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most recent 30 from 2018-03-22T14:26:38Z

Do pseudo-differential operators form a sheaf of algebras?

Mon, 02/19/2018 - 14:14

Let $M$ be a smooth manifold.

I have been trying to figure out from the literature I know whether (any flavor of) pseudo-differential operators form a sheaf of algebras (w.r.t. the usual topology on $M$). Sadly the best results I could find showed at best that pseudo-differential form a sheaf of left $C^{\infty}$-modules without treating the question of whether the multiplication is well defined or associative. So my question is rather simple:

Do pseudo-differential operators form a sheaf of (associative) $\mathbb{C}$-algebras on $M$? If not, what fails?

I'm pretty sure that if one considers pseudo-differential operators modulo smothing operators than these do form a sheaf of algebras (However a precise reference here would be welcome). As for the entire space of pseudo-differential operators this seems rather non-trivial to me (I'm actually not entirely sure whether this should be true or not). I apologize if this is too elementary for this site.

Mysterious symmetry - in search for a bijection

Mon, 02/19/2018 - 14:13

I have a mysterious symmetry that I have not managed to prove. First some definitions (see picture below)

Fix a partition that fit in a staircase shape with $n$ rows. There are $Catalan(n)$ such shapes. We can represent this with a diagram $D$, as below, where the gray squares is the partition. The yellow squares are enumerated $1,\dotsc,n$ from top to bottom, and thus any permutation in $S_n$ is seen as a labeling of the yellow squares.

Let $a+b=n$. Given a permutation in $S_n$ seen as a labeling of the yellow squares, the first block is the squares with labels $1,\dotsc,a$ and the second block is the remaining $b$ squares.

A permutation $\sigma \in S_n$ is called $(D,a,b)$-good if the following holds:

  • The smallest label in each of the two blocks appear lowest in its block.
  • If $i$ and $i+1$ are in the same block, and $i$ below $i+1$, then the square in the same row as $i+1$ and same column as $i$ must be white.

In the diagram, the permutation $342615$ is shown, and one can verify that it is $(D,4,2)$-good. The white squares that has to be white due to the second condition has been marked with bullets.

Let $Good(D,a,b)$ denote the set of $(D,a,b)$-good permutations.

Warmup exercise

Show that $|Good(D,a,b)|=|Good(D,b,a)|$.

Finally, we define the ascent, $asc_D$-statistic on permutations as follows: For every white square $S$, we let $S_1$ be the index of the yellow square in the same row, and $S_2$ be the index of the yellow square in the same column. Then $asc_D(\sigma)$ is the number of white squares $S$, such that $\sigma(S_1)<\sigma(S_2)$. In our diagram, $asc_D(342615) = 1+1+2+0+1 = 5$, where the terms are contributions from each row.

My problem

Show (bijectively) that for every diagram $D$ and choice of $a+b=n$, $$ \sum_{\sigma \in Good(D,a,b)} q^{asc_D(\sigma)} =\sum_{\sigma \in Good(D,b,a)} q^{asc_D(\sigma)}. $$

For the diagram $D$ here, we have that $|Good(D,4,2)|=|Good(D,2,4)|=20$, and that both sums above become $$1 + 3 q + 4 q^2 + 4 q^3 + 4 q^4 + 3 q^5 + q^6.$$

Comments: For some diagrams $D$, it is straightforward to produce a bijection, in particular the case when $D$ has no gray squares. One would hope that a bijection would the number ascents 'within blocks', that is, ascents where $S_1$ and $S_2$ belong to the same block. However, this cannot be done for general $D$.

One can generalize the problem to permutations with more than two blocks, but the 2-block case implies the general case.

I am quite confident this result follows (non-bijectively) from a result by C. Athanasiadis, but it requires several messy steps.


This is related to the $p_\lambda$-expansion of certain LLT polynomials.

If $X$ and $Y$ are homotopy equivalent, then are $X \times \mathbb{R}^{\infty}$ and $Y \times \mathbb{R}^{\infty}$ homeomorphic?

Mon, 02/19/2018 - 14:13

Let $X$ and $Y$ be reasonable spaces. Since $\mathbb{R}^{\infty}$ is contractible, $$ X \times \mathbb{R}^{\infty} \cong Y \times \mathbb{R}^{\infty} \;\;\; \implies \;\;\; X \simeq Y. $$

Is the converse also true?

My vague intuition: the factors of $\mathbb{R}^{\infty}$ provide so much extra room that there will never be a geometric obstruction to producing a homeomorphism. Evidently, there is no homotopy-theoretic obstruction, so maybe the converse is true.

On the other hand, I really have no idea and could be missing something basic. For example, the plane with two punctures is homotopy equivalent to a wedge of two circles. However, I do not know about a homeomorphism $$ (\mathbb{C} - \{0, 1\}) \times \mathbb{R}^{\infty} \overset{?}{\cong} (S^1 \vee S^1) \times \mathbb{R}^{\infty}. $$

Annihilators of indecomposable representations in the BGG category $\mathcal O$ over semisimple Lie algebra

Mon, 02/19/2018 - 12:21

Let $\mathfrak g$ be a finite-dimensional (complex) semisimple Lie algebra. Then we consider the BGG category $\mathcal O$.

For a given indecomposable module $M$ in $\mathcal O$. And assume that we know all composition factors $L(\lambda_i)$ (simple height weight module of weight $\lambda_i$) of $M$, for $i =1,\ldots, m$.

${\bf My ~question}$: Is there any method to detect the annihilator of $\text{Ann}M$? Can we say something about $\text{Ann} M$ in terms of annihilators $\text{Ann}L(\lambda_i)$, $i=1,\ldots ,m$. (In general I think $\text{Ann}M \neq \bigcap_{i=1}^m\text{Ann}L(\lambda_i)$, is this correct?) Thanks!

*$\text{Ann} N = \{x\in \mathfrak g|~xN=0\}$, for all $N \in \mathcal O$.

Could the sequence A287326 be generalized in order to receive expansion of natural power n>3?

Mon, 02/19/2018 - 11:55

The sequence - is Binomial distributed triangular array, that shows us necessary items to expand perfect cube $n^3$. Summation of $n$-th row of Triangle A287326 from $0$ to $n-1$ returns $n^3$. But is it exist simillar patterns in order to receive expansion of power $n>3$, where $n$ - positive integer?

$$ \begin{matrix} & & & & & 1\\ & & & & 1 & & 1\\ & & & 1 & & 7& & 1\\ & & 1 & & 13& & 13& & 1\\ & 1 & & 19& & 25& & 19& & 1\\ \end{matrix} $$ Figure 1. Triangle A287326.

It derived by means of identity $$ x^3=\sum\limits_{m=0}^{x-1}3!\cdot mx-3!\cdot m^2+1 $$

For detailed info on derivation, please, reffer to links below. Thank you !

Tate-Shafarevich group over number fields

Mon, 02/19/2018 - 10:55

Let $A$ be an abelian variety over a number field $K$, $\text{Sha}(A/K)$ its Tate-Shafarevich group, $\ell$ a prime.

Is it known that the $\ell$-primary torsion subgroup $\text{Sha}(A/K)\{\ell\}$ is trivial for almost all primes $\ell$?

Is every endomorphism of the sheaf of holomorphic functions on a disk a differential operator?

Mon, 02/19/2018 - 01:33

Let $D= \{z\in \mathbb{C}:|z| < 1\}$ be the unit disk. And consider the sheaf of holomorphic functions $\mathcal{O}_{D}$.

Question (?) : Is there a sheaf endomorphisms $\phi : \mathcal{O}_D \to \mathcal{O}_D$ which is not a (possibly infinite order) differential operator. I.e. not of the form:

$$\phi=\Sigma_{n=0}^{\infty} b_n(z) \partial^n$$

Where $\partial =\frac{d}{d z}$ and $b_n \in \mathcal{O}_D$ .

EDIT: Suppose I require that $\phi$ be continuous w.r.t. to the natural frechet topology on $\mathcal{O}_D$ coming from uniform convergence on compact subsets, does the answer change?

Haar measure on $\mathrm{SL}_3(\mathbb{Z}) \backslash \mathrm{SL}_3(\mathbb{R}) / \mathrm{SO}_3(\mathbb{R})$

Mon, 02/19/2018 - 00:05

The bi-invariant Haar measure on the quotient $\mathrm{SL}_2(\mathbb{Z}) \backslash \mathrm{SL}_2(\mathbb{R}) / \mathrm{SO}_2(\mathbb{R})$ represents the moduli space of rank two real lattices modulo rotation and is easy to write down. An element of the quotient can be written (using the Iwasawa decomposition) as: \[\begin{pmatrix} y^{1/2} & xy^{-1/2} \\ 0 & y^{-1/2} \end{pmatrix}\] with $z=x+iy$ in a fundamental domain of the action of $\mathrm{SL}_2(\mathbb{Z})$ over the upper-half plane, such as $D = \{(x,y) : x^2+y^2 \geq 1,|x| \leq 1/2,y > 0\}$. In these coordinates, the invariant measure is a scale of the hyperbolic measure, namely \[\frac{3}{\pi} \frac{dx \, dy}{y^2}.\] Is there a similar nice parametrisation for the rank 3 case (namely explicating the bi-invariant Haar measure of $\mathrm{SL}_3(\mathbb{Z}) \backslash \mathrm{SL}_3(\mathbb{R}) / \mathrm{SO}_3(\mathbb{R})$?

What is this quotient of the triangle 2-3-7 group?

Sun, 02/18/2018 - 23:51

I have been working with Hurwitz groups, and I came across the group $G := \langle a, b \ | \ a^2, b^3, (ab)^7, ([a,b]^2ab)^6 \rangle$. I'm trying to figure out exactly what this group is. I know it contains two copies of ${\rm PSL}(2,13)$, and there seems to be a very large 2-group in there as well, of order $2^{28}$ at least. Is this group even finite, and if so, what is the order? Does it contain any other simple groups?

$K[[X_1,...]]$ is a UFD (Nishimura's Theorem)

Sun, 02/18/2018 - 23:18

Let us define the infinitely-many-variable formal power series ring

$$ K[[X_1,\ldots]] \colon= \underset{m \geq 1}{\varprojlim}\,K[[X_1,\ldots,X_m]]. $$

$K[[X_1,\ldots]]$ is known to be a UFD by a theorem of Nishimura (c.f. On the unique factorisation theorem for formal power series, Journal Math. Kyoto. Univ. Vol 7. No 2. 1967, 151-160).

Now let us choose an irreducible element $f \in K[[X_1,\ldots]]$ and consider the image $f_m \in K[[X_1,\ldots,X_m]]$ of $f$ by the natural quotient ring homomorphism $K[[X_1,\ldots]] \twoheadrightarrow K[[X_1,\ldots,X_m]]$.

Q. Is $f_m$ also irreducible for $m \gg 0$?

Is every true statement independent of $PA$ equivalent to some consistency statement?

Sat, 02/17/2018 - 17:15

Most true statements independent of PA that I know of is equivalent to some consistency statement. For example

  • Con(PA), Con(PA + Con(PA)), Con(PA + Con(PA) + Con (PA + Con(PA)), $\dots$
  • Goodstein's theorem is equivalent to Con(PA)
  • Any conjunction or disjunction of the above.

Is every true statement independent of PA equivalent to some consistency statement?

By "equivalent to some consistency statement", I mean that $PA \vdash S \iff Con(T)$, for some theory $T$. Also, $T$ should be either finite, or specified by a Turing machine that outputs its axioms (and such that PA proves that the Turing machine never stops outputting statements), so that the description of $T$ doesn't throw PA off.

EDIT: In particular, are there are $\Pi^0_1$ examples?

Fast Comparing of the Volume of Simplices Defined by Sidelengths

Sat, 02/17/2018 - 01:06

I have a problem, that requires sorting a set of simplices, that are defined via their sidelengths, according to volume; the value of the individual volumes isn't relevant in my problem.


are there faster methods of comparing the volumes of two simplices (that are defined via their sidelengths), than comparing the absolute values of their Cayley Menger determinants?

  • if no, where can I find details about the initial proof?
  • if yes, what are relevant algorithms for that problem?

Estimating the critical probabilities $\mathrm{P_{c1}}$ and $\mathrm{P_{c2}}$ mathematically for the infinite system case

Sat, 02/17/2018 - 00:47

Suppose I have an $\mathrm{N\times N}$ square matrix consisting of only $0$'s and $1$'s. The probability of a certain element being $1$ is $\mathrm{p}$. At that certain probability $\mathrm{p}$, we count the number the number of white clusters and number of black clusters in the matrix. Any two elements who share a side (i.e. any one of them lies to the left, right, top or bottom of the other) or share a vertex are considered to belong to the same cluster. BUT, there's one special case i.e. if anywhere in the matrix there occurs a situation like this:

0 | 1 1 | 0


1 | 0 0 | 1

That is two $1$'s share a vertex (diagonally connected) and two $0$'s also share a vertex as shown, the $50\%$ of the times, one should consider the $1$'s to belong to same cluster and other $50\%$ of the times one should consider the $0$'s to belong to the same cluster.

Suppose I am gradually increasing $\mathrm{p}$ from $0$ to $100$ in steps of $0.001$ and counting the number of black and white clusters for each $\mathrm{p}$. My aim is to find that critical probability $\mathrm{P_{c1}}$ at which number of white clusters increases from $1$ to a number greater than $1$ as $\mathrm{N}\to\infty$. Also I need to find $\mathrm{P_{c2}}$ at which number of black clusters decreases from a number greater than $1$ to $1$.

I wrote a program for the $1000\times 1000$ case, and averaged the critical probability over $10$ iterations. (The program basically had a loop which ran from $\mathrm{p}=0$ to $\mathrm{p}=100$ and I outputted those $\mathrm{p}$'s at which the number of white clusters increased from $1$ and number of black clusters decreased to $1$) The value I got for $\mathrm{P_{c1}}$ was $0.05$ and for $\mathrm{P_{c2}}$ it was $0.96$. However, I don't know if any purely mathematical technique exists to find the convergence value of these two critical probabilities for the infinite matrix case (i.e. $\mathrm{N}\to \infty$). Around $3$ decimal places of accuracy will be enough for my purpose. Any help will be appreciated.


I realize that my question wasn't very clear. I'm adding an example.

$$\begin{bmatrix} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{bmatrix}$$ Here, $f$ and $k$ share a vertex and so do $g$ and $j$.

If $f$ and $k$ were both $1$'s and $g$ and $j$ were both $0$'s, then you should either consider the two $1$'s ($f$ and $k$) to belong to the same "black" cluster (i.e. connected along the vertex) and $0$'s ($g$ and $j$) to belong to the same "white" cluster. BUT, suppose if you did consider the two $1$'s to belong to the same cluster (i.e. connected along the vertex), then you should not consider the two $0$'s to be connected along the vertex.

Decomposition of injective modules over Noetherian rings

Sat, 02/17/2018 - 00:42

Let $A=\mathbb{C}[x_1,\ldots,x_n]$ be a polynomial algebra over the complex numbers. I am interested in injective modules over $A$.

Since $A$ is projective over itself, the $\mathbb{C}$-dual module $A^\ast=\mathbb{C}[[x_1,\cdots,x_n]]$ is known to be injective and the generatings $x_i$ of $A$ act on $A^\ast$ by partial derivatives $\partial/\partial x^i$. As the ring $A$ is Noetherian, $A^\ast$ splits in the direct sum of indecomposable injective modules. My question is about the structure of this decomposition.

As I can see, for any $y\in \mathbb{C}^n$ the module $A^\ast$ has the indecomposible $A$-submodules $$M_y=\{p(x)e^{yx}| p(x)\in A\}\,,$$ where $yx$ stands for the inner product of vectors in $\mathbb{C}^n$. Furthermore, it seems that $M_y$ is isomorphic to the injective hull $E_A(\mathbb{C})$ of the trivial $A$-module $\mathbb{C}$ on which $p(x)\in A$ acts by multiplication by $p(y)\in \mathbb{C}$.

Is it true that

$$ A^\ast\simeq \bigoplus_{y\in \mathbb{C}^n} M_y \quad? $$ This would be rather strange, as any finite sum of functions of the form $p(x)e^{yx}$ for some $p$'s and $y$'s can't produce a divergent power series from $A^\ast$.

What is a good reference on the subject?

Log-concavity of the maximum of gaussians

Sat, 02/17/2018 - 00:23

Let $Z_1,\ldots, Z_n$ be independent gaussian random variables. Is it true that $X=\max\{Z_1,\ldots,Z_n\}$ has a log-concave distribution function?

Does exist a Kahler-Einstein metric on the blow-up of $\mathbb{P}^3$ along a smooth plane cubic?

Sat, 02/17/2018 - 00:18

This might be well known for the experts but I am not able to find a reference. I was wondering if there exists a Kahler-Einstein metric on the Fano threefold given by blow-up of $\mathbb{P}^3$ along a smooth plane cubic or not. I believe that the answer shoud be "no" by Matsushima's criterion saying that if the automorphism group of the variety is not reductive then there is no such metric.

I was thinking that the following might be a (sketch of) proof: The automorphisms of $\mathbb{P}^3$ fixing a plane (say $x_3=0$) are of the form $$\begin{pmatrix} * & * & * & * \\ * & * & * & * \\ * & * & * & * \\ 0 & 0 & 0 & * \end{pmatrix} $$
This group has projective dimension 12 and since 9 points determines a cubic in $\mathbb{P}^2$ we have that the automorphism group of the blow-up is 3-dimensional. On the other hand, the 3-dimensional unipotent subgroup $$\begin{pmatrix} 1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & * \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ acts trivially on the plane $x_3=0$, so in particular fixes the 9 points determining the cubic and hence lifts to the automorphism group of the blow-up. Since they have the same dimension (but the latter might be non connected), the automorphism group if a finite extension of a unipotent group, hence not reductive.

Am I right? Sorry for being sketchy! Thanks a lot in advance for any comment, I just started to introduce myself to the subject.

PS: By the way, does the fact that $\operatorname{Aut}(X)$ is reductive implies that the connected component of the identity $\operatorname{Aut}^0(X)$ is reductive as well?

Graphs of groups with homomorphisms not necessarily injective

Fri, 02/16/2018 - 23:44

I'm wondering if there is any literature on graphs of groups where the maps $G\to H$ from an edge group $G$ to its endpoint group $H$ are not necessarily $\pi_1$-injective. Or is this just too general to actually say anything meaningful? Are there any results on this subject?

Numerical Methods

Fri, 02/16/2018 - 22:26

While using the adaptive step size Runge Kutta method to approximate the solution of ordinary differential equations, what must generally be the tolerance to compare the error with?

Specialization map étale cohomology

Fri, 02/16/2018 - 21:59

Let $R$ be a henselian dvr, $s,\eta\in\text{Spec}(R)$ the closed and generic points, and $f : X\to \text{Spec}(R)$ a proper smooth scheme.

For a prime $\ell$ invertible on $R$, is there a specialization map

$$sp^i_{\eta,s} : R^if_*(\mu_{\ell^n})_{\eta}\to R^if_*(\mu_{\ell^n})_{s}$$

(with $\eta$ and $s$, not the geometric points over them)?

Is it an isomorphism, injective, surjective?

Amortization differences [on hold]

Fri, 02/16/2018 - 19:44

not a math guru but work in the mortgage field and have a question I am hoping someone can answer.

Scenario: one loan for $200,000 at 5pct over 30 years has a monthly principle and interest payment of $1074.

If broken up into two loans, each for $100,000, one loan with a 4% interest rate and the other with a 6pct interest rate, the payments are $477 and $600 respectively. Totaling $1077.

Considering that the combination scenario has an average rate equal to 5% and the amortization schedules each end at 360 months...... why the difference in total payments?

This payment difference increases with larger loan amounts and bigger swings in interest rates (e.g. 2% and 8%) for the same average rate.

Thanks in advance!