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https://mathoverflow.net/questions/435659 | 10 | Assume that $\frac{1}{p}+\frac{1}{q}=1$ for two positive real numbers $p,q$.
For what kind of $C^\*$ algebras $A$ does the following hold:
$$\frac{ab+ba}{2}\leq \frac{a^p}{p}+\frac{b^q}{q}$$ $\forall a,b\in A^+$?
As a second question we pose the trace version of this question:
Let $A$ be a $C^\*$ algebra with a faithful trace.
Is it true to say that for every two positive elements $a$, $b$ we have the following inequality?
$$\DeclareMathOperator\trace{trace}\trace(ab)\leq \frac{\trace(a^p)}{p} +\frac{\trace(b^q)}{q}$$
| https://mathoverflow.net/users/36688 | For what kind of $C^*$ algebras does the inequality $\frac{(ab+ba)}{2}\leq\frac{ a^p}{p} +\frac{b^q }{q}$ hold for $a,b>0$? | Let me expand slightly on the comments I made above, and give the most general solution.
Clearly the inequality $\frac{ab + ba}{2} \leq \frac{a^2}{2} + \frac{b^2}{2}$ holds for all positive elements $a,b$ in a $C^\ast$-algebra since $a-b$ is self-adjoint and therefore $0\leq (a-b)^2 = a^2 + b^2 - ab - ba$.
Also (just to get it out of the way before we get technical), the paper [Farenick, Douglas R., Manjegani, S. Mahmoud, Young's inequality in operator algebras. J. Ramanujan Math. Soc. 20 (2005), no. 2, 107–124] shows that for all $p,q >1$ with $\tfrac{1}{p} + \tfrac{1}{q} = 1$ all $C^\ast$-algebras $A$ with a faithful tracial state $\tau$, the inequality $\tau(ab) \leq \frac{\tau(a^p)}{p} + \frac{\tau(b^q)}{q}$ holds for all $a,b\in A\_+$.
Here comes the technical part:
**Proposition:** Let $A$ be a $C^\ast$-algebra and let $p,q>1$ such that $\tfrac{1}{p} + \tfrac{1}{q} = 1$ and $p\neq q$. Then the inequality $\frac{ab+ba}{2} \leq \frac{a^p}{p} + \frac{b^q}{q}$ holds for all $a,b\in A\_+$ if and only if $A$ is commutative.
**Proof:** If $A$ is commutative, then $A \cong C\_0(X)$ for some locally compact Hausdorff space $X$, and the inequality in question holds by Young's inequality since the order relation $\leq$ in $C\_0(X)$ is the pointwise $\leq$ of complex numbers.
Conversely, suppose $A$ is non-commutative. A classical result (which I think might go back to Kadison (I have added a proof below)) shows that $A$ contains a $C^\ast$-subalgebra which surjects onto $M\_2(\mathbb C)$. As positive elements in quotients of $C^\ast$-algebras lift to positive elements (by functional calculus), it suffices to witness the failure of our desired inequality in $M\_2(\mathbb C)$.
Assume without loss of generality that $p>2$. Let $a(\epsilon) = \left(\begin{array}{cc} \epsilon & 0 \\ 0 & 0\end{array}\right)$ for $\epsilon>0$ and $b = \left(\begin{array}{cc} 1/2 & 1/2 \\ 1/2 & 1/2 \end{array}\right)$. We will show that $a(\epsilon)$ and $b$ fail the inequality for sufficiently small $\epsilon$. Note that $a(\epsilon)^p = a(\epsilon^p)$ and that $b^q = b$ (since $b$ is a projection). So we should verify that the
\begin{equation}
\frac{a(\epsilon^p)}{p} + \frac{b}{q} - \tfrac{1}{2}(a(\epsilon)b + ba(\epsilon)) = \left( \begin{array}{cc} \tfrac{\epsilon^p}{p} + \tfrac{1}{2q} - \tfrac{\epsilon}{2} & \tfrac{1}{2q} - \tfrac{\epsilon}{4} \\ \tfrac{1}{2q} - \tfrac{\epsilon}{4} & \tfrac{1}{2q} \end{array} \right)
\end{equation}
is not positive semidefinite for small $\epsilon$. We do this by showing that it has a negative determinant. In fact, the determinant of the above matrix is
\begin{equation}
\frac{\epsilon^p}{2pq} + \frac{1}{4q^2} -\frac{\epsilon}{4q} - \frac{1}{4q^2} - \frac{\epsilon^2}{16} + \frac{\epsilon}{4q} = \epsilon^2 (\frac{\epsilon^{p-2}}{2pq} - \frac{1}{16})
\end{equation}
and this is clearly negative for sufficiently small $\epsilon >0$. QED.
ADDON: Upon request I will add the following which I think is due to Kadison (the driving force is Kadison transitivity). I have added most details, but some minor elementary things are left for the reader.
**Proposition**: Let $A$ be a $C^\ast$-algebra. The following are equivalent:
(i) $A$ is non-commutative;
(ii) there exists an irreducible representation $\pi \colon A \to \mathcal B(H)$ where $\mathrm{dim}(H) \geq 2$;
(iii) $A$ contains a (non-zero) nilpotent element;
(iv) there exists a $C^\ast$-subalgebra $B \subseteq A$ which contains a two-sided closed ideal $J$ such that $B/J \cong M\_2(\mathbb C)$.
**Proof**:
(iv) $\Rightarrow$ (i) is obvious, (i) $\Rightarrow$ (ii) is standard (irreducible representations separate points, so if every irreducible representation is 1-dimensional, then all commutators $ab-ba$ in $A$ vanish).
(ii) $\Rightarrow$ (iii): Fix an irreducible representation $\pi \colon A \to \mathcal B(H)$ where $\mathrm{dim}(H) \geq 2$. Let $\xi, \eta \in H$ be orthogonal unit vectors. By Kadison transitivity there are elements $a,b\in A$ such that
\begin{equation}
\pi(a) \xi = \tfrac{1}{2}\xi , \quad \pi(a) \eta = \eta, \quad \pi(b) \xi = \eta \quad \pi(b) \eta = \xi.
\end{equation}
Replacing $a$ with $|a|$ we may assume $a\geq 0$. Let $f,g\in C\_0((0,\|a\|])$ such that $fg = 0$ and $f(1/2) = 1$ and $g(1) = 1$. I claim that $x= f(a) b g(a) \in A$ is nilpotent of order 2. In fact, clearly $x^2 = 0$ since $g(a) f(a) = 0$ (by functional calculus), so we should show that $x\neq 0$. Since $f(\pi(a)) \xi = \xi$ and $g(\pi(a)) \eta = \eta$ we have
\begin{equation}
\pi(x) \eta = f(\pi(a)) \pi(b) g(\pi(a)) \eta = f(\pi(a)) \pi(b) \eta = f(\pi(a)) \xi = \xi
\end{equation}
so $x\neq 0$.
(iii) $\Rightarrow$ (iv): Represent $A\subseteq \mathcal B(H)$ faithfully (not the same $H$ as above), and let $x\in A$ be a nilpotent element of order 2. We may assume $\|x\| =1$. Let $x = u|x|$ be the polar decomposition of $x$ in $\mathcal B(H)$. Then $u$ is a partial isometry with orthogonal range and source projections, and thus $C^\ast(u) \cong M\_2(\mathbb C)$. Let $y= |x| + |x^\ast|$ (which is the sum of two orthogonal positive elements, so $\|y\|=1$). Then $yu = uy$ and $u^\ast y =y u^\ast$ and thus there is a unique $\ast$-homomorphism $\psi \colon C\_0((0,1]) \otimes M\_2(\mathbb C) \to \mathcal B(H)$ such that $\psi(f\otimes e\_{1,2}) = f(y) u$. The image of $\psi$ takes values in $A$ since $y^k u = y^{k-1} x \in A$, so $f(y)u \in A$ for any polynomial $f$ with trivial constant term. Let $B = \psi(C\_0((0,1])\otimes M\_2(\mathbb C))$ (which is equal to $C^\ast(x)$ but this is irrelevant) and $J = \psi(C\_0(0,1) \otimes M\_2(\mathbb C))$. Then $B/J \cong M\_2(\mathbb C)$.
| 13 | https://mathoverflow.net/users/126109 | 435780 | 176,141 |
https://mathoverflow.net/questions/435751 | 4 | Let $n\geq 3$ be a positive integer and $\kappa=(k\_1, \dots, k\_n)\in \mathbb{Z}^n$. Denote by $B\_n$ the braid group on $n$ strings. Consider the braid on $n+1$ strings $\sigma\_\kappa:=\sigma\_1^{k\_1}\dots \sigma\_n^{k\_n}$, where $\sigma\_i=\sigma\_i^+$ is the generator taking the $i$-th string to the $(i+1)$-st string and $(i+1)$-st string to the $i$-th string in a single overlap (as in Fig. 1.9 on p. 16 of Kassel and Turaev, Braid groups, volume 247, Springer science and Business media).
On closing up the braid $\sigma\_\kappa$ one obtains a link, which we call $L\_\kappa$, with associated Alexander polynomial $\Delta\_\kappa(t)\in \Lambda$, where $\Lambda=\mathbb{Z}[t^{\pm 1}]$, which is well defined up to a unit in $\Lambda$. A well known formula for the Alexander polynomial $\Delta\_b(t)$ of a link associated to a braid $b$ states that
$\Delta\_{b}(t)=\frac{(1-t)}{(1-t^n)}\operatorname{det}\left(\operatorname{I}\_n-\bar{\psi}\_n(b)\right)$, where $\bar{\psi}\_n: B\_n\rightarrow \operatorname{GL}\_{n-1}(\Lambda)$ is the reduced Burau representation (c.f. Chapter 3 of loc. cit.), which arises from interpreting $B\_n$ as the mapping class group of a disk with $n$ marked points.
Using this formula, I was able to show for $n=3, 4$ that the Alexander polynomial $\Delta\_\kappa(t)$ is $\prod\_{i=1}^n F\_{k\_i}(t)$, where $F\_r(t):=1-t+t^2+\dots+(-1)^{r-1}t^{r-1}=\frac{1-(-1)^r t^r}{1+t}$. When $n$ gets larger, the size of the matrices get larger and in general it is not clear to me if a formula similar to this should hold for all $n$? I wasn't able to find this computation appear in the literature, but it seems like something that should be known. Also, I was wondering if this family of braids has any special significance.
| https://mathoverflow.net/users/470091 | Alexander polynomials for a certain family of closed braids | The closure of the braid $\sigma\_\kappa$ is a connected sum of torus links $T(2,k\_i)$ (which are closures of 2-braids). Since the Alexander polynomial is multiplicative with respect to connected sums, $\Delta\_\kappa = \prod\_i \Delta\_{T(2,k\_i)}$.
You can see more generally that if you have a braid $\beta = \beta\_{\rm top}\beta\_{\rm bot} \in B\_n$ which is a product of two braids where $\beta\_{\rm top} \in \langle \sigma\_1,\dots,\sigma\_m\rangle$ and $\beta\_{\rm bot}\in \langle \sigma\_{m+1},\dots,\sigma\_{n-1}\rangle$, then the closure of $\beta$ is the connected sum of the closures of $\beta\_{\rm top}$ and $\beta\_{\rm bot}$ (viewed as an $(m+1)$-braid and an $(n-m)$-braid, respectively). The circle running (mostly) between the $(m+1)^{\rm st}$ and the $(m+2)^{\rm nd}$ strands exhibits the connected sum.
| 3 | https://mathoverflow.net/users/13119 | 435781 | 176,142 |
https://mathoverflow.net/questions/435784 | 7 | A function $f \in L^1(\mathbb R^n)$ is said to be of *bounded variation* if there exists a constant $C \geq 0$ such that
$$
\int\_{\mathbb R^n} f(x) \operatorname{div} \phi(x) \; dx
\leq
C \sup\_{ x \in \mathbb R^n } \lvert\phi(x)\rvert
$$
for all compactly supported differentiable vector fields $\phi : \mathbb R^n \rightarrow \mathbb R^n$. The smallest such constant $C$ can be called the total variation of $f$.
I am wondering about the requirement that $f \in L^1(\mathbb R^n)$. For example, we could replace the requirement “$f \in L^1(\mathbb R^n)$” by “$f \in L^\infty(\mathbb R^n)$” and still obtain a reasonable theory.
For example, compactly support smooth functions are essentially bounded and have total variation. The total variation of constant functions is zero, and constant functions are clearly in $L^\infty$ but not $L^1$.
Hence, if consider the subspace of functions in $L^p(\mathbb R^n)$ with bounded variation (defined analogously to above), then at the very least we get non-trivial vector spaces for any $1 \leq p \leq \infty$, and there is no obvious inclusion between those spaces.
**Question:** Why is $L^1$ usually required? Are there exceptions to this discussed in the literature?
I am aware of the functions of *local bounded variation*, where the function $f$ is merely expected to be Lebesgue integrable. But that big generalization already seems like a step way too far.
Among the possible reasons I can imagine are:
* Authors working on the subject are only interested in analysis over bounded domains, in which case $f \in L^1$ suffices.
* The (historical) application in elasticity actually prefers $L^1$.
* There are actual mathematical reasons to require $f \in L^1$ that are much less obvious than the simple musings above.
| https://mathoverflow.net/users/2082 | What happens if we consider functions of bounded variation that are not in $L^1$? | The main historical reason for which the requirement $f\in L^1$ enters in the definition of $BV$ is that functions of bounded variation (*tout court*) of several variables were introduced by Lamberto Cesari building upon previous work by Leonida Tonelli in order to solve the problem of characterization of (hyper)surfaces of finite (Lebesgue) area (see reference [1] for a complete description of their work). When measuring the surface area (i.e. the perimeter) of a bounded domain $G\in\Bbb R^n$, then $f=\chi\_G$ (where $\chi\_G$ is the characteristic function of the domain $G$), thus the requirements of simple absolute integrability of $f$ suit very naturally in that context.
Another, more practical reason is that, by simply dropping the requirement $f\in L^1$ while keeping the other one, you get the space of functions of locally finite variation $BV\_\text{loc}$: this space is at once more general than $BV$ and $BV^p$, $p>1$ (as are called the spaces defined by requiring $f\in L^p$, $p>1$): even if it is not Banach, this space is not extremely difficult to manage, thus working in it allows more generality without the need of a really cumbersome technical machinery.
**Notes: brief (scant...) literature survey**
* The space $BV^2$ as described in [3], chapter 5, §3.1-3.7 pp. 214-227, is not defined by using functions $f\in L^2$ but functions $f\in BV$ such that $\partial\_if\in L^2$ whose (inward) trace on their domain of definition $G$ (this implies that the theory must be developed on bounded domains of finite perimeter) is square summable too. Then, from this starting point the Authors go on by defining a scalar product, proving its completeness and the compactness of the embedding operator and a few other results. However, I am not aware of other applications of this space than the ones described in this book.
* The only fields where $BV^p$ functions currently play an active role seem to be approximation theory and analysis of nonlinear integral equations (for functions of a single variable). A nice annotated bibliography, updated up to 1998 and which includes also works on functions of bounded $\phi$-variation, has been published in [2], part VI, pp. 241-272.
**References**
[1] Lamberto Cesari, *Surface area* (English), Annals of Mathematics Studies No. 35, Princeton: Princeton University Press, pp. X+595 (1956), [MR0074500](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0074500), [Zbl 0073.04101](https://zbmath.org/?q=an%3A0073.04101).
[2] Richard M. Dudley, Rimas Norvaiša, *Differentiability of six operators on nonsmooth functions and $p$-variation. With the collaboration of Jinghua Qian* (English), Lecture Notes in Mathematics, 1703, Berlin: Springer, pp. viii+277 (1999), ISBN 3-540-65975-7, [MR1705318](https://mathscinet.ams.org/mathscinet-getitem?mr=MR1705318), [Zbl 0973.46033](https://zbmath.org/?q=an%3A0973.46033).
[3] Aizik Isaakovich Vol'pert and Sergei Ivanovich Hudjaev (1985), *[Analysis in classes of discontinuous functions and equations of mathematical physics](https://books.google.com/books?id=lAN0b0-1LIYC&printsec=frontcover&dq=%22Analysis+in+classes+of+discontinuous+functions%22)*, Mechanics: analysis, 8, Dordrecht–Boston–Lancaster: Martinus Nijhoff Publishers, pp. xviii+678, ISBN 90-247-3109-7, [MR 0785938](https://www.ams.org/mathscinet-getitem?mr=0785938), [Zbl 0564.46025](https://zbmath.org/?q=an%3A0564.46025).
| 7 | https://mathoverflow.net/users/113756 | 435792 | 176,145 |
https://mathoverflow.net/questions/435749 | 1 | Let $(X,\tau)$ be a topological space. A *retraction* is a continuous map $r:X\to X$ such that $r$ is the identity on $\text{im}(r)$. We call $S\subseteq X$ a *retract* of $X$ if there is a retraction $r:X\to X$ such that $\text{im}(r) = S$.
We say that $(X,\tau)$ is rc if all retracts are closed. It turns out that all Hausdorff spaces have this property.
Is the class of rc-spaces closed under topological products?
| https://mathoverflow.net/users/8628 | Is the class of rc-spaces closed under products? | Take $X$ to be an [RC space which isn't $T\_2$](https://topology.pi-base.org/spaces?q=rc%2B%7E%24T_2%24) such as the [one-point compactification of the rationals](https://topology.pi-base.org/spaces/S000029). We will show $X^2$ is not RC. Note that it is not $T\_2$ as its factors are not $T\_2$.
First we will note that the diagonal $\{\langle x,x\rangle:x\in X\}\subseteq X^2$ is a retract: send $\langle x,y\rangle \mapsto\_f \langle x,x\rangle$. This is the composition of the (continuous) projection map onto the first coordinate and the (homeomorphic) map $x\mapsto \langle x,x\rangle$, and thus continuous. Since this map restricted to the diagonal is the identity, it is a retraction.
As noted in [this answer](https://mathoverflow.net/q/191020), a non-$T\_2$ space has a non-closed diagonal: choose $x\not=y$ for which no open neighborhoods are disjoint. Then any basic neighborhood $U\times V$ of $\langle x,y\rangle$ must have $z\in U\cap V$ and therefore $\langle z,z\rangle\in U\times V$, showing $\langle x,y\rangle$ is a limit point of the diagonal.
Thus the diagonal is a retract of $X^2$ which is not closed, showing $X^2$ is not $RC$.
| 3 | https://mathoverflow.net/users/73785 | 435793 | 176,146 |
https://mathoverflow.net/questions/434947 | 0 | I'm asking here instead of the economics stackexchange because I'm interested more in the applied mathematics part, instead of just the economics; I'm interested in seeing what new research is being published, and what are the ongoing trends and currents in this area. In that sense, would anyone have interesting journal suggestions for someone interested in applied mathematics, but with a touch of economics, or applied mathematics aimed at economics?
(cross-posted here from MSE given the lack of answers)
| https://mathoverflow.net/users/204581 | Journals of applied mathematics with an economics bent? | <https://www.resurchify.com/impact/category/Applied-Mathematics>
Journal of Econometrics is a journal covering the categories related to Applied Mathematics (Q1); Economics and Econometrics (Q1); History and Philosophy of Science (Q1). It is published by Elsevier BV. The overall rank of Journal of Econometrics is 462. ISSN of this journal is/are 3044076.
| 1 | https://mathoverflow.net/users/493315 | 435802 | 176,148 |
https://mathoverflow.net/questions/435767 | 3 | I start with a thesis: the natural notion of equality is additional data (paths/morphisms), not a binary relation (the fact that they exist). So, in particular, with such a constructivization (replacing property $\to$ structure):
* sets $\to$ $\infty$-groupoids
* categories $\to$ $\infty$-categories
At the same time, it is somewhat unsatisfactory that the concepts on the right have much more cumbersome, technical definitions. The natural answer to this would be: definitions are given in terms of sets i.e. from a 1-world perspective and one would expect the concept of a $\infty$-category have a simple and natural definition in the $\infty$-world.
I know this is an [important open problem in homotopy type theory](https://mathoverflow.net/questions/145770/how-do-you-define-infinity-1-categories-in-homotopy-type-theory), but homotopy type theory is the internal language of a fairly large class of $\infty$-categories (including all $\infty$-topoi anyway). Thus, it is poorer than the $\infty\text{-}\rm{Groupoid}$ internal language (the most expressive $\infty$-topos?).
Questions
1. What is the description of the $\infty\text{-}\rm{Groupoid}$ internal language?
2. Is there a natural definition of the $\infty$-category in this language?
P.S. I don't mean that I see specific reasons why moving from HoTT to the internal language of $\infty\text{-}\rm{Groupoid}$ should help (on the contrary: in 1-world the concept of a category is interpreted in any finitely complete category, no advantages from the expressive means of toposes, much less $\rm{Set}$ is not), but I still can't be sure otherwise.
| https://mathoverflow.net/users/148161 | Does the concept of a $\infty$-category have a natural definition in the $\infty$-world? | Homotopy type theory (HoTT) gives a natural internal language for studying $\infty$-groupoids. [Riehl and Shulman](https://arxiv.org/abs/1705.07442) give an extension of HoTT which gives an analogous internal language for studying $\infty$-categories. Essentially, the Riehl-Shulman framework is able to work because a bit of extra strictness of equality is encoded in the expanded language of their type theory.
However, there is no known way to even define the notion of an $\infty$-category in plain HoTT, without something like the Riehl-Shulman framework extending the basic language. This is closely related to the open problem of defining the type of semi-simplicial types in plain HoTT, and [also](https://homotopytypetheory.org/2014/03/03/hott-should-eat-itself/) to the open problem of defining HoTT internally to the language of HoTT.
I think this is a fundamentally important question. As mentioned by the OP, many practitioners of $\infty$-category theory are very attracted to the idea that $\infty$-groupoids, with their weak notion of equality, should be thought of as *more fundamental* than sets, with their strict notion of equality. But the fact is, we still do not know how to do all the mathematics we want starting purely from the principles of HoTT and weak equality. Both possible resolutions would yield striking pictures:
* If it's possible, then the vision of a fundamentally homotopical world would be vindicated.
* If it's not, then there should be interesting obstructions to study -- we should be able to understand strictness of equality as a "resource" [1] which a logical theory may have to greater or lesser degrees -- on a par with the "resource" of consistency strength. Theories with access to more strictness would be fundamentally "stronger" than theories without such access.
HoTT is still a young field. One day surely we will know which of these possibilities obtain!
[1] I know I got this terminology from someone else, but I can't remember who. Apologies for the lack of attribution!
| 6 | https://mathoverflow.net/users/2362 | 435805 | 176,149 |
https://mathoverflow.net/questions/435758 | 9 | The [nLab page on $\infty$-categories](https://ncatlab.org/nlab/show/infinity-category) splits the known definitions of $\infty$-categories into two types:
* [Algebraic $\infty$-categories](https://ncatlab.org/nlab/show/algebraic+definition+of+higher+categories), in which composition is expressed "externally", e.g. as a some kind of map "$X\_1\times\_{X\_0}X\_1\to X\_1$";
* [Geometric $\infty$-categories](https://ncatlab.org/nlab/show/geometric+definition+of+higher+categories), in which composition is expressed "internally", being defined by means of existential assertions with unicity of composition holding only up to a contractible space of choices.
From what I understand (i.e. very little), the current known definitions of globular $\infty$-categories are all algebraic, including e.g. Batanin $\infty$-categories and Grothendieck–Maltsiniotis $\infty$-categories.
Part of the problem, as I understand it, is that the (reflexive or not) globe category $\mathbb{G}$ is not a test category. From what I gather, this means that we don't have a straightforward way to put a model structure on $\mathsf{Fun}(\mathbb{G}^\mathrm{op},\mathsf{Set})$ modelling the homotopy theory of $\infty$-groupoids, as described in the nLab page [*model structure on presheaves over a test category*](https://ncatlab.org/nlab/show/model+structure+on+presheaves+over+a+test+category).
1. Does this rule out $\mathsf{Fun}(\mathbb{G}^\mathrm{op},\mathsf{Set})$ admitting *any* model structure modelling $\infty$-categories or $\infty$-groupoids?
2. Even if this is so, would it possible (at least in principle) to nevertheless develop a "geometric definition"¹ of globular $\infty$-categories without making use of model categories altogether², and then show that the resulting theory is equivalent to the usual homotopy theory of $\infty$-categories in some appropriate sense (other than "Quillen equivalent")?
¹Here I have in mind something like finding a family of globular sets that fulfills in the globular theory a similar role as to what the horns $\Lambda^n\_k$ do for Kan complexes, quasicategories, and $(\infty,2)$-categories. A further requirement is that this family should be "convenient": it has to be small enough to be reasonably easy to work with, so e.g. globular versions of associahedra don't count.
²I've heard that we have a similar situation for [Lurie's new model of $(\infty,2)$-categories](https://kerodon.net/tag/01W4), in the sense that there isn't a model structure on $\mathsf{sSet}$ recovering them as its fibrant objects. (It is this assertion that made me wonder about this question in the first place, but is it indeed true?)
| https://mathoverflow.net/users/130058 | Is there a "geometric definition" of globular $\infty$-groupoids/categories? | In short there isn't: the problem is that if you just have globular sets - and if you want $k$-cells to model $k$-arrows following the globular structure - then globular sets have no way of expressing the idea that some cell $f$ is the composite of two cells $g$ and $h$. You can only express that two cells $g$ and $h$ are composable and $f$ is parallel to what their composite should be, but you have no way of saying that $f$ is equivalent to that composite, like you could do with simplicial or cubically shaped higher cells.
So this prevent you to define composition in a "geometric way" as a mere lifting property. Composition has to be an additional structure on the globular set.
This is closely related to the reason why Globular sets are not a test category: If you look at the type of spaces that are obtained as geometric realization of a globular set, you only get some very specific spaces (Exercice: they are all wedge sums of spheres) hence you can get all spaces this way.
Now there is I think a lot of comments one can add to this, and answer to your subquestions:
* The fact that globes are not a (weak) test category doesn't completely rule out the existence of a model structure on globular sets. What it rules out is the existence of a model structure where the cofibrations are monomorphisms and the globes are contractible.
* Lurie's model for $(\infty,2)$ has a model structure - it justs not on simplicial sets but on "simplicial sets with a class of marked 2-cells" (the "thin" cell) the problem is that his definition of $(\infty,2)$-category involved the notion of thin cell - it justs so happen that once a "simplicial sets with a notion of thin cell" is an $(\infty,2)$-category then the thin cell can be uniquely characterized, so that as long as we only look at the $(\infty,2)$-categories only then the functor forgetting these marked cells is fully faithful.
* There is a fairly canonical extension of the category of globular sets that still feel globular and is rich enough to encode compositions: This is [Joyal category $\Theta$](https://ncatlab.org/nlab/show/Theta+category) and it has model structure (due to [Ara](https://arxiv.org/abs/1206.4354) following a conjecture of Joyal and Cisinski) that models $(\infty,n)$-category (in a "geometric way"). It is probably the simplest way to do this.
* The definition of "algebraic" vs " geometric" model isn't completely clear cut. I would be tempted to think about it in this way: A geometric model is one where you have a model category where all objects are cofibrants and only the vibrant objects should be thought of as $\infty$-something (the general objects are "presentations"). Morphisms between fibrant objects corresponds directly to "weak functor" already. An algebraic model structure is when you have a model structure where all objects are fibrants, are all considered as "$\infty$-something", but morphisms are some sort of "strict functor". Cofibrant objects corresponds to nice objects $X$ (typically, freely generated in some sense) which have the property that any "weak functor" $X \to Z$ is equivalent to a strict morphism $X \to Z$. In this point of view, [Nikolaus construction](https://arxiv.org/abs/1003.1342) is a general process that turn a geometric model into an algebraic model. Its dual version due to [Ching and Riehl](https://arxiv.org/abs/1403.5303) in the simplicial case, and [Bourke and myself](https://arxiv.org/abs/2005.05384) in the general case) take an algebraic model and turn it into a geometric model. But the problem is that if you apply to algebraic Globular $\infty$-groupoids (even if you assume their canonical model structure exists) then the model you get isn't globular anymore: It will have cells of fairly complex shapes that can encoded compoisiton.
| 5 | https://mathoverflow.net/users/22131 | 435807 | 176,150 |
https://mathoverflow.net/questions/435801 | 11 | Given an initial integer $x\_0>0$, one can consider the first prime of the recursive sequence $x\_i=1+2x\_{i-1}$.
Naïvely such a prime should exist for $x\_0$ arbitrary since the sequence $\log(x\_i)$ is asymptotically
an arithmetic progression. Sometimes it takes however some
time: for $x\_0=147$ my Maple algorithm stops when hitting a 771-digit number
labelled as prime by Maple (Maple does not use a primality proof but
some strong primality tests if I am not mistaken).
Starting with $x\_0=658$ I lost patience: No prime among the first
$56000$ iterates leading to numbers with almost $17000$ digits.
I tried to find an easy reason: If $x\longmapsto 2x+1$
is $k$-periodic for $k$ prime numbers $p\_0,\dotsc,p\_{k-1}$ such that $x\_i\equiv 0\pmod{p\_i}$ and $x\_i>p\_i$ then there is obviously no prime in this sequence. Such an easy argument fails for all small values of $k$ for the sequence $x\_0=658,\dotsc$.
Perhaps my patience ran out a bit early and the sequence will hit
eventually a prime:
*Does an integral sequence given by $x\_0>0$, $x\_i=1+2x\_{i-1}$ necessarily contain a prime number?*
| https://mathoverflow.net/users/4556 | First prime of the form $x_i$ for $x_0=658$ and $x_i=1+2x_{i-1}$ | The $n$th term of your sequence is $x\_{n} = 659 \cdot 2^{n} - 1$. People have long searched for prime values of numbers $k \cdot 2^{n} - 1$ for small $k$, with the goal of proving that $k = 509203$ is the *smallest* positive integer for which $k \cdot 2^{n} - 1$ is composite for all $n$. (This is called the Riesel problem.)
According to the MathWorld entry on [Riesel numbers](https://mathworld.wolfram.com/RieselNumber.html), in 2004 Dave Linton discovered that $659 \cdot 2^{800516} - 1$ is prime, and that $800516$ is the smallest $n$ so that $659 \cdot 2^{n} - 1$ is prime. A lot of patience would have been required indeed.
Currently, the smallest $k$ for which it is unknown if $k \cdot 2^{n} - 1$ is ever prime is $k = 23669$ and a distributed computing project related to this is run by [PrimeGrid](https://www.primegrid.com/stats_trp_llr.php). For $k = 23669$ all $n < 13968972$ have been tested.
| 23 | https://mathoverflow.net/users/48142 | 435808 | 176,151 |
https://mathoverflow.net/questions/428627 | 1 | Here is the definition of a space $X$ to be Rothberger, if for each sequence $(\mathcal{U}\_{n})\_{n\in\mathbb{N}}$ of open covers of $X$, there exists a sequence $(U\_{n})\_{n\in\mathbb{N}}$ where $U\_{n}\in\mathcal{U}\_{n}$ and $X=\bigcup\_{n\in\mathbb{N}} U\_{n}$.
Definition: A subset $A$ of $X$ is said to be semi-open if $A\subset cl(int(A))$ where $cl$ and $int$ are closure and interior operator, respectively.
The space $X$ is said to be s-Rothberger if we can find a sequence for each semi-open covers sequence of $X$.
Let $\mathbb{R}$ be set of real numbers and $p\in\mathbb{R}$ be fixed. Consider $\mathbb{R}$ with the Fort space $$\tau=\{U\subset\mathbb{R}: p\notin U \text{ or if } p\in U \text{ then } \mathbb{R}\setminus U \text{ is finite}\}.$$ Since this space is compact and scattered, it is Rothberger. Is this space also s-Rothberger? It actually may be hard to determine all semi-open sets of this space, but any help or hint will greatly be appreciated.
| https://mathoverflow.net/users/489732 | Rothberger property and semi-open sets | First let's identify the semi-open sets. We first note all open sets are open. A simpler definition of the open sets are (assuming $p=0$),
$$\tau=\{U\subseteq\mathbb R:0\in U\Rightarrow\mathbb R\setminus U\text{ finite}\}.$$
If $0\not\in S$, then $S$ is open and thus semi-open.
If $0\in S$ and $S$ is open, then $S$ is infinite. If $S$ is not open, then $int(S)=S\setminus\{0\}$. Then $0\in cl(int(S))$ provided $S$ is infinite. Thus the semiopen sets are exactly
$$\tau\_{semi}=\{S\subseteq\mathbb R:0\in S\Rightarrow S\text{ infinite}\}.$$
So $\{\mathbb Q\}\cup\{\{x\}:x\in\mathbb R\setminus\mathbb Q\}$ is an uncountable partition of $\mathbb R$ into semi-open sets. Thus there exists no countable subcover, violating s-Lindelof (every semi-open cover has a countable subcover) and thus s-Rothberger.
| 0 | https://mathoverflow.net/users/73785 | 435813 | 176,153 |
https://mathoverflow.net/questions/435797 | 1 | In a commutative ring $R$, when does the assumption $r\_i\mid r$ for $1\le i\le n$ imply $\prod\_{1\le i\le n} r\_i\mid r$ (when $r\_i$ are fixed)?
Does there exist any criterion for this implication that is related to regular sequences?
| https://mathoverflow.net/users/2191 | When an element of a ring that is divisible by a finite set of elements is necessarily divisible by their product? | This is just an expansion of my comment. Question is local, so we may assume that the ring is local. If $r\_1,\ldots, r\_n$ is a regular sequence, then they are so in any order and any subset forms a regular sequence. So, assume $r\_i|r$ for all $i$ and assume we have shown $r=ar\_1r\_2\cdots r\_i$ for some $i<n$, the case of $i=1$ being given.
Then $r\_1,\ldots, r\_i$ is a regular sequence modulo $r\_{i+1}$. Then $r\_1\cdots r\_i$ is also a regular sequence modulo $r\_{i+1}$ and then since $r\_{i+1}|r$ implies $r\_{i+1}|a$ and by induction we are done.
| 2 | https://mathoverflow.net/users/9502 | 435815 | 176,154 |
https://mathoverflow.net/questions/435803 | 3 | Recently, [Ching and Salvatore](https://arxiv.org/abs/2002.03878 "Koszul duality for topological E_n-operads") have proven that the $E\_n$ operad is Koszul self dual. While thinking about the analogous question for the framed $E\_n$ operad, I realized there is an obvious first question: does the spectrum $\Sigma^\infty\_+ O(n)^\vee$ have an $A\_\infty$-structure? Perhaps worth noting, since $\Sigma^\infty\_+ O(n)$ is self dual, it suffices to show $\Sigma^{-n(n-1)/2}\_+ O(n)$ has an $A\_\infty$ structure.
Edit: I realized this question is very basic; the Spanier-Whitehead dual of any space is canonically $E\_\infty$ via the diagonal.
| https://mathoverflow.net/users/134512 | Is $\Sigma^\infty_+ O(n)^\vee$, the Spanier-Whitehead dual of the orthogonal group, an $A_\infty$-ring spectrum? | The question as stated probably requires clarification. If
$X$ is a space, then the S-dual $D\_+(X)$ (i.e., functions from $X\_+$ to the sphere) is always an $E\_\infty$-ring spectrum. In particular, it will also be an $A\_\infty$-ring.
Perhaps what is being asked is whether $D\_+(G)$ is an **$A\_\infty$-coalgebra** when $G$ is a topological group. The answer is yes, because $G$ is an $A\_\infty$-space.
| 6 | https://mathoverflow.net/users/8032 | 435825 | 176,156 |
https://mathoverflow.net/questions/435841 | 5 | Let $E/ \Bbb{C}$ be an elliptic curve which has complex multiplication over a number field $K$.
Then it is widely known that $j(E) \in \overline { \Bbb{Z}}$.
What is the known generalization of this statement to abelian varieties of arbitrary dimension?
| https://mathoverflow.net/users/144623 | Generalization of $j(E) \in \overline { \Bbb{Z}}$ to abelian varieties of arbitrary dimension | The $j$ invariant gives an isomorphism over $\mathbb Z$,
$$j:\mathcal A\_1\to\mathbb A^1,$$
of the moduli space of elliptic curves. So $j(E)\in\overline{\mathbb Z}$ can be interpreted as saying that $\langle E\rangle\in \mathcal A\_1(\overline{\mathbb Z})$, where I've written $\langle E\rangle$ for the isomorphism class of $E$. As noted by Dror, there are Igusa invariants which (largely) do the same thing for $\mathcal A\_2$. But in general, probably the right interpretation is that one should look at the moduli space $\mathcal A\_g/\mathbb Z$, and then it is a theorem that an abelian variety $A$ that has complex multiplication satisfies
$$ \langle A\rangle \in \mathcal A\_g(\overline{\mathbb Z}). $$
Another useful interpretation is that the elements of $\mathcal A\_g(\overline{\mathbb Z})$ have potential good reduction, so in particular this is true for CM abelian varieties. There is also the criterion of Neron-Ogg-Shafarevich that you might find relevant; see
*Serre, Jean-Pierre; Tate, J.*, [**Good reduction of abelian varieties**](http://dx.doi.org/10.2307/1970722), Ann. Math. (2) 88, 492-517 (1968). [ZBL0172.46101](https://zbmath.org/?q=an:0172.46101).
| 7 | https://mathoverflow.net/users/11926 | 435854 | 176,162 |
https://mathoverflow.net/questions/435853 | -1 | Given a non-zero high-dimensional vector, $v\in (\mathbb{R} \setminus \{0\}) ^ d$, and a random sign vector $s \in \{-1,1\}^d$ (i.e., each entry is a rademacher random variable).
Empirically, I find that the distribution of $s \cdot v$ seems to be $\mathcal{N}(0, \frac{||v||\_2^2}{d})$, or at least I can' find an example when it isn't as $d \to \infty $.
(If I would have added that the coordinates are independent and iid, this would be just the Central Limit Theorem.)
Is there some known form of CLT that explains why this would work for a general vector multiplied by random signs?
(or a counterexample that I'm missing)
Thank you!
| https://mathoverflow.net/users/131785 | The distribution of the sum of a non-zero vector with random signs | Of course, your empirically motivated conjecture will not hold in general. E.g., it will not hold if $v\_1^2$ is much greater than $\sum\_{i\ge2}v\_i^2$, where the $v\_i$'s are the coordinates of $v$.
On the other hand, by (say) the [Berry--Esseen inequality](https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem#Non-identically_distributed_summands), the dot product $s\cdot v$ of $s$ and $v$ will be approximately normal if $\sum\_i|v\_i|^3$ is much less than $(\sum\_i v\_i^2)^{3/2}$, which will in turn follow if e.g. $\max\_i v\_i^2$ is much less than $\sum\_i v\_i^2$. (I guess the condition that $\sum\_i|v\_i|^3$ is much less than $(\sum\_i v\_i^2)^{3/2}$ held in your empirical studies.)
| 3 | https://mathoverflow.net/users/36721 | 435856 | 176,163 |
https://mathoverflow.net/questions/435429 | 9 | *[Cross-posted from MSE.](https://math.stackexchange.com/questions/4580731)*
Following [Bankston - The total negation of a topological property](https://doi.org/10.1215/ijm/1256048236), a topological space is called *anticompact* if all its compact subsets are finite. The linked MSE post above has two examples:
**Example 1:** The topology on $\mathbb{R}$ generated by the usual Euclidean topology together with the cocountable topology is an uncountable space that is connected anticompact and Hausdorff.
**Example 2:** Given a free ultrafilter $\mathfrak{F}$ on $\mathbb{N}$, take the ultrafilter topology $\tau = \mathfrak{F} \cup \{\emptyset\}$ on $\mathbb{N}$. This is a countably infinite, connected anticompact space. It is $T\_1$, but not $T\_2$.
Does there exist a countably infinite connected anticompact space that is Hausdorff?
---
Note that such an example would have to be totally path disconnected. Also it cannot be sequential, as a consequence of [this answer](https://math.stackexchange.com/a/4567778) to [Is a space where only finite subsets are compact sets always discrete?](https://math.stackexchange.com/questions/517411/is-a-space-where-only-finite-subsets-are-compact-sets-always-discrete). In particular, it cannot be first countable.
| https://mathoverflow.net/users/458159 | Is there a connected Hausdorff anticompact space that is countably infinite? | Such an example has been constructed by [Banakh and Stelmakh](https://arxiv.org/abs/2211.12579).
More precisely, they constructed an anticompact countable connected Hausdorff space which is Brown and strongly rigid.
| 3 | https://mathoverflow.net/users/61536 | 435864 | 176,167 |
https://mathoverflow.net/questions/435523 | 1 | In the proof of Lemma 2.1 in
*Ershov, Mikhail; He, Sue*, [**On finiteness properties of the Johnson filtrations**](https://doi.org/10.1215/00127094-2018-0005), [ZBL06904638](https://zbmath.org/?q=an:06904638),
the authors claim the following (without proof).
*Let $G$ be a finitely generated group, and let $g\_1,\ldots,g\_k\in G$ be pairwise commuting elements, such that the union of the centralizers $S:=\bigcup\_{i=1}^k\mathrm{Cent}\_G(g\_i)$ generates $G$. Then it is possible to find a finite generating set for $G$ contained in $S$.*
I can't see why this should be true in general, but I don't have any counter-example either. Can anybody help me?
| https://mathoverflow.net/users/47274 | A group-theoretic lemma in a paper by Ershov and He | Here is @AndyPutman's [comment](https://mathoverflow.net/questions/435523/a-group-theoretic-lemma-in-a-paper-by-ershov-and-he#comment1122032_435523) as an answer (so that it can be accepted), made CW to avoid reputation. If @AndyPutman prefers to post the answer, then I will delete this.
>
> This only needs the fact that $G$ is finitely generated and $S$ is a generating set. If $T$ is a finite generating set for $G$, then since $S$ is another generating set each $t$ in $T$ can be written as a word in $S$. Let $S'$ contain the letters that appear in such words as $t$ ranges over $T$. Then $S'$ is a finite generating set.
>
>
>
| 1 | https://mathoverflow.net/users/2383 | 435867 | 176,168 |
https://mathoverflow.net/questions/435754 | 5 | Let $(M,g)$ be a (connected, paracompact, $C^{\infty}$-smooth) Riemannian manifold with Riemannian metric $g$. The exponential map is defined for each point $p \in M$ to be the map $\exp\_p : T\_p M \to M$ that sends a tangent vector $v \in T\_p M$ to the endpoint of the unique geodesic $\gamma$ satisfying $\gamma(0)=p$ and $\gamma'(0)=v$.
In particular, the notion of a geodesic allows us to define an exponential map. My question concerns the reversal of this construction:
>
> Suppose that for each $p \in M$, we have a map $\sigma\_p : T\_p M \to M$ with some desirable properties. Can one define a notion of $\sigma$-geodesic that coincides with notion of a geodesic when $\sigma =\exp$?
>
>
>
If this can be done, I suspect that it is well-known, but I've never come across such a construction.
| https://mathoverflow.net/users/105103 | Can the exponential map be used to define geodesics (and hence, generalisations of geodesics)? | Exponential map in your definition is closely related to the smooth family of smooth curves smoothly depending on the position such that in every point in every direction there exists precisely one curve at this point in this direction. In literature, such a family is sometimes called path structure.
There are two differences between your structure and path structure: the first one is that you curves are parameterised, and the curves of path structure are not. But this an additional structure, so my answer below will has sense also for your setup.
The other difference is that for a path structure there exists precisely one, or, in the case of irreversible path structures, at most two curves from the family. You definition allows that (for three points A,B,C) the curves from A and B to C come to C with the same velocity vector which is not possible for geodesics. Or, even more wild, take 3 points A, B, C and consider the trajectory $\gamma\_{ABC} $
of your exponential map which starts from A and then passes through first B and then C (assume such trajectory exists, i.e., choose B and C on a trajectory from A). Your definition does not require that the segment of $\gamma\_{ABC}$ from $B$ to $C$ is a trajectory of your exponential map starting from $B$ and going to $C$, though this property may be essential for geodesics.
But, returning to path structures, most path structures do not come from an affine connection (known since at least Cartan for dim 2, essentially the same proof works for all dimensions, see <https://arxiv.org/abs/1101.2069>). In dimension 2, every reversible path structure is locally Finsler-metrisable <https://arxiv.org/abs/1002.0243>, the question whether every irreversible path structures is metrisable is still open.
In higher dimensions most path structures are, even microlocally, not Finsler-metrisable, see (the open access paper) <https://link.springer.com/article/10.1007/s10714-022-03006-2> .
| 7 | https://mathoverflow.net/users/14515 | 435870 | 176,169 |
https://mathoverflow.net/questions/435863 | 3 | I'm looking at the linear PDE in 3+1 dimensions,
$$
\left[ -(\partial\_t - \xi)^2 - \partial\_k \partial\_k \right] \phi(t,x) = 4\pi^2 \delta(t)\delta(x)\label{1} \tag{1}
$$
Where $\xi$ is generally a complex parameter and $k=1,2,3$. So I'm after a particular Green's function $\phi(t,x) = \phi(t,x,\xi)$ and would like to study the $\xi$-dependence. The solution is required to be $SO(3)$ invariant and should fall off in all directions at infinity.
Now for purely imaginary parameter $\xi = i \alpha$ where $\alpha$ is real it's easy to see that a solution is
$$
\phi(t,x) = \frac{e^{it\alpha}}{t^2 + r^2}\label{2}\tag{2}$$
with $r = \sqrt{x\_1^2 + x\_2^2 + x\_3^2}$. So far so good :)
Now if $\xi$ is real, or has a non-zero real part, we would have
$$
\phi(t,x) = \frac{e^{\xi t}}{t^2 + r^2}\label{3}\tag{3}
$$
as the analytic continuation of \eqref{2} but this blows up for $t\to\pm\infty$, depending on the sign of $\xi$. So this analytically continued solution is not okay.
The solution which does satisfy the boundary conditions can still be found and is
$$\phi(t,x) = \frac{\cos(\xi r) + \frac{t}{r} \sin(\xi r)}{t^2 + r^2}\label{4}\tag{4}$$
which does fall off in all directions at infinity. Notice that if $\xi$ is purely imaginary the analytic continuation of the solution \eqref{4} will behave badly and will blow up for large $r$.
I was wondering if it's possible to construct one of the solutions if the other is known. In other words if I know the Green's function on the imaginary line, can I somehow guess or construct the Green's function everywhere on the complex plane? Or if I know it everywhere except the imaginary axis, can I find the form of the Green's function along it?
Or one just have to consider both problems totally independently and work hard twice? :)
| https://mathoverflow.net/users/41312 | Linear PDE, analytic continuation, Green's function and boundary conditions | **Q:** *Do I have to consider both problems (real $\xi$ or imaginary $\xi$) totally independently and work hard twice?.*
**A:** A single calculation suffices, you could just do the inverse Fourier transform of $[(\omega+i\xi)^2+k^2]^{-1}$ for complex $\xi=i\alpha+\beta$ to arrive at the solution that works for both real and imaginary $\xi$,
$$\phi(t,x) = e^{i\alpha t}\frac{r\cos\beta r + t\sin\beta r}{r(t^2 + r^2)}.$$
This is not an analytic function of $\xi$, so you cannot reconstruct it by analytic continuation from the real or imaginary axis.
| 8 | https://mathoverflow.net/users/11260 | 435876 | 176,172 |
https://mathoverflow.net/questions/435826 | 9 | Let $g$ be a piecewise smooth, zero average, function over $[0,1]$ such that $\min g^2>0$. I would like to show that
$$
\int\_0^1 g\sqrt{1-r/g^2}\int\_0^1 \frac{1}{g\sqrt{1-r/g^2}} \leq 1
$$
for all $r \in \mathopen[-1,\min g^2\mathclose[$. I don't know that this is true but I am persuaded that it is, based on intuition (see motivation below), numerical tests and on a couple of, admittedly trivial, particular cases (e.g., $g^2$ is constant).
If it helps: note that second integral is almost the derivative of the first with respect to $r$.
Also, one could rescale $g$ to get rid of $r$. Say $0<r<\min g^2$. Let $g\_r=g/\sqrt{r}$ so that $\min g\_r^2>1$. Then, we're after
$$
\int\_0^1 g\_r\sqrt{1-1/g\_r^2}\int\_0^1 \frac{1}{g\_r\sqrt{1-1/g\_r^2}} \leq 1.
$$
We should be able to do something similar for $r<0$.
I am aware of an inequality $E(X)E(1/X)\geq 1$ for $X$ positive. Here, $g$ has zero average meaning it can't be positive.
Motivation: I am trying to show that a certain linkage (something like a carpenter's ruler) grows in span. The problem boils down to the above inequality. With that in mind, I've tried to apply something like Cauchy's arm lemma, to no avail.
The question is [cross-posted](https://math.stackexchange.com/questions/4587970/integral-inequality-prove-int-01-f-int-01-1-f-leq-1-for-some-function-f).
**For future generations:** Inequality disproven thanks to @fedja's answer. To construct a counter example consider the case where $g$ takes three values $g\_{1,2,3}$ over three intervals of lengths $p\_{1,2,3}$. Suppose $g\_{1,2}>0$ and $g\_3<0$. This case is enough by @Iosif's observation. Also, let $f=\sqrt{g^2-1}$. Now consider the limit $p\_1\to 0$, $g\_1\to 1$. To disprove the inequality, we would also like for $p\_1/f\_1$ to go to infinity. Thus, take $g\_1=1+p\_1^3$ for instance. Then, the inequality reads, to leading order, $(p\_2f\_2-p\_3f\_3)p\_1/f\_1<1$. Disproving the inequality amounts to finding $p\_{2,3}$ and $g\_{2,3}$ such that $p\_2+p\_3=1$ and $p\_2g\_2+p\_3g\_3=0$ and such that $p\_2f\_2-p\_3f\_3>0$. It suffices to take $g\_2>-g\_3>1$ and $p\_2=-g\_3/(g\_2-g\_3)$ and $p\_3=g\_2/(g\_2-g\_3)$. For instance, $g\_2=3$, $g\_3=-2$, $p\_1=0.004$ appear to do the trick.
| https://mathoverflow.net/users/486571 | Integral inequality: Prove $\int_0^1 f\int_0^1 1/f \leq 1$ for a certain function $f$ | Matematika, shmatematika: any minimally decent CAS should immediately detect and tell the human operator that, for $r=1$, the inequality reads
$$
(\int\_X F-\int\_Y G)(\int\_X 1/F-\int\_Y 1/G)\le 1
$$
where $F=\sqrt{f^2-1}, G=\sqrt{g^2-1}$, $f,g> 1$, $\mu(X)+\mu(Y)=1$, and $\int\_X f=\int\_Y g$. Now it becomes obvious that the inequality can easily fail: Take any $f,g$ with the expression in the first parentheses not $0$. That first difference doesn't feel small moving of the distribution of $f$ or $g$ but the second one is extremely unstable: if you move a tiny portion of values of $f$ or $g$ sufficiently close to $1$ without changing the full integral, you can get arbitrarily large size and any sign you want.
The second part (negative r) is true and equally trivial:
Let $r=-1$ (scaling) and $F=\sqrt{f^2+1}, G=\sqrt{g^2+1}$. Then
$|F-f|,|G-g|\le 1$, so the first difference is at most $\mu(X)+\mu(Y)=1$ in absolute value (after subtracting integrals of $f$ and $g$ that compensate each other) and the second one is trivially bounded by the same quantity (since the denominators are at least $1$). So we get the bound $1$ for the absolute value of the product.
I would *love* to see an *intelligent* computer program for solving elementary inequalities and do not see any principal obstacles to creating it and even have some ideas about how this task could be possibly accomplished, but, alas, what is around is still nowhere close :-(.
| 6 | https://mathoverflow.net/users/1131 | 435877 | 176,173 |
https://mathoverflow.net/questions/435868 | 23 | This question is inspired by [On moments of inertia of planar and 3D convex bodies](https://mathoverflow.net/questions/435361/on-moments-of-inertia-of-planar-and-3d-convex-bodies).
Let $f:{\mathbb R}^3\setminus\{0\}\to{\mathbb R}$ be an even homogeneous ($f(kx)=f(x)$ for all real $k\neq 0$) continuous function.
Can one always find an orthogonal basis $(x\_1,x\_2,x\_3)$ such that
$f(x\_1)=f(x\_2)=f(x\_3)$ ?
Remark. Even without the condition that $f$ is even, I do not know a counterexample. This condition comes from the application that I mention in the first sentence.
Remark 2. Thanks to @Aleksei Kulikov, who found the reference: this was a problem posed by Rademacher and solved by Kakutani in 1942 (Annals of math., 43 (1942) 739-741). Most interestingly, the reviewer
in MR0007267 wrote that the corresponding problem in $R^n, n\geq 4$ was unsolved in 1942.
I found that the $n$-diensonal generalization was obtained by H. Yamabe and Z. Yujobô,
Osaka Math. J. 2 (1950), 19–22.
| https://mathoverflow.net/users/25510 | A property of even continuous functions on the sphere | It seems there aren't any counterexamples, even if $f$ is homogeneous but not even.
If there is some counterexample $f$ to the question, and letting $X=\mathbb{R}^3\setminus\{x\_1=x\_2=x\_3\}$, we can consider the map $F:SO(3)\to X$, given by $F(u,v,w)=(f(u),f(v),f(w))$ (we will sometimes represent elements $r\in SO(3)$ by $(r(e\_1),r(e\_2),r(e\_3))$, $(e\_i)\_{i=1}^3$ being the usual basis of $\mathbb{R}^3$).
As $\pi\_1(SO(3))\cong\mathbb{Z}\_2$ and $\pi\_1(X)\cong\mathbb{Z}$, we will have $F\_\*(\gamma)=0\;\forall\gamma\in\pi\_1(SO(3))$.
Now consider the loop $\gamma:t\mapsto\gamma\_t=r\_{2\pi t}(v)$ given by rotations of angle $2\pi t$ around $v:=(1,1,1)$. Note that $\gamma\_{\frac{1}{3}}(e\_1)=e\_2,\gamma\_{\frac{1}{3}}(e\_2)=e\_3$ and $\gamma\_{\frac{1}{3}}(e\_3)=e\_1$. So if $\gamma\_t=(u,v,w)$, then $\gamma\_{t+\frac{1}{3}}=(\gamma\_{t+\frac{1}{3}}(e\_1),\gamma\_{t+\frac{1}{3}}(e\_2),\gamma\_{t+\frac{1}{3}}(e\_3))=(\gamma\_{t}(e\_2),\gamma\_{t}(e\_3),\gamma\_{t}(e\_1))=(v,w,u)$.
So the path $\alpha=F\_\*\gamma\in\pi\_1(X)$ satisfies that if $\alpha(t)=(x\_1,x\_2,x\_3)$, then $\alpha(t+\frac{1}{3})=(x\_2,x\_3,x\_1)$. From this we will deduce that $\alpha$ is not trivial, which is a contradiction. To deduce that $\alpha$ is not trivial, we can first deformation retract $X$ to a circumference perpendicular to the line $\{x\_1=x\_2=x\_3\}$, so that identifying this circumference with $\mathbb{S}^1$, the rotation $(x\_1,x\_2,x\_3)\mapsto(x\_2,x\_3,x\_1)$ becomes the rotation $r\_{\frac{2\pi}{3}}$ of $\mathbb{S}^1$.
So we have to prove that any loop $\beta:[0,1]\to\mathbb{S}^1$ satisfying $\beta(t+\frac{1}{3})=e^{\frac{2\pi i}{3}}\beta(t)$ is not trivial. This is true because as $\beta(\frac{1}{3})=e^{\frac{2\pi i}{3}}$, the path $\beta|\_{[0,\frac{1}{3}]}$ rotates a total angle of $2\pi(n+\frac{1}{3})$ for some $n\in\mathbb{Z}$, so the winding number of $\beta$ is $3(n+\frac{1}{3})\neq0$.
| 21 | https://mathoverflow.net/users/172802 | 435882 | 176,174 |
https://mathoverflow.net/questions/435855 | 6 | The following is true:
1. The counit components $\epsilon : X^A\times A\to X$ of the cartesian closed structure of $Set$ are the components of the initial cowedge; in other words, $X$ is the coend $\int^A X^A\times A$. Size issues apart, the reason is that $\int^A X^A\times A$ is isomorphic to the value at $X$ of $\text{Lan}\_11$, and the latter is the identity functor.
But what about the same statement in a generic cartesian closed category $\cal C$, with enough colimits if needed?
2. The counit components $\epsilon : X^A\times A\to X$ are the components of the initial cowedge in $\cal C$; in other words, $X$ is the coend $\int^A X^A\times A$.
And what about the most general statement in this direction:
3. Let $L\_a\dashv R\_a$ be a parametric adjunction -give by functors $L : A\times C\to D$ and $R : A^{op}\times D \to C$. Then the counit $\epsilon : L\_aR\_ad \to d$ is a cowedge, and $d\cong\int^aL\_aR\_ad $; dually, the unit $c\to R\_aL\_ac$ is a wedge, and $c\cong\int\_a R\_aL\_ac$.
I think 3. is false in such a generality, but then
* What is an instructive counterexample?
* Under which conditions it is true, considering that it is *sometimes* true?
| https://mathoverflow.net/users/7952 | The coend of a parametric counit | Unfortunately statement 2 is not true in general. First, notice that
$$\hom\Bigl(\int^A X^A \times A,Y\Bigr) = \int\_A \hom(X^A,Y^A) = \hom(X^{(-)},Y^{(-)}).$$
So the question is if the functor
$$\mathcal{C} \to [\mathcal{C}^{\mathrm{op}},\mathcal{C}], ~ X \mapsto X^{(-)}$$ is fully faithful. The canonical map
$$\hom(X,Y) \to \hom(X^{(-)},Y^{(-)})$$
has a retraction: we can evaluate a natural transformation at $1 \in \mathcal{C}$. So the composition $\hom(X,Y) \to \hom(X^{(-)},Y^{(-)}) \to \hom(X,Y)$ is always the identity. But I don't see a reason why the composition $\hom(X^{(-)},Y^{(-)}) \to \hom(X,Y) \to \hom(X^{(-)},Y^{(-)})$ should be the identity. It holds when $1$ is a [**separator**](https://ncatlab.org/nlab/show/separator) of $\mathcal{C}$. I am too lazy to write down the proof, but it is an easy exercise.
But now consider the topos of $G$-sets $\mathcal{C} = {}\_G \mathbf{Set}$ for some group $G$ with non-trivial center $Z(G)$ (maybe another argument works for all non-trivial groups). It is well-known that for two $G$-sets $X,A$ the $G$-set $X^A$ consists of all maps of the underlying sets $\alpha : U(A) \to U(X)$, and the $G$-action is $(g \cdot \alpha)(x) := \alpha(g^{-1} \cdot x)$. Any element $g \in Z(G)$ induces an element in the center $Z(\mathcal{C})$ and hence a natural transformation $X^{(-)} \to X^{(-)}$, namely $\alpha \mapsto g \cdot \alpha$. Evaluating this at $1 \in \mathcal{C}$ always gives the identity, so we get a counterexample when $g \neq 1$.
| 3 | https://mathoverflow.net/users/2841 | 435886 | 176,175 |
https://mathoverflow.net/questions/435859 | 9 | Let $\Sigma \subset \mathbb{R}^3$ be a compact embedded surface with boundary $\partial \Sigma$ and $i:\Sigma\setminus \partial\Sigma \to \mathbb{R}^3 \setminus \partial\Sigma$ the inclusion.
Is the following true?
If $i\_\*(\pi\_1(\Sigma\setminus \partial \Sigma))=0$, then $\Sigma$ is orientable.
| https://mathoverflow.net/users/64231 | Links and non-orientable surfaces | Yes, the surface is orientable. To simplify the LaTex and the exposition, I will change the notation and setting a small amount.
---
Suppose that $F$ is a compact connected embedded surface in three-sphere. Suppose that the image of $\pi\_1(F)$ in $\pi\_1(S^3 - \partial F)$ is trivial. We must show that $F$ is orientable.
In the base case, where $F$ has no boundary, this follows from Alexander's theorem.
Suppose that $\alpha$ is a boundary component of $F$. Let $\alpha'$ be a curve embedded in $F$ which is disjoint from, but is isotopic to (in $F$), the boundary component $\alpha$. Thus $\alpha$ and $\alpha'$ cobound an annulus in $F$. Since the image of $\pi\_1(F)$ is trivial in $\pi\_1(S^3 - \partial F)$, we have that $\alpha'$ bounds an immersed disk in $S^3 - \partial F$. By Dehn's lemma (that is, by a version of the Disk Theorem) $\alpha'$ bounds an embedded disk $D$ in $S^3 - \partial F$. Thus $\alpha$ is an unknot.
This holds for all boundary components of $F$. In fact, since all of the boundary components bound disks in $S^3 - \partial F$, we deduce that $\partial F$ is a split link. Surgering along separating spheres reduces us to the case where $F$ has only one boundary component. [There is some work here.]
Now consider the annulus $A$ between $\alpha$ and $\alpha'$. Note that $A$ is two-sided. So we may and do isotope $D$ slightly to make it transverse to $F$ and disjoint from the interior of $A$. Suppose that $\beta$ (perhaps equal to $\alpha'$) is an innermost curve of $F \cap D$. Let $D' \subset D$ be the subdisk bounded by $\beta$. Let $B \subset F$ be a small annulus neighbourhood of $\beta$.
We surger $F$ along $D'$ to obtain a new surface $F'$. That is, we form $F - B$ and glue on a pair of disks, both parallel to $D'$. If $F'$ is non-orientable, then it still has trivial $\pi\_1$-image and has lower complexity (either has no boundary, has lower genus, or meets $D$ in a simpler way) than $F$. This is a contradiction.
We deduce that $F'$ is orientable and so is two-sided. We also deduce that the two boundaries of $B$ are attached to opposite sides of $F'$. Thus there is a (orientation reversing) curve $\gamma$ in $F$ that meets $\beta$ exactly once. We deduce that $\gamma$ has linking number one with $\alpha$. Thus $\gamma$ is non-trivial in the image of $\pi\_1(F)$, a contradiction.
---
I think that the condition can be reduced to "$H\_1$-image is trivial". The above argument does not immediately work (because surgery along an orientable surface can cause genus to increase).
| 4 | https://mathoverflow.net/users/1650 | 435887 | 176,176 |
https://mathoverflow.net/questions/435837 | 4 | To each pair $(S,\mathcal{X})$ where $S=(s\_i)\_{i\in\mathbb{N}}$ is a decreasing sequence of positive real numbers and $\mathcal{X}\subseteq\mathbb{R}$, we can associate the **alternation game** $A\_S(\mathcal{X})$ as follows:
* Players $1$ and $2$ jointly build an increasing sequence of natural numbers $$a\_0<b\_0<a\_1<b\_1<...,$$ subject to the rule that $$\sum\_{i<a\_0}s\_i>\sum\_{a\_0\le i<b\_0}s\_i> ... >\sum\_{b\_k\le i< a\_{k+1}}s\_i>\sum\_{a\_{k+1}\le i<b\_{k+1}}s\_i>...$$
* Player $1$ wins iff the alternating series $$\left(\sum\_{i<a\_0}s\_i\right)-\left(\sum\_{a\_0\le i<b\_0}s\_i\right)+ ... +\left(\sum\_{b\_k\le i< a\_{k+1}}s\_i\right)-\left(\sum\_{a\_{k+1}\le i<b\_{k+1}}s\_i\right)+...$$ converges to an element of $\mathcal{X}$.
An [old question](https://mathoverflow.net/questions/426794/a-game-of-harmonic-seriess) of mine asked about the behavior of these games when we specifically take $S$ to be the harmonic sequence $s\_i={1\over i}$, but turned out to be difficult to attack. I'd like to get a better understanding of *why* this sort of question might be hard, and this seems like a good starting point:
>
> Is it consistent with $\mathsf{ZFC}$ that *some* $A\_S(\mathcal{X})$ is undetermined?
>
>
>
A positive answer to this question would help explain potential wildness, but would also seem to require us to be able to "code into" the alternation games (a la the [Banach game](https://mathoverflow.net/questions/281240/is-the-banach-game-quantifier-intractable-beckers-guess)), and at present I don't see how to do that.
| https://mathoverflow.net/users/8133 | Can these alternating series games be undetermined? | If we take $s\_i = 2^{-i}$, we should even get that $\mathrm{ZFC}$ proves the existence of some $\mathcal{X}$ with undetermined $A\_{(2^{-i})\_{i \in \mathbb{N}}}(\mathcal{X})$. The key parts are that different plays yield different reals (as nothing in the tail can overcome a difference in a prefix), and that each strategy is compatible with continuumsly many plays (as a move here doesn't constrain the future at all).
We should thus be able to do the standard trick of well-ordering both players strategies with the least possible order type, and then building $\mathcal{X}$ in stages to foil any particular winning strategy.
| 5 | https://mathoverflow.net/users/15002 | 435906 | 176,180 |
https://mathoverflow.net/questions/435676 | 4 | Let $\mathbb{A}$ be the ring of algebraic integers. Consider a sequence $(d\_i)\_{i \in I}$, with $I$ a finite set and $d\_i \in \mathbb{A} \cap \mathbb{R}\_{\ge 1}$, such that $$d\_i d\_j = \sum\_{k \in I} n\_{i,j,k} d\_k,$$ for all $i,j \in I$, with $n\_{i,j,k} \in \mathbb{Z}\_{\ge 0}$.
A subset $J \subset I$ is called a *subsystem* if $\forall i,j \in J$ and $\forall k \not \in J$ then $n\_{i,j,k} = 0$. Let $\langle i \rangle$ be the smallest subsystem containing $i$.
**Question**: Let $i,j \in I$ such that $\langle i \rangle = \langle j \rangle$. Is it true that $\mathbb{Z}[d\_i] = \mathbb{Z}[d\_j]$?
*Observation*: The vector $v = (d\_k)\_{k \in I}$ is a common eigenvector for the matrices $M\_i = (n\_{i,j,k})\_{j,k \in I}$, with eigenvalue $d\_i$, because: $$ M\_i v = (\sum\_{k \in I} n\_{i,j,k} d\_k)\_{j \in I} = (d\_i d\_j)\_{j \in I} = d\_i v.$$ By Frobenius-Perron Theorem (stated below), the eigenvalue $d\_i$ of $M\_i$ is its [spectral radius](https://en.wikipedia.org/wiki/Spectral_radius).
*[Frobenius-Perron Theorem](https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem)*: A square matrix $M$ with nonnegative real entries has a non-negative real eigenvalue. If moreover $M$ has an eigenvector $v$ with strictly positive entries, then the eigenvalue of $v$ is the largest non-negative real eigenvalue and is the spectral radius of $M$.
| https://mathoverflow.net/users/34538 | Positive system of algebraic integers | No, here are infinitely many counter-examples for $I = \{1,2\}$.
Let $n,m \in \mathbb{Z}\_{\ge 1}$ such that $m | n^2$, $n | 2m^2$ and $n \neq 2m$ (e.g. $n=m=1$).
Take $d\_1 = n(1+\sqrt{2})$, $d\_2 = m(3+2\sqrt{2})$. Here are the matrices $M\_i = (n\_{i,j,k})\_{j,k \in I}$:
$$\left(\begin{matrix}0&\frac{n^2}{m}\\m&2n \end{matrix}\right), \
\left(\begin{matrix}m&2n\\\frac{2m^2}{n}&5m\end{matrix}\right)$$
Now $\langle 1 \rangle = \langle 2 \rangle = I$, but $\mathbb{Z}[d\_1] = \mathbb{Z}[n \sqrt{2}] \neq \mathbb{Z}[2m \sqrt{2}] = \mathbb{Z}[d\_2]$, because $n \neq 2m$.
---
Here is a counter-example which is also a *fusion ring* (the initial motivation), with $I = \{1,2,3\}$.
Take $d\_1 = 1$, $d\_2 = 3+2\sqrt{2}$, $d\_3 = 4+3\sqrt{2}$. Here are the matrices $(M\_i)$:
$$\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1 \end{matrix}\right), \
\left(\begin{matrix}0&1&0\\1&0&4\\0&4&3 \end{matrix}\right), \
\left(\begin{matrix}0&0&1\\0&4&3\\1&3&6 \end{matrix}\right)$$
Now $\langle 2 \rangle = \langle 3 \rangle = I$, but $\mathbb{Z}[d\_2] = \mathbb{Z}[2\sqrt{2}] \neq \mathbb{Z}[3\sqrt{2}] = \mathbb{Z}[d\_3]$.
For the notion of **fusion ring**, see Definition 3.1.7 in the following reference:
P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik; Tensor categories. Mathematical Surveys and Monographs (2015) 205.
---
*Conclusion*: We should replace $\mathbb{Z}[d\_i] = \mathbb{Z}[d\_j]$ by $\mathbb{Q}(d\_i) = \mathbb{Q}(d\_j)$ in the question.
| 2 | https://mathoverflow.net/users/34538 | 435917 | 176,183 |
https://mathoverflow.net/questions/434790 | 5 | Kechris in his *Classical Descriptive Set Theory* book gives the following definition (Definition 24.1) and characterization (Theorem 24.15) of Baire class $1$ functions:
>
> **Definition.** Let $X,Y$ be metrizable spaces. A function $f:X\rightarrow Y$ is of Baire class $1$ if $f^{-1}(U)$ is $F\_\sigma$ for every open set $U\subseteq Y$.
>
>
> **Theorem.** (Baire) Let $X$ be a Polish space, $Y$ separable metrizable and $f:X\rightarrow Y$. Then the following are equivalent:
>
>
> 1. $f$ is of Baire class $1$.
> 2. $f{\upharpoonright} K$ has a point of continuity for every compact $K\subseteq X$.
>
>
>
When I first read this theorem in his book, I suspected its statement differed from Baire's original formulation, but I also thought it was the current and commonly accepted one. But reading some recent papers (mainly in real analysis and general topology) dealing with this class of functions, I discovered that there are *a lot* of different definitions and characterization theorems in circulation.
For example (I'm leaving out many others) I've seen Baire class $1$ functions being defined as pointwise limits of sequences of continuous functions (this definition and Kechris' are equivalent if $X$ is $0$-dimensional or $Y=\mathbb{R}$), and also I've also read a variety of "Baire's characterization" theorems in which $Y$ is assumed to be some kind of topological vector space. Lastly, I've noticed that every time Baire's characterization theorem is stated differently from Kechris' presentation, it's either without a reference or with a more than hundred years old' one.
Is there a paper giving some semblance of order to all these definitions and theorems going under the same name?
Thanks!
| https://mathoverflow.net/users/141146 | Baire class $1$ functions and Baire's characterization theorem | I had a similar problem some time ago and I had the impression that a general review on Baire classes is in fact missing, and would be useful.
Probably you already saw it, but in [this](https://www.tau.ac.il/%7Etsirel/Courses/MeasCategory/lect5.pdf) chapter there is at least a discussion of how the definition through $F\_\sigma$ sets originated from the classical definition as pointwise limit, and it covers the characterization theorem.
Other titles I found useful:
* A.C.M. Van Rooij & W.H. Schikhof, A Second Course On Real
Functions, Cambridge University Press, 1982.
(If I remember well, the only place where I found an elementary proof that real functions with countably many discontinuities are Baire 1).
* S. Fung, Functions Of Baire Class One, University of California San
Diego, Department of Mathematics, May 2013.
* D.M. Bressoud, A Radical Approach to Lebesgue's Theory of Integration
(especially pages 109-119, but in general an interesting reading from a historical point of view).
| 3 | https://mathoverflow.net/users/167834 | 435918 | 176,184 |
https://mathoverflow.net/questions/435919 | 31 | In 2002, the discovery of the AKS algorithm proved that it is possible to determine whether an integer is prime in polynomial time deterministically. However, it is still not known whether there is an algorithm for factoring an integer in polynomial time.
To me, this is the most counter-intuitive observation in mathematics. If one can know for certain that a given integer is composite, why is it apparently so difficult to find its factors? Why doesn’t knowing that something exists give one a recipe for determining what that something is?
One way to resolve this problem would be to find a polynomial time algorithm to factor integers. However, if this were possible, it appears that a completely new idea would be needed to do so.
My question is is there an example of a problem similar to integer factorization in which it has been proven that an algorithm can determine the existence of a certain entity in polynomial time but there is no algorithm that computes that entity in polynomial time?
| https://mathoverflow.net/users/7089 | Why is integer factoring hard while determining whether an integer is prime easy? | What I think you're asking for are examples of **search** problems that seem to be hard, while a corresponding **decision** problem is solvable in polynomial time (but not totally trivial). It is true that such problems do not arise in practice very often; typically, an efficient decision procedure can be turned into an efficient search. For example, if you can determine whether or not an arbitrary SAT instance is satisfiable, then you can find satisfying assignments easily, just by taking each variable in turn, trying the two possible settings in turn (TRUE or FALSE), and asking if the smaller instance is satisfiable. Or, for an optimization problem, if you can solve the decision problem ("is there a solution with cost at most $k$?") then you can find the optimum value by performing a binary search on $k$.
You might find examples of what you're looking for on the CS Theory StackExchange: [Easy decision problem, hard search problem](https://cstheory.stackexchange.com/questions/1848/easy-decision-problem-hard-search-problem). But perhaps none of the examples there is as convincing as factorization.
It should be pointed out, however, that "the decision version of factorization" is (arguably) not primality testing, but the following problem:
>
> Given a positive integer $n$ and a bound $k$, does there exist $p$ ($1 < p < k$) such that $p\mid n$?
>
>
>
A fast algorithm for this decision problem would indeed yield a fast algorithm for factoring. So arguably, what's special about factoring is that there is "a" decision problem (primality testing) that looks very close to "the" decision problem for factoring, but which seemingly cannot be parlayed into a solution to "the" decision problem. Stated this way, it's perhaps less surprising that there appears to be a computational gap between the two decision problems. An analogy here might be [subgraph isomorphism](https://en.wikipedia.org/wiki/Subgraph_isomorphism_problem), which is $\mathsf{NP}$-hard, while graph isomorphism appears to be much easier.
| 53 | https://mathoverflow.net/users/3106 | 435925 | 176,188 |
https://mathoverflow.net/questions/434800 | 3 | Let $K$ be a finite extension of the $p$-adic numbers with valuation ring $\mathcal{R}$ and uniformizer $\pi$. Consider a smooth and connected rigid $K$-variety $X=Sp(A)$ and assume that the affine formal model $\mathfrak{X}=Spf(A^{\circ})$ is normal (i.e. $A^{\circ}$ is a normal integral domain). My question is whether the fact that $X$ admits a normal affine formal model is stable under finite extensions of the ground field. That is, let $L$ be a finite extension of $K$ (separable as $K$ is of characteristic zero) with valuation ring $\mathcal{R}\_{L}$ and uniformizer $\omega$. Is it true that $X\_{L}=Sp(A\otimes\_{K}L)$ admits a normal formal model? Letting $A\_{L}=A\otimes\_{K}L$, is $A\_{L}^{\circ}$ a normal ring in general? Under which conditions does it hold that $A^{\circ}\_{L}= A^{\circ}\otimes\_{\mathcal{R}}\mathcal{R}\_{L}$?
Context: By the reduced fiber theorem there is a finite extension $L$ of $K$ such that $A^{\circ}\_{L}$ has geometrically reduced special fiber. As this process introduces roots of $\pi$, the induced morphism $\mathcal{R}\rightarrow \mathcal{R}\_{L}$ will not be étale, just finite flat.
Thus, the morphism $A^{\circ}\rightarrow A^{\circ}\otimes\_{\mathcal{R}}\mathcal{R}\_{L}$ is not neccesarily étale, and $A^{\circ}\otimes\_{\mathcal{R}}\mathcal{R}\_{L}$ is not necessarily normal.
I am interesting in finding out if for every smooth irreducible rigid $K$-variety with a normal affine formal model with geometrically irreducible special fiber there is a finite extension $L$ such that $X\_{L}$ admits an affine formal model with integral geometric fiber.
| https://mathoverflow.net/users/476832 | On the stability of having a normal formal model under finite extensions of the base field | As for your first question, $X\_L$ indeed admits a normal formal model by virtue of normalisation. Whether $A^{\circ}\otimes\_R R\_L$ is already normal (which is equivalent to $A^{\circ}\otimes\_R R\_L=(A\_L)^{\circ}$, as the latter is normal) is less transparant and is related to issues of wild ramification.
Here is an example to show such base change may fail to be normal. The elliptic curve $y^2=x^3+2$ over $\mathbb{Q}\_2$ has bad reduction, but has good reduction over the wild Kummer extension $\mathbb{Q}\_2(\sqrt{2})$. Over $\mathbb{Q}\_2$ the minimal normal crossings model contains a component of multiplicity 2, above which the base change has a non-normal component. So to obtain a counterexample we can take formal completion and let $\mathrm{Spf}(A)$ be some affine piece containing this multiplicity 2 component.
Using logarithmic geometry one may prove there is no obstruction in certain 'tame' situations: suppose $A^{\circ}/R$ is log regular (for instance coming from a snc degeneration as in the above example), where we equip $A^{\circ}$ and $R$ with the standard log structure of nonzero elements. Then if $A^{\circ}$ is log smooth or $R\_L/R$ is log étale (i.e. $L/K$ is tame), we have that $A^{\circ}\otimes\_R R\_L$ is log regular and hence normal. Section 3.1 of [this paper](https://arxiv.org/pdf/1403.5538.pdf) may be relevant. Whenever nice normal crossings models are available, components with multiplicity prime-to-$p$ are log smooth and so above these there is no problem. In particular, if you're only concerned with pieces of semistable formal models of algebraic varieties, all is good.
| 1 | https://mathoverflow.net/users/93776 | 435938 | 176,191 |
https://mathoverflow.net/questions/435322 | 1 | $G$ is a finite solvable group. Let $\{P\_{1}, P\_{2}, \dotsc , P\_{s}\}$ be a Sylow basis of $G$. We have that $G=P\_{1}P\_{2}\dotsm P\_{s}$. Set
\begin{equation}
\begin{aligned}
%% The alignment is never used ….
T=\prod\limits\_{t=1}^{s-1}P\_t,
H=\prod\limits\_{k\neq3}^sP\_k,
K=\prod\limits\_{r\neq2}^sP\_r.\nonumber
\end{aligned}
\end{equation}
Suppose that $T$ is nilpotent (i.e. $T=P\_{1}\times P\_{2}\times P\_{3}\times \dotsb \times P\_{s-1}$), $N\_H(P\_s)=P\_s$ and $N\_K(P\_s)=P\_s$.
Can we get that $N\_G(P\_s)=P\_s$?
| https://mathoverflow.net/users/478453 | The property of self-normalizing subgroup | Yes. Nilpotency of $T$ is not needed. Assume that $N\_G(P\_s)>P\_s$. Then for some $i<s$, $N\_G(P\_s)$ has a
Sylow $p\_i$-subgroup $Q\_i\ne 1$. Assume without loss that $p\_i$
divides $|H|$. Then $P\_iP\_s$ is a Hall $\{p\_i,p\_s\}$-subgroup of $H$
and $G$. By Hall's extension of Sylow's theorems to solvable groups,
$(Q\_iP\_s)^g\le P\_iP\_s$ for some $g\in G$. Then $P\_s$ and $P\_s^g$ are
Sylow $p\_s$-subgroups of $P\_iP\_s$, so for some $k\in P\_iP\_s$,
$P\_s^{gk}=P\_s$. Now $Q\_i^{gk}P\_s=(Q\_iP\_s)^{gk}\le P\_iP\_s\le H$. But
$Q\_i$ normalizes $P\_s$, so $Q\_i^{gk}$ normalizes
$P\_s^{gk}=P\_s$. Therefore $Q\_i^{gk}$ is a non-identity $p\_i$-subgroup
of $N\_H(P\_s)=P\_s$, an impossibility.
| 2 | https://mathoverflow.net/users/99221 | 435943 | 176,193 |
https://mathoverflow.net/questions/435941 | 1 | I am very confused by a sum I have been trying to solve analytically/ numerically for a long time. It comes from the idea of a physical problem where the observation is made that has a combined response of a number of entities. For example, I want to evaluate the mathematical sum at a observation point $\omega$ that looks like the following when $M \to \infty$ .
$$ L(\omega) = \sum\_{n = 0}^{N-1}\sum\_{m = 1}^{M} \exp\left(i \left( (\omega\_m - \omega)n + \beta\_m \right) \right) $$
Where all $\omega\_m$s are random draws from a normal distribution
$$ \omega\_m \sim \mathcal{N}(\mu, \sigma^2) $$
and the $\beta\_m$ are the normal draws from a uniform distribution
$$ \beta\_m \sim \mathcal{U}[-\pi, +\pi] $$
Let's try to solve it first with the sum with respect to $m$ and then $n$. The sum with respect to $m$ can be approximated to an infinite integral when $M \to \infty$.
$$ \mathbb{E}(L(\omega)) \approx M \sum\_{n = 0}^{N-1} \int\_{-\infty}^{+\infty} \int\_{-\pi}^{+\pi} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\exp(i(x - \omega)n)\frac{1}{2\pi} \exp(i\beta) d\beta dx $$
This integral is $0$ because of the integral $\int\_{-\pi}^{+\pi}\exp(i\beta) d\beta$.
Let's approach this sum first with respect to $n$ and then $m$. The function has a closed form with respect to the sum with $n$.
$$ L(\omega) = \sum\_{m = 1}^{M} \frac{\sin\left( \frac{N (\omega\_m-\omega)}{2} \right)}{\sin\left( \frac{(\omega\_m-\omega)}{2} \right)} \exp\left(i \left( (1-N)\frac{\omega\_m-\omega}{2} - \beta\_m \right) \right)$$
If I take the expectation here with an integral approximation, it also becomes $0$. However, I proceeded with finding the expression for the absolute value of $L(\omega)$ from the above expression.
$$ |L(\omega)|^2 = \sum\_{m = 1}^{M} \left| \frac{\sin\left( \frac{N (\omega\_m-\omega)}{2} \right)}{\sin\left( \frac{(\omega\_m-\omega)}{2} \right)} \right|^2 + \sum\_{p = 1}^{P} \sum\_{q = 1}^{Q} \frac{\sin\left( \frac{N (\omega\_p-\omega)}{2} \right)}{\sin\left( \frac{(\omega\_p-\omega)}{2} \right)} \frac{\sin\left( \frac{N (\omega\_q-\omega)}{2} \right)}{\sin\left( \frac{(\omega\_q-\omega)}{2} \right)} \cos\left( (1 - N) \frac{(\omega\_p-\omega\_q)}{2} + \beta\_q - \beta\_p \right) $$. The sum with $p$ and $q$ are similar to $m$. The second term is clearly $0$ based on the same type of approximations with expected value when $M \to \infty$. So, taking only the first term, the expected value becomes,
$$ |L(\omega)|^2 \approx M \int\_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) \left| \frac{\sin\left( \frac{N (x-\omega)}{2} \right)}{\sin\left( \frac{(x-\omega)}{2} \right)} \right|^2 dx $$
Can I further reduce this to a closed form or a form that can only depend on $N$ numerically?
Let's take a simpler form by choosing $\mu = 0$, $\sigma = 1$ and find the integral at $\omega = 0$.
$$ |L(0)|^2 \approx M \int\_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{(x)^2}{2}\right) \left| \frac{\sin\left( \frac{N (x)}{2} \right)}{\sin\left( \frac{(x)}{2} \right)} \right|^2 dx $$
| https://mathoverflow.net/users/489481 | A double sum with complex numbers having stochastic variables | Your final integral can be readily evaluated by expanding the fraction of sines into sums of exponentials $e^{ikx/2}$ with integer $k$, and integrating term by term with the Gaussian weight, to arrive at
$$|L(0)|^2 \equiv\frac{M}{\sqrt{2\pi}} \int\_{-\infty}^{+\infty} e^{-x^2/2} \frac{\sin^2\left(Nx/2 \right)}{\sin^2\left( x/2 \right)} \, dx=$$
$$=MN+2M e^{-N^2/2} \sum \_{k=1}^{N-1} k e^{\frac{1}{2} \left(2 k N-k^2\right)}.$$
| 1 | https://mathoverflow.net/users/11260 | 435951 | 176,196 |
https://mathoverflow.net/questions/435947 | 7 | I don't know whether this is the right place to discuss a part of someone's thesis or not. If it is wrong, let me know; I will delete my post.
I am reading this [thesis](https://trace.tennessee.edu/cgi/viewcontent.cgi?article=8804&context=utk_graddiss).
Corollary 4.1.15. on page 63 says that if the number of ends (see Definition 4.1.6. of the above thesis) of Freudenthal space (i.e., a $\sigma$-compact, locally compact, connected, and locally connected space) is infinite, then its space of ends is homeomorphic to the Cantor set.
As far as know, the space of ends of a non-compact surface can be any closed subset of the Cantor set; see Theorem 2 of [this](https://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf). For example, the space of ends of $\Bbb R^2\setminus \Bbb N$ is homeomorphic to the space $\{1/n\}\cup\{0\}$, which is an infinite set but certainly not homeomorphic to the Cantor set.
So, my question is the following:
>
> Are these two notions of the space of ends (one coming from
> Freudenthal compactification and another, as mentioned in Ian Richard's
> paper) different?
>
>
>
| https://mathoverflow.net/users/363264 | If the number of ends of Freudenthal space is infinite, then its space of ends is homeomorphic to the Cantor set? | Of course, yourself and YCor already answered this in the comments, but since I see you tagged geometric group theory, maybe you will be interested in this explicit answer about bi-Holder homeomorphisms : The end boundary of an accessible infinitely-ended group is bi-Hölder equivalent to the standard Cantor ternary set if and only if it is virtually free. This is Corollary 1.9 of this paper : <https://arxiv.org/abs/2010.07671>. The involved distance is the visual distance with respect to a base-point.
| 2 | https://mathoverflow.net/users/111917 | 435961 | 176,197 |
https://mathoverflow.net/questions/435858 | 4 | The context for this question comes from [this arxiv preprint](https://arxiv.org/abs/2210.07753). Specifically, a remark in the final proof of the paper. To make the question more self-contained, I'll phrase this question in a slightly more general setting.
Fix a category $A$ and a combinatorial model category $\mathcal{M}$ where all objects are cofibrant [1]. Given a pair objects $X : \mathcal{M}$ and $a : A$, we define $a \otimes X$ to be the left Kan extension of $X$ along the inclusion $\{a\} \to A$. Explicitly then,
$$
(a \otimes X)(a') = \hom(a,a') \cdot X
$$
**The question:** Endowing $\mathbf{Fun}(A,\mathcal{M})$ with the projective model structure, can an arbitrary functor $F : A \to \mathcal{M}$ be built out of an (iterated) homotopy colimit of $a \otimes X$s?
As a first step, we can use the classical Yoneda formula to realize $F$ as $\int^{a : A} a \otimes F(a)$, so $F$ can certainly built as the coequalizer of coproducts of presheaves of the form $a \otimes X$. I'm unsure how to argue that this is the homotopy colimit of some functor. It seems natural enough to extend this coequalizer formula to a simplicial object:
$$
G(n) = \coprod\_{a\_0 \to \dots \to a\_n} a\_0 \otimes F(a\_n)
$$
This then appears to be a cofibrant diagram with respect to the Reedy model structure on simplicial objects of $\mathbf{Fun}(A, \mathcal{M})$ whose colimit is $F$. However, my understanding is that this alone does *not* imply that $\mathsf{hocolim}\,G = F$ since $\Delta : \mathbf{Fun}(A, \mathcal{M}) \to \mathbf{Fun}(\Delta^\mathsf{op}, \mathbf{Fun}(A, \mathcal{M}))$ is not necessarily a right Quillen functor.
Is it still the case that $\mathsf{hocolim}\,G = F$? Or do I need to search for a different $G$ entirely? I asked a similar question on a different forum and was adviced to use the Bousfield-Kan formula. This certainly appears to be quite related but I'm not quite sure how to join the pieces together.
[1]: For the actual example, $\mathcal{M}$ is $\mathbf{sSet}^\sharp\_{/B}$ with the Cartesian model structure.
| https://mathoverflow.net/users/76636 | Decomposing a $\mathcal{M}$-valued presheaf into a homotopy colimit of representables | The diagram $G$ is a projectively cofibrant diagram because
nondegenerate simplices split off as a coproduct summand in every degree
and every simplicial level is a coproduct of (enriched) representables.
The homotopy colimit of a projectively cofibrant diagram can be computed as its colimit.
Therefore, the homotopy colimit of $G$ can be computed as $F$.
| 2 | https://mathoverflow.net/users/402 | 435965 | 176,199 |
https://mathoverflow.net/questions/434451 | 23 | I want to follow the discussion from [here](https://mathoverflow.net/questions/191016/all-retracts-are-closed-as-separation-axiom) concerning about the strength of the separation "all retract subspaces are closed".
(A retract subspace of a topological space $X$ is a subspace $A$ where there exists continuous $f: X\to A$ such that $f|\_A = \mathrm{id}\_A$.)
Write $\mathrm{KC}$ as "all compact subsets are closed", and $\mathrm{RC}$ as "all retract subspaces are closed". We have $T\_2\Rightarrow \mathrm{KC}$, $T\_2\Rightarrow \mathrm{RC}$ and neither of the reversed implications. The cocountable topology over $\mathbb{R}$ is a $\mathrm{KC}$ example not $\mathrm{RC}$: $f(x) = |x|$ is a continuous function from $\mathbb{R}$ with cocountable topology to its subspace $[0,+\infty)$ since the preimages of countable sets are countable, but $[0,+\infty)$ is certainly not closed.
So I would like to ask if $\mathrm{RC}$ implies $\mathrm{KC}$, because the only $\mathrm{RC}$ non-Hausdorff spaces I know are those compact $\mathrm{KC}$ spaces (note that compact $\mathrm{KC}$ implies $\mathrm{RC}$). Any help appreciated.
| https://mathoverflow.net/users/494268 | "All retracts are closed" and "all compacts are closed" | RC does not imply KC: in [this paper](https://arxiv.org/abs/2211.12579) Banakh and Stelmakh construct a semi-Hausdorff countable Brown space $X$ which is strongly rigid (and hence $X$ has $RC$) and contains a non-closed compact subset (so, $X$ fails to have $KC$).
This example also shows that $KC$ does not follow from the semi-Hausdorff property, which is intermediate between $T\_1$ and $T\_2$.
| 6 | https://mathoverflow.net/users/61536 | 435973 | 176,204 |
https://mathoverflow.net/questions/435893 | 3 | Let $\varphi := \frac{\sqrt{5}-1}{2}$ be the golden ratio, and $H(x):=-x\log\_2 x -(1-x) \log\_2(1-x)$ be the binary entropy function for a Bernoulli random variable.
Show that for all $\delta > 0$, one can choose sufficiently small $\gamma > 0$, such that if $\mu$ is a probability measure on $[0, 1]$, with
\begin{align\*}
\mathbb{E}\_{p \sim \mu}(p) \geq \varphi - \gamma
\end{align\*}
and
\begin{align\*}
\mathbb{E}\_{(p, q) \sim \mu \times \mu} H(pq) < \mathbb{E}\_{p \sim \mu} H(p),
\end{align\*}
then
\begin{align\*}
\mu(p \leq \frac12) < \delta.
\end{align\*}
This is true when $\gamma$ is allowed to be $0$: in that case, it has been [shown](https://arxiv.org/abs/2211.11504) that $\mu = \delta\_\varphi$.
This comes up as a key estimate in Will Sawin's [proposed proof](https://arxiv.org/abs/2211.11504) that the lower bound of Frankl's union closed set conjecture can be pushed to $\frac{3 - \sqrt{5}}{2} + \delta$ for some small $\delta > 0$; see also Justin Gilmer's [original breakthrough paper](https://arxiv.org/abs/2211.09055).
Update: I believe (could be wrong) following the linearization approach [here](https://arxiv.org/abs/2211.11504), it suffices to show the result for $\mu$ of the form $u \delta\_0 + v \delta\_{1/2} + w \delta\_y$, with $y > \varphi - \gamma$ and $u + v + w = 1$. However I cannot get any quantitative estimate on $\gamma$ for a given $\delta$.
| https://mathoverflow.net/users/4923 | A variational estimate related to the union closed set conjecture | Actually I don't see what your trouble is then. Switching to natural $\log$ (which is just scaling of $H$), you have the key inequality
$$
\varphi H(x^2)\ge xH(x)
$$
(ask Zachary Chase if you want to see a reasonably short and almost computation-free complex analysis proof of it) with equality only for $x=0,\varphi,1$.
Now I want to upgrade it to
$$
2\varphi H(xy)\ge xH(y)+yH(x)\,.
$$
Then, taking the expectations, we will get what we want if we take into account that $xy$ is far away from $\varphi^2$ with probability $\delta^2$ (I'm not shooting for the best dependence at this point).
To get this extended version, fix $xy$. Then the RHS is
$$
xy\log\frac 1y+yx\log\frac 1x+x(1-y)\log\frac 1{1-y}+y(1-x)\log\frac 1{1-x}
$$
The sum of the first two terms is $xy\log \frac 1{xy}$, so it is completely determined by the product $xy$.
Now, expand
$$
y(1-x)\log \frac 1{1-x}=y(1-x)\left(x+\tfrac 12 x^2+\tfrac 13 x^3+\tfrac 14 x^4+\dots\right)
\\
=yx\left(1-\frac 1{1\cdot 2}x-\frac 1{2\cdot 3}x^2-\dots\right)
$$
and similarly for the other term.
Then the sum of the two troublesome terms is
$$
xy\left(1-\sum\_{k=1}^\infty \frac 1{k(k+1)}(x^k+y^k)\right)
$$
but, for fixed product $xy$, the sum $x^k+y^k$ is minimized with $x=y$, so the two-variable inequality immediately follows from the one-variable one.
*Edit: the full endgame*
The main idea in analyzing the stability in the inequalities is the following. Suppose we have some continuous function $F(x)$ on a compact set (of any nature) that is non-negative and vanishes only at finitely many points $x\_1,\dots,x\_n$. If we invent *any* continuous function $G(x)$ ("gain") that also vanishes at these points and such that $G(x)/F(x)$ stays bounded in a small neighborhood of each of the points, then $G/F$ is bounded on the entire compact, i.e., there exists $c>0$ with $F\ge cG$ everywhere.
Now I want to write the chain of inequalities
$$
2\varphi H(XY)\ge 2\sqrt{XY}H(\sqrt{XY})\ge XH(Y)+YH(X)\,.
$$
Denoting $Z=\sqrt{XY}$, we can easily analyze $F(Z)=\varphi H(Z^2)-ZH(Z)$ at its only zeroes $0,\varphi,1$. We have
$$
F(Z)=(2\varphi-1)Z^2\log\frac 1Z+O(Z^2), \quad Z\to 0;
\\
F(Z)=(2\varphi-1)(1-Z)\log\frac 1{1-Z}+O(1-Z), \quad Z\to 1;
\\
F(Z)=C(Z-\varphi^2)^2+O(|Z-\varphi|^3), \quad Z\to\varphi
$$
(I'm too lazy to compute $C>0$ explicitly, but it is not required: the proof of the key inequality you can obtain from Zachary (or, possibly, even from me if I still have it in the mess on my desk) shows that $\varphi$ is a regular zero of the second order and the function is globally non-negative, so $C>0$ is automatic). Thus, if I invent absolutely anything that is bounded and matches these asymptotic behaviors up to a constant factor, I will have it as a lower bound with some coefficient. So, I just write
$$
G(Z)=Z^2(1-Z)(Z-\varphi^2)[\log\tfrac 1Z+\log\tfrac 1{1-Z}]
$$
and have
$$
2\varphi H(XY)\ge 2\sqrt{XY}H(\sqrt{XY})+aG(\sqrt{XY})
$$
with some numeric constant $a>0$.
Now passing from one variable to two is even simpler. Note that when we discussed why the diagonal was the worst possible case, we said that $x^k+y^k$ is minimized for fixed $xy$ for $x=y$, i.e., for $k=1$ used the inequality $x+y\ge 2\sqrt{xy}$. Using (for $k=1$ only) the identity
$x+y=2\sqrt{xy}+(\sqrt x-\sqrt y)^2$, we arrive at
$$
2\sqrt{XY} H(\sqrt{XY})\ge XH(X)+YH(Y)+\frac 12XY(\sqrt X-\sqrt Y)^2\.
$$
Thus, in the whole chain, we can claim the gain
$$
\widetilde G(X,Y)=XY(1-\sqrt{XY})(\sqrt{XY}-\varphi)^2[\log \tfrac 1{\sqrt{XY}}+\log \tfrac 1{1-\sqrt{XY}}]+XY(\sqrt X-\sqrt Y)^2.
$$
Now fix $\delta>0$ and let $\gamma$ be very small. Then we can write
$$
2\varphi H(XY)\ge XH(Y)+YH(X)+2a\widetilde G(X,Y)
$$
with a numeric $a>0$. Taking the expectations, we obtain
$$
2\varphi EH(XY)\ge 2\varphi EH(X)-2\gamma EH(X)+a E\widetilde G(X,Y)\,.
$$
Thus, it suffices to show that under your assumptions,
$$
E\widetilde G(X,Y)\ge a(\delta)E H(X)
$$
with $a(\delta)$ depending on $\delta$, but not on $\gamma$ for $\gamma<\gamma(\delta)$.
Now, your assumptions imply that $P(Y>\varphi+b\delta)>b\delta$ for some explicit numeric $b>0$. Indeed, otherwise we would have
$$
EY\le \frac 12\delta+\varphi(1-\delta)+2 b\delta
$$
so if $3b<\varphi-\tfrac 12$, we'll have a $b\delta$ drop from $\varphi$ for which $\gamma$ would not be able to compensate.
We have for $Y>\varphi+b\delta$
$$
\widetilde G(X,Y)\ge
\begin{cases}
{\rm const} X\log\frac 1X, & X<\frac 14 \\
{\rm const}\delta^2, & \frac 14\le X\le \varphi+\frac b2\delta\\
{\rm const}\delta^2 (1-X)\log\frac 1{1-X}, & \varphi+\frac b2\delta\le X\le 1
\end{cases}
$$
(in the middle case the first term in $\widetilde G$ is useless and the whole estimate comes from the second term; otherwise the second term is ignored).
So, we have at least ${\rm const} \delta^2 H(X)$ in all cases and
$$
E\widetilde G(X,Y)\ge {\rm const} \delta^2 EH(X)P(Y>\varphi+b\delta)\ge {\rm const} \delta^3 EH(X)
$$
and we are done.
Some estimates here can be improved, but you requested just a qualitative result, so I didn't bother to chase the best constants and powers: you and Will will easily figure that stuff now once you know that the desired result is possible.
| 5 | https://mathoverflow.net/users/1131 | 435977 | 176,208 |
https://mathoverflow.net/questions/435572 | 5 | The following question comes from a typo in an old notebook of mine (I changed what I was calling my forcing notion partway through writing the definition of properness):
Say that a forcing $\mathbb{P}$ is **pseudoproper** iff for all sufficiently large $\theta$, all countable $M\prec H\_\theta$, and all $\mathbb{P}$-generic-over-$V$ filters $G$, there is some $\mathbb{Q}\in M$ and some $\mathbb{Q}$-generic-over-$M$ filter $\hat{G}$ such that $M[\hat{G}]=M[G\cap M]$.
For example, the (definitely non-proper) Levy collapse $Col(\omega,\omega\_1)$ is pseudoproper. Roughly, given $M$ appropriate, let $\mathbb{Q}$ be the forcing consisting of all (in the sense of $M$) finite partial maps $p:\omega\rightarrow\omega\_1\cup\{\perp\}$ which never take on the same ordinal value twice but are allowed to re-use $\perp$. If $G$ is $Col(\omega,\omega\_1)$-generic over $V$, then $G\cap M$ gives a partial map $\omega\rightarrow\omega\_1$, and "filling in the gaps with $\perp$" gives a $\mathbb{Q}$-generic-over-$M$ (indeed, over $V$) filter $\hat{G}$ which is appropriately equivalent to $G\cap M$.
It's easy to see that pseudoproperness is not preserved by equivalence of forcing notions; a more "annotated" version of the Levy collapse fails to be pseudoproper. This suggests the following strong antithesis of pseudoproperness. Say that a forcing $\mathbb{P}$ is **rude** if there is no pseudoproper $\hat{\mathbb{P}}$ such that $(i)$ forcing with $\mathbb{P}$ adds a $\hat{\mathbb{P}}$-generic and $(ii)$ forcing with $\hat{\mathbb{P}}$ adds a $\mathbb{P}$-generic.
>
> **Question**: Do rude forcings exist?
>
>
>
| https://mathoverflow.net/users/8133 | Highly improper forcings | If we allow only for separative forcings $\hat{\mathbb{P}}$ in the definition of rudeness then all non-trivial separative $\sigma$-closed forcings are rude. This does not answer your question literally, but hopefully in spirit.
EDIT: This restriction is not necessary after all, see below.
Suppose that $\mathbb P$ is some non-trivial separative $\sigma$-closed forcing and that $\hat{\mathbb{P}}$ is any separative forcing which is forcing equivalent to $\mathbb P$.
>
> Claim: There is an embedding $\tau\colon 2^{<\omega}\rightarrow\hat{\mathbb{P}}$ so that the pointwise image of any branch through $2^{<\omega}$ has a lower bound in $\hat{\mathbb{P}}$.
>
>
>
This is straightforward to arrange for $\mathbb P$ but to do it for $\hat{\mathbb{P}}$ requires a little bit of care. First as both $\hat{\mathbb{P}}$, $\mathbb P$ are separative forcings, we can consider them as dense subsets $\hat{\mathbb{P}}\subseteq \mathbb B$, $\mathbb P\subseteq\mathbb C$ of complete Boolean algebras. Now $\mathbb B$ and $\mathbb C$ are forcing equivalent so that there are $b\in\mathbb B$ and $c\in\mathbb C$ and an isomorphism $\pi\colon\mathbb B\upharpoonright b\rightarrow\mathbb C\upharpoonright c$. Now choose some $p\_\emptyset\in \hat{\mathbb{P}}$, $p\_\emptyset\leq b$. If $s\in 2^{<\omega}$ and $p\_s$ is defined, then find incompatible $q\_{s^\frown 0}, q\_{s^\frown 1}\in\mathbb P$ so that $q\_{s^\frown i}\leq \pi(p\_s)$ and then further find $p\_{s^\frown i}\in\hat{\mathbb{P}}$ with $\pi(p\_{s^\frown i})\leq q\_{s^\frown i}$ for $i<2$. This completes the construction. Set $\tau(s)=p\_s$. Clearly if $b$ is any cofinal branch through $2^{<\omega}$ then $\pi[\tau[b]]$ is interlaced with a descending sequence in $\mathbb P$ and thus has a lower bound.
It follows that $\hat{\mathbb{P}}$ cannot be pseudoproper: Suppose $\theta$ is sufficiently large and $M\prec H\_\theta$ is countable with $\mathbb P,\tau\in M$. Pick a real $x\in 2^{\omega}$ which codes an ordinal above the ordertype of $M\cap\mathrm{Ord}$ and let $p\in\mathbb P$ be a lower bound of $\tau[\{x\upharpoonright n\mid n<\omega\}]$. $p$ is not in $M$, but if $G$ is $\hat{\mathbb{P}}$-generic with $p\in G$ then $M[G\cap M]$ can define $x$. But $x$ is not definable over $M[\hat{G}]$ whenever $\mathbb Q\in M$ and $\hat{G}$ is $\mathbb Q$-generic-over-$M$ (assuming that $H\_\theta$ is a model of sufficiently much of $\mathrm{ZFC}$).
EDIT: Now lets show that there is no pseudoproper $\hat{\mathbb P}$ forcing equivalent to $\mathbb P$ at all: Suppose $\hat{\mathbb P}$ were such a forcing. There always is a some separative $\hat{\mathbb P}'$ and some surjection $\rho:\hat{\mathbb P}\rightarrow \hat{\mathbb P}'$ which preserves the order and so that $p ,q $ are compatible in $\hat{\mathbb P}$ iff $\rho(p), \rho(q)$ are compatible in $\hat{\mathbb P}'$. As before let $M\prec H\_\theta$ with $\hat{\mathbb P}, \hat{\mathbb P}', \mathbb P, \rho, \tau\in M$, $x$ be a complicated real, $p\in\mathbb P'$ below $\tau[\{x\upharpoonright n\mid n<\omega\}]$ and $G'$ $\hat{\mathbb P}'$-generic with $p\in G'$. Then $G:=\rho^{-1}[G']$ is $\hat{\mathbb P}$-generic and $M[G\cap M]$ can compute $G'\cap M=\rho[G\cap M]$, so we get a contradiction as before.
| 4 | https://mathoverflow.net/users/125703 | 435998 | 176,213 |
https://mathoverflow.net/questions/435798 | 2 | Let $K$ be a number field, $\mathfrak{p}$ be a prime of it, and $L=K(\mathfrak{p}^n)$ be the ray class field of $K$ with finite conductor $\mathfrak{p}^n$ (we do not care about the infinite part of the conductor).
1. Is it true that $L/K$ is totally ramified at $\mathfrak{p}$?
2. (Here we may assume $L/K$ is cyclic.) Is it true that the order of the zeroth Tate cohomology group $\smash{\hat{H}}^0(\operatorname{Gal}(L/K), U\_L)$ is equal to $2^t$, where $U\_L$ is the group of units of the ring of integers of $L$ and $t$ is the number of infinite places of $K$ which ramified in $L$?
| https://mathoverflow.net/users/149579 | Ramification of primes and order of $\smash{\hat{H}}^0$ in ray class fields with one finite prime divisor | The answer to Question 1 is negative. Take $K = {\mathbb Q}(\sqrt{-6})$
and let ${\mathfrak p} = (2,\sqrt{-6})$ denote the prime ideal above $2$.
The class number of $K$ is $2$, its maximal unramified (abelian) extension
is $L = {\mathbb Q}(\sqrt{-3},\sqrt{2})$. The ray class number formula shows that $h\{\mathfrak m\} = h \cdot \Phi(\mathfrak m)/(E:E^{(1)})$; here $\Phi$ is Euler's Phi function in $K$. In the present case, $E$ is generated by $-1$, hence the index in the denominator is $2$ for all ideals with norm $> 4$, and we simply have $h\{\mathfrak m\} = \Phi(\mathfrak m)$. We now compute the ray class numbers $h\{\mathfrak p^m\}$:
$$ \begin{array}{c|c}
m & h\{\mathfrak p^m\} \\ \hline
1 & 2 \\
2 & 4 \\
3 & 4 \\
4 & 8
\end{array} $$
Observe that $\Phi(\mathfrak p^m) = \Phi(\mathfrak p) \cdot N(\mathfrak p)^{m-1} = 2^{m-1}$ in our case.
This shows that the ray class field defined modulo $\mathfrak p^2 = (2)$ has conductor $(2)$ (because the ray class field defined modulo $\mathfrak p$ is strictly smaller) and that it contains the Hilbert class field of $K$ with conductor $(1)$.
In Question 2 there are conditions missing. The order $2^t$ is something I would expect for quadratic extensions.
| 2 | https://mathoverflow.net/users/3503 | 436003 | 176,216 |
https://mathoverflow.net/questions/435849 | 4 | Is it true that
$$(v-u)^2+(w-u)^2+(w-v)^2 \\
+\left(\sqrt{\frac{1+u^2}{1+v^2}}
+\sqrt{\frac{1+v^2}{1+u^2}}\right) (w-u)(w-v) \\
-\left(\sqrt{\frac{1+u^2}{1+w^2}}+\sqrt{\frac{1+w^2}{1+u^2}}\right) (w-v)(v-u) \\
-\left(\sqrt{\frac{1+w^2}{1+v^2}}+\sqrt{\frac{1+v^2}{1+w^2}}\right) (v-u) (w-u)>0$$
for any real $u,v,w$ such that $u<0<v<w$?
Certain numerical evidence suggests it is true.
This inequality arose in the [previous answer](https://mathoverflow.net/a/435844/36721).
| https://mathoverflow.net/users/36721 | An algebraic inequality in three real variables | Rewrite the inequality in terms of $a,b,c>1$ via $u=(\frac{1}{a}-a)/2$, $v=(b-\frac{1}{b})/2$, and $w=(c-\frac{1}{c})/2$. Note that $\sqrt{1+u^2}=\frac{1+a^2}{2a}$ and so on. Then the conjectured inequality is equivalent to
\begin{equation}
((c - b)(a + c)(a + b))^2(a^2b^2c^2-a^2bc+ab^2c+abc^2+ab+ac-bc+1)>0.
\end{equation}
Note that the second factor equals
\begin{equation}
a^2bc(bc-1) + abc(b+c-1) + ab + ac +1,
\end{equation}
which is positive since $b,c>1$.
| 4 | https://mathoverflow.net/users/18739 | 436004 | 176,217 |
https://mathoverflow.net/questions/436009 | 3 | Let $T(n,k)$ be a triangle of coefficients such that $T(n,k)\geqslant0$ for $n>0$, $0<k\leqslant n$, $0$ otherwise. Also
$$T(2n+1,1)=\frac{1}{2n+1}, T(2n,1)=0$$
$$T(n,k)=\frac{1}{n}(T(n-1,k-1)+(n-2)(T(n-2,k)+\frac{T(n-2,k-1)}{n-1}))$$
Let
$$P(n,m)=m\sum\limits\_{k=1}^{m}n^{k-1}T(m,k)(-1)^{m+k}$$
I conjecture that
$$P(n,m)=2^{n-1}, 0<n\leqslant m$$
To verify it one may use this pari prog:
```
T(n)=my(v, v1, v2); v=vector(n, i, 0); v[1]=1; v1=v; if(n>1, v[1]=0; v[2]=1/2); v2=v; for(i=3, n, v[1]=if(i%2,1/i); for(j=2, i, v[j]=(v2[j-1]+(i-2)*(v1[j]+v1[j-1]/(i-1)))/i); v1=v2; v2=v); v
P(n, m)=my(A=T(m)); m*sum(k=1, m, n^(k-1)*A[k]*(-1)^(m+k))
```
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Powers of $2$ up to $2^{m-1}$ from a polynomial of degree $m-1$ | This is a fun problem! You start out by describing the conjectured coefficients of $P(n,m)$, but presumably you started out by computing the polynomial which interpolates $2^{n-1}$ and then noticed a pattern in the coefficients. So I'll start in that order: Let $Q(n,m)$ be the unique degree $m-1$ polynomial in $n$ which obeys
$$Q(n,m) = 2^{n-1} \ \text{for} \ 0 < n \leq m.$$
For any $n>0$, we know that $2^{n-1}$ is the sum of the $(n-1)$-st row of Pascal's triangle:
$$2^{n-1} = \sum\_{k=0}^{n-1} \binom{n-1}{k} =\sum\_{k=0}^{n-1} \frac{(n-1)(n-2) \cdots (n-k)}{k!}.$$
If $n \leq m$, then the terms with $k \geq m$ are zero, so we have
$$2^{n-1} = \sum\_{k=0}^{m-1} \frac{(n-1)(n-2) \cdots (n-k)}{k!} \quad 0 < n \leq m. \quad (\ast)$$
The right hand side of $(\ast)$ is clearly a polynomial in $n$ of degree $m-1$, so it is $Q(n,m)$. I'm going to switch the variable $n$ to $x$ for clarity. So
$$Q(x,m) = \sum\_{k=0}^{m-1} \frac{(x-1)(x-2) \cdots (x-k)}{k!}.$$
So
$$Q(x,m) - Q(x,m-1) = \frac{(x-1)(x-2) \cdots (x-m+1)}{(m-1)!}$$
and
$$Q(x,m) - Q(x,m-1) = \frac{x-m+1}{m-1} \left( Q(x,m-1) - Q(x,m-2) \right). \quad (\dagger)$$
Now, put
$$Q(x,m) = \sum\_{k=0}^{m-1} U(m,k) x^k.$$
Comparing the coefficient of $x^{\ell}$ on both sides of $(\dagger)$, we get
$$U(m,\ell) - U(m-1, \ell) = $$
$$\frac{1}{m-1} \left( U(m-1, \ell-1) - U(m-2, \ell-1) \right) - \left( U(m-1, \ell) - U(m-2, \ell) \right).$$
Presumably, it shouldn't be hard to rearrange this into your recurrence; I'm going to stop here.
| 6 | https://mathoverflow.net/users/297 | 436012 | 176,220 |
https://mathoverflow.net/questions/435971 | 8 | Can we prove the consistency of $\mathsf{ZF+SVC}$ + "There is a Reinhardt cardinal?" (Preferably from the consistency of $\mathsf{ZF}$ with a Reinhardt cardinal, but using a stronger assumption is also okay.)
Here $\mathsf{SVC}$ means the *Small Violation of Choice*, claiming the axiom of choice is forcible by a set forcing.
Here are some easy observations.
* Let $\kappa$ be a critical point of $j\colon V\to V$. Then no forcing $\mathbb{P}\in V\_\kappa$ can force $\mathsf{AC}$. Similarly, no forcing $\mathbb{P}\in V\_{j^\omega(\kappa)}$ can force $\mathsf{AC}$ (as $j^n(\kappa)$ is also a critical point of some elementary embedding $V\to V$.)
* If there is a super Reinhardt cardinal $\kappa$, then $\mathsf{SVC}$ fails: if there is a $\mathbb{P}$ forcing $\mathsf{AC}$, then we can pull it back to $V\_\kappa$ by using $j\colon V\to V$ satisfying $j(\kappa)>\operatorname{rank} \mathbb{P}$.
| https://mathoverflow.net/users/48041 | Compatibility of $\mathsf{SVC}$ and Reinhardtness | No, a Reinhardt cardinal implies SVC is false.
First, if there is a Reinhardt cardinal, then by Woodin's proof of the Kunen inconsistency theorem, for sufficiently large regular cardinals $\delta$, the set $S^\delta\_\omega$ of ordinals of cofinality $\omega$ cannot be partitioned into $\delta$-many disjoint stationary sets.
On the other hand, SVC implies that for all sufficiently large regular cardinals $\delta$, every stationary subset of $\delta$ can be partitioned into $\delta$-many stationary sets: to see this, let $\mathbb P$ be a partial order that forces choice, and let $\delta$ be a regular cardinal such that $|\mathbb P| < \delta$ in $V^{\mathbb P}$. Fix any stationary subset $S$ of $\delta$. In $V^\mathbb P$, $S$ remains stationary (by a standard argument, a club in the extension contains a club in the ground model) and there is a partition of $S$ into stationary sets $\langle S\_\alpha\rangle\_{\alpha < \delta}$ by Solovay's theorem. For each $p\in \mathbb P$, let $S^p\_\alpha = \{\xi < \delta : p\Vdash \xi\in S\_\alpha\}$ (fixing a name, etc, etc). We claim there is some $p\in \mathbb P$ such that $A\_p = \{\alpha < \delta : S^p\_\alpha\text{ is stationary}\}$ has cardinality $\delta$. If not, $|\bigcup\_{p\in \mathbb P} A\_p| < \delta$, so there is some $\alpha\in \delta\setminus \bigcup\_{p\in \mathbb P} A\_p$. But then in $V^{\mathbb P}$, $S\_\alpha \subseteq \bigcup\_{p\in \mathbb P} S^p\_\alpha$ is a stationary set which is contained in the union of fewer than $\delta$ nonstationary sets, which is a contradiction. So fix $p$ such that $|A\_p| = \delta$. Then in $V$, we have partitioned $S$ into $\delta$-many stationary sets of the form $S\_\alpha^p$.
| 9 | https://mathoverflow.net/users/102684 | 436013 | 176,221 |
https://mathoverflow.net/questions/435949 | 9 | Let $\mathcal{C}$ be a category with finite limits and a (parameterized) natural numbers object $(N,0,s)$. Let $1$ denote the terminal object of the category. It's easy to show that the following is a coproduct diagram.
$$ 1 \xrightarrow{0} N \xleftarrow{s} N $$
**My question is:** is this coproduct diagram necessarily universal?
---
Background
----------
A coproduct $A+B$ is said to be *universal* if pulling it back along any morphism into $A+B$ gives a coproduct diagram. (See [Introduction to extensive and distributive categories](https://www.sciencedirect.com/science/article/pii/002240499390035R) by Carboni and Walters, definition 2.10.)
Often, natural numbers objects are considered in the context of extensive categories, in which all coproducts are universal (see lemma 2.11 of the same paper). However, I'm curious whether this coproduct is universal even if the ambient category is not assumed to be extensive. I suspect it is not true, since I can't find a simple proof, but I'm not well versed enough to come up with a counter-example.
One reason to ask whether the coproduct is universal is because it allows the definition of certain functions "by cases". If we have an arrow $u : X \rightarrow N$, then we might want to define an arrow $f : X \rightarrow Y$ in a way that $f(x)$ is defined by cases, depending on whether $u(x) = 0$ or not. Doing this requires that we can write $X$ as the coproduct of $X\_0$ and $X\_{>0}$, where $X\_0$ is the equalizer of $u$ and $0$, and $X\_{>0}$ is the equalizer of $1{\dot -}u$ and $0$. It's not too hard to show that this is possible for any $u$ iff the coproduct $N=1+N$ is universal.
| https://mathoverflow.net/users/128139 | Is the coproduct $N=1+N$ universal? | The answer is no in general - but this is a fairly subtle issue. First, let's go over why this is "almost true".
Given $e:X \to N$ I denote by $X\_0$ and $X\_{>0}$ the pullback of $\{0\}$ and $N\_{>0}$ along $e$.
So, given $N$ a parametrized NNO, and $f,g :X \to Y$ two functions, what you can easily do using the NNO property, is construct a function $h:X \times N \to Y$ such that $h(x,0)= f(x)$ and $h(x,Sn)=g(x)$.
And if on top of this you have a map $e:X \to N$ then you can define $h'(x)=h(x,e(x))$ which morally is defined $h'(x) = f(x)$ if $e(x)=0$ and $h'(x)=g(x)$ if $e(x)>0$. More precisely, its restriction to $X\_0$ and $X\_{>0}$ will coincide respectively with the restriction of $f$ and $g$.
But that doesn't quite give you a pushout because the maps $f$ and $g$ are defined on the whole of $X$ and not on $X\_0$ and $X\_{>0}$. If you have a way to somehow extend a map only defined on $X\_0$ and $X\_{>0}$ to a map defined on the whole of $X$, then this construction would show the result (at least the existence part). For example, I claim that this method shows:
**Proposition:** Let $e:X \to N$. Assume there are two maps $u\_0:X \to X\_0$ and $u\_{>0} :X \to X\_{>0}$. Then $X = X\_0 \coprod X\_{>0}$
For example, if both $X\_0$ and $X\_{>0}$ have global sections (map from $1$) then the proposition can be applied with $u\_0$ and $u\_{>0}$ two constant maps.
The general idea of the proof is that using the discussion above you can modify $u\_0$ and $u\_{>0}$ into maps $\tilde{u\_0}$ and $\tilde{u\_{>0}}$ so that they restrict to the identity on $X\_0$ and this provide you with a canonical way of extending functions on $X\_0$ and $X\_{>0}$ to the whole of $X$ to apply the discussion above. This only proves the existence part of the coproduct property, not the uniqueness - but there is a neat trick to deduce uniqueness:
**Lemma :** Let $i:A \to C$ and $j:B \to C$ and assume that the induced natural transformation $Hom(C,Y) \to Hom(A,Y) \times Hom(B,Y)$ admit a section, which is natural in $Y$ and sends the pair $(i,j)$ to the identity of $C$. Then $C$ is the coproduct of $A$ and $B$ (with the maps $i$ and $j$).
Indeed, under the assumption of the lemma, the functor $Hom(A,Y) \times Hom(B,Y)$ is a retract of $C$, by a certain map $P: Hom(C,Y) \to Hom(C,Y)$, which by the Yoneda lemma corresponds to a map $P:C \to C$, but the second assumption give you that $P$ is the identity.
To deduce the proposition, just observe that the way we build a map $X \to Y$ out of two maps $X\_0 \to X $ and $X\_{>0} \to X$ is natural in $Y$, and gives the identity of $X$ if one start form the two inclusion map (for this we need to use the precise way the function $\tilde{u\_0}$ and $\tilde{u\_{>0}}$ were defined using the NNO).
**Now, a counterexample.**
So the general idea is we want to start from a situation where one of the two fiber has no map from 1 - that's not quite sufficient but this is necessary. I'm starting from the following assumption. Let $C$ be a category such that:
* $C$ has all finite limits and is extensive (all coproducts are universal and disjoint). I don't really need extensivity, I think I only need a (maybe strict) initial object - but most categories with NNO are extensive anyway.
* $C$ has a parametrized NNO.
* There exists a function $f: N \to N$ such that $f$ is not the constant equal to zero function but for all map $p: 1 \to N$ the composite $fp$ is equal to $0$.
I would recommend just assuming these exist. But a typical example of this situation (and to be honest the only ones I know) is to take $C$ to be the free topos (with NNO) or the free (extensive) cartesian closed category with NNO, or the free extensive category with NNO. In each case you can take $f$ to be a primitive recursive function which sends an integer $n$ to $1$ if $n$ is a code of a proof in the theory of elementary topos with NNO there exists a map $ \* \to \emptyset$ - of course replacing elementary topos with NNO, by extensive (cartesian closed) with NNO in the other two cases.
In these categories, the only maps $1 \to N$ are the standard integer, so as such a proof of course does not exist, all global section are sent to $0$. But the point is that because of Gödel's theorem (applied to the recursively enumerable theory of say elementary topos with NNO), that function $f$ isn't the function equal to $0$.
Ok, so of course $C$ is extensive, so it isn't our counterexample on the nose.
Take $P$ to be the full subcategory of $C$ of objects $X$ that are either the initial object or admit at least one function $1 \to X$.
I claim that $P$ is a coreflective subcategory of $C$. Indeed, for any object $Y$ of $C$, then either it has no map $1 \to Y$ - in which case there is only one map $X \to Y$ with $X \in P$ given by the unique map from the initial object - Or there are some maps $1 \to Y$ in which case $Y \in P$. So in both case, there is a universal map $X \to Y$ with $X \in P$.
In particular, the category $P$ has all finite limits (by computing them in $C$ and applying the reflection). It is closed under product and coproduct in $C$ and contains the NNO of $C$, so it has a parametrized NNO.
Now consider our function $f:N \to N$, we assumed above. The fiber of $f$ over $N\_{>0}$ has no map from $1$, so in $P$ it becomes the initial object. The fiber of $f$ over $0$ cannot be isomorphic (by the canonical mao) to $N$ because otherwise, we would have $f=0$. So the coproduct of the two is simply the fiver over $0$ - which is not isomorphic to $N$ (well - in fact, it will likelt be isomorphic to $N$ - but I mean not isomorphic by the canonical map).
| 4 | https://mathoverflow.net/users/22131 | 436017 | 176,222 |
https://mathoverflow.net/questions/435997 | 2 | $\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\GL{GL}$Let $V, W$ be finite dimensional complex vector spaces and $M\in \Hom(V, W)$ a full rank linear map. I want to see if there exists a Lie group $G$ and representations $\pi: G \to \GL(V)$ and $\rho: G \to \GL(W)$ s.t. $\rho(g) M = M \pi(g)$. The problem is that I don't know what the group $G$ is. I'm looking for an algorithm that will find the group $G$ and the representations $\rho, \pi$ for any $M$. In case it makes things easier, I mostly care about finding unitary symmetries of $M$, meaning that $V,W$ have inner products and the representations are of the form $\pi: G \to U(V)$ and $\rho: G \to U(W)$.
Two questions:
* Is there a well defined notion of the maximal symmetry group of $M$? Obviously, the trivial group will always be a symmetry of $M$, I care about finding all the symmetries.
* Does there exist an algorithm to solve this problem? I'm imagining that the group $G$ could be constructed by finding generators. Starting with Lie algebra generators $X\in \mathfrak{gl}(V), Y\in \mathfrak{gl}(W)$ s.t. $YM = MX$. The process would start with the trivial Lie algebra and would extend it step by step until no more extensions can be found. Then, an analogous process should find all the discrete symmetries. The result should be the maximal symmetry group of $M$ together with the representations $\pi$ and $\rho$.
| https://mathoverflow.net/users/265376 | Algorithm for finding the symmetries of a linear operator | Note that the identity $\rho M = M \pi$ (I'm omitting the $g$ argument here) implies that $\rho$ has to preserve the image of $M$ and can do anything it likes the cokernel of $M$, while $\pi$ has to preserve the kernel of $M$ and it can do anything it likes on the complement. Writing $V = \ker M \oplus Z$ and $W = Z \oplus \operatorname{coker} M$, where the subspaces $\operatorname{im} M = Z = \operatorname{coim} M$ are identified by putting $M$ in reduced row echelon form, the obvious most general parametrization of $\rho$ and $\pi$ in an adapted basis are
$$
\rho = \begin{bmatrix} \zeta & \* \\ 0 & GL(\operatorname{coker}M) \end{bmatrix}, \quad
\pi = \begin{bmatrix} GL(\ker M) & \* \\ 0 & \zeta \end{bmatrix}, \quad \text{with} \quad
M = \begin{bmatrix} 0 & I \\ 0 & 0 \end{bmatrix} ,
$$
where $\zeta \in GL(Z)$, $GL(-)$ schematically denotes any elelement of the corresponding group, and $\*$ denotes an arbitrary matrix block of the right dimension.
If you want to restrict $\rho$ and $\pi$ to be unitary, then it's a little bit more complicated. Instead of using reduced row echelon form, you have to use the singular value decomposition (SVD) to bring $M$ into positive diagonal form $\Sigma$ on $Z$. In an adapted basis, $\rho$ and $\pi$ are parametrized as
$$
\rho = \begin{bmatrix} \kappa & 0 \\ 0 & U(\operatorname{coker}M) \end{bmatrix}, \quad
\pi = \begin{bmatrix} U(\ker M) & 0 \\ 0 & \lambda \end{bmatrix}, \quad \text{with} \quad
M = \begin{bmatrix} 0 & \Sigma \\ 0 & 0 \end{bmatrix} ,
$$
where $\kappa,\lambda \in U(Z)$ and $\kappa \Sigma = \Sigma \lambda$. The standard solution of the last equality is $\kappa = \lambda$ with the matrices in block-diagonal form, each diagonal block being unitary, the blocks corresponding to the groups of equal singular values in $\Sigma$.
In each case, you can reconstruct the group $G$ and the multiplication rule in it from the above parametrizations.
| 3 | https://mathoverflow.net/users/2622 | 436022 | 176,225 |
https://mathoverflow.net/questions/425214 | 6 | It is known that there is a representation of the affine Lie algebra $\widehat{\mathfrak{sl}\_q}$ (over $\mathbb{Z}$) on the algebra of symmetric functions, where the action of the Chevalley generators $E\_i,F\_i$, $i=0, \ldots q-1$ on the basis of Schur polynomials $s\_\lambda$ is given by
$$E\_i s\_\lambda= \sum\_{\lambda\setminus \mu \text{ is an } i-\text{cell}} s\_\mu$$
$$F\_i s\_\lambda= \sum\_{\mu\setminus\lambda \text{ is an } i-\text{cell}} s\_\mu$$
where a cell $(x,y)$ in Young diagram is called an $i$-cell if $x-y\equiv i\pmod{q}$.
Using the fact that this action commutes with multiplication by the power sum symmetric functions $p\_{qk}$ for all $k\in \mathbb{Z}\_{\ge 1}$, one can show that the $\mathbb{Q}$-vector space generated by the action of $\widehat{\mathfrak{sl}\_q}\otimes \mathbb{Q}$ on $s\_\emptyset$ is $\mathbb{Q}\left[p\_n\ |\ q\nmid n\right]$.
I am interested, for $q$ prime, in the $\mathbb{Z}\_{(q)}$-lattice generated by the action of $\widehat{\mathfrak{sl}\_q}\otimes \mathbb{Z}\_{(q)}$ on $s\_\emptyset$, where $\mathbb{Z}\_{(q)}$ is the subring of $\mathbb{Q}$ consisting of fractions $\frac{r}{s}$ with $s$ not divisible by $q$. From Darij Grinberg's [answer](https://mathoverflow.net/a/403672/160416) to a previous question of mine, one can see that this is contained in $\mathbb{Z}\_{(q)}\left[p\_r\ |\ q\nmid r\right]$. Computations suggest that equality should hold for $q>2$, but for $q=2$ the situation is different.
I observed that the $\mathbb{Z}\_{(2)}$-lattice generated by the action of $\widehat{\mathfrak{sl}\_2}\otimes \mathbb{Z}\_{(2)}$ on $s\_\emptyset$ seems to be $\mathbb{Z}\_{(2)}\left[p\_1,2p\_3,4p\_5,8p\_7, \ldots\right]$. It's not clear to me why this particular lattice should be stable under the operators $E\_0,E\_1,F\_0,F\_1$ and why the prime $2$ behaves differently here. I am therefore looking for an explanation of this phenomenon.
I am not very familiar with affine Lie algebras and their representations, so please excuse me if I said anything wrong or if this follows easily from well-known facts about highest-weight modules.
| https://mathoverflow.net/users/160416 | Action of $\widehat{\mathfrak{sl}_2}$ on symmetric functions with $\mathbb{Z}_{(2)}$ coefficients | I ended up finding a proof of this phenomenon using the vertex operator description of the basic representation of $\widehat{\mathfrak{sl\_2}}$. It can be found in the following preprint: <https://arxiv.org/abs/2211.10584>.
| 2 | https://mathoverflow.net/users/160416 | 436023 | 176,226 |
https://mathoverflow.net/questions/436029 | 1 | Let $X$ be a complex smooth cubic threefold and $C$ be a smooth twisted cubic, then $C\subset Y\subset X$ for a unique cubic surface $Y$ in $X$ (or equivalently a hyperplane section of $X$). When $Y$ is smooth, we have $27$ lines on $Y$.
What can we say about the position of the a line $L$ and the twisted cubic $C$ (or say $\mathcal{O}\_L(C)$)?
I will ask a question about the singular $Y$ separately.
| https://mathoverflow.net/users/nan | twisted cubic in a smooth hyperplane section of a cubic threefold | Although this is not necessarily, let me assume that $Y$ is a smooth cubic surface. Then the linear system of a twisted cubic curve $C \subset Y$ generates a linear system that induces a morphism
$$
\pi \colon Y \to \mathbb{P}^2
$$
which is a blowup of 6 points, say $P\_1,\dots,P\_6$, and so that $\mathcal{O}\_Y(C)$ is the pullback of $\mathcal{O}\_{\mathbb{P}^2}(C)$. In other words, $C$ is the preimage of a line under $\pi$.
If $C$ is irreducible, the corresponding line does not pass through the points $P\_i$. Therefore, it doesn't intersect 6 lines on $Y$ (the exceptional divisors over the $P\_i$), intersect ones 15 lines (the strict transforms of lines through $P\_i$ and $P\_j$, $i \ne j$), and intersect twice 6 lines (the strict transforms of conics).
| 1 | https://mathoverflow.net/users/4428 | 436030 | 176,227 |
https://mathoverflow.net/questions/436020 | 3 | For prime $p$ let $E\_p[\dots]$ and $P\_p[\dots]$ be the external and polynomial $\mathbb{Z}\_p$--algebras.
It is known that for $n\geqslant 1$ and odd $p$ where is an isomorphism
of primitively generated Hopf algebras
$H\_\*(\Omega^2S^{2n+1};\mathbb{Z}\_p)=E\_p[x\_0,x\_1,x\_2,\dots]\otimes P\_p[y\_1,y\_2,\dots]$
where $\deg(x\_i)=2p^in-1$ and $\deg(y\_i)=2p^in-2$.
For $p=2$ where is an isomorphism
of primitively generated Hopf algebras
$H\_\*(\Omega^2S^{2n+1};\mathbb{Z}\_2)=P\_2[z\_0,z\_1,z\_2,\dots]$
where $\deg(z\_i)=2^{i+1}n-1$.
$\mathbf{Question:}$ How does a similar description of $H\_\*(\Omega^3S^{2n+1};\mathbb{Z}\_p)$ looks like for $n\geqslant 2$ ?
| https://mathoverflow.net/users/494199 | Homology of iterated loop spaces on odd--dimensional spheres | For the case where $n=dp$, the calculation of $H\_\*(\Omega^3S^{2n+1};\mathbb{Z}/p)$ is reviewed in Section 4.2 of the paper [The triple loop space approach to the telescope conjecture](https://people.math.rochester.edu/faculty/doug/mypapers/obit.pdf) by Mahowald, Ravenel and Shick. I don't think that the assumption that $n=dp$ makes any real difference, although it will take a little work to straighten out all the indices for the general case. All the real work is done in the book [Homology of iterated loop spaces](http://www.math.uchicago.edu/%7Emay/BOOKS/homo_iter.pdf) by Cohen, Lada and May; the paper of Mahowald, Ravenel and Shick just spells out the simplifications that occur in the case at hand.
| 5 | https://mathoverflow.net/users/10366 | 436035 | 176,230 |
https://mathoverflow.net/questions/435890 | 3 | Let $B$ denote the $n$-dimensional unit ball. Assume $u\in\bigcap\_{1\le p<2} W\_0^{1,p}(B)$ satisfies $$
\int a\_{ij}\partial\_j u \partial\_iv =0,$$ for any $v\in C\_c^{\infty}(B)$, where we assume $a\_{ij}(x)$ are just uniformly elliptic, i.e. there exists $\lambda,\Lambda>0$ such that $$\lambda \lvert\xi\rvert^2\le a\_{ij}(x) \xi\_i\xi\_j \le \Lambda \lvert\xi\rvert^2$$ for any $x\in B$. Under these conditions, can we show that $u$ is identically $0$?
If we just assume $u\in W^{1,1}\_0(B)$, the answer is negative, see the post [Uniqueness of solutions to elliptic PDE in $W^{1,1}$](https://math.stackexchange.com/questions/4181319/uniqueness-of-solutions-to-elliptic-pde-in-w1-1).
And when $a\_{ij}=1$ ($i=j$), we can show $u\equiv 0$ by Poisson integral formula, while in general case this approach may not be generalized.
| https://mathoverflow.net/users/442224 | $W^{1,p}$ ($1\le p<2$) uniqueness of elliptic equations | The answer is **yes** if and only if you have optimal elliptic regularity in $W^{-1,p'}(B)$ for some $p>2$ for the adjoint differential operator, i.e., the one given by the transpose matrix $A^\top$.
Consider your question from a functional-analytic point of view: You are essentially asking whether the (bounded linear) operator $$-\nabla \cdot A\nabla \colon W^{1,p}\_0(B) \to W^{-1,p}(B), \qquad \langle -\nabla \cdot A \nabla u, v\rangle := \int\_B A\nabla u \cdot \nabla v$$ is injective for some $p<2$, where of course $A = (a\_{ij})$ and $W^{-1,p}(B)$ is the dual space of $W^{1,p'}\_0(B)$.
For this it is sufficient (and in this case equivalent, since the Sobolev scale is reflexive, at least if we stay away from $p=1$) that the adjoint operator of $-\nabla \cdot A \nabla$ is surjective.
This adjoint operator happens to be $-\nabla \cdot A^\top \nabla \colon W^{1,p'}\_0(B) \to W^{-1,p'}(B)$ which is still uniformly elliptic. Since $p' > 2$ and $B$ is bounded, we thus already know from Lax-Milgram that for every $f \in W^{-1,p'}(B)$ there is a unique $w \in W^{1,2}\_0(B)$ such that $-\nabla \cdot A^\top \nabla w = f$.
So the question of surjectivity for $-\nabla \cdot A^\top \nabla$ as before becomes the question whether $w \in W^{1,p'}\_0(B)$. But that is a very question of **optimal elliptic regularity** for $-\nabla \cdot A^\top\nabla$ in $W^{-1,p'}(B)$ for *some* $p' > 2$.
Such questions are classical and answered in many articles and textbooks. The property in question is true for the ball, and also for, say, (weak) Lipschitz domains. You can also have mixed boundary conditions within this class. (And also beyond, but that seems not to be the scope of this question.) I would like to point to the classical work [G89] which covers this question in some generality and also points to the historical development.
---
[G89] *Gröger, Konrad*, [**A $W^{1,p}$-estimate for solutions to mixed boundary value problems for second order elliptic differential equations**](http://dx.doi.org/10.1007/BF01442860), Math. Ann. 283, No. 4, 679-687 (1989). [ZBL0646.35024](https://zbmath.org/?q=an:0646.35024).
| 4 | https://mathoverflow.net/users/85906 | 436037 | 176,231 |
https://mathoverflow.net/questions/436028 | -1 | In this [paper](https://arxiv.org/abs/1206.2251), if we denote the $k$ largest eigenvalues by $\lambda\_N,\lambda\_{n-1},··· ,\lambda\_{N-k+1}, $ then for Gaussian ensembles the joint distribution function of rescaled eigenvalues has the limit:
$$
\lim\_{N\to\infty}P(N^{2/3}(\lambda\_N-2)\le s\_1,\dots, N^{2/3}(\lambda\_{N-k+1}-2)\le s\_k)=F\_{\beta, k}(s\_1,\dots, s\_k)
$$
Can we say the gap $\lambda\_N-\lambda\_{N-k+1}=O\_p(N^{-2/3})$ by the continuous mapping theorem?
| https://mathoverflow.net/users/168083 | Can we apply the continuous mapping theorem for the limiting joint distribution of the Tracy-Widom law? | $\newcommand\la\lambda$The answer is: yes, of course.
Indeed, let $X\_{N,i}:=N^{2/3}(\la\_i-2)$. By the limit theorem you cited and (say) [Example 2.3, p. 18](https://archive.org/details/convergenceofpro0000bill), the $k$-tuple $(X\_{N,N},\dots,X\_{N,N-k+1})$ converges in distribution to some $k$-tuple $(Y\_1,\dots,Y\_k)$ of random variables. So, by continuous mapping theorem, $N^{2/3}(\la\_N-\la\_{N-k+1})=X\_{N,N}-X\_{N,N-k+1}$ converges in distribution to $Y\_1-Y\_k$.
So, $N^{2/3}(\la\_N-\la\_{N-k+1})=O\_P(1)$, that is, $\la\_N-\la\_{N-k+1}=O\_P(N^{-2/3})$, as desired.
| 0 | https://mathoverflow.net/users/36721 | 436038 | 176,232 |
https://mathoverflow.net/questions/436036 | 2 | I found this question: [Chernoff style concentration bound for ratio of variables](https://mathoverflow.net/questions/420837/chernoff-style-concentration-bound-for-ratio-of-variables).
I want to ask if we get similar thing for the ratio of the sum and the one Gaussian variable.
Given i.i.d. Gaussian random variables $X\_1,\dots, X\_k$ with $N(0, 1)$. Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that
$$
P\left(\frac{X\_1^2+X\_2^2+\dots+X\_k^2}{X\_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ?
$$
Can we find such $k$?
---
In [these 2020 slides by Andrew Nobel](https://nobel.web.unc.edu/wp-content/uploads/sites/13591/2020/10/Probability_Inequalities.pdf),
for $Y\sim \chi\_k^2$ where $(Y=\sum\_{I=1}^k X\_i^2)$, for $t\in (0,1)$
$$
P(Y\ge (1+\epsilon)k)\le \exp(-k(t^2-t^3)/4).
$$
| https://mathoverflow.net/users/168083 | Can we find such $k$ so that the following inequality holds? | $\newcommand\ep\epsilon $In the clever answer by Fedor Petrov, it was shown that
\begin{equation\*}
Q:=P\Big(\frac{X\_1^2+\dots+X\_k^2}{X\_1^2}<C\Big)\le C/k, \tag{1}\label{1}
\end{equation\*}
where $C:=1/\ep^2>0$ and the $X\_i$'s are any iid random variables.
Let us show that for the standard normal $X\_i$'s as in the OP, the upper bound $C/k$ in \eqref{1} can be replaced by a bound decreasing exponentially in $k$.
For $C\le1$, $Q=0$. So, without loss of generality (wlog) $C>1$. Also, wlog $k\ge2$.
For
\begin{equation\*}
c:=C-1>0, \tag{2}\label{2}
\end{equation\*}
any $x\in [0,\sqrt{(k-1)/c}\,]$, and
$$h:=\frac{k-1-cx^2}{4(k-1)},$$
we have
\begin{equation\*}
Q=P\Big(\sum\_{i=2}^k X\_i^2<cX\_1^2\Big)\le Q\_1+Q\_2, \tag{3}\label{3}
\end{equation\*}
where
\begin{equation\*}
Q\_1:=P(X\_1^2\ge x^2)\le e^{-x^2/2}=:R\_1 \tag{4}\label{4}
\end{equation\*}
and
\begin{equation\*}
\begin{aligned}
Q\_2&:=P\Big(\sum\_{i=2}^k X\_i^2<cx^2\Big) \\
&=P\Big(\sum\_{i=2}^k(1-X\_i^2)>k-1-cx^2\Big) \\
&\le\exp\{-h(k-1-cx^2)+(k-1)\ln Ee^{h(1-X\_1^2)}\} \\
&=\exp\{-h(k-1-cx^2)+(k-1)(h-\tfrac12\,\ln(1+2h))\} \\
&\le\exp\{-h(k-1-cx^2)+2(k-1)h^2\} \\
&=\exp-\frac{(k-1-cx^2)^2}{8(k-1)}=:R\_2.
\end{aligned}
\tag{5}\label{5}
\end{equation\*}
Choosing now $x$ to be the positive root of the equation $R\_1=R\_2$, from \eqref{3}, \eqref{4}, and \eqref{5} we get
\begin{equation}
Q\le 2\exp-\frac{k-1}{2(1+\sqrt C)^2}, \tag{6}\label{6}
\end{equation}
which is the promised bound, decreasing exponentially in $k$.
One may also note that typically $k$ and $C$ will be large, and then the exponent in the bound in \eqref{6} will be about $-k/(2C)$ -- compare this with the bound in \eqref{1}.
| 2 | https://mathoverflow.net/users/36721 | 436051 | 176,237 |
https://mathoverflow.net/questions/436032 | 13 | *Below, all sentences/formulas are first-order and in the language of arithmetic. For simplicity, we conflate numbers and numerals, and conflate sentences/formulas and their Godel numbers.*
Given a formula $\varphi(x)$ and a sentence $\theta$, say that $\theta$ **asserts its own $\varphi$-ness** iff $\mathsf{PA}\vdash\theta\leftrightarrow\varphi(\theta).$ Let $\mathsf{SR}(\theta)$ be the set of $\varphi$ such that $\theta$ asserts its own $\varphi$-ness. Bounded truth predicates show that $\mathsf{SR}(\theta)$ is never empty. I'm curious how much $\mathsf{SR}(\theta)$ could actually depend on $\theta$; in particular, if there's a lot of potential variety here, this might give a meaningful notion of "degree of self-referentiality" of a sentence.
Here's one way to make this precise:
>
> Are there sentences $\theta,\theta'$ such that $\mathsf{SR}(\theta)\not\cong\mathsf{SR}(\theta')$ as partial orders?
>
>
>
The partial ordering I have in mind is provability: $\sigma\le\rho$ iff $\mathsf{PA}\vdash\forall x[\sigma(x)\rightarrow\rho(x)]$. By considering bounded truth predicates, $\mathsf{SR}(\theta)$ always contains a copy of $\mathbb{Z}$: briefly, look at $\tau\_n^+(x)=$ "$x$ is a true $\Sigma\_n$ sentence" and $\tau\_n^-(x)=$ "$x$ is not a false $\Sigma\_n$-sentence" for $n$ sufficiently large. Another "canonical" element is the formula $\varphi(x)\equiv\theta$ (basically, "ignore input, output $\theta$"). Finally, $\mathsf{SR}(\theta)$ is always a distributive lattice. Beyond this, however, I don't see anything useful.
| https://mathoverflow.net/users/8133 | Are there different "levels" of self-referentiality in arithmetic? | If $\varphi$ is not required to behave the same way on Gödel codes of equivalent sentences or any such thing, then $\mathsf{SR}(\theta)$ is always equivalent to the preorder on all arbitrary formulas, by defining in PA a bijection $f$ between $\mathbb{N} - \{\theta\}$ and $\mathbb{N}$, and noting that any formula $\varphi$ gives rise to a formula $\varphi' \in \mathsf{SR}(\theta)$ via $\varphi'(\theta) = \theta$ and $\varphi'(n) = \varphi(f(n))$ for $n \neq \theta$.
We have that $\varphi \leq \psi$ iff $\varphi' \leq \psi'$, and that every formula $\varphi \in \mathsf{SR}(\theta)$ is equivalent to some $\psi'$ (specifically, take $\psi(n) = \varphi(f^{-1}(n))$). Thus, the map $\varphi \mapsto \varphi'$ is an equivalence from the preorder of arbitrary formulas to the preorder $\mathsf{SR}(\theta)$.
| 8 | https://mathoverflow.net/users/3902 | 436053 | 176,238 |
https://mathoverflow.net/questions/436052 | 9 | We believe there is always a prime in the interval $[x,x+\sqrt{x})$, for $x$ sufficiently large, but proving this is inaccessible, even under RH.
What if we just wanted a sequence of integers free of multiples of previous elements? These sequences are called primitive. Can we find a primitive sequence with this property?
Defining $a\_1 = 6$ and then recursively defining $a\_{n+1} = \max\{i: a\_n<i<a\_n+\sqrt{a\_n}, \ a\_j \nmid i, 1\leq j \leq n\}$, (if such an integer exists) we obtain a sequence
6,8,10,13,15,17,21,25,29,33,38,44,49,55,62,69,77,83,92,101,111,121,131,142,151,163,173,185,197,211,223,237,251,265,281,295,311,327,341,359,373,389,407,427,447,467,487,...
which appears to be infinite, and (although it doesn't look like it at first) far sparser than the primes.
Is there any hope of proving that it is infinite, or rather that there is some starting point that results in an infinite sequence? (Note, this process terminates quickly if $a\_1$ is chosen to be any smaller than 6.) Could we do so with $\sqrt{x}$ replaced with smaller powers of $x$?
| https://mathoverflow.net/users/495982 | Primitive sequences with elements in every interval $[x, x + \sqrt x)$ | One can obtain such a sequence after some finite starting point. Let $k\_0$ be large, and for each $k \geq k\_0$ we can partition the interval $[2^k,2^{k+1})$ into about $100 \cdot 2^{k/2}$ disjoint intervals of length about $\frac{1}{100} \cdot 2^{k/2}$. In each such interval $I$, use sieve theory to select a number $a\_I$ of the form $a\_I = p\_k b\_I$, where $p\_k \sim k \log k$ is the $k^{th}$ prime and $b\_I$ is some number with no prime factors less than $p\_k$ (so in particular the smallest prime factor of $a\_I$ is $p\_k$). It is then a simple case analysis to check that none of the $a\_I$ can divide each other (note that no two distinct elements of $[2^k, 2^{k+1})$ can divide each other, whereas for $k < k'$ all the numbers chosen in $[2^k,2^{k+1})$ have the prime factor $p\_k$ which is not present in any number chosen in $[2^{k'}, 2^{k'+1})$).
Probably with some computer assistance and some effective sieve theory one can in fact start the sequence at some quite small value such as $6$.
| 6 | https://mathoverflow.net/users/766 | 436064 | 176,240 |
https://mathoverflow.net/questions/436060 | 1 | It is easy to see that for any $x$ and $y$ on the unit sphere of a Hilbert space $H$ there exists a surjective isometry $U$ such that $Ux=y$. Does something more general also hold? That is, given two pairs $(x\_1, y\_1)$ and $(x\_2, y\_2)$ of vectors on the unit sphere such that $||x\_1-x\_2||=||y\_1-y\_2||$, can we find a surjective isometry $U$ such that $Ux\_1=y\_1$ and $U x\_2=y\_2$?
| https://mathoverflow.net/users/69275 | Isometries of Hilbert space | For complex scalars, this fails in $\mathbb{C}^2$. Take $x\_1 = y\_1 = (1,0)$, $x\_2 = (i,0)$, $y\_2 = (0,1)$. No (complex linear) isometry can take the one-dimensional subspace spanned by $x\_1$ and $x\_2$ onto the two-dimensional subspace spanned by $y\_1$ and $y\_2$.
For real scalars, it is true. It suffices to find an isometry from ${\rm span}(x\_1,x\_2)$ to ${\rm span}(y\_1,y\_2)$ which takes $x\_1$ to $y\_1$ and $x\_2$ to $y\_2$. This just means that you have two points on the unit circle in $\mathbb{R}^2$ and I have two points on the unit circle, and the distance between your two points equals the distance between my two points, and you want an isometry of $\mathbb{R}^2$ to itself that takes your points to mine. Which is easy.
| 3 | https://mathoverflow.net/users/23141 | 436066 | 176,241 |
https://mathoverflow.net/questions/436071 | 11 | Let $M\_1,M\_2$ be two simply connected, connected, compact smooth manifolds without boundary and of the same dimension. Assume that $\mathfrak{X}(M\_1)\cong \mathfrak{X}(M\_2)$ as Lie algebras.
>
> **Question.** Are $M\_1$ and $M\_2$ diffeomorphic?
>
>
>
This seems like a basic question but I did not find anything on Google.
| https://mathoverflow.net/users/32135 | Does the Lie algebra of vector fields $\mathfrak{X}(M)$ determine the diffeomorphism class of a manifold $M$? | The answer is *yes*, and the assumptions "simply connected" and "compact" are actually unnecessary.
In fact, it is possible to reconstruct any smooth manifold $M$, up to diffeomorphisms, by using the subalgebra $\mathfrak{X}\_0(M)$ of vector fields with compact support. See
*Shanks, M. E.; Pursell, Lyle E.*, [**The Lie algebra of a smooth manifold**](http://dx.doi.org/10.2307/2031961), Proc. Am. Math. Soc. 5, 468-472 (1954). [ZBL0055.42105](https://zbmath.org/?q=an:0055.42105), [JSTOR](https://www.jstor.org/stable/2031961#metadata_info_tab_contents).
| 11 | https://mathoverflow.net/users/7460 | 436078 | 176,246 |
https://mathoverflow.net/questions/436080 | 13 | In my research, I've recently started to play with Voronoi tessellations. I currently have a Python code that creates the tessellation and I am trying to color the polygonal regions according to their regularity. The best-case scenario would be to have a color scale (let's say from red to blue) where regions in red would be the "most irregular" (in some sense) and regions in blue would be the "most regular" (in some sense). It is not strictly necessary to have a scale; I would happily settle to color regions by range, e.g., if the region is "very irregular" I color it red, if it is "sort of regular" I color it yellow and if it is "regular" I color it green.
What I am looking for is a way to associate a number to say how regular a polygon is (for now, I am more interested in hexagons, but it wouldn't hurt to have something for a general polygon).
For example, the first idea that came to mind was taking the average of the angles and measuring the difference from this average to the expected internal angle of the hexagon (which is $2\pi/3$), but this does not work because the average is always $2\pi/3$ since the polygons I'm working with are always convex.
Some ideas that I have discussed with my advisor are
* looking at $\bar{D}/\sigma(D)$, where $D$ is the set of distances from the vertices of the polygon to it's center, $\bar{D}$ is the average of $D$ and $\sigma(D)$ is the standard deviation of $D$. This ratio ranges from 0 ("maximum" irregularity) to $\infty$ ("maximum" regularity);
* looking only at $\sigma(D)$, which ranges from 0 ("maximum" regularity) to $\infty$ ("maximum" irregularity).
If possible, I would like to work only with angles due to the way the code I have produces the Voronoi tessellation. However, my intuition says that either
1. I'll have to forget angles altogether and look only at distances;
2. I'll have to look at angles and at distances.
Therefore, my question is: is there such a metric for characterizing the regularity of a polygon that involves only the internal angles? If not, is there a more interesting metric that involves distances than the ones I listed above?
I would be more than happy to share more information about the problem if need be.
**Edit:** [this picture it a prototype to what I am trying to do.](https://i.stack.imgur.com/rv5mC.png)
The green regions are hexagons (regular or not), the white ones are the infinite regions of the tessellation and the red ones are the finite regions that are not hexagons.
| https://mathoverflow.net/users/168737 | How to characterize the regularity of a polygon? | Internal angles are not enough to determine the regularity of a polygon. E.g., angles of $2\pi/3$ between sides of length $1,1,4,1,1,4$ make an irregular hexagon.
For a metric of regularity, I suggest $$A \ / \ \sum d\_i^2$$ where $A$ is the area of the hexagon and the $d\_i$ are the side lengths.
Using this metric:
* regular hexagons have a regularity of $\sqrt{3}/4 \sim 0.433$;
* irregular hexagons have lower regularity;
* the hexagon with six unit lengths along a $1\times2$ rectangle has a regularity of $1/3$;
* the hexagon with six unit lengths along a $0\times 3$ rectangle has a regularity of $0$.
The area is easy to calculate as $\sum (x\_i y\_{i+1}-x\_{i+1}y\_i)/2$. So the metric should be good for getting a stable optimization, since the numerator and denominator are both simple quadratic functions of the coordinates, not requiring any square roots.
| 13 | https://mathoverflow.net/users/nan | 436088 | 176,250 |
https://mathoverflow.net/questions/425280 | 3 | I'm interested in studying a certain generalization of determinental varieties as defined here:
<https://en.wikipedia.org/wiki/Determinantal_variety>
To be more specific, I must first lay out a few definitions.
Consider the variety
$X=Mat\_{n\_1, n\_2}(\mathbb{C})\times Mat\_{n\_2, n\_3}(\mathbb{C})\times... Mat\_{n\_{k-1}, n\_k}(\mathbb{C}).$
There exists an action of $(g\_1, ..., g\_k)\in GL\_{n\_1}(\mathbb{C})\times... \times GL\_{n\_k}(\mathbb{C})$ on $X$ given by
$$(g\_1, ..., g\_k)\cdot(x\_1, x\_2, ..., x\_{k-1})=(g\_1x\_1g\_2^{-1}, ..., g\_{k-1}x\_{k-1}g\_k^{-1}).$$
Letting $r\_{ij}$ denote the rank of $x\_ix\_{i+1}... x\_j$, the orbits are completely determined by all valid choices for the values of $r\_{ij}$. One can see from this definition that when $k=2$, choosing any $0\leq r\_{12}\leq \min\{n\_1, n\_2\}$ precisely gives us the case of determinental varieties of rank $r\_{12}$.
In this case, it is known that the closure is given by all of those elements of $M\_{n\_1, n\_2}(\mathbb{C})$ or rank *at most* $r\_{12}$. As indicated on the wiki page, these closures are singular, and have well known resolutions.
I believe that in the more general case I have outlined above, the closures of the orbits will also be singular, where the "=$r\_{ij}$"'s determining the orbits, will be replaced with all "$\leq r\_{ij}$" for the closures, and that these closures will, in general, be singular.
**Question:** I would like to know if these varieties have been studied in general by algebraic geometers, if so, what is their name? Are there known resolutions of the closures, and bonus points if we have a good understanding of their fibers, or rather, the cohomology of the fibers.
Note, that is this paper of Zelevinsky:
[http://www.math.utah.edu/~ptrapa/math-library/zelevinsky/zelevinski\_tworemarks.pdf](http://www.math.utah.edu/%7Eptrapa/math-library/zelevinsky/zelevinski_tworemarks.pdf)
though dressed up differently, the varieties I'm interested in studying are embedded inside Schubert varieties. I think the implication being that one might know where to go from here?
Thanks.
| https://mathoverflow.net/users/88516 | Searching for resolutions of generalized determinental varieties | These are "type A quiver cycles" (a name chosen so as not to collide with type A quiver varieties, which involve taking quotients; here the quotient would be a point). Your guess for the closure is correct. We compute the equivariant cohomology classes of these varieties in [Four positive formulae for type A quiver polynomials](https://arxiv.org/abs/math/0308142), and give non-small resolutions of them (also in types D,E) in [Kempf collapsing and quiver loci](https://arxiv.org/abs/math/0608327). You can find the K-classes in the latter paper, though we weren't first on that.
There are similar results available if some of the arrows in the type A quiver are flipped, in [Three combinatorial formulas for type A quiver polynomials and K-polynomials](https://arxiv.org/abs/1503.05880).
| 2 | https://mathoverflow.net/users/391 | 436092 | 176,252 |
https://mathoverflow.net/questions/436102 | 3 | Let $X$ be a scheme over a field $k$. (Feel free to assume that $X$ is an algebraic variety, if needed.) Also, let $M^\bullet$ be a complex in the derived category of quasi-coherent sheaves $\mathsf{D}\_\text{qc}(\mathcal{O}\_X)$. (Feel free to assume $M^\bullet$ bounded if needed. But I would rather not suppose it coherent.)
**I want to understand what kind of information we can get from $M^\bullet$ if we know its derived fibers $\mathsf{L}x^\ast M^\bullet$ for every $x\in X(k)$.** Let me be concrete and give two precise questions:
1. Is it true that $M^\bullet=0$ if $\mathsf{L}x^\ast M^\bullet=0$ for every $x\in X(k)$? (This is the content of [this answer](https://mathoverflow.net/questions/12992/when-does-a-quasicoherent-sheaf-vanish), if we allow $x$ to be every point of $X$, not only the $k$-points.)
2. Let $\mathsf{D}^b\_\text{lf}(\mathcal{O}\_X)$ be the full subcategory of $\mathsf{D}\_\text{qc}(\mathcal{O}\_X)$ whose objects $M^\bullet$ satisfy that $\mathscr{H}^i(M^\bullet)$ is locally free and zero for $i$ large enough. Then, is it true that $M^\bullet\in \mathsf{D}^b\_\text{lf}(\mathcal{O}\_X)$ if, for every $x\in X(k)$, $\mathsf{L}\_i x^\ast M^\bullet$ is finite-dimensional and zero for $i$ large enough?
| https://mathoverflow.net/users/131975 | What do we know about a sheaf $M$ if we know its derived fibers $\mathsf{L}x^* M$, for $x\in X(k)$? | No to both questions:
1. Take $M = K(X)$, the field of rational functions on $X$.
2. Take $M$ to be the structure sheaf of any regular point on $X$ (assuming $\dim(X) > 0$).
| 4 | https://mathoverflow.net/users/4428 | 436103 | 176,256 |
https://mathoverflow.net/questions/435840 | 0 | von-Neumann entropy
===================
I know von-Neumann entropy on density matrix $S=-{\rm Tr}(\rho \ln\rho)$ is similar to Shannon entropy $S=-\sum\_i p\_i\ln p\_i$ in classical mechanics. And I want to get Bose-Einstein distribution and Fermi-Dirac distribution with the principle of maximum entropy. I work with some key points shown in this document [http://userpage.fu-berlin.de/~marekgluza/ASM2\_16/sheet06.pdf](http://userpage.fu-berlin.de/%7Emarekgluza/ASM2_16/sheet06.pdf).
Statement of Problem
====================
What I was confused is "Among all density matrices with the same diagonal (in some basis), the matrix with all entries outside the diagonal equal to zeros has the largest von-Neumann entropy". How can we get this conclusion?
Schur's theorem
===============
The document also says it's a consequence of Schur's theorem. Does Schur's theorem say von-Neumann entropy is concave? And this problem [Bound on the von Neumann entropy of a positive, positive semi-definite, and unit-trace density matrix?](https://mathoverflow.net/questions/68249/bound-on-the-von-neumann-entropy-of-a-positive-positive-semi-definite-and-unit) also talks about Schur-concave functions. But how can we know density matrix with off-diagonal elements equal to zero has maximum entropy?
| https://mathoverflow.net/users/495865 | Dose density matrix with off-diagonal elements equal to zero has maximum von-Neumann entropy? | The following observation allows to reduce the problem to the classical, commuting case.
I will assume finite dimensionality although generalizations are possible.
Let the Hamiltonian have the following spectral decomposition, $H = \sum\_k E\_k \Pi\_k $, where $E\_k, \Pi\_k$ are respectively the eigenvalues, spectral projectors.
Define the following dephasing map:
\begin{align}
D\_H(X) &:= \lim\_{T\to \infty} \frac{1}{T} \int\_0^T \!\! dt \, e^{-it H} X e^{itH} \\
& = \sum\_k \Pi\_k X \Pi\_k .
\end{align}
Note that the first line is also valid in infinite dimensions. It is easy to see that $D$ is positive, trace preserving and unital (in fact it is completely positive as one has a Kraus representation).
Let the entropy of a state $\rho$ be given by
$$S(\rho) = -\operatorname{Tr} [\rho \ln \rho]. $$
The observation is the following. It turns out that, for any quantum state $\rho$
$$S (D\_H (\rho)) \ge S(\rho) . $$
Since we want to maximize the entropy we can do so among the fixed points of $D\_H$ which are states that commutes with $H$. From here on you can follow the classical proof with Lagrange multipliers.
The observation is part of the characterization of entropy increasing channels. In particolar, if $T$ is positive, trace preserving and unital, and $h: \mathbb{R} \rightarrow \mathbb{R}$ is a concave function, then
$$
\operatorname{Tr} [h(T(\rho))] \ge \operatorname{Tr}[h(\rho)]$$
for all density operators $\rho$.
For a proof, see Theorem 8.8 and corollary 8.1 of *Quantum channels & operations guided tour* by M. Wolf ([freely available here](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwiG5ODx8ef7AhXlzIUKHbxGCK0QFnoECA4QAQ&url=https%3A%2F%2Fwww-m5.ma.tum.de%2Ffoswiki%2Fpub%2FM5%2FAllgemeines%2FMichaelWolf%2FQChannelLecture.pdf&usg=AOvVaw0ifu39ST08JdtnUx7QZI5s)).
| 0 | https://mathoverflow.net/users/74539 | 436105 | 176,257 |
https://mathoverflow.net/questions/436082 | 0 | I take $F$ from $\Omega\subset \mathbb C^n$ to $\mathbb C^n$ to be a holomorphic function such that
$$| \det(J\_F)|\leq 1,$$
where $J\_F$ is the Jacobian matrix of $F$.
My question: Is there any classification of functions of this type?
| https://mathoverflow.net/users/151216 | Holomorphic function on $\mathbb C^n$ | This function has constant Jacobian by Liouville. Then it is
map of Jacobian 1 composed with a homothety or its differential is degenerate everywhere. The constant Jacobian biholomorphisms are subject of much research, see for example
*Rosay, Jean-Pierre,
Automorphisms of Cn, a survey of Andersén-Lempert theory and applications. Complex geometric analysis in Pohang (1997), 131–145,
Contemp. Math., 222, Amer. Math. Soc., Providence, RI, 1999.*
| 0 | https://mathoverflow.net/users/3377 | 436113 | 176,260 |
https://mathoverflow.net/questions/435649 | 8 | Let $k$ be a local field of characteristic $p$ and $\omega \in k$ a uniformiser. Consider the Artin-Schreier extension $L\_n = k[x]/(x^p - x - \omega^n)$ for each $n \in \mathbb{Z}$.
Is there an explicit description for the image of the norm map $N\_{L\_n/k}: L\_n^\times \to k^\times$?
If $n < 0$ it seems that the extension is wildy ramified and I'm not sure how to calculate the image of the norm group in this case.
| https://mathoverflow.net/users/5101 | Image of the norm map for Artin-Schreier extensions | Local class field theory has something to say about the norm groups.
To set up notation, let's put $k = \kappa((t))$ with uniformiser $\omega = t$.
I'll assume $n < 0$ is not divisible by $p$ in the following (without loss).
The element $x \in L\_n$ satisfying $x^p - x = t^n$ has valuation $v(x) = n/p$ (where I always normalise the valuation to satisfy $v(t) = 1$). For integers $a, b$ satisfying $ap + bn = 1$, the element $\pi := t^a x^b$ is now a uniformiser of $L\_n$. If $\sigma$ is the generator of $G := \operatorname{Gal}(L\_n/k)$ sending $x$ to $x+1$, a calculation shows that $v(\sigma(\pi) - \pi) = (1-n)/p$.
Hence for the ramification groups in lower numbering we have $G\_{-n} =G$, $G\_{-n+1} = 1$. For the upper numbering we still have $G^{-n} = G$, $G^{-n+1} = 1$ (the renumbering is trivial up to the relevant point). (See for instance Section II.10 of Neukirch's Algebraische Zahlentheorie for the definition of these groups.)
It follows that the ideal $(t)^{1-n}$ is precisely the conductor of the extension $L\_n/k$, and so the subgroup $1 + (t)^{1-n} \leq k^\times$ is contained in $N\_{L\_n/k}(L\_n^\times)$, but $1 + (t)^{-n}$ is not. (See again Neukirch, Sections 1 and 6 of Chapter V.)
If $\kappa = \mathbb{F}\_p$ is the prime field, then $1 + (t)^{1-n}$ has index $p$ in $1 + (t)^{-n}$, showing that the only elements of $1 + (t)^{-n}$ which are norms from $L\_n$ are the ones in $1 + (t)^{1-n}$. Note that this matches Lubin's answer for $n=-1$, and satisfies the desideratum (from the comments to the question) of finding many non-norms in $k$.
I do not know if one can easily identify the norm group more precisely than this. Serre's Corps Locaux has some more material, starting in V.3. The upshot there seems to be that one has a good handle on the norm in the graded components of $\kappa[[t]]^\times$, where $\kappa[[t]]^\times$ is filtered by the subgroups $1 + (t)^m$. This gives little more information than the above, however (it is precisely the above for $\kappa = \mathbb{F}\_p$). Piecing things together for an ungraded version is presumably going to be tedious.
Edit: In the comments, Daniel very reasonably asks to characterise the elements of $1 + (t)^{-n}$ which do occur as norms, in the case of a general finite residue field $\kappa$. Here the graded idea does help (see Proposition 5(iii) in V.3 in Serre): Restricting to the graded component $(1 + (t)^{-n})/(1 + (t)^{-n+1})$ of $k^\times$ and identifying it with the additive group $\kappa$ (sending $c \in \kappa$ to the class of $1 + ct^{-n}$), the image of the norm map is the image of an additive polynomial $\alpha X^p + \beta X \in \kappa[X]$. However, we know that this polynomial must in fact be defined over $\mathbb{F}\_p$ (since everything we do arises via base change from $\mathbb{F}\_p$, see also Serre's discussion in V.4) and not injective on $\mathbb{F}\_p$. Hence it is a constant multiple of the Artin-Schreier polynomial $X^p - X$, and can be written as $(\beta X)^p - \beta X$ with $\beta \in \mathbb{F}\_p^\times$. This confirms Daniel's suspicion that an element $1 + ct^{-n} + ... \in k^\times$ is a norm if and only if $c$ has the form $d^p - d$ for some $d \in \kappa$.
Edit, later: I somehow missed that Serre actually gives an explicit description of the norms later, in XIV.5. Specifically, he defines a local symbol $[a, b)\_v \in \mathbb{Z}/p$ for $a \in k$, $b \in k^\times$, which is essentially equivalent to computing the local Artin symbol in the cyclic Artin-Schreier extension corresponding to $a$.
Relevant for us is the fact that $[t^n, b)\_v = 0$ iff $b$ is a norm of the extension $L\_n/k$. Further, Serre gives the explicit formula (Corollaire to Proposition 15) $[a, b)\_v = \operatorname{Tr}\_{\kappa/\mathbb{F}\_p}(\operatorname{Res}(a \cdot \mathrm{d}b/b))$, where the residue of a differential is defined in the usual way (write the differential as $f \mathrm{d}t$ for some Laurent series $f$, and extract its $t^{-1}$-coefficient.)
From this formula we see that an element $b = 1 + ct^n + \dots$ is a norm in $L\_n/k$ iff $\operatorname{Tr}\_{\kappa/\mathbb{F}\_p}(c) = 0$; the latter condition is equivalent to $c$ having the form $d^p - d$ by additive Hilbert 90.
The advantage of the formula over what's written above is of course that it also allows to treat a general element $b \in k^\times$.
| 4 | https://mathoverflow.net/users/496038 | 436126 | 176,264 |
https://mathoverflow.net/questions/435426 | 2 | ### Background
Throughout, let $X$ be a smooth complex manifold.
1. It is a classical fact that a coherent analytic sheaf admits a **local** resolution by locally free sheaves (also known as a local syzygy). Griffiths and Harris' *Principles of Algebraic Geometry* (p. 696) gives a nice proof of this: by definition, at any point $z\_0$ we have
$$\mathscr{O}^p\to\mathscr{O}^q\to\mathscr{F}$$
on some open neighbourhood $U$ of $z\_0$, and applying Oka's lemma gives
$$\mathscr{O}^r\to\mathscr{O}^p\to\mathscr{O}^q\to\mathscr{F}$$
on some possibly smaller neighbourhood $U'\subseteq U$ of $z\_0$, and we can repeat this process finitely many times, eventually terminating with an exact sequence, since the syzygy theorem tells us that eventually the stalk of the kernel at $z\_0$ will be free.
2. It is natural to ask if this generalises to **complexes** of coherent sheaves. One answer to this is given in [SGA 6, §I, Corollarie 5.10 & Exemples 5.11], which states that
$$D^\mathrm{b}(X)\_\mathrm{coh}\simeq D^\mathrm{b}(X)\_\mathrm{perf}$$
or, in (vague) words, that complexes of coherent sheaves are perfect (i.e. locally quasi-isomorphic to a bounded complex of locally free sheaves). This is proved by what might fairly be called "general abstract methods" (in particular, it is proved in much more generality than just for smooth complex manifolds).
### Question
Is there a generalisation of the proof method of 1 to the setting of 2? That is, is there a nice manual construction of a local syzygy for a **complex** of coherent analytic sheaves?
| https://mathoverflow.net/users/73622 | Resolving complexes of coherent analytic sheaves | If I interpret your question correctly, then I believe there is indeed such a construction.
The construction relies first of all on the existence of local resolutions as in your point 1. Secondly, it relies on the fact that vector bundles are projective objects over Stein domains, for example over any ball in some local coordinates. This fact follows from the local to global spectral sequence of Ext. It follows from the second point that one has a *"Horseshoe lemma"* over any Stein domain, cf., i.e., Weibel, Homological Algebra, Lemma 2.2.8.
Then, one may construct a local *Cartan-Eilenberg resolution* $P\_{\bullet,\bullet}$ of any bounded complex $\mathcal{F}\_\bullet$ of coherent sheaves. This construction is based on taking local resolutions of each $\mathcal{F}\_k$, $B\_k(\mathcal{F})$, $H\_k(\mathcal{F})$, and using the Horseshoe lemma repeatedly in an explicit way. The Cartan-Eilenberg resolution is a double complex satisfying various nice properties. In particular, there exists an explicit quasi-isomorphism from the total complex $\mathrm{Tot}\_\bullet(P\_{\bullet,\bullet})$ of $P$ to $\mathcal{F}\_\bullet$, see i.e., Weibel, Section 5.7.
| 2 | https://mathoverflow.net/users/49151 | 436128 | 176,265 |
https://mathoverflow.net/questions/436134 | 6 | For a positive integer $n$, the prime omega function value $\Omega(n):=\sum\_{p\mid n}{\nu\_p(n)}$ counts the number of prime divisors of $n$ with multiplicities. A result of Hardy and Wright, [1, Theorem 430 on p. 472], implies that $\frac{1}{x}\sum\_{n\leq x}{\Omega(n)}\sim\log\log{x}$ as $x\to\infty$.
**Question:**
Letting the variable $q$ range over prime powers, is it true that
$\left(\sum\_{q\leq x}{1}\right)^{-1}\cdot\sum\_{q\leq x}{\Omega(q-1)}\sim\log\log{x}$
as $x\to\infty$?
This question is motivated by some work in progress concerning certain algorithms over finite fields $\mathbb{F}\_q$. An affirmative answer would imply that these algorithms are efficient for "most" finite fields. In fact, it would suffice if
$\left(\sum\_{q\leq x}{1}\right)^{-1}\sum\_{q\leq x}{\operatorname{mpe}(q-1)}\in O(\log\log{x})$
where $\operatorname{mpe}(n):=\max\_{p\mid n}{\nu\_p(n)}\leq\Omega(n)$. The following table provides some computational evidence, namely the values of $f(x):=\left(\log\log{x}\sum\_{q\leq x}{1}\right)^{-1}\sum\_{q\leq x}{\Omega(q-1)}$ for $x=10^n$ with $n\in\{5,6,7,8,9\}$.
| $x$ | $10^5$ | $10^6$ | $10^7$ | $10^8$ | $10^9$ |
| --- | --- | --- | --- | --- | --- |
| $f(x)$ | 1.91446 | 1.86387 | 1.82433 | 1.7924 | 1.76574 |
Reference:
[1] G.H. Hardy and E.M. Wright, *An Introduction to the Theory of Numbers. Edited and revised by D.R. Heath-Brown and J.H. Silverman. With a foreword by Andrew Wiles*, Oxford University Press, Oxford, 6th edn. 2008.
| https://mathoverflow.net/users/57975 | Average value of the prime omega function $\Omega$ on predecessors of prime powers | Yes, this is true.
First, let us observe that replacing prime powers with primes cannot make a difference, and the same goes to replacing $\Omega$ with $\omega$.
Erdős, in "On the normal number of prime factors of $p-1$ and some related problems concerning Euler’s $\phi$-function (Quart. Journ. of Math. 6, 205-213 (1935)) proved a Hardy--Ramanujan result for $\omega(p-1)$, namely that $\omega(p-1) \sim \log \log p$ for almost all $p$.
C. B. Haselgrove, in "Some theorems in the analytic theory of numbers" (J. Lond. Math. Soc. 26, 273-277 (1951)), proved that
$$\frac{\sum\_{p \le x} \omega(p-1)}{\sum\_{p \le x}1}\sim \log \log x.$$
In fact, he computed any fixed moment.
H. Halberstam in "On the distribution of additive number-theoretic functions. III." (J. London Math. Soc. 31 (1956), 14–27) extended Haselgrove's result to $\omega(f(p))$ for any irreducible $f \in \mathbb{Z}[x]$.
A central limit theorem can be proved for $\omega(p-1)$, and this is due to Barban (according to Elliott's book mentioned below). Some modern references for this:
1. Peter D. T. A. Elliott, "Probabilistic number theory. II", Mathematischen Wissenschaften 240. Springer-Verlag, Berlin-New York, 1980.
2. Krishnaswami Alladi, "Moments of additive functions and the sequence of shifted primes", Pacific J. Math. 118 (1985), no. 2, 261–275.
3. Andrew Granville and Kannan Soundararajan, "Sieving and the Erdős-Kac theorem", Equidistribution in number theory, an introduction, 15–27, NATO Sci. Ser. II Math. Phys. Chem., 237, Springer, Dordrecht, 2007.
| 10 | https://mathoverflow.net/users/31469 | 436137 | 176,271 |
https://mathoverflow.net/questions/436122 | 0 | Following this question:[Can we find such $k$ so that the following inequality holds?](https://mathoverflow.net/questions/436036/can-we-find-such-k-so-that-the-following-inequality-holds/436040#436040).
Consider a sequence of independent $n-$dimensional random vectors $u, v\_1, v\_2,\dots, v\_k$ uniformly distributed on the unit sphere on $\mathbb{R}^n$. Define $X\_i=\sqrt{n}(u\cdot v\_i)$ for $i=1,\dots, k$. Note that by central limit theorem, we have $X\_i\to N(0,1)$ as $n\to \infty$ for $i=1,\dots, k$. Can we still prove this similarly inequality?
>
> Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that
> $$
> \lim\_{n\to \infty}\mathbb{P}\left(\frac{X\_1^2+X\_2^2+\dots+X\_k^2}{X\_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ?
> $$
>
>
>
| https://mathoverflow.net/users/168083 | Can we find the following $k$ so that the following inequality holds for asymptotic normal? | $\newcommand\ep\epsilon\newcommand{\R}{\mathbb R}$By the spherical symmetry, conditionally on $u$, the $X\_i$'s are iid random variables (r.v.'s) each with a conditional distribution not depending on $u$. So, even unconditionally, the $X\_i$'s are iid r.v.'s each equal $X:=\sqrt n\,e\_1\cdot v$ in distribution, where $v:=v\_1$ and $e\_1$ is the first vector in the standard basis of $\R^n$.
Next, the random vector $v$ equals $\frac{(G\_1,\dots,G\_n)}{\sqrt{G\_1^2+\dots+G\_n^2}}$ in distribution, where $(G\_1,\dots,G\_n)$ is a standard Gaussian random vector in $\R^n$.
So, $X^2$ equals $\frac{G\_1^2}{(G\_1^2+\dots+G\_n^2)/n}$ in distribution. So, by the law of large numbers, $X^2$ converges to $G^2$ in distribution as $n\to\infty$, where $G:=G\_1$.
Because the $X\_i$'s are independent copies of $X$, it follows that
\begin{equation}
\frac{X\_1^2+\dots+X\_k^2}{X\_1^2}\to\frac{G\_1^2+\dots+G\_k^2}{G\_1^2}
\end{equation}
in distribution as $n\to\infty$.
So, by the [previous answer](https://mathoverflow.net/a/436051/36721),
\begin{equation}
L:=\lim\_{n\to\infty}P\Big(\frac{X\_1^2+\dots+X\_k^2}{X\_1^2}<C\Big) \\
=P\Big(\frac{G\_1^2+\dots+G\_k^2}{G\_1^2}<C\Big)
\le 2\exp-\frac{k-1}{2(1+\sqrt C)^2},
\end{equation}
where $C:=1/\ep^2>1$.
Similarly one can get a better upper bound on $L$ by using the [other previous answer](https://mathoverflow.net/a/436068/36721).
| 2 | https://mathoverflow.net/users/36721 | 436140 | 176,272 |
https://mathoverflow.net/questions/436104 | 2 | If stumbled accross [self-avoiding walk](https://en.wikipedia.org/wiki/Self-avoiding_walk)s. They seem to be deterministically generated, but from a quick google search term seem to be randomly generated variants.
However, as far as I can see they are only defined on a lattice (like $\mathbb Z^d$) (which is clearly not dense in $\mathbb R^d$). Now I wondered whether we can also stochastically generate a self-avoiding walk in continuous space, say (for simplicity) on $[0,1)^2$, such that the whole space is visited if the walk goes on forever (i.e. some kind of denseness is achieved).
As a further step, I would like to know whether this process can be constructed in a way such that it has the Markov property (but this seems to be pretty hard).
I'm thankful for any adivce and/or reference on that.
| https://mathoverflow.net/users/91890 | Is there something like a "self-avoiding Markov chain" on a continuous space? | As the question is asked, the answer is "no": if a continuous curve $\gamma:\mathbb{R}\_{\geq 0}\to [0,1)^2$ is self-avoiding, i.e., injective, then the image $\gamma(\mathbb{R}\_{\geq 0})$ is nowhere dense in $[0,1)^2$. Indeed, the images $\gamma([0,T])$ are compact, hence closed. Also, compactness implies the inverse map $\gamma^{-1}:\gamma([0,T])\to[0,T]$ is continuous, i.e., the image $\gamma([0,T])$ is homeomorphic to an interval. Hence, it cannot contain any balls, for if it did, removing a point from it would make the fundamental group non-trivial, while removing a point from an interval cannot do so.
Baire's theorem now implies that $\gamma(\mathbb{R}\_{\geq 0})=\cup\_{T\in\mathbb{N}}\gamma([0,T])$ is nowhere dense.
**Update:** one way to relax the self-avoidance property: we say that a curve is non-self-crossing if it belongs to the closure of the set of injective curves with respect to the metric
$$
d(\gamma\_1,\gamma\_2)=\inf\_\sigma \sup\_t|\gamma\_1(\sigma(t))-\gamma\_2(t)|.
$$
where the infimum is over all parametrizations $\sigma$. Such a curve may intersect its past trace, but then it bounces back on the same side it came from. The $SLE\_\kappa$ curves are like that (but non-simple) for $\kappa>4$, and space filling for $\kappa\geq 8$. It is conjectured that self-avoiding walk on a lattice $\epsilon \mathbb{Z}^2$ (uniformly chosen from all walks staying in a domain $\Omega$ and with given endpoints) converges in the scaling limit $\epsilon\to 0$ to the $SLE\_{8}$ process. See Duminil-Copin, H., Kozma, G., & Yadin, A. (2014). Supercritical self-avoiding walks are space-filling. In Annales de l'IHP Probabilités et statistiques (Vol. 50, No. 2, pp. 315-326).
| 2 | https://mathoverflow.net/users/56624 | 436149 | 176,274 |
https://mathoverflow.net/questions/436130 | 2 | Why do we ask Shimura datum to have Hodge weight $(-1,1),(0,0),(1,-1)$?
I know it's related to the decomposition of a complex Lie algebra $\frak{g}\_{\mathbb{C}}=\frak{t}\oplus\frak{p}^{+} \oplus \frak {p} ^{-}$ but I'm having trouble finding what in Shimura's varieties theory needs this assumption.
Does anyone have an enlightening explanation?
| https://mathoverflow.net/users/169282 | Why do we ask Shimura datum to have Hodge weight $(-1,1),(0,0),(1,-1)$? | This is the condition which gives the Shimura variety the (almost) complex structure, which is obviously necessary if you want to view it as a variety over $\mathbb C$. This is explained in Theorem 1.21 of [Milne's notes](https://www.jmilne.org/math/xnotes/svi.pdf#X.1.21) (in slightly different language, but all the equivalences are established later in the text), but let me summarize here.
Let $(G,X)$ be a Shimura datum, where $X$ is the conjugacy class of maps $h:\mathbb S\to G\_{\mathbb R}$. At the level of real points, $h$ gives a map $h:\mathbb C^\times\to G(\mathbb R)$, which via the adjoint action gives an action of $\mathbb C^\times$ on the tangent space of $G(\mathbb R)$ at the identity. The subspace on which this subgroup is trivial coincides with the subspace tangent to the compact subgroup $K$ stabilizing $h$, which gives an action of $\mathbb C^\times$ on the tangent space $T\_hX$. Now the condition on weights comes in, and tells us that the only characters of $\mathbb C^\times$ occurring in the complexification $(T\_hX)\_{\mathbb C}$ are $z$ and $\bar z$, and $T\_hX$ can be identified with the subspace of the complexification on which the character is $z$. This equips $T\_hX$ with a complex vector space structure, and by homogeneity, $X$ receives an almost complex structure.
| 1 | https://mathoverflow.net/users/30186 | 436151 | 176,275 |
https://mathoverflow.net/questions/436124 | 2 | Can one supply related references or detailed proofs of the following two explicit formulas?
$$
{}\_2F\_1\biggl(2\alpha+1,2;\alpha+3;\frac{1}{2}\biggr)
=2\frac{\alpha B(1/2,\alpha)-1-\alpha}{(1-\alpha) \alpha B(2,\alpha +1)}
$$
and
$$
{}\_2F\_1(2\alpha+1,\alpha+1;\alpha+3;-1)
=\frac{1}{2^{2\alpha}}\frac{\alpha B(1/2,\alpha)-1-\alpha}{(1-\alpha)\alpha B(2,\alpha+1)},
$$
where the Beta function is denoted and defined by
$$
B(z,w)=\int\_0^1t^{z-1}(1-t)^{w-1}\textrm{d}t
=\int\_0^\infty\frac{t^{z-1}}{(1+t)^{z+w}}\textrm{d}t
$$
for $\Re(z),\Re(w)>0$ and the Gauss hypergeometric function ${}\_2F\_1$ is defined by
$$
{}\_pF\_q(\alpha\_1,\dotsc,\alpha\_p;\beta\_1,\dotsc,\beta\_q;z)
=\sum\_{n=0}^\infty\frac{(\alpha\_1)\_n\cdots(\alpha\_p)\_n} {(\beta\_1)\_n\cdots(\beta\_q)\_n}\frac{z^n}{n!}
$$
for $\alpha\_i\in\mathbb{C}$, $\beta\_i\in\mathbb{C}\setminus\{0,-1,-2,\dotsc\}$, $p,q\in\mathbb{N}$, and $z\in\mathbb{C}$, in terms of the rising factorial
$$
(z)\_n=\prod\_{\ell=0}^{n-1}(z+\ell)
=
\begin{cases}
z(z+1)\dotsm(z+n-1), & n\in\mathbb{N};\\
1, & n=0.
\end{cases}
$$
| https://mathoverflow.net/users/147732 | Ask for references or proofs of two explicit formulas for special Gauss hypergeometric functions | These formulas are special cases of what I like to call
extendable evaluations: set for simplicity $F={}\_2F\_1$. If one
has an explicit formula for $F(a,b;c;z)$ and (for example) also for
$F(a+1,b;c;z)$, then using the contiguity relations, it is immediate
to find explicit formulas for $F(a+k,b+l;c+m;z)$ for any
$(k,l,m)\in {\mathbb Z}^3$. For $z=-1$ there is the classical Kummer evaluation
$$F(a,b;a-b+1;-1)=\dfrac{2^{-a}\Gamma(1/2)\Gamma(a-b+1)}{\Gamma(a/2+1/2)\Gamma\
(a/2-b+1)}\;,$$
and the less classical (sorry I do not remember a good reference)
\begin{align\*}&F(a+1,b;a-b+1;-1)=2^{-a-1}\Gamma(1/2)\Gamma(a-b+1)\ \cdot\\
&\phantom{=}\cdot\left(\dfrac{1}{\Gamma(a/2+1/2)\Gamma(a/2-b+1)}+\dfrac{1}{\Gamma(a/2+1)\Gamma(a/2-b+1/2)}\right)\;,\end{align\*}
so this is extendable and in particular immediately gives your formula for
$z=-1$. Similarly, there are two extendable two-parameter families
for $z=1/2$ and two for $z=2$.
| 4 | https://mathoverflow.net/users/81776 | 436154 | 176,277 |
https://mathoverflow.net/questions/436157 | 6 | $\DeclareMathOperator\Spec{Spec}\newcommand\Ring{\mathrm{Ring}}\newcommand\op{^\text{op}}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Sh{Sh}$In the category of schemes the objects of the form $\Spec(K)$ with $K$ a field can be characterized as follows. They are precisely the non-empty schemes which have no proper subobjects in $Sch$.
Consider now the category $\Ring\op$ with its various Grothendieck topologies (Zariski, étale, fpqc, etc.). From those we get various big sheaf topoi which contain the category of schemes as a full subcategory. As a functor, $\Spec(K)$ is the representable presheaf $\Hom\_{\Ring}(K,-)$.
Even though $\Spec(K)$ has no proper subschemes when $K$ is a field, it can have non-trivial sub-pre-sheaves. Can it have non-trivial Zariski sub-sheaves? If yes, can we switch to one of the finer topologies to prevent this? Is one of the topologies on $\Ring\op$ fine enough, so that the sheaves of the form $\Spec(K)$ with $K$ a field are precisely the objects with a trivial subobject lattice in $\Sh(\Ring\op,\text{sth})$?
| https://mathoverflow.net/users/219922 | Subsheaves of Spec K, K a field | There is no hope for this in any subcanonical topology coarser than the fppf topology, or more generally, any subcanonical topology in which morphisms $\operatorname{Spec} C \to \operatorname{Spec} K$ are not automatically covers when $K$ is a field and $C$ is non-trivial.
But it is true with the fpqc topology.
Indeed, if $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$ is any morphism that is not a cover and $F$ is the sheaf image, then $F$ is not the top subsheaf of $\operatorname{Spec} K$; if we further assume that $A$ is not trivial then $F$ is also not the bottom subsheaf of $\operatorname{Spec} K$.
(We have $\alpha \in F (A)$ if and only if there exist a cover of $\operatorname{Spec} A$ by affines $\operatorname{Spec} \beta\_i : \operatorname{Spec} B\_i \to \operatorname{Spec} A$ such that each $\beta\_i \circ \alpha : K \to B\_i$ factors through $\phi : K \to C$.
So $\textrm{id}\_K \in F (K)$ if and only if $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$ is a cover.)
Conversely, by the above argument, in any subcanonical topology such that $\operatorname{Spec} K$ only has the top and bottom subsheaves, it must be that every morphism $\operatorname{Spec} C \to \operatorname{Spec} K$ is either a cover or has $C$ trivial.
Nothing in the argument assumes that $K$ is a field, but if we assume that covers are faithfully flat and $K$ is non-trivial, it will follow that $K$ is a field: because $K$ is non-trivial, there exist a field $L$ and a ring homomorphism $K \to L$, and $\operatorname{Spec} L \to \operatorname{Spec} K$ is a cover so $K \to L$ is faithfully flat, hence $K$ is an integral domain with a unique prime ideal, i.e. a field.
Thus, with the fpqc topology, $\operatorname{Spec} K$ has only the top and bottom subsheaves if and only if $K$ is a field.
| 8 | https://mathoverflow.net/users/11640 | 436161 | 176,279 |
https://mathoverflow.net/questions/419383 | 16 | I study algebraic geometry / number theory and from time to time I stumble upon 2-categorical (co)limits. I have two main examples in mind:
>
> Example 1) In étale cohomology, the (triangulated) derived category of $\overline{\mathbb{Q}\_\ell}$-sheaves is defined as a 2-colimit of the derived categories of sheaves with coefficients in finite extensions of $\mathbb{Q}\_\ell$.
>
>
>
>
> Example 2) The fact that $\textsf{QCoh}\to \textsf{Sch}$ is a stack should mean that, given a scheme $X$ and a covering $\{U\_i\to X\}$, the category $\textsf{QCoh}(X)$ is a 2-limit of the $\textsf{QCoh}(U\_i)$.
>
>
>
While the second example is somewhat straightforward, given that we may describe the limit as a category of descent data, the first one is awkward.
For sure, I shouldn't need to understand a lot of 2-category theory to make sense of these (and all related) examples. There are a lot of intricacies in the 2-categorical world... For example, should we consider (co)limits in the (2,0)-category of categories or on the (2,1)-category of categories? Should we consider lax 2-functors or strict 2-functors? What changes with those choices? (Bear in mind that I know very little of all of this.)
All in all, how should an algebraic geometer approach these kinds of statements? Also, is there a quick reference for all of this?
| https://mathoverflow.net/users/131975 | 2-categories for the working algebraic geometer | A standard reference for category theory is *Categories for the Working Mathematician* (which I assume the OP knew about based on the question title). The closest reference I know to "2-categories for the working mathematician" is Steve Lack's *[A 2-categories companion](https://arxiv.org/abs/math/0702535v1)*, which presents the essential parts of the theory (including 2-limits, in Section 6) in only 73 pages. Lack is a famous 2-category theorist in Sydney, Australia, and throughout this monograph he emphasizes lax vs strict, pseudo vs strong, bicategories vs 2-categories, etc. He also discusses situations where the 2-cells are invertible. I mention this because of the comment telling the OP to look at (2,1)-categories.
Another nice (and much newer) reference is *[Elements of $\infty$-category theory](https://emilyriehl.github.io/files/elements.pdf)* by Emily Riehl and Dominic Verity. It's 606 pages, which might seem daunting, but Appendix B is an introduction to 2-category theory and is only 22 pages. An earlier draft of this book had the title *$\infty$-categories for the working mathematician*, and the book seems to be written with the goal of making the machinery easier to use, including to folks in algebraic geometry. Riehl and Verify are also world-class category theorists, and much of their approach to $\infty$-categories is 2-categorical in nature.
| 4 | https://mathoverflow.net/users/11540 | 436163 | 176,280 |
https://mathoverflow.net/questions/434699 | 10 | In a survey article [Algebraic geometry in mixed characteristic](http://arxiv.org/abs/2112.12010v1), B. Bhatt writes
For instance, given a commutative ring $R$ with a finitely generated ideal $I$,
* the assignment carrying $R$ to the $\infty$-category $D\_\text{$I$−comp}(R)$ of derived $I$-complete $R$-complexes forms a stack for the flat topology (or even a suitably defined $I$-completely flat topology), unlike the corresponding assignment at the triangulated category level.
What would be a reference for the above sort of statement? By this I mean I don't need a reference for that exact fact, although that would be fine too. I know what the words individually mean, but I have never seen a place where I can get an idea for how to go about proving this.
This question is very similar to the following question, which at the time of writing has no answer:
[2-categories for the working algebraic geometer](https://mathoverflow.net/questions/419383/2-categories-for-the-working-algebraic-geometer)
| https://mathoverflow.net/users/80739 | Reference request: infinity categories for the commutive algebraist/algebraic geometer | I don't want this question to hang around forever on the "unanswered queue," so let me add an answer, even though I think the comments largely answer it. My motivation here is to advertise a few other sources beyond those mentioned in the comments.
First, as the comments point out, Lurie has written a ton about this kind of statement, and especially *Spectral Algebraic Geometry* seems the most user-friendly to commutative algebraists and algebraic geometers (as of today, [the most recent update](https://www.math.ias.edu/%7Elurie/papers/SAG-rootfile.pdf) is from Feb of 2018, SAG is 2319 pages, and Lurie writes that it's 67% complete). I'm sure it was a big part of what got Bhargav Bhatt excited enough to write the survey the OP cites.
Bhatt also cites Toen and Vezzosi's papers on homotopical algebraic geometry (well, Homotopical Algebraic Geometry II is more of a book than a paper, at 228 pages), and I've always found these to have a bit more of an algebraic geometry focus relative to Lurie's writings (which often strike me as more topological). In addition to HAG I and HAG II, I recommend Toen's "global overview" (from 2009) and Toen's "derived algebraic geometry" survey from 2014. Still, I could not find in these sources the statement Bhatt cites, so I think [Dylan Wilson is probably right](https://mathoverflow.net/questions/434699/reference-request-infinity-categories-for-the-commutive-algebraist-algebraic-ge#comment1119784_434699) that the best way to get this statement is to deduce it from Appendix D of SAG.
Also, the comment pointing out work of Positselski is worth following. His paper "[Remarks on derived complete modules and complexes](https://arxiv.org/abs/2002.12331v5)" discusses different definitions of $D\_{I-comp}(R)$, a comparison between them, and their formal properties. See also [his slides](https://users.math.cas.cz/%7Epositselski/derived-adic-complete.pdf), which include a discussion of the badly-behaved non-derived case that Bhatt mentioned.
Turning to sources not mentioned in the comments or in Bhatt's article, I recommend:
1. The stacks project, e.g., [this entry on derived completion](https://stacks.math.columbia.edu/tag/091N) following work of Greenlees and May. Even though this does not explicitly mention $\infty$-categories, much of the story for $\infty$-categories is inspired by work that came earlier using things like model categories.
2. Kiran Kedlaya's notes on prismatic cohomology, especially [this entry about derived completion](https://kskedlaya.org/prismatic/sec_derived-complete.html). Reading this, it seems Kedlaya has the same philosophical approach to the subject that Bhatt has. This makes sense, because Kedlaya writes that these lecture notes were inspired by [2018 notes from a course Bhatt taught](http://www-personal.umich.edu/%7Ebhattb/teaching/prismatic-columbia/), and it seems plausible to me that the statement the OP asks about would be contained in those notes. Similarly, [this paper by Bhatt and Scholze](https://arxiv.org/pdf/1905.08229.pdf) seems like required reading.
3. As a homotopy theorist, I cannot help but advertise [recent work by Barthel, Heard, and Valenzuela](https://arxiv.org/pdf/1808.00895.pdf), which revisits the work of Greenlees and May in (1), works out the theory of derived completion for comodules, and [puts both pieces together](https://arxiv.org/pdf/1511.03526.pdf) in the setting of stable $\infty$-categories, in a way I really like, conceptually.
For a commutative algebraist or algebraic geometer, I suppose (2) would be the easiest way to get deeper into the types of results Bhatt mentioned.
| 3 | https://mathoverflow.net/users/11540 | 436166 | 176,282 |
https://mathoverflow.net/questions/436108 | 4 | $\newcommand\Spt{\mathit{Spt}}\newcommand\GrAb{\mathit{GrAb}}$Let $A$ be a ring spectrum. Suppose that $A$ has a Künneth theorem — i.e. the homology theory $A\_\ast : \Spt \to \GrAb$ is a strong monoidal functor [1].
**Question:** Does it follow that $A$ is a module over Morava $K$-theory $K(h)$ for some prime $p$ and some $0 \leq h \leq \infty$?
An affirmative answer would be a variation on the theorem that every ring spectrum which is a *field* is a module over some $K(h)$. See e.g. Lurie's notes [Uniqueness of Morava $K$-theory](https://people.math.harvard.edu/%7Elurie/252xnotes/Lecture24.pdf).
[1] I have learned [here](https://mathoverflow.net/questions/435685/if-pi-ast-a-is-graded-commutative-then-is-a-ast-a-lax-monoidal-functor "If pi_* A is graded-commutative, then is A_* a lax monoidal functor?") that it's not so straightforward to say exactly what it means to "have a Künneth theorem" — right now I'm just assuming that there is *some* way to make $A\_\ast$ into a strong monoidal functor, but it seems in general there need not a canonical way to make $A\_\ast$ into a lax monoidal functor, unless $A$ is homotopy commutative. I think for a start, I'd be happy with an answer which assumes that $A$ is homotopy commutative, and assumes that the strong monoidal structure comes from the lax monoidal structure which exists canonically in this case.
| https://mathoverflow.net/users/2362 | If $A_\ast$ has a Künneth theorem, then is $A$ a module over Morava $K$-theory? | If you want $K(h)\_\*$ to be strong monoidal, then you need the target category to be the category of graded $K(h)\_\*$-modules, not the category of graded abelian groups. Thus, I assume you are really asking for conditions under which $A\_\*(-)$ gives a strong symmetric monoidal functor to the category of graded $A\_\*$-modules.
Recall that the wedge of spectra is the same as the categorical product. It follows that the wedge also gives the categorical product in the category of homotopy commutative ring spectra. Also, it is easy to see that $\text{GrMod}\_{A\_\*\times B\_\*}\simeq\text{GrMod}\_{A\_\*}\times\text{GrMod}\_{B\_\*}$ as symmetric monoidal categories. Thus, if $A$ and $B$ each have a Künneth isomorphism, so does $A\vee B$. In particular, any ring spectrum of the form $K(i)\vee K(j)$ has a Künneth isomorphism, but it is not a $K(h)$-module for any $h$ unless $i=j$.
| 7 | https://mathoverflow.net/users/10366 | 436168 | 176,284 |
https://mathoverflow.net/questions/433532 | 4 | Why can't the monoidal Dold–Kan correspondence be extended to non-connected CDGAs over a field of characteristic 0?
I understand that there is a technical problem with the original proof due to Quillen given in ["Rational Homotopy Theory" (Remark on p.223)](https://people.math.rochester.edu/faculty/doug/otherpapers/quillen-rational.pdf). However, I don't understand what is the conceptual reason for this.
Edit: In the initial post I mistakenly used the term connective to denote CDGAs that have the ground field in dim 0 and 0 in negative degrees. It was pointed out by Ben Wieland in the [comments](https://mathoverflow.net/questions/433532/monoidal-dold-kan-correspondence-for-non-connected-cdga#comment1116597_433532) that these algebras should be called connected instead.
| https://mathoverflow.net/users/143549 | Monoidal Dold–Kan correspondence for non-connected CDGA | The answer to the question:
>
> Why can't the monoidal Dold–Kan correspondence be extended to non-connected CDGAs over a field of characteristic 0?
>
>
>
is that in fact the Dold-Kan correspondence *can* be extended to this setting.
The classical Dold-Kan correspondence is an equivalence of categories between the category $ch^+$ of non-negatively graded chain complexes of abelian groups, and the category $sAb$ of simplicial abelian groups. The normalization functor $N: sAb \to ch^+$ is monoidal. So is the inverse functor $\Gamma: ch^+\to sAb$, thanks to the Alexander-Whitney map. The natural isomorphism $N\Gamma \cong Id\_{Ch^+}$ is monoidal, but the natural isomorphism $\Gamma N \cong Id\_{sAb}$ is not, as explained in the paper *[Equivalences of monoidal model categories](https://arxiv.org/pdf/math/0209342.pdf)* by Schwede and Shipley.
The Dold-Kan equivalence can be defined for any commutative ring $R$, analogously to the classical case $R = \mathbb{Z}$, leading to a Quillen equivalence between non-negatively graded $R$-modules and simplicial $R$-modules (where $R$ is regarded as a monoid in the category of simplicial abelian groups). This Quillen equivalence is weak monoidal rather than strong monoidal. The monoidality is enough to induce an adjoint pair on categories of $O$-algebras, for any operad $O$ (e.g., the commutative monoid operad).
Let $k$ be a field of characteristic zero. Then the Quillen equivalence $(N,\Gamma)$ is *nice* in the sense of Definition 3.5.5 of the paper [Homotopical Adjoint Lifting Theorem](https://link.springer.com/article/10.1007/s10485-019-09560-2) by me and Donald Yau. This is the content of Example 3.5.6. Consequently, Corollary 5.4.1 proves that the induced adjunction between CDGAs over $k$ and simplicial commutative $k$-algebras is a Quillen equivalence. Note that we place no restriction at all on the values of the CDGAs in degree zero, so we're in the "non-connected" setting of the question.
For completeness, let me unpack why this setting is "nice":
1. In $\mathcal{N} = ch^+\_k$ (with the projective model structure, which here coincides with the injective model structure), for any symmetric group $\Sigma\_n$, every object of $\mathcal{N}^{\Sigma\_n}$ is projectively cofibrant (essentially, $\Sigma\_n$ acts freely due to the characteristic being zero). So all the conditions of niceness are automatic for $\mathcal{N}$.
2. Let $\mathcal{M}$ denote the category of simplicial $k$-modules. Whenever $X\in \mathcal{M}^{\Sigma\_n}$ is cofibrant, so is the object of $\Sigma\_n$-coinvariants $X\_{\Sigma\_n}$. Furthermore, the domains of the generating cofibrations of $\mathcal{M}$ are cofibrant.
3. For cofibrant objects $W$ and $X$, the map $N^2: N(W\otimes X)\to NW \otimes NX$ is a weak equivalence and remains a weak equivalence after taking $\Sigma\_n$-coinvariants (if $W$ and $X$ have $\Sigma\_n$-actions), again, because $\mathcal{N}$ is so nicely behaved.
4. For any $X\in \mathcal{M}^{\Sigma\_n}$ cofibrant in $\mathcal{M}$, if $f$ is a (trivial) cofibration, then so is $X\otimes\_{\Sigma\_n} f^{\Box n}$, where $f^{\Box n}$ is the $n$-fold iterated pushout product. This condition guarantees that the homotopy theory of $\mathcal{M}$ plays nicely with symmetric powers, so that commutative monoids (and algebras over more general operads) inherit transferred model structures. It holds because we're in the characteristic zero setting.
As for the proof of Corollary 5.4.1, it comes down to Theorem 4.2.1, which works by a transfinite induction over the $Sym$ functor, where $Sym^n(X) = X^{\otimes n}/\Sigma\_n$. Because each of these functors preserves weak equivalences between cofibrant objects (again using the characteristic zero assumption), the transfinite induction works and proves the relevant morphism is a weak equivalence, to know that the induced adjoint pair is a Quillen equivalence.
| 2 | https://mathoverflow.net/users/11540 | 436171 | 176,285 |
https://mathoverflow.net/questions/435524 | 1 | Let $D$ be a Weil divisor on a normal toric variety $X$ with fan $\Sigma$ that is invariant under the action by the torus $T$. Then Proposition 4.3.2 of the textbook *Toric Varieties* by Cox, Little and Schenck says that the vector space of global sections of the invertible sheaf $\mathcal{O}\_X(D)$ associated to $D$ is given by:
$$\Gamma(X, \mathcal{O}\_X(D)) \cong \bigoplus\_{\operatorname{div}(\chi^m) + D \ge 0} \mathbb{C} \cdot \chi^m,$$
where $\operatorname{div}(\chi^m)$ is the divisor associated to a character $\chi^m$ of $T$. The toric varieties in their textbook are defined over the complex numbers.
Does an analogous theorem holds for real toric varieties? Namely, if $Y$ is a normal real toric variety and $D$ is a torus invariant Weil divisor on $Y$, is there an isomorphism:
$$\Gamma(Y, \mathcal{O}\_X(D)) \cong \bigoplus\_{\operatorname{div}(\chi^m) + D \ge 0} \mathbb{R} \cdot \chi^m?$$
| https://mathoverflow.net/users/136356 | Explicit description of the space of global sections of a torus invariant Weil divisor over a real toric variety | Yes, since the complex global sections are isomorphic to the tensor product of the real global sections with $\mathbb C$ by flat base change, and this isomorphism is equivalent for the action of the real torus, so when we write the complex global sections as a sum of eigenspaces this also writes the real global sections as a sum of eigenspaces.
| 1 | https://mathoverflow.net/users/18060 | 436176 | 176,286 |
https://mathoverflow.net/questions/417735 | 2 | Levy’s characterisation theorem for Brownian motion states that for a local martingale $X$ with $X\_0 = 0$, $X$ is a Brownian motion if and only if it has quadratic variation $\langle X, X \rangle\_t = t$.
The usual proofs of this fact use the characteristic function of the normal distribution. I am seeking an alternate proof in order to improve my intuition about the theorem.
**Question:** Is there a proof of this that does not go through the characteristic function?
| https://mathoverflow.net/users/173490 | Alternate proof of Levy’s characterisation of Brownian motion | Fix $A>0$; we will be proving that $X\_t\stackrel{\mathcal D}= B\_t$ on $[0;A].$ Let $T^{(m)}$ be an a.s. increasing sequence of stopping times such that $T^{(m)}\to\infty$ almost surely, and for each $m$, $X\_{n\wedge T^{(m)}}$ is a martingale. Then, $Y\_n=X\_{\epsilon n\wedge T^{(m)}}$, $n=0,1,\dots,$ is a discrete-time martingale satisfying $$\mathbb{E}((Y\_{n+1}-Y\_n)^2|\mathcal{F}\_{\epsilon n})=\mathbb{E}(\epsilon\wedge(T^{(m)}-\epsilon n)\_+|\mathcal{F}\_{\epsilon n})=:\epsilon\_n.$$
Put $t\_n:=\sum\_{i=0}^{n-1}\epsilon\_n$. Note that $0\leq\epsilon\_n\leq \epsilon$ almost surely, hence $\epsilon n-t\_n$ is non-negative and increasing. We infer that for $N=[A\epsilon^{-1}]$, $$\mathbb{P}(\max\_{n\leq N}|\epsilon n-t\_n|\geq \delta)=\mathbb{P}(\epsilon N-t\_N\geq \delta)\leq \delta^{-1}\mathbb{E}(\epsilon N-t\_N)=\delta^{-1}\mathbb{E}((N\epsilon-T^{(m)})\_+)\leq\delta^{-1}\mathbb{E}((A+1)-T^{(m)})\_+).$$ By monotone convergence, the right-hand side tends to zero as $n\to\infty,$ note that it also does not depend on $\epsilon.$ Therefore, given $\delta>0$, we can choose $m$ such that for all $\epsilon$,
$$
\mathbb{P}(\max\_{n\leq N}|\epsilon n-t\_n|\geq \delta)\leq \delta.
$$
By Skorokhod embedding theorem *[A. V. Skorohod, Studies in the theory of random processes, Addison-Wesley, Reading, Mass.,1965]*, there exists a Brownian motion $\{B\_t\}\_{t\geq 0}$, which we can take independent of $\{X\_t\}\_{t\geq 0}$, and a sequence of stopping times $0=\tau\_0<\tau\_1<\tau\_2<\dots$ such that $\{B\_{\tau\_n}\}\_{n\geq 0}$ has the same distribution as $\{Y\_n\}\_{n\geq 0}.$ Since both $B\_t$ and $X\_t$ are continuous, it is enough to show that $\max\_{n\leq A\epsilon^{-1}}|\tau\_n - n\epsilon|\to 0$ in probability as $\epsilon\to 0$ and then $m\to\infty$. In view of the above discussion, it is enough to show that for any fixed $m$, $\max\_{n\leq A\epsilon^{-1}}|\tau\_n - t\_n|\to 0$ in probability as $\epsilon\to 0$.
Recall that the construction in Skorokhod embedding theorem runs iteratively: conditionally on $\mathcal{F'\_n}=\sigma(\mathcal{F}\_{\epsilon n},\tau\_n, \{B\_{t}\}\_{0\leq t\leq\tau\_n})$, we construct $\tau\_{n+1}-\tau\_{n}$ as a stopping time for the Brownian motion $B\_{\tau\_n+t}-B\_{\tau\_n}$ such that $B\_{\tau\_{n+1}}-B\_{\tau\_n}$ has the same distribution as the conditional distribution of $Y\_{n+1}-Y\_{n}$. This implies:
$$\mathbb{E}(\tau\_{n+1}-\tau\_n|\mathcal{F}'\_n)=\mathbb{E}(B\_{\tau\_{n+1}}^2-B\_{\tau\_n}^2|\mathcal{F}'\_n)=\mathbb{E}(Y\_{n+1}^2-Y\_{n}^2|\mathcal{F}'\_n)=\epsilon\_n.$$ That is, $\tau\_n-t\_n$ is in fact a martingale. In particular, by Doob's inequality, we have $$\mathbb{P}(\max\_{n\leq N}|\tau\_n-t\_n|>\delta)\leq\delta^{-2}\mathbb{E}(\tau\_N-t\_N)^2,$$ which we will apply to $N=[A\epsilon^{-1}]$. In fact, Skorokhod embedding theorem guarantees that
$\mathbb{E}(\tau\_{n+1}-\tau\_n)^2\leq C\mathbb{E}(Y\_{n+1}-Y\_{n})^4$, thus
$$
\mathbb{E}(\tau\_{N}-t\_N)^2\leq \sum\_{n=0}^{N-1}\mathbb{E}(\tau\_{n+1}-\tau\_n)^2\leq \sum\_{n=0}^{N-1}\mathbb{E}(Y\_{n+1}-Y\_{n})^4.
$$
It remains to show that the right-hand side tends to zero as $\epsilon\to 0$. This is well-known for bounded, continuous local martingale, see Karatzas-Shreve, Lemma 5.10. We can reduce to this case by localization: since $X\_t$ is a.s. bounded on any finite interval, we have that $\hat{T}^{(m)}:=\min\{t:|X\_t|\geq m\}\to \infty$ almost surely, we can replace $T^{(m)}$ by $\hat{T}^{(m)}\wedge T^{(m)}$.
| 1 | https://mathoverflow.net/users/56624 | 436187 | 176,289 |
https://mathoverflow.net/questions/436045 | 3 | Let $A$ be a (unital) $C^\*$-algebra and $X,Y$ right Hilbert $A$-modules which are finitely generated and projective. It seems to be well-known that if $T: X \to Y$ is an $A$-linear map, then $T$ is necessarily adjointable. I could not find a proof though. Can someone give a reference or proof of this little fact?
| https://mathoverflow.net/users/216007 | Linear map between projective finitely generated Hilbert modules is adjointable | Looking at the proof of Lemma 6.21 in the notes of de Commer that you're reading (<https://arxiv.org/pdf/1604.00159.pdf>, per the comments), it seems like the relevant property of the modules is that of having compact identity operator. Suppose that a Hilbert $A$-module $X$ has this property. Since finite sums of the form $\sum\_{i} |x\_i\rangle\,\langle y\_i|$ are dense in the compact operators, some such sum must be very close to the identity, and hence invertible. Let $K=\left( \sum\_i |x\_i\rangle\,\langle y\_i|\right)^{-1}$. Then for every Hilbert module $Y$ and every $A$-linear map $T:X\to Y$ we have
\begin{equation\*}
T = T\circ \operatorname{id}\_{X} = \sum\_i |TKx\_i\rangle\,\langle y\_i|,
\end{equation\*}
which is compact and therefore adjointable.
To tie this to the actual question as asked: for Hilbert modules over unital $C^\*$-algebras, the property of having compact identity operator is equivalent to being finitely generated and projective in the purely algebraic sense, and it is also equivalent to being the image of an orthogonal projection on the Hilbert module $A^n$ for some $n$. These equivalences are explained in Wegge-Olsen's book on $K$-theory.
| 5 | https://mathoverflow.net/users/85913 | 436195 | 176,291 |
https://mathoverflow.net/questions/436194 | 5 | Let $C\_{lsc}(\mathbb{R}^n)$ be the space of lower semicontinuous convex functions $\mathbb{R}^n \to \mathbb{R}$. The Legendre-Fenchel (LF) transform of $f \in C\_{lsc}(\mathbb{R}^n)$ is:
$$ f^\*(y) := \sup\_{x \in \mathbb{R}^n} (\langle x, y \rangle - f(x)) $$
It is known that the LF transform is continuous and an involution on $C\_{lsc}(\mathbb{R}^n)$ (Wijsman 1963). I want to know if the LF transform is Lipschitz. That is, given $f$ and $g$ lsc, is there a way to bound $\|f^\* - g^\*\|$ by $\|f-g\|$ (under a standard norm)?
I haven't found any sources that do this, so my suspicion is that the LF transform is not Lipschitz. Does anyone know if this is true, or if not a simple counterexample?
It is worth noting that Attouch and Wets ("Isometries of the Legendre-Fenchel transform") constructed norms under which the LF transform is an isometry -- however these norms are not particularly useful for me. I am looking for any results using standard norms (any of the p-norms).
| https://mathoverflow.net/users/136732 | Is the Legendre transform as an operator Lipschitz? | This is basically true for sup norm by Fenchel's inequality. Indeed, for all $y$,
$$
f^\*(y) = \sup\_x\left( \langle x,y\rangle - f(x) \right) \leq \sup\_x\left( g(x)+g^\*(y) - f(x) \right) \leq \|f-g\|\_{\infty} + g^\*(y).
$$
The same is true when the roles of $f,g$ are reversed, giving essentially what you want. One should take care, though, in writing the final conclusion because it is possible for $f^\*(y) = g^\*(y) = +\infty$, despite $\|f-g\|\_{\infty}=0$ (e.g., take $f=g=0$). That said, if $\|f-g\|\_{\infty}<\infty$, then it is clear from above that
$$
\{y : f^\*(y) = +\infty\} = \{y : g^\*(y) = +\infty\},
$$
so it is reasonable to write
$$
\|f^\*-g^\*\|\_{\infty} \leq \|f-g\|\_{\infty},
$$
provided you adopt the convention $|(+\infty)-(+\infty)| = 0$ for handling the indeterminate form that can appear in defining the sup norm on the LHS.
| 6 | https://mathoverflow.net/users/99418 | 436199 | 176,293 |
https://mathoverflow.net/questions/436173 | 2 | Following this question: [Can we apply the continuous mapping theorem for the limiting joint distribution of the Tracy-Widom law?](https://mathoverflow.net/questions/436028/can-we-apply-the-continuous-mapping-theorem-for-the-limiting-joint-distribution).
We know that
>
> $$
> \lim\_{N\to\infty}P(N^{2/3}(\lambda\_N-2)\le s\_1,\dotsc, N^{2/3}(\lambda\_{N-k+1}-2)\le s\_k)=F\_{\beta, k}(s\_1,\dotsc, s\_k).
> $$
>
>
>
Can we say that for any $\epsilon>0$, there exists a constant $c>0$ such that
$$
P(N^{2/3}(\lambda\_N-\lambda\_{N-1})\le c)\ge 1-\epsilon?
$$
Or assume that $(N^{2/3}(\lambda\_N-2),\dotsc, N^{2/3}(\lambda\_{N-k+1}-2))\to (Y\_1,\dotsc, Y\_{N-k+1})$ in distribution.
The question becomes that
for any $\epsilon>0$, there exists a constant $c>0$ such that
$$
P(Y\_N-Y\_{N-1}\le c)\ge 1-\epsilon?
$$
---
We can say $N^{2/3}(\lambda\_N-\lambda\_{N-1})=O\_p(1)$.
| https://mathoverflow.net/users/168083 | Can we get that $ P(N^{2/3}(\lambda_N-\lambda_{N-1})\le c)\ge 1-\epsilon$? | The probability distribution of the spacing $\delta\_N=\lambda\_N-\lambda\_{N-1}$ of the eigenvalues $\lambda\_N$ and $\lambda\_{N-1}$ at the edge of the spectrum decays exponentially for $\delta\_N\gg N^{-2/3}$, with a decay rate that is independent of $N$. So no matter how small $\epsilon$, you can always find a $c$ such that $P(N^{2/3}\delta\_N\le c)\ge 1-\epsilon$.
I agree with the ChatGPT bot in the deleted answer that $c$ will depend on $\epsilon$, but I do not agree that $c$ will depend on $N$. The scaling with $N$ is fully contained in the mean level spacing $N^{-2/3}$ for large $N$ (Tracy-Widom law).
| 3 | https://mathoverflow.net/users/11260 | 436211 | 176,297 |
https://mathoverflow.net/questions/436209 | 3 | Suppose a function $f(x): \mathbb R^d \mapsto \mathbb R^D$, and its stochastic approximator, $g(x; W): \mathbb R^d \mapsto \mathbb R^D$. Here $W$ is some random variable. Then $g(x; W)$ is unbiased in the sense that
$$\mathbb E\_W [g(x;W)] = f(x),$$
for any $x$.
I think the following two are not equal, but how to prove it?
$$\mathbb E\_W\left[\frac{dg(x;W)}{dx}\right] \text{ vs. } \frac{df(x)}{dx}.$$
| https://mathoverflow.net/users/34972 | Expected gradient vs. gradient of expectation | Indeed, the equality will not hold in general. For counterexamples, see [this](https://mathoverflow.net/a/105773/36721) or [this](https://mathoverflow.net/a/172898/36721).
For sufficient conditions for the equality when $d=1$, see e.g. [Folland, Theorem 2.27](http://home.ustc.edu.cn/%7Eluke2001/pdf/realfolland.pdf) or the more general [Lemma 2.3](https://projecteuclid.org/journals/annals-of-probability/volume-43/issue-5/Exact-Rosenthal-type-bounds/10.1214/14-AOP942.full). This immediately extends to any $d$ if the derivatives are understood in the [Gateaux](https://en.wikipedia.org/wiki/Gateaux_derivative) sense.
As seen from the discussion of Theorem 2.27 in [Folland](http://home.ustc.edu.cn/%7Eluke2001/pdf/realfolland.pdf), the extension to $d>1$ is somewhat more problematic if the derivatives are understood in the [Fréchet](https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative) sense. [This answer](https://mathoverflow.net/a/105775/36721) may also be of interest to you.
| 5 | https://mathoverflow.net/users/36721 | 436224 | 176,304 |
https://mathoverflow.net/questions/428448 | 2 | Assume that $B$ is a complete boolean algebra endowed with a Hausdorff topology, with respect to which all operations on $B$ are continuous, $0$ has a base of full sets (recall that $A\subset B$ is full if $b\le a\in A\Rightarrow b\in A$), and for any net $(b\_i)$ in $B$ which decreases to $0$ we have $b\_i\to 0$.
>
> What can be said about $B$? Is $B$ a measure algebra? Can existence of such a topology characterized in terms of $B$?
>
>
>
Generally, what is a good references on topologies on Boolean algebras?
| https://mathoverflow.net/users/53155 | Is a Boolean algebra with an order continuous topology a measure algebra? | It is not true that $B$ is necessarily a measure algebra. The counterexample is due to Michel Talagrand, who constructed a Maharam algebra that is not a measure algebra.
>
> *Maharam, D.*, [**An algebraic characterization of measure algebras**](http://dx.doi.org/10.2307/1969222), Ann. Math. (2) 48, 154-167 (1947). [ZBL0029.20401](https://zbmath.org/?q=an:0029.20401).
>
>
>
>
> *Talagrand, Michel*, [**Maharam’s problem**](http://dx.doi.org/10.4007/annals.2008.168.981), Ann. Math. (2) 168, No. 3, 981-1009 (2008). [ZBL1185.28002](https://zbmath.org/?q=an:1185.28002).
>
>
>
The relevant material can also be found in Fremlin's textbook *Measure Theory*, specifically in [Chapter 39](https://www1.essex.ac.uk/maths/people/fremlin/chap39.pdf).
First I'll give some background, which I think will be helpful for your more general question, as well as relating Talagrand's example to your more specific question. A *submeasure* on a Boolean algebra $A$ is a function $A \rightarrow [0,\infty]$ that:
1. $\mu(0) = 0$
2. $\mu$ is monotone
3. $\mu$ is subadditive: $\mu(a \lor b) \leq \mu(a) + \mu(b)$.
A *finite* submeasure is one where $\mu(1) < \infty$, and a *positive* (sometimes called *strictly positive*) submeasure is one such that $\mu(a) = 0$ implies $a = 0$.
Proofs about finitely-additive measures that only use axioms 1 to 3 without using additivity for disjoint elements go through for submeasures. In particular, if $\mu$ is finite we have a pseudometric on $A$:
$$
d(a,b) = \mu(a \triangle b)
$$
which is a metric iff $\mu$ is positive.
It is not difficult to prove that for all $\mu$, the functions $\lnot, \land, \lor$ and $\mu$ are uniformly continuous with respect the metric $d$. The monotonicity of $\mu$ implies that an $\epsilon$-ball around $0$ is down-closed (a "full set" in your terminology) so $0$ has a base of full sets.
If we try out the submeasure that has $\mu(0) = 0$ and $\mu(a) = 1$ for all $a \neq 0$, then $d$ is the discrete metric with distance $1$ between distinct elements. This shows we need an extra assumption to get your third requirement on the topology on $A$. In fact we only need to add a version of it for sequences:
A submeasure is *continuous* (called *Maharam* in Fremlin 393A) if for each non-increasing sequence $(a\_n)\_{n \in \mathbb{N}}$ with $\bigwedge\_{n=0}^\infty a\_n = 0$ we have $\lim\_{n \to \infty}\mu(a\_n) = 0$.
If $A$ is $\sigma$-complete and $\mu$ is finite, strictly positive and continuous, then $A$ is c.c.c. (Fremlin 393C), which is to say a disjoint set of non-zero elements is countable. It follows that $\mu$ preserves infima of downward directed nets (see *e.g.* [Fremlin 316F(c)](https://www1.essex.ac.uk/maths/people/fremlin/chap31.pdf), though try your own proof using Zorn's lemma or the well-ordering theorem). Therefore your third condition is also satisfied, and also $A$ is a complete Boolean algebra.
So a (sigma-)complete Boolean algebra with a continuous positive finite submeasure (known as a *Maharam algebra*) will satisfy your criteria. By Theorem 1.2 of Talagrand's paper, there are such algebras that are not measure algebras, in fact admitting no non-zero finite measures at all.
You can find many interesting proofs and references to the literature for results characterizing both Maharam algebras and measure algebras in Fremlin's §393, e.g. 393E, as well as in Talagrand's article.
| 1 | https://mathoverflow.net/users/61785 | 436249 | 176,308 |
https://mathoverflow.net/questions/436261 | 1 | I’m looking for a version of the Cameron Martin theorem for the Brownian motion under random shifts. Here is the precise statement:
Let $\mathbb P$ be Wiener measure on $\Omega := C[0, 1]$. Given a $C[0, 1] $ valued random variable $F$, define the translation map $T\_F: \Omega \to \Omega$ by $T\_F (\omega) = \omega + F(\omega)$, and denote the pushforward of $\mathbb P$ under this map by $T\_F^{\ast} \, \mathbb P $.
Denote by $W^{1, 2}$ the space of absolutely continuous functions on $[0, 1]$ with derivative in $L^2$.
>
> **Theorem:** $\mathbb Q$ is equivalent to $\mathbb P$ if and only if $\mathbb Q = T\_F ^\ast \, \mathbb P$ for some $F$ such that $F(0) = 0$ and $F \in W^{1, 2}$ almost surely.
>
>
>
My questions are two-fold:
1. Is this statement true?
2. If so, where can I find a proof?
So far the statements I could find were only for deterministic
shifts $F$, these are what is usually called the Cameron Martin theorem
for Brownian motion. On the other hand, statements of Girsanov’s theorem deal with random shifts, but typically are concerned with the Radon Nikodym derivative, and are silent on the interpretation as shifts by $W^{1, 2}$ processes.
| https://mathoverflow.net/users/173490 | Full version of Cameron Martin theorem for Brownian motion | Let $\varphi$ be a smooth function such that $\varphi(x)=0$ for $x\leq 0$ and $\varphi(x)=1$ for $x\geq 1$, and $0\leq\varphi(x)\leq 1$ else. If $\tau\_1$ and $\tau\_2$ are is the first times $B\_t$ hits $1,2$ respectively, and $M$ is a maximum of the $B\_t$ on $[0,1]$, put $F\_t=-M\varphi(\frac{t-\tau\_1}{\tau\_2-\tau\_1})$. Then, $\sup (B\_t+F\_t)\leq 2$ almost surely, hence $\mathbb{P}$ is not absolutely continuous with respect to $\mathbb{Q}$.
| 3 | https://mathoverflow.net/users/56624 | 436265 | 176,314 |
https://mathoverflow.net/questions/436244 | 5 | First, some motivation. Let $X$ be a complex manifold, and $A$ a Hermitian connection on some complex vector bundle $E$ over $X$. It is known that the existence of $A$ such that the $(0,2)$-part of the curvature form $F\_A$ vanishes implies the existence of a holomorphic structure on $E$.
Now let $M$ be an almost complex manifold, and $A$ a Hermitian connection on the tangent bundle $TM$.
>
> Does $F\_{A}^{0,2}=0$ imply the existence of a complex structure on $M$?
>
>
>
After discussing this with my professor, he believed that this might not be the case, but I've yet to find a counterexample. The only place one would know to look so far is on an almost complex $4$-manifold that does not admit a complex structure. Otherwise, if the result turns out to be true, is there a standard reference on this matter?
| https://mathoverflow.net/users/143629 | Does $F_{A}^{0,2}=0$ for a connection $A$ on $TM$ almost complex give a complex structure? | Here is an example to think about: Let $S^6 = \mathrm{G}\_2/\mathrm{SU}(3)$ be the $6$-sphere endowed with its $\mathrm{G}\_2$-invariant almost Hermitian structure. There is a $\mathrm{G}\_2$-invariant (special) Hermitian connection $A$ on the tangent bundle of $S^6$ whose curvature is of type $(1,1)$, but the only $\mathrm{G}\_2$-invariant almost-complex structures on $S^6$ (and there are exactlty two) are not integrable.
The point is that the (special) Hermitian connection $A$ has *torsion* of type $(0,2)$, and that is what prevents the almost-complex structure from being integrable.
Of course, there might still exist *some* complex structure on $S^6$ (whether this is so is not currently known), but the existence of an Hermitian connection on its tangent bundle with curvature of type $(1,1)$ has proved of no help, so far, in finding a complex structure on $S^6$.
Here is an actual counterexample: S. T. Yau has shown that there are compact parallelizable 4-manifolds $M^4$ that support no complex structure. See S. T. Yau, *Parallelizable manifolds without complex structure*, Topology **15** (1976), 51–53. Since the tangent bundle is trivial, there is an almost-complex structure $J$ on $M$ for which the tangent bundle of $M$ is trivial as a complex vector bundle. Hence, the tangent bundle of $M$ supports a flat $J$-complex connection, i.e., a connection $A$ for which the complex structure is parallel and $F\_A = 0$. *A fortiori*, we have $F^{0,2}\_A = 0$, but $M$ does not admit a complex structure. Of course, the torsion of $A$ is non-zero and, in fact, must have a nontrivial $(0,2)$-piece.
| 6 | https://mathoverflow.net/users/13972 | 436274 | 176,317 |
https://mathoverflow.net/questions/436275 | 2 | For $X$ compact metric spaces and $f:X\to X$ continuous, is there a nice characterization of the systems $(X,f)$ for which, for every pair of points $x,y\in X$ with disjoint orbits, we have $\omega(x)\cap\omega(y)=\emptyset$?
| https://mathoverflow.net/users/167834 | Dynamical systems with disjoint $\omega$-limits of single points | This seems to be true if and only if every $x \in X$ is eventually periodic.
The reverse direction is obvious; if all points of $X$ are eventually periodic and $x, y$ have disjoint orbits, then each has $\omega$-limit set equal to a different periodic orbit, which must be disjoint.
For the forward direction, assume your condition. For every $x \in X$, $\omega(x)$ is a closed $f$-invariant set, and therefore contains a minimal subsystem $M$. If the orbit of $x$ does not eventually enter $M$, you already contradict your condition; for arbitrary $m \in M$, $\omega(m) = M$. Then $x$ and $m$ are in disjoint orbits, but $\omega(x) \cap \omega(m) = M \neq \varnothing$.
If $M$ does not consist of a single orbit, then it contains $x \neq y$ in different orbits, but by minimality, $\omega(x) = \omega(y) = M$, again contradicting your condition. So $M$ consists of a single orbit, say $O(m)$. If $m$ is not periodic, then $M$ is a countable closed set in a complete metric space, therefore it has an isolated point $f^n m$. But then $O(f^{n+1} m)$ is not dense in $M$, contradiction to minimality.
So, every minimal subsystem $M$ is a periodic orbit, and every $x \in X$ has orbit eventually in one of these periodic orbits, therefore every $x$ is eventually periodic.
Throughout I've not assumed invertibility; if $(X, f)$ were invertible, then you remove the word "eventually."
| 6 | https://mathoverflow.net/users/116357 | 436277 | 176,318 |
https://mathoverflow.net/questions/436278 | 8 | I would like to know if there is a special name for the following concept, papers that feature something similar or a general reference. Let $\mathcal{C}$ be a category and $\mathcal{D}$ a subcategory (or more generally a subclass of objects and morphisms). For a diagram $F:J\to\mathcal{C}$, say that the $\mathcal{D}$-limit of $F$ is a cone in $\mathcal{C}$ that satisfy a universal property respect to all cones whose vertex and morphisms to the vertex are in $\mathcal{D}$. $\mathcal{D}$-colimit would be the same concept but with cocones. It is clear that a $\mathcal{C}$-limit is precisely a usual limit, while any cone is a $\emptyset$-limit.
I am particularly interested in understanding in greater generality the behaviour of the examples in the following questions:
-[Do colimits of manifolds coincide with underlying colimits as topological spaces?](https://mathoverflow.net/questions/396340/do-colimits-of-manifolds-coincide-with-underlying-colimits-as-topological-spaces)
-[Are GIT's good categorical quotients just locally ringed space coequalizers?](https://mathoverflow.net/questions/59812/are-gits-good-categorical-quotients-just-locally-ringed-space-coequalizers)
-[Colimits of manifolds](https://mathoverflow.net/questions/38575/colimits-of-manifolds)
[M-complete category](https://ncatlab.org/nlab/show/M-complete+category) is certainly related but too specific.
| https://mathoverflow.net/users/494777 | Reference for certain categorical limits | This is such a natural notion, that I feel it must be written about somewhere, but I don't think it's in any of my usual sources like Borceux's books, Mac Lane's *Categories for the working mathematician*, or Emily Riehl's book *Category Theory in Context*.
It seems to me that such a notion, if previously invented, might be called a "relative limit" or "relative colimit." It feels like the sort of thing Grothendieck could have written about. Googling led me to writings of Lurie definining the analogous notion for $\infty$-categories, e.g., in *Higher Topos Theory* (4.3.1) and at [this link in Kerodon](https://kerodon.net/tag/02KF).
The theory of Mahavier limits is also related, as [this paper by Ittay Weiss spells out](https://pure.port.ac.uk/ws/files/8457717/MahavierLimitsUni.pdf).
| 4 | https://mathoverflow.net/users/11540 | 436288 | 176,320 |
https://mathoverflow.net/questions/436290 | 6 | (This is in a sense a follow-up to my [earlier question on a geometric definition of globular $\infty$-groupoids](https://mathoverflow.net/questions/435758/is-there-a-geometric-definition-of-globular-infty-groupoids-categories))
---
We know by [Scholie 8.4.14 of Cisinski's thesis](http://www.numdam.org/item/?id=AST_2006__308__R1_0) that the globe category $\mathbb{G}$ is not a weak test category. Thus, there is no model structure on $\mathsf{Fun}(\mathbb{G}^\mathsf{op},\mathsf{Set})$ such that:
* This model structure models $\infty$-groupoids;
* The cofibrations in this model structure are precisely the monomorphisms;
* The $n$-globes $G\_n:=\mathrm{Hom}\_{\mathbb{G}}(-,[n])$ are contractible.
Nevertheless, there could still in theory be a model structure on $\mathsf{Fun}(\mathbb{G}^\mathsf{op},\mathsf{Set})$ modelling $\infty$-groupoids, even though it might be really bad behaved, and some of the constructions in it would be very ad hoc¹.
¹In particular we would need to somehow construct a geometric realisation functor without using the abstract procedure in [nLab, nerve and realization](https://ncatlab.org/nlab/show/nerve+and+realization), as that recovers only those homotopy types which are wedge sums of spheres (as Simon Henry [explained to me here](https://mathoverflow.net/a/435807)).
**Question.** Is the following statement true?
* There exists no model structure on $\mathsf{Fun}(\mathbb{G}^\mathsf{op},\mathsf{Set})$ that is Quillen equivalent to the Kan–Quillen model structure on simplicial sets.
| https://mathoverflow.net/users/130058 | Inexistence of a Kan–Quillen model structure on globular sets | Zhen Lin points out below that I've been way too cavalier with transferring model structures along a reflection. So the following answer is not clearly correct. I will leave this up as community wiki because I still think it addresses the spirit of the question, showing that spaces can be "modeled" in some sense by globular sets (or even just by graphs).
---
Contrary to my guess in the comments, the answer is *no*: there does exist a model structure on globular sets (reflexive or otherwise) which is Quillen equivalent to the Kan-Quillen model structure on spaces.
To see this, note that if $\mathcal A$ is a reflective subcategory of $\mathcal B$, and if $\mathcal A$ has a model structure, then the model structure transfers to $\mathcal B$, and the resulting adjunction is a Quillen equivalence.
Now, as mentioned in the comments, the category $Gph$ of graphs (reflexive or otherwise) is a reflective subcategory of $Glob$. Moreover, the category $Pos$ of posets is a reflective subcategory of $Gph$. Thus $Pos$ is reflective subcategory of $Glob$. So it will suffice to find a model structure on $Pos$ which is Quillen equivalent to topological spaces. This is proven by [Raptis](https://dx.doi.org/10.4310/HHA.2010.v12.n2.a7), by transferring the [Thomason model structure](https://ncatlab.org/nlab/show/Thomason+model+structure) on $Cat$.
| 1 | https://mathoverflow.net/users/2362 | 436295 | 176,321 |
https://mathoverflow.net/questions/435698 | 4 | For an algebraic number field $K$, let $\zeta\_K(s)$ be the Dedekind zeta function associated to $K$, and let $\zeta\_K'(s)$ be its derivative.
I believe that the following statement is true:
$$\zeta\_K\left(\dfrac12\right)\neq 0 \implies \zeta\_K'\left(\dfrac12\right)\neq0.$$
I maganged to prove this for quadratic and cubic number fields, but I'm unsure how to prove the general case rigorously. Any thoughts or insight on how one could go about proving it would be really appreciated.
Thank you!
| https://mathoverflow.net/users/175440 | Proving $\zeta_K\left(\frac12\right)\neq 0 \implies \zeta_K'\left(\frac12\right)\neq0?$ | If $\zeta\_K(1/2) \neq 0$ then $\zeta'\_K(1/2) = 0$ if and only if
$$
\log |D\_K| = (\log(8\pi) + \gamma) n + \frac\pi2 r\_1,
$$
where $D\_K$ is the discriminant of $K$, and
$\gamma = 0.5772156649\ldots$ is Euler's constant.
We expect that this is impossible because such an equality would give
a closed form for $\gamma$, which should not exist.
Moreover, unless some $(r\_1,r\_2)$ yields
$|D\_K|$ surprisingly close to an integer, we expect to be able to prove
that $[K:{\bf Q}] \leq N$ is impossible in time polynomial in $N$;
for instance it takes only a few seconds in **gp** to do this for $N=200$
(the nearest is about $3.567 \cdot 10^{-5}$, for $(r\_1,r\_2) = (28,36)$).
But I doubt that we can expect to prove this for all $r\_1$ and $r\_2$
in the foreseeable future.
One might hope instead to prove that $\zeta'\_K(1/2) \neq 0$ by showing that
$|D\_K|$ would have to be too small for a number field of given $r\_1,r\_2$.
Unfortunately this is not known either, even under the Generalized Riemann
Hypothesis (GRH): the best we can prove is that if $\zeta\_K$ satisfies GRH then
$$
\log |D\_K| > (\log(8\pi) + \gamma - o(1)) n + (\frac\pi2 - o(1)) r\_1
$$
as $n \to \infty$.
(The appearance of the same linear combination of $r\_1$ and $r\_2$
is not coincidental; see for instance these lecture notes on the
discriminant bound:
[https://abel.math.harvard.edu/~elkies/M229.19/disc.pdf](https://abel.math.harvard.edu/%7Eelkies/M229.19/disc.pdf).)
To derive the condition on $D\_K$,
start from the functional equation $\xi\_K(s) = \xi\_K(1-s)$ where
$$
\xi\_K(s) = \Gamma(s/2)^{r\_1} \Gamma(s)^{r\_2}
(4^{-r\_2} \pi^{-n} |D\_K|)^{s/2} \zeta\_K(s).
$$
(This was already suggested in the comments by **edward cornfoot** and **reuns**.)
Thus $\xi'\_K(1/2) = 0$. Hence if $\zeta\_K(1/2) \neq 0$ then
$\xi\_K(1/2) \neq 0$, so the logarithmic derivative $\xi'\_K / \xi\_K$
also vanishes at $s = 1/2$. But this logarithmic derivative is
$$
\frac{r\_1}{2}( \psi(1/4) - \log\pi) + r\_2 (\psi(1/2) - \log 2\pi) + \frac12 \log |D\_K| + \frac{\zeta'\_K(1/2)}{\zeta\_K(1/2)}
$$
where $\psi$ is the logarithmic derivative of the Gamma function.
Therefore $\zeta'\_K(1/2) = 0$ if and only if
$$
\log|D\_K| = r\_1 (\log \pi - \psi(1/4) ) + 2r\_2 (\log 2\pi - \psi(1/2)).
$$
Our condition then follows from the known special values
$$
\psi(1/2) = -\log 4 - \gamma, \quad \psi(1/4) = -\frac\pi2 - \log 8 - \gamma.
$$
| 17 | https://mathoverflow.net/users/14830 | 436297 | 176,322 |
https://mathoverflow.net/questions/436296 | 1 | I was reading an article on Probabilistic Number Theory by M.Kac where I am not able to understand why a particular equation mentioned [here](https://projecteuclid.org/journalArticle/Download?urlId=bams%2F1183513946) in page $657$ equation $(7.7)$ is true?
I do understand that $\frac{(\nu(m)-\log\log n)^2}{\log\log n}<\omega^2$ and that it follows that $\frac{(\nu(m)-\log\log n)^2}{n\log\log n}<\frac{\omega^2}{n}$. But why after summing over $m$ we are getting a integral of $\omega^2$? Specifically, why did the inequality changed into an equality?
If somebody could explain this to me I would really appreciate it.
| https://mathoverflow.net/users/483436 | Why is $\sum_{m=1}^{n}\frac{(\nu(m)-\log\log n)^2}{n\log\log n}=\int_{-\infty}^{\infty}\omega^2\, \mathrm{d}\sigma_n(\omega)$? | Denote $f\_m=(\nu(m)-\log\log n)(\log\log n)^{-1/2}$ and define $\rho(\omega)=\sum\_{m=1}^n\delta(\omega-f\_m)$, with $\delta(x)$ the [Dirac delta function](https://en.wikipedia.org/wiki/Dirac_delta_function). Because of the identity $\int g(\omega)\delta(\omega-\omega\_0)\,d\omega=g(\omega\_0)$, one has
$$n^{-1}\sum\_{m=1}^n f\_m^2=n^{-1}\int \omega^2\rho(\omega)\,d\omega.$$
The function $\sigma(\omega)$ introduced by Kac is given by$^\ast$
$$\frac{d}{d\omega}\sigma(\omega)=n^{-1}\rho(\omega),$$
so one may equivalently write
$$n^{-1}\sum\_{m=1}^n f\_m^2=\int \omega^2\,d\sigma(\omega),$$
which is the formula desired by the OP.
$^\ast$ The relation between $\sigma(\omega)$ and $\rho(\omega)$ follows from the identity $d\theta(\omega)/d\omega=\delta(\omega)$, with $\theta(\omega)$ the unit step function.
| 4 | https://mathoverflow.net/users/11260 | 436299 | 176,323 |
https://mathoverflow.net/questions/436310 | 4 | Are there infinitely many pairs of positive integers $(a,b)$ such that $2(6a+1)$ divides $6b^2+6ab+b-6a^2-2a-3$? That is, if there are infinitely many different $a$ and for which at least one value of $b$ can be found for a given $a$. Some $a$ values are $0,2,3,7,11,17....$
I think that the answer is yes but I have no idea how to show this.
If the answer is yes, are there any polynomial parametric solutions?
| https://mathoverflow.net/users/265714 | Division problem | If $6a+1$ is a prime and a quadratic residue modulo $17$ (which is true for infinitely many values of $a$), then there are infinitely many positive integers $b$ with the required property.
First observe that $b$ is good if and only if $$f(a,b):=6b^2+6ab+b-6a^2-2a-3$$ is even and divisible by $6a+1$. Hence $b$ must be odd: $b=2c+1$. Now we need to find infinitely many positive integers $c$ such that $f(a,2c+1)$ is divisible by $6a+1$. The identity
$$6f(a,2c+1) + (6a-5-12c)(6a + 1)=(12c+6)^2-17$$
shows that $c$ is good if and only if $(12c+6)^2-17$ is divisible by $6a+1$. So we need to guarantee that the congruence
$$(12c+6)^2\equiv 17\pmod{6a+1}$$
has a solution. This is equivalent to $17$ being a quadratic residue modulo $6a+1$. By quadratic reciprocity, this is equivalent to $6a+1$ being a quadratic residue modulo $17$, and we are done.
| 11 | https://mathoverflow.net/users/11919 | 436319 | 176,331 |
https://mathoverflow.net/questions/436321 | 15 | There is a (most likely folklore) theorem - if in a category every morphism has a right inverse then that category is a groupoid. The proof is an honest oneliner: for $x:A\to B$ find $x':B\to A$ with $xx'=\operatorname{id}\_B$; now for $x'$ find $x'':A\to B$ with $x'x''=\operatorname{id}\_A$. Then $x''=xx'x''=x$, so $x'x=\operatorname{id}\_A$ too.
I wonder if this admits a generalization showing that only part of the horn filling conditions for Kan complexes suffice. Let me recall that a simplicial set $X\_\bullet$ is a Kan complex iff any morphism $\Lambda\_i[n]\to X\_\bullet$ extends along the inclusion $\Lambda\_i[n]\hookrightarrow\Delta[n]$, for all $0\leqslant i\leqslant n$. Here $\Lambda\_i[n]$ is the boundary of the generic $n$-simplex $\Delta[n]$ with the $i$th facet removed.
So for a starter, let me ask this. Suppose I require on $X\_\bullet$ all those extension conditions except for $\Lambda\_2[2]$. Is $X\_\bullet$ still a Kan complex?
Looks like this should be true but I cannot finish the argument even for the particular case of $\Lambda\_2[2]$ when $d\_1$ is degenerate. Thus for a $1$-simplex $x$ with $d\_0(x)=A$, $d\_1(x)=B$ we have to find (using all $\Lambda\_i[n]$ except for $\Lambda\_2[2]$) a $2$-simplex $\tau$ with $d\_0(\tau)=x$ and $d\_1(\tau)=s\_0(A)$.
I begin imitating the groupoid statement. Using $\Lambda\_0[2]$, I find a $2$-simplex $\sigma$ with $d\_1(\sigma)=s\_0(B)$ and $d\_2(\sigma)=x$; let $x'=d\_0(\sigma)$. Again using $\Lambda\_0[2]$, find another $2$-simplex $\rho$ with $d\_1(\rho)=s\_0(A)$ and $d\_2(\rho)=x'$. Let $x''=d\_0(\rho)$. Now using $\Lambda\_1[2]$, find $\pi$ with $d\_0(\pi)=x$ and $d\_2(\pi)=x'$, and let $i=d\_1(\pi)$. The problem is that there is no reason for $i$ to be $s\_0(A)$. We certainly can use $\Lambda\_0[2]$ again to find $\iota$ with $d\_2(\iota)=i$, $d\_1(\iota)=s\_0(A)$, but so what?
I feel like there is lots of reserve unused as I have all the higher Kan conditions available but still cannot figure out how to do it.
And in fact, I would like to omit each of the $\Lambda\_n[n]$ if possible...
| https://mathoverflow.net/users/41291 | Is there a higher analog of "category with all same side inverses is a groupoid"? | Yes, this is possible. The following is a classical result of the theory of quasi-categories (You'll find it in the early part of Lurie's Higher topos theory or in Joyal notes on quasi-categories - where they discuss equivalences in quasi-categories):
**Proposition:** A quasi-category $X$ is a Kan complex if and only its homotopy category is a groupoid.
So if you have the lifting property against the $\Lambda\_i[n] \to \Delta[n]$ for $0<i<n$ and the lifting property against $\Lambda\_0[2] \to \Delta[2]$ then you can show that every arrow in the homotopy category has a one sided inverse using this last lifting property, and hence the homotopy category is a groupoid, which shows that $X$ is a Kan complex by the proposition.
| 21 | https://mathoverflow.net/users/22131 | 436322 | 176,332 |
https://mathoverflow.net/questions/436316 | 11 | Let $f\colon\mathbb R^2\to\mathbb R$ be a measurable function such that
\begin{equation\*}
F(t):=\int\_{\mathbb R}dx\,f(t,x)
\end{equation\*}
exists and is finite for all real $t$. Suppose that
\begin{equation\*}
f\_t(t,x):=\frac{\partial f(t,x)}{\partial t}
\end{equation\*}
exists and is finite for all real $t,x$, and also suppose that
\begin{equation\*}
\int\_{\mathbb R}dx\, f\_t(t,x)
\end{equation\*}
exists and is finite for all real $t$.
Then, under certain additional conditions,
\begin{equation\*}
F'(t)=\int\_{\mathbb R}dx\, f\_t(t,x) \tag{1}\label{1}
\end{equation\*}
for all real $t$; see e.g. [Folland, Theorem 2.27](http://home.ustc.edu.cn/%7Eluke2001/pdf/realfolland.pdf) or the more general [Lemma 2.3](https://projecteuclid.org/journals/annals-of-probability/volume-43/issue-5/Exact-Rosenthal-type-bounds/10.1214/14-AOP942.full).
Among counterexamples to \eqref{1} are [this](https://mathoverflow.net/a/105773/36721) and [this](https://mathoverflow.net/a/172898/36721). However, in those counterexamples the function $f$ is not continuous.
A [question](https://mathoverflow.net/questions/436209/expected-gradient-vs-gradient-of-expectation/436224#comment1123964_436224) arose whether there is a counterexample to \eqref{1} with a continuous function $f$. Such an example will be given in the answer below.
| https://mathoverflow.net/users/36721 | Counterexamples to differentiation under integral sign, revisited | A simple example is given by
$$
f(t,x)=\cases{\exp(-(x-t^{-2})^2),&$t\neq 0$,\\0,&$t=0.$}
$$
For each fixed $t\neq 0$, $\int f(t,x)\,dx$ is a Gaussian integral equal to $\sqrt{\pi}$, while for $t=0$, the integral equals to zero. Therefore, $F(t)$ is not even continuous at $0$, let alone differentiable.
On the other hand, at $t\neq 0$, any partial derivative of $f$ has a form $P(t^{-1},x)\exp(-(x-t^{-2})^2)$ for some polynomial $P$. This tends to zero as $t\to 0$, uniformly in $x$ in compacts. Therefore, $f$ is smooth with all partial derivatives vanishing on at $\mathbb{R}$. Also, any expression of this form, and in particular $\partial\_t f(t,x)$, is clearly integrable over $x$ for all $t$.
**Update:** The above example may feel like cheating, since in essence the discontinuity of $f$ has just been moved to infinity. Here's a better one in that respect:
$$
f(t,x)=\cases{\mathrm{sign}(t)|t|^\frac{1}{2}\exp\left(-\frac{(x-\sqrt[5]{t})^2}{|t|}\right),&$t\neq 0$,\\0,&$t=0.$}
$$
This has $F(t)=\sqrt{\pi}t$ while still $\partial\_t f(t,x)|\_{t=0}\equiv 0$. It has an additional feature that
$$
\int\_I\partial\_t f(t,x)\,dx\neq\partial\_t \int\_I f(t,x)\,dx
$$
for any non-empty interval $I$ containing the origin. It has smooth restrictions onto vertical and horisontal lines, but it is not overall smooth or even continuously differentiable: indeed, if it were, it would have its derivatives bounded on compacts, thus $\int\_I$ and $\partial\_t$ would commute for any compact interval $I$.
| 10 | https://mathoverflow.net/users/56624 | 436333 | 176,336 |
https://mathoverflow.net/questions/436329 | 8 | Consider the following martingale: $X\_1 \sim \mathcal{N}(0, 1)$, and for any $n > 1$, $X\_n \sim \mathcal{N}(X\_{n-1}, X\_{n-1}^2)$ (notice, this is a conditional distribution given $X\_{n-1}$).
I am looking for concentration bounds for $X\_n$, which I suspect exist, based on numerical simulation.
**Additional information:**
It is easy to see that $\operatorname{Var}(X\_n) = 2^{n-1}$, by induction over $n$.
The $n=1$ case is trivial. Since $X\_n$ is unbiased, for any $n > 1$, $$\underset{X\_n}{\mathbb{E}} \left[X\_n^2 \mid X\_{n-1} \right] = \underset{X\_n}{\mathbb{E}} \left[(X\_{n-1} + (X\_n - X\_{n-1}))^2 \,|\, X\_{n-1} \right] = X\_{n-1}^2 + \underset{X\_n}{\mathbb{E}} \left[(X\_n - X\_{n-1})^2 \mid X\_{n-1} \right] = 2 X\_{n-1}^2.$$
Using this fact we get, $$\operatorname{Var}(X\_n) = \underset{X\_1, \ldots, X\_n}{\mathbb{E}} \left[X\_n^2 \right] = \underset{X\_1, \ldots, X\_{n-1}}{\mathbb{E}} \left[ \underset{X\_n}{\mathbb{E}} \left[X\_n^2 \mid X\_{n-1} \right] \right] = \underset{X\_1, \ldots, X\_{n-1}}{\mathbb{E}} \left[2 X\_{n-1}^2 \right] = 2 \operatorname{Var}(X\_{n-1}).$$
At first glance, this might lead to the assumption that the distribution becomes less concentrated as $n$ grows, but running some simulations, it seems like the distribution actually gets more concentrated around $0$, and the growth of the variance results from the tails becoming heavier.
It is possible to state this question in a more general form, by defining $X\_n \sim \mathcal{N}(X\_{n-1}, f(X\_{n-1}))$ for any non negative function $f$. A similar question was presented in [this thread](https://mathoverflow.net/questions/227556/adaptive-version-of-the-azuma-hoeffding-inequality), but the answers do not apply to this example.
This question has close ties to the method of mixtures (see e.g., Theorem 2.7 in [De la Pena et al.](https://arxiv.org/pdf/0709.2233.pdf)), but as far as I understand, none of their results can be used to answer the question, the way it was formalized here.
| https://mathoverflow.net/users/496182 | Concentration bounds for martingales with adaptive Gaussian steps | Observe that $X\_n=X\_{n-1}(1+Z\_n)$ where $\{Z\_k\}\_{k \ge 1}$ are i.i.d. standard normal. Hence to analyze the asymptotic distribution of $|X\_n|$, pass to logarithms, to get $$\log(|X\_n|)= \log(|X\_1|) +\sum\_{k=2}^n \log(|1+Z\_k|) \tag{$\*$} \,.$$
According to Wolfram alpha,
\begin{align}
& \mu:=E\bigl[ \log(|1+Z\_k|) \bigr] \\[8pt] = {} & \frac1{\sqrt{2\pi}}\int\_{-\infty}^\infty \log(|1+z|)e^{-z^2/2} \, dz\in [-0.2085,-0.2084] \,.
\end{align}
Thus the law of large numbers yields that for large $n$, with high probability,
$$ \log(|X\_n|) <-n/5 \quad \text{whence } |X\_n|<e^{-n/5} \,.$$
This confirms the prediction of the O.P.
More precise information can be extracted from $(\*)$ via the asymptotic distribution theory for sums of i.i.d. variables [1].
[1] Petrov, Valentin V. "Sums of independent random variables." De Gruyter, 2022.
| 11 | https://mathoverflow.net/users/7691 | 436338 | 176,338 |
https://mathoverflow.net/questions/436326 | 9 | Let $V$ be an irreducible finite dimensional complex representation of the product of groups $G\times H$. Is it necessarily isomorphic to a tensor product of irreducible representation of $G$ and $H$? If not what is a counter-example, and under what extra assumptions this is known to be true?
Remark. I think for continuous representations of compact groups this is true.
| https://mathoverflow.net/users/16183 | Irreducible representations of a product of two groups | If your groups are finite, then Andy’s [answer](https://mathoverflow.net/a/436339/2383) is perfectly fine. If you want to do topological groups you must be slightly (but not much) more careful. First since we are dealing with finite dimensional representations, algebraic and topological reducibility are the same.
It is enough to prove that your irreducible representation decomposes as a tensor product in the case of discrete groups. For if it is isomorphic to one of the form $U\otimes V$, then since the linear isomorphism is continuous, it is enough to observe that restricting the original representation to $G$ gives $\dim V$ copies of $U$ and restricting to $H$ gives $\dim U$ copies of $V$ and so $U$ and $V$ must give continuous representations of $G$, $H$.
So we may assume that $G$, $H$ are discrete and we are dealing with no topology. Then replacing $G$, $H$ with their group algebras and $\mathbb C$ by algebras over an algebraically closed field, it suffices to show that if $A$, $B$ are $K$-algebras with $K$ algebraically closed, then any finite dimensional irreducible representation of $A\otimes B$ is equivalent to a tensor product of irreducible representations of $A$ and $B$.
But the image of $A$ and $B$ under any finite dimensional representation is a finite dimensional algebra and so without loss of generality we may assume that $A$, $B$ are finite dimensional.
The crux of the matter, which is basically Andy’s proof in different words, is the special case that $A,B$ are matrix algebras over $K$. Then $M\_n(K)\otimes M\_m(K)\cong M\_{nm}(K)$ and the isomorphism intertwines the actions on $K^n\otimes K^m\cong K^{nm}$. Thus in the case $A$, $B$ are matrix algebras then the unique $A\otimes B$ irreducible rep is the tensor product of the unique irreducible $A$ and $B$ reps. Since arbitrary semisimple algebras over an algebraically closed field are finite direct products of matrix algebras and tensor product distributes over direct product, this handles the case $A$, $B$ are semisimple and also shows that $A\otimes B$ is semisimple in this case.
$\DeclareMathOperator\rad{rad}$If $A$, $B$ are general finite dimensional $K$-algebras with $K$-algebraically closed, then $A/{\rad(A)}\otimes B/{\rad(B)}$ is semisimple by the above. Moreover, the kernel of the natural map $A\otimes B\to A/{\rad(A)}\otimes B/{\rad(B)}$ is easily checked to be $A\otimes \rad(B)+\rad(A)\otimes B$, which is a nilpotent ideal. Thus $(A\otimes B)/{\rad(A\otimes B)}\cong A/{\rad(A)}\otimes B/{\rad(B)}$ and so the desired result that irreducible reps are tensor products follows since $C$ and $C/{\rad(C)}$ have the same irreducible reps for any $K$-algebra $C$.
**Simpler proof**
Here is an argument along the line of Andy's that works for arbitrary groups or even algebras that avoids the radical and just uses Burnside's theorem that a finite dimensional representation of a $K$-algebra over an algebraically closed field $K$ is irreducible if and only if it is surjective.
Let $A,B$ be $K$-algebras (not necessarily finite dimensional) with $K$-algebraically closed (they could be group algebras) and let $W$ be a finite dimensional irreducible $A\otimes B$-module. Let $U$ be an irreducible $A$-subrepresentation of $W$. Then the sum $S$ of all irreducible $A$-submodules of $W$ isomorphic to $U$ is invariant under any $A$-enomorphism of $W$ and hence under $B$ as the $B$-action commutes with $A$. Hence $S$ is $A\otimes B$-invariant and so $S=W$ by irreducibility. Thus $W\cong U^m$ for some $m$ as an $A$-module (this is a standard argument). Thus, up to isomorphism, we may assume that $W=U\otimes V$ with $V$ a vector space of dimension $m$ and where $A$ acts via matrices of the form $\rho(a)\otimes 1$ where $\rho$ is the irreducible representation associated to $U$. By Burnside, $\rho$ is onto $End\_k(U)$. But since $End\_K(U\otimes V)= End\_k(U)\otimes End\_k(V)$, it follows that the centralizer of $End\_K(U)\otimes 1$ in $End\_K(W)$ is $1\otimes End\_k(V)$. Since the action of $B$ commutes with that of $A$, as a representation of $B$, we have $W$ is of the form $b\mapsto 1\otimes \psi(b)$ for some representation $\psi$ of $B$ on $V$. Clearly, if $V$ is not irreducible, then neither is $U\otimes V$ and so we are done.
| 6 | https://mathoverflow.net/users/15934 | 436343 | 176,341 |
https://mathoverflow.net/questions/436346 | 47 | It is often said that instead of proving a great theorem a mathematician's fondest dream is to prove a great lemma. Something like Kőnig's tree lemma, or Yoneda's lemma, or really anything from [this list](https://en.wikipedia.org/wiki/List_of_lemmas).
When I was first learning algebra, one of the key lemmas we were taught was [Zorn's lemma](https://en.wikipedia.org/wiki/Zorn%27s_lemma). It was almost magical in its power and utility. However, I can't remember the last time Zorn's lemma appeared in one of my papers (even though I'm an algebraist). In pondering why this is, a few reasons occurred to me, which I'll list below. I don't want to lose my old friend Zorn, and so my question is:
>
> What are some reasons to keep (or, perhaps in line with my thoughts, abandon) Zorn's lemma?
>
>
>
**Edited to add**: One purpose to this question is to know whether or not I should be rewriting my proofs to use Zorn's lemma, instead of my usual practice of using transfinite recursion, if there is a mathematical reason to prefer one over the other. Hopefully this clarifies the mathematical content of this question.
---
To motivate the discussion, let me give an example of how I would now teach ungraduates a result that was taught to me using Zorn's lemma.
**Theorem**: Every vector space $V$ has a basis.
*Proof*: First, fix a well-ordering for $V$. We will recursively work our way through the ordering, deciding whether to keep or discard elements of $V$. Suppose we have reached a vector $v$; we keep it if it is linearly independent from the previously kept vectors (equivalently, it is not in their span), otherwise we discard it. If $B$ is the set of kept vectors we see it is a basis as follows. Any vector $v\in V$ is in the span of $B$, because it is either in $B$ or in the span of the vectors previously kept. On the other hand, the elements of $B$ are linearly independent because a nontrivial combination $c\_1 v\_1 + \dotsb +c\_k v\_k=0$, where $v\_1<v\_2<\dotsb<v\_k$ and $c\_k\neq 0$ can be rearranged so $v\_k$ is a linear combination of the previous vectors, so $v\_k$ cannot belong to $B$ after all. $\quad\square$
Here are some of the benefits I see for this type of proof over the usual Zorn's lemma argument.
**1. The use of choice is disentangled from the other parts of the proof.**
When applying Zorn's lemma, it is difficult to see exactly how the axiom of choice is being used to reach the conclusion of a maximal element. One way to visualize its use is that Zorn's lemma lets us recursively build a maximal chain through the poset. This chain must have a greatest element. However, this construction is hidden behind the magic words "Abracadabra Zornify".
Is it a historical artifact that choice is hidden this way?
**2. We can more easily see whether or not to use a choice principle.**
In the proof above, if $V$ is already well-orderable (without AC), then we don't need to ever use the axiom of choice.
**3. Zorn's lemma is no easier than transfinite recursion.**
Each part of transfinite recursion already (implicitly) occurs in most Zorn's lemma arguments. The base case of the recursion corresponds, roughly, to showing that that the poset is nonempty (i.e., has some starting point). The successor ordinal step often occurs at the end; after asserting that some maximal element of the poset exists, we show that this maximal element has some claimed property by working by contradiction, and then passing to a slightly bigger element of the poset (i.e., the next successor). The limit ordinal step occurs when we show that chains have upper bounds.
**4. Zorn's lemma often includes unnecessary complications.**
In the proof I gave above, there is no need to define a complicated set, together with a poset relation. We can use strong induction, to avoid differentiating between the zero, successor, and nonzero limit steps. We don't need to combine the contradiction at the end with any successor step; they are entirely separated.
**5. Transfinite recursion is a more fundamental principle.**
As a matter of pedagogy, shouldn't we teach students about transfinite induction before we teach them a version of it that is also combined with AC, and that requires the construction of a complicated poset?
**6. Transfinite recursion applies to situations where Zorn's lemma does not.**
To give just one example: There are some recursions that continue along all of the ordinals (for a proper class amount of time). Zorn's requires, as a hypothesis, an end.
| https://mathoverflow.net/users/3199 | Zorn's lemma: old friend or historical relic? | I agree with almost everything in your post. But still, I believe I know why people use Zorn's lemma.
**My answer.** Zorn's lemma encapsulates succinctly many of the consequences of AC via transfinite recursion, but without requiring any involvement of the ordinals or knowledge of transfinite recursion to be used.
To those who are deeply familiar with transfinite recursion, of course, every use of Zorn's lemma can be seen as sumblimating the underlying construction, which achieves the maximal elements by a transfinite process that simultanesously explains why they exist. To appeal to Zorn seems to hide this essential explanatory underlying mechanism.
And yet, the alternative perspective is that Zorn's lemma abstracts away from the recursive process, producing in the end a simpler argument that relies only on the core consequences of the recursive process, which do not rely on any explicit engagement with ordinals or recursion. And precisely because of that feature, Zorn's lemma arguments can be undertaken and understood by mathematicians who are unfamiliar with the ordinals and transfinite recursion.
In the vector space example, to show every vector space has a basis, one can mount a transfinite recursive process: you pick an element, and then if it doesn't span, you pick another, and so on transfinitely until you have a basis. (My view of this example is a little different from how you described it, since I view the choice function as more primitive than the well order — I would build the basis by choosing amongst the element not yet in the span — indeed I prefer to view WOP itself as the outcome of recursively choosing elements.)
With Zorn's lemma, however, there is no need for ordinals or transfinite recursion, and the Zorn's lemma argument instead encapsulates abstractly their effects — the partial order consists in effect of partial undertakings of the recursive process. In this sense, the Zorn argument is simpler, abstracting away from the transfinite constructive "process".
I find the situation to be analogous to Martin's axiom and forcing. Martin's axiom is the poor mathematician's forcing, just as Zorn's lemma is the poor mathematician's choice+transfinite recursion.
My personal view is that the ordinals and transfinite recursion are one of the wonders of mathematics, a sublime achievement of the intellect resulting in many beautiful arguments and constructions. I tend to prefer the transfinite recursive arguments as providing a deeply explanatory account of the consequences of Zorn's lemma. (Even the well-order theorem seems fundamentally less mysterious when explained via the transfinite recursion — pick any element as the least element, and now pick a next element, and a next, and so on transfinitely.)
Further, although I recognize that many mathematicians have little involvement or experience with the ordinals and transfinite recursion, I also believe that their mathematical life would be improved by knowing more of them.
| 57 | https://mathoverflow.net/users/1946 | 436348 | 176,343 |
https://mathoverflow.net/questions/390681 | 2 | The Routh-Hurwitz criterion explicitly specifies a finite set of inequalities on the coefficients of a polynomial, necessary and sufficient that all zeros lie in the unit circle or in the left half complex plane.
Is there a similar set of explicit inequalities on the coefficients of a (real or complex) matrix, necessary and sufficient that all eigenvalues lie in the unit circle or in the left half complex plane? I am not looking for a decision algorithm (one could just compute the eigenvalues...). Instead, I'd like to have explicit inequalities for 2 x 2 and 3 x 3 matrices (with an elegant proof), and a scheme to generate the inequalities for larger matrices.
(The related page
[Routh-Hurwitz for eigenvalues](https://mathoverflow.net/questions/58701/routh-hurwitz-for-eigenvalues)
is very old and less specific. It gives no explicit inequalities, hence does not answer my question.)
| https://mathoverflow.net/users/56920 | Routh-Hurwitz criterion for matrices | Hmm, a "similar set". Is the following similar enough? I do not know. But it has to be pointed out that there are of course very nice and largely tractable numerical tests for stability of matrices (and much better than attempting to compute the eigenvalues). Let $I$ denote the identity matrix. The spectrum of a (real or complex) matrix $A$ is in the left half plane, if and only if there is a positive definite, Hermitian $P$ such that
$$A^\* P + PA = -I.$$
For the unit circle the respective equation is
$$ A^\*PA - P = -I.$$
These are the so-called Lyapunov equations. The answer is a bit tongue in cheek, but still. First you have a bunch of linear equalities for the entries of $P$ using the coefficients in $A$. Then you need to check positive definiteness. To do this numerically, as an example Cholesky factorization has been [proposed](https://www.sciencedirect.com/science/article/pii/0024379588902236).
| 2 | https://mathoverflow.net/users/85570 | 436360 | 176,350 |
https://mathoverflow.net/questions/436369 | 8 | This question is naive, but I didn't get an answer at [MSE](https://math.stackexchange.com/questions/4587217/is-there-a-measure-theory-for-proper-classes): Is it straightforward to extend measure theory to proper classes?
Of course when one tries to define measures on "large sets" problems of non-measurability arise (e.g. non-Lebesgue measurable sets, Ulam's theorem, etc.). But setting those aside (since they arise already for measures over sets), are there any special problems that dealing with proper classes would raise?
To be a little bit more precise (and I'm afraid I can't do any better at the moment), let $\Omega$ be a proper class. If you prefer that we have something concrete, then let $\Omega$ be the proper class of all ordinals, for instance. Is there anything wrong with saying that a measure is a real-valued function with the usual properties (countable additivity, etc.) on a collection of sub-classes of $\Omega$?
In other words, can the usual definition of a measure be extended straightforwardly to proper classes or are there problems lurking here?
If this isn't entirely straightforward and you know of any references, I would appreciate them.
| https://mathoverflow.net/users/96899 | Is there a measure theory for proper classes? | As a [sledgehammer](https://www.collinsdictionary.com/us/dictionary/english/a-sledgehammer-to-crack-a-nut), working in Higher Order Set Theory (HOST) as proposed in
* *Rhea, Alec*, An Axiomatic Approach to Higher Order Set Theory. (2022)
<https://doi.org/10.48550/arXiv.2206.10060>
you can just go ahead and collect up proper classes inside other proper classes, no need to move higher up the hierarchy or inevitable collapse downwards due to membership issues. I designed this theory to work with things like topology, analysis, measure theory, etc. over the surreals without having to worry about anything too subtle, and it works equally well for measure theory on arbitrary proper classes -- no subtleties lurking about.
This theory is equiconsistent with ZFC plus a countable collection of inaccessible cardinals, however, so it is significantly stronger in consistency strength than most standard set theories. If there's anything you'd like me to elaborate on please don't hesitate to ask.
| 6 | https://mathoverflow.net/users/92164 | 436373 | 176,354 |
https://mathoverflow.net/questions/436372 | 8 | Let $u$ and $v$ be two vectors in $\mathbb{C}^n$.
Define a permutation of a vector $v':=\sigma(v)$ by $v'\_j = v\_{\sigma(j)}$ for any $\sigma \in S\_n$.
It is easy to show the following for $u,v \in \mathbb{C}^n$ by considering $\sigma$ being a swap:
>
> $u = c\vec{1}$ or $v = c\vec{1}$ for some $c \in \mathbb{C} \iff $ $\langle u,\sigma(v) \rangle$ is invariant for any $\sigma \in S\_n$.
>
>
>
I am wondering if the following is also true:
>
> Let $n \geq 3$. $u = c\vec{1}$ or $v = c\vec{1}$ for some $c \in \mathbb{C} \iff $ $\left|\langle u,\sigma(v) \rangle \right|$ is invariant for any $\sigma \in S\_n$.
>
>
>
Note that for $n=2$ there is a simple counterexample $u=v=(1,-1)$. However I was unable to find any counterexample for $n \geq 3$.
**Update.** The conjecture is false in light of Matt's answer. However the known counterexamples are all highly "structural". Is there some way to identify all counterexamples?
| https://mathoverflow.net/users/22954 | Use random inner product to test if at least one vector is uniform | This (and specifically the $\Leftarrow$) is false: let
$$u = (-1,0,1)$$
$$v = (1, \omega, \omega^2)$$
where $\omega$ is a cube root of unity.
| 5 | https://mathoverflow.net/users/nan | 436376 | 176,355 |
https://mathoverflow.net/questions/436374 | 2 | Recently I have read many papers which focusing on the image enhancement. The description,
>
> $D$ are the Toeplitz matrices from the discrete gradient with forward difference
>
>
>
occurs many times.
An example is shown below:
To minimize an object function,
$$ F(T)=\sum\_x\left((T(x)-L(x))^2 + \lambda\sum\_{d\in\lbrace h,v\rbrace}A\_d(x)\left(\partial\_dT(x)\right)^2\right), $$
where $L(x)$ and $A\_d(x)$ is known. So to minimize this function, we differentiating it and setting the derivative to 0, the problem is simplified to:
$$t=\left(I + \sum\_{d\in\lbrace h, v\rbrace}D\_d^TDiag(a\_d)D\_d\right)^{-1}l,$$
where $m$ is the vectorized version of $M$ and the operator $Diag(m)$ is to construct a diagonal matrix using vector $m$, and **$D$ are the Toeplitz matrices from the discrete gradient with forward difference** .
So my question is, what on earth is $D$ ? How to calculate this $D$ ? And what does the transpose (or adjoint) of this $D^T$ mean ?
| https://mathoverflow.net/users/475095 | What is the "Toeplitz matrices from the discrete gradient operators with forward difference" | A Toeplitz matrix is a band matrix in which each descending diagonal from left to right has constant elements. The derivative of a function can be represented in a finite-difference calculation as the product of a Toeplitz matrix and equally spaced values of $f$. One distinguishes forward differences and backwards differences, the forward difference is given by
$$D[f](x)=f(x+\delta x)-f(x),$$
corresponding to a matrix with elements (in the $4\times 4$ case)
$$D=\begin{pmatrix}
-1&1&0&0\\
0&-1&1&0\\
0&0&-1&1\\
1&0&0&-1
\end{pmatrix}
$$
The backward difference is given by $-D^\top$, minus the transpose of the forward difference,
$$-D^\top[f](x)=f(x)-f(x-\delta x).$$
| 2 | https://mathoverflow.net/users/11260 | 436378 | 176,356 |
https://mathoverflow.net/questions/436366 | 4 | In Walter Michaelis' paper [*Lie Coalgebras*](https://doi.org/10.1016/0001-8708(80)90056-0), he gives on page 9 an explicit example of a Lie coalgebra which is not the union of its finite-dimensional Lie subcoalgebras. In fact, Michaelis' example has exactly two finite-dimensional Lie subcoalgebras: the zero coalgebra and a certain one-dimensional subcoalgebra.
What I want to know is whether one can do even better: is there a Lie coalgebra $C\neq0$ (over any field) whose only finite-dimensional Lie subcoalgebra is $0$? Ideally, I'd like an explicit example.
| https://mathoverflow.net/users/126183 | Lie coalgebra with no finite-dimensional subcoalgebras | Consider the Lie algebra of vector fields on the formal disk, $\mathfrak g=k[[t]]d/dt$, where $k$ is a field of characteristic zero and $k[[t]]$ is the $k$-algebra of formal Taylor power series in the variable $t$. Then $k[[t]]$ is naturally a topological $k$-algebra with a pro-finite-dimensional ( = pseudocompact = linearly compact) topology, and $\mathfrak g$ is a topological Lie algebra with a pro-finite-dimensional topology.
Consequently, there exists a unique coassociative coalgebra $\mathcal C$ such that $k[[t]]\simeq \mathcal C^\*$ as a topological associative algebra, and there exists a unique Lie coalgebra $\mathcal L$ such that $\mathfrak g = \mathcal L^\*$ as a topological Lie coalgebra. Now, of course, $\mathcal C$ is the union of its finite-dimensional subcoalgebras. However, $\mathcal L$ has no nonzero finite-dimensional Lie subcoalgebras. In fact, $\mathcal L$ has no nonzero proper Lie subcoalgebras at all.
To prove these assertions about $\mathcal L$, one can translate them back into the topological Lie algebra language, where they become more intuitively clear and can be checked explicitly by hand. The claims to prove are that the topological Lie algebra $\mathcal g$ has no proper open Lie ideals, and in fact, it does not even have any nonzero proper closed Lie ideals.
The topological Lie algebra $\mathfrak g$ has a topological basis $L\_{-1}$, $L\_0$, $L\_1$, $L\_2$, $\dots$, where $L\_n = t^{n+1}d/dt$. The commutation relations are $[L\_i,L\_j]=(j-i)L\_{i+j}$. Suppose $\mathfrak h$ is a nonzero closed Lie ideal in $\mathfrak g$. Considering the action of $L\_0$ in $\mathfrak h$, one shows that $\mathfrak h$ is homogeneous, i.e., if it contains an infinite linear combination of $L\_n$ with nonzero coefficient at some $L\_j$, then it also contains the element $L\_j$ itself. Now one can use the action of $L\_{-1}$ and $L\_n$ with $n\gg0$ in $\mathfrak h$ to prove that $\mathfrak h=\mathfrak g$ if $\mathfrak h\ne0$.
Exercises: pinpoint a three-dimensional closed Lie subalgebra in $\mathfrak g$ isomorphic to $\mathfrak{sl}\_2(k)$, and also an infinite family of pro-nilpotent open Lie subalgebras in $\mathfrak g$ forming a base of the topology of $\mathfrak g$. Realize that these Lie subalgebras are not Lie ideals in $\mathfrak g$. Conclude that the Lie coalgebra $\mathcal L$ is a union of its finite-dimensional Lie coideals, but these Lie coideals are not Lie subcoalgebras.
| 2 | https://mathoverflow.net/users/2106 | 436385 | 176,359 |
https://mathoverflow.net/questions/436381 | 18 | It may seem silly to ask *"Why are there three types of non-Archimedean geometry?"*, that would be like asking why there are three (and even more) different Weil cohomologies. So I have to clarify my question.
Let $X$ be a scheme on a $p$-adic ring (i.e. an extension of $\mathbb{Z}\_{p}$).
**-** What are the relationships between the rigid, berkovich, and adic analytification of $X$?
**-** For each of them, are the GAGA statements *exactly* the same as in the complex case or are there (subtle) changes?
**-** What are the cohomologies on these analytification? Are they Weil cohomologies? Are there comparison isomorphisms?
**-** Finally, a question that may be too general, how do you know in the context if you have to work with one analytification more than another?
Feel free to answer even if you don't have the answer to all my questions :)
| https://mathoverflow.net/users/169282 | Why are there three kinds of non-archimedean geometry? | Tate's rigid-analytic geometry was the first theory of (global) nonarchimedean geometry to have been devised, and in some sense could be seen as a "proof of concept" that such a theory can exist, despite some general skepticism which was present at the time, including from Grothendieck. For a while this theory was the only one, and when alternatives were introduced, they didn't present enough benefits for people to switch over to Berkovich or adic settings. You can see [this question of mine](https://mathoverflow.net/q/393797/30186) where this is discussed in the answers.
Now, rigid analytic geometry is somewhat limited - you should think of it as analogous to the classical theory of varieties over fields. We would like to have a theory which is more akin to that of schemes, which could admit a greater range of spaces one can study. The fact rigid spaces are also not "honest" topological spaces is also often undesireable. Berkovich was the first to get around this, and his theory has a few advantages: firstly, one has a real topological space, and even a rather "nice" one - almost of "Euclidean" type, which makes it amenable to the usual topological study. Secondly, it is not strictly tied to the nonarchimedean world - one can apply it to any Banach space, including archimedean ones, which leads among others to the theory of Berkovich spaces over $\mathbb Z$ as developed by Poineau. Berkovich's original motivation, though, was that the bigger set of points gives a susbtitute of generic points, making it useful in developing the theory of etale cohomology.
Adic spaces were also developed by Huber largely with refining the theory of etale cohomology in mind. They are more tied to nonarchimedean algebras, and for those they can be seen as a further refinement of Berkovich spaces. This class of spaces permits you to use a much bigger class of rings, most famously perfectoid algebras, as shown by Peter Scholze.
Now to try and answer your questions:
* Every rigid space has both a Berkovich and an adic analytification, which are compatible in the obvious way with analytification of schemes of finite type over a nonarchimedean field. There isn't an analytification from Berkovich to adic spaces in general, but there should be one for spaces topologically of finite type over a field, again compatible in the obvious way.
* The statements should be the same, but the notion of coherent sheaf itself is significantly more subtle, especially in the rigid setting. This is discussed in Bosch's "Lectures on Formal and Rigid Geometry".
* As I mentioned, studying etale cohomology has driven Berkovich and Huber to describe their theories. I believe this cohomology makes sense in all settings, and would expect suitable comparison theorems, but I'm not certain on that. In a different direction, let me also mention rigid cohomology, which is a cohomology theory of varieties in characteristic $p$ which is defined in terms of rigid spaces. It can also be defined using the other types of spaces.
* Let me refer you once again to my [other question](https://mathoverflow.net/q/393797/30186) on the use of rigid spaces - my general impression is that it can be still useful conceptually and pedagogically, and has the biggest backlog of literature written in this language, but can overall be replaced by the other theories. As I mentioned Berkovich spaces are convenient if you want to mix in the archimedean setting, and they have also found uses in dynamics and tropical geometry (IIRC in some sense, Berkovich analytification is the inverse limit over all tropical models of a variety). Finally, adic spaces are probably the best fit for most applications dealing with "nonclassical" nonarchimedean spaces, especially non-noetherian ones like perfectoids.
By the way, on those and related topics, I highly recommend the set of notes by Brian Conrad on [Several approaches to non-archimedean geometry](https://math.stanford.edu/%7Econrad/papers/aws.pdf).
| 19 | https://mathoverflow.net/users/30186 | 436389 | 176,361 |
https://mathoverflow.net/questions/436391 | 1 | I wonder if one can write the following matrix in the form $A = \begin{pmatrix} 0 & B \\ B^\* & 0 \end{pmatrix}.$
The matrix I have is of the form
$$ C = \begin{pmatrix} 0 & a & b & 0 & 0 & 0 \\
\bar a & 0 & 0 &b & 0& 0\\
\bar b & 0 & 0 & a & f & 0 \\
0 & \bar b & \bar a & 0 & 0 &f \\
0 & 0 & \bar f & 0 & 0 & a\\
0 & 0 & 0 & \bar f & \bar a & 0
\end{pmatrix}.$$
The reason I believe it should be possible is that the eigenvalues of $A$ are symmetric with respect to zero $\pm \vert a \vert, \pm \sqrt{ \vert a \vert^2+ \vert b \vert^2 + \vert f \vert^2 \pm \vert a \vert^2( \vert b \vert^2 + \vert f \vert^2)}$ where in the latter case all sign combinations are allowed.
Hence, I wonder if there exists $U$ such that
$$A = UCU^{-1}$$
| https://mathoverflow.net/users/119875 | Transforming matrix to off-diagonal form | The general recipe to accomplish the block off-diagonalization is as follows. The matrix $C$ has eigenvalues $\pm\lambda\_1,\pm\lambda\_2,\ldots \pm\lambda\_3$. Define $\Lambda=\text{diag}\,(\lambda\_1,\lambda\_2,\lambda\_3)$, and decompose
$$C=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast,$$
with $U$ the unitary matrix of eigenvectors of $C$;
Then the matrix product
$$\Omega^\ast U^\ast CU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$
with $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$, has the desired form.
The explicit form of $U$ is complicated for arbitrary complex numbers $a,b,f$.
| 2 | https://mathoverflow.net/users/11260 | 436397 | 176,366 |
https://mathoverflow.net/questions/436353 | 8 | Let $X\_0$ be a compact complex algebraic surface with an isolated singularity and let $X\_t$ be a smoothing of $X\_0$ over the disc. How can we compute the fundamental group of $X\_t$ say in terms of the topology of a minimal resolution and some local information of the singularity? If it helps we can assume the singularity is rational and that the smoothing is $\mathbb{Q}$-Gorenstein.
| https://mathoverflow.net/users/12402 | Fundamental group of a smoothing of a complex surface | Non surprisingly, this usually involves Seifert-Van Kampen theorem, but the actual computation can be a tricky one. However, since you have just one singularity, life will be probably easier.
For an example where the fundamental group turns out to be trivial, you can look at the celebrated paper by Lee and Park, in which they construct a simply-connected Campedelli surface via $\mathbb{Q}$-Gorenstein smoothing of a family of rational surfaces with isolated $T$-singularities:
*Lee, Yongnam; Park, Jongil*, [**A simply connected surface of general type with (p\_g=0) and (K^2=2)**](http://dx.doi.org/10.1007/s00222-007-0069-7), Invent. Math. 170, No. 3, 483-505 (2007). [ZBL1126.14049](https://zbmath.org/?q=an:1126.14049).
| 5 | https://mathoverflow.net/users/7460 | 436398 | 176,367 |
https://mathoverflow.net/questions/436405 | 7 | One model of Robinson arithmetic which is obviously not our usual integers is $\mathbb{Z}[X]^+$, that is the set containing 0 and also all polynomials with coefficients in $\mathbb{Z}$ with positive lead coefficient. This ring also satisfies a bit more, since it satisfies distributivity, commutativity, and associativity, and that no number is its own successor, which Robinson arithmetic only can prove are true for specific natural numbers.
One basic property that the natural numbers have which $\mathbb{Z}[X]^+$ lacks if a division algorithm. That is, the property that for any $a, b \in \mathbb{N}$, and $b \neq 0$, there is a $q$ and $r$ such that $a=bq+r$, and where $r < b$. (Here we are for simplicity including 0 as a natural number.) We can see that $\mathbb{Z}[X]^+$ fails this by considering for example elements like $a=X$ and $b=2$. However, despite this, $\mathbb{Z}[X]^+$ has many of the properties we expect out of the natural numbers, including unique factorization; and a major reason we care about having a division algorithm in the natural numbers is to use it to prove unique factorization.
**Question:** Is there a nice model of Robinson arithmetic which satisfies commutative, associative, and distributive properties, no number is its own successor, and also the division algorithm and is not just the natural numbers?
Nice here is left vague, but at an absolute minimum requires being computable. I'm hoping to see something like some subset of some ring that an algebraist would recognize as a reasonable object.
The obvious thing to try to use that the nonnegative part of a discrete ordered ring is always a model for Robinson Arithmetic, and there are a lot of examples of things one can construct there, but I have not been able to find one that works for this purpose.
| https://mathoverflow.net/users/127690 | Systems intermediate in strengthen between Robinson arithmetic and PA | Yes. For example, you can take the nonnegative part of the ring of polynomials in $\mathbb Q[X]$ with integer constant coefficient (i.e., $\mathbb Q[X]X+\mathbb Z$).
For a more sophisticated example due to Shepherdson [2], the nonnegative part of the ring of *Puiseux polynomials* $\sum\_{i\le n}a\_iX^{i/k}$ over a real-closed field such as $\tilde{\mathbb Q}\cap\mathbb R$ with $a\_0\in\mathbb Z$ is a nonstandard computable model of the theory IOpen = Robinson’s arithmetic + induction for open (= quantifier-free) formulas in the language of ordered semirings. IOpen proves the existence of division with remainder and much more.
Computable nonstandard models exist for extensions of IOpen by some mild algebraic axioms, in particular GCD; see Mohsenipour [1]. However, these constructions get rather messy, and I’m not sure “an algebraist would recognize them as a reasonable object”. On the other hand, there is no computable nonstandard model of Peano arithmetic by Tennenbaum’s theorem [3], and this has been extended to models of $IE\_1$ (= Robinson’s arithmetic + induction for bounded existential formulas) by Wilmers [4].
### References
[1] Shahram Mohsenipour: *A recursive nonstandard model for open induction with GCD property and confinal primes*, Logic in Tehran (Ali Enayat et al., eds.), Lecture Notes in Logic no. 26, Association for Symbolic Logic, 2006, pp. 227–238.
[2] John C. Shepherdson: *A nonstandard model for a free variable fragment of number theory*, Bulletin de l’Académie Polonaise des Sciences, Série des sciences mathématiques, astronomiques et physiques 12 (1964), no. 2, pp. 79–86.
[3] Stanley Tennenbaum: *Non-archimedean models for arithmetic*, Notices of the American Mathematical Society 6 (1959), p. 270.
[4] George Wilmers: *Bounded existential induction*, Journal of Symbolic Logic 50 (1985), no. 1, pp. 72–90.
| 16 | https://mathoverflow.net/users/12705 | 436406 | 176,368 |
https://mathoverflow.net/questions/436393 | 7 | Let $F : \mathbb C \to \mathbb C$ be an entire function of finite order. Since the zeros of $F$ are countable there exists a constant $c \in \mathbb R$ such that $F$ is zero-free on the line $e^{ic} \mathbb R$. I'm wondering of the following stronger statement holds true: there exists a $c \in \mathbb R$ and a $d>0$ such that
$$
|F(z)| \geq d
$$
for all $z \in e^{ic} \mathbb R$, that is, $F$ is uniformly bounded from below on some line $e^{ic} \mathbb R$. Above I wrote the assumption that $F$ is of finite order. I feel like such an assumption might be needed since it gives control over the density of the zeros of $F$.
| https://mathoverflow.net/users/170539 | Are entire functions uniformly bounded from below on a line through the origin? | The answer is negative, and here are two ways to construct counterexamples.
1. Let $G$ be a bounded simply connected region whose closure does not contain zero, but $G$ intersects every ray from the origin. For example, $G$ can be a neighborhood of a sufficiently long compact piece of a logarithmic spiral.
Consider the open set $D=\cup\_n t\_nG$, where $t\_n>0, t\_n\to\infty$ sufficiently fast, so that $t\_nG$ are disjoint. Then one can find an entire function $f$ such that $f(z)\to 0$ as $z\to\infty, z\in D$. Such a function
can be easily constructed using Runge's approximation theorem. Moreover, by taking the sequence $t\_n$ of sufficiently fast growth, one can control the growth of $f$, so that $f$ will be of finite order, and even of zero order. For details of the order control, one may consult, for example
MR1040926 Eremenko, A. E.; Sodin, M. L. On the behavior of an entire function on a sequence of concentric circles. Complex Variables Theory Appl. 12 (1989), no. 1-4, 267–276, or
MR0545054 Golʹdberg, A. A.; Eremenko, A. E. Asymptotic curves of entire functions of finite order. Mat. Sb. (N.S.) 109(151) (1979), no. 4, 555–581, 647 (there is an English translation).
2. One can use the construction of Balashov,
MR0324032
Balašov, S. K.
Functions with completely regular growth along curves of regular rotation.
Dokl. Akad. Nauk SSSR 209 (1973), 525–528 (There is an English translation).
MR0328073 Balašov, S. K. Entire functions of finite order with roots on curves of regular rotation. Izv. Akad. Nauk SSSR Ser. Mat. 37 (1973), 603–629.
(There is an English translation).
(See also his papers MR0333171, MR0409807). Functions constructed in these papers are of any finite order $>1/2$ and have
zeros regularly distributed on spirals. They have controlled asymptotics on spirals, in particular one can arrange that $f(z)\to 0$ on some logarithmic spiral.
| 9 | https://mathoverflow.net/users/25510 | 436410 | 176,369 |
https://mathoverflow.net/questions/436411 | 1 | Let $X$ be a smooth Fano threefold with a finite group $G$ action. Assume that the orbit space $X/G$ is smooth. Is it true that $J(X/G)\cong J(X)^G$ As an abelian variety? Here, $J(X)^G$ is the $G$-invariant part of $J(X)$.
I am particular interested in the case that $G$ is $\mathbb{Z}\_2=\langle 1,\tau\rangle$ and $\tau$ is an involution.
| https://mathoverflow.net/users/41650 | Intermediate Jacobian under group action | No, because there is no reason for $J(X)^G$ to be connected. Here is a silly example: consider a smooth elliptic curve $C\subset \mathbb{P}^3$ given by $\sum x\_i^2=\sum a\_ix\_i^2=0$; take for $X$ $\mathbb{P}^3$ with $C$ blown up, and let $\tau $ be the involution of $\mathbb{P}^3$ which changes the sign of one coordinate. We have a $G$-equivariant isomorphism $J(X) \cong JC $, and $\tau $ acts on $JC$ ($\cong C$)
with 4 fixed points, while $J(X/G)=0$.
| 2 | https://mathoverflow.net/users/40297 | 436419 | 176,372 |
https://mathoverflow.net/questions/436218 | 11 | $\newcommand{\Z}{\mathbb Z/p\mathbb Z}$
Can one partition a group of prime order as $A\cup(A-A)$ where $A$ is a subset of the group, $A-A$ is the set of all differences $a'-a''$ with $a',a''\in A$, and the union is disjoint?
As stated, the answer is "yes", at least if the order of the group is $p\equiv 2\pmod 3$, in which case one can take $A$ to be an appropriately located interval of an appropriate length: namely, $A=[n,2n-1]$ where $n=(p+1)/3$. One also can dilate $A$ replacing intervals with arithmetic progressions.
Are there any other examples where $A$ and $A-A$ partition the whole group?
>
> Suppose that $A\cup(A-A)$ is a partition of a group of prime order; does it follow that $A$ is an arithmetic progression?
>
>
>
**Added two days later.** A set found by Peter Mueller in the comments can be generalized to produce infinitely many counterexamples, essentially answering the original question. Specifically, for a prime $p\equiv 5\pmod 8$ let $m:=(p+3)/8$ and define $A:=[-(2m-1),-m]\cup[m,2m-1]$. It is easily verified that then $A-A=[-(m-1),m-1]\cup[2m,p-2m]$, so that $A-A$ is disjoint from $A$, and $A\cup(A-A)$ is the whole group, while $A$ is not an arithmetic progression.
| https://mathoverflow.net/users/9924 | $\mathbb Z/p\mathbb Z=A\cup(A-A)$? | *This is a previous comment which was moved to [chat](https://chat.stackexchange.com/rooms/141236/discussion-on-question-by-seva-mathbb-z-p-mathbb-za-cupa-a) by Ben Webster:* In fact for every prime $p$ if $A=[-(2m-1),-m] \cup [m, 2m-1]$ for $(p+3)/8\le m<(p+3)/6$, then $\mathbb Z/p\mathbb Z$ is a disjoint union of $A$ and $A-A$, and in addition this set $A$ is not an arithmetic progression if and only if $m\ne(p+1)/6$. There are many more examples though. One peculiar sporadic one happens for $p=41$ with $A$ the multiplicative subgroup of order $10$ of $(\mathbb Z/p\mathbb Z)^\star$.
| 5 | https://mathoverflow.net/users/18739 | 436422 | 176,373 |
https://mathoverflow.net/questions/436428 | 4 | Call a topological space $X$ standard Borel if $X$ is standard Borel as a measurable space (equipped with its Borel $\sigma$-algebra), i.e. if there is a Borel isomorphism between $X$ and a Polish space.
Clearly all Polish spaces are standard Borel by definition, but the converse is not true (any Borel subset of a Polish space is standard Borel, but it is Polish under the subspace topology if and only if it is $G\_\delta$).
>
> Is there a purely topological characterization of which topological spaces are standard Borel?
>
>
>
If an exact characterization is hopeless, I am interested in sufficient or necessary topological conditions that are weaker than Polish. For example, is any standard Borel topological space second-countable? Metrizable? Homeomorphic to a Borel subset of a Polish space?
| https://mathoverflow.net/users/160416 | Which topological spaces have a standard Borel $\sigma$-algebra? | Here are two examples showing that none of your candidate notions work.
First, we can observe that every Quasi-Polish space (<https://doi.org/10.1016/j.apal.2012.11.001>) admits a Baire class 1 isomorphism to a Polish space, and thus has a standard Borel $\sigma$-algebra. However, take e.g. the Scott domain $\mathcal{O}(\mathbb{N})$, with underlying set $\mathcal{P}(\mathbb{N})$ and the topology generated by $\{U \subseteq \mathbb{N} \mid n \in U\}$. This space is not Hausdorff, so clearly not metrizable and not isomorphic to any subspace of a Polish space.
For our second example, let us consider the space $\mathbb{R}[X]$ of polynomials of the reals. It is topologized as the limit of the compact Polish space of polynomials of degree up to $n$ and coefficients bounded by $n$. This space is not second-countable, but it is separable, so again, it is not metrizable. As there is a $\Delta^0\_2$-bijection between $\mathbb{R}[X]$ and the Polish space $\mathbb{R}^\*$, it again has a standard Borel $\sigma$-algebra.
| 5 | https://mathoverflow.net/users/15002 | 436430 | 176,375 |
https://mathoverflow.net/questions/435800 | 16 | Consider the (continuous, injective, abelian group homomorphism) map $\Phi \colon \mathbb{R} \to \prod\_{a>0} (\mathbb{R}/a\mathbb{Z})$ (where the target is given the product topology) taking $x\in \mathbb{R}$ to the family $(x \mod a) \_ {a>0}$; depending on your inclination, you may prefer to view it as $\Phi \colon \mathbb{R} \to (\mathbb{S}^1)^{\mathbb{R} \_ {>0}}$ taking $x$ to $(\exp(2i\pi b x)) \_ {b>0}$ (with $b=1/a$).
I'd like to know more about the following two topological spaces, indeed topological groups:
* The image of $\Phi$ with the subspace topology, call it $R$, or, equivalently, the coarsest topology on $\mathbb{R}$ making $\Phi$ continuous (this is strictly coarser than the usual topology).
* The closure of the image of $\Phi$ with the subspace topology, call it $S$ (since the target of $\Phi$ is compact, so is $S$).
There seems to be a lot of interesting things to say about $R$ and $S$ in the line of “examples and counterexamples in general topology”; for example, it turns out that [if $(x\_n)$ is a real sequence such that $\Phi(x\_n) \to 0$ then in fact $x\_n \to 0$ in the reals](https://math.stackexchange.com/questions/4590203/if-x-n-to-0-pmoda-for-every-a0-does-it-follow-that-x-n-to-0), so $R$ has the same convergent sequences as $\mathbb{R}$ even though it does not have the same topology (so $R$ is not metrizable). I have a million more questions about $R$, about $S$, and also about the map $\beta\mathbb{R} \to S$ (and maybe about $\beta R \to S$ while I'm at it), but rather than ask a million questions now, let me start by asking what has already been written about them out there in the literature:
**Question:** Do $R$ and $S$ have standard names? Are they described in a paedagogical fashion in some textbook on general topology or such text?
A presentation in the spirit of Steen & Seebach's classic *Counterexamples in Topology* would be a good start (I was unable to find these spaces in the book in question but, of course, I may have missed them). A name would help with the search (this is somewhat related to solenoids, but solenoids are metrizable whereas $R$ and $S$ are not, as I mention above).
Of course, if you have some favorite facts to share about $R$ and $S$, they're welcome as well.
| https://mathoverflow.net/users/17064 | Where can I learn more about the topology on $\mathbb{R}$ induced by the map $\mathbb{R} \to \prod_{a>0} (\mathbb{R}/a\mathbb{Z})$? | Short answer: $S$ is known as the *Bohr compactification* of $\mathbf R$ (often denoted $S=b\mathbf R$, see [1], 26.11), and $R$ is known as $\mathbf R$ with the *Bohr topology* (often denoted $R=\mathbf R^+$, see e.g. [2, 3, 4]).
For a longer answer, write $\chi\_b(x)=\exp(2\pi ibx)$ so that your $\Phi(x)=(\chi\_b(x))\_{b>0}$. Then $R$ is $\mathbf R$ with the weakest topology making $\chi\_b$ continuous for all $b>0$. Since $\chi\_0$ is constant and $\chi\_{-b}$ is $\chi\_b$ composed with complex conjugation, we see that it makes no difference to use $(\chi\_b(x))\_{b\in\mathbf R}$ instead. That shows that your definition coincides with those in *loc. cit.*
More generally if $G$ is any locally compact abelian group with dual $\hat G$ and we write $$\Phi(g)=(\chi(g))\_{\chi\in\hat G},$$ these references define $bG$ as the closure of $\Phi(G)$ in $\mathbf T^{\hat G}$, and $G^+$ as $G$ with its relative topology in $bG$. Favorite facts about $G^+$ are that it is sequentially closed in $bG$ (no sequence in $G^+$ can converge to a point of $bG\smallsetminus G^+$ [5]), and Glicksberg's theorem that $G^+$ has exactly the same compact sets as $G$ [6]. The map $\beta G\to bG$ is also studied in [7].
Finally, *since you ask*, my own favorite facts about the Bohr topology $G^+$ occur for $G=\mathbf R^n$, $n>1$: namely, a parabola is Bohr dense in the plane, and more generally the image of any polynomial map $\mathbf R^m\to \mathbf R^n$ is Bohr dense in its affine hull [8].
---
[1] *Hewitt, E.; Ross, K. A.*, [**Abstract harmonic analysis. Vol. I.**](//archive.org/details/in.ernet.dli.2015.134308/page/n382/mode/1up) Berlin-Göttingen-Heidelberg: Springer-Verlag. VIII, 519 p. (1963). [ZBL0115.10603](https://zbmath.org/?q=an:0115.10603).
[2] *Comfort, W. Wistar; Hernández, Salvador; Trigos-Arrieta, F. Javier*, [**Relating a locally compact Abelian group to its Bohr compactification**](https://doi.org/10.1006/aima.1996.0042), Adv. Math. 120, No. 2, 322-344 (1996). [ZBL0863.22004](https://zbmath.org/?q=an:0863.22004).
[3] *Hernández, Salvador*, [**The dimension of an LCA group in its Bohr topology**](https://doi.org/10.1016/S0166-8641(97)00126-0), Topology Appl. 86, No. 1, 63-67 (1998). [ZBL0935.22006](https://zbmath.org/?q=an:0935.22006).
[4] *Hernández, Salvador; Remus, Dieter; Javier Trigos-Arrieta, F.*, [**Contributions to the Bohr topology by W.W. Comfort**](https://doi.org/10.1016/j.topol.2019.02.018), Topology Appl. 259, 28-39 (2019). [ZBL1414.22012](https://zbmath.org/?q=an:1414.22012).
[5] See [MR 40:4685](//ams.org/mathscinet-getitem?mr=40:4685).
[6] *Glicksberg, I.*, [**Uniform boundedness for groups**](https://doi.org/10.4153/CJM-1962-017-3), Can. J. Math. 14, 269-276 (1962). [ZBL0109.02001](https://zbmath.org/?q=an:0109.02001).
[7] *Zlatoš, Pavol*, [**The Bohr compactification of an abelian group as a quotient of its Stone-Čech compactification**](https://doi.org/10.1007/s00233-020-10121-6), Semigroup Forum 101, No. 2, 497-506 (2020). [ZBL1471.22001](https://zbmath.org/?q=an:1471.22001).
[8] [Zbl 0795.22003](//zbmath.org/0795.22003), [Zbl 1355.37048](//zbmath.org/1355.37048)
| 16 | https://mathoverflow.net/users/19276 | 436436 | 176,379 |
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