Recent MathOverflow Questions

scale factors in the contest of differential operators

Math Overflow Recent Questions - Tue, 01/02/2018 - 14:33

With the original aim of properly understand the ellipsoidal coordinates $(\lambda, \mu, \nu)$, I would like you to help me understanding the concept of "scale factors" in the contest of differential operators. Particularly, I would like you to provide me a relatively fast and intuitive explanation of such concept for someone like me who has a knowledge of differential calculus mostly coming from the courses of calculus and electromagnetism.

In order to make you understand which is my issue, I am going to copy and paste here a text from Wikipedia (

" Scale factors and differential operators

For brevity in the equations below, we introduce a function

\begin{eqnarray} S(\sigma) = (a^2 + \sigma)(b^2 + \sigma)(c^2 + \sigma) \end{eqnarray} where $\sigma$ can represent any of the three variables $(\lambda, \mu, \nu)$. Using this function, the scale factors can be written

\begin{eqnarray} h_{\lambda} & = & \frac{1}{2}\sqrt{\frac{(\lambda - \mu)(\lambda - \nu)}{S(\lambda)}}\\ h_{\mu} & = & \frac{1}{2}\sqrt{\frac{(\mu- \lambda)(\mu- \nu)}{S(\mu)}}\\ h_{\nu} & = & \frac{1}{2}\sqrt{\frac{(\nu- \lambda)(\nu - \mu)}{S(\nu)}} \end{eqnarray}

Hence. the infinitesimal volume element equals \begin{eqnarray} dV = \frac{(\lambda - \mu)( \lambda - \nu )( \lambda - \nu )}{8\sqrt{ - S(\lambda)S(\mu)S(\nu)}}d\lambda d\mu d\nu \end{eqnarray} "

What I personally expect as coefficient in the previous formula is the Jacobian determinant and not the product of three different "scale factors". Is there anybody who can provide me an intuitive explanation of this scale factors and how they are related with the Jacobian determinant?

Does anyone have a copy of Salce's paper "Cotorsion theories for abelian groups"?

Math Overflow Recent Questions - Tue, 01/02/2018 - 14:26

The paper "Cotorsion theories for abelian groups" by L. Salce, was published in 1979 in Symposia Math. 21, pages 1-21. According to Google Scholar, it's been cited 233 times, and I keep seeing citations of this paper as fundamental to cotorsion pairs. But I can't find it anywhere online, or in my library. Any help would be appreciated!

Yang-Mills connection on circle bundle

Math Overflow Recent Questions - Tue, 01/02/2018 - 14:16

Let $(M,g)$ be a connected, oriented, compact Riemann surface with positive constant curvature $K_g$, and let $P\to M$ be a principal $S^1$-bundle on $M$. Can we find a Yang-Mills connection $A$ on $P$ such that the norm of the curvature $2$-form $F_A\in A^2(M,\mathbb R)$ is equal to $K_g$ (i.e. $|F_A|_g=K_g$)?

Defining morphisms/rational maps between coarse moduli spaces

Math Overflow Recent Questions - Tue, 01/02/2018 - 14:05

Frequently in the theory of moduli I see map constructed between moduli spaces simply by saying where geometric points go, and no justification at all for why the defined map is actually a morphism of varieties/schemes.

For example, consider $M_{1,3}$. Given a curve $C$ of genus $1$ with three marked points $p_i$, consider $L=\mathbb{O}_C(p_1+p_2+p_3)$. Since $h^0(L)=3$, we get an embedding of $C$ in $\mathbb{P}^2$ as a cubic.

By considering the images of the $p_i$ too, this gives a point in $X:=\mathbb{P}(\text{cubics in $3$ variables})\times (\mathbb{P}^2)^3$, which is not well defined (it depends on the isomorphism $\mathbb{P}(H^0(L))$ with $\mathbb{P}^2$). There is an action of $PGL_2$ on $X$, and the image of $(C,p_1,p_2,p_3)$ identifies one orbit of this action. We may take a rational quotient $X/PGL_2$ by Rosenlicht's theorem, and the claim is now that this gives a rational map from $M_{1,3}$ to $X/PGL_2$.

Things like this are usually asserted without justification. The only way that I can prove this is by providing a (rational) map from the stack $\mathcal{M}_{1,3}$ to $X/PGL_2$, i.e. for any family of curves, I produce a certain $PGL_2$-torsor. However, this is long and technical, and certainly not the kind of thing that I would say nothing about in a paper. Furthermore, this stuff is done in classical geometry all the time, with no reference to stacks. Is there a direct method to see this that I am not aware of?

Prikry forcing and Cohen generic

Math Overflow Recent Questions - Tue, 01/02/2018 - 13:50

Let $\kappa$ be a measurable cardinal and let $\mathcal{U}$ be a normal measure on $\kappa$. Let $\mathbb{P}$ be the standard Prikry forcing using $\mathcal{U}$. Let $\mathbb{Q} = \text{Add}(\kappa, 1)$ be Cohen forcing for adding a new subset to $\kappa$ using partial functions from $\kappa$ to $2$ of size ${<}\kappa$.

Question: Is there a projection from $\mathbb{P}$ to $\mathbb{Q}$?

A set-family where in each coloring, at least one set is single-colored

Math Overflow Recent Questions - Tue, 01/02/2018 - 13:41

Given an integer $k$, we would like a family of $k$-sized sets such that, in any coloring of the items with two colors, at least one set is single-colored. What is the smallest number of sets we need?

Obviously $s(1)=1$, and it is easy to prove that $s(2)=3$. What about $k\geq 3$?

As a side-question: I often run into similar questions regarding set-families, and I feel I am missing some background to approach them. What is a good textbook to study about such problems?

Happy new semiprime after prime year!

Math Overflow Recent Questions - Tue, 01/02/2018 - 13:05

After the change of the year i realized ,us everyone did, that $2018=2\times1009$ and of course $1009$ is a prime number. $2017$ is also a prime number. Furthermore $2019=3\times 673$ and $673$ is also a prime number. So here is the conjecture.

For every number $n\in \mathbb{N}$ there exists a prime number $p$ such that

$p+1=2\times p_1$,

$p+2=3\times p_2$,

$p+3=4\times p_3$,

$p+4=5\times p_4$,

. . .

$p+n=(n+1)\times p_n$

where $p_1,p_2,p_3,.....,p_n$ are some prime numbers.

One can notice that this is an optimal situation in the sense that in every 2 numbers at least one should be divided by $2$ in every 3 numbers at least one should be divided by $3$ and in every n numbers at least one should be divided by $n$. Of course the conjecture is too hard in the sense that it implies the open conjecture that there are infinitely many primes such that $2p-1$ is also prime, or you can ask the same question for the oposite direction meaning:

For every number $n\in \mathbb{N}$ there exists a prime number $p$ such that

$p-1=2\times p_1$,

$p-2=3\times p_2$,

. . .

$p-n=(n+1)\times p_n$

where $p_1,p_2,p_3,.....,p_n$ are some prime numbers that implies that the Sophie Germain Primes are infinite, or both directions simultaneously,etc.

I dont know if this is a known question,i tried to find related ones but I didn't made it.


Thinking to this direction one can make much more general conjectures that implies these ones. Par example:

For any number $n\in \mathbb{N}$ and for every prime $p'<n$ and every power $i$ for each one of them there exists a $p$ such that

every number $p+1,...,p+n$ is divided by:

1) some(or none) primes $p'<n$ to some powers $i$ in any legal way meaning that none of this ways should force $p'|p$ or for some $n'$ $n'|p+j, n'|p+k$ and $n'$ does not divide $k-j$ or some $n''$ not to divide $n''$ consecutive numbers.

2) at most one more prime number.

Of course this would imply much more than the twin prime conjecture...

approaching the general philosophy that "With prime numbers (or natural numbers in general) everything that is possible to happen happens somewhere"

Happy New Year!

How to compute the index of a given weight?

Math Overflow Recent Questions - Tue, 01/02/2018 - 11:04

I am learning Borel Weil Bott theorem in rational homogeneous varieties. Working on them, I need to know two questions: 1. How to judge if a weight is singular? 2. How to compute the index of a given weight?

To make my questions precise, let $G$ be a semisimple Lie group over $\mathbb C$. Choose a base of simple roots $\alpha_i, i=1,\ldots, n$, we form a weight lattice. Let $\lambda_i,i=1,\ldots, n$ be the corresponding fundamental weight. Now I have a weight $\lambda=\sum_{i=1}^n m_i\lambda_i$, my question is 1. How to determine if $\lambda$ is singular? Definition: a weight $\lambda$ is called singular, if there is a root $\alpha$ such that $\left<\lambda, \alpha\right>=0$. 2. If $\lambda$ is nonsingular, how to calculate the index of $\lambda$? Definition: let $w$ be a Weyl group element, the length of $w$ is the number of shortest letters of reflections. The index of $\lambda$ is the shortest length of such $w$ that takes $\lambda$ to the fundamental chamber.

For example, let $G=B_n$ and $\lambda=\lambda_{n-2}-(1+n)\lambda_{n-1}+2\lambda_n$. How can I determine whether $\lambda$ is singular? If $\lambda$ is not singular, how should I compute the index of $\lambda$?

Do actions of BS(1,n) on finite sets factor through abelian quotients?

Math Overflow Recent Questions - Tue, 01/02/2018 - 11:03

Suppose $BS(1,n)$ is the Baumslag-Solitar group and $S_m$ is the symmetric group. If $\Phi: BS(1,n) \to S_m$ is a homomorphism, must the image of $\Phi$ be abelian?

Where do the (Akizuki)-Nakano Identities First Appear

Math Overflow Recent Questions - Tue, 01/02/2018 - 10:49

The answers to this M.O. question give a history of the Kaehler identities. The identities can be extended to the vector bundle-valued setting, and play a central role in the proof of the Kodaira vanishing theorem. Where do these identities first appear? Moreover, is it more common to refer to these identities as the Nakano identities or the Akizuki-Nakano identities?

A consequence of these identities is the Bochner-Kodaira-Nakano identity, or as Demailly refers to it, the Bochner-Calabi-Kodaira-Nakano identity. Where does this first appear, and how are the contributions of Bochner, Calabi, Kodaira, and Nakano related?

Does $\int_{0}^{\infty}e^{-xz}\sum_{n=0}^{\infty}a_{n}\frac{x^n}{n!}dx$ converge for $z>0$ with $a_{n} > n! $, for $ n>1$? [on hold]

Math Overflow Recent Questions - Tue, 01/02/2018 - 09:32

Let $g$ be exponential generating function such that $g(x)= \sum_{n=0}^{+\infty}a_{n}\frac{x^n}{n!}$ extended by analytic continuation along $\mathbb{R+}$ and has a positive radius of convergence. We assume $a_{n }$ is increasing sequence such that $a_{n }> n! $ for $n>1$. My question here is to ask about convergence of the integral below in general under the given conditions.

Question: Does $\displaystyle\int_{0}^{\infty}e^{-xz}\sum_{n=0}^{\infty}a_{n}\frac{x^n}{n!}dx$ converge for $z>0$ with $a_{n} > n!$ , with $ n>1 $?. $z, x $ are real numbers.

Note 1: I found this integral as a condition for summability of Exponential Generating function in this paper definition 3 page 17.

Note 2: The motivation of this question to know more about sufficient condition for summability .

Chow group and base change

Math Overflow Recent Questions - Tue, 01/02/2018 - 07:43

Let $k$ be a field with algebraic closure $\overline{k}$. Let $f\colon X\to k$ be a smooth projective variety(geometrically connected) over $k$.

Is the base change map $$\phi_i\colon \mathrm{CH}^i(X)\to\mathrm{CH}^{i}(X_{\overline{k}})$$ always injective?

(If $i=1$, $\mathrm{CH}^1(X)=\mathrm{Pic}(X)$, the Hochschild-Serre spectral sequence gives $$0\to H^1(\mathrm{Gal}(\overline{k}/k),f_{\overline{k},*}\mathbb{G}_m)\to\mathrm{Pic}(X)\to\mathrm{Pic}_{X_\overline{k}/\overline{k}}(k)\to H^2(\mathrm{Gal}(\overline{k}/k),f_{\overline{k},*}\mathbb{G}_m)$$ The condition implies $f_{\overline{k},*}\mathbb{G}_m=\mathbb{G}_m$, so $H^1(\mathrm{Gal}(\overline{k}/k),f_{\overline{k},*}\mathbb{G}_m)=0$, and $\mathrm{Pic}_{X_{\overline{k}}/\overline{k}}(k)\to\mathrm{Pic}_{X_{\overline{k}}/\overline{k}}(\overline{k})$ is injective, we know $\phi_1$ is injective.

If $X$ is quasi-projective, then $\phi_1$ is not injective, for example $X=\mathbb{P}^1_{\mathbb{R}}-\{\pm i\}$, then $\mathcal{O}(1)$ is a nontrivial element in $\mathrm{ker}(\phi_1)$.)

Bilinear product of two summable families

Math Overflow Recent Questions - Tue, 01/02/2018 - 06:42

Consider the following statement, which I suspect is false as written:

Let $E,F,G$ be (Hausdorff) topological vector spaces (over $\mathbb{R}$), let $\varphi\colon E\times F\to G$ be continuous and bilinear, and let $(x_i)_{i\in I}$ and $(y_j)_{j\in J}$ be summable families in $E$ and $F$ respectively with $\sum_{i\in I} x_i = x$ and $\sum_{j\in J} y_j = y$. Then $(\varphi(x_i,y_j))_{(i,j)\in I\times J}$ is summable with $\sum_{(i,j)\in I\times J} \varphi(x_i, y_j) = \varphi(x,y)$.

(Lest there be any doubt, “$(z_k)_{k\in K}$ summable with $\sum_{k\in K} z_k = z$” means that for every neighborhood $V$ of $z$ there is $K_0\subseteq K$ finite such that for any $K_0 \subseteq K_1 \subseteq K$ finite we have $\sum_{k\in K_1} z_k \in V$”.)

I am interested both in “nice” counterexamples to the statement above and in strengthenings of the hypotheses which would make it true—or basically any information regarding variations of this statement (I know essentially nothing except for the pretty much trivial fact that when $E,F,G$ are finite-dimensional it is correct). Since the rules of MO are to ask one specific question, and since I am mostly interested in counterexamples, let me ask:

Question: Is there a counterexample to the above statement with “nice” spaces $E,F,G$ (e.g., locally convex, complete, metrizable… or even Banach spaces)?

—but again, any information concerning it is welcome.

Comments (added 2018-01-04):

  • It is clear that, under the hypotheses of the statement above, $\sum_{(i,j)\in I_1\times J_1} \varphi(x_i, y_j)$ converges to $\varphi(x,y)$ where $I_1$, $J_1$ range over the finite subsets of $I$ and $J$ respectively; what is to be proven is that $\sum_{(i,j)\in K_1} \varphi(x_i, y_j)$ converges to $\varphi(x,y)$ where $K_1$ ranges over the finite subsets of $I\times J$. The subtlety, of course, is that $K_1$ can fail to be a rectangle.

  • The following result is found in Seth Warner's book Topological Rings (1993), theorem 10.15: if $E,F,G$ be are Hausdorff commutative topological groups, and $\varphi\colon E\times F\to G$ is continuous and $\mathbb{Z}$-bilinear, and $(x_i)_{i\in I}$ and $(y_j)_{j\in J}$ summable families in $E$ and $F$ respectively with $\sum_{i\in I} x_i = x$ and $\sum_{j\in J} y_j = y$, then provided $(\varphi(x_i,y_j))_{(i,j)\in I\times J}$ is summable, it sum is $\sum_{(i,j)\in I\times J} \varphi(x_i, y_j) = \varphi(x,y)$. So the crucial question is the summability of $(\varphi(x_i,y_j))$, not the equality with $\varphi(x,y)$. Even with the very weak hypothesis that $E,F,G$ are commutative topological groups, I still don't have a counterexample!

  • The following possibly related result is found in Kamal Kant Jha's 1972 paper “Analysis of Bounded Sets in Topological Tensor Products” (corollary 3.3): If $(x_i)$ is a totally summable family in a locally convex space $E$ [meaning that there exists $L\subseteq E$ closed, absolutely convex and bounded, such that $\{x_i\}\subseteq L$ and $\sum_i p_L(x_i) < +\infty$ for $p_L$ the gauge of $L$] and ditto for $(y_j)$ in $F$, then $(x_i \otimes y_j)$ is totally summable in $E \mathbin{\otimes_\varepsilon} F$.

Question about actions of full transformation monoids

Math Overflow Recent Questions - Tue, 01/02/2018 - 05:12

[Reposted from math.stackexchange]

Consider a monoid $M$ acting on a set $X$, where $M$ is the full transformation monoid on some set $A$.

Say that $B\subseteq A$ fixes $x\in X$ iff, for all $m\in M$, if $m(b) = b$ for all $b\in B$, then $mx=x$.

Say that $B\subseteq A$ pins down $x\in X$ iff, for all $m,n\in M$, if $m(b)=n(b)$ for all $b\in B$, then $mx=nx$.

[Apologies if there's more standard terminology for these notions]

Question 1: If $B$ fixes $x$, does it follow that $B$ pins down $x$?

Question 2: If $B$ and $B'$ both pin down $x$, does it follow that $B\cap B'$ pins down $x$?

Are the inverses of a set of quadratic polynomials linearly independent?

Math Overflow Recent Questions - Mon, 01/01/2018 - 22:29

Is a collection of reciprocals of monic reducible quadratic polynomials, that is functions of the form

$$ \{ \left( (x-a_i)(x-b_i) \right)^{-1} \}_{i=1}^{k}, $$

linearly independent over a finite field? This can be seen for the reciprocals of linear functions from the invertibility of Cauchy matrices.

Edit: To address ABX's very nice observation/obstruction, assume that $k$ is very small compared to the characteristic of the field.

Lefschetz fixed-point theorem for the Frobenius map

Math Overflow Recent Questions - Mon, 01/01/2018 - 20:24

Where can one find a proof of Lefschetz fixed-point theorem for the Frobenius map on elliptic curves over algebraic closures of $F_{p}$ ?

This could immediately follow if their coholomogies (for the sheaf of regular functions) were Weil cohomologies. But the proof of this is also hard to find.

Yet, there are references to this fact in connection with the use of Picard-Fuchs equation and counting rational points on such curves.

Is the order on repeated exponentiation the Dyck order?

Math Overflow Recent Questions - Mon, 01/01/2018 - 15:55

The Catalan numbers $C_n$ count both

  1. the Dyck paths of length $2n$, and
  2. the ways to associate $n$ repeated applications of a binary operation.

We call the latter magma expressions; we will explain below.

Dyck paths, and their lattice structure

A Dyck path of length $2n$ is a sequence of $n$ up-and-right strokes and $n$ down-and-right strokes, all having equal length, such that the sequence begins and ends on the same horizontal line and never passes below it. A picture of the five length-6 Dyck paths is shown here:

A: B: C: D: E: /\ / \ /\/\ /\ /\ / \ / \ / \/\ /\/ \ /\/\/\

There is an order relation on the set of length-$2n$ Dyck paths, given by comparing their heights; we call it the height order, though in the title of the post, we called it "Dyck order". For $n=3$ it gives the following lattice:

A | B / \ C D \ / E

For any $n$, one obtains a poset structure on the set of length-$2n$ Dyck paths using height order, and in fact this poset is always a Heyting algebra (it represents the subobject classifier for the topos of presheaves on the twisted arrow category of $\mathbb{N}$, the free monoid on one generator; see this mathoverflow question).

Magma expressions and the "exponential evaluation order"

A set with a binary operation, say •, is called a magma. By a magma expression of length $n$, we mean a way to associate $n$ repeated applications of the operation. Here are the five magma expressions of length 3:

A: B: C: D: E: a•(b•(c•d)) a•((b•c)•d) (a•b)•(c•d) (a•(b•c))•d ((a•b)•c)•d

It is well-known that the set of length-$n$ magma expressions has the same cardinality as the set of length-$2n$ Dyck paths: they are representations of the $n$th Catalan number.

An ordered magma is a magma whose underlying set is equipped with a partial order, and whose operation preserves the order in both variables. Given an ordered magma $(A,$•$,\leq)$, and magma expressions $E(a_1,\ldots,a_n)$ and $F(a_1,\ldots,a_n)$, write $E\leq F$ if the inequality holds for every choice of $a_1,\ldots,a_n\in A$. Call this the evaluation order.

Let $P=\mathbb{N}_{\geq 2}$ be the set of natural numbers with cardinality at least 2, the logarithmically positive natural numbers. This is an ordered magma, using the usual $\leq$-order, because if $2\leq a\leq b$ and $2\leq c\leq d$ then $a^c\leq b^d$.

Question: Is the exponential evaluation order on length-$n$ expressions in the ordered magma $(P,$^$,\leq)$ isomorphic to the height order on length-$2n$ Dyck paths?

I know of no a priori reason to think the answer to the above question should be affirmative. A categorical approach might be to think of the elements of $P$ as inhabited sets, choose an arbitrary element of each, and use them to define functions between the various Hom-sets (e.g. a map $\mathsf{Hom}(c,\mathsf{Hom}(b,a))\to\mathsf{Hom}(\mathsf{Hom}(c,b),a)$), do so in a recursively-definable way, hope to prove they are injective. However, while defining these maps seems to be doable in an ad hoc manner, I don't see how to generalize it. I also don't see how one should use the assumption that the sets are not just inhabited but have cardinality at least two.

However, despite the fact that I don't know where to look for a proof, I do have evidence to present in favor of an affirmative answer to the above question.

Evidence that the orders agree

It is easy to check that for $n=3$, these two orders do agree:

a^(b^(c^d)) A := A(a,b,c,d) | | a^((b^c)^d) B / \ / \ (a^b)^(c^d) (a^(b^c))^d C D \ / \ / ((a^b)^c)^d E

This can be seen by taking logs of each expression. (To see that C and D are incomparable: use a=b=c=2 and d=large to obtain C>D; and use a=b=d=2 and c=large to obtain D>C.) Thus the evaluation order on length-3 expressions in $(P,$^$,\leq)$ agrees with the height order on length $6$ Dyck paths.

(Note that the answer to the question would be negative if we were to use $\mathbb{N}$ or $\mathbb{N}_{\geq 1}$ rather than $P=\mathbb{N}_{\geq2}$ as in the stated question. Indeed, with $a=c=d=2$ and $b=1$, we would have $A(a,b,c,d)=2\leq 16=E(a,b,c,d)$.)

It is even easier to see that the orders agree in the case of $n=0,1$, each of which has only one element, and the case of $n=2$, where the order $(a^b)^c\leq a^{(b^c)}$ not-too-surprisingly matches that of length-4 Dyck paths:

/\ /\/\ ≤ / \

Indeed, the order-isomorphism for $n=2$ is not too surprising because there are only two possible partial orders on a set with two elements. However, according to the OEIS, there are 1338193159771 different partial orders on a set with $C_4=14$ elements. So it would certainly be surprising if the evaluation order for length-4 expressions in $(P,$^$,\leq)$ were to match the height order for length-8 Dyck paths. But after some tedious calculations, I have convinced myself that these two orders in fact do agree for $n=4$! Of course, this could just be a coincidence, but it is certainly a striking one.


Existence of double eigenvalue

Math Overflow Recent Questions - Mon, 01/01/2018 - 09:18

Let $A$ and $B$ be complex $4\times 4$ matrices. Assume both are Hermitian, and that they are linearly independent.

Must there exist a nonzero real linear combination $aA + bB$ which has a repeated eigenvalue?

Projectives in the category of quasi coherent sheaves Qch(X)

Math Overflow Recent Questions - Sat, 12/30/2017 - 17:36

X is a non singular projective variety over an infinite field k. How to prove there are no projectives with a surjective map to the structure sheaf O_X in Qch(X) and Coh(X). Coh(X) is the category of coherent sheaves on X.

Is there a function $f$ satisfy :$B(f+f^{-1})=B(f)+B(f^{-1})$ with $B$ is a Borel transform?

Math Overflow Recent Questions - Sat, 12/30/2017 - 15:27

Let $y(x) = \sum_k y_k x^k$ be a formel power series and ,Assume that is

invertible then $y^{-1}$ exist ,Also assum that $y$ ,$y^{-1}$ are Borel summable

then my question here is :

Question: Is there a function $f$ satisfy :$B(f+f^{-1})=B(f)+B(f^{-1})$ with $B$ is a Borel transform ?


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