Recent MathOverflow Questions

A nonempty proper subset with positive measure

Math Overflow Recent Questions - Sun, 11/12/2017 - 05:06

Does a measurable set with positive measure have a nonempty proper subset with positive measure?

Tensor Product of vector spaces

Math Overflow Recent Questions - Sun, 11/12/2017 - 04:25

I have this problem.

Let $V$ be a vector space over a field. I'm working with a certain linear operator $F$ of $V \otimes V$ that satisfies a certain equation A.

Now let $M$ be a nonempty set and I denote with the symbol $\mathbb{C}^M$ the space of mappings from $M$ to $\mathbb{C}$. For each map $f: M \times M \rightarrow M \times M$ I associate its pullback $F:\mathbb{C}^{M\times M} \rightarrow \mathbb{C}^{M\times M}$

$F(\phi)=\phi(f)$, with $\phi:M \times M \rightarrow \mathbb{C}$.

Then in the article, the author apply the equation A mentioned above with the pullback $F$ . I dont'unterstand how because $A$ is a tensor equation. (page 4)

What is the largest hyper-graph class solved ( have a quasi-polynomial ) Babai's new algorithm?

Math Overflow Recent Questions - Sun, 11/12/2017 - 04:07

I am currently working on survey on graph isomorphism, the best known result for graph isomorphism is $exp((\log{n})^c)$ given by Babai.

Now let me define a bounded rank (maximum cardinality of any of the edge) hyper graph class, it is a collection of all hyper-graphs such that rank is bounded (at max $k$)

There exist a simple reduction which reduces hyper-graph instance into a graph and then apply a babai's new algorithm.

Question : Is there any hyper-graph class bigger than bounded $k$ rank hyper-graph such babai's new result is still applicable to that class?

A length decreasing homotopy of a closed curve in a simply connected manifold

Math Overflow Recent Questions - Sun, 11/12/2017 - 04:05

Assume that $M$ is a simply connected Riemannian manifold and $\gamma$ is a simple smooth closed curve with bases point $p\in M$ and length $l(\gamma)$.

Is there a homotopy of smooth closed curves $\gamma_t,\;\;t\in [0,\;1]$ based at $p$ with $\gamma_0=\gamma$, $\gamma_1=p$ such that $\forall t\in [0,\;1]$ we have $l(\gamma_t)\leq l(\gamma)$? Moreover can we choose the homotopy $\gamma_t$ of smooth closed curves based at point $p$ such that $l(\gamma_t)$ is a decreasing function of $t$?

Internal operations on uncomputable functions

Math Overflow Recent Questions - Sun, 11/12/2017 - 03:14
  1. Is there know set of operations for which uncomputable functions are, let's name it down-unclosed? I mean a set of operations which takes two ( or more) uncomputable functions and return computable function? It is obvious it depends on certain property, like rates of growth, so are this hypothetical internal operations structured like onion, being internal only inside shells of similar growth? So what is known for other classes of functions? non primitive recursive (NPR)? For primitive recursive answer is known.

  2. Or relaxing a bit, it returns an uncomputable ( or NPR) function, which grows much slower?

  3. Is there any general notion describing such relationships ( "down-unclosure" under set of operations in bigger set which has built in order?) Maybe category abstract nonsense is ready defined?

Example regarding not primitive recursive function:

Let A be Ackerman function, so it is not primitive recursive ( it grows faster than any primitive recursive function). Obviously A + A is another NPR function, equivalent to original one, and similar for multiplication or exponents etc. But $$A/A$$ looks like quite normal computable function, an probably there are variants like linear combinations fraction or maybe even complicated functional composition on arguments ( for example $$A/(A°g)$$ where g is linear function and ° means function composition, so I mean Ackerman fraction of two functions, one of them with argument shifted $$x \to ax+b$$).


Example of down-unclosure is visible in various areas. There are a proof techniques based on lowering some parameters in the bounded from below set ( for example in natural numbers, contraction of volumes, or more generally measure of sets etc ), various analytical techniques based on control of remainder growth, or even set theoretic techniques based on "almost all" elements which may be seen as certain types of such "down-unclosure" trick.

Is there a "universal" connected compact metric space?

Math Overflow Recent Questions - Sat, 11/11/2017 - 20:08

Fact 1. The Cantor set $K$ is "universal" among nonempty compact metric spaces in the following sense: given any nonempty compact metric space $X$, there exists a continuous surjection $f\colon K \to X$.

Fact 2. The closed interval $I$ has a similar "universal" property among nonempty compact connected and locally connected metric spaces: given any such space $X$, there exists a continuous surjection $f \colon I \to X$.

This makes me wonder: is there a compact connected metric space $J$ such that for any nonempty compact connected metric space $X$, there exists a continuous surjection $f \colon J \to X$?

Such a space $J$, if it exists, would be 'intermediate' between $I$ and $K$: there would need to be continuous surjections

$$ K \to J \to I $$

Fact 1 is sometimes called the Alexandroff–Hausdorff theorem, since appeared in the second edition of Felix Hausdorff’s Mengenlehre in 1927 and also in an article by Pavel Alexandroff published in Mathematischen Annalen in the same year. Fact 2 was proved by Hans Hahn in 1914 and reproved by him more nicely in 1928. For a nice history of these results, see:

One may rightly complain that "universal" is the wrong word above, since we're not claiming there exists a unique continuous surjection, and indeed there's usually not. A better term is versal. There can be two non-homeomorphic spaces having the same versal property. For example, $I^2$ would work just as well as $I$ in Fact 2, thanks to the existence of space-filling curves.

Nonetheless we can create a category in which these versal properties become universal, by a cheap trick. Let $\mathrm{CompMet}$ be the collection of all homeomorphism classes of nonempty compact metric spaces, and put a partial order on this where $[X] \ge [Y]$ iff there exists a continuous surjection $f \colon X \to Y$. The homeomorphism class of the Cantor set is the top element of the poset $\mathrm{CompMet}$. My question asks if the subset of $\mathrm{CompMet}$ coming from connected compact metric spaces has a top element.

I'd also appreciate any interesting information on this poset $\mathrm{CompMet}$.

For example, I think that there's a map sending each element of $\mathrm{CompNet}$ to its number of connected components, and I think that this is an order-preserving map from $\mathrm{CompMet}$ to the cardinals less than or equal to the continuum. But there also seems to be an order-preserving map sending each element of $\mathrm{CompNet}$ to its number of path-connected components. Are there other interesting maps like this?

Anti-trace applied on the trace zero subgroup and distortion maps

Math Overflow Recent Questions - Sat, 11/11/2017 - 12:43

I was reading Pairings for Beginners by Craig Costello and he talks about the base field subgroup $\mathcal{G}_1 = E[r] \cap \ker(\pi - 1)$ and the trace zero subgroup $\mathcal{G}_2 = E[r] \cap \ker(\pi - q)$. He then defines the $Tr$ map and $aTr$ map on the entire $r$-torsion. See section 4.1 regarding the $r$-torsion.

I understood that $Tr$ moves everything in $r$-torsion in $\mathcal{G}_1$ except for $\mathcal{G}_2$ which sends the elements there into $\mathcal{O}$ (which is also in $\mathcal{G}_1$). Also $aTr$ sends "everything" from $r$-torsion into $\mathcal{G}_2$.

First question: Does $aTr$ sends also $\mathcal{G}_1$ into $\mathcal{G}_2$ or only in $\mathcal{O}$ ? I am asking because $Tr$ does not send $\mathcal{G}_2$ into entire $\mathcal{G}_1$ but only in $\mathcal{O}$.

Also, he talks about distortion maps and searching on google it seems that a distortion map is defined as a monomorphism $$\phi : {\mathcal{G}_1} \longrightarrow E[r] \setminus {\mathcal{G}_1}$$ But the Pairings for Beginners does not define the distortion map at all, even more it allows that the distortion map to be defined over subgroups other than $\mathcal{G}_1$ (see Figure 4.5, 4.6), so here is my second question.

Second question: It seems that distorsion map when is defined over $\mathcal{G}_2$ moves the elements out of $\mathcal{G}_2$ also. Is it true ?

Also, regarding distorsion maps, I come with the third question.

Third question: It seems that $\mathcal{O}$ could be mapped via the distorsion map out of the torsion. Is this desired in general or should be avoided ?

Convex hull of geodesic triangles in Cat(0) spaces

Math Overflow Recent Questions - Sat, 11/11/2017 - 09:19

Let $(X,d)$ be a Cat(0) space of dimension 2 homeomorphic to the plane. Following the notation of "Metric spaces of non-positive curvature" by Bridson and Haefliger, I define:

  • a geodesic triangle $\Delta\subset X$ consists of three points $p,q,r\in X$, its vertices, and a choice of three geodesic segments joining them, its sides.

  • A subset $C$ of $X$ is said to be convex if every pair of points $x,y\in C$ can be joined by a geodesic in $X$ and the image of every such geodesic is contained in $C$

  • the (closed) convex hull of a subset $A$ of a geodesic space $X$ is the intersection of all (closed) convex subspaces of $X$ containing $A$.

Since $(X,d)$ is of dimension 2, it makes sense to talk about the interior of a geodesic triangle $\Delta$: it is the bounded subset of $X$ delimited by $\Delta$. Is it true that the convex hull of every geodesic triangle $\Delta$ always consists of the union of $\Delta$ with its interior? If not, could you show a counterexample?

Characterization of Determinants, Decomposition of O(n), Vector Space Orientations

Math Overflow Recent Questions - Sat, 11/11/2017 - 07:01

Is there a way to characterize the determinant in arbitrary finite-dimensional vector spaces without either:

  • Coming up rather arbitrarily with some axioms for a 'nice function we would like to have' and then proving that this 'nice' function is unique

  • Assuming we know (before defining the determinant in first place and decomposing $O(n)$ to $SO(n)$ and $O_{\_}(n)$ ) that a vector space has exactly two orientations, and then constructing $\Lambda^p(V)$ (whereby the exactness of two possible orientations is built-in as the anti-symmetry property to the wedge product) ?

As a simpler first attempt, in the Euclidian case, one possible way I can think of is to

  1. Prove without determinants that $O(n)$ has exactly 2 connected components $A,B$.

  2. Characterize 'orientation-preserving' mappings $T$ satisfying $T|_A = 1$ and $T|_B = -1$.

  3. Somehow extend the results to $GL(n)$ and show that what we get is unique.

For the first step, the Cartan-Dieudonne Theorem seems quite helpful (there is a nice constructive proof of it for $R^n$ here that doesn't use determinants); it should be possible to prove that isometries consisting of at most even vs. odd reflections across hyperplanes form two disjoint classes. (How to prove this?)

For the second step, is there a well-known characterization of such mappings? (edit: reasked here as a standalone question).

Is there a Lie group decomposition that could help?

As a whole, are there any other known characterizations of the determinant that don't rely on definitions based on implicit assumptions about the nature of orientations?

Is there a calibre $\aleph_1$ Moore space which is not separable

Math Overflow Recent Questions - Sat, 11/11/2017 - 01:31

A topological space has calibre $\aleph_1$ if for every uncountable sequence $\langle U_\alpha\mid\alpha\lt\aleph_1\rangle$ of nonempty open sets $U_\alpha\subset X$, there is an uncountable subfamily $\Lambda\subset\aleph_1$ with $\bigcap_{\alpha\in\Lambda}U_\alpha\neq\emptyset$.

Is there a calibre $\aleph_1$ Moore space which is not separable?

(Under CH, the answer is no, since every first countable calibre $\aleph_1$ space is separable.)


Long term behavior of Hamilton-Jacobi PDE for moving interface

Math Overflow Recent Questions - Thu, 11/09/2017 - 18:04

Suppose $u: \mathbb{R}^m \times [0,\infty) \to \mathbb{R}$ and $u_0,g: \mathbb{R}^m \to \mathbb{R}$ where $m$ is 2 or 3. Further, suppose that $u_0$ and $g$ are both smooth. Let's say $u(x,t)$ is a solution (maybe in the sense of viscosity solution) to the initial value problem:

\begin{align} u_t = g\|Du\|, \;\; & (x,t) \in \mathbb{R}^m \times (0, \infty) \\ u(x,t) = u_0(x), \;\; & (x,t) \in \mathbb{R}^m \times \{0\} \end{align}

Intuitively, because $u$ is non-decreasing in $t$ where $g > 0$ and non-increasing where $g < 0$, it seems that we should have $$ \text{sign}(u) \to \text{sign}(g), \;\; t \to \infty $$ so long as (where defined) $\|Du\| \nrightarrow 0$, or, if the gradient norm does go to zero, at least not too quickly.

Is there literature on whether the this conjecture is true under these conditions, or perhaps either more strict or more general ones?

A trace formula for $\mathrm{GSp(4)}$

Math Overflow Recent Questions - Thu, 11/09/2017 - 06:28

The Arthur trace formula and its variations provide general results for reductive groups, however to the extent of my knowledge only few specific instances of the formula have been really worked out in details, i.e. with precise decomposition of the spectrum, as for $\mathrm{GL}(n)$.

Is there such a formula for $\mathrm{GSp(4)}$ or instances of use of the general trace formula in that setting? Or should I rather come back to the classification of representations of $\mathrm{GSp(4)}$ (perhaps as in Robert and Schmidt)?

Thanks for any idea!

How are constants/functions named after their discoverer?

Math Overflow Recent Questions - Thu, 11/09/2017 - 06:16

In general, when a paper references an object discovered/defined in another paper by author X, it goes something along the lines of:

"Let $\tau$ be the constant defined by X in 1999 [1]$\ldots$",


"Let $f_{\mu}$ denote the function that generalizes the case $\ldots$ (X, [1])".

At what point does the literature start talking about "X's constant" or "The X function"?

Who/what determines that an object discovered by somebody deserves to take his name?

Deforming a non-positively curved Riemannian manifold into a negatively curved one

Math Overflow Recent Questions - Thu, 11/09/2017 - 06:08

Cheeger deformations can be used to deform some non-negatively curved Riemannian manifolds into positively curved manifolds (e.g., sectional curvatures strictly positve), see What is a Cheeger deformation?.

I would like to ask: Is there a similar, or at least somehow comparable method of deforming a non-positively curved Riemannian manifold into a negatively curved one?

In particular, I would be very interested in such a method for compact Riemannian locally symmetric spaces of Rank $>1$, which are non-positively curved but not negatively curved.

Calculating a certain parameter for an abstract simplicial complex

Math Overflow Recent Questions - Thu, 11/09/2017 - 05:46

Let $\mathcal{C}$ be an abstract simplicial complex on some finite set $\Omega$. I say that a subset $\Lambda\subset\Omega$ is minimally non-simplicial if it is not a simplex, but all of its subsets are. Write $h(\mathcal{C})$ for the size of the largest minimally non-simplicial subset of $\Omega$.

I need to calculate $h(\mathcal{C})$ for a particular family of abstract simplicial complexes. I'm wondering if this parameter can be seen somehow using algebraic topology -- via some (co)homology calculation or suchlike. Also, if this parameter has been studied somewhere in the literature, then I would be very interested to know about that.

I apologise if my question is vague/ ill-posed/ inappropriate -- I am not at all an expert on abstract simplicial complexes. For those who are interested, this family of complexes arose when studying some properties of permutation groups.

What are the primitive elements in a polynomial Hopf algebra with primitive indeterminates?

Math Overflow Recent Questions - Thu, 11/09/2017 - 05:23

Asked on math.stackexchange but didn't get response (in fact got a negative vote without any comment), so trying here.

Is it true that in any polynomial Hopf algebra K[X1,X2,...] over a field K with indeterminates primitive, the primitive elements are precisely the linear homogeneous polynomials? (Perhaps with some additional assumptions like characteristic of K is 0?). If so, could someone kindly give me a reference? A paper I am reading says (without citation) that it is well-known.

When we have L^infinite(G) is subset C_0(G)?

Math Overflow Recent Questions - Thu, 11/09/2017 - 05:06

Let $G$ be locally compact group. Question:When we have $L^\infty(G)$ is subset $C_0(G)$?

Aymptotical formula of partition function with finite length

Math Overflow Recent Questions - Thu, 11/09/2017 - 05:05

Given $n,k$ with $n \ge k$. Let $p(n,k)$ denotes the number of partitions with length no more than $k$.

What is the approximate formula of $p(n,k)$?

Non-homogeneous space $X$ such that $X\cong X\setminus \{x\}$ for all $x\in X$

Math Overflow Recent Questions - Thu, 11/09/2017 - 04:38

What is an example of a topological space $(X,\tau)$ with the properties that

  1. $X\cong X\setminus \{x\}$ for all $x\in X$, and
  2. $(X,\tau)$ is not topologically homogeneous


Does positive semi-definiteness of a multivariate trigonometric polynomial imply that the constant term is positive?

Math Overflow Recent Questions - Thu, 11/09/2017 - 04:02

Let $f(\theta_1,...,\theta_d)=\sum _{m \in \Omega} a_m \cos(\sum _{i=1}^d m_i \theta_i) + b_m \sin(\sum _{i=1}^d m_i \theta_i)$ be a trigonometric polynomial in $d$ variables, where the $a_m$ and $b_m$ are real numbers and $\Omega \subseteq \mathbb{N}^d$ is a finite subset. Let the constant term of $f$ vanish ($a_0=b_0=0$ or just $0 \notin \Omega$) and $f\neq 0$. Can $f$ be nonnegative on $\mathbb{R}^d$?


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