Math Overflow Recent Questions


Coend formula for derived enriched left Kan extension

Tue, 03/13/2018 - 10:42

Let $f: \mathbf{C} \to \mathbf{D}$ be a functor between small categories and $\mathbf{M}$ a cocomplete category. Then we have the coend formula $$ (f_! X)(d) = \int^{c \in \mathbf{C}} \mathbf{D} (f(c),d) \otimes X(c) \quad (*) $$ for the left Kan extension of a functor $X : \mathbf{C} \to \mathbf{M}$ along $f$.

Now assume that all categories and functors involved here are enriched over simplicial sets (for example). Then the same formula holds for the enriched left Kan extension when interpreted correctly (in terms of tensoring over simplicial sets).

In the next step assume that $\mathbf{C}, \mathbf{D}$ and $\mathbf{M}$ are actually a simplicial model categories. This is the setup discussed in:

I would like to see a formula like $(*)$ for the $derived\ enriched$ left Kan extension, preferrably using the derived hom and maybe a replacement of $X$?

For the case where $\mathbf{C}$ and $\mathbf{D}$ are just enriched over simplicial sets, but not treated as model categories, I found a model for the derived enriched left Kan extension in Riehl's $Categorical\ Homotopy\ Theory$ using the bar construction, but can we maybe express it directly using the appropriate homotopical analogon of $(*)$?

Thank you for any hints.

On complexity of a combinatorial number theoretic problem?

Tue, 03/13/2018 - 08:39

Given the matrix $\begin{bmatrix} r_{11}&\dots&r_{1n}\\ \vdots&\ddots&\vdots\\ r_{m1}&\dots&r_{mn} \end{bmatrix}\in\Bbb Z^{m\times n}$ with $0<r_{ij}<2^n$ and $a,q\in\Bbb Z$ with $|a|,|q|<2^n$ what is the complexity of deciding if there is a product of form $r_{1j_1}\cdot\dots\cdot r_{mj_m}\equiv a\bmod q$?

Measurability of $T \to \Pi_{ker T}$ w.r.t. SOT

Tue, 03/13/2018 - 06:30

I came across the following technical question, to which I could not - after some time of thinking - find an answer:

Let $\mathcal{U},\mathcal{H}$ be two real (in general infinite dimensional) separable Hilbert spaces. For some linear subspace $\bar{U} \subseteq \mathcal{U}$, let $\Pi_U$ denote the orthogonal projection on this subspace. The question is:

Is the mapping $T \mapsto \Pi_{\text{ker}T}$ measurable from $L(\mathcal{U},\mathcal{H})$ to $L(\mathcal{U})$ when both spaces are endowed with the strong operator topology (and measurability is meant w.r.t. the Borel-$\sigma$-algebra of the SOT on both sides)?

Any hints and thoughts on this are more than appreciated!

Group Action from $\mathbb{C}^*$ and Hodge decomposition

Tue, 03/13/2018 - 04:58

In Claire Voisin's book (Hodge Theory and complex algebraic geometry), at the first exercise of chapter 6, the author claims that if we have a continuous action from $\mathbb{C}^*$ on $H_\mathbb{C}$ (here $H_\mathbb{C}$ is the complexified of a real vector space), satisfying conditions, then the latter vector space can be decomposed as followed : $$H_\mathbb{C}=\bigoplus_{p+q=k} H^{p,q}$$ with $\overline{H^{p,q}}=H^{q,p}$.

The $k$ appearing here is given in the conditions of our action : $$\rho : \mathbb{C}^* \to GL(H_{\mathbb{C}})$$ satisfies : $\forall t\in \mathbb{R}^*$, $\rho(t)=t^kId$

and $\rho(\overline{z})=\overline{\rho(z)}$ where the conjugation on $GL(H_\mathbb{C})$ is defined as follow : $g\in GL(H_\mathbb{C})$ then $\overline{g}(u)=\overline{g(\overline{u})}$.

I think I succeed to show most of the results here but I am now confused to show the assertion $\overline{H^{p,q}}=H^{q,p}$.

Here's how I proceed :

Given any $z$ in $S^1$ we can show that there exists a base and a sequence of integers $n_i$ such that $\rho(z)$ is diagonalizable with $z^{n_i}$ as eigen values. Then we can show that there exists a base which diagonalizes at the same time all the $\rho(z)$ for $z$ in $S¹$ and moreover that the sequence of integers is independant from the $z$ chosen in $S^1$.

Using the conditions of restriction on real numbers, we then show that there exists a sequence of integers such that for every non null complex $z=|z|e^{i\theta}$, $\rho(z)$ can be written as a diagonal matrix in which the diagonal terms are $|z|^ke^{i n_j\theta}$.

Denoting that the base used for the diagonalization is the same for any $z$, the eigenspaces don't not depend of the $z$ chosen : So we have : $E_{|z|^ke^{i n_j\theta}}=E_{|z'|^ke^{i n_j\theta'}}$ Now let $\chi$ be any character of $\mathbb{C}^*$. We define vector spaces as follows :

If $\chi : z \mapsto |z|^ke^{i n_j\theta}$, then $H_\chi:= E_{|z|^ke^{i n_j\theta}}$ and $H_\chi:=0$ if not. So we can decompose our space as follows : $$H_\mathbb{C}=\bigoplus_{\chi} H_{\chi}$$

Now for any $n_j$ appearing in the matrix we define $p$ and $q$ as $p:=\frac{k+n_j}{2}$ and $q:=\frac{k-n_j}{2}$ So we have $|z|^ke^{i n_j\theta}=z^p\overline{z}^q$. Each character appearing in the decomposition above is now of the form : $z \mapsto z^p\overline{z}^q$ and we can denote $H^{p,q}$ the corresponding eigenspace. We notice that if $H^{p,q}$ appears in the decomposition then $H^{q,p}$ also appears ( as the eigenvalues are conjugated each one to another).

Then I tried to use the condition $\rho(\overline{z})=\overline{\rho(z)}$ to show $\overline{H^{p,q}}=H^{q,p}$ but with no success. Explicitely, I take an eigen vector in $H^{p,q}$, meaning that for all $z$ $$\rho(z)(u)=z^p\overline{z}^qu$$ and I want to show that for all z we have : $$\rho(z)(\overline{u})=z^q\overline{z}^p\overline{u}$$ but did not succeed it.

Thanks in advance for your help.

What is the Proof for "the maximum number of n-vectors that can form a linearly independent set is 'n'? [on hold]

Tue, 03/13/2018 - 04:34

I have tried to prove it this way.

Let the number of n-vectors be 'n'.

The set formed should have a maximum number of vectors. In this case, it is 'n'.

The set formed is linearly independent, which means that it just includes all the vectors and it is equal to 'n'.

It will be great if someone can verify.

Thank you

If $n=x^k+y^k$ then also $n=a^2+b^2=c^3+d^3=\ldots =x^k+y^k$

Tue, 03/13/2018 - 04:29

Are there infinitely many positive integers with the property:
If $n$ is a sum of two $k$th powers then it is also the sum of two $k-1$th powers, the sum of two $k-2$th powers, ... , the sum of two squares?
For example let $n=2^5+1^5=31$. It is the sum of two 5th powers but not the sum of two squares. I am not sure if this has been studied before.
Here we suppose that $k\ge 3$.

Minimum space needed to compute a binomial coefficient

Tue, 03/13/2018 - 04:26

What is the minimum space required by an algorithm that, given as input two integers $n \geq k \geq 0$, performing only integer operations of addition, subtraction, multiplication, and division, returns as output the binomial coefficient $\binom{n}{k}$ ?

Clearly, since $$\max_{0 \leq k \leq n} \binom{n}{k} =\binom{n}{[n/2]} \sim \frac{2^{n}}{\sqrt{\pi n / 2}},$$ at least $n + o(n)$ bits are required (in the worst case).

Analog of Tenenbaum's theorem for EFA

Tue, 03/13/2018 - 04:09

EFA can prove the exponential function to be total, but it cannot prove the superexponential function to be total. Is there an analog of Tenenbaum's theorem (which states the PA has no recursive non-standard models) for EFA which states that EFA has no "sub-superexponential" non-standard models? (Here "sub-superexponential" should mean something like "only finitely iterated exponential" space complexity.) Can one say even more, like that EFA has no primitive recursive non-standard models?

Embeddings of fields and rational points

Tue, 03/13/2018 - 03:58

Let $S$ be a irreducible scheme over a field $k$ (for example a smooth projective curve over algebraically closed field). Denote by $k(S)$ its field of fractions. Let $K$ be a(n algebraically closed) field. Given an embedding $k(S) \hookrightarrow K$ we can construct a $K$-point of $S\otimes K$ in the following way: choose an affine open set in $S$ with the coordinate ring $R$. Then we can restrict the embedding to the embedding $R \hookrightarrow K$. After tensor product with $K$ it becomes a $K$-point in $S\otimes K$.

My question is: which points arise as the image of this map?

Does the lattice of all topologies embed into the lattice of $T_1$-topologies?

Tue, 03/13/2018 - 03:41

Let $\kappa$ be an infinite ordinal, and let $\text{Top}(\kappa)$ be the lattice of all topologies on $\kappa$, ordered by $\subseteq$. Let $\text{Top}^{T_1}(\kappa)$ be the lattice of all $T_1$-topologies on $\kappa$.

Is there an injective lattice homomorphism $\varphi: \text{Top}(\kappa)\to \text{Top}^{T_1}(\kappa)$?

Given a polynomial system, determine a simple set containing all its solutions

Tue, 03/13/2018 - 03:06

Suppose we have a system of complex polynomials $f = (f_1, \ldots, f_n)$, where each $f_i$ can be viewed as a function $\mathbb{C}^n \to \mathbb{C}$. The solutions of $f$ are the points $x \in \mathbb{C}^n$ such that $f(x) = 0$.

There are results bounding the magnitude of the zeros of univariate polynomials. One result for instance is given here. With this result we know that all zeros of a univariate polynomial $p$ are inside a complex disk or radius $R$ centered at the origin. The value $R$ can be determined using only the coefficients of $p$.

In the title, by simple set I mean a set like the above, which is a disk. What I mean by simple is a set which is determined explicitly by the coefficients of the $f_i$'s and has some simple determining rule. I'm looking for a way to know where the solutions are in advance. Sharp results are welcome.

I know there are some results relating this systems with Newton polyhedra, but I only saw results about counting solutions. If there is a connection between Newton polyhedra and the location of the solutions I would be glad to know. If anyone know some relevant article or book please share it here so I can read it.

Thanks you very much.

Binarily universal members of $[0,1]$

Tue, 03/13/2018 - 02:38

Let $r\in[0,1]$. We look at the binary represenation of $r$ and say that $r$ is binarily universal if every finite binary string appears in at least one place in the binary representation of $r$. Let $U$ be the set of binarily universal members of $[0,1]$.

Is $U$ a Borel set? If yes, what is its Borel measure?

Does the boundary of immediate basin contain a fixed point?

Tue, 03/13/2018 - 02:27

Let $f$ be a rational map of degree $d\geq 2$, and $B$ is a simply connected immediate basin of an supper-attracting fixed point of $f$. I want to know whether there exists a fixed point of $f$ contained in $\partial B$. Any hint will be welcome! Or some similar result can be given!

Are holomorphic vector bundles over Kähler manifolds Kähler

Tue, 03/13/2018 - 01:28

Let $X$ be a Kähler manifold and $E\to X$ a holomorphic vector bundle. Is there a Kähler structure on $E$ compatible with its complex structure?

A conjecture: Given $nD$ finite measures that bisects

Mon, 03/12/2018 - 22:01

Given $nD$ finite measures $µ_1, . . ., µ_{nD}$ in $\mathbb{R}^n$ does there exists a set of at most D hyperplanes that bisect each of the measures? How far is this problem solved?

That is closely related to a conjecture in L. Barba and P. Schnider, Sharing a pizza: bisecting masses with two cuts, Proceedings of the 29th Canadian Conference on Computational Geometry, CCCG’17.

Definition: A Hyperplane bisects $\mu$ iff the value of $\mu$ on two sides of the hyperplane are equal.

Etale Homotopy (Artin Mazur) reference

Mon, 03/12/2018 - 18:52

In the book of Artin and Mazur, there is an appendix to pro-categories, specifically in A.2.2 one can find the following:

It can be shown ([25]) that morphisms between the functors associated to pro-objects X,Y are in 1-1 correspondence with elements of Hom(X,Y)

In the bibliography [25] is Quillen's Homotopical Algebra.

Question: I am searching for that proof in Quillen's work, but I can't find it. Someone can help?

Edit: As John suggested, I exlpain in details what is the question.

Let $C$ be a category, we can associate to $C$ a pro-category of pro-objects (see pro-object). Let $X$ be a pro-object indexed by $I$, then we have a a functor $C \rightarrow Set$ which to $T \in Ob_C \mapsto \varprojlim_I Hom(X_i, T)$, on morphisms it is defined in the natural way. Let this functor be $hX$.

The question is: It can be shown ([25]) that morphisms between the functors associated to pro-objects $X,Y$ are in 1-1 correspondence with elements of $Hom(X,Y)$ as pro-objects.

Direct proof of the equivalence of symmetric monoidal $K$-theory and exact sequence $K$-theory?

Mon, 03/12/2018 - 17:08

When all exact sequences split in $C$, we have $\Omega B C \simeq K(C):=\Omega Q(C)$. Heuristically, this is because the space of upper-triangular matrices is contractible. Can this be made precise? I don't quite see how the standard proof comes down to this idea.

Let me elaborate. Let $C$ be a Quillen exact category. We may perform two constructions:

  • On the one hand, $C$, and hence also the groupoid $\iota C$ of isomorphisms in $C$, is symmetric monoidal under $\oplus$. We may deloop $\iota C$ with respect to $\oplus$ to obtain $BC$, and then the "symmetric monoidal K-theory" of $C$ is $\Omega B C$.

  • On the other hand, we may take the $Q$ construction or the $S_\bullet$ construction of $C$, and the "exact sequences K-theory" of $C$ is $K(C) = \Omega QC$ or $K(C) = \Omega S_\bullet C$.

The theorem (found in Grayson, Higher Algebraic K-Theory II, last theorem on p. 11 [1]) is that when every exact sequence splits, these two constructions are equivalent: $\Omega BC \simeq \Omega Q C$.

It seems to me the idea of the comparison should be as follows. In the bar construction $BC$, an $n$-simplex consists of an object $c \in C$ equipped with a decomposition as the direct sum of $n$ objects $c = c_{0,1} \oplus c_{1,2} \dots \oplus c_{n-1,n}$. Whereas in $S_\bullet C$, an $n$-simplex consists of an object $c \in C$ equipped with the structure of an $n$-step filtration in $C$ $c_{0,1} \hookrightarrow c_{0,2} \dots \hookrightarrow c_{0,n} = c$ (the indices here are chosen to agree with the $S_\bullet$ construction). In fact, there's a natural map $S_\bullet C \to B C$ sending a filtration to its associated graded. The point should be that the fiber of this map is contractible when every exact sequence splits. Assuming exact sequences split, the fiber over a given "associated graded" is connected, and its automorphisms are automorphisms $c \to c$ which are "upper triangular" (with $1$'s on the diagonal) with respect to the direct sum decomposition. Hence the heuristic.


This heuristic argument seems compelling to me, but again, I don't quite see how to read Grayson's proof as a formalization of this line of thinking. Is there somewhere where this heuristic argument is made precise?

If I'm totally off-base, I'd appreciate any conceptual illumination of this theorem. It's also possible that glossing over the difference between the $S^{-1} S$ construction and the $\Omega B S$ construction, and the difference between the $Q$-construction and the $S_\bullet$ construction, are getting me into trouble here.

Edit: I'm now convinced I don't know a relevant sense in which the group of upper-triangular matrices is contractible to begin with -- just the fact that this would be true if we were working in the the Euclidean topology over $\mathbb R$. So maybe getting a handle on that is the first step.

[1] It may appear from the structure of the paper that Grayson's argument relies on a homology computation and comparison to the plus construction, but this is not the case -- the comparison to the plus construction separates out cleanly in the logic of the paper, and the comparison I'm talking about is in fact reasonably direct, just not quite as conceptually clear as I would dream it could be. Specifically, the theorem in question really proves that $S^{-1} S \simeq \Omega QC$ (the part I'm interested in) and combines this with the homology computation showing that $S^{-1} S = K_0R \times BGL(R)^+$.

Periods in the trivial extension algebra of the incidence algebra of the divisor lattice

Mon, 03/12/2018 - 16:45

Definition of $C_L$ for people who like number theory:

Let $m$ be a number with prime factorisation $m=p_1^{n_1} ... p_r^{n_r}$ with $n_i>0$. Define $I_m$ to be the incidence algebra of the divisor lattice of $m$. Up to isomorphism this just depence on the $n_i$ and not on the primes so lets define with $L=[n_1+1,...,n_r+1]$, $I_L$ as the incidence algebra of the divisor lattice of a number $m$ with prime factorisation $m=p_1^{n_1} ... p_r^{n_r}$. Define $C_L$ as the trivial extension algebra of $I_L$. For the definition of trivial extension algebra see for example (together with a proof that this is always a Frobenius algebras) or the textbook on Frobenius algebras I by Skowronski and Yamagata.

Definition of $C_L$ for people who like Nakayama algebras:

For a natural number $n \geq 2$ denote by $A_n$ the hereditary Nakayama algebra with dimension $n$ given by quiver and relations over a field $K$ (that is the Nakayama algebra with Kupisch series $[n,n-1,...,2,1]$). For a list of integers $L=[n_1,...,n_r]$. Define the algebra $C_L$ as the trivial extension of the algebra $A_{n_1} \otimes_K \cdots \otimes_K A_{n_r}$.


Is $C_L$ isomorphic to a known/studied algebra from the literature? Experiments with small lists suggest that the algebra is always periodic or at least all simple modules are periodic with the same period. (Recall that a module $M$ is called periodic with period $n$ when $n$ is the smallest integer such that $\Omega^n(M) \cong M$. An algebra is called periodic in case the regular module is periodic as a bimodule.) The period seems to behave in a strange way. What could it be (in case such a finite periodic exists)?

Here some examples, where after the list $L$ the number is the period of a simple module of $C_L$. Sadly even with a computer it takes long to do such calculations.

$L=[t]: 2t$ for $t \geq 1$ (this is well known)

$L=[2,2]: 5$

$L=[3,2]: 22$


$L=[5,2]: 12$ (this was surprising)

$L=[6,2]: 43$


$L=[3,3]: 8$

$L=[2,2,2]: 6$

$L=[3,2,2]: 26$

Guesses what the period might be for general $L$ are also welcome. I should say that it is not too good tested yet as even for those small lists it takes several minutes to do the calculations with a computer.

More general, one might ask what those periods are (in case they exist) for the trivial extension algebra of tensor products of hereditary path algebras of Dynkin type. For example the trivial extension of $k D_4 \otimes_K k A_2$ has the property that all simple modules have period equal to 6.

Structure of Perfect Matchings in edge-colored Graphs

Mon, 03/12/2018 - 16:16

I am dealing with a problem arising from some quantum mechanical calculations, related to Graph Theory. The aim of this post is to get some literature references or suggestions and ideas how to think about this problem.

I am interested in whether an undirected edge-colored graph with a certain structure of perfect matchings can exist. I give two examples; for the first, I know how to solve my question, for the second question I don't know whether it exists (computer searches were unsuccessful).

The edge-color can have two values: one for the first vertex, one for the second. In all of my examples, there are only two colors (white and red), so an edge can be white-white, white-red, red-white and red-red.

Example 1: One red vertex

I want a graph $G=(V,E)$, with an even number of vertices, such that

  1. $\#_\text{pm}(G) = \lvert V\rvert$,
  2. every perfect matching contains exactly one half-red edge,
  3. in each perfect matching, a vertex has a red edge connected to it exactly once.

For $\lvert V\rvert=4$ and $\lvert V\rvert=6$, there is a simple solution:

The perfect matchings for $\lvert V\rvert=4$ are shown here:

It has four perfect matchings (axiom 1), every perfect matching contains only one half-red edge (axiom 2), and every vertex has one half-red edge incident with it (indicated by red letters; axiom 3).

The above solution can be scaled to every even number of vertices. I 3D-printed a few, and colored them. These are solutions for $\lvert V\rvert=12$ and $\lvert V\rvert=28$:

Example 2: Two red vertices

I want a graph $G=(V,E)$ for even number of vertices such that

  1. $\#_{\text{pm}}(G)=\binom{\lvert V\rvert}{2}$,
  2. every perfect matching contains exactly two half-red edges,
  3. in all perfect matchings, a vertex has a red edge connected to it exactly $\lvert V\rvert-1$ times.

I can solve this case for $\lvert V\rvert=4$:

However, I am not able to solve it for any higher even number of vertices. I used a computer search for $\lvert V\rvert=6$, which has $\binom{6}{2}=15 $, but the computer didn't find any solution (of course, I was able to search only a tiny fraction of all possible graphs due to the enormous number of possibilties).

My questions:

  1. Is there any noteworthy alternative way of looking at this problem than in terms of edge-colored graphs?
  2. Are related problems investigated in graph theory? Any references to such problems would probably help, as I am quite stuck and don't know how to search the existing mathematical literature for such questions.
  3. With which techniques could the existence of graph like those mentioned above be proven or disproven?
  4. Can one prove (without exhaustive search) that a graph satisfying the axioms of example 2, and with $\lvert V\rvert=2n$ with $n>2$ cannot exist?

Any hint will be very much appreciated!

Glue morphism of schemes defined over irreducible components

Mon, 03/12/2018 - 16:10

Let $k$ be an algebraically closed field and $X, Y$ be quasi-projective $k$-scheme. Let $X_1, X_2$ be two irreducible components of $X$ and $f_i:X_i \to Y$ be morphims such that $f_1|_{X_1 \cap X_2}=f_2|_{X_1 \cap X_2}$. Then, does the morphisms $f_1$ and $f_2$ glue to a morphism $f:X \to Y$ such that its restriction to $X_i$ is $f_i$?