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curious little things - aggregated feedsenMath Overflow Recent Questions: Constant Gaussian curvature surfaces in 3-space containing lines
https://mathoverflow.net/questions/284098/constant-gaussian-curvature-surfaces-in-3-space-containing-lines
<p>How can one construct surfaces in $\mathbb R^3$ of constant negative Gaussian curvature containing a line in $\mathbb R^3$? (this question is inspired by <a href="https://math.stackexchange.com/q/2478117/72694">this MSE post</a>).</p>Sun, 22 Oct 2017 03:49:38 -0600Math Overflow Recent Questions: How to prove this by extension group
https://mathoverflow.net/questions/284094/how-to-prove-this-by-extension-group
<p><strong>Let R be a commutative ring.$r\in R$.Let M and N are R-modules such that $rM=0$ and there is a short exact sequence $0\rightarrow N\xrightarrow r N\rightarrow N/rN \rightarrow 0$.</strong></p>
<p>if X is arbitrary R-module,$X\xrightarrow r X$ is R module morphism by multiplication,then we can get the induced map $Hom_R(M,r)$ is zero.</p>
<p>by above,it is not difficult to see $Hom_R(M,N)=0$.and we have exact sequence:$0\rightarrow Hom_R(M,N/rN)\rightarrow Ext^1_R(M,N)\xrightarrow {Ext^1(M,r)} Ext^1_R(M,N)$.By the same argument,we use the definition of derived functor $Ext^1$,take a injective resolution of $N$,then we can prove $Ext^1(M,r)=0$. So $Hom_R(M,N/rN)\cong Ext^1_R(M,N)$.</p>
<blockquote>
<p><strong>My question</strong>:How to prove this by extension group?</p>
</blockquote>
<p>That is:for any s.e.s. $\delta:0\rightarrow N\rightarrow W\rightarrow M\rightarrow 0$,take pushout with $\delta$ and morphism $N\xrightarrow rN$.We get a new s.e.s.,How to prove this new s.e.s. splits?</p>
<p>Thank you in advance!</p>Sun, 22 Oct 2017 03:04:39 -0600Math Overflow Recent Questions: Does this Algorithm on my Matrix Converge
https://mathoverflow.net/questions/284092/does-this-algorithm-on-my-matrix-converge
<p>I have an algorithm for an $\infty\times n$ nonnegative matrix described by Ethan Bolker, and I am asking if there are any theorems that may help prove convergence to a matrix with desired row and column sums. </p>
<p><strong>Summary of Algorithm</strong>: Start with a matrix with desired row and column sums, set some entries to 0 (this will change some of the column and row sums), multiply affected columns by a constant to reattain the correct column (this will also affect the row sums), multiply affected rows by a constant to reattain the correct row sum, and continue multiply to columns and rows by constants until the matrix converges to a matrix with the desired row and column sums.</p>
<p>Consider an $\infty\times n$ matrix with columns $j=1,2,\ldots, n$ and rows labelled by integers $i$ whose prime factor contains primes at most $n$: $i=2^{a_2}3^{a_3}\cdots p_n ^{a_n}$, where $p_n\le n$ and $a_k\in \mathbb{Z}_{\ge 0}.$</p>
<p>(1) <strong>Starting Point:</strong> Start with the matrix whose $i,j$ entry is $$a_{i,j}=\frac{\prod_{p\le n, p \text{ prime}}\frac{p-1}{p}}{in}$$
As a consequence, each column has sum $1/n$, each row has sum $$r_i:=\frac{\prod_{p\le n, p \text{ prime}}\frac{p-1}{p}}{i},$$ and the entries along a given row are the same.</p>
<p>(2) <strong>Make some Zeros:</strong> Set $a_{i,j}$ to $0$ if $\frac{j}{\gcd(i,j)}$ is composite. This will create infinitely zeros in the matrix, and the zeros only occur in columns $j$ where $j$ is composite. These zeros do not disconnect the matrix in the following sense: Construct the bipartite graph with one vertex set for the rows and one for the columns. Join two vertices when their intersection is not one of the entries that must be 0. I've shown this graph is connected.</p>
<p>(3) Multiply each column by a constant to reattain column sum $1/n$.</p>
<p>(4) Multiply each row by a constant to reattain row sum $r_i$ for row $i$.</p>
<p>Then I'd like to repeat steps (3) and (4) until the matrix converges to a matrix with column sum $1/n$ and row $i$ sum $r_i$ for each $i$. </p>
<p>Are there any theorems that will allow me to show convergence? When I work out examples by hand for small $n$, the multipliers needed to correct the row and column sums seem to be getting closer to $1$.</p>Sun, 22 Oct 2017 02:07:53 -0600Math Overflow Recent Questions: Distribution of elements in ideal class groups of imaginary quadratic orders
https://mathoverflow.net/questions/284091/distribution-of-elements-in-ideal-class-groups-of-imaginary-quadratic-orders
<p>Let $X$ be a large positive number. Put $h(d)$ for the size of the ideal class group of the quadratic order of discriminant $d$, or equivalently, </p>
<p>$$\displaystyle h(d) = \#\{\operatorname{SL}_2(\mathbb{Z})\text{-orbits of integral binary quadratic forms of discriminant } d\}.$$</p>
<p>Put $C(d)$ for the ideal class group of the quadratic order of discriminant $d$, and for each positive integer $m$ put $C_m(d)$ to be the subgroup of $C(d)$ consisting of those elements of order dividing $m$. Consider the homomorphism $\rho$ of $C(d)$ to itself defined by $w \mapsto w^4$. The kernel of this map is precisely $C_4(d)$. </p>
<p>We now suppose $d > 0$, and suppose that each class $w \in C(-d)$ is parametrized by a reduced form $f_w = a_w x^2 + b_w xy + c_w y^2$, so that $|b_w| \leq a_w \leq c_w, a_w, c_w > 0$. </p>
<p>Is the set </p>
<p>$$\displaystyle R(X) = \{(a,b,c) \in \mathbb{Z}^3 : |b| \leq a \leq c, a, c \geq 1, 4ac - b^2 \leq X, a < X^{1/4}\}$$</p>
<p>well-distributed with respect to $\rho$? More precisely, for a given $d > 0$ we identify the subset of $C(-d)$ given by $C^\dagger(-d) = \{w \in C(d) : f_w \in R(X)\}$ and ask whether or not $\# (C^\dagger(-d)/C_4(-d)) \approx \#C^\dagger(-d)/\# C_4(-d)$ (here the left hand side should read "cardinality of $C^\dagger(-d)$ modded out by $C_4(-d)$"), at least on average as $X$ tends to infinity.</p>Sun, 22 Oct 2017 02:07:15 -0600Math Overflow Recent Questions: Can an exponential family MRF lose its exponentiality when adding CPDT constraints?
https://mathoverflow.net/questions/284089/can-an-exponential-family-mrf-lose-its-exponentiality-when-adding-cpdt-constrain
<p>I am reading the <a href="https://people.eecs.berkeley.edu/~wainwrig/Papers/WaiJor08_FTML.pdf" rel="nofollow noreferrer">book</a> "Graphical Models, Exponential Families, and Variational Inference" by Wainwright and Jordan. In there, they show that any Markov Random Field (<a href="https://en.wikipedia.org/wiki/Markov_random_field" rel="nofollow noreferrer">MRF</a>) with exclusively finite random variables can be written as a member of an <a href="https://en.wikipedia.org/wiki/Exponential_family" rel="nofollow noreferrer">exponential family</a>, just by using indicator functions for all the cliques.</p>
<p>I now wonder, how would one incorporate constraints on the conditional probability distribution tables (CPDT) into this scheme. For instance, Wainwright and Jordan mention hidden Markov models (<a href="https://en.wikipedia.org/wiki/Hidden_Markov_model" rel="nofollow noreferrer">HMMs</a>) as an example of exponential family MRFs. But most of the time, HMMs are homogeneous, i.e. the CPDTs are time independent, which is a constraint on their CPDTs.</p>
<p>So, if homogeneous HMMs also belong to an exponential family, how do the sufficient statistics look like?</p>
<p>More generally: Can adding constraints on the CPDTs of an exponential family MRF make it "non-exponential", or is there always a way to adapt the sufficient statistics so that there will be an exponential family containing it? What are some relevant references?</p>Sun, 22 Oct 2017 01:22:55 -0600Math Overflow Recent Questions: On sum of squares?
https://mathoverflow.net/questions/284088/on-sum-of-squares
<p>If $(a,b)=1$ and $2|b$ then $p$ is prime and $p|a^2+b^2\implies p\not\equiv3\bmod4$.</p>
<ol>
<li><p>For any other $k\in\Bbb N_{>2}$, is there a polynomial that represents an odd prime $p$ if and only if $p\not\equiv(2^k-1)\bmod2^k$?</p></li>
<li><p>Is there such a polynomial for infinitely many $k\in\Bbb N$?</p></li>
</ol>Sun, 22 Oct 2017 00:41:28 -0600Math Overflow Recent Questions: Non-isotropic Schur root
https://mathoverflow.net/questions/284087/non-isotropic-schur-root
<p>Let $Q$ be a quiver, maybe with cycles and multiple arrows.</p>
<p>Define $Q$ to be and wild if the Tits form is indefinite.</p>
<p>If $Q$ is wild, is it true that it always admits an imaginary non-isotropic Schur root, i.e. a dimension vector $d$ such that there exists a representation $M$ over a field $K$ with $\operatorname{End}_KM=K$ and such that $q(d)<0$?</p>
<p>If I remember correctly this is true, but I cannot find a reference.</p>Sun, 22 Oct 2017 00:25:39 -0600Math Overflow Recent Questions: What is the good notion of supervariety?
https://mathoverflow.net/questions/284084/what-is-the-good-notion-of-supervariety
<p>The principal (I think) difference between the notions of manifold in differential (including complex analytic) topology and in algebraic (or especially arithmetic) geometry is that for the former the "local models" are the same while for the latter they may vary.</p>
<p>This somehow suggests that there may be a version of the notion of supervariety in algebraic/arithmetic geometry, which has variable local models too. Except that I have no idea what these local models can be.</p>
<p>Has this been investigated?</p>Sat, 21 Oct 2017 22:38:38 -0600Math Overflow Recent Questions: premise and decidability [on hold]
https://mathoverflow.net/questions/284083/premise-and-decidability
<p>Regarding the 17 camels riddle question that was asked lately. Why does the story of 17 camels decides that the man who helped the three sons inherit what's actually not thiers (by the will of the father) as a wise man?</p>Sat, 21 Oct 2017 22:08:23 -0600Math Overflow Recent Questions: What is the most useful rationality criterion of surfaces?
https://mathoverflow.net/questions/284082/what-is-the-most-useful-rationality-criterion-of-surfaces
<p>The motivation for this question is that I would like to extract some information from derived category of surfaces to conclude the rationality of surface. There is a well known rationality criterion called Castelnuovo's criterion. I recall it as follows: Let $X$ be a smooth projective surface and if $q:=h^1(O_X)=0, P_2(X)=h^0(2K_X)=0$, then $X$ is rational surface. Where $K_X$ is canonical divisor of $X$. I found that $q=0$ is somewhat relatively easier to obtain from the derived category: say if $D^b(X)$ admits a full exceptional collection or stronger assumption, say if $D^b(X)$ admits a full exceptional collection of line bundles, then, $q=h^1(O_X)=0$. On the other hand, I found $h^0(2K_X)=0$ is really hard to obtain. There is some very special case where I can obtain such equality, say if a surface admit a cyclic strong exceptional collection of line bundles, then I can conclude that $-K_X$ is strictly effective by some techniques. Then $-2K_X$ is strictly effective, thus $h^0(2K_X)=0$. Thus, I can conclude that $X$ is rational surface. But in general, there are many rational surfaces with $-K_X$ not effective at all. So in general, I do not expect such method will work.</p>
<p>There is another classical way to conclude $X$ is rational, it uses the classification of surfaces: one can first obtain that $X$ is uniruled, or equivalently, the kodaira dimension $kod(X)=-\infty$, then $X$ is a ruled surface and $q=0$ implies that $X$ is rational. For example, in the work of M.Brown and I.Shipman: <a href="https://arxiv.org/abs/1312.3918" rel="nofollow noreferrer">https://arxiv.org/abs/1312.3918</a></p>
<p>They proved that the existence of full strong exceptional collection of line bundles on $X$ implies that rationality of surfaces. Basically, they showed that there exists some free rational curve on $X$ ($C^2\geq 0, p_a(C)=0$), then $X$ is covered by rational curves(or equivalently, $kod(X)=-\infty$), together with $q=0$, the $X$ is rational surface. It looks like this way of showing the rationality is more flexible, it uses some techniques coming from higher dimensional birational geometry, say the concept of free rational curves.</p>
<p>Also, of course the classification results for rational surfaces: any rational surface is blow up of $\mathbb{P}^2, \mathbb{F}_n,n\neq 1$ is also used sometimes. In our recent work: <a href="https://arxiv.org/abs/1710.03972" rel="nofollow noreferrer">https://arxiv.org/abs/1710.03972</a></p>
<p>we showed the existence of cyclic strong exceptional collection of line bundles imply the surfaces is weak del pezzo ($-K_X$ is big and nef). We used the alternative definition of weak del pezzo, i.e: it is either $\mathbb{F}_0,\mathbb{F}_2$ or blow up of $\mathbb{P}^2$ at most $8$ points in so called almost general position. We can extract information from cyclic strong exceptional collection of line bundles to showed this. </p>
<p>So my question is
<strong>Do you know some other way(criterion) to conclude that $X$ is a rational surface?</strong> Since I am not an expert in higher dimensional algebraic geometry, I am very curious to know some techniques from there (because maybe we can understand things deeper by observing things from general perspective). For example, some concepts like uniruled, unirational, rational connected might be very different from each other in higher dimensional case, but in the case of surfaces, maybe they coincide or very close to each other, then maybe some criterion of them in higher dimensional geometry will help. </p>Sat, 21 Oct 2017 20:58:44 -0600