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https://mathoverflow.net/questions/439235 | 3 | I have been playing with the following function:
$$
f(x)=\frac{\pi x (1-x^2)}{\sin\pi x}\prod\_{k=2}^\infty \frac{\sin(\pi x/k)}{\pi x/k}
$$
It is hard to get correct numerical values. I'll start with the basic. Can you confirm that (1) we have absolute convergence, (2) the limit exists if $x$ is an integer, (3) if $x$ is a composite number then $f(x)=0$, and if $x$ is prime, $f(x)\neq 0$?
For simplicity, let's consider the absolute value of $f(x)$. Now the interesting part. If $x$ is not too close to a composite number, it sounds like
$$
|f(x)|\sim \exp(-\lambda x)
$$
for some $\lambda >0$, possibly $\lambda\approx 4.5$. Is there an asymptotic formula that can be easily derived? Since $f(x)=0$ if $x$ is composite, that formula would be valid only for some values of $x$, for instance if the fractional part is within some range, or if $x$ is prime, which seems to be where the formula works best.
If this was true, you could approximately compute the number of primes $<n$ as
$$\pi(n)\approx \sum\_{k=2}^n e^{\lambda k} |f(k)|.
$$
The formula is useless for computational purposes, but I am wondering if it might have some theoretical interest. Anyway, my question is this: can you get some asymptotic formula as $x\rightarrow\infty$ depending on the fractional part of $x$ or if $x$ is prime? Mine might not be correct. Even better, what is the value of $\lambda$ assuming it ever exists?
| https://mathoverflow.net/users/140356 | Curious infinite product, convergence, connection to prime numbers | I believe that the doubly infinite product $\prod \_{k=1} ^\infty \prod\_{m=1} ^\infty \left( 1 - \frac {x^2} {k^2 m^2} \right)$ is absolutely convergent and equals $f(x) \frac {\sin^2 (\pi x)} {(\pi x)^2 (1 - x^2)}$. Is that wrong? I am just using $\frac {\sin(\pi x)} {\pi x} = \prod\_{m=1} ^\infty \left(1 - \frac {x^2} {m^2} \right)$ here.
| 5 | https://mathoverflow.net/users/494014 | 439255 | 177,405 |
https://mathoverflow.net/questions/439249 | 1 | Let $f : \mathbb R\_+ \to \mathbb C$ be a measurable and integrable function where $\mathbb R\_+ = [0,\infty)$. The Laplace transform of $f$ is given by
$$
Lf(s) = \int\_0^\infty f(x)e^{-sx} \, dx.
$$
A classical Theorem due to Lerch [M. Lerch, Sur un point de la théorie des fonctions génératrices d’Abel, Acta Math. 27, 339 -351 (1903)] states that if
$$
Lf(\delta n) = 0, \quad n \in \mathbb N
$$
for some $\delta>0$ then $f=0$ almost everywhere. Suppose now that we replace the Laplace transform by the two-sided Laplace transform,
$$
Tf(s) = \int\_{\mathbb R} f(x)e^{-sx} \, dx. \quad (\*)
$$
Suppose that $Tf$ defines an entire function (clearly, we need a decay assumption on the negative axis so that $Tf$ is entire).
**Question:** Are there Lerch-type results for the two-sided Laplace transform? That is, suppose that $f$ belongs to a certain function class $C \subset L^1(\mathbb R)$. Can we find discrete sets $A \subset \mathbb R$, so that $Tf(a)$ for all $a \in A$ implies $f=0$ almost everywhere? In particular, I'm interested in function classes $C$ so that $A$ can be chosen to be uniformly discrete, $\inf\_{a \neq b, a,b \in A} > 0$.
| https://mathoverflow.net/users/223636 | Discrete uniqueness sets for the two-sided Laplace transform? | Of course, if $A$ has accumulation points then
your statement is correct, since you assume the function $Lf$ to be entire.
Otherwise, there are no restrictions on zeros of such functions $Lf$ (unless you somehow restrict your class $C$ of functions $f$). For example, let $f$ be an infinite sum of $\delta$-functions sitting at integers. Then your integral becomes a sum, and changing the variable to $z=e^{-s}$ we obtain an arbitrary Laurent series in $\mathbf{C}\backslash\{0\}$. This can have
arbitrary sequence of zeros.
Of course a sum of delta-functions is not integrable, but it is easy to modify this example, to make it integrable by integrating by prts:
$$\sum\_{-\infty}^\infty a\_ke^{-sk}=\int\_{-\infty}^\infty e^{-sx}dn(x)=-s\int\_{-\infty}^\infty n(x)e^{-sx}dx,$$
where $n$ is a step function jumping by $a\_k$ at $k$.
The integral in the RHS have the same zeros as the LHS, except at $s=0$, and $n(x)$ is an integrable (step) function. By integrating few more times, you can make your $f$ arbitrarily smooth.
The main condition of Lerch's theorem is that $f(x)=0$ for $x<0$, which ensures that $LF$ is bounded in the right half-plane. If we allow an arbitrary support of $f$, no conclusion about zeros can be made, except, of course that this is a discrete set.
| 2 | https://mathoverflow.net/users/25510 | 439263 | 177,408 |
https://mathoverflow.net/questions/439266 | 5 | Traditionally in dependent type theory with axiom K or uniqueness of identity proofs, every type $A$ is 0-truncated, and thus the type of lists on $A$, $\mathrm{List}(A)$, is 0-truncated and the free monoid on $A$. However, in homotopy type theory, not every type is 0-truncated; for example, the circle type $\mathbb{S}^1$ is a 1-truncated type which is provably not 0-truncated, and in general, types represent infinity-groupoids rather than sets. Thus, it makes sense to speak of the type of lists on the circle type, $\mathrm{List}(\mathbb{S}^1)$, or the type of lists on an arbitrary untruncated type $B$, $\mathrm{List}(B)$. Since neither $\mathbb{S}^1$ nor $B$ are provably 0-truncated, one cannot prove that the type of lists on $\mathbb{S}^1$ or $B$ is a monoid. However, in higher algebra, there is a structure which generalizes monoids from sets to infinity-groupoids, called $A\_\infty$-spaces. So is the list of types on $B$, $\mathrm{List}(B)$, the free $A\_\infty$-space on $B$?
| https://mathoverflow.net/users/483446 | Are lists in homotopy type theory free $A_\infty$-spaces? | In an informal sense, the answer "should be yes", in the sense that if one ignore type theory and work with an $\infty$-topos one can make sense of the construction $List(A)$ either by the usual universal property of list objects or as $List(A) = \coprod\_\mathbb{N} A^n$ (both definitions can be shown to be equivalent) and $List(A)$ is indeed the free $A\_\infty$-algebra on $A$.
However, the big problem is that we don't even know how to phrase the precise question you are asking in terms of Homotopy type theory, and it is not possible to make sense of what I just said (at least at the present time) within homotopy type theory. The problem is that we do not know how to define what is an $A\_\infty$-algebra within homotopy type theory. Being able to talk about this type of higher algebraic structure in HoTT, or show that it is impossible in some precise sense, is one of the biggest open problem in HoTT.
There are extensions of HoTT (like cubical HoTT, or maybe the simplicial HoTT of Riehl and Shulman with some additional work) that can/should be able to talk about higher structures like $A\_\infty$-algebras, and where this problem might have a solution, but I don't think it has been worked out in details.
The best one might hope to do in standard "book HoTT" (at least with present technology) is for each **external** (that is metatheoretical) integer $n$, define what is an $n$-truncated $A\_\infty$-algebra and show that for an $n$-truncated type, $List(A)$ is the free $n$-truncated $A\_\infty$-algebra on the type $A$. I even suspect we would only know how to do this for small values of $n$, though I might be wrong.
| 7 | https://mathoverflow.net/users/22131 | 439267 | 177,409 |
https://mathoverflow.net/questions/439269 | 6 | The word "local" in category theory does not seem to have a precise definition in itself but it often appears as part of other terminology. To my understanding, it is then used in the following ways, which AFAIU are quite unrelated:
* A category is locally $P$ if all of its slice categories are $P$.
Example: [locally cartesian closed categories](https://ncatlab.org/nlab/show/locally+cartesian+closed+category).
This is applied to functors in two ways:
+ A functor $F : \mathcal C \to \mathcal D$ is locally $P$ if its restriction to slice categories $F/x : \mathcal C / x \to \mathcal D / Fx$ is $P$.
Example: [locally cartesian closed functor](https://ncatlab.org/nlab/show/locally+cartesian+closed+functor).
+ A functor $F : \mathcal C \to \mathcal D$ is locally $P$ if its restriction $F/\top : \mathcal C \to \mathcal D / F \top$ (where $\top$ is the terminal object) is $P$.
Example: [local right adjoint](https://ncatlab.org/nlab/show/parametric+right+adjoint) functor.
* For (2-)categories and (2-)functors, locally $P$ means that $P$ holds on Hom-sets/categories.
Examples: locally small, [locally fully faithful](https://ncatlab.org/nlab/show/locally+fully+faithful+2-functor), [local adjunction](https://ncatlab.org/nlab/show/local+adjunction), ...
* Locally $P$ means that $P$ holds after localization/sheafification (or similar).
Examples: localization, local epimorphism, local object, local equivalence, ...
Am I correct that these uses are fairly unrelated? If so, are there synonyms that I can use to disambiguate? The current situation makes it difficult to come up with good terminology for things. For example, I am looking for names for the following concepts:
1. A functor $F : \mathcal C \to \mathcal D$ whose restriction $F/\top : \mathcal C \to \mathcal D / F \top$ is (a) full / (b) faithful / (c) fully faithful,
2. A functor $F : \mathcal C \to \mathcal D$ whose restrictions to slice categories $F/x : \mathcal C / x \to \mathcal D / Fx$ are all (a) full / (b) faithful / (c) fully faithful.
I don't care that 1b is equivalent to 2b and 1c to 2c. What I'm looking for is a general way to come up with good names for concepts. Currently, the best names I can come up with are:
1. (a) locally full / (b) locally faithful / (c) locally fully faithful,
2. (a) locally full / (b) locally faithful / (c) locally fully faithful, (yes, that's the same).
Unfortunately, it turns out that [locally fully faithful](https://ncatlab.org/nlab/show/locally+fully+faithful+2-functor) is already the term used for 2-functors that are fully faithful on Hom-categories. So I have 6 concepts, for which I can come up with only 3 terms which moreover are already taken.
| https://mathoverflow.net/users/66017 | Overloading of the word "local" in category theory | The terminology situation is certainly unfortunate. There is an nLab page for [locally](https://ncatlab.org/nlab/show/locally), which lists the different usages of the term (it doesn't currently include those for "local" rather than "locally"). The page suggests "slice-wise" to disambiguate the first case, and "hom-wise" to disambiguate the second case.
| 12 | https://mathoverflow.net/users/152679 | 439270 | 177,410 |
https://mathoverflow.net/questions/439264 | 1 | Let $\mathcal{F}$ be the space of all functions that uniformly and independently map the alphabet $\mathcal{X}$ to the set $\{1,2,\ldots,A\}$. Let $p(x|y)$ be an arbitrary conditional probability distribution. Does the following inequality hold?
\begin{align}
\mathbb{E}\_\mathcal{F}\left[\sqrt{\sum\_{x\in\mathcal{X}}p^2(x|y)\mathbf{1}[\mathcal{F}(x)=1]}\right]&\overset{?}{\le}\sqrt{\sum\_{x\in\mathcal{X}}p^2(x|y)}\mathbb{E}\_\mathcal{F}\left[\mathbf{1}[\mathcal{F}(x)=1]\right]\\
&=\frac{1}{A}\sqrt{\sum\_{x\in\mathcal{X}}p^2(x|y)},
\end{align}
where $\mathbf{1}[\cdot]$ is the indicator function.
I know that using Jensen's inequality for $\sqrt{x}$ we have:
\begin{align}
\mathbb{E}\_\mathcal{F}\left[\sqrt{\sum\_{x\in\mathcal{X}}p^2(x|y)\mathbf{1}[\mathcal{F}(x)=1]}\right]&\leq\sqrt{\sum\_{x\in\mathcal{X}}p^2(x|y)\mathbb{E}\_\mathcal{F}\left[\mathbf{1}[\mathcal{F}(x)=1]\right]}\\
&=\frac{1}{\sqrt{A}}\sqrt{\sum\_{x\in\mathcal{X}}p^2(x|y)}.
\end{align}
| https://mathoverflow.net/users/68835 | Does the following expectation-based inequality hold? | No, the factor $\frac1{\sqrt A}$ is the best you can get.
Indeed, letting $\mathcal X=[n]:=\{1,\dots,n\}$ and $a\_x:=p^2(x|y)$, we have
$$R\_n:=\frac{E\sqrt{\sum\_{x\in\mathcal{X}}p^2(x|y)1[\mathcal{F}(x)=1]}}{\sqrt{\sum\_{x\in\mathcal{X}}p^2(x|y)}}
=E\sqrt{\frac
{\sum\_{i\in[n]} a\_i Y\_i}
{\sum\_{i\in[n]} a\_i} },$$
where the $Y\_i$'s are iid Bernoulli random variables such that $P(Y\_1=1)=\frac1A=1-P(Y\_1=0)$.
If now, say, $a\_i=a>0$ for all $i\in[n]$, then, by the law of large numbers and the Fatou lemma,
$$\liminf\_{n\to\infty}R\_n=\liminf\_{n\to\infty}E\sqrt{\frac
{\sum\_{i\in[n]} Y\_i}n }\ge\sqrt{EY\_1}=\frac1{\sqrt A}.$$
Thus, the factor $\frac1{\sqrt A}$ is the best you can get, as claimed.
---
(Of course, for your question to make sense, you should have said "Let $\mathcal{F}$ be a uniformly distributed random element of the space of all functions that map the alphabet $\mathcal{X}$ to the set $\{1,2,\ldots,A\}$" instead of "Let $\mathcal{F}$ be the space of all functions that uniformly and independently map the alphabet $\mathcal{X}$ to the set $\{1,2,\ldots,A\}$".)
| 1 | https://mathoverflow.net/users/36721 | 439273 | 177,412 |
https://mathoverflow.net/questions/439238 | 1 | Let $V$ be a real vector space. Given a subset $A \subseteq V$, say that a point $x \in A$ lies in the *algebraic interior* of $A$ if every affine line $\ell$ that passes through $x$ has the property that $x \in (\ell \cap B)^\circ$. Here $\ell \cap B$ is a subinterval of $\ell \cong \mathbb{R}$, so we defined its interior $(\ell \cap B)^\circ$ by equipping $\ell$ with the Euclidean topology.
Say that a subset $U \subseteq V$ is *algebraically open* if the algebraic interior of $U$ is $U$ itself. We thereby get a topology on $V$.
Any affine functional $\pi: V \to \mathbb{R}$ is continuous with respect to this topology. Is this topology defined by the algebraically open subsets the coarsest topology such that every affine functional on $V$ is continuous map?
I had asked this question on [Math Stack Exchange](https://math.stackexchange.com/q/4622368/1141296) earlier.
| https://mathoverflow.net/users/136356 | abstract description of the topology on a real vector space defined by the algebraically open sets | This is not true, already in $\mathbf R^2$. Indeed, if $V$ is finite-dimensional, then the Euclidean topology is the coarsest topology for which all linear¹ maps $V \to \mathbf R$ are continuous: choosing a basis $e\_1,\ldots,e\_n$ and taking the corresponding coordinate projections $\pi\_i \colon V \to \mathbf R$ shows that all boxes $(a\_1,b\_1) \times \cdots \times (a\_n,b\_n)$ have to be open, and these generate the Euclidean topology.
So it suffices to find a set that is open in your sense but not in the Euclidean topology.
**Example.** Let $V = \mathbf R^2$, set $v\_0 = (1,0)$, and inductively choose points $v\_1, v\_2, \ldots$ such that $\lVert v\_i \rVert = 2^{-i}$ and $v\_i$ does not lie on the (finitely many) lines $\overline{v\_jv\_k}$ for $j,k < i$. Define $Z = \{v\_0,v\_1,\ldots\}$ and $U = V \setminus Z$. Then $U$ is not open in the Euclidean topology because the $v\_i$ converge to $0 \not \in Z$. But for every line $\ell \subseteq V$, the intersection $\ell \cap Z$ contains at most $2$ points by construction, so $\ell \cap U$ is open.
(For the algebraic geometers in the room, this construction is remarkably similar to an example I laid out in [this answer](https://mathoverflow.net/a/423827/82179): being closed cannot be checked on each curve.)
---
¹It doesn't matter if you say "linear" or "affine linear" here, because any affine linear map differs from a linear one by a translation, and translation $\mathbf R \to \mathbf R$ is continuous.
| 2 | https://mathoverflow.net/users/82179 | 439274 | 177,413 |
https://mathoverflow.net/questions/439192 | 4 | I wonder if there are regular cardinals $\lambda$ and $\kappa$ such that $\kappa < \lambda \leq 2^\kappa$ and such that, consistently, the following holds:
Let $c: [\lambda]^2 \to \kappa$ be such that for $\alpha < \beta < \gamma$, we have $c(\alpha, \gamma) \leq \max \lbrace c(\alpha, \beta) , c(\beta, \gamma) \rbrace$. The there is a cofinal $H \subseteq \lambda$ such that $range(c \restriction [H]^2) \subseteq \eta$ for some $\eta < \kappa$.
Now, the second part is obviously true, even without the max condition, in case $\lambda $ is weakly compact, but then we lose $\lambda \leq 2^\kappa$ since $\lambda$ is strong limit.
I am particularly interested in the case where $\kappa = \aleph\_0$, or more generally a regular cardinal.
| https://mathoverflow.net/users/495743 | Ramsey-like property with order condition | A coloring $c:[\kappa]^2\rightarrow\theta$ is *subadditive of the first kind* if for all $\alpha<\beta<\gamma<\kappa$, $c(\alpha,\gamma)\le\max\{c(\alpha,\beta),c(\beta,\gamma)\}$. It is *subadditive of the second kind* if for all $\alpha<\beta<\gamma<\kappa$, $c(\alpha,\beta)\le\max\{c(\alpha,\gamma),c(\beta,\gamma)\}$. The combination of both is called *subadditive*, and is the subject matter of the following paper: <http://www.assafrinot.com/paper/36>
The Ramsey-theoretic results you are after may be found as Lemma 3.38(2), Corollary 3.42(2), and Theorem 3.45(2) of the above paper. In particular, assuming PFA, for every regular cardinal $\kappa>2^{\aleph\_0}$ that is not the successor of a singular cardinal of countable cofinality, every coloring $c:[\kappa]^2\rightarrow\omega$ that is subadditive of the first kind admits a cofinal subset $A\subseteq\kappa$ such that $c``[A]^2$ is finite.
| 5 | https://mathoverflow.net/users/20033 | 439275 | 177,414 |
https://mathoverflow.net/questions/439277 | 1 | I have read the first three chapters of Hartshorne and was wondering what are the applications of the notions presented in number theory or arithmetic geometry. I already know that the notion of scheme is of utmost importance in number theory but what about the cohomology of sheaves ? Why is étale cohomology useful in number theory for instance ?
Note : About the étale cohomoloy groups, I already know that they are very important examples of representations of absolute Galois groups.
| https://mathoverflow.net/users/170999 | Sheaf cohomology in number theory | This is slightly too log to be a comment, but certainly too short to be a defintive answer. Apologees in advance.
I would suggest reading the volume 2 book ("*Cohomology of algebraic varieties*", by Danilov) of the EMS series "*Algebraic Geometry*". You will learn a lot about étale cohomology, in a non-technical way in this book. The basic motivation for Grothendieck to develop étale cohomology was to micmick an argument of Serre, who proved some analogues of the Weil conjectures over the field of complex numbers (see [*Analogues K"ahleriennes de certaines conjecture de Weil*](https://www.jstor.org/stable/1970088)).
Grothendieck noticed early on that if we had a cohomology theory for varieties over finite fields having nice properties (like 6 operations, Lefschetz trace formula and so on) then the Weil conjectures "*would open like a nutshell when the time is ripe : hand pressure is enough, the shell opens like a perfectly ripened avocado!*" (the quote is from *Recoltes et Semailles*, page 552-3)
It was realized, only a few years later, that an analogue of the Hodge conjecture in étale cohomology could be formulated in terms of Galois actions on étale cohomology groups of a smooth projective variety over a finite field. This is the famous Tate cojecture. This conjecture is really remarkable as it would, for instance, imply most of the Grothendieck's standard conjectures. In the aftermath of this conjecture, some area of representation theory merged with arithmetic in order to study in more details the etale cohomology groups as representation spaces.
| 3 | https://mathoverflow.net/users/37214 | 439292 | 177,419 |
https://mathoverflow.net/questions/438021 | 1 | It is [well known](https://math.stackexchange.com/a/509018/656606) that an open subscheme of an affine scheme is not necessarily an affine one. But what are (if possible the most general) sufficient conditions for its affinity? And is it known how, in such cases, to directly describe its ring in terms of its corresponding radical ideal?
(I'm just starting to learn algebraic geometry, so any of your answers will most likely be useful to me)
| https://mathoverflow.net/users/148161 | Under what conditions is an open subscheme of an affine scheme affine and what ring corresponds to it? | Here is the best general result I'm aware of:
**Lemma.** *Let $A$ be a commutative ring with an ideal $I$, let $X = \operatorname{Spec} A$ with closed subset $Z = V(I)$ and open complement $U = D(I)$.*
1. *If $Z$ is a Cartier divisor, then $U$ is affine and dense in $X$.*
2. *If $A$ is Noetherian and $U$ is affine and dense in $X$, then $Z$ is a Weil divisor.*
The assumption in (1) means that $I$ is locally generated by one element (i.e. there exist elements $f\_1,\ldots,f\_n \in A$ generating the unit ideal such that each $I\_{f\_i} \subseteq A\_{f\_i}$ is a principal ideal). The conclusion in (2) means that every minimal prime containing $I$ has height $1$, i.e. $Z$ has pure codimension $1$ in $X$.
Every Cartier divisor is a Weil divisor (this is Krull's principal ideal theorem), and the converse is true if $X$ has mild singularities. For this problem, you can take $X$ to be locally factorial (or locally $\mathbf Q$-factorial if you don't mind replacing your ideal by a suitable power); in particular it is true when $A$ is regular. This gives:
**Corollary.** *Suppose $A$ is Noetherian and locally $\mathbf Q$-factorial, and that $U \subseteq X$ is dense. Then $U$ is affine if and only if every component of $Z$ has codimension $1$.* $\square$
There is also something to say about the case where $Z$ contains an entire irreducible component of $X$, but let me not attempt this here. At any rate, algebraic geometers often stick to irreducible schemes, where any nonempty open subset is dense.
*Proof of Lemma* (sketch). Statement (1) is relatively easy if you know some scheme-theory. The slogan is that the property of a morphism being affine can be checked locally (Hartshorne, Exercise II.5.17). Let me do it in a way that also computes the ring $B = \Gamma(U,\mathcal O\_U)$ in terms of $I$. By general stuff we get a morphism of schemes $\phi \colon U \to \operatorname{Spec} B$ (see e.g. Hartshorne, Exercise II.2.4), which we will show is an isomorphism.
Choose $f\_1,\ldots,f\_n \in A$ generating the unit ideal and elements¹⁾ $g\_1,\ldots,g\_n \in A$ such that $I\_{f\_i} = (g\_i)\_{f\_i} \subseteq A\_{f\_i}$. Then
$$U \cap D(f\_i) = D(g\_i) \cap D(f\_i) = D(f\_ig\_i) \cong \operatorname{Spec} A\_{f\_ig\_i},$$
and these cover $U$ since the $D(f\_i)$ cover $X$. The sheaf condition for $\mathcal O\_U$ gives an exact sequence
$$0 \to B \to \bigoplus\_{i=1}^n A\_{f\_ig\_i} \to \bigoplus\_{i<j} A\_{f\_ig\_if\_jg\_j},\label{1}\tag{1}$$
where the second map takes $(a\_i)\_i$ to $(a\_i-a\_j)\_{i<j}$. Localising \eqref{1} at $f\_i$ on the one hand computes $B\_{f\_i}$ by exactness of localisation, but on the other hand computes $\Gamma(U \cap D(f\_i),\mathcal O\_U) = A\_{f\_ig\_i}$ via the sheaf condition on $U \cap D(f\_i)$. If $Y = \operatorname{Spec} B$, we see that both $D(f\_i) \subseteq Y$²⁾ and its inverse image $U \cap D(f\_i) \subseteq U$ are affine, with isomorphic coordinate rings, so $\phi$ is an isomorphism above $D(f\_i) \subseteq Y$. Since these cover $Y$, we conclude that $\phi$ is an isomorphism, i.e. $U$ is affine. (See also Hartshorne, Exercises II.2.16, II.2.17, and II.5.17.)
Statement (2) is a little harder, and relies on the observation that for a Noetherian *local* ring $(R,\mathfrak m)$, the punctured spectrum $\operatorname{Spec} R \setminus\{\mathfrak m\}$ is affine if and only if $\dim R \leq 1$. See [Tag [0BCQ](https://stacks.math.columbia.edu/tag/0BCQ)] for a proof. Statement (2) then follows by taking $(R,\mathfrak m)$ to be the localisation of $A$ at a minimal prime $\mathfrak p$ containing $I$. If $U$ is affine, then so is $U \times\_X \operatorname{Spec} A\_{\mathfrak p} = \operatorname{Spec} A\_{\mathfrak p} \setminus \{\mathfrak p\}$ since a fibre product of affine schemes is affine. This forces $\operatorname{ht} \mathfrak p \leq 1$, i.e. every irreducible component of $Z$ has codimension $\leq 1$. The assumption that $U$ is dense means that every component of $Z$ has codimension exactly $1$. $\square$
---
¹⁾ A priori you only get $I\_{f\_i} = \big(\tfrac{g\_i}{f\_i^r}\big)\_{f\_i}$ for some $g\_i \in A$ and $r\in \mathbf N$, but then $g\_i$ also generates the ideal in $A\_{f\_i}$ since $f\_i$ is invertible.
²⁾ This is abuse of notation: we should really take the ring homomorphism $\psi \colon A \to B$ and talk of $D(\psi(f\_i)) \subseteq Y$. Then $B\_{f\_i}$ (localisation as $A$-modules) agrees with $B\_{\psi(f\_i)}$ (localisation as $B$-modules), etcetera. I hope this does not lead to confusion.
| 3 | https://mathoverflow.net/users/82179 | 439295 | 177,421 |
https://mathoverflow.net/questions/438983 | 1 | Let $\mathscr{C}:=\{\gamma : \mathbb{R}\_+\rightarrow\mathbb{R}^n \mid \gamma \ \text{ continuous}\}$ be the set of all $\mathbb{R}^n$-valued paths over $[0,\infty)$. Endow $\mathscr{C}$ with the $\sigma$-algebra $\mathfrak{C}$ generated by all projections $\gamma \mapsto \gamma\_t$ (for $t\geq 0$ fixed). Let further
$$\phi : \mathscr{C}\rightarrow\mathscr{C} \ \ \text{ be given by } \ \phi(\gamma):=(\gamma\_{t+1})\_{t\geq 0} \ \text{ (left-shift),}$$ and call *$\phi$-invariant* any event $S\in\mathfrak{C}$ with $\phi^{-1}(S)=S$.
Do you know of examples for $\mathscr{C}$-valued stochastic processes $Y$ that attain $\phi$-invariant events with trivial probability, i.e. are such that $\mathbb{P}\_Y(S)\in\{0,1\}$ whenever $S\in\mathfrak{C}$ is $\phi$-invariant?
| https://mathoverflow.net/users/160714 | Shift-ergodic stochastic processes in continuous time | Do you not also want that $\mathbb{P}\_Y$ is $\phi$-invariant?
In any case, yes there are extremely many continuous-time continuous-path real-valued stochastic processes whose law is ergodic under the time-$1$-shift map. Of course, the most trivial example would be where
$$ \mathbb{P}(Y\_t=c \ \ \forall t \geq 0) = 1 $$
for some $c \in \mathbb{R}$. An almost-as-trivial case would be
$$ Y\_t = p(2\pi ft + \Phi) $$
where $p(\cdot)$ is a $1$-periodic continuous function, $f$ is an irrational number, and $\Phi$ is a random variable with $\Phi \sim \mathrm{Uniform}(0,2\pi)$. (I don't know if this would count as a "non-trivial example class" under your bounty statement!)
Perhaps the simplest "non-trivial" example would be the *Ornstein-Uhlenbeck* process:
(I hope there are no typos in the following!) Fix two parameters $\theta,\sigma>0$. Starting off with a two-sided Wiener process $(W\_t)\_{t \in \mathbb{R}}$ (meaning that $(W\_t)\_{t \geq 0}$ and $(W\_t)\_{t \leq 0}$ are independent Wiener processes), we can define a stochastic process $(Y\_t)\_{t \in \mathbb{R}}$ where for almost every sample path of the Wiener process, for all $t \in \mathbb{R}$,
\begin{align\*}
Y\_t &= \sigma \! \int\_{-\infty}^t e^{\theta(s-t)} \, dW\_s \quad\quad\quad\quad\quad\quad\ \, \text{(Riemann-Stieltjes integral)} \\
&= \sigma \left( W\_t - \theta \! \int\_{-\infty}^t e^{\theta(s-t)} W\_s \, ds \right) \quad \text{(classical integral, obtained by integration by parts)}.
\end{align\*}
This is the unique two-sided-time strong solution of the stochastic differential equation
$$ dY\_t = -\theta Y\_t \, dt + \sigma dW\_t, $$
or equivalently (as a random differential equation)
$$ \frac{d(Y\_t - \sigma W\_t)}{dt} = -\theta Y\_t. $$
$\text{[}$The one-sided-time solution $(Y\_t^{(y\_0)})\_{t \geq 0}$ of this equation starting at any $y\_0 \in \mathbb{R}$ would be
\begin{align\*}
Y\_t^{(y\_0)} &= e^{-\theta t} \left( y\_0 + \sigma \! \int\_0^t e^{\theta s} \, dW\_s \right) \\
&= \sigma W\_t + e^{-\theta t} \left( y\_0 - \sigma\theta \! \int\_0^t e^{\theta s} W\_s \, ds \right). \text{]}
\end{align\*}
A one-sided-time stochastic process with the law of $(Y\_t)\_{t \geq 0}$ (or a two-sided-time stochastic process with the law of $(Y\_t)\_{t \in \mathbb{R}}$) is called an Ornstein-Uhlenbeck process.
*Sketch-proof that OU process is shift-ergodic.* Firstly, working in two-sided-time,
* one can check that replacing $(W\_t)\_{t \in \mathbb{R}}$ with $(W\_{t+1}-W\_1)\_{t \in \mathbb{R}}$ transforms $(Y\_t)\_{t \in \mathbb{R}}$ to $(Y\_{t+1})\_{t \in \mathbb{R}}$;
* since the Wiener process has memoryless stationary increments, one can show that the law of $(W\_t)\_{t \in \mathbb{R}}$ is mixing and hence ergodic with respect to the map $(\gamma\_t) \mapsto (\gamma\_{t+1}-\gamma\_1)$;
and so the law of $(Y\_t)\_{t \in \mathbb{R}}$ is ergodic with respect to the map $(\gamma\_t) \mapsto (\gamma\_{t+1})$. Finally, having ergodicity in two-sided time guarantees ergodicity in the one-sided-time restriction.
---
In general, given any $\phi$-invariant probability measure $\mathbb{P}$ on $\mathscr{C}$, one can express $\mathbb{P}$ as a weighted average of ergodic $\phi$-invariant probability measures on $\mathscr{C}$. This is essentially the *ergodic decomposition theorem*, which holds for any Borel-measurable dynamical system on a Polish space; in this case, the $\sigma$-algebra $\mathfrak{C}$ is precisely the Borel $\sigma$-algebra of the topology of uniform convergence on compact sets, which is a Polish topology.
| 1 | https://mathoverflow.net/users/15570 | 439298 | 177,423 |
https://mathoverflow.net/questions/439278 | 8 | In appendix A.2 of the orange book, Ravenel defines a ring spectrum $E$ to be *flat* if $E\wedge E$ is equivalent to a coproduct of suspensions of $E$. (Call this definition (1).) I've seen this definition used in talks and the like as well, but I'm confused about where it comes from. I know two meanings of the word "flat" in this context. Firstly, Lurie's version (2): a spectrum $E$ is flat (over the sphere spectrum) if $\pi\_0E$ is flat as an abelian group and $\pi\_0E\times\pi\_n\mathbb{S}\to\pi\_nE$ is an isomorphism for all $n$. Second, the version defined in e.g. the nLab page on the Adams spectral sequence (3): a ring spectrum $E$ is flat if $E\_\*E$ is flat over $\pi\_\*E$.
As far as I can tell, these three definitions are all different, though I guess (1) implies (3). What I would like to know is, why did Ravenel call this condition "flat", and what is its precise relationship to the other two definitions?
| https://mathoverflow.net/users/158123 | Why did Ravenel define a ring spectrum to be flat if its smash-square splits into copies of itself? | Here is a little context to maybe complement Tom's and Nick's answers.
---
The definition (2) in terms of being flat over $\pi\_\* \Bbb S$ is new - it's a specialization of a definition of flatness over $R$ that comes from derived / spectral algebraic geometry. Over $\Bbb S$ this is so rare that most examples are either in characteristic zero, or they are tailor-made to satisfy it (eg: localizations). The main utility would be to get a Kunneth theorem, identifying $E\_\* X$ with $E\_\* \otimes\_{\pi\_\* \Bbb S} \pi\_\* X$. It has very specific goals, such as making base-change and descent effectively calculable.
---
As Nick says, Ravenel's interest is in the general Adams-Novikov spectral sequence, and specifically getting better than the $E\_1$-term. (In any case where the version (2) definition applies, the Adams-Novikov spectral sequence is pretty degenerate.) Slightly more explicitly, the $E\_1$-term starts with the homotopy groups of $E \wedge X$, $E \wedge E \wedge X$, and so on, and produces a spectral sequence for the homotopy groups of $X$. We would ideally like to understand $\pi\_\*(E \wedge E \wedge \dots \wedge E \wedge X)$ in terms of $E\_\* X$.
This leads to definitions (1) and (3). Both of these are geared at getting a Kunneth isomorphism
$$\pi\_\*(E \wedge E \wedge Y) \cong \pi\_\*(E \wedge E) \otimes\_{\pi\_\* E} E\_\* Y$$
which we can then apply inductively to $X$. Definition (1) gets at this Kunneth isomorphism explicitly, because the splitting of $E \wedge E$ tells you that
$$
E \wedge E \wedge Y \simeq \bigvee\_i \Sigma^{n\_i} (E \wedge Y).
$$
Definition (3), by contrast, gets at this isomorphism with the Kunneth spectral sequence. You re-express
$$
E \wedge E \wedge Y \simeq (E \wedge E) \wedge\_E (E \wedge Y)
$$
and this has a spectral sequence
$$
Tor\_{\*\*}^{E\_\*} (E\_\* E, E\_\* Y) \Rightarrow \pi\_\*((E \wedge E) \wedge\_E (E \wedge Y)).
$$
Definition (3) implies that this degenerates to a Kunneth isomorphism.
---
So why would we choose version (1), when version (3) is usually strictly stronger? There are two reasons.
* Maybe $E$ isn't good enough. Version (3) has an assumption - it only applies if $E$ is a highly structured (associative) ring spectrum, so that modules over $E$ and smash products over $E$ make sense. This is a problem if you're working with a ring spectrum that doesn't admit that nice of a multiplication (or you simply don't know that it does). By contrast, version (1) applies to not-associative-but-homotopy-associative things -- for example, the mod-$p$ Moore spectrum when $p$ is a prime greater than 3.
* More seriously, maybe it is 1992, when the orange book is being published. In 1992, categories of module spectra over highly structured ring spectra don't really exist, and there is no high-powered Kunneth spectral sequence available to you. Version (1) is what you've got.
| 8 | https://mathoverflow.net/users/360 | 439299 | 177,424 |
https://mathoverflow.net/questions/438839 | 4 | Consider the following situation, let $\mathcal{R}$ be a discrete valuation ring with uniformizer $\pi$ (say the valuation ring of a finite extension $K$ of $\mathbb{Q}\_{p}$. Let $\{ A\_{n}\}\_{n\in\mathbb{N}}$ be a projective family of $\mathcal{R}$ algebras equipped with maps $A\_{n+1}\rightarrow A\_{n}$, considering their $\pi$-adic completions, we get an associated projective family $\{ \widehat{A\_{n}}\}\_{n\in\mathbb{N}}$, along with short exact sequences:
\begin{equation}
0\rightarrow A\_{n}\rightarrow \widehat{A\_{n}}\rightarrow \widehat{A\_{n}}/A\_{n}\rightarrow 0
\end{equation}
together with transition maps between the sequences. We will denote $B\_{n}=\widehat{A\_{n}}/A\_{n}$ for simplicity. Assume that the morphism $\varprojlim\_{n}A\_{n}\rightarrow \varprojlim\_{n}\widehat{A\_{n}}$ is an isomorphism and that all $\widehat{A\_{n}}$ are compact with respect to the $\pi$-adic topology. I want to be able to conclude that if this is the case, then for sufficiently $n$ the map $A\_{n}\rightarrow \widehat{A\_{n}}$ is already an isomorphism. For this, it would be enough to show two things:
1. The first right derived functor $R^{1}\varprojlim\_{n}A\_{n}= R^{1}\varprojlim\_{n}A\_{n}$.
2. If $\varprojlim\_{n} B\_{n}=0$ then $B\_{n}=0$ for sufficiently high $n$.
Notice that all morphisms considered here are continuous when all modules are endowed with the $\pi$-adic topology. Furthermore, the $\mathcal{R}$-modules $B\_{n}$ are compact, multiplication by $\pi$ is an isomorphism and the closure of $0$ is the whole $B\_{n}$. I do not expect complete answers to any of these questions, but I seem to lack the tools to tackle these kinds of problems, so suggestions of references where something like this might come up are very much appreciated.
Context: If $X=\mathrm{Sp}(A)$ is a smooth connected affinoid space over $K$ and $f\in A$, I would like to show that for $X(\frac{\pi^{n}}{f})=\mathrm{Sp}(C\_{n})$ the rings $C^{\circ}\_{n,A}$ defined in [MO Question: Noetherianess of some subalgebras of an affinoid algebra](https://mathoverflow.net/questions/432242/on-the-noetherianess-of-some-subalgebras-of-an-affinoid-algebra), agree with $A^{\circ}$ for sufficiently high $n$. In this setting the inverse limit
$\varprojlim\_{n}C^{\circ}\_{n,A}\rightarrow \varprojlim\_{n}\widehat{C^{\circ}\_{n,A}}=A^{\circ}$ by
theorem 2.6 in the reference.
*Hansen, David*, [**Vanishing and comparison theorems in rigid analytic geometry**](http://dx.doi.org/10.1112/S0010437X19007371), Compos. Math. 156, No. 2, 299-324 (2020). [ZBL1441.14085](https://zbmath.org/?q=an:1441.14085).
| https://mathoverflow.net/users/476832 | On inverse limits of $\pi$-adically complete algebras | Not sure about the main question, but regarding the "context" question, one can show the following:
If $A$ is any affinoid, and $f \in A$ any element with nowhere-dense zero locus, then $A^\circ = A\times\_{A\langle \pi^n/f \rangle} {A\langle \pi^n/f \rangle}^\circ$ for all sufficienly large $n$.
The point is that $A \to {A\langle \pi^n/f \rangle}$ is strict for the supermum seminorms as soon as the image of $\mathrm{Spa} {A\langle \pi^n/f \rangle} \to \mathrm{Spa} A$ contains the Shilov boundary of $\mathrm{Spa} A$, which it will for all sufficiently large $n$ because the Shilov boundary is finite and (by our assumption on $f$) disjoint from $V(f)$.
Hope this helps!
| 1 | https://mathoverflow.net/users/496798 | 439311 | 177,428 |
https://mathoverflow.net/questions/439310 | 2 | Let $S$ be a simplicial complex and let $Bary(S)$ denote its barycentric subdivision.
Of course, the geometric realizations of $S$ and $Bary(S)$ are homeomorphic.
However, one issue that arises in practical computation utilizing barycentric subdivisions is the size of $Bary(S)$ is *much* larger than $S$, and so when $S$ is large, then $Bary(S)$ is HUGE!
Has there been any research on "smaller" (i.e., having less simplices) simplicial models of $Bary(S)$ for general $S$ that maintain isomorphic homology groups, i.e., are homotopy equivalent?
I believe this would have much relevance to many areas of persistent homology and computational geometry.
| https://mathoverflow.net/users/408316 | Do there exist smaller simplicial models of barycentric subdivisions? | Consider an arbitrary finite simplicial complex $\ S.\ $ First, look at it purely combinatorially. Thus, let $\ \{a\ b\}\ $ be a 1-simplex of $\ S.\ $ Then define a subdivided simplicial complex $\ S(a\ b\ c),\ $ where $\ c\ $ is a new fixed vertex that didn't belong to $\ S.\ $ The simplexes that don't contain $\ \{a\ b\}\ $ stay the same. Every simplex
$$ \{a\ b\}\cup A $$
of $\ S\ $ that does contain $\{a\ b\} $
gets replaced by $\ \{a\ c\}\cup A\ $ and
$\ \{c\ b\}\cup A.$
Now we can iterate this easy construction.
---
---
In turn, let's look at a geometric implementation $\ |S|,\ $ where we will have $\ c:=\frac{a+b}2.$
We may like to have iterations that have smaller and smaller mesh (mesh of a complex is the maximal length of 1-dimensional simplexes of that complex) so that mesh approaches zero.
We achieve this effect of mesh approaching zero by each time selecting 1-simplex $\ \{a\ b\}\ $ that has the maximal diamater among all present 1-simplexes (i.e. this diameter would be the current mesh). The mesh will approach zero slowly but it will.
| 3 | https://mathoverflow.net/users/110389 | 439312 | 177,429 |
https://mathoverflow.net/questions/439320 | 3 | I am trying to observe the behavior of $x\_n \in (0,1)$ defined such that the function
\begin{equation}
f\_n(x):=e^{-1/x}\Bigl(1+\frac{1}{n^2 x^n} \Bigr)
\end{equation}
attains its maximum inside the interval $(0,1)$ at $x=x\_n$.
Upon using the Wolfram alpha, I have found out that as $n \to \infty$ it seems that $x\_n \to 0^+$ and
\begin{equation}
\int\_0^1 f\_n(x) dx \leq 2 \int\_0^{2x\_n} f\_n(x)dx.
\end{equation}
That is, as $n \to \infty$, the graph of $f\_n$ on $(0,1)$ is sufficiently localized around its maximum value.
Now my question are the following two:
1. I am trying to estimate the rate at each $x\_n \to 0^+$ as $n \to \infty$. But I cannot find a nice way to do so.
2. Is the above estimate for the integral correct? How one can prove it? Could anyone please help me with it as well?
This kind of analysis is quite new to me, so I am a bit stuck. I deeply appreciate any help.
| https://mathoverflow.net/users/56524 | Locating the maximum point $x_n$ of $f_n(x):=e^{-1/x}\Bigl(1+\frac{1}{n^2 x^n} \Bigr)$ in $(0,1)$ | The maximum $x\_n$ of
$$f\_n(x):=e^{-1/x}\Bigl(1+\frac{1}{n^2 x^n} \Bigr)$$
is the smallest solution in $(0,1)$ of the equation
$$x=n x^n+\frac{1}{n}.$$
For $n\gg 1$ this gives $x\_n\rightarrow 1/n$.
The integral is given by
$$\int\_0^1 f\_n(x)dx=\text{Ei}(-1)+n^{-2}\,\Gamma (n-1,1)+1/e$$
$$\qquad\rightarrow \sqrt{2 \pi } e^{-n} n^{n-\frac{7}{2}}\;\;\text{for}\;\;n\gg 1.$$
Here is a comparison of the exact integral (gold data points) and the asymptote (blue) --- the difference is hardly noticeable for $n>10$.
| 6 | https://mathoverflow.net/users/11260 | 439321 | 177,430 |
https://mathoverflow.net/questions/439316 | 22 | It is well known that some problems in functional analysis and in general topology are independent from ZFC: to name a few, Kaplansky's conjecture, the existence of outer automorphisms of the Calkin algebra, the existence of a Suslin line etc. This is not too surprising, since many results in this fields require a nontrivial amount of set theory to be proved. On the other hand, differential geometry seems to be "higher up" in mathematical complexity (i.e., further away from set theoretical questions) and so it seems reasonable to me that no "natural" statement in diff. geom. is independent from ZFC.
Is that the case? That is, are there some statements which are independent from ZFC (or are conjectured to be)?
Edit: many comments (and the only, at the moment, answer) focused on computably undecidable problems. While this does indeed answer the question as formulated (because at least one instance of the problem must be independent), it is not exactly what I had in mind in that it does not offer an explicit example of an independent statement, or at least I don't see how to get one. So a more precise reformulation of my question is (but I am also very interested in more examples of computably indecidable problems): is there an explicit (in some sense of the word) statement in differential geometry that is (or is conjectured to be) independent from ZFC?
| https://mathoverflow.net/users/493244 | Statements in differential geometry independent from ZFC | [Using the comments for context on undecidability/independence of ZFC]
A computably undecidable problem is whether or not a homology sphere has a metric of positive scalar curvature [Page 79 of *[Computers, Rigidity, and Moduli](https://www.google.com/books/edition/Computers_Rigidity_and_Moduli/Q5sAEAAAQBAJ)*].
| 19 | https://mathoverflow.net/users/1847 | 439326 | 177,432 |
https://mathoverflow.net/questions/437521 | 5 | Let $F\_n^{(k)}(x)= \sum\_j {\binom{n+(k-1)j}{kj} x^j}$ and $G\_n^{(k)}(x)= \sum\_j {\binom{n+j}{kj} x^j}.$
I am interested in the coefficients ${a\_{n,k,j}}$ such that
$$G\_n^{(k)}(x)=\sum\_{j\geq0 }{a\_{n,k,j}} F\_j^{(k)}((-1)^ { k}x).$$
Computations suggest the following:
Let $z\sum\_{j\geq 0}C\_{1,j}^{(k)}z^j$ be the inverse series of $\sum\_{j=1}^k (-1)^{j-1}z^j$ and let $(\sum\_{j\geq 0}C\_{1,j}^{(k)}z^j)^m=\sum\_{j\geq 0}C\_{m,j}^{(k)}z^j.$
Then $$ a\_{n,k,j}=(-1)^{k j}C\_{kj+1,n-(k-1)j}^{(k-1)}.$$
Any idea how to prove this?
**Remark:** $C\_{1,n}^{(2)}=C\_n$ are the Catalan numbers and
$( C\_{1,n}^{(3)})\_{n\geq 0}=(1,1,1,0,-4,-14,-30,-33,\dots).$ (cf. OEIS, A103779).
**Edit:** Perhaps the following observation may be useful.
A matrix inversion theorem of Gould and Hsu implies a similar result:
The coefficients $c\_{n,k,j}$ which give $$\sum\_j {c\_{n,k,j}} F\_j^{(k)}(x)=x^n$$ are $(-1)^{n-j} A\_{n-j,k,kj+1}$ where $A\_{n,k,r}=\frac{r}{kn+r}\binom{kn+r}{n}$ are Fuss-Catalan numbers.
Here we have $\sum\_{n}A\_{n,k,r}x^n=(\sum\_{n}A\_{n,k,1}x^n)^r $ and $ x\sum\_{n}A\_{n,k,1}x^{(k-1)n}$ is the inverse series of $y-y^k.$
Is there a combinatorial or other reason for the appearance of these special inverse series?
| https://mathoverflow.net/users/5585 | A polynomial identity related to Catalan numbers | These assertions can be proved using (formal) generating functions.
Using that for $j\geq 0, k\geq 1$
\begin{align\*}
\sum\_{n\geq 0} {n-j+kj \choose kj} t^n &=\frac{t^j}{(1-t)^{kj+1} }\;\;\mbox{ and }\\
\sum\_{n\geq 0}{n+j \choose kj} t^n&=\frac{t^{kj-j}}{(1-t)^{kj+1}}\;\;\;,
\end{align\*}
gives that
\begin{align\*}
\sum\_{n\geq 0} t^n F\_n^{(k)}(x) &=\frac{1}{1-t}\,\frac{1}{1-\frac{xt}{(1-t)^k}}\;\;\mbox{ and }\\
\sum\_{n\geq 0} t^n G\_n^{(k)}(x) &=\frac{1}{1-t}\,\frac{1}{1-\frac{xt^{k-1}}{(1-t)^k}}\;\;\;,
\end{align\*}
(I) consider first the (simpler) Gould-Hsu case. Here $c\_{n,k,j}= (-1)^{n-j} [t^{n-j} ]\, C\_k(t)^{kj+1}$ where
$C\_k(t)$ denotes the $k$-ary tree function, which is defined by
$$C\_k(t)=1+tC\_k(t)^k\;\;\;.$$
Thus \begin{align\*}
\sum\_{j\geq 0} c\_{n,k,j} F\_j^{(k)}(x)&=\sum\_{j\geq 0} F\_j^{(k)}(x)(-1)^{n-j}[t^{n-j}]C\_k(t)^{kj+1}\\
&=\sum\_{j\geq 0} F\_j^{(k)}(x)(-1)^{n-j}[t^{n}] t^j C\_k(t)^{kj+1}\\
&=(-1)^n [t^n] \sum\_{j\geq 0} F\_j^{(k)}(x)(-1)^{j}t^j C\_k(t)^{kj+1}\\
&=(-1)^n [t^n] \frac{C\_k(t)}{1+tC\_k(t)^k}\frac{1}{1+\frac{xtC\_k(t)^k}{(1+tC\_k(t)^k)}^k}\\
&=(-1)^n [t^n] \frac{1}{1+xt}=x^n\\
\end{align\*}
(II) now to your case above. Here $a\_{n,k,j}=(-1)^{kj}[t^{n-(k-1)j}] A\_{k-1}(t)^{kj+1}$ where $yA\_k(y)$ is the inverse of $y(z)=\sum\_{j=1}^k (-1)^{j-1} z^j$ .
A similar computation as above here gives
\begin{align\*}
\sum\_{j\geq 0} a\_{n,k,j} F\_j^{(k)}((-1)^kx)&=[t^n] \frac{A\_{k-1}(t)}{1-T(t)} \frac{1}{1- (-1)^k\frac{T(t)x}{(1-T(t))^k}}
\end{align\*}
where $T(t):=(-1)^kt^{k-1}A\_{k-1}(t)^k$. This will simplify to the generating function for the $G\_n^{(k)}$ if
$$\frac{A\_{k-1}(t)}{1-T(t)}=\frac{1}{1-t}\;\;.$$
And this in turn follows (for $k\geq 2$) with simple steps after substituting $x=tA\_{k-1}(t)$ in the equality $$t=\frac{x+(-1)^{k-1}x^k}{1+x}$$.
(III) The Ansatz $b\_{n,k,j}=[t^n] T(t)^j Z(t)$ leads to the generating function
\begin{align\*}
\frac{Z(t)}{1-T(t)} \frac{1}{1- \frac{T(t)x}{(1-T(t))^k}}
\end{align\*}
for $R\_n(x):=\sum\_{j \geq 0} b\_{n,k,j} F\_j^{(k)}(x)$. One will expect this to be a simple function of $t$ only if $\frac{Z(t)}{1-T(t)}$ and $\frac{T(t)}{(1-T(t))^k}$ simplify
to simple functions of $t$, i.e. can be "solved" for $t$. The targetet generating functions more or less require that $Z=C\_k, T=-tC\_k^k $ in case (I), resp. that
$Z=A\_{k-1}, T=-t^{k-1}A\_{k-1}$ in case (II), this explains the appearance of these special inverse series.
| 3 | https://mathoverflow.net/users/48831 | 439349 | 177,438 |
https://mathoverflow.net/questions/416724 | 2 | Let $M$ be a 1-skeleton of a triangulation of a sphere with $V$ vertices and $E$ edges.
**Definition 1** A polyhedron is a map $M\to \mathbb R^3$ that is affine on edges (and non-degenerate on faces). The space of polyhedra is identified with (apparantly an open subset of) $\mathbb R^{3V}$.
**Definition 2** Two polyhedra are called isometric if they induce the same metric on $M$. We will denonote this fact by $[P]=[Q]$
**Definition 3** A polyhedron $P$ is called infinitesimally rigid if the kernel (at $P$) of the Jacobian of the natural map $f:\mathbb R^{3V}\to\mathbb R^E$ which sends $P$ to the tuple of square-lenghts of its edges has the minimal possible dimension equal to 6. In other words $P$ has no nontrivial infinitesimal deformations (or more algebraically: Zariski tangent space to the scheme $f^{-1}f(P)$ has dimension $6$ at $P$).
It is a well known result of Gluck that the set of infinitesimally rigid polyhedra $M\to \mathbb R^3$ form an open and dense set in $\mathbb{R}^{3V}$. However it can happen that in the set [P] of polyhedra isometric to a given infinitesimally rigid polyhedron $P$ there is a non-rigid representative. Is this phenomenon non-typical?
In other words, is the following true:
For a generic $P\in \mathbb R^{3V}$ (you may chose suitable definition) $[P]$ is 6-dimensional (that is, *all* polyhedra isometric to $P$ are infinitesimally rigid). To put it differently, after quotienting out by isometries of the euclidean space we have that the moduli space of a generic metric polyhedron is a finite union of points (without nilpotents).
| https://mathoverflow.net/users/13842 | Generic infinitesimal rigidity of polyhedra | In short, no, this cannot happen.
To be more specific. A standard definition for ``generic'' is that the coordinates of $P \in \mathbb{R}^{3V}$ form an algebraically independent set. It is a nice exercise proving that: (i) almost all (in the measure theory sense) points in $\mathbb{R}^{3V}$ are generic, and (ii) every generic polyhedra is infinitesimally rigid.
Bob Connelly proved in his paper *Generic global rigidity* (see Proposition 3.3) the following: given a generic $P \in \mathbb{R}^{3V}$ and $Q \in [P]$, if $P$ is infinitesimally rigid then so too is $Q$. (His result is stated in the more general setting of bar-joint frameworks and so is more general.) The result you need now just follows from basic differential geometry.
| 3 | https://mathoverflow.net/users/122002 | 439361 | 177,443 |
https://mathoverflow.net/questions/439116 | 3 | (This is a natural continuation of a [previous post](https://mathoverflow.net/questions/438594/).)
**I. Quintic method**
Given the Lehmer quintic,
$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1 = 0$$
From [this post](https://mathoverflow.net/questions/438701/), we saw its roots $x\_i$ can be ordered such that,
$$(x\_1^4\, x\_2^3\, x\_3^2\, x\_4)^{1/5} + (x\_2^4\, x\_3^3\, x\_4^2\, x\_5)^{1/5} + \dots + (x\_5^4\, x\_1^3\, x\_2^2\, x\_3)^{1/5} = 0$$
Or equivalently,
$$\frac1{x\_1}-\frac1{x\_1 x\_2}+\frac1{x\_1 x\_2 x\_3}-\frac1{x\_1 x\_2 x\_3 x\_4}+\frac1{x\_1x\_2x\_3x\_4x\_5} = 0$$
This ordering is *useful*, since using the **same** order of roots, it turns out that,
$$(x\_1\,x\_2^4\,x\_3^2\,x\_4^7)^{1/11} + (x\_2\,x\_3^4\,x\_4^2\,x\_5^7)^{1/11} + \dots + (x\_5\,x\_1^4\,x\_2^2\,x\_3^7)^{1/11} = z\_1$$
$$(x\_1^2\,x\_2\,x\_3^7\,x\_4^4)^{1/11} + (x\_2^2\,x\_3\,x\_4^7\,x\_5^4)^{1/11} + \dots + (x\_5^2\,x\_1\,x\_2^7\,x\_3^4)^{1/11} = z\_2$$
$$(x\_1^4\,x\_2^7\,x\_3\,x\_4^2)^{1/11} + (x\_2^4\,x\_3^7\,x\_4\,x\_5^2)^{1/11} + \dots +(x\_5^4\,x\_1^7\,x\_2\,x\_3^2)^{1/11} = z\_3$$
$$(x\_1^7\,x\_2^2\,x\_3^4\,x\_4)^{1/11} + (x\_2^7\,x\_3^2\,x\_4^4\,x\_5)^{1/11} + \dots +(x\_5^7\,x\_1^2\,x\_2^4\,x\_3)^{1/11} = z\_4$$
where the $z\_i$ are now roots of **four** different $11$-deg equations.
Notice that $z\_1$ and $z\_4$ are "complementary", with the $x\_1, x\_4$ of their starting terms just swapping exponents, likewise with the $x\_2, x\_3$. (The same can be said for $z\_2$ and $z\_3$).
---
**II. Example**
Let $n=-1$ and we have
$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$
with roots $x\_i = 2\cos\frac{2\pi k}{11}$ in the same order $k= 1,4,5,2,3.$ Using the four expressions above, these yield the four quintics,
\begin{align}
-1 + 529 y + 5361 y^2 + 756 y^3 - 377 y^4 &+ y^5\\
-1 - 439 y - 45701 y^2 - 5536 y^3 - 135 y^4 &+ y^5\\
-1 - 4806 y - 5771 y^2 - 1543 y^3 - 47 y^4 &+ y^5\\
-1 + 408 y + 2215 y^2 + 1724 y^3 - 740 y^4 &+ y^5
\end{align}
such that,
$$y\_1^{1/11}+y\_2^{1/11}+y\_3^{1/11}+y\_4^{1/11}+y\_5^{1/11} = z\_i$$
are roots of the four $11$-degree equations,
\begin{align}
-16954 - 2387 z + 3762 z^2 + 2200 z^3 + 704 z^4 & + 187 z^5 - 110 z^6 - 33 z^7 + 22 z^8 + z^{11}\\
9908 + 8987 z + 7029 z^2 + 4499 z^3 + 1188 z^4 & - 418 z^5 - 352 z^6 - 33 z^7 + 22 z^8 + z^{11}\\
7213 - 572 z - 1804 z^2 + 2563 z^3 + 1430 z^4 & - 418 z^5 - 352 z^6 - 33 z^7 + 22 z^8 + z^{11}\\
-18043 + 7777 z + 253 z^2 + 3168 z^3 - 385 z^4 & - 418 z^5 - 110 z^6 - 33 z^7 + 22 z^8 + z^{11}
\end{align}
Incidentally, the unexplained phenomenon of shared coefficients (also found in the cubic method) appears again.
---
\*\*III. Questions
1. For the quintic with one root "fixed", there are now $(n-1)! = 24$ permutations, creating 24 different quintics. Why is it there are only **four** useful ones (*I checked*) that can solve an $11$-deg equation?
2. And what would be one ($a,b,c,d)$ such that $(x\_1^a\,x\_2^b\,x\_3^c\,x\_4^d)^{1/p}$ and its Galois conjugates can be used to solve $p=31$?
| https://mathoverflow.net/users/12905 | Using the Lehmer quintic to solve $11$-degree equations and higher? | **Question 1:**
An irreducible Lehmer quintic $L(n)$ has Galois group ${\bf Z} / 5 {\bf Z}$.
Let $\sigma$ be a generator, and order the roots $x\_i$ ($i \bmod 5$) so that
$x\_{i+1} = \sigma(x\_i)$ for each $i$.
Note that $\prod\_i x\_i = -1$ because $L(n)$ has constant coefficient $1$.
Thus $\prod\_i x\_i^{a\_i} = (-1)^c \prod\_i x\_i^{a\_i+c}$ for any integer $c$.
You chose $c$ that makes $a\_0 = 0$, but other choices may be more useful.
For sums of (1/11)th powers, taking $c=6$ converts the exponents $(4,2,7,0,1)$
to $(10,8,2,6,7)$, proportional to the 5th roots of unity $\bmod 11$
(namely $c\_i \equiv -3^i \bmod 11$ for $i=0,1,2,3,4$).
The four choices of exponents correspond to the four isomorphisms from
the Galois group to the group of 5th roots of unity $\bmod 11$.
**Question 2:**
For (1/31)st powers we likewise need 5th roots of unity $\bmod 31$, which are
$(1,2,4,8,16)$ (or ($0,1,3,7,15$) if you insist on having a zero exponent).
Again there are four choices, again with "shared coefficients" including
some zero coefficients, such as
```
x^31 + 744*x^26 - 620*x^25 - 4960*x^24 - 16058*x^23 + 27528*x^22 + 186372*x^21 + 13392*x^20 + 57908*x^19 - 2151400*x^18 + 6700526*x^17 + 6967374*x^16 + 33644796*x^15 - 86990712*x^14 - 125974204*x^13 - 244592976*x^12 + 559706116*x^11 + 1329731980*x^10 + 1937968348*x^9 + 760060790*x^8 - 615794292*x^7 - 539362428*x^6 + 376540942*x^5 + 474549116*x^4 + 105617372*x^3 - 15563860*x^2 - 3550678*x + 645413
```
(found numerically by computing to high precision and applying GP's **aldgep**).
| 3 | https://mathoverflow.net/users/14830 | 439364 | 177,446 |
https://mathoverflow.net/questions/439324 | 2 | I played around with the Fourier series of the Eisenstein series resp. divisor sums and did some calculations, see below. Although the deduction is ***not rigorous / wrong*** (as the power series for the Bernoulli numbers is not convergent), I wonder, why the final result - at least in the area $0<Im(z)\leq \frac{1}{2}$ near the imaginary axis - is still so good? (Even the correction term in case of $E\_2$, i.e. for $m=1$, is matched!).
\begin{equation}
\boxed{\sum\_{n=1}^{\infty}{\sigma\_m(n)\,e(nz)}\approx\frac{B\_{m+1}}{2(m+1)}\left(1-\frac{1}{z^{m+1}}\right)+\left[m=1\right]\frac{1}{2}\frac{1}{2\pi iz}}
\end{equation}
*Can someone give a reason for this behavior, or can someone calculate the error term explicitly?*
For odd $m\in\mathbb{N}$:
\begin{equation\*}
\begin{aligned}
\sum\_{n=1}^{\infty}{\sigma\_m(n)\,e(nz)}&=\sum\_{n=1}^{\infty}{\left(\zeta(m+1)\,n^m\sum\_{q=1}^{\infty}{\frac{c\_q(n)}{q^{m+1}}}\right)e(nz)}\\
&=\zeta(m+1)\sum\_{q=1}^{\infty}{\frac{1}{q^{m+1}}\sum\_{n=1}^{\infty}{\sideset{}{^\star}\sum\limits\_{a\leq q}{n^m\,e\left(z+\frac{a}{q}\right)^n}}}\\
&=\zeta(m+1)\sum\_{q=1}^{\infty}{\frac{1}{q^{m+1}}\sideset{}{^\star}\sum\limits\_{a\leq q}\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\left(\frac{1}{e\left(z+\frac{a}{q}\right)-1}\right)}\\
&=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum\_{q=1}^{\infty}{\frac{1}{q^{m+1}}\frac{1}{2\pi iz}\sum\_{d|q}{\mu\left(\frac{q}{d}\right)}\frac{2\pi i dz}{e(dz)-1}}\\
&=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum\_{q=1}^{\infty}{\frac{1}{q^{m+1}}\frac{1}{2\pi iz}\sum\_{n=0}^{\infty}{\frac{B\_n}{n!}\left(2\pi i z\right)^n
\sum\_{d|q}{\mu\left(\frac{q}{d}\right)d^n}}}\\
&=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum\_{q=1}^{\infty}{\frac{1}{q^{m+1}}\frac{1}{2\pi iz}\sum\_{n=0}^{\infty}{\frac{B\_n}{n!}\left(2\pi i z\right)^n
J\_n(q)}}\\
&=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum\_{n=0}^{\infty}{\frac{B\_n}{n!}\left(2\pi i z\right)^{n-1}\sum\_{q=1}^{\infty}{\frac{J\_n(q)}{q^{m+1}}}}\\
&=\zeta(m+1)\left(-\frac{1}{2\pi i}\right)^m\frac{\operatorname{d}^m}{\operatorname{d}z^m}\sum\_{n=0}^{\infty}{\frac{B\_n}{n!}\left(2\pi i z\right)^{n-1}\frac{\zeta(m+1-n)}{\zeta(m+1)}}\\
&=-\sum\_{n=0}^{\infty}{\left(\frac{B\_n}{n!}\left(2\pi i z\right)^{n-1-m}\zeta(m+1-n)\prod\_{k\leq m}{\left(n-k\right)}\right)}\\
&=\frac{m!}{\left(2\pi iz\right)^{m+1}}\zeta(m+1)-\frac{B\_{m+1}}{(m+1)!}\zeta(0)m!+\left[m=1\right]\frac{1}{2}\frac{1}{2\pi iz}\\
&=\frac{B\_{m+1}}{2(m+1)}\left(1-\frac{1}{z^{m+1}}\right)+\left[m=1\right]\frac{1}{2}\frac{1}{2\pi iz}
\end{aligned}
\end{equation\*}
Notation: $c\_q(n)$ is [Ramanujan's sum](https://en.wikipedia.org/wiki/Ramanujan%27s_sum), $J\_n(q)$ [Jordan's totient function](https://en.wikipedia.org/wiki/Jordan%27s_totient_function), $B\_n$ the [Bernoulli numbers](https://en.wikipedia.org/wiki/Bernoulli_number), $\sideset{}{^\star}\sum\limits\_{a\leq q}{}$ means $\sum\limits\_{\substack{a\leq q\\\gcd(a,q)=1}}{}$, $\left[~\right]$ is the [Iverson bracket](https://en.wikipedia.org/wiki/Iverson_bracket) and $e(z)=e^{2\pi iz}$ as usual.
Numerical Approximations:
Set $f\_m(z):=\sum\_{n=1}^{\infty}{\sigma\_m(n)\,e(nz)},~~~$ $g\_m(z):=\frac{B\_{m+1}}{2(m+1)}\left(1-\frac{1}{z^{m+1}}\right)+\left[m=1\right]\frac{1}{2}\frac{1}{2\pi iz}$
\begin{equation\*}
\begin{aligned}
f\_1\left(\frac{i}{2}\right)&=0.04916444073\\
g\_1\left(\frac{i}{2}\right)&=0.04917839024\\~\\
f\_3\left(\frac{i}{3}\right)&=0.3333338608\\
g\_3\left(\frac{i}{3}\right)&=0.3333333333\\~\\
f\_5\left(\frac{i}{3}-\frac{1}{10}\right)&=-0.1956996617 - 1.099266745\*I\\
g\_5\left(\frac{i}{3}-\frac{1}{10}\right)&=-0.1957162803 - 1.099271842\*I\\~\\
f\_7\left(\frac{i}{\sqrt{7}}+\frac{1}{5}\right)&=-1.365047865 - 1.273191433\*I\\
g\_7\left(\frac{i}{\sqrt{7}}+\frac{1}{5}\right)&=-1.363026857 - 1.272859570\*I\\~\\
\end{aligned}
\end{equation\*}
Another surprising fact is that $g\_1(z),\,g\_3(z)$ and $g\_5(z)$ satisfy [Ramanujan's differential equations](https://en.wikipedia.org/wiki/Eisenstein_series#Ramanujan_identities) for $L(q),\,M(q),\,N(q)$ (as usual $q=e(z)$) as well!
By the way, the corresponding result for Eisenstein series is
\begin{equation}
\boxed{\sum\_{\substack{n,k\in\mathbb{Z}\\\left(n,k\right)\neq\left(0,0\right)}}^{\infty}{\frac{1}{\left(nz+k\right)^{m+1}}}\approx-\left(\frac{2\pi i}{z}\right)^{m+1}\frac{B\_{m+1}}{(m+1)!}+\left[m=1\right]\frac{1}{2z}}
\end{equation}
Best,
M.
| https://mathoverflow.net/users/108459 | Fourier series of Eisenstein series — elegant and very good approximation | As Paul Garrett says, this reflects (for $k \geq 4$ even) the modularity of the Eisenstein series $E\_k(z) = -\frac{B\_k}{2k} + \sum\_{n=1}^{\infty} \sigma\_{k-1}(n) e(nz)$ with respect to the matrix $S = (\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix})$, which sends $z$ to $-1/z$ (so it preserves the imaginary axis and swaps $0$ and $i\infty$). For $m \geq 3$ odd, writing $E\_{m+1} |\_{m+1} S = E\_{m+1}$ you get an asymptotic expansion of $E\_{m+1}(z)$ at $z=0$, which agrees with what you wrote. For $m=1$ the Eisenstein series $E\_2$ is only quasi-modular, but you can also work out the asymptotic expansion.
The integer $m$ should be odd, because $E\_k(z)$ has no modularity property whatsoever for $k$ odd.
| 2 | https://mathoverflow.net/users/6506 | 439368 | 177,447 |
https://mathoverflow.net/questions/439352 | 5 | I read that, with the Scott topology, suprema of sequences are topological limits (See page 1 of [this article](https://akjournals.com/view/journals/10473/88/1-2/article-p35.xml)).
Let $(X, \le)$ be a DCPO, and $D$ be a directed subset of $X$.
I can easily see that the identity net on $D$ converges to $\bigvee D$.
How about the other direction? If the identity net on $D$ converges to $x$, does it imply that $x = \bigvee D$?
| https://mathoverflow.net/users/55915 | Scott topology: Suprema of sequences are topological limits | $\newcommand\LL{\mathcal L}$**Brief answer:** The Scott topology is not Hausdorff, and therefore we have to deal here with the set of limits, rather than with the (unique) limit. Here, for the set $\LL\_D$ of limits of the identity net on $D$ we have
\begin{equation\*}
\LL\_D=L\_s:=\{y\in X\colon x\le s\},
\end{equation\*}
where
\begin{equation\*}
s:=s\_D:=\sup D=\bigvee D.
\end{equation\*}
One can similarly see that the statement "Suprema of sequences are also topological limits, i.e., $\sup x\_n=x \iff \lim x\_n=x$" in the paper linked in your post is incorrect in general (and almost always). E.g., if $X=(-\infty,0]$ with the usual order and $x\_n=0$ for all $n$, then $\sup x\_n=0$, whereas the set of limits of $(x\_n)$ in the Scott topology is the entire set $X=(-\infty,0]$.
---
**Details:**
First, a review of definitions.
Let $X$ be a set with a partial order $\le$.
* A point $b\in X$ is an *upper bound* of a subset $A$ of $X$ if $x\le b$ for all $x\in A$.
* A nonempty subset $D$ of $X$ is *directed* if any finite subset of $D$ has an upper bound in $D$.
* A point $s\in X$ is the *supremum* of a subset $A$ of $X$ and denoted by $\sup A$ or $\bigvee A$ if $s$ is the least upper bound of $A$, that is, if $s$ is an upper bound of $A$ and for any upper bound $b$ of $A$ we have $s\le b$.
* A subset $A$ of $X$ is an *upper set* if for any $x\in A$ we have $U\_x\subseteq A$, where $U\_x:=\{y\in X\colon x\le y\}$.
* A subset $A$ of $X$ is a *lower set* if for any $x\in A$ we have $L\_x\subseteq A$, where $L\_x:=\{y\in X\colon y\le x\}$.
* The pair $(X,\le)$ of a set $X$ with a partial order $\le$ is a *directed-complete partial order (dcpo)* if each directed subset $D$ of $X$ has a supremum. In what follows, $(X,\le)$ will be a dcpo.
* A subset $G$ of $X$ is *open in the Scott topology*, or *Scott-open*, if $G$ is an upper set and for any directed subset $D$ of $X$ such that $\sup D\in G$ we have $D\cap G\ne\emptyset$. Equivalently, a subset $F$ of $X$ is *closed in the Scott topology*, or *Scott-closed*, if $F$ is a lower set and for any directed subset $D$ of $F$ we have $\sup D\in F$.
* The identity net $I\_D$ on a directed subset $D$ of $X$ is the map $D\ni x\mapsto I\_D(x):=x\in X$.
* The set $\LL\_D$ of limits of the identity net $I\_D$ is defined by the following condition on $x\in X$:
\begin{equation\*}
\begin{aligned}
x\in\LL\_D\iff &\text{for any Scott-open $G$ such that $x\in G$ } \\
&\text{there is some $z\_G\in D$ such that $U\_{z\_G}\subseteq G$.}
\end{aligned}
\end{equation\*}
---
Let now $D$ be any directed subset of $X$.
Suppose that $x\in\LL\_D$.
We want to show that then $x\in L\_s$. Suppose the contrary. Then $x\in V:=X\setminus L\_s$, and the set $V$ is Scott-open (since the set $L\_x$ is Scott-closed for any $x\in X$). On the other hand, $z\_V\in D$ and $s=\sup D$, so that $z\_V\le s$, that is, $s\in U\_{z\_V}$. Also, by the definition of $\LL\_D$, we have $U\_{z\_V}\subseteq V$. So, $s\in U\_{z\_V} \subseteq V=X\setminus L\_s$, which contradicts the obvious fact that $s\in L\_s$.
We see that $x\in L\_s$ for any $x\in\LL\_D$, that is,
\begin{equation\*}
\LL\_D\subseteq L\_s. \tag{1}\label{1}
\end{equation\*}
Let us now show that $s\in\LL\_D$. Take any any Scott-open $G$ such that $s\in G$. Since $G$ is Scott-open and $s=\sup D\in G$, we have $D\cap G\ne\emptyset$. Take now any $z\_G\in D\cap G$. Since $z\_G\in G$ and the Scott-open set $G$ is an upper set, we get $U\_{z\_G}\subseteq G$. Since $z\_G\in D$, we see that indeed $s\in\LL\_D$.
Note also that for any $x\in\LL\_D$ we have $L\_x\subseteq\LL\_D$. Therefore and because $s\in\LL\_D$, we see that $L\_s\subseteq\LL\_D$.
Now, in view of \eqref{1}, we conclude that
\begin{equation\*}
\LL\_D=L\_s=L\_{\sup D},
\end{equation\*}
as claimed.
| 7 | https://mathoverflow.net/users/36721 | 439376 | 177,450 |
https://mathoverflow.net/questions/439372 | 11 | Let $A$ be a set of cardinality $4n$. We define a pairing in $A$ to be a partition of $A$ into sets of cardinality $2$. What is the minimum number of pairings in $A$ such that every subset of $A$ of cardinality $4$ is the union of two pairs from at least one pairing?
This question is motivated by the computational problem of producing such a set of pairings. An answer may be a proof that the minimum number of pairings with this property is or is not $\mathcal O (n^2)$.
| https://mathoverflow.net/users/66833 | Minimum number of pairings that make all quadruples | It is a nice exercise in number theory.
Let $p$ be a prime slightly above $4n$. Let $1\le x<y<z<t\le 4n$.
Note that $0<(y+t)-(x+z)<4n<p$.
Now, there exists $a\in \mathbb Z\_p$ such that $(x+a)(z+a)=(y+a)(t+a)$ in $\mathbb Z\_p$ (this equation only pretends to be quadratic; in fact it is linear in $a$ with non-zero (in $\mathbb Z\_p$) coefficient $(y+t)-(x+z)$ at $a$). Also, we cannot have both parts $0$ simultaneously. Thus, if we consider $p^2-p$ pairings $P\_{a,b}$ in $\mathbb Z\_p\setminus\{-a\}$ given by $u\sim v\Longleftrightarrow (u+a)(v+a)=b$, we will have $x\sim z$ and $y\sim t$ simultaneously for some $a\in\mathbb Z\_p, b\in\mathbb Z\_p^\*$. To reduce $P\_{a,b}$ to $[1,4n]$, just make all pairs in $P\_{a,b}$ for which both $u,v$ are in the range and then pair the remaining set of numbers of even cardinality in any way you want.
That seems pretty economical. If $N=4n$, then it is essentially $N^2$ while the trivial lower bound is roughly speaking $(N^4/24)/((N/2)^2/2)=N^2/3$. You can try to fight for this $3$, of course, but without me :-)
| 18 | https://mathoverflow.net/users/1131 | 439377 | 177,451 |
https://mathoverflow.net/questions/438711 | 2 | I'm searching for some references about groups of invariance of the Painlevé property other than the book of Robert Conte or more generally birational transformation of Riemann surfaces.
| https://mathoverflow.net/users/497573 | References for group of invariance of the Painlevé property | I assume "the book of Robert Conte" refers to *The Painlevé Handbook* by Robert Conte and Micheline Musette (Springer, 2008); the "groups of invariance of the Painlevé property" are briefly described in Appendix A of that book.
The book by Harold T. Davis *Introduction to Nonlinear Differential and Integral Equations* (Dover, 1962) provides further background.
In this connection, the book by Einar Hille, *Ordinary Differential Equations in the Complex Domain* (Wiley, 1976; Dover reprint 1997) gives a different treatment and also provides historical references.
| 0 | https://mathoverflow.net/users/106467 | 439382 | 177,454 |
https://mathoverflow.net/questions/439220 | 1 | Let $A$ be a unital $C^{\*}$-algebra and $a \in A$ be normal, with spectrum $\sigma(a)$. Let $B = C^{\*}(a)$ be the $C^{\*}$-algebra generated by $1$ and $a$, which is abelian. Let $\hat{B}$ be the space of all homomorphisms $\chi: B \to \mathbb{C}$ with the weak\* topology, which makes it a compact Hausdorff topological space.
Gelfand Isomorphism Theorem states that $\varphi: B \to C(\hat{B})$ is an isometric $\*$-isomorphism. Moreover, $\hat{a}: \hat{B} \to \sigma(a) \subset \mathbb{C}$ is a homeomorphism, so that $\hat{B} \cong \sigma(a)$.
On one hand, given a positive functional $\omega: C(\hat{B}) \to \mathbb{C}$, Riesz Representation Theorem states that there exists a unique regular Borel measure $\mu$ on $\sigma(a)$ such that:
$$\omega(f) = \int\_{\hat{B}}f(\chi)d\mu\_{\omega}(\chi).$$
On the other hand, given a function $g \in C(\sigma(a))$, there is a natural identification $g \mapsto g\circ \hat{a}$ from $C(\sigma(a))$ to $C(\hat{B})$. Hence, each $\mu\_{\omega}$ induces a new measure $\nu\_{\omega}$ on $\sigma(a)$ by using $\hat{a}^{-1}$ as a push-forward.
By the abstract change of variables formula, it must hold:
$$\int\_{\hat{B}} (g\circ \hat{a})(\chi)d\mu\_{\omega}(\chi) = \int\_{\sigma(a)}g(z)d\nu\_{\omega}(z) \tag{1}\label{1}$$
My question is: can we extend this analysis to $A$ instead of $B$ (and $\hat{A}$ instead of $\hat{B}$ and so on) when $A$ is abelian? I don't think so because, in this case, $\hat{a}$ is not a homeomorphism between $\hat{A}$ and $\sigma(a)$, but I couldn't tell if a general version of formula (\ref{1}) holds from the literature I know on the topic.
| https://mathoverflow.net/users/152094 | Spectral theorem for unital $C^{*}$-algebras | To see that (1) continues to hold when B is replaced by A,
it suffices (by linearity and density of step functions) to consider only the case when g is the characteristic function of some measurable subset S of σ(a).
Indeed, in this case the right side equals the ν-measure of S, whereas the left side equals the μ-measure of the â-preimage of S. Since ν was defined as the â-pushforward of μ, this equality holds by definition.
| 1 | https://mathoverflow.net/users/402 | 439388 | 177,456 |
https://mathoverflow.net/questions/424813 | 0 | Let $B(t, \omega)$ be a Brownian motion defined on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, adapted to a filtration $\{\mathcal{F}\_t\}$. Let $\phi(t, \omega)$ be a $\{\mathcal{F}\_t\}$-adapted quadratic variation process such that the Ito integral
$$
Z(t, \omega) := \int\_0^t \phi(s, \omega) d B(s, \omega)
$$
is a local martingale.
My question: is the (path) distribution/pushforward measure $\mathbb{P}\circ Z^{-1}$ absolutely continuous with respect to the distribution $\mathbb{P} \circ B^{-1}$ of the Brownian motion? If so, how to prove? If not, do we have a concrete counter example of $\phi$?
| https://mathoverflow.net/users/170508 | Distribution of local martingale is absolutly continuous to that of the Brownian motion? | Let $ B $ be a Brownian Motion, $\mu\_{B}=\mathsf{P}\circ B^{-1} $ and $ Z $ be a continuous local martingale, $ \mu\_{Z}=\mathsf{P}\circ Z^{-1}$.
Then $ \mu\_{Z}\ll \mu\_{B} $ if and only if $ \mu\_{Z}=\mu\_{B} $.
In this case, if $ Z $ may be expressed as Ito stochastic integral of a predicatble $ \phi $ with respect to $ B $:
\begin{equation\*}
Z(t,\omega)=\int\_0^t \phi(s,\omega)\,\mathrm{d}B(s,\omega)
\end{equation\*}
Then processes $ |\phi|=(|\phi(s,\omega)|,s\ge 0) $ and 1 are indistinguishable, i.e., for almost all $ \omega $ trajectories $ \phi(\cdot,\omega)\equiv 1 $.
Now we prove above facts. Since $ Z $ and $ B $ are continuous local martingale, for its predictable variation process $ \langle Z,Z\rangle$ and $\langle M,M\rangle $ the following facts hold, (cf. D. Revuz, M. Yor, *Continuous Martingales and Brownian Motion*, Corrected 3rd Ed. Springer, 2005, p124. Th.4.1.8.)
\begin{gather\*}
\textrm{pr}-\lim\_{n\to\infty}\sum\_{k=1}^{2^n}\Big[Z\Big(\frac{kt}{2^n}\Big)- Z\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2=\langle Z,Z\rangle\_t,\qquad \forall t>0.\tag{1}\\
\textrm{pr}-\lim\_{n\to\infty}\sum\_{k=1}^{2^n}\Big[B\Big(\frac{kt}{2^n}\Big)- B\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2=t,\qquad \forall t>0. \tag{2}
\end{gather\*}
(2) is equivalent to the following,
\begin{equation\*} \lim\_{n\to\infty}\mu\_{B}\Big(\Big|\sum\_{k=1}^{2^n}\Big[x\Big(\frac{kt}{2^n}\Big)- x\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2 - t \Big|>\epsilon\Big)=0,\quad \forall \epsilon >0,\quad \forall t>0. \tag{3}
\end{equation\*}
Due to $ \mu\_{Z}\ll \mu\_{B} $ and (3), as $ n\to\infty $,
\begin{align\*}
&\mathsf{P}\Big(\Big|\sum\_{k=1}^{2^n}\Big[Z\Big(\frac{kt}{2^n}\Big)- Z\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2-t \Big|>\epsilon\Big)\\
&\quad =\mu\_{Z}\Big(\Big|\sum\_{k=1}^{2^n}\Big[x\Big(\frac{kt}{2^n}\Big)- x\Big(\frac{(k-1)t}{2^n}\Big)\Big]^2 - t \Big|>\epsilon\Big)\\
& \quad \longrightarrow 0,\quad \forall \epsilon >0,\quad \forall t>0. \tag{4}
\end{align\*}
comparing (1) and (4), get
\begin{gather\*}
\mathsf{P}(\langle Z,Z\rangle\_t=t)=1, \quad\forall t>0,\\
\mathsf{P}(\langle Z,Z\rangle\_t=t, \forall t>0 )=1 \tag{5}
\end{gather\*}
Now from the Lévy's characterization of Brownian Motion, $Z$ is a Brownian motion and
$\mu\_Z=\mu\_{B} $.
Remark: The conclusion in following book is also useful: C. Dellacherie & P. Meyer, *Probabilities and Potential B*, volume 72 of *North-Holland Mathematics Studies*.
North-Holland, Amsterdam, 1982.
| 3 | https://mathoverflow.net/users/103256 | 439389 | 177,457 |
https://mathoverflow.net/questions/439366 | 2 | Does there exist a sequence of strongly regular graphs with parameters $(n,d,\lambda,\mu)$ (so every pair of adjacent vertices have $\lambda$ common neighbours, and every pair of non-adjacent ones have $\mu$ common neighbours), with $d - \mu = \Omega(n)$ and $\mu - \lambda = \Omega(n)$?
| https://mathoverflow.net/users/141963 | Strongly regular graphs with certain parameters | I think, the answer is negative (I expect that by $\Omega(n)$ you mean something positive of order $n$). Denote $\mu-\lambda=k,d-\mu=\ell$.
The eigenvalues of the adjacency matrix of such a graph $G$ are $d$ and $t\_{1,2}$ which are the roots of a quadratic equation $t^2+kt-\ell=0$. The multiplicities $n\_{1},n\_2$ enjoy the equations $n\_1+n\_2=n-1$, $n\_1t\_1+n\_2t\_2=-d$. Consider two cases.
1. $n\_1=n\_2$. Then $n\_1=n\_2=(n-1)/2$ and $-d=(t\_1+t\_2)\frac{n-1}2=-k\frac{n-1}2$ which is too large (of order $n^2$).
2. $n\_1\ne n\_2$. Then from $t\_1+t\_2=-k$ and $n\_1t\_1+n\_2t\_2=-d$ we get $t\_2(n\_2-n\_1)=kn\_1-d$ and so $t\_2$ is rational, i.e., the discriminant $k^2+4\ell$ of our quadratic equation is a perfect square, let it be equal to $k^2+4\ell=(k+2x)^2$, then $x^2+kx=\ell$. It yields that $x$ is bounded. So, the roots are $t\_2=x$ and $t\_1=-k-x$. And $n\_1t\_1+n\_2t\_2=-d$ reads as $n\_2x+(n-1-n\_2)(-k-x)=-d$, so $n\_2(k+2x)=(n-1)(k+x)-d$ and $n\_2=(n-1)-(n-1)\cdot\frac{x}{k+2x}-\frac{d}{k+2x}=n-O(1)$. In other words, the matrix $A-xI$ has bounded rank. But by Ramsey theorem it contains a large minor with $-x$ along diagonal and either 0's or 1's outside diagonal. This minor does not have full rank only if $x\in \{0,-1\}$ which is impossible.
| 4 | https://mathoverflow.net/users/4312 | 439390 | 177,458 |
https://mathoverflow.net/questions/439359 | 8 | Let $(\mathbb{R}^2,g)$ be a complete Riemannian manifold. Let $K\subset \mathbb{R}^2$ be a compact, connected set, and let $\text{conv}(K)$ be its convex hull, i.e., the intersection of all geodesically-convex sets containing $K$.
Is $\text{conv}(K)$ bounded? Compact?
If it helps, I am interested in cases where the curvature of $g$ has both signs, but is flat outside of $K$.
| https://mathoverflow.net/users/97528 | Convex hulls of compact sets in a 2-manifold | *The convex hull of a compact set $K \subset M^2$ in a complete manifold need neither be bounded, nor closed.*
Both counterexamples are rotationally symmetric, and the second has a Euclidean metric outside of a compact region.
To prove that a convex hull need not be bounded, consider a sphere with an infinitely long and thin spike attached at the north pole. (This is diffeomorphic to $\mathbf{R}^2$, and can be done with a complete metric.) For our compact set $K$ we take a loop $\gamma$ enclosing the spike, say at height $z = 10$. Any convex set containing $K$ also contains the strip $\{ 10 \leq z < 10 + \epsilon \}$. However, no region of the form $\{ 10 \leq z < h \}$ with $h > 10$ is convex. Therefore any convex set containing $K$ must also contain the whole spike, making it unbounded.
To find an example where $\operatorname{conv} K$ is not closed we construct a rotationally symmetric metric on $M = \mathbf{R}^2 \subset \mathbf{R}^3$ by gluing in a 'mushroom' with a thin waist $\gamma\_1$ at height $z = 1$, of radius one for example, above which there lies a larger sphere, say of radius five. Outside of the unit disc $D \subset \mathbf{R}^2$, the Euclidean metric is unchanged.
*Claim.* The convex hull of $K = \{ z = 1/2 \}$ is $\{ 1/2 \leq z < 1 \}$.
*Proof.*
Let $C$ be the convex hull of the set. This contains $\{ 1/2 \leq z < 1/2 + \epsilon \}$ for some small $\epsilon > 0$. None of the sets $\{ 1/2 \leq z < h \}$ with $h < 1$ is geodesically convex, so $\{ 1/2 \leq z < 1 \} \subset C$.
To show that $\{ 1/2 \leq z < 1 \}$ is geodesically convex, take two points $x,y$ lying in it, and let $\gamma$ be a minimizing geodesic connecting the two. This has length at most $4 \pi$, say. By preservation of angular momentum—Clairaut's relation—, if $\gamma$ crossed the waist, then it would cross into the large sphere, thus making it longer than allowed.
Still by Clairaut's relation, $\gamma$ has angular momentum strictly larger than that of the waist $\gamma\_1$. (It cannot have angular momentum equal to it, because that would make it an unbounded geodesic converging to $\gamma\_1$.) Therefore $\max z(\gamma) < 1$. Q.E.D.
| 5 | https://mathoverflow.net/users/103792 | 439391 | 177,459 |
https://mathoverflow.net/questions/439283 | 1 | Let $M$ be a surface with boundary and let $f\_t: M \to M, t \in [0,1]$ be a differentiable family of diffeomorphisms (I think this is usually called a diffeotopy). Suppose I have a Liouville form $\lambda\_0 \in \Omega\_1(M)$ such that $f\_0$ is exact, i.e. $f\_0^\*\lambda\_0 - \lambda\_0 = dF$ for some function $F: M \to \mathbb{R}$. Is it always possible to find a differentiable family of Liouville forms $\lambda\_t$ such that $f\_t$ is exact for all $t \in [0,1]$?
| https://mathoverflow.net/users/173545 | Deformation of a Liouville form with a diffeotopy | The statement is not true. For a function to be exact, it needs to preserve the symplectic form $d\lambda$. This puts strong conditions on the function, for instance it's differential should have determinant $1$ at fixed points. This is clearly not preserved in a diffeotopy.
| 0 | https://mathoverflow.net/users/173545 | 439401 | 177,462 |
https://mathoverflow.net/questions/439397 | 2 | Let $X$ be a compact metric space and $\mathcal{M}(X)$ be the space of variational-bounded, signed Borel measures equipped with the Kantorovich–Rubinshtein norm, cf. [Section 8.3, 1]:
$$||\mu||\_0:= \sup\Bigg\{\int\_X fd \mu: f \in \mathrm{Lip}\_1(X), \sup\_{x \in X}|f(x)|\leq 1 \Bigg\}$$
where $\mathrm{Lip}\_1(X)$ deontes the set of Lipschitz functions on $X$ with Lipschitz constant smaller or equal to $1$.
Let $K\subseteq X$ be closed (with nonemtpy interior) and $M>0$ be a constant. Are the sets
$$\{\mu \in \mathcal{M}(X) \mid \mu(K)< M\} \text{ and } \{\mu \in \mathcal{M}(X) \mid |\mu|(K)< M\}$$
open in $\mathcal{M}(X)$?
Remark: Bogachev shows that the topology induced by the KR-norm on the set of all nonnegative Borel-measures $\mathcal{M}\_+(X)$ coincides with the topology of weak convergence. Hence, one can verify that the above sets restricted to $\mathcal{M}\_+(X)$ are open in $\mathcal{M}\_+(X)$.
References: [1]
*Bogachev, V. I.*, Measure theory. Vol. I and II, Berlin: Springer (ISBN 978-3-540-34513-8/hbk). xvii, 500 p./v.1; xiii, 575 p./v.2. (2007). [ZBL1120.28001](https://zbmath.org/?q=an:1120.28001).
| https://mathoverflow.net/users/498220 | Open sets in the space of signed measures equipped with the Kantorovich–Rubinshtein norm | No, those sets are not open. Indeed, take any non-isolated point $x$ in $X$ and a sequence $(x\_n)\_{n\in\omega}$ that converges to $x$. For every $n$, consider the sign measure $\mu\_n=\delta\_{x}-\delta\_{x\_n}$, where $\delta\_p$ is the Dirac measures at a point $p\in X$. The definition of the Kantorovich-Rubinshtein norm guarantees that $\|\mu\_n\|\_0$ tends to zero as $n\to\infty$. For $M=\frac 12$ and $K=\{x\}$, the zero measure belongs to the sets $\{\mu\in\mathcal M(X):|\mu|(K)<M\}\subseteq\{\mu\in\mathcal M(X):\mu(K)<M\}$ but for every $n\in\omega$ the measure $\mu\_n$ does not belong to those two sets, which implies that these sets are not open in $\mathcal M(X)$.
| 6 | https://mathoverflow.net/users/61536 | 439404 | 177,463 |
https://mathoverflow.net/questions/439350 | 0 | Let $R$ be a finite-dimensional irreducible representation of $U(N)$, with the set of weights $W\_R$. Each element of $W\_R$ is a vector of length $N$ with integer entries.
Firstly, I would like to know if it is true that $$\sum\_{\mu\in W\_R}m\_\mu\mu=A(R)(1,\ldots,1),$$ where $m\_\mu$ is the multiplicity of the weight $\mu$ and $A(R)$ is some integer that varies with the representation. Note that we are considering $U(N)$ rather $SU(N)$, because I think the sum of weights of an irrep vanishes for $SU(N)$.
Secondly, I would like to know if there is a simple formula for $A(R)$, or if there is an existing name for it?
| https://mathoverflow.net/users/122036 | Sum of weights of an irreducible representation of $U(N)$ | As discussed in the [comments](https://mathoverflow.net/questions/439350/sum-of-weights-of-an-irreducible-representation-of-un#comment1133244_439350), your sum is a Weyl-fixed character, so trivial for $G = \operatorname{SU}(N)$ and a multiple of $\det = (1, \dotsc, 1)$ for $\operatorname U(N)$.
To be concrete, as I guessed in the [comments](https://mathoverflow.net/questions/439350/sum-of-weights-of-an-irreducible-representation-of-un#comment1133257_439350), one sees that your sum $\sum\_\mu m\_\mu\mu$ (the sum taking place in the character lattice $X^\*(T)$ of the (implicitly chosen) maximal torus $T$, not in $\mathbb C[X^\*(T)]$ as in the Weyl character formula) is precisely the character of $\det \circ R$. Again, since $\det \circ R$ is Weyl invariant and $\det$ spans the Weyl-invariant part of the character lattice, your integer $A(R)$ is precisely the integer $n$ such that we have $\det \circ R = \det(\cdot)^n$.
Inspired by [your comment](https://mathoverflow.net/questions/439350/sum-of-weights-of-an-irreducible-representation-of-un#comment1133302_439350), I realise we can be a little more explicit. (On having written this, I realise that it was actually *exactly* what you were saying; I mistook your $\lambda\_i$ for an indexing of weights, rather than a component of $\lambda$. Oops, sorry!)
Let $\lambda$ be the highest weight of $R$. Then all weights $\mu$ of $R$ agree on the centre of $\operatorname U(N)$ with $\lambda$. Specifically, they all act as $z I\_N \mapsto z^\ell$ for some integer $\ell$. (If we think of $\lambda$ as an element of $\mathbb Z^N$, then $\ell$ is the sum of the components.) Thus, we have that $R$ agrees on the centre with $z I\_N \mapsto z^\ell I\_{\dim(R)}$, so $\det \circ R$ agrees on the centre with $z I\_N \mapsto \det(z^\ell I\_{\dim(R)}) = z^{\ell\dim(R)} = \det(z I\_N)^{\ell\dim(R)/N}$, so $A(R)$ equals $\ell\dim(R)/N$. As you [point out](https://mathoverflow.net/questions/439350/sum-of-weights-of-an-irreducible-representation-of-un#comment1133302_439350), we can use the [Weyl dimension formula](https://en.wikipedia.org/wiki/Weyl_character_formula#Weyl_dimension_formula) to compute $\dim(R)$ in terms of $\lambda$ if desired.
| 2 | https://mathoverflow.net/users/2383 | 439408 | 177,465 |
https://mathoverflow.net/questions/439400 | 0 |
>
> Let $f \in H\_{0}^{1}(0,1)$ and $\lambda >0$ big enough. Consider $0 <\alpha < 1$ and some $k > 0$. I would like to show the following inequality
> $$
> \int\_{\lambda^{-k}}^{1}|f(x)|^{2}dx \leq C\lambda^{-p}\int\_{\lambda^{-k}}^{1}x^{\alpha}|f^{\prime}(x)|^{2}dx
> $$
> for some $p> 0$. Here, $C> 0$ is constant.
>
>
>
* I want to use the following Hardy's Inequalities:
Let $-\infty \leq a \leq b \leq \infty$, $g(x) \geq 0$,$h(x)\geq0$. Then the following statements are equivalent:
$$
\bigg(\int\_{a}^{b}|Qu(x)|^{2}g(x)^{2}dx\bigg)^{\frac{1}{2}} \leq C\bigg(\int\_{a}^{b}|u(x)|^{2}h(x)^{2}dx\bigg)^{\frac{1}{2}},
$$
$$
K = \sup\_{x \in [a,b]}\bigg(\int\_{a}^{x}[g(t)]^{2}dt\bigg)^{\frac{1}{2}}\bigg(\int\_{x}^{b}[h(t)]^{-2}dt\bigg)^{\frac{1}{2}} < \infty ,
$$
where $Qu(x) = \int\_{x}^{b}u(s)ds$. Moreover, the best constant $C$ satisfies $K \leq C \leq 2K$.
* How $f(1) = 0$, because $f \in H\_{0}^{1}(0,1)$, then
$$
|f(x)| = |f(1) - f(x)|= \bigg|\int\_{x}^{1}f^{\prime}(s)ds\bigg|= |Qf^{\prime}(x)|.
$$
Consider $h(x) = x^{\alpha/2}$, $g(x) =1$, $a=\lambda^{-k}$ and $b=1$.Thus, according to the inequality above, we would have to show that
$$
\sup\_{x \in [\lambda^{-k},1]}\bigg(\int\_{\lambda^{-k}}^{x}dt\bigg)^{\frac{1}{2}}\bigg(\int\_{x}^{1}t^{-\alpha}dt\bigg)^{\frac{1}{2}} < \infty
$$
But it is clear that the supreme of the above product is finite. My big question is:
After applying Hardy's inequality, how do I get $\lambda^{-p}$ to appear in front of the integral.
| https://mathoverflow.net/users/481556 | Perhaps an application of Hardy's inequality | $\newcommand\la\lambda\newcommand\al\alpha$If $C$ is allowed to depend on $\lambda$, just take $C=2\lambda^p$.
If $C$ is not allowed to depend on $\lambda$, take any nonzero $f\in H\_{0}^{1}(0,1)$ and let $\lambda\to\infty$. Then the left-hand side of your desired inequality will go to $\int\_0^1|f(x)|^2\,dx>0$ whereas its right-hand side will go to $0$, so that your desired inequality will fail to hold.
---
Let $f(x)=x(1-x)$ and let $\la\to\infty$. Then
$$\int\_{\la^{-k}}^1|f(x)|^2\,dx\to\frac1{30}$$
and
$$\int\_{\la^{-k}}^1 x^\al|f'(x)|^2\,dx\to h(\al):=\frac{\alpha ^2+\alpha +2}{\alpha ^3+6 \alpha ^2+11 \alpha +6},$$
so that the constant factor $L$ in the inequality
$$\int\_{\la^{-k}}^1|f(x)|^2\,dx
\le L\,\int\_{\la^{-k}}^1 x^\al|f'(x)|^2\,dx$$
must be $\ge\dfrac1{30h(\al)}$. So, $L$ cannot go to $0$ as $\la\to\infty$.
| 2 | https://mathoverflow.net/users/36721 | 439409 | 177,466 |
https://mathoverflow.net/questions/439308 | 6 | I am trying to understand a proof in [this paper](https://arxiv.org/abs/1512.01669) (specifically theorem 5.4). In it, a fact is used that every element of the $W^\*$-algebra $A$ is a linear combination of exponential unitaries.
I've tried to reason this fact out, but I don't follow where this fact comes from or why it necessarily holds true. I do know that any element can be written as a sum of four unitaries (as explained [here](https://math.stackexchange.com/questions/2256498/prove-that-every-element-a-of-a-c-algebra-a-is-a-finite-linear-combination)), but I can't seem to find any information about being a sum of specifically exponential unitaries, which is important to the logic of the proof.
Why is this true, and is this known specifically about $W^\*$-algebras or is it a more general fact? Any information on this would be appreciated.
| https://mathoverflow.net/users/498367 | Every element of a $W^*$-algebra is a linear combination of exponential unitaries? | It is true in any unital $C^\ast$-algebra. Any element $x$ can be written as a linear combination of self-adjoints, by $x=\tfrac{1}{2}(x+x^\ast)+\tfrac{1}{2i}i(x−x^\ast)$. Moreover, if $y$ is a self-adjoint element with $\| y\| \leq 1$, and $\arccos\colon [−1,1]\to [0,π]$, then $u=\exp(i \arccos(y))$ is an exponential unitary such that $y=\tfrac{1}{2}(u+u^\ast)$. By combining these two results one gets the desired fact.
| 7 | https://mathoverflow.net/users/126109 | 439420 | 177,474 |
https://mathoverflow.net/questions/439313 | 1 | In the Lucas–Lehmer test with $ \quad p \quad $ an odd prime.
we know that $ \quad S\_0=4 \quad $ and $ \quad S\_i=S\_{i-1}^2-2 \quad $ for $\quad i>0 \quad$
$M\_p=2^p-1 \quad$ is prime if $ \quad S\_{p-2} \equiv 0 \bmod {(2^p-1)}$
after some observations i found a link with Triangle of coefficients of Chebyshev's OEIS [A053122](https://oeis.org/A053122)
$a(n,k)=(-1)^{n-k}\left(\begin{matrix} n+1+k \\ 2 \cdot k+1 \end{matrix}\right)$
from which
$S\_{i+1} \cdot S\_{i+2}\cdot \quad \ldots \quad \cdot S\_{i+m} = \sum\limits\_{k=0}^{2^m-1}{(-1)^{k+1}\left(\begin{matrix} 2^m+k \\ 2 \cdot k+1 \end{matrix}\right) \cdot (S\_i^2)^k }$
can anyone help me find any reference or pointer on how to derive this result?
| https://mathoverflow.net/users/140242 | Lucas–Lehmer test and triangle of coefficients of Chebyshev's | By the [composition property](https://en.wikipedia.org/wiki/Chebyshev_polynomials#Composition_and_divisibility_properties) of Chebyshev polynomials $T\_m(T\_n(x))=T\_{mn}(x)$. Since $x^2-2 = 2T\_2(\tfrac{x}2)$, we have $S\_i = 2T\_{2^i}(2)$ for all $i\geq 0$.
Furthermore, since $T\_{2^k}(x) = \frac{U\_{2^{k+1}-1}(x)}{2U\_{2^{k}-1}(x)}$, we have
\begin{split} S\_{i+1}\cdot S\_{i+1}\cdots S\_{i+m} &= 2^m T\_{2^{i+1}}(2)\cdot T\_{2^{i+2}}(2)\cdots T\_{2^{i+m}}(2) \\
&=\frac{U\_{2^{m+i+1}-1}(2)}{U\_{2^{i+1}-1}(2)} \\
&= U\_{2^m-1}(T\_{2^{i+1}}(2))\\
&= \frac{U\_{2^{m+1}-1}(T\_{2^i}(2))}{2T\_{2^i}(2)}\\
&= \frac{U\_{2^{m+1}-1}(\tfrac{S\_i}2)}{S\_i}.
\end{split}
and it remains to use an [explicit expression](https://en.wikipedia.org/wiki/Chebyshev_polynomials#Explicit_expressions) for $U\_{2^{m+1}-1}(x)$ to obtain the formula in question.
| 3 | https://mathoverflow.net/users/7076 | 439421 | 177,475 |
https://mathoverflow.net/questions/439431 | 0 | Let $(X\_t)\_{t \ge 0}$ be a Lévy process on $\mathbb R$ with Lévy measure $\nu$.
Define the jump counting measure $N(t, A) = \lvert\{s \in [0, t] \mathrel: \Delta X\_s \in A\}\rvert$
where $\Delta X\_s$ denotes the jump height at time $s$.
For a fixed Borel set $A$ such that $0 \notin A$, let
$N\_t = N(t,A)$.
How can I prove that $E[N\_t] = t\nu(A)$?
| https://mathoverflow.net/users/495513 | Lévy measure and jump behaviour of the corresponding Lévy process | This follows immediately from (say) the following statements in [Schilling - An Introduction to Lévy and Feller Processes](https://arxiv.org/abs/1603.00251):
* parts b) and c) of Lemma 9.4, stating that $N(\cdot,A)$ is a Poisson process of intensity $\nu(A):=EN(1,A)$, where $A$ is any Borel subset of $\mathbb R^d\setminus\{0\}$
* Corollary 9.13, stating that the intensity measure $\nu$ coincides with the Lévy measure.
| 2 | https://mathoverflow.net/users/36721 | 439438 | 177,479 |
https://mathoverflow.net/questions/439286 | 8 | Here's an idea for a knot-based Diffie–Hellman exchange:
* Public: random (oriented) knot $P$.
* Private: random (oriented) knots $A$ and $B$.
* Exchange: Alice sends (randomized or canonical representation of) $A'=A\oplus P$, Bob sends (randomized or canonical representation of) $B'=P\oplus B$. Here $\oplus$ is knot connected sum.
* Shared key: (invariant of) $A\oplus P\oplus B=A\oplus B'=A'\oplus B$.
Questions:
* Why is this a good/bad idea? References?
* What is the complexity of solving $P\oplus X=X'$ for $X$ given $P$ and $X'$? Is factoring knots difficult?
* What are good digital representations of knots? How to efficiently generate and randomize knot representations? (E.g., Gauss codes and Reidemeister moves?) "Diffusing" the sums $A'$, $B'$ (randomly or to some canonical form) would be important for security, assuming factoring is hard in the first place.
* What is a good choice of invariant in the last step? Should be efficient to compute for Alice and Bob but infeasible knowing $P$, $A'$, and $B'$.
* If this makes sense, can it be done in reasonable space/time (e.g. $n\log n$ diagram storage for $n$ crossings, obfuscation/diffusion, final invariant computation)?
TL;DR: Could the knot monoid be useful for public key cryptography? Is connected sum of knots one-way, even knowing one of the factors? Do random planar projections obfuscate knot factorizations?
---
EDIT: To make this question more pointed, given a "random" planar projection of a "random" knot with $n$ crossings, I would like to know:
* First, what are good notions of random above?
* What is the distribution of factors of such a random knot, e.g. is it mostly filled with trefoils and other small knots?
* What computational methods are used to factor knots?
* What are some bounds for space/time complexity of such methods (as a function of $n$)?
* What is the "average" complexity (at least intuitively or in practice)?
* For the envisioned cryptographic purposes, $n$ on the order of 1000 or 10000 might be of reasonable size. Where does this stand in regards to current computational capabilities?
Abandoning uniformly random instances, what is the state of producing hard instances? (One of the answers states that this is currently infeasible.)
| https://mathoverflow.net/users/129839 | Knot Diffie–Hellman | Here I assume that by “addition” of knots you mean the usual connect sum, as defined [here](https://en.wikipedia.org/wiki/Connected_sum#Connected_sum_of_knots). With that said, I think you correctly ask the relevant question: “Is factoring knots difficult?”
In favour of your idea is the fact that we do not (yet?) have a polynomial-time algorithm to factor knots. Against your idea is the “issue” that there are methods (normal surface theory, sutured manifolds) that seem to work very well in practice. Furthermore, we (well, three-manifold topologists) do not know how to produce “hard instances”. Thus your proposal is probably a “bad idea”. :(
---
This should be compared to the situation of factoring numbers. There we really don’t have good general techniques (disregarding quantum algorithms…). Also, we have so many hard instances that your web browser is using one right now!
| 8 | https://mathoverflow.net/users/1650 | 439442 | 177,480 |
https://mathoverflow.net/questions/439402 | 14 | This is in some sense a follow up to the question asked here [Properties of the category of compact Hausdorff spaces](https://mathoverflow.net/questions/382348/properties-of-the-category-of-compact-hausdorff-spaces)
To clarify: The category $\text{Prof}$ of profinite sets sits inside the category $\text{CHaus}$ of compact Hausdorff spaces in a nice way. Every compact Hausdorff space $X$ can be covered by a profinite set (specifically the stone Czech compactification of the underlying set of $X$). $\text{Prof}$ can be characterized by a universal property without reference to the category of topological spaces: It is the free completion of the category $\text{Fin}$ of finite sets under cofiltered limits.
Since $\text{CHaus}$ is, unlike $\text{Top}$, a category that is quite nicely behaved (it is for example the category of algebras for the ultrafilter monad on sets), it seems natural to ask: Can it also be characterized as a category by a universal property similar to $\text{Prof}$?
| https://mathoverflow.net/users/76299 | Is there a universal property characterizing the category of compact Hausdorff spaces? | I definitely expect that there is much more than one good answer. But, here is one that one can get easily by just patching together several classical facts:
1. The category of compact Hausdorf topological space is the category of algebras for the ultrafilter monad.
2. the Ultrafilter monad is the codensity monad for the inclusion of finite set into sets. If I'm not mistaken, it follows that it is terminal for monads on sets such that $ S \to M(S)$ is an isomorphism for all finite sets $S$.
3. the category of monads on $Set$ is equivalent to the opposite of category of monadic right adjoint functors $C \to Set$, or equivalent to the category of monadic left adjoint functor $Set \to C$
So that we get:
**Theorem:** The category of compact Hausdorff spaces, together with its forgetful functor to the category of set is initial in the category of Monadic right adjoint functor $U: C \to Set$ such that the left adjoints $L: Set \to C$ is fully faithful when restricted to finite sets.
**Theorem:** The category of compact Hausdorff spaces, together with the Ultrafilter space functor $\beta: Set \to C$ is terminal in the category of monadic left adjoint functor that are fully faithful when restricted to finite sets.
There are many way to twist this, for example using that $U$ is $Hom(1,\\_)$, or writing the universal property of the codensity monad some other way (the way of writing I chose is maybe a bit atypical...) but it depends what kind of universal property you like
| 19 | https://mathoverflow.net/users/22131 | 439451 | 177,486 |
https://mathoverflow.net/questions/439195 | 9 | I am currently looking into how to construct a **skew-Hadamard matrix** of order 292. Where can I find such construction?
According to multiple papers (e.g. [Koukouvinos and Stylianou - On skew-Hadamard matrices](https://doi.org/10.1016/j.disc.2006.06.037) and [Seberry and Yamada - Amicable Hadamard matrices and amicable orthogonal designs](https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=d42da0293057242ecdd9ce5654c7e4d6fbf5a10e)) this matrix has been constructed. However, none of the papers I have found gives a reference to the actual construction that is being used to obtain it.
From what I have seen, it looks like the first paper which mentions its existence was published in 1978 ([Seberry - On skew Hadamard matrices](https://ro.uow.edu.au/infopapers/988)), with the first (*non-skew*) construction of an Hadamard matrix of order 292 being published in 1975 ([Spence - Skew-Hadamard Matrices of the Goethals–Seidel Type](https://doi.org/10.4153/CJM-1975-066-9)).
---
The idea is to make constructions for various types of Hadamard matrices available in [SageMath](https://sagemath.org). We currently have quite a few already (and discovered gaps in literature along the way).
| https://mathoverflow.net/users/498306 | Construction of skew-Hadamard matrix of order 292 | The skew Hadamard matrix of order 292 can be constructed from skew supplementary difference sets (kindly supplied to us by Prof. Djokovic). This same construction is used, for example, in [Djokovic - Skew-Hadamard Matrices of Orders 436,580, and 988 Exist](https://arxiv.org/abs/0706.1973).
An implementation of this can now be found in [SageMath](https://trac.sagemath.org/ticket/34848).
Note that $H=\{1,2,4,8,16,32,64,55,37\}$ is the order 9 subgroup of $Z\_{73}$, and let
$$
J\_1 = \{5,9,11,25\}\\
J\_2 = \{11,13,17,25\}\\
J\_3 = \{5,9,13,17\}\\
J\_4 = \{1,3,13\}\\
$$
Then, we can obtain supplementary difference sets $(X\_1,X\_2,X\_3,X\_4)$ with parameters $(73; 36,36,36,28; 63)$ and with block $X\_1$ skew as follows:
$$
X\_1 = \bigcup\_{x\in J\_1} xH\\
X\_2 = \bigcup\_{x\in J\_2} xH\\
X\_3 = \bigcup\_{x\in J\_3} xH\\
X\_4 = \{0\} \cup \bigcup\_{x\in J\_4} xH\\
$$
From each set $X\_i$, let $a\_i = (a\_{i,0}, a\_{i,1},...,a\_{i,72})$ be the {±1}-row vector such that $a\_{i,j} = −1$ *iff* $j \in X\_i$.
Then, we can construct four circulant matrices $A\_1, A\_2, A\_3, A\_4$ where $a\_i$ is the first row of $A\_i$. We have that $A\_1$ is skew, and by plugging them in the Goethals-Seidel array we obtain a skew Hadamard matrix of order **292**:
$$
H = \left(\begin{array}{rrrr}
A\_1 & A\_2R & A\_3R & A\_4R \\
-A\_2R & A\_1 & -A\_4^TR & A\_3^TR \\
-A\_3R & A\_4^TR & A\_1 & -A\_2^TR \\
-A\_4R & -A\_3^TR & A\_2^TR & A\_1
\end{array}\right)
$$
Here $R$ denotes the matrix having ones on the back-diagonal and all other entries zero.
| 5 | https://mathoverflow.net/users/498306 | 439452 | 177,487 |
https://mathoverflow.net/questions/439437 | 0 | Let $p(u,x):=(4 \pi u)^{-1/2}e^{-\frac{x^2}{4u}},u>0,x \in \mathbb{R}.$
Let $\mathcal{E}:=\{\phi \in C\_c^\infty (\mathbb{R}),\operatorname{supp}(\phi) \subset B(0,1),\|\phi\|\_\infty \leq 1\}.$
Prove or disprove that for all $U>0,\beta>0,$ there exist $\epsilon>0,C>0$ such that for all $\lambda \in \left]0,1\right],u,v \in [0,U],$ $$\sup\_{x \in \mathbb{R}} \sup\_{\phi \in \mathcal{E}}\left(\int\_0^{|v-u|} \int\_{\mathbb{R}} \left(\int\_{\mathbb{R}} \phi\_x^\lambda(y\_1)p(r,y\_1-y\_2) \, dy\_1 \right)^2 \,dy\_2 \, dr\right)^{1/2}\leq C|v-u|^\varepsilon \lambda^{1/2-\beta},$$
where $\phi\_x^\lambda(y) = \lambda^{-1} \phi(\lambda^{-1}(y-x)).$
| https://mathoverflow.net/users/138491 | Integral with inequality | $\newcommand\EE{\mathcal E}\newcommand\la\lambda\newcommand\R{\mathbb R}\newcommand\ep\varepsilon$What you wanted us to prove is not true.
Indeed, take any $\phi\in\EE$ such that $\phi\ge1\_{[-1/2,1/2]}$. Write $A\gg B$ for $A\ge cB$, where $c$ is a universal positive real constant.
Then, for $w:=x-y\_2$,
\begin{equation}
\begin{aligned}
&\int\_\R \phi\_x^\la(y\_1)p(r,y\_1-y\_2)\,dy\_1 \\
&\gg\frac1\la\,\int\_\R dy\_1\, 1(|y\_1-x|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{(y\_1-y\_2)^2}{4r} \\
&=\frac1\la\,\int\_\R dz\, 1(|z|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{(w+z)^2}{4r} \\ &\ge\frac1\la\,\int\_\R dz\, 1(|z|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{w^2+z^2}{2r} \\
&\ge\exp\Big(-\frac{\la^2}{8r}\Big)\frac1{\sqrt r}\,\exp-\frac{w^2}{2r}. \end{aligned}
\end{equation}
So,
\begin{equation}
\begin{aligned}
&\int\_\R dy\_2\,\Big(\int\_\R \phi\_x^\la(y\_1)p(r,y\_1-y\_2)\,dy\_1\Big)^2 \\
&\gg \exp\Big(-\frac{\la^2}{4r}\Big)
\int\_\R dw\,\frac1r\,\exp-\frac{w^2}r \\
&\gg \frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big)
\end{aligned}
\end{equation}
and hence
\begin{equation}
\begin{aligned}
I&:=\int\_0^{|v-u|}dr\,\int\_\R dy\_2\,\Big(\int\_\R \phi\_x^\la(y\_1)p(r,y\_1-y\_2)\,dy\_1\Big)^2 \\
&\gg \int\_0^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \\
&\ge \int\_{|v-u|/2}^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \\
&\ge \int\_{|v-u|/2}^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{2|v-u|}\Big) \\
&\gg |v-u|^{1/2}\exp\Big(-\frac{\la^2}{2|v-u|}\Big).
\end{aligned}
\end{equation}
Letting now, for instance, $U=1$, $v=1$, and $u=0$, for all $\la\in(0,1]$ we get
\begin{equation}
I\gg1.
\end{equation}
So, if $\beta<1/2$, then there is no real $\ep>0$ and $C>0$ such that $I^{1/2}\le C|v-u|^{\ep} \la^{1/2-\beta}$ for all $\la\in(0,1]$. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 439456 | 177,489 |
https://mathoverflow.net/questions/439259 | 4 | Sarnak and Strömbergsson studied [Epstein zeta function](https://encyclopediaofmath.org/wiki/Epstein_zeta-function) $\zeta(L,s)=\sum\limits\_{0\neq v\in L}\langle v,v\rangle^{-s}$ of a number of highly symmetric lattices in their [Inventiones Math. paper](http://www2.math.uu.se/%7Eastrombe/papers/minima.pdf).
In particular, they show (Thm 1 in [loc.cit.]) that in dimensions $n=4,8,24$ one has $\zeta(L,s)$, with $s>0$, reaching a strict local minimum
$\zeta(L\_n,s)$ on $L\_4=D\_4,$ $L\_8=E\_8$ and the Leech lattice $L\_{24}=\Lambda\_{24}$, respectively.
Their analysis has two parts - first they need to show that $\zeta(L\_n,s)<0$ for $\frac{n}{2}>s>0$, and $L\_n$ as above. Then they study polynomial invariants of the automorphism groups of $L\_n$ (acting on bilinear forms), and establish that these only start to differ from the corresponding invariants of the full orthogonal group $O(n)$ at high enough degrees, allowing for an analytic argument to complete the proof.
It turns out that for Barnes-Wall lattice $\Lambda\_{16}$ the needed property of the invariant ring holds, too - so their Thm 1 could apparently be extended to the case $n=16$ if $\zeta(\Lambda\_{16},s)<0$ for $8>s>0$ holds.
So my question is whether $\zeta(\Lambda\_{16},s)$ can be found in the literature. Note that for $n=4,8,24$ the authors were able to use information in [Conway-Sloane](https://link.springer.com/book/10.1007/978-1-4757-6568-7) book (I'm struggling to fill in details there, though) - but the info on $\Lambda\_{16}$ there appears to be less complete.
| https://mathoverflow.net/users/11100 | Epstein zeta function of Barnes-Wall and related lattices | It turns out this question, and more, is basically answered, for all Barnes-Wall lattices $\Lambda\_{n}$, in affirmative in [Spherical designs and zeta functions of lattices](https://arxiv.org/abs/math/0611735) - the published version [here](https://doi.org/10.1155/IMRN/2006/49620) with the extra assumption $s>n/2$.
It's not clear from this text whether for $n=16$ the assumption $s>8$ is needed, but it's remarked that it's needed for $n$ big enough.
| 2 | https://mathoverflow.net/users/11100 | 439459 | 177,490 |
https://mathoverflow.net/questions/439458 | 1 | Let $G$ be a linearly reductive algebraic group and $X$ be an affine $G$-variety over an algebraically closed field $\mathbb{K}$. Let $Y\subset X$ be a (closed) affine subvariety of $X$ which is also $G$-stable.
>
> Is there any "nice" description of the invariant ring $\mathbb{K}[Y]^G$ in terms of the invariant ring $\mathbb{K}[X]^G$? Maybe something like- it's a quotient or a subring of $\mathbb{K}[X]^G$.
>
>
>
For my situation, $X$ is an affine space.
| https://mathoverflow.net/users/338456 | Invariant ring of the subvariety | As we discussed in the comments, by linear reductivity of $G$, there is a $G$-module splitting of the surjection $\mathbb K[X] \to \mathbb K[Y]$; so the restriction map $\mathbb K[X]^G \to \mathbb K[Y]^G$ is a surjection. Thus $\mathbb K[Y]^G$ is a quotient of $\mathbb K[X]^G$, as both a $G$-module and an algebra; and is a $G$-submodule of $\mathbb K[X]^G$, but (even if $G$ is trivial) need not be a subalgebra.
| 5 | https://mathoverflow.net/users/2383 | 439464 | 177,492 |
https://mathoverflow.net/questions/439467 | 0 | Let $p(u,x):=(4 \pi u)^{-1/2}e^{-\frac{x^2}{4u}},u>0,x \in \mathbb{R}.$
Let $\mathcal{E}:=\{\phi \in C\_c^\infty (\mathbb{R}),\operatorname{supp}(\phi) \subset B(0,1),\|\phi\|\_\infty \leq 1\}.$
Prove that for all $U>0,\beta>0,$ there exist $\epsilon>0,C>0$ such that for all $\lambda \in \left]0,1\right],u,v \in [0,U],$ $$u\leq v \implies \sup\_{x \in \mathbb{R}} \sup\_{\phi \in \mathcal{E}}\left(\int\_u^{v} \int\_{\mathbb{R}} \left(\int\_{\mathbb{R}} \phi\_x^\lambda(y\_1)p(v-r,y\_1-y\_2) \, dy\_1 \right)^2 \,dy\_2 \, dr+\int\_0^u\int\_{\mathbb{R}}\left(\int\_{\mathbb{R}}\phi\_x^\lambda(y\_1)(p(v-r,y\_1-y\_2)-p(u-r,y\_1-y\_2))dy\_1\right)^2dy\_2dr\right)\leq C|v-u|^\varepsilon \lambda^{1-2\beta},$$
where $\phi\_x^\lambda(y) = \lambda^{-1} \phi(\lambda^{-1}(y-x)).$
How can we prove this inequality?
| https://mathoverflow.net/users/138491 | Integral and inequality | At least when (say) $U=1$, $v=1$, and $u=0$, the integral
$$\int\_u^{v} \int\_{\mathbb{R}} \left(\int\_{\mathbb{R}} \phi\_x^\lambda(y\_1)p(v-r,y\_1-y\_2) \, dy\_1 \right)^2 \,dy\_2 \, dr$$
coincides (in view of the variable change $r\leftrightarrow v-r$) with the integral
$$\int\_0^{|u-v|} \int\_{\mathbb{R}} \left(\int\_{\mathbb{R}} \phi\_x^\lambda(y\_1)p(r,y\_1-y\_2) \, dy\_1 \right)^2 \,dy\_2 \, dr$$
in [your previous post](https://mathoverflow.net/q/439437/36721).
So, in view of the [previous answer](https://mathoverflow.net/a/439456/36721), your current inequality is not true either.
| 1 | https://mathoverflow.net/users/36721 | 439469 | 177,495 |
https://mathoverflow.net/questions/439463 | 0 | An issue from 3D tessellated geometry: Find the direction vector of a plane that minimizes the silhouette of a set of triangles. To say it another way, find the direction vector that is most perpendicular to a set of triangles. Each triangle area is half of the length of the normal vector of any two sides—call it $n\_i$. The projected area is simply the dot product of these face vectors with the target plane normal, $a$. Some of these could be negative, so we square and sum—this is a minimal least squares problem: minimize.
$$
f(a)=\sum\_i (a \cdot n\_i)^2 \text{ over all } a \in \mathbb{S}\_2.
$$
This could be done with constrained optimization by creating a Lagrangian, but this seems to lead to a set of nonlinear equations.
Instead, use the spherical angles and redefine $a$ as:
$$
a = [\cos\phi \sin\theta, \sin\phi \sin\theta, \cos\theta]
$$
The derivative of the gradient of $f$ could be set to zero and we solve for the angles: $\phi$ and $\theta$.
Here are those two equations:
$$
\frac{df}{d\phi} =2(x\_{ni} \cos\phi \sin\theta+y\_{ni}\sin\phi \sin\theta +z\_{ni}\cos\theta)(y\_{ni}\sin\theta \cos\phi-x\_{ni}\sin\theta sin\phi)=0
$$
$$
\frac{df}{d\theta}=2(x\_{ni} \cos\phi \sin\theta+y\_{ni}\sin\phi \sin\theta +z\_{ni}\cos\theta)((x\_{ni}\cos\phi+y\_{ni}\sin\phi)\cos\theta-z\_{ni}\sin\theta)=0
$$
Now, here's where I get confused! I essentially have two terms on the left hand sides that could be zero in each equation. Given that the first big term $(x\_{ni} \cos\phi \sin\theta+y\_{ni}\sin\phi \sin\theta +z\_{ni}\cos\theta)$ is the same in both equations, this being zero won't help us since our equations will reduce to one. and we need two to solve for the two variables ($\phi$ and $\theta$).
The latter two parenthetical terms then make for some nice equations that we can reduce to:
$$\phi = \arctan(y\_{ni} / x\_{ni})$$
and $$\theta = \arctan((x\_{ni}\cos\phi+y\_{ni}\sin\phi) / z\_{ni})$$
Gee, I'm smart! I code this up and give it a try where the sum of normals results in:
$x\_{ni} = 500; y\_{ni}= -800; z\_{ni} = -0.06$. Clearly the answer should be mostly in the $z$-direction but as you can see from these last two equations, it is clearly not!
Where did I go wrong?
| https://mathoverflow.net/users/498458 | Trigonometry/spherical angles/minimum-least-squares | If you use constrained optimization, you do get linear equations.
I set $g(x)=||x||^2$. For every $x \in \mathbb{R}^3$,
$$\nabla f(x) = \sum\_i 2(x \cdot n\_i)n\_i \text{ and } \nabla g(x) = 2x.$$
You look at (unit) vectors $x$ such that $\nabla f(x) = \lambda \nabla g(x)$ for some real number $\lambda$, namely you look at eigenvectors of the symmetric endomorphism $u$ given by
$$u(x) = \sum\_i (x \cdot n\_i)n\_i.$$
Actually, using an orthogonal basis of eigenvectors of $u$, you get that the minimum of $f(x) = x \cdot u(x)$ over all $x \in \mathbb{S}\_2$ is achieved when $x$ is an eigenvector associated to the least eigenvalue of $u$.
This can be proved directly. Call $\lambda\_1 \le \lambda\_2 \le \lambda\_3$ the eigenvalues of $u$ and call $(e\_1,e\_2,e\_3)$ an orthonormal basis of associated eigenvectors. Then for every $x = \xi\_1e\_1+\xi\_2e\_2+\xi\_1e\_3 \in \mathbb{R}^3$,
$$f(x) = x \cdot u(x) = \sum\_{i=1}^3 \lambda\_i\xi^2 \ge \sum\_{i=1}^3 \lambda\_1\xi^2 = \lambda\_1||x||^2,$$and equality holds when $x$ is a multiple of $e\_1$.
| 0 | https://mathoverflow.net/users/169474 | 439476 | 177,499 |
https://mathoverflow.net/questions/439403 | 4 | $\newcommand{\R}{\mathbb R}$
Let $\Omega\subset \R^d$ be a smooth, bounded open set and fix $p\geq 1$.
* **Fact 1**: the usual Lebesgue differentiation theorem says that, if $u\in L^p(\Omega)$, then
$$
u(x)=\lim\limits\_{r\to 0}\frac{1}{|B\_r(x)|}\int\_{B\_r(x)}u(y)dy
$$
for Lebesgue-a.e. $x\in \Omega$.
* **Fact 2**: for $u\in W^{1,p}(\Omega)$ (the usual Sobolev space) it is well known that the boundary trace $v:= tr(u)$ is well-defined as an element of $W^{1-1/p,p}(\partial\Omega)$.
>
> **Question:** is it true that
> $$
> v(x)=\lim \limits\_{r\to 0}\frac{1}{|B\_r(x)\cap \Omega|}\int\_{B\_r(x)\cap \Omega}u(y)dy
> $$
> for $\mathcal H^{d-1}$-a.e. $x$ in the boundary?
>
>
>
This seems a very natural question to ask and I hope that the answer is already written somewhere out there. I suspect that the notion of $p$-capacity should play a role here (being the precise representative of a $W^{1,p}$ function $p$-quasicontinuous), but I'm not too familiar with capacity theory so I'd rather avoid appealing to such raffinate machinery and reinventing the wheel, if possible.
(I am also aware that the trace can be recovered as the continuous limit along any transversal curve, but somehow I did not manage to conclude from this.)
Actually any reference to either a positive or negative statement would make me happy, I just need a black-box theorem that I could apply.
| https://mathoverflow.net/users/33741 | Lebesgue differentiation theorem at boundary points for Sobolev traces | It is true - this is Theorem 5.7 in Evans and Gariepy’s *Measure Theory and Fine Properties of Functions* (2015 version).
Note that the theorem is stated for BV functions, but Sobolev functions are BV, so it holds also for your case.
| 6 | https://mathoverflow.net/users/173490 | 439479 | 177,500 |
https://mathoverflow.net/questions/439483 | 0 | (Asked previously in [MSE](https://math.stackexchange.com/questions/4625673/find-a-probability-density-from-the-moments))
Suppose a probability distribution $p(x)$ has moments $m\_n=\int p(x)x^ndx$ given by $m\_0=1$, $m\_1=1$, $m\_2=2$ and, for $n>0$,
$$m\_{n+1}={2n \choose n}.$$
The moment generating function exists everywhere and is
$$f(t)=1+\sum\_{n=0}^\infty \frac{(2n)!}{n!^2(n+1)!}t^{n+1}=1+te^{2t}\left(I\_0(2t)-I\_1(2t)\right),$$
in terms of Bessel functions. An inverse Fourier transform $\int e^{-itx}f(it)dt$ then gives the function
$$ \rho(x)=\frac{\delta(x)}{2\pi}+\frac{1}{\pi x^{3/2}\sqrt{4-x}}.$$
This function has all the correct moments, except for the norm, $m\_0$, because it is in fact not normalizable.
Is this weird? If the moment generating function converges everywhere, a probability distribution is uniquely determined (right?), but this $\rho$ is not a probability distribution. Can I conclude that no random variable can possibly have those moments?
| https://mathoverflow.net/users/83671 | Probability distribution with shifted central binomial moments | The function $\mathbb R\ni t\mapsto g(t):=f(it)$ is not the characteristic function of any probability distribution.
Indeed, if it were, then we would have $|g(t)|\le1$ for all real $t$. However, in fact, $|g(10)|=1.316\ldots>1$.
So, $f$ is not the moment generating function of any probability distribution, and hence there is no probability distribution with your "moments" $m\_0,m\_1,\dots$.
---
You wrote
>
> Can I conclude that no random variable can possibly have those moments?
>
>
>
It is now shown that, indeed, "no random variable can possibly have those moments".
You also wrote
>
> If the moment generating function converges everywhere, a probability distribution is uniquely determined (right?), but this $\rho$ is not a probability distribution.
>
>
>
The uniqueness result is actually this: If two probability distributions (say on the real line) have the same moment generating function which is finite in a neighborhood of $0$, then the two distributions are the same.
However, in our case it has been shown that $f$ is not the moment generating function of any probability distribution. So, no contradiction whatsoever here.
---
Finally, concerning this:
>
> $$ \rho(x)=\frac{\delta(x)}{2\pi}+\frac{1}{\pi x^{3/2}\sqrt{4-x}}.$$
>
>
> This function has all the correct moments, except for the norm, $m\_0$, because it is in fact not normalizable.
>
>
>
Let
$$p(x):=x\rho(x)=\frac{1}{\pi x^{1/2}\sqrt{4-x}}$$
for $x\in(0,4)$, with $p(x):=0$ for real $x\notin(0,4)$. Then $p$ is a true probability density, with moments
$$\mu\_n:=\int\_{\mathbb R}x^n p(x)\,dx=m\_{n+1}.$$
That is, your almost-moments $m\_1,m\_2,\dots$ are just shifted true moments $\mu\_0,\mu\_1,\dots$, respectively -- exactly with no place for your $m\_0$. This completes the resolution of the puzzle.
| 2 | https://mathoverflow.net/users/36721 | 439489 | 177,503 |
https://mathoverflow.net/questions/439490 | 4 | I want to show that a certain Lie subgroup (i.e. generated by the exponential of elements in some Lie subalgebra) of a Lie group is closed. My knowledge of the subject of Lie groups is rudimentary, and I would really appreciate any pointers towards relevant results.
My question
-----------
**Is the following statement true?**
>
> Let $G$ be a connected finite-dimensional real Lie group, and $H$ a Lie subgroup (i.e. generated by $\exp(\mathfrak{h})$ for some sub-lie algebra $\mathfrak{h}$ of the lie algebra $\mathfrak{g}$ of $G$). If the center of $H$ (i.e. $Z(H) = \lbrace x \in H: xh = hx\text{ for all }h \in H\rbrace$) is a closed subset of $G$, then $H$ is a closed subset of $G$.
>
>
>
If true I would love a reference. If for some reason it is true only for semi-simple $G$ with finite center, that would be great as well.
Some context
------------
[The closed subgroup theorem](https://en.wikipedia.org/wiki/Closed-subgroup_theorem) says that any connected subgroup of $G$ which is closed is in fact a Lie group. On the wikipedia page there are [some criteria](https://en.wikipedia.org/wiki/Closed-subgroup_theorem#Conditions_for_being_closed) which allow one to deduce that the group associated to some lie algebra $\mathfrak{h} \subset \mathfrak{g}$ is closed, but they don't seem to apply in my case of interest.
I found an [encyclopedia of math article](https://encyclopediaofmath.org/wiki/Lie_group,_solvable) where the claim is made that any connected subgroup of a simply connected solvable real Lie group is closed. This is false as noted in the answers to [this stackoverflow question](https://math.stackexchange.com/questions/3237589/subgroups-of-solvable-lie-groups-are-closed). However for Lie subgroups it seems to be true. The article gives [a paper of Malcev](https://mathscinet.ams.org/mathscinet-getitem?mr=13165) as a reference, there is an [an errata for the paper](https://mathscinet.ams.org/mathscinet-getitem?mr=18651), and also some errors are pointed out by Chevalley in the mathscinet review (where he implies that the particular result that interests me was not new at the time but, alas, does not give a reference). I can use this to solve some, but not all, of the cases that interest me.
Finally, I found [some papers](https://mathscinet.ams.org/mathscinet-getitem?mr=56613) by Morikuni Goto which seem likely to contain results I could piece together to either get my statement, or something similarly useful.
He gives conditions on the Lie algebra of a connected Lie group $H$ that guarantee that every isomorphic embedding has a closed image. In particular if $Z(H)$ is compact then any continuous isomorphic image of $H$ in a Lie group is closed.
However, the subgroups $H$ that interest me have non-compact center, which happens to be closed in the larger group (this means some isomorphisms could send $H$ to a non-closed subgroup inside some other group).
| https://mathoverflow.net/users/7631 | Is a Lie subgroup whose center is closed, a closed subgroup itself? | No. Consider a group of the form $G=V\rtimes K$ where $V$ is a Euclidean group and $K$ is a compact 2-torus and $K$ acting faithfully on $D$, with no nonzero invariant vector. (For instance $G=(\mathbf{R}^2\rtimes\operatorname{SO}(2))^2$.) Let $D$ be a dense line in $K$: then $V\rtimes D$ is dense and non-closed in $G$, and has a trivial center, hence its center is closed in $G$.
(However, I think that the statement is true when the subgroup is semisimple: in a connected Lie group, an immersed Lie subgroup that is semisimple with the assumption that its center is closed in $G$, has to be closed — I'm not sure of a reference for this.)
| 7 | https://mathoverflow.net/users/14094 | 439497 | 177,505 |
https://mathoverflow.net/questions/439502 | 1 | I am reading the paper "P.Sjolin, Convolution with Oscillating Kernels, Indiana University Mathematics Journal Vol. 30, No. 1 (1981), pp. 47-55" where $L^p-L^p$ boundedness of the operator
$$Tf(x)=\int \frac{e^{i|x-y|^{a}}}{|x-y|^{\alpha}}
f(y)dy$$
is studied. Here $0<\alpha<n$ and $a>0$, $a\neq 1$.
The author repeatedly utilizes the claim that:
if the inequality
$$\|Tf\|\_{p}\leq C \|f\|\_{p}\qquad (1)$$
necessitates that $p\geq p\_{0}$ for some $1< p\_{0}\leq 2$ then (1) also necessitates that that $p\leq p\_{0}^{\prime}$, the dual exponent of $p\_{0}$. That is if (1) is false for all $p<p\_{0}$ for some $1<p\_{0}\leq 2$
then (1) is also false for all $p>p\_{0}^{\prime}$.
This seems to have something to do with duality but I can't make that precise. Where does this statement come from ?
| https://mathoverflow.net/users/116555 | Why does failure of boundedness of this operator for $p<q$ implies its failure for $p>q^{\prime}$? | This is a standard property of self-adjoint (including convolution) operators, following from duality.
Let $Tf = k\*f$ and assume that $||Tf||\_{p} \leq C ||f||\_{p}$.
Then $$||Tf||\_{p'} = \sup\_{\substack{g \in L^p \\ ||g||\_{p}} =1} \left<Tf,g\right> =\sup\_{\substack{g \in L^p \\ ||g||\_{p}} =1} \left<f,Tg\right>.$$
From Holder we then have that this is
$$\leq \sup\_{\substack{g \in L^p \\ ||g||\_{p}} =1} ||Tg||\_{p} ||f||\_{p'} \leq C ||f||\_{p'}.$$
| 2 | https://mathoverflow.net/users/630 | 439506 | 177,508 |
https://mathoverflow.net/questions/439265 | 1 | The principle of induction over identity families, do not prohibit instances different from `refl: x == x` but its computation rule is only defined for this instance, i.e. `ind(C,c,x,x,refl) :≡ c(x)`.
If a function defined by path induction receives arguments different from the expected ones for the computation rule e.g. Using the higher inductive type "Interval": `ind(D, d, i0, i1, seg) :≡ ?` where `i0, i1: Interval`, `seg: i0 == i1`:
* The computation of the function is undefined?
* This is similar to the case of `inr` and `inl` that are irreducible, i.e. `inr(a) :≡ inr(a)`?
| https://mathoverflow.net/users/495133 | Computation over non-reflexivity | After analyzing, I have come with a conclusion: the path induction principle is not different from other axioms/rules that assert the existence of a function without providing an explicit definition, e.g. $succ: N \rightarrow N, inl: A \rightarrow A + B$.
As axioms, these functions do not need a definition, since a definition should be a proof of the axiom and therefore no longer an axiom. The specific case of $inr(a) :≡ inr(a)$ is due to $inr$ been not defined but rather taken as an axiom.
Path induction have the same behavior, $ind(C,c,x,y,p) :≡ ind(C,c,x,y,p)$ but have an exception when the arguments are $(x,x,refl\_x)$, in this case is reducible to $c(x)$.
This seems to be where the so called non-trivial paths come, since the cases when $p \neq relf\_x$ are the ones that cannot be reduced to the reflexive case $c(x)$ but exists even if they are irreducible.
e.g.
In the case of $S^1$ we have elements:
$base: S^1, loop: base = base$
We can define the inverse of loop by path induction
$loop^{-1} :≡ ind((x,y,p) \Rightarrow y = x, x \Rightarrow refl\_x, base,
base, loop)$
Since $loop \neq refl\_{base}$ (Lemma 6.4.1 of the HoTT book) the definition of $loop^{-1}$ is irreducible and different from loop itself.
The proof that this new element is indeed different from loop can be seen with the equivalence $\Omega(S^1) \simeq Z $ (described on HoTT book section 8.1) that also shows that using this same irreducible non-reflexive paths we can build other new different non-reflexive paths.
| 1 | https://mathoverflow.net/users/495133 | 439507 | 177,509 |
https://mathoverflow.net/questions/439293 | 6 | Let $X$ be a complex compact manifold, and write $\mathcal{O}\_X$ for the sheaf of holomorphic functions on $X$. Let $\mathcal{O}\_X^{\times}$ be the subsheaf consisting of holomorphic functions. These are both sheaves of abelian groups. If we identify $H^1(X, \mathcal{O}\_X^{\times})$ with the Picard group $\text{Pic}(X)$, then we can regard classes of $H^1(X, \mathcal{O}\_X^{\times})$ geometrically as iso classes of line bundles on $X$.
Surely, one way to obtain from an abstract class $[\alpha] \in H^1(X, \mathcal{O}\_X^{\times})$ the line bundle back is via the Cech cocycle data containing the whole glueing information about the associated line bundle up to isomorphism.
Recently I saw a different construction associating to an abstract class $[\alpha] \in H^1(X, \mathcal{O}\_X^{\times})$ a representing line bundle and I would like to learn more about the background of this construction and why this gives finally a line bundle which coincides with that one obtained from Cech cycle. Does it has a name under which it can be looked up in literature? What is it's geometrical origin? Is there some construction from (differential) topology motivating it?
Here the construction: We take some injective resolution $1 \to \mathcal{O}\_X^{\times} \to \mathcal{I}^0 \to \mathcal{I}^1 \to ... $ Then a $[\alpha] \in H^1(X, \mathcal{O}\_X^{\times})$ is represented by construction by a section $\alpha \in H^0(X, \mathcal{I}^1)$ becoming zero in $H^0(X, \mathcal{I}^2)$.
We consider the bundle $\mathcal{M}$ obtained as fibre product
$$
\require{AMScd}
\begin{CD}
\mathcal{M} @>{} >> \mathbb{Z}\_X \\
@VVV @VV{\alpha}V \\
\mathcal{I}^0 @>{}>> \mathcal{I}^1
\end{CD}
$$
where $\mathbb{Z}\_X \to \mathcal{I}^1$ is induced by $\alpha$ via $1 \vert \_U \to \alpha \vert \_U $. Since $ \mathcal{I}^0 $ contains $\mathcal{M}$ and $\mathcal{O}\_X^{\times}$ as abelian subsheaves, there is a action $\mathcal{O}\_X^{\times} \times \mathcal{M} \to \mathcal{M}$ by $\mathcal{O}\_X^{\times}$, which restricts on $\mathcal{O}\_X^{\times}$ to the usual multiplication $\mathcal{O}\_X^{\times} \times \mathcal{O}\_X^{\times} \to \mathcal{O}\_X^{\times}$.
It's well defined because for $U \subset X$ the sections $\mathcal{M}(U)$ are given as sections in $\mathcal{I}^0(U)$ mapping to $\alpha \vert \_U$ and $\mathcal{O}\_X^{\times} \to \mathcal{I}^0 \to \mathcal{I}^1$ is exact.
Then we obtain an invertible sheaf $\mathcal{L}$ as $(\mathcal{M} \times \mathcal{O}\_X)/\mathcal{O}\_X^{\times}$ by moding out the diagonal action.
The claim is that this coincides with the line bundle obtained from class $[\alpha]$ via Cech cycle. Why? The construction reminds me on algebraic analoga of topological [associated bundle](https://ncatlab.org/nlab/show/associated+bundle). The aim behind the construction is obviously to exchange the fiber $\mathcal{O}\_X^{\times}$ by the "right" fiber $\mathcal{O}\_X$ giving the resulting object the desired structure of a line bundle. (This fiber replacing business is essentially for what the construction of associated bundle good for. Morally one wants to keep the same base space & in certain sense the same twisting behaviour, but replace the fiber)
But I not see from this construction why this one gives up to isomorphism the same line bundle one obtains from the Cech cocycle data.
| https://mathoverflow.net/users/108274 | Construction of a line bundle from a class $[\alpha] \in H^1(X, \mathcal{O}_X^{\times})$ as $\mathcal{O}_X^{\times}$-Torsor | Let $\mathcal M$ be the sub-sheaf of $\mathcal I^0$ of sections mapping to (the restrictions of) $\alpha$. (I suppose that's what the OP has intended anyways.) That's easily seen to be an $\mathcal{O}\_X^\times$-torsor. By [Stacks Project Tag 0A6G](https://stacks.math.columbia.edu/tag/0A6G), if $X=\bigcup\_i U\_i$ is an open cover trivialising this torsor, say via sections $s\_i\in\mathcal M(U\_i)$, then the Čech cocycle defined by the $s\_{ij}\in\mathcal O\_{U\_{ij}}^\times$ such that $s\_{ij}(s\_i|\_{U\_{ij}})=s\_j|\_{U\_{ij}}$ does indeed represent $[\alpha]\in H^1(\mathcal O\_X^\times)$.
Moreover, let $\mathcal L$ be the quotient sheaf of the action of $\mathcal O\_X^\times$ on $\mathcal M\times \mathcal O\_X$ which acts on sections as $s.(m,z)=(sm,s^{-1}z)$ – that's not quite the diagonal action, but exactly what you should be familiar with from the associated bundle construction.
$\mathcal L$ is trivialised over the same open cover by the classes $[s\_i,1]$ of the sections $(s\_i,1)\in \mathcal M(U\_i)\times\mathcal O\_X(U\_i)$ and by construction, $s\_{ij}[s\_i,1]=[s\_i,s\_{ij}]=[s\_{ij}s\_i,1]=[s\_j,1]$, i.e., the cocycle yields the transition functions of the associated line bundle.
By the way, the construction works quite generally, cf. [Stacks Project Tag 040E](https://stacks.math.columbia.edu/tag/040E).
| 2 | https://mathoverflow.net/users/15782 | 439527 | 177,516 |
https://mathoverflow.net/questions/439518 | 2 | Consider $X$ a smooth cubic surface in $\mathbb{P}^3$, and let $l\_1,...,l\_6$ be six disjoint lines contained in $X$.
What is the linear system giving the blow-down map $X \to \mathbb{P}^2$, so that the lines $l\_k$ are contracted to points ?
The other way round is well-known : if $p\_1,\dots,p\_6$ are six points in general position, the rational map $\mathbb{P}^2 \to \mathbb{P}^3$ obtained by the linear system of cubic containing the six points has image a cubic surface.
| https://mathoverflow.net/users/21030 | what is the linear system on a cubic surface giving the blow-down map to the plane | As you probably know, if a representation of $X$ as blowup is given,
$$
K\_X = -3h + \sum l\_i,
$$
where $h$ is the line class. Consequently, the linear system
$$
|-K\_X + \sum l\_i|
$$
gives the required contraction.
| 3 | https://mathoverflow.net/users/4428 | 439533 | 177,519 |
https://mathoverflow.net/questions/439474 | 1 | We say that a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ has *property ${\bf B}$* if there is $S\subseteq V$ such that for all $e\in E$ with $|e|\geq 2$ we have $$(e\cap S) \neq \emptyset \neq (e\cap (V\setminus S)).$$
If $k\in \omega$, a hypergraph $H=(V,E)$ is $k$-*uniform* if all elements of $E$ have cardinaliy $k$.
Suppose the hypergraph $(\omega,E)$ is $k$-uniform for some $k\geq 4$, and we have that $$2\cdot |e\_0\cap e\_1| < k$$ whenever $e\_0\neq e\_1\in E$. Does this necessarily imply that $(\omega,E)$ has property ${\bf B}$?
| https://mathoverflow.net/users/8628 | Uniform hypergraphs with small edge intersections and propery ${\bf B}$ | There are counterexamples for every integer $k\ge3$.
In fact, if $2\le k\lt\omega$, there is a $k$-uniform hypergraph $H=(V,E)$ such that $|V|=\aleph\_0$, $\{e\_1,e\_2\}\in\binom E2\implies|e\_1\cap e\_2|\le1$, and $H$ has chromatic number $\chi(H)=\aleph\_0$.
Namely, let $V=\binom{\mathbb N}{k-1}$ and $E=\{\binom X{k-1}:X\in\binom{\mathbb N}k\}$.
It follows from Ramsey's theorem that $\chi(H)\gt n$ for each $n\lt\omega$; the other properties are obvious. Moreover, $H$ has a finite subhypergraph $H\_n$ with $\chi(H\_n)\gt n$.
| 2 | https://mathoverflow.net/users/43266 | 439540 | 177,521 |
https://mathoverflow.net/questions/439526 | 5 | Let $\mathbb{G}$ be a compact quantum group (in the sense of Woronowicz) with discrete dual $\widehat{\mathbb{G}}$ which we view as a von Neumann algebraic locally compact quantum group (in the sense of Vaes-Kustermans). Let us denote its function algebra by $(\ell^\infty(\widehat{\mathbb{G}}), \hat{\Delta})$.
Consider a von Neumann algebra $M$ and a unital completely isometric (normal) map
$$\alpha: M \to M \overline{\otimes} \ell^\infty(\widehat{\mathbb{G}})$$ satisfying the coaction property
$$(\alpha\otimes \iota) \alpha = (\iota\otimes \hat{\Delta})\alpha.$$
Is it true that $\alpha$ is automatically multiplicative?
Of course, if $\Gamma$ is a discrete group, then a completely isometric map
$$\alpha: M \to M\overline{\otimes}\ell^\infty(\Gamma)= \prod\_{g\in \Gamma} M$$
is automatically multiplication-preserving because the map $\alpha$ is then a direct product of unital completely isometric maps $\alpha\_g: M \to M$ which are automatically $C^\*$-isomorphisms (by a result by Choi).
I tried to apply the same trick in this case: $$\ell^\infty(\widehat{\mathbb{G}})\cong \prod\_{\gamma \in \operatorname{Irr}(\mathbb{G})} B(H\_\gamma)$$
and the map $\alpha$ then breaks down as a collection of maps
$$\alpha\_\gamma: M \to M\_{n\_\gamma}(M): m \mapsto [u\_{ij}^\gamma \rhd m]$$
where $U^\gamma = [u\_{ij}^\gamma]$ is the irreducible representation $\gamma$ and $\rhd: \mathcal{O}(\mathbb{G})\odot M \to M$ the induced left module structure. If we can show that these maps are multiplication-preserving, then we are done. This, in turn, is equivalent with showing that their images are $C^\*$-algebras, but neither of these claims are clear to me. On the level of left $\mathcal{O}(\mathbb{G})$-modules, the multiplicativity means
$$g\rhd (mn)= (g\_{(1)}\rhd m)(g\_{(2)}\rhd n)$$
or in terms of matrix coefficients
$$u\_{ij}^\gamma\rhd (mn) = \sum\_{k=1}^{n\_\gamma} (u\_{ik}^\gamma\rhd m)(u\_{kj}^\gamma\rhd n).$$
Is the multiplicativity of the coaction $\alpha$ somehow automatic? I am starting to believe this isn't true, but I was not able to find a counterexample. Thanks in advance for your help!
| https://mathoverflow.net/users/470427 | Completely isometric coaction of discrete quantum group is multiplicative? | Yes, such a map $\alpha$ is automatically multiplicative and thus defines an action of $\widehat{\mathbb{G}}$ on $M$.
As in the question, denote by $\alpha\_\gamma : M \to M \otimes B(H\_\gamma)$ the components of $\alpha$, for any irreducible unitary representation $\gamma$ of $\mathbb{G}$. Fix an irreducible representation $\gamma$. It suffices to prove that $\alpha\_\gamma$ is multiplicative.
Since $\alpha$ is unital completely isometric, $\alpha$ is also completely positive. Thus, all $\alpha\_\gamma$ are unital completely positive (ucp).
We first prove that $\alpha\_\varepsilon(x) =x$ for all $x \in M$. By the coaction property, $\alpha\_\gamma \circ \alpha\_\varepsilon = \alpha\_\gamma$ for all $\gamma$. So, if $\alpha\_\varepsilon(x)=0$, it follows that $\alpha(x)=0$ and thus $x=0$ because $\alpha$ is supposed to be isometric. Since $\alpha\_\varepsilon(\alpha\_\varepsilon(x)-x) = 0$ for all $x \in M$, it follows that $\alpha\_\varepsilon(x) =x$ for all $x \in M$.
Let $\rho$ be the contragredient of $\gamma$ and choose morphisms $t \in \operatorname{Mor}(\varepsilon,\rho \otimes \gamma)$ and $s \in \operatorname{Mor}(\varepsilon,\gamma \otimes \rho)$ such that $t^\* t = 1$ and $(s^\* \otimes 1)(1 \otimes t) = 1$. Define the ucp map
$$\theta : M \otimes B(H\_\gamma) \to M : \theta(x) = (1 \otimes t^\*)(\alpha\_\rho \otimes \text{id})(x) (1 \otimes t) \; .$$
By the coaction property and the fact that $\alpha\_\varepsilon = \text{id}$ proven above, $\theta(\alpha\_\gamma(x)) = x$ for all $x \in M$. Fix a unitary $u \in \mathcal{U}(M)$. Since $\theta(\alpha\_\gamma(u)) = u$ is a unitary and $\|\alpha\_\gamma(u)\| \leq 1$, we find that $\alpha\_\gamma(u)$ belongs to the multiplicative domain of $\theta$. We have that $\alpha\_\gamma(u)^\* \alpha\_\gamma(u) \leq \alpha\_\gamma(u^\*u) = 1$. Applying $\theta$ and using that $\alpha\_\gamma(u)$ belongs to the multiplicative domain of $\theta$, we find that
$\theta(1-\alpha\_\gamma(u)^\* \alpha\_\gamma(u)) = 0$. Below I will prove that $\theta$ is faithful. So, we conclude that $\alpha\_\gamma(u)^\* \alpha\_\gamma(u) = 1$ for every unitary $u \in \mathcal{U}(M)$. This implies that $\alpha\_\gamma$ is multiplicative.
It remains to prove that $\theta$ is faithful. Assume that $x \in M \otimes B(H\_\gamma)$ such that $\theta(x^\* x) = 0$. Then, $(\alpha\_\rho \otimes \text{id})(x)(1 \otimes t) = 0$. Apply $\alpha\_\gamma \otimes \text{id} \otimes \text{id}$ to conclude that
$$(1 \otimes s^\* \otimes 1) ((\alpha\_\gamma \otimes \text{id})\alpha\_\rho \otimes \text{id})(x) (1 \otimes 1 \otimes t) = 0 \; .$$
Using the coaction property of $\alpha$ and the fact that $\alpha\_\varepsilon = \text{id}$ as proven above, the left hand side of the above expression equals
$$x (1 \otimes s^\* \otimes 1)(1 \otimes 1 \otimes t) = x \; .$$
So $x = 0$ and the faithfulness of $\theta$ is proven.
| 5 | https://mathoverflow.net/users/159170 | 439543 | 177,522 |
https://mathoverflow.net/questions/439551 | 1 | Let $\mathcal{L}(E)$ be the algebra of all bounded linear operators on a complex Hilbert space $E$.
On $\mathcal{L}(E)^2$, we have two equivalent norms:
\begin{eqnarray\*}
N\_1(A\_1,A\_2)
&=&\sup\left\{\|A\_1x\|^2+\|A\_2x\|^2,\;x\in E,\;\|x\|=1\;\right\},
\end{eqnarray\*}
and
$$N\_2(A\_1,A\_2)=\sup\left\{|\langle A\_1x,y\rangle|^2+|\langle A\_2x,y\rangle|^2,\;x,y\in E,\;\|x\|=\|y\|=1\;\right\}.$$
>
> Assume that $A\_1A\_2=A\_2A\_1$ and $A\_1$ et $A\_2$ are normal operators on $E$. How to show that
> $$N\_1(A\_1,A\_2)= N\_2(A\_1,A\_2)?$$
>
>
>
**My attempt:**
Notice that by the Cauchy–Schwarz inequality we have always $N\_2(A\_1,A\_2)\leq N\_1(A\_1,A\_2)$.
Now we aim to prove that the converse inequality holds when $A\_1A\_2=A\_2A\_1$ and $A\_1$ and $A\_2$ are normal operators on $E$. I tried to apply the spectral theorem.
Since $A\_1$ and $A\_2$ are commuting normal operators, il is well known that there exists a suitable measure space $(X,\mu)$; $\mu(X)<\infty$,
two functions $\varphi\_1,\varphi\_2\in L^\infty(\mu)$ and a unitary operator $U:E\longrightarrow L^2(\mu)$, such that each $A\_k$ is unitarily equivalent to multiplication by $\varphi\_k$, $k=1,2$. i.e.
$$UA\_kU^\*f=\varphi\_kf,\;\forall f\in E,\,k=1,2.$$
So, we can write
$$A\_kf=\varphi\_kf,\;\forall f\in L^2(\mu),\,k=1,2.$$
Hence,
$$\langle A\_kf\;,\;g\rangle=\langle \varphi\_kf\;,\;g\rangle=\int\_X\varphi\_k f\bar{g}d\mu,$$
and
$$\|A\_kf\|^2=\langle A\_kf\;,\;A\_kf\rangle=\langle \varphi\_kf\;,\;\varphi\_kf\rangle=\int\_X|\varphi\_k|^2|f|^2d\mu.$$
I am trying to solve the following question, but I did not reach to any answer, I would be so glad if anyone could help me on that.
| https://mathoverflow.net/users/113054 | Prove that $N_1(A_1,A_2)= N_2(A_1,A_2)$ when $A_1A_2=A_2A_1$ and $A_1$ and $A_2$ are normal operators | I will follow the OP's initial observation and notation. From the formula
$$\|A\_1 f\|^2+\|A\_2 f\|^2=\int\_X\left(|\varphi\_1|^2+|\varphi\_2|^2\right)|f|^2$$
it is clear that
$$N\_2(A\_1,A\_2)\leq N\_1(A\_1,A\_2)\leq\mathrm{ess}\sup\left(|\varphi\_1|^2+|\varphi\_2|^2\right).$$
Hence it suffices to show that, for any $\varepsilon>0$, we have
$$N\_2(A\_1,A\_2)>\mathrm{ess}\sup\left(|\varphi\_1|^2+|\varphi\_2|^2\right)-\varepsilon.$$
Let $C$ denote the essential supremum on the right-hand side. By definition, there is a set $U\subset X$ of positive measure on which $|\varphi\_1|^2+|\varphi\_2|^2$ is at least $C-\varepsilon/2$ (pointwise). This set has a subset $V\subset U$ of positive measure such that both $\varphi\_1(V)\subset\mathbb{C}$ and $\varphi\_2(V)\subset\mathbb{C}$ can be covered by a disk of radius $\sqrt{\varepsilon}/4$. Then,
$$\frac{1}{\mu(V)}\int\_V|\varphi\_k|^2-\left|\frac{1}{\mu(V)}\int\_V\varphi\_k\right|^2=\frac{1}{\mu(V)}\int\_V\left|\varphi\_k-\frac{1}{\mu(V)}\int\_V\varphi\_k\right|^2\leq\varepsilon/4$$
shows that
$$\left|\frac{1}{\mu(V)}\int\_V\varphi\_1\right|^2+\left|\frac{1}{\mu(V)}\int\_V\varphi\_2\right|^2\geq\frac{1}{\mu(V)}\int\_V\left(|\varphi\_1|^2+|\varphi\_2|^2\right)-\varepsilon/2\geq C-\varepsilon.$$
Choosing both $f$ and $g$ to be the $L^2$-normalized indicator function $\mathbf{1}\_V/\sqrt{\mu(V)}$, the previous inequality becomes
$$|\langle A\_1f,g\rangle|^2+|\langle A\_2f,g\rangle|^2\geq C-\varepsilon.$$
The proof is complete.
| 2 | https://mathoverflow.net/users/11919 | 439559 | 177,530 |
https://mathoverflow.net/questions/439344 | 3 | Let $X$ be a real analytic vector field defined on $\mathbb{R}^n$. Assume the origin $0 \in \mathbb{R}^n$ is a zero of $X$. Assume, furthermore, that we know that the center-stable manifold (in the sense of [Al Kelley - The stable, center-stable, center, center-unstable, unstable manifolds](https://doi.org/10.1016/0022-0396(67)90016-2)) of $X$ at $0$ is actually *stable*. In this context, this means that every trajectory starting in the center-stable manifold close enough to $0$ converges to $0$. The center-stable manifold being stable implies that it is unique (also in Kelley's article above).
>
> Do we know in this situation whether the center-stable manifold is analytic?
>
>
>
If not,
>
> Are there any known sufficient conditions that allow one to conclude that the center-stable manifold is analytic?
>
>
>
| https://mathoverflow.net/users/43097 | Analyticity of central stable manifolds | Quick answer to the first question: no, there is no reason why it should be analytic. Take *e.g.* the parametric vector field (written as a Lie derivative)$$X(x,y):=-x^3\partial\_x+(y+\alpha x)\partial\_y~~~,~\alpha\in\mathbb{R}.$$ Here the center manifold is a $C^\infty$ regular curve through the origin, tangent there to the vector $\left[1,-\alpha\right]$. If I understood correctly what the center-stable manifold being stable means, then $X$ satisfies that property.
If $\alpha=0$, then clearly the center manifold is the line $\left\{y=0\right\}$, hence analytic. Yet for every nonzero value of $\alpha$ it cannot be analytic. The easiest way to see this is to express the center manifold as the graph $\left\{y=-\alpha x+f(x)\right\}$, and this can be done explicitly by solving the underlying (affine) differential equation $x^3y'=y+\alpha x$. On the other hand, one computes formally the Taylor series of $f$ at $0$ by looking for a $\hat{f}(x)=\sum\_{n>1} a\_n x^n\in\mathbb{R}[[x]]$ solving the same differential equation. Both objects are unique and, if $f$ were to be analytic, then the Taylor series $\hat{f}$ would be convergent and its sum would coincide with $f$. However it is not difficult to show that $a\_n$ diverges like $\mathrm{cst}\sqrt{n!}$, hence the Taylor series at $0$ has a null convergence radius.
| 2 | https://mathoverflow.net/users/24309 | 439561 | 177,531 |
https://mathoverflow.net/questions/439187 | 4 | I am self-studying branched coverings. I read it from B. Maskit's Kleinian groups book. I want some more references for reading branched covers. In particular, I want to understand how to create branched covers of a given topological space or more precisely branched covers of surfaces.
If anyone shears related references like books or articles it will be a great help.
| https://mathoverflow.net/users/490039 | Books for learning branched coverings | Montesinos wrote several papers defining the meaning of *branched coverings* and proving basic properties(not just between manifolds, but for general topological spaces):
*Montesinos-Amilibia, José María*, Branched folded coverings and 3-manifolds, Castrillón López, Marco (ed.) et al., Contribuciones matemáticas en honor a Juan Tarrés. Madrid: Universidad Complutense de Madrid, Facultad de Ciencias Matemáticas (ISBN 978-84-695-4421-1). 295-315 (2012). [ZBL1297.57008](https://zbmath.org/?q=an:1297.57008).
Despite the title, he works in much greater generality than 3-manifolds. In the paper he also refers to his earlier work (which I cannot access):
*Montesinos-Amilibia, José María*, Branched coverings after Fox, Bol. Soc. Mat. Mex., III. Ser. 11, No. 1, 19-64 (2005). [ZBL1104.57002](https://zbmath.org/?q=an:1104.57002).
You should be able to get it through your interlibrary loan (I am not sufficiently motivated to do so myself).
Here is another paper, which you can find online and has basic definitions and properties. One advantage is that the paper also has a definition of a PL branched covering, which is much easier to digest:
*Montesinos-Amilibia, José-María*, [**Open 3-manifolds and branched coverings: a quick exposition**](https://eudml.org/doc/228624), Rev. Colomb. Mat. 41, No. 2, 287-302 (2007). [ZBL1149.57300](https://zbmath.org/?q=an:1149.57300).
Everybody else appear to be happy to work with a much more limited definition which I explain [here](https://math.stackexchange.com/questions/722017/a-more-general-definition-of-branched-covering/724717#724717).
Note, however, that there is a disagreement even about the meaning of the word "covering" which many traditionalists among complex geometers interpret simply as a locally biholomorphic map, something a topologist would find abhorrent.
| 4 | https://mathoverflow.net/users/39654 | 439567 | 177,532 |
https://mathoverflow.net/questions/439521 | 2 |
>
> Is there a general $\alpha$-tuple implementation that is of height $2$, that both doesn't require infinity of the naturals, and is at the same time stable under lack of Extensionality?
>
>
>
My own try to solve this question depends on a modified Holmes ordered pairs.
*Define:* $\langle x,y,z,..,s \rangle = \{ (x,1), (y,2),(z,3),..,(s,n) \}$
Where $1,2,3,..,n$ are the usual von Neumann naturals; and where $(,)$ is defined after Holmes (see [page 221](https://randall-holmes.github.io/head.pdf)), but with slight modification, as:
$$(x,y)= \{\{x',0,1\},\{x',2,3\} ,\{y',4,5\} ,\{y',6,7\}\mid \\x' \in x,y' \in y \} \cup \{x \mid \not \exists k \ (k\in x) \} \\ \cup \{y \mid \not \exists k \ (k\in y) \}$$
This tuple can work even if Extensionality fails, can be of any ordinal length and it doesn't beg having infinitely many naturals (as Quine-Rosser pairs demand), and it is just $2$-type higher than its projections, which I think it's the minimal height a tuple can have if it's to meet these criteria. However, it does have a pretty much complex definition.
>
> Are there simpler general $\alpha$-tuple implementations that can meet the above mentioned three criteria?
>
>
>
| https://mathoverflow.net/users/95347 | Are there known general tuple implementations that are 2 types high and withstand absence of extensionality and infinity? | This is a partial answer to this question:
I was always under the impression that a type-level pair must depend on having some kind of well ordered infinite class of objects, well at least this was the experience I found with the Quine-Rosser pair, and in addition it should presuppose Extensionality (like Quine-Rosser pairs, or even Holmes's $1$-high type pairs). However, it's only today when I came to realize that this is not necessarily correct.
I think I can cook up a pair that fulfills the aforementioned criteria.
Lets work in $\sf ZFA \neg C$.
Now, let $F$ be a total injective function that sends sets to nonempty sets of nonempty sets not having sets $1,2$ among their elements. That is: $$ F(x)=y \to \\ \exists z\, (z \in y) \land \forall z (z \in y \to \exists u \, (u \in z) \land 1 \not \in z \land 2 \not \in z)$$
Now, we define an "inserter" function $I\_\alpha$ that inserts $\alpha$ to all elements of a set, that is:
$$I\_\alpha(x)=\{y \cup \{\alpha\} \mid y \in x \}$$
Now, we define the following pair:
$$(x,y) = I\_1 (F(x)) \cup I\_2 (F(y))$$
We can easily retrieve the $\alpha^{th}$ projection of $(x,y)$: We take the set of all elements of $(x,y)$ that have $\alpha$ among their elements, apply the de-inserter function $I\_\alpha^{-1}$ on it, then apply $F^{-1}$ and we get the $\alpha^{th}$ projection of $(x,y)$.
We can extend that to any $\lambda$-tuple: $$(x\_1,x\_2,...)^\lambda = \underset {\alpha < \lambda} \bigcup I\_\alpha(F(x\_\alpha))$$
Of course, to answer the above question, we can simply take the tuple to be: $$\{\{(x\_1,x\_2,...)^\lambda\}\}$$
The main drawback is that this definition is not that general, for instance it doesn't work in the usual known models of $\sf NFU$, where we have strictly more empty objects than nonempty. In such models I only know of pairs with height of at least $2$ that can do the job.
| 1 | https://mathoverflow.net/users/95347 | 439569 | 177,534 |
https://mathoverflow.net/questions/439545 | 6 | Let $F$ be a number field and $\mathbb{A}$ its adele ring. $G$ be a classical group and $
\pi$ be a unitary cuspidal automorphic representation of $G(\mathbb{A})$.
Then I am wondering whether there is a known theorem on the local component of $\pi$. For example, the statement I am expecting is $\pi\_v$ is supercuspidal, square-integrable, tempered or generic etc for all places $v$.
Any comments are welcome!
| https://mathoverflow.net/users/29422 | Local component of cuspidal automorphic representation | Let me work in the category of $L^2$-automorphic representations. Assuming your global representation $\pi$ is irreducible, about the only thing you can say about an arbitrary local component $\pi\_v$ is that it is an irreducible smooth representation of $G\_v$. For almost all $v$, you can also say $\pi\_v$ will be spherical.
I don't know exactly which groups you include in classical groups (unitary groups? non-quasi-split forms?) but if you allow compact groups then the trivial representation is cuspidal, and in positive rank it is locally non-generic everywhere.
For (split) SO(5) you have Saito-Kurokawa lifts, which are locally non-tempered. Also, many cuspidal representation of SO(5) are not generic everywhere. See references on Siegel modular forms, SO(5) or GSp(4).
As mentioned in comments, $\pi\_v$ can typically only be supercuspidal or discrete series at a finite number of places (e.g., if G is GL($n$) with $n > 1$).
| 8 | https://mathoverflow.net/users/6518 | 439574 | 177,536 |
https://mathoverflow.net/questions/374863 | 0 | If $(M,g)$ is a smooth Riemannian manifold and $c : [a,b] \to M$ is a smooth embedded simple curve on $M$, it is always possible to choose locally a Riemannian metric $g\_0$ on $M$ for which $c$ is a geodesic for $g\_0$. As I understand, this can be done by pulling back a tubular neighborhood of $c$ to a disc bundle along $c$. There is always a flat Riemannian metric for which the $0$-section ($c$ itself) is a smooth geodesic. My question is: is it possible to locally realize $c$ as a geodesic to a metric such that the tube is positively curved? For instance, one of constant sectional curvature? I mean, can $g\_0$ be positively curved? If it helps, $M$ can be assumed to be a surface.
| https://mathoverflow.net/users/94097 | Discs bundles along a curve and positive curvature | It appears to me that your question does not involve the metric $g$ at all. Any connected embedded curve in a smooth $n$-manifold has a tubular neighborhood diffeomorphic to $T = I\times D$,
where $I$ is an interval or a circle and $D$ is the unit $(n-1)$-dimensional disk. You can map $T$ into the unit sphere such that $I\times \{0\}$ maps into a geodesic and and $T$ onto a tubular neighborhood of that geodesic. If $g\_0$ is the pullback of the standard metric on the sphere, then it has positive sectional curvature $1$.
| 3 | https://mathoverflow.net/users/613 | 439578 | 177,538 |
https://mathoverflow.net/questions/437768 | 1 | Suppose we have a function $f:\mathbb{R}^n\to\mathbb{R}$ that is at least three times differentiable. Clearly, there is a relationship between the symmetric trilinear form $$T\_1=\nabla^3f(x),$$ and the $n^2 \times n$ matrix $$T\_2=\frac{\partial \operatorname{vec}(\nabla^2f(x))}{\partial x^T}.$$ That relationship being a particular reshape operation that can take you from one to the other. In other words, they both contain the same information, but in different forms. According to "Matrix Differential Calculus with Applications in Statistics and Econometrics" (Magnus and Neudecker), at least in my understanding, $T\_2$ is the true derivative, as it is the matrix of all third order partial derivatives of $f$. Although, the text I have just cited doesn't consider third-order differentials extensively.
Let's define $\|\cdot\|$ as the norm induced by the vector 2-norm, so that $\|T\_2\|$ is the largest singular value of $T\_2$, and the following holds for $T\_1$: $$\|T\_1\|=\max\_{h:\|h\|\_2\leq 1}|\nabla^3f(x)[h,h,h]|.$$
Now, my question is as follows: is there any connection between the norms of the two operators For example, if $\|T\_1\|=m$, then does that imply $\|T\_2\|=m$?
Alternatively, would a Lipschitz bound on $\nabla^2f(x)$ provide a bound on the norm of $T\_2$ in a similar way as it does for $T\_1$? If $\nabla^2f(x)$ is $m$-Lipschitz, then my understanding is that $\|T\_1\|\leq m$, but this seems to imply that $T\_1$ is the correct derivative.
| https://mathoverflow.net/users/484618 | Third order matrix differential norm | First, we will denote the $n^2\times n$ matrix of third-order derivatives as $\mathbf{K}$, which has the following structure:
$$
\mathbf{K}=\frac{\partial\operatorname{vec}(\mathbf{H})}{\partial\mathbf{x}^T}=\begin{bmatrix}
\frac{\partial f}{\partial \mathbf{x}\_1\mathbf{x}\_1\mathbf{x}\_1} & \cdots & \frac{\partial f}{\partial \mathbf{x}\_1\mathbf{x}\_1\mathbf{x}\_{n-1}} & \frac{\partial f}{\partial\mathbf{x}\_1\mathbf{x}\_1\mathbf{x}\_n} \\
\frac{\partial f}{\partial \mathbf{x}\_1\mathbf{x}\_2\mathbf{x}\_1} & \cdots & \frac{\partial f}{\partial \mathbf{x}\_1\mathbf{x}\_2\mathbf{x}\_{n-1}} & \frac{\partial f}{\partial \mathbf{x}\_1\mathbf{x}\_2\mathbf{x}\_n} \\
\vdots & \ddots & \vdots & \vdots \\
\frac{\partial f}{\partial \mathbf{x}\_n\mathbf{x}\_n\mathbf{x}\_1} & \cdots & \frac{\partial f}{\partial\mathbf{x}\_n\mathbf{x}\_n\mathbf{x}\_{n-1}} & \frac{\partial f}{\partial \mathbf{x}\_n\mathbf{x}\_n\mathbf{x}\_n}
\end{bmatrix}
$$
and we note that the tri-linear operator is defined as follows:
$$
\nabla^3f(\mathbf{x})[\mathbf{a},\mathbf{b},\mathbf{c}]=\sum\_{i,j,k}^n\frac{\partial f}{\partial\mathbf{x}\_i\partial\mathbf{x}\_j\partial\mathbf{x}\_k}\mathbf{a}\_{i}\mathbf{b}\_{j}\mathbf{c}\_{k}
$$
We will show the following relationship holds:
$$
\|\nabla^3f(\mathbf{x})\|\leq\|\mathbf{K}\|\leq \sqrt{n}\|\nabla^3f(\mathbf{x})\|
$$
**Proof**:
I claim that $\nabla^3f(\mathbf{x})[\mathbf{a},\mathbf{b},\mathbf{c}]=(\mathbf{a}\otimes \mathbf{b})^T\mathbf{K}\mathbf{c}$ -- that is to say $\mathbf{K}$ can perform the same action as the tri-linear operator. This can be seen by simply writing out the RHS and comparing it to the form of the tri-linear operator. Taking the absolute value and maximizing over all vectors with norm less than 1 gives us the following:
$$ \|\nabla^3f(\mathbf{x})\|=\max\_{\|\mathbf{a},\mathbf{b},\mathbf{c}\|\leq 1}|\nabla^3f(\mathbf{x})[\mathbf{a},\mathbf{b},\mathbf{c}]|=\max\_{\|\mathbf{a},\mathbf{b},\mathbf{c}\|\leq 1}|(\mathbf{a}\otimes \mathbf{b})^T\mathbf{K}\mathbf{c}|
$$
This immediately implies the LHS of the bound in question, as the RHS of the preceding equation is bounded above by $\|\mathbf{K}\|$. Now we can bound it below by choosing appropriate vectors. In particular, take $\mathbf{c}$ to be the left-most right singular vector of $\mathbf{K}$, and let $\mathbf{u}$ be the left-most left singular vector of $\mathbf{K}$. For now, we leave $\mathbf{a}$ and $\mathbf{b}$ unspecified. Computing the matrix-vector product $\mathbf{K}\mathbf{c}$ gives the following result:
$$
\|\mathbf{K}\|\cdot|(\mathbf{a}\otimes \mathbf{b})^T\mathbf{u}|=|(\mathbf{a}\otimes \mathbf{b})^T\sigma\_{1}\mathbf{u}|\leq\|\nabla^3f(\mathbf{x})\|
$$
where $\sigma\_1$ is the largest singular value of $\mathbf{K}$. Now, we will rewrite the LHS as $\|\mathbf{K}\|\cdot|\mathbf{a}^T\operatorname{mat}(\mathbf{u})\mathbf{b}|$. We know that $\|\operatorname{mat}(\mathbf{u})\|\_F=1$ as the vector $\mathbf{u}$ has unit norm. Thus we can bound the largest singular vector of $\operatorname{mat}(\mathbf{u})$ below by $1/\sqrt{n}$.
Then let $\mathbf{b}$ be the left-most right singular vector of $\operatorname{mat}(\mathbf{u})$, and let $\mathbf{a}$ be the left-most left singular vector of the same matrix. We can now write the following:
$$
\|\mathbf{K}\|\cdot|\mathbf{a}^T\operatorname{mat}(\mathbf{u})\mathbf{b}|\geq\frac{1}{\sqrt{n}}\|\mathbf{K}\|
$$
Rearranging gives us the upper bound of the result in question.
| 0 | https://mathoverflow.net/users/484618 | 439585 | 177,540 |
https://mathoverflow.net/questions/439494 | 0 | Let $P$ be a closed polygon defined by the sequence
$p\_0,\,\dots,\,p\_{n-1},p\_0$ of points.
>
> **Question:**
>
>
> how can one construct, with straightedge and compass alone, another sequence of points $q\_0,\,\dots,q\_{n-1}$ such that:
>
>
> * $q\_i$ lies on the bisector of $p\_i$ and $p\_{i+1}$
> * $q\_i,p\_{i+1}$ and $q\_{i+1}$ are collinear
>
>
>
| https://mathoverflow.net/users/31310 | Constructing a polygon from another with collinearity constraints | Let $L\_i$ be the bisector of $p\_i$ and $p\_{i+1}$, and let $f\_i \colon L\_i \to L\_{i+1}$ be the central projection through $p\_{i+1}$. This is a projective transformation with constructible coefficients. Your question is about the fixed points of the projective transformation $f\_n \circ \cdots \circ f\_1$. The coefficients of this projective transformation are also constructible. The fixed points are found by solving a quadratic equation, which can be done by compass and ruler.
It follows also that the number of solutions is among $0, 1, 2$ or $\infty$, although it is not clear whether there are examples with no solutions.
| 2 | https://mathoverflow.net/users/98590 | 439611 | 177,547 |
https://mathoverflow.net/questions/312876 | 6 | I, ask my question as a comment [in this post](https://mathoverflow.net/questions/311630/nash-isometric-embedding-for-noncompact-manifolds). Without answer I post a more detailed version.
I am looking for a reference about $C^\infty$ Nash isometric embedding for non compact manifold.
My question is what are exactly the hypothesis needed on a complete manifold $M$ in order to be properly isometrically embedded into some $\mathbb{R}^n$ (I am not very interested by the optimal dimension $n$) **and** which admits a nice projection (or equivalently a tubular neighborhood of fixed width). Any modern reference will appreciated.
Thx in advance
| https://mathoverflow.net/users/9253 | Nash embedding for complete manifold | Your nice projection is usually called *positive reach*.
In order to have it, one has to have both curvature bounds and a lower bound on injectivity radius.
Plus the volume growth must be at most polynomial.
Say the Lobachevsky plane does not admit a smooth embedding positive reach; see this question: [Does a Riemannian manifold with bounded geometry..](https://mathoverflow.net/q/124840).
I suspect that there are more conditions.
| 6 | https://mathoverflow.net/users/1441 | 439613 | 177,549 |
https://mathoverflow.net/questions/439576 | 0 | Let $X\sim \mathcal{N}(\mu,\sigma^2)$ be a Gaussian random variable with random mean $\mu\sim {\sf Bernoulli}(p)$, i.e., $\mu=1$ with probability $p$ and $\mu=0$ with probability $1-p$. In other words, to draw a realization of $X$, first one flips a coin to decide whether $\mu=0$ or $\mu=1$ and then, one samples from the Gaussian distribution with chosen mean.
Consider the family $\mathcal{M}$ of Bernoulli random variables taking on the values $0$ or $1$.
Let $\mathcal{I}\left(Y,X\right)$ be the mutual information between $Y$ and $X$ (a natural definition can be found, e.g., [in this mathoverflow post](https://mathoverflow.net/questions/321364/mutual-information-between-continuous-and-discrete-variables-from-numerical-data)).
My intuition tells that a global solution for the optimization problem
$$\max\_{Y\in\mathcal{M}} \mathcal{I}(Y,X)$$
is given by $\mu$ itself, i.e.,
$$\mathcal{I}(\mu,X) = \max\_{Y\in\mathcal{M}} \mathcal{I}(Y,X).$$
Any reference or ideas on how to prove or disprove this claim would be appreciated.
| https://mathoverflow.net/users/138242 | Maximal mutual information between a continuous and a discrete random variables | $\newcommand{\de}{\delta}\newcommand{\M}{\mathcal M}
\newcommand{\ep}{\varepsilon}
\newcommand{\thh}{\theta}\newcommand\I{\mathcal I}\newcommand{\Si}{\Sigma}$Here is the (slightly edited) definition of the mutual information given in the [answer](https://mathoverflow.net/questions/321364/mutual-information-between-continuous-and-discrete-variables-from-numerical-data) linked in the OP:
>
> Let $D$ be any discrete random variable (r.v.) with distinct values $d\_i$ taken with probabilities $p\_i=P(D=d\_i)>0$ for $i\in I$, where $I$ is a denumerable (that is, at most countable) set. Let $X$ be any r.v. (defined on the same probability space as $D$), with values in any nonempty set $S$ (given also some sigma-algebra $\Sigma$ over $S$, to make $S$ a measurable space). Let $\nu$ be the probability distribution of $X$, so that $\nu(B)=P(X\in B)$ for all $B\in\Sigma$. For each $i\in I$ and each $B\in\Sigma$, let
> \begin{equation\*}
> \nu\_i(B):=P(D=d\_i,X\in B).
> \end{equation\*}
> Then $\nu\_i$ is a (sub-probability) measure absolutely continuous with respect to $\nu$, so that we can consider a Radon--Nikodym density
> \begin{equation\*}
> \rho\_i:=\frac{d\nu\_i}{d\nu}
> \end{equation\*}
> of the measure $\nu\_i$ with respect to $\nu$, so that the values of $\rho\_i$ are in $[0,1]$.
>
>
> Then the mutual information between $D$ and $X$ is defined as follows:
> \begin{equation\*}
> \I(D,X):=\sum\_{i\in I}\int\_S d\nu\;\rho\_i\ln\frac{\rho\_i}{p\_i}.
> \end{equation\*}
>
>
>
---
Let us now answer the current question.
Much more generally than in the OP, let $X$ be any r.v. as in the highlighted text above. Let $\nu$ denote the distribution of $X$.
Let $\M$ denote the set of all Bernoulli r.v.'s $Y$ (defined on the same probability space as $X$) with $P(Y=1)\in(0,1)$;
we suppose that the probability space is rich enough so that for any coupling $\gamma$ of any Bernoulli distribution and the distribution $\nu$ of $X$ there is a r.v. $Y$ such that the joint distribution of the pair $(Y,X)$ is $\gamma$.
In view of the highlighted definition, for any $Y\in\M$,
\begin{equation\*}
\I(Y,X)=\int\_S d\nu\;
\Big(\rho\ln\frac{\rho}{p}+(1-\rho)\ln\frac{1-\rho}{1-p}\Big),
\end{equation\*}
where
$p:=P(Y=1)$, $\rho:=\frac{d\nu\_1}{d\nu}$, and $\nu\_1(B):=P(Y=1,X\in B)$ for all $B\in\Si$; we are assuming the convention that $0\times\text{anything}:=0$.
We want to maximize $\I(Y,X)$ over all $Y\in\M$. This is very easy to do, noting that $u\ln\frac{u}{p}+(1-u)\ln\frac{1-u}{1-p}$ is convex in $u\in[0,1]$. So, $\I(Y,X)$ is maximized when $\rho=1\_A$ for some $A\in\Si$, and then $p=\nu(A)$ and $\I(Y,X)=\int\_A d\nu\;
\ln\frac1{p}=p\ln\frac1p$. Maximizing the latter expression in $p\in(0,1)$, we see that, contrary to the conjecture in the OP,
\begin{equation}
\max\_{Y\in\M}\I(Y,X)=\frac1e,
\end{equation}
for any r.v. $X$ whatsoever provided that $X$ is defined on a rich enough probability space.
| 1 | https://mathoverflow.net/users/36721 | 439618 | 177,550 |
https://mathoverflow.net/questions/439624 | 2 | Let $\gamma: S^1 \to \partial B \subset \mathbf{R}^3$ be a smooth, simple closed curve in the boundary of the unit ball. Suppose that $\gamma$ intersects every horizontal plane $\Pi\_t = \{ z = t\}$ at most twice:
\begin{equation}
\# \gamma(S^1) \cap \Pi\_t \leq 2 \quad \text{for all $t$.}
\end{equation}
>
> Does $\gamma$ bound a unique minimal disk (resp. minimal surface)?
>
>
>
| https://mathoverflow.net/users/103792 | A geometric criterion for uniqueness in the Plateau problem? | The answer is no.
Take two parallel circles of unit radius in $z=\pm \epsilon$ with $\epsilon$ small. Tilt the two circles very slightly toward one another. This satisfies your hypotheses. There are clearly at least three minimal surfaces with this boundary the tilted flat disks and a stable and unstable annulus (obtained by perturbing the appropriate pieces of a catenoid).
If you want non-uniqueness in the class of disks then form a small bridge between the two closest points on the circle. This boundary also satisfies your hypotheses. The [bridge principle](https://link.springer.com/article/10.1007/BF01192091) implies there is a stable minimal surfaces with this boundary that should look like the two flat disks joined by a small bridge. However, it is pretty clear the area minimizing disk is the "ribbon" solution going around the outside. Hence, you have two distinct minimal disks.
| 5 | https://mathoverflow.net/users/127803 | 439628 | 177,553 |
https://mathoverflow.net/questions/216338 | 7 | Let $f,g:(D^2,j\_\mathrm{std})\to(B^{2n}(1),J)$ be two pseudo-holomorphic maps. The following unique continuation result is well-known (it may be proved using either Aronszajn's Lemma or the Carleman similarity principle as in Floer--Hofer--Salamon):
**Lemma:** Suppose $f=g$ over a neighborhood of $0\in D^2$. Then $f=g$ everywhere.
Is there a similar unique continuation result for pseudo-holomorphic *submanifolds* (instead of *maps*)? Here is a precise formulation of what I expect might be true:
**Conjecture**: Let $U,V\subseteq(D^2)^\circ$ be two closed topological disks with smooth boundary, and let $\phi:U^\circ\to V^\circ$ be a biholomorphism (which necessarily extends continuously to a homeomorphism $U\to V$). Suppose $f=g\circ\phi$ over $U$. Then $\phi$ extends holomorphically to an open neighborhood of $U$ (and hence $f=g\circ\phi$ over this neighborhood by the unique continuation result for maps).
Is some statement along these lines known?
| https://mathoverflow.net/users/35353 | Does pseudo-holomorphic *submanifolds* satisfy unique continuation? | The desired statement is in fact a well known property of pseudo-holomorphic maps. Lemma 1.3.1 in [Singularities and positivity of
intersections of J-holomorphic curves](https://mathscinet.ams.org/mathscinet-getitem?mr=1274930) by Dusa McDuff says that for a pair of pseudo-holomorphic maps $u,v:(D^2,0)\to(M,p)$ with $du(0)\ne 0$, either $u(0)=v(0)$ is an isolated point of intersection, or $v$ factors through $u$. This (combined with, say the fact that critical points of pseudo-holomorphic maps are isolated, see Lemma 1.2.1 in the same paper) gives the desired result.
| 2 | https://mathoverflow.net/users/35353 | 439651 | 177,559 |
https://mathoverflow.net/questions/439649 | 1 | Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $H=(H\_t, t\ge 0)$ be a stochastic process with continuous trajectories. Fix $T>0$. For $n \ge 1$, we define
$$
H\_{s,n} := \sum\_{i=1}^{2^n} H\left(\frac{(i-1) T}{2^n}\right) 1\_{ \left (\frac{(i-1) T}{2^n}, \frac{i T}{2^n} \right]}(s) \quad \forall s \in [0, T].
$$
Then for $(\omega, n) \in \Omega \times \mathbb N$, the map $s \mapsto H\_{s, n} (\omega)$ is a step function. Because $H$ has continuous trajectories, we have
$$
H\_{s, n} (\omega) \underset{n \rightarrow \infty}{\longrightarrow} H\_{s} (\omega) \quad \forall (\omega, s) \in \Omega \times [0, T].
$$
---
It is mentioned at page 35 of this [note](https://ipgold.epfl.ch/%7Eleveque/Lecture_Notes/sc1_final.pdf) that
>
> **Preliminary fact** If $\mathbb E[ \int\_0^T H\_s^2 \mathrm d s] < \infty$, then
> $$
> \mathbb{E}\bigg[\int\_0^T (H\_{s,n}-H\_s)^2 \mathrm d s\bigg] \underset{n \rightarrow \infty}{\longrightarrow} 0.
> $$
>
>
>
I have tried to verify this statement but got stuck. Could you elaborate on how to prove it?
---
**My attempt**
* Fix $\omega \in \Omega$. We define $f\_n:[0, T] \to \mathbb R$ by $f\_n (s) := H\_{s,n} (\omega)$. We define $f:[0, T] \to \mathbb R$ by $f (s) := H\_{s} (\omega)$. Then $\sup\_n \|f\_n\|\_\infty \le \|f\|\_\infty < \infty$. Also, $f\_n-f$ converges to $0$ pointwise. By dominated convergence theorem, we have
$$
\int\_0^T (f\_n-f)^2 \mathrm d s \underset{n \rightarrow \infty}{\longrightarrow} 0.
$$
* We define $Y\_n:\Omega \to \mathbb R$ by
$$
Y\_n (\omega) := \int\_0^T (H\_{s,n} (\omega)-H\_s(\omega))^2 \mathrm d s.
$$
As proved above, $Y\_n \underset{n \rightarrow \infty}{\longrightarrow} 0$ almost surely.
---
I posted [this](https://math.stackexchange.com/questions/4628234/how-is-mathbb-e-int-0t-h-s2-mathrm-d-s-infty-important-for-this-clai) question on MSE but have not received any answer so far.
| https://mathoverflow.net/users/477203 | How is $\mathbb E[ \int_0^T H_s^2 \mathrm d s] < \infty$ important for this claim? | This statement is false in general.
E.g., let $T=1$ and $p\_k:=2^{-k}$ for integers $k=1,2,\dots$, so that $\sum\_{k=1}^\infty p\_k=1$. Let $A\_1,A\_2,\dots$ be pairwise disjoint events with respective probabilities $p\_1,p\_2,\dots$.
Let
\begin{equation}
H\_t:=\sum\_{k=1}^\infty 1\_{A\_k}\,2^k\,\sum\_{j=0}^{2^k}\Big(1-8^k\Big|t-\frac j{2^k}\Big|\Big)\_+,
\end{equation}
where $u\_+:=\max(0,u)$.
Then
\begin{equation}
E\int\_0^1 H\_t^2\,dt=\sum\_{k=1}^\infty p\_k\; (2^k)^2\,2^k \int\_0^1 \Big(1-8^k\Big|t-\frac 1{2^k}\Big|\Big)\_+^2\,dt
=\sum\_{k=1}^\infty p\_k\; (2^k)^2\,2^k \frac23\,\frac1{8^k}=\frac23<\infty.
\end{equation}
On the other hand, for all $t\in(0,1]$ we have $H\_{t,n}\ge2^n$ on the event $A\_n$, which has probability $p\_n=\frac1{2^n}$. So,
\begin{equation}
E\int\_0^1 H\_{t,n}^2\,dt\ge(2^n)^2\frac1{2^n}\to\infty
\end{equation}
and hence
\begin{equation}
E\int\_0^1 (H\_{t,n}-H\_t)^2\,dt\to\infty\ne0
\end{equation}
as $n\to\infty$. $\quad\Box$
---
The problem here is, of course, that the process $(H\_t)$ is not bounded (by a nonrandom constant). A standard construction of the stochastic integral -- see e.g. Proposition 2.6 in Ch. 3 of Karatzas, Ioannis; Shreve, Steven (1991), Brownian Motion and Stochastic Calculus, 2nd ed. -- is done a bit differently: first, the integrand process is truncated to obtain a bounded process, and then the truncated, bounded process is approximated by simple, piecewise-constant processes, and thus the original integrand process is approximated by simple ones (in $L^2(\Omega\times[0,T])$).
| 2 | https://mathoverflow.net/users/36721 | 439654 | 177,560 |
https://mathoverflow.net/questions/439572 | 1 | I am working on $\mathbb{R}^4$ with the sign convention $(1,-1,-1,-1)$.
I wonder if there is Schwartz function $f(x)$ on $\mathbb{R}^4$ such that the support satisfies the condition $0<x^2 < 4m^2$ for some given positive constant $m$.
Here $x^2$ is of course with respect to the above metric, and $x=(x\_0,x\_1,x\_2,x\_3)$.
The first idea that comes to me is a function of the form
\begin{equation}
f(x)=g(x^2)
\end{equation}
where $g : \mathbb{R} \to \mathbb{R}$ is the smooth function defined by $g(t):=e^{-\frac{1}{4m^2-t}-\frac{1}{t}}$ for $0<t<4m^2$ and $0$ otherwise.
However, such $f$ does not show decay in the case $\lvert x\_1 \rvert \to \infty$ while $x^2=2m^2$ and as a result $\sup\_{x \in \mathbb{R}^4} \lvert x\_1 f(x) \rvert = \infty$.
I wonder if there exists a Schwartz function with causal support. Moreover, it would be better if $f(\Lambda x) =f(x)$ for all elements $\Lambda$ of the restricted Lorentz group.
Could anyone please provide an example?
| https://mathoverflow.net/users/56524 | Is there an example of a causally supported Schwartz function on $\mathbb{R}^4$ invariant under the Lorentz transform? | If Lorentz invariance is not required:
Let $\phi$ be any smooth bump function $\phi:\mathbb{R}\to\mathbb{R}$ that is non-zero precisely on $(0,4m^2)$ (including the one you used in the question statement).
Let $f:\mathbb{R}^4\to \mathbb{R}$ be given by
$$ f(x\_0, x\_1, x\_2, x\_3) = \phi(x\_0^2 - x\_1^2 - x\_2^2 - x\_3^2) \exp(-x\_0^2 - x\_1^2 - x\_2^3 - x\_3^3) $$
Since the Gaussian term is nowhere vanishing, $f$ has the desired support property.
Since $\phi$ is a smooth bump function, there is a sequence of numbers $M\_k$ such that $|\phi^{(k)}| \leq M\_k$. This shows that the $k$th order partial derivatives of $f$ are uniformly bounded by a $k$th degree polynomial in $(x\_0, x\_1, x\_2, x\_3)$ times the Gaussian, which shows that $f$ is Schwartz.
| 3 | https://mathoverflow.net/users/3948 | 439655 | 177,561 |
https://mathoverflow.net/questions/439635 | 14 | Suppose we have a diffeomorphism $f:{\mathbb{S}}^n\_{+}\to\mathbb{S}^n$ of class $C^1$ of the closed upper hemisphere onto a submanifold of $\mathbb{S}^n$ with boundary.
>
> **Question.** Is it possible to extend it to a diffeomorphism $F:\mathbb{S}^n\to\mathbb{S}^n$?
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I believe that when $n=2$ this should follow from the smooth Schoenflies theorem, but still some work is necessary. I think I know how to do, but I did not write a rigorous proof.
From what I understand when $n\geq 3$, the generalized Schoenflies theorem allows us to extend $f$ to a homeomorphism, but that is much less than extending to a diffeomorphism.
I expect that the answer might depend on $n$ ($n\geq 3$, $n=4$, $n\geq 5$).
I would greatly appreciate it if you could provide references where I cold find relevant theorems (if there are any), including the case $n=2$. I need references for a proper citation in my paper.
**Edit:** This question is strictly related to another post: [Gluing two diffeomorphisms together](https://mathoverflow.net/q/38498/121665)
| https://mathoverflow.net/users/121665 | Extending diffeomorphisms | The answer is positive and follows from Corollary 2 in
*Palais, Richard S.*, [**Extending diffeomorphisms**](http://dx.doi.org/10.2307/2032968), Proc. Am. Math. Soc. 11, 274-277 (1960). [ZBL0095.16502](https://zbmath.org/?q=an:0095.16502).
(A caveat: Palais is not entirely clear about the degree of smoothness he allows, he just says "differentiable." However, I think, it works for $C^k$-smooth map for every $k>0$.)
Applying Corollary 2 in the case of maps of closed $n$-balls $B^n$ to $S^n$, one obtains that if $\phi, \psi: B^n \to S^n$ are smooth embeddings, then there exists a diffeomorphism $F: S^n\to S^n$ such that $\phi=F\circ \psi$ on $B^n$. Now, take $\psi$ to be the identity embedding $B^n\to S^n$ (where $B^n$ is a hemisphere). Then it follows that $F$ is the desired extension of $\phi$.
PS: For some reason I had Palais' paper open in my browser for a week before you posted the question. :)
| 11 | https://mathoverflow.net/users/39654 | 439658 | 177,563 |
https://mathoverflow.net/questions/439071 | 5 | I am studying the enlightening article "The Picard Group of $\mathcal{M}\_{1, 1, S}$", written by Fulton and Olsson, but I have some problems with a proof.
**Setting**
Let $\mathcal{M}\_{1, 1, k}$ denote the stack of elliptic curves over an algebraically closed field $k$ with $char(k)=3$. We have
\begin{equation\*}
V = Spec(k[\lambda][1/(\lambda(\lambda - 1))]) \rightarrow \mathcal{M}\_{1, 1, k} \rightarrow \mathbb{A}^{1}\_{k}
\end{equation\*}
where the first map (which is étale) is the Legendre family, hence induced by the ellptic curve $E\_{V} \rightarrow V$
\begin{equation\*}
E\_{V} \colon Y^{2}Z = X(X-Z)(X - \lambda Z)
\end{equation\*}
and the second map is the morphism into the coarse moduli space induced by the $j$-invariant. The composition is induced by the ring map sending $j \mapsto \frac{2^{8}(\lambda^{2}-\lambda+1)^{3}}{\lambda^{2} (1 - \lambda)^{2}}$.
Recall that in $char(k)=3$ the elliptic curves have automorphism group $\mathbb{Z}/2\mathbb{Z}$ except that for the curve with $j$-invariant equal to $0$ (which in the family corresponds to $\lambda = -1$) in which case the automorphism group is isomorphic to $G = \mathbb{Z}/4\mathbb{Z} \ltimes \mathbb{Z}/3\mathbb{Z}$.
**The problem**
I want to understand how the group $G$ acts on the localization in $\mathcal{M}\_{1, 1, k}$ at the point corresponding to the curve whith automorphism group $G$. Since $V \rightarrow \mathcal{M}\_{1, 1, k}$ is étale, the problem is equivalent to understanding how it acts on the localization at $\lambda = -1$ in V.
**What it is said in the article**
For comfort, we can translate and consider $\mu = \lambda + 1$ so the localization is at $\mu = 0$ and we have the morphism $Spec(k[[\mu]]) \rightarrow Spec(k[[j]])$ induced by $j \mapsto \frac{\mu^{6}}{(\mu^{4}-1)}$ (since we hare in $char(k)=3$).
Now the group $G$ acts on $k[[\mu]]$. The group sits in an exact sequence
\begin{equation\*}
1 \rightarrow \{ \pm 1 \} \rightarrow G \rightarrow S\_{3} \rightarrow 1
\end{equation\*}
and the action on $k[[\mu]]$ factors through the action of $S\_{3}$ on $k[[\mu]]$ given by the two automorphisms
\begin{equation\*}
\alpha \colon \mu \mapsto -\mu
\end{equation\*}
and
\begin{equation\*}
\beta \colon \mu \mapsto \mu/(1-\mu) = \mu(1 + \mu + \mu^{2} + \ldots).
\end{equation\*}
**My question is**: How the authors of the article determine the actions of $\alpha$ and $\beta$?
**My attempt at a solution**
I think to know how $\alpha$ acts in such a way.
First of all, if we consider a model for an elliptic curve with $j$ invariant equal to 0, like $y^{2}=x^{3} + a\_{4}x + a\_{6}$, with $a\_{4} \ne 0$, the morphism of elliptic curves are of the form
\begin{equation\*}
\begin{split}
x &\mapsto u^{2}x' + r \\
y &\mapsto u^{3}y'
\end{split}
\end{equation\*}
and imposing the equality between the curve in $(x, y)$ and the curve in $(x', y')$, i.e. imposing the morphism being an automorphism, we obtain $u^{4}=1$ and $r^{3} + a\_{4}r + (1-u^{2})a\_{6}=0$. The $12$ elements $(u, r)$ which satisfy the equations are the group elements of $G$. In particular $\alpha = (i, 0)$.
To see how $\alpha$ acts:
let $Spec(k[x, y])/(y^{2} = x^{3} + a\_{2}x^{2} + a\_{4}x + a\_{6})) \rightarrow Spec(k)$ be a generic elliptic curve. On the coefficients $(a\_{2}, a\_{4}, a\_{6})$, the transformation $(u, r)=(i, 0)$ acts via $(i, 0) \star (a\_{2}, a\_{4}, a\_{6}) = (i^{2}a\_{2}, i^{4}a\_{4}, i^{6}a\_{6})$. Since we are interested at the point $(0, 1, 1)$ we consider the curve $y^{2} = x^{3} + \mu x^{2} + x + 1$ on which the action is $(i, 0) \star (\mu) = i^{2} \mu = - \mu$ and taking the localization at $\mu=0$ is exactly what we want to see the action of $\alpha= (i, 0)$ at the point $\lambda = -1$ (i.e. $\mu=0$).
My problem is that the previous argument cannot be applied to $\beta$.
Can somone help me? (maybe also with a simpler way to compute the action of $\alpha$)
Thanks
| https://mathoverflow.net/users/498211 | How the automorphism group of an elliptic curve acts at the localization of the stack $\mathcal{M}_{1, 1, k}$ at the corresponding point | From Section 3 of the paper, we see that the moduli stack of elliptic curves (over a ring $A$) has a quotient stack presentation $$ \mathcal{M}\_{1,1,A} \simeq [U/G] $$ where $U = \operatorname{Spec} A[a\_{1},a\_{3},a\_{2},a\_{4},a\_{6},\Delta^{-1}]$ and $G = \operatorname{Spec} A[u^{\pm},r,s,t]$. The group law on $G$ and the action of $G$ on $U$ come from making the change-of-variable \begin{equation} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix} \mapsto \begin{bmatrix} u^{2} & & r \\ u^{2}s & u^{3} & t \\ & & 1 \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix} \tag{1}\label{eqn01} \end{equation} in the Weierstrass equation $Y^{2}Z+a\_{1}XYZ+a\_{3}YZ^{2} = X^{3} + a\_{2}X^{2}Z + a\_{4}XZ^{2} + a\_{6}Z^{3}$ (see Silverman, *Arithmetic of Elliptic Curves*, Section III.1, Table 3.1).
Suppose $2 \in A^{\times}$. Given $a\_{1},a\_{3},a\_{2},a\_{4},a\_{6} \in A$ such that $\Delta \in A^{\times}$, we can change variables so that $a\_{1} = a\_{3} = 0$ (which implies $s = t = 0$), and furthermore we can replace $A$ by an etale cover $A \to A'$ and change variables so that \begin{equation} \begin{aligned} a\_{6} &= 0 \\ a\_{2}+a\_{4}+1 &= 0 \end{aligned} \tag{2}\label{eqn02} \end{equation} (i.e. write the equation in (dehomogenized) Legendre normal form \begin{equation} y^{2} = x(x-1)(x-\lambda) \tag{3}\label{eqn03} \end{equation} for $\lambda = a\_{4}$). The formula \eqref{eqn01} becomes $(x,y) \mapsto (u^{2}x+r,u^{3}y)$. The equation \eqref{eqn03} is transformed into \begin{equation} \textstyle y^{2} = (x+\frac{r}{u^{2}})(x+\frac{r-1}{u^{2}})(x+\frac{r-\lambda}{u^{2}}) \tag{4} \label{eqn04} \end{equation} and imposing the condition \eqref{eqn02} gives \begin{equation} \begin{aligned} r(r-1)(r-\lambda) &= 0 \\ \textstyle (1+\frac{r}{u^{2}})(1+\frac{r-1}{u^{2}})(1+\frac{r-\lambda}{u^{2}}) &= 0 \end{aligned} \tag{5} \label{eqn05} \end{equation} respectively.
Let $k$ be an algebraically closed field of $\operatorname{char} k = 3$, and set $A := k[[\mu]]$ and $\lambda = \mu-1$. The stabilizer $\Gamma\_{\overline{x}}$ can be identified with the set of pairs $(u,r) \in A^{\times} \times A$ satisfying \eqref{eqn05}, and the action $\Gamma\_{\overline{x}} \times A \to A$ is defined by \begin{equation} (u,r) \cdot \lambda := \textstyle \frac{1}{u^{4}}(r(r-1) + r(r-\lambda) + (r-1)(r-\lambda)) \end{equation} (namely the coefficient of $x$ in \eqref{eqn04}, the new $a\_{4}$). You can check that the solutions are \begin{equation} (u,r) = (\pm 1 , 0), (\pm \sqrt{\lambda},0), (\pm \sqrt{-1},1), (\pm \sqrt{\lambda-1},1), (\pm \sqrt{-\lambda},\lambda),(\pm \sqrt{1-\lambda},\lambda) \end{equation} which sends $\mu$ to \begin{equation} \mu,\frac{\mu}{\mu-1},-\mu,\frac{\mu}{\mu+1},\frac{-\mu}{\mu-1},\frac{-\mu}{\mu+1} \end{equation} respectively. It seems the authors don't specify what $\alpha,\beta$ are but if they're interested in checking a certain condition for a set of generators of $S\_{3}$, then it is enough to choose elements of order 2 and 3.
| 2 | https://mathoverflow.net/users/15505 | 439663 | 177,564 |
https://mathoverflow.net/questions/438926 | 2 | To the language of set theory add a primitive unary predicate $\operatorname {Universe}$ and a primitive total unary function $j$. Add all axioms of $\sf ZF$ in the language of this theory, i.e. the new primitives allowed to be used in instances of Separation and Replacement, I'll refer to this simply as $\sf ZFj$.
We add the following axioms:
**Cumulative:** $\forall X: \operatorname {Universe}(X) \to \exists \lambda: X=V\_\lambda$
**Modeling:** if $\psi$ is an axiom of $\sf ZFj$, and $\psi^X$ is the $``\in X"$ bounded form of $\psi$; then: $$\forall X: \operatorname {Universe}(X) \to \psi^X$$
**Elementarity:** if $\phi(x\_1,..,x\_n)$ is a formula in signature $\{=,\in\}$, having $``x\_1,..,x\_n\! \!"$ as its sole free variables, and none of them occur bound, then: $$\forall X: \operatorname {Universe}(X) \to\forall x \in X \, (j(x) \in X) \land \\\forall x\_1,..,x\_n \in X \\ (\phi(x\_1,..,x\_n) \iff \phi(j(x\_1),..,j(x\_n))) \\ \land \exists x \in X: j(x) \neq x$$
The intention is to render the restriction of $j$ to any Universe $X$ a non-trivial elementary embedding from $X \to X$.
**Reflection:** if $\phi$ is a formula [defined functions and predicates allowed] not using the symbol $X$, and $\phi^X$ is the $``\in X"$ bounded form of $\phi$; then: $$\forall \vec{p} \, (\phi \to \exists X: \operatorname {Universe} (X) \land \phi^X)$$
>
> Is this consistent? If so, what's its consistency strength?
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> In particular does it manage to interpret Reinhardt cardinals at each universe? Would we have a club of Reinhardt cardinals?
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| https://mathoverflow.net/users/95347 | Can we interpret Reinhardt cardinals this way? | It is inconsistent. Call an ordinal b an independent critical point if for every for every Universe X,
if c∈X then there is a function f with domain X and an ordinal α, such that "f is an elementary
embedding from X to Vα" holds, for all x∈X f(x)∈X, and b is the least ordinal with f(b)≠b.
Note that the critical point of j is an independent critical point because the restriction of j to any
Universe X has the above property of f. Let c be the least independent critical point. By the
axiom schema of Reflection, there is a universe K such that "c is the least independent critical
point" holds relativized to K. By the definition of independent critical point, there is a function
g with domain K and ordinal γ such tha g is an elementary embedding from K to Vγ, and for all x∈K g(x)∈K
and c is the least ordinal with g(c)≠c. Let F(x) be a formula expressing
"x is the least independent critical point". Define a sequence s by
s0=c and s(n+1) is the least ordinal α such that α is greater than g(s(n)) and for K, Vα reflects all
subformulas of F. Let t=U{sn|n∈}. Then g(t)=t, and F(c) holds in V(t). By elementarity,F(g(c)) holds
in V(g(t)). That is F(g(c)) holds in V(t). But this is impossible.
| 3 | https://mathoverflow.net/users/133981 | 439682 | 177,569 |
https://mathoverflow.net/questions/439206 | 5 | By Neumann's addition theorem, we know that the following identity holds (including for complex $\alpha$):
$$J\_0(u)J\_0(v)+2\sum\_{n=1}^\infty J\_n(u)J\_n(v) \cos(n\alpha) = J\_0 \left( \sqrt{u^2+v^2-2uv \cos\alpha} \right)$$
What is the corresponding identity for
$$\sum\_{n=1}^\infty J\_n(u)J\_n(v)\sin(n\alpha)=\text{ ?}$$
I know that the sum vanishes when it runs over all $n$ from $-\infty$ to $+\infty$ (<https://dlmf.nist.gov/10.23>) but I haven't been able to derive an answer for the sum over positive $n$ only. Any help is appreciated!
**EDIT:**
I managed to obtain an integral representation of the sum using the Hilbert transform on the circle:
$$2\sum\_{n=1}^\infty J\_n(u)J\_n(v)\sin(n\alpha)=\frac{1}{2\pi}\int J\_0 \left(\sqrt{u^2+v^2-2uv\cos\psi}\right)\cot\left(\frac{\alpha-\psi}{2} \right) \, d\psi+iJ\_0 \left(\sqrt{u^2+v^2-2uv\cos\alpha}\right)$$
Does anyone know how to evaluate the above integral in closed form?
| https://mathoverflow.net/users/7154 | Sum over Bessel functions: what is $\sum_{n=1}^\infty J_n(u)J_n(v)\sin(n\alpha)$? | I don't think a "closed form" expression exists, I tried several approaches. I guess the best one can do is to use a modified version of the OP's integral representation.
Denote
$$
w(\psi) = \sqrt{u^2+v^2-2uv\cos\psi},\tag{1}
$$
then
\begin{align}
S &= 2\sum\_{n=1}^\infty J\_n(u)J\_n(v)\sin(n\alpha) \tag{2a}\\
&= \frac{\sin\alpha}{\pi} \int\_0^\pi \mathrm d\psi
\frac{J\_0[w(\psi)] - J\_0[w(\alpha)]}{\cos\psi-\cos\alpha}.\tag{2b}
\end{align}
Note that the subtraction of $J\_0[w(\alpha)]$ removes the simple pole at $\psi=\alpha$ (if $\alpha\in\mathbb R$), such that this integral is along the real axis, while the integral in the OP's edit was shifted infinitesimally to the upper complex plane, leading to the imaginery correction term. The identity (2) also holds for complex $u,v,\alpha$.
Changing the integration variable from $\psi$ to $\omega=w(\psi)$ is possible, but leads to additional square-root denominators, e.g.,
\begin{align}
S = \frac{4u\sin\alpha}{v\pi} \int\_{u-v}^{u+v} \mathrm d\omega
\frac{J\_0[\omega] - J\_0[w(\alpha)]}{\sqrt{[1-(u+\omega)^2][(u-\omega)^2-1]}\,[\omega^2-w(\alpha)^2]}\tag{3}
\end{align}
if $u\geq v > 0$.
| 2 | https://mathoverflow.net/users/90413 | 439685 | 177,570 |
https://mathoverflow.net/questions/439665 | 6 | Question about curve stabilisers acting on annular curve graphs, plus context since I'm interested in being fact-checked.
**Definition:** let the group $G$ act by isometries on a metric space $(X,d)$. The *stable translation length* of $g\in G$ is $\tau(g) = \lim\_{n\rightarrow\infty} d(x,g^nx)/n$, (for any $x\in X$).
**Context:** if $S$ is a connected, oriented, hyperbolic surface of sufficiently high finite complexity, and $C(S)$ is its curve graph, then the action of the mapping class group $MCG(S)$ on $C(S)$ is acylindrical, and this implies the existence of $K>0$, depending on $S$, such that $\tau(g)\geq K$ for $g$ pseudo-Anosov (Theorem 1.3 and Lemma 2.2 in [Bowditch](https://homepages.warwick.ac.uk/%7Emasgak/papers/bhb-tight.pdf)).
My question is about **annular** subsurfaces of $S$. Fix an isotopy class $\gamma$ of essential simple closed curves, and let $G < MCG(S)$ be the stabiliser of $\gamma$.
Let $C(\gamma)$ be the curve graph of the annulus in $S$ with core curve $\gamma$, defined in Section 2.4 of [Masur and Minsky](https://arxiv.org/pdf/math/9807150.pdf), where it is explained that $C(\gamma)$ is quasi-isometric to $\mathbb R$, and $G$ acts by isometries, and the Dehn twist $t$ satisfies $\tau(t) =1$ for this action.
**Question:** Does there exist $K>0$ (allowed to depend on $S$) such that the $G$--action on $C(\gamma)$ has the property that for all $g\in G$, either $\tau(g)=0$ or $\tau(g)>K$?
I tentatively guess "no" but I don't know an argument.
On the other hand, I don't think one can use the above result about acylindricity to say "yes": I believe it's well-known (and seems fairly straightforward) that the image $\bar G$ of $G$ in $Isom(C(\gamma))$ does not in general act acylindrically on $C(\gamma)$. Actually, I also struggle to find a reference for non-acylindricity of $\bar G$ on $C(\gamma)$; is there one or is this MCG folklore?
(My coauthors and I have a result which is "sharp" given a "no" to the question, we're writing the introduction to the paper, and want to comment on MCGs.)
| https://mathoverflow.net/users/76590 | Translation length on annular curve graphs | I think that the answer is “yes, such a constant $K$ exists.”
Suppose that $g$ is the given mapping class in the stabiliser of $\gamma$. Replace $g$ by a large power (bounded by the topology of $S$) so that it fixes the curves of its canonical reducing system (and also their orientations and their sides). In each component of the complement of the reducing system, this new $g$ acts as either the identity, or as a pseudo-Anosov map.
If we only have the former, $g$ is a product of powers of disjoint Dehn twists, and we are done.
If we have some of the latter, then we take a further power so that $g$ fixes the switches of its stable train-tracks. In particular, we can control the hyperbolic geometry of a neighbourhood of $\gamma$. The action of $g$ on arcs of $C(\gamma)$ sends endpoints into the “cusps” of the stable track, and otherwise does a power of a Dehn twist. The motion of the endpoints (once close enough to the cusps) is sufficiently slow that it does not contribute to the stable translation length, and we are done.
| 2 | https://mathoverflow.net/users/1650 | 439686 | 177,571 |
https://mathoverflow.net/questions/439681 | 3 | I see many mathematicians conflating the definitions of traveling waves and solitons, and I am unable to understand, from a mathematical point of view, the differences between these two types of solutions for a nonlinear dispersive PDE. All I know is the following:
Consider for example a nonlinear dispersive PDE which is [completely integrable](https://en.wikipedia.org/wiki/Integrable_system), i.e. has infinite conservation laws, ( I think the property of complete integrability will not probably add something to the question)
* The **traveling waves** are solutions of the form $u\_0(x+ct)$ where $u\_0$ is the initial data and $c\in\mathbb{R}$.
* The **Solitons** are subset of the traveling waves, that remain with the same shape even after colliding with another soliton. The phenomena of solitons appear after a cancelation between the dispersive effects and the nonlinearity of the equation.
So here are my questions:
1. How do we know if a nonlinear PDE has solitons as solutions, knowing that it has traveling wave solutions ? On other words, how do we prove mathematically that a traveling wave is a soliton (without using simulation).
2. For example, solutions of the form $u(t,x)=e^{i(x-t)}$ or $u(t,x)=\frac{1}{1-\frac12 e^{i(x-ct)}}$, $x\in \mathbb{T}:=\mathbb{R}/(2\pi\mathbb{Z}),$ can be considered as solitons?
3. Does a traveling wave that is almost periodic solution, *i.e. the set $\{u(\cdot+\tau), \tau\in \mathbb{R}\}$ is relatively compact*, can lead to the fact that it is a soliton ?
| https://mathoverflow.net/users/498602 | Mathematical difference between solitons and traveling waves for a non-linear dispersive PDE | A necessary requirement for a traveling wave $u(x,t)=f(x-ct)$ to be a "solitary wave" or "soliton" is that the two limits $\lim\_{s\rightarrow\pm\infty}f(s)=\alpha\_\pm$ exist. This is the condition of *shape invariance* and *localisation*. The stability under collision may or may not be added as extra condition, but in much of the literature any shape-invariant localised wave is called a soliton. One further distinguishes homoclinic and heteroclinic solutions, depending on whether $\alpha\_+$ is equal to $\alpha\_-$ or not.
In response to Q2, the waves $u(t,x)=e^{i(x-t)}$ or $u(t,x)=\frac{1}{1-\frac12 e^{i(x-ct)}}$ are no solitons, because they are not localised.
Concerning Q1, without the "stability upon collision" condition, one way to identify solitonic solutions of a second order wave equation $f''(s)=F[f(s)]$ is to plot the flow lines in the *f-g* plane of the two coupled equations $f'(s)=g(s)$, $g'(s)=F[f(s)]$. [Homoclinic](https://en.wikipedia.org/wiki/Homoclinic_orbit) or [heteroclinic](https://en.wikipedia.org/wiki/Heteroclinic_orbit) orbits then correspond to solitonic solutions.
| 6 | https://mathoverflow.net/users/11260 | 439694 | 177,572 |
https://mathoverflow.net/questions/439696 | 4 | A well known off-diagonal Ramsey result says that every $C\_3$-free graph $G$ on $N$ vertices has an independent set of size $\Omega(\sqrt{N\log N})$.
It is a conjecture of Erdos that every $C\_4$-free graph $G$ on $N$ vertices has an independent set of size $\Omega(N^{1/2+c})$ for some absolute constant $c>0$.
My question is: is there some $g$ where we know that any $N$-vertex graph $G$ without any copies of $C\_3,C\_4,\dots,C\_{g-1}$ has an independent set of size $\Omega(N^{1/2+c})$ for some $c>0$?
| https://mathoverflow.net/users/130484 | Independent sets in graphs with girth $\ge g$ | Yes, and for any $c<1/2$. Namely, for $\delta>0$ assume that our graph does not have an independent set of size $N^{1-\delta}$. In particular this yields that the chromatic number is at least $N^{\delta}$, thus there is an induced subgraph $H$ with all degrees at least $N^{\delta}$ (proof: remove vertices of smaller degree while they exist, this does not affect the large chromatic number). Then if you consider the neighborhood of arbitrary vertex $v$, you get that it has at least $N^{\delta}$ neighbours, they in turn have at least $N^\delta(N^\delta-1)$ neighbours, and all these guys are distinct if $G$ does not contain $C\_3$ and $C\_4$, they have at least $N^\delta(N^\delta-1)^2$ new neighbours which are all distinct if the graph does not contain also $C\_5$ and $C\_6$, etc. On step about $1/\delta$ you get too many vertices.
| 4 | https://mathoverflow.net/users/4312 | 439698 | 177,573 |
https://mathoverflow.net/questions/439700 | 1 | Let $S$ be a metric space and denote the set of probability measures on $S$ by $\mathcal{P}(S)$. Fix $\mu\in \mathcal{P}(S)$ and denote the law of $N\geq 1$ i.i.d samples $X=(X\_1,\ldots,X\_N)$ from $\mu$ as $\mu\_N$. Do we have the following for $p\geq 1$:
\begin{align}
\inf\_{\nu \in \mathcal{P}(S)} \mathbb{E}\_{X\sim \mu\_N}[\mathcal{W}\_p(\frac{1}{N}\sum\_{i=1}^N \delta\_{X\_i}, \nu)] = \mathbb{E}\_{X\sim \mu\_N}[\mathcal{W}\_p(\frac{1}{N}\sum\_{i=1}^N \delta\_{X\_i}, \mu)] \,?
\end{align}
In plain words: Is the empirical distribution, on average, closest to its underlying distribution?
So far, I did neither succeed in proving this nor in finding some reference, even though there are many articles studying the asymptotics of the r.h.s.. Does anyone know whether this is true, at least in some simplified scenario ($S=[0,1], p=1$ e.g.)?
| https://mathoverflow.net/users/176220 | Is the Wasserstein distance to the empirical measure minimized by the underlying distribution? | No. E.g., let $N=1$ and suppose that $X:=X\_1$ has a nondegenerate zero-mean distribution $\mu$ such that $E|X|^p<\infty$. Let $Y$ be an independent copy of $X$.
Then the expected $\mathcal W\_p$-distance from the empirical distribution to $\mu$ is
$$E\mathcal W\_p(\delta\_X,\mu)^p=E|X-Y|^p>E|X|^p=E\mathcal W\_p(\delta\_X,\delta\_0)^p;$$
the inequality here is an instance of a strict version of Jensen's inequality, which holds because the distribution $\mu$ is nondegenerate.
| 1 | https://mathoverflow.net/users/36721 | 439704 | 177,574 |
https://mathoverflow.net/questions/439711 | 1 |
>
> Suppose that $Y$ is an independent copy of a random variable (r.v.) $X$ with a zero-mean nondegenerate distribution. Is it then always true that $E|X-Y|>E|X|$?
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>
>
To get the non-strict version of this inequality, condition on $X$ and then apply Jensen's inequality to the zero-mean random variable $Y$.
If $p\in(1,\infty)$ and $E|X|^p<\infty$, then $E|X-Y|^p>E|X|^p$ -- because then the function $|\cdot|^p$ is strictly convex.
In view of the well-known expression of the absolute moments in terms of the characteristic function (c.f.), the highlighted question can be restated as follows:
>
> Suppose that $X$ is a r.v with a zero-mean nondegenerate distribution and c.f. $f$. Is it then always true that
> $$\int\_0^\infty\frac{dt}{t^2}\,(\Re f(t)-|f(t)|^2)>0\,\text{?}$$
>
>
>
| https://mathoverflow.net/users/36721 | A strict inequality for the $L^1$-norm of a symmetrized zero-mean random variable | No, a symmetric random sign gives equality ($X$ is $\pm 1$ with probability $1/2$).
| 2 | https://mathoverflow.net/users/120845 | 439713 | 177,576 |
https://mathoverflow.net/questions/439710 | 3 | The [original post](https://math.stackexchange.com/questions/4158432/reference-request-for-a-riemannian-fokker-planck-equation) is in StackExchange but no one has answered it yet. I personally think it is more related to the research area so I put it in MathOverflow. Below is the question in the original post:
I am looking for any reference that states, and proves, a Fokker-Planck equation for Riemannian manifolds.
In particular, if $dX\_t = \mu(X\_t) dt + \sigma(X\_t)dB\_t$ is a stochastic differential equation on a manifold, I want to relate $\mu$ and $\sigma$ to the time evolution of the density of $X\_t$, just like the Euclidean Fokker-Planck equation. It would be great if there is a global description of the time evolution, but a local coordinate expression would be okay too.
| https://mathoverflow.net/users/480283 | Reference request for a Riemannian Fokker-Planck equation | An early reference is [Coordinate-independent formulation of the Langevin equation](https://doi.org/10.1063/1.527396) (1986).
>
> A diffusion process on a compact Riemannian manifold is considered,
> and a coordinate-invariant Fokker-Planck equation is formulated. A
> covariant form of the Langevin equation is also derived, and the
> formalism is applied to the stochastic quantization of lattice gauge
> theories.
>
>
>
| 2 | https://mathoverflow.net/users/11260 | 439718 | 177,577 |
https://mathoverflow.net/questions/439716 | 3 | Consider the (non-unital) $\mathbb{C}$-algebra (point-wise multiplication) of $\mathcal{S}$ of Schwartz functions on $\mathbb{R}$.
**Question:** Does there exist some character (non-zero multiplicative functional to $\mathbb{C}$) $\omega$ of $\mathcal{S}$ that is *not* the evaluation map at any point in $\mathbb{R}$, i.e. does $\mathcal{S}$ as an algebra, admit characters other than the functionals given by Dirac measures?
Some remarks:
1. This can not happen if we require $\omega$ to be continuous with respect to the $L^\infty$ norm (one can extend $\omega$ to $C\_0(\mathbb{R})$, which is a commutative $C^\ast$-algebra, then use Gelfand-Naimark), my question does not pose any continuity restriction on $\omega$, and can be seen of an algebraic nature.
2. This amounts to the question whether the ideal $C\_c^\infty(\mathbb{R})$ of compactly supported smooth functions is contained in a codimension $1$ ideal of $\mathcal{S}$.
| https://mathoverflow.net/users/40789 | Characters of algebra of Schwartz functions | Let $m$ be a multiplicative functional. Let $A={\mathbb C}\oplus\mathcal S$ be the algebra $\mathcal S$ extended by the constant functions. This algebra is unital. Setting $m(f+\lambda)=m(f)+\lambda$ extends $m$ to a multiplicative functional on $A$.
For $f\in A$ let $s(f)=\sup\_{x\in\mathbb R}|f(x)|$. Assume $s(f)<1$. Then we claim that $\frac1{1-f}$ lies in $A$.
We have $\frac1{1-f}-1=\frac f{1-f}$ and any derivative of the latter is of a quotient of a polynomial in the derivatives of $f$ divided by a power of $1-f$. One concludes that $\frac f{1-f}$ lies in $\mathcal S$, hence $1-f$ is invertible in $A$.
For $\lambda\ne 0$ we have $f-\lambda=\lambda(\frac f\lambda -1)$, so if $|\lambda|>s(f)$, the latter is invertible. Now we have $m(f-m(f))=0$, hence $f-m(f)$ is not invertible, hence $|m(f)|\le s(f)$. This means that $m$ is continuous in the sup-norm. Hence it extends to a continuous functional on $C\_c({\mathbb R})$, i.e., a Radon measure. For this to be multiplicative, it must be a point measure.
| 4 | https://mathoverflow.net/users/473423 | 439722 | 177,578 |
https://mathoverflow.net/questions/439717 | 0 |
>
> Suppose that $Y$ is an independent copy of a random variable (r.v.) $X$ with a zero-mean nondegenerate distribution. Is there a non-tautological, preferably simple characterization of the cases when
> $$E|X-Y|=E|X|\,\text{?} \tag{1}\label{1}$$
>
>
>
This question is a modification of this [previous one](https://mathoverflow.net/q/439711/36721).
The inequality $E|X-Y|\ge E|X|$ always holds. To get it, condition on $X$ and then apply Jensen's inequality to the zero-mean random variable $Y$.
If $p\in(1,\infty)$ and $E|X|^p<\infty$, then $E|X-Y|^p>E|X|^p$ -- because then the function $|\cdot|^p$ is strictly convex.
In view of the well-known expression of the absolute moments in terms of the characteristic function (c.f.), equality \eqref{1} can be restated as
$$\int\_0^\infty\frac{dt}{t^2}\,(\Re f(t)-|f(t)|^2)=0.$$
| https://mathoverflow.net/users/36721 | Equality cases in a certain case of Jensen's inequality | Since the distribution of the r.v. $X$ is zero-mean, $X$ integrable (hence also the copy $Y$).
The equality is attained iff the distribution of $X$ is carried by at most two points.
Indeed, by independence of $X$ and $Y$,
$$E\big(|X-Y|\big|X\big) = f(X) \text{ where } f(x) := E|x-Y|.$$
Hence
$$f(X) =E\big(|X-Y|\big|X\big) \ge |X| \text{ a.s.}$$
Looking at the expectations, one deduces that
\begin{equation}
\begin{aligned}
E|X-Y| = E|X| &\iff f(X) = |X| \text{ a.s.} \\
&\iff \text{ for $P\_X$ a.e. } x, \quad f(x) = |x| \\
&\iff \text{ for $P\_X$ a.e. } x, E|x-Y| = |E(x-Y)| \\
&\iff \big(\text{ for $P\_X$ a.e. } x, Y-x \ge 0 \text{ a.s. or } Y-x \le 0 \text{ a.s.}\big) \\
&\iff \big( \text{ for $P\_X$ a.e. } x, \mathrm{ess}\inf Y \ge x \text{ or } \mathrm{ess}\sup Y \le x \big).
\end{aligned}
\end{equation}
Since $X$ and $Y$ have the same distribution, the last condition means that the support of $P\_X$ contains at most two points (namely the essential inf and the essential sup).
| 1 | https://mathoverflow.net/users/169474 | 439727 | 177,580 |
https://mathoverflow.net/questions/439728 | 1 | $\newcommand\Fl{\mathrm{Fl}}$Let $G$ be a connected, reductive algebraic group over $\mathbb{C}$. Fix a maximal torus $T$ and Borel subgroup $B$. Let $L$ be a generalized Levi ($L = Z\_G(s)^\circ$, for some semisimple element $s \in T$). Then $L$ is also a reductive group, and $B \cap L$ is a Borel subgroup in $L$. Therefore there is a closed embedding of flag varieties
$$ i \colon \Fl\_L = L/B \cap L \hookrightarrow \Fl\_G = G/B. $$
I am wondering if this embedding respects the Schubert stratification in the following sense: There is an inclusion of Weyl groups $j\colon W\_L \hookrightarrow W\_G$. For any $w \in W\_L$, there is a Schubert cell $C\_w \subset \Fl\_L$, and similarly for $G$. Then is
$$ i(C\_w) = C\_{j(w)}? $$
If $L$ is an actual Levi subgroup, I believe then the answer is yes. Therefore I am wondering about the slight generalization to generalized Levis.
| https://mathoverflow.net/users/492133 | Inclusion of flag varieties and Schubert decomposition | As you suspect, the equality of inflated Schubert cells does hold for actual Levis $L$. Let $N$ be the unipotent radical of the parabolic subgroup of $G$ containing $B$ that has $L$ as a Levi componnet. Then $w^{-1}N w$ is contained in $B$ for all $w \in W\_L$, and $(B \cap L)w(w^{-1}N w)$ equals $B w$.
However, the equality need not hold for generalised Levis $L$. In fact, I think for every non-Levi $L$ it will fail for certain $w \in W\_L$, and my example below will probably suggest how you would check this.
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}$Consider $L = \SL\_2 \times \SL\_2$ in $\Sp\_4$ and $w = (w\_0, w\_0)$, where $w\_0$ is the long element in the Weyl group of $\SL\_2$. Let $\alpha\_1$ and $\alpha\_2$ be the positive roots of the two $\SL\_2$'s, numbered so that $\beta \mathrel{:=} \tfrac1 2(\alpha\_1 - \alpha\_2)$ is positive. We have that $U\_w \times B \to U w B$ is an isomorphism of varieties, where $U\_w = U \cap w U^- w^{-1}$, and analogously for $U\_{L, w} \times B \to U\_L w B$, with hopefully obvious notation. Then $U\_\beta w B$ is obviously contained in $B w B$, but does not contain any element of $U\_{L, w}w B = U\_L w B$.
| 2 | https://mathoverflow.net/users/2383 | 439730 | 177,581 |
https://mathoverflow.net/questions/400891 | 2 | Let $M$ be a $n$-dimensional compact Riemannian manifold, and $N$ a smooth submanifold of $M$ of dimension strictly less than $n$.
Denote by $N\_{\varepsilon}$ the $\varepsilon$-neighbourhood of $N$ - that is, the set $\{p \in M \ | \ d(x, N) < \varepsilon\}$.
Here $d$ denotes the Riemannian distance.
Let $\text{Vol}$ denote the Riemannian volume measure on $M$.
**Question:**
Define the function $f: [0, \infty) \to \mathbb R\_+$ by $f(\varepsilon) := \text{Vol}(N\_{\varepsilon})$.
Is it true that
$$\lim\_{\varepsilon \to 0+} f’’(\varepsilon)$$
exists? If so, can we find an expression for it in terms of the Riemannian metric on $M$ and the embedding of $N$ in $M$?
| https://mathoverflow.net/users/173490 | Second derivative of the volume of the $\varepsilon$-neighbourhood of a submanifold | I guess we are interested in closed submanifolds; otherwise the question does not have much sense.
Suppose $k=\mathrm{codim}\, N$.
Note that $f(\varepsilon)=O(\varepsilon^k)$, so $f''(0)=0$ if $k\ge 3$.
It remains to consdier two cases $k=2$ and $k=1$.
Given a normal vector $v$ to $N$ denote by $\rho(v)$ the jacobian of the expontential map from the normal bundle $E$ to $N$ to $M$.
Note that
$$f(\varepsilon)=\int\limits\_{x\in N}\ \int\limits\_{v\in B\_\varepsilon\subset E\_x} \rho(v).$$
Note that $\rho$ is smooth and $\rho(0)=1$.
If $k=2$, then it follows that $f''(0)=2\cdot\pi\cdot \mathrm{vol}\,N$.
For $k=1$, we get $f''(0)=0$,
| 3 | https://mathoverflow.net/users/1441 | 439732 | 177,582 |
https://mathoverflow.net/questions/439729 | 1 | Is it possible to have a properly embedded annulus $A$ in a genus two handlebody $V$ such that the two boundary curves of $A$ in $\partial V$ represent different isotopy classes of curves on $\partial V$? (I should also mention that in the case I care about, the curves bounding $A$ cannot be meridians, i.e. cannot bound embedded disks in $V$). My intuition says that this is not possible, but maybe I'm just not visualizing things correctly. And I haven't yet thought up a good way to prove this either way.
If the answer is yes, such an annulus exists, can we say anything about whether the boundary curves of $A$ in $\partial V$ are separating/non-separating? For instance, could you have both boundaries be non-separating (as is possible in genus 3)? Or could you have one boundary separating and one non-separating?
I should mention that I mean the following by "properly embedded": $X$ is properly embedded in $Y$ via $f:X \to Y$ if $f(\partial X) = f(X) \cap \partial Y$, and $f(X)$ is transverse to $\partial Y$ at any point of $f(\partial X)$.
| https://mathoverflow.net/users/156387 | Properly embedded annuli in genus two handlebody? | Choose a non-trivial loop $a\subset \partial V$ and a disjoint meridian $b\subset \partial V$. Let $D\subset V$ be a disk with $\partial V = b$ and let $A'$ be an annulus in $V$ with $\partial A'$ equal to two parallel copies of $a$. Define $A$ to be the result of taking a boundary connected sum of $A'$ and $D$ parallel to an embedded arc $p\subset \partial V$ joining $a$ to $b$. Then one boundary component of $A$ is homologous (in $\partial V$) to $a$ and the other boundary component is homologous to $a \sqcup b$.
If $a$ is separating and $b$ is non-separating then one boundary component of $A$ is separating and the other is not.
| 2 | https://mathoverflow.net/users/284 | 439733 | 177,583 |
https://mathoverflow.net/questions/430449 | 0 | I've edited (ten days ago) a question on Physics Stack Exchange, this [*Mathematical characterization of gravitational geons*](https://physics.stackexchange.com/questions/726281/mathematical-characterization-of-gravitational-geons), post with identifier **726281** the users of the site were kind adding in the comment thread links about papers related to gravitational geons. And a question on Meta of MathOverflow, identifier **5432** of this [post.](https://meta.mathoverflow.net/questions/5432/asking-if-a-post-about-a-mathematical-characterization-of-gravitational-geons-ca) I think that I should to delete both posts if it is edited an answer for this post.
>
> **Question (Updated after the interesting comments of professor in comments).** Please, can you identify an article and the corresponding equations/formulation (equation and page from the article) characterizing a gravitational geon? Secondly the post has more focus on electromagnetic fields and classical gravity, but I'm interested in the discussion of possible singularities in the solution. Can you tell us, from your knowleges or as a reference request, the properties of these solutions (the gravitational geons)?**Many thanks.**
>
>
>
The companion question is from an advice that was given from kind professors on Meta: (in my words) summarize the more important properties of gravitational geons from your knowledges or from the literature answering my post as a reference request.
I'm not able to read the article but I'm motivated to study your answer.
In Wikipedia is edited the article [*Geon (physics)*](https://en.wikipedia.org/wiki/Geon_(physics)) of an hypothetical object in theoretical physics, that studied the physicist John Archibald Wheeler. In the post of Physics Stack Exchange were added from kind users comments about references for the literature.
| https://mathoverflow.net/users/142929 | Mathematical characterization of gravitational geons as reference request, and their properties as main question | A gravitational geon is a space-time configuration that is bounded (asymptotically flat at spatial infinity) and stable (held together for all times by its own gravitational attraction). No such object is believed to exist, a geon should be radiating gravitational waves, lose energy, and decay for long times. No-go theorems exist, see for example [On Derrick's theorem in curved spacetime](https://arxiv.org/abs/1906.00702).
A simple explanation why the non-linear Klein-Gordon equation in Minkowski 3+1 dimensional spacetime does not support stable, time-independent solutions of finite energy is given in [lecture notes](https://indico.cern.ch/event/951466/contributions/4009710/attachments/2109820/3548883/Aveiro_Set_2020_Lecture_4_final_print.pdf) by Herdeiro (page 7-10).
There are ways to avoid the no-go theorem. Periodic (rather than time-independent) solutions are one possibility ([Q-balls](https://en.wikipedia.org/wiki/Q-ball)). The obstruction does not apply to 1+1 dimensions, see [Gravitational geons in 1+1 dimensions.](https://arxiv.org/abs/0807.0611)
| 2 | https://mathoverflow.net/users/11260 | 439734 | 177,584 |
https://mathoverflow.net/questions/439632 | 2 | I have come across the following problem. Let $d\in\mathbb{N}$. Let $G$ be any $k$-regular connected directed graph with $n$ vertices, no parallel edges and no 2-cycles. For a vertex $v\in G$, let $e\_v$ denote the union of $\{v\}$ and the end vertices of edges starting at $v$. I would like to assure that there are sequences of vertices $\{v\_i\}\_{i=1}^j$ and $\{v'\_i\}\_{i=1}^j$ for a graph $G$ such that $v\_{i+1}\in S\_i\cap e\_{v'\_{i+1}}$ where $S\_0=G$ and $S\_{i+1}=S\_i\setminus e\_{v'\_{i+1}}$ with $$j\geq \frac{kn}{d}-C$$ for a constant $C>0$ that does not depend on the graph $G$. The problem would be to show that we can create sequences long enough. The constant $C$ depend only on $d$ and $k$ while the length of the sequences $j$ can vary for different graphs of this type.
I am only able to prove this for $k(k+1)\leq d$, but I do not think this is optimal. It is easy to see that for $k=1$ it is possible to have inequality with $d=1$ and $C=1$. I would like to know if there are similar theorems already proven or any insight that might help prove or disprove this conjecture.
| https://mathoverflow.net/users/494777 | Bound for a sequence of vertices in a graph | Let $q$ be a prime power and let $P$ be a projective plane of order $q$. It has $q^2+q+1$ points and $q^2+q+1$ lines. Each point lies on $q+1$ lines, and each line has $q+1$ points. Each pair of lines has exactly 1 common point. Since the point-line incidence graph is regular and bipartite, there is a bijection $L$ from points to lines such that $L(v)$ is a line through $v$ for every point $v$.
Now construct a graph $G$. The vertex set $V$ is the set of points of $P$. The out-neighbourhood of a vertex $v$ is $L(v)-v$, i.e. $e\_v=L(v)$. So $n=q^2+q+1$ and $k=q$.
Now consider distinct $v'\_1,\ldots,v'\_{q+1}$. Since distinct lines have one common point, $|L(v'\_1)\cup\cdots\cup L(v'\_t)|\ge \sum\_{i=1}^{q+1} (q-i+2)=\frac12(q+2)(q+3)>\frac12n+2q$. Thus, after $q+1=O(n^{1/2})$ steps, already less than half the vertices are available for the $\{v\_i\}$ sequence. Even if all the remaining vertices can be chosen (most unlikely), in total less than half the vertices can be chosen.
I suspect the last part of this argument is unnecessarily weak and that the real bound is a lot smaller than $n/2$.
This example doesn't strictly violate the conjecture as stated since $C$ is allowed to be a function of $(k,d)$ which is a function of $n$.
| 1 | https://mathoverflow.net/users/9025 | 439751 | 177,589 |
https://mathoverflow.net/questions/439329 | 2 | Consider that $u\in H^1(\Omega)$ with $\Delta u\in L^2(\Omega)$ (in the distributional sense) such that for some $\lambda>0$ we have that:
$$\begin{cases} \Delta u(x)=\lambda u(x), & x\in\Omega\\ \dfrac{\partial u}{\partial\nu} (x)=0, & x\in\partial\Omega\end{cases}$$
We assume $\Omega\subseteq\mathbb{R}^2$ to be an open, connected, bounded and has a uniform Lipschitz boundary.
**1)** Is it true that $u\in H^2(\Omega)$? If this is not true:
**2)** Is it true that $u\in C(\overline{\Omega})$? If this is not true:
**3)** How can we prove that $u\in L^{\infty}(\Omega)$?
I know that (3) is valid from the inequality posted here: [Contractivity of Neumann Laplacean](https://mathoverflow.net/questions/402066/contractivity-of-neumann-laplacean)
But I do not know how to prove that inequality. Maybe it can be done in an easier way...
**I wonder if there is an estimate of the form:**
$\Vert u\Vert\_{\infty}\leq c\lambda^{\alpha}$, **where $c$ is a constant depending on $\Omega$?**
I know that such estimates hold for for the Dirichlet laplacian.
I found some references about the problem here:
<https://math.stackexchange.com/questions/2309436/regularity-of-laplacian-eigenfunctions-in-convex-polygon>
but they do not represent counter-examples for any of my questions.
P.S. I found that (2) might also be true from that post: [Is the linear span of the Neumann eigenfunctions dense in $C(\overline{D})$](https://mathoverflow.net/questions/143842/is-the-linear-span-of-the-neumann-eigenfunctions-dense-in-c-overlined) but I did not understand the argument. Is there any clear way of getting more information about the regularity of the eigenfunction $u$ of $\lambda$?
| https://mathoverflow.net/users/61629 | Are Neumann Laplacian eigenfunctions in $C(\overline{\Omega})$? | For (3), this follows from De-Giorgi-Nash-Moser so long as you are OK with dependence of the constant on the $C^{0, 1}$ character of the domain, not just e.g. the size.
This can be checked by running through the De Giorgi proof (all the energy inequalities are valid for the Neumann problem), or, with some extra care, flattening the boundary. There is a book by Gary Lieberman, "Oblique Derivative Problems for Elliptic Equations," which is a good reference for the regularity theory in general.
There are some tricks that will give a power dependence on $\lambda$ like stated in the question for some large, explicit $\alpha$. First treat $\lambda u$ as a zero-order term and run a finite number of Moser iterations (can be skipped in 2D to get $\alpha = 1$) to show that $\|u\|\_{L^p} \leq C \lambda^\alpha \|u\|\_{L^2}$ for a $p > \frac{n}{2}$, then treat $\lambda u = f$ in $- \Delta u = f$ and show that $\|u\|\_{L^\infty} \leq C \|f\|\_{L^p}$. This will generally not give the optimal value $\alpha$.
This also answers (2); De Giorgi-Nash-Moser gives that $u \in C^{0, \beta}$ for some $\beta$ which will depend on the Lipschitz constant of the domain.
(1) is false, as can be checked on non-convex polygons in the plane by blowing up at the corners. The blow-up limit is essentially explicit (like $z^\alpha$ for some $\alpha > \frac{1}{2}$, but tending to $\frac{1}{2}$ as the corner becomes more oblique), so locally this has $|D^2 u|^2 \approx |z|^{2\alpha - 4}$, which fails to be integrable. While obviously not a proof, this is the correct heuristic here.
It's unclear to me whether or not (3) remains true with $c=c(|\Omega|)$ only, at least specifically in 2D.
| 1 | https://mathoverflow.net/users/378654 | 439755 | 177,590 |
https://mathoverflow.net/questions/439719 | 0 | Let $X$ and $Y$ be two integral separated Noetherian Gorenstein schemes over a base field $k$ of arbitrary characteristic whose local rings are unique factorization domains and $f: X\to Y$ an étale map between them. Since we assumed these to be Gorenstein, their canonical bundles $\omega\_X$, resp. $\omega\_Y$ exist and are line bundles / invertible sheaves.
The map $f$ induces via pullback the map $f^\*: \text{Pic}(Y) \to \text{Pic}(X)$ between Picard groups and the rather natural question which arrises at this point is how $\omega\_X$ and $f^\*\omega\_Y$ are related to each other?
Since $X$ and $Y$ where moreover assumed to be integral separated and locally factorial and therefore the Cartier divisors coincide with Weil-divisors, this question question can be equivalently stated in terms of Weil divisors and canonical classes: how $K\_X$ and $f^\*K\_Y$ are related to each other? Is there any explicitly formula known?
If we think of étale maps as algebro geometric pendants to topological coverings, then in case $X$ and $Y$ smooth ($\simeq$ manifolds in topological sense) one could expect/hope that it might hold $\omega\_X= f^\*\omega\_Y$, because in topological setting the the canonical bundles are given locally as determinant bundles of holomorphic $n$-forms and coverings are local isomorphisms.
If that's not the case in full generality for the assumptions of $X,Y$ and $f$ as above, is there at least an explicit formula à la Hurwitz for smooth curves/Riemann surfaces known relating $\omega\_X$ and $ f^\*\omega\_Y$ to each other?
If yes, how general this formula is? Does it only hold for smooth $Y,X$? Depend it on characteristic of base field $k$?
I'm pretty sure that if we drop the étaleness assumption and so allow some even rather tame ramifications then it is nearly hopeless to expect the existence of such formula relating $\omega\_X$ and $ f^\*\omega\_Y$ in such generality, so I hoped that maybe the étaleness assumption might provide the right amount of price which we are ready to pay to have such formula.
| https://mathoverflow.net/users/108274 | Relation between canonical bundles under étale maps | The usual definition of $\omega\_X^\bullet$ is $\pi\_X^!(k[0])$, where $\pi\_X \colon X \to \operatorname{Spec} k$ is the structure morphism. If $f \colon X \to Y$ is a map of $k$-schemes, we therefore get an isomorphism $f^! \omega\_Y^\bullet \stackrel\sim\to \omega\_X^\bullet$; see for instance [Tag [0ATX](https://stacks.math.columbia.edu/tag/0ATX)]. If $f$ is étale, then $f^! \cong f^\*$ [Tag [0FWI](https://stacks.math.columbia.edu/tag/0FWI)], so we get
$$f^\* \omega\_Y^\bullet \stackrel\sim\to \omega\_X^\bullet.$$
The next best thing is when $f$ is a perfect morphism (e.g. flat, or l.c.i., or when $Y$ is regular). In this case, we get an isomorphism $Lf^\*(-) \otimes^{\mathbf L}\_{\mathcal O\_X} \omega\_{X/Y}^\bullet \stackrel\sim\to f^!(-)$ [Tag [0B6U](https://stacks.math.columbia.edu/tag/0B6U)], giving
$$Lf^\*\omega\_Y^\bullet \overset{\mathbf L}{\underset{\mathcal O\_X}\otimes} \omega\_{X/Y}^\bullet \stackrel\sim\to \omega\_X^\bullet.$$
If $Y$ is Gorenstein, then $\omega\_Y^\bullet$ is invertible (in particular flat), so this simplifies to
$$f^\*\omega\_Y^\bullet \underset{\mathcal O\_X} \otimes \omega\_{X/Y}^\bullet \stackrel\sim\to \omega\_X^\bullet.$$
So it remains to compute $\omega\_{X/Y}^\bullet = f^! \mathcal O\_Y$. For a finite morphism, this is characterised by
$$f\_\*f^! \mathcal O\_Y = R\mathscr Hom\_{\mathcal O\_Y}(f\_\*\mathcal O\_X,\mathcal O\_Y),$$
viewed naturally as $f\_\*\mathcal O\_X$-module [Tag [0AX2](https://stacks.math.columbia.edu/tag/0AX2)]. I'm not sure if there is a simpler description of this, but the subtleties of the Riemann–Hurwitz formula for wild ramification show that this computation can become a bit involved.
(Also, while quasi-finite maps of smooth curves are always flat, I'm not sure if a quasi-finite map of Gorenstein schemes is always perfect, so it is possible that the second half of this answer doesn't apply in complete generality. Maybe there are better ways to compute $f^!\omega\_Y^\bullet$.)
| 5 | https://mathoverflow.net/users/82179 | 439756 | 177,591 |
https://mathoverflow.net/questions/439753 | 1 | The [hook length formula](https://en.wikipedia.org/wiki/Hook_length_formula) give a simple product expression for the number of standard Young tableaux of a given shape $\lambda$, where $\lambda$ is an integer partition, or equivalently, the number of ways to build the Ferrers diagram of $\lambda$ from the empty partition by adding boxes one at a time.
I'm aware that the hook length formula has analogues for certain ('$d$-complete') posets, I have a rather more pedestrian question: is there a hook length formula for plane partitions?
Here what I mean by 'hook-length formula for plane partitions' is the following. Consider the plane partition $\Lambda$ as a union of $1 \times 1 \times 1$ cubes, the $3D$ analogue of the Ferrers diagram of a partition, and let $SYT(\Lambda)$ be the number of sequences $\emptyset \subset \Lambda^{(1)} \subset \ldots \subset \Lambda^{(N)} = \Lambda$, where each $\Lambda^{(i)}$ is a valid plane partition and $\Lambda^{(i)},\Lambda^{(i+1)}$ differ by a single $1 \times 1 \times 1$ cube (such sequences are just the $3D$ analogues of standard Young tableaux). My question is, is there a simple product formula for the number $|SYT(\Lambda)|$? Or is there good reason to think such a formula should or should not exist?
| https://mathoverflow.net/users/76764 | hook length formula for plane partitions | I'll convert my comments to an answer.
You are asking about the number of linear extensions of a poset $P$ which is a finite order ideal (downwards closed set) in $\mathbb{N}^3$. With [SageMath](https://www.sagemath.org/) I was easily able to compute that the number of linear extensions of $P=[3]\times[3]\times[3]$ is $6405442434150 = 2 \times 3 \times 5^2 \times 607 \times 70350823$. This number has a huge prime factor in it, which rules out the possibility of any product formula like you had in mind.
In general, there are many two-dimensional arrays of numbers which have remarkable enumerative properties. Depending on your viewpoint, these remarkable properties are either the result of connections to representation theory or to planar statistical mechanical models. But at any rate, their naive 3D analogs fail to behave nicely from an enumerative point of view. For instance, you were asking about 3D Standard Young Tableaux. On the other hand, 100 years ago, MacMahon considered "3D plane partitions" (also called "solid partitions"), i.e., 3D arrays of numbers that are weakly decreasing in all directions. He conjectured a simple product formula for their generating function, but his conjecture was wrong already in the $x^6$ coefficient. See for instance the [Wikipedia page on solid partitions](https://en.wikipedia.org/wiki/Solid_partition) and see also this [prior MathOverflow question about 3D analogs of tableaux, et cetera](https://mathoverflow.net/questions/125846).
[ Nevertheless, higher-dimensional partitions have been investigated to some degree. For example, there are some nontrivial asymptotic results, as mentioned on the Wikipedia page linked above. Also, very recently there have been some interesting enumerative results related to higher-dimensional partitions when you "change the question you are asking": see, e.g., the preprint ["MacMahon's statistics on higher-dimensional partitions"](https://arxiv.org/abs/2009.00592) by Amanov and Yeliussizov, or the preprint ["Fully complementary higher dimensional partitions"](https://arxiv.org/abs/2301.12272) by Aigner that was just posted to the arXiv today! ]
| 6 | https://mathoverflow.net/users/25028 | 439757 | 177,592 |
https://mathoverflow.net/questions/439687 | 1 | There is
>
> a unique simplicial set with a unique non-degenerate simplex in each dimension, (updated) and such that all faces of the non-degenerate simplex are non-degenerate.
>
>
>
Does it have a name, and what can be said about it ? What is it geometrically ? Can one think of it as a "fat" point ?
I have seen it referred to as the dunce's hat but I cannot find a reference.
An explicit construction is as follows: $ S\_n $ is the set of equivalence relations on $\{0,1,\dotsc,n\}$ such that each equivalence class is an interval (i.e. of form $[a,a+1,\dotsc,b]$ for some $a$, $b$). In another notation, $S\_n := \bigsqcup\_{m\leq n} \operatorname{Hom}\_\text{surj}(n,m)$ is the set of order-preserving surjections from $n$ to $m$, $m\leq n$ (i.e. $\operatorname{Hom}\_\text{surj}(n,m)$ is the set of all order-preserving surjections from the linear order $n$ to linear order $m$).
(Update). By the answer of Dmitri Pavlov $S$ is weakly contractible. A comment by Reid Barton suggested $S$ is something like an "ordered" classifying space (BG) for the standard representation of the group of automorphisms of a dense linear order. Namely, $$S\_n := Hom\_{orders} (n^\leq, \Bbb Q^\leq)/Aut(\Bbb Q^\leq)$$ That is, (a representative of) an $n$-simplex is an ordered tuple $x\_0\leq ... \leq x\_n $, and two tuples $(x\_0\leq ... \leq x\_n)$ and $(y\_0\leq...\leq y\_n)$ are considered equal iff there is an order preserving map $g:\Bbb Q\to \Bbb Q$ such that $y\_i=gx\_i$,$0\leq i \leq n$. (Such a $g$ exists iff both tuples give rise to the same equivalence relation $i\approx j$ iff $x\_i=x\_j$.)
Does this construction have a name ? What is a reference and correct terminology for the classifying spaces of group representations (actions) ?
A related question:
>
> let $\mathrm{Eq}\_\bullet$ be the simplicial set where $\mathrm{Eq}\_n$ is the set of equivalence relations on $\{0,1,\dotsc,n\}$. Equivalently, it is something like a classifying space for the standard representation of the symmetric group of an infinite set $X$:
> $$Eq\_n := Hom\_{Sets}(n, X)/Aut(X)$$
>
>
>
Where can I read about this simplicial set ?
Note it classifies equivalence relations in the following sense:
to give an equivalence relation on a set $X$ is the same as to give a morphism to $\mathrm{Eq}\_\bullet$ from the simplicial set represented by $X$.
| https://mathoverflow.net/users/494312 | The simplicial set with a unique non-degenerate simplex in each dimension | As already pointed out in the comments, such a simplicial set is highly nonunique. For example, in addition to the simplicial set $S$ described in the second paragraph one could take the wedge of simplicial spheres, with one sphere for every dimension.
If we concentrate our attention on the simplicial set $S$ described in the second paragraph, it is easy to show that $S$ is weakly contractible.
For example, it has a single vertex, therefore it is connected and by writing down an explicit presentation for the fundamental group we see that the fundamental group is trivial.
Next, computing the chain complex of normalized simplicial chains on $S$ with coefficients in an abelian group $A$, we get the chain complex
$$A←A←A←A←⋯,$$
where the differentials alternate between zero and identity on $A$.
This chain complex is contractible, which proves that $S$ is weakly contractible.
| 1 | https://mathoverflow.net/users/402 | 439767 | 177,597 |
https://mathoverflow.net/questions/438367 | 8 | Let $G$ be a finitely generated group.
I am trying to count the number of permutation representations on $n$ elements, i.e. homomorphisms from $G$ to the symmetric group $S\_n$.
Equivalently this is the number of ways that $G$ can act on the set $\{1,2,\ldots,n\}$.
Let $a\_{G,n}$ be the number of homomorphisms from $G$ to $S\_n$.
Let $\alpha(G) = \lim\_n \frac{\log(a\_{G,n})}{n\log n}$.
How can we determine $\alpha(G)$?
For example, if $G = \mathbb Z$ we have $a\_{G,n} = n!$, since a homomorphism $\mathbb Z \to S\_n$ is determined by the image of its generator. By Stirling's approximation we have $\alpha(\mathbb Z) = 1$.
More generally, for $G = F\_s$ the free group on $s$ generators, we get $\alpha(F\_s) = s$.
For $G = \mathbb Z/2$, a morphism $G\to S\_n$ can be seen as an order 2 element in $S\_n$.
We can create such a permutation by dividing the elements of $\{1,2,\ldots,n\}$ in to pairs, which can be done in $\frac{n!}{(n/2)!2^{n/2}}$ ways.
This is asymptotically $\exp(\frac12 n\log n)$.
There are some more permutations of order 2 that have fixed points, but not enough to change the asymptotic result.
So $\alpha(\mathbb Z/2) = \frac12$.
More generally, for any finite group $G$ we have $\alpha(G) = 1-\frac1{|G|}$.
How can we determine $\alpha(G)$ for an arbitrary group?
| https://mathoverflow.net/users/470870 | Asymptotic number of permutation representations of a given group | In general it is not known whether $\alpha(G)$ exists as a limit (as opposed to limsup). For virtually solvable groups of finite rank (whose subgroup growth is at most polynomial) the relation between subgroup growth and representation growth given in the comments shows that $\alpha(G) \le 1$ (and equals it if the group is residually finite).
For finite groups $\alpha(G) = 1 - \tfrac 1{|G|}$ (this is likely provable by elementary means, much more precise results on the asymptotic behaviour of $a\_{G, n}$ in this case are given by Müller <https://zbmath.org/0862.20019>).
An opposite situation is that of so-called large groups (which have a finite-index subgroup surjecting onto the free group $F\_2$). For these $\alpha(G) > 0$ if it exists, and it has been computed exactly for a few families, for example:
* Fuchsian groups (e.g. surface groups) where $\alpha(G) = -\chi(G)$ (see <https://zbmath.org/1059.20021>, <https://zbmath.org/1068.20052>)
* Right-angled Artin groups where $\alpha(G)$ equals the independence number of the defining graph (see <https://zbmath.org/1481.20094>)
There also are relevant sporadic examples in another paper of Müller--Schlage-Puchta (<https://zbmath.org/1127.20022>). In particular they observe that the 2-generator 1-relator group
$$ G = \langle a, b | a^mb^n \rangle $$
admits the free product $C\_m \* C\_n$ as a quotient and hence it satisfies that $\alpha(G) \ge 2 - \tfrac 1n - \tfrac 1m$ (more generally you can get 1-relator $d$-generated groups with $\alpha(G)$ arbitrarily close to $d$).
| 5 | https://mathoverflow.net/users/32210 | 439768 | 177,598 |
https://mathoverflow.net/questions/438671 | 4 | The holomorphic function
$$F(\tau)=-\frac{1}{\vartheta\_4(\tau)}\sum\_{n\in\mathbb Z}\frac{(-1)^nq^{\frac{n^2}{2}-\frac 18}}{1-q^{n-\frac12}}=2q^{\frac38}(1+3q^{\frac12}+7q+14q^{\frac32}+\dots),$$
is a mock modular form for the congruence subgroup $\Gamma^0(4)$ of $\text{SL}(2,\mathbb Z)$ with shadow $\eta^3$. Here, $q=e^{2\pi i\tau}$, $\vartheta\_4(\tau)=\sum\_{n\in\mathbb Z}(-1)^n q^{n^2/2}$ is a Jacobi theta function and $\eta$ is the Dedekind eta function. The $q$-series is the OEIS sequence A256209.
As a consequence, the sum of $F$ and a non-holomorphic period integral
$$ \hat F(\tau,\bar\tau)=F(\tau)-\frac i2\int\_{-\bar\tau}^{i\infty}\frac{\eta(w)^3}{\sqrt{-i(w+\tau)}}\mathrm dw. $$
transforms as a non-holomorphic modular form of weight $(\tfrac12,0)$ for $\Gamma^0(4)$, and it is clear that $\partial\_{\bar\tau}\hat F(\tau,\bar\tau)=-\frac{i}{2\sqrt{2y}}\overline{\eta(\tau)^3}$.
I am interested in special values of $F$ and $\hat F$, such as at the elliptic points of $\text{SL}(2,\mathbb Z)$. For instance, numerically I find that
$$\hat F(i,\bar i)=\frac{\vartheta\_4(i)}{2^{\frac54}}, \\
\hat F(\alpha,\bar\alpha)=\frac{e^{-\frac{\pi i}{4}}\vartheta\_4(\alpha)}{2\sqrt3},$$
where $\alpha=e^{2\pi i/3}$, while $\hat F(i+1,\overline{i+1})=0$. The latter can be proven by realising that $\hat F$ is more precisely a non-holomorphic modular form for $\Gamma^0(2)$, whose multiplier system evaluated at the elliptic fixed point $1+i$ of $\Gamma^0(2)$ is $-1$.
My questions:
* Are explicit values of mock modular forms and their completions at elliptic points known in the literature?
* Is there a general strategy to calculate series and period integrals, such as those involved in $\hat F(i,\bar i)$ and $\hat F(\alpha,\bar\alpha)$?
| https://mathoverflow.net/users/321953 | Evaluation of mock modular forms at elliptic points | The holomorphic function $F$ is related to the $q$-series $H^{(2)}$ of Mathieu moonshine in a simple way. We have
\begin{equation}
\begin{aligned}
H^{(2)}(\tau)&=2\frac{\vartheta\_2(\tau)^4-\vartheta\_4(\tau)^4}{\eta(\tau)^3}-\frac{24}{\vartheta\_3(\tau)}\sum\_{n\in\mathbb Z}\frac{q^{\frac{n^2}{2}-\frac18}}{1+q^{n-\frac12}}\\
&=2q^{-\frac18}\left(-1+45q+231q^2+770q^3+2277q^4+\dots\right)
\end{aligned}
\end{equation}
The $q$-series is the OEIS sequence A169717, and the Fourier coefficients are sums of dimensions of irreducible representations of the sporadic group $M\_{24}$. It relates to $F$ as
\begin{equation}\label{FH\_relation}
F(\tau)=\frac{1}{24}\left( H^{(2)}(\tau)+2\frac{\vartheta\_2(\tau)^4+\vartheta\_3(\tau)^4}{\eta(\tau)^3}\right).
\end{equation}
Since both $H^{(2)}$ and $F$ are mock modular forms with shadow $\eta^3$, their non-holomorphic completions $\widehat H^{(2)}$ and $\widehat F$ are simply related by
\begin{equation}\label{FhatHhat}
\widehat F(\tau)=\frac{1}{24}\left( \widehat H^{(2)}(\tau)+2\frac{\vartheta\_2(\tau)^4+\vartheta\_3(\tau)^4}{\eta(\tau)^3}\right).
\end{equation}
Unlike $ \widehat F$, $\widehat H^{(2)}$ is a non-holomorphic modular function for $\text{SL}(2,\mathbb Z)$, and transforms under the generators as
\begin{equation}
\begin{aligned}
\widehat H^{(2)}(\tau+1,\bar\tau+1)&=e^{-\frac{\pi i}{4}}\widehat H^{(2)}(\tau,\bar\tau), \\
\widehat H^{(2)}(-1/\tau,-1/\bar\tau)&=-\sqrt{-i\tau}\widehat H^{(2)}(\tau,\bar\tau).
\end{aligned}
\end{equation}
From the phases of the transformations under the elliptic elements of $\text{SL}(2,\mathbb Z)$ stabilising the elliptic fixed points $i$ and $\alpha$, we easily find
\begin{equation}\begin{aligned}
\widehat H^{(2)}(i,\bar i)&=0, \\
\widehat H^{(2)}(\alpha,\bar \alpha)&=0.
\end{aligned}\end{equation}
Thus at $\tau=i$ and $\tau=\alpha$ the completion $ \widehat H^{(2)}(\tau,\bar\tau)$ evaluates to the modular "difference" between $ \widehat H^{(2)}$ and $\widehat F$, which can be determined using the Chowla-Selberg formula.
| 1 | https://mathoverflow.net/users/321953 | 439774 | 177,599 |
https://mathoverflow.net/questions/439794 | 4 | In my course on probabilistic graphical models, my professor made a claim which I find a little sus. In discussing the equivalence between Markov Random Fields and Factor Graphs, the following example was given:
An MRF over 3 variables was drawn as a triangle graph (3-cycle). This admits density function like $\psi(x, y, z)$. Naturally, an equivalent factor graph can be drawn by having only one factor connected to all three variables, $p(x,y,z) \propto f(x, y, z)$. However the example where a factor graph with three factors, one along each edge, was also considered, which admits a density like $p(x,y,z) \propto f\_1(x,y) \cdot f\_2(y,z) \cdot f\_3(x,z)$.
The question was posed that all three graphical models are equivalent for the appropriate choice of factors. Formally, the claim is as follows:
>
> For any nonnegative function $f(x,y,z)$, there exists nonnegative functions $u, v, w$ such that $f(x,y,z)=u(x,y)\cdot v(y,z)\cdot w(x,z)$. We can assume that $x,y,z \in \{0,1\}$.
>
>
>
My guess at a counterexample would be a function along the lines of $f(x,y,z) = xy + yz + xz$, but I'm stuck at finding a way to *prove* that this cannot be written as a product of pairwise factors.
Intuitively, for a binary alphabet, $f(x,y,z)$ can be encoded using 8 values, where each of $u,v,w$ need 4 values, leaving 8 degrees of freedom versus 12. But this intuition seems weak to me.
| https://mathoverflow.net/users/123034 | Can nonnegative functions $f(x,y,z)$ be written as a product of pairwise functions $u(x,y) v(y,z) w(x, z)$? | Let $f(x,y,z)=x^2+y^2+z^2$ for $(x,y,z) \in \mathbb{R}^3$. The only zero of $f$ is $(0,0,0)$.
If we had three functions $u,v,w$ such that $f(x,y,z)=u(x,y)v(y,z)w(z,x)$ for every $(x,y,z) \in \mathbb{R}^3$, at least one of the functions $u,v,w$ would vanish at $(0,0)$, so $f$ would vanish on a whole line.
This argument also works on $\{0,1\}^3$.
| 17 | https://mathoverflow.net/users/169474 | 439799 | 177,603 |
https://mathoverflow.net/questions/439783 | 2 |
>
> **Question:** Does there exist an isomorphic predual of $\ell^1$, which does not have a quotient isomorphic to $c\_0$?
>
>
>
Thanks in advance.
---
Edit: The answer is no. Let $X$ be a Banach space such that $X^\*$ is isomorphic to $\ell^1$. If there didn't exist any surjective bounded $T:X\to c\_0$, then $X$ would be a Grothendieck space. Being separable, $X$ would be reflexive.
It is awkward to answer your own question in a few hours after you post it. Please accept my apologies, I will delete this post in a couple of days.
| https://mathoverflow.net/users/164350 | $\ell^1$ predual with no $c_0$ quotient? | In my weak$^\*$ basic sequences paper with Rosenthal we proved that if $\ell\_1$ embeds into $X^\*$ and $X$ is separable, then $c\_0$ is isomorphic to a quotient of $X$.
| 2 | https://mathoverflow.net/users/2554 | 439801 | 177,605 |
https://mathoverflow.net/questions/439792 | 17 | Let $T\subset \mathbb{R}^n$ be a fixed simplex, $H\subset \mathbb{R}^n$ be a variable affine hyperplane. Is it true that the maximal area (i.e. the $(n-1)$-dimensional volume) of $T\cap H$ is attained when $H$ contains a facet of $T$?
| https://mathoverflow.net/users/4312 | Is a facet always a maximal area section of a simplex? | No, there is a $5$-dimensional simplex with a hyperplane section which is larger than any of its facets, see [Walkup, A simplex with a large cross-section, Am. Math. Monthly, January 1968](https://www.jstor.org/stable/pdf/2315102.pdf).
The idea is to squeeze a regular 5-simplex along the common perpendicular of two opposite 2-dimensional faces. The volume of the mid-section between these faces does not change, but it turns out that all facets become smaller than this cross-section if the scaling coefficient is small enough.
For more information see also the [PhD thesis of Dirksen](https://d-nb.info/1082537128/34).
| 23 | https://mathoverflow.net/users/98590 | 439818 | 177,610 |
https://mathoverflow.net/questions/439803 | 1 | I am trying to prove whether the space $L(H,K)$ has martingale type 2 for Hilbert spaces $H,K$. It is known that Hilbert spaces have martingale type 2, so I was wondering whether the space of bounded linear operators also enherit the property. We recall the definition here
>
> **Definition.** Let $X$ be a Banach space and $(M\_n)\_{n=0}^N$ be a martingale with values in $X$. We say $X$ has martingale type $p\in [1,2]$ if for some $m\in (1,\infty)$, there exists a constant $C=C(m,p,X)$ such that
> \begin{align\*}
> \mathbb E[\Vert M\_N\Vert\_X^m]\leq C \operatorname{\mathbb E}\left[ \left\vert \Vert M\_0\Vert\_X^p+ {\sum}\_n^N\Vert \Delta M\_n\Vert\_X^p\right\vert^{m/p}\right],
> \end{align\*}
> where $\Delta M\_n$ is the martingale difference.
>
>
>
Since $H$ and $K$ both have type $2$ a first line of applying that to linear operators leads to the following problem:
>
> **Problem.** Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space and $H,K$ be separable Hilbert spaces. Consider a random bounded linear operator $A:\Omega\to L(H,K)$. Assume for any integer $m\geq 2$ there exists $C=C(m,K,H)$ such that for all $0\neq x\in H$ we have
> \begin{align\*}\mathbb{E}[\Vert Ax\Vert\_K^m] \leq C \Vert x\Vert\_H^m \end{align\*}
> There exists $\Gamma=\Gamma(m,K,H)$ such that
> $$\mathbb E[\Vert A\Vert\_{L(H,K)}^m]\leq C\Gamma.$$
>
>
>
The only attempt I currently have is to show the statement through Kolmogorov continuity criterion for Hilbert spaces. I do not know whether such statement exists, but I was willing to use the paper [sample properties of random fields](https://digitalcommons.lsu.edu/cgi/viewcontent.cgi?article=1112&context=cosa) to get the result. Note that if I believe Kolmogorov continuity holds for Hilbert spaces, then I get that the Holder seminorm $\vert A\vert\_{C^\alpha}$ bounded in $L^m(\mathbb P)$ which together with the fact that $A$ is linear bounded operator gives the result. This is too overkill.
**Context.**
| https://mathoverflow.net/users/425509 | The space of linear operators between Hilbert spaces has martingale type 2 | As noted by Mikael de la Salle, $L(H,K)$ contains an isometric copy of $\ell^\infty$, which is not of martingale type $p$ for any $p\in(1,2]$, and hence $L(H,K)$ is not of martingale type $p$ for any $p\in(1,2]$.
---
The answer to your highlighted **Problem** is also negative. Indeed, suppose that $H=K=\ell^2$ and $A$ is a random linear operator such that $P(A=A\_k)=p\_k$ for integers $k\ge2$, where
$$A\_k e\_j=b\_j\,1(j=k)$$
for $j=1,2,\dots$, the $e\_j$'s are the standard basis vectors of $\ell^2$, $b\_j:=\ln j$, $p\_k=\frac c{k\ln^2k}\,1(k\ge2)$, and $c:=1/\sum\_{k\ge2}\frac 1{k\ln^2k}$, so that $\sum\_{k\ge2}p\_k=1$.
Then for any $x=(x\_1,x\_2,\dots)\in H$ and any integer $m\ge2$
$$E\|Ax\|^m=\sum\_{k\ge2}p\_k\|A\_kx\|^m=\sum\_{k\ge2}p\_k b\_k^m |x\_k|^m
\le C\_m\|x\|^m,$$
where
$$C\_m=\max\_{k\ge2}p\_k b\_k^m=\max\_{k\ge2}\frac c{k\ln^{2-m}k}<\infty.$$
On the other hand, $\|A\_k\|=b\_k$ and hence
$$E\|A\|^m=\sum\_{k\ge2}p\_k\|A\_k\|^m=\sum\_{k\ge2}p\_k b\_k^m
=\sum\_{k\ge2}\frac c{k\ln^{2-m}k}=\infty$$
for all $m\ge1$.
| 1 | https://mathoverflow.net/users/36721 | 439819 | 177,611 |
https://mathoverflow.net/questions/439816 | 0 | Recall that a probability $\mu$ measure on a metric space $(X,d)$ is called *Ahlfors $q$-regular* if there are $0<c\le C$ such that: for $\mu$-a.e.\ $x\in X$ one has
$$
cr^q \le \mu(B(x,r)) \le Cr^q,
$$
for any $0\le r\le \operatorname{diam}(X,d)$.
If $(X,d)$ is a finite metric space, then can it support an Ahlfors regular measure? If so, what are concrete examples of Ahlors $q$-regular measures on discrete metric spaces; for $q>1$?
| https://mathoverflow.net/users/491352 | Can a measure on a finite metric space be Alhfors regular? | Take any $x\in X$ with $m:=\mu(\{x\})\in(0,\infty)$ (since $X$ is finite and $\mu$ is a probability measure, such a point $x$ exists). Let $R:=\min\{d(y,x)\colon y\in X\setminus\{x\}\}$. Then $R\in(0,\infty)$ and $B(x,r)=\{x\}$ for $r\in(0,R)$.
Letting now $r\downarrow0$, from $cr^q \le \mu(B(x,r)) \le Cr^q$ we get $m\le 0$ if $q>0$ and $\infty\le m$ if $q<0$. So, we get a contradiction with $m\in(0,\infty)$ unless $q=0$.
So, $\mu$ can be Ahlfors $q$-regular only if $q=0$.
On the other hand, any probability measure over a finite set is clearly Ahlfors $0$-regular (assuming the convention that $r^q=1$ if $r=0=q$).
Thus, any probability measure over a finite set is Ahlfors $q$-regular iff $q=0$.
| 2 | https://mathoverflow.net/users/36721 | 439824 | 177,613 |
https://mathoverflow.net/questions/439347 | 7 | A common technique for constructing objects (sheaves) and morphisms in algebraic geometry is faithfully flat descent. Roughly speaking this consists on constructing an object or a morphism "locally" in a fpqc covering of a scheme, verifying that the local constructions form a descent datum, and using the theorem that says that fpqc descent data are effective to show that there is a global object or morphism defined over the original scheme.
Usually, a lot of constructions from algebraic geometry can be easily done in the complex geometric setting. Moreover, usual "analytic" opens of a complex variety are "smaller" than Zariski opens, so most constructions work better in the analytic topology. For example, to develop a good theory of principal bundles on the algebraic setting, one needs to work with étale-locally trivial objects, rather than with Zariski-locally trivial ones.
My question is if there is a "canonical" way to translate an argument or construction in the algebraic setting using fpqc descent to an argument in complex geometry.
Roughly speaking my question would be if, given a construction over a complex algebraic variety using fpqc descent, one could do the same construction over the analytification by, for example, gluing local data over a covering by usual opens.
| https://mathoverflow.net/users/143492 | Faithfully flat descent in complex analytic geometry | There will not be a way to directly translate faithfully flat descent into gluing over analytic opens; the former is genuinely stronger than the latter in the analytic category. However basically any faithfully flat descent statement that holds for schemes and makes sense for complex analytic spaces (including those involving coherent sheaves) does hold. See Kiehl's paper Äquivalenzrelationen in analytischen Räumen, for example.
The greater flexibility of analytic opens does manifest itself in improvements to effective descent results (or, more generally, effectivity of quotients). The aforementioned paper more or less shows that every separated flat equivalence relation is effective, which is certainly not true for schemes (you'd get an algebraic space in general).
| 3 | https://mathoverflow.net/users/109356 | 439830 | 177,615 |
https://mathoverflow.net/questions/303687 | 3 | Hairer in his [spdes notes](http://www.hairer.org/notes/SPDEs.pdf) on pg.6, says that GFF is the stationary solution of $u\_{t}(z)=\Delta u(z)+\xi(z,t)$, where
$\xi$ is the space-time white noise
$$\xi(x,t)=\sum \sqrt{\lambda\_{k}} B\_{k}(t)e\_{k}(x)$$
for iid Brownian motions $B\_{k}$ and L2 basis $e\_{k}$. So that means we take
$$u(x,t)=\Delta^{-1}\xi(x,t). $$
and that at each fixed time it is a GFF
$$h(x)=\sum \frac{1}{\sqrt{\lambda\_{k}}} B\_{k}(t)e\_{k}(x).$$
In many of the standard spde textbooks, we can find existence and uniqueness of invariant measure for parabolic pdes (eg. using the Bismut-Elworthy-Li formula). But I want to know if for SHE we have stronger results eg. on the rate.
>
> Q: Does SHE in a bounded domain with zero boundary converge to a steady state? So if $u(x,0)=GFF$ or $=0$, do we have some asymptotic results? Any rates? What is the largest function space over which it makes sense to take limits?
>
>
>
In Walsh's spdes book pg.418 there is a computation for SHE on infinite domains with $d\geq 3$ showing that one obtains the Green function covariance. But is there a clean treatment for bounded domains, at least for reference purposes.
**Weak convergence**
First we show that covariances agree. For bounded domains D the formula is
$$u(x,t)=e^{t\Delta }u(x,0)+\int\_{[0,t]}\int\_{D}H(t-s,x-y)dW(s,y),$$
where the second term is a Wiener integral with Heat kernel H for domain D.
We will compute the covariance for the bounded domain for zero initial data and $\xi(x,t):=\sum B\_{k}(t) e\_{k}(x)$. We have
$$u(x,t)=\int\_{[0,t]}\int\_{D}H(t-s,x-y)dW(s,y)$$
$$=\sum\_{k\geq 1}\int\_{0}^{t}e^{-\lambda\_{k}(t-s)}dB\_{k}(s) e\_{k}(x).$$
Therefore, by Ito isometry we indeed obtain the Green function:
$$E[u(t,x)u(t,y)]=\frac{1}{2}\sum\_{k\geq 1}\frac{e\_{k}(x)e\_{k}(y)}{\lambda\_{k}}(1-e^{-\lambda\_{k}t})\to G(x,y).$$
**Function space weak limit**
We have that the SHE $u\in C^{-\varepsilon}(\mathbb{R}\_{+},D)$ and the GFF $h\in H^{-\varepsilon}(D)$ for all $\varepsilon>0$. By Morrey's inequality
$$ H^{2}(D)\subset C^{0,\varepsilon}(D)=C^{\varepsilon}(D), $$
where $H^{2}(D)$ is the Sobolev space where second weak derivatives are also square integrable and so we also have $H^{2}(D)\subset H^{\varepsilon}(D)$ for $\varepsilon\leq 2$. So we will work with functions $f\in H^{2}(D)$.
By the covariance computation above we also obtain
$$\left \langle u(\cdot,t),f \right\rangle\_{H^{2}} \stackrel{law}{\to} \left \langle h(\cdot),f \right\rangle\_{H^{2}}.$$
| https://mathoverflow.net/users/99863 | Gaussian free field limiting distribution of additive Stochastic heat eqn bounded domain | Yes, indeed, one nice reference is ["An introduction to singular stochastic PDEs: Allen-Cahn equations, metastability and regularity structures"](https://arxiv.org/abs/1901.07420) section 2.5.1 Gaussian free field.
>
> The stochastic heat equation (2.3.1) does not admit an invariant measure, since we have seen that its zeroth Fourier mode performs a Brownian
> motion. This problem, however, is easily cured by considering the equation
> $$\partial\_{t}\phi(t, x) = \Delta \phi(t, x) − \phi(t, x) +2 \sqrt{\epsilon}\xi(t, x) , (2.5.1)$$
> where we have reintroduced the parameter $\epsilon$ to keep track of its effect.
>
>
>
Then the Gaussian measure $\mu\_{\epsilon,GFF}$ of $\sqrt{\epsilon}\phi\_{GFF}$ i.e. the GFF with covariance $\epsilon(-\Delta+1)^{-1}$ is invariant for the (2.5.1) i.e.
$$E\_{\mu\_{\epsilon,GFF}}[f(\phi(t,\cdot))]=E\_{\mu\_{\epsilon,GFF}}[f(\sqrt{\epsilon}\phi\_{GFF}(\cdot))]$$
for any integrable functional $f$.
| 1 | https://mathoverflow.net/users/99863 | 439839 | 177,617 |
https://mathoverflow.net/questions/439809 | 5 | Given a complete graph with $n$ nodes, if we want to use $n$ complete subgraphs to cover the graph and ask what is the minimum possible size of each complete subgraph, the answer is $\Theta(\sqrt{n})$: there are $\Theta(n^2)$ edges and each clique can cover at most $O(n)$ edges, so the lower bound for the complete subgraph size is $\Omega\left(\sqrt{\frac{n^2}{n}}\right)=\Omega\left(\sqrt{n}\right)$. This lower bound is also tight: we can partition vertices into $\sqrt{n}$ sets with $\sqrt{n}$ nodes in each set, and any pair of sets form a complete subgraph.
I want to ask for a generalization of the above question when the graph is not a complete graph: when the graph has $m$ edges, is it always possible to cover the graph by $n$ cliques with size $\sqrt{\frac{m}{n}}$? (**the clique can contain non-edge**). To be precise, I want to find the smallest possible $k(n,m)$ such that for any graph $G$ with $n$ vertices and $m$ edges, there exists $n$ vertex sets $S\_1,S\_2,...,S\_n\subseteq V$ each with size $k(n,m)$ and every edge $(u,v)$ has $u,v$ both contained in some $S\_i$.
I searched for clique edge covering but most of the results only consider the case where cliques cannot cover non-edge. I wonder if there exists any similar research in the setting when cliques can cover non-edge.
| https://mathoverflow.net/users/498695 | Cover a graph with small size complete graphs | Here is an example showing that the clique size $\sqrt{m/n}$ does not suffice.
Graphs on $n$ vertices without 4-cycles have less than $O(n^{3/2})$ edges, and there is a series of examples where this asymptotic bound is attained. If one wishes to cover all edges of such a dense graph with $n$ cliques of equal size, then these cliques must contain $\Theta(n^{1/2})$ edges on average, and are still 4-cycle-free. Thus the clique size is at least $n^{1/3}$, while $\sqrt{m/n} = \Theta(n^{1/4})$.
| 5 | https://mathoverflow.net/users/98590 | 439844 | 177,619 |
https://mathoverflow.net/questions/439843 | 0 | Let $R$ be an integral domain, $S$ be a finitely presented $R$ algebra. Then for a flat $R$ module $M$ which is also a finitely generated $S$ module I need to show that $M \otimes\_{R}T$ is a fintely presented $S\otimes\_{R}T$ module where $T$ is the quotient field of $R$.
My attempt :
Since $M$ is finitely generated $S$ module then we have an $S$ module surjection $$S^{n\_1} \rightarrow M.$$ Since tensor product is right exact we would have a surjection $$S^{n\_1}\otimes\_{R}T \rightarrow M \otimes\_{R}T$$, now if I can show that the kernel of this map is finitely generated I will be done and here is what I feel like finitely presented $R$ algebra property of $S$ might be used , but I am unable to progress further from here.
Any hint or suggestion is much appreciated, thank you.
| https://mathoverflow.net/users/443060 | Proving finite presentation | I am following the notation in the comments. I do not have privilege to make a comment, do I should write and answer. Also from the comments I understand that you are working just with commutative rings.
Notice that $R[x\_1,\dots, x\_n]/(f\_1,\dots,f\_m)\otimes T$ is isomorphic to a quotient of $T[x\_1,\dots,x\_n]$ so it is a noetherian ring (because T is a field). Then any finitely generated module is finitely presented.
What I do not understand is why you neet M\_R to be flat.
| 1 | https://mathoverflow.net/users/490959 | 439851 | 177,621 |
https://mathoverflow.net/questions/424313 | 3 | I am trying to derive the critical coupling strength for synchronisation in a network of phase oscillators with noisy input.
I am following the steps outlined in
Sakaguchi, Hidetsugu. ["Cooperative phenomena in coupled oscillator systems under external fields."](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.967.3027&rep=rep1&type=pdf) Progress of theoretical physics 79.1 (1988): 39-46.
However I am stuck in an expansion and I would be grateful if someone had an insight to help.
I have a problem to derive the expansion in Eq. 25 but i will also copy here the equations for convenience.
We consider phase oscillators with all-to-all coupling in the thermodynamic limit, and thereby have expressed the mean-field Fokker-Planck for the phase distribution
$$
\frac{\partial n(\psi ; \omega)}{\partial t}=-\frac{\partial}{\partial \psi}\{(\omega-K \sigma \sin \psi) n(\psi ; \omega)\}+D \frac{\partial^{2}}{\partial \psi^{2}} n(\psi ; \omega),
$$
where $\psi$ denotes the phase and $\omega$ the natural frequency, $\sigma$ the synchronisation order parameter, and $K$ the coupling strength.
We compute the stationary solution:
$$
\begin{aligned}
n(\psi ; \omega)=& \exp \left(\frac{-K \sigma+\omega \psi+K \sigma \cos \psi}{D}\right) n(0 ; \omega) \\
& \times\left\{1+\frac{\left(e^{-2 \pi \omega / D}-1\right) \int\_{0}^{\phi} e^{(-\omega \phi-K \sigma \cos \phi) / D} d \phi}{\int\_{0}^{2 \pi} e^{(-\omega \phi-K \sigma \cos \phi) / D} d \phi} \right\},
\end{aligned}
$$
and we are about to write a self-consistent equation for the synchronisation order parameter $\sigma$. We have from previously (Eq.9)
$$
\begin{aligned}
\sigma \exp \left(i \phi\_{0}\right)
&=\int\_{-\infty}^{\infty} d \omega g(\omega) \int\_{0}^{2 \pi} d \psi n(\psi ; \omega) \exp (i \psi),
\end{aligned}
$$
where $g(\omega)$ denotes the natural frequency distribution, and $\phi\_0$ stands for the mean phase.
We write the self consistent equation for the order parameter
$$
\sigma=\int\_{-\infty}^{\infty} d \omega g\left(\omega+\omega\_{0}\right) \int\_{0}^{2 \pi} d \psi n(\psi ; \omega) \exp (i \psi),
$$
and to find the critical coupling strength we proceed to expand the expression for the order parameter in powers of $K \sigma/D$
$$
\begin{aligned}
\sigma=& K \sigma\left[\frac{1}{2} \int\_{-\infty}^{\infty} g\left(D \omega+\omega\_{0}\right) \frac{d \omega}{\omega^{2}+1}-\frac{K^{2} \sigma^{2}}{4 D^{2}} \int\_{-\infty}^{\infty} g\left(D \omega+\omega\_{0}\right)\left\{\frac{1}{\omega^{2}+4}-\frac{\omega}{\left(\omega^{2}+1\right)^{2}}\right\} d \omega\right.\\
&\left.+O\left(\left(\frac{K \sigma}{D}\right)^{4}\right)\right].
\end{aligned}
$$
So my question is how can I derive this expansion. The author mentions that they consider that $g(\omega+\omega\_0)$ is symmetric about 0, and that the imaginary part of the self-consistent equation for the order parameter is zero. So the expansion is only for the real part.
I would be grateful for any insight!
| https://mathoverflow.net/users/483817 | Derivation of a series expansion | To simplify the notation I denote $\tilde{K}=K\sigma/D$ and $\tilde{\omega}=\omega/D$.
First expand
\begin{align}
&n(\psi , \omega)= \exp \bigl(-\tilde{K} +\tilde{\omega} \psi+\tilde{K} \cos \psi\bigr) n(0 , \omega)\nonumber\\
&\times\biggl[1+\frac{\bigl(e^{-2 \pi \tilde{\omega}}-1\bigr) \int\_{0}^{\phi} e^{(-\tilde{\omega} \phi-\tilde{K} \cos \phi)} d \phi}{\int\_{0}^{2 \pi} e^{(-\tilde{\omega} \phi-\tilde{K} \cos \phi)} d \phi} \biggr]
\end{align}
to fourth order in $K$ and evaluate
$$\int\_0^{2\pi}n(\psi,\omega)\cos(\psi)\,d\psi=\pi n(0,\omega)\bigl[\frac{\tilde{K}}{1+\tilde{\omega}^2}-\frac{\tilde{K}^2}{(1+\tilde{\omega}^2)^2}
+\frac{ \tilde{K}^3 \bigl(3 \tilde{\omega}^4+2 \tilde{\omega}^2+5\bigr)}{2 \bigl(\tilde{1+\omega}^2\bigr)^3 \bigl(4+\tilde{\omega}^2\bigr)}+{\cal O}(K^4)\bigr].$$
The function $n(0,\omega)$ is determined by the normalisation
$$\int\_0^{2\pi}n(\psi,\omega)\,d\psi=1,$$
which gives to fourth order in $K$ the equation
$$n(0,\omega)=\frac{1}{2 \pi }+\frac{\tilde{K}}{2 \pi \bigl(\tilde{\omega}^2+1\bigr)}-\frac{\tilde{K}^2}{4\pi}\frac{\tilde{\omega}^2-2}{(\tilde{\omega}^2+1)(\tilde{\omega}^2+4)}-\frac{\tilde{K}^3}{4\pi}\frac{ \tilde{\omega}^4-17 \tilde{\omega}^2+6}{ \bigl(\tilde{\omega}^2+1\bigr)^2 \bigl(\tilde{\omega}^2+4\bigr) \bigl(\tilde{\omega}^2+9\bigr)}+{\cal O}(K^4).$$
We thus arrive at
$$\int\_0^{2\pi}n(\psi,\omega)\cos(\psi)\,d\psi=\tfrac{1}{2}\tilde{K}\frac{1}{1+\tilde{\omega}^2}+\tfrac{1}{4}\tilde{K}^3\frac{2\tilde{\omega}^2-1}{(1+\tilde{\omega}^2)^2(4+\tilde{\omega}^2)}+{\cal O}(K^4).$$
This then gives the equation for $\sigma$,
$$
\sigma= \tfrac{1}{2} K \sigma\int\_{-\infty}^{\infty} g\left(D \omega+\omega\_{0}\right) \frac{d \omega}{\omega^{2}+1}$$
$$\qquad-\frac{K^{3} \sigma^{3}}{4 D^{3}} \int\_{-\infty}^{\infty} g\left(D \omega+\omega\_{0}\right)\left\{\frac{1}{\omega^{2}+4}-\frac{\color{red}{\omega^2}}{\left(\omega^{2}+1\right)^{2}}\right\} d \omega+{\cal O}(K^4).
$$
The term of order $K$ agrees with the formula in the OP, the term of order $K^3$ is slightly different, the OP has $\omega$ instead of the red $\omega^2$. I think this is a typo (the integrand is an even function of $\omega$).
| 3 | https://mathoverflow.net/users/11260 | 439860 | 177,622 |
https://mathoverflow.net/questions/439862 | 6 | So, I know that the existence of a scale (that is, a linear cofinal set in $({}^\omega \omega , \leq^\ast )$, where $\leq^\ast $ is eventual domination, is equivalent to $\mathfrak{b} = \mathfrak{d}$, and that we can obtain the latter through Martin's Axiom (see Hechler forcing).
My question is, what about cofinal trees of a certain height and width? A scale is a tree of width $1$ and height $\mathfrak{d}$. If $\mathfrak{b} < \mathfrak{d}$, can I at least force the existence of a cofinal tree of width $< \mathfrak{d}$?
| https://mathoverflow.net/users/495743 | Cofinal trees in $({}^\omega \omega , \leq^\ast )$ | Hechler proved that you can force the existence of a cofinal subset of $\omega^\omega$ having any shape you like (subject to one or two necessary restrictions):
**Theorem:** (Hechler) Suppose that $(P,\leq)$ is a partially ordered set with the property that every countable subset of $P$ has a strict upper bound in $P$. Then there is a ccc forcing notion which adds a cofinal subset of $(\omega^\omega,\leq^\*)$ that is order isomorphic to $(P,\leq)$.
The result can be found in:
*Hechler, Stephen H.*, On the existence of certain cofinal subsets of ${}^\omega\omega$, Axiom. Set Theor., Proc. Symp. Los Angeles 1967, 155-173 (1974). [ZBL0326.02047](https://zbmath.org/?q=an:0326.02047).
**EDIT:**
In my original answer to this question, I said that Hechler's result implies that, yes, any reasonable-looking tree can be the order type of a cofinal subset of $\omega^\omega$. K. P. Hart pointed out to me in an email that this is not in fact the case.
The poset $(\omega^\omega,\leq^\*)$ is (countably) directed, which means that it cannot contain two incompatible elements. The same is true for any cofinal subset of this order. Thus the only trees that are allowed as the order type of cofinal subsets of $(\omega^\omega,\leq^\*)$ are the unbranching trees with uncountable cofinality: i.e., the type of $\mathfrak{d}$ when $\mathfrak{b} = \mathfrak{d}$. (Or at least the type of some ordinal $\kappa$ with cofinality $\mathfrak{d}$, which can be attained by packing extra functions into a scale in certain models of $\mathfrak{c} > \mathfrak{d} = \mathfrak{b}$.)
But that being said, if $(T,\leq)$ is any tree-like poset such that every branch has uncountable cofinality, then $T$ can be used to generated a countably directed poset $(P,\leq)$, namely the poset of downward-closed subsets of $T$ that do not contain a branch, ordered by inclusion. (I'm envisioning an upward-growing tree when I say "downward closed" here -- feel free to reverse it according to your own preferences.) And this poset can, in a Hechler-style forcing extension, be isomorphic to the order type of a cofinal subset of $(\omega^\omega,\leq^\*)$.
| 6 | https://mathoverflow.net/users/70618 | 439865 | 177,623 |
https://mathoverflow.net/questions/439744 | 2 | Let $X,Y$ be algebraic subsets of $\mathbb A^n.$ I would like to show that if $X$ and $Y$ intersect "transversely" then $I(X)+I(Y)$ is radical (so $I(X\cap Y)=I(X)+I(Y)$).
How to prove it?
"transversely" can mean that $T\_p\, X+ T\_p\, Y=T\_p\, \mathbb A^n$ for all $p\in X\cap Y$ and $X$ and $Y$ are smooth at all intersection points, but I hope this condition can be relaxed somewhat. (Though I realize limitations. For example, for $X=V(u^2-v^2)$ and $Y=V(v)$, $I(X)+I(Y)=(u^2,v)$ is not radical, even though the tangent spaces at $(0,0)$ span $T\_{(0,0)}\,\mathbb A^2$.)
This statement was conjectured in this [math stackexchange answer](https://math.stackexchange.com/questions/322872/conditions-for-sqrt-mathfraka-b-sqrt-mathfraka-sqrt-mathfrak) (see the top answer), but not proved, so I am hoping to find a proof here.
| https://mathoverflow.net/users/23935 | When a sum of the ideals is radical | (Adapting earlier comment into an answer.) Assume for simplicity that $X$ and $Y$ are irreducible, or in general take an irreducible component of each. Let $Z$ be an irreducible component of the intersection $X \cap Y$ and let $p \in Z$. Let $a = \dim X$, $b = \dim Y$.
Since $X$ and $Y$ are smooth at $p$, $\dim T\_p X = a$ and $\dim T\_p Y = b$. By transversality, $\dim T\_p X \cap T\_p Y = a+b-n$.
On the one hand, $T\_p Z \subseteq T\_p X \cap T\_p Y$. So $\dim T\_p Z \leq a+b-n$.
On the other hand, $\dim Z \geq a+b-n$, see for example <https://math.stackexchange.com/a/3419739/343280>.
It follows that $\dim T\_p Z = \dim Z$. So $Z$ is reduced at the point $p$.
(I make no claim of originality in this argument, but I can't remember where I might have read it. If anyone knows where this might be found - somewhere in Fulton's Intersection Theory maybe? - I'll add a citation to this answer.)
| 2 | https://mathoverflow.net/users/88133 | 439873 | 177,625 |
https://mathoverflow.net/questions/439869 | 3 | I have heard the expression recently that one should be careful when constructing cones in the homotopy category - namely, that this is not functorial. However, when working through some examples in cochain complexes, I was playing around with the following construction which I haven't been able to understand in relation to the above.
For simplicity let $k$ be a field, and let $Ch^\*\_h(k)^{\rightarrow}$ be the category with
* Objects cochain maps $f : C \to D$
* Morphisms given by triples $(\rho,\sigma,h)$ that fit into homotopy commutative squares
$$\require{AMScd}
\begin{CD}
C @>{f}>> D ;\\
@V{\rho}VV @VV{\sigma}V \\
C' @>{f'}>> D';
\end{CD}
$$ where the key point is that homotopy is specified.
The identity map is given by $(1\_C, 1\_D, 0)$ and composition of squares
$$\require{AMScd}
\begin{CD}
C @>{f}>> D ;\\
@V{\rho}VV @VV{\sigma}V \\
C' @>{f'}>> D';
\end{CD}
$$
with homotopy $h$ and
$$\require{AMScd}
\begin{CD}
C' @>{f'}>> D' ;\\
@V{\rho'}VV @VV{\sigma'}V \\
C'' @>{f''}>> D'';
\end{CD}
$$ and homotopy $h'$ is given by
$$ (\rho', \sigma', h') \circ (\rho, \sigma, h) = (\rho' \rho, \sigma' \sigma, \sigma'h + h' \rho)$$ where $\sigma'h + h' \rho$ is indeed a homotopy between $\sigma'\sigma f$ and $f'' \rho' \rho$.
By my calculations this induces a functor $\mathsf{Cone} : Ch^\*\_h(k)^{\rightarrow} \to Ch^\*(k)$ into the category of cochain complexes where
* On objects $\mathsf{Cone}(f)$ is sent to the usual cone of a morphism
* On morphisms $(\rho,\sigma,h)$ from $f$ to $f'$ the map is given by the cochain map
$$\begin{bmatrix} \rho & 0 \\ h & \sigma \end{bmatrix} : \mathsf{Cone}(f) \to \mathsf{Cone}(f').$$
That this agrees with composition follows from matrix multiplication
$$\begin{bmatrix} \rho' \rho & 0 \\ \sigma'h + h' \rho & \sigma' \sigma \end{bmatrix} = \begin{bmatrix} \rho' & 0 \\ h' & \sigma' \end{bmatrix} \begin{bmatrix} \rho & 0 \\ h & \sigma \end{bmatrix} : \mathsf{Cone}(f) \to \mathsf{Cone}(f') \to \mathsf{Cone}(f'').$$
Now let $K(Ch^\*(k))$ be the homotopy category of cochain complexes. If I have understood correctly, the difference between this construction and the one on the homotopy category is that the objects in $K(Ch^\*(k))^{\rightarrow}$ are homotopy classes of morphisms and representative homotopies are not specified.
Taking the Kan extension of $\mathsf{Cone} : Ch^\*\_h(k)^{\rightarrow} \to Ch\_\*(k)$ along the functor $\Pi : Ch^\*\_h(Q)^{\rightarrow} \to K(Ch^\*(k))$ into the homotopy category (by sending everything to its homotopy class) yields a functor
$$ L : K(Ch^\*(k))^{\rightarrow} \to Ch^\*(k).$$
where there is a natural transformation from $L \Pi$ to $\mathsf{Cone}$, and we get a square
$$\require{AMScd}
\begin{CD}
Ch\_h^\*(k)^{\rightarrow} @>{\mathsf{Cone}}>> Ch^\*(k) ;\\
@V{\Pi}VV @VV{\Pi'}V \\
K(Ch^\*(k))^{\rightarrow} @>{\Pi' L}>> K(Ch^\*(k));
\end{CD}$$ which commutes up to natural transformation and where $\Pi'$ sends morphisms to their homotopy classes.
My question: this functor $\Pi' L$ looks like a good candidate for a functorial cone. My intuition is that something must go wrong but I've been unable to figure out what it is so far. I also want to know if this kind of thing is what people are really asking for when they ask for a functorial cone, or whether they mean something different.
| https://mathoverflow.net/users/171987 | Why does this construction not give a functorial cone in the homotopy category of cochain complexes? | The main issue you run into is that the projection
$\Pi: Ch^\*\_h(k)^{\to} \to K(Ch^\*(k))^{\to}$
is not fully faithful, and so the Kan extension $L$ is not automatically an "extension": it does not satisfy $L \Pi(f) \simeq \mathrm{Cone}(f)$.
(This occurs prominently when the map $f$ is the map $0 \to k$. If $g = k[-1] \to 0$, there are several maps $g \to f$ in $Ch^\*\_h(k)^{\to}$ but only one map in the homotopy category, and they all induce different maps $\mathrm{Cone}(g) \to \mathrm{Cone}(f)$.)
Another small issue is that $K(Ch^\*(k))^{\to}$ is a large category, and so you need some work to ensure that such a Kan extension exists.
| 2 | https://mathoverflow.net/users/360 | 439874 | 177,626 |
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