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https://mathoverflow.net/questions/434962 | 3 | The question is the following: given a matrix
$$A=\begin{pmatrix}
1& 2 & & & & \\
1& 0& 1 & & & \\
& 1& 0& 1 & &\\
& & \ddots & \ddots & \ddots & \\
& & & 1& 0 & 1\\
& & & & 1 &0
\end{pmatrix}.$$
Is it possible to give analytic expressions for the eigenvalues and eigenvectors of $A$?
Wang et. al [[1](https://www.sciencedirect.com/science/article/pii/S0024379522002920?via%3Dihub)] show that if the elements on the main diagonal are all 0, the eigenvalues and eigenvectors of $A$ can be expressed in trigonometric functions.
Thanks for your answer.
**References**
[[1](https://www.sciencedirect.com/science/article/pii/S0024379522002920?via%3Dihub)] W. Wang, C. M. Wang and S. L. Guo, [On the walk matrix of the Dynkin graph $D\_n$](https://www.sciencedirect.com/science/article/pii/S0024379522002920?via%3Dihub), Linear Algebra Appl. 653 (2022) 193-206.
| https://mathoverflow.net/users/495150 | Eigenvalues and eigenvectors of non-symmetrical tridiagonal matrix | Defining the $2 \times 2$ transfer matrix
\begin{align}\tag{1}
Q = \begin{pmatrix} -\lambda & 1 \\ -1 & 0 \end{pmatrix},
\end{align}
the characteristic polynomial (CP) of the $M \times M$ matrix $A\_M$ is given by
\begin{align}
P\_M(\lambda) &= \det(A\_M -\lambda \, I)\tag{2a}\\
&=\langle 1{-}\lambda, 2| \, Q^{M-1} \,|1,0\rangle \tag{2b}\\
&= 2 T\_M\left(-\frac \lambda 2 \right)
+ U\_{M-1}\left(-\frac \lambda 2 \right) \tag{2c}\\
&= 2\cos(M \varphi) + \frac{\sin(M \varphi)}{\sin(\varphi)}\tag{2d},
\end{align}
with Chebyshev polynomials $T\_M,U\_M$, and with $\lambda=-2\cos\varphi$.
The unnormalized right eigenvectors $A\_M x\_\mu=\lambda\_\mu x\_\mu$ have the elements
\begin{align}
x\_{\mu,m} = \langle 1, 0| \, (-Q)^m \, |1,0\rangle,\quad m=0,\ldots,M-1.\tag{3}
\end{align}
The eigenvector normalization can be related to the derivative $P\_M'(\lambda)$, see, e.g., <https://arxiv.org/abs/2103.10776> for details.
Regarding to your question, I don't think that a closed form expression exist for $\lambda\_\mu$ if $M>5$, as due to the left boundary term, the CP does not factorize in this case (up to one trivial eigenvalue $\lambda=\pm1$ if $M=3n\pm1$). However, it is often not necessary to explicitly calculate the eigenvalues, see the cited paper, as the CP together with the eigenvectors contains enough, or even more, information.
**Note added** (22.11.22,11:22):
If $(A\_M)\_{11}=a\_0$, then
\begin{align}
P\_M(\lambda) &= 2 T\_M\left(-\frac \lambda 2 \right)
+ a\_0 \, U\_{M-1}\left(-\frac \lambda 2 \right) \tag{4a} \\
&=2\cos(M \varphi) +a\_0 \frac{\sin(M \varphi)}{\sin(\varphi)}\tag{4b},
\end{align}
such that for $a\_0=0$ the eigenvalues are the well known zeroes of the Chebyshev polynomial of the first kind $T\_M$. This is the case in the paper [1] cited by the OP.
| 3 | https://mathoverflow.net/users/90413 | 435026 | 175,885 |
https://mathoverflow.net/questions/434798 | 2 | Let $M^n$ be a complete simply connected Riemannian manifold with $\operatorname{sec}\_M \leq 0$ (i.e. a Hadamard manifold) and assume that there is a constant $a \geq 0$ such that $\operatorname{sec}\_M \geq -a^2$. Do you know whether this implies an upper bound on the volume growth of area minimizing submanifolds i.e. whether there exists a function $f: [0,\infty) \rightarrow [0,\infty)$ such that for any area minizing submanifold $\Sigma^k \subseteq M$ and any $p\in M$ it holds that $\mathcal{H}^k(B(p,r)\cap \Sigma) \leq f(r)$ for every $r\geq 0$. Here $\mathcal{H}^k$ is the $k$-dimensional Hausdorff measure.
I would be very grateful for a reference or a counterexample in the case when $M = \mathbb{R}^n$.
| https://mathoverflow.net/users/313861 | Upper bound on volume growth of area minimizers | Consider the complex curve $w = z^k$ in $\mathbb{C}^2\simeq\mathbb{R}^4$, which is calibrated and therefore area-minimizing. The area of the part of this curve that lies inside the polydisk $\max\{|z|,|w|\}\le 1$ is $(k{+}1)\pi$. (The reason is that, because the standard Kähler form $\Omega = \tfrac{i}{2}(\mathrm{d}z\wedge\mathrm{d}\overline{z} + \mathrm{d}w\wedge\mathrm{d}\overline{w})$ calibrates this curve, the total area of this part of the curve is the sum of the areas of the projections, counted with multiplicities, onto the $z$- and $w$-axes. The projection onto the $z$-axis is $1$-to-$1$ onto the unit disk and the projection onto the $w$-axis is $k$-to-$1$ onto the unit disk (except for $w=0$).
Thus, there is no upper bound on the area of an area-minimizing surface inside a fixed ball in $\mathbb{R}^4$.
| 4 | https://mathoverflow.net/users/13972 | 435040 | 175,890 |
https://mathoverflow.net/questions/435025 | 4 | Let $f^{\*}$ be the usual decreasing rearrangement function of a measurable function $f$ on a measure space $(X, \mu)$. Let $1<p<n$ and set $$p'=\frac{pn}{n-p}.$$ Also, let $g$ be a positive function on $\mathbb{R}\_{+}$. Is it true that $$\left(\int\_{0}^{\infty}(f^{\*}(s))^{p'}(g(s))^{-p'}ds\right)^{p/p'}\leq c\int\_{0}^{\infty}(f^{\*}(s))^{p}(g(s))^{-p}s^{\frac{p}{p'}-1}ds$$ for some positive constant $c>0$ (**later added**: possibly depending on function $g$)?
Maybe it helps that, for $p, q\in (0, +\infty)$, the *Lorentz space* is defined via the quasi-norm $$||f||\_{p, q}=\left(\int\_{0}^{\infty}(f^{\*}(s))^{q}s^{\frac{q}{p}-1}ds\right)^{1/q}$$ and we have the Hoelder inequality $$||fg||\_{p, q}\leq C||f||\_{p\_{1},q\_{1}}||g||\_{p\_{2},q\_{2}},$$ where $\frac{1}{p}=\frac{1}{p\_{1}}+\frac{1}{p\_{2}}$ and $\frac{1}{q}=\frac{1}{q\_{1}}+\frac{1}{q\_{2}}$.
Any help is appreciated!
In the case when $g$ is non-decreasing, see [here](https://mathoverflow.net/questions/435128/inequality-with-decreasing-rearrangement-and-non-decreasing-function).
| https://mathoverflow.net/users/163368 | Inequality with decreasing rearrangement function | No (if $c$ cannot depend on $f^\*$ or $g$). Indeed, let $h:=f^\*/g$. Then $h$ can be any positive function and the inequality in question can be rewritten as
$$lhs:=\Big(\int\_0^\infty h(s)^{p'}ds\Big)^{p/p'}
\le rhs:=c\int\_0^\infty h(s)^p s^{p/p'-1}\,ds. \tag{1}\label{1}$$
Note that $p'>p>0$.
To obtain a contradiction, suppose that \eqref{1} holds for all positive functions $h$. Then it holds for all nonnegative functions $h$. Take now any real $k>0$ and let $h:=1\_{(k,k+1)}$. Then $lhs=1$ while
$rhs<ck^{p/p'-1}\to0$ as $k\to\infty$. So, we get $1\le0$, a contradiction. $\quad\Box$
---
The OP has changed the question, thus invalidating the above answer.
After the change, the answer is still no. Indeed, let us leverage the idea in the above answer as follows. Recall that $p'>p>0$, so that
$$a:=\frac2{1-p/p'}>2.$$
Let $(k\_j)\_{j\ge0}$ be a strictly increasing sequence of natural numbers such that
$$k\_j\sim j^a$$
(as $j\to\infty$). Let $(f\_n)\_{n\ge0}$ be any sequence of strictly positive real numbers decreasing to $0$. Let
$$f^\*:=\sum\_{n\ge0}f\_n\,1\_{[n,n+1)},\quad
g:=\sum\_{j\ge0}f\_{k\_j}\,\big(1\_{[k\_j,k\_j+1)}
+(j+1)^{(a+2)/p}\,1\_{[k\_j+1,k\_{j+1})}\big).$$
Then
$$lhs\ge\Big(\sum\_{j\ge0}\int\_{[k\_j,k\_j+1)}1\Big)^{p/p'}=\infty,$$
whereas
$$rhs<c\sum\_{j\ge0}\big(k\_j^{p/p'-1}+(j+1)^{-(a+2)}k\_{j+1}\big)<\infty,$$
since $k\_j^{p/p'-1}\sim j^{-2}$ and $k\_{j+1}\sim j^a$. So, in general the inequality $lhs\le rhs$ cannot hold for any real $c>0$, even if $c$ is allowed to depend on both $f^\*$ and $g$. $\quad\Box$
| 6 | https://mathoverflow.net/users/36721 | 435042 | 175,891 |
https://mathoverflow.net/questions/435050 | 10 | $\newcommand{\orb}{\mathrm{orb}}$Let $T$ ($K$) be the torus (Klein bottle) with one cone point of order $q\geq 2$. The presentation of their orbifold fundamental groups are easy to find. Namely,
$$\pi\_1^{\orb}(T)=\{a,b\ |\ (aba^{-1}b^{-1})^q=1\}.$$
$$\pi\_1^{\orb}(K)=\{a,b\ |\ (aba^{-1}b)^q=1\}.$$
Want to know the centre of these two groups. Are they trivial?
| https://mathoverflow.net/users/126243 | Centre of orbifold fundamental group of torus (Klein bottle) with one cone point | These groups have trivial centers. As one proof, they are both fuchsian and so embed in $\mathrm{PSL}(2, \mathbb{R})$ (well, the orientation preserving subgroups do). However, elements of $\mathrm{PSL}$ that commute must have common fixed points on the "circle at infinity" of the hyperbolic plane.
| 9 | https://mathoverflow.net/users/1650 | 435051 | 175,893 |
https://mathoverflow.net/questions/435070 | 3 | In IZF, we can easily prove there is a minimal cauchy complete field extending the rationals: the dedekind reals are cauchy complete, so just intersect all of its cauchy complete subfields.
CZF can still prove that there are dedekind reals, and that they are cauchy complete, but intersecting all of its cauchy complete subfields is impredicative (i.e. CZF doesn't have the powerset axiom). Also keep in mind that, without countable choice, you can't prove the cauchy reals are cauchy complete (but if they were they'd be minimal).
So, can CZF prove that there is a minimal extension of the rationals that is cauchy complete?
---
Another approach I thought of is taking the set of dedekind reals that can be formed using a well founded tree of cauchy sequences whose leaves are rational numbers. But I'm pretty sure the property of being a well founded tree has unbounded quantifiers, so we can't form a subset using it.
A set that I think is minimal is if we take the set of trees as above, but instead of requiring it to be well founded, we just require that the values at the nodes are determined by the leaves (i.e. for every other tree that has the same structure and leaves, it will also have the same internal nodes). But I couldn't figure out how to prove that this gives the minimal subfield we are looking for.
| https://mathoverflow.net/users/65915 | Does CZF prove there is a minimal cauchy completion of the rationals? | You can prove this from the regular extension axiom $\mathbf{REA}$ using the general theory of inductive definitions. See e.g. Theorem 5.11 in Aczel & Rathjen, [*Notes on Constructive Set Theory*](https://ncatlab.org/nlab/files/AczelRathjenCST.pdf). I suspect that $\mathbf{REA}$ is strictly necessary, although I don't think anyone has proved that specifically. A relevant paper on this kind of thing is Curi & Rathjen, [*Formal Baire Space in Constructive Set Theory*](https://doi.org/10.1515/9783110324921.123).
$\mathbf{REA}$ is usually seen as a safe axiom to assume if you need it for something. It isn't strictly predicative, but sits proof theoretically at the same level as inductive types regularly used in type theory. It's sometimes referred to as "semi-predicative."
| 3 | https://mathoverflow.net/users/30790 | 435072 | 175,903 |
https://mathoverflow.net/questions/435024 | 4 | Let $\mathbb{G}$ be a compact quantum group, $B$ be a $C^\*$-algebra together with a right action
$$\beta: B \to B\otimes C(\mathbb{G})$$ which is a non-degenerate $\*$-homomorphism satisfying $(\beta \otimes \iota)\beta = (\iota \otimes \Delta)\beta$ and the Podles density condition. A right $B$-Hilbert module $X$ is called $\mathbb{G}$-equivariant if it is equipped with a coaction
$$\delta: X \to X\otimes C(\mathbb{G})$$ such that
* $\beta(\langle x,y\rangle\_X) = \langle\delta(x), \delta(y)\rangle\_{X\otimes C(\mathbb{G})}$
* $(\delta \otimes \iota)\delta = (\iota \otimes \Delta)\delta$
* $[\delta(X)(1\otimes C(\mathbb{G}))]= X\otimes C(\mathbb{G})$.
---
Example: Let $U\in B(H\_U)\otimes C(\mathbb{G})$ be a finite-dimensional representation of $\mathbb{G}$. Then the right $B$-Hilbert module $H\_U\otimes B$ becomes $\mathbb{G}$-equivariant for
$$\delta(\xi\otimes b) = U\_{13} (\xi \otimes \beta(b)).$$
We denote this $\mathbb{G}$-equivariant module by $B\times U.$
---
I'm looking for references/proofs of the following facts:
(1) If $X$ is finitely generated as a right Hilbert module and $\mathbb{G}$-equivariant, then $X$ embeds equivariantly in a finite direct sum $\bigoplus\_{i=1}^n (B\times U\_i)$ where the $U\_i$ are irreducible representations of $\mathbb{G}$.
(2) If $\beta$ is a homogeneous action, i.e. $B$ is unital and $\{b\in B: \alpha(b) = b\otimes 1\}= \mathbb{C}1\_B$, then every $\mathbb{G}$-equivariant right Hilbert $B$-module decomposes as a direct sum of equivariant irreducible submodules.
---
Thanks in advance for your help.
| https://mathoverflow.net/users/470427 | Reference request: decomposability of $\mathbb{G}$-Hilbert modules | Let's assume that $G$ is a reduced compact quantum group, that is, the Haar state on $C(G)$ is faithful.
(1): A direct reference is [1, Lemma 4.2]. You can also get this from a careful study of [2, Théorème 3.2].
(2): This (for countably generated modules) follows from [2, Théorème 3.2] and [3, Proposition 4.6].
1. *Neshveyev, Sergey; Tuset, Lars*, [**Hopf algebra equivariant cyclic cohomology, (K)-theory and index formulas**](https://doi.org/10.1023/B:KTHE.0000031399.40342.7d), (K)-Theory 31, No. 4, 357-378 (2004). [ZBL1067.19002](https://zbmath.org/?q=an:1067.19002).
2. Roland Vergnioux, KK-théorie équivariante et opératour de Julg-Valette pour les groupes quantiques, Ph.D. thesis, Universit ́e Paris Diderot-Paris 7, 2002.
3. *De Commer, Kenny; Yamashita, Makoto*, [**Tannaka-Krein duality for compact quantum homogeneous spaces. I: General theory**](https://www.emis.de/journals/TAC/volumes/28/31/28-31abs.html), Theory Appl. Categ. 28, 1099-1138 (2013). [ZBL1337.46045](https://zbmath.org/?q=an:1337.46045).
| 3 | https://mathoverflow.net/users/9942 | 435074 | 175,904 |
https://mathoverflow.net/questions/435028 | 2 | ##### How many prime numbers of $b$ bits are there?
Beyond the prime number theorem, one can give explicit bounds on the number of primes below some integer $n$, or in a given interval. For instance, Rosser and Schoenfeld [RS62, Corollary 3] prove:
>
> *For $λ≥20.5$, $\frac{3}{5}λ/\lnλ < π(2λ) - π(λ) < \frac{7}{5}λ/\ln(λ)$ where $π(x)$
> denotes the number of primes $≤x$.*
>
>
>
From this, one immediately derives that
>
> *For $b≥5$, the number of $b$-bit primes is between $\frac{3}{5\ln 2} 2^b/b \simeq 0.865⋅2^b/b$ and $\frac{7}{5\ln 2} 2^b/b\simeq 2.02⋅2^b/b$.*
>
>
>
Experimentally for small values of $b$, one gets these values for $α = \frac{π(2^{b+1})-π(2^b)}{2^b/b}$:
$$\begin{array}{l|llllllllllllllllllllllllllllll}
b& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30\\
α & 1.00 & 1.00 & 0.750 & 1.25 & 1.09 & 1.22 & 1.26 & 1.34 & 1.32 & 1.34 & 1.37 & 1.36 & 1.38 & 1.38 & 1.39 & 1.39 & 1.39 & 1.40 & 1.40 & 1.40 & 1.41 & 1.41 & 1.41 & 1.41 & 1.41 & 1.41 & 1.41 & 1.41 & 1.42 & 1.42
\end{array}$$
My question is about explicit (and proven!) bounds that would be closer to the experimental data:
* It *seems* that for $b≥4$, the value of $α$ is always $≥1$: Is it true and proved?
* ~~The value of $α$ *seems* to increase with $b$ from $b = 5$: Is it true and proved?~~ False, *cf.* $b = 9$ or $b=12$ for instance.
* Rosser and Schoenfeld prove that $α ≤ 2.02$: Are there improvements? Is $α ≤ 1.5$ true?
[RS62]: J. Barkley Rosser and Lowel Schoenfeld. Approximate formulas for some function of prime numbers. Illinois J. Math. 6(1): 64-94 (March 1962). DOI: [10.1215/ijm/1255631807](http://doi.org/10.1215/ijm/1255631807).
| https://mathoverflow.net/users/16178 | Explicit bounds on number of primes of given size | Dusart proved in his [thesis](https://www.unilim.fr/laco/theses/1998/T1998_01.pdf) that
$$\frac{x}{\log x-1}<\pi(x)<\frac{x}{\log x-1.1},\qquad x\geq 60184.$$
It follows after a bit of calculation that
$$0.975\frac{x}{\log x}<\pi(2x)-\pi(x)<\frac{x}{\log x},\qquad x\geq 2^{31}.$$
Hence for $b\geq 31$, the value of $\alpha$ lies between $0.975/\log 2\approx 1.4066$ and $1/\log 2\approx 1.4427$.
| 8 | https://mathoverflow.net/users/11919 | 435077 | 175,906 |
https://mathoverflow.net/questions/435085 | 9 | Let $^na$ denote the $n$th tetration of $a$, so that $^0a=1$ and
$$^{n+1}a=a^{^na}$$
for $n=0,1,\dots$. (For complex $x$ and $y$, here we use the definition $x^y:=e^{y\ln x}$, where $\ln$ is the principal branch of the logarithm.)
It appears that $^9(-\sqrt2)$ is very close to $1$, but not exactly $1$ — so that the sequence $\big(^n(-\sqrt2)\big)$ is almost(?) periodic:
$$^9(-\sqrt2)-1\approx(4.99+1.51\, i)\times10^{-45}.$$
Is this a mere coincidence or is there an explanation for this?
| https://mathoverflow.net/users/36721 | The $9$th tetration of $-\sqrt2$ | This is not a huge coincidence: the idea is that the sequence $a\_n={}^{n}(-\sqrt{2})$ has small norm until $n=6$, then it gets out of hand for $n=7$ ($a\_7\sim-33+29i$), so that $a\_8=e^{a\_7\ln(-\sqrt{2})}$ is almost $0$ and $a\_9$ is almost $1$.
In general when sequences of this type get out of hand (increase exponentially in norm), it seems likely that at some point you will raise $e$ to a number with a big negative real part so a similar phenomenon will happen.
In fact, as you say the sequence is almost periodic, in the sense that $a\_{9n}$ converges to some point close to $1$ when $n$ goes to infinity. Let $k=\ln(-\sqrt{2}):=\pi i+\frac{1}{2}\ln(2)$ and let $f(x)=e^{kx}$, so that $a\_{n+1}=f(a\_n)$. Then $f'(x)=kf(x)$, and by the chain rule we get that $(f^n)'(x)=k^n\prod\_{i=1}^nf^i(x)$.
Now let $g(x)=f^9(x)$. We know that $|g(1)-1|<10^{-40}$, so by the proof of the [Banach fixed point theorem](https://en.wikipedia.org/wiki/Banach_fixed-point_theorem), to check that $g^n(1)$ converges to some point close to $1$ it is enough with the following proposition:
**Proposition:** If $x\in B(1,10^{-30})$, then $|g'(x)|<\frac{1}{2}$.
To prove it, note that $|g'(x)|=|k|^9\lvert\prod\_{i=1}^9f^i(x)\rvert\leq10^6\prod\_{i=1}^9|f^i(x)|$, so we just have to estimate $|f^i(x)|$. It is also easy to check that $|f^i(x)|<1.49$ for $i=1,\dots,6$. Also, for any $x$ with $|x|<1.5$ we have $|f'(x)|=|kf(x)|\leq4|f(x)|\leq4e^{3.5\cdot1.5}<1000$.
This implies by induction that $|f^i(x)-f^i(1)|\leq1000|f^{i-1}(x)-f^{i-1}(1)|$ for $i\leq6$, so $|f^6(x)-f^6(1)|\leq10^{-15}$, so $|f^7(x)-f^7(1)|<10^{-10}$ and $|f^8(x)|=|e^{kf^7(x)}|\sim |e^{-103-95i}|<|10^{-40}|$. So, finally, $|g'(x)|\leq10^6\prod\_{i=1}^9|f^i(x)|\leq10^6\cdot1.49^7\cdot100\cdot10^{-40}\leq\frac{1}{2}$, as we wanted.
| 22 | https://mathoverflow.net/users/172802 | 435088 | 175,908 |
https://mathoverflow.net/questions/435004 | 12 | Let $M$ be a manifold. Then is the ring of smooth functions $C^\infty(M,\mathbb{R})$ [formally smooth](https://en.wikipedia.org/wiki/Formally_smooth_map) over $\mathbb{R}$?
If it helps, feel free to assume that $M$ is compact.
---
(This is not a joke question. And yes, I know about $C^\infty$-rings and topological algebras, but I'm still interested in the above as stated.)
| https://mathoverflow.net/users/27013 | Are algebras of smooth functions formally smooth? | It seems $C^\infty(M;\mathbb{R})$ is not formally smooth for any positive-dimensional manifold. (The following argument came up in a discussion with Thomas Nikolaus, we later also found it [in this MO question](https://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials/9723#9723)):
Let's start by discussing the case $M=\mathbb{R}$. Our strategy will be to show that $\Omega^1\_{C^\infty(M;\mathbb{R})/\mathbb{R}}$ is not a projective $C^\infty(M;\mathbb{R})$-module. For a projective module $P$ over some ring $R$, different elements $x,x'\in P$ can always be separated by homomorphisms $P\to R$ (e.g. by embedding $P$ into a free module and using coordinate functions), equivalently the map $P\to P^{\vee\vee}$ into the double dual is injective. So it suffices to exhibit distinct elements of $\Omega^1\_{C^\infty(M;\mathbb{R})/\mathbb{R}}$ which cannot be separated by homomorphisms to $C^\infty(M;\mathbb{R})$.
Claim 1: In $\Omega^1\_{C^\infty(\mathbb{R};\mathbb{R})/\mathbb{R}}$, $de^x \neq e^x dx$. To detect this, it suffices to find a derivation $\partial: C^\infty(\mathbb{R};\mathbb{R})\to M$ with $\partial e^x \neq e^x \partial x$. Let $K$ be the fraction field of the local ring of stalks of smooth functions on $\mathbb{R}$ at $0$. Then $e^x$ and $x$ are algebraically independent in $K$, so we may find a transcendence basis of $K$ over $\mathbb{R}$ consisting of $e^x$, $x$ and other elements $a\_i$. It follows that $\Omega^1\_{K/\mathbb{R}}$ is a free module on $de^x$, $dx$ and the $da\_i$, in particular we find a derivation $K\to K$ taking $de^x$ to $1$ and $dx$ to $0$. Precomposing with the map $C^\infty(\mathbb{R};\mathbb{R})\to K$ yields the claim.
Claim 2: Every homomorphism $\Omega^1\_{C^\infty(\mathbb{R};\mathbb{R})}\to C^\infty(\mathbb{R};\mathbb{R})$ takes $de^x$ and $e^x dx$ to the same element. Indeed, such homomorphisms correspond to derivations $C^\infty(\mathbb{R};\mathbb{R})\to C^\infty(\mathbb{R};\mathbb{R})$, which are given by smooth vector fields. But any smooth vector field $f\cdot \frac{\partial}{\partial x}$ takes $de^x$ to $f\cdot\frac{\partial}{\partial x} e^x$, and $e^x dx$ to $f\cdot e^x \frac{\partial}{\partial x} x$, which agree.
So $\Omega^1\_{C^\infty(\mathbb{R};\mathbb{R})/\mathbb{R}}$ cannot be projective, and thus $C^\infty(\mathbb{R};\mathbb{R})$ is not formally smooth.
A zoomed-out version of the above argument is that the $C^\infty(M;\mathbb{R})$-linear dual of $\Omega^1\_{C^\infty(M;\mathbb{R})/\mathbb{R}}$ is always vector fields, and so the double dual is always $1$-forms. Hence the above observation tells us that the map from "algebraic $1$-forms" $\Omega^1\_{C^\infty(M;\mathbb{R})/\mathbb{R}}$ into its double dual identifies with the map to "smooth $1$-forms" $\Omega^1(M;\mathbb{R})$, and we have used algebraically independent functions with a linear dependence between their derivatives to show that map is not injective.
For $M=\mathbb{R}^n$, the same argument works for $de^{x\_1}$ and $e^{x\_1}dx\_1$ with the coordinate function $x\_1$. Finally, on a general manifold, take a coordinate ball around an arbitrary point, and extend the functions $e^{x\_1}$, $x\_1$ in any way. If $\psi$ is a function supported on our ball (and $1$ in a smaller neighbourhood of $x$), then we still have that $\psi \cdot d\widetilde{e^{x\_1}}$ and $\psi\cdot \widetilde{e^{x\_1}}d \widetilde{x\_1}$ agree as "smooth $1$-forms", but not as "algebraic $1$-forms", as witnessed by going into the fraction field of stalks around our point.
EDIT: For a short proof that the stalks of $e^x$ and $x$ at $0$ are algebraically independent, first observe that since these are analytic functions, any polynomial relation $f(e^x,x)=0$ which holds in a neighbourhood of $0$ holds on all of $\mathbb{R}$, and in fact on all of $\mathbb{C}$. Since every nonzero $a$ is attained infinitely often as value of $e^x$, each of the polynomials $f(a,x)$ for fixed nonzero $a$ has infinitely many zeros in $x$, thus vanishes. So the coefficient polynomials in $f(y,x) = \sum g\_i(y) x^i$ each have infinitely many zeros, thus vanish, and so $f=0$ as polynomial, proving algebraic independence.
EDIT2: In the zero-dimensional case, if $M$ is finite, $C^\infty(M;\mathbb{R})=\mathbb{R}^M$ is clearly formally smooth over $\mathbb{R}$. In the infinite case, they shouldn't be (following the comment by Martin Brandenburg). Indeed, a variant of the argument works. We still have that $C^\infty(M;\mathbb{R})$-valued derivations are vector fields, hence trivial. However, there exist nontrivial derivations on $\mathbb{R}^M$. Identify $M=\mathbb{N}$, fix a nonprincipal ultrafilter on $\mathbb{N}$, and let $K$ be the corresponding ultraproduct (which is automatically a field). The function $f: n\mapsto n$ is algebraically independent from $1$ in $K$, and so we find a derivation $K\to K$ taking $df$ to $1$. In particular, $df\neq 0$ in $\Omega^1\_{\mathbb{R}^\mathbb{N}/\mathbb{R}}$. (If your manifolds are not second-countable, they still contain $\mathbb{N}$ as retract and you can still pull this nonzero element back.)
| 10 | https://mathoverflow.net/users/39747 | 435099 | 175,909 |
https://mathoverflow.net/questions/420365 | 4 | $\newcommand{\real}{\mathrm{real}}$I am having trouble with understanding the axiom (OS3) in [this book](https://link.springer.com/book/10.1007/978-1-4612-4728-9) by Glimm and Jaffe.
It defines
\begin{equation}
\mathcal{A} = \left \{ A(\phi) = \sum\_{j = 1}^N c\_j \exp \left( \phi(f\_j) \right) \; \Big \vert \; c\_j \in \mathbb{C}, f\_j \in \mathcal{D} \right \}
\end{equation}
where $\mathcal{D}$ is supposed to be the space of compactly supported, smooth functions on $\mathbb{R}^d$ and $\phi \in \mathcal{D}'(\mathbb{R}^d)$.
Now, this is notationally somewhat messed up (imho) but it is also stated earlier in the text, that $\mathcal{D}'(\mathbb{R}^d)$ should refer to *real* distributions hence, I assume that $\mathcal{D}$ should also refer to *real* functions - but I have not found this statement anywhere.
They then go on by defining the time-reflection operator $\theta : \mathcal{A} \to \mathcal{A}$ and consider a special supspace $\mathcal{A}\_+ \subset \mathcal{A}$ and give the axiom:
\begin{equation}
0 \le \int\_{\mathcal{D}'(\mathbb{R}^d)} \left( \theta A \right)^- A \mathrm{d} \mu
\end{equation}
where the superscript $-$ is supposed to mean complex conjugation.
This might sound okay but then they state that this is equivalent to the positive definiteness of matrices of the form
\begin{equation}
M\_{i j} = \int\_{\mathcal{D}'(\mathbb{R}^d)} \exp \left[ i \phi \left( f\_i - \theta f\_j \right) \right] \mathrm{d} \mu \left( \phi \right)
\end{equation}
with $f\_j$ in the new space $\mathcal{D}\_{\real}(\mathbb{R}^d)$.
I feel that it is completely unclear whether they are talking about $\mathcal{D}$ as a space of real or complex functions.
Furthermore, I cannot see the equivalence of the two statements unless $\mathcal{A}$ was defined with an additional $i$ in the exponent and with $\mathcal{D} \supset \mathcal{D}\_{\real}(\mathbb{R}^d)$ as a space of *real* functions.
[nlab](https://ncatlab.org/nlab/show/Osterwalder-Schrader+theorem) also gives the almost identical definition as the book but without stating anything about real or complex spaces.
| https://mathoverflow.net/users/18936 | Understanding the Osterwalder-Schrader conditions as formulated by Glimm and Jaffe | Since this has confused me multiple times, I write this answer in the hope that it might help others.
First, recall that reflection positivity as formulated by [Osterwalder and Schrader](https://link.springer.com/article/10.1007/BF01645738) states that
\begin{equation}
\sum\_{n,m} G\_{n+m} \left( \theta f\_n^\* \otimes f\_m \right) \ge 0
\end{equation}
for all finite sequences $(f\_n)\_{n \in \mathbb{N}}$ of **complex** Schwartz functions on $(\mathbb{R}^d)^n$ with supports in $(\mathbb{R}\_{\ge 0} \times \mathbb{R}^{d-1})^n$.
Here,
\begin{equation}
\left( \theta f \right) \left( x^1\_1, \dots, x^1\_d, x^2\_1, \dots x^d\_d \right)
=
f \left( -x^1\_1, x^1\_2 \dots, x^1\_d, -x^2\_1, x^2\_2 \dots x^2\_d, \dots \right)
\end{equation}
for all $x^1, \dots x^d \in \mathbb{R}^d$.
Second, the whole idea of having a measure $\mu$ on a space of distributions is to be able to write the $G\_n$s as moments of that measure.
From here on $\mu$ will be a corresponding measure on the space of **real** distributions.
Also, let $Z$ denote the moment-generating function of $\mu$ (assuming it exists) and $\hat{\mu}$ its characteristic function.
Now, for any finite sequences $(\phi\_n, c\_n)\_{n \in \mathbb{N}}$ of **real** Schwartz functions on $\mathbb{R}^d$ with supports on $\mathbb{R}\_{\ge 0} \times \mathbb{R}^{d-1}$ and complex numbers, define
\begin{equation}
f\_a = \frac{1}{a!} \sum\_n c\_n \phi\_n^{\otimes a}
\end{equation}
for all $a \in \mathbb{N}$.
Then a simple calculation (assuming everything is well-defined) shows that
\begin{equation}
\sum\_{m, n} c\_n^\* c\_m Z \left( \theta \phi\_n + \phi\_m \right)
=
\sum\_{a, b = 0}^\infty G\_{a+b} \left( \theta f\_a^\* \otimes f\_b \right)
=
\lim\_{N \to \infty} \sum\_{a, b = 0}^N G\_{a+b} \left( \theta f\_a^\* \otimes f\_b \right)
\ge 0 \, .
\end{equation}
Similarly, we obtain a connection with the characteristic function by setting
\begin{equation}
g\_a = \frac{\left(-i\right)^a}{a!} \sum\_n c\_n \phi\_n^{\otimes a}
\end{equation}
for all $a \in \mathbb{N}$.
Then
\begin{equation}
\sum\_{m, n} c\_n^\* c\_m \hat{\mu} \left( \theta \phi\_n - \phi\_m \right)
=
\sum\_{a, b = 0}^\infty G\_{a+b} \left( \theta g\_a^\* \otimes g\_b \right)
=
\lim\_{N \to \infty} \sum\_{a, b = 0}^N G\_{a+b} \left( \theta g\_a^\* \otimes g\_b \right)
\ge 0 \, .
\end{equation}
Observe that we need either $Z( \theta \phi\_n + \phi\_m )$ or $\hat{\mu}( \theta \phi\_n - \phi\_m )$.
| 1 | https://mathoverflow.net/users/18936 | 435109 | 175,912 |
https://mathoverflow.net/questions/435110 | 41 | Today, somebody posted on the nLab a link to Kirti Joshi's preprint on the arXiv from last month: <https://arxiv.org/abs/2210.11635>
In that preprint, Kirti Joshi claims that
* he agrees with Scholze and Stix that Mochizuki's proof of ABC is incomplete,
* Scholze and Stix's rigidity claim in Remark 9 of their paper "[Why abc is still a conjecture](https://ncatlab.org/nlab/files/why_abc_is_still_a_conjecture.pdf)" is wrong
* "This paper provides the first proof of Mochizuki’s non-redundancy claim by establishing that the isomorphs are of distinct arithmetic-geometric provenance (and even continuous families of isomorphs exist) and therefore are non-redundant"
If these results are confirmed, what are the consequences of this preprint on the validity of IUT as a theory and Mochizuki's proof of the ABC conjecture?
| https://mathoverflow.net/users/483446 | Consequences of Kirti Joshi's new preprint about p-adic Teichmüller theory on the validity of IUT and on the ABC conjecture | To give a simple answer: There would be no direct implications. The paper doesn't claim a proof of Corollary 3.12, the ABC conjecture, or any other Diophantine inequalities. I'm pretty sure that, if Joshi had a proof of one of these, he would say it.
| 34 | https://mathoverflow.net/users/18060 | 435117 | 175,914 |
https://mathoverflow.net/questions/435115 | 8 | Is there a predicate $P(x)$ such that $\mathrm{ZF} \vdash \exists! x. P(x)$, and $\mathrm{ZF} \vdash \forall x. P(x) \to (x \subseteq \mathbb R)$, but $\mathrm{ZF} \nvdash \forall x. P(x) \to \mathsf{Borel}(x)$?
Since ZF is consistent with the statement that every real subset is Borel, we have $\mathrm{ZF} \nvdash \forall x. P(x) \to \neg \mathsf{Borel}(x)$. So this is asking whether there is a definable set in ZF such that ZF can neither prove or disprove it is Borel.
I think there is no easy, round-the-corner proof here, but the general direction is to prove that if a model has a non-Borel set then it is non-definable.
| https://mathoverflow.net/users/136535 | Definable set in ZF that cannot be proved to be Borel | There is a common view amongst mathematicians that uses of the axiom of choice must somehow be inherently undefinable, and that if one sticks with definable sets only, then everything will be very nicely behaved.
The question here is an instance of this perspective, since it is asking whether every definable set of reals must be Borel.
But unfortunately, the view is simply mistaken. One way to see this is to observe that the constructible universe L of Gödel has a [definable well ordering of the set-theoretic universe](https://mathoverflow.net/a/6600/1946). Thus, in $L$ all uses of the axiom of choice can be undertaken in a definable manner, since one can always make the choices by picking the least element with respect to the definable well order.
For example, in $L$ there is an $L$-least non-Borel set. So we can take $Px$ to assert: $x$ is the least non-Borel set in $L$, using the definable $L$-order. This defines a unique set in any model of set theory, a set of reals, and ZF cannot prove that it is Borel, because it is not Borel if $V=L$. But it could be Borel in other models of set theory, since by forcing we can make every set of reals in $L$ countable and hence Borel.
Similarly, Asaf mentions another example in his comment, namely, the set of reals of the constructible universe $\newcommand\R{\mathbb{R}}\R^L$. This is definable in $V$, but it is consistent that this set is not Borel. This definition has very low complexity in the descriptive set-theoretic hierarchy, $\Sigma^1\_2$. That is, it is defined by a formula of the form $\exists y\in\R\forall z\in\R\ \varphi(x,y,z)$, where the quantifiers range over $\R$ and $\varphi$ involves only quantification over integers.
More generally, any set at all can become definable in a forcing extension of the universe. Indeed, there is a single definition $P$, such that for any model of set theory $M$ and any object $a\in M$, there is a forcing extension $M[G]$ in which $P$ defines $a$ in $M[G]$. For example, let $Px$ hold when $x$ is the set coded (in one of the canonical manners) by the class of ordinals defined by the GCH pattern on the regular cardinals, with some default value if the coding fails. Since by Easton's theorem we can control the GCH pattern as we wish on the regular cardinals, we can use this method to make any desired set definable by this same definition. And we can arrange the forcing to not add reals or sets of reals, so the concept of Borel will be the same in $M$ as in $M[G]$.
Another universal definition arises with the [Universal finite set](https://arxiv.org/abs/1711.07952) and its generalization to [The universal $\Sigma\_1$ definition](https://arxiv.org/abs/1909.09100), which shows how to make any set you like definable on the universal finite sequence.
Thus, there is no necessary implication from definable-in-ZF to always-having-nice-regularity-properties. Even badly behaved sets can be definable, and indeed, any set at all can become definable.
| 23 | https://mathoverflow.net/users/1946 | 435118 | 175,915 |
https://mathoverflow.net/questions/435128 | 2 | This question is a continuation of the question [here](https://mathoverflow.net/questions/435025/inequality-with-decreasing-rearrangement-function/435042?noredirect=1#comment1120917_435042).
Let $f^{\*}$ be the usual decreasing rearrangement function of a measurable function $f$ on a measure space $(X, \mu)$. Let $1<p<n$ and set $$p'=\frac{pn}{n-p}.$$ Also, let $g$ be a positive, non-decreasing function on $\mathbb{R}\_{+}$. Is it true that $$\left(\int\_{0}^{\infty}(f^{\*}(s))^{p'}(g(s))^{-p'}ds\right)^{p/p'}\leq c\int\_{0}^{\infty}(f^{\*}(s))^{p}(g(s))^{-p}s^{\frac{p}{p'}-1}ds$$ for some positive constant $c>0$ (possibly depending on function $g$)?
Any help is appreciated!
| https://mathoverflow.net/users/163368 | Inequality with decreasing rearrangement and non-decreasing function | $\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$The answer to this question is yes.
Indeed,
let $h:=f^\*/g$. Then $h$ is a nonincreasing function and the inequality in question can be rewritten as
\begin{equation\*}
\Big(\int\_0^\infty h(s)^{p'}ds\Big)^{p/p'}
\le c\int\_0^\infty h(s)^p s^{p/p'-1}\,ds. \tag{1}\label{1}
\end{equation\*}
Note that $p'>p>0$.
Letting
\begin{equation\*}
a:=p'/p>1\quad\text{and}\quad u:=h^p,
\end{equation\*}
we see that it is enough to show that
\begin{equation\*}
L\overset{\text{(?)}}\le \frac1a\,R, \tag{2}\label{2}
\end{equation\*}
where
\begin{equation\*}
L:=\Big(\int\_0^\infty u(s)^a\,ds\Big)^{1/a},\quad
R:=\int\_0^\infty u(s)s^{1/a-1}\,ds,
\end{equation\*}
and $u$ is a nonnegative nonincreasing function.
By approximation, without loss of generality (wlog) the function $u$ is piecewise constant, with just a finite number of discontinuities and with $u(s)=0$ for all large enough $s>0$. So, wlog
\begin{equation}
u(s)=\int\_{(s,\infty)}\mu(dt)
\end{equation}
for some finite measure $\mu$ with a finite support $S\_\mu\subseteq(0,\infty)$ and all real $s\ge0$. Then
\begin{equation}
R=\int\_0^\infty ds\,s^{1/a-1}\,\int\_{(s,\infty)}\mu(dt)
=\int\_{(0,\infty)}\mu(dt)\,\int\_0^t ds\,s^{1/a-1}
=a\int\_{(0,\infty)}\mu(dt)\,t^{1/a}
=a\int\_{(0,\infty)}\nu(dt),
\end{equation}
where $\nu(dt):=\mu(dt)\,t^{1/a}$, and
\begin{equation\*}
L:=\Big(\int\_0^\infty ds\,\Big(\int\_{(s,\infty)}\nu(dt)\,t^{-1/a}\Big)^a\Big)^{1/a}.
\end{equation\*}
Since $a>1$, $L^a$ is convex in $\nu$ (actually, by Minkowski's inequality, even $L$ itself is convex in $\nu$), whereas $R$ is affine in $\nu$.
Note that the support of the measure $\nu$ is finite and,
by homogeneity, wlog $\nu$ is a probability measure.
So, we have the following:
Given any value of $R$, the maximum of $L$ over all probability measures $\nu$ with support $S\_\nu$ in a given compact interval $I$ (say of the form $[0,N]$) and with the cardinality of $S\_\nu$ not exceeding a given natural number is attained at a Dirac measure $\de\_z$ supported on a singleton set $\{z\}\subseteq I$. So, wlog $\nu=\de\_z$, and then
\begin{equation}
L^a=\int\_0^\infty ds\,\Big(\int\_{(s,\infty)}\de\_z(dt)\,t^{-1/a}\Big)^a
=\int\_0^\infty ds\,z^{-1}\,1(z>s)=1
\end{equation}
and
\begin{equation}
R=a\int\_{(0,\infty)}\de\_z(dt)=a.
\end{equation}
Thus, \eqref{1} is proved.
$\quad\Box$
We also see that the constant factor $\frac1a$ in \eqref{2} is the best possible one. So, the best possible constant factor in \eqref{1} is $c=\frac1a=\frac p{p'}$.
| 2 | https://mathoverflow.net/users/36721 | 435130 | 175,917 |
https://mathoverflow.net/questions/360856 | 8 | In the Euclidean plane, for a closed smooth curve of length $\ell$ whose curvature is bounded above by $\epsilon$ we have the inequality
$$ \ell \ge 2\pi \epsilon^{-1} $$
which follows from the fact that the total curvature is $2|k|\pi$ with $k \not= 0$ the winding number.
Is there a known generalisation of this to closed curves in a CAT(0) (simply connected with negative sectional curvature) Riemannian manifold? That is given such a manifold $X$, is there a function $f : [0, 1] \to [0, +\infty[$ (say) so that for any closed smooth curve in $X$ of length $\ell$ and curvature bounded above by $\epsilon$ we have $\ell \ge f(\epsilon)$? (of course we want $\lim\_{\epsilon\to 0} f(\epsilon) = 0$ and for pinched negative curvature I would expect it to go faster than $1/\epsilon$).
We can also ask a purely metric version of this question: given a CAT(0)-space $X$, is there a function $f : [0, 1] \to [0, +\infty[$ (say) so that for any closed curve in $X$ of length $\ell$ which is a local $(1+\epsilon)$-quasi-geodesic we have $\ell \ge f(\epsilon)$?
(By "local $(1+\epsilon)$-quasi-geodesic" I mean a curve $\gamma$ such that for any two points $x, y \in X$ such that $d(x, y) \le C$ and $x, y$ lie on $\gamma$, if $a$ is the length of the (shortest) arc between $x$ and $y$ on $\gamma$ then $a - d(x, y) \le \epsilon$. Here $C$ is a constant depending on $X$.)
**Edit following comments**: in Gromov-hyperbolic spaces it seems that the condition of being a "local-quasi-geodesic" implies (for sufficiently small $\epsilon$) that the curve is a global quasi-geodesic, in particular it cannot close. I think the proof of Theorem 1.13, p. 405 of Bridson--Haefliger can be immediately adapted to do this (and for this particular problem points (1) and (2) of the theorem are sufficient).
So we can take $f = +\infty$ in a neighbourhood of 0 (depending on the hyperbolicity constant). The comment by shurtados shows that for the hyperbolic plane we can take the neighbourhood to be $[0, 1[$.
As noted in Ycor's comment the "purely metric" version is not optimal for CAT(0) spaces (as opposed to hyperbolic) and for these a version that would include non-Riemannian $X$ and singular curves should probably involve some sort of "curvature measure" on the curve whose integral would be computable. This question seems interesting even for euclidean space.
| https://mathoverflow.net/users/32210 | Length and curvature for closed curves in negatively curved spaces | The Reshetnyak majorization theorem (see [9.56](https://arxiv.org/pdf/1903.08539v5.pdf)) states that any closed rectifiable curve $\alpha$ in a CAT(0) length space $U$ can be *majorized* by a convex plane figure $F$; that is, there is *short* (= 1-Lipschitz) map (= majorization) $m\colon F\to U$ such that $m|\_{\partial F}$ is the arc-length parametrization of $\alpha$.
Note that majorization does not decrease the curvature; that is, curvature of $\alpha$ cannot be smaller than curvature of $\partial F$ at the corresponding point.
Therefore, the inequality
$$ \ell \ge 2\cdot \pi\cdot \varepsilon^{-1}$$
holds in CAT(0) spaces as well.
| 4 | https://mathoverflow.net/users/1441 | 435142 | 175,924 |
https://mathoverflow.net/questions/434970 | 13 | As shown in Simpson's excellent [Subsystems of Second Order Arithmetic](https://www.personal.psu.edu/t20/sosoa), the ‘big five’ system ATR$\_0$ from second-order reverse mathematics is equivalent to the following principle:
*For arithmetical $\varphi$ such that $(\forall n)(\exists \text{ at most one } X)\varphi(X, n)$, there is $Z$ such that $(\forall m)(m\in Z\leftrightarrow (\exists X)\varphi(X,m))$.*
My question is whether this ‘at most one’ comprehension principle is also equivalent to the following ‘at most finitely many’ principle, where $w^{1^\*}$ is a finite sequence of sets of length $\lvert w\rvert$.
*For arithmetical $\varphi$ such that
$$
(\forall n)(\exists w^{1^\*})(\forall X)\Bigl[\varphi(X, n)\rightarrow (\exists i< \lvert w\rvert)(X=w(i))\Bigr],\tag{\*}\label{star}
$$
there is $Z$ such that $(\forall m)(m\in Z\leftrightarrow (\exists X)\varphi(X,m))$.*
Note that the condition \eqref{star} guarantees that there are only finitely many $X$ satisfying $\varphi(X,n)$, for fixed $n$.
| https://mathoverflow.net/users/33505 | "At most one" versus "at most finitely many" | The answer is positive, assuming extra induction, and a sketch is as follows.
Let $\varphi(X,n)$ be as in (\*).
1. Define an analytic code $A\_n$ as follows $X\in A\_n\leftrightarrow \varphi(X, n)$.
2. Use induction (say for $\Sigma\_2^1$-formulas) to show that $A\_n$ can be enumerated (as a finite sequence).
3. Now use V.4.10 from [Subsystems of Second-order Arithmetic](https://www.personal.psu.edu/t20/sosoa) to show that $\cup\_{n\in\mathbb{N}}A\_n$ can be enumerated, say by $(X\_m)\_{m\in \mathbb{N}}$.
4. Then for all $n\in \mathbb{N}$, we have
$$
(\exists X\subset \mathbb{N})\varphi(X, n)\leftrightarrow (\exists m\in \mathbb{N})\varphi(X\_m, n),
$$
and we are done. Note that step 3. makes use of ATR$\_0$.
| 4 | https://mathoverflow.net/users/33505 | 435159 | 175,931 |
https://mathoverflow.net/questions/434423 | 8 | This is something which I'm sure is well known to experts which I would appreciate some information about. In his paper [1], Scholze proves (e.g. Theorem 8.4, Theorem 8.8) that on a proper adic space $X$ over $\textrm{Spa}(k,\mathcal{O}\_{k})$, with $k$ an appropriate non-archimedean field, the natural map $H^{i}(X\_{\textrm{pet}}, \hat{\mathbb{Z}}\_{p}) \otimes B\_{\textrm{dR}} \to H^{i}(X\_{\textrm{pet}}, \mathbb{B}^{+}\_{\textrm{dR}})$ is an isomorphism, where I use the notation "pet" for the pro-etale site. I think a similar statement should be true for $X$ the adic space associated to a smooth algebraic variety over $k$, but I cannot find the appropriate reference.
I am aware of work [2] of Diao, Lan, Liu and Zhu which works with a general smooth variety, but I am really interested in the "intermediate" comparison between cohomology wich coefficients in $\hat{\mathbb{Z}}\_{p}$ and $\mathbb{B}^{+}\_{\textrm{dR}}$ rather than the actual comparison isomorphism of $p$-adic Hodge theory, and it seems to me they only handle this intermediate fact in the proper case. The related work [3] seems to do something like what I want in Proposition 3.3.4, but in a compactly supported setting which isn't really what I'm interested in.
If it helps I am really just looking to understand the maps $H^{i}(X\_{\textrm{pet}}, \hat{\mathbb{Z}}\_{p}) \otimes B\_{\textrm{dR}} \to H^{i}(X\_{\textrm{pet}}, \mathbb{B}^{+}\_{\textrm{dR}})$ when $X = \mathbb{G}\_{m}^{w}$ is the $w$-fold product of the multiplicative group, so an explicit description of this map would also be appreciated.
Edit: I may have found what I want in Theorem 7.14 of [4]; I will take a look at the paper and report back.
1. <https://www.math.uni-bonn.de/people/scholze/pAdicHodgeTheory.pdf>
2. <https://arxiv.org/pdf/1803.05786.pdf>
3. <https://arxiv.org/pdf/1912.13030.pdf>
4. <https://arxiv.org/pdf/1801.01779.pdf>
| https://mathoverflow.net/users/113933 | $p$-adic comparison of cohomology with coefficients in $\mathbb{Z}_{p}$ and $\mathbb{B}_{\textrm{dR}}$ on general smooth algebraic varieties | The result is false in the open case.
If true, a long exact sequence would show that also
$$H^i(X\_{\mathrm{proet}},\hat{\mathbb Z}\_p)\otimes k\to H^i(X\_{\mathrm{proet}},\hat{\mathcal O}\_X)$$
is an isomorphism, where I assume $k$ is algebraically closed (which I think is also implicit in the question, or otherwise one should base change $X$ to an algebraic closure before taking cohomology). But this can't even be true for $i=0$ in general, as the right-hand side receives a map from $H^0(X,\mathcal O\_X)$ -- if $X$ is affine, this is infinite-dimensional.
One key issue is that if $X$ is open, it is not quasicompact as a rigid space, and hence taking cohomology does not commute with filtered colimits. Thus, it is not even clear that the map
$$ H^i(X\_{\mathrm{proet}},\hat{\mathbb Z}\_p)\otimes \mathbb Q\_p\to H^i(X\_{\mathrm{proet}},\hat{\mathbb Q}\_p)$$
is an isomorphism. In fact, it is not an isomorphism even for $X=\mathbb A^1$ and $i=1$, where the left-hand side vanishes but right-hand side is infinite-dimensional, see Corollary 3.10 of [Le Bras' Thesis](https://arxiv.org/abs/1801.00422).
| 5 | https://mathoverflow.net/users/6074 | 435163 | 175,933 |
https://mathoverflow.net/questions/434794 | -2 | This posting has been Edited. The edited material shall be noted.
[The projectively extended real line](https://en.wikipedia.org/wiki/Projectively_extended_real_line) $\hat {\mathbb R}= \mathbb R \cup \{\infty\}$ is one system which allows division by zero! Yet it has many [undefined](https://en.wikipedia.org/wiki/Projectively_extended_real_line#Arithmetic_operations_that_are_left_undefined) arithmetic operations.
If we weaken the requirement of the operators being binary *functions* to being ternary ***relations***, and so can relate many values to the same arguments; then those undefined operations can be settled in the sense of getting some kind of closure under them.
I'll adopt the notation "$a \* b \to c$" to mean: the operator $\*$ is a ternary relation that sends the ordered pair $(a,b)$ to $c$; for convience, it can be read as: $c$ is a result of $a \* b$.
Now we may coin a closure notion over $\hat {\mathbb R}$ under an operator $\*$ as:
$ \hat {\mathbb R} \text { closed under } \* \iff \forall x,y \in \hat{\mathbb R} \exists z \in \hat {\mathbb R}: x \* y \to z$
I'll revert to the functional expression (i.e. $x \* y = z$) only when $z$ is unique per $(x,y)$
The following is a possible extension:
$\infty + \infty = \infty \\ \infty - \infty \to r \\ \infty \times 0 \to r \\ 0 \times \infty \to r \\ \infty / \infty \to r \\ 0 /0 \to r$
Where $r$ is any element of $\hat{\mathbb R}$.
>
> [EDIT]: the above was the original suggestion, which was motivated by defining subtraction and division as the reciprocal relation of addition and multiplication respectively, i.e.:
>
>
>
>
> $a - b \to c \iff c + b \to a \\ a / b \to c \iff c \times b \to a$
>
>
>
>
> But, ($\infty + \infty \to \infty$) doesn't conform to that! We have $r - \infty = \infty$ so this results in $\infty + \infty
> \to r$, so this to be corrected to ($\infty + \infty \to r$); for all $r \in \hat{\mathbb R}$.
>
>
>
>
> Additional motivation is to have multiplication by zero be equivalent with self subtraction.
>
>
>
Accordingly $\hat{\mathbb R}$ would be closed under operators "$+,-,\times, /$" in the sense defined here.
Perhaps a similar move can be done for the Riemanian sphere $\mathbb C^\*$, to gain closure under more operators.
>
> Is this system consistent relative to $\sf ZFC$? More specifically, what are the particulars of what could be viewed to constitute somehwat natural interpretation of this system in $\sf ZFC$?
>
>
>
| https://mathoverflow.net/users/95347 | Is this extension of the projectively extended real line, consistent? | This is an account on the particulars of an interpretation of the original system [before the edit] presented in this question in set theory.
First we define an extended kind of rationals to suit adding a rational that is higher than all other rationals, the latter would correspond to $\infty$. The set of all those rationals shall be denoted by $\mathbb Q^\*$
$\text {Define}: r \in \mathbb Q^\* \iff r \subseteq \mathbb R \times \mathbb R \land [\operatorname {image}(r) = \mathbb R \lor \operatorname {preimage}(r) = \mathbb R] \land \\\exists a,b \in \mathbb Z : r= \{\langle a \times x, b \times x \rangle \mid x \in \mathbb R \}$
In the above definition $r$ is meant to represent the rational number $a/b$ for integers $a,b$.
A strict smaller than relation $<$ can be defined on $\mathbb Q^\*$ as:
$r < s \iff \\ r \neq s \land \exists x \exists a \, \exists b \,( a < b \land \langle a, x \rangle \in r \land \langle b, x \rangle \in s)$
As usual elements of $\mathbb Q^\*$ strictly below $0/1$ are negatively signed, those above except $\infty$ are positively signed, while the rest (i.e.; $0, \infty$) are unsigned, this can also be captured in terms of sets as the signed rationals being those which have both their images and preimages being $\mathbb R$.
So we define: $\infty = \{ \langle x,0 \rangle \mid x \in \mathbb R \}$, i.e. the reciprocal of $0/1 = \{\langle 0,x\rangle \mid x \in \mathbb R \}$
**Addition** of any two extended rationals $r,s$ is given by:
$r + s =\{\langle a+b,c \rangle \mid \langle a,c \rangle \in r\land \langle b,c \rangle \in s\}$
This gives: $r + \infty = \infty + r= \infty$ for all $r \in \mathbb Q$
**Subtraction over** $\mathbb Q^\*$:
$r - s \to q \iff \\ q+s = r \lor q= \{\langle a-b,c \rangle \mid \langle a,c \rangle \in r\land \langle b,c \rangle \in s\} $
This gives: $r - \infty = \infty-r = \infty$, for all $r \neq \infty$
and $ \infty - \infty \to r $, forall $r \in \mathbb Q^\*$.
**Multiplication over** $\mathbb Q^\*$:
$r \neq 0, s \neq 0 \\ r \times s = \{\langle k \times x,a \rangle , \langle t \times h, m^2 \rangle \mid (\langle t,m \rangle \in r [s] \land \langle h,m \rangle \in s [r]) \land (\langle k,1 \rangle \in r [s] \land \langle x,a \rangle \in s [r]) \} $
Where: $(z \in r[s] \land u \in s[r])$, is short for: $(z \in r \land u \in s) \lor (z \in s \land u \in r)$
if $r =0 \lor s=0 \implies r \times s \to (r-r)+(s-s) $
So, we get: $r \times \infty = \infty \times r= \infty$, for all $r \neq 0$
and, $0 \times \infty \to r; \infty \times 0 \to r $, forall $r \in \mathbb Q^\*$
Now, that we defined addition and multiplication of the extended rationals, we can define extended reals as Dedekind cuts over $\mathbb Q^\*$
A Dedekind cut shall be defined here as a binary partition on $\mathbb Q^\*$, with one block being an initial segment (i.e.; closed under $<$) of $\mathbb Q^\*$ that is open upwardly. So, for example $\{\mathbb Q^\* \setminus \{\infty\}, \{\infty\}\}$ is a Dedekind cut, and it is taken to represent the real number that corresponds to the extended rational number $\infty$, and so it'll be denoted by "$[\infty]$". The elements of a Dedekind cut are to be termed *initial*, *terminal* abbreviated as *init,term*, the former is the one closed dowardly, the latter is the one closed upwardly.
We can define a total order $<$ on Dedekind cuts themselves, this is given by:
$K < L \iff \operatorname {init}(K) \subsetneq \operatorname {init}(L)$
That said, then clearly $[\infty]$ is strictly greater than all other cuts. Formally this is: $$ r \neq [\infty] \implies r < [\infty]$$, for every extended real $r$.
The set of extended reals to be designated by $\hat {\mathbb R}$
As a terminology if $S$ and $C$ are nonempty subsets of $\mathbb Q^\*$, then:
$S \ \* \ C = \{a \ \* \ b \mid a \in S \land b \in C \} $
Where "$\*$" is some arithmetic operator.
Define: $X=\{S,-\} \iff X=\{S, \mathbb Q^\* \setminus S \}$
**Addition of extended reals:**
$K+L = \{ \operatorname {init}(K) + \operatorname {init}(L), - \} $
This yields: $r + [\infty] = [\infty]$, for all $r \in \hat{\mathbb R}$
**Subtraction of extended reals:**
$ K-L \to X \iff \\ X + L=K \lor \\K < [\infty] \land L=[\infty] \land X=[\infty]$
Another definition is:
$K-L \to X \iff \\ X + L=K \lor \\ X = \{\operatorname {init} (K) - \{x \in \operatorname {term}(L) \mid L \neq [\infty] \to x \neq \infty \}, - \}$
This gives: $ r - [\infty]= [\infty]-r = [\infty] $, for all $r \neq [\infty]$,
and: $[\infty] -[\infty] \to r $, for all $ r \in \hat {\mathbb R}$.
Define: $\operatorname {Comp} (S) = S \cup \{\operatorname {Inf}(S)\} $
$\operatorname {Comp}$ is read as the *completion* set of.
**Multiplication of extended reals:**
$K > 0 \land L > 0: \\K \times L= \{\operatorname {Comp} (\operatorname {term}(K) \times \operatorname {term}(L)) , -\}$
$ K > 0 \land L < 0: K \times L = 0 - (K \times (0 - L))$
$ K < 0 \land L < 0: K \times L = (0-K) \times (0-L)$
$ K = 0 \lor L = 0:\\ K \times L \to X \iff (K-K) + (L-L) \to X$
Accordingly: $r \times [\infty] = [\infty]$, if $r \neq 0$;
and: $0 \times [\infty] \to r, [\infty] \times 0 \to r $; for all $r \in \hat {\mathbb R}$
**Division of extended reals:**
$ K/L \to X \iff X \times L \to K $
Yielding: $[\infty]/r = [\infty], r/[\infty]=0$, for all $r \neq [\infty] $;
and: $r/ 0 = [\infty], [\infty]/[\infty] \to r, 0/0 \to r $, for all $r \in \hat{\mathbb R}$.
This would establish the interpretation of the projectively extended real line in $\sf ZFC$, and also provides an answer to the undetermined expressions mentioned in the Wikipedia page, along the lines mentioned in this question. So, it proves the consistency of this system.
| 1 | https://mathoverflow.net/users/95347 | 435165 | 175,934 |
https://mathoverflow.net/questions/435167 | 1 | It is known that the Minimum Spanning Tree (MST) of a finite set of points in the Euclidean plane is contained in the point set's Delaunay triangulation, but is that all that can be said about their relation?
>
> **Question:**
>
>
> can the *longest* side of a triangle in the Delaunay triangulation of a planar point set be an edge of that point set's MST or can these edges be unconditionally ruled out as MST edges?
>
>
>
| https://mathoverflow.net/users/31310 | Relation of MSTs in the Euclidean plane to Delaunay triangulations | It cannot be for any planar triangulation. Say we have a triangle with vertices $x, y, z$ and $xy$ is the longest edge.
Consider a run of [Prim’s MST algorithm](https://en.m.wikipedia.org/wiki/Prim%27s_algorithm) which at each step adds an edge to a growing tree if it is minimum length among all those edges that have exactly one endpoint already in the tree.
Say without loss of generality that $x$ is added to the tree before $y$. If $xy$ gets added to the tree at some step, then it is minimum weight among all edges with exactly one endpoint in the tree.
Because it is longer than $xz$, this means that $z$ must already be in the tree, otherwise we would have chosen to add $xz$ instead since it’s shorter than $xy$.
But if $z$ is already in the tree, then at this step we would choose to add $zy$ over $xy$ because it is shorter.
Thus it’s impossible that we add $xy$ at any step of the algorithm, so it’s not in the MST.
| 2 | https://mathoverflow.net/users/97414 | 435168 | 175,935 |
https://mathoverflow.net/questions/435138 | 2 | I've been working on the spectrum of the closure of the operator $J: \mathcal{D}(J)= \mbox{span}\{ e\_n: n \in \mathbb{Z}\} \subseteq \ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$ defined for $x=(x\_n)\_{n \in \mathbb{Z}} \in \mathcal{D}(J)$ by
$$(Jx)\_{n} =i((2n+1)x\_{n+1}-(2n-1)x\_{n-1}).$$
I know that $J$ is essentially self-adjoint and I have shown that if $\lambda$ is an eigenvalue of $\overline{J}$, then $-\lambda$ is also an eigenvalue. But I don't know another approach to study $\sigma(\overline{J})$. Can you give me any help?
| https://mathoverflow.net/users/142048 | Spectrum of $(Jx)_n =i((2n+1)x_{n+1}-(2n-1)x_{n-1})$ on $\ell^2(\mathbb{Z})$ | Under the Fourier series isomorphism $\ell^2(\mathbb{Z}) \cong L^2(-\pi,\pi)$, $u(t) = \sum\_{n\in\mathbb{Z}} x\_n e^{int}$, the operator becomes
$$\begin{aligned}
(Ju)(t) &= 4i\sin(t) u'(t) + 2i\cos(t) u(t) \\
&= \begin{cases}
+4i\left|\sin(t)\right|^{1/2} \partial\_t (\left|\sin(t)\right|^{1/2} u(t)) & t\in(0,\pi) \\
-4i\left|\sin(t)\right|^{1/2} \partial\_t (\left|\sin(t)\right|^{1/2} u(t)) & t\in(-\pi,0)
\end{cases} .
\end{aligned}$$
Solving the eigenvalue equation $Ju = \lambda u$ as an ODE, gives two independent weak solutions
$$
u\_{\lambda,\pm}(t) = \frac{\left|\tan(t/2)\right|\_\pm^{-i\lambda/4}}{\left|\sin(t)\right|\_\pm^{1/2}} ,
$$
where $\left|A\right|\_\pm = \left|A\right| \Theta(\pm A)$. The explicit expressions tells us that $u\_{\lambda,\pm} \not\in L^2(-\pi,\pi)$ for any complex $\lambda$. However, for $\Im\lambda > 0$ we have $u\_{\lambda,\pm}$ in $L^2\_{\text{loc}}$ near $t=0$, while for $\Im\lambda < 0$ we have $u\_{\lambda,\pm}$ in $L^2\_{\text{loc}}$ near $t=\pi$.
Thus, for $\lambda \in \mathbb{C} \setminus \mathbb{R}$ we can adapt the variation of constants formula to define the resolvents (hopefully getting all the factors correct)
$$\begin{aligned}
((J-\lambda)^{-1}v)(t) = \begin{cases}
\frac{\Theta(t)}{4i} \int\_t^\pi u\_{\lambda,+}(t) u\_{-\lambda,+}(s) v(s)\, ds + \frac{\Theta(-t)}{4i} \int\_{-\pi}^{t} u\_{\lambda,-}(t) u\_{-\lambda,-}(s) v(s)\, ds
& \Im\lambda > 0 \\
-\frac{\Theta(t)}{4i} \int\_0^t u\_{\lambda,+}(t) u\_{-\lambda,+}(s) v(s)\, ds - \frac{\Theta(-t)}{4i} \int\_{t}^{0} u\_{\lambda,-}(t) u\_{-\lambda,-}(s) v(s)\, ds
& \Im\lambda < 0
\end{cases} .
\end{aligned}$$
The resolvent is symmetric, $((J-\lambda)^{-1})^\* = (J-\bar{\lambda})^{-1}$ and is well-defined for any $v\in L^2(-\pi,\pi)$. Moreover, it did not require any boundary conditions other than being defined from $L^2$ to $L^2$. Hence, $J$ is essentially self-adjoint, with the unique self-adjoint extension corresponding to the above resolvent. As a function of $\lambda$, $(J-\lambda)^{-1}$ is discontinuous across $\mathbb{R}\subset \mathbb{C}$ (the solutions $u\_{\lambda,\pm}$ switch the location of their $L^2\_{\text{loc}}$ behavior as $\lambda$ crosses $\mathbb{R}$), hence $\sigma(J) = \mathbb{R}$, with generalized eigenfunctions given by $u\_{\lambda,\pm}(t)$.
**NB:** In the comments, Giorgio Metafune outlined essentially the same argument.
| 3 | https://mathoverflow.net/users/2622 | 435174 | 175,937 |
https://mathoverflow.net/questions/435187 | 3 | Let $P\subset \mathbb{R}^2$ be a set of positive Lebesgue measure. Is it always true that a suitable rotation and translation of $P$ always contains a set of the form $\{re^{i\theta}:r\in E, \theta\in [0,2\pi)\}$ or $A×B,$ where $E,A,B$ are sets of positive Lebesgue measure in $\mathbb{R}?$
Note: I can show that the two types of positive measure sets, mentioned above, are different (in the sense that no one type contains the other type always).
| https://mathoverflow.net/users/483450 | Property of sets of positive Lebesgue measure in $\mathbb{R}^2$ | Firstly, a set $P$ of positive measure need not contain anything of the form $A\times B$, for example consider for some $k\in\mathbb{R}\setminus\{0\}$ the set $P=\{(x,y)\in\mathbb{R}^2;x-ky\not\in\mathbb{Q}\}$. Then the complement of $P$ is a null set, and any set $A\times B$ where $A,B$ have positive measure contains some point $(x,y)$ with $x-ky\in\mathbb{Q}$. This is because the set $A-kB$ [contains an open interval](https://math.stackexchange.com/questions/596273/sum-of-positive-measure-set-contains-an-open-interval). We can substitute $\mathbb{Q}$ by any other set dense in $\mathbb{R}$.
Now let $P=\{(x,y)\in\mathbb{R}^2;x\not\in\mathbb{Q},y\not\in\mathbb{Q},x-y\not\in\mathbb{Q}\}$. Any rotation/translation of $P$ cannot contain a set $A\times B$, with $A,B$ of positive measure, because then $A\times B$ would not intersect some set of parallel lines dense in $\mathbb{R}^2$ and with non zero slope, and we get a contradiction as in the previous case.
The same set $P$ also doesn't contain sets $\{re^{i\theta}:r\in E, \theta\in F\}$, where $E,F$ have positive measure. To check that, it will be enough to check that every point $p\_0\in\mathbb{R}^2\setminus\{0\}$ with polar coordinates $(r\_0,\theta\_0)$ has a neighborhood $U$ such that $U$ doesn't contain sets $\{re^{i\theta}:r\in E, \theta\in F\}$, where $E,F$ have positive measure.
This is true because near $p\_0$, any set of parallel lines dense in $\mathbb{R}^2$ can be expressed as $\{r(k\cos(\theta)+l\sin(\theta))\not\in Q\}$, for $k,l$ not both $0$ and $Q$ some dense set in $\mathbb{R}$. We can suppose that $p\_0\not\in\{k\cos(\theta)+l\sin(\theta)=0\}$, so after changing $Q$ by $-Q$ if necessary, we can take logarithms and the equation of the parallel lines becomes $f(r)+g(\theta)\not\in\{\ln(q);q\in Q\cap\mathbb{R}^+\}$, where $f(r)=\ln(r),g(\theta)=\ln(k\cos(\theta)+l\sin(\theta))$. However, if $E,F$ have positive measure, then $f(E)$ and $g(F)$ have positive measure, so $f(E)+g(F)$ contains some open interval, so the same reasoning of before works.
| 2 | https://mathoverflow.net/users/172802 | 435188 | 175,943 |
https://mathoverflow.net/questions/435200 | 0 | The subject of this question are perfect matchings of a complete undirected graph $G(V,E), n:=\mathrm{card}(V)=2k$, without self-loops or parallel edges and $n=2k$ vertices.
The objective is to determine a perfect matching $\mathrm{M}\_{\text{opt}}$ of minimal weight.
Consider now the greedy heuristic that determines a perfect matching $\mathrm{M}\_{\text{app}}$ by starting with $\mathrm{M}\_{\text{app}}\,=\,\emptyset$ and keeps adding the shortest edge that is not adjacent to any edge in $\mathrm{M}\_{\text{app}}$ until it is a perfect matching.
>
> **Question:**
>
>
> Is it true, that the longest edge in $\mathrm{M}\_{\text{opt}}$ can't be strictly longer than the longest edge in $\mathrm{M}\_{\text{app}}$?
>
>
>
I am convinced that that must be true for the following reason: if $\mathrm{M}\_{\text{opt}}$ contains a longer edge, then it must also contain an edge that compensates for the weight difference and is not contained in $\mathrm{M}\_{\text{app}}$.
As the shortest edge is already contained in $\mathrm{M}\_{\text{app}}$ it can't be the edge that compensates for the weight difference; therefore it must be adjacent to some shorter edge in $\mathrm{M}\_{\text{app}}$, which completes the proof; is that actually the case?
If the longest edge of $\mathrm{M}\_{\text{opt}}$ can't indeed be longer than the longest edge of $\mathrm{M}\_{\text{app}}$ the performance of matching algorithms could in some cases be improved substantially by reducing the number of candidate edges
| https://mathoverflow.net/users/31310 | Edge-length constraints from greedy matching | Let the weights of the edges in a 6-cycle in $K\_6$ be $1,2,5,6,5,2$ (in cyclic order), and let other weights be large. Then the optimal matching will be $2,2,6$, while the greedy one will consist of $1,5,5$.
A similar construction works with longer cycles.
| 2 | https://mathoverflow.net/users/17581 | 435201 | 175,945 |
https://mathoverflow.net/questions/435129 | 4 | Let $(Y, \Sigma,\mu)$ be measure space and $X$ a Polish space endowed with its Borel $\sigma$-algebra. Suppose that $f:Y\times X\to \mathbb R$ is a Carathéodory function (i.e. continuous in $x\in X$ for each $y\in Y$, measurable and bounded by a $L^1$ function that does not depend on $x$). Let ${\Sigma}\_0$ be sub $\sigma$-algebra of $\Sigma$, and let $g(\cdot,x)=E(f(\cdot,x)|\Sigma\_0)$ denote the conditional expectation with respect to ${\Sigma}\_0$. Does $g$ have a version that is a Carathéodory function as well?
Thanks!
| https://mathoverflow.net/users/470906 | Is the conditional expectation of a Caratheodory function a Caratheodory function? | Here is a positive answer for the case that $\Sigma\_0$ is generated by a random variable with values in a Polish space, so that we can use regular conditional probabilities and for some kernel $\kappa:Y\to\Delta(Y)$, we can let
$$\mathbb{E}(f(\cdot,x)|\Sigma\_0)\_y=\int f(,\cdot,x)~\mathrm d\kappa\_y.$$
Then continuity follows from the assumption that there is a dominating integrable function and the dominated convergence theorem.
| 1 | https://mathoverflow.net/users/35357 | 435217 | 175,947 |
https://mathoverflow.net/questions/435232 | 4 | For a set $A\subseteq \omega$ we let the *upper density* of $A$ be defined as $d^+(A) := \lim\sup\_{n\to\infty}\frac{|A\cap(n+1)|}{n+1}$. Let $\text{FrU}(\omega)$ be the collection of [free ultrafilters](https://en.wikipedia.org/wiki/Ultrafilter) on $\omega$.
[Asaf Karagila](https://mathoverflow.net/users/7206/asaf-karagila) provided a convincing argument that there is a free ultrafilter ${\cal U}$ on $\omega$ such that $d^+(U) > 0$ for all $U\in {\cal U}$.
**Question.** What is $$\sup\big\{\inf \{d^+(U): U \in {\cal U}\}: {\cal U} \in \text{FrU}(\omega)\big\}\;?$$
| https://mathoverflow.net/users/8628 | Supremum of infimum of measure of members of a free ultrafilter | The answer is: zero.
The reason is that every ultrafilter has zero as the infimum of the upper density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive upper density, then we can split $U$ in half $U=A\sqcup B$ each with half the upper density (just take every other element of $U$ into $A$, the others into $B$). One of these sets will be in the ultrafilter, and so we will have a set in $\mathcal{U}$ with half the upper density of $U$. By iterating this, we can make the upper density of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.
| 9 | https://mathoverflow.net/users/1946 | 435233 | 175,950 |
https://mathoverflow.net/questions/435195 | 0 | I'm now solving an LP that has a few coupling rows (as in Dantzig-Wolfe decomposition) and a few coupling columns (as in Benders decomposition) simultaneously; other rows and columns are block-angular. Is there an algorithm that decomposes such LP? Thank you very much!
| https://mathoverflow.net/users/494373 | Combining Dantzig-Wolfe and Benders decomposition | You could apply Benders as the main algorithm and use Dantzig-Wolfe for the subproblem. Alternatively, you could apply Dantzig-Wolfe as the main algorithm and use Benders for the subproblem. For LP, Benders and Dantzig-Wolfe are equivalent if you take duals, so you could also either apply Benders as the main algorithm and Benders for the dual of the subproblem or Dantzig-Wolfe as the main algorithm and Dantzig-Wolfe for the dual of the subproblem.
| 0 | https://mathoverflow.net/users/141766 | 435235 | 175,951 |
https://mathoverflow.net/questions/435241 | 34 | Does there exist a polynomial $f \in \mathbb{Z}[x,y]$ such that
$$\displaystyle f(a,b) > 0 \text{ for all } a,b \in \mathbb{Z}$$
and
$$\displaystyle \liminf\_{(x,y) \in \mathbb{R}^2} f(x,y) = -\infty?$$
In other words, does there exist a polynomial $f$ which takes on positive values at every integer point, but still there exists a sequence $(x\_k, y\_k)$ of real pairs such that $\lim\_{k \rightarrow \infty} f(x\_k, y\_k) = -\infty$?
Note that if such a sequence exists, the norm of its elements must tend to infinity. This is because $f$ is continuous, and therefore the image of any compact set under $f$ is necessarily compact, and thus in particular must be bounded.
| https://mathoverflow.net/users/10898 | Ruling out the existence of a strange polynomial | The polynomial $f(x,y)=(x^2+1)(5y^2+5y+1)\in\mathbb{Z}[x,y]$ is an example. Note that $5y^2+5y+1>0$ for $y\in\mathbb{Z}$, but $5y^2+5y+1<0$ at $y=-\frac{1}{2}$.
| 52 | https://mathoverflow.net/users/95685 | 435244 | 175,953 |
https://mathoverflow.net/questions/435251 | 9 | This is a refinement of my [question asked earlier](https://mathoverflow.net/questions/435241/ruling-out-the-existence-of-a-strange-polynomial), which is answered beautifully in the negative by Thomas Browning. The example he gave was geometrically reducible. Now I want to ask the same question, but with the extra assumptions that $f(0,0) = 0$ and $f$ is geometrically irreducible for all $c \in \mathbb{R}$.
For convenience, the question is: does there exist a polynomial $f \in \mathbb{Z}[x,y]$ such that $f(0,0) = 0$ and $f$ is geometrically irreducible, and such that
$$\displaystyle f(a,b) > 0 \text{ for all } a,b \in \mathbb{Z}$$
and
$$\displaystyle \liminf\_{\mathbb{R}^2} f(x,y) = -\infty?$$
| https://mathoverflow.net/users/10898 | Ruling out the existence of a strange polynomial II | Just a slight modification of the previous [example](https://mathoverflow.net/a/435244/36721):
$$f(x,y):=\left(5 y^2+5 y+1\right) \left(x^2+y^2\right)+\left(10 y^2+10 y+1\right) y^2.$$
---
Your conditions $f(0,0) = 0$ and
$$f(a,b) > 0 \text{ for all } a,b \in \mathbb{Z}$$
contradict each other. So, I was assuming that you meant
$f(0,0) = 0$ and
$$f(a,b) > 0 \text{ for all } (a,b) \in \mathbb{Z}^2\setminus\{(0,0)\}.$$
| 20 | https://mathoverflow.net/users/36721 | 435252 | 175,955 |
https://mathoverflow.net/questions/433196 | 1 | The origin question: Let $\Omega \subset \mathbb{H}^2$ be a domain of the hyperbolic plane $\mathbb{H}^2$. Let $u: \Omega \to \mathbb{H}^2$ be injective and an isometry from $\Omega$ to its image. Does there exist a Mobius transformation $\gamma\in \text{PSL}(2,\mathbb{R})$ such that $u=\gamma\mid\_\Omega$?
The modified question: Let $\Omega \subset \mathbb{H}^2$ be a connected domain of the hyperbolic plane $\mathbb{H}^2$. Let $u: \Omega \to \mathbb{H}^2$ be an orientation-reserving $C^1$ isometry from $\Omega$ to its image. Does there exist a Mobius transformation $\gamma\in \text{PSL}(2,\mathbb{R})$ such that $u=\gamma\mid\_\Omega$?
Thanks for all comments and answers. I have found the answer from "Dierkes, Ulrich; Hildebrandt, Stefan; Tromba, Anthony J. Global analysis of minimal surfaces", on page 273, Lemma 1, which reads as follows:
Lemma: Let $f: U \to \mathbb{H}^2$ be a $C^1$ isometry on an open connected subset $U$ of the hyperbolic plane. Then
$$
f(w)=\frac{Aw+B}{Cw+D}, \, A,B,C,D \in \mathbb{R},
$$
and $AD-BC=1$.
| https://mathoverflow.net/users/161514 | Is a local isometry of the hyperbolic plane the restriction of a global isometry? | The hyperbolic plane has this property, as does the Euclidean plane.
If $E$ is any subset of $ \mathbb{H}^2$, and $u : K \to\mathbb{H}^2$ is an isometry, then there is an extension of $u$ which is an isometry of $\mathbb{H}^2$ onto itself. I use "isometry" in the sense: $d(u(x),u(y)) = d(x,y)$ for all $x,y \in E$, where $d$ is the distance in $\mathbb{H}^2$.
So the OP reduces to: is an isometry of $\mathbb{H}^2$ onto itself necessarily a Möbius transformation?
| 3 | https://mathoverflow.net/users/454 | 435259 | 175,958 |
https://mathoverflow.net/questions/435231 | 5 | Let $a$ be a strictly positive $C^\infty$ smooth function on the unit interval. Does there exist a strictly positive $C^\infty$ smooth function $f$ on $I$ such that
$$ f’’(x) \leq 0\quad \text{and} \quad (af)’’(x) \leq 0$$
for all $x \in I$?
| https://mathoverflow.net/users/50438 | On existence of a concave function | Such a function doesn't exist for some choices of $a$. For notational purposes I will change the unit interval by $[-2,2]$.
Consider a $C^\infty$ function $a:[-2,2]\to\mathbb{R}$ such that $a(0)=3$, $a(1)=1$ and $a(x)=a(-x)\forall x$.
Suppose that we have a convex function $f:[-2,2]\to\mathbb{R}$ such that $f''\leq0$, $(af)''\leq0$. We can assume $f'(0)\leq0$ (if not change $f(x)$ by $f(-x)$). So $f$ is decreasing in $[0,2]$ and we have $f(0)\geq f(1)$. Now apply Jensen's inequality to $af$ to obtain a contradiction:
$$f(1)=(af)(1)\geq\frac{1}{2}((af)(0)+(af)(2))\geq\frac{1}{2}(af)(0)=\frac{3}{2}f(0)\geq\frac{3}{2}f(1)>f(1)$$
| 4 | https://mathoverflow.net/users/172802 | 435264 | 175,962 |
https://mathoverflow.net/questions/435268 | 2 | This is a variant of the Nash equilibrium. Let's say that there are 3 prizes: A Ferrari, a diamond watch, and a new boat. There are 6 players. 3 players with a motive while 3 players with another motive, but all 6 players are playing for themselves: 3 players, for example, are wearing red shirts. The people wearing red shirts have an objective: to pick the prize that they think no red shirt or purple shirt will pick. The people wearing purple shirts have a different objective: to pick the same prize as someone wearing a red shirt so they can claim their chosen prize
The Scenarios:
* If a Purple Shirt picks a prize (Ex. diamond watch) no one else picks, they can't claim the prize (Ex. diamond watch)
* If a Purple Shirt picks a prize (Ferrari, for example) that at least one other Purple Shirt picks (Ferrari, for example), they collide and the Purple Shirts that picked the prize(Ferrari. for example) go home with nothing, regardless of how many Red Shirts picked the prize (Ferrari, for example).
* If a Purple Shirt picks a prize (Boat, for example) that at least one Red Shirt picks(Boat, for example), then that means the Purple Shirt wins the prize assuming that the Purple Shirt is the only Purple Shirt picks the prize(Boat, for example).
* If a Red Shirt picks a prize (Ferrari, for example) picks a prize that no other Red Shirt chooses(Ferrari, for example), and assuming that the amount of Purple Shirts that choose the prize(Ferrari, for example) follows the interval (2, ∞), then the Red Shirt rightfully claims the prize(Ferrari, for example). This means that if at least one other Red Shirt picks the same prize as the Red Shirt did, the Red Shirt goes home with nothing, and regardless of how many Purple Shirts picked the same prize.
* When it comes to collisions, purple shirts collide first(Which means that if 2 or more Purple Shirts pick the same prize, they go home with nothing), then Red Shirts(Which means that if 2 or more Red Shirts picked the same prize, that means that the Red Shirts come home with nothing.) We also assume that we're using whole non-negative numbers and not decimals
We assume that they have folders and write their answers down on paper, and can't communicate with anyone during the game. They hand the paper to the judge/referee. The following questions are:
The formula for increasing the amount of players?
The formula for increasing the amount of prizes(How many prizes, not their value)?
The formula for both increasing the amount of players and prizes?
This is my first question on mathoverflow.net, so it may contain errors.
Citations
Nash Equilibrium(mathoverflow.net)
Monty Hall Problem(Einstein's Riddle by Jeremy Stangroom)
[Simple(?) game theory](https://mathoverflow.net/questions/41679/simple-game-theory)
| https://mathoverflow.net/users/495433 | Nash equilibrium at another level | With the given parameters, the Nash equilibria are exactly those situations where the three purple players pick each a different price. This ensures that the red players can't win anyway, so they'll do whatever.
To see this, we first observe that a purple player would never change to a prize another purple player already picked. If there is a price no purple player has chosen, but a red player, a purple player not currently winning would change to it. If there is a price no player has currently chosen, a red player not currently winning would change to it.
If we go for 4 prizes, and still 3 players of each colour, there is no Nash equilibrium. There is at least one prize not chosen by a purple player. Such prizes are the only options for a red player to win, so there will be at least one red player picking one of them. This in turn means at least one purple player is losing, and would thus switch over to the unclaimed-by-purple prize.
This generalizes as follows: As long as there are at least as many purple players as prizes, there is a Nash equilibrium, and every prize being chosen by at least one purple player suffices to be a Nash equilibrium. If there are at least as many red players as prizes, there also is a Nash equilibrium, and having red and purple players "spread out as much as possible" works. If there are fewer purple players than prizes, and at least one red player but fewer than prizes, there is no Nash equilibrium.
| 3 | https://mathoverflow.net/users/15002 | 435271 | 175,963 |
https://mathoverflow.net/questions/435272 | 1 | The general approach here is a follow up of the approach outlined in a prior posting on [extending the projectively extended real line](https://mathoverflow.net/questions/434794/is-this-extension-of-the-projectively-extended-real-line-consistent). In particular arithmetic operators break down to ternery relations that may sometimes admit multivaluation. Here, it'll be applied to extend the affinely extended real line.
Working in $\sf ZFC$, We start with any model $(\mathbb R, =, <, +, -, \times, /)$ satisfying the customary sentences of arithmetic of reals. Now we take any two distinct sets $+\infty, -\infty$ that are not elements of $\mathbb R$ and adjoin them to $\mathbb R$ to get $\hat {\mathbb R}$. That is, we have: $$ \hat{\mathbb R} = \mathbb R \cup \{+\infty, -\infty\}$$
We define $=^\*$ by the restriction of $=$ to $\hat {\mathbb R}$, so it is the set $= \restriction\_\hat{\mathbb R}$
Similarly $\neq^\*$ is the set $\neq \restriction\_\hat{\mathbb R}$
Of course we have $\langle +\infty, -\infty \rangle \in \, \neq^\*$
Note: Whenever "$\times$" appears as an infix then it designates Cartesian product, otherwise if "$\times$" appears as a constant then it stands for the customary multiplicaiton relation over reals.
Now we extend $<$ to an extended strict less than relation $<^\*$ over $\hat{\mathbb R}$, as:
$<^\* = \, (< \cup \ (\hat {\mathbb R} \times \{+\infty\}) \cup (\{-\infty\} \times \hat {\mathbb R} )) \ \cap \neq^\*$
Then we extend the set $+$ to $+^\*$ defined as:
Let: $+' = + \ \cup \\ \hat{\mathbb R} \times \{+\infty\} \times \{+\infty\} \ \cup \\ \hat{\mathbb R} \times \{-\infty\} \times \{-\infty\} \ \cup \\ \{+\infty\} \times \{-\infty\} \times \hat {\mathbb R} $
Define: $+^\* = +' \cup \{\langle b,a,c \rangle \mid \langle a,b,c \rangle \in +' \}$
We define $-^\*$ as the reciprocal of $+^\*$:
$-^\* = \{ \langle a,b,c \rangle \mid \langle c,b,a \rangle \in +^\* \}$
Let $\hat{\mathbb R}^+; \hat{\mathbb R}^-$ be the sets of all extended reals (i.e., elements of $\hat{\mathbb R}$), strictly above or strictly below zero, respectively.
We extend the set $\times$ to $\times^\*$ defined as:
Let: $\times' = \times \ \cup \\ \hat{\mathbb R}^+ \times \{+\infty\} \times \{+\infty\} \ \cup \\ \hat{\mathbb R}^+ \times \{-\infty\} \times \{-\infty\} \ \cup \\ \hat{\mathbb R}^- \times \{+\infty\} \times \{-\infty\}\ \cup \\ \hat{\mathbb R}^- \times \{-\infty\} \times \{+\infty\} \ \cup \\ \{+\infty\} \times \{0\} \times \hat{\mathbb R} \ \cup \\\{-\infty\} \times \{0\} \times \hat{\mathbb R} $
Define: $\times^\* = \times' \cup \{\langle b,a,c \rangle \mid \langle a,b,c \rangle \in \times' \}$
We define $/^\*$ as the reciprocal of $\times^\*$:
$/^\* = \{ \langle a,b,c \rangle \mid \langle c,b,a \rangle \in \times^\* \}$
So $(\hat{\mathbb R}, =^\*, <^\*, +^\*, -^\*, \times^\*, /^\*)$ is our model of the affinely extended real line extended to be closed under all above-mentioned arithmetic operators.
So, we get all defined expressions mentioned in the Wikipedia page on arithmetic of [affinely extending the real line](https://en.wikipedia.org/wiki/Extended_real_number_line), and also settle all mentioned undefined expressions to:
$ 0/0 \to r \in \hat{\mathbb R}\\ \pm\infty + (\mp \infty) \to r \in \hat{\mathbb R} \\ \pm\infty - (\pm \infty) \to r \in \hat{\mathbb R} \\0 \times \pm \infty \to r \in \hat{\mathbb R} \\ \pm\infty /\pm\infty \to r \geq 0 \\ \pm \infty / \mp \infty \to r \leq 0 \\ r/0 \to \pm \infty$
Unlike [extending the projectively extended real line](https://mathoverflow.net/questions/434794/is-this-extension-of-the-projectively-extended-real-line-consistent), here the operational multi-valuation is not that trivial. And if we demand closure over more operators, like extending this method to the complex number line and demand closure under exponentiation\rooting; then one expect more different multi-valuation patterns.
>
> Is there an effective way that governs extending this approach as to cover all iterative $+$ operators and their reciprocals?
>
>
>
By iterative $+$ operators I mean multiplication, exponentiation, hyperexponentiation, etc..
| https://mathoverflow.net/users/95347 | Is there an effective way to generalize this approach of affinely extending the number line? | First of all, mentioning set-theoretic background is really just making this more complicated than it needs to be, so I'll ignore it. Additionally, I'll use "$\leadsto$" in place of your "$\rightarrow$" since I also want to talk about conventional limits.
It sounds like you want $t\leadsto c$ whenever $t$ "has a form" that equals or converges to $c$ (for $t,c$ appropriate terms). This can be made straightforwardly precise, but I don't think it's particularly useful. Specifically, given an $n$-ary partial function $f:\mathbb{R}^n\rightarrow\mathbb{R}$, we can consider its multivalued-and-partial extension $\hat{f}$ to $\hat{\mathbb{R}}^n$ by setting $\hat{f}(a\_1,...,a\_n)\leadsto b$ iff there are sequences of (non-extended!) reals $(\alpha^1\_i)\_{i\in\mathbb{N}}\rightarrow a\_1,..., (\alpha^n\_i)\_{i\in\mathbb{N}}\rightarrow a\_n$ such that for each $i$ the expression $\beta\_i:=f(\alpha\_i^1,...,\alpha\_i^n)$ is defined and $(\beta\_i)\_{i\in\mathbb{N}}\rightarrow b$. Here, "$\rightarrow$" refers to convergence in $\hat{\mathbb{R}}$ in the usual sense.
This gives all the examples you include, as well as things like $$\sin(\pm\infty)\leadsto x\iff x\in[-1,1]$$ and $C(x)\leadsto y$ for all $x,y\in\hat{\mathbb{R}}$ where $C$ is [Conway's base $13$ function](https://en.wikipedia.org/wiki/Conway_base_13_function). Additionally, if $f$ is total on $\mathbb{R}$ then $\hat{f}$ is total on $\hat{\mathbb{R}}$ and extends $f$ (for the latter point just take $\alpha^k\_i=a$), and all "value sets" $$\hat{f}[x]:=\{y: \hat{f}(x)\leadsto y\}$$ will always be **closed** for any $f$ whatsoever.
This last observation points to the right way to think about what's going on here, in my opinion: we're really just taking the topological closure of the graph of $f$ as a subset of (the appropriate power of) $\hat{\mathbb{R}}$ and declaring that to be the graph of $\hat{f}$. But I don't think we actually get anything interesting this way.
| 3 | https://mathoverflow.net/users/8133 | 435273 | 175,964 |
https://mathoverflow.net/questions/435191 | 8 | This is about a rather concrete problem that occurs in the middle of a lecture by Scholze. First I'll refer to the lecture, but then I'll state the problem.
In <https://www.youtube.com/watch?v=q6Tv2vJJShg> , at around the 22 minute mark, Peter Scholze claims that if you have a simplicial hypercover of a profinite set by profinite sets, then you can write it as a cofiltered limit of hypercovers of finite sets by finite sets. After a lot time of grappling with this technicality, I have finally given up and am hoping someone can help me.
**Question:** $S$ is a profinite set. $T\_{\bullet}\to S$ is a simplicial hypercover of $S$ by profinite sets $T\_i.$ Can we find some cofiltered poset $J,$ and some simplicial hypercovers $T\_{\bullet, j} \to S\_j,$ where $S\_j$ is finite and each $T\_{n,j}$ is finite, such that the cofiltered limit of the hypercovers $T\_{\bullet, j} \to S\_j$ is precisely our original hypercover $T\_{\bullet}\to S$?
---
Currently, I can prove that $T\_{\bullet}\to S$ is a cofiltered limit of certain simplicial objects; however, I do not know that these objects are hypercovers. My main difficulty is that the hypercover condition, namely that the map $T\_{n+1, j} \to (\operatorname{cosk}\_n\operatorname{sk}\_n T\_{\bullet, j})\_{n+1}$ is surjective, seems to be very hard to keep true when constructing these sets $T\_{n, j}.$
To elaborate a little more, let's think just about the case of $T\_{1, j} \to (\operatorname{cosk}\_0 \operatorname{sk}\_0T\_{\bullet, j})\_1 = T\_{0,j}\times\_{S\_j} T\_{0, j}$ is surjective.
Assume that there is some index set $I$ so that $S = \lim\_i S\_i$ and $T\_0 = \lim\_i T\_{0, i}.$ For an index $i,$ one might get the idea to try producing a simplicial hypercover of $S\_i.$ A natural choice would be to first assume each projection map $S \to S\_i$ and $T\_0 \to T\_{0, i}$ is surjective (assuming this loses no generality, up to changing how we present $S$ and $T\_0$ as limits). Then surjectivity of $T\_0\to S$ implies that $T\_0\to S\_i$ is surjective, which implies that there's some index $i'$ so that $T\_0\to S\_i$ factors through $T\_{0, i'}.$
So, we try building a hypercover of $S\_i$ by starting with $T\_{0, i'} \to S\_i.$ To go to the next stage, there's some list of conditions we want to satisfy; the first is that whatever object $X$ we use as our next term, there should be a surjection $X \to T\_{0,i'} \times\_{S\_i} T\_{0, i'}.$ I cannot seem to easily construct such an $X,$ mainly because the map $T\_0 \times\_S T\_0 \to T\_{0,i'} \times\_{S\_i} T\_{0, i'}$ is not always surjective! (Imagine for example that $S\_i = \{\*\}$ is a singleton and $T\_0 = T\_{0, i'}$; then the latter fiber product is just $T\_0 \times T\_0,$ but the fiber product $T\_0 \times\_S T\_0$ might end up being some proper subset of $T\_0\times T\_0$ is $S$ is bigger than $S\_i$).
| https://mathoverflow.net/users/318125 | A hypercover of profinite sets as a limit of hypercovers of finite sets | I'm sorry for being cryptic.
The subtle point in the construction is that the maps $T\_n\to T\_{n,j}$ are not all surjective, i.e. one cannot construct this pro-system as a system of quotients.
By induction on $n$, one constructs a cofiltered system of $n$-skeletal simplicial hypercovers $T\_{\bullet,j}^{(n)}\to S\_j$ (of finite sets by finite sts) whose limit is the $n$-skeleton of $T\_\bullet\to S$, where each inductive step is refining the previous one; in the end, one takes the limit over $n$ of the induced coskeletal simplicial hypercovers.
In the first step, one needs to write the surjection $T\_0\to S$ as a limit of surjections of finite sets; this can be done by simply taking the cofiltered system of compatible finite quotients. Next, one needs to write $T\_1$ as a cofiltered limit of surjections $T\_{1,j}\to T\_{0,j}\times\_{S\_j} T\_{0,j}$. For this, we prove
>
> Lemma. Let $X=\mathrm{lim}\_i X\_i$ be a profinite set written as a limit of finite sets $X\_i$ along a cofiltered category $I$. Let $Y\to X$ be a surjection of profinite sets. Then there is a cofiltered category $J$ with a functor $g: J\to I$ and (functorial) surjections of finite sets $Y\_j\to X\_{g(j)}$ whose limit is $Y\to X$.
>
>
>
Admitting the lemma for the moment, one can find the desired $T\_{1,j}\to T\_{0,j}\times\_{S\_0} T\_{0,j}$. But for higher $n$, one is in the similar situation, for the surjection $T\_n\to (\mathrm{cosk}\_{n-1} \mathrm{sk}\_{n-1} T\_\bullet)\_n$ where the target already has a given presentation as a cofiltered limit of finite sets.
Thus, it remains to prove the lemma. For this, one considers the category $J$ whose objects $j\in J$ consist of an object $i\in I$ and a factorization of $Y\to X\to X\_i$ into $Y\to Y\_j\to X\_i$ where $Y\_j$ is finite and $Y\_j\to X\_i$ is surjective. A morphism $j\to j'$ is given by a map $i\to i'$ and a map $Y\_j\to Y\_{j'}$ making the obvious diagrams (from $Y$, and to $X\_i$) commute.
One needs to see that $J$ is cofiltered, and that $Y$ is the limit of the $Y\_j$'s. Given two $j,j'\in J$, one can first arrange via cofilteredness of $I$ that they map to the same $i\in I$. Then we have two maps $Y\to Y\_j\to X\_i$ and $Y\to Y\_{j'}\to X\_i$ where $Y\_j\to X\_i$ and $Y\_{j'}\to X\_i$ are surjective. We get a similar factorization over $Y\_j\times\_{X\_i} Y\_{j'}$, which is still finite and surjective over $X\_i$, so we find some $j''$ dominating $j$ and $j'$. Now, given two morphisms $f,g: j\to j'$, we need to find some $j''$ mapping to $j$ equalizing $f$ and $g$. Again, we can assume that the diagram $f,g: j\to j'$ gets contracted to $i\in I$ (using that $I$ is cofiltered). Then we get two maps $Y\_j\to Y\_{j'}$ over $X\_i$, and $Y$ factors over the equalizer $Z\subset Y\_j$. This is finite, but not necessarily surjective over $X\_i$! Still, as $Y\to X$ is surjective, there is some $i'\to i$ such that $Z\times\_{X\_i} X\_{i'}\to X\_{i'}$ is surjective. This defines the desired object $j''\in J$.
It remains to see that $Y\to \mathrm{lim}\_j Y\_j$ is bijective. Injectivity is easy to see (for any finite quotient $Y'$ of $Y$, $Y\to Y'\times X\_i\to X\_i$ defines an object of $J$). For surjectivity, the argument at the end of the last paragraph applies (one needs to see that for any $j$ there is some $j'$ such that $Y\_{j'}\to Y\_j$ factors over the image of $Y\to Y\_j$, which was proved there).
| 9 | https://mathoverflow.net/users/6074 | 435277 | 175,966 |
https://mathoverflow.net/questions/435210 | 3 | **Question.** Let $u: B^3 \to \mathbf{R}$ be a harmonic function with $u(0) = 0$, $Du(0) = 0$, where its homogeneous harmonic blow-up is a polynomial $p = p(x,y)$ in two variables, so independent of $z$; in other words $p$ is a non-zero homogeneous harmonic polynomial so that
\begin{equation}
u(x,y,z) = p(x,y) + o( \lvert (x,y,z) \rvert^m),
\end{equation}
where $2 \leq m = \operatorname{deg} p$. Must $u$ be translation-invariant with respect to $z$? Can the origin be isolated in the singular set $u^{-1}(0) \cap \lvert Du \rvert^{-1}(0)$?
| https://mathoverflow.net/users/103792 | A harmonic function degenerate in one direction | The questions have been answered in the comments, I am just recording them here: Alexandre Eremenko pointed out that *no*, the function $u$ need not be translation-invariant, because the dependencies on $z$ could be 'hidden' inside a polynomial of higher degree, say
\begin{equation}
u = p(x,y) + q(x,y,z),
\end{equation}
with $\operatorname{deg} q > \operatorname{deg} p$.
This also gives a hint for the second question: the answer is *yes*, there exist examples of such $u$ that only have isolated singularity at the origin. The example given below $u$ is basically of the form above—with $q$ picked so as to have an isolated singularity at the origin—, except for the fact that one multiplies $q$ by a small constant to avoid introducing new singular points.
Specifically, pick a constant $\delta \in (0,1/3)$ and define
\begin{equation}
u(x,y,z) = x^2 - y^2 + \delta(2x^3 - 3xy^2 - 3xz^2).
\end{equation}
Then
\begin{equation}
Du(x,y,z) = (2x + \delta( 6x^2 - 3y^2 - 3z^2),-2y - 6\delta xy,-6\delta xz).
\end{equation}
At a critical point $(x,y,z)$:
* from $D\_y u = 0$ one finds that $y(1 + 3\delta x) = 0$. As $\delta < 3$, the second factor never vanishes if $\lvert x \rvert < 1$, so $y = 0$;
* from $D\_z u = 0$ one finds that either $x = 0$ or $z = 0$.
* from $D\_x u = 0$, if $x = 0$ then immediately $z = 0$. If instead $z = 0$ then $0 = D\_x u = 2x + 6\delta x^2 = 2x(1 + 3 \delta x)$. Again, our choice of a sufficiently small $\delta$ means that $1 + 3 \delta x > 0$ on $B^3$, so $x = 0$.
Therefore $Du(x,y,z) = 0$ is equivalent to $(x,y,z) = 0$. Obviously $u(0,0,0) = 0$, so this is indeed the unique singular point.
| 3 | https://mathoverflow.net/users/103792 | 435278 | 175,967 |
https://mathoverflow.net/questions/435248 | 0 | Consider a smooth hypersurface $X\subset\mathbb{P}^{n+1}$ of degree $d$ over a nice field (such as $\mathbb{C}$), we know that the cone $C(\operatorname{id}:\mathcal{O}\_X\rightarrow\mathcal{O}\_X)=0$. Under the isomorphism
$$\hom(\mathcal{O}\_X,\mathcal{O}\_X)\cong\hom(\mathcal{O}\_X,\mathcal{O}\_X(d-n-2)[n])^\vee\cong\mathbb{C}$$
one can identify $\operatorname{id}$ with a complex map $$f：\mathcal{O}\_X\rightarrow\mathcal{O}\_X(d-n-2)[n].$$
Can we compute the cone $E$ of this map in $D^b(X)$? Explicitly, what are the cohomology sheaves of $E$?
| https://mathoverflow.net/users/nan | Compute the cone of $\mathcal{O}_X\rightarrow\mathcal{O}_X(d-n-2)[n]$ | The cohomology sheaves are easy to see from the long sequence of cohomology sheaves of the triangle
$$
\mathcal{O}\_X \to \mathcal{O}\_X(d-n-2)[n] \to E.
$$
If $n \ge 2$ it gives
$$
\mathcal{H}^i(E) =
\begin{cases}
\mathcal{O}\_X(d-n-2), & i = -n,\\
\mathcal{O}\_X, & i = -1,\\
0, & \text{otherwise.}
\end{cases}
$$
And if $n = 1$ there is a single cohomology sheaf in the degree $-1$, and it is an extension of $\mathcal{O}\_X$ by $\mathcal{O}\_X(d-3)$.
| 3 | https://mathoverflow.net/users/4428 | 435280 | 175,968 |
https://mathoverflow.net/questions/435283 | 3 | Consider the following fragment from the book "Compact quantum groups and their representation categories" by Neshveyev-Tuset (p72, in section 2.5):
>
> $\ \ \ $ Assume $\mathscr{C}$ is a category having all the properties of a strict $\scr C^\*$-tensor category except existence of direct sums and subobjects. First complete it with respect to direct sums. For this, consider the category $\mathscr{C}'$ consisting of $n$-tuples $(U\_1,\ldots,U\_n)$ of objects in $\mathscr{C}$ for all $n\geqslant 1$. Morphisms are defined by $$\operatorname{Mor}\big((U\_1,\ldots,U\_n),(V\_1,\ldots,V\_m)\big)=\oplus\_{i,j}\operatorname{Mor}(U\_i,V\_j).$$
> Composition and involution are defined in the obvious way. Note that the norm is uniquely determined by the $\rm C^\*$-condition $\|T\|^2=\|T^\*T\|$, and for its existence we do need condition (ii) (c) in Definition 2.1.1 to be satisfied in $\mathscr{C}$. The tensor product of $(U\_1,\ldots,U\_n)$ and $(V\_1,\ldots,V\_m)$ is defined as the $nm$-tuple consisting of objects $U\_i\otimes V\_j$ ordered lexicographically. The category $\mathscr{C}'$ has direct sums: they can be defined by concatenation.
>
>
>
As I understand it, the morphism spaces between objects $(U\_1, \dots, U\_n)$ and $(V\_1, \dots, V\_m)$ just consist of matrices of morphisms between these objects, and composition and involution are defined as the natural matrix operations.
Tensoring morphisms corresponds to the formal Kronecker product of matrices.
What I do not understand however, is how these morphism spaces become Banach spaces (which is a requirement for these kinds of categories). There is a sentence in the text that mentions something about this: "Note that the norm .... to be satisfied in $\mathscr{C}$". However, I do not understand what the author is saying here.
Can someone clarify?
| https://mathoverflow.net/users/216007 | Adding finite direct sums to a C*-tensor category | If you care about completeness, you just want to observe that the norm defined this way restricts to the original norms on the direct summands $\operatorname{Mor}(U\_i, V\_j)$. Since there are only finitely many summands, a sequence in $\bigoplus\_{i,j} \operatorname{Mor}(U\_i, V\_j)$ is a Cauchy sequence if and only if the components in the summands are so.
The remark about Definition 2.1.1 is about checking the C$^\*$-condition for the norm defined this way. You define $\|T\|$ as the square root of ‘the norm’ of $T^\* T$, but how do you determine this?
One way to do is to treat $\bigoplus\_{i, j} \operatorname{Mor}(U\_i, U\_j)$ as a right Hilbert C$^\*$-module over the C$^\*$-algebra $\bigoplus\_i \operatorname{Mor}(U\_i, U\_i)$. (You want to use that condition to obtain a Hilbert module.) Then $T^\* T$ is an adjointable endomorphism of this Hilbert C$^\*$-module, hence you can take its operator norm.
Now, according to this convention, $\|T^\* T\|$ should be defined as the square root of the operator norm of $(T^\* T)^\* (T^\* T)$ acting on the same module. You can then use the fact that adjointable endomorphisms form a C$^\*$-algebra with respect to the operator norm, hence you have $\|T\|^2 = \|T^\* T\|$.
| 5 | https://mathoverflow.net/users/9942 | 435289 | 175,971 |
https://mathoverflow.net/questions/435216 | 8 | For the symmetric group on $n$ objects $S\_n$ the question of how to write its longest element $w\_0$ as a reduced decomposition is an important combinatorical problem. As example, in this [question](https://mathoverflow.net/questions/370333/number-of-reduced-decompositions-of-the-longest-element-of-the-weyl-group) the number of such decompositions is discussed.
Following [Stembridge](https://link.springer.com/article/10.1023/A:1022452717148), we define a relation $∼$ on the set of reduced expressions for $w\_0$. Let $\mathbf{w}$ and $\mathbf{w}'$ be two reduced expressions for $w\_0$ and define $\mathbf{w} \sim \mathbf{w}'$ if we can obtain $\mathbf{w}'$ from $\mathbf{w}$ by applying a single commutation. Now, define the equivalence relation $\simeq$ by taking the reflexive transitive
closure of $\sim$. Each equivalence class under $\simeq$ is called a commutation class.
The question of the number of commutation classes (of reduced expressions for $w\_0 \in S\_n$) is still an open problem. However, is there an formula for the number of equivalence classes containing just one element?
| https://mathoverflow.net/users/491434 | One element commutation classes of reduced decompositions of the longest element of the Weyl group | The reduced words that are in their own commutation classes are:
* $s = [123\cdots(n-2)(n-1)(n-2)\cdots321][23\cdots(n-3)(n-2)(n-3)\cdots32][3\cdots(n-4)(n-3)(n-4)\cdots3]\cdots$
* the reversal of $s$ (writing the word in reverse order)
* the complement of $s$ (mapping each letter $i$ to $n-i$)
* the reverse complement of $s$
This is just one word in $S\_2$ and two words in $S\_3$ (no big surprises there!). Otherwise it is four words. For example, in $S\_6$, they are
* 123454321234323
* 323432123454321
* 543212345432343
* 343234543212345
On a separate note, to add to the list of related topics, you might want to look at Fishel et al. (<https://doi.org/10.1016/j.ejc.2018.07.002>) about how commutation classes and braid classes interact.
| 12 | https://mathoverflow.net/users/495435 | 435293 | 175,973 |
https://mathoverflow.net/questions/433762 | 6 | I am a little confused on the $p$-adic regulator on elliptic curves and what happens when you switch to different Weierstrass models. Restrict to ell. curves over $\mathbb Q$ for simplicity.
From my understanding, to define the p-adic height (and more precisely the $p$-adic sigma function), you need an integral Weierstrass model minimal at $p$. As long as that condition is cleared, the $p$-adic regulator should be independent of the model.
However, when I try this in practice, I get different results.
For example, (and this is the case with most examples), take the rank 1 curves on Sage:
```
E=EllipticCurve('143a1')
E.is_p_minimal(7)
E1=E.short_weierstrass_model()
E1.is_p_minimal(7)
```
It's easy to check, they're both minimal at $p=7$ and that $p=7$ is a good ordinary non-anomalous prime. (The model of $E$ is automatically The minimal Weierstrass model).
However, their regulators are completely different.
```
E.padic_regulator(7)
2*7^2 + 5*7^3 + 7^4 + 5*7^5 + 3*7^6 + 5*7^7 + 7^8 + 5*7^9 + O(7^10)
E1.padic_regulator(7)
3*7 + 3*7^2 + 4*7^3 + 3*7^4 + 5*7^6 + 7^8 + 3*7^9 + 5*7^11 + 7^12 +O(7^13)
```
Not only are they different, but their $p$-valuation as you can see is also different, which, without going through details, would imply in the first case that the Iwasawa $\lambda$ invariant would be greater than 1, whereas in the second case it would imply that $\lambda\_p(E)= 1$, which is absurd since $\lambda\_p(E)$ does not depend on the model. The correct regulator is the first one, of The minimal curve, since in this example one can check $\lambda\_p(E) >1$.
At first I was convinced it was an implementation error by Sage, so I posted pretty much the same question on [Ask Sage](https://ask.sagemath.org/question/64632/p-adic-regulator-on-different-weierstrass-models/) with the assumption that it was a bug, but for various reasons I could go through, now I have started to doubt that it was due to implementation errors, which is why I also decided to post the question here.
So in short, am I misunderstanding something? Shouldn't both the regulators be the same, since both models are minimal at $p$?
| https://mathoverflow.net/users/124772 | Does the $p$-adic regulator depend on Weierstrass model? | Here a long comment to settle this question.
This is really a bug in the implementation of $p$-adic heights in sagemath. I have announced it [as a bug here on the sage trac list](https://trac.sagemath.org/ticket/34790). I hope to add the code to fix it soon after this post.
The current implementation is based on [1], which in turn is a refinement of the method by Mazur-Stein-Tate mentioned before.
First of all the $p$-adic height of a point on an elliptic curve must be independent of the chosen model. The formula used looks like $\hat{h}\_p(P) = \log\_p(\sigma\_p(P)/d(P))$ up to a chosen constant factor, where $\sigma\_p$ is the canonical $p$-adic $\sigma$-function for this model and $d(P)$ is the square root of the denominator of the $x$-coordinate of $P$. However this formula only holds if $P$ has good reduction at all places for the given model and trivial reduction at $p$. Instead of figuring out the correction factors introduced when the point has bad reduction in the model, one uses the fact that $\hat{h}\_p$ is quadratic. First a certain multiple $Q=nP$ has good reduction everywhere. Then a multiple $mQ$ has trivial reduction at $p$.
The algorithm implemented took $n=\operatorname{lcm}(c\_v)$, the lowest common multiple of all Tamagawa numbers, but this is only valid if the model is minimal. Instead for primes $p$ at which the model is not minimal, one needs to correct this.
As explained in [1], David Harvey has optimised the implementation by a clever trick to avoid the next multiplication by $m$ which brings the point into the formal group. As one only needs a $p$-adic approximation of $d(mQ)$ and of $\sigma(mQ)$. This follows the idea explained in the other answer.
I would avoid to try and extend this trick to the first multiplication $nP$ which aims at getting the point to have good reduction at all primes. The reason is that there will be cancellation in the fraction obtained by the formula using the $n$-th division polynomial $f\_n$. When the point $P$ has good reduction at all primes then $d(nP) = f\_n(P)\cdot d(P)^{n^2}$ as explained in Proposition 1 [here](https://www.maths.nottingham.ac.uk/plp/pmzcw/download/pheight.pdf). Otherwise this formula does not holds as such.
[1] MR2395362 Harvey, David
[Efficient computation of p-adic heights](https://arxiv.org/abs/0708.3404).
LMS J. Comput. Math. 11 (2008), 40–59.
| 2 | https://mathoverflow.net/users/5015 | 435294 | 175,974 |
https://mathoverflow.net/questions/435205 | 4 | I suppose that what I look for is known, but I can't find it.
Let $\left\lbrace I\_n=[a\_n,b\_n)\right\rbrace$ and $\left\lbrace J\_n=[b\_n,c]\right\rbrace$ ($n\in\mathbb{N}$) be two countable families of intervals in the unit circle $S^1$. Notice that $I\_n$ and $J\_n$ are adjacent for every $n$, and that the extremum $c$ is fixed. Assume that the total length of every pair goes to 0, that is $$\lim\_{n\to\infty}|c-a\_n|= 0.$$ Assume also that the length of the "left" interval is a higher order infinitesimal than the length of the "right" one, that is $$\lim\_{n\to\infty}\frac{|b\_n-a\_n|}{|c-b\_n|}= 0.$$
Now let $R:S^1\to S^1$ be an irrational rotation. I say that $x\in S^1$ ultimately first visits $J\_n$ if, for every sufficiently large $n$, $$\min\left\lbrace k:R^k(x)\in J\_n\right\rbrace<\min\left\lbrace k:R^k(x)\in I\_n\right\rbrace.$$
Now my question: is it possible to take the families $\{I\_n\}$ and $\{J\_n\}$ such that every point in $S^1$ ultimately first visits $J\_n$?
| https://mathoverflow.net/users/167834 | First visit of intervals for an irrational rotation | No, it is not possible. In the following I will use $I\_n=(a\_n,b\_n)$ instead of $[a\_n,b\_n)$ (this is not a problem, you can just increase $a\_n$ a bit so that the statement with $I\_n=(a\_n,b\_n)$ is stronger).
For fixed $n$, we say that $x$ first visits $J\_n$ if $\min\left\lbrace k:R^k(x)\in J\_n\right\rbrace<\min\left\lbrace k:R^k(x)\in I\_n\right\rbrace$. So the set of points that first visit $J\_n$ is $A\_n:=\bigcup\_{m=1}^\infty\left(R^{-m}J\_n\setminus\bigcup\_{k=1}^{m-1}R^{-k}I\_n\right)$.
Then the set of points that ultimately first visit $J\_n$ is $X:=\bigcup\_{N=1}^\infty\bigcap\_{n=N}^\infty A\_n$. What we want is $X=\mathbb{S}^1$, or equivalently, $\varnothing=\mathbb{S}^1\setminus X=\bigcap\_{N=1}^\infty\bigcup\_{n=N}^\infty(\mathbb{S}^1\setminus A\_n)$. Moreover, $\mathbb{S}^1\setminus A\_n$ is the set of points that first visit $I\_n$, that is, $B\_n:=\bigcup\_{m=1}^\infty\left(R^{-m}I\_n\setminus\bigcup\_{k=1}^{m-1}R^{-k}J\_n\right)$.
So, we want $\bigcap\_{N=1}^\infty\bigcup\_{n=N}^\infty B\_n=\varnothing$. However, for each $N$, the set $\bigcup\_{n=N}^\infty B\_n$ is open and dense: it is open because $B\_n$ is open for all $n$. To check that it is dense, we will prove that for every $\varepsilon>0$ there is some $n$ such that $B\_n$ is [$\varepsilon$-dense](https://en.wikipedia.org/wiki/Dense_set#Related_notions) in $\mathbb{S}^1$. Indeed, let $z$ be the complex number such that $R(x)=zx$ for all $x$. Now let $k$ be such that $\{z,z^2,\dots,z^{k}\}$ is $\varepsilon$-dense in $\mathbb{S}^1$, and let $n$ be so big that the distance between any two points of $\{z,z^2,\dots,z^{k}\}$ is bigger than $|c-a\_n|$. This implies that $R^{-a}I\_n$ does not intersect $R^{-b}J\_n$ if $a,b<k$ (this is obvious if $a=b$, and if not we use that $d(z^{-a},z^{-b})>|c-a\_n|$, so as $I\_n$ and $J\_n$ are contained in an interval of length $|c-a\_n|$, $z^{-a}I\_n$ doesn't intersect $z^{-b}J\_n$). So $B\_n$ contains $\bigcup\_{m=1}^k\left(R^{-m}I\_n\setminus\bigcup\_{k=1}^{m-1}R^{-k}J\_n\right)=\bigcup\_{m=1}^kR^{-m}I\_n$, which is $\varepsilon$-dense in $\mathbb{S}^1$ as we wanted.
So as $\bigcup\_{n=N}^\infty B\_n$ is open and dense for all $N$, by Baire's theorem $\bigcap\_{N=1}^\infty\bigcup\_{n=N}^\infty B\_n$ is nonempty.
| 2 | https://mathoverflow.net/users/172802 | 435298 | 175,975 |
https://mathoverflow.net/questions/16347 | 7 | $\newcommand\Box{\mathrm{Box}}\newcommand\Set{\mathrm{Set}}\newcommand\op{^\text{op}}\DeclareMathOperator\Hom{Hom}$A cubical set $\Box\op \to \Set$ is a model for a homotopy type, via Grothendieck and Cisinski (here $\Box$ is the box category with objects the natural numbers and arrows generated by face and degeneracy maps ‘as usual’). A typical example is the singular cubical set of a space, $n \mapsto \Hom(I^n,X)$. The homotopy groups of $X$ can be recovered from this cubical set as it satisfies a property analogous to that of Kan complexes (horns have fillers). In general, do the homotopy groups of a cubical set satisfying this ‘Kan’ condition agree with that of the homotopy type it represents?
| https://mathoverflow.net/users/4177 | Homotopy groups of cubical sets | I think a reference for this would be [Theorem 3.24](https://arxiv.org/pdf/2202.03511.pdf#page=26) of
>
> Homotopy groups of cubical sets, Daniel Carranza, Chris Kapulkin, 2022. arXiv:2202.03511, <https://doi.org/10.48550/arxiv.2202.03511>
>
>
>
| 4 | https://mathoverflow.net/users/130058 | 435306 | 175,978 |
https://mathoverflow.net/questions/435324 | 9 |
>
> Has anyone ever attempted to write down axioms capturing the behaviour of ${\bf Rel}$, the category of relations?
>
>
>
Lawvere's [ETCS](https://ncatlab.org/nlab/show/ETCS) attempts to axiomatize the behaviour of the subcategory ${\bf Set}$ of ${\bf Rel}$ and ends up with a theory equiconsistent with $ZF-{\sf Replacement}$; I'm curious if axiomatizing ${\bf Rel}$ leads to a category where we can always define a subcategory corresponding to ${\bf Set}$, or if this is a second-order property not necessarily possible in 'ETCR'. If it is possible this would give ETCR at least the consistency strength of ETCS, and I'm curious if this is an upper bound on its consistency strength as well or if it's strictly stronger.
| https://mathoverflow.net/users/92164 | Elementary theory of the category of relations | As Sam has pointed out in the comments, [SEAR](https://ncatlab.org/nlab/show/SEAR) is close in spirit to this sort of theory. The difference is that SEAR has "elements" as a basic notion in addition to sets and relations, whereas an axiomatization of Rel would contain only sets and relations. One could in theory translate the axioms of SEAR into statements about Rel only, but the result would be rather klunky.
Probably a better thing to call "ETCR" would be to write down the axioms of a [power allegory](https://ncatlab.org/nlab/show/allegory) (also mentioned by Andreas in the comments), which is known to be equivalent to an elementary topos, and then add a suitable "well-pointed" axiom. In such a theory you would certainly always be able to isolate the "functions" from among the relations and define a category satisfying ETCS, and vice versa --- although these would of course be metatheoretic model constructions or syntactic translations rather than definitions internal to the theories.
In particular, a thus-formulated ETCR would have the same consistency strength as ETCS, namely "bounded Zermelo" a.k.a. "Mac Lane set theory". By the way, I don't think calling this "ZF $-$ Replacement" is quite right. The notion of "subtracting" an axiom (schema) from a theory is not well-defined since a theory can be axiomatized in many equivalent ways, and the result of "subtracting" some axioms depends on the whose axiomatization. In particular, if your axiomatization of "ZF" includes the full separation schema in addition to replacement, then "ZF $-$ Replacement" will be stronger than ETCS. Whereas if your axiomatization of "ZF" doesn't include any separation explicitly (instead deriving it from replacement), then "ZF $-$ Replacement" won't have any separation axiom and hence should be weaker than ETCS. There is an axiomatization of ZF such that "ZF $-$ Replacement" is equiconsistent with ETCS, namely if you include Bounded Separation explicitly in addition to the full Replacement schema, but this is not one of the usual axiomatizations of ZF.
The fact that SEAR is equiconsistent with ZF is just a choice about the default meanings of names. By default, "ETCS" does not include a replacement schema or a full separation schema, whereas by default "SEAR" does include such a schema. But you can easily add a replacement schema to ETCS to get a theory ETCS+R that's equiconsistent with ZF, or remove the replacement (actually, collection) schema from SEAR to get a theory equiconsistent with ETCS. The same would be true for any ETCR. There's a fairly extensive discussion of replacement-style axioms for categorical set theories in my paper [Comparing material and structural set theories](https://arxiv.org/abs/1808.05204v2).
| 14 | https://mathoverflow.net/users/49 | 435330 | 175,985 |
https://mathoverflow.net/questions/435336 | 9 | In mathematical contexts the term *spline* essentially refers to interpolating or approximating piecewise functions with continuity constraints.
According to [the history of mathematical splines](https://en.wikipedia.org/wiki/Spline_(mathematics)#History)
>
> In the foreword to (Bartels et al., 1987), Robin Forrest describes "lofting", a technique used in the British aircraft industry during World War II to construct templates for airplanes by passing thin wooden strips (called "splines") through points laid out on the floor of a large design loft, a technique borrowed from ship-hull design. used in aircraft design
>
>
>
and also remarks that
>
> The word "spline" was originally an East Anglian dialect word.
>
>
>
Today however I came across a different meaning of the term "spline" in an article headlined [Design parameters for spline connections](https://www.geartechnology.com/articles/22611-design-parameters-for-spline-connections):
>
> Splines are machine elements that connect a shaft with a rotor.
>
>
>
That leads to further questions regarding the "genealogy" of the term "spline" that is nowadays used in the context of piecewise functions:
* Are the "wooden-strip splines" named after the "shaft-connection splines", or is the equality of the term "spline" in these cases a mere coincidence?
* A more linguistic question: is the word "spline" a true dialect word that corresponds to a different "official" word, or is it rather a proper English word that was in broader use locally, maybe because of a higher concentration of machines, resp. machine-design activities?
| https://mathoverflow.net/users/31310 | True origin of the term "Spline" | The Oxford English Dictionary doesn't necessarily give the earliest uses of a word. But spline with the meaning "A long, narrow, and relatively thin piece or strip of wood, metal, etc.; a slat." is quoted from 1756, much earlier than "flexible strip of wood or hard rubber used by draftsmen in laying out broad sweeping curves" (1891) and "A rectangular key fitting into grooves in a shaft and wheel or other attachment so as to allow longitudinal movement of the latter." (1864)
OED says "Originally East Anglian dialect: perhaps for splind (compare older Danish splind , North Frisian *splinj*) and related to splinder", where "splinder" is a splinter with uses noted as early as c1440: "Þe splyndre or speele þerof schal entre into hys hoond".
In the Philosophical Transactions of the Royal Society of 1803 I found: "This word is, perhaps, not universally understood: it is the technical translation of the French *cheville*. When there is a gap, chink, or rift, in the wainscot, which a carpenter is employed to fill up, he cuts a lath to the length of the aperture, planes it to the right width, and inserts it. Such inserted bits of wood, contrived to fit a given vacancy, are called *splines*, or *chevilles*...
| 11 | https://mathoverflow.net/users/9025 | 435337 | 175,988 |
https://mathoverflow.net/questions/435316 | 2 | Nancy Dykes says in the proof of Theorem 3.4 in her article [Generalizations of realcompact spaces](https://msp.org/pjm/1970/33-3/pjm-v33-n3-p05-s.pdf) that by a result of
John Mack, if for every $p\in \beta X\setminus X$ there exists a nonnegative
upper semicontinuous function $f$ on $\beta X$ such that $f$ is positive on $X
$ and $f\left( p\right) =0$, then $X$ is realcompact.
I looked at both of John Mack's articles in the references but couldn't find
this result. How can I prove this result?
| https://mathoverflow.net/users/86099 | A question about a realcompact space and upper semicontinuous function | The following characterisation is well known. It can be found in Engelking's book as Theorem 3.11.10.
>
> **Theorem**: A Tychonoff space $X$ is realcompact if and only if for each $p\in\beta X\setminus X$ there is a continuous function $\varphi:\beta X\rightarrow [0,\infty)$ such that $\varphi|\_X>0$ and $\varphi(p)=0$. $\;\blacksquare$
>
>
>
Thus suppose the weaker condition, that for $p\in\beta X\setminus X$ there is an upper semicontinuous function $f:\beta X\rightarrow[0,\infty)$ with $f|\_X>0$ and $f(p)=0$.
>
> **Theorem (Katětov-Tong)**: A space $Y$ is normal (not necessarily $T\_1$) if and only if for any upper semicontinuous function $f:Y\rightarrow \mathbb{R}$ and any lower semicontinuous function $g:Y\rightarrow \mathbb{R}$ satisfying $f(y)\leq g(y)$ for each $y\in Y$ there is a continuous function $\varphi:Y\rightarrow\mathbb{R}$ with $f(y)\leq\varphi(y)\leq g(y)$ for all $y\in Y$. $\;\blacksquare$
>
>
>
Of course $Y=\beta X$ is normal. Since $\beta X$ is compact, every upper semicontinuous function defined on it has an upper bound which it attains. We can assume the upper semicontinuous function $f:\beta X\rightarrow[0,\infty)$ is bounded above by $1$. Take $g:\beta X\rightarrow\mathbb{R}$ to be the characteristic function of $\beta X\setminus\{p\}$, so that $g(p)=0$ and $g(q)=1$ if $p\neq q$. Since $\beta X\setminus\{p\}$ is open, $g$ is lower semicontinuous.
We have $f\leq g$ everywhere, so from the Katětov-Tong Theorem we obtain a continuous function $\varphi:\beta X\rightarrow [0,\infty)$ with $\varphi|\_X>0$ and $\varphi(p)=0$. Thus we see the sufficiency of this condition for the realcompactness of $X$.
| 2 | https://mathoverflow.net/users/54788 | 435339 | 175,990 |
https://mathoverflow.net/questions/435340 | 1 | Consider the partial differential equation
$$\psi\_t(t,x)=i\kappa \psi\_{xx}(t,x) ~\mbox{for}~ 0<(t,x)\in\mathbb{R}\times\mathbb{R}$$
with boundary conditions
$$\psi(0,x)=0 ~\mbox{for}~ x>0,$$
$$\psi(t,0)=\psi\_0(t) ~\mbox{for}~ t\ge0.$$
Are these equation uniquely solvable whenever $\psi\_0$ is sufficiently smooth?
Can one give an explicit expression for the linear operator mapping $\psi\_0$ to the solution?
| https://mathoverflow.net/users/56920 | Schrödinger equation with nonstandard boundary conditions | The equation under consideration is uniquely solvable in $H^s(\mathbb{R}^+)$, as soon as $\psi\_0\in H^{(2s+1)/4}(\mathbb{R}^+)$, and there is a fairly explicit expression for the propagator. You can find these results, for example, in
J. L. Bona, S. Sun, B. Zhang,
Nonhomogeneous boundary-value problems for one-dimensional nonlinear Schrödinger equations, <https://doi.org/10.1016/j.matpur.2017.11.001>,
where also the nonlinear problem is investigated.
| 1 | https://mathoverflow.net/users/169603 | 435348 | 175,993 |
https://mathoverflow.net/questions/435344 | 2 | I need to find the following paper:
“K. Urbanik and WA Woyczynski, A random integral and Orlicz spaces, Bulletin de l'Académie Polonaise des Sciences, Série des sciences mathématiques, astronomiques et physiques, 15 (1967), p. 161-169.”
It is possible to find it on the internet?
Thanks
| https://mathoverflow.net/users/495513 | Reference request: “A random integral and Orlicz spaces” | I could not find it on the internet so I uploaded it here: <https://www.transfernow.net/dl/20221126nnUfCto7>
| 2 | https://mathoverflow.net/users/54263 | 435354 | 175,994 |
https://mathoverflow.net/questions/435368 | 3 | $\DeclareMathOperator\Var{Var}\DeclareMathOperator\Motives{Motives}$Let us assume for the moment that we have a "nice" category of motives, that is for fields $k$ we have a contravariant functor
$$\Var(k)\to \Motives(k).$$
Now for any field extension $l/k$, we have a natural forgetful functor
$$\Var(l)\to\Var(k).$$
Would we expect to have an extension of this functor to motives, that is a corresponding functor
$$\Motives(l)\to \Motives(k)$$
compatible with the forgetful functor of schemes? I've only found references for the "other direction", that is restriction of scalars.
| https://mathoverflow.net/users/152554 | Functor between categories of motives | This would be a motivic analogue of the induced representation functor, and should be adjoint to the restriction functor, at least for finite field extensions.
This is just because the cohomology of a variety obtained by the forgetful functor is the induced representation of the cohomology.
It should be possible to construct this straightforwardly in most existing theories of motives.
For infinite field extensions it may depend on exactly which category of motives you mean and which finiteness conditions it might have - it certainly doesn't exist for Chow motives, say.
| 6 | https://mathoverflow.net/users/18060 | 435369 | 175,999 |
https://mathoverflow.net/questions/435371 | 1 | Assume that for each $n\in\mathbb{N}$, there's a stochastic function $f\_n$ of type $\mathbb{R}^{m}\to\Delta\mathbb{R}^{m}$, and for each $x\in\mathbb{R}^{m}$, the distributions $\frac{f\_n(x)-x}{\frac{1}{n}}$ will weakly converge as $n$ limits to $\infty$, s.t. the n'th distribution is about $O(\frac{1}{n^2})$ away from the limiting distribution w.r.t. the usual earthmover distance. Also, the function which maps a given starting point $x$ to its (rescaled) limiting distribution $\lim\_{n\to\infty}\frac{f\_n(x)-x}{\frac{1}{n}}$ is known to be Lipschitz.
The particular thing I'm trying to prove is that, for any finite interval of time $[0,I]$, $\epsilon,\delta$, then in the $n\to\infty$ limit, all but $\delta$ of the trajectories you get from applying $f\_n$ $In$ times to a starting point are $\epsilon$-close to the trajectory you get from the differential equation $\frac{dx}{dt}=\lim\_{n\to\infty}\mathbb{E}\left[\frac{f\_n(x)-x}{\frac{1}{n}}\right]$ (ie, in a tiny time interval, the point just moves deterministically to the average of its next positions, instead of moving randomly)
Now, there's a strong informal intuition for why it would work out this way. Because the function which maps a given starting point $x$ to its (rescaled) limiting distribution $\lim\_{n\to\infty}\frac{f\_n(x)-x}{\frac{1}{n}}$ is Lipschitz, that means that for large $n$, it should be be possible to iterate the function a whole bunch of times in a row without the distribution on next-positions changing much. Splitting the application of the function into a deterministic "drift" term that goes to the average next position, and a random term with a mean of 0, the resulting distribution should look something like "sum up all the drift terms, and by Martingale Central Limit Theorem shenanigans, the random terms will mostly cancel themselves out and look like a Gaussian with the same covariance matrix."
But since the "spread" of the Gaussian is about $\sqrt{\text{number of iterations}}\cdot\text{step-size}$, and the step size scales as $\frac{1}{\text{total number of iterations}}$, the "spread" after $n$ steps would be about $\sqrt{n}\cdot\frac{1}{n}=\frac{1}{\sqrt{n}}$ and so in the $n\to\infty$ limit, the Gaussian sharpens up to a single point, a dirac-delta distribution. The variance is decreasing too fast as $n$ ramps up, and so, in the limit, the differential equation is deterministic, not stochastic, and only the "drift" terms survive from the repeated application of $f\_n$ for large $n$.
The issue is, although the informal argument is quite clear, I don't know what theorems would be involved in rigorously proving this conjecture. The stochastic differential equation theorems I've looked at seem to primarily be about how to analyze stochastic differential equations once you already have them, not about rigorously showing that a given sequence of discrete stochastic processes have a certain SDE as their limit. A response like "if you prove this bound and that bound and this other bound on your stochastic function we can directly apply Theorem 3.24 from BlahBlah to show your desired result" would be ideal.
| https://mathoverflow.net/users/148137 | How to rigorously prove that this sequence of stochastic processes converges to a deterministic process? | I am guessing in "The particular thing I'm trying to prove is that,..." you are talking about the convergence of discrete generator to continuous one. The natural topology for these questions is Skorokhod. See here for some ideas and references/keywords: <https://math.stackexchange.com/questions/4225750/continuous-limit-of-a-discrete-stochastic-process>
>
> We denote by $P\_n$ the transition kernel of $X^n$ given by
> $$P\_nf(x) = \mathbb{E}\left[f\left(X\_{t\_1}^n\right)\left| X\_0 = x\right.\right], $$
> for functions $f$ of a suitable class (we will take that class to be $C^2\left(\left[0,1\right]\right)$). We will show that the discrete generators of $X^n$,
> $$\mathcal{L}\_n = \frac{P - I}{\Delta t^n}$$
> where $I$ is the identity operator (i.e. $If = f$ for any function), converges to $\mathcal{L}$.
>
>
>
As mentioned there some good references are:
* Billingsley's Convergence of Probability Measures.
* Ethier and Kurtz's Markov Processes.
If you are dealing with SPDEs, you can also check out the literature in Wong-Zakai type theorems.
| 1 | https://mathoverflow.net/users/99863 | 435373 | 176,000 |
https://mathoverflow.net/questions/435377 | 2 | Every non-singular complex projective cubic surface has $27$ lines. Many such surfaces contain points where three lines intersect (called Eckardt points). There are even surfaces with many Eckardt points, like the Fermat cubic, which has $18$. Is there any such non-singular complex projective cubic surface where four, five, or six lines intersect at a point?
[Post on Math StackExchange](https://math.stackexchange.com/questions/4585682/is-there-a-non-singular-cubic-surface-that-has-a-point-where-four-lines-intersec)
| https://mathoverflow.net/users/493889 | Is there a non-singular cubic surface that has a point where four lines intersect? | No, this is not possible. If p is a smooth point on any surface S, and is
contained in a line l on S, then l is contained in the tangent plane at p,
call it T\_p. Now if S is a cubic then it intersects T\_p in a cubic curve
(with some singularity at p, even though S is smooth at p); and a cubic curve
can contain at most three lines.
| 12 | https://mathoverflow.net/users/14830 | 435379 | 176,002 |
https://mathoverflow.net/questions/428454 | 6 | A famous corollary of [Matiyasevich's theorem](https://en.wikipedia.org/wiki/Diophantine_set#Matiyasevich%27s_theorem) is that there exists a Diophantine equation such that it is undecidable (under some recursively axiomatizable theory $T$, such as ZFC) whether that equation has any natural-number solutions. I will somewhat sloppily describe such a Diophantine equation as "undecidable" (with respect to $T$) for brevity.
I've never actually seen an explicit example of an undecidable Diophantine equation. The closest that I've seen is in [this answer](https://mathoverflow.net/a/81986/95043) to another Math Overflow question, which provides an example that's *almost* explicit, but contains one unspecified parameter $K$, whose value depends on the recursively axiomizable theory $T$. However, that answer claims that $K$ can be effectively computed from the axioms of $T$.
A. Are there any simpler explicit Diophantine equations that are known to be undecidable with respect to some common axiomatic system (ideally ZFC, or else ZF or ZFC + CH or something)?
B. For, say, ZFC (or a similar theory, as described in question A), which of the following mutually exclusive statements best captures how well we can actually calculate the value of $K$?
1. An explicit numerical value of $K$ has been reported in the literature.
2. We know how to calculate a valid choice of $K$ and could actually do it in the real world if we really wanted to, but doing so would be too tedious for anyone to have actually bothered to do it so far.
3. We know how to calculate a valid choice of $K$ in principle, but doing so would be impossible in practice - e.g. it would require many, many years of labor, or more computational resources than are currently available with existing supercomputers.
4. We know how to calculate a valid choice of $K$, but it isn't clear whether doing so would be feasible in practice (so it isn't clear whether case #2 or #3 above is correct).
5. We know that $K$ can be effectively computed from the axioms of $T$, but we don't know how to actually do so, so we cannot find a valid choice of parameter $K$ with our current mathematical knowledge.
| https://mathoverflow.net/users/95043 | How constructive is Matiyasevich's theorem? | Turning the proof that there exists a Diophantine equation encoding eg the consistency of $\mathrm{ZFC}$ into a program that actually computes the polynomial would be a bit tedious, but there should not be any significant obstacles. Whether running said program would actually yield an answer, some kind of overflow error, or whether we'd simply get too bored waiting for it to do anything at all is less clear to me.
What I am certain of is that if we do get an answer, it would be sufficiently monstrous that we couldn't gain any insight from it. (And I assume that people generally agreeing with me is the reason why people don't really try it out.)
Getting a "small" undecidable Diophantine equation will require much more work. At any stage of the encodings, one should look for ways to be more efficient.
To better understand the boundary between decidable and undecidable Diophantine equations, it is probably more promising to look at properties of the polynomials that allow for decision procedures vs those properties polynomials obtained from the Turing machine translations have.
| 2 | https://mathoverflow.net/users/15002 | 435391 | 176,006 |
https://mathoverflow.net/questions/435389 | 3 | In $G=\operatorname{PGL}(4,5)$ there are two elementary abelian $2$-subgroups of order $16$ denoted by $E\_{1}$ and $E\_{2}$ with $N\_{G}(E\_{1})=E\_{1}.\operatorname{Sp}(4,2)$ and $N\_{G}(E\_{2})=E\_{2}.(2^{3}:S\_{3})$.
$N\_{G}(E\_{1})$ is a maximal subgroup of $G$ and $(2^{3}:S\_{3})$ is a point stabilizer of a nonidentity element of $E\_{2}$ in $\operatorname{Sp}(4,2)$ (viewing $E\_{2}$ as a symplectic basis of $\operatorname{Sp}(4,2)$).
By comparison, $H=\operatorname{PGL}(4,3)$ has $N\_{H}(F\_{1})=F\_{1}.M\_{1}$ and $N\_{H}(F\_{2})=F\_{2}.M\_{2}$ where $F\_{i}$ are again elementary abelian $2$-subgroups of order $16$ and $M\_{i}$ are, by some Magma computation, two maximal subgroups of $\operatorname{Sp}(4,2)$ of orders $72$ and $120$, respectively. I see that $2^{4}.\operatorname{Sp}(4,2)$ is not a subgroup of $H$ any more, in contrast. (It appears in the corresponding finite unitary group now.) Also $M\_{1}$ and $M\_{2}$ are stabilizers in $\operatorname{Sp}(4,2)$ of $f\_{1}\in F\_{1}$ and $f\_{2}\in F\_{2}$ which have orbit sizes $10$ and $6$.
Is there an explanation for this phenomenon? What are these orbits of sizes $10$ and $6$ now? I hope to generalizer this. For instance, in $S=\operatorname{PGL}(8,3)$, what are the analogous orbit sizes in $2^6$?
I've only just found out that $M\_{1}\cong \operatorname{SO}^{+}\_{4}(2)$ and $M\_{2}\cong \operatorname{SO}^{-}\_{4}(2)$. They do have orbit sizes 1,5,10 and 1,9,6….
| https://mathoverflow.net/users/488802 | Normalisers and stabilisers in classical groups $\operatorname{PGL}_{4}$ | The difference between the examples arises principally because $5 \equiv 1 \bmod 4$ and $3 \equiv 3 \bmod 4$.
For $q \equiv 1 \bmod 4$, $G := {\rm GL}(4,q)$ has centre $Z$ divisible by $4$, and contains a group $S$ of symplectic type with $N\_G(S) = ZS.{\rm Sp}(4,2)$. The group $S$ maps onto your group $E\_1$ in ${\rm PGL}(4,q)$.
As you pointed out, you get similar behaviour in the unitary group when $q \equiv 3 \bmod 4$
When $q \equiv 3 \bmod 4$, $G := {\rm GL}(4,q)$ does not have an element of order $4$ in its centre. It has extraspecial subgroups $S^+$ and $S^-$ of two different types, with normalizers $ZS^+ {\rm GO}^+(4,2)$ and $ZS^- {\rm GO}^+(4,2)$ The groups $S^+$ and $S^-$ map onto your groups $F\_1$ and $F\_2$.
The inverse image of your group $E\_2$ when $q=5$ has centre of order $4$ and derived group of order $2$ (as for $E\_1$) but, unlike the inverse image of $E\_1$, it has exponent $8$ and a smaller normalizer.
| 4 | https://mathoverflow.net/users/35840 | 435402 | 176,008 |
https://mathoverflow.net/questions/435401 | 5 | I am trying to prove the following. (I posted this on [math.se](https://math.stackexchange.com/questions/4568521/initially-horizontal-geodesic-is-always-horizontal) with no success)
>
> Let $E,B$ be Riemannian manifolds. Suppose
> $\pi: E\to B$ is a Riemannian submersion.
>
>
> For each $x\in E$, define $V\_x E = \ker \pi\_{\*}$ and $H\_xE = (V\_x
> E)^{\perp}$. $W\in T\_x E$ is said to be *horizontal* if $W\in H\_x E$.
>
>
> Let $\gamma:[0,1]\to E$ be a geodesic curve such that $\gamma'(0)$ is
> horizontal. Prove that $\gamma'(t)$ is horizontal for all $t\in
> [0,1]$.
>
>
>
**My attempt**
I mostly followed the proof [here](https://www.ime.usp.br/%7Egorodski/teaching/mat5771-2021/fabricio-submersions.pdf)(Theorem 3.6).
Firstly, for any smooth curve $\gamma:[0,1]\to B$, there is always a locally defined *horizontal lift* to $E$. That is, for any $p\in \pi^{-1}(\gamma(0))$, there is $\epsilon>0$ and a smooth curve $\gamma\_E:[0,\epsilon) \to E$ such that $\pi\circ \gamma\_E = \gamma$, $\gamma\_E(0)=p$ and $\gamma\_E'$ is always horizontal.
So, we take a geodesic on $B$ with initial condition $\gamma\_B(0)=\pi\circ\gamma(0), \gamma\_B'(0)=\pi\_{\*}\gamma'(0)$. We can do so (at least locally) through the uniqueness and existence of geodesic.
Then, take a horizontal lift $\gamma\_E:I\to E$ with $\gamma\_E(0)=\gamma(0)$. Here, $I$ is an open subset of $[0,1]$ containing $0$.
We can prove that $\gamma\_E$ is a geodesic. (This argument is not the main topic of this question.)
Then, by the uniqueness of geodesic, we have $\gamma\_E(t) = \gamma(t)$ for $t\in I$.
**Here is the question.** It seems that we need to prove that $I$ is a closed(and therefore clopen) set of $[0,1]$, too. Then since $[0,1]$ is connected we have $I=[0,1]$. But I do not know how to do it.
For example, [this book](https://www.google.com/books/edition/Riemannian_Geometry/wEcBBQAAQBAJ?hl=en&gbpv=1&dq=initially+horizontal++submersion&pg=PA101&printsec=frontcover) on the proof of Proposition 2.109, (ii),
(Here $\tilde{c}$ corresponds to our $\gamma$, and $c$ corresponds to $\gamma\_B$)
>
> ...Hence the set of parameters where the geodesic $\tilde{c}$ is
> horizontal, and where it is a lift of $c$ is an open set containing
> $0$. These two conditions being also closed, they are satisfied on the
> maximal interval of definition of $\tilde{c}$.
>
>
>
Any help?
| https://mathoverflow.net/users/220580 | Initially horizontal geodesic is always horizontal | After choosing local coordinates, by the implicit function theorem (I'm omitting a bunch of technical computations) there is a smooth function $\varphi:TE \to TE$ such that $\varphi(x,-): T\_x E \to T\_x E$ is the projection to $V\_x E$.
Let $I\subseteq [0,1]$ be the set on which $\gamma'$ is horizontal: this is the set on which $\varphi\circ \gamma' = 0$ (the zero section). As a composition of two smooth functions ($\gamma'$ is a smooth function from $[0,1]\to TE$ and $\varphi$ is smooth), the function $\varphi\circ\gamma'$ is continuous, and hence $(\varphi\circ\gamma')^{-1}(0)$ is closed (the zero section is a closed subset of $TE$).
| 7 | https://mathoverflow.net/users/3948 | 435405 | 176,009 |
https://mathoverflow.net/questions/420558 | 1 | Both cohomology and homotopy groups capture global topological information of a manifold $X$. It is interesting to ask if they can be computed from local data. A triangulation $T$ is a natural presentation of a manifold.
The cohomology by definition can be computed from $T$. However, the problem seems much harder for homotopy groups. In fact, humans struggle even for the simplest case $X=S^n$, at least for $\pi\_{k>n}(X)$.
Hope is not lost, as we know the answer to $k \leq n$ for spheres. Therefore my question:
**Question:** Given a triangulated (compact) manifold $X$ of dimension $n$, is there an algorithm that computes $\pi\_{k \leq n}(X)$ (up to isomorphism, or in the form of generators and relations)?
| https://mathoverflow.net/users/124549 | (Lower) homotopy groups from triangulations | Being a manifold or the dimension restriction $k\leq n$ doesn't matter, the following applies to finite simplicial complexes in general:
As others have explained, if the fundamental group is not finite, the higher homotopy groups might be infinitely generated, so in that case even the format of the answer is unclear. Also, it is known that deciding whether a group given by a finite presentation (the simplicial complex gives you such a presentation for $\pi\_1$) is actually finite is not possible algorithmically (the halting problem can be reduced to this problem, so if this is problem would admit an algorithmic solution, the halting problem would, too). This problem arises even if you restrict attention to manifolds: Any finite presentation of a group can be turned into an explicit manifold of dimension $4$ or greater via handlebodies.
However, there are algorithms that will produce the list of elements in finite time if $\pi\_1$ happens to be finite (and simply never terminate otherwise). From this, it is possible to produce a finite simplicial complex describing the universal cover. For simply-connected finite simplicial complexes, the problem of computing any given homotopy group is known to be be algorithmic, in fact, there exist [actual implementations](https://www-fourier.ujf-grenoble.fr/%7Esergerar/Kenzo/). (I do think they scale pretty badly with degree though).
EDIT: For the $\pi\_1$ statement, let me give you an easy but horribly inefficient algorithm. For a finitely presented group $G$, and a finite group $H$ given by a multiplication table, a map $G\to H$ can be specified by finite data: Pick an image for each generator, such that the relations are satisfied. Conversely, a map $H\to G$ can be described by choosing for each element of $H$ a word in the generators of $G$, and for each pair of elements $h\_1,h\_2\in H$, a finite sequence of rewritings using the relations of $G$, identifying the word associated to $h\_1h\_2$ with the product of words associated to $h\_1,h\_2$. For two such maps, a proof witnessing that the composite $G\to H\to G$ is the identity can similarly be specified in a finite amount of data: For each generator, a finite sequence of rewrites identifying it with its image. Similarly for the other composite. So a proof identifying a finitely presented group with a concrete finite group $H$ can be given as a finite bundle of data, now just enumerate all such bundles by size until you found one.
| 3 | https://mathoverflow.net/users/39747 | 435424 | 176,012 |
https://mathoverflow.net/questions/435383 | 3 | Let $\sigma(n)$ be the sum of the divisors of $n$. Is it always true that if $n$ is odd, that $$n\mid\sum\_{k=1}^{\frac{n-1}{2}}k^2\sigma(k)\sigma(n-k)?$$
I have checked this up to $n=100$, and I suspect there's some simple argument for why this should be true, probably using Eisenstein series identities but I'm not seeing it.
The motivation from this is from the old Mathoverflow question [Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers](https://mathoverflow.net/questions/372476/van-der-pols-identity-for-the-sum-of-divisors-and-a-quartic-polynomial-equation), where this term appears as the constant in the polynomials generated which for an integer $n$, $n$ is an odd perfect number if and only if $n$ is a root of the $n$th polynomial there. It seems unlikely that the method used there will work as any sort of viable attack on that problem, since they have a very tough set of quartic polynomials they would need to prove are irreducible, and one has very little control over the behavior of the coefficients. But if the above relationship is true, then this at least can be replaced with a similar looking cubic polynomial which still seems unlikely to be helpful, but at least marginally more plausible as an angle of productive attack since then if there is an integer root, the other two roots have to be quadratics. And the coefficient of the $x^2$ will always be $-1$, so the sum of the roots is pretty restricted.
| https://mathoverflow.net/users/127690 | Divisibility relation with a specific sum of divisors | From the paper of [Touchard](https://oeis.org/A000385/a000385.pdf) that is linked in the question on we get the relation $$3nS\_0(n)-\frac{n(n-1)\sigma(n)}{6}=\frac{10}{n}S\_2(n)
....(1)$$
here $S\_i(n)=\sum\_{k=1}^{n-1}k^i\sigma(k)\sigma(n-k)$.
Now we can prove $n\nmid S\_2(n)$ in general when $n$ is even. For $n$ odd, first take the case when $5\nmid n$.
We need to prove $6|n(n-1)\sigma(n)$. If $3|n$ we are done ($2$ divides $n-1$). Otherwise: we have $n=\prod p\_i^{a\_i}, 3\nmid p$.
Now $n \not\equiv 1 (\mod 3)$ when odd number primes like $p\_k \equiv -1 (\mod 3)$ have powers $a\_k$ that are odd. But $\sigma(n)=\prod\_{i} \frac{p\_i^{a\_i+1}-1}{p\_i-1}$. So, for those primes $p\_k \equiv -1 (\mod 3)$, $p\_k^{a\_k+1}\equiv 1 (\mod 3)$ as $a\_k$ is odd and $3 \nmid p\_k-1$.
So, we get that $6|n(n-1)\sigma(n)$.
Now, for the case $5\mid n$,we divide both sides by $5$. We are left to prove that $30\mid n(n-1)\sigma\_(n)$. As $5\mid n$, we need to prove $6|m(5^am-1)\sigma(5^am)$. Here, $n=5^am, 5\nmid m$. Again, as $2|(n-1)$ we need to prove $3|m(5^am-1)\sigma(5^am-1)$.
We focus on the case when $3 \nmid m$, because the other case is trivial. Again $5^am \not\equiv 1 (\mod 3) $ when odd number of primes $p\_k$ s.t $p\_k \equiv -1 (\mod 3)$ (including $5$) of $m$ have odd powers $a\_k$, so there's at least such prime. We do the same thing as previous case and find that $3|\sigma(5^am)$ which proves that $n|S\_2(n)$.
| 2 | https://mathoverflow.net/users/156029 | 435433 | 176,016 |
https://mathoverflow.net/questions/435411 | 3 | Nine different types of singularities are possible on a cubic surface, according to [Wikipedia.](https://en.wikipedia.org/wiki/Cubic_surface) How exactly is the "type" of singularity defined? I know that the number corresponding to the singularity is the number of degrees of freedom removed, but how can we say that the two different surfaces with a $D\_4$ singularity have the same type of singularity while the surface with an $A\_4$ singularity has another type? What properties can tell the difference between the two singularities?
| https://mathoverflow.net/users/493889 | How do we define the type of a singularity on a cubic surface? | All the singularities involved in this classification are Rational Double Points. These singularities are taut, in other words, their analytic type is uniquely determined by the configuration of curves in their minimal resolution.
Such a resolution is a finite set of (-2)-curves whose dual diagram is a Dynkin diagram of the same type of the singularity.
In particular, since the Dynkin diagram of type $A\_4$ is non-isomorphic to the Dynkin diagram of tipe $D\_4$, this allows you to distinguish the two singularities.
| 3 | https://mathoverflow.net/users/7460 | 435439 | 176,019 |
https://mathoverflow.net/questions/435438 | 0 | Can someone explain the Palm distribution? Or provide some information about Palm distribution. The
article called 《A tutorial on Palm distributions for spatial point processes》 is hard to understand.
| https://mathoverflow.net/users/495598 | About Palm distribution | C. Palm's theory of spatial point processes relies heavily on measure theory in an abstract setting. A more gentle introduction is given in the lecture notes [Conditioning in spatial point processes](https://data.math.au.dk/publications/csgb/2015/math-csgb-2015-14.pdf). Section 3, in particular, defines the Palm distribution on a finite set of points. This definition involves only elementary concepts.
---
I now notice that these lecture notes are the same source as mentioned in the OP (without a link, so I did not notice earlier). I do think this is the most accessible exposition of the Palm distribution, by considering the finite case first.
| 1 | https://mathoverflow.net/users/11260 | 435442 | 176,021 |
https://mathoverflow.net/questions/433698 | 4 | A Riemannian manifold $(M, g)$ is said to be an almost Ricci soliton if there exists a complete vector field $X \in \Gamma(TM)$ and a smooth function $\lambda: M \to \mathbb{R}$ such that
$$\operatorname{Ric} + \frac{1}{2}\mathscr{L}\_{X} g = \lambda g$$ When this vector field is the gradient of a smooth function $f: M \to \mathbb{R}$, we say $M$ is a gradient almost Ricci soliton, and this equation becomes:
$$\operatorname{Ric} + \operatorname{Hess}(f) = \lambda g$$
Obviously, any Einstein manifold is a Ricci soliton and hence an almost Ricci soliton (gradient as well, trivially), so these are trivial examples.
If $M$ satisfies: $$\operatorname{div}({\operatorname{Rm}}) = 0$$
we then say $M$ has harmonic curvature (notice this happens if and only if $M$ has harmonic Weyl curvature and constant scalar curvature. I think that part of some work I've been doing with some other people shows that any gradient almost Ricci soliton with harmonic curvature satisfies the property that for any $p \in M$, there is a neighborhood $U\_p \ni p$ such that $U\_p$ has constant sectional curvature (and is therefore necessarily Einstein) (**EDIT AT NOVEMBER 27:** this supposes the dimension is $\geq 4$. Also, I've come to realize since the initial writing of this post that the Einstein examples might not be exhaustive).
As a sanity check, I'm looking for some explicit examples of nontrivial (i.e, not Einstein and with nonconstant $\lambda$) gradient almost Ricci solitons (preferably of dimension $\geq 5$) with harmonic curvature. Can anyone here provide some examples? I'd appreciate any help. Thanks in advance!
| https://mathoverflow.net/users/119418 | What are some explicit examples of nontrivial gradient almost Ricci solitons with harmonic curvature? | **I'm revising my answer to shorten it, since there is a much simpler way to describe these solutions more fully.**
Let $(N^n,h)$ be a metric of constant sectional curvature $k$ and consider the quadratic form
$$
g = \frac{\mathrm{d}u^2}{k-a\,u^2+ b\,u^{1-n}} + u^2\,h
$$
on $M^{n+1} = \mathbb{R}^+\times N$, where $a$ and $b$ are constants and $u>0$ is the coordinate on $\mathbb{R}^+$. If $I\subset\mathbb{R}^+$ is an interval on the $u$-line on which $k-a\,u^2+ b\,u^{1-n} >0$, then $g$ is a Riemannian metric on $I\times N$ that is conformally flat and has constant scalar curvature $S = n(n{+}1)a$. Hence it has harmonic curvature. The Ricci curvature is
$$
\mathrm{Ric}(g) = \bigl(n\,a - \tfrac{1}{2}\,b\,u^{-n-1}\bigr)\,g
+ \frac{(n^2{-}1)b\,\mathrm{d}u^2}{2\bigl(b\,u^2+k\,u^{n+1}-a\,u^{n+3}\bigr)},
$$
so $g$ is Einstein if and only if $b=0$.
Moreover, it is now easy to construct (by quadrature) a function $f = f(u)$ on $I$ such that $\mathrm{Ric}(g) + \mathrm{Hess}\_g(f) = \lambda\,g$ for some function $\lambda$. When $b\not=0$, $\lambda$ will not be constant. Thus, this gives a completely explicit $3$-parameter family of non-trivial almost Ricci solitons with harmonic curvature.
If $I = (r\_1,r\_2)$ where $r\_2>r\_1>0$ are simple roots of $k-a\,u^2+ b\,u^{1-n}=0$, then the curve $v^2 = k-a\,u^2+ b\,u^{1-n}$ in the $uv$-plane has a smooth circle component $C$ between the lines $u=r\_1$ and $u=r\_2$. In this case, the metric $g$ extends to a smooth complete metric on $C\times N$ (assuming that $(N,h)$ is complete). In this way, one can construct many complete or compact examples of such metrics. However, when $b\not=0$, the functions $f$ and $\lambda$ will only be locally defined unless one passes to the simply-connected cover of $C$, so that $M = \mathbb{R}\times N$. On this covering space, $f$ (and $\lambda$) can be globally defined.
*Remark:* In dimension $3$, it turns out that *every* conformally flat metric $(M^3,g)$ with constant scalar curvature (i.e., every metric in dimension $3$ with harmonic curvature) that admits a 'Ricci potential', i.e., a function $f$ such that $\mathrm{Ric}(g)+\mathrm{Hess}\_g(f) = \lambda\,g$ for some function $\lambda$, is locally of the above form for some $(N^2,h)$ with constant curvature. It was after I worked out this fact, via an exterior differential system analysis, that I realized that the above construction could be used to produce examples in any dimension.
In dimensions above $3$, it is not likely that every metric with harmonic curvature that admits a Ricci potential in the above sense is of the form given above, but I don't know a classification, even when $n=4$.
| 4 | https://mathoverflow.net/users/13972 | 435448 | 176,023 |
https://mathoverflow.net/questions/435403 | 4 | Let $M$ be a Riemannian $n$-manifold and $x \in M$. For the metric tensor $g\_{ij}$ of geodesic normal coordinates at $x$, there is a formula $g\_{ij}(y) = \delta\_{ij} + \frac13 R\_{kijl} y^k y^l + O(\|y\|^3)$. Assume that the injectivity radius of $M$ is positive.
**Question.** Are there constants $c\_1, c\_2>0$ (that only depends on $M$) such that the following holds, regardless the choice of $x$ at which the geodesic normal coordinates are taken?
$$ \|y\| < c\_1 \implies | g\_{ij}(y) - \delta\_{ij} | < c\_2 \cdot \|y\| $$
Due to higher order terms, $C$ would have to be larger than $\frac13 R\_{kijl}$, but what exactly should it be? I'd like to express it in terms of simple intrinsic quantities of $M$.
The full Taylor expansion of $g\_{ij}$ needs computer algebra if written in terms of derivatives of the Riemann tensor, see Equation 11.1 in [here](https://arxiv.org/pdf/0903.2087.pdf).
Related: [Riemann's formula for the metric in a normal neighborhood](https://mathoverflow.net/questions/185527/riemanns-formula-for-the-metric-in-a-normal-neighborhood?rq=1)
| https://mathoverflow.net/users/156792 | Bounds for metric in normal coordinate | As mentioned by Deane Yang in the comments and his (deleted) answer, one can estimate the components of the metric in normal coordinates using a transport ODE (I know it from [Dolgov-Khriplovich (1983)](https://doi.org/10.1007%2fBF00760057) Eq.(34), and also in more implicit form from [Florides-Synge (1971)](https://doi.org/10.1098%2frspa.1971.0085), but perhaps there is a more canonical reference). The way I see the question, it does not ask to estimate the derivatives of the metric, only its pointwise deviation from the Euclidean metric $\delta\_{ij}$. So I see this strategy as completely viable.
The radial ODE satisfied by the metric $g\_{ij}(y)$ in normal coordinates is
$$\begin{aligned}
r \partial\_r (r\partial\_r +1) (g(ry) - \delta)\_{ij}
&= \frac{1}{2} g^{kl}(ry) [r\partial\_r(g(ry) - \delta)\_{ik}] [r\partial\_r (g(ry)-\delta)\_{jl}]
- 2r^2 y^k y^l R\_{ikjl}(ry) \\
&= \frac{1}{2} g^{kl}(ry) [r\partial\_r g\_{ik}(ry)] [r\partial\_r g\_{jl}(ry)]
- 2r^2 y^k y^l R\_{ikjl}(ry)
=: r^2 F\_{ij}(r, g(ry)-\delta, \partial\_r g(ry)) ,
\end{aligned}$$
where $y=(y^i)$ is any fixed coordinate vector. The linear differential operator has the following forward Green function/fundamental solution:
$$
G(s,r) = \Theta(s-r) \frac{1}{r} \left(1-\frac{r}{s}\right) .
$$
To solve the equation with initial condition $g\_{ij}(0) = \delta\_{ij}$, we can adapt the [usual argument](https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem) using Picard iteration and the Banach fixed point theorem to the integral equation
$$\begin{aligned}
(g(sy)-\delta)
&= \int\_0^s dr\, r\left(1-\frac{r}{s}\right) \, F(r, g(ry)-\delta, \partial\_r g(ry))
=: \Gamma[g(ry)-\delta] , \\
\partial\_s g(sy)
&= \int\_0^s dr \, \frac{r^2}{s^2} F(r,g(ry)-\delta,\partial\_r g(ry)) ,
\end{aligned}$$
where the second equation is a consequence of the first, which we'll need shortly. Some uniform estimates on $|F|$, coupled with the integrals $\int\_0^s dr\, r(1-r/s) = s^2/6$ and $\int\_0^s dr\, r^2/s^2 = s/3$, will allow us to find lower bounds on the interval of existence of the solution to the transport ODE.
The original question didn't specify what norm $|g(y) - \delta|$ refers to. I'll just suppose that it is the Frobenius norm with respect to $\delta\_{ij}$, $|h| = (h\_{ik} h\_{jl} \delta^{ij}\delta^{kl})^{1/2}$, which is quite natural for dealing with tensors in normal coordinates. To get anywhere with estimating $|F|$, it helps to assume some a priori uniform bounds on $|g(ry) - \delta| \le D < 1$, $|\partial\_r g(ry)| \le C$. The constants $C$, $D$ are for now arbitrary, but they can be tuned to specific values later. The first useful uniform estimate is
$$
|g^{kl}(ry) [r\partial\_r g\_{ik}(ry)] [r\partial\_r g\_{jl}(ry)]| \le \frac{C^2}{1-D} .
$$
Next, the curvature term satisfies
$$
|y^k y^l R\_{.k.l}(ry)| \le |R(ry)| \|y\|^2 ,
$$
where $\|y\|^2 = |yy| = y^k y^l \delta\_{kl}$ and
$$
|R(ry)| = (R\_{ijkl}(ry) R\_{i'k'j'l'}(ry) \delta^{ii'} \delta^{kk'} \delta^{jj'} \delta^{ll'})^{1/2}
$$ is the Frobenius norm of the endomorphism $h^{kl} \mapsto h^{kl} R\_{ikjl}(ry)$. It would be nice to express the bound on the Riemann tensor in terms of some geometric quantity, but $|R(ry)|$ only has meaning inside the normal coordinate chart. However, replacing each $\delta^{ij}$ by $g^{ij}(ry)$ we get an invariant curvature scalar and we can essentially bound one with the other. Writing $\delta^{ij} = g^{ij}(ry) - (g^{ij}(ry) - \delta^{ij})$, noting that $|g^{..}(ry)-\delta^{..}| \le \frac{D}{1-D}$, and repeatedly using the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
|R(ry)| &\le \frac{\|R\|(ry) \|y\|^2}{(1-D)^2} \le \frac{\mathcal{R} \|y\|^2}{(1-D)^2} , \\
\|R\| &= \max\{R^{(1)}, R^{(2)}, R^{(3)}, R^{(4)}\} , \\
R^{(1)} &= (R\_{ijkl} R^{ijkl})^{1/2} , \\
R^{(2)} &= (R\_{ijkl} R^{i'jkl} R\_{i'j'k'l'} R^{ij'k'l'})^{1/4} , \\
R^{(3)} &= (R\_{ijkl} R^{kli'j'} R\_{i'j'k'l'} R^{k'l'ij})^{1/4} , \\
R^{(4)} &= (R\_{ikjl} R^{ki'lj'} R\_{i'k'j'l'} R^{k'il'j})^{1/4} , \\
\mathcal{R} &= \sup\_{x\in M} \|R\|(x) .
\end{aligned}$$
The global geometric invariant $\mathcal{R}$ of $(M,g)$ is what we will eventually use to determine the constants sought in the original question.
The main upper bound on the size of the interval $s\in [0,t]$ is that the iteration must map $(g(ry)-\delta) \mapsto \Gamma(g(ry)-\delta)$ to the space satisfying the same a priori uniform bounds. Using the above estimates on $|F|$ these inequalities are expressed as
$$\begin{aligned}
\frac{t^2}{6} \left(\frac{1}{2} \frac{C^2}{1-D} + 2\frac{\mathcal{R \|y\|^2}}{(1-D)^2}\right) &\le D , \\
\frac{t}{3} \left(\frac{1}{2} \frac{C^2}{1-D} + 2\frac{\mathcal{R} \|y\|^2}{(1-D)^2}\right) &\le C ,
\end{aligned}$$
which has a consistent solution $D=1/4$, $C=(2\mathcal{R}\|y\|^2/3)^{1/2}$, $t=(3\mathcal{R}\|y\|^2/8)^{-1/2}$ (no claim about the optimality of $t$). So, the integral equation $(g(ry)-\delta) = \Gamma[g(ry)-\delta]$ and the above parameters for the a priori estimates gives use the uniform bound
$$
|g(ry)-\delta| \le \frac{r^2}{6} \left(\frac{1}{2} \frac{C^2}{1-D} + 2\frac{\mathcal{R} \|y\|^2}{(1-D)^2}\right) \le \frac{2}{3} \mathcal{R} \|ry\|^2
$$
for $\|ry\| \le (3\mathcal{R}/8)^{-1/2}$. [This section](https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem#Optimization_of_the_solution%27s_interval) on Wikipedia explains how to go from here prove the contraction property of a sufficiently high power of $\Gamma$ to prove existence.
Since now we can just replace the normal coordinate vector $ry^i$ by $y^i$ we get what was hopefully the desired estimate valid for any normal coordinate chart on $M$:
$$
\|y\| \le \min\left\{\sqrt{\frac{8}{3\mathcal{R}}}, \rho\_I\right\}
\implies
|g(y) - \delta| \le \frac{2}{3} \mathcal{R} \|y\|^2 ,
$$
where $\rho\_I$ is the injectivity radius of $(M,g)$ (positive by assumption), which has to be there because some normal coordinate charts cannot exceed the radius $\rho\_I$ anyway.
As expected, we get a larger coefficient $2/3 > 1/3$ than in the Taylor formula. It might be possible to optimize this coefficient and the bound on $\|y\|$ a bit more, if needed. Also changing the Frobenius norm to a different one could also affect the constant and the definition of $\mathcal{R}$ in terms of global geometric invariants of $(M,g)$.
| 7 | https://mathoverflow.net/users/2622 | 435449 | 176,024 |
https://mathoverflow.net/questions/434592 | 4 | Let $X,Y$ be metric spaces. Let $f,g : X \to Y$ be two maps and $x\_0 \in X$. Let us say that $f$ and $g$ are *tangent at* $x\_0$ if the following condition is satisfied: For every $\epsilon > 0$ there is some $\delta > 0$ such that for all $x \in X$ we have
$$d(x,x\_0) \leq \delta \implies d(f(x),g(x)) \leq \epsilon \cdot d(x,x\_0).$$
In this case, let's write $f \sim\_{x\_0} g$. An equivalent characterization is
$$
f(x\_0)=g(x\_0) \quad \text{and} \quad \lim\_{x \to x\_0,\, x \neq x\_0} \frac{d(f(x),g(x))}{d(x,x\_0)} = 0.$$
This relation appears (in the special case of normed vector spaces) in Dieudonné's "Foundations of Modern Analysis" to give a conceptual definition of the Fréchet derivative.
**Question.** What are good references for the properties of this tangent relation? Is it, perhaps, studied under a different name (touching, closeness, etc.)? In particular, I would like to know if the following results about the compatibility with composition are already in the literature?
(1) Let $X,Y,Z$ be metric spaces, $x\_0 \in X$, $y\_0 \in Y$. Let $g\_1,g\_2 : Y \to Z$ be functions with $g\_1 \sim\_{y\_0} g\_2$, and let $f : X \to Y$ be a function which is Lipschitz continuous at $x\_0$ (def. [here](https://math.stackexchange.com/questions/3916802/)) with $f(x\_0)=y\_0$. Then also $g\_1 \circ f \sim\_{x\_0} g\_2 \circ f$.
(2) Let $X,Y,Z$ be metric spaces, $x\_0 \in X$. Let $f\_1, f\_2 : X \to Y$ be functions with $f\_1 \sim\_{x\_0} g\_2$. If $g : Y \to Z$ is Lipschitz continuous, then also $g \circ f\_1 \sim\_{x\_0} g \circ f\_2$.
They are not hard to prove, but what makes them interesting is that they offer a very conceptual (new?) proof of the chain rule of the Fréchet derivative.
| https://mathoverflow.net/users/2841 | Reference request: "Tangent relation" in metric spaces | One reference:
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> Elisabeth Burroni and Jacques Penon, [A metric tangential calculus](http://tac.mta.ca/tac/volumes/23/10/23-10abs.html). *Theory and Applications of Categories* 23 (2010), 199–220.
>
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>
The first sentence of the paper says that a fuller account of their work can be found in a longer (99-page) arXiv paper:
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> Elisabeth Burroni and Jacques Penon, [Elements for a metric tangential calculus](https://arxiv.org/abs/0912.1012). ArXiv:0912.1012, 2009.
>
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>
Remark 1.6(2) of the first of these papers looks like it might cover the two specific properties you mention, but I haven't checked carefully.
| 2 | https://mathoverflow.net/users/586 | 435456 | 176,026 |
https://mathoverflow.net/questions/435428 | 7 | I am trying to understand the intuition for Luna's Étale Slice Theorem in the affine setting over $\mathbb{C}$.
Here is the setup. Let $X$ be an affine algebraic variety and $G$ a reductive group acting on $X$. Moreover, let $x\in X$ be a closed point and $G\_{x}$ its stabilizer under the $G$ action where $G\_{x}$ is also a reductive group, and $V\subseteq X$ a $G\_{x}$ invariant locally closed affine variety containing $x$.
The group action $G\times V\to X$ by $(g,v)\mapsto g\cdot v$ is $G\_{x}$ invariant if the action of $G\_{x}$ on $G\times V$ is by $h\cdot (g,v)\mapsto (gh^{-1},hv)$. Indeed, one can check that $gh^{-1}\cdot h\cdot v=gh^{-1}h\cdot v=g\cdot v\in X$ but if $g\in G\_{x}$ then $g\cdot v\in V$ since $V$ is $G\_{x}$ invariant.
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> What is the geometric intuition for the quotient $G\times\_{G\_{x}}V= (G\times V)// G\_{x}$?
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Formally, it seems we are identifying points $(g,v)\sim (g',v')$ if there exists some $h\in G\_{x}$ such that $gh^{-1}\cdot hv$ and $g'\cdot v'$ map to the same point in $X$. Namely $v,v'\in V$ lie in the same $G$ orbit. So $G\times\_{G\_{x}} V$ is just the set of $G$-orbits in $V$?
This induces a morphism $\psi: G\times\_{G\_{x}} V\to X$. Say $V$ is an étale slice if $\psi$ is an étale morphism. Luna's theorem shows the existence of such étale slices.
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> What is the structure of $G\times\_{G\_{x}} V$ as a variety?
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> What is the correct way to think about the morphism $\psi$?
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| https://mathoverflow.net/users/495308 | Intuition for Luna's Étale Slice Theorem | Luna Étale's Slice Theorem is probably the most powerful result we have to understand the local structure of a moduli space.
Moduli spaces (say of semi-stable vector bundles on a smooth projective curve) may be constructed by taking the geometric quotient of a Quot-scheme by a reductive group $G$. The stabilizer of a point $x$ is usually included in $\mathrm{Hom}(E\_x, E\_x)$, where $E\_x$ the vector bundle corresponding to $x$. In case $E\_x$ is semi-stable, there are many techniques available to compute $\mathrm{Hom}(E\_x, E\_x)$. If you are in a favorable situation, the Quot-scheme is smooth at $x$ and Luna Étale's Slice tells you that the singularity of the moduli space at $x$ is locally isomorphic to $\mathbb{C}^n/G\_x$.
I would advise reading Seshadri's notes *Fibrés vectoriels sur les courbes algébriques* (written by Drézet) where all of this is very clearly explained. And with many examples.
| 3 | https://mathoverflow.net/users/37214 | 435458 | 176,027 |
https://mathoverflow.net/questions/435457 | 6 | This is a cross post from MSE of
<https://math.stackexchange.com/questions/4562196/normalizer-of-maximal-torus-is-maximal>
Let $ T $ be a maximal torus in a compact connected simple Lie group $ K $. For which groups $ K $ is the normalizer $ N(T) $ maximal among the proper closed subgroups of $ K $?
I know this is true for the infinite families of compact connected simple groups $ SU\_n, n \geq 2 $ and $ SO\_{2n}, n\geq 3 $, see
<https://arxiv.org/abs/math/0605784>
table 5 fourth row $ p=1 $ case for $ SU\_n, n \geq 2 $ and table 7 first row $ p=2 $ case for $ SO\_{2n}, n \geq 3 $.
| https://mathoverflow.net/users/387190 | When is the normalizer of the maximal torus maximal? | $\def\fg{\mathfrak{g}}\def\ft{\mathfrak{t}}\def\long{\text{long}}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SU{SU}$The point of this answer, as discussed in comments, is to note that this is NOT true in types BCFG. In comments, Mikhail Borovoi said he could prove the result is true in types ADE (now an [answer](https://mathoverflow.net/a/435501)), so collectively this resolves the claim.
Let $G$ be a compact connected simple Lie group with Lie algebra $\fg$. Choose a torus $T$ and with corresponding Lie algebra $\ft$, and let
$$\fg = \ft \oplus \bigoplus\_{\beta} \fg\_{\pm \beta}$$
be the root decomposition, where $\beta$ runs over positive roots and each $\fg\_{\pm \beta}$ is two dimensional. (Since I'm not tensoring by $\mathbb{C}$, I don't get to go all the way down to one dimensional subspaces.)
Now, suppose that $\fg$ has two lengths of roots, "long" and "short". A case by case check of the root systems shows that the sum of two long roots is always either $0$, long or not a root. Therefore,
$$\fg\_{\long} = \ft \oplus \bigoplus\_{\beta\ \long} \fg\_{\pm \beta}$$
is a subalgebra; let $G\_{\long}$ be the corresponding subgroup.
Then $T \subsetneq G\_{\long}$ and, because the only choice in defining $G\_{\long}$ is the choice of $T$, the normalizer $N(T)$ also normalizes $G\_{\long}$. So $G\_{\long} N(T)$ is a subgroup containing $N(T)$ of dimension $\dim G\_{\long} > \dim T$.
Explicitly, we have:
* In type $B\_n = \SO(2n+1)$, the group $G\_{\long}$ is $\SO(2n)$ and $G\_{\long} N(T) = O(2n)$.
* In type $C\_n = \Sp(2n)$, the group $G\_{\long}$ is $\Sp(2)^n$ and I think that $G\_{\long} N(T) = \Sp(2)^n \rtimes S\_n$.
* In type $F\_4$, the group $G\_{\long}$ is $\SO(8)$, or possibly one of the other compact real forms of $D\_4$. I'm not sure what $G\_{\long} N(T)$ is explicitly; it might depend on which compact form of $F\_4$ we are using.
* In $G\_2$, the group $G\_{\long}$ is $\SU(3)$. I think that $G\_{\long} N(T)$ is $SU(3) \rtimes C\_2$, with $C\_2$ acting by an outer automorphism.
These examples, and a little bit of thought about the $F\_4$ case, lead me to conjecture: $G\_{\long} N(T)$ sits in a short exact sequence $1 \to G\_{\long} \to G\_{\long} N(T) \to \operatorname{Out}(G\_{\long}) \to 1$. (I'm not ready to guess whether or not the sequence is always semidirect.)
| 10 | https://mathoverflow.net/users/297 | 435474 | 176,032 |
https://mathoverflow.net/questions/435469 | 1 | Let $(E, d)$ be a metric space and $\mathcal F$ a collection of real-valued functions on $E$. Assume that for all $x,x\_n \in E$ with $n\in \mathbb N$,
$$
x\_n \to x \iff [f(x\_n) \to f(x) \quad \forall f \in \mathcal F]. \quad (\star)
$$
Let $\tau$ be the metric topology on $E$ induced by $d$ and $\tau'$ the initial topology induced by $\mathcal F$. Clearly, $\tau' \subset \tau$. It's [possible](https://mathoverflow.net/a/36390/477203) that $\tau' \neq \tau$.
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> **My question** Are there some conditions on $\mathcal F$ that ensures $\tau' = \tau$?
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Thank you so much for your elaboration!
| https://mathoverflow.net/users/477203 | Conditions that ensure the metric topology of $E$ coincides with the initial topology induced by a collection of real-valued functions on $E$ | If $\mathcal{F}$ is countable, $\tau'$ is metrizable with a compatible metric $\rho$ given by
$$\rho(x,y)=\sum\_n 2^{-n}~|f\_n(x)-f\_n(y)|\wedge1$$
for some enumeration of $\mathcal{F}$.
Since the topology of a metric space is determined by the convergent sequences, this is a sufficient condition.
| 2 | https://mathoverflow.net/users/35357 | 435477 | 176,033 |
https://mathoverflow.net/questions/435459 | 3 | Let $k$ be a field, $A$ a finite dimensional $k$-Algebra, $\text{mod}\,A$ the category of finite dimensional left $A$-modules and $\text{rad}\_A$ the collection of radical morphisms in $\text{mod}\,A$. For a natural number $n$ we denote by $\text{rad}\_A^n$ the collection of $n$ compositions of morphisms in $\text{rad}\_A$ and set $\text{rad}\_A^{\omega} := \bigcap\_{n\in \mathbb{N}} \text{rad}\_A^n$. It is well-known that if $A$ is of infinite representation type, then $\text{rad}\_A^{\omega}\neq 0$. However, in every example it seems to be that we always have $(\text{rad}\_A^{\omega})^2 \neq 0$ aswell.
**Question:** Is it always true that if $A$ is of infinite representation type, then $(\text{rad}\_A^{\omega})^2 \neq 0$?
| https://mathoverflow.net/users/145920 | For a finite dimensional $k$-Algebra $A$ does infinite representation type imply $(\text{rad}_A^{\omega})^2 \neq 0$? | This is proven in the paper (as the title suggests):
*Coelho, Flávio U.; Marcos, Eduardo N.; Merklen, Héctor A.; Skowroński, Andrzej*, [**Module categories with infinite radical square zero are of finite type**](https://doi.org/10.1080/00927879408825084), Commun. Algebra 22, No. 11, 4511-4517 (1994). [ZBL0812.16019](https://zbmath.org/?q=an:0812.16019).
| 5 | https://mathoverflow.net/users/18756 | 435484 | 176,035 |
https://mathoverflow.net/questions/433927 | 3 | According to [Wikipedia](https://en.wikipedia.org/wiki/Doubling_space),
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> However, many results from classical harmonic analysis and computational geometry extend to the setting of metric spaces with doubling measures.
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My question is: what are some examples of these results, and where can a more thorough explanation of why this is the case be found?
| https://mathoverflow.net/users/408316 | Results in computational geometry utilizing doubling dimension of a metric space | The question is a bit unclear, so I'm not sure this is what you seek.
But here is a connection to computational geoemtry.
There is considerable literature on
[geometric spanners](https://en.wikipedia.org/wiki/Geometric_spanner) and [doubling metric spaces](https://en.wikipedia.org/wiki/Doubling_space).
Here is a sampling of references from 2006, 2008, 2019, 2022. And from their references you can locate many more.
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> Damian, Mirela, Saurav Pandit, and Sriram Pemmaraju. "Distributed spanner construction in doubling metric spaces." In *International Conference on Principles of Distributed Systems*, pp. 157-171. Springer, Berlin, Heidelberg, 2006.
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> Gottlieb, Lee-Ad, and Liam Roditty. "An optimal dynamic spanner for doubling metric spaces." In *European Symposium on Algorithms*, pp. 478-489. Springer, Berlin, Heidelberg, 2008.
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> Borradaile, Glencora, Hung Le, and Christian Wulff-Nilsen. "Greedy spanners are optimal in doubling metrics." In *Proceedings of the Thirtieth Annual ACM-SIAM Symposium on Discrete Algorithms*, pp. 2371-2379. Society for Industrial and Applied Mathematics, 2019.
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> Bartal, Yair, Ora Nova Fandina, and Ofer Neiman. "Covering metric spaces by few trees." *Journal of Computer and System Sciences* 130 (2022): 26-42.
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| 2 | https://mathoverflow.net/users/6094 | 435485 | 176,036 |
https://mathoverflow.net/questions/435394 | 3 | In the article titled Tilting Exercises (See <http://arXiv.org/abs/math/0301098v3>) the authors define a notion of tilting perverse sheaves on an algebraic vareity $X$ with respect to stratification $\lbrace X\_{\nu} \rbrace$. It appears that §1.3 onwards they assume that all the morphisms $i\_{\nu} : X\_{\nu} \hookrightarrow X$ are affine. They further assume that all the strata $X\_{\nu}$ have no higher nontrivial cohomology. They obtain an equivalence of bounded homotopy category of tilting sheaves with the full subcategory $D$ (of complexes of sheaves constant along the strata of $X$) of $D(X)$ (=the bounded derived category of constructible sheaves). This is §1.5, Proposition of loc. cit.
It is certainly clear that there are spaces of interest where the above assumptions on the strata are not satisfied.
Q1. Has the theory of perverse tilting sheaves been explored out of the context of the paper cited above? For example consider the algebraic variety given by compactification of the moduli of principally polarized abelian varieties or loosely speaking the minimal or some toroidal compactification of locally symmetric space associated to $\mathrm{Sp}(2g)$ and some congruence subgroup. Is it possible to understand how big or small the subcategory of tilting perverse sheaves will be?
Q2. If the notion of tilting perverse sheaves has not been explored in more general context then I would like to ask if there are certain philosophical reasons that it may not be so meaningful to pursue the theory of tilting sheaves in more generality?
Any comments or suggestions are welcome. Thank you.
| https://mathoverflow.net/users/58056 | Perverse tilting sheaves | At @random123's request, I'm will try to argue that looking for tilting perverse sheaves takes one very close to a setting in which [BBM] work.
Suppose that we have a suitably stratified variety
\begin{equation}
X = \bigsqcup\_{\lambda \in \Lambda} X\_\lambda
\end{equation}
and we wish to understand $D = D^b\_\Lambda(X,k)$, the derived category of $\Lambda$-constructible sheaves of $k$-vector spaces.
Often we try to understand derived categories by finding a tilting generator $T$, i.e. an object $T$ which generates $D$ and has no higher extensions. In this case $D$ is equivalent to a suitable derived category of modules over $End(T)$ ("tilting theory").
In general, describing all tilting complexes in $T$ is hopeless. (In examples I am familiar with, there are more than I can possibly comprehend, and classifying them is hopeless -- it is somewhat analogous to classifying all $t$-structures on $D$.)
However, the theory of highest weight categories (aka quasi-hereditary algebras) leads one to look for tilting complexes $T$ which are perverse and may be written as a direct sum $T = \bigoplus T\_\lambda$ in a way that is "compatible with the stratification", more precisely:
1. $T\_\lambda$ is supported on $\overline{X}\_\lambda$.
2. The set $\{ T\_\mu \; | \; \mu \le \lambda \}$ generates the full subcategory of $D$ generated consisting of complexes supported on $\overline{X}\_\lambda$.
Firstly, note that "one $T\_\lambda$ per strata" implies that there are no local systems on strata, in other words that $\pi\_1(X\_\lambda) = \{ 1 \}$. (Or at least that fundamental groups have no finite-dimensional representations.)
Next, it is a nice exercise to show that these assumptions imply that (if one denotes by $j : X\_\lambda \hookrightarrow \overline{X}\_\lambda$ the inclusion, one has a surjection $Hom^i(T\_\lambda, T\_\lambda) \to Hom^i(j^\*T\_\lambda, j^\*T\_\lambda)$. The latter group is equal to $H^i(X\_\lambda, k)$, because I hope I convinced you in the previous paragraph that $j^\*T\_\lambda$ is a shift of a constant sheaf on $X\_\lambda$.
In other words, we are forced to take a stratification with strata which look a lot like they are contractible. Thus we are very close to the setting in which [BBM] work.
(Please see comments following question for more specialized comments.)
*Aside:* I understand that the title "tilting exercises" refers to the medieval practice of riding towards each other carrying long jousting poles. When reading this beautiful paper, I was mystified as to why they also cite Oliver Twist. Only recently did I understand, but leave this as a tilting exercise for the interested reader.
| 4 | https://mathoverflow.net/users/919 | 435492 | 176,040 |
https://mathoverflow.net/questions/435483 | 6 | Let $K$ be a $p$-adic local field with uniformizer $\pi \in \mathcal{O}\_{K}$ and residue field $k = \mathcal{O}\_{K}/\pi$. Let $G$ be a Lubin-Tate formal $\mathcal{O}\_{K}$-module and $G\_{0}$ its reduction to $k$.
Let $G\_{0}[\pi]$ denote the $\pi$-torsion subgroup of $G\_{0}$, this is a finite group scheme over $k$. The action of $\mathcal{O}\_{K}$ on $G\_{0}$ yields an action of the multiplicative monoid $\Gamma = (k, \cdot)$ on $G\_{0}[\pi]$ and hence on its coordinate ring $\mathcal{O}\_{G\_{0}[\pi]}$.
**Question:** Does there always exist an element $g \in \mathcal{O}\_{G\_{0}[\pi]}$ which generates the coordinate ring as a module over the monoid ring $k[\Gamma]$? How to find such an element?
**Example:** Suppose that $K = \mathbb{Q}\_{p}$, $\pi = p$ and $G = \mathbb{G}^{\wedge}\_{m}$ the formal multiplicative group, so that
$\mathcal{O}\_{G[p]} \simeq \mathbb{F}\_{p}[x]/((1+x)^{p}-1)$.
Then an element $[\gamma] \in \Gamma = (\mathbb{Z}/p, \cdot)$ acts on $g = x+1$ via
$[\gamma]g = [\gamma](x)+[\gamma](1) = (x+1)^{\gamma}-1+1 = (x+1)^{\gamma}$
and hence the translates of $g$ along $0, \cdots, p-1 \in \mathbb{F}\_{p}$ are of the form
$(x+1)^{0}, (x+1)^{1}, \cdots, (x+1)^{p-1}$
This is a $\mathbb{F}\_{p}$-basis of $\mathcal{O}\_{G\_{0}[\pi]}$, so $g = x+1$ gives an element which generates the whole coordinate ring as needed.
**Some thoughts:** 1) In the case of $\mathbb{G}\_{m}^{\wedge}$, where $\mathbb{G}\_{m}[p] = \mu\_{p}$, another way to find an element $g$ is to notice that Cartier duality gives an isomorphism
$\mathcal{O}\_{\mu\_{p}} \simeq \mathcal{O}\_{\mathbb{Z}/p}^{\*} \simeq Fun(\mathbb{Z}/p, \mathbb{F}\_{p})^{\*} \simeq \mathbb{F}\_{p}[\mathbb{Z}/p]$,
(where $\*$ denotes the dual vector space) so that the coordinate ring is just the regular representation of the multiplicative monoid $\Gamma = \mathbb{Z}/p$ and hence must be generated by a single element.
2. Also in the case of $\mathbb{G}\_{m}^{\wedge}$, the invariant differential is given by $\omega(x) dx = \frac{1}{1+x} dx$. So perhaps it is possible to relate a generating element $g$ to invariant differentials?
3. We have verified that such an element also exists when $G\_{0}$ is the Honda formal group law of height 2 over $\mathbb{F}\_{4} \simeq W(\mathbb{F}\_{4})/p$, and can be taken to be
$g = 1 + x + x^{2} + x^{3} \in \mathcal{O}\_{G\_{0}[p]} \simeq \mathbb{F}\_{4}[x]/(x^{4})$
| https://mathoverflow.net/users/16981 | Generating the coordinate ring of the Lubin-Tate formal group | I'll assume that $\pi=p$; I guess that the general case is the same but I have not checked. We now have $k=\mathbb{F}\_{p^n}$ and $\mathcal{O}\_K=W\mathbb{F}\_{p^n}$. The group of roots of unity in $W\mathbb{F}\_{p^n}$ is cyclic of order $p^n-1$, with generator $\omega$ say, and maps by an isomorphism to $\mathbb{F}\_{p^n}^\times$. Put $f(x)=px+x^{p^n}$. Lubin-Tate theory then says that there is a unique formal group law $F$ with $f(F(x,y))=F(f(x),f(y))$ and that this has $[p](x)=f(x)$ and $[\omega^i](x)=\omega^ix$. The ring $R=\mathcal{O}\_{G\_0[\pi]}$ is just $W\mathbb{F}\_{p^n}[[x]]/f(x)$, which has basis $B=\{x^i:0\leq i<p^n\}$ over $W\mathbb{F}\_{p^n}$. Put $a=1+x+\dotsb+x^{p^n-1}\in R$. I claim that this is a generator of the required type. More explicitly, let $b$ be the image of $a$ under the substitution $x\mapsto [0](x)=0$ i.e. $b=1$, and let $c\_i$ be the image of $a$ under the substitution $x\mapsto[\omega^i](x)=\omega^ix$. The claim is then that the list $b,c\_0,c\_1,\dotsc,c\_{p^n-2}$ is a basis for $R$ over $W\mathbb{F}\_{p^n}$. If we write this list in terms of the monomial basis, we get a matrix $M$ of shape $p^n\times p^n$, which we must prove to be invertible. The top row (corresponding to $b$) is just $(1,0,0,\dotsc,0)$, and the bottom right block of size $(p^n-1)\times(p^n-1)$ is the Vandermonde matrix with entries $\omega^{ij}$. The Vandermonde determinant formula shows that this is invertible.
| 2 | https://mathoverflow.net/users/10366 | 435512 | 176,043 |
https://mathoverflow.net/questions/435506 | 2 | If $X$ is any set, we let $[X]^2:=\big\{\{x,y\}:x\neq y\in X\big\}$. If $G=(V,E)$ is an undirected graph and $v\in V$, we define $N\_G(v) = \{w\in V:\{v,w\}\in E\}$.
For $i =1,2$, let $G\_i=(V\_i,E\_i)$ be non-empty, finite, simple, undirected graphs with $V\_1\cap V\_2 = \emptyset$. Suppose that $E \subseteq [V\_1\cup V\_2]^2$ has $E\cap [V\_i]^2=E\_i$ for $i=1,2$, and in addition has the following property:
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> there is $k\in\mathbb{N}$ such that $|N\_G(v\_1) \cap V\_2| = k$ for all $v\_1\in V\_1,$
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where $G := (V\_1\cup V\_2, E)$.
Is there a "global constant" $c\_k\in\mathbb{N}$ depending on $k$ only such that we have $$\chi(G) \leq \max\{\chi(G\_1), \chi(G\_2)\} + c\_k,$$ no matter what the initial choice of $G\_1, G\_2$ and $E$ (adhering to the property above) was?
| https://mathoverflow.net/users/8628 | "Combined" chromatic number of $2$ graphs glued together with $2$ edges per vertex | This is not true even for $k=1$. Assume that $(V\_1,E\_1)$ is a complete $d$-partite graph with each part having $d$ vertices, and so is $(V\_2,E\_2)$. Take a bijection $f\colon V\_1\to V\_2$ and add edges between $v$ and $f(v)$ for all $d^2$ vertices $v\in V\_1$. $f$ is chosen so that for each part in $V\_1$, its vertices are joined with different parts in $V\_2$ (for example, we may join $i$-th vertex of the $j$-th part in $V\_1$ with $j$-th vertex of $i$-th part in $V\_2$).
Assume that this graph has a proper coloring with $d+c$ colors (where $c=c\_1$ is a constant). Note that each color may be used for the vertices of at most one part in $V\_1$, and analogously for $V\_2$. Therefore at least $d-c$ parts in $G\_1$ are monochromatic, and so are at least $d-c$ parts in $G\_2$. Thus, if $d>3c$, there exists a color (say, white) such that there is a white part in $V\_1$ and white part in $V\_2$. But then there is an edge with white endpoints.
| 5 | https://mathoverflow.net/users/4312 | 435515 | 176,045 |
https://mathoverflow.net/questions/435496 | 1 | Suppose you're a shopkeeper in the business of selling Items. An "Item" is a thing whose only property is that the quantity that can be bought by a purchaser must be a positive integer; all Items are identical.
A book whose identity I do not recall (I'll post it here maybe tomorrow?) seems to hold that the way to model the number $Y$ of Items that your customers will buy today is as a negative binomial distribution:
\begin{align}
& \Pr(Y=n) = \binom{-r}{\phantom{+}n} (-q)^n p^r \text{ for } n=0,1,2,3,\ldots \\[8pt]
\text{where } & 0<p<1,\, q = 1-p,\, r>0, \\[8pt]
\text{and } & \binom {-r}{\phantom{+}n} = \frac{\overbrace{(-r)(-r-1)(-r-2)\cdots (-r-n+1)}^{\text{$n$ factors}}}{n!}.
\end{align}
Supposing the number $N$ of purchasers today has a Poisson distribution (which makes sense for well known reasons under some plausible assumptions) that would mean the number of Items purchased by each purchaser has a logarithmic series distribution:
$$
\Pr(X=x) = \frac 1 {-\log(1-u)}\cdot \frac{u^x} x \text{ for } x = 1,2,3,\ldots
$$
where $0<u<1$ and $\dfrac1{-\log(1-u)} \cdot \dfrac u{1-u}$ is the average number of Items purchased by each purchaser.
(To prove this, use moment-generating functions. I tried to do it by more straightforward combinatorial reasoning and it's a mess.) (**Postscript some days later:** ok, I now see that there's at least one other fairly convenient way to prove this.)
**My question is:** Is there some "intuitive" argument for the logarithmic series distribution to occur here that is independent of any knowledge that the negative binomial distribution occurs here, so that one can deduce from that intuitive argument that the negative binomial distribution should occur here? Similar to the well known "intuitive" arguments that would show the Poisson distribution should occur above? ("Intuitive" and "intuition" are words that mathematicians use too much, without examining the various different meanings for which they use it. This practice exhibits Augean stability.)
| https://mathoverflow.net/users/6316 | Why should the logarithmic series distribution model the number of "Items" bought? | $\newcommand\la\lambda\newcommand\La\Lambda\newcommand{\Ga}{\Gamma}\newcommand{\R}{\mathbb R}$**Some preliminaries:** To make it explicit that the distribution of $Y$ depends on the parameter $r$, let $Y\_r:=Y$, so that
\begin{equation\*}
\Pr(Y\_r=j) = \binom{-r}j (-q)^j p^r \tag{1}\label{1}
\end{equation\*}
for $j=0,1,\dots$. We know that the number $N$ of purchasers has the Poisson distribution with some parameter $\mu$.
We have
\begin{equation\*}
Y\_r=\sum\_{i=1}^N X\_i,
\end{equation\*}
where $X\_i$ is the number of Items bought by the $i$th purchaser. Suppose that the purchasers act independently and are statistically interchangeable, so that the $X\_i$'s are iid. We also suppose that the number $N$ of purchasers is independent of the numbers $X\_i$ of Items bought by purchasers. It follows that the distribution of $Y\_r$ is a [compound Poisson distribution](https://en.wikipedia.org/wiki/Compound_Poisson_distribution):
\begin{equation\*}
P\_{Y\_r}=\sum\_{k=0}^\infty \Pr(N=k)\,P\_X^{\*k}
=\sum\_{k=0}^\infty \frac{\mu^k}{k!}\,e^{-\mu}\,P\_X^{\*k}, \tag{2}\label{2}
\end{equation\*}
where $X:=X\_1$, $P\_Z$ denotes the distribution of a random variable (r.v.) $Z$, and $Q^{\*k}:=\underbrace{Q\*\cdots\*Q}\_{k}$.
**We have to plausibly/"intuitively" recover the distribution $P\_X$ of $X$ given \eqref{2}.** In view of the infinite divisibility of the negative binomial distribution of $Y\_r$ and by \eqref{2}, we have
\begin{equation\*}
\lim\_{n\to\infty}P\_{Y\_{r/n}}^{\*n}=P\_{Y\_r}
=\lim\_{n\to\infty}\big(P\_0+\tfrac\mu n\,(P\_X-P\_0)\big)^{\*n}.
\end{equation\*}
So, it is plausible that for large $n$
\begin{equation\*}
(1-\tfrac\mu n)P\_0+\tfrac\mu n\,P\_X=P\_0+\frac\mu n\,(P\_X-P\_0)
\approx P\_{Y\_{r/n}}.
\end{equation\*}
Therefore and because $X\ge1$, we see that $P\_X$ is the limit as $n\to\infty$ of the conditional distribution of $Y\_{r/n}$ given $Y\_{r/n}\ne0$. So, for integers $j\ge1$, in view of \eqref{1},
\begin{equation\*}
\Pr(X=j)=\lim\_{s\downarrow0}\frac{\Pr(Y\_s=j)}{1-\Pr(Y\_s=0)} \\
=\lim\_{s\downarrow0}\frac{(j-1+s)(j-2+s)\cdots(1+s)s\; q^j p^s/j!}{1-p^s}
=\frac{q^j/j}{-\ln p},
\end{equation\*}
as desired.
$$\mbox{\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*}$$
Answer to the changed question
------------------------------
In the initial version of the question, the negative binomial distribution of $Y\_r$ appeared to be assumed, and the question appeared to be asking for a more intuitive derivation of the logarithmic series distribution for $X$ based on that "negative binomial" assumption.
After the above answer was posted, the OP changed the question, which now calls for a derivation of the logarithmic series distribution for $X$ without assuming the negative binomial distribution of $Y\_r$.
Let us try to answer the changed question as well. Here I will be following lines of [Fisher's paper](https://www.jstor.org/stable/1411#metadata_info_tab_contents) (pp. 54--58).
Let $Z$ denote the random number of books purchased by a **potential** buyer (not necessarily an actual purchaser). Assume that $Z$ depends on some "hidden" positive r.v. $\La$ so that the conditional distribution of $Z$ given $\La$ is the Poisson distribution with mean $\La$:
\begin{equation\*}
\Pr(Z=z\mid\La)=\frac{\La^z}{z!}\,e^{-\La}\,1(z=0,1,\dots). \tag{10}\label{10}
\end{equation\*}
In turn, assume that $\La$ has the gamma distribution with positive parameters $a$ and $b$, so that
\begin{equation\*}
\Pr(\La\in d\la)=\frac1{\Ga(a)b^a}\,\la^{a-1}e^{-\la/b}\,1(\la>0)\,d\la. \tag{20}\label{20}
\end{equation\*}
Then
\begin{align}
\Pr(Z=z)&=\int\_\R\Pr(\La\in d\la)\Pr(Z=z\mid\La) \notag \\
&=\frac{\Ga(a+z)}{\Ga(a)}\frac1{z!}\,\frac{1}{b^a (1+1/b)^{a+z}}\,1(z=0,1,\dots) \notag \\
&=\frac{a(a+1)\cdots(a+z-1)}{z!}\,p^a(1-p)^z\,1(z=0,1,\dots), \tag{30}\label{30}
\end{align}
where
\begin{equation\*}
p:=\frac1{1+b}.
\end{equation\*}
The "Poisson" assumption \eqref{10} seems natural.
Unfortunately, just as in the mentioned paper by Fisher, I cannot offer a reason to model the mixing distribution of $\La$ as a gamma distribution, except that then it is easy to get \eqref{30}. One may note here, though, that the gamma distribution is the [conjugate prior distribution](https://en.wikipedia.org/wiki/Conjugate_prior#Table_of_conjugate_distributions) with respect to the Poisson family of distribution -- a fact prized by Bayesians.
Anyhow, once this modeling is accepted, we next note that the expected number of purchases by a potential buyer is $EZ=EE(Z\mid\La)=E\La=ab$, which may be reasonably assumed small. If $b>0$ is fixed (that is, if $p=\frac1{1+b}\in(0,1)$ is fixed), then it follows that $a>0$ is small.
So, the distribution of $Z$ will be close to the Dirac distribution supported on the singleton set $\{0\}$. However, since we are interested in those potential buyers who are actually purchasers, we consider the conditional distribution of $Z$ given $Z>0$. In view of \eqref{30}, the probability mass function (pmf) -- say $f\_{a,p}$ -- of the just mentioned conditional distribution is given by the formula
\begin{align\*}
& f\_{a,p}(z)=\frac{\Pr(Z=z)\,1(z>0)}{1-\Pr(Z=0)} \\[6pt]
= {} & \frac{a(a+1)\cdots(a+z-1)}{z!}\,\frac{p^a(1-p)^z}{1-p^a} \, 1(z=1,2,\dots).
\end{align\*}
Since $a$ is small, it is natural to consider the limit
\begin{equation\*}
f\_{0+,p}(z):=\lim\_{a\downarrow0}f\_{a,p}(z)
=\frac1{-\ln p}\frac{(1-p)^z}z\,1(z=1,2,\dots).
\end{equation\*}
So, $f\_{0+,p}$ is the pmf of the logarithmic series distribution, as desired.
| 4 | https://mathoverflow.net/users/36721 | 435530 | 176,049 |
https://mathoverflow.net/questions/428913 | 1 | I'm curious about if the following question is in the literature or what work can be done about it.
Denote the number of distinct primes dividing an odd perfect number $N$ with the arithmetic function $\omega(N)$. Wikipedia has this section dedicated to near-square primes from the article [*Landau's problems.*](https://en.wikipedia.org/wiki/Landau%27s_problems#Near-square_primes)
Under the assumption that there exists an odd perfect number $N$, we denote the number of distinct near-squares primes dividing $N$ as $\omega\_{\text{nsp}}(N)$, thus $\omega\_{\text{nsp}}(N)\leq \omega(N)$.
Here I add as general reference that a near-square prime is a prime number of the form $p=n^2+1$, that's the sequence *A002496* from The On-Line Encyclopedia of Integer Sequences.
>
> **Question.** I would like to know if we can improve the previous lower bound and/or the upper bound, in this way $$\text{lower bound}<\omega\_{\text{nsp}}(N)<\text{upper bound}\tag{1}$$
> being these bounds functions of $N$, that's $\text{lower bound}=\text{lower bound}(N)$ and $\text{upper bound}$ denotes a function $\text{upper bound}(N)$. **Many thanks.**
>
>
>
I can to deduce obvious improvements for particular cases of Touchard's theorem, or when I consider Euler's theorem for odd perfect numbers. Also I know bounds for the radical of an odd perfect number, bounds for the Euler's totient function $\varphi(N)$, and I know the product formula representation for these arithmetic functions.
| https://mathoverflow.net/users/142929 | Number of distinct near-squares primes dividing an odd perfect number | In general, very few prime factors in an odd perfect number can be of the form $n^2+1$.
In particular, if N is an odd perfect number then $\frac{\sigma(N)}{N}=2$, and for any $m$ (perfect or not), $\frac{\sigma(n)}{n} \leq \prod\_{p|n}\frac{p}{p-1}$ with equality if and only if $n=1$.
Thus, for any perfect number, one needs $ \prod\_{p|N}\frac{p}{p-1}>2$. However,
$$\prod\_{p, p =i^2+1}\frac{p}{p-1} < \prod\_{i=1}^{\infty} \frac{(2i)^2+1}{(2i)^2} < 1.4$$
So, only a small fraction of the primes can be of the form $n^2+1$. Turning this into a bound on $\omega(N)$ directly is going to be very difficult, because one cannot rule out that one has an odd perfect number with a few primes not of this form, and then a lot of very big primes of the form $n^2+1$ that contribute very little to the product.
Note that similar remarks would apply to almost any class of primes of the form $p=Q(n)$ for some polynomial $Q$. There's very little content here dealing with the fact that we're looking at $n^2+1$ at all.
| 3 | https://mathoverflow.net/users/127690 | 435552 | 176,060 |
https://mathoverflow.net/questions/435500 | 1 | The purity of Brauer group states that for a smooth (quasi-)projective variety $X$ over a field $k$, removing a closed subscheme $Z \subset X$ of codimension at least $2$ ensures that the restriction map $H^2(X,\mathbb{G}\_m) \rightarrow H^2(X-Z,\mathbb{G}\_m)$ is an isomorphism. This means that the result cannot be extended to curves. Does anyone know of a simple counterexample, preferably when the smooth projective curve has genus at least $2$, if it matters?
| https://mathoverflow.net/users/172132 | Counterexample to purity of Brauer group for curves | The results of [this answer](https://mathoverflow.net/a/380825/82179) carry over (mutatis mutandis) to curves over finite fields. This shows that if $C$ is a (smooth, geometrically integral) curve over $\mathbf F\_q$ with function field $K$, then there is an exact sequence
$$0 \to \operatorname{Br}(C) \to \operatorname{Br}(K) \to \bigoplus\_{v \in C^{\text{cl}}} \mathbf Q/\mathbf Z,$$
obtained by taking the residues in $\operatorname{Br}(K\_v) \cong \mathbf Q/\mathbf Z$ for each completion $K\_v = \operatorname{Frac}\big(\widehat{\mathcal O\_{C,v}}\big)$ at closed points $v \in C^{\text{cl}}$. (That is, Brauer classes on $C$ are Brauer classes on $K$ that are unramified at all closed points of $C$.)
Thus if $\bar C$ is the smooth compactification of $C$, as soon as there exists a Brauer class on $C$ that is ramified at a point $v \in \bar C \setminus C$, then the restriction $\operatorname{Br}(\bar C) \to \operatorname{Br}(C)$ is not an isomorphism (it is always injective since $\operatorname{Br}(\bar C) \to \operatorname{Br}(K)$ is injective).
But by class field theory we can actually compute $\operatorname{Br}(C)$: by the [other post](https://mathoverflow.net/a/380825/82179) it sits in an exact sequence
$$0 \to \operatorname{Br}(C) \to \bigoplus\_{v \in \bar C \setminus C} \operatorname{Br}(K\_v) \to \mathbf Q/\mathbf Z \to 0.$$
If $\lvert \bar C \setminus C\rvert \geq 2$, we can produce classes whose invariants sum to $0$ in $\mathbf Q/\mathbf Z$, which therefore come from $\operatorname{Br}(C)$. Applying the same sequence to $\bar C$ shows $\operatorname{Br}(\bar C) = 0$. $\square$
Note: a direct computation of triviality of $\operatorname{Br}(\bar C)$ (not using class field theory) is given in [Brauer3, Rmq. 2.5(b)]. Presumably those texts contain many more useful computations and examples; for instance I expect that the passage from a curve to a dense open might be discussed somewhere as well.
**Remark.** On the other hand, if $C$ is a curve over an *algebraically closed* field $k$, then $\operatorname{Br}(C) \hookrightarrow \operatorname{Br}(K) = 0$ by Tsen's theorem, so $\operatorname{Br}(C) = 0$. So the Brauer classes really have to come from the nontriviality of $H^2(\kappa(v),\mathbf Z) \cong H^1(\kappa(v),\mathbf Q/\mathbf Z)$ for closed points $v \in \bar C \setminus C$, i.e. we need the residue fields $\kappa(v)$ not to be algebraically closed.
---
**References.**
[Brauer3] A. Grothendieck, *Le groupe de Brauer. III: Exemples et compléments.* Dix Exposés sur la Cohomologie des Schémas, Adv. Stud. Pure Math. **3**, 88-188 (1968). [ZBL0198.25901](https://zbmath.org/?q=an:0198.25901).
| 1 | https://mathoverflow.net/users/82179 | 435560 | 176,064 |
https://mathoverflow.net/questions/435556 | 1 | Recently, I came across the notion of [quasi-isometries](https://en.wikipedia.org/wiki/Quasi-isometry), while thinking of *"discrete* spaces which are surrogates for approximate continuous ones".
What (metric)/geometric properties are preserved by quasi-isometries? Also, are there good references on the topic?
---
**Direction/Angle:** Concretely, as an example of the direction I'm thinking in, I am interested in graph approximations to compact smooth manifolds (e.g. [in this post](https://mathoverflow.net/questions/288909/does-there-exist-a-notion-of-discrete-riemannian-metric-on-graph)). In that post, the described graphs are quasi-isometrically embedded into the target manifolds.
| https://mathoverflow.net/users/36886 | What properties are preserved by quasi-isometries | One can think of quasi-isometric spaces as spaces which look the same when seen from far away. Examples of properties preserved under quasi-isometries are for example Gromov-hyperbolicity (for geodesic metric spaces), growth types of Dehn functions and various notions of "rank". As a reference I would recommend Buyalo-Schroeder: *Elements of Asymptotic Geometry*, Bridson-Haefliger: *Metric spaces of non-positive curvature* and Burago-Burago-Ivanov: *A course in metric spaces*.
| 2 | https://mathoverflow.net/users/313861 | 435561 | 176,065 |
https://mathoverflow.net/questions/434711 | 1 | Given $X$=$l^1$ and its dual space $X^\*=l^\infty$. Now take $f=(1, 1/2, 2/3, 3/4,...) \in X^\*$. Then clearly $\|f\|\_\infty = 1$. I have found that $H=\ker f$ is a proximinal hyperplane in $X$.
Note: A subspace $Y$ of a normed linear space $X$ is called proximinal if for all $x \in X$, $P\_Y(x) \neq \emptyset$, where $P\_Y(x)= \{ y \in Y : \|x-y\| = d(x, Y) \}$. Here $y\in P\_Y(x)$ will be called a best approximation; the distance function is defined as $d(x, Y)= \inf \|x-y\_0\|$, $\forall y\_0 \in Y$. I need to find the set of best approximations $B\_Y(x)$ of $x$ to $B\_Y$, where $B\_Y$ is the closed unit ball in $Y$ defined as $B\_Y=\{y\_0 \in Y: \|y\_0\| \leq 1\}$. I am stuck at the confusion that whether $B\_Y$ exists or not and if it exists, what would $P\_{B\_Y}(x)$ be? Kindly help me. Thank you in advance.
| https://mathoverflow.net/users/494605 | Finding the set of best approximation | Similar to $P\_Y(x)$, there is no such ready formula for evaluating $P\_{B\_Y}(x)$, when $Y=\ker (f)$, and so is for $d(x,B\_Y)$. In some cases, for instance when $d(x,Y)=d(x,B\_Y)$, it is easier to understand $P\_{B\_Y}(x)$. In fact in few cases one can write $P\_{B\_Y}(x)=B\_Y\cap P\_Y(x)$, when $d(x,Y)=d(x,B\_Y)$.
Now the point $f=(1,1/2,2/3,\ldots)\in\ell\_\infty$ is a smooth point of $\ell\_\infty$ (in fact Frechet smooth) and $f (e\_1)=1$. Hence $J\_X (f)\neq\emptyset$ and a singleton. It now follows that $P\_Y(e\_1)=\{0\}$ and clearly $P\_{B\_Y}(e\_1)=\{0\}$ as $d(e\_1,Y)=d(e\_1,B\_Y)$. Here $Y=\ker (1,1/2,2/3,\ldots)$ and $J\_X(f)=\{x\in S\_X: f(x)=1\}$.
Now $Y$ is a strongly proximinal subspace of $\ell\_1$ as $(1,1/2,2/3,\ldots)$ is an SSD point of $\ell\_\infty$. This follows that $d(e\_i,B\_Y)>d(e\_i,Y)$, whenever $i>1$. So in these cases one needs to evaluate separately $d(e\_i,B\_Y)$ and figure out whether $P\_{B\_Y}(e\_i)\neq\emptyset$. See (Proposition 2.10, Indumathi, Lalithambigai, J. Convex Analysis, {vol 18}(2), 353--366, (2011)).
For a point $x\in\ell\_1$, suppose $\inf\{\|y\|:y\in P\_Y(x)\}\leq 1$.
Then $d(x,Y)=d(x,B\_Y)$ if and only if $P\_{B\_Y}(x)=P\_Y(x)\cap B\_Y$.
| 0 | https://mathoverflow.net/users/76412 | 435574 | 176,068 |
https://mathoverflow.net/questions/435573 | 2 | Let $f \in W^{1,p}(U)$, then how to prove that $|f| \in W^{1,p}(U)$, where $W$ means the sobolev space over some open subset $U \in \mathbb{R}^n$.
In Lieb's Analysis he prove that Let $f$ be in $W^{1, p}(\Omega)$. Then the absolute value of $f$, denoted by $|f|$ and defined by $|f|(x)=|f(x)|$, is in $W^{1, p}(\Omega)$ with $\nabla|f|$ being the function
$$
(\nabla|f|)(x)= \begin{cases}\frac{1}{|f|(x)}(R(x) \nabla R(x)+I(x) \nabla I(x)) & \text { if } f(x) \neq 0 \\ 0 & \text { if } f(x)=0\end{cases}
$$
here $R(x),I(X)$ denote the real part and imaginary part of $f(x)$.
In the proof he uses the inequality $|\frac{1}{|f|(x)}(R(x) \nabla R(x)+I(x) \nabla I(x))|^2 \leq |\nabla R(x)|^2+|\nabla I(x)|^2$ and chain rule to show that $\nabla|f| \in L^p$, but how it implies $\partial |f| \in L^p(U)$, do we have $|\frac{1}{|f|(x)}(R(x) \partial R(x)+I(x) \partial I(x))|^2 \leq |\partial R(x)|^2+|\partial I(x)|^2$ ?
In all, I have three question:
1. How to prove that $|\frac{1}{|f|(x)}(R(x) \nabla R(x)+I(x) \nabla I(x))|^2 \leq |\nabla R(x)|^2+|\nabla I(x)|^2$
2. How to prove that $\partial |f| \in L^p(U)$.
3. I wonder whether or not $|f| \in W\_0^{1,p}(U)$ when $f \in W\_0^{1,p}(U)$. I can't find a counterexample
| https://mathoverflow.net/users/494763 | Derivative of the absolute value | Part 1 follows from the Cauchy-Schwartz inequality, applied to the two vectors $(R(x), I(x))$, $(\nabla R(x), \nabla I(x))$.
Part 2 follows from the simple inequality $|\partial\_j R(x)| \leq \sqrt {\sum\_i (\partial\_i R(x))^2 } = |\nabla R(x)|$ for all $j$ (and similarly for $I$).
I am not sure about part 3 myself.
| 2 | https://mathoverflow.net/users/173490 | 435580 | 176,071 |
https://mathoverflow.net/questions/311877 | 18 | I consider *convex polytopes* $P\subseteq\Bbb R^d$ (convex hull of finitely many points) which are arc-transitive, i.e. where the automorphism group acts transitively on the 1-flags (incident vertex-edge pairs). Especially, this includes that the polytope is vertex- and edge-transitive. The *graph* of a polytope is the graph isomorphic to its 1-skeleton.
>
> **Question**: Are there arc-transitive polytopes, where the graph has more symmetries than the polytope?
>
>
>
When weakening the requirement of arc-transitivity, there are examples:
* A rhombus is edge- but not vertex-transitive. However, its graph is vertex-transitive. (Thanks to Henrik for the comment)
* There are vertex-transitive neighborly polytopes other than a simplex, but none of these can be edge-transitive. Their graphs are complete and are therefore edge-transitive.
| https://mathoverflow.net/users/108884 | Can the graph of a symmetric polytope have more symmetries than the polytope itself? | The answer is *No*, the graph of an arc-transitive polytope cannot have more symmetries than the polytope.
The polytope and its graph have the exact same symmetries!
---
In the article [*"Capturing Polytopal Symmetries by Coloring the Edge-Graph"*](https://arxiv.org/abs/2108.13483) I prove the following:
>
> (Corollary 5.3.) Let $P\subset \Bbb R^d$ be a polytope with edge-graph $G$. Consider a coloring $G^c$ of the edge-graph where two vertices/edges have the same color if and only if they are in the same orbit w.r.t. the symmetry group $\mathrm{Aut}(P)\subseteq\mathrm O(\Bbb R^d)$. Then $\mathrm{Aut}(G^c)\simeq\mathrm{Aut}(P)$.
>
>
>
In particular, if $P$ is vertex- and edge-transitive, then each vertex/edge of $G^c$ has the same color, that is, the graph is uncolored. Thus $\mathrm{Aut}(G)\simeq\mathrm{Aut}(P)$.
At the core of the proof is a result by Ivan Izmestiev from
>
> [I. Izmestiev (2010), *"The Colin de Verdière Number and Graphs of Polytopes"*](https://link.springer.com/article/10.1007/s11856-010-0070-5)
>
>
>
(summarized as Theorem 4.1 in my article), which roughly states that polytope skeleta are *spectral embeddings* of the edge-graph, when assigning certain vertex/edge weights.
These weights can be used as colors, and it then follows from basic spectral graph theory that every combinatorial symmetry of the colored graph extends to a geometric symmetry of the polytope.
| 5 | https://mathoverflow.net/users/108884 | 435585 | 176,073 |
https://mathoverflow.net/questions/435471 | 1 | Let $(E, d)$ be a Polish space and $\mathcal C\_b(E)$ the space of all real-valued bounded continuous functions on $E$. Let $p \in [1, \infty)$ and $\mathcal P\_p (E)$ the space of all Borel probability measures on $E$ with finite $p$-th moments. We define the [Wasserstein metric](https://en.wikipedia.org/wiki/Wasserstein_metric#Definition) $W\_p$ on $\mathcal P\_p (E)$ by
$$
W\_p (\mu, \nu) := \inf\_{\gamma \in \Pi(\mu, \nu)} \left [ \int\_{E \times E} (d(x,y))^p \mathrm d \gamma (x, y) \right ]^{1/p} \quad \forall \mu, \nu \in \mathcal P\_p (E).
$$
Here $\Pi(\mu, \nu)$ is the set of all Borel probability measures on $E\times E$ whose marginals are $\mu, \nu$ respectively. Fix some $a \in E$. Then we have from Villani's *Optimal Transport: Old and New*,
>
> **Theorem 6.9** Let $\mu\_n, \mu \in \mathcal P\_p (E)$ with $n\in \mathbb N$. The following statements are equivalent.
>
>
> * 1. $W\_p (\mu\_n, \mu) \to 0$.
> * 2. $\int f \mathrm d \mu\_n \to \int f \mathrm d \mu$ for all $f \in \mathcal C\_b(E)$ and $\int (d(x, a))^p \mathrm d \mu\_n \to \int (d(x, a))^p \mathrm d \mu$.
>
>
>
We define $g\_p:E \to \mathbb R, x \mapsto (d(x, a))^p$ and $\mathcal F\_p := \mathcal C\_b(E) \cup \{g\_p\}$. **Theorem 6.9** gives me a feeling that the metric topology of $\mathcal P\_p (E)$ is the initial topology induced by $\mathcal F\_p$. This would be true if **Theorem 6.9** holds not only for sequences but also for **nets**.
Because a metric space is a [sequential space](https://en.wikipedia.org/wiki/Sequential_space), $[(1) \implies (2)]$ holds for nets.
>
> Could you elaborate on if the direction $[(2) \implies (1)]$ of **Theorem 6.9** holds if we replace a sequence $(\mu\_n)\_{n \in \mathbb N}$ with a net $(\mu\_d)\_{d\in D}$?
>
>
>
Many thanks you so much for your explanation.
| https://mathoverflow.net/users/99469 | Does the topology of Wasserstein space $(\mathcal P_p (E), W_p)$ coincide with the initial topology induced by $\mathcal C_b(E) \cup \{g_p\}$? | A topology generated by countably many point-separating real functions [is metrizable](https://mathoverflow.net/a/435477/35357). To apply this here, it suffices to show that there is a countable family $\mathcal{G}$ of bounded real functions on $E$ such that the topology of weak convergence of measures is generated by the functions $\mu\mapsto g~\mathrm d\mu$ with $g$ in $\mathcal{G}$. That this is possible is
Theorem 6.6 in "Probability measures on metric spaces" by Parthasarathy. The possibly first proof of this fact is in
>
> Varadarajan, Veeravalli S. "Weak convergence of measures on separable
> metric spaces." *Sankhyā: The Indian Journal of Statistics*
> (1933-1960) 19.1/2 (1958): 15-22.
>
>
>
| 1 | https://mathoverflow.net/users/35357 | 435604 | 176,077 |
https://mathoverflow.net/questions/435609 | 3 | I have a math/stat problem where I need to show the convergence of the average of a sequence of experiments to an interval. The sequence of experiments is not i.i.d., hence the standard law of large number does not apply. However, the framework satisfies some assumptions which might facilitate the convergence proof. I think the question can fit this advanced forum because it seems to go beyond standard applications of probability results.
Suppose we have a sequence of random experiments $(a\_n)\_{n\in \mathbb{N}}$. In particular, each $a\_n$ is a random draw from a probability distribution $P\_n: B\rightarrow [0,1]$, where $B$ is a finite set.
The probability distributions $P\_n$ are potentially different across $n$. However, for each $b\in B$ and $n\in \mathbb{N}$, we know that $P\_n(b)\in [\nu\_\ell(b), \nu\_u(b)]$, where the latter interval does not vary across $n$.
Let $x\_N(b):=\frac{1}{N}\sum\_{n=1}^N \mathbb{1}(a\_n=b)$ for a finite $N\in \mathbb{N}$, where $\mathbb{1}(a\_n=b)$ takes value 1 if $a\_n=b$ and 0 otherwise.
I would like to show that, as $N\rightarrow \infty$, $x\_N(b)$ falls in $[\nu\_\ell(b), \nu\_u(b)]$.
Could you help me to do that? If you think the statement is wrong, can you explain why?
**Note: the draws may not be independent.**
| https://mathoverflow.net/users/42412 | A convergence problem | I assume that $(a\_n)\_{n \ge 1}$ are random variables taking values on a finite subset $B$, and that $\nu\_l(b) \le P[a\_n = b|a\_1,\ldots,a\_{n-1}] \le \nu\_u(B)$ almost surely for every $n \ge 1$ and $b \in B$.
If yes, then for each $b \in B$, the formula
$$M\_n(b) := \sum\_{k=1}^n\frac{1}{k}\big(1\_{[a\_k=b]}-P[a\_k = b|a\_1,\ldots,a\_{k-1}]\big)$$
defines a square-integrable martingale. This martingale has orthogonal increments and is bounded in $L^2(P)$, since
$$E\Big[\frac{1}{k^2}\big(1\_{[a\_k=b]}-P[a\_k = b|a\_1,\ldots,a\_{k-1}]\big)^2\Big] \le \frac{1}{4k^2}.$$
Hence it converges almost surely and in $L^2$.
We deduce that the averages
$$\frac{S\_n(b)}{n} := \frac{1}{n}\sum\_{k=1}^n \big(1\_{[a\_k=b]}-P[a\_k = b|a\_1,\ldots,a\_{k-1}]\big)$$
converge almost surely to $0$, by Cesàro lemma since
$$S\_n(b) = \sum\_{k=1}^n k(M\_k(b)-M\_{k-1}(b)),$$
$$S\_n(b) = \sum\_{k=1}^n kM\_k(b) - \sum\_{k=1}^n kM\_{k-1}(b)),$$
$$S\_n(b) = \sum\_{k=0}^n kM\_k(b) - 0 - \sum\_{k=0}^{n-1} (k+1)M\_k(b)),$$
$$S\_n(b) = nM\_n(b) - \sum\_{k=0}^{n-1} M\_k(b),$$
$$\frac{S\_n(b)}{n} = M\_n(b) - \frac{1}{n}\sum\_{k=0}^{n-1}M\_k(b).$$
As a result, the averages $\frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k = b]}$ and $\frac{1}{n}\sum\_{k=1}^n P[a\_k = b|a\_1,\ldots,a\_{k-1}]$ have the same limit points as $n \to +\infty$, which belong to $[\nu\_l(b),\nu\_u(b)]$.
---
ADDENDUM (answers to the questions added by the OP)
Step 1. $|M\_n(b)| \le \sum\_{k=1}^n 1/k$. Therefore $M\_n(b)$ is in $L^2(P)$.
Step 2. On $L^2(\Omega,\mathcal{A},P)$, the conditional expectation $E[\cdot|\mathcal{F\_n}]$ coincides with the orthogonal projection on $L^2(\Omega,\mathcal{F\_n},P)$. Hence $M\_{n+1}(b)-M\_n(b)$ is orthogonal to $L^2(\Omega,\mathcal{F\_n},P)$, therefore to $M\_0(b),\ldots,M\_n(b)$.
Step 3. Do not confuse $E[M\_n^2]$ finite for every $n$ and $E[M\_n^2]$ bounded independently on $n$. The last statement follows from Pythagore equality (write $N\_n$ as the sum of the pairwise orthogonal random variables $M\_1-M\_0,\ldots,M\_n-M\_{n-1}$) and from the convergence of the series $\sum\_k 1/k^2$.
Step 4. The theorem applied here is the martingale convergence theorem, for martingales which are bounded in $L^2(P)$. Convergence in $L^2(P)$ can also be proved simply bu using Cauchy lemma and Pythagore theorem, thanks to the pairwise orthogonalality of the random variables $M\_n-M\_{n-1}$ and the convergence of the series $\sum\_k 1/k^2$.
Step 5. No question on this step.
Step 6. Two sequences $(u\_n)$ and $(v\_n)$ of real numbers whose difference converges to $0$ have the same limit points: remind that the limit points are the limit of convergent subsequences. Because of the assumption $u\_n-v\_n \to 0$, for every increasing map $\phi$ from $\mathbb{N}$ to $\mathbb{N}$, and every real number $\ell$, $u\_{\phi(n)} \to \ell$ if and only if $v\_{\phi(n)} \to \ell$.
| 4 | https://mathoverflow.net/users/169474 | 435611 | 176,079 |
https://mathoverflow.net/questions/435617 | 4 | Let $\mathcal{E}$ be a coherent sheaf on an irreducible scheme $S$ ($S$ can be pretty nice, say noetherian of finite type), and let $\mathbf{P}(\mathcal{E}):=\mathrm{Proj}(\mathrm{Sym}(\mathcal{E}))$ be the projective bundle associated to $\mathcal{E}$. Is it true that if the structural map $\mathbf{P}(\mathcal{E})\to S$ is dominant (or even surjective), then $\mathbf{P}(\mathcal{E})$ is irreducible?
| https://mathoverflow.net/users/14143 | Basic question on projective bundles | This can fail when $\mathscr E$ has torsion:
**Example.** Let $k$ be a field, let $R = k[t]$ with maximal ideal $\mathfrak m = (t)$, and let $S = \operatorname{Spec} R$. Take $\mathscr E = R\oplus R/\mathfrak m = Rx \oplus Ry/(ty)$. Then
$$\operatorname{Sym}^\*(\mathscr E) = R[x,y]/(ty),$$
so taking $\mathbf{Proj}$ gives the locus $V(ty) \subseteq \mathbf P^1\_R = \mathbf P^1\_{[x:y]} \times \mathbf A^1\_t$. This has two components: a horizontal component $y=0$ mapping isomorphically down to $S = \mathbf A^1$, and a vertical component $t=0$ living entirely above the origin $0 \in \mathbf A^1$.
**Remark.** I do not know a clean criterion for irreducibility of $\mathbf P\_S(\mathscr E)$ for a coherent sheaf $\mathscr E$ on a locally Noetherian scheme $S$. Clearly a necessary criterion is that $\operatorname{Supp}(\mathscr E)$ is irreducible, since this is the image of $\mathbf P\_S(\mathscr E) \to S$ (of course the image is closed, either since $\mathbf P\_S(\mathscr E) \to S$ is proper or since it equals the support). So, as in your question, let me restrict to the case $S$ irreducible and $\mathbf P\_S(\mathscr E) \to S$ surjective. For irreducibility questions, we may replace $S$ by $S\_{\text{red}}$ to assume $S$ is integral.
Then there is a positive result if all $\operatorname{Sym}^n\mathscr E$ all torsion-free. In fact, in this case $\mathbf P\_S(\mathscr E)$ and even $\mathbf A\_S(\mathscr E)$ is integral: if $\eta \in S$ is the generic point, then torsion-freeness means that the natural map $\operatorname{Sym}^\*\_{\mathcal O\_S}(\mathscr E) \to \operatorname{Sym}^\*\_{\kappa(\eta)}(\mathscr E\_\eta)$ is injective, so $\operatorname{Sym}^\*(\mathscr E)$ embeds into a polynomial algebra over a field. Conversely, if $\mathbf A\_S(\mathscr E)$ is integral, clearly we need all $\operatorname{Sym}^n \mathscr E$ to be torsion-free.
However, this turns out to be a subtle condition on $\mathscr E$. For instance, when $\mathscr E = \mathcal I$ is a sheaf of ideals, then there is a natural map $\operatorname{Sym}^n \mathcal I \twoheadrightarrow \mathcal I^n$, which is an isomorphism if and only if $\operatorname{Sym}^n\mathcal I$ is torsion-free. This is a well-studied problem; for instance if $\mathcal I = \mathfrak m$ is the ideal of a closed point, then $\operatorname{Sym}^\* \mathfrak m$ is torsion-free if and only if $R$ is regular [Vas, Cor. 2.1.4].
**Example.** This gives a very transparent set of examples: if $S = \mathbf A^n$ and $I = \mathfrak m$ is the ideal of the origin, then $\operatorname{Sym}^\* I = \bigoplus\_n I^n$ is the Rees algebra by the discussion above, so $\mathbf P(I)$ is the blowup of $\mathbf A^n$ in the origin (which of course we know is integral).
(This also very clearly shows the issue with the wrong argument in an earlier version of this answer, as pointed out in the comments below.)
On the other hand, if $I = \mathfrak m$ is the ideal of a singular point, then $\mathbf P(I)$ does not agree with the Rees algebra, and at least $\mathbf A(I)$ is not integral.
Two things I couldn't figure out:
* If $S$ is integral and $\mathbf P\_S(\mathscr E)$ is irreducible, then is $\mathbf P\_S(\mathscr E)$ integral?
* If $S$ is integral and $\mathbf P\_S(\mathscr E)$ is integral, then is $\mathbf A\_S(\mathscr E)$ integral, i.e. are all symmetric powers of $\mathscr E$ torsion-free?
I suspect the answer to both questions is negative, but I have not yet found counterexamples. Maybe [Sasha's answer](https://mathoverflow.net/a/435703/82179) gives an example where $\mathbf P\_S(\mathscr E)$ is integral (since it is irreducible, generically reduced, and Cohen–Macaulay) but $\mathbf A\_S(\mathscr E)$ is not.
Here is an example showing that $\mathbf P\_S(\mathscr E)$ is not always reduced if we drop the irreducibility assumption:
**Example.** Let $k$ be a field, $R = k[t]$ with maximal ideal $\mathfrak m = (t)$, and let $S = \operatorname{Spec} R$. Take $\mathscr E = R \oplus R/\mathfrak m^2 = Rx \oplus Ry/(t^2y)$. Then
$$\operatorname{Sym}^\*(\mathscr E) = R[x,y]/(t^2y),$$
whose $\mathbf{Proj}$ is not reduced. However, as in the first example, it is also not irreducible.
---
**References.**
[Vas] W. V. Vasconcelos, *Arithmetic of blowup algebras*. London Mathematical Society Lecture Note Series **195**. Cambridge University Press (1994). DOI:[10.1017/CBO9780511574726](https://doi.org/10.1017/CBO9780511574726).
| 8 | https://mathoverflow.net/users/82179 | 435619 | 176,080 |
https://mathoverflow.net/questions/435628 | 1 | Let $k \in \mathbb{Z}^+$.
Is it possible to prove that, for some given
$m \in \{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23\}$,
there are only finitely many $k$ such that the closed interval $[k, k+99]$ contains (exactly) $m$ prime numbers?
Backstory: years ago, we wrote an informative article, in Italian, entitled "Quanti numeri primi in 100 interi consecutivi?" (see Matematicamente.it Magazine, Vol. 25(228), 2015 [1]), easily showing the existence of only a finite number of $k$ such that there are more than $23$ primes in $[k, k+99]$ (i.e., if $m \geq 24$, then $2 \leq k \leq 17$) and confirming the existence of infinitely many $k$ such that $m=0$ and $m=1$ (see Lemma 4 from [1]).
Moreover, it seemed reasonable to us to conjecture the existence of infinitely many $k$-tuples such that $m \in [2, 23]$.
**Question:** Is it possible, after more than $7$ years, to prove/disprove (at least) the conjecture above?
Lastly, in [1], we noted that it would be sufficient to prove just a single case of Polignac's conjecture up to a prime gap of $98$ (starting from a prime gap of $2$, then moving on $4$, and so forth, up to $98$) in order to confirm the conjecture, and we also wrote that we were inspired by prof. Tao's article entitled "Polymath8b: Bounded intervals with many primes, after Maynard" (19 Nov. 2013).
| https://mathoverflow.net/users/481829 | Infinitely many $k \in \mathbb{N}$ such that the closed interval $[k, k+99]$ contains from $2$ to $23$ prime numbers | It is compatible with current knowledge that all prime gaps are greater than 100\*, so it's not possible currently to show that for any of m in {2, ..., 23} that there are infinitely many such intervals (though it is surely true). Cases 0 and 1 follow from PNT (or even just Chebyshev's theorem on prime density) and Dirichlet's theorem.
\* The best current bound is 246.
| 3 | https://mathoverflow.net/users/6043 | 435632 | 176,086 |
https://mathoverflow.net/questions/435636 | 0 | Consider an ambient metric space $(\mathcal{X},\Vert\cdot\Vert\_\infty)$. Let $\mathcal{B}\_1 = \mathcal{B}\_{\Vert\cdot\Vert\_K}(0,1)\subseteq\mathcal{X}$ be the closed unit ball with respect to some norm $\Vert\cdot\Vert\_K$. Denote the $\varepsilon$-covering number of $\mathcal{B}\_1$ with respect to $\Vert\cdot\Vert\_\infty$ by $\mathcal{N}(\varepsilon, \mathcal{B}\_1,\Vert\cdot\Vert\_\infty)$. That is, we can find a set of points $\{x\_1,\dots,x\_n\}\subseteq\mathcal{X}$ with $n = \mathcal{N}\left(\varepsilon, \mathcal{B}\_1,\Vert\cdot\Vert\_\infty\right)$ such that for all $x\in\mathcal{B}\_1$, there exists $i\in[n]$ with
\begin{equation}
\Vert x - x\_i\Vert\_\infty\leq\varepsilon
\end{equation}
From the observation that the ball is closed and the covering number is defined with $\leq$ instead of $<$ (see equation above), I am tempted to assume that the covering number is a **right-continuous** function of $\varepsilon$. Is this assumption correct?
| https://mathoverflow.net/users/495129 | Right-continuity of covering number | Without additional assumptions on the metric space, it may appear that for every $\varepsilon>1$ the covering number equals 1, but for $\varepsilon=1$ it is infinite. For example, let positive integers be the points and the distance between $n$ and $m>n$ be equal $1+1/m$.
For compact metric space, it is right-continuous as you may take a convergent subsequence of cover sets.
| 1 | https://mathoverflow.net/users/4312 | 435638 | 176,089 |
https://mathoverflow.net/questions/434685 | 2 | Let $a,b$ be two positive integers. Let the sequence $\{s\_n\}\_n$ be the set of all possible sums of squares $a^2+b^2$, such that they are in *ascending order*
\begin{align\*}
& n=1 & s\_1=1^2+1^2=2 \\
& n=2 & s\_2=1^2+2^2=5 \\
& n=3 & s\_3=2^2+1^2=5 \\
& n=4 & s\_4=2^2+2^2=8
\end{align\*}
etc. Here $n$ serves as a counting index for the sequence and does not serve any other purpose. I have found that the number of identical values that each sum can have is equal to the [number of lattice points lying on a quarter of circumference of a circle of radius $n$](https://math.stackexchange.com/questions/2512236/number-of-lattice-points-on-a-circle). For example, $s\_2=s\_3=5$ we have two values, but there are other cases, such as
\begin{align\*}
& n=31 & s\_{31}=1^2+7^2=50 \\
& n=32 & s\_{32}=5^2+5^2=50 \\
& n=33 & s\_{33}=7^2+1^2=50
\end{align\*}
To the best of my knowledge, there are no references to the asymptotic form of $\{s\_n\}\_n$ as $n\to \infty$, if indeed it exists. A brute-force approach with $n\le10^6$ led to the expression
$$
\{s\_n\}\_n\sim \beta n,\qquad\beta\approx1.276\ldots.
$$
Therefore my question is: is there any known closed form of the asymptotics of this particular succession? A more profound and interesting question is, why does it seem to be linear?
| https://mathoverflow.net/users/148012 | Asymptotic analysis of a peculiar sum of squares sequence | The answer has been provided by [Noam D. Elkies](https://mathoverflow.net/users/14830/noam-d-elkies), for which I'll copy here his comment.
Usually this is studied in the nearly equivalent form of asking for the number of $(x,y)$ such that $x^2+y^2\le N$.
The answer is asymptotic to the area of this circle, which is $\pi N$. Since you're using a quarter-circle, we replace the factor $\pi$ by $\pi/4$.
Then the $n$-th sum of squares should be asymptotic to
$$
\frac{4}{\pi}n
$$
Indeed your $\beta$ nearly equals $4/\pi=1.273\ldots$
| 0 | https://mathoverflow.net/users/148012 | 435643 | 176,091 |
https://mathoverflow.net/questions/435644 | 26 | Typical courses on real integration spend a lot of time defining the Lebesgue measure and then spend another lot of time defining the integral with respect to a measure. This is sometimes criticized as being inefficient or roundabout (see, e.g., the question [“Why isn't integral defined as the area under the graph of function?”](https://mathoverflow.net/questions/321916/why-isnt-integral-defined-as-the-area-under-the-graph-of-function)) and one might seek a way to define the Lebesgue integral directly, without mentioning measures at all (the Lebesgue measure can then retrospectively be defined as the integral of the characteristic function, making its properties obvious if those of the integral are correctly obtained).
Now some years ago, I taught a course on real analysis (which I hadn't myself written, conceived or organized) using the following definition of the Lebesgue integral on $\mathbb{R}$:
* First, define a “step function” $\mathbb{R} \to \mathbb{R}$ as a finite linear combination of characteristic functions of intervals, and the integral of such a function as the linear form which takes the characteristic function of $I$ to the length of $I$.
* Next, we say that $f\colon \mathbb{R} \to \mathbb{R}$ is *integrable* iff there exists a series $(\Sigma f\_n)$ of step functions such that $\sum\_{n=0}^{+\infty} \int|f\_n| < +\infty$ and such that $f(x) = \sum\_{n=0}^{+\infty} f\_n(x)$ for every $x$ for which the RHS converges (**edit**: see below) *absolutely*; and when this is the case, we define $\int f := \sum\_{n=0}^{+\infty}\int f\_n$.
This provides a very short definition of what a Lebesgue-integrable function is, without going through the roundabout route of defining the measure first. Now of course it's not all rosy either: one has to check that this definition makes sense, and that it satisfies the usual properties of the Lebesgue integral. (And even if one knows in advance what the Lebesgue integral is, it's not quite obvious that this definition reconstructs it, because it's not trivial that one can construct a series $(\Sigma f\_n)$ of step functions that converges to $f(x)$ at *every* point where it converges.)
(I also mentioned this definition in passing in the question [“Can the Riemann integral be defined through a closure/completion process?”](https://mathoverflow.net/questions/288327/can-the-riemann-integral-be-defined-through-a-closure-completion-process))
But anyway, **my question is**: who came up with this definition? Has anyone else seen it? What is its history? And are there any prominent courses in real integration that use it?
**Edit / correction:** Following Willie Wong's comment to Kostya\_I's answer, I realize I had misremembered the definition, it's “$f(x) = \sum\_{n=0}^{+\infty} f\_n(x)$ for every $x$ for which the RHS converges absolutely” (i.e., $\sum\_{n=0}^{+\infty} |f\_n(x)| < +\infty$) rather than just “…for which the RHS converges”, so it appears that Jan Mikusiński is indeed the author of the definition I *meant* to write. But this raises the question of whether the definition I had *actually* written (with “converges” instead of “converges absolutely”) is different or whether this is irrelevant: if someone wants a crack at it, let them do so!
| https://mathoverflow.net/users/17064 | What is the origin/history of the following very short definition of the Lebesgue integral? | This definition is due to Jan Mikusiński, see Mikusiński, Jan,
The Bochner integral. Basel, Stuttgart: Birkhauser, 1978.
Mikusiński has co-authored another book on integration with Hartman in 1961, where a standard exposition of Lebesgue integration is given. So we may infer that Mikusiński's definition was invented between 1961 and 1978.
| 21 | https://mathoverflow.net/users/56624 | 435651 | 176,092 |
https://mathoverflow.net/questions/435479 | 5 | $\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\GL{GL}$I am looking to classify the homomorphisms from the group $\PSL(2,p)=\Aut(\mathbb{P}^1\mathbb{F}\_p)$ to $\PGL(n,2)=\GL(n,2)=\Aut(\mathbb{F}\_{2^n})$ when $p$ is a Mersenne prime, i.e. $2^n=p+1$.
There is a bijection $\mathbb{P}^1\mathbb{F}\_p \rightarrow \mathbb{F}\_{2^n} \cong \mathbb{F}\_2[x]/g(x)$ given by $k \mapsto x^\infty+x^k$ where $x^\infty=0$. Given a Möbius transformation $f$, define the homomorphism $T$ by $T\_f(x^\infty+x^k)=x^{f(\infty)}+x^{f(k)}$.
When $n=3$, $p=7$, $T$ is an isomorphism (see [Brown and Loehr - Why is $\PSL(2, 7) \cong \GL(3, 2)$?](https://personal.math.vt.edu/brown/doc/PSL(2,7)_GL(3,2).pdf)). What is $\operatorname{im}(T)\subset \PGL(n,2)$?
| https://mathoverflow.net/users/15133 | How to classify homomorphisms from $\operatorname{PSL}(2,p)$ to $\operatorname{PGL}(n,2)$ when $2^n=p+1$? | The map $T \colon \mathrm{PSL}\_2(p) \to \operatorname{Sym}(\mathbb{F}\_{2^n}) \colon f \mapsto T\_f$ does not have its image in $\mathrm{GL}\_n(2)$ for other Mersenne primes $p = 2^n - 1$, unlike the case $p = 7 = 2^3 - 1$.
For instance, let $p = 31$ and consider $\mathbb{F}\_{32} \cong \mathbb{F}\_2[x] / (g)$ with $g(x) = x^5 + x^2 + 1$, as in the question, and let $f \colon \mathbb{P}\_1(\mathbb{F}\_p) \to \mathbb{P}\_1(\mathbb{F}\_p) \colon a \mapsto -a^{-1}$ (so $f \in \mathrm{PSL}\_2(p)$). The map $T\_f$ is then the permutation of $\mathbb{F}\_{32}$ mapping each $x^k$ to $1 + x^{f(k)}$. In particular,
\begin{align\*}
T\_f(1) &= 1, \\
T\_f(x^2) &= 1 + x^{15}, \\
T\_f(x^5) &= 1 + x^{6}.
\end{align\*}
However, $x^5 + x^2 + 1 = 0$ in $\mathbb{F}\_{32}$, but
$$ T\_f(x^5) + T\_f(x^2) + T\_f(1) = 1 + x^6 + x^{15} = 1 + x^2 + x^4 \neq 0 ,$$
so $T\_f$ cannot be linear, i.e., it is not contained in $\mathrm{GL}\_n(2)$.
| 3 | https://mathoverflow.net/users/12858 | 435664 | 176,097 |
https://mathoverflow.net/questions/435662 | 7 | Let $k$ be a global field and $C$ a smooth projective curve over $k$ which is not isotrivial. Then there is a well-known trichotomy:
* If $g(C) = 0$ and $C(k) \neq \emptyset$, then $C \cong \mathbb{P}^1$. In particular $C(k)$ is infinite.
* If $g(C) = 1$ and $C(k) \neq \emptyset$, then $C(k)$ is a finitely generated abelian group.
* If $g(C) \geq 2$ then $C(k)$ is finite (Mordell conjecture).
However, there exist projective *regular* curves $C$ over $k$ whenever $k$ is imperfect, which need not be smooth.
What happens with $C(k)$ in this case? Is there a similar structure theorem given by geometric invariants, which determine when $C(k)$ can be infinite?
| https://mathoverflow.net/users/5101 | Rational points on regular curves over global fields | *If the (geometrically integral, projective) curve $C$ over the global field $k$ is regular but not smooth over $k$, then $C(k)$ is finite*:
First of all $C$ is smooth over $k$ if and only if also the base change $C\_{\bar{k}}$ is regular.
This is equivalent to $C$ being conservative, i.e. $g(C)=g(C\_{\bar{k}})$, where $g$ is denotes the arithmetic genus of the normalization (or in function field terminology the genus of $k(C)/k$), see e.g. Theorem 2.5.1 in [1](https://arxiv.org/abs/2102.06941).
Finally, it is a consequence of a theorem of Jeong [2] that
for global fields $k$ (in fact for all $k$ finitely generated and of transcendence degree $1$ over a perfect field), $g(C)>g(C\_{\bar{k}})$ implies that $C(k)$ is finite, see Proposition 5.7 in [3].
[1](https://arxiv.org/abs/2102.06941) B. Poonen, Rational Points on Varieties
[2] S. Jeong. Rational points on algebraic curves that change genus. J. Number Theory 67(2):170–181, 1997
[3] <https://arxiv.org/abs/2102.06941>
| 7 | https://mathoverflow.net/users/50351 | 435667 | 176,098 |
https://mathoverflow.net/questions/375655 | -1 | Let $H$ be a Hilbert space and $B(H)$ be the space of all bounded operators on $H$.
The Wold decomposition says that: an operator $x$ in $B(H)$ is an isometry if and only if $x=x\_u\oplus x\_s$ where $x\_u$ and $x\_s$ are unitary and unilateral shift respectively. Indeed $x\_u$ is the restriction of $x$ to $H\_u=\bigcap x^nH$ and $x\_s$ is the restriction of $x$ to $H\_s$ where,
$$H\_s:=H\ominus H\_u=\bigoplus\_{n\geq0} (x^nH\ominus x^{n+1}H)$$
For $x\in B(H)$, let us denote $e\_x$ by the projection onto $xH$.
Q. I am looking for a commuting pair of isometries $(x\_1,x\_2)$ in $B(H)$ such that all the following hold:
i) $e\_y$ commutes with all projections $e\_{x^n}$ for $n\geq1$.
ii) $H\_s^{(x)}:=H\ominus (\cap x^nH)$ is not $y$-invariant.
| https://mathoverflow.net/users/84390 | A commuting pair of isometries | Such a pair $(X,Y)$ is constructed as follows. Consider a Hilbert space $M$ with an orthonormal basis $\{e\_n:n\in\mathbb Z\}$ and the bilateral shift $U$ on $M$ such that $Ue\_n=e\_{n+1}$. Denote by $S$ the restriction of $U$ to the space $N$ generated by $\{e\_n:n\geq 0\}$, and construct the space $H=N\oplus M\oplus M\oplus\cdots$, that is, the space of square summable sequences $(m\_0,m\_1,\dots)$ such that $m\_0\in N$ and $m\_j\in M$ for $j>0$. Now we define $X,Y$ by setting
$$X(m\_0,m\_1,\dots)=(Sm\_0,Um\_1,Um\_2,\dots),$$ and
$$Y(m\_0,m\_1,\dots)=(0,m\_0,m\_1,\dots).$$
The range projections are easily calculated: $$YY^\*(m\_0,m\_1,\dots)=(0,P\_Nm\_1,m\_2,\dots),$$ and
$$X^nX^{\*n}(m\_0,m\_1,\dots)=(S^nS^{\*n}m\_0,m\_1,\dots).$$
It is trivial to verify that $YY^\*$ commutes with $X^nX^{\*n}$ for every $n$, but $Y$ does not leave the space
$$H\ominus\left(\bigcap\_n X^nH\right)=\{(m\_0,0,0,\dots):m\_0\in N\}$$
invariant.
This construction is based on an idea suggested by Ronald Douglas and formalized in *Canonical models for bi-isometries* ([actual paper](https://doi.org/10.1007/978-3-0348-0221-5_7), [MR review](https://mathscinet.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=JOUR&pg8=ET&review_format=html&s4=bercovici%20and%20douglas&s5=&s6=&s7=&s8=All&sort=Newest&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=1&mx-pid=2931928)).
| 1 | https://mathoverflow.net/users/12465 | 435671 | 176,099 |
https://mathoverflow.net/questions/435598 | 1 | I have a cube $X\in \mathbb R^{N\times N\times N}$ such that no matter how the cube is rotated by $90^\circ$ along any of the axes, the result is unchanged. What is the maximum number of distinct entries in the cube if $N$ is odd?
Here is my answer. I am sure that it is less than or equal to the true answer.
Let $n=(N+1)/2$. There exists a subcube of shape $\mathbb R^{n\times n\times n}$ that contains all of the distinct entries. This subcube must be on the corner of the cube, so I can imagine the indices of the entries as being in $\mathbb N\_n^3,\mathbb N\_n:=1,...,n$.
The claim I am unsure of: the indices of any one of the distinct entries are the same up to a permutation, e.g., the $(1,2,3), (1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)$ entries must all be the same. This is sufficient for the cube to be rotation invariant, but it may not be necessary, which would allow for the final answer to be greater.
Assuming that this claim is true, it's enough to just count the number of sets of indices that are equal up to a permutation. That is the same as unordered sampling with replacement, and thus, the number of distinct entries is $\binom{n+3-1}{3} = \binom{n+2}{3}$
---
Results following @MaartenHavinga:
I attempted to follow Maarten's instructions to generate the $(0,0,0)$ corner of the cube (the first dimension is the row, second is the column, third is the depth, i.e. the "upper left front" part of the cube).
When $n=3$, the results are good:
```
Group 0:
[[0 0 0]]
Group 1:
[[0 0 1]
[0 1 0]
[1 0 0]]
Group 2:
[[0 0 2]
[0 2 0]
[2 0 0]]
Group 3:
[[0 1 1]
[1 0 1]
[1 1 0]]
Group 4:
[[0 1 2]
[0 2 1]
[1 0 2]
[1 2 0]
[2 0 1]
[2 1 0]]
Group 5:
[[0 2 2]
[2 0 2]
[2 2 0]]
Group 6:
[[1 1 1]]
Group 7:
[[1 1 2]
[1 2 1]
[2 1 1]]
Group 8:
[[1 2 2]
[2 1 2]
[2 2 1]]
Group 9:
[[2 2 2]]
array([[[ 1.62, -0.61, -0.53],
[-0.61, -1.07, 0.87],
[-0.53, 0.87, -2.3 ]],
[[-0.61, -1.07, 0.87],
[-1.07, 1.74, -0.76],
[ 0.87, -0.76, 0.32]],
[[-0.53, 0.87, -2.3 ],
[ 0.87, -0.76, 0.32],
[-2.3 , 0.32, -0.25]]])
```
However, when $n=4$, the results are slightly off:
```
Group 0:
[[0 0 0]]
Group 1:
[[0 0 1]
[0 1 0]
[1 0 0]]
Group 2:
[[0 0 2]
[0 2 0]
[2 0 0]]
Group 3:
[[0 0 3]
[0 3 0]
[3 0 0]]
Group 4:
[[0 1 1]
[1 0 1]
[1 1 0]]
Group 5:
[[0 1 2]
[1 2 0]
[2 0 1]]
Group 6:
[[0 2 1]
[1 0 2]
[2 1 0]]
Group 7:
[[0 1 3]
[0 3 1]
[1 0 3]
[1 3 0]
[3 0 1]
[3 1 0]]
Group 8:
[[0 2 2]
[2 0 2]
[2 2 0]]
Group 9:
[[0 2 3]
[0 3 2]
[2 0 3]
[2 3 0]
[3 0 2]
[3 2 0]]
Group 10:
[[0 3 3]
[3 0 3]
[3 3 0]]
Group 11:
[[1 1 1]]
Group 12:
[[1 1 2]
[1 2 1]
[2 1 1]]
Group 13:
[[1 1 3]
[1 3 1]
[3 1 1]]
Group 14:
[[1 2 2]
[2 1 2]
[2 2 1]]
Group 15:
[[1 2 3]
[2 3 1]
[3 1 2]]
Group 16:
[[1 3 2]
[2 1 3]
[3 2 1]]
Group 17:
[[1 3 3]
[3 1 3]
[3 3 1]]
Group 18:
[[2 2 2]]
Group 19:
[[2 2 3]
[2 3 2]
[3 2 2]]
Group 20:
[[2 3 3]
[3 2 3]
[3 3 2]]
Group 21:
[[3 3 3]]
array([[[ 1.62, -0.61, -0.53, -1.07],
[-0.61, 0.87, -2.3 , -0.76],
[-0.53, 1.74, 0.32, -0.25],
[-1.07, -0.76, -0.25, 1.46]],
[[-0.61, 0.87, 1.74, -0.76],
[ 0.87, -2.06, -0.32, -0.38],
[-2.3 , -0.32, 1.13, -1.1 ],
[-0.76, -0.38, -0.17, -0.88]],
[[-0.53, -2.3 , 0.32, -0.25],
[ 1.74, -0.32, 1.13, -0.17],
[ 0.32, 1.13, 0.04, 0.58],
[-0.25, -1.1 , 0.58, -1.1 ]],
[[-1.07, -0.76, -0.25, 1.46],
[-0.76, -0.38, -1.1 , -0.88],
[-0.25, -0.17, 0.58, -1.1 ],
[ 1.46, -0.88, -1.1 , 1.14]]])
```
I will refer to positions starting from index 1. In the 2nd depth position, the entry in the 3rd row and 4th column (`-1.1`) is distinct from the entry in the 4th row and 3rd column (`-0.17`). This is because the indices are `(3, 4, 2)` and `(4, 3, 2)` which are an odd permutation.
However, in order for the cube to be rotation invariant to rotations about the row and column axes (i.e., around the entry `-0.88` in the second depth dimension), these two entries should be equal.
---
Results following @MaartenHavinga's corrected answer for `n=4`:
```
Group 0:
[[0 0 0]]
Group 1:
[[0 0 1]
[0 1 0]
[1 0 0]]
Group 2:
[[0 0 2]
[0 2 0]
[2 0 0]]
Group 3:
[[0 0 3]
[0 3 0]
[3 0 0]]
Group 4:
[[0 1 1]
[1 0 1]
[1 1 0]]
Group 5:
[[0 1 2]
[1 2 0]
[2 0 1]]
Group 6:
[[0 2 1]
[1 0 2]
[2 1 0]]
Group 7:
[[0 1 3]
[0 3 1]
[1 0 3]
[1 3 0]
[3 0 1]
[3 1 0]]
Group 8:
[[0 2 2]
[2 0 2]
[2 2 0]]
Group 9:
[[0 2 3]
[0 3 2]
[2 0 3]
[2 3 0]
[3 0 2]
[3 2 0]]
Group 10:
[[0 3 3]
[3 0 3]
[3 3 0]]
Group 11:
[[1 1 1]]
Group 12:
[[1 1 2]
[1 2 1]
[2 1 1]]
Group 13:
[[1 1 3]
[1 3 1]
[3 1 1]]
Group 14:
[[1 2 2]
[2 1 2]
[2 2 1]]
Group 15:
[[1 2 3]
[1 3 2]
[2 1 3]
[2 3 1]
[3 1 2]
[3 2 1]]
Group 16:
[[1 3 3]
[3 1 3]
[3 3 1]]
Group 17:
[[2 2 2]]
Group 18:
[[2 2 3]
[2 3 2]
[3 2 2]]
Group 19:
[[2 3 3]
[3 2 3]
[3 3 2]]
Group 20:
[[3 3 3]]
array([[[ 1.62, -0.61, -0.53, -1.07],
[-0.61, 0.87, -2.3 , -0.76],
[-0.53, 1.74, 0.32, -0.25],
[-1.07, -0.76, -0.25, 1.46]],
[[-0.61, 0.87, 1.74, -0.76],
[ 0.87, -2.06, -0.32, -0.38],
[-2.3 , -0.32, 1.13, -1.1 ],
[-0.76, -0.38, -1.1 , -0.17]],
[[-0.53, -2.3 , 0.32, -0.25],
[ 1.74, -0.32, 1.13, -1.1 ],
[ 0.32, 1.13, -0.88, 0.04],
[-0.25, -1.1 , 0.04, 0.58]],
[[-1.07, -0.76, -0.25, 1.46],
[-0.76, -0.38, -1.1 , -0.17],
[-0.25, -1.1 , 0.04, 0.58],
[ 1.46, -0.17, 0.58, -1.1 ]]])
```
| https://mathoverflow.net/users/495707 | Number of distinct entries in a rotation invariant cube | Since rotations give only even permutations, the 3 distinct index triples $\{a,b,c\}$ with $a \neq b \neq c \neq a$ give twice as much distinct entries as you gave: entries at $(a,b,c)$ and $(b,a,c)$ may be distinct in this case. This adds $n$ choose $3$ distinct entries.
However, for entries $\{a,b,n\}$ a 90 degree rotation can interchange $b$ and $n$ because $N$ is odd and $n=(N+1)/2=1-n$ modulo $N$, which is an odd permutation identifying $\binom{n-1}{2}$ unique entry pairs.
That gives a total of $\binom{n+2}{3} + \binom{n}{3} - \binom{n-1}{2}$ which is $\binom{n+2}{3} + \binom{n-1}{3}$ distinct entries/dimensions in your tensor space. Check for instance for $n=3, N=5$: giving $10$ distinct entries.
| 1 | https://mathoverflow.net/users/490128 | 435680 | 176,101 |
https://mathoverflow.net/questions/335437 | 6 | A famous result of Sullivan (closely related to work of Wilkerson) says that the group of isotopy classes of diffeomorphisms of a simply-connected closed smooth manifold of dimension $\geq 5$ is commensurable with an arithmetic group. Sullivan gave an outline of its proof, and several later papers filled in more details treating some additional cases. These in particular give more of a hint about the required results from the theory of algebraic groups.
One such paper, Triantafillou's [The Arithmeticity of Groups of Automorphisms of Spaces](http://www.ams.org/books/conm/231/3367/conm231-3367.pdf) contains the following statement regarding a useful general result. It concerns maps from a short exact sequence of groups
$$1 \to A \to A' \to A'' \to 1$$
to a short exact sequence of groups obtained by taking $\mathbb{Q}$-points of an extension of algebraic groups
$$1 \to G \to G' \to G'' \to 1.$$
>
> To complete the proof one needs an argument to the effect that
> the middle vertical map of a diagram of short exact sequences involving the $\mathbb{Q}$-points of algebraic groups is arithmetic if the outer maps
> are arithmetic and the kernel of the algebraic extension is unipotent.
> Here a map from a group into an algebraic group over $\mathbb{Q}$ is called
> *arithmetic* if its kernel is finite and its image is arithmetic. It appears
> that no such result exists in print nor is it folklore in the subject.
> Upon our inquiry, A. Borel furnished a proof of this algebraic fact
> which involves basic structure theorems of algebraic groups and the
> behaviour of lattices under the maps $\exp$ and $\log$ between the Lie
> algebra and the unipotent group.
>
>
>
However, she does not provide Borel's proof and instead uses a weaker result sufficient for her purposes. Is there a proof of this statement available in the literature somewhere?
| https://mathoverflow.net/users/798 | A result of Borel on extensions of arithmetic groups | A proof appears in Section 2.3 of my paper [Mapping class groups of manifolds with boundary are of finite type](https://arxiv.org/abs/2204.01945). As pointed out in the comments, it is crucial that $G$ is unipotent.
| 1 | https://mathoverflow.net/users/798 | 435683 | 176,102 |
https://mathoverflow.net/questions/376729 | 9 | Derived mapping spaces between little $d$-disks operads $E\_d$ play an important role in embedding calculus. For example, [Dwyer-Hess](https://projecteuclid.org/euclid.gt/1513732411) expresses the homotopy of framed long knots as loop spaces such mapping spaces, a result which was generalized by [Boavida de Brito-Weiss](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/topo.12048).
When stating these results, the $E\_d$-operad is usually *unitary*: its space of 0-ary operations is contractible (in fact a point). There is also a *non-unitary* version $E\_d^{nu}$, where replace the $0$-ary operations by an empty set. There is a natural forgetful map from the derived mapping space of unitary operads to that of non-unitary operads
$$Map^h\_{\mathsf{Op}}(E\_m,E\_n) \to Map^h\_{\mathsf{Op}}(E\_m^{nu},E\_n^{nu}).$$
>
> Is this a weak equivalence?
>
>
>
This is known if we assume $m=n$ *and* we replace the operads by their rationalizations, by Section 7 of [Fresse-Willwacher](https://arxiv.org/abs/2003.02939).
| https://mathoverflow.net/users/798 | Maps between unitary little disks operads and non-unitary little disks operads | A positive answer is the main theorem of my paper with Krannich and Horel, [Two remarks on spaces of maps between operads of little cubes](https://arxiv.org/abs/2211.00908). The proof uses a result of Haugseng and Kock to reduce it to a theorem of Lurie.
| 3 | https://mathoverflow.net/users/798 | 435684 | 176,103 |
https://mathoverflow.net/questions/435647 | 2 | I have been looking into bootstrapping lately and although I believe to have understood the basic process somewhat, I am fuzzy on the mathematical details. I will begin with my understanding of what bootstrapping is and then my understanding of the mathematics going on in the background. I might very well be mistaken on either one.
**Bootstrapping** (as I understand it):
The idea behind (most) bootstrapping is to study a random variable $X$ and its (unknown) distribution by resampling a sample $x=(x\_1, ..., x\_n)$ of $X$. Meaning, we choose randomly $n$ times an entry of $x$ to create a new sample $x^{(1)}=(x\_1^{(1)}, ..., x\_n^{(1)})$. This way, we create a bunch of samples $(x^{(i)})\_i$.
For each of these samples, we can now compute, for example, the mean and plot these means into a histogram. This histogram can be normalized to give us a distribution of the mean across all new samples. This process could of course be applied to any other statistic we might want to compute for any of the samples, giving us some histogram each time.
**Mathematical model** (as I understand it):
Let $X$ be a random variable with unknown distribution $f\_X$. We have a sample $x=(x\_1, ..., x\_n)$ of $X$. This sample defines a (discrete) empirical distribution $f\_x$. We can now define $n$ independent variables $Y\_1, ..., Y\_n \sim f\_x$. Calculating the mean of a resample of $x$ would now be equivalent to sampling the random variable $X' = \frac{1}{n}(Y\_1 + ... + Y\_n)$. Sampling $X'$ a lot, producing and normalizing a histogram would tend towards the distribution of $\bar X = \frac{1}{m}(X'\_1 + ... + X'\_m)$ as $m$ tends to infinity where $X'\_1, ..., X'\_m$ are independent and distributed as $X'$. The CLT now tells us that this would be a Gaussian distribution.
Barring any mistakes I have made here, I have two questions:
1. Am I correct here? Does the histogram of the mean of bootstrapped samples really resemble a Gaussian distribution? (In the sense that we can normalize the histogram and take the limit.)
2. What if we replace the mean by any function $g$ on $X\_1, ..., X\_n$ that produces different $X'$, e.g. $g=\min$? Do we still get something Gaussian? My instinct says no but I cannot see why the CLT would not work the same.
| https://mathoverflow.net/users/151546 | Bootstrapping and the central limit theorem | $\newcommand{\X}{\mathbf X}\newcommand{\x}{\mathbf x}\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$Your understanding of the purposes of the bootstrap is largely incorrect.
Here is what the [foundational paper by Efron](https://projecteuclid.org/journals/annals-of-statistics/volume-7/issue-1/Bootstrap-Methods-Another-Look-at-the-Jackknife/10.1214/aos/1176344552.full) that introduced the bootstrap method says:
>
> We discuss the following problem: given a random sample $\X=(X\_1,X\_2,\dots,X\_n)$ from an unknown probability distribution $F$, estimate the sampling distribution of some prespecified random variable $R(\X,F)$, on the
> basis of the observed data $\x$.
>
>
>
We see that here the (say real-valued) random variable (r.v.)
\begin{equation\*}
Y:=R(\X,F)
\end{equation\*}
in question is, **not a function of just the sample $\X$, but also of the unknown probability distribution $F$ (of each $X\_i$).**
To estimate the distribution of the r.v. $Y=R(\X,F)$ by the bootstrap method, we obtain (usually by computer simulation) a large number, say $B$, of (desirably/approximately) independent random samples
\begin{equation\*}
\x^\*\_1=(x^\*\_{1,1},\dots,x^\*\_{1,n}),\dots,\x^\*\_B=(x^\*\_{B,1},\dots,x^\*\_{B,n})
\end{equation\*}
from the empirical distribution
\begin{equation\*}
F\_\x=\frac1n\sum\_{i=1}^n\de\_{x\_i}
\end{equation\*}
corresponding to the observed sample $\x=(x\_1,\dots,x\_n)$, where $\de\_x$ is the Dirac delta measure supported on the singleton set $\{x\}$. So, the $Bn$ r.v.'s $x^\*\_{j,i}$ with $j=1,\dots,B$ and $i=1,\dots,n$ are (desirably/approximately) iid each with the distribution $F\_\x$.
Here $n$ is large enough so that the empirical distribution $F\_\x$ be close enough to the true but unknown distribution $F$, and $B$ is very large -- which is affordable because of the large computer power we have nowadays.
Then, for each $j=1,\dots,B$, in the formula $Y=R(\X,F)$ we replace the r.v. $\X$ and the unknown distribution $F$ by the known $\x^\*\_j$ and $F\_\x$, respectively, to get
\begin{equation\*}
y^\*\_1:=R(\x^\*\_1,F\_\x),\dots,y^\*\_B:=R(\x^\*\_B,F\_\x).
\end{equation\*}
Because the empirical distribution $F\_\x$ is close enough to the true but unknown distribution $F$, the empirical distribution
\begin{equation\*}
\frac1B\sum\_{j=1}^B\de\_{y^\*\_j} \tag{1}\label{1}
\end{equation\*}
will be somewhat close to the desired unknown distribution of $Y=R(\X,F)$ -- if the function $R$ is continuous in an appropriate sense. The empirical distribution \eqref{1} is called the *bootstrap distribution of $Y=R(\X,F)$*.
The simplest example of the function $R$ in Efron's paper is given by
\begin{equation\*}
R(\X,F)=\bar X-\int\_\R x\,F(dx),
\end{equation\*}
where $\bar X:=\frac1n\sum\_{i=1}^n X\_i$ and $F$ is the Bernoulli distribution with an unknown parameter.
As noted by Efron, clearly in this case one does not need any simulation to find/estimate the mean and the variance of the distribution of $R(\x^\*\_1,F\_\x)$ -- they are obviously $0$ and $\bar x(1-\bar x)$, respectively, where, of course, $\bar x:=\frac1n\sum\_{i=1}^n x\_i$.
---
Of course, neither the true distribution of $Y=R(\X,F)$ nor its bootstrap distribution will be even approximately normal in general. However, recall that the bootstrap distribution is random (or, at least, pseudo-random), even for a given realization $\x$ of the random sample $\X$ -- because the bootstrap distribution depends on the (pseudo-)random bootstrap samples $\x^\*\_1,\dots,\x^\*\_B$. Therefore, for each realization $\x$ of $\X$, the bootstrap distribution of $R(\X,F)$ (being the empirical distribution \eqref{1} (based on $y^\*\_1,\dots,y^\*\_B$)) will satisfy a central limit theorem for empirical measures -- see e.g. [Dudley](https://projecteuclid.org/journals/annals-of-probability/volume-6/issue-6/Central-Limit-Theorems-for-Empirical-Measures/10.1214/aop/1176995384.full) and subsequent papers.
| 1 | https://mathoverflow.net/users/36721 | 435687 | 176,105 |
https://mathoverflow.net/questions/435679 | 4 | The motivation for this question comes from Theorem 3.3 of the 1995 paper [*Tilings of Triangles*](https://core.ac.uk/download/pdf/82351337.pdf) by M. Laczkovich, which states:
>
> Let $x$ and $y$ be non-zero integers such that $x+2y\neq 0\neq y+2x$. Then there is a positive integer $k$ such that the equilateral triangle can be dissected into $n=|xy(x+2y)(y+2x)k^2|$ congruent triangles.
>
>
>
I am curious about the realizable squarefree parts of such integers, hereafter $s(x,y)$. Since we focus on the squarefree part, it suffices to consider the case where $\gcd(x,y)=1$.
In the range $|x|,|y|<10,000$, the realizable values of $s(x,y)$ are
$$1, 5, 6, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 29, 30, 33,\ldots$$
of which the squarefree integers missing are
$$2, 3, 7, 26, 31, 38, 43, 51, 53,\ldots$$
This sequence is not in OEIS, but I am very far from confident that it is complete; the smallest solution for $s(x,y)=19$ is given by $x=578,y=-225$ and for $s(x,y)=37$ it is $x=5929,y=648$. However, if there are any squarefree integers missing from the image of $s$ then no compatible OEIS sequences exist, even allowing for an initial $0$ term.
I would be curious to see a proof that specific values like $7$ are not realizable by this function, even if it does not generalize to a complete characterization. Alternatively, pointers to open problems that render solving this very difficult or conjectures that would imply certain results here would also be welcome.
---
As an aside, here is a sketch of a rough heuristic argument (not a proof!) that the equation $s(x,y)=k$ should have only finitely many solutions for a fixed $k$ with $x$ and $y$ coprime:
Consider the squarefree parts of $x$, $y$, $2x+y$, and $x+2y$, hereafter $a,b,c,d$ respectively. No prime greater than $3$ can be shared among two of $a,b,c,d$ without violating coprimality, so we have finitely many ways to partition the factors of $k$ among the four values times at most $4^4$ ways to assign them all an additional factor of $1,2,3,$ or $6$. So we are seeking a solution to one of a finite number of equation systems of the form $x=\pm a\cdot p^2, y=\pm b\cdot q^2, x+2y=\pm c\cdot r^2, 2x+y=\pm d\cdot s^2$.
Let's estimate the expected number of solutions to one such equation system with $\frac z2 < |x|,|y|\le z$. We have $O(\sqrt{z})$ choices for each of $x$ and $y$ in this range, hence $O(z)$ choices for the pair. For each such choice of $x$ and $y$, if we model an integer $n$ of having an $O(1/n)$ chance to be square, and assume independence, we'll get an $O(1/z^2)$ chance of success (ie that both $c(x+2y)$ and $d(2x+y)$ ended up being perfect squares) for every such $x,y$ we pick. Adding up, we have (heuristically) an expected $O(1/z)$ total solutions for valid $x,y$ in this range. If we sum this over the ranges $[\frac z2=1,z=2],[\frac z2=2,z=4],[4,8],[8,16],\ldots$ we find that the sum converges. Hence, the expected number of solutions across all the equations, and so the expected number of solutions to $s(x,y)=k$, is finite, though the expected *size* of a solution if one exists is infinite. This accords fairly well with the data so far, including the isolated very large solutions.
(Of course, there might well be substantial correlations or patterns in the factors of $x+2y$ and $2x+y$ that make this model of the terms as independent random draws extremely
bad! But absent reasons to suspect such patterns in one direction or another, I'd guess this is roughly the behavior of this function.)
*Cross-posted from Math StackExchange [here](https://math.stackexchange.com/questions/4558701/squarefree-parts-of-integers-of-the-form-xyx2yy2x).*
| https://mathoverflow.net/users/89672 | Squarefree parts of integers of the form $xy(x+2y)(y+2x)$ | If a given squarefree integer $s$ is given, determining whether there is a pair of integers $(x,y)$ for which $s(x,y) = s$ is a problem about rational points on elliptic curves. In particular, there is a bijection to solutions to
$$
xy(x+2y)(y+2x) = sz^{2}
$$
and rational points on the elliptic curve $E\_{s} : Y^{2} = X^{3} + 5sX^{2} + 4s^{2} X$, where $X = \frac{2sx}{y}$ and $Y = \frac{2s^{2} z}{y^{2}}$. This elliptic curve always has four points on it (namely $(-4s,0)$, $(-s,0)$, $(0,0)$ and the point at infinity) that form a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. If $s = 1$, the torsion subgroup is larger, but if $s > 1$, the existence of an additional rational point (and hence a value of $(x,y)$ with $s(x,y) = s$ with $x + 2y \ne 0$ and $y + 2x \ne 0$) is equivalent to at least one of $E\_{s}$ or $E\_{-s}$ having positive rank.
These curves each have rank zero for $s \in \{ 2, 3, 7, 26, 31, 38, 51 \}$ and hence it is impossible to solve $s(x,y) = s$ for these values of $s$. For $s = 43$, $E\_{-43}$ has positive rank and one rational point has $X = \frac{926171883}{29604481}$. This produces $x = -21538881$ and $y = 59208962$ for which
$$
xy(x+2y)(y+2x) = -43 \cdot 215287751668080^{2}.
$$
A solution for $s = 53$ can be found with $x = -28322$ and $y = 223925$ which gives
$$
xy(x+2y)(y+2x) = -53 \cdot 2897871340^{2}.
$$
The most complicated solution I found (within the range of [Cremona's database of elliptic curves](https://www.lmfdb.org/EllipticCurve/Q/)) was with $s = 139$ where it is necessary to take $x = -4659047399556225$ and $y = 15599782505124098$.
It is worth noting that the heuristic you propose at the end of your question is actually not correct, and if $E\_{s}$ or $E\_{-s}$ has positive rank, then there will be infinitely many solutions to $s(x,y) = s$, but it's also the case that elliptic curves "just barely" have infinitely many rational points.
Finally, all of these elliptic curves are quadratic twists of $E\_{1} : Y^{2} = X^{3} + 5X^{2} + 4X$, and for this reason, there is the possibility of an analogue of [Tunnell's theorem](https://en.wikipedia.org/wiki/Tunnell%27s_theorem) that would tell you (conditional on the Birch and Swinnerton-Dyer conjecture) for precisely which values of $s$ there is a solution and for which there isn't.
| 7 | https://mathoverflow.net/users/48142 | 435688 | 176,106 |
https://mathoverflow.net/questions/435589 | 1 | This is a follow-up to [this question of mine](https://mathoverflow.net/questions/428242/isometric-embeddings-of-c-0-into-metric-spaces).
It is well-known that the Banach space $\ell\_1$ does not contain any *isomorphic* copies of $c\_0$. One can even go further and show that $\ell\_1$ does not contain any *bilipschitz* copy of $c\_0$. Is it however known whether $\ell\_1$ does not contain any bilipschitz copy of the unit ball $B\_{c\_0}$ of $c\_0$?
| https://mathoverflow.net/users/15860 | Bilipschitz embedding of the unit ball of $c_0$ into $\ell_1$ | Yes, this is known. Raynaud showed that $B\_{c\_0}$ does not uniformly (in particular, bilipschitz) embed into any stable Banach space. $\ell\_1$ is stable.
* [Yves Raynaud, Espaces de Banach superstables, distances stables et homeomorphismes uniformes](https://link.springer.com/article/10.1007/BF02763170)
**Added later.**
In the comments Bill pointed out an easier proof. Here is an elementary argument that doesn't use differentiation theory.
Suppose $f:B\_{c\_0}\to \ell\_1$ with
$$\frac1C\|x-y\|\le \|f(x)-f(y)\|\le C\|x-y\|$$
Let $(s\_i)$ be the summing basis of $c\_0$. For $\vec{n}=(n\_1, \ldots, n\_k)$, let $x(\vec{n})=\frac1k \sum\_{i=1}^k s\_{n\_i}$.
By standard stabilization argument you can find $M\subseteq \mathbb N$ so that for all $\vec{n}\in M^k$, $f(x(\vec{n}))\approx h\_0+\sum\_{i=1}^k h\_{n\_i}$ where $h\_i$'s are successive blocks in $\ell\_1$ and the support of $h\_i$'s go to infinity as $n\_i\to \infty$ (except $h\_0$). Use Ramsey to stabilize the norms of $h\_i$'s. Considering $\vec{n}<\vec{m}$, $\|x(\vec{n})-x(\vec{m})\|\_{c\_0}=1$ we see that
$$\frac1C\le \|f(x(\vec{n}))-f(x(\vec{m}))\|\approx \|\sum\_{i=1}^k h\_{n\_i}-\sum\_{i=1}^k h\_{m\_i}\|=\sum\_{i=1}^k (\|h\_{n\_i}\|+\|h\_{m\_i}\|).$$
Now consider $n\_1<m\_1<\ldots<n\_k<m\_k$ then $\|x(\vec{n})-x(\vec{m})\|\_{c\_0}=\frac1k$. The blocks $h\_{n\_1}<h\_{m\_1}<\ldots h\_{n\_k}<h\_{m\_k}$ are interlacing but the norm of the sum of the difference in $\ell\_1$ is the same as above. Then
$$\sum\_{i=1}^k (\|h\_{n\_i}\|+\|h\_{m\_i}\|)=\|\sum\_{i=1}^k h\_{n\_i}-\sum\_{i=1}^k h\_{m\_i}\|\approx\|f(x(\vec{n}))-f(x(\vec{m}))\|\le \frac{C}{k}.$$
which yields a contradiction for large $k$.
| 7 | https://mathoverflow.net/users/3675 | 435692 | 176,108 |
https://mathoverflow.net/questions/435690 | 2 | $\DeclareMathOperator\GL{GL}$Let $G$ be a $2$-generator pro-$p$-group of finite rank, i.e. it is isomorphic to a closed subgroup of $\GL\_d(\mathbb{Z}\_p)$ for some integer $d$. Assume that $G$ is torsion-free. Recall that the dimension $\dim(G)$ of $G$ as a $p$-adic analytic group can be described as $d(U)$ where $U$ is any uniform open pro-$p$-subgroup of $G$ and $d(U)$ denotes the minimal cardinality of a topological generating set for $U$.
**Question:** Is $\dim(G)=2$?
| https://mathoverflow.net/users/492970 | The dimension of a torsion-free $p$-adic analytic group generated by two generators | A pro-$p$ group $G$ has $\textit{lower rank}$ $r$ if $r$ is minimal such that every open subgroup of $G$ contains an open subgroup generated by at most $r$ elements. Lubotzky and Mann showed that the lower rank of a $p$-adic analytic pro-$p$ group is the number of generators of its associated Lie algebra. On the other hand, its dimension is the dimension of the Lie algebra. Thus, for instance, every open subgroup of $SL\_d(\mathbb{Z}\_p)$ (including torsion free ones) has lower rank $2$ and dimension $d^2-1$.
| 3 | https://mathoverflow.net/users/5034 | 435696 | 176,109 |
https://mathoverflow.net/questions/435705 | 1 | I am looking for a formula giving the asymptotic expansion of the renewal equation when there is exponential growth (for lack of better terms).
Consider the renewal equation, for an unknown $M(t)$:
\begin{equation}
M(t)=a\int\_0^\infty M(t-\tau)f(\tau)\mathrm{d}\tau
\end{equation}
where $f(\tau)$ is a probability density and $a>0$.
When $a=1$, $M(t)$ represents the average value of a renewal process, and it is possible to obtain an asymptotic expansion of it as $t\to \infty$:
\begin{equation}
\lim\_{t\to\infty}\left [ M(t) - \frac{t}{\mu}\right]=\frac{\sigma^2-\mu^2}{2\mu^2}
\end{equation}
See for example Chap 6. A first course in Stochastic Process by S. Karlin and HM. Taylor.
However, when $a>1$ the arguments used to derive this expansion can no longer be used (conditioning on the first renewal event), because the quantity is growing exponentially.
What would be the asymptotic expansion of $M(t)$ when $a>1$?
| https://mathoverflow.net/users/420641 | Asymptotic expansion of the renewal function for an exponential growing population | For any $\theta>0$, if you substitute $M(t)=e^{\theta t}N(t),$ then the equation becomes
$$
N(t)=\int\_0^\infty N(t-\tau)ae^{-\theta\tau}f(\tau)d\tau.
$$
If $a>1$, then there is, by continuity, exactly one $\theta$ such that $ae^{-\theta \tau}f(\tau)$ is a probability density. For this $\theta$, the problem is reduced to the $a=1$ case.
| 0 | https://mathoverflow.net/users/56624 | 435712 | 176,112 |
https://mathoverflow.net/questions/424595 | 5 | I'm reading Casselman's notes "[Introduction to the theory of admissible representations of p-adic reductive groups](https://personal.math.ubc.ca/%7Ecass/research/pdf/p-adic-book.pdf)". In chapter 4 "The asymptotic behavior of matrix coefficients", the main result of this chapter states that for some torus elements $a$ (in a sense, those who are "close to zero"), the matrix coefficients $\left\langle\pi(a)v,\tilde{v}\right\rangle$ equlas to $\left\langle\pi\_N(a)u,\tilde{u}\right\rangle\_N$, where $\pi$ is an admissible representation, $\pi\_N$ is its Jacquet module with respect to the unipotent radical $N$, $v\in V\_\pi$, $\tilde{v}\in \widetilde{V}\_\pi$, and $u$ (respectively, $\tilde{u}$) is the image of $v$ (respectively, $\tilde{v}$) in $V\_{\pi,N}$ (respectively, $\widetilde{V}\_{\pi,N}$).
In his notes "[Remarks on Macdonald’s book on p-adic spherical functions](https://personal.math.ubc.ca/%7Ecass/research/pdf/Macdonald.pdf)", Casselman refers to this result (Theorem 6.7) and then makes the following remark: "This result says that any matrix coefficient is asymptotically equal to an $A$-finite expression".
Apparently, Casselman means that $\left\langle\pi\_N(a)u,\tilde{u}\right\rangle\_N$ is a sum of $A$-finite functions. I can't figure out why this is true.
This question is related to the old question [Reference request - Jacquet module and asymptotic of matrix coefficients](https://mathoverflow.net/questions/90367/reference-request-jacquet-module-and-asymptotic-of-matrix-coefficients). I would very much appreciate a rather detailed explanation or a good reference for both the p-adic case and the Archimedean case.
| https://mathoverflow.net/users/123363 | Asymptotic behavior of matrix coefficients | For the $p$-adic case, the idea is as follows (thanks to [Elad](https://mathoverflow.net/users/103908/darkl) for pointing out the direction).
Recall (from the unpublished notes by Casselman, [*Introduction to the theory of admissible representations of p-adic reductive groups*](https://personal.math.ubc.ca/%7Ecass/research/pdf/p-adic-book.pdf)) that $k$ is a non-Archimedean locally compact field, $G$ is a group of $k$-rational points of a reductive algebraic group defined over $k$, and $P$ is a parabolic subgroup of $G$ with Levi decomposition $P=MN$.
We note that it is sufficient to show that the function $x\mapsto\left\langle \pi\_{N}(x)u,\tilde{u}\right\rangle \_{N}$ is an $A\_M$-finite function ($A\_M$ is the center of the Levi part $M$). Indeed, From Jacquet and Langlands (Lemma 8.1 in [*Automorphic Forms on $\operatorname{GL}(2)$: Part I, volume 114.*](https://doi.org/10.1007/BFb0058988) Springer, 2006), one can deduce that the space of continuous finite functions on the locally compact abelian group $A\_M$ is spanned by functions of the form
\begin{equation}
\prod\_{i=1}^{r}\chi\_i(a\_i)\left\lvert a\_i\right\rvert^{p'\_i}\log\_q ^{p\_i}\left\lvert a\_i\right\rvert,
\end{equation}
where $r$ is such that $A\_M\cong k^r$, $\left(p'\_{1},\dotsc,p'\_{r}\right)\in \mathbb{R}^r$, $\left(p\_{1},\dotsc,p\_{r}\right)\in \mathbb{Z}^r\_{\ge 0}$, and for all $1\leq i\leq r$, $\chi\_i:k^\times \to \mathbb{C}^\times$ are unitary characters.
Now, in order to prove that the function $x\mapsto\left\langle \pi\_{N}(x)u,\tilde{u}\right\rangle \_{N}$ is an $A\_M$-finite function, we use the following technical lemma.
* **Lemma.** Let $R$ be a group with center $Z\left(R\right)\cong K\times\mathbb{Z}^{r}$, where $K$ is a compact group. Let $\left(H,\sigma\right)$ be a (complex) smooth $R$-module of finite length and let $v\in H$. Then, the $Z\left(R\right)$-module generated by $v$ is finite dimensional.
The Jacquet module is a smooth $G$-module of finite length (See Theorems 3.3.1 and 6.3.10 in the abovementioned unpublished notes by Casselman). Hence, we apply the lemma with $R=M$ (with $Z(R)=A\_M$), $H=V\_N$, $\sigma=\pi\_N$, and $v=u\in V\_{N}$. This gives that $U:=\left\{ \pi\_{N}\left(a\right)u|\ a\in A\_{M}\right\} $
is of finite dimension. Let $\left\{ \pi\_{N}\left(b\_{1}\right)u,\dotsc, \pi\_{N}\left(b\_{\ell}\right)u\right\}$
be a basis of $U$. Then,
\begin{equation}
\pi\_{N}\left(a\right)u=\sum\_{i=1}^{\ell}c\_{i}(a)\pi\_{N}\left(b\_i\right)u.
\end{equation}
Therefore,
\begin{equation}
\left\langle \pi\_{N}\left(ma\right)u,\tilde{u}\right\rangle \_{N}=\sum\_{i=1}^{\ell}c\_{i}\left(a\right)\left\langle \pi\_{N}\left(m b\_i\right)u,\tilde{u}\right\rangle \_{N}.
\end{equation}
For more details, as well as the proof of the lemma, see [Hazan - A Note on the Asymptotic Expansion of Matrix Coefficients over $p$-adic Fields](https://arxiv.org/abs/2211.15822).
For the Archimedean case, one can find an asymptotic expansion of the matrix coefficients, as a finite sum of finite functions, in Casselman's paper *Jacquet modules for real reductive groups* (see the Lemma in Section 5, [Proceedings of the International Congress of Mathematicians (Helsinki, 1978)](https://www.mathunion.org/fileadmin/ICM/Proceedings/ICM1978.2/ICM1978.2.ocr.pdf)).
| 2 | https://mathoverflow.net/users/123363 | 435724 | 176,117 |
https://mathoverflow.net/questions/435723 | 3 | The [Dirac delta function](https://en.wikipedia.org/wiki/Dirac_delta_function) appears in the Sokhotsky formula,
$$\text{Im}\lim\_{\epsilon\to 0^+} \frac{1}{x-i\epsilon} = \pi\delta(x),$$
to be understood in the integral sense
$$\text{Im}\lim\_{\epsilon\to 0^+} \int \frac{f(y)}{y-x-i\epsilon}dy=\pi f(x),$$
for a real valued function $f(x)$.
I stumbled on an identity that has a similar flavour,
$$\lim\_{\epsilon\to 0^+}\int\_x^b \frac{\epsilon f(y)}{(y-x)^{1-\epsilon}} dy=f(x).\label{1}\tag{$\ast$}$$
The upper limit $b>x$ of the integral is arbitrary, one may send it to infinity if $f(x)$ has compact support.
A corollary is
$$
\lim\_{\epsilon\to 0^+}\int\_a^b \frac{\epsilon f(x)}{[(b-x)(x-a)]^{1-\epsilon}}\,dx=\frac{f(a)+f(b)}{b-a}.$$
All of this can be interpreted as a delta function representation in terms of the unit step function $\theta(x)$,
$$\lim\_{\epsilon\to 0^+} \frac{\epsilon\theta(x)}{x^{1-\epsilon}}=\delta(x),\tag{$\ast\ast$}$$
acting on compactly supported functions.
---
**Q:** One can readily check the formula \eqref{1} for polynomial functions $f(x)$. Is there a more comprehensive derivation? Is this representation of the delta function known?
| https://mathoverflow.net/users/11260 | Representation of the Dirac delta function | As usual in such examples, there is no need to integrate against a test function. One can simply use the fact that if a sequence (or net) of distributions converges in the distributional sense, then so does the one obtained by differentiating term by term. In particular, this applies when the sequence consists of functions which converge in pretty well any sensible classical sense, e.g., locally $L^1$ as in the case in point, that of the functions which are defined as $x^{\epsilon}$ on the positive real axis, and $0$ elsewhere. They converge to the Heaviside function and we can differentiate to obtain the required result.
Most of the examples of $\delta$-sequences in the literature can be verified in this way: consider the terms’ primitives and show that they converge to the Heaviside function. The result then follows as above.
The first example (Sokhotsky) in the question can be proved in one line, after integrating $\dfrac{\epsilon}{x^2+\epsilon^2}.$
| 12 | https://mathoverflow.net/users/495655 | 435730 | 176,120 |
https://mathoverflow.net/questions/435729 | 1 | The sum I am looking for is the following sum as $M \to \infty$:
$$ L(\omega) = \sum\_{m = 1}^{M} \frac{\sin\left( N \frac{\omega\_m - \omega}{2} \right)}{\sin\left( \frac{\omega\_m - \omega}{2} \right)} \cos\left(N \frac{\omega\_m - \omega}{2} + \beta\_m \right) $$
where
* $\omega\_m$ is a random number from a Gaussian distribution having the parameters (mean $\mu$ and variance $\sigma^2$).
$$ \omega\_m \sim \mathcal{N}(\mu, \sigma^2) $$
* the $\beta\_m$ are random numbers drawn from a uniform distribution from $-\pi$ to $+\pi$,
$$ \beta\_m \sim \mathcal{U}[-\pi, +\pi] $$
| https://mathoverflow.net/users/489481 | Is it possible to sum this analytically in any way? | Since the average over $\beta$ will give a vanishing expectation value of $L$, let me omit it for now and set $\beta=0$. I will also simplify the question by setting $\omega=\mu=0$ and $\sigma=1$. Then the expectation value of $L$ has a compact expression
$$\mathbb{E}[L(0)]=M\;\int\_{-\infty}^\infty \frac{dx}{\sqrt{2\pi}}e^{-x^2/2} \frac{\sin\left( N x/2 \right)}{\sin\left( x/2\right)} \cos\left(N x/2 \right)=Me^{-\frac{1}{8} (2 N-1)^2} \sum \_{j=1}^N e^{-\frac{1}{2} (j-1) (j-2 N)}.$$
| 2 | https://mathoverflow.net/users/11260 | 435734 | 176,123 |
https://mathoverflow.net/questions/435704 | 0 | Given the function
$$
E(M) = \sum\_{i=1}^N \sum\_{a=1}^K \left( M\_{ia} \cdot \left\lVert\sum\_{i=1}^N M\_{ia}\cdot x\_i\right\rVert\_2^2 \right)
$$
$x$ is a given constant matrix, $x\_i$ is a the $n\_\text{th}$ column of that.
$M\_{ia} \in \{0,1\}$ and $\sum\_{i=1}^N M\_{ia}=\frac{N}{K} $, and we also have $\sum\_{a=1}^N M\_{ia}=1 \\\\$.
The question is, can $E$ of $M$ be further simplified to a form using only quadratic and linear terms of $M$?
Again, thank you so much.
| https://mathoverflow.net/users/495775 | Can this function be simplified to use only quadratic, linear terms of M with given conditions? | Yes, $$E(M) = \frac{N}{K} \lVert X M \rVert\_F^2$$
writing $X$ for the matrix $\begin{bmatrix} x\_1 & \dots & x\_N \end{bmatrix}$.
Note $\sum\_{i=1}^N M\_{ia}x\_i = X M e\_a$ where $e\_a$ is the $a$th standard basis vector. The constraints imply $M$ is a binary $N \times K$ matrix with a single nonzero per row and each column having the same number $m = N/K$ of nonzeros. Let the single nonzero in row $i$ be in column $j\_i$. Only the term with $a=j\_i$ is nonzero in the sum over $a$ so that $E(M) = \sum\_{i=1}^N \lVert X M e\_{j\_i} \rVert\_2^2$. Finally note that each column index $j$ will be visited exactly $m$ times in the sum over $i$, so that $E(M) = m \sum\_{j=1}^K \lVert X M e\_j \rVert\_2^2 = m \lVert X M \rVert\_F^2$.
| 0 | https://mathoverflow.net/users/70005 | 435740 | 176,128 |
https://mathoverflow.net/questions/414286 | 3 | I find the $d$-separation criterion (see, e.g., [Theorem 2 here](https://ftp.cs.ucla.edu/pub/stat_ser/r130-reprint.pdf); note however the preceding definition, which basically means we are treating discrete random variables) a really useful sufficient criterion for conditional independence of random variables. However, I am struggling to find a general version for arbitrary random variables, say with values in a Polish space (where conditional independence is defined via the respective disintegrations).
**My question:** Is there a version of $d$-separation that treats arbitrary Polish spaces valued random variables, or at least arbitrary $\mathbb{R}^d$ valued random variables?
*To make the post self-contained:* Let $\pi \in \mathcal{P}(S\_1\times S\_2 \times \dots \times S\_N)$ be a Borel probability measure on the polish space $S\_1 \times ... \times S\_N$. Assume there is a directed acyclic graph with nodes $\{1, 2, \dots, N\}$, where without loss of generality the nodes are increasing in topological order (i.e., there is no edge pointing from $k$ to $l$ for $k > l$). Denote by $pa(k)$ the set of parents of a node $k$. Assume $\pi$ has a regular disintegration of the form
$$
\pi = \pi\_{1} \otimes \pi\_{2, pa(2)} \otimes \pi\_{3, pa(3)} \otimes \dots \otimes \pi\_{N, pa(N)},
$$
where $\pi\_{k, pa(k)}$ is a stochastic kernel mapping from $(\otimes\_{j \in pa(k)} S\_j)$ to $\mathcal{P}(S\_k)$.
Note that this disintegration naturally encodes certain conditional independencies. Indeed, if $(X\_1, X\_2, \dots, X\_N) \sim \pi$ is a random variable with distribution $\pi$, the given disintegration means that $X\_k$ is conditionally independent of $\{X\_j : j \in \{1, ..., k-1\}\backslash pa(k)\}$ given $\{X\_j : j \in pa(k)\}$.
On the other hand, this disintegration also implies many other conditional independences between the variables, and $d$-separation is a criterion to identify all of them (atleast for discrete spaces). The definition of $d$-separation is as follows: Take disjoint $A, B, C \subset \{1, \dots, N\}$. We say that $C$ $d$-separates $A$ and $B$, if there are no paths from a node $i \in A$ to a node $j \in B$ which is not blocked by a node in $C$. A path between nodes is blocked if either of the following criteria is satisfied:
1. there is a chain element $a \rightarrow b \leftarrow c$ contained in the path such that $b \not \in C$ and none of the descendants of $b$ is in $C$.
2. there is a chain element $a \rightarrow b \rightarrow c$ or $a \leftarrow b \rightarrow c$ contained in the path such that $b \in C$.
Then the fundamental result we want is that if $C$ $d$-separates $A$ and $B$, then $\{X\_j : j\in A\}$ and $\{X\_j : j \in B\}$ are conditionally independent given $\{X\_j : j \in C\}$ (all under $\pi$, of course).
| https://mathoverflow.net/users/106046 | General version of $d$-separation | Yes, there are more general versions of the $d$-separation criterion, in particular also for standard Borel spaces (i.e. Polish spaces).
For completeness, the classical version of $d$-separation (i.e. on discrete random variables) has been generalized several times to the best of my knowledge.
### 1. $d$-separation for absolute continuous probability distributions
Lauritzen et al. (see [Independence Properties of Directed Markov Fields](https://onlinelibrary.wiley.com/doi/epdf/10.1002/net.3230200503)) showed that the $d$-separation criterion holds for absolute continuous probability distributions with respect to a product measure. More specifically, let $\mathcal{H}\_1, \ldots, \mathcal{H}\_n$ be measurable spaces and $P$ a probability distribution on $\prod\_{i=1}^{n} \mathcal{H}\_i$ such that
$$P(X\_1 \in E\_1, \ldots, X\_n \in E\_n) = \int\_{x\_1 \in E\_1} \cdots \int\_{x\_n \in E\_n} f(x\_1,\ldots, x\_n) \ \mu\_1(\mathrm{d} x\_1) \otimes \cdots \otimes \mu\_n(\mathrm{d} x\_n)$$
where $E\_i$ is measurable, $f$ being a density and $\mu\_i$ being a measure on $\mathcal{H}\_i$. In this situation, a distribution $P$ is called **compatible** with a DAG $G$ (with vertices $\{1, \ldots, n\}$) if the density $f$ factorizes as
$$f(x\_1, \ldots, x\_n) = \prod\_{i=1}^{n} f(x\_i | x\_{\textsf{pa}(i)}) \quad \mu\textrm{-a.e.}$$
where $\textsf{pa}(i)$ is the set of parents of vertex $i$ and $f(x\_i | x\_{\textsf{pa}(i)})$ is the conditional density of $X\_i$ given its parents.
Theorem 1 in the reference shows that the following two statements are equivalent
* $P$ is compatible with $G$
* $P$ satisfies the global Markov property, i.e. for every disjoint triple of subsets $A,B,C \subseteq \{1, \ldots, n\}$ such that $C$ d-separates $A$ and $B$ in $G$, we have that
$$f(x\_A, x\_B, x\_C) = f(x\_A | x\_C) \cdot f(x\_B | x\_C) \cdot f(x\_C) \quad \mu\textrm{-a.e.},$$
i.e. $\{X\_i: i\in A\}$ is cond. independent of $\{X\_j: j \in B\}$ given $\{X\_k: k\in C\}$.
### 2. $d$-separation for standard Borel spaces (i.e. Polish spaces)
A version of the $d$-separation criterion (with several other equivalent conditions) for standard Borel spaces has been proven in by P. Forré and J. Mooij ([Markov Properties for Graphical Models with Cycles and Latent Variables](https://arxiv.org/abs/1710.08775)). Your desired version is stated in Theorem 3.2.1. ($3. \Longleftrightarrow 6.$)
In this setting a probability distribution on standard Borel spaces $\prod\_{i=1}^n \mathcal{H}\_i$ given by
$$P(X\_1 \in E\_1, \ldots, X\_n \in E\_n) = \int\_{x\_1 \in E\_1} \cdots \int\_{x\_n \in E\_n} P(\mathrm{d} x\_1, \ldots, \mathrm{d} x\_n)$$
with $E\_{i} \in \mathcal{H}\_i$ is **compatible** with a DAG $G$ if
$$P(\mathrm{d} x\_1, \ldots, \mathrm{d} x\_n) = \prod\_{i = 1}^n P(\mathrm{d} x\_i | x\_{\textsf{pa}(i)})$$
where $P(\mathrm{d} x\_i | x\_{\textsf{pa}(i)})$ is the regular conditional probability distribution of $X\_i$ given its parents in $G$.
In Theorem 3.2.1. ($3. \Longleftrightarrow 6.$) it is proven that the following two statements are equivalent:
* $P$ is compatible with $G$
* $P$ satisfies the global Markov property, i.e. for every disjoint triple of subsets $A, B, C \subseteq \{1,\ldots, n\}$ such that $A$ is $d$-separated from $B$ by $C$, we have that $\{X\_i: i\in A\}$ is cond. independent of $\{X\_j: j \in B\}$ given $\{X\_k: k\in C\}$ where conditional independence is defined via regular conditional probabilities.
### 3. $d$-separation on Markov categories
As already mentioned in the question, conditional independence is defined via regular conditional probability distributions in all examples considered.
Recently, Tobias Fritz and me showed the $d$-separation criterion in the more abstract setting of Markov categories (see [The $d$-Separation Criterion in Categorical Probability](https://arxiv.org/abs/2207.05740)). There we have shown that whenever you can do an operation that behaves like Bayesian disintegration (i.e. whenever conditionals exist), the $d$-separation criterion applies.
This result includes the $d$-separation criterion on standard Borel spaces and finite probability distributions as specific instances. But it also includes non-probabilistic settings, where $d$-separation applies.
For example, it implies that the $d$-separation criterion also applies to finite possibilistic networks, i.e. networks where a conditional $f(x | a)$ specifies whether $x \in X$ is possible or impossible given $a \in A$ by setting $f(x | a) = 1$ or $f(x|a) = 0$.
| 2 | https://mathoverflow.net/users/493730 | 435741 | 176,129 |
https://mathoverflow.net/questions/435685 | 5 | Let $A$ be a homotopy ring spectrum. Then the homology theory $A\_\ast : Spectra \to GrAb$ lifts to a homology theory valued in $GrMod(\pi\_\ast A)$. If $A$ is homotopy commutative, then this functor $A\_\ast : Spectra \to GrMod(\pi\_\ast A)$ is lax monoidal. But it makes sense to ask for a lax monoidal structure on this functor even if $A$ is not homotopy commutative, so long as $\pi\_\ast A$ is graded-commutative. And the answer can be *yes* --e.g. at the prime 2, Morava $K$-theory $K(n)$ is not homotopy commutative, but $K(n)\_\ast$ is strong monoidal.
**Question:** Let $A$ be a homotopy ring spectrum such that $\pi\_\ast A$ is graded-commutative. Then is the functor $A\_\ast : Spectra \to GrMod\_{\pi\_\ast A}$ lax monoidal? Even lax symmetric monoidal?
| https://mathoverflow.net/users/2362 | If $\pi_\ast A$ is graded-commutative, then is $A_\ast$ a lax monoidal functor? | As pointed out in the comments, the functor $A\_\*$ cannot in general be lax symmetric monoidal without making some alterations.
Here is an incomplete discussion of when $A\_\*$ can be lax monoidal.
The first observation is that, for any homotopy associative ring spectrum $A$, the functor $A\_\*$ naturally takes values in $\pi\_\* A$-bimodules. The left and right actions of $A$ on itself produce natural left and right actions of $A$ on $A \otimes X$ for any $X$, and lax monoidality of $\pi\_\*$ show that $A\_\* X$ is then a $\pi\_\* A$-bimodule.
The second observation is that this makes the functor $A\_\*$ a lax monoidal functor to the category of $\pi\_\* A$-bimodules. The homotopy associativity of $A$ ensures that the two composites $A \otimes A \otimes A \to A \otimes A \to A$ are homotopic, and so for any $X$ and $Y$ the two composites
$$
(A \otimes X) \otimes A \otimes (A \otimes Y) \to (A \otimes X) \otimes (A \otimes Y) \to A \otimes (X \otimes Y)
$$
are homotopic. The lax monoidality of $\pi\_\*$ then tells us that the two composites
$$
A\_\* X \otimes \pi\_\* A \otimes A\_\* Y \to A\_\*(X \otimes Y)
$$
are equal, establishing $\pi\_\* A$-bilinearity. However, tracking which maps actually appear on the spectrum level, this specifically uses the right $\pi\_\* A$-module structure on $A\_\* X$ and the left $\pi\_\* A$-module structure on $A\_\* Y$.
We now assume that the coefficient ring $\pi\_\* A$ is graded-commutative. We would like to show that, for any $X$, the left and right module structures "coincide": $a \cdot x = \pm x \cdot a$ for any $a \in \pi\_\* A$ and $x \in A\_\* X$ according to the Koszul sign rule. If we can do this, then bimodule bilinearity collapses to module bilinearity.
Suppose $x \in A\_d X$ comes from a map $S^d \to A \otimes X$. We can express $A$ as a filtered hocolim of finite spectra $A\_i$, and get a lift to a map $S^d \to A\_i \otimes X$, with an adjoint map $DA\_i \otimes S^d \to X$ using Spanier-Whitehead duality. This gives us a lift of $x$ to a factorization
$$
S^d \to \xrightarrow{\eta} A \otimes (DA\_i \otimes S^d) \to A \otimes X.
$$
Therefore, it suffices to check this in the "universal" cases where $x \in A\_\* (DA\_i)$ comes from the canonical map $\eta: S^0 \to A \otimes DA\_i$.
This lifts to the limit over $i$, however: the canonical unit map $S^0 \to map(A,A)$ to the function spectrum. In these terms, we are asking if the canonical element $1 \in [A,A]$ is sent to the same element under post-multiplication by $a$ on either the left or the right; or equivalently, if "multiply by $a$ on the left" and "multiply by $a$ on the right" are always homotopic maps $A \to A$, rather than merely giving equal maps $\pi\_\* A \to \pi\_\* A$.
---
An example of this can be constructed from the following differential graded algebra. Start with the discrete ring $\Bbb Z[x]$. We first form a bimodule with a single generator $y$, annihilated by $x$ on the left and right; we form a second bimodule with a single generator $z$, annihilated by $x^2$ on the left and $x$ on the right (so it has a basis $\{z,xz\}$). Form the mapping cone $M$ of the map $z \mapsto 2y$: it has $H\_0 = \Bbb Z/2$, generated by $y$, and $H\_1 = \Bbb Z$, generated by $xz$. Let $A$ be the square-zero extension $\Bbb Z[x] \times M$ by this differential graded bimodule $M$. The coefficient ring is graded-commutative because almost all products are zero. However, in mod-$2$ homology we see the element $z$ which no longer commutes with the generator $x$ from $A\_\*$.
| 6 | https://mathoverflow.net/users/360 | 435747 | 176,131 |
https://mathoverflow.net/questions/435715 | 3 | Let us consider a Lie group $G$ with Lie algebra $\mathfrak{g}$ and let $L\mathfrak{g} = C^\infty(S^1, \mathfrak{g})$ the Lie algebra of the loop group $LG$.
My question is about continuous Lie algebra 2-cocycles on $L\mathfrak{g}$.
It is well-known (see, e.g., Prop. 4.2.4 in Pressley-Segal "Loop groups") that if $G$ is semisimple and compact, the only continuous $G$-invariant 2-cocycles are of the form
$$\omega(X, Y) = \int\_{S^1} b(X(t), Y^\prime(t))dt, $$
where $b$ is some (necessarily symmetric) $G$-invariant bilinear map on $\mathfrak{g}$.
If I understand correctly, the proof in Pressley-Segal does not use the compactness assumption (which is a general assumption throughout the entire chapter there), but the compactness is used in the proof that all continuous 2-cocycles on $L\mathfrak{g}$ are cohomologous to a $G$-invariant one (this follows from averaging over $G$, which only works in the compact case).
**Question: What is known if $G$ is a non-compact semisimple Lie group?**
Still, all continuous $G$-invariant cocycles are of the form given above, but now there may be non-trivial cohomology classes that are not represented by a $G$-invariant one.
**More precisely: What are examples of non-trivial classes in $H^2\_c(L\mathfrak{g}, \mathbb{R})$ that are not represented by a $G$-invariant one, where $G$ is some semisimple Lie group?**
| https://mathoverflow.net/users/16702 | Non-invariant forms on loop Lie algebra of semisimple Lie group | Thanks to Yves Cornulier, for suggesting to look at the paper of Neeb and Wagemann. After reading Example 6.2 of that paper (arxiv version), I think the answer to my question is that in fact all 2-cycles have a representative of the form in my original post. In other words, the assumption of $G$-invariance in the proposition from the book of Pressley-Segal can be dropped (at least up to replacing the cocycle with a cohomologous one).
Explicitly, set $A = C^\infty(S^1)$.
By the results of Neeb and Wagemann, any cocycle $\omega$ on $L \mathfrak{g} = A \otimes \mathfrak{g}$ is defined by two continuous linear maps, $$f\_1 :\Lambda^2 (A) \otimes \mathrm{Sym}^2(\mathfrak{g}) \to \mathbb{R}, \qquad f\_2 : A \otimes Z\_2(\mathfrak{g}) \to \mathbb{R}.$$
(In general, there is also $f\_3$, but that does not occur in this special case.)
Moreover, we can also ignore $f\_2$, because $H^2(\mathfrak{g}) = 0$ in the semisimple case, hence the $f\_2$ part corresponds to a coboundary.
The map $f\_1$ corresponds to a map
$$ \tilde{f}\_1: A \times A \to \mathrm{Sym}(\mathfrak{g})^{\mathfrak{g}}$$
and the corresponding cocycle $\omega$ is then given by
$$ \omega(X, Y) = \sum\_{ij=1}^n \tilde{f}\_1(X^i, Y^j)(b\_i, b\_j),$$
where $b\_1, \dots, b\_n$ is a basis for $\mathfrak{g}$, and we expanded $X = \sum\_i X^i b\_i$, $Y = \sum\_j Y^j b\_j$.
Now, the condition on $f\_1$ is that there exist a continuous linear map $f\_1^\flat : \Omega^1(S^1) \to \mathrm{Sym}^2(\mathfrak{g})^{\mathfrak{g}}$ with
$$ \tilde{f}\_1(a, b) = f\_1^\flat(a \,db - b \,da) $$
and
$$ \Gamma(f\_1^\flat(da)) = 0$$
for all $a \in A$ (in general there is a term depending on $f\_2$ on the right hand side and such cocycles are called "coupled", but this term vanishes in our case).
Here $\Gamma: \mathrm{Sym}^2(\mathfrak{g})^{\mathfrak{g}} \to Z\_3(\mathfrak{g})$ is the Koszul map, which is injective for semisimple Lie algebras, hence $f\_1^\flat(da) = 0$.
Hence we are looking for 1-currents $T$ on $S^1$ that are co-closed, $\partial T = 0$, and it is well-known that the space of such is one-dimensional; up to a scalar, they are of the form
$$ T(\alpha) = \int\_{S^1} \alpha. $$
Hence $f\_1^\flat = T \otimes b$ for some $b \in \mathrm{Sym}^2(\mathfrak{g})^{\mathfrak{g}}$ and assembling $\omega$ for this $f\_1^\flat$ yields the cocycle from the original post.
| 1 | https://mathoverflow.net/users/16702 | 435748 | 176,132 |
https://mathoverflow.net/questions/433646 | 3 | Consider a simple closed curve $\gamma$ in $\mathbb R^3$. Suppose that $\gamma$ has length $\ell$ and contains a line segment $s$ of length $k<\ell/2$. Let $\Sigma$ be a surface with boundary $\gamma$ of minimum area. I'm interested in knowing what such curve $\gamma$ causes the area of $\Sigma$ to be largest possible.
I suspect that the area of $\Sigma$ is largest possible when $\gamma$ is the union of $s$ and a circular arc of length $\ell-k$, in which case $\Sigma$ is a *circular segment*. Is this known, or does anyone have an idea how it can be (dis)proved?
If $\gamma$ is not required to contain a line segment (i.e., the case $k=0$), it is well known that $\Sigma$ has maximum area when $\gamma$ is a circle, so that agrees with my claim.
| https://mathoverflow.net/users/493914 | Isoperimetric inequality for minimal surfaces bounded by space curves containing a line segment | The area minimizing disc $\Sigma$ with boundary $\gamma$ has nonpositive curvature in the sense of Alexandrov.
Applying Reshetnyak majorization theorem (see [9.56](https://arxiv.org/pdf/1903.08539v5.pdf)), we get a convex plane figure $F$ and a length-nonincreasing map $m\colon F\to \Sigma$ such that the restriction $m|\_{\partial F}$ is an arc-length parametrization of $\gamma$.
In particular,
$$\textrm{area}\,F\geqslant\textrm{area}\,\Sigma.$$
Observe that the line segment $s$ corresponds to a line segment in $\partial F$.
This construction reduces you question to the convex plane figures, where you know the answer already.
In addition one need to use that for a plane curve area minimizing surface is a disc.
| 3 | https://mathoverflow.net/users/1441 | 435753 | 176,133 |
https://mathoverflow.net/questions/435728 | 4 | In a paper that I am reading the author quotes the following result about harmonic functions. According to him this should be "easy to show" but I don't seem to be able to do so.
Let $u:\overline{B^n}\to \mathbb{R}$ be harmonic, where $\overline{B^n}\subset\mathbb{R}^n$ is the closed unit ball. I would like to prove that
$$
\int\_{B\_1} |Du|^2\leq \int\_{\partial B\_1} |\partial\_{\tau}u|^2,
$$
where $\partial\_{\tau}u$ is the tangential derivative of $u$. I tried to use both the Gauss-Green theorem and to write the Laplacian in spherical coordinates, but I always get stuck.
---
Here are some correct but apparently useless computations.
Using the Gauss-Green formula we get
$$
\begin{equation}
0=\int\_{B\_1} u\Delta u =-\int\_{B\_1}|Du|^2+\int\_{\partial {B\_1}}u\frac{\partial u}{\partial \nu}.
\end{equation}
$$
On the other hand, the expression of the Laplacian in spherical coordinates is
$$
\begin{equation}
\begin{aligned}
0=\Delta u=\frac{\partial^2u}{\partial r^2} +\frac{n-1}{r}\frac{\partial u}{\partial r}+ \frac{1}{r^2}\Delta\_{\partial B\_1} u\\
=\frac{1}{r^{n-1}}\frac{\partial}{\partial r}(r^{n-1}\frac{\partial u}{\partial r})+ \frac{1}{r^2}\Delta\_{\partial B\_1} u,
\end{aligned}
\end{equation}
$$
and integrating this against $u$ and noting that $\partial\_{\tau}u=1/r\partial\_{\theta}u$ we find the same expression as before (obviously)
$$
\begin{equation}
\tag{2}
\begin{aligned}
0=-\int\_{\partial B\_1} \int\_0^1|\frac{\partial u}{\partial r}|^2 r^{n-1}+\int\_{\partial B\_1}u \frac{\partial u}{\partial \nu}-\int\_0^1r^{n-3}\int\_{\partial B\_1}|\partial\_{\theta}u|^2(r\theta)\\
=-\int\_{B\_1} |\frac{\partial u}{\partial r}|^2 +\int\_{\partial B\_1}u \frac{\partial u}{\partial \nu}-\int\_{ B\_1}|\partial\_{\tau}u|^2\\
=-\int\_{B\_1}|Du|^2+\int\_{\partial {B\_1}}u\frac{\partial u}{\partial \nu}.
\end{aligned}
\end{equation}
$$
| https://mathoverflow.net/users/351083 | An inequality for harmonic functions | Consider first the $d=2$ case. Then, $u$ is a real part of an analytic function. We can write $$u(z)=\frac12\sum\_{n=0}^{\infty}(a\_nz^n+\overline{a}\_n\overline{z}^n)$$ and $$\partial\_\nu u(z)=\frac12\sum\_{n=0}^{\infty}(a\_nnz^n+\overline{a}\_nn\overline{z}^n)$$ for $|z|=1$. When multiplying them out and integrating over the circle, we get terms of the form $c\cdot z^m=c\cdot e^{im\theta}$ for some $m\in\mathbb{Z}$, which integrage to zero unless $m=0$. This yields
$$
\int\_{\partial B\_1} u\partial\_\nu u=\pi\sum\_{n=0}^\infty n|a\_n|^2\leq \pi\sum\_{n=0}^\infty n^2|a\_n|^2=\int\_{\partial B\_1} |\partial\_\nu u|^2.
$$
The generalization to higher dimensions is straightforward, for one can still write an arbitrary harmonic function as $f(x)=\sum\_{n,j}a\_{n,j}r^n Y\_{n,j}(x/r)$, where $Y\_{n,j}$ are spherical harmonics (hence mutually orthogonal).
**Update:** I misread the question and the above proof proves a different inequality - with $\partial\_\nu$ instead of $\partial\_\tau$ in the right-hand side - I'm assuming $\partial\_\tau$ stands for the spherical gradient.
However, the same idea works in the case of OP's inequality. Note that the spherical harmonics are an orthonormalgonal basis of $L^2(S\_1)$, and also $Y\_{n,j}$ is an eigenfunction of the Laplace-Beltrami operation on the sphere with eigenvalue $n(n+d-2)$. In particular, they are orthogonal with respect to the Dirichlet form:
$$
\int\_{S\_1}\partial\_\tau Y\_{n,j}\cdot\partial\_\tau Y\_{\hat{n},\hat{j}}= \int\_{S\_1}\Delta\_{S\_1} Y\_{n,j} Y\_{\hat{n},\hat{j}}=n(n+d-2)\delta\_{n,\hat{n}}\delta\_{j,\hat{j}}.
$$ This means that if $u(x)=\sum\_{n,j}a\_{n,j}r^n Y\_{n,j}(x/r)$, then, as explained above,
$$
\int\_{S\_1} u\partial\_\nu u=\sum\_{n,j} a^2\_{n,j}n\leq\sum\_{n,j} a^2\_{n,j}n(n+d-2)=\int\_{S\_1} |\partial\_\tau u|^2.
$$
| 3 | https://mathoverflow.net/users/56624 | 435757 | 176,134 |
https://mathoverflow.net/questions/435642 | 15 | Lately I have become interested in solid $F$-modules where $F$ is some discrete field. Ideally, one would want a category that is as nicely behaved as solid abelian groups or solid $\mathbb{F\_p}$-modules.
I have managed to show that a discrete field is a solid module over itself, so we get that the usual $\prod\_I F$ are all solid, which seems like a promising start. However, I have been unsuccessful in trying to show that one gets an abelian category.
If one tries to mimic the proof that solid abelian groups are an abelian category, the main obstruction we run into is $\operatorname{RHom}\_F (\prod\_I F, F)$. This is quite a difficult computation, and the main tool used to do this with condensed abelian groups is the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{R}/\mathbb{Z} \to 0$ so that we can use cohomology, but we don't have a similar sequence for $F$-modules.
The proof strategy for the analogue statement in $\mathbb{F}\_p$-modules is heavily reliant on the fact that $\mathbb{F}\_p$ is finite.
Essentially, the proof one needs is that $F\_\blacksquare$ is an analytic ring. I know from Scholze's notes that it is suspected not all discrete rings form an analytic ring but I was hoping that this might be true for fields.
So my question is: should we expect a good theory of solid $F$-modules and why/why not?
EDIT: Some more details about my question. I am interested in whether $F\_\blacksquare$ is analytic, where $F[S]^\blacksquare = \varprojlim F[S\_i]$.
The motivation for this is a bit convoluted, but I'm in a situation where having the $\prod\_I F$ as compact projective generators of an abelian category would be particularly convenient in order to extend certain functors only defined on these compact projective generators.
| https://mathoverflow.net/users/170467 | Is there a good theory of solid vector spaces? | I will prove that the result is true if $F$ is a finitely generated field, but fails if $F$ is countably generated field that is not finitely generated.
Let me first discuss the case $F=\mathbb Q$. For $F=\mathbb Q$, one has the idempotent solid $\mathbb Z$-algebra $\hat{\mathbb Z}=\mathrm{lim}\_n \mathbb Z/n\mathbb Z$ (where $n$ runs over nonzero integers). One can form the Bousfield localization of $D(\mathbb Z\_\blacksquare)$ that kills all $\hat{\mathbb Z}$-modules. As a Bousfield localization, this can also be described as a full subcategory of $D(\mathbb Z\_\blacksquare)$ in terms of the corresponding local objects; and one computes that the localization of $\mathbb Z\_\blacksquare[S] = \mathrm{lim}\_i \mathbb Z[S\_i]$ is $\mathrm{lim}\_i \mathbb Q[S\_i]$ (as the quotient $\mathrm{lim}\_i \mathbb Q/\mathbb Z[S\_i]$ is a module over $\hat{\mathbb Z}$, while $\hat{\mathbb Z}$ has no maps to $\mathbb Q$, and thus no maps $\mathrm{lim}\_i \mathbb Q[S\_i]$). This easily implies that $\mathbb Q$ with $\mathbb Q\_\blacksquare[S] = \mathrm{lim}\_i \mathbb Q[S\_i]$ defines an analytic ring.
If $F$ is any finitely generated field, write it as the field of fractions of some domain $R$ that is a finitely generated $\mathbb Z$-algebra. Then $R\_\blacksquare$ exists as an analytic ring, and one can form the idempotent $R\_\blacksquare$-algebra $\hat{R}=\mathrm{lim}\_f R/f$, where $f$ runs over nonzero elements of $R$. Passing to the corresponding Bousfield localization, the localization of $\mathrm{lim}\_i R[S\_i]$ is $\mathrm{lim}\_i K[S\_i]$, from which one gets the desired result.
Now assume that $F$ is not finitely generated, but still countably generated. I claim that
$$\mathrm{Ext}^1\_F(\prod\_{\mathbb N} F,F)\neq 0,$$
where the $\mathrm{Ext}^1$ is computed in condensed $F$-modules; but if $F\_\blacksquare$ was an analytic ring then such Ext-groups would have to be zero.
Write $F$ as a sequential colimit of finitely generated fields $F\_n$. By writing $R\mathrm{Hom}\_F$ as $R\mathrm{lim}\_n R\mathrm{Hom}\_{F\_n}$ and a lim-lim^1-sequence, it suffices to show that
$$ \mathrm{lim}^1 \mathrm{Hom}\_{F\_n}(\prod\_{\mathbb N} F,F)\neq 0.$$
But one can show that any morphism of condensed abelian groups $\prod\_{\mathbb N} F\to F$ factors over a finite product, hence it suffices that
$$ \mathrm{lim}^1 (\bigoplus\_{\mathbb N} \mathrm{Hom}\_{F\_n}(F,F))\neq 0.$$
But for any tower $(X\_n)\_n$ of abelian groups, one has $\mathrm{lim}^1(\bigoplus\_{\mathbb N} X\_n)\neq 0$ as long as $(X\_n)\_n$ does not satisfy the Mittag-Leffler condition (see I. Emmanouil. Mittag-Leffler condition and the vanishing of lim^1. Topology, 35(1):267–271, 1996). But if the $F\_n$ are strictly increasing, $\mathrm{Hom}\_{F\_n}(F,F)$ forms a strictly decreasing chain of abelian groups.
It is likely that the result fails for any $F$ that is not finitely generated; I did not try to think about this.
| 12 | https://mathoverflow.net/users/6074 | 435760 | 176,136 |
https://mathoverflow.net/questions/435774 | 3 | Recall a prior posting titled [Is there an effective way to generalize this approach of affinely extending the number line?](https://mathoverflow.net/questions/435272/is-there-an-effective-way-to-generalize-this-approach-of-affinely-extending-the), and especially the accepted [answer](https://mathoverflow.net/a/435273/95347) given to it. So we are working in $\sf ZFC$ with the definitions of operators given in that answer.
The issue I'm asking about here is if we can always figure out when algebra of the presented extended reals departs from alegbra of the reals. There are of course obvious violations, like the real rule "$x + 1 > x$" no longer holds generally over $\hat{\mathbb R}$, but $x +1 \geq x$ does! Another example, $0^0$ seem to confom to $0/0$, i.e. $$0^0 \leadsto k \iff 0/0 \leadsto k$$, also we do have: $\infty^0 \leadsto \infty/\infty; (-\infty)^0 \leadsto -\infty/-\infty$
So, the rule $x^0 \leadsto x/x$, is preserved.
Also $$0^{-\infty}\leadsto 1/0^\infty \leadsto 1/0 \leadsto -\infty, \infty $$
However we don't have: $$ \sqrt[\infty]{0} \leadsto0^{1/\infty} \leadsto 0^0$$, since $\sqrt[\infty]{0} \leadsto r \iff r \in [-1,1]$,
while $0^0 \leadsto r \iff r \in \hat{\mathbb R}$
So here the rule $$k=1/x \implies \sqrt[x]{y} \leadsto y^k$$; no longer applies generally over $\hat{\mathbb R}$.
>
> Given an arithmetical expression $\psi$ that holds over $\mathbb R$, is it always decidable whether this also holds over $\hat{\mathbb R}$?
>
>
>
| https://mathoverflow.net/users/95347 | Can we always know if an algebraic rule over the reals is preserved over the extended reals or not? | Of course this depends on the signature (= choice of basic functions) used, but for a wide variety of such the answer will be **yes**. This is because $(\hat{\mathbb{R}};\overline{\hat{F}})$ is interpretable in $({\mathbb{R}};\overline{{F}})$ for any tuple of functions $\overline{F}$ containing at least $+$ and $\times$ (these let us define ordering). Basically, the definition of $\hat{f}$ I gave in terms of sequences can be reformulated in terms of limit-style language.
The point, then, is that for many choices of $\overline{F}$ the structure $(\mathbb{R};{\overline{F}})$ is decidable and so anything interpretable in it is also decidable. E.g. for $\overline{F}=(+,\times)$ this is due to Tarski.
An interesting near example is $\overline{F}=(+,\times,\mathit{exp})$; the decidability of the resulting structure is [an old question of Tarski](https://en.wikipedia.org/wiki/Tarski%27s_exponential_function_problem) which is still open. Macintyre and Wilkie showed that it has an affirmative answer assuming Schanuel's conjecture
| 4 | https://mathoverflow.net/users/8133 | 435775 | 176,140 |
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