Recent Questions - MathOverflowmost recent 30 from www.4124039.com2019-03-24T16:12:13Zhttp://www.4124039.com/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://www.4124039.com/q/3262160Systems of $n$ divisibility conditions on $n$ prime variablesNellhttp://www.4124039.com/users/1369322019-03-24T16:07:07Z2019-03-24T16:07:07Z
<p>Consider a system of <span class="math-container">$n$</span> divisibility conditions on <span class="math-container">$n$</span> prime variables:
<span class="math-container">$$p_i|a_{i,1} p_1 + \dotsc + a_{i,n} p_n,\;\;\;\;\;1\leq i\leq n,$$</span>
where <span class="math-container">$a_{i,j}$</span> are bounded integers. How many solutions are there with all <span class="math-container">$p_i$</span> prime and in a range <span class="math-container">$[N_0,N_1]$</span>? (We can assume that <span class="math-container">$N_0$</span> is quite a bit larger than all <span class="math-container">$a_{i,j}$</span>.) Or, put otherwise: under what conditions are there very few solutions?</p>
<hr>
<p>Some thoughts: if <span class="math-container">$\lbrack N_0,N_1\rbrack$</span> is dyadic (that is, of the form <span class="math-container">$\lbrack N, 2N\rbrack$</span>) then the divisibility conditions can be replaced by a system of equations <span class="math-container">$c_i p_i = a_{i,1} p_1 + \dotsc + a_{i,n} p_n$</span>, <span class="math-container">$c_i$</span> bounded. That system should not in general have non-trivial solutions (meaning: it will not have solutions, as <span class="math-container">$0$</span> is not a prime), though of course the fact that there are <span class="math-container">$C^k$</span> possibilities for <span class="math-container">$c_i$</span> can be annoying to say the least.</p>
<p>Of course, <em>some</em> conditions are needed on <span class="math-container">$a_{i,j}$</span> for the system to have few solutions: if <span class="math-container">$a_{i,j}=0$</span> for all <span class="math-container">$(i,j)$</span>, the system does have very many solutions!</p>
http://www.4124039.com/q/3262150Simple One-relator GroupsCarl-Fredrik Nyberg Broddahttp://www.4124039.com/users/1209142019-03-24T16:02:54Z2019-03-24T16:02:54Z
<p>One-relator groups, that is, groups which admit a finite presentation <span class="math-container">$\langle A \: | \: w=1 \rangle$</span> for some <span class="math-container">$w \in A^\ast$</span>, are well studied objects in combinatorial group theory. Many abstract properties of groups have been studied for these groups; for example, Wise showed that all one-relator groups with torsion are residually finite, a result of Baumslag says that whenever <span class="math-container">$w$</span> is a positive word, i.e. involving no negative powers, the group is residually soluble. </p>
<p>My first question is the following: </p>
<p><strong>Are there any examples of (results regarding) simple one-relator groups?</strong> </p>
<p>It is clear that any example must be non-residually finite, as no residually finite group is simple, and so it must be torsion-free. There exists examples of non-residually finite one-relator groups, such as the Baumslag-Solitar groups <span class="math-container">$BS(m, n) = \langle a, t \: | \: t^{-1}a^mt = a^n \rangle$</span> unless <span class="math-container">$|m| = 1, |n| = 1$</span>, or <span class="math-container">$|m| = |n|$</span>. It should be noted that by a result of Sapir and ?pakulová, almost all (in a well-defined way) one-relator groups with three or more generators are residually finite. Hence, almost all one-relator groups are not simple.</p>
<p><strong>Does there exist an algorithm for deciding whether or not a given one-relator group is simple?</strong></p>
<p>It should be noted that while simplicity of groups is a Markov property, from which it follows that it is undecidable in general to decide whether a group is simple or not, one-relator groups are in many ways easier to deal with than general groups; as an example they have decidable word problem.</p>
<p>A naturally related question is the following:</p>
<p><strong>What is the smallest (with respect to the number of defining relations) presentation known for a simple group?</strong></p>
<p>For the finite case, <span class="math-container">$A_5$</span> admits a presentation with two generators and three defining relations. For the infinite case (which, admittedly, I find more interesting, personally) Thompson's group <span class="math-container">$V$</span> was recently shown by Bleak and Quick to admit a presentation with two generators and seven defining relations. I would be surprised if either of these turn out to be the smallest known examples.</p>
http://www.4124039.com/q/3262141Identifying elements in the kernel of an explicit endomorphism of a Jacobian varietyuser1158945http://www.4124039.com/users/910232019-03-24T16:00:55Z2019-03-24T16:09:19Z
<p>I hope this question fits here.</p>
<p>Let <span class="math-container">$H/k$</span> be a genus <span class="math-container">$2$</span> curve and <span class="math-container">$J$</span> its Jacobian variety. Since <span class="math-container">$J(k)\cong \text{Pic}^0(H)(k)$</span> we have that its generic point looks like <span class="math-container">$[(x_1,y_1)+(x_2,y_2)-2\infty]\in J$</span>. In Mumford coordinates we can see it as <span class="math-container">$g:=\langle x^2
-Ax + B,Cx+D\rangle:=\langle u(x),v(x)\rangle\in J$</span> with <span class="math-container">$u(x_i)=0$</span> and <span class="math-container">$v(x_i)=y_i$</span>. This means that <span class="math-container">$A:=x_1+x_2, B:=x_1x_2, C:=\tfrac{y_1-y_2}{x_1-x_2}, D:=\tfrac{x_2y_1-x_1y_2}{x_1-x_2}$</span>. </p>
<p>I calculated explicitly an element <span class="math-container">$\gamma\in\text{End}_k(J)$</span> using Mumford coordinates for the generic point, that is <span class="math-container">$\gamma(g)=<x^2 + \tfrac{A_1(g)}{A_2(g)}x + \tfrac{B_1(g)}{B_2(g)}, \tfrac{C_1(g)}{C_2(g)}x + \tfrac{ D_1(g)}{D_2(g)}>$</span> where <span class="math-container">$A_1,A_2,B_1,B_2,C_1,C_2,D_1,D_2\in k[J]$</span>. </p>
<p>My question is, since this endomorphism works fine for the generic point, and <span class="math-container">$J$</span> has dimension <span class="math-container">$2$</span>, we have that if for some <span class="math-container">$D\in J(k)$</span>, its image under <span class="math-container">$\gamma$</span>, namely <span class="math-container">$\gamma(D)$</span> is of the form <span class="math-container">$[(x_1,y_1)-\infty]$</span>, then <span class="math-container">$\gamma$</span> won't be defined for <span class="math-container">$D$</span>, since some of the denominators will be zero. </p>
<p>I want to distinguish when <span class="math-container">$\gamma(D)$</span> is 0 or when it is non-zero non-generic of the form <span class="math-container">$[(\tilde x,\tilde y)-\infty]$</span>. </p>
<p>I have noted that when the ALL the denominators are 0, it looks like the image 0 in fact, but when it lies in the "theta divisor", (the image is a point of the form <span class="math-container">$[(x_1,y_1)-\infty]$</span>), some of the denominators are non-zero. However I do not know how to distinguish this formally or maybe my examples are just "lucky" examples.</p>
<p>Is there a way with this information to distinguish when <span class="math-container">$\gamma(D)$</span> is exactly [0] ? </p>
<p><strong>What I did</strong></p>
<p>I tried to calculate the formula of <span class="math-container">$\gamma$</span> using MAGMA via the function field of <span class="math-container">$J$</span>, using the usual relations for the Jacobian of <span class="math-container">$H$</span> plus the denominators of <span class="math-container">$\gamma$</span> as relations, but the computation does not finish and eats all my memory eventually.</p>
<p>I just need to know if a point in the image is 0 or non-generic when <span class="math-container">$\gamma(D)$</span> has 0's in the denominators using the information that I have, or maybe using an element of the function field of <span class="math-container">$J$</span> that can distinguish if a divisor is 0 or if it has the point at infinity with multiplicity 1.</p>
http://www.4124039.com/q/3262121Journal losing indexing servicesPietro Paparellahttp://www.4124039.com/users/1046332019-03-24T15:49:05Z2019-03-24T16:05:39Z
<p>I recently had a paper accepted by a journal. When I looked it up on the AMS¡¯ <em><a href="https://mathscinet.ams.org/mathscinet/index.html" rel="nofollow noreferrer">Mathematical Reviews</a></em>, I noticed that it was previously indexed by the service but, at present, is it not. The journal is not pay-for-play, and one of the previous editors was a very famous mathematician.</p>
<p>I asked the handling editor about this and this person informed me that the journal could not ¡°ensure regular periodicity over the year¡±. This person also informed me that the journal was removed by zbMATH.</p>
<p>My questions:</p>
<ol>
<li>What possible reasons are there for a journal to be stripped of indexing?</li>
<li>Is it necessarily a sign of (lack of) quality that a journal is stripped of its indexing?</li>
</ol>
http://www.4124039.com/q/3262111Constructivist defininition of linear subspaces of $\mathbb{R}^n$?RBega2http://www.4124039.com/users/1278032019-03-24T15:32:06Z2019-03-24T15:32:06Z
<p>Let me preface this by saying I'm not someone who has every studied mathematical logic or philosophy of math, so I may be mangling terminology here (and the title is a little tongue in cheek).</p>
<p>I (and probably most of us) would define a linear subspace, <span class="math-container">$W$</span> of <span class="math-container">$\mathbb{R}^n$</span> to be a non-empty set of vectors that is closed under vector addition and scalar multiplication. This definition has the ``drawback" of not having a constructive (as far as I can tell) way to choose a basis of <span class="math-container">$W$</span>. </p>
<p>Alternatively, one could define a subspace <span class="math-container">$W$</span>, of <span class="math-container">$\mathbb{R}^n$</span> to just be a set which is the span of some finite collection of vectors, i.e. <span class="math-container">$W=\mathrm{span}\{ w_1, \ldots, w_N\}$</span>. This is a much clunkier definition (in my opinion) but does have the advantage of allowing one to construct a basis by using Gaussian elimination. It has the ``drawback" that showing that the kernel of a matrix is a subspace is slightly non-trivial (though still constructive).</p>
<p>Naively, the second definition defines a smaller set of objects than the first, but it seems that in most philosophies of math they actually define the same set of objects. </p>
<p>My questions: </p>
<p>1) Is the second definition actually how a constructivist would define subspace?</p>
<p>2) Are there any system of axioms where the two definitions are really different?</p>
<p>3) Are there any reasons to prefer the first definition to the second definition besides taste?</p>
<p>This question comes out of something that always bothers me when I teach the service course in Linear Algebra. Namely, I usually define (as does the textbook) a subspace using the first definition, but in practice only ever have students consider subspaces that are defined by the second one (or equally concretely as kernels of matrices). Pedagogically, I sometimes wonder how much is being gained by using this first definition (as it doesn't fit in with the extremely explicit nature of the rest of the course).</p>
http://www.4124039.com/q/3262041Artin¡¯s theorem on induced representations and the kernelinequalitynoob2http://www.4124039.com/users/1374412019-03-24T13:02:21Z2019-03-24T13:02:21Z
<p>Let <span class="math-container">$G$</span> be a finite group, and let <span class="math-container">$X$</span> be a family of subgroups of <span class="math-container">$G$</span> closed under conjugation and under passage to subgroups. Suppose further that <span class="math-container">$G$</span> is the union of the elements of <span class="math-container">$X,$</span> and denote by <span class="math-container">$R(G)$</span> the representation ring. By Artin¡¯s theorem, we have a surjective map
<span class="math-container">$$Ind: \oplus_{H \in X} \mathbb{Q} \otimes R(H) \rightarrow \mathbb{Q} \otimes R(G),$$</span> given by taking the induced character and extending by linearity. I will now define two special classes of elements in the source of this map. </p>
<p><strong>Elements of Class 1:</strong> Let <span class="math-container">$H,H¡¯ \in X, H¡¯ \subset H,$</span> and let <span class="math-container">$\psi¡¯ \in R(H¡¯).$</span> Let <span class="math-container">$\psi = Ind_{H¡¯}^H \psi¡¯.$</span> Then elements of class <span class="math-container">$1$</span> are of the form <span class="math-container">$\psi-\psi¡¯.$</span> </p>
<p><strong>Elements of class 2:</strong>
Let <span class="math-container">$H \in X, s \in G$</span> and set <span class="math-container">$H^s = sHs^{-1}.$</span> Let <span class="math-container">$\psi \in R(H),$</span> and define <span class="math-container">$\psi^s \in R(H^s)$</span> by <span class="math-container">$\psi^s(shs^{-1})= \psi(h).$</span> Then elements of class <span class="math-container">$2$</span> are of the form <span class="math-container">$\psi-\psi¡¯.$</span> </p>
<p>It is easy to see that all elements of class <span class="math-container">$1$</span> and <span class="math-container">$2$</span> lie in the kernel of the map <span class="math-container">$Ind.$</span> I want to prove that in fact, the kernel is generated by elements of class <span class="math-container">$1$</span> and <span class="math-container">$2.$</span> The hint I have seem is that one should base change go <span class="math-container">$\mathbb{C},$</span> and then use duality. It is then claimed that one is reduced to proving that if for every <span class="math-container">$H \in X$</span> one has a class function <span class="math-container">$f_H$</span> satisfying restriction snd conjugation conditions as above, then there exists a class function <span class="math-container">$f$</span> on <span class="math-container">$G$</span> restricting to <span class="math-container">$f_H$</span> for all <span class="math-container">$X.$</span>. </p>
<p>I supppse we should use Frobenius reciprocity in some clever way, but I do not see how the statement reduces to what I just claimed, so I would appreciate help, or full solutions.</p>
http://www.4124039.com/q/3262031Unique way to topologise finite algebra over Huber ringNibhttp://www.4124039.com/users/1370022019-03-24T12:28:32Z2019-03-24T12:28:32Z
<p>Let me start with the following Lemma.</p>
<blockquote>
<p><span class="math-container">$\textbf{Lemma}$</span> Let <span class="math-container">$A$</span> be a Tate ring, and let <span class="math-container">$f\colon A\to B$</span> be a finite <span class="math-container">$A$</span>-algebra. Then there is a unique way to topologise <span class="math-container">$B$</span> turning it into a Huber ring, making the structure morphism <span class="math-container">$A\to B$</span> continuous and such that there exists rings of definition <span class="math-container">$A_0$</span>, <span class="math-container">$B_0$</span> of <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, respectively, with <span class="math-container">$f(A_0)\subset B_0$</span> such that the induced map <span class="math-container">$A_0\to B_0$</span> is finite.</p>
<p><span class="math-container">$\textbf{Proof.}$</span> <span class="math-container">$\textit{Existence}$</span>: Give <span class="math-container">$B$</span> the canonical <span class="math-container">$A$</span>-module topology, i.e. give it the quotient topology for any surjection <span class="math-container">$A^n\twoheadrightarrow B$</span> of <span class="math-container">$A$</span>-modules (this does not depend on the choice). Let <span class="math-container">$A_0$</span> be a ring of definition and let <span class="math-container">$\pi\in A_0\cap A^\times$</span> be a pseudo-uniformizer. Choose a finite number of generators <span class="math-container">$\{b_i\}$</span> of <span class="math-container">$B$</span> as an <span class="math-container">$A$</span>-module s.t. <span class="math-container">$1$</span> is one of them, with each <span class="math-container">$b_i$</span> being integral over <span class="math-container">$A_0$</span>. Denote by <span class="math-container">$M$</span> the <span class="math-container">$A_0$</span>-submodule of <span class="math-container">$B$</span> generated by <span class="math-container">$\{b_i\}$</span>. Then the integrality allows to choose <span class="math-container">$m\gg 0$</span> such that <span class="math-container">$B_0:=A_0[\pi^mb_1,\ldots,\pi^m b_n]\subset M$</span>. Since also <span class="math-container">$\pi^m M\subset B_0$</span>, the subspace topology on <span class="math-container">$B_0$</span> is the <span class="math-container">$\pi$</span>-adic topology. Thus, <span class="math-container">$B$</span> is a Huber ring with ring of definition <span class="math-container">$B_0$</span>. Clearly the structure morphism is continuous and the induced map <span class="math-container">$A_0\to B_0$</span> is finite as <span class="math-container">$B_0$</span> is integral and of finite type over <span class="math-container">$A_0$</span>. </p>
<p><span class="math-container">$\textit{Uniqueness}$</span>: Since <span class="math-container">$B_0$</span> is finite over <span class="math-container">$A_0$</span>, there exists <span class="math-container">$m\gg 0$</span> such that <span class="math-container">$\pi^m B_0\subset M$</span>. On the other hand, since the structure morphism is continuous, <span class="math-container">$\pi$</span> is topologically nilpotent in <span class="math-container">$B$</span> and so <span class="math-container">$\pi^N M\subset B_0$</span> for <span class="math-container">$N\gg 0$</span>. This shows that the topology on <span class="math-container">$B$</span> is the one uniquely determined by making <span class="math-container">$M$</span> with the <span class="math-container">$\pi$</span>-adic topology an open subgroup, which is precisely the canonical topology.</p>
</blockquote>
<ol>
<li><p>I am wondering if the condition on the existence of <span class="math-container">$A_0,B_0$</span> making <span class="math-container">$A_0\to B_0$</span> finite is automatically satisfied knowing that the structure morphism is continuous and finite étale. Namely, in <a href="https://webusers.imj-prg.fr/~matthew.morrow/Espaces/Perf%20spaces%20IMJ-PRG%20final.pdf" rel="nofollow noreferrer">Remark 7.2 of these notes</a> by Matthew Morrow, the uniqueness in case <span class="math-container">$A\to B$</span> being finite étale is stated without any finiteness condition on the level of rings of definition. </p></li>
<li><p>In Huber's book "Étale Cohomology of Rigid Analytic Varieties and Adic Spaces" it says in <span class="math-container">$(1.4.2)$</span> that for a (complete) Huber ring (it does not say Tate) and <span class="math-container">$A\to B$</span> a finite algebra, the canonical <span class="math-container">$A$</span>-module topology on <span class="math-container">$B$</span> turns it into a Huber ring. My proof above clearly makes use of the existence of a topologically nilpotent unit, and I would like to know what the general argument looks like. Huber refers to <span class="math-container">$(3.12.10)$</span> of his "Bewertungsspektrum und rigide Geometrie", which is unfortunately not available to me.</p></li>
</ol>
http://www.4124039.com/q/3262024Sufficient conditions for the coefficients of a generating function to dominate those of its squareVince Vatterhttp://www.4124039.com/users/26632019-03-24T12:26:27Z2019-03-24T14:20:21Z
<p>Let <span class="math-container">$f(z)$</span> be a generating function (so in particular, its power series coefficients are nonnegative). I am interested in conditions which would ensure that for every <span class="math-container">$n$</span>, the coefficient of <span class="math-container">$z^n$</span> in <span class="math-container">$f$</span> is greater than or equal to the coefficient of <span class="math-container">$z^n$</span> in <span class="math-container">$f^2$</span>. Equivalently, I would like the generating function <span class="math-container">$f-f^2$</span> to have nonnegative coefficients.</p>
<p>One necessary condition is that <span class="math-container">$f$</span> have no constant term, so <span class="math-container">$f(0)=0$</span>. Another necessary condition is that <span class="math-container">$f$</span> have infinitely many terms; polynomials will never satisfy this condition. But these necessary conditions are very far from sufficient.</p>
<p>For an example of a generating function that does satisfy this condition, let <span class="math-container">$f$</span> denote the generating function for the shifted Catalan numbers,
<span class="math-container">$$
f=\frac{1-\sqrt{1-4z}}{2}=z+z^2+2z^3+5z^4+14z^5+\cdots.
$$</span>
Then we have that <span class="math-container">$f-f^2=z$</span>, which satisfies the condition.</p>
<p>However the condition seems to be quite fragile. If we instead take
<span class="math-container">$$
f=\frac{1-2z-\sqrt{1-4z}}{2z}=z+2z^2+5z^3+14z^4+42z^5+\cdots
$$</span>
then we have <span class="math-container">$f-f^2=z+z^2+z^3-6z^5-\cdots$</span>, which does not satisfy it.</p>
<p>The only other necessary condition I know of is that the dominant singularity of <span class="math-container">$f$</span> cannot be a pole, because if it were a pole then (by Pringsheim's Theorem) there would be a real number <span class="math-container">$z_0>0$</span> such that <span class="math-container">$f(z_0)>1$</span>, and thus we would have <span class="math-container">$f^2(z_0)>f(z_0)$</span>, so the condition could not hold.</p>
<p>Other than a few specific examples (e.g., the shifted Catalan numbers above), I know of no <em>sufficient</em> conditions.</p>
http://www.4124039.com/q/3262001Singularity of a hypersurfaceuser130022http://www.4124039.com/users/1300222019-03-24T12:07:17Z2019-03-24T12:07:17Z
<p>Alexander-Hirschowitz Theorem: Fix <span class="math-container">$r \ge 2$</span> and
<span class="math-container">$d \ge 2$</span>, and consider the linear system
<span class="math-container">$L = L^{(r )}_d(-\sum_{i=1}^n 2p_i)$</span>
consisting of hypersurfaces
of degree at most <span class="math-container">$d$</span> in <span class="math-container">$r$</span> variables that are singular at <span class="math-container">$n$</span> general points <span class="math-container">$\{p_i\}$</span>.Then</p>
<p>(a) For <span class="math-container">$d = 2$</span>, the linear system <span class="math-container">$L$</span> is special if and
only if <span class="math-container">$2 \le n \le r$</span>.</p>
<p>(b) For <span class="math-container">$d \ge 3$</span>, the linear system <span class="math-container">$L$</span> is special if and
only if the triple <span class="math-container">$(r,d,n)$</span> is one of the following: <span class="math-container">$(2, 4, 5), (3, 4, 9), (4, 4, 14), (4, 3, 7)$</span>.</p>
<p>Thus if we consider degree <span class="math-container">$5$</span> hypersurfaces in <span class="math-container">$\mathbb{P}^3$</span> having singularity at <span class="math-container">$15$</span> general points, then <span class="math-container">$(n, d, r)$</span> is not as in the list of the above Theorem. Therefore by above theorem, <span class="math-container">$L$</span> is non special and therefore having expected dimension which is negative. In other words, there does not exist any degree <span class="math-container">$5$</span> hypersurface in <span class="math-container">$\mathbb{P}^3$</span> which is singular along <span class="math-container">$15$</span> general points. Please correct me if i have made any mistake in understanding the statement of the Theorem.</p>
http://www.4124039.com/q/3261994Open problems concerning all the finite groupsSebastien Palcouxhttp://www.4124039.com/users/345382019-03-24T11:55:41Z2019-03-24T11:55:41Z
<p>What are the open problems concerning <em>all</em> the finite groups?</p>
<p>The references will be appreciated. Here are two examples: </p>
<ul>
<li><p><strong>Aschbacher-Guralnick conjecture</strong> (<a href="https://doi.org/10.1016/0021-8693(84)90183-2" rel="nofollow noreferrer">AG1984</a> p.447): the number of conjugacy classes of maximal subgroups of a finite group is at most its class number (i.e. the number of conjugacy classes of elements, or the number of irreducible complex representations up to equiv.).</p></li>
<li><p><strong>K.S. Brown's problem</strong> (<a href="https://doi.org/10.1006/jabr.1999.8221" rel="nofollow noreferrer">B2000</a> Q.4; <a href="https://doi.org/10.1016/j.aim.2015.10.018" rel="nofollow noreferrer">SW2016</a> p.760): Let <span class="math-container">$G$</span> be a finite group, <span class="math-container">$\mu$</span> be the Möbius function of its subgroup lattice <span class="math-container">$L(G)$</span>. Then the sum <span class="math-container">$\sum_{H \in L(G)}\mu(H,G)|G:H|$</span> is nonzero.</p></li>
</ul>
<p>There are two types of problems, those involving an upper/lower bound (like Aschbacher-Guralnick conjecture) and those "exact", involving no bound (like K.S. Brown's problem). I guess the first type is much more abundant than the second, so for the first type, please restrict to the main problems.</p>
http://www.4124039.com/q/3261982Lie structure over $R$-modulejustanothermathstudenthttp://www.4124039.com/users/1270692019-03-24T11:11:28Z2019-03-24T13:14:51Z
<p>In Higgins' paper <em>Baer invariants and the Birkhoff-Witt theorem</em> (J. Algebra <strong>11</strong> (1969) 469¨C482, doi:<a href="https://doi.org/10.1016/0021-8693%2869%2990086-6" rel="nofollow noreferrer">10.1016/0021-8693(69)90086-6</a>) the following definition is given:</p>
<p>A <em>Lie structure</em> over the <span class="math-container">$R$</span>-module <span class="math-container">$M$</span> is a <span class="math-container">$T(M)$</span>-bimodule <span class="math-container">$A$</span> together with a bilinear function <span class="math-container">$M\otimes M\to A$</span> taking <span class="math-container">$x \otimes y \mapsto \langle x,y\rangle$</span>, satisfying </p>
<ul>
<li><p><span class="math-container">$\langle x,x\rangle = 0$</span>;</p></li>
<li><p><span class="math-container">$\langle x,y\rangle t(uv-vu)=(xy-yx)t\langle u,v\rangle$</span>, for all <span class="math-container">$x,y,u,v \in M$</span> and <span class="math-container">$t \in T(M)$</span>; and</p></li>
<li><p><span class="math-container">$(\langle x,y\rangle z-z\langle x,y\rangle)+(\langle y,z\rangle x-x\langle y,z\rangle)+(\langle z,x\rangle y-y\langle z,x\rangle) = 0$</span>, for <span class="math-container">$x,y,z \in M$</span>.</p></li>
</ul>
<p>My question is, how does this generalize the case of a Lie algebra over a field? And what is the motivation behind the second condition? Why can't we simply define a Lie structure over a ring to be an alternating bilinear form which satisfies the Jacobi identity?</p>
http://www.4124039.com/q/326196-4Name for a ratio expressed as a decimal [on hold]Rich Shttp://www.4124039.com/users/1374372019-03-24T10:37:56Z2019-03-24T10:37:56Z
<p>Given the ratio 3/4, we can express this in different ways.</p>
<p><code>75% - Percent
3/4 - Fraction
3:4 - Ratio</code></p>
<p>However, I was looking for a word to describe a ratio expressed just as a simple decimal value, or a number represented as per-ten or per-dec. </p>
<p><code>0.75 - per-unit
7.5 - per-dec</code></p>
<p>But is there an official term for these?</p>
http://www.4124039.com/q/3261900Model in $\mathsf{ZFC}$ with neither $P$- nor $Q$-pointsDominic van der Zypenhttp://www.4124039.com/users/86282019-03-24T09:10:40Z2019-03-24T12:08:43Z
<p>A <span class="math-container">$P$</span>-point is an ultrafilter <span class="math-container">$\scr U$</span>on <span class="math-container">$\omega$</span> such that for every function <span class="math-container">$f:\omega\to\omega$</span> there is <span class="math-container">$x\in {\scr U}$</span> such that the restriction <span class="math-container">$f|_x$</span> is either constant, or injective.</p>
<p>A <span class="math-container">$Q$</span>-point is an ultrafilter <span class="math-container">$\scr U$</span>on <span class="math-container">$\omega$</span> such that for every function <span class="math-container">$f:\omega\to\omega$</span> with the property that <span class="math-container">$f^{-1}(\{m\})$</span> is finite for each <span class="math-container">$m\in \omega$</span>, there is <span class="math-container">$x\in {\scr U}$</span> such that the restriction <span class="math-container">$f|_x$</span> is injective.</p>
<p><span class="math-container">$P$</span>-points need not exist, and <span class="math-container">$Q$</span>-points need not exist.</p>
<p><strong>Question.</strong> Is it possible that neither <span class="math-container">$P$</span>- nor <span class="math-container">$Q$</span>-points exist?</p>
http://www.4124039.com/q/3261831The norm of isotropic sub-Gaussian random vector may not be sub-Gaussianzbh2047http://www.4124039.com/users/1203022019-03-24T03:56:40Z2019-03-24T16:01:48Z
<p>Suppose <span class="math-container">$X$</span> is a isotropic sub-Gaussian <span class="math-container">$n$</span>-dimensional random vector (i.e. <span class="math-container">$EXX^T=I_n$</span>, and for any unit vector <span class="math-container">$u$</span>,<span class="math-container">$\|\left<X,u\right>\|_{\psi_2}\le K$</span>). It is said that <span class="math-container">$\|X\|_2-\sqrt n$</span> may not be sub-Gaussian with bounded norm <span class="math-container">$CK$</span> which does not depend on <span class="math-container">$n$</span>. But I havn't found a counter example. </p>
<p>When <span class="math-container">$X$</span> is a uniform ball distribution or a uniform hypercube distribution, it can both be proved that <span class="math-container">$\|X\|_2-\sqrt n$</span> is sub-Gaussian. Moreover, if <span class="math-container">$X_i$</span> are independent, the proposition is also true.</p>
<p>Can someone show a counter example? Thank you!</p>
http://www.4124039.com/q/3261733Theorem on formal functions and cohomological flatnessmartinhttp://www.4124039.com/users/1374002019-03-23T21:08:22Z2019-03-24T14:43:36Z
<p>Let <span class="math-container">$f:X\rightarrow S$</span> be a proper morphism of schemes with Noetherian target. The theorem on formal functions says that for any point <span class="math-container">$s\in S$</span> there is an isomorphism between inverse limits of <span class="math-container">$(f_*O_X)_s/\mathfrak{m}_s^n (f_*O_X)_s$</span> and <span class="math-container">$\Gamma(X_s, O_X\otimes_{O_S}O_S/\mathfrak{m}_s^n O_S)$</span>, if I understand correctly. If
<span class="math-container">$f$</span> happens to be flat and cohomologically flat in degree 0, then we know the isomorphism between inverse limits is actually an isomorphism at <span class="math-container">$n$</span>-th stage for any <span class="math-container">$n>0$</span>. </p>
<ul>
<li>Is there an example of a morphism <span class="math-container">$f$</span> such that the induced comparison morphism is an isomorphism at <span class="math-container">$n$</span>-th stage (and thus isomorphism at <span class="math-container">$m$</span>-th stage for every <span class="math-container">$0<m<n$</span>) and is not an isomorphism at any later stage? </li>
<li>Is there a description of <span class="math-container">$(f_*O_X)_s$</span> in terms of fiberwise information? </li>
</ul>
http://www.4124039.com/q/3261711Power series rings and the formal generic fibreRinmyakuhttp://www.4124039.com/users/1133772019-03-23T20:39:19Z2019-03-24T14:19:02Z
<p>Let <span class="math-container">$S = K[[S_1,\ldots,S_n]]$</span> and consider <span class="math-container">$d$</span> elements
<span class="math-container">\begin{equation*}
f_1,\ldots,f_d \in S[[X_1,\ldots,X_d]]
\end{equation*}</span>
and the prime ideal <span class="math-container">${\frak P} \colon\!= (f_1,\ldots,f_d)$</span> generated by them. </p>
<p>Suppose that <span class="math-container">${\frak P}$</span> satisfies the following three conditions<span class="math-container">$\colon$</span></p>
<ol>
<li><p><span class="math-container">${\frak P} \cap S = 0$</span>.</p></li>
<li><p><span class="math-container">${\mathrm{ht}}({\frak P}) = d$</span>. </p></li>
<li><p>Neither of <span class="math-container">$\overline{f_1},\ldots,\overline{f_d} \in K[[X_1,\ldots,X_d]]$</span> is zero. </p></li>
</ol>
<p>(The condition 3. is equivalent to that every <span class="math-container">$f_i$</span> contains at least one term <span class="math-container">$c_{e_1,\ldots,e_d}X_1^{e_1} \ldots X_d^{e_d}$</span> such that <span class="math-container">$c_{e_1,\ldots,e_d} \notin (S_1,\ldots,S_n)$</span>.)</p>
<h2>Q. Is the ring <span class="math-container">$S[[X_1,\ldots,X_d]]/{\frak P}$</span> necessariy finite over <span class="math-container">$S$</span>?</h2>
http://www.4124039.com/q/3261622Injection of Besov spaces in $L^p$Bazinhttp://www.4124039.com/users/219072019-03-23T17:46:42Z2019-03-24T11:34:53Z
<p>I believe that for <span class="math-container">$p\ge 2$</span>, we have the continuous injection (for <span class="math-container">$p=2$</span>, it is an equality),
<span class="math-container">$$
B^0_{p,2}(\mathbb R^n)\subset L^p(\mathbb R^n),
$$</span>
where <span class="math-container">$B^0_{p,2}(\mathbb R^n)$</span> is the Besov space.
Is there a straightforward proof?</p>
http://www.4124039.com/q/3261571Dense filter and selective ultrafilterar.grighttp://www.4124039.com/users/1183662019-03-23T17:02:44Z2019-03-24T15:04:35Z
<p>We say that <span class="math-container">$\rho(A)=\lim_{n\to\infty}\frac{|A\cap n|}{n}$</span> is the density of subset <span class="math-container">$A\subset\omega$</span> if the limit exists. Let us define the filter <span class="math-container">$\mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}$</span>. </p>
<p><strong>Question:</strong> Does there exist (in ZFC & CH) selective ultrafilter <span class="math-container">$\mathcal{U}$</span> and a bijection <span class="math-container">$\varphi:\omega\to\omega$</span> such that
<span class="math-container">$\varphi(\mathcal{F_1})\subset\mathcal{U}$</span> ?</p>
<p><strong>Remark:</strong> Someone had downvoted two similar questions. Please, explain what is wrong if something is wrong</p>
http://www.4124039.com/q/3261523Supremum of certain modified zeta functions at 1Davide Cesare Venianihttp://www.4124039.com/users/439512019-03-23T16:08:33Z2019-03-24T11:06:28Z
<p>Let <span class="math-container">$D$</span> be an integer number and let <span class="math-container">$\chi$</span> be the Dirichlet character defined by
<span class="math-container">$$\chi(m) = 0 \text{ if $m$ even, } \chi(m) = (D/m) \text{ if $m$ odd,}$$</span>
where <span class="math-container">$(D/m)$</span> denotes the Jacobi symbol. Let us denote by <span class="math-container">$\zeta_D(s)$</span> its L-series:
<span class="math-container">$$ \zeta_D(s) = L(s,\chi) = \sum_{n = 1}^\infty \frac{\chi(n)}{n^s} = \prod_p \frac{1}{1-(D/p)p^{-s}},$$</span>
where <span class="math-container">$(D/p) = 0$</span> if <span class="math-container">$p | 2d$</span>.</p>
<p>For <span class="math-container">$s > 1$</span> there are the following bounds:
<span class="math-container">$$\frac{\zeta(2s)}{\zeta(s)} = \prod_p \frac{1}{1+p^{-s}} \leq \zeta_D(s) \leq \prod_p \frac{1}{1-p^{-s}} = \zeta(s).$$</span>
Both of them become trivial for <span class="math-container">$s = 1$</span>. Still, <span class="math-container">$\zeta_D(1)$</span> is a finite positive number for <span class="math-container">$D < 0$</span>.</p>
<blockquote>
<p>Consider the sequence
<span class="math-container">$$ \zeta_{-d}(1),\, d = 1,2,3,\ldots$$</span>
Is this sequence bounded above or below? If yes, what is an upper or lower bound?</p>
</blockquote>
http://www.4124039.com/q/3261420Quantifier elimination and where is this quantified convex program in the polynomial hierarchy?VS.http://www.4124039.com/users/1365532019-03-23T13:28:54Z2019-03-24T10:42:23Z
<p>I have a quantified convex program of the form that I need to solve</p>
<p><span class="math-container">$$\exists(x_{1,1},\dots,x_{1,n})\in\mathbb R^n\quad\forall(x_{2,1},\dots,x_{2,n})\in\mathbb R^n$$</span>
<span class="math-container">$$\vdots$$</span>
<span class="math-container">$$\exists(x_{2t-1,1},\dots,x_{2t-1,n})\in\mathbb R^n\quad\forall(x_{2t,1},\dots,x_{2t,n})\in\mathbb R^n$$</span>
<span class="math-container">$$\phi_1(x_{1,1},\dots,x_{2t,n})\leq a_1\wedge\dots\wedge\phi_r(x_{1,1},\dots,x_{2t,n})\leq a_r$$</span>
where <span class="math-container">$\phi_1,\dots,\phi_r$</span> are either linear degree <span class="math-container">$d=1$</span> or quadratic degree <span class="math-container">$d=2$</span> convex polynomials with <span class="math-container">$O(n)$</span> terms each with all polynomial coefficients and <span class="math-container">$a_1,\dots,a_r$</span> in <span class="math-container">$\mathbb Z$</span> and having at most <span class="math-container">$m$</span> bits each.</p>
<p>I feel quantifier elimination over reals will help here. However how to go about it is unclear.</p>
<ol>
<li><p>Does quantifier elimination tell anything about the time complexity of the program and if so what is the complexity and is it possible to use quantifier elimination to reduce the number of quantifications and if so how does the parameters change?</p></li>
<li><p>Is it in <span class="math-container">$\Sigma_{k}$</span> in the polynomial hierarchy where <span class="math-container">$k=O(t)$</span> and independent of <span class="math-container">$m,n,r$</span>?</p></li>
</ol>
http://www.4124039.com/q/3261171First order partial differential equationDaaanhttp://www.4124039.com/users/1359362019-03-23T05:24:28Z2019-03-24T13:40:57Z
<p>I know there is a solution to this pde</p>
<p><span class="math-container">$$\partial_{t} f(t,x)= \partial_{x}(v(x)f(t,x))$$</span>
<span class="math-container">$$ f(0,x)=g(x)$$</span>
( Where <span class="math-container">$v$</span> and <span class="math-container">$g$</span> are known functions)
which is given by
<span class="math-container">$$ f(t,x)=\frac{1}{v(x)} h(t+\int \frac{1}{v(x)})$$</span>
where <span class="math-container">$h(x)$</span> is determined by initial condition <span class="math-container">$g(x)$</span>.</p>
<p>The question is if I use the method of characteristic would it give me the same solution with this initial condition?</p>
http://www.4124039.com/q/3261103Non-abelian cohomologiesKiuhttp://www.4124039.com/users/214912019-03-23T03:25:05Z2019-03-24T11:07:25Z
<p>Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients? </p>
<p>If a topological group <span class="math-container">$G$</span> acts on a space <span class="math-container">$X$</span> one can construct its equivariant cohomology ring <span class="math-container">$H^*_G(X)$</span> (say with coefficients in <span class="math-container">$\mathbb{R}$</span>). Is there a notion of equivariant cohomology for <span class="math-container">$(X, G)$</span> with coefficients in a non-commutative algebra <span class="math-container">$A$</span>? </p>
http://www.4124039.com/q/3260525On a problem for determinants associated to Cartan matrices of certain algebrasMarehttp://www.4124039.com/users/619492019-03-22T13:39:05Z2019-03-24T13:53:02Z
<p>This is a continuation of <a href="http://www.4124039.com/questions/324534/classification-of-algebras-of-finite-global-dimension-via-determinants-of-certai/325637?noredirect=1#comment813583_325637">Classification of algebras of finite global dimension via determinants of certain 0-1-matrices</a> but this time with a concrete conjecture and using the simplification suggested by user esg to make the problem more compact.</p>
<p>Let <span class="math-container">$n \geq 4$</span> and <span class="math-container">$w \in \{3,4,...,n-1 \}$</span>.
Let <span class="math-container">$Z$</span> be the matrix of the cyclic shift (the companion matrix of <span class="math-container">$X^n-1$</span>), and for non-zero <span class="math-container">$\mathbf{v}\in \{0,1\}^n$</span> let
<span class="math-container">$\mathrm{diag}(\mathbf{v})$</span> be the diagonal matrix with <span class="math-container">$\mathbf{v}$</span> on the diagonal,
and <span class="math-container">$M_\mathbf{v}:=I + Z+ \ldots + Z^{w-1}-\mathrm{diag}(\mathbf{v})$</span>.<br>
For fixed <span class="math-container">$n$</span>, call the tuple <span class="math-container">$(w,v)$</span> perfect in case <span class="math-container">$\det(M_\mathbf{v})=(-1)^{(w-1)(n-1)}$</span>.
(here I used the reformulation obtained by user esg for my problem. I refer to the previous thread for a motivation. Roughly stated, perfect pairs correspond to certain algebras with finite global dimension. )</p>
<p>Define <span class="math-container">$G_n := \{ w \in \{3,4,...,n-1\} | $</span>there exists a nonzero <span class="math-container">$\mathbf{v}\in \{0,1\}^n$</span> with <span class="math-container">$(w,v)$</span> perfect <span class="math-container">$\}$</span>.</p>
<p>It is best to picture the <span class="math-container">$v$</span> as two-colored necklaces (with colours corresponding to 1 and 0), so a cyclic shift just means rotating the necklace. </p>
<p>It is an interesting question what the set <span class="math-container">$G_n$</span> is explicitly but my first guess was wrong and it seems that <span class="math-container">$G_n$</span> is complicated to describe for large <span class="math-container">$n$</span>.</p>
<p>But here are two conjectures that would be nice in case they are true:</p>
<blockquote>
<p>Conjecture 1: Maximum(<span class="math-container">$G_n$</span>)=<span class="math-container">$\frac{n+2}{2}$</span> in case <span class="math-container">$n$</span> is even and Maximum(<span class="math-container">$G_n$</span>)=<span class="math-container">$\frac{n+1}{2}$</span> in case <span class="math-container">$n$</span> is odd.</p>
<p>Conjecture 2: a) For <span class="math-container">$n$</span> even the number of perfect tuples <span class="math-container">$(w,v)$</span> with <span class="math-container">$w=(n+2)/2$</span> is equal to <span class="math-container">$\frac{3^{n/2-1}+1}{2}$</span> when we identify two tuples <span class="math-container">$(w,v_1)$</span> and <span class="math-container">$(w,v_2)$</span> when <span class="math-container">$v_1$</span> is a cyclic shift of <span class="math-container">$v_2$</span>.</p>
<p>b) For <span class="math-container">$n$</span> odd the number of perfect tuples <span class="math-container">$(w,v)$</span> with <span class="math-container">$w=(n+1)/2$</span> is equal to <span class="math-container">$n-1$</span> when we identify two tuples <span class="math-container">$(w,v_1)$</span> and <span class="math-container">$(w,v_2)$</span> when <span class="math-container">$v_1$</span> is a cyclic shift of <span class="math-container">$v_2$</span>.</p>
</blockquote>
<p>The conjectures are tested with the computer for <span class="math-container">$n \leq 20$</span>.</p>
<p>Here two examples:</p>
<p>For <span class="math-container">$n=13$</span> and <span class="math-container">$w=7$</span>, the <span class="math-container">$v$</span> up to cyclic shift with <span class="math-container">$(w,v)$</span> perfect are:</p>
<p>[ [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 ], </p>
<p>[ 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1 ], </p>
<p>[ 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1 ], </p>
<p>[ 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1 ], </p>
<p>[ 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1 ], </p>
<p>[ 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1 ], </p>
<p>[ 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1 ], </p>
<p>[ 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1 ], </p>
<p>[ 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1 ], </p>
<p>[ 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1 ], </p>
<p>[ 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1 ], </p>
<p>[ 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ] ]</p>
<p>For <span class="math-container">$n=8$</span> and <span class="math-container">$w=5$</span>, the perfect <span class="math-container">$(w,v)$</span> up to cyclic shifts are:</p>
<p>[ [ 0, 0, 0, 1, 0, 0, 0, 1 ], </p>
<p>[ 0, 0, 0, 1, 0, 0, 1, 1 ], </p>
<p>[ 0, 0, 0, 1, 0, 1, 0, 1 ], </p>
<p>[ 0, 0, 0, 1, 0, 1, 1, 1 ], </p>
<p>[ 0, 0, 0, 1, 1, 0, 0, 1 ], </p>
<p>[ 0, 0, 0, 1, 1, 0, 1, 1 ], </p>
<p>[ 0, 0, 0, 1, 1, 1, 0, 1 ],</p>
<p>[ 0, 0, 0, 1, 1, 1, 1, 1 ], </p>
<p>[ 0, 0, 1, 0, 0, 1, 1, 1 ], </p>
<p>[ 0, 0, 1, 0, 1, 0, 1, 1 ], </p>
<p>[ 0, 0, 1, 0, 1, 1, 1, 1 ],</p>
<p>[ 0, 0, 1, 1, 0, 1, 0, 1 ], </p>
<p>[ 0, 0, 1, 1, 1, 1, 0, 1 ],</p>
<p>[ 0, 1, 0, 1, 1, 0, 1, 1 ] ]</p>
<p>edit: It might be also a good idea to think about conjecture 1 in terms of representation theory/homological algebra. Here is the non-elementary formulation of conjecture 1:</p>
<p>Let <span class="math-container">$A$</span> be a selfinjective (connected) Nakayama algebra with <span class="math-container">$n$</span> simple modules and Loewy length <span class="math-container">$w$</span> with <span class="math-container">$(n+2)/2 < w<n$</span>. Then a generator <span class="math-container">$M$</span> with every non-projective indecomposable summand being simple has the property that <span class="math-container">$End_A(M)$</span> has infinite global dimension.</p>
<p>(equivalently, one can look at generators <span class="math-container">$M$</span> with every non-projective indecomposable summand being a radical of an indecomposable projective module).</p>
http://www.4124039.com/q/3260357Smoothness of the moduli space of Drinfeld modulesbad mathematicianhttp://www.4124039.com/users/1373542019-03-22T08:19:22Z2019-03-24T13:55:40Z
<p>I'm studying the proof of Thm 1.5.1. in Laumon's "Cohomology of Drinfeld Modular Varieties". Notation: <span class="math-container">$\mathfrak{m}$</span> is a square zero ideal of <span class="math-container">$\mathcal{O}$</span> and <span class="math-container">$k=\mathcal{O}/\mathfrak{m}$</span>. Laumon shows that the obstruction to the existence of a lift of a Drinfeld module <span class="math-container">$\phi: A \rightarrow k[\tau]$</span> to a Drinfeld module <span class="math-container">$\phi':A \rightarrow \mathcal{O}[\tau]$</span> lies in the second Hochschild cohomology <span class="math-container">$HH^2(A,\mathfrak{m}[\tau])$</span>.</p>
<p>I get that this is an obstruction to the existence of a lift of <span class="math-container">$\phi$</span> to a ring morphism <span class="math-container">$\phi':A \rightarrow \mathcal{O}[\tau]$</span>. However, I don't see how the degree condition is controlled by the Hochschild cohomology. I mean just because there exists a lift as a ring morphism <span class="math-container">$\phi'$</span>, this does not need to be of the same rank as <span class="math-container">$\phi$</span>.</p>
<p>Laumon does not explain this, and neither do Blum and Stuhler in "Drinfeld Modules and Elliptic Sheaves". Can some clarify this for me? Thanks.</p>
http://www.4124039.com/q/3260219More mysterious properties of Gram matrixDaniil Rudenkohttp://www.4124039.com/users/216202019-03-22T00:51:35Z2019-03-24T13:19:30Z
<p>This is another question related to the mysterious properties of the Gram matrix in dimension <span class="math-container">$4$</span>. Here's the <a href="http://www.4124039.com/q/325830/91764">previous question</a>.</p>
<p>The following fact could be extracted from <a href="https://arxiv.org/abs/math/0402087" rel="nofollow noreferrer">0402087</a>:</p>
<blockquote>
<p>For any <span class="math-container">$a_i\neq 0$</span> a polynomial </p>
<p><span class="math-container">$$(t+a_1a_2a_3)(t+a_2a_4a_6)(t+a_1a_5a_6)(t+a_3a_4a_5)-$$</span></p>
<p><span class="math-container">$$(t-a_1a_2a_4a_5)(t-a_1a_3a_4a_6)(t-a_2a_3a_5a_6)(t-1)$$</span> </p>
<p>has a nonzero double root if and only if Gram determinant</p>
<p><span class="math-container">$$
\begin{vmatrix}
1 & -\frac{a_1+\frac{1}{a_1}}{2}& -\frac{a_2+\frac{1}{a_2}}{2}&-\frac{a_3+\frac{1}{a_3}}{2} \\
-\frac{a_1+\frac{1}{a_1}}{2} & 1 &-\frac{a_6+\frac{1}{a_6}}{2}&-\frac{a_5+\frac{1}{a_5}}{2}\\
-\frac{a_2+\frac{1}{a_2}}{2} & -\frac{a_6+\frac{1}{a_6}}{2} & 1&-\frac{a_4+\frac{1}{a_4}}{2}\\
-\frac{a_3+\frac{1}{a_3}}{2} & -\frac{a_5+\frac{1}{a_5}}{2} & -\frac{a_4+\frac{1}{a_4}}{2} & 1\\
\end{vmatrix}
$$</span></p>
<p>vanishes.</p>
</blockquote>
<p><strong>Question:</strong> How to prove it? I was able only to verify it in Maple.</p>
http://www.4124039.com/q/32596325Why is so much work done on numerical verification of the Riemann Hypothesis?Tomhttp://www.4124039.com/users/1191142019-03-21T13:52:38Z2019-03-24T10:47:32Z
<p>I have noticed that there is a huge amount of work which has been done on numerically verifying the <a href="https://en.wikipedia.org/wiki/Riemann_hypothesis" rel="nofollow noreferrer">Riemann hypothesis</a> for larger and larger non-trivial zeroes.</p>
<p>I don't mean to ask a stupid question, but is there some particular reason that numerical verifications give credence to the truth of the Riemann hypothesis or some way that the computations assist in proving the hypothesis (as we know, historically hypotheses and conjectures have had numerical verification to the point where it seemed that they must be true but the conjectures then turned out to be false, especially hypotheses related to prime numbers and things like that).</p>
<p>Is there something special about this hypothesis which makes this kind of argument more powerful than normal? Would one be able to use these arguments somewhere in the case for a proof of the hypothesis or would they never be used in the proof at all (and yes, until it is proven we cannot know that, sure).</p>
http://www.4124039.com/q/3259161Is this problem in $NP$?VS.http://www.4124039.com/users/1365532019-03-21T00:25:34Z2019-03-24T14:26:26Z
<p>Where is </p>
<blockquote>
<p>Given two <span class="math-container">$n$</span> many algebraically independent homogeneous system of polynomials (hence zero-dimensional system) in <span class="math-container">$\mathbb Z[x_1,\dots,x_n]$</span> with degree <span class="math-container">$2$</span> with absolute value of coefficients bound by <span class="math-container">$2^{n^c}$</span> size at a fixed <span class="math-container">$c\geq1$</span> do all their integer roots with <span class="math-container">$\infty$</span>-norm at most <span class="math-container">$2^{n}$</span> agree?</p>
</blockquote>
<p>in the polynomial hierarchy?</p>
<p>It is in <span class="math-container">$coNP$</span>.</p>
<ol>
<li><p>Is this in <span class="math-container">$NP$</span>?</p></li>
<li><p>Is there a polynomial-time solution?</p></li>
</ol>
http://www.4124039.com/q/3251082Why control a continuous approximation of stochastic gradient descent instead of just the SGD?Stefan Perkohttp://www.4124039.com/users/786502019-03-10T20:08:15Z2019-03-24T16:08:07Z
<p>In "<a href="https://arxiv.org/abs/1511.06251" rel="nofollow noreferrer">Stochastic modified equations and adaptive stochastic gradient algorithms</a>" (Li et. al 2015) the authors approximate stochastic gradient descent, as in</p>
<p><span class="math-container">$$x_{k+1} = x_k - \eta u_k \nabla f_{\gamma_k}(x_k),$$</span></p>
<p>by an SDE, a so called <em>stochastic modified equation</em> (SME)</p>
<p><span class="math-container">$$d X_t = -u_t \nabla f(X_t)dt + u_t \sqrt{\eta \Sigma(X_t)}dW_t,$$</span></p>
<p>in the sense of weak approximation of order 1 by viewing the SGD update as the application of the Euler scheme to the SDE.</p>
<p>Here</p>
<ul>
<li><span class="math-container">$\eta > 0$</span> is the <em>maximum</em> learning rate</li>
<li><span class="math-container">$T > 0$</span></li>
<li><span class="math-container">$u : [0,T] \to [0,1]$</span> controls the learning rate</li>
<li><span class="math-container">$(\gamma_k)_k$</span> are iid RV with values in <span class="math-container">$\{1,\dots, n\}$</span> and <span class="math-container">$\mathbb P(\gamma_k = i) = 1/n$</span>, which represent random samples from a training set (e.g.)</li>
<li><span class="math-container">$f(x) = \frac 1 n \sum_{i=1}^n f_i(x)$</span> is the objective</li>
<li><span class="math-container">$(W_t)_t$</span> is Brownian motion independent of the <span class="math-container">$\gamma_k$</span></li>
</ul>
<p>and</p>
<p><span class="math-container">$$\Sigma(x) := \frac 1 n \sum_{i=1}^n (\nabla f(x) - \nabla f_i(x))(\nabla f(x) - \nabla f_i(x))^T.$$</span></p>
<p>Of course, many conditions are left implicit here, s.t. the <span class="math-container">$f_i$</span> are continuously differentiable or that the coefficients of the SME satisfy growth conditions that guarantee the existence of a unique solution.</p>
<p>Now, the goal is to solve a control problem for the SME, such as</p>
<p><span class="math-container">$$\min_{u} \mathbb E(f(X_T)),$$</span></p>
<p>and use the solution to control the learning rate of the original SGD.
What I wonder is</p>
<blockquote>
<p>Why can't we just control the original SGD directly? What can we gain by passing to this continuous problem?</p>
</blockquote>
<p>Among other things, I think this is important because the difference between the real <span class="math-container">$x_k$</span> and its continuous counterpart ("<span class="math-container">$|X_{\eta k} - x_k|$</span>") can be quite large since the approximation is only weak. I think this prevents us from letting <span class="math-container">$u$</span> be dependent on <span class="math-container">$X_t$</span> as well, because then the solution cannot be meaningfully used for the original SGD. </p>
http://www.4124039.com/q/3236167Is there any physical or computational justification for non-constructive axioms such as AC or excluded middle??_?http://www.4124039.com/users/569382019-02-20T02:38:00Z2019-03-24T13:47:20Z
<p>I became interested in mathematics after studying physics because I wanted to better understand the mathematical foundations of various physical theories I had studied such as quantum mechanics, relativity, etc. </p>
<p>But after years of post-graduate study in mathematics, I've become a bit disillusioned with the way mathematics is done, and I often feel that much of it has become merely a logical game with objects that have no meaning outside of the confines of the game; e.g., existence of enough injectives/projectives, existence of bases for any vector space, etc. </p>
<p>I am not really an applied person---I love abstract theories such as category theory---but I would like to feel that the abstractions have some meaning outside of the logical game. To me it seems that using non-constructive proof techniques divorces the theory from reality.</p>
<p>Essentially, my question boils down to the following:</p>
<ol>
<li><p>If I adopt a constructive foundation (like say an intuitionistic type theory), is there any (apparently) insurmountable difficulty in modelling the physical universe and the abstract processes that emerge from it (quantum mechanics, relativity, chess, economies, etc.). </p></li>
<li><p>Is there any physical/computational justification for assuming excluded middle or some sort of choice axiom? </p></li>
</ol>
http://www.4124039.com/q/31428019Gluing hexagons to get a locally CAT(0) spaceDylan Thurstonhttp://www.4124039.com/users/50102018-10-31T19:41:44Z2019-03-24T12:13:12Z
<p>I believe that there are four ways to glue (all) the edges of a regular Euclidean hexagon to get a locally CAT(0) space:
<a href="https://i.stack.imgur.com/LTbvQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LTbvQ.png" alt="Gluings of a hexagon"></a>
The first two give the torus and the Klein bottle, respectively. What are the last two? In particular, do their fundamental groups have another name? Do they have the same fundamental group?</p>
<hr>
<p>EDIT: HJRW points out that I missed the several ways to glue a hexagon to get a non-orientable surface of Euler characteristic -1. I count 8 possibilities up to symmetries.</p>
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