Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2024-03-19T01:14:33Zhttps://mathoverflow.net/feedshttps://creativecommons.org/licenses/by-sa/4.0/rdfhttps://mathoverflow.net/q/467287-1A Near Closed-Form Expression of Strict Partition Function Inquiryjableshttps://mathoverflow.net/users/5249412024-03-18T23:28:47Z2024-03-19T00:53:32Z
<p>I am an independent researcher working in various fields of mathematics and sciences. I am working on a strict partition problem. I believe I have found a very fast exact solution that is a near-closed-form expression to the number of strict partition triplets that sum to any Natural n. Is this something that has already been discovered? Or is this something worth publishing? And if so, does anyone have any helpful advice on how to go about getting published? I am also working on a near-closed-form expression for the strict partition function of quadruplets and also a generalization to a closed-form expression of the strict partition function for any sized tuple. All help is much appreciated. Thank you.</p>
https://mathoverflow.net/q/4672860Solving system of linear diophantine equations over the integersuser1868607https://mathoverflow.net/users/828392024-03-18T23:26:16Z2024-03-18T23:26:16Z
<p>In general, solving a system of linear diophantine equations over the integers is polynomial time solvable on the size of the coefficients of the equations.</p>
<p>I am interested in an extension of this result that would handle exponential sized coefficients of the special form <span class="math-container">$2^{2^n}$</span> where <span class="math-container">$n$</span> is a constant written in unary (or equivalently, it is of linear size in the size of the system).</p>
<p>Is it possible to get polynomial time complexity for such restricted exponentially sized constants?</p>
https://mathoverflow.net/q/4672840Are the coefficients in the stationary phase approximation computed explicitly somewhereMedohttps://mathoverflow.net/users/1165552024-03-18T22:47:49Z2024-03-18T22:47:49Z
<p>In Stein's "Harmonic analysis" book, page 334, one can find
the asymptotic expansion</p>
<p><a href="https://i.stack.imgur.com/5SHA6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5SHA6.png" alt="enter image description here" /></a></p>
<p>An instructive proof is given for the case <span class="math-container">$k=2$</span>. It is clear enough to generalize to the cases <span class="math-container">$k\geq 3$</span>. But the coefficients are not computed explicitly. It is not a matter of merely expanding the phase and amplitude. For instance, some integrals are used in obtaining the expansion. These integrals are given in terms of their asymptotics.</p>
<p><strong>Are the coefficients <span class="math-container">$a_{j}$</span> (the first few hopefully) computed explicitly somewhere ?</strong></p>
https://mathoverflow.net/q/467282-4Goldbach's conjecture with Claude 3 [closed]Sylvain JULIENhttps://mathoverflow.net/users/136252024-03-18T22:11:53Z2024-03-18T22:14:30Z
<p>I wanted to test the mathematical abilities of Claude 3, so I typed a prompt leading to the following conjecture:</p>
<p>Assume Goldbach's conjecture and denote by <span class="math-container">$r_{0}(n):=\inf\{r\geq 0\mid(n-r,n+r)\in\mathbb{P}^{2}\}$</span>. Then for all integer <span class="math-container">$k>0$</span>, there exists <span class="math-container">$N_{k}$</span> such that whenever <span class="math-container">$m>N_k$</span>, <span class="math-container">$r_{0}(m)<m/k$</span>.</p>
<p>Numerically it seems one can take <span class="math-container">$N_{k}:=k^{3}$</span>.</p>
<p>Can the latter be proven under the only assumption that Goldbach's conjecture is true?</p>
https://mathoverflow.net/q/4672810Find transseries from difference equationopfromthestarthttps://mathoverflow.net/users/1668432024-03-18T21:57:55Z2024-03-18T22:43:06Z
<p>I want to find a method to solve equations of the form
<span class="math-container">$f(x+1)=f(x)+g(x)$</span> for a given function <span class="math-container">$g$</span> and <span class="math-container">$f(x)=0$</span>.
The paper <a href="https://core.ac.uk/download/pdf/232380616.pdf" rel="nofollow noreferrer">here</a> has solutions for <span class="math-container">$f(x+1)=\lambda(x)f(x)+g(x)$</span>, which is more general than I need and I was not entirely able to understand it. It also had the requirement that <span class="math-container">$g(x)=O(x^{-2})$</span>, which I would like to remove. Are there any cases of finding a solution of this for transseries or any other formal series form?</p>
https://mathoverflow.net/q/4672780Is this reduction process always commutative?Snaredhttps://mathoverflow.net/users/5148092024-03-18T21:25:53Z2024-03-18T21:25:53Z
<p>There was a question posted on this website earlier today by a new Account "Nick". He asked,</p>
<pre><code>The numbers 1,2,..,100 are written in a blackboard.
A move is defined as follows:
- select two numbers a,b
- erase them and instead write ab+a+b.
- After 99 moves exactly one number is left.
Find its possible values.
</code></pre>
<p>I found the problem to be fascinating upon reading it, so I enquired further about what he had tried, or where he had seen or how he came up with this problem, as he (or she of course) did not add appropriate details to it. They then quickly deleted the problem as if they were only looking for a solution and didn't know how to interact with someone who didn't know exactly already.</p>
<p>So, my question is, is this a well-studied problem? How many different integers can be made using this process, and what is known about this problem?</p>
<p>My try: Consider how the process works on just <span class="math-container">$\{x,y,z\}$</span>. For the first product, without loss of generality, choose <span class="math-container">$x,y$</span> to give smaller set <span class="math-container">$\{xy+x+y, z\}$</span>, which naturally each converts to <span class="math-container">$(xy+x+y)z + xy+x+y + z$</span>, or <span class="math-container">$x+y+z + xy + xz + yz + xyz$</span>, so it did not matter if we first chose to contract (x,y) or (y,z) or (x,z). When three elements remain, the result is entirely determined.</p>
<p>Motivated by that, I hope that the answer will always be 1 and we can give it a shot: Suppose that <span class="math-container">$\xi(n)$</span> is the predicate that the number of possible outcomes with a set of <span class="math-container">$n$</span> unique integers under this process is one. We know that <span class="math-container">$\xi(3)$</span>, so all we need is to show that <span class="math-container">$\xi(n) \implies \xi(n+1)$</span>, by induction. So assume that we have <span class="math-container">$n$</span> elements, and that maps to the powerset of the n elements, where each subset within that powerset gets producted with the corresponding elements, and all of those get summed up. Now, what happens when we add another element <span class="math-container">$w$</span> to <span class="math-container">$S$</span>, so that we are looking at <span class="math-container">$|S \cup w| = n+1$</span>? Well, there are two cases:</p>
<p>Case 1: The next contraction does not at all involve <span class="math-container">$w$</span>.</p>
<p>This case is trivial since now the size of <span class="math-container">$S$</span> has been reduced to <span class="math-container">$n$</span>, and therefore the induction hypothesis applies, and then the size 2 and size <span class="math-container">$n$</span> powerset combine in a very straightforward fashion to create the <span class="math-container">$n+1$</span> powerset as we were looking for.</p>
<p>Case 2: The next contraction does involve <span class="math-container">$w$</span>, let's say <span class="math-container">$x$</span> and <span class="math-container">$w$</span>. Then we get <span class="math-container">$wx+w+x$</span>, and a size <span class="math-container">$n$</span> set, which again, by the induction hypothesis is all combinations of everything else, so again, it nicely composes with the <span class="math-container">$2$</span>and <span class="math-container">$n$</span> case giving that <span class="math-container">$\xi(n) \implies \xi(n+1)$</span>.</p>
<p>Therefore, there can only be one value attained by this process. Is this proof correct, and has it been discovered before? Are there related areas of mathematics to this, where one takes for example <span class="math-container">$a,x \mapsto ax+\lambda a + (1-\lambda)x$</span> instead and proves something? I am trying to figure out if this is an entire category of problems that have been worked on and solved for centuries, or if this is just some trivial one-off question that the guy deleted because it truly was absolutely meaningless and worthless?</p>
<p>Thanks in advanced for the responses and please make sure to include citation to actual research papers.. Thanks.</p>
https://mathoverflow.net/q/4672768Proofs of the valence formula that avoid tricky contours?Terry Taohttps://mathoverflow.net/users/7662024-03-18T19:58:09Z2024-03-19T00:47:03Z
<p><span class="math-container">$\DeclareMathOperator\ord{ord}\DeclareMathOperator\Im{Im}$</span>The valence formula for a modular form asserts that if <span class="math-container">$f: \mathbf{H} \to \mathbf{C}$</span> is a modular form of weight <span class="math-container">$k$</span> on the upper half-plane <span class="math-container">${\mathbf H} := \{ z: \Im z > 0 \}$</span> in the sense that
<span class="math-container">$$ f\left(\frac{az+b}{cz+d}\right) = (cz+d)^k f(z)$$</span>
for all <span class="math-container">$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2({\bf Z})$</span>, and <span class="math-container">$f$</span> is holomorphic on all of <span class="math-container">${\mathbf H}$</span> and also bounded as <span class="math-container">$\Im z \to \infty$</span> then (if <span class="math-container">$f$</span> is not identically zero) one has the identity
<span class="math-container">$$ \sum_\rho \ord_\rho(f) + \ord_\infty(f) + \frac{1}{2} \ord_i(f) + \frac{1}{3} \ord_{e^{\pi i/3}}(f) = \frac{k}{12}$$</span>
where <span class="math-container">$\rho$</span> ranges over the zeroes of <span class="math-container">$f$</span> away from the (partially) fixed points <span class="math-container">$i, e^{\pi i/3}, \infty$</span> of the action, with each orbit of <span class="math-container">$\operatorname{SL}_2(\mathbf{Z})$</span> avoiding these fixed points being represented precisely once in this sum, and the order of vanishing at infinity defined in terms of the (squared) nome variable <span class="math-container">$q = e^{2\pi i z}$</span>.</p>
<p>The standard proof of this identity proceeds by integrating the logarithmic derivative of <span class="math-container">$f'/f$</span> on a somewhat complicated contour designed to avoid zeroes, and which looks something like this (image taken from <a href="https://rtca2023.github.io/pages_Lyon/eberl.pdf" rel="nofollow noreferrer" title="Eberl - Some new tricks for formalising advanced mathematics">this source</a>):</p>
<p><a href="https://i.stack.imgur.com/uJ6V2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uJ6V2.png" alt="complicated contour" /></a></p>
<p>The claim then follows from a routine application of the residue theorem, after estimating the contribution of the various components of the contour appropriately.</p>
<p>It turns out that this argument, while straightforward, is somewhat difficult to formalize in proof assistant languages; see the discussion in the slides <a href="https://rtca2023.github.io/pages_Lyon/eberl.pdf" rel="nofollow noreferrer" title="Eberl - Some new tricks for formalising advanced mathematics">linked previously</a>. Are there other proofs of this formula that do not rely on a tricky contour integration? I have experimented with a Green's formula type approach in which <span class="math-container">$f'/f$</span> is integrated against a suitable cutoff function, say on the unit disk in the (squared) nome variable <span class="math-container">$q = e^{2\pi i z}$</span>, but the computations are quite complicated. One can also obtain this formula from using the <span class="math-container">$j$</span>-invariant to coordinatize the modular curve, though this is somewhat circular as often the valence formula is needed to establish properties of this invariant. I also considered trying to use general Riemann surface tools such as the Riemann–Hurwitz formula, but among other things I ran into the need to triangularize a Riemann surface, which also would be complicated to formalize I think.</p>
https://mathoverflow.net/q/467275-6Is the Square-free Test A Polynomial Time Algorithm?user524928https://mathoverflow.net/users/5249282024-03-18T19:09:23Z2024-03-18T19:09:23Z
<p>The paper</p>
<p>"MILLER’s PRIMALITY TEST",
Volume 8, number 2 INFORMATION PROCESSING LETTERS February 1979
by H.W. LENSTRA, Jr.,</p>
<p>claims to have a square-free algorithm of polynomial time complexity, see Theorem 2, this claim seems to be very wrong. A more recent paper on the same topic</p>
<p>"Detecting squarefree numbers", arXiv:1304.6937,</p>
<p>by Andrew R. Booker, Ghaith A. Hiary, Jon P. Keating</p>
<p>contradicts this claim, on the very first page it comments about the square-free test being a difficult problem, not a polynomial time algorithm as Lenstra's result claims.</p>
<p>The Lenstra's result is repeated in books and papers, for example, Corollary 2 in the paper</p>
<p>"On the Divisibility of Fermat Quotients", Michigan Math. J. 59 (2010), 313-328,</p>
<p>by Jean Bourgain, Kevin Ford, Sergei V. Konyagin, & Igor E. Shparlinski.</p>
<p>Our question: Please comment on the correctness of Mr. Lenstra's claim. If this claim is erroneous, should this paper be retracted? There is no easy correction, we think it is best to retract it to prevent further proliferation of this error in the literature and to avoid confusing the young students, researchers.</p>
https://mathoverflow.net/q/4672741Classification of complex irreducible representations of $\mathrm{GL}_n(\mathbb{F}_q)$ [duplicate]asvhttps://mathoverflow.net/users/161832024-03-18T18:56:27Z2024-03-18T20:21:57Z
<p>Is there a classification of complex irreducible representations of the group <span class="math-container">$\operatorname{GL}_n(\mathbb{F}_q)$</span>, where <span class="math-container">$\mathbb{F}_q$</span> is a finite field with <span class="math-container">$q$</span> elements?</p>
https://mathoverflow.net/q/4672720On partitions into distinct parts and binaryNotamathematicianhttps://mathoverflow.net/users/2319222024-03-18T17:41:14Z2024-03-18T17:41:14Z
<p>Let <span class="math-container">$a(n)$</span> be <a href="https://oeis.org/A000009" rel="nofollow noreferrer">A000009</a> (i.e., number of partitions of <span class="math-container">$n$</span> into distinct parts or number of partitions of <span class="math-container">$n$</span> into odd parts).</p>
<p>Let</p>
<p><span class="math-container">$$
b(n) = \sum\limits_{i=0}^{n} a(i)
$$</span></p>
<p>Let</p>
<p><span class="math-container">$$
\ell(n) = \left\lfloor\log_2 n\right\rfloor
$$</span></p>
<p>Let <span class="math-container">$\operatorname{tr}(n)$</span> be <a href="https://oeis.org/A007814" rel="nofollow noreferrer">A007814</a> (i.e., number of trailing zeros in the binary expansion of <span class="math-container">$n$</span>).</p>
<p>Let <span class="math-container">$c(n)$</span> be an integer sequence such that we start with <span class="math-container">$A = n, B = 2^{\operatorname{tr}(A)}, C = B$</span> and while <span class="math-container">$A\ne B$</span> consecutively apply <span class="math-container">$A := 3\cdot2^{\ell(\frac{A}{B})} - \frac{A}{B} - 1, B = 2^{\operatorname{tr}(A)}, C := B + C$</span>. Then <span class="math-container">$c(n)$</span> is the value of <span class="math-container">$C$</span> after the whole transformation.</p>
<p>Let <span class="math-container">$d(n)$</span> be an integer sequence of numbers <span class="math-container">$\frac{c(2k)}{2}$</span> such that we take these numbers in order if their appearance.</p>
<p>I conjecture that <span class="math-container">$$d(b(n)) = 2^n.$$</span></p>
<p>I also conjecture that values of <span class="math-container">$d(k)$</span> for <span class="math-container">$k$</span> from <span class="math-container">$b(n-1)$</span> to <span class="math-container">$b(n)-1$</span> interpreted as its binary expansion and then converted to positions of bits (starting from the right side) gives partitions of <span class="math-container">$n$</span> into distinct parts.</p>
<p>Here is the PARI/GP program to check it numerically:</p>
<pre><code>a(n) = if( n<0, 0, polcoeff( prod( k=1, n, 1 + x^k, 1 + x * O(x^n)), n))
b(n) = sum(i=0, n, a(i))
c(n) = my(A = n, B = 1 << valuation(A, 2), C = B, D); while(!(A == B), D = A/B; A = 3*(1 << logint(D, 2)) - D - 1; B = 1 << valuation(A, 2); C += B); C
d_upto(n) = my(A = 0, v1, v2); v1 = vector(n, i, 0); v2 = []; for(i=1, n, until(!setsearch(v2, c(2*A)/2), A++); B = c(2*A)/2; v1[i] = B; v2 = setunion(v2, [B])); v1
v1 = d_upto(2^8)
test(n) = v1[b(n)] == 2^n
parts(n) = select(x -> (x > 0), Vecrev(binary(v1[n])), 1)
</code></pre>
<p>Is there a way to prove it?</p>
https://mathoverflow.net/q/4672684Prime differences and zero multiplicityFelixsonhttps://mathoverflow.net/users/3142362024-03-18T16:59:17Z2024-03-18T23:11:25Z
<p>Paul Erdős conjectured, for consecutive primes, that:</p>
<p><span class="math-container">$$\sum_{p_n < x} (p_n -p_{n-1})^2 = O(x \log x)$$</span></p>
<p>Let's call this hypothesis EH. Assuming the Riemann hypothesis (RH), Selberg showed that a result follows that differs from EH merely by a logarithmic factor.</p>
<p>Does EH imply RH, or RH EH? What are the state of the art results on EH?</p>
<p>By assuming that all the zeros of <span class="math-container">$\zeta(s)$</span> are simple (or other deep vertical fact tied to the critical strip), EH might also follow. Any comments are welcomed.</p>
https://mathoverflow.net/q/4672661Is the irreducible $ \mathrm{SU}(3) $ subgroup of $ \mathrm{SU}(6) $ maximal?Ian Gershon Teixeirahttps://mathoverflow.net/users/3871902024-03-18T16:19:59Z2024-03-18T18:56:37Z
<p><span class="math-container">$\DeclareMathOperator\SO{SO}\DeclareMathOperator\O{O}\DeclareMathOperator\SU{SU}\DeclareMathOperator\U{U}\DeclareMathOperator\S{S}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PSU{PSU}\newcommand{\irr}{\mathrm{irr}}\DeclareMathOperator\T{T}\DeclareMathOperator\A{A}\DeclareMathOperator\N{N}$</span>Is the <span class="math-container">$6$</span>-dimensional <span class="math-container">$ (2,0) $</span> irrep of <span class="math-container">$ \SU(3) $</span> maximal in <span class="math-container">$ \SU(6) $</span>?</p>
<p>For those of you who are interested in context, I started wondering this the other day when I tried to write down the maximal subgroups of <span class="math-container">$ \SU(6) $</span>. My guess so far is that the full list of maximal (proper closed) subgroups of <span class="math-container">$ \SU(6) $</span> is:</p>
<p>Type I (normalizer of maximal connected subgroup)
<span class="math-container">\begin{align*}
& \U(5) \cong \S(\U(5) \times \U(1)) \\
& \S(\U(4) \times \U(2)) \\
& \S(\U(3) \times \U(3))\rtimes \S_2 \\
& 6 \circ_2 \Sp(3) \\
& 6 \circ_2 \SO(6) \\
& 6 \circ_3 SU(3)_{\irr}
\end{align*}</span>
Type II (finite maximal closed subgroup, for the last 2 groups GAP subscripts are used to label the center and the outer automorphisms when multiple groups of this structure description exist)
<span class="math-container">\begin{align*}
& 6.\A_7
\\
&6.\PSL(3,4).2_1
\\
&6_1.\PSU(4,3).2_2
\end{align*}</span>
Type III (normalizer of a subgroup which is connected but not maximal connected)
<span class="math-container">\begin{align*}
& \N(\T^6)=\S(\U(1) \times \U(1) \times \U(1) \times \U(1) \times \U(1) \times \U(1)) \rtimes \S_6\\
&\S( \U(2) \times \U(2) \times \U(2) ) \rtimes \S_3\\
\end{align*}</span></p>
<p>Note on notation. <span class="math-container">$ \rtimes $</span> means split extension (semidirect product). <span class="math-container">$ \cdot $</span> means nonsplit extension. <span class="math-container">$ \circ $</span> denotes central product, in most cases here we have <span class="math-container">$ 6 \circ_2 H $</span>, which is just the group generated by <span class="math-container">$ H $</span> and <span class="math-container">$ \zeta_6I $</span> but that group is not a direct product since already <span class="math-container">$ -I \in H $</span>, we get a central product essentially with three <span class="math-container">$ H $</span> components. Similar idea for <span class="math-container">$ 6 \circ_3 \SU(3)_{\irr} $</span> having two components.</p>
<p>Here <span class="math-container">$ \N $</span> denotes normalizer. Recall that a positive dimensional (type I and type III above) maximal subgroup of a simple Lie group equals the full normalizer of its identity component.</p>
<p><a href="https://arxiv.org/abs/math/0605784" rel="nofollow noreferrer">The paper</a>
classifies all maximal closed subgroups of <span class="math-container">$ \SU(n) $</span> whose identity component is not simple (here trivial counts as simple). According to table 5 the maximal closed subgroups of <span class="math-container">$ \SU(4) $</span> of this type are:</p>
<p>The normalizer of the maximal torus (row 4 table 5, <span class="math-container">$ \ell=6, p=1 $</span>)
<span class="math-container">$$
\N(\T)=\S(\U(1) \times \U(1) \times \U(1) \times \U(1)) \rtimes \S_6
$$</span>
and (row 4 of table 5, <span class="math-container">$ \ell=3, p=2 $</span>)
<span class="math-container">$$
\S( \U(2) \times \U(2) \times \U(2) ) \rtimes \S_3
$$</span>
As well as (row 1 table 5, <span class="math-container">$ p=5,q=1 $</span> )
<span class="math-container">$$
\S(\U(5) \times \U(1) )\cong \U(5)
$$</span>
and (row 1 table 5, <span class="math-container">$ p=4,q=2 $</span> )
<span class="math-container">$$
\S(\U(4) \times \U(2) )
$$</span>
and the normalizer of <span class="math-container">$ \S(\U(3) \times \U(3))= \{\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}:A,B\in U(3),\det(A)\det(B)=1 \} $</span> which is a split extension (row 1 table 5 <span class="math-container">$ p=q=3 $</span>)
<span class="math-container">$$
\langle \S(U(3) \times \U(3)),SWAP_{\oplus}\rangle \cong \S(\U(3) \times \U(3)) \rtimes \S_2
$$</span>
where the normalizing matrix <span class="math-container">$ SWAP_{\oplus}=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $</span> swaps the two blocks in the direct sum.</p>
<p>Next, we consider maximal closed subgroups with nontrivial simple connected component.</p>
<p>By dimension, such a subgroup would be isogeneous to <span class="math-container">$ \SU(2)$</span>, <span class="math-container">$\SU(3)$</span>, <span class="math-container">$\Sp(2)$</span>, <span class="math-container">$G_2$</span>, <span class="math-container">$\SU(4)$</span>, <span class="math-container">$\SO(7)$</span>, <span class="math-container">$\Sp(3)$</span>, <span class="math-container">$\SU(5)$</span>, <span class="math-container">$\SO(8) $</span> of dimensions <span class="math-container">$ 3$</span>, <span class="math-container">$8$</span>, <span class="math-container">$10$</span>, <span class="math-container">$14$</span>, <span class="math-container">$21$</span>, <span class="math-container">$21$</span>, <span class="math-container">$24$</span>, <span class="math-container">$28 $</span> respectively.</p>
<p>Of these the only one with 6d irreps are: 6d irrep of <span class="math-container">$ \SU(2) $</span>, the <span class="math-container">$ (2,0) $</span> 6d irrep of <span class="math-container">$ \SU(3) $</span>, fundamental irrep of <span class="math-container">$ \Sp(3) $</span>,</p>
<p>Of these only
<span class="math-container">$$
6 \circ_2 \Sp(3)=\langle\zeta_6 I,\Sp(3)\rangle
$$</span>
is maximal subgroup of <span class="math-container">$ \SU(6) $</span>.</p>
<p>Even dimensional irreps of <span class="math-container">$ \SU(2) $</span> are always symplectic so all <span class="math-container">$ \SU(2) $</span> subgroups of <span class="math-container">$ \SU(6) $</span> are contained in a conjugate of <span class="math-container">$ \Sp(3) $</span>. See <a href="https://math.stackexchange.com/questions/4536082/understanding-the-4-dimensional-irrep-of-su-2/4536205#4536205">this MathSE question</a>.</p>
<p>Finally we consider subgroups with trivial connected component. These are finite since <span class="math-container">$ \SU(6) $</span> is compact. To be maximal they must at least be primitive. For example there is a very large <span class="math-container">$ 6 \circ_2 2.J_2 $</span> subgroup of <span class="math-container">$ SU(6) $</span> but it is not maximal because it is not even Lie primitive: it is contained in <span class="math-container">$ 6 \circ_2 \Sp(3) $</span>. Also there is an <span class="math-container">$ \A_7 $</span> subgroup that is not Lie primitive, it is contained in <span class="math-container">$ \SO(6) $</span> since it is the standard <span class="math-container">$ \A_{n+1} $</span> subgroup of <span class="math-container">$ \SO(n) $</span> arising from the deleted permutation representation.</p>
<p>Even Lie primitive subgroups may not be maximal if they are contained in another larger Lie primitive finite subgroup. For example there is a subgroup <span class="math-container">$ 3.\A_7 \subset 6.\PSU(4,3) \subset \SU(6) $</span> which is Lie primitive but not maximal.</p>
<p>A maximal finite subgroup which is irreducible in the adjoint representation is always a maximal closed subgroup. This includes the following subgroups
The central product
<span class="math-container">$$
6.\A_7
$$</span>
of order <span class="math-container">$ 6(2,520)=15,120 $</span> (maximal closed since it is maximal finite and a 2-design)
<span class="math-container">$$
6.\PSL(3,4).2_1
$$</span>
of order <span class="math-container">$ 6(20,160)2 $</span> (maximal closed since it is maximal finite and a 3-design).
<span class="math-container">$$
6_1.\PSU(4,3).2_2
$$</span>
of order <span class="math-container">$ 6(3,265,920)2 $</span> (maximal closed since it is maximal finite and a 3-design).</p>
<p>For references on designs and maximality see <a href="https://math.stackexchange.com/questions/4272017/finite-maximal-closed-subgroups-of-lie-groups/4477296#4477296">this MathSE question</a></p>
<p>This is consistent with the fact
that a maximal <span class="math-container">$ 2 $</span>-design group is maximal closed ( all <span class="math-container">$ 3 $</span> designs are <span class="math-container">$ 2 $</span> designs).</p>
<p>This question <a href="https://math.stackexchange.com/questions/4828244/is-the-irreducible-su3-subgroup-of-su6-maximal">cross posted from MSE</a></p>
https://mathoverflow.net/q/4672641Irreducible subspaces in the space of functions on Grassmannian acted by $\mathrm{GL}_n(\mathbb{F}_q)$asvhttps://mathoverflow.net/users/161832024-03-18T16:06:06Z2024-03-18T17:34:16Z
<p><span class="math-container">$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Gr{Gr}$</span>Let <span class="math-container">$\mathbb{F}_q$</span> be a finite field with <span class="math-container">$q$</span> elements. Let <span class="math-container">$\Gr_{i,n}(\mathbb{F}_q)$</span> denote the Grassmannian of linear <span class="math-container">$i$</span>-dimensional subspaces in <span class="math-container">$\mathbb{F}_q^n$</span>.</p>
<p>The group <span class="math-container">$\GL_n(\mathbb{F}_q)$</span> acts on the space of complex valued functions on <span class="math-container">$\Gr_{i,n}(\mathbb{F}_q)$</span>. By Proposition 5.1 in <a href="https://arxiv.org/abs/1811.08675" rel="nofollow noreferrer">this</a> paper it is multiplicity free and has the length <span class="math-container">$1+\min\{i,n-i\}$</span>.</p>
<p><strong>Is there an explicit description of irreducible subspaces?</strong> The simplest unknown to me case is <span class="math-container">$n=4,i=2$</span>.</p>
https://mathoverflow.net/q/46725816A conjecture about inclusion–exclusionM.Monethttps://mathoverflow.net/users/4709392024-03-18T13:55:12Z2024-03-18T16:43:42Z
<p><span class="math-container">$\newcommand\calF{\mathcal{F}}
\def\cupdot {\stackrel{\bullet}{\cup}}
\def\minusdot {\stackrel{\bullet}{\setminus}}$</span>This post presents a conjecture that we have with some colleagues. It is about reordering inclusion–exclusion to use only the intersections that do not cancel and the operations of disjoint union and subset complement. We are posting it here in case someone has an idea or know combinatorialists that might be interested (we are from TCS).</p>
<p>I write <span class="math-container">$[n] = \{1,\dotsc,n\}$</span>. Let <span class="math-container">$\calF = \{S_1,\dotsc,S_n\}$</span> be a
finite family of pairwise incomparable (for inclusion) finite sets. For <span class="math-container">$T \subseteq
[n]$</span>, <span class="math-container">$T\neq \emptyset$</span>, define <span class="math-container">$S_T = \bigcap_{j\in T} S_j$</span>.
By <a href="https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="noreferrer">inclusion–exclusion</a> we have
<span class="math-container">$$\left\vert\bigcup_{i=1}^n S_i\right\vert = \sum_{T\subseteq [n],T\neq \emptyset} (-1)^{|T|+1} |S_T|.$$</span></p>
<p>It can happen that <span class="math-container">$S_T = S_{T'}$</span> for <span class="math-container">$T \neq T'$</span>, so that some of the
intersections disappear from the above sum. Call such an intersection <em>cancelling</em>,
the other intersections being <em>non-cancelling</em>.</p>
<p><strong>Example:</strong></p>
<p>Take <span class="math-container">$\calF = \{S_1,S_2,S_3\}$</span> with <span class="math-container">$S_1=\{a,b,d\}$</span>, <span class="math-container">$S_2=\{a,b,c,e\}$</span>,
<span class="math-container">$S_3=\{a,c,f\}$</span>. Then:
<span class="math-container">$$ \begin{align}
|S_1 \cup S_2 \cup S_3| ={} &|S_1| + |S_2| + |S_3| \\\\
& - (|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3|) \\\\
& + |S_1 \cap S_2 \cap S_3|.
\end{align} $$</span></p>
<p>Since we have <span class="math-container">$S_1 \cap S_3 = S_1 \cap S_2 \cap S_3$</span>, we obtain
<span class="math-container">\begin{align*}
|S_1 \cup S_2 \cup S_3| &= |\{a,b,d\}| + |\{a,b,c,e\}| + |\{a,c,f\}|- |\{a,b\}| - |\{a,c\}|.
\end{align*}</span>
The non-cancelling intersections are the ones that remain, i.e., <span class="math-container">$\{a,b,d\}$</span>, <span class="math-container">$\{a,b,c,e\}$</span>, <span class="math-container">$\{a,c,f\}$</span>, <span class="math-container">$\{a,b\}$</span>, and <span class="math-container">$\{a,c\}$</span>, while
<span class="math-container">$\{a\}$</span> (<span class="math-container">$=S_1 \cap S_3 = S_1 \cap S_2 \cap S_3$</span>) is a cancelling term.</p>
<p>The conjecture is that we can always express the union <span class="math-container">$\bigcup_{i=1}^n S_i$</span> (not its size, but <em>the set itself</em>)
from the non-cancelling intersections, using only the operations of <em>disjoint
union</em> and <em>subset complement</em>. Formally, for two sets <span class="math-container">$A$</span>, <span class="math-container">$B$</span> such that
<span class="math-container">$A\cap B = \emptyset$</span>, define the <em>disjoint union</em> <span class="math-container">$A\cupdot B = A
\cup B$</span>. For two sets <span class="math-container">$A,B$</span> such that <span class="math-container">$B\subseteq A$</span>, define the <em>subset
complement</em> <span class="math-container">$A\minusdot B = A \setminus B$</span>.</p>
<p><strong>Conjecture:</strong>
For any finite family of pairwise incomparable finite sets <span class="math-container">$\calF =
\{S_1,\dotsc,S_n\}$</span>, we can express <span class="math-container">$\bigcup_{i=1}^n S_i$</span> using only the
non-cancelling intersections and the operations of disjoint union and subset
complement.</p>
<p><strong>Example:</strong>
Continuing the example, we can express~<span class="math-container">$S_1 \cup S_2 \cup S_3 =
\{a,b,c,d,e,f\}$</span> with <span class="math-container">$ \bigl[\bigr((\{a,b,d\} \minusdot \{a,b\}) \cupdot
\{a,b,c,e\}\bigr) \minusdot \{a,c\}\bigr] \cupdot \{a,c,f\}$</span>: the reader can
easily check that each <span class="math-container">$\cupdot$</span> (resp., each <span class="math-container">$\minusdot$</span>) is a valid
disjoint union (resp., subset complement), and that we have only used the
non-cancelling intersections. Note that this is not the only valid
expression, for instance we can also obtain the union with the expression
<span class="math-container">$[\{a,b,d\} \minusdot \{a,b\}] \cup [\{a,b,c,e\} \minusdot \{a,c\}] \cup
\{a,c,f\}$</span>.</p>
<p><strong>What we know:</strong>
We have tried on millions of examples by a bruteforce approach and this always seems to be true. We have some partial results in <a href="http://arxiv.org/abs/2401.16210" rel="noreferrer" title="Amarilli, Monet, and Suciu - The Non-Cancelling Intersections Conjecture">a note on arXiv</a> (calling this conjecture the “Non-Cancelling Intersections Conjecture”) concerning a reformulation of this conjecture, but we do not have a general solution so far.</p>
https://mathoverflow.net/q/4672560How to distinguish birth and death bifurcations?Azurhttps://mathoverflow.net/users/1679362024-03-18T13:09:11Z2024-03-18T16:35:31Z
<p>Let <span class="math-container">$f : \mathbb{R} \to \mathbb{R}$</span> have a <strong>degenerate</strong> critical point at <span class="math-container">$x = 0 \, ($</span><em>ie</em>, <span class="math-container">$f(0) = f'(0) = f''(0) = 0)$</span>.</p>
<p>Perturbing <span class="math-container">$f$</span> locally around <span class="math-container">$0$</span> may cause multiple scenarios:</p>
<ul>
<li><strong>Birth:</strong> the critical point splits into multiple (non-degenerate) critical points. Example: <span class="math-container">$f(x) = x^3$</span> with perturbation <span class="math-container">$f(x) - \varepsilon x$</span> (for <span class="math-container">$\varepsilon > 0$</span>).</li>
<li><strong>Death:</strong> the critical point completely disappears. Example: <span class="math-container">$f(x) = x^3$</span> with perturbation <span class="math-container">$f(x) + \varepsilon x$</span>.</li>
<li><strong>Neither</strong>. The critical point is just slightly perturbed, and then no-longer degenerate, but it neither splits nor dies. An example can be found in <a href="https://math.stackexchange.com/a/4881715/656302">this answer</a>.</li>
</ul>
<hr />
<p><strong>Overarching question:</strong> Consider the three scenarios of perturbation I described above (birth/death of critical points, or neither). Given a function <span class="math-container">$f$</span> and a degenerate critical point <span class="math-container">$x$</span> of <span class="math-container">$f$</span>, is there a way to distinguish which scenarios could happen, and which could not?</p>
<p>I would be perfectly happy with references treating this topic; as most results concerning birth-death bifurcation (that I could find) take a lot of statements for granted, as most of this is now part of the "mathematical folklore". (This is the reason I'm asking this on MO, and not MSE; for this is for research purposes; and the original question I asked on MSE on a similar topic took a long time to gain an answer).</p>
<hr />
<p><em>Incomplete attempt at answering this question:</em> the Hopf index.</p>
<p><em>Edit: as was pointed out in the comments, a post is technically only allowed to contain one question; so whichever comes next is NOT a question. It is my attempt at the problem.</em></p>
<p>As explained in <a href="https://math.stackexchange.com/a/4882275/656302">this answer</a>: to each (isolated) critical point <span class="math-container">$x$</span> of <span class="math-container">$f$</span>, one can associate a Hopf index, <span class="math-container">$\lambda(x)$</span>. The sum of all these indices (over all critical points of <span class="math-container">$f$</span>) is a topological invariant by the Poincaré-Hopf theorem; and hence, in particular:</p>
<p>If <span class="math-container">$x$</span> can be killed by some small perturbation of <span class="math-container">$f$</span>, then <span class="math-container">$\lambda(x) = 0$</span>.</p>
<p>Hence, my first question is:</p>
<p><strong>1a)</strong> is the converse true? <em>ie</em>, if <span class="math-container">$\lambda(x) = 0$</span>, there exists a perturbation of <span class="math-container">$f$</span> that kills <span class="math-container">$x$</span>?</p>
<p>If this is true, then this would give us a criterion that identifies when a critical point can die (it would be necessary and sufficient that <span class="math-container">$\lambda(x) = 0$</span>). If not:</p>
<p><strong>1b)</strong> Is there a criterion that allows us to determine whether the critical point <span class="math-container">$x$</span> can die or not? (when submitted to an arbitrary perturbation of <span class="math-container">$f$</span>)</p>
<p>Either way, the question I am mostly interested in is the following:</p>
<p><strong>2)</strong> In either case: are death bifurcations <strong>rare</strong>? In other words: given a generic <span class="math-container">$f$</span> with a degenerate critical point; is it true that this critical point cannot die? (If <strong>1a)</strong> holds, then this reduces to asking that for a generic <span class="math-container">$f$</span> and <span class="math-container">$x$</span> (degenerate), then <span class="math-container">$\lambda(x) \ne 0$</span>).</p>
<p>The reason I care about this question is that, in my set-up, I would like to exclude all cases which lead to a death bifurcation. If these are extremely rare, then I can do so sensibly. I am, however, not so sure that this is true (it is very easy to construct degenerate critical points which die through perturbation, simply by choosing odd-degree polynomials). I am also uncertain whether we can have birth without death, or if both come hand in hand?</p>
<hr />
<p>Finally, this is a bit unrelated from questions <strong>1)</strong> and <strong>2)</strong>, but still related to the overarching question:</p>
<p><strong>3)</strong> Say we have already ruled out deaths. So our degenerate critical point can either bifurcate into <em>multiple</em> critical points (birth scenario), or it will not split, and small perturbations of <span class="math-container">$f$</span> will still have only one critical point. Is there a formal way/criterion to distinguish between these two cases?</p>
https://mathoverflow.net/q/4672202Is there a version of 3-SAT that is NP-complete but grows like $2^n$ instead of $2^{n \choose 3}$?Logan https://mathoverflow.net/users/5248742024-03-17T19:21:46Z2024-03-18T16:32:25Z
<p>If I have <span class="math-container">$n$</span> variables and I want to write down all 3-SAT problems, the number of problems is <span class="math-container">$2^{8{n \choose 3}}$</span>, since each clause has 3 variables and each variable can be negated or not.</p>
<p>But empirically with SAT solvers, the hardest SAT problems have far fewer terms than this (if you have too few or too many terms SAT is easy to solve).</p>
<p>I suspect there is a subclass of 3-SAT that grows like <span class="math-container">$2^{(c_0+c_1 n)}$</span> and is still NP-complete.</p>
<p>Does anyone know of such a class or have suggestions on how to construct it?</p>
<p>(update)</p>
<p>As pointed out, if we allow n clauses but each of the clauses can choose from whatever variable they want, we get <span class="math-container">$O(n^2)^n=2^{2 n log n+o(n log n)}$</span> possible problems.</p>
<p>It seems like we need some kind of windowed-3-SAT where there are <span class="math-container">$n*k$</span> clauses and clauses in the span <span class="math-container">$[k*i,k*(i+1)]$</span> can choose from variables <span class="math-container">$x_{i-w}$</span> to <span class="math-container">$x_{i+w}$</span>. This gives us a finite number of choices for each clause and hence exponential growth in the number of problems (in n).</p>
<p>Is there some obvious reason this is/isn't NP-complete? (for W=0 it's obvious to me that it's not.And I could probably convince myself for w=1 it also isn't).</p>
<p>Trying out random problems with minisat, it looks like k=4..8 w=100+ is the regime where these problems are non-trivial. (-1 indicates a timeout of more than 1 second to solve). This is with n=100</p>
<p><a href="https://i.stack.imgur.com/9gbhG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9gbhG.png" alt="enter image description here" /></a></p>
<p>Here are some solution times for random problems with k=4 w=100. For smaller w, the solution times look linear in n, but there's definitely something non-linear going on for w=100. I suspect that if there is true NP-hard behavior going on you can get it with a much smaller w but that for random problems the larger w makes it more likely that we find the hard problems.</p>
<p><a href="https://i.stack.imgur.com/l9uYs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l9uYs.png" alt="enter image description here" /></a></p>
https://mathoverflow.net/q/4672140Convex sets via fixed point equationsrimuhttps://mathoverflow.net/users/1752802024-03-17T16:58:27Z2024-03-18T19:18:34Z
<p>I have an equation of the general form
<span class="math-container">$$ X = S \cup T X $$</span>
where <span class="math-container">$S \subset \mathbb R^n$</span> is a convex polytope (given by its bounding hyperplanes), <span class="math-container">$T\colon \mathbb R^n \to \mathbb R^n$</span> is a linear map, and the largest <span class="math-container">$X \subset \mathbb R^n$</span> must be found.</p>
<p>Is there an easy way to solve this and similar problems?</p>
https://mathoverflow.net/q/4671683Integral involving Legendre polynomialZurab Silagadzehttps://mathoverflow.net/users/323892024-03-16T17:36:30Z2024-03-18T21:40:48Z
<p>In a physics problem the following integral shows up <span class="math-container">$$\int\limits_0^{2\pi}P_m(\cos{(\theta-\alpha)})\,\cos^{m+2}{(n\alpha)}\;d\alpha,$$</span> where <span class="math-container">$P_m$</span> is the Legendre polynomial and <span class="math-container">$n,m$</span> are integer numbers. How this integral can be evaluated?</p>
https://mathoverflow.net/q/4670382Why is this polynomial factorizable? [closed]LichenSDUhttps://mathoverflow.net/users/1143622024-03-14T08:29:05Z2024-03-18T21:15:09Z
<p>I met a curious problem on factorizing a homogenerous polynomial of degree 9.</p>
<p><strong>Problem: Show that the following polynomial can be divided by <span class="math-container">$(a_1+a_2+a_3)$</span>:</strong></p>
<p><span class="math-container">\begin{align}
&\quad\left|
\begin{array}{ccc}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \\
\end{array}
\right|^3 \\ &
+ \left(
a_1^3 \left(b_2 c_3-b_3 c_2\right)
+a_2^3 \left(b_3 c_1-b_1 c_3\right)
+a_3^3 \left(b_1 c_2-b_2 c_1\right)
\right)\left(b_1+b_2+b_3\right)^2\left(c_1+c_2+c_3\right)^2 \\
&- \left(
b_1 \left(c_2 a_3-c_3 a_2\right)^3
+b_2 \left(c_3 a_1-c_1 a_3\right)^3
+b_3 \left(c_1 a_2-c_2 a_1\right)^3
\right)
\left(b_1+b_2+b_3\right)^2 \\
&- \left(
c_1 \left(a_2 b_3-a_3 b_2\right)^3
+c_2 \left(a_3 b_1-a_1 b_3\right)^3
+c_3 \left(a_1 b_2-a_2 b_1\right)^3
\right)
\left(c_1+c_2+c_3\right)^2
\end{align}</span></p>
<p>The problem looks quite peculiar as the expression contains both <span class="math-container">$(b_1+b_2+b_3)$</span> and <span class="math-container">$(c_1+c_2+c_3)$</span>, while <span class="math-container">$(a_1+a_2+a_3)$</span> does not appear. Mathematica tells me it is true(one factor is <span class="math-container">$(a_1+a_2+a_3)$</span> while the other factor consists of more than 200 terms), but I can not figure out a human proof.</p>
https://mathoverflow.net/q/4669453Sum of three square is a square and sum of their product taken two at a time is also a squareGuruprasadhttps://mathoverflow.net/users/4944062024-03-13T08:01:19Z2024-03-18T22:49:47Z
<p>Let <span class="math-container">$a^2 + b^2 + c^2 = X^2$</span> and</p>
<p><span class="math-container">$$(ab)^2 + (ac)^2 + (bc)^2 = Y^2$$</span><br />
Such that <span class="math-container">$a,b,c,x,y$</span> are all Integers</p>
<p>How to find All non trivial solutions ?</p>
<p>Is there any parametrization which gives many solutions ?</p>
<p>What is the smallest solution for above equation ?</p>
https://mathoverflow.net/q/4667704Do maximal compact logics exist?Noah Schweberhttps://mathoverflow.net/users/81332024-03-10T19:43:37Z2024-03-18T17:53:20Z
<p>By "logic" I mean regular logic in the sense of abstract model theory (see e.g. the last section of <a href="https://link.springer.com/book/10.1007/978-1-4757-2355-7" rel="nofollow noreferrer">Ebbinghaus/Flum/Thomas' book</a>). My question is simple:</p>
<blockquote>
<p>Is there a logic <span class="math-container">$\mathcal{L}$</span> which is fully compact and such that no properly-stronger logic is fully compact?</p>
</blockquote>
<p>It seems like there should be an easy negative answer as long as the logics are required to be set-sized, but I can't quite get the details to work; the <em>relativizability</em> property of regular logics makes everything rather tedious. And if we allow class-sized logics (e.g. <span class="math-container">$\mathcal{L}_{\infty,\omega}$</span>, although of course that's not fully compact) then this idea goes out the window and I have no intuition.</p>
<p>I'm happy to add any definability criteria which would help, but I do <em>not</em> want to add Lowenheim-Skolem-type conditions. In particular, the results of <a href="https://link.springer.com/article/10.1007/s00153-004-0212-8" rel="nofollow noreferrer">Shelah/Vaananen (<em>A note on extensions of infinitary logic</em>)</a> don't seem to apply here.</p>
https://mathoverflow.net/q/4665662Frobenius theorem and the size of integral manifoldGeorgehttps://mathoverflow.net/users/5008592024-03-07T11:26:51Z2024-03-18T21:30:36Z
<p>Let <span class="math-container">$X =(X_0,X_1)\in \mathbb{R}^2$</span> and <span class="math-container">$Y=(Y_0,Y_1)\in \mathbb{R}^2$</span> be two vector fields on <span class="math-container">$\mathbb{R}^2$</span> such that <span class="math-container">$X,Y$</span> are independent on each tangent plane and
<span class="math-container">$[X,Y]:=XY-YX=0$</span>.<br />
Then by Frobenius theorem, the partial differential equation on <span class="math-container">$\mathbb{R}^2$</span> given by
<span class="math-container">$\frac{d}{ds}f=X_0(f(s,t),g(s,t)),\frac{d}{ds}g=X_1(f(s,t),g(s,t))$</span>,<br />
<span class="math-container">$\frac{d}{dt}f=Y_0(f(s,t),g(s,t)),\frac{d}{dt}g=Y_1(f(s,t),g(s,t))$</span>,<br />
<span class="math-container">$ (f(0,0),g(0,0))=(0,0)$</span><br />
has a solution in <span class="math-container">$-\epsilon<s<\epsilon,-\epsilon<t<\epsilon$</span> for some positive real number <span class="math-container">$\epsilon$</span>.</p>
<p>My question is if <span class="math-container">$f,g$</span> are maximal solutions (i.e <span class="math-container">$f,g$</span> cannot be extended) then is the image of <span class="math-container">$(f(s,t),g(s,t))$</span> equal to the whole <span class="math-container">$\mathbb{R}^2$</span>?</p>
<p>edit<br />
I'm especially curious about the case when <span class="math-container">$X_0,X_1,Y_0,Y_1$</span> are polynomials.</p>
https://mathoverflow.net/q/4658433Continuity of eigenvector of zero eigenvaluemuddyhttps://mathoverflow.net/users/1557032024-02-24T20:26:22Z2024-03-18T22:51:16Z
<p>Wonder whether anyone has an idea on showing the following or to point out that it is not true:</p>
<p>Let <span class="math-container">$A(t) \in \Re^{n \times n}$</span> be differentiable over an interval <span class="math-container">$I$</span>, and it has a zero eigenvalue for all <span class="math-container">$t \in I$</span>. Then, there exists an eigenvector <span class="math-container">$v(t)$</span> corresponding to the zero eigenvalue of <span class="math-container">$A(t)$</span> for <span class="math-container">$t \in I$</span> such that <span class="math-container">$v(t)$</span> is continuous a.e. on <span class="math-container">$I$</span>.</p>
https://mathoverflow.net/q/4658417Goldman symplectic form vs Weil–Petersson symplectic formAMath91https://mathoverflow.net/users/1277392024-02-24T19:28:59Z2024-03-18T22:09:16Z
<p>I'm confused about the exact multiplicative factor that relates Goldman symplectic form on the <span class="math-container">$\operatorname{SL}(2,\mathbb R)$</span>-character variety and the Weil–Petersson symplectic form on Teichmüller space.</p>
<p>One way to define Goldman symplectic form is the following:
<span class="math-container">$$
\omega_{h} = \int_{S} \operatorname{trace} (\dot{\nabla} \wedge \smash{\dot{\nabla}}')
$$</span>
where <span class="math-container">$\dot{\nabla}$</span> and <span class="math-container">$\smash{\dot{\nabla}}'$</span> are variations of the Levi-Civita connection of the hyperbolic metric <span class="math-container">$h$</span>. If those variations are induced by divergence-free, <span class="math-container">$h$</span>-self-adjoint, traceless endomorphisms <span class="math-container">$\dot{J}$</span> and <span class="math-container">$\dot{J}'$</span>, then it is easy to see that
<span class="math-container">$$
\dot{\nabla} = -\frac{1}{2} J\nabla\dot{J} \ .
$$</span>
Now, Goldman claims that this form is <span class="math-container">$-8$</span> times the Weil-Petersson symplectic form, but, integrating by parts, I get instead a multiplicative factor of <span class="math-container">$+4$</span>. What am I doing wrong?</p>
https://mathoverflow.net/q/4585332Global duality theorem for 2-partDualityhttps://mathoverflow.net/users/1446232023-11-16T12:28:59Z2024-03-18T17:41:59Z
<p><span class="math-container">$\DeclareMathOperator\coker{coker}\DeclareMathOperator\Sha{Sha}$</span>Let <span class="math-container">$K$</span> be a number field.
Let <span class="math-container">$E/K$</span> be an elliptic curve over <span class="math-container">$K$</span>.
Suppose finiteness of <span class="math-container">$\Sha(E/K)$</span>.</p>
<p>According to Global duality theorem (cf. J.S Milne, Arithmetic duality theorems, 6.14(b)
<a href="https://www.jmilne.org/math/Books/ADTnot.pdf" rel="nofollow noreferrer">https://www.jmilne.org/math/Books/ADTnot.pdf</a>)</p>
<p>states</p>
<p><span class="math-container">$$0\to \Sha(E/K)\to H^1(G_K,E)\to \bigoplus_{v\in M_K}H^1(G_{K_v},E)\to \widehat{E(K)}^*\to 0$$</span> is an exact sequence,
where <span class="math-container">$\widehat{E(K)}$</span> is a completion with respect to finite index subgroups, and <span class="math-container">$*$</span> is Pontryagin dual.</p>
<p>My question is, are there known result for <span class="math-container">$Sha(E/K)[2]$</span> ?</p>
<p><span class="math-container">$$0\to \Sha(E/K)[2]\to H^1(G_K,E)[2]\to \bigoplus_{v\in M_K}H^1(G_{K_v},E)[2]\to ?\to 0:\text{exact}$$</span></p>
<p>In other words, can we find <span class="math-container">$?$</span> in explicit form like in the above sequence ?</p>
<p>It is ok to exchange <span class="math-container">$H^1(G_K,E)[2]$</span> and <span class="math-container">$\bigoplus_{v\in M_K}H^1(G_{K_v},E)$</span> to some appropriate other groups.</p>
https://mathoverflow.net/q/44553516Recommendations to learn about the use of toposes in logic?huurdhttps://mathoverflow.net/users/1356282023-04-26T08:06:29Z2024-03-18T17:37:45Z
<p>I'd like to learn about the use of toposes in logic. The "logic" side I know quite well, but of the "topos" side I am totally ignorant.</p>
<p>Which books/articles (formal and/or casual) would you recommend to start with ?</p>
<p>Is it usefull to learn first about Grothendieck toposes in order to contextualize the topic ?</p>
<p>Is it usefull to trace back to Lawere and Tierney's work in the 70's, or can I start directly with more recent accounts like Caramello and others ?</p>
https://mathoverflow.net/q/4196558Is $\frac{1}{L(1+it)}$ unbounded?Holomorphic manifoldhttps://mathoverflow.net/users/4799892022-04-05T07:57:03Z2024-03-18T22:09:15Z
<p>Let <span class="math-container">$\chi$</span> be a Dirichlet character and <span class="math-container">$L(s, \chi)$</span> be the corresponding L-function. Is <span class="math-container">$$\frac{1}{L(1+it, \chi)}$$</span> unbounded for <span class="math-container">$t \in \mathbb{R}$</span>? I'm aware that this is true if <span class="math-container">$L=\zeta$</span>, but I'm not sure of general <span class="math-container">$L$</span>.</p>
https://mathoverflow.net/q/4049740Approximation of the product $(\bar{z} - a)^{-1} \cdot (z - b)^{-1}$Mathieuhttps://mathoverflow.net/users/3899852021-09-27T22:17:14Z2024-03-19T01:02:00Z
<p><span class="math-container">$\def\zbar{\smash{\overline z}\vphantom z}$</span>I would like to construct an approximation of the product
<span class="math-container">\begin{equation}
f(z) = \frac{1}{\zbar-a} \frac{1}{z-b},
\end{equation}</span>
where <span class="math-container">$a, b \in \mathbb{C}$</span>, and <span class="math-container">$|{a}/{z}|, |{b}/{z}| <1$</span>.</p>
<p>More precisely, I would like to obtain a separable expression of the form:
<span class="math-container">\begin{equation}
f(z) = \sum_{k=0}^\infty g_k(a,b) h_k(z, \overline{z}),
\end{equation}</span>
where <span class="math-container">$g_k$</span> depends only on <span class="math-container">$a$</span> and <span class="math-container">$b$</span>, and <span class="math-container">$h_k$</span> depends only on <span class="math-container">$z$</span> and <span class="math-container">$\overline{z}$</span>. We can rewrite the product <span class="math-container">$f(z)$</span> as a double sum:
<span class="math-container">\begin{equation}
\frac{1}{\zbar-a} \frac{1}{z-b} = \frac{1}{|z|^2}\left(\sum_{j=0}^{\infty} \left(\frac{a}{\overline{z}}\right)^j \right) \left(\sum_{k=0}^\infty \left(\frac{b}{z}\right)^k \right) = \frac{1}{|z|^2} \sum_{j,\;k} \left(\frac{a}{\overline{z}}\right)^j \left(\frac{b}{z}\right)^k.
\end{equation}</span></p>
<p>Unfortunately, the function <span class="math-container">$f$</span> is not analytic so we cannot use a <a href="https://en.wikipedia.org/wiki/Cauchy_product" rel="nofollow noreferrer">Cauchy product</a> to obtain the desired form.</p>
<p><strong>Question:</strong> Is there a technique to obtain the desired factorization of <span class="math-container">$f$</span> when <span class="math-container">$|{a}/{z}|, |{b}/{z}| <1$</span>?</p>
https://mathoverflow.net/q/2727454Linear convergence rate of proximal point algorithmyonhttps://mathoverflow.net/users/445522017-06-21T21:51:51Z2024-03-18T20:04:40Z
<p>For $T : R^n \to P({R^n})$ maximally monotone, the proximal point algorithm (step size $c>0$)
$$
x^{k+1} = (I + c T)^{-1} x^k,
$$
converges linearly with rate $\kappa = \frac{1}{1 + c \sigma}$ if $T$ is strongly monotone with parameter $\sigma > 0$.</p>
<p>I'm interested in analyzing the linear convergence rate in case of matrix-valued step sizes, i.e., $C \succ 0$,
$$
x^{k+1} = (I + C T)^{-1} x^k.
$$
I could only manage to prove a bound depending on $\lambda_{\text{min}}(C)$, while in practice I numerically observe that the convergence rate depends on the whole spectrum of $C$. </p>
<p>It seems like such a basic algorithm, so I am surprised that I could not find classic literature (e.g. by Rockafellar) on this topic.</p>
<p><strong>Background:</strong> many proximal algorithms for solving problems of the form
$$
\min_x \max_y~G(x) - F(y) + \langle Kx,y \rangle
$$
such as Douglas-Rachford, ADMM or Chambolle-Pock fit the above setting of proximal point algorithms given a special choice of $C$. In case $G$ and $F$ are both strongly convex, $T$ is strongly monotone and my goal is to connect the linear convergence rate to the choice of metric/step size. </p>
https://mathoverflow.net/q/887322Differential structures on unit-root Frobenius modulesAnonymoushttps://mathoverflow.net/users/131732012-02-17T15:56:57Z2024-03-19T01:02:46Z
<p>Let <span class="math-container">$\mathcal{E}$</span> be the ring of series <span class="math-container">$f(T)=\sum_{n \in \mathbb{Z}} a_nT^n$</span>, <span class="math-container">$a_n \in \mathbb{Q}_p$</span> such that the <span class="math-container">$\{a_n\}$</span> are bounded and <span class="math-container">$a_{-n}\rightarrow 0$</span> as <span class="math-container">$n \rightarrow +\infty$</span>. This is the usual coefficient ring in Fontaine's theory of <span class="math-container">$(\varphi, \Gamma)$</span>-modules. This ring comes equipped with a Frobenius endomorphism <span class="math-container">$\varphi$</span> and a derivation <span class="math-container">$\frac{d}{dT}$</span>. I recently read that any finite free module over <span class="math-container">$\mathcal{E}$</span> equipped with a unit-root <span class="math-container">$\varphi$</span>-structure automatically admits a unique compatible differential module structure. Why is this true?</p>
<p>Conversely, if we take a finite free module <span class="math-container">$M$</span> over <span class="math-container">$\mathcal{E}$</span> equipped with a differential module structure can we find a compatible unit-root Frobenius structure on <span class="math-container">$M$</span>? Furthermore, if we can find one, to what extent is it unique?</p>