Recent MathOverflow Questions

Traces in associative algebras

Math Overflow Recent Questions - Tue, 04/03/2018 - 05:54

Are there some books or papers about the general definition of traces:

If $\mathscr{A}$ is an associative algebra over $K$ then the space of traces is the set of all linear functionals $\tau:\mathscr{A}\to K$ which satisfy $\tau(\mathcal{A}\mathcal{B})=\tau(\mathcal{B}\mathcal{A})$ for all $\mathcal{A},\mathcal{B}\in\mathscr{A}$.


Uniformly convex, uniformly smooth Banach space which is not convex of power type

Math Overflow Recent Questions - Tue, 04/03/2018 - 05:51

It is well known that every uniformly convex Banach space $X$ admits an equivalent norm which such that it is convex of power type, i.e. the modulus of convexity with respect to the new norm satisfies $\delta_X(\varepsilon)\geq C\varepsilon^p$ for some $p$ but the same need not be true for the original norm.

Is there an example of Banach space which is uniformly convex and uniformly smooth which is not convex of power type?

Is the existence of double complement of a set provable in Intuitionistic ZF?

Math Overflow Recent Questions - Tue, 04/03/2018 - 05:39

In Powell's article [1] he introduces the axiom of double complement, which says a double complement $\{x : \lnot\lnot(x\in A)\}$ is a set for any set $A$.

I can't find similar axiom from other references, even in the Friedman's article [2] on double negation over set theories. Hence it is natural to ask the relation between his axiom and other axioms of IZF. (Note: Powell consider the axiom of collection rather than the replacement in his article, but I will consider full IZF.)

I have made some attempts on this problem: if $f(\beta):=\sup\{\alpha\in\mathrm{On} : \lnot\lnot(\alpha<\beta)\}$ exists for each ordinal $\beta$, then the axiom of double complement holds: then $\{x\in V_{f(\operatorname{rank}(A))} : \lnot\lnot(x\in A)\}$ would be the double complement of $A$. (Here $V_\alpha := \bigcup_{\beta\in\alpha} \mathcal{P}(V_\beta)$ is a von Neumann hierarchy. Axiom of power set is necessary in my argument.) However checking $f(\beta)$ is a set is at least as hard as checking the axiom of double complement.

Forcing or realizability seems not helpful to me. This is because we need to generate a set whose double complement is proper class to prove the independence of the axiom of double complement, and it seems to need a proper-class sized name. However both methods just deal with set-sized names.

My question is: Is the axiom of double complement provable from full IZF? If not, is it indenpendent from IZF? Is the axiom of double complement related to the law of excluded middle? I would appreciate any help.


[1] Powell, William C. "Extending Gödel's negative interpretation to ZF." The Journal of Symbolic Logic 40.2 (1975): 221-229.

[2] Friedman, Harvey. "The consistency of classical set theory relative to a set theory with intuitionistic logic 1." The Journal of Symbolic Logic 38.2 (1973): 315-319.

Existence of Arithmetic Progression from density inequality

Math Overflow Recent Questions - Tue, 04/03/2018 - 05:39

Let $A\subset \{0,1,\dots, N-1\}$ such that $$|A\cap [0,N/3)|\geq \left(\delta+\dfrac{\delta}{8}\right)\cdot \dfrac{N}{3},$$ where $\delta\in (0,1]$. Prove that exists arithmetic progression $P$ with $|P|\ge N/3$ such that $|A\cap P|\geq \left(\delta+\dfrac{\delta}{8}\right)|P|$.

Remark: By $[0,N/3)$ I denote the set of integers from this interval.

If $N$ is divisible by $3$ then we can take $P:=[0,N/3)$ which is also a progression and $|P|=N/3$ we have $|A\cap P|\ge \left(\delta+\dfrac{\delta}{8}\right)|P|$.

If $N$ is not divisible by $3$ and if we we take $P:=[0,N/3)$ which is also progression and has $\lfloor\frac{N}{3}\rfloor+1$ elements. However, I am not able to derive the needed inequality.

Would be very grateful if anyone can explain how to approach in the case when $3\nmid N$?

P.S. This question refers to the proof of Roth's theorem.

About existence of Characters of $B(X)$

Math Overflow Recent Questions - Tue, 04/03/2018 - 04:55

Let $X$ be a Banach space. Let $B(X)$ be the space of all bounded linear operators on $X$. Does $B(X)$ have an empty character space for any $X$? I know that $M_n(\mathbb{C})$ has no characters and also $B(H)$ where $H$ is a Hilbert space.

Is there any simple expression for $(\dot{F}\phi_u)\times(\frac{\partial \dot{F}}{\partial u} \phi_v)$

Math Overflow Recent Questions - Tue, 04/03/2018 - 04:14

$F$ is an isometry of two dimensional regular surfaces $A$ to $B$. $\phi$ and $F\circ\phi$ are the coordinate chart for $A$ and $B$ respectively.$\phi_u$ and $\phi_v$ are unit tangent vector of $A$ at a point $P$. I am trying to simplify $(\dot{F}\phi_u)\times(\frac{\partial \dot{F}}{\partial u} \phi_v)$. where $\frac{\partial \dot{F}}{\partial u}$ is a matrix obtained by differentiating each elements of $\dot{F}$ partially with respect to $u$.
Is there any simple expression for $(\dot{F}\phi_u)\times(\frac{\partial \dot{F}}{\partial u} \phi_v)$.
Thank you

Lower bound on the eigenvalues of the Laplacian

Math Overflow Recent Questions - Tue, 04/03/2018 - 02:12

I am looking for a graph for which $2 d_{i} < \mu_{i}$, for some index $i$, where $\mu_{1} \leq \mu_{2} \leq \dots\leq \mu_{n}$ are the eigenvalues of the Laplacian matrix $L(G)$ and $d_{1} \leq d_{2} \leq \dots \leq d_{n}$ are the node degrees.

According to the literature and existing upper/lower bounds on the eigenvalues of the Laplacian as a function of node degrees, it seems there is a graph with this property. However, I was not able to find such a graph by generating all graph with $4, 5, \dots, 10$ vertices. Any help or advice for a possible hypothesis confirmation or rejection would be appreciated.

Related references:

First order decidability of limit of gradient flow?

Math Overflow Recent Questions - Mon, 04/02/2018 - 08:10

Let $f: \mathbb{R}^n\to\mathbb{R}$ be a polynomial function, and let $p$ be a critical point. Consider the ascending manifold $A_p$ consisting of all points whose limit under the gradient flow of $f$ is equal to $p$. Is $A_p$ a semialgebraic set? If not, is there an additional assumption on $f$ that makes it so?

This is an offshoot of a previous question of mine that did not get much attention.

What’s the etiquette on using diagrams that need color to be understood?

Math Overflow Recent Questions - Sun, 04/01/2018 - 11:29

I’m working on a paper that makes heavy use of colorful diagrams to supplement the text. For most of these it would probably not be possible to create grayscale versions that convey the same information as effectively. I’m a bit worried about this because (1) I imagine that some people like to print out papers to read them but these people might not want/be able to print in color, and (2) some readers may be colorblind.

What are the expectations on an author in my situation? Will it be considered rude to leave as-is, so long as the diagrams are not technically necessary to verify the arguments? Am I expected to include a description in the captions (“this region is red, this region is blue...”)? Or most stringently, would I be expected to post a “colorblind version” somewhere that tries to recreate the diagrams in grayscale as best as I can?

I’m interested in all opinions, but especially those of people who would have trouble with color for whatever reason.

Integral domain $R$ with fraction field $K$ such that for every $u \in K$, the subring $R[u]$ of $K$ is flat $R$-module

Math Overflow Recent Questions - Sun, 04/01/2018 - 10:46

Let $R$ be an integral domain with fraction field $K$. If for every $u \in K$, the subring $R[u]$ of $K$ is a flat $R$-module, then is it true that $R$ is a Prufer domain ?

If $R$ were moreover a GCD domain, then I could prove that $R$ is a Prufer domain; but without the GCD assumption, I don't see what would happen. If every overring of $R$ i.e. subring $S$ of $K$ with $R \subseteq S \subseteq K$, then also $R$ would be Prufer (this is standard). But I am considering only very special type of overrings.

Points on Sphere whose image, under symmetric positive definite matrix, is contained in cube

Math Overflow Recent Questions - Sat, 03/31/2018 - 14:42

Let $\Sigma \in \mathbb{R}^{n \times n}$ be a symmetric, positive definite matrix and let $\mu_r$ denote surface measure on the sphere in $\mathbb{R}^n$ with radius $r$. Let $$ R = \{x \in \mathbb{R}^n : |x_i| \leq a_i, \ i = 1, \dots, n\} $$ denote an $n$-dimensional rectangle that centers the origin.

Many geometric properties are known when considering $\Sigma$ as a linear map from $\mathbb{R}^n$ to itself, in particular the map will transform a sphere to an ellipse, rotate the standard basis and scale them proportional to the eigenvalues, etc. As such there are many reformulations of my question, which is the following:

Ideally, I would like $\mu_r(\Sigma^{-1}R)$ as explicitly as possible given in terms of $r$ and the elements of $\Sigma$ and the $a_i.$

Geometrically, this is the surface area of the points on the sphere with corresponding points on the ellipsoid (under $\Sigma$) contained in $R.$

Other comments or answers to related problems are welcome!

References on Gerbes [on hold]

Math Overflow Recent Questions - Sat, 03/31/2018 - 10:19

Some excerpt of comments in some questions

  • Thanks a lot! Is there any chance you know of a reference where I could find the proofs of these facts? and the response is The standard reference is Giraud's book Cohomologie non-abelienne. This book is unreadable in the strongest possible meaning of the word "unreadable". So... no. Another users response is Depends on taste, André, I find most of the contemporary articles in this area, which are often nonsystematic in terminology and notation, plus wave hands and use jargon on most issues, much less readable than Giraud's book.
  • See Giraud's book on nonabelian cohomology. Response from another user is I don't think telling someone to see a dense 470 page book in French on non-abelian cohomology is a helpful comment.

Another excerpt from some notes by Breen and Messing is We now describe in more detail the content of the present text. While we have placed ourselves firmly within the context of algebraic geometry, in which the concepts of gerbes and stacks have been to date most fully developed.

Only references I am familiar with are

Out of these, only first article is more or less readable. The other two by Lawrence Breen are really not readable for me.

I am getting demotivated and irritated by lack of notes on gerbes and even in math over flow there are not so much to see. Is this out of fashion now? Are there any one else who read/work on these? I am not looking for something in Physics.

Are rationals everywhere equally dense?

Math Overflow Recent Questions - Sat, 03/31/2018 - 08:44

I would like to know is there any notion of density over the rationals with which we could determine are rationals everywhere equally dense on the real line, because, for example, I am not sure would all the rationals in some $\delta>0$ neghborhood of, for example $\frac {1}{2}$, would be equally dense as in some $\delta>0$ neghborhood of zero, that is, would that definition of density give same results for rationals in $[\frac {1}{2}- \delta, \frac {1}{2}+\delta]$ as for those in $[0-\delta,0+\delta]$.

Because, although rationals are everywhere dense in the sense that between any two of them there is an infinite number of them it is probably true that irrationals are not self-similar in their positioning over the intervals on the real line, so if irrationals do not have that kind of self-similarity neither do rationals have, right?

So it would maybe be enough to define that notion of density over the irrationals and then study irrationals to know how exactly in the sense of that density the rationals are placed.

There probably have been ideas like this one before so someone can tell something about this task, I believe that this can be put on a firm ground.

What is known about this?

Is this a characterization of cocompleteness?

Math Overflow Recent Questions - Sat, 03/31/2018 - 08:33

If $A$ is a cocomplete category and $C$ is small, we can say that for all functor $C \stackrel{l}{\to} A$ there exists the Kan Extension Lan$_{y_C}(l): \hat{C} \to A$. Moreover, the Kan extension is pointwise.

Is it known if the other implication is true? Namely:

Conj: If for all functor $C \stackrel{l}{\to} A$ there exists the Kan Extension Lan$_{y_C}(l): \hat{C} \to A$ then $A$ is cocomplete.

Modular forms, Maass forms and Automorphic representations

Math Overflow Recent Questions - Sat, 03/31/2018 - 07:52

I am beginning to learn about automorphic forms, and stay perplex concerning the two languages of "forms" versus "representations" often used at the same time. As far as I understand,

  • a modular/Maass form generates an automorphic representation (by considering the space of its right translations)
  • conversely, an automorphic representation gives a unique automorphic form (as a newvector, i.e. a nontrivial element in $\pi^K$ which is one dimensional for a suitable (?) compact subgroup $K$)

I would like to understand, where does the distinction between modular form and Maass form appear in the representation language?

Thanks in advance for any reference or explanation

Distributional + $C^0$ convergence

Math Overflow Recent Questions - Sat, 03/31/2018 - 07:47

Let $C^0(\mathbb{R}^d)$ the space of bounded continuous functions taken with the norm $\|f\|_{C^0}=\sup_{x\in\mathbb{R}^d}|f(x)|$. And let $\mathcal{S}'$ be the space of tempered distributions.

Consider a family of Fourier multipliers $\{\psi_j(D)\}_{j\in\mathbb{N}}$ such that $$f=\sum_{j\in\mathbb{N}}\psi_j(D)f$$ holds in $\mathcal{S}'$.

Suppose that for a given $f\in\mathcal{S}'$ the sum $\sum_{j\in\mathbb{N}}\psi_j(D)f$ converges also in $C^0$. How can it be proven that, in such case, we must have $f\in C^0$?

Conjugacy classes of non-normal subgroups of a finite $p$-group

Math Overflow Recent Questions - Sat, 03/31/2018 - 07:12

Let $G$ be a finite $p$-group of derived length $d$ and nilpotency class $c$. Suppose that $G$ is not a Dedekind group (i.e., possesses at least one non-normal subgroup). Suppose that $G^{(d-1)}$ has order p and is the unique normal subgroup of order $p$. Let $s$ be the number of conjugacy classes of non-trivial subgroups of $G$ that do not contain $G^{(d-1)}$ (thanks to the assumptions on $G$, they are automatically non-normal).

Is it true that $s\geq c-1$?

By an inspection with GAP, I could check this inequality for such $p$-groups of small order in GAP library which I checked. Does there exist a method (may be induction) to prove this inequality? Any comment or answer will be appreciated!

Normal form of two tangent symplectic surfaces in $\mathbb R^4$

Math Overflow Recent Questions - Sat, 03/31/2018 - 06:53

Suppose $\Sigma_1$ and $\Sigma_2$ are two distinct smooth symplectic surfaces in $(\mathbb R^4,\omega)$. Suppose they both pass through $0$ and their tangent spaces coincide at $0$. Suppose moreover, that there is a smooth $J$-holomorphic structure for which both $\Sigma_1$ and $\Sigma_2$ are $J$-holomophic.

Question. Is it true that locally close to $0$ one can introduce any holomorphic coordinates $(z,w)$ such that $C_1$ is given by $w=0$ and $C_2$ by $w=z^n$, $n\ge 1$?

Or at least can one make so that $C_1$ is given by $w=0$ and $C_2$ by $w=z^n+O(|z^{n+1}|)$?

Note that this question is an identical reformulation of my previous question: Two smooth tangent almost complex curves in a $4$-manifold , I ask it here to remove ambiguities about its interpretation.

Seeking for results and references about location of zeros of complex polynomials

Math Overflow Recent Questions - Sat, 03/31/2018 - 06:25

Suppose we have a complex polynomial of a complex variable $P(z)=\sum_{k=0}^n c_kz^k$.

I am interested in all what is known of location of zeros as a function of the coefficients, that is, are there some known functions $w_1,...w_n$ so that $w_l(c_0....c_n)$ gives us some set in the plane for every $l=1,...,n$, and it would be nice if every $w_l$ would give us some disk or a square but even other simply connected sets would do the job.

Do you know of some free books and/or articles that discuss these problems?

If there are no results of this nature what are the results known about zeros of complex polynomials as a functions of coefficients?

Algebraic independence of the composition of functions

Math Overflow Recent Questions - Sat, 03/31/2018 - 06:10

Is my statement below true, and if yes, how is this provable?

$f_1\colon Z_1\ open\ \subseteq\mathbb{C}\to\mathbb{C}$

$f_2\colon Z_2\ open\ \subseteq\mathbb{C}\to\mathbb{C}$

$g\colon Z_3\ open\ \subseteq\mathbb{C}\to\mathbb{C}$, g is transcendental over $\mathbb{C}$.

If $f_1$ and $f_2$ are algebraically independent of each other over $\mathbb{C}$, the functions $$F_1\colon Z\ open\ \subseteq\mathbb{C}\to\mathbb{C}, z\mapsto f_1(g(z)),$$ $$F_2\colon Z\ open\ \subseteq\mathbb{C}\to\mathbb{C}, z\mapsto f_2(g(z))$$ are algebraically independent of each other over $\mathbb{C}$.

Assume at first all the functions are elementary functions according to Liouville an Ritt. But the statement shall be extended to other functions.

Maybe there is a proof for meromorphic functions, but with another proof without differentiation, the statement can be extended to other functions.


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