</p>\n<p>Also, if you wish, you can contact the Josef and Anni Albers Foundation who provided this figure to the book authors.</p>\n<p>Ebook is available at <a href=\"https://bookstore.ams.org/HMATH/46\" rel=\"nofollow noreferrer\">https://bookstore.ams.org/HMATH/46</a></p>\n</div>",
"comments": [
{
"comment_id": 1313351,
"author_name": "testaccount",
"author_id": 510150,
"author_url": "https://mathoverflow.net/users/510150/testaccount",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The picture appears in the preface, which is freely available. See page 10 (viii) of the PDF <a href=\"https://www.ams.org/books/hmath/046/hmath046-endmatter.pdf\" rel=\"nofollow noreferrer\">ams.org/books/hmath/046/hmath046-endmatter.pdf</a></span>",
"created_date": "2025-11-16 02:59:36Z"
}
]
}
]
},
{
"question_id": 507298,
"title": "Explicit derivation of the Artin-Schreier equation from a Kummer extension in non-discretely valued complete fields",
"link": "https://mathoverflow.net/questions/507298/explicit-derivation-of-the-artin-schreier-equation-from-a-kummer-extension-in-no",
"score": 6,
"view_count": 270,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "nt.number-theory;field-extensions;valuation-rings",
"creation_date": "2026-03-04 05:57:07Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Let <span class=\"math-container\">$(K, |\\cdot|)$</span> be a complete ultrametric field of characteristic <span class=\"math-container\">$0$</span> with residue field <span class=\"math-container\">$k$</span> of characteristic <span class=\"math-container\">$p > 0$</span>. Assume <span class=\"math-container\">$K$</span> contains a primitive <span class=\"math-container\">$p$</span>-th root of unity <span class=\"math-container\">$\\zeta$</span>. Consider a Kummer extension <span class=\"math-container\">$L = K(a^{1/p})$</span> where <span class=\"math-container\">$a \\in O_K^\\times$</span>. Suppose the extension of residue fields <span class=\"math-container\">$l/k$</span> is of degree <span class=\"math-container\">$p$</span> and separable. It is well-known that <span class=\"math-container\">$l/k$</span> is an Artin-Schreier extension of the form <span class=\"math-container\">$z^p - z = \\bar{\\gamma}$</span>. How can we find <span class=\"math-container\">$\\gamma$</span> from <span class=\"math-container\">$a$</span>?</p>\n<p>More precisely, for the case where <span class=\"math-container\">$|.|$</span> is not discret, how can we found <span class=\"math-container\">$\\gamma\\in O_K$</span> such that <span class=\"math-container\">$a=(1+X)^p(1+p(\\zeta-1)\\gamma)$</span>?</p>\n<p>For the case where <span class=\"math-container\">$|.|$</span> is discret and <span class=\"math-container\">$\\pi$</span> is the uniformizer, we can construct an algorithm such that if <span class=\"math-container\">$|a_i-1|>|p(\\zeta-1)|$</span> we can construct <span class=\"math-container\">$u_i$</span> with <span class=\"math-container\">$a_i=u_i^p a_{i+1}$</span> and <span class=\"math-container\">$|a_{i+1}-1|\\leq |\\pi(a_i-1)|$</span>. Therefore at some stage we obtain <span class=\"math-container\">$|a_i-1|\\leq |p(\\zeta-1)|$</span>.\nIn the general case, however, we have\n<span class=\"math-container\">$|a_{i+1}| \\leq q_i \\, |a_i - 1|$</span> for some <span class=\"math-container\">$q_i \\in (0,1)$</span>.\nThere is no concrete information about the constants <span class=\"math-container\">$q_i$</span>.\nIf we assume that\n<span class=\"math-container\">$|a_i - 1| > |p(\\zeta - 1)|$</span> for all <span class=\"math-container\">$i$</span>,\nwe can only prove that\n<span class=\"math-container\">$|a_i - 1| \\to |p(\\zeta - 1)|$</span>,\nand so far I have not been able to derive a contradiction.</p>\n</div>",
"comments": [
{
"comment_id": 1321394,
"author_name": "Lubin",
"author_id": 11417,
"author_url": "https://mathoverflow.net/users/11417/lubin",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">In future, please have pity on an aged mathematician with failing vision, and avoid using “$a$” and “$\\alpha$” in the same paragraph.</span>",
"created_date": "2026-01-23 01:13:37Z"
}
],
"answers": [
{
"answer_id": 507322,
"question_id": 507298,
"score": 7,
"is_accepted": false,
"author_name": "Will Sawin",
"author_id": 18060,
"author_url": "https://mathoverflow.net/users/18060/will-sawin",
"author_reputation": 166969,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-01-22 17:05:38Z",
"last_edited_date": "2026-01-22 21:03:36Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>If <span class=\"math-container\">$a$</span> has the special form <span class=\"math-container\">$a = 1 + p(\\zeta-1) b$</span> for <span class=\"math-container\">$b \\in \\mathcal O_K^\\times$</span> then we can expand <span class=\"math-container\">$(1+(\\zeta-1) x)^p= a$</span> using the binomial theorem as</p>\n<p><span class=\"math-container\">$$ p (\\zeta-1) x + \\dots + (\\zeta-1)^p x^p = p (\\zeta-1) b $$</span></p>\n<p>where the omitted terms are divisible by <span class=\"math-container\">$p(\\zeta-1)^2$</span>. From the minimal polynomial\n<span class=\"math-container\">$$\\frac{ (y+1)^p -1}{y} = y^{p-1} + \\binom{p}{1} y^{p-2} + \\dots + \\binom{p}{2} y+ \\binom{p}{1} $$</span> of <span class=\"math-container\">$\\zeta-1$</span>, we see that <span class=\"math-container\">$$(\\zeta-1)^{p-1} \\equiv - p \\pmod {p (\\zeta-1)},$$</span> so\n<span class=\"math-container\">$$(\\zeta-1)^p \\equiv - p (\\zeta-1) \\pmod{ p (\\zeta-1)^2}.$$</span></p>\n<p>Thus, dividing by <span class=\"math-container\">$p(\\zeta-1)$</span> and modding out by <span class=\"math-container\">$(\\zeta-1)$</span> gives</p>\n<p><span class=\"math-container\">$$x - x^p \\equiv b \\pmod {(\\zeta-1)} $$</span>\nwhich is the desired Artin–Schreier form.</p>\n<p>For the general case, one can observe that every <span class=\"math-container\">$a$</span> for which the Kummer extension is unramified can be expressed as a <span class=\"math-container\">$p$</span>th power times a number of this special form. To do this, one just has to check that <span class=\"math-container\">$a_0 = 1 + p(\\zeta-1) b_0$</span> for <span class=\"math-container\">$b_0$</span> 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 <span class=\"math-container\">$p$</span> unramified extension, and there is only one such extension, so every such <span class=\"math-container\">$a$</span> must be a power of <span class=\"math-container\">$a_0$</span> times a <span class=\"math-container\">$p$</span>th power by a general uniqueness statement in Kummer theory.</p>\n</div>",
"comments": [
{
"comment_id": 1322041,
"author_name": "AZZOUZ Tinhinane Amina",
"author_id": 117853,
"author_url": "https://mathoverflow.net/users/117853/azzouz-tinhinane-amina",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks. My real problem is to find a way to recover $a_0$ from $a$. In other words, I want to find $u$ such that $a = u^p a_0$. I tried, without success, to construct an iterative algorithm to find $u$. Obviously, we cannot construct a convergent sequence $(u_n)$ converging to $u$; the only possible approach seems to be to find a suitable stopping condition for the algorithm. Is there anything in the literature that could help with this issue?</span>",
"created_date": "2026-01-29 07:49:54Z"
},
{
"comment_id": 1322042,
"author_name": "AZZOUZ Tinhinane Amina",
"author_id": 117853,
"author_url": "https://mathoverflow.net/users/117853/azzouz-tinhinane-amina",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">More generally, given a finite extension $(L, |\\cdot|)$ of a complete valued field $(K, |\\cdot|)$, how can one find a generator of the ring of integers $O_L$ of $L$ as an $O_K$-module?</span>",
"created_date": "2026-01-29 07:50:27Z"
},
{
"comment_id": 1322090,
"author_name": "Will Sawin",
"author_id": 18060,
"author_url": "https://mathoverflow.net/users/18060/will-sawin",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@AZZOUZTinhinaneAmina Constructing an iterative algorithm seems not so difficult to me. Consider some $u_i$ such that $a/ u_i^p$ has $p$-adic valuation $v_i$. If $v_i $ is at least the $p$-adic valuation of $p(\\zeta-1)$ then we are done. Otherwise we need to adjust $u_i$ by an element of valuation $\\frac{v_i}{p}$. We can try such elements until we bring $u_i^p$ closer to $a$.</span>",
"created_date": "2026-01-29 13:59:23Z"
},
{
"comment_id": 1325050,
"author_name": "AZZOUZ Tinhinane Amina",
"author_id": 117853,
"author_url": "https://mathoverflow.net/users/117853/azzouz-tinhinane-amina",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">This is possible if $K$ has a discrete valuation. Indeed, if $\\pi$ is a uniformizer, then $|a/u_i^p - 1| < |\\pi| \\, |a/u_{i-1}^p - 1|$. Therefore, at some stage we obtain $|a/u_i^p - 1| \\le |p(\\zeta - 1)|$. In the general case, however, we have $|a/u_i^p - 1| < q_i \\, |a/u_{i-1}^p - 1|$ for some $q_i \\in (0,1)$. There is no concrete information about the constants $q_i$. If we assume that $|a/u_i^p - 1| > |p(\\zeta - 1)|$ for all $i$, we can only prove that $|a/u_i^p - 1| \\to |p(\\zeta - 1)|$, and so far I have not been able to derive a contradiction.</span>",
"created_date": "2026-02-27 07:55:52Z"
}
]
}
]
},
{
"question_id": 25630,
"title": "Major mathematical advances past age fifty",
"link": "https://mathoverflow.net/questions/25630/major-mathematical-advances-past-age-fifty",
"score": 73,
"view_count": 44000,
"answer_count": 43,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "soft-question;ho.history-overview",
"creation_date": "2026-03-04 05:39:20Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>From 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.\"</p>\n<p>Have matters improved for the elderly mathematician? Please answer with major discoveries made by mathematicians past 50.</p>\n</div>",
"comments": [
{
"comment_id": 54004,
"author_name": "Pietro Majer",
"author_id": 6101,
"author_url": "https://mathoverflow.net/users/6101/pietro-majer",
"score": 72,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Rmk: Hardy suffered of depression, and was living not exactly in the most suitable environment for that. Unfortunately, this wrong idea of \"mathematics is a young man's game\" had an incredible success. </span>",
"created_date": "2010-05-23 08:10:43Z"
},
{
"comment_id": 54061,
"author_name": "BCnrd",
"author_id": 3927,
"author_url": "https://mathoverflow.net/users/3927/bcnrd",
"score": 38,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Cliff Taubes (b. 1954) recently solved Weinstein conjecture, Gopal Prasad (b. 1945) has done multiple great things (separately with J-K. Yu, A. Rapinchuk, & S-K. Yeung) on buildings, Zariski-dense and arithmetic subgroups of ss groups over number fields, classification of \"fake\" projective spaces, etc., Serre turned 50 in 1976 (e.g., his precise modularity conjecture published in 1986 exerted vast influence over number theory ever since), and Jean-Marc Fontaine (b. 1944) is as dominant as ever in $p$-adic Hodge theory (e.g., Colmez-Fontaine thm. in 2000, recent work with L. Fargues, etc.)</span>",
"created_date": "2010-05-23 13:04:44Z"
},
{
"comment_id": 56856,
"author_name": "paul Monsky",
"author_id": 6214,
"author_url": "https://mathoverflow.net/users/6214/paul-monsky",
"score": 19,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">This isn't exactly what you were asking for, but Littlewood himself, after overcoming depression at age 72, did good mathematics throughout his 80's--it's hardly a young man's game.</span>",
"created_date": "2010-06-01 23:50:48Z"
},
{
"comment_id": 56860,
"author_name": "Victor Protsak",
"author_id": 5740,
"author_url": "https://mathoverflow.net/users/5740/victor-protsak",
"score": 12,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Re \"Littlewood himself\": Of course it was well known that Littlewood was the name Hardy used to publish his lesser results (cf \"A mathematician's miscellany\").</span>",
"created_date": "2010-06-02 00:09:39Z"
},
{
"comment_id": 270526,
"author_name": "Tom Leinster",
"author_id": 586,
"author_url": "https://mathoverflow.net/users/586/tom-leinster",
"score": 15,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">What's really odd is when Abel and Galois are wheeled out in support of the view that mathematics is a young person's game. Spot the logical flaw. </span>",
"created_date": "2012-08-20 18:25:28Z"
}
],
"answers": [
{
"answer_id": 25631,
"question_id": 25630,
"score": 82,
"is_accepted": false,
"author_name": "David Hansen",
"author_id": null,
"author_url": "",
"author_reputation": 13308,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-23 07:12:56Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Roger Apery was 62 when he proved the irrationality of $\\zeta(3)$.</p>\n</div>",
"comments": [
{
"comment_id": 54025,
"author_name": "Gerry Myerson",
"author_id": 3684,
"author_url": "https://mathoverflow.net/users/3684/gerry-myerson",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I think disregarded is a bit too strong - more like, met with considerable skepticism, on the grounds that it was not expected that such a venerable problem would be solved by such low-tech methods. The community took the proof seriously enough to go through it in detail and then acknowledged that it was valid. </span>",
"created_date": "2010-05-23 10:22:56Z"
},
{
"comment_id": 54074,
"author_name": "Ben Webster",
"author_id": 66,
"author_url": "https://mathoverflow.net/users/66/ben-webster",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Victor, read the article \"A proof that Euler missed\" <a href=\"http://www.maths.mq.edu.au/~alf/45.pdf\" rel=\"nofollow noreferrer\">maths.mq.edu.au/~alf/45.pdf</a> This provides some historical context around the announcement of Apery's proof. Skepticism toward proofs the prover has not produced the details of is healthy, not snubbing.</span>",
"created_date": "2010-05-23 14:08:36Z"
},
{
"comment_id": 54278,
"author_name": "Junkie",
"author_id": 5267,
"author_url": "https://mathoverflow.net/users/5267/junkie",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">\"Upon re-reading it and the article that Wadim linked it became clear that the so-called \"community\" acted in a worst possible manner. It was only thanks to the determination of a few outstanding mathematicians that he got the recognition that he deserved.\" I would say this differently. Thru the determination of others Apery's ideas went from a convoluted multi-100 page work to a 3-page note. Has anyone bothered to see if his original manuscript did in fact <i>prove</i> bounded denominators? That's the crux, and vDP's cheeky \"utterly compelling\" numerically (so is 1.2020569..., no?) is a nip off.</span>",
"created_date": "2010-05-24 02:43:43Z"
},
{
"comment_id": 54499,
"author_name": "S. Carnahan",
"author_id": 121,
"author_url": "https://mathoverflow.net/users/121/s-carnahan",
"score": 10,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The following is my (probably flawed) recollection of part of Cohen's lecture at the Lenstra Truerfeest a little over 7 years ago: \"Apéry gave a shameful talk. He explained almost nothing, and many of his formulas didn't make any sense. One of his sums seemed to have zeroes in the denominator of every term. But there was one formula that he wrote that looked interesting and new, and Hendrik was sitting next to me with a calculator. I asked him to check the first few terms on his calculator, and they matched very well. After the talk we were able to use it to reconstruct a proof.\"</span>",
"created_date": "2010-05-25 03:36:24Z"
},
{
"comment_id": 54500,
"author_name": "S. Carnahan",
"author_id": 121,
"author_url": "https://mathoverflow.net/users/121/s-carnahan",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I think between \"Hendrik was sitting next to me with a calculator.\" and \"I asked him to check the first few terms\", Cohen made some remark about calculators being rare and expensive back then.</span>",
"created_date": "2010-05-25 03:49:24Z"
}
]
},
{
"answer_id": 146035,
"question_id": 25630,
"score": 63,
"is_accepted": false,
"author_name": "Ian Morris",
"author_id": null,
"author_url": "https://mathoverflow.net/users/1840",
"author_reputation": 6277,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-02-16 21:25:26Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>An 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.</p>\n</div>",
"comments": [
{
"comment_id": 376442,
"author_name": "Federico Poloni",
"author_id": 1898,
"author_url": "https://mathoverflow.net/users/1898/federico-poloni",
"score": 45,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">For the lazy, 2013-1955=58.</span>",
"created_date": "2013-10-27 12:44:51Z"
},
{
"comment_id": 1325586,
"author_name": "Gerry Myerson",
"author_id": 3684,
"author_url": "https://mathoverflow.net/users/3684/gerry-myerson",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">However, neither $2013$ nor $1955$ is a prime, so this doesn't count as two primes differing by less then $70$ million.</span>",
"created_date": "2026-03-04 06:13:21Z"
}
]
},
{
"answer_id": 25635,
"question_id": 25630,
"score": 62,
"is_accepted": false,
"author_name": "arun s",
"author_id": null,
"author_url": "",
"author_reputation": 525,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-23 07:32:13Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Weierstrass approximation theorem was proved by Karl Weierstrass when he was 70 years old</p>\n</div>",
"comments": []
},
{
"answer_id": 25840,
"question_id": 25630,
"score": 62,
"is_accepted": false,
"author_name": "John Stillwell",
"author_id": null,
"author_url": "",
"author_reputation": 12688,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-25 03:11:23Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Since 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.</p>\n<p>In 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.</p>\n</div>",
"comments": [
{
"comment_id": 54565,
"author_name": "Halfdan Faber",
"author_id": 1320,
"author_url": "https://mathoverflow.net/users/1320/halfdan-faber",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">John, Great answer! I don't think there should be any limit on the number of good answers from any one contributor, for a question like this.</span>",
"created_date": "2010-05-25 07:47:15Z"
}
]
},
{
"answer_id": 25672,
"question_id": 25630,
"score": 57,
"is_accepted": false,
"author_name": "Victor Miller",
"author_id": null,
"author_url": "https://mathoverflow.net/users/2784",
"author_reputation": 4818,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-02-15 23:53:28Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><a href=\"https://en.wikipedia.org/wiki/Kurt_Heegner\" rel=\"nofollow noreferrer\">Kurt Heegner</a> 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.</p>\n</div>",
"comments": [
{
"comment_id": 73131,
"author_name": "The Mathemagician",
"author_id": 3546,
"author_url": "https://mathoverflow.net/users/3546/the-mathemagician",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">GREAT example,Victor.A sad story,too. </span>",
"created_date": "2010-07-19 06:07:35Z"
}
]
},
{
"answer_id": 25632,
"question_id": 25630,
"score": 54,
"is_accepted": false,
"author_name": "Tony Huynh",
"author_id": null,
"author_url": "https://mathoverflow.net/users/2233",
"author_reputation": 32855,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-01-06 16:19:28Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><a href=\"https://en.wikipedia.org/wiki/Leonhard_Euler\" rel=\"noreferrer\">Leonhard Euler</a>. According to the Wikipedia page, he still managed to produce one paper per <em>week</em> in the year 1775 (at age 68), despite deteriorating eyesight. As a concrete example, at age 65 he proved that <span class=\"math-container\">$2^{31} − 1$</span> is a Mersenne prime, which may have remained the largest known prime for the next 95 years.</p>\n</div>",
"comments": [
{
"comment_id": 54028,
"author_name": "Franz Lemmermeyer",
"author_id": 3503,
"author_url": "https://mathoverflow.net/users/3503/franz-lemmermeyer",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">A more precise version of the history on M_31 can be found here: <a href=\"http://primes.utm.edu/notes/by_year.html\" rel=\"nofollow noreferrer\">primes.utm.edu/notes/by_year.html</a> He certainly did not do the necessary calculations himself at a time when he was completely blind.</span>",
"created_date": "2010-05-23 10:58:05Z"
}
]
},
{
"answer_id": 25679,
"question_id": 25630,
"score": 49,
"is_accepted": false,
"author_name": "jeebus",
"author_id": null,
"author_url": "",
"author_reputation": 1,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-23 16:04:52Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><strong>Marina Ratner</strong> (b. 1938) proved Ratner's Theorems around 1990. They are some of the biggest advances in ergodic theory for quite a long time.</p>\n</div>",
"comments": []
},
{
"answer_id": 25649,
"question_id": 25630,
"score": 45,
"is_accepted": false,
"author_name": "John Stillwell",
"author_id": null,
"author_url": "",
"author_reputation": 12688,
"license": "CC BY-SA 2.5",
"is_edited": true,
"created_date": "2010-05-24 22:36:42Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>P. 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.</p>\n</div>",
"comments": [
{
"comment_id": 54054,
"author_name": "Wadim Zudilin",
"author_id": 4953,
"author_url": "https://mathoverflow.net/users/4953/wadim-zudilin",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">But Sergey Ivanovich (Adian) was much younger at the time of solving the Burnside problem! +1 for recalling a remarkable family of results.</span>",
"created_date": "2010-05-23 12:46:31Z"
},
{
"comment_id": 54213,
"author_name": "John Stillwell",
"author_id": 1587,
"author_url": "https://mathoverflow.net/users/1587/john-stillwell",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Another in the same family of results is the theorem of A. A. Markov that the homeomorphism problem is unsolvable for manifolds of dimension $\\ge 4$, proved in 1958 when Markov was 55.</span>",
"created_date": "2010-05-23 23:17:59Z"
}
]
},
{
"answer_id": 25638,
"question_id": 25630,
"score": 42,
"is_accepted": false,
"author_name": "Victor Protsak",
"author_id": null,
"author_url": "https://mathoverflow.net/users/5740",
"author_reputation": 14697,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-01-06 16:18:56Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>There are many examples of people doing significant work into their 60s and 70s, but fewer great discoveries. Here are a couple of my favorites:</p>\n<ol>\n<li><p>August Ferdinand Möbius discovered the <b>Möbius band</b> in 1858 at age 68 (the date referenced in <a href=\"https://en.wikipedia.org/wiki/Moebus_strip\" rel=\"noreferrer\">Wikipedia</a>). 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.</p>\n</li>\n<li><p>Julius 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 <b>\"line geometry\"</b> (it is possible that the roots of this discovery go back to his 1846 monograph). The first volume of his book <em>Neue Geometrie des Raumes</em> 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.</p>\n</li>\n</ol>\n<hr/>\nThe question has been closed, but perhaps the following recent example deserves mention: \n<ol start=\"3\">\n<li>As described in a Quanta magazine <a href=\"https://www.quantamagazine.org/20170328-statistician-proves-gaussian-correlation-inequality/\" rel=\"noreferrer\">article</a>, a retired German statistician Thomas Royen proved the <a href=\"https://en.wikipedia.org/wiki/Gaussian_correlation_inequality\" rel=\"noreferrer\"><b>Gaussian correlation inequality</b></a> (GCI) in 2014 at the age of 67. GCI was a major conjecture at the interface of probability and convex geometry that remained open for more than 40 years. An additional twist to the story is that the proof went virtually unnoticed for almost 2 years.</li>\n</ol>\n</div>",
"comments": [
{
"comment_id": 54009,
"author_name": "John Stillwell",
"author_id": 1587,
"author_url": "https://mathoverflow.net/users/1587/john-stillwell",
"score": 11,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">At an even later age, Möbius in 1863 discovered the classification of closed orientable surfaces by genus.</span>",
"created_date": "2010-05-23 08:51:09Z"
},
{
"comment_id": 54016,
"author_name": "Victor Protsak",
"author_id": 5740,
"author_url": "https://mathoverflow.net/users/5740/victor-protsak",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thank you, I didn't remember that! On the other hand, Kolmogorov and Yushkevich (\"Mathematics of the 19th century\", vol 2) indicate that Plücker's idea of line geometry originated already in his 1846 \"System des Geometrie des Raumes\". I don't have Klein's \"Lectures on the development of mathematics in the 19th century\" close at hand to clarify this point, but apparently Plücker switched to doing physics around that time (1846) due to strained relations with German mathematicians and unfavorable reception of his analytic methods.</span>",
"created_date": "2010-05-23 09:13:00Z"
},
{
"comment_id": 270617,
"author_name": "David Corwin",
"author_id": 1355,
"author_url": "https://mathoverflow.net/users/1355/david-corwin",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">\"The idea of using higher-dimensional objects as points\" - What does this mean? Could it mean that Plucker's idea led to generic points in algebraic geometry (and more generally, non-closed points of schemes)?</span>",
"created_date": "2012-08-21 01:47:40Z"
},
{
"comment_id": 271048,
"author_name": "Victor Protsak",
"author_id": 5740,
"author_url": "https://mathoverflow.net/users/5740/victor-protsak",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">No, just to the idea of more abstract spaces, such as homogeneous spaces $G/H$ and the simplest examples of moduli spaces. </span>",
"created_date": "2012-08-23 01:34:23Z"
}
]
},
{
"answer_id": 27378,
"question_id": 25630,
"score": 33,
"is_accepted": false,
"author_name": "brondell",
"author_id": null,
"author_url": "https://mathoverflow.net/users/6639",
"author_reputation": 1,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-01-06 16:10:19Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>According to wiki, Mihailescu got his PhD at the age of 42; and then proved <a href=\"https://en.wikipedia.org/wiki/Catalan%27s_conjecture\" rel=\"noreferrer\">Catalan's conjecture</a> in 2002, age 47, so almost 50.</p>\n</div>",
"comments": []
},
{
"answer_id": 25634,
"question_id": 25630,
"score": 26,
"is_accepted": false,
"author_name": "Angelo",
"author_id": null,
"author_url": "",
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"created_date": "2010-05-23 07:30:04Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Zariski 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.</p>\n</div>",
"comments": []
},
{
"answer_id": 25671,
"question_id": 25630,
"score": 26,
"is_accepted": false,
"author_name": "Victor Miller",
"author_id": null,
"author_url": "",
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"created_date": "2010-05-23 14:59:15Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Louis de Branges solved the Bieberbach conjecture in 1985 when he was 53.</p>\n</div>",
"comments": [
{
"comment_id": 54206,
"author_name": "Victor Protsak",
"author_id": 5740,
"author_url": "https://mathoverflow.net/users/5740/victor-protsak",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Was that his first or his last attempt at the solution? I understand that he worked on it for quite some time.</span>",
"created_date": "2010-05-23 22:41:18Z"
},
{
"comment_id": 54293,
"author_name": "Victor Miller",
"author_id": 2784,
"author_url": "https://mathoverflow.net/users/2784/victor-miller",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Depending on how one counts it might have been his second or third. de Branges is no shrinking violet. He's also announced solutions to the Poincare conjecture (before Perlman) and the Riemann Hypothesis.</span>",
"created_date": "2010-05-24 03:36:31Z"
},
{
"comment_id": 54385,
"author_name": "Deane Yang",
"author_id": 613,
"author_url": "https://mathoverflow.net/users/613/deane-yang",
"score": 5,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I'm not aware that de Branges ever claimed a proof of the Poincare conjecture, which is a bit far from de Branges' area of research. He has been working relentlessly on the Riemann hypothesis for quite a while now.</span>",
"created_date": "2010-05-24 15:25:45Z"
},
{
"comment_id": 73132,
"author_name": "The Mathemagician",
"author_id": 3546,
"author_url": "https://mathoverflow.net/users/3546/the-mathemagician",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Deane A few months ago,he claimed at the archive to actually have the proof-but so far,no one's been able to verify it. I hope someone really takes a good hard look at it-if only to stop him from saying it.</span>",
"created_date": "2010-07-19 06:10:00Z"
},
{
"comment_id": 1047197,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@VictorMiller I suspect you're confusing the Poincare conjecture with the invariant subspace problem. <a href=\"https://www.lrb.co.uk/the-paper/v26/n14/karl-sabbagh/the-strange-case-of-louis-de-branges\" rel=\"nofollow noreferrer\">Karl Sabbagh</a> quotes de Branges as saying, \"The first case in which I made an error was in proving the existence of invariant subspaces for continuous transformations in Hilbert spaces. This was something that happened in 1964, and I declared something to be true which I was not able to substantiate. And the fact that I did that destroyed my career. My colleagues have never forgiven it.\"</span>",
"created_date": "2021-11-11 17:39:52Z"
}
]
},
{
"answer_id": 25689,
"question_id": 25630,
"score": 26,
"is_accepted": false,
"author_name": "Deane Yang",
"author_id": null,
"author_url": "",
"author_reputation": 28230,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-23 16:46:23Z",
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"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Although 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.</p>\n<p>I 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.</p>\n<p>And is any of this that different from other fields?</p>\n</div>",
"comments": [
{
"comment_id": 54250,
"author_name": "Victor Protsak",
"author_id": 5740,
"author_url": "https://mathoverflow.net/users/5740/victor-protsak",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">He told me several anecdotes about Hardy, but he presented each story in a sarcastic tone. “Hardy’s opinion that mathematics is a young man’s game is nonsense,” he said. (Goro Shimura, André Weil As I Knew Him, <a href=\"http://www.ams.org/notices/199904/shimura.pdf\" rel=\"nofollow noreferrer\">ams.org/notices/199904/shimura.pdf</a>)</span>",
"created_date": "2010-05-24 01:37:40Z"
},
{
"comment_id": 139021,
"author_name": "Thierry Zell",
"author_id": 8212,
"author_url": "https://mathoverflow.net/users/8212/thierry-zell",
"score": 6,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">How are other fields different? Well, I always picture experimental sciences as being much more hierarchical, with the (older) team leader being credited for the work of the whole team. Thus, one's impressive achievements would come later in life. <b>Disclaimer:</b> I don't know what I'm talking about, as I wrote it's the image I got, but not necessarily from very reliable sources.</span>",
"created_date": "2011-02-15 16:56:14Z"
}
]
},
{
"answer_id": 25644,
"question_id": 25630,
"score": 26,
"is_accepted": false,
"author_name": "Anweshi",
"author_id": null,
"author_url": "https://mathoverflow.net/users/2938",
"author_reputation": 7582,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-01-06 16:10:33Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><a href=\"https://en.wikipedia.org/wiki/Theorema_Egregium\" rel=\"nofollow noreferrer\">Theorema Egregium</a> was published by Gauss in 1828. Since <a href=\"https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss\" rel=\"nofollow noreferrer\">Gauss was born in 1777</a>, he ought to have been a little over 50 then.</p>\n<p>Ref: Disquisitiones generales circa superficies curva (1828)</p>\n</div>",
"comments": [
{
"comment_id": 270621,
"author_name": "Turbo",
"author_id": 10035,
"author_url": "https://mathoverflow.net/users/10035/turbo",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Gauss usually keeps his stuff in his sleeve for a while. So you never knew when he had the insight.</span>",
"created_date": "2012-08-21 04:20:08Z"
}
]
},
{
"answer_id": 25650,
"question_id": 25630,
"score": 23,
"is_accepted": false,
"author_name": "Franz Lemmermeyer",
"author_id": null,
"author_url": "",
"author_reputation": 33319,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-23 10:54:30Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Furtwängler proved the principal ideal theorem when he was almost 60. No small feat given that Artin and Schreier simultaneously were working on it. </p>\n</div>",
"comments": []
},
{
"answer_id": 25700,
"question_id": 25630,
"score": 22,
"is_accepted": false,
"author_name": "Kristal Cantwell",
"author_id": null,
"author_url": "https://mathoverflow.net/users/1098",
"author_reputation": 6627,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-01-06 16:17:47Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Paul Erdős continued to do work in many fields including combinatorics after his 50th birthday. Some of his papers are <a href=\"https://www.renyi.hu/%7Ep_erdos/Erdos.html\" rel=\"noreferrer\">here</a></p>\n</div>",
"comments": []
},
{
"answer_id": 25824,
"question_id": 25630,
"score": 20,
"is_accepted": false,
"author_name": "Michael Greenblatt",
"author_id": null,
"author_url": "",
"author_reputation": 551,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-24 23:52:34Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Something 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.</p>\n</div>",
"comments": [
{
"comment_id": 1179468,
"author_name": "Daniel Asimov",
"author_id": 5484,
"author_url": "https://mathoverflow.net/users/5484/daniel-asimov",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">But Levinson was an expert in complex analysis, so analytic number theory may have been right up his alley.</span>",
"created_date": "2023-09-30 05:15:19Z"
}
]
},
{
"answer_id": 25636,
"question_id": 25630,
"score": 18,
"is_accepted": false,
"author_name": "DoubleJay",
"author_id": null,
"author_url": "",
"author_reputation": 2473,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-23 07:51:28Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Christos 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.</p>\n</div>",
"comments": []
},
{
"answer_id": 25772,
"question_id": 25630,
"score": 16,
"is_accepted": false,
"author_name": "Zoran Skoda",
"author_id": null,
"author_url": "",
"author_reputation": 5322,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-24 14:34:02Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>This 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). </p>\n<p>If 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. </p>\n</div>",
"comments": [
{
"comment_id": 54377,
"author_name": "Deane Yang",
"author_id": 613,
"author_url": "https://mathoverflow.net/users/613/deane-yang",
"score": 6,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">well, even your formulation is a bit extreme. I don't see the need to require \"no major research before 50\". How about just a well-known mathematician whose best work was after the age of 50? Although this is less common, I don't think it is any more rare than it is in any other field.</span>",
"created_date": "2010-05-24 14:38:36Z"
},
{
"comment_id": 54607,
"author_name": "gowers",
"author_id": 1459,
"author_url": "https://mathoverflow.net/users/1459/gowers",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">As I understand it, Apery, who has already been mentioned, remains a good example even if you restrict the question in this way.</span>",
"created_date": "2010-05-25 12:54:19Z"
},
{
"comment_id": 73129,
"author_name": "The Mathemagician",
"author_id": 3546,
"author_url": "https://mathoverflow.net/users/3546/the-mathemagician",
"score": 5,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Deane,gowers: Thank you,gentleman.This old wives' tale of Hardy's has probably prevented many a late bloomer from pursuing thier dreams. And that's tragic. </span>",
"created_date": "2010-07-19 06:04:07Z"
},
{
"comment_id": 173823,
"author_name": "Zoran Skoda",
"author_id": 35833,
"author_url": "https://mathoverflow.net/users/35833/zoran-skoda",
"score": 11,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Andrew, a person who does mathematics in order to be famous or \"major mathematician\" and not to enjoy path of curiosity is tragic in the first place, whatever be his/her achievements. The present day celebrity counterculture, statistics based on \"official results\", meaningless linear lists of comparison and so on aggravate the situation. Among the worst is the funding agency habit that they award more future support to those who got in past more support, else param. equal. That is, if you achieved the same with more spending in past they consider you more efficient, what is outrageously wrong.</span>",
"created_date": "2011-06-26 08:59:40Z"
},
{
"comment_id": 1059421,
"author_name": "H A Helfgott",
"author_id": 398,
"author_url": "https://mathoverflow.net/users/398/h-a-helfgott",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">People who do what is clearly their best work (in the strict sense, not just \"work that is as good as what they did before\") around 50 or later do exist - see, e.g., Marina Ratner (what would be some other indisputable examples?) - but they are rare - rare enough that people remark repeatedly on it.</span>",
"created_date": "2022-01-06 16:32:49Z"
}
]
},
{
"answer_id": 25813,
"question_id": 25630,
"score": 15,
"is_accepted": false,
"author_name": "Steve D",
"author_id": null,
"author_url": "",
"author_reputation": 4494,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-24 22:31:50Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Philip 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).</p>\n<p>Steve</p>\n</div>",
"comments": []
},
{
"answer_id": 25658,
"question_id": 25630,
"score": 14,
"is_accepted": false,
"author_name": "Leonid Petrov",
"author_id": null,
"author_url": "",
"author_reputation": 1785,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-23 11:55:32Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Karl 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).</p>\n</div>",
"comments": []
},
{
"answer_id": 32441,
"question_id": 25630,
"score": 13,
"is_accepted": false,
"author_name": "Péter Komjáth",
"author_id": null,
"author_url": "",
"author_reputation": 7175,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-07-19 04:51:13Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Tibor Rado introduced the busy beaver function and proved its noncomputablity at the age of 67. </p>\n</div>",
"comments": []
},
{
"answer_id": 25683,
"question_id": 25630,
"score": 13,
"is_accepted": false,
"author_name": "The Mathemagician",
"author_id": null,
"author_url": "https://mathoverflow.net/users/3546",
"author_reputation": 487,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-01-06 16:16:48Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>And 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.</p>\n</div>",
"comments": [
{
"comment_id": 73130,
"author_name": "The Mathemagician",
"author_id": 3546,
"author_url": "https://mathoverflow.net/users/3546/the-mathemagician",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">And we can add John Milnor to that list as well with his late contributions to dynamics.</span>",
"created_date": "2010-07-19 06:05:57Z"
}
]
},
{
"answer_id": 27320,
"question_id": 25630,
"score": 12,
"is_accepted": false,
"author_name": "Jérôme Poineau",
"author_id": null,
"author_url": "",
"author_reputation": 4317,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-06-07 07:05:00Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>When Khare and Wintenberger proved Serre's conjecture, Wintenberger was older than fifty.</p>\n</div>",
"comments": []
},
{
"answer_id": 25675,
"question_id": 25630,
"score": 10,
"is_accepted": false,
"author_name": "Abtan Massini",
"author_id": null,
"author_url": "",
"author_reputation": 1826,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2010-05-23 15:22:07Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Connes has initiated whole new areas of mathematics since turning 50: spectral triples, and his novel approach to the Riemann hypothesis, for example.</p>\n</div>",
"comments": [
{
"comment_id": 54099,
"author_name": "BCnrd",
"author_id": 3927,
"author_url": "https://mathoverflow.net/users/3927/bcnrd",
"score": 23,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">But has his approach to RH really yielded progress? My impression from one top expert in analytic number theory is that it ultimately isn't novel (after stripping away the fancy-looking language) and hasn't shed any light on any key issues (after quite some years now). </span>",
"created_date": "2010-05-23 15:36:09Z"
},
{
"comment_id": 54277,
"author_name": "Junkie",
"author_id": 5267,
"author_url": "https://mathoverflow.net/users/5267/junkie",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">One \"idea\" of the Connes reformulation is that one can \"see\" how a dynamical system of primes could prove RH, if one ignores issues like renormalisation and dealing with infinitely many primes rather than finitely many (he proves the S-local analogue of the trace formula). His later programme with Marcolli/Consani has used evermore iffy language and analogues, IMO. On the mathematical end, Meyer had a nice paper on some of the function space constructs, though he doesn't actually try to get RH to appear via his re-working. <a href=\"http://projecteuclid.org/euclid.dmj/1113847338\" rel=\"nofollow noreferrer\">projecteuclid.org/euclid.dmj/1113847338</a> </span>",
"created_date": "2010-05-24 02:35:47Z"
},
{
"comment_id": 54378,
"author_name": "Zoran Skoda",
"author_id": 35833,
"author_url": "https://mathoverflow.net/users/35833/zoran-skoda",
"score": 6,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">With my admiration for our guru in noncommutative geometry, we must be realistic: his Fields medal-winning discovery of classification of factors of type III and his a bit later single-handed introduction of major characters in noncommutative geometry like the introduction of cyclic homology 30 years ago, while not as perfect as some modern ramifications seem historically far deeper and more striking discoveries than the more synthetic mature (and more collaborative) works at present. </span>",
"created_date": "2010-05-24 14:46:40Z"
}
]
},
{
"answer_id": 105104,
"question_id": 25630,
"score": 10,
"is_accepted": false,
"author_name": "Harun Šiljak",
"author_id": null,
"author_url": "",
"author_reputation": 1544,
"license": "CC BY-SA 3.0",
"is_edited": false,
"created_date": "2012-08-20 18:02:24Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>A 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).</p>\n</div>",
"comments": [
{
"comment_id": 270555,
"author_name": "Joseph O'Rourke",
"author_id": 6094,
"author_url": "https://mathoverflow.net/users/6094/joseph-orourke",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Here is a link to the <i>PNAS</i> paper: <a href=\"http://www.pnas.org/content/early/2012/05/16/1206569109.abstract\" rel=\"nofollow noreferrer\">pnas.org/content/early/2012/05/16/1206569109.abstract</a></span>",
"created_date": "2012-08-20 20:10:09Z"
}
]
},
{
"answer_id": 25640,
"question_id": 25630,
"score": 10,
"is_accepted": false,
"author_name": "Anweshi",
"author_id": null,
"author_url": "https://mathoverflow.net/users/2938",
"author_reputation": 7582,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-01-06 16:10:55Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The <a href=\"https://en.wikipedia.org/wiki/Fermat_number\" rel=\"nofollow noreferrer\">Fermat number</a> <span class=\"math-container\">$F_6$</span> was shown to have nontrivial factorization, by Landry at the age of 82. And apparently it was Landry's only mathematical publication.</p>\n<p>(Source: Ribenboim, Prime number records(the smaller book).)</p>\n<p>This 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.</p>\n</div>",
"comments": []
},
{
"answer_id": 55520,
"question_id": 25630,
"score": 8,
"is_accepted": false,
"author_name": "user4245",
"author_id": null,
"author_url": "",
"author_reputation": 819,
"license": "CC BY-SA 2.5",
"is_edited": false,
"created_date": "2011-02-15 14:40:25Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Andre Weil lay the modern foundation of \"theta series\" in Acta math. (1964/65) when he was almost 60 years old!</p>\n</div>",
"comments": []
},
{
"answer_id": 26748,
"question_id": 25630,
"score": 8,
"is_accepted": false,
"author_name": "Franz Lemmermeyer",
"author_id": null,
"author_url": "https://mathoverflow.net/users/3503",
"author_reputation": 33319,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-01-06 16:15:49Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>This \"almost\" answers Zoran Škoda's question: <a href=\"https://www.mathi.uni-heidelberg.de/~roquette/manu.html#gruen\" rel=\"nofollow noreferrer\">Otto Grün</a> (his theorems in group theory are still well known) published his first paper at the age of 46.</p>\n</div>",
"comments": []
},
{
"answer_id": 105122,
"question_id": 25630,
"score": 5,
"is_accepted": false,
"author_name": "Alain Valette",
"author_id": null,
"author_url": "",
"author_reputation": 11431,
"license": "CC BY-SA 3.0",
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"created_date": "2012-08-20 21:57:47Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Burnside proved the $p^aq^b$ theorem at age 53.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 379866,
"title": "Conceptual proof of braid group actions on quantum groups",
"link": "https://mathoverflow.net/questions/379866/conceptual-proof-of-braid-group-actions-on-quantum-groups",
"score": 7,
"view_count": 637,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "rt.representation-theory;qa.quantum-algebra;quantum-groups;geometric-representation-theory",
"creation_date": "2026-03-04 20:25:05Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Roughly 1990, Lusztig wrote a series of papers on quantum groups. Perhaps the result that the braid groups acts on <span class=\"math-container\">$U_q(\\mathfrak{g})$</span> is the proof which is least conceptual.</p>\n<p>The original paper contains a case by case check for <span class=\"math-container\">$m=2,3,4,6$</span> as far as I understand.</p>\n<p>If it is not a coincidence, then it needs an explanation. My question is</p>\n<blockquote>\n<p>Any conceptual proof up to now is known?</p>\n</blockquote>\n<p>At least not one by one.</p>\n<ul>\n<li><p>For example, can we define quantum groups for the type <span class=\"math-container\">$I_n$</span>? (The naive version is to work in the field of Puiseux series.)</p>\n</li>\n<li><p>Perhaps we have very limited 'conceptual' understanding on quantum groups up to now. There is a lot on the negative part <span class=\"math-container\">$\\mathfrak{f}$</span> (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?</p>\n</li>\n<li><p>The 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?)</p>\n</li>\n<li><p>For example, the simplest case, in the springer realization of <span class=\"math-container\">$U_q(\\mathfrak{sl}_3)$</span>, does the braid group action have any geometric meaning?</p>\n</li>\n</ul>\n</div>",
"comments": [
{
"comment_id": 964605,
"author_name": "Paul Gustafson",
"author_id": 91709,
"author_url": "https://mathoverflow.net/users/91709/paul-gustafson",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Have you seen <a href=\"https://mathoverflow.net/a/115252\">mathoverflow.net/a/115252</a> ? In particular, Drinfeld's ICM talk has an explicit formula for the universal 𝑅-matrix.</span>",
"created_date": "2020-12-28 01:40:43Z"
},
{
"comment_id": 964623,
"author_name": "Noah Snyder",
"author_id": 22,
"author_url": "https://mathoverflow.net/users/22/noah-snyder",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Do you need this for the Lusztig integral form? Or would the h-adic version be enough? In the latter case I thought you could get it pretty directly from the classical r-matrix following Drinfeld, as Paul has suggested. If you really want the integral form done carefully then you probably just need to do what Lusztig does, but the h-adic version tells you what formulas you should use.</span>",
"created_date": "2020-12-28 04:19:04Z"
},
{
"comment_id": 964647,
"author_name": "Pavel Safronov",
"author_id": 18512,
"author_url": "https://mathoverflow.net/users/18512/pavel-safronov",
"score": 5,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I think the question is about the quantum Weyl group (denoted by $T_{i, e}', T_{i, e}''$ in Lusztig's book) rather than about the quantum $R$-matrix.</span>",
"created_date": "2020-12-28 11:08:47Z"
}
],
"answers": [
{
"answer_id": 508754,
"question_id": 379866,
"score": 0,
"is_accepted": false,
"author_name": "Dat Minh Ha",
"author_id": 143390,
"author_url": "https://mathoverflow.net/users/143390/dat-minh-ha",
"author_reputation": 1728,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-04 20:25:05Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>For 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 <a href=\"https://arxiv.org/abs/2401.06402\" rel=\"nofollow noreferrer\">https://arxiv.org/abs/2401.06402</a>.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508755,
"title": "Hanner's unpublished example of a non-fiber bundle",
"link": "https://mathoverflow.net/questions/508755/hanners-unpublished-example-of-a-non-fiber-bundle",
"score": 10,
"view_count": 120,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "gn.general-topology;fibre-bundles;principal-bundles",
"creation_date": "2026-03-04 20:26:04Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Given a topological group <span class=\"math-container\">$G$</span> and a closed subgroup <span class=\"math-container\">$H \\le G$</span>, one can ask when the projection map <span class=\"math-container\">$G \\to G/H$</span> 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 <em>local cross-section</em>.</p>\n<p>He refers to an unpublished example of Hanner of a compact abelian group of infinite dimension <span class=\"math-container\">$G$</span> and a zero-dimensional subgroup <span class=\"math-container\">$H$</span> which fails this condition. Mostert similarly hints at this result in 1952, and Karube gives a similar but supposedly different example in 1958.</p>\n<p>Does anyone know the precise folklore construction of Hanner?</p>\n</div>",
"comments": [],
"answers": []
},
{
"question_id": 487398,
"title": "Niveau of the Hodge structure of an hypersurface in $\\mathbb{P}^n$",
"link": "https://mathoverflow.net/questions/487398/niveau-of-the-hodge-structure-of-an-hypersurface-in-mathbbpn",
"score": 1,
"view_count": 351,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "ag.algebraic-geometry;hodge-theory;intersection-theory;hypersurfaces",
"creation_date": "2026-03-04 21:07:35Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Assume <span class=\"math-container\">$X:=X_d$</span> is an hypersurface of degree <span class=\"math-container\">$d$</span> in <span class=\"math-container\">$\\mathbb{P}^n$</span>. Assume in addition that <span class=\"math-container\">$d<n+1$</span>, hence <span class=\"math-container\">$X$</span> is a Fano variety.</p>\n<p>The hyperplane section theorem of Lefschetz ensures that the only interesting part of the cohomology is <span class=\"math-container\">$H^{n-1}(X,\\mathbb{Q})$</span>, and as <span class=\"math-container\">$X$</span> is Fano, it is well known that <span class=\"math-container\">$H^{n-1,0}(X)=0$</span> : the Hodge structure is of coniveau <span class=\"math-container\">$1$</span>.</p>\n<p>Recall that an hodge structure <span class=\"math-container\">$H$</span> is of niveau <span class=\"math-container\">$\\alpha$</span> if <span class=\"math-container\">$H^{p,q}=0$</span> for <span class=\"math-container\">$\\mid p-q \\mid > \\alpha$</span>.</p>\n<p>Now for very low degree hypersurfaces, the niveau is bigger. I observed the following facts with direct computations :</p>\n<ul>\n<li>If <span class=\"math-container\">$d\\leq 2$</span> then <span class=\"math-container\">$X_d$</span> is cellular, and hence its cohomology is of niveau 0.</li>\n<li>For <span class=\"math-container\">$d=3$</span> and <span class=\"math-container\">$n\\leq 6$</span>, the cohomology is of niveau <span class=\"math-container\">$2$</span>.</li>\n</ul>\n<p>Furthermore 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 <span class=\"math-container\">$n$</span>.</p>\n<p>So 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 :</p>\n<p><strong>With intersection theory</strong></p>\n<p>We can easily compute the Chern classes of <span class=\"math-container\">$X$</span> as it is an hypersurface, and so Riemann-Roch formula gives an explicit element <span class=\"math-container\">$P\\in \\mathbb{Q}(x,y)$</span> such that in a formal power extension, the coefficient of <span class=\"math-container\">$x^py^q$</span> is exactly the vanishing part of <span class=\"math-container\">$h^{p,q}$</span>.</p>\n<p>However it seems really annoying to expend this fraction for a general couple <span class=\"math-container\">$(d,n)$</span>.</p>\n<p><strong>With vanishing of the pole</strong></p>\n<p>Let <span class=\"math-container\">$S=\\mathbb{C}[x_0,\\cdots,x_n]$</span>, <span class=\"math-container\">$f$</span> the homogenous polynomial defining <span class=\"math-container\">$X$</span> and <span class=\"math-container\">$J_f$</span> the homogenous ideal generated by the jacobian of <span class=\"math-container\">$f$</span>. Finally, let <span class=\"math-container\">$R=S/J_f$</span>. Then a vanishing of the pole argument gives an isomorphism</p>\n<p><span class=\"math-container\">$$R^{kd-n-1}\\simeq H^{n-k,k-1}_{van}(X)$$</span></p>\n<p>where <span class=\"math-container\">$R^m$</span> is the part of degree <span class=\"math-container\">$m$</span>, and <span class=\"math-container\">$van$</span> denotes the vanishing part of the cohomology.</p>\n<p>Now if I choose <span class=\"math-container\">$f= \\sum_{i=0}^n x_i^d$</span>, then the jacobian is really easy to compute, so we can find the degree where <span class=\"math-container\">$R^m$</span> vanishes. More precisely, we have <span class=\"math-container\">$R^m=0$</span> exactly when</p>\n<p><span class=\"math-container\">$$m>(n+1)(d-2).$$</span></p>\n<p>So <span class=\"math-container\">$H^{n-k,k-1}_{van}(X)=0$</span> for <span class=\"math-container\">$k>(n+1)(d-1)/d$</span>. It is exactly the niveau bound I was looking for !</p>\n<p><strong>Conclusion of the argument and question</strong></p>\n<p>I 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 <span class=\"math-container\">$(d,n)$</span>, so my bound is true for any hypersurface of degree <span class=\"math-container\">$d$</span>.</p>\n<p>So 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.</p>\n<p>Also it is rather disappointing to prove the result only for a precise choice of <span class=\"math-container\">$f$</span>, and then use the independance of the choice. I would rather prove the result for any choice of <span class=\"math-container\">$f$</span>, or directly with the computation of Chern numbers.</p>\n<p>Do you have any idea to find a better argument ?</p>\n</div>",
"comments": [
{
"comment_id": 1270805,
"author_name": "Jason Starr",
"author_id": 13265,
"author_url": "https://mathoverflow.net/users/13265/jason-starr",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Are you asking how to extend the Griffiths residue calculus to complete intersections in generalized flag varieties (not necessarily of type $A_n$)? There is an article of Carlsson and Toledo with a quite general form of the Griffiths residue calculus.</span>",
"created_date": "2025-02-07 21:10:27Z"
},
{
"comment_id": 1270809,
"author_name": "Littlebird",
"author_id": 484878,
"author_url": "https://mathoverflow.net/users/484878/littlebird",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Yes it could be a possibility. Which one are you talking about ? And can we hope to find a generalization without the flag assumption ?</span>",
"created_date": "2025-02-07 21:27:10Z"
},
{
"comment_id": 1270852,
"author_name": "abx",
"author_id": 40297,
"author_url": "https://mathoverflow.net/users/40297/abx",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">You'll find a precise formula (with proof, and more generally for complete intersections) in SGA 7, Exposé XI, Corollaire 2.8.</span>",
"created_date": "2025-02-08 05:27:41Z"
},
{
"comment_id": 1325662,
"author_name": "Enrico",
"author_id": 52811,
"author_url": "https://mathoverflow.net/users/52811/enrico",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Since the partial derivatives (for a smooth projective hypersurface of degree $d$ in $\\mathbf{P}^n$) form a regular sequence, you might compute the Hilbert-Poincaré series of the Jacobian algebra (from $\\frac{(1-t^{d-1})^{n+1}}{(1-t)^{n+1}}$). This would give you an immediate generalization to the (quasi)smooth hyper surfaces in weighted projective spaces. You may also take a look at <a href=\"https://arxiv.org/abs/1801.09586\" rel=\"nofollow noreferrer\">arxiv.org/abs/1801.09586</a> for a practical way of computing Hodge numbers and extracting useful information.</span>",
"created_date": "2026-03-04 22:48:43Z"
}
],
"answers": [
{
"answer_id": 487402,
"question_id": 487398,
"score": 0,
"is_accepted": false,
"author_name": "Sasha",
"author_id": 4428,
"author_url": "https://mathoverflow.net/users/4428/sasha",
"author_reputation": 42266,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-02-07 19:30:16Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Since <span class=\"math-container\">$H^{p,q}(X) = H^q(X,\\Omega^p_X)$</span> and since for <span class=\"math-container\">$p + q \\ne 1$</span> the only nontrivial cohomology of <span class=\"math-container\">$\\Omega^p(X)$</span> is <span class=\"math-container\">$H^p(X, \\Omega^p_X) = H^{p,p}(X)$</span>, and it is 1-dimensional, one has\n<span class=\"math-container\">$$\nH^{p,q}(X) = 0 \n\\iff\n\\chi(\\Omega^p_X) = (-1)^p.\n$$</span>\nOn the other hand, <span class=\"math-container\">$\\chi(\\Omega^p_X)$</span> can be computed by Riemann--Roch.</p>\n</div>",
"comments": [
{
"comment_id": 1270792,
"author_name": "Will Sawin",
"author_id": 18060,
"author_url": "https://mathoverflow.net/users/18060/will-sawin",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">This idea was already mentioned in the question, in the first bolded section, the difficulty being how to do this computation in a nice way and find a simple expression for when the cohomology vanishes.</span>",
"created_date": "2025-02-07 19:33:20Z"
}
]
}
]
},
{
"question_id": 371129,
"title": "Where can I read about Veblen functions / klammersymbols beyond the large Veblen ordinal?",
"link": "https://mathoverflow.net/questions/371129/where-can-i-read-about-veblen-functions-klammersymbols-beyond-the-large-veblen",
"score": 3,
"view_count": 398,
"answer_count": 2,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "set-theory;lo.logic;ordinal-numbers",
"creation_date": "2026-03-05 06:41:04Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>So, I'm not sure to what extent this is a thing. John Baez mentions in <a href=\"https://johncarlosbaez.wordpress.com/2016/07/07/large-countable-ordinals-part-3/\" rel=\"nofollow noreferrer\">this blog post</a> 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.</p>\n<p>But, 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 <span class=\"math-container\">$\\omega_1^{\\mathrm{CK}}$</span>.)</p>\n<p>Now 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?</p>\n<p>Thank you all!</p>\n</div>",
"comments": [
{
"comment_id": 938434,
"author_name": "Fedor Pakhomov",
"author_id": 36385,
"author_url": "https://mathoverflow.net/users/36385/fedor-pakhomov",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">If one takes Veblen-like construction as an informal concept, then it would be hard to justify some $\\alpha<\\omega_1^{\\mathsf{CK}}$ to be the ultimate Veblen ordinal (the suprema of all ordinals obtained by Veblen-style constructions). The reason for this is that it wouldn't be clear why one shouldn't be allowed to make the next step of diagonalization and overcome this particular ordinal.</span>",
"created_date": "2020-09-08 09:51:53Z"
},
{
"comment_id": 938435,
"author_name": "Fedor Pakhomov",
"author_id": 36385,
"author_url": "https://mathoverflow.net/users/36385/fedor-pakhomov",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">However, I think that it is possible to give Bachmann-Howard ordinal as a (non-least) upper bound for reasonable constructions of this kind. The motivation here is that it is reasonable to expect that this constructions should be formalizable in Kripke-Platek set theory with infinity $\\mathsf{KP}\\omega$ and the proof-theoretic ordinal of $\\mathsf{KP}\\omega$ is Bachmann-Howard ordinal.</span>",
"created_date": "2020-09-08 09:52:06Z"
},
{
"comment_id": 938503,
"author_name": "SSequence",
"author_id": 112385,
"author_url": "https://mathoverflow.net/users/112385/ssequence",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I think a link to original paper (of howard I think?) was posted on fom (if I am re-calling correctly). I don't think it was that long ago (maybe around an year or bit more). I will try to find and post the link (in few days perhaps). Other than that, one could probably find a number of relevant topics by using a search or following the links in some of the topics under \"Related\".</span>",
"created_date": "2020-09-08 14:35:19Z"
},
{
"comment_id": 938659,
"author_name": "SSequence",
"author_id": 112385,
"author_url": "https://mathoverflow.net/users/112385/ssequence",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Here is the link I was referring to in comment above: <a href=\"https://arxiv.org/abs/1903.04609\" rel=\"nofollow noreferrer\">arxiv.org/abs/1903.04609</a></span>",
"created_date": "2020-09-09 04:54:54Z"
},
{
"comment_id": 938909,
"author_name": "Fedor Pakhomov",
"author_id": 36385,
"author_url": "https://mathoverflow.net/users/36385/fedor-pakhomov",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@HarryAltman I think that it is very unlikely that there is some concrete point of failure. But in my opinion the reasoning that either there is a concrete point of failure or we could reach unlimitely large ordinals isn't valid. This is because we are dealing with an informal notion here. Furthermore, it is rather unlikely that any particular formalization would be completely satisfactory (precisely because then it would lead to a concrete point of failure, which we would be able to diagonalize against).</span>",
"created_date": "2020-09-10 10:23:23Z"
}
],
"answers": [
{
"answer_id": 371199,
"question_id": 371129,
"score": 2,
"is_accepted": false,
"author_name": "Andreas Weiermann",
"author_id": 62695,
"author_url": "https://mathoverflow.net/users/62695/andreas-weiermann",
"author_reputation": 471,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2020-09-08 15:04:44Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Here is some relevant information by a grandmaster on the subject:\n<a href=\"http://www.mathematik.uni-muenchen.de/%7Ebuchholz/articles/jaegerfestschr_buchholz3.pdf\" rel=\"nofollow noreferrer\">http://www.mathematik.uni-muenchen.de/~buchholz/articles/jaegerfestschr_buchholz3.pdf</a></p>\n</div>",
"comments": [
{
"comment_id": 948877,
"author_name": "Harry Altman",
"author_id": 5583,
"author_url": "https://mathoverflow.net/users/5583/harry-altman",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">OK, taking a look at this I'm afraid I don't understand this. He explains how to construct $\\Gamma_\\alpha$ in the preliminaries, but I'm having trouble following it beyond that. His $\\phi$ only has <i>one</i> subscript, but (going by the equation before Lemma 1.6) it acts like multiple (i.e., acts like Klammersymbols) if you go into uncountables? I don't really understand how to get beyond that though, especially if merely getting beyond (the equivalent of) a single subscript requires getting to $\\Omega$! I'd hope for each step to be somehow expressible in terms of things constructed so far...</span>",
"created_date": "2020-10-19 02:12:49Z"
}
]
},
{
"answer_id": 508772,
"question_id": 371129,
"score": 0,
"is_accepted": false,
"author_name": "Harry Altman",
"author_id": 5583,
"author_url": "https://mathoverflow.net/users/5583/harry-altman",
"author_reputation": 2747,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-05 06:24:38Z",
"last_edited_date": "2026-03-05 06:41:04Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>OK, these people define the \"dimensional Veblen\" which they show can approach the Bachmann-Howard ordinal: <a href=\"https://arxiv.org/abs/2310.12832\" rel=\"nofollow noreferrer\">https://arxiv.org/abs/2310.12832</a></p>\n<p>So, 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 <span class=\"math-container\">$\\omega$</span>\"; there will always be a maximum, finite, indexing rank in each relation even if you allow non-uniformity. So this might be the answer!</p>\n<p>However, 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.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508777,
"title": "Please help me with this AWESOME Goldbach Conjecture idea! [closed]",
"link": "https://mathoverflow.net/questions/508777/please-help-me-with-this-awesome-goldbach-conjecture-idea",
"score": 0,
"view_count": 37,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "prime-numbers;open-problems;factorization;goldbach-type-problems",
"creation_date": "2026-03-05 10:02:07Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<div>\n<aside class=\"s-notice s-notice__info post-notice js-post-notice mb16\" role=\"status\">\n<div class=\"d-flex fd-column fw-nowrap\">\n<div class=\"d-flex fw-nowrap\">\n<div class=\"flex--item wmn0 fl1 lh-lg\">\n<div class=\"flex--item fl1 lh-lg\">\n<div>\n Requests to check work for correctness and announcements of results are off-topic on MathOverflow.\n \n </div>\n</div>\n</div>\n</div>\n</div>\n<hr class=\"my12 outline-none baw0 bb bc-blue-400\"/>\n<div class=\"fw-nowrap fc-black-500\">\n<div class=\"d-flex fd-column lh-md\">\n<div class=\"mb0 d-flex\">\n<div class=\"flex--item mr8\">\n<svg aria-hidden=\"true\" class=\"svg-icon iconLightbulb\" height=\"18\" viewbox=\"0 0 18 18\" width=\"18\"><path d=\"M15 6.38A6.5 6.5 0 0 0 7.78.04h-.02A6.5 6.5 0 0 0 2.05 5.6a6.3 6.3 0 0 0 2.39 5.75c.49.39.76.93.76 1.5v.24c0 1.07.89 1.9 1.92 1.9h2.75c1.04 0 1.92-.83 1.92-1.9v-.2c0-.6.26-1.15.7-1.48A6.3 6.3 0 0 0 15 6.37M4.03 5.85A4.5 4.5 0 0 1 8 2.02a4.5 4.5 0 0 1 5 4.36 4.3 4.3 0 0 1-1.72 3.44c-.98.74-1.5 1.9-1.5 3.08v.1H7.2v-.14c0-1.23-.6-2.34-1.53-3.07a4.3 4.3 0 0 1-1.64-3.94M10 18a1 1 0 0 0 0-2H7a1 1 0 1 0 0 2z\"></path></svg>\n</div>\n<p> If this is not the first question of the OP which has been closed for the same reason, then consider flagging for moderator attention.</p>\n</div>\n<div class=\"mb0 mt6 d-flex\">\n<p class=\"ml24 pl2\">Closed <span class=\"relativetime\" title=\"2026-03-05 10:23:50Z\">15 mins ago</span>.</p>\n</div>\n<div class=\"ml24 pl2\">\n</div>\n</div>\n</div>\n<div class=\"mt24 d-flex gsx gs8\">\n<a class=\"s-btn s-btn__outlined flex--item js-post-notice-edit-post\" href=\"/posts/508777/edit\">\n Improve this question\n </a>\n</div>\n</aside>\n</div>\n<p>Hello MathOverflow Community!</p>\n<p>I 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.</p>\n<p>Also 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!</p>\n<p>I’ve been pondering the Goldbach Conjecture for several years now</p>\n<p>(which states that every even number greater than 2 is the sum of two primes)</p>\n<p>I 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.</p>\n<p>This is the function</p>\n<p>For any even number ‘N’ equal to or greater than 8,</p>\n<p>First subtract any arbitrary prime number that is both</p>\n<ol>\n<li>Less than N</li>\n<li>Not a prime factor of N</li>\n</ol>\n<p>If this produces a prime number, congratulations the function halts/terminates and it has found two unique prime numbers that sum to N.</p>\n<p>If 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.</p>\n<p>This 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.</p>\n<p>I 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.</p>\n<p>There 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.</p>\n<p>I don’t have a specific question about this function, but here’s a few general questions that come to mind:</p>\n<ol>\n<li>Is this function already known? I have searched the internet thoroughly and have not found anything close. Although honestly given my limited knowledge in mathematics I may not even know what to look for.</li>\n<li>Is this function basically just as difficult to prove as the original Goldbach conjecture, or does this provide any meaningful progress?</li>\n<li>Can anyone that’s more programming savvy than me test this for much larger numbers to find a potential counterexample? I have little to no programming knowledge and asked AI to run this function which it seemed to only be able to validate up to 50,000- with 0 counterexamples found.</li>\n</ol>\n<p>I 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.</p>\n<p>Any 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!</p>\n<ul>\n<li>Christopher Oesterling</li>\n</ul>\n</div>",
"comments": [],
"answers": []
},
{
"question_id": 502120,
"title": "Examples for the use of AI and especially LLMs in notable mathematical developments",
"link": "https://mathoverflow.net/questions/502120/examples-for-the-use-of-ai-and-especially-llms-in-notable-mathematical-developme",
"score": 56,
"view_count": 11000,
"answer_count": 20,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "big-list;computer-science;examples;big-picture;experimental-mathematics",
"creation_date": "2026-03-05 09:46:23Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<blockquote>\n<h3>The purpose of this question is to collect examples where large language models (LLMs) like ChatGPT have led to notable mathematical developments.</h3>\n</blockquote>\n<p>The emphasis in this question is on LLMs, but answers about other machine-learning tools are also welcome.</p>\n<p>This question complements two questions that I asked before: <a href=\"https://mathoverflow.net/questions/12085/experimental-mathematics-leading-to-major-advances\">Experimental mathematics leading to major advances</a> (January 2010) and <a href=\"https://mathoverflow.net/questions/396470/the-use-of-computers-leading-to-major-mathematical-advances-ii\">The use of computers leading to major mathematical advances II</a> (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.</p>\n<p>I relaxed the threshold from \"major\" (in the previous two questions) to \"notable\" to allow more answers.</p>\n<p>A related question specifically about Deep Mind is this: <a href=\"https://mathoverflow.net/questions/463937/what-mathematical-problems-can-be-attacked-using-deepminds-recent-mathematical\">What mathematical problems can be attacked using DeepMind's recent mathematical breakthroughs?</a> ; Another related question referring to deep learning is <a href=\"https://mathoverflow.net/questions/390174/what-are-possible-applications-of-deep-learning-to-research-mathematics;\">What are possible applications of deep learning to research mathematics?</a> See also <a href=\"https://mathoverflow.net/questions/463937/what-mathematical-problems-can-be-attacked-using-deepminds-recent-mathematical\">What mathematical problems can be attacked using DeepMind's recent mathematical breakthroughs?</a> .</p>\n</div>",
"comments": [
{
"comment_id": 1309848,
"author_name": "Sam Hopkins",
"author_id": 25028,
"author_url": "https://mathoverflow.net/users/25028/sam-hopkins",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">There have been many similar questions on MO to this about the use of AI/machine learning in research math; see, e.g., <a href=\"https://mathoverflow.net/questions/463937\">mathoverflow.net/questions/463937</a> and other questions linked there.</span>",
"created_date": "2025-10-26 17:43:38Z"
},
{
"comment_id": 1309850,
"author_name": "Jochen Glueck",
"author_id": 102946,
"author_url": "https://mathoverflow.net/users/102946/jochen-glueck",
"score": 8,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I haven't voted on the question (in either way), but I consider it likely that answers - if you get some - will lead to a lot of discussion regarding how significant the LLM contribution actually was.</span>",
"created_date": "2025-10-26 17:48:35Z"
},
{
"comment_id": 1310829,
"author_name": "LSpice",
"author_id": 2383,
"author_url": "https://mathoverflow.net/users/2383/lspice",
"score": 42,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">My instinct is to downvote the question, though I don't have any better justification than that I hate the intrusion of AI into every sphere, and would rather not see t here; but that's unreasonable personal bias, so I just won't vote. But it does seem nonsensioal to me that the question would be at 9 – 7 while both answers, reasonable as far as I can tell, are at 0 – 2. I hope downvoters will consider leaving a comment about what they think is an appropriate answer.</span>",
"created_date": "2025-10-26 20:27:21Z"
},
{
"comment_id": 1310957,
"author_name": "Gil Kalai",
"author_id": 1532,
"author_url": "https://mathoverflow.net/users/1532/gil-kalai",
"score": 9,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I think that there should be a special badge for controversial questions :).</span>",
"created_date": "2025-10-27 19:59:04Z"
},
{
"comment_id": 1311032,
"author_name": "LSpice",
"author_id": 2383,
"author_url": "https://mathoverflow.net/users/2383/lspice",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\"><a href=\"https://mathoverflow.net/posts/comments/1310837\">Re</a>, just to be clear, I meant my rant to express dissatisfaction with the ubiquity of AI, not with you or this question; I hope I gave no offence. <a href=\"https://mathoverflow.net/posts/comments/1310957\">Re</a>, I thought there was, but searching just turned up a post <a href=\"https://meta.stackexchange.com/questions/34543\">Can we have a badge for controversy?</a> which seems to indicate that the answer to the titular question is, or 15 years ago was, \"no.\"</span>",
"created_date": "2025-10-28 14:57:16Z"
}
],
"answers": [
{
"answer_id": 503160,
"question_id": 502120,
"score": 33,
"is_accepted": false,
"author_name": "Lior Silberman",
"author_id": null,
"author_url": "",
"author_reputation": 2873,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-10-27 23:16:37Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Boris Alexeev and Dustin Mixon posted last week their paper <a href=\"https://arxiv.org/abs/2510.19804\" rel=\"noreferrer\">Forbidden Sidon subsets of perfect difference sets, featuring a human-assisted proof</a>, 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.</p>\n<p>The 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).</p>\n<p>My 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.</p>\n</div>",
"comments": [
{
"comment_id": 1311449,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 7,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">This is a very interesting paper. But I think it's important to point out that this use of ChatGPT was a mixed success. They do cite one instance where one of their intermediate results (Proposition 20) was formally proved by the LLM autonomously. On the other hand, they also say that their efforts at vibe coding the nearly trivial result that if f is a fixed point–free involution on a finite set S, then S has even cardinality was \"a multi-day struggle.\"</span>",
"created_date": "2025-10-31 14:32:08Z"
},
{
"comment_id": 1311450,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">IMO, what Alexeev and Mixon did was closer to \"autoformalization\" than to automated discovery of new theorems. Another impressive example of an autoformalization effort is the development by Math.Inc of a tool called Gauss, which helped them complete a challenging formalization project that Tao and Kontorovich had proposed but had not completed.</span>",
"created_date": "2025-10-31 14:39:16Z"
},
{
"comment_id": 1312277,
"author_name": "NooneAtAll3",
"author_id": 519775,
"author_url": "https://mathoverflow.net/users/519775/nooneatall3",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">it wasn't $1000 problem - it was a strong statement that <i>would</i> have proven $1000 problem had it been true. But even Erdos said this formulation was most likely false</span>",
"created_date": "2025-11-07 02:12:26Z"
},
{
"comment_id": 1312378,
"author_name": "Kevin Buzzard",
"author_id": 1384,
"author_url": "https://mathoverflow.net/users/1384/kevin-buzzard",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Note that in this particular case, the statement of the main theorem had already been formalized in a Lean repository of Erdos problems, maintained by Google DeepMind. In particular, in this case the statement had already been inspected by experts. You are absolutely right that in general people probably won't be so lucky.</span>",
"created_date": "2025-11-08 01:57:39Z"
},
{
"comment_id": 1312513,
"author_name": "aorq",
"author_id": 1079,
"author_url": "https://mathoverflow.net/users/1079/aorq",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The paper quotes Erdős clearly stating \"I offer a thousand dollars for a proof or disproof of this conjecture.\"</span>",
"created_date": "2025-11-09 13:32:48Z"
}
]
},
{
"answer_id": 503135,
"question_id": 502120,
"score": 22,
"is_accepted": false,
"author_name": "Sudipta Roy",
"author_id": null,
"author_url": "",
"author_reputation": 170,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-10-27 12:06:23Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Here is an example <a href=\"https://arxiv.org/abs/2510.20013v1\" rel=\"noreferrer\">Counterexample to majority optimality in NICD with erasures</a></p>\n<p>From the abstract:</p>\n<p>We 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.</p>\n</div>",
"comments": []
},
{
"answer_id": 503112,
"question_id": 502120,
"score": 16,
"is_accepted": false,
"author_name": "Zach Teitler",
"author_id": null,
"author_url": "",
"author_reputation": 6361,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-10-26 19:40:32Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>This paper</p>\n<p>Sergey Avvakumov, Roman Karasev,\n<em>Tensor rank of the determinant and periodic triangulations of <span class=\"math-container\">$\\mathbb{R}^n$</span></em></p>\n<p><a href=\"https://arxiv.org/abs/2509.22333\" rel=\"noreferrer\">https://arxiv.org/abs/2509.22333</a></p>\n<p>includes 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.\"</p>\n</div>",
"comments": [
{
"comment_id": 1310831,
"author_name": "Gil Kalai",
"author_id": 1532,
"author_url": "https://mathoverflow.net/users/1532/gil-kalai",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks, Zach! I knew the paper and I met Sergey today, but did not know about the role of ChatGPT :)</span>",
"created_date": "2025-10-26 20:35:06Z"
}
]
},
{
"answer_id": 503575,
"question_id": 502120,
"score": 14,
"is_accepted": false,
"author_name": "Sam Hopkins",
"author_id": null,
"author_url": "https://mathoverflow.net/users/25028",
"author_reputation": 26357,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2025-12-04 18:00:42Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>At the request of Gil Kalai, I'm converting a comment to an answer.</p>\n<p>The 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: <a href=\"https://arxiv.org/abs/2511.02864\" rel=\"nofollow noreferrer\">https://arxiv.org/abs/2511.02864</a>.</p>\n<p>Below is the abstract of the paper.</p>\n<blockquote>\n<p>AlphaEvolve 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.</p>\n</blockquote>\n<p>EDIT:</p>\n<p>Some further developments are in the paper \"New Nikodym set constructions over finite fields\" by Terence Tao (<a href=\"https://arxiv.org/abs/2511.07721\" rel=\"nofollow noreferrer\">https://arxiv.org/abs/2511.07721</a>) whose abstract reads</p>\n<blockquote>\n<p>For any fixed dimension <span class=\"math-container\">$d \\geq 3$</span> we construct a Nikodym set in <span class=\"math-container\">$\\mathbb{F}_q^d$</span> of cardinality <span class=\"math-container\">$q^d - \n(\\frac{d-2}{\\log 2} +1+o(1)) q^{d-1} \\log q$</span> in the limit <span class=\"math-container\">$q \\to \\infty$</span>, when <span class=\"math-container\">$q$</span> is an odd prime power. This improves upon the naive random construction, which gives a set of cardinality\n<span class=\"math-container\">$q^d - (d-1+o(1)) q^{d-1} \\log q$</span>, and is new in the regime where <span class=\"math-container\">$\\mathbb{F}_q$</span> has unbounded characteristic and <span class=\"math-container\">$q$</span> not a perfect square. While the final proofs are completely human generated, the initial ideas of the construction were inspired by output from the tools <em>AlphaEvolve</em> and <em>DeepThink</em>. We also give a new construction of Nikodym sets in <span class=\"math-container\">$\\mathbb{F}_q^2$</span> for <span class=\"math-container\">$q$</span> a perfect square that match the existing bounds of <span class=\"math-container\">$q^2 - q^{3/2} + O(q \\log q)$</span>, assuming that <span class=\"math-container\">$q$</span> is not the square of a prime <span class=\"math-container\">$p \\equiv 3 \\pmod{4}$</span>.</p>\n</blockquote>\n<p>And also in the paper \"Sum-difference exponents for boundedly many slopes, and rational complexity\" again by Terence Tao (<a href=\"https://arxiv.org/abs/2511.15135\" rel=\"nofollow noreferrer\">https://arxiv.org/abs/2511.15135</a>) whose abstract reads</p>\n<blockquote>\n<p>The dimension of Kakeya sets can be bounded using sum-difference exponents <span class=\"math-container\">$\\mathrm{SD}(R;s)$</span> for various sets of rational slopes <span class=\"math-container\">$R$</span> and output slope <span class=\"math-container\">$s$</span>; the arithmetic Kakeya conjecture, which implies the Kakeya conjecture in all dimensions, asserts that the infimum of such exponents is <span class=\"math-container\">$1$</span>. The best upper bound on this infimum currently is <span class=\"math-container\">$1.67513\\dots$</span>. In this note, inspired by numerical explorations from the tool <em>AlphaEvolve</em>, we study the regime where the cardinality of the set of slopes <span class=\"math-container\">$R$</span> is bounded. In this regime, we establish that these exponents converge to <span class=\"math-container\">$2$</span> at a rate controlled by the <em>rational complexity</em> of <span class=\"math-container\">$s$</span> relative to <span class=\"math-container\">$R$</span>, which measures how efficiently <span class=\"math-container\">$s$</span> can be expressed as a rational combination of slopes in <span class=\"math-container\">$R$</span>.</p>\n</blockquote>\n</div>",
"comments": [
{
"comment_id": 1312286,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">In a nutshell, the idea is to solve a combinatorial optimization problem by evolving <i>code for generating combinatorial objects</i> rather than evolving the <i>combinatorial objects themselves</i>. To do this, one needs to be able to make small random perturbations of the code while still having the code compile; this is where LLMs come in, since writing code is one thing LLMs are good at.</span>",
"created_date": "2025-11-07 05:05:22Z"
},
{
"comment_id": 1312320,
"author_name": "Sam Hopkins",
"author_id": 25028,
"author_url": "https://mathoverflow.net/users/25028/sam-hopkins",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@TimothyChow I believe the approach used to find better cap sets by DeepMind (discussed at <a href=\"https://mathoverflow.net/questions/463937\">mathoverflow.net/questions/463937</a>) was along the same lines.</span>",
"created_date": "2025-11-07 13:17:53Z"
},
{
"comment_id": 1312323,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Yes, AlphaEvolve is \"FunSearch 2.0\".</span>",
"created_date": "2025-11-07 13:50:51Z"
},
{
"comment_id": 1312699,
"author_name": "Gil Kalai",
"author_id": 1532,
"author_url": "https://mathoverflow.net/users/1532/gil-kalai",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks, Sam. There is a blogpost about the paper here: <a href=\"https://terrytao.wordpress.com/2025/11/05/mathematical-exploration-and-discovery-at-scale/\" rel=\"nofollow noreferrer\">terrytao.wordpress.com/2025/11/05/…</a></span>",
"created_date": "2025-11-11 09:37:48Z"
}
]
},
{
"answer_id": 503229,
"question_id": 502120,
"score": 13,
"is_accepted": false,
"author_name": "littleO",
"author_id": null,
"author_url": "https://mathoverflow.net/users/38114",
"author_reputation": 146,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2025-10-29 18:27:52Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>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:</p>\n<blockquote>\n<p>The 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.</p>\n</blockquote>\n<p><a href=\"https://arxiv.org/abs/2510.23513\" rel=\"noreferrer\">https://arxiv.org/abs/2510.23513</a></p>\n<p>See also this discussion by Damek Davis that helps put the result in perspective: <a href=\"https://x.com/damekdavis/status/1982529760505782510?s=46\" rel=\"noreferrer\">https://x.com/damekdavis/status/1982529760505782510?s=46</a></p>\n</div>",
"comments": []
},
{
"answer_id": 503147,
"question_id": 502120,
"score": 10,
"is_accepted": false,
"author_name": "JoshuaZ",
"author_id": null,
"author_url": "https://mathoverflow.net/users/127690",
"author_reputation": 8600,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2025-10-28 02:56:49Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Scott Aaronson Phillip Harris, Freek Witteveen have a <a href=\"https://arxiv.org/abs/2509.21131\" rel=\"noreferrer\">recent paper on the bounds of amplification of QMA (quantum Merlin-Arthur)</a>. A critical part of the paper involved a linear algebra trick suggested by GPT5. See <a href=\"https://scottaaronson.blog/?p=9183\" rel=\"noreferrer\">Aaronson's blog entry here</a>.</p>\n</div>",
"comments": []
},
{
"answer_id": 504664,
"question_id": 502120,
"score": 9,
"is_accepted": false,
"author_name": "Timothy Chow",
"author_id": null,
"author_url": "https://mathoverflow.net/users/3106",
"author_reputation": 91241,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2025-12-05 04:29:08Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>There have been some notable recent examples of LLMs playing an important role in solving certain Erdős problems, e.g., <a href=\"https://leanprover.zulipchat.com/#narrow/channel/219941-Machine-Learning-for-Theorem-Proving/topic/Erdos.20124.20AI.20Solution/with/560981950\" rel=\"noreferrer\">Problem 124</a> and <a href=\"https://x.com/llllvvuu/status/1995596080042721740\" rel=\"noreferrer\">Problem 481</a> (although I think the latter turned out to be implied by a result of Klarner in 1982), which were purportedly solved <a href=\"https://x.com/SebastienBubeck/status/1994946303546331508\" rel=\"noreferrer\">entirely by Aristotle</a>, and <a href=\"https://mathstodon.xyz/@tao/115591487350860999\" rel=\"noreferrer\">Problem 367</a>, 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.</p>\n</div>",
"comments": [
{
"comment_id": 1315639,
"author_name": "Dustin G. Mixon",
"author_id": 29873,
"author_url": "https://mathoverflow.net/users/29873/dustin-g-mixon",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Relevant: <a href=\"https://xenaproject.wordpress.com/2025/12/05/formalization-of-erdos-problems/\" rel=\"nofollow noreferrer\">xenaproject.wordpress.com/2025/12/05/…</a></span>",
"created_date": "2025-12-05 18:11:38Z"
},
{
"comment_id": 1321557,
"author_name": "Carlo Beenakker",
"author_id": 11260,
"author_url": "https://mathoverflow.net/users/11260/carlo-beenakker",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\"><a href=\"https://www.erdosproblems.com/728\" rel=\"nofollow noreferrer\">Erdős Problem 728</a> has now been solved by the LLM's GPT-5.2 Pro and Aristotle, operated by Kevin Barreto: <a href=\"https://arxiv.org/abs/2601.07421\" rel=\"nofollow noreferrer\">arxiv.org/abs/2601.07421</a> ; see the discussion by <a href=\"https://mathstodon.xyz/@tao/115891256726420022\" rel=\"nofollow noreferrer\">Terence Tao</a></span>",
"created_date": "2026-01-24 22:34:38Z"
},
{
"comment_id": 1322642,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Another paper about Aletheia and Erdős problems: <a href=\"https://arxiv.org/abs/2601.22401\" rel=\"nofollow noreferrer\">Semi-Autonomous Mathematics Discovery with Gemini: A Case Study on the Erdős Problems</a>.</span>",
"created_date": "2026-02-04 00:48:31Z"
},
{
"comment_id": 1323638,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Terry Tao just gave a <a href=\"https://www.youtube.com/watch?v=zJvuaRVc8Bg\" rel=\"nofollow noreferrer\">talk at IPAM</a> on this general topic, with an emphasis on solving Erdős problems with LLMs and Lean (although AlphaEvolve does get a mention). He gives some examples that are not already listed above.</span>",
"created_date": "2026-02-14 04:01:32Z"
},
{
"comment_id": 1324947,
"author_name": "Sam Hopkins",
"author_id": 25028,
"author_url": "https://mathoverflow.net/users/25028/sam-hopkins",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Here is another interesting Erdős problem example, where apparently the AI solved the whole thing: <a href=\"https://arxiv.org/abs/2602.21275\" rel=\"nofollow noreferrer\">arxiv.org/abs/2602.21275</a></span>",
"created_date": "2026-02-26 03:10:08Z"
}
]
},
{
"answer_id": 503113,
"question_id": 502120,
"score": 8,
"is_accepted": false,
"author_name": "Marco Ripà",
"author_id": null,
"author_url": "",
"author_reputation": 2151,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-10-26 20:08:28Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Not exactly a <em>notable</em> result, but in my recent preprint <a href=\"https://www.researchgate.net/publication/394419253_Evaluation_of_GPT-5_on_an_Advanced_Extension_of_Kashihara%27s_Problem\" rel=\"noreferrer\">Evaluation of GPT-5 on an Advanced Extension of Kashihara's Problem</a> I describe how GPT-5 has been able to improve the general version of an extended combinatorial problem I originally solved in 2010.</p>\n</div>",
"comments": []
},
{
"answer_id": 504142,
"question_id": 502120,
"score": 8,
"is_accepted": false,
"author_name": "Timothy Chow",
"author_id": null,
"author_url": "",
"author_reputation": 91241,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-11-21 04:08:54Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The abstract of\n<a href=\"https://cdn.openai.com/pdf/4a25f921-e4e0-479a-9b38-5367b47e8fd0/early-science-acceleration-experiments-with-gpt-5.pdf#page8\" rel=\"noreferrer\">Early science acceleration experiments with GPT-5</a> 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.\"</p>\n</div>",
"comments": [
{
"comment_id": 1314051,
"author_name": "Sam Hopkins",
"author_id": 25028,
"author_url": "https://mathoverflow.net/users/25028/sam-hopkins",
"score": 14,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I found this passage on pg. 29 really interesting: \"Our experience illustrates a pitfall in using AI: although GPT-5 possesses enormous internal knowledge and the capability to locate even more using the internet, it may not always report the original information sources accurately. This has the potential to deceive even seasoned researchers into thinking their findings are novel. We expect that our experience is not unique, and urge others to take special care in attribution when working with LLM-assisted proofs.\"</span>",
"created_date": "2025-11-21 04:11:30Z"
},
{
"comment_id": 1314059,
"author_name": "Paata Ivanisvili",
"author_id": 50901,
"author_url": "https://mathoverflow.net/users/50901/paata-ivanisvili",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I agree. Even without AI, humans often believe they’ve discovered something new only to learn it was proved earlier by someone else. What’s interesting now is to understand how the rate of such misattributions from LLM-assisted work compares to the natural rate of human rediscovery</span>",
"created_date": "2025-11-21 07:16:00Z"
},
{
"comment_id": 1314078,
"author_name": "Yemon Choi",
"author_id": 763,
"author_url": "https://mathoverflow.net/users/763/yemon-choi",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@SamHopkins I find the quote telling, although as someone who studied history as one of their subjects in the UK's 16-18 high-school specialisms, I find the apparent surprise of a lot of scientists and mathematicians in this regard to be a bit depressing. (Any historian worth their salt will know the distinction between primary and secondary sources and have done some basic training of source analysis, etc)</span>",
"created_date": "2025-11-21 14:23:12Z"
},
{
"comment_id": 1314083,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@YemonChoi Yes. It shouldn't even require any formal training (just common sense) to know that you shouldn't just blindly copy a reference from someone else's bibliography without checking its accuracy, but of course many scientists and mathematicians have been doing this since time immemorial.</span>",
"created_date": "2025-11-21 15:19:43Z"
}
]
},
{
"answer_id": 503560,
"question_id": 502120,
"score": 7,
"is_accepted": false,
"author_name": "Timothy Chow",
"author_id": null,
"author_url": "",
"author_reputation": 91241,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-11-06 14:57:04Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I 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.</p>\n<p>As reported in <a href=\"https://www.scientificamerican.com/article/inside-the-secret-meeting-where-mathematicians-struggled-to-outsmart-ai/\" rel=\"noreferrer\">Scientific American</a>, 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 <span class=\"math-container\">$\\mathbb{Q}$</span>?\" To Ono's amazement, the AI autonomously solved the problem. You can read <a href=\"https://www.facebook.com/photo/?fbid=9758357980879433&set=a.164126083636052\" rel=\"noreferrer\">Ono's account on his Facebook page</a> (also reproduced below), or listen to him talk about it <a href=\"https://www.youtube.com/watch?v=vGmrcm1aD6g&t=790s\" rel=\"noreferrer\">here</a>.</p>\n<p>Even 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 <a href=\"https://x.com/littmath/status/1931403214613598252\" rel=\"noreferrer\">tweet by Daniel Litt</a>, who was also one of the invited mathematicians but was not too impressed when he read over the chat log.\n<a href=\"https://i.sstatic.net/LR9D59gd.jpg\" rel=\"noreferrer\"><img alt=\"enter image description here\" src=\"https://i.sstatic.net/LR9D59gd.jpg\"/></a></p>\n</div>",
"comments": [
{
"comment_id": 1312191,
"author_name": "Sam Hopkins",
"author_id": 25028,
"author_url": "https://mathoverflow.net/users/25028/sam-hopkins",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">A similar project, but on a smaller scale and led by Christian Stump, for using PhD-level mathematics problems to benchmark AI is: <a href=\"https://math.science-bench.ai/\" rel=\"nofollow noreferrer\">math.science-bench.ai</a></span>",
"created_date": "2025-11-06 15:04:20Z"
},
{
"comment_id": 1323667,
"author_name": "Sam Hopkins",
"author_id": 25028,
"author_url": "https://mathoverflow.net/users/25028/sam-hopkins",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Here is another, somewhat similar project \"First Proof\" (<a href=\"https://1stproof.org/\" rel=\"nofollow noreferrer\">1stproof.org</a>): \"This project represents our preliminary efforts to develop an objective and realistic methodology for assessing the capabilities of AI systems to autonomously solve research-level math questions. After letting these ideas ferment in the community, we hope to produce a more structured benchmark.\"</span>",
"created_date": "2026-02-14 14:37:28Z"
},
{
"comment_id": 1324948,
"author_name": "Sam Hopkins",
"author_id": 25028,
"author_url": "https://mathoverflow.net/users/25028/sam-hopkins",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">A paper from \"Aletheia\" (Google DeepMind) was just posted about solving 4/10 of the First Proof problems: <a href=\"https://arxiv.org/abs/2602.21201\" rel=\"nofollow noreferrer\">arxiv.org/abs/2602.21201</a>. Maybe I should post this as a separate answer.</span>",
"created_date": "2026-02-26 03:20:26Z"
},
{
"comment_id": 1325180,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@SamHopkins Yes, First Proof is probably worth an answer, even though the problems were specifically chosen to be \"already solved.\" Daniel Litt has a very interesting <a href=\"https://www.daniellitt.com/blog/2026/2/20/mathematics-in-the-library-of-babel\" rel=\"nofollow noreferrer\">blog post</a> that discusses First Proof in some detail, and how it has nudged him slightly closer to \"believer\" and further from \"skeptic.\"</span>",
"created_date": "2026-02-28 13:44:57Z"
}
]
},
{
"answer_id": 504668,
"question_id": 502120,
"score": 7,
"is_accepted": false,
"author_name": "Vamsi",
"author_id": null,
"author_url": "",
"author_reputation": 3453,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-12-05 06:27:12Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>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 <a href=\"https://arxiv.org/pdf/2511.06849\" rel=\"noreferrer\">https://arxiv.org/pdf/2511.06849</a> 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.</p>\n</div>",
"comments": []
},
{
"answer_id": 503193,
"question_id": 502120,
"score": 6,
"is_accepted": false,
"author_name": "prof-g",
"author_id": null,
"author_url": "",
"author_reputation": 1,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-10-28 17:25:14Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>here is a paper on bottleneck duality in flow networks with lattice coefficients from fall 2024.</p>\n<p><a href=\"https://arxiv.org/abs/2410.00315\" rel=\"noreferrer\">https://arxiv.org/abs/2410.00315</a></p>\n<p>the 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.</p>\n</div>",
"comments": [
{
"comment_id": 1311053,
"author_name": "darij grinberg",
"author_id": 2530,
"author_url": "https://mathoverflow.net/users/2530/darij-grinberg",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">This is funny! I once tried in vain to detropicalize max-flow-min-cut (you can find <a href=\"https://mathoverflow.net/questions/297015\">some traces of that on MO</a>), while you have managed to tropicalize it even further (+ becomes max) and then extend it to distributive lattices :)</span>",
"created_date": "2025-10-28 17:54:26Z"
}
]
},
{
"answer_id": 508759,
"question_id": 502120,
"score": 5,
"is_accepted": false,
"author_name": "Timothy Chow",
"author_id": null,
"author_url": "",
"author_reputation": 91241,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-04 20:58:44Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The \"notability\" here may be less because of the problem itself, and more because Donald Knuth was involved. <a href=\"https://cs.stanford.edu/~knuth/papers/claude-cycles.pdf\" rel=\"noreferrer\">Knuth wrote</a>:</p>\n<blockquote>\n<p>Shock! 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.<br/><br/>\nHere’s the problem, which came up while I was writing about directed Hamiltonian cycles for a future\nvolume of <i>The Art of Computer Programming</i>:<br/><br/>\nConsider the digraph with <span class=\"math-container\">$m^3$</span> vertices <span class=\"math-container\">$ijk$</span> for <span class=\"math-container\">$0 \\le i, j, k < m$</span>, and three arcs from\neach vertex, namely to <span class=\"math-container\">$i^+jk$</span>, <span class=\"math-container\">$ij^+k$</span>, and <span class=\"math-container\">$ijk^+$</span>, where <span class=\"math-container\">$i^+ = (i+1) \\bmod m$</span>. Try to find\na general decomposition of the arcs into three directed <span class=\"math-container\">$m^3$</span>-cycles, for all <span class=\"math-container\">$m > 2$</span>.</p>\n</blockquote>\n<p>Knuth then went on to to describe how Filip Stappers used Claude interactively to discover a conjectural solution, which was then rigorously proved.</p>\n</div>",
"comments": [
{
"comment_id": 1325654,
"author_name": "darij grinberg",
"author_id": 2530,
"author_url": "https://mathoverflow.net/users/2530/darij-grinberg",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Note that, despite his fame as a combinatorialist, Don Knuth is no stranger to heuristical and non-deterministic methods; there are sections about genetic algorithms and simulated annealing in TAoCP.</span>",
"created_date": "2026-03-04 21:11:23Z"
}
]
},
{
"answer_id": 503191,
"question_id": 502120,
"score": 4,
"is_accepted": false,
"author_name": "Piyush Grover",
"author_id": null,
"author_url": "",
"author_reputation": 1839,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-10-28 16:54:01Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Using 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).</p>\n<p>Paper here: <a href=\"https://arxiv.org/abs/2509.14185\" rel=\"nofollow noreferrer\">https://arxiv.org/abs/2509.14185</a>\nArticle: <a href=\"https://deepmind.google/discover/blog/discovering-new-solutions-to-century-old-problems-in-fluid-dynamics/\" rel=\"nofollow noreferrer\">https://deepmind.google/discover/blog/discovering-new-solutions-to-century-old-problems-in-fluid-dynamics/</a></p>\n</div>",
"comments": [
{
"comment_id": 1311071,
"author_name": "Geordie Williamson",
"author_id": 919,
"author_url": "https://mathoverflow.net/users/919/geordie-williamson",
"score": 6,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Note that this is an application of neural networks, but not LLMs. (I think they tried to use AlphaEvolve, but this wasn't the main ingredient in the paper...)</span>",
"created_date": "2025-10-28 21:42:59Z"
}
]
},
{
"answer_id": 503192,
"question_id": 502120,
"score": 4,
"is_accepted": false,
"author_name": "Jalaj",
"author_id": null,
"author_url": "",
"author_reputation": 1,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-10-28 17:09:11Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>A 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.</p>\n</div>",
"comments": []
},
{
"answer_id": 504786,
"question_id": 502120,
"score": 4,
"is_accepted": false,
"author_name": "JoS",
"author_id": null,
"author_url": "",
"author_reputation": 711,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2025-12-09 07:22:35Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The paper <a href=\"https://arxiv.org/abs/2511.18828\" rel=\"nofollow noreferrer\">Solving a Research Problem in Mathematical Statistics with AI Assistance</a> by Edgar Dobriban documents how GPT-5 Pro helped close a gap between upper and lower bounds in robust density estimation under Wasserstein contaminations.</p>\n<p>The mathematical results appear in version 3 of <a href=\"https://arxiv.org/abs/2308.01853\" rel=\"nofollow noreferrer\">Minimax Statistical Estimation under Wasserstein Contamination</a> (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.</p>\n<p>The key AI contributions were:</p>\n<ul>\n<li>For the <strong>upper bound</strong>: suggesting a sharper analysis of the contamination bias via the optimal transport map and interpolation along geodesics</li>\n<li>For the <strong>lower bound</strong>: proposing the dynamic Benamou–Brenier formulation of optimal transport, a technique the authors were unfamiliar with</li>\n</ul>\n<p>The meta-paper provides detailed transcripts and a balanced discussion of limitations (hallucinated references, glossed-over details requiring days to verify).</p>\n<p>See also the author's <a href=\"https://x.com/EdgarDobriban/status/1998101290027507892\" rel=\"nofollow noreferrer\">Twitter/X thread</a> summarizing the experience.</p>\n</div>",
"comments": [
{
"comment_id": 1316041,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">\"over the last 15 years, the present author has acquired a good amount of experience with research in the mathematical sciences (reading and writing proofs, literature search), writing papers, and a good working knowledge of basic optimal transport, at the level of Figalli and Glaudo (2021). These skills took many years to develop, starting from the school level, through doctoral training, and continuing at the level of a faculty member. Without such skills, we feel that it would be effectively impossible to properly use current AI tools for advanced mathematical research.\"</span>",
"created_date": "2025-12-10 02:29:45Z"
}
]
},
{
"answer_id": 507767,
"question_id": 502120,
"score": 4,
"is_accepted": false,
"author_name": "Timothy Chow",
"author_id": null,
"author_url": "",
"author_reputation": 91241,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-02-03 18:41:27Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I 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.\" <a href=\"https://meta.mathoverflow.net/questions/6348/use-of-llms-in-notable-mathematical-developments#comment35131_6357\">Ravi Vakil commented</a> 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.</p>\n<p>The abstract of <a href=\"https://arxiv.org/abs/2601.07222\" rel=\"nofollow noreferrer\">The motivic class of the space of genus 0 maps to the flag variety</a> 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).\"</p>\n</div>",
"comments": []
},
{
"answer_id": 504641,
"question_id": 502120,
"score": 3,
"is_accepted": false,
"author_name": "JoS",
"author_id": null,
"author_url": "https://mathoverflow.net/users/69630",
"author_reputation": 711,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2025-12-04 17:45:40Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The computational complexity paper <a href=\"https://arxiv.org/abs/2512.02808\" rel=\"noreferrer\">\"Search versus Decision for <span class=\"math-container\">$S_2^P$</span>\"</a> by Lance Fortnow writes in the acknowledgements:</p>\n<blockquote>\n<p>While 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.</p>\n</blockquote>\n<p>EDIT (additional context):\nThe author further elaborated on <a href=\"https://x.com/fortnow/status/1996257537310486803\" rel=\"noreferrer\">Twitter/X</a>: 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.\".</p>\n</div>",
"comments": [
{
"comment_id": 1315462,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">That acknowledgement reads to me like it says the AI was <i>not</i> used in the mathematical development itself. It only helped write the paper.</span>",
"created_date": "2025-12-04 12:19:49Z"
},
{
"comment_id": 1315473,
"author_name": "JoS",
"author_id": 69630,
"author_url": "https://mathoverflow.net/users/69630/jos",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I agree that the acknowledgement could be interpreted this way. However, the author <a href=\"https://x.com/fortnow/status/1996257537310486803\" rel=\"nofollow noreferrer\">clarified on Twitter/X</a>: 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.\".</span>",
"created_date": "2025-12-04 14:08:12Z"
},
{
"comment_id": 1315480,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">You should edit your answer to clarify and include this then. I'm happy to withdraw my downvote if you do so.</span>",
"created_date": "2025-12-04 14:57:24Z"
},
{
"comment_id": 1315510,
"author_name": "JoS",
"author_id": 69630,
"author_url": "https://mathoverflow.net/users/69630/jos",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thank you, I have edited the answer accordingly!</span>",
"created_date": "2025-12-04 17:46:05Z"
}
]
},
{
"answer_id": 506692,
"question_id": 502120,
"score": 3,
"is_accepted": false,
"author_name": "Sam Hopkins",
"author_id": null,
"author_url": "",
"author_reputation": 26357,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-01-06 03:43:18Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>(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.)</p>\n<p>In \"Bruhat intervals that are large hypercubes\" (<a href=\"https://arxiv.org/abs/2601.01235\" rel=\"nofollow noreferrer\">https://arxiv.org/abs/2601.01235</a>), Jordan Ellenberg, Nicolas Libedinsky, David Plaza, José Simental, and Geordie Williamson find large hypercubes (of order <span class=\"math-container\">$n \\log n$</span>, in particular, superlinear) inside of Bruhat graphs of the symmetric group <span class=\"math-container\">$S_n$</span>, 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 <span class=\"math-container\">$n$</span>, and then were able to give a natural \"interpretation\" to what the machine was doing (at least in the case of <span class=\"math-container\">$n=2^m$</span> 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.</p>\n<p>The abstract of their paper is</p>\n<blockquote>\n<p>We study the question of finding big Bruhat intervals that are poset hypercubes in the symmetric group <span class=\"math-container\">$S_n$</span>. 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 <span class=\"math-container\">$n$</span> tested, and which we show works well for general <span class=\"math-container\">$n$</span>. When <span class=\"math-container\">$n$</span> is a power of 2 we exhibit a hypercube of dimension <span class=\"math-container\">$O(n\\log n)$</span>, 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 <em>dyadically well-distributed</em> 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 <span class=\"math-container\">$S_n$</span> gives a lower bound (and possibly is equal to) the maximal possible coefficient of the second-highest degree term in the Kazhdan--Lusztig <span class=\"math-container\">$R$</span>-polynomial in <span class=\"math-container\">$S_n$</span>. As a surprising consequence, we obtain a new lower bound of order <span class=\"math-container\">$n\\log n$</span> for the maximal number of frozen variables appearing in the cluster algebras attached to the open Richardson varieties in <span class=\"math-container\">$S_n$</span>, and a similar result for moduli spaces of embeddings of Bruhat graphs.</p>\n</blockquote>\n</div>",
"comments": [
{
"comment_id": 1319598,
"author_name": "Sam Hopkins",
"author_id": 25028,
"author_url": "https://mathoverflow.net/users/25028/sam-hopkins",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I will say that I find this sentence in the acknowledgments quite cryptic: \"This paper involves essential contributions from Adam Zsolt Wagner, who cannot be named as a coauthor for technical reasons.\"</span>",
"created_date": "2026-01-06 17:43:01Z"
},
{
"comment_id": 1322606,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I'd guess it's some kind of technicality regarding funding or conflict of interest.</span>",
"created_date": "2026-02-03 18:30:48Z"
},
{
"comment_id": 1322644,
"author_name": "Geordie Williamson",
"author_id": 919,
"author_url": "https://mathoverflow.net/users/919/geordie-williamson",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Yes, Google DeepMind requires GDM affiliated authors to get permission to be named as authors for papers where GDM resources were used. Adam made an essential contribution to this work.</span>",
"created_date": "2026-02-04 01:13:06Z"
}
]
},
{
"answer_id": 508776,
"question_id": 502120,
"score": 2,
"is_accepted": false,
"author_name": "liuyao",
"author_id": null,
"author_url": "",
"author_reputation": 730,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-05 09:46:23Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>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.</p>\n<p><a href=\"https://arxiv.org/abs/2511.13391\" rel=\"nofollow noreferrer\">https://arxiv.org/abs/2511.13391</a></p>\n<p>The <a href=\"https://en.wikipedia.org/wiki/Kissing_number#Some_known_bounds\" rel=\"nofollow noreferrer\">table on wikipedia</a> and the one maintained by <a href=\"https://cohn.mit.edu/kissing-numbers/\" rel=\"nofollow noreferrer\">Henry Cohn</a> have been updated.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508737,
"title": "Ordinals embeddable into the collection of strictly increasing functions $f:\\omega\\to\\omega$",
"link": "https://mathoverflow.net/questions/508737/ordinals-embeddable-into-the-collection-of-strictly-increasing-functions-f-ome",
"score": 6,
"view_count": 236,
"answer_count": 2,
"is_answered": true,
"has_accepted_answer": true,
"accepted_answer_id": 508740,
"tags": "set-theory;lo.logic;order-theory;lattice-theory;ordinal-numbers",
"creation_date": "2026-03-05 09:24:58Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Let <span class=\"math-container\">$\\text{I}(\\omega)$</span> denote the collection of strictly increasing functions <span class=\"math-container\">$f:\\omega\\to\\omega$</span> and consider the poset <span class=\"math-container\">$\\big(\\text{I}(\\omega),\\leq_\\omega\\big)$</span> where <span class=\"math-container\">$f \\leq_\\omega g$</span> for <span class=\"math-container\">$f, g \\in \\text{I}(\\omega)$</span> if and only if <span class=\"math-container\">$f(n) \\leq g(n)$</span> for all <span class=\"math-container\">$n\\in \\omega$</span>.</p>\n<p>What is the smallest ordinal not embeddable into <span class=\"math-container\">$\\text{I}(\\omega)$</span>?</p>\n</div>",
"comments": [],
"answers": [
{
"answer_id": 508740,
"question_id": 508737,
"score": 12,
"is_accepted": true,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"author_reputation": 34953,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-04 11:07:34Z",
"last_edited_date": "2026-03-05 08:47:55Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The answer is <span class=\"math-container\">$\\omega_1$</span>, the least uncountable ordinal.</p>\n<p>Indeed, suppose <span class=\"math-container\">$f_\\alpha,\\alpha<\\omega_1$</span> is an increasing sequence in <span class=\"math-container\">$I(\\omega)$</span>. I claim that for each <span class=\"math-container\">$n$</span>, the value <span class=\"math-container\">$f_\\alpha(n)$</span> eventually stabilizes. Indeed, if not, let <span class=\"math-container\">$\\alpha_k$</span> be the least such that <span class=\"math-container\">$f_{\\alpha_k}(n)>k$</span>. Since <span class=\"math-container\">$\\omega_1$</span> is regular, there is a countable ordinal <span class=\"math-container\">$\\alpha>\\alpha_k$</span> for all <span class=\"math-container\">$k$</span>. But then <span class=\"math-container\">$f_\\alpha(n)>k$</span> for all <span class=\"math-container\">$k$</span>, which is impossible.</p>\n<p>Therefore for each <span class=\"math-container\">$n$</span> there is an ordinal <span class=\"math-container\">$\\beta_n$</span> such that <span class=\"math-container\">$f_\\alpha(n)=f_{\\beta_n}(n)$</span> for all <span class=\"math-container\">$\\alpha>\\beta_n$</span>. But using regularity once again and taking <span class=\"math-container\">$\\beta>\\beta_n$</span> for all <span class=\"math-container\">$n$</span>, we find <span class=\"math-container\">$f_\\alpha=f_\\beta$</span> for all <span class=\"math-container\">$\\alpha>\\beta$</span>, which is a contradiction.</p>\n<p>Conversely, for any countable ordinal <span class=\"math-container\">$\\eta$</span> there is an order-preserving embedding <span class=\"math-container\">$c:\\eta\\to\\Bbb Q$</span>, and we can assume <span class=\"math-container\">$c(\\alpha)>1$</span> for all <span class=\"math-container\">$\\alpha\\in\\eta$</span>. Setting <span class=\"math-container\">$f_\\alpha(n)=\\lfloor c(\\alpha)n\\rfloor$</span> we get an order-embedding <span class=\"math-container\">$\\eta\\to I(\\omega),\\alpha\\mapsto f_\\alpha$</span>.</p>\n</div>",
"comments": [
{
"comment_id": 1325643,
"author_name": "Noah Schweber",
"author_id": 8133,
"author_url": "https://mathoverflow.net/users/8133/noah-schweber",
"score": 5,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">+1, and a footnote to the OP/other readers: if we weaken our partial ordering to <i>eventual domination</i> ($f\\le g$ iff $\\exists n\\forall m>n(f(m)\\le g(m))$) then the argument breaks down and in fact a very different situation occurs: any countable set of functions has a common upper bound in this weaker sense (given $\\{f_i:i\\in\\omega\\}$ consider $g:x\\mapsto\\sum_{i<x}f_i(x)$), and so we definitely <b>can</b> embed $\\omega_1$. Indeed, at this point we wind up in the setting of <b>cardinal characteristics of the continuum</b>. (The same occurs with $\\mathcal{P}(\\omega)$ versus $\\mathcal{P}(\\omega)/fin$).</span>",
"created_date": "2026-03-04 17:22:10Z"
},
{
"comment_id": 1325647,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@NoahSchweber Indeed, I have in the past investigated that question (see an archive of my <a href=\"https://web.archive.org/web/20201202092556/https://wojowu.students.wmi.amu.edu.pl/2017/10/30/long-well-ordered-chains-of-functions-under-eventual-domination/\" rel=\"nofollow noreferrer\">old blog</a>). I have shown that, letting $\\mathfrak b$ be the bounding number, the supremum of embeddable ordinals is at least $\\mathfrak b^+$. I was considering mentioning it in the post, but it was going a bit far afield.</span>",
"created_date": "2026-03-04 17:57:41Z"
}
]
},
{
"answer_id": 508756,
"question_id": 508737,
"score": 8,
"is_accepted": false,
"author_name": "bof",
"author_id": 43266,
"author_url": "https://mathoverflow.net/users/43266/bof",
"author_reputation": 15434,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-04 20:33:11Z",
"last_edited_date": "2026-03-05 09:24:58Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><strong>Theorem.</strong> A linearly ordered set <span class=\"math-container\">$A$</span> is embeddable in <span class=\"math-container\">$I(\\omega)$</span> if and only if <span class=\"math-container\">$A$</span> is embeddable in <span class=\"math-container\">$\\mathbb R$</span>.</p>\n<p><strong>Proof.</strong> As in Wojowu's answer we can embed <span class=\"math-container\">$[1,\\infty)$</span> into <span class=\"math-container\">$I(\\omega)$</span> by setting <span class=\"math-container\">$f_t(n)=\\lfloor nt\\rfloor$</span> for <span class=\"math-container\">$1\\le t\\lt\\infty$</span> and <span class=\"math-container\">$n\\lt\\omega$</span>. For the converse, choose an injection <span class=\"math-container\">$\\pi:\\omega\\times\\omega\\to\\omega$</span> and define a map <span class=\"math-container\">$\\varphi:I(\\omega)\\to\\mathbb R$</span> by setting\n<span class=\"math-container\">$$\\varphi(f)=\\sum_{n\\lt f(m)}2^{-\\pi(m,n)}.$$</span>\nSince <span class=\"math-container\">$f\\lt_\\omega g\\implies\\varphi(f)\\lt\\varphi(g)$</span>, every chain in <span class=\"math-container\">$I(\\omega)$</span> is embeddable in <span class=\"math-container\">$\\mathbb R$</span>.</p>\n<p><strong>Corollary 1.</strong> Every countable linear order is embeddable in <span class=\"math-container\">$I(\\omega)$</span>.</p>\n<p><strong>Proof.</strong> By a theorem of Cantor every countable linear order is embeddable in <span class=\"math-container\">$\\mathbb Q$</span> and therefore in <span class=\"math-container\">$\\mathbb R$</span>. Alternatively, if <span class=\"math-container\">$A$</span> is a linearly ordered set and <span class=\"math-container\">$i:A\\to\\omega$</span> is injective, then the map\n<span class=\"math-container\">$$a\\mapsto\\sum_{x\\lt a}2^{-i(x)}$$</span>\nis an order-embedding of <span class=\"math-container\">$A$</span> into <span class=\"math-container\">$\\mathbb R$</span>.</p>\n<p><strong>Corollary 2.</strong> <span class=\"math-container\">$\\omega_1$</span> is not embeddable in <span class=\"math-container\">$I(\\omega)$</span>.</p>\n<p><strong>Proof.</strong> <span class=\"math-container\">$\\omega_1$</span> is not embeddable in <span class=\"math-container\">$\\mathbb R$</span>.</p>\n<p><strong>Remark.</strong> For posets <span class=\"math-container\">$P$</span> and <span class=\"math-container\">$Q$</span> let us write <span class=\"math-container\">$P\\le Q$</span> if there is a map <span class=\"math-container\">$\\varphi:P\\to Q$</span> such that <span class=\"math-container\">$x\\lt y\\implies\\varphi(x)\\lt\\varphi(y)$</span>. The theorem was proved by showing that <span class=\"math-container\">$\\mathbb R\\le I(\\omega)$</span> and <span class=\"math-container\">$I(\\omega)\\le\\mathbb R$</span>. Hence, for any poset <span class=\"math-container\">$P$</span>, we have <span class=\"math-container\">$P\\le I(\\omega)$</span> if and only if <span class=\"math-container\">$P\\le\\mathbb R$</span>.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508774,
"title": "More conjectural formulas for Riemann's zeta function (II)",
"link": "https://mathoverflow.net/questions/508774/more-conjectural-formulas-for-riemanns-zeta-function-ii",
"score": 0,
"view_count": 43,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "nt.number-theory;sequences-and-series;riemann-zeta-function;binomial-coefficients;combinatorial-identities",
"creation_date": "2026-03-05 09:22:33Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Motivated by Zeilberger's series\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty\\frac{21k-8}{k^3\\binom{2k}k^3}=\\zeta(2)$$</span>\nand Questions <a href=\"https://mathoverflow.net/questions/508743\">508743</a> and <a href=\"https://mathoverflow.net/questions/508768\">508768</a>\nI have discovered several identities involving higher-order derivatives of the function\n<span class=\"math-container\">$$Z(k):=\\frac{(21k-8)\\Gamma(k+1)^6}{k^3\\Gamma(2k+1)^3}.$$</span>\nNamely,\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty Z'(k)=-6\\zeta(3),\\tag{1}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty Z''(k)=\\frac{57}2\\zeta(4),\\tag{2}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty Z^{(3)}(k)=18\\pi^2\\zeta(3)-324\\zeta(4),\\tag{3}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty Z^{(4)}(k)=\\frac{1959}2\\zeta(6)-432\\zeta(3)^2,\\tag{4}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty Z^{(5)}(k)=19440\\zeta(2)\\zeta(5)-540\\zeta(3)\\zeta(4)-31860\\zeta(7),\\tag{5}$$</span>\nand\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty Z^{(6)}(k)=31104\\zeta(5,3)+38880\\zeta(2)\\zeta(3)^2-77760\\zeta(3)\\zeta(5)-\\frac{84807}4\\zeta(8),\\tag{6}$$</span>\nwhere <span class=\"math-container\">$$\\zeta(5,3):=\\sum_{k_1>k_2>0}\\frac1{k_1^5k_2^3}=0.0377\\ldots.$$</span></p>\n<p>Note that <span class=\"math-container\">$(1)$</span> is equivalent to a formula obtained by K. C. Au in 2022.</p>\n<p><strong>QUESTION</strong>. Are the identities <span class=\"math-container\">$(2)-(6)$</span> new? If some of them are new, how to prove them?</p>\n<p>Your comments are welcome!</p>\n</div>",
"comments": [],
"answers": []
},
{
"question_id": 508768,
"title": "More conjectural formulas for Riemann's zeta function",
"link": "https://mathoverflow.net/questions/508768/more-conjectural-formulas-for-riemanns-zeta-function",
"score": 2,
"view_count": 115,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "nt.number-theory;sequences-and-series;riemann-zeta-function;binomial-coefficients;combinatorial-identities",
"creation_date": "2026-03-05 09:21:58Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Motivated by <a href=\"https://mathoverflow.net/questions/508743\">Question 508743</a> and the known identities</p>\n<p><span class=\"math-container\">$$\\sum_{k=1}^\\infty\\frac{(-1)^k}{k^3\\binom{2k}k}=-\\frac25\\zeta(3)\\ \\ \\text{and}\\ \\ \\sum_{k=1}^\\infty\\frac1{k^4\\binom{2k}k}=\\frac{17}{36}\\zeta(4),$$</span>\nI found the following identities involving higher-order derivatives of the functions\n<span class=\"math-container\">$$f(k)=\\frac{e^{\\pi ik}}{k^3\\binom{2k}k}=\\frac{e^{\\pi ik}\\Gamma(k+1)^2}{\\Gamma(2k+1)}\n\\ \\ \\text{and}\\ \\ g(k)=\\frac1{k^4\\binom{2k}k}=\\frac{\\Gamma(k+1)^2}{k^4\\Gamma(2k+1)}.$$</span></p>\n<p><strong>Conjecture 1</strong>. We have\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty f'(k)=\\frac{9}{5}\\zeta(4)-\\frac{2}5\\pi\\zeta(3)i.\\tag{I1}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty f''(k)=\\frac{\\pi^5}{25}i+\\frac{8}{15}\\pi^2\\zeta(3)-\\frac{52}5\\zeta(5).\\tag{I2}$$</span></p>\n<p><strong>Conjecture II</strong>. We have\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty g'(k)=-\\frac{22}9\\zeta(5),\\tag{II1}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty g''(k)=\\frac{3073}{216}\\zeta(6),\\tag{II2}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty g^{(3)}(k)=\\frac{176\\zeta(2)\\zeta(5)-1423\\zeta(7)}{12},\\tag{II3}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty g^{(4)}(k)=\\frac{752537\\zeta(8)+25344\\zeta(5,3)}{1080},\\tag{II4}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty g^{(5)}(k)=660\\zeta(4)\\zeta(5)+\\frac{7115}3\\zeta(2)\\zeta(7)-\\frac{283010}{27}\\zeta(9),\\tag{II5}$$</span>\nwhere\n<span class=\"math-container\">$$\\zeta(5,3):=\\sum_{k_1>k_2>0}\\frac1{k_1^5k_2^3}=0.0377\\ldots.$$</span></p>\n<p><strong>QUESTION</strong>. Are the conjectural identities known? If some of them are new, how to prove them?</p>\n</div>",
"comments": [
{
"comment_id": 1325677,
"author_name": "Zhi-Wei Sun",
"author_id": 124654,
"author_url": "https://mathoverflow.net/users/124654/zhi-wei-sun",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">(II1) is equivalent to (3.15) of my 2015 paper available from <a href=\"http://maths.nju.edu.cn/~zwsun/165s.pdf\" rel=\"nofollow noreferrer\">maths.nju.edu.cn/~zwsun/165s.pdf</a> which was confirmed by J. Ablinger in his 2017 paper available from <a href=\"https://www.tandfonline.com/doi/full/10.1080/10586458.2015.1116028\" rel=\"nofollow noreferrer\">tandfonline.com/doi/full/10.1080/10586458.2015.1116028</a> .</span>",
"created_date": "2026-03-05 02:15:34Z"
},
{
"comment_id": 1325683,
"author_name": "Deyi Chen",
"author_id": 287674,
"author_url": "https://mathoverflow.net/users/287674/deyi-chen",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">$$\\sum_{k=1}^\\infty g^{(4)}(k)=\\frac{352 \\zeta(3) \\zeta(5)}{15} + \\frac{727193}{1080} \\zeta(8) - \\frac{352}{15} \\zeta(3, 5). $$</span>",
"created_date": "2026-03-05 04:02:44Z"
},
{
"comment_id": 1325685,
"author_name": "Deyi Chen",
"author_id": 287674,
"author_url": "https://mathoverflow.net/users/287674/deyi-chen",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">$$\\sum_{k=1}^\\infty g^{(6)}(k)=\\frac{29703837}{560} \\zeta(10) + \\frac{5730 \\zeta(5)^2}{7} + \\frac{14230 \\zeta(7, 3)}{7} - 704 \\zeta(2) \\zeta(5, 3).$$</span>",
"created_date": "2026-03-05 04:21:46Z"
},
{
"comment_id": 1325686,
"author_name": "Zhi-Wei Sun",
"author_id": 124654,
"author_url": "https://mathoverflow.net/users/124654/zhi-wei-sun",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks. I did also find (II4) and have just added it.</span>",
"created_date": "2026-03-05 04:26:59Z"
},
{
"comment_id": 1325698,
"author_name": "eddy ardonne",
"author_id": 58345,
"author_url": "https://mathoverflow.net/users/58345/eddy-ardonne",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Zhi-WeiSun But you forgot to update the lower bound, the summation should read $k_1 > k_2 > 0$.</span>",
"created_date": "2026-03-05 07:29:11Z"
}
],
"answers": []
},
{
"question_id": 508775,
"title": "Characterization of spacelike simplices in $1+n$-dimensional Minkowski",
"link": "https://mathoverflow.net/questions/508775/characterization-of-spacelike-simplices-in-1n-dimensional-minkowski",
"score": 2,
"view_count": 13,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "mg.metric-geometry;convex-geometry;simplicial-stuff;special-relativity;minkowski-space",
"creation_date": "2026-03-05 09:21:22Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Let <span class=\"math-container\">$\\mathbb M^n = \\mathbb R^{1,n-1}$</span> be <span class=\"math-container\">$n$</span>-dimensional Minkowski space and <span class=\"math-container\">$\\eta\\colon \\mathbb M^n \\times \\mathbb M^n \\to \\mathbb R $</span> the corresponding indefinite inner product.</p>\n<p>How do I see for a given <span class=\"math-container\">$n$</span>-simplex <span class=\"math-container\">$\\Delta \\subseteq \\mathbb R^{1,n-1}$</span> whether it is space like. More precisely, let <span class=\"math-container\">$x_1, \\dots, x_n \\in \\mathbb R^{1,n-1}$</span> and let <span class=\"math-container\">$\\Delta :=\\operatorname{Conv}(\\{x_1,\\dots,x_n\\})$</span> be the convex hull. How do I know <span class=\"math-container\">$\\Delta$</span> is spacelike, i.e., <span class=\"math-container\">$\\forall x,y \\in \\Delta\\colon \\eta(x-y, x-y) \\geq 0$</span> (using signature <span class=\"math-container\">$(-+\\dots+)$</span>).</p>\n<p>What I am looking for, is an easy way to see whether this is the case given the points <span class=\"math-container\">$x_1,\\dots, x_n$</span>.</p>\n<p>For <span class=\"math-container\">$n = 2$</span> it is sufficient that <span class=\"math-container\">$\\eta(x_1-x_2, x_1-x_2)\\geq 0$</span>. But for higher dimensions it is not sufficient to show, that the points <span class=\"math-container\">$\\{x_1,\\dots, x_n\\}$</span> are pairwise spacelike. E.g., the points\n<span class=\"math-container\">$$\\begin{pmatrix}\n0\\\\2\\\\0\n\\end{pmatrix},\\begin{pmatrix}\n0\\\\-2\\\\0\n\\end{pmatrix},\n\\begin{pmatrix}\n1\\\\0\\\\0\n\\end{pmatrix}\n\\in \\mathbb R^{1,3}\n$$</span>\nare clearly pairwise spacelike but there convex hull contains a timelike direction.</p>\n</div>",
"comments": [
{
"comment_id": 1325714,
"author_name": "Dmitrii Korshunov",
"author_id": 13842,
"author_url": "https://mathoverflow.net/users/13842/dmitrii-korshunov",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Pick any vertex $x_1$ of your simplex, then it is a problem to decide if the linear space spanned by $x_i-x_1$ is positive definite.</span>",
"created_date": "2026-03-05 10:32:51Z"
}
],
"answers": []
},
{
"question_id": 508621,
"title": "Primes of the form $x^2+ny^2$ when the class group is $C_2C_4$",
"link": "https://mathoverflow.net/questions/508621/primes-of-the-form-x2ny2-when-the-class-group-is-c-2c-4",
"score": 3,
"view_count": 201,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "nt.number-theory;algebraic-number-theory;class-field-theory",
"creation_date": "2026-03-05 07:44:40Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>In my <a href=\"https://mathoverflow.net/questions/507906/primes-of-the-form-p-x2ny2-when-the-class-number-is-8\">previous</a> 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.</p>\n<p>Specificially, I would like to know a proof (or counterexample) to the conjectures:</p>\n<p>Conjecture 1: If <span class=\"math-container\">$q$</span> and <span class=\"math-container\">$r$</span> are <span class=\"math-container\">$5$</span>(mod <span class=\"math-container\">$8$</span>), with <span class=\"math-container\">$q=a^2+b^2$</span> and <span class=\"math-container\">$r=c^2+d^2$</span>, with <span class=\"math-container\">$a+bi$</span> and <span class=\"math-container\">$c+di$</span> primary (<span class=\"math-container\">$b,d>0$</span> are even and <span class=\"math-container\">$a+b$</span> and <span class=\"math-container\">$c+d$</span> are congruent to <span class=\"math-container\">$1$</span> mod <span class=\"math-container\">$4$</span>) and <span class=\"math-container\">$h(-qr)=8$</span>, then <span class=\"math-container\">$p=x^2+qry^2$</span> if and only if <span class=\"math-container\">$q, r,$</span> and <span class=\"math-container\">$(a+bi)(c+di)$</span> are quadratic residues mod <span class=\"math-container\">$p$</span>.</p>\n<p>For an example: <span class=\"math-container\">$q=5$</span> and <span class=\"math-container\">$r=13$</span>. The corresponding primary Gaussian primes are <span class=\"math-container\">$-1+2i$</span> and <span class=\"math-container\">$3+2i$</span>. Multiplying them you get <span class=\"math-container\">$-7+4i$</span>. This is what must be a quadratic residue mod <span class=\"math-container\">$p$</span> (along with <span class=\"math-container\">$q$</span> and <span class=\"math-container\">$r$</span>) so that <span class=\"math-container\">$p=x^2+qry^2$</span></p>\n<p>EDIT: The main thing that concerns me is dealing with the <span class=\"math-container\">$C_4$</span> part of the form class group. The remaining cases can be split with congruence conditions. For example, for <span class=\"math-container\">$n=65$</span>, as Will Jagy pointed out, the main cases to distinguish are <span class=\"math-container\">$x^2+65y^2$</span> and <span class=\"math-container\">$9x^2+8xy+9y^2$</span>. I conjecture that the quadratic residuacity of <span class=\"math-container\">$(a+bi)(c+di)$</span> is enough to distinguish them. This is what I'm asking for a proof of.</p>\n</div>",
"comments": [
{
"comment_id": 1325151,
"author_name": "Will Jagy",
"author_id": 3324,
"author_url": "https://mathoverflow.net/users/3324/will-jagy",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I'm fiddling with $x^2 + 65 y^2.$ What do you say happens? Oh, not cyclic, all have exponent 4.</span>",
"created_date": "2026-02-28 02:54:29Z"
},
{
"comment_id": 1325152,
"author_name": "Thomas Blok",
"author_id": 109590,
"author_url": "https://mathoverflow.net/users/109590/thomas-blok",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">$qr=65 or 145$ were my test cases. I managed to verify for low primes that if $7+4i$ and $5$ are quadratic residues mod $p$, then $p=x^2+65y^2$. Similarly, for $qr=145$, $1+12i$ needs to be a quadratic residue. It's weird, you have to multiply the primary $a+bi$ and $c+di$, otherwise you get the wrong answer (e.g. for $qr=65$, using $8+i$ fails).</span>",
"created_date": "2026-02-28 03:03:37Z"
},
{
"comment_id": 1325154,
"author_name": "Will Jagy",
"author_id": 3324,
"author_url": "https://mathoverflow.net/users/3324/will-jagy",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Alright. The competition (to $x^2 + 65 y^2$) is $9 x^2 + 8xy + 9 y^2.$ The first few such primes are $ 29, 61, 181, 521.$ What happens to your various modular quantities for these?</span>",
"created_date": "2026-02-28 03:15:27Z"
},
{
"comment_id": 1325171,
"author_name": "Emil Jeřábek",
"author_id": 12705,
"author_url": "https://mathoverflow.net/users/12705/emil-je%c5%99%c3%a1bek",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Does $C_2C_4$ mean $C_2\\times C_4$?</span>",
"created_date": "2026-02-28 11:13:49Z"
},
{
"comment_id": 1325227,
"author_name": "Thomas Blok",
"author_id": 109590,
"author_url": "https://mathoverflow.net/users/109590/thomas-blok",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@WillJagy, $5$ and $13$ are quadratic residues modulo $p$ for all those numbers. But $7+4i$ isn't a quadratic residue modulo any of them. In other words, the polynomial $x^4-14x^2+65$ doesn't split completely for the $9x^2+8xy+9y^2$ primes, but does for the $x^2+65y^2$ primes.</span>",
"created_date": "2026-02-28 20:31:55Z"
}
],
"answers": []
},
{
"question_id": 508743,
"title": "Some series related to $\\zeta(3),\\zeta(4),\\zeta(5),\\zeta(6),\\zeta(7)$",
"link": "https://mathoverflow.net/questions/508743/some-series-related-to-zeta3-zeta4-zeta5-zeta6-zeta7",
"score": 11,
"view_count": 356,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "nt.number-theory;sequences-and-series;riemann-zeta-function;combinatorial-identities",
"creation_date": "2026-03-05 07:19:54Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Let\n<span class=\"math-container\">$$f(k)=\\frac{1}{k^2\\binom{2 k}{k}},$$</span>\nand\n<span class=\"math-container\">$$f^{(n)}(k)=\\frac{d^n}{dk^n} f(k).$$</span>\nInspired by <a href=\"https://mathoverflow.net/questions/508718\">Question 508718</a>, I discovered the following identities</p>\n<p><span class=\"math-container\">$$\\sum_{k=1}^\\infty f^{(1)}(k)=-\\frac{4}{3}\\zeta \\! \\left(3\\right),\\tag{1}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty f^{(2)}(k)=\\frac{31}{6}\\zeta(4),\\tag{2}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty f^{(3)}(k)=-38 \\zeta \\! \\left(5\\right)+8 \\zeta \\! \\left(2\\right) \\zeta \\! \\left(3\\right),\\tag{3}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty f^{(4)}(k)=-32 \\zeta \\! \\left(3\\right)^{2}+\\frac{979}{6}\\zeta \\! \\left(6\\right),\\tag{4}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty f^{(5)}(k)=760 \\zeta \\! \\left(2\\right) \\zeta \\! \\left(5\\right)+360 \\zeta \\! \\left(3\\right) \\zeta \\! \\left(4\\right)-2465 \\zeta \\! \\left(7\\right)\n.\\tag{5}$$</span></p>\n<p>Since <span class=\"math-container\">$f^{(1)}(k)=-\\frac{2}{k^{3}{\\binom{2 k}{k}}}-2\\frac{H_{2k}-H_{k}}{k^2\\binom{2k}k}$</span> , we can deduce that <span class=\"math-container\">$(1)$</span> is a direct corollary of <a href=\"https://www.sciencedirect.com/science/article/pii/0022314X85900198\" rel=\"noreferrer\">I. J. Zucker's series (2.9)</a> <span class=\"math-container\">$\\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 $</span> and <a href=\"https://mathoverflow.net/questions/508718\">Question 508718</a>.</p>\n<p>My questions:</p>\n<ol>\n<li>Are the identities (2)-(5) known?</li>\n<li>How can they be proven?</li>\n<li>What is the exact value of <span class=\"math-container\">$\\sum_{k=1}^\\infty f^{(n)}(k)$</span> for <span class=\"math-container\">$n\\geq 6$</span>?</li>\n</ol>\n</div>",
"comments": [
{
"comment_id": 1325658,
"author_name": "Henri Cohen",
"author_id": 81776,
"author_url": "https://mathoverflow.net/users/81776/henri-cohen",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">$$\\sum_{k=1}^\\infty f^{(6)}(k)=(110483/12)\\zeta(8)-5088\\zeta(3)\\zeta(5)+960\\zeta(2)\\zeta(3)^2-672\\zeta(3,5)$$ where $\\zeta(3,5)$ is a multizeta value. I suspect all are linear combinations of MZVs. Of course, I assume as yours, this is numerical.</span>",
"created_date": "2026-03-04 22:11:20Z"
},
{
"comment_id": 1325659,
"author_name": "eddy ardonne",
"author_id": 58345,
"author_url": "https://mathoverflow.net/users/58345/eddy-ardonne",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Henri, do you have some intuition for when a multizeta value is needed? For $n = 7$, this is not the case.</span>",
"created_date": "2026-03-04 22:16:28Z"
},
{
"comment_id": 1325663,
"author_name": "Henri Cohen",
"author_id": 81776,
"author_url": "https://mathoverflow.net/users/81776/henri-cohen",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">This has been thoroughly explored by the experts (not me) who can, at least conjecturally, say how many primitive MZVs must be added.</span>",
"created_date": "2026-03-04 22:57:59Z"
},
{
"comment_id": 1325681,
"author_name": "Deyi Chen",
"author_id": 287674,
"author_url": "https://mathoverflow.net/users/287674/deyi-chen",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">$$\\sum_{k=1}^\\infty f^{(8)}(k)=\\frac{13510461 \\zeta(10)}{10} - 967680 \\zeta(3) \\zeta(7) + 322560 \\zeta(2) \\zeta(3) \\zeta(5) + 120960 \\zeta(4) \\zeta(3)^2 - 618480 \\zeta(5)^2 - 37632 \\zeta(2) \\zeta(5, 3) + 49200 \\zeta(7, 3) $$</span>",
"created_date": "2026-03-05 03:13:49Z"
}
],
"answers": [
{
"answer_id": 508761,
"question_id": 508743,
"score": 5,
"is_accepted": false,
"author_name": "eddy ardonne",
"author_id": 58345,
"author_url": "https://mathoverflow.net/users/58345/eddy-ardonne",
"author_reputation": 673,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-04 22:06:53Z",
"last_edited_date": "2026-03-05 07:19:54Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>This is a partial answer to question 3., that does not fit as a comment. Given the form of the relations for <span class=\"math-container\">$1\\leq n \\leq 5$</span>, it is natural to write the following ansatz for <span class=\"math-container\">$n=6$</span>\n<span class=\"math-container\">$$\n\\sum_{k=1}^\\infty f^{(6)}(k) = c_0 \\zeta(8) + c_1 \\zeta(5)\\zeta(3) + c_2 \\zeta(3)^2 \\zeta(2)\n$$</span>\nand find the appropriate rational <span class=\"math-container\">$c_i$</span>. This problem can be solved by an <a href=\"https://en.wikipedia.org/wiki/Integer_relation_algorithm\" rel=\"nofollow noreferrer\">en.wikipedia.org/wiki/Integer_relation_algorithm</a>, as was suggested to me in this <a href=\"https://mathoverflow.net/questions/409365/finding-elements-in-a-real-extension-of-mathbbq-that-are-close-to-some-numb#comment1050189_409365\">comment</a> to this <a href=\"https://mathoverflow.net/q/409365/58345\">question (409365)</a>.</p>\n<p>I tried this using Mathematica, which implements an integer relation algorithm via LatticeReduce[]. To my surprise, I did not find a solution for the case <span class=\"math-container\">$n=6$</span>, even when using 300 digit precision. However, for <span class=\"math-container\">$n=7$</span>, I found the following relation (for which 100 digit precision suffices):\n<span class=\"math-container\">$$\n\\sum_{k=1}^\\infty f^{(7)}(k) = -\\frac{974470}{3} \\zeta(9) + 103530 \\zeta(7) \\zeta(2) + 33180 \\zeta(6) \\zeta(3) + 71820 \\zeta(5) \\zeta(4) - 4480 \\zeta(3)^3 \\ .\n$$</span></p>\n<p>These results suggest that at least for <span class=\"math-container\">$n=6$</span>, the structure for the sum <span class=\"math-container\">$\\sum_{k=1}^\\infty f^{(n)}(k)$</span> as suggested by the results of the OP is not general enough. I did not (yet) try cases <span class=\"math-container\">$n\\geq 8$</span>, so the general picture is not clear at the moment.</p>\n<p><em>Edit</em> As pointed out by Henri Cohen in <a href=\"https://mathoverflow.net/questions/508743/some-series-related-to-zeta3-zeta4-zeta5-zeta6-zeta7#comment1325658_508743\">this comment</a>, in general one needs the <a href=\"https://en.wikipedia.org/wiki/Multiple_zeta_function\" rel=\"nofollow noreferrer\">en.wikipedia.org/wiki/Multiple_zeta_function</a>.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508468,
"title": "Availability of Medusa algorithm implementations",
"link": "https://mathoverflow.net/questions/508468/availability-of-medusa-algorithm-implementations",
"score": 3,
"view_count": 228,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "complex-dynamics;mathematical-software",
"creation_date": "2026-03-05 06:56:33Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<div>\n<aside class=\"s-notice s-notice__info post-notice js-post-notice mb16 js-bounty-notification\" role=\"status\">\n<div class=\"d-flex fd-column fw-nowrap\">\n<div class=\"d-flex fw-nowrap\">\n<div class=\"flex--item mr8\">\n<svg aria-hidden=\"true\" class=\"svg-icon iconClock\" height=\"18\" viewbox=\"0 0 18 18\" width=\"18\"><path d=\"M9 17c-4.36 0-8-3.64-8-8s3.64-8 8-8 8 3.64 8 8-3.64 8-8 8m0-2c3.27 0 6-2.73 6-6s-2.73-6-6-6-6 2.73-6 6 2.73 6 6 6M8 5h1.01L9 9.36l3.22 2.1-.6.93L8 10z\"></path></svg>\n</div>\n<div class=\"flex--item wmn0 fl1 lh-lg\">\n<div class=\"flex--item fl1 lh-lg\">\n<div>\n<b>The <a href=\"https://mathoverflow.net/help/bounty\">bounty</a> expires <span title=\"2026-03-05 12:13:11Z\">in 1 hour</span></b>. Answers to this question are eligible for a <span class=\"s-badge s-badge__bounty d-inline px4 py2 ba bc-transparent bar-sm fs-caption va-middle\">+50</span> reputation bounty.\n <a dir=\"auto\" href=\"/users/575237/oscarthegrumpygrouch\">OscarTheGrumpyGrouch</a> wants to <b>draw more attention</b> to this question.\n </div>\n</div>\n</div>\n</div>\n</div>\n</aside>\n</div>\n<p>I 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) <a href=\"https://pi.math.cornell.edu/%7Edynamics/Matings/index.html\" rel=\"nofollow noreferrer\">Cornell dynamics webpage</a>.</p>\n<p>The original programs were written in C++ for Unix systems, with a minimal interface, and the source code is available as Medusa.tar.gz.</p>\n<p>Are 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 <strong>modern open-source equivalents for experimenting with polynomial matings</strong> that are compatible with typical desktop environments?</p>\n<p>Thanks</p>\n</div>",
"comments": [
{
"comment_id": 1324993,
"author_name": "Michael Engelhardt",
"author_id": 134299,
"author_url": "https://mathoverflow.net/users/134299/michael-engelhardt",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">macOS is a Unix system.</span>",
"created_date": "2026-02-26 14:59:54Z"
},
{
"comment_id": 1325053,
"author_name": "Peter Taylor",
"author_id": 46140,
"author_url": "https://mathoverflow.net/users/46140/peter-taylor",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The Windows Subsystem for Linux gives a way to execute programs written for *nix on a Windows machine. It's the recommended way to run Sage on a Windows machine nowadays, so the <a href=\"https://doc.sagemath.org/html/en/installation/index.html\" rel=\"nofollow noreferrer\">Sage installation instructions</a> up to but not including the <code>curl</code> 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 <code>make</code>.</span>",
"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": "<span class=\"comment-copy\" itemprop=\"text\">I tried to compile the source code but failed; it's using obsolete C++ headers and memory‑management functions, requiring significant updates.</span>",
"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": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I have uploaded a set of patches at <a href=\"https://github.com/LegionMammal978/medusa-matings\" rel=\"noreferrer\">https://github.com/LegionMammal978/medusa-matings</a> to allow the original source code to be built on modern systems. I have tested this workflow on Windows 11:</p>\n<ol>\n<li>Download and install <a href=\"https://www.msys2.org/\" rel=\"noreferrer\">MSYS2</a>, and open the UCRT64 shell.</li>\n<li>Install the necessary build dependencies with <code>pacman -S git make mingw-w64-ucrt-x86_64-gcc patch</code>.</li>\n<li>Download and build the tools with these commands:\n<pre><code>git clone https://github.com/LegionMammal978/medusa-matings.git\ncd medusa-matings\n./download.sh\ncd Medusa\nmake all\n</code></pre>\n</li>\n<li>Use the tools via <code>./mate_interact</code>, <code>./draw_pb_sphere</code>, <code>./drawpullback</code>, or <code>./find_matings</code> according to their documentation.</li>\n<li>To view the <code>.eps</code> files generated by <code>draw_pb_sphere</code> and <code>drawpullback</code>, you can install Ghostscript with <code>pacman -S mingw-w64-ucrt-x86_64-ghostscript</code> and convert to PDF with <code>ps2pdf image.eps image.pdf</code>. Alternatively, you can use any other PostScript tool available for Windows.</li>\n</ol>\n<p>Using this, I have been able to reproduce the five example pictures available from <a href=\"https://pi.math.cornell.edu/%7Edynamics/Matings/Pics/index.html\" rel=\"noreferrer\">https://pi.math.cornell.edu/~dynamics/Matings/Pics/index.html</a>, the first one nearly exactly and the other four exactly. Here are the parameters I deduced:</p>\n<ul>\n<li>\"1/255 Mate 1/255\": 1/255, 1/255, 400 iterations with <code>mate_interact</code>; 144 dpi with <code>draw_pb_sphere</code>.</li>\n<li>\"1/7 Mate 10/63\": 1/7, 10/63, 50 iterations with <code>mate_interact</code>; 144 dpi with <code>draw_pb_sphere</code>.</li>\n<li>\"1/5 Mate 1/5\": 1/5, 1/5, 38 iterations with <code>mate_interact</code>; 144 dpi with <code>draw_pb_sphere</code>.</li>\n<li>\"1/6 Mate 5/14\": 1/6, 5/14, 15 iterations with <code>mate_interact</code>; 144 dpi with <code>draw_pb_sphere</code>.</li>\n<li>\"1/4 Mate 1/6\": 1/4, 1/6, 16 iterations with <code>mate_interact</code>; 144 dpi with <code>draw_pb_sphere</code>.</li>\n</ul>\n</div>",
"comments": [
{
"comment_id": 1325694,
"author_name": "Carlo Beenakker",
"author_id": 11260,
"author_url": "https://mathoverflow.net/users/11260/carlo-beenakker",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">thank you for this service to the community!</span>",
"created_date": "2026-03-05 07:07:44Z"
}
]
}
]
},
{
"question_id": 508771,
"title": "Quandle and homeomorphism of spaces",
"link": "https://mathoverflow.net/questions/508771/quandle-and-homeomorphism-of-spaces",
"score": 0,
"view_count": 20,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "quandles",
"creation_date": "2026-03-05 06:23:58Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Does a homeomorphism between pairs of spaces induce an isomorphism of the fundamental quandle?</p>\n</div>",
"comments": [
{
"comment_id": 1325692,
"author_name": "Jim Conant",
"author_id": 9417,
"author_url": "https://mathoverflow.net/users/9417/jim-conant",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">AFAIK, the fundamental quandle is not defined for an arbitrary space, just for knot complements.</span>",
"created_date": "2026-03-05 06:33:51Z"
},
{
"comment_id": 1325693,
"author_name": "good bye",
"author_id": 566232,
"author_url": "https://mathoverflow.net/users/566232/good-bye",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Sorry, here I assume that the pair of spaces satisfies the conditions required to define the fundamental quandle.</span>",
"created_date": "2026-03-05 06:36:46Z"
},
{
"comment_id": 1325701,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Do you mean as opposed to being isotopic knot complements? If the fundamental quandle of a knot complement is a functor on the full subcategory of Top on the spaces where they are defined, then the answer is yes.</span>",
"created_date": "2026-03-05 08:24:46Z"
},
{
"comment_id": 1325703,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I'm not sure if Cattabriga, A., Horvat, E. Knot Quandle Decompositions. Mediterr. J. Math. 17, 98 (2020). <a href=\"https://doi.org/10.1007/s00009-020-01530-6\" rel=\"nofollow noreferrer\">doi.org/10.1007/s00009-020-01530-6</a> helps, but maybe. The abstract starts <i>We show that the fundamental quandle defines a functor from the oriented tangle category to a suitably defined quandle category.</i></span>",
"created_date": "2026-03-05 08:27:01Z"
}
],
"answers": []
},
{
"question_id": 508707,
"title": "Existence of limit intersection in an $\\omega_1$-chain of almost decreasing subsets of $\\omega$",
"link": "https://mathoverflow.net/questions/508707/existence-of-limit-intersection-in-an-omega-1-chain-of-almost-decreasing-subs",
"score": 10,
"view_count": 216,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": true,
"accepted_answer_id": 508741,
"tags": "set-theory;infinite-combinatorics",
"creation_date": "2026-03-04 21:18:35Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Is it consistent that there exists a <span class=\"math-container\">$\\subseteq^*$</span>-decreasing chain <span class=\"math-container\">$(X_\\alpha)_{\\alpha<\\omega_1}$</span> of infinite subsets of <span class=\"math-container\">$\\omega$</span> such that for all ordinals <span class=\"math-container\">$\\alpha_0 < \\alpha_1 < \\cdots < \\alpha_\\omega$</span> with <span class=\"math-container\">$\\alpha_\\omega = \\sup_{n<\\omega} \\alpha_n$</span>, <span class=\"math-container\">$\\bigcap_{\\gamma \\leq \\omega} X_{\\alpha_\\gamma}$</span> is finite?</p>\n<p>Note that if <span class=\"math-container\">$(X_\\alpha)_{\\alpha<\\omega_1}$</span> has a pseudo-intersection (i.e. some infinite <span class=\"math-container\">$Y \\subseteq \\omega$</span> such that <span class=\"math-container\">$Y \\subseteq^* X_\\alpha$</span> for all <span class=\"math-container\">$\\alpha$</span>), then this chain doesn't have such a sequence of ordinals. Therefore, <span class=\"math-container\">$\\omega_1 < \\mathfrak{t}$</span> implies the non-existence of such a chain.</p>\n</div>",
"comments": [
{
"comment_id": 1325511,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">What is $\\subseteq^*$?</span>",
"created_date": "2026-03-03 14:06:43Z"
},
{
"comment_id": 1325512,
"author_name": "Clement Yung",
"author_id": 146831,
"author_url": "https://mathoverflow.net/users/146831/clement-yung",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@Wojowu Almost containment. $X \\subseteq^* Y$ if $X \\setminus Y$ is finite.</span>",
"created_date": "2026-03-03 14:07:48Z"
},
{
"comment_id": 1325518,
"author_name": "Will Brian",
"author_id": 70618,
"author_url": "https://mathoverflow.net/users/70618/will-brian",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">If you had said $\\gamma < \\omega$ rather than $\\gamma \\leq \\omega$ in the intersection at the end, then I would be able to prove the answer is no. I'm guessing that maybe you know this already, and that's why you phrased the question exactly as you did.</span>",
"created_date": "2026-03-03 15:23:11Z"
}
],
"answers": [
{
"answer_id": 508741,
"question_id": 508707,
"score": 8,
"is_accepted": true,
"author_name": "Jonathan Schilhan",
"author_id": 103802,
"author_url": "https://mathoverflow.net/users/103802/jonathan-schilhan",
"author_reputation": 483,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-04 11:34:30Z",
"last_edited_date": "2026-03-04 21:18:35Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>No.</p>\n<p>If given <span class=\"math-container\">$(X_\\alpha)_{\\alpha < \\omega_1}$</span>, then there is a ccc forcing <span class=\"math-container\">$\\mathbb{P}$</span> that adds a pseudointersection <span class=\"math-container\">$Y$</span>. In the forcing extension, the sets <span class=\"math-container\">$A_n = \\{ \\alpha < \\omega_1 : Y \\setminus X_\\alpha \\subseteq n \\}$</span> form a countable cover of <span class=\"math-container\">$\\omega_1$</span> (which is still the real <span class=\"math-container\">$\\omega_1$</span>), so one of them is stationary. Say <span class=\"math-container\">$A_m$</span> is stationary. This means that there is <span class=\"math-container\">$\\alpha \\in A_m \\cap \\lim A_m$</span> and in particular there is an increasing sequence <span class=\"math-container\">$\\langle \\alpha_n : n \\leq \\omega \\rangle$</span> completely contained within <span class=\"math-container\">$A_m$</span> with <span class=\"math-container\">$\\alpha = \\alpha_\\omega = \\sup \\alpha_n$</span> and <span class=\"math-container\">$\\bigcap_{n \\leq \\omega} X_{\\alpha_n}$</span> is infinite, as it contains <span class=\"math-container\">$Y \\setminus m$</span>.</p>\n<p>Now 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 <span class=\"math-container\">$(p_n)$</span> in <span class=\"math-container\">$\\mathbb{P}$</span> deciding more and more of the <span class=\"math-container\">$\\alpha_n$</span>'s, ensuring these become cofinal in what <span class=\"math-container\">$\\alpha_\\omega < \\omega_1$</span> has been decided to be, and deciding bigger and bigger naturals that should be in the intersection of these <span class=\"math-container\">$X_{\\alpha_n}$</span>, <span class=\"math-container\">$n \\leq \\omega$</span>.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 507483,
"title": "Proving bounds for function set with similarities to Legendre polynomials",
"link": "https://mathoverflow.net/questions/507483/proving-bounds-for-function-set-with-similarities-to-legendre-polynomials",
"score": 2,
"view_count": 210,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "real-analysis;ca.classical-analysis-and-odes;lower-bounds;legendre-polynomials;upper-bounds",
"creation_date": "2026-03-05 05:53:08Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<div>\n<aside class=\"s-notice s-notice__info post-notice js-post-notice mb16 js-bounty-notification\" role=\"status\">\n<div class=\"d-flex fd-column fw-nowrap\">\n<div class=\"d-flex fw-nowrap\">\n<div class=\"flex--item mr8\">\n<svg aria-hidden=\"true\" class=\"svg-icon iconClock\" height=\"18\" viewbox=\"0 0 18 18\" width=\"18\"><path d=\"M9 17c-4.36 0-8-3.64-8-8s3.64-8 8-8 8 3.64 8 8-3.64 8-8 8m0-2c3.27 0 6-2.73 6-6s-2.73-6-6-6-6 2.73-6 6 2.73 6 6 6M8 5h1.01L9 9.36l3.22 2.1-.6.93L8 10z\"></path></svg>\n</div>\n<div class=\"flex--item wmn0 fl1 lh-lg\">\n<div class=\"flex--item fl1 lh-lg\">\n<div>\n<b>The <a href=\"https://mathoverflow.net/help/bounty\">bounty</a> expires <span title=\"2026-03-12 05:53:08Z\">in 7 days</span></b>. Answers to this question are eligible for a <span class=\"s-badge s-badge__bounty d-inline px4 py2 ba bc-transparent bar-sm fs-caption va-middle\">+100</span> reputation bounty.\n <a dir=\"auto\" href=\"/users/112616/f-jatpil\">F. Jatpil</a> wants to <b>draw more attention</b> to this question.\n </div>\n</div>\n</div>\n</div>\n</div>\n</aside>\n</div>\n<p>I search to prove (<span class=\"math-container\">$n\\in \\mathbb{N}_0$</span>)</p>\n<p><span class=\"math-container\">$$ |f_n(x)| \\le 1 \\text{ for } x\\in[0,1]$$</span></p>\n<p>where</p>\n<p><span class=\"math-container\">$$\nf_n(x) = 1 + \\sum_{j=1}^n k_{n,j}x^{1/j}\n\\quad\n\\text{ with }\n\\quad\nk_{n,j}=(-1)^{n+j+1} \\binom{n}{j} \\binom{n+j}{j}.$$</span></p>\n<p>I would be happy with any <span class=\"math-container\">$n$</span>-independent bound. There is a strong numerical evidence the bounding is true. Two remarks:</p>\n<ol>\n<li>Functions <span class=\"math-container\">$f_n$</span> fulfill <span class=\"math-container\">$\\int_0^1 f_n(x^{m+1})dx = 0$</span> for <span class=\"math-container\">$m<n$</span> (<span class=\"math-container\">$m,n\\in \\mathbb{N}_0)$</span> of which I have a proof.</li>\n<li>The coefficients are those of shifted Legendre polynomials.</li>\n</ol>\n<p>The 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).</p>\n<p>Update:</p>\n<p>First, I want to lower my expectations, I would by happy with any bound that rises slower than a power, e.g. logarithmic <span class=\"math-container\">$|f_m (x)|<K\\log(m)$</span>, but valid for all <span class=\"math-container\">$x$</span> from the closed interval <span class=\"math-container\">$[0,1]$</span>. Quickly one realizes the problem comes from the neighborhood of <span class=\"math-container\">$x=0$</span> ,where oscillations accumulate.</p>\n<p>Here are some possibly useful equations (established by AI, checked by me):</p>\n<p><span class=\"math-container\">$$\nf_{n}(x)-f_{n-1}(x)=nx\\left[f'_{n}(x)+f'_{n-1}(x)\\right]\\text{ for }n\\ge2. \n$$</span>\n<span class=\"math-container\">$$\nxf'_{n}(x)=(-1)^{n}+\\frac{1}{n}f_{n}(x)+\\sum_{k=1}^{n-1}(-1)^{n-k}\\frac{2k+1}{k(k+1)}f_{k}(x)\n$$</span>\n<span class=\"math-container\">$$\nf_{n}(x)=-f_{n-1}(x)+\\frac{2}{n}\\,x^{1/n}\\int_{x}^{1}t^{-1-1/n}\\,f_{n-1}(t)\\,dt\n$$</span>\n<span class=\"math-container\">$$\n\\int_{0}^{1}f_{n}^{2}(x)dx+\\frac{n}{2}\\int_{0}^{1}\\bigl(f_{n}(x)+f_{n-1}(x)\\bigr)^{2}dx=\\int_{0}^{1}f_{n-1}^{2}(x)dx\n$$</span>\n<span class=\"math-container\">$$\n\\int_{0}^{1}f_{n}^{2}(x)dx\\le\\int_{0}^{1}f_{1}^{2}(x)dx=\\frac{1}{3}.\n$$</span>\n<span class=\"math-container\">$$\n\\left|f_{n}(x)\\right|\\le C_{\\varepsilon}(1+\\ln n) \\text{ for } x\\in[\\varepsilon,1]\n$$</span></p>\n<p>Update 2:</p>\n<p>The functions <span class=\"math-container\">$(-1)^nf_n(x^\\alpha)$</span> with some appropriate <span class=\"math-container\">$\\alpha$</span> (denoted as <span class=\"math-container\">$R_m$</span> and <span class=\"math-container\">$r_m$</span> 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:</p>\n<pre><code>kill(all);\nL(n,x) := legendre_p(n,2*x-1);\nk[n,j] := (-1)^(j+n+1)*binomial(n+j,j)*binomial(n,j)<span class=\"math-container\">$\nR(n,x) := 1 + sum(k[n,j]*x^(1/j), j, 1, n)$</span>\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]);\n</code></pre>\n<p>And here are pictures</p>\n<p><a href=\"https://i.sstatic.net/65SI6RQB.png\" rel=\"nofollow noreferrer\"><img alt=\"<span class=\" math-container\"=\"\" src=\"https://i.sstatic.net/65SI6RQB.png\"/>$r_5$ and <span class=\"math-container\">$L_5$</span>\" /></a></p>\n<p><a href=\"https://i.sstatic.net/Ix7RiuEW.png\" rel=\"nofollow noreferrer\"><img alt=\"<span class=\" math-container\"=\"\" src=\"https://i.sstatic.net/Ix7RiuEW.png\"/>$r_6$ and <span class=\"math-container\">$L_6$</span>\" /></a></p>\n</div>",
"comments": [
{
"comment_id": 1321797,
"author_name": "mathworker21",
"author_id": 129185,
"author_url": "https://mathoverflow.net/users/129185/mathworker21",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">you doubt which proof is feasible?</span>",
"created_date": "2026-01-27 09:19:53Z"
},
{
"comment_id": 1321798,
"author_name": "F. Jatpil",
"author_id": 112616,
"author_url": "https://mathoverflow.net/users/112616/f-jatpil",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">A doubt technical algebraic proof (re-arranging sum, factorials, factorizing terms, ect..) is feasible. Looking for a similar proof for Legendre polynomials I could not find.</span>",
"created_date": "2026-01-27 09:25:06Z"
},
{
"comment_id": 1322667,
"author_name": "Carlo Beenakker",
"author_id": 11260,
"author_url": "https://mathoverflow.net/users/11260/carlo-beenakker",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">note that $f_n(1)=(-1)^n$, so it is sufficient to prove that $|f_n(x)|\\leq |f_n(1)|$ for $x\\in[0,1]$.</span>",
"created_date": "2026-02-04 09:15:45Z"
},
{
"comment_id": 1323409,
"author_name": "Alex Ravsky",
"author_id": 43954,
"author_url": "https://mathoverflow.net/users/43954/alex-ravsky",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\"><a href=\"https://math.stackexchange.com/questions/5122014/proving-bounds-for-function-set-with-similarities-to-legendre-polynomials\">Crossposted</a> at Mathematics.SE.</span>",
"created_date": "2026-02-11 21:34:25Z"
}
],
"answers": []
},
{
"question_id": 508770,
"title": "Classification of symmetric bilinear forms over Laurent polynomial ring under congruence and Schur complement",
"link": "https://mathoverflow.net/questions/508770/classification-of-symmetric-bilinear-forms-over-laurent-polynomial-ring-under-co",
"score": 1,
"view_count": 21,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "ac.commutative-algebra;polynomials;finite-fields;quadratic-forms;schur-complement",
"creation_date": "2026-03-05 04:46:20Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Consider the ring <span class=\"math-container\">$R$</span> of Laurent polynomials in <span class=\"math-container\">$n$</span> variables over the finite field of two elements, <span class=\"math-container\">$R= \\mathbb F_2[x_0,x_0^{-1},x_1,x_1^{-1},\\ldots]$</span>, with the standard involution by inverting monomials. Concretely, I'm interested in a small number of variables, like <span class=\"math-container\">$n\\in\\{1,2,3,4\\}$</span>. Consider \"hermitian\" <span class=\"math-container\">$R$</span>-valued matrices <span class=\"math-container\">$M$</span>, i.e., <span class=\"math-container\">$M=M^\\dagger$</span> where <span class=\"math-container\">$\\dagger$</span> denotes both transposition and ring involution. I'm interested in the equivalence classes of such matrices under both of the following two equivalences:</p>\n<p>(1) Congruence, that is, <span class=\"math-container\">$M\\sim A^\\dagger MA$</span> for some invertible <span class=\"math-container\">$R$</span>-valued matrix <span class=\"math-container\">$A$</span>.</p>\n<p>(2) Schur complement, that is, if <span class=\"math-container\">$M$</span> has block form,\n<span class=\"math-container\">$$M=\\begin{pmatrix}X&Y\\\\Y^\\dagger&Z\\end{pmatrix}\\;,$$</span>\nand <span class=\"math-container\">$Z$</span> has an inverse <span class=\"math-container\">$Z^{-1}$</span>, then <span class=\"math-container\">$M\\sim X-YZ^{-1}Y^\\dagger$</span>.\nI believe it may be sufficient to consider this equivalence for <span class=\"math-container\">$Y=1$</span> and <span class=\"math-container\">$Y=\\begin{pmatrix}0&1\\\\1&0\\end{pmatrix}$</span>.</p>\n<p><strong>Question</strong>: 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 <span class=\"math-container\">$R$</span>) 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?</p>\n<p>I 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 <span class=\"math-container\">$n\\leq 3$</span>). On the other hand, this also suggests that the classification is not too simple.</p>\n</div>",
"comments": [],
"answers": []
},
{
"question_id": 508769,
"title": "Conjectural series for $L(s,\\genfrac(){}{}{-3}\\cdot)$",
"link": "https://mathoverflow.net/questions/508769/conjectural-series-for-ls-genfrac-3-cdot",
"score": 0,
"view_count": 49,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "nt.number-theory;sequences-and-series;binomial-coefficients;l-functions;combinatorial-identities",
"creation_date": "2026-03-05 03:39:47Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><span class=\"math-container\">$\\newcommand\\Ksymb{\\genfrac(){}{}}$</span>For <span class=\"math-container\">$s=1,2,3,\\dotsc$</span>, let us define\n<span class=\"math-container\">$$L_{-3}(s):=L\\left(s,\\Ksymb{-3}\\cdot\\right)=\\sum_{k=1}^\\infty\\frac{\\Ksymb{-3}k}{k^s}=\\sum_{n=0}^\\infty\\left(\\frac1{(3n+1)^s}-\\frac1{(3n+2)^s}\\right),$$</span>\nwhere <span class=\"math-container\">$\\Ksymb{-3}k$</span> is the Kronecker symbol which coincides with the Legendre symbol <span class=\"math-container\">$\\genfrac(){}{}k3$</span>.</p>\n<p>It is well known that <span class=\"math-container\">$$\\sum_{k=1}^\\infty\\frac1{k\\binom{2k}k}=\\frac{\\pi}{3\\sqrt3}=L_{-3}(1).\\tag{0}\\label{508769_0}$$</span>\nMotivated by this and Questions <a href=\"https://mathoverflow.net/questions/508743\" title=\"Some series related to \\$\\zeta(3),\\zeta(4),\\zeta(5),\\zeta(6),\\zeta(7)\\$\">508743</a> and <a href=\"https://mathoverflow.net/questions/508768\" title=\"More conjectural formulas for Riemann's zeta function\">508768</a>, I make the following conjecture involving higher-order derivatives of the function\n<span class=\"math-container\">$$h(k)=\\frac1{k\\binom{2k}k}=\\frac{\\Gamma(k+1)^2}{k\\Gamma(2k+1)}.$$</span></p>\n<p><strong>Conjecture</strong>. We have\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty h'(k)=-\\frac32L_{-3}(2),\\tag{1}\\label{508769_1}$$</span>\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty h''(k)=3L_{-3}(3)=\\frac{4\\pi^3}{27\\sqrt3},\\tag2\\label{508769_2}$$</span>\nand\n<span class=\"math-container\">$$\\sum_{k=1}^\\infty h'''(k)=\\frac34\\left(2\\pi^2L_{-3}(2)-27L_{-3}(4)\\right).\\tag{3}\\label{508769_3}$$</span></p>\n<p><strong>QUESTION</strong>. Are the three conjectural identities known? If some of them are new, how to prove them?</p>\n<p>Your comments are welcome!</p>\n</div>",
"comments": [],
"answers": []
},
{
"question_id": 508739,
"title": "Numbers between twin primes",
"link": "https://mathoverflow.net/questions/508739/numbers-between-twin-primes",
"score": 6,
"view_count": 475,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "nt.number-theory;prime-numbers",
"creation_date": "2026-03-05 02:25:33Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I 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. <span class=\"math-container\">$\\ n\\ $</span> such that <span class=\"math-container\">$\\ n\\pm1\\ $</span> are both prime), are abundant?</p>\n</div>",
"comments": [
{
"comment_id": 1325608,
"author_name": "Wojowu",
"author_id": 30186,
"author_url": "https://mathoverflow.net/users/30186/wojowu",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Primes themselves are not abundant, so it seems the question is about the density of abundant numbers. As stated on <a href=\"https://en.wikipedia.org/wiki/Abundant_number\" rel=\"nofollow noreferrer\">Wikipedia</a>, abundant numbers have a natural density, 0.247. I don't understand how this question is really about \"numbers between (twin) primes\", so let me know if this answers it.</span>",
"created_date": "2026-03-04 11:13:09Z"
},
{
"comment_id": 1325611,
"author_name": "Gerry Myerson",
"author_id": 3684,
"author_url": "https://mathoverflow.net/users/3684/gerry-myerson",
"score": 12,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The number between twin primes is always a multiple of six, so (except for six itself) is always abundant.</span>",
"created_date": "2026-03-04 11:41:17Z"
},
{
"comment_id": 1325616,
"author_name": "Timothy Chow",
"author_id": 3106,
"author_url": "https://mathoverflow.net/users/3106/timothy-chow",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@GerryMyerson Why not add a couple of extra sentences of explanation, and post your comment as an answer?</span>",
"created_date": "2026-03-04 12:55:03Z"
},
{
"comment_id": 1325619,
"author_name": "Euro Vidal Sampaio",
"author_id": 547394,
"author_url": "https://mathoverflow.net/users/547394/euro-vidal-sampaio",
"score": 9,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Happy birthday in advance, Recaman.</span>",
"created_date": "2026-03-04 13:35:59Z"
},
{
"comment_id": 1325646,
"author_name": "Daniel Asimov",
"author_id": 5484,
"author_url": "https://mathoverflow.net/users/5484/daniel-asimov",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I would like the question to explain exactly what the phrase \"numbers between primes\" means, and why it omits the word \"twin\".</span>",
"created_date": "2026-03-04 17:50:28Z"
}
],
"answers": [
{
"answer_id": 508748,
"question_id": 508739,
"score": 6,
"is_accepted": false,
"author_name": "Elias Panholzer",
"author_id": 586494,
"author_url": "https://mathoverflow.net/users/586494/elias-panholzer",
"author_reputation": 140,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-04 15:07:37Z",
"last_edited_date": "2026-03-04 17:19:07Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Every multiple of <span class=\"math-container\">$6$</span> greater than <span class=\"math-container\">$6$</span> is abundant. Every twin prime pair greater than <span class=\"math-container\">$(3, 5)$</span> is of the form <span class=\"math-container\">$(6n-1, 6n+1)$</span>. So every number between a prime pair except <span class=\"math-container\">$4$</span> and <span class=\"math-container\">$6$</span> is abundant.</p>\n</div>",
"comments": [
{
"comment_id": 1325633,
"author_name": "LSpice",
"author_id": 2383,
"author_url": "https://mathoverflow.net/users/2383/lspice",
"score": 4,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Also mentioned by @GerryMyerson in the <a href=\"https://mathoverflow.net/posts/comments/1325611\">comments</a>.</span>",
"created_date": "2026-03-04 16:01:16Z"
},
{
"comment_id": 1325682,
"author_name": "Wlod AA",
"author_id": 110389,
"author_url": "https://mathoverflow.net/users/110389/wlod-aa",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">------- C'mon ... --------</span>",
"created_date": "2026-03-05 03:22:09Z"
}
]
}
]
},
{
"question_id": 416763,
"title": "Intersection of the kernel with the interpolation space",
"link": "https://mathoverflow.net/questions/416763/intersection-of-the-kernel-with-the-interpolation-space",
"score": 1,
"view_count": 414,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "fa.functional-analysis;sobolev-spaces;interpolation-spaces",
"creation_date": "2026-03-05 02:03:35Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><span class=\"math-container\">$\\DeclareMathOperator\\Ker{Ker}$</span>Given two Banach spaces <span class=\"math-container\">$X$</span> and <span class=\"math-container\">$Y$</span> with a continuous inclusion <span class=\"math-container\">$X\\subset Y$</span>, and another couple <span class=\"math-container\">$X’ \\subset Y’$</span> with the same properties. Take <span class=\"math-container\">$f : Y \\longrightarrow Y’$</span> linear continuous, such that <span class=\"math-container\">$f_{\\mid X}$</span> induces a linear continuous map from <span class=\"math-container\">$X$</span> to <span class=\"math-container\">$X'$</span>. My question is for <span class=\"math-container\">$0<s<1$</span> and <span class=\"math-container\">$p>1$</span>, is the following true:</p>\n<p><span class=\"math-container\">$$(X\\cap \\Ker(f),\\Ker(f))_{s,p}=\\Ker(f) \\cap (X,Y)_{s,p}\\;\\;?$$</span></p>\n<p>Here I'm considering the K-method for the interpolation.</p>\n<p>The inclusion <span class=\"math-container\">$(X\\cap \\Ker(f),\\Ker(f))_{s,p}\\subset \\Ker(f) \\cap (X,Y)_{s,p}$</span> follows directly from the definition. My problem is the other inclusion.</p>\n<p>It's clear that if we have <span class=\"math-container\">$Z\\subset Y$</span> then in general the following is not true:</p>\n<p><span class=\"math-container\">$$(X\\cap Z,Z)_{s,p}=Z \\cap (X,Y)_{s,p},$$</span></p>\n<p>one can take <span class=\"math-container\">$X=H^{2}(U)$</span>, <span class=\"math-container\">$Z=H^{1}(U)$</span>, <span class=\"math-container\">$Y=L^{2}(U)$</span>, <span class=\"math-container\">$s=\\frac{1}{2}$</span>, and <span class=\"math-container\">$p=2$</span>. But this does not contradict our case (because <span class=\"math-container\">$Z$</span> here is not even closed in <span class=\"math-container\">$Y$</span>).</p>\n<p>If what I'm asking is not true in general, is it true under the following assumptions:</p>\n<ul>\n<li><p><span class=\"math-container\">$f(X)$</span> is closed in <span class=\"math-container\">$X'$</span>, and <span class=\"math-container\">$f(Y)$</span> is closed in <span class=\"math-container\">$Y'$</span>, and</p>\n</li>\n<li><p><span class=\"math-container\">$f$</span> is open onto <span class=\"math-container\">$f(Y)$</span>, and <span class=\"math-container\">$f_{\\mid X}$</span> is open onto <span class=\"math-container\">$f(X)$</span>.</p>\n</li>\n</ul>\n</div>",
"comments": [
{
"comment_id": 1069517,
"author_name": "Willie Wong",
"author_id": 3948,
"author_url": "https://mathoverflow.net/users/3948/willie-wong",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Is your counterexample really a counterexample? What norms are you putting on $X\\cap Z$ and on $Z$? In some sense the \"correct\" norm should be the $H^2$ norm on the former (consider $X\\cap Z$ as a subspace of $X$) and the $L^2$ norm on the latter (considering $Z$ as a subspace of $Y$). By thinking of $Z = H^1(U)$ with its own norm, you are introduce additional data that is not available in the $\\ker$ case.</span>",
"created_date": "2022-02-22 22:07:48Z"
},
{
"comment_id": 1069549,
"author_name": "M.Oud",
"author_id": 477563,
"author_url": "https://mathoverflow.net/users/477563/m-oud",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Of course just saying $Z \\subset Y$ , isn't enough, we assume everything fits the interpolation functor. And here we mean that $Z$ is a Banach space with contnuous inclusion in $Y$, and $X \\cap Z $ is given the max norm so that we have a continuous inclusion of $X \\cap Z$ in $Z$. In the exemple, this coïncide with the standard sobolev norms. Is it clear now?</span>",
"created_date": "2022-02-23 02:04:47Z"
},
{
"comment_id": 1069553,
"author_name": "Willie Wong",
"author_id": 3948,
"author_url": "https://mathoverflow.net/users/3948/willie-wong",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">You miss my point entirely, so let me ask a different way: in your analogy, what is the norm on $\\ker f$? My guess is it is the induced norm as a closed subspace of $Y$. But then your \"counterexample\" is different in that your $Z$ has a finer topology.</span>",
"created_date": "2022-02-23 03:22:14Z"
}
],
"answers": [
{
"answer_id": 416820,
"question_id": 416763,
"score": 0,
"is_accepted": false,
"author_name": "Willie Wong",
"author_id": null,
"author_url": "https://mathoverflow.net/users/3948",
"author_reputation": 42242,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2022-02-23 17:14:50Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>(Making CW as this is an extended comment.)</p>\n<p>Let's generalize slightly<sup>1</sup>: let <span class=\"math-container\">$X\\hookrightarrow Y$</span> be an embedding of Banach spaces, and let <span class=\"math-container\">$S\\subset Y$</span> be a closed subspace with the induced norm.</p>\n<p>Your question essentially boils down to comparing the <span class=\"math-container\">$K$</span> functionals <span class=\"math-container\">$K$</span> and <span class=\"math-container\">$K^{(S)}$</span>, where\n<span class=\"math-container\">$$ K(t,z) = \\inf_{z = x + y} \\|x\\|_X + \\|y\\|_Y $$</span>\nand <span class=\"math-container\">$K^{(S)}$</span> is analogously defined, except that instead of allowing arbitrary decompositions <span class=\"math-container\">$z = x+y$</span> with <span class=\"math-container\">$x\\in X$</span> and <span class=\"math-container\">$y\\in Y$</span>, you are looking at decompositions with <span class=\"math-container\">$x \\in X\\cap S$</span> and <span class=\"math-container\">$y\\in S$</span>. (Obviously here <span class=\"math-container\">$z\\in S$</span> as well.)</p>\n<p>Clearly <span class=\"math-container\">$K(t,z) \\leq K^{(S)}(t,z)$</span> for <span class=\"math-container\">$z\\in S$</span>.</p>\n<p>Your question will have a positive answer if <span class=\"math-container\">$K(t,z) \\gtrsim K^{(S)}(t,z)$</span> for some implicit constant independent of <span class=\"math-container\">$t$</span> and <span class=\"math-container\">$z$</span>.</p>\n<p>(This is why I objected to your \"<span class=\"math-container\">$H^1$</span> counterexample; the <span class=\"math-container\">$K$</span>-functional for the <span class=\"math-container\">$(X\\cap Z,Z)$</span> interpolation has almost nothing to do with the functional for the <span class=\"math-container\">$(X,Y)$</span> interpolation in that case. Compared to this case where it is essentially a geometry question about how <span class=\"math-container\">$S$</span> is situated in <span class=\"math-container\">$X$</span> and in <span class=\"math-container\">$Y$</span>.)</p>\n<p>A sufficient condition is therefore that there exists a projection <span class=\"math-container\">$\\Pi:Y\\to S$</span> with the property that <span class=\"math-container\">$\\Pi(X) \\subseteq X\\cap S$</span> and the operator norms <span class=\"math-container\">$\\|\\Pi\\|_{Y\\to S}$</span> and <span class=\"math-container\">$\\|\\Pi\\|_{X\\to X\\cap S}$</span> are both bounded.</p>\n<p>(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.<sup>2</sup>)</p>\n<p>This would be the case if, for example, <span class=\"math-container\">$X, Y$</span> were Hilbert, and <span class=\"math-container\">$S$</span> is such that whenever <span class=\"math-container\">$x\\in X$</span> is orthogonal to <span class=\"math-container\">$S$</span> w.r.t. the <span class=\"math-container\">$X$</span> inner product, it is also orthogonal w.r.t. the <span class=\"math-container\">$Y$</span> inner product.</p>\n<hr/>\n<p>As Hannes noted in a comment below: the result alluded to above is given as Theorem 1 in Section 1.17.1 of <a href=\"https://www.sciencedirect.com/bookseries/north-holland-mathematical-library/vol/18\" rel=\"nofollow noreferrer\">Interpolation Theory, Function Spaces, Differential Operators</a> (ed. Hans Triebel, North-Holland (1978)).</p>\n<hr/>\n<p><sup>1</sup>: When <span class=\"math-container\">$Y$</span> is separable, every closed subspace is a kernel, so in this case it is not more general. <a href=\"https://arxiv.org/abs/1811.02399\" rel=\"nofollow noreferrer\">https://arxiv.org/abs/1811.02399</a></p>\n<p><sup>2</sup>: 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 <a href=\"https://en.wikipedia.org/wiki/Complemented_subspace\" rel=\"nofollow noreferrer\">complemented subspaces</a>.</p>\n</div>",
"comments": [
{
"comment_id": 1069575,
"author_name": "Hannes",
"author_id": 85906,
"author_url": "https://mathoverflow.net/users/85906/hannes",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Chapter 1.17 in the Interpolation book of Triebel supports your complemented-subspace/projection suggestion. (And possibly worth a read for OP in any case.)</span>",
"created_date": "2022-02-23 08:58:00Z"
},
{
"comment_id": 1069590,
"author_name": "M.Oud",
"author_id": 477563,
"author_url": "https://mathoverflow.net/users/477563/m-oud",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thank you @Willie Wong. Its very useful,</span>",
"created_date": "2022-02-23 10:32:00Z"
},
{
"comment_id": 1069618,
"author_name": "Willie Wong",
"author_id": 3948,
"author_url": "https://mathoverflow.net/users/3948/willie-wong",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@OUDRANE a quick note, however: the complemented subspace argument is sufficient, but not necessary. An example: the Hardy spaces $H^p(\\mathbb{T})$ can be described as those $L^p$ functions (necessarily $\\subset L^1$) whose Fourier transforms have no negative modes. $H^1$ is not complemented in $L^1$, but for $p\\in (1,\\infty)$ we do have $H^p$ is complemented in $L^p$. However the sort of interpolation result you are asking for is true for Hardy spaces. (In this case you can write $H^p$ as the kernel of some $f:L^1\\to \\ell^\\infty$.)</span>",
"created_date": "2022-02-23 14:20:58Z"
},
{
"comment_id": 1069619,
"author_name": "Willie Wong",
"author_id": 3948,
"author_url": "https://mathoverflow.net/users/3948/willie-wong",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The proof of that last statement as given in <a href=\"http://www.numdam.org/article/AIF_1992__42_4_875_0.pdf\" rel=\"nofollow noreferrer\">numdam.org/article/AIF_1992__42_4_875_0.pdf</a> is based on relaxing the conditions on $\\Pi$: instead of looking for a single projection operator, the author constructs a family of uniformly bounded operators (not necessarily projections!) $\\Pi_f: L^1 \\to H^1$, indexed by $f\\in H^1$, such that $\\Pi_f(f) = f$.</span>",
"created_date": "2022-02-23 14:30:26Z"
},
{
"comment_id": 1069751,
"author_name": "M.Oud",
"author_id": 477563,
"author_url": "https://mathoverflow.net/users/477563/m-oud",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thank you a lot @Willie Wong , I think just the condition of the existence of projections is enough for me ( for my original probleme ). But this is very interesting!. And also I found here a paper devoted to this probleme <a href=\"https://www.researchgate.net/publication/313846656_Interpolation_of_subspaces\" rel=\"nofollow noreferrer\">researchgate.net/publication/…</a> . They try to go more general on the conditions that allow the interpolation functor to be compatible with the Kernel.</span>",
"created_date": "2022-02-24 06:53:55Z"
}
]
}
]
},
{
"question_id": 508763,
"title": "Intersection of centralizers in $S_n$",
"link": "https://mathoverflow.net/questions/508763/intersection-of-centralizers-in-s-n",
"score": 1,
"view_count": 92,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "gr.group-theory;finite-groups;symmetric-groups;centralisers",
"creation_date": "2026-03-05 01:53:05Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Let <span class=\"math-container\">$Z(g_1)$</span> and <span class=\"math-container\">$Z(g_2)$</span> be the centralizers of two permutations <span class=\"math-container\">$g_1$</span> and <span class=\"math-container\">$g_2$</span> in the symmetric group <span class=\"math-container\">$S_n$</span>. Is there an algorithm which calculates the intersection <span class=\"math-container\">$Z(g_1) \\cap Z(g_2)$</span> as a subgroup of <span class=\"math-container\">$S_n$</span>? Any references will be highly appreciated.</p>\n</div>",
"comments": [
{
"comment_id": 1325664,
"author_name": "LSpice",
"author_id": 2383,
"author_url": "https://mathoverflow.net/users/2383/lspice",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">What does it mean to calculate it as a subgroup of $S_n$? To enumerate the elements?</span>",
"created_date": "2026-03-04 23:20:20Z"
},
{
"comment_id": 1325666,
"author_name": "Derek Holt",
"author_id": 35840,
"author_url": "https://mathoverflow.net/users/35840/derek-holt",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@LSpice The standard meaning of computing a subgroup of $S_n$ is finding generators of that subgroup. When computing with large groups you work with generators and avoid enumerating elements.</span>",
"created_date": "2026-03-04 23:26:03Z"
},
{
"comment_id": 1325667,
"author_name": "jessica rhoades",
"author_id": 587098,
"author_url": "https://mathoverflow.net/users/587098/jessica-rhoades",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@LSpice, to give a description in terms of known subgroups of $S_n$. For example $Z(g_1)$ is a product of wreath products of $\\mathbb{Z}_k$ with $S_N$. Can the intersection be computed as a product of known groups.</span>",
"created_date": "2026-03-04 23:28:00Z"
},
{
"comment_id": 1325668,
"author_name": "YCor",
"author_id": 14094,
"author_url": "https://mathoverflow.net/users/14094/ycor",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The centralizer of the set of left translations in a group $G$ is the set of right translations. Hence, every 2-generated group of order $n$ appears as centralizer of some pair in $S_n$, so you won't find something as explicit as for centralizer of individual elements.</span>",
"created_date": "2026-03-04 23:34:54Z"
},
{
"comment_id": 1325671,
"author_name": "jessica rhoades",
"author_id": 587098,
"author_url": "https://mathoverflow.net/users/587098/jessica-rhoades",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@YCor, thanks that is very insightful. It seems there is no nice structural result related the intersection of the centralizers. One cannot expect these intersections to be products of wreath products?</span>",
"created_date": "2026-03-05 00:18:20Z"
}
],
"answers": [
{
"answer_id": 508764,
"question_id": 508763,
"score": 6,
"is_accepted": false,
"author_name": "Derek Holt",
"author_id": 35840,
"author_url": "https://mathoverflow.net/users/35840/derek-holt",
"author_reputation": 38726,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-04 23:32:22Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The intersection of the centralizers of <span class=\"math-container\">$g_1$</span> and <span class=\"math-container\">$g_2$</span> is just the centralizer of the subgroup <span class=\"math-container\">$\\langle g_1,g_2 \\rangle$</span> of <span class=\"math-container\">$S_n$</span>. Yes there is an efficient algorithm for that problem. It can be done in polynomial time.</p>\n<p>It is described in detail in Section 6.1.2 of the book \"Permutation Group Algorithms\" by Akos Seress.</p>\n</div>",
"comments": [
{
"comment_id": 1325670,
"author_name": "jessica rhoades",
"author_id": 587098,
"author_url": "https://mathoverflow.net/users/587098/jessica-rhoades",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Thanks! In theorem 6.1.6 when they say the centralizer can be computed, the algorithm outputs an abstract group or just the generators in $S_n$.</span>",
"created_date": "2026-03-05 00:17:06Z"
},
{
"comment_id": 1325702,
"author_name": "Derek Holt",
"author_id": 35840,
"author_url": "https://mathoverflow.net/users/35840/derek-holt",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">It outputs generators (because that is the conventional meaning of \"computing a subgroup\" in computational group theory), but the theory described in that section of the book tells you the structure of the group. You can also use other standard algorithms to compute the structure of a given group. Note that as $n$ gets large, the probability that $g_1$ and $g_2$ generate $A_n$ or $S_n$ approaches $1$, so for most pairs of elements the centralizer will be trivial.</span>",
"created_date": "2026-03-05 08:25:33Z"
}
]
}
]
},
{
"question_id": 508746,
"title": "Multiplicativity conjecture for odd-index derangements",
"link": "https://mathoverflow.net/questions/508746/multiplicativity-conjecture-for-odd-index-derangements",
"score": 0,
"view_count": 81,
"answer_count": 0,
"is_answered": false,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "co.combinatorics",
"creation_date": "2026-03-04 23:43:11Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<blockquote>\n<p><strong>Conjecture.</strong> Let <span class=\"math-container\">$D_n$</span> denote the number of derangements on <span class=\"math-container\">$n$</span> and <span class=\"math-container\">$S$</span> denote the set of all <span class=\"math-container\">$k$</span> such that <span class=\"math-container\">$k \\nmid D_{2n-1}$</span> for all <span class=\"math-container\">$1 < n \\leq k$</span>. Then we have the following: (a) if <span class=\"math-container\">$a \\in S$</span> and <span class=\"math-container\">$p \\mid a$</span> is prime then <span class=\"math-container\">$p \\in S$</span>, (b) if <span class=\"math-container\">$a,b \\in S$</span> and <span class=\"math-container\">$9 \\nmid ab$</span> then <span class=\"math-container\">$ab \\in S$</span>.</p>\n</blockquote>\n<p>For reference, this is the set <span class=\"math-container\">$S$</span> up to <span class=\"math-container\">$n = 100$</span>: <span class=\"math-container\">$$\\{1,3,5,7,15,17,19,21,23,25,29,35,43,49,51,57,59,61,69,71,73,75,85,87,95\\}.$$</span></p>\n<p>Now I can prove (b) under the stronger assumption that <span class=\"math-container\">$\\gcd(a,b) = 1$</span> and <span class=\"math-container\">$a,b \\in S$</span> implies that <span class=\"math-container\">$ab \\in S$</span>.</p>\n<p>It can be shown that <span class=\"math-container\">$$D_{n+k} = (-1)^kD_n \\pmod{k} \\implies D_{n+2k} = D_n \\pmod{k}.$$</span> This means that <span class=\"math-container\">$$D_{2(n+k)-1} = D_{(2n-1)+2k} = D_{2n-1} \\pmod{k}$$</span> and so the sequence <span class=\"math-container\">$x_n = D_{2n-1} \\pmod{k}$</span> has period <span class=\"math-container\">$k$</span>.</p>\n<p>Now we show that if <span class=\"math-container\">$k \\in S, k \\mid D_{2n-1} \\implies n = 1 \\pmod{k}$</span>.</p>\n<p>This is since <span class=\"math-container\">$$D_{2n-1} = D_{2\\alpha-1} \\pmod{k}$$</span> where <span class=\"math-container\">$\\alpha \\equiv n \\mod{k}$</span> and <span class=\"math-container\">$\\alpha \\in \\{1,\\ldots,k\\}$</span> and so <span class=\"math-container\">$k \\mid D_{2\\alpha-1}$</span>. But this must mean that <span class=\"math-container\">$n = 1 \\pmod{k}$</span> since <span class=\"math-container\">$k \\nmid D_{2\\alpha-1}$</span> for each <span class=\"math-container\">$1 < \\alpha \\leq k$</span>.</p>\n<p>Let <span class=\"math-container\">$a,b \\in S$</span> with <span class=\"math-container\">$\\gcd(a,b) = 1$</span>. To show <span class=\"math-container\">$ab \\in S$</span> suppose for sake of contradiction that <span class=\"math-container\">$ab \\not \\in S$</span>. Then there exists <span class=\"math-container\">$n$</span> such that <span class=\"math-container\">$1 < n \\leq ab$</span> and <span class=\"math-container\">$ab \\mid D_{2n-1}$</span> and so <span class=\"math-container\">$a \\mid D_{2n-1}$</span> and <span class=\"math-container\">$b \\mid D_{2n-1}$</span>. Now by the last fact this means that <span class=\"math-container\">$n = 1 \\pmod{a}$</span> and <span class=\"math-container\">$n = 1 \\pmod{b}$</span> and since <span class=\"math-container\">$\\gcd(a,b) = 1$</span> it means that <span class=\"math-container\">$n = 1 \\pmod{ab}$</span>, which is a contradiction if <span class=\"math-container\">$1 < n \\leq ab$</span>.</p>\n</div>",
"comments": [],
"answers": []
},
{
"question_id": 508758,
"title": "Topology on the set of smooth sections of a smooth fiber bundle",
"link": "https://mathoverflow.net/questions/508758/topology-on-the-set-of-smooth-sections-of-a-smooth-fiber-bundle",
"score": 1,
"view_count": 88,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "dg.differential-geometry;differential-topology;frechet-manifold",
"creation_date": "2026-03-04 22:34:56Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Let <span class=\"math-container\">$p: B \\to M$</span> be a smooth fiber bundle. Denote <span class=\"math-container\">$\\Gamma(B)$</span> the set of smooth sections. What is the natural topology on this set?</p>\n<p>It 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?</p>\n<p>Edit: if possible, give citations, please.</p>\n</div>",
"comments": [
{
"comment_id": 1325652,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">If M is compact then yes, it's a Fréchet manifold.</span>",
"created_date": "2026-03-04 21:04:54Z"
},
{
"comment_id": 1325653,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">BTW, giving charts is enough to specify the topology.</span>",
"created_date": "2026-03-04 21:11:22Z"
},
{
"comment_id": 1325690,
"author_name": "Pedro Lauridsen Ribeiro",
"author_id": 11211,
"author_url": "https://mathoverflow.net/users/11211/pedro-lauridsen-ribeiro",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">It might help having a look at <a href=\"https://mathoverflow.net/a/349234/11211\">mathoverflow.net/a/349234/11211</a> and the technical appendix to <a href=\"https://physics.stackexchange.com/a/317839/16767\">physics.stackexchange.com/a/317839/16767</a> . I also second David Roberts's point that a manifold topology is uniquely determined by a compatible atlas which doesn't even need to be maximal, you don't need to specify the former separately. This still holds true when the manifold is modeled after a possibly infinite-dimensional (say, locally convex Hausdorff) topological vector space.</span>",
"created_date": "2026-03-05 05:39:38Z"
},
{
"comment_id": 1325704,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">For what it's worth, this textbook has a lot of detail <a href=\"https://arxiv.org/abs/2112.08114\" rel=\"nofollow noreferrer\">arxiv.org/abs/2112.08114</a> if not exactly the example you want</span>",
"created_date": "2026-03-05 08:29:25Z"
}
],
"answers": [
{
"answer_id": 508762,
"question_id": 508758,
"score": 1,
"is_accepted": false,
"author_name": "Pierre PC",
"author_id": 129074,
"author_url": "https://mathoverflow.net/users/129074/pierre-pc",
"author_reputation": 4203,
"license": "CC BY-SA 4.0",
"is_edited": false,
"created_date": "2026-03-04 22:34:56Z",
"last_edited_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>People usually consider one of the two <a href=\"https://en.wikipedia.org/wiki/Whitney_topologies\" rel=\"nofollow noreferrer\">Whitney topologies</a>.</p>\n<p>The 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 <span class=\"math-container\">$|\\partial^\\alpha f_\\phi|$</span> over <span class=\"math-container\">$K$</span>, where <span class=\"math-container\">$K$</span> is compact inside an open set <span class=\"math-container\">$U$</span> for a trivializing chart <span class=\"math-container\">$\\phi:\\pi_B^{-1}U\\to V\\times\\mathbb R^r$</span>, <span class=\"math-container\">$f_\\phi$</span> is the section expressed in those coordinates, and <span class=\"math-container\">$\\alpha\\in\\mathbb R^d$</span> is a multi-index.</p>\n<p>The 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 <span class=\"math-container\">$\\sup_{\\phi,\\alpha,K}|\\partial^\\alpha f_\\phi|$</span> with the argument like before, over some collection of <span class=\"math-container\">$(\\phi,\\alpha,K)$</span> such that the domains <span class=\"math-container\">$U$</span> of the trivializations <span class=\"math-container\">$\\phi$</span> are locally finite, and <span class=\"math-container\">$\\alpha$</span> takes only finitely many values (equivalently, the partial derivatives have bounded order).</p>\n</div>",
"comments": [
{
"comment_id": 1325672,
"author_name": "andrey safin",
"author_id": 587092,
"author_url": "https://mathoverflow.net/users/587092/andrey-safin",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I consider your answer to be rather unclear. In such a construction it is better to point out what is meant by \"induced\". In the first paragraph you basically gave topology on sections on trivializing open sets. Which coincides with standard Frechet topology (and btw i can't get the difference with the second paragraph except for finite orders). And what's next? What is the base of the topology in the whole set of sections, for instance?</span>",
"created_date": "2026-03-05 00:28:04Z"
},
{
"comment_id": 1325699,
"author_name": "David Roberts",
"author_id": 4177,
"author_url": "https://mathoverflow.net/users/4177/david-roberts",
"score": 1,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@andreysafin a subset of the set of all smooth sections is open iff its intersection with all charts is an open subset of that chart. You don't even need to talk about the base of the topology, you can describe the open sets directly.</span>",
"created_date": "2026-03-05 07:29:54Z"
}
]
}
]
},
{
"question_id": 508750,
"title": "Minimum degree in a graph defined by sumsets",
"link": "https://mathoverflow.net/questions/508750/minimum-degree-in-a-graph-defined-by-sumsets",
"score": 2,
"view_count": 124,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": true,
"accepted_answer_id": 508760,
"tags": "co.combinatorics;graph-theory;additive-combinatorics;sumsets",
"creation_date": "2026-03-04 22:13:57Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I have the following conjecture.</p>\n<p><strong>Conjecture.</strong> <em>Let <span class=\"math-container\">$A$</span> be a finite set in some abelian group <span class=\"math-container\">$G$</span>, and let <span class=\"math-container\">$A = B\\cup B'$</span> be a partition of <span class=\"math-container\">$A$</span> into two disjoint nonempty sets. We define a graph on the vertex set <span class=\"math-container\">$B'$</span> by placing an edge between <span class=\"math-container\">$b_1'$</span> and <span class=\"math-container\">$b_2'$</span> if and only if <span class=\"math-container\">$b_1' + b_2' \\notin A+B$</span>. My conjecture is that there always exists some <span class=\"math-container\">$b'\\in B'$</span> such that</em>\n<span class=\"math-container\">$$\\deg(b') \\le {|A+B|\\over |B|}.$$</span></p>\n<p>Here's some vague reasoning why the inequality might be true. If <span class=\"math-container\">$B$</span> is very small, then the right-hand side is close to <span class=\"math-container\">$|A|$</span>. If <span class=\"math-container\">$B$</span> is very large, then <span class=\"math-container\">$B'$</span> is very small and the degree on the left-hand side, which is bounded above by <span class=\"math-container\">$|B|-1$</span>, is small.</p>\n<p>The trouble arises when <span class=\"math-container\">$A+B$</span> is small. In this case the inequality might seem more far-fetched, and indeed there are cases where some of the <span class=\"math-container\">$b'$</span> have degree <span class=\"math-container\">$> |A+B|/|B|$</span>, but still I've always been able to find <em>some</em> <span class=\"math-container\">$b'$</span> with small degree. (And I've never actually found any cases where equality holds for the minimiser <span class=\"math-container\">$b'$</span>, it's always a strict inequality.)</p>\n<p>I 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 <span class=\"math-container\">$G={\\bf Z}$</span> 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 <span class=\"math-container\">$A$</span>, like arithmetic progressions and geometric rogressions, combinations of the two, etc.)</p>\n<p>Counterexamples over other groups would be interesting to me as well, as they would rule out many of the basic \"Ruzsa calculus\" techniques.</p>\n</div>",
"comments": [],
"answers": [
{
"answer_id": 508760,
"question_id": 508750,
"score": 4,
"is_accepted": true,
"author_name": "Mark Wildon",
"author_id": 7709,
"author_url": "https://mathoverflow.net/users/7709/mark-wildon",
"author_reputation": 12112,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-04 21:45:49Z",
"last_edited_date": "2026-03-04 22:13:57Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>Here is a family of counterexamples using elementary abelian <span class=\"math-container\">$2$</span>-groups, i.e. vector spaces over <span class=\"math-container\">$\\mathbb{F}_2$</span>.</p>\n<p>Take <span class=\"math-container\">$G = \\mathbb{F}_2^n$</span> and let <span class=\"math-container\">$G = B' \\oplus \\langle v \\rangle$</span>, so <span class=\"math-container\">$B'$</span> is a codimension <span class=\"math-container\">$1$</span> hyperplane in <span class=\"math-container\">$G$</span> with a complement spanned by <span class=\"math-container\">$v$</span>. Let <span class=\"math-container\">$C$</span> be a codimension <span class=\"math-container\">$1$</span> hyperplane in <span class=\"math-container\">$B'$</span> and let <span class=\"math-container\">$B = C + v$</span>. Thus <span class=\"math-container\">$|B'| = 2^{n-1}$</span>, <span class=\"math-container\">$|B| = |C| = 2^{n-2}$</span> and since <span class=\"math-container\">$B \\cap B' = \\varnothing$</span>, we have <span class=\"math-container\">$|A| = |B \\cup B'| = |B| + |B'| = 3 \\times 2^{n-2}$</span>.</p>\n<p>Observe that</p>\n<p><span class=\"math-container\">$$A+ B = (B \\cup B') + B = (B + B) \\cup (B' + B) = C \\cup (B' + v),$$</span></p>\n<p>where we used <span class=\"math-container\">$B + B = (C+ v) + (C+ v) = C$</span> and, since <span class=\"math-container\">$C \\subseteq B'$</span>,</p>\n<p><span class=\"math-container\">$$B' + B = B' + C + v = B' + v.$$</span></p>\n<p>Thus <span class=\"math-container\">$|A+B| = 3 \\times 2^{n-2}$</span> and the ratio in the question is <span class=\"math-container\">$|A+B|/|B| = 3$</span>.</p>\n<p>For the graph we have an edge <span class=\"math-container\">$\\{b_1', b_2'\\}$</span> for <span class=\"math-container\">$b_1', b_2' \\in B'$</span> if and only if</p>\n<p><span class=\"math-container\">$$b_1' + b_2' \\not\\in A + B = C \\cup (B' + v).$$</span></p>\n<p>Since <span class=\"math-container\">$B'$</span> is a hyperplane, we never have <span class=\"math-container\">$b_1' + b_2' \\in B'+ v$</span>. Hence there is an edge <span class=\"math-container\">$\\{b_1', b_2'\\}$</span> if and only if <span class=\"math-container\">$b_1'+b_2' \\not\\in C$</span>, the codimension <span class=\"math-container\">$1$</span> hyperplane inside <span class=\"math-container\">$B'$</span>. This is the case if and only if <span class=\"math-container\">$b_1', b_2'$</span> come from different cosets of <span class=\"math-container\">$C$</span> in the union <span class=\"math-container\">$B' = C \\cup (B' \\backslash C)$</span>. Therefore the graph is is the complete bipartite graph with bipartition <span class=\"math-container\">$\\{C, B' \\backslash C\\}$</span>. In particular, every vertex has degree <span class=\"math-container\">$|C| = |B' \\backslash C| = 2^{n-2}$</span>. Therefore we get a counterexample if and only if <span class=\"math-container\">$2^{n-2} \\not\\le 3$</span>. The smallest such <span class=\"math-container\">$n$</span> is <span class=\"math-container\">$n= 4$</span>.</p>\n</div>",
"comments": []
}
]
},
{
"question_id": 508710,
"title": "What is the Joyal model structure actually useful for?",
"link": "https://mathoverflow.net/questions/508710/what-is-the-joyal-model-structure-actually-useful-for",
"score": 16,
"view_count": 593,
"answer_count": 1,
"is_answered": true,
"has_accepted_answer": false,
"accepted_answer_id": null,
"tags": "ct.category-theory;higher-category-theory;simplicial-stuff;model-categories;infinity-categories",
"creation_date": "2026-03-04 05:32:13Z",
"last_activity_date": "",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>I'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 <span class=\"math-container\">$\\infty$</span>-categories (or, rather, the theory of quasi-categories as a model of <span class=\"math-container\">$\\infty$</span>-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:</p>\n<blockquote>\n<p>What is the Joyal model structure actually useful for?</p>\n</blockquote>\n<p>In fact, I don't even have a good understanding of what the existence of the Joyal model structure really <em>means</em>, so a related question (which may share the same answer) would be:</p>\n<blockquote>\n<p>What is the principal mathematical content in the existence of the Joyal model structure?</p>\n</blockquote>\n<p>Or to be really concrete:</p>\n<blockquote>\n<p>Name 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.</p>\n</blockquote>\n<p>I guess that's three questions, but they're basically all asking the same thing.</p>\n</div>",
"comments": [
{
"comment_id": 1325522,
"author_name": "Tim Campion",
"author_id": 2362,
"author_url": "https://mathoverflow.net/users/2362/tim-campion",
"score": 3,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">I’m not sure what you mean. What’s an example of a model structure which you do find useful?</span>",
"created_date": "2026-03-03 16:06:33Z"
},
{
"comment_id": 1325530,
"author_name": "Tyler Lawson",
"author_id": 360,
"author_url": "https://mathoverflow.net/users/360/tyler-lawson",
"score": 6,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Here are two examples that I find difficult to do without establishing much of the Joyal model structure. (a) Suppose $f: C \\to D$ is a functor between quasicategories, $A \\subset C$, and we have an equivalence $\\phi: f|_A \\to g$ of functors. Then there exists an extension of $g$ to a functor $g': C \\to D$ and an extension of $\\phi$ to an equivalence $\\phi': f \\to g'$. (b) Suppose $K \\to L$ is an equivalence of (marked) simplicial sets (eg: the inclusion of the \"spine\" of $B\\Bbb N$). Then for any quasicategory C, the map of functor categories $Fun(L,C) \\to Fun(K,C)$ is an equivalence.</span>",
"created_date": "2026-03-03 16:24:12Z"
},
{
"comment_id": 1325531,
"author_name": "Tyler Lawson",
"author_id": 360,
"author_url": "https://mathoverflow.net/users/360/tyler-lawson",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">(More generally, a model structure's role is often to establish \"well-definedness\" of many constructions.)</span>",
"created_date": "2026-03-03 16:27:38Z"
},
{
"comment_id": 1325532,
"author_name": "Tyler Lawson",
"author_id": 360,
"author_url": "https://mathoverflow.net/users/360/tyler-lawson",
"score": 2,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">The biggest foundational application of the Joyal model structure, though, is probably the establishment of an equivalence between quasicategories and categories enriched in spaces. We already cared about many topics from the latter, and quasicategories provide a technical advance over them.</span>",
"created_date": "2026-03-03 16:31:55Z"
}
],
"answers": [
{
"answer_id": 508716,
"question_id": 508710,
"score": 14,
"is_accepted": false,
"author_name": "Tim Campion",
"author_id": 2362,
"author_url": "https://mathoverflow.net/users/2362/tim-campion",
"author_reputation": 67580,
"license": "CC BY-SA 4.0",
"is_edited": true,
"created_date": "2026-03-03 16:25:00Z",
"last_edited_date": "2026-03-04 05:32:13Z",
"body_html": "<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>If 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. <em>should</em> work, and these considerations are typically of a model categorical flavor.</p>\n<p>For 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 <em>should</em> be possible (and before that presumably allowed Joyal and Lurie some hope that it might be possible).</p>\n<p>Some 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 <em>per se</em>. But the model categorical philosophy still feels evident in motivating why you write one set of axioms and not another.</p>\n<hr/>\n<p>I think I’ve been addressing primarily your first question above.</p>\n<p>For your second question, I think I will refer back to my <a href=\"https://mathoverflow.net/questions/508710/what-is-the-joyal-model-structure-actually-useful-for/508716#comment1325522_508710\">comment</a> 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.</p>\n<p>For your third question, I think the examples given by <a href=\"https://mathoverflow.net/questions/508710/what-is-the-joyal-model-structure-actually-useful-for/508716#comment1325530_508710\">Tyler Lawson</a> and <a href=\"https://mathoverflow.net/questions/508710/what-is-the-joyal-model-structure-actually-useful-for/508716#comment1325535_508716\">Phil Tosteson</a> in the comments are great. Phil’s is particularly punchy — you should not take for granted that a functor category <span class=\"math-container\">$Fun(A,B)$</span> 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 <span class=\"math-container\">$A$</span>…</p>\n<p>(Tyler’s “(b)” gives a great example of how things are even better since in fact <span class=\"math-container\">$A$</span> 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 <span class=\"math-container\">$\\mathbb N$</span> is Joyal-equivalent to its Hasse diagram (viewed as a 1-skeletal simplicial set).)</p>\n</div>",
"comments": [
{
"comment_id": 1325533,
"author_name": "Tim Campion",
"author_id": 2362,
"author_url": "https://mathoverflow.net/users/2362/tim-campion",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Perhaps relatedly— trying to understand the join of simplicial categories appears to be a bit of a nightmare.</span>",
"created_date": "2026-03-03 16:34:34Z"
},
{
"comment_id": 1325535,
"author_name": "Phil Tosteson",
"author_id": 52918,
"author_url": "https://mathoverflow.net/users/52918/phil-tosteson",
"score": 11,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">Nice answer! To summarize, the reason quasicategories are an especially nice model for infinity categories is that they are fibrant objects in a model category where every object is cofibrant. So unlike simplicial categories or topological categories, we don't need to perform complicated cofibrant replacements to compute mapping spaces, and this provides an a priori justification for many definitions.</span>",
"created_date": "2026-03-03 16:42:58Z"
},
{
"comment_id": 1325536,
"author_name": "Tim Campion",
"author_id": 2362,
"author_url": "https://mathoverflow.net/users/2362/tim-campion",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">@PhilTosteson Yeah! I guess for those mapping spaces it’s also important that it’s a (cartesian) monoidal model category. The fact that the join is also basically a Quillen bifunctor is a cherry on top. OTOH I don’t know how to conceptualize why left/right and co/cartesian fibrations work so well strictly in terms of the Joyal model structure, though it seems telling that various marked model structures on slice categories can be developed in such close analogy to Joyal. I suppose Cisinski might have something to say about it.</span>",
"created_date": "2026-03-03 16:50:40Z"
},
{
"comment_id": 1325537,
"author_name": "Tim Campion",
"author_id": 2362,
"author_url": "https://mathoverflow.net/users/2362/tim-campion",
"score": 0,
"body_html": "<span class=\"comment-copy\" itemprop=\"text\">For that matter— the fact that the Kan-Quillen model structure is cartesian monoidal (ie cartesian products are “homotopically correct” and adjointly, you can compute mapping spaces of simplicial sets in a straightforward way) is already something not to be taken lightly— plain cubical sets don’t have this property, which is a large part of why people like Kan moved away from them to the simplicial land in the first place.</span>",
"created_date": "2026-03-03 16:57:53Z"
}
]
}
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] |