Recent Questions - MathOverflow most recent 30 from mathoverflow.net 2021-05-17T15:05:04Z https://mathoverflow.net/feeds https://creativecommons.org/licenses/by-sa/4.0/rdf https://mathoverflow.net/q/393007 0 Representing $\forall x\in \mathbb{N}$ via ZFC Mister Orange https://mathoverflow.net/users/226047 2021-05-17T14:52:14Z 2021-05-17T14:52:14Z <p>given just some arbitrary predicate <span class="math-container">$\phi(x)$</span> (e.g. <span class="math-container">$\phi(x)="x$</span> is uneven&quot; or &quot;x is prime&quot;), I was wondering if one can <em>formally</em> make sense out of statements like <span class="math-container">\begin{equation*}{\forall x\in \mathbb{N}: \phi(x)}\end{equation*}</span> by only using axioms from ZFC. By this I mean that I would like to have the above statement as a predicate logic formula without the <span class="math-container">$\ldots\in\mathbb{N}$</span> part, i.e. only using <span class="math-container">$\forall x:$</span> or <span class="math-container">$\exists x:$</span> parts and where <span class="math-container">$\forall x$</span> means <em>all sets</em> and not all natural numbers x).</p> <p>To put it in another way: I would be nice to get the above statement as a predicate logic formula where the discourse universe is all sets (as in ZFC).</p> <p>What I already know:</p> <ul> <li><a href="https://math.stackexchange.com/questions/123831/how-to-express-the-set-of-natural-numbers-in-zfc">https://math.stackexchange.com/questions/123831/how-to-express-the-set-of-natural-numbers-in-zfc</a> : One can define natural numbers in various ways easily, e.g. by defining <span class="math-container">$0:=\emptyset$</span>, <span class="math-container">$1:=\{0,\emptyset\}$</span>, <span class="math-container">$2:=\{\ldots\}$</span>.</li> </ul> <p>Thanks a lot!</p> <p>Thomas</p> https://mathoverflow.net/q/393004 -1 Bad proofs history [duplicate] Gergely https://mathoverflow.net/users/39422 2021-05-17T14:23:04Z 2021-05-17T14:23:04Z <p>I searched for bad proofs in mathematics and this Wikipedia page came through:</p> <p><a href="https://en.wikipedia.org/wiki/Mathematical_fallacy" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Mathematical_fallacy</a></p> <p>But mathematical fallacies are bad proofs on intention.</p> <p>Then on Stack Exchange I have found</p> <p><a href="https://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs">Examples of interesting false proofs</a></p> <p>and</p> <p><a href="https://math.stackexchange.com/questions/348198/best-fake-proofs-a-m-se-april-fools-day-collection">https://math.stackexchange.com/questions/348198/best-fake-proofs-a-m-se-april-fools-day-collection</a></p> <p>These give a number of interesting bad proofs, but I am interested in the history of sometimes long accepted proofs in published mathematical research that later turned out to be false.</p> <p>Is there a historical account of this?</p> https://mathoverflow.net/q/393003 2 Littelmann Path model and RSK e and f operators Mathprof https://mathoverflow.net/users/140532 2021-05-17T14:14:31Z 2021-05-17T14:20:37Z <p>The <a href="https://en.wikipedia.org/wiki/Littelmann_path_model" rel="nofollow noreferrer">Littelmann path model</a> defines <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> operators which correspond in type A to the <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> operators on semistandard Young tableaux (i.e. since they both are different ways of looking at Type A crystal.)</p> <p>In the context of semistandard Young tableaux, we can think of the <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> operators as being described by a paired parenthesis operation, and that operation commutes with RSK insertion thus the <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> operators are often thought of as acting on words, not just semistandard Young tableaux.</p> <p>In the context of the Littelmann path model, there are many paths to the same point, all equally compatible with the same <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> operators. Some of those paths, particularly those that are a sequence of various length one steps from the set <span class="math-container">$\{e_i-v\}_{i=1\cdots,n}$</span> (where <span class="math-container">$v=\frac{1}{n}(1,1,\cdots,1)$</span> so the vector sums to zero) might reasonably be considered an analogue of a word, with <span class="math-container">$e_i- v$</span> corresponding to i in the word (with the end of the word corresponding to the start of the path at the origin). Call these path &quot;special&quot; paths for purposes of this post.</p> <p>Since every highest weight path of the same weight generators the same crystal, there's a sense in which order doesn't matter to the <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> operators in the Littelmann path model. In contrast, not every word of the same type RSK inserts to the same semistandard Young Tableaux, so special paths must not be fully compatible with RSK and the <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> paired parenthesis operators on words. Is there a natural subset of special paths that is consistent on which the paired parenthesis <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> operators are consistent with the <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> operators of the Littelmann path model? For example, on special paths which correspond to reading words of Semistandard Young tableaux, will the <span class="math-container">$e_i$</span> and <span class="math-container">$f_i$</span> operators in the Littelmann path model result in paths which are also the reading word of semistandard Young tableaux?</p> <p>I suspect this is too much to ask for, although it seemed to work on a couple of very small examples.</p> https://mathoverflow.net/q/392999 0 Completed Hochschild (co)homology FunctionOfX https://mathoverflow.net/users/111070 2021-05-17T13:58:56Z 2021-05-17T14:59:40Z <p>Let <span class="math-container">$A$</span> be a <span class="math-container">$\mathbb{C}[[h]]$</span> algebra (not necessarily commutative). The Hochschild homology is then defined via a bar construction and that <span class="math-container">$HH_0(A)=A/[A,A]$</span>. Note that each <span class="math-container">$HH_i(A)$</span> is a <span class="math-container">$\mathbb{C}[[h]]$</span>-module. We can define <span class="math-container">$\overline{HH_0}(A):=A/\overline{[A,A]}$</span>, where the overline means taking the <span class="math-container">$h$</span>-adic completion. I want to ask if there is a version of the bar resolution that produces the 'completed' version of <span class="math-container">$HH$</span> and how <span class="math-container">$\overline{HH_i}$</span> and <span class="math-container">$HH_i$</span> are related and when they are equal.</p> <p>Edit: More assumptions on <span class="math-container">$A$</span> might be needed to make the question 'nice'.</p> https://mathoverflow.net/q/392998 0 In practice, how is the lebesgue measure usually generalized? exfret https://mathoverflow.net/users/115247 2021-05-17T13:51:37Z 2021-05-17T13:51:37Z <p><strong>The general question</strong></p> <p>It is easy to find on the <a href="https://en.wikipedia.org/wiki/Lebesgue_measure#Relation_to_other_measures" rel="nofollow noreferrer">wikipedia page</a> for Lebesgue measure that Haar measure is a common generalization that preserves the idea of &quot;invariance under some group action&quot;. While wondering about the &quot;most natural&quot; way of defining a measure on lines of <span class="math-container">$\mathbb{R}^2$</span> (see below for more information), it struck me that it's not always obvious what the &quot;most natural&quot; measure is for certain spaces is like in <span class="math-container">$\mathbb{R}^n$</span>. I was wondering how this problem is usually tackled, if it even comes up.</p> <p>Is finding an &quot;appropriate&quot; Haar measure usually easy to find? Is it unique in some sense or usually dependent on the use case? What can we do in the case of the space of lines in <span class="math-container">$\mathbb{R}^2$</span> (or general hyperplanes in <span class="math-container">$\mathbb{R}^n$</span>)?</p> <p><strong>The specific case of lines in <span class="math-container">$\mathbb{R}^2$</span></strong></p> <p>To give an example, and some motivation, consider lines in <span class="math-container">$\mathbb{R}^2$</span>. Is there a &quot;unique&quot;, &quot;most natural&quot; measure on this space? By this, I mean some invariance under isometries, plus some extra properties that satisfy my intuition of what it should look like. My primary motivation is whether there is actually a &quot;most natural&quot; way to solve <a href="https://en.wikipedia.org/wiki/Bertrand_paradox_(probability)" rel="nofollow noreferrer">Bertrand's paradox</a>.</p> <p>Specifically something satisfying the properties below would be nice. In what follows, <span class="math-container">$S$</span> is a measurable set of lines, <span class="math-container">$m(S)$</span> is the measure of <span class="math-container">$S$</span>, and if we have an operation <span class="math-container">$f:\mathbb{R}^2\to \mathbb{R}^2$</span> we can consider <span class="math-container">$f(S)$</span> to be the set of <span class="math-container">$f(L)$</span> for each <span class="math-container">$L$</span> in <span class="math-container">$S$</span> (where <span class="math-container">$f(L)$</span> for some line <span class="math-container">$L$</span> is just the set of <span class="math-container">$f(p)$</span> for <span class="math-container">$p$</span> in <span class="math-container">$L$</span>).</p> <ol> <li>If <span class="math-container">$f$</span> is a rotation, <span class="math-container">$m(S)=m(f(S))$</span></li> <li>If <span class="math-container">$f$</span> is a reflection, <span class="math-container">$m(S)=m(f(S))$</span></li> <li>If <span class="math-container">$f$</span> is a translation, <span class="math-container">$m(S)=m(f(S))$</span></li> <li>If <span class="math-container">$f$</span> is a dilation with factor <span class="math-container">$C$</span> (i.e.- two points <span class="math-container">$d$</span> apart get dilated to <span class="math-container">$C\cdot d$</span> apart), then <span class="math-container">$C^2\cdot m(S)=m(f(S))$</span></li> <li>If <span class="math-container">$M$</span> is a compact subset of <span class="math-container">$\mathbb{R}^2$</span>, then the set of lines <span class="math-container">$S$</span> that intersect <span class="math-container">$M$</span> is measurable and <span class="math-container">$0&lt;m(S)&lt;∞$</span></li> </ol> <p>I see three cases: Either there are no measures satisfying the above, in which case perhaps there is a weakening of the conditions to give some meaningful content (we might be able to at least satisfy a chosen subset of the above); there is exactly one measure satisfying the above (up to perhaps some relatively trivial modifications) and I am happy; or there are many and perhaps we need some more conditions specifying what a &quot;natural&quot; measure should be.</p> <p>Of course, this is a relatively open-ended question, so I am satisfied with any related comments. I simply don't want to wade through a couple courses in measure theory just for a chance of learning the answer. :)</p> https://mathoverflow.net/q/392996 1 Some questions on a paper of Baumslag and Solitar user101010 https://mathoverflow.net/users/99414 2021-05-17T13:27:32Z 2021-05-17T14:42:27Z <p>I was recently trying taking a look at the paper &quot;Some two-generator one-relator non-hopfian groups&quot; by Baumslag and Solitar where they introduce the groups now known as the Baumslag-Solitar groups given by the presentation <span class="math-container">$$G = \left&lt; a,b \mid a^{-1}b^la = b^m \right&gt;$$</span> The paper is only 3-pages long and a bit skimpy on details (for me as someone without much group-theory know-how). I have a few questions about some of the claims in the paper.</p> <p>In considering the case where there is a prime <span class="math-container">$p$</span> that divides <span class="math-container">$l$</span> but not <span class="math-container">$m$</span>, the authors claim that the homomorphism given by <span class="math-container">$a \mapsto a, b \mapsto b^p$</span> has a nontrivial kernel, and namely, that the element <span class="math-container">$$[b^{l/p}, a]^p b^{l-m}$$</span> which is in the kernel of the above map is actually nontrivial in <span class="math-container">$G$</span> (here the commutator convention is <span class="math-container">$[g_1,g_2] = g_1^{-1}g_2^{-1}g_1g_2$</span>). With a little fussing, we can see that this map is surjective, and thus, if we see that that element is nontrivial, we have a non-hopfian group (which was a punchline/motivation in the above paper). My first question is, why is this element nontrivial?</p> <p>My second question is with regards to a claim made on the first page. Taking <span class="math-container">$l = 2, m = 3$</span> we can see that <span class="math-container">$a$</span> and <span class="math-container">$b^4$</span> generate <span class="math-container">$G$</span>. The claim is that for the surjection <span class="math-container">\begin{align*} F\langle x,y\rangle &amp;\to G \\ x &amp;\mapsto a \\ y &amp;\mapsto b^4 \end{align*}</span> the kernel is finitely normally generated, but not by a single element. Why is this kernel finitely normally generated but not by a single element?</p> <p>I bit of playing around yields some elements of the kernel, and I imagine that there is some folding-style way of seeing if a given element of a free group is in the normal closure of some finite list of other elements, but I don't know how to do that. If I did, I could probably show that the kernel is not normally generated by a single element.</p> https://mathoverflow.net/q/392994 0 Factoring integers of the form $n=p q^2$ using elliptic curves joro https://mathoverflow.net/users/12481 2021-05-17T12:39:57Z 2021-05-17T12:39:57Z <p>We got argument and strong experimental support that integers of the form <span class="math-container">$n=p q^2$</span> can be factored using elliptic curves easier than general integers</p> <blockquote> <p>Q1 Is this known?</p> </blockquote> <blockquote> <p>Q2 Is this worth publishing?</p> </blockquote> <blockquote> <p>Q3 Can the method be improved for general integers?</p> </blockquote> <p>Assume <span class="math-container">$E$</span> is elliptic curve modulo <span class="math-container">$n$</span> with known point <span class="math-container">$P$</span> of the order of <span class="math-container">$P$</span> over <span class="math-container">$\mathbb{F}_p$</span> is <span class="math-container">$\rho = S U$</span> where <span class="math-container">$S$</span> is <span class="math-container">$B_1$</span> smooth. We can find the factor <span class="math-container">$p$</span> working over the curve modulo <span class="math-container">$n$</span> in time <span class="math-container">$O(B_1 + \sqrt{U})$</span> and constant memory. It is an open problem if this can be generalized to arbitrary <span class="math-container">$n$</span>.</p> <p>Stage 1:<br /> This is the first stage of the elliptic curve factoring algorithm. Set <span class="math-container">$Q=P$</span> and for primes <span class="math-container">$r$</span>, <span class="math-container">$r \le B_1$</span> let <span class="math-container">$Q=r^{\log_r{n}} Q$</span>.</p> <p>If we reach the point at infinity modulo <span class="math-container">$p$</span> we are done and the algorithm stops with factor <span class="math-container">$p$</span>.</p> <p>Stage 2: This is based on the birthday paradox and Pollard's rho algorithm for discrete logarithms in groups.</p> <p>Let <span class="math-container">$\mathrm{kronecker}(a,n)$</span> denote the kronecker symbol and for a point <span class="math-container">$a$</span> on an elliptic curve <span class="math-container">$X(a)$</span> denote the <span class="math-container">$X$</span> coordinate of <span class="math-container">$a$</span>.</p> <p>For points <span class="math-container">$a,b$</span> on <span class="math-container">$E$</span> define the function <span class="math-container">$f(a,g)$</span>:</p> <p>If <span class="math-container">$\mathrm{kronecker}(X(a),n)=1,f(a,g)=2 a + 5 g$</span>, else <span class="math-container">$f(a,g)=3 a + g$</span>.</p> <p>For <span class="math-container">$Q$</span> from stage 1, set <span class="math-container">$a=Q,b=Q,g=Q$</span>.</p> <p>For <span class="math-container">$n$</span> from <span class="math-container">$1$</span> to <span class="math-container">$B_2$</span> set <span class="math-container">$a=f(a,g),b=f(b,g),b=f(b,g)$</span>.</p> <p>If <span class="math-container">$\gcd(X(a)-X(b),n) \bmod n&gt;1$</span> return the factor.</p> <p>Stage 2 finds multiple of <span class="math-container">$U Q$</span> in time <span class="math-container">$B_2$</span> while <span class="math-container">$U \sim B_2^2$</span>.</p> <p>Since the kronecker symbol is multiplicative, we have <span class="math-container">$\mathrm{kronecker}(a,n)=\mathrm{kronecker}(a,p)$</span> for <span class="math-container">$a$</span> coprime to <span class="math-container">$n$</span>.</p> <p>Iterates of <span class="math-container">$f^k(a,g)$</span> are deterministic random walk modulo <span class="math-container">$p$</span>, so <span class="math-container">$f^N(a,g)=f^{2N}(a,g)$</span> happens for <span class="math-container">$N = O(\sqrt{U}) \sim B_2$</span>.</p> https://mathoverflow.net/q/392993 1 On Markoff-type diophantine equation Bogdan https://mathoverflow.net/users/31472 2021-05-17T12:22:43Z 2021-05-17T13:46:56Z <p>Do there exist integers <span class="math-container">$x,y,z$</span> such that <span class="math-container">$$x^2+y^2-z^2 = xyz -2 \quad ?$$</span></p> <p>Why this is interesting? First, this equation arose in an answer to the previous Mathoverflow question <a href="https://mathoverflow.net/questions/316708/what-is-the-smallest-unsolved-diophantine-equation">What is the smallest unsolved diophantine equation?</a> but was not asked explicitly as a separate question. The context is that, in a well-defined sense for the notion of &quot;smallness&quot;, the equation above is the &quot;smallest&quot; open Diophantine equation.</p> <p>Second, this equation is one of the simplest non-trivial representative of the family of equations <span class="math-container">$ax^2+by^2+cz^2=dxyz+e$</span>, which generalises a well-known Markoff equation <span class="math-container">$x^2+y^2+z^2=3xyz$</span>. The well-known methods for the former (Vieta jumping) has been extended to the general case if <span class="math-container">$a,b,c$</span> are all natural numbers and are divisors of <span class="math-container">$d$</span> (see, for example, Fine, Benjamin, et al. &quot;On the Generalized Hurwitz Equation and the Baragar–Umeda Equation.&quot; Results in Mathematics 69.1-2 (2016): 69-92). The question seems to be much more challenging when <span class="math-container">$a,b,c$</span> have different signs. The simplest case with different signs is <span class="math-container">$a=b=d=1$</span> and <span class="math-container">$c=-1$</span>, which leads to the family of equations <span class="math-container">$x^2+y^2-z^2=xyz+e$</span>. The equation above is the first non-trivial example from this family.</p> https://mathoverflow.net/q/392991 -1 Examples of transitive geodesic flows that are not ergodic alvarezpaiva https://mathoverflow.net/users/21123 2021-05-17T12:12:53Z 2021-05-17T12:43:45Z <p>What would be an easy example of a transitive geodesic flow (defined as: there is a geodesic whose velocity vectors are dense on the unit tangent bundle) that is not ergodic?</p> https://mathoverflow.net/q/392990 5 What is the "Prikry–Silver collapse" when CH fails? Asaf Karagila https://mathoverflow.net/users/7206 2021-05-17T11:31:37Z 2021-05-17T13:34:12Z <p>We all know and love Cohen reals, and we can (and often do) define the Cohen forcing as partial functions <span class="math-container">$p\colon\omega\to 2$</span> with finite domain. The Prikry–Silver forcing is defined as partial functions <span class="math-container">$p\colon\omega\to 2$</span> with co-infinite domain.</p> <p>These two couldn't be any more different. For example, Cohen reals are aggressively non-minimal, whereas Prikry–Silver reals are minimal.</p> <p>We can look at a similar situation with other forcings that are given by finite conditions. For example <span class="math-container">$\operatorname{Col}(\omega,\omega_1)$</span> is a forcing notion whose conditions are finite partial functions <span class="math-container">$p\colon\omega\to\omega_1$</span>. We can ask what would be the Prikry–Silver analogue of this forcing, then. That is, <span class="math-container">$\{p\colon\omega\to\omega_1\mid\operatorname{dom} p\text{ is co-infinite}\}$</span>.</p> <p>Interestingly, assuming CH this is the same as the standard collapsing forcing. This follows from the fact that the cardinality of the partial order is <span class="math-container">$2^{\aleph_0}$</span>, which under CH is just <span class="math-container">$\aleph_1$</span>, and we know that any forcing of size <span class="math-container">$\aleph_1$</span> which collapses <span class="math-container">$\omega_1$</span> is equivalent to <span class="math-container">$\operatorname{Col}(\omega,\omega_1)$</span>.</p> <blockquote> <p><strong>Question.</strong> Is this so-called &quot;Prikry–Silver collapse&quot; provably equivalent to the standard collapsing forcing?</p> </blockquote> https://mathoverflow.net/q/392985 0 Sum involving consequtive prime numbers Andrej Leško https://mathoverflow.net/users/169583 2021-05-17T09:50:53Z 2021-05-17T11:49:59Z <p>Let <span class="math-container">$p_i$</span> be the <span class="math-container">$i$</span>th prime number, <span class="math-container">$q$</span> any prime number and <span class="math-container">$m$</span> a positive integer <span class="math-container">$\geq 1$</span>. For real positive <span class="math-container">$x$</span> define the set: <span class="math-container">$$T_m(x,q)=\left\{p_i\leqslant x;\textrm{frac}\left(\frac{p_{i+1}-1}{q^m}\right) &lt; \textrm{frac}\left(\frac{p_i-1}{q^m}\right)\right\}$$</span> <strong>Question:</strong> how to estimate the sum <span class="math-container">$T(x,q)=\sum_{m\geq1}\vert{T_m(x,q)\vert}$</span>, as <span class="math-container">${x\to\infty}$</span>, where <span class="math-container">$\textrm{frac}(u)$</span> is the fractional part of <span class="math-container">$u$</span> and <span class="math-container">$\vert{T_m(x,q)\vert}$</span> the cardinality of the set <span class="math-container">$T_m(x,q)$</span>?</p> https://mathoverflow.net/q/392980 3 Is the Sato-Tate conjecture known for Bianchi modular forms? Barinder Banwait https://mathoverflow.net/users/5744 2021-05-17T08:50:33Z 2021-05-17T12:48:40Z <p>Originally formulated for elliptic curves, the Sato-Tate conjecture regarding the equidistribution of Frobenius trace values according to the Haar measure on a certain compact group (the Sato-Tate group) was generalised in 1994 by Serre to any motive over a number field - see 13.5? in . As is now well known, the Sato-Tate conjecture has been proven for elliptic curves over totally real fields  as well as for all Hilbert modular forms .</p> <p>More recently, in 2018, the authors in  prove modularity lifting theorems for Galois representations over CM fields which yield several attractive corollaries including the Sato-Tate conjecture for elliptic curves over CM fields.</p> <p>My main question is the following:</p> <h3>Does it follow from the work of  that the Sato-Tate conjecture is known for some class of cuspidal automorphic forms for <span class="math-container">$\mbox{GL}_2$</span> over CM fields?</h3> <p>I am chiefly concerned with the case of the CM field being an imaginary quadratic field, in which case the automorphic forms are often called <em>Bianchi modular forms</em>, hence the question in the title. A concise account of cusp forms for <span class="math-container">$\mbox{GL}_2$</span> associated to more general CM fields may be found in Section 2 of .</p> <p>If I may hazard a guess at my question above, I would think that it is currently known only for automorphic <span class="math-container">$\mbox{GL}_2$</span> newforms over CM fields of parallel weight 2 with rational Fourier coefficients. But I am not an expert in this area, hence my question on this forum.</p> <p>I am also curious to know the following:</p> <h3>What is the precise formulation of the Sato-Tate conjecture for general cuspidal automorphic forms for <span class="math-container">$\mbox{GL}_2$</span> over CM fields?</h3> <p>I think  also establishes the Generalised Ramanujan Conjecture (GRC) in the parallel weight 2 setting. For general parallel weight <span class="math-container">$k$</span>, does GRC imply that there is an inequality</p> <p><span class="math-container">$$\left|\frac{c(\frak{p})}{\mbox{Nm}({\frak{p}})^{\frac{k-1}{2}}}\right| &lt; 2,$$</span></p> <p>where the <span class="math-container">$c(\frak{p})$</span> are the Fourier coefficients of the (cuspidal new <span class="math-container">$\mbox{GL}_2$</span>) automorphic form at the integral prime ideals of the CM field <span class="math-container">$K$</span>? If so, then I can begin to see how one can at least start talking about equidistribution.</p> <p>References:</p> <p>: P. B. Allen, F. Calegari, A. Caraiani, T. Gee, D. Helm, B. V. Le Hung, J. Newton, P. Scholze, R. Taylor, J.A. Thorne. <em>Potential automorphy over CM fields</em>. <a href="https://arxiv.org/abs/1812.09999" rel="nofollow noreferrer">arXiv:1812.09999</a> (2018)</p> <p>: T.Barnet-Lamb, T. Gee, D.Geraghty. <em>The Sato-Tate cojecture for Hilbert Modular forms</em>. JAMS (2011)</p> <p>: T. Barnet-Lamb, D. Geraghty, M. Harris and R. Taylor . <em>A family of Calabi-Yau varieties and potential automorphy II</em>. P.R.I.M.S. 47 (2011)</p> <p>: E. Ghate. <em>Critical values of twisted tensor L-functions over CM-Fields</em>. Proc. Symp. Pure Math. (1999)</p> <p>: J.P. Serre. <em>Propriétés conjecturales des groupes de Galois motiviques et des représentations <span class="math-container">$\ell$</span>-adiques</em>. Proc. Symp. Pure Math. (1994)</p> https://mathoverflow.net/q/392979 5 Intersection of $\mathrm{PGL}_2(q_0)$'s in $\mathrm{PGL}_2(q_0^3)$ Nick Gill https://mathoverflow.net/users/801 2021-05-17T08:36:03Z 2021-05-17T12:08:51Z <p><span class="math-container">$\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\GL{GL}$</span>I would like an explanation for some strange numerology which I encountered when studying intersections of subfield subgroups in <span class="math-container">$\PGL_2(q)$</span>. Here's the situation...</p> <p>Let <span class="math-container">$G=\PGL_2(q)$</span> and let <span class="math-container">$H=\PGL_2(q_0)$</span> with <span class="math-container">$q=q_0^a$</span> for <span class="math-container">$a$</span> an odd integer at least <span class="math-container">$3$</span>. Let <span class="math-container">$H_1$</span> be a distinct conjugate of <span class="math-container">$H$</span> in <span class="math-container">$G$</span>. I believe that the possible intersections of <span class="math-container">$H$</span> and <span class="math-container">$H_1$</span> are isomorphic to one of the following: <span class="math-container">$$\{1\}, \,\, C_{q_0-1} \,\, C_{q_0+1}\,\, \textrm{or} \,\, [q_0].$$</span> Note that, provided <span class="math-container">$q_0&gt;3$</span>, there is one conjugacy class of groups of each of these sizes in <span class="math-container">$H$</span>.</p> <p>Suppose that <span class="math-container">$L$</span> is a subgroup of <span class="math-container">$H$</span> of one of the above isomorphism types. Define <span class="math-container">$$x(L) = \{ H_3 \mid H_3 \in H^g, \, H_3 \cap H=L\}.$$</span> Now assume that <span class="math-container">$q=q_0^3$</span> with <span class="math-container">$q_0&gt;3$</span>. Under this assumption we have the following remarkable numerology:</p> <ol> <li>If <span class="math-container">$L\cong C_{q_0-1}$</span>, then <span class="math-container">$|x(L)|=q_0^2+q_0 = |H: L|$</span>.</li> <li>If <span class="math-container">$L\cong C_{q_0+1}$</span>, then <span class="math-container">$|x(L)|=q_0^2-q_0 = |H:L|$</span>.</li> <li>If <span class="math-container">$L\cong [q_0]$</span>, then <span class="math-container">$|x(L)|=q_0^2-1 = |H:L|$</span>.</li> </ol> <p>Note that if <span class="math-container">$L\cong \{1\}$</span>, then <span class="math-container">$|x(L)|=q_0^6-q_0^3-q_0^2+q_0+1$</span> which is not such an interesting number. One could extend the definition of <span class="math-container">$L$</span> to include <span class="math-container">$\PGL_2(q_0)$</span> itself, in which case <span class="math-container">$|x(L)|=1=|H:L|$</span> but whatever. There does not seem (to me) be an obvious reason for these sets to have size <span class="math-container">$|H:L|$</span> in each case. It certainly depends upon the fact that <span class="math-container">$q=q_0^3$</span> -- it would not be true if <span class="math-container">$q=q_0^a$</span> for <span class="math-container">$a\neq 3$</span>. Perhaps there is some group isomorphic to <span class="math-container">$H$</span> acting on the set <span class="math-container">$x(L)$</span>?</p> <p>Some notes:</p> <ol> <li>If we want to calculate <span class="math-container">$|x(L)|$</span> for <span class="math-container">$L$</span> non-trivial, then there is an easy method. We observe that any pair of <span class="math-container">$G$</span>-conjugates of <span class="math-container">$L$</span> in <span class="math-container">$H$</span> are, in fact <span class="math-container">$H$</span>-conjugate. It is then easy enough to prove that the number of conjugates of <span class="math-container">$H$</span> that contain <span class="math-container">$L$</span> is <span class="math-container">$|N_G(L)|/|N_H(L)|$</span>. Since the possible intersections are so restrictive, one then obtains that <span class="math-container">$$|x(L)|=\frac{|N_G(L)|}{|N_H(L)|} - 1.$$</span> So I am asking why, in all cases when <span class="math-container">$L$</span> is non-trivial, we should have <span class="math-container">$$\frac{|N_G(L)|}{|N_H(L)|} - 1= |H:L|.$$</span></li> <li>I ran some GAP code to check one possibility: suppose that <span class="math-container">$G=\PGL_2(5^3), H=\PGL_2(5)$</span> and <span class="math-container">$L$</span> is a cyclic subgroup of <span class="math-container">$H$</span> of order <span class="math-container">$4$</span>. Consider the corresponding set <span class="math-container">$x(L)$</span>, a set of size 30 whose elements are conjugates of <span class="math-container">$H$</span>. I wondered if this set might be an orbit of some group <span class="math-container">$H_2$</span>, a conjugate of <span class="math-container">$H$</span> in <span class="math-container">$G$</span>, acting on <span class="math-container">$H^G$</span> by conjugation. But computer says NO.</li> <li>If you prefer to think in <span class="math-container">$\GL_2(q)$</span> rather than <span class="math-container">$\PGL_2(q)$</span>, then please crack on.</li> </ol> https://mathoverflow.net/q/392978 7 Completely positive maps on Coxeter groups - the general case worldreporter14 https://mathoverflow.net/users/64444 2021-05-17T08:21:11Z 2021-05-17T12:22:06Z <p>In the reference below Bozejko and Speicher showed the following (for the full statement see the remark on page 9):</p> <blockquote> <p>Let <span class="math-container">$(W,S)$</span> be a Coxeter system, let <span class="math-container">$\mathcal{H}$</span> be a Hilbert space and denote the group algebra of <span class="math-container">$W$</span> by <span class="math-container">$\mathbb{C}[W]$</span>. Let further <span class="math-container">$T_s \in \mathcal{B(H)}$</span>, <span class="math-container">$s\in S$</span> be self-adjoint, contractive operators which satisfy for all <span class="math-container">$s,t \in S$</span> the Braid relation <span class="math-container">$T_s T_t ... =T_tT_s ...$</span> where the number of factors coincides with the coefficient <span class="math-container">$m_{s,t}$</span>. Then, if <span class="math-container">$W$</span> is amenable, the quasi-multiplicative map <span class="math-container">$\varphi: \mathbb{C}[W]\rightarrow \mathcal{B(H)}$</span> defined by <span class="math-container">$\varphi(e)=1$</span>, <span class="math-container">$\varphi(s)=T_s$</span> is completely positive.</p> </blockquote> <p>On page 10 it is asked whether the statement above holds for all infinite (possibly non-amenable) Coxeter groups. I'm wondering what the status of this question is. Is it still open? Bozejko and Speicher announced a proof for the case where <span class="math-container">$\mathcal{H}=\mathbb{C}$</span>, but I couldn't find the corresponding paper. Was it ever published?</p> <p><em>Bożejko, Marek; Speicher, Roland</em>, <a href="http://dx.doi.org/10.1007/BF01450478" rel="nofollow noreferrer"><strong>Completely positive maps on Coxeter groups, deformed commutation relations, and operator spaces</strong></a>, Math. Ann. 300, No. 1, 97-120 (1994). <a href="https://zbmath.org/?q=an:0819.20043" rel="nofollow noreferrer">ZBL0819.20043</a>.</p> https://mathoverflow.net/q/392974 2 An integral transform and the Stone-Weierstrass theorem Jun https://mathoverflow.net/users/110835 2021-05-17T06:24:12Z 2021-05-17T11:01:33Z <p>For a bounded function <span class="math-container">$\operatorname{F}: \mathbb{R}_{\,\ge\ 0} \to \mathbb{R}$</span> (<em>not necessarily non-negative</em>), if <span class="math-container">$$\int_{0}^{\infty}\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad \forall s &gt; 0$$</span> where <span class="math-container">$k \in \mathbb{N}$</span> is a positive constant, is it true that <span class="math-container">$$\int_{0}^{\infty}\left(\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\right)^{h} \, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad \forall s &gt; 0$$</span> where <span class="math-container">$h \in \mathbb{N}$</span>?</p> <p>This question is inspired by a comment to the answer in <a href="https://math.stackexchange.com/questions/3996738/condition-for-an-integral-to-be-zero">https://math.stackexchange.com/questions/3996738/condition-for-an-integral-to-be-zero</a> (which required to check the assumptions of the Stone-Weierstrass theorem).</p> <p>More generally, ignoring the question above, my main concern is this:</p> <ul> <li>Use Stone-Weierstrass to prove that <span class="math-container">$$\int_{0}^{\infty}\frac{x^{k}\,s}{(s^{2} + x^{2})^{\left(k + 3\right)/2}\,\,}\, \operatorname{F}\left(x\right)\,{\rm d}x = 0\quad \forall s &gt; 0$$</span> if and only if <span class="math-container">$F \equiv 0$</span>.</li> </ul> https://mathoverflow.net/q/392973 10 Problems concerning subspaces of $M_{n}(\mathbb{Q})$ Sugata Mandal https://mathoverflow.net/users/215016 2021-05-17T06:07:56Z 2021-05-17T10:42:48Z <p>Let <span class="math-container">$M_{n}(\mathbb{Q})$</span> denote the <span class="math-container">$n$</span> times <span class="math-container">$n$</span> matrices over the rational number field. <span class="math-container">$N$</span> be a subspace of <span class="math-container">$M_{n}(\mathbb{Q})$</span>. Then if all the non-zero matrices in <span class="math-container">$N$</span> are invertible, what is the maximum the dimension of <span class="math-container">$N$</span> can be?</p> <p>We already know that if we take <span class="math-container">$M_{n}(\mathbb{R})$</span> instead of <span class="math-container">$M_{n}(\mathbb{Q})$</span> then the answer is <span class="math-container">$\rho(n)$</span>. where <span class="math-container">$\rho(n)$</span> is Radon Hurwitz number i.e if <span class="math-container">$n = (2a + 1 ) 2^{c+4d}$</span> where <span class="math-container">$0 \leq c \leq 3$</span> then <span class="math-container">$\rho(n) = 2^{c} + 8d$</span> .</p> https://mathoverflow.net/q/392961 2 Random walk on $n$-dimensional cube gks https://mathoverflow.net/users/171475 2021-05-17T01:21:54Z 2021-05-17T14:10:53Z <p>Consider a symmetric random walk along the edges of an <span class="math-container">$n$</span>-dimensional unit cube. At each time step, a particle located at a particular vertex <span class="math-container">$(a_1, \ldots, a_n)$</span> moves to an adjacent neighbor each with equal probability.</p> <p>For an arbitrary starting point <span class="math-container">$(a_1, \ldots, a_n)$</span>, how can I compute the probability that we will hit <span class="math-container">$(1, 1, \ldots, 1)$</span> before hitting <span class="math-container">$(0, 0, \ldots, 0)$</span>? This will be a function of <span class="math-container">$n$</span>.</p> <p>I thought about making a Markov chain with <span class="math-container">$2^{n}$</span> states, but I'm not entirely sure if this is the right approach. Under this representation, I'm pretty sure some state <span class="math-container">$u = (a_1, \ldots, a_n)$</span> can move to some other state if and only if we can retrieve the second state by turning a bit off in <span class="math-container">$u$</span> or turning a bit on (that is not already on) in <span class="math-container">$u$</span>. I think this may be the best approach, but I'm a bit stuck.</p> <p>I wrote an expression for the probability of reaching the state with all ones prior to the state with all zeros for each different starting state, and I also use each temr in the expressions for others (so we need to solve for the variables), but I see no easy way to finish.</p> <p>A solution is provided here <a href="https://arxiv.org/pdf/0711.2675.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/0711.2675.pdf</a>, but it uses circuits and a less mathematical approach to derive the answer.</p> <p>Any help is appreciated.</p> https://mathoverflow.net/q/392932 2 An information-theoretic upper-bound on prime gaps Aidan Rocke https://mathoverflow.net/users/56328 2021-05-16T15:53:57Z 2021-05-17T14:08:24Z <h2>An upper-bound that is consistent with Cramer's conjecture:</h2> <p>While going over a <a href="https://mathoverflow.net/questions/384109/information-theoretic-derivation-of-the-prime-number-theorem">related question</a>, it occurred to me to define the expected information gained from observing large primes that are consecutive:</p> <p><span class="math-container">\begin{equation} \Delta I_N = \mathbb{E}[K_U([1,P_{N+1}])] - \mathbb{E}[K_U([1,P_{N}])] \geq 0 \tag{1} \end{equation}</span></p> <p>where the expectation is calculated over prefix-free Universal Turing Machines <span class="math-container">$U \in \{U_n\}_{n \in \mathbb{N}}$</span>. By the Invariance theorem for Kolmogorov Complexity , we may choose any computable probability distribution <span class="math-container">$P$</span> over <span class="math-container">$\{U_n\}_{n \in \mathbb{N}}$</span> without affecting the asymptotics of <span class="math-container">$\Delta I_N$</span>.</p> <p>Now, given Chaitin's theorem that almost all integers are incompressible, for large <span class="math-container">$N$</span> we have:</p> <p><span class="math-container">\begin{equation} \mathbb{E}[K_U([1,P_{N}])] \sim P_N \cdot \ln P_N - P_N \tag{2} \end{equation}</span></p> <p><span class="math-container">\begin{equation} \mathbb{E}[K_U([1,P_{N+1}])] \sim P_{N+1} \cdot \ln P_{N+1} - P_{N+1} \tag{3} \end{equation}</span></p> <p>which gives us useful asymptotic bounds:</p> <p><span class="math-container">\begin{equation} 0 \leq \Delta I_N \leq P_{N+1} \cdot \ln P_{N+1} - P_{N} \cdot \ln P_{N} \tag{4} \end{equation}</span></p> <p><span class="math-container">\begin{equation} 0 \leq P_{N+1} - P_N \leq P_{N+1} \cdot \ln P_{N+1} - P_{N} \cdot \ln P_{N} \tag{5} \end{equation}</span></p> <p>Considering (5), we may observe that:</p> <p><span class="math-container">\begin{equation} P_N \sim \pi(N) \cdot \ln P_N = N \cdot \ln P_N \tag{6} \end{equation}</span></p> <p>which allows us to deduce:</p> <p><span class="math-container">\begin{equation} P_N \cdot \ln P_N \sim N \cdot (\ln P_N)^2 \tag{7} \end{equation}</span></p> <p>and as Bertrand's postulate implies <span class="math-container">$\ln P_N \leq \ln P_{N+1} &lt; \ln P_{N+1} + 1$</span> we also have:</p> <p><span class="math-container">\begin{equation} P_{N+1} \cdot \ln P_{N+1} \sim (N+1) \cdot (\ln P_N)^2 \tag{8} \end{equation}</span></p> <p>By combining (5), (7) and (8) we obtain the following upper-bound on prime gaps that is consistent with Cramer's conjecture :</p> <p><span class="math-container">\begin{equation} 0 \leq P_{N+1} - P_N \leq (\ln P_N)^2 \tag{9} \end{equation}</span></p> <h2>Question:</h2> <p>This analysis makes me wonder whether such investigations into the distribution of prime gaps have been pursued in greater detail using tools from algorithmic information theory. My intuition suggests that information-theoretic tools may be useful in the analysis of the average case besides the worst case(considered here).</p> <hr /> <h2>Proof that <span class="math-container">$K_U([1,N]) \sim N\cdot \ln N -N$</span>:</h2> <p>If we define the computable function <span class="math-container">$f$</span>:</p> <p><span class="math-container">\begin{equation} \forall A \subset \mathbb{N}, f \circ A = \prod_{i=1}^{|A|} a_i \tag{10} \end{equation}</span></p> <p>then for a fixed prefix-free UTM <span class="math-container">$U$</span>:</p> <p><span class="math-container">\begin{equation} K_U(f \circ A) \leq K_U(A) + \text{Cst} \tag{11} \end{equation}</span></p> <p>where <span class="math-container">$K_U(f) = \text{Cst}$</span>.</p> <p>Now, an integer <span class="math-container">$N$</span> is incompressible if <span class="math-container">$K_U(N) \sim \ln(N)$</span> so we may deduce the following upper and lower bounds on <span class="math-container">$K_U([1,N])$</span>:</p> <p><span class="math-container">\begin{equation} K_U(N!) - \text{Cst} \leq K_U([1,N]) \leq N \cdot \ln N - N \tag{12} \end{equation}</span></p> <p>where <span class="math-container">$K_U(N!)\sim N \cdot \ln N - N$</span> and so it follows that for large <span class="math-container">$N$</span>:</p> <p><span class="math-container">\begin{equation} K_U([1,N]) \sim N \cdot \ln N - N \tag{13} \end{equation}</span></p> <h2>An important Corollary:</h2> <p>If the prime encoding <span class="math-container">$X_N = \{x_n\}_{n=1}^N$</span> is defined such that <span class="math-container">$x_n =1$</span> if <span class="math-container">$n$</span> is prime and <span class="math-container">$x_n = 0$</span> otherwise, then due to the derivation in  the information gained from observing each prime number in <span class="math-container">$[1,N]$</span> is given by:</p> <p><span class="math-container">\begin{equation} \mathbb{E}[K_U(X_N)] \sim \pi(N) \cdot \ln N \sim N \tag{14} \end{equation}</span></p> <p>so if we define the difference in information gained:</p> <p><span class="math-container">\begin{equation} \Delta I_N' = \mathbb{E}[K_U(X_{P_{N+1}})] - \mathbb{E}[K_U(X_{P_{N}})] \sim P_{N+1} - P_N \tag{15} \end{equation}</span></p> <p>an important consequence of (4) and (9) is that:</p> <p><span class="math-container">\begin{equation} \max (\Delta I_N, \Delta I_N') \leq (\ln P_N)^2 \tag{16} \end{equation}</span></p> <p>I think this is a key information-theoretic insight.</p> <h2>References:</h2> <ol> <li>A. N. Kolmogorov. Three approaches to the quantitative definition of information. Problems of Information and Transmission, 1(1):1–7, 1965</li> <li>G. J. Chaitin, A theory of program size formally identical to information theory, Journal of the ACM 22, (1975) 329–340.</li> <li>Ming Li &amp; Paul Vitányi. Kolmogorov Complexity and its Applications. Handbook of Theoretical Computer Science, Volume A: Algorithms and Complexity. 1990.</li> <li>Peter Grünwald and Paul Vitányi. Shannon Information and Kolmogorov Complexity. 2010.</li> <li>Cramér, H. &quot;On the Order of Magnitude of the Difference Between Consecutive Prime Numbers.&quot; Acta Arith. 2, 23-46, 1936.</li> <li>Granville, A. &quot;Harald Cramér and the Distribution of Prime Numbers.&quot; Scand. Act. J. 1, 12-28, 1995.</li> <li>K. Ford, B. Green, S. Konyagin, and T. Tao, Large gaps between consecutive prime numbers. Ann. of Math. (2) 183 (2016), no. 3, 935–974</li> <li>Aidan Rocke (<a href="https://mathoverflow.net/users/56328/aidan-rocke">https://mathoverflow.net/users/56328/aidan-rocke</a>), information-theoretic derivation of the prime number theorem, URL (version: 2021-04-08): <a href="https://mathoverflow.net/q/384109">information-theoretic derivation of the prime number theorem</a></li> </ol> https://mathoverflow.net/q/392922 0 Bounding the $n$-th term of a sequence, given a non-linear recursive bound UserA https://mathoverflow.net/users/105925 2021-05-16T13:45:23Z 2021-05-17T14:23:06Z <p>I asked the <a href="https://math.stackexchange.com/questions/4136248/bounding-the-nth-term-of-a-sequence-given-a-recursive-non-linear-bound">following</a> question in MSE:</p> <blockquote> <p>Let <span class="math-container">$a,b\in\mathbb{R}^+$</span>. Suppose that <span class="math-container">$\{x_n\}_{n=0}^\infty$</span> is a sequence satisfying <span class="math-container">$$|x_n|\leq a|x_{n-1}|+b|x_{n-1}|^2,$$</span> for all <span class="math-container">$n\in\mathbb{N}$</span>. How can we bound <span class="math-container">$|x_n|$</span> with a number <span class="math-container">$M_n$</span> depending on <span class="math-container">$n$</span>, <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, and <span class="math-container">$x_0$</span>?</p> </blockquote> <p>However, I did not get an answer. It seems that there should be some obvious observation to give a naive upper bound? Although it would be nice to have an asymptotically tight bound in closed form, if possible of course!</p> https://mathoverflow.net/q/392901 2 On the equation $[U, V] - V_x = C(x)$ JPwin https://mathoverflow.net/users/170939 2021-05-16T05:08:03Z 2021-05-17T10:37:17Z <p>While considering the zero curvature equation <span class="math-container">$U_t - V_x + [U, V] = 0$</span>, I developed a similar problem, albeit one that discards time dependence entirely. For a given <span class="math-container">$U(x)$</span> and <span class="math-container">$C(x)$</span>, find <span class="math-container">$V(x)$</span> such that <span class="math-container">$[U, V] - V_x = C(x)$</span> holds. After considering this <a href="https://math.stackexchange.com/questions/1307098/is-there-any-inverse-commutator-for-matrices">question</a>, I believe that the system would either have a single solution or an infinite class of solutions depending on the choice of <span class="math-container">$C$</span> and <span class="math-container">$U$</span>. In the <a href="https://math.stackexchange.com/questions/1307098/is-there-any-inverse-commutator-for-matrices">question</a>, I found the method involving the Kronecker product to be more suited for matrix functions as I ultimately hope to prove that the compatibility condition is isospectral. (I do <em>not</em> want to assume that the eigenvalues are constant across space, as I believe that it would produce a circular argument.)</p> <p>I had also wondered if the system could be transformed by finding <span class="math-container">$\gamma(x)$</span> and <span class="math-container">$\eta(x)$</span> such that <span class="math-container">$[U + \gamma, V + \eta] = [U,V] - V_x$</span>. Ultimately, I realized such an attempt would reduce to solving <span class="math-container">$[\eta, \gamma] + [V, \gamma] + [\eta, U] = V_x$</span>, which is no simpler than the original equation. Most of the difficulty arises from the lack of commutativity of the matrix product.</p> <p><strong>Question:</strong> For a given <span class="math-container">$U(x)$</span> and <span class="math-container">$C(x)$</span>, can one solve for <span class="math-container">$V(x)$</span> such that <span class="math-container">$[U, V] - V_x = C(x)$</span> holds?</p> https://mathoverflow.net/q/392890 15 Which unfoldings of the $d$-dimensional hypercube tile $(d{-}1)$-space? Joseph O'Rourke https://mathoverflow.net/users/6094 2021-05-15T23:08:08Z 2021-05-17T14:35:38Z <p>A six year old question, <a href="https://mathoverflow.net/q/199097/6094">Which unfoldings of the hypercube tile <span class="math-container">$3$</span>-space?</a>, has just been answered by <a href="https://mathoverflow.net/a/392828/6094">Moritz Firsching</a>: All <span class="math-container">$261$</span> unfoldings tile space! So now we know:</p> <ul> <li>For <span class="math-container">$d=2$</span>, the unfolding of the square tiles <span class="math-container">$\mathbb{R}$</span>.</li> <li>For <span class="math-container">$d=3$</span>, each of the <span class="math-container">$11$</span> unfoldings of the <span class="math-container">$\mathbb{R}^3$</span>-cube tile <span class="math-container">$\mathbb{R}^2$</span>.</li> <li>For <span class="math-container">$d=4$</span>, each of the <span class="math-container">$261$</span> unfoldings of the <span class="math-container">$\mathbb{R}^4$</span>-hypercube tile <span class="math-container">$\mathbb{R}^3$</span>.</li> </ul> <p>The natural next question (also posed at <a href="https://matheducators.stackexchange.com/a/20883/511">MESE</a>) is:</p> <blockquote> <p><em><strong>Q</strong></em>. Is it true that, for every <span class="math-container">$d$</span>, each of the unfoldings of the <span class="math-container">$d$</span>-dimensional cube tiles <span class="math-container">$\mathbb{R}^{d-1}$</span>? If not, up to which <span class="math-container">$d$</span> does this hold?</p> </blockquote> <p>Brute-force calculation may be limited by the combinatorics: <a href="https://mathoverflow.net/q/300713/6094">Number of hypercube unfoldings</a>, <a href="https://oeis.org/A091159/internal" rel="nofollow noreferrer">OEIS:A091159</a>: <span class="math-container">$$1,11,261,9694,502110,33064966,2642657228 ,\ldots$$</span></p> <p><em>Added</em>. I should add that recently Satyan Devadoss and his students proved that no unfolding of a <span class="math-container">$d$</span>-dimensional hypercube self-overlaps, i.e., each forms a <a href="https://en.wikipedia.org/wiki/Net_(polyhedron)" rel="nofollow noreferrer">net</a>.</p> <blockquote> <p>DeSplinter, Kristin, Satyan L. Devadoss, Jordan Readyhough, and Bryce Wimberly. &quot;Nets of higher-dimensional cubes.&quot; <em>Canad. Conf. Comput. Geom.</em> 2020, pp.114-120. <a href="https://www.cccg.ca/" rel="nofollow noreferrer">CCCG proceedings link</a>.</p> </blockquote> https://mathoverflow.net/q/392813 3 Bounding the max-loaded bin using${m \choose k} \|A\|_k^k$ usul https://mathoverflow.net/users/29697 2021-05-15T02:11:41Z 2021-05-17T14:46:27Z <p>Throw <span class="math-container">$m$</span> balls into <span class="math-container">$n$</span> bins independently, each ball selecting a bin from the distribution <span class="math-container">$A \in \Delta_n$</span>. This question is about <strong>lower-bounding the max-loaded bin</strong>.</p> <p><strong>Background.</strong> In <a href="https://mathoverflow.net/a/323756/29697">this MO answer</a> I wrote about an upper bound based on collisions. Let <span class="math-container">$Z_k$</span> be all subsets of <span class="math-container">$k$</span> distinct balls. For <span class="math-container">$S \in Z_k$</span>, let <span class="math-container">$1_S$</span> be the indicator that all balls in <span class="math-container">$S$</span> land in the same bin. Then <span class="math-container">$\mathbb{E} [1_S] = \sum_{i=1}^n A_i^k = \|A\|_k^k$</span>. Let <span class="math-container">$C_k = \sum_{S \in Z_k} 1_S$</span>, the number of <span class="math-container">$k$</span>-way collisions. Then by Markov's inequality, <span class="math-container">$$\Pr[ \text{max-loaded bin } \geq k] = \Pr[ C_k \geq 1] \leq \mathbb{E}[C_k] = {m \choose k} \|A\|_k^k .$$</span></p> <p>For example, in the classic case of <span class="math-container">$m=n$</span> and uniform <span class="math-container">$A$</span>, where <span class="math-container">$\|A\|_k^k = n^{1-k}$</span>, we can use Stirling to get a bound closely approaching <span class="math-container">$\frac{n}{k^k}$</span>, as expected. And this bound can be very tight: if we throw half as many balls (cut <span class="math-container">$m$</span> in half), the probability decreases by a factor of about <span class="math-container">$2^k$</span>.</p> <p><strong>Question.</strong></p> <blockquote> <p>Is there a lower-bound on the max-loaded bin that matches this upper bound, in a sense? Or, what is the tightest non-asymptotic lower bound you know for this setting?</p> </blockquote> <p><strong>Motivation.</strong> First, notice that if we pretended each collision <span class="math-container">$1_S$</span> were independent, we would obtain <span class="math-container">$$\Pr[ \text{max-loaded bin} &lt; k ] = \Pr[ C_k = 0] = \Pr[1_S = 0 ~ (\forall S \in Z_k)] \leq \left(1 - \|A\|_k^k\right)^{m \choose k} \leq \exp\left(- {m \choose k} \|A\|_k^k\right) .$$</span></p> <p>That would be such a cool result, matching the upper bound so neatly. Unfortunately, the collisions are positively associated, not negatively: <span class="math-container">$\Pr[1_S = 1 \mid 1_{S'} = 1] \geq \Pr[1_S = 1]$</span>. So I don't know of techniques to prove this. (Yet simulations suggest something not too far away could hold, at least for <span class="math-container">$A$</span> uniform...)</p> <p><strong>What else I've tried.</strong> Well, if <span class="math-container">$X_i$</span> is the number of balls in bin <span class="math-container">$i$</span>, then the <span class="math-container">$X_i$</span> are negatively associated, so I think we can get a bound by pretending the <span class="math-container">$X_i$</span> are independent Binomials, but it doesn't match. The standard approach would be to bound the variance of <span class="math-container">$C_k$</span> and use Chebyshev. I was able to get an only-somewhat-horrible expression for <span class="math-container">$\text{Var}(C_k)$</span>, but I had trouble pushing it through to get a tight bound here. It might work. Finally, I'll mention that <a href="http://www.dblab.ntua.gr/%7Egtsat/collection/scheduling/Balls%20into%20Bins%20A%20Simple%20and%20Tight%20Anal.pdf" rel="nofollow noreferrer">Raab and Steger</a> is only asymptotic, whereas I'm hoping with this approach to get a concrete bound for any given <span class="math-container">$m,n,k$</span>.</p> https://mathoverflow.net/q/392596 3 Cutting convex polygons into triangles of same diameter Nandakumar R https://mathoverflow.net/users/142600 2021-05-12T15:15:07Z 2021-05-17T11:31:25Z <p>This question continues from: <a href="https://mathoverflow.net/questions/375536/cutting-convex-regions-into-equal-diameter-and-equal-least-width-pieces">Cutting convex regions into equal diameter and equal least width pieces</a></p> <p><strong>Definitions:</strong> The diameter of a convex region is the greatest distance between any pair of points in the region. The least width of a 2D convex region can be defined as the least distance between any pair of parallel lines that touch the region. In above post, it was also noted that for any n, any convex planar region allows partition into n convex pieces all of same diameter (these pieces could have different numbers of edges).</p> <p><strong>Question:</strong> How can any given convex n-vertex polygon be partitioned into the least number of triangular regions all with the same diameter?</p> <p><strong>Note 1:</strong> One observes that any convex quadrilateral can be cut into 2 triangles of same diameter (if the diameter of the quad is one of its diagonals) or 3 triangles of same diameter (if the diameter of the quad is one of its sides). What we ask above is a lower bound on the number on the number of equal diameter triangles into which any given convex n-gon can be cut as a function of n (there is also the possibility that some convex n-gons <em>might not</em> allow partition into any number of equal diameter triangles).</p> <p><strong>Note 2:</strong> For non-convex polygons - even quadrilaterals - it is easy to see that there is no lower bound on the number of equal perimeter triangles into which the polygon can be cut.</p> <p><strong>Note 3:</strong> A similar question can be asked about cutting a convex n-gon into triangles all of same least width - and higher dimensional analogs are conceivable for these questions.</p> <p><strong>Note 4:</strong> Questions of cutting a given convex n-gon into equal area or equal perimeter triangles are more natural. The equal area case is introduced at <a href="https://en.wikipedia.org/wiki/Equidissection" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Equidissection</a> and the equal perimeter case discussed in the specific case of the n-gon being a square is treated here: <a href="https://math.stackexchange.com/questions/2822589/dissect-square-into-triangles-of-same-perimeter">https://math.stackexchange.com/questions/2822589/dissect-square-into-triangles-of-same-perimeter</a></p> https://mathoverflow.net/q/392303 6 $\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$ Evans Gambit https://mathoverflow.net/users/157738 2021-05-09T10:58:08Z 2021-05-17T11:26:15Z <p>I have been reading <a href="https://reader.elsevier.com/reader/sd/pii/S0001870809000450" rel="noreferrer">Asok and Doran - <span class="math-container">$\mathbb A^1$</span>-homotopy groups, excision, and solvable quotients</a>. In the Example 2.17, page 1155, there is a claim that for <span class="math-container">$m&gt;1$</span> and <span class="math-container">$n\geq 1$</span>, <span class="math-container">$\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$</span>.</p> <p>Let me explain the notation quickly:</p> <p><span class="math-container">$\mu_n$</span> is the group scheme of <span class="math-container">$n^\text{th}$</span> roots of unity acting on <span class="math-container">$\mathbb{A}^m-0$</span> by diagonal action. So <span class="math-container">$\mathbb{A}^m-0/\mu_n$</span> is a smooth scheme. <span class="math-container">$\mathbb{G}_m/\mathbb{G}_m^n$</span> is the cokernel of the <span class="math-container">$n^\text{th}$</span>-power map <span class="math-container">$\mathbb{G}_m\xrightarrow{(-)^n} \mathbb{G}_m$</span> in the category of Nisnevich sheaves of abelian groups.</p> <p>My question is how to see the isomorphism <span class="math-container">$\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$</span>?</p> <p>Comments are most welcome!</p> https://mathoverflow.net/q/392282 12 Smoothness of distance function to a compact set James_T https://mathoverflow.net/users/154189 2021-05-08T23:35:11Z 2021-05-17T14:06:55Z <p>Fix a non-empty compact subset <span class="math-container">$K\subseteq \mathbb{R}^n$</span> and let <span class="math-container">$d_K(x):=\min_{z \in K} \,\|z-x\|$</span> be the map sending any <span class="math-container">$x\in \mathbb{R}^n$</span> to its distance from <span class="math-container">$K$</span>.</p> <p>Suppose that:</p> <ul> <li><span class="math-container">$K$</span> is <em>regular</em> : it has a non-empty interior <span class="math-container">$\overset{\circ}{K}$</span>, and the closure of <span class="math-container">$\overset{\circ}{K}$</span> is <span class="math-container">$K$</span>; in particular <span class="math-container">$K$</span> has co-dimension <span class="math-container">$0$</span>.</li> <li><span class="math-container">$K$</span> has a <span class="math-container">$C^{k+1}$</span> boundary.</li> <li><span class="math-container">$K$</span> is convex.</li> </ul> <p>Then:</p> <ul> <li><p><strong>Question 1:</strong> Is there some exponent <span class="math-container">$k+1&lt;p&lt;\infty$</span> such that <span class="math-container">$d_K^p$</span> is also <span class="math-container">$k$</span> times continuously differentiable on <span class="math-container">$\mathbb{R}^n$</span>? Note, this is known to be true locally, on some open neighbourhood of <span class="math-container">$K$</span> (see <a href="https://link.springer.com/book/10.1007/978-3-642-61798-0" rel="nofollow noreferrer">Lemma 14.16</a> for instance). But the result does not require convexity so maybe it is possible to get smoothness on all of <span class="math-container">$\mathbb{R}^n$</span> by incorporating this assumption.</p> </li> <li><p><strong>Question 2:</strong> Are there reasonable &quot;geometric&quot; conditions on <span class="math-container">$K$</span> which will guarantee that <span class="math-container">$d_K^p$</span> is <span class="math-container">$C^k$</span> on <span class="math-container">$\mathbb{R}^n$</span>, for some <span class="math-container">$0&lt;p&lt;\infty$</span>?</p> </li> </ul> https://mathoverflow.net/q/392169 9 The double cover in the classical limit of $U_q(\mathfrak{sl}_2)$ Quin Appleby https://mathoverflow.net/users/160055 2021-05-07T13:13:16Z 2021-05-17T13:10:30Z <p>I am trying to learn about Drinfeld–Jimbo quantum groups and I am having trouble with the classical limit of <span class="math-container">$U_q(\mathfrak{sl}_2)$</span>. When properly expressed the limit makes sense as <span class="math-container">$q\to 1$</span> — see for example this <a href="https://mathoverflow.net/questions/126461/quantized-enveloping-algebras-at-q-1">question</a>.</p> <p>But we get <span class="math-container">$U(\mathfrak{sl}_2) \otimes \mathbb{Z}/2$</span>. This fact goes on to create many problems, such as the effective doubling of the number of representations, half of which are generally regarded as &quot;uninteresting&quot;.</p> <p>It seems to me that finding a presentation of <span class="math-container">$U_q(\mathfrak{sl}_2)$</span> would be of great benefit. However I can't seem to find any discussion of this anywhere. Can anybody explain what is going on here? Is the double cover somehow a 'necessary consequence' of quantization? Is there some reason why people felt it best to stick with the double cover approach?</p> https://mathoverflow.net/q/343783 -8 Can we define a probability measure $\mu$ that follows all my requirments? Arbuja https://mathoverflow.net/users/87856 2019-10-13T18:30:41Z 2021-05-17T14:53:46Z <p><strong>I posted a similar post on <a href="https://math.stackexchange.com/questions/4141241/can-we-define-a-probability-measure-mu-that-follows-all-my-requirments">math stack exchange</a> in case it's not suitable for this site.</strong></p> <p>Suppose we have <span class="math-container">$f:A\to B$</span> and <span class="math-container">$S\subseteq A$</span>.</p> <p>If <span class="math-container">$A'=A^1$</span> is the <a href="https://en.wikipedia.org/wiki/Derived_set_(mathematics)" rel="nofollow noreferrer">derived set</a> of <span class="math-container">$A$</span>, such that <span class="math-container">$A=A^0$</span> and <span class="math-container">$A^{k+1}=\left(A^{k}\right)'$</span>, then if</p> <p><span class="math-container">$$\omega(A)=\begin{cases}\min\left\{k:A^k=\emptyset\right\} &amp; \exists(k)(A^k=\emptyset)\\ \infty &amp; \forall(k)\left(A^{k}\neq \emptyset\right)\end{cases}$$</span></p> <p>Then I wish to define probability measure <span class="math-container">$\mu(S,A)$</span> such that <span class="math-container">$\mu$</span>:</p> <ul> <li>Is additive and monotonic</li> <li>Is zero when <span class="math-container">$\omega(A)&gt;\omega(S)$</span></li> <li>Is <span class="math-container">$1$</span> when <span class="math-container">$S=A$</span></li> <li>Is <span class="math-container">$\lambda(S)/\lambda(A)$</span> when <span class="math-container">$\lambda$</span> is the the <a href="https://en.wikipedia.org/wiki/Derived_set_(mathematics)" rel="nofollow noreferrer">Lebesgue Measure</a> and <span class="math-container">$\lambda(A)&gt;0$</span>.</li> <li>Is zero for countable <span class="math-container">$S$</span> and uncountable <span class="math-container">$A$</span>, regardless of whether <span class="math-container">$\lambda(A)$</span> is positive or zero.</li> </ul> <p>I presume <em>most cases</em> can be accomplished by doing the following:</p> <ul> <li>Intervals which cover <span class="math-container">$S$</span> and <span class="math-container">$A$</span> should have two lengths: <span class="math-container">$0$</span> and <span class="math-container">$\epsilon$</span>.</li> <li>If <span class="math-container">$A$</span> is uncountable, the total number of intervals of covering <span class="math-container">$S$</span> and <span class="math-container">$A$</span> must be <span class="math-container">$+\infty$</span></li> <li>If <span class="math-container">$A$</span> is countable, the <em>only</em> intervals that can cover <span class="math-container">$S$</span> and <span class="math-container">$A$</span> must have <span class="math-container">$\epsilon$</span> length.</li> </ul> <p>Then we compute <span class="math-container">$\mu$</span>, by finding the minimum value of the sum of the length of intervals covering <span class="math-container">$S$</span>, divided by the sum of the length of intervals covering <span class="math-container">$A$</span>.</p> <p><strong>Here is my attempt at mathematically defining <span class="math-container">$\mu$</span>. (I don't know if it would meet my requirements)</strong></p> <blockquote> <p>If <span class="math-container">$f:A\to B$</span> and <span class="math-container">$S\subseteq A$</span>, then:</p> <ul> <li>If <span class="math-container">$\ell$</span> is the length of an interval</li> <li><span class="math-container">$(I_{j,1})_{j=1}^{n_1}$</span> and <span class="math-container">$(I_{j,2})_{j=1}^{n_2}$</span> are a sequence of open intervals covering <span class="math-container">$S$</span> such that <span class="math-container">$\ell(I_{j,1})=\epsilon$</span> and <span class="math-container">$\ell(I_{j,2})=0$</span></li> <li><span class="math-container">$\left(J_{k,1} \right)_{k=1}^{m_1}$</span> and <span class="math-container">$\left(J_{k,2}\right)_{k=1}^{m_2}$</span> is a sequence of open intervals covering <span class="math-container">$A$</span> such that <span class="math-container">$\ell(J_{k,1})=\epsilon$</span> and <span class="math-container">$\ell(J_{k,2})=0$</span></li> </ul> <p><span class="math-container">$$\delta(A)=\begin{cases} m_1+m_2=+\infty &amp; A \; \text{is uncountable} \\ m_2=0 &amp; A \; \text{is countable} \end{cases}$$</span></p> <p><span class="math-container">$$\mu^{*}(S,A)=\inf\limits_{m_1,m_2\in\mathbb{N}\cup\{\infty\}}\left\{\frac{\left\{\sum\limits_{j=1}^{n_1}\ell(I_{j,1})+\sum\limits_{j=1}^{n_2}\ell(I_{j,2}):S\subseteq\bigcup\limits_{r=1}^{2}\bigcup\limits_{j=1}^{n_r}I_{j,r}\right\}}{\left\{\sum\limits_{k=1}^{m_1}\ell(J_{k,1})+\sum\limits_{k=1}^{m_2}\ell(J_{k,2}):\delta(A),A\subseteq\bigcup\limits_{r=1}^{2}\bigcup\limits_{k=1}^{m_r}J_{k,r}\right\}}\right\}$$</span></p> </blockquote> <p>The problem is if <span class="math-container">$S$</span> and <span class="math-container">$A$</span> countable and <span class="math-container">$\omega(S)$</span> and <span class="math-container">$\omega(A)=\infty$</span>; then my definition of <span class="math-container">$\mu$</span> isn't additive.</p> <p>For example, if function <span class="math-container">$f$</span> is:</p> <p><span class="math-container">$$\begin{cases} 2 &amp; S_1=\ln(\mathbb{Q}_{&gt;0})\cap[0,1]\\ 1 &amp; S_2=\mathbb{Q}\cap[0,1]\\ 0 &amp; S_3=\frac{1}{2\mathbb{N}+1}\\ \end{cases}$$</span></p> <p>I assume (but can't prove), <span class="math-container">$S_3$</span> has measure zero and <span class="math-container">$S_1$</span> and <span class="math-container">$S_2$</span> have measure <span class="math-container">$1$</span>. However, <span class="math-container">$0+1+1\neq 1$</span> and thus, <span class="math-container">$\mu$</span> is not additive.</p> <p><strong>Therefore, I need an alternative. Is there another probability measure which gives everything I listed but is also additive for countable <span class="math-container">$A$</span> and <span class="math-container">$\omega(A)=\infty$</span>?</strong></p> https://mathoverflow.net/q/226299 10 The Picard number of the Kummer surface of an abelian surface guest31 https://mathoverflow.net/users/84225 2015-12-16T22:33:39Z 2021-05-17T14:44:16Z <p>Let $A$ be an abelian surface and $\text{Km}(A)$ be the Kummer surface of $A$. If I remember correctly, the Picard number $\rho(\text{Km}(A))$ is equal to $16+\rho(A)$. </p> <blockquote> <p>Does anyone know any reference or proof for this fact?</p> </blockquote> https://mathoverflow.net/q/217557 3 Is the sumset or the sumset of the square set always large? Mark Lewko https://mathoverflow.net/users/630 2015-09-06T05:02:38Z 2021-05-17T14:13:03Z <p>Let A be a finite subset of $\mathbb{N}$, $\mathbb{R}$, or a sufficiently small subset of $\mathbb{F}_{p}$.</p> <p>Do we have a lower bound of the form $|A|^{1+\delta}$ on the following quantity:</p> <p>$$\max (|\{a+b : a,b \in A\}|, |\{a^2+b^2 : a,b \in A\}| ) ?$$</p> <p>In other words, is either the sumset of $A$ or the sumset of the square set of $A$ guaranteed to be large? </p> <p>This is very similar to the sum-product problem (which is formally connected to the variant question of lower bounding $\max(|2A|, |2A^2|)$). My hope is that this problem might be easier than the sum-product problem and better bounds may be available.</p> https://mathoverflow.net/q/109027 35 Applications of Frobenius theorem and conjecture Mikko Korhonen https://mathoverflow.net/users/10146 2012-10-06T23:10:39Z 2021-05-17T14:20:01Z <p>A theorem of Frobenius states that if $n$ divides the order of a finite group $G$, then the number of solutions to $x^n = 1$ in $G$ is a multiple of $n$. Frobenius conjectured that if the number of solutions is exactly $n$, then the set of solutions form a characteristic subgroup of $G$. The conjecture was eventually proved in the 90's, and the full proof uses the classification of finite simple groups.</p> <p>The theorem feels a bit isolated for me.. I'm not sure how the conjecture fits into a wider context, either. What is their importance, if any? Are there any good examples of applications of the theorem or the conjecture? If I'm interested in finite groups, why should I care about the theorem or the conjecture, other than that they are kind of neat?</p> <p>One example I know is that if $G$ has every Sylow subgroup cyclic, then with Frobenius theorem we can show that the Sylow subgroup corresponding to the largest prime divisor of $G$ is normal. Also (this one is too easy, but I like it) for any prime $p$, the number of elements satisfying $x^p = 1$ in the symmetric group $S_p$ is $(p-1)! + 1$, so Frobenius theorem implies $(p-1)! \equiv -1 \mod p$. </p>