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https://mathoverflow.net/questions/420538 | 3 | I'm trying to show that manifolds are affine, i.e. $\text{Man}(M,N)\cong \mathbb{R}\text{-Alg}(C^\infty(N),C^\infty(M)) $. If I could show this for $N=\mathbb{R}^n$, then I know how to do the rest using the strong Whitney's embedding theorem. But it seems hard to show the surjectivity of $ \text{Man}(M,\mathbb{R}^n)\hookrightarrow \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M)):f\mapsto f^\*$.
I found this post [Are all manifolds are affine?](https://mathoverflow.net/questions/88986/are-all-manifolds-affine), but it didn't mention any explicit material about this, and I failed finding anything other than that post. I learned that $\text{Man}(M,\mathbb{R}^n)\cong C^\infty\text{-Ring}(C^\infty(\mathbb{R}^n),C^\infty(M))$ from *Models for Smooth Infinitesimal Analysis*, but I don't know how to show that $C^\infty\text{-Ring}(C^\infty(\mathbb{R}^n),C^\infty(M))\cong \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))$.
Thanks in advance for any solution or material about this.
**EDIT:** I had a further investigation (but no success) and I'm putting it here with supplementary details of my question.
The images of the coordinates functions on $\mathbb{R}^n$ consist to give a map $ \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))\to \text{Man}(M,\mathbb{R}^n)$, which tells the injectivity of $ \text{Man}(M,\mathbb{R}^n)\hookrightarrow \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))$. To show the desired surjectivity, we still need to verify that the composition $$\mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))\to \text{Man}(M,\mathbb{R}^n)\hookrightarrow \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))$$
is the identity.
I find that it suffices to show that, for any $\mathbb{R}$-algebra homomorphism $\varphi:C^\infty(\mathbb{R}^n)\to C^\infty(M)$, if $\rho\in C^\infty(\mathbb{R}^n)$ satisfies $\rho|\_V=0$ for some open subset $V\subset\mathbb{R}^n$, then $\varphi(\rho)|\_{f^{-1}(V)}=0$, where $f:M\to \mathbb{R}^n$ is given by the images of the coordinate functions on $\mathbb{R}^n$ under $\varphi$. If this is true, then we can define the restriction of $\varphi$ to any open subset $U\subset \mathbb{R}^n$,
$$ \varphi|\_U :C^\infty(U)\to C^\infty(f^{-1}(U)),$$
by putting, for any $g\in C^\infty(U)$, $\varphi|\_U(g)$ to be the unique smooth function such that
$$\varphi|\_U(g)|\_{f^{-1}(V)}=\varphi(G)|\_{f^{-1}(V)} $$
for any open set $V\subset U$ and $G\in C^\infty(\mathbb{R}^n)$ such that $G|\_V=g|\_V$; to show that $\varphi(G)|\_{f^{-1}(V)} $ does not depend on the choice of $G$ requires exactly the property I claimed above.
If the restrictions of $\varphi$ can be defined, then we obtain a sheaf morphism. Investigating the germs and we can find that $\varphi$ is indeed the pullback by $f$.
| https://mathoverflow.net/users/167862 | How to show that $\text{Man}(M,\mathbb{R}^n)\cong \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^n),C^\infty(M))$? | The proof of this fact is available in modern textbooks.
For example, see Theorem 7.16 in Jet Nestruev's *Smooth Manifolds and Observables* (Second Edition, 2020).
In fact, the cited book contains a lot of material that explains how to pass between differential geometric objects and the corresponding algebraic objects, e.g., vector bundles and modules, differential forms and Kähler differentials, etc.
| 10 | https://mathoverflow.net/users/402 | 420552 | 171,130 |
https://mathoverflow.net/questions/420547 | 4 | I'm looking for the proof that $\mathtt{PSP}$, the statement that every uncountable subset of the the Baire space $\mathbb{N}^\mathbb{N}$ contains an homeomorphic copy of the Cantor space $2^\mathbb{N}$, implies the consistency of inaccessible cardinals.
I was able to track down Mycielski's paper *On the axiom of determinateness* in which, building on Specker's *Zur Axiomatik der Mengenlehre*, this result is proven (read also this [question](https://mathoverflow.net/questions/275216/inaccessible-cardinals-and-the-perfect-set-property-for-coanalytic-sets) on this account), but I don't read german and Mycielski's article does not report Specker's full argument (and it's also pretty hard to read due to painful notation).
Is there a "newer" account on this proof? Is there a book or thesis where it is the full proof is reported?
Thanks!
| https://mathoverflow.net/users/141146 | $\mathtt{PSP}$ implies the consistency of inaccessible cardinals | See Proposition 11.5 (and the discussion leading up to it) in Kanamori's book **The Higher Infinite**. Note that Kanamori states the result a bit more optimally: if $M\models\mathsf{ZF}$ + "$\omega\_1$ is regular" + $\mathsf{PCP}$ then $\omega\_1^M$ is inaccessible in $L^M$ (and in fact $M\models$ "$\omega\_1$ is inaccessible to reals," that is, $\omega\_1^M$ is inaccessible in $L[a]^M$ for every $a\in\mathbb{R}^M$).
---
Since it's pretty short, I'll sketch the proof here for completeness:
Suppose $M$ satisfies the necessary conditions; from now on, all statements are made within $M$.
First, we prove an auxiliary result: that $\omega\_1^{L[a]}$ being countable for each real $a$ implies that $\omega\_1$ is inaccessible to reals. We prove the contrapositive. Suppose $\omega\_1$ is not inaccessible in $L[a]$ for some real $a$. By the usual condensation argument (relativized) we have $L[a]\models\mathsf{GCH}$, so we must have $\omega\_1=\rho^{+^{L[a]}}$ for some $L[a]$-cardinal $\rho$. Now $\rho$ is countable in reality, hence coded by a real; let $b$ be a real coding a well-ordering of $\omega$ isomorphic to $\rho$. Then $a\oplus b$ is a real with $\omega\_1^{L[a\oplus b]}\ge\omega\_1$.
Now suppose $\omega\_1$ is not inaccessible to reals. By the above, this means that $\omega\_1^{L[a]}=\omega\_1$ for some real $a$. But per Godel we know that $L[a]\models\omega\_1\le 2^{\aleph\_0}$, or put another way there is an injection from $\omega\_1^{L[a]}$ into the reals. Sicne $\omega\_1^{L[a]}=\omega\_1$, this means there is an injection from $\omega\_1$ into the reals, whence (by Bernstein) $\mathsf{PSP}$ fails.
| 4 | https://mathoverflow.net/users/8133 | 420553 | 171,131 |
https://mathoverflow.net/questions/420501 | 1 | As the title says, How to find the probability of vectors a, b, c, on some unit sphere, all lies on same side of some hyperplane passing through the origin. Information present are the angles between vectors a and b, b and c, c and a.
I am trying to bound the area of sphere in which the normal of hyperplane can reside but I feel stuck. Any hints or directions?
| https://mathoverflow.net/users/480694 | Probability that three vectors of a unit sphere lie on one side of a hyperplane if angle between the vectors are given | $\newcommand\al\alpha\newcommand\be\beta\newcommand\ga\gamma$It appears that the question is as follows: Given unit vectors $a,b,c$ with angles
$$\al:=\cos^{-1}(b\cdot c),\quad \be:=\cos^{-1}(a\cdot c),\quad \ga:=\cos^{-1}(b\cdot a)$$
(where $\cdot$ denotes the dot product), find the probability, say $p$, that the vectors $a,b,c$ are to the same side of a hyperplane through the origin chosen uniformly at random.
The answer is
$$p=\frac{2\pi-\al-\be-\ga}{2\pi}. \tag{1}\label{1}$$
Indeed,
$$p=P(a\cdot U>0,b\cdot U>0,c\cdot U>0)
+P(a\cdot U<0,b\cdot U<0,c\cdot U<0)
=2P(a\cdot U>0,b\cdot U>0,c\cdot U>0),$$
where $U$ is a random vector uniformly distributed on the unit sphere. Next, the random vector $U$ equals $G/|G|$ in distribution, where $G$ is a standard Gaussian random vector and $|G|$ is the Euclidean norm of $G$. So,
$$p=2P(X>0,Y>0,Z>0),$$
where
$$X:=a\cdot G,\quad Y:=b\cdot G,\quad Z:=c\cdot G,$$
so that $X,Y,Z$ are zero-mean, unit-variance jointly normal random variables with correlations
$$\rho\_{X,Y}=a\cdot b=\cos\ga,\quad
\rho\_{Y,Z}=b\cdot c=\cos\al,\quad
\rho\_{X,Z}=a\cdot c=\cos\be.$$
Now \eqref{1} follows from the known formula
$$P(X>0,Y>0,Z>0)=\frac{\cos^{-1}(-\rho\_{X,Y})+\cos^{-1}(-\rho\_{Y,Z})+\cos^{-1}(-\rho\_{X,Z})-\pi}{4\pi}\tag{2}\label{2}$$
-- see e.g. the second display in Section 6 on p. 355 of [Plackett](https://www.jstor.org/stable/2332716).
---
Note that \eqref{1} holds for any dimensions $\ge3$.
---
Another way to derive \eqref{2} and hence \eqref{1} is, of course, to note that the numerator of the ratio in \eqref{2} is the area of the spherical triangle on the unit sphere with angles $\pi-\al,\pi-\be,\pi-\ga$. This area can be expressed as a double integral in spherical coordinates. Yet other ways to find this area can be found e.g. on [this page](https://math.stackexchange.com/questions/110075/deriving-the-surface-area-of-a-spherical-triangle). The formula for this area is [Girard's theorem](https://www.theoremoftheday.org/GeometryAndTrigonometry/Girard/TotDGirard.pdf).
| 5 | https://mathoverflow.net/users/36721 | 420559 | 171,133 |
https://mathoverflow.net/questions/403291 | 1 | Throughout, by "logic" I mean regular logic (in the sense of Ebbinghaus–Flum–Thomas) whose sentences are coded by elements of $\mathsf{HC}$. Say that $\mathcal{L}$ is **Barwise compact** iff whenever $\mathbb{A}$ is a countable admissible set and $X\subseteq\mathbb{A}$ is a $\mathbb{A}$-c.e. $\mathcal{L}$-theory each of whose $\mathbb{A}$-finite subtheories is satisfiable, then $X$ is satisfiable. Finally, say that a logic $\mathcal{L}$ has the **single-sentence DLS property** iff every satisfiable $\mathcal{L}$-sentence has a countable model.
It's easy to show that $\mathcal{L}\_{\omega\_1,\omega}$ has the single-sentence DLS property *(favorite silly proof: apply Mostowski absoluteness to $\mathit{Col}(\omega,\kappa)$ for large enough $\kappa$)*, and Barwise showed that $\mathcal{L}\_{\omega\_1,\omega}$ is Barwise compact. Moreover, Lindstrom's characterization (see [Väänänen, *Lindstrom's Theorem*](http://www.math.helsinki.fi/logic/opetus/lt/lindstrom_theorem1.pdf)) of first-order logic can be [tweaked](https://mathoverflow.net/questions/350778/improving-a-lindstrom-y-fact-about-mathcall-omega-1-omega) to show that no logic with the above two properties can be *much stronger than* $\mathcal{L}\_{\omega\_1,\omega}$. However, there's still a large gap here, and I'm curious whether anything is known about it. In particular:
>
> Is there an already-introduced logic which is properly stronger than $\mathcal{L}\_{\omega\_1,\omega}$, has the single-sentence DLS property, and is Barwise compact?
>
>
>
I'd also be interested in what happens if we weaken the definition of Barwise compactness to allow replacing "admissible set" with "transitive model of $T$" for some first-order $T\subseteq \mathit{Th}(\mathsf{HC})$.
| https://mathoverflow.net/users/8133 | Natural strong logic with Barwise compactness property | A very nice family of examples is provided by Harrington's 1980 paper *[Extensions of countable infinitary logic which preserve most of its nice properties](https://link.springer.com/content/pdf/10.1007/BF02021129.pdf)*. Harrington shows that if we expand $\mathcal{L}\_{\omega\_1,\omega}$ by any infinitary propositional connective satisfying an appropriate tameness property - and there are $2^{2^{\aleph\_0}}$ of these - we preserve multiple tameness properties, including Barwise compactness and dLS for individual sentences (the latter basically trivially).
| 1 | https://mathoverflow.net/users/8133 | 420561 | 171,134 |
https://mathoverflow.net/questions/389952 | 2 | I'd like to know if there is a concentration inequality for the sample covariance matrix that don't assume the knowledge of the true mean.
---
**Background.**
Given a probability distribution $\mu$ on $\mathbb R^d$, the covariance matrix of $\mu$ is defined as follows:
$$\Sigma := \mathbb E [(x - \bar \mu)(x -\bar \mu)^\top] $$
where $x \sim \mu$ and $\bar \mu = \mathbb E [x]$.
If $X = (x\_1, \cdots x\_m)$ is an i.i.d. sample drawn from $\mu$, then we can define two estimators:
\begin{align\*}
& \hat \Sigma\_1 := \frac1m \sum\_{i=1}^m (x\_i - \bar \mu)(x\_i - \bar \mu)^\top, \text{ where } \bar \mu = \mathbb E\_{x \sim \mu} [x] \\
& \hat \Sigma\_2 := \frac1{m-1} \sum\_{i=1}^m (x\_i - \bar x)(x\_i - \bar x)^\top, \text{ where } \bar x = \frac1m (x\_1 + \cdots x\_m)
\end{align\*}
They both satisfy $\mathbb E\_X \hat \Sigma\_1 = \mathbb E\_X \hat \Sigma\_2 = \Sigma$.
The second estimator $\hat \Sigma\_2$ is of interest because $\bar \mu$ is often not known in practice.
---
**Question.**
I'm interested in the concentration of $\hat \Sigma\_2$ to $\Sigma$ as $m \rightarrow \infty$. More precisely, given a number $t > 0$, I'd like to know whether there exists a constant $A>0$ and a term $\alpha \in (0,1)$ that depend on $\mu$ and $t$ such that
$$\text{Prob}(\| \Sigma - \hat \Sigma\_2 \| \ge t) \le A \cdot \alpha^m$$
where $\|\cdot \|$ is the spectral norm, also known as the 2-norm. (The Frobenius norm is also fine, since for any $d \times d$ matrix $A$, $\|A\| \le \|A\|\_F \le \sqrt{d} \|A\|$)
In the case of the difference $\|\Sigma - \hat \Sigma\_1\|$, such an answer can be obtained using the [matrix Bernstein inequality](https://arxiv.org/abs/1501.01571). However, I'm less sure about $\|\Sigma - \hat \Sigma\_2\|$. I have an idea, which is to use the fact that:
$$\hat \Sigma\_1 - \hat \Sigma\_2 = \frac1{m(m-1)} \sum\_{i\neq j} (x\_i-\bar\mu) (x\_j-\bar\mu)^\top$$
which follows from:
\begin{align\*}
\hat \Sigma\_2 =& \frac1m \sum\_i x\_i x\_i^\top - \frac1{m(m-1)} \sum\_{i\neq j} x\_i x\_j^\top \\
=& \frac1m \sum\_i (x\_i-\bar\mu) (x\_i-\bar\mu)^\top - \frac1{m(m-1)} \sum\_{i\neq j} (x\_i-\bar\mu) (x\_j-\bar\mu)^\top \\
=& \hat \Sigma\_1 - \frac1{m(m-1)} \sum\_{i\neq j} (x\_i-\bar\mu) (x\_j-\bar\mu)^\top
\end{align\*}
But now I'm not sure how to control the sum of the quantities $(x\_i-\bar\mu) (x\_j-\bar\mu)^\top$, which are *not independent*.
This should be a fairly standard question with a standard answer, but I couldn't find an answer to this. A [similar question](https://math.stackexchange.com/questions/2374810/concentration-inequality-for-covariance)'s only answer wasn't addressing my question; it was addressing the case for $\hat \Sigma\_1$.
| https://mathoverflow.net/users/156792 | Concentration inequality for the sample covariance matrix | A variant of this (taking $\frac1m$ instead of $\frac1{m-1}$ for the empirical covariance) is answered on Proposition 2.6, page 10 of <https://arxiv.org/pdf/2110.06357.pdf> .
| 2 | https://mathoverflow.net/users/156792 | 420569 | 171,137 |
https://mathoverflow.net/questions/229088 | 6 | There exists a Latin square of order $8$ which can be partitioned into $2 \times 4$ subrectangles:
$$
\begin{bmatrix}
\color{red} 1 & \color{red} 2 & \color{red} 3 & \color{red} 4 & \color{purple} 5 & \color{purple} 6 & \color{purple} 7 & \color{purple} 8 \\
\color{red} 2 & \color{red} 3 & \color{red} 4 & \color{red} 1 & \color{purple} 6 & \color{purple} 7 & \color{purple} 8 & \color{purple} 5 \\
7 & 8 & \color{blue} 1 & \color{blue} 2 & \color{blue} 3 & \color{blue} 4 & 5 & 6 \\
8 & 5 & \color{blue} 2 & \color{blue} 3 & \color{blue} 4 & \color{blue} 1 & 6 & 7 \\
\color{pink} 5 & \color{pink} 6 & \color{pink} 7 & \color{pink} 8 & \color{green} 1 & \color{green} 2 & \color{green} 3 & \color{green} 4 \\
\color{pink} 6 & \color{pink} 7 & \color{pink} 8 & \color{pink} 5 & \color{green} 2 & \color{green} 3 & \color{green} 4 & \color{green} 1 \\
\color{orange} 3 & \color{orange} 4 & \color{brown} 5 & \color{brown} 6 & \color{brown} 7 & \color{brown} 8 & \color{orange} 1 & \color{orange} 2 \\
\color{orange} 4 & \color{orange} 1 & \color{brown} 6 & \color{brown} 7 & \color{brown} 8 & \color{brown} 5 & \color{orange} 2 & \color{orange} 3 \\
\end{bmatrix}
$$
If we take the row-symbol parastrophe of this Latin square (i.e., replace each entry $(i,j,l\_{ij})$ with $(l\_{ij},j,i)$), then the entry colors define a decomposition of $K\_{8,8}$ into $2$-regular spanning subgraphs of $K\_{4,4}$. I would like to generalize this.
*Question*: For which divisors $a$ and $b$ of $n$ does there exist a Latin square of order $n$ that can be partitioned into $a \times b$ subrectangles?
(*Note*: We assume $a \leq b$. Subrectangles must have $b$ symbols. We don't assume that the boundaries of the subrectangles align.)
Observations:
* It's trivially possible when
+ $a$ divides $b$ (construct a Latin square with blocks that are $b \times b$ subsquares)
+ $a=1$ or $b=n$.
* The first non-trivial case is when $a=2$, $b=3$, and $n=6$. If my code is correct, then it is impossible (by exhaustive search). (I'm tempted to think this is just because the parameters are too small.) The best possible is $4$ subrectangles, which is straightforward to construct.
* My code found a random Latin square which gave an $a=2$, $b=5$, and $n=10$ example:
$$
\begin{bmatrix}
\color{red} 7 & \color{red} 1 & 9 & 5 & \color{red} {10} & 4 & 8 & \color{red} 2 & 6 & \color{red} 3 \\
\color{red} 3 & \color{red} {10} & 5 & 8 & \color{red} 2 & 6 & 4 & \color{red} 1 & 9 & \color{red} 7 \\
\hline
\color{blue} 9 & \color{blue} 6 & 4 & 7 & 1 & \color{blue} 2 & \color{blue} {10} & \color{blue} 3 & 5 & 8 \\
\color{blue} 6 & \color{blue} 3 & 7 & 1 & 5 & \color{blue} 9 & \color{blue} 2 & \color{blue} {10} & 8 & 4 \\
\hline
\color{pink} 4 & 5 & 1 & \color{pink} 2 & \color{pink} 8 & 3 & 9 & \color{pink} 7 & \color{pink} {10} & 6 \\
\color{pink} 2 & 9 & 6 & \color{pink} {10} & \color{pink} 4 & 1 & 3 & \color{pink} 8 & \color{pink} 7 & 5 \\
\hline
\color{purple} 1 & 2 & \color{purple} 8 & 3 & 7 & \color{purple} {10} & 6 & \color{purple} 5 & 4 & \color{purple} 9 \\
\color{purple} 8 & 4 & \color{purple} {10} & 6 & 3 & \color{purple} 5 & 7 & \color{purple} 9 & 2 & \color{purple} 1 \\
\hline
\color{brown} 5 & 8 & \color{brown} 2 & 9 & 6 & 7 & \color{brown} 1 & 4 & \color{brown} 3 & \color{brown} {10} \\
\color{brown} {10} & 7 & \color{brown} 3 & 4 & 9 & 8 & \color{brown} 5 & 6 & \color{brown} 1 & \color{brown} 2 \\
\end{bmatrix}
$$
| https://mathoverflow.net/users/48278 | For which divisors $a$ and $b$ of $n$ does there exist a Latin square of order $n$ that can be partitioned into $a \times b$ subrectangles? | I eventually co-authored a paper which includes this topic. The non-trivial results are:
>
> * There exists a Latin square of order $n$ which decomposes into $2 \times (n/2)$ subrectangles for all even $n \not\in \{2,6\}$.
> * There exists a Latin square of order n which decomposes into $3 \times (n/3)$ subrectangles if and only if $3$ divides $n$ and $n > 9$.
> * For sufficiently large n, there exists a Latin square of order $n$ which partitions into $d \times (n/d)$ subrectangles if and only if $d$ divides $n$.
>
>
> Akbari, Marbach, Stones, Wu, *Balanced equi-n-squares* Electron. J. Combin, 2020. ([pdf](https://www.combinatorics.org/ojs/index.php/eljc/article/download/v27i4p8/pdf/))
>
>
>
The smallest cases we didn't resolve are
* $20 \times 20$ Latin squares into $4 \times 5$ subrectangles;
* $28 \times 28$ Latin squares into $4 \times 7$ subrectangles; and
* $30 \times 30$ Latin squares into $5 \times 6$ subrectangles.
| 5 | https://mathoverflow.net/users/48278 | 420575 | 171,138 |
https://mathoverflow.net/questions/420567 | 1 | Is there study on polynomial-like functions of the following kind?
$$f(z) = c\_0 + a\_1z+b\_1\bar{z} + a\_2z^2+b\_2\bar{z}^2 + ...+ a\_nz^n+b\_n\bar{z}^n$$
My reason for studying it is polynomials are analytic so that they can't approximate non-analytic complex functions with arbitrary precision. While the above function is non-analytic so they might do the approximation.
Any pointer is appreciated. Thanks!
| https://mathoverflow.net/users/151671 | Complex polynomial-like functions with conjugate terms | Yes, there are studies under the title harmonic maps of the plane. In general, a harmonic map is of the form $f+\overline{g}$ where $f,g$ are analytic. Most of this literature is about univalent harmonic maps (in a region) but there are some papers on the general case, including polynomials and entire functions in the whole plane. Here is a small sample:
P. Duren, Harmonic mappings in the plane, Cambridge Tracts in Mathematics, 156. Cambridge Univ. Press, Cambridge, 2004.
A. Lyzzaik, Local properties of light harmonic mappings, Canadian J.
Math., 44 (1992) 135–153.
D. Khavinson and G. Swiatek, On the maximal number of zeros of cer- ´
tain harmonic polynomials, Proc. Amer. Math. Soc., 131 (2003) 409–414.
W. Bergweiler and A. Eremenko, On the number of solutions of a transcendental equation arising in the theory of gravitational lensing, CMFT, Comput. Methods Funct. Theory 10 (2010), No. 1, 303--324.
S. Nakane, D. Schleicher,
On multicorns and unicorns. I. Antiholomorphic dynamics, hyperbolic components and real cubic polynomials.
Internat. J. Bifur. Chaos Appl. Sci. Engrg. 13 (2003), no. 10, 2825–2844.
| 3 | https://mathoverflow.net/users/25510 | 420588 | 171,140 |
https://mathoverflow.net/questions/417973 | 6 | It is known that if a finite group $G$ admits a faithful *topological* action on the 3-sphere $S^3$, then $G$ admits a faithful action on $S^3$ by *isometries*. (Pardon proved that a topological action implies a smooth action, and Dinkelbach & Leeb proved that a smooth action implies an isometric one.) I wonder if this extends to infinite groups acting on $R^3$:
**Question:** Let $G$ be a finitely generated group that admits a faithful, co-compact, topological action on $R^3$, such that no orbit has an accumulation point. Must $G$ admit an action by isometries on one of Thurston’s geometries, preserving the above properties (i.e. faithful, co-compact, accumulation-free)?
**Update:** The comments below suggest that the answer is negative in this generality (an "official" answer with references and explanation would be welcome). **What if G is assumed to be Gromov-hyperbolic?** I'm most interested in the 1-ended case, anticipating an isometric action on $\mathbb{H}^3$. (1-endedness excludes $\mathbb{S}^2 \times \mathbb{R}$.) This is partly motivated by Cannon's conjecture.
By topological action I mean an action by homeomorphisms.
**Update:** instead of just assuming that no orbit has an accumulation point, I'm happy with stronger discreteness conditions such as proper discontinuity
| https://mathoverflow.net/users/69681 | From topological actions on $\mathbb{R}^3$ to isometric actions | I believe (though have not checked carefully) that the argument in my paper proves:
>
> If $\Gamma$ (discrete) acts continuously and properly discontinuously on a smooth three-manifold $M$, then that action can be uniformly approximated by a smooth action.
>
>
>
The point is simply that each step in the argument is local on the quotient space $M/\Gamma$ (which is a reasonable topological space given proper discontinuity).
Here is a (sketched) better argument, which proves the indented statement above as a consequence of my paper. Fix $x\in M$, and consider the stabilizer $\Gamma\_x\leq\Gamma$, which is finite. Choose coset representatives $g\_i\in\Gamma/\Gamma\_x$, so $\Gamma x=\{g\_ix\}\_i$. Fix a $\Gamma\_x$-invariant open neighborhood $U$ of $x$ whose translates $g\_iU$ are all disjoint (should exist by proper discontinuity). Now smooth the action of $\Gamma\_x$ on $U$ using my paper, and smooth the homeomorphisms $g\_i:U\to g\_iU$ using Bing--Moise. This determines a smoothing of the action of $\Gamma$ on $\Gamma U\subseteq M$. By making the approximations sufficiently $C^0$-close, we ensure that this smoothed action of $\Gamma$ on $\Gamma U\subseteq M$ splices together with the original action of $\Gamma$ on $M\setminus\Gamma U$ to define a new action of $\Gamma$ on $M$, which is now smooth over $\Gamma U$. Now iterate a (locally) finite number of times to cover all of $M$.
| 2 | https://mathoverflow.net/users/35353 | 420591 | 171,142 |
https://mathoverflow.net/questions/420102 | 6 | Let $G \neq 1$ be a finite perfect group which is not simple.
Is it true that $G$ necessarily has a maximal subgroup whose derived subgroup
has nontrivial core in $G$?
*Remark 1:* This holds for all such $G$ of order less than 100000.
*Remark 2:* In case the answer is negative, I would mainly be interested
in a counterexample with nontrivial Frattini subgroup.
| https://mathoverflow.net/users/128342 | Does a non-simple perfect group always have a maximal subgroup whose derived subgroup has nontrivial core? | The answer to the title question is 'No.' Following YCor's comment, an example is furnished by $S=J\_1$, the smallest Janko sporadic group, and its complex irreducible character $\chi$ of degree 76 (in Atlas notation, 76a). I have checked by hand (hopefully correctly), using the Atlas, that $(\chi\downarrow M,1\_M)>0$ for all maximal subgroups $M$ of $J\_1$. For any prime $p$ not dividing $|J\_1|=2^3.3.5.7.11.19$, let $N\_p$ be an $F\_pS$-module affording the mod-$p$ reduction of $\chi$. Then, as YCor points out, the semidirect product $N\_pS$ has the property that for every maximal subgroup $H\le N\_pS$, $[H,H]$ does not contain $N\_p$.
However, this example has trivial Frattini subgroup.
| 6 | https://mathoverflow.net/users/99221 | 420592 | 171,143 |
https://mathoverflow.net/questions/420583 | 2 | Given a solvable Lie algebra $\frak{a}$ and a semisimple Lie algebra $\frak{g}$ we can take their semidirect product $\frak{a} \rtimes \frak{g}$, with respect to a Lie algebra map $\frak{g} \to \mathrm{Der}(\frak{a})$. For the zero map $\frak{g} \to \mathrm{Der}(\frak{a})$ we get the usual direct product. But can there exist more than one non-zero map $\frak{g} \to \frak{a}$, giving two non-isomorphic semi-direct product? What is a motivating example? I will be interested most in a example in the context of Levi's theorem.
| https://mathoverflow.net/users/176218 | Non-isomorphic direct products of a solvable and a semisimple Lie algebra | Let $K$ be the ground field of characteristic zero.
For $n\ge 1$, let $\mathfrak{v}\_n$ denote an $n$-dimensional irreducible representation of $\mathfrak{sl}\_2(K)$ over $K$ ($(n-1)$-th symmetric power of the standard 2-dimensional representation). Then $\mathfrak{v}\_3\rtimes\mathfrak{sl}\_2(K)$ and ($\mathfrak{v}\_1\oplus\mathfrak{v}\_2)\rtimes\mathfrak{sl}\_2(K)$ are non-direct semidirect products and are not isomorphic (the first is perfect and not the second).
Note: however, in general (say over the complex or real numbers), there are only finitely many homomorphisms from $\mathfrak{g}$ to $\mathrm{Der}(\mathfrak{a})$ up to conjugacy, and hence there are only finitely many isomorphism classes of semidirect products.
| 2 | https://mathoverflow.net/users/14094 | 420595 | 171,145 |
https://mathoverflow.net/questions/420551 | 4 | A real matrix is called totally positive if all of its minors are positive. Of course, to check that a given matrix is totally positive it is not necessary to check all the minors: for example, it suffices to check that all initial minors are positive. A minor $A\begin{pmatrix}I\\J\end{pmatrix}$ of $A$, corresponding to the sets of row indices $I$ and column indices $J$ is called initial if both $I$ and $J$ consist of consecutive indices and $1\in I\cup J$ (so every entry of $A$ is the lower right entry of a unique initial minor). In fact, there are many distinct families of minors such that their positivity implies total positivity, parametrized by "double wiring diagrams" (see <https://link.springer.com/article/10.1007/BF03024444>).
There are similar tests for total non-negativity. But all such tests seem to optimize the number of minors to be checked and not the total computational cost. One would imagine that checking more smaller minors could in some cases be done faster then checking fewer larger minors. Has any work been done in this direction?
| https://mathoverflow.net/users/5018 | Total positivity tests: optimal in the number of minors vs. the computational cost | There are at least a few places looking at complexity for total positivity counting arithmetic operations. I don't know if people have looked at trying to find smaller minors. These use initial minors but try to compute smartly.
>
> Cryer, Colin W. Some properties of totally positive matrices. Linear Algebra Appl. 15 (1976), no. 1, 1--25. [link](https://doi.org/10.1016/0024-3795(76)90076-8)
>
>
>
In Remark 4.2 of the above, the complexity of checking total positivity with matrix factorization is analyzed and found to be cubic.
>
> Gasca, M.; Peña, J. M. Total positivity and Neville elimination. Linear Algebra Appl. 165 (1992), 25--44. [link](https://doi.org/10.1016/0024-3795(92)90226-Z)
>
>
>
After Theorem 5.4 in the above, another cubic method for checking total positivity is discussed. (Here STP is ''strictly totally positive'' and TP is ''totally positive'' which is often called ''totally nonnegative.'')
>
> Daniel Carter, Charles Johnson. The Complexity of Checking Partial Total Positivity. [arXiv:2103.08742 [cs.CC]](https://arxiv.org/abs/2103.08742)
>
>
>
This recent preprint gives another cubic algorithm that makes use of Dodgson condensation.
| 5 | https://mathoverflow.net/users/51668 | 420597 | 171,146 |
https://mathoverflow.net/questions/420600 | 2 | Assume factorization of $N$ is unknown. What is the best complexity we know to find roots of the irreducible equation $$ax^2+bx+c\equiv0\bmod N?$$
Is this problem equivalent to any hardness results?
| https://mathoverflow.net/users/10035 | On roots of irreducible quadratics modulo composites | Ability to find all roots leads to finding factorization of $N$. For example, if $N=pq$ is a product of two odd primes, and we find a root $x'$ of $x^2-1\equiv 0\pmod N$ such that $x'\equiv 1\pmod p$ and $x'\equiv -1\pmod q$, then we can compute $p$ as $\gcd(N,x'-1)$.
| 6 | https://mathoverflow.net/users/7076 | 420604 | 171,148 |
https://mathoverflow.net/questions/420605 | 3 | I sincerely apologize if MathOverflow is not the appropriate place to ask this question. I also tried consulting M.SE but it seems that this question gained little to no interest .
---
Consider a Heegaard splitting $M = V\cup\_F W$ of a $3$-manifold $M$ with splitting surface $F = \partial V = \partial W$. Suppose there is a meridian pair $\{D\_1,D\_2\}$, that is, $D\_1$ is a meridian disk in $V$ and $D\_2$ a meridian disk in $W$. Both disks intersect transversaly so that their boundary curves $\partial D\_1$ and $\partial D\_2$ intersect in exactly one point.
Now I would like to verify that $V\setminus N(D\_1)$ is ambient isotopic to $V\cup N(D\_2)$ where $N(\cdot)$ denotes a regular neighbourhood. But I'm somewhat new to geometric topology and I am not sure on how to proceed.
---
My attempt:
-----------
What I know is that both disks $D\_i$ are compression disks in $V$ respectively $W$, thus we can compress along $D\_1$ in $V$, that is, we cut at the regular neighbourhood $N(D\_1)$ of $D\_1$ in order to obtain $V\setminus N(D\_1)$.
Now the union of $N(D\_1)$ and $N(D\_2)$ along the square in which they intersect is a $3$-ball and so we can isotope "through" that ball.
But now I'm unsure if this is the correct attempt and if this already shows that $V\setminus N(D\_1)$ is ambient isotopic to $V\cup N(D\_2)$.
Another Idea was to somehow construct the deired isotopy in a "pedestrian way" by considering a collar of the boundary $\partial V$ and somehow extending an isotopy from the boundary to the entire handlebody $V$ but so far I haven't had much success with this approach either.
Any feedback would be highly appreciated! Thanks in advance.
| https://mathoverflow.net/users/136505 | Given a Heegaard splitting $M = V\cup_F W$, then $V\setminus N(D_1)$ is ambient isotopic to $V\cup N(D_2)$ for a meridian pair $\{D_1,D_2\}$ | Yes, you are correct that the two handlebodies ($V$ cut along $D$ and $V$ plus a neighbourhood of $E$) are isotopic. This is discussed when defining *stabilisation* and the inverse operation *destabilisation* of Heegaard splittings. I don’t have a copy of Hempel’s book with me, but I am fairly sure it is discussed there.
| 5 | https://mathoverflow.net/users/1650 | 420606 | 171,149 |
https://mathoverflow.net/questions/420607 | 5 | Many connected [vertex-transitive](https://en.wikipedia.org/wiki/Vertex-transitive_graph) graphs $G=(V,E)$ have the property that some of their automorphisms other than the identity have fixed points. To point out two simple examples:
* If $G = K\_3$ then the automorphism swapping the points of an edge and leaving the remaining point intact has $1$ fixed point.
* For $G = C\_4$, the "mirror map" along one of the diagonals has those diagonal points as fixed points.
But for both graphs, there are automorphisms that do not have any fixed points. (For both examples, consider a rotation map.)
Is there a connected vertex-transitive graph $G=(V,E)$ with $|V| \geq 2$ such that every automorphism has a fixed point?
| https://mathoverflow.net/users/8628 | Connected vertex-transitive graph with the fixed-point property | Let me convert my comments into an answer.
There is no such **[EDIT:] finite** graph $G$. Indeed, something stronger can be said. Suppose that a group $\Gamma$ acts transitively by permutations on a finite set $X$ , with $\#X \geq 2$. Then there is some $\gamma \in \Gamma$ for which $\gamma\colon X \to X$ has no fixed points. The proof is a simple application of Burnside's Lemma, see e.g. <https://math.stackexchange.com/questions/106158/every-transitive-permutation-group-has-a-fixed-point-free-element>.
For infinite graphs, see this other question: [Infinite vertex-transitive graph where every automorphism has a fixed vertex](https://mathoverflow.net/questions/420668/infinite-vertex-transitive-graph-where-every-automorphism-has-a-fixed-vertex)
| 9 | https://mathoverflow.net/users/25028 | 420611 | 171,150 |
https://mathoverflow.net/questions/420264 | 2 | I got this question because of [this post](https://mathoverflow.net/questions/415806/limit-of-the-extremal-process-of-i-i-d-gaussians-see-from-the-tip), discussing the limiting distribution of extreme point process of i.i.d. Gaussians seen from the tip. For notations please refer that post.
At first, @Iosif Pinelis gave an answer suggesting the number of points in a certain interval $[-A,0]$ is geometrically distributed, which sounds weird to me since the extreme point process seen from the critical speed is roughly Poisson point process, so how can it be so different when seen from the tip? At that time I just accepted it since the derivation is indeed correct.
But today I realized we can get the limiting distribution of the extreme point process seen from the tip, directly from the above Poisson point process, by noticing the random shift between these two point processes converges to some Gumbell distribution. This allows me to represent the geometric stuff by a Poisson point process with **random** intensity measure.
So it just came to me naturally:
>
> Is it the case that **any** point process can be represented as a Poisson point process with a random intensity measure, if it satisfies the independence condition for PPP? (or do we need other necessary constraints?)
>
>
>
>
> Suppose we have a description of a point process, like that @Iosif's geometric-type Laplace functional for $\bar{\mathcal{E}}$, the limit extreme point process seen from the tip, how to identify the random intensity measure to give that representation?
>
>
>
I may fail to organise my questions clearly, but basically I just find it extremely interesting that the geometric representation and the random intensity measure representation are actually the same. So I wonder whether it is universal, and if yes, practically how to achieve that.
| https://mathoverflow.net/users/174600 | How to identify a point process as Poisson point process with (possibly) random intensity measure? | $\bullet$ The answer to the first question is affirmative, but the notion of "independence" needs to be modified if we allow for the point process to be directed by an external random variable.
A Poisson point process with a random intensity is known as a [Cox process.](https://en.wikipedia.org/wiki/Cox_process) The "independence" property is defined in terms of a "thinning" procedure, see [Lectures on the Poisson Process:](https://www.math.kit.edu/stoch/~last/seite/lectures_on_the_poisson_process/media/lastpenrose2017.pdf)
Given a point process $N$ on $\mathbb{R}^+$ and $p\in(0,1]$, the thinned process $N\_p$ is obtained by retaining every point of the process with a probability $p$ and deleting it with probability $1-p$. Then $N$ is called the $p$-inverse of $N\_p$.
*Theorem:* A point process is a Cox process if and only if, for
every $p\in(0,1]$ there exists a point process which is the $p$-inverse of the process.
$\bullet$ Concerning the second question: For each integer $k$, the $k$-th [factorial moment measure](https://en.wikipedia.org/wiki/Factorial_moment_measure) of the Cox process equals the $k$-the [ordinary moment measure](https://en.wikipedia.org/wiki/Moment_measure) of the random intensity. So point process and random intensity have the same mean, and the variance of the random intensity equals the variance of the point process minus the mean.
This implies a simple necessary criterion for a point process to be Cox process: the variance must be greater than the mean ("bunching"). If the variance is smaller than the mean ("anti-bunching") the point process cannot be a Poisson process with a random intensity.
| 1 | https://mathoverflow.net/users/11260 | 420624 | 171,154 |
https://mathoverflow.net/questions/420510 | 6 | Propositional logic can be presented as in Mendelson’s book, with the sole inference rule of modus ponens, and with the following three axioms:
$$B \Rightarrow (C \Rightarrow B)$$
$$(B \Rightarrow (C \Rightarrow D)) \Rightarrow ((B \Rightarrow C) \Rightarrow (B \Rightarrow D))$$
$$((\neg C) \Rightarrow (\neg B)) \Rightarrow (((\neg C) \Rightarrow B) \Rightarrow C)$$
This is a sound and complete theory, as are several other theories for propositional logic.
I have questions about similar theories for propositional logic:
1. Given a complete set of logical connectives and a finite set of sound axioms and inference rules, can we algorithmically determine if the resulting theory is complete?
2. Given a finite set of axioms and inference rules $X$ (not necessarily sound) and a formula $\alpha$ in the underlying language of propositional connectives, can we algorithmically determine if $\alpha$ can be derived from $X$?
| https://mathoverflow.net/users/419791 | Decidability of completeness in propositional logic | It is undecidable, because it is even undecidable to recognize whether a finite set of axioms together with the rule of modus ponens axiomatizes exactly classical propositional logic by the Post-Linial theorem. This was shown in 1948 by Linial and Post, see their [announcement (p. 50)](https://www.ams.org/journals/bull/1949-55-01/S0002-9904-1949-09162-0/S0002-9904-1949-09162-0.pdf), but the first published [proof](https://doi.org/10.1305/ndjfl/1093957737) is by Yntema. There are many similar results for other propositional logics.
| 6 | https://mathoverflow.net/users/58913 | 420630 | 171,157 |
https://mathoverflow.net/questions/420608 | 11 | Suppose $d≥0$, $m≥0$, $n≥0$, and $\def\R{{\bf R}} f\colon \R^m→\R^n$ is a smooth map
whose rank at any point of $\R^m$ is at most $d$.
Here and below, smooth means infinitely differentiable.
Can we find an open cover $\{U\_i\}\_{i∈I}$ of $\R^m$
such that for any $i∈I$ the restriction of $f$ to $U\_i$
is a smooth map $U\_i→\R^n$ that factors through $\R^d$
as the composition of smooth maps $U\_i→\R^d→\R^n$?
If the rank of $f$ is constant, this follows immediately from the constant rank theorem,
so the interesting case is when the rank is not constant.
The question is nontrivial only when $0<d<\min(m,n)$
and the simplest nontrivial case appears to be $d=1$, $m=n=2$.
| https://mathoverflow.net/users/402 | Does every smooth map of rank at most d factor through a d-manifold? | There is a counterexample with $d=1$ and $m=n=2$. Here is one way to construct such an example: Let $g:\mathbb{R}\to\mathbb{R}$ be a smooth function such that $g'(t)>0$ for $t\not=0$ and $g^{(k)}(0) = 0$ for $k\ge 0$. Now, define a smooth mapping $f:\mathbb{R}^2\to\mathbb{R}^2$ by the rules
$$
f(x,y) = \begin{cases}\bigl(0,yg(y^2{-}x^2)\bigr) & y^2>x^2 \\ \bigl(0,0\bigr) & y^2 = x^2\\ \bigl(xg(x^2{-}y^2),0\bigr) & x^2>y^2\end{cases}
$$
Note that the differential of $f$ vanishes identically on the locus $x^2=y^2$, but it has rank $1$ everywhere else. The image of $f$ lies in the union of the $x$ and $y$ axes. In fact, note that the mapping $(x,0)\mapsto f(x,0) = (xg(x^2),0)$ is injective, with nonvanishing derivative except at $x=0$, with a similar statement for $(0,y)\mapsto f(0,y) = (0,yg(y^2))$.
I claim that the point $(0,0)$ does not have an open neighborhood $U$ on which there exists a smooth map $\pi:U\to\mathbb{R}$ and a smooth map $\iota:\pi(U)\to\mathbb{R}^2$ such that $f(x,y) = \iota\bigl(\pi(x,y)\bigr)$ for all $(x,y)\in U$. To see this, assume that $\pi$ and $\iota$ exist with the specified properties and note that we can assume without loss of generality that $\pi(0,0) = 0$ and that $U$ is the open disk $x^2+y^2 < r^2$ for some $r>0$ and $\pi(U) = (-s,s)$ for some $s>0$. Because of the above noted behavior of $f$ restricted to the $x$ or $y$ axis, it follows that $\iota$ must be a injection that is an immersion except at $0\in\pi(U)$. From this it is easy to see that $\iota(\pi(U))$ for $r$ sufficiently small must be contained in both the $x$-axis and the $y$-axis, which is clearly impossible.
| 10 | https://mathoverflow.net/users/13972 | 420646 | 171,161 |
https://mathoverflow.net/questions/420641 | 5 | Let $X$ be a compact Kähler manifold with Kähler form $\omega$, then from Kodaira & Spencer's paper on deformations [III](https://www.jstor.org/stable/1969879?casa_token=Lm4RJWHwjh4AAAAA%3Ahj6Mji5mA0uzR5pq4cNuPWCRZOUp24xKCZ50qXtMxR7_N3CaUW0ltkqwtMlvuoSb3oRx_D8wgolZsQPjd0UfEmFPzNk8jTXurtrtgmoObNCQLU-0MFokMg&seq=1), p.75, the authors state that the Kähler form satisfies the Laplace equation $\Delta\omega=0$.
As we know, $\Delta\omega=0\Leftrightarrow d\omega=0,d^\*\omega=0$. By definition, every Kähler form satisfies $d\omega=0$, but it is not obvious why $d^\*\omega$ should also be 0. Actually, from Kodaira & Morrow's book *Complex Manifolds* p.115, they have a proof of $\Delta\omega=0$ by showing $\bar\partial\omega=0$ and $\bar\partial^\*\omega=0$, then it's a result of the famous Kähler identities $\Delta=2\Delta\_{\bar\partial}$, but their proof seems too complicated as it involves covariant differentiation. Does anybody have a simple reason why $d^\*\omega$ should be $0$, i.e. $\omega\in \ker\Delta$?
| https://mathoverflow.net/users/99826 | Does the Kähler form $\omega$ satisfy $d^*\omega=0$? | Recall that $d^\* = -\ast d\ast$ and $\ast\omega = \frac{1}{(n-1)!}\omega^{n-1}$, see Example 1.2.32 of Huybrechts' *Complex Geometry: An Introduction* for example. Therefore
$$d^\*\omega = -\ast d\ast\omega = -\ast d\left(\frac{1}{(n-1)!}\omega^{n-1}\right) = 0$$
as $\omega$ is closed.
| 13 | https://mathoverflow.net/users/21564 | 420658 | 171,165 |
https://mathoverflow.net/questions/420656 | 3 | A subset $A\subseteq \mathbb{R}$ is said to be a $Q$-set if every subset $B\subseteq A$ is $F\_\sigma$ wrt the subspace topology on $A$. For example $\mathbb{Q}$ is a $Q$-set. The first time I have seen this definition is in:
*Balogh, Zoltán*, [**There is a $Q$-set space in ZFC**](http://dx.doi.org/10.2307/2048543), Proc. Am. Math. Soc. 113, No. 2, 557-561 (1991).
Now my questions are:
1. Can we prove in $\mathtt{ZF}+\mathtt{DC}$, where $\mathtt{DC}$ stands for dependent choice, that every co-analytic $Q$-set is countable?
2. Is the descriptive complexity (with connections with set-theoretic hypothesis) of these sets been studied in any paper/book/thesis?
Thanks!
| https://mathoverflow.net/users/141146 | Co-analytic $Q$-sets | I do not know about full references but if you haven't seen it yet there is a nice discussion of this topic in Miller's [https://people.math.wisc.edu/~miller/res/dstfor.pdf](https://people.math.wisc.edu/%7Emiller/res/dstfor.pdf). (Incidentally this may be the most fun math book to read). See in particular sections 2-5.
In particular, Theorem 5.1 states that Martin's Axiom implies every second countable Hausdorff space of cardinality less than the continuum is a Q-set. If $V=L$ there are $\Pi^1\_1$ sets $A \subseteq \mathbb R$ of size $\aleph\_1$ so that in any $\aleph\_1$-preserving forcing remain uncountable and no new elements are added. Forcing martin's Axiom over $L$ with ccc forcing (so the standard way) will therefore give a consistent example of an $\aleph\_1$-sized co-analytic Q-set.
Also Corollary 3.2 can be used to answer your first question. This corollary states that any second countable space which contains a perfect set is not Q (actually it uses a weaker hypothesis to make a stronger conclusion). Staring at the proof of the relevant results towards proving this corollary you need to convince yourself that DC is enough. Given this, if every set $A \subseteq \mathbb R$ contains a perfect set (for instance in the Solovay model) then no uncountable set of reals can be a Q-set.
Putting these two results together also we get (modulo an inaccessible) that the existence of a projective uncountable Q-set is independent of ZFC.
I do not know if anyone has thought about these definability issues further, though I would check the references of the Miller book to start.
| 7 | https://mathoverflow.net/users/114946 | 420660 | 171,166 |
https://mathoverflow.net/questions/420663 | 13 | Historically, the theory of ultracategories was invented [by Makkai](https://www.sciencedirect.com/science/article/pii/000187088790020X) to prove a strong conceptual completeness theorem for first-order logic, roughly: if $T$ and $S$ are two first-order theories such that the ultracategory of models of $T$ is equivalent to the ultracategory of models of $S$, then the theories $T$ and $S$ are "essentially the same" (their pretopos completions are equivalent).
There is some more recent material about ultracategories by Lurie, see [here](https://www.math.ias.edu/~lurie/278x.html) and [here](https://www.math.ias.edu/~lurie/papers/Conceptual.pdf), which uses a simplified definition of the notion of an ultracategory (which is, for the purposes of proving a strong conceptual completeness theorem, equivalent to Makkai's original definition). The [definition](https://www.math.ias.edu/~lurie/278xnotes/Lecture25X-Ultracategories.pdf) (see Definition 7) is still technical, though. So it is natural to ask if one can formulate this notion in more familiar terms when considering special cases of the notion of a category. Indeed, [an ultraset turns out to be the same as a compact Hausdorff space](https://www.math.ias.edu/~lurie/278xnotes/Lecture26X-Ultrasets.pdf) (Theorem 9).
I wonder if the notion of an ultracategory simplifies to something more familiar if we restrict to categories which are not a set, but a *monoid* or a *preordered set*:
* What is an ultracategory with exactly one object? (Call these things *ultramonoids*.)
* What is an ultracategory with at most one morphism between any pair of objects?
Coming back to the original motivation of ultracategories, an ultracategory usually consists of [structures in the sense of model theory](https://en.wikipedia.org/wiki/Structure_(mathematical_logic)). Let $\mathbf A$ be a structure. Can we equip, say, the endomorphism monoid $\mathrm{End}(\mathbf A)$ of $\mathbf A$ or the monoid $\mathrm{End}'(\mathbf A)$ of elementary embeddings $\mathbf A\hookrightarrow\mathbf A$ with the structure of an ultramonoid?
More generally, if $\mathcal M$ is any ultracategory and $M\in\mathcal M$ an object, is $\mathrm{End}(M)$ an ultramonoid? If $F\colon\mathcal M\to\mathcal N$ is an ultrafunctor, is for each $M\in\mathcal M$ the induced map $\mathrm{End}(M)\to \mathrm{End}(F(M))$ an ultrafunctor between ultramonoids?
| https://mathoverflow.net/users/480841 | Ultracategories with one object | $\newcommand{\cat}{\mathrm}
\newcommand{\St}{\cat{Stone}^\cat{fr}}
\newcommand{\Cat}{\cat{Cat}}
\newcommand{\Cart}{\cat{Cart}}
\newcommand{\Fun}{\cat{Fun}}
\newcommand{\Mon}{\cat{Mon}}
\newcommand{\Set}{\cat{Set}}
\newcommand{\Po}{\cat{Poset}}$
EDIT : I misread the definition of ultracategory fibration, apparently only *certain* locally cartesian edges are closed under composition. After a thought about it, what I said for monoids remains true, but not for posets. I just need to modify the arguments.
I would argue that a monoid is not the same as a category with one object, namely a monoid is the same as a *pointed* category with one object. This is not really relevant if you look at the category of categories as a $1$-category (because then $\hom\_\Cat(BM, BM')$ is indeed isomorphic to $\hom\_\Mon(M,M')$), but it does if you more naturally view it as a $(2,1)$-category ($\Fun(BM,BM')$ has nontrivial morphisms !)
In particular, I'll interpret your question as :
>
> What is an ultracategory $\mathcal M$ with a morphism from the terminal ultracategory "$\*$" such that for all $I$, $\*\_{\beta I}\to \mathcal M\_{\beta I}$ is essentially surjective ?
>
>
>
Of course, because $\*\_{\beta I}\simeq \prod\_I\*$ in a way compatible with $\mathcal M\_{\beta I}\simeq \prod\_I\mathcal M\_\*$, this is equivalent to $\*\to \mathcal M\_\*$ being essentially surjective.
The first question is : What is a pointed ultracategory ?
In my original post, I made a mistake because I had misread the definition of ultrcategory. In particular, a morphism $\St\to \mathcal M$ is more than just a point in $\mathcal M\_\*$ : it's a point with a certain property. Namely, (see definition 1 in Lurie's notes) we ask that the comparison morphisms for base-change functors associated to *all* compositions $\beta I\to \beta J\to \beta K$ be isomorphisms when applied to this point (while in the definition of an ultracategory, one only requires this if $\beta I \to \beta J$ comes from a map $I\to J$).
But, in particular, if this point is the only point of $\mathcal M\_\*$ (and thus of $\mathcal M\_{\beta I}$ for all $I$), this amounts to asking that cartesian morphisms actually be closed under composition ! In other words,
>
> An ultramonoid is an ultracategory which is a *cartesian* fibration $\mathcal M\to \St$ with a section $\St\to \mathcal M$ sending cartesian edges to cartesian edges, such that on fibers over $\*$, the map $\*\to \mathcal M\_\*$ is essentially surjective.
>
>
>
Now that this is clarified, the rest of my answer essentially goes through without changes. But note that there was initially a mistake, and also note that this will not work for posets !
The second thing to observe is that there is a functor $\Mon\to\Cat\_\*$ which implements the equivalence of $(2,1)$-categories I've mentioned above.
This functor is fully faithful (in fact it has a right adjoint - given by $(C,x)\mapsto \cat{End}\_C(x)$, this will be relevant later - and the unit map $M\to \cat{End}\_{BM}(\bullet)$ is an isomorphism), and therefore the functor $\Fun((\St)^\cat{op}, \Mon)\to \Fun((\St)^\cat{op},\Cat)$ is also fully faithful.
By what was clarified above, ultramonoids can therefore be viewed as a full subcategory of $\Fun((\St)^\cat{op},\Cat\_\*)$, via the Grothendieck construction (while this is not the case for general ultracategories ! for them it would be certain pseudo-functors). Precisely, it is the full subcategory spanned by functors $(\St)^\cat{op}\to\Cat\_\*$ that send the diagrams $(\{i\}\to \beta I)\_{i\in I}$ to product diagrams. Piecing things together, using the fact that $\Mon\to\Cat\_\*$ and $\Cat\_\*\to\Cat$ both reflect and preserve products, we find :
>
> The category of ultramonoids is equivalent to the full subcategory of $\Fun((\St)^\cat{op},\Mon)$ spanned by those functors that send the diagrams $(\{i\}\to \beta I)\_{i\in I}$ to product diagrams.
>
>
>
But we can be more concrete. Indeed, $\Fun((\St)^\cat{op},\Mon)\simeq \Mon(\Fun((\St)^\cat{op},\Set))$, and $\Mon \to \Set$ preserves and reflects products, so that :
>
> The category of ultramonoids is equivalent to the category of monoids in the category of ultrasets.
>
>
>
(note that for ultrasets the subtlety disappears: the fibrations must be cartesian, and not only locally cartesian)
Ultrasets are compact Hausdorff topological spaces, so finally:
**Corollary** : The category of ultramonoids is equivalent to the category of monoids in compact Hausdorff topological spaces.
Now, if $\mathcal M$ is an ultracategory with a point $x : \*\to \mathcal M\_\*$, there is, in general, no way to make an ultramonoid out of it because of this cartesian vs locally cartesian business. However, if it is a "nice" point, i.e. if it is a point for which the comparison morphisms for general compositions $\beta I\to \beta J\to \beta K$ are isomorphisms, then we can view it as a morphism $\St\to \mathcal M$ of ultracategory fibrations. In particular we can take the full subcategory $\mathcal M\_x$ of $\mathcal M$ spanned by the image of $\St$, and this should still be an ultracategory fibration. But now this is a pointed ultracategory fibration with an essentially surjective point, so it is an ultramonoid, as explained above.
We therefore find tha a sufficiently nice object $x$ (in more details: a point $\St\to \mathcal M$) in an ultracategory $\mathcal M$ has an ultramonoid of endomorphisms, $\cat{End}\_\mathcal M(x)$, i.e. a compact Hausdorff topological space of endomorphisms.
(For posets, I don't know a satisfying answer. I'm not sure you will get a more satisfying answer than to the question "what is an ultracategory ?", but I would be interested in seeing one)
EDIT to answer the questions in a comment below (rather than as a long sequence of further comments):
1- I used "pointed category" to mean something more naive than in the nLab, namely a category $C$ with a point $\*\to C$. Morphisms of such are functors $f : C\to D$ that preserve the point, and $2$-morphisms are natural transformations that are the identity on the point (this is an unfortunate clash of terminology: one can define a "pointed object" in an arbitrary category, and "pointed objects of $Cat$" do not coincide with the nLab's pointed categories). With this definition, the (usual) $1$-category of monoids is equivalent to the $(2,1)$-category of pointed categories for which the point $\*\to C$ is essentially surjective.
2- "which question": I was adressing the question about ultramonoids (at first I thought also posets, but in the end, no). I don't know if I'm using the same definition of ultramonoid as you are, because it's unclear to me what "an ultracategory with one object" means in the same way that it's unclear to me what "a category with one object" means.
If you mean in the strictest sense and using Definition 7, then no I am not using the same definition (a choice which my first paragraph tries to explain). There are 2 reasons I did not use the same definition: a- what I explained in the first paragraph of this (partial) answer and b- the strict definition is not invariant under the several presentations of "ultracategories" and not invariant under equivalence of ultracategories. I know that this means I'm not technically answering your question (in any case, I wasn't answering all of it, this was just a contribution ! I'm sorry if I made it sound like it was supposed to be everythin). But maybe in the situation you're interested in, the definition I used might be more relevant ?
3- The notation $\*\_{\beta I}\to \mathcal M\_{\beta I}$ refers to $\*^I\to \mathcal M^I$ in the language of definition 7, specifically the map that picks out this one object. Sorry again, Definition 1 was clearer to me, which is why I phrased my answer in this language (note that the two definitions are equivalent in a suitable sense).
4- I define the $(2,1)$-category of ultramonoids as a full subcategory of the category of pointed ultracategories (in the sense I described earlier). In particular, morphism *categories* (rather than morphism sets) are categories of ultrafunctors and, I guess but am not sure about the terminology, ultra natural transformations.
5- A "point" of a category is an object therein, equivalently a functor $\*\to C$. I used this latter perspective (which may seem unnecessarily pedantic) to define a point of an ultracategory, as an ultrafunctor from the trivial ultracategory $\*$ to $\mathcal M$. That's the content of the paragraph following "The first question is..." - I was trying to identify those more concretely.
6- I'm not sure what the nice objects are in the case of "models of a first order theory". Essentially these are the models $M$ such that for any two sets $J,K$, function $f: J\to \beta K$ and ultrafilter $\mathcal U$ on $J$, the canonical morphism $(\prod\_K M)/\mathcal V \to \prod\_J ((\prod\_K M)/\mathcal f(j))/\mathcal U$ is an isomorphism, where $\mathcal V = \lim\_\mathcal U f(j)$ in $\beta K$, equivalently $\mathcal U\wr \mathcal V\_\bullet$ where $\mathcal V\_\bullet = f$.
It looks like this map is always injective by definition of $\mathcal U\wr \mathcal V\_\bullet$, but most likely almost never surjective. For instance take $K =\*$, then this is $\mathcal M\to (\prod\_J \mathcal M)/\mathcal U$, which is almost never surjective.
7- I'm sorry, I misunderstood that these were your main questions. So, for the question of which structures, this amounts to answering 6-. The answer seems to be "almost never" - my guess would be "if and only if $M=\*$". It makes sense to a certain extent : what would be the compact Hausdorff topology on the set of elementary embeddings/morphisms ?
For the second question however, the way everything is set up makes it very easy to answer : a good point in $\mathcal M$ is an ultrafunctor $\*\to \mathcal M$. This is clearly stable under composition of ultrafunctors : if $F:\mathcal{M\to N}$ is an ultrafunctor, so is $\*\to \mathcal M\to\mathcal N$, and in particular, $F$ preserves "nice" objects. A corollary of how everything is set up is also (for free) that you get a morphism of ultramonoids (in the sense I described, so a morphism of compact Hausdorff topological monoids) $End(M)\to End(F(M))$.
8- By strongly connected I meant "only one isomorphism class". I'm not sure there is a standard terminology and what it is, if there is.
| 11 | https://mathoverflow.net/users/102343 | 420671 | 171,168 |
https://mathoverflow.net/questions/420657 | 7 | Let $G$ be a finite group and let $M$ be a representation of $G$ over a field $k$. Suppose that, for every cyclic subgroup $C$ of $G$, we have $M|\_{C} \cong k[C]^{\oplus [G:C]}$. Can we conclude that $M \cong k[G]$?
In characteristic zero, this is immediately true by character theory, but I don't know what happens in finite characteristic.
**Motivation:** I was thinking about the normal basis theorem in Galois theory, and I remembered that many sources do the cyclic case separately. Suppose that $M/k$ is a Galois extension with Galois group $G$. Then, for each cyclic subgroup $C$, we know that $M/\text{Fix}(C)$ is a Galois extension with Galois group $C$ so, if we knew the normal basis theorem for cyclic extensions, we would know that $M \cong \text{Fix}(C)[C] \cong k[C]^{\oplus [G:C]}$ as a $k[C]$-module. I was wondering whether this might already be enough to finish the proof with no more field theory input.
| https://mathoverflow.net/users/297 | Is a $k[G]$-module which is free on every cyclic subgroup free? | Suppose that $E = C\_p \times \dots \times C\_p$ be an elementary abelian $p$-group, say of rank $r$, and let $k$ be a field of characteristic $p$. Then the group algebra $kE$ is isomorphic to $k[x\_1, \dots, x\_r]/(x\_1^p, \dots, x\_r^p)$. If $v$ is in the vector space spanned by the $x\_i$, then $(1+v)^p=1$, so $1+v$ generates a subgroup of the group of units of $kE$. This is called a *cyclic shifted subgroup* of $kE$.
A result of Dade originally, then independently later proved by Carlson and Avrunin-Scott, says the following: A $kE$-module $M$ is free if and only if the restriction of $M$ to each cyclic shifted subgroup is free.
(Dade: Lemma 11.8 in Endo-Permutation Modules over p-Groups II, <https://www.jstor.org/stable/1971169>. Carlson: Theorem 4.4 in The varieties and the cohomology ring of a module, <https://www.sciencedirect.com/science/article/pii/0021869383901217.>)
I believe that there are examples of modules which restrict to free modules over every honest cyclic subgroup but not over every cyclic shifted subgroup, but I don't have one at my fingertips. There should be an example with $G = C\_2 \times C\_2$, I think.
| 10 | https://mathoverflow.net/users/4194 | 420672 | 171,169 |
https://mathoverflow.net/questions/420678 | 8 | Let $\Lambda \subset\mathbb{R}^n$ be a lattice satisfying $\|x-y\|\_2^2 \in \mathbb{Z}$ for all $x,y\in\Lambda$. How small can $\text{vol}(\Lambda)=\det(\Lambda)$ be?
For example, in dimension $2$, the hexagonal lattice has smallest volume of any integral lattice, equal to $\frac{\sqrt{3}}{2}$. In higher dimensions, rescaling an even unimodular lattice by a factor of $\sqrt{2}$ yields a lattice with integral distances with $\text{vol}(\Lambda) = 2^{-\frac{n}{2}}$.
What is the smallest volume possible as $n\rightarrow\infty$?
| https://mathoverflow.net/users/12176 | The smallest volume possible for a lattice with integer distances? | After scaling your lattice by $\sqrt{2}$, the Gram matrix has integral entries, so the absolute value of its determinant, being a nonzero integer, is at least $1$. This gives a lower bound of $2^{-n/2}$.
| 10 | https://mathoverflow.net/users/40821 | 420682 | 171,172 |
https://mathoverflow.net/questions/420679 | 4 | Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be a bounded analytic function such that its derivative is also bounded. What kind of bound can we get on the higher order derivatives of $f$? Does it follow that they are bounded as well?
| https://mathoverflow.net/users/78173 | Bounded real analytic function with bounded derivative and its higher order derivatives | No, $f(x)=\int\_0^x \sin t^2\, dt$ is a counterexample.
| 10 | https://mathoverflow.net/users/48839 | 420686 | 171,175 |
https://mathoverflow.net/questions/420706 | 0 | Let $\left( S\_{n}^{1}\right) $ and $\left( S\_{n}^{2}\right) $ two sequences
of operators in $\mathcal{L}(E\_{1},F\_{1})$ and $\mathcal{L}(E\_{2},F\_{2})$
where $E\_{i},F\_{i},i=1,2$ are Hilbert spaces such that $\left\Vert
S\_{n}^{1}x-S^{1}x\right\Vert \_{F\_{1}}\underset{n\rightarrow \infty }{%
\rightarrow }0,$ $\forall x\in E\_{1}$ for some $S^{1}\in \mathcal{L}%
(E\_{1},F\_{1})$ and such that
$$
\ker S\_{n}^{1}=\{0\}\Leftrightarrow \ker S\_{n}^{2}=\{0\},\text{ }\forall
n\geq 0.
$$
Assume that $\ker S^{1}=\{0\}.$ Does this implies that $\ker S^{2}=\{0\}?.$
Thank you in advance.
| https://mathoverflow.net/users/106804 | Kernels of sequences of operators | No, even if $\|S^2\_n x-S^2x\|\_{F\_2}\to0$ for some $S^2\in\mathcal L(E\_2,F\_2)$ and all $x\in E\_2$.
Indeed, let e.g. $E\_1=F\_1=E\_2=F\_2=H:=\ell^2$, $S^1\_n=I$, $S^1=I$, $S^2=0$, and $S^2\_n x=(e^{-|k-2n|}x\_k)\_{k=1}^\infty$ for $x=(x\_k)\_{k=1}^\infty\in\ell^2$. Then all the conditions hold. In particular,
$$\|S^2\_n x\|\_H^2\le e^{-n}\sum\_{k=1}^n|x\_k|^2+\sum\_{k=n+1}^\infty|x\_k|^2\to0$$
as $n\to\infty$.
However, the conclusion $\text{ker}S^2=\{0\}$ does not hold.
| 2 | https://mathoverflow.net/users/36721 | 420707 | 171,181 |
https://mathoverflow.net/questions/308196 | 4 | Consider the following Diophantine equation $$x^2+x+1=(a^2+a+1)(b^2+b+1)(c^2+c+1).$$ Assume also that $x,a,b,c,a^2+a+1,b^2+b+1, c^2+c+1$ are all primes. We'll call such a quadruplet $(x,a,b,c)$ a *triple threat*.
I'd like to be able to show that there are no triple threats. I've been able to prove the following weaker result that if $(x,a,b,c)$ is a triple-threat and $x \equiv 1$ (mod 5), then none of $a,b$ or $c$ can be $1$ (mod 5). For my purposes, this is useful, but I can get a stronger result with a simpler proof if I can get that there are no triple threats.
Let me briefly sketch the idea of the proof: Most of what I have been able to do has been from looking at the more general equation
$$x^2+x+1=(a^2+a+1)(b^2+b+1)p$$ where $x,a,b,p,a^2+a+1$ and $b^2+b+1$ are all prime. Note that this equation does solutions (and probably has infinitely many). Call this an almost-triple threat.
The key approach is showing the following sort of result: If $a \leq b$, and we have some specific modulo restriction on $x,a,b,p$ then one must have $a^2+a+1<p$. If one then applies to the original triple threat equation one can in some circumstances get a contradiction by forcing that $a^2+1<c^2+c+1<b^2+b+1<a^2+a+1$.
Let me briefly sketch how in most contexts I've obtained that $a^2+a+1<p$.
My primary method of attack on this equation has been to rewrite it as
$$(x-a)(x+a+1)=((b^2+b+1)p-1)(a^2+a+1) $$ and
$$(x-b)(x+b+1)=((a^2+a+1)p-1)(b^2+b+1). $$
Since $a^2+a+1$ and $b^2+b+1$ are assumed to be prime, one has that either the pair of relations $a^2+a+1|x-a$ and $x+a+1|(b^2+b+1)p-1$, or the pair of relations $a^2+a+1|x+a+1$ and $x-a|(b^2+b+1)p-1$. One gets similar pairings using the second version above with $b^2+b+1$. One has then four cases:
I) $a^2+a+1|x-a$ and $b^2+b+1|x-b$
II) $a^2+a+1|x-a$ and $b^2+b+1|x+b+1$
III) $a^2+a+1|x+a+1$ and $b^2+b+1|x-b $.
IV) $a^2+a+1|x+a+1$ and $b^2+b+1|x+b+1$.
It is not too hard to show in Cases II and III that one always has $a^2+a+1<p$. The difficulty is primarily in Cases I and IV. In Case I one gets with a little algebra that $x+a+1|p(a+b+1) + \frac{x-b}{b^2+b+1}$. One then has $k(x+a+1)= p(a+b+1) + \frac{x-b}{b^2+b+1}$. For many choices of $x,a,b$ mod 5 one can force $k$ to be large. Say $k \geq 3$. Then one has $$a+x+1 \leq \frac{ p(a+b+1) + \frac{x-b}{b^2+b+1}}{3},$$ and one can get that $a^2+a+1<p$ from this inequality and a little work. (I can expand on this part if necessary.)
The main difficulty right now is dealing with Case IV. I can handle Case IV under some small modulus assumptions, but I cannot prove that $a^2+a+1<p$ in Case IV unconditionally.
In Case IV one has with a little work that
$x-a|p(a+b+1)-\frac{x+b+1}{b^2+b+1}$ and $x-b|p(a+b+1)-\frac{x+a+1}{a^2+a+1}$ but it seems tough to use these to get a tight enough bound to conclude what is wanted without having some extra modulus restriction on the variables. Any thoughts on either clearing out Case IV, or any other method of showing that triple threats don't exist via some other method would be appreciated.
| https://mathoverflow.net/users/127690 | A specific Diophantine equation restricted to prime values of variables. | Pace Nielsen and Cody Hansen just [put this preprint on the Arxiv](https://arxiv.org/abs/2204.08971) which shows that no triple threats exist.
| 3 | https://mathoverflow.net/users/127690 | 420712 | 171,183 |
https://mathoverflow.net/questions/420711 | 2 | I posted this on MathStackExchange, but it hasn't even got 10 views, so probably it is better to post here. I hope it is not inappropriate.
I am reading a paper of Brezis and Oswald about existence and uniqueness of positive solutions to sublinear elliptic equations:
\begin{equation}
- \Delta u = f(x, u), \ u \geq 0, \ u \not\equiv 0 \text{ in } \Omega, \qquad u = 0 \text{ on } \partial \Omega.
\end{equation}
As usual, $\Omega$ is a bounded smooth domain of $\mathbb R^N$ and there are certain hypotheses on $f$, but I think they are not important for my question.
They prove in the paper that any solution of the problem (given the suitable hypotheses on $f$) satisfy
$$
(\*) \qquad u > 0 \text{ in } \Omega, \qquad \frac{\partial u}{\partial \nu} < 0 \text{ on } \partial \Omega,
$$
where $\nu$ is the outer unit normal. Then they claim that from $(\*)$ it follows that if $u\_1, u\_2$ are two solutions of the problem, then $u\_1/u\_2 \in L^\infty(\Omega)$. In the book "Qualitative Analysis of Nonlinear Elliptic Partial Differential Equations" Radulescu also presents this result and explains that it follows by L'Hospital's rule, but I also don't understand this.
>
> I) Why does $u\_1/u\_2 \in L^\infty(\Omega)$?
>
>
>
>
> II) Does I) still hold if $u$ is a solution of the problem
> $$
> - \Delta u = f(x, u) \text{ in } \Omega, \qquad u = 0 \text{ on } \Gamma, \qquad \frac{\partial u}{\partial \nu} = 0 \text{ on } \Gamma\_1,
> $$
> where $\partial \Omega = \Gamma \cup \Gamma\_1$, both parts being smooth? It is known that $(\*)$ still holds in this case with $\Gamma$ in the place of $\partial \Omega$.
>
>
>
The paper: [https://sites.math.rutgers.edu/~brezis/PUBlications/110-journal.pdf](https://sites.math.rutgers.edu/%7Ebrezis/PUBlications/110-journal.pdf)
| https://mathoverflow.net/users/113406 | Quotient of solutions of a semilinear Dirichlet problem is $L^\infty$ | The fact that both $u\_j$ solve the PDE is not an issue. What matters is that both $u\_j$ are smooth over the closure $\overline\Omega$, positive in $\Omega$, vanish on $\partial\Omega$ and their normal derivatives don't vanish. This is where you may invoque L'Hopital's rule, if you like this tool.
What is more important is that we can with little effort derive an estimate of the ratio $w=\frac{u\_2}{u\_1}$.
The paper assumes the important property that $u\mapsto f(x,u)$ is decreasing, which I shall use together with the maximum principle. This function satisfies the PDE
$$-\Delta w-2\nabla\log u\_1\cdot\nabla w=\frac1{u\_1}(f(x,wu\_1)-wf(x,u\_1)).$$
If $w(\bar x)>1$ for some $\bar x\in\Omega$, then the assumption on $f$ ensures that $f(x,wu\_1)-wf(x,u\_1)\le0$ and thus $-\Delta w-2\nabla\log u\_1\cdot\nabla w\le0$. By the maximum principle, $w$ cannot achieve a local maximum at $\bar x$. This shows that either $w\le1$ everywhere, or it achieves its maximum at the boundary. In the latter case, $w$ coincides (L'Hopital) with the ratio of normal derivatives. In conclusion, we have
$$\max\_\Omega\frac{u\_2}{u\_1}\le\max\left\{1,\max\_{\partial\Omega}\frac{\partial u\_2/\partial\nu}{\partial u\_1/\partial\nu}\right\}.$$
| 3 | https://mathoverflow.net/users/8799 | 420720 | 171,184 |
https://mathoverflow.net/questions/420717 | 2 | It is known that every Lipschitz function $f \colon [-1,1] \to \mathbb R$ can be expressed as a series in the Chebyshev polynomials $$f = \sum\_{n = 0}^\infty a\_n T\_n $$ which is absolutely convergent under the $\lVert \, \cdot \, \rVert\_\infty$ norm (see Approximation Theory and Approximation Practice by Lloyd Trefethen, 2018, p.19-20). I'm wondering whether the series converges under the Lipschitz norm, i.e. $$ \Big\lVert \sum\_{n = N}^\infty a\_n T\_n \Big\rVert\_\text{Lip} \xrightarrow{N \to \infty} 0 $$
(and if so, is convergence unconditional? absolute?) where the Lipschitz norm is defined by $$\lVert f \rVert\_\text{Lip} := |f(-1)| + \sup\_{x \neq y}\bigg|\frac{f(y)-f(x)}{y-x} \bigg|$$
I'm interested in this since $(\text{Lip}[-1,1], \lVert \, \cdot \, \rVert\_\text{Lip})$ is a Banach space, while $(\text{Lip}[-1,1], \lVert \, \cdot \, \rVert\_\infty)$ is not. Thank you.
| https://mathoverflow.net/users/47757 | Are Chebyshev polynomials a Schauder basis of $\mathrm{Lip}[-1,1]$? | No, the space of Lipschitz functions on an infinite metric space is non-separable so it can't have a Schauder basis.
| 6 | https://mathoverflow.net/users/15129 | 420721 | 171,185 |
https://mathoverflow.net/questions/420585 | 1 | Let $X=(X\_n)\_{n\in\mathbb N}$ be a stochastic process in $\{0,1\}^{\mathbb N}$ with distribution $\mu$. I do not at first make any assumptions about $X$ being stationary or having any kind of correlations or associations. Let $Y=(Y\_n)\_{n\in\mathbb N}$ be an iid Bernoulli process with distribution product measure $\pi\_p$, i.e. $\mathbb P(Y\_n=1)=p$ for all $n$ and independently of all other $Y\_j$.
I want to study the conditions under which $\pi\_p\preceq \mu$, that $X$ stochastically dominates $Y$. The order $X(\omega)\geq Y(\omega)$ uses the natural ordering of their sample paths for each $\omega\in\Omega$ a common probability space upon which they are constructed, for which I just say $X\geq Y$.
From Lemma 1.1 in Liggett et al. (1997), I can conclude that a sufficient condition for $\pi\_p\preceq \mu$ is that
$$
\mathbb P(X\_n=1\mid X\_{s\_j}=\epsilon\_{s\_j}, \text{ for all } j=1,2,\ldots,m)\geq p
\tag{1}\label{eq1}
$$
for any $n$ and any $\{s\_1,s\_2,\ldots,s\_m\}\subset \{1,2,\ldots,n-1\}$ and any $\epsilon=(\epsilon\_n)\_{n\in\mathbb N}$. Of course we must be conditioning on an event of positive probability, so I always assume that.
My question is on why this condition isn't also necessary.
I know that $\pi\_p\preceq \mu$ if and only if there exists a coupling of $X$ and $Y$ such that $\mathbb P(X\geq Y)=1$. (I hope it is clear that I am now talking about coupled "copies of" $X$ and $Y$ with the appropriate marginals.) Such a coupling must also satisfy (\ref{eq1}) yes?
I use $\mathbb P^\epsilon$ to refer to the conditional joint distribution of $(X,Y)$, conditioned on $X\_{s\_j}=\epsilon\_{s\_j}$ for $j=1,2,\ldots,m$:
$$\mathbb P^\epsilon(X\_n=1)=\mathbb P^\epsilon(X\_n=1,Y\_n=0)+\mathbb P^\epsilon(X\_n=1,Y\_n=1)$$
Since the coupling satisfies $\mathbb P(X\geq Y)=1$, we also get $\mathbb P^\epsilon(X\geq Y)=1$ (assuming we condition on an event of positive probability). Hence $\mathbb P^\epsilon(X\_n=1,Y\_n=1)=\mathbb P^\epsilon(Y\_n=1)=p$ (since $\{X\_n=0,Y\_n=1\}$ has probability zero under $\mathbb P$ and $\mathbb P^\epsilon$), therefore we clearly have
$\mathbb P^\epsilon(X\_n=1)\geq p$ as desired.
So I have (\ref{eq1}, for all $n,\epsilon,m$) if and only if $\pi\_p\preceq\mu$.
Another way I look at it is from a simulation point of view. Assuming I have complete access to the distribution of $X$ (simulating $Y$ is easy), I can simulate coupled $(X\_1,Y\_1)$ with $\mathbb P(X\_1=1)\geq p$. Then simulate $X\_2$ conditioned on the realized value of $X\_1$ (which obviously has positive probability for nontrivial $p$), etc. At every step we must have the probability that $X\_n=1$ is at least $p$ (conditioned on an event of positive probability). Of course, maybe the structure of $X$ could possibly prevent this type of simulation? Though as long as this is possible, then (\ref{eq1}), for all $n,m,\epsilon$, should be necessary and sufficient for $\pi\_p\preceq \mu$.
Am I misunderstanding something?
Liggett & Steif (2006) explores the situation when $X$ has certain associations, and there are many other papers exploring special cases for certain types of processes as well.
Primary References:
Liggett et al (1997) <https://www.jstor.org/stable/2959530>
Liggett & Steif (2006) <https://dx.doi.org/10.1016/j.anihpb.2005.04.002>
(and follow any references from those and papers that cite them)
| https://mathoverflow.net/users/68851 | Stochastic process on $\{0,1\}^{\mathbb N}$ domination of product measures, necessary and sufficient conditions | Ok, here is a counterexample with the index set $\{1,2\}$ (you can easily extend it to whole $\mathbb N$ if you wish).
Let $(Y\_1,Y\_2)$ be independent Bernoulli($1/2$) and set
$$
(X\_1,X\_2) = \begin{cases}
(Y\_1,Y\_2), & Y\_1 + Y\_2>0,\\
(0,1), & Y\_1 = Y\_2 = 0.
\end{cases}
$$
In this case $$
\mathrm P(X\_1 = 1\mid X\_2 = 1) = \frac{P(X\_1 = 1, X\_2 = 1)}{P(X\_2 = 1)} = \frac{1/4}{3/4} = \frac13<\frac12.
$$
| 2 | https://mathoverflow.net/users/8146 | 420728 | 171,186 |
https://mathoverflow.net/questions/420732 | 5 | I've copied over this question from [what I asked on Mathematics Stack Exchange](https://math.stackexchange.com/q/4422568/976076), in the hope that some experts here can help me find a way to check the residual finiteness of this group.
Consider the group $G=K\rtimes \mathbb{Z}$ defined as follows:
The subgroup $K$ is generated by elements $x\_i,y\_k$ with $i,k \in {\mathbb Z}$ and $k > 0$, and it has defining relations
\begin{eqnarray\*}
x\_i^2 &=& y\_j^2= 1\ \mbox{for all}\ i,j,\\
[x\_j,x\_i] &=& y\_{j-i}\ \mbox{for}\ j>i,\\
[y\_k,x\_i] &=& 1\ \mbox{for all}\ i,k,
\end{eqnarray\*}
The action of $({\mathbb Z},+)$ on $K$ is defined by the automorphism $1 \in {\mathbb Z}$ maps $x\_i$ to $x\_{i+1}$ for all $i \in {\mathbb Z}$.
**Question**: Is group $G$ residually finite?
**The progress**: My idea is to check if $K$ is residually finite first (because if $K$ is not residually finite, then $G$ can't be). So far, if a word $w$ from $K$ satisfies the following condition, then there is a homomorphism from $K$ to a finite group that doesn't send $w$ to the identity.
* if there exists $x\_i$ in $w$, and the total power of $x\_i$ is odd. (we can map $K$ to some direct product of $\mathbb{Z}\_2$)
* if $w= y\_j$ and $j$ is odd. (We can map $K$ to the Heisenberg group over $\mathbb{Z}\_2$)
I am not sure how to show such homomorphism exists for any general word. (e.g. a string of $y\_i$'s).
| https://mathoverflow.net/users/479955 | Is this semi-direct product residually finite? | Yes, it's residually finite.
This group maps onto the residually finite wreath product $C\_2\wr\mathbf{Z}$ and the kernel is central, free over the $y\_k$, $k\ge 0$, as 2-elementary abelian group.
So one needs to show that for every non-empty subset $J$ of the set $J$ of positive integers, the element $y\_J=\prod\_{j\in J}y\_j$ survives in some finite quotient.
Let $t$ be the generator of $\mathbf{Z}$ and kill $t^n$. This kills $y\_n$ and identifies $y\_i$ to $y\_{i+n}$ for all $n$ (and identifies $x\_i$ to $x\_{i+n}$. At the level of the quotient, the central kernel of the homomorphism onto $C\_2\wr C\_n$ is freely generated by the $y\_i$, $0<i<n$, as $2$-elementary abelian group. In particular, $y\_J$ is not trivial in this quotient.
| 6 | https://mathoverflow.net/users/14094 | 420735 | 171,187 |
https://mathoverflow.net/questions/420695 | 6 | Let $M$ be a simply connected, (orientable), non-compact, 3-manifold without boundary. Must $M$ be homeomorphic with a topological subspace of $\mathbb{R}^3$?
| https://mathoverflow.net/users/69681 | Does every simply connected, orientable, non-compact, 3-manifold embed in $\mathbb{R}^3$? | When the manifold is the universal cover of a compact $3$-manifold $M$ (to begin with, lets say without boundary) then you construct the embedding by hands, using geometrization. In your question let's replace $\Bbb R^3$ with $S^3$ and we will see the embedding into $\Bbb R^3$ comes for free in the situation you are interested in.
An instructive case comes from the case where your manifold is a connect-sum of lens spaces. This has been the subject of a few recent threads:
[Universal covers of non-prime 3-manifolds](https://mathoverflow.net/questions/419294/universal-covers-of-non-prime-3-manifolds/)
The basic idea is that you choose a collection of reducing spheres for the connect sum decomposition, call them $\Sigma$. Then $M \setminus \Sigma$ is a disjoint union of punctured lens spaces. Each of these have universal covers diffeomorphic to punctured spheres, so they embed in $S^3$. You then glue the embedded punctured spheres together (in $S^3$) so that the appropriate boundary spheres are glued together. If you view the embedded punctured spheres from the perspective of their complements in $S^3$ we are essentially doing the canonical construction of a Cantor set. There is the exceptional case of $\Bbb RP^3 \# \Bbb RP^3$ where we are constucting the standard embedding $S^0 \to S^3$.
But this is the basic idea. The remaining geometric 3-manifolds have universal covers that are also subsets of $S^3$, so you similarly glue these together along the (lifted) torus decomposition or sphere decompositions. The tori (being incompressible) will lift to copies of $\Bbb R^2$, so those gluings are a little easier to visualize.
| 4 | https://mathoverflow.net/users/1465 | 420738 | 171,188 |
https://mathoverflow.net/questions/420684 | 11 | Recall the notation $(z;q)\_n=(1-z)(1-zq)(1-zq^2)\cdots(1-zq^{n-1})$.
My earlier [MO question](https://mathoverflow.net/questions/420603/congruence-modulo-2-for-q-series) did not find enough interest or yield an answer. Perhaps the modulo $2$ part might have thrown people off. So, I now can write another question which, if proved, would certainly capture the modular problem immediately.
So, I would like to propose:
>
> **QUESTION.** Are these two $q$-series equal to each other?
> $$\sum\_{n\geq1}\frac{q^{\binom{n+1}2}}{(q;q)\_n}\sum\_{m=1}^n\frac{q^m}{1-q^m}
> =\sum\_{n\geq1}\frac{q^{\frac{n(3n+1)}2}(1+q^{2n+1})(-q;q)\_n}{(q;q)\_n}
> \sum\_{j=1}^n\frac{1+q^{2j}}{1-q^{2j}}.\tag1$$
>
>
>
**Postscript.** This is a response to Vladimir Dotsenko's request. For the [LHS](https://mathoverflow.net/questions/420603/congruence-modulo-2-for-q-series) of my earlier post use (1) above, for the [RHS](https://mathoverflow.net/questions/420603/congruence-modulo-2-for-q-series) use the identity
$$\sum\_{k\geq0}x^{k+2}q^{k+1}\prod\_{j=1}^k(1-xq^j)=\sum\_{n\geq0}(-1)^n\left(x^{3n+2}q^{\frac{(n+1)(3n+2)}2}+x^{3n+3}q^{\frac{(n+1)(3n+4)}2}\right)$$
and replace $x\rightarrow q^{-1}, q\rightarrow q^4$.
| https://mathoverflow.net/users/66131 | Equality of two $q$-series. Proof? | I take it from looking at the previous problem that you are familiar with the [Dyson rank](https://en.wikipedia.org/wiki/Rank_of_a_partition) on partitions with distinct parts. Let's denote by $Q(r,n)$ the number of partitions of $n$ into distinct parts that have rank $r$. The following expression for the generating function of $Q$ is well known:
$$F(x,q)\mathrel{\mathop:}= \sum\_{r,n} Q(r,n)x^rq^n=\sum\_{n\geq 0}\frac{q^{\binom{n+1}{2}}}{(xq;q)\_n}.$$
This comes from grouping the partitions according to the number of parts (or equivalently the largest staircase partition that fits in the diagram).
Another expression from the generating function comes from grouping the partitions according to the largest pentagonal partition that fits in the diagram (the ones that appear as the fixed points of Franklin's involution mentioned by Vladimir Dotsenko in the comments). More specifically this means that for any partition with distinct parts we find the unique $n$ such that the partition either has:
* $n$ parts that are $\geq n+1$ and all other parts are $\le n$
* $n$ parts that are $\geq n+2$, one part $=n+1$ and all other parts are $\le n$
The contribution of the first group to the generating function of $Q$ is given by $\frac{(-x^{-1}q;q)\_n}{(xq;q)\_n}x^nq^{\frac{n(3n+1)}{2}}$ and the contribution of the second group is given by $\frac{(-x^{-1}q;q)\_n}{(xq;q)\_n}x^n q^{\frac{(n+1)(3n+2)}{2}}$. So putting everything together we get
$$F(x,q)=\sum\_{n\geq 0} x^n q^{\frac{n(3n+1)}{2}}(1+q^{2n+1})\frac{(-x^{-1}q;q)\_n}{(xq;q)\_n}.$$
Using these two expressions, you can obtain your identity by evaluating $$\frac{d}{dx}F(x,q)\Bigg\rvert\_{x=1}$$
in two ways.
| 18 | https://mathoverflow.net/users/2384 | 420742 | 171,189 |
https://mathoverflow.net/questions/419776 | 3 | Note: Here all processes take values in $[0, 1]$.
Let $W$ be a standard one dimensional Brownian motion, and $\sigma > 0$ a constant.
Let $X$ be the solution to the SDE
$$dX\_t = \sigma X\_t \, dW\_t$$
with $X\_0 = 1$ a.s.
For every $\varepsilon > 0$, let $A\_\varepsilon$ be the event $\{\text{max}\_{0 \leq t \leq 1} X\_t \geq \frac{1}{\varepsilon}\}$, and let $\mathbb P^\varepsilon$ be the probability measure defined by
$$\mathbb P^\varepsilon (E) =\frac{ \mathbb P(E \cap A\_\varepsilon)}{\mathbb P(A\_\varepsilon)}.$$
for all events $E$.
Define also for each $\varepsilon > 0$ the process $Y^\varepsilon$ by
$$Y^\varepsilon\_t := X\_t^{-C(\varepsilon)}.$$
where
$$C(\varepsilon) := \frac{\log \frac{1}{\varepsilon} +\frac{\sigma^2}{2}}{\sigma}.$$
**Question:** Is it true that
$$\lim\_{\varepsilon \to 0} \mathbb E\_{\mathbb P^\varepsilon} \left [\int\_{0}^1 \lvert Y\_t^\varepsilon - e^t \rvert \, dt \right ] = 0?$$
Where $\mathbb E\_{\mathbb P^\varepsilon}$ denotes the expectation under $\mathbb P^\varepsilon$.
*Remark:* It may be useful to adapt the method given by Yuval Peres in the answer [here](https://mathoverflow.net/questions/417015/a-large-noise-limit), however I do not know how to deal with the additional integral of $\sigma X\_t$ against the Brownian bridge.
| https://mathoverflow.net/users/173490 | Another large noise limit | $\newcommand{\si}{\sigma}\newcommand{\ep}{\varepsilon}\newcommand\num{\operatorname{num}}\newcommand\den{\operatorname{den}}\newcommand{\R}{\mathbb R}
\newcommand{\vpi}{\varphi}$The conjecture is not true in general.
The limit depends on $\si$. In particular, let us show that the limit in question is, not $0$, but $\infty$ if
\begin{equation\*}
\si>2 + \sqrt3; \tag{-2}\label{-2}
\end{equation\*}
(also see the heuristics at the end of this answer).
Indeed, let $P:=\mathbb P$, $P\_\ep:=\mathbb P\_\ep$, $E\_\ep:=\mathbb E\_{P\_\ep}$,
\begin{equation\*}
m:=\ln\frac1\ep\to\infty,\quad l:=m+\si^2/2,\quad\mu:=-\frac\si2, \quad r:=\frac m\si,
\end{equation\*}
\begin{equation\*}
M\_t:=\max\_{s\in[0,t]}(W\_s+\mu s),
\end{equation\*}
\begin{equation\*}
B\_t:=\{M\_t\ge r\}.
\end{equation\*}
Note that $(X\_t)$ is a geometric Brownian motion, so that
\begin{equation\*}
X\_t=\exp(\si W\_t-\si^2 t/2),
\end{equation\*}
whence
\begin{equation\*}
Y\_t:=Y^\ep\_t=X\_t^{-C(\ep)}=e^{\si l t/2}e^{-l(W\_t+\mu t)} \tag{-1}\label{-1}
\end{equation\*}
and
\begin{equation\*}
A\_\ep=\{M\_1\ge r\}\supseteq B\_t;
\end{equation\*}
here and in the sequel, $t\in(0,1)$.
It follows that
\begin{equation\*}
E\_\ep Y\_t\ge\frac\num\den, \tag{0}\label{0}
\end{equation\*}
where
\begin{equation\*}
\num:=Ee^{-l(W\_t+\mu t)}1\_{B\_t},\quad \den:=P(A\_\ep).
\end{equation\*}
Formula 1.4.8(1) on p. 256 in Handbook of Brownian Motion - Facts and Formulae, Second Edition, by Borodin and Salminen can be rewritten as
\begin{equation\*}
P(M\_t<u,W\_t+\mu\_t\in dz) \\
=\vpi\Big(\frac{z-\mu t}{\sqrt t}\Big)\frac{dz}{\sqrt t}
-e^{2\mu u}\vpi\Big(\frac{z-2u-\mu t}{\sqrt t}\Big)\frac{dz}{\sqrt t}
\tag{1}\label{1}
\end{equation\*}
for $z<u$, where $\vpi$ is the standard normal pdf.
Using \eqref{1} (and noting that $M\_t\ge W\_t+\mu\_t$), one can find
\begin{equation\*}
\begin{gathered}
\num=\int\_\R P(M\_t\ge r,W\_t+\mu\_t\in dz)e^{-lz} \\
=
\frac{1}{2} \left(\text{erf}\left(\frac{\si t \left(2 m+\si ^2+\si \right)-2 m}{2 \sqrt{2} \si
\sqrt{t}}\right)+1\right) \\
\times \exp \left(\frac{1}{8} \left(\frac{4 m^2 (\si t-4)}{\si }+4 m (\si +1)
(\si t-2)+(\si +2) \si ^3 t\right)\right) \\
+\frac{1}{2} e^{\frac{1}{8} t \left(2 m+\si ^2\right)
(2 m+\si (\si +2))} \text{erfc}\left(\frac{\si t \left(2 m+\si ^2+\si \right)+2 m}{2
\sqrt{2} \si \sqrt{t}}\right)
\end{gathered}
\end{equation\*}
and
\begin{equation\*}
\begin{gathered}
\den=P(M\_1\ge r) \\
=
1-\frac{1}{2} e^{-m}
\left(\text{erfc}\left(\frac{\frac{\si }{2}-\frac{m}{\si }}{\sqrt{2}}\right)-2\right)-\frac{1}{2}
\text{erfc}\left(-\frac{\frac{m}{\si }+\frac{\si }{2}}{\sqrt{2}}\right).
\end{gathered}
\end{equation\*}
If now $\si>2+\sqrt3$, then the interval $I\_\si:=(\frac{4\si-1}{\si^2},\min(1,\frac4\si))$ is nonempty and contained in the interval $(0,1)$. Moreover, for any $\si>2+\sqrt3$ and any $t\in I\_\si$, we have $\frac\num\den\to\infty$ (as $m\to\infty$), and hence, by \eqref{0}, $E\_\ep Y\_t\to\infty$. Thus, by Fatou's lemma, the limit in question is $\infty$. $\quad\Box$
---
Let me offer two competing heuristics to explain this result:
**Heuristic I: The large-deviation effect:** The large-deviation event $A\_\ep=\{M\_1\ge r\}=\{M\_1\ge m/\si\}$ (with $m\to\infty$) implies that $W\_t\approx mt/\si$. (In this case, this follows, for instance, from the independence of $W\_1$ from the Brownian bridge $B\_\cdot$, where $B\_t:=W\_t-tW\_1$ for $t\in[0,1]$.) So, on the event $A\_\ep$ we have $X\_t\approx\exp((m-\si^2/2)t)$ and hence
\begin{equation\*}
Y\_t\approx\exp\Big(-\frac{m^2}\si\,(1+o(1))t\Big), \tag{2}\label{2}
\end{equation\*}
so that we may expect $\int\_0^1 Y\_t\,dt$ to be somewhat small on the event $A\_\ep$, on the order of $\si/m^2$. The smaller $\si$ is, the more pronounced this effect should be. I think we will indeed have $E\_\ep\int\_0^1 Y\_t\,dt\to0$ if \eqref{-2} does not hold, but I have not checked all the details here.
**Heuristic II: The counterbalancing effect of a re-weighting exponential factor:** However, if $\si$ is large enough, then the large-deviation effect of Heuristic I may be overshadowed by the factor $e^{-lW\_t}$ in the representation of $Y\_t$ in \eqref{-1}. Indeed, this exponential factor can be very large for negative values of $W\_t$ and negligible for positive values of $W\_t$, since $l\sim m\to\infty$. So, even though negative values of $W\_t$ are somewhat suppressed by the large-deviation condition $M\_1\ge m/\si$, this suppression may be counterbalanced by the re-weighting exponential factor $e^{-lW\_t}$, which greatly "favors" negative values of $W\_t$. This counterbalancing effect will be more successful when the large-deviation effect is less strong, that is, when the spread/diffusion coefficient $\si$ of the Brownian motion is large enough. In this case,
the conditional expectation of $Y\_t$ given $A\_\ep$ may resemble much more
the unconditional expectation of $Y\_t$, which is
\begin{equation\*}
e^{l^2t/(2+o(1))}=e^{m^2t/(2+o(1))},
\end{equation\*}
which is very, very large (as $m\to\infty$).
Heuristic II is absent in the [previous setting](https://mathoverflow.net/questions/417015/a-large-noise-limit), where we do not have a very influential re-weighting exponential factor, such as the just considered factor $e^{-lW\_t}$.
| 1 | https://mathoverflow.net/users/36721 | 420752 | 171,192 |
https://mathoverflow.net/questions/420734 | 3 | I asked this over on Math.SE but it remained completely silent for over a week so I've deleted it and am reposting it here (I'm not really sure which site it fits better). The question itself is somewhat vague since I don't know precisely what I'm looking for and I'm hoping someone can point me in the right direction. Let me also preface this by saying that while I am familiar with the basic definitions of groupoids and sheaves, that is basically where my knowledge stops, so I am very much feeling my way in the dark here. I think a concrete example would best illustrate the situation.
Consider the open sets in $\mathbb{C}$ (or any Riemann surface, really) as the objects of a category. Let $\mathrm{Hom}(U,V)=\{\text{holomorphic maps }U\longrightarrow V\}$; composition of morphisms is just composition of functions. If we keep only the isomorphisms in this category, we get the titular "conformal" groupoid $\mathrm{Conf}$.
Now consider some sheaf of functions $\mathcal{O}$ on $\mathbb{C}$, e.g. smooth or holomorphic or whatever. $\mathrm{Conf}$ "acts" on $\mathcal{O}$ the sense that any $\phi\in\mathrm{Hom}(U, V)$ can be used to transfer sections between $\mathcal{O}(U)$ and $\mathcal{O}(V)$ in a reasonable way: if $f\in\mathcal{O}(V)$, then we can define $f\circ\phi\in\mathcal{O}(U)$.
But note that this action doesn't really involve the whole structure of $\mathcal{O}$ since I haven't mentioned the restriction maps of $\mathcal{O}$. And indeed $\mathrm{Conf}$ has more structure than just a groupoid, there are also "restriction operations": if $W\subset U$ and $\phi\in\mathrm{Hom}(U, V)$, we can form $\phi\vert\_W\in\mathrm{Hom}(W, \phi(W))$. The action from before behaves naturally with respect to this: if again $f\in\mathcal{O}(V)$, then
$$
(f\circ\phi)\vert\_W = f\vert\_{\phi(W)}\circ\phi\vert\_W
$$
The reason for the CFT tag should be clearer now, but let me say some more on this. If we look at something with different symmetry, like a standard Poincaré-covariant field theory on Minkowski space, we could play a similar game, but it doesn't seem as interesting: once we know that a map $\phi:U\longrightarrow V$ between open subsets preserves the metric, it has (I believe) a unique extension to a global Poincaré transformation, so somehow the transformations are too rigid to admit an interesting local structure (but I may be glossing over something here, I don't know). On the other hand, the same is emphatically not the case for conformal transformations: here it is much more important to compare not just global field configurations but also individual subsystems, i.e. field configurations on open subsets. [This post](https://physics.stackexchange.com/q/108472) goes into more detail on this, also working with the groupoid of conformal maps and its corresponding algebroid.
As I have said, this is all somewhat vague, and I haven't even made my question explicit. I suppose it would be something like, how do we tie all of this together neatly? That is, what algebraic structure captures this kind of groupoid-with-restriction and how do we describe its action on a given sheaf more precisely? Whatever the answers are, given that the question originated in CFT, they would be particularly satisfying if it followed that $\mathrm{Conf}$ bore some strong relation to the Witt algebra.
| https://mathoverflow.net/users/480683 | Conformal groupoid |
>
> That is, what algebraic structure captures this kind of groupoid-with-restriction and how do we describe its action on a given sheaf more precisely?
>
>
>
This structure is well known and has many equivalent incarnations:
[inductive groupoid](http://www-users.york.ac.uk/%7Evarg1/esnslidesf.pdf), [inverse semigroup](https://ncatlab.org/nlab/show/inverse%20semigroup),
[etale groupoid](https://ncatlab.org/nlab/show/%C3%A9tale+groupoid),
[etale stack](https://arxiv.org/abs/1212.2282).
Indeed, the bicategories of these objects turn out to be equivalent
once one imposes certain reasonable conditions (such as requiring inverse semigroups to be complete and distributive)
and equips them with an appropriate notion of 1-morphisms and 2-morphisms.
This equivalence is the main result of the [PhD thesis of Nilan Manoj Chathuranga](https://ttu-ir.tdl.org/handle/2346/88013) (soon to be on arXiv).
These languages also provide a very satisfactory answer to your second question: the action of Conf on a given sheaf is now simply a sheaf of sets (appropriately defined) on the inductive groupoid or inverse semigroup associated to Conf.
Equivalently, it is an etale map to the etale stack associated to Conf, i.e., an object in the slice category Et/Conf, where Et denotes the category of etale stacks and etale maps.
| 3 | https://mathoverflow.net/users/402 | 420755 | 171,194 |
https://mathoverflow.net/questions/420722 | 5 | **Question 1:** Suppose that two hyperbolic 3-manifolds $M\_1$ and $M\_2$ with finitely generated fundamental group satisfy the property that for every closed geodesic in $M\_1$, there is a closed geodesic in $M\_2$ with the same length. Are $M\_1$ and $M\_2$ isometric?
**Question 2:** The same as question one, but suppose that we also know the corresponding geodesics have the same rotational holonomy.
We can rephrase these questions in the language of Kleinian groups, as whether the multipliers of elements of a Kleinian group determine the group.
For finite volume manifolds, it should suffice to prove that $M\_1$ and $M\_2$ have the same volume.
| https://mathoverflow.net/users/479962 | Does length spectrum determine a hyperbolic 3-manifold? What if we also know holonomies? | I think your questions are answered by [this](https://arxiv.org/abs/math/0606343) paper of Leininger, McReynolds, Neumann and Reid. In their terminology, your first question is asking about manifolds with equal *length sets*, and your second question is asking about manifolds with equal *complex length sets*. Their Theorem 1.3 constructs many pairs of hyperbolic 3-manifolds with equal length sets but different volumes, and in §6.2 they note that these manifolds actually have equal complex length sets.
| 5 | https://mathoverflow.net/users/1463 | 420758 | 171,195 |
https://mathoverflow.net/questions/420751 | 5 | Let $G$ be an elementary abelian group, so that $G = (\mathbb{Z}/p)^k$ for some $k$. We can then compute the Morava $K$-theory of $BG = (BZ/p)^k$ pretty easily: $K(n)^\*(BG) = K(n)^\*[[x]]/[p](x) (x)\_{K(n)^\*} ... (x)\_{K(n)^\*} K(n)^\*[[x]]/[p](x)$, the tensor being taken $k$ times. We know that $[p](x) = x^{p^n}$. What is this cohomology $K(n)^\*(BG)$ as a representation of $GL\_k(\mathbb{F}$\_p)$?
Maybe we can compute the topological cyclic homology via a slice spectral sequence computation.
| https://mathoverflow.net/users/480262 | Computation of cohomology of Morava $K$-Theory | Bjorn Schuster and I looked at this problem in our article `On the $GL(V)$-module structure of $K(n)^\*(BV)$' Math Proc Cambridge Phil Soc vol 122 (1997) pp73--89. In particular, we show that in at least some cases it is not a permutation module. If you don't have access to the journal, there is a preprint that started off on Dave Benson's preprint server and was put on ArXiv in 2007: <https://doi.org/10.48550/arXiv.0711.5016>
Before our article, Nick Kuhn had shown that $K(n)^\*(BV)$ always has the same Brauer character as a permutation module for $GL(V)$ in `Morava $K$-theories of some classifying spaces' Trans Amer Math Soc vol 304 (1987) 193--205, and this motivated our work.
| 6 | https://mathoverflow.net/users/124004 | 420761 | 171,197 |
https://mathoverflow.net/questions/420759 | 2 | I'm almost not at all knowledgable in either Freidlin-Wentzel theory or Kramers' escape problem as it is known in the physics community, so please excuse some of my naivety.
One can use Freidlin-Wentzel theory to study the problem of a particle with position given by the SDE
$$dX\_t = V(X\_t)\,dt + \sqrt{\epsilon}\,dW\_t (\*)$$
escaping from a local minimum over some barrier, or more precisely derive the mean first passage time over that barrier. To this end one has to introduce the concept of large deviation principles, prove Schilder's theorem, then transfer the LDP for the Brownian motion $W$ to $X$ via the Ito map. Then it seems one can *start* studying the escape problem.
What suprises me that in the physics/chemistry/etc. community they study the same (?) problem, but their derivation of the mean passage time makes no reference at all to Large deviations and proceeds more or less elementary using the Fokker-Planck equation associated with (\*) (cf. e.g. [these lectures notes](http://physics.gu.se/%7Efrtbm/joomla/media/mydocs/LennartSjogren/kap8.pdf)).
How do these two approaches relate to each other? The physics approachs seems so much easier that it makes me question the merit of studying Freidlin-Wentzel theory to solve escape problems.
| https://mathoverflow.net/users/78650 | Kramers' escape problem: statistical physics vs. Large deviations | The Kramers theory is limited to reversible processes in equilibrium, while the Freidlin-Wentzel theory generalises this to irreversible processes out of equilibrium.
The distinction appears in the stochastic differential equation
$$dx\_t=f(x\_t)dt+ \sqrt{2\epsilon}dW\_t,$$
in the Kramers theory it is assumed that $f(x)=-\nabla U$ is the gradient of a potential $U:\mathbb{R}^d\rightarrow\mathbb{R}$, and the escape rate is given in terms of this potential. The system is in equilibrium with probability distribution $p(x)\propto e^{-U/\epsilon}$. The Freidlin-Wentzel theory applies to more general functions $f$, and can describe systems out of equilibrium
| 2 | https://mathoverflow.net/users/11260 | 420762 | 171,198 |
https://mathoverflow.net/questions/420354 | 6 | In Lurie's "Higher Algebra", Remark 5.4.5.2 towards the end, there is the following statement: "It follows that $\mathbb{E}\_M$ can be identified with the colimit of a diagram of $\infty$-operads parametrized by $M$, each of which is equivalent to $\mathbb{E}\_k$". I do not see how this follows from the preceding discussion there, but let me give a bit of context:
Lurie introduces the $\mathbb{E}\_M$-operad of a topological manifold $M^n$ by forming the pullback $\operatorname{BTop}(n)^\otimes \times\_{\operatorname{BTop}(n)^\coprod} B\_M^\coprod$, where $B\_M$ is defined as the category $\operatorname{BTop}(n)\_{/M}$ with $\operatorname{BTop(n)}$ the full sub-$\infty$-category on $\mathbb{R}^n$ of the category of topological manifolds and embeddings. Note that the operadic structures are the coCartesian symmetric monoidal structure $\coprod$, and disjoint unions, respectively (i.e. $\operatorname{Mul}\_{\operatorname{BTop}(n)^\otimes}(\mathbb{R}^n,\mathbb{R}^n ; \mathbb{R}^n) = \operatorname{Emb}(\mathbb{R}^n \sqcup \mathbb{R}^n, \mathbb{R}^n)$ for example). To put it bluntly, $\mathbb{E}\_M$ describes embeddings of (disjoint unions of) disks into $M$.
In the previously mentioned remark, Lurie shows that $B\_M \simeq \operatorname{Sing}(M)$ so that, since $\operatorname{BTop}(n)^\otimes$ has only one object $\mathbb{R}^n$, the underlying $\infty$-category of $\mathbb{E}\_M$ is equivalent to $\operatorname{Sing}(M)$. It is also not very hard to see that if we take the full sub-operad of $\mathbb{E}\_M$ on one object, i.e. one point in $M$, it is equivalent to $\mathbb{E}\_n$. However, I do not see how the statement about colimits above follows from these observations, as Lurie claims - I however find the statement very interesting, as it is in my eyes much more explicit then the assembly of $\mathbb{E}\_M$ by copies of $\mathbb{E}\_k$ that Lurie proves later.
| https://mathoverflow.net/users/156537 | $\mathbb{E}_M$ as colimit of little cubes operads | I agree this is sort of scattered around as remarks (e.g. HA.2.3.3.4, say). Let's try to spell it out more cleanly.
1. I claim that, if $X$ is a groupoid, then the category of $X$-families of operads is the same as the category of functors $\mathsf{Fun}(X, \mathsf{Op})$. Indeed, $\mathsf{Op}$ is a (non-full) subcategory of $\mathsf{Cat}\downarrow \mathsf{Fin}\_\*$, so begin by observing that $\mathsf{Fun}(X,\mathsf{Cat}\downarrow \mathsf{Fin}\_\*)$ is equivalent, by the Grothendieck construction, to the category of diagrams $X \leftarrow \mathcal{E} \to \mathsf{Fin}\_\*$ where $\mathcal{E} \to X$ is a cocartesian fibration. But if $X$ is a groupoid, every functor with target $X$ is a cocartesian fibration (here I am using the 'homotopy invariant notions', otherwise you have to say that we restrict to maps which are categorical fibrations, etc.) So in fact, $\mathsf{Fun}(X, \mathsf{Cat}\downarrow \mathsf{Fin}\_\*)\simeq \mathsf{Cat}\downarrow(X \times \mathsf{Fin}\_\*)$. But now the definition of an $X$-family of operads corresponds exactly to the requirement that the functor from $X$ to $\mathsf{Cat}\downarrow \mathsf{Fin}\_\*$ factors through the subcategory of operads.
2. Great. Let $\mathcal{E}$ be a generalized operad and $\mathcal{O}$ be an operad. By definition, a map $\mathrm{Assem}(\mathcal{E}) \to \mathcal{O}$ is the same as a map of generalized operads $\mathcal{E} \to \mathcal{O}$. This, in turn, is the same as a map of generalized operads over $\mathcal{E}\_{\langle 0\rangle}\times \mathsf{Fin}\_\*$ from $\mathcal{E}$ to $\mathcal{E}\_{\langle 0\rangle}\times \mathcal{O}$. By HA.2.3.2.13, this is the same as a map of families of operads over $\mathcal{E}\_{\langle 0\rangle}$ from $\mathcal{E} \to \mathcal{E}\_{\langle 0\rangle} \times \mathcal{O}$. Finally, if $\mathcal{E}\_{\langle 0\rangle}$ happens to be a groupoid, then by (1) this is the same as a map from the diagram of operads classifying $\mathcal{E}$ to the constant diagram at $\mathcal{O}$. This completes the proof that $\mathrm{Assem}(\mathcal{E})$ is the colimit of the corresponding diagram (when $\mathcal{E}\_{\langle 0\rangle}$ happens to be a groupoid.)
| 2 | https://mathoverflow.net/users/6936 | 420770 | 171,202 |
https://mathoverflow.net/questions/420737 | 4 | Given a group $G$, the outer automorphism group $Out(G)$ acts on the center by $Z(G)$ by lifting an outer automorphism to an actual automorphism and evaluating this on elements of $Z(G)$. What is classified by the degree three group cohomology $H^3(Out(G),Z(G))$ ?
Given an algebra $A$, say finite-dimensional over a field, the outer automorphism group $Out(A)$ acts on the group of central units $Z(A)^\times$ by lifting an outer automorphism to an actual automorphism and then evaluating. The action is trivial if $A$ is a central algebra. What is classified by the degree three group cohomology $H^3(Out(A),Z(A)^\times)$ ?
My motivation for this question is the following. If $A$ is my algebra, then we have a *crossed module*
$$
A^\times \to Aut(A)
$$
given by inner automorphisms and the evaluation action of $Aut(A)$ on $A^\times$. The homotopy groups of this crossed module are $\pi\_0=Out(A)$ and $\pi\_1=Z(A)^\times$, and the usual action of $\pi\_0$ on $\pi\_1$ is the one described above. Crossed modules have a so-called *k-invariant*, which is precisely a class $\xi\in H^3(\pi\_0,\pi\_1)$.
Baez and Lauda have shown that crossed modules are classified in a sense by triples $(\pi\_0,\pi\_1,\xi)$. The classification is saying that the crossed module - viewed as a monoidal category - is equivalent to the usual monoidal category constructed from the 3-cocycle $\xi$.
*Baez, John C.; Lauda, Aaron D.*, [**Higher-dimensional algebra. V: 2-Groups**](http://www.emis.de/journals/TAC/volumes/12/14/12-14abs.html), Theory Appl. Categ. 12, 423-491 (2004). [ZBL1056.18002](https://zbmath.org/?q=an:1056.18002).
It is also [known](https://arxiv.org/abs/2204.03900) that if $A$ and $B$ are Picard-surjective and Morita equivalent, then their crossed modules are equivalent and so their classes coincide. Moreover, if $A$ is Picard-surjective and central-simple, then its class vanishes.
Summarizing, associated to every algebra is a class in $H^3(Out(A),Z(A)^\times)$. What is the intrinsic meaning of this class, apart from classifying some crossed module?
For groups instead of algebras it is basically the same story, and the question is analogous.
| https://mathoverflow.net/users/3473 | 3-cocycles on outer automorphism groups | Among other things, the third cohomology contains an invariant for the existence of group-graded algebras whose degree-1-piece is $A$ / group extension with $G$ as normal subgroup. This is a theorem of Schreier. If I'm not mistaken, the cohomology class $\xi$ that comes from crossed modules is precisely Schreier's invariant.
There is a generalisation of both of these with more abstract nonsense sprinkled in (2-groups among them) see [this earlier question of mine](https://mathoverflow.net/questions/401964/connection-between-the-classifications-of-group-extensions-and-group-graded-alge)
| 2 | https://mathoverflow.net/users/3041 | 420772 | 171,203 |
https://mathoverflow.net/questions/420765 | 11 | I am embarrassed to be stuck on this seemingly simple question.
Suppose that $X,Y$ are mean-zero real-valued random variables and $\tilde X,\tilde Y$ are their "independent copies": $\tilde X,\tilde Y$ are mutually independent and independent of $(X,Y)$, and $\tilde X$ (resp., $\tilde Y$) is distributed identically to $X$ (resp., $Y$).
Here is the inequality I am trying to prove/disprove:
for some universal constant $c>0$,
$$
\mathbb{E}|\tilde X-\tilde Y|
\le
c\left(
\mathbb{E}|X-Y|
+
\sqrt{|\mathbb{E}XY|}
\right).
$$
Update. Note that the related inequality,
$$
\mathbb{E}|\tilde X-\tilde Y|^2
\le
\mathbb{E}|X-Y|^2
+
2|\mathbb{E}XY|
,
$$
is trivially true.
| https://mathoverflow.net/users/12518 | Decoupling inequality/counterexample | [**EDITED** *to ensure that the random variables have expectation 0*]
I think the answer is no.
Let $Z$ be a random variable taking the values $\pm 1$ with probability $p$ each and 0 with probability $1-2p$; and let $N$ be a standard normal independent of $Z$.
Set $X=ZM+(1-|Z|)N$ and $Y=-ZM+(1-|Z|)N$, where $M=\sqrt{(1-2p)/(2p)}$.
Now $\mathbb E|X-Y|=2pM=O(\sqrt p)$ and $\mathbb EXY=-2pM^2+(1-2p)=0$.
On the other hand, $\mathbb E|\tilde X-\tilde Y|=\Omega(1)$.
| 13 | https://mathoverflow.net/users/11054 | 420773 | 171,204 |
https://mathoverflow.net/questions/420776 | 10 | Let $A\in \mathbb{S}^{N\times N}$ be a symmetric, real and stable matrix, i.e., $\rho(A)<1$, where $\rho(A)$ stands for the spectral radius of $A$. Then, $$\sum\limits\_{i=0}^{\infty} A^{2i}=\left( I-A^2 \right)^{-1}.\tag{$\star$}\label{star}$$
Now, let $A$ be asymmetric. It is not clear whether the power series $$F(A):=\sum\_{i=0}^{\infty} A^i (A^i)^{\top}$$ admits a *compact closed-form* expression as in equation \eqref{star} (when it converges). If the matrix $A$ is normal — i.e., it commutes with its transpose — and stable, then we have that $F(A)=\left(I-AA^{\top}\right)^{-1}$. In general, that is not the case.
>
> **Framework.** Let $A$ be a random matrix generated as follows. Let $\widetilde{A}\in \mathcal{G}(N,p)$ be the adjacency matrix associated with the realization of a random graph over $N$ nodes with probability $p$ of arrow drawing, i.e., the entries $\widetilde{A}\_{ij}$ of $\widetilde{A}$ are i.i.d. $\mathsf{Bernoulli}(p)$ for all $i\neq j$ — this is often referred to as a binomial random graph. Set $\widetilde{A}\_{ii}=0$ for all $i$. Let $A$ be obtained from $\widetilde{A}$ by: **i) [off-diagonal]** normalizing the entries of $\widetilde{A}$ as $A\_{ij}:=\alpha\_1 \widetilde{A}\_{ij}/d\_{\max}$, for all $i\neq j$, where $d\_{\max}$ is the maximum *in-flow* degree of the random graph and $0<\alpha\_1<1$ is some constant; **ii) [diagonal]** setting the diagonal terms as $A\_{ii}:=\alpha-\sum\_{k\neq i} \widetilde{A}\_{ik}$, where $0<\alpha\_1\leq\alpha<1$. In other words, the rows of $A$ sum to $\alpha$ and its support is given by the realization of the binomial random graph on $N$ nodes.
>
>
>
**Question.** Is it true that $$\mathbb{P}\left(\lVert\sum\_{i=0}^{\infty}A^i \left(A^i\right)^{\top} - \left(I-AA^{\top}\right)^{-1}\rVert\_\text{max}>\varepsilon\right)\overset{N\rightarrow \infty}\longrightarrow 0?$$
Is there any known result in this direction? Any reference would be appreciated. The specifics of the random nature of $A$ are not important other than that of $A$ being stable and randomly generated (so that it is more natural to consider the limit $N\rightarrow\infty$).
Some numerical experiments are pointing in the direction of a *concentration* of $\sum\_{i=0}^{\infty} A^i \left(A^i\right)^{\top}$ about $\left(I-AA^{\top}\right)^{-1}$ as the dimension $N$ grows large (I may attach them later, if necessary) under the referred framework.
**Context.** $F(A)$ is the covariance matrix associated with the time-series reflecting the state-evolution of certain linear stochastic dynamical systems. Having a closed form expression for the covariance in terms of $A$ is helpful for model identification purposes.
| https://mathoverflow.net/users/138242 | (Asymmetric) matrix power series in closed form: $\sum_{i=0}^{\infty} A^i \left(A^i\right)^{\top}={?}$ | Let me answer the first question:
**Q:** Does $F=\sum\_{i=0}^{\infty} A^i (A^i)^{\top}$ for non-symmetric $A$ have a closed-form solution analogous to the solution $\sum\_{i=0}^{\infty} A^{2i}=\left( I-A^2 \right)^{-1}$ for a symmetric $A$?
**A:** Yes, in terms of the [vectorization operation](https://en.wikipedia.org/wiki/Vectorization_(mathematics)):
$$\operatorname{vec}(F)=(I-{A} \otimes A)^{-1}\cdot\operatorname{vec}(I).\tag{$\ast$}\label{ast}$$
This does require the inversion of an $N^2\times N^2$ matrix, rather than the inversion of an $N\times N$ matrix as in the case of a symmetric $A$.
The vectorization formula follows because $F$ solves the [Lyapunov equation](https://en.wikipedia.org/wiki/Lyapunov_equation)
$$AFA^\top=F-I.$$
I do not have an answer to the second question, whether \eqref{ast} converges for large $N$ to $(I-AA^\top)^{-1}$.
| 7 | https://mathoverflow.net/users/11260 | 420782 | 171,207 |
https://mathoverflow.net/questions/420744 | 14 | Let $\mathbb{S}=\{x\in\mathbb{R}^n|x\_1^2+\ldots +x\_n^2=1\}$ be the unit sphere in $\mathbb{R}^n$, $\mathbb{C}[x]=\mathbb{C}[x\_1,\ldots ,x\_n]$ the complex-valued polynomial functions on $\mathbb{R}^n$, and $\Delta=\partial\_1^2+\ldots +\partial\_n^2$ the Laplacian. Let $H\_n=\ker(\Delta)$ be the space of harmonic polynomials. Consider the restriction to the sphere
$$\rho:\mathbb{C}[x\_1,\ldots ,x\_n]\to\mathbb{C}[x\_1,\ldots ,x\_n]/(x\_1^2+\ldots +x\_n^2-1)$$
and the image $\bar{H}\_n=\rho(H\_n)\subseteq\mathbb{C}[x\_1,\ldots ,x\_n]/(x\_1^2+\ldots +x\_n^2-1)$.
Now, there is the curious fact that $\bar{H}\_n$ is in fact the whole of $\mathbb{C}[x\_1,\ldots ,x\_n]/(x\_1^2+\ldots +x\_n^2-1)$. In other words, for every $f\in\mathbb{C}[x\_1,\ldots ,x\_n]$, the polynomial restriction $f|\_{\mathbb{S}}$ on the sphere has a harmonic representative $h\in H\_n$ such that $f|\_{\mathbb{S}}=h|\_{\mathbb{S}}$.
This follows from a $\sum(\mathrm{radial})\cdot (\mathrm{harmonic})$ decomposition
**Theorem.** Let $f$ be a polynomial. Then $f(x)=\sum\_i f\_i(x\_1^2+\ldots +x\_n^2)\cdot h\_i(x)$ where $h\_i\in H\_n$ and $f\_i\in\mathbb{C}[t]$.
Indeed: $f|\_\mathbb S =(\sum\_i f\_i(x\_1^2+\ldots+x\_n^2) \cdot h\_i)|\_\mathbb S=\sum\_i f\_i(1)\cdot h\_i|\_\mathbb S = (\sum\_i f\_i(1)h\_i)|\_\mathbb S$ and now $h:=\sum\_i f\_i(1)h\_i$ is harmonic.
In particular, harmonic restrictions on the sphere form an algebra (equal to the algebra of all polynomial restrictions). From this, I think, one can prove the spherical harmonics are a Hilbert basis on the sphere by using Stone-Weierstrass (and density of continuous functions in $L^2$) without proving that they form a basis of eigenfunctions of the Laplace-Beltrami operator $\Delta\_\mathbb{S}$ on $\mathbb S$ (which by the way they do, and it's kinda interesting in itself).
>
> **Q.** Is this stuff part of some theory, or just a coincidence for the usual Laplacian?
>
>
>
For example, what if $D$ is a differential operator with symbol $\sigma\_D(x)$ and $X\_D\subseteq\mathbb{R}^n$ the variety $\{x|\sigma\_D(x)=1\}$. Is there sometimes a polynomial decomposition
$$f=\sum\_i f\_i(\sigma\_D(x))\cdot h\_i(x)$$
with $Dh\_i=0$? Or an "intrinsic" operator $P$ on $X\_D$, related to $D$ in a similar way as $\Delta\_\mathbb{S}$ is related to $\Delta$, such that the homogeneous $h\in\ker D$ are eigenfunctions of $P$?
| https://mathoverflow.net/users/4721 | Harmonic polynomials on the sphere | I view this as a concatenation of two facts:
---
**Fact 1:** Let $k$ be a field, let $I$ be an ideal of $k[x\_1, \ldots, x\_n]$ and let $J$ be associated graded ideal of $I$, meaning that a degree $d$ homogenous polynomial $g(x)$ is in $J$ if and only if there is an element of $I$ of the form $g(x) + (\text{terms of degree $< d$})$. Let $\{ p\_a \}$ be a set of homogenous polynomials which maps to a basis of $k[x]/J$. Then $\{ p\_a \}$ also maps to a basis of $k[x]/I$.
The proof of Fact 1 is just an upper triangularity argument.
Apply Fact 1 with $I$ the ideal $\langle \sum x\_i^2 -1 \rangle$ and $J$ the ideal $\langle \sum x\_i^2 \rangle$. So this reduces us to the question of why harmonic polynomials are a basis for $\mathbb{R}[x]/\langle \sum x\_i^2 \rangle$.
---
Turn $\mathbb{R}[x\_1, x\_2, \ldots, x\_n]$ into a module over itself by
$$g(x\_1, \ldots, x\_n) \ast f(x\_1, \ldots, x\_n) := g(\tfrac{\partial}{\partial x\_1}, \ldots, \tfrac{\partial}{\partial x\_n}){\big(} f(x\_1, \ldots, x\_n) {\big)}$$
To be clear, this is a differential operator, encoded by $g$, acting on the polynomial $f$.
This action induces an inner product on the degree $d$ homogenous polynomials by $\langle g,f \rangle := g \ast f$, since $g \ast f$ is a degree $0$ polynomial and can be thought of an element of $\mathbb{R}$.
Let $J \subseteq \mathbb{R}[x\_1, \ldots, x\_n]$ be any graded ideal of $\mathbb{R}[x\_1, \ldots, x\_n]$. We write $J = \bigoplus J\_d$ for its composition into graded pieces. Let $J^{\perp} = \bigoplus J\_d^{\perp}$, where the orthogonal complement is taken with the above inner product on $\mathbb{R}[x\_1, \ldots, x\_n]\_d$. By generalities of inner products, the map $J^{\perp} \hookrightarrow R[x\_1, \ldots, x\_n] \twoheadrightarrow R[x\_1,\ldots, x\_n]/J$ is an isomorphism.
**Fact 2** We have $$J^{\perp} = \{ f : g \ast f=0 \ \forall g \in J \}.$$ Moreover, if $g\_1$, $g\_2$, ..., $g\_N$ is a list of generators for $J$, it is sufficient to impose that $g\_i \ast f=0$ for each generator $g\_i$.
Thus, the set of polynomials with $\sum (\tfrac{\partial}{\partial x\_i})^2 f =0$ is a basis for $\mathbb{R}[x]/(\sum x\_i^2)$.
---
I learned this by reading Mark Haiman's papers on "diagonal harmonics", and trying to figure out why he was calling them harmonic functions. I'll try to find some better references.
| 6 | https://mathoverflow.net/users/297 | 420784 | 171,208 |
https://mathoverflow.net/questions/420781 | 2 | Let $(X,\mu)$ be a measure space and let $1<p<\infty$.
>
> **Question.** Is the space $L^p(X,\ell^p)$,
> $$
> \lVert f\rVert\_p=\Bigl(\int\_X\sum\_{i=1}^\infty \lvert f\_i\rvert^p\, dx\Bigr)^{1/p},
> \qquad
> f=(f\_i)\_{i=1}^\infty,
> $$
> uniformly convex?
>
>
>
If the answer if "yes" I would appreciate a reference to the statement/proof.
I know that $L^p(X,\ell^p\_M)$, where $\ell^p\_M$ is finite dimensional $\ell^p$ space
$$
\lVert f\rVert\_p=\Bigl(\int\_X\sum\_{i=1}^M \lvert f\_i\rvert^p\, dx\Bigr)^{1/p},
\qquad
f=(f\_1,\dotsc,f\_M)
$$
is uniformly convex. This result was proved by Clarkson in [Uniformly convex spaces](https://doi.org/10.2307/1989630) where he introduced the notion; see also [Uniformly convex Banach spaces](https://mathoverflow.net/q/420011/121665). I haven't checked whether the argument applies to the case of values into $\ell^p$.
| https://mathoverflow.net/users/121665 | Is $L^p(X,\ell^p)$, $1<p<\infty$, uniformly convex? | By Proposition 1.2.24 in T. Hytonen, J. Van Neerven, M. Veraar and L. Weis, Analyis in Banach spaces Vol I, Springer, the spaces $L^p(X; \ell^p)$ and $L^p(X\times \mathbb N)$, $X \times \mathbb N$ with the product measure, are isometric (at least when $\mu$ is $\sigma$-finite). Then the uniform convexity of $L^p(X; \ell^p)$ follows from that of $L^p$-spaces.
| 8 | https://mathoverflow.net/users/150653 | 420791 | 171,211 |
https://mathoverflow.net/questions/420785 | 5 | Consider dihedral Galois extensions $L/\mathbb{Q}$ of degree $n$ (and we know they exist thanks to Shafarevich), can we show there always exists an extension $L/\mathbb{Q}$ unramified at $p$, for all $p \mid 2n$?
| https://mathoverflow.net/users/477248 | Dihedral extension unramified at primes dividing order of group? | $\DeclareMathOperator\Gal{Gal}$The answer is "yes", and this is an easy exercise in class field theory: if, for example, $q$ is a prime number that is $1\pmod n$, and $F$ is a quadratic number field in which $q$ splits, then there is a quotient of the ray class group of $F$ with modulus $q$ that has order $n$ and on which $\Gal(F/\mathbb{Q})$ acts by $-1$, so that the corresponding ray class field $L/F$ is Galois over $\mathbb{Q}$ with dihedral Galois group and is unramified (over $F$) outside of $q$. See, for example, Section 3.1 in <https://arxiv.org/abs/0805.1231> for the details of this computation. So if you also arrange $F$ to be unramified at all $p|2n$, then $L/\mathbb{Q}$ is as required.
| 4 | https://mathoverflow.net/users/35416 | 420793 | 171,212 |
https://mathoverflow.net/questions/231770 | 6 | The following question arises when I attempt to understand the modular parameterization of the elliptic curve $$E:y^2-y=x^3-x$$
In [Mazur-Swinnerton-Dyer](https://eudml.org/doc/142281) and [Zagier](http://people.mpim-bonn.mpg.de/zagier/files/doi/10.1007/BFb0084592/fulltext.pdf)'s construction, a theta function associated with a positive definite quadratic form is induced:
$$\theta(q)=\sum\_{x\in\mathbb{Z}^4}q^{\frac{1}{2}x^{T}Ax}$$
where $$A=\left(\begin{matrix}2 & 0 & 1 & 1\\
0 & 4 & 1 &2\\
1 & 1 & 10 & 1\\
1 & 2 & 1 & 20
\end{matrix}\right)$$
$A$ is a positive definite matrix of determinant $37^2$, and we have $37A^{-1}=K^TAK$ where $K$ is an integral matrix of determinant $\pm 1$.
**Question:** Suppose $A$ is a positive definite $4\times 4$ matrix with integral entries. All diagonal entries are even numbers. The determinant of $A$ is a square number $N^2$. Is it true that for every $N=p$ ($p>2$ is a prime number), there is *at least one* $A$ that $NA^{-1}=K^TAK$, where $K$ is an integral matrix of determinant $\pm 1$?
| https://mathoverflow.net/users/18286 | Integral quaternary forms and theta functions | The answer is true, using the following construction.
Let $B$ be the quaternion algebra of discriminant $p$ and let $O$ be a maximal order with an element $x$ satisfying $x^2 = -p$. The reduced norm is a quadratic form on $O$, with positive definite Gram matrix $A$ of determinant $p^2$. The matrix $A^{-1}$ then represents the reduced norm on the dual lattice $O^{\sharp}$. Recall that $\text{nrd}(\text{diff}(O)) = \text{discrd}(O) = p$ and since $O$ is maximal, $\text{diff}(O)$ is invertible and $O = \text{diff}(O) O^{\sharp}$ (see e.g.
*Voight, John*, [**Quaternion algebras**](http://dx.doi.org/10.1007/978-3-030-56694-4), [ZBL07261776](https://zbmath.org/?q=an:07261776). Section 16.8), hence
$$
p O^{\sharp} \subseteq \text{diff}(O) O^{\sharp} = O
$$
which implies that $p \cdot (O^{\sharp} / O) = 0$ (therefore $O^{\sharp} / O$ is an abelian group of exponent $p$ and size $p^2$, so isomorphic to $(\mathbb{Z} / p \mathbb{Z})^2$). Next, we note that the matrix $A^{-1}$ is also the change of basis matrix between the chosen basis of $O$ and its dual, therefore $pA^{-1}$ is integral.
This is not enough, but so far we have not used the element $x$. The matrix $pA^{-1}$ is the matrix representing the norm form on the ideal $xO^{\sharp}$, since $x^2 = -p$. But $nrd(xO^{\sharp})=p$ is a maximal order with an element $x$ such that $x^2 = -p$. By a theorem of Ibukiyama (see reference below), it is isomorphic (hence isometric as lattices) to $O$.
This could also be seen directly (referring to some of the answers above - it is possible to do it using a finite number of cases) by explicitly writing down the maximal orders. I will list below explicit constructions, which are based on
*Ibukiyama, Tomoyoshi*, [**On maximal orders of division quaternion algebras over the rational number field with certain optimal embeddings**](http://dx.doi.org/10.1017/S002776300002016X), Nagoya Math. J. 88, 181-195 (1982). [ZBL0473.12012](https://zbmath.org/?q=an:0473.12012). -
If $p \equiv 3 \bmod 4$, then $B = (-p,-1)$, and $O = \mathbb{Z}<(1+i)/2,j>$. In this case, we have
$$
A = \left( \begin{array}{cccc}
2 & 1 & 0 & 0 \\
1 & \frac{p+1}{2} & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 1 & \frac{p+1}{2}
\end{array} \right)
,
pA^{-1} = \left( \begin{array}{cccc}
\frac{p+1}{2} & -1 & 0 & 0 \\
-1 & 2 & 0 & 0 \\
0 & 0 & \frac{p+1}{2} & -1 \\
0 & 0 & -1 & 2
\end{array} \right),
K = \left( \begin{array}{cccc}
-1 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & -1 & 1 \\
0 & 0 & 1 & 0
\end{array} \right)
$$
If $p \equiv 1 \bmod 4$, then choosing $q \equiv 3 \bmod 4$ such that
$\left( \frac{q}{p} \right) = -1$, we can find $c$ such that
$c^2 \equiv -p \bmod q$, and then $B = (-p,-q)$ and
$$
O = \mathbb{Z} \oplus \mathbb{Z} \frac{1+j}{2} \oplus
\mathbb{Z} \frac{i(1+j)}{2} \oplus \mathbb{Z} \frac{(c+i)j}{q}
$$
In this case, we compute that
$$
A = \left( \begin{array}{cccc}
2 & 1 & 0 & 0 \\
1 & \frac{q+1}{2} & 0 & c \\
0 & 0 & \frac{p(q+1)}{2} & p \\
0 & c & p & \frac{2(p+c^2)}{q}
\end{array} \right)
,
pA^{-1} = \left( \begin{array}{cccc}
\frac{(q+1)(c^2+p)}{2q} & -c-\frac{c^2+p}{q} & -c & \frac{c(q+1)}{2} \\
-c-\frac{c^2+p}{q} & 2(c^2 + \frac{c^2+p}{q}) & 2c & -c(q+1) \\
-c & 2c & 2 & -q \\
\frac{c(q+1)}{2} & -c(q+1) & -q & \frac{q(q+1)}{2}
\end{array} \right)
$$
and if $\{e\_1, e\_2, e\_3, e\_4 \}$ is the above basis for $O$, then we see that $ \{e\_2 e\_4, ce\_1 - e\_4, e\_1, j e\_2 \} $ is a basis for the resulting module, which written in terms of the original basis yields the matrix
$$
K = \left( \begin{array}{cccc}
0 & c & 1 & -\frac{q+1}{2} \\
-c & 0 & 0 & 1 \\
-1 & 0 & 0 & 0 \\
\frac{q+1}{2} & -1 & 0 & 0
\end{array}
\right).
$$
Referring to Will Jagy's wondering in the first answer, the reason that this does not work for most lattices is that they do not correspond to a maximal order (as in the answer by few\_reps, the quotient of the lattices is actually cyclic) and there are only one or two (depending on $p \bmod 4$ isomorphism classes of maximal orders which contain a root of $-p$. For example, for $p = 37$, there are only two isomorphism classes of maximal orders in the quaternion algebra, and only one of them contains a root of $-p$.
| 4 | https://mathoverflow.net/users/74819 | 420800 | 171,216 |
https://mathoverflow.net/questions/420799 | 3 | Let $Pic\_n^0$ denote the even part of the $K(n)$-local Picard group, and let $Pic\_n^\*$ denote $Hom(Pic\_n^0, W(\mathbb{F}\_{p^n})^x)$. Denote by $L$ the profinite group ring $\mathbb{Z}\_p[[Pic\_n^\*]] $. Let $\lambda$ be an element of $Pic\_n$; this gives a map $Pic\_n^\* \to W(\mathbb{F}\_{p^n})^x$, given by evaluation at $\lambda$. This induces a ring map $L \to W(\mathbb{F}\_{p^n})$. We can consider the $Pic\_n$-graded homotopy group of a ($K(n)$-local) spectrum $X$. What we want is a "universal" $L$-module $P(X)$ such that $P (x)\_{L}W(\mathbb{F}\_{p^n})$, where we regard $W(\mathbb{F}\_{p^n})$ as a $L$-module via $\lambda$.
At height $1$, this problem is not too hard: we know that $Pic\_1^0 = \mathbb{Z}\_p^x$, so that $Pic\_1^\* = \mathbb{Z}\_p^x$. If $\psi$ is any topological generator of $\mathbb{Z}\_p$, then $L$ is isomorphic to $p-2$ copies of $\mathbb{Z}\_p[[T]]$, where $T$ corresponds to $\psi-1$. Let $P$ denote $\mathbb{Z}\_p$, with the trivial $Pic\_1^\*$-action. The other action on $\mathbb{Z}\_p$ coming from an element $\lambda$ is given by $g\*x = g(\lambda) x$; here, we are thinking of $End(\mathbb{Z}\_p^x)$ as $\mathbb{Z}\_p^x$. In this case, for a fixed lambda with $\lambda(k) = k^m, P (x)\_L \mathbb{Z}\_p = \mathbb{Z}\_p[[T]] = \mathbb{Z}\_p/(k^m-1)$. But $k^m-1$ is a unit unless $(p-1)|m$, and in that case, $k^{(p-1)\*q}-1 = u\*p\*q$ where $u$ denotes the unit. It follows that $\pi\_{(p-1)\*q} L\_K(1) S = \mathbb{Z}/(p\*q)$, as standard computations tell us.
Can this connection with Iwasawa theory be studied via some geometry on the Lubin-Tate stack?
| https://mathoverflow.net/users/480262 | Studying a connection between Iwasawa theory and the $K(n)$-local Picard group by some geometry on the Lubin-Tate stack | A long time ago there were various attempts to make this sort of approach work, but they were not very successful. If you want to try again then you should make sure that you are familiar with the phenomena observed by computation in the height two case (which are quite complex). The computations are due to Shimomura, Yabe and their collaborators. One perspective on the answer is explained in Section 15.2 of the memoir *Morava $K$-theories and localisation* (by Mark Hovey and myself), and another in the paper *The homotopy groups of the $E(2)$ local sphere at $p > 3$ revisited* (by Mark Behrens).
| 6 | https://mathoverflow.net/users/10366 | 420804 | 171,219 |
https://mathoverflow.net/questions/420816 | 8 | Usually, having more comprehension axiom means the more you can prove. We wonder if the converse can be the case.
Is there a natural problem $\mathsf{P}$ so that $\mathsf{P}+\neg(\Gamma-\mathsf{ComprehensionAxiom})$ implies
$\mathsf{Q}$, but $\mathsf{P}$ does not implies $\mathsf{Q}$.
Also I'd like to exclude some trivial case such as $\mathsf{Q}$ is $\neg(\Gamma-\mathsf{ComprehensionAxiom})$. We can accept $\mathsf{Q}$ to be some Comprehension Axiom, but not the negation of them.
The implication is not necessary over $\mathsf{RCA}$ and $\Gamma$ can be any set of formulas.
| https://mathoverflow.net/users/74918 | Comprehension axiom that helps in the opposite direction | David Belanger's work is relevant.
The principle $\mathsf{WKL\_0}$ (= "Every infinite subtree of $2^{<\omega}$ has an infinite path") is not literally a comprehension principle, so its negation can be used as a choice of $\mathsf{Q}$. Belanger [showed](https://www.jstor.org/stable/43303769) that over $\mathsf{RCA\_0}$, the principle $(\star)\equiv$ "There is a complete theory with finitely many models and a single nonprincipal type" is equivalent to $\mathsf{ACA\_0\vee\neg WKL\_0}$. Consequently, if we take as our base theory $\mathsf{RCA\_0+(\star)}$, we get $\neg\mathsf{ACA\_0}\implies\mathsf{Q}$.
What if we *do* want to consider $\mathsf{WKL\_0}$ as a kind of comprehension principle (it's one of the "Big Five" after all), so that we can't set $\mathsf{Q}=\neg\mathsf{WKL\_0}$? Well, in this case another result of Belanger is relevant: that a rather complicated model-theoretic principle (see Theorem 2.23 [here](https://www.sciencedirect.com/science/article/pii/S0168007215000330?via%3Dihub)), which I'll refer to as $(\dagger)$ here, implies over $\mathsf{RCA\_0}$ the disjunction $\mathsf{WKL\_0\vee I\Sigma\_2}$. Consequently, over $\mathsf{RCA\_0}$ the negation of $\mathsf{WKL\_0}$ (which we're now considering a comprehension principle!), plus the principle $(\dagger)$, implies $\mathsf{I\Sigma\_2}$ - which is not itself a consequence of $\mathsf{RCA\_0}+(\dagger)$ so this is nontrivial.
| 10 | https://mathoverflow.net/users/8133 | 420818 | 171,223 |
https://mathoverflow.net/questions/420756 | 6 | I was wondering if it is well understood under what circumstances say three univariate polynomials $f(x),g(x),h(x)$ have a common root.
In this situation, I can see that the resultant of each pair must vanish but that only ensures that each pair has a common root. Is there a way to generate a finite set of polynomials in the coefficients of $f,g,h$ which tells you when all 3 share at least one common root?
Would be interested in an answer for the more general (more than 3 polynomials) case too.
EDIT: After thinking about it a tad more here is a possibly interesting observation. If you have $n$ polynomials of degree at most $n$ then you can write it as a linear system in $x,x^2,...,x^n$. Using determinants, minors, etc you will be able to get $n-1$ necessary and sufficient relations between the coefficients which tells you when the polynomials share a common root. I would suspect that means in general if you have $k$ polynomials it might be possible to give $k-1$ polynomials in the coefficients which will be necessary and sufficient conditions for having a common root.
| https://mathoverflow.net/users/164946 | When do multiple polynomials have a common root? | Assume that $f\_n$ is monic. Then for indeterminates $u\_1,\ldots,u\_{n-1}$, all coefficients of the polynomial $Res\_x(f\_n,u\_1f\_1+\ldots+u\_{n-1}f\_{n-1})\in k[u\_1,\ldots,u\_{n-1}]$ vanish if and only if $f\_1,\ldots,f\_n$ have a common root [expanding out this resultant then gives the list of polynomials].
This is how elimination theory works - if $f\_1(x\_1,\ldots,x\_m)=0,\ldots, f\_n(x\_1,\ldots,x\_m)=0$ set-theoretically define a variety $V$, $f\_n$ is monic in $x\_m$ (which one ensure by applying a Noether normalization coordinate change), and $\pi:\mathbb{A}^m\to \mathbb{A}^{m-1}$ is the projection away from the last coordinate, then $\pi(V)$ is a variety in $\mathbb{A}^{m-1}$ defined by the coefficients of $Res\_x(f\_n,u\_1f\_1+\ldots+u\_{n-1}f\_{n-1})$. Repeatedly eliminating variables like this eventually gives you a finite morphism surjecting $V$ to $\mathbb{A}^{\dim(V)}$.
| 5 | https://mathoverflow.net/users/3404 | 420821 | 171,224 |
https://mathoverflow.net/questions/420825 | 7 | I'll phrase this in terms of spectral AG, but I'm curious about the same question in the classical context.
We define a nonconnective spectral Deligne-Mumford stack to be a spectrally-ringed topos which admits a cover by the étale spectra of $E\_{\infty}$-rings. The definition of a nonconnective spectral *scheme*, on the other hand, is a spectrally-ringed *space* which admits a cover by the Zariski spectra of $E\_{\infty}$-rings. Suppose we instead look at spectrally-ringed *topoi* which are locally isomorphic to the Zariski topos of an $E\_{\infty}$-ring. Clearly, these generalize nonconnective spectral schemes; and, on the other hand, these objects admit a natural functor to nonconnective spectral DM-stacks, which should be a fully faithful embedding extending the inclusion of schemes as schematic stacks.
My question is this: are these objects actually more general than nonconnective spectral schemes, or is the underlying topos always generated by a topological space? If the former, how much more general are they? (For instance, a nonconnective spectral DM-stack is schematic iff it admits a cover by $(-1)$-truncated affine objects. Is there a similar characterization of these "generalized schematic stacks"?)
| https://mathoverflow.net/users/158123 | If we replace the spectrally ringed space in the definition of a spectral scheme with an arbitrary infinity-topos, what objects do we get? | Yes, they are more general. This is in fact already the case with ordinary rings. Let's call a classically-ringed $\infty$-topos which is locally the Zariski $\infty$-topos of an affine scheme an *$\infty$-scheme*. A classical scheme is then the same as a $0$-localic $\infty$-scheme. To construct an $\infty$-scheme which is not classical, let $F$ be any object in the Zariski $\infty$-topos $\mathrm{Shv}(X)$ of a classical scheme $X$. Then the slice $\infty$-topos $\mathrm{Shv}(X)\_{/F}$ with the restricted sheaf of rings is an $\infty$-scheme (it is covered by open subschemes of $X$), which is classical iff $F$ is $0$-truncated. This same construction works to define spectral $\infty$-schemes that are not $0$-localic.
A theorem of Lurie (Theorem 2.3.13 in [DAG V](https://www.math.ias.edu/%7Elurie/papers/DAG-V.pdf)) implies that every $\infty$-scheme is such a slice over a $1$-localic $\infty$-scheme. I do not know if there exist $1$-localic examples that are not slices over classical schemes. If one works with the étale topology, however, this theorem implies that every DM $\infty$-stack is a slice over a classical DM stack.
$\infty$-schemes can also be identified with the full subcategory of $\mathrm{Shv}\_{\mathrm{Zar}}(\mathrm{CRing}^\mathrm{op})$ consisting of those Zariski sheaves (of spaces) $F$ that admit an effective epimorphism from a coproduct of representable sheaves $R\_i$ such that each $R\_i\to F$ is "representable by slice $\infty$-topoi" (see Proposition 2.4.17(6) in loc. cit.). Unlike with classical schemes, however, such sheaves do not usually satisfy descent with respect to the étale topology, because higher Zariski and étale cohomology generally disagree.
**ETA:** This last remark implies that the natural functor from $\infty$-schemes to DM $\infty$-stacks is *not* fully faithful. If it were, then the functor of points of any $\infty$-scheme would be the same as the functor of points of the associated DM $\infty$-stack, which is an étale sheaf.
| 13 | https://mathoverflow.net/users/20233 | 420829 | 171,225 |
https://mathoverflow.net/questions/419672 | 3 | I am having trouble finding any results concerning real interpolation of vector-valued Sobolev spaces. Namely, I would like to know if a continuous embedding of the type,
$$
L^p(0,T;X\_1)\cap W^{1,p}(0,T;X\_0) \hookrightarrow \mathcal{C}([0,T]; (X\_0, X\_1)\_{\theta, p})
$$
can hold for some $\theta \in (0,1)$, where $X\_1$ and $X\_0$ are two Banach spaces that are compatible for (real) interpolation and $T > 0$.
Any reference or anything I could grab on to takle this will be greatly appreciated. Thank you very much.
Disclaimer : I am not very familiar with interpolation so I would prefer rather accessible sources if possible.
**EDIT** : I change the ambitions of my question which involved an inclusion inside a vector-valued fractional Sobolev space which seemed a little bit two complex to first tackle.
| https://mathoverflow.net/users/146986 | Real interpolation for vector-valued Sobolev spaces | The desired embedding is indeed correct for $\theta = 1-1/p$. This is a classical result in interpolation theory and the theory of evolution equations. See for example the book of Amann [2], Theorem III.4.10.2.$^1$
In fact, the real interpolation space $$X:= \bigl(X\_0,X\_1\bigr)\_{1-1/p,p}$$ can be *defined* (equivalently to several other methods) as the space of point evaluations of zero of functions in the intersection space. This is called the (Lions-Peetre) **trace method** which you will find in nearly any book on interpolation theory. (For instance in Bergh/Löfström [1], Corolllary 3.12.3) From there, one uses e.g. continuity of shift semigroups to obtain the desired embedding.
If, as quite often when dealing with the spaces in the question, $X\_1 \hookrightarrow X\_0$, then the embedding is true for all $0 \leq \theta \leq 1-1/p$ by factoring through $\theta=1-1/p$ and a natural embedding of interpolation spaces. ("Take more of the bigger space $X\_0$".) However, if you are content with $\theta < 1-1/p$, then you can in fact do better and even have Hölder-continuity in time of order $\alpha = 1-1/p-\theta$.
Since I also checked the previous version of your question, let me mention that results as asked there are indeed also known. A common name is *mixed derivative theorem*. (Which is unfortunate, since web searches will not be very effective.) But naturally they become much more involved considering noninteger differentiability scales. Right now I know that such results (and more or less rigorous proofs) are considered e.g. in another work of Amann [3] and by Denk and Kaip; maybe rather see the accessible version for $X\_0 = L^p$ in [4], Proposition 4.3. But there are surely more.
---
[1] *Bergh, Jöran; Löfström, Jörgen*, Interpolation spaces. An introduction, Grundlehren der mathematischen Wissenschaften. 223. Berlin-Heidelberg-New York: Springer-Verlag.
[2] *Amann, Herbert*, Linear and quasilinear parabolic problems. Vol. 1: Abstract linear theory, Monographs in Mathematics. 89. Basel: Birkhäuser.
[3] *Amann, Herbert*, Compact embeddings of vector-valued Sobolev and Besov spaces, Glas. Mat., III. Ser. 35, No. 1, 161-177 (2000).
[4] *Tolksdorf, Patrick*, [**On the $\mathrm {L}^p$-theory of the Navier-Stokes equations on three-dimensional bounded Lipschitz domains**](http://dx.doi.org/10.1007/s00208-018-1653-4), Math. Ann. 371, No. 1-2, 445-460 (2018).
$^1$ I am not quite sure whether this qualifies as an accessible source, since Amann's works are notoriously hard to read, but I hope this particular result should be accessible enough..
| 3 | https://mathoverflow.net/users/85906 | 420840 | 171,228 |
https://mathoverflow.net/questions/420837 | 2 | Suppose I have two random variables $X\_1$ and $X\_2$. $X\_1,X\_2$ are both sums of random variables, and I can find Chernoff bounds for both variables independently. That is to say, I have
$$Pr[\mid X\_i - \mathbb{E}X\_i \mid \geq \delta \mathbb{E}X\_i] \leq 2\exp(-\frac{\delta^2 \mathbb{E}X\_i}{3}) $$ for both $i=1,2$.
Now, I have a new variable $$Y = \frac{X\_1 - X\_2}{X\_1 + X\_2}$$ I know how I can get concentration bounds using the union bound for both the numerator and denominator. But is there a way that I can get some kind of bound for the fraction itself? Here is what I attempted.
First, get the bounds for the numerator. Then holding the lower and upper bounds of the numerator constant, try to get concentration bounds for the (lower(or upper) bound/denominator random variable) term, i.e., get lower bounds for $\frac{LB(X\_1-X\_2)}{X\_1+X\_2}$, where LB is taking the lower bound we get from the numerators concentration bound. Then try to find probability and lower bound for this new $\frac{constant}{X\_1+X\_2}$ type term. Similarly try to find probability and upper bound using the numerator upper bound term. But the calculation got messy and I couldn't finish it.
Is there a method for solving problems like this?
| https://mathoverflow.net/users/480998 | Chernoff style concentration bound for ratio of variables | $\newcommand{\de}{\delta}$You have the bounds
\begin{equation}
P(|X\_i-m\_i|\ge \de\_i m\_i]\le2e^{-\de\_i^2 m\_i/3}
\end{equation}
for $i=1,2$ and $\de\_i>0$, where
\begin{equation}
m\_i:=EX\_i.
\end{equation}
Clearly, these bounds are useful only if $m\_i>0$ for $i=1,2$, which will be henceforth assumed. Assume also that $0<\de\_i<1$ for $i=1,2$.
Then on the event
\begin{equation}
\begin{aligned}
A&:=\{|X\_1-m\_1|<\de\_1 m\_1,|X\_1-m\_1|<\de\_2 m\_2\} \\
&=\{m\_1(1-\de\_1)<X\_1<m\_1(1+\de\_1),m\_2(1-\de\_2)<X\_2<m\_2(1+\de\_2)\}
\end{aligned}
\end{equation}
we have $X\_1>0$ and $X\_2>0$, which implies that
\begin{equation}
Y = \frac{X\_1-X\_2}{X\_1+X\_2}
\end{equation}
is increasing in $X\_1$ and decreasing in $X\_2$, so that the event
\begin{equation}
B:=\Big\{\frac{m\_1(1-\de\_1)-m\_2(1+\de\_2)}{m\_1(1+\de\_1)+m\_2(1-\de\_2)} <Y<\frac{m\_1(1+\de\_1)-m\_2(1-\de\_2)}{m\_1(1-\de\_1)+m\_2(1+\de\_2)}\Big\}
\end{equation}
occurs. That is, $A\subseteq B$ and hence $1-P(B)\le1-P(A)$.
So, by the union bound, we get the concentration result:
\begin{equation}
1-P(B)\le 2e^{-\de\_1^2 m\_1/3}+2e^{-\de\_2^2 m\_2/3}.
\end{equation}
One can now (quasi-)optimize the choices of $\de\_1$ and $\de\_2$, depending on one's objectives.
| 1 | https://mathoverflow.net/users/36721 | 420851 | 171,230 |
https://mathoverflow.net/questions/420857 | 1 | Let $P\in\mathbb{F}\_p[T]$ (not supposed irreducible). All roots $\xi$ of $P$ have a certain order $k$ such that $\xi^k=1$.
**Question:** is it possible to know the order of the roots of the given polynomial $P$, or at least a upper bound of the order?
It is clear that if $k$ is an order of (a root of) $P$ then by the description of factors of cycltomic polynomials $\Phi\_k$ then
$$ \text{ord}\_k(p)\leq \deg(P) $$
where $\text{ord}\_k(p)$ is the order of $p$ in $(\mathbb{Z}/k\mathbb{Z})^\*$ and that basically
$$ \log\_p(k)\leq \text{ord}\_k(p) $$
because for $a$ to be the order of $p$ we must have $p^a\geq k$.
So we have
$$ \log\_p(k)\leq\deg(P) $$
and hence
$$ k\leq p^{\deg(P)} $$
This is a very big upper bound and for algorithms I would like to have a smaller bound or better a list of possible orders.
Thanks for your help!
| https://mathoverflow.net/users/34066 | Order of roots for a polynomial $P\in\mathbb{F}_p[T]$ | Let $d = \deg P$. The list of possible orders is simply the list of divisors of $p^d-1$.
Proof that every such order appears: Let $n$ divide $p^d-1$. Let $\alpha$ be an element of $\mathbb F\_{p^d}$ of order $n$, which exists since the multiplicative group of $\mathbb F\_{p^d}$ is cyclic. Then $\alpha$ generates a subfield $\mathbb F\_{p^e} \subseteq \mathbb F\_{p^d}$ for some divisor $e$ of $d$. The minimal polynomial of $\alpha$ has degree $e$, so the $d/e$th power of it does the trick.
Proof that these are the only such orders that appear. Let $P$ be such a polynomial, and let $e$ be the least natural number such that $k$ divides $p^e-1$. Then every element of $\overline{\mathbb F\_p}$ of order $k$ lies in $\mathbb F\_{p^e}$, and none of them lie in any smaller finite field, so they all have minimal polynomials of degree $e$. Since $P$ has roots only these elements, $P$ is a product of these minimal polynomials, so $\deg P$ is a multiple of $e$. Thus $p^{ d}-1$ is a multiple of $p^e-1$ and hence is a multiple of $k$.
So your upper bound is sharp, but despite this, there is a list of size $p^{ o(d)}$ (by bounds for the divisor function).
| 3 | https://mathoverflow.net/users/18060 | 420859 | 171,233 |
https://mathoverflow.net/questions/420839 | 6 | I was wandering if there was a book, thesis or some notes where Shelah's argument for
1. $\mathtt{ZF}+\mathtt{DC}+$"All sets of reals are Lebesgue measurable" is equiconsistent with $\mathtt{ZFC} + \exists \kappa$ inaccessible
2. $\mathtt{ZF}+\mathtt{DC}+$"All sets of reals have the Baire property" is equiconsistent with $\mathtt{ZFC}$
contained in [Can you take Solovay's inaccessible away?](http://dx.doi.org/10.1007/BF02760522) is explained in a newer and/or "more digestible" way. Is there?
Thanks!
| https://mathoverflow.net/users/141146 | Shelah's "Can you take Solovay's inaccessible away?" | Chapter 9.5 of the book by Bartoszynski-Judah presents Raisonnier's proof (answering your question 1):
Assume that $\aleph\_1$ is not inaccessible in $L$, hence a successor in $L$. So there is a real $x$ which knows that the $L$-predecessor is countable, hence $\aleph\_1^{L[x]}=\aleph\_1$, so $X:=\mathbb R \cap L[x]$ is uncountable.
Now assume that all $\Sigma^1\_2(x)$-sets are Lebesgue measurable. Then (this needs some work) the "Raisonier filter" $F\_X$ built from $X$, which is a $\bf \Sigma^1\_3$-set, is a rapid filter and therefore (again some work, known earlier) not measurable.
| 9 | https://mathoverflow.net/users/14915 | 420867 | 171,235 |
https://mathoverflow.net/questions/420870 | 4 | I am reading a very nice paper of Newton and Thorne, [*Symmetric power functoriality for holomorphic modular forms*](https://doi.org/10.1007/s10240-021-00127-3), and there is an argument concerning the (Zariski-closure of) image of certain $p$-adic Galois representations that I do not fully understand. In case it is helpful, I am talking about Lemma 3.5 of the paper, but I doubt that the context is too relevant — I think I have understood the necessary $p$-adic Hodge theory arguments, and the only thing left is something routine with algebraic groups.
Let $\rho: G\_{\mathbf{Q}\_p} \to \mathrm{GL}\_2(\overline{\mathbf{Q}}\_p)$ be a continuous irreducible representation. Then (essentially by definition of what a reductive group is), the Zariski closure of the image of $\rho$ (call it $H$) is a reductive subgroup of $\mathrm{GL}\_2$. From the context of the paper ($\rho$ comes from a classical modular form of weight $k \geq 2$), $\rho$ is Hodge–Tate with distinct Hodge–Tate weights. In particular, the Sen operator is regular semisimple, so a theorem of Sen implies that $H$ has Lie algebra base-changing to $\mathbf{C}\_p$ to something containing a regular semisimple element. This in turn implies that $H$ is NOT a finite group, as its Lie algebra is nontrivial (could be convenient if the approach ends up being to use some classification theorem for all algebraic subgroups of $\mathrm{GL}\_2$ but I was hoping to avoid that route and understand something more conceptual). It also implies that the maximal torus of $H$ is a regular torus in $\mathrm{GL}\_2$.
Somebody told me that we could do a $p$-adic Hodge theory argument to argue that $H$ is connected, but in fact since the modular form $\rho$ comes from could have $p$ in the level, there is no crystalline assumption to be had, and I don't think that we can actually do that.
Anyhow, it seems that using just this information ($H$ an infinite reductive subgroup of $\mathrm{GL}\_2$ whose maximal torus is regular in $\mathrm{GL}\_2$), it is supposed to follow that if $H$ does not contain $\mathrm{SL}\_2$, then it is contained in the normalizer of a maximal torus of $\mathrm{GL}\_2$. Perhaps I could get this out of the general classification of algebraic subgroups of $\mathrm{GL}\_2$ (by hopefully ruling out a bunch of cases due to not being reductive), which I don't know how to prove but was able to find on Google. Is this the way to go? Or is there a more conceptual way of proving what is used here? It would be nice, of course, to have something that generalizes to higher-dimensional representations.
| https://mathoverflow.net/users/165625 | Reductive subgroups of $\mathrm{GL}_2$ over an algebraically closed field of characteristic zero | Let $T\_H$ be a maximal torus in $H$.
Suppose that $H^\circ$ is not a torus. Then $T\_H$ has non-trivial roots on $\operatorname{Lie}(H)$, and the subgroup of $H$ generated by opposite root groups is $\operatorname{SL}\_2$.
Thus, if $H$ does not contain $\operatorname{SL}\_2$, then its identity component equals $T\_H$; and you have indicated that that torus is regular in $G = \operatorname{GL}\_2$, which I take to mean that its centraliser $T\_G \mathrel{:=} C\_G(T\_H)$ in $G$ is a maximal torus in $G$. Thus, $H$ normalises the maximal torus $T\_G = C\_G(H^\circ)$ in $G$.
| 6 | https://mathoverflow.net/users/2383 | 420875 | 171,237 |
https://mathoverflow.net/questions/420874 | 13 | Let $M$ be a Riemannian manifold and let $\Delta$ be its Laplacian operator. There is a large literature on a **spectral gap** for such a $\Delta$, that is, finding an interval $(0,c)$ which does not contain any eigenvalues of $\Delta$. Why are people interested in producing such spectral gaps? What interesting information do they give about the manifold? Are there interesting applications in the `real work', i.e. in physics?
| https://mathoverflow.net/users/438034 | Why are we interested in spectral gaps for Laplacian operators | A spectral gap gives information on geometry of the manifold via Cheeger's inequality, <https://en.wikipedia.org/wiki/Cheeger_constant> See also Buser's inequality discussed there. More directly, a spectral gap for the Laplacian yields exponential decay for the heat kernel and determines the asymptotic rate of decay. The heat kernel represents transition probabilities for Brownian motion on the manifold, so a spectral gap means the motion mixes rapidly (in the finite volume case) and dissipates quickly (in the infinite volume case).
<https://en.wikipedia.org/wiki/Heat_kernel>
| 7 | https://mathoverflow.net/users/7691 | 420879 | 171,239 |
https://mathoverflow.net/questions/420894 | 6 | [Bhargava 2021](https://arxiv.org/abs/2111.06507) proves van der Waerden's conjecture about Galois groups of random integer polynomials: over all $x^n + a\_{n-1} x^{n-1} + \cdots + a\_0 = 0$ with $a\_k \in \{-H, \ldots, H\}$, the number of polynomials $E\_n(H)$ with Galois group not equal to the full symmetric group $S\_n$ is $O(H^{n-1})$.
* First (and this is a very basic question), am I correct that the $O$ here is in terms of $H$ only, so that in detail we have $E\_n(H) < c\_n H^{n-1}$ for some constants $c\_n$ that depend on $n$?
* Second, is there a conjectured growth rate for $c\_n$?
From [Bhargava 2021](https://arxiv.org/abs/2111.06507) it seems like in the worst case $c\_n$ might have a factor related to the number of nonisomorphic groups of order $n$ (possibly with other factors), but presumably most of the count concentrates on fewer of the groups.
| https://mathoverflow.net/users/22930 | Is there a conjectured dependence on $n$ in van der Waerden's conjecture? | There are two different questions you could ask here.
One is the dependence on $n$ in the conjectured asymptotic best constant, i.e. the $c\_n$ such that we have $E\_n (H) < c\_n H^{n-1} + o(H^{n-1})$ with the little $o$ depending on $n$. This one has a precise answer, because we do not need to worry about all the groups that are proven or expected to occur with frequency $o(H^{n-1})$. So we just need to deal with the reducible polynomials, where it was computed by Chela that we can take
$$ c\_n = 2^n (\zeta(n-1) - 1/2) + 2 k\_n $$
where $k\_n$ is the volume of the region in $\mathbb R^{n-1}$ consisting of tuples $x\_1,\dots, x\_{n-1}$ with $$-1 \leq x\_1,\dots, x\_{n-1}, \sum\_{i=1}^{n-1} x\_i \leq 1 .$$
It is straightforward to check that this has growth rate proportional to $2^n$ in $n$.
The other is the dependence on $n$ of the conjectured best constant valid for all $H$. This is a somewhat annoying question because it will depend heavily on small values of $H$. For example, we must take $c\_n \geq 3^{n-1}$ to deal with the case where $H$ is $1$ (and the constant coefficient vanishes). It's possible that we can take $c\_n \leq 3^n$ - I don't see a reason against it.
| 10 | https://mathoverflow.net/users/18060 | 420895 | 171,242 |
https://mathoverflow.net/questions/420906 | 0 | Given two primes $p,q\in[T,2T]$, how many integers $m$ of size $O(T^{3/2+\epsilon})$ are there such that the residues $m\bmod p$ and $m\bmod q$ are both $O(polylog(T))$? **Looking for an answer**
Is it possible to construct any such in $polylog(T)$ time without integer programming? *Answered below*
| https://mathoverflow.net/users/10035 | Constructing an integer with small residues for two distinct primes in polynomial time | If such $m$ exists, then $m=up+a=vq+b$ for some $u,v\in O(T^{1/2+\epsilon})$ and $a,b\in O(\mathrm{polylog}(T))$. Then $up-vq=b-a$ and thus $\frac{p}q - \frac{v}u=\frac{b-a}{uq}$, implying that $\frac{v}u$ represents a rational approximation to $\frac{p}q$, which so good that it must be a convergent. So, it enough to compute a continuous fraction for $\frac{p}q$ and search for $\frac{v}u$ among its convergents.
| 3 | https://mathoverflow.net/users/7076 | 420907 | 171,245 |
https://mathoverflow.net/questions/420900 | 0 | Find the first derivative of a SUM of all elements of an INVERSE of a square matrix (whose elements are functions of $z$) at $z=1$ knowing that all of the matrix' elements evaluate to $1$ at $z=1$.
This is needed in Statistics to find the expected number of trials to generate one of several preselected patterns (such as 3 consecutive sixes when rolling a die, etc.)
I believe the answer is: $-$ the reciprocal of the SUM of all elements of the inverse of the matrix of first derivatives (evaluated at $z=1$) but I don't know how to prove it (I have verified it up to 5 by 5 - quite difficult to go any higher).
P.S. What is the second derivative?
| https://mathoverflow.net/users/141969 | Derivative involving a singular matrix | I first rephrase the question using formulas and then will provide a proof.
Start from the $n\times n$ matrix $A(z)$, such that $[A(1)]\_{ij}=1$ for all $i,j\in\{1,2,\ldots n\}$. The matrix $B(z)$ is the elementwise derivative, $B\_{ij}=dA\_{ij}/dz$.
Now the question which the OP asks is whether
$$\lim\_{z\rightarrow 1}\frac{d}{dz}\sum\_{i,j=1}^n [A^{-1}(z)]\_{ij}=-\left(\lim\_{z\rightarrow 1}\sum\_{i,j=1}^n[B^{-1}(z)]\_{ij}\right)^{-1}\qquad\qquad(\ast)$$
For a proof I decompose $A=(z-1)B+v^\top v$, where $v=(1,1,1,\ldots 1)^\top$. I apply the [Woodbury identity](https://en.wikipedia.org/wiki/Woodbury_matrix_identity),
$$A^{-1}=[(z-1)B+v^\top v]^{-1}=(z-1)^{-1}B^{-1}\left[1-v^\top \bigl(z-1+vB^{-1}v^\top\bigr)^{-1}vB^{-1}\right]$$
$$\Rightarrow\sum\_{i,j=1}^n [A^{-1}]\_{ij}=vA^{-1}v^\top=\frac{vB^{-1}v^\top}{vB^{-1}v^\top+z-1}$$
$$\Rightarrow\lim\_{z\rightarrow 1}\frac{d}{dz}\sum\_{i,j=1}^n [A^{-1}]\_{ij}=-\frac{1}{vB^{-1}v^\top}=-\left(\sum\_{i,j=1}^n[B^{-1}]\_{ij}\right)^{-1},$$
which is the equation ($\ast$).
---
The OP also asks about second derivatives. Define $C\_{ij}=d^2A\_{ij}/dz^2$ and decompose $A=(z-1)B+\tfrac{1}{2}(z-1)^2C+v^\top v$. The Woodbury identity gives
$$vA^{-1}v^\top=\frac{vX^{-1}v^\top}{vX^{-1}v^\top+z-1},$$
$$X=B+\tfrac{1}{2}(z-1)C\Rightarrow X^{-1}=B^{-1}-\tfrac{1}{2}(z-1)B^{-1}C B^{-1}+{\cal O}(z-1)^2.$$
Thus we obtain the second derivative from
$$\lim\_{z\rightarrow 1}\frac{d^2}{dz^2}\sum\_{i,j=1}^n [A^{-1}]\_{ij}=\frac{2-vB^{-1}C B^{-1}v^\top}{(vB^{-1}v^\top)^2}.$$
| 2 | https://mathoverflow.net/users/11260 | 420909 | 171,246 |
https://mathoverflow.net/questions/420896 | 31 | Do there exist integers $x,y,z$ such that
$$
xy(x+y)=7z^2 + 1 ?
$$
The motivation is simple. Together with Aubrey de Grey, we developed a computer program that incorporates all standard methods we know (Hasse principle, quadratic reciprocity, Vieta jumping, search for large solutions, etc.) to try to decide the solvability of Diophantine equations, and this equation is one of the nicest (if not the nicest) cubic equation that our program cannot solve.
| https://mathoverflow.net/users/89064 | Is equation $xy(x+y)=7z^2+1$ solvable in integers? | There is **no solution**.
It is clear that at least one of $x$ and $y$ is positive and that neither is divisible by 7. We can assume that $a := x > 0$. The equation implies that there are integers $X$, $Y$ such that
$$ X^2 - 7 a Y^2 = a (4 + a^3) $$
(with $X = a (a + 2y)$ and $Y = 2z$).
First consider the case that $a$ is odd. Then $4 + a^3$ is also odd (and positive), so we can consider the Jacobi symbol
$$ \left(\frac{7a}{4+a^3}\right) \,. $$
One of the two numbers involved is ${} \equiv 1 \bmod 4$, so
by quadratic reciprocity,
$$ \left(\frac{7a}{4+a^3}\right)
= \left(\frac{4+a^3}{7}\right) \left(\frac{4+a^3}{a}\right)
= \left(\frac{4+a^3}{7}\right) $$
($4 + a^3$ is a square mod $a$). Since $7 \nmid a$, we have
$4 + a^3 \equiv 3$ or $5 \bmod 7$, both of which are nonsquares
$\bmod 7$, so the symbol is $-1$. This implies that there is
an odd prime $p$ having odd exponent in $4 + a^3$ and such that
$7a$ is a quadratic nonresidue $\bmod p$. This gives a
contradiction (note that $p \nmid a$).
Now consider the case $a = 2b$ even; write $b = 2^{v\_2(b)} b'$.
Then we have that $4 + a^3 = 4 (1 + 2 b^3)$ and
$$ \left(\frac{7a}{1 + 2b^3}\right)
= \left(\frac{14b}{1 + 2b^3}\right)
= \left(\frac{2}{1 + 2b^3}\right)^{1+v\_2(b)}
\left(\frac{7b'}{1 + 2b^3}\right) \,. $$
If $b$ is odd, then this is
$$ \left(\frac{2}{1 + 2b^3}\right)
(-\left(\frac{-1}{b}\right))
\left(\frac{1 + 2b^3}{7}\right)
\left(\frac{1 + 2b^3}{b}\right) \,, $$
which is always $-1$ (the product of the first two factors is $1$;
then conclude similarly as above). We obtain again a contradiction.
Finally, if $b$ is even, then
$$ \left(\frac{2}{1 + 2b^3}\right)^{1+v\_2(b)}
\left(\frac{7b'}{1 + 2b^3}\right)
= \left(\frac{1 + 2b^3}{7}\right)
\left(\frac{1 + 2b^3}{b'}\right) = -1$$
again (the first symbol is $1$, and quadratic reciprocity holds
with the positive sign), and the result is the same.
---
Here is an alternative proof using the product formula for
the quadratic Hilbert symbol.
If $(a,y,z)$ is a solution (with $a > 0$), then for all
places $v$ of $\mathbb Q$, we must have
$(7a, a(4+a^3))\_v = 1$.
We can rewrite the symbol as follows.
$$ (7a, a(4+a^3))\_v
= (-7, a (4 + a^3))\_v (-a, a)\_v (-a, 4+a^3)\_v
= (-7, a(4 + a^3))\_v $$
(the last two symbols in the middle expression are $+1$).
So it follows that
$$ (-7, a)\_v = (-7, 4 + a^3)\_v \,.$$
When $v = \infty$, the symbols are $+1$, since $a > 0$.
When $v = 2$, the symbols are $+1$, since $-7$ is a
$2$-adic square.
When $v = p \ne 7$ is an odd prime, one of the symbols is
$+1$ (and therefore both are), since $a$ and $4 + a^3$
have no common odd prime factors.
Finally, when $v = 7$, the symbol on the right is
$$ (-7, 4 + a^3)\_7 = \left(\frac{4 + a^3}{7}\right) = -1 $$
as in the first proof.
Putting these together, we obtain a contradiction to the
product formula for the Hilbert symbol.
| 36 | https://mathoverflow.net/users/21146 | 420918 | 171,248 |
https://mathoverflow.net/questions/420899 | 6 | Let $f(x)\in K((x))$ be an algebraic formal Laurent series and let $P(x,y)\in K(x)[y]$ be its minimal polynomial. Is $P(x,y)$
always separable? An example of non separable polynomial comes
from Puiseux series: the polynomial $y^2-x$ has a double root
$y=\sqrt{x}$. So I wonder what happen if we restrict to Laurent series.
Thank you in advance.
| https://mathoverflow.net/users/122378 | Is the minimal polynomial of an algebraic formal Laurent series always separable? | Yes. Let $p=char(K)$ and $\alpha\in \overline{K(x)}\cap K((x))$ assumed to be inseparable over $K(x)$.
* Let $L= K^{1/p^\infty}$ which is perfect. If $\alpha$ is inseparable over $L(x)$ then $\alpha$'s monic $L(x)$-minimal polynomial is $$f(y)=g(y^p)=h(y)^p$$ with $h(y)\in L(x)^{1/p}[y]=L(x^{1/p})[y]$ so $$[L(x^{1/p},\alpha):L(x)] = p \deg(h) =[L(x,\alpha):L(x)]$$ ie. $x^{1/p}\in L(x,\alpha)$.
This gives that $x^{1/p}\in L((x))[\alpha]= L((x))$ which is a contradiction.
>
> Whence $\alpha$ is separable over $L(x)$ $\implies f(y)$ is separable.
>
>
>
* There is a finite extension $E/K$ such that $f(y)\in E[y]$. Take a basis $E=\bigoplus\_{j=1}^q b\_j K$ with $b\_1=1$. We get that
$$E(x)=\bigoplus\_{j=1}^q b\_j K(x),\qquad
E((x))=\bigoplus\_{j=1}^q b\_j K((x))$$ $$f(y)=\sum\_{j=1}^q b\_j f\_j(y), \qquad f\_j(y)\in K(x)[y]$$
$f(\alpha)=\sum\_{j=1}^q b\_j f\_j(\alpha) = 0$ in $E((x))$ gives that $$f\_1(\alpha)=0$$
$f\_1(y)\in K(x)[y]$ being monic and of degree $= \deg f$ it must be that $f(y)=f\_1(y)$.
$f\_1$ is in $K(x)[y]$, separable, irreducible, which proves that $\alpha$ is in fact separable over $K(x)$.
| 4 | https://mathoverflow.net/users/84768 | 420919 | 171,249 |
https://mathoverflow.net/questions/420887 | 5 | Given a multi-variable polynomial $F$, denote the number of monomials by $N(F)$. Take for instance, \begin{align\*}N(x(x+y)+(x+y)^2-(x-y)^2)=N(x^2+5xy)&=2 \qquad \text{and} \\
N((x+z)(x+y)^2)=N(x^3 + 2x^2y + x^2z + xy^2 + 2xyz + y^2z)&=6.\end{align\*}
Denote the symmetric group on $n$ letters $\{1,2,\dots,n\}$ by $\mathfrak{S}\_n$. Define the action of $\mathfrak{S}\_n$ on a function $F(x\_1,\dots,x\_n)$ in a natural way: given $w\in\mathfrak{S}\_n$, then $w\cdot F(x\_1,\dots,x\_n)=F(x\_{w(1)},\dots,x\_{w(n)})$. Introduce the (multi-variable) rational functions
$$G(\mathbf{x},\mathbf{z})=\prod\_{k=1}^n\frac{x\_1z\_1+x\_2z\_2+\cdots+x\_kz\_k}{x\_k-x\_{k+1}}.$$
Assuming that the symmetric groups act only on the $x$-variables, let's compute the polynomial
$$G\_{\mathfrak{S}\_{n+1}}=\sum\_{w\in\frak{S}\_{n+1}}w\cdot G.$$
I would like to ask:
>
> **QUESTION.** Let $C\_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Is there a combinatorial proof that $N(G\_{\mathfrak{S}\_{n+1}})=C\_n$?
>
>
>
| https://mathoverflow.net/users/66131 | Counting monomials and the Catalan numbers | Any monomial $P:=\prod x\_i^{c\_i}$ of degree $\sum c\_i=n$ maps to a non-zero constant after symmetrization
$$
P\to \Phi(P):=G\_{\mathfrak{S}\_{n+1}}\frac{P}{(x\_1-x\_2)(x\_2-x\_3)\ldots (x\_n-x\_{n+1})}.
$$
Indeed, $\Phi(P)$ is a constant by a degree consideration, to prove that this constant is non-zero you may use, for example, Theorem 2 [here](https://arxiv.org/pdf/1512.07136.pdf).
Thus any monomial $\prod (x\_iz\_i)^{c\_i}$ maps to a non-zero constant times $\prod z\_i^{c\_i}$.
Thus, you simply count the number of monomials which may arise, when you multiply the linear forms $\prod\_{k=1}^n(x\_1z\_1+x\_2z\_2+\cdots+x\_kz\_k)$. The only condition is that $c\_1+\ldots+c\_i\geqslant i$ for all $i$. Such sequences are indeed enumerated by Catalan numbers, a bijection with lattice paths below diagonal is straightforward.
| 9 | https://mathoverflow.net/users/4312 | 420923 | 171,252 |
https://mathoverflow.net/questions/420700 | 3 | Crosspost from Math.SE as I did not receive an answer there:
In Lang's book *Elliptic Functions*, he shows how to generate the ray class field with conductor $N$ of an imaginary quadratic number field $k$ using the $j$-invariant of an elliptic curve $A/\mathbb{C}$ with $\mathrm{End}(A)\cong\frak{o}\_\text{$k$}$ and the values of the Weber function at $N$-torsion points of $A$. Namely, Theorem 2 of Chapter 10 reads as follows:
>
> Let $A$ be an elliptic curve whose ring of endomorphisms is the ring of algebraic integers $\frak{o}\_\text{$k$}$ in an imaginary quadratic number field $k$, and $A$ is defined over $k(j\_A)$. Let $h$ be the Weber function on $A$, giving the quotient of $A$ by its group of automorphisms. Then $k(j\_A, h(A\_N))$ is the ray class field of $k$ with conductor $N$.
>
>
>
However, his proof starts out by saying "[l]et $K$ be the smallest Galois extension of $k$ containing $j\_A=j(\frak{a})$ and all coordinates $h(A\_N)$" and then he goes on to prove that $K$ is the ray class field of $k$ with conductor $N$. After that, he concludes that "[t]his proves Theorem 2", without ever mentioning the fact that we have not yet proved $K=k(j\_A, h(A\_N))$, i.e. we do not yet know that $k(j\_A, h(A\_N))$ is Galois. Is there an obvious reason I am missing here?
Notice that he does a similar thing when he proves that the Hilbert class field of $k$ is $k(j(\frak{a}))$ with $\frak{a}$ some fractional ideal of $k$: He starts by defining $K$ as the smallest Galois extension of $k$ containing all $j(\frak{a}\_\text{$i$})$, where the $\frak{a}\_\text{$i$}$ are a set of representatives for the ideal class group, proves that $K$ is the Hilbert class field of $k$ and then decides to be done. However, here I can conclude the argument: It was shown in the course of the proof that all $j(\frak{a}\_\text{$i$})$ are conjugate, hence $j(\frak{a})$ has at least degree $h\_k$ over $k$ and since this is also the degree of the Hilbert class field of $k$ over $k$, the Hilbert class field must already be $k(j(\frak{a}))$. Maybe something similar is possible for the ray class field?
| https://mathoverflow.net/users/480786 | Lang's proof concerning ray class fields of imaginary quadratic number fields | I have found out what's going on here and it is so trivial that I wonder why this did not occur to me earlier: The conclusion of the proof is that $K$ is the ray class field of $k$ modulo $N$ - but this means that $K$ is abelian over $k$, so the intermediate field $k(j\_A, h(A\_N))$ must be Galois over $k$ because all intermediate extensions of abelian extensions are Galois! Duh ...
| 6 | https://mathoverflow.net/users/480786 | 420937 | 171,258 |
https://mathoverflow.net/questions/420913 | 2 | Let $p$ be large prime in $[T,2T]$ where $T$ is a parameter.
1. Can we have an integer pair $a,b$ satisfying $ab\equiv c\bmod p$ such that $|c|$ is of size $O(\operatorname{polylog}(T))$ and $a,b$ are of size $O(T^{1/2 +\varepsilon})$?
2. If $1/2$ is impossible what is the best rational in exponent we can get?
3. Given $p$ how many such minimal generically positioned pairs are possible?
| https://mathoverflow.net/users/10035 | Small near-reciprocals | There is a nice article of Heath-Brown in the Mathematical Intelligencer, called Arithmetic applications of Kloosterman sums, where this problem is discussed under the heading, "An elementary problem." He derives exponent $3/4$ and remarks that it is open to improve on it. I am not aware of any subsequent improvements.
| 4 | https://mathoverflow.net/users/2627 | 420941 | 171,260 |
https://mathoverflow.net/questions/420081 | 5 | Let $G$ be a group and $F$ a field. I am particularly interested in the case where $G$ is a uniform lattice in a Lie group and $F=\mathbb{F}\_2$, or in finite groups $G$ where
$\operatorname{char} F$ divides
$|G|$, but the discussion applies to general $G$ and $F$.
Let $\rho:G\to \mathrm{GL}(V)$ be a representation of $G$ on a finite-dimensional $F$-vector space $V$. Letting $G$ act trivially on $F$, we may form the cohomology ring $H^\*(G,F)$, which acts via the cup product on the graded module $H^\*(G,V)$. In particular, the cup product induces a linear map
$$
p: H^1(G,F)\otimes\_F H^1(G,V)\to H^2(G,V).
$$
My question is whether there are $G$ and $V$ for which:
* $p$ is injective,
* $\dim H^1(G,V) \geq 1$,
* $\dim H^1(G,F)\geq 2$.
More generally, is there a recipe for constructing such examples? I am not aware of even a single example of this kind.
The injectivity of $p$ is the real issue. For example, if we take $V=F$, then the cup product $\cup :H^1(G,F)\times H^1(G,F)\to H^2(G,F)$ is well-known
to be anti-symmetric, which means that
$\dim \ker p\geq \frac{1}{2}\dim H^1(G,F)(\dim H^1(G,F)-1)$ if the characteristic of $F$ is $2$, or $\dim \ker p\geq \frac{1}{2}\dim H^1(G,F)(\dim H^1(G,F)+1)$ if $F$ is not of characteristic $2$, so $\ker p$ would not be trivial when $\dim H^1(G,F)\geq 2$.
(This special example is studied in a paper of [Hillman](https://www.cambridge.org/core/journals/bulletin-of-the-australian-mathematical-society/article/kernel-of-the-cup-product/93B684246E1E69E5E75A002CE8B4EC3D).)
| https://mathoverflow.net/users/86006 | Examples of a group $G$ and an $F$-representation $V$ where $\cup:H^1(G,F)\otimes H^1(G,V)\to H^2(G,V)$ is injective | $\newcommand{\bZ}{\mathbb{Z}}$Let $G$ be a finite group of order divisible by $p:=\mathrm{char}\, F$ such that $\dim\_F \mathrm{Hom}(G,F)\geq 2$ (e.g. $G=\bZ/p\times\bZ/p$). Take $V$ to be the kernel of the augmentation map $e(\sum a\_g\cdot g)=\sum a\_g$ from $F[G]$ to $F$. Since $F[G]$ is an injective $G$-module, the short exact sequence $0\to V\to F[G]\xrightarrow{e} F\to 0$ yields isomorphisms $H^i(G, F)\simeq H^{i+1}(G,V)$ for $i\geq 0$ (this is automatic for $i>0$ and for $i=0$ follows from the fact that $e:F[G]^G\to F$ is the zero map because the order of $G$ is divisible by $p$).
These isomorphisms are compatible with the cup-product in the sense that the following diagram commutes for every $i$ (this follows from the associativity of cup-product using the observation that the isomorphism $H^i(G,F)\to H^{i+1}(G,V)$ is obtained by cupping with a class in $H^1(G,V)$)
$\require{AMScd}$
\begin{CD}
H^1(G,F)\otimes H^{i+1}(G,V) @>\cup >> H^{i+2}(G,V)\\
@| @|\\
H^1(G,F)\otimes H^{i}(G,F) @>>\cup > H^{i+1}(G,F)
\end{CD}
For $i=0$ the bottom map is obviously an isomorphism, hence the top map is an isomorphism, as desired.
| 3 | https://mathoverflow.net/users/39304 | 420945 | 171,262 |
https://mathoverflow.net/questions/420927 | 3 | Let $X\_0, X\_1, X\_2, \ldots$ be a sequence of i.i.d. real-valued random variables on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with continuous CDF $F(x)$ and define a sequence of empirical CDFs $F\_n(x) = \frac{1}{n} \sum\_{i = 1}^n \mathbf{1}\_{\{X\_i \leq x\}}(x)$ ($\mathbf{1}\_A$ is an indicator function). It is well known that
$$\sup\_{x \in \mathbb{R}} \left|F\_n(x) - F(x)\right| \to 0 \text{ a.s.}$$
(This is the fundamental theorem of mathematical statistics, or FTMS, supposedly.) Are we then safe to say that
$$F\_n(X\_0) \to U \text{ in law}$$
where $U$ follows a standard uniform distribution? Or could we make the even stronger claim that we can redefine the random variables into a probability space such that the convergence happens almost surely?
It seems like this would be the case. FTMS suggests we can write $F\_n(X\_0) = F(X\_0) + o(1)$ (holding almost surely) and the distribution of $F(X\_0)$ is a standard uniform distribution. While $F\_n$ is affected by the asymptotics, $X\_0$ is not. The probability $X\_0$ takes a value that is not covered by the convergence is zero thanks to FTMS being a uniform convergence result. Is it that easy? Am I overthinking this?
| https://mathoverflow.net/users/113992 | For a random sequence $X_0, X_1, X_2, \ldots$ and $F_n$ the empirical CDF, does $F_n(X_0)$ converge to a uniform random variable? | $\newcommand\Om\Omega\newcommand\om\omega\newcommand\R{\mathbb R}$This is indeed straightforward. Spelling out
$$\sup\_{x\in\R}|F\_n(x)-F(x)|\to0 \text{ a.s.}, $$
we see that for some subset $N$ of $\Om$ of outer probability $0$ and all $\om\in\Om\_0:=\Om\setminus N$ we have
$$\sup\_{x\in\R}\Big|\frac1n\sum\_{j=1}^n 1(X\_j(\om)\le x)-F(x)\Big|\to0$$
and hence
$$\frac1n\sum\_{j=1}^n 1(X\_j(\om)\le X\_0(\om))\to F(X\_0(\om)).$$
So, indeed we have
$$F\_n(X\_0)\to F(X\_0)$$
a.s. and hence in law. Also, if $F$ is continuous, then the random variable $F(X\_0)$ has the standard uniform distribution.
| 2 | https://mathoverflow.net/users/36721 | 420951 | 171,264 |
https://mathoverflow.net/questions/420947 | 1 | I am wondering can we say something about the cover time $T$ for a box, eg. $[-n,n]^2\cap \mathbb{Z}^d$, by the simple symmetric random walk on $\mathbb{Z}^2$ starting from zero?
For example, the expected time, the variance, or some scale $a\_n$ such that $T\leq a\_n$ with high probability?
(I tried search on web but found nothing. Maybe I am not searching with the correct things, but...)
| https://mathoverflow.net/users/174600 | Cover time of a box by SRW in $\mathbb{Z}^2$? | "Cover times for Brownian motion
and random walks in two dimensions": Annals Math 160 (2004), 433-464,
By Dembo, Peres, Rosen, Zeitouni.
See Theorem 1.4
| 5 | https://mathoverflow.net/users/35520 | 420954 | 171,265 |
https://mathoverflow.net/questions/420939 | 0 | how to show that non-decision problem is NP-hard?
So far I could find out that problems which are NP-hard do not have to be decision problems.
But how to show a non-decision problem is NP-hard?
Is it possible to show problem L is NP-hard by showing that another NP-hard problem B is reducible to L?
Does B have to be NP-hard or NP-complete and why is it like this?
| https://mathoverflow.net/users/480697 | NP-hardness of non-decision problems | A definition of NP-hard is: if the problem can be solved in polynomial time, then every problem in NP can be solved in polynomial time. This definition works for function problems.
Example: Consider the optimization problem "find the length of the shortest path in an instance of the traveling salesman problem." This is an optimization problem, not a decision problem. It is NP-hard because, in particular, if we can solve it then we can use that solution to solve the decision version of TSP, which is NP-complete. And by definition of NP-complete, if we can solve decision-TSP in polynomial time then we can solve any problem in NP in polynomial time.
Usually the reductions used here are "Cook" reductions which work as follows: We assume we have an oracle for problem B, and show a polynomial-time algorithm (where an oracle call happens in one time step) for solving A. This reduces A to B.
| 3 | https://mathoverflow.net/users/29697 | 420957 | 171,266 |
https://mathoverflow.net/questions/420949 | 6 | Context:
--------
In celebrating the centenary of Ramanujan's birth, Freeman Dyson presented the following career advice for talented young physicists [1]:
>
> My dream is that I will live to see the day when our young physicists, struggling to bring the predictions of superstring theory into correspondence with the facts of nature, will be led to enlarge their analytic machinery to include not only theta-functions but mock theta-functions … But before this can happen, the purely mathematical exploration of the mock- modular forms and their mock-symmetries must be carried a great deal further. —Freeman Dyson
>
>
>
Question:
---------
Was Freeman Dyson guided by physical intuitions that could have convinced top-notch quantum field theorists of his generation, such as Richard Feynman? Though I am aware that Freeman Dyson and Richard Feynman collaborated on Feynman's approach to quantum field theory, the precursor to string theory, I doubt that Feynman would have advanced the hypothesis that Ramanujan's work had any important consequences for theoretical physics.
Complementary insights:
-----------------------
In parallel, I wonder whether it may not be equally sensible to reconcile quantum theory with the facts of probabilistic number theory where probabilistic events are of a deterministic and frequentist nature. Upon closer inspection, this would be a complementary effort but I don't know of a systematic research program aimed at this particular objective although a large number of physicists appear to have a strong interest in the pair correlation conjecture which emerged from a tea-time discussion between Freeman Dyson and Hugh Montgomery.
These are related observations, which may be relevant for a couple reasons: (1) The theory of modular forms potentially enters mathematical physics via the analysis of the Pair-Correlation conjecture. (2) By John Bell's own admission, his 1964 theorem known as Bell's theorem was motivated by the super-deterministic theory proposed by De Broglie and Bohm.
Furthermore, I suspect that Erdős is often quoted saying:
>
> God may not play dice with the universe, but something strange is
> going on with the prime numbers.
>
>
>
because all mathematical systems may be constructed from Peano Arithmetic, and the prime numbers are the atomic units of the integers, so the distribution of the prime numbers may be viewed as fundamental scientific data. Based on a recent discussion with Max Tegmark [7], who believes that a physicist can only understand the mathematical relations between things, this perspective is worth consideration if we assume that the mathematical structure of the Universe emerged from an information-theoretic singularity(i.e. Big Bang Cosmology).
**Note:** Contrary to those who are voting to close this question, I believe that if there are fundamental physical insights which motivated Freeman Dyson's hypothesis then this question is of interest to the MathOverflow community.
References:
-----------
1. Jeffrey A. Harvey. Ramanujan’s influence on string theory, black holes and moonshine. 2019.
2. Hardy, G. H.; Ramanujan, S. “The normal number of prime factors of a number n”, Quarterly Journal of Mathematics. 1917.
3. Erdős, Paul; Kac, Mark. “The Gaussian law of errors in the theory of additive number theoretic functions”. American Journal of Mathematics. 1940.
4. Montgomery, Hugh L. "The pair correlation of zeros of the zeta function", Analytic number theory, Proc. Sympos. Pure Math. 1973.
5. Bell, J.S.“On the Einstein-Podolsky-Rosen paradox,” Physics. 1964.
6. Tegmark, Max. "The Mathematical Universe". Foundations of Physics. Arxiv. 2008.
7. Email discussion with Max Tegmark on tabletop experiments for the Mathematical Universe Hypothesis via Probabilistic Number Theory. Dec 18 2021.
| https://mathoverflow.net/users/56328 | Freeman Dyson's approach to string theory | Dyson's [A walk through Ramanujan's garden](https://www.semanticscholar.org/paper/A-walk-through-Ramanujan%E2%80%99s-garden-Dyson/eb6d4ec50b6666a07521157c22c14c81380ce424) gives the background of this comment: He explains that the "seeds from Ramanujan's garden have been
blowing on the wind and have been sprouting all over the landscape. Some of the seeds even blew over into physics." He then writes that he received a preprint from a superstring theorist entitled *Atkin-Lehner symmetry*, in which modular forms entered physics in ways that mathematicians never dreamt of, and concludes that "Perhaps we may one day see a preprint written by a physicist with the title *Mock Atkin-Lehner Symmetry*."
Dyson also indicates in the same text that he knows little about superstring theory, so my answer to the question: *"Was Freeman Dyson guided by physical intuitions"* is: No, he was guided by his experience that fundamental math often makes it into physics in unexpected ways.
| 22 | https://mathoverflow.net/users/11260 | 420958 | 171,267 |
https://mathoverflow.net/questions/420959 | 7 | This question was asked at the french ENS oral examination. I do not really know how to approach it. I think the answers no.
What I've gathered so far :
Lets call $T$ the subset of $\mathbb{R}^2$ in the title (for obvious reasons). If the union exists, by Baire's theorem it must be uncountable. Let $E$ be the set of maps $T \to \mathbb{R}^2$ of which the images disjointly cover $\mathbb{R}^2$.
Since the domain is compact every map is uniformly continuous : by uncountability, for all $\epsilon > 0$ there is an $\eta$ such that an uncountably infinite number of maps $f$ verify the property $|x - y| \leq \eta \implies |f(x) - f(y)| < \epsilon$.
By uncountability, I also think an uncountably infinite number of the maps above should also have their image in a well chosen compact region of the plane, denoted $K$.
Let $E'$ be an uncoutably infinite subset of $E$ where all the maps verify the two properties ("uniform" uniform continuity and image in a given compact set $K$). For any finite subset of $T$, by using successive extractions I should be able to find a sequence of maps $(f\_n)$ in $E'$ such that for any point $x$ in this finite subset the images $f\_n(x)$ converges to a point $y \in K$. By using uniform uniform continuity I hoped to find that the maps $f\_n$ would be arbitrarily close to one another and necessarily cross.
However, the more I think about this approach the less likely I think it is to work.
Would anyone have an idea ?
| https://mathoverflow.net/users/466576 | Can $\mathbb{R}^2$ be covered by disjoint sets homeomorphic to the union of the segments $[(0,0), (0,1)], [(0,0), (1,1)], [(0,0), (1,0)]$? | Such sets are called *triods*. R. L. Moore (Concerning triods in the plane and the junction points of plane continua, Proceedings of the National Academy of Sciences USA, vol. 14, 1928, pp. 85-88) proved that every set of pairwise disjoint triods in the plane is countable.
<https://www.pnas.org/doi/abs/10.1073/pnas.14.1.85>
| 7 | https://mathoverflow.net/users/43266 | 420965 | 171,269 |
https://mathoverflow.net/questions/420914 | 5 | Taken from Math Stack Exchange.
>
> Let $\mathcal{F}$ be a set of $\mathcal{L}\_\in$-formulae, $\kappa$ be a cardinal and $A \subset \textrm{Ord}$. Then, $\kappa$ is called $\mathcal{F}$-Mahlo if $A \cap \kappa$ intersects every club definable in $H\_\kappa$ by a formula $\varphi \in \mathcal{F}$. $\kappa$ is $\mathcal{F}$-Mahlo if it is $\mathcal{F}$-Mahlo onto $\textrm{Reg}$.
>
>
>
This has some interesting properties. For example, if we let $\Pi$ denote the standard Levy hierarchy, then every $\Pi\_1$-Mahlo cardinal is a weakly inaccessible limit of weakly inaccessible cardinals, i.e. weakly 2-inaccessible. Now, a well-known result is that $\kappa$ is $\Pi^1\_n$-indescribable iff it is $\Sigma^1\_{n+1}$-indescribable (a similar thing applies to reflecting ordinals). Does this apply to Mahloness? In other words, is $\kappa$ $\Pi\_n$-Mahlo iff it is $\Sigma\_{n+1}$-Mahlo? Also, does any kind of similar equivalence apply to $\Delta\_n$-Mahloness?
| https://mathoverflow.net/users/473200 | Equivalences of $\mathcal{F}$-Mahloness | The paper ["Small Definably-large Cardinals" by Roger Bosch](https://doi.org/10.1007/3-7643-7692-9_3) proves that an inaccessible cardinal is $\Sigma\_{n+1}$-Mahlo if and only if it is $\Pi\_n$-Mahlo except for $n=1$ (I'm referring to the boldface hierarchy, not the lightface hierarchy which is less relevant to your question) and that an inaccessible cardinal is $\Sigma\_2$-Mahlo if and only if it is $\Delta\_2$-Mahlo. Whether $\Pi\_1$-Mahlo cardinals are always $\Delta\_2$-Mahlo (and thus $\Sigma\_2$-Mahlo) an open problem as far as I know.
| 6 | https://mathoverflow.net/users/352898 | 420971 | 171,270 |
https://mathoverflow.net/questions/420822 | 4 | A lower bound of the number of conjugacy classes in the automorphism group of a simple Lie algebra $\mathfrak{s}$, of finite dimension over an arbitrary field $\mathbb{F}$, can be the size of the image of the automorphism group by the trace.
I think this map $\operatorname{tr}:\operatorname{Aut}(\mathfrak{s})\longrightarrow\mathbb{F}$ must be surjective. However, I have no clue. (For splitable Lie algebras, there is an explicit way).
| https://mathoverflow.net/users/235856 | Conjugacy classes in the automorphism group of a simple Lie algebra | 1. If $\mathfrak{s}$ is $K$-anisotropic, where $K$ is a real or $p$-adic field (this is equivalent to $\mathfrak{s}$ not containing $\mathfrak{sl}\_2$, and also to the corresponding group be compact), then the trace is bounded, hence non-surjective.
2. Suppose that $\mathrm{Aut}(\mathfrak{s})^0$ is Zariski-dense in the underlying algebraic group (this is automatic if $K$ is perfect, by Rosenlicht's theorem). In this case, by Zariski density, if $\mathrm{Aut}(\mathfrak{s})^0$ has infinitely many traces (resp. characteristic polynomials), then the same holds over the algebraic closure. So we can reduce (in this case) to the algebraically closed case.
3. In the algebraically closed case, if the number of characteristic polynomials of $\mathrm{Aut}(\mathfrak{s})^0$ is finite, it is (by connectedness) reduced to one, and hence it follows that every element in $\mathrm{Aut}(\mathfrak{s})^0$ is unipotent. In characteristic zero, this is not possible. I'm not sure about the modular case.
The same conclusion holds for the number of traces, although the characteristic zero is then used in a stronger way.
4. If $\mathfrak{s}$ is the Lie algebra of a simple algebraic group, the restrictions on the characteristic should be dropped (still needing Rosenlicht), to infer that the number of characteristic polynomials is infinite.
| 4 | https://mathoverflow.net/users/14094 | 420976 | 171,272 |
https://mathoverflow.net/questions/420993 | 2 | Let us consider the space $M\_n(\mathbb{C})$. By a unitary matrix $U=(u\_{ij})$ we mean that $U^{-1}=(\overline{u\_{ji}})$.
>
> Q. Let $U$ be a unitary matrix. I am looking for the pairs of matrices $(D,A)$ satisfying the following conditions: (1) $D$ is a diagonal matrix in $M\_n(\mathbb{C})$, (2) $UAU^{-1}=D$, (3) $U$ and $A$ have the same eigen-vectors. In general case, does there exist such a pair? How can we find a formula to produce all such of the pairs?
>
>
>
| https://mathoverflow.net/users/84390 | On the eigen vectors of a diagonalizable matrix | I [mentioned](https://mathoverflow.net/questions/420993/on-the-eigen-vectors-of-a-diagonalizable-matrix#comment1081449_420993) the silly example $U = A = D$ offhand, but, in fact, it's essentially the only example. In general, note that the columns of $U^{-1}$ are the eigenvectors of $A$. So we're asking for, at least, a unitary matrix $U$ that admits the columns of $U^{-1}$ as eigenvectors. But $U U^{-1} = I$, so this means that $U$ must be diagonal.
| 2 | https://mathoverflow.net/users/2383 | 420994 | 171,277 |
https://mathoverflow.net/questions/421012 | -2 | Which extension of $\sf ZFC$ prove that
$$ {\sf ZFC} \not \vdash \exists x \, ( \operatorname {CH}(x) \land x \neq \emptyset \land x \neq 1)$$
Where $\operatorname {CH}(x) \iff \neg \exists \kappa \, (|x| < \kappa < |P(x)|) $
In English: $\sf ZFC$ doesn't prove the continuum hypothesis of any set other than the empty set and the singleton of the empty set.
I know that "$\sf ZFC + CH$ fails everywhere", can prove that, which is too strong. But I'm asking if this can be proved in a much less strong theory, some theory whose consistency strength is just a little above $\sf ZFC$.
| https://mathoverflow.net/users/95347 | Which extension of ZFC proves that ZFC can only prove CH satisfied by the first two sets? | The statement you propose is equivalent to consistency of GCH failing for all (infinite) cardinals. This problem is well-studied. Of course the theory ZFC+Con(ZFC+"GCH fails everywhere") gives an answer to your problem, but this is completely tautological. It is standard in such situations to calibrate the consistency strength against the large cardinal hierarchy: as is discussed in [this answer](https://math.stackexchange.com/a/3186971/127263), the precise equiconsistency statement is not easy to state in ZFC, but it amounts to existence of stationarily many measurable cardinals of high Mitchell rank. It is between measurable cardinals and strong cardinals.
Let me also mention that your claim in the last paragraph is false: ZFC+"GCH fails everywhere" doesn't prove Con(ZFC+"GCH fails everywhere"), by Godel's incompleteness.
| 5 | https://mathoverflow.net/users/30186 | 421038 | 171,288 |
https://mathoverflow.net/questions/420999 | 6 | The simplest case of the Fundamental Theorem of Projective Geometry states that, if $f: \mathbb{R}^2\to\mathbb{R}^2$ is a bijection that preserves lines – in the sense that if $L\subseteq\mathbb{R}^2$ is a line then so is $f[L]$ – then $f$ is an affine transformation. [1]
The condition that $f$ be a bijection is stronger than necessary, and it is sufficient to ask for $f$ to be injective. (*Proof*: let $L\_1$ and $L\_2$ be two parallel lines in $\mathbb{R}^2$; then $f[L\_1]$ and $f[L\_2]$ are also parallel lines: they are lines since $f$ preserves lines, and parallel since $f$ is injective. Take any point $p\in\mathbb{R}^2$, and take a line $L$ through $p$ not parallel to $f[L\_1]$ and $f[L\_2]$. This line intersects $f[L\_1]$ in some point, say $f(p\_1)$, and it intersects $f[L\_2]$ in some point, say $f(p\_2)$. The image under $f$ of the line through $p\_1$ and $p\_2$ must be the line through $f(p\_1)$ and $f(p\_2)$, which contains $p$, and therefore $p$ is in the image of $f$.)
**My question is**: can the condition be weakened further? For example, if we assume only that $f: \mathbb{R}^2\to\mathbb{R}^2$ preserves lines, must $f$ be an affine transformation?
---
**Added later**: Thanks to @Wojowu in the comments for explaining how to construct a function $\mathbb{R}^2\to\mathbb{R}$ that maps every line to the whole of $\mathbb{R}$, which answers the second question I asked.
So: **what if we also insist that there be three non-collinear points in the image of $f$?** It follows that there must be two distinct lines in the image of $f$, and then the argument above (more or less) shows that $f$ is surjective. But must it be *in*jective?
---
[1] For a proof of the general case see e.g. Andrew Putman, [The fundamental theorem of projective
geometry](https://www3.nd.edu/%7Eandyp/notes/FunThmProjGeom.pdf).
| https://mathoverflow.net/users/8217 | Functions $\mathbb{R}^2\to\mathbb{R}^2$ that preserve lines | Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a function that maps lines to lines, and suppose that there are three non-collinear points in the image of $f$.
**Lemma 1:** If $f[\ell\_1]$ and $f[\ell\_2]$ are distinct parallel lines, then $\ell\_1$ and $\ell\_2$ are distinct parallel lines.
**Proof:** Suppose otherwise. Then $\ell\_1$ and $\ell\_2$ intersect at a point $x$, and $f(x)$ must lie on both $f[\ell\_1]$ and $f[\ell\_2]$; this is clearly impossible since the lines are parallel.
**Lemma 2:** Let $a, b, c, d \in \mathbb{R}^2$ be four points such that $f(a), f(b), f(c), f(d)$ form the vertices of a non-degenerate parallelogram (in that order). Then $a, b, c, d$ also form the vertices of a non-degenerate parallelogram in the same order, so in particular $a + c = b + d$.
**Proof:** Apply Lemma 1 twice, once to each pair of opposite edges.
**Lemma 3:** If four points in the image of $f$ form the vertices of a non-degenerate parallelogram, then they each have one preimage under $f$.
**Proof:** Suppose otherwise. Then there are two points $a, a' \in \mathbb{R}^2$ with $f(a) = f(a')$; let $f(b), f(c), f(d)$ be the other three vertices of a non-degenerate parallelogram (in that order). Then we have $a + d = b + c$ and $a' + d = b + c$ by applying Lemma 2, whence it follows that $a = a'$.
**Lemma 4:** $f$ is surjective.
**Proof:** Let $x,y,z$ be three points such that $f(x), f(y), f(z)$ are non-collinear points. It follows that $x,y,z$ are also non-collinear. The line through $x,y$ maps to the line through $f(x),f(y)$. Given any point $P$ in the plane of $f(x), f(y), f(z)$, we can take the line through $f(z)$ and $P$ and let it intersect the line through $f(x)$ and $f(y)$ at a point $Q$, which is in the image of $f$ and is therefore $f(w)$ for some point $w$. The line through $w$ and $z$ maps to the line through $f(z)$ and $f(w)$, which contains $P$, so $P$ is indeed in the image of $f$.
**Lemma 5:** $f$ is injective.
**Proof:** Since $f$ is surjective by Lemma 4, every point $f(a)$ in the image of $f$ is the vertex of a non-degenerate parallelogram with its other three vertices in the image of $f$. By Lemma 3, there cannot be a point $a' \ne a$ with $f(a) = f(a')$. The result follows.
**Corollary:** $f$ is a bijection, so is an affine transformation by the claim at the beginning of the OP.
Can we weaken the condition even further? In particular, if we weaken 'maps lines to lines' to 'preserves collinearity' (the difference being that a line can map to a proper subset of a line), then does the result still hold? Lemma 2 holds (and so does Lemma 3), so the difficult part is proving surjectivity.
| 3 | https://mathoverflow.net/users/39521 | 421042 | 171,291 |
https://mathoverflow.net/questions/421006 | 2 | One of the most intriguing things I've read about over the last few years is [Diaconescu's theorem](https://en.wikipedia.org/wiki/Diaconescu%27s_theorem), which says that, in some forms of constructivist/intuitionistic set theory, even if the law of the excluded middle is not presupposed in the background logic, the LEM is still a theorem under a set-theoretic principle, the axiom of choice. Hence, to remain constructive, that kind of set theory has to reject the AC, which is not necessarily appealing to constructivists, at least one of whom (Bishop) has claimed that the AC is constructively plausible, as he said something like "a choice is presupposed by the very concept of existence".
Now, as far as I can tell, the law of noncontradiction is not quite an axiom in normal modern logic, but instead a theorem of the argument from explosions. I'm wondering, though, **if there is a set-theoretic, or at least broadly "mathematical," principle that does the same for the LNC as the AC does for the LEM?** This is all I've come up with so far:
>
> Let $$\mathfrak{j}(S) = a$$ be the *justification function* on sentences *S*, where = the *justification value* (c.f. Fregean truth values) of *S*. Assume the following:
>
>
> 1. $$\mathfrak{j}(A \land B) = \mathfrak{j}(A) + \mathfrak{j}(B)$$
> 2. $$(\mathfrak{j}(A) > 0) \rightarrow (\mathfrak{j}(\neg A) < 0)$$ also $$(\mathfrak{j}(A) = x) \rightarrow (\mathfrak{j}(\neg A) = -x)$$
> 3. $$\Box(\mathfrak{j}(B) = 0) \rightarrow \neg B$$
>
>
>
>
> Then $$\mathfrak{j}(A \land \neg A) = \mathfrak{j}(A) + \mathfrak{j}(\neg A) = x + (-x) = x - x = 0$$
>
>
>
> >
> > ["Modalized"] Then $$\Box(\mathfrak{j}(A \land \neg A) = 0)$$ i.e. the LNC is (via (3)) true. QED
> >
> >
> >
>
>
>
When using (3), one must proceed with caution. Taking the concept of justification in play to be stated in terms of something like abstract [evidentialism](https://iep.utm.edu/evidenti/), (3) is perilously close to [Fitch's paradox of knowability](https://plato.stanford.edu/entries/fitch-paradox/), which says that if all truths *can* be known, then all of them *are*, wherefore it follows that there apparently must be unknowable truths (or else we would be [hyperlogically omniscient](https://plato.stanford.edu/entries/logic-epistemic/#LogiOmni)). So for now, I would recommend limiting the domain of the justification function as such, to a closed-off set-theoretic or category-theoretic mathematical universe.
Another problem I'm having with how I've formulated the argument is that (2) seems to involve double-negation introduction, and anyway the whole idea has it that negative numbers are more elementary than they are in normal set theory (e.g. if we style the logic along the lines of a Boolean algebra augmented by a negative unit of antijustification value). IIRC, constructivists/intuitionists reject DN elimination, but not necessarily DN introduction. So if I was trying to maneuver around the ~LEM crowd, here, maybe I pulled the maneuver off with respect to (2), as it resembles DNI more than DNE. Still, I've never felt that DNI/E were far off from the LNC, either; I didn't see too much difference between the following:
1. $$\neg(x = \neg x)$$
2. $$x ≠ \neg x$$
3. $$\neg\neg x = x$$
4. $$x = \neg\neg x$$
According to one common and at least moderately strong [argument](https://plato.stanford.edu/entries/dialetheism/#ArguNega), the semantic content/identity of the negation operator is equivalent to the LNC. I would say, though, that DNI/E, which "look like" the law of identity, are the content/identity in question. By contrast, the LNC seems like the identity law applied to the negation *and* conjunction operators together (and the LEM "looks like" *identity* + *negation* + *disjunction*).
**Are there any set/category/other such theories, in which a mathematical principle generates the LNC?** I tried Googling "law of noncontradiction is a theorem" but the only seemingly relevant result I remember seeing had something to do with Hegel-style dialectical logic or whatever.
| https://mathoverflow.net/users/147890 | The LNC as a mathematical theorem | Non-contradiction can only be a mathematical theorem, rather than a logical assumption, against a background of some logic which does not already contain it. One of the few well-motivated systems with that property is relevant logic.
So the best answer to your question would be some mathematical theory, like a relevant theory of arithmetic $T$, using relevant logic, for which $T\vdash PA$. Then non-contradiction for arithmetic statements would follow from the logic and arithmetic together but not from the logic alone.
One place to start looking for such a theory is Friedman and Meyer’s 1971 paper [*Whither Relevant Arithmetic*](https://www.jstor.org/stable/2275433). Their theory $RA$ of relevant arithmetic does not imply all the theorems of $PA$, and they considered bridging the gap with an $\omega$-rule.
You could look for arithmetical axioms to bridge the gap also, motivated by the [arithmetical results](https://mathoverflow.net/a/252573/44143) left unproved by their theory $RA$. Any such axioms would give a positive answer to your question.
| 5 | https://mathoverflow.net/users/nan | 421048 | 171,293 |
https://mathoverflow.net/questions/421005 | 4 | Suppose $M$ is a cusped finite-volume hyperbolic $3$-manifold, say with a single cusp for simplicity. Following [NZ, Section 4] we can parametrize deformations of the hyperbolic structure with a complex paramter $u$, chosen so that $u = 0$ gives the complete hyperbolic structure on $M$. We can identify $u$ with the derivative of the holonomy of a meridian acting on the cusp torus. There is a second parameter $v$ associated with the longitude. A point $u$ is compatible with $p/q$ Dehn surgery on $M$ if $pu + qv = 2\pi i$.
I want to understand the complex length
$$
\operatorname{L}\_{\mathbb C} (\gamma) = \operatorname{length}(\gamma) + i \operatorname{torsion} \gamma
$$
of the geodesic core $\gamma$ of the torus added in the surgery. If we choose $r,s \in \mathbb{Z}$ with $ps - qr = 1$, then $(p,q)$ and $(r,s)$ are a basis for the homology of the torus near the cusp, and $(p,q)$ is the curve sent to zero by the surgery, so $\gamma$ corresponds to $(r,s)$ up to orientation and we conclude that
$$
\operatorname{L}\_{\mathbb C} (\gamma) = \pm(ru + sv).
$$
[NZ] determine that the right sign is $-(ru + sv)$. Furthermore they observe that
$$
\operatorname{length}(\gamma) = -\Re(ru + sv) = -\frac{1}{2\pi} \Im(u \overline v)
= \frac{1}{2\pi} \Re(i u \overline v)
$$
In particular, the last expression $\Re(i u \overline v)/2\pi$ makes sense on all of generalized Dehn surgery space, not just at the points $u$ with $pu + qv = 2\pi i$ for coprime integers $p,q$.
The obvious conjecture is that the complex length is also a function
$$
\operatorname{L}\_{\mathbb C} (\gamma) = \frac{1}{2\pi} i u \overline v \tag{1}
$$
that makes sense on all of Dehn surgery space. However, I don't see an obvious way to prove this from the method of [NZ]. **My question is:** Is formula (1) (or something closely related) true? How could one prove it?
[NZ]
W. D. Neumann and D. Zagier, *Volumes of hyperbolic three-manifolds.* Topology 24, 307--332 [DOI 10.1016/0040-9383(85)90004-7](https://doi.org/10.1016/0040-9383(85)90004-7)
| https://mathoverflow.net/users/113402 | Complex length of geodesic added in hyperbolic Dehn surgery | After more thought, I realize this can't possibly be true, by an easy extension of the methods in [NZ]. We can re-write our two equations $pu + qv = 2\pi i$ and $\operatorname{L}\_{\mathbb C} (\gamma) = -(ru + sv)$ as the matrix equation
$$
\begin{pmatrix}
p & q \\
r & s
\end{pmatrix}
\begin{pmatrix}
\Re u & \Im u \\
\Re v & \Im v
\end{pmatrix}
=
\begin{pmatrix}
0 & 2 \pi \\
- \operatorname{length}(g) & - \operatorname{torsion}(g)
\end{pmatrix}.
$$
Taking the determinant is what gives $\operatorname{length}(\gamma) = \frac{1}{2\pi} \Re(i u \overline v)$. If we subtract $\left( \begin{smallmatrix} 2\pi & - 2\pi \\ 0 & 0 \end{smallmatrix}\right)$ from both sides and take the determinant again we get
$$
(\Re(u) + 2 \pi s)( \Im (v) + 2 \pi r) - (\Re(v) - 2 \pi r) (\Im(u) + 2 \pi s) = -2\pi \operatorname{torsion}(g).
$$
which is *not* the imaginary part of $i u \overline v$.
| 2 | https://mathoverflow.net/users/113402 | 421065 | 171,302 |
https://mathoverflow.net/questions/421068 | 6 | The singular cohomology with integer coefficients of a projective variety is isomorphic to the Čech cohomology of the constant sheaf of integers on this variety.
If the above statement is correct then consider the following example:
Look at $\mathbb{P}^1$ and the standard open cover $U\_0,U\_1$. Then the Čech complex for any sheaf $\mathcal{F}$ on $\mathbb{P}^1$ would be
$$
0 \rightarrow \mathcal{F}(U\_0)\oplus\mathcal{F}(U\_1) \rightarrow \mathcal{F}(U\_0\cap U\_1) \rightarrow 0.
$$
If we consider the constant sheaf of integers this becomes
$$
0 \rightarrow \mathbb{Z}\oplus\mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0
$$
with non-trivial map $(a,b)\mapsto a-b$.
But the singular cohomology of $\mathbb{P}^1$ is
$$H^i\mathbb{P}^1=\begin{cases} \mathbb{Z},\quad i=0,2 \\
0,\quad \text{else.}\end{cases}$$
Where am I going wrong?
| https://mathoverflow.net/users/481166 | Čech cohomology is isomorphic to singular cohomology | The Cech complex only computes cohomology if the subspaces have vanishing cohomology themselves: you need
$$
H^i(U\_0; \mathcal{F}) \cong
H^i(U\_1; \mathcal{F}) \cong
H^i(U\_0 \cap U\_1; \mathcal{F}) \cong 0
$$
for all $i > 0$. In the example you've written, this is not true for $U\_0 \cap U\_1$. In general there is actually a Mayer-Vietoris long exact sequence relating the cohomology of $X$, $U\_0$, $U\_1$, and $U\_0 \cap U\_1$, or more generally a Mayer-Vietoris spectral sequence if the cover has more open sets.
(As a remark, you may have seen this constraint pushed under the rug because it is automatically true in special cases. If $X$ is a scheme and the subspaces $U\_0$, $U\_1$, and $U\_0 \cap U\_1$ are affine---the last automatic if $X$ is separated---then Serre's vanishing theorem says that these higher cohomologies vanish whenever $\mathcal{F}$ is a quasicoherent sheaf of $\mathcal{O}\_X$-modules.)
| 23 | https://mathoverflow.net/users/360 | 421069 | 171,303 |
https://mathoverflow.net/questions/421062 | 3 | **Motivation.** [Swiss license plates](https://en.wikipedia.org/wiki/Vehicle_registration_plates_of_Switzerland) consist of $2$ letters indicating the region, followed by a number, such that the pairing (region, number) is unique by car. In the small town where I live, I saw two cars today, both from my region, having license plate numbers differing by $1$. This motivated the following.
**Question.** Let $n\geq 2$ be an integer. We say that $S\subseteq \{1,\ldots,n\}$ *contains neighbors* if there is $k\in S$ such that $k+1\in S$. Let $M(n)$ denote the minimum integer $\leq n$ such that at least half of the sets $S\subseteq \{1,\ldots, n\}$ with $|S| = M(n)$ contain neighbours. What is the value of $$\lim\inf\_{n\to\infty}\frac{M(n)}{n}?$$
| https://mathoverflow.net/users/8628 | Probability of picking neighbors in $\{1,\ldots, n\}$ | **Claim** The number of sets $S$ of cardinality $m$ with no neighbors is precisely $\binom{n+1-m}{m}$.
**Proof** Encode a subset $S$ of $[n]$ as a binary string $x\_0 x\_1 \ldots x\_n$ where $x\_0=0$ and $x\_j = 1$ if and only if $j \in S$. Then the sets we want correspond to strings with $m$ ones and no consecutive $11$'s, starting with $0$. A string starting with $0$ has no $11$'s if and only if it is a concatenation of $0$'s and $01$'s. If $x\_0 x\_1 \ldots x\_n$ has $m$ ones then it must be made of $m$ copies of $01$ and $n+1-2m$ copies of $0$; the number of ways to order these is $\binom{n+1-m}{m}$. $\square$
We want to understand when $\binom{n+1-m}{m} \geq \tfrac{1}{2} \binom{n}{m}$. Dividing both sides by $\tfrac{n^m}{m!}$, we want
$$\prod\_{j=m-1}^{2m-2} (1-j/n) \geq \tfrac{1}{2} \prod\_{k=0}^{m-1} (1-k/n)$$
or
$$\sum\_{j=m-1}^{2m-2} \log (1-j/n) \geq - \log 2 - \sum\_{k=0}^{m-1} \log (1-k/n)$$
$$ - \left( \tfrac{3m^2}{2} + O(m) \right) \tfrac{1}{n} + O(m^3/n^2) \geq - \log 2- \left( \tfrac{m^2}{2} + O(m) \right) \tfrac{1}{n} + O(m^3/n^2).$$
So
$$m^2/n = \log 2 + O(m/n) + O(m^3/n^2)$$
and the threshold is at $m = \sqrt{(\log 2) n}$.
---
This makes sense intuitively: If we choose a random set of size $m \approx \sqrt{(\log 2) n}$, then the probability that it contains $k$ and $k+1$ is $\approx \sqrt{(\log 2) n}^2/n^2 \approx (\log 2)/n$. Nonrigourously, the probability that $S$ does not contain $k$ and $k+1$ for any $k$ should be about $(1-(\log 2)/n)^{n-1} \approx e^{- \log 2} = 1/2$.
| 7 | https://mathoverflow.net/users/297 | 421070 | 171,304 |
https://mathoverflow.net/questions/421059 | 6 | Let $G$ be a compact Lie group with a compatible biinvariant metric $d$. The hyperspace $K(G)$ of nonempty compact *subsets* of $G$ is a compact metric space with the Hausdorff metric, and it is easy to check that *subgroups* of $G$ form a closed subspace in $K(G)$, hence we may talk about the (compact) **space of closed subgroups of $G$**. Let us denote this space by $\mathbf{K}(G)$.
General question:
(1) Does anyone know any **source** that may help exploring spaces of the form $\mathbf{K}(G)$?
I have a conjecture:
(2) For a compact connected Lie group $G$ the following are equivalent:
a) $G$ is a limit point in $\mathbf{K}(G)$ (that is, it can be approximated by proper closed subgroups).
b) The circle group is a quotient of $G$.
Is it true? ( b)$\implies$a) is easy, take inverse images of finite subgroups of the circle group by the quotient map.)
For (1) I have found only the papers of Fischer and Gartside: [On the space of subgroups of a compact group I](https://www.researchgate.net/publication/229347536_On_the_space_of_subgroups_of_a_compact_group_I)
and
[On the space of subgroups of a compact group II](https://doi.org/10.1016/j.topol.2008.11.004).
They mostly deal with arbitrary compact $G$ or profinite $G$, not Lie groups.
For (2) I found the MO question [Approximating Lie groups by finite groups](https://mathoverflow.net/questions/190705/approximating-lie-groups-by-finite-groups), which says that only compact abelian Lie groups can be approximated by *finite* subgroups (it refers to a paper of A. M. Turing, [Finite Approximations to Lie Groups](https://doi.org/10.2307/1968716)).
| https://mathoverflow.net/users/479121 | Hausdorff distance in compact Lie groups | Assume a sequence of subgroups $G\_n$ converges to $G$. Up to extraction, we can assume that $\mathrm{Lie}(G\_n)$ converges to a Lie subalgebra $\mathrm{Lie}(H)$. Since the adjoint action of $G\_n$ preserves $\mathrm{Lie}(G)$, by passing to the limit we get that $\mathrm{Lie}(H)$ is an ideal of $\mathrm{Lie}(G)$.
Now a bit of (elementary ?) Lie theory should give you that $\mathrm{Lie}(G\_n) = \mathrm{Lie}(H)$ for $n$ large enough. Let $H$ be the connected subgroup with Lie algebra $\mathrm{Lie}(H)$. One concludes that $G\_n/H$ is a sequence of discrete groups approximating $G/H$. By your second reference, $G/H$ is abelian, which proves your conjecture.
| 2 | https://mathoverflow.net/users/173096 | 421076 | 171,306 |
https://mathoverflow.net/questions/421028 | 3 | Galois revealed that an algebraic equation $f(x)=0$ with coefficients in a field $K$ of zero characteristic is solvable by radicals if and only if the Galois group of $f(x)$ over $K$ is solvable. However, many mathematicions actually expect that one could give a criterion for solvability by radicals simply by coefficients.
Joint with Qing-Hu Hou at Tianjin University, we formulate the following conjecture based on our computation.
**Conjecture.** Suppose that $f(x)=ax^n+bx+c$ is irreducible over $\mathbb Q$ (the field of rational numbers), where $n,a,b,c\in\mathbb Z$, $n>0$ and $a\not=0$. Provided that $\gcd(b,nac)=1$,
the Galois group of $f(x)$ over $\mathbb Q$ is isomorphic to the symmetric group $S\_n$,
and hence the equation $f(x)=0$ is not solvable by radicals if $n\ge5$.
Via an internet search, we note that the conjecture in the case $a=1$ and $\gcd(n-1,c)=1$ is known to be true, see, e.g., [Osaka's JNT paper](https://doi.org/10.1016/0022-314X(87)90029-1).
**QUESTION.** Is the conjecture true? Can one prove it completely?
| https://mathoverflow.net/users/124654 | A criterion for the equation $ax^n+bx+c=0$ not solvable by radicals via $a,b,c$ and $n$ | No, the conjecture is false at least for $n = 5$. The irreducible quintic trinomial $f(x) = 85x^{5} - 4x + 1$ satisfies $\gcd(b,5ac) = \gcd(-4,5 \cdot 85 \cdot 1) = 1$. However, the Galois group of $f(x)$ is solvable.
This can be found as follows. The family of solvable quintic trinomials is
$$
f(x) = (4u^{2} + 16)x^{5} + (5u^{2} - 5) x + (4u^{2} + 10u + 6)
$$
with $u \in \mathbb{Q}$. (Apparently, this dates back to Weber from 1898, although there are many other modern sources.) Taking $u = -19/21$ and then scaling the polynomial gives rise to the example above.
| 13 | https://mathoverflow.net/users/48142 | 421081 | 171,308 |
https://mathoverflow.net/questions/420539 | 3 | **Definition** Let $E\_{1} \xrightarrow{\pi\_{1}} M\_{1}$, $E\_{2} \xrightarrow{\pi\_{2}} M\_{2}$ be two vector bundles over $M\_{1}$ and $M\_{2}$ with fibers $V\_{1}$, $V\_{2}$ respectively. The exterior tensor product $E\_{1} \boxtimes E\_{2}$ is defined as the vector bundle over $M\_{1} \times M\_{2}$, whose fiber over $(x, y) \in M\_{1} \times M\_{2}$ is $E\_{1 x} \otimes E\_{2 y}$, where $E\_{1 x}$ is the fibre of $E\_{1}$ over $x$ and $E\_{2 y}$ is the fibre of $E\_{2}$ over $y$.
In the article [Brouder, Dang, Laurent-Gengoux, and Rejzner - Properties of field functionals and
characterization of local functionals](https://hal.archives-ouvertes.fr/hal-01654120) they have the so called fundamental result on the projective tensor product of sections of a vector bundle
>
> Proposition III.8. Let $\Gamma(M, B)$ be the space of smooth sections
> of some smooth finite rank vector bundle $B \rightarrow M$ on a
> manifold $M$. Then $\Gamma(M, B)^{\hat{\otimes}\_{\pi}
> k}=\Gamma\left(M^{k}, B^{\boxtimes k}\right)$.
>
>
>
Where $\boxtimes$ means exterior tensor product and $\otimes\_\pi$ is the projective tensor product.
I am trying to prove this theorem. My first step is to prove the algebraic isomorphism.
This is how I am trying to prove the algebraic isomorphism by following the same steps as in [A nonlinear theory of generalized tensor fields
on Riemannian manifolds](https://core.ac.uk/download/pdf/11592565.pdf) by Eduard Nigsch at page 89:
Let $V$ be the fiber of the **trivial vector bundle** $B \xrightarrow{\pi} M$.
We know there is a bilinear map from $\alpha :V\times V \rightarrow V\otimes V$ with $$\alpha(v\_1,v\_2)=v\_1 \otimes v\_2$$
where $v\_1,v\_2 \in V$.
Now let $\psi :B \rightarrow U\times V$ be a local trivialization map with $U \subset M$.
Now we define the local sections $\sigma\_i$ by
$\sigma\_i(x) =\psi^{-1}(x,v\_i)$.
We also define the local sections $\gamma\_{ij}$ by
$\gamma\_{ij} =\phi^{-1}(x,y,v\_i\otimes v\_j)$
where $\phi :B{\boxtimes}B \rightarrow U^2 \times V\otimes V$ is a local trivialization map of the bundle $B{\boxtimes}B \rightarrow M \times M$.
Define the map
$g :\Gamma\left(M, B\right) \times \Gamma\left(M, B\right) \rightarrow \Gamma\left(M\times M, B{\boxtimes}B\right)$ by
$$g=\sum\_{i, j} m \circ\left(i d \times m\left(\cdot, \gamma\_{i j}\right)\right) \circ\left(\sigma\_{i}^{\*} \times \sigma\_{j}^{\*}\right)$$
where $m: C^{\infty}(M) \times \Gamma(B\otimes B)\rightarrow \Gamma(B\otimes B) $ is the module multiplication on $\Gamma(B\otimes B)$ and $\sigma\_{i}^{\*}$ is the dual of $\sigma\_{i}$.
By the properties of the tensor product there is a bilinear map $f: \Gamma(M, B)\otimes \Gamma(M, B)\rightarrow \Gamma\left(M\times M, B{\boxtimes}B\right)$ such that if $g$ is the map $h:\Gamma\left(M, B\right) \times \Gamma\left(M, B\right) \rightarrow \Gamma(M, B)\otimes \Gamma(M, B)$
we have that $$g=f \circ h.$$
For the inverse define $h^{-1}: \Gamma\left(M\times M, B{\boxtimes}B\right)\rightarrow \Gamma(M, B)\otimes \Gamma(M, B)$
$h^{-1}(s)=\sum\_{i, j} \gamma\_{i j}^{\*}(s) \alpha\_{i} \otimes \beta\_{j}$ for $s \in \Gamma\left(M\times M, B{\boxtimes}B\right)$
Now we have that $$
\begin{aligned}
&h^{-1}\left(f\left(t \otimes u\right)\right)=h^{-1}\left(h \left(t^{i} \alpha\_{i}, u^{j} \beta\_{j}\right)\right)=h^{-1} \left(t^{i} u^{j} \gamma\_{i j}\right)=t^{i} u^{j} \alpha\_{i} \otimes \beta\_{j}\\
&=t^{i} \alpha\_{i} \otimes u^{j} \beta\_{j}=t \otimes u \text { and }\\
&h(h^{-1}(s))=h\left(s^{i j} \alpha\_{i} \otimes \beta\_{j}\right)=s^{i j} g\left(\alpha\_{i}, \beta\_{j}\right)=s^{i j} \gamma\_{i j}=s .
\end{aligned}
$$
Is my proof correct?
| https://mathoverflow.net/users/138482 | Fundamental result on the projective tensor product of sections of a vector bundle | The tensor product (either one) is symmetric $A \otimes B \cong B \otimes A$, associative $(A\otimes B) \otimes C \cong A \otimes (B \otimes C)$ and distributive over direct sums $A \otimes (B \oplus C) \cong (A\otimes B) \oplus (A\otimes C)$. Spaces of sections preserve direct sum decompositions $\Gamma(M, E\_1 \oplus E\_2) \cong \Gamma(M,E\_1) \oplus \Gamma(M,E\_2)$. Any vector bundle $E\to M$ can be realized as a summand of a trivial vector bundle $(V\times M \to M) \cong (E\oplus E' \to M)$, where we can realize the sub-bundles $E$ and $E'$ and the image and kernel of a fiber-wise projection $P$ on $V\times M$ ([it is sufficient](https://en.wikipedia.org/wiki/Vector_bundle#Operations_on_vector_bundles) that $M$ is contractible to a compact space). The completed tensor product satisfies the identity $C^\infty(M\_1) \otimes C^\infty(M\_2) \cong C^\infty(M\_1 \times M\_2)$ (for details see Trèves *opological Vector Spaces, Distributions and Kernels* 1970).
With the above background, the proof is straightforward. Start with the observation that for a trivial bundle $\Gamma(M, V\times M) \cong C^\infty(M) \otimes V$. Hence $$\Gamma(M\_1,V\_1\times M\_1) \otimes \Gamma(M\_2,V\_2\times M\_2)\cong C^\infty(M\_1\times M\_2) \otimes (V\_1\otimes V\_2) \cong \Gamma(M\_1\times M\_2, (V\_1 \times M\_1) \boxtimes (V\_2 \times M\_2)).$$
Applying direct sum decompositions to both sides gives
$$
(\Gamma(M\_1,E\_1) \oplus \Gamma(M\_1,E'\_1)) \otimes (\Gamma(M\_2,E\_2) \oplus \Gamma(M\_2,E'\_2))
\cong \Gamma(M\_1\times M\_2, (E\_1\oplus E'\_1) \boxtimes (E\_2\oplus E'\_2)) .
$$
Finally, the desired isomorphism
$$
\Gamma(M\_1,E\_1) \otimes \Gamma(M\_2,E\_2)
\cong \Gamma(M\_1\times M\_2, E\_1 \boxtimes E\_2)
$$
follows because both sides coincide with the image of the projection $P\_1\otimes P\_2$ (interpreted as acting fiber-wise or on spaces of sections as needed).
| 1 | https://mathoverflow.net/users/2622 | 421098 | 171,313 |
https://mathoverflow.net/questions/421106 | 3 | Let $M$ be an $n$-dimensional complete Riemannian manifold and $r$ is the distance function to a fixed point.
The Hessian comparison theorem says that if the sectional curvature of $M$ is bounded (precisely $k\le \operatorname{sec}\le K$), then the Hessian of $r$ is bounded by the Hessians of the distance function for the space form with constant sectional curvature $k$ and $K$ (precisely $\operatorname{Hess}\_Kr\le\operatorname{Hess}r\le \operatorname{Hess}\_kr$).
My question: What is the difference between the Hessian of $r$ and the Hessian of the distance function for the space form with constant sectional curvature 0? It's better to have a Taylor expansion.
Following the notation in Petersen's Riemannian geometry, the Hessian of the distance function for the space form with constant sectional curvature $k$ is $\operatorname{Hess}\_kr=\frac{\operatorname{sn}\_k'(r)}{\operatorname{sn}\_k(r)}g\_r$ where $g\_r=g-(\operatorname{d}r)^2$ and $\frac{\operatorname{sn}\_k'(r)}{\operatorname{sn}\_k(r)}=\frac{1}{r}-\frac{k}{3}r+O(r^2)$. Note that $\operatorname{Hess}\_0r=\frac{1}{r}g\_r$. Based on the Hessian comparison theorem, I guess that $\operatorname{Hess}r-\operatorname{Hess}\_0r=-\frac{\operatorname{sec}}{3}rg\_r+O(r^2)$.
---
I originally guessed that the difference $\operatorname{Hess}r-\operatorname{Hess}\_0r$ is a multiplication of the Ricci tensor, big thanks to MySheperd who excluded this possibility and suggested that the difference is $O(r)$ in the answer below. Then I started to ask the above question.
Thank you!
| https://mathoverflow.net/users/102515 | Hessian of the distance function--comparison with the space form with constant sectional curvature 0 | $dr$ is defined regardless of the metric. If $g$ and $g\_{0}$ are two metrics, and $\nabla^{g}$ and $\nabla^{g\_{0}}$ are their corresponding Levi-Civita connections, then the connection difference yields a tensorial operation $D:T^\*M\rightarrow T^\*{}M\otimes T^{\*}M$ such that $\mathrm{Hess}\,r-\mathrm{Hess}\_{0}\,r=\nabla^{g}dr-\nabla^{g\_{0}}dr=D(dr)$. If one now assumes that $g\_{0}$ is locally-flat, i.e., that its curvature tensor has 0-constant sectional cruvature, then one can pick parallel coordinates for this metric which makes $(\nabla^{g\_{0}}dr)\_{ij}=\partial\_{i}\partial\_{j}r$. Therefore, in these coordiantes $D(dr)\_{ij}=-\Gamma\_{ij}^{k}\partial\_{k}r$, where $\Gamma\_{ij}^{k}$ are the chirstoffel symbols of the metric $g$. This includes only first derivatives of the metric, hence in general it won't be e.g. a multipication of the Ricci-Tensor.
In fact, if one picks $r$ to be a distance function, and picks polar coordinates with respect to this function, then an example of a locally-flat metric is $g\_{0}=dr\otimes dr+\sum\_{i=1}^{d-1}dx\_{i}\otimes dx\_{i}$. These polar coordinates are then parallel coordinates for $g\_{0}$, and $D(dr)\_{ij}=-\delta\_{k}^{r}\Gamma\_{ij}^{k}$. Since in polar coordinates $\Gamma\_{ij}^{k}=o(r)$, this would imply that the difference vanishes as r goes to zero. This example also shows that the expression would depend on the choice of the particular locally-flat metric $g\_{0}$.
---
**Edit:** Regarding @Borromean follow up question, the difference does not have to be $o(r)$ either. Let us take a unit disc $\mathcal{D}\subset \mathbb{R^{2}}$ and pick for it the usual polar coordinates $(r,\theta)$. Write two different metrics: $g=dr\otimes dr+r^{2}d\theta\otimes d\theta$ and $g\_{0}=d\theta\otimes d\theta+\theta^{2}dr\otimes dr$. Note how both of these metrics are flat: they are simply eucliden metrics written in polar coordinates, with the roles of $r$ and $\theta$ exchanged. As such, both have constant sectional curvature 0. Note how $r$ is a distance function in one metric but not in the other.
However, $\mathrm{Hess}\,r=\nabla^{g} dr=-\Gamma\_{\theta\theta}^{r}d\theta\,\otimes d\theta=-\frac{1}{r} d\theta\,\otimes d\theta$ while $\mathrm{Hess}\_{0}r=\nabla^{g\_{0}}dr=-\frac{1}{\theta}(dr\otimes d\theta+d\theta\otimes dr)$. So in this case, the difference will explode as $r\rightarrow 0$. Moreover, since the curvature of both metrics vanishes, the coefficents in the expansion of this difference in either $r$ or $\theta$ will include no expressions involving the curvature, as all curvature ingredients vanish identically.
| 5 | https://mathoverflow.net/users/144247 | 421118 | 171,318 |
https://mathoverflow.net/questions/421104 | 3 | If $K/\mathbb{Q}$ is an infinite algebraic extension, define as usual the class group $Cl\_K$ by the direct limit via the natural (conorm) map $Cl\_K := \lim\limits\_{\rightarrow} Cl\_F$,
where $F$ runs over all finite extensions of $\mathbb{Q}$. (I think this definition coincides with the Picard group of the ring of integers in $K$.)
For example, it is easy to see $Cl\_{\overline{\mathbb{Q}}}=0$. From Iwasawa theory if $K$ is a $\mathbb{Z}\_p$-extension of some finite extension of $\mathbb{Q}$, we know that the $p$-primary part of the class group (I will call it $p$-class group for short):
$$Cl\_K(p)=(\mathbb{Q}\_p/\mathbb{Z}\_p)^{\lambda} \times \text{ some p-group of bounded exponent}$$
If moreover $K$ is abelian over $\mathbb{Q}$, Ferrero-Washington proved that the $p$-group on the right hand side is $0$.
Now let $K=\mathbb{Q}(\mu\_{\infty})$, the field joining all the roots of unity, which is the maximal abelian extension of $\mathbb{Q}$.
**What is** $Cl\_K$ or what is $Cl\_K(p)$?
I think $Cl\_K$ has subgroups of forms, for example $H=\mathbb{Q}\_{37}/\mathbb{Z}\_{37}$, because $H$ is the $37$-class group of $F:=\mathbb{Q}(\mu\_{37^\infty})$ and one can show that the natural map from $Cl\_{F}(p)$ to $Cl\_K$ is injective, using the fact that $H$ is in the minus part (the complex conjugatation acts by $-1$) of $Cl\_F$.
| https://mathoverflow.net/users/95241 | What's the class group of $\mathbb{Q}^{\mathrm{ab}}$? | From Armand Brumer, [*The class group of all cyclotomic integers*](https://www.sciencedirect.com/science/article/pii/0022404981900864):
>
> As an abelian group, $\mathrm{Pic}(O\_\infty)$ is isomorphic to a countable direct sum of copies of $\mathbb Q/\mathbb Z$.
>
>
>
Here $O\_\infty$ is the ring of cyclotomic integers, and $\mathrm{Pic}$ is the Picard group - for a number ring it coincides with the class group, while for a union of an increasing tower of such it coincides with the direct limit, as in your definition.
| 4 | https://mathoverflow.net/users/30186 | 421121 | 171,320 |
https://mathoverflow.net/questions/421115 | 3 | I am an amateur in $K$-theory, I have just started reading from "The K-book" by Charles Weibel. I have only read the definition of $K\_1$ which is stated as a quotient of $GL(R)$. The union of the sequence $R^{ \times} = GL\_1(R) \subset GL\_2(R) \subset ...\subset GL\_n(R) \subset GL\_{n+1}(R) \subset..$ is called $GL(R)$ and $K\_1(R)$ is defined to be $GL(R)/[GL(R),GL(R)].$
In one of the examples he has proved that if $R$ is any commutative ring then $K\_1(R) \cong R^{\times} \oplus SK\_1(R)$. Where $SK\_1(R)$ is denoted as the kernel of a surjective map from $K\_1(R) \rightarrow R^{\times}$ which is induced from the determinant map $GL(R) \rightarrow R^{\times}$. I understand the splitting. But when I am trying to compute $SK\_1(R)$ for $R = \mathbb{Z}\_4$ or $R = \mathbb{Z}\_4[t]$ explicitly I am at a loss.
If you could please guide me to the right direction I would be grateful.
| https://mathoverflow.net/users/443060 | $K_1(\mathbb{Z}_4)$ and $K_1(\mathbb{Z_4}[t])$ | As mentioned in Example 1.3.5, the group $SK\_1$ vanishes over euclidean domains. The reason is that then you can transform any invertible matrix into diagonal form using elementary row and column operations, by the Smith normal form algorithm. (Note that we are using more than just the abstract elementary divisor theorem for PIDs, we really need that the transformation matrices are built from elementary matrices. In fact, there are examples of PIDs with nonzero SK\_1.)
Example 1.3.5 is stated only for euclidean *domains*, but you don't need to be without zero-divisors, you just need a euclidean algorithm. For example, the same argument also applies to local rings such as $\mathbb{Z}/4$, and you can set up a similar argument for $\mathbb{Z}/4[t]$.
| 5 | https://mathoverflow.net/users/39747 | 421122 | 171,321 |
https://mathoverflow.net/questions/421123 | 2 | Let $f:X\rightarrow Y$ be a proper and surjective morphism of projective $k$-varieties, where $k$ is a field of characteristic $0$. Let $L$ be a line bundle over $X$. Assume that $L$ is nef and generically ample ( here generically ample means $L$ is ample on the generic fiber of $f$). Does there exists an line bundle $A$ on $Y$ such that $L\otimes f^\*A$ is big?
| https://mathoverflow.net/users/297229 | Nef and generically ample line bundles | This holds for *any* ample line bundle $A$ on $Y$. Indeed, then $f^\*A$ is nef [Laz, Ex. 1.4.4(i)], hence so is $D= L + f^\*A$. Setting $n = \dim X$ and $m = \dim Y$, we see that $D^n \geq 0$ [Laz, Thm. 1.4.9] and $D$ is big if and only if $D^n > 0$ [Laz, Thm. 2.2.16].
For positivity of $D^n$, we may replace $(L,A)$ by $(dL,dA)$ for $d \gg 0$ to assume $A$ is very ample. There is a dense open $U \subseteq Y$ with preimage $V = f^{-1}(U)$ such that $f \colon V \to U$ is flat (and proper) of relative dimension $r = n-m$ and $L|\_V$ is relatively $f$-ample [Laz, Thm. 1.2.17]. General hyperplanes $H\_1,\ldots,H\_m \in \lvert A\rvert$ intersect transversally in a finite (reduced) scheme $S \subseteq U$. We find
$$L^r \cdot (f^\*A)^m = L^r \cdot f^\*S = \sum\_{s \in S} \big(L\big|\_{X\_s}\big)^r > 0.$$
Since $f^\*A$ and $L$ are nef, we get $L^i \cdot (f^\*A)^{n-i} \geq 0$ for all $i$ [Laz, Ex. 1.4.16], hence
$$D^n = \big(L + f^\*A\big)^n = \sum\_{i=0}^n {n \choose i}L^i \cdot \big(f^\*A\big)^{n-i} \geq {n \choose r} L^r \cdot \big(f^\* A\big)^m > 0.\tag\*{$\square$}$$
---
**References.**
[Laz] R. Lazarsfeld, *Positivity in algebraic geometry I*. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge **48**, Springer, Berlin (2004). [ZBL1093.14501](https://zbmath.org/?q=an:1093.14501).
| 4 | https://mathoverflow.net/users/82179 | 421129 | 171,323 |
https://mathoverflow.net/questions/420984 | 0 | Causality seems to play an important role in physics. There also seems to be a close parallel between "$P$ causes $Q$" and "if $P$ then $Q$." Mathematical logic studies logical inference; has there been any formal mathematical study of causal inference?
| https://mathoverflow.net/users/14024 | Has there been any mathematical study of causality? | I am converting my comments into an answer.
Setting aside the alleged parallel between causation and inference for a moment, there has indeed been some mathematical investigation of cause and effect. The Stanford Encyclopedia of Philosophy article on [causal models](https://plato.stanford.edu/entries/causal-models/) is a good starting point for reading about this topic. One of the theories mentioned there has been described in some detail in semi-popular terms in [*The Book of Why*](https://en.wikipedia.org/wiki/The_Book_of_Why) by Judea Pearl and Dana Mackenzie. This theory lies more in the domain of statistics than mathematics proper, and tries to address that perennial problem that statistical correlations do not in themselves prove causation. Pearl has also written a more technical monograph, [*Causality: Models, Reasoning, and Inference*](https://en.wikipedia.org/wiki/Causality_(book)).
Returning to the word "inference," let us note there are several interpretations of what that word means. In mathematics, the flavors that most commonly arise are the *material conditional* $P \Rightarrow Q$, which is equivalent to $(\neg P) \vee Q$, and the *provability* relation $T \vdash \phi$. Though both of these relations might superficially resemble the relation "$P$ causes $Q$," most people who have thought about the analogy have concluded that causality more closely resembles a different kind of conditional statement, namely a *counterfactual* conditional. Roughly speaking, "$P$ caused $Q$" seems akin to the statement, "If $P$ had not occurred then $Q$ would not have occurred." A counterfactual conditional is a very different beast from the material conditional. In everyday speech, the material conditional rarely comes up, except when someone half-jokingly says something like, "If Chris is a good cook then I'm the king of England!" Conditionals in everyday speech are much more likely to be counterfactual, and nowadays, counterfactuals are usually analyzed using [possible world semantics](https://plato.stanford.edu/entries/possible-worlds/) and modal logic. Again the Stanford Encyclopedia of Philosophy has a good article on [David Lewis's attempt to analyze causation in terms of counterfactuals](https://plato.stanford.edu/entries/causation-counterfactual/). Though the mathematics of modal logic is rigorous, the question of whether Lewis has successfully used it to analyze causality is a philosophical one, and is controversial.
Finally, although you might think that causality plays an essential [role in modern physics](https://en.wikipedia.org/wiki/Causality_(physics)), the truth is more subtle. The fundamental equations of physics make no explicit mention of causality, and are in fact time-reversible. Intuitively, causality involves an [arrow of time](https://en.wikipedia.org/wiki/Arrow_of_time), which is a notoriously difficult concept to explain in terms of modern physics. Causality does get mentioned sometimes in physics, e.g., in special relativity, but not in a way that suggests a direct connection with provability or the material conditional.
To summarize, while your intuition might suggest that mathematical logic should be intimately related to causality, closer inspection reveals that the relationship is not as tight as you might have expected.
| 9 | https://mathoverflow.net/users/3106 | 421141 | 171,325 |
https://mathoverflow.net/questions/421113 | 1 | Given a random variable $X$, satifying $P(0\leq X \leq 1)=1$, and $\mathsf{E}[X^2] = \alpha$. We know its maximum variance $\text{Var}(X) = \alpha(1-\alpha)$ achived by a binary random variable $P(X =x) = \begin{cases} &1-\alpha, &x=0 \\ &\alpha, &x=1 \end{cases}$.
Now my problem is given a random vector $\boldsymbol{X}$, and $\text{supp}\boldsymbol{X} = [0,1]^n$, and $\mathsf{E}\boldsymbol{X}={\boldsymbol{\alpha}}$. After a linear transformation $\boldsymbol{H} (\boldsymbol{H} \succ \boldsymbol{0})$, I want to know whether the maximum trace of the covaraince matrix $\text{cov}(\boldsymbol{H}\boldsymbol{X})=\mathsf{E}[(\boldsymbol{H}\boldsymbol{X}-\mathsf{E}[\boldsymbol{H}\boldsymbol{X}])(\boldsymbol{H}\boldsymbol{X}-\mathsf{E}[\boldsymbol{H}\boldsymbol{X})^\text{T}]$ can be achived by a discrete random vector whose support $\text{supp}\boldsymbol{X}=\{0,1\}^n$.
The trace can be expanded as $\text{tr}(\text{cov}(\boldsymbol{H}\boldsymbol{X}))=\sum\_{k=1}^{n} h\_{i,k}^2 \mathsf{E}{\bigl(X\_k-\mathsf{E}{X\_k}\bigr)^2}
+ \sum\_{k=1}^{n} \sum\_{\substack{\ell=1\\\ell\neq k}}^{n}
2h\_{i,k} h\_{i,\ell} \bigl( \mathsf{E}{X\_{k} X\_{\ell}} -
\mathsf{E}{X\_{k}}\mathsf{E}{X\_{\ell}}\bigr)$. If we use the similar method as in the random variable case, we can maximize the first term of RHS of above equation, be the change of second term of RHS cannot be determined.
| https://mathoverflow.net/users/472652 | The maximum trace of a covariance can be achieved by a discrete random vector? | The answer is yes. Indeed, let $X:=\boldsymbol{X}$, $H:=\boldsymbol{H}$, and $a:=\boldsymbol{\alpha}$. By approximation and compactness, without loss of generality (wlog), the matrix $H$ is nonsingular, so that the trace in question is
$$\sum\_{i=1}^n Var\,l\_i (X),$$
where the $l\_i$'s are linearly independent linear functionals determined by the matrix $H$. By compactness, the maximum of this trace over all random vectors $X$ with mean $EX=a$ and $P(X\in[0,1]^n)=1$ is attained at some maximizing $X$. In what follows, let $X$ be such a maximizer.
To obtain a contradiction, suppose that there is some point $x\in S\_X\cap[0,1]^n\setminus\{0,1\}^n$, where $S\_X$ is the support of the distribution $\mu\_X$ of the random vector $X$. Then (i) $\{x-h,x+h\}\subset[0,1]^n$ for some nonzero vector $h$ and (ii) $\mu\_X(U\_x)>0$ for some set $U\_x$ relatively open in $[0,1]^n$ and such that $x\in U\_x$. Wlog, the length of the vector $h$ and the diameter of the set $U\_x$ are small enough so that $U\_{x-h}\cup U\_{x+h}\subseteq[0,1]^n$.
Moving now half of the mass $\mu\_X(U\_x)$ by the parallel translation by vector $h$ and moving the remaining half of the mass $\mu\_X(U\_x)$ by the parallel translation by vector $-h$, we obtain a new probability distribution of some random vector $Y$ with values in $[0,1]^n$ such that $EY=a=EX$ and $Var\,l\_i (Y)\ge Var\,l\_i (X)$ for **all** $i$, with at least one of these inequalities being strict (namely, strict for all $i$ with $l\_i(h)\ne0$). This contradicts the assumption that $X$ is a maximizer. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 421143 | 171,326 |
https://mathoverflow.net/questions/421130 | 5 | The $L$ genus can be expressed as combinations of the Pontryagin classes with the first few terms as follows:
$$L\_1=\frac{1}{3}p\_1,$$
$$L\_2=\frac{1}{45}(7p\_2-p\_1^2),$$
$$L\_3=\frac{1}{945}(62p\_3-13p\_1p\_2+2p\_1^3)$$
Let $a\_k$ denote the coefficient of $p\_k$ in $L\_k$. Is there any estimate of $a\_k$? Is it true that $a\_k$ is positive and decreasing?
Checking the first 14 $a\_k$ using the formula in <http://www.mctague.org/carl/blog/2014/01/05/computing-L-polynomials/>, it is true that $a\_k$ is decreasing (and exponentially fast). I am wondering if this is true in general.
| https://mathoverflow.net/users/40517 | Coefficient of the top Pontryagin class in $L$-genus | The coefficient of $p\_k$ is given by
$$2^{2n}(2^{2n-1}-1)\frac{B\_n}{(2n)!} = \zeta(2n)\frac{2^{2n}-2}{\pi^{2n}},$$
see e.g. Appendix A of this older version of [Weiss](https://arxiv.org/pdf/1507.00153v3.pdf) (warning: for Weiss, the convention is that the Bernoulli numbers $B\_n$ are positive for all positive $n$). From this it is clear that it is always positive. You can presumably obtain growth estimate from growth estimates for the zeta function (or equivalently, Bernoulli numbers).
In fact, all coefficients of the $L$-polynomial can be expressed in terms of multiple zeta values, by the results of [Berglund-Bergström](https://link.springer.com/article/10.1007/s00208-018-1647-2) (and by their Corollary all $a\_k$ are positive).
| 11 | https://mathoverflow.net/users/798 | 421144 | 171,327 |
https://mathoverflow.net/questions/419301 | 15 |
>
> Are the weak topologies of $\ell\_1$ and $L\_1$ homeomorphic?
>
>
>
Strangely may it sound, the question seeks contrasts between norm and weak topologies of Banach spaces from the non-linear point of view. And indeed, no linear homomorphism between weak topologies of non-isomorphic spaces exists as prevented by the Closed Graph Theorem. Note that the fact that $\ell\_1$ is a Schur space yet $L\_1$ is not is *a priori* not helpful, unless I am missing something trivial.
However, all infinite-dimensional, separable Banach spaces are homeomorphic. For weak topologies this is not so, as the weak topology of a reflexive space is $\sigma$-compact whereas this does not hold for $L\_1$, say. We could then ask:
>
> Is there a pair of non-isomorphic, non-reflexive Banach spaces whose weak topologies are homeomorphic?
>
>
>
31.03.2022: As [noticed](https://mathoverflow.net/questions/419301/distinguishing-topologically-weak-topologies-of-banach-spaces#comment1076935_419301) by Jerzy, the weak topologies of $\ell\_1$ and $C[0,1]$ are not homeomorphic.
| https://mathoverflow.net/users/15129 | Distinguishing topologically weak topologies of Banach spaces | The weak topologies of the spaces $\ell\_1$ and $L\_1$ are not homeomorphic because of the following
**Theorem.** *Assume that $X,Y$ are two Banach spaces whose weak topologies are homeomorphic. If $X$ has Shur's property, then $Y$ has Shur's property, too.*
Proof. For a topological space $T$ let $T\_s$ be the sequential coreflection of $T$, which is $T$ endowed with the topology consisting of sequentially open sets. A subset $U\subseteq T$ is *sequentially open* if for any convergent sequence $(x\_n)\_{n\in\omega}$ in $T$ with $\lim x\_n\in U$ there exists $m\in\omega$ such that $x\_n\in U$ for all $n\ge m$.
Assume that $h:X\_w\to Y\_w$ is a homeomorphism between the Banach spaces $X,Y$ endowed with the weak topologies. The homeomorphism $h$ remains a homeomorphism betweem the sequential coreflections $X\_s$ and $Y\_s$ of the spaces $X\_w$ and $Y\_w$, respectively. Since $X$ has Shur's property, the sequential coreflection $X\_s$ of $X\_w$ coincides with $X$. Since $X=X\_s$ is a Baire space, so is the space $Y\_s$. The Baireness of the space $Y\_s$ implies that the closed unit ball $B\_Y$ of $Y$ has non-empty interior in $Y\_s$. Let $y\_\*$ be an interior point of the set $B\_Y$ in $Y$. Assuming that $Y$ fails to have Shur's property, we can find a null sequence $(y\_n)\_{n\in \omega}$ in $Y\_w$ consisting of vectors of norm $\|y\_n\|=3$. Then $(y\_n+y\_\*)\_{n\in\omega}$ is a sequence in $Y\_s\setminus B$ that converges to $y\_\*$, which is not possible as $y\_\*$ is an interior point of $B$ in $Y\_s$. This contradiction implies that the Banach space $Y$ has Shur's property.
---
The proof of the above theorem suggests the following characterization of the Shur property.
**Characterization.** A Banach space $X$ has Shur's property if and only if the sequential coreflection $X\_s$ of the weak topology of $X$ is a Baire space if and only if $X\_s=X$.
**Remark.** Concerning the second question, it can be shown that any reflexive separable infinite-dimensional Banach spaces endowed with the weak topologies are sequentially homeomorhic (= the sequential coreflections of their weak topologies are homeomorphic), see this [paper](https://arxiv.org/abs/1908.09115) for more information in this direction.
| 3 | https://mathoverflow.net/users/61536 | 421149 | 171,329 |
https://mathoverflow.net/questions/421154 | 1 |
>
> Let $A \in M(n)$, let $\lambda \in \mathbb{R}$, let $V\_{\lambda} := \ker(\lambda I - A)$ and let $x:\mathbb{R} \to \mathbb{R}^{n}$ be a solution of $\dot x= Ax$ such that $x(t\_0) \in V\_{\lambda}$ for some $t\_0 \in \mathbb{R}$, then $x(t) \in V\_{\lambda}$ for every real $t \in \mathbb{R}$.
>
>
>
---
Since $x(t\_0) \in V\_{\lambda}$, then $\dot x(t\_0)= \lambda x(t\_0)$, given that $\dot x(t\_0) = Ax(t\_0)$. But I don't know how to go on!
| https://mathoverflow.net/users/481556 | $V_{\lambda}$ is invariant under $A$ | Define $y(t) = \exp(\lambda(t - t\_0)) x(t\_0)$. Verify that $y' = Ay$ and that $y(t) \in V\_\lambda, \forall t \in \mathbb R$. Uniqueness of solutions gives $y = x$ and, therefore, the desired result.
| 2 | https://mathoverflow.net/users/115302 | 421156 | 171,332 |
https://mathoverflow.net/questions/421138 | 6 | Homomorphisms $B\_n \to B\_{2n}$ and $B\_n \to S\_{2n}$ have been classified in [Chen–Kordek–Margalit - Homomorphisms between braid groups](https://people.math.gatech.edu/%7Edmargalit7/papers/bn2bn.pdf) and [Lin - Braids and permutations](https://arxiv.org/abs/math/0404528) respectively. I am interested in the corresponding question for symmetric groups: what are the homomorphisms $S\_n \to S\_{n+k}$ (up to conjugation)?
I know from [Explicit description of all morphisms between symmetric groups.](https://mathoverflow.net/questions/36452/explicit-description-of-all-morphisms-between-symmetric-groups) that such a classification is difficult in full generality—I am willing to restrict my attention to suitably small $k$. The answer to the linked question discusses the classification of maximal subgroups of $S\_m$ isomorphic to $S\_n$, but I do not see how this is directly useful, as the image of a map $S\_n \to S\_{n+k}$ needn't be maximal.
When $k = 1$ and $n + 1 \neq 6$, the only index $n+1$ subgroups of $S\_{n+1}$ are point stabilizers. Thus, any non-cyclic map $S\_n \to S\_{n+1}$ is conjugate to the obvious inclusion.
When $k > 1$, we may combine\* the identity map $S\_n \to S\_n$ with a sign representation $S\_n \to \mathbb{Z}\_2 \to S\_{k}$ to produce a map $S\_n \to S\_{n+k}$ not conjugate to an inclusion. Following Chen–Kordek–Margalit, we might hope that these are all such homomorphisms (for small $k$).
Is anything known about these homomorphisms?
\*In case it wasn't clear: this combination is found by letting $S\_n$ act on $[1,\ldots, n]$ and $S\_k$ act on $[n+1,\ldots,n+k]$
| https://mathoverflow.net/users/153494 | Is there a classification of homomorphisms $S_n \to S_{n+k}$ for small $k$? | As spin observed in a comment below, if you know all (conjugacy classes of) subgroups of $S\_n$ of index up to $m$, then you can determine the equivalence classes of homomorphisms $\phi:S\_n \to S\_m$, because the subgroups tell you the possible actions on the orbits of the image of $\phi$. The subgroups of index up to $m=n^2$ are listed explicitly in Mikko Korhonen's answer to the the MSE post [Large subgroups of Symmetric Group](https://math.stackexchange.com/questions/3536888).
As an illustration, I will answer the question on the assumption that $k<n$.
For $n \ge 5$, a homomorphism from $S\_n$ that is not injective has image of order $1$ or $2$, so we can restrict attention to injective maps $\phi:S\_n \to S\_{n+k}$.
From the MSE post referred to above we find that, for $n>6$, the only subgroups of index less than $2n$ are $S\_n$, $A\_n$, and the point stabilizers, which are isomorphic to $S\_{n-1}$.
So if $S\_n$ is acting faithfully on the set $\Omega := \{1,2,\dotsc,n+k\}$ with $k<n$, then there must be a single orbit $\Delta$ of length $n$ on which $S\_n$ acts faithfully.
Since we are interested in classifying maps up to conjugation, we can assume that $\Delta = \{1,2,\dotsc,n\}$ and that the image of $\phi(g)$ restricted to $\Delta$ is $g$ for all $g \in S\_n$.
Furthermore, the image of $\phi$ restricted to $\Omega \setminus \Delta$ has order $1$ or $2$.
So, $\phi(g)\_{\Omega \setminus \Delta} = 1$ for $g \in A\_n$, and for $g \in S\_n \setminus A\_n$, we have, up to conjugation,
$\phi(g)\_{\Omega \setminus \Delta}$ can be
$1$, or $(n+1,n+2)$, or $(n+1,n+2)(n+3,n+4)$, etc., which gives a total of $\lfloor \frac{k+2}{2} \rfloor$ equivalence classes of injective homomorphisms $\phi$.
| 7 | https://mathoverflow.net/users/35840 | 421166 | 171,336 |
https://mathoverflow.net/questions/421164 | 1 | Let $A$ and $B$ be two, possibly dependent, random variables, and let $X$ be a random variable independent of $(A,B)$. For simplicity, let's concern ourselves with discrete random variables. Is the following inequality always true?
$$I(A+X : B+X) \geq I(A:B) \label{eq:conj} \tag{\*}$$
This is clearly true when $A$ and $B$ are independent, as the RHS is then $0$. At the other extreme, it is also true when $A = B$ since the RHS is $H(A)$ while the LHS is $H(A+X)$. And, $H(A+X) \geq H(A+X~|~X) = H(A~|~X) = H(A)$, where the last equality uses $A\bot X$. On the other hand, I cannot see a proof even when $B = f(A)$ for some deterministic function $f$.
It is not too hard to see if we added $X$ to only one of $A$ or $B$, the mutual information inequality would be flipped. That is, $I(A+X: B) \leq I(A:B)$. Intuitively this makes sense: a random variable plus noise gives less information about another random variable than without the noise. However, when we add (the same) $X$ to both $A$ and $B$ and ask for the mutual information between them, I have no good intuition.
The setting in which \eqref{eq:conj} arose, $X$ is a Bernoulli, and $A$ and $B$ are sums of iid Bernoullis with common elements. More precisely, $X\_1, \ldots, X\_n$ are iid Bernoullis, and $A = \sum\_{i\in S} X\_i$ and $B = \sum\_{j\in T} X\_j$ where $S,T$ are (potentially intersecting) subsets of $\{1,2,\ldots, n\}$. I experimentally verified \eqref{eq:conj} for small $n$.
Any help/pointers would be appreciated.
| https://mathoverflow.net/users/436290 | A question about mutual information | This is not true in general. E.g., let each of the random variables $A,B,X$ take values in the set $\{1,2\}$. Let
the matrix $(p\_{a,b}\colon a=1,2,\,b=1,2)$ of the probabilities $p\_{a,b}:=P(A=a,B=b)$ be the following matrix:
$$\frac1{10^4}\left(
\begin{array}{cc}
1456 & 3987 \\
4533 & 24 \\
\end{array}
\right);$$
in particular, $p\_{1,1}=\dfrac{1456}{10^4}$.
Let
$$P(X=1)=\frac{8201}{10000}=1-P(X=2).$$
Then
$$I(A+X:B+X)=0.335\ldots\not\geq 0.342\ldots=I(A:B).$$
| 3 | https://mathoverflow.net/users/36721 | 421170 | 171,337 |
https://mathoverflow.net/questions/421163 | 1 | Suppose $d\in \{3,4,\dotsc\}$ and $A\subseteq \mathbb{R}^d$ is non-empty, open and connected with its complement $A^c$ connected too and $\text{int}(A^c)\neq \emptyset$. Its boundary $S:=\partial A$ is then [connected too](https://math.stackexchange.com/questions/170337/connectedness-of-the-boundary?noredirect=1&lq=1). Let $A^{(2)}$ be the set of 2-dimensional affine planes within the affine space $\mathbb{R}^d$. For any $x\in S$, define
\begin{gather\*}
S\_x:=\bigcup\_{n=1}^\infty S\_x^{(n)} \\
S\_x^{(1)}:=\{y\in S\mid\exists H\in A^{(2)}\mathrel:\text{ $x$ and $y$ are in the same connected component of }H\cap S\},\\
S\_x^{(n+1)}:=\{y\in S|\,\exists z\in S\_x^{(n)}:\,y\in S\_z^{(1)}\}.
\end{gather\*}
Is $S\_x=S$ for all $x \in S$? If not, is $\{S\_x\mid x\in S\}$ a countable partition of $S$?
| https://mathoverflow.net/users/33927 | Can one explore a surface along ‘piecewise planar’ curves? | Consider the set
$S=\{\,(t,t^2,t^3)\in\mathbb R^3\mid\,t\in\mathbb R\}$.
Clearly $S$ is connected, and so is its complement $A=\mathbb R^3\setminus S$.
Note that each plane has at most 3 points of intersection with $S$.
It follows that $S\_x=\{x\}$ for any $x\in S$.
The only problem is the set $A^c=S=\partial A$ has empty interior, but it is easy to fix by removing from $A$ a closed ball centered at a point in $S$.
| 2 | https://mathoverflow.net/users/1441 | 421178 | 171,341 |
https://mathoverflow.net/questions/421034 | 1 | I am reading one lecture note [Dynamics for Spherical Models of Spin-Glass and Aging](https://doi.org/10.1007/978-3-540-40908-3_5) by Alice Guionnet. On page 124, it says that
for some $\alpha>0$,
$$
e^L=P\left(\exp(\alpha\sup\_{|s-t|\le\delta}\frac{|B\_s-B\_t|^2}{|s-t|})<\infty\right)
$$
Moreover, how to show that
$$
E\left[\exp(\alpha\sup\_{|s-t|<\delta}\frac{|B\_s-B\_t|^2}{|s-t|})\right]\le e^L
$$
---
I do not know what is $L$ here. I check the the reference and found that
I just found a useful result: for $\alpha-$Holder continuity Brownian motion, there exists $C=C(\alpha)>0$ s.t. $0<\epsilon\le 1$,
$$
-C\epsilon^{-\frac{2}{1-2\alpha}}\le \log P(\sup\_{|s-t|<\delta}\frac{|B\_s-B\_t|}{|s-t|^\alpha}\le \epsilon)\le -C^{-1}\epsilon^{-\frac{2}{1-2\alpha}}
$$
| https://mathoverflow.net/users/168083 | For some $\alpha>0$, $ e^L=P\left(\exp(\alpha\sup_{|s-t|\le\delta}\frac{|B_s-B_t|^2}{|s-t|})<\infty\right) $? | First, there is a typo there - should be E instead of P, as in the second display in your question. Second, the argument as written is not quite right, but it can be rescued. Indeed, $\sup\_{t<\delta} (B\_t/\sqrt{t})=\infty$, and therefore that expectation is $\infty$. The solution is not to divide by $|t-s|$ but rather by $|t-s|^{\alpha}$ for some $\alpha$ close to $1$- then the expectation will be finite (using arguments from [28] to control the mean of the sup, and Borel's lemma to control the deviations). The only difference is that instead of $1/M\delta$ you will have $1/M\delta^\alpha$, which is still fine for the rest of the argument.
| 2 | https://mathoverflow.net/users/35520 | 421179 | 171,342 |
https://mathoverflow.net/questions/421066 | 5 | I posted the following question on MSE originally because, not being research-level, it seemed more appropriate for that site. However, there was no activity for it on MSE and I feel that it certainly can be answered on MO, so I am now posting it here.
---
A number of conditions that are equivalent to the statement that a Banach space $X$ has Radon-Nikodym property (RNP) are found on page 217 of ``Vector Measures" by Diestel & Uhl. One such condition is that: a Banach space $X$ has the RNP iff every function $f:[0,1]\to X$ that is of bounded variation is weakly differentiable off a fixed set of measure zero.
To me, the condition that $f:[0,1]\to X$ is ``weakly differentiable off a fixed set of measure zero" means that for a.e. $t\in(0,1)$, there exists a vector $x\_{t}\in X$ such that $\lim\_{h\to 0}\frac{(\varphi\circ f)(t+h)-(\varphi\circ f)(t)}{h}=\varphi(x\_{t})$ for every $\varphi\in X^{\*}$. Note that that this interpretation of weak differentiability at a point $t\in(0,1)$ matches the definition found in [this paper](https://link.springer.com/content/pdf/10.1007/s00009-015-0656-6.pdf) at the bottom of page 2802 (the second page of the introduction).
**First question:** Is this interpretation of the condition that $f:[0,1]\to X$ is ``weakly differentiable off a fixed set of measure zero" correct? If not, what is the correct interpretation?
For example, one might reasonably also say that $f:[0,1]\to X$ is weakly differentiable if there exists a function $g:[0,1]\to X$ such that, for every smooth test function $\varrho:[0,1]\to\mathbb{R}$ with compact $\text{supp}(\varrho)\subset(0,1)$, it follows that $\int\_{0}^{1}\varrho'(t)f(t)dt=-\int\_{0}^{1}\varrho(t)g(t)dt$. This is, in turn, equivalent (see [here](https://www.math.ucdavis.edu/%7Ehunter/pdes/ch6A.pdf), Prop. 6.36) to the dual condition that $\int\_{0}^{1}\varrho'(t)(\varphi\circ f)(t)dt=-\int\_{0}^{1}\varrho(t)(\varphi\circ g)(t)dt$ for every $\varphi\in X^{\*}$. Clearly, if $g:[0,1]\to X$ is defined for a.e. $t\in[0,1]$ by $g(t)=x\_{t}$ (where $x\_{t}\in X$ is as stated above the first question) then the dual condition holds at least formally.
**Second question:** Is the dual condition for weak differentiability equivalent to the interpretation of weak differentiability above the first question?
| https://mathoverflow.net/users/165007 | Radon-Nikodym property in Diestel & Uhl: a definition clarification | Diestel & Uhl can only mean the first interpretation you gave for two reasons:
1. For the second interpretation, the term “off a fixed set of measure zero” makes no sense.
2. Even in case $X=\mathbb R$ (which certainly has the RNP) the condition is not satisfied if interpreted in the second way: Cantor's stairway function $f$ is monotone and thus of bounded variation, and for $f$ the only candidate for $g$ in the second interpretation is the $0$-function.
The latter shows also that the answer to your second question is negative.
| 6 | https://mathoverflow.net/users/165275 | 421185 | 171,345 |
https://mathoverflow.net/questions/421114 | 1 | In this question, the term “word” implies a binary word, i.e. a sequence of bits.
Let $W(w)$ denote the number of non-zero bits in a word $w$.
Assuming that $l \geq 2$ is even, an $l$-bit word $w$ is *balanced* if and only if $W(w) = l/2.$
Assuming that $w$ is an $l$-bit word and $0 \le k < l$, let $R(w, k)$ denote the result of the left bitwise rotation (i.e. the left [circular shift](https://en.wikipedia.org/wiki/Circular_shift)) of $w$ by $k$ bits. For example, if $w = 0100110001110000$, then $l = 16$ and $$\begin{array}{l}
R(w,0) = w = {\text{0100110001110000}},\\
R(w,1) = {\text{1001100011100000}},\\
R(w,2) = {\text{0011000111000001}},\\
\ldots \\
R(w,15) = {\text{0010011000111000}}.
\end{array}$$
Let $w\_0 \oplus w\_1$ denote the result of the bitwise [“exclusive or”](https://en.wikipedia.org/wiki/Exclusive_or) operation for two words. For example, $$0100110001110000 \oplus 1010010001000010 = 1110100000110010.$$
Assuming that $w\_0$ and $w\_1$ are two words of length $l$, let $f(w\_0, w\_1)$ denote the minimal element (the smallest number) in the tuple $$\begin{array}{l}
(W(w\_0 \oplus w\_1),\\
W(w\_0 \oplus R(w\_1, 1)),\\
W(w\_0 \oplus R(w\_1, 2)),\\
\ldots \\
W(w\_0 \oplus R(w\_1, l - 1))).
\end{array}$$
Given an arbitrary natural number $n \geq 4$ which is a multiple of $4$, I want to generate a set $T$ (if it exists) such that:
1. Cardinality of $T$ is $2$;
2. Each element of $T$ is a balanced $2^n$-bit word;
3. For any pair $(w\_0, w\_1)$ of elements of $T$ we have $f(w\_0, w\_1) \geq n$.
In other words, I need two balanced $2^n$-bit words $(w\_0, w\_1)$ such that $f(w\_0, w\_1) \geq n$. Is there an efficient algorithm to solve the problem? The word “efficient” here means that it should take *significantly* less than $2^{2n}$ operations to generate a single element of $T$: for example, if $n=32$, it is infeasible to perform $2^{64}$ operations for a single word.
| https://mathoverflow.net/users/122796 | Is there an efficient generalized algorithm to generate a set of binary words satisfying a particular cross-correlation property? | If $m = 2^a b$ where $b$ is odd, the words $(0^{2^k} 1^{2^k})^{2^{a-k-1}b}$ for $k < a$ have cross-correlation $2^{a-1}b$. This appears to give a (potentially one of many) set of maximum size which achieves the upper bound on cross-correlation (since the average Hamming distance over all rotations of two balanced words of the same length is half the length), and is far stronger than required by the current version of the question.
| 1 | https://mathoverflow.net/users/46140 | 421189 | 171,346 |
https://mathoverflow.net/questions/419705 | 3 | This question is a cross post from the Math StackExchange since it got no attention at all there: <https://math.stackexchange.com/questions/4414601/maps-that-preserve-winding-numbers>
I am looking for a characterisation of the continuous maps on some subset of $A\subseteq \mathbb{C}$ that preserve the winding numbers of all closed curves in $A$, i.e. if $\gamma$ is a closed curve that lies in $A$ and $x\in A$ is a point not lying on $\gamma$, then $$\text{ind}\_{f\circ \gamma}(f(x)) = \text{ind}\_\gamma(x)\ .$$
Translations clearly satisfy this. Multiplications with a non-zero complex number do as well. $\mathbb{R}$-linear maps with positive determinant probably as well. $\mathbb{R}$-linear maps with negative determinant on the other hand will flip the sign of the winding number.
Another example is the inverse function, which has this property on any region not containing 0.
Is there any good classification of functions that have that kind of property?
| https://mathoverflow.net/users/32355 | Maps that preserve winding numbers | I found an answer to this question: the question of how a continuous function changes the winding number of a closed curve can be studied quite generally. The important concept here is the [degree of the function](https://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping) (also known as the Brouwer degree).
If a curve $\gamma$ has winding number $n$ around a point $z$ (say w.l.o.g. $z = 0$) then it is homotopic to a circle that winds $n$ times around $z$. So if we want to know the winding number of $f(\gamma)$ around $f(0)$, it is enough to look at how $f$ acts on a simple counter-clockwise circle around $0$, since and that is its degree and we have $$\text{ind}\_{f(\gamma)}(0) = \text{deg}\_f(f(0)) \cdot \text{ind}\_\gamma(0)$$
In fact, the idea of a degree generalises the concept of winding number.
The degree of a particular function $f$ that is real-differentiable can also be easily computed using the formula
$$\text{deg}\_f(f(0)) = \sum\_{y\in f^{-1}(0)} \text{sgn}\ \text{det}(Df(y))$$
where $Df(y)$ is the Jacobian of $f$ at $y$.
In particular, a real-differentiable function that is one-to-one and whose Jacobian has positive determinant at $0$ preserves the winding number at $0$.
At least I think this is the basic idea; I am not 100% clear on the details here, so I apologise for any mistakes in this answer. But perhaps this already helps someone who's struggling like I was!
| 0 | https://mathoverflow.net/users/32355 | 421194 | 171,347 |
https://mathoverflow.net/questions/132878 | 16 | I'm looking for small concrete examples of non-pivotal finite tensor categories to do some calculations with.
Here a finite tensor category is, according to [Etingof-Ostrik](http://arxiv.org/abs/math/0301027), a rigid monoidal category whose underlying category is equivalent to the category of finite dimensional modules over a finite dimensional algebra. A tensor category is pivotal if there's an isomorphism of tensor functors between the identity functor and the double dual functor.
The example I was able to find in the literature is the category of representations of a 72-dimensional Hopf algebra in Remark 2.11 of [Andruskiewitsch-Angiono-Iglesias-Torrecillas-Vay](http://arxiv.org/abs/1204.5807). (In fact, they give three such examples.) But I was hoping for a smaller example.
Bonus question: What I would really like is an example of a category where not only is the double dual functor nontrivial, but there's no invertible object X such that the double dual is isomorphic as a tensor functor to conjugation by X. (Note that the 72 dimensional Hopf algebras mentioned above do not give counterexamples, because their duals are pivotal.)
Easier question: I'd also love to hear about *any* other examples of non-pivotal finite tensor categories beyond the three from AAITV, even if they're not smaller.
| https://mathoverflow.net/users/22 | Are there small examples of non-pivotal finite tensor categories? | Monoidal categories $\mathcal{C}$ which admit a monoidal natural isomorphism $\Phi\_X:X^{\*\*} \rightarrow \beta\otimes X\otimes \beta^\*$ are called $\textit{quasi-pivotal}$ . For $\mathcal{C}=$Rep$(H)$ with $H$ a finite dimensional Hopf algebra, quasi-pivotal structures on $\mathcal{C}$ are in bijection with pairs $(l,b)$ where $l$, $b$ are group-like elements in $H$, $H^\*$ respectively satisfying $$S^2(h)=b(h\_3)\;b^{-1}(h\_1)\; l\;h\_2 \; l^{-1} \;\;\;\; \text{for all}\;\; h\in H.$$
Such a pair $(l,b)$ is called a $\textit{pair in involution}$. The paper [Generalized Taft algebras and pairs in involution](https://arxiv.org/pdf/1908.10750.pdf), constructs a family of Hopf algebras that do not admit a pair in involution, thereby providing an answer to the bonus question.
---
"Book Hopf algebras" provide an example of Hopf algebras that admit a quasi-pivotal structure but don't always admit a pivotal structure. See [this paper](https://arxiv.org/pdf/1810.03350.pdf) for further details.
---
Furthermore, by a result of [Shimizu](https://arxiv.org/pdf/1608.05905.pdf), a quasi-pivotal structure on $\mathcal{C}$ yields a pivotal structure on $\mathcal{Z}(\mathcal{C})$. Thus, the Drinfeld doubles $D(H)$ of the Hopf algebras admitting a pair in involution (discussed above) provide examples of pivotal Hopf algebras.
| 3 | https://mathoverflow.net/users/58202 | 421197 | 171,349 |
https://mathoverflow.net/questions/404064 | 3 | While I don't work on the regularity theory for the optimal transport map, I was curious about the open problem 1.28 listed in Ambrosio and Gigli's *User's guide*: the problem to determine whether we have $T\in W^{1,1}$ for the transport map $T$ is still open?
More precisely, assume $\Omega\_1,\Omega\_2\subset \mathbb{R}^d$ are two bounded and connected open sets, $\mu=\rho \mathcal{L}^d |\_{\Omega\_1}, \nu=\eta \mathcal{L}^d |\_{\Omega\_2}$ with $0<c\leqslant \rho,\eta \leqslant C$ for some $c,C\in\mathbb{R}$. Assume also that $\Omega\_2$ is convex, then the optimal transport map $T\in W^{1,1}(\Omega\_1)$?
| https://mathoverflow.net/users/137743 | Open problem 1.28: $W^{1,1}$ regularity for optimal transport map | The problem has indeed been solved by De Philippis-Figalli. The reference is:
<https://link.springer.com/article/10.1007/s00222-012-0405-4>
the paper linked above is a subsequent, related, development of the theory and concerns stability of optimal maps.
| 2 | https://mathoverflow.net/users/58975 | 421200 | 171,350 |
https://mathoverflow.net/questions/419722 | 4 | This question is inspired by the [recent answer](https://mathoverflow.net/q/419324), where @RobPratt proposed to use integer linear programming (ILP) for solving a number-theoretic optimization problem. I will consider a very similar problem described by sequence [OEIS A061057](https://oeis.org/A061057), which for a given integer $n$ asks to find a split of $n!$ into two co-factors with the smallest difference, i.e. to find
$$\min\_{d\,\mid\, n!}\quad \left|\frac{n!}{d} - d\right|.$$
Equivalently, it can be posed as finding the largest divisor $d\mid n!$ such that $d\leq \sqrt{n!}$.
---
This problem has a simple ILP formulation based on the prime factorization
$$n! = p\_1^{m\_1} p\_2^{m\_2} \cdots p\_k^{m\_k}$$
with integer variables $x\_1,\dots,x\_k$, representing the exponents in the prime factorization of $d=p\_1^{x\_1}p\_2^{x\_2}\cdots p\_k^{x\_k}$, as follows:
$$\begin{cases}
\sum\_{i=1}^k x\_i\log p\_i\ \longrightarrow\ \max,\\
\sum\_{i=1}^k x\_i\log p\_i\leq \log \sqrt{n!},\\
0\leq x\_i\leq m\_i,\quad i\in\{1,2,\dots,k\}.
\end{cases}$$
---
Existing ILP solvers are pretty happy with this problem formulation and solve it for $n\leq 40$ in a matter of seconds. The issue, however, is that their solutions quickly become incorrect.
I've tried to solve this problem with 3 solvers: [Gurobi](https://www.gurobi.com), [CPLEX](https://www.ibm.com/analytics/cplex-optimizer), and [GLPK](https://www.gnu.org/software/glpk/), with default parameters from Sage. The code using GLPK can be even [run online at SageMathCell](https://sagecell.sagemath.org/?z=eJxtkl9PwjAUxd9J-A7X-EArsAiJL8S98GKMmPDgA4lRUtjdUrO1tS1zavju9s82MbKEjd6e--s5N80whx0auzWq5JYIuhgOhgNwT2UghYzvLcnZ3krdfjgrnYpSgEtoK1_McilA5iCAmdDj1kx_RlB8Lx1NaZkR9VpBLjWoSQVcuGMSbrEyhNKEm3dtSUCn8BwW4oK-9I54qdzGI28wuxcWC9QrLpDptZaFZhUBI8sadTq6W60fRhOoWMOr1l76pA8IQCOqcRyPSwR-bGvmUu1KJDxCg3QCQgqBhWuuMVRo58MP5jr-PRtkEfc6ywnLsu1eCmM1cycQaJ7VC9w6A52bCB2ncecKSlkQRU9j_2MYD_C6Je0oQWfQbuXuDffetpf9xYQBkb62cUm-QS08sh5fJze0TVSHRL6jcMCalQc0pKFdQjj2s5i1iI23HprjMIDnEGrHVjhvhaSa-jo9PzvfFgW_Z2i0By3i7RHt7RGx0cz72wPT84pZr5i4hfvNPVfpMMaTqz9z4ekPZNHmXQ==&lang=sage&interacts=eJyLjgUAARUAuQ==). As an example, it computes the difference (= 928) for $n=15$.
I've compared their results to the known values from A061057. To my surprise, the celebrated commercial solvers Gurobi and CPLEX showed worst and quite similar performance by producing incorrect results for $n\geq 18$ and $n\geq 19$ respectively. The freeware GLPK did much better job by failing only for $n=27$ and $n\geq 31$.
Gurobi has an extensive [discussion on numerical issues](https://www.gurobi.com/documentation/9.5/refman/guidelines_for_numerical_i.html), which does not seem to be much relevant here, especially given that GLPK's performance was significantly better.
---
So, I have a few related questions:
>
> Q1. What is the cause of failure for ILP solvers in this problem?
>
>
>
>
> Q2. Is there a way to alleviate the issue and extend the range of correctly soluble $n$?
>
>
>
>
> Q3. Assuming that the correct answer is unknown, how must trust we can put in the result produced by an ILP solver and/or how we can verify it?
>
>
>
| https://mathoverflow.net/users/7076 | Reliability of ILP approach to number-theoretic optimization | Well, the question is broad, but let us address at least some of it.
>
> Q1. What is the cause of failure for ILP solvers in this problem?
>
>
>
Let's concentrate in the GLPK failure with $n=27$. [Here is a SageMath cell](https://sagecell.sagemath.org/?q=lfikax) for demonstrating what happens, based on your original code with some extra outputs. And here is the output.
```
we were using absolute tolerance 1e-05
we were using relative tolerance 1e-07
objective should <= 32.278769313503155
objective is 32.27878369954806
absolute exceedance 1.4386044902892081e-05
relative exceedance 4.4568133199780874e-07
(-3002360256, {2: 6, 3: 6, 5: 6, 7: 1, 11: 2, 13: 2}, {2: 17, 3: 7, 7: 2, 17: 1, 19: 1, 23: 1})
```
We note that the offered solution is invalid; the first factor is $2^6 3^6 5^6 7^1 11^2 13^2 = 104351247000000$, which is *more* than $\sqrt{27!}$.
Numerically, we were trying to maximize the objective value $\sum\_i x\_i \log p\_i$, while simultaneously constraining it to be at most $\log \sqrt{27!}$, represented as a float. Observe that the resulting objective *exceeds* the constraint. The solver seems to be happy because the constraint is *almost* satisfied. (It seems like it is slightly outside the tolerance, but it is pretty near, and I did not read enough of the documentation to be 100% sure that these are the only tolerance parameters affecting the behaviour.)
I think this is an instructive example: The failure is *not* caused by rounding errors when converting the original inputs into floats (although they could be a problem too). It is caused by the relatively loose-looking *tolerance* in the solver. Such tolerance values are in fact quite common in numerical solvers. And they might come as a big surprise if you are expecting exact results; especially since the user did not *ask* for such tolerance, it was just the solver's default.
You might hope that you could simply reduce the tolerance. But the GLPK manual warns: "(Do not change this parameter without detailed understanding its purpose.)"
---
>
> Q2. Is there a way to alleviate the issue and extend the range of correctly soluble n?
>
>
>
I would try either of the following options:
1. Use stricter tolerance *and* higher precision floats, if the solver allows this. Then read the solver's manual very carefully, to find what (if anything) it actually guarantees for the solutions. Then double-check the solutions (e.g. are they even feasible, in the exact mathematical formulation of the problem). But even if we rigorously validate that the solver's numerical solution is *feasible*, do we really know it was *optimal*? Well, we can hope...
2. Use an exact solver (like PPL in my comments). Since the problem involves logarithms of rational numbers, we must approximate them with rationals, within some precision. How much precision to needed, is patently problem-specific, but at least in this problem it should be doable. Fix some precision and work out the *worst possible error* that the rounding could cause, and if this is small enough not to change the integer results, we are happy. At least if we have confidence on the solver...
Detailed solution for this problem
----------------------------------
In this particular problem (factorial-splitting), there are only two places where we must approximate reals with rationals: the upper bound $\log\sqrt{n!}$ and the logarithms of the prime factors. We can simply make sure that the former is approximated upwards and the latter are approximated downwards. Then we will find a rational solution with objective value *at least* as big as the true optimal value. We might find something where the objective value in fact exceeds the true upper bound, but we can check that afterwards. If that happens, increase the precision until we find a feasible solution. Then we *know* we got the true optimum (assuming, of course, that we trust the computer and the ILP solver).
[Here is a SageMath cell](https://sagecell.sagemath.org/?q=ikgnlc) demonstrating the exact rational solution with $20!$, and what happens if the precision is not enough: we get an error message instead of an incorrect solution. So if a solution is found, I would be fairly confident it is correct. Here are the outputs with 10, 20 and 30 bits precision:
```
using 10-bit approximation
done in 1.243 s
Error: First factor too big
None
using 20-bit approximation
done in 1.042 s
(20, 800640, {2: 7, 3: 2, 5: 3, 7: 2, 13: 1, 17: 1}, {2: 11, 3: 6, 5: 1, 11: 1, 19: 1})
using 30-bit approximation
done in 1.061 s
(20, 800640, {2: 7, 3: 2, 5: 3, 7: 2, 13: 1, 17: 1}, {2: 11, 3: 6, 5: 1, 11: 1, 19: 1})
```
And here are the results for $n=40,\ldots,45$. OEIS has them up to $n=41$ and they match.
```
(40, 470500040794291200, {2: 10, 3: 15, 5: 2, 7: 1, 11: 1, 13: 3, 19: 1, 23: 1, 29: 1, 31: 1, 37: 1}, {2: 28, 3: 3, 5: 7, 7: 4, 11: 2, 17: 2, 19: 1})
(41, 2323929740464193400, {2: 35, 3: 3, 5: 2, 7: 4, 11: 1, 17: 1, 19: 1, 23: 1, 31: 1, 41: 1}, {2: 3, 3: 15, 5: 7, 7: 1, 11: 2, 13: 3, 17: 1, 19: 1, 29: 1, 37: 1})
(42, 20720967220237197312, {2: 26, 3: 13, 7: 3, 13: 3, 17: 1, 23: 1, 29: 1, 41: 1}, {2: 13, 3: 6, 5: 9, 7: 3, 11: 3, 17: 1, 19: 2, 31: 1, 37: 1})
(43, 69638496398882611200, {2: 13, 3: 10, 5: 7, 7: 3, 11: 1, 13: 1, 17: 2, 19: 2, 31: 1, 41: 1}, {2: 26, 3: 9, 5: 2, 7: 3, 11: 2, 13: 2, 23: 1, 29: 1, 37: 1, 43: 1})
(44, 61690805562507264000, {2: 20, 3: 5, 5: 6, 7: 6, 11: 2, 13: 2, 17: 2, 19: 1, 31: 1}, {2: 21, 3: 14, 5: 3, 11: 2, 13: 1, 19: 1, 23: 1, 29: 1, 37: 1, 41: 1, 43: 1})
(45, 416497216789463040000, {2: 24, 3: 2, 5: 6, 7: 3, 11: 1, 13: 3, 17: 1, 23: 1, 29: 1, 31: 1, 37: 1, 43: 1}, {2: 17, 3: 19, 5: 4, 7: 3, 11: 3, 17: 1, 19: 2, 41: 1})
```
---
>
> Q3. Assuming that the correct answer is unknown, how must trust we can put in the result produced by an ILP solver and/or how we can verify it?
>
>
>
If the solver is exact, and if you have verified that your input encoding does not cause spurious results (see above), the remaining suspects are the solver itself (both algorithm and implementation), and hardware (random bit flips in the memory, design errors like the infamous [Pentium FDIV bug](https://en.wikipedia.org/wiki/Pentium_FDIV_bug)). Some of these suspicions could be alleviated by independent runs (different hardware, different solver, different parameters).
Another prospect is **certificates** of optimality. You could hope that the ILP solver could output such a certificate and it could be independently verified. I'm not aware of any common ILP solver that does this, but there is some research towards this direction. Here are just a few hits from googling:
*Cheung, Kevin K. H.; Moazzez, Babak*, [**Certificates of optimality for mixed integer linear programming using generalized subadditive generator functions**](http://dx.doi.org/10.1155/2016/5017369), Adv. Oper. Res. 2016, Article ID 5017369, 11 p. (2016). [ZBL1387.90153](https://zbmath.org/?q=an:1387.90153).
*Cheung, Kevin K. H.; Gleixner, Ambros; Steffy, Daniel E.*, [**Verifying integer programming results**](http://dx.doi.org/10.1007/978-3-319-59250-3_13), Eisenbrand, Friedrich (ed.) et al., Integer programming and combinatorial optimization. 19th international conference, IPCO 2017, Waterloo, ON, Canada, June 26–28, 2017. Proceedings. Cham: Springer. Lect. Notes Comput. Sci. 10328, 148-160 (2017). [ZBL1418.90176](https://zbmath.org/?q=an:1418.90176).
| 3 | https://mathoverflow.net/users/171662 | 421204 | 171,353 |
https://mathoverflow.net/questions/421107 | 6 | $\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Gal{Gal}\newcommand{\ab}{\mathrm{ab}}$Let $G(\mathbb Q) = \Gal(\overline{\mathbb Q} / \mathbb Q)$ be the absolute Galois group. It's well-known that the abelianization $G(\mathbb Q)^{\ab}$ of $G(\mathbb Q)$ is isomorphic to $\Aut(\mathbb Q / \mathbb Z) = \widehat {\mathbb Z}^\times$, and that the fixed field $\mathbb Q^{\ab}$ of the commutator subgroup of $G(\mathbb Q)$ may be constructed by adjoining all roots of unity to $\mathbb Q$.
Abelian Galois groups are generalized by solvable Galois groups, or more generally hypoabelian Galois groups (recall that a group is hypoabelian if its derived series stabilizes at the trivial group, possibly after transfinitely many steps).
**Question 1:** What is the hypoabelianization of $G(\mathbb Q)$?
**Question 2:** What is the fixed field of the maximal perfect normal subgroup of $G(\mathbb Q)$?
(Recall that in general, the maximal perfect normal subgroup is the subgroup at which the derived series stabilizes, possibly after transfinitely many steps; the hypoabelianization of a group is its quotient by its maximal perfect normal subgroup.)
| https://mathoverflow.net/users/2362 | What is $\mathbb Q^{\mathrm{hypoab}}$? | My [comment](https://mathoverflow.net/questions/421107/what-is-mathbb-q-mathrmhypoab?noredirect=1#comment1081832_421107), Wojowu's [answer](https://mathoverflow.net/a/421125/82179), YCor's [comment](https://mathoverflow.net/questions/421107/what-is-mathbb-q-mathrmhypoab?noredirect=1#comment1081902_421125), and Z. M's [comment](https://mathoverflow.net/questions/421107/what-is-mathbb-q-mathrmhypoab?noredirect=1#comment1081944_421107) already contain everything we need. Let me provide a little more detail here. I will shift the indices by $1$ for reasons that will become apparent:
**Definition.** Set $K\_0 = \mathbf Q$ and let $K\_1 = \mathbf Q(\boldsymbol \mu)$ be the field obtained by adjoining the roots of unity $\boldsymbol \mu \subseteq \bar{\mathbf Q}$. Inductively define $K\_{i+1}=K\_i\big(\sqrt[\infty\ \ ]{K\_i^\times}\big)$, and set $K\_\infty = \underset{\substack{\longrightarrow \\ i}}{\operatorname{colim}} K\_i$.
We claim that this is the extension we're after. We first introduce some notation:
**Definition.** Given a profinite group $G$, its (*profinte*) *derived series* is the transfinite chain of *closed* subgroups
$$G = G^{(0)} \trianglerighteq G^{(1)} \trianglerighteq \cdots \trianglerighteq G^{(\alpha)} \trianglerighteq \cdots$$
defined by $G^{(\alpha+1)} = \overline{[G^{(\alpha)},G^{(\alpha)}]}$ and $G^{(\beta)} = \bigcap\_{\alpha < \beta} G^{(\alpha)}$ for any limit ordinal $\beta$ (which is already closed as each $G^{(\alpha)}$ is closed). One could alter the notation to distinguish it from the abstract derived series, but I will never use the latter (the same goes for the Kronecker–Weber theorem: it computes the *topological* abelianisation, not the abstract one!). Note that for any continuous surjective homomorphism $G \to H$ of profinite groups, the image of $G^{(\alpha)}$ is $H^{(\alpha)}$.
**Lemma.** *Let $G$ be a profinite group. Then $G^{(\omega + 1)} = G^{(\omega)}$, and this group is trivial if and only if $G$ is pro-soluble¹.*
*Proof.* For any finite group $G$, the descending chain $G^{(i)}$ stabilises after finitely many steps, so $G^{(\omega + 1)} = G^{(\omega)}$. The same statement for profinite groups follows since any closed normal subgroup $H \trianglelefteq G$ is the intersection of the open normal subgroups $U \trianglelefteq G$ containing it. Similarly, $G^{(\omega)} = 1$ if and only if the same holds in every finite quotient $G/U$, i.e. if and only if all $G/U$ are soluble. $\square$
Let's denote $G^{(\omega)}$ by $G^{(\infty)}$. For $n \in \mathbf N \cup \{\infty\}$, we will say that $G$ is *$n$-soluble* if $G^{(n)} = 1$, and we write $G^{n\text{-}\!\operatorname{sol}} = G/G^{(n)}$ for its maximal $n$-soluble quotient (in which we omit $n$ if $n = \infty$). For instance, $G$ is $1$-soluble if and only if it is abelian, and $\infty$-soluble if and only if it is pro-soluble (equivalently, hypoabelian as profinite group).
**Theorem.** *Let $\Gamma = \Gamma\_{\mathbf Q}$ be the absolute Galois group of $\mathbf Q$.*
1. *For $n \in \mathbf N \cup \{\infty\}$, the fixed field of $\Gamma^{(n)}$ is $K\_n$ (i.e. $K\_n$ is the maximal pro-soluble extension of derived length $\leq n$);*
2. *For $n \in \mathbf N \setminus \{0\}$, the Galois group $\operatorname{Gal}(K\_{n+1}/K\_n) = \Gamma^{(n)}/\Gamma^{(n+1)}$ is isomorphic to*
$$\operatorname{Hom}\_{\operatorname{cont}}\!\big(K\_n^\times,\hat{\mathbf Z}(1)\big),$$
*where $K^\times$ has the discrete topology and $\hat{\mathbf Z}(1) = \lim\_m \boldsymbol \mu\_m$ is the Tate module of $\bar{\mathbf Q}^\times$.*
*Proof.* Statement (1) is trivial for $n=0$, and is the Kronecker–Weber theorem for $n=1$. Statements (1) and (2) for finite $n \geq 2$ follow inductively by Kummer theory (see the corollary below). Finally, statement (1) for $n = \infty$ follows from the statement at finite levels, since $K\_\infty = \bigcup\_n K\_n$ and $G^{(\infty)} = \bigcap\_n G^{(n)}$. $\square$
Note also that the Galois group $\operatorname{Gal}(K\_1/K\_0)$ is isomorphic to $\operatorname{Aut}(\boldsymbol \mu) = \hat{\mathbf Z}^\times$. However, explicitly computing $\operatorname{Gal}(K\_{n+1}/K\_n)$ in a meaninful way is pretty hard, let alone saying anything about how the various pieces fit together.
---
**Edit:** After writing this answer, I became aware of the following two striking results:
**Theorem** (Iwasawa). *The Galois group $\operatorname{Gal}(K\_\infty/K\_1) = \Gamma^{(1)}/\Gamma^{(\infty)}$ is a* free *pro-soluble group $\widehat{F\_\omega}^{\operatorname{sol}}$ on countably infinitely many generators.*
So we know that $\Gamma^{\operatorname{sol}}$ sits in a short exact sequence
$$1 \to \widehat{F\_\omega}^{\operatorname{sol}} \to \Gamma^{\operatorname{sol}} \to \hat{\mathbf Z}^\times \to 1.$$
I find it hard to imagine that this sequence splits as a semi-direct product (but I am more optimistic about the derived length $\leq 2$ situation).
**Theorem** (Shafarevich). *Any finite soluble group $G$ occurs as a quotient of $\operatorname{Gal}(K\_\infty/\mathbf Q) = \Gamma^{\operatorname{sol}}$.*
A modern reference is Neukirch–Schmidt–Wingberg's *Cohomology of number fields*, Corollary 9.5.4 (Iwasawa) and Theorem 9.6.1 (Shafarevich). (This is a truly great book, but even at $>800$ pages it can be a bit terse at times.)
---
We used the following general result:
**Lemma** (Kummer theory). *Let $m \in \mathbf Z\_{>0}$, and $K$ be a field of characteristic not dividing $m$ that contains $\boldsymbol \mu\_m$.*
1. *The maximal abelian extension of exponent $m$ of $K$ is $L=K\big(\sqrt[m\ \ ]{K^\times}\big)$;*
2. *The map*
\begin{align\*}
\operatorname{Gal}(L/K) = \Gamma\_K^{\operatorname{ab}}/m &\to \operatorname{Hom}\_{\operatorname{cont}}\!\big(K^\times,\boldsymbol \mu\_m\big) = \left(K^\times/(K^\times)^m\right)^\vee \\
\sigma &\mapsto \left(a \mapsto \frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right)
\end{align\*}
*is an isomorphism of profinite groups, where $K^\times/(K^\times)^m$ has the discrete topology and $A^\vee$ denotes the Pontryagin dual of a locally compact abelian group $A$.*
We avoid the notation $\widehat A$ for Pontryagin duals, since it clashes with the notation for profinite completions. (Note that Z. M's [comment](https://mathoverflow.net/questions/421107/what-is-mathbb-q-mathrmhypoab?noredirect=1#comment1081944_421107) uses $(-)^\vee$ for a linear dual, which differs from my notation by a Tate twist.)
Because it's not very hard, let's include a proof.
*Proof.* For (2), by Pontryagin duality it suffices to show that the dual map
\begin{align\*}
K^\times/(K^\times)^m &\to \operatorname{Hom}\!\big(\Gamma\_K,\boldsymbol \mu\_m\big) = \left(\Gamma\_K^{\operatorname{ab}}/m\right)^\vee \\
a &\mapsto \left(\sigma \mapsto \frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right)
\end{align\*}
is an isomorphism. Note that it is well-defined since any two $m$-th roots of $a$ differ (multiplicatively) by an element of $\boldsymbol \mu\_m \subseteq K$, on which $\sigma$ acts as the identity. Since $\boldsymbol \mu\_m \subseteq K$, the $\Gamma\_K$-module $\boldsymbol \mu\_m$ has trivial action, so $\operatorname{Hom}\_{\operatorname{cont}}(\Gamma\_K,\boldsymbol \mu\_m) = H^1(K,\boldsymbol \mu\_m)$. The Kummer sequence
$$1 \to \boldsymbol \mu\_m \to \mathbf G\_m \stackrel{(-)^m}\to \mathbf G\_m \to 1$$
and Hilbert's theorem 90 compute $K^\times/(K^\times)^m \stackrel\sim\to H^1(K,\boldsymbol \mu\_m)$ via the map above. Now (1) follows since $\sigma \in \Gamma\_K$ is in the kernel of $\Gamma\_K \to \big(K^\times/(K^\times)^m\big)^\vee$ if and only if $\sigma$ fixes all $m$-th roots of elements in $K$. $\square$
**Corollay.** *Let $K$ be a field of characteristic $0$ containing $\boldsymbol \mu$.*
1. *The maximal abelian extension of $K$ is $L=K\big(\sqrt[\infty\ \ ]{K^\times}\big)$.*
2. *The map*
\begin{align\*}
\operatorname{Gal}(L/K) &\to \operatorname{Hom}\_{\operatorname{cont}}\!\big(K^\times,\hat{\mathbf Z}(1)\big) \\
\sigma &\to \left(a \mapsto \left(\frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right)\_{m \in \mathbf Z\_{>0}}\right)
\end{align\*}
*is an isomorphism of profinite groups.*
*Proof.* Take inverse limits over all $m \in \mathbf Z\_{>0}$ in the lemma above, noting that the inverse limit pulls out of $\operatorname{Hom}(K^\times,-)$. $\square$
---
¹Linguistic footnote: *soluble* and *solvable* mean the same thing. I used to think that this is one of those BrE vs AmE things (for instance, my Oxford Advanced Learner's dictionary does not contain the word *solvable* at all). But I think some folks in the UK also use *solvable*, so it's not entirely clear to me.
| 8 | https://mathoverflow.net/users/82179 | 421232 | 171,358 |
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