Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2023-12-25T01:10:26Zhttps://mathoverflow.net/feedshttps://creativecommons.org/licenses/by-sa/4.0/rdfhttps://mathoverflow.net/q/4609880Comparing the Stacks Project Homotopy limit with limits in the $\infty$-categoryuser141099https://mathoverflow.net/users/4969412023-12-25T00:58:12Z2023-12-25T00:58:12Z
<p>In the Stacks project Tag 08TC, there is a definition of a homotopy limit in a derived category, and I expect it to compare with a limit in the <span class="math-container">$\infty$</span>-categorical enhancement. I guess this is also hinted at in the beginning of the proof of
Proposition 1.2.1.19 in Lurie's Higher Algebra. However, I have trouble to write down the details. Could someone clarify this for me please? Also, is there a reference available?</p>
<p>Alternatively, is there a way to see this through the Bousfield-Kan formula?</p>
https://mathoverflow.net/q/4609870Is $\sum\limits_{k=1}^nk^i=S_3(n)\times\frac{P_{i-3}(n)}{N_i}$ for odd $i>1,\sum\limits_{k=1}^nk^i=S_2(n)\times\frac{P_{i-2}'(n)}{N_i}$ for even $i$?piehttps://mathoverflow.net/users/5055322023-12-25T00:47:16Z2023-12-25T01:02:53Z
<p>I asked this question <a href="https://math.stackexchange.com/questions/4828000/conjecture-sum-limits-k-1nki-s-3n-times-fracp-i-3nn-i-for-odd">here</a></p>
<hr />
<p>When I was in high school, I was fascinated by <span class="math-container">$\displaystyle\sum\limits_{k=1}^n k= \frac{n(n+1)}{2}$</span> so I tried to find the general solution for <span class="math-container">$\displaystyle\sum\limits_{k=1}^n k^i$</span> s.t <span class="math-container">$i \in \mathbb{N}$</span>.</p>
<p>I was able to to find the sum up to <span class="math-container">$i=6$</span>. Here, I tried to search for a pattern to find the general solution of <span class="math-container">$\sum\limits_{k=1}^n k^i $</span> s.t <span class="math-container">$i \in \mathbb{N}$</span> which I failed to do, but I noticed this pattern:<br />
<span class="math-container">$\\[3pt]$</span>
<span class="math-container">$$S_1(n):={\sum\limits_{k=1}^n k= \frac{n(n+1)}{2}}$$</span></p>
<p><span class="math-container">$$S_2(n):={\sum\limits_{k=1}^n k^2= \color{blue}{\frac{n(2n+1)(n+1)}{6}}}$$</span></p>
<p><span class="math-container">$$S_3(n):={\sum\limits_{k=1}^n k^3= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}}}$$</span></p>
<p><span class="math-container">$$S_4(n):={\sum\limits_{k=1}^n k^4=\color{blue}{\frac{n(2n+1)(n+1)}{6}} \times \frac{3n^2+3n-1}{ 5}}$$</span></p>
<p><span class="math-container">$$S_5(n):={\sum\limits_{k=1}^n k^5= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}} \times\frac{2n^2+2n-1}{ 3}}$$</span></p>
<p><span class="math-container">$$S_6(n):={\sum\limits_{k=1}^n k^6= \color{blue}{\frac{n(2n+1)(n+1)}{6}} \times \frac{3n^4+6n^3-3n+1}{ 7 }}$$</span></p>
<p><span class="math-container">$$S_7(n):={\sum\limits_{k=1}^n k^7= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}} }\times\frac{3n^4+6n^3-n^2-4n+2}{ 6}}$$</span></p>
<p><span class="math-container">$$S_8(n):={\sum\limits_{k=1}^n k^8= \color{blue}{\frac{n(2n+1)(n+1)}{6}} \times \frac{5n^6+15n^5+5n^4-15n^3-n^2+9n+3}{ 15}}$$</span></p>
<p><span class="math-container">$$S_9(n):={\sum\limits_{k=1}^n k^9= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}} \times\frac{(n^2+n-1)(2n^4 +4n^3-n^3-3n^2+3)}{ 5}}$$</span></p>
<p><span class="math-container">$$S_{10}(n):={\sum\limits_{k=1}^n k^{10}= \color{blue}{\frac{n(2n+1)(n+1)}{6}} \times \frac{ 3 n^8+ 12 n^7+ 8 n^6 - 18 n^5- 10 n^4+ 24 n^3 + 2 n^2 - 15 n +5}{ 11}}$$</span></p>
<p><span class="math-container">$$S_{11}(n):={\sum\limits_{k=1}^n k^{11}= \color{red}{\frac{\color{red}{n^2(n+1)^2}}{\color{red}{4}}} \times\frac{2n^8 +8n^7+4n^6-16n^5-5n^4+26n^3-3n^2-20n+10}{ 6}}$$</span></p>
<p><span class="math-container">$\\[3pt]$</span>
I noticed that:</p>
<p>For odd <span class="math-container">$i>1$</span>, <span class="math-container">$\displaystyle\sum\limits_{k=1}^n k^i= \color{red}{\frac{{n^2(n+1)^2}}{{4}}} \times \frac{P_{i-3}(n)}{N_i}$</span> s.t <span class="math-container">$P_{i-3}(n)$</span> is an <span class="math-container">$i-3$</span> polynomial with integer coefficients <span class="math-container">$ \{a_{i-3},\dots a_1,a_0 \}$</span> such that <span class="math-container">$\gcd \{a_{i-3},\dots a_1,a_0 \}=1$</span>, <span class="math-container">$N_i\in \mathbb {N}$</span>.</p>
<p>For even <span class="math-container">$i$</span>, <span class="math-container">$\displaystyle\sum\limits_{k=1}^n k^i= \color{blue}{\frac{n(2n+1)(n+1)}{6}}\times \frac{P_{i-2}'(n)}{N_i}$</span> s.t <span class="math-container">$P_{i-2}'(n)$</span> is an <span class="math-container">$i-2$</span> polynomial with integer coefficients <span class="math-container">$\{ a_{i-2},\dots a_1,a_0 \}$</span> such that <span class="math-container">$\gcd \{ a_{i-2},\dots a_1,a_0 \}=1$</span>, <span class="math-container">$N_i\in \mathbb {N}$</span>.</p>
<p>When I was in high school I couldn't prove this pattern, and I remembered this observation that I had totally forgotten about. Now, after two years from my first attempt, I tried to prove this pattern again, but I couldn't.</p>
<hr />
<p>This question has been partly answered <a href="https://mathoverflow.net/questions/444681/maybe-faulhaber-polynomial-s-kx-0-have-only-rational-roots-0-frac12">here</a> (The answer shows that <span class="math-container">$\displaystyle\sum\limits_{k=1}^n k^i$</span> is divisible by <span class="math-container">${n^2(n+1)^2}$</span> for odd <span class="math-container">$i>1$</span> and
<span class="math-container">$\displaystyle\sum\limits_{k=1}^n k^i$</span> is divisible by <span class="math-container">${n(2n+1)(n+1)}$</span> for even <span class="math-container">$i$</span>) the only missing part is to show that the denominator is a multiple of <span class="math-container">$4$</span> if <span class="math-container">$i\in 2\mathbb{N}+1$</span>, and the denominator is a multiple of <span class="math-container">$6$</span> if <span class="math-container">$i \in 2\mathbb{N}$</span>.</p>
https://mathoverflow.net/q/460986-4Examiner si vous savez ça sinon laisser [closed]Hamma Hamahttps://mathoverflow.net/users/5196172023-12-25T00:40:04Z2023-12-25T00:40:04Z
<p>Bonjour, SMIC ou Solution Maths Institute Clay</p>
<p>19+19=38</p>
<p>38+38=76</p>
<p>380+380=760</p>
<p>20×38=760</p>
<p>2×38=76</p>
<p>38=2×20</p>
<p>10X=11</p>
<p>10111=23</p>
<p>23=10123</p>
<p>1)10.868673767=(2.713+2.718)×2</p>
<p>2)10.868673767=2.713+2.718+2.718281828+2.719391939</p>
<p>e=2.718281828</p>
<p>=3-p</p>
<p>=3-b</p>
<p>=3-></p>
<p>8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160]=1.500.000+3288</p>
<p>[8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160+8300+8300+1660+9080+6080+1516+7080+4080+1116+101080+101160]=1.500.000+1644</p>
<p>1644-1640=4
830660€=943660<span class="math-container">$
943.660$</span>-830.660€=113.000
113.000/2=56.500
1.000.000$-943.660=56 340
56.500-56.340=160</p>
<p>Cordialement</p>
<p>Ahmadou Mbacké Niane
Diplômé Génie Électrique et Informatique Industrielle
Diplômé Géomètre-TOPOGRAPHE
Ayant fréquenté le Master GER EPT en 2018</p>
<p>CECI N'EST PLUS BASIQUE QUAND ON AJOUTE 20 ET 2 ON A 760 ET 76</p>
<p>Cher M. AHMADOU MBACKE NIANE</p>
<p>Je pense qu'il n'y a rien d'utile que je puisse dire en réponse à ce que vous avez écrit. Le problème P v NP est un défi profond concernant la nature des algorithmes, pas une question de manipulation de symboles de cette manière basique.</p>
<p>Bien cordialement,
Martin Bridson</p>
<hr />
<p>Martin R Bridson FRS (he/him)
President, Clay Mathematics Institute
<a href="http://www.claymath.org" rel="nofollow noreferrer">http://www.claymath.org</a></p>
<p>CMI President's Office
Andrew Wiles Building
Radcliffe Observatory Quarter
Woodstock Road
Oxford, OX2 6GG
United Kingdom</p>
https://mathoverflow.net/q/4609851Show me that I have not simplified the proof of the Adian-Rabin theoremPerry Bleiberghttps://mathoverflow.net/users/529692023-12-24T23:54:37Z2023-12-25T00:12:32Z
<p>I am not a mathematics researcher but I am concerned that this question, posed with slightly different wording on math.stackexchange, may be too esoteric for that forum since it concerns the details of a proof that I don't think is standard in the graduate level curriculum. At least, nobody has answered it so far.</p>
<p>Let <span class="math-container">$G$</span> be a group with presentation <span class="math-container">$<x_1,x_2...,x_m|R>$</span> and let <span class="math-container">$G'=G \ast<y_0>$</span>.</p>
<p>Now define <span class="math-container">$y_i=y_{0}x_i$</span>. Notice that <span class="math-container">$G'=<y_0,y_1,...y_m|R'>$</span> for appropriately defined <span class="math-container">$R'$</span> and that <span class="math-container">$<y_i>$</span> is free on <span class="math-container">$y_i$</span> for all <span class="math-container">$i$</span>.</p>
<p>Consider the group</p>
<p><span class="math-container">$$H = <y_0,y_1,...y_m,t_1,t_2,...t_m,|R',t_iy_it^{-1}_i=y_{i}^2>$$</span></p>
<p>Either take my word for it that <span class="math-container">$G'$</span> embeds (isomorphically) into <span class="math-container">$H$</span>, or notice that, since <span class="math-container">$<y_i>$</span> is free on <span class="math-container">$y_i$</span> for all <span class="math-container">$i$</span>, <span class="math-container">$H$</span> is an iterated HNN extension on <span class="math-container">$G'$</span> and therefore embeds <span class="math-container">$G'$</span> by <span class="math-container">$m$</span> applications of Britton's Lemma.</p>
<p>Now let <span class="math-container">$w$</span> be a nonidentity element in G. Define <span class="math-container">$H^{'}$</span> as <span class="math-container">$H$</span> subject to the relation <span class="math-container">$(w^{-1}y_{0}^{-1}wy_0)t_i(w^{-1}y_{0}^{-1}wy_0)^{-1} = t_{i}^2$</span> for each <span class="math-container">$t_i$</span>.</p>
<p>I can show that <span class="math-container">$w^{-1}y_{0}^{-1}wy_0$</span> is infinite order since it is cyclically reduced word not containing the identity and length <span class="math-container">$\geq 2$</span>. <span class="math-container">$t_i$</span> is also infinite order in <span class="math-container">$H$</span> since it is the pivot letter in an HNN extension. So it would seem that I'm not introducing any torsion for the elements <span class="math-container">$y_0,t_i$</span>.</p>
<p>I also don't think I'm introducing any torsion for elements of <span class="math-container">$G$</span> since it does not reference them at all.</p>
<p>In particular it seems like <span class="math-container">$G'$</span> embeds into <span class="math-container">$H'$</span>. I have no idea how to prove this, but I have done pages of calculations in this group to try and disprove this.</p>
<p>+++++++++++++++++</p>
<p>Is this true? Here's why I think no, in spite of my best efforts to (dis)prove it:</p>
<p>In the proof of the Adian-Rabin theorem, <span class="math-container">$G$</span> is the free product of some group with unsolvable word problem <span class="math-container">$U$</span> with a group <span class="math-container">$G_{-}$</span>, which has the property that any group it embeds into does not satisfy some property <span class="math-container">$M$</span>. (A <span class="math-container">$G_{-}$</span> with such a property is given by hypothesis).</p>
<p>The proof constructs <span class="math-container">$H$</span>, and then uses <span class="math-container">$H$</span> in the construction of another group, call it <span class="math-container">$K$</span>. <span class="math-container">$K$</span> has the property that it embeds <span class="math-container">$G_{-}$</span> iff <span class="math-container">$w\neq1$</span>, and is trivial otherwise. Proving that <span class="math-container">$K$</span> depends on <span class="math-container">$w$</span> in this way is the crux of the theorem. However, I think that my group <span class="math-container">$H'$</span> also has the desired property w.r.t. <span class="math-container">$w$</span>, which would mean that the proof could be simplified somewhat.*</p>
<p>All proofs of Adian-Rabin that I have seen go as far as constructing <span class="math-container">$K$</span>, they don't stop at <span class="math-container">$H$</span> or construct <span class="math-container">$H'$</span>, so there must be something about my construction that makes it not necessarily embed <span class="math-container">$G_{-}$</span>. I cannot figure out what that could be, for reasons outlined above.</p>
<p>*NB: In <span class="math-container">$H'$</span>, <span class="math-container">$w=1$</span> implies <span class="math-container">$w^{-1}y_{0}^{-1}wy_0=1$</span> implies the <span class="math-container">$t_i$</span>'s and thus the <span class="math-container">$y_i$</span>'s all equal 1. The proof that <span class="math-container">$K$</span> is trivial goes exactly like this too, except it involves the extra relations used to construct <span class="math-container">$K$</span>.</p>
https://mathoverflow.net/q/4609830Estimate for the operator $A A_D^{-1}$Yulia Meshkovahttps://mathoverflow.net/users/1266232023-12-24T23:04:41Z2023-12-24T23:04:41Z
<p>Let <span class="math-container">$O\subset\mathbb{R}^d$</span>
be a bounded domain of the class <span class="math-container">$C^{1,1}$</span>
(or <span class="math-container">$C^2$</span>
for simplicity). Let the operator <span class="math-container">$A_D$</span>
be formally given by the differential expression <span class="math-container">$A=-\mathrm{div}g(x)\nabla$</span>
with the Dirichlet boundary condition. (<span class="math-container">$A$</span>
means the expression, and <span class="math-container">$A_D$</span>
is the corresponding operator.) Here <span class="math-container">$g(x)>0$</span>, <span class="math-container">$g,g^{−1}\in L_\infty$</span>.
The precise definition is given via the quadratic form on <span class="math-container">$H^1_0(O)$</span>.</p>
<p>If the matrix of coefficients <span class="math-container">$g$</span>
and the boundary are smooth enough, we can define <span class="math-container">$A_D$</span>
by the differential expression <span class="math-container">$-\mathrm{div}g(x)\nabla$</span>
on <span class="math-container">$H^2(O)\cap H^1_0(O)=\mathrm{Ran}A_D^{-1}$</span>, so for the expression <span class="math-container">$A$</span>
acting on the resolvent <span class="math-container">$A^{−1}_D$</span>
we have
<span class="math-container">$$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=\Vert A_D A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}=1.$$</span></p>
<p><strong>Question:</strong></p>
<p>Is there are any hope to prove the estimate
<span class="math-container">$$\Vert A A_D^{-1}\Vert _{L_2(\mathbb{R}^d)\rightarrow L_2(\mathbb{R}^d)}\leqslant \mathrm{const}$$</span>
for the operator with non-smooth coefficients in the domain of class <span class="math-container">$C^{1,1}$</span>
(or <span class="math-container">$C^2$</span>
)? (Here the expression <span class="math-container">$A$</span> is initially considered as operator from <span class="math-container">$H^1$</span>
to <span class="math-container">$H^{−1}$</span>
.) E. g., by using of mollification of coefficients.</p>
<p>(I am also interested in this question for strongly elliptic systems in a divergent form: <span class="math-container">$A=b(D)^∗g(x)b(D)$</span>
, <span class="math-container">$b=\sum _{l=1}^d b_lD_l$</span>
, <span class="math-container">$b_l$</span>
are constant matrices, the symbol of <span class="math-container">$b(D)$</span>
has maximal rank.)</p>
https://mathoverflow.net/q/460981-1Why do we define independence for zero-probability events?Matthiashttps://mathoverflow.net/users/1779492023-12-24T22:27:09Z2023-12-24T23:03:08Z
<p>I am learning about probability and the definition of pairwise independence is given as <span class="math-container">$P(AB) = P(A)P(B)$</span>. My textbook motivates this definition as one to capture the intuition where the knowledge of one event occurring doesn't change the your knowledge of the likelihood of the other event occurring. I understand that this definition implies the two conditional probability statements that are closer to that above intuition, given that both events in question have non-zero probability.</p>
<p>But it seems to me that there are examples where this definition does not match intuition, when one or more of the events have zero probability. Consider an experiment where you pick a random number <span class="math-container">$R$</span> uniformly between 0 and 1. The event where that random number <span class="math-container">$R$</span> is equal to 0.5 has probability 0. So, by the standard definition of independence, it is trivially independent with the event that your number <span class="math-container">$R$</span> is in <span class="math-container">$[0.25, 0.75]$</span>. But intuitively, the knowledge of <span class="math-container">$R = 0.5$</span> occurring should make you 100% confident that <span class="math-container">$R \in [0.25, 0.75]$</span>.</p>
<p>I guess my question is why is independence defined for zero-probability events when the definition does not capture the intuition of independence? What is the utility of it?</p>
https://mathoverflow.net/q/4609793Psychological test for Euclidean geometryAnton Petruninhttps://mathoverflow.net/users/14412023-12-24T22:04:34Z2023-12-24T22:31:24Z
<p>There is the so-called <a href="https://en.wikipedia.org/wiki/Force_Concept_Inventory" rel="nofollow noreferrer">FCI test</a>.
It contains a list of questions such that anyone who can speak will have an opinion.
Based on the answers one can determine if the person knows elementary mechanics.
I learned about this test from the <a href="https://www.youtube.com/watch?v=WwslBPj8GgI" rel="nofollow noreferrer">lecture of Eric Mazur</a>, which is an interesting lecture,
and it describes an application of such a list.</p>
<p>It seems that there are no analogous tests in mathematics.
Or did I miss something?</p>
<p>Sergei Tabachnikov shared a couple of questions that might work:</p>
<ul>
<li><p>One needs to peel potato. What is faster and why: peeling a pound of large or small potatoes?</p>
</li>
<li><p>Why does the reflection in the mirror interchanges left and right, but not up and down?</p>
</li>
<li><p>Two persons are walking down an escalator (that is moving down), counting their steps.
Who will count more steps, the one who is moving faster or slower?</p>
</li>
</ul>
<blockquote>
<p>Could you share more questions of this type; please make one question per answer.</p>
</blockquote>
https://mathoverflow.net/q/4609770Commutative/ symmetric second covariant derivativeKhaled T.https://mathoverflow.net/users/5196112023-12-24T20:47:23Z2023-12-24T21:18:34Z
<p>Consider a smooth manifold <span class="math-container">$M$</span> together with an affine connection (or covariant derivative) <span class="math-container">$\nabla$</span> on the tangent bundle <span class="math-container">$TM$</span>.</p>
<p>Is it possible to have an affine connection, possibly with non-zero torsion, such that <span class="math-container">$\nabla_{X,Y}^2 = \nabla_{Y,X}^2$</span>? I.e. the second covariant derivative is symmetric/commutative, while the covariant derivative itself is not (since it might have torsion)?</p>
<p>How are such connections or manifolds called? If anyone can refer me to an article or reference, that would be great.</p>
<p>PS. Note that the connection is not necessarily curvature-free in the case that <span class="math-container">$\nabla_{X,Y}^2 = \nabla_{Y,X}^2$</span>, but instead all the curvature is due to torsion. That is, <span class="math-container">$R(X,Y)Z = \nabla_{T(X,Y)}Z$</span>.</p>
https://mathoverflow.net/q/460976-1Possibility of upper bounding $\|f\|_{L^2}$ by the weighted $L^2$ norm of its Laplace transformFei Caohttps://mathoverflow.net/users/1634542023-12-24T20:35:07Z2023-12-24T20:35:07Z
<p>Assume that <span class="math-container">$f$</span> is a smooth and bounded function defined on <span class="math-container">$[0,\infty)$</span> such that its mean value is zero (i.e., <span class="math-container">$\int_{\mathbb{R}_+} xf(x) \mathrm{d}x = 0$</span>). We denote by <span class="math-container">$$\hat{f}(\xi) = \int_{\mathbb{R}_+} f(x)\mathrm{e}^{-x\xi} \mathrm{d}x$$</span> to be the Laplace transform of <span class="math-container">$f$</span>. I am wondering if it is possible to upper bound <span class="math-container">$\|f\|^2_{L^2}$</span> by some functionals involving <span class="math-container">$$\int_{\mathbb{R}_+} \frac{|\hat{f}(\xi)|^2}{\xi^2} \mathrm{d}\xi \label{1}\tag{1}$$</span> I admit that it is not possible to bound the <span class="math-container">$L^2$</span> norm of <span class="math-container">$f$</span> by the <span class="math-container">$L^2$</span> norm of its Laplace transform, but will such bound be possible with the modification considered in \eqref{1} ? Any help is greatly appreciated!</p>
https://mathoverflow.net/q/4609750Connection of eigenspace of finite Hilbert matrix and its continuous operator counterpartMatteo Saccardihttps://mathoverflow.net/users/5031182023-12-24T20:14:37Z2023-12-24T20:14:37Z
<p>I am trying to understand the connection between the eigenspace of the continuous operator
<span class="math-container">$$
H(x,y) = \frac{1}{x+y}
$$</span>
which is nothing but the square of the Laplace operator, and its discrete counterpart, i.e. the Hilbert matrix
<span class="math-container">$$
H_{i,j} = \frac{1}{i+j}\,, i,j=1,\dots,N \,.
$$</span>
In particular, <a href="https://www2.math.upenn.edu/%7Ecle/papers/laplce_rev2.pdf" rel="nofollow noreferrer">it can be proven</a> that the generalized eigenfunctions of the continuous operator are
<span class="math-container">$$
u_s(x) = \frac{1}{\sqrt{2\pi}} x^{-\frac{1}{2}+is}\,, \int_0^\infty dy \, H(x,y) u_s(y) = \vert \lambda_s \vert^2 u_s(x)
$$</span>
where
<span class="math-container">$$
\lambda_s = \Gamma(\frac{1}{2}+is)\,, \quad \vert \lambda_s \vert^2 = \frac{\pi}{\cosh{\pi s}} \,.
$$</span>
Intuitively, the discretized spectrum and eigenvectors should converge to their continuous values, but I do not fully understand nor find in the literature a discussion about the two distinct effects of</p>
<ol>
<li>discretization, i.e. <span class="math-container">$i,j=1,2,\dots$</span>, possibly up to <span class="math-container">$\infty$</span>, and</li>
<li>finite, but continuous extension (i.e. having <span class="math-container">$\int_0^X$</span> rather than <span class="math-container">$\int_0^\infty$</span>).</li>
</ol>
<p>I believe their effects need to be taken separately into account, since considering and infinite amount of points but not continuously leads to some contradictions, e.g.
<span class="math-container">$$
\mathrm{tr}{H} = \int_{-\infty}^\infty ds \, \vert \lambda_s \vert^2 = \pi < \sum_{n=1}^\infty \frac{1}{n+n} = \infty \,.
$$</span></p>
<p>I was wondering whether there was a way to link finite and discrete (i.e. numerical) eigenvalues of the Hilbert matrix to their continuous operator counterpart. Thank you all!</p>
https://mathoverflow.net/q/460971-5short exact sequence of groups [closed]Sajjad Mohammadihttps://mathoverflow.net/users/5007562023-12-24T18:45:34Z2023-12-24T19:19:57Z
<p>We have a short exact sequence as
<span class="math-container">$$0 \rightarrow \mathbb{Z}_2\rightarrow G \rightarrow \mathbb{Z}_2\rightarrow 0,$$</span>
can we conclude that the group <span class="math-container">$G$</span> is isomorphic to <span class="math-container">$\mathbb{Z}_2 + \mathbb{Z}_2$</span> or <span class="math-container">$\mathbb{Z}_4$</span>?
Also, Let the group <span class="math-container">$G$</span> be isomorphic to the direct sum of finite numbers of free abelian groups finitely generated such as <span class="math-container">$\mathbb{Z}_n$</span>, then is the map <span class="math-container">$G\rightarrow \mathbb{Z}$</span> trivial map?</p>
https://mathoverflow.net/q/4609700Does there exist an $L$-function for any subset of $\mathbb{N}$?martinhttps://mathoverflow.net/users/450572023-12-24T18:32:38Z2023-12-24T19:29:15Z
<p>Consider the following prime sum:</p>
<p><span class="math-container">\begin{aligned}
\sum _{p}{\frac {\cos(x\log p)}{p^{1/2}}}
\end{aligned}</span></p>
<p><a href="https://i.stack.imgur.com/Tt23O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tt23O.png" alt="enter image description here" /></a></p>
<p>whose spikes appear at the Riemann <span class="math-container">$\zeta$</span> zeros as shown <a href="https://mathoverflow.net/questions/460561/does-this-partial-sum-over-primes-spike-at-all-zeta-zeros">here</a>.</p>
<p>Taking these detected spikes (up to some partial sum), and denoting these approximated zeros as <span class="math-container">$t'$</span>, the primes can be recovered with the sum:</p>
<p><span class="math-container">\begin{aligned}
\sum _{n \in t'}{\frac {\cos(t'\log x)}{t'^{1/2}}}
\end{aligned}</span></p>
<p><a href="https://i.stack.imgur.com/qh9zL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qh9zL.png" alt="enter image description here" /></a></p>
<p>albeit with added noise. Compare with the plot</p>
<p><span class="math-container">\begin{aligned}
\sum _{n \in t}{\frac {\cos(t\log x)}{t^{1/2}}}
\end{aligned}</span></p>
<p><a href="https://i.stack.imgur.com/Hp0oQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hp0oQ.png" alt="enter image description here" /></a></p>
<p>where <span class="math-container">$t$</span> is the imaginary part of the actual Riemann zeta zeros.</p>
<p>It turns out that it appears that this method can be used for any arbitrary integer sets</p>
<p><strong>e.g.</strong></p>
<p><em>the set of twin primes:</em></p>
<p><a href="https://i.stack.imgur.com/Cufhe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Cufhe.png" alt="enter image description here" /></a></p>
<p><em>the square numbers:</em></p>
<p><a href="https://i.stack.imgur.com/1ZtdJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1ZtdJ.png" alt="enter image description here" /></a></p>
<p>Does this imply that there exists an <span class="math-container">$L$</span>-function that corresponds to <em>any</em> subset of <span class="math-container">$\mathbb{N}$</span>?</p>
https://mathoverflow.net/q/4609691Two spectral sequences arising from a simplicial spectrumBrian Shinhttps://mathoverflow.net/users/946242023-12-24T18:07:49Z2023-12-24T19:19:21Z
<p>Let <span class="math-container">$X_\bullet$</span> be a simplicial spectrum, and let <span class="math-container">$X = |X_\bullet|$</span> be the geometric realization.
Let's assume each <span class="math-container">$X_n$</span> is connective.</p>
<p>From this situation, we can form two filtrations on <span class="math-container">$X$</span>: the <em>skeletal</em> filtration and the <em>levelwise Postnikov</em> filtration.
You can probably guess what these will be, but let me spell it out for my own sake.</p>
<p>The skeletal filtration
<span class="math-container">$$\mathrm{Fil}^\mathrm{sk}_*(X) = \{ 0 \to \mathrm{sk}^0(X_\bullet) \to \mathrm{sk}^1(X_\bullet) \to \mathrm{sk}^2(X_\bullet) \to \cdots \}$$</span>
is an increasing filtration with graded pieces <span class="math-container">$\mathrm{gr}^\mathrm{sk}_s(X)= \mathrm{Fil}^\mathrm{sk}_s(X)/\mathrm{Fil}^\mathrm{sk}_{s-1}(X) = \Sigma^s X_s$</span>.
The resulting spectral sequence has signature
<span class="math-container">$$E^1_{n,s}(\mathrm{sk}) = \pi_{n-s}(X_s) \implies \pi_n(X)$$</span>
with differentials <span class="math-container">$d^r : E^r_{n,s} \to E^r_{n-1,s-r}$</span>.
The second page is quite nice: we find that <span class="math-container">$E^2_{n,s}(\mathrm{sk}) = \mathrm{H}_s(\pi_{n-s}(X_\bullet))$</span>, where for each <span class="math-container">$j \in \mathbb{Z}$</span> we're viewing <span class="math-container">$\pi_j(X_\bullet)$</span> as a chain complex via Dold-Kan.</p>
<p>On the other hand, the levelwise Postnikov filtration
<span class="math-container">$$\mathrm{Fil}^\mathrm{LP}_*(X) = \{ \cdots \to |\tau_{\geq 2} (X_\bullet)| \to |\tau_{\geq 1} (X_\bullet)| \to |\tau_{\geq 0} (X_\bullet)| = X\}$$</span>
is a decreasing filtration with graded pieces <span class="math-container">$\mathrm{gr}^\mathrm{LP}_s(X) = \mathrm{Fil}^\mathrm{LP}_s(X)/\mathrm{Fil}^\mathrm{LP}_{s+1}(X) = \Sigma^s|\pi_s (X_\bullet)|$</span>.
The resulting spectral sequence has signature
<span class="math-container">$$E^1_{n,s}(\mathrm{LP}) = \mathrm{H}_{n-s}(\pi_s(X_\bullet)) \implies \pi_n(X)$$</span>
with differentials <span class="math-container">$d^r : E^r_{n,s} \to E^r_{n-1,s+r}$</span>.
Note in particular that the <span class="math-container">$E^1$</span> page here is the <span class="math-container">$E^2$</span> page coming from the skeletal filtration, up to a shift <span class="math-container">$s \mapsto n-s$</span>.</p>
<p>Finally, the</p>
<blockquote>
<p><strong>Question.</strong> What exactly is the relationship between the skeletal filtration and the levelwise Postnikov filtration?</p>
</blockquote>
<p>In particular, I'm looking for an explanation for the essentially matching spectral sequence pages.</p>
<p>I suspect the key is hidden in the fact that the two filtrations are two "edges" of a single diagram <span class="math-container">$\mathbb{N} \times \mathbb{N}^\mathrm{op} \to \mathrm{Spt}$</span>, <span class="math-container">$(a,b) \mapsto \mathrm{sk}_a(\tau_{\geq b} (X_\bullet))$</span>.
Restricting to <span class="math-container">$b=0$</span> yields the skeletal filtration, while taking the colimit over <span class="math-container">$a$</span> yields the levelwise Postnikov filtration.
I imagine this gives some way to interpolate between the two spectral sequences.</p>
<p>More generally, I imagine the various spectral sequences one can extract from a bilfiltered spectrum <span class="math-container">$M \in \mathrm{Fun}(\mathbb{Z},\mathrm{Fun}(\mathbb{Z},\mathrm{Spt}))$</span> have some sort of relationship, though maybe not as straightforward as shifting and shearing pages.
In general, it seems our understanding of the various spectral sequences arising from a bifiltered spectrum is not quite at a satisfactory state.
See <em>e.g.</em> <a href="https://mathoverflow.net/q/293725/94624">this other question</a>.
I am very much interested in any developments on this front.</p>
https://mathoverflow.net/q/4609680Lattice not contained in any connected subgroup is not contained in any positive dimensional subgroupIan Gershon Teixeirahttps://mathoverflow.net/users/3871902023-12-24T17:21:34Z2023-12-24T17:21:34Z
<p>Let <span class="math-container">$ G $</span> be a simple Lie group and let <span class="math-container">$ \Gamma $</span> be a lattice in <span class="math-container">$ G $</span>. If <span class="math-container">$ \Gamma $</span> is not contained in any connected subgroup of <span class="math-container">$ G $</span> does that imply that <span class="math-container">$ \Gamma $</span> is not contained in any positive dimensional subgroup of <span class="math-container">$ G $</span>?</p>
<p>Context:
I was thinking about the lattice <span class="math-container">$ SL(2,\mathbb{Z}) $</span> in <span class="math-container">$ SL(2,\mathbb{C}) $</span>. <span class="math-container">$ SL(2,\mathbb{Z}) $</span> is not contained in any proper connected subgroup of <span class="math-container">$ SL(2,\mathbb{C}) $</span>. And it is also true that <span class="math-container">$ SL(2,\mathbb{Z}) $</span> is not contained in any positive dimensional proper subgroup.</p>
<p>Cross-posted <a href="https://math.stackexchange.com/questions/4828889/lattice-not-contained-in-any-connected-subgroup-is-not-contained-in-any-positive">https://math.stackexchange.com/questions/4828889/lattice-not-contained-in-any-connected-subgroup-is-not-contained-in-any-positive</a> its been up on MSE for over a week with 1 upvote and no comments or answers so I thought MO might be better</p>
https://mathoverflow.net/q/4609671How does the behaviour of a hyperderived functor of many variables change if you use $\prod$-totalisation instead of $\oplus$-totalisation?FShrikehttps://mathoverflow.net/users/3200402023-12-24T16:02:03Z2023-12-24T16:02:03Z
<p><span class="math-container">$\newcommand{\tot}{\operatorname{Tot}}\newcommand{\A}{\mathscr{A}}\newcommand{\L}{\mathbb{L}}\newcommand{\R}{\mathbb{R}}$</span>Say <span class="math-container">$T$</span> is is a functor <span class="math-container">$\A_1\times\A_2\times\cdots\times\A_n\to\A$</span> of Abelian categories, additive in each variable separately. Assume the Abelian categories <span class="math-container">$\A_\bullet$</span> are as nice as we could wish, maybe even that they are module categories.</p>
<p>When defining the left or right hyperderived functors of <span class="math-container">$T$</span> (which Cartan-Eilenberg call instead the (co)homology invariants of <span class="math-container">$T$</span>) we take, for some tuple <span class="math-container">$X_1,X_2,\cdots,X_n$</span>, left or right Cartan-Eilenberg resolutions <span class="math-container">$Q_1,Q_2,\cdots,Q_n$</span> - which are upper half plane double complexes of (co)homological type - and according to some sign conventions also explained in the text of Cartan-Eilenberg we can form a <span class="math-container">$2n$</span>-complex <span class="math-container">$T(Q_1,Q_2,\cdots,Q_n)$</span>. Then we would declare the hyperderived functor of <span class="math-container">$T$</span> at <span class="math-container">$X_1,\cdots,X_n$</span> to be the homology of the totalisation of this <span class="math-container">$2n$</span>-complex.</p>
<p>In the text of Cartan-Eilenberg it seems the convention is to always define <span class="math-container">$\L_\ast T(X_1,\cdots,X_n)=H_\ast(\tot^{\oplus}(T(Q_1,\cdots,Q_n))),\,\R_\ast T(X_1,\cdots,X_n)=H^\ast(\tot^{\oplus}(T(Q_1,\cdots,Q_n)))$</span> taking the <span class="math-container">$\oplus$</span>-totalisation in both instances. However, in the text of Weibel (which only discusses the single variable case) the convention is to define <span class="math-container">$\L_\ast T$</span> by <span class="math-container">$\oplus$</span>-totalisation and <span class="math-container">$\R_\ast T$</span> by <span class="math-container">$\prod$</span>-totalisation; <span class="math-container">$\R_\ast T(X_1,\cdots,X_n):=H^\ast(\tot^{\prod}(T(X_1,\cdots,X_n)))$</span>.</p>
<p>Either choice results in well-defined functors. The two notions only disagree on unbounded complexes. <strong>Are there any reasons to prefer one totalisation convention over the other, in this case?</strong> My gut tells me <span class="math-container">$\R_\ast$</span> is best defined with <span class="math-container">$\prod$</span>-totalisation simply because <span class="math-container">$\oplus$</span> sits on the "left" and <span class="math-container">$\prod$</span> on the "right" when thinking about universal properties. I cannot point to any formal property of one definition over the other that would inform this choice. I suppose another benefit is that under the <span class="math-container">$\prod$</span>-totalisation definition, <span class="math-container">$\R_\ast$</span> is formally dual to <span class="math-container">$\L_\ast$</span>. But, Cartan and Eilenberg must have had a reason for doing it their way, right?</p>
<p>One difficulty in pinpointing reasons to prefer one definition over the other is that I am not aware of anywhere where these notions are studied in generality. The Cartan-Eilenberg texts lays out some definitions. I know of a detailed study of <span class="math-container">$\L_\ast T$</span> where <span class="math-container">$T$</span> is the tensor product in Grothendieck's EGA 3, but this is a special case.</p>
https://mathoverflow.net/q/4609663What exactly is a Tannakian subcategory?David Corwinhttps://mathoverflow.net/users/13552023-12-24T15:59:31Z2023-12-24T15:59:31Z
<p>I've searched all the standard references (Deligne--Milne, Saavedra-Rivano) and cannot find a definition of <em>Tannakian subcategory</em>. What I find is many authors who discuss the Tannakian subcategory generated by an object, which should correspond by Tannaka duality to the image of the Tannakian fundamental group in the automorphisms of the fiber functor applied to that object. Even Deligne--Milne mention this on p.61 without defining the notion. They do define a tensor</p>
<p>Question: What is the definition of a Tannakian subcategory of a Tannakian category?</p>
<p>More specifically: Is a Tannakian subcategory necessarily full?</p>
<p>(*) I would assume it is at least an abelian subcategory closed under tensor products and duals, and containing the unit object. Note the former means that a sequence is exact in the subcategory iff it is in the larger category.</p>
<p>But that leaves open the following possibilities:</p>
<ol>
<li><p>Let <span class="math-container">$G$</span> be a pro-algebraic group over <span class="math-container">$\mathbb{C}$</span> and <span class="math-container">$G \twoheadrightarrow H$</span> a quotient. Then <span class="math-container">$\mathrm{Rep}_{\mathbb{C}}(H) \subseteq \mathrm{Rep}_{\mathbb{C}}(G)$</span> is a (full) Tannakian subcategory by any definition.</p>
</li>
<li><p>Let <span class="math-container">$\mathcal{C}'$</span> the category of <span class="math-container">$\mathbb{C}$</span>-representations of <span class="math-container">$C_2$</span> (let's say a skeletal version), and let <span class="math-container">$\mathcal{C}$</span> denote the category whose objects are those of <span class="math-container">$\mathcal{C}'$</span> and whose morphisms are all maps of <span class="math-container">$\mathbb{C}$</span>-vector spaces. Then <span class="math-container">$\mathcal{C}'$</span> is a non-full subcategory of <span class="math-container">$\mathcal{C}$</span> that satisfies (*), and <span class="math-container">$\mathcal{C}$</span> is equivalent to the category of representations of the trivial group. But surely this is not what people mean, right? (In particular, this example is especially evil; the embedding depends on the isomorphism class of the category <span class="math-container">$\mathcal{C}$</span>.)</p>
</li>
<li><p>Now let <span class="math-container">$\mathcal{C}'$</span> be the category of <span class="math-container">$\mathbb{C}$</span>-linear representations of <span class="math-container">$S_3$</span>, and let <span class="math-container">$\mathcal{C}$</span> be the category whose objects are those of <span class="math-container">$\mathcal{C}$</span> and whose morphisms are those that are <span class="math-container">$C_2=\{\mathrm{id},(12)\}$</span>-equivariant. Then (assuming we don't do anything too evil), we can view <span class="math-container">$\mathcal{C}'\subseteq \mathcal{C}$</span> as an inclusion of subcategories. This inclusion satisfies (*) but it is not full: for example, there is an inclusion from either the trivial representation or the sign representation into the two-dimensional irreducible representation of <span class="math-container">$S_3$</span>. Somehow, this example seems more allowable than 2., because the inclusion <span class="math-container">$C_2 \to S_3$</span> is an epimorphism in the category of (algebraic) groups.</p>
</li>
</ol>
<p>Notice that in all cases, the inclusion functor is a morphism of Tannakian category (it's an exact tensor functor). But it's not clear that it should be called a Tannakian subcategory.</p>
https://mathoverflow.net/q/4609640Is the integer factorization into prime numbers normally distributed?mathoverflowUserhttps://mathoverflow.net/users/1659202023-12-24T14:28:06Z2023-12-24T23:20:30Z
<p>Let <span class="math-container">$P_1(n) := 1$</span> if <span class="math-container">$n=1$</span> and <span class="math-container">$\max_{q|n, \text{ }q\text{ prime}} q$</span> otherwise, denote the largest prime divisor of <span class="math-container">$n$</span>.</p>
<p>Let us define some rooted trees <span class="math-container">$T_{n,m}$</span> for <span class="math-container">$1 \le m \le n$</span> by:</p>
<ul>
<li><span class="math-container">$T_{n,m}$</span> has as root the number <span class="math-container">$m$</span>.</li>
<li>If there are primes <span class="math-container">$p$</span> such that <span class="math-container">$P_1(m) \le p \le \frac{n}{m}$</span>, then the children of <span class="math-container">$m$</span> are <span class="math-container">$T_{n,mp}$</span>, otherwise <span class="math-container">$m$</span> is a leaf.</li>
</ul>
<p>We set <span class="math-container">$T_n := T_{n,1}$</span>. Here is an image of <span class="math-container">$T_{20}$</span>:</p>
<p><a href="https://i.stack.imgur.com/JtNzl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JtNzl.png" alt="T_20" /></a></p>
<p>We can encode these trees as parentheses <span class="math-container">$()$</span> or binary words <span class="math-container">$w_n$</span> <span class="math-container">$10$</span> where <span class="math-container">$(=1$</span> and <span class="math-container">$)=0$</span>. Here are some words <span class="math-container">$n$</span> and <span class="math-container">$w_n$</span>:</p>
<pre><code>1
10
2
1100
3
110100
4
11100100
5
1110010100
</code></pre>
<p>From these words only it is possible to reconstruct all prime factorizations of all numbers <span class="math-container">$1 \le m \le n$</span>, without to have to factor any number <span class="math-container">$m$</span>, only a check is needed if a number is prime or not.</p>
<p>Observing those words we see that they seem to be kind of random.
To make this more precise, I suggest separating the <span class="math-container">$1$</span>s and <span class="math-container">$0$</span>s from each other by permutations <span class="math-container">$\sigma$</span> from <span class="math-container">$S_{2n}$</span>, where <span class="math-container">$|w_n| = 2n$</span>. This gives us a measure of its randomness, where we define the string for example:</p>
<p><span class="math-container">$$0000000000111111$$</span></p>
<p>as being separated (the zeros than the ones), whereas the string:</p>
<p><span class="math-container">$$010101010101010101$$</span></p>
<p>is not separated. In <a href="https://statweb.stanford.edu/%7Ecgates/PERSI/papers/77_04_spearmans.pdf" rel="nofollow noreferrer">Diaconis & Graham & Knuth</a> the following distance measure for permutations is suggested:</p>
<p><span class="math-container">$$T(\pi,\sigma) = n - C(\pi^{-1} \sigma), \pi,\sigma \in S_n$$</span></p>
<p>where <span class="math-container">$C$</span> counts the number of distinct cyclces of the permutation. Using this, one can define the separation measure as for a arbitrary binary word <span class="math-container">$w$</span> as:</p>
<p><span class="math-container">$$Z(w) = \max_{ \sigma \in S_{|w|}, \sigma \text{ separates } w} T(1,\sigma)$$</span></p>
<p>The standardized version <span class="math-container">$z(w_n):=z_n$</span> in the tree case, is given by Diaconis and Graham as :</p>
<p><span class="math-container">$$z_n := \frac{Z(w_n)-\mu_n}{\sqrt{\operatorname{Var(n)}}} = \frac{|w|-C^*(w)-|w|+\log(|w |)}{\sqrt{\log(|w|)}}$$</span>
which is equal to:</p>
<p><span class="math-container">$$z_n=\sqrt{\log(2n)}-\frac{C^*(w_n)}{\sqrt{\log(2n)}}$$</span></p>
<p>where we have set: <span class="math-container">$C^*(w_n) := \max_{ \sigma \in S_{|w|}, \sigma \text{ separates } w}C(\sigma)$</span></p>
<p><strong>Question: Is it true that for <span class="math-container">$n \rightarrow \infty$</span> we have <span class="math-container">$z_n \approx N(0,1)$</span> is distributed as the standard normal variable?</strong></p>
<p><a href="https://sagecell.sagemath.org/?z=eJyVU02P2yAQva-0_wE5qgSJv_BxVV97zqGntazIsklChTEFskn313fAYCe7K1X1xcAb3rx5MwzsiPYUS_Ly_ITg40ck65qGnfs0sxctEZ1PmDDsMzh2N6w0H9lh4G_cTNoAI5nDnp8GyMFkPw0MyxQZ22lb05iwP3MxaCZRjRoPbRU6ThopxCXynAaDQA-RVBaFX-0oaRfBgkkcaUhdl5_1JbRMvtR_hbRJkv-auMSLxFlhT7yO3ulY2L9gTnbXXeLo50L38s7NEBS8URyQbFmb39o6m0gho1M9yClRcM2tHatSzhsoV4iwMPB3eaqMgiILO3qHOtXSqdadPDFM02ox--psjnWGYkCPtB9217jdBDSFnlo24AaghrcphwZ4e_iaqNpK0qaavTFtWP1TX1gcgb3Ty_R4sZ3lk8TNraGtv35z1_-b-0cHXSSoDfSvQA8BmZuDfd7_6QU72IsSMDlRwTuE4NcMonZiOnk6UvgGLNt1_r_Rsnwco-jCey4XSqXyTikmB7wAAREiIvfxm0ejN74XH4eOrtGu6_8YnG31QO3qBwszpTL6ktGWFPMJXFUFrvJyLXMDdV4kB5dHXKbg9_carUGh8o1_n7uarne4OXgZ-DFoWIOiVUMKD8gPsoLHZQ1WiuTmPF1xOMZnbux00t2IhSAR-wurdTiM&lang=sage&interacts=eJyLjgUAARUAuQ==" rel="nofollow noreferrer">Here</a> is an image of the histogram for <span class="math-container">$1 \le n \le 2000$</span>:</p>
<p><a href="https://i.stack.imgur.com/Jrdsi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jrdsi.png" alt="histogram_2000" /></a></p>
<p>(Notice that in the image above I have not really computed <span class="math-container">$\max$</span> over all separating permutations but just taken one permutation given by the sorting algorithm.)</p>
https://mathoverflow.net/q/4609452Computing the Dieudonné module of $\mu_p$ from Fontaine's Witt CovectorHJKhttps://mathoverflow.net/users/4904352023-12-24T06:40:06Z2023-12-24T16:33:51Z
<p>In <em><a href="http://www.numdam.org/item/AST_1977__47-48__1_0/" rel="nofollow noreferrer">Groupes <span class="math-container">$p$</span>-divisibles sur les corps locaux</a></em>, Fontaine introduced a uniform construction of Dieudonné modules through the definition of the Witt covector. Consider a perfect field <span class="math-container">$k$</span> of characteristic <span class="math-container">$p$</span> and a finite <span class="math-container">$k$</span>-algebra <span class="math-container">$R$</span>. Fontaine presented a group functor</p>
<p><span class="math-container">$$\operatorname{CW}(R) = \{(\dotsc, a_{-3}, a_{-2}, a_{-1}) \mid a_i\in R\},$$</span></p>
<p>where for some (negative) integer <span class="math-container">$r$</span>, the ideal generated by <span class="math-container">$\{a_n \mid n < r\}$</span> is nilpotent. The addition is the usual Witt vector addition, which is well-defined due to the nilpotency condition. This functor is obviously represented by an affine group scheme, so pro-represented by a formal <span class="math-container">$p$</span>-group. He gave the following theorem:</p>
<blockquote>
<p><strong>Theorem (Fontaine).</strong> The Dieudonné module of a finite group scheme <span class="math-container">$G$</span> is given by
<span class="math-container">$M(G) = \operatorname{Hom}_{\text{formal-groups}}(G, \operatorname{CW}).$</span></p>
</blockquote>
<p>For a constant group scheme <span class="math-container">$\mathbb{Z}/p\mathbb{Z}$</span>, the homomorphisms correspond to group homomorphisms on <span class="math-container">$k$</span>-points. If <span class="math-container">$G = \alpha_p = \operatorname{Spec} k[x]/(x^p)$</span>, then these homomorphisms are represented by <span class="math-container">$(\dotsc, 0, 0, ax)$</span> for some <span class="math-container">$a \in k$</span>. For <span class="math-container">$G = \mu_p$</span>, we know that <span class="math-container">$\operatorname{Hom}_{\text{formal-groups}}(\mu_p, \operatorname{CW}) = k$</span>. However, I cannot find a single explicit homomorphism.</p>
<blockquote>
<p><strong>Question 1.</strong> How to represent the homomorphisms from <span class="math-container">$\mu_p=\operatorname{Spec} k[x]/(x^p-1)$</span> to <span class="math-container">$\operatorname{CW}$</span> in terms of the coordinates
<span class="math-container">$\operatorname{CW}(\mu_p) = \{(\dotsc, a_{-3}, a_{-2}, a_{-1}) \mid a_i\in k[x]/(x^p-1)\}$</span>?</p>
</blockquote>
<p>There is an exact sequence</p>
<p><span class="math-container">$$ 0 \rightarrow \varinjlim \mu_{p^n} \rightarrow \operatorname{CW} \xrightarrow{1-V} \operatorname{CW} \rightarrow 0, $$</span></p>
<p>which appears to be analogous to the Artin–Schreier sequence</p>
<p><span class="math-container">$$ 0 \rightarrow \mathbb{Z}_p/\mathbb{Q}_p \rightarrow \operatorname{CW} \xrightarrow{1-F} \operatorname{CW} \rightarrow 0. $$</span></p>
<p>Let <span class="math-container">$A$</span> denote the kernel of the map <span class="math-container">$\operatorname{CW} \xrightarrow{1-V} \operatorname{CW}$</span>. By definition, <span class="math-container">$V$</span> acts trivially on <span class="math-container">$A$</span>, so <span class="math-container">$F$</span> acts on <span class="math-container">$A$</span> as multiplication by <span class="math-container">$p$</span>. Thus, every element of <span class="math-container">$A[p^n](R)$</span> can be represented as <span class="math-container">$(\dotsc, a, a, a)$</span> for some <span class="math-container">$a\in R$</span> satisfying <span class="math-container">$a^{p^n}=0$</span>.</p>
<p>Therefore, <span class="math-container">$A[p^n]$</span> is represented by <span class="math-container">$k[x]/(x^p)$</span>. Considering the actions of <span class="math-container">$V$</span> and <span class="math-container">$F$</span>, it becomes evident that <span class="math-container">$A[p^n]$</span> is isomorphic to <span class="math-container">$\mu_{p^n}$</span>.</p>
<blockquote>
<p><strong>Question 2.</strong> How can we explicitly represent the isomorphism <span class="math-container">$\mu_p$</span> with <span class="math-container">$A[p] = \ker(\operatorname{CW} \xrightarrow{1-V} \operatorname{CW})[p]$</span> in terms of the coordinates <span class="math-container">$A[p](\mu_p) = \{(\dotsc, a, a, a) \mid a \in k[x]/(x^p-1)\}$</span>?</p>
</blockquote>
https://mathoverflow.net/q/4609370Solve NP-hard type problems with linear programmingJuan Carloshttps://mathoverflow.net/users/5195882023-12-24T01:36:24Z2023-12-24T15:46:03Z
<p>I would like to know if there is any way to solve an NP-hard type problem, for example, the TSP, sum of subsets or knapsack problem, by using linear programming and not by brute force.</p>
<p>I ask this because I made a program which uses a linear programming algorithm and it solves them, but the execution time does not vary specifically from the input size, so I can't define its asymptotic behavior <span class="math-container">$O(n)$</span> yet.</p>
<p>For example, for the TSP with an input size <span class="math-container">$n=20$</span>, depending on the cases and tested on an ordinary computer, it takes between <span class="math-container">$30$</span> seconds and <span class="math-container">$3$</span> minutes to solve the problem.</p>
<p>Greetings.</p>
https://mathoverflow.net/q/4609280On fifth powers forming a Sidon setSayan Duttahttps://mathoverflow.net/users/3113662023-12-23T21:21:18Z2023-12-24T18:36:51Z
<p>We call a set of natural numbers <span class="math-container">$\mathcal S$</span> to be a <a href="https://en.wikipedia.org/wiki/Sidon_sequence" rel="nofollow noreferrer"><em>Sidon Set</em></a> if <span class="math-container">$a+b=c+d$</span> for <span class="math-container">$a,b,c,d\in \mathcal S$</span> implies <span class="math-container">$\{a,b\}=\{c,d\}$</span>. In other words, all pairwise sums are distinct.</p>
<p>Erdős conjectured that there exists a nonconstant integer-coefficient polynomial whose values at the natural numbers form a Sidon sequence. Specifically, he asked if the set of fifth powers is a Sidon set.</p>
<p>Ruzsa came close to this by showing that there is a real number <span class="math-container">$c$</span> with <span class="math-container">$0<c<1$</span> such that the range of the function
<span class="math-container">$$f(x)=x^5+\left\lfloor cx^4\right\rfloor$$</span>
is a Sidon sequence.</p>
<p>I mainly have two (related) questions :-</p>
<ol>
<li><p>How much (if any) progress have been made on Ruzsa'a result? Are there any other almost-polynomials that have been found which may be able to do the job?</p>
</li>
<li><p>How far has the computers checked the second conjecture? Related to this is the question : what algorithms do we know of (better that brute force) to check whether a given set is a Sidon set or not? Checking all the cases (ie., brute forcing) won't be a good idea since <span class="math-container">$100^5=10^{10}$</span> is already quite large.</p>
</li>
</ol>
<hr />
<p>Gerry Myerson points out OEIS A046881, which gives some really useful results. But, it is through here that I came to know that Euler had a proof that there are infinitely many numbers that can be written as the sum of two fourth powers in two different ways (source<a href="https://math.hawaii.edu/home/talks/resume-fev2011.pdf" rel="nofollow noreferrer">https://math.hawaii.edu/home/talks/resume-fev2011.pdf</a>). But, I couldn't find Euler's proof. Can somebody please refer me to the proof?</p>
https://mathoverflow.net/q/460914-1Maximizing a function involving minimums with variable parameterslznhttps://mathoverflow.net/users/5195672023-12-23T14:26:07Z2023-12-24T15:44:27Z
<p>I am working on a problem where I have a set of <span class="math-container">$m$</span> data points, each represented as a triplet <span class="math-container">$(a_i, b_i, c_i)$</span> for <span class="math-container">$ i = 1, 2, \dots, m $</span>. I am interested in a target function defined as:</p>
<p><span class="math-container">$f(t) = 2a + 2 \min(t, b) + 2\lfloor(t+\min(t, b + c))/2\rfloor $</span></p>
<p>where <span class="math-container">$ t $</span> is a variable, and I need to calculate the maximum value of <span class="math-container">$ f(t) $</span> for each <span class="math-container">$ t = 1, 2, \dots, n $</span> across all the given triplets. In other words, for each value of <span class="math-container">$t$</span>, I need to identify which triplet <span class="math-container">$(a_i, b_i, c_i)$</span> yields the maximal <span class="math-container">$f(t)$</span>.</p>
<p>My challenge is to find an efficient method or algorithm to calculate these maximum values for each <span class="math-container">$ t $</span>. I am currently unsure how to approach this problem, especially considering the presence of minimum functions in <span class="math-container">$ f(t) $</span>.</p>
<p>I would appreciate any insights or suggestions on methods to tackle this optimization problem. Specifically, I am interested in:</p>
<ol>
<li>Strategies for dealing with the minimum functions within the target function.</li>
<li>Possible optimization algorithms or techniques that could be employed.</li>
<li>Any relevant mathematical theories or concepts that might be helpful in solving this problem.</li>
</ol>
<p>Thank you for any assistance you can provide!</p>
https://mathoverflow.net/q/4608874Is there an elliptic curve analogue to the 4-term exact sequence defining the unit and class group of a number field?Snacchttps://mathoverflow.net/users/5195432023-12-23T01:12:08Z2023-12-24T14:38:37Z
<p>Let <span class="math-container">$K$</span> be a number field. One has the following exact sequence relating the unit group and ideal class group <span class="math-container">$\text{cl}(K)$</span>:
<span class="math-container">$$1\to \mathcal{O}_K^\times\to K^\times \to J_K\to \text{cl}(K)\to 1$$</span>
where <span class="math-container">$J_K$</span> is the group of fractional ideals of <span class="math-container">$K$</span>. This is essentially a restatement of the definition of unit group and class group respectively. However, there is an alternate way to construct <span class="math-container">$\text{cl}(K)$</span>. One can consider the group of units of the algebraic integers <span class="math-container">$\mathcal{O}_{\bar{K}}^\times$</span>, and this has a natural galois action. You can consider it's Tate-Shafaravich group
<span class="math-container">$$\text{Sha}(K,\mathcal{O}_{\bar{K}}^\times)=\text{ker}\left(H^1(K,\mathcal{O}_{\bar{K}}^\times)\to \prod_{\mathfrak{p}}H^1(K_{\mathfrak{p}},\mathcal{O}_{\bar{K_{\mathfrak{p}}}}^\times)\right)$$</span>
You can prove (see <a href="https://www.ma.imperial.ac.uk/%7Ebuzzard/maths/research/notes/why_is_an_ideal_class_group_a_tate_schaferevich_group.pdf" rel="nofollow noreferrer">here</a>) that there is a canonical isomorphism <span class="math-container">$\text{Sha}(K,\mathcal{O}_{\bar{K}}^\times)\simeq \text{cl}(K)$</span> Using the behavior of number fields as analogy, is it possible to create an analogous exact sequence for elliptic curves? I.e, one of the following form:</p>
<p><span class="math-container">$$1\to E(K)\to X \to Y \to \text{Sha}(E/K)\to 1$$</span></p>
<p>for some "natural" objects <span class="math-container">$X,Y$</span>.</p>
https://mathoverflow.net/q/4608625Expressing an invertible sparse matrix as a product of few elementary matricesJohn Pardonhttps://mathoverflow.net/users/353532023-12-22T15:52:45Z2023-12-24T16:16:00Z
<p>Let <span class="math-container">$M$</span> be an <span class="math-container">$n \times n$</span> matrix with integer entries. Suppose that <span class="math-container">$M$</span> is invertible (over the integers) and that <span class="math-container">$M$</span> has at most <span class="math-container">$An$</span> nonzero entries, each of which is less than <span class="math-container">$B$</span> in absolute value, for some constants <span class="math-container">$A,B<\infty$</span>.</p>
<blockquote>
<p>Is <span class="math-container">$M$</span> necessarily the product of <span class="math-container">$Cn$</span> elementary matrices whose single off-diagonal entries are less than <span class="math-container">$D$</span> in absolute value, for some constants <span class="math-container">$C$</span> and <span class="math-container">$D$</span> depending on <span class="math-container">$A$</span> and <span class="math-container">$B$</span>?</p>
</blockquote>
<p>A simple Gaussian elimination does not appear to come anywhere close to showing this. Sparsity is lost rapidly (the number of nonzero entries can grow exponentially in the number of steps in the elimination process), and so there will be order <span class="math-container">$n^2$</span> elementary matrices, with off-diagonal entries of order <span class="math-container">$2^{n^2}$</span> times <span class="math-container">$B$</span> (or something like that). Perhaps the bound I'm asking for is too optimistic, but are there any significant improvements on the "trivial" Gaussian elimination approach (which doesn't take much advantage of sparsity)?</p>
<p>One could ask the same question over the rationals, where instead of bounding the absolute value of the entries, we bound their height (absolute value of numerator plus absolute value of denominator, when written in lowest terms). I presume this is more or less equivalent to the problem over the integers (and I would be happy with an answer in either setting).</p>
https://mathoverflow.net/q/4608203Adjunctions and inverse limits of derived categoriesuser141099https://mathoverflow.net/users/4969412023-12-21T22:58:49Z2023-12-25T01:05:47Z
<p>Consider a tower <span class="math-container">$\dots\to A_{2}\to A_{1}$</span> of rings. This gives rise to a diagram <span class="math-container">$\mathbb{N}^{\text{op}}\to\text{Cat}_{\infty}$</span> of <span class="math-container">$\infty$</span>-categories (confusing <span class="math-container">$\mathbb{N}^{\text{op}}$</span> with its nerve), sending <span class="math-container">$t$</span> to the derived category <span class="math-container">$D(A_{t})$</span> and <span class="math-container">$t+1\to t$</span> to <span class="math-container">$A_{t}\otimes_{A_{t+1}}^{\text{L}}-$</span>, etc.</p>
<p>Now let <span class="math-container">$A:=\varprojlim_{t}A_{t}$</span>. Then there is a functor <span class="math-container">$\text{L}\colon D(A) \to \varprojlim_{t}D(A_{t})$</span> which, roughly speaking, sends a complex <span class="math-container">$M$</span> of <span class="math-container">$A$</span>-modules to the data of the complexes <span class="math-container">$A_{t}\otimes_{A}^{\text{L}}M$</span> for all <span class="math-container">$t$</span>. If I am not mistaken, Proposition 5.5.3.13 in Lurie's Higher Topos Theory implies that <span class="math-container">$\text{L}$</span> is a left adjoint. Fix a right adjoint <span class="math-container">$\text{R}$</span>.</p>
<p>Here is my question: How do we describe <span class="math-container">$\text{R}$</span> explicitely? I understand that intuitively, we should take a limit of a given system <span class="math-container">$M_{\bullet}=\left( M_{t}\right)_{t}$</span> of complexes of <span class="math-container">$A_{t}$</span>-modules; but one always takes the limit of a <em>diagram</em>. How precisely do we produce a suitable diagram <span class="math-container">$\mathbb{N}^{\text{op}}\to D(A)$</span> from a given object <span class="math-container">$M_{\bullet}\in\varprojlim_{t}D(A_{t})$</span>?</p>
https://mathoverflow.net/q/4608094Local triviality of torsors for relative reductive groupsC.D.https://mathoverflow.net/users/1241852023-12-21T19:17:05Z2023-12-24T14:26:56Z
<p>Let <span class="math-container">$X \to S$</span> be a relative (smooth proper) curve, and <span class="math-container">$G \to X$</span> a reductive group scheme. The following two results are well-known:</p>
<ol>
<li>(Drinfeld-Simpson) For arbitrary <span class="math-container">$S$</span>, if <span class="math-container">$G$</span> is defined over <span class="math-container">$S$</span>, then any <span class="math-container">$G$</span>-bundle on <span class="math-container">$X$</span> becomes Zariski-locally trivial after an étale base change <span class="math-container">$S' \to S$</span>.</li>
<li>(Steinberg-Borel-Springer -- actually an older result due to Lang, thanks Jason) If <span class="math-container">$S = \operatorname{Spec} k$</span> for a field, then writing <span class="math-container">$F$</span> for the function field of <span class="math-container">$X_{\overline{k}}$</span>, we have <span class="math-container">$H^1(F, G_F) = 0$</span>. It follows that every <span class="math-container">$G$</span>-bundle on <span class="math-container">$X$</span> is Zariski-locally trivial after an étale base change <span class="math-container">$S' \to S$</span>.</li>
</ol>
<p>Note the differences: in the first result, <span class="math-container">$G$</span> is ``constant" relative to the morphism <span class="math-container">$X \to S$</span>. In the second result, <span class="math-container">$G$</span> is allowed to be a relative group scheme over <span class="math-container">$X$</span>, but <span class="math-container">$S$</span> is restricted to be a field, and a cohomological vanishing argument is being used to descend.</p>
<p>These two results suggest a possible common strengthening:</p>
<ol start="3">
<li>For arbitrary <span class="math-container">$S$</span> and <span class="math-container">$G$</span> a reductive group scheme over <span class="math-container">$X$</span>, any <span class="math-container">$G$</span>-bundle on <span class="math-container">$X$</span> becomes Zariski-locally trivial after an étale base change <span class="math-container">$S' \to S$</span>.</li>
</ol>
<p><strong>Is this true?</strong> Does it follow from these two results?</p>
https://mathoverflow.net/q/4606302Is the Fortissimo space on discrete $\omega_1$ radial?Steven Clontzhttps://mathoverflow.net/users/737852023-12-18T18:21:17Z2023-12-24T23:29:12Z
<p>Let <span class="math-container">$\omega_1$</span> have the discrete topology. Its Fortissimo space is <span class="math-container">$X=\omega_1\cup\{\infty\}$</span> where neighborhoods of <span class="math-container">$\infty$</span> are co-countable.</p>
<p>A space is radial provided for every subset <span class="math-container">$A$</span> and point in its closure, there exists a transfinite sequence within <span class="math-container">$A$</span> converging to it (see <a href="https://zbmath.org/1059.54001" rel="nofollow noreferrer">d-4 Pseudoradial spaces</a>).</p>
<p>Then <span class="math-container">$\infty\in cl(A)$</span> for any uncountable <span class="math-container">$A$</span>, and given the "transfinite sequence" <span class="math-container">$a_\alpha=\min\{a\in A:\beta<\alpha\Rightarrow a>a_\beta\}$</span> for <span class="math-container">$\alpha<\omega_1$</span>, we have that every neighborhood of <span class="math-container">$\infty$</span> contains a final subsequence. Since <span class="math-container">$\infty$</span> is the only possible point added by the closure operator, this shows <span class="math-container">$X$</span> is radial.</p>
<p>The problem is: is <span class="math-container">$a_\alpha$</span> actually a transfinite sequence? This mapping from the ordinal space <span class="math-container">$\omega_1$</span> to <span class="math-container">$X$</span> is not continuous as its image is discrete. In fact, <a href="https://math.stackexchange.com/questions/4828939">every convergent continuous image of an infinite limit ordinal in <span class="math-container">$X$</span> is eventually constant</a>. Under this interpretation, the space is not radial.</p>
<hr />
<p>After more careful review of <a href="https://zbmath.org/1059.54001" rel="nofollow noreferrer">my reference</a> and JDH's comment, I think it's clear that the first interpretation is correct, and the space is in fact radial.</p>
<p>However, does the more restrictive concept appear in the literature anywhere?</p>
https://mathoverflow.net/q/4598776A cubic equation, and integers of the form $a^2+32b^2$Bogdan Grechukhttps://mathoverflow.net/users/890642023-12-06T16:18:37Z2023-12-24T17:16:06Z
<p>I am trying to determine whether there are any integers <span class="math-container">$x,y,z$</span> such that
<span class="math-container">$$
1+2 x+x^2 y+4 y^2+2 z^2 = 0. \quad\quad\quad (1)
$$</span>
It is clear that <span class="math-container">$x$</span> is odd. We can consider this equation as quadratic in <span class="math-container">$(y,z)$</span> with parameter <span class="math-container">$x$</span>. After multiplying by <span class="math-container">$16$</span>, we can rewrite the equation as
<span class="math-container">$$
(8y+x^2)^2+32z^2=x^4-32x-16.
$$</span>
Writing <span class="math-container">$x=2t+1$</span> and denoting <span class="math-container">$s=8y+x^2$</span>, we obtain
<span class="math-container">$$
s^2+32z^2=P(t), \quad\quad\quad (2)
$$</span>
where <span class="math-container">$P(t)=(2t+1)^4-32(2t+1)-16$</span>. If this equation has no integer solutions, then so is the original one.</p>
<p>With <span class="math-container">$Z=2z$</span> we can rewrite it as
<span class="math-container">$$
s^2+8Z^2=P(t) \quad\quad\quad (3)
$$</span>
with <span class="math-container">$Z$</span> even. It is known that every prime congruent to <span class="math-container">$1$</span> modulo <span class="math-container">$8$</span> is of the form <span class="math-container">$s^2+8Z^2$</span>. <span class="math-container">$P(t)$</span> is always equal to <span class="math-container">$1$</span> modulo <span class="math-container">$8$</span>, and takes infinitely many prime values by Bunyakovsky conjecture. By finding <span class="math-container">$t$</span> such that <span class="math-container">$P(t)$</span> is prime, we can generate as many solutions to (3) as we want, but <span class="math-container">$Z$</span> happens to be always odd in all these solutions up to a large bound.</p>
<p>Unfortunately, there seems to be no coprime <span class="math-container">$a,b$</span> such that all primes equal to <span class="math-container">$a$</span> modulo <span class="math-container">$b$</span> are of the form <span class="math-container">$s^2+32z^2$</span>, so we cannot easily apply the same method directly to (2).</p>
<p>So, are there any integers <span class="math-container">$x,y,z$</span> satisfying (1)?</p>
https://mathoverflow.net/q/45652813Four new series for $\pi$ and related identities involving harmonic numbersZhi-Wei Sunhttps://mathoverflow.net/users/1246542023-10-16T04:17:08Z2023-12-24T19:46:38Z
<p>Recently, I discovered the following four new (conjectural) series for <span class="math-container">$\pi$</span>:
<span class="math-container">\begin{align}\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^k\binom{3k}k}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}&=\frac{3\pi}2,\tag{1}
\\[8pt]\sum_{k=1}^\infty\frac{(35k^2-29k+6)3^k\binom{3k}k}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}&=\sqrt3\,\pi,\tag{2}
\\[8pt]\sum_{k=1}^\infty\frac{(40k^2-20k+3)2^k\binom{4k}{2k}}
{k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}}&=\frac{\pi}2,\tag{3}
\\[8pt]\sum_{k=1}^\infty\frac{(64k^2-48k+7)3^k\binom{4k}{2k}}
{k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}}&=\frac{4\pi}{3\sqrt3}.\tag{4}
\end{align}</span>
The series in <span class="math-container">$(1)$</span>–<span class="math-container">$(4)$</span> have converging rates <span class="math-container">$27/32$</span>, <span class="math-container">$81/256$</span>, <span class="math-container">$2/27$</span> and <span class="math-container">$1/9$</span> respectively, and so it is easy to check the identities <span class="math-container">$(1)$</span>–<span class="math-container">$(4)$</span> numerically.</p>
<p>I also consider some variants of <span class="math-container">$(1)$</span>–<span class="math-container">$(4)$</span> involving the harmonic numbers
<span class="math-container">$$H_n=\sum_{0<k\le n}\frac1k\ \ (n=0,1,2,\ldots).$$</span>
Namely, motivated by <span class="math-container">$(1)$</span>-<span class="math-container">$(4)$</span>, I conjecture the following identities:
<span class="math-container">\begin{align}\sum_{k=1}^\infty\frac{\binom{3k}k8^k((5k^2-4k+1)(2H_{2k-1}-3H_{k-1})-8k+2)}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}
&=3\pi\log2-12G,\tag{5}
\\\sum_{k=1}^\infty\frac{\binom{3k}k8^k((5k^2-4k+1)(6H_{4k-1}-7H_{2k-1})+22k-10)}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}
&=3\pi\log2+24G,\tag{6}
\end{align}</span>
<span class="math-container">\begin{align}
&\sum_{k=1}^\infty\frac{\binom{3k}k8^k\left((5k^2-4k+1)(2H_{3k-1}-2H_{2k-1}-H_{k-1})+\frac{2(219k^3-249k^2+87k-10)}{3k(3k-1)}\right)}{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}
\\&\qquad\qquad=\frac{9}4\pi^2,\tag{7}\end{align}</span>
<span class="math-container">\begin{align}
&\sum_{k=1}^\infty\frac{\binom{3k}k3^k\left((35k^2-29k+6)(H_{2k-1}-H_{k-1})-\frac{4(4k-1)(7k-3)}{5(2k-1)}\right)}
{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}
\\&\qquad=\frac{3}{10}(2\pi\sqrt3\log3-9K),\tag{8}
\end{align}</span>
<span class="math-container">\begin{align}&\sum_{k=1}^\infty\frac{\binom{3k}k3^k\left((35k^2-29k+6)(H_{4k-1}-H_{2k-1})+\frac{2(42k^2-36k+7)}{5(2k-1)}\right)}
{k(3k-1)(3k-2)\binom{2k}k\binom{4k}{2k}}
\\&\qquad=\frac{99}{10}K-\frac{\pi}5\sqrt3\log3,\tag{9}
\end{align}</span>
<span class="math-container">\begin{align}
&\sum_{k=1}^\infty\frac{\binom{4k}{2k}2^k((40k^2-20k+3)(2H_{6k-1}-H_{3k-1}-H_{k-1})-32k+4)}
{k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}}
\\&\qquad=2G+\frac{\pi}2\log2,\tag{10}
\end{align}</span>
<span class="math-container">\begin{align}
&\sum_{k=1}^\infty\frac{\binom{4k}{2k}3^k\left((64k^2-48k+7)(5H_{2k-1}-4H_{k-1})-\frac{2(6k-1)(24k-13)}{2k-1}\right)}
{k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}}
\\&\qquad=\frac43\pi\sqrt3\log3-9K,\tag{11}
\end{align}</span>
and
<span class="math-container">\begin{align}
&\sum_{k=1}^\infty\frac{\binom{4k}{2k}3^k\big((64k^2-48k+7)(10H_{6k-1}-5H_{3k-1}-3H_{k-1})-\frac{8(96k^2-78k+17)}{6k-3}\big)}
{k(4k-1)(4k-3)\binom{3k}k\binom{6k}{3k}}
\\&\qquad=\frac49\pi\sqrt3\log3+32K,\tag{12}
\end{align}</span>
where
<span class="math-container">$$G:=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}
\ \ \text{and}\ \ \ K:=\sum_{k=1}^\infty\frac{(\frac k3)}{k^2}=\sum_{n=0}^\infty\left(\frac1{(3n+1)^2}-\frac1{(3n+2)^2}\right).$$</span></p>
<p><strong>QUESTION.</strong> Can one prove the identities <span class="math-container">$(1)$</span>-<span class="math-container">$(12)$</span> via known tools (including the WZ pairs and hypergeometric transformation formulas)?</p>
<p>Your commments are welcome!</p>
https://mathoverflow.net/q/3766982Is this lower bound on the singular values of the sum of two matrices correct?Gabriele Olivahttps://mathoverflow.net/users/1688612020-11-17T16:16:21Z2023-12-24T23:34:52Z
<p>Equation 7 of <a href="http://www.m-hikari.com/imf-2011/29-32-2011/ulukokIMF29-32-2011.pdf" rel="nofollow noreferrer">this paper (Ramazan Türkmen, Zübeyde Ulukök, <em>Inequalities for Singular Values of Positive Semidefinite Block Matrices</em>, International Mathematical Forum, Vol. 6, 2011, no. 31, 1535 - 1545)</a> says that it is well known that, given two complex-valued matrices <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, it holds</p>
<p><span class="math-container">$$\sigma_j(A+B) \geq \sigma_j(A)+\lambda_n(B)$$</span></p>
<p>where <span class="math-container">$\sigma_j$</span> is the <span class="math-container">$j$</span>-th singular value and <span class="math-container">$\lambda_n$</span> is the <span class="math-container">$n$</span>-th eigenvalue</p>
<ol>
<li><p>Is this correct? Do we need some hypotheses on <span class="math-container">$A$</span>, <span class="math-container">$B$</span>?</p>
</li>
<li><p>Can somebody point me to a reference?</p>
</li>
</ol>
<p>Thank you.</p>
https://mathoverflow.net/q/2617733Lower estimate on sectional curvature of the boundaryasvhttps://mathoverflow.net/users/161832017-02-09T13:20:04Z2023-12-24T21:44:53Z
<p>Let $M^n$ be an $n$-dimensional smooth compact Riemannian manifold with boundary. Assume that the sectional curvature of $M$ is at least $\kappa$, the diameter is at most $D$, and the second fundamental form of the boundary is at least $\lambda$. </p>
<p><strong>Question.</strong> Does there exist a lower bound on the sectional curvature of the boundary $\partial M$ in terms of $n,\kappa,\lambda$ and (possibly) $D$?</p>
<p>Even partial results under extra assumptions might be helpful.</p>