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https://mathoverflow.net/questions/440648 | 6 | I attended a talk recently and the speaker said, essentially, that gauge theory invariants are expected to never be able to detect exotic 4-spheres because they always vanish, for a reason related to something called "neck cutting." I'm not sure what this means specifically, but I'm aware that Seiberg-Witten theory, for example, would not detect an exotic 4-sphere. My question is: how do we know that other potential gauge theories (SW with two spinors, Vafa-Witten, etc.) would never produce a nontrivial invariant for homotopy 4-spheres?
| https://mathoverflow.net/users/314845 | "Neck cutting" and why gauge theory doesn't work on homotopy 4-spheres | We cannot go as far as to say all of SW theory is insufficient, but the invariants that exist now are insufficient, via neck 'stretching' and 'pinching'. If you're aware of the proof (such as [No homotopy 4-sphere invariants using ECH=SWF](https://msp.org/agt/2021/21-5/p11.xhtml)), then I believe similar gauge theories fall to the same logic if they have some sort of "independence of metric and perturbation to equations".
To slightly clarify, there are multiple factors: the trivial 2nd cohomology (vaguely 2nd cohomology "rules all" in 4d gauge theories), the lack of stability of moduli of gauge-theoretic solutions when equations are perturbed, and being able to stretch (using metrics) a standard ball out of the sphere (to consider "Floer homology cobordism maps").
In contrast, w.r.t. Gluck's construction to attempt to get exotic spheres from special subsets of the standard sphere, it fails with SW (and tentatively other gauge theories). This latter fact (applied to 4-manifolds with $b^2\_+>0$) hasn't been documented in the literature I believe but here is my proof:
On any 4-manifold with $b^2\_+>0$ *the Gluck construction does not alter the SW invariants,* because the induced $\mathbb{Z}$-action on monopole Floer homology of $S^1\times S^2$ (the boundary of the tubular neighborhood of the surgery 2-sphere) is trivial. Indeed, Gluck's action $R(\theta,x)=\big(\theta,r\_\theta(x)\big)\in\text{Diff}\_+(S^1\times S^2)$ preserves the unique torsion spin-c structure on $S^1\times S^2$ and $R\_\*=\text{id}\_\*$ on $H\_\*(S^1\times S^2;\mathbb{Z})$, so we're done by Propositions 9.7.1+36.1.3 of Kronheimer--Mrowka's bible.
| 3 | https://mathoverflow.net/users/12310 | 440679 | 177,933 |
https://mathoverflow.net/questions/336494 | 26 | For $1\leq p,q \leq \infty$ such that $\frac1p +\frac1q\geq 1$, Young's inequality states $\|f\star g\|\_r\leq \|f\|\_p\|g\|\_q$ (we work on $\mathbf{R}^d$ here), where $1+\frac1r = \frac1p+\frac1q$. Equivalently
\begin{align\*}
\|f\|\_p=\|g\|\_q=\|h\|\_{r'}=1\Rightarrow \int\_{\mathbf{R}^d}\int\_{\mathbf{R}^d}f(x)g(y)h(x+y)\,\mathrm{d}x\,\mathrm{d}x \leq 1.
\end{align\*}
The most elementary proof that I know is based on the (generalized) Hölder inequality on $\mathbf{R}^d\times\mathbf{R}^d$ (for three functions), applied on three "mixing" functions $\varphi(x,y)^a \psi(x,y)^b$ where $\{\varphi,\psi\}$ runs over the possible pairs of $\{(x,y)\mapsto f(x); (x,y)\mapsto g(y) ; (x,y)\mapsto h(x+y)\}$ and $a$ and $b$ are adequately chosen.
There is of course a way to guess the correct exponents, but I find this proof a bit tedious and, when it comes to teach it, a bit artificial ("consider these three functions and ... the magic happens").
Instead, I am wondering if it is possible to prove it in a different way, remaining at the same level of knowledge.
The relation between $p,q,r$ rewrites $\frac{1}{r'} = \frac{1}{p'}+\frac{1}{q'}$. This, together with Hölder inequality, proves that any element in $L^{r'}(\mathbf{R}^d)$ is the (pointwise) product of two elements respectively in $L^{p'}(\mathbf{R}^d)$ and $L^{q'}(\mathbf{R}^d)$.
Can we use this to prove (something like)
\begin{multline\*}
\sup\_{\|h\|\_{r'}=1} \int\_{\mathbf{R}^d}\int\_{\mathbf{R}^d} f(x)g(y) h(x+y)\,\mathrm{d} x\,\mathrm{d} y \\
\leq \sup\_{\|\varphi\|\_{q'}=1,\|\psi\|\_{p'}=1} \int\_{\mathbf{R}^d}\int\_{\mathbf{R}^d} f(x)g(y) \varphi(x)\psi(y)\,\mathrm{d} x\,\mathrm{d} y\quad ?
\end{multline\*}
I did not succeed but still feel that the correspondance between the exponents in the convolution and poncutal products is not a coincidence.
Note that using (a bit of) interpolation theory (I did not check in details) :
* Young's inequality can be obtained by Fourier transform (precisely using $\widehat{f\star g}=\widehat{f}\widehat{g}$), at least for exponents in $[1,2]$ and then all the other ones by a duality argument.
* The case $\{p,q\}=\{1,\infty\}$ is straightforward and by a duality argument it is possible to recover then $\{p,q\}=\{1,r\}$, and then an interpolation argument should recover some intermediate exponents.
However, I'd really much appreciate a proof without interpolation.
| https://mathoverflow.net/users/27767 | Proofs of Young's inequality for convolution | I found the proof given by Hormander in the Analysis of Linear Partial Differential Operators to be quite clear and instructive. It can be found in the first volume, pages 116-117, and for convenience I will reproduce it here.
>
> $|u\_1 \* u\_2 \* \dots \* u\_k(0)| \leq \|u\_1\|\_{p\_1}\dots\|u\_k\|\_{p\_k}$ if $\sum 1/p\_i = k - 1$ and $1 \leq p\_i \leq \infty$.
>
>
>
When for some $p\_i$ we have $p\_i = \infty$, then every other $p\_j = 1$, and thus by Fubini, we have the assertion.
The other cases can be reduced to this case by using a convexity argument: letting $v\_i = |u\_i|^{p\_j}$, we have to prove $v\_1^{t\_1} \*\dots \*v\_k^{t\_k}(0) \leq 1$ assuming that $\int v\_i = 1$, $0 \leq v\_i$, $0 \leq t\_i \leq 1$ and $\sum t\_i = k - 1$.
However $v\_1^{t\_1} \* \dots \*v\_k^{t\_k}(0)$ as a function of $t$ is convex, since the integrand as a function of $t$ is convex.
Then, $(t\_1, \dots t\_k) = \sum (1 - t\_i) f\_i$ where $f\_i$ is the vector with $i$-th component $0$ and everywhere else $1$, and this reduces the assertion to the first case.
Now letting $u\_1 \*\dots \* u\_k = u$, $|u \* v(0)| \leq \|u\_1\|\_{p\_1} \dots \|u\_k\|\_{p\_k} \|v\|\_{q'}$, where $1/p\_1 + \dots + 1/p\_k = k - 1 + 1/q$, and $q'$ is the holder conjugate of $q$.
Viewing $(u\*v)(0)$ as a functional on $L^{q'}$, we have by duality that $\|u\|\_q \leq \|u\_1\|\_{p\_1} \dots \|u\_k\|\_{p\_k}$.
The case $k = 2$ is the desired case.
| 3 | https://mathoverflow.net/users/nan | 440680 | 177,934 |
https://mathoverflow.net/questions/440685 | 2 | Let $A/\mathbb{Q}$ be an abelian variety with good reduction at a prime $p$. Assume $\mathcal{A}/\mathbb{Z}\_{(p)}$ is an integral model at $p$(hence proper smooth).
For any number field $K$ and any prime ideal $\mathfrak{p}$ of $K$ over $p$ with residue field $\kappa$, by valuation criterion on $\mathcal{O}\_{K,\mathfrak{p}}$, any $K$-point of $A$ extends uniquely to an $\kappa$-point of $\mathcal{A}$'s special fiber.
By fixing a prime ideal of $\overline{\mathbb{Z}}$ lying over $p$ and taking injective limit, we get a well defined reduction map\begin{equation} A(\overline{\mathbb{Q}})\rightarrow \mathcal{A}\_{\mathbb{F}\_p}(\overline{\mathbb{F}\_p}). \end{equation}
Does the above map induce isomprhisms on $\ell$-primary parts of both groups? Notice that both of their $\ell$-primary subgroups are isomorphic to $(\mathbb{Q}\_{\ell}/\mathbb{Z}\_{\ell})^{\oplus 2g}$.
| https://mathoverflow.net/users/177957 | The reduction map on the $\ell$-primary torsion of abelian variety | This is true. One reference for this is Theorem C.1.4 in Hindry-Silverman's "Diophantine Geometry: An Introduction" which says that for an abelian variety $A$, number field $K$ and any prime $\mathfrak p$ above $p$, the reduction map $A(K)\to A(O\_K/\mathfrak p)$ is injective on prime-to-$p$-torsion. The result you ask for follows since $\overline{\mathbb Q}$ is a union of number fields.
One way to prove the required injectivity follows from the description of the kernel of the reduction map over the completion $K\_{\mathfrak p}$, which can be identified with the formal group associated to $A$. Formal groups over $p$-adic fields have no prime-to-$p$ torsion.
| 3 | https://mathoverflow.net/users/30186 | 440687 | 177,938 |
https://mathoverflow.net/questions/376603 | 13 | The following result is Proposition 2.4.3 in [1]:
>
> **Theorem.** Let $K\subset\mathbb{R}^n$ be a bounded convex set with the non-empty interior. Then $\partial K\in C^{1,1}$ if and only if
> there is $r>0$ such that $K$ is the unioin of balls of radius $r$.
>
>
>
**Question.** Do you know who is the author of this result?
Hörmander does not provide any reference.
**Edit.** I am still quite puzzled about the result. The two answers below show that the result was proved in an unpublished PhD from 1957, it was mentioned without a proof or reference in a paper by Kiselman and the first published proof I am aware of appears in Hormander's book. The result is in my opinion very beautiful not entirely trivial so I expect there should be other references.
**I am still waiting for more answers showing other references to published proofs.**
[1] **L. Hörmander**,
*Notions of convexity.*
Progress in Mathematics, 127. Birkhäuser Boston, Inc., Boston, MA, 1994.
| https://mathoverflow.net/users/121665 | Regularity of convex sets in $\mathbb{R}^n$ | Theorem 1.3 in:
<https://arxiv.org/pdf/1304.4179.pdf>
and the historical discussion below it is relevant. The result (in a form that applies without the convexity assumption) seems to first appear in 1957.
| 4 | https://mathoverflow.net/users/112954 | 440697 | 177,939 |
https://mathoverflow.net/questions/440613 | 8 | Let $B\_t$ be the classic Brownian motion. I understand that, if $s>1/2$, almost surely $B\_t$ is nowhere $s$-Hölder continuous i.e. almost surely for no point $x$ it happens that $B\_t\in C^s(x)$.
Moreover I read a claim that said the same about any translation by a continuous function: given $s>1/2$ and $f$ continuous then almost surely $f+B\_t$ is nowhere $s$-Hölder continuous. Is this true and if so why?
| https://mathoverflow.net/users/39180 | Regularity of translations for Brownian motion | The result holds for any bounded function $f$, in the following sense: for any real $s>1/2$,
\begin{equation}
P^\*(A)=0,
\end{equation}
where
\begin{equation}
A:=\Big\{\exists t\_0\in[0,1]\ \limsup\_{t\to t\_0}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}<\infty\Big\},
\end{equation}
$P^\*$ is the outer probability, $\limsup\_{t\to t\_0}:=\limsup\_{t\to t\_0,t\in[0,1]}$,
\begin{equation}
W\_f:=W+f,
\end{equation}
and $W$ is a standard Wiener process.
The proof is obtained by a straightforward adaptation of the proof of the Paley--Wiener--Zygmund theorem on the almost sure nowhere differentiability of the Brownian motion.
Indeed, since $W$ and $f$ are bounded,
\begin{equation}
A\subseteq B=\bigcup\_{M=1}^\infty B\_M,
\end{equation}
where
\begin{equation}
B:=\Big\{\exists t\_0\in[0,1]\ \sup\_{t\in[0,1]}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}<\infty\Big\},
\end{equation}
\begin{equation}
B\_M:=\Big\{\exists t\_0\in[0,1]\ \sup\_{t\in[0,1]}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}\le M\Big\}.
\end{equation}
Next,
\begin{equation}
B\_M:=B\_{M,1}\cup B\_{M,2},
\end{equation}
where
\begin{equation}
B\_{M,1}:=\Big\{\exists t\_0\in[0,1/2]\ \sup\_{t\in[0,1]}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}\le M\Big\},
\end{equation}
\begin{equation}
B\_{M,2}:=\Big\{\exists t\_0\in[1/2,1]\ \sup\_{t\in[0,1]}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}\le M\Big\}.
\end{equation}
Let $r$ be any integer such that
\begin{equation}
r>1/(s-1/2).
\end{equation}
Let then $n$ be any integer $\ge n\_r$, where $n\_r$ is the smaller integer $q$ such that $2^{q-1}>r$.
Assuming the event $B\_{M,1}$ occurs, let $K$ be a random integer in the set $\{1,\dots,2^{n-1}\}$ such that $t\_0\in[\frac{K-1}{2^n},\frac K{2^n}]$. Then for $j=1,\dots,r$
\begin{equation}
\begin{aligned}
&\Big|W\_f\Big(\frac{K+j}{2^n}\Big)-W\_f\Big(\frac{K+j-1}{2^n}\Big)\Big| \\
&\le\Big|W\_f\Big(\frac{K+j}{2^n}\Big)-W\_f(t\_0)\Big|
+\Big|W\_f\Big(\frac{K+j-1}{2^n}\Big)-W\_f(t\_0)\Big| \\
&\le M\Big|\frac{K+j}{2^n}-t\_0\Big|^s
+M\Big|\frac{K+j-1}{2^n}-t\_0\Big|^s\le\frac{2Mj^s}{2^{sn}} \le\frac{2Mr^s}{2^{sn}}
\end{aligned}
\end{equation}
So,
\begin{equation}
B\_{M,1}\subseteq\bigcap\_{n\ge n\_r}\bigcup\_{k=1}^{2^{n-1}}C\_{n,k},
\end{equation}
where
\begin{equation}
C\_{n,k}:=\bigcap\_{j=1}^r \Big\{\Big|W\_f\Big(\frac{k+j}{2^n}\Big)-W\_f\Big(\frac{k+j-1}{2^n}\Big)\Big|
\le\frac{2Mr^s}{2^{sn}}\Big\}.
\end{equation}
By the independence of the increments of the Wiener process and because the pdf of the normally distributed random variable $W\_f\big(\frac{k+j}{2^n}\big)-W\_f\big(\frac{k+j-1}{2^n}\big)$ is $\le2^{n/2-1}$, we have
\begin{equation}
\begin{aligned}
P(C\_{n,k})&=\prod\_{j=1}^r P\Big(\Big|W\_f\Big(\frac{k+j}{2^n}\Big)-W\_f\Big(\frac{k+j-1}{2^n}\Big)\Big|
\le\frac{2Mr^s}{2^{sn}}\Big) \\
& \le\Big(2^{n/2}\frac{2Mr^s}{2^{sn}}\Big)^r=\frac C{2^{(1+a)n}},
\end{aligned}
\end{equation}
where $C:=(2Mr^s)^r$ and $a:=r(s-1/2)-1>0$. So,
\begin{equation}
P\Big(\bigcup\_{k=1}^{2^{n-1}}C\_{n,k}\Big)\le2^{n-1}\frac C{2^{(1+a)n}}\le\frac C{2^{an}}
\end{equation}
and
\begin{equation}
P^\*(B\_{M,1})\le\lim\_{n\to\infty}\frac C{2^{an}}=0.
\end{equation}
So, $P^\*(B\_{M,1})=0$. Similarly, $P^\*(B\_{M,2})=0$. So, $P^\*(B\_M)=0$, for all $M$.
Thus,
\begin{equation}
P^\*(A)\le P^\*(B)\le\sum\_{M=1}^\infty P^\*(B\_M)=0,
\end{equation}
as claimed. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 440698 | 177,940 |
https://mathoverflow.net/questions/440695 | 0 | This is a follow-up question from [my previous question](https://mathoverflow.net/questions/439653/maximum-number-of-vectors-with-bounds-on-inner-products/440643?noredirect=1#comment1136719_440643).
Suppose there are (2n+1) vectors $\{m\_1,m\_2,...,m\_n\}$, $\{p\_1,p\_2,...,p\_n\}$ and $p^\*$ in $R^{k+1}$. $m\_i$ are weakly positive vectors. $p\_i$ and $p^\*$ are probability vectors (whose coordinates are weakly positive and sum to 1).
The vectors satisfy the following properties:
$$ m\_i p\_j \geq m\_i p^\* = 1 \quad \forall i\neq j,$$
$$ m\_i p\_i = 0 \quad \forall i.$$
The question is: what is the maximum number of $n$?
| https://mathoverflow.net/users/498587 | Maximum number of vectors with bounds on inner products (follow up question) | This is not an answer to the question, but here are some upper/lower bounds. Firstly, if we let $A\_i\subseteq\{0,1,\dots,k+1\}$ be the non zero coordinates of $m\_i$, then we can't have $A\_i\subseteq A\_j$ for $i\neq j$, because then we would have $m\_ip\_j=0$ (you already mentioned this in the other question I think). This means that $(A\_i)\_i$ are all different and form an antichain of $\{1,\dots,k+1\}$, thus $n\leq\binom{k+1}{\left\lfloor\frac{k+1}{2}\right\rfloor}$ by [Sperner's theorem](https://en.wikipedia.org/wiki/Sperner%27s_theorem).
For a superpolynomial lower bound, suppose that $\sqrt{k+1}$ is an even integer for simplicity. Let $p^\*=(\frac{1}{k+1},\dots,\frac{1}{k+1})$, and for each $A\subseteq\{1,\dots,\sqrt{k+1}\}$ of size $\frac{\sqrt{k+1}}{2}$, let $p\_A$ have coordinates $(p\_i)\_j=\frac{2}{\sqrt{k+1}}$ when $j\in A$ and $0$ if not, and let $m\_A$ have coordinates $2\sqrt{k+1}$ in $\{1,\dots,\sqrt{k+1}\}\setminus A$ and $0$ else.
The number of vectors $p\_A$, $m\_A$ is then $\binom{\sqrt{k+1}}{\frac{\sqrt{k+1}}{2}}$, which grows faster than polynomially.
| 1 | https://mathoverflow.net/users/172802 | 440700 | 177,941 |
https://mathoverflow.net/questions/440705 | 11 | A key construction in Thurston's proof of the existence of hyperbolic structures on Haken manifolds is the so-called "skinning map" associated to a 3-manifold $M$ with boundary whose interior admits a hyperbolic metric. The map
$$\sigma\_M:\mathrm{Teich}(\partial M) \to \mathrm{Teich}(\overline{\partial M})$$ takes a conformal structure on the boundary and produces a new one. Conformal structures on $\partial M$ are in bijection with geometrically finite hyperbolic metrics on $M$. See [here](https://people.math.harvard.edu/%7Ectm/papers/home/text/papers/rs/rs.pdf) for a quick reference.
The problem is, I'm having trouble understanding what exactly this map is. The most I've gathered is that one can lift a hyperbolic structure on $M$ to the covering space $\tilde M$ of the interior corresponding to the image of $\pi\_1(\partial M)\to\pi\_1(M)$, and that this induces a new conformal structure, but how? In particular, how can the covering space $\tilde M$ be described as a manifold? Is it just $\partial M\times \mathbb{R}$?
| https://mathoverflow.net/users/499323 | Definition of Thurston's skinning map | Let’s simplify to the case where $M$ has exactly one boundary component, say $\partial M = S$. So the hyperbolic structures on $M$ are parametrised by the conformal structures on $S$. Fix one such conformal structure $\rho$ and lift the resulting hyperbolic structure on $M$ to the cover corresponding to $\pi\_1(S)$. As you suggest, this cover is homeomorphic to $S \times \mathbb{R}$. So it has two ends, each giving a conformal structure on $S$. One of these is $\rho$, the conformal structure we started with. The other is the (orientation reverse of) $\rho'= \sigma\_M(\rho)$ - that is, the image of $\rho$ under the skinning map.
I believe that $\sigma\_M$ is called the "skinning map" because $\rho \sqcup \rho'$ gives the "skin at infinity" of the lifted hyperbolic structure on $S \times \mathbb{R}$.
| 13 | https://mathoverflow.net/users/1650 | 440707 | 177,942 |
https://mathoverflow.net/questions/439697 | 6 | I'm posting this question in hopes that someone more familiar with the literature will be able to point me in the right direction (or give an obvious answer).
Let $M^{d-2} \hookrightarrow \mathbb{R}^d$ be a smooth embedding and let $g$ be the metric induced on $M$ from the flat metric on $\mathbb{R}^d$. Under what conditions can $M$ be viewed (locally) as an isometrically-embedded hypersurface of $\mathbb{R}^{d-1}$?
In the case where $d=3$, it is fairly clear: for a curve in $\mathbb{R}^3$, if the torsion of the spacecurve vanishes, then the spacecurve is a planecurve. What is not clear to me is how this notion generalizes to higher dimensions.
I have considered two directions. My first guess is that there must exist a covariantly constant normal vector field in the normal bundle of $M$. This generalizes the notion that the direction of the binormal of a spacecurve is fixed when the spacecurve is actually a planecurve.
My second instinct is that there must exist a tensor $II \in \Gamma(\odot^2 T^\* M)$ such that
$$R\_{abcd} = II\_{ac} II\_{bd} - II\_{ad} II\_{bc}\,,$$
in line with the Gauss equation. Clearly, this condition is necessary but it is not obvious if it sufficient.
If anyone knows of literature on this subject, I'd love to be pointed in that direction. This seems like a rudimentary-enough question that it must have been studied. The literature I have found discusses the embeddability of surfaces in $\mathbb{R}^3$ - this is always possible by the Burstin-Cartan-Janet Theorem, but it's a special case because $3$ is the Janet dimension for surfaces, and this coincidence does not hold for $d \geq 5$. The Nash embedding theorem seems to give upper bounds, but that's not exactly the kind of result I'm looking for.
| https://mathoverflow.net/users/104933 | Existence of local isometric embedding of smooth $(M^{d-2},g)$ in $\mathbb{R}^{d-1}$ | It's hard to answer the OP's question satisfactorily without considering what would actually constitute an answer. The most obvious answer is tautological: $M^{d-2}\hookrightarrow\mathbb{R}^d$ lies in a hyperplane if and only if there is a nonzero affine function $\ell:\mathbb{R}^d\to\mathbb{R}$ such that $\ell(M^{d-2}) = \{0\}$. It seems clear that the OP wouldn't consider that criterion an answer though, because it requires one to test for the existence of something for which it may not be clear how to test. Instead, it seems that the OP wants an answer of the form "If certain differential-geometric invariant quantities defined on $M^{d-2}\subset\mathbb{R}^d$ vanish, then $M^{d-2}$ lies in a hyperplane. One might also hope for a 'converse' that said that if $M^{d-2}\subset\mathbb{R}^d$ lies in a hyperplane, then those quantities do vanish.
Here is a cautionary example worth considering: Let $f:\mathbb{R}\to\mathbb{R}$ satisfy $f(t) = 0$ for $t\le0$ and $f(t) = e^{-1/t}$ for $t>0$. Now consider the smooth space curves $\alpha,\beta:\mathbb{R}\to\mathbb{R}^3$ defined by
$$
\alpha(t) = \bigl(t,f(t),f(-t)\bigr)\quad\text{and}\quad
\beta(t) = \bigl(t,f(t)+f(-t),0\bigr)
$$
All of the differential invariants of $\alpha$ and $\beta$ are equal (in particular, they both have vanishing torsion), but $\alpha(\mathbb{R})$ does not lie in a plane while $\beta(\mathbb{R})$ clearly does.
The issue is that if an immersed space curve $\gamma:\mathbb{R}\to\mathbb{R}^3$ lies in a plane, then, $\gamma',\gamma'',\gamma'''$ must be linearly dependent, so in particular, the third order polynomial differential invariant $I\_3 = (\gamma'\times\gamma'')\cdot\gamma'''\,(\mathrm{d}t)^6$ must vanish identically. However the vanishing of $I\_3$, by itself, is not enough to guarantee that $\gamma'''$ is a linear combination of $\gamma'$ and $\gamma''$ with smooth coefficients (which would indeed imply that $\gamma(\mathbb{R})$ lay in a plane). To get this, one needs an additional nondegeneracy condition, usually taken to be the condition that $\gamma'$ and $\gamma''$ be linearly independent. (It turns out, though, that it suffices to impose the weaker nondegeneracy condition that the second order polynomial differential invariant $I\_2 = (\gamma'\times \gamma'')\cdot(\gamma'\times \gamma'')\,(\mathrm{d}t)^6$ only vanish to finite order at any point of $\mathbb{R}$.) [Note that, with regard to an element of arc-length $\mathrm{d}s$, defined up to a sign by $I\_1 = (\gamma'\cdot\gamma')\,(\mathrm{d}t)^2 = (\mathrm{d}s)^2$, one has $I\_2 = \kappa^2\,(\mathrm{d}s)^6$ and $I\_3 = \tau\kappa^2(\mathrm{d}s)^6$, where $\kappa$ and $\tau$ have their usual meanings.] In the above pair of examples, $\alpha$ and $\beta$ have the same invariants $I\_k$ for $k=1,2,3$, but $I\_2$ vanishes to infinite order at $t=0$.
In higher dimensions, something similar has to be done, i.e., one has to identify conditions on the polynomial differential invariants of $M^{d-2}\subset\mathbb{R}^d$ that hold when $M^{d-2}$ lies in a hyperplane and then impose nondegeneracy conditions before the vanishing of these invariants suffices to imply that $M^{d-2}$ does, in fact, lie in a hyperplane.
The most obvious condition (as mentioned by Yang and Petrunin) is that the rank $r$ of the second fundamental form $I\!I$ of $M$ should be at most $1$ at every point. This is equivalent to the vanishing of a second order polynomial differential invariant that is asection of $\Lambda^2(S^2(T^\*M))$ that we can write informally as $I\_2 = {I\!I}\wedge{I\!I}$. It's easy to see that the variety of second fundamental forms $I\!I$ at a point that satisfy this condition has codimension $\tfrac12 d(d{-}3)$, so this is already a fairly stringent condition as soon as $d>3$.
Another important invariant is what some authors call the *nullity* $\nu(p)$ of $I\!I$ at $p\in M$, i.e., the dimension of the subspace $N\_p\subset T\_pM$ consisting of the vectors $v\in T\_pM$ such that ${I\!I}\_p(v,u)=0$ for all $u\in T\_pM$. Note that bounding the nullity from above is an *open* condition on $I\!I$, so one can regard it as a nondegeneracy condition.
Using these terms, one has the following special case of a classic consequence of the structure equations:
**Theorem:** If $M^{d-2}\subset\mathbb{R}^d$ has its second fundamental form of rank $r\le 1$ at every point and nullity at most $d{-}4$ at every point, then $M^{d-2}$ lies in a hyperplane.
Note that the above theorem is vacuous for $d=3$, but, for $d\ge 4$, it gives a sufficient criterion for $M$ to lie in a hyperplane that depends only on *second-order* information. As we have seen, though, when the nullity is allowed to be as large as $d{-}3$, one has to go to third order information and, even then, put some condition on the locus where the nullity is as large as $d{-}2$ (i.e., where the second fundamental form vanishes) in order to conclude that $M$ lies in a hyperplane.
| 5 | https://mathoverflow.net/users/13972 | 440714 | 177,944 |
https://mathoverflow.net/questions/440615 | 1 | I don't know if it is a well-known problem, but I have been struggling to come up with an algorithm.
I have a set of linear constraints $Ax\le b$, $b\ge 0$ ($b$ and $A$ are given, $x$ is a variable). These constraints designate the domain for variable $x$. Imagine I have one new constraint $cx\le d$, which may or may not further constrain the original domain.
Now I can modify the right hand side of the original domain with variable $y$, $0\leq y \leq b$. The problem is to find $y$ such that $$Ax\le b-y$$ makes sure that the domains $Ax\le b-y$ and $Ax\le b-y \cup cx\leq d$ are the same (so in other words that adding a new constraint $cx\leq d$ does not change the domain. In fact I want to find minimal $y$ (for instance minimum of $\sum\_{i} y\_i$) that provides this feature.
| https://mathoverflow.net/users/499253 | Adding linear constraint to the domain | By LP duality, the new constraint $cx \le d$ is redundant iff there exists $u \ge 0$ such that
\begin{align}
u A &= c \tag1\label1 \\
u (b - y) &\le d \tag2\label2
\end{align}
To see the easy direction of the iff, note that \eqref{1} and \eqref{2} imply
$$cx = u A x \le u (b - y) \le d$$
You want to minimize $\sum\_i y\_i$ subject to \eqref{1}, \eqref{2} and $0 \le y \le b$, and this is a quadratically constrained linear programming (QCLP) problem.
| 1 | https://mathoverflow.net/users/141766 | 440728 | 177,948 |
https://mathoverflow.net/questions/440692 | 2 | **Question.** Does there exist an entire function $h$ satisfying three following assertions:
* $h$ belongs to the $H^2$ Hardy space in every horizontal upper half-plane;
* $zh - 1$ belongs to $H^2(\mathbb{C}\_+)$, where $\mathbb{C}\_+ = \{\text{Im}(z) > 0\}$;
* $h$ has infinitely many zeroes in some horizontal strip $\{\text{Im}(z) \in [y\_1, y\_2] \}$?
Does the answer change if we additionally assume that $h$ is of finite order?
**On the first assertion.** Assume that the function $H$ belongs to the Hardy space in every upper horizontal half-plane. and let $G = \mathcal{F}H\in L^2[0, +\infty)$. In terms of $G$ the first assertion is equivalent to the fact that $G e^{\delta x}\in L^2[0, \infty)$ for every $\delta > 0$ and $H(z) = \int\_0^{\infty} G(x)e^{izx}dx$ for all $z\in \mathbb{C}$.
In this situation $G$ can be expressed as $G = G\_1 + G\_2$ where $G\_1\in C^1[0,\gamma]$ for some $\gamma > 0$ and $\|G\_2\|\_{L^1[0,\infty)} < \varepsilon$. Via this decomposition one can show that $H(z)$ tends to $0$ as $|z|\to \infty$ uniformly in every horizontal strip (consequently $H(z) = w$ has only finitely many roots for all $\omega\in \mathbb{C}\setminus\{0\}$).
**Construction attempts.** To fulfil the first two assertions one can take function $H$ from the previous paragraph and put $h = (1 + H)/(z - z\_0)$ where $H(z\_0)= -1$. However in this situation $h(z) = 0 \Longleftrightarrow H(z) = -1$, which has only a finite number of roots in any horizontal half-plane.
Furthermore, by the Paley-Wiener theorem, the third assertion cannot hold if $\mathcal{F}h$ has a compact support.
| https://mathoverflow.net/users/498423 | Existence of the special entire Hardy space function with infinitely many zeros in the strip | Let
$$f(z)=\frac{e^{iz}-1}{iz}.$$
This function is in the Hardy class for any upper half-plane,
and has these properties: $f(0)=1,$ $f(2\pi n)=0$,
$$|f(z)|\leq C\frac{e^y+1}{|z|+1},$$
(this evidently holds for large and small $|z|$, therefore there is a constant $C$ so that this holds everywhere).
Since the $L^2$ norm has the property $\| f(kx)\|^2=\| f(x)\|^2/k,$ and the norm is invariant under real translations, the following function $g$ belongs to the Hardy spaces for upper half-planes:
$$g(z)=\sum\_{n=0}^\infty f(n^2(z-a\_n)),$$
where $a\_n\in\{2\pi k: k\in {\mathbf{Z}}\}$.
Moreover, it is entire, if the sequence $a\_n$ grows fast enough. (It follows from the estimate for $f$ that the series is uniformly convergent on any compact subset of the plane).
Now $g(a\_n)=1$, so $h(z)=(g(z-i)-1)/(z-i)$, where has all properties that you required.
Some extra labor is needed to show that $h$ can be made of finite order.
| 1 | https://mathoverflow.net/users/25510 | 440729 | 177,949 |
https://mathoverflow.net/questions/438836 | 1 | Let $(X\_n)\_{n\in\mathbb{N}}$ be a sequence of i.i.d. real-random variables. Let further $0<c<1$. I'm interested in the asymptotic properties of
$$
\sum\_{k=1}^n c^k X\_k.
$$
I can prove that this converges a.s. for $n\to\infty$ iff $\mathbb{E}(\max(0,\log(|X\_1|)<\infty$.
To be more specific:
* Are there known conditions for $\limsup\limits\_{n\to\infty}\sum\_{k=1}^n c^k X\_k=\infty$ a.s.?
* As far as I know, the asymptotic behaviour of $c^n\sum\_{k=1}^nX\_k$ is fairly well understood (c.f. ["A note on Fellers strong law of large numbers" by Chow, and Zhang (1986)](https://projecteuclid.org/journals/annals-of-probability/volume-14/issue-3/A-Note-on-Fellers-Strong-Law-of-Large-Numbers/10.1214/aop/1176992464.full)). Is there a known connection (or differences) between the asymptotic behaviour of $c^n\sum\_{k=1}^nX\_k$ and $\sum\_{k=1}^n c^k X\_k$? E.g. is it possible that $\lim\limits\_{n\to\infty}c^n\sum\_{k=1}^nX\_k=\infty$ a.s., but $\lim\limits\_{n\to\infty}\sum\_{k=1}^n c^k X\_k<\infty$ a.s.? I can only proof that if $\lim\limits\_{n\to\infty}c^n\sum\_{k=1}^nX\_k=\infty \text{ a.s.} \Rightarrow \lim\limits\_{n\to\infty}\sum\_{k=1}^n c^k X\_k$ diverges a.s.
* I'm having trouble constructing examples, where $\lim\limits\_{n\to\infty}\sum\_{k=1}^n c^k X\_k=\infty$ a.s. holds. Does anybody know a way to construct such examples?
I would be very grateful for any advice or reference!
| https://mathoverflow.net/users/473107 | Asymptotic properties of weighted random walks / infinite convolutions of random variables | Let $(X\_n)\_{n\in\mathbb{N}}$ be i.i.d. real random variables and let $0<c<1$.
Then the following are equivalent:
(a) There exists $r>0$ such that $P(|X\_k|>e^{rk} \; \, \text{infinitely often} )=0$.
(b) $\mathbb{E}(\max(0,\log(|X\_1|)<\infty$.
(c) For all $r>0$, we have $P(|X\_k|>e^{rk} \; \, \text{infinitely often} )=0$.
(d) The series $ \sum\_{k=1}^n c^k X\_k $ converges a.s.
(e) $\lim \_n c^n S\_n =0 \;$ a.s., where $S\_n:=\sum\_{k=1}^n X\_k$.
**Proof:** For a random variable $Y \ge 0$, we have
$$\sum\_{k=1}^\infty {\bf 1}\_{\{Y \ge k\}} \le Y \le \sum\_{k=0}^\infty {\bf 1}\_{\{Y \ge k\}} \,,$$
so taking expectations gives the well-known inequalities
$$\sum\_{k=1}^\infty P(Y \ge k) \le \mathbb{E}(Y) \le \sum\_{k=0}^\infty P(Y \ge k) \,.$$
Thus by the Borel-Cantelli lemma, if $Y\_k \ge 0$ are i.i.d. and $r>0$, then
$$\mathbb{E}(Y\_1) <\infty \Longleftrightarrow \mathbb{E}( Y\_1/r) <\infty \Longleftrightarrow $$ $$ \sum\_k P(Y\_k \ge {rk}) <\infty \Longleftrightarrow P(Y\_k \ge {rk} \; \, \text{infinitely often} )=0 \, . \tag{\*}$$
Applying this to $Y\_k=\max(0,\log(|X\_k|)$ shows that
$ \;(a) \Longrightarrow(b) \Longrightarrow (c)$. Obviously $(c) \Longrightarrow (a)$.
Clearly $\; (c) \Longrightarrow (d) \Longrightarrow (a)$ and
$\; (c) \Longrightarrow (e)$.
Finally, suppose (e) holds. Then
$c^n X\_n =(c^n S\_n) - c(c^{n-1}S\_{n-1}) \to 0$ a.s.,
and (a) follows.
QED
| 1 | https://mathoverflow.net/users/7691 | 440730 | 177,950 |
https://mathoverflow.net/questions/440744 | 6 | The vector space dimension of the cohomology group of the $2$-plane Grassmannian $\mathrm{Gr}\_{2,n}$ is given by the number of tuples $(\lambda\_1,\lambda\_2)$ satisfying
$$
n - 2 \geq \lambda\_1 \geq \lambda\_2 \geq 0.
$$
Explicitly this is given by
$$
\binom{n}{2}.
$$
This also happens to be the dimension of $V\_{\pi\_2}$ the second fundamental representation of $\frak{sl}\_n$. I am guessing this is not an accident, especially since the $2$-plane Grassmannian corresponds (in the usual way) to $V\_{\pi\_2}$.
Does this extend to the general identity
$$
\mathrm{dim}(H^{\*}(\mathrm{Gr}\_{d,n})) = \mathrm{dim}(V\_{\pi\_d})?
$$
If it does, then what is a conceptual explanation for this?
EDIT: Since $V\_{\pi\_d}$ is isomorphic to the exterior power
$$
\Lambda^d(V\_{\pi\_1})
$$
and $V\_{\pi\_1}$ is of dimension of $n$, we see that the RHS of the claimed identity is the binomial coefficient
$$
\binom{n}{d}.
$$
It follows from the general formula given in this [answer](https://mathoverflow.net/questions/196546/hard-lefschetz-theorem-for-the-flag-manifolds) that the LHS is the same binomial coefficient. Thus the identity does indeed extend from $2$-planes to $d$-planes. So the question is if there is a conceptual reason for this . . .
| https://mathoverflow.net/users/491434 | The dimension of the Grassmannian cohomology ring $H^*(\mathrm{Gr}_{n,d})$ and the fundamental $\frak{sl}_n$-representation $V_{\pi_d}$ | This is very standard. For a compact complex variety admitting a cell decomposition, the (co)homology is the free Abelian group generated by the cells (over $\mathbb{C}$ there is no room for the differential in the cellular complex). In the cellular decomposition of the Grassmannian, the (Schubert) cells are indexed by possible reduced row echelon forms of a $d\times n$-matrix, that is by possible positions of pivots. To the cell having pivots in positions $i\_1,\ldots,i\_d$ you can assign the basis element $e\_{i\_1}\wedge\cdots\wedge e\_{i\_d}$ in $\Lambda^d(V\_{\pi\_1})$.
| 10 | https://mathoverflow.net/users/1306 | 440753 | 177,954 |
https://mathoverflow.net/questions/440770 | 1 | Is there an example of a multivalued maximal monotone operator that is not the convex subdifferential of a proper convex lower semicontinuous? Besides, among these type of operators, are there any physically important? (describing any non-smooth dynamics of the real world). Thank you!
| https://mathoverflow.net/users/60556 | Any example of a multi-valued monotone maximal operator without subdifferential? | My prime example of such an operator comes from saddle point problems of the form
$$
\min\_x\max\_y F(x) + \langle Kx,y\rangle - G(y)
$$
with $F,G$ being two proper, convex, lower-semicontinuous functions defined on Hilbert spaces $X$ and $Y$, respectively, and $K:X\to Y$ linear and bounded. The Fenchel-Rockafellar optimality system is
$$
0\in \begin{pmatrix} \partial F(x) & -K^\*y\\ Kx & \partial G(y)\end{pmatrix}.
$$
The operator on the right hand side is indeed monotone as the sum of a subdifferential and a skew-symmetric one and it's also maximally so, since he both are and the second one is defined everywhere. It is not a subgradient of any function on $X\times Y$ since it's not symmetric.
| 1 | https://mathoverflow.net/users/9652 | 440777 | 177,959 |
https://mathoverflow.net/questions/440771 | 2 | Consider the matrix $$D=\begin{pmatrix}1&0\\0&e^{i\theta}\end{pmatrix}.$$ For the commonly used norms $\|\cdot\|$ on $\mathbb{C}^2$ or for $\theta=0$ the associated subordinate norm is $1$. Is it always true ? can a subordinate norm be strictly bigger than one ?
| https://mathoverflow.net/users/126690 | Existence of weird complex norms | Consider the norm $\Vert\cdot\Vert$ on $\mathbb{C}^2$ defined by $\Vert z \Vert^2 := |z\_1|^2+|z\_1+z\_2|^2$. Then
$$\left\Vert \left(\begin{array}{c}1 \\ -1 \end{array}\right) \right\Vert^2 = 1,$$
$$\left\Vert D\left(\begin{array}{c}1 \\ -1 \end{array}\right) \right\Vert^2 = \left\Vert \left(\begin{array}{c}1 \\ -e^{i\theta} \end{array}\right) \right\Vert^2 = 1 + |1-e^{i\theta}|^2.$$
If $e^{i\theta} \ne 1$, we derive that the subordinate norm of $D$ is $>1$.
| 6 | https://mathoverflow.net/users/169474 | 440782 | 177,960 |
https://mathoverflow.net/questions/440758 | 1 | Given a probability space $(\Omega, \mathcal {A}, P)$, what are the minimum and maximum of the quantity
$$
P(A\_1 \cap \cdots \cap A\_n) - P(A\_1) \cdots P(A\_n)
$$
over $A\_1, \ldots, A\_n \in \mathcal {A}$, $n \geq 1$?
When $n = 2$, it is easily seen, from the Cauchy-Schwarz inequality (since
$$
P(A\_1 \cap A\_2) - P(A\_1) P(A\_2 ) = E ((1\_{A\_1} -P(A\_1)) (1\_{A\_2} -P(A\_2))) \,
$$
and $ E ((1\_{A\_i} -P(A\_i))^2) = P(A\_i) - P(A\_i)^2 \leq \frac 14$, $i=1,2$),
that $-\frac 14$ and $\frac 14$ are lower and upper bounds,
achieved on simple examples (on $[0,1]$ with $P$ the Lebesgue measure e.g.).
But now for arbitrary $n \geq 3$?
| https://mathoverflow.net/users/498800 | Gap to independence | The suggestions in the [comment](https://mathoverflow.net/questions/440758/gap-to-independence#comment1137007_440758) by usul are correct.
Indeed, let
\begin{equation}
p:=P(B),\quad B:=\bigcap\_1^n A\_j,\quad p\_j:=P(A\_j).
\end{equation}
We want to find the extreme values of
\begin{equation}
d:=p-\prod\_1^n p\_j.
\end{equation}
Clearly, $p\_j\ge p$ for all $j$ and hence
\begin{equation}
d\le p-p^n\le\max\_{0\le p\le1}(p-p^n)=r-r^n,
\end{equation}
where $r:=1/n^{1/(n-1)}$. On the other hand, if $A\_1=\dotsb=A\_n$ and $p\_j=r$ for all $j$, then $d=r-r^n$. So,
\begin{equation}
\max d=r-r^n=\frac{n-1}{n^{n/(n-1)}}
\end{equation}
(so that $\max d\to1$ as $n\to\infty$).
Next,
\begin{equation}
B^c=\bigcup\_1^n A\_j^c,
\end{equation}
where $^c$ denotes the complement. So,
\begin{equation}
1-p=P(B^c)\le\sum\_1^n P(A\_j^c)=n-\sum\_1^n p\_j,
\end{equation}
so that $\sum\_1^n p\_j\le n-(1-p)$ and hence, by the AM–GM inequality,
\begin{equation}
d\ge p-\Bigl(\frac{n-(1-p)}n\Bigr)^n=p-\Bigl(1-\frac{1-p}n\Bigr)^n=:f(p).
\end{equation}
Since $f(p)$ is increasing in $p\in[0,1]$, we have $f(p)\ge f(0)=-\bigl(1-\frac1n\bigr)^n$. So,
$d\ge-\bigl(1-\frac1n\bigr)^n$. On the other hand, if the sets $A\_1^c,\dotsc,A\_n^c$ form a partition of $\Omega$ and $P(A\_j^c)=\frac1n$ for all $j$, then $p=0$ and $p\_j=1-\frac1n$ for all $j$, whence $d=-\bigl(1-\frac1n\bigr)^n$.
So,
\begin{equation}
\min d=-\Bigl(1-\frac1n\Bigr)^n
\end{equation}
(so that $\min d\to-1/e$ as $n\to\infty$).
| 2 | https://mathoverflow.net/users/36721 | 440790 | 177,962 |
https://mathoverflow.net/questions/440797 | 3 | Let $A\in M\_n(\mathbb{R})$ be a matrix and $\|\cdot\|$ be a norm on $\mathbb{C}^n$. When we look at the operator norm of $A$ with respect to $\|\cdot\|$ we can either consider the inclusion of $M\_n(\mathbb{R})$ in $M\_n(\mathbb{C})$ or the restriction of $\|\cdot\|$ to $\mathbb{R}^n$. Are these points of view equivalent?
In other words do we always have:
$$
\sup\_{x\in\mathbb{R}^n\setminus\lbrace0\rbrace}\frac{\|Ax\|}{\|x\|}=\sup\_{x\in\mathbb{C}^n\setminus\lbrace0\rbrace}\frac{\|Ax\|}{\|x\|}
$$
for $A\in M\_n(\mathbb{R})$.
I hoped that we could use as an intermediary step [this other question](https://mathoverflow.net/questions/440771/existence-of-weird-complex-norms) however, I was a bit optimistic.
| https://mathoverflow.net/users/126690 | Real and complex operator norms | $\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No. E.g., let
$$A=\begin{bmatrix}1&0\\0&0 \end{bmatrix}$$
and
$$\left\|\begin{bmatrix}z\_1\\z\_2 \end{bmatrix}\right\|=|z\_1|+|z\_1+iz\_2|.$$
Then the real norm of $A$ is $1$ and the complex norm of $A$ is $2$.
---
Indeed, if $x=\begin{bmatrix}x\_1\\x\_2\end{bmatrix}\in\R^2$ with $x\_1\ne0$, then
$$\frac{\|Ax\|}{\|x\|}=\frac{|x\_1|+|x\_1+0i|}{|x\_1|+|x\_1+ix\_2|}
\le\frac{2|x\_1|}{|x\_1|+|x\_1|}=1,$$
with $\frac{\|Ax\|}{\|x\|}=1$ if $x\_2=0$. So, the real norm of $A$ is $1$.
If now
$z=\begin{bmatrix}z\_1\\z\_2\end{bmatrix}\in\C^2$ with $z\_1\ne0$, then
$$\frac{\|Az\|}{\|z\|}=\frac{2|z\_1|}{|z\_1|+|z\_1+iz\_2|}
\le2,$$
with $\frac{\|Az\|}{\|z\|}=2$ if $z\_2=iz\_1$. So, the complex norm of $A$ is $2$.
| 2 | https://mathoverflow.net/users/36721 | 440802 | 177,965 |
https://mathoverflow.net/questions/440791 | 0 | I am trying to figure out if the infinite product $$\omega=\frac{5\sqrt{3}}{12}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
p\ge 13}}\left(\frac{p-2}{p-1}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
p\ge 13}}\left(\frac{p}{p-1}\right)$$ is asymptotically equal to the infinite product $$c=\frac{5775}{2592\pi}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
p\ge 13}}\left(\frac{p(p-2)}{(p-1)^2}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
p\ge 13}}\left(\frac{p^2}{p^2-1}\right)$$.
So, I reformulated the products as $$\omega(x)=\frac{5\sqrt{3}}{12}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
13\leq p\leq x}}\left(\frac{p-2}{p-1}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(\frac{p}{p-1}\right)$$ and $$c(x)=\frac{5775}{2592\pi}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
13\leq p\leq x}}\left(\frac{p(p-2)}{(p-1)^2}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(\frac{p^2}{p^2-1}\right)$$.
Then, taking the quotient we get
$$\frac{\omega(x)}{c(x)}=\frac{216\sqrt{3}\pi}{1155}\prod\limits\_{\substack{p\equiv 1\pmod3 \\
13\leq p\leq x}}\left(1-\frac{1}{p}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(1+\frac{1}{p}\right)$$ $$=\frac{216\sqrt{3}\pi}{1155}\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(1-\frac{1}{p^2}\right)\prod\limits\_{\substack{p\equiv 1\pmod3 \\
13\leq p\leq x}}\left(1-\frac{1}{p}\right)\prod\limits\_{\substack{p\equiv 2\pmod3 \\13\leq p\leq x}}\left(\frac{1}{1-\frac{1}{p}}\right).$$
As $x\to\infty$, from [A. Languasco's paper](https://core.ac.uk/reader/81188410), we observe that the last two products in the quotient above are asymptotically equal. So, the limit of the quotient depends on the product $$\frac{216\sqrt{3}\pi}{1155}\prod\limits\_{\substack{p\equiv 2\pmod3 \\
13\leq p\leq x}}\left(1-\frac{1}{p^2}\right)$$, but I don't know if this converges to $1$ or not as $x\to\infty$. I can see that the last product (except the constant) has something to do with $\frac{1}{\zeta(2)}$ but I don't know if it's actually smaller or greater than that.
I am essentially trying to check if eventually the products would be same or not?
I would really appreciate it if somebody could provide some hints or ideas on how to progress from here.
| https://mathoverflow.net/users/483436 | Asymptotic equivalence of two infinite products of prime numbers in residue classes | I don't know why you are restricting the products to $p \geq 13$ or where the factor $5\sqrt{3}/12$ is coming from. I am going to ignore that and discuss the following product over all primes $p$:
$$
C = \prod\_{p}\left(1- \frac{\chi(p)}{p-1}\right)
$$
where the terms in the product are in order of increasing $p$ and $\chi$ is the nontrivial Dirichlet character mod $3$, so $\chi(p) = 1$ if $p \equiv 1 \bmod 3$, $\chi(p) = -1$ if $p \equiv 2 \bmod 3$, and $\chi(3) = 0$.
When $p \equiv 1 \bmod 3$ we have
$$
1- \frac{\chi(p)}{p-1} = 1 - \frac{1}{p-1} = \frac{p-2}{p-1}
$$
and when $p \equiv 2 \bmod 3$ we have
$$
1- \frac{\chi(p)}{p-1} = 1 + \frac{1}{p-1} = \frac{p}{p-1}
$$
When $p = 3$, $\chi(p) = 0$, so
$$
1- \frac{\chi(p)}{p-1} = 1.
$$
Therefore the factors in $C$ are the same as what you wrote, but you broke up the product into separate products over $p \equiv 1 \bmod 3$ and $p \equiv 2 \bmod 3$ (I ignore the condition $p \equiv 13$). That is a subtle issue because *your products don't converge*. I am multiplying the terms at all primes in $C$ together in increasing order, not using separate products depending on $p \bmod 3$.
As an analogue, consider the alternating harmonic series $S = \sum\_{n \geq 1} (-1)^{n-1}/n$. This converges (and equals $\log 2$), but you can't write
$S$ with two separate sums over odd and even $n$:
$$
S \not= \sum\_{{\rm odd } \ n} \frac{1}{n} - \sum\_{{\rm even } \ n} \frac{1}{n}
$$
because those separate sums individually do not converge.
Returning to $C$, we want to insert factors into each term to improve the convergence (to make it absolutely convergent). To do that, write
$$
1 - \frac{\chi(p)}{p-1} = 1-\frac{\chi(p)/p}{1-1/p} =
1 - \frac{\chi(p)}{p} + \frac{\chi(p)}{p^2} + O\left(\frac{1}{p^3}\right)
$$
by expanding $1/(1-1/p)$ into a geometric series in powers of $1/p$. Since $1-\chi(p)/p \sim 1$ as $p \to \infty$, dividing by $1-\chi(p)/p$ tells us
$$
\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p} = 1 + O\left(\frac{1}{p^2}\right),
$$
so the product
$$
\prod\_{p} \left(\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p}\right)
$$
is absolutely convergent.
Now we can rewrite $C$:
$$
C = \prod\_p \left(1 - \frac{\chi(p)}{p-1}\right) = \prod\_p \left(1-\frac{\chi(p)}{p}\right)\left(\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p}\right),
$$
where the product is in order of increasing $p$. Can we split apart this product?
Yes! When ${\rm Re}(s) > 1$, the $L$-function of $\chi$ has the absolutely convergent Euler product representation
$$
L(s,\chi) = \prod\_{p} \frac{1}{1-\chi(p)/p^s}
$$
and it can be shown, with some nontrivial work, that this product remains valid on the line ${\rm Re}(s) = 1$ *provided* the product is taken in order of increasing $p$ (on that line the product is no longer absolutely convergent). Taking $s = 1$ and reciprocating,
$$
\prod\_p \left(1 - \frac{\chi(p)}{p}\right) = \frac{1}{L(1,\chi)}.
$$
Feeding that into the formula for $C$ above,
$$
C = \frac{1}{L(1,\chi)}\prod\_p \frac{1-\chi(p)/(p-1)}{1-\chi(p)/p}.
$$
It can be shown that $L(1,\chi) = \pi/(3\sqrt{3})$, so
$$
C = \frac{3\sqrt{3}}{\pi}\prod\_p \frac{1-\chi(p)/(p-1)}{1-\chi(p)/p},
$$
where the $p$-th factor in this product is $1+O(1/p^2)$.
What is the $p$-th factor here? When $\chi(p) = 1$,
$$
\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p} = \frac{1-1/(p-1)}{1-1/p} = \frac{p(p-2)}{(p-1)^2}
$$
and when $\chi(p) = -1$,
$$
\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p} = \frac{1+1/(p-1)}{1+1/p} = \frac{p^2}{p^2-1}.
$$
And when $\chi(p) = 0$ (namely when $p = 3$,
$$
\frac{1-\chi(p)/(p-1)}{1-\chi(p)/p} = 1.
$$
Because these terms are $1 + O(1/p^2)$, the order of multiplication now does not matter and we can split apart this product depending on $p \bmod 3$:
$$
C = \frac{3\sqrt{3}}{\pi}\prod\_{p\equiv 1 \bmod 3}\frac{p(p-2)}{(p-1)^2} \prod\_{p \equiv 2 \bmod 3} \frac{p^2}{p^2-1}.
$$
On the MO page [here](https://mathoverflow.net/questions/31150/calculating-the-infinite-product-from-the-hardy-littlewood-conjecture-f?rq=1) I discuss such a product with $\chi$ replaced by a more general Legendre symbol.
| 3 | https://mathoverflow.net/users/3272 | 440806 | 177,967 |
https://mathoverflow.net/questions/440804 | 3 | Consider the functor $\mathcal{F}: \text{Sch}/\mathbb{Q}\longrightarrow \text{Sets}$, defined by sending a scheme $X$ with coordinate ring $\mathcal{O}(X)$ to the set of orbits $B(\mathcal{O}(X))\setminus SL\_2(\mathcal{O}(X))$, where $B$ is the Borel subgroup of upper triangular matrices.
My question is if this functor is a sheaf.
What I've tried: it boils down to showing that for any open $U$ and a cover $\cup\_i U\_i = U$ by opens $U\_i$, the following sequence is exact
$$0\longrightarrow \mathcal{F}(U)\longrightarrow \prod\_i\mathcal{F}(U\_i)\rightrightarrows\prod\_{i,j}\mathcal{F}(U\_i\cap U\_j).$$
In [this](https://math.stackexchange.com/questions/86509/reference-for-it-is-enough-to-specify-a-sheaf-on-a-basis) stack exchange thread, I saw that it is enough to prove that the above sequence is exact for a basis of the topology, so it is enough to prove the statement when $U$ is affine and the $U\_i$ are basic opens.
I managed to prove the injectivity of $\mathcal{F}(U)\longrightarrow \prod\_i\mathcal{F}(U\_i)$ without much trouble, but I am not sure I understand how to prove that this functor admits descent, i.e. that the sequence is exact in the middle.
For example, even when there are two open subsets $U\_1,U\_2$ with $U = U\_1\cup U\_2$. Cosets of $B(\mathcal{O}(X))\setminus SL\_2(\mathcal{O}(U\_i))$ are parameterized by pairs $(c\_i,d\_i)\in \mathcal{O}(U\_i)$. Denote by $f\_i\in \mathcal{O}(U)$ the element for which $U\_i = D(f\_i)$, then possibly replacing the representative $(c\_i,d\_i)$ by multiplying with a power of $f\_i$, we may assume $(c\_i,d\_i)$ are in $\mathcal{O}(U)$.
Ideally, I would like to say that $(c\_1,d\_1)$ is a global section that lifts $s = \{(c\_1,d\_1)|\_{U\_1}, (c\_2,d\_2)|\_{U\_2}\}$, however, for this I would need to find a $b\in \mathcal{O}(U)^{\times}$ such that $bc\_1 = c\_2, bd\_1 = d\_2$, however, the only thing that follows from the fact that the section $s$ equalizes, is that there exists a $b\in \mathcal{O}(U\_1\cap U\_2)^{\times}$ such that $b(c\_1,d\_1) = (c\_2,d\_2)$. Write $b = b'/(f\_1f\_2)^m$, then this only means that there exists a power of $f\_1f\_2$ such that
$$(f\_1f\_2)^n(b'c\_1 - (f\_1f\_2)^mc\_2) = 0$$
in $\mathcal{O}(U)$. From this equality, I can only deduce that there exists an invertible $b'\in\mathcal{O}(U)$ for which
$$b'(f\_1f\_2)^nc\_1 = (f\_1f\_2)^{m+n}c\_2.$$
Had I known that $f\_1f\_2$ is not a zero divisor, then I could conclude that $f\_1f\_2$ is invertible in $\mathcal{O}(U)$ and complete the argument, but I don't. And generally, one cannot choose an irreducible basis of opens for a scheme unless it is irreducible to begin with.
In summary: don't know how to prove that the sequence is exact in the middle nor how to find a counter example. Either of which would be appreciated.
| https://mathoverflow.net/users/174655 | Is this functor $\mathcal{F}: \text{Sch}/\mathbb{Q}\longrightarrow \text{Sets}$ a sheaf? | No, this isn't true. Take $X = \mathbb{P}^1$ with $U\_1$ and $U\_2$ being the standard affine cover of $\mathbb{P}^1$. Let the coordinate rings of $U\_1$ and $U\_2$ be $k[t]$ and $k[t^{-1}]$, so the coordinate ring of $U\_1 \cap U\_2$ is $k[t, t^{-1}]$.
Map $U\_1$ to $B \backslash \text{SL}\_2$ by $\begin{bmatrix} 1&0 \\ t&1 \\ \end{bmatrix}$ and map $U\_2$ to $B \backslash \text{SL}\_2$ by $\begin{bmatrix} 0&-1 \\ 1&t^{-1} \\ \end{bmatrix}$.
We have
$$\begin{bmatrix} 1&0 \\ t&1 \\ \end{bmatrix}=
\begin{bmatrix} t^{-1} & 1 \\ 0&t \end{bmatrix}
\begin{bmatrix} 0&-1 \\ 1&t^{-1} \\ \end{bmatrix},$$
so these maps coincide on $U\_1 \cap U\_2$. But $\text{SL}\_2$ is affine and $\mathbb{P}^1$ is connected and projective, so $\text{SL}\_2(\mathbb{P}^1)$ contains only constant maps.
---
The point here is that $B \backslash \text{SL}\_2$ IS just $\mathbb{P}^1$, and the map $\text{SL}\_2 \to \mathbb{P}^1$ has local sections but no global sections.
| 6 | https://mathoverflow.net/users/297 | 440807 | 177,968 |
https://mathoverflow.net/questions/440781 | 9 | $\newcommand{\A}{\mathcal{A}}\newcommand{\T}{\mathcal{T}}$The notions of *continuous map* of topological spaces and *measurable function* of measurable spaces are very similar:
* A map of topological spaces from $(X,\T\_X)$ to $(Y,\T\_Y)$ is **continuous** if for each $V\in\T\_Y$, we have $f^{-1}(V)\in\T\_X$.
* A map of measureable spaces from $(X,\A\_X)$ to $(Y,\A\_Y)$ is **measurable** if for each $V\in\A\_Y$, we have $f^{-1}(V)\in\A\_X$.
Similarly, a map from $(X,\T\_X)$ to $(Y,\T\_Y)$ is **open** if for each $U\in\T\_X$, we have $f(U)\in\T\_Y$.
Does the analogue of openness for measurable spaces have a name? Is there some place where I can read more about it and its properties?
| https://mathoverflow.net/users/130058 | Analogue of open/closed maps for measurable spaces | There are at least three different answers that can be given to this question, and in all three interpretations the answer essentially states that all maps are “open”, for the appropriate analogue of “open”.
In the first interpretation, we may ask: under what conditions is the image of a measurable set measurable?
Of course, in measure theory we typically work with equivalence classes of measurable maps modulo equivalence almost everywhere, i.e., up to a set of measure 0 (alias negligible set), so first we need to give a more precise definition of an image that interacts nicely with negligible sets.
Below, I reuse the notation of the main post.
**Definition.** The *measurable image* of a mesurable map $\def\cA{{\cal A}}f\colon X→Y$ is a subset $V∈\cA\_Y$ such that $f^{-1}(Y∖V)$ is negligible and, furthermore, for any $W∈\cA\_Y$ such that $W⊂V$ and $f^{-1}(W)$ is negligible, the set $W$ itself is negligible. The *measurable image* of $U∈\cA\_X$ is defined as the measurable image of the map $f|\_U\colon U→Y$, where $U$ is equipped with the induced structure of an enhanced measurable space.
Of course, this definition requires the additional data of σ-ideals of negligible sets $\def\cN{{\cal N}}\cN\_X⊂\cA\_X$ and $\cN\_Y⊂\cA\_Y$, which we therefore include in the given data. Adopting the terminology of [arXiv:2005.05284](https://arxiv.org/abs/2005.05284), we refer to a triple $(X,\cA\_X,\cN\_X)$ as an *enhanced measurable space*. If we want composition of measurable maps to respect the equivalence relation of equality almost everywhere, we must also require that preimages of negligible sets are negligible, in complete analogy to the existing requirement that preimages of measurable sets are measurable. Below, a *morphism* of enhanced measurable spaces is an equivalence class of such maps modulo equality almost everywhere; in the uncountably separated case some care must be exercised when defining equality almost everywhere, see Definition 4.13 in [arXiv:2005.05284](https://arxiv.org/abs/2005.05284).
The following theorem is due to Fremlin (Lemma 451Q in his *Measure Theory*). An exposition that matches the presentation of this post can be found in Proposition 4.65 of [arXiv:2005.05284](https://arxiv.org/abs/2005.05284).
**Theorem (Fremlin, 2003).** The measurable image exists whenever $(X,\cA\_X,\cN\_X)$ is Marczewski-compact and $(Y,\cA\_Y,\cN\_Y)$ is strictly localizable. The equivalence class (up to a negligible set) of the measurable image is uniquely defined.
Marczewski-compactness is essentially the abstract reformulation of a Radon measure (not relying on the structure of a topological space) and strict localizability amounts to being isomorphic to a disjoint union of σ-finite spaces. In particular, with the exception of trivial counterexamples constructed for a specific purpose, all measure spaces used in standard textbooks are compact and strictly localizable, since Radon measure spaces are automatically compact and all σ-finite measure spaces are strictly localizable. See the sources cited above for more details.
In the second interpretation we may ask for a more direct relation between measurable maps of measurable spaces and open maps of topological spaces.
Here is a theorem that establishes such a relation.
**Theorem (Theorem 1.1 in [arXiv:2005.05284](https://arxiv.org/abs/2005.05284)).** The category of compact strictly localizable enhanced measurable spaces is equivalent to the category of hyperstonean topological spaces and open continuous maps.
In one direction, the equivalence sends an enhanced measurable space $(X,\cA\_X,\cN\_X)$ to the Gelfand spectrum of the von Neumann algebra of bounded measurable functions on $X$, or, equivalently, the Stone spectrum of the Boolean algebra $\cA\_X/\cN\_X$. In the other direction, the equivalence sends a hyperstonean topological space $\def\cT{{\cal T}}(X,\cT\_X)$ to the enhanced measurable space $(X,\cA\_X,\cN\_X)$, where $\cN\_X$ is the set of all meager subsets of $X$ (alias sets of first category) and $\cA\_X$ is the set of all subsets of $X$ with the Baire property (i.e., symmetric differences of open subsets and meager subsets).
Thus, morphisms of enhanced measurable spaces correspond precisely to open maps of hyperstonean topological spaces.
Finally, if you are willing to go a little bit beyond point-set topological spaces in the setting of [locales](https://ncatlab.org/nlab/show/locale) (which are a rather mild adjustment of the usual notion of a topological space), then a third answer is possible.
**Theorem (Theorem 1.1 in [arXiv:2005.05284](https://arxiv.org/abs/2005.05284)).** The category of compact strictly localizable enhanced measurable spaces is equivalent to the category of *measurable locales*.
Here the category of measurable locales is a full subcategory of locales whose underlying frames are Boolean algebras admitting a completely additive measure.
It is in this precise sense that we can consider measure theory to be a part of general topology (understood here as including the study of locales). Having nonspatial locales (i.e., locales that do not come from a point-set topological space) is essential, since almost all measurable locales are not spatial. The notion of an open map makes sense for locales, and can be defined in a formally similar way: the image of an open sublocale must be open. With this definition, it is easy to prove (Lemma 2.44 in [arXiv:2005.05284](https://arxiv.org/abs/2005.05284)) that any morphism of measurable locales is an open map of locales.
| 5 | https://mathoverflow.net/users/402 | 440814 | 177,971 |
https://mathoverflow.net/questions/440743 | 12 | Decades ago, Voevodsky constructed the six-functor formalism in motivic homotopy theory [[Ayoub's thesis](https://user.math.uzh.ch/ayoub/PDF-Files/THESE.PDF)]. This construction seems very technical, long and "hard".
Very recently [[Mann's thesis](https://arxiv.org/abs/2206.02022)], the six-functor formalism has been defined to be a lax symmetric monoidal functor $D:Corr(C,E)\rightarrow Cat\_\infty$ such that the induced functors $\otimes,f^\*$ and $f\_!$ have right adjoints. This construction is concise and short.
Question: Are the two constructions "equivalent"? Would Mann's definition surprise Voevodsky or would he just say: "this is exactly what I meant."?
| https://mathoverflow.net/users/173315 | Voevodsky's six functor formalism VS Lucas Mann's | There may be some confusion in this question about what exactly Voevodsky/Ayoub and Mann do, as they do very different things.
* Mann's thesis constructs a formalism of six operations in the setting of rigid-analytic geometry, using some $\infty$-categorical construction techniques developed for this purpose by Liu and Zheng. Along the way he gives an abstract definition of what a "formalism of six operations" is using categories of spans, but this definition was certainly well known and already appears (in a more complete form, see below) in the book of Gaitsgory and Rozenblyum.
* Ayoub's thesis, based on Voevodsky's unpublished ideas, explains how one gets for free a formalism of six operations on the category of schemes out of some very simple axioms (what he calls a "homotopy 2-functor"). These axioms are of a geometric rather than categorical nature. The output of Ayoub's theorem (combined with the $\infty$-categorical construction techniques of Gaitsgory-Rozenblyum or of Liu-Zheng) is in particular a formalism of six operations in the sense of Mann.
Note also that the definition of a formalism of a six operations in Mann's thesis is far from capturing everything. A more complete definition would be a right-lax symmetric monoidal functor from an $(\infty,2)$-category in which both the 1-morphisms and the 2-morphisms are spans (Gaitsgory and Rozenblyum work with an intermediate 2-category, probably for simplicity's sake, in which the 1-morphisms are spans and the 2-morphisms are just morphisms). These 2-morphisms encode the functoriality of bivariant homology, i.e., the isomorphisms $f\_!\simeq f\_\*$ for $f$ proper and $f^!\simeq f^\*$ for $f$ étale. But even this does not capture all the structure that we have in the examples, such as the contravariance of bivariant homology with respect to quasi-smooth morphisms.
| 21 | https://mathoverflow.net/users/20233 | 440819 | 177,975 |
https://mathoverflow.net/questions/440754 | 1 | It is known that the closed unit ball of $L\_{\infty}(\mu)$ is weakly compact in $L\_{1}(\mu)$. A natural question arises in the case of spaces of Bochner integral functions:
Question. Let $X$ be a Banach space. In what cases the closed unit ball of $L\_{\infty}(\mu,X)$ is weakly compact in $L\_{1}(\mu,X)$ ?
I am not sure that this question is interesting or remains open.
| https://mathoverflow.net/users/41619 | Weak compactness of the closed unit ball of $L_{\infty}(\mu,X)$ in $L_{1}(\mu,X)$ | As Jochen commented, you need $X$ to be reflexive, and this is sufficient. It is enough to show that the unit ball of $L\_\infty(X)$ is closed in the reflexive space $L\_2(X)$. But the injection from $L\_\infty(X)$ into $L\_2(X)$ is the adjoint of the injection from $L\_2(X)$ into $L\_1(X)$, so this is automatic. (Recall that $L\_p(X)^\* = L\_{p'}(X^\*)$ when $X$ is reflexive and $p <\infty$. See Diestel-Uhl section IV.1.)
| 4 | https://mathoverflow.net/users/2554 | 440822 | 177,977 |
https://mathoverflow.net/questions/440824 | 5 | A classical result from Young in 1936 says that if $f\in C^\alpha$ and $g\in C^\beta$ with $\alpha+\beta>1$ then $\int f \, dg$ exists as a Riemann-Stieltjes integral.
However, I am interested in the converse. Clearly if the support of $f$ is disjoint with the support of $g$ then they can have as bad of analytic properties as you'd want.
However, can we have $f\in C^\alpha$ and $f$ nowhere locally $C^{\alpha'}$ for any $\alpha'>\alpha$, $g\in C^\beta$ and $g$ nowhere locally $C^{\beta'}$ for any $\beta'>\beta$, with $\alpha+\beta<1$, and have the Riemann Stieltjes integral converge?
| https://mathoverflow.net/users/479223 | Converse to Young's classical result on Riemann-Stieltjes integration | $\newcommand\al\alpha\newcommand\be\beta$Yes, of course. Take any $\al>0$ and $\be>0$ such that $\al+\be<1$. For $x\in[0,1]$, let
$$f(x):=\sum\_{j=1}^\infty 2^{-j}(x-r\_j)\_+^\al$$
and
$$g(x):=\sum\_{j=1}^\infty 2^{-j}(x-r\_j)\_+^\be,$$
where $(r\_1,r\_2,\dots)$ is an enumeration of the rational numbers in $[0,1)$ and $u\_+:=\max(0,u)$.
For any real $u,v$ one has $u\_+^\al-v\_+^\al\le(u-v)\_+^\al$ and hence for any $x,y$ in $[0,1]$
$$f(x)-f(y)\le\sum\_{j=1}^\infty 2^{-j}(x-y)\_+^\al=(x-y)\_+^\al,$$
so that $f\in C^\al[0,1]$. Also, for any $j$
$$\liminf\_{x\downarrow r\_j}\frac{f(x)-f(r\_j)}{(x-r\_j)^\al}\ge2^{-j}>0$$
and hence for any real $\al'>\al$
$$\liminf\_{x\downarrow r\_j}\frac{f(x)-f(r\_j)}{(x-r\_j)^{\al'}}=\infty,$$
so that $f$ is nowhere $C^{\al'}$ on $[0,1]$.
Similarly, $g\in C^\be[0,1]$ but $g$ is nowhere $C^{\be'}$ on $[0,1]$ for any real $\be'>\be$.
Finally,
$$\int\_0^1 f\,dg=\sum\_{j=1}^\infty\sum\_{k=1}^\infty 2^{-j}2^{-k}\,I\_{j,k},$$
where
$$I\_{j,k}:=\int\_0^1(x-r\_j)\_+^\al\be(x-r\_k)\_+^{\be-1}\,dx
\le\int\_0^1\be(x-r\_k)\_+^{\be-1}\,dx\le1.$$
So, $\int\_0^1 f\,dg\in[0,1]\subset\mathbb R$. $\quad\Box$
| 6 | https://mathoverflow.net/users/36721 | 440829 | 177,979 |
https://mathoverflow.net/questions/440772 | 3 | Let $p \in [1, \infty)$. Let $\mathcal P\_p(\mathbb R^d)$ be the space of all Borel probability measures on $\mathbb R^d$ with finite $p$-th moments. Let $D\_p$ be the collection of all Borel measurable functions $f:\mathbb R^d \to \mathbb R\_{\ge 0}$ such that $\int\_{\mathbb R^d} f (x) \, \mathrm d x = 1$ and $\int\_{\mathbb R^d} |x|^p f (x) \, \mathrm d x < \infty$. So $f \in D\_p$ if and only if $f$ is a density of some $\mu \in \mathcal P\_p(\mathbb R^d)$.
Let $F:D\_p \to \mathcal P\_p(\mathbb R^d)$ that sends a density $f$ to its corresponding $\mu \in \mathcal P\_p(\mathbb R^d)$. We endow $\mathcal P\_p(\mathbb R^d)$ with the $L\_p$-Wasserstein metric $W\_p$. We endow $D\_p$ with the norm $[\cdot]$ defined by
$$
[f-g]\_p := \int\_{\mathbb R^d} |f(x)-g(x)| \cdot |x|^p \, \mathrm d x \quad \forall f,g \in D\_p.
$$
>
> Is $F$ Lipschitz? Are there some special properties about $F$?
>
>
>
Thank you so much for your elaboration?
| https://mathoverflow.net/users/477203 | Is this map (from the space of probability densities to the Wasserstein space) Lipschitz? | $\newcommand{\vpi}{\varphi}\newcommand\R{\mathbb R}$
>
> **Claim 1:** The map $F$ is not Lipschitz if $p>1$.
>
>
>
>
> **Claim 2:** The map $F$ is $1$-Lipschitz if $p=1$: For all $f,g$ in $D\_p$,
> \begin{equation\*}
> W\_1(F(f),F(g))\le[f-g]\_1. \tag{1}\label{1}
> \end{equation\*}
>
>
>
*Proof of Claim 1:* Take any real $a>0$ and a small $h>0$. Let $f$ and $g$ be the densities of the uniform distributions over the cubes $[0,h]^d$ and $[a,a+h]\times[0,h]^{d-1}$. Then $[f-g]\_p\to a^p$ whereas $W\_p(F(f),F(g))\to a$ as $h\downarrow0$. Letting now $a\downarrow0$, we complete the proof of Claim 1. $\quad\Box$
*Proof of Claim 2:* By (say) Theorem 1.17 (Duality) in [A user's guide to optimal transport](https://www.math.umd.edu/%7Eyanir/OT/AmbrosioGigliDec2011.pdf) (and the last paragraph on p. 8 there),
\begin{equation\*}
W\_1(F(f),F(g))=\int\vpi\,d\mu+\int\psi\,d\nu, \tag{2}\label{2}
\end{equation\*}
for some $\vpi\in L^1(\mu)$ and $\psi\in L^1(\nu)$ such that $\vpi=\psi^\*$ and $\psi=\vpi^\*$, where
\begin{equation\*}
\theta^\*(y):=\inf\_{x\in\R^d}(|x-y|-\theta(x)) \tag{3}\label{3}
\end{equation\*}
for $y\in\R^d$; note that the function $\theta^\*$ is $1$-Lipschitz, as the infimum of $1$-Lipschitz functions $y\mapsto|x-y|-\theta(x)$.
So, $\vpi=\psi^\*$ is $1$-Lipschitz. Using \eqref{3} again, we see that $\psi(y)=\vpi^\*(y)\le|y-y|-\vpi(y)$ for all $y$, so that $\psi\le-\vpi$. So, by \eqref{2},
\begin{equation\*}
\begin{gathered}
W\_1(F(f),F(g))\le\int\vpi\,d\mu-\int\vpi\,d\nu
=\int\vpi\,d(\mu-\nu)=\int(\vpi-\vpi(0))\,d(\mu-\nu) \\
\le\int|\vpi-\vpi(0)|\,d|\mu-\nu|
=\int|\vpi(x)-\vpi(0)|\,|f(x)-g(x)|\,dx \\
\le\int|x|\,|f(x)-g(x)|\,dx=[f-g]\_1,
\end{gathered}
\end{equation\*}
which completes the proof of Claim 2 (the latter inequality follows because $\vpi$ is $1$-Lipschitz). $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 440836 | 177,983 |
https://mathoverflow.net/questions/440206 | 13 | Recently I've been learning more about differential geometry, and I came upon the notion of a [diffeological space](https://en.wikipedia.org/wiki/Diffeology), which encompasses a number of already known extensions of smooth manifolds or related notions, like Banach and Frechét manifolds, complex and analytic manifolds, but also includes a number of other constructions (like quotients and mapping spaces), making the category of diffeological spaces quite well-behaved and nice to work with.
However, I couldn't find much information about applications of diffeology to "ordinary" differential geometry, and would love to hear about some results in this vein. Have diffeological spaces been used to obtain meaningful results about ordinary manifolds (smooth, complex, analytic, p-adic, etc.), specially for cases in which there's no known proof that does not use diffeology?
One example would be something like using the de Rham cohomology of (or other constructions involving) the diffeological space of diffeomorphisms/symplectomorphisms/smooth maps to prove results about ordinary manifolds.
| https://mathoverflow.net/users/130058 | Applications of diffeological spaces to ordinary differential geometry | As far as I am aware, you can find some applications of diffeology "merely" in differential geometry of manifolds in the following list (I am not sure this list is exhaustive):
* **The (internal) tangent space of the diffeomorphism group of a compact manifold is the space of its vector fields**, See [Hector G. Géométrie et topologie des espaces difféologiques. Analysis and geometry in foliated manifolds (Santiago de Compostela, 1994). 1995 Nov 17:55-80.](https://books.google.com/books?hl=en&lr=&id=Fec7DwAAQBAJ&oi=fnd&pg=PA55&dq=Hector%20G.%20G%C3%A9om%C3%A9trie%20et%20topologie%20des%20espaces%20diff%C3%A9ologiques.%20Analysis%20and%20geometry%20in%20foliated%20manifolds%20(Santiago%20de%20Compostela,%201994).%201995%20Nov%2017:55-80.&ots=dErQjDNNko&sig=dSc43L8CJ8ZTDCrPBBJnUy0VSNQ#v=onepage&q&f=false)**or**[[J.D. Christensen, E. Wu, Tangent spaces and tangent bundles for diffeological spaces, Cah. Topol. G'{e}om. Diff'{e}r. Cat'{e}g. 57(1) (2016), 3-50.]](http://cahierstgdc.com/wp-content/uploads/2017/11/ChristensenWu.pdf).
* **The diffeomorphism group of a Lie foliation**, See [Hector G, Macías-Virgós E, Sotelo-Armesto A. The diffeomorphism group of a Lie foliation. InAnnales de l'Institut Fourier 2011 (Vol. 61, No. 1, pp. 365-378).](http://www.numdam.org/item/10.5802/aif.2605.pdf)
* **De Rham cohomology of diffeological spaces and foliations**, See [Hector G, Macías-Virgós E, Sanmartín-Carbón E. De Rham cohomology of diffeological spaces and foliations. Indagationes Mathematicae. 2011 Aug 1;21(3-4):212-20.](https://www.sciencedirect.com/science/article/pii/S0019357711000140)
* **The basic de Rham complex of a singular foliation**, See [Miyamoto D. The Basic de Rham Complex of a Singular Foliation. International Mathematics Research Notices.](https://academic.oup.com/imrn/article-abstract/2023/8/6364/6540673?redirectedFrom=PDF)
* **Basic forms and orbit spaces: a diffeological approach**, See [Karshon Y, Watts J. Basic forms and orbit spaces: a diffeological approach. SIGMA. Symmetry, Integrability and Geometry: Methods and Applications. 2016 Mar 8;12:026.](https://www.emis.de/journals/SIGMA/2016/026/sigma16-026.pdf)
* **The orientation-preserving diffeomorphism group of $\mathbb{S}^2$ deforms to SO (3) smoothly**, See [Li J, Watts JA. The orientation-preserving diffeomorphism group of $\mathbb{S}^2$ deforms to SO (3) smoothly. Transformation Groups. 2011 Jun;16(2):537-53.](https://link.springer.com/article/10.1007/s00031-011-9130-0)
* **Smooth Lie group actions are parametrized diffeological subgroups**, See [Iglesias-Zemmour P, Karshon Y. Smooth Lie group actions are parametrized diffeological subgroups. Proceedings of the American Mathematical Society. 2012 Feb;140(2):731-9.](https://www.ams.org/journals/proc/2012-140-02/S0002-9939-2011-11301-7/S0002-9939-2011-11301-7.pdf)
* **Differential forms on manifolds with boundary and corners** See [Gürer S, Iglesias-Zemmour P. Differential forms on manifolds with boundary and corners. Indagationes Mathematicae. 2019 Sep 1;30(5):920-9.](https://www.sciencedirect.com/science/article/abs/pii/S0019357719300436)
* **The Geodesics of the 2-Torus** See [here](http://math.huji.ac.il/%7Epiz/documents/DBlog-Rmk-TGOT2T.pdf).
* **Every symplectic manifold is a (linear) coadjoint orbit** See [Donato P, Iglesias-Zemmour P. Every symplectic manifold is a (linear) coadjoint orbit. Canadian Mathematical Bulletin. 2022 Jun;65(2):345-60.](https://www.cambridge.org/core/journals/canadian-mathematical-bulletin/article/abs/every-symplectic-manifold-is-a-linear-coadjoint-orbit/60EEA4D0397E72FAE554ADEE9EC800B0)
* **De Rham cohomology for the complement of a (dense) irrational flow on the torus** See [this MO post and Patrick I-Z's answer](https://mathoverflow.net/questions/427637/can-one-make-sense-of-de-rham-cohomology-for-the-complement-of-a-dense-irratio/449827#449827)
I also believe that even if one is interested in smooth manifolds, it is much easier and conceptual (or perhaps natural) to define "smooth" manifolds through diffeological spaces, just like the definition of topological manifolds. Actually, we can say that a smooth manifold is a diffeological space that is locally Euclidean: every point in the space has an open neighborhood (with respect to the D-topology) which is diffeomorphic to an open subset of a fixed Euclidean space. After that, one can add Hausdorfness and second-countability requirements, if needed. In this situation, there is no need to talk about the smooth compatibility of charts, because it is automatically verified. Notice that diffeological spaces are an extension of Euclidean spaces in the first place.
| 10 | https://mathoverflow.net/users/131015 | 440841 | 177,984 |
https://mathoverflow.net/questions/440783 | 2 | I am reading the research article *"The Hausdorff dimension of the boundary of Mandelbrot set and Julia sets"* by Shishikura. The following two definitions are given without any examples in this paper. Therefore to understand the following definitions, I need examples of these two definitions:
Let $f$ be rational map on $\mathbb{C}\_{\infty}$. A closed subset $X$ of $\mathbb{C}\_{\infty}$ is called a hyperbolic subset for $f$ if
1. $f(X) \subset X$ and
2. there exist a positive constant $c$ and $\kappa$ $> 1$ such that $\lVert (f^n)^{'} \rVert \geq c \kappa^{n}$ on $X$ for $n \geq 0$. Here $\rVert . \rVert$ denotes the norm of derivative with respect to the spherical metric on $\mathbb{C}\_{\infty}$.
Let $\Lambda$ be open set of $\mathbb{C}$. A family $\{f\_\lambda : \lambda \in \Lambda \}$ of rational maps is $J$-Stable at $\lambda\_0 \in \Lambda$, if there exists a continuous map $h : \Lambda ^{'} \times J(f\_{\lambda\_{0}}) \rightarrow \mathbb{C}\_{\infty}$, such that $\Lambda ^{'}$ is neighborhood of $\lambda\_0$ in $\Lambda$, $h\_{\lambda} \equiv h(\lambda,.)$ is conjugacy from $(J(f\_{\lambda\_{0}}),f\_{\lambda\_{0}})$ to $(J(f\_{\lambda}), f\_{\lambda})$ and $h\_{\lambda\_{0}} = \mathrm{id}\_{J(f\_{\lambda\_{0}})}$.
Where $J(f)$ denotes the Julia set of $f$ and in both definitions $\mathbb{C}\_{\infty}$ denotes the Riemann sphere.
To get the example of second definition I tried to work with the quadratic family $P(z) = z^2 + c$. But didn't get anything.
Note that I am not aware about hyperbolic dynamics. After studying some general theory of complex dynamics and some great examples in it. I am studying this research article to understand the proof that Hausdorff dimension of the boundary of the Mandelbrot set is $2$.
| https://mathoverflow.net/users/499397 | Examples of hyperbolic set and J-stable sets | Hyperbolic functions - for example, quadratic polynomials with an attractive periodic point - are examples of maps that are J-stable. The notion of J-stability arises from the famous article of Mañe, Sad and Sullivan. It is discussed in McMullen's book on renormalisation.
The trivial examples of hyperbolic sets are repelling periodic orbits. You can think of hyperbolic sets as a generalisation of this to more general repelling subsets of the Julia set. The Julia set of a hyperbolic map is also a hyperbolic set. More generally, any compact forward-invariant set disjoint from the closure of the critical orbits is a hyperbolic set.
There is a connection between the two notions : A hyperbolic set will move holomorphically in a neighbourhood in parameter space. This is why they are relevant to Shishikuras proof: The main point is that there are parameters for which there are hyperbolic sets of dimension arbitrarily close to 2. Then you find corresponding sets of parameters via the corresponding holomorphic motions.
| 1 | https://mathoverflow.net/users/3651 | 440849 | 177,987 |
https://mathoverflow.net/questions/440726 | 7 | Let $\Sigma\_n$ be a genus $n$ surface, let $\mathcal{H}\_n$ be a genus $n$ handle body, and let $F\_n$ be a free group of rank $n$. Fix an identification of $\pi\_1(\mathcal{H}\_n)$ with $F\_n$. I know several proofs of the following result:
**Theorem**: Let $\phi\colon \pi\_1(\Sigma\_n) \rightarrow F\_n$ be a surjection. Then there exists an orientation-preserving homeomorphism $\psi\colon \Sigma\_n \rightarrow \partial \mathcal{H}\_n$ such that $\phi$ factors as
$$\pi\_1(\Sigma\_n) \stackrel{\psi\_{\ast}}{\longrightarrow} \pi\_1(\partial \mathcal{H}\_n) \longrightarrow \pi\_1(\mathcal{H}\_n) = F\_n.$$
However, I do not know any references for it, nor who to attribute it to. Does anyone know any references, preferably the original one?
| https://mathoverflow.net/users/499341 | Surjections from genus $n$ surface group to free group of rank $n$ | I think the first formal proof is due to Zieschang, Stallings probably knew it:
<https://mathscinet.ams.org/mathscinet/search/publdoc.html?pg1=INDI&s1=187195&sort=Newest&vfpref=html&r=101&mx-pid=161901>
There is a discussion at the end of the paper that refers to a correspondence with Lyndon. There it is mentioned that the claim is implicit in Satz 2.
| 4 | https://mathoverflow.net/users/69797 | 440851 | 177,988 |
https://mathoverflow.net/questions/440852 | 12 | Let $X$ be a smooth projective variety over a number field $K$. Then there are two cohomology groups we can attach to $X$: the algebraic de Rham cohomology group
$H^k\_{\text{dR}}(X/K), $
which is a finite dimensional $K$-vector space, and the singular cohomology group
$H^k\_{\text{sing}}(X(\mathbf{C}), \mathbf{Q}), $
which is a finite dimensional $\mathbf{Q}$-vector space. De Rham's theorem gives us an isomorphism between these two cohomology groups:
$$\sigma: H^k\_{\text{dR}}(X/K) \otimes\_{K} \mathbf{C} \xrightarrow{\sim} \,H^k\_{\text{sing}}(X(\mathbf{C}), \mathbf{Q}) \otimes\_{\mathbf{Q}} \mathbf{C}.$$
The two groups in this isomorphism both have a rational structure. The de Rham cohomology group $H^k\_{\text{dR}}(X/K) \otimes\_{K} \mathbf{C}$ has a $K$-lattice inside it given by $H^k\_{\text{dR}}(X/K)$. And the singular cohomology group $H^k\_{\text{sing}}(X(\mathbf{C}), \mathbf{Q}) \otimes \mathbf{C}$ has a $\mathbf{Q}$ lattice inside it given by $H^k\_{\text{sing}}(X(\mathbf{C}), \mathbf{Q})$.
**My question is:** what is the relation between these two lattices under the isomorphism $\sigma$? For example, if $K = \mathbf{Q}$, then does the $K$-lattice on the de Rham side map to the $\mathbf{Q}$-lattice on the singular cohomology side? In general, if $K \neq \mathbf{Q}$, is there any relation between these two lattices we can speak of?
| https://mathoverflow.net/users/394740 | Comparing singular cohomology with algebraic de Rham cohomology | This is the subject of *periods*: recall that the de Rham isomorphism between $H^k\_{\text{dR}}(X/K) \otimes\_K \mathbf C = H^k\_{\text{dR}}(X\_{\mathbf C}/\mathbf C)$ and $H^k\_{\text{sing}}(X(\mathbf C),\mathbf Q) \otimes\_{\mathbf Q} \mathbf C = H^k\_{\text{sing}}(X(\mathbf C),\mathbf C)$ is defined by integrating $k$-forms $\eta \in H^k\_{\text{dR}}(X\_{\mathbf C}/\mathbf C)$ along $k$-cycles $\Delta^k \to X(\mathbf C)$.
This shows that the two lattices should *not* be the same: the integral of a $\mathbf Q$-valued $k$-form along an integral $k$-cycle $\Delta^k \to X(\mathbf C)$ need not be a rational number. For instance, if $X = \mathbf G\_m = \mathbf A^1 \setminus \{0\}$, then $H\_1(X(\mathbf C),\mathbf Z)$ is generated by a loop around the origin, and $H^1\_{\text{dR}}(X/\mathbf Q) = H^0(X,\Omega\_X^1)$ is generated by $\tfrac{\text{d}z}{z}$. The integral is $2 \pi i$, which is not a rational number. Thus, the lattices do not agree. You can see an incarnation of this in Hodge theory, where Tate twists show up by multiplying a lattice by $2\pi i$.
There are similar examples in the projective case, for instance on elliptic curves (but they become harder to compute). The question is which complex numbers are periods is one that is heavily studied. A great introduction is the book [*Periods and Nori motives*](https://doi.org/10.1007/978-3-319-50926-6) by Annette Huber and Stefan Müller-Stach, Parts III and IV (Chapters 11–16).
| 16 | https://mathoverflow.net/users/82179 | 440854 | 177,989 |
https://mathoverflow.net/questions/440405 | 7 | On affine space, a sufficiently smooth continuous-time [Hamiltonian dynamic system](https://en.wikipedia.org/wiki/Hamiltonian_system) $\dot p = \nabla\_q H, \dot q = -\nabla\_p H$ conserves $H$, preserves the volume form (e.g. if we are looking for ergodicity), and lends itself to a limited physical intuition of position-momentum.
But on the other hand any flow $\dot p = f(p)$ can be seen as the projection of a Hamiltonian system with $H(p, q) = q^t f(p) + g(p)$, where the "gauge" $g$ is any differentiable scalar function.
The resulting Hamiltonian flow is
$$
\begin{align}
\dot p &= f(p) \\
-\dot q &= \left[\nabla f(p)\right] q+ \nabla g(p)
\end{align}
$$
A quick example shows how the $p, q$ system conserves volume in phase space. Let $f$ be multiplication by a negative matrix $A$. Then $p \to 0$ at roughly the same rate at which $q \to \infty$.
Is there a name for this construction, and is there a physical meaning for the variable $q$ and the function $g$? Are there any interesting choices for $g$? (My first instinct is to set $\dot q = 0$, but I am doubt that the "directional derivative" $p \mapsto \left[\nabla f(p)\right] q$ is a conservative function that doesn't depend on $q$.)
| https://mathoverflow.net/users/140723 | Hamiltonian-ization of a dynamic system | This construction is somewhat similar to the Martin-Siggia-Rose formalism for writing expectation values over solutions of a stochastic differential equation
$$
\dot{p}(t)=f(p,t)+\xi(p,t)
$$
with $\xi$ a Gaussian noise with correlation function
$$
\mathbb{E}[\xi(p,t)\xi(p',t')]=G(p,t,p',t')
$$
as a path integral
$$
\mathbb{E}[O[p]]=\int\mathcal{D}p\mathcal{D}q\,O[p]\,e^{-S[p,q]}
$$
with action
$$
S[p,q]=\int \mathrm{d}t\, q\left(\dot{p}-f(p)\right)+\frac{1}{2}\int \mathrm{d}t\mathrm{d}t'\,G(q(t),t,q(t'),t')q(t)q(t').
$$
In your situation $G\equiv 0$, so there is no stochastic aspect, and the Lagrangian is just $L=q\dot{p}-qf(p)$, where $p$ is really a generalized position coordinate, whence the Hamiltonian is
$$
H=P\dot{p}-L=(P-q)\dot{p}+qf(p)
$$
where now $P$ is the conjugate momentum to $p$, and identifying $P=q$ yields your ansatz.
As to a physical interpretation of $q$, I don't think there is an obvious one (particularly since $g(q)$ is arbitrary and influences the evolution of $q$).
| 0 | https://mathoverflow.net/users/45250 | 440855 | 177,990 |
https://mathoverflow.net/questions/440869 | 2 | Forgive me for asking what is undoubtedly an elementary question.
The Weil representation (defined below) of the metaplectic group $\operatorname{Mp}\_2(\mathbb{Z})$ can be defined in terms of the generators traditionally denoted $(T,1)$ and $(S,\sqrt{\tau})$. On a superficial level the images of these generators seems to depend *only* on the discriminant form $(L'/L,Q)$ of a lattice $(L,Q)$ (defined below), however, much of the theory seems to depend on the lattice itself (in particular, its signature). For example, in her [thesis](https://tuprints.ulb.tu-darmstadt.de/4458/) Alfes-Neumann works with theta functions defined with respect to a lattice of signature $(1,2)$, and in [later work](https://arxiv.org/abs/1607.02701) Bruinier and his students [Klein and Kupka](https://pubmed.ncbi.nlm.nih.gov/34840518/) use lattices of signature $(1,0)$ while citing Alfes-Neumann's formulas for the lattice of signature $(1,2)$.
While this isn't stated in the papers, it seems that the reason this can be justified is because there is an isomorphism of discriminant groups which preserves the quadratic form. I should also mention the lattices both satisfy $b\_--b\_+=\pm1$, where the signature of the lattice is $(b\_+,b\_-)$ (there is a duality between a lattice of signature $(b\_-,b\_+)$ and a lattice of signature $(b\_+,b\_-)$ that we can use).
**The question is then: Does the Weil representation $ \rho\_L $ depend only on the discriminant form $ (L'/L,Q) $?** That is, if $ M $ is another lattice with quadratic form $ R $ and there exists a group isomorphism $ \phi:L'/L\to M'/M $ with $ R(\phi(x)) = Q(x) $ for all $ x $ in $ L'/L $, is it the case that the Weil representations $ \rho\_L $ and $ \rho\_M $ are equivalent?
It seems trivial that this should be the case, but then why use the more complicated lattice of Alfes' thesis instead of a simple one? Moreover, there are other results in the literature which work for lattices of certain signature. Do these results extend to any lattice with an appropriate isomorphism?
---
Recall a lattice is a finitely generated $\mathbb{Z}$-module $L$ together with a quadratic form $Q$. Specifying a quadratic form on $L$ is equivalent to specifying a symmetric bilinear form on $L$. An even lattice is one for which $Q(\lambda)\in \mathbb{Z}$ for all $\lambda$. The dual lattice of $L$ is the set $\\{y\in L\otimes\_\mathbb{Z}\mathbb{Q}:(x,y)\in\mathbb{Z}\text{ for all }x\in L\\}$. Here $Q$ is extended in the natural way.
Let $ (L,Q) $ be an even nondegenerate lattice of signature $ (b\_+,b\_-). $ The finite dimensional group algebra $ \mathbb{C}[L'/L] $ has its standard basis vectors denoted by $ \mathfrak{e}\_h, $ $ h\in L'/L. $
There is a unitary representation $ \rho\_L $ of $ \operatorname{Mp}\_2(\mathbb{Z})$ in $\operatorname{GL}(\mathbb{C}[L'/L])$ defined on the generators by
$$\rho\_L(T,1)\mathfrak{e}\_h = e(Q(h))\mathfrak{e}\_h$$
$$\rho\_L(S,\sqrt{z})=\frac{\sqrt{i}^{b\_--b\_+}}{\sqrt{|L'/L|}}\sum\_{h'\in L'/L}e((-h,h'))\mathfrak{e}\_{h'}.$$
This is the Weil representation associated to $ L. $
| https://mathoverflow.net/users/167073 | Does the Weil representation depend only on the discriminant group? | The Weil representation depends only on the discriminant form, as you already observed.
The thesis of Alfes-Neumann and the paper that you cite use various theta lifts, which do *not* depend only on the discriminant form. The lattice $L$ needs to be given, not because the Weil representation depends specifically on $L$ rather than $L'/L$ (it does not), but because theta lifts involve integrating against kernels (essentially, a theta function) that depends crucially on $L$ itself.
| 3 | https://mathoverflow.net/users/499465 | 440870 | 177,993 |
https://mathoverflow.net/questions/440848 | 7 | Simon Plouffe found experimentally a series for $\pi$ that can be written as
$$\frac{\pi}{24} = \sum\_{n=1}^\infty \frac{1}{n} \left( \frac{3}{e^{\pi n}-1} -\frac{4}{e^{2\pi n}-1} +\frac{1}{e^{4\pi n}-1}\right) $$
A related series for $\log(2)$, easily found with lindep in PARI, is
$$\frac{\log(2)}{8} = \sum\_{n=1}^\infty \frac{1}{n} \left( \frac{2}{e^{\pi n}-1} -\frac{3}{e^{2\pi n}-1} +\frac{1}{e^{4\pi n}-1}\right) $$
Linas Vepstas proves several formulas similar to these in his paper [On Plouffe's Ramanujan Identities](https://arxiv.org/pdf/math/0609775.pdf), for constants of the form $\zeta(4m-1)$ , $\zeta(4m+1)$ and related powers of $\pi$.
However, regarding $\pi$ and $\log(2)$, the closest expression found in the reference is
$$S\_1(2\pi)+T\_1(2\pi) = \frac{\pi}{6} - \frac{3}{4}\log(2)$$
where
$$S\_1(x)=\sum\_{n=1}^\infty \frac{1}{n(e^{xn}-1)}$$
and
$$T\_1(x)=\sum\_{n=1}^\infty \frac{1}{n(e^{xn}+1)}$$
Therefore, there seems to be another similar equation that would allow for proving the above two series for $\pi$ and $\log(2)$, by eliminating for one constant or the other.
>
> What is this second equation involving $\pi$ and $\log(2)$ ?
>
>
>
| https://mathoverflow.net/users/85758 | Proving a series for $\pi$ by Plouffe | It is immediate to show that
$S\_1(x)=-x/24-\log(\eta(ix/(2\pi)))$
and $T\_1(x)=S\_1(x)-2S\_1(2x)$, so all these formulas are immediate
consequences of the properties of the Dedekind $\eta$ function.
| 3 | https://mathoverflow.net/users/81776 | 440881 | 177,998 |
https://mathoverflow.net/questions/439672 | 3 | There are well-known results about nilpotent and solvable (=virtually nilpotent) Kähler groups coming from the work of (to name a few) Campana, Carlson-Toledo, Arapura-Nori, Delzant...
1. Are there any interesting known restrictions on *amenable* Kähler groups?
2. What about interesting known examples?
More generally,
**Definition.** A group $G$ is a **von-Neumann group** if it does not contain a non-abelian free group.
It is clear that any amenable group is a von-Neumann group.
3. What is known about Kähler groups which are also von-Neumann group?
| https://mathoverflow.net/users/105615 | Kahler groups with no non-abelian free groups? | For 1: beyond the case of nilpotent/solvable groups, I know of no restrictions on amenable kähler groups.
For 3: this is an interesting question but again I know of no restriction. This is probably due to our lack of ideas or to the lack of any suitable technology.
For 2: besides the 2-step nilpotent Kähler groups (not virtually Abelian) built in the 90s (Campana, Carlson-Toledo), I know of no new examples. In particular I think that it is still unknown whether there can be some 3-step nilpotent examples (not virtually 2-step nilpotent).
| 2 | https://mathoverflow.net/users/61960 | 440884 | 178,000 |
https://mathoverflow.net/questions/440756 | 1 | I'm interested in the following Cauchy problem for a linear diffusion equation
$$
\begin{cases}
{^C}\!D^{a}\_tu(t,x) = \sigma\Delta u(t,x),\\
u(0)=u\_0\in X.
\end{cases}
$$
where ${^C}\!D^{\sigma}\_t$
denotes the Caputo fractional derivative of order $a\in (0,1)$, and $X$ is a Banach space.
I wonder if there are any well-posedness results and characterizations of the $C\_0$ semigroup and its properties. Any reference would be great. Thanks.
| https://mathoverflow.net/users/98050 | Fractional reaction-diffusion with Caputo derivative | In $L^2(\Omega)$, see
>
> K. Sakamoto and M. Yamamoto, Initial value/boundary value problems for fractional diffusion-wave equations and applications to some inverse problems, J. Math. Anal. Appl.,382 (2011), 426–447.
>
>
>
For a general theory in Banach spaces, see
>
> E. G. Bajlekova, Fractional evolution equations in Banach spaces, Thesis, Eindhoven, 2001.
>
>
>
Note that solutions to fractional evolution equations do not satisfy the semigroup property!
| 1 | https://mathoverflow.net/users/124904 | 440888 | 178,002 |
https://mathoverflow.net/questions/440411 | 1 | I do not understand the following proof in the paper [Abelian varieties](http://van-der-geer.nl/%7Egerard/AV.pdf) by Edixhoven, van der Geert, and Moonen:
**(1.12) Rigidity Lemma.** Let $X$, $Y$ and $Z$ be algebraic varieties over a field $k$. Suppose that $X$ is complete. If $f: X \times Y \to Z$ is a morphism with the property that, for some $y \to Y(k)$, the fibre $X \times \{y\}$ is mapped to a point $z \in Z(k)$ then $f$ factors through the projection $\operatorname{pr}\_Y : X \times Y \to Y$.
Proof. We may assume that $k = \bar k$. Choose a point $x\_0 \in X(k)$, and define a morphism $g: Y \to Z$ by $ g(y) = f(x\_0,y)$ (here is the passage to to algebraic closed field involved, since otherwise we may not find such point $x\_0$…).
*Question:* Why $ k$ can be assumed to be algebraically closed? Assuming the lemma has been proved for fibre products $X\_{\bar{k}}$, $Y\_{\bar{k}}$, $Z\_{\bar{k}}$, how can we derive the statement for schemes over not algebraically closed $k$? It appears that there should be a little lovely diagram chase involved but I do not know how finally to construct the morphism $ Y \to Z$.
Finally there is the machinery of fppf-descent which justifies immediately the reduction to $k=\overline{k}$. But I would like to know if this can be also showed with elemenary methods — presumably a (tricky?) diagram chase. It looks rather similar to the problem [Proof of rigidity lemma](https://mathoverflow.net/questions/440404/proof-of-rigidity-lemma)
I posted recently, and there it turned out that the reduction to $k=\overline{k}$ can be justified by simple diagram chase. Is it here also possible to argue in similar way without ‘deep’ methods?
| https://mathoverflow.net/users/108274 | Reduction step to $k=\bar{k}$ in the proof of rigidity lemma | One way to argue is as follows: given a morphism $f \colon X \times Y \to Z$, consider its graph $\Gamma\_f \subseteq X \times Y \times Z$. Let $W$ be the scheme-theoretic image of $\Gamma\_f$ under the projection $\pi\_{Y \times Z} \colon X \times Y \times Z \to Y \times Z$, and let $X \times W \subseteq X \times Y \times Z$ be the preimage of $W$ under $\pi\_{Y \times Z}$.
Then we have $\Gamma\_f \subseteq X \times W$, and the statement that $f$ factors via $\pi\_Y \colon X \times Y \to Y$ is equivalent to the statement that $\Gamma\_f = X \times W$. Indeed, if $f$ factors through a morphism $g \colon Y \to Z$, then $W = \Gamma\_g$ and $\Gamma\_f = X \times W$. Conversely, if $\Gamma\_f = X \times W$, then the projection $\pi\_Y \colon W \to Y$ is an isomorphism since the same holds after multiplying with $X$ (this is a form of fppf descent, I suppose ― if you want you can choose a closed point in $X$ to reduce to the case of base change along a finite field extension). Thus $W$ is the graph of a morphism (namely $Y \underset{\pi\_Y}{\stackrel\sim\leftarrow} W \underset{\pi\_Z}\to Z$).
All of the above constructions are defined over $k$. Formation of scheme-theoretic image commutes with flat base change (on the level of sheaves, it is given by image factorisation $\mathcal O\_{Y \times Z} \twoheadrightarrow \mathcal O\_W \hookrightarrow \pi\_{Y \times Z,\*} \mathcal O\_{X \times Y \times Z}$), and checking whether an inclusion of subschemes is an equality can be done over $\bar k$. $\square$
| 1 | https://mathoverflow.net/users/82179 | 440890 | 178,003 |
https://mathoverflow.net/questions/440892 | 3 | I am trying to prove the following assertion: Let $B$ be a simply-connected set, and let $B' \subsetneq B$ be a proper open connected subset. Then, there exists a point $x \in \partial B'$ of the boundary such that there is a basis of neighborhoods of $x$ in $B$ whose intersections with $B'$ are connected.
This seems like it should be true. For example, in the one-dimensional case, a proper open subinterval of an interval always has an endpoint satisfying this property, whereas for a circle, one can take the complement of a point $x \in S^1$, and the boundary point $x$ does not have the property in question.
I am happy if one takes the special case of $B$ a manifold, or even a product of intervals.
| https://mathoverflow.net/users/68565 | Boundaries of subsets of simply-connected domains | It seems if you take $B=\mathbb{R}^2$ and $B'$ the complement of the closure of $\Big\{\big(x,\sin\big(\frac{1}{x}\big)\big);x\in(0,\infty)\Big\}$ this is a counterexample. (Added bonus: $B'$ is also simply connected)
| 9 | https://mathoverflow.net/users/172802 | 440893 | 178,004 |
https://mathoverflow.net/questions/440897 | 5 | Let $M$ be a compact complex manifold of dimension three.
Let us say that a divisor $D$ on $M$ is big if there is a constant $C>0$ such that
$$ h^0(M, \mathcal O\_M(nD)) > C n^3 $$
for sufficiently large $n \in \mathbb N$.
Assume that $M$ has a big divisor. My question is
>
> Is $M$ projective?
>
>
>
If the answer for the above question is no, can adding some topological conditions to $M$ change the answer?
What if $M$ is simply-connected and $h^{2,0}=h^{0,2}=0$ with trivial canonical class?
| https://mathoverflow.net/users/69559 | Big divisors and projectivity | The existence of a big divisor implies that $M$ is Moichizon so the "only obstruction" to projectivity is that $M$ may not be Kahler. In fact, this can happen even for nonprojective varieties.
Hironaka's construction gives for any projective $3$-fold $X$ equipped with curves $C, C' \subset X$ intersecting transversally at two points $P, Q$ a birational modification $f : \tilde{X} \to X$ where $\tilde{X}$ is not projective (see Hartshorne appendix B).
However, we can choose $X$ such that it is,
(1) simply connected,
(2) has $h^{2,0} = h^{0,2}$ and,
(3) has a big divisor $D$ disjoint from $C, C'$ (which consequently cannot be ample).
Then,
$$ H^0(\tilde{X}, \mathcal{O}\_X(n f^\* D)) = H^0(\tilde{X}, f^\* \mathcal{O}\_X(n D)) = H^0(X, \mathcal{O}\_X(n D) \otimes f\_\* \mathcal{O}\_{\tilde{X}})
\\
\supset H^0(X, \mathcal{O}\_X(n D)) $$
and since $D$ is big, this grows as a degree $3$ polynomial. Since the topological properties are birational invariants we see that $\tilde{X}$ satisfies the required properties but is not projective.
Concretely, choose $X$ to be the blowup of $\mathbb{P}^3$ at a point $P \in \mathbb{P}^3$. The exceptional fiber is $\mathbb{P}^2$ which indeed contains two suitable curves $C, C' \subset X$. Choose $D$ to be the pullback of any ample hyperplane $\mathbb{P}^2 \subset \mathbb{P}^3$ which does not contain the special point $P$. Furthermore, $X$ is rational so $h^{2,0} = h^{0,2} = 0$ and $\pi\_1(X) = 0$.
| 5 | https://mathoverflow.net/users/154157 | 440900 | 178,007 |
https://mathoverflow.net/questions/440877 | 2 | $\begin{align}x \text { is predicatively} &\text{ definable } \iff \exists x\_1,..,x\_n \exists \varphi:\\ & \rho(x\_1) < \rho(x) ,.., \rho(x\_n) < \rho(x) \ \land \\&\forall y \, (y \in x \iff V\_{\rho(x)} \models \varphi(y,x\_1,..,x\_n))\end{align} $
Where $\rho(x)$, the rank of $x$, is defined as the ordinal index of the first stage $V\_\alpha$ of which $x$ is a subset.
Now, let $\sf HPD$ be the class of all hereditarily predicatively definable sets.
>
> What is the maximal fragment of $\sf ZFC$ satisfied by $\sf HPD$?
>
>
>
| https://mathoverflow.net/users/95347 | Which fragment of ZF does the class of all hereditarily predicatively definable sets capture? | HPD satisfy extensionality and regularity trivially.
It satisfy Pairing, as if $x,y\in HPD$ we can explicitly write down the definition of $(x,y)$ using only $x,y$ as parameters (and they have strictly smaller rank).
It satisfy infinity (it has all of the Ordinals and it computes Ordinals correctly).
It has powerset, $V\_{\alpha+2}$ calculate $HPD\cap V\_{\alpha+1}$ correctly.
It doesn't have $\Delta\_0$ separation even when restricted to formulaes without parameters:
Note that $HPD(\omega)$.
Externally we know that $(2^{|\omega|})^{HPD}$ is countable
I claim that $HPD$ also see that, this is because $V\_{\omega+2}$ has a canonical well-ordering for $(\mathcal P(\omega))^{HPD}$ of ordertype $\omega$ (it is even without parameters), simply by intertwining Ackermann coding for parameters with Godel encoding for formulaes (and noting that every $HPD$ subset of $\omega$ must be seen from $V\_{\omega+1}$), so $HPD\models |\omega|=|\mathcal P(\omega)|$
Like @GabeGoldberg stated, for every $X\subseteq \omega$ we have $X'=X\cup\{\omega\}\in HPD$, but most such $X=\{a\in X'\mid a\ne\omega\}\notin HPD$, and $\omega$ is definable without parameters so we don't have $\Delta\_0$-separation w/o parameters.
---
I'm not quite sure about choice and union, but I would imagine that union fails in a strong sense (at every infinite successor level, note that the union of HPD sets with limit rank are closed under union, but at successor ranks we can use parameters from one rank bellow, which causes a problem)
| 2 | https://mathoverflow.net/users/113405 | 440906 | 178,009 |
https://mathoverflow.net/questions/440905 | 2 | Let us assume $<$ is some class relation without minimal elements, meaning $\forall a, \exists b, b< a$. This means that for every $n\in\omega$, one can build a decreasing function $f$ with domain $n+1$, meaning that $f(k+1)<f(k)$ for every $k\in n$. This is a very simple inductive argument.
My question is, how can we show the existence of a decreasing function whose domain is all $\omega$?
If I had access to a global choice function $\tau$, I could do the following. For every $n\in\omega$, inductively show the existence of functions with domain $n+1$ such that $f(0)=\varnothing$ and $$f(k+1)=\tau\{x\in V\_\alpha|x<f(k)\}$$ for $k\in n$, where $\alpha$ is the least ordinal such that $V\_\alpha$ contains some set less than $f(k)$, as in Scott's trick. Then I could simply use replacement and union over all of these to get the desired set.
Since ZFC with global choice is a conservative extension of ZFC, I understand that there should be some way to prove this result without invoking global choice, at least for a fixed class relation $<$. But how?
| https://mathoverflow.net/users/147705 | Infinite decreasing sequence for class relation without minimal elements | Why do you need a global choice function, though? Yes, it's neater, but unnecessary.
Define, for each $x\in V\_\alpha$ an ordinal $\alpha\_x$ such that it is the minimal for which $V\_{\alpha\_x}$ contains witnesses that $x$ is not minimal in $<$. Now let $\alpha\_0$ be defined for witnesses that $\varnothing$ is not minimal; then $\alpha\_{n+1}=\sup\{\alpha\_x\mid x\in V\_{\alpha\_n}\}$. Finally, $\alpha=\sup\alpha\_n$.
Now all you need is to have a choice function for $V\_\alpha$. This can be modified, of course, to start with any set, not just $\varnothing$.
Alternatively, use Reflection to find some $V\_\alpha$ such that $(V\_\alpha,<\restriction V\_\alpha)$ reflects the "no minimal elements", which is essentially what we did before, and then work with a choice function for that specific $V\_\alpha$.
| 4 | https://mathoverflow.net/users/7206 | 440907 | 178,010 |
https://mathoverflow.net/questions/440898 | 5 | In this [paper](https://statweb.stanford.edu/%7Ecgates/PERSI/papers/zabell82.pdf) by Diaconis and Zabell from 1982, Theorem 2.1 and the remark after essentially stated that
>
> Given two probability measures $P$ and $Q$ on the same probability space $\Omega$. If $Q\ll P$ and their
> RN-derivative $\frac{dQ}{dP} \in L^\infty$ , then $Q$ can be obtained from $P$ by conditioning.
>
>
>
Their proof is very easy and the assumption $\frac{dQ}{dP} \in L^\infty$ is essential for it. However I was wondering if it is possible to relax the assumption $\frac{dQ}{dP} \in L^\infty$ or if this is a necessary condition, is it? I was looking at concrete cases of this problem, for example between $\text{Exp}(a)$ and $\text{Exp}(b)$.
I know very little about the literature on this topic, any suggestion is also appreciated.
| https://mathoverflow.net/users/49551 | Radon-Nikodym derivative and conditional probability | Theorem 2.1 in the quoted paper of Diaconis and Zabell actually states that boundedness of the ratios
$Q(\omega)/P(\omega)$ is a **necessary and sufficient condition** for obtaining $Q$ from $P$ by conditioning in the discrete case. This is true in the general case as well.
Indeed, **conditioning** here means that there are a probability space $(\widetilde \Omega,\widetilde P)$ that covers the original space $(\Omega,P)$ and an event $E\subset \widetilde\Omega$ such that $Q$ is the image of the conditional measure $\widetilde P|E$ under the projection $\widetilde\Omega\to\Omega$. Since $\widetilde P|E \le \widetilde P/\widetilde P(E)$, the same inequality holds for the image measures as well, which means that $dQ/dP\le 1/\widetilde P(E)$.
| 2 | https://mathoverflow.net/users/8588 | 440920 | 178,013 |
https://mathoverflow.net/questions/440914 | 2 | For $A\subseteq \omega$ we let the *lower and upper density* be defined as $$\mu^-(A):= \lim\inf\_{n\to\infty}\frac{|A\cap n|}{n+1} \text{ and } \mu^+(A):= \lim\sup\_{n\to\infty}\frac{|A\cap n|}{n+1}$$ respectively.
Let $s:\omega\to\{0,1\}$ be an infinite binary string, $n\in\omega\setminus\{0\}$ a positive integer and $t\in\{0,1\}^n$ a finite binary string of length $n$. Then we define the *set of starting points of $t$ in $s$* by $$\text{start}(t, s) = \{k\in \omega: (\forall i\in n)\; t(i) = s(k+i)\}.$$ We say that $s$ is *uniform for* $t\in\{0,1\}^n$ if $$\mu^-\big(\text{start}(t,s)\big) = \mu^+\big(\text{start}(t,s)\big) = 1/2^n.$$
(The $1/2^n$ part is motivated by the fact that there are $2^n$ binary strings of length $n$.) We say that $s:\omega\to\{0,1\}$ is *strongly uniform* if for all positive integers $n$ and for all $t\in\{0,1\}^n$ we have that $s$ is uniform for $t$.
It is not clear to me whether strongly uniform infinite binary strings exist. A candidate could be the *[Champernowne binary string](https://mathworld.wolfram.com/BinaryChampernowneConstant.html)* which is obtained by concatenating the binary representations of the integers: $$0\; 1 \; 10\; 11\; 100\; 101\;\ldots.$$
**Question.** Is the the Champernowne binary string strongly uniform? If not, is there a strongly uniform infinite binary string?
| https://mathoverflow.net/users/8628 | Strongly uniform infinite binary strings | Yes, the Champernowne binary string is [normal](https://en.wikipedia.org/wiki/Champernowne_constant#Properties) (or strongly uniform, in your terms); see also [here](https://en.wikipedia.org/wiki/Normal_number#Properties_and_examples).
[Almost all infinite binary string are normal](https://en.wikipedia.org/wiki/Normal_number#Properties_and_examples), in the sense that the set of the corresponding real numbers in $[0,1]$ is of Lebesgue measure $1$.
| 6 | https://mathoverflow.net/users/36721 | 440922 | 178,014 |
https://mathoverflow.net/questions/440903 | 8 | Let $M\_1,M\_2 \subset \mathbf{S}^n$ be two smoothly embedded, connected hypersurfaces of the round sphere, which are realized as the zero sets of two homogeneous polynomials $P\_1,P\_2$ in $\mathbf{R}^{n+1}$:
\begin{equation}
M\_i = \{ P\_i = 0 \} \cap \mathbf{S}^n.
\end{equation}
>
> Is there an upper bound for the number of connected components of $M\_1 \setminus M\_2$ in terms of $\operatorname{deg} P\_1$ and $\operatorname{deg} P\_2$?
>
>
>
I apologize for the elementary question; my impression is that it might follow from Bezout's theorem, perhaps combined with Mayer–Vietoris, but I'm not an algebraic geometer by trade, so I can't say for certain. (I'd be happy to look up references myself if you have a suggestion!)
| https://mathoverflow.net/users/103792 | Intersection of two hypersurfaces via... Bezout's theorem? | In Proposition 4.13 of Coste’s introduction to semi-algebraic geometry, a bound of $d(2d-1)^{s+k-1}$ is given for the number of connected components of a system of $s$ real polynomial equations and inequations of degree at most $d\ge 2$ in $k$ variables. In your case, $k=n+1$ and $s=3$.
| 15 | https://mathoverflow.net/users/3404 | 440924 | 178,015 |
https://mathoverflow.net/questions/440954 | 3 | Let $\sigma=(\sigma\_1,...,\sigma\_m)$ be i.i.d. uniform binary 0-1 valued variables.
I'm trying to figure out what is the order of $E[||\sigma||\_p]$ with respect to $m$.
Jensen's inequality gives an upper bound of $(m/2)^{1/p}$, but how do I get a lower bound? (I'm hoping for a lower bound of the same order.)
| https://mathoverflow.net/users/61472 | Expected p-norm of binary vector | $\newcommand\si\sigma$
For any real $p>0$,
$$m^{-1/p}\,E\|\si\|\_p=E\Big(\frac1m\sum\_{j=1}^m\si\_j\Big)^{1/p}.$$
By the law of large numbers, $\frac1m\sum\_{j=1}^m\si\_j\to2^{-1}$ in probability (as $m\to\infty$). So, by Fatou's lemma,
$$\liminf\_{m\to\infty}m^{-1/p}\,E\|\si\|\_p\ge2^{-1/p}$$
and hence
$$E\|\si\|\_p\ge(1-o(1))(m/2)^{1/p}.$$
Thus, the upper bound $(m/2)^{1/p}$ is asymptotically exact for large $m$.
---
Let us now get an asymptotically exact explicit lower bound on $E\|\si\|\_p$. Since you said "Jensen's inequality gives an upper bound of $(m/2)^{1/p}$" and "$p$-norm" is mentioned in the title of your post, one should conclude that $p\ge1$ and hence $1/p\in(0,1]$ (for $p>0$).
Note that
$$E\|\si\|\_p=\mu\_{1/p},$$
where $\mu\_q:=ES^q$ and $S:=\sum\_{j=1}^m\si\_j$, so that $S$ has the binomial distribution with parameters $m$ and $1/2$ and hence $\mu\_1=m/2$ and $\mu\_2=(m/2)^2+m/4$. Note also that $\mu\_q$ is log convex in $q>0$. So, for $t:=\frac{p-1}{2p-1}\in[0,1)$ we have $\mu\_1\le\mu\_{1/p}^{1-t}\mu\_2^t$, whence
$$\mu\_{1/p}\ge\mu\_1^{1/(1-t)}\mu\_2^{-t/(1-t)}
=\Big(\frac m2\Big)^{2-1/p}
\Big(\Big(\frac m2\Big)^2+\frac m4\Big)^{-1+1/p}
=\Big(\frac m2\Big)^{1/p}\Big(1+\frac1m\Big)^{-1+1/p}.$$
Thus,
$$E\|\si\|\_p\ge L\_{p,m}:=\Big(\frac m2\Big)^{1/p}\Big(1+\frac1m\Big)^{-1+1/p}.$$
Clearly, $L\_{p,m}\sim(m/2)^{1/p}$ (as $m\to\infty$). Thus, we have the explicit lower bound, $L\_{p,m}$, on $E\|\si\|\_p$ such that
$$E\|\si\|\_p\sim L\_{p,m}.$$
| 3 | https://mathoverflow.net/users/36721 | 440956 | 178,025 |
https://mathoverflow.net/questions/440945 | 2 | Let $X$ be a finite set and let $\emptyset\neq \mathcal{H}\subseteq \{ 0,1 \}^{\mathcal{X}}$. Let $\{(X\_n,L\_n)\}\_{n=1}^N$ be i.i.d. random variables on $X\times \{0,1\}$ with law $\mathbb{P}$. Without knowing more of $\mathcal{H}$, what is the best risk-bounds available via VC-dimension or Rademacher-complexity theory on the *true risk*
$$
\sup\_{h\in \mathcal{H}}\,\mathbb{E}\_{(X,L)\sim \mathbb{P}}[I(h(x)=L)] - \frac1{N}\sum\_{n=1}^N\, I(h(X\_n)=L\_n)?
$$
| https://mathoverflow.net/users/491352 | VC-based risk bounds for classifiers on finite set | $\newcommand\HH{\mathcal H}\newcommand\ep\varepsilon$Let
\begin{equation}
D\_N:=\sup\_{h\in\HH}\Big(E\,I(h(X)=L) - \frac1{N}\sum\_{n=1}^N\, I(h(X\_n)=L\_n)\Big).
\end{equation}
It was shown in (the proof of) [Long's main result, 1999](https://link.springer.com/article/10.1023/A:1007666507971) (which he ascribed to Talagrand) that
\begin{equation}
P(|D\_N|>\ep)\le\exp\Big(d\_\HH-cN\ep^2)
\end{equation}
for some real $c>0$, all $N$, and all real $\ep>0$, where $d\_\HH$ is the VC dimension of $\HH$.
Similar results, but in terms of certain covering numbers for $\HH$ rather than the VC dimension, were obtained by [Talagrand, 1994](https://projecteuclid.org/journals/annals-of-probability/volume-22/issue-1/Sharper-Bounds-for-Gaussian-and-Empirical-Processes/10.1214/aop/1176988847.full).
At least as recently as in October 2019, these results by Talagrand and Long still [seemed to be the state of the art](https://projecteuclid.org/journals/annals-of-statistics/volume-47/issue-5/Exact-lower-bounds-for-the-agnostic-probably-approximately-correct-PAC/10.1214/18-AOS1766.full).
| 3 | https://mathoverflow.net/users/36721 | 440959 | 178,027 |
https://mathoverflow.net/questions/440939 | 2 | I was looking for a detailed explanation of a standard construction involving the projective tangential variety but I'm not able to find it anywhere, so maybe here some expert can enlight me on this topic.
Let me briefly sketch the setting: $X^n \subset \mathbb{P}^N$ is a projective variety of dimension $n$ (we can assume the required properties for $X$ ad libitum) and $TX=\bigcup\_{x \in X}\mathbb{T}\_{x}X$ the projective tangential variety, where $\mathbb{T}\_xX$ is the projective tangent space. Attached to $X$ we can also consider its tangent bundle $\mathcal{T}\_X$. The essential fact is that $(\mathcal{T}\_X)\_x$ is a vector space of dimension $n$ while $\mathbb{T}\_xX$ is a projective space of dimension $n$. Of course if we denote with $\hat{X}$ the affine cone over $X$ then $(\mathcal{T}\_X)\_x=T\_\hat{x}\hat{X}/\hat{x}$.
1. **First question**
With these in mind I want to "construct" a projective bundle $\mathbb{P}(\mathcal{E})$ over $X$ with some notion of "natural" fiber such that $\mathbb{P}(\mathcal{E}\_x)=\mathbb{T}\_xX$. My first guess would be something of the form $\mathbb{P}(\mathcal{T}\_X \oplus \mathcal{O}\_X(i))$ (for projective dimensional reasons) for some $i$ but I'm not sure if this is the right way to think about it.
2. **Second question**
If such a projective bundle $\mathbb{P}(\mathcal{E})$ exists, what is the linear system $W$ and line bundle $\mathcal{L}$, with $W \subset H^0(\mathbb{P}(\mathcal{E}),\mathcal{L})$, that gives me the natural map $\phi\_W:\mathbb{P}(\mathcal{E}) \rightarrow TX \subset \mathbb{P}^N$?
Thanks in advance for all the responses.
| https://mathoverflow.net/users/146431 | Linear system giving the projective embedding of the tangential variety | Consider the Euler exact sequence
$$0\rightarrow \mathcal{O}\_{\mathbb{P}}\rightarrow \mathcal{O}\_{\mathbb{P}}(1)^{N+1}\rightarrow \mathcal{T}\_{\mathbb{P}}\rightarrow 0\,.$$Restrict to $X$, and pull-back by the inclusion $ \mathcal{T}\_X\subset \mathcal{T}\_{\mathbb{P}|X}$. This gives an exact sequence
$$0\rightarrow \mathcal{O}\_{X}\rightarrow \tilde{\mathcal{T}}\_{X}\rightarrow \mathcal{T}\_X\rightarrow 0$$with an injective map $\tilde{\mathcal{T}}\_{X}\hookrightarrow \mathcal{O}\_{X}(1)^{N+1}$. The tangential variety is $\mathbb{P}(\tilde{\mathcal{T}}\_{X})$; it maps to $\mathbb{P}^N$ through $\mathbb{P}(\tilde{\mathcal{T}}\_{X})\hookrightarrow \mathbb{P}(\mathcal{O}\_{X}(1)^{N+1})=X\times \mathbb{P}^N\rightarrow \mathbb{P}^N$.
| 1 | https://mathoverflow.net/users/40297 | 440966 | 178,028 |
https://mathoverflow.net/questions/440909 | 1 | I asked a simillar question with the weaker restriction:
[On the equation $a^4+b^4+c^4=2d^4$ in positive integers $a\lt b\lt c$ such that $a+b\ne c$](https://mathoverflow.net/questions/438196/on-the-equation-a4b4c4-2d4-in-positive-integers-a-lt-b-lt-c-such-that)
.
---
I couldn't find any solution to this equation. And if $a^2+2ab+b^2=c^2$, then $c>d=a+b$.
Main question: Find some solutions to the equation $a^4+b^4+c^4=2d^4$ in natural numbers with $a<b<c<d$.
Thanks for advance.
| https://mathoverflow.net/users/nan | On the equation $a^4+b^4+c^4=2d^4$ in natural numbers with $a<b<c<d$ | If we take a point $3Q(m)$ then we get a solution as follows.
For more details, please see [$a^4+b^4+c^4=2d^4$](https://mathoverflow.net/questions/438196/on-the-equation-a4b4c4-2d4-in-positive-integers-a-lt-b-lt-c-such-that).
```
a = 100954906225546184690686373445232988785377384105455647295649619
847550243590542235300309485123275004949554849358738273723252717
90197500095135824990543003573711100267705489
b = 182720084009424134346291581847691617723750250933518665717300613
005234908926888631952883207666518927841277103892927263065478117
85836937759879470109480968700349292383907888
c = 186724255151010815056751706873402208434522449763207787663872587
945588542980120170792502069164146307065641765086510884090297462
19862671046394577805380183600483828969086905
d = 186780052966035701609386026218590575760779098084873786620653552
576571949525574948356153636564402836837404166574064647539252512
98760140535473903764903746388941211178351357
```
| 4 | https://mathoverflow.net/users/150249 | 440969 | 178,030 |
https://mathoverflow.net/questions/440913 | 3 | **Problem:** Let $M$ be a semifinite von Neumann algebra with a faithful semifinite normal trace $\tau$. Let $m$ be a positive element in $M$ and let $e\_{(0,\infty)}(m)$ be the spectral projection of $m$ on $(0,\infty)$. How can we show that $e\_{(0,\infty)}(m)m^{-1/2}$ is $\tau$-measurable?
I got stuck with this problem while reading $\tau$-measurable operators from the book '[Lectures on Selected Topics in von Neumann Algebras](https://drive.google.com/file/d/1H6ki8gjUPY5dsWE-CrGn9WFNhMEfEkIv/view?usp=sharing)' by Hiwi. Here I recall the definition of $\tau$-measurable operator.
**Definition 1:** For each $\epsilon,\delta>0$, define $$\mathscr{O}(\epsilon,\delta)=\{m\text{ affiliated to } M:eH\subseteq \mathcal{D}(m),\,\|me\|\leq \epsilon \text{ and }\tau(1-e)\leq\delta \text{ for some } e\in Proj(M)\}.$$
Let $m$ be a densely defined closed operator such that $m$ is affiliated to $M$. We say that $m$ is $\tau$-measurable if for any $\delta >0$, there exists an $\epsilon >0$ such that $m\in\mathscr{O}(\epsilon,\delta)$. We denote by $\widetilde{M}$ the set of such $\tau$-measurable operators.
Thanks in advance for any help or suggestion.
| https://mathoverflow.net/users/477204 | $\tau$-measurable operator | I think that this is simply not true. Take $M = \ell^\infty(\mathbb{N})$ with the semifinite trace $\tau(F) = \sum\_n F(n)$. When $p \in M$ is a nonzero projection, we have $\tau(p) \geq 1$. So the only $\tau$-measurable operators are the elements of $M$ itself. Taking $m \in M$ given by $m(k) = 1/k$, the element $m^{-1/2}$ is not $\tau$-measurable.
| 3 | https://mathoverflow.net/users/159170 | 440990 | 178,038 |
https://mathoverflow.net/questions/440992 | 1 | I am working with Ash's Probability and Measure Theory, Second Edition, specifically on theorem 6.2.1 (some convergence criteria for a random variable sequence).
We are given a sequence $(X\_i)\_{i \ge 1}$of random variables, and:
1. $\sum\_{i=1}^\infty \text{Var} X\_i$ converges.
2. Wlog, $\forall i.\:E(X\_i)=0$, (so assumption 1 becomes $\sum\_{i=1}^\infty X\_i^2$ converges).
3. $S\_n=\sum\_{i=1}^n X\_i$.
4. $\mu$ is the probability measure.
The proof says at a certain point that
$$\forall \epsilon>0.\:\mu\Bigl[\bigcup\_{k=1}^\infty\{|S\_{m+k}-S\_m|\geq\epsilon\}\Bigr]\to0\text{ as }m\to \infty$$
implies:
$$\forall \epsilon>0.\:\mu\Bigl[\bigcup\_{j,k\geq n}\{|S\_j-S\_k|\geq\epsilon\}\Bigr]\to0\text{ as }n\to \infty$$
I have been butting my head against exactly how to prove this for ages. Does anyone have a proof of this?
| https://mathoverflow.net/users/499564 | Equivalence of unions in probability theory | Let $\epsilon>0$ and $n \ge 1$. Then
$$\bigcap\_{k=1}^\infty\{|S\_{n+k}-S\_n|<\epsilon = \bigcap\_{j \geq n}\{|S\_j-S\_n|<\epsilon\} \subset \bigcap\_{j,k\geq n}\{|S\_j-S\_k|<2\epsilon\} .$$
Hence, taking complements,
$$\bigcup\_{j,k\geq n}\{|S\_j-S\_k|\geq 2\epsilon\} \subset \bigcup\_{k=1}^\infty\{|S\_{n+k}-S\_n|\geq\epsilon\}.$$
| 3 | https://mathoverflow.net/users/169474 | 440993 | 178,040 |
https://mathoverflow.net/questions/440988 | 1 | Let $z>0$ be fixed. Consider the function $p\_a: \mathbb R^2\_+\to\mathbb R\_+$ given as
$$
p\_a(t,x):=\frac{1}{\sqrt{2\pi N\_a(t)}}\left[\exp\left(-\frac{(x-z)^2}{2N\_a(t)}\right)-\exp\left(-\frac{(x+z)^2}{2N\_a(t)}\right)\right],
$$
where $N\_a:\mathbb R\_+\to\mathbb R\_+$ is defined by
$$N\_a(t):=\int\_0^t\frac{ds}{(1+a(s))^2}$$
and $a:\mathbb R\_+\to [0,1]$ is some measurable function. I can show there exists a unique function $a^\*$ (which is also decreasing) to the equation
$$a^\*(t)=\int\_0^\infty p\_{a^\*}(t,x)dx\equiv \text{Erf}\left(\frac{z}{\sqrt{2N\_{a^\*}(t)}}\right),\quad \forall t>0,$$
where $\text{Erf}$ is the Gauss error function (<https://en.wikipedia.org/wiki/Error_function>). Is there (efficient) numerical scheme to compute/approximate $a^\*$?
| https://mathoverflow.net/users/493556 | Numerical solution to some functional equation | $\newcommand\erf{\operatorname{erf}}\newcommand\R{\mathbb R}$The functional equation in question is
\begin{equation\*}
a=F(a) \tag{1}\label{1}
\end{equation\*}
on $(0,\infty)$, where $a$ is in the closed convex set, say $A$, of all nonincreasing functions from $(0,\infty)$ to $[0,1]$ with norm $\|\cdot\|:=\|\cdot\|\_\infty$ and
\begin{equation\*}
F(a)(t):=\erf\frac{z}{\sqrt{2N\_a(t)}}
\end{equation\*}
for real $t>0$.
For any $a\in A$, any function $h$ from $[0,\infty)$ to $\R$ such that $a+uh\in A$ for all small enough $u>0$, and all such $u$, let $g\_{a,h}(u):=F(a+uh)$. Then for all real $t>0$
\begin{equation\*}
g'\_{a,h}(0+):=\lim\_{u\downarrow0}\frac{g\_{a,h}(u)-g\_{a,h}(0)}u \\
=\frac2{\sqrt\pi}\,\exp\Big(-\frac{z^2}{2N\_a(t)}\Big)
\frac{-z}{2\sqrt2\,N\_a(t)^{3/2}} \int\_0^t\frac{-2ds\,h(s)}{(1+a(s))^3}
\end{equation\*}
and $\big|\int\_0^t\frac{-2ds\,h(s)}{(1+a(s))^3}\big|\le2N\_a(t)\|h\|$,
so that, with $y:=\frac z{\sqrt{2\,N\_a(t)}}>0$
\begin{equation\*}
|g'\_{a,h}(0+)|\le\frac2{\sqrt\pi}\,e^{-y^2}y\,\|h\|\le r\|h\|,
\end{equation\*}
where $r:=\sqrt{\frac2{\pi e}}\in(0,1)$.
So, the map $F$ is a contraction. So, there is a unique solution $a^\*\in A$ of \eqref{1}, and the iterations $a\_{n+1}=F(a\_n)$ with any $a\_0\in A$ converge to $a\_\*$ uniformly on $(0,\infty)$.
| 1 | https://mathoverflow.net/users/36721 | 441004 | 178,042 |
https://mathoverflow.net/questions/441005 | 3 | Let $V(\pi\_1)$ be the usual vector/matrix representation of the Lie algebra $\frak{sl}\_n$, for $n > 2$. A basic fact is the tensor product $V(\pi\_1) \otimes V(\pi\_1)$ decomposes as
$$
V(\pi\_1) \otimes V(\pi\_1) \simeq V\_{2\pi\_1} \oplus V\_{\pi\_2}.
$$
For higher tensor powers
$$
V(\pi\_1)^{\otimes k}
$$
does there exist a formula to describe the decomposition into irreducibles? Thinking about this in Young diagram terms, it seems (more or less) clear that the $k^{\text{th}}$-tensor power will have summands whose diagrams contain $k$-boxes, and any Young diagram with $k$-boxes will appear. Hence the question reduces to the question of multiplicities.
| https://mathoverflow.net/users/499575 | Decomposition of tensor powers of the vector representation of $\frak{sl}_n$ | The multiplicity of the Young diagram of shape $\lambda$ is the number of standard Young tableaux of shape $\lambda$, which can be computed with the [hook length formula](https://en.wikipedia.org/wiki/Hook_length_formula). This can be deduced from [Schur-Weyl duality](https://en.wikipedia.org/wiki/Schur%E2%80%93Weyl_duality).
| 7 | https://mathoverflow.net/users/297 | 441006 | 178,043 |
https://mathoverflow.net/questions/440853 | 2 | It could be a naive question. Probably, it is not true.
However, this question makes sense in the setting of function spaces.
For example, for $L\_\infty (0,1)$, we have $L\_p(0,1)\supset L\_\infty (0,1)$
and $L\_p(0,1)$ is reflexive when $p>1$.
On the other hand, $L\_1(0,1)$ itself is weakly sequentially complete and any rearrangement-invariant function spaces on $(0,1)$ is a subset of $L\_1(0,1)$.
I am wondering whether we have such a result in the setting of general Banach spaces.
BTW: Davis, Figiel, Johnson and Pełczyński's construction gives us a smaller reflexive space.
| https://mathoverflow.net/users/91769 | For a Banach space $X$, can we find a reflexive (or weakly sequentially complete) space $Y$ such that $X\subset Y$? | Let $K$ be a compact scattered space and $X=C(K)$ the space of continuous functions on $K$. We want to show that there is no injective bounded linear $T:X\to Y$ into a weakly sequentially complete (w.s.c.) Banach space $Y$ if $K$ has *large*(see below) cardinality.
Let $T:X\to Y$ be as above. $C(K)$ has Pelczynski property (V) and $Y$ is w.s.c., so $T$ is weakly compact. Since $C(K)$ has Dunford-Pettis property, then $T$ is completely continuous. Lastly $C(K)$ contains no copy of $\ell^1$ since $K$ is scattered. Thus, $T$ is compact.
Since $T$ is compact, its range $T(X)$ is separable, so $card(T(X))\leq\mathfrak{c}=card(\mathbb{R})$. $T$ is injective, so $card(X)=card(T(X))$. Together, we have
$$ card(K)\leq card(X)=card(T(X))\leq\mathfrak{c}.$$
However, this leads to a contradiction when $card(K)>\mathfrak{c}$. For example, take $K=[0,\alpha]$ for some ordinal $\alpha$ with the order topology, where the cardinality of $\alpha$ is strictly greater than $\mathfrak{c}$.
---
Note (2023-02-18): Following the footsteps in this post, we can actually make a slightly more general statement. If $X$ is a Banach space with property (V), $X^\*$ has Schur property, and $card(X)>\mathfrak{c}$, then there exists no injective bounded $T:X\to Y$ into a w.s.c. $Y$.
| 5 | https://mathoverflow.net/users/164350 | 441010 | 178,044 |
https://mathoverflow.net/questions/441012 | 2 | I am reading this paper <https://arxiv.org/abs/1608.04797>
Let $\Theta$ be the stack $\mathbb A^1/{\mathbb G\_m}$. Let $X$ be a smooth projective curve of genus $g$ over a field $k$. Let $Coh\_P$ be the stack of coherent sheaves parameterized by $X$ and with Hilbert polynomial $P$.
The paper says : groupoid of maps $f$ from $\mathbb A^1/{\mathbb G\_m}$ to $Coh\_P$ is equivalent to the groupoid whose objects consist of a coherent sheaf $f(1)=[E] \in Coh\_P $ and a $\mathbb Z$ weighted filtration of $E$. Maps $\*/ \mathbb G\_m \to Coh\_P$ correspond to $\mathbb Z$ graded coherent sheaves and restriction of a map $f$ from $\Theta \to Coh\_P$ to $0/ \mathbb G\_m$ classify gr(E).
I am unable to make sense of this.
>
> 1. What does the stack $\Theta$ look like and what does f(1) mean? More precisely what does 1 mean here?
>
>
>
>
> 2. Where do the filtrations and gradations appear from?
>
>
>
| https://mathoverflow.net/users/111666 | Maps from $\mathbb A^1/ \mathbb G_m$ to Coherent sheaves | Maps $\mathrm{B} G \to \mathcal{X}$ correspond to an object of $\mathcal{X}(k)$ along with a $G$-action. Indeed, the map $\* \to \mathrm{B} G \to \mathcal{X}$ selects an object $x$ and for each test scheme $T$ we get a natural map,
$$ \{ T \text{-torsors} \} \to \mathcal{X}(T) $$
so that the trivial torsor maps to $x$. This is determined by the induced $G$-action on $x$ since, after a cover, each torsor becomes trivial and the gluing maps pass to gluing data for an object of $\mathcal{X}$.
Maps $[E/G] \to \mathcal{X}$ correspond to "$G$-equivariant objects over $E$" meaning the data of $x \in \mathcal{X}(E)$ with a $G$-action in $\mathcal{X}$ which is compatible with the $G$-action on $E$ along the functor $\mathcal{X} \to \mathrm{Sch}\_k$.
First, $\mathbb{G}\_m$-equivariant coherent sheaves are exactly coherent sheaves with a $\mathbb{Z}$-grading. In some sense, this is an incarnation of the fact that the actions of tori split up into a weight space decomposition. In the affine case, this is very clear, we need a comultiplication map,
$$ M \to M \otimes k[t, t^{-1}] $$
which gives a decomposition of $M$ into the sum over $M\_n = \{ m \mid m \mapsto m \otimes t^n\}$. The (co)associativity of the action shows that this is a $\mathbb{Z}$-grading.
Now we need to consider $\mathbb{G}\_m$-equivariant sheaves over $\mathbb{A}^1$. This means a coherent sheaf $\mathcal{F}$ on $X \times \mathbb{A}^1$ flat over $\mathbb{A}^1$ with a $\mathbb{G}\_m$-action over $\mathbb{A}^1$. We get an actual sheaf $\mathcal{F}|\_1$ taking the fiber over $1 \in \mathbb{A}^1$ which is what "$f(1)$" should mean. Consider this situation affine-locally. We have a $A[t]$-module $M$ which is $\mathbb{Z}$-graded compatibly with the $\mathbb{Z}$-grading on $A[t]$ (although this grading on $A[t]$ is trivial in negative degrees this need not be true of $M$ e.g. $M = A[t, t^{-1}]$. We think of $f(1) = M/(t-1)$ as the underlying object which messes up the grading but preserves the decreasing filtration
$$ M\_{\ge n} = \bigoplus\limits\_{k \ge n} M\_n $$
because the action of $(t-1)$ preserves this filtration but not preserve the ascending filtration. Therefore, we get a $\mathbb{Z}$-filtered $A$-module $M/(t-1)$.
This is analogous to the "dynamical description of Parabolic, Levi, and Cartan subgroups" where we are asking for the limit of a $\mathbb{G}\_m$-parametrized sheaf "as $t \to 0$". This is the sort of condition that gives a parabolic so it may not be so surprising that it recovers filtered objects.
| 2 | https://mathoverflow.net/users/154157 | 441016 | 178,046 |
https://mathoverflow.net/questions/441002 | 0 | Suppose $V\subset \mathbb{R}^3$ be non-empty and at least twice differentiable (Smooth) and let $S$ be the surface that encloses $V$ (for example a sphere). Let $\textbf{F}\in \mathbb{R}^3$ be a smooth vector field for all space. Let $\textbf{n}$ denote the normal to the surface $S$. Does the surface integral over $S$ preserve the curl operation with respect to the vector field $\textbf{F}$. In other words, Does the surface integral of $\textbf{n}\times\textbf{F}$ commute with the curl operation $$\textbf{curl}\biggl(\oint\_{S}^{}{\textbf{n}\times\dfrac{\textbf{F}(\textbf{r}^\prime)}{|\textbf{r}-\textbf{r}^\prime|} ~ds}\biggr) = \oint\_{S}^{}{\textbf{n}\times \dfrac{ \textbf{curl} (\textbf{F}) (\textbf{r}^\prime)}{|\textbf{r}-\textbf{r}^\prime|}~ds}?$$ Here the surface intergrals are evaluated with respect to the position $\textbf{r}^\prime$ and produce vector fields.
| https://mathoverflow.net/users/353746 | Does surface integral preserve the curl operation? | $\renewcommand\r{\mathbf r}\newcommand\n{\mathbf n}\newcommand\F{\mathbf F}\newcommand\0{\mathbf 0}\newcommand\curl{\operatorname{\mathbf{curl}}}$No. E.g., let $S$ be the unit sphere and let $\F:=(1,0,0)$, so that $\curl\F=\0$ and hence the right-hand side of the identity in question is $\0$.
On the other hand, the left-hand side of the identity in question is
\begin{equation}
\begin{aligned}
\curl\oint\_{S}ds(\r')\,\n(\r')\times\dfrac{\F(\r')}{|\r-\r'|}
&=\oint\_{S}ds(\r')\,\curl\Big(\n(\r')\times\dfrac{\F(\r')}{|\r-\r'|}\Big),
\end{aligned}
\end{equation}
where the latter $\curl$ is of course with respect to $\r$.
For $\r=(1,0,0)$, the first coordinate of the latter (vector) integral (rewritten in the spherical coordinates) is
\begin{equation}
\int\_0^\pi d\phi\int\_0^{2\pi}d\theta\, g(\phi,\theta),
\end{equation}
where
\begin{equation}
g(\phi,\theta):=-\frac{\cos^2\phi+\sin^2\phi\;\sin^2\theta}
{(2-2 \cos \theta \sin\phi)^{3/2}} \, \sin\phi,
\end{equation}
which is manifestly $<0$ for $(\phi,\theta)\in(0,\pi)\times(0,2\pi)$.
So, the left-hand side of the identity in question is not $\0$, and thus
the identity does not hold in general. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 441018 | 178,047 |
https://mathoverflow.net/questions/441008 | -2 | Consider $$\left\|2\sum\_{i<j}L\_{ij}+4\sum\_i \operatorname{diag}e\_i \right\|,$$ where
(1) $L\_{ij}=\operatorname{diag}e\_i+\operatorname{diag}e\_j-e\_ie\_j^T-e\_je\_i^T$
(2) $e\_i$ denotes $n$-by-$1$ vector with only $i$-th element equals to $1$ and others are $0$
(3) $\operatorname{diag}e\_i$ is a $n$-by-$n$ diagonal matrix with only $i$-th diagonal element equals to $1$ and others are $0.$
I am wondering is it possible to obtain the spectral norm of this matrix, or if it is not possible, can we obtain the upper bound of it?
| https://mathoverflow.net/users/494410 | How to compute the spectral norm of this matrix | Making through the terrible notations, we see that
$$M\_n:=2\sum\_{i<j}L\_{ij}+4\sum\_{i}\text{diag}\,e\_i=(2n+4)I\_n-2\,1\_n1\_n^\top,$$
where $I\_n$ is the $n\times n$ identity matrix and $1\_n$ is the $n\times1$ column matrix of $1$'s.
The eigenvectors of the symmetric matrix $M\_n$ are the nonzero multiples of $1\_n$ and the nonzero vectors orthogonal to $1\_n$. The eigenvalue corresponding to the eigenvector $1\_n$ is $4$, and eigenvalue corresponding to the eigenvectors orthogonal to $1\_n$ is $2n+4>4\ge0$. Thus, $\|M\_n\|=2n+4$.
| 2 | https://mathoverflow.net/users/36721 | 441021 | 178,048 |
https://mathoverflow.net/questions/440997 | 3 | Is there a lower bound on the crossing number of a knot (resp., link) with braid index $b$?
For knots, I believe the smallest crossing number for braid index 2 is 3, the smallest crossing number for braid index 3 is 4, the smallest crossing number for braid index 4 is 6, the smallest crossing number for braid index 5 is 8, the smallest crossing number for braid index 6 is 10, and the smallest crossing number for braid index 7 is 12. The urge to extrapolate is strong.
There is no upper bound, of course -- already with braid index 2 there are knots with arbitrarily large (odd) crossing number.
| https://mathoverflow.net/users/6043 | Bounds for the crossing number in terms of the braid index? | As in the comment of dvitek, as for the relation of the braid index and the crossing number, Ohyama proved $c(L) \geq 2b(L)-2$ in [On the Minimal Crossing Number and the Braid Index of Links](https://doi.org/10.4153/CJM-1993-007-x).
Here I add three additional information.
**(a)** A simpler proof of $c(L) \geq 2b(L)-2$ (based on other results, but it adds additional insight)
Let $\alpha(L)$ be the arc index of a link. It is not hard to see that
$2b(L) \leq \alpha(L)$
(just try to change a minimum grid diagram into a braid diagram -- see [Grid diagrams, braids, and contact geometry](http://gokovagt.org/proceedings/2008/ggt08-ngthurston.pdf) for example).
On the other hand, $\alpha(L) \leq c(L)+2$ for a prime link, and equality happens if and only if $L$ is alternating [(An upper bound of arc index of links)](https://doi.org/10.1017/S0305004100004576).
Combining these two inequalities we get
$2b(L) \leq c(L)+2$. In particular, when $L$ is prime, the equality occurs if and only if $L$ is alternating and $\alpha(L)=2b(L)$.
**(b)** Though many results are stated for non-split links, a similar conclusion $2b(L) \leq c(L)+2m$ holds for split links if $L$ is split union $L= L\_1\sqcup \cdots \sqcup L\_m$ of $m$ non-split links $L$, because $c(L)=c(L\_1) + \cdots + c(L\_m)$ and $b(L)=b(L\_1) + \cdots + b(L\_m)$ hold (see [The crossing number of composite knots](https://doi:10.1112/jtopol/jtp028), [Studying links via closed braids IV: composite links and split links](https://doi.org/10.1007/BF01233423))
**(c)**
If you allow to use one additional natural quantity, the maximum euler characteristic $\chi(L)$, we can get lower and upper bounds
$-\chi(L)+b(L) \leq c(L)\leq (2b(L)-5)(-\chi(L)+b(L))$
[(A quantitative Birman–Menasco finiteness theorem and its application to crossing number)](https://londmathsoc.onlinelibrary.wiley.com/doi/10.1112/topo.12259)
| 3 | https://mathoverflow.net/users/193957 | 441026 | 178,049 |
https://mathoverflow.net/questions/441040 | 3 | Following the terminology of
*Drozd, Yuriy A.*, [**Derived tame and derived wild algebras**](https://doi.org/10.48550/arXiv.math/0310171), Algebra Discrete Math. 2004, No. 1, 57-74 (2004). [ZBL1067.16028](https://zbmath.org/1067.16028).
let $A$ and $R$ be algebras over a field $k$. A strict family of $A$-complexes based on $R$ is a complex $P^\* \in K^-(A\otimes R^{\rm{op}})$ of finitely generated projective $A-R$-bimodules such that
* For every indecomposable finite dimensional $R$-module $L$, $P^\*\otimes\_R L$ is indecomposable.
* For every pair of finite dimensional $R$-modules $L$ and $L'$, if $P^\* \otimes L \cong P^\*\otimes L'$ then $L \cong L'$.
An algebra $A$ is called derived wild if there exists a strict family for every finite dimensional algebra $R$.
My question is: Does anyone have an explicit description of such a strict family for a (preferably simple) derived wild algebra $A$ and some wild algebra $R$, say $R = k\langle x,y \rangle$ for example.
| https://mathoverflow.net/users/157483 | Explicit proof that algebra is derived wild | A few such examples are constructed in
*Bekkert, Viktor; Drozd, Yuriy; Futorny, Vyacheslav*, [**Derived tame local and two-point algebras**](https://doi.org/10.1016/j.jalgebra.2009.05.023), J. Algebra 322, No. 7, 2433-2448 (2009). [ZBL1191.16017](https://zbmath.org/1191.16017).
| 0 | https://mathoverflow.net/users/157483 | 441044 | 178,052 |
https://mathoverflow.net/questions/441043 | 6 | Let $G$ be a compact Lie group and $G/K$ a connected homogeneous space. A **homogeneous vector bundle** over $G/K$ is a vector bundle is one that is isomorphic to a vector bundle of the form
$$
G \times\_{\rho} V \to G/K, ~~~ (g,v) \mapsto [g],
$$
where $(V,\rho)$ is a $K$-module. Can there exist line bundles over $G/K$ that are not homogeneous? What are some typical examples?
| https://mathoverflow.net/users/499575 | Non-homogeneous line bundles over a homogeneous space | Yes. This happens whenever $G$ admits nontrivial vector bundles $E$ which can be equipped with an equivariant structure for the $K$-action. Then $E$ descends in the same way to $G \times\_{\rho} E \to G / K$. The homogenous ones are exactly the case that $E$ is a trivial $G$-bundle.
A trivial example is $G = S^1$ and $K = 1$ then take $E$ to be the Mobius bundle. This gives the Mobius bundle on $G/K$ which is not homogeneous according to this presentation. Of course, it is homogeneous if identify $G$ with $G / \{ \pm 1 \}$ and equip the trivial line bundle with the flip $\{ \pm 1 \}$-action.
Less trivial examples arise from the observation that if $K$ is connected then $\pi\_1(G) \to \pi\_1(G/K)$ is a surjection and therefore $H^1(G/K, \mathbb{Z}/2\mathbb{Z}) \to H^1(G, \mathbb{Z}/2 \mathbb{Z})$ is injective. These are the spaces parametrizing real line bundles via the first Stiefel–Whitney class. Its injectivity means that if $K$ is compact and $G/K$ admits any nontrivial line bundle, it stays nontrivial after pulling back to $G$ so cannot be homogenous.
For example, take $G = U(n)$ and $K = SU(n)$ then $G/K = S^1$ then the Mobius bundle on $S^1$ is not homogenous.
However, in a better sense, every line bundle is homogeneous if you choose the right presentation. Every line bundle $L$ defines via its Stiefel–Whitney class, a map $w\_1(L) : \pi\_1(G) \to \mathbb{Z}/2\mathbb{Z}$ giving a finite index normal subgroup. This defines a $\mathbb{Z}/2\mathbb{Z}$-covering group $G' \to G$ so that $L$ is trivialized on $G'$. Then $L$ is homogeneous for the pair $(G', \mathbb{Z} / 2 \mathbb{Z})$. Hence for each line bundle $L$ on $G/K$ we can form $(G', K')$ with $K'$ the preimage under $G' \to G$ such that $L$ is homogeneous for the new pair $(G', K')$ noting that $G'/K' = G/K$.
Now if you meant *complex* line bundle the story is more interesting. Complex line bundles $L$ are classified by their first Chern class $c\_1(L) \in H^2(X, \mathbb{Z})$. Just as before we are interested in whether the class is sent to zero under $H^2(G/K, \mathbb{Z}) \to H^2(G, \mathbb{Z})$. It is an interesting fact that every compact Lie group has $\pi\_2(G) = 0$. Therefore, if $\pi\_1(G) = 0$ then $H^2(G, \mathbb{Z}) = 0$ so every complex line bundle on the homogeneous space will again be homogeneous.
| 9 | https://mathoverflow.net/users/154157 | 441045 | 178,053 |
https://mathoverflow.net/questions/441052 | 1 | Let random matrix $\mathbf{X} \in \mathbb{C^{\mathrm{m} \times \mathrm{n}}}$ and random vector $\mathbf{y} \in \mathbb{C^{\mathrm{m} \times 1}}$ are unknown distributed, but their covariance and correlation are known, $\mathbf{C}\_{\mathbf{X}}$, $\mathbf{C}\_{\mathbf{y}}$, and $\mathbf{C}\_{\mathbf{Xy}}$. The question is whether there is any bound for the following expression, (like Cauchy-Schwarz or Jensen's)
$?\le \mathbb{E}[(\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1} \mathbf{X}^{\mathrm{H}} \mathbf{y}] \le ?$
Also, I have to mention $\le$ here means an element-wise inequality since the corresponding expectation is an $\mathrm{n} \times 1$ vector.
| https://mathoverflow.net/users/496172 | Bound for expectation of random matrix | No, of course not.
Indeed, consider a simplest case when $m=n=1$, and $X:=\mathbf X$ and $y:=\mathbf y$ are iid standard normal random variables. Then
$$E(\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1} \mathbf{X}^{\mathrm{H}} \mathbf{y}=E\frac yX$$
does not even exist (because here $\frac yX$ has [the standard Cauchy distribution](https://en.wikipedia.org/wiki/Cauchy_distribution#Properties)).
| 2 | https://mathoverflow.net/users/36721 | 441053 | 178,055 |
https://mathoverflow.net/questions/441062 | 1 | Let $z>0$ be fixed and $A$ be the set of non-increasing functions from $\mathbb R\_+$ to $[0,1]$ with norm $\|\cdot\|:=\|\cdot\|\_\infty$. Define by $F$ the operator on $A$ by
\begin{equation\*}
F(a)(t):=\text{Erf}\left(\frac{z}{\sqrt{2N\_a(t)}} \right),\quad \forall t\ge 0,
\end{equation\*}
where $\text{Erf}$ is the Gauss error function and $N\_a:\mathbb R\_+\to\mathbb R\_+$ is defined by
$$N\_a(t):=\int\_0^t\frac{ds}{(1+a(s))^2}.$$
Iosif has shown in [Numerical solution to some functional equation](https://mathoverflow.net/questions/440988/numerical-solution-to-some-functional-equation) that $F$ maps $A$ to $A$ and is a contraction map. I'm looking for a numerical approximation of the fixed point $a^\*$ of $F$. Let $a\_0\equiv 0$ and $a\_{n}:=F(a\_{n-1})$ for all $n\ge 1$. Then it follows that
$$\|a\_n-a^\*\|\le Cr^n,\quad \forall n\ge 1,$$
where $C>0, r\in (0,1)$ are some constants. How could we implement (code in computer) the iteration $a\_{n}:=F(a\_{n-1})$, and obtain a numerical approximation of $a\_N$ for any $N$?
PS : Here the numerical approximation of $a\_N$ means that: For any fixed $T>0$, any $\epsilon>0$ and any subset $\{0=t\_0<t\_1<\cdots<t\_M=T\}$, the computer yields a output $\{y\_0, y\_1,\ldots, y\_M\}$ s.t.
$$\max\_k |y\_k-a\_N(t\_k)|\le \epsilon.$$
Of course, any other numerical method (different from the above iteration) is highly appreciated.
| https://mathoverflow.net/users/493556 | Implementable numerical scheme for the equation $a=\text{Erf}\big(z/\sqrt{2N_{a}}\big)$ | $\newcommand\erf{\operatorname{erf}}\newcommand\R{\mathbb R}\newcommand{\de}{\delta}\newcommand\ep\epsilon
$In the [previous answer](https://mathoverflow.net/questions/440988/numerical-solution-to-some-functional-equation), it was shown that the operator $F$ on $A$ is $r$-Lipschitz for a certain universal constant $r\in(0,1)$ with respect to the norm $\|\cdot\|\_\infty$. The same proof holds for the corresponding operator $F\_T\colon A\_T\to A\_T$, where $A\_T$ is the set of all non-increasing functions from $(0,T]$ to $[0,1]$ with norm given by $\|a\|:=\sup\_{t\in(0,T]}|a(t)|$ for $a\in A\_T$ and
\begin{equation\*}
F\_T(a)(t):=\text{Erf}\left(\frac{z}{\sqrt{2N\_a(t)}} \right)
\end{equation\*}
for $a\in A\_T$ and $t\in(0,T]$; this follows because for any $t\in(0,T]$ the value of $N\_a(t)$ depends only on the values of $a$ on $(0,T]$.
The additional ingredient needed to answer the current question is the observation that $F\_T(a)(t)$ is $L$-Lipschitz in $t$ with the same $L$ for all $a\in A\_T$. Indeed, for any $a\in A\_T$ and $t\in(0,T]$, with $u:=z/\sqrt{N\_a(t)}$,
\begin{equation\*}
F\_T(a)'(t):=\frac d{dt}\,F\_T(a)(t)
=-\frac2{\sqrt\pi}\,e^{-u^2/2}u^3\,\frac1{2\sqrt2\,z^2}\frac1{(1+a(t))^2},
\end{equation\*}
so that
\begin{equation\*}
|F\_T(a)'(t)|\le L:=c/{z^2},
\end{equation\*}
where $c:=(3/e)^{3/2}/\sqrt{2\pi}$.
As requested, take now any real $\ep>0$, any natural $K$, and any $t\_0,\dots,t\_K$ such that $0=t\_0<t\_1<\cdots<t\_K=T$. For each $n=0,1,\dots$, we want a computer to yield $y\_{n,0},\ldots,y\_{n,K}$ s.t.
\begin{equation\*}
M\_n:=\max\_{k\in[K]}|a\_n(t\_k)-y\_{n,k}|\le\ep, \tag{$\*$}\label{1}
\end{equation\*}
where, as usual, $[K]:=\{1,\dots,K\}$.
Take any real $\de>0$. By refining the "partition" $0=t\_0<t\_1<\cdots<t\_K=T$, assume without loss of generality that
\begin{equation\*}
\max\_{k\in[K]}|t\_k-t\_{k-1}|\le\de.
\end{equation\*}
Let (say) $a\_0:=1$ and $a\_n:=F\_T(a\_{n-1})$ for $n\ge1$.
For $k\in[K]$, let $y\_{0,k}:=0$ and for $n\ge1$ let (a computer compute)
\begin{equation\*}
y\_{n,k}:=F\_T(Y\_{n-1})(t\_k),
\end{equation\*}
where
\begin{equation\*}
Y\_n(t):=\sum\_{k\in[K]}y\_{n,k}\,1(t\_{k-1}<t\le t\_k).
\end{equation\*}
Then, recalling that $F\_T$ is $r$-Lipschitz, for $n\ge1$ we have
\begin{equation\*}
\begin{aligned}
M\_n&=\max\_{k\in[K]}|F\_T(a\_{n-1})(t\_k)-F\_T(Y\_{n-1})(t\_k)| \\
&\le\|F\_T(a\_{n-1})-F\_T(Y\_{n-1})\| \\
&\le r\|a\_{n-1}-Y\_{n-1}\| \\
&=r\sup\_{t\in(0,T]}|a\_{n-1}(t)-Y\_{n-1}(t)| \\
&=r\max\_{k\in[K]}\sup\_{t\in(t\_{k-1},t\_k]}|a\_{n-1}(t)-y\_{n-1,k}|;
\end{aligned}
\end{equation\*}
also, $|a\_{n-1}(t)-y\_{n-1,k}|\le|a\_{n-1}(t\_k)-y\_{n-1,k}|+L\de$, because ($F\_T(a)(t)$ is $L$-Lipschitz in $t$ and hence) $a\_{n-1}$ is $L$-Lipschitz.
It follows that for all $n\ge1$
\begin{equation\*}
M\_n\le rM\_{n-1}+rL\de
\end{equation\*}
and hence (by induction on $n$, with $M\_0=0$)
\begin{equation\*}
M\_n\le C\de,
\end{equation\*}
where $C:=rL/(1-r)$. Finally, choosing $\de=\ep/C$, we get \eqref{1}. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 441069 | 178,058 |
https://mathoverflow.net/questions/441064 | 2 | Let $n=2m$. What is the order of the following permutation $\sigma$?
$$1\leq k\leq m \Rightarrow \sigma(k)=2k , \quad \sigma(m+k)=2k-1$$
| https://mathoverflow.net/users/84390 | If $n=2m$, what is the order of the permutation $\sigma(k)=2k , \quad \sigma(m+k)=2k-1$ | By adding a fixed point at $0$ (which preserves the order), the permutation $\sigma$ considered is just the multiplication by $2$ modulo $2m+1$. Thus, for $k \ge 0$, $\sigma^k$ is the identity map if and only if it fixes $1$, namely if and only if $2m+1$ divides $2^k-1$.
Hence, the order of $\sigma$ is the order of $2$ in $(\mathbb{Z}/(2m+1)\mathbb{Z})^\times$. I do not think that there are formulas for this, although the order necessarily divides $\phi(2m+1)$ by Lagrange theorem.
| 8 | https://mathoverflow.net/users/169474 | 441073 | 178,060 |
https://mathoverflow.net/questions/218855 | 9 | All sheaf topoi have [W-types](http://ncatlab.org/nlab/show/W-type) and in fact there's an explicit construction given by [Benno van den Berg & Ieke Moerdijk](http://www.phil.cmu.edu/projects/ast/Papers/vdBM_Wtypes.pdf), but the construction is quite involved.
I would like to know whether the inverse image part of a geometric morphism always preserves W-types for more general reasons or pointers to references detailing the conditions on functors so they preserve W-types.
Inverse image functors always preserve the natural numbers object, which is a particular kind of W-type, but the proof of this fact is very specific to the NNO so this might suggest that they don't all preserve W-types, but I have not found anything about this stated anywhere.
| https://mathoverflow.net/users/10875 | W-types and inverse image functor | We have a canonical map in one direction, namely $f^\*(W(p)) \to W(f^\*(p))$, but this map can fail to be an isomorphism. Here is an explicit counterexample.
Let $X$ be the set of countably-brancing trees, so $X = W(p)$ where $p : \mathbb{N} \to \mathbf{2}$ is a constant map. A tree is either a leaf or a node with countably many children.
Let $f$ be the canonical geometric morphism from the classifying topos $\mathcal{E}$ of enumerations of $X$ to $\mathrm{Set}$. (Incidentally, $f$ is surjective and open.)
As the definition of $p$ uses only geometric constructions, $f^\*(p)$ is again a constant map $\mathbb{N} \to \mathbf{2}$. So $W(f^\*(p))$ is again the set of countably-branching trees, just in $\mathcal{E}$ instead of $\mathrm{Set}$.
But $f^\*(X)$ does not coincide with $W(f^\*(p))$: As internally in $\mathcal{E}$ there is an enumeration of all elements of $f^\*(X)$, we can build a certrain tree, an element of $f^\*(X)$, which has all elements of $f^\*(X)$ as children. Hence $f^\*(X)$ contains an infinite path and is thus not well-founded.
| 2 | https://mathoverflow.net/users/31233 | 441081 | 178,062 |
https://mathoverflow.net/questions/441080 | 3 | Suppose $V\subset \mathbb{R}^3$ be non-empty and at least twice differentiable (Smooth) and let $S$ be the surface that encloses $V$ (for example a sphere). Let $\textbf{F}\in \mathbb{R}^3$ be a smooth vector field for all space. Let $\textbf{n}$ denote the normal to the surface $S$. It is well known using Helmhotz decompostion that we can decompose $\textbf{F}$ into two vector fields in $V$; $$\textbf{F} = \nabla \sigma + \nabla \times \Gamma$$ where $\nabla \sigma$ is called the irrotatioal part and $\nabla \times \Gamma$ the selonodial part of $\textbf{F}$. This decomposositon is known not to be unique. I want to know if the class of all solutions are related. For example take $\sigma\_1$, $\sigma\_2$, $\Gamma\_1$, and $\Gamma\_2$ so that $$\textbf{F} = \nabla \sigma\_1 + \nabla \times \Gamma\_1 = \nabla \sigma\_2 + \nabla \times \Gamma\_2$$ and then $$ \nabla (\sigma\_1- \sigma\_2) + \nabla \times (\Gamma\_1- \Gamma\_2) = \textbf{0}.$$ This should happen only if both these vector fields vanish. Or $$ \nabla \times (\Gamma\_1- \Gamma\_2) = \textbf{0}$$ $$ \nabla (\sigma\_1- \sigma\_2) = \textbf{0}.$$ Is there anything we can say about the solutions (i,e are they linear independent or multiples of one another).
| https://mathoverflow.net/users/353746 | Are all Helmholtz decompositions related? | **Q:** *How are two Helmholtz decompositions related?*
**A:** The scalar fields differ by a *harmonic function*.
Starting from a first decomposition $\sigma\_1,\Gamma\_1$, you can construct a second one by adding to $\sigma\_1$ a harmonic function $h$,
$$\sigma\_2=\sigma\_1+h,\;\;\text{with}\;\;\nabla^2 h=0.$$
Then the matching $\Gamma\_2$ is determined by the equation
$$\nabla\times(\Gamma\_1-\Gamma\_2)=\nabla h.$$
| 7 | https://mathoverflow.net/users/11260 | 441083 | 178,063 |
https://mathoverflow.net/questions/441084 | 1 | Let $f: X \to Y$ be a map of CW-complexes with continuous maps as morphisms.
How would one show that homotopy pullback $\mathcal D/Y → \mathcal D/X$ is right adjoint?
Here $\mathcal D$ is the derived category (edit, I mean the category of CW-complexes with continuous maps).
| https://mathoverflow.net/users/30211 | Homotopy pullback is right adjoint in the derived category | There is no such functor $\mathcal D/Y\to \mathcal D/X$. It's clear what is meant to be on objects, but it is not well-defined on morphisms.
Let $f$ be the inclusion of a point $p$ into a circle $C$. Let $g$ and $h$ be the inclusions of two points, say $q$ and $r$, into $C$. Regard $g$ and $h$ as objects of $\mathcal D/C$. There is a unique isomorphism between them. The homotopy pullback of $g$ along $f$ is the space of paths from $p$ to $q$ in $C$. The homotopy pullback of $h$ along $f$ is the space of paths from $p$ to $r$ in $C$. What map is the isomorphism between $g$ and $h$ meant to give from the one homotopy pullback to the other?
| 6 | https://mathoverflow.net/users/6666 | 441088 | 178,064 |
https://mathoverflow.net/questions/441082 | 1 | Before stating the question, I would like to first use an example for the type of formulation that I'm interested in.
Suppose we consider the continuity equation $\partial\_t \rho + \mathrm{div}( \rho v ) = 0$ with boundary conditions $\rho(0) = \rho\_0$ and $\rho(1) = \rho\_1$.
Now, if we were to test this equation with function $f$, we could arrive at the following equation (assuming zero boundary terms):
$$\int\_\Omega f ( \rho\_1 - \rho\_0 ) \ \mathrm{d}x = \int\_0^1 \int\_\Omega \nabla f \cdot (\rho v) \ \mathrm{d}x \ \mathrm{d}t.
$$
On the left side of the equation, we have values of $f$, while on the right, its gradient.
By considering all $v$ and $\rho$ satisfying the continuity equation we could, in some sense, think that we are trying to "retrieve" $\nabla f$ back.
---
Question
--------
I would like to ask whether there are PDEs which could be used for "retrieving" the hessian $\nabla^2 f$ back in the above sense. (Or just in which hessian and not, for example, laplacian, appears explicitly)
| https://mathoverflow.net/users/170491 | Are there PDEs in which Hessian appears in the weak formulation | Integration by parts of $\nabla^{2} f$ against a symmetric field $\sigma=(\sigma\_{ij})$ yields eventually the formula,
$$
\int\_{\Omega}(\nabla^{2}f,\sigma)dx=\int\_{\Omega}f\,\nabla^{\*}\nabla^{\*}\sigma\,dx+\int\_{\partial\Omega}\partial\_nf\,(\sigma,n\otimes n)\,dS+\int\_{\partial\Omega}f\,T\sigma\,dS
$$
where $(\nabla^{\*}\sigma)\_{j}=\sum\_{i}\partial\_{i}\sigma\_{ij}$, $\nabla^{\*}\nabla^{\*}\sigma=\sum\_{i,j}\partial\_{i}\partial\_{j}\sigma\_{ij}$, and $T$ is a first order boundary operator mixing $\nabla^{\*}\sigma$ and the derivatives of $\sigma\cdot n$ on the boundary. Hence, requiring zero boundary conditions, you could cook up a "second order" continuity equation of the form $\partial\_{t}\rho-\nabla^{\*}\nabla^{\*}(\rho\sigma)=0$ in order to obtain,
$$
\int\_{0}^{1}\int\_{\Omega}(\nabla^{2}f,\rho\sigma)\,dx \,dt=\int\_{0}^{1}\int\_{\Omega}f\,\nabla^{\*}\nabla^{\*}(\rho\sigma)\,dx\,dt=\int\_{0}^{1}\int\_{\Omega}f\,\partial\_{t}\rho\,dx\,dt=\int\_{\Omega}f(\rho(1)-\rho(0))dt
$$
assuming that $f$ is either time independent or satisfies $f(0)=0$ and $f(1)=0$.
| 1 | https://mathoverflow.net/users/144247 | 441097 | 178,068 |
https://mathoverflow.net/questions/441099 | 3 | Let $X$ be a locally compact Hausdorff space. Denote $C\_c(X)$ and $C\_0(X)$ the space of continuous functions with compact support and vanishing at infinity respectively. By Riesz representation theorem, the dual space $C\_0(X)^\*$ is isomorphic to the space of finite Radon measures $M(X)$. Since $C\_c(X)$ is dense in $C\_0(X)$, we actually have
$$ C\_c(X)^\* = C\_0(X)^\* = M(X) $$
In probability textbooks, the vague topology on $M(X)$, depending on the author, is defined as the weak\* topology induced by either $C\_c(X)$ or $C\_0(X)$. However, these two topologies are in general not the same.
I've heard somewhere that the reason why probabilists don't care about this ambiguity is, because $C\_c(X)$ and $C\_0(X)$ induce the same vague topology on a closed ball in $M(X)$, and probabilists only care about measures of norm $\leq 1$ anyway.
So is the statement true? And how to prove it?
| https://mathoverflow.net/users/49284 | Vague Topologies induced by $C_c$ and $C_0$ are the same on a closed ball of finite Radon measures? | The dual unit ball of a normed space $E$ is weakly compact à la Alaoglu and hence there is no strictly coarser Hausdorff topology. Applying this to $E=C\_0(X)$ you get that the topologies $\sigma(M(X),C\_0(X))$ and $\sigma(M(X),C\_c(X))$ coincide on all bounded sets of $M(X)$.
| 4 | https://mathoverflow.net/users/21051 | 441100 | 178,069 |
https://mathoverflow.net/questions/440994 | 12 | Here is a naive idea for a forcing $\mathbb A(\kappa)$, for an inaccessible cardinal $\kappa$. Conditions are pairs $(P,p)$, where $P \in V\_\kappa$ is a partial order and $p \in P$. We define the ordering $(Q,q) \leq (P,p)$ to hold when $P$ is a regular suborder of $Q$, and $q \leq\_Q p$.
The Amoeba's body grows larger and its nucleus gets wiser.
It is easy to see that whenever $G \subseteq \mathbb A(\kappa)$ is generic and $(P,p) \in G$, then $G$ induces a generic filter $G\_P \subseteq P$. Also, every cardinal below $\kappa$ is collapsed to $\omega$.
**Question:** Is $\kappa$ preserved? Does $\mathbb A(\kappa)$ add any bounded subsets of $\kappa$ that aren't added by some $G\_P$ as above?
| https://mathoverflow.net/users/11145 | Amoeba collapse | $\kappa$ is preserved, and moreover all reals are added by the small generics.
Let $(P\_0,p\_0)$ be a condition and let $\sigma$ be a name for a real. First, enumerate the elements of $P\_0$ below $p\_0$ as $\langle p\_i : 0<i< \lambda \rangle$. Let $(Q\_1,q\_1) \leq (P\_0,p\_1)$ decide $\sigma(0)$. Then let $(Q\_2,q\_2) \leq (Q\_1,p\_2)$ also decide $\sigma(0)$. Note that we are enlarging the partial order but going below $p\_2$ instead of $q\_1$. This makes sense because $P\_0 \lhd Q\_1$. Consider what happens at stage $\omega$. We have an increasing sequence of posets $P\_0 \lhd Q\_1 \lhd Q\_2 \lhd \dots$. If we let $Q\_\omega$ be the union, then this is a regular superorder of each $Q\_n$, since this is expressible as the first-order property, for all $q \in Q\_\omega$ (which is in some $Q\_m$ for $m \geq n$), there is $r \in Q\_n$ such that all $s \leq r$ are compatible with $q$.
So we continue transfinitely until we reach a poset $Q\_\lambda$ that is a regular superorder of $P\_0$ and each $Q\_i$, $0<i<\lambda$. It has the property that for all $p \in P\_0$ below $p\_0$, there is $r \in Q\_\lambda$ below $p$ such that $(Q\_\lambda,r)$ decides $\sigma(0)$. Now repeat process $\omega$-times until we reach some poset $R\_1$ such that for all $p \in P\_0$ below $p\_0$ and all $n<\omega$, there is $r \leq p$ in $R\_1$ such that $(R\_1,r)$ decides $\sigma(n)$.
Next, repeat this whole process with respect to $R\_1$ and iterate, reaching a closure point $R\_\omega \in V\_\kappa$. We will have that for all $r \in R\_\omega$ below $p\_0$, and all $n<\omega$, there is $r' \leq r$ in $R\_\omega$ such that $(R\_\omega,r')$ decides $\sigma(n)$. In other words, for each $n$, the set of $r \in R\_\omega$ such that $(R\_\omega,r)$ decides $\sigma(n)$ is dense below $p\_0$ in $R\_\omega$.
Now let us explain the claim in the OP that the generic for $\mathbb A$ adds a generic $G\_P$ for all posets $P$ appearing in $G$. Fix $(P,p) \in \mathbb A$, and suppose $D$ is a dense subset of $P$. For any $(Q,q) \leq (P,p)$ there is $q' \leq q$ in $Q$ such that $q' \leq d$ for some $d \in D$, since $D$ is predense in $Q$. Thus $(P,1)$ forces that the set $G\_P := \{ p \in P : (P,p) \in G \}$ is generic.
So if we force below $(R\_\omega,p\_0)$, then for each $n \in \omega$, the generic $G$ will have some element of the form $(R\_\omega,r)$ deciding $\sigma(n)$. This means that $\sigma^G$ will be an element of $V[G\_{R\_\omega}]$. By the arbitrariness of $(P\_0,p\_0)$ and $\sigma$, the desired conclusion follows.
| 6 | https://mathoverflow.net/users/11145 | 441103 | 178,071 |
https://mathoverflow.net/questions/441102 | 2 | I have a really stupid question that I don't seem to know the answer to and have been too embarassed to ask. In some number theory papers, I encounter sums of the form $$\sum\_{\substack{{x \asymp B}\\P(x) =0}}1$$ to count solutions "of size B" to some quadratic form $P$ (often $x \in \mathbb{Z}^n$ but let us ignore that for now). Alternately, I see statements like let us consider a set of $k$ $\textbf{primes of size $N\log N$}$.
In Vinogradov notation, if $f(x) \ll g(x)$ and $g(x)\ll f(x)$, we write $f(x) \asymp g(x)$, but what does it mean in this context?
For example, when is a prime of size $N\log N$? Vinogradov notation suggests that for a prime
$p \asymp N\log N$, there exist constants $C, C' >0$ such that
$$CN\log N\leq p \leq C'N\log N.$$
In the same vein, $x\asymp B$ should probably mean that there are such positive $k, k'$ such that
$kB\leq |x| \leq k'B$.
My question is: how small and big can these $C$, $C'$, $k$, $k'$ be?
Also, let $\mathcal{P} = \{p \textrm{ prime}: p \asymp N \log N\}$. What does the set $\mathcal{P}$ look like?
| https://mathoverflow.net/users/nan | What does it mean to have a number of size $B$? | This is a good question, and the answer is that writing $x\asymp B$ under a sum **is sloppy notation without further clarification**. It can mean that $c\_1 B<x<c\_2 B$ for any fixed constants $c\_1$ and $c\_2$ (and then the sum will depend on those constants), or it can mean the same for some constants $c\_1$ and $c\_2$ fixed explicitly at the beginning of the paper. For example, the author might say in the notation section that $x\asymp B$ will mean $B<x<2B$, although some authors would denote this relation by $x\sim B$ to distinguish it from $x\asymp B$.
| 6 | https://mathoverflow.net/users/11919 | 441104 | 178,072 |
https://mathoverflow.net/questions/441122 | 4 | In the book " Number Theory IV Transcendental Numbers" written by Parsin and Shafarevich ([book](https://link.springer.com/book/10.1007/978-3-662-03644-0), page 104) it is asserted that to explicit a transcendence measure of a complex number $w$, it is sufficient to prove that there is an infinite sequence of polynomials $P\_m(x)\in\mathbb Z[x]$ of fixed degree such that
$$0<H(P\_m)\le c^m\qquad e^{-\lambda\_1m}\le|P\_m(w)[\le e^{-\lambda\_2m}$$
where $c$, $\lambda\_1$, and $\lambda\_2$ are constants, such that $c > 1, \lambda\_l > \lambda\_2 > 0$ with $H(P\_m)=\max\{|\text{coefficients of $P\_m$}|\}$. I do not know why. Can anyone give the argument behind this assertion or a reference?
This result is used to give a transcendence measure of $\ln(r)$ ($r$ is a rational)
Thanks in advance
| https://mathoverflow.net/users/33128 | Transcendence measure: of $\ln(a/b)$ | Warning: this is for irrationaity measure, not transcendence measure.
Let $a/b$ be an approximation of $w$ such that $|w-a/b|=b^{-\kappa}$. Then $$P\_m(a/b)=P\_m(w)+(w-a/b)P\_m'(\theta)$$
for certain $\theta$ between $a/b$ and $w$. Note that $P\_m(a/b)$ is either 0 or at least $b^{-d}$ in absolute value, where $\deg P\_m\leqslant d$. Choose $m$ such that $b^{-d}\geqslant 2e^{-\lambda\_2 m}$, say, $m=\lceil \frac{\log 2+d\log b}{\lambda\_2}\rceil$.
Then, if $|P\_m(a/b)|\geqslant b^{-d}$, we get $$C(d,w)\cdot c^m b^{-\kappa}\geqslant |b^{-\kappa}P\_m'(\theta)|=|(w-a/b)P\_m'(\theta)|=|P\_m(a/b)-P\_m(w)|\geqslant e^{-\lambda\_2 m},$$
thus $\kappa\log b\leqslant m(\log c+\lambda\_2)+O(1)$ and $\kappa\leqslant d(1+\frac{\log c}{\lambda\_2})+o(1)$.
If $P\_m(a/b)=0$, then analogously
$$C(d,w)\cdot c^m b^{-\kappa}\geqslant |b^{-\kappa}P\_m'(\theta)|=|(w-a/b)P\_m'(\theta)|=|P\_m(w)|\geqslant e^{-\lambda\_1 m},$$
so $\kappa\log b\leqslant m(\log c+\lambda\_1)+O(1)$, and $\kappa\leqslant d\cdot \frac{\log c+\lambda\_1}{\lambda\_2}+o(1)$.
So, in both cases we may conclude that the irrationality measure of $w$ does not exceed $$d\cdot \frac{\log c+\lambda\_1}{\lambda\_2}.$$
| 1 | https://mathoverflow.net/users/4312 | 441123 | 178,077 |
https://mathoverflow.net/questions/441114 | 6 | According to some authors, it is built in *A.A.Beilinson "Higher regulator of modular curves"* a class $\mathbf{Eis}\_{\phi}$ in the motivic cohomology of the modular curve where $\phi$ is a Schwartz function over the finite adeles. Since the modular curve is only quasi-projective, I assume it is mixed-motivic cohomology? I am right ?
I say "according to some authors" because I haven't read this article by Beilinson (I can't find it anywhere). This document is a bit old and sometimes Beilinson is hard to read so my question is
**Does anyone know of a reference where this construction is done with the technical details? If not how can I find this article of Beilinson?**
| https://mathoverflow.net/users/169282 | References for the construction of Beilinson's motivic Eisenstein classes | The Eisenstein classes $\mathrm{Eis}^k\_\phi$ live in the motivic cohomology $H^{k+1}\_{\mathcal{M}}(E^k, \mathbf{Q}(k+1))$, where $E \to Y(N)$ is the universal elliptic curve over the open modular curve $Y(N)$. For example the classes $\mathrm{Eis}^0\_\phi$ are the Siegel modular units. At the time, Beilinson defined motivic cohomology as a graded piece of algebraic $K$-theory.
Beilinson's construction is hard to read. Scholl gave an account of the Eisenstein symbol in [An introduction to Kato's Euler systems](https://www.dpmms.cam.ac.uk/%7Eajs1005/preprints/euler.pdf), in the survey [The Beilinson conjectures](https://www.dpmms.cam.ac.uk/%7Eajs1005/preprints/d-s.pdf) (with Deninger), and in his book in preparation [$L$-functions of modular forms and higher regulators](https://www.dpmms.cam.ac.uk/%7Eajs1005/mono/index.html). Doran and Kerr also explained Beilinson's construction in [Algebraic $K$-theory of toric hypersurfaces](https://arxiv.org/abs/0809.4669), Section 5.
| 5 | https://mathoverflow.net/users/6506 | 441126 | 178,080 |
https://mathoverflow.net/questions/441128 | 3 | This posting is related to a recent question asked in [MSE](https://math.stackexchange.com/q/4640948/121671): Suppose $(X,\mathscr{B},\mu)$ is a $\sigma$-finite measure space. If $\nu$ is another measure on $\mathscr{B}$, $\nu(X)=\mu(X)$, and $\nu\ll\mu$, is there a measurable map $T:(X,\mathscr{B})\rightarrow(X,\mathscr{B})$ such that $\nu=\mu\circ T^{-1}$? (Here $\mu\circ T^{-1}$ is the push-forward of $\mu$ by $T$, that is $(\mu\circ T^{-1})(A):=\mu(T^{-1}(A))$).
The answer is yes for Borel spaces $X$ when $(X,\mathscr{B},\mu)$ is isomorphic to the standard Lebesgue space $((0,1),\mathscr{B}((0,1)),\lambda)$ ( $\lambda$ is Lebesgue measure restricted to the unit interval with the Borel $\sigma$-algebra) regardless of whether $\nu\ll\mu$ or not. Indeed, on the standard Lebesgue space, one can use the quantile function $Q\_\nu(q):=\inf\{x: \nu(0,x]\geq q\}$ to get $\nu=\lambda\circ Q^{-1}$.
My question is whether the statement holds another measure $\mu$
on $((0,1),\mathscr{B}(0,1))$ that is not equivalent to $\lambda$ (i.e. either $\mu\not\ll\lambda$ or $\lambda\not\ll\mu$)?
| https://mathoverflow.net/users/78591 | A type of coupling problem I | $\newcommand\de\delta$The answer to your first question is no.
E.g., let $\mu:=2\de\_{1/3}+2\de\_{2/3}$ and $\nu:=\de\_{1/3}+3\de\_{2/3}$, where $\de\_a$ is the Dirac measure with support $\{a\}$. Then all the conditions imposed on $\mu$ and $\nu$ hold. However, $\nu(\{1/3\})=1$ is an odd integer, whereas all the values of the measure $\mu$ are even integers. So, $\nu\ne\mu T^{-1}$ for any $T$.
---
MO posts should not have multiple questions. So, I suggest you post the other two questions separately.
| 2 | https://mathoverflow.net/users/36721 | 441132 | 178,081 |
https://mathoverflow.net/questions/441096 | -4 | **Nested Selection:** For every infinite set $G$ of pairwise disjoint infinite sets such that any two distinct elements $x,y$ of $G$ either "$y$ is a set of proper supersets of elements of $x$ and each element of $x$ has a proper superset of it in $y$" or "$x$ is a set of proper supersets of elements of $y$ and each element of $y$ has a proper superset of it in $x$"; there exists a set $cG$ that is a choice set on $G$ (i.e. $cG$ is a subset of $\bigcup G$ that has exactly one element from each element of $G$, among its elements) and such that for any two elements $a,b$ of $cG$ we have $a$ subset of $b$ or $b$ subset of $a$.
>
> Is Nested selection equivalent to $AC$?
>
>
>
The formal capture is a little bit messy. It is:
$\forall G: \operatorname {infinite} (G) \land \forall h \in G (\operatorname {infinite}(h)) \land \\ \forall k,l \in G (k \neq l \to k \cap l = \emptyset) \land \\ \forall x,y \in G (\forall z \in y [x] \exists u \in x [y] (z \supsetneq u) \land \\ \forall v \in x[y] \exists w \in y[x] (w \supsetneq v) ) \\ \implies \\ \exists cG: cG \subseteq \bigcup G \land \\\forall g \in G \exists! m (m \in g \land m \in cG) \land \\\forall a,b \in cG: a \subseteq b \lor b \subseteq a$
>
> If the above nested selection is too particular theme to pace with Choice.Then is the following general form equivalent to choice?
>
>
>
Define $\begin{align} Y \text { is } \Phi\text{-image of } X \iff &\forall a \in X \exists b \in Y: \Phi(a,b) \land \\ &\forall b \in Y \exists a \in X: \Phi(a,b)\end{align}$
**General Nested Selection:** If $\Phi$ is a transitive asymmetric binary relation, then if $G$ is an infinite set of pairwise disjoint infinite sets such that every two distinct sets $X,Y \in G$ either $ Y \text { is } \Phi\text{-image of } X $ or $ X \text { is } \Phi\text{-image of } Y$; then there exists a choice set $C$ on $G$ such that for any two distinct elements $a,b \in C$ either $ \Phi(a,b)$ or $\Phi(b,a)$.
By $C$ being a choice set on $G$ it means that $C \subseteq \bigcup G$ and $C$ shares exactly one element with each element of $G$.
| https://mathoverflow.net/users/95347 | Is Nested Selection equivalent to AC? | For a counterexample to Nested Selection, construct sets $A\_\alpha\subseteq\omega$ ($\alpha\lt\omega\_1$) so that, for $\alpha\lt\beta\lt\omega\_1$, we have $|A\_\alpha\setminus A\_\beta|\lt\aleph\_0=|A\_\beta\setminus A\_\alpha|$. Let $\mathcal S\_\alpha=\{X\subseteq\omega:|X\triangle A\_\alpha|\lt\aleph\_0\}$ and let $G=\{\mathcal S\_\alpha:\alpha\lt\omega\_1\}$. A choice set for $G$ which was totally ordered by inclusion would be a chain of length $\omega\_1$ in $\mathcal P(\omega)$, which doesn't exist.
**P.S.** To construct the sets $A\_\alpha$, first partition $\omega$ into infinitely many disjoint infinite sets $N\_i$ ($i\lt\omega$), and then define $A\_\alpha$ recursively for $\alpha\lt\omega\_1$ so that $|A\_\alpha\cap N\_i|\lt\aleph\_0$ for each $i\lt\omega$ and $|A\_\beta\setminus A\_\alpha|\lt\aleph\_0=|A\_\alpha\setminus A\_\beta|$ for each $\beta\lt\alpha$.
| 3 | https://mathoverflow.net/users/43266 | 441138 | 178,084 |
https://mathoverflow.net/questions/440984 | 6 | In [1] Fedorcuk, using diamond, proved that there is a hereditarily separable compact space of cardinality $2^{2^\omega}$.To my best knowledge, Kunen created a humanly digestible proof, but he has not published it (and he passed away). Can I find Kunen's proof somewhere? Or do you know any other proof of the theorem of Fedorcuk?
[1] Fedorčuk, V. V.
The cardinality of hereditarily separable bicompacta. (Russian)
Dokl. Akad. Nauk SSSR 222 (1975), no. 2, 302–305.
| https://mathoverflow.net/users/71011 | How to construct large, hereditarily separable compact spaces? | I was informed that the following joint paper of Dzamoja and Kunen contains the proof I was looking for.
[Dz̆amonja, Mirna; Kunen, Kenneth Measures on compact HS spaces.
Fund. Math. 143 (1993), no. 1, 41–54](http://matwbn.icm.edu.pl/ksiazki/fm/fm143/fm14314.pdf).
| 3 | https://mathoverflow.net/users/71011 | 441158 | 178,088 |
https://mathoverflow.net/questions/441161 | 3 | Let $T$ be a maximal torus of a compact Lie group $K$,
and let $\Psi \subset {\mathfrak t}$ be the (finite) set of coroots for $(K,T)$, where $\mathfrak t$ is the Lie algebra of $T$.
Assume now that $K'$ is another compact Lie group that admits a homomorphic embedding $i:K\hookrightarrow K'$ verifying the condition that $i(Z\_K)\subset Z\_{K'}$, where $Z\_K, Z\_{K'}$ are the centers of $K$ and $K'$, respectively. Let $T'$ be a maximal torus of $K'$ such that $i(T)\subset T'$. This gives a vector space inclusion ${\mathfrak t} \subset {\mathfrak t}'$, into the Lie algebra of $T'$.
Question: Is it true that
$$ \Psi = {\mathfrak t} \cap \Psi' \quad ?$$
where $\Psi' \subset {\mathfrak t}'$ are the coroots of $(K',T')$.
If this is not always true, are there some simple conditions under which it becomes true?
| https://mathoverflow.net/users/409881 | Sub-coroot systems | This can fail. Consider the natural embedding $\operatorname{Sp}\_4(\mathbb C) \to \operatorname{SL}\_4(\mathbb C)$, where $\operatorname{Sp}\_4$ is taken with respect to the symplectic form $(x, y) \mapsto x\_1 y\_4 + x\_2 y\_3 - x\_3 x\_2 - x\_4 y\_1$. (This is a map of complex Lie groups, but it carries maximal compact subgroups to maximal compact subgroups, so you can work in that setting if you prefer.) If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}\_4(\mathbb C)$ by $\alpha\_1, \alpha\_2, \alpha\_3$, in the obvious fashion, then the restrictions of $\alpha\_1$ and $\alpha\_3$ to the diagonal torus in $\operatorname{Sp}\_4(\mathbb C)$ are the short simple (with respect to the upper-triangular Borel) root $a$, and the restriction of $\alpha\_2$ is the long simple root $b$. Then $a^\vee$ equals $\alpha\_1^\vee + \alpha\_3^\vee$, which is not a coroot of $\operatorname{SL}\_4(\mathbb C)$.
(As discussed in [comments](https://mathoverflow.net/questions/441161/sub-coroot-systems/441172#comment1138119_441172), the original version of the question did not include the condition on centres, and so admitted any proper, maximal-rank subgroup as a counterexample. This example does satisfy the centre condition.)
| 6 | https://mathoverflow.net/users/2383 | 441172 | 178,091 |
https://mathoverflow.net/questions/441168 | 6 | Given a simplicial set $X\_\bullet$, define its **powerset simplicial set** $\mathcal{P}\_\bullet(X)$ as the composition
$$\Delta^\mathsf{op}\xrightarrow{X\_\bullet}\mathsf{Set}\xrightarrow{\mathcal{P}}\mathsf{Set},$$
where $\mathcal{P}$ is the covariant powerset functor.
**How homotopically well-behaved is the powerset simplicial set construction?**
1. If $X\_\bullet$ is a Kan complex, must $\mathcal{P}\_\bullet(X)$ also be one?
2. Given a Kan complex $X\_\bullet$, are the homotopy groups of $X\_\bullet$ related to those of $\mathrm{Ex}^\infty\_\bullet(\mathcal{P}\_\bullet(X))$?
| https://mathoverflow.net/users/130058 | Homotopical properties of powersets of simplicial sets | The first question has a negative answer, given by the simplicial set $\def\Exi{{\sf Ex}^{\sf\infty}}X=\Exi Y$, where $Y$ is a simplicial set generated by vertices $a,b,b',c,c',d,d'$, 1-simplices $ab,ab',ac,ac',ad,ad'$, 2-simplices $abc,acd,abd,ab'c',ac'd',ab'd',abc'$, and 3-simplices $abcd,ab'c'd'$.
We specify a 3-horn $Λ^3\_0→P(X)$ by setting its 1st face to $\{acd,ac'd'\}$, 2nd face to $\{abd,ab'd'\}$, 3rd face to $\{abc,ab'c',abc'\}$.
Any 3-simplex of $P(X)$ that fills in this horn must in particular contain a 3-simplex of $X$ whose 3rd face is $abc'$. This forces its 1st face to be $ac'd'$ and its 2nd face to be $abd$, which means that the 1-simplex connecting 0th and 3rd vertices would have to be simulataneously $ad$ and $ad'$, which is impossible.
| 2 | https://mathoverflow.net/users/402 | 441182 | 178,095 |
https://mathoverflow.net/questions/441180 | 0 | [This is a sequel to the previous question [sub-coroot systems](https://mathoverflow.net/questions/441161/sub-coroot-systems), that has been answered! :-) ]
Let $T$ be a maximal torus of a compact Lie group $K$,
and let $\Lambda \subset {\mathfrak t}$ be the coroot lattice for $(K,T)$, where $\mathfrak t$ is the Lie algebra of $T$.
Assume now that $K'$ is another compact Lie group that admits a homomorphic embedding $i:K\hookrightarrow K'$ verifying the condition that $i(Z\_K)\subset Z\_{K'}$, where $Z\_K, Z\_{K'}$ are the centers of $K$ and $K'$, respectively. Let $T'$ be a maximal torus of $K'$ such that $i(T)\subset T'$. This gives a vector space inclusion ${\mathfrak t} \subset {\mathfrak t}'$, into the Lie algebra of $T'$.
Question: Is it true that
$$ \Lambda = {\mathfrak t} \cap \Lambda'$$
where $\Lambda' \subset {\mathfrak t}'$ is the coroot lattice of $(K',T')$?
If this is not always true, are there some simple conditions under which it becomes true?
| https://mathoverflow.net/users/409881 | Sub-coroot lattices | As for your [other question](https://mathoverflow.net/questions/441161/sub-coroot-systems), I will work with complex Lie groups, but you can pass to maximal compact subgroups if you prefer.
Consider the natural embedding $\operatorname{SO}\_4(\mathbb C) \to \operatorname{SL}\_4(\mathbb C)$, where $\operatorname{SO}\_4$ is taken with respect to the quadratic form $(x\_1, x\_2, x\_3, x\_4) \mapsto x\_1 x\_4 + x\_2 x\_3$. If we denote the simple (with respect to the upper-triangular Borel) roots of the diagonal torus in $\operatorname{SL}\_4(\mathbb C)$ by $\alpha\_1, \alpha\_2, \alpha\_3$, in the obvious fashion, then the common restriction of $\alpha\_1$ and $\alpha\_3$ to the diagonal torus in $\operatorname{SO}\_4(\mathbb C)$ is a simple (with respect to the upper-triangular Borel) root $a$, and the common restriction of $\alpha\_1 + \alpha\_2$ and $\alpha\_2 + \alpha\_3$ is a simple root $b$. Then $a^\vee$ equals $\alpha\_1^\vee + \alpha\_3^\vee$, $b^\vee$ equals $\alpha\_1^\vee + 2\alpha\_2^\vee + \alpha\_3^\vee$, and $\Lambda$ equals $\mathbb Z a^\vee + \mathbb Z b^\vee$, but $(\Lambda' \cap \mathfrak t)/\Lambda$ has order $2$, generated by the image of $\alpha\_2^\vee$.
| 1 | https://mathoverflow.net/users/2383 | 441185 | 178,096 |
https://mathoverflow.net/questions/441194 | 2 | The following is inspired from the most recent [riddle of the week](https://www.spiegel.de/karriere/fuenf-baelle-und-fuenf-eimer-raetsel-der-woche-a-6b30bd21-3487-4ecf-aa34-bcd7efe66509) of the German news magazine [Der Spiegel](https://www.spiegel.de).
For any positive integer $n\in\mathbb{N}$, let $[n]$ denote the set $\{1,\ldots,n\}$. Let $E\_n$ be the expected value of the size of the image of any function $f:[n]\to[n]$, or more explicitly, $$E\_n = \frac{1}{n^n}\sum\_{f:[n]\to[n]}|\text{im}(f)|.$$
What is the value of $\lim\_{n\to\infty}E\_n/n$, or, if the limit does not exist, what are the values of $\lim\inf\_{n\to\infty}E\_n/n$ and $\lim\sup\_{n\to\infty}E\_n/n$, respectively?
| https://mathoverflow.net/users/8628 | Approximate size of the image of functions $f:[n]\to[n]$ | For each $i \in [n]$, the probability that $i$ is in the image of a random function $f: [n] \to [n]$ is $1 - \frac{(n-1)^n}{n^n}$. By linearity of expectation, the expected size of the image of $f$ is
$$n-\frac{(n-1)^n}{n^{n-1}}.$$
Thus, $$\lim\_{n \to \infty} \frac{E\_n}{n}= \lim\_{n \to \infty} 1 - \frac{(n-1)^n}{n^n}=1-\frac{1}{e}.$$
| 15 | https://mathoverflow.net/users/2233 | 441196 | 178,099 |
https://mathoverflow.net/questions/439835 | 2 | I've tried to find counterexamples or results in this direction, but I haven't found what I'm after (except for the $\mathbb{R}^2$ case).
Allard's regularity theorem guarantees that $(\Lambda,r\_0)$-perimeter minimisers are $C^{1,\frac{1}{2}-\epsilon}$ for all $0<\epsilon<\frac{1}{2}$. But as I understand it, Allard's regularity theorem can never prove a set is $C^{1,1}$. The sets I'm interested in are calibrable, but they do not minimise any obvious variational equations (e.g. the Cheeger set for a square is a square with rounded edges, but I'm not sure which variational equation that minimises).
I want to apply Ketterer's [Heintze-Karcher inequality for metric measure spaces](https://doi.org/10.1017/CBO9781139108133.001) to a $(\Lambda,r\_0)$-perimeter minimising set $E$, but this requires a certain outer curvature to be finite everywhere, corresponding to an interior ball condition. The interior and exterior ball conditions together are satisfied if and only if $E$ has $C^{1,1}$ boundary. Some way of proving an interior ball condition for calibrable sets would also be greatly appreciated.
| https://mathoverflow.net/users/106263 | Are $(\Lambda,r_0)$-perimeter minimising sets $C^{1,1}$? | No.
---
Let $f(x,y):=(x^2-y^2)\log(x^2+y^2)$. Then $f$ has bounded mean curvature on bounded sets, and $f\in C^{1,1-\epsilon}$ for all $\epsilon>0$, but $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are not Lipschitz.
| 2 | https://mathoverflow.net/users/106263 | 441205 | 178,102 |
https://mathoverflow.net/questions/441190 | 4 | The papers *Periods of integrals on algebraic manifolds* by Griffiths is often quoted as the first instance where the Hodge ring of a smooth projective hypersurface (say defined by the homogenous polynomial $f$) is related to the jacobian ring of $f$.
However, this paper is divided in three parts of equal length (about 70 pages each, so more than 200 pages in total) and the references I have don't give the precise section nor a set of pages where the result might be found. I have tried to flip through the paper but with this superficial approach, I haven't been able to locate the computation of the Hodge ring of a smooth projective hypersurface in terms of the Jacobian ring of the homogeneous polynomial defining it.
Does anyone know the precise location of this computation in Griffth's papers?
Thanks in advance.
| https://mathoverflow.net/users/37214 | Precise reference in Griffiths' papers : computation of the Hodge theory of a smooth projective hypersurface | As mentioned in the post pointed out by Jason, the correct reference is *On the Periods of Certain Rational Integrals II* by
P. Griffiths,
Ann. Math. 90, no. 3 (1969), pp. 496-541. The best place to look at is §10, where the results are explained using sheaf cohomology.
| 2 | https://mathoverflow.net/users/40297 | 441210 | 178,104 |
https://mathoverflow.net/questions/441063 | 23 | Is the Salem prize discontinued? On the [relevant Wikipedia entry](https://en.wikipedia.org/wiki/Salem_Prize), I don't see anyone since 2018 on there. Why was it discontinued?
| https://mathoverflow.net/users/499631 | What happened to the Salem prize? | I inquired with the IAS. The Salem prize has not been discontinued, but the pandemic has interrupted operations. It should reopen for nominations later this year.
You can email me for the full message I received, which includes some personal details.
| 30 | https://mathoverflow.net/users/11260 | 441212 | 178,105 |
https://mathoverflow.net/questions/441130 | 5 | $\newcommand{\Aut}{\operatorname{Aut}}$Let $G$ be an abelian group.
It seems to be a well-known fact (for example [here](https://mathoverflow.net/questions/349880/classifying-space-for-fibrations-with-eilenberg-maclane-space-fibers-and-nontriv?noredirect=1&lq=1)) that $B\Aut(K(G,1))$, the classifying space of the topological monoid of (unbased) self homotopy equivalences of the Eilenberg-Maclane space $K(G,1)$, has nontrivial homotopy groups $\pi\_1 = \Aut(G),\pi\_2 = G$.
There is a fibration of topological monoids
$$\Aut\_\*(K(G,1)) \to \Aut(K(G,1)) \to K(G,1)$$
where the first map is the inclusion of the fiber, and the second map is evaluating at the basepoint $e$ ($\Aut\_\*$ is the topological monoid of *based* homotopy equivalences).
One can readily show that $\Aut\_\*(K(G,1))$ is homotopy equivalent to $\Aut(G)$ endowed with the discrete topology. Taking classifying spaces we get a fibration sequence
$$K(\Aut(G),1) \to B\Aut(K(G,1))\to K(G,2)$$
which induces the LES in homotopy
$$\cdots \to 0 \to \pi\_2(X) \to G \overset{\phi}{\to} \Aut(G) \to \pi\_1(X) \to 0 \to \cdots$$
where $X$ is $B\Aut(K(G,1))$.
To get the stated result, it suffices to show that the map $\phi: G \to \Aut(G)$ sends $g$ to conjugation by $g$. Why is this true? I appreciate any references.
| https://mathoverflow.net/users/141291 | Computing the homotopy type of $B\operatorname{Aut}(K(G,1))$ using a fibration sequence: why is $G \to \text{Aut(G)}$ given by conjugation? | This argument is rather elementary. Maybe we should later move this to MathStackExchange. Anyway:
As mentioned above in comments, $K(G,1)$ is not a topological monoid and $K(G,2)$ doesn't exist, so the initial setting is the fibration
$Aut\_\*(K(G,1))\hookrightarrow Aut(K(G,1))\twoheadrightarrow K(G,1)$
where $Aut(K(G,1))$ is the topological monoid of self-homotopy equivalences and $Aut\_\*(K(G,1))$ is the submonoid of pointed ones, for $K(G,1)$ has a base-point $x\_0$. This base-point can be taken as the unique vertex of a $CW$-model in order to apply the HEP (homotopy extension property). We obviously identify $\pi\_1(K(G,1),x\_0)=G$.
We have an isomorphism $\pi\_0(Aut\_\*(K(G,1)))\cong Aut(G)$ which sends a pointed self-equivalence $f\colon K(G,1)\to K(G,1)$ to $\pi\_1(f,x\_0)\colon G\to G$.
We now want to compute the morphism
$$\partial\colon G=\pi\_1(K(G,1),x\_0)\longrightarrow \pi\_0(Aut\_\*(K(G,1)))\cong Aut(G)$$
which is part of the long exact sequence of homotopy groups induced by the previous fibration.
Looking at the definition of this morphism in any basic text dealing with homotopy groups, we can proceed as follows:
1. Pick $g\in G$.
2. Choose a representing closed path at $x\_0$ $$\gamma\_g\colon [0,1]\longrightarrow K(G,1).$$
3. Lift it to a non-closed path starting at the base-point of $Aut(K(G,1))$, which is the identity in $K(G,1)$, $$\tilde{\gamma}\_g\colon [0,1]\longrightarrow Aut(K(G,1)).$$
4. $\partial(g)$ is the connected component of $\tilde{\gamma}\_g(1)$.
Now let's proceed wisely in our example:
If we apply the HEP to the pair given by $\gamma\_g$ and the identity in $K(G,1)$, we obtain an unbased homotopy
$$H\colon [0,1]\times K(G,1)\longrightarrow K(G,1)$$
such that $H(0,-)$ is the identity in $K(G,1)$ and $H(-,x\_0)=\gamma\_g$.
All maps $H(t,-)$ are homotopic to the identity in $K(G,1)$, hence they are self-homotopy equivalences, therefore we can choose the lifting in 3 as $$\tilde{\gamma}\_g(t)=H(t,-).$$
Then $\partial(g)$, regarded as an automorphism of $G$, is precisely the morphism induced in $\pi\_1(-,x\_0)$ by the based map $H(1,-)\colon K(G,1)\to K(G,1)$. This map is based because $\tilde{\gamma}\_g$ lifts a closed path.
Let us compute $\pi\_1(H(1,-),x\_0)$. Let $\mu\colon [0,1]\to K(G,1)$ be a closed path based at $x\_0$ representing an element $h\in G$. The homotopy $H(-,\mu)$ is an unbased homotopy between $H(0,\mu)=\mu$ and $H(1,\mu)$. Since $H(-,x\_0)$ is the path $\gamma\_g$ we have $H(1,\mu)\simeq \gamma\_g\*\mu\*\gamma\_g^{-1}$. Therefore $\partial(g)$ maps $h$ to $ghg^{-1}$.
| 3 | https://mathoverflow.net/users/12166 | 441226 | 178,107 |
https://mathoverflow.net/questions/441234 | 11 | For $A,B\in{\cal P}(\omega)$ we say $A\subseteq^\* B$ if $A\setminus B$ is finite (that is, $A$ is "almost contained" in $B$). We write $A\simeq\_{\text{fin}} B$ if $A\subseteq^\* B$ and $B\subseteq^\* A$ (that is, the sets $A, B$ are "almost the same set" except for finitely many elements). It is easy to see that $\simeq\_{\text{fin}}$ is an equivalence relation on ${\cal P}(\omega)$.
We denote the collection of equivalence classes on ${\cal P}(\omega)$ with respect to $\simeq\_{\text{fin}}$ by ${\cal P}(\omega)/(\text{fin})$. Using $\subseteq^\*$ on representatives of equivalence classes, it is easy to see that we can make ${\cal P}(\omega)/(\text{fin})$ into a partially ordered set.
In the Noah Schweber's [beautiful post](https://mathoverflow.net/a/440492/8628) inspiring this question, we learn that there is no order-embedding from $\omega\_1$ into ${\cal P}(\omega)$, but $\omega\_1$ *can* be order-embedded in ${\cal P}(\omega)/(\text{fin})$. So within the Continuum Hypothesis ${\sf (CH)}$ we get that $2^{\aleph\_0} = \omega\_1$ can be order-embedded in ${\cal P}(\omega)/(\text{fin})$.
**Question.** Can the cardinal $2^{\aleph\_0}$ (well-ordered by $\in$) be order-embedded in ${\cal P}(\omega)/(\text{fin})$ even if $\neg{\sf(CH)}$? If not: can every member of $2^{\aleph\_0}$ be order-embedded in ${\cal P}(\omega)/(\text{fin})$?
| https://mathoverflow.net/users/8628 | Can the cardinal $2^{\aleph_0}$ be order-embedded in ${\cal P}(\omega)/(\text{fin})$? | Yes! The fact that this is consistent is originally due to Laver. Later, Baumgartner, Frankiewicz, and Zbierski strengthened Laver's result to the following theorem:
**Theorem:** Is it consistent that $\mathsf{MA}\_{\sigma\text{-linked}}$ holds and that every Boolean algebra of size $\mathfrak{c}$ can be order-embedded in $\mathcal P(\omega)/\mathrm{fin}$.
The theorem is proved in
*Baumgartner, J.; Frankiewicz, R.; Zbierski, P.*, [**Embedding of Boolean algebras in P((\omega) )/fin**](https://doi.org/10.4064/fm-136-3-187-192), Fundam. Math. 136, No. 3, 187-192 (1990). [ZBL0718.03039](https://zbmath.org/?q=an:0718.03039).
On the other hand, Kunen proved in his thesis that in the Cohen model, there is no order-preserving embedding of $\omega\_2$ into $\mathcal P(\omega)/\mathrm{fin}$. So the statement "every linear order of size $\mathfrak{c}$ can be order-embedded into $\mathcal P(\omega)/\mathrm{fin}$" is independent of ZFC.
I believe it is still an open problem whether the "$\mathsf{MA}\_{\sigma\text{-linked}}$" in the theorem above can be strengthened to simply "MA". I seem to remember hearing at one point that Woodin had proved something about this, but I'm not sure it was ever published and I don't remember exactly what he proved.
| 13 | https://mathoverflow.net/users/70618 | 441237 | 178,110 |
https://mathoverflow.net/questions/441167 | 2 | I have two smoothly embedded orientable surfaces $S\_1,S\_2\subset S^3 \times [0,1]$ with boundary such that
$(i)$ $S\_1\cap S\_2$ is a smoothly embedded surface without boundary and
$(ii)$ $\overline{S\_1}\cap \overline{S\_2}=\overline{S\_1\cap S\_2}$
Now I want to prove that $S\_1 \cup S\_2$ is a smoothly embedded surface, by showing that each point in it has an open neighborhood $U \subset S\_1 \cup S\_2$ completely contained in either $S\_1$ or $S\_2$.
I tried to do this by doing a case distinction depending on where the point lies. The cases where I can't show this are when the point lies in $ int(S\_1) \cap \partial S\_2$, $\partial S\_2 \cap \partial S\_1$ or $\partial S\_1 \cap int(S\_2)$ (the last case is of course equivalent to the first).
Here are my questions:
1. How can one show that there is an open neighborhood in $S\_1 \cup S\_2$ completely contained in $S\_1$ or $S\_2$ in those last three cases? (Somehow I feel that one should make use of the tubular neighborhood theorem...)
2. Is $S\_1 \cup S\_2$ even guaranteed to be a smooth submanifold with boundary under my conditions?
3. What are alternative conditions which could guarantee that $S\_1 \cup S\_2$ is a submanifold when $S\_1 \cap S\_2 \neq \emptyset$?
According to the accepted answer in this mathoverflow question which is very similar to mine ([When is the union of embedded smooth manifolds a smooth manifold?](https://mathoverflow.net/questions/78733/when-is-the-union-of-embedded-smooth-manifolds-a-smooth-manifold)), a sufficient condition would be that $S\_1 \cap S\_2$ is a n-submanifold and $\overline{S\_1}\cap S\_2=\overline{S\_1\cap S\_2}\cap S\_2$.
I think this is not true (under the assumption that submanifolds with boundary are considered submanifolds, else there was another counterexample in the comments of the answer): Consider $S\_1=[-1,1] \subset \mathbb{R}^2$ and $S\_2 \subset \mathbb{R}^2$ to be the graph of the smooth function $f:[-1,1] \to \mathbb{R},t\mapsto \begin{cases} 0 \ \ \ \ \ \ \ \ \ if \ \ t \in [-1,0] \\ e^{-t^{-2}} \ \ \ if \ \ t>0\end{cases}$. Then the previous conditions hold, but obviously $S\_1 \cup S\_2$ does not define a submanifold (removing the point (0,1) produces 3 components). But correct me if I am wrong here, since this would provide me with an alternative sufficient condition I could use to show that $S\_1 \cup S\_2$ is a submanifold.
Most of the ideas I had so far, did not use the fact that I deal with surfaces inside $S^3 \times [0,1]$, and I believe that this should work in every dimension. But if someone knows how to make use of this assumption, I would be happy because that is the specific case I am interested in the most.
Thank you for your time.
| https://mathoverflow.net/users/171941 | Sufficient condition for the union of two submanifolds to be a submanifold | In the meantime, a very similar question of mine has been answered here <https://math.stackexchange.com/a/4642619/857154> , which answers these questions aswell. Moishe Kohan has provided a counterexample to my claim for 1-manifold which most likely carries over to surfaces. Therefore 1) cannot be shown, the answer to 2) is no, and an alternative condition would be the following:
(a) For every sequence $(x\_i)$ in $S\_2$ converging to a point $x\in S\_1$, for all sufficiently large $i$, $x\_i\in S\_1$.
(b) Same as (a) with the roles of $S\_1, S\_2$ swapped.
| 0 | https://mathoverflow.net/users/171941 | 441239 | 178,111 |
https://mathoverflow.net/questions/441106 | 5 | **Problem:** Let $M\subseteq B(H)$ be a finite von Neumann algebra with a faithful tracial state $\tau$. Let $\widetilde{M}$ be the $\tau$-measurable operators on $M$ (recalled below). Extend the trace $\tau$ on $\widetilde{M}\_+$ by $\tau(a):=\int\_0^\infty\lambda\tau(e\_\lambda)$ where $a=\int\_0^\infty\lambda\,de\_\lambda$ is the spectral decomposition (See Equation 4.6 of Hiwi). If $\{x\_n\}\_{n\in\mathbb{N}}$ be a sequence of positive elements from $M$, which converges strongly to an element $x$ of $\widetilde{M}\_+$, then can we say that $\{\tau(x\_n)\}\_{n\in\mathbb{N}}$ converges to $\tau (x)$?
Or, can we at least say that if $\{x\_n\}\_{n\in\mathbb{N}}$ be a sequence of positive elements from $M$, which converges strongly to an element $x$ of $\widetilde{M}\_+$, then $\{x\_n\}\_{n\in\mathbb{N}}$ converges to $x$ in the measure topology?
**(A positive answer of any one of the above two questions would be sufficient for me.)**
I got stuck with this problem while reading $\tau$-measurable operators from the book '[Lectures on Selected Topics in von Neumann Algebras](https://drive.google.com/file/d/1H6ki8gjUPY5dsWE-CrGn9WFNhMEfEkIv/view?usp=sharing)' by Hiwi. Here I recall the definition of $\tau$-measurable operator.
**Definition 1:** For each $\epsilon,\delta>0$, define $$\mathscr{O}(\epsilon,\delta)=\{m\text{ affiliated to } M:eH\subseteq \mathcal{D}(m),\,\|me\|\leq \epsilon \text{ and }\tau(1-e)\leq\delta \text{ for some } e\in Proj(M)\}.$$
Let $m$ be a densely defined closed operator such that $m$ is affiliated to $M$. We say that $m$ is $\tau$-measurable if for any $\delta >0$, there exists an $\epsilon >0$ such that $m\in\mathscr{O}(\epsilon,\delta)$. We denote by $\widetilde{M}$ the set of such $\tau$-measurable operators.
**Theorem 2:** (Theorem 4.12 of Hiwi) The $\widetilde{M}$ is a complete metrizable Hausdorff topological \*-algebra with $\{\widetilde{M}\cap\mathscr{O}(\epsilon,\delta):\epsilon,\delta >0\}$ as a neighborhood basis of $0$. Moreover, $M$ is dense in $\widetilde{M}$.
Thanks in advance for any help or suggestion.
| https://mathoverflow.net/users/477204 | Continuity of the extension of a tracial state with respect to the strong operator topology | With the specific definition of strong convergence in the comment (namely, a sequence $x\_n \in M\_+$ is said to converge strongly to $x \in \widetilde{M}\_+$ if and only if $x\_n \xi \to x \xi$ for all $\xi \in D(x)$), both properties indeed hold.
Take such a sequence $x\_n \in M\_+$ converging strongly to $x \in \widetilde{M}\_+$. We prove that $\tau(x\_n) \to \tau(x)$ and $x\_n - x \to 0$ in measure.
First assume that $\tau(x)=+\infty$. Choose $\kappa > 0$. Since $\tau(x) = +\infty$, we can choose a spectral projection $p = e\_{[0,\lambda]}(x)$ such that $\tau(xp) > \kappa + 1$. By assumption, $x\_n p \to x p$ strongly in the usual sense. By the uniform boundedness principle, the sequence $x\_n p$ is bounded in operator norm and $\tau(x\_n p) \to \tau(x p)$. We can thus take $n\_0$ such that $|\tau(x\_n p) - \tau(xp)| < 1$ for all $n \geq n\_0$. Thus, $\tau(x\_n p) > \kappa$ for all $n \geq n\_0$. Since
$$\tau(x\_n) \geq \tau(x\_n^{1/2} p x\_n^{1/2}) = \tau(x\_n p) > \kappa \; ,$$
we conclude that $\tau(x\_n) > \kappa$ for all $n \geq n\_0$. Thus, $\tau(x\_n) \to + \infty$.
Next assume that $\tau(x) < +\infty$. We then have a well-defined normal positive functional $\omega$ on $M$ satisfying
$$\omega(a) = \tau((1+x)^{1/2} a (1+x)^{1/2}) \quad\text{for all $a \in M$.}$$
By assumption $x\_n (1+x)^{-1} \to x (1+x)^{-1}$ strongly in the usual sense. Again by uniform boundedness,
$$\omega(x\_n (1+x)^{-1}) \to \omega(x (1+x)^{-1}) \; .$$
This precisely says that $\tau(x\_n) \to \tau(x)$.
To prove that $x\_n - x \to 0$ in measure, first note the following standard result: if $y\_n$ is a sequence in $M$ such that $y\_n \to 0$ strongly, then $y\_n \to 0$ in measure. Indeed, given $y \in M$ and $\delta > 0$, consider the spectral projection $e = e\_{[0,\delta]}(y^\* y)$. Since $y^\* y \geq \delta (1-e)$, we find that
$$\delta \, \tau(1-e) \leq \tau(y^\* y) \quad\text{and}\quad \|y e \| \leq \delta \; .$$
Writing $\|y\|\_2 = \sqrt{\tau(y^\* y)}$, taking $\delta = \|y\|\_2$ and using the notation $\mathcal{O}(\cdot,\cdot)$ for the basic neighborhoods of $0$ in the measure topology as in the question, it follows that
$$y \in \mathcal{O}(\|y\|\_2,\|y\|\_2) \quad\text{for all $y \in M$.}$$
When $y\_n \to 0$ strongly, also $\|y\_n\|\_2 \to 0$ and thus $y\_n \to 0$ in measure.
We now return to the sequence $x\_n - x$. Fix $\varepsilon > 0$. Take a spectral projection $p$ of $x$ such that $xp$ is bounded and $\tau(1-p) < \varepsilon/2$. Since $(x\_n - x) p \to 0$ strongly in the usual sense, also $(x\_n - x)p \to 0$ in measure. So, for all $n$ large enough, we have that $(x\_n - x)p \in \mathcal{O}(\varepsilon,\varepsilon/2)$. It follows that $x\_n - x \in \mathcal{O}(\varepsilon,\varepsilon)$ for all $n$ large enough. Thus, $x\_n \to x$ in measure.
| 2 | https://mathoverflow.net/users/159170 | 441242 | 178,112 |
https://mathoverflow.net/questions/441224 | 2 | I noticed [this](https://mathoverflow.net/questions/380070/is-a-mixture-of-real-analytic-functions-again-analytic) post. But still I'd like to follow up with a specific case I have in mind. Say $p(x| \theta)$ is the density of a Gaussian distribution on $\mathbb{R}^n$ with mean $\theta$ and known covariance $\Sigma$. Let
$$f(x) = \int\_{\Theta} p(x| \theta) \Lambda(d\theta),$$
where $\Theta$ is a proper subset of $\mathbb{R}^n$, and $\Lambda$ is an arbitrary probability measure over $(\Theta, \mathcal{B}(\Theta))$, not necessarily continuous w.r.t. the Lebesgue measure. Let $g(x)$ be the density of another Gaussian on $\mathbb{R}^n$ with mean $\mu$ and known covariance $\Sigma + \Psi$. Is it possible for $g(x)$ and $f(x)$ to agree on a set of positive Lebesgue measure?
| https://mathoverflow.net/users/157159 | Mixture of gaussian density agree with another gaussian on positive measure | $\newcommand\R{\mathbb R}\renewcommand\th{\theta}\newcommand{\Th}{\Theta}\newcommand{\Si}{\Sigma}
\newcommand\La\Lambda\newcommand{\C}{\mathbb C}$The answer is: This will be so if (and only if) $\Lambda$ itself is a (possibly degenerate) Gaussian distribution.
Let us prove the "only if" part. Here it does not really matter whether it is a priori assumed that $\Th$ is a proper subset of $\R^n$. Also, by substitutions $x\leftrightarrow\Si^{1/2}x$ and $\th\leftrightarrow\Si^{1/2}\th$, without loss of generality $\Si=I\_n$, the identity matrix. So,
\begin{equation\*}
f(x)=(2\pi)^{-n/2}\int\_{\R^n}\exp\Big(-\frac12\sum\_{j=1}^n(x\_j-\th\_j)^2\Big)\,\La(d\th)
\end{equation\*}
for $x=(x\_1,\dots,x\_n)\in\R^n$. Since
\begin{equation\*}
\sum\_{j=1}^n(x\_j+iy\_j-\th\_j)^2=\sum\_{j=1}^n(x\_j-\th\_j)^2-\sum\_{j=1}^n y\_j^2
+2i\sum\_{j=1}^n(x\_j-\th\_j)y\_j, \tag{2}\label{2}
\end{equation\*}
the function $f$ can be extended to a holomorphic function on $\C^n$. Similarly, the function $g$ can be extended to a holomorphic function on $\C^n$.
So, $f$ and $g$ are real-analytic on each (straight) line in $\R^n$.
Suppose now $f=g$ on some subset $A$ of $\R^n$ of Lebesgue measure $|A|>0$. Without loss of generality, the set $A$ is bounded.
Take any $x\in\R^n$. Then
\begin{equation\*}
0<|A|=|A-x|=\int\_{S^{n-1}}du\,\int\_0^\infty r^{n-1}\,dr\,1(ru\in A-x),
\end{equation\*}
where $\int\_{S^{n-1}}du$ is the integral with respect to the surface measure on the unit sphere $S^{n-1}$ in $\R^n$. So, by the Tonelli theorem, there is some $u\_x\in S^{n-1}$ such that $\int\_0^\infty r^{n-1}\,dr\,1(ru\_x\in A-x)>0$ and hence
\begin{equation\*}
|A\cap(x+\R\_+ u\_x)|=\int\_0^\infty dr\,1(x+ru\_x\in A)=\int\_0^\infty dr\,1(ru\_x\in A-x)>0.
\end{equation\*}
So, the set $A\cap(x+\R\_+ u\_x)$ is an infinite and bounded subset of the line $x+\R u\_x$, and $f=g$ on this infinite bounded subset of a line. Recalling that $f$ and $g$ are real-analytic on each line in $\R^n$, we conclude that $f=g$ on the line $x+\R u\_x$, and hence $f(x)=g(x)$.
So, $f=g$ (on the entire space $\R^n$), which can be rewritten as the identity
\begin{equation\*}
e^{-\|x\|^2/2}\int\_{\R^n}\exp\Big(\sum\_{j=1}^n \th\_j x\_j\Big)\,L(d\th)
=C\exp\big(-x^\top Bx+x^\top\mu\big)
\end{equation\*}
for all $x\in\R^n$,
where $\|\cdot\|$ is the Euclidean norm,
$$L(d\th):=e^{-\|\th\|^2/2}\La(d\th),$$
$B$ is some positive-definite $n\times n$ real matrix, $\mu$ is some vector in $\R^n$, and $C:=L(\R^n)=(2\pi)^{n/2}f(0)$. So, for (joint) moment generating function (mgf) $M\_{L/C}$ of the probability measure $L/C$, some symmetric matrix $R$, and all $x\in\R^n$ we have
\begin{equation\*}
M\_{L/C}(x)=\exp\big(-x^\top Rx+x^\top\mu\big)
\end{equation\*}
Any mgf is log convex. So, $R$ must be positive semidefinite. Thus, $M\_{L/C}$ is the mgf of a (possibly degenerate) Gaussian distribution. So, $L/C$ is a (possibly degenerate) Gaussian distribution. So, $\La$ is a (possibly degenerate) Gaussian distribution.
Vice versa, if $\La$ is a (possibly degenerate) Gaussian distribution, then $f$ is clearly a Gaussian distribution. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 441256 | 178,118 |
https://mathoverflow.net/questions/441231 | 1 | I have been recently trying to look at substitution tilings with finite local complexity by examining their admissible patch\pattern atlas, which is sometimes called their language. I have also seen the term dictionary used.
To the best of my understanding, given a finite collection of proto-tiles $\mathcal{P}'$ and a substitution rule $S$ defined on these proto-tiles, an admissible patch\pattern $Q$ is a partial tiling that occurs in $S^n(P)$ for some $P\in \mathcal{P'}$ and $n\in \mathbb{N}$. This is at least the case when the substitution is symbolic and the prototiles are the letters of the alphabet.
From what I searched so far, I have not seen many give the way to compute these atlases\langauges, although I have seen people write them partially in some cases. Therefore I was wondering about a method or algorithm that people are using implicitly without mentioning.
I have only found two references for the computation of a language, for symbolic substitutions in $1$-dimension, in algorithm 3 of [Grout: A 1-Dimensional Substitution Tiling Space Program](https://arxiv.org/abs/1512.00398) and in section 3.2 of the paper [Computations for symbolic substitutions](https://arxiv.org/abs/1705.11130), by Dan Rust and Scott Balchin.
My question is whether a variation of this algorithm is what people are using implicitly when giving admissible patches for general substitutions? For example in the book by Baake and Grimm, [Aperiodic order, Volume 1: A Mathematical Invitation](http://www.aperiodicorder.org/), when they talk about block substitutions in section 4.9 and inflation tilings in section 6?
I would appreciate any input on this question, and also corrections to any incorrect statements I may have made.
| https://mathoverflow.net/users/143153 | Computing admissible patches of a substitution | The higher dimensional situation isn't very different to the one-dimensional situation. Of course, there are probably quicker ways to do it than the following, but this at least works and is reasonably fast for most purposes. Also, keep in mind that this method is to check a single word/patch to see if it's legal. If you want to build the set of *all* legal words/patches of a particular size, then there are faster methods that 'build the language up' and let you use the fact that you know all patches of size at most $N$ to build the patches of size $N+1$.
**Method**
I think the method that Scott and I gave in that paper is a bit naïve and very much brute-force. There are faster methods.
In one dimension, for any legal word $u$, there must exist a partition $u\_1u\_2 = u$ and a power $p$ such that $u\_1$ is a suffix of a word of the type $S^p(a\_1)$ and $u\_2$ is a prefix of a word of the type $S^p(a\_2)$ (of course $u\_1$ or $u\_2$ may be empty). This just comes from the definition of a word being legal.
There are obviously only finitely many pairs of letters to consider and there are only finitely many prefix/suffix pairs of words of the form $S^p(a)$, so this is a finite check. How finite? Well bounds exist, and they come from the fact that for a primitive substitution $S$, there are constants $c, C$ so that for all $a \in \mathcal{A}$, $$c \lambda^k < |S^k(a)| < C \lambda^k,$$ where $\lambda$ is the Perron--Frobenius eigenvalue of the substitution matrix $M$ (and these constants can be calculated in an algorithmic fashion if required). The lower bound means that we just need to pick $k$ so that $c\lambda^k > |u|$ and then check the pairs $S^{k+i}(a)$ where $i \in \{0,1,\ldots, \#\mathcal{A}-1\}$ (in the worst case, you may have to go all the way to $i = \#\mathcal{A}-1$).
**Higher dimensions**
The same kind of argument works in higher dimensions, you just need to consider partitions into more than just a pair of patches. For instance, in a block substitution in 2d, maybe your legal patch comes from the meeting of four corners of large supertiles, so you need to partition your patch into four rectangles. This is where the geometry plays a role in higher dimensions. Thankfully though, for true inflation rules (not weird ones where some tiles don't grow under substitution), all supertiles are eventually larger than any particular ball, and so you just need to consider how tiles can surround a vertex/edge/face/etc., for which there are only finitely many possibilities (assuming FLC).
If you know that your substitutions is recognisable (which by Mossé/Solomyak is true for any aperiodic primitive substitution), that can also help immensely, as recognisability implies local recognisability, so up to the boundary of your patch, if your patch is legal, then there is a unique way to partition your patch into super-tiles, then super-super-tiles, etc., and so the only difficult work then is checking the boundaries, which can possibly introduce some choices that all need to be checked. In practice, this is the fastest method.
**Example**
For instance, let's check if the word $u = bbaababbabaababbaabbabaa$ is legal for the Thue-Morse substitution $S \colon a\mapsto ab, b\mapsto ba$. The Thue-Morse substitution is recognisable, so let's start cutting up into super-tiles then super-supertiles... (notice that we get more boundary problems to account for at every iteration)
$b(ba)(ab)(ab)(ba)(ba)(ab)(ab)(ba)(ab)(ba)(ba)a$ is the only legal way to cut $u$ into super-tiles.
$b(baab)(abba)(baab)(abba)(abba)(ba)a$ is the only legal way to cut $u$ into super-super-tiles.
$b(baababba)(baababba)(abba)(ba)a$ is the only legal way to cut $u$ into super${}^3$-tiles
We can't cut it up any more as there doesn't exist an internal super${}^4$-tile. What we can do now though is use the fact that we just need to find a four-letter legal word $v$ that has $u$ as a subword of its third substitutive image and we also know that the middle two letters have to be $b$s because $baababba = S^3(b)$. Well, there's only one such four-letter legal word and that's $abba$. And indeed, if we check $S^3(abba)$ we get $abbabaa |bbaababbabaababbaabbaba |ab$, which contains our word $u$ (between the bars).
If a word is not legal, then some part of the above process would have failed.
In higher dimensions, the process works exactly the same.
| 3 | https://mathoverflow.net/users/21271 | 441276 | 178,124 |
https://mathoverflow.net/questions/441260 | 6 | I am trying to find a reference for the following well-known result on the functoriality of the Leray spectral sequence:
Let $\pi:X\to Y$ and $\pi':X'\to Y'$ be morphisms of schemes and denote by $E\_2^{p,q}(\mathcal{F})=\mathrm{H}^q(Y,R^q\pi\_\*\mathcal{F})\implies \mathrm{H}^{p+q}(X,\mathcal{F})$ and $(E')\_2^{p,q}(\mathcal{G})\implies \mathrm{H}^{p+q}(X',\mathcal{G})$ their associated Leray spectral sequences (in étale cohomology for some étale sheafs $\mathcal{F},\mathcal{G}$). Assume that there are $f:X'\to X$ and $g:Y'\to Y$ such that the obvious square commutes. Then there is a morphism of spectral sequences $E\_r^{p,q}(\mathcal{F})\to (E')^{p,q}\_r(g^{-1}\mathcal{F})$ induced by the natural pullback maps on cohomology $E\_2^{p,q}(\mathcal{F})\to (E')\_2^{p,q}(g^{-1}\mathcal{F})$ such that the map on the infinity-pages agrees with the canonical pullback morphism $\mathrm{H}^n(X,\mathcal{F})\to \mathrm{H}^n(X',f^{-1}\mathcal{F})$.
It would be amazing if someone knows a reliable source for this frequently used result. Thanks in advance!
| https://mathoverflow.net/users/492820 | Leray spectral sequence and pullbacks | I apologize for the self promotion, but page 570 of my article *The Leray spectral sequence is motivic* has a very brief discussion of the functoriality of Leray.
**Added** In a bit more detail, here are the key points:
* To every object $(C, F)$ in the (bounded below, biregularly filtered) filtered derived category, one has a spectral sequence
with $E\_1^{pq}= H^{p+q}(Gr^p\_FC)$. This construction is functorial. See Deligne, Théorie de Hodge II, sect. 1; III, sect. 7.
* With your notation, the spectral sequence associated to
$(\mathbb{R} \pi\_\* \mathcal{F},\tau)$ becomes the Leray spectral sequence after reindexing to transform $E\_1$ to $E\_2$ [loc. cit.]. Here $\tau$ is the canonical filtration, given by truncations.
* Let $\mathcal{F}'= f^{-1}\mathcal{F}$. Then
we have a morphism in the filtered derived category
$$ (\mathbb{R} \pi\_\* \mathcal{F},\tau)\to \mathbb{R}g\_\*(\mathbb{R} \pi'\_\* \mathcal{F}',\tau)$$
This will induce a morphism of Leray spectral sequences.
| 8 | https://mathoverflow.net/users/4144 | 441288 | 178,128 |
https://mathoverflow.net/questions/440785 | 6 | The following question originates from a Physics problem, so I apologize if I am not using a suitable mathematical jargon.
The original system involves $N$ massless electric charges at position $\boldsymbol{r}\_1$, $\boldsymbol{r}\_2$, ..., $\boldsymbol{r}\_N$ which can move on a plane pierced by a uniform and constant transverse magnetic field $\boldsymbol{B}$. The motion equation governing the dynamics of the $j$-th charge is:
\begin{equation}
\label{eq:Motion\_eq}
\boldsymbol{0}=q\_j\,\dot{\boldsymbol{r}}\_j\times\boldsymbol{B}\, +\, q\_j\boldsymbol{E}(\boldsymbol{r}\_j)
\end{equation}
where $q\_j$ is the charge of the $j$-th electric charge and $\boldsymbol{E}(\boldsymbol{r}\_j)$ is the electric field generated at position $\boldsymbol{r}\_j$ by all the remaining charges. Moreover, at any time, the electric field at position $\boldsymbol{E}(\boldsymbol{r}\_j)$ can be written as
$$
\boldsymbol{E}(\boldsymbol{r}\_j) = \sum\_{i=1\\i\neq j}^N q\_i\frac{\boldsymbol{r}\_j-\boldsymbol{r}\_i}{|\boldsymbol{r}\_j-\boldsymbol{r}\_i|^3}.
$$
The motion equation above is manifestly of *first* order in time and the only initial conditions one needs are the initial positions of the massless charges, i.e. $\boldsymbol{r}\_1(t=0)$, $\boldsymbol{r}\_2(t=0)$, ..., $\boldsymbol{r}\_N(t=0)$.
Now let us switch from *massless* to *massive* charges. This means that the aforementioned motion equation transforms as:
\begin{equation}
M\_j \ddot{\boldsymbol{r}}\_j=q\_j\,\dot{\boldsymbol{r}}\_j\times\boldsymbol{B}\, +\, q\_j\boldsymbol{E}(\boldsymbol{r}\_j)
\end{equation}
We assume that all the masses $M\_j$ are very small. As far as I understand, the introduction of a non-zero mass constitutes a *singular perturbation* as it manifestly alters the order of the differential equations. Accordingly, the number of initial conditions which one should set doubles.
My question is: if one knows the solution $\{\boldsymbol{r}\_1(t),\,\boldsymbol{r}\_2(t),\,\dots,\,\boldsymbol{r}\_N(t)\}$ of the *massless* problem (i.e. of the system of first-order differential equations), what can be said about the solution of the *massive* problem (i.e. of the system of second-order differential equations)?
For simplicity we can assume $M\_j=M\,\forall j$ and we can also make additional assumptions, if needed. E.g. that the energy $$H\_\mathrm{massive}=T+V=\sum\_{j=1}^N\frac{1}{2}M\_j\dot{\boldsymbol{r}}\_j^2 + V$$ of the massive system is limited and does not significantly depart from that of the massless system $$H\_\mathrm{massless}=V=\frac{1}{2}\sum\_{j=1}^N\sum\_{i\neq j}^N \frac{q\_iq\_j}{|\boldsymbol{r}\_i-\boldsymbol{r}\_j|}\,.$$ Another possible and reasonable assumption would be that the initial velocities of the 2nd-order problem deviate only little from the fixed initial velocities of the 1st order problem.
I guess that a term $\mathcal{O}(M^{-1})$ will show up in the solutions of the perturbed system. Is this correct? Apart from this term, is it possible to write a sort of Taylor expansion involving powers of $M$? Is there a general framework to approach this singular-perturbation problem?
| https://mathoverflow.net/users/101308 | Solution of an ODE upon singular perturbation | Let me change the notations to fit the mathematical literature. I will denote by $x(t) \in \mathbb{R}^{3N}$ the positions of the particles, by $y(t) \in \mathbb{R}^{3N}$ their velocities and by $0 < \varepsilon \ll 1$ their common small mass. Your *massive* problem can be written as a first order ODE as:
$$
\begin{cases}
\dot{x} = f(x,y), \\
\varepsilon \dot{y} = g(x,y),
\end{cases}
$$
with $f(x,y) = y$ and $g(x,y) = q y \times B + q E(x)$. The question now becomes whether solutions to the degenerate system $\dot{x} = f(x,y)$ and $g(x,y) = 0$ correctly approximate solutions with small $\varepsilon > 0$.
This question has been studied by many authors, probably starting with Tikhonov's paper:
>
> Andrei Nikolaevich Tikhonov. "*Systems of differential equations
> containing small parameters in the derivatives.*" Mat. Sb.(NS) 31.73
> (1952): 575-586.
>
>
>
As a starting point, I would suggest the book:
>
> Wolfgang Wasow, *Asymptotic expansions for ordinary differential
> equations.* Reprint of the 1976 edition. Dover Publications, Inc., New
> York, 1987. x+374 pp. ISBN: 0-486-65456-7
>
>
>
In particular, Section 39 contains an introduction to the topic (and surveys Tikhonov's results), and Section 40 discusses series expansions (with respect to $\varepsilon$), which correspond to the framework you are looking for.
Eventually, let me mention two difficulties:
1. As already mentioned in the comments, a key difficulty in such singularly perturbed problems lies in the discrepancy between the boundary conditions, which can lead to the presence of boundary layers in the solutions with $\varepsilon > 0$ (in your case, boundary layers in time, located near $t = 0$). The study of boundary layers is a whole field by itself. If you are not familiar with it, Wikipedia's page on [singular perturbations](https://en.wikipedia.org/wiki/Singular_perturbation) already features a nice example with the ODE $\varepsilon \ddot{x}(t) + \dot{x}(t) = - e^{-t}$.
2. In your physical problem, there is another source of singularity, since the electric field becomes singular as two particles become very close to each other. This difficulty is typically not included in the above mentioned literature, which always requires some kind of regularity on $f$ and $g$ (at least continuous for instance). Note that this difficulty is somehow independent, since it is already present with $\varepsilon = 0$ or $\varepsilon = 1$. In order to extend the known results for the general setting explained above to your case, you would first typically have to derive *a priori* estimates proving that the particles won't be collapsing together on the considered time interval.
---
Following your request, here are two (hopefully more "practical") examples:
a) Concerning boundary layers, let us look at $\dot{x} = 1$ on $\mathbb{R}$, with initial condition $x(0) = 0$. Then the solution is $x(t) = t$. Note that, in this massless case, $\dot{x}(0) = 1$.
Now the massive problem is $\varepsilon \ddot{x}\_\varepsilon + \dot{x}\_\varepsilon = 1$, with boundary conditions $x\_\varepsilon(0) = 0$ and $\dot{x}\_\varepsilon(0) = y^0$ given. The explicit solution is
$$
x\_\varepsilon(t) = t + \varepsilon (y^0 - 1) (1-e^{-t/\varepsilon}).
$$
In particular, when $y^0 = 1$ (the initial speed according to the massless problem), one has $x\_\varepsilon(t) = t$, so the massive solution is (in this particular case) exactly the same.
On the contrary, when $y^0 \neq 1$, there is a short period of time, say $t \in [0,5\varepsilon]$ where the solution $x\_\varepsilon(t)$ does not resemble the massless solution $t$. The particle quickly transitions from its initial speed to the one predicted by the massless model. After this initial period called "boundary layer", one has $x\_\varepsilon(t) \approx t + \varepsilon (y^0-1)$ so resembles the massless solution $t$, with a slight correction due to the initial phase.
b) Concerning expansions, let us consider the massless model $\dot{x} + x = 0$, with initial condition $x(0) = 1$, so that $x(t) = e^{-t}$ is the solution, with $\dot{x}(0) = -1$ the initial speed in this model. Now consider the massive version $\varepsilon \ddot{x}\_\varepsilon + \dot{x}\_\varepsilon + x\_\varepsilon = 0$ with $x\_\varepsilon(0) = 1$ and let us use $\dot{x}\_\varepsilon(0) = -1$ as the initial condition matching the one of the massless problem, to avoid the boundary layers mentioned above.
You can then indeed look (at least formally) for a solution under the form
$$
x\_\varepsilon(t) = \sum\_{k=0}^{+\infty} \varepsilon^k x\_k(t)
$$
with $\dot{x}\_0 + x\_0 = 0$, $x\_0(0) = 1$ and, for $k \geq 0$,
$$
\dot{x}\_{k+1} + x\_{k+1} = - \ddot{x}\_k.
$$
Here $x\_0$ is the solution to the massless problem so $x\_0(t) = e^{-t}$, and you can solve iteratively using the variation of constant formula.
For example, $x\_1(t) = - t e^{-t}$, and so on.
In this case, you can prove that if you truncate the series, you have a solution which approximates $x\_\varepsilon$ at any precision.
| 2 | https://mathoverflow.net/users/50777 | 441296 | 178,130 |
https://mathoverflow.net/questions/441207 | 5 | Let $\mathbb{S}\_n$ denote the set of $n \times n$ symmetric positive semidefinite matrices. I am trying to figure out whether $k: \mathbb{S}\_n \times \mathbb{S}\_n \to \mathbb{R}\_+$ defined as:
$$k(A, B) = \text{Tr}[(A^{1/2} B A^{1/2})^{1/2}]$$
is a positive definite kernel. Can anyone find a counterexample showing it is not? Or prove it is indeed a positive definite kernel?
**Some analysis so far:** Having failed to find a counterexample numerically, I am trying to show:
$$\sum\_{ij} k(A\_i, A\_j) x\_i x\_j \geq 0$$
holds for any $A\_1, \dots, A\_m \in \mathbb{S}\_n$ and $x\_1, \dots, x\_m \in \mathbb{R}$.
Without loss of generality we can assume that $x\_1, \dots, x\_m \in \{-1, +1\}$ since the change of $x\_i \mapsto \text{sign}(x\_i)$ and $A\_i \mapsto x\_i^2 A\_i$ preserves the value on the left hand side above and all matrices remain positive semidefinite after this change of variables.
Let $\mathcal{P}$ denote the set of indices where $x\_i x\_j = +1$ and let $\mathcal{N}$ denote the set of indices where $x\_i x\_j = -1$.
Then we just need to show the following inequality holds:
$$\sum\_{i=1}^m \text{Tr}[A\_i] + 2 \sum\_{(i,j)\in\mathcal{P}} k(A\_i, A\_j) \geq 2 \sum\_{(i,j)\in\mathcal{N}} k(A\_i, A\_j)$$
This is where I'm stuck, or maybe a different approach is needed.
| https://mathoverflow.net/users/59128 | Is $k(A, B) = \text{Tr}[(A^{1/2} B A^{1/2})^{1/2}]$ a positive definite kernel? | Counterexample for $n = 2$ :
Let $A\_k$ be the orthonormal projection on the span of $$(\cos(2 \pi (k-1) / 5), \sin(2 \pi (k-1) / 5))^\mathsf{T} , \quad k = 1...5.$$
Then $k(A\_k,A\_l) = \vert \cos(2 \pi (k-l) / 5) \vert$ .
The corresponding matrix has the eigenvalue $-0.11803398874989484820458683436563811772$ with multiplicity $2$ .
| 3 | https://mathoverflow.net/users/17261 | 441297 | 178,131 |
https://mathoverflow.net/questions/441131 | 2 | **Theorem.** *Let $m$ be an integer and $P\_m$ the vector space of degree $m$ polynomials in one real variable. There is a constant $C$ such that, for all $a<b$ and $p \in P\_m$,
$$\|p\|\_{L^\infty(a,b)} \leq {C \over b-a} \|p\|\_{L^1(a,b)}$$*
*Proof.* Since $P\_m$ is finite-dimensional, there is a norm equivalence $\|\cdot\|\_{L^\infty(0,1)} \leq C \|\cdot\|\_{L^1(0,1)}$. The corresponding inequality follows by change of variable $x = (b-a)y+a$. □
I'd like to generalize this as much as possible, but I'm not sure how much is possible. One possible generalization is to all measurable sets $E$:
$$\|p\|\_{L^\infty(E)} \leq {C \over |E|} \|p\|\_{L^1(E)}?$$
The given Theorem is the special case where $E$ is a single interval. Maybe it's necessary to impose that $E$ is in the interval, to avoid some trivialities.
I think the following generalization might be false though. You could allow $\mu$ to vary over probability measures and ask for:
$$\|p\|\_{L^\infty(d\mu)} \leq C \|p\|\_{L^1(d\mu)}?$$
I tried to look in this direction, thinking of the Banach-Alaoglu theorem.
I'm marking this as "Complex Variables" because reverse Hölder inequalities appear in complex analysis textbooks.
| https://mathoverflow.net/users/73890 | Most general reverse Hölder inequality for polynomials | Although Robert Israel's answer is completely accurate, I was able to find the generalization I was looking for. In case someone is looking for this exact thing in the future, I'm writing it up here.
**Theorem.** *Let $m$ be an integer and $P\_m$ the space of polynomials of degree $m$ in one real variable. Let $\kappa > 0$. There is a constant $C$ such that the following holds. If $a<b$ and $d\mu = g(x) dx$ is a Radon probability measure with density function $g(x)$ on $[a,b]$, such that $\|g\|\_{L^{\infty}(a,b)} \leq {\kappa \over b-a}$, and $p \in P\_m$, then
\begin{align}
\|p\|\_{L^{\infty}(a,b)} \leq C \|p\|\_{L^1(d\mu)}.
\end{align}*
**Proof.** We begin with the case $a=0$ and $b=1$. By Alaoglu's theorem, the set $Q = \{ g \in L^{\infty}(a,b) \; : \; \|g\|\_{L^{\infty}(a,b)} \leq \kappa \text{ and } g \geq 0 \text{ and } \int\_a^b g \,dx = 1 \}$ is compact in the weak-\* topology of $L^{\infty}(a,b)$ as the dual of $L^1(a,b)$. Put
\begin{align}
\phi(f,\mu) & = \int |f| \, d\mu.
\end{align}
Denote by $S \subset P\_m$ the unit sphere defined by $\|p\|\_{L^{\infty}(0,1)} = 1$. On the set $S\times Q$, we impose the topology $(\text{uniform}) \times (\text{weak}-\*)$.
We show that $\phi$ is continuous on $S \times Q$ in this topology. To that end, assume that $\mu\_k \to \mu\_{\infty}$ in the weak-\* topology, and $p\_k \to p\_{\infty}$ uniformly. Then,
\begin{align}
|\phi(p\_k,\mu\_k) - \phi(p\_{\infty},\mu\_{\infty})| & \leq
|\phi(p\_\infty,\mu\_k) - \phi(p\_\infty,\mu\_{\infty})|
+|\phi(p\_\infty,\mu\_k) - \phi(p\_k,\mu\_k)|
\end{align}
On the right-hand-side, the first term $\to 0$ by weak-\* convergence of $\mu\_k$, and the second term is bounded by
\begin{align}
|\phi(p\_\infty,\mu\_k) - \phi(p\_k,\mu\_k)|
& = \left|
\int |p\_{\infty}| - |p\_k| \, d\mu\_k
\right|
\leq
\int |p\_{\infty} - p\_k| \, d\mu\_k
\leq \|p\_{\infty} - p\_k\|\_{L^{\infty}(0,1)},
\end{align}
which also $\to 0$ as $k \to \infty$. Thus, $\phi$ is continuous on the compact set $S\times Q$ and hence it attains a minimum, say at $(p\_0,\mu\_0)$. If $\phi(p\_0,\mu\_0) = 0$ then $\|p\_0\|\_{L^1(d\mu\_0)} = \int |p\_0(x)| g\_0(x) \, dx = 0$, so that $p\_0(x)g\_0(x) = 0$ a.e. Because $d\mu\_0 = g\_0(x) \, dx$ is a probability density function, it must be that $\int g\_0(x) \, dx = 1$, and hence $g\_0(x) \neq 0$ on a set of positive measure, thus $p\_0(x)=0$ on a set of positive measure. In particular, $p\_0(x) = 0$ for infinitely many values of $x$. Since $p\_0 \in P\_m$, it must be that $p\_0=0$, contradicting $\|p\_0\|\_{L^{\infty}(0,1)} = 1$. Thus, it must be that $\phi(p,\mu) \geq \epsilon>0$ for all $(p,\mu) \in S \times Q$. By homogeneity in $p$, we find that
\begin{align}
\|p\|\_{L^{\infty}(0,1)} \leq \epsilon^{-1}\phi(p,\mu) = \epsilon^{-1} \|p\|\_{L^1(d\mu)}.
\end{align}
so that $C = \epsilon^{-1}$.
For generic $a<b$ and probability measure $d\nu = h(y) \, dy$ on $[a,b]$, we perform the substitution $p(x) = q((x-a)/(b-a)) = q(y)$ and $(b-a)h(y)dy = g(x)dx$ and use the scalings
\begin{align}
\|q\|\_{L^{\infty}(a,b)} = \|p\|\_{L^{\infty}(0,1)} \leq C \|p\|\_{L^1(d\mu)} = C\|q\|\_{L^1(d\nu)}.
\end{align}
□
Maybe the following corollary is more aesthetically pleasing though:
\begin{align}
\|p\|\_{L^{\infty}(d\mu)} \leq C \|p\|\_{L^1(d\mu)}.
\end{align}
However, I find this version confusing because it obfuscates the interval $[a,b]$, and that interval cannot be dispensed with, essentially because of Robert Israel's example.
| 1 | https://mathoverflow.net/users/73890 | 441312 | 178,133 |
https://mathoverflow.net/questions/441306 | 5 | A well-known theorem of Atiyah and Bott states that given a finite dimensional oriented manifold $M$ with circle action, the $S^1$-equivariant cohomology of $M$ (with $\mathbb{Q}$ coefficients) is isomorphic to the $S^1$-equivariant cohomology of the fixed points after inverting the Euler class of the normal bundle $\nu$ of $i: Fix(M)\hookrightarrow M$ (and shifting degree by the rank of $\nu$). In particular, the isomorphisms are the equivariant pushforward $i\_\ast$ in one direction and $i^\ast/eul(\nu)$ in the other. Notice that the $S^1$-equivariant cohomology of a space is a $\mathbb{Q}[u]$-module and $eul(\nu)$ is a multiple of a power of $u$. This means that we just need to localize at $u$ to obtain the theorem.
Now, if we let $X$ be a smooth manifold and $LX$ its free loop space, $S^1$ acts trivially on the former and by loop rotation on the latter. The inclusion map $i:X\hookrightarrow LX$ sending $X$ to the trivial loops is exactly the inclusion of the fixed points. We cannot apply the theorem above for dimensional reasons.
**Question 1** Is it still true that ${{H\_{S^1}}(LX)}\_{(u)} \cong{H\_{S^1}(X)}\_{(u)}$?
The same could be rephrased as an isomorphism between the localized cyclic homology (see Jones) of $\Omega^\ast(X)$, namely $H\hat{C}\_{-\ast}(\Omega^\ast (X))$ and $H^\*(X)\otimes \mathbb{Q}[u,u^{-1}]$.
On one hand it seems natural enough for this isomorphism to hold, on the other hand, I have not seen anything stating any connection between them, not even in the paper of Jones in which he defines localized cyclic homology.
**Question 2** Is the corresponding statement for localized cyclic homology true? That is, given a cochain complex $S$, is it true that
$H\hat{C}\_{-\*}(S)\cong H^n(S)\otimes\mathbb{Q}[u,u^{-1}]$?
*Atiyah, Michael F.; Bott, Raoul*, [**The moment map and equivariant cohomology**](https://doi.org/10.1016/0040-9383(84)90021-1), Topology 23, 1-28 (1984). [ZBL0521.58025](https://zbmath.org/?q=an:0521.58025).
*Jones, John D. S.*, [**Cyclic homology and equivariant homology**](https://doi.org/10.1007/BF01389424), Invent. Math. 87, 403-423 (1987). [ZBL0644.55005](https://zbmath.org/?q=an:0644.55005).
| https://mathoverflow.net/users/148223 | Is the localised $S^1$-equivariant cohomology of the free loop space of a space $X$ isomorphic to that of $X$ itself? | I don't know what you would mean by the normal bundle of $M$ in $LM$, but the answer to one question is "no": When the $\mathbb Q[u]$-module $H^\ast\_{S^1}(LM)$ is localized by inverting $u$, it does not become the same as the cohomology of $M$.
In fact, in my paper in Topology (1985) "Cyclic homology, derivations, and the free loopspace" I prove a very different and incompatible general statement: this localized cohomology depends only on the fundamental group of $M$. More precisely, for a $2$-connected map $X\to Y$ the relative equivariant cohomology of $LX\to LY$ is a torsion $\mathbb Q[u]$-module.
There are related statements involving cyclic homology of differential graded algebras.
| 4 | https://mathoverflow.net/users/6666 | 441344 | 178,141 |
https://mathoverflow.net/questions/441334 | 4 | Let $\textbf{F}\in \mathbb{R}^3$ be a smooth vector field for all space. It is well known using [Helmholtz decomposition](https://en.wikipedia.org/wiki/Helmholtz_decomposition) that we can decompose $\textbf{F}$ into two vector fields in $V$: $$\textbf{F} = \nabla \sigma + \nabla \times \Gamma,$$ where $\nabla \sigma$ is called the irrotational part and $\nabla \times \Gamma$ the solenoidal part of $\textbf{F}$. If both these fields are non-zero, is it true that $\nabla \sigma$ and $\nabla \times \Gamma$ are linearly independent?
| https://mathoverflow.net/users/353746 | Are the irrotational and solenoidal parts of a smooth vector field linearly independent? | $\newcommand\R{\mathbb R}\newcommand\na{\nabla}\newcommand\om{\boldsymbol{\omega}}\newcommand\si{\sigma}\newcommand\Ga{\Gamma}\newcommand\F{\mathbf F}\newcommand\x{\mathbf x}\newcommand\0{\mathbf 0}$The answer is no, in general. E.g., take any nonzero $\om\in\R^3$ and let
$$\si(\x):=\om\cdot\x\quad\text{and}\quad\Ga(\x):=\om\times\x$$
for $\x\in\R^3$, with $\F:=\na\si+\na\times\Ga$. Then $\na\si(\x)=\om$ and $\na\times\Ga(\x)=2\om$ for all $\x\in\R^3$, so that $\na\si$ and $\na\times\Ga$ are nonzero but linearly dependent.
---
On a positive note, suppose that $\na\si$ and $\na\times\Ga$ are nonzero and linearly dependent, so that $\na\si=c\na\times\Ga$ for some nonzero real $c$. Then $\na^2\si=\na\cdot(\na\si)=c\na\cdot(\na\times\Ga)=\0$, so that $\si$ is a harmonic scalar field. If we assume that $\si$ is bounded above or below, then [Liouville's theorem](https://en.wikipedia.org/wiki/Harmonic_function#Liouville%27s_theorem) will imply that $\si$ is constant. [Moreover](https://math.stackexchange.com/a/1685017/96609), if $\si(\x)=o(|\x|)$ as $|\x|\to\infty$, then $\si$ is constant. Furthermore, similarly it is easy to show that, if $\si$ is bounded above or below by a function $\si\_\*$ such that $\si\_\*(\x)=o(|\x|)$ as $|\x|\to\infty$, then $\si$ is constant -- and hence the fields $\na\si$ and $\na\times\Ga$ are zero, a contradiction.
Thus, the answer to your question becomes yes given that $\si$ is bounded above or below by a function $\si\_\*$ such that $\si\_\*(\x)=o(|\x|)$ as $|\x|\to\infty$.
On the other hand, the counterexample in the first part of the answer shows that the sublinear growth condition, $\si\_\*(\x)=o(|\x|)$ as $|\x|\to\infty$, cannot be relaxed to the condition that $\si\_\*(\x)=O(|\x|)$ as $|\x|\to\infty$.
| 9 | https://mathoverflow.net/users/36721 | 441347 | 178,142 |
https://mathoverflow.net/questions/441354 | 8 | It is known that every orientable 3-manifold has a spin structure, because its tangent bundle is trivial. Also it is known that if a manifold $X$ has a spin structure, then the number of distinct spin structures is equal to the order of $H^1(X;\Bbb Z\_2)$. In particular, the real projective space $\Bbb RP^3$ has exactly two different spin structures, say $s\_1,s\_2$. Is there a self diffeomorphism of $\Bbb RP^3$ that interchanges these two spin structures, i.e. is there a diffeomorphism $f:\Bbb RP^3\to \Bbb RP^3$ satisfying $f^\* s\_1=s\_2$?
| https://mathoverflow.net/users/164671 | Two different spin structures of the real projective space $\Bbb RP^3$ | I'm editing this to reflect the discussion.
You might think that there cannot be an automorphism of ${\mathbb R}P^3$ that acts non-trivially on the degree one cohomology. But the bijection between the spin structures and degree one cohomology is not canonical, it's a torsor. So it may be possible for an automorphism to swap the two spin structures.
To fix conventions, a spin structure is a structure on a manifold with a given orientation. The space ${\mathbb R}P^3$ is orientable because $3$ is odd, and there is an orientation reversing diffeomorphism which gives a bijection between the two spin structures for one and the two spin structures for the other. So we fix an orientation.
Now in Section IV.B.(i) of Dabrowski and Trautman, "Spinor structures on spheres and projective spaces", in the second to last paragraph they give a topological way to distinguish the two spin structures, so there cannot be an orientation preserving diffeomorphism swapping them.
They then go on to make the claim that the two spin structures are related by an orientation reversing isometry. This I do not understand, and it misled me to give my original answer.
| 5 | https://mathoverflow.net/users/460592 | 441365 | 178,144 |
https://mathoverflow.net/questions/440378 | 1 | Let $\mathbf{X}$ be $(n \times p)$-dimensional data matrix ($n > p$) whose rows $\mathbf{x}\_i$ are i.i.d. with some finite moments:
$$
\mathbf{X}^\top = [\mathbf{x}\_1, \ldots \mathbf{x}\_n]^\top.
$$
By the singular decomposition, we can obtain
$$
\mathbf{X} = \mathbf{U} \mathbf{D} \mathbf{V}^\top = \mathbf{U}\_{:r} \mathbf{D}\_{:r} \mathbf{V}\_{:r}^\top,
$$
where $r$ is the rank of $\mathbf{X}$ which may $r<p$. Notice that here I used a Matlab colon notation ":" to denote submatrices
\begin{align\*}
\mathbf{U}\_{:r} &:= [\mathbf{u}\_1, \mathbf{u}\_2, \ldots, \mathbf{u}\_r], \\
\mathbf{D}\_{:r} &:= \operatorname{diag}([d\_1,d\_2,\ldots,d\_r]), \\
\mathbf{V}\_{:r} &:= [\mathbf{v}\_1, \mathbf{v}\_2, \ldots, \mathbf{v}\_r],
\end{align\*}
(Notice that $\mathbf{U}\_{:r}$ is not a square matrix, so $\mathbf{U}\_{:r}^\top \mathbf{U}\_{:r} = \mathbf{I}$, but $\mathbf{U}\_{:r}\mathbf{U}\_{:r}^\top \neq \mathbf{I}$). Then the question is
>
> **Question.**
>
>
> As $n \to \infty$ while $p$ is fixed, what can we know about the asymptotic property of $\mathbf{U}\_{:r}$, for example, the limit of $n^{-1} \mathbf{y}^\top \mathbf{U}\_{:r} \mathbf{U}\_{:r}^\top \mathbf{y}$, where $\mathbf{y}$ is a random vector whose elements are i.i.d. and independent to $\mathbf{X}$?
>
>
>
As an example, we can know about the asymptotic properties of $\mathbf{D}\_{:r}$ and $\mathbf{V}\_{:r}$: by the law of large numbers, as $n \to \infty$ (while $p$ is fixed),
$$
\frac{1}{n} \mathbf{X}^\top \mathbf{X} = \frac{1}{n} \mathbf{V}\_{:r} \mathbf{D}\_{:r}^2 \mathbf{V}\_{:r}^\top \xrightarrow{\mathbb{P}} \mathbb{E} \big[ \mathbf{x}\_i \mathbf{x}\_i^\top \big],
$$
so $\mathbf{V}\_{:r}$ and $\mathbf{D}\_{:r}^2$ converges to the eigenvectors and eigenvalues of $\mathbb{E} \big[ \mathbf{x}\_i \mathbf{x}\_i^\top \big]$.
Of course that since the number of rows of $\mathbf{U}$ goes to infinity, we cannot directly say what is the limit of $\mathbf{U}\_{:r}$, but we may know (or at least the existence) the limit of $n^{-1} \mathbf{y}^\top \mathbf{U}\_{:r} \mathbf{U}\_{:r}^\top \mathbf{y}$.
Here we assume that $\mathbf{X}$ and $\mathbf{y}$ have finite moments (and sub-Gaussianity).
If this question is too elementary, I really apologize for that, but this question is very important to prove the asymptotic properties of my statistical estimator. It would be really appreciated if you give any help.
Thanks,
| https://mathoverflow.net/users/159685 | Asymptotic property of the left singular vectors of i.i.d. data matrix | This is for the self-reference.
>
> **Claim.**
> There exists some $C>0$ such that
> $$\frac{1}{n} \mathbf{y}^\top \mathbf{U}\_{:r} \mathbf{U}\_{:r}^\top \mathbf{y} \xrightarrow{\mathbb{P}} C.$$
>
>
>
*Proof.*
The main idea of the proof is that linear approximation of the left singular vectors $\mathbf{U}\_{:r}$ using the limit of $\mathbf{V}\_{:r}$ and $\mathbf{D}\_{:r}$.
*(Step 1: Limit of $\mathbf{V}\_{:r}$ and $\mathbf{D}\_{:r}$).* Since we know that $n^{-1} \mathbf{X}^\top \mathbf{X} \xrightarrow{\mathbb{P}} \mathbb{E}[\mathbf{x}\_i \mathbf{x}\_i^\top]$,by the Davis-Kahan theorem and Weyl's inequality, we have
\begin{align\*}
&\mathbf{V}\_{:r} = \mathbf{V}\_0 \mathbf{O}^\top + O\_{\mathbb{P}}\bigg(\frac{1}{\sqrt{n}}\bigg), &\frac{\mathbf{D}\_{:r}^2}{n} = \frac{\mathbf{D}\_{0}^2}{n} + O\_{\mathbb{P}}\bigg(\frac{1}{\sqrt{n}}\bigg),
\end{align\*}
where $\mathbf{V}\_0$ and $\mathbf{D}\_0^2/n$ are the first $r$ eigenvectors and eigenvalues of $\mathbb{E}[\mathbf{x}\_i\mathbf{x}\_i^\top]$, respectively, and $\mathbf{O}$ is some orthogonal matrix which depends on $\mathbf{V}\_{:r}$ and $\mathbf{V}\_0$.
*(Step 2: Approximation of $\mathbf{U}\_{:r}$).*
By using the previous result, we can approximate the left singular vectors $\mathbf{U}\_{:r}$ as follows:
\begin{align\*}
\mathbf{U}\_{:r} = \mathbf{X} \mathbf{V}\_{:r} \mathbf{D}\_{:r}^{-1} = \mathbf{X} \mathbf{V}\_0 \mathbf{O}^\top \mathbf{D}\_0^{-1} + \mathbf{X} \mathbf{R} \approx \mathbf{X} \mathbf{V}\_0 \mathbf{O}^\top \mathbf{D}\_0^{-1},
\end{align\*}
where
\begin{align\*}
\mathbf{R} &:= \mathbf{V}\_{:r} \mathbf{D}\_{:r}^{-1} - \mathbf{V}\_0 \mathbf{O}^\top \mathbf{D}\_0^{-1} \\
&= (\mathbf{V}\_{:r} - \mathbf{V}\_0 \mathbf{O}^\top) \mathbf{D}\_{:r}^{-1} + \mathbf{V}\_0 \mathbf{O}^\top (\mathbf{D}\_{:r}^{-1} - \mathbf{D}\_0^{-1}) \\
&= O\_{\mathbb{P}}\bigg(\frac{1}{n}\bigg).
\end{align\*}
Therefore, we have
\begin{align\*}
\frac{1}{n} \mathbf{y}^\top \mathbf{U}\_{:r} \mathbf{U}\_{:r}^\top \mathbf{y} &= \frac{1}{n} \mathbf{y}^\top \mathbf{X} \mathbf{V}\_0 \mathbf{O} \mathbf{D}\_0^{-2} \mathbf{O}^\top \mathbf{V}\_0^\top \mathbf{X}^\top \mathbf{y} + O\_{\mathbb{P}} \bigg(\frac{1}{\sqrt{n}}\bigg) + O\_{\mathbb{P}} \bigg(\frac{1}{n}\bigg) \\
&= \frac{1}{n} \mathbf{y}^\top \mathbf{X} \mathbf{V}\_0 \mathbf{D}\_0^{-2} \mathbf{O}^\top \mathbf{V}\_0^\top \mathbf{X}^\top \mathbf{y} + O\_{\mathbb{P}} \bigg(\frac{1}{\sqrt{n}}\bigg) \\
&\xrightarrow{\mathbb{P}} \mathbb{E}[y\_i \mathbf{x}\_i^\top] \mathbf{V}\_0 \mathbf{D}\_0^{-2} \mathbf{V}\_0^\top \mathbb{E}[\mathbf{x}\_i y\_i] = C,
\end{align\*}
so we have the result. $\square$
| 0 | https://mathoverflow.net/users/159685 | 441368 | 178,146 |
https://mathoverflow.net/questions/441230 | 0 | I have a question about Theorem 3.7.25. of *Computational commutative algebra I* by M. Kreuzer and L. Robbiano.
Let $K$ be a perfect field, $I \subseteq K[x\_1, \ldots, x\_n]$, be a zero dimensional radical ideal in normal $x\_n$ position, let $g\_n \in K[x\_n]$ be the monic generator of the elimination ideal $I \cap K[x\_n]$, and let $d = \deg(g\_n)$.
Then,
the reduced Groebner basis of the ideal $I$ with respect to lex is of the
form $\{x\_1 − g\_1,\ldots , x\_{n−1} − g\_{n−1}, g\_n\}$, where $g\_1,\ldots, g\_{n−1} \in K[x\_n]$.
My question is if we have a zero-dimensional ideal $I$ with the reduced Groebner basis of the form $\{x\_1 − g\_1,\ldots, x\_{n−1} − g\_{n−1}, g\_n\}$, can we claim that it's a radical ideal in general? If not, what are the conditions we need to be sure $I$ is radical?
| https://mathoverflow.net/users/152308 | Are zero dimensional ideals radical? | From the comments, I got my answer and I will write it here for future reference.
In general, if $ I = (x\_1 - g\_1, \ldots, x\_{n-1}-g\_{n-1}, g\_n)$ with $g\_1, \ldots, g\_n \in K[x\_n]$, then $K[x\_1, \ldots, x\_n]/I \cong K[x\_n]/(g\_n)$. So $I$ is radical iff $K[x\_n]/(g\_n)$ is reduced iff $g\_n$ is separable.
Also, from the same book, Proposition 3.7.15 (Seidenberg’s Lemma):
Let $K$ be a field, let $P = K[x\_1, \ldots, x\_n]$, and let $I \subseteq P$ be a zero-dimensional ideal. Suppose that, for every $i \in \{1, \ldots, n\}$, there exists a non-zero polynomial $g\_i \in I \cap K[x\_i]$ such that $gcd(g\_i , g\_i^{\prime} ) = 1$. Then $I$ is a radical ideal.
| 2 | https://mathoverflow.net/users/152308 | 441375 | 178,148 |
https://mathoverflow.net/questions/441362 | 2 | Consider the following exponential of matrices $\exp(X+\delta Y)$, where $\delta$ is a smaller number, and $X,Y$ are non-commuting matrices. I am interested in expanding it in such a way that
$$
\exp(X+\delta Y) = e^Xe^{\delta Y}e^{\delta A\_1}e^{\delta^2 A\_2}e^{\delta^3 A\_3}...,
$$
namely that organizing the terms of the expansion according to the order of $\delta$.
From [wikipedia](https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#:%7E:text=this%20Hopf%20algebra.-,Zassenhaus%20formula,-%5Bedit%5D), we have
$$
e^{t(X+Y)} = e^{tX}~ e^{tY} ~e^{-\frac{t^2}{2} [X,Y]} ~
e^{\frac{t^3}{6}(2[Y,[X,Y]]+ [X,[X,Y]] )} ~
e^{\frac{-t^4}{24}([[[X,Y],X],X] + 3[[[X,Y],X],Y] + 3[[[X,Y],Y],Y]) } \cdots
$$
hence it is naturally to guess that
$$
A\_1 = -\frac{1}{2}[X,Y]+\frac{1}{6}[X,[X,Y]]... = \sum\_{n=2}^\infty\frac{(-1)^{n+1}}{n!}[X,[X,[X,...[X,Y]].
$$
This seems to be correct for the particular example I have in mind, but I don't have proof for that. Also I couldn't figure out the higher order terms $A\_{2,3,...}$. Any help is really appreciated.
| https://mathoverflow.net/users/476103 | Reorganizing the terms in the Baker–Campbell–Hausdorff formula (or Zassenhaus formula) for $\exp(X+\delta Y)$ for small $\delta$ | This expansion is derived by K. Kumar in [On Expanding the Exponential](https://doi.org/10.1063/1.1704742), see equation (9) (with $t=1$) and section 6.
| 2 | https://mathoverflow.net/users/11260 | 441380 | 178,149 |
https://mathoverflow.net/questions/441363 | 1 | $E(i+1)=(I-AT)E(i)+1/2(AT)^2$
How to find the maximum value of $E$ in this expression without using the iterative method? An approximate estimation is also acceptable. Only the $E$ vector is unknown, and the others are known matrix vectors.
It would be better if we could get a formula of $E$ about $T$.
| https://mathoverflow.net/users/499742 | How to find the maximum value of the following difference equation without using iterative method？ | Concerning your request "It would be better if we could get a formula of $E$":
By induction on $i$,
$$E(i)=B^i E(0)+\sum\_{j=0}^{i-1} B^jC \tag{1}\label{1}$$
for $i=0,1,\dots$, where $B:=I-AT$ and $C:=1/2(AT)^2$.
If $(I-B)^{-1}$ exists, then $\sum\_{j=0}^{i-1} B^j=(I-B^i)(I-B)^{-1}$ and hence \eqref{1} can be rewritten as
$$E(i)=B^i (E(0)-(I-B)^{-1}C)+(I-B)^{-1}C. \tag{2}\label{2}$$
---
(As for your other request, "How to find the maximum value of $E$", it is incomprehensible to me, as $E(i)$ apparently is a *vector* function of $i$.)
| 0 | https://mathoverflow.net/users/36721 | 441391 | 178,153 |
https://mathoverflow.net/questions/440856 | 5 | We recall two definitions. Let $A$ be a linearly topologized ring which is complete and Hausdorff.
>
> We say that $A$ is *pseudo-compact* if, for every open ideal $I\subset A$, the ring $A/I$ is artinian. We say that $A$ is *admissible* if $A$ has an ideal of definition (an open ideal $I\subset A$ such that every neighbourhood of $0$ contains $I^n$ for some $n$).
>
>
>
Both seem to be used in the definition of formal schemes in algebraic geometry. **I wonder if one of them implies the other.**
| https://mathoverflow.net/users/131975 | What's the relation between pseudo-compact and admissible rings? | Neither of these properties implies the other:
**There is an admissible algebra that is not pseudo-compact**
If $A$ is a ring with the discrete topology that is not artinian then it is admissible but not pseudocompact.
Less degenerately if $A$ is a noetherian ring and $I$ is an ideal such that $A/I$ is not artinian then the $I$-adic completion of $A$ is admissible but not pseudocompact --- the first example is simply the case $I=0$.
**There is a pseudo-compact ring that is not admissible**
If $G$ is any profinite group then the complete group algebra $K[[G]]$ over an algebraically closed field $K$ of characteristic zero, in the sense of <https://www.ams.org/journals/bull/1966-72-02/S0002-9904-1966-11513-6/S0002-9904-1966-11513-6.pdf>, is pseudocompact.
I guess you are only interested in commutative rings but if we take $G$ to be the profinite completion of the integers then $K[[G]]$ is commutative.
Now for every prime $p$, there is a quotient $C\_p$ of $G$ that is cyclic of order $p$. Moreover there is non-trivial homomorphism $\theta\_p\colon K[[G]]\to K$ that factors through $C\_p$. For each $p$, $\ker \theta\_p$ is an open (maximal) ideal. Moreover if $p\neq q$ then $\ker \theta\_p\neq \ker \theta\_q$.
If an open ideal $I$ has the property that for every $p$ there is some $n$ such that $I^n\subset \ker \theta\_p$, then since each $\ker \theta\_p$ is maximal, $I$ is contained in the intersection of all the $\ker \theta\_p$. It follows that $K[[G]]/I$ is not artinian since it has infinitely many maximal ideals.
Thus $K[[G]]$ is pseudo-compact but not admissible.
| 4 | https://mathoverflow.net/users/345 | 441398 | 178,158 |
https://mathoverflow.net/questions/441399 | 2 | I am trying to better understand the straightening-unstraightening equivalence of Lurie in the $\infty$-categorical setting. In the case that I am interested in, this equivalence states that
$$
\mathrm{LFib}(\mathcal{C}) \simeq \mathrm{Fun}(\mathcal{C}, \mathrm{Spaces}),
$$
where $\mathcal{C}$ is an $\infty$-category and $\mathrm{LFib}(\mathcal{C})$ is *some* $\infty$-subcategory of $(\mathrm{Cat}\_\infty)\_{/\mathcal{C}}$ spanned by the left fibrations with codomain $\mathcal{C}$. What I don't understand and can't quite find in the literature is whether the morphisms in $\mathrm{LFib}(\mathcal{C})$, which I take to be functors $\mathcal{D} \to \mathcal{E}$ sitting over $\mathcal{C}$, are themselves left fibrations. I.e., if I have a natural transformation $\eta : D \to E$ of functors $D, E : \mathcal{C} \to \mathrm{Spaces}$, does it straighten to a left fibration $\mathcal{D} \to \mathcal{E}$ between the straightenings of $D$ and $E$?
| https://mathoverflow.net/users/322094 | Morphisms in category of left fibrations | The answer is yes, but $\mathrm{LFib}(\mathcal C)$ is *also* the full subcategory of $(\mathrm{Cat}\_\infty)\_{/\mathcal C}$, it just so happens that you can prove that any morphism between such is a left fibration (this does not remain true in the case of cocartesian fibrations, though).
Here is a proof: Let $f:\mathcal{D\to E}$ be a morphism of left fibrations over $\mathcal C$, which I'll denote by $p,q$ respectively. By the dual of HTT.2.4.1.3.(3), if an edge $\alpha$ in $\mathcal D$ is such that $f(\alpha)$ is $q$-coCartesian, then $\alpha$ is $f$-coCartesian if and only if it is $p$-coCartesian.
But we are in left fibrations, so all edges are $p$-coCartesian. In particular, $\alpha$ is $f$-coCartesian if and only if $f(\alpha)$ is $q$-coCartesian, the latter being a void condition: every edge is $f$-coCartesian.
So we have seen that any edge is $f$-coCartesian, and therefore we just need a sufficient supply of (arbitrary) edges along $f$: let $\alpha: e\_0\to e\_1$ be an edge in $\mathcal E$, with a lift $d\_0$ of $e\_0$ along $f$, i.e. $f(d\_0) = e\_0$. We then push things down: $q(\alpha): q(e\_0) \to q(e\_1)$ is an edge in $\mathcal C$, with source $q(e\_0) = p(d\_0)$, so it has a lift $\tilde \alpha: d\_0\to d\_1$. Now $f(\tilde\alpha): f(d\_0)\to f(d\_1)$ is some map lifting $q(\alpha)$, and it is $q$-coCartesian, just like $\alpha$ (and any map in $\mathcal E$), so there is an equivalence $e\_1\simeq f(d\_1)$ with appropriate $2$-simplices showing that have found a lift of $\alpha$.
| 1 | https://mathoverflow.net/users/102343 | 441404 | 178,159 |
https://mathoverflow.net/questions/403450 | 6 | There's a theory of algebraic geometry over $\mathbb{Z}\_2$-graded commutative rings, often called "[algebraic supergeometry](https://arxiv.org/abs/2008.00700)" or the theory of [superschemes](https://ncatlab.org/nlab/show/super-scheme). From what I understand, there's also a variant theory of $\mathbb{Z}$-graded algebraic geometry, for rings whose multiplication is $\mathbb{Z}$-graded commutative, satisfying $ab=(-1)^{\deg(a)\deg(b)}ba$.
Now, many structures arising in algebraic topology are not commutative, but some are instead *graded-commutative*―for instance, this is the case for the [cohomology ring](https://en.wikipedia.org/wiki/Cohomology_ring) of any space.
>
> **Question.** Can one use the theory of $\mathbb{Z}$-graded algebraic geometry to say something useful about some of the graded-commutative structures found in algebraic topology, such as e.g. cohomology rings?
>
>
>
One thing I imagine one could do is say take the $\mathrm{Spec}$ of a cohomology ring, and then study it algebro-geometrically as a scheme in the $\mathbb{Z}$-graded setting. Has this sort of strategy ever been successfully carried out?
(Of course there's DAG/SAG, which work wonderfully for the purposes of homotopy theory, but I'm nevertheless curious about this question considered from the point of view of graded-commutative algebraic geometry.)
| https://mathoverflow.net/users/130058 | Applications of $\mathbb{Z}$-graded algebraic geometry to algebraic topology | Lars Hesselholt and Piotr Pstrągowski have since posted a paper to the arXiv doing *exactly this*!
>
> Hesselholt–Pstrągowski, *Dirac geometry I: Commutative algebra*. [[arXiv]](https://arxiv.org/abs/2207.09256)
>
>
>
In their paper, they develop a theory of $\mathbb{Z}$-graded-commutative algebraic geometry in the sense of schemes built from $\mathbb{Z}$-graded rings satisfying $ab=(-1)^{\deg(a)\deg(b)}ba$, which they call **Dirac rings**.
My (limited) understanding of it is that the main example and motivation for such a theory is that the $\pi\_\*$ of a commutative algebra in spectra is a Dirac ring ([Example 2.2](https://arxiv.org/pdf/2207.09256.pdf#page=9) there).
Here's the abstract from the arXiv:
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> **Abstract.** The homotopy groups of a commutative algebra in spectra form a commutative algebra in the symmetric monoidal category of graded abelian groups. The grading and the Koszul sign rule are remnants of the structure encoded by anima as opposed to sets. The purpose of this paper and its sequel is to develop the geometry built from such algebras. We name this geometry Dirac geometry, since the grading exhibits the hallmarks of spin. Indeed, it is a reflection of the internal structure encoded by anima, and it distinguishes symmetric and anti-symmetric behavior, as does spin. Moreover, the coherent cohomology, which we develop in the sequel admits half-integer Serre twists.
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**Edit:** There's now also a sequel paper:
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> Hesselholt–Pstrągowski, *Dirac geometry II: Coherent cohomology* [[arXiv]](https://arxiv.org/abs/2303.13444)
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| 6 | https://mathoverflow.net/users/130058 | 441426 | 178,166 |
https://mathoverflow.net/questions/441420 | 1 | Define: $\operatorname {wo}^n(x) \iff \forall y (y \in^n x \to \operatorname {wo} (y))$
Where $\operatorname {wo}(y)$ refers to $y$ being well orderable.
Where $y \in^0 x \iff y=x \\ y \in^{n+1} x \iff \exists z (z \in^n x \land y \in z)$
**n-well ordered choice:** for $n=0,1,2,...$, for every set $x$ of nonempty sets, if $\operatorname{wo}^n(x)$, then $x$ admits a choice function on it.
If we add this schema to axioms of $\sf ZF$, would it entail axiom of choice?
| https://mathoverflow.net/users/95347 | Does n-well ordered choice schema imply the axiom of choice? | $2$-well ordered choice is enough to imply AC.
Let $α$ be any ordinal, and look at $\mathcal P^2(α)\setminus\{\emptyset\}$.
We have $x\in^2 \mathcal P^2(α)\setminus\{\emptyset\}⇒∃z⊆\mathcal P(α)\;(x\in z)⇒x\subseteq α⇒x\text{ is well orderable}$.
A choice function on $\mathcal P^2(α)\setminus\{\emptyset\}$ induce a well ordering on $\mathcal P(α)$.
"powerset of well-orderable set is well-orderable" is famously equivalent to the axiom of choice.
| 5 | https://mathoverflow.net/users/113405 | 441427 | 178,167 |
https://mathoverflow.net/questions/441441 | 0 | The following expression arises in the study of hierarchical models. I suspect that the sum of the underlined $4$ terms become constant as $\alpha\rightarrow \infty$. Mathematica agrees when prompted with 'toy' versions, but I'm having some difficulty seeing how it generalizes.
I would greatly appreciate any help or observations.
\begin{align}
\log \text{P}\propto
\underbrace{
\log \Gamma(k \alpha)
- \log \Gamma(k\alpha + length(\ell)) -
k \log \Gamma(\alpha) + \sum^k\_{i=1} \log \Gamma(c\_{i} + \alpha)} \ + \ f(\text{other parameters})
\end{align}
**Where:** $\ell$ is a list of items; Each item in $\ell$ must be assigned to exactly one bin, so $c\_{i}$ is the count of items in $\ell$ assigned to bin $i$, and there are a total of $k$ bins.
Current thinking:
It is well known that as $\alpha\rightarrow \infty$, the argument of the sum in the rightmost term $\log \Gamma(c\_{i}+a)$ increases only linearly in its arguments rather than superlinearly. This is because $\displaystyle \lim\_{\alpha\rightarrow\infty} \big(\frac{\partial^2}{\partial\alpha^2}\log \Gamma(c\_{i} + \alpha)\big) = \lim\_{\alpha\rightarrow\infty} \big(\frac{\partial^2}{\partial c\_{i}^2}\log \Gamma(c\_{i} + \alpha)\big) = \displaystyle \lim\_{\alpha\rightarrow\infty} \big(\frac{\partial}{\partial c\_{\ell,i}}\Psi(c\_{i} + \alpha)\big)=0$, where $\Psi$ is the Digamma function. Therefore any partition of $\ell$ (i.e. any choices of the different $c\_{i}$) will cause this term to sum to the same constant.
The other terms are all constant for fixed $\alpha$, $\ell$, and $k$.
| https://mathoverflow.net/users/105727 | Infinite limit of sums of gamma functions is constant? | For each $k>0$, $c\in\mathbb{C}$ it holds that
$$\lim\_{\alpha\rightarrow\infty}\frac{\Gamma(\alpha+c)}{\Gamma(\alpha)\alpha^c}=1\Rightarrow\lim\_{\alpha\rightarrow\infty} \left(\log\frac{\Gamma(k\alpha+c)}{\Gamma(k\alpha)}-c\log k\alpha\right)=0.$$
Apply this to
$$I= - \log \frac{\Gamma(k\alpha + L)}{\Gamma(k\alpha)} + \sum^k\_{i=1} \log\frac{\Gamma(\alpha+c\_i )}{\Gamma(\alpha)}
$$
and you find (using $\sum\_{i=1}^k c\_i=L$) that
$$\lim\_{\alpha\rightarrow\infty} I=-L\log k\alpha+\sum\_{i=1}^k c\_i\log\alpha=-L\log k.$$
So indeed, the function $\log P$ approaches an $\alpha$-independent limit for $\alpha\rightarrow\infty$.
| 2 | https://mathoverflow.net/users/11260 | 441442 | 178,172 |
https://mathoverflow.net/questions/441000 | 7 | Let $Y=\Sigma(\alpha\_{1},\dots,\alpha\_{n})$ be a Seifert fibered homology 3-sphere, let $\pi:Y\to\Sigma$ denote the projection to the orbifold surface, and let $T\subset Y$ be the pre-image under $\pi$ of a small disk neighborhood of the singular point of order $\alpha\_{i}$, which we identify with a standard fibered solid torus $T(\alpha\_{i},p\_{i})\cong S^{1}\times D^{2}$.
Here, $T(\alpha\_{i},p\_{i})$ is such that a regular fiber is identified with the subset $(t,e^{2\pi it p\_{i}/\alpha\_{i}}\cdot z)\subset S^{1}\times D^{2}$ for $z\in D^{2}\setminus\{0\}$, $p\_{i}$ is such that $1\le p\_{i}<\alpha\_{i}$, $p\_{i}\beta\_{i}\equiv 1\pmod{\alpha\_{i}}$, and $(b,(\alpha\_{1},\beta\_{1}),\dots,(\alpha\_{n},\beta\_{n}))$ are the Seifert invariants of $Y$, normalized so that $1\le \beta\_{i}<\alpha\_{i}$.
Now let $F\_{i}\subset T(\alpha\_{i},p\_{i})$ denote the core of the Seifert fibered solid torus, corresponding to the exceptional fiber in $Y$ over the singular point of order $\alpha\_{i}$ in $\Sigma$. Given this setup, we obtain a distinguished framing of $F\_{i}$ given by $S^{1}\times \{z\}\subset S^{1}\times D^{2}\cong T(\alpha\_{i},p\_{i})$ for $z\in D^{2}\setminus\{0\}$, which we'll denote by $\lambda\_{T}$. On the other hand, since $F\_{i}$ is null-homologous there is also a canonical framing $\lambda\_{S}$ of $F\_{i}$ induced by any Seifert surface $S$ with $\partial S=F\_{i}$.
My question is: how do the framings $\lambda\_{T}$ and $\lambda\_{S}$ compare? Furthermore, is the Seifert framing $\lambda\_{S}$ determined locally by the tuple $(\alpha\_{i},\beta\_{i})$ corresponding to this fiber, or does it depend in an essential way on the global topology of $Y$? Thanks!
| https://mathoverflow.net/users/133991 | Calculating the Seifert framing for an exceptional fiber in a Seifert-fibered integer homology 3-sphere | Here's an answer to my own question (thanks to Matt Hedden for the approach):
It will be helpful to fix a presentation of $\pi\_{1}(Y)$. Let $T\_{1},\dots,T\_{n}$ be regular neighborhoods of the exceptional fibers. We can write $Y'=Y\setminus(\cup\_{j}T\_{j})$ as a circle bundle over $S^{2}\setminus(\cup\_{i}D\_{i})$ with Euler class $b$. Let $\ell\in\pi\_{1}(Y')$ correspond to the singular fiber, and for $j=1,\dots, n$ let $m\_{j}\in\pi\_{1}(Y')$ correspond to the meridians around the deleted disks, oriented so that $m\_{j}=\partial D\_{j}$. With respect to these generators we can write
$$\pi\_{1}(Y')=\langle m\_{j},\ell\;|\;[m\_{j},\ell]=m\_{1}\cdots m\_{n}\ell^{-b}=1\rangle.$$
After Dehn filling to obtain $Y$, we obtain the presentation
$$\pi\_{1}(Y)=\langle m\_{j},\ell\;|\;[m\_{j},\ell]=m\_{1}\cdots m\_{n}\ell^{-b}=m\_{j}^{\alpha\_{j}}\ell^{\beta\_{j}}=1\rangle.$$
The meridians $\mu\_{j}$ and longitudes $\lambda\_{T\_{j}}$ of the filling fibered solid tori $T\_{j}$ are given by $\mu\_{j}=\alpha\_{j}m\_{j}+\beta\_{j}\ell$ and $\lambda\_{T\_{j}}=-p\_{j}m\_{j}+q\_{j}\ell$, where $p\_{j},q\_{j}$ are the unique integers which satisfy $q\_{j}\alpha\_{j}+p\_{j}\beta\_{j}=1$ and $1\le p\_{j}<\alpha\_{j}$.
Now fix $1\le i\le n$ and write the Seifert longitude of the exceptional fiber $K\_{i}$ as
$$\lambda\_{S\_{i}}=\lambda\_{T\_{i}}+N\_{i}\mu\_{i}\in H\_{1}(\partial T\_{i}).$$
The Seifert longitude is characterized by the property that it maps to zero under the inclusion $H\_{1}(\partial T\_{i})\hookrightarrow H\_{1}(Y\setminus T\_{i})$. The above fundamental group presentation for $Y$ induces a natural presentation
$$H\_{1}(Y\setminus T\_{i})=\mathbb{Z}\langle m\_{j},\ell\rangle/(\alpha\_{j}m\_{j}+\beta\_{j}\ell=0\text{ for all }j\neq i, \sum\_{j}m\_{j}=b\ell).$$
Solving for $\lambda\_{S\_{i}}=0$, we see that
$$0=\lambda\_{S\_{i}}=\lambda\_{T\_{i}}+N\_{i}\mu\_{i}=(N\_{i}\alpha\_{i}-p\_{i})m\_{i}+(N\_{i}\beta\_{i}+q\_{i})\ell$$
$$=(N\_{i}\alpha\_{i}-p\_{i})(b\ell-\sum\_{j\neq i}m\_{j})+(N\_{i}\beta\_{i}+q\_{i})\ell$$
$$=(N\_{i}\alpha\_{i}-p\_{i})(b\ell+\sum\_{j\neq i}\frac{\beta\_{j}}{\alpha\_{j}}\ell)+(N\_{i}\beta\_{i}+q\_{i})\ell$$
$$\implies (N\_{i}\alpha\_{i}-p\_{i})(b+\sum\_{j\neq i}\frac{\beta\_{j}}{\alpha\_{j}})+(N\_{i}\beta\_{i}+q\_{i})=0.$$
Rearranging, we obtain
$$N\_{i}=\frac{-q\_{i}+p\_{i}b+p\_{i}\sum\_{j\neq i}\frac{\beta\_{j}}{\alpha\_{j}}}{\beta\_{i}+\alpha\_{i}b+\alpha\_{i}\sum\_{j\neq i}\frac{\beta\_{j}}{\alpha\_{j}}}.$$
Finally using the relation $\sum\_{j\neq i}\frac{\beta\_{j}}{\alpha\_{j}}=-\Big(\frac{\beta\_{i}}{\alpha\_{i}}+b+\frac{1}{\alpha\_{1}\cdots\alpha\_{n}}\Big)$ and rearranging, we obtain the explicit formula
$$N\_{i}=\frac{\alpha\_{1}\cdots\alpha\_{n}}{\alpha\_{i}}(q\_{i}-bp\_{i})-p\_{i}\sum\_{j\neq i}\frac{\alpha\_{1}\cdots\alpha\_{n}}{\alpha\_{i}\alpha\_{j}}\beta\_{j}.$$
| 2 | https://mathoverflow.net/users/133991 | 441464 | 178,178 |
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