Recent MathOverflow Questions

Constant Gaussian curvature surfaces in 3-space containing lines

Math Overflow Recent Questions - Sun, 10/22/2017 - 03:49

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 this MSE post).

How to prove this by extension group

Math Overflow Recent Questions - Sun, 10/22/2017 - 03:04

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 $0\rightarrow N\xrightarrow r N\rightarrow N/rN \rightarrow 0$.

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.

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 $Hom_R(M,N/rN)\cong Ext^1_R(M,N)$.

My question:How to prove this by extension group?

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?

Thank you in advance!

Does this Algorithm on my Matrix Converge

Math Overflow Recent Questions - Sun, 10/22/2017 - 02:07

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.

Summary of Algorithm: 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.

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 $a_k\in \mathbb{Z}_{\ge 0}.$

(1) Starting Point: 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.

(2) Make some Zeros: 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.

(3) Multiply each column by a constant to reattain column sum $1/n$.

(4) Multiply each row by a constant to reattain row sum $r_i$ for row $i$.

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 $i$.

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 $1$.

Distribution of elements in ideal class groups of imaginary quadratic orders

Math Overflow Recent Questions - Sun, 10/22/2017 - 02:07

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,

$$\displaystyle h(d) = \#\{\operatorname{SL}_2(\mathbb{Z})\text{-orbits of integral binary quadratic forms of discriminant } d\}.$$

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 $C_4(d)$.

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 $|b_w| \leq a_w \leq c_w, a_w, c_w > 0$.

Is the set

$$\displaystyle R(X) = \{(a,b,c) \in \mathbb{Z}^3 : |b| \leq a \leq c, a, c \geq 1, 4ac - b^2 \leq X, a < X^{1/4}\}$$

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.

Can an exponential family MRF lose its exponentiality when adding CPDT constraints?

Math Overflow Recent Questions - Sun, 10/22/2017 - 01:22

I am reading the book "Graphical Models, Exponential Families, and Variational Inference" by Wainwright and Jordan. In there, they show that any Markov Random Field (MRF) with exclusively finite random variables can be written as a member of an exponential family, just by using indicator functions for all the cliques.

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 (HMMs) 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.

So, if homogeneous HMMs also belong to an exponential family, how do the sufficient statistics look like?

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?

On sum of squares?

Math Overflow Recent Questions - Sun, 10/22/2017 - 00:41

If $(a,b)=1$ and $2|b$ then $p$ is prime and $p|a^2+b^2\implies p\not\equiv3\bmod4$.

  1. For any other $k\in\Bbb N_{>2}$, is there a polynomial that represents an odd prime $p$ if and only if $p\not\equiv(2^k-1)\bmod2^k$?

  2. Is there such a polynomial for infinitely many $k\in\Bbb N$?

Non-isotropic Schur root

Math Overflow Recent Questions - Sun, 10/22/2017 - 00:25

Let $Q$ be a quiver, maybe with cycles and multiple arrows.

Define $Q$ to be and wild if the Tits form is indefinite.

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 $q(d)<0$?

If I remember correctly this is true, but I cannot find a reference.

What is the good notion of supervariety?

Math Overflow Recent Questions - Sat, 10/21/2017 - 22:38

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.

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.

Has this been investigated?

premise and decidability [on hold]

Math Overflow Recent Questions - Sat, 10/21/2017 - 22:08

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?

What is the most useful rationality criterion of surfaces?

Math Overflow Recent Questions - Sat, 10/21/2017 - 20:58

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.

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:

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.

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:

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.

So my question is Do you know some other way(criterion) to conclude that $X$ is a rational surface? 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.

Detecting stability in solutions to coupled nonlinear equations in aerodynamics,

Math Overflow Recent Questions - Sat, 10/21/2017 - 19:45

I'm studying a quasi-steady force model (published in fluid dynamics journal) that consists of coupled, nonlinear ODEs that describe unsteady aerodynamics -- recently, my advisor and I have found some solutions to the system that appear to be stable-ish.

My question is: how can I proceed to confirm the stability of such solutions? I'm having a hard time thinking of a way to linearize the models for lift and drag forces, and then setting up an eigenvalue problem.

One way I've thought of is to manually perturb the state variables. That is, write functions that stop my solver when a steady state solution is detected, look at the final values of all of the state variables, then proceed to make a 2nd call to the solver, using these final values as the new initial values. And, changing any of the initial values before making the 2nd call to the solver is then effectively perturbing a state variable from its steady state. If after this small perturbation, the solutions gotten from the 2nd call to the solver again settle back into the same set of solutions (i.e. a steady-state solution for a fixed set of parameters), then this would indicate that the solutions could be stable.

Is this a good approach?

If so, should I manually perturb all of the state variables -- or only the state variables that I am interested in focusing on?

By what percentage should I perturb the state variables, in order to be able to conclude that the solutions are stable? Is there a common benchmark / standard to follow in perturbation / stability theory?

If this is not a good approach, can you recommend another approach?


Waring rank vs tensor rank of symmetric tensors?

Math Overflow Recent Questions - Sat, 10/21/2017 - 19:27

Suppose we work in an algebraically closed field. Then, do the Waring rank (symmetric tensor rank) and tensor rank of a symmetric tensor coincide in general? Recall that tensor rank is rank with respect to the Segre variety and Waring rank is rank with respect to the Veronese variety.

Can we approximate any open set by sub-domains with smooth boundary?

Math Overflow Recent Questions - Sat, 10/21/2017 - 19:06

In some books, mainly about PDEs, I read that any open set can be approximated by sub-domain with smooth boundary (not just piecewise smooth). In 2 dimensional case, this seemly to be quite trivial: for any subdomain, use small open balls to cover its boundary and then mollify the connection parts. But in the higher dimensional case, I think this is not that obvious.

So the first question is: how can we approximate any open set by sub-domain with smooth boundary?

And the second question is: In what meaning the approximation is? Pointwise, i.e., we can find subdomain $D_n$ with smooth boundary such that $D_n\uparrow A$? uniformly pointwise? Or in the Lebesgue measure sense? etc.

Here "$D_n\uparrow A$" uniformly pointwise" means means that $\partial D_n\subset A\cap A_n^c$, where $A_n:=\{ x \in A:d(x,\partial A)\geq \frac1n \}$,.

alternating and symmetric powers of the standard representation of the symmetric group

Math Overflow Recent Questions - Sat, 10/21/2017 - 13:36

Let $n \geq 7$ and $V = \mathbb{C}^n$ be the standard representation for $S_{n+1}$, the symmetric group of cardinal $(n+1)!$

Let $k$ be an integer such that $2 \leq k \leq n$. Is it true or false that $\bigwedge^k V$ does not appear as an irreducible sub-representation of $\mathrm{Sym}^k V$? I am looking for a reference which would either prove or give counter-examples to this.

Thanks in advance!

EDIT : Is there an online software (analogous to LIE) that would compute the decompositions into irreducible representations of the tensor product of two ireps for the symmetric group?

Why is the definition of the higher homotopy groups the "right one"?

Math Overflow Recent Questions - Sat, 10/21/2017 - 11:39

If someone asked me the question for the fundamental group, I would answer as follows:

  1. The connection to classification of covering spaces.
  2. The fundamental group of many spaces is an object of independent interest. For instance, for an Elliptic curve over the complex numbers, the fundamental group is the lattice defining the curve or equivalently, it is related to the torsion points on the curve.
  3. Related to 1, the arithmetic fundamental group is closely related and the arithmetic fundamental group is itself very important. For instance, the Galois groups of fields is an example.
  4. The fundamental group offers very natural proofs of fundamental theorems like the fundamental theorem of algebra.

However, for the higher homotopy groups, the best answer I could give would be something along the lines of the the long exact sequence of homotopy groups]1 for fibrations. Maybe the Hurewicz theorem is also an answer to my question except that I think the Hurewicz theorem is usually used to get information about the homotopy groups from the homology groups. If this is not true, that would be an interesting answer too.

I am almost sure this is entirely due to my background (in arithmetic geometry) and lack of formal training in algebraic topology and that the higher homotopy groups are indeed a natural object to study.

Ideally, I would appreciate answers that either connect the higher homotopy groups to important invariants of spaces that were already studied (1,2, 3 above) or proofs of statements not about the higher homotopy groups that however use the higher homotopy groups in an essential way (4 above and I guess the long exact sequence comes under here).

Isotrivial factors of Jacobian

Math Overflow Recent Questions - Sat, 10/21/2017 - 11:32

Let $k$ be an algebraically closed field of positive characteristic that it is not the algebraic closure of a finite field. Fix a smooth proper $k$-curve $C$ and write $J_C$ for its Jacobian abelian variety.

Assume that one of the isogeny factors of $J_C$ is isogenous to an isotrivial abelian variety. Is it true that $J_C$ has a factor isogenous to an isotrivial Jacobian?

Are there any other finite summation formulas for the series ${}_4\phi_3$

Math Overflow Recent Questions - Sat, 10/21/2017 - 08:22

Some analysis shows that there are only $3$ essentially different classes of terminating ${}_4\phi_3$ summations: $$ {}_4\phi_3\left[{q^{-n},a,b,-q^{1-n}/ab \atop q^{1-n}/a,q^{1-n}/b,-ab};q,q\right]=\begin{cases}0,& n=2m+1,\\ \large{\frac{(ab;q)_n(q,a^2,b^2;q^2)_m }{\left(a,b;q\right)_n(a^2b^2;q^2)_m}},&n=2m,\end{cases}\tag{1} $$

$$ {}_4\phi_3\left[{ q^{-n},a,b,q^{\frac12-n}/ab \atop q^{1-n}/a,q^{1-n}/b,abq^{\frac12}};q,q\right]=q^{n/2}\frac{(-q^{\frac12},a,b;q^{\frac12})_n(ab;q)_n}{(ab;q^{\frac12})_n(a,b;q)_n},\tag{2} $$

$$ {}_4\phi_3\left[{q^{-n},-q^{-n},a,b \atop \sqrt{qab},-\sqrt{qab},q^{-2n}};q,q\right]=\frac{(q a,qb;q^2)_n }{\left(q,qab;q^2\right)_n},\tag{3} $$ with $2$ continuous parameters $a$ and $b$ (besides $q$) and $n$ a non-negative integer. $(1),(2)$ and $(3)$ are due to Bailey, W. N. (1941), Jackson, F. H. (1941), and Andrews, G.E. (1976), respectively.

The classification is based on the following principles. The series of the type ${}_4\phi_3\left[{q^{-n},aq^r,... \atop a,...};q,q\right]$, where $r=0,1,2,...$ does not depend on $n$, are not included in the classification; however if $r=O(n)$, like $r=n$, then it might be included. Two summations fall under the same class if

(i) some of the parameters in the numerator or denominator of the series ${}_4\phi_3$ differ from each other by a factor of $q^{\pm r},~r=0,1,2,...$; e.g. if there is a summation for ${}_4\phi_3\left[{q^{-n},-q^{-n},aq,b/q \atop \sqrt{qab},-\sqrt{qab},q^{-2n}};q,q\right]$ then it is in class $(3)$.

(ii) they are related by Sears' transformation formula $$ {}_4\phi_3\left[{q^{-n},a,b,c \atop d,e,f};q,q\right]=\frac{(e/a,f/a;q)_n}{(e,f;q)_n}{}_4\phi_3\left[{q^{-n},a,d/b,d/c \atop d,aq^{1-n}/e,aq^{1-n}/f};q,q\right], $$ where $abc=defq^{n-1}$.

Q1: Are there any other classes of summation formulas for ${}_4\phi_3$ with $2$ continuous parameters that are essentially different from $(1-3)$ ?

There is an invariant preserved under Sears' transformation that can distinguish the class $(2)$:

Let $q>0$, and for ${}_4\phi_3\left[{q^{-n},a_1,a_2,a_3 \atop a_4,a_5,a_6};q,q\right]$, where $a_i\in\mathbb{R}$ define $$ I=\text{sign}(a_1a_2a_3). $$ With this definition we have for $(1),(2),(3)$ $$ I_1=-1,~I_2=+1,~I_3=-1. $$ However the invariant $I$ can not distinguish between classes $(1)$ and $(3)$.

Q2: How to rigorously prove that the classes $(1)$ and $(3)$ are distinct?

This question was motivated by the paper Andrews, G.E. (2011). On Shapiro’s Catalan convolution, Adv. Appl. Math. 46, 15-24 where the author writes

"It turns out that (1.10) and (1.12) are, in fact, specializations of a new q-hypergeometric series summation. Indeed, this is one of the most interesting aspects of this paper. It is seldom that a new q-hypergeometric summation turns up. The summation in question is"

and then he gives formula $(2)$ found by Jackson $70$ years earlier. I looked papers citing Andrews' paper and didn't find any that stated that the result was not in fact new. Therefore such classification as above might be useful.

Bailey, W. N. (1941). A note on certain q-identities, Quart. J. Math. 12, 173-175.

Jackson, F. H. (1941). Certain q-identities, Quart. J. Math. 12, 167-172.

Andrews, G.E. (1976). On q-analogues of the Watson and Whipple summation, SIAM J. Math. Anal., 7, pp. 332-336.

A bounded operator associated with a principal bundle

Math Overflow Recent Questions - Sat, 10/21/2017 - 03:45

Assume that $(X, B, G)$ is a $G$- principal bundle where $G$ is a compact topological or Lie group. The normalized Haar measure of (each fiber of ) $X$ is denoted by $\mu$. The space of continuos complex valued function on $X$ is denoted by $C(X)$ which is equiped with the sup norm and the standard operation.

To this principal bundle, we associate a bounded linear operator $T: C(X) \to C(B)$ with $T(f)(b)=\int_{q^{-1}(b)} fd\mu$.

The above linear map $T$ has a right invers $S_q :C(B) \to C(X)$ with $S_q (g)= g\circ q$. Moreover $S_q$ is a linear isometry.

Question: Is it true that every isometric right inverse of $T$ is a multiplicative operator?If yes,are there some right inverse $S_p$ for some continuous $p:X \to B$ but $p$ is not homotopic to $q$?In particular what is the answer to the above question for the hopf fibration $S^3 \to S^2$?

Base change for Borel-Moore homology

Math Overflow Recent Questions - Fri, 10/20/2017 - 18:06

For a seperated scheme of finite type $X$ over $\mathbf{C}$, let $H_*(X)$ denote its Borel-Moore homology, which is defined by $$ H_k(X) = R^{-k}\Gamma(X, \omega_X) $$ where $\omega\in D_c(X, \mathbf{C})$ is a dualising object in the derived category of constructible $\mathbf{C}$-sheaves on $X$. It is immediate from the definition and the six operations for constructible sheaves that $H_*$ is covariant with respect to proper morphisms and contravariant with respect to smooth morphisms. Let $f_*$ and $f^*$ denote respectively the direct and inverse image maps.


If we are given a cartesian diagramme of schemes over $\mathbf{C}$ $\require{AMScd}$ \begin{CD} Y' @>\tilde g>> Y \\ @VVf'V @VVfV \\ X' @>g>> X \end{CD} where $f$ is proper and $g$ is smooth, do we have an equality of the maps between homology groups $f'_*\tilde g^* = g^*f_*: H_*(Y)\to H_*(X')$?

In the book [N. Chriss & V. Ginzburg, Representation theory and complex geometry, 8.3.34] , this is shown in the case where $g$ is locally a trivial fibration, by reducing it to the commutativity of the following diagramme of derived functors

\begin{CD} f_*f^! @>>> \mathrm{id}_X @>>> g_*g^* \\ @VVV @. @AAA \\ f_*\tilde g_*\tilde g^*f^! @. = @. g_* f'_*{f'}^! g^* \end{CD}

in which all the morphisms are adjunction morphisms except for the base change morphism $\tilde g^*f^! \cong {f'}^! g^*$ in the bottom line, whose definition can be found in [SGA4, Exposé 18,]. One gets the result by apply the diagramme to $\omega_X$ and then take global sections.

The commutativity of the diagramme above can be easily checked in the case where $g: X'\to X$ is locally a trivial fibration. While I don't know how to do it if $g$ is only assumed to be smooth, I still suspect the commutativity to remain true.

Let me also add that there is such an equality $f'_*\tilde g^* = g^*f_*: A_*(Y)\to A_*(X')$ for Chow groups and for $f$ proper and $g$ flat, see for example [Fulton, Intersection theory, Prop. 1.7].

The Largest Piece of Circumference

Math Overflow Recent Questions - Fri, 10/20/2017 - 13:33

We put $n$ lines on the unit circle. Each line is chosen independently, we randomly choose two points from anywhere in the circle and connect them. We continue doing this $n$ times.

After $n$ random lines have been chosen what is the probability that the largest piece of circumference on the circle is of length $\pi/2$ or longer?


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