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https://mathoverflow.net/questions/424142 | 1 | We work over $\mathbb{C}$. Let $X$ be a normal projective irreducible variety, and let $\mathbb{C}^\*$ act nontrivially on $X$. The fixed point locus of $X$, namely $X^{\mathbb{C}^\*}$, can be decomposed into a disjoint union of connected fixed point components, let us call them $F\_1,\ldots,F\_s$. Moreover, for any $k=1,\ldots,s$, we can define the *plus* cell
as
$$X^+(F\_k)=\{p\in X\mid \lim\_{t\to 0} t\cdot p \in F\_k\}$$
and similarly also the *minus* cell, by considering the limit at $\infty$.
If $X$ is smooth, then the celebrated theorem of Bialynicki-Birula tell us that
\begin{equation}
X=\bigsqcup\_{i=1\ldots k} X^+(F\_k)\tag{\*}\label{star}
\end{equation}
(and similarly for the minus cell), and moreover this is an affince cell decomposition.
**Question:** If $X$ is only normal, do we still have a decomposition as in \eqref{star} (but in this case the cell will not be afine, as explained in [Białynicki-Birula decomposition for singular projective variety](https://mathoverflow.net/questions/356995/bia%C5%82ynicki-birula-decomposition-for-singular-projective-variety))?
I've tried to search on the literature, but what I've found is usually much more deep of what I'm asking and I have a hard time trying to translate that result in my specific case. Any help, or reference, would be much appreciated. Thanks in advance!
| https://mathoverflow.net/users/481375 | Plus and minus Białynicki-Birula decomposition for normal variety | Yes, \eqref{star} is always a disjoint union (that's obvious). Moreover, each set $X^+(F\_k)$ is locally closed and the map $x\mapsto\lim\_{t\to0}t\cdot x$ induces an affine morphism $\pi\_k:X^+(F\_k)\to F\_k$. In general, the morphism $\pi\_k$ is not a fiber bundle anymore.
These assertions can be easily reduced to the smooth case by using a theorem of Sumihiro (Sumihiro, Hideyasu: [Equivariant completion](https://doi.org/10.1215/kjm/1250523277). J. Math. Kyoto Univ. 14 (1974), 1–28) according to which $X$ can be embedded equivariantly into a projective space.
The assertions are more generally valid for any normal complete variety. The paper "Konarski, Jerzy: Decompositions of normal algebraic varieties determined by an action of a one-dimensional torus.
Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 26 (1978), 295–300" seems to be a good reference.
| 5 | https://mathoverflow.net/users/89948 | 424186 | 172,338 |
https://mathoverflow.net/questions/424189 | 1 | How can I evaluate ( estimate ) the sum $s(n)=\sum\_{k=1}^{n-1}\mu ^{2}(k(n-k))$ ($\mu$ is the Möbius function). Trivial estimate
$s(n)<\varphi (n)$ follows from the fact that $\mu (k(n-k))$ is zero if $k$ is not prime to $n$. It seem that in the case of $n$ - primorial, $s(n)$ is pretty close to $\varphi (n)$
| https://mathoverflow.net/users/169583 | Möbius function summation | Let $f\_n(x)=x(n-x)$. You count squarefree values of $f\_n(k)$ for $k$ between $1$ and $n$.
Given a prime $p$ let $a\_{p,n}$ be the number of solutions $k\in \mathbb{Z}/p^2\mathbb{Z}$ to the congruence $f\_n(k)\equiv 0\bmod p^2$.
The heuristic asymptotic answer is $n$ times the following infinite product over primes
$$\prod\_{p} \left(1-\frac{a\_{p,n}}{p^2} \right),$$
where the $p$th term stands for the probability that $f\_n(k)$ is indivisible by $p^2$. This answer can be established unconditionally for quadratic polynomials $f$; this is due to Ricci (1933). Reference and proofs of special cases are given in these lecture notes of Rudnick:
[http://www.math.tau.ac.il/~rudnick/courses/sieves2015/squarefrees.pdf](http://www.math.tau.ac.il/%7Erudnick/courses/sieves2015/squarefrees.pdf)
The mentioned result of Ricci deals with a fixed polynomial, while your $f\_n$ varies with $n$. However, the arguments can be made uniform in your choice of $f\_n$.
Let us investigate the above product. The value of $a\_{p,n}$ is $2$ if $n$ is indivisible by $p$ (corresponding to $k\equiv 0,n\bmod p^2$) and is equal to $p$ otherwise (corresponding to $k\equiv 0,p,2p,\ldots\bmod p^2$). Hence the product is $A\_n B\_n$ where
$$A\_n=\prod\_{p\nmid n} \left(1-\frac{2}{p^2}\right),$$
$$B\_n=\prod\_{p\mid n} \left(1-\frac{1}{p}\right)=\frac{\phi(n)}{n}.$$
Then we see that your sum, divided by $\phi(n)$, is asymptotic to $A\_n$. The constant $A\_n$ is asymptotic to $1$ if $n$ is the product of all primes up to $x$, because we can estimate it naively as follows, if we take its logarithm:
$$A\_n=\exp\left(O\left(\sum\_{p>x}\frac{1}{p^2}\right)\right)=\exp\left(O\left(\frac{1}{x}\right)\right)$$.
| 4 | https://mathoverflow.net/users/31469 | 424193 | 172,340 |
https://mathoverflow.net/questions/424171 | 4 | In Hirschhorn's "Model Categories and Their Localizations", section 15.7, there's the following corollary to the preceding Proposition:
$\mathbb{Corollary}$ 15.7.2 If $\mathfrak{M}$ is a cellular model category and $\mathfrak{C}$ is a Reedy category, then the cofibrations of the Reedy model category on $\mathfrak{M}^{\mathfrak{C}}$ are monomorphisms.
I think we can use this to prove the following:
If we've got a Reedy model category $\mathfrak{A}$ and a category $\mathcal{B}$ in which all cofibrations are monomorphisms and weak equivalences are pointwise (i.e. just a rough way to say that it's equipped with an injective model structure) then in the category $\text{Psh}(\mathfrak{A},\text{Psh}(\mathcal{B},\textit{Ssets}))$ all Reedy cofibrations are monomorphisms.
Is there a way to have the converse work for some types of Reedy model categories? (By converse I mean all monomorphisms being Reedy cofibrations)
If I'm not mistaken, this will imply some sort of relation between model structures.
Edit:
Big thanks to Simon for pointing out the vague, unexplicit nature of the question.
| https://mathoverflow.net/users/482932 | Can Reedy cofibrations be monomorphisms? | I believe what you are after is the notion of "[elegant Reedy category](https://ncatlab.org/nlab/show/elegant+Reedy+category)"
This sort of things isn't true for a general Reedy category, but for an elegant one $R$ (see the link for the definition) if $\mathcal{E}$ is a Grothendieck topos (for e.g. simplicial presheaves on something) in which the cofibration are the monomorphisms, then in $\mathcal{E}^{R^{op}}$ the Reedy cofibration are exactly the monomorphism. This is explained on the nLab page linked and you'll also find reference about this there.
The last section of the nLab page gives many exemples of elegant Reedy category. In particular, Reedy categories that satisfies an analogue of the Eilenberg-Zilber lemma are elegant Reedy category (most exemple come from this).
| 4 | https://mathoverflow.net/users/22131 | 424195 | 172,342 |
https://mathoverflow.net/questions/424199 | 4 | Given a $\delta$-hyperbolic group $G$, I have been told that the Rips $n$-complex of $G$ is contractible for high enough $n$. The only proof I have found for this statement is in an [expository essay](https://warwick.ac.uk/fac/sci/maths/people/staff/lopezdegamiz/hyperbolic_groups.pdf) by Jone Lopez de Gamiz. I don't believe the proof there is independent, but I can't tell where it's from. Can someone direct me to a more original source for this theorem?
| https://mathoverflow.net/users/14257 | Looking for a citation: the Rips $n$-complex of a $\delta$-hyperbolic group is contractible for high enough $n$ | In his monograph *Hyperbolic groups* (1987), Gromov states and proves:
**Lemma 1.7.A.** *Let $X$ be a $\delta$-hyperbolic space
such that every $x\in X$ can be joined by a segment with
a fixed reference point $x\_0 \in X$. Then the polyhedron
$P\_d(X)$ is contractible for all $d \geq 4 \delta$.*
Here, the polyhedron $P\_d(X)$ is the Rips complex whose simplices are the finite collections of points in $X$ that are pairwise at distance $\leq d$.
This is probably the first published reference where the statement appears, even though the result is attributed to E. Rips. In the late 1980s, several texts have been dedicated to hyperbolic groups in general, and the the result can be found there too:
* E. Ghys, P. de la Harpe, *Sur les groupes hyperboliques d'après Mikhael Gromov* (1989).
* E. Ghys, *Les groupes hyperboliques*, [Séminaire Bourbaki](http://perso.ens-lyon.fr/ghys/articles/groupeshyperboliquesbourbaki.pdf) (1989-1990).
* Coornaert, Delzant, Papadopoulos, *Géométrie et théorie des groupes* (1990).
* E. Ghys, A. Haefliger, A. Verjovsky, *Group theory from a geometrical viewpoint* (1991).
But the result can also be found in more modern references, such Bridson and Haefliger's book *Metric spaces of non-positive curvature*, as already mentioned; or Drutu and Kapovich's book *Geometric group theory*.
| 9 | https://mathoverflow.net/users/122026 | 424202 | 172,343 |
https://mathoverflow.net/questions/424210 | 4 | Consider the quantum anharmonic oscillator, with Hamiltonian $H=p^2/2+q^2/2+gq^4$ for some real $g\geq 0$, with $p$ and $q$ obeying the usual Heisenberg commutation relations. For $g=0$, the ground energy is equal to $1/2$. Suppose $g>0$ and $g$ is algebraic. I would guess that the ground state energy is then not an algebraic number. Are there any results along these lines?
| https://mathoverflow.net/users/83174 | Ground state energy of anharmonic oscillator: algebraic or transcendental? | There is no exact expression for the ground state energy $E\_0$ for any nonzero $g$, but there are upper and lower bounds: for $g=1/2$ the upper bound for $2E\_0$ is 1.3923516415302918570 and the lower bound for $2E\_0$ is 1.3923516415302918502 , see [Upper and lower bounds of the ground state energy of anharmonic oscillators using renormalized inner projection.](https://doi.org/10.1063/1.529452) There is no indication that $E\_0$ can be expressed as the root of a polynomial, for all we know it's a transcendental number.
| 2 | https://mathoverflow.net/users/11260 | 424212 | 172,347 |
https://mathoverflow.net/questions/424216 | 2 | Given a real $n \times n$ matrix $A$ (feel free to assume its entries are non-negative, if it helps), what is known about the problem of computing the quantity
$$
\max\_{\sigma \in S\_n} \left\{\sum\_{j=1}^na\_{j,\sigma(j)}\right\}?
$$
(Here $S\_n$ is the symmetric group, so we are maximizing over all permutations of the columns of $A$: if you like you can write the sum as $\mathrm{trace}(AP\_\sigma)$, where $P\_\sigma$ is the permutation matrix associated with the permutation $\sigma$).
For example, is it known to be NP-hard, or is there a known polynomial-time algorithm for computing it (maybe just for certain special classes of matrices)? Does it have a name and/or is it equivalent to a well-known problem? It feels vaguely graph isomorphism-ish to me, but I'm having trouble pinning it down.
As one minor note, numerical examples show that the greedy algorithm (i.e., pick the largest entry of $A$, then forget about that row and column and pick the largest entry of the remaining $(n-1) \times (n-1)$ submatrix, and repeat) does not always produce the maximum value even if $A$ is doubly stochastic.
| https://mathoverflow.net/users/11236 | Maximum permuted row/column sum of a matrix | This is the maximum weight matching problem in the weighted complete bipartite graph $K\_{n,n}$, also known as the [assignment problem](https://en.wikipedia.org/wiki/Assignment_problem). It does have a few polynomial-time algorithmic solutions.
| 10 | https://mathoverflow.net/users/7076 | 424217 | 172,348 |
https://mathoverflow.net/questions/424165 | 6 | I am currently reading Rolfsen's "Knots and Links". At page 82 Whitehead manifold $W$ is defined and an exercise asking to show that $W\times \mathbb{R}\cong \mathbb{R}^4$ is left. Reference in Wiki says that this can be proved using Generalized Schonflies theorem, can anyone give any hint or material about this? Appreciation to your help!
| https://mathoverflow.net/users/483712 | How to prove the product of Whitehead manifold and $\mathbb{R}$ is homeomorphic to $\mathbb{R}^4$? | I would suggest a classical proof showing that the one-point compactification of $W$ is a manifold factor. See [Wild wild whitehead manifold](https://www.ams.org/journals/notices/201904/rnoti-p581.pdf). The proof is originally due to J. Andrews and L. Rubin, Bull. Amer. Math. Soc. 71(1965), 675-677. Once you understand that, you will see why Whitehead manifold cross R is homeomorphic to $R^4$. The key is the resulting manifold is simply connected at infinity. In general, the claim is true for any open contractible manifold (by combined forces).
For example,
J. Glimm, Two Cartesian products which are Euclidean spaces, Bull. Soc. Math. France 88 (1960), 131–135.
D. R. McMillan, Jr., Cartesian products of contractible open manifolds, Bull. Amer. Math. Soc. 67 (1961), 510–514.
J. Stallings, The piecewise-linear structure of Euclidean space, Proc. Cambridge Philos. Soc. 58 (1962), 481–488.
E. Luft, On contractible open topological manifolds, Invent. Math. 4 (1967), 192–
201.
E. Luft, On contractible open 3-manifolds, Aequationes Math. 34 (1987), 231–
239.
M. H. Freedman, The topology of four-dimensional manifolds, J. Differential
Geom. 17 (1982), 357–453.
However, the proofs are mainly focused on the study of the fundamental group at infinity (which is unlike the "unknotting" technique used in Andrews-Rubin's proof).
| 4 | https://mathoverflow.net/users/114032 | 424221 | 172,350 |
https://mathoverflow.net/questions/424214 | 4 | Let $S$ be a compact complex surface. It is well-known that the following two facts are equivalent
1. $c\_1^2(S) = 3 c\_2(S)$ and $S \neq \mathbb{CP}^2$
2. The universal cover of $S$ is biholomorphic to the unit ball in $\mathbb{C}^2$.
The unit ball in $\mathbb{C}^2$ is biholomorphic the complex hyperbolic plane $\mathbf{H}^2\_{\mathbb{C}}$ and its group of biholomorphisms is $\mathrm{PU}(1,2)$. Consequently, a compact complex surface is the same thing as the datum of a torsion-free, co-compact lattice in $\mathrm{PU}(1,2)$.
My question is : are there constructions of such surfaces that do not transit through lattices in $\mathrm{PU}(1,2)$? For instance, are there constructions of surfaces satisfying $c\_1^2(S) = 3 c\_2(S)$ coming from algebraic geometry, or from ramified covering-type of construction?
Thanks!
| https://mathoverflow.net/users/25511 | Constructions of complex surfaces covered by the ball of $\mathbb{C}^2$ | Below are two book references, both originating in the 1983 paper by Hirzebruch:
*Hirzebruch, Friedrich*, Arrangements of lines and algebraic surfaces, Arithmetic and geometry, Pap. dedic. I. R. Shafarevich, Vol. II: Geometry, Prog. Math. 36, 113-140 (1983). [ZBL0527.14033](https://zbmath.org/?q=an:0527.14033).
*Barthel, Gottfried; Hirzebruch, Friedrich; Höfer, Thomas*, Geradenkonfigurationen und algebraische Flächen. (Configurations of lines and algebraic surfaces). Eine Veröffentlichung des Max-Planck-Instituts für Mathematik, Bonn, Aspects of Mathematics, D4. Braunschweig/Wiesbaden: Friedr. Vieweg & Sohn. XII, 308 S.; (1987). [ZBL0645.14016](https://zbmath.org/?q=an:0645.14016).
*Tretkoff, Paula*, [**Complex ball quotients and line arrangements in the projective plane. With an appendix by Hans-Christoph Im Hof**](http://dx.doi.org/10.1515/9781400881253), Mathematical Notes (Princeton) 51. Princeton, NJ: Princeton University Press (ISBN 978-0-691-14477-1/pbk; 978-1-400-88125-3/ebook). ix, 215 p. (2016). [ZBL1342.14001](https://zbmath.org/?q=an:1342.14001).
In this approach, ball quotients are constructed via a combination of blow-ups and (finite) ramified coverings of $P^2$.
| 6 | https://mathoverflow.net/users/39654 | 424246 | 172,360 |
https://mathoverflow.net/questions/424271 | 1 | Let $\Omega \subset \mathbb{R}^2$ be a domain which is "well behaved" (has all "wishable" properties), so as its boundary. For every $u \in C^\infty(\Omega,\mathbb{R})$, I would like to express the following integral
$$\int\_{\partial \Omega} \frac{((Du)^\perp \cdot n)^2}{|Du|^3} dS,$$
where $dS$ is the Hausdorff measure of $\partial \Omega$ and $n$ its unit normal, as an integral over $\Omega$ with the standard Lebesgue measure. Here, $(Du)^\perp=(\partial u/\partial x\_2, - \partial u/\partial x\_1)$ and $|Du|=\sqrt{(\partial u/\partial x\_1)^2 + (\partial u/\partial x\_2)^2 }$.
The presence of the normal might indicate that the divergence theorem could be of use, but the fact that the term is quadratic makes it hard to handle. Any ideas are appreciated!
| https://mathoverflow.net/users/121671 | Expressing the integral over boundary of a domain as an integral over the domain | $\newcommand{\Om}{\Omega}$Let
\begin{equation\*}
I\_u(\Om):=\int\_{\partial\Om} \frac{((Du)^\perp\cdot n)^2}{|Du|^3}\,dS.
\end{equation\*}
Then
\begin{equation\*}
I\_u(\Om)=\int\_\Om \frac{I\_u(\Om)}{|\Om|}\,dz
\end{equation\*}
if $|\Om|:=\int\_\Om dz>0$.
This trivial representation of $I\_u(\Om)$ as "an integral over $\Omega$ with the standard Lebesgue measure" is probably not what you had in mind when asking your question.
Apparently, you wanted the integrand to not depend on $\Om$ and you wanted the integral representation of $I\_u(\Om)$ to hold for all nice enough domains $\Om$. So then, what you wanted seems to be a formula of the form
\begin{equation\*}
I\_u(\Om)=\int\_\Om F(u)(z)\,dz \tag{1}\label{1}
\end{equation\*}
for some operator $F$ acting on nice enough functions $u$ and for all nice enough domains $\Om$.
However, such a representation of $F$ would be additive in the sense that, if $|\Om\_1\cap\Om\_2|=0$ for some nice enough domains $\Om\_1$ and $\Om\_2$, then $I\_u(\Om\_1\cup\Om\_2)=I\_u(\Om\_1)+I\_u(\Om\_2)$. But this is false for "almost all" nice functions $u$ and, say, $\Om\_1=[0,1]\times[0,1]$ and $\Om\_1=[1,2]\times[0,1]$ -- because in general the integrand $\frac{((Du)^\perp\cdot n)^2}{|Du|^3}$ is $>0$ almost everywhere on $\{1\}\times[0,1]$.
So, a representation of the form \eqref{1} is impossible.
| 2 | https://mathoverflow.net/users/36721 | 424275 | 172,367 |
https://mathoverflow.net/questions/424269 | 2 | In the end of the Abstract of the paper [Minsky and Papert - Unrecognizable Sets of Numbers](https://doi.org/10.1145/321328.321337), the authors write "…for every
infinite regular set $A$ there is a nonregular set $A'$ for which
$$ \lvert\pi\_A(n)-\pi\_{A'}(n)\rvert\leq 1\text{",} $$
where $\pi\_A(n)$ is the counting function for $A$.
But I don't find a reference in the paper. Also I want to know if the following statement is true or not: "…for every
infinite nonregular set $B$ there is a regular set $B'$ for which
$$ \lvert\pi\_B(n)-\pi\_{B'}(n)\rvert\leq 1\text{."} $$
If I understand right the "regular set" in this paper means "automatic set".
| https://mathoverflow.net/users/159935 | A question on regular sets | The second question has a negative answer. The asymptotic behavior of $\pi\_B(n)$ and $\pi\_{B'}(n)$ would be the same, and if $\pi\_B(n)$ satisfies any of the criteria for non-regularity on page 283 of the paper by Minski and Papert, then $\pi\_{B'}(n)$ would satisfy the same, and thus, cannot be regular.
For the first question, the relevant portion of the paper is Section 5, titled "Impossibility of a Converse".
| 2 | https://mathoverflow.net/users/483601 | 424278 | 172,369 |
https://mathoverflow.net/questions/424279 | 3 | I have a multivariate polynomial $P$ which is a product of $M$ low degree polynomials $p\_i$
$$P(x\_1, x\_2, \dotsc, x\_n) = \prod\_{i=1}^M p\_i(x\_1, x\_2, \dotsc, x\_n)$$
where the maximum degree of each $p\_i$ can be $4$. For example:
$$P(x\_1, x\_2, x\_3) = (1+x\_1x\_2^2x\_3)(x\_3^3+x\_1x\_2^3).$$
I need to find out whether a specific term $x\_1^{m\_1}x\_2^{m\_2}\dotsm x\_n^{m\_n}$ appears in $P$ or not. Can I find the answer without expanding the products explicitly?
| https://mathoverflow.net/users/402086 | Checking presence of a specific term in product polynomial | You can solve the problem via integer linear programming as follows.
Let $a\_{ijk}$ be the power of $x\_j$ in term $k$ of $p\_i$, that is, $p\_i = \sum\_k \prod\_j x\_j^{a\_{ijk}}$. Let binary decision variable $y\_{ik}$ indicate whether term $k$ of $p\_i$ is used to form the target term $\prod\_j x\_j^{m\_j}$. The target is attainable if and only if the following linear system is feasible:
\begin{align}
\sum\_k y\_{ik} &= 1 &&\text{for $i\in\{1,\dotsc,M\}$} \tag1\label1 \\
\sum\_{i,k} a\_{ijk} y\_{ik} &= m\_j &&\text{for $j\in\{1,\dotsc,n\}$} \tag2\label2
\end{align}
Constraint \eqref{1} selects exactly one term in each $p\_i$.
Constraint \eqref{2} forces the product of the selected terms to match the desired powers.
Note that this formulation works even if the powers $a\_{ijk}$ and $m\_j$ are not nonnegative integers.
| 4 | https://mathoverflow.net/users/141766 | 424282 | 172,370 |
https://mathoverflow.net/questions/413189 | 14 | Fixing a dimension $n \ge 4$, is the class of closed hyperbolic $n$-manifolds recursively enumerable?
Since hyperbolic manifolds are triangulable I can reformulate this in the following more explicit way: say $M\_1, \ldots, M\_m,\ldots$ is an enumeration of all triangulations of $n$-manifolds. Is there a Turing machine which outputs a sequence $i\_1, \ldots, i\_m, \ldots$ such that each $M\_{i\_k}$ is hyperbolic, every hyperbolic $n$-manifold occurs as one of the $M\_{i\_k}$ and no two of them are homeomorphic to each other?
This problem reduces to that of deciding whether a triangulated manifold is hyperbolic, since the homeomorphism problem for hyperbolic manifolds is decidable. This is well-known to be possible when $n$ is $2$ or $3$ but i haven't found any references for $n \ge 4$.
**Added later**: ~~actually it seems that it is not possible to list all triangulations of manifolds in dimensions $6$ and higher. The question still makes sense for n=4, 5.~~ While it may not be possible to get a complete list of all triangulation of all smooth manifolds it is possible (as outlined by HJRW in the comments) to get a list of triangulations which include all smooth manifolds at least once. So the question makes sense for all dimensions again.
| https://mathoverflow.net/users/32210 | Are hyperbolic $n$-manifolds recursively enumerable? | The class of closed hyperbolic manifolds is recursively enumerable. I’ll describe a terrible algorithm which nevertheless gives an enumeration.
A couple of basic facts: a hyperbolic $n$-manifold $M$ admits a triangulation by geodesic simplices, and the representation of the fundamental group into $PO(n,1)$ may be conjugated to have matrices with algebraic entries. The former follows from taking a dirichlet domain and subdividing, the latter from [Mostow rigidity](https://en.wikipedia.org/wiki/Mostow_rigidity_theorem?wprov=sfti1).
By taking the preimage of the triangulation in $\mathbb{H}^n$ (realized eg as one sheet of a hyperboloid in $\mathbb{R}^{n,1}$), we have a triangulation of $\mathbb{H}^n$ invariant under the action of $\pi\_1(M)$ acting by algebraic matrices. We now perturb the vertices of the triangulation equivariantly to be points with algebraic coordinates (this is why we use simplices rather than more complicated polyhedra, since they are stable under small perturbations).
We may recover $M$ from a finite amount of this information: choose one simplex from each orbit, and for each pair of faces that are glued together, there will be a matrix in $PO(n,1)$ with algebraic entries gluing the faces together by an isometry.
Conversely, if we have a collection of simplices in $\mathbb{H}^n$ with algebraic coordinates, and the faces are paired by isometries, then we may check that they give rise to an $n$-manifold using the Poincaré polyhedron theorem. For each codimension two face of the simplices, we check that the angles about the faces sum to $2\pi$. This is possible since all of the coordinates of the vertices are algebraic. What I have in mind here is a version of Poincaré’s theorem due to Seifert; see [Rob Riley’s paper.](https://www.jstor.org/stable/2007537)
To enumerate hyperbolic n-manifolds, recursively enumerate simplices in $\mathbb{H}^n$ with algebraic coordinates, together with pairings between the faces. Throw out the ones that don’t satisfy Poincaré’s theorem. Then throw out repeats eg by using [Sela’s algorithm](https://doi.org/10.2307/2118520) to eliminate manifolds with the same fundamental group.
(Let me know if you’d like more details on any aspect of this algorithm, I realize it’s very sketchy. )
| 5 | https://mathoverflow.net/users/1345 | 424283 | 172,371 |
https://mathoverflow.net/questions/424254 | 2 | Assume that $Q=(q\_{ij})$ is a $k\times k$ with $q\_{ij}\in \{0, 1\}.$ The two side subshift of finite type associated to the matrix $Q$ is a left shift map $T:\Sigma\_{Q}\rightarrow \Sigma\_{Q}$, where
$$\Sigma\_{Q}:=\{x=(x\_{i})\_{i\in \mathbb{Z}} : x\_{i}\in \{1,...,k\} \hspace{0.2cm}\textrm{and} \hspace{0.1cm}Q\_{x\_{i}, x\_{i+1}}=1 \hspace{0.2cm}\textrm{for all}\hspace{0.1cm}i\in \mathbb{Z}\}.$$
It is known $\Sigma\_{Q}$ is compact and $T$-invariant. We denote by $\mathcal{L}$ the set of admissible words. We also denote by $\mathcal{L}\_{n}$ the set of admissible words with length $n.$
Denote $X:=\cup\_{i=1}^{m} \mathcal{L}\_{i}$ for some $m \in \mathbb{N}.$ Is $T':X^{\mathbb{Z}} \to X^{\mathbb{Z}}$ a two sided subshift of finite type?If so, what is the relation between $T$ and $T'$?
| https://mathoverflow.net/users/127839 | Union of admissible words are subshift of finite type | So, I probably did not initially understand you correctly. Let me analyze four interpretations of your construction; the first is what I thought first, the second gives something uninteresting, the third gives something uninteresting, the fourth is now my best guess of what you meant (you may want to jump there first to check). Definitions of words used can be found in [1].
**Interpretation 1.**
We take $X$ to be the words of *exactly* length $n$ rather than words of length *up to* $n$, and by $X^{\mathbb{Z}}$ we mean all concatenations of words from $X$ in the usual sense of formal language theory. This is what I guessed at first, but this interpretation *is quite unlikely to be what you mean*, because it is simply not what you wrote. With this interpretation your question contains a (slightly) nontrivial mathematical problem, which is probably why I initially guessed this one. I rewrite slightly more carefully (and more correctly) what I wrote in the comments.
In this interpretation, you take $X^{\mathbb{Z}}$ to mean all $x \in \{1,\ldots,k\}^{\mathbb{Z}}$ such that for some $P \subset \mathbb{Z}$, and choices $w\_p \in X$ for $p \in P$, you have for $p<q$ consecutive in $P$ that the length $|w\_p| = |q - p|$, and $x\_{p+i} = (w\_p)\_i$ for $i < |w\_p|$ (where the word $w\_p$ is indexed starting from $0$). So literally concatenations of words from $X$ (but you can't tell where words start!)
With this interpretation, what we get is that $X^{\mathbb{Z}}$ is a *renewal system* (just by definition; the definition of a renewal system is you take any finite set of words $X$ and consider $X^{\mathbb{Z}}$ in the above sense).
In general, renewal systems are always sofic (meaning factors of subshifts of finite type, equivalently sets of labels of bi-infinite paths in edge-labeled finite directed graphs), and they are topologically mixing.
They need not in general be of finite type, and indeed in you construction, you can get a proper sofic one. Namely, if $\Sigma\_Q$ is the golden mean shift meaning it comes from the matrix $Q = \left(\begin{smallmatrix} 1 & 1 \\ 1 & 0 \end{smallmatrix}\right)$ as a vertex shift (what you refer to as a subshift of finite type is usually called a vertex shift), setting $m = 2$ we would get $X = \{11, 12, 21\}$, and now for all $n$, $X^{\mathbb{Z}}$ has admissible word $12211^{2n}1221$ but not $12211^{2n+1}1221$, so it is not of a subshift of finite type.
You can get a "formula" for the renewal system easily in the sense that there is a simple algorithm that produces a presentation. Of course $X$ itself is a reasonable presentation of a renewal system, but to get it in a form that more explicitly represents a sofic shift, just take a graph where $|X|$ cycles of length $m$ intersect in a single vertex, and edge-label the $i$th cycle by the $i$th word in $X$. (You can then simplify this presentation if you like.)
(If you insist on representing sofic shifts as matrices, I think there are things to say about that but I'll stop here.)
**Interpretation 2.**
We take the same interpretation of $X^{\mathbb{Z}}$, but $X$ is what you wrote, i.e. words up to length $m$. This interpretation doesn't give anything interesting: $X^{\mathbb{Z}}$ is equal to $A^{\mathbb{Z}}$ where $A = \{w \in X \;|\; |w| = 1\}$. I.e. it's just a full shift over the alphabet $\{1,\ldots,k\}$ (or a subset of that in case $Q$ was not essential).
**Interpretation 3.**
You take $X^{\mathbb{Z}}$ to just mean $A^{\mathbb{Z}}$ where the alphabet $A$ happens to be the set $X$, so the action of $T'$ shifts "by one word" on each time step. Then $X^{\mathbb{Z}}$ is just the full shift on $|X|$ letters.
**Interpretation 4.**
We take $X$ to be what you wrote (so words up to length $m$), but we take $X^{\mathbb{Z}}$ to mean "marked" concatenations, so e.g. let's say the letters in $Q$ are "lowercase", and define $Y$ as $X$ but capitalizing each word. So e.g. in the golden mean example with $m = 2$, renaming the letters as $a = 1, b = 2$, we would have $X = \{a, b, aa, ab, ba\}$ and $Y = \{A, B, Aa, Ab, Ba\}$. (If you don't find the capitalization natural, you can also imagine markers (of length zero) between the words.) Then take $X^{\mathbb{Z}}$ to mean $Y^{\mathbb{Z}}$ (and then that is interpreted as in Interpretation 1, i.e. as the set of bi-infinite concatenations).
This $Y^{\mathbb{Z}}$ is always a topologically mixing subshift of finite type (now subshift of finite type is in the general sense that it's defined by a finite set of forbidden words), and there is a rather simple formula for it.
First of all, its isomorphism type (up to topological conjugacy) is completely determined by how many words of length $n$ there are in $\Sigma\_Q$: you can tell where words start and you can tell them apart, so you can just as well rename them to be "word 1 of length 1, word 2 of length 1, ..., word $i$ of length $n$, ...". Let $N\_n$ be the number of words of length $n$.
So now, $X^{\mathbb{Z}}$ is, up to conjugacy, obtained by simply taking $N\_n$ cycles of length $n$ that meet in a single vertex, and taking (unlabeled!) bi-infinite paths in that graph. In the polynomial matrix presentation explained at least in [2], you can represent this very compactly: it is simply the the $1$-by-$1$ matrix with the unique entry $\sum\_{n = 1}^m N\_n t^n$. In this representation, $t^n$ represents a path of length $n$ between two vertices, and in our case we just want to have paths loop around a single vertex. I didn't try to work out nice formula for an vertex/edge shift presentation in terms of $Q$, there might be a very simple formula for that too. (It is explained in [2] how a polynomial matrix is turned into an edge or vertex shift, but you don't necessarily get what you'd call a "formula" for the resulting integer matrix.)
(Side note: If we used edge shifts instead of vertex shifts there is a very nice formula for $N\_i$; not sure there is one with vertex shifts. In the edge shift presentation, $Q$ is a $k$-by-$k$ matrix over natural number entries ($\mathbb{N} = 0,1,2,\ldots$), and gives rise to a graph with $k$ vertices and $Q\_{a,b}$ edges from vertex $a$ to vertex $b$, and $\Sigma\_Q$ is just the bi-infinite paths in this graph. Now the number of admissible words of length $n$ is simply $N\_n = |Q^n|\_1$, the $1$-norm = sum of entries in the graph. Edge shifts are the same as vertex shifts are the same as general subshifts of finite type *up to topological conjugacy*, but I don't know if you can *exactly* mimic the word counts of a vertex shift with an edge shift; all edge shifts can directly be seen as vertex shifts, but not vice versa.)
[1] *Lind, Douglas; Marcus, Brian*, [**An introduction to symbolic dynamics and coding**](http://dx.doi.org/10.1017/9781108899727), [ZBL07279890](https://zbmath.org/?q=an:07279890).
[2] *Boyle, M.; Wagoner, J. B.*, Positive algebraic (K)-theory and shifts of finite type, Brin, Michael (ed.) et al., Modern dynamical systems and applications. Dedicated to Anatole Katok on his 60th birthday. Cambridge: Cambridge University Press (ISBN 0-521-84073-2/hbk). 45-66 (2004). [ZBL1148.37303](https://zbmath.org/?q=an:1148.37303).
| 1 | https://mathoverflow.net/users/123634 | 424290 | 172,374 |
https://mathoverflow.net/questions/424200 | 6 | Let $X$ be a Hausdorff compact space, and let $\mathrm {Ba}$, $\mathrm {Bo}$ be its Baire, respectively, Borel, $\sigma$-algebras. Let $\mu:\mathrm {Ba}\to[0,+\infty)$ be a finite Baire measure: it is well-known that it extends uniquely to a *regular* Borel measure $\tilde\mu:\mathrm {Bo}\to[0,+\infty)$ (since $\mu$ induces a positive linear functional on $C(X)$, this follows from the Riesz-Markov representation theorem, as e.g. stated in Rudin's *Real and Complex analysis*, thm 2.14). It follows from the regularity that the measure $\tilde\mu:\mathrm {Bo}\to[0,+\infty)$ also satisfies:
>
> **1.** For any $B\in \mathrm {Bo}$ there exists $A\in \mathrm {Ba}$ such that $\tilde\mu(A\Delta B)=0$.
>
>
>
***Question.*** Is the following stronger statement true, i.e.
>
> **2.** For any $B\in \mathrm {Bo}$ there exists $A\in \mathrm {Ba}$ such that $B\subset A$ and $\tilde\mu(A\setminus B)=0$.
>
>
>
In other words: *Is it the case that the completion of the measure space $(X,\mathrm {Ba},\mu)$ is an extension of $(X,\mathrm {Bo},\tilde\mu)$?*
In case the answer is ***yes***, I would be especially happy with an elementary proof independent from the R-M theorem of the fact that $\mathrm {Bo}$ is included in the $\mu$-completion of $\mathrm {Ba}$.
In case the answer is ***no***: Is there an abstract measure-theoretic construction, for abstract measure spaces $(X,\mathcal A,\mu)$, producing an extension into a complete measure space $(X,\mathcal{\tilde A},\tilde\mu)$ verifying the analog of **(1)**, that is: For any $B\in\mathcal{\tilde A}$ there exists $A\in \mathcal A$ such that $\tilde\mu(A\Delta B)=0$, and such that in the case of a finite Baire measure $\mu$ it gives $\tilde{\mathrm{Ba}}\supset{\mathrm{Bo}} $?
***(Edit June 9, 2022)*** Since the answer is no, let me elaborate on point 2. What if one does the Carathéodory construction starting from the measure space $(X,\text{Ba},\mu)$? That is, consider $\mu^\*:P(X)\to[0,+\infty)$, the maximum exterior measure which is below $\mu$ on $\text{Ba}$, i.e. $\mu^\*(E)=\inf\{\mu(B):E\subset B\in\text{Ba}\}$, and the $\sigma$-algebra $\mathcal C$ of all Carathéodory $\mu^\*$-measurable sets. At least for the given examples, $\mathcal C\supset \text{Ba}$. Is this always true? By topological reasons (normality of $X$), $\mu^\*$ is $\sigma$-additive on the family of closed sets; moreover (by regularity of $\tilde\mu$), it must coincide with the regular Borel extension $\tilde\mu$ on closed sets.
| https://mathoverflow.net/users/6101 | Extending a finite Baire measure to a regular Borel measure | Let $X=\{0,1\}^{\aleph\_1}$ and $x\in X$, and let $\mu$ be the Borel probability measure such that $\mu(\{x\})=1$. Since $\{x\}$ itself is not a Baire set, there is no Baire set $A\subseteq \{x\}$ such that $\mu(A)=1$. By taking the complements, we get answer **no** for the question.
For the second question: The usual way of extending a Baire measure to the Borel sets relies on the measure being $\tau$-additive, which is a topological concept. So it is likely that an abstract version would need some kind of property with a topological flavour.
| 6 | https://mathoverflow.net/users/95282 | 424295 | 172,376 |
https://mathoverflow.net/questions/424314 | 0 | Let $L/\mathbb{Q}$ be a finite extension and $f\_{1},\dotsc,f\_{n}\in L[x\_{1},\dotsc,x\_{k}]$ be degree $d$ homogeneous polynomials. Is there a way to find homogenous degree $d’$ polynomials $g\_{1},\dotsc,g\_{n}\in L[x\_{1},\dotsc,x\_{k}]$ such that $f\_{1}g\_{1}+\dotsc f\_{n}g\_{n}\in \mathbb{Q}[x\_{1},\dots,x\_{k}]$ and $d’<d$?
| https://mathoverflow.net/users/483823 | Changing base field for sum of polynomials | Not always. Take $k=3$, $n=2$, $f\_1=\sqrt{2}x\_1^2+x\_2^2$, $f\_2=x\_3^2$.
| 1 | https://mathoverflow.net/users/4312 | 424315 | 172,383 |
https://mathoverflow.net/questions/424187 | 1 | Let $Z$ be a finite set and $\mathcal S$ be a finite set of vectors in $Z^n$.
For each $z\in Z$ let $x\_z$ be a formal variable. A **distribution** (please excuse abuse of language) $d \in \mathbf N^n$ is a vector with nonnegative integer entries, not all zero, such that the set $\{\sum\_{i=1}^n d\_i x\_{s\_i} | s\in \mathcal S\}$ consists of a single element.
For example, if
$$\mathcal S = \{(1, 2, 2, 3), (1, 2, 3, 2), (2, 1, 3, 1)\},$$ then $d=(1,1,0,0)$ is a distribution, and so is $(1,0,1,1)$. If $$\mathcal S = \{(1,1,2,3),(1,2,1,3),(2,3,3,1)\},$$ then $(1,1,1,2)$ is a distribution.
A distribution $d$ is **minimal** if there is no distribution $d'$ such that $d'\_i \leq d\_i$ for all $i\in\{1,\dots,n\}$ and $d' \neq d$.
The distributions in the examples above are minimal.
I am looking for a set $\mathcal S$ with the following properties:
* there are no two positions $1\leq i < j\leq n$ with $s\_i = s\_j$ for all $s\in\mathcal S$.
* there are two minimal distributions $d\neq d'$ on $\mathcal S$ such that $d\_i > 0$ if and only if $d'\_i > 0$ for all $i\in\{1,\dots,n\}$.
Note that, without the first requirement, there are trivial examples of such sets.
For example, let $$\mathcal S = \{(1,1,2,2,2,1), (1,1,2,2,1,2), (1,1,2,1,2,2), (1,1,1,2,2,2), (2,2,1,1,1,1)\}.$$ Then $d = (2,1,1,1,1,1)$ and $d'=(1,2,1,1,1,1)$ are minimal distributions. The first condition is violated, because $s\_1 = s\_2$ for all $s\in\mathcal S$.
| https://mathoverflow.net/users/3032 | are minimal subdistributions determined by their support? | Extracted from the comments:
there are examples of the following type: $Z=\{0,1\}$, the set $\mathcal{S} $ consists of sequences $(t\_1,\ldots,t\_n)$ satisfying two different relations of the form $\sum c\_i t\_i=C$, where $c\_i$ are positive integers and $C$ is a positive integer (it is natural to choose it close to $\frac12 \sum c\_i$). Then $(c\_1,\ldots,c\_n)$ is a distribution, and it is easy to come up with example when two such distributions are both minimal and other conditions hold.
| 1 | https://mathoverflow.net/users/4312 | 424321 | 172,385 |
https://mathoverflow.net/questions/424339 | 1 | Let $\lambda>0$ be given. Define
$$G\_{\lambda}(\xi) = \chi\_{\_{\lbrace |\xi|^{2} \leq \lambda \rbrace }}.
$$
and
$$
E\_{0}(\lambda)f = \mathcal{F}^{-1}[G\_{\lambda}(|\xi|^{2})\mathcal{F}(f)], \ \ f \in L^{2}(\mathbb{R}^{n})
$$
How do I show that
$$(E\_{0}(\lambda)f|f) = \|E\_{0}(\lambda)f\|^{2}\_{L^{2}}, \ \ f \in L^{2}(\mathbb{R}^{n})\qquad ?$$
| https://mathoverflow.net/users/481556 | Spectral family associated with the Laplacian operator in $L^{2}(\mathbb{R}^{n})$ | Your identity amounts to
$$\frac12(E\_0(\lambda)^\*+E\_0(\lambda))=E\_0^\*(\lambda) E\_0(\lambda).$$
Since ${\cal F}^\*=\cal F$, this is equivalent to saying that $$\frac12(G\_\lambda+\overline{G}\_\lambda)=|G\_\lambda|^2,$$
which is true because $G\_\lambda(\xi)$ equals either $0$ or $1$.
**Edit**. To explain the first equality, let us define the bounded linear operator
$$L:=E\_0(\lambda)-E\_0^\*(\lambda) E\_0(\lambda).$$
Your assumption is that $(Lf\mid f)=0$ for every $f$, which means that $L^\*+L=0$.
| 2 | https://mathoverflow.net/users/8799 | 424341 | 172,392 |
https://mathoverflow.net/questions/424340 | 1 | I have the following question, which I am thinking about for days now and can't get the answer right. I have a sequence of elements in this order $x\_{1},x\_{2},...,x\_{2n}$, $n \ge 1$ and then I perform a permutation $\pi$ on this set, so it becomes $x\_{\pi(1)},x\_{\pi(2)},...,x\_{\pi(2n)}$. Let us denote the sign of this permutation by $\varepsilon\_{\pi}$.
Take the element $x\_{1}$ and some arbitrary other element $x\_{i}$ on the set $\{x\_{2},x\_{2},...,x\_{2n}\}$. This is what happens now: after the permutation $\pi$, I erase the elements $x\_{1}$ and $x\_{i}$ of the new ordered sequence, so I end up with something like $x\_{\pi(1)},x\_{\pi(2)},\cdots, \hat{x}\_{1},...,\hat{x}\_{i},...,x\_{\pi(2n)}$, where the hat $\hat{x}\_{l}$ means to omit the element $x\_{l}$. This new sequence has now $2n-2$ elements and I want to bring it to its original form where the subindices are ordered in increasing order, only now the elements $x\_{1}$ and $x\_{i}$ are obviously not present anymore. Thus I need another permutation, say $\sigma$, which permuts $2n-2$ elements, not $2n$ anymore.
So my question is: how can I relate the sign of $\sigma$ with the sign of $\pi$? Is there any relation between these two?
| https://mathoverflow.net/users/235802 | Sign of the permutation which brings a subsequence back to its original form | Imagine a bubble sort where you bring each element to its original position. $x\_1$ would have taken $\pi^{-1}(1) - 1$ transpositions to bring it back to its original position. Let's perform those transpositions, even though we'll discount them in a moment. $x\_i$ would have taken $\lvert\pi^{-1}(i) - i\rvert$ transpositions to bring it back to its original position if we had left $x\_1$ in place. Your notation $x\_{\pi(1)}, \dotsc, \hat x\_1, \dotsc, \hat x\_i, \dotsc, x\_{\pi(2n)}$ suggests you want to assume that $\pi^{-1}(i)$ is greater than $\pi^{-1}(1)$, but I assume that's just a quirk. In any case, with $x\_1$ moved back to its original position, we require $\lvert\pi^{-1}(i) - i\rvert - 1$ transpositions to move $i$ back to its proper position if $\pi^{-1}(1)$ is between $\pi^{-1}(i)$ and $i$, and $\lvert\pi^{-1}(i) - i\rvert$ otherwise.
Thus, discounting these transpositions, the sign of the permutation required to re-sort the permuted list after removing the re-placed $x\_1$ and $x\_i$ is $\varepsilon\_\pi\cdot(-1)^{\pi^{-1}(1) + \lvert\pi^{-1}(i) - i\rvert}$ if $\pi^{-1}(1)$ is between $\pi^{-1}(i)$ and $i$, and $\varepsilon\_\pi\cdot(-1)^{\pi^{-1}(1) - 1 + \lvert\pi^{-1}(i) - i\rvert}$ otherwise.
| 4 | https://mathoverflow.net/users/2383 | 424344 | 172,393 |
https://mathoverflow.net/questions/424351 | 1 | I am reading Engelen´s paper and have trouble with this proof of Lemma 2.1 (a) (link is below).
>
> It is easily seen that any non-empty open subspace $U$ of $\mathbb{Q}^\infty$ can be
> written as an infinite disjoint union of non-empty basic clopen
> subsets; hence, $U = \mathbb{N} \times \mathbb{Q}^\infty \simeq \mathbb{Q}^\infty$.
>
>
>
In particular, I don't see **how the $U$ can be written as the said disjoint union and how that implies that $U \simeq \mathbb{Q}^\infty$**.
Can anyone please help with this or provide any reading source? I would be really grateful.
**Definition:**
$\mathbb{Q}^\infty$ is defined as a set of all rational sequences, endowed with the standard product topology.
**Source:** Engelen - [Countable Product of zero-dimensional absolute $F\_{\sigma \delta}$ spaces](https://ur.booksc.me/book/8624681/d91b4b), Lemma 2.1 (a).
| https://mathoverflow.net/users/421149 | Why can any open subset $U$ of $\mathbb{Q}^\infty$ be written as disjoint union of basic clopen subsets? | We can first express the open set $U$ as a countable union of basic clopen subsets $A\_n=\{(q\_k)\_{k\in\mathbb{N}};i\_{n,k}<q\_k<j\_{n,k} \text{ for $k=1,\dots,n$}\}$, where $i\_{n,k}$ and $j\_{n,k}$ are irrational numbers. Some $A\_n$ may be empty.
To answer the question it is enough to express $A\_n\setminus\bigcup\_{i=1}^{n-1}A\_i$ as a finite union of disjoint basic clopens. To do it, we can work in $\mathbb{Q}^n$ instead of $\mathbb{Q}^\infty$ due to the construction of the sets $A\_n$.
Writing $\mathbb{Q}^n=\mathbb{Q}\_1\times\dots\times\mathbb{Q}\_n$, notice that for each $k\leq n$, the irrationals $i\_{1,k},\dots,i\_{n,k},j\_{1,k},\dots,j\_{n,k}$ give a partition of $\mathbb{Q}\_k$ into finitely many clopen subsets. Taking a product of these partitions, we obtain a partition of $\mathbb{Q}^n$ into disjoint clopens such that any of the $A\_i$, with $i\leq n$, is a finite union of these clopens. So $A\_n\setminus\bigcup\_{i=1}^{n-1}A\_i$ is also a finite union of the clopens.
If we pick the $A\_n$ so that no finite union of them covers $U$, we also get infinitely many non empty clopens.
This implies that $U\equiv\mathbb{Q}^\infty$ because any basic clopen set like the ones above is homeomorphic to $\mathbb{Q}^\infty$, which can be deduced using that any clopen subset of $\mathbb{Q}$ is homeomorphic to $\mathbb{Q}$ by [Sierpinski's theorem](https://math.wvu.edu/%7Ekciesiel/Other/ElectronicReprints/141.SierpinskiThmOnQReprint.pdf).
| 4 | https://mathoverflow.net/users/172802 | 424361 | 172,396 |
https://mathoverflow.net/questions/424298 | 0 | Every weakly null sequence in a Banach space, as a subset, is clearly relatively weakly compact. To quantify the elementary fact, we need the following quantities:
$$\delta\_{0}((x\_{n})\_{n}):=\sup\_{x^{\*}\in B\_{X^{\*}}}\limsup\_{n}|\langle x^{\*},x\_{n}\rangle |$$ for a bounded sequence $(x\_{n})\_{n}$ of a Banach space $X$, where $B\_{X^{\*}}$ is the closed unit ball of $X^{\*}$.
For a bounded subset $A$ of $X$ we set:
$$\operatorname{wck}\_{X}(A)=\sup\{\textrm{d}(\textrm{clust}\_{X^{\*\*}}((x\_{n})\_{n}),X)\colon (x\_{n})\_{n}\text{ is a sequence in }A\}, $$
where $\textrm{clust}\_{X^{\*\*}}((x\_{n})\_{n})=\cap\_{n=1}^{\infty}\overline{\{x\_{m}:m>n\}}^{w^\*}$ is the set of all weak$^{\*}$-cluster points of $(x\_{n})\_{n}$ in $X^{\*\*}$. It follows from the Eberlein-Smulyan theorem that $\operatorname{wck}\_{X}(A)=0$ if and only if $A$ is relatively weakly compact.
Let $A$ and $B$ be non-empty subsets of a Banach space $X$, we set
$$\textrm{d}(A,B)=\inf\{\|a-b\|\colon a\in A,b\in B\},$$
$$\widehat{\textrm{d}}(A,B)=\sup\{\textrm{d}(a,B)\colon a\in A\}.$$
$\textrm{d}(A,B)$ is the ordinary distance between $A$ and $B$, and $\widehat{\textrm{d}}(A,B)$ is the (non-symmetrised) Hausdorff distance from $A$ to $B$. When $A$ is a bounded subset of $X$, we set
$$\textrm{wk}\_{X}(A)=\widehat{\textrm{d}}\big(\overline{A}^{\sigma(X^{\*\*},X^{\*})},X\big). $$
It is a direct consequence of the Banach-Alaoglu theorem that $A$ is relatively weakly compact if and only if $\textrm{wk}\_{X}(A)=0$.
Question 1. $\operatorname{wck}\_{X}(\{x\_{n}:n=1,2,\cdots\})\leq \delta\_{0}((x\_{n})\_{n})$ for every bounded sequence $(x\_{n})\_{n}$ of a Banach space $X$ ?
Question 2. $\operatorname{wk}\_{X}(\{x\_{n}:n=1,2,\cdots\})\leq \delta\_{0}((x\_{n})\_{n})$ for every bounded sequence $(x\_{n})\_{n}$ of a Banach space $X$ ?
Thank you !
| https://mathoverflow.net/users/41619 | Weakly null sequences in Banach spaces | I answer Question 1 by myself and I am sure my proof is correct.
Let $A=\{x\_{n}:n=1,2,\cdots\}$ and let $0<c<\operatorname{wck}\_{X}(A)$ be arbitrary. Then there exists a sequence $(z\_{n})\_{n}$ in $A$ so that $\textrm{d}(\textrm{clust}\_{X^{\*\*}}((z\_{n})\_{n}),X)>c$. Let $Y=\overline{\textrm{span}}\{x\_{n}\colon n=1,2,\ldots\}$ and $i\_{Y}\colon Y\rightarrow X$ be the inclusion map. Since $i\_{Y}^{\*\*}\colon Y^{\*\*}\rightarrow X^{\*\*}$ is an isometric embedding, we get $$\textrm{d}(\textrm{clust}\_{Y^{\*\*}}((z\_{n})\_{n}),Y)\geqslant \textrm{d}(\textrm{clust}\_{X^{\*\*}}((z\_{n})\_{n}),X)>c.$$ Let $\varepsilon>0$.
Take any $y^{\*\*}\_{0}\in \textrm{clust}\_{Y^{\*\*}}((z\_{n})\_{n})$ and let $d=\textrm{d}(y^{\*\*}\_{0},Y)$. By the Hahn-Banach theorem, there exists $y^{\*\*\*}\_{0}\in S\_{Y^{\*\*\*}}$ so that $\langle y^{\*\*\*}\_{0},y^{\*\*}\_{0}\rangle=d$ and $\langle y^{\*\*\*}\_{0},y\rangle=0$ for all $y\in Y$. We let
$$C=B\_{Y^{\*}}\cap \{y^{\*\*\*}\in Y^{\*\*\*}\colon |\langle y^{\*\*\*},y^{\*\*}\_{0}\rangle-d|<\varepsilon\}.$$ By Goldstine's theorem, $y^{\*\*\*}\_{0}\in \overline{C}^{\sigma(Y^{\*\*\*},Y^{\*\*})}$. Since $\langle y^{\*\*\*}\_{0},y\rangle=0$ for all $y\in Y$, we get $0\in \overline{C}^{\sigma(Y^{\*},Y)}$. Since $Y$ is separable, there exists a weak$^{\*}$ null sequence $(f\_{m})\_{m}$ in $C$. By passing to a subsequence, we may assume that the limit $\lim\limits\_{m}\langle y^{\*\*}\_{0},f\_{m}\rangle$ exists, which is denoted by $a$. By the definition of $C$, $|a-d|\leqslant \varepsilon$. Since $y^{\*\*}\_{0}\in \textrm{clust}\_{X^{\*\*}}((z\_{n})\_{n})$, we get a subsequence $(y\_{n})\_{n}$ of $(z\_{n})\_{n}$ so that $|\langle y^{\*\*}\_{0}-y\_{n},f\_{m}\rangle|<\frac{1}{n}$ for $m=1,2,\ldots,n$. This implies that $\lim\limits\_{n\to\infty}\langle f\_{m},y\_{n}\rangle=\langle y^{\*\*}\_{0},f\_{m}\rangle$ for each $m$ and then $\lim\limits\_{m\to\infty}\lim\limits\_{n\to\infty}\langle f\_{m},y\_{n}\rangle=a$.
Claim. $|a|\leq \delta\_{0}((x\_{n})\_{n})$.
Case 1. the set $\{y\_{n}:n=1,2,\cdots\}$ is finite.
In this case, there exists a subsequence $(y\_{k\_{n}})\_{n}$ and $y\_{0}\in Y$ so that $y\_{k\_{n}}=y\_{0}$ for all $n$. Hence $$a=\lim\_{m}\lim\_{n}\langle f\_{m},y\_{k\_{n}}\rangle=\lim\_{m}\langle f\_{m},y\_{0}\rangle=0.$$ The claim holds trivially.
Case 2. the set $\{y\_{n}:n=1,2,\cdots\}$ is infinite.
In this case, we get two strictly increasing sequences $(k\_{i})\_{i},(l\_{i})\_{i}$ of positive integers so that $y\_{k\_{i}}=x\_{l\_{i}}(i=1,2,\cdots)$. Hence, for each $m$, we get
$$|\langle y^{\*\*}\_{0},f\_{m}\rangle|=\lim\limits\_{i\to\infty}|\langle f\_{m},y\_{k\_{i}}\rangle|\leq \delta\_{0}^{Y}((x\_{n})\_{n}):=\sup\_{y^{\*}\in B\_{Y^{\*}}}\limsup\_{n}|\langle y^{\*},x\_{n}\rangle|.$$ By the Hahn-Banach theorem, we get $|a|\leq \delta\_{0}((x\_{n})\_{n})$.
By Claim, we get $d\leq \delta\_{0}((x\_{n})\_{n})+\varepsilon$ and so is $c$. Letting $\varepsilon\rightarrow 0$, we get $c\leq \delta\_{0}((x\_{n})\_{n})$. As $c$ was arbitrary, the proof is complete.
| 0 | https://mathoverflow.net/users/41619 | 424383 | 172,401 |
https://mathoverflow.net/questions/424403 | 0 | Let $(a\_{n})\_{n}$ be a bounded real-valued sequence. Suppose that $(b\_{n})\_{n}$ is a sequence (not necessarily a subsequence) in the set $A:=(a\_{n})\_{n}$. Assume that the limit $\lim\limits\_{n}b\_{n}$ exists. Moreover, assume that $(b\_{n})\_{n}$ is not eventually constant (we say that a sequence $(c\_{n})\_{n}$ is eventually constant if there exists a positive integer $N$ so that $c\_{n}=c\_{N+1}$ for all $n>N$). My question is the following:
Question. $\lim\limits\_{n}b\_{n}\leq \limsup\limits\_{n}a\_{n}$ ?
Thank you!
| https://mathoverflow.net/users/41619 | the limit of a scalar sequence in a sequence of scalars | Yes. Denote $B=\{b\_n\colon n=1,2,\ldots\}$. If $B$ is finite, then, since $\lim b\_n$ exists, the sequence $(b\_n)$ is eventually constant. If $B$ is infinite, there exists a subsequence $n\_1<n\_2<\ldots$ such that all $b\_{n\_j}$ are distinct. Denote $b\_{n\_j}=a\_{m\_j}$. All $m\_j$ are distinct, thus $\lim\_j a\_{m\_j}\leqslant \limsup a\_n$.
| 0 | https://mathoverflow.net/users/4312 | 424404 | 172,404 |
https://mathoverflow.net/questions/424415 | 5 | If we violate the partition principle and add to $\sf ZF$ the axiom that there exists a set $X$ that has a partition on it that is greater in cardinality than the set of singleton subsets of $X$.
>
> Can $X$ be a standard stage $V\_\alpha$ of the cumulative hierarchy?
>
>
>
>
> Can $X$ be a non-standard stage $V\_{\alpha'}$ of the cumulative hierarchy?
>
>
>
About the second question what I mean is there is a non-well founded model of $\sf ZF$ such that $\alpha'$ is seen as an ordinal internally but externally it is a transitive set of transitive sets but has subsets of it that are infinite descending membership chains.
| https://mathoverflow.net/users/95347 | Can a stage of the cumulative hierarchy violate the partition principle? | If you are asking whether or not a $V\_\alpha$ could violate the partition principle, the answer is easily yes.
As we all know, it is always the case that $\Bbb R$ can be partitioned into $\aleph\_1$ parts; but it is consistent that there is no injection from $\omega\_1$ into the reals. Next, observe that $V\_{\omega+1}$ is exactly the same cardinality as $\Bbb R$.
| 4 | https://mathoverflow.net/users/7206 | 424417 | 172,406 |
https://mathoverflow.net/questions/424410 | 4 | Let $H^1$ be the Hardy space on $\mathbb{R}^n$, defined e.g. as the set of $u\in L^1$ such that $Ru\in L^1$, where $R$ is the Riesz transform on $\mathbb{R}^n$. It seems to me that simple functions with bounded support and average 0 are dense in $H^1$ (simple functions = finite linear combinations of characteristic functions). But I grew suspicious since I can not find any mention of this fact in the literature. If true this would be very useful, e.g. it implies a short proof that the complex interpolation between $H^1$ and $L^\infty$ is an $L^p$ space.
So, can anyone confirm/disprove/point at the relevant literature?
| https://mathoverflow.net/users/7294 | Dense subspaces of the Hardy space $H^1$ | It seems to be true. Due to the atomic decomposition, it is enough to approximate in $H^1$ any "$(1,\infty)$"-atom, that is, a function $a$ such that $$\mathrm{supp}\, a \subset Q,\quad \|a\|\_{L^\infty}\leq |Q|^{-1}, \quad \int\_Q a=0$$
for a cube $Q$.
In order to do this, we can cut $Q$ into $N=k^n$ smaller equal cubes $Q\_1, Q\_2, \ldots, Q\_N$ for a large $k$. We put
$$
b\_N(x)=\sum\_{j=1}^N \Big( \frac{1}{|Q\_j|}\int\_{Q\_j}a(y)\,dy\Big) \chi\_{Q\_j}(x).
$$
Put $g\_N=a-b\_N$. Due to Lebesgue differentiation theorem we have
$$
\lim\limits\_{N\to \infty}\int |g\_N(x)|^2dx = 0.
$$
Then for big $N$ the following holds:
$$
\mathrm{supp}\, g\_N\subset Q, \quad \|g\_N\|\_{L^2}\leq \varepsilon |Q|^{-1/2}, \quad \int\_Q g\_N = 0.
$$
It means that $\varepsilon^{-1} g\_N$ is a $(1,2)$-atom. Hence, $\|g\_N\|\_{H^1}\leq C\varepsilon$.
The atomic decomposition in $H^1$ is described in the book "Weighted norm inequalities and related topics" by Garcia-Querva and Rubio de Francia. Another reference is "Modern Fourier Analysis" by Grafakos.
| 3 | https://mathoverflow.net/users/69086 | 424419 | 172,407 |
https://mathoverflow.net/questions/424374 | 9 | This question is a request for assistance in surveying the existing literature on applications of Lawvere's Fixed Point Theorem (LFPT).
Yanofsky [0] has demonstrated several applications of LFPT to prove various limitative results, in particular Goedel's First Incompleteness Theorem, and alludes to its applicability for proofs of the Second Incompleteness Theorem:
>
> Goedel’s second incompleteness theorem about the unprovability within
> arithmetic of the consistency of arithmetic. This theorem is a simple
> consequence of the first incompleteness theorem. However Kreisal has
> a direct model theoretic proofs that uses a diagonal method (see, e.g.,
> page 860 of Smorynski’s article in [1].) This proof seems amenable to our
> scheme.
>
>
>
Has there been any work conducted to further this vein of inquiry? Can the LFPT be used to prove Goedel's Second Incompleteness Theorem?
[0] <https://arxiv.org/abs/math/0305282>
[1] Handbook of mathematical logic. North-Holland Publishing Co., Amsterdam, 1977. Edited by Jon Barwise, With the cooperation of H. J. Keisler,
K. Kunen, Y. N. Moschovakis and A. S. Troelstra, Studies in Logic and the
Foundations of Mathematics, Vol. 90.
| https://mathoverflow.net/users/483796 | Has Goedel's Second Incompleteness Theorem been proven using Lawvere's Fixed Point Theorem? | As suggested by the OP, I'm turning my comment into an affirmative answer. WARNING: The presentation of Joyal's proof in the paper I cited contains an incorrect conclusion about Joyal's sentence (that it is undecidable from mere consistency). I have modified my answer accordingly.
Both Gödel's first and second incompleteness theorems have been proven by André Joyal mimicking the arithmetization of metamathematics in Gödel's original proof with internal reasoning inside the initial arithmetic universe. This latter is none other than (the pretopos completion of) the syntactic category, in coherent logic, of the sequents expressing the axioms of [primitive recursive arithmetic](https://en.wikipedia.org/wiki/Primitive_recursive_arithmetic) ($PRA$); in particular the type theoretic treatment of arithmetic universes is not really needed in the proof. Joyal gives an alternative construction of the initial arithmetic universe by defining the category of primitive recursive predicates, which corresponds to taking a different site of definition of the classifying topos (Coste's construction).
The proof has been very recently made available on arxiv in [the paper of van Dijk/Oldenziel](https://arxiv.org/abs/2004.10482), although it's been part of categorical folklore since the seventies. Once an internal initial arithmetic universe has been shown to exist inside the initial arithmetic universe, its externalization provides the Gödel numbering of formulas, and one can define functorially the provability predicate $Prov(x)$. Then the proof of the incompleteness theorems proceeds by building a self-referential sentence similar to Gödel's sentence "I am not provable", exactly as Gödel did, using that this sentence is the fixed point of $\neg Prov(x)$, and an argument essentially equivalent to applying Lawvere's fixed point theorem provides its construction. The sentence is however slightly changed to "I am provably false" (which I call Joyal's sentence, the fixed point of $Prov(\neg x)$). It is well known that a fixed point for $\neg Prov(x)$ is equivalent to $\neg Prov(\bot)$, i.e., to $Con(PRA)$, and similarly Joyal also proves categorically that Joyal's sentence is equivalent to $Incon(PRA)$.
The second incompleteness then follows from the observation that Joyal's sentence cannot be provably false (though for certain consistent but $\omega$-inconsistent recursive extensions of PRA it can very well be provably true; here's what goes wrong with the conclusion of the cited paper).
While Lawvere's fixed point theorem is quite trivial, Gödel's incompleteness theorems are much deeper, more complex and subtle than just the fixed point lemma, since they contain the arithmetization of syntax, which Joyal has shown to correspond precisely to internal reasoning inside the arithmetic universe (it is certainly not trivial at all to internally construct an arithmetic universe inside an arithmetic universe). His proof thus deserves to be much widely known than it is.
| 12 | https://mathoverflow.net/users/12976 | 424422 | 172,408 |
https://mathoverflow.net/questions/424413 | 1 | If there are some square free numbers, a finite amount of them $\{t\_1,t\_2,...t\_k\}$ and we define the set $\mathcal{N}\_L=\{l\in\mathbb{N}/L-l^2\notin\bigcup t\_j\mathbb{Z}^2\}$, where $n\notin\bigcup t\_j\mathbb{Z}^2$ means that if we write $n$ as the product of its square part and square free part as $n=tm^2$ then $t\notin\{t\_1,t\_2,...t\_k\}$.
It seems intuitive to me the fact that as $L\rightarrow\infty$ then $\#\mathcal{N}\_L\rightarrow\infty$, being $\#$ the amount of elements in the set.
However, I don't know how to prove it rigorously as I have almost no control on the explicit form of the elements of the set. Thank you for any help.
| https://mathoverflow.net/users/411616 | Justify that a certain set depending on a parameter is large | Let $p\_1,\dots,p\_s$ be the distinct primes in $t\_1,\dots,t\_k$.
We claim that, for any prime $p\_i$, there is $a\_i$ such that whenever $\ell \equiv a\_i \pmod{p\_i}$, we have $L - \ell^2 \not\equiv 0 \pmod{p\_i}$. This is clear, since there are at least two different squares modulo $p\_i$ (in other words, $a\_i$ can always be chosen to be either 0 or 1).
By the Chinese Reminder Theorem, the system $\ell \equiv a\_i \pmod{p\_i}$, for $i=1,\dots,s$, has a solution.
If $\ell$ is a solution to the system of congruences above, then $L - \ell^2$ is not divisible by any of the primes $p\_i$ so it cannot have its square-free part equal to any of the numbers $t\_1,\dots,t\_k$.
The solution of the system has the form $\ell \equiv a \pmod{p\_1p\_2 \dots p\_s}$, for some $a<p\_1p\_2\dots p\_s$. Therefore the number of choices for $\ell$ in the interval $[0,L]$ is at least $\lfloor L/(p\_1\dots p\_s) \rfloor$, so it grows without a bound.
| 1 | https://mathoverflow.net/users/483601 | 424425 | 172,409 |
https://mathoverflow.net/questions/423865 | 6 | Suppose we have a simplicial complex / poset / small category without loops $X$ equipped with a functor $F$ into the category of posets / small categories without loops. Suppose further that for each arrow / edge $e$ of $X$, the functor $Fe$ is an equivalence of categories. If we perform the Grothendieck construction on this situation, we end up with a new poset / small category without loops $Y$ equipped with a functor $\pi \colon Y \to X$. Under what circumstances does $\pi$ induce a fibration on geometric realizations?
| https://mathoverflow.net/users/135175 | When is the Grothendieck / category of elements construction a fibration on geometric realizations? | Just so this question doesn't sit around forever with no "answer" (even though it's perfectly answered in the comments), I am writing a CW expansion to Cisinski's comment, to help folks who don't have Quillen's lecture notes from 1973 on hand.
Definition: a commutative square of categories is **homotopy cartesian** if the corresponding square of classifying spaces is. Recall that this means the map from $E'$ to the homotopy pullback of $h$ and $g$ is a homotopy equivalence in:
$$ \begin{array}{rrrr}
E' & \stackrel{h'}{\rightarrow} & E\\\
g'\downarrow & & \downarrow g \\\
B' & \stackrel{h}{\rightarrow} & B
\end{array}
$$
The following result is general, but applies of course to the OP's original situation where $Y$ is a Grothendieck construction over $X$.
Corollary to Quillen's Theorem B: Let $f: Y\to X$ be a prefibred functor such that for every morphism $e: d\_1\to d\_2$ in $X$, the base-change functor $u\_\*: f^{-1}(d\_1)\to f^{-1}(d\_2)$ is a homotopy equivalence. Then for any $d\in X$, the category $f^{-1}(d)$ is homotopy equivalent to the homotopy fiber of $f$ over $X$, meaning the following square is homotopy cartesian (where $i$ is the inclusion functor):
$$ \begin{array}{rrrr}
f^{-1}(d) & \stackrel{i}{\rightarrow} & Y\\\
\downarrow & & \downarrow f \\\
pt & \stackrel{d}{\rightarrow} & X
\end{array}
$$
This is the same as saying "the fibers are all homotopy fibers."
Consequently, for any $c$ in $f^{-1}(d)$, we have a long exact sequence
$$ \dots \to \pi\_{i+1}(X,d)\to \pi\_i(f^{-1}(d),c)\to \pi\_i(Y,c)\to \pi\_i(X,d)\to \dots$$
This is the notion of $f$ being "like a fibration" that you get from Quillen's Theorem B.
| 4 | https://mathoverflow.net/users/11540 | 424429 | 172,411 |
https://mathoverflow.net/questions/424427 | 12 | Let R be a noetherian ring and V a valuation ring with maximal ideal $\mathfrak{m}\_V$. Does every morphism of rings $\varphi: R \rightarrow V$ factor through a discrete valuation ring?
One may consider the homomorphic image $S$ of $\varphi$ in V. This is a noetherian subring and we may even assume it to be local by localizing at $\mathfrak{m}\_S :=\mathfrak{m}\_V \cap S$. Then we can find by a standard procedure a discrete valuation ring $T$ dominating $S$ in $L= Frac(S)$:
We can consider the blowup $\tilde{S}$ of $S$ in its closed point $\mathfrak{m}\_S$ and pick a generic point $\mathfrak{n}$ of an irreducible component of the exceptional divisor.
Then $\mathcal{O}\_{\tilde{S},\mathfrak{n}}$ is a 1-dimensional local ring with field of fractions $L$ which dominates $S$.
Normalizing $\mathcal{O}\_{\tilde{S},\mathfrak{n}}$ in $L$ yields the desired DVR $T$.
I guess the question is if we may choose $\mathfrak{n}$ above in a way such that $T \subset V \subset Frac(V)$. I would be also happy with extending $V$ such that this statement holds.
Is a statement like this known to be true or false?
| https://mathoverflow.net/users/111259 | Does every map from a noetherian ring to a valuation ring factor through a DVR? | The answer is no. I give two examples, which are standard non-dvr points on the Riemann-Zariski space of the plane.
(1) Let $R=k[x,y]$ and let $V\subseteq k(x,y)$ be the subring consisting of rational functions $f(x,y)$ which have non-negative valuation along $x=0$, and such that the restriction of $f$ to $x=0$ (which is a well-defined element of $k(y)$) has non-negative valuation along $y=0$. This is a valuation ring of rank two constructed from two dvrs by "concatenation". Clearly $V$ contains $R$.
Concretely, a monomial $x^ay^b$ belongs to $V$ if $a>0$ or $a=0$ and $b\geq 0$. So $x$ is infinitely divisible by $y$. The value group is $\mathbf{Z}^2$ with lexicographic ordering.
Suppose that $R\to V$ factors through a dvr $\mathcal{O}$. Let $a$ and $b$ be the valuations of the images of $x$ and $y$ in $\mathcal{O}$, respectively. Clearly $a,b>0$ since otherwise either $x$ or $y$ is invertible in $\mathcal{O}$ and hence also in $R$. So some power of $y$ is divisible by $x$ in $\mathcal{O}$ and hence also in $V$, which does not happen.
(2) Let again $R=k[x,y]$, pick an irrational number $\alpha>1$, and let $|\cdot|$ be the nonarchimedean norm on $R$ with $|x|=e^{-1}$ and $|y|=e^{-\alpha}$. Concretely, the norm of a polynomial $f=\sum a\_{mn} x^my^n$ is the supremum of $|e^{-m-\alpha n}|$ over all $(m,n)$ with $a\_{mn}\neq 0$. The completion $V$ of $R$ with respect to this norm is a valuation ring of rank one with value group $\Gamma=\mathbf{Z}+\alpha\cdot \mathbf{Z}$ (it looks a bit like $k[[\Gamma\_+]]$), with $x$ of valuation one and $y$ of valuation $\alpha$.
If $R\to V$ factors through a dvr $\mathcal{O}$, then there exist positive integers $a, b$ such that $x^a=uy^b$ where $u$ is a unit in $\mathcal{O}$. But this cannot happen in $V$, since $\alpha$ is irrational!
In Huber's classification of points on the adic unit disc (which is closely related), example (1) corresponds to a Type 5 point, and example (2) to a Type 3 point.
| 19 | https://mathoverflow.net/users/3847 | 424434 | 172,413 |
https://mathoverflow.net/questions/424447 | 2 | This question is related to my other question [Sign of the permutation which brings a subsequence back to its original form](https://mathoverflow.net/questions/424340/sign-of-the-permutation-which-brings-a-subsequence-back-to-its-original-form). Suppose I have a complete ordered set $\{a\_{1},\dotsc,a\_{2n}\}$ and take $\pi$ to be a permutation of $\{1,\dotsc,2n\}$ such that $a\_{\pi(1)}\le a\_{\pi(2)}\le \dotsb \le a\_{\pi(2n)}$. If we throw away the element $a\_{1}$ and some other element, say, $a\_{j}$, we end up with a complete ordered set which has now $2n-2$ elements. Suppose $\pi'$ is the permutation which brings it to an increasing order again, that is $a\_{\pi'(2)}\le a\_{\pi'(3)}\le\dotsb \le a\_{\pi'(2n-2)}$. What is the relation between the sign of $\pi$ and the sign of $\pi'$?
| https://mathoverflow.net/users/235802 | Given $\pi$ permutation on $\{1,\dotsc,n\}$, what is the sign of a permutation of $\{2,\dotsc,\hat\jmath,\dotsc,n\}$? | $\DeclareMathOperator\sgn{sgn}$This is almost the same as your previous question, just with the order of the operations switched—whether you think of $\pi$ as ordering or disordering is just a matter of taking the inverse.
You previously needed $\pi^{-1}(1) - 1$ transpositions to bring the element $a\_1$ to its new position $a\_{\pi(\pi^{-1}(1))}$. Instead, simply throw away $a\_1$. Separately, you would have needed $\lvert\pi^{-1}(j) - j\rvert$ transpositions to bring the element $a\_j$ to its new position $a\_{\pi(\pi^{-1}(j))}$. This is unaffected by throwing away $a\_1$, unless $\pi^{-1}(j)$ equals $1$, in which case it should be reduced by $1$. Thus, throwing away both $a\_1$ and $a\_j$ gives you a new permutation $\pi'$ whose sign is $\sgn(\pi') = \sgn(\pi)\cdot(-1)^{(\pi^{-1}(1) - 1) + \lvert\pi^{-1}(j) - j\rvert}$ unless $\pi^{-1}(j) = 1$, in which case it is $\sgn(\pi') = \sgn(\pi)\cdot(-1)^{\pi^{-1}(1) + \lvert\pi^{-1}(j) - j\rvert}$.
| 5 | https://mathoverflow.net/users/2383 | 424449 | 172,416 |
https://mathoverflow.net/questions/424412 | 2 | Let $G$ be a finite abelian etale group scheme over a number field $k$. Let $E$ be an elliptic curve over $k$ and $C := E\backslash \{O\}$ its affine model of the same equation.
Recall that for a variety $X$, the pointed (etale cohomology) set $H^1(X,G)$ classifies all $X$-torsors under $G$.
Given a torsor $Y \rightarrow E$ under $G$, the pull-back by the canonical inclusion $C \rightarrow E$ gives us an element whose class is in $H^1(C,G)$. Thus we have a map $$H^1(E,G) \rightarrow H^1(C,G)$$ defined by pull-backs.
I want to know if this map has any chance of being an isomorphism. There isn't much in the literature about the cohomology groups $H^i(E,-)$, usually $E$ is on the right, and therefore I hope to get references understanding $E$-torsors under some group rather than some $X$-torsors under $E$.
| https://mathoverflow.net/users/172132 | Torsors over elliptic curves | Indeed this map is an isomorphism. There is a diagram of five-term exact sequences, arising from the Leray spectral sequence for the maps $E\to \text{Spec}(k), C\to \text{Spec}(k)$, from
$0\to H^1(k, H^0(E\_{\overline{k}}, G))\to H^1(E, G)\to H^0(k, H^1(E\_{\overline{k}}, G))\to H^2(k, H^0(E\_{\overline{k}}, G))\to H^2(E, G)$
to the analogous sequence with $C$
$0\to H^1(k, H^0(C\_{\overline{k}}, G))\to H^1(C, G)\to H^0(k, H^1(C\_{\overline{k}}, G))\to H^2(k, H^0(C\_{\overline{k}}, G))\to H^2(C, G)$.
By comparison with the complex-analytic setting, the map $H^i(E\_{\overline{k}}, G)\to H^i(C\_{\overline{k}},G)$ is an isomorphism for $i=0,1$. Hence this map of five-term exact sequences is an isomorphism on the first, third, and fourth terms.
We would like to argue that it is an isomorphism on the second term. But this is immediate from the five lemma.
| 2 | https://mathoverflow.net/users/6950 | 424452 | 172,419 |
https://mathoverflow.net/questions/424444 | 11 | This question now has two sequels, [Pointless groups II](https://mathoverflow.net/questions/424456/pointless-groups-ii) (to which @R.vanDobbendeBruyn gave a [counterexample](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091038_424456) for an infinite, imperfect field) and [Pointless groups III](https://mathoverflow.net/questions/424489/pointless-groups-iii), both using revised wording suggested by [@DanielLitt](https://mathoverflow.net/questions/424444/pointless-groups#comment1091016_424453).
Fix a field $k$—not algebraically closed, and, in fact, this question is obviously of interest only if $k$ is finite. (EDIT: @R.vanDobbendeBruyn pointed out in the [comments](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091035_424456) to [Pointless groups II](https://mathoverflow.net/questions/424456/pointless-groups-ii) that my memory that the $k$-points are Zariski-dense in the infinite case is only guaranteed true for reductive groups.) Then "group" means "smooth, connected, affine algebraic group scheme of finite type over $k$".
Is it possible to have a (smooth, connected) group $G$, and a proper (smooth, connected) subgroup $H$, such that $G(k) = H(k)$? If so, then can this even happen for $H$ the trivial subgroup?
I have a small store of counterexamples showing that certain naïve statements about rational points of algebraic groups can fail for small fields $k$. (For example: $C\_G(S)(k) \ne C\_G(S(k))$ for $S$ a maximal torus in $\operatorname{SL}\_2$ when $k$ has $2$ elements.) None of them provides an example of this sort of behaviour, but, then again, it is a very small testbed. If this question is too elementary for MO, then I have no objection to migrating it to MSE.
| https://mathoverflow.net/users/2383 | Pointless groups | Perhaps I'm missing something, but it seems to me that $k=\mathbb{F}\_2$, $G=\mathbb{G}\_{m, k}$ works, where $H$ is the trivial subgroup.
| 13 | https://mathoverflow.net/users/6950 | 424453 | 172,420 |
https://mathoverflow.net/questions/424456 | 7 | This question is a sequel to [Pointless groups](https://mathoverflow.net/questions/424444/pointless-groups), where I asked for a certain kind of counterexample. @DanielLitt [produced](https://mathoverflow.net/a/424453) an elegant and easy-to-understand counterexample, but also [suggested](https://mathoverflow.net/questions/424444/pointless-groups#comment1091016_424453) a sense in which it might be the only counterexample.
At the suggestion of [@R.vanDobbendeBruyn](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091038_424456) and [@YCor](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091103_424456), this question has yet another sequel: [Pointless groups III](https://mathoverflow.net/questions/424489/pointless-groups-iii).
Fix a field $k$ — not algebraically closed, and, in fact, this question is obviously of interest only if $k$ is finite. (EDIT: @R.vanDobbendeBruyn points out in the [comments](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091035_424456) that my memory that the $k$-points are Zariski-dense in the infinite case is only guaranteed true for reductive groups, and gives a [counterexample](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091038_424456) for an infinite, imperfect field.) Then "group" means "smooth, connected, affine algebraic group scheme of finite type over $k$".
If $G$ is a (smooth, connected) group, and $G(k)$ is trivial, then does that imply that $k = \mathbb F\_2$ and $G$ is a split torus?
| https://mathoverflow.net/users/2383 | Pointless groups II | One other type of example is constructed in Conrad–Gabber–Prasad's *Pseudo-reductive groups*, Example 11.3.2. The construction is kind of classic (and possibly predates the book), so let me recall it here:
**Example.** Let $k = \bar{\mathbf F}\_p(t)$, and let $G \subseteq \mathbf G\_a \times \mathbf G\_a$ be the subgroup
$$\left\{(x,y) \in \mathbf G\_a \times \mathbf G\_a\ \big|\ y^p = x-tx^p\right\}.$$
It is smooth and becomes isomorphic to $\mathbf G\_a$ over $k(t^{1/p})$ (replace $y$ by $y+t^{1/p}x$ and eliminate $x$).
**Lemma.** *If $p > 2$, then $G(k) = \{(0,0)\}$.*
*Proof.* Let $(x,y) \in G(k)$. If $x = 0$, then $y = 0$ as well, so we may assume $x \neq 0$. Then $x$ cannot be a $p$-th power, for otherwise $t$ is a $p$-th power. Thus there exists $a \in \bar{\mathbf F}\_p$ such that $v:=v\_a(x)$ is not divisible by $p$. The equation $y^p = x-tx^p$ gives
$$(y+ax)^p = x-(t-a)x^p.$$
The ultrametric triangle inequality implies that if $\alpha+\beta = \gamma$, then the set of valuations $\{v\_a(\alpha),v\_a(\beta),v\_a(\gamma)\} \subseteq \mathbf Z \cup \{\infty\}$ has at most $2$ elements.
Since $v\_a((y+ax)^p)$ is divisible by $p$ but neither $v=v\_a(x)$ nor $v\_a((t-a)x^p)=pv+1$ is, we conclude that $v=pv+1$, i.e. $(p-1)v=-1$, which is impossible since $p > 2$ and $v \in \mathbf Z$. $\square$
For $p = 2$ there is also the point $(1/t,0)$, and again Conrad–Gabber–Prasad show that this is the only other point.
**Remark.** On the other hand, if $k$ is an infinite field and either $k$ is perfect or $G$ is reductive, then $G$ is unirational, hence $G(k) \subseteq G$ dense. See for instance [Milne's notes](https://www.jmilne.org/math/CourseNotes/iAG200.pdf), Corollaries 19.21 and 19.22. Thus, the only examples are among imperfect fields and finite fields.
| 9 | https://mathoverflow.net/users/82179 | 424462 | 172,423 |
https://mathoverflow.net/questions/424438 | 7 | The bar construction is a functor $A\mapsto Bar(A)$ from the category of augmented differential graded algebras over a commutative ring $R$ to the category of chain complexes of $R$-modules. It sends an algebra $A$ to the derived tensor product $R\otimes^{L}\_AR$.
Assume that I have a map of augmented algebras $A\to B$ which induces a quasi-isomorphism $Bar(A)\to Bar(B)$. This does not necessarily implies that the original map was a quasi-isomorphism but I believe it does if some connectivity or co-connectivity assumptions are made on $A$ and $B$. Philosophically speaking, the bar construction is the algebraic analogue of the loop space construction for based spaces and this construction is conservative if one restricts to connected based spaces.
I am using the homological grading convention. My guess is that the Bar construction is conservative when restricted to simply coconnected algebras (i.e. whose homology is concentrated in degrees $\leq -2$ apart from the unit in degree zero). The analogy with spaces suggests that this might actually also be true for coconnected algebras but I'm not so sure about that. In any case, I'm interested in finding a reference for this fact.
| https://mathoverflow.net/users/10707 | Under which conditions is the bar construction a conservative functor? | $\vphantom{0pt}$ Hi Geoffroy. The natural way to approach this is by finding conditions under which the cobar functor preserves quasi-isomorphisms. If $A \to A'$ is a morphism of dg algebras, then you get a zig-zag
$$ A \leftarrow \Omega BA \to \Omega BA' \to A'.$$
The two counit maps here are always quasi-isomorphisms, without connectivity/coconnectivity hypotheses. The cobar functor preserves quasi-isomorphisms of coalgebras concentrated in homological degree $\geq 2$, and also of coalgebras concentrated in degree $\leq 0$. (I hope I didn't mess up any degree shifts just now.) So natural hypotheses which imply that the bar construction reflects quasi-isomorphisms is that $A$ and $A'$ are either both connected or both coconnected.
Off the top of my head I don't have a good reference for the above, but let me tell you how the proofs go. With no assumption at all, it is true that the cobar functor takes quasi-isomorphisms to filtered quasi-isomorphisms, where the cobar construction is filtered by its natural "adic" filtration. A filtered quasi-isomorphism is a quasi-isomorphism if the filtration is complete and exhaustive. The reason the cobar construction fails to preserve quasi-isomorphisms in general is that the adic filtration is typically not complete, but if your algebra is concentrated in degrees as above then the adic filtration becomes bounded below in every homological degree, in particular complete, and you win.
| 5 | https://mathoverflow.net/users/1310 | 424467 | 172,426 |
https://mathoverflow.net/questions/424461 | 0 | This question is based on [this Math.SE answer](https://math.stackexchange.com/questions/4182260/on-the-function-n-mapsto-a-n-frac-1n-for-a-given-power-series-sum-n/4445420#4445420), so let's recall a few concepts dealt with there. If $\{a\_n\}\_{n\in\Bbb N}$ is the sequence of coefficients of a power series $\sum\_{n=0}^\infty a\_nz^n$ with unit convergence radius, then
$$
\limsup\_{n\to+\infty} \ln\lvert a\_n\rvert/n=0\iff \limsup\_{n\to+\infty} \lvert a\_n\rvert^{1\over n}=1.
$$
Shmuel Agmon was able to prove (see [1], chapter 1, §§ 1.1 and 1.2 pp. 264-265, but for a clearer exposition [2], §3 p. 497) that under these assumptions there exists a unique *smallest concave majorant* or *envelope* $\{C(n)\triangleq c\_n\}\_{n>0}$ of the sequence $\{\ln\lvert a\_n\rvert\}\_{n>0}$ (see the references or the cited answer for relevant definitions). In [1], chapter 1, §1.2 p. 265 he states that the function $C(x)$ was introduced by Georges Valiron for the study of entire functions.
**Question**. I wasn't able to locate a precise reference, so does anyone know where Valiron introduces the function $C(x)$ ($C(n)=C(x)|\_{\Bbb N\setminus\{0\}})$?
**Note**. I had a quick look to Valiron's monograph [3] but I was not able to find a definition.
**References**
[1] Shmuel Agmon, "[Sur les séries de Dirichlet](https://doi.org/10.24033/asens.971)" (French), Annales Scientifiques de l’École Normale Supérieure, Troisième (III) Série 66, 263-310 (1949), [MR0033352](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0033352), [Zbl 0034.34602](https://www.zbmath.org/?q=an%3A0034.34602).
[2] Shmuel Agmon, "[Functions of exponential type in an angle and singularities of Taylor series](https://doi.org/10.2307/1990611)" (English), Transactions of the American Mathematical Society 70, 492-508 (1951), [MR0041222](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0041222), [Zbl 0045.34902](https://www.zbmath.org/?q=an%3A0045.34902).
[3] Georges Valiron, *[Lectures on the general Theory of integral functions](https://archive.org/details/lecturesonthegen033156mbp/mode/2up)*, translated by E. F. Collingwood, with a preface by W. H. Young (English) Cambridge: Deighton, Bell and Co., pp. XII+208 (1923), [JFM 50.0254.01](https://www.zbmath.org/?q=an%3A50.0254.01).
| https://mathoverflow.net/users/113756 | Reference(s) on the smallest concave majorant for the sequence of coefficients of a given power series? | A better reference on Valiron is
George Valiron, *Fonctions analytiques*, Presses universitaires de France, 1954, [MR0061658](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0061658), [Zbl 0055.06702](https://zbmath.org/?q=an%3A0055.06702).
This smallest concave majorant is nothing but the Newton polygon generalized to power series. Valiron also calls it Hadamard's polygon.
He has a picture of it in Section 67.
Valiron studies it in the case of entire functions ($\log|a\_n|/n\to-\infty$), but also mentions the case of finite radius of convergence.
Its existence is almost evident: since $\log|a\_n|/n$ is bounded from above, there is a constant majorant. So the class of concave majorants is non-empty, therefore there is a unique smallest concave majorant, since the minimum of any set of concave functions bounded from below is concave.
**Edit**. Valiron refers to his 2 papers in the *Annales scientifiques de l'École Normale Supérieure* (Série 3):
* "[Les théorèmes généraux de M. Borel dans la théorie des fonctions entières](https://doi.org/10.24033/asens.724)", 37, 219-253 (1920), [JFM 47.0301.01](https://zbmath.org/?q=an%3A48.1218.01), and
* "[Recherches sur le théorème de M. Picard](https://doi.org/10.24033/asens.736)", 38, 389-429 (1921), [JFM 48.1218.01](https://zbmath.org/?q=an%3A48.1218.01).
| 1 | https://mathoverflow.net/users/25510 | 424485 | 172,430 |
https://mathoverflow.net/questions/424471 | 4 | Let $\chi$ denote the Legendre symbol of conductor $q$. A Siegel zero for the $ L $ series associated to $ \chi $, which we denote by $ L(s,\chi) $ is a real zero $ \sigma $ satisfying $ 1-\frac{c}{\log |q|} < \sigma < 1$ for some constant $c$.
I have read in many places (see for example the second page of the article by Ajit Bhand and Ram Murty available [here](https://mast.queensu.ca/%7Emurty/class-groups_final_revised.pdf)) that the non-existence of Siegel zeros for $ L(s,\chi) $ can be used to prove the bound
$$
h(d) > c\_1 \frac{\sqrt{d}}{\log(d)},
$$
where $ h(d)$ is the class number of the associated imaginary quadratic field and $c\_1$ is another *effective constant* which can be calculated depending on $c$.
>
> **My Question :** How to explicitly compute $c\_1$ from $c$?
>
>
>
If anyone can direct me to a proof of the above statement, I think that would also suffice.
| https://mathoverflow.net/users/148866 | Calculating the explicit constant – Siegel zeros and class numbers | One place to find this worked out in detail is the paper ["On the Siegel-Tatuzawa theorem"](https://www.impan.pl/shop/en/publication/transaction/download/product/102670) by Jeffrey Hoffstein (published in 1980 in Acta Arithmetica). Lemma 1 of that paper states that if $\chi$ is a quadratic Dirichlet character with conductor $d > 10^{6}$ and if $L(s,\chi)$ is nonzero on $(\beta,1)$ and $1-\beta$ is small (specifically $(1 - \beta)^{-1} < 11.657 \log(d)$), then
$$
L(1,\chi) > 1.507 (1 - \beta).
$$
Also given in Lemma 1 is a lower bound on $L(1,\chi)$ under the assumption that $L(s,\chi)$ doesn't vanish on $(0,1)$.
Combining this with the Dirichlet class number formula $L(1,\chi) = \frac{2 \pi h(d)}{w\_{d} \sqrt{d}}$ gives the result you seek with $c\_{1} = \frac{1.507}{\pi} c$ provided $c < \frac{1}{11.657}$ and $d > 4$.
| 8 | https://mathoverflow.net/users/48142 | 424488 | 172,431 |
https://mathoverflow.net/questions/424489 | 12 | This question is a sequel to [Pointless groups](https://mathoverflow.net/questions/424444/pointless-groups), to which @DanielLitt produced an elegant and easy-to-understand [counter-example](https://mathoverflow.net/a/424453), and [Pointless groups II](https://mathoverflow.net/questions/424456/pointless-groups-ii), where @R.vanDobbendeBruyn [pointed out](https://mathoverflow.net/questions/424456/pointless-groups-ii?noredirect=1#comment1091035_424456) that my implicit claim about Zariski density of rational points of groups over infinite fields is only guaranteed for *reductive* groups (or perfect fields), and gave a [counterexample](https://mathoverflow.net/a/424462) (still elegant, though less easy for me to understand than a split torus over $\mathbb F\_2$!) for an infinite but imperfect field.
At the suggestion of [@R.vanDobbendeBruyn](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091038_424456) and [@YCor](https://mathoverflow.net/questions/424456/pointless-groups-ii#comment1091103_424456), I am engaging in the extreme bad manners of once again modifying my search for counterexamples to rule out the counterexamples already provided. The revised question is @DanielLitt's [wording](https://mathoverflow.net/questions/424444/pointless-groups#comment1091016_424453).
So, 3rd time's a charm, hopefully … fix a *finite* field $k$. Then "group" means "smooth, connected, affine algebraic group scheme of finite type over $k$".
If $G$ is a (smooth, connected) group, and $G(k)$ is trivial, then does that imply that $k = \mathbb F\_2$ and $G$ is a split torus?
| https://mathoverflow.net/users/2383 | Pointless groups III | Yes.
Let $|k| = q$. Let $T$ be an $n$-dimensional torus defined over $k$ and let $\overline{T}$ be the base change of $T$ to $\overline{k}$. Then the character group $\text{Hom}(\overline{T}, \mathbb{G}\_m)$ is isomorphic to $\mathbb{Z}^n$ and the Frobenius acts on $\text{Hom}(\overline{T}, \mathbb{G}\_m)$ by a matrix $F$. If $T$ splits over $\mathbb{F}\_{q^n}$, then $F$ obeys $F^n = q^n \text{Id}$. The isomorphism class of the torus $T$ is determined by the $\text{GL}\_n(\mathbb{Z})$ conjugacy class of $F$, and we have $\# T(\mathbb{F}\_q) = \det(F - \text{Id})$. So we are reduced to the following linear algebra problem:
>
> If $F$ is an integer matrix obeying $F^n = q^n \text{Id}$ and $\det(F - \text{Id}) = 1$, can we deduce that $q=2$ and $F = 2 \text{Id}$?
>
>
>
The answer is "yes". The condition that $F^n = q^n \text{Id}$ means that the eigenvalues of $F$ are of the form $q \zeta$ for $\zeta$ a root of unity. So we have $\det(F - \text{Id}) = \prod\_i (q \zeta\_i - 1) = 1$ where the $\zeta\_i$ are various roots of unity.
But $|q \zeta - 1| \geq 1$ for any prime power $q$ and any root of unity $\zeta$, and strictly greater unless $q=2$ and $\zeta=1$, so the only solution is $q=2$ and $F = \text{Id}$. $\square$
| 13 | https://mathoverflow.net/users/297 | 424493 | 172,432 |
https://mathoverflow.net/questions/424501 | 2 | I can deduce the following simple proposition, the definitions for $\sigma(x)$ the sum of divisors functions and $\varphi(x)$ the Euler totient function are assumed. After I present a conjecture that I've tested for the square $1000\times 1000$. (Please add feedback in comments if you think that the question can be improved before downvote it.) The sequence from the OEIS related to the post is *A023194*.
**Claim.** *Let $B\geq1$ and $C\geq 1$ be positive integers with $B$ a prime numbers satisfying $$\sigma(C)=B,\tag{1}\label{1}$$
then $\varphi(B)$ has the factorization
$$\varphi(B)=\operatorname{rad}(C)\cdot (B-C),\tag{2}\label{2}$$
where $\operatorname{rad}(n)=\prod\_{\substack{p\mid n\\p\text{ prime}}}p$ denotes the product of distinct primes dividing an integer $n> 1$.*
One also can deduce under the same assumptions of **Claim** the following identities $\varphi(C)=C\left(1-\frac{1}{\operatorname{rad}(C)}\right)$, $\frac{\varphi(C)}{C}=\frac{C-1}{B-1}$ and $\sigma(C^{1+\lambda})=\frac{C^{\lambda}-1}{\operatorname{rad}(C)-1}+BC^{\lambda}$, for $\lambda\geq 1$ integer.
**Conjecture.** *If \eqref{1} and \eqref{2} hold for some integers $B$ and $C$ strictly greater than $1$, then $B$ is a prime number.*
>
> **Question.** I would like to know if it is possible to prove previous conjecture, or **what work can be done** about it.
>
>
>
I was inspired in an edited question (on Mathematics Stack Exchange, now closed with identificator [**4423186**](https://math.stackexchange.com/questions/4423186/on-the-equation-sigma-left-square-right-textprime-propositions-that-can)) that I'm going to delete from my profile on MSE.
**As application (that I evoke)** I tried, and I invite to it if you think that it is interesting, to combine these identities for an odd perfect number of the form (for example) $BCM^2$, for pairwise coprime integers $B$, $C$ and $M$ under the previous assumptions.
| https://mathoverflow.net/users/142929 | A conjecture concerning the equation $\sigma\left(\square\right)=\text{prime}$ | Your conjecture is true. Notice first that it is enough to show that $C=p^k$ for some prime $p$. Indeed, in this case
$$
B=\sigma(C)=\frac{p^{k+1}-1}{p-1},
$$
hence
$$
\mathrm{rad}(C)(B-C)=p\frac{p^k-1}{p-1}=B-1.
$$
This means that $\varphi(B)=B-1$, so $B$ is prime.
To show that $C$ must be a prime power, assume the contrary. Let $p$ be the least prime factor of $C$. Since $C$ has at least one larger prime factor (otherwise $C=p^k$) , we have
$$
\mathrm{rad}(C)\geq p(p+1)=p^2+p.
$$
On the other hand,
$$
B\geq \varphi(B)=\mathrm{rad}(C)(B-C)\geq (p^2+p)(B-C).
$$
Therefore,
$$
C\geq B\left(1-\frac{1}{p^2+p}\right).
$$
Next, if $C=p\_1^{k\_1}\ldots p\_l^{k\_l}$, then
$$
\sigma(C)=\prod\_i \frac{p\_i^{k\_i+1}-1}{p\_i-1}
$$
and
$$
\frac{\sigma(C)}{C}=\prod\_i\frac{p\_i-p\_i^{-k\_i}}{p\_i-1}.
$$
Every factor in the product above is $\geq 1$, so we can bound the product from below by any one factor. In particular, for some $k\geq 1$
$$
\frac{\sigma(C)}{C}\geq \frac{p-p^{-k}}{p-1}.
$$
So
$$
B=\sigma(C)\geq C\frac{p-1/p}{p-1}=C\frac{p+1}{p}.
$$
Substituting into the previous inequality, we obtain
$$
C\geq B\left(1-\frac{1}{p^2+p}\right)\geq C\left(1-\frac{1}{p^2+p}\right)\frac{p+1}{p}.
$$
Dividing by $C$, we arrive at
$$
1\geq \left(1-\frac{1}{p^2+p}\right)\frac{p+1}{p}=1+\frac{1}{p}-\frac{1}{p^2}>1,
$$
which is a contradiction.
| 7 | https://mathoverflow.net/users/101078 | 424507 | 172,440 |
https://mathoverflow.net/questions/424495 | 6 | For a hobby software project I am working with exact rational arithmetic, as it happens this produces numbers $\frac{n}{k}$ of huge size even after reducing them, I am searching for an efficient algorithm to "simplify" these rational with a specified error tolerance.
In my own working out the "simplicity function" I was trying to maximize was $S(\frac{n}{k}) = |\frac{1}{k}|$ because it seemed the simplest.
In the following I will focus on finding a rational with minimal denominator, but if you have a solution with a different definition of "simple rational" that still decreases $\log(n) + \log(k)$ it would still solve my problem.
So given 2 rational numbers $p,q \in \mathbb{Q}$ with $p\lt q$, and a $k \in \mathbb{N}^+$ we can find a rational $\frac{n}{k}$ s.t. $p \le \frac{n}{k} \le q$ iff $\lceil kp\rceil \le\lfloor kq\rfloor$ by choosing $n \in \left[\lceil kp\rceil,\lfloor kq\rfloor\right]\cap\mathbb{Z}$.
Also if $k \ge \frac{1}{q-p}$ then the set $\left[\lceil kp\rceil,\lfloor kq\rfloor\right]\cap\mathbb{Z}$ is never empty.
Is there a simply way to find the minimal $k$ such that $\lceil kp\rceil \le\lfloor kq\rfloor$?
In a sense this could be more mathematically formulated as finding a global maximum of a function like
\begin{equation}
f(n,k) =
\left\{
\begin{array}{lll}
\frac{1}{k} & \text{if} & p \le \frac{n}{k} \le q \\
0 & \text{if} & \text{otherwise}
\end{array}
\right.
\end{equation}
Or
\begin{equation}
f(n,k) =
\left\{
\begin{array}{lll}
\frac{1}{\log(|n|)+\log(|k|)} & \text{if} & p \le \frac{n}{k} \le q \\
0 & \text{if} & \text{otherwise}
\end{array}
\right.
\end{equation}
But since it came from an algorithmic context I preferred to keep it in a semi-algorithmic formulation.
| https://mathoverflow.net/users/38349 | How can I efficiently find the "simplest" rational in an interval? | Here's a version of what has been said in the comments that's reformulated into an explicit algorithm, first assuming that $p$ and $q$ are *irrational* to avoid the edge case:
* Lazily compute the continued fraction expansions of $p = [a\_0; a\_1, a\_2, a\_3,\ldots]$ and $q = [b\_0; b\_1, b\_2, b\_3,\ldots]$: stop at the first $i$ such that $a\_i \neq b\_i$ (namely $a\_i < b\_i$ if $i$ is even, and $a\_i > b\_i$ if $i$ is odd, all this is for $p<q$).
* The sought-after rational is then $[a\_0; a\_1, a\_2, a\_3,\ldots, a\_{i-1},\min(a\_i,b\_i)+1]$.
Indeed, this is clear from the paths taken in the [Stern-Brocot tree](https://en.wikipedia.org/wiki/Stern%E2%80%93Brocot_tree): the branch toward $p$ from the root proceeds by going $a\_0$ steps to the right, then $a\_1$ to the right, then $a\_2$ to the left, and so on; and similarly for $q$ with $b\_j$ instead of $a\_j$. The properties of the Stern-Brocot tree make it clear that the sought-after rational is the node of first divergence between these two branches; now the continued fraction expansion of the rational node in question is obtained by tacking a final $1$ to the common sequence, so it's $[a\_0; a\_1, a\_2, a\_3,\ldots, a\_{i-1},\min(a\_i,b\_i), 1]$ which is more standardly written as $[a\_0; a\_1, a\_2, a\_3,\ldots, a\_{i-1},\min(a\_i,b\_i) + 1]$.
If $p$ and/or $q$ are rational the same procedure applies provided we take the continued fraction expansion of the form $[a\_0;a\_1,\ldots,a\_r,1]$ and/or $[b\_0;b\_1,\ldots,b\_s,1]$, because again this tells us that $p$ is reached in the Stern-Brocot tree by going first $a\_0$ steps to the left, then $a\_1$ to the right, and ending in $a\_r$ steps in the final direction (and stop there):
* If there is $i\leq\min(r,s)$ such that $a\_i\neq b\_i$, this is exactly as above.
* If $s\geq r$ and $b\_i = a\_i$ for all $0\leq i\leq r$, then the sought-after rational is $p = [a\_0;a\_1,\ldots,a\_r,1]$ itself (it is an ancestor of $q$ in the Stern-Brocot tree).
* If $r\geq s$ and $a\_i = b\_i$ for all $0\leq i\leq s$, then the sought-after rational is $q = [b\_0;b\_1,\ldots,b\_r,1]$ itself (it is an ancestor of $p$ in the Stern-Brocot tree).
**Example:** Suppose I want to find the rational with smallest denominator between $1.728 = [1; 1, 2, 1, 2, 10, 1]$ and $1.729 = [1; 1, 2, 1, 2, 4, 2, 2, 1, 1, 1]$: the algorithm in question shows that it's $[1; 1, 2, 1, 2, 4, 1] = 102/59$; and of course, this rational is also the rational with smallest denominator between itself and any of the two numbers I started with, which exemplifies the edge cases.
| 9 | https://mathoverflow.net/users/17064 | 424509 | 172,441 |
https://mathoverflow.net/questions/424364 | 2 | Consider the SDE
$$dX\_t =b(t)dt + a(t)dW\_t,\quad \forall t>0,$$
with $X\_0>0$ has a density function $\rho:\mathbb R\_+\to\mathbb R\_+$. Consider the probability $g(t):=\mathbb P[\inf\_{0\le s\le t}X\_s>0]$. Can we prove $g'$ exists and
$$-\frac{C}{\sqrt{t}}\le g'(t)\le 0,\quad \forall t>0,$$
where $C$ depends only on $b,a,\rho$. If it helps, we may assume any "rational" condition satisfied by $b,a,\rho$.
| https://mathoverflow.net/users/nan | Is $g(t)=\mathbb P[\inf_{0\le s\le t}X_s>0]$ differentiable with respect to $t$? | **Yes**, $g(t)$ is differentiable with respect to $t$ under the assumption that $a(s) > 0 $ for $s \in [0,t]$.
---
*Proof*. Introduce the time-change $\tau(s) := \int\_0^s a(u)^{2} du$ and set $c(s) := b(\tau^{-1}(s)) a(\tau^{-1}(s))^{-2}$ for $s \in [0,t]$. In terms of this time change, the process $X\_t$ can be written equivalently (in law) as $$
X\_t \overset{d}{=} X\_0 + \int\_0^{\tau(t)} c(r) dr + W\_{\tau(t)} \;.
$$ By Girsanov's Theorem, \begin{align\*}
& g(t) = E\left[ 1(\inf\_{0 \le s \le \tau(t)} \left( \int\_0^s c(r) dr + W\_s \right) \ge -X\_0 ) \right] \\
&= E\left[ 1(\inf\_{0 \le s \le \tau(t)} W\_s \ge -X\_0 ) e^{\left(\int\_0^{\tau(t)} c(s) dW\_s - \frac{1}{2} \int\_0^{\tau(t)} c(s)^2 ds \right)} \right] \\
&= E\left[ 1(\sup\_{0 \le s \le \tau(t)} (-W\_s) \le X\_0 ) e^{\left(\int\_0^{\tau(t)} c(s) dW\_s - \frac{1}{2} \int\_0^{\tau(t)} c(s)^2 ds \right)} \right] \\
&= 1 - E\left[ 1(\sup\_{0 \le s \le \tau(t)} W\_s \ge X\_0) e^{\left(-\int\_0^{\tau(t)} c(s) dW\_s - \frac{1}{2} \int\_0^{\tau(t)} c(s)^2 ds \right)} \right]
\end{align\*}
where in the last step we used that $-\inf\_{0 \le s \le \tau(t)} W\_s = \sup\_{0 \le s \le \tau(t)} (-W\_s)$ together with the reflection symmetry of Brownian motion, i.e., $-W\_s$ is also a standard BM.
Let $T = \inf\{ t \ge 0 \mid W\_t = X\_0 \}$. By the strong Markov property of BM, \begin{equation}
g(t) = 1 - E\left[ 1(T \le \tau(t)) e^{\left(-\int\_0^{T} c(s) dW\_s - \frac{1}{2} \int\_0^{T} c(s)^2 ds \right)} \right]
\tag{1} \label{g\_T}
\end{equation}
Moreover, since the process $\{ W\_s \}\_{s \in [0,T]}$ is a Brownian bridge from $(0,0)$ to $(T,X\_0)$, we have $$
W\_s = X\_0 \frac{s}{T} + (T-s) \int\_0^s \frac{1}{T-r} dB\_r \;,
$$ where $B$ is a standard BM. Hence, \begin{align\*}
\int\_0^{T} c(s) dW\_s = \frac{X\_0}{T} \int\_0^T c(s) ds + \int\_0^T \left( c(s) - \frac{1}{T-s} \int\_s^T c(r) dr \right) d B\_s
\end{align\*}
Inserting this into \eqref{g\_T} yields, \begin{align\*}
&g(t) = \\
& \quad 1 - E\left[ 1(T \le \tau(t)) e^{- \int\limits\_0^{T} \left(-\frac{X\_0}{T} c(s)-\frac{1}{2} c(s)^2 + \frac{1}{2} (c\_s - \frac{1}{T-s} \int\limits\_s^T c\_r dr)^2 \right) ds} \right] \end{align\*} Since the distribution of the hitting time $T$ is inverse gamma with parameters $1/2$ and $(1/2) X\_0^2$, \begin{equation} \begin{aligned}
&g(t) = \\
& \quad 1 - \int\limits\_0^{\tau(t)} \frac{X\_0 e^{- \frac{X\_0^2}{2 z}- \int\limits\_0^{z} \left(-\frac{X\_0}{z} c(s)-\frac{1}{2} c(s)^2 + \frac{1}{2} (c\_s - \frac{1}{z-s} \int\limits\_s^z c\_r dr)^2 \right) ds}}{\sqrt{2 \pi} z^{3/2}} dz
\end{aligned} \tag{$\star$} \label{final\_g}\end{equation}
which is differentiable with respect to $t$ without any additional assumptions. $\Box$
---
This proof also answers this MO question: [Probability that a drifted Gaussian process does not hit zero](https://mathoverflow.net/q/409414/64449)
| 0 | https://mathoverflow.net/users/64449 | 424512 | 172,442 |
https://mathoverflow.net/questions/424405 | 4 | In Mike Shulman's article [Brouwer’s fixed-point theorem in real-cohesive homotopy type theory](https://www.cambridge.org/core/journals/mathematical-structures-in-computer-science/article/abs/brouwers-fixedpoint-theorem-in-realcohesive-homotopy-type-theory/8270C2EAC4EE5D5CDBA17EEB3FF6B19E), the fundamental axiom adopted for his real-cohesive homotopy type theory (axiom $\mathbb{R}\flat$), which states that the only (continuous) functions from the Dedekind real numbers to every discrete cohesive type $A$ are the constant functions to $A$, requires defining the Dedekind real numbers. However, in predicative mathematics, without assuming propositional resizing or an alternative like replacement or subset collection in the type theory (making all sets of Dedekind real numbers equivalent to each other), the Dedekind real numbers are
1. no longer small relative to the propositions from which the Dedekind cuts are defined from, as the set of Dedekind cuts (which are pairs of subsets/predicates of a dense strict linear order) is not small, but only locally small, relative to the propositions
2. no longer unique, as a result of point 1., there exists a set of Dedekind real numbers for every internal universe in the ambient universe, and we can no longer speak of *the* Dedekind real numbers. If there are no internal universes, we cannot even define a set of Dedekind real numbers
3. assuming there is at least one internal universe in the ambient universe, no set of Dedekind real numbers is even guaranteed to be terminal in the ambient universe, depending upon the internal universe hierarchy in the ambient universe: Suppose one has an internal hierarchy of universes indexed by the natural numbers, where for natural numbers $n$ and $m$, $U\_n$ embeds into $U\_m$ for $n \leq m$. Then for every set of Dedekind real numbers $\mathbb{R}\_n$ defined from propositions in $U\_n$ there is a larger set of Dedekind real numbers $\mathbb{R}\_m$ defined from propositions in $U\_m$, where $\mathbb{R}\_n$ embeds in $\mathbb{R}\_m$, and thus there is no such terminal Archimedean ordered field in the ambient universe.
In such a case, how would one even go about even defining axiom $\mathbb{R}\flat$? Do we simply get multiple inequivalent versions of axiom $\mathbb{R}\flat$ of varying strengths? Suppose one simply selects a random set of Dedekind real numbers $\mathbb{R}$ for axiom $\mathbb{R}\flat$, how exactly does that particular axiom $\mathbb{R}\flat$ interacts with sets of Dedekind real numbers $\mathbb{R}'$ where $\mathbb{R}$ embeds into $\mathbb{R}'$ but it isn't possible to prove (without propositional resizing/subset collection/etc) that $\mathbb{R}$ is equivalent to $\mathbb{R}'$, as well as sets of Dedekind real numbers $\mathbb{R}^\dagger$ where $\mathbb{R}^\dagger$ embeds into $\mathbb{R}$?
| https://mathoverflow.net/users/483446 | Predicativity and axiom $\mathbb{R}\flat$ in real cohesive homotopy type theory | My suggestion would be to start from your intended semantics. First ask what kind of "predicative topos" you have in mind where the theory you're asking about would have a model, and construct a particular such category that you're interested in. Then once you have a model, ask what axioms that model satisfies. That's the motivation for the original axiom $\mathbb{R}\flat$: it holds in an intended model of [Euclidean-topological infinity-groupoids](https://www.ncatlab.org/nlab/show/D-topological+infinity-groupoid).
| 2 | https://mathoverflow.net/users/49 | 424520 | 172,445 |
https://mathoverflow.net/questions/424479 | 0 | Let $(x\_{n})\_{n}$ be a sequence in a Banach space $X$. Assume that the set $\{x\_{n}:n=1,2,\cdots\}$ is finite. Let $(f\_{m})\_{m}$ be a weak\*-null sequence in $X^{\*}$ satisfying the following conditions:
(1) the limit $a\_{m}:=\lim\limits\_{n}\langle f\_{m},x\_{n}\rangle$ exists for each $m$;
(2) the limit $a:=\lim\limits\_{m}a\_{m}$ exists.
Question. $a=0$ ?
Thank you !
| https://mathoverflow.net/users/41619 | Weak*-null sequences in dual spaces | Let's write $\{x\_{n}:n=1,2,\cdots\}=\{y\_{1},y\_{2},\cdots,y\_{N}\}$. For each $i=1,2,\cdots,N$, we set $A\_{i}=\{n:x\_{n}=y\_{i}\}$. Then $\cup\_{i=1}^{N}A\_{i}=\mathbb{N}$. Then there exists $1\leq i\_{0}\leq N$ so that $A\_{i\_{0}}$ is infinite. Write $A\_{i\_{0}}=\{k\_{n}:n=1,2,\cdots\}$. Hence $x\_{k\_{n}}=y\_{i\_{0}}$ for all $n$. So we get
$$a=\lim\_{m}\lim\_{n}f\_{m}(x\_{k\_{n}})=\lim\_{m}f\_{m}(y\_{i\_{0}})=0.$$
| 0 | https://mathoverflow.net/users/41619 | 424521 | 172,446 |
https://mathoverflow.net/questions/424518 | 2 | We all know that the famous Hypergeoemtric function $\_2F\_1$ has an integral form as follows:
$$\_2F\_1(a,b,c;z)=-\frac{e^{-i\pi c} \Gamma(c)}{4\Gamma(b)\Gamma(c-b)\sin\pi b\sin\pi(c-b)}\int\_P t^{b-1}(1-t)^{c-b-}(1-zt)^{-a}dt,$$
where $c\neq0,-1,-2,...$ and $P$ is Pochhammer contour. This integral form can give an analytic continuation of Hypergeometric function from $|z|<1$ to the whole $z$-plain except the slit $[1,\infty]$.
Recently, I read a paper -
* W. Becken, P. Schmelcher, *The analytic continuation of the Gaussian hypergeometric function $\_2F\_1(a,b,c;z)$ for arbitrary parameters*, Journal of Computational and Applied Mathematics **126** (2000) 449–478. <https://doi.org/10.1016/S0377-0427(00)00267-3>, <https://core.ac.uk/download/pdf/82108003.pdf>
The paper gave us a conclusion that Gaussian hypergeometric function can be analytically continuated to the whole complex $z$-plain excluding only $e^{\pm i\pi/3}$. If the conclusion is true, it will help me a lot since the continuation domain covers nearly a whole plain, and I can regard the hypergeometric function as an entire function except two exact points. Can I directly use this conclusion? Is this conclusion verified formally in maths world? Is the continuated function single-valued?
| https://mathoverflow.net/users/483996 | Does any Hyper-geometric function can be analytically continuated to the whole complex plain except $e^{\pm i\pi/3}$ | Continuation to the whole $z$-plane? As stated, not true. Example:
$$
{}\_2F\_1\left(-\frac12,1;1;z\right) = (1-z)^{1/2},\quad |z|<1,
$$
cannot be extended analytically to any neighborhood of $z=1$.
| 3 | https://mathoverflow.net/users/454 | 424542 | 172,452 |
https://mathoverflow.net/questions/415346 | 6 | I'm quite a newbe in the field of motives & A1 homotopy theory,
so please forgive me if the question is too elementary:
In the intro from [wikipedia on Nisnevish topology](https://en.wikipedia.org/wiki/Nisnevich_topology)
is remarked that it's founder Yevsey Nisnevich was originally motivated
by the theory of adeles.
Could somebody elaborate which features/ nuances in
the development of Nisnevich topology
reflect the striking influence on it
by the theory of adeles as stated there?
| https://mathoverflow.net/users/108274 | Nisnevich topology inspired by Adeles | He used the topology to give a cohomological interpretation of the class group/set $c(G)$ of an affine algebraic group (which is defined in terms of adeles, see below), and then to partially prove the Grothendieck-Serre conjecture. I quote Nisnevich's summary of section 2 of his thesis.
2.1. The principal role in the proofs of all results of the thesis is played by a direct relationship between the two kinds of invariants mentioned above: the arithmetic and cohomological invariants for a general affine group scheme $G$.
The idea of such relationships already appeared in the classical isomorphism $$H^{1}(X\_{Zar},\mathbb{G}\_{m,X})\simeq\mathrm{Pic}(\mathbb{G}\_{m,X})=c(\mathbb{G}\_{m,X}),$$ which was generalized by Voskresenky and Harder to some other groups $G$. Unfortunately, in general [the] Zariski topology on $X$ is too weak to give a description of $c(G)$ and other arithmetic invariants and properties of $G$. To describe them one needs a stronger topology on $X$. For these purposes an appropriate topology is in introduced in Ch. I of the thesis. To use the topology in Ch. IV-V we define it in the following more general situation.
2.2. Let $R$ be a one dimensional regular ring, $k$ the field of fractions of $R$, $X=\mathrm{Spec}\,R$, $A$ and $A(X)$ the ring of adeles and $X$-integral adeles of $X$, $G$ an affine group scheme over $X$ with smooth general fibre $G\otimes\_{R}k$, $c(G)=G(A(X))\backslash G(A)/G(k)$ the set of double cosets of $G(A)$, or in the case when $c(G)$ is a group, the class group of $G$.
2.3 Theorem-definition. There exists a Grothendieck topology $X\_{cd}$ on $X$ (we shall call it the completely decomposed topology) for which one has a canonical bijection $$\alpha(G)\colon H^{1}(X\_{cd},G)\simeq c(G).$$
| 6 | https://mathoverflow.net/users/nan | 424543 | 172,453 |
https://mathoverflow.net/questions/420071 | 7 | Given a general topological space $X$ does the category $\mathbf{coShv}(X,\mathbf{Mod}\_R)$ have enough projectives ? I know that under some conditions this is true, for example if $X$ is a cell complex equipped with the Alexandrov topology and $R$ is a field (in the case of cellular sheaves of vector spaces). In this case one can define cosheaf homology as the left derived functor of the global section functor. I am wondering if one can always define cosheaf homology in this way ? I read that the problem with cosheaves is that cosheafification does not exist in general. Is this problem connected with the problem of defining cosheaf homology ?
| https://mathoverflow.net/users/299054 | Does the category of cosheaves have enough projectives? | The answer is yes. A reference for the claim about Alexandrov spaces and cellular (co)sheaves is [Justin Curry's thesis](https://arxiv.org/pdf/1303.3255.pdf). When he proves Claim 7.1.9 he points out that the statement is true for all spaces $X$. It's also not essential that $R$ is a field. You can see directly that Justin's proof does not use any assumptions on $R$. The main point is that, given a cosheaf $F$, there is a projective cosheaf $P$ surjecting onto $F$. Once you have $P$, you can build the full projective resolution as usual. To define $P$ all you need is direct sums in the category of cosheaves of $R$-modules, and this follows from the bicompleteness of the category of $R$-modules, as observed in Theorem 2.14 of Prasolov's 2018 paper [Cosheaves](https://arxiv.org/pdf/1804.07988.pdf).
Now, just because you have projective resolutions does not mean you should expect your proof about sheaves (via injective resolutions) to immediately dualize. As Bredon writes in Cosheaves and Homology (1968) "In our opinion one cannot expect to find anything like a complete duality with sheaf theory and one must be prepared, from the start, to dispense with some basic properties."
In particular, to actually compute cosheaf homology, it's often more useful to use a resolution by flabby cosheaves, as Bredon does, but the existence of such a resolution requires X to be homologically locally connected (see page 35 of Bredon's *Sheaf Theory* book). The existence of cosheaf homology was never in doubt, and is [discussed in this mathoverflow thread from ten years ago](https://mathoverflow.net/questions/99969/cosheaf-homology-and-a-theorem-of-beilinson-in-a-paper-on-mixed-tate-motives).
Also, without some restriction on $X$, we don't even know how to define constant cosheaves. The constant precosheaf is easy (just send everything to a fixed $R$-module $M$) but constant cosheaf is harder and traditionally required $X$ to be locally connected.
It's also true that in Bredon's time, cosheafification was not known to exist in general, as the OP wrote. However, as pointed out in [this mathoverflow thread](https://mathoverflow.net/questions/120729/cosheafification), cosheafification does exist in your setting of $R$-modules, because that's a locally presentable and accessible category. A published reference is the [2016 paper of Prasolov](https://arxiv.org/pdf/1605.01555.pdf).
I think the reason cosheafification was considered hard in Bredon's time was the lack of cosmall objects. But this argument via local presentability gets around that. I don't think it's the same problem with the existence of resolutions. I should point out that Prasolov's 2018 paper develops quite a lot of theory that might help you, but it's about cosheaves of Pro(R)-modules. In that setting you do NOT have enough projectives but you have enough quasi-projectives; see Prasolov's appendix A to learn more.
| 0 | https://mathoverflow.net/users/11540 | 424546 | 172,454 |
https://mathoverflow.net/questions/424545 | 20 | I sent a paper to an elite journal (the top in the field). Two weeks later I got a decision "reject" but the editors added that "we believe it should deserve a good publicity and publication".
The paper was described by the associate editor as "of very good quality" and the reason for the rejection is that "the techniques used look rather far to me from the *journal* readership".
It should be mentioned that the main results of the paper are very much related to the scope of the journal. Moreover, this journal has already published more than 5 papers on the subject with results similar to those of mine (but in rather special cases, and according to few experts in the field no doubt that my new result is a significant step forward).
My questions are:
1. Is it common that a paper that contains results of high enough quality that are well in line with the journal scope, is rejected because the techniques are not familiar to the readership?
2. If it is common, could anyone explain the reason behind this policy? To me it seems odd, as in mathematics applying tools from one subject to solve problems in another subject, as long as it is done correctly, is considered to be a good development.
| https://mathoverflow.net/users/161837 | Rejection for a seemingly odd reason | When you submit to an elite journal, expect a rejection most of the time. Then submit to a less-prestigious journal. It is a waste of your time to attempt an analysis of the reasons given for rejection.
1. Yes, it is common for journals that receive far more submissions than they can publish to reject most of them — sometimes for boilerplate reasons, sometimes for no reason at all.
| 48 | https://mathoverflow.net/users/454 | 424547 | 172,455 |
https://mathoverflow.net/questions/424549 | 8 | We all take for granted the theorem that every ordinal $\alpha > 0$ has a Cantor normal form, and there are plenty of proofs of it, some of which are on this site. However, where was it proved? Was it actually proved by Cantor in his 1883 paper introducing ordinals? Or somewhere else?
| https://mathoverflow.net/users/473200 | Where was the Cantor normal form theorem first proved? | Cantor proved the normal form theorem in his 1895 paper [Beiträge zur Begründung der transfiniten Mengenlehre](https://eudml.org/doc/157768), which was his last paper on transfinite set theory.
| 12 | https://mathoverflow.net/users/11260 | 424550 | 172,456 |
https://mathoverflow.net/questions/424539 | 1 | I asked this question fourteen days ago on MathStackexchange (see [here](https://math.stackexchange.com/questions/4460652/star-autonomous-categories-are-categorifications-of-boolean-algebras)). I have not received any answers or comments until now. It seems to me that on MathStackexchange not many people are familiar with star-autonomous categories. Following [Nick Champion's advice](https://meta.mathoverflow.net/questions/5012/can-i-ask-a-question-on-mathoverflow-and-also-on-another-site/5013#5013), I therefore have decided to cross-post the question on this site.
**1. Question**
The [n-Lab article](https://ncatlab.org/nlab/show/Chu+construction) on the Chu-construction says:
>
> "Armed with just this much knowledge, and knowledge of how star-autonomous categories behave (as **categorified versions of Boolean algebras**, or perhaps better Boolean rigs), the star-autonomous structure on $\operatorname{Chu}(C,d)$ can pretty much be deduced (or strongly guessed) […]."
>
>
>
How do star-autonomous categories behave as categorified versions of Boolean algebras or Boolean rigs? In what way is the term categorification used here?
**2. Wikipedia says**
One explanation might be given on [wikipedia](https://en.wikipedia.org/wiki/*-autonomous_category#Examples):
>
> "A degenerate example [of a star-autonomous category] (all homsets of cardinality at most one) is given by any Boolean algebra (as a partially ordered set) made monoidal using conjunction for the tensor product and taking 0 as the dualizing object."
>
>
>
I suppose the internal hom of two objects $a,b$ in this category is $\neg a \lor b $, correct? The dual functor is the complement?
Edit for future readers: The quoted statement on the nLab seems to have been inaccurate. The nLab entry has now been changed to: "Armed with just this much knowledge, and knowledge of how star-autonomous categories behave (*as categorified versions of linear logic*), the star-autonomous structure on $\operatorname{Chu}(C,d)$ can pretty much be deduced (or strongly guessed).“
| https://mathoverflow.net/users/160778 | Star-autonomous categories are categorifications of Boolean algebras? | The starting point for decategorification is the observation that a category in which any parallel arrows are equal must necessarily be a preorder.
Restricting to skeletal categories makes it a poset.
Thus, we might as well ask: what is a \*-autonomous category whose underlying category is a poset?
For the specific case when the poset is a Boolean algebra,
we set $A→B=A⊸B$, and $A⊢B$ means there is a morphism $A→B$.
We have $A⊗X≅(A⊸X^\*)^\*$, which means $A⊗X=A∧B$.
Most axioms of propositional logic are now straightforward to verify.
In particular, we can identify the global dualizing object $⊥$:
if for all objects $A$ the canonical map $A→(A⊸⊥)⊸⊥$
is an isomorphism, then $⊥$ must be the initial object,
since $B⊢(A→B)$ in our case.
More generally, \*-autonomous posets that are not Boolean algebras provide models for [linear logic](https://ncatlab.org/nlab/show/linear%20logic). So in the strict sense \*-autonomous categories do not categorify Boolean algebras (or rigs), but rather the corresponding algebraic structure for linear logic.
| 4 | https://mathoverflow.net/users/402 | 424554 | 172,457 |
https://mathoverflow.net/questions/424457 | 3 | Do we have any examples of non-modular elliptic curves over number fields $K \neq \mathbb{Q}$?
In particular, I came across a paper by Freitas, Le Hung, and Siksek, "[Elliptic curves over real quadratic fields are modular](https://doi.org/10.1007/s00222-014-0550-z)", which shows that there are at most finitely many non-modular elliptic curves over a fixed totally real number field. Is there any known example of $E/K$ where $E$ is a non-modular elliptic curve and $K$ is a totally real number field?
| https://mathoverflow.net/users/478525 | Non-modular elliptic curves | It is a widely believed conjecture that all elliptic curves, over any number field $K$, are modular (in the sense that there exists an automorphic representation [\*] $\pi$ of $\operatorname{GL}\_2 / K$ whose $L$-function is the same as that of $E$). No counterexamples are known, and it would be *extremely* big and disturbing news for number theory if somebody stumbled across one.
([\*] If you are surprised not to see the word "cuspidal" here, then there's a reason for that: you need to take non-cuspidal $\pi$ if $E$ has CM, and the field it has CM by is a subfield of $K$. In all other cases $\pi$ will be cuspidal. In particular, if you are only looking at totally-real $K$, then you can validly write "cuspidal" there.)
| 9 | https://mathoverflow.net/users/2481 | 424562 | 172,458 |
https://mathoverflow.net/questions/424551 | 3 | **The main references** for this question are
**1** : V.Voevodsky's paper [Triangulated categories of motives over a field](https://www.math.ias.edu/vladimir/sites/math.ias.edu.vladimir/files/s5.pdf)
**2** : the book "Lecture notes in motivic cohomology" written by Carlo Mazza, Vladimir Voevodsky and Charles Weibel.
**Context:** in both **1** and **2** we can find a possible definition of motivic cohomology using $\mathit{Hom}$ in the category of geometric motives over $k$. I begin to recall the main ideas of this.
This category looks to be in **1** the pseudo-abelianisation of a localisation of the bounded complexes of smooth schemes over a field $k$, and it is in **2** the same construction, but now on the category of sheaves (of $R$-algebra) with transfers for the Nisnevich site (**questions:** why this particular site? Do we obtain the same category if $R=\mathbb{Z}$?). Then the category of effective Chow motives over $k$ embeds through a functor $M$ into effective geometric motives over $k$ (see proposition 20.1 of **2**). Now, given a smooth scheme $X$ we define as in [**2**, definition **14.16**] the motivic cohomology (for a ring $R$) to be
$$H^{n,i}\_{\text{mot}}(X,R) \overset{\text{def}}{=}\mathit{Hom}\_{\text{geo. motives}}(M(X),R(i)[n]).$$
**My question is:** Are we able to generalize the above construction, replacing $R$ by a sheaf. In particular, I am interested in the case where $R=M$ with $M$ a Chow motive over $X$ seen as a locally constant sheaf over $X$. If the answer is "Yes and it is exactly the same construction" then is there any reference which deals with it? More generally do you have any reference for motivic cohomology of a scheme $X$ with coefficient in a Chow motive over $X$?
| https://mathoverflow.net/users/169282 | Motivic cohomology as $\mathit{Hom}$ in the category of geometric motives, with coefficient in a Chow motive | You can also define the motivic cohomology of $X$ as a $\mathrm{Hom}$ in the category $\mathrm{DM}(X)$ of motives over $X$, as defined by Cisinski-Déglise – see Example 11.2.3 in their book Triangulated categories of mixed motives (Springer, 2019).
Once in $\mathrm{DM}(X)$, you can use "motivic sheaves on $X$", for example using relative Chow motives. The category of relative Chow motives exists for reasonable base, see the work of Corti-Hanamura and the article "Borel–Moore motivic homology and weight structure on mixed motives" by Fangzhou Jin.
| 2 | https://mathoverflow.net/users/6506 | 424563 | 172,459 |
https://mathoverflow.net/questions/424455 | 6 | As discussed here, [Are monads monadic?](https://mathoverflow.net/questions/19906/are-monads-monadic), in "On the monadicity of finitary monads" by Steve Lack, the following is shown, the forgetful functor from $Mnd\_f(C) \rightarrow Endo\_f(C)$ is monadic, note the finitarity restrictions on both domain and codomain. In the same paper, Steve Lack, also shows a generalization for operads, but those recover only cartesian monads as discussed here [Monad arising from operad](https://mathoverflow.net/questions/66313/monad-arising-from-operad/66324#66324). Are there any results known, which generalize this to the forgetful functor $Mnd(C) \rightarrow Endo(C)$? Note the lack of finitarity restrictions.
Edit: Assume that C supports enough constructions to have internal left Kan extensions up to universes, thus guaranteeing some form of free monads. Note that for example on the universe in type theory we can take a left kan extension and then take its free monad to obtain a free monad, which exists in a higher universe. This ensures existence of arbitrary free monads at the cost of having to think about a universe.
Edit 2: To explain existence of free monad under those conditions and stop derailing the discussion away from the issue of infinitary monads, I would like to point out the construction and then write a quick argument.
```
data Lan {l1 l2 l3 : _} (G : Set l1 -> Set l2) (A : Set l3) : Set (lsuc l1 ⊔ l2 ⊔ l3) where
FMap : {X : Set l1} -> G X -> (X -> A) -> Lan G A
```
```
data Freer {l1 l2 l3 : _} (F : Set l1 -> Set l2) (A : Set l3) : Set (lsuc l1 ⊔ l2 ⊔ l3) where
Pure : A -> Freer F A
Impure : Lan F (Freer F A) → Freer F A
```
Here Lan means left kan extension of F along identity.
Note that this always exists even when the naive free monad of F does not.
Further more by algebraic-freeness of existing free monads in type theory we obtain $Talg (Freer \: F) \cong Falg (Lan \: F)$. Further by dualizing a result of Hinze we can obtain $Falg (Lan \: F) \cong Falg \: F$, for all $F$. We can actually obtain this result even when $F$ is not a strong endofunctor. (see page 2133 (26 of 52) in [Adjoint folds and unfolds—An extended study](https://www.cs.ox.ac.uk/people/ralf.hinze/publications/SCP-78-11.pdf))
Edit 3: for proofs feel free to assume any background that would be compatible with predicative HoTT or MLTT/MLTT+K and would make this property hold if it's known to turn out consistent.
| https://mathoverflow.net/users/129807 | Are infinitary monads monadic? | These question of existence of free monad are not "derailling" the discusion. They are the whole point of the discusion. Let me clarify :
If I'm not mistaken, we have the following:
**Theorem:** Let $V$ be a monoidal category. Let $A$ be the category of monoids in $V$, then the forgetfull functor $U: A \to V$ is monadic if and only if it admits a left adjoint.
**Sketch of the Proof:** Just check the other two conditions of Beck criterion. The forgetfull functor is conservative. So we need to check that it create U-split coequalizer. Given a U-split coequalizer diagram $ X \rightrightarrows Y$ in $A$, and let $Z$ be its colimit in $V$ ( which exists by assumption). Then $Z$ comes with a monoid structure because the colimit defining it is split, so preserved by any functor, in particular the tensor product, so that $Z \otimes Z $ is the coequalizer of $X \otimes X$ and $ Y \otimes Y$ which you can use to construct an operaion $ Z \otimes Z \to Z$ making the relevant maps morphism of monoids.
With a little bit of additional work, you can conclude that this makes $Z$ a colimit in the category of monoids. $\square$
It follows that :
**Corollary:** Given $C$ any category, and $S \subset End(C)$ a full subcategory of endofunctors such that, $S$ contains the identity, is closed under composition, and every element of $S$ generates a free monads which is also in $S$, then the category of monads that are in $S$ is monadic over $S$.
**Proof:** Apply the previous result to the monoidal category $S$ (for the composition of endofunctors). $\square$
So the all point of the question is to find nice class of endofunctor for which free monads construction are available. For example :
**Proposition :** If $C$ is a locally presentable then the category of accessible monads on $C$ is monadic over the category of accessible endofunctor on $C$.
Though I have to insist that I do not know any exemple of category for which free monad construction on $C$ exists unconditionally. In fact, I'm relatively convinced this is impossible. To give an example :
**Proposition:** It is inconsistent with ZF (even with IZF) that every endofunctor of the category of sets admit a free monads.
**proof:** Consider the power-set endofunctor $X \mapsto \mathcal{P}(X)$, if the free monad on $\mathcal{P}$ existed, then as endomorphism monad exists in Set, the category of algebras for this monad would be the algebras for the endofunctor $\mathcal{P}$. In partiular the category of $\mathcal{P}-algebra$ would have an initial object, but by a well known theorem of Lambek, an initial algebra for an endofunctor $\mathcal{P}$ is always such that $\mathcal{P}(X) \simeq X$, but this is impossible by the diagonal argument. $\square$
So, unless the type theory you are using is inconsistent with ZFC, or you are using a notion of "endofunctor of sets" that does not corresponds to the usual notion, you won't be able to show that any endofunctor of the category of sets admit a free monad.
I don't quite understand what you are trying to explain with universes, but I think the problem you'll run into is that if you start form an endomorphism of the category of U-sets, then you might be able to construct something that look like a free monad on the category of V-sets for V a larger universe, but that's no longer a monad on U-set, so the forgetfull functor don't take you back to an endofunctor on U-set, but on V-set, but if you start with an endofunctor on $V$-set, then you need to go to an even larger universe W to get your free monad, but for endofunctors on W you'll need to get to a larger one and so one... And if your theory is consistent with ZFC, you can never find this way a setting where both the free monad and the forgetfull functor are simultenously defined, so that you can have a well defined "free monad monad" that acts on a category.
What will work (and maybe this is essentially what you are doing) is that if $\kappa$ is an innaccesible cardinal ( any regular cardinal actually), then any endofunctor of the category of $\kappa$-small sets can be extended into a $\kappa$-accessible endofunctor of the category of all sets, and then amongst $\kappa$-accessible endofunctors free monad constructions are available, and you can get a monadicity result by working with $\kappa$-accessible endofunctor.
| 6 | https://mathoverflow.net/users/22131 | 424564 | 172,460 |
https://mathoverflow.net/questions/422925 | 3 | Let $\rho(A)$ denote the spectral radius of a square matrix $A$. Let $r,d$ be positive integers. Let $X\_1,\dots,X\_r$ be $d\times d$-real matrices. Then do we necessarily have $$\rho(X\_1\dots X\_r)^{2/r}\leq \frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r)?$$
If $X\_1,\dots,X\_r$ are $d\times d$-real matrices with $d\leq r$ and
$$\rho(X\_1\dots X\_r)^{2/r}=\frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r),$$
then is $\text{rank}(X\_j)=1$ for $1\leq j\leq r$?
If $1\leq d\leq r$, then do there always exist $d\times d$-real matrices $X\_1,\dots,X\_r$ with
$$\rho(X\_1\dots X\_r)^{2/r}=\frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r)?$$
I have used a gradient descent algorithm to try to find counterexamples of this inequality, and the gradient descent algorithm was good at finding examples where
$$\rho(X\_1\dots X\_r)^{2/r}\approx\frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r),$$ but in all these examples, we have $\text{rank}(X\_j)=1$ for all $j$, and the gradient descent algorithm was unable to find any cases where $$\rho(X\_1\dots X\_r)^{2/r}>\frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r).$$
| https://mathoverflow.net/users/22277 | Is $\rho(X_1\dots X_r)^{2/r}\leq \frac{d}{r}\cdot\rho(X_1\otimes X_1+\dots+X_r\otimes X_r)$ for $d\times d$-real matries $X_1,\dots,X_r$? | The answer to all these questions is **Yes**. We shall answer this question in the more general case when $X\_1,\dots,X\_r$ are complex matrices (the inequality for this question looks a little bit different in the complex case since we need to apply conjugations and transposes). This answer shall be in the context of quantum channels. I recommend [The Theory of Quantum Information](https://cs.uwaterloo.ca/%7Ewatrous/TQI/) by John Watrous (2018) for more information on quantum information theory. Much of content of this answer is similar to the answers to [this related question](https://mathoverflow.net/q/423510/22277); fedja remarked that the idea behind the answer to that previous question also applies to this question, so fedja could have answered at least most of this question before I did.
A function $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ is said to be positive if $\mathcal{E}(P)$ is positive semidefinite whenever $P$ is positive semidefinite. If $W$ is a finite dimensional vector space, then let $L(W)$ be the set of all homomorphisms from $W$ to $W$. We say that $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ is completely positive if whenever $V$ is a finite dimensional complex Hilbert space, the mapping $\mathcal{E}\otimes 1\_V:L(\mathbb{C}^d\otimes V)\rightarrow L(\mathbb{C}^d\otimes V)$ is positive. We say that a linear mapping $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ is trace preserving if $\text{Tr}(\mathcal{E}(A))=\text{Tr}(A)$ whenever $A\in M\_d(\mathbb{C})$. We say that a linear mapping $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ is a quantum channel if it is both completely positive and trace preserving.
If $X\_1,\dots,X\_r$ are complex matrices, then define a mapping $\Phi(X\_1,\dots,X\_r):M\_{d}(\mathbb{C})\rightarrow M\_{d}(\mathbb{C})$ by letting
$$\Phi(X\_1,\dots,X\_r)(X)=X\_1XX\_1^\*+\dots+X\_rXX\_r^\*.$$ The completely positive mappings from $M\_d(\mathbb{C})$ to $M\_d(\mathbb{C})$ are precisely the mappings of the form
$\Phi(X\_1,\dots,X\_r)$, and the quantum channels $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ are precisely the mappings of the form $\Phi(X\_1,\dots,X\_r)$ where $X\_1^\*X\_1+\dots+X\_r^\*X\_r=1\_d$.
If $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$; quantum channels are a lot like stochastic matrices.
I claim that $\rho(X\_1\dots X\_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(X\_1,\dots,X\_r))$ whenever $X\_1,\dots,X\_r$ are $d\times d$-complex matrices.
Let $O\_{d,r}$ be the set of all $(X\_1,\dots,X\_r)\in M\_d(\mathbb{C})^r$ such that
$\Phi(X\_1,\dots,X\_r)$ is not nilpotent. Let $E\_{d,r}$ be the set of all $(X\_1,\dots,X\_r)\in M\_d(\mathbb{C})^r$ such that
$\Phi(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})$ is a quantum channel for some complex $\lambda\neq 0$ and invertible matrix $B$. Then by [my answer to my other question](https://mathoverflow.net/a/423803/22277), $E\_{d,r}$ is a dense subset of $O\_{d,r}$, and clearly $O\_{d,r}$ is dense in $M\_d(\mathbb{C})^r$.
Observe that $$\frac{\rho(Y\_1\dots Y\_r)^{2/r}}{\rho(\Phi(Y\_1,\dots,Y\_r))}=\frac{\rho(\lambda BY\_1B^{-1}\dots \lambda BY\_rB^{-1})^{2/r}}{\rho(\Phi(\lambda BY\_1B^{-1},\dots,\lambda BY\_rB^{-1}))}$$ whenever $\lambda\neq 0$, $B$ is invertible, and $\Phi(Y\_1,\dots,Y\_r)$ is not nilpotent. Therefore, if $\rho(Z\_1\dots Z\_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(Z\_1,\dots,Z\_r))$ whenever $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel, then $\rho(Y\_1\dots Y\_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(Y\_1,\dots,Y\_r))$ whenever $(Y\_1,\dots,Y\_r)\in E\_{d,r},$
so because $E\_{d,r}$ is dense in $M\_{d}(\mathbb{C})^{r}$, we will be able to conclude that $\rho(X\_1\dots X\_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(X\_1,\dots,X\_r))$ whenever $X\_1,\dots,X\_r\in M\_d(\mathbb{C})^r$.
If $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel, then $$d=\text{Tr}(1\_d)=\text{Tr}(Z\_1^\*Z\_1+\dots+Z\_r^\*Z\_r)=\|Z\_1\|^2\_2+\dots+\|Z\_r\|\_2^2$$ where $\|\cdot\|\_2$ denotes the Frobenius norm.
Therefore, by the arithmetic-geometric mean inequality, we have our main inequality, namely
$$\rho(Z\_1\dots Z\_r)^{2/r}\leq\|Z\_1\dots Z\_r\|\_\infty^{2/r}\leq(\|Z\_1\|\_\infty^2\dots\|Z\_r\|\_\infty^2)^{1/r}\leq(\|Z\_1\|\_2^2\dots\|Z\_r\|\_2^2)^{1/r}\leq\frac{\|Z\_{1}\|\_2^2+\dots+\|Z\_r\|\_2^2}{r}
=\frac{d}{r}=\frac{d}{r}\rho(\Phi(Z\_1,\dots,Z\_r)).$$
Furthermore, if $\rho(Z\_1\dots Z\_r)^{2/r}=\frac{d}{r}\rho(\Phi(Z\_1,\dots,Z\_r))$,
then $\|Z\_j\|\_\infty^2=\|Z\_j\|^2\_2=d/r$ for all $j$. However, The only way to get $\|Z\_{j}\|\_\infty=\|Z\_j\|\_2$ is if $\text{Rank}(Z\_j)=1$ since $\|W\|\_{\infty}=\|W\|\_2$ precisely when $\text{Rank}(W)=1$.
Therefore, for all $j$, there are column vectors $u\_j,v\_j$ such that $Z\_j=u\_jv\_j^\*$.
In this case, $$\|Z\_j\|\_2^2=\text{Tr}((u\_jv\_j^\*)(u\_jv\_j^\*)^\*)=\text{Tr}(u\_jv\_j^\*v\_ju\_j^\*)=
\text{Tr}(u\_j^\*u\_jv\_j^\*v\_j)=\|u\_j\|^2\cdot\|v\_j\|^2.$$ Therefore, $\|Z\_j\|\_2=\|u\_j\|\cdot\|v\_j\|=\sqrt{d/r}$.
$$\rho(Z\_1\dots Z\_r)=|\text{Tr}(Z\_1\dots Z\_r)|=|\text{Tr}(u\_1v\_1^\*\dots u\_rv\_r^\*)|
=|\text{Tr}(v\_1^\*u\_2\dots v\_r^\*u\_1)|$$
$$=|\langle v\_1,u\_2\rangle|\dots|\langle v\_r,u\_1\rangle|.$$
Since $\rho(Z\_1\dots Z\_r)=\|Z\_1\|\_\infty\dots\|Z\_r\|\_\infty$, we have
$$|\langle v\_1,u\_2\rangle|\dots|\langle v\_r,u\_1\rangle|=\|u\_1\|\cdot\|v\_1\|\dots \|u\_r\|\cdot\|v\_r\|.$$
This means that there are constants $\gamma\_1,\dots,\gamma\_r$ where
$v\_j=\overline{\gamma\_j}\cdot u\_{j+1}$. Therefore,
$Z\_j=u\_jv\_j^\*=u\_j\gamma\_j u\_{j+1}^\*=\gamma\_j u\_ju\_{j+1}^\*$.
Now, set $w\_j=\|u\_j\|\cdot v\_{j}=\|u\_j\|\cdot\overline{\gamma\_j}u\_{j+1}.$
Then $$1\_d=\sum\_{j=1}^{r}Z\_j^\*Z\_j=\sum\_{j=1}^{r}(u\_jv\_j^\*)^\*u\_jv\_j^\*=\sum\_{j=1}^{r}v\_ju\_j^\*u\_jv\_j^\*=\sum\_{j=1}^{r}\|u\_{j}\|^2\cdot v\_jv\_j^\*=\sum\_{j=1}^{r}w\_jw\_j^\*.$$
Observe that $1\_d=\sum\_{j=1}^{r}w\_jw\_j^\*$ precisely when $x=\sum\_{j=1}^{r}w\_{j}\cdot\langle x,w\_j\rangle$. We say that a tuple of vectors $(h\_1,\dots,h\_r)$ is a normalized tight frame if $1\_d=\sum\_{j=1}^{r}h\_jh\_j^\*$. We say that a normalized tight frame is equal-norm if $\|h\_1\|=\dots=\|h\_r\|$. Observe that $\|w\_j\|^2=d/r$ for all $j$, so $(w\_1,\dots,w\_r)$ is an equal-norm tight frame.
One can also obtain tuples $(Z\_1,\dots,Z\_r)$ where $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel and where $\rho(Z\_1,\dots,Z\_r)^{2/r}=d/r$ from equal-norm tight frames. Suppose that $(w\_1,\dots,w\_r)\in(\mathbb{C}^d)^r$ is an equal-norm tight frame. Then we necessarily have $\|w\_j\|^2\_2=d/r$ for all $j$. Now, suppose that $|\alpha\_j\beta\_j|^2=r/d$ for $1\leq j\leq r$. Let $v\_j=\alpha\_jw\_j,u\_{j+1}=\beta\_{j+1}w\_j$, and as always, set $Z\_j=u\_jv\_j^\*$ for $1\leq j\leq r$. Then the reader can verify that $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel with $\rho(Z\_1,\dots,Z\_r)^{2/r}=\frac{d}{r}$ and every quantum channel $\Phi(Z\_1,\dots,Z\_r)$ with $\rho(Z\_1,\dots,Z\_r)^{2/r}=\frac{d}{r}$ is of this form.
The existence of equal-norm tight frames $(w\_1,\dots,w\_r)\in(\mathbb{R}^d)^r$ is already a known result. For example, the existence of such frames is Corollary 7.1 in [An Introduction to Finite Tight Frames](https://www.math.auckland.ac.nz/%7Ewaldron/Preprints/Frame-book/draft26-7.17.pdf) by Shayne F. D. Waldron.
By applying continuity, we can actually show that $\text{Rank}(X\_j)=1$ even if we drop the assumption that $\Phi(X\_1,\dots,X\_r)$ is a quantum channel.
Let $f:O\_{d,r}\rightarrow\mathbb{R}$ be the function where
$f(X\_1,\dots,X\_r)=\frac{\rho(X\_1\dots X\_r)^{2/r}}{\rho(\Phi(X\_1,\dots,X\_r))}.$
Let $g^-:M\_d(\mathbb{C})\rightarrow\mathbb{R}$ be the function where
$g^-(X)=|\lambda\_1\lambda\_2|$ where $\lambda\_1,\dots,\lambda\_d$ are the eigenvalues of $X$ arranged in an order so that $|\lambda\_1|\geq|\lambda\_2|\geq\dots\geq|\lambda\_d|$.
Define $g:O\_{d,r}\rightarrow\mathbb{R}$ by letting $$g(X\_1,\dots,X\_r)=\frac{g^-(X\_1)+\dots+g^-(X\_r)}{r\cdot\rho(\Phi(X\_1,\dots,X\_r))}.$$
Suppose that $(X\_1,\dots,X\_r)\in O\_{d,r}$ and $f(X\_1,\dots,X\_r)=\frac{d}{r}$. Then for each $\delta>0$, there is a neighborhood $U$ of $(X\_1,\dots,X\_r)$ where if
$(Y\_1,\dots,Y\_r)\in U$, then $f(Y\_1,\dots,Y\_r)>\frac{d}{r}-\delta$. Now, let $(Y\_1,\dots,Y\_r)\in U\cap E\_{d,r}$. Then there is some $\lambda$ and invertible $B$ where if we set $Z\_j=\lambda BY\_jB^{-1}$ for all $j$, then $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel. Now, let $\lambda\_{j,1},\dots,\lambda\_{j,d}$ be the eigenvalues of $Z\_j$ ordered in a way so that $|\lambda\_{j,1}|\geq|\lambda\_{j,2}|\geq\dots\geq|\lambda\_{j,d}|$. Let $\sigma\_{j,1}\geq\dots\geq\sigma\_{j,d}$ denote the singular values of $Z\_j$.
Then
$$\frac{d}{r}-\delta<f(Y\_1,\dots,Y\_r)=f(Z\_1,\dots,Z\_r)=\rho(Z\_1\dots Z\_r)^{2/r}
\leq(\sigma\_{1,1}^2\dots\sigma\_{r,1}^2)^{1/r}$$
$$\leq\frac{\sigma\_{1,1}^2+\dots+\sigma\_{r,1}^2}{r}
\leq\frac{(\sigma\_{1,1}^2+\sigma\_{1,2}^2)+\dots+(\sigma\_{r,1}^2+\sigma\_{r,2}^2)}{r}$$
$$\leq\frac{\|Z\_1\|\_2^2+\dots+\|Z\_r\|\_2^2}{r}=\frac{d}{r}.$$
We conclude that $\sigma\_{j,1}^2\leq\frac{d}{r}$ and $\sigma\_{j,2}^2\leq\delta$. Therefore, by Weyl's inequality, we have
$$|\lambda\_{j,1}\lambda\_{j,2}|\leq\sigma\_{i,1}\sigma\_{j,2}\leq\sqrt{\frac{d\delta}{r}}.$$
Therefore, we have
$$g(Y\_1,\dots,Y\_r)=g(Z\_1,\dots,Z\_r)\leq\sqrt{\frac{d\delta}{r}}.$$
We therefore conclude that $g(X\_1,\dots,X\_r)=0$ which means that
$\text{Rank}(X\_j)\leq 1$ for $1\leq j\leq r$.
| 0 | https://mathoverflow.net/users/22277 | 424572 | 172,462 |
https://mathoverflow.net/questions/424568 | 2 | Consider an $n \times n$ unitary $U$, drawn from the Haar measure. I'm trying to find the distribution for $U^{2}$. Is it true that $U^{2}$ is also Haar random?
Note that for any fixed unitary $V$, $VU$ and $UV$ are both Haar random unitaries, by the translational invariance of the Haar measure. But our case is slightly different where the matrix we are multiplying $U$ by also "depends" on $U$.
| https://mathoverflow.net/users/166840 | Question about squaring a Haar random unitary | The probability distribution of $U^p$ for $U$ uniformly distributed with the Haar measure in $\text{U}(n)$ has been calculated by Eric Rains in [Images of eigenvalue distributions under power maps](https://arxiv.org/abs/math/0008079).
For $p=2$ and $n$ even the eigenvalue distribution of $U^2$ is obtained by taking the union of the eigenvalues of two *independent* matrices $U\_1$ and $U\_2$, uniformly distributed in $\text{U}(n/2)$. For $n$ odd the two independent matrices are taken from $\text{U}((n+1)/2)$ and $\text{U}((n-1)/2)$.
So $U^2$ is only Haar random in the trivial case $n=1$, but not for $n>1$. For $n=2$, in particular, the two eigenvalues of $U^2$ are independent, in contrast to the two eigenvalues of $U$.
To see how squaring the $2\times 2$ unitary $U$ removes the correlations, use that the joint probability distribution of the two eigenvalues $\lambda\_1$ and $\lambda\_2$ of $U$ is
$$P(\lambda\_1,\lambda\_2)\propto|\lambda\_1-\lambda\_2|^2=2-2\,{\rm Re}\,\lambda\_1\bar{\lambda}\_2.$$
Squaring $U$ identifies $\pm\lambda\_i$, so the contributions from the correlator ${\rm Re}\,\lambda\_1\bar{\lambda}\_2$ cancel and $P(\lambda\_1^2,\lambda\_2^2)$ becomes independent of $\lambda\_i$.
| 5 | https://mathoverflow.net/users/11260 | 424573 | 172,463 |
https://mathoverflow.net/questions/424566 | 10 | Suppose $A=(a\_{ij})$ is a symmetric (0,1)-matrix with 1's along the diagonal, and let $A\_{ij}$ be the matrix obtained by removing the $i$-th row and $j$-th column.
Based on substantial numerical experimentation it seems that the bound $$\operatorname{Perm}(A) \geq 2 \operatorname{Perm}(A\_{ij})$$
holds for any $(i,j)$, $i \neq j$. I would be interested in any bound of the form $\operatorname{Perm}(A) \geq c \operatorname{Perm}(A\_{ij})$ for a constant $c>0$. (The bound $\operatorname{Perm}(A) \geq \operatorname{Perm}(A\_{ij})$ is trivial if $a\_{ij}=1$, but not so clear if $a\_{ij}=0$.)
My numerical experimentation suggests that this is a specific case of a far more general fact:
**Conjecture:** If $A$ is any matrix with nonnegative entries, $i \neq j$ and $\operatorname{Perm}(A)>0$, then $$\operatorname{Perm}(A)^2 \geq 4 a\_{ii}a\_{jj} \operatorname{Perm}(A\_{ij})\operatorname{Perm}(A\_{ji}). $$
It seems that there are various families of matrices $A$ that achieve equality.
| https://mathoverflow.net/users/126170 | A bound for the permanent of a nonnegative matrix | Please check the details.
Below I denote $i=1,j=2$.
Expand ${\rm Perm}\,A$ as a polynomial in $a\_{11}$ and $a\_{22}$ as $a\_{11}a\_{22}X+a\_{11}Y+a\_{22}Z+T$. Then $$({\rm Perm}\,A)^2\geqslant 4a\_{11}a\_{22}(XT+YZ).$$
I claim that $XT+YZ$ is coefficient-wise not less than ${\rm Perm}\, A\_{12} {\rm Perm}\, A\_{21}$.
For this sake, consider the weighted complete bipartite graph $K\_{nn}$, with parts $M=\{m\_1,\ldots,m\_n\}$ and $W=\{w\_1,\ldots,w\_n\}$, and weight of the edge $m\_iw\_j$ equal to $a\_{ij}$. Then the permanent of matrix $A$ is the weighted sum of perfect matchings (as usual, the weight of a set of edges is defined as the product of all weights of edges in this set).
The permanent of $A\_{12}$ is the weighted sum of matchings between $M\setminus \{m\_1\}$ and $W\setminus \{w\_2\}$. Call them blue matchings. Analogously, the permanent of $A\_{21}$ is the weighted sum of matchings between $M\setminus \{m\_2\}$ and $W\setminus \{w\_1\}$. They are called red matchings. Thus, $a\_{11} a\_{22}{\rm Perm}\, A\_{12} {\rm Perm}\, A\_{21}$ is a weighted sum of such multisets of edges: ($m\_1w\_1,m\_2w\_2$, blue matching, red matching). Call it a special multiset. Any special multiset $S$ of edges is a regular multigraph of degree 2, thus it is partitioned onto cycles. Thus, $S$ is partitioned onto two perfect matchings by $2^p$ ways, where $p$ is the number of long cycles (=cycles containing more than 2 vertices.) Denote by $q$ the
number of long cycles not containing the edges $m\_1w\_1,m\_2w\_2$.
I claim that the coefficient of weight of $S$ in $a\_{11}a\_{22}(XT+YZ)$ is at least $2^q$ (in $a\_{11} a\_{22}{\rm Perm}\, A\_{12} {\rm Perm}\, A\_{21}$ it is exactly $2^q$).
Indeed, partition $S$ onto two matchings. Remove the edges $m\_1w\_1$ and $m\_2w\_2$ from these matchings. If they are removed from different matchings, this corresponds to $YZ$. If they are removed from the same matching, it corresponds to $XT$.
| 2 | https://mathoverflow.net/users/4312 | 424593 | 172,465 |
https://mathoverflow.net/questions/424567 | 2 | Let $N=\{1,\dots,n\}$. Let $M$ be the set of all binary trees $B$ formed from all elements of $N$ (i.e of size $n$). Let $P$ be a numeric property defined for all $B\in M$. Let $O\in M$ be the optimal $B$ that minimizes $P$.
Example: Let $P$ be the sum of the heights of all nodes of $B$. This measures the "balancing" of $B$, and for $n=2^m-1$ $O$ is even unique.
Task: Given $P$, find (some - ties are OK) $O$.
Attempt 1: Stupid but surefire. Generate all permutations $\Pi$ of $N$. Generate the canonic tree (start empty, insert the elements of $\Pi$ in their order). Compute $P$, pick best $O$.
Attempt 2: If it is possible to derive a simple algorithm that takes a Hamilton walk through the "rotahedron" $G$ of $M$ ($B$ and $B'$ are connected in $G$ if a left rotation move turns $B$ into $B'$) this would at least reduce the load from $n!$ to Catalan($n$). *My question*: Is such an algorithm known?
Note: If $P$ is "very local" such that $G$, turned to a directed graph by pointing the arrow to the smaller $P$, is a DAG, this would speed up things enormously (start anywhere, follow the arrows), but even then, you could end up in a local minimum. ("My" $P$, BTW, would be again the sum of the heights, but dropping the first $k$ elements of $\Pi$.)
| https://mathoverflow.net/users/11504 | Finding the best binary tree with a general property | There are many ways to generate the binary trees systematically. An older example is in [this paper](https://watermark.silverchair.com/290171.pdf%5DADCCAqcGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMrKjgZkHACgY74gsDAgEQgIICeMZ44ylFTEpgx8fPGdZC0E8nZO4Pmd0nc0uOXM3dRDI03jqH5OdyxRr6xUxItzDL8H88_Dxae1YoAzHc4UynsczKNba1D3k7jk7VGuk293X-4zKk0gYt8RBXF0zo8kDqORhgYlaT5jeROsI0p1RAwxgoSclskVQIxrWQfleFRGbOP5P61HpxD0KzztfGbFks-nn92Ws2xB73dNi8N-gGyJDvHqyYJmTxqLRQixJvvuHnLl6imFlCmHvR5se5ozkoRi7rsrfTKUOdB7FuP1K6Y__8i6KRCucJ1WbFNXm22JsYr2OHH074uyXnoswSF9SG1Pd6Ffsh4VFIQaLsQpHp-FqaQ4lWRaF_ZEyLta74b-C_CTEynpvYfv1C16mU_lOc03IFLcP5UvzdAy65rPihALMDVtnhspuYRMNW2oKbpCNP-sGeQb0TZkpJgfoziOOwoV9ZcGqM-hoI4an0dBTJhFOKVFjwgfg8DpSIJYqu1yGCvvxeKSo4I-Ouiw7r1ZlKMdGZ6XUk8oAN5scfH_Ky2dEC0pG4PhUJzMxjmYSGe6mSR_YhNVc691DMVb_G9akVvLBIgiM1VULPeizNsIxp6yhv9vwzDgyGV3kK2iNXDO0NBZUs_FjY1p-QD35SXPPDoR8tPcT4_YmJdCZUOzUsdkyczmCO6fAVDOI3QAPKKBXzJQLFDXPCN8grdaekEFXScAYIU57-_It_sQdEk3AylucTi6jaf1_NT1_-c_H6UYgcxIJcew9fB5iXge2N_D6L0YIMIhAnGjxetn57nxEWFvuo1KgbYU-TGTOD0Hl8KffPMywYJaJil4Bf6eP6lwAri5H_tEjD5haA).
Another one, perhaps more efficient for your purposes, is [here](https://doi.org/10.1016/0196-6774(90)90030-I).
| 1 | https://mathoverflow.net/users/9025 | 424597 | 172,466 |
https://mathoverflow.net/questions/424602 | 3 | $\require{AMScd}$
Related to [this](https://mathoverflow.net/questions/424558/in-luries-higher-topos-theory-lemma-4-3-2-7), I have a question about the proof given in [Kerodon](https://kerodon.net/) of the following result:
>
> [**Proposition 7.3.7.1**](https://kerodon.net/tag/030V): Let $C$ be an $\infty$-category, let $\bar{F} : C^\rhd \to D$ be a functor of $\infty$-categories, and let $U : D \to E$ be another functor of $\infty$-categories. Assume that $F = \bar{F}\_{\vert C}$ is U-left Kan extended from a full subcategory $C\_0 \subseteq C$. Then $\bar{F}$ is a U-colimit diagram if and only if the composite map $C\_0^\rhd \to C^\rhd \to D$ is a $U$-colimit diagram.
>
>
>
The proof proceeds by observing that $\bar{F}$ is a $U$-colimit diagram iff for each $Y : D$, the following commutative diagram is a homotopy pullback square (regarding $D$ as a constant functor):
$$
\begin{CD}
\hom(\bar{F},D) @>{}>> \hom(F,D)\\
@VVV @VVV \\
\hom(U \circ \bar{F},U(D)) @>>> \hom(U \circ F,U(D))
\end{CD}
$$
Likewise, the composite is a $U$-colimit diagram if and only if for each $Y : D$ the following is a homotopy pullback square:
$$
\begin{CD}
\hom(\bar{F}\_{\vert C\_0^\rhd},D) @>{}>> \hom(F\_{\vert C\_0},D)\\
@VVV @VVV \\
\hom(U \circ \bar{F}\_{\vert C\_0^\rhd},U(D)) @>>> \hom(U \circ F\_{\vert C\_0},U(D))
\end{CD}
$$
Accordingly, it suffices to show that the first diagram is a pullback square if and only if the second is. Restriction maps assemble these two squares into a cube, with the first as the back face and the second as the front face. Therefore, it suffices to argue that each restriction map is a homotopy equivalence. To this end, Kerodon uses [another lemma](https://kerodon.net/tag/030F) to observe that the assumption that $\bar{F}$ is $U$-Kan extended from $F$ yields two pullback squares (the left and right faces of the cube):
$$
\begin{CD}
\hom(\bar{F},D) @>{}>> \hom(\bar{F}\_{\vert C\_0^\rhd},D)\\
@VVV @VVV \\
\hom(U \circ \bar{F},U(D))
@>>> \hom(U \circ \bar{F}\_{\vert C\_0^\rhd},U(D))
\end{CD}
$$
$$
\begin{CD}
\hom(F,D) @>{}>> \hom(F\_{\vert C\_0},D)\\
@VVV @VVV \\
\hom(U \circ F,U(D)) @>>> \hom(U \circ F\_{\vert C\_0},U(D))
\end{CD}
$$
**Question**
I fail to see how the existence of pullback squares shows that the four restriction maps are homotopy equivalences, as Kerodon indicates. It would now suffice to argue that only
$\hom(U \circ F,U(D)) \to \hom(U \circ F\_{\vert C\_0},U(D))$ and $\hom(U \circ \bar{F},U(D)) \to \hom(U \circ \bar{F}\_{\vert C\_0^\rhd},U(D))$, but I cannot show this. As is, using 2-for-3 with pullbacks we can conclude that the front face being a pullback square implies that the back face is a pullback square, so the "if" is already easily shown. Is an alternative approach needed for "only if"?
| https://mathoverflow.net/users/76636 | Question about the proof of Kerodon tag 030V (Proposition 7.3.7.1) | I think you are right that the cube is not a levelwise equivalence - I saw you already put a comment on the Kerodon page so you should get an answer soon.
Here's how I would fix the proof: call your first square $S$ and your second square $S\_0$ (for "restriction to $C^0$). Restriction is a map $S\to S^0$. Draw this as a cube where arrows are facing a- towards the screen, b- down and c- towards the right. I strongly recommend actually drawing the cube to follow what I say below.
Then the left face has equivalences as screen-facing arrows. This left face is the following square:
$$\require{AMScd}\begin{CD}\hom(\overline F,D) @>>> \hom(\overline F\_{\mid (C^0)^\triangleright},D) \\
@VVV @VVV \\
\hom(U\circ \overline F,U(D)) @>>> \hom(U\circ \overline F\_{\mid (C^0)^\triangleright},U(D))\end{CD}$$
Here the horizontal maps (which are facing the screen in my cube) are equivalences : indeed writing $\star$ for the point of a cone, in $\text{Fun}(X^\triangleright, Y)$, $\hom(G,y) \simeq \hom(G(\star), y)$, and here $\star$ is preserved by $(C^0)^\triangleright \to C^\triangleright$.
The right face of the cube does *not* have equivalences as screen-facing arrows, it is the following:
$$\begin{CD}\hom( F,D) @>>> \hom( F\_{\mid C^0},D) \\
@VVV @VVV \\
\hom(U\circ F,U(D)) @>>> \hom(U\circ F\_{\mid C^0},U(D))\end{CD}$$
It is, however, a pullback square by definition of $U$-left Kan extension (or by some previous lemma).
In any case, the left face having equivalences as mentioned above means that the big cube can be seen as an equivalence between two squares: the rectangle that covers the back face and the right face on the one hand, and the front face on the other hand.
In more detail, the left face having equivalences as mentioned above implies that the outer square in the following:
$$\begin{CD}\hom(\overline F, D) @>>>\hom(F,D) @>>> \hom( F\_{\mid C^0},D) \\
@VVV @VVV @VVV \\
\hom(U\circ \overline F,U(D)) @>>> \hom(U\circ F, U(D)) @>>> \hom(U\circ F\_{\mid C^0},U(D))\end{CD}$$
is equivalent to the front face, namely:
$$\begin{CD}\hom(\overline F\_{\mid C^0}, D) @>>>\hom(F\_{\mid C^0},D)\\
@VVV @VVV \\
\hom(U\circ \overline F\_{\mid C^0},U(D)) @>>> \hom(U\circ F\_{\mid C^0}, U(D))\end{CD}$$
Now in the first rectangle, the rightmost square is a pullback, so by the pullback lemma, the outer rectangle is a pullback if and only if the leftmost square is.
The outer rectangle is a pullback if and only if the equivalent one is.
So the leftmost square is a pullback if and only if the last square I drew is one. I think this is what you wanted.
| 4 | https://mathoverflow.net/users/102343 | 424606 | 172,469 |
https://mathoverflow.net/questions/424585 | 2 | The spherical derivative of a meromorphic function is defined as
$$f^\#(z):=\frac{|2f'(z)|}{1+|f(z)|^2}.$$
The motivation is that given a piecewise smooth curve $\gamma$ in the complex plane, the length of $f\circ \gamma$ in the Riemann sphere is given by $\int\_{\gamma} f^\#(z)d|z|$. When $f=az$, where $a$ is a nonzero complex number, then $f^\#(z)$ is radially symmetric and is strictly decreasing along the radial direction. Also, it is well known that $f\_1^\#=f\_2^\#$ is equivalent to the existence of complex numbers $p, q$ with $|p|^2+|q|^2=1$ such that $f\_1=\frac{pf\_2-\bar{q}}{qf\_2+\bar{p}}$.
---
I did know from literature (see for example Lehto's 1961 paper) that the behavior of $f^\#$ near an essential singularity of $f$ can reflect certain value distributional properties of $f$ near the singularity, but I'm trying to understand how the property of $f^\#$ can determine certain rigidity property of $f$.
For example, here is a standard **question**: if generally $f^\#(z)$ is strictly decreasing along radial directions, is it ture that $f$ must be a Mobius transform? Are there any other possibilities?
| https://mathoverflow.net/users/51546 | On a rigidity question related to spherical derivative of meromorphic functions | The other possibility is that $f(z)=e^z$. Explicit computation gives
$$\frac{d}{dr}f^\#(re^{i\phi})=2e^{r\cos\phi}\cos\phi(1-e^{2r\cos\phi}),$$
which is $\leq 0$ everywhere.
On the other hand, the exponential is the only exception among entire functions.
Indeed, if $f^\#$ is decreasing on each radius, then $f^\#$ has no zeros,
this implies that $f'$ has no zeros. On the other hand, $f^\#(z)\leq f^\#(0)$ must be bounded, and a theorem of Clunie and Hayman implies that $f$ has at most exponential type (that is at most order 1, normal type). Then $f'$ has at most exponential type, and since it has no zeros, it must be $e^{az+b}$ by the Hadamard factorization theorem.
Refs. The original paper of Clunie and Hayman is
MR0192055
Clunie, J.; Hayman, W. K.
The spherical derivative of integral and meromorphic functions.
Comment. Math. Helv. 40 (1966), 117–148.
Their proof was much simplified (and generalized) in
MR2869124
Barrett, Matthew; Eremenko, Alexandre
Generalization of a theorem of Clunie and Hayman.
Proc. Amer. Math. Soc. 140 (2012), no. 4, 1397–1402.
Edit. One can describe all meromorphic functions with your property: there is probably nothing except $L(e^{az})$,
where $L$ is linear-fractional transformation.
Sketch of the proof for meromorphic functions. As before, we have
that $f$ has no critical points (that is $f'(z)\neq 0$ and all poles are simple), but now $f$ is of order at most 2, normal type. (By definition of the order of a meromorphic function in terms of Nevanlinna characteristic). Now we use the theorem of R. Nevanlinna which describes all meromorphic functions of finite order without
critical points. All such functions are ratios $f=w\_1/w\_2$ of two
linearly independent solutions of the differential equation
$$w''=Pw,$$
where $P$ is a polynomial. The order of the function is $(\deg P+2)/2$. The case $\deg P=0$ corresponds to an exponential, so it remains to consider orders $3/2$ (Airy functions) and $2$ (Weber functions).
Now there is an asymptotic theory
of these differential equations which gives a very precise asymptotic formula for solutions as $z\to\infty$. (In physics, this formula is called the WKB approximation). It implies that for orders $>1$, the spherical derivative of $f$ is unbounded. (This proof is independent of the previous proof for entire functions but uses deeper tools, like Nevanlinna theory).
Refs.
R. Nevanlinna, Uber Riemannsche Flachen mit endlich vielen Windungspunkten, Acta Math. 58 (1932) 295–373.
Expositions in English:
A. Eremenko, Entire and meromorphic solutions of ordinary differential equations, Chapter 6 in the book: Complex Analysis I, Encyclopaedia of Mathematical Sciences, vol. 85; Springer, NY, 1997, 141-153.
G. Gundersen, J. Heittokangas and A. Zemirni, Asymptotic integration theory for $f''+P(z)f=0$, Expositiones math., 40, 1 (2022) 94-126.
See also A. Eremenko, A Toda lattice in dimension 2 and Nevanlinna theory, J. Math. Phys., Anal. Geom., 31 (2007) 39-46 for a generalization of this theorem of Nevanlinna.
| 5 | https://mathoverflow.net/users/25510 | 424609 | 172,470 |
https://mathoverflow.net/questions/424577 | 5 | I'm reading a paper in which the author use $(p,q)$-forms and currents on a complex analytic space.
My question is how to define $(p,q)$-current on complex space? Does it have similar properties like closeness, positivity and cohomology as in the smooth case? Could we define intersection theory or wedge product in this case?
| https://mathoverflow.net/users/167083 | How to define a current on a complex analytic space | The definition can be found in for example Section 3.3 of T. Bloom, M. Herrera: De Rham Cohomology of an Analytic Space (Inventiones math. 7, 275-296 (1969)) and Section 4.2 of M. Herrera, D. Lieberman: Residues and Principal Values on Complex Spaces (Math. Ann. 194, 259-294 (1971)) (in the more general setting of semianalytic sets).
If $M \subseteq U$, where $M$ is an analytic subset of an open set $U \subseteq \mathbb{C}^n$, and $\mathcal{E}\_U$ denotes the sheaf of smooth forms on $U$, then one may first define $N\_{M,U} \subseteq \mathcal{E}\_U$ as the sheaf of smooth forms on $U$ whose pullback to the regular part of $M$ vanishes. Then one defines the sheaf of smooth forms on $M$ as $\mathcal{E}\_M := \mathcal{E}\_U/N\_{M,U}$. This construction then globalizes to define smooth forms on an analytic space $\mathcal{E}\_X$ which on a local model $M \subseteq U$ coincides with $\mathcal{E}\_M$ defined above.
Then, one equips the space $\Gamma\_c(U,\mathcal{E}\_U)$ of sections with compact support with the usual appropriate (locally convex inductive) topology induced by uniform convergence of the coefficients of the form and of their derivatives. Since the kernel of the projection $\Gamma\_c(U,\mathcal{E}\_U) \to \Gamma\_c(M,\mathcal{E}\_M)$ is closed, one may equip $\Gamma\_c(M,\mathcal{E}\_M)$ with the quotient Hausdorff topology. Finally, for a general analytic space $X$, one equips $\Gamma\_c(X,\mathcal{E}\_X)$ with the appropriate topology which induces the given topology on the local models.
Regarding the question of what properties these currents have, this seems to be way broad question to answer here, but would have to be checked on a case by case basis.
One word of caution is that the definition in Bloom-Herrera is actually stated in two ways, claimed to be equivalent to each other, where the other definition is that $N\_{M,U}$ is alternatively defined as the space of smooth forms $\alpha$ on $U$ such that for any smooth map $g : W \to U$, with $W$ a manifold and $g(W) \subseteq M$, $g^\*\alpha = 0$. As far as I am aware, the equivalence of these two definitions is only addressed recently in M. Andersson, H. Samuelsson Kalm: A note on smooth forms on analytic spaces (Math. Scand. 127, 521–526 (2021)). With this alternative definition, it is straight-forward that a smooth map $f : N \to M$ of analytic spaces induces a pullback map $f^\*$ of smooth forms, and thus a push-forward of smooth proper maps of currents, while with the simpler definition of smooth forms I used above, this is not so immediate.
---
The definition in the local models also has a more concrete version, unless I am mistaken about some topological subtlety:
Using the notation above, the push-forward of a current $T' := i\_\*T$ on $M$ under the inclusion $i : M \to U$ yields a current $T'$ on $U$ which vanishes on the ideal of test-forms in $N\_{M,U}$, and conversely, any current $T'$ with this property should be the push-forward of a unique current on $M$, and one could thus take this as the definition.
The global case should then be possible to describe through a set of compatible currents in the local models.
| 7 | https://mathoverflow.net/users/49151 | 424610 | 172,471 |
https://mathoverflow.net/questions/424612 | 2 | Let $X$ be a smooth projective variety and $E$ a vector bundle of rank $3$ over $X$. Moreover let $L \in Pic(X)$ be a line bundle and $$q:S^2E \rightarrow L$$ a $L-$valued quadratic form.
Then we can consider the subvariety $C\_q \subset \mathbb{P}(E)$, where $\pi:\mathbb{P}(E) \rightarrow X$ is the projective bundle associated to $E$, given as the zero locus of the section $s\_{C\_q} \in H^0(\mathbb{P}(E), \mathcal{O}\_{\mathbb{P}(E)}(2) \otimes \pi^\*L)$ corresponding to $q$. With the restriction projection $\pi:C\_q \rightarrow X$ the scheme $C\_q$ has a structure of a conic bundle over $X$. Call $D \subset X$ the discriminant divisor, i.e. the locus of points $x \in X$ such that $\pi^{-1}(x)$ is a singular conic in the fiber $\mathbb{P}^2\_x$.
How can we express $D \in H^2(X,\mathbb{Z})$ as a function of Chern classes of $E$ and $L$?
| https://mathoverflow.net/users/146431 | Class of the discriminant of a conic bundle | The map $q$ induces a morphism
$$
E \to E^\vee \otimes L
$$
and the discriminant is the zero locus of its determinant
$$
\det(E) \to \det(E^\vee) \otimes L^{\otimes 3}.
$$
Thus
$$
[D] = 2c\_1(E^\vee) + 3c\_1(L).
$$
| 4 | https://mathoverflow.net/users/4428 | 424614 | 172,472 |
https://mathoverflow.net/questions/303637 | 4 | In 2003 E. S. Croot [Ann. of Math. 157(2)(2003), 545-556] proved the Erdos-Graham Conjecture which states that if $\{2,3,\ldots\}$ is partitioned into finitely many subsets then one of the subsets contains finitely many distinct integers $x\_1,\ldots,x\_m$ satisfying $\sum\_{k=1}^m1/x\_k=1$.
Here I ask for the density version of this result.
QUESTION: Let $A$ be a subset of $\{2,3,\ldots\}$ having positive lower (or upper) asymptotic density. Are there finitely many distinct elements $a\_1 <\ldots< a\_m$ of $A$ with $\sum\_{k=1}^m1/a\_k = 1$?
Clearly, the set $\{3,5,7,\ldots\}$ has asymptotic density $1/2$, and it is known that
$$\frac1 3 + \frac1 5 +\frac 1 7 +\frac 1 9 +\frac 1 {11} +\frac 1 {33} +
\frac1{35} + \frac1 {45} + \frac 1 {55} + \frac1 {77} + \frac1 {105} = 1.$$
Motivated by Szemeredi's theorem, in 2007 I formulated the above question and conjectured that it has a positive answer. It seems that Croot's proof of the Erdos-Graham conjecture could not be adapted to answer my question.
Any comments are welcome!
| https://mathoverflow.net/users/124654 | Density version of the Erdos-Graham conjecture | The answer is yes (even to the strong upper density version), as proved in this preprint <https://arxiv.org/abs/2112.03726>.
(I can give this answer with a high degree of confidence, since the proof is now completely formalised in Lean! This was a joint project by myself and Bhavik Mehta - see <https://github.com/b-mehta/unit-fractions> for the code)
| 9 | https://mathoverflow.net/users/385 | 424616 | 172,473 |
https://mathoverflow.net/questions/424605 | 5 | *Note: This question is closely related to an earlier question: [A large noise limit](https://mathoverflow.net/questions/417015/a-large-noise-limit).*
Let $W$ be a standard one dimensional Brownian motion.
For every $\varepsilon > 0$, let $A\_\varepsilon$ denote the event
$$\{\underset{0 \leq t \leq 1}{\text{max}} W\_t \geq \frac{1}{\varepsilon}\} \;, $$
and let $\mathbb P^\varepsilon$ be the probability measure given by
$$P^\varepsilon (E) = \frac{\mathbb P(E \cap A\_\varepsilon)}{\mathbb P(A\_\varepsilon)} \;, $$
for all measurable events $E$.
We denote by $\mathbb E\_{\mathbb P^\varepsilon}$ the expectation under $\mathbb P^\varepsilon$.
**Question:** Is it true that
$$\lim\_{\varepsilon \to 0} \mathbb E\_{\mathbb P^\varepsilon} \big [\lvert \varepsilon W\_1 - 1 \rvert \big ]= 0?$$
| https://mathoverflow.net/users/173490 | Endpoint of Brownian motion conditional on high maxima | $\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\de}{\delta}$Yes, this is true:
By the reflection principle (see e.g. [Proposition 2](https://ocw.mit.edu/courses/15-070j-advanced-stochastic-processes-fall-2013/aca1518a09539a09ddd37428ab0d0268_MIT15_070JF13_Lec7.pdf), for $M:=\max\_{0\le t\le1}W\_t$,
\begin{equation}
\begin{aligned}
&P(M\ge1/\ep,W\_1\in dx) \\
&=P(W\_1\in2/\ep-dx)1(x<1/\ep)+P(W\_1\in dx)1(x>1/\ep) \\
&=\vpi(2/\ep-x)1(x<1/\ep)dx+\vpi(x)1(x>1/\ep)dx,
\end{aligned}
\end{equation}
where $\vpi$ is the standard normal pdf.
So,
\begin{equation}
P(A\_\ep)=P(M\ge1/\ep)=2G(1/\ep)\sim2\ep\vpi(1/\ep),
\end{equation}
where $1-G$ is the standard normal cdf, and
\begin{equation}
\begin{aligned}
&E1(M\ge1/\ep)|\ep W\_1-1| \\
&=\int\_{\mathbb R}P(M\ge1/\ep,W\_1\in dx)|\ep x-1| \\
&=2\big(\ep\vpi(1/\ep)-G(1/\ep)\big)=o(\ep\vpi(1/\ep))
\end{aligned}
\end{equation}
(as $\ep\downarrow0$).
Hence,
\begin{equation}
\de(\ep):= E\_{P\_\ep}|\ep W\_1-1|=\frac{E1(M\ge1/\ep)|\ep W\_1-1|}{P(A\_\ep)}\to0. \tag{1}\label{1}
\end{equation}
---
The expression for $\de(\ep)$ in \eqref{1} can be rewritten as follows:
\begin{equation}
\de(\ep)=\frac\ep{r(1/\ep)}-1, \tag{2}\label{2}
\end{equation}
where $r:=G/\vpi$ is the Mills ratio. Then one can use known asymptotic expansions of and bounds on the Mills ratio (see e.g. [this paper](https://eudml.org/doc/122588)) to get asymptotic expansions of and bounds on $\de(\ep)$.
For instance, from Propositions 1.3 and the simplest two cases of formula (1.8) (with $m=1,2$) of that paper we get
\begin{equation}
\de(\ep)=\ep^2 - 2 \ep^4 + 10 \ep^6 - 74 \ep^8 + 706 \ep^{10}+O(\ep^{12})
\end{equation}
(as $\ep\downarrow0$) and
\begin{equation}
\frac{\ep^2}{1 + 2 \ep^2}<\frac{\ep^2 + 7 \ep^4}{1 + 9 \ep^2 + 8 \ep^4}
<\de(\ep)<\frac{\ep^2 + 3 \ep^4}{1 + 5 \ep^2}<\ep^2.
\end{equation}
One may note that the difference between the upper and lower bounds $\dfrac{\ep^2 + 3 \ep^4}{1 + 5 \ep^2}$ and $\dfrac{\ep^2 + 7 \ep^4}{1 + 9 \ep^2 + 8 \ep^4}$ on $\de(\ep)$ is $\dfrac{24 \ep^8}{1 + 14 \ep^2 + 53 \ep^4 + 40 \ep^6}$, which is $<0.00000024$ if $\ep=0.1$.
| 7 | https://mathoverflow.net/users/36721 | 424617 | 172,474 |
https://mathoverflow.net/questions/424619 | 4 | Let $SL(n)$ be algebraic group defined over finite field $\mathbb{F}\_{p^n}$, $B$ be Borel subgroup consist of upper triangular matrices and $T$ be maximal torus consist of diagonal matrices. Let $W$ be Weyl group, defined by $N\_{G}T/T$. Let $G/B=\cup\_{w\in W} B\cdot e\_{w}$ be the Bruhat decomposition, where $e\_{w}=wB$. Denote $X:=B\cdot e\_{w}$ be Schubert cell and $O^{-}$ be opposite big cell i.e.$B^{-}\cdot e\_{id}$, where $B^{-}$ is the opposite Borel.
My question is how to calculate the number of $\mathbb{F}\_{p^n}$-points of $Y:=X\cap O^{-}$. Is this one polynomial count?(i.e. there exists a polynomial $f$ with $\mathbb{Z}$-coefficients not depend on $n$, such that the number of $\mathbb{F}\_{p^n}$-points is $f(p^n)$)
| https://mathoverflow.net/users/147080 | May Schubert cell intersection with opposite big cell polynomial count? | Yes, this is polynomial point count. An intersection of a Bruhat cell and an opposite Bruhat cell is called a "Richardson variety", and Richardson varieties come with decompositions known as Deodhar decompositions. Each piece of the Deodhar decomposition is of the form $\mathbb{G}\_m^{N-2k} \times \mathbb{A}^k$ where $N$ is the dimension of the Richardson, so it has $(q-1)^{N-2k} q^k$ points over $\mathbb{F}\_q$, and adding them all up gives a polynomial in $q$.
To give the first nontrivial example, in $\text{GL}\_3$, the big Schubert cell is flags which can be written as
$$\begin{bmatrix} 1&0&0 \\ x&1&0 \\ z&y&1 \\ \end{bmatrix} B.$$
The intersection with the opposite cell is the open locus $z(xy-z) \neq 0$.
We can decompose this as the union of the pieces where $x\neq 0$ and where $x = 0$, so we get
$$\left\{ \begin{bmatrix} 1&0&0 \\ x&1&0 \\ z&\tfrac{w+z}{x}&1 \\ \end{bmatrix} B : wxz \neq 0 \right\} \sqcup \left\{ \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ z&y&1 \\ \end{bmatrix} B : z \neq 0 \right\},$$
so we have
$$(q-1)^3 + (q-1)q$$
points.
The Deodhar stratification of $(B\_+ w B\_+ \cap B\_- u B\_+)/B\_+$ depends on a choice of reduced word for $w$. Deodhar's original paper is
*Deodhar, Vinay V.*, [**On some geometric aspects of Bruhat orderings. I: A finer decomposition of Bruhat cells**](http://dx.doi.org/10.1007/BF01388520), Invent. Math. 79, 499-511 (1985). [ZBL0563.14023](https://zbmath.org/?q=an:0563.14023).
For very explicit formulas, you might also like
*Marsh, R. J.; Rietsch, K.*, [**Parametrizations of flag varieties**](http://dx.doi.org/10.1090/S1088-4165-04-00230-4), Represent. Theory 8, 212-242 (2004). [ZBL1053.14057](https://zbmath.org/?q=an:1053.14057).
In your particular case, you want $w = w\_0$ and $u = e$. If you take the reduced word $(s\_1 s\_2 \cdots s\_{n-1}) (s\_1 s\_2 \cdots s\_{n-2}) \cdots (s\_1 s\_2) (s\_1)$, Allen Knutson pointed out to me that the Deodhar stratification has a very explicit form -- it corresponds to taking a lower triangular matrix as above and specifying the ranks of the left-justified submatrices in consecutive rows. (In the example above, the two strata depend on whether the submatrix in row $2$ and column $1$ has rank $1$ or $0$.) I don't know whether Allen has written up this fact.
| 8 | https://mathoverflow.net/users/297 | 424626 | 172,476 |
https://mathoverflow.net/questions/422249 | 1 | Suppose that $V$ is a finite dimensional complex Hilbert space. Let $L(V)$ denote the collection of all linear mappings from $V$ to $V$. Let $A\_1,\dots,A\_r:V\rightarrow V$ be linear operators. Then define a completely positive superoperator $\Phi(A\_1,\dots,A\_r):L(V)\rightarrow L(V)$ by letting $\Phi(A\_1,\dots,A\_r)(X)=A\_1XA\_1^\*+\dots+A\_rXA\_r^\*$, and define the $L\_2$-spectral radius of $A\_1,\dots,A\_r$ by letting
$\rho\_2(A\_1,\dots,A\_r)=\rho(\Phi(A\_1,\dots,A\_r))^{1/2}$.
The Cauchy-Schwarz inequality holds for the $L\_2$-spectral radius. If $V,W$ are finite dimensional complete Hilbert spaces, and $A\_i:V\rightarrow V,B\_i:W\rightarrow W$ for $1\leq i\leq r$, then $$\rho(A\_1\otimes B\_1+\dots+A\_r\otimes B\_r)\leq\rho(\Phi(A\_1,\dots,A\_r))^{1/2}\rho(\Phi(B\_1,\dots,B\_r))^{1/2}.$$
Now, define the $L\_{2,d}$-spectral radius of $A\_1,\dots,A\_r$ by letting
$$\rho\_{2,d}(A\_1,\dots,A\_r)$$
$$=\sup\{\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}\mid \rho(\Phi(X\_1,\dots,X\_r))\neq 0,X\_1,\dots,X\_r\in M\_d(\mathbb{C})\}.$$
From the Cauchy-Schwarz inequality, we observe that $$\rho\_{2,d}(A\_1,\dots,A\_r)\leq\rho\_{2,g}(A\_1,\dots,A\_r)\leq\rho\_{2,\dim(V)}(A\_1,\dots,A\_r)=\rho\_2(A\_1,\dots,A\_r)$$
whenever $1\leq d\leq g$. One should therefore think of $\rho\_{2,d}$ as an approximation to $\rho\_{2}$.
Suppose that $1\leq d<\dim(V)$. Then is there some $r$ along with linear operators $A\_1,\dots,A\_r:V\rightarrow V$ such that $\rho\_{2,d}(A\_1,\dots,A\_r)<\rho\_{2,d+1}(A\_1,\dots,A\_r)?$
Example 0: $\rho\_{2,d}(A)=\rho\_{d}(A)=\rho(A).$ Therefore, $\rho\_{2,d},\rho\_{2}$ should be considered to be a sort of spectral radius but for multiple linear operators.
Example 1: $$\rho\_{2,1}(A\_1,\dots,A\_r)=\sup\{\rho(\alpha A\_1+\dots+\alpha\_r A\_r):|\alpha\_1|^2+\dots+|\alpha\_r|^2=1\}.$$
Example 2: Observe that if $d\geq\dim(V)$, then $$\rho\_{2,d}(A\_1,\dots,A\_r)=\rho(\Phi(A\_1,\dots,A\_r))^{1/2}.$$
Example 3: $$\rho\_{2,1}(\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix})=1/\sqrt{2}<1=\rho\_{2,2}(\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}).$$
Example 4: Let $C\_1,\dots,C\_r$ be the $r\times r$-matrices, where $(C\_i)\_{i,i+1\mod r}=1$ and where all the other entries in $C\_i$ are zero. Then
$\rho(C\_1\otimes X\_1+\dots+C\_r\otimes X\_r)=\rho(X\_1\dots X\_r)^{1/r}$. Therefore,
$$\rho\_{2,d}(C\_1,\dots,C\_r)=\sup\{\frac{\rho(X\_1\dots X\_r)^{1/r}}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}:\rho(\Phi(X\_1,\dots,X\_r))\neq 0,X\_1,\dots,X\_r\in M\_d(\mathbb{C})\}.$$
Now, suppose that $j\_1,\dots,j\_r$ are natural numbers with $1\leq j\_1\leq\dots\leq j\_r=d$ and where $j\_{i+1}-j\_{i}\in\{0,1\}$ for $1\leq i\leq r$. For $1\leq i\leq r$, let $X\_i$ be the $d\times d$ matrix where $(X\_i)\_{j\_{i},j\_{i+1}}=1$ (here $i$ is taken modulo $r$) and where all the other entries in $X\_i$ are zero. Then $\rho(X\_1\dots X\_r)=1$. Let $n\_{1},\dots,n\_{d}$ be the natural numbers defined by letting $n\_{j}=\{i\mid 1\leq i\leq r,j\_{i}=j\}$. Then $r=n\_1+\dots+n\_d$. Suppose now that $n\_j=r/d$ for all $j$. Then $\rho(\Phi(X\_1,\dots,X\_r))=\frac{r}{d}.$ Therefore, $$\rho\_{2,d}(C\_1,\dots,C\_r)\geq\sqrt{\frac{d}{r}}$$ whenever $d$ is a factor of $r$.
| https://mathoverflow.net/users/22277 | Are these $L_2$-spectral radii approximations strictly increasing? | **Yes.** If $C\_1,\dots,C\_r$ are the matrices in Example 4, then by [my answer to my other question](https://mathoverflow.net/a/424572/22277), $$\rho\_{2,d}(C\_1,\dots,C\_r)=\sqrt{d/r}$$
whenever $1\leq d\leq r$.
| 0 | https://mathoverflow.net/users/22277 | 424628 | 172,477 |
https://mathoverflow.net/questions/424613 | 19 | Mark Hovey maintains a [list of open problems in model category theory](https://www-users.cse.umn.edu/%7Etlawson/hovey/model.html). I think this list is quite old, and I don't know if Hovey is still updating it or not.
>
> My question is:
>
>
>
>
> i) which of the 13 problems in that list are still
> considered open and still important in homotopy theory?
>
>
> ii) or dually, which of the problems are considered settled or are perceived as no longer relevant given the current status of homotopy theory?
>
>
>
| https://mathoverflow.net/users/139854 | Mark Hovey's open problems in the theory of model categories | I am a former student of Mark Hovey's, and during grad school, I wrote a document giving an update on the status of the 13 problems (as of 2012 or 2013, I guess). I just briefly went through it a moment ago to give some updates, but it's still in rough shape. I apologize for what I'm sure will be many omissions, as so much work has been done in this area over the past 20 years. Nevertheless, here you go (using Hovey's words to set up each of the problems in case his website goes down again):
1. The safest sort of problem to work on with model categories is building one of interest in applications. The essential idea is: whenever someone uses the word homology, there ought to be a model category around. I like this idea a great deal, and it might lead to expansion of algebraic topology into many different areas. The simplest example that I personally do not understand is complexes of (quasi-coherent?) sheaves over a scheme. There is certainly a model structure here, and it is probably even known. But I think it would be good to find this out, and find out how the model structure is built. I believe this should be a symmetric monoidal model category. I also have the impression that one can not generalize the usual model structure on chain complexes over a ring, because you won't have projectives. But these two impressions sort of contradict each other, since the second one would lead you to generalize the injective model structure on chain complexes over a ring, but this model structure is not symmetric monoidal. So there is something for me at least to learn here.
**The machinery to resolve this question was developed by Mark Hovey (2007) in “[Cotorsion pairs, model category structures, and representation theory](https://link.springer.com/article/10.1007/s00209-002-0431-9),” which provides a method for building an abelian model structure on a bicomplete abelian category with prescribed classes of acyclic, cofibrant, and fibrant objects $(W,C,F)$ such that $W$ is thick and the pairs $(C,F \cap W)$ and $(C\cap W, F)$ are complete cotorsion pairs. An excellent survey is Hovey’s “[Cotorsion Pairs and Model Categories](http://homepages.math.uic.edu/%7Ebshipley/hovey.pdf),” where this appears as Theorem 2.5. Conditions are also given to make this model structure monoidal (Theorem 4.2). Hovey used this theory to build a model structure on unbounded chain complexes of quasi-coherent sheaves over a (nice) scheme where weak equivalences are quasi-isomorphisms and fibrations are dimensionwise split surjections with dimensionwise injective kernel (see Hovey “[Model Category Structures on Chain Complexes of Sheaves](https://www.ams.org/journals/tran/2001-353-06/S0002-9947-01-02721-0/S0002-9947-01-02721-0.pdf)” Theorem 4.4). He also built a model structure on chain complexes of modules over a nice ringed space (Theorem 5.2) and found conditions to make this model structure satisfy the pushout product and monoid axioms (5.7, 5.9, 5.10). Jim Gillespie then proved a theorem which allows one to build a model structure on Ch(A) where A is an abelian model category with a nice cotorsion pair (Theorem 7.9). Gillespie’s work generalizes the above and can be found in “[The Flat Model Structure on Complexes of Sheaves](https://www.ams.org/journals/tran/2006-358-07/S0002-9947-06-04157-2/S0002-9947-06-04157-2.pdf)” and “[Cotorsion Pairs and Degreewise Homological Model Structures](https://www.intlpress.com/site/pub/files/_fulltext/journals/hha/2008/0010/0001/HHA-2008-0010-0001-a012.pdf).” Gillespie continued to generalize and hone this theory, resulting in [more than 30 papers on the topic](https://scholar.google.com/citations?user=V8c2YscAAAAJ&hl=en). A great survey is his paper "[Hereditary abelian model categories](https://scholar.google.com/citations?view_op=view_citation&hl=en&user=V8c2YscAAAAJ&citation_for_view=V8c2YscAAAAJ:IjCSPb-OGe4C)." These kinds of model structures, that come from cotorsion pairs, are called *abelian model structures* and nowadays a compatible pair of cotorsion pairs (the type that give rise to an abelian model structure) is known as a *Hovey triple*. Many other authors have also worked in this area, including Sergio Estrada, Daniel Bravo, Jan Stovicek, Sinem Odabasi, Hanno Becker, and probably many more people I should mention.**
2. A scheme is a generalization of a ring, in the same way that a manifold is a generalization of R^n. So maybe there is some kind of model structure on sheaves over a manifold? Presumably this is where de Rham cohomology comes from, but I don't know. It doesn't seem like homotopy theory has made much of a dent in analysis, but I think this is partly due to our lack of trying. Floer homology, quantum cohomology--do these things come from model structures?
**These problems appear to still be open as stated. There are good reasons why one cannot have a model category of manifolds, and these are discussed in my expository paper “[On Colimits in Various Categories of Manifolds](http://personal.denison.edu/%7Ewhiteda/files/Expository/On%20Colimits%20in%20Various%20Categories%20of%20Manifolds.pdf)” among other places. I don’t know whether or not there are similar obstructions for sheaves over a manifold. If so then the methods of getting around this obstruction mentioned in the expository paper may apply in those settings as well (i.e., using Voevodsky style enlargement or using Diffeologic Spaces). The note advertises Dan Dugger's nice paper "[Sheaves and Homotopy Theory](https://pages.uoregon.edu/ddugger/cech.html)." I should disclaim that I wrote this paper early in my grad student career, so it's probably badly written, naive, and might even have errors.**
**The years since Hovey originally wrote his problem list have seen a massive development in the theory of infinity categories, and a partial answer can be given in that language. Consider the category of smooth $\infty$-groupoids, which contains the categories of smooth manifolds and Lie groupoids. Making use of the global model structure on simplicial presheaves and the fact that BG is a fibrant object therein, one can recover the de Rham cohomology of a smooth manifold as $\pi\_0$ of a mapping space in the category of smooth $\infty$-groupoids. Similarly one can recover Cech hypercohomology. A nice survey of these results can be found at the nLab page on “smooth infinity-groupoid structures.” So, although it remains unknown whether there is a model category of sheaves over a manifold, at least de Rham cohomology can be recovered from model category theoretic considerations. There does not appear to have been any work done to recover Floer homology or quantum cohomology from a model category.**
**UPDATE: Tyler Lawson has done nice work on homotopy theory for Floer homology. And I still think a model structure in this context is possible. It’s something I’d like to work on someday. Mark Hovey and I used to talk about this and we had some ideas I hope to work out one day.**
3. Every stable homotopy category I know of comes from a model category. Well, that used to be true, but it is no longer. Given a flat Hopf algebroid, Strickland and I have constructed a stable homotopy category of comodules over it. This clearly ought to be the homotopy category of a model structure on the category of chain complexes of comodules, but we have been unable to build such a model structure. My work with Strickland is still in progress, so you will have to contact me for details.
**This was solved by Mark Hovey in “[Homotopy Theory of Comodules over a Hopf Algebroid](https://arxiv.org/abs/math/0301229)” (published in contemp. math.) where he constructs for a given Hopf algebroid $(A,\Gamma)$ a model structure on the category of unbounded chain complexes of $\Gamma$-comodules with weak equivalences the homotopy isomorphisms (not the homology isomorphisms) and the cofibrations are the degreewise split monomorphisms whose cokernel is a complex of relative projectives with no differential. See Theorems 2.1.1, 2.1.3, 5.1.4, and (for the monoidal structure) 5.1.5.**
**Regarding the general philosophy behind this question, it is interesting to note that examples have been given for triangulated categories which do not arise as the homotopy category of a model category. The most well-known appears in “[Triangulated Categories without Models](https://arxiv.org/abs/0704.1378)” by Muro, Schwede, and Strickland. Note however that the definition of triangulated category used in the book Model Categories is different from the standard definition. Hovey’s definition of triangulated category T requires T to come with an action of Ho(sSet). Such triangulated categories are the ones which come up in homotopy theory, but they have not been studied systematically other than in Hovey’s book. So, technically, you could ask if every triangulated category in Hovey's sense comes from a model category. And the answer is probably "no." Since the triangulated category community kinda rejected Hovey's definition of "triangulated category", I don't think an explicit counterexample would generate much interest.**
4. Given a symmetric monoidal model category C, Schwede and Shipley have given conditions under which the category of monoids in C is again a model category (with underlying fibrations and weak equivalences). On the other hand, the category of commutative monoids seems to be much more subtle. It is well-known that the category of commutative differential graded algebras over Z cannot be a model category with underlying fibrations and weak equivalences (= homology isos). On the other hand, the solution to this is also pretty well-known--you are supposed to be using E-infinity DGAs, not commutative ones. Find a generalization of this statement. Here is how I think this should go, broken down into steps. The first step: find a model structure on the category of operads on a given model category. (Has this already been done? Charles Rezk is the person I would ask). We probably have to assume the model category is cofibrantly generated.
**This has pretty much been solved. [My thesis](https://arxiv.org/abs/1403.6759) gave conditions under which a category of commutative monoids has a transferred model structure, and also under which it’s equivalent to E-infinity algebras. Also, there is a model structure on categories of operads, due to [Berger and Moerdijk (2003)](https://arxiv.org/abs/math/0206094), if M satisfies some conditions, like having a nicely behaved interval object. For more general M, say just cofibrantly generated, there’s a semi-model structure on operads [due to Spitzweck](https://arxiv.org/abs/math/0101102) (a published reference is [Fresse’s book on operads and modules](https://link.springer.com/book/10.1007/978-3-540-89056-0)).**
5. The second step: show that the category of algebras over a cofibrant operad admits a model structure, where the fibrations and weak equivalences are the underlying ones. Show that a weak equivalence of cofibrant operads induces a Quillen equivalence of the categories of algebras. Show that an E-infinity operad is just a cofibrant approximation to the commutative ring operad. (This latter statement is probably known, since to me it seems to be the whole point of E-infinity).
**Let M be a monoidal model category. If M is nice (like, simplicial sets, equivariant topological spaces, chain complexes over a field of characteristic zero, symmetric spectra, equivariant orthogonal spectra, etc) then algebras over any operad have a transferred model structure. If $M$ is less nice, you have a transferred semi-model structure. Many authors did work in this direction, including Spitzweck, Berger–Moerdijk, Elmendorf–Mandell (for spectra), Fresse, Harper, Hess, Casacuberta, Gutierrez, Moerdijk, Vogt, Caviglia, Harper, Hornbostel (in a motivic setting), Muro, and Pavlov-Scholbach. I give some history in my papers with Donald Yau, and also what we consider to be the most general approach with the weakest hypotheses on $M$. Let's focus on full model structures instead of semi-model structures. [Our first paper](https://arxiv.org/abs/1503.06720) proves that all operads are admissible (have a transferred full model structure) in chain complexes, spaces, and symmetric spectra, and [our second paper](https://arxiv.org/abs/1609.03635) covered equivariant spaces and spectra, simplicial abelian groups, the category of small categories, the stable module category, etc. If M is only cofibrantly generated, then you have a semi-model structure on algebras over a Sigma-cofibrant operad, again by Spitzweck and Fresse. My first paper with Donald Yau generalizes this to a wider class of operads (just entrywise cofibrant), and [our third paper](https://arxiv.org/abs/1606.01803) handles rectification (a weak equivalence of operads inducing a Quillen equivalence of categories of algebras) in greater generality. I should point out that many authors had rectification results for $\Sigma$-cofibrant operads (including Berger-Moerdijk and Fresse), and Pavlov-Scholbach had admissibility and rectification results under different assumptions on $M$.**
6. Find conditions under which algebras over a noncofibrant operad admit a model structure that generalize the monoid axiom of Schwede-Shipley. This would include the case where everything is fibrant, for example. Show that, under some more conditions, a weak equivalence of operads induces a Quillen equivalence of the algebra categories. Thus, sometimes you can use commutative, sometimes you can't, but you can always use E-infinity. And using E-infinity will not hurt you when you can use commutative.
**I did this in my thesis, inventing the Commutative Monoid Axiom to get a model structure on commutative monoids. This also included the situation of rectification with E-infinity (indeed, under some more conditions, because it’s not true in the monoidal model category of compactly generated topological spaces). I generalized this in my work with Donald Yau. In our first paper, we get conditions on a model category M so that all operads are admissible (meaning: have a transferred model structure), or weaker conditions so that entrywise cofibrant operads are semi-admissible (have a transferred semi-model structure), and doing rectification in our third paper. To respond to Dmitri Pavlov's answer, let me remark that Jacob Lurie also had a result that ends up with a model structure on commutative monoids, but under much stronger conditions on $M$. Mark and I were not aware of Lurie's approach until after mine was done, and back then there was actually an error in Lurie's approach that wiped out the applications (it was only applicable to chain complexes over a field of characteristic zero). Since my approach worked for all known examples (where commutative monoids have a transferred model structure) plus some new ones, we decided to call my condition the "commutative monoid axiom" instead of giving Lurie's condition that name.**
7. Let A be a cofibrant operad as above. Use the above results to construct spectral sequences that converge to the homotopy groups of the space of A-algebra structures on a given object X, and to the homotopy groups of the mapping space of A-algebra maps between two given A-algebras. These spectral sequences for the A-infinity operad are the key formal ingredients to the Hopkins-Miller proof that Morava E-theory admits an action by the stabilizer group.
**This has been done. Rezk got the program started in part 2 of [his thesis](https://dspace.mit.edu/handle/1721.1/41793), setting up a spectral sequence to compute the homotopy groups of the moduli space of $A$-algebra structures on $X$, when $A$ is a cofibrant operad (in simplicial sets). [Vigleik Angeltveit did it](https://arxiv.org/pdf/0810.5032.pdf) in the context of spectra. A great paper by Niles Johnson and Justin Noel "[Lifting homotopy T-algebra maps to strict maps](https://arxiv.org/abs/1301.1511)" sets up a Bousfield-Kan spectral sequence that converges to the homotopy groups of the space of $A$-algebra maps between two spaces. Here $A$ is a simplicial monad and $M$ is a simplicial model category. Possibly there is room here for generalization to non-simplicial cases, but I doubt that Hovey was asking for an answer in that level of generality.**
8. My general theory is that the category of model categories is not itself a model category, but a 2-model category. Weak equivalences of model categories are Quillen equivalences, and weak equivalences of Quillen functors are natural weak equivalences. Define a 2-model category and show the 2-category of model categories is one. Note that the homotopy 2-category at least makes sense (in a higher universe): we can just invert the Quillen equivalences and the natural weak equivalences. This localization process for an n-category has been studied by Andre Hirschowitz and Carlos Simpson in descent pour les n-champs, on xxx.
**It will be debatable whether or not this problem has been satisfactorily solved. I’ve got a [recent paper with Boris Chorny](https://arxiv.org/abs/1805.05378) that can be thought of as seeking the internal hom of the “2-model category of model categories” (meaning, a model structure on a category of functors between two model categories). Boris [has done a lot of work](https://arxiv.org/search/math?searchtype=author&query=Chorny%2C%20B) in the direction of this problem and would not think it’s solved as of now. However, I think there’s a strong argument that [Reid Barton’s thesis](https://arxiv.org/abs/2004.12937) essentially solves this problem, or rather a slight but necessary weakening of what Hovey was asking for. Barton makes the argument himself, in section 1 of his thesis, laying out why he thinks this is a solution to what Hovey was asking for, and why what Hovey was asking for was literally impossible.**
9. The 2-category of simplicial model categories is supposed to be (according to me) 2-Quillen equivalent to the 2-category of model categories. Even without having all the definitions one can try to find out if this is true. For example, Dan Dugger has shown that every model category (with some hypotheses--surely cofibrantly generated at least) is Quillen equivalent to a simplicial model category. Understand his result in the context of the preceding two problems. That is, does Dugger's construction in fact give a 2-functor from model categories to simplicial model categories? Does it preserve enough structure to make it clear that it will induce some kind of equivalences on the homotopy 2-categories?
**Again, people will debate if this has been solved. But if you restrict attention to combinatorial model categories then essentially it has been. Dugger proved that [every combinatorial model category is Quillen equivalent to a simplicial combinatorial model category](https://arxiv.org/abs/math/0007068) (Batanin and I [recently proved the same](https://arxiv.org/abs/2001.03764) for combinatorial semi-model categories). Back when Clark Barwick was writing [a paper about “partial model categories”](https://arxiv.org/abs/1102.2512) I convinced myself that the collection of them was equivalent to partial simplicial model categories. I have a hazy recollection that Lennart Meier did [some](https://arxiv.org/abs/1408.2743) [work](https://arxiv.org/abs/1503.02036) related to this in 2015. Barton’s thesis also has Theorem 1.3.4 which is like a 2-model category of simplicial combinatorial premodel categories.**
10. Is every monoidal model category Quillen equivalent to a simplicial monoidal model category? This would remove the loose end in my book on model categories, where I am unable to show that the homotopy category of a monoidal model category is a central algebra over the homotopy category of simplicial sets. The centrality is the problem, and I can cope with this problem for simplicial monoidal model categories.
**Essentially yes. The relevant paper here is "[Admissible replacements for simplicial monoidal model categories](https://arxiv.org/abs/2008.00515)" by Bayinder and Chorny. Also, the centrality problem was resolved. I learned one solution from Jerome Scherer, who told me it's in his 2008 paper (with Chacholski) [Representations of Spaces](https://arxiv.org/abs/math/0511577). Denis-Charles Cisinski points out that [he also solved it in 2002](https://conf.math.illinois.edu/K-theory/0602).**
11. Charles Rezk has a homotopy theory of homotopy theories. This is just a category, though it is large. The objects are generalizations of categories where composition is not associative on the nose--that is, they are some kind of simplicial spaces. Understand the relationship between Rezk's point of view and mine on the 2-category of model categories. They should be equivalent in some sense.
**Again, modulo the fact that what Hovey had in mind doesn’t exactly exist, this problem has been solved. There are [many, many models for the "homotopy theory of homotopy theories"](https://ncatlab.org/nlab/show/%28infinity%2C1%29-category), including quasi-categories, relative categories, simplicial categories, topological categories, Segal categories, Segal spaces, complete Segal spaces, $A\_\infty$-categories, etc, etc. Each has a model structure and these model structures are all Quillen equivalent, even in a coherent way. See work of [Toen](https://hal.archives-ouvertes.fr/hal-00772983/document) and [Barwick and Schommer-Pries](https://arxiv.org/pdf/1112.0040.pdf). There are also models for $\infty$-categories with extra structure, like the category of cofibration categories (whose homotopy theory was worked out beautifully [by Karol Szumilo](https://arxiv.org/abs/1411.0303)) or the category of fibration categories. There's also a model-free approach due to Riehl and Verity, called [$\infty$-cosmoi](https://ncatlab.org/nlab/show/infinity-cosmos).**
**However, the theory of model categories is not equivalent to the theory of $(\infty,1)$-categories, because not every infinity category comes from a model category. If an infinity category comes from a model category, it must have all limits and colimits.**
**There are two ways to proceed. You can weaken what you mean by "model category" or you can look for an equivalence with $(\infty,1)$-categories with extra structure. Both approaches work. For example, if you weaken from "model category" to, say, "partial model category", then Barwick and Kan’s paper on the subject (linked above) shows how the theory of partial model categories is equivalent to the theory of $\infty$-categories. The same is true if you weaken from "model category" to "relative category". As for $(\infty,1)$-categories with extra structure, the right notion is "presentable $(\infty,1)$-categories." The theory of combinatorial model categories [is equivalent in a strong way](https://arxiv.org/abs/2110.04679) to the theory of presentable infinity categories.**
12. In the appendix to my book on model categories, I said maybe what we are doing in associating to a model category its homotopy category is the wrong thing. Maybe we should be associating to a model category C the homotopy categories of all the diagram categories C^I, together with all the adjunctions induced by functors I --> J. This would make homotopy limits and colimits part of the structure. Does this viewpoint have any value?
**I think what Mark had in mind here was basically the [theory of derivators](https://en.wikipedia.org/wiki/Derivator), first conceived by Grothendieck in 1983, and worked out by many authors over the past fifteen years or so. I would say it’s widely accepted that the viewpoint of derivators does have value and is an acceptable way to do homotopy theory. I don’t claim that it's equivalent in a formal way to the theory of model categories or infinity categories.**
13. Find a model category you can prove is not cofibrantly generated. This is just an annoyance, not a very significant problem, but it has been bugging me for a while. The obvious candidate for this is the simplest nontrivial model category, the one on chain complexes where weak equivalences are chain homotopy equivalences. Mike Cole is, so far as I know, the first to write down a desciption of this model category, though one certainly has the feeling that Quillen must have known about it. But how do you prove something is not cofibrantly generated?
**Many examples have now been given. I’ll list all the ones I know:**
* George Raptis proves in Remark 4.7 of “[Homotopy Theory for Posets](https://projecteuclid.org/journals/homology-homotopy-and-applications/volume-12/issue-2/Homotopy-theory-of-posets/hha/1296223882.pdf)” that Strom’s model structure on Top is not cofibrantly generated. This answers a question which was implicit in Hovey’s book when he discusses this model structure, but which he never explicitly asked.
* Christensen and Hovey's paper ["Quillen model structures for relative homological algebra"](https://arxiv.org/abs/math/0011216) prove that the absolute model structure on relative chain complexes is not cofibrantly generated (see section 5).
* [Isaksen's model structure](https://arxiv.org/abs/math/0106152) on pro-simplicial sets is not cofibrantly generated, and the [category of simplicial sets is not fibrantly generated](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.235.7756&rep=rep1&type=pdf). For many more examples see Scott Balchin's recent book [A Handbook of Model Categories](https://link.springer.com/book/10.1007/978-3-030-75035-0)
* [Chorny showed](https://arxiv.org/abs/math/0112176) that the model category of maps of spaces or simplicial sets (i.e. the arrow category) is not cofibrantly generated. This might have been the first example known.
* [Adamek, Herrlich, Rosicky, and Tholen](https://link.springer.com/article/10.1023/A:1015270120061) produce a model structure on the category of small categories that is not cofibrantly generated.
* Chorny-Rosicky “[class combinatorial model categories](https://arxiv.org/abs/1110.4252)” includes examples of Pro spaces and Ind spaces, and about Fun(M,N), that are not confibrantly generated in the classical sense for set-theoretic reasons.
* There are several examples in [Emily Riehl’s book](https://emilyriehl.github.io/files/cathtpy.pdf). Chapters 12 and 13 discuss “category cofibrantly generated” when the left class is a category not a set. The left class must be defined as a retract-closure, but the right is automatically closed under retracts. Examples include Hurewicz models. She also covers "enriched cofibrantly generated" more generally.
* [Barthel and Riehl](https://arxiv.org/abs/1204.5427): Cole’s
mixed model structure – this is an example like the above for
topological model categories.
* [Barthel May Riehl](https://arxiv.org/abs/1310.1159) – this is an
example like the above for chain enriched.
* [Lack’s 2007 paper](https://arxiv.org/abs/math/0607646) has a non-cofibrantly generated model structure on Arr(Cat), but it is cofibrantly generated in Riehl's sense.
* [Bourke and Garner](https://www.sciencedirect.com/science/article/abs/pii/S002240491500170X) have notion of "cofibrantly generated by a double category",
plus an example that is not cofibrantly generated as a category (because the right class is
not closed under retracts).
**UPDATE: In my paper with Donald Yau about arrow categories, we list several examples of monoidal model categories that are not cofibrantly generated.**
| 38 | https://mathoverflow.net/users/11540 | 424641 | 172,481 |
https://mathoverflow.net/questions/424408 | 20 | **Question P.** *Can a polynomial $P(x)=\sum\_{n=0}^ma\_nx^n$ with coefficients $a\_n\in\{-1,1\}$ (and $P(1)=0$) have a multiple root in the interval $(\tfrac12,1)$?*
Also I am interested in a similar question for analytic functions.
**Question A.** *Let $f(x)=\sum\_{n=0}^\infty a\_nx^n$ a series with coefficients $a\_n\in\{-1,1\}$ such that $\sup\_{m\in\mathbb N}|\sum\_{n=0}^ma\_i|<\infty$. Can the analytic function $f$ have a multiple root in the interval $(0,1)$?*
| https://mathoverflow.net/users/61536 | Multiple roots of polynomials with coefficients $\pm 1$ |
>
> **Question P.** *Can a polynomial $P(x)=\sum\_{n=0}^ma\_nx^n$ with coefficients $a\_n\in\{-1,1\}$ (and $P(1)=0$) have a multiple root in the interval $(\tfrac12,1)$?*
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>
>
Yes. The following four Littlewood polynomials:
* $z^{27} + z^{26} + z^{25} + z^{24} + z^{23} - z^{22} - z^{21} + z^{20} + z^{19} + z^{18} - z^{17} - z^{16} - z^{15} - z^{14} - z^{13} - z^{12} - z^{11} - z^{10} + z^9 + z^8 - z^7 + z^6 - z^5 + z^4 - z^3 + z^2 + z - 1 = (z^{18} + z^{16} + 2z^{15} + 2z^{13} + z^{12} + 2z^{11} + 3z^{10} + 3z^8 + 2z^7 + z^6 + 2z^5 + 2z^3 + 1)(z^2 + 1)(z - 1)(z^3 + z^2 - 1)^2$
* $z^{27} + z^{26} + z^{25} - z^{24} - z^{23} - z^{22} + z^{21} - z^{20} - z^{19} + z^{18} + z^{17} - z^{16} - z^{15} + z^{14} + z^{13} - z^{12} - z^{11} - z^{10} - z^9 - z^8 + z^7 + z^6 - z^5 + z^4 - z^3 + z^2 + z - 1 = (z^{21} - z^{20} + 2z^{19} - 2z^{18} + z^{17} + z^{16} - 3z^{15} + 3z^{14} - 2z^{13} + 2z^{11} - 4z^{10} + 4z^9 - 2z^8 - z^7 + 3z^6 - 4z^5 + 2z^4 - z^3 - z^2 + z - 1)(z^3 + z^2 - 1)^2$
* $z^{27} + z^{26} + z^{25} + z^{24} + z^{23} - z^{22} - z^{21} + z^{20} - z^{19} - z^{18} - z^{17} + z^{16} - z^{15} + z^{14} + z^{13} - z^{12} - z^{11} - z^{10} - z^9 - z^8 + z^7 + z^6 - z^5 + z^4 - z^3 + z^2 + z - 1 = (z^{18} + z^{16} + 2z^{15} + 2z^{13} + z^{12} + 2z^{11} + z^{10} + 3z^8 + z^6 + 2z^5 + 2z^3 + 1)(z^2 + 1)(z - 1)(z^3 + z^2 - 1)^2$
* $z^{27} + z^{26} + z^{25} - z^{24} - z^{23} - z^{22} + z^{21} + z^{20} + z^{19} + z^{18} - z^{17} + z^{16} - z^{15} - z^{14} - z^{13} + z^{12} - z^{11} - z^{10} - z^9 - z^8 + z^7 + z^6 - z^5 + z^4 - z^3 + z^2 + z - 1 = (z^{18} + z^{16} + 2z^{13} - z^{12} + 2z^{11} + z^{10} + 3z^8 + z^6 + 2z^5 + 2z^3 + 1)(z^2 + 1)(z - 1)(z^3 + z^2 - 1)^2$
all have repeated factor $z^3 + z^2 - 1$ with root $z \approx 0.75488$.
| 14 | https://mathoverflow.net/users/46140 | 424650 | 172,484 |
https://mathoverflow.net/questions/424646 | 6 | Let $V$ be a (complex, finite-dimensional) vector space. Suppose that two (finite) groups $A$ and $B$ act on $V$. Furthermore, suppose that these actions commute, so that the direct product $A \times B$ also acts on $V$.
As a representation of $A$, we can decompose $V$ as the direct sum $$ V = \bigoplus\_i V\_i ,$$ where each $V\_i$ is an irreducible representation. And we can do the same by considering $V$ as a representation of $B$, say
$$V = \bigoplus\_j V\_j' .$$
Also, it is known that every irreducible representation of a direct product of groups $A \times B$ is of the form $V\_i \otimes V\_j'$, where $V\_i$ is irreducible for $A$ and $V\_j'$ is irreducible for $B$.
My question is : If I know the two decompositions of $V$ as both a representation of $A$ and of $B$, can I understand its decomposition $$ V = \bigoplus\_{i,j} V\_i \otimes V\_j' $$ into irreducible representations of $A \times B$ in a simple way, or is it a hard problem in general?
| https://mathoverflow.net/users/73667 | If I know how to decompose a vector space in irreducible representations of two groups, can I understand the decomposition as a rep of their product? | To be totally clear: no, the decomposition as a representation of $A$ and the decomposition as a representation of $B$ separately don't determine the decomposition as a representation of $A \times B$, because this is not enough information by itself to determine which irreducibles of $A$ pair with which irreducibles of $B$ in general.
The smallest counterexample is $A = B = C\_2$ acting on a $2$-dimensional vector space $V$ such that, as a representation of either $A$ or $B$, $V$ decomposes as a direct sum of the trivial representation $1$ and the sign representation $-1$. This means that $V$ could be either $1 \otimes 1 + (-1) \otimes (-1)$ or $1 \otimes (-1) + (-1) \otimes 1$ (the $+$ here is a direct sum but I find writing direct sums and tensor products together annoying to read) and you can't tell which. You can construct a similar counterexample out of any pair of groups $A, B$ which both have non-isomorphic irreducibles of the same dimension.
What you can do instead is the following. If you understand the action of $A$, then you get a canonical decomposition of $V$ as a direct sum
$$V \cong \bigoplus\_i V\_i \otimes \text{Hom}\_A(V\_i, V)$$
where $V\_i$ are the irreps of $A$ and $\text{Hom}\_A(V\_i, V)$ is the *multiplicity space* of $V\_i$. If instead of considering the multiplicity space we just write a direct sum of a bunch of copies of $V\_i$ that decomposition is not canonical and not unique. The point of doing this canonically is that if the action of $B$ commutes with the action of $A$ then the action of $B$ naturally descends to a family of actions on each multiplicity space $\text{Hom}\_A(V\_i, V)$. Once you decompose each of these actions into irreps of $B$ then you've understood the entire action of $A \times B$; this is how you prove that result you cite about direct products. Of course you could also start with $B$ instead.
| 13 | https://mathoverflow.net/users/290 | 424654 | 172,486 |
https://mathoverflow.net/questions/424430 | 10 | Consider the $N$ by $N$ matrix
$$M\_N= \begin{pmatrix} 1+3\lambda & -1-2\lambda & - \lambda & 0 & 0 &0 &0\\
-1-2\lambda & 2+3\lambda & -1 & -\lambda & 0 & 0 & 0\\
-\lambda & -1 & 2(1+\lambda) &-1 & -\lambda & 0& 0 \\
0 & \ddots & \ddots & \ddots & \ddots& \ddots & 0 \\
0 & 0 & -\lambda & -1 & 2(1+\lambda) & -1 & -\lambda \\
0&0 & 0& -\lambda & -1& 2+3\lambda & -1-2\lambda \\
0 & 0 & 0 &0 & -\lambda & -1-2\lambda& 1+3\lambda \\
\end{pmatrix}.
$$
I want to show that there exists $\varepsilon>0$ such that for all $\lambda \in (-\varepsilon,\varepsilon)$ **and independent of $N$**, the matrix $M\_N$ does not have an eigenvector of the form $u=(0,u\_2,...,u\_N)$,i.e. where the first component vanishes.
Numerically it seems to be true up to $\varepsilon =1/4,$ but it is hard to understand this 5 term recurrence for me.
| https://mathoverflow.net/users/150549 | Support of eigenvectors | This looks to be true for $\varepsilon=1/4$.
First of all, the space of symmetric ($u\_i=u\_{N+1-i}$ for all $i=1,\ldots, N$) vectors is invariant for $M\_N$, and so is the space of antisymmetric ($u\_i=-u\_{N+1-i}$ for all $i=1,\ldots, N$) vectors. Since the sum of these two subspaces consists of all vectors, we conclude that for any eigenvalue there exists either a symmetric or antisymmetric eigenvector. At first, we consider the case when this vector has zero first coordinate (thus zero last coordinate aswell). For a vector $u=(u\_1,\ldots,u\_N)$ we attribute a polynomial $p\_u:=\sum u\_i t^{N+1-i}$. Then $M\_Nu$ corresponds to a polynomial $p\_u(—\lambda/t^2—1/t+2+2\lambda-t-\lambda t^2)+R$, where $R$ is a remainder term which depends on $u\_1,u\_2,u\_{N-1},u\_N$. If $u$ is a symmetric or antisymmetric eigenvector with $u\_1=u\_N=0$ and eigenvalue $\mu$, the remainder $R$ equals $-\lambda u\_2(t-1)^2(t^{n-2}\pm 1/t)$, and we see that $\lambda\ne 0$ and all roots of $—\lambda/t^2—1/t+2+2\lambda-t-\lambda t^2-\mu$ must lie on the unit circle. If they are $e^{\pm ia}, e^{\pm ib}$, then $2\cos a, 2\cos b$ must be roots of $-\lambda x^2-x+2+4\lambda-\mu=0$, and by Vieta theorem $|1/\lambda|=2|\cos a+\cos b|\leqslant 4$.
So, it remains to consider the case when the same eigenvalue has both a symmetric and antisymmetric eigenvector. In natural coordinates the corresponding matruces differ only by few elements, that may possibly yield some interlacing type claim.
| 2 | https://mathoverflow.net/users/4312 | 424655 | 172,487 |
https://mathoverflow.net/questions/424272 | 16 | Main Question
-------------
This question is about finding *explicit*, *calculable*, and *fast* error bounds when approximating continuous functions with polynomials to a user-specified error tolerance.
---
EDIT (Apr. 23): See also a [revised version of this question](https://mathoverflow.net/questions/442057/explicit-and-fast-error-bounds-for-approximating-continuous-functions) on *MathOverflow*.
Also: The other answer to this question claims that the result it mentions "should be easy to map ... to [0, 1]". I don't see how, especially since not every function that meets the result's assumptions on [0, 1] can be extended to one that does so on [-1, 1]. In addition, does the result on the polynomial interpolant relate to the "exact" Chebyshev coefficients, or the "approximate" and "easier-to-compute" coefficients? (These and other unsure points on the Chebyshev method have made me prefer other methods to approximate functions — especially those that don't introduce transcendental functions, and these other methods include linear combinations and iterated Boolean sums of Bernstein polynomials.)
In addition, a comment to that answer mentions choosing "rational nodes close to the Chebyshev nodes" as a way of avoiding transcendental functions. But are there results on the use of such rational nodes to approximate continuous functions and, if so, do they provide explicit error bounds (with no hidden constants) at the rate $O(1/n^{r/2})$ for the class of functions with continuous $r$-th derivative on [0, 1]?
---
In this question:
* A polynomial $P(x)$ is written in *Bernstein form of degree $n$* if it is written as $P(x)=\sum\_{k=0}^n a\_k {n \choose k} x^k (1-x)^{n-k},$ where $a\_0, ..., a\_n$ are the polynomial's *Bernstein coefficients*.
* For the *Bernstein polynomial* of $f(x)$ of degree $n$, $a\_k = f(k/n)$.
Let $f:[0,1]\to [0,1]$ be continuous and polynomially bounded (both $f$ and $1-f$ are bounded below by min($x^n$, $n(1-x)^n$) for some integer $n$), let $r\ge 3$, and denote the Bernstein polynomial of degree $n$ of a function $g$ as $B\_n(g)$.
Given that $f$ has a continuous $r$-th derivative (or has a Lipschitz continuous $(r-1)$-th derivative), are there results that give a sequence of polynomials $P\_n$ with the following error bound?
$$| f(x) - P\_n(f)(x) | \le \epsilon(f, n, x) = O(1/n^{r/2}),$$ where:
1. $\epsilon(f, n, x)$ is a fully determined function, with all constants in the expression having a **known exact value or upper bound**.
2. $P\_n(f)(x)$ is an approximating polynomial of degree $n$ that can be readily rewritten to a polynomial in Bernstein form with coefficients in $[0, 1]$. Preferably, $P\_n(f)$ has the form $B\_n(W\_n(f))$ where $W\_n(f)$ is easily computable from $f$ using rational arithmetic only (see "Remarks", later).
One way to answer this (see [this question](https://math.stackexchange.com/questions/3904732/what-are-ways-to-compute-polynomials-that-converge-from-above-and-below-to-a-con)) is to find a sequence $W\_n(f)$ and an explicit and tight upper bound on $C\_1>0$ such that, for each integer $n\ge 1$ that's a power of 2— $$\max\_{0\le k\le 2n}\left|\left(\sum\_{i=0}^k \left(W\_n\left(\frac{i}{n}\right)\right) {n\choose i}{n\choose {k-i}}/{2n \choose k}\right)-W\_{2n}\left(\frac{k}{2n}\right)\right|\le \frac{C\_1 M}{n^{r/2}},$$ where $M$ is the maximum absolute value of $f$ and its derivatives up to the $r$-th derivative (or, respectively, the maximum of $|f|$ and the Lipschitz constants of $f$ and its derivatives up to the $(r-1)$-th derivative).
Then $| f(x) - B\_n(W\_n(f))(x) | \le \frac{C\_1}{1-\sqrt{2/2^{r+1}}}\frac{M}{n^{r/2}}=O(1/n^{r/2})$ (see Lemma 3 in "[Proofs for Polynomial-Building Schemes](https://peteroupc.github.io/bernsupp.html#Proofs_for_Polynomial_Building_Schemes)), although this is only guaranteed to work for power-of-2 values of $n$. For example, $W\_n$ can be $2f-B\_n(f)$ and $r$ can be 3 or 4, or $W\_n$ can be $B\_n(B\_n(f))+3(f-B\_n(f))$ and $r$ can be 5 or 6.
### Motivation
My motivation for this question is to implement "[*approximate Bernoulli factories*](https://peteroupc.github.io/bernsupp.html#Approximate_Bernoulli_Factories)", or algorithms that toss heads with a probability equal to a polynomial in Bernstein form that comes within $\epsilon$ of a continuous function $f(x)$. This involves finding a reasonably small degree $n$ of that polynomial, then the algorithm works as follows:
1. Flip the coin $n$ times, count the number of heads as $h$.
2. With probability equal to the $h$-th Bernstein coefficient, return heads; otherwise tails.
Note that the algorithm requires finding only one Bernstein coefficient per run. And for ordinary Bernstein polynomials, finding it is trivial — $f(h/n)$ — but the degree $n$ can be inordinate due to Bernstein polynomials' slow convergence; for example, if $\epsilon=0.01$ and $f$ is Lipschitz with constant 1, the required polynomial degree is 11879.
* Approximating $f$ with a rational function is also interesting, but is outside the scope of this question.
* *Exact* Bernoulli factories require a slightly different approach to finding the polynomials; see [another question of mine](https://math.stackexchange.com/questions/3904732/what-are-ways-to-compute-polynomials-that-converge-from-above-and-below-to-a-con).
### Polynomials with faster convergence than Bernstein polynomials
As is known since Voronovskaya (1932), the Bernstein polynomials converge uniformly to $f$, in general, at a rate no faster than $O(1/n)$, regardless of $f$'s smoothness, which means that it won't converge in a finite expected running time. (See also a [related question](https://math.stackexchange.com/questions/3889382/bound-on-approximation-error-of-bernstein-polynomial) by Luis Mendo on ordinary Bernstein polynomials.)
But Lorentz (1966, "The degree of approximation by polynomials with positive coefficients") has shown that if $f(x)$ is positive (the case that interests me) and has $k$ continuous derivatives, there are polynomials with non-negative Bernstein coefficients that converge to $f$ at the rate $O(1/n^{k/2})$ (and thus can be faster than the $O(1/n^{2+\epsilon})$ needed for a finite expected running time, depending on $f$).\*
Thus, people have developed alternatives, including *iterated Bernstein polynomials*, to improve the convergence rate. These include:
* Micchelli, C. (1973). The saturation class and iterates of the Bernstein polynomials. Journal of Approximation Theory, 8(1), 1-18.
* Guan, Zhong. "[Iterated Bernstein polynomial approximations](https://arxiv.org/pdf/0909.0684)." arXiv preprint arXiv:0909.0684 (2009).
* Güntürk, C. Sinan, and Weilin Li. "[Approximation with one-bit polynomials in Bernstein form](https://arxiv.org/pdf/2112.09183)" arXiv preprint arXiv:2112.09183 (2021).
* The "Lorentz operator": Holtz, Olga, Fedor Nazarov, and Yuval Peres. "[New coins from old, smoothly](https://link.springer.com/article/10.1007/s00365-010-9108-5)" Constructive Approximation 33, no. 3 (2011): 331-363.
* Draganov, Borislav R. "On simultaneous approximation by iterated Boolean sums of Bernstein operators." Results in Mathematics 66, no. 1 (2014): 21-41.
Usually, papers like those express a bound on the error when approximating a function with polynomials as follows: $$| f(x) - P\_n(f)(x) | \le c\_n \epsilon(f, n, x),$$ where $\epsilon(f, n, x)$ is a fully determined function, $c\_n>0$ is a constant that may depend on $n$, and $P\_n(f)(x)$ is an approximating polynomial of degree $n$.
There are results where the error bound $\epsilon(.)$ is in $O(1/n^{k/2})$, but in all those results I've seen so far (e.g., Theorem 4.4 in Micchelli; Theorem 5 in Güntürk and Li), $c\_n$ is unknown, and no upper bound for $c\_n$ is given by the results in the papers above, so that the error bound is unimplementable and there is no way of knowing beforehand whether $P\_n$ will come close to $f$ within a user-specified error tolerance. (There is also a separate matter of rewriting the polynomial to its Bernstein form, but this is much more manageable, especially with the Lorentz operator.)
### Remarks
* I prefer approaches that involve only rational arithmetic and don't require transcendental or trigonometric functions to build the Bernstein-form polynomials.
+ Unlike with rational arithmetic (where arbitrary precision is trivial thanks to Python's `fractions` module), transcendental and trig. functions require special measures to support arbitrary accuracy, such as constructive/recursive reals — floating point won't do for my purposes.
+ In general, "rounding" a polynomial's coefficients or "nodes" to rational numbers will add a non-trivial error that, for my purposes, has to be accounted for in any error bound.
---
\* If the polynomials are not restricted in their coefficients, then the rate $O(1/n^k)$ is possible (e.g., DeVore and Lorentz 1993). But this result is not useful to me since my use case (approximate Bernoulli factories) requires the polynomials to have Bernstein coefficients in $[0, 1]$.
| https://mathoverflow.net/users/171320 | Explicit and fast error bounds for polynomial approximation | If $f, f', \dotsc, f^{(\nu-1)}$ are all absolutely continuous on $[-1, 1]$, and $f^{(\nu)}$ has bounded variation $V$ on $[-1, 1]$, where $\nu \ge 1$, then the polynomial interpolant $p\_n$ of degree $n > \nu$ through the Chebyshev points of the second kind,
$$ p\_n(x\_j) = f(x\_j), \quad x\_j = \cos \tfrac{j\pi}{n}, \quad j=0,\dotsc,n, $$
satisfies the bound
$$ \sup\_{-1 \le x \le 1} | f(x) - p\_n(x)| \le \frac{4V}{\pi \nu (n - \nu)^\nu}. $$
See Theorem 7.2, equation (7.5) of [1].
It should be easy to map the result to [0, 1]. If all you know is that $f$ has $k$ continuous derivatives, then you can take $\nu = k-1$, provided $k \ge 2$. The convergence rate $O(n^{-(k-1)})$ is faster than you asked for when $k > 2$ (and equal at $k=2$).
I am unclear exactly what you mean by being "readily rewritten" to the Bernstein polynomial basis. One explicit possibility: you can use a discrete Fourier transform (specifically, the FFT) to recover the coefficients of the Chebyshev series of $p\_n$ from the values at the Chebyshev points above (again, see [1]), then use an explicit expansion of the Chebyshev polynomials in the Bernstein basis (e.g., the one on the [Wikipedia page for the Bernstien polynomials](https://en.wikipedia.org/wiki/Bernstein_polynomial)).
[1] *Trefethen, Lloyd N.*, Approximation theory and approximation practice, Other Titles in Applied Mathematics 128. Philadelphia, PA: Society for Industrial and Applied Mathematics (SIAM) (ISBN 978-1-611972-39-9/pbk). 305 p. (2013). [ZBL1264.41001](https://zbmath.org/?q=an:1264.41001).
| 5 | https://mathoverflow.net/users/70005 | 424657 | 172,488 |
https://mathoverflow.net/questions/424683 | 9 | In a closed (say differentiable) Riemannian manifold you see only continuous features when looking at small neighbourhoods of points. From afar,
discrete features appear ((co)homology, closed geodesics, eigenvalues of the laplacian and so on).
In physics, I have the impression that more or less the opposite is going on: seen from
afar you get (general) relativity, differential equations etc.
Zooming in you have particles, quantum mechanics, discrete eigenvalues, ....
An obvious explanation might be that the 'manifold-picture' is
not a good model for our universe.
Is there an obvious mathematical object which behaves
like the universe: local properties are mainly discrete and
global properties tend to be continuous? (Coming up with the
adele-stuff is kind of cheating, I think, since 'locality' in
this context is mainly a convention. The other obvious 'explanation', string-theory, seems also to be somewhat controversial.)
**Which mathematical objects behave like the
universe with respect to smoothness/discreteness?**
Motivation: I have no real motivation for this question other than curiosity: It is quite striking that one of the most studied objects of mathematics, manifolds, has smoothness/discreteness inversed with respect to physics (in my limited understanding of both areas).
| https://mathoverflow.net/users/4556 | Why are discreteness and smoothness in physics inversed with respect to geometry? | The "manifold picture" can be applied to physics in the context of the [Brillouin zone](https://en.wikipedia.org/wiki/Brillouin_zone), see for example [On Brillouin Zones](https://www.math.stonybrook.edu/preprints/ims98-7.pdf). The reason that discreteness and smoothness appear inverted, is that the Brillouin zone describes [reciprocal space](https://en.wikipedia.org/wiki/Reciprocal_lattice). The distinction is not fundamental, one can equivalently describe a crystal in real space, where discrete features appear on short distance, or in reciprocal space, where discrete features appear at large distance.
So to answer the specific question in the OP: I don't think there is a need to abandon the manifold picture to describe physical matter, you just want to apply it to reciprocal space rather than to real space.
| 6 | https://mathoverflow.net/users/11260 | 424690 | 172,494 |
https://mathoverflow.net/questions/424686 | 2 | Let $(C, \|\cdot\|)$ be the Banach space of continuous paths $x: [0,1]\rightarrow\mathbb{R}^d$ starting at zero with sup-norm $\|\cdot\|$.
Let further $B\subset C$ be the subspace of $0$-started continuous paths of bounded variation, i.e.
$$B = \big\{x\in C \ \big| \ \|x\|\_1:=\sup\nolimits\_{(t\_\nu)}{\textstyle\sum\_\nu}|x\_{t\_\nu} - x\_{t\_{\nu-1}}|<\infty \ \text{ and } \ x(0)=0\big\}$$
where the above sup runs over all partitions $(t\_\nu)$ of $[0,1]$.
The set $B$ can be endowed with both the (uniform) norm $\|\cdot\|$ and the (1-variation) norm $\|\cdot\|\_1$.
Do you know if the Borel $\sigma$-algebras on $(B, \|\cdot\|)$ and $(B, \|\cdot\|\_1)$ coincide?
Any references or hints are welcome.
*Edit:* As pointed out by Gerald Edgar, we narrowed $B$ down to paths started at zero to make the question more meaningful.
*Edit 2:* As Gerald Edgar pointed out in the comments to his answer below, the answer to this question is **no, the Borel $\sigma$-algebras cannot coincide** because there are more $\|\cdot\|\_1$-open sets in $B$ than there are $\|\cdot\|$-Borel sets in $C$.
| https://mathoverflow.net/users/472548 | Borel $\sigma$-algebras on paths of bounded variation | I think there is a problem with a mere semi-norm. The constant functions have variation distance $0$ from each other. Any variation-open set contains either all the constants, or none of them. Therefore, any variation-Borel set contains either all the constants, or none of them.
Choose $\mathbf u \in \mathbb R^n \setminus \{0\}$. Take the subset $\{0,\mathbf{u}\} \subseteq B$ of two constant functions. It is sup-norm closed, so it is a uniform-Borel set. But it is not a variation-Borel set.
---
Maybe a better question would restrict to the subset of $B$ with $x(0) = 0$. Then at least we have a norm.
---
rmcerafl pointed out a problem with the following. For now, I will leave it here for reference.
>
> Let $C\_0 = \{x \in C : x(0)=0\}$ with the uniform norm. Let
> $B\_0 = \{x \in B : x(0)=0\}$ with the variation norm. Both of these are complete separable metric spaces. The inclusion map $i : B\_0 \to C\_0$ is continuous and injective. Then citing a result or two from descriptive set theory\*, we get: $i(B\_0)$ is a Borel subset of $C\_0$ and $i : B\_0 \to i(B\_0)$ is a Borel-isomorphic map. This shows, in particular, the variation-Borel subsets of $B\_0$ coincide with the uniform-Borel subsets of $i(B\_0)$.
>
>
>
---
${}^\*$ Proposition 8.3.5 and Theorem 8.3.7 in:
*Cohn, Donald L.*, Measure theory, Boston, Basel, Stuttgart: Birkhäuser. IX, 373 p. DM 42.00 (1980). [ZBL0436.28001](https://zbmath.org/?q=an:0436.28001).
| 2 | https://mathoverflow.net/users/454 | 424692 | 172,495 |
https://mathoverflow.net/questions/424694 | 20 | Let $p$ be a prime, and consider $$S\_p(a)=\sum\_{\substack{1\le j\le a-1\\(p-1)\mid j}}\binom{a}{j}\;.$$
I have a rather complicated (15 lines) proof that $S\_p(a)\equiv0\pmod{p}$. This must be
extremely classical: is there a simple direct proof ?
| https://mathoverflow.net/users/81776 | Analogue of Fermat's "little" theorem | Let $$P(x)=(1+x)^a-1-x^a=\sum\_{1 \le j \le a-1} \binom{a}{j}x^j.$$ Working in a field $F$ where $|\{\mu \in F: \mu^{p-1}=1\}|=p-1$ (roots of unity of order $p-1$ exist), we have
$$ \frac{1}{p-1}\sum\_{\mu^{p-1}=1}P(x \mu) = \sum\_{\substack{1 \le j \le a-1\\(p-1)\mid j}} \binom{a}{j}x^j.$$
We now specialize the field to $\mathbb{F}\_p$, and let $x=1$:
$$ -\sum\_{\mu \in \mathbb{F}\_p^{\times}}P( \mu) = \sum\_{\substack{1 \le j \le a-1\\(p-1)\mid j}} \binom{a}{j}.$$
To conclude, observe that $P(0)=0$ and $|\mathbb{F}\_p| = p$, so $$\sum\_{\mu \in \mathbb{F}\_p^{\times}} P(\mu) = \sum\_{\mu \in \mathbb{F}\_p} P(\mu) = \sum\_{\mu \in \mathbb{F}\_p} ((1+\mu)^a - \mu^a)=0,$$
because $\mu\mapsto \mu+1$ is a permutation of $\mathbb{F}\_p$.
---
A slightly more general congruence is due to Glaisher (1899), as I've found in a survey by Granville, see [equation (11) here](https://dms.umontreal.ca/%7Eandrew/Binomial/sumsofbcs.html). The precise reference is Glaisher, J. W. L., "A congruence theorem relating to sums of binomial-theorem coefficients.", Quart. J. 30, 150-156, 349-360, 361-383 (1899). See [here for the zbMath review](https://zbmath.org/?q=an%3A29.0152.05) in German. There is no entry in MathSciNet.
Namely, Glaisher proved that for any $0 \le j\_0 \le p-1$ we have
$$\sum\_{\substack{1 \le j \le a \\ j \equiv j\_0 \bmod (p-1)}} \binom{a}{j} \equiv \binom{k}{j\_0} \bmod p$$
where $k$ is any positive integer with $k \equiv a \bmod p$ Applying this with $j\_0=0$ and $k=a$, and observing that the term $j=a$ contributes $1 \bmod p$, your result follows. Clicking on equation (11) in the link above leads to a [detailed proof](https://dms.umontreal.ca/%7Eandrew/Binomial/pfeq11.html) which utilizes Lucas' theorem.
| 38 | https://mathoverflow.net/users/31469 | 424696 | 172,496 |
https://mathoverflow.net/questions/424699 | 2 | Let $X$ be a variety over $k = \mathbb{C}$ with an action of a finite group $G$. According to [this paper](https://arxiv.org/pdf/1711.07909.pdf) (Section 4), the induced action of $G$ on the cohomology of $X$ respects the mixed Hodge structure, allowing to define a $G$-equivariant $E$-polynomial
$$
e^G(X) = \sum\_{k, p, q} (-1)^k [H\_c^{k; p, q}(X; \mathbb{Q})] u^p v^q \in Z[u, v] \otimes\_\mathbb{Z} R(G) ,
$$
where $R(G)$ is the representation ring of $G$.
The coefficients of $e^G(X)$ can be computed from the $E$-polynomials of the quotients $e(X / H)$, where $H$ is a subgroup of $G$. For example, the group $G = S\_3$ has three irreducible representations: $T$ (trivial), $S$ (sign) and $D$ (standard), so we can write $e^{S\_3}(X) = a \otimes T + b \otimes S + c \otimes D$. Using that $e(X) = a + b + 2c$, $e(X / S\_3) = a$, $e(X / \langle \tau \rangle) = a + c$ and $e(X / \langle \rho \rangle) = a + b$, where $\tau = (1 \; 2)$ and $\rho = (1 \; 2 \; 3)$, we find that
$$
a = e(X / S\_3), \quad b = e(X) - 2 \cdot e(X / \langle \tau \rangle) + e(X / S\_3), \quad c = e(X / \langle \tau \rangle) - e(X / S\_3) .
$$
However, note that $S\_3$ has more subgroups than representations, so there is a non-trivial relation between the $E$-polynomials that always holds:
$$
e(X) - 2 \cdot e(X / \langle \tau \rangle) - e(X / \langle \rho \rangle) + 2 \cdot e(X / S\_3) = 0
$$
Now, a more refined invariant than the $E$-polynomial, is the class of a variety in the Grothendieck of varieties $\textrm{K}(\textbf{Var}\_k)$, that is, there is a ring morphism $\textrm{K}(\textbf{Var}\_k) \to \mathbb{Z}[u, v]$ sending $[X]$ to $e(X)$.
My question is: does the analogue relation hold in $\textrm{K}(\textbf{Var}\_k)$ as well? That is,
$$
[X] - 2 \cdot [X / \langle \tau \rangle] - [X / \langle \rho \rangle] + 2 \cdot [X / S\_3] = 0
$$
and similar relations for other finite groups $G$. Or, does anyone know a counterexample where this relation fails?
| https://mathoverflow.net/users/484150 | Representation-induced relations in the Grothendieck of varieties | No.
Let $X$ be a curve of genus $6g+1$ for some $g>0$ with a free action of $S\_3$.
None of $X, X/\langle \tau \rangle$, $X/ \langle \rho \rangle$, and $X/ S\_3$ will be stably birational to each other, since they have genus, respectively, $6g+1$, $3g+1$, $2g+1$, $g+1$ by Riemann-Hurwitz and thus all have different genuses.
By Larsen-Lunts, it follows that there is no $\mathbb Z$-linear relation between $X, X/\langle \tau \rangle$, $X/ \langle \rho \rangle$, and $X/ S\_3$.
So, in particular, the relation fails in this case.
| 5 | https://mathoverflow.net/users/18060 | 424700 | 172,497 |
https://mathoverflow.net/questions/424672 | 3 | I'm trying to understand Berman's classic paper on the subject ("Local Nondeterminism and Local Times of Gaussian Processes"). In order to define local nondeterminism, he considers the ratio
$$
\frac{Var(X\_{t\_m}-X\_{t\_{m-1}}|X\_{t\_{m-1}}, X\_{t\_{m-2}}, \ldots X\_{t\_1})}{Var(X\_{t\_m}-X\_{t\_{m-1}})}.
$$
Then he says that this is always between 0 and 1. Apparently this is an almost sure statement, because the top is a r.v., however I think I heard somewhere that with a Gaussian process (which $X$ is here) a conditional variance is actually a deterministic function. Am I right about this? I can't find a reference for it, though. I think that the conditional variance is not bounded a.s. by the variance in general, so it must be something to do with being Gaussian. Can someone help with this, please?
Thanks,
Greg
| https://mathoverflow.net/users/148035 | Local nondeterminism | If $X, Y\_1,\ldots Y\_m$ are jointly Gaussian variables over a probability space $\Omega$, think of them as vectors in the Hilbert space $L^2(\Omega)$. WLOG all these variables have mean zero, otherwise subtract the means without affecting variances. We can decompose $$X=Y+Z$$
where $Y\!\!\in $span$(Y\_1,\ldots Y\_m)$ and $Z$ is orthogonal to this span, so
$${\rm Var}(X)={\rm Var}(Y)+{\rm Var}(Z)\,.$$
Since uncorrelated Gaussian variables are independent,
$${\rm Var} (X\, |\, Y\_1,\ldots Y\_m)={\rm Var}\, (Y+Z\, |\, Y\_1,\ldots Y\_m) ={\rm Var} ( Z ) \,.$$
| 4 | https://mathoverflow.net/users/7691 | 424706 | 172,498 |
https://mathoverflow.net/questions/424715 | 3 | Let $F$ be a free group and $w\in F$ a cyclically reduced word. Let $v$ be a non-trivial proper subword of $w$. Is it true that $v\notin \langle w^F\rangle$?
| https://mathoverflow.net/users/10482 | Is is true that a proper subword cannot lie in the normal closure of a word? | This is true. See Theorem 2 of [ON RELATORS AND DIAGRAMS FOR GROUPS WITH ONE DEFINING RELATION
BY
C. M. WEINBAUM](https://projecteuclid.org/journalArticle/Download?urlId=10.1215%2Fijm%2F1256052287&referringURL=https%3A%2F%2Fwww.google.com%2F&isResultClick=False).
| 5 | https://mathoverflow.net/users/15934 | 424716 | 172,502 |
https://mathoverflow.net/questions/424722 | 2 | Throughout, we denote by $\mu$ the Lebesgue measure on $[0, 1]$.
Let $f \in L^1([0, 1])$ be a nowhere zero function, that is, the set of all $x \in [0, 1]$ such that $f(x) = 0$ is empty.
Suppose there exists some $\varepsilon$ with $0 < \varepsilon < 1$ such that for every open subset $E$ of $[0, 1]$ with Lebesgue measure $\varepsilon$, we have
$$\int\_E f \, d\mu > 0.$$
**Question:** Is it true that $\mu(\{f > 0\}) > 1 - \varepsilon$?
| https://mathoverflow.net/users/173490 | Is this function positive on a set of large measure? | I'm guessing your question means what it says literally, that $\int\_U f>0$ for every open set $U$ of measure *exactly* $\epsilon$. Is this right? (The answer even with this restriction is still positive).
Let $N=\{x\colon f(x)<0\}$ and suppose for a contradiction that $\mu(N)\ge\epsilon$. Let $N\_1$ be a subset of $N$ of measure exactly $\epsilon$ and
let $\delta<\frac 12\int\_{N\_1}|f|\,d\mu$. Let $0<\eta<\epsilon$ be such that $\int\_A |f|<\delta$ whenever $\mu(A)\le\eta$. Now let $N\_2$ be a measurable subset of $N\_1$ of measure $\epsilon-\eta$. By regularity of Lebesgue measure, there exists an open set $U$ containing $N\_2$ of measure at most $\epsilon$. By the choice of $\eta$, we see $\int\_{U}f<-\delta$.
Now let $s$ be chosen so that $\mu([0,s)\setminus U)=\epsilon-\mu(U)$ (so that in particular, $\mu([0,s)\setminus U)<\eta$. Now we see $\mu([0,s)\cup U)=\epsilon$ and $\int\_{[0,s)\cup U}f<0$, giving the required contradiction.
| 3 | https://mathoverflow.net/users/11054 | 424724 | 172,503 |
https://mathoverflow.net/questions/424721 | 0 | Suppose $W$ is a proper von Neumann Algebra contained in $B(H)$ and the identity in $W$ is the identity mapping of $H$ (namely, $W$ does not have non-trivial null space).
1. Given a self-adjoint $T\in W$, does there exists a projection $P$ that is not in $W$ such that one of $TP, PT, PTP$ is not in $W$? (we definitely do not want all of them to be zero) Can we find a sufficient condition for this to happen?
2. Conversely, given a self-adjoint $T$ that is not in $W$, does there exists a projection $Q$ that is in $W$ such that one of $TQ, QT, QTQ$ is in $W$? (again we do not want all of them to be zero) Can we find a sufficient condition for this to happen?
In general, if $T$ is seld-adjoint, then there exists a projection in $W^\*(T)$ (the von Neumann Algebra generated by $T$) such that the range of that projection is the closure of the range of $T$ and that projection corresponds to the characteristic function of $\sigma(T)$. So far, this is the only thing I know about this question. Besides, given $T$ self-adjoint and not in $W$, can $W^\*(T)\cap W\neq\emptyset$? If we have, say $f(T)\in W^\*(T)\cap W$, can we always recover the entire $W^\*(T)$ using $f(T)$ (even when $f$ is not the characteristic function of $\sigma(T)$)?
| https://mathoverflow.net/users/151332 | "Project" an operator outside of a von Neumann Algebra into it | These aren't research-level questions, but anyway. Let $W$ be a von Neumann algebra unitally contained in $B(H)$ and let $T \in W$ be nonzero (doesn't have to be self-adjoint). Suppose for every projection $P \in B(H)$ we have $PT \in W$. Taking linear combinations and weak\* limits, we get $ST \in W$ for all $S \in B(H)$.
Since $T \neq 0$, we can find nonzero vectors $v,w \in H$ with $Tv = w$. Then for any $u \in H$, taking $S = u\otimes w$ so that $ST$ is a scalar multiple of $u \otimes v$, we get $u \otimes v \in W$. Taking adjoints, we also have $v\otimes u \in W$ for all $u \in H$. Then $(u\otimes v)(v\otimes u') = u\otimes u' \in W$ for all $u,u' \in H$ and this implies that $W = B(H)$. Conclusion: if $W \neq B(H)$ then there is some projection $P \in B(H)$ with $PT \not\in W$.
The second question trivially fails with $W = \mathbb{C}\cdot I$, as long as ${\rm dim}(H) > 1$.
| 1 | https://mathoverflow.net/users/23141 | 424728 | 172,504 |
https://mathoverflow.net/questions/424734 | 2 | Let $ \Omega = \mathbb{T}^d (1 \leq d \leq 3)$ be the $d$ dimensional torus and $ u \in H^2(\Omega) $ be a complex valued function. For some $ 0 < \alpha < 1 $, let $ g(u) = |u|^\alpha u $.
**My question is**: do we have some estimates for $ g(u) $ which can ensure that it belongs to some fractional Sobolev space $ H^s(\Omega) $ with $ 1<s<2 $, possibly depending on $\alpha$?
This question occurs in the analysis of nonlinear Schrödinger equation, where we have to handle some terms like
$$ \| (e^{i \tau \Delta} - I) g(u) \|\_{L^2} \leq C \tau^{\gamma} \| g(u) \|\_{H^{2\gamma}} ,\quad 0 < \gamma \leq 1. $$
Of course, we can differentiate $ g(u) $ once, and get
$$ \nabla g(u) = |u|^\alpha \nabla u + |u|^{2\alpha-2}u^2 \overline{\nabla u}. $$
Then it seems not clear how to obtain the fractional estimates of these terms.
It may be related that when $ d=1 $, the power function $ |x|^\alpha (0 < \alpha < 1) $ is in $ H^s $ for any $ s<\alpha+1/2 $. Are there similar results for general $ |u|^\alpha $ with $ u $ sufficiently smooth?
**Update**:
To be precise, the question is about how does the non-smoothness of the power nonlinearity, i.e. $|u|^\alpha (0<\alpha<1)$, affect the regularity of $g(u)$ provided that $ u $ is sufficiently smooth. I wonder what's the best one could expect for $ g(u) \in H^s $ with $ 1 <s<2 $ since $ g(u) $ is at least in $ H^1 $ by the Sobolev embedding of $ H^2 \hookrightarrow L^\infty $ when $ 1 \leq d \leq 3 $.
As one can see, for $0<\alpha<1$, if we differentiate $g(u)$ twice, there will be some term with negative power and, in general, $ g(u) $ is in $ H^1 $ but not in $ H^2 $ even though $ u \in H^2 $. While in the smooth case, saying the cubic case ($\alpha = 2$), one has
$$ \| |u|^2 u \|\_{H^s} \leq C \| u \|\_{H^s}^3, \quad s > \frac{d}{2}, $$
which implies that we will not lose any regularity of $g(u)$ for sufficiently smooth $ u $ if the nonlinearity is smooth.
**Some attempts in 1D:**
Use the follwoing definition of the fractional Sobolev space $H^s$: $ u \in H^s $ for some $ 0 < s < 1 $ if and only if
$$ u \in L^2(\Omega) \text{ and } \int\_\Omega \int\_\Omega \frac{|u(x) - u(y)|^2}{|x-y|^{2s+d}} dx dy < \infty. $$
By the inequality
$$ |x^{\alpha} - y^{\alpha}| \leq |x - y|^\alpha, \quad x, y \in \mathbb{R}, \quad 0 < \alpha < 1, $$
and the Sobolev embedding into Holder space in 1D, one has
$$ ||u(x)|^\alpha - |u(y)|^\alpha| \leq |u(x) - u(y)|^\alpha \leq C | x - y|^\alpha. $$
Then by the definition above, one has $ |u|^\alpha \in H^s $ for any $ 0\leq s<\alpha $. Then $ \nabla g(u) \in H^s $ since $ \nabla u \in H^1 $ and $|u|^\alpha \in H^s $ and thus $g(u) \in H^{1+s}$ for any $ 0 \leq s<\alpha $.
| https://mathoverflow.net/users/484187 | $H^s$ norm of non-integer power of functions | In Christ-Weinstein, JFA 100 (1991) 87-109 you can find the fractional chain rule (Proposition 3.1)
$$
\|F(u)\|\_{\dot H^s\_r}\le C
\|F'(u)\|\_{L^p}\|u\|\_{\dot H^s\_q}
$$
where $s\in(0,1)$, $p,q,r\in(1,\infty)$, $1/r=1/p+1/q$, $F\in C^1$, and $\dot H^s\_r$ is the homogeneous Sobolev space with norm $\||D|^su\|\_{L^r}$. You can use this inequality to estimate the first derivative of $g(u)$.
| 3 | https://mathoverflow.net/users/7294 | 424739 | 172,510 |
https://mathoverflow.net/questions/424747 | 1 |
>
> Is the topological group $(\mathbf{Q}\_p/\mathbf{Z}\_p)^{\oplus k}$, $k\ge 1$, a $\sigma$-compact topological group when endowed with its natural $p$-adic topology?
>
>
>
More generally, I'm looking for a criterion for locally compact Hausdorff topological groups to not contain a nested exhaustive sequence of compact subgroups (i.e. locally compact Hausdorff topological groups $G$ with a countable family of nested compact subsets $K\_n$ such that $\bigcup\_{n\ge 0}K\_n = G$).
**Example** Real and complex Lie groups are $\sigma$-compact exactly when they are compact.
**Example** The same group $(\mathbf{Q}\_p/\mathbf{Z}\_p)^{\oplus k}$, endowed with the discrete topology. It is uncountable, so it's not $\sigma$-compact when endowed with the discrete topology, if I'm not missing anything.
| https://mathoverflow.net/users/nan | $\sigma$-compactness of some locally compact Hausdorff topological groups | I agree with @MatthewDaws ... for example the real Lie group $(\mathbb R,+)$ with the usual topology is locally compact and sigma-compact but not compact.
Take any locally compact group $G$. There is a neighborhood $V$ of $e$ with compact closure. The subgroup $H$ generated by $V$ is an open, locally compact, sigma-compact subgroup of $G$. The coset space $G/H$ is discrete.
---
The example. The group $\mathbb Q\_p$ of $p$-adic numbers is locally compact. The subgroup $\mathbb Z\_p$ of $p$-adic integers is an open, compact, subgroup. The quotient $\mathbb Q\_p/\mathbb Z\_p$ is countable and discrete.
Indeed,
$$
\mathbb Q\_p = \bigcup\_{n=0}^\infty p^n \mathbb Z\_p .
$$
For a fixed $k \in \mathbb N$, also $(\mathbf{Q}\_p/\mathbf{Z}\_p)^{\oplus k}$ is countable and discrete.
| 1 | https://mathoverflow.net/users/454 | 424752 | 172,513 |
https://mathoverflow.net/questions/424746 | 2 | Let $A$ be a linear map over the finite-field vector space $(\mathbb F\_2)^n$, i.e., an $\mathbb F\_2$-valued $n\times n$ matrix, not necessarily symmetric. I'm interested in the sum
$$Z(A) = \sum\_{X\in \mathbb F\_2^n} (-1)^{X^T A X}\;,$$
where
$$x\rightarrow (-1)^x$$
should be thought of as a function from $\mathbb F\_2$ to $\mathbb Z$ (or $\mathbb R$ or $\mathbb C$).
Is there a way to efficiently compute $Z(A)$ for large matrices $A$? Can one say anything interesting about for which $A$ we have $Z(A)=0$? For example, one can easily see that $Z(A)=0$ if $A=1\oplus B$ where $1$ is a $1\times 1$ matrix.
The motivation behind this question comes from physics. $Z(A)$ is the partition function of a discrete path integral, $X$ is are the different configurations of degrees of freedom which are summed over, and $X^T A X$ is a quadratic action.
Reposting [this mathematics stackexchange question](https://math.stackexchange.com/q/4454008/487705) here since I didn't get any answers there.
| https://mathoverflow.net/users/115363 | Sum over exponentiated bilinear form in finite-field vector space | This is a multidimensional Gauss sum, and can be handled by the same methods used to handle Gauss sums.
$Z(A)=0$ if and only if $X^T A X$ is nonzero for some $X \in \ker (A + A^T)$, and, if $Z(A) \neq 0$, then $$Z(A) = \pm 2^{ \frac{n + \dim ( ker (A + A^T))}{2}}$$ which implies your divisibility claim.
To prove this, just note that
$$Z(A)^2 = \sum\_{X\_1, X\_2 \in \mathbb F\_2^n} (-1)^{X\_1^T A X\_1 + X\_2^T A X\_2} = \sum\_{X,Y\in \mathbb F\_2^n} (-1)^{ (X+Y)^T A (X+Y) + X^T A X } $$
and the exponent satisfies
$$(X+Y)^T A (X+Y) + X^T A X = X^T A X + X^T A Y + Y^T A X + Y^T A Y + X^T A X$$ $$ = X^T A Y + Y^T A X + Y^T A Y = X^T A Y + X^T A^T Y + Y^T A^T Y $$ so
$$Z(A)^2 = \sum\_{Y \in \mathbb F\_2^n} (-1)^{Y^T A Y} \sum\_{X\in \mathbb F\_2^n} (-1)^{ X^T (A + A^T ) Y } $$
Now the inner sum $\sum\_{X\in \mathbb F\_2^n} (-1)^{ X^T (A + A^T ) Y } $ vanishes unless $(A+A^T) Y =0$, i.e. $Y \in \ker (A+A^T)$, and equals $2^n$ in that case.
Restricted to $\ker (A +A^T)$, $Y^T A Y$ is actually a linear form, so the outer sum vanishes unless it is identically zero on $\ker (A + A^T)$ and is $2^{ \dim (A+A^T)}$ otherwise, giving
$$Z(A)^2 =2^{ n + \dim (\ker(A+A^T))}$$
unless $Y^T A Y$ is nonzero for some $Y \in \ker (A +A^T)$ and $$Z(A)^2= 0$$ if $Y^T A Y$ is nonzero for some $Y \in \ker (A +A^T)$, and thus the claim above.
| 5 | https://mathoverflow.net/users/18060 | 424769 | 172,518 |
https://mathoverflow.net/questions/424774 | 4 | I recall that there was a theorem mimicking the change of variables' integral formula. Surprisingly, I can't find it on the Fosco Loregian book. The change of variables formula states that, if $f: E \to B$ is a measurable mapping between measurable spaces and $\mu$ is a measure on $E$, then for all $g: B \to \mathbb{R}$
$$ \int\_E f(g(x)) d \mu(x) = \int\_B g(x) d f\_\* \mu(x) $$
Analogously, I want to understand how to reduce the coend over a big category if the bifunctor depends on a smaller category.
In other words, let $p: E \to B$ be a functor of categories and a bifunctor $\alpha: B \times B^{op} \to D$. Define $p^\*F: E\times E^{op} \to D$ as the composition $F(p \times p^{op})$. In case $p$ has set-fibers and is a surjection, I think one should have
$$ \int^{y \in E} (p^\*F) (y,y) \simeq \int^{x \in B} E\_{x} \cdot F(x,x )$$
where $\cdot$ is the copower and $E\_x$ is the fiber of $E \to B$ in $x$.
>
> Is there a general formula for $$ \int^{y \in E} (p^\*F) (y,y) $$ and what is a reference to the proof?
>
>
>
| https://mathoverflow.net/users/140013 | Change of coordinates for coends | $\require{AMScd}$First of all, I suspect that "$p$ has set fibers and it is surjective" means that it is a discrete fibration. In that case, $p$ corresponds to a functor $G\_p : C \to Set$ (covariant if $p$ is an opfibration, contravariant if it's a fibration) defined precisely as $Gx = E\_x$. So, a better way to rephrase your question would be the following
>
> **Conjecture**. Let $F : C^{op}\times C \to D$ a functor with codomain a cocomplete category; let $P : C^{op}\to Set$ be a presheaf. Then
> $$\int^X PX^-\cdot F(X^-,X^+)\cong \int^{W\in Elts(P)} F(\Sigma\_PW,\Sigma\_PW) \tag{$\heartsuit$}$$ where
>
>
> * the "sign rule" is a convention on dummy variables that says that all $X^-$ vary simultaneouly, and contravariantly, and all $X^+$ vary
> simultaneously, and covariantly;
> * $Elts(P)$ denotes the category of elements of $P$ and $\Sigma\_P$ is the associated forgetful functor, which is a discrete fibration.
>
>
> Without loss of generality, all discrete fibrations are of this form.
>
>
> Dually, if $Q : C\to Set$ is a copresheaf, $$\int^X QX^+\cdot
> F(X^-,X^+)\cong \int^{W\in Elts(P)} F(\Sigma\_PW,\Sigma\_PW) $$
>
>
>
But coends have nothing special against ends, and by properly dualising the statement, we get that for a presheaf $P$ and copresheaf $Q$,
$$\begin{gather\*}
\int\_X PX^+\pitchfork F(X^-,X^+)\cong \int\_{W\in Elts(P)} F(\Sigma\_PW,\Sigma\_PW) \\ \int\_X QX^-\pitchfork F(X^-,X^+)\cong \int\_{W\in Elts(P)} F(\Sigma\_PW,\Sigma\_PW)
\end{gather\*}$$ (or probably the opposite of $Elts(P)$?)
This is meant to address the problem that the integral $\int^X PX\cdot F(X,X)$ depends on three variables, and what *exactly* is the coend you're performing?
The precise sense of the conjecture is that for every fibration $p : E\to C$ there is a coequalizer diagram
$$
\sum\_{x\to y} E\_y \cdot F(y,x) \rightrightarrows \sum\_{x\in X} E\_x\cdot F(x,x) \to \textstyle\int p^\*F
$$
or more precisely that $\int p^\*F$ is the coend of the functor
$$\begin{CD}
C^{op}\times C \\
@V\Delta\times C VV\\
C^{op}\times C^{op} \times C\\
@VG\_p\times C^{op}\times C VV\\
Set \times C^{op} \times C\\
@VSet\times FVV\\
Set\times D\\
@V\\_\cdot\\_VV\\
D
\end{CD}$$
In "stupid" examples, this works:
* If $P$ is constant, say $PX\equiv A$ for a set $A$, then $$\int^X PX\cdot F(X,X) \cong A\cdot \int^X F(X,X)$$ which, since $Tw(C\times A)=Tw(C)\times A$ when $A$ is discrete, seems to be precisely the RHS of $(\heartsuit)$.
* If $P$ is representable and $F$ is mute, it reduces to a "Yoneda-like" argument and actually this is enough to say that the conjecture is true for *any* $P$ and $F$ mute, because $\\_\cdot\\_$ is separately cocontinuous in both variables; alternatively, one can argue that the conjecture reduces to the fact that the colimit of $F$ weighted by $P$ is the coend $\int^X PX\cdot FX$. Incidentally, this suggests that if you want to state a similar conjecture in an enriched setting, it will only work for the bases of enrichment that have a meaningful Grothendieck construction, e.g. for the class of categories of [this paper](https://arxiv.org/abs/1804.03829).
**Remark**. By the abovementioned cocontinuity of $\\_\cdot\\_$ it might be sufficient to prove $(\heartsuit)$ in the case where $P$ is representable. But one must then find a way to circumvent the following problem: let's compute with the LHS of $(\heartsuit)$, and get
$$\begin{align\*}
\int^X PX\cdot F(X,X) &= \int^X \big(\text{colim}\_\alpha C(X,A\_\alpha)\big)\cdot F(X,X) \\
&= \text{colim}\_\alpha\int^X C(X,A\_\alpha) \cdot F(X,X) \\
&=\text{colim}\_\alpha\int^{(X,u) \in C/A\_\alpha} F(\Sigma (X,u),\Sigma (X,u))
\end{align\*}$$ from which it is "evident" (?!) that the sense in which this is just $\int^{W \in Elts(P)} F(\Sigma W, \Sigma W)$ is the same sense in which $Elts(P)\cong \int^X C/X\times PX$. Yet, I don't see a formal manipulation ensuring that this is true, and I find a direct, hands-on proof pretty daunting.
Probably it's a good point to stop and find counterexamples, so that at least we have a clear idea of when the conjecture is false. We're in the same ballpark of [this question](https://mathoverflow.net/questions/356473/category-in-a-coend-depends-on-the-integration-variable-of-an-outer-coend) I asked years ago, so I suspect that a similar condition as the one stated in the comments, to "compute colimits fiberwise" is needed.
A different path that one might take is more belligerent: find a cowedge $F(\Sigma\_P(X,a), \Sigma\_P(X,a)) \to \int^X PX\cdot F(X,X)$ and prove that it is initial. But I don't trust the back-of-the-envelope computation that seems to say $(\heartsuit)$ is true in general.
---
PS: join the [ItaCa](https://progetto-itaca.github.io) gang!
| 3 | https://mathoverflow.net/users/7952 | 424781 | 172,519 |
https://mathoverflow.net/questions/424775 | 1 | Suppose $p(x\_1,...,x\_n)$ is a polynomial in $n$ real variables whose Hessian is not identically zero. Can we say that there is a set $Z \subset {\mathbb R}^n$ of measure zero and a constant $M$ such that for all real numbers $c\_1,...,c\_n$ the cardinality of the set of $(x\_1,...,x\_n) \in {\mathbb R}^n - Z$ for which ${\partial p \over \partial x\_i}(x\_1,...,x\_n) = c\_i$ for each $i$ is at most $M$? It's possible I might be able to exclude some $(c\_1,...,c\_n)$ but I'd have be careful what I was excluding.
I've looked around online at resources on systems of polynomial equations and nothing really applied. Also, the above might generalize to systems $p\_i(x\_1,...,x\_n) = c\_i$ for $(p\_1,...,p\_n)$ whose Jacobian is not identically zero, but I only need it in the above situation.
Basically, if this follows from something well-known, or someone has pointers on where to look I'd be appreciative.
| https://mathoverflow.net/users/149955 | Is there a uniform bound on the number of solutions to ${\partial p \over \partial x_i} (x_1,...,x_n) = c_i$ outside a set of measure zero? | This follows from basic facts about polynomials, to your generalization of a polynomial map $F: \mathbb R^n \to \mathbb R^n$. Since the Jacobian determinant of $F$ does not vanish identically, its zero set $Z$ (the critical points of $F$) has measure 0, since $F$ is a polynomial. Away from critical values, the inverse image is a dimension 0 manifold, i.e. a finite set. Finally, the size of this finite set is bounded by the product of the degrees of the $F\_i$'s by Bezout, since it is the intersection of the $F\_i^{-1}(c\_i)$, each of which is a hypersurface of degree $\deg F\_i$
| 3 | https://mathoverflow.net/users/5279 | 424782 | 172,520 |
https://mathoverflow.net/questions/424445 | 0 | **Inequality 1**
\begin{align}
\mathbb{P}\left(\frac{1}{n} \sum\_{i=1}^{n}\left(f\left(x\_{i}\right)-\mathbb{E}[f]\right) z\_{i} \geq \frac{\epsilon}{8}\right) \leq 2 \exp \left(-\frac{\epsilon^{2} n d}{9^{4} c L^{2}}\right)
\end{align}
Since we assumed that the range of the functions is in $[-1,1]$ we have $\mathbb{E}[f] \in[-1,1]$ and hence:
**Inequality 2**
$$
\mathbb{P}\left(\exists f \in \mathcal{F}: \frac{1}{n} \sum\_{i=1}^{n} \mathbb{E}[f] z\_{i} \geq \frac{\epsilon}{8}\right) \leq \mathbb{P}\left(\left|\frac{1}{n} \sum\_{i=1}^{n} z\_{i}\right| \geq \frac{\epsilon}{8}\right)
$$
**By Hoeffding's inequality, the above quantity is smaller than $2 \exp \left(-n \epsilon^{2} / 8^{3}\right)$ (recall that $\left.\left|z\_{i}\right| \leq 2\right)$.**
Thus we obtain with an union bound:
$$
\begin{aligned}
\mathbb{P}\left(\exists f \in \mathcal{F}: \frac{1}{n} \sum\_{i=1}^{n} f\left(x\_{i}\right) z\_{i} \geq \frac{\epsilon}{4}\right) & \leq|\mathcal{F}| \cdot \mathbb{P}\left(\frac{1}{n} \sum\_{i=1}^{n}\left(f\left(x\_{i}\right)-\mathbb{E}[f]\right) z\_{i} \geq \frac{\epsilon}{8}\right)+\mathbb{P}\left(\left|\frac{1}{n} \sum\_{i=1}^{n} z\_{i}\right| \geq \frac{\epsilon}{8}\right) \\
& \leq 2|\mathcal{F}| \cdot \exp \left(-\frac{\epsilon^{2} n d}{9^{4} c L^{2}}\right)+2 \exp \left(-\frac{n \epsilon^{2}}{8^{3}}\right)
\end{aligned}
$$
---
I am not getting how Union bound is getting happened using Ineq 1 and 2. Can anyone help me with that how they able to reach last inequality?
| https://mathoverflow.net/users/nan | Union Bound of two events? | By the triangle inequality, the event
$$A=\left\{\exists f \in \mathcal{F}: \frac{1}{n} \sum\_{i=1}^{n} f\left(x\_{i}\right) z\_{i} \geq \frac{\epsilon}{4}\right\}$$
is contained in the union of the two events
$$B=\left\{\exists f \in \mathcal{F}: \frac{1}{n} \sum\_{i=1}^{n}\left(f\left(x\_{i}\right)-\mathbb{E}[f]\right) z\_{i} \geq \frac{\epsilon}{8}\right\}$$
and
$$D=\left\{\exists f \in \mathcal{F}: \ \frac{1}{n} \sum\_{i=1}^{n} \mathbb{E}[f] z\_{i} \geq \frac{\epsilon}{8}\right\} \,.
$$
Moreover, $B=\bigcup\_{f \in \mathcal{F}}B\_f$ where
$$B\_f=\left\{ \frac{1}{n} \sum\_{i=1}^{n}\left(f\left(x\_{i}\right)-\mathbb{E}[f]\right) z\_{i} \geq \frac{\epsilon}{8}\right\} \,,
$$
so
$$A\subset \left(\bigcup\_{f \in \mathcal{F}} B\_f \right) \cup B \,$$
To this relation the union bound is applied, together with inequality 2 to bound $P(D)$.
| 0 | https://mathoverflow.net/users/7691 | 424786 | 172,522 |
https://mathoverflow.net/questions/424771 | 2 | Consider a random matrix $A \in \mathbb{R}^{m\times n}$ with i.i.d. entries, with mean zero and variance 1 and $m <n $. I am interested in the expectation of $$E\_{A}(\mathrm{Tr}( (A^T A + \lambda \mathrm{Id})^{-1})).$$ This question is similar to [Trace of inverse of random positive-definite matrix in high dimension?](https://mathoverflow.net/a/332889/483386), except $A$ is non-symmetric. I am unfamiliar with Marchenko-Pastur distribution, so any references would be great!
| https://mathoverflow.net/users/483386 | Trace inverse of random PSD matrix? | I think the product $A^\top A$ in the OP should read $AA^\top$, to avoid a trivial contribution from zero eigenvalues (assuming $A\in\mathbb{R}^{m\times n}$ and $m<n$).
For $m,n\gg 1$, and $m/n\equiv r\in (0,1)$ fixed, an integration over the [Marchenko–Pastur distribution](https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution) gives the answer (with $x\_\pm=(1\pm\sqrt{r})^2$)
$$\lim\_{m,n\rightarrow\infty}\mathbb{E}[m^{-1}\mathrm{Tr}\,(n^{-1}AA^\top + \lambda I)^{-1})]=\int\_{x\_-}^{x\_+} \frac{1}{x+\lambda}\frac{\sqrt{\left(x\_+-x\right) \left(x-x\_-\right)}}{2 \pi r x}\,dx$$
$$\qquad=\frac{1}{2\lambda r}\left(\sqrt{\lambda^2+2 \lambda (1+r)+(1-r)^2}-\lambda+r-1\right).$$
The rescaling of $AA^\top$ by a factor $1/n$ is needed for a $\lambda$-dependent answer in the large $n$ limit. If you do not rescale, then
$$\lim\_{m,n\rightarrow\infty}\mathbb{E}[\mathrm{Tr}\,(AA^\top + \lambda I)^{-1})]=\frac{r}{1-r}.$$
This diverges for $r=1$, in that case the trace grows as $\sqrt{n}$, see <https://mathoverflow.net/a/332889/11260>
| 1 | https://mathoverflow.net/users/11260 | 424789 | 172,524 |
https://mathoverflow.net/questions/422922 | 3 | Tyma Gaidash has recently [posted](https://math.stackexchange.com/a/4442541/2513) solutions to some quintics in terms of Inverse Beta Regularized function. He also [found](https://math.stackexchange.com/questions/46934/what-is-the-solution-of-cosx-x/4389007#4389007) the closed form for the equation $\cos x=x$ using the same Inverse Beta Regularized function.
So, the question is: can a general quintic be solved in terms of this function?
| https://mathoverflow.net/users/10059 | Can a general quintic be solved using inverse beta regularized function? | Yes, it seems, it can. Any general quintic can be reduced to the [Bring-Jerrard form](https://mathworld.wolfram.com/Bring-JerrardQuinticForm.html) $x^5+ax+b=0$.
Then Tyma Gaidash found a solution for the Bring-Jerrard form via Inverse Beta Regularized function:
$x=\frac{5b}{4a\left(\text I^{-1}\_{\frac{3125b^4}{256a^5}+1}(2,4)-1\right)}$.
In computer algebra systems and spreadsheets like Excel the function is defined only for the subscript argument in the range $[0,1]$. But there is nothing that prevents to define this function on a bigger domain either by functional equations or analytic continuation.
Tyma Gaidash gives only one solution, but other roots can be obtained by factoring and solving general quartic.
Moreover, in his recent finds he found solutions for equations of other orders, including non-integer ones:
$x^r+ax+b=0\implies x=\frac{br}{a(1-r)}\text I^{-1}\_\frac{ b^{r-1}(r-1)}{a^r\left(\frac1r-1\right)^r}(-r,2)$
and
$x^r+ax+b=0\implies x=\frac{b r}{a(1-r)\text I^{-1}\_\frac{b^{r-1}(r-1)}{a^r\left(\frac1r-1\right)^r}(r-1,2)}$
So, **yes**. A solution of general quintic and some equations of higher orders can be expressed in terms of inverse beta regularized function granted the function is defined on a greater range than its usual range of definition.
As a side note: Tyma Gaidash managed to express using inverse beta regularized function besides polynomial roots also such things as Dottie number, elliptic and trigonometric and hyperbolic functions, logarithms and exponents, Lambert W-function, inverse functions of $\text{erf}(x)$ and $\text{erfc}(x)$, logarithmic and exponential integrals $\text{Ei}(x)$ and $\text{li}(x)$.
| 2 | https://mathoverflow.net/users/10059 | 424790 | 172,525 |
https://mathoverflow.net/questions/424731 | 1 | Let $k=\mathbb F\_q\left(\left(\frac1T\right)\right)$, $\overline k$ be an algebraic closure of $k$ and $K$ be the completion of $\overline k$ for the $\frac1T$ valuation. Consider the morphism $\sigma:k\to L=K\left(\left(\frac1Z\right)\right)$ defined by $\sigma(\sum\_{n\ge m}a\_n\left(\frac1T\right)^n)=\sum\_{n\ge m}a\_n\left(\frac1Z\right)^n$. Let $\alpha\in \overline k$. Does there exist a continuous morphism $\tilde\sigma: k(\alpha)\to L$ ($L$ is endowed with the topology induced by the $\frac1Z$-valuation) whose restriction to $k$ is $\sigma$?
Thanks in advance for any answer.
| https://mathoverflow.net/users/33128 | Extension of morphisms in function fields | Not if $\alpha = \sqrt{T}$ since $\alpha^2 = T$ is sent to $Z$ which does not have a square root in $L$.
| 1 | https://mathoverflow.net/users/18060 | 424792 | 172,526 |
https://mathoverflow.net/questions/424756 | 1 | Let $X\neq \emptyset$ be a set. By $\text{Part}(X)$ we denote the set of all partitions of $X$ not containing $\emptyset$ as an element.
First, note that $\bigcup{\frak P}$ is the collection of subsets of $X$ that are a member of at least one partition in ${\frak P}$. We say that ${\frak P}\subseteq \text{Part}(X)$ is *admissible* if it is closed under [refinement](https://en.wikipedia.org/wiki/Partition_refinement) and if for all $b\in \bigcup {\frak P}$ there is $b\_0\in \bigcup{\frak P}$ such that $b \subseteq b\_0$, and $b\_0$ is maximal in $\bigcup{\frak P}$ with respect to set inclusion. Moreover, we say that ${\frak P}$ has the *optimality property* if there is $P\_0\in {\frak P}$ such that $|P\_0\setminus P| \leq |P \setminus P\_0|$ for all $P\in {\frak P}$.
Given ${\frak P}\subseteq \text{Part}(X)$ as well as a non-empty set $S\subseteq X$, we let ${\frak P}|\_S$ be ${\frak P}$ "cut down to $S$", or more formally, $${\frak P}|\_S := \{P \in \text{Part}(S): (\exists Q \in \text{Part}(X))(\forall
b\in P)(\exists b' \in Q) b = b'\cap S\}.$$
Let $X$ be non-empty and ${\frak P}\subseteq \text{Part}(X)$ be admissible. Suppose that ${\cal C}$ is a collection of subsets of $X$ such that for $C, C'\in {\cal C}$ we have $C\subseteq C'$ or $C'\subseteq C$, and also we have $\bigcup {\cal C} = X$. Suppose further that for every set $C\in {\cal C}$, the collection ${\frak P}|\_C \subseteq \text{Part}(C)$ has the optimality property.
**Question.** Does this imply that $\frak P$ itself has the optimality property?
**Note.** Without the maximality requirement for admissibility, there would be an easy negative answer for the question. Let $X = \omega$, and let ${\frak P}$ consist of the partitions of $\omega$ such that every block (= member of the partition) is finite. Let ${\cal C}$ consist of the sets of the form $n := \{0, \ldots, n-1\}$ for all positive integers $n$. Then the union of ${\cal C}$ equals $\omega$, and ${\frak P}|\_n$ trivially satisfies the optimality property, but ${\frak P}$ itself does not.
| https://mathoverflow.net/users/8628 | Optimal partitions amongst a given set of partitions | Modifying your note at the end slightly, try the following.
Let $X=\{0,1,2,\ldots\}$ and let $P=\{\{0,1\},\{2,3\},\{4,5\},\ldots\}$. Let $\mathfrak{P}$ be all partitions of $X$ which refine $P$ and moreover contain only finitely many of the blocks in $P$ (thus for instance $P$ itself does not belong). Cover $X$ by sets of the form $\{0,1,\ldots,n\}$, like in your note. Now I think $\mathfrak{P}$ is admissible but still fails the optimality property.
| 2 | https://mathoverflow.net/users/25028 | 424793 | 172,527 |
https://mathoverflow.net/questions/424776 | 2 | Let $\Omega\subset\mathbb{C}^n$ be any $C^{\infty}$ bounded domain and let $H^2(\partial\Omega)$ denotes a holomorphic Hardy space which is a $L^2(\partial\Omega)$ closure of $A^{\infty}(\Omega)(=\mathscr{O}(\Omega)\cap C^{\infty}(\overline{\Omega})).$ Because $H^2(\partial\Omega)$ is a closed subspace of $L^2(\partial\Omega)$ there exists a countable dense subset {$\phi\_j$} where $ j\in\mathbb{N}$ such that
$$
f=\sum\_{j=1}^{\infty}\langle f,\phi\_j\rangle\phi\_j\qquad
\forall f\in H^2(\partial\Omega)
$$
**My Question**: can we choose $ \forall j\in\mathbb{N},\;\; \phi\_j\in A^{\infty}(\Omega)?$
| https://mathoverflow.net/users/482747 | Regarding basis of holomorphic Hardy space | One should be able to find a countable linearly independent subset of $A^\infty(\Omega)$ whose linear span is dense in $H^2(\partial\Omega)$ (it would suffice to show that the orthogonal complement of this subset inside $H^2$ is zero).
Once you have such a subset, order it in any ways as a sequence and then apply the Gram--Schmidt orthogonalization process.
---
Some more details on the second step. Suppose that we have a sequence $(f\_n)\_{n=1}^\infty$ in an inner product space, which is linearly independent. Equivalently, if we define $V\_n$ to be the linear span of the set $\{f\_1,\dots, f\_n\}$, then we require that $0 \subset V\_1 \subset V\_2 \subset \dots$ where each inclusion is strict.
The Gram-Schmidt orthogonalization process takes this sequence and produces an orthonormal sequence $(u\_n)\_{n=1}^\infty$, with the property that the span of $\{u\_1,\dots, u\_n\}$ is equal to $V\_n$ for each $n\geq 1$. In particular, if $\bigcup\_{n=1}^\infty V\_n$ is dense in the original inner product space, then we have produced an orthonormal basis.
In particular, if we start in the inner product space that is $A^\infty(\Omega)$ with the $L^2$-inner product, then we automatically get an orthonormal basis for $H^2(\partial \Omega)$ that is contained in $A^\infty(\Omega)$.
| 1 | https://mathoverflow.net/users/763 | 424800 | 172,528 |
https://mathoverflow.net/questions/424744 | 7 | Work in a theory with (deep breath) a countable number of primitives denoted with capital letters from the end of the alphabet with numerical subscripts $\{X\_n,Y\_n,Z\_n,\dots\}\_{n<\omega}$ indicating which kind of primitive they are together with primitive relations $\{\in\_n\}\_{n<\omega}$ with numerical subscripts indicating which kind of primitive they are a relation on, such that each $n$-primitive is also an $n+1$-primitive for all $n<\omega$ and such that $\in\_n$ is $\in\_{n+1}$ restricted to $n$-primitives for all $n<\omega$, and such that each $\in\_n$ is *left closed* so primitives to the left of $\in\_n$ are at most $n$-primitives. Call $0$-primitives *sets*, and suppress subscripts when they are obvious from context.
For axioms assume $Z$ plus foundation (denoted $Z^+$) for each level of primitives, so we can pair/union/powerset/separate/etc. at each primitive stage. Note that we can pair etc. $n$- and $m$-primitives for $n<m$ since all $n$-primitives are also $m$-primitives. For all $n<\omega$, say that a predicate $\phi$ in the language of this theory is *safe above $n$* iff no $m$-primitives for $m>n$ occur in $\phi$, and let $\Phi\_n$ denote the set of all predicates safe above $n$. Add
>
> **Class Building Axioms**. For all $n$, if $\phi$ is a predicate in the language of this theory safe above $n$, then there exists an $n+1$-primitive $X\_{n+1}$ whose members are precisely the $n$-primitives $Y\_n$ sayisfying $\phi$. $$\forall\phi\in\Phi\_n\exists X\_{n+1}\big(Y\_n\in X\_{n+1}\iff\phi(Y\_n)\big).$$ We denote the $n+1$-primitive guaranteed by this axiom together with a predicate $\phi$ safe above $n$ by $$\{Y\_n:\phi(Y\_n)\}\_{n+1}.$$
>
>
>
Define functions as subsets of arbitrary Cartesian products as usual. For all $n<\omega$, define the *$n+1$-primitive of all $n$-primitives* by $$\widehat{V\_{n+1}}=\{X\_n:X\_n=X\_n\}\_{n+1}.$$
>
> **Replacement**. If $F$ is a function and $dmnF$ is an $n$-primitive and $F(X)$ is an $n$-primitive for all $X\in dmnF$, then $rngF$ is an $n$-primitive. $$\forall n\forall F(F\ \text{is a function}\wedge dmnF\in \widehat{V\_{n+1}}\wedge\forall X\in dmnF(F(X)\in\widehat{V\_{n+1}})\implies rngF\in\widehat{V\_{n+1}}).$$
>
>
>
This gives us replacement at each stage, since (for example) we can form the $1$-primitive $$X=\{\mathcal{P}^n(\emptyset):n\in\omega\}$$ by class building and form the surjective function $$\langle\mathcal{P}^n(\emptyset):n<\omega\rangle\subset\omega\times X$$ to prove that $X$ is in fact a set by replacement, then take $\mathcal{P}^\omega(\emptyset)=\bigcup X$ and proceed to define $\mathcal{P}^{\omega+\omega}(\emptyset)$ and higher stages of the cumulative hierarchy as usual.
>
> **Question**. What is the consistency strength of this theory?
>
>
>
It's bounded above trivially by $ZFC$ plus the existence of a countable number of inaccessible cardinals $\{\kappa\_i\}\_{i<\omega}$ with $\kappa\_i<\kappa\_{i+1}$ for all $i<\omega$, where we take $\widehat{V\_{n+1}}=V\_{\kappa\_n}$ for all $n<\omega$ together with usual membership. Is this also a lower bound?
---
**Edit**: Replacement as originally phrased made the theory inconsistent, since for all $n$ we have that $\{(0,\widehat{V\_n})\}$ is trivially a function whose domain is a set, so $\{\widehat{V\_n}\}$ would be a set for all $n$ and this together with unions yields obvious inconsistencies. I believe the fix proposed above avoids this issue and still allows for use of the axiom in all desired situations, since it correctly describes the kind of replacement we have in the inaccessible cardinal situation.
| https://mathoverflow.net/users/92164 | Consistency strength of an attempt at higher order set theory | I believe $ZFC$ plus the existence of a countable collection of strictly inaccessible cardinals is also a lower bound on the consistency strength of this theory, and am posting this as a CW answer to close the question.
For a proof sketch, define ordinals to be hereditarily membership transitive primitives, define rank $\rho$ as usual for all primitives, and for each $n<\omega$ define $$Ord\_n=\{X:X\ \text{is an ordinal}\}\_{n+1}.$$ In particular, each $Ord\_n$ is a proper $n+1$ primitive (it is not also an $n$-primitive) since it is also an ordinal and would thusly be a member of itself by definition if it were an $n$-primitive, contradicting foundation. Further, defining stages of the cumulative hierarchy $V\_\alpha$ for ordinals $\alpha$ using rank in the usual way we have that $$V\_{Ord\_n+1}\models MK$$ for all $n<\omega$. To see this, observe that for all $n<\omega$ we have that $$V\_{Ord\_n}=\widehat{V\_{n+1}}\models ZFC$$ and if we have two $n+1$-primitives $X,Y\in V\_{Ord\_n+1}$ (viewed as 'classes over $V\_{Ord\_n}$') with $X\in Y$ then $$\rho(X)<\rho(Y)\leq Ord\_n\implies X\in V\_{Ord\_n}=\widehat{V\_{n+1}},$$ so a 'class $X$ over $V\_{Ord\_n}$' becomes a 'set in $V\_{Ord\_n}$' as soon as it is a member of another 'class over $V\_{Ord\_n}$'. Using the [characterization](https://neugierde.github.io/cantors-attic/Inaccessible) of strictly inaccessible cardinals as those ordinals $\alpha$ such that $V\_{\alpha+1}\models MK$, we see that $Ord\_n$ is strictly inaccessible for all $n<\omega$.
| 1 | https://mathoverflow.net/users/92164 | 424802 | 172,529 |
https://mathoverflow.net/questions/424783 | 3 | Let $k$ be a field, $A$ a $k$-algebra and $X$ a finite dimensional indecomposable $A$-module. Then $\text{End}\_A (X)$ is a local ring. Let $m$ be its maximal ideal. Can we say anything about the $k$-dimension of $\text{End}\_A (X)/m$? It always equals $1$ if $k$ is algebraically closed. Can we bound it if $k$ is not algebraically closed?
| https://mathoverflow.net/users/145920 | Dimension of division rings coming from indecomposable modules | Even if $A$ is a finite dimensional $k$-algebra, there may be no bound on the dimension of $\text{End}\_A(X)/m$.
Let $Q$ be the Kronecker quiver (i.e., the quiver with two vertices and two arrows from the first vertex to the second), and let $A=kQ$ be its path algebra over $k$.
Suppose $k(\alpha)$ is a simple algebraic field extension of $k$. Let $X$ be the representation of $Q$ over $k$ with $k(\alpha)$ at both vertices, the first arrow acting as the identity, and the second arrow acting as multiplication by $\alpha$. Then $\text{End}\_A(X)\cong k(\alpha)$.
So as long as $k$ has simple algebraic extensions of unbounded degree, there is no bound on $\dim\_k\left(\text{End}\_A(X)/m\right)$ for finite dimensional indecomposable $A$-modules $X$.
| 7 | https://mathoverflow.net/users/22989 | 424815 | 172,530 |
https://mathoverflow.net/questions/424799 | 2 | Let $x\_1,...,x\_N$ be $N$ mutually distinct real numbers.
I wonder: How does the expression
$$ f\_x(N)=\sup\_{\lambda \in \mathbb R}\min\_{i \neq j} \min\_{n \in \mathbb Z}\left\vert \lambda (x\_i-x\_j)-n\right\rvert$$
depend on $N$? I acknowledge the answer might depend on $x$, but I am looking for a lower bound on $\inf\_{x; all x\_i distinct}f\_x(N)$
Let's say we have two numbers, then we can choose $\lambda (x\_1-x\_2)=1/2.$ Hence, the optimal value in that case is $1/2.$
| https://mathoverflow.net/users/457901 | Maximal distance of set to integers | If I understood correctly, your question is a form of the [Lonely runner conjecture](https://en.wikipedia.org/wiki/Lonely_runner_conjecture). You can see several results and a summary on that Wikipedia link. An important special case is when the difference of the $x\_i$'s are irrational, in which case you can spread them so that you can get arbitrarily close to $1/N$.
| 8 | https://mathoverflow.net/users/955 | 424816 | 172,531 |
https://mathoverflow.net/questions/424704 | 3 | I am looking for a reference. I hope that what follows is in some textbook.
Let $q$ be an odd prime power and let $\ell$ be a positive integer. Now, let $\mathfrak{q}:\mathbb{F}\_{q^\ell}^2\to\mathbb{F}\_{q^\ell}$ be a non-degenerate quadratic form. If we compose $\mathfrak{q}$ with the trace mapping $\mathrm{Tr}:\mathbb{F}\_{q^\ell}\to\mathbb{F}\_q$, we obtain a non-degenerate quadratic form $Q:\mathbb{F}\_{q^\ell}^2\to\mathbb{F}\_q$, where $\mathbb{F}\_{q^\ell}^2$ is viewed as a $2\ell$-dimensional vector space over $\mathbb{F}\_q$.
Using this quadratic forms we obtain an embedding, $\mathrm{O}\_2^\varepsilon(q^\ell)\le \mathrm{O}\_{2\ell}^\varepsilon(q)$. Actually, we obtain an embedding, $\mathrm{SO}\_2^\varepsilon(q^\ell)\le \mathrm{SO}\_{2\ell}^\varepsilon(q)$.
Now, consider the groups $\Omega\_2^\varepsilon(q^\ell)$ and $\Omega\_{2\ell}^\varepsilon(q)$: these are the kernels of the spinor norm homomorphisms $\mathrm{SO}\_2^\varepsilon(q^\ell)\to\mathbb{F}\_{q^\ell}^\ast/(\mathbb{F}\_{q^\ell}^\ast)^2$ and $\mathrm{SO}\_{2\ell}^\varepsilon(q)\to\mathbb{F}\_{q}^\ast/(\mathbb{F}\_{q}^\ast)^2$.
In the embedding above,
is it true that $\mathrm{SO}\_2^\varepsilon(q^\ell)\nleq \Omega\_{2\ell}^\varepsilon(q)$?
This does not seem entirely obvious because a reflection of $\mathrm{O}\_2^\varepsilon(q^\ell)$ (when viewed as an element of $\mathrm{O}\_{2\ell}^\varepsilon(q)$) is no longer a reflection. Therefore it seems cumbersome to use the spinor norms above.
| https://mathoverflow.net/users/45242 | Inclusions among finite orthogonal groups over finite fields | The answer will depend on $q$, $\epsilon$ and potentially also $\ell$. Instead of looking at spinor norms you can find an element $x$ such that $\mathrm{SO}\_2^\epsilon(q^\ell)=\langle \Omega\_2^\epsilon(q^\ell),x\rangle$. You then lift $x$ to $\mathrm{O}^\epsilon\_{2\ell}(q)$ and see what sort of element it is. When you can choose $x$ to be an involution you can then use the knowledge of which involutions of $\mathrm{O}^\epsilon\_{2\ell}(q)$ lie in $\Omega^\epsilon\_{2\ell}(q)$ to see if $\mathrm{SO}\_2(q^\ell)$ is contained in $\Omega^\epsilon\_{2\ell}(q)$. For some values of $q$ and $\ell$ you can take $x=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ or some other involution but for others you will have to choose a different element.
This in done in Lemma 5.3.4 of my book with Tim Burness for the inclusion $\mathrm{O}\_{n/2}(q^2)^\epsilon\leqslant \mathrm{O}^\epsilon\_{n}(q)$. When $q$ is even you have $\mathrm{O}^\epsilon\_{n/2}(q^2)\leqslant\Omega^\epsilon\_n(q)$.
| 2 | https://mathoverflow.net/users/3214 | 424826 | 172,533 |
https://mathoverflow.net/questions/424777 | 7 | For a free group $F$, an element $w$ is primitive if it is part of some free basis for $F$.
Let $\pi:F[x\_0,x\_1,...,x\_n]\rightarrow F[x\_1,x\_2,...,x\_n]$ be defined $\pi (x\_0)=1$ and $\pi (x\_i)=x\_i$ for $i\geq 1$.
My question is:
Is there an example of a cyclically reduced primitive element $w$ in $F[x\_0,x\_1,...,x\_n]$ satisfying $\pi(w)=1$ whose length is greater than 1?
[Added for clarification]
Note: if $w$ is an example, then there must be more than one occurrence of $x\_0$ in $w$ and the exponent sum of $x\_0$ in $w$ must be $\pm 1$.
| https://mathoverflow.net/users/484234 | Primitive elements in a free group with trivial projection | No, there cannot be a primitive element $w \in \ker \pi$ that is not conjugate to $x\_0$ or $x\_0^{-1}$.
The map $\pi$ factors through $F[x\_0, x\_1, \dots, x\_n] / \langle \langle w \rangle \rangle$, which is isomorphic to $F\_n$ since $w$ is primitive. As the free group is Hopfian, this means the induced surjection $F[x\_0, x\_1, \dots, x\_n] / \langle \langle w \rangle \rangle \to F\_n$ is in fact an isomorphism, so $w$ and $x\_0$ have the same normal closure in $F\_{n+1}$. A theorem of Magnus from 1931, citing from Lyndon and Schupp's book "Combinatorial Group Theory", states:
**Proposition II.5.8.** If two elements $r\_1$ and $r\_2$ in a free group $F$ have the same normal closure in $F$, then $r\_1$ is conjugate to $r\_2$ or to $r\_2^{-1}$.
| 10 | https://mathoverflow.net/users/24447 | 424830 | 172,534 |
https://mathoverflow.net/questions/424827 | 10 | Consider the minimal class of (simple, undirected) connected graphs (strictly speaking, isomorphism classes of connected graphs) which contains a single vertex $K\_1$, and is closed under following operations:
1. add a copy $\tilde{v}$ of a vertex $v$ (neighbours of the new vertex are the same as of $v$) and do not join it with $v$ (the new graph must be connected, in other words, you are not allowed to start with this operation applied to $K\_1$);
2. add a copy $\tilde{v}$ of a vertex $v$ (neighbours of the new vertex are the same as of $v$) and join it with $v$;
3. glue two graphs by a vertex.
Alternative description of this class (I hope that we have a proof, it is not very short) is the following: any simple cycle with at least 5 vertices must have two disjoint chords.
Was such class studied?
| https://mathoverflow.net/users/4312 | "Gluing and copy" graphs | This is the same as the class of distance-hereditary graphs, which have received a fair amount of attention (see <https://en.wikipedia.org/wiki/Distance-hereditary_graph>).
| 11 | https://mathoverflow.net/users/385 | 424833 | 172,535 |
https://mathoverflow.net/questions/424829 | 3 | Let $L$ be a Galois extension of a number field $K$ with the Galois group $G$. Let $N$ be the smallest integer with the following property: For any conjugacy class $C$ of $G$ there exists an unramified prime $\mathfrak{p}$ of $K$ such that $\text{Frob}\_\mathfrak{p}\in C$ and $\text{Norm}\_{K/\mathbb{Q}}\mathfrak{p}\le N$.
Is there any good bound of $N$, with respect to $|G|,[L:\mathbb{Q}]$ and $\text{disc}\_{L/\mathbb{Q}}$? If $K=\mathbb{Q}$ and $L/K$ is abelian, this is done in Chapter 18 of [this book](https://bookstore.ams.org/coll-53). I'll be glad to know about the $K=\mathbb{Q}$ case.
| https://mathoverflow.net/users/95601 | A variant of effective Chebotarev theorem | There are bounds on $N$, but whether they are good or not I leave up to you to decide.
First, conditional on GRH, Lagarias and Odlyzko proved a bound on $N$ in their 1977 paper ''Effective versions of the Chebotarev density theorem''. This was made explicit by Bach and Sorenson in 1996 (in "Explicit bounds for primes in residue classes'' published in Mathematics of Computation). Their bound is
$$ N \leq (4 \log({\rm disc}\_{L/\mathbb{Q}}) + 2.5 [L : \mathbb{Q}] + 5)^{2}. $$
Unconditional results are substantially less good. In 2017, Zaman proved (in "Bounding the least prime ideal in the Chebotarev density theorem") that $N \ll ({\rm disc}\_{L/\mathbb{Q}})^{40}$. A more complicated bound that is better in some situations was proven by Thorner and Zaman in their 2017 Algebra and Number Theory paper "An explicit bound for the least prime ideal in the Chebotarev density theorem".
To state Thorner and Zaman's result, choose an abelian subgroup $A$ of ${\rm Gal}(L/K)$ such that $A \cap C$ is nonempty, let $M = L^{A}$ be the fixed field of $A$ (so that $L/M$ is abelian). Let $Q$ be the maximum value of the norm (from $M$ to $\mathbb{Q}$) of the conductor of characters of $A$. Then
$$ N \ll ({\rm disc}\_{M/\mathbb{Q}})^{694} Q^{521} + ({\rm disc}\_{M/\mathbb{Q}})^{232} Q^{367} [M : \mathbb{Q}]^{290 [M : \mathbb{Q}]}. $$
| 7 | https://mathoverflow.net/users/48142 | 424837 | 172,537 |
https://mathoverflow.net/questions/424603 | 3 | Let $K$ be a compact subset of $\mathbb{R}^n$ and $\mathcal{U}$ be a finite collection of open subsets covering $K$ satisfying the minimality property: *for every $U\in \mathcal{U}$, the sub-collection $\mathcal{U}-\{U\}$ does not cover $K$.*
*Notation:* Given any subset $A\subseteq K$ denote the relative complement $K-A:=\{x\in K:\, x\not\in A\}$. Given a point $x\in K$ and a subset $A\subseteq K$ let $\|x-A\|:=\inf\_{a\in A} \|x-a\|$.
Is there a condition on the open cover $\mathcal{U}$ such that the partition of unity $\{\psi\_U\}\_{U \in \mathcal{U}}$ is such that each
$$
\psi\_U(x):= \frac{\|x-(K-U)\|^2}{\sum\_{V \in \mathcal{U}}\,\|x-(K-V)\|^2}
$$
is Lipschitz?
If so, what are explicit bounds on the Lipschitz constant of the functions $\{\psi\_U\}\_{U\in \mathcal{U}}$?
| https://mathoverflow.net/users/469470 | Lipschitz-regularity of partition of unity | For convenience write
$$ \phi\_U(x) = \| x - (K-U)\|. $$
Let $N = |\mathcal{U}|$ the number of open sets in the cover.
Define
$$ f\_1(x) = \frac{1}{N} \sum \phi\_U(x), \qquad f\_2(x) = \left( \frac1N \sum \phi\_U^2(x) \right)^{1/2} $$
Note that since $\mathcal{U}$ is a cover we have
$$ 0 < f\_1(x) \leq f\_2(x) \leq \sqrt{N} f\_1(x) $$
Let
$$ \delta\_1 = \inf f\_1, \quad \delta\_2 = \inf f\_2 $$
and we also have $0 < \delta\_1 \leq \delta\_2 \leq \sqrt{N} \delta\_1 $. Note also that $\delta\_1$ is a Lebesgue number of the covering $\mathcal{U}$.
Now, you've defined
$$ \psi\_U(x) = \frac{\phi\_U^2(x)}{N f\_2^2(x)} $$
so
$$ N(\psi\_U(x) - \psi\_U(y)) = \frac{\phi\_U^2(x) f\_2^2(y) - \phi\_U^2(y) f\_2^2(x) }{f\_2(x) f\_2(y)} $$
$$ = \frac{(\phi\_U(x) - \phi\_U(y))(\phi\_U(x) + \phi\_U(y))}{f\_2^2(x)} + \frac{\phi\_U^2(y) (f\_2^2(y) - f\_2^2(x))}{f\_2(x)^2 f\_2(y)^2}. $$
So we have the following estimate
$$ N |\psi\_U(x) - \psi\_U(y)| \leq \left|\frac{(\phi\_U(x) - \phi\_U(y))(\phi\_U(x) + \phi\_U(y))}{f\_2^2(x)}\right| + N \left| \frac{f\_2^2(y) - f\_2^2(x)}{f\_2(x)^2} \right| $$
Now, it suffices to consider only those $x,y$ with $\|x-y\| \leq \delta \leq \delta\_1$ for a fixed $\delta$. This is because
1. We know that $0 \leq \psi\_U(x) \leq 1$ and hence if $\|x-y\| > \delta$, then the corresponding difference quotient is $\leq \frac{1}{\delta}$.
2. On the other hand, if we define
$$ \epsilon = \inf\_{U\in \mathcal{U}} d( U \setminus \cup (\mathcal{U} \setminus \{U\}), K \setminus U ) $$
then essentially the [same argument of Pietro Majer](https://mathoverflow.net/questions/424603/lipschitz-regularity-of-partition-of-unity#comment1091500_424603) shows that the Lipschitz constant of the family is at least $1/\epsilon$. (Note: minimality of the covering is used to get that $U\setminus \cup(\mathcal{U}\setminus \{U\}) \neq\emptyset$.) Note that $\epsilon \geq \delta\_1$. So we don't expect a much better upper bound estimate of the Lipschitz constant than $1/\delta$.
Now, $\phi\_U(x)$ is well-known to be $1$-Lipschitz. So we have that $$\phi\_U(x) + \phi\_U(y) \leq 2 \phi\_U(x) + \delta \leq (2N+1) f\_1(x)$$
So
$$ \left|\frac{(\phi\_U(x) - \phi\_U(y))(\phi\_U(x) + \phi\_U(y))}{f\_2^2(x)}\right| \leq \frac{2N+1}{\delta\_2} \|x - y\| $$
Similarly, we have
$$ |f^2\_2(x) - f^2\_2(y)| \leq \frac1N \sum |\phi\_U(x) - \phi\_U(y)| |2\phi\_U(x) + \delta| \leq \|x-y\| \cdot (\delta + 2 f\_1(x)) $$
so
$$ N \left| \frac{f\_2^2(y) - f\_2^2(x)}{f\_2(x)^2} \right| \leq \frac{3N}{\delta\_2} \|x-y\|$$
So putting everything together we find that the Lipschitz constant of the family $\psi\_U$ are uniformly bounded by $6 / \delta\_2$.
### Summary
1. For the finite collection $\mathcal{U}$, each function in the corresponding partition of unity is necessarily Lipschitz.
2. The uniform Lipschitz constant can be controlled by a quantity similar to the Lebesgue number of the covering.
| 4 | https://mathoverflow.net/users/3948 | 424862 | 172,541 |
https://mathoverflow.net/questions/421155 | 2 | Let $V$ be a complex finite dimensional inner product space. If $A\_{1},\dots,A\_{n}:V\rightarrow V$ are linear operators, then let $\Phi(A\_{1},\dots,A\_{n}):L(V)\rightarrow L(V)$ be the superoperator defined by letting $\Phi(A\_{1},\dots,A\_{n})(X)=A\_1XA\_1^\*+\dots+A\_nXA\_n^\*$. The completely positive superoperators from $L(V)$ to $L(V)$ are precisely the mappings of the form $\Phi(A\_{1},\dots,A\_{n})$.
If $A$ is an operator, then let $\rho(A)$ denote the spectral radius of $A$. Then
$$\rho(A\_{1}\otimes B\_{1}+\dots+A\_{n}\otimes B\_{n})\leq\rho(\Phi(A\_{1},\dots,A\_{n}))^{1/2}\rho(\Phi(B\_{1},\dots,B\_{n}))^{1/2}.$$
Is there a characterization of the systems of linear operators $((A\_{1},\dots,A\_{n}),(B\_{1},\dots,B\_{n}))$ for which
$$\rho(A\_{1}\otimes B\_{1}+\dots+A\_{n}\otimes B\_{n})=\rho(\Phi(A\_{1},\dots,A\_{n}))^{1/2}\rho(\Phi(B\_{1},\dots,B\_{n}))^{1/2}?$$
| https://mathoverflow.net/users/22277 | When does the Cauchy-Schwarz inequality for spectral radii of tensor products become equality? | **Yes.** We have a necessary and sufficient characterization for when the Cauchy-Schwarz inequality becomes equality.
For this post, $U,V,W$ shall denote finite dimensional complex Hilbert spaces. Suppose that $A\_1,\dots,A\_n:U\rightarrow U,B\_1,\dots,B\_n:V\rightarrow V$ are linear operators. Then define an operator $\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n):L(V,U)\rightarrow L(V,U)$ by letting $$\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n)(X)=A\_1XB\_1^\*+\dots+A\_nXB\_n^\*.$$
Observe that $\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$ is similar to $A\_1\otimes \overline{B\_1}+\dots+A\_n\otimes\overline{B\_n}$ (Here, $\overline{C}=(C^{T})^{\*}=(C^\*)^T$, so $\overline{C}$ is the matrix obtained by replacing every entry in $C$ with its complex conjugate), so the spectral radius Cauchy-Schwarz inequality
$$\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))
\leq\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\cdot\rho(\Phi(B\_1,\dots,B\_n))^{1/2}$$
hold in all cases. It is not too hard to prove the spectral radius Cauchy-Schwarz inequality by using the conventional Cauchy-Schwarz inequality and the characterization of $\rho(\Phi(A\_1,\dots,A\_n))^{1/2}$ given in the 1998 paper The $p$-norm joint spectral radius for even integers by Ding-Xuan Zhou.
Observe that if there is a $\lambda$ and an invertible $B$ with $B\_j=\lambda BA\_jB^{-1}$ for $1\leq j\leq n$, then
$$\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))
=\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\cdot\rho(\Phi(B\_1,\dots,B\_n))^{1/2}.$$
I claim that when $(A\_1,\dots,A\_n)$ and $(B\_1,\dots,B\_n)$ have no invariant subspaces, this is the only way in which the spectral radius Cauchy-Schwarz inequality becomes equality. By decomposing $(A\_1,\dots,A\_n)$ and $(B\_1,\dots,B\_n)$ according to their invariant subspaces, we obtain necessary and sufficient conditions for when the Cauchy-Schwarz inequality is actually an equality.
Recall that a linear operator $\mathcal{E}:L(U)\rightarrow L(V)$ is called positive if $\mathcal{E}(P)$ is positive semidefinite whenever $P$ is positive semidefinite. A linear operator $\mathcal{E}:L(U)\rightarrow L(V)$ is called completely positive if
$\mathcal{E}\otimes 1\_W:L(U\otimes W)\rightarrow L(V\otimes W)$ is positive whenever $W$ is a finite dimensional Hilbert space. We say that an operator $\mathcal{E}:L(U)\rightarrow L(V)$ is trace preserving if $\text{Tr}(\mathcal{E}(X))=\text{Tr}(X)$ for all $X\in L(U)$. An operator $\mathcal{E}:L(U)\rightarrow L(V)$ is said to be a quantum channel if $\mathcal{E}$ is both trace preserving and completely positive.
The completely positive mappings from $L(V)$ to $L(V)$ are precisely the mappings of the form $\Phi(A\_1,\dots,A\_n)$. Observe that any linear mapping from $L(V,U)$ to $L(V,U)$ is of the form $\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$. It is not too hard to show using the Frobenius inner product that the mapping $\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$ is trace preserving if and only if $A\_1^\*B\_1+\dots+A\_n^\*B\_n=1\_V$. In particular, the quantum channels are precisely the mappings of the form $\Phi(A\_1,\dots,A\_n)$ where $A\_1^\*A\_1+\dots+A\_n^\*A\_n=1\_V$.
Observe that if $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$.
Let $E\_{U,n}$ be the collection of tuples $(A\_1,\dots,A\_n)\in L(U)^n$ such that there is some complex number $\lambda$ and invertible $B$ such that if $B\_j=\lambda BA\_jB^{-1}$ for $1\leq j\leq n$, then $\Phi(B\_1,\dots,B\_n)$ is a quantum channel. [By this answer](https://mathoverflow.net/a/423803/22277), $L(U)^n\setminus E\_{U,n}$ is a quite small set whenever $n>1$. In particular, if $(A\_1,\dots,A\_n)$ has no-invariant subspace, then $(A\_1,\dots,A\_n)\in E\_{U,n}$.
Theorem: Suppose that $(A\_1,\dots,A\_n)\in L(U)^n,(B\_1,\dots,B\_n)\in L(V)^n$ have no invariant subspace. Then $$\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))=\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\cdot\rho(\Phi(B\_1,\dots,B\_n))^{1/2}$$ if and only if there is some invertible matrix $C$ along with some complex number $\eta\neq 0$ where
$A\_j=\eta CB\_jC^{-1}$ for $1\leq j\leq n$.
Proof: Since $(A\_1,\dots,A\_n),(B\_1,\dots,B\_n)$ have no invariant subspace, there are non-zero complex numbers $\mu,\nu$ along with invertible matrices $A,B$ where if we set $R\_j=\mu AA\_jA^{-1},S\_j=\nu BB\_jB^{-1}$ for $1\leq j\leq n$, then $\mathcal{E}=\Phi(R\_1,\dots,R\_n),\mathcal{F}=\Phi(S\_1,\dots,S\_n)$ are quantum channels. Let $W$ be a complex inner product space with orthonormal basis $(e\_1,\dots,e\_n).$ Now, let $R,S\in L(V,V\otimes W)$ be the operators defined by letting
$R=\sum\_{j=1}^{n}R\_j\otimes e\_j,S=\sum\_{j=1}^{n}S\_j\otimes e\_j$. Then
$$\mathcal{E}(X)=\text{Tr}\_{W}(RXR^\*),\mathcal{F}(X)=\text{Tr}\_{W}(SXS^\*).$$
Define a mapping $\mathcal{G}$ by setting $$\mathcal{G}(X)=\text{Tr}\_{W}(RXS^\*)=\sum\_{k=1}^nR\_kXS\_k^\*.$$
Since $R\_1^\*R\_1+\dots+R\_n^\*R\_n=S\_1^\*S\_1+\dots+S\_n^\*S\_n=1\_V$, the mappings $R,S$ are isometries.
Now, let $\lambda$ be an eigenvalue of $\mathcal{G}$. Then there is some
eigenvector $X$ with $\mathcal{G}(X)=\lambda X$. Now perform a polar decomposition of $X$ to write $X=HP$ where $H$ is an isometry and $P$ is a positive semidefinite matrix. Therefore, we have
$$\lambda HP=\lambda X=\mathcal{G}(X)=\text{Tr}\_{W}(RXS^\*)=\text{Tr}\_W(RHPS^\*).$$ Now, set $T=(H^\*\otimes 1\_W)RH$. Then
$$\lambda P=\lambda H^\*HP=H^\*\text{Tr}\_W(RHPS^\*)=\text{Tr}\_W((H^\*\otimes 1\_W)RHPS^\*)=\text{Tr}\_W(TPS^\*).$$ Now let $P=\sum\_{k=1}^{n}\sigma\_ke\_ke\_k^\*$. Then $\text{Tr}(\lambda P)=\lambda\cdot\sum\_{k=1}^n\sigma\_k$. Therefore, we have
$$\text{Tr}(\lambda P)=\text{Tr}(\text{Tr}\_W(TPS^\*))=\text{Tr}(TPS^\*)=\sum\_{k=1}^{n}\sigma\_k\text{Tr}(Te\_ke\_k^\*S^\*)=\sum\_{k=1}^n\sigma\_k\text{Tr}(Te\_k(Se\_k)^\*)=\sum\_{k=1}^n\sigma\_k\langle Te\_k,Se\_k\rangle.$$
Therefore, since $$\lambda\cdot\sum\_{k=1}^n\sigma\_k=\sum\_{k=1}^n\sigma\_k\langle Te\_k,Se\_k\rangle,$$ we know that $|\lambda|\leq 1$. Furthermore, if $|\lambda|=1$, then we know that $Te\_k=\lambda Se\_k$ whenever $\sigma\_k>0$.
Therefore, we have $T|\_{\text{Im}(P)}=\lambda\cdot S|\_{\text{Im}(P)}$. In this case, we have $\lambda P=\text{Tr}\_W(TPS^\*)=\text{Tr}\_W(\lambda SPS^\*)$, so $P=\text{Tr}\_W(SPS^\*)$. Since
$P=\text{Tr}\_W(SPS^\*)$ and since $(S\_1,\dots,S\_n)$ has no invariant subspace, we know that $P$ is positive semidefinite. Therefore, $(H^\*\otimes 1\_W)RH=T=\lambda S$.
Now, if $v\in V$, then $\|RH v|=\|v\|=\|\lambda Sv\|=\|(H^\*\otimes 1\_W)RHv\|$. Therefore, if $v\in V$, then $RHv\in\text{Im}(H)\otimes 1\_W$. Therefore, since $\text{Im}(H)$ is a non-trivial invariance subspace of $U$, we know that $\text{Im}(H)=U$. Therefore, $H$ is a unitary operator.
Thus, $H^\*R\_jH=\lambda S\_j$ for $1\leq j\leq n$. Thus, $H^\*\mu AA\_jA^{-1}H=\lambda \nu BB\_jB^{-1}$. We conclude that
$$A\_j=\mu^{-1}\lambda\nu A^{-1}HBB\_jB^{-1}H^{-1}A=\mu^{-1}\lambda\nu A^{-1}HBB\_j(A^{-1}HB)^{-1}.$$
Q.E.D.
Theorem: Suppose that $A\_1,\dots,A\_n:U\rightarrow U,B\_1,\dots,B\_n:V\rightarrow V$ be linear operators. Then assign $U,V$ bases so that
$$A\_j=\begin{bmatrix}
A\_{j,1,1}&\dots&A\_{j,1,u}\\
\vdots&\ddots&\vdots\\
A\_{j,u,1}&\dots&A\_{j,u,u}
\end{bmatrix}$$
and
$$B\_j=\begin{bmatrix}
B\_{j,1,1}&\dots&B\_{j,1,v}\\
\vdots&\ddots&\vdots\\
B\_{j,v,1}&\dots&B\_{j,v,v}
\end{bmatrix}$$
and where for each $\alpha,\beta$ each of the $n$ matrices $A\_{1,\alpha,\beta},\dots,A\_{n,\alpha,\beta}$ have the same dimensions,
for each $\alpha,\beta$, each of the $n$ matrices $B\_{1,\alpha,\beta},\dots,B\_{n,\alpha,\beta}$ have the same dimensions, and where $A\_{j,\alpha,\beta}=0$ whenever $\alpha>\beta$, and where $B\_{j,\alpha,\beta}=0$ whenever $\alpha>\beta$, and where for $1\leq\alpha\leq u$, the matrices
$(A\_{1,\alpha,\alpha},\dots,A\_{n,\alpha,\alpha})$ have no non-trivial invariant subspace, and where if $1\leq\beta\leq v$, the matrices
$(B\_{1,\beta,\beta},\dots,B\_{n,\beta,\beta})$ have non-trivial no invariant subspace either.
Then $\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))=\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\cdot\rho(\Phi(B\_1,\dots,B\_n))^{1/2}$ if and only if there are $\alpha,\beta$ with $1\leq\alpha\leq u,1\leq\beta\leq v$ and where
1. $\rho(\Phi(A\_1,\dots,A\_n))=\rho(\Phi(A\_{1,\alpha,\alpha},\dots,A\_{n,\alpha,\alpha}))$
2. $\rho(\Phi(B\_1,\dots,B\_n))=\rho(\Phi(A\_{1,\beta,\beta},\dots,A\_{n,\beta,\beta}))$, and
3. There is an invertible matrix $C$ and some complex number $\lambda\neq 0$ such that $A\_{j,\alpha,\beta}=\lambda CB\_{j,\alpha,\beta}C^{-1}$ for $1\leq j\leq n$.
**A few observations:**
In this answer, we actually have a couple of different proofs of the spectral radius Cauchy-Schwarz inequality. To prove the Cauchy-Schwarz inequality, we need to prove that $\rho(\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n))\leq 1$ whenever $\Phi(R\_1,\dots,R\_n),\Phi(S\_1,\dots,S\_n)$ are quantum channels, and the general case will follow from the fact that $E\_{V,n}$ is dense in $L(V)^n$ whenever $n>1$. We have already shown that $\rho(\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n))\leq 1$, but there is another way of showing this using the induced trace norm.
If $\mathcal{H}:L(V)\rightarrow L(V)$, then define the induced trace norm of $\mathcal{H}$ to be $\|\mathcal{H}\|\_1=\max\{\|\mathcal{H}(X)\|\_1:\|X\|\_1\leq 1\}.$ Recall that if $R,S$ are isometries and $A$ is a complex matrix, then $\|RAS^\*\|\_1=\|A\|\_1$ whenever the matrix multiplication exists. Also, recall that $\|\text{Tr}\_{V}(X)\|\_1\leq\|X\|\_1$ whenever this inequality makes sense.
We have
$$\|\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n)(X)\|\_1=\|\text{Tr}\_{W}(RXS^\*)\|\_1\leq\|RXS^\*\|\_1=\|X\|\_1.$$
Therefore, $\|\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n)\|\_1\leq 1$, so
$\rho(\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n))\leq 1$ since the induced trace norm is submultiplicative.
**Empirical verification**
I have empirically verified using computer calculations that the conclusions that we have made are reasonable. Define a fitness function $F:M\_d(\mathbb{C})^n\times M\_d(\mathbb{C})^n\rightarrow\mathbb{R}$ by letting
$$F(A\_1,\dots,A\_n;B\_1,\dots,B\_n)=\frac{\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))}{\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\rho(\Phi(B\_1,\dots,B\_n))^{1/2}}.$$
One can maximize the value of $F(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$ using gradient ascent to obtain examples of tuples $(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$ with
$F(A\_1,\dots,A\_n;B\_1,\dots,B\_n)\approx 1$, but in each of these examples, we always have a complex number $\lambda$ along with some invertible $B$ where
$B\_j\approx \lambda BA\_jB^{-1}$ for $1\leq j\leq n$.
| 1 | https://mathoverflow.net/users/22277 | 424866 | 172,543 |
https://mathoverflow.net/questions/424760 | 8 | Disclaimer: This is a crosspost (see [MathStackexchange](https://math.stackexchange.com/questions/4466696/cartesian-monoidal-star-autonomous-categories)). Apologies if cross-posting is frowned upon. However, it seems that on Stackexchange there are not many people familiar with star-autonomous categories.
**1. Question**
Any rigid cartesian monoidal category is trivial (see [here](https://math.stackexchange.com/questions/1042126/is-there-a-special-name-or-any-research-on-cartesian-compact-closed-categories)). Star-autonomy is a generalization of rigidity. Are there (non-trivial) examples of star-autonomous cartesian monoidal categories? Neither Zhen Lin's nor Martin Brandenburg's arguments for the rigid case seem to be easily adaptable to star-autonomous categories.
**2. Additional remarks**
I use the following definition:
A monoidal category $(C, \otimes,I)$ carries a *star-autonomous structure* if there exists an equivalence of categories $S: C^{op} \xrightarrow{\sim} C$ with inverse $S’$ such that there there are bijections $\phi\_{X,Y,Z}: \operatorname{Hom}\_C(X \otimes Y,SZ) \xrightarrow{\sim} \operatorname{Hom}\_C(X, S(Y \otimes Z))$ natural in $X,Y,Z.$
I tried the category **Pos** of posets and monotone maps. This category is cartesian closed. The product is the product order. The set of monotone maps between two posets becomes a poset by setting $$f \leq g \text{ for } f,g: X \rightarrow Y \text{ if } f(x)\leq g(x) \text{ for all } x\in X.$$ Does this category admit a star-autonomous structure? Is it even self-dual? If on objects one sets $S(X):=X^{op}$ with $X^{op}$ the dual poset, it is not clear to me how to define the mapping on morphisms in order to obtain a duality functor. In the category of sup-lattices such a definition is possible, but it uses the existence of joins.
| https://mathoverflow.net/users/160778 | Cartesian monoidal star-autonomous categories | [I'm going to assume $S' \cong S$, which holds in every symmetric monoidal $\*$-autonomous category. (See e.g. Lemma 5.6 of [this paper](http://arxiv.org/abs/1108.6020v2).) This applies here since cartesianness implies symmetry. Part of the proof also works with weaker assumptions, such as merely that the monoidal unit is terminal. Hat tip to Todd Trimble for indicating what this answer really should be.]
**Proposition:** The cartesian monoidal $\*$-autonomous categories are exactly those that are equivalent to Boolean algebras (with $S$ being negation and $\land$ as monoidal structure).
**Proof:**
Every Boolean algebra, considered as a [posetal category](https://en.wikipedia.org/wiki/Posetal_category), is a $\*$-autonomous cartesian monoidal category, since both $x \land y \le \lnot z$ and $x \le \lnot(y \land z)$ are equivalent to $x \land y \land z = \bot$.
For the other direction, let us first show that such a category $C$ must be a preorder; this is also known as [Joyal's Lemma](https://arxiv.org/abs/0910.2401).
Taking $Z := SI$ produces an isomorphism
$$1 \cong \mathrm{Hom}\_C(X \otimes Y,I) \cong \mathrm{Hom}\_C(X \otimes Y,SSI) \cong \mathrm{Hom}\_C(X,S(Y \otimes SI))$$
natural in $X$ and $Y$. This yields an isomorphism $S(Y \otimes S'I) \cong I$ natural in $Y$ by Yoneda, or equivalently $Y \otimes S'I \cong SI$. Hence the dualizing object $SI$ is "absorbing" for the monoidal structure, and since $S$ is a contravariant equivalence, we also know that it is initial.
But now $SI \otimes SI \cong SI$ together with initiality shows that the two product projections $SI \otimes SI \to SI$ are equal. Hence any two parallel morphisms $f,g : A \to SI$ are equal for any $A$. But then since
$$
\mathrm{Hom}\_C(X,Y) \cong \mathrm{Hom}\_C(X\otimes SY,SI),
$$
we can conclude that any two parallel morphisms are equal, and $C$ is equivalent to a poset.
In the poset case, the cartesian monoidal structure must be $\land$, so we have a meet-semilattice, and $S$ is an antitone involution with
$$
x \land y \le Sz \quad \Longleftrightarrow \quad x \le S(y \land z).
$$
But then it follows that $C$ is a Heyting algebra, $Sx = (x \to \bot)$ is negation, and every element is equal to its double negation. Therefore $C$ is a Boolean algebra. $\Box$
---
By the way, $\mathbf{Pos}$ is not contravariant equivalent to itself, for example because it has a strict initial object but not a strict terminal object.
| 6 | https://mathoverflow.net/users/27013 | 424871 | 172,544 |
https://mathoverflow.net/questions/424856 | 2 | Let $p$ be an odd prime and $F={\mathbb Z}/p{\mathbb Z}$. With $a,b,c\in F$, let $Q(x,y)=ax^2+bxy+cy^2$ where $p\nmid b^2-4ac$. I wish to prove that the number of solutions $(x,y)\in F^2$ of
$$ax^2+bxy+cy^2=u$$ is the same for all units $u\in F$.
For example, when $p=5$, the equation $x^2-2xy+3y^2=u$ has one solution when $u=0$ and 6 solutions for each of $u=1,2,3,4$.
| https://mathoverflow.net/users/47804 | Are the number of solutions to $ax^2+bxy+cy^2\equiv u\pmod{p}$, $(x,y)\in\{0,\dotsc,p-1\}$, the same for all units $u$? | Let $F$ be a finite field, such as $\mathbf Z/(p)$ but it could be a more general finite field (including of characteristic $2$, and not necessarily $\mathbf Z/(2)$, so I won't make a change of linear variables using division by $2$ as in another answer). If both $a = 0$ and $c = 0$ then the equation becomes $bxy = u$, so necessarily $b \not= 0$ since $b^2 - 4ac = b^2$ and you said $b^2 - 4ac \not= 0$. A solution $(x,y)$ has both coordinates not equal to $0$, with $y = u/(bx)$. Thus
the solution set for each $u \in F^\times$ is $\{(x,u/(bx)) : x \in F^\times\}$, whose size is $|F| - 1$.
If $a \not= 0$ or $c \not= 0$ then by symmetry (swapping the roles of $x$ and $y$ in the equation) we can suppose $a \not= 0$. Then
solving $ax^2 + bxy + cy^2 = u$ is the same as solving $x^2 + (b/a)xy + (c/a)y^2 = u/a$, and as $u$ runs over $F^\times$ so does $u/a$. Thus we can suppose $a = 1$: we want to count solutions of $x^2 + bxy + cy^2 = u$ in $F^2$. Set $R = F[t]/(t^2 + bt + c)$, a finite ring. The norm map ${\rm N}\_{R/F} \colon R \to F$ is multiplicative, and using the basis $\{1,t\}$ we can realize your expression for $Q(x,y)$ as a norm value: for $x, y \in F$,
$$
{\rm N}\_{R/F}(x + yt) =
\det\begin{pmatrix}x&-cy\\y&x-by\end{pmatrix} = x^2 - bxy + cy^2 = (-x)^2 + b(-x)y + cy^2.
$$
Therefore the equation $x^2 + bxy + cy^2 = u$ is the same as ${\rm N}\_{R/F}(-x+yt) = u$ for $x, y \in F$. On units the norm map ${\rm N}\_{R/F} \colon R^\times \to F^\times$ is a group homomorphism, so as with all homomorphisms between finite groups, all values are taken on an equal number of times.
Thus it remains to show the norm map ${\rm N}\_{R/F} \colon R^\times \to F^\times$ is surjective.
Case 1: $t^2 + bt + c$ is irreducible in $F[t]$. Then $R$ is a field, so $R^\times$ is cyclic and the norm map $R^\times \to F^\times$ on the nonzero elements of finite fields is onto (if $|F| = q$ then $|R| = q^2$ and a generator of $R^\times$ is mapped to a generator of $F^\times$).
Case 2: $t^2 + bt + c$ is reducible in $F[t]$. Write it as $(t-r)(t-s)$.
Since $b^2 - 4c = (r-s)^2$, from $b^2 - 4c \not= 0$ we have
$r \not= s$. Then $R \cong F[t]/(t-r) \times F[t]/(t-s) \cong F \times F$, and in the basis $\{(1,0),(0,1)\}$, the norm mapping
has the formula ${\rm N}\_{R/F}(x,y) = xy$, which maps $R^\times = F^\times \times F^\times$ onto $F^\times$.
You definitely want $b^2 - 4ac \not= 0$, since if $b^2 - 4ac = 0$ your desired conclusion can fail. As a simple example, consider $Q(x,y) = x^2 + 2xy + y^2 = (x+y)^2$. Then the equation $Q(x,y) = u$ has solutions in $F$ if $u$ is a nonzero square in $F$ and it has no solutions in $F$ if $u$ is a nonzero nonsquare in $F$. When
$F = \mathbf Z/(p)$ for odd $p$ (or more generally $F$ is a finite field of odd characteristic), half of $F^\times$ is squares and half is nonsquares, so the equation $(x+y)^2 = u$ for $u \in F^\times$ has solutions for half the values of $u$.
| 5 | https://mathoverflow.net/users/3272 | 424877 | 172,548 |
https://mathoverflow.net/questions/424876 | 2 | Given a directed acyclic graph $G=(V,E)$ with a source node $s$ and a sink node $t$, and we have a weight function that is defined on $E\times E$, $f:E\times E\to R^{+}$. We want to find a $s$-$t$ path $P$ that maximizes the sum $\sum\_{\forall e\_i,e\_j\in P} f(e\_i,e\_j)$.
Does this problem have a polynomial solution?
| https://mathoverflow.net/users/481313 | Longest path on directed acyclic graph when the weight is defined on the pair of edges | The *quadratic shortest path problem* (QSSP) can be reduced to the problem. Because QSSP is NP-hard [1], the problem in the question has no polynomial-time solution unless P = NP.
To remove the restriction of the acyclicity of the graph, we can use the standard technique of using $V \times \{0,\dots,|V|\}$ as the new vertex space where the integer in a pair represents the length of the current path, so an edge $(u,v)$ is converted to $((u,k), (v,k+1))$.
Furthermore, by adding edges $((t,k), (t,k+1))$ corresponding a loop at the destination, we ensure all paths have the same path length.
To convert the minimization problem to the maximization problem, let $W$ be a number larger than all weights and then convert weights as $W - f(e\_1,e\_2)$. Because all paths have the same length in the new graph, the ordering of the solutions is exactly reversed.
* [1] Rostami, Borzou, et al. "On the quadratic shortest path problem." International Symposium on Experimental Algorithms. Springer, Cham, 2015. <https://hal.inria.fr/hal-01251438/file/QSPPaperHal.pdf>.
| 2 | https://mathoverflow.net/users/476793 | 424887 | 172,551 |
https://mathoverflow.net/questions/424879 | 12 | Consider the two matrices with some parameter $s \in \mathbb R$
$$A\_1= \begin{pmatrix} s& -1 &0& 0 \\1&0 &0&0 \\ 0&0&1&0 \\0&0&0&1 \end{pmatrix}$$
and
$$A\_2= \begin{pmatrix} s& -1 &-1& 0 \\1&0 &0&0 \\ -1&0&s&-1 \\0&0&1&0 \end{pmatrix}.$$
I then noticed that the eigenvalues of arbitrary products of $A\_1$ and $A\_2$, i.e. e.g. $A\_1A\_2A\_1$ and $A\_1A\_1A\_2A\_1$ etc. all have eigenvalues $\lambda\_1,1/\lambda\_1$ and $\lambda\_2, 1/\lambda\_2.$
It is clear that the product of eigenvalues is equal to one, as both matrices are in $\text{SL}(4,\mathbb R)$, but I don't see why they have to come in two pairs that multiply up to one, respectively.
| https://mathoverflow.net/users/150564 | Eigenvalues come in pairs | This follows from the identities
$$A\_1^{-1}=UA\_1U^{-1},\;\;A\_2^{-1}=UA\_2U^{-1},$$
$$A\_1^{\top}=VA\_1V^{-1},\;\;A\_2^{\top}=VA\_2V^{-1},$$
with
$$U=U^{-1}=\left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
\end{array}
\right),\;\;V=V^{-1}=\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1 \\
\end{array}
\right).$$
Hence for any string of products $M=A\_1^{n\_1}A\_2^{n\_2}A\_1^{n\_3}A\_2^{n\_4}\cdots A\_1^{n\_N-1}A\_2^{n\_N}$ of the two matrices $A\_1$ and $A\_2$ one has
$$M^\top = VU M^{-1} (VU)^{-1}.$$
It follows that if $\lambda$ is an eigenvalue of $M$, then also $1/\lambda$ is an eigenvalue:
$${\rm det}\,(\lambda-M)={\rm det}\,(\lambda-M^\top)={\rm det}\,(\lambda-VU M^{-1}(VU)^{-1})={\rm det}\,(\lambda-M^{-1}),$$
which implies that
$${\rm det}\,(\lambda-M)=0\Leftrightarrow {\rm det}\,(\lambda^{-1}-M)=0.$$
The case $\lambda=0$ is excluded because $A\_1$ and $A\_2$ are nonsingular for any $s$.
| 13 | https://mathoverflow.net/users/11260 | 424896 | 172,554 |
https://mathoverflow.net/questions/424895 | 5 | Consider the Sierpinski space $\text{S} = (\{0,1\}, \tau)$ where $\tau = \big\{\emptyset,\{0\}, \{0,1\}\big\}$. Endow $\text{S}^\omega$ with the product topology.
If $X, Y$ are topological spaces with $\text{S}^\omega \cong X \times Y$, does this imply that there are $\alpha, \beta \in \big(\omega\cup \{\omega\}\big)$ such that $X\cong \text{S}^\alpha$ and $Y \cong \text{S}^\beta$?
(Note that $\text{S}^0 = \text{S}^\emptyset$ is the one-point space.)
| https://mathoverflow.net/users/8628 | Decomposing $\{0,1\}^\omega$ endowed with the Sierpinski topology | I think the answer is yes. I try to prove the stronger statement that every such homeomorphism is of the form $S^{\mathbb{N}}=S^A\times S^B$ for a disjoint union $\mathbb{N}=A\amalg B$.
Let me switch the roles of 0,1. This makes the following more easily readable.
We can introduce a order on $S^w$ by defining $x\le y$, iff each open set containing $x$ also contains $y$. We can do the same for $X,Y$ and note that the homeomorphism induces an order preserving map.
Furthermore $S^w$ is a lattice, e.g. any subset of it has a least upper bound and a greatest lower bound. If a product of two orders is a lattice, so are the two factors.
Especially $S^w,X,Y$ have a unique minimal element 0. Let $f:X\times Y\rightarrow S^w$ denote the homeomorphism. Write $[a,b]$ for $\{c\mid a\le c\le b\}$. For products of orders we have $[(a\_1,b\_1),(a\_2,b\_2)]=[a\_1,a\_2]\times [b\_1,b\_2]$. Thus if $e\_i$ is the characteristic function of $i\in \mathbb{N}$, we have $[0,e\_i]$ has only the two elements $0,e\_i$. Pick $(x,y)$ with $f(x,y)=e\_i$. We then have:
$[0,e\_i]=[f(0,0),f(x,y)]=f([0,x]\times [0,y])$ and hence exactly one of $[0,x]$ $[0,y]$ consists of two elements, and the other one of one element.
If $[0,x]$ consists of two elements, put $i$ into $A$ otherwise, put it into $B$.
$S^A=f(X\times 0)$ follows by writing any element in $S^A$ as a least upper bound of $e\_i's$ and using that $f$ is compatible with least upper bounds. Analogously $S^B=f(0\times Y)$ follows.
Last we have to show that for an arbitrary element $g\in S^{\mathbb{N}}$
we have $g=f(x,y)$, where $f(x,0)=g\_A,f(0,y)=g|\_B$ and $g|\_A$ is the function which agrees with $g$ on $A$ and is zero otherwise. This again follows since $f$ is compatible with least upper bounds and the least upper bound of $g|\_A$ and $g|\_B$ is $g$, and the least upper bounds of $(0,x)$ and $(0,y)$ is $(x,y)$.
| 2 | https://mathoverflow.net/users/3969 | 424918 | 172,558 |
https://mathoverflow.net/questions/424898 | 5 | Let $G$ be a finitely generated group, and let $F\_1, F\_2$ be two subgroups of $G$ which are free of finite rank at least 2. I am wondering what conditions can be placed on $G$ so that $F\_1\cap F\_2$ is finitely generated. For example:
* It is a classical result of Howson that if $G$ is itself free then $F\_1\cap F\_2$ will be finitely generated.
* I. Kapovich (Subgroup properties of fully residually free groups. Trans. Amer. Math. Soc. 354 (2002), no. 1, 335–362.) extended Howson's result to limit groups and certain hyperbolic groups. However, in these cases *all* finitely generated subgroups have pairwise intersections which are finitely generated. So this is stronger than the situation I am asking about.
On the other hand, if we set $G$ to be a free product of two free groups amalgamated across a non-finitely generated subgroup, so $G=F\_1\ast\_CF\_2$ where $C$ is not finitely generated, then $F\_1\cap F\_2=C$ is not finitely generated. ($G=F(a, b)\times\mathbb{Z}$ gives a finitely presented example of this phenomenon - see [wikipedia](https://en.wikipedia.org/wiki/Howson_property#Examples_and_non-examples).)
I'm interested in any class of groups where for all $F\_1, F\_2$ free, $F\_1\cap F\_2$ is finitely generated, but I was specifically wondering:
1. Let $G$ be hyperbolic. Is $F\_1\cap F\_2$ necessarily finitely generated? (Rips' construction gives finitely generated $H, K<G$ with $H\cap K$ not finitely generated, but the subgroups given by Rips are not in free.)
2. Let $G$ be a one-relator group. Is $F\_1\cap F\_2$ necessarily finitely generated? (The above example $F\_1\ast\_C F\_2$ has cohomological dimension 2, so possibly can embed into a one-relator group.)
| https://mathoverflow.net/users/6503 | Finite generation of intersections of free subgroups | The answer to both 1. and 2. is no.
Burns and Brunner showed that free-by-cyclic groups do not have the "finitely generated intersection property" (a.k.a. Howson property), which was recently generalized by Bamberger and Wise in [Failure of the finitely generated intersection property for ascending HNN extensions of free groups](https://arxiv.org/abs/2201.12580). In fact, it is proved that there are finitely generated *free* subgroups with infinitely generated intersection. A random 2-generator 1-relator group will be hyperbolic. That it is free-by-cyclic with positive probability is Theorem 6.1 in:
*Dunfield, Nathan M.; Thurston, Dylan P.*, [**A random tunnel number one 3-manifold does not fiber over the circle**](http://dx.doi.org/10.2140/gt.2006.10.2431), Geom. Topol. 10, 2431-2499 (2006). [ZBL1139.57018](https://zbmath.org/?q=an:1139.57018).
*Burns, R. G.; Brunner, A. M.*, [**Two remarks on the group property of Howson**](http://dx.doi.org/10.1007/BF01673500), Algebra Logic 18, 319-325 (1980); translation from Algebra Logika 18, 513-522 (1979). [ZBL0448.20030](https://zbmath.org/?q=an:0448.20030).
| 9 | https://mathoverflow.net/users/24447 | 424920 | 172,559 |
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