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https://mathoverflow.net/questions/442150 | 3 | Does there exist a modal formula $φ$ with the following properties?
1. for every finite Kripke frame $F$ there is some ground substitution $\sigma$ such that for every point $w \in F$ we have $F, w \Vdash \sigma(φ)$,
2. there exists an infinite Kripke frame $F$ such that for every ground substitution $\sigma$ there is some point $w \in F$ with $F, w \nVdash \sigma(φ)$.
A ground substitution is a map $σ$ from the set of variables to the set of formulas without variables (also called "ground formulas"). Their domain is naturally extended to the set of all modal formulas, by forcing $σ$ to be a homomorphism. This precisely formalises "uniformly substituting" all variables in a formula by formulas without variables.
This question arose when studying unifiability in the modal logic K.
I know the following "in between" property:
If there is a finite (intransitive) tree $F$ of depth $m$ and a ground substitution $σ$ such that
* for all variables $p$, we have $md(σ(p)) ≤ m - md(φ)$
* $F, w \Vdash \sigma(\varphi)$ for all $w \in F$
then $\sigma(\varphi)$ is valid and condition 2. above can not be fulfilled. Here $md$ stands for the modal degree of a formula. I.e. the $σ$ which exist in condition 1. have to be "very complicated" in comparison to $F$ and $φ$.
| https://mathoverflow.net/users/484622 | Existence of certain formulas in modal logic K | $\def\eq{\leftrightarrow}\def\sset{\subseteq}$An example of such a formula is
$$\phi(p,q)=(\Box p\to p)\land(p\to(q\eq\neg\Box q)).$$
**Lemma.** The formula $\phi$ satisfies condition 1.
**Proof:**
Let $(F,R)$ be a finite frame. For all $i\in\omega$, let $W\_i=\{w\in F:w\vDash\Box^i\bot\}$, where $\Box^i=\underbrace{\Box\cdots\Box}\_i$. Since $W\_0\sset W\_1\sset W\_2\sset\cdots$ and $F$ is finite, there is an $n$ such that $W\_n=\bigcup\_{i\in\omega}W\_i$. Note that $W\_n$ is the largest converse well-founded generated subframe of $F$. By well-founded recursion, there exists a unique $X\sset W\_n$ such that
$$x\in X\iff\exists y\,(x\mathrel Ry\land y\notin X)$$
for all $x\in W\_n$. Then $F\vDash\phi$ under the valuation
$$\begin{align\*}w\vDash p&\iff w\in W\_n,\\w\vDash q&\iff w\in X.\end{align\*}$$
$W\_n$ is defined in $F$ by the ground formula $\Box^n\bot$. To see that $X$ is also definable by a ground formula, put $\xi\_0=\bot$ and $\xi\_{i+1}=\neg\Box\xi\_i$ for each $i$ (that is, $\xi\_{2i}=(\Diamond\Box)^i\bot$ and $\xi\_{2i+1}=(\Diamond\Box)^i\Diamond\top$). By induction on $i$, we can show
$$\forall x\in W\_i\:(x\in X\iff F,x\vDash\xi\_i),$$
hence $X$ is definable by the formula $\Box^n\bot\land\xi\_n$. Thus, the substitution $\sigma(p)=\Box^n\bot$, $\sigma(q)=\xi\_n$ satisfies $F\vDash\sigma(\phi)$. **QED**
To see that $\phi$ satisfies 2, note that condition 2 is equivalent to “$\phi$ is not unifiable”. Since Löb’s rule $\Box\alpha\to\alpha\mathrel/\alpha$ is admissible in **K**, any unifier of $\phi$ also unifies $p$, hence $q\eq\neg\Box q$; but the latter is not unifiable, as it is not satisfiable in nonempty reflexive frames. Let me spell out a more explicit argument for completeness:
**Lemma.** The formula $\phi$ satisfies condition 2.
**Proof:**
Let $(F,R)=(\mathbb N,\{(n+1,n):n\in\mathbb N\})$ be the infinite descending irreflexive intransitive chain, and assume for contradiction that $F\vDash\sigma(\phi)$ for a ground substitution $\sigma$. Then $F\models\sigma(\Box p\to p)$ implies $n\models\sigma(p)$ for all $n\in\mathbb N$ by induction on $n$, hence also $F\models(\sigma(q)\eq\neg\Box\sigma(q))$. Thus,
$$n+1\models\sigma(q)\iff n+1\nvDash\Box\sigma(q)\iff n\nvDash\sigma(q).$$
On the other hand, for every ground formula $\alpha$, we can prove by induction on the complexity of $\alpha$ that $\{n\in\mathbb N:n\models\alpha\}$ or its completement is finite, thus for all sufficiently large $n\in\mathbb N$, we have
$$n+1\models\sigma(q)\iff n\models\sigma(q).$$
This is a contradiction. **QED**
The argument actually works for all logics between $\mathbf K$ and $\mathbf{Alt\_1=K}\oplus\Diamond p\to\Box p$.
| 4 | https://mathoverflow.net/users/12705 | 442159 | 178,425 |
https://mathoverflow.net/questions/442160 | 5 | Let $f: [0, 1] \to \mathbb R$ be a function of bounded variation. We say that $g$ is a $C^0$ reparametrization if $g = f \circ s$ for $s$ a continuous increasing bijection from a finite interval $I$ to $[0, 1]$.
**Question:** Does a continuous function of bounded variation always admit a $C^0$ reparametrization that is continuously differentiable?
*Remark: Reparametrizing by the (inverse of) the variation of $f$ gives a $1$-Lipschitz function, which is at least differentiable a.e.*
| https://mathoverflow.net/users/173490 | Can a continuous bounded variation function be $C^0$-reparametrized to be continuously differentiable? | Non-constant $C^1$ functions have intervals of monotonicity: if $f'(x\_0)\neq 0$ then $f$ is monotone on some interval $(x\_0-\epsilon,x\_0+\epsilon)$; this follows from the inverse function theorem. Evidently monotone continuous reparameterizations preserve this property. But a function of
bounded variation does not have to have such intervals. See examples in the comments.
| 12 | https://mathoverflow.net/users/25510 | 442164 | 178,426 |
https://mathoverflow.net/questions/442155 | 9 | Let us assume that there is a non-trivial elementary embedding $j \colon L\_\gamma \to L\_\gamma$ and $\gamma \geq \omega\_1^V$. Can we conclude that $0^{\#}$ exists?
In general, it is known that if there is an elementary embedding $j \colon L\_\alpha \to L\_\beta$ such that $\mathrm{crit}\, j = \delta$, $(\delta^{+})^L \leq \alpha$ and $\delta \geq \omega\_2^V$ then one can conclude that $0^{\#}$ exists, but can we get something better by starting with the assumption that the embedding is from $L\_\gamma$ to itself?
| https://mathoverflow.net/users/41953 | $0^{\#}$ and self embeddings of $L_\gamma$ | No, such embeddings can consistently exist in $L$. In the following, $\omega\_1$ and $\omega\_2$ will denote these cardinals as computed in $V$.
Let us assume that $0^\sharp$ exists in $V$. There is an increasing sequence $\langle \xi\_n\mid n<\omega\rangle\in V$ of order indiscernibles over the structure $(L\_{\omega\_2};\in, \alpha\mid\alpha<\omega\_1)$, for example take the first $\omega$-many Silver indiscernibles $\geq\omega\_1$. In $L$, we can define the tree $T$ which attempts to find such a sequence. As $T$ is illfounded in $V$, $T$ is illfounded in $L$, i.e. there is such a sequence $\langle \xi\_n\mid n<\omega\rangle$ in $L$. Now let $X=\mathrm{Hull}^{L\_{\omega\_2}}(\omega\_1\cup\{\xi\_n\mid n<\omega\})$ and let $\pi\colon X\rightarrow L\_\gamma$ be the transitive collapse of $X$. Then $\omega\_1<\gamma$. Let us write $\bar \xi\_n=\pi(\xi\_n)$. We now have
* $L\_\gamma=\mathrm{Hull}^{L\_\gamma}(\omega\_1\cup\{\bar \xi\_n\mid n<\omega\})$ and
* the $\langle \bar\xi\_n\mid n<\omega\rangle$ are order indiscernible over $(L\_\gamma;\in,\alpha\mid\alpha<\omega\_1)$.
This allows us to build elementary embeddings $j\colon L\_\gamma\rightarrow L\_\gamma$ similar to how one builds elementary embeddings $j\colon L\rightarrow L$ from Silver indiscernibles. Here is the construction:
Let $i\colon\omega\rightarrow\omega$ be any strictly increasing function. We can define an elementary $j\colon L\_\gamma\rightarrow L\_\gamma$, $j\in L$, by
$j(\tau^{(L\_\gamma;\in, \alpha\mid\alpha<\omega\_1)}(\bar\xi\_0,\dots,\bar\xi\_n))=\tau^{(L\_\gamma;\in,\alpha\mid\alpha<\omega\_1)}(\bar\xi\_{i(0)},\dots,\bar\xi\_{i(n)})$ for any term $\tau(x\_0,\dots, x\_n)$ in the language with the $\in$-relation and a constant for every ordinal below $\omega\_1$. It is not difficult to check that $j$ is well-defined, elementary and, in case $i\neq\mathrm{id}\_{\omega}$, even non-trivial.
| 10 | https://mathoverflow.net/users/125703 | 442166 | 178,428 |
https://mathoverflow.net/questions/442189 | 0 | For countable models, elementary equivalence is not equivalent to isomorphism.
For example, let $\frak{A}= \omega+\Bbb{Z}\*\omega$ and $\frak{B}= \omega+\Bbb{Z}\*\omega^\*$ ($ω^∗$ is the reverse of $ω$). Then $\frak{A}\equiv\frak{B}$ but $\frak{A}\not\cong\frak{B}$ for $\frak{A}$ and $\frak{B}$ are not atomic. (For two countable atomic models if $\frak{A}\equiv\frak{B}$, then $\frak{A}\cong\frak{B}$.)
I am not sure how to prove $\frak{A}\equiv\frak{B}$ exactly even though I know Ehrenfeucht–Fraïssé game can do it. Also, I am not sure how to prove they are not atomic.
*Note:* This question has been posted on the Math StackExchange website, but can not get an answer.
| https://mathoverflow.net/users/nan | Need proof on a model being elementarily equivalent but non-isomorphic | To show that the models are not isomorphic assume an isomorphism $f:\frak{A}\to\frak{B}$, take $(a\_n)\_{n\in\omega}$ be an increasing sequence such $a\_i,a\_j$ are from a different copy of $\Bbb Z$ for each $i\ne j$.
Now look at $f(a\_i),f(a\_j)$. If they are from the same copy of $\Bbb Z$ then there exists some natural number $n$ that is their difference, and so there is a first order formula that represent this ("There exists exactly $n-1$ different elements that are between $f(a\_i)$ and $f(a\_j)$), because $f$ is an isomorphism it would imply that there is exactly $n-1$ elements between $a\_i,a\_j$ which is a contradiction.
Now for each $a\_i$ let $b\_i$ be the unique $k\in\omega^\*$ such that $f(a\_i)∈{\Bbb Z}×\{k\}$, this result with $(b\_i)\_{i\in\omega}$ being an increasing sequence in $ω^\*$, contradiction.
---
To see that they are elementary equivalent we want to look at the EF game of length $n$, $G\_n(\frak A,B)$.
When the Spoiler pick an element from $ω$ of one of the structures, the Duplicator will choose the same element from $\omega$ in the other one.
Next if the Spoiler choose an element not from $\frak A\setminus\omega$, then the Duplicator will choose any element from ${\Bbb Z}×\{-n^n\}$ in $\frak B$, and dually if the Spoiler choose from $\frak B\setminus\omega$.
Next if the Spoiler chooses an element that is a finite distance from a chosen element, the Duplicator will just duplicate this distance with the corresponded element in the other model.
If the Spoiler chooses a new $\Bbb Z$, the Duplicator will duplicate the $\Bbb Z$ but will cap the distance between the index of the chosen $\Bbb Z$ to the index of existing chosen indexes to have $k^k$ where $k$ is the number of steps left to the game. This will leave enough room to full out any kind of structure the Spoiler will try to create.
---
To see that they are not atomic, let for each natural $n$ $φ\_n(x)$ be "$x$ is at least $n$" and let $τ(x)$ be the partial type consistent of all $\phi\_n$. Note that for each $a\in\frak A$ we have $a\notin ω⇔τ(a)$, in particular $τ⊆\operatorname{tp}(a)$.
Now let $ψ(x)$ be any formula (that does not contain $z\ne z$ for some variable $z$), $ψ(x)$ can only describe what kind of finite linear orders exists/don't exists bellow $x$ and what kind of finite linear orders above $x$ and the distance of those orders from $x$ *if it is finite* and their distance from $0$ *if it is finite*
We don't care about what happens above $x$ because the end segment of $a$ and the end segment of $b$ are isomorphic for all $a,b$.
Now take some $a\notin ω$ and assume that $\operatorname{tp}(a)$ is isolated by $ψ(x)$. Note that any finite linear order exists bellow $a$, in particular $ψ(x)$ only describe what kind of orders **do** exists bellow $x$ and the distance of those orders if it is finite.
Take a natural number $m$ that is bigger than the sum of the length of the orders that $ψ$ describe plus the distance from $x$ of those orders plus the distance from $0$ of those orders, and any element $b∈ω\setminus m$ will satisfy $ψ$, in particular $ψ(x)\;\not\!\!\!\implies φ\_b(x)$ and hence does not isolate $\operatorname{tp}(a)$
| 1 | https://mathoverflow.net/users/113405 | 442190 | 178,433 |
https://mathoverflow.net/questions/442181 | 3 | Consider a set $N$ with elements $n\_1, n\_2, \dots, n\_k$ which are distinct integers. Introduce the notation $N\_{i=1,2,\dots,s}$ for the $s$ blocks of a set partition of $N$. Consider a supplementary integer $n\_{k+1}$.
We have the identity
$(\sum\_{i=1}^{k+1} n\_i -k)^{k-1}= \sum\_{\text{set partitions } N\_i \text{ of } N } \prod\_{l=1}^s ((\sum\_{n\_j \in N\_l} n\_j)−|\_l|+1)^{|N\_l|-1}(\_{k+1}−1)!/(\_{k+1}−)!$
I have proven the identity for $=1,2,…,8$ analytically (using a symbolic manipulation program) and for $=9,10,11$
numerically. On the right hand side, we have expanded the polynomial in $n\_{k+1}$
into falling factorials according to the number $s$ of blocks in the set partition $N\_i$. The number $k$ is a positive integer.
The identity arises in the context of the theory of conjugacy class sums of the symmetric group.
I have proven parts of this identity, e.g. for particular values of $n\_{k+1}$, and so on. I have tried various inductions (on the number of variables $k$ as well as on the number of parts s in the partition) and derivations, though not exhaustively.
Is there a known or unknown analytic proof of this identity ? Is there a standard and/or efficient strategy to prove the identity ?
(Edited for clarity following a first comment and fixed a typo indicated in a second comment.)
| https://mathoverflow.net/users/70000 | Proof of a combinatorial identity for a sum over partitions of sets giving rise to a symmetric polynomial? | The notation and framing of the problem are not optimal, because the $n\_i$'s being integers is a bit of a distraction.
Let $[k]:=\{1,\ldots,k\}$, and let ${\rm Part}\_k$ be the set of set partitions of $[k]$. We also use $|A|$ to denote the cardinality of a set $A$.
Let $x\_1,\ldots,x\_k$ and $y$ be formal variables.
The problem is to prove
$$
(1+y+x\_1+\cdots+x\_k)^{k-1}=\sum\_{\Theta\in{\rm Part}\_k}
y(y-1)\cdots(y-|\Theta|+2)\times\prod\_{J\in\Theta}
\left(1+\sum\_{i\in J}x\_i\right)^{|J|-1}\ .
$$
I think this is true because the matrices made of [Stirling numbers](https://en.wikipedia.org/wiki/Stirling_number) of first and second kinds are inverses of each other.
To recover the original notations set $x\_i=n\_i-1$ and $y=n\_{k+1}-1$.
---
**Update:** Here is a proof.
Use the matrix-tree/Cayley formula for counting trees according to specified vertex-degrees (see, e.g., [these notes](https://math.berkeley.edu/~mhaiman/math172-spring10/matrixtree.pdf) by Mark Haiman), which reads
$$
z\_1\cdots z\_n(z\_1+\cdots+z\_n)^{n-2}=\sum\_{T}\prod\_{\{i,j\}\in T}z\_i z\_j\ ,
$$
where the sum is over trees $T$ on the set $[n]$ which connect all vertices.
Edges are treated as subsets $\{i,j\}$ of $[n]$ of cardinality 2.
Take $n=k+1$, $z\_1=1$, $z\_2=x\_1,\ldots,z\_n=x\_k$.
Then we get a sum over rooted forests
$$
(1+x\_1+\ldots+x\_k)^{k-1}=\sum\_{F,R}\frac{\prod\_{\{i,j\}\in F}x\_i x\_j}{\prod\_{i\in[k]\backslash R} x\_i}\ ,
$$
where $F$ is the set of edges and $R$ is the set of roots (exactly one per connected component).
Now instead of using it once, globally on $[k]$, use the previous identity *locally* in each $J$. This gives
$$
\prod\_{J\in\Theta}\left(1+\sum\_{i\in J}x\_i\right)^{|J|-1}
=\sum\_{\Pi\le \Theta}\sum\_{F,R|\Pi}\frac{\prod\_{\{i,j\}\in F}x\_i x\_j}{\prod\_{i\in[k]\backslash R} x\_i}\ ,
$$
where the sum is restricted to forests whose components form the partition $\Pi$ which refines the fixed partition $\Theta$.
**Remark 1:**
The last step we will need is related to Möbius inversion in the partition lattice.
For $y=-1$ the multiplying factor
$y(y-1)\cdots(y-|\Theta|+2)$ is the Möbius function $\mu(\Theta,\mathbf{1})$. For reference, the partition lattice is the set ${\rm Part}\_k$ together with the partial order $\Theta\_1\le\Theta\_2$ when the blocks of $\Theta\_1$ are contained in blocks of $\Theta\_2$, i.e., $\Theta\_1$ is refinement or further subdivision of $\Theta\_2$.
The minimal element is $\mathbf{0}$, the partition made of singletons. The maximal element is $\mathbf{1}$, the partition made the single block $[k]$. The möbius matrix $(\mu(\Theta\_1,\Theta\_2))\_{\Theta\_1,\Theta\_2}$ with rows and columns indexed by set partitions, is the inverse of the matrix given by the indicator function of the statement $\Theta\_1\le \Theta\_2$.
Now note that the right-hand side of the identity to be established is
$$
RHS=\sum\_{\Theta\in{\rm Part}\_k}
\sum\_{\Pi\le\Theta}
y(y-1)\cdots(y-|\Theta|+2)
\times\left(\sum\_{F,R|\Pi}\frac{\prod\_{\{i,j\}\in F}x\_i x\_j}{\prod\_{i\in[k]\backslash R} x\_i}\right)\ ,
$$
$$
=\sum\_{\Pi\in{\rm Part}\_k}
\sum\_{\Theta\ge\Pi}y(y-1)\cdots(y-|\Theta|+2)
\times\left(\sum\_{F,R|\Pi}\frac{\prod\_{\{i,j\}\in F}x\_i x\_j}{\prod\_{i\in[k]\backslash R} x\_i}\right)\ ,
$$
$$
=\sum\_{\Pi\in{\rm Part}\_k}\left(
\sum\_{F,R|\Pi}\frac{\prod\_{\{i,j\}\in F}x\_i x\_j}{\prod\_{i\in[k]\backslash R} x\_i}\right)\times
\sum\_{\Theta\ge\Pi}y(y-1)\cdots(y-|\Theta|+2)\ .
$$
Now, we have
$$
\sum\_{\Theta\ge\Pi}y(y-1)\cdots(y-|\Theta|+2)=
\frac{1}{y+1}\sum\_{j=1}^{|\Pi|}S(|\Pi|,j)\ (y+1)y(y-1)\cdots((y+1)-j+1)\ ,
$$
where the $S(\cdot,\cdot)$ are the Stirling numbers of the second kind. We also ``contracted'' $\Pi$ so blocks become points.
By a [well known property of the Stirling numbers](https://en.wikipedia.org/wiki/Stirling_number#Expansions_of_falling_and_rising_factorials),
$$
\sum\_{j=1}^{|\Pi|}S(|\Pi|,j)\ (y+1)y(y-1)\cdots((y+1)-j+1)=(y+1)^{|\Pi|}\ ,
$$
so
$$
RHS=\sum\_{\Pi\in{\rm Part}\_k}\left(
\sum\_{F,R|\Pi}\frac{\prod\_{\{i,j\}\in F}x\_i x\_j}{\prod\_{i\in[k]\backslash R} x\_i}\right)\times (y+1)^{|\Pi|-1}\ ,
$$
$$
=\frac{1}{x\_1\cdots x\_k(y+1)}
\sum\_{\Pi\in{\rm Part}\_k}
\sum\_{F,R|\Pi}(y+1)^{|\Pi|}\prod\_{i\in R}x\_i\prod\_{\{i,j\}\in F}x\_ix\_j\ .
$$
This clearly is a sum over trees on a set with $n=k+1$ elements. Using the matrix tree/Cayley formula again (but in the reverse direction) with $z\_1=y+1$, $z\_2=x\_1,\ldots,z\_n=x\_k$, and cleaning up, we get the desired result $(1+y+x\_1+\cdots+x\_k)^{k-1}$.
**Remark 2:** Since this problem showed up in a study of cycle structures of permutations, one could mention the following. Forests give set partitions by taking connected components. Permutations give set partitions by taking the supports of the disjoint cycles of the permutation. So there is nothing surprising in forests featuring in the proof above. However, there is also a *direct* relation between forests and permutations. Writing a permutation as a minimal product of transpositions gives a forest. The correspondence is not 1:1 of course, and this has been studied by Dénes and other, see my previous answer
<https://math.stackexchange.com/questions/3535217/in-how-many-ways-can-a-permutation-cycle-be-decomposed-as-a-product-of-transposi/3536747#3536747>
| 5 | https://mathoverflow.net/users/7410 | 442192 | 178,434 |
https://mathoverflow.net/questions/442163 | 20 | One motivation for studying infinity categories is that the derived category does not satisfy Zariski descent, although the infinity categorical version does.
I would like to see an example of Zariski descent failing for the (one categorical) derived category of coherent sheaves on a scheme.
That is I would like an example of an (preferably affine) algebraic variety $X$, an open cover $\{U\_i \hookrightarrow X\}$ and a descent datum in the derived category which does not glue uniquely to an element of the derived category of $X$.
| https://mathoverflow.net/users/153310 | The derived category does not satisfy descent - example | Consider the canonical functor $H$ from the homotopy category of homotopy coherent descent data in the ∞-category of coherent sheaves
to the category of descent data in the derived category of coherent sheaves.
Nonuniqueness of gluing amounts to $H$ not being a full functor
and is observed for covers of cardinality 3 and higher.
Nonexistence of gluing amount to $H$ not being essentially surjective
and is observed for covers of cardinality 4 and higher.
For covers of cardinality 0, 1, and 2, the functor $H$ is an equivalence.
Let's analyze first the case of covers of cardinality 3.
Given three opens $U\_1$, $U\_2$, $U\_3$,
an object in the domain of $H$
will have chain complexes of coherent sheaves $F\_i$ over $U\_i$,
isomorphisms $t\_{i,j}:F\_i→F\_j$ of their restrictions to $U\_i∩U\_j$ ($1≤i<j≤3$, setting $t\_{j,i}=t\_{i,j}^{-1}$),
and a choice of a homotopy $h$ (or rather its homotopy class) from the composition $t\_{3,1}t\_{2,3}t\_{1,2}$ to the identity map on the restriction of $F\_1$ to $U\_1∩U\_2∩U\_3$.
In particular, objects with the same data of $F\_i$ and $t\_{i,j}$, but nonhomotopic $h$ are nonisomorphic.
An object in the codomain of $H$ has a similar description, except that $h$ is merely required to exist, but its data is not included.
It is now clear how to show that $H$ is not full: pick two objects in the domain of $H$ with the same data of $F\_i$ and $t\_{i,j}$,
but different (nonhomotopic) $h$.
Their images under $H$ will see the data of $h$ discarded, so the two objects become isomorphic in the codomain of $H$,
despite not being isomorphic in the domain of $H$.
Therefore, to construct a concrete example, one needs to specify $U\_i$, $F\_i$, and $t\_{i,j}$
such that the identity map $p$ on $F\_1$ restricted to $U\_1∩U\_2∩U\_3$
admits a homotopy $h$ from $p$ to itself
that is not homotopic to the identity homotopy from $p$ to itself.
That is to say, we want $\def\Aut{\mathop{\sf Aut}\nolimits}\def\End{\mathop{\sf End}\nolimits} π\_1(\Aut(F\_1))$ to be nontrivial.
For example, we can take $\def\cO{{\cal O}} F\_1=\cO[0]⊕\cO[1]$, then $$\End(F\_1)=\cO[-1]⊕\cO[0]⊕\cO[0]⊕\cO[1],$$ and a nontrivial 1-cocycle in $\Aut(F\_1)$ yields such a homotopy $h$.
The case of covers of cardinality 4 is similar.
As before, for an object in the domain of $H$ we have $U\_i$ and $F\_i$ ($1≤i≤4$), $t\_{i,j}$ ($1≤i<j≤4$), $h\_{i,j,k}$ ($1≤i<j<k≤4$),
and now on $⋂\_i U\_i$ there is a homotopy $q$ from a certain composition of homotopies $h$ to the identity homotopy on the identity map on the restriction of $F\_1$ to $⋂\_i U\_i$.
The functor $H$ discards the data of $h$ and $q$.
In the codomain of $H$, some maps $h$ must exist (but their choice is not given to us),
but $q$ need not exist at all.
Indeed, by modifying the above example appropriately, we can construct descent data for the derived category
in which the composition of homotopies $h$ is not homotopic to the identity homotopy,
which means that the existence of gluing is violated for the derived category.
| 11 | https://mathoverflow.net/users/402 | 442194 | 178,435 |
https://mathoverflow.net/questions/442183 | 9 | This is a reference question:
Let $\psi(x)$ be the psi-Chebyshev function. Is there any unconditional result in the literature that proves that there exists $0<a<2$ such that
$$
\int\_2^x (\psi(y)-y)^2 \mathrm dy =O(x^{a})
?$$
| https://mathoverflow.net/users/9232 | $\psi(x)-x$ on average | Impossible as this would imply that $\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1}$ is analytic on $\Re(s)\ge 1/2$.
Your bound
$$\int\_1^x |\psi(y)-y|^2 dy = O(x^a)$$ for some $a < 2$ implies (with Cauchy-Schwarz inequality) that $$\int\_x^{2x} (\psi(y)-y)dy = O(x^{a/2+1/2}) \implies\int\_1^2 (\psi(xt)-xt)dt = O(x^{a/2-1/2})
$$
i.e.
$$s\int\_1^\infty\int\_1^2 (\psi(tx)-tx)dt x^{-s-1}dx= \left(\frac{-\zeta'(s)}{\zeta(s)}-\frac{s}{s-1}\right)\int\_1^2 t^s dt-f(s)$$
extends analytically from $\Re(s) > 1$ to $\Re(s) > a/2-1/2$ which is absurd.
(where $f(s)=s\int\_1^2 t^s\int\_1^t x^{-s}dxdt$ is an irrelevant entire term)
| 8 | https://mathoverflow.net/users/84768 | 442209 | 178,440 |
https://mathoverflow.net/questions/442198 | 3 | In my setting, $\mu$ and $\nu$ are probability measures on $\mathbb{R}^{2}$ with compact support. For any function $f\in{C^{2}\_{b}(\mathbb{R}^{2})}$, I have the estimate:
$$
|\mathbb{E}\_{\mu}(f)-\mathbb{E}\_{\nu}(f)|\leq{a\|Df\|\_{L^{\infty}}+b\|D^{2}f\|\_{L^{\infty}}}
$$
for some coefficients $a,b>0$. Does the estimate above imply an estimate for the Wasserstein distance of the form:
$$
W\_{2}(\mu,\nu)\leq{\eta(a,b)}
$$
where $\eta$ is a function of $a$ and $b$ such that $\eta(a,b)\rightarrow{0}$ as $(a,b)\rightarrow{(0,0)}$?
| https://mathoverflow.net/users/80052 | Getting Wasserstein closeness from a derivative estimate | Assuming the support of both measures is fixed, it is enough to bound $W\_1(\mu,\nu)$. That is, we want to bound $\vert \int f d\mu -d\nu \vert$ for any $1$-Lipschitz function $f$.
Take a triangulation of the plane with equilateral triangles of size $\sqrt b$ and let $f\_1$ be the linear interpolation of $f$. We have $\vert f - f\_1 \vert = O(\sqrt b)$. Also the derivative of $f\_1$ is $O(1)$. Take $f\_2$ to be the convolution of $f\_1$ with a Gaussian of width $\sqrt b$. It is not too hard to see that the second derivative of $f\_2$ is $O(1/\sqrt b)$, and its first derivative is $O(1)$, implying that $\vert \int f\_2 d\mu -d\nu \vert = O(a+\sqrt b)$. Now $\vert f\_2-f\_1\vert = O(\sqrt b)$ because the derivative of $f\_1$ is $O(1)$. Hence $\vert f-f\_2\vert $ is also $O(\sqrt b)$, meaning that $\vert \int f d\mu -d\nu \vert = O(a+\sqrt b) + O(\sqrt b)$.
| 2 | https://mathoverflow.net/users/112954 | 442210 | 178,441 |
https://mathoverflow.net/questions/442203 | 11 | The category of condensed sets is the colimit of the toposes of $\kappa$-condensed sets over all cardinals $\kappa$, or equivalently the category of "small sheaves" on the large site of all compact Hausdorff spaces. As such, it is not itself a topos, although it is an infinitary pretopos. I have encountered claims that it is additionally locally cartesian closed, but I have not found a proof written down.
*Is the category of condensed sets (locally) cartesian closed?*
Based on general results about small (pre)sheaves that I have found, I suspect that this ought to be equivalent to some statement such as "For any compact Hausdorff spaces $A,B$ there exists a set of compact Hausdorff spaces $C\_i$ and maps $e\_i : A\times C\_i \to B$ such that for any extremally disconnected compact Hausdorff space $X$, any map $A\times X\to B$ factors through a map $X\to C\_i$ for some $i$." But I have not checked the details.
Of course, the category of $\kappa$-condensed sets is locally cartesian closed, since it is a topos. If the category of all condensed sets *is* also locally cartesian closed, I would also like to know under what conditions the embedding of $\kappa$-condensed sets in condensed sets is a locally cartesian closed functor. My guess would be that this is some kind of closure property of $\kappa$ that ought to be true for arbitrarily large values of $\kappa$.
| https://mathoverflow.net/users/49 | Are condensed sets (locally) cartesian closed? | Condensed sets are indeed locally cartesian closed. On the other hand, for no cardinal $\kappa$ (no matter how inaccessible) the functor from $\kappa$-condensed sets to condensed sets preserves all internal Hom's. Indeed, consider the internal Hom from a discrete set $I$ towards the discrete set $\{0,1\}$: This is evidently $\prod\_I \{0,1\}$, and within all condensed sets it is represented by this profinite set. But if $|I|>\kappa$, this is not a $\kappa$-condensed set.
To prove that internal Hom's exist, one uses the following criterion. See for example Proposition 2.1.9 of [Lucas Mann's Thesis](https://arxiv.org/abs/2206.02022). (Some incomplete discussion along those lines is also after Lecture 2 of [Condensed Mathematics](https://people.mpim-bonn.mpg.de/scholze/Condensed.pdf).) I should mention that we define a profinite set to be $\kappa$-small if the corresponding Boolean algebra is $\kappa$-small. (This is in general slightly different than asking about the size of underlying set of the profinite set. If $\kappa$ is a strong limit, the distinction disappears, but for regular $\kappa$ it is relevant.)
>
> Lemma. Let $X$ be a functor from profinite sets to sets that satisfies the hypersheaf condition, but is not necessarily small. Let $\kappa$ be a regular cardinal. Then $X$ is a $\kappa$-condensed set if and only if $X$ takes $\kappa$-cofiltered limits (of profinite sets) to $\kappa$-filtered colimits (of sets), i.e.
> $$
> X(\varprojlim\_i S\_i) = \varinjlim\_i X(S\_i)
> $$
> for $\kappa$-cofiltered diagrams of profinite sets $S\_i$.
>
>
>
In particular, $X$ is a condensed set if and only if the functor
$$
X: \mathrm{ProFin}^{\mathrm{op}} = \mathrm{Ind}(\mathrm{Fin}^{\mathrm{op}})\to \mathrm{Set}
$$
is accessible, i.e. commutes with $\kappa$-filtered colimits for some large enough $\kappa$.
Now one checks that this applies to the internal Hom $\mathrm{Hom}(X,Y)$ between any condensed sets $X$ and $Y$. More precisely, note first that this criterion implies that condensed sets admit arbitrary small limits (as any $\lambda$-small limit commutes with $\kappa$-filtered colimits for $\kappa>\lambda$). Resolving $X$ by disjoint unions of profinite sets, this reduces the existence of internal Hom's to the case that $X=S$ is a profinite set. In that case, the internal Hom $\mathrm{Hom}(S,Y)$ is given by the functor taking any profinite set $T$ to $Y(S\times T)$. But if $T\mapsto Y(T)$ sends $\kappa$-cofiltered limits to $\kappa$-filtered colimits, then so does $T\mapsto Y(S\times T)$. See also Corollary 2.1.10 and Remark 2.1.12 in Lucas Mann's Thesis.
Essentially the same argument works in a slice (over some profinite set, and then in general), showing that condensed sets are locally cartesian closed.
[**Edit:** In the comments, Mike Shulman asked whether the internal Hom commutes with the inclusion of $\lambda$-small $\kappa$-condensed sets into all condensed sets. And indeed it does when $\kappa$ is regular and $\lambda<\kappa$, and in fact this is what the argument above proves. Here, a condensed set is $\lambda$-small $\kappa$-condensed if it is a $\lambda$-small colimit of $\kappa$-small profinite sets. Even better, see Z.M's question in the comments, the proof shows that if $\kappa$ is regular and $\lambda<\kappa$, and $X$ is a $\lambda$-small condensed set (of any condensedness) and $Y$ is a $\kappa$-condensed set (but of any cardinality) then $\mathrm{Hom}(X,Y)$ is a $\kappa$-condensed set.]
| 13 | https://mathoverflow.net/users/6074 | 442211 | 178,442 |
https://mathoverflow.net/questions/442212 | 9 | Let $\mathcal{C}$ denote a category with a zero object. Let $\mathrm{N}(X)$ denote the norm of an object $X$ of a category $\mathcal{C}$, defined as $\mathrm{N}(X) = \,|\mathrm{Hom}(X,X)\,|$. From this terminology, we say a non-zero object $X$ is *finite* if $\mathrm{N}(X)$ is finite. Let $\mathrm{P}(\mathcal{C})$ denote the isomorphism classes of all finite simple objects and $\mathrm{M}(\mathcal{C})$ denote the isomorphism classes of all finite semisimple objects. For the former, letting $[X] \in \mathrm{P}(\mathcal{C})$, the norm $\mathrm{N}([X])$ is well-defined so we choose a representative of an isomorphism class $[X]$ so we have $\mathrm{N}([X]) = \mathrm{N}(X)$. According to [1], the *categorical zeta function* of our category $\mathcal{C}$ is defined as
$$\zeta\_\mathcal{C}(s) = \prod\_{[X] \in \mathrm{P}(\mathcal{C})} \frac{1}{1-\mathrm{N}(X)^{-s}}.$$
As an example, if $\mathcal{C}$ denotes the category of $\mathbb{Z}$-modules, we have $\zeta\_{\mathcal{C}}(s) = \zeta(s)$, the Riemann zeta function. Generalizing this to $O\_K$-modules, where $O\_K$ denotes the ring of integers of a number field $K$, we have $\zeta\_{\mathcal{C}}(s) = \zeta\_K(s)$, the Dedekind zeta function. From the definition of a semisimple object, it seems to me we ought to have the relationship
$$\prod\_{[X] \in \mathrm{P}(\mathcal{C})} \frac{1}{1-\mathrm{N}(X)^{-s}} = \sum\_{[X] \in \mathrm{M}(\mathcal{C})} \frac{1}{\mathrm{N}(X)^s},$$
since we can write semisimple objects into a coproduct (direct sum, in this case) of simple objects but I'm not quite sure if I can exactly treat $\mathrm{N}$ as if it was multiplicative. I'm not finding other sources that consider the above.
>
> **Question:** Is it true, for any category with a zero object, we have the relationship $$\prod\_{[X] \in \mathrm{P}(\mathcal{C})} \frac{1}{1-\mathrm{N}(X)^{-s}} = \sum\_{[X] \in \mathrm{M}(\mathcal{C})} \frac{1}{\mathrm{N}(X)^s} \;?$$
>
>
>
**References:**
1.) N. Kurokawa. Zeta Functions of Categories. *Proc. Japan Acad.* **Vol 72A** No. 10, 221-222 (1996).
| https://mathoverflow.net/users/70508 | Properties of categorical zeta function | Consider the category $\def\Set{\mathbf{Set}}\Set\_\*$ of pointed sets.
Kurokawa defines the simple objects to be those nonzero $X$ for which the only morphisms $X\to Y$ are monomorphisms or zero, so the only simple object of $\Set\_\*$ is the two-element set $P = \{\*, 0\}$, whose norm is $N(P) = 2$.
Therefore,
$$
\zeta\_{\Set\_\*}(s) = \frac1{1-2^{-s}}
$$
On the other hand, the finite pointed sets are the semisimples, and $N(X) = (n+1)^n$ if $|X|=n+1$.
Therefore,
$$
\sum\_{[X]\in M(\Set\_\*)}\frac1{N(X)^s} = \sum\_{n=0}^\infty\frac1{(n+1)^{ns}}
$$
which does not agree with the zeta function for $\Set\_\*$.
---
I imagine this even fails when $\def\cC{\mathcal{C}}\cC = \def\Z{\mathbb{Z}}\def\Mod{\mathbf{Mod}}\Z\Mod$.
The finite semisimple $\Z$-modules are precisely the direct sums of cyclic groups of prime order, hence there is a unique semisimple $\Z$-module of every finite order.
The issue is that the norm of such a module is not its order, unlike with the simples.
Indeed, even consider $\Z/3\oplus\Z/3$.
In this case, the endomorphism ring has order $3^4$ (rather than the $3^2$ you would want).
In fact, you can prove the following "almost multiplicity" of the norm for an abelian category:
**Claim 1.** If $\cC$ is abelian, then for simple objects $X$ and $Y$, we have
$$
N(X\times Y) = \begin{cases} N(X)\times N(Y), & \text{if $X\not\cong Y$} \\ N(X)^4, & \text{otherwise} \end{cases}
$$
>
> *Proof.* Finite products and coproducts coincide for abelian categories, so
> $$
> \begin{align\*}
> N(X\times Y) &= |\def\Hom{\operatorname{Hom}}\Hom(X\times Y, X\times Y)| \\
> &= |\Hom(X\sqcup Y, X\times Y)| \\
> &= |\Hom(X, X)|\times|\Hom(X, Y)|\times|\Hom(Y, X)|\times|\Hom(Y, Y)|
> \end{align\*}
> $$
> Now, by Schur's lemma, $\Hom(X, Y) = \{0\}$ if and only if $X\not\cong Y$ for simples $X$ and $Y$.
> Therefore, in one case, you get $N(X)\times N(Y)$, and in the other case, you get $N(X)^4$. $\blacksquare$
>
>
>
In particular, the series $\sum\_{[X]\in M(\Z\Mod)}\frac1{N(X)^s}$ would not agree with $\sum\_{n=1}^\infty\frac1{n^s}$.
---
**Edit.** ($\times2$) After some thought, I believe there is a good "fix" for the norm that should still be satisfying.
**Definition.** For a more general object $Y$ of $\cC$, define its **norm** to be
$$
\bar N(Y) := \prod\_{[X]\in P(\cC)}|\Hom(Y, X)|
$$
In the context of a category where Schur's lemma applies, this will be an extension of the original definition of a norm for simple objects---in that $N(X) = \bar N(X)$ for $X$ simple---but this definition has the advantage of being multiplicative:
**Claim 2.** In an abelian category, for all semisimple objects $X$ and $Y$, we have $\bar N(X\oplus Y) \cong \bar N(X)\times \bar N(Y)$.
The proof of the above claim is similar to Claim 1, but now with this multiplicative norm, one can show (analogous to the usual Euler product identity for the Riemann zeta function) that
$$
\zeta\_\cC(s) = \sum\_{[X]\in M(\cC)}\frac1{\bar N(X)^s}
$$
for $\cC$ an abelian category.
---
I edited my answer to map *into* simples because this actually makes this identity true also for $\Set\_\*$!
We have for any pointed set $X$ of cardinality $n+1$ that $\bar N(X) = |\Hom(X, \{\*, 1\})| = 2^n$, and so
$$
\sum\_{[X]\in M(\Set\_\*)}\frac1{\bar N(X)^s} = \sum\_{n=0}^\infty\frac1{2^{ns}} = \frac1{1-2^{-s}} = \zeta\_{\Set\_\*}(s)
$$
In fact, we can now prove the following
**Claim 3.** In any category $\cC$ with a zero object and finite coproducts such that finite simple objects satisfy Schur's lemma,
$$
\prod\_{[X]\in P(\cC)}\frac1{1-N(X)^{-s}} = \sum\_{[X]\in M(\cC)}\frac1{\bar N(X)^s}
$$
where "Schur's lemma" is the statement that any nonzero map $X\to Y$ between finite simple objects is an isomorphism (which holds also for $\Set\_\*$), and $M(\cC)$ is the set of isomorphism classes of finite coproducts of finite simple objects.
| 11 | https://mathoverflow.net/users/160838 | 442220 | 178,446 |
https://mathoverflow.net/questions/442230 | 1 | Let $W$ be a one-dimensional Brownian motion. Consider the stochastic differential equation (SDE)
$$dX\_t = C(t)(1-X\_t)dW\_t,\quad \forall t\ge 0,$$
where $C$ is a continuous and bounded function. Under which condition (on $C$),
1. the above SDE has an explicit solution?
2. any solution $X$ satisfies $\lim\_{t\to\infty} X\_t=1$ almost surely?
| https://mathoverflow.net/users/493556 | On a martingale defined via some SDE | If $X\_0$ and $C$ are deterministic (or independent of $W$), then
$$X\_t = 1-(1-X\_0)\exp\left(-\int\_0^tC(s)\mathrm dW\_s-\frac12\int\_0^tC(s)^2\mathrm ds\right).$$
This will go to 1 if the argument of the exponential goes to $-\infty$. I claim that this is equivalent to the condition $I:=\int\_0^\infty C(s)^2\mathrm ds=+\infty$.
If $I<\infty$, then the argument of the exponential converges in $\mathrm L^2$ to the same integral for $t=\infty$, which has distribution $\mathcal N(-I/2,I)$; this makes it impossible for it to converge almost surely to $-\infty$.
If $I=\infty$, then $M:t\mapsto\int\_0^t C(s)\mathrm dW\_s$ is a continuous martingale. By the Dambis-Dubins-Schwarz theorem [RY, Theorem V-(1.6) in my edition], there exists a Brownian motion $B$ on the same probability space such that $M\_t=B\_{\langle M,M\rangle\_t}$ for all $t\geq0$. In these terms, the argument of the exponential is $B\_{\langle M,M\rangle\_t}-\langle M,M\rangle\_t/2$. Since almost surely we have $\lim\_{\tau\to\infty}(B\_\tau-\tau/2)=-\infty$ and $\lim\_{t\to\infty}\langle M,M\rangle\_t=I=+\infty$, the exponential goes to zero and $X$ goes to 1 almost surely.
---
[RY] D. Revuz and M. Yor, *[Continuous Martingales and Brownian Motion](https://doi.org/10.1007/978-3-662-06400-9),* third edition (1993).
| 2 | https://mathoverflow.net/users/129074 | 442231 | 178,448 |
https://mathoverflow.net/questions/442188 | 5 | Let $\mathscr{P}$ be a polyhedral decomposition of a real vector space $V$. By that I mean that $\mathscr{P}$ is a finite set of polyhedra in $V$ satisfying the following three properties:
1. The union of the polyhedra in $\mathscr{P}$ is equal to $V$.
2. If $P\in \mathscr{P}$ and $Q$ is a face of $P$, then $Q\in\mathcal{P}$.
3. If $P,Q\in\mathscr{P}$, then $P\cap Q$ is a (possibly empty) face of both $P$ and $Q$.
Question: Is the union of the bounded elements of $\mathscr{P}$ contractible?
If $\mathscr{P}$ consists of the set of faces of an essential hyperplane arrangement, then a positive answer is stated in Exercise 1.7.a of Stanley's lecture notes [https://www.cis.upenn.edu/~cis6100/sp06stanley.pdf](https://www.cis.upenn.edu/%7Ecis6100/sp06stanley.pdf), though I don't know the reference for a proof. Note that, when $\mathscr{P}$ does not come from a hyperplane arrangement, then the bounded complex need not be pure.
| https://mathoverflow.net/users/10273 | The bounded complex of a polyhedral decomposition | Below I construct a deformation retract of $V$ onto the bounded sub-complex of $\mathscr P$, proving that the latter is contractible.
For a polyhedral complex $\mathscr P$ resp. a polyhedron $P\in\mathscr P$ I write $\mathscr P^b\subseteq\mathscr P$ resp. $P^b\subseteq P$ for its bounded sub-complex.
Also, by *a retract* of a polyhedron $P$ (or polyhedral complex $\mathscr P$) I mean a [strong deformation retract](https://en.wikipedia.org/wiki/Retraction_(topology)) $\phi\_P: P\times[0,1]\to P^b$ from $P$ onto $P^b\subseteq P$.
Such clearly exist if $P$ is bounded (just use the identity) or if $P$ is a half-infinite edge. But below I explain how to define such for all polyhedra, in such a way that they are compatible on their boundaries and so can be glued together to form a retract for all of $\mathscr P$. The latter is therefore homotopy equivalent to $\bigcup\_{P\in\mathscr P} P=V$, hence contractible.
*Proof.*
* Each $P\in\mathscr P$ can be written as $P=P^b+C\_P$, where $C\_P$ is the [recession cone](https://en.wikipedia.org/wiki/Recession_cone) of $P$. Fix a choice of *generalized barycentric coordinates* (e.g. mean value coordinates, Wachspress coordinates, ...), i.e. for each point $x\in P$ we have a canonical decomposition $x=x\_b+x\_C$ with $x\_b\in P^b$ and $x\_C\in C\_P$ that only depends on the shape of $P$ and the location of $x$ relative to it (see also below). Then set $\phi\_P(x,t):=x\_b+(1-t)x\_C$.
* For $P,Q\in\mathscr P$ the above definition ensures that $\phi\_P(x,t)=\phi\_Q(x,t)$ for all $x\in P\cap Q$, so that we can extend the retract to all of $\mathscr P$.
$\square$
---
**Some notes on "the canonical choice"**
As aluded to in the comments, it is sufficient to construct a section $s\_P:P\to P^b\times C\_P$ for the projection
$$\pi\_P:P^b\times C\_P\to P,(x\_b,x\_C)\mapsto x\_b+x\_C,$$
(i.e., $\pi\_P\circ s\_P=\mathrm{id}\_P$) that matches with prescribed sections $s\_\sigma:\sigma\to\sigma^b\times C\_\sigma$ on the faces $\sigma\subseteq P$. Here is one idea for how to do this.
The recession cones of polyhedra in $\mathscr P$ form a full fan $\mathscr F:=\{C\_P\mid P\in\mathscr P\}$ of $V$. For each cone $C\_P\in\mathscr F$ choose a ray $r\_P\subseteq C\_P$.
Now, for a point $x\in P$ let $x\_b'$ be the intersection of $\partial P$ with the translated and inverted ray $-r\_P+x$ (which exists since the linearity space of $C\_P$ is empty). In particular, $x\_b'$ is in some face $\sigma\subseteq \partial P$. Then set $x\_b:=s\_\sigma(x\_b')$ and $x\_C:= x-x\_b$.
| 1 | https://mathoverflow.net/users/108884 | 442236 | 178,450 |
https://mathoverflow.net/questions/442202 | 2 | Let $(P,\leq)$ be a finite poset that contains a (global) minimal element $0$ and a (global) maximal element $1$. We say that a subset $U \subset P$ is *upward closed* if $x \in U$ and $y \geq x$ forces $y \in U$. Given an arbitrary subset $A \subset P$, let $A^+$ be the smallest upward closed subset of $P$ containing $A$. We say that $P$ satisfies Property (X) if the following holds:
>
> Let $U \subset P$ be any nonempty upward closed subset whose complement $D := P - U$ is also nonempty. For every subset $A \subset D$, we have the inequality
> $$ \frac{|A|}{|D|} \leq \frac{|A^+ \cap U|}{|U|}, $$
> where $|\bullet|$ denotes cardinality.
>
>
>
Has anybody encountered this, or an equivalent property before? I don't know what to search for or where to look. There is a more specific question here:
>
> let $L$ be the poset $\{1,2,\ldots,\ell\}$ with the usual ordering, and for any integer $k > 0$ consider $P = L^k$ with the product partial order. Property X holds trivially for $k = 1$ and I have a clunky proof for $k = 2$. Does Property (X) hold for higher $k$? And if so, is there a slick proof?
>
>
>
| https://mathoverflow.net/users/18263 | Posets with cardinality bounds on upward-closed subsets | We have $A\subset A^+\cap D$ and $(A^+\cap D)^+=A^+$, so we can assume $A=A^+\cap D$, let $V=A^+$ the problem turn into $\frac{|V|-|V\cap U|}{|P|-|U|}=\frac{|V\cap(P-U)|}{|P|-|U|}\leq\frac{|V\cap U|}{|U|}\Leftrightarrow\frac{|V|}{|P|}\leq\frac{|V\cap U|}{|U|}\Leftrightarrow|V||U|\leq|P||V\cap U|$
for all upward closed subset $U,V\subset P$.
Let $U\lor V=\{u\lor v,u\in U,v\in V\},U\land V=\{u\land v,u\in U,v\in V\}$, if $U,V$ be a upward closed subset then $U\lor V=U\cap V$. If $P$ is finite distributive lacttice, like $L^k$, then by [Ahlswede-Daykin inequality](https://en.wikipedia.org/wiki/Ahlswede%E2%80%93Daykin_inequality), with four functions are identically $1$, we have: $|V||U|\leq|V\land U||V\lor U|\leq|P||V\cap U|$ as we want.
| 4 | https://mathoverflow.net/users/432274 | 442239 | 178,451 |
https://mathoverflow.net/questions/442171 | 3 | Let $G$ be a $p$-group contained in $S\_{p^m}$ for some $m\in \mathbb{N}$, this is, the symmetric group of degree $p^m$. Assume that the lower central series, LCS for short, of $G$ is well understood.
Is there anything known about the LCS of $W(G):=G\wr C\_p$, where $C\_p$ is a cyclic group of order $p$, and $\wr$ is the (right regular) permutation action of $C\_p$ on $G\times \dotsb\times G$. That is, if $(g\_1,g\_2,\dotsc, g\_p)\in G\times \dotsb\times G$, then the action of $C\_p=\langle x\rangle$ on the base group $G\times \dots\times G$ is defined by
$(g\_1,g\_2,\dotsc, g\_p)^x=(g\_p,g\_1,\dotsc,g\_{p-1})$.
I am particularly interested in the indices $|\gamma\_i(W(G)):\gamma\_{i+1}(W(G))|$ for $i \in \mathbb{N}$.
I have been searching for some bibliography or papers on the topic but I have not been able to find anything.
| https://mathoverflow.net/users/500544 | Is there anything known about the lower central series of a group $G\wr C_p$? | Here is a solution in the special case in which each lower central factor $\gamma\_i(G) / \gamma\_{i+1}(G)$ is elementary abelian.
First consider the case in which $G$ is elementary abelian, written additively, so $G \cong \mathbf F\_p^d$ for some $d$. Let $x$ be a generator of $C\_p$. Then for $g \in G^p$ we have $[g,x] = \Delta(g)$, where $\Delta = \sigma-1$ and $\sigma : G^p \to G^p$ is the linear map defined by $\sigma(g\_1, \dots, g\_p) = (g\_p, g\_1, \dots, g\_{p-1})$. Therefore
$$\gamma\_i(G \wr C\_p) = \Delta^i(G^p) \qquad (i \ge 2).$$
By the $\mathbf{F}\_p[X]$ polynomial identity
$(X-1)^{p-1} = 1 + X + \cdots + X^{p-1}$
we have
$$\Delta^{p-1} = 1 + \sigma + \cdots + \sigma^{p-1}.$$
Therefore
$$\Delta^{p-1}(g) = (s, \dots, s), \qquad \text{where}~s=g\_1 + \cdots + g\_p.$$
Therefore $\gamma\_{p-1}(G \wr C\_p) = \Delta^{p-1}(G^p)$ is just the diagonal copy of $G$ in $G^p$, which is also the kernel of $\Delta$. It follows that the lower central factors of $G \wr C\_p$ are $G \times C\_p, G, \dots, G$.
Now consider the general case. Let $G = \Gamma\_1 \ge \cdots \ge \Gamma\_k = 1$ be the lower central series of $G$. Then $C\_p$ acts on each factor $\Gamma\_i^p / \Gamma\_{i+1}^p$, and the lower central series of $\Gamma\_i / \Gamma\_{i+1} \wr C\_p$ is as described above. By lifting to $G$ we get a long central series:
$$G \wr C\_p = \Gamma\_1^p C\_p \ge \Delta(\Gamma\_1^p) \Gamma\_2^p \ge \Delta^2(\Gamma\_1^p) \Gamma\_2^p \ge \cdots \ge \Delta^{p-1}(\Gamma\_1^p) \Gamma\_2^p \ge \Gamma\_2^p \ge \Delta(\Gamma\_2^p) \Gamma\_3^p \ge \cdots.$$
I claim this is in fact the lower central series. To see this it suffices to observe that $\Delta^i(\Gamma\_j^p)\Gamma\_{j+1}^p$ (for $i \le p-1$) contains the preimage $D$ in $\Gamma\_j^p$ of the diagonal subgroup of $\Gamma\_j^p/\Gamma\_{j+1}^p$, and $[D, G^p] = \Gamma\_{j+1}^p$.
In general if $\gamma\_i(G)/\gamma\_{i+1}(G)$ is not elementary abelian it seems more complicated. For example $C\_{p^2} \wr C\_p$ seems to have lower central factors $C\_{p^2} \times C\_p, C\_p, C\_p, \dots, C\_p$.
| 2 | https://mathoverflow.net/users/20598 | 442240 | 178,452 |
https://mathoverflow.net/questions/441993 | 9 |
>
> Question 1:Is there a reference that lists all possible finite subgroups and their orders of $\mathrm{SO}(n)$ and $\mathrm{O}(n)$ for $n=4$ or even higher $n$ over the real numbers?
>
>
>
I can only find incomplete lists on wikipedia, mathoverflow or other sources in the internet.
>
> Question 2: Is there a computer algebra package that can list those finite subgroups up to some order? Maybe a package of GAP?
>
>
>
| https://mathoverflow.net/users/61949 | Finite subgroups of $\mathrm{SO}(n)$ and $\mathrm{O}(n)$ | Finite subgroups of $\operatorname{SO}(4)$ and $\operatorname{O}(4)$ are enumerated in Chapter 4 of the book:
*Conway, John H.; Smith, Derek A.*, **On quaternions and octonions: their geometry, arithmetic, and symmetry**, Natick, MA: A K Peters (ISBN 1-56881-134-9/hbk). xii, 159 p. (2003). [ZBL1098.17001](https://zbmath.org/?q=an:1098.17001).
There may be results for some $n>4$, but since every finite group has faithful orthogonal representations, this gets hopeless at some point.
| 10 | https://mathoverflow.net/users/10266 | 442248 | 178,455 |
https://mathoverflow.net/questions/442243 | 0 | Currently, I'm reading the paper "Towards the fixed point property for superreflexive spaces" by Andrzej Wiśnicki. In this article, given $X\_1,\dots,X\_n$ Banach spaces, he defines $(X\_1\oplus\dots\oplus X\_n)\_{\infty}$ as the product $X\_1\times\dots\times X\_n$ with the norm
\begin{equation\*}
\lVert (x\_1,\dots,x\_n)\rVert\_{\infty}=\max(\lVert x\_1\rVert\,\dots,\lVert x\_n\rVert)
\end{equation\*}
With this, in the next proposition, he uses the fact that, given a free ultrafilter on $\mathbb{N}$, $\mathcal{U}$, the following relation is satisfied:
$((X\_1\oplus\dots\oplus X\_n)\_{\infty}) \_{\mathcal{U}}=((X\_1) \_{\mathcal{U}}\oplus\dots\oplus (X\_n) \_{\mathcal{U}}) \_{\infty}$,
where $(X)\_{\mathcal{U}}$ denotes the ultrapower of $X$ respect to $\mathcal{U}$. Why this equality holds?
In addition, if you have access to the paper, I would like to ask one more question. In this same proposition, after using the first fact I asked about, they say that it is sufficient to prove that $(X\_1\oplus\dots\oplus X\_n)\_{\infty}$ has property $(S)$. Why? They have to prove it for $((X\_1) \_{\mathcal{U}}\oplus\dots\oplus (X\_n) \_{\mathcal{U}}) \_{\infty}$ in case they want to use the relation above, don't they?
Thank you very much!!
| https://mathoverflow.net/users/490148 | The ultrapower of the direct sum is the direct sum of ultrapowers | I'll attempt to record a careful proof here for your first question. My apologies for any errors or inefficiencies, as I am a bit rusty. I will write $\|\,\|\_j$ for the norm on $X\_j$ and $\|\,\|$ for the max-norm on $(X\_1\oplus\cdots\oplus X\_n)\_\infty$.
Writing $\Pi\_0$ for the space of bounded sequences in $(X\_1\oplus\cdots\oplus X\_n)\_\infty$, we have two seminorms on $\Pi\_0$: for $x\in\Pi\_0$, define
$$\|x\|\_{1,s}=\lim\_{i\to\mathcal{U}}\|x\_i\|$$
and
$$\|x\|\_{2,s}=\max\_{1\leq j\leq n}(\lim\_{i\to\mathcal{U}}\|x\_i^j\|\_j)$$
We have used subscripts to denote entries along the sequence direction, superscripts to denote the summand direction; e.g. $x\_i^j$ is the $i$th element of the sequence in $X\_j$. Observe that $\|\,\|\_{1,s}$ and $\|\,\|\_{2,s}$ essentially differ by commuting a $\lim\_{i\to\mathcal{U}}$ with a $\max\_{1\leq j\leq n}$. Thus we may write $\|\,\|\_{1,s}$ as
$$\|x\|\_{1,s}=\lim\_{i\to\mathcal{U}}(\max\_{1\leq j\leq n}\|x\_i^j\|\_j)$$
But this is just $\|x\|\_{2,s}$; indeed, if we write $c^j=\lim\_{i\to\mathcal{U}}\|x\_i^j\|\_j$, then for each $\epsilon>0$ and each $j$ we have
$$\{i:|\|x\_i^j\|\_j-c^j|<\epsilon\}\in\mathcal{U}$$
so by taking finite intersections
$$\{i:|\|x\_i^j\|\_j-c^j|<\epsilon\,\forall j\}\in\mathcal{U}$$
which in particular implies
$$\{i:\max\_{1\leq j\leq n}c^j-\epsilon\leq\max\_{1\leq j\leq n}\|x\_i^j\|\_j\leq\max\_{1\leq j\leq n}c^j+\epsilon\}\in\mathcal{U}$$
so (by arbitrariness of $\epsilon>0$) we have $$\lim\_{i\to\mathcal{U}}\max\_{1\leq j\leq n}\|x\_i^j\|=\max\_{1\leq j\leq n}c^j=\max\_{1\leq j\leq n}\lim\_{i\to\mathcal{U}}\|x\_i^j\|\_j$$
and $\|\,\|\_{1,s}=\|\,\|\_{2,s}$ on $\Pi\_0$. Finally, observe that $((X\_1\oplus\cdots\oplus X\_n)\_\infty)\_{\mathcal{U}}$ is the quotient of $((X\_1\oplus\cdots\oplus X\_n)\_\infty,\|\,\|\_{1,s})$ by the nullset $\|\,\|\_{1,s}=0$, and $((X\_1)\_{\mathcal{U}}\oplus\cdots\oplus(X\_n)\_{\mathcal{U}})\_\infty$ is the quotient of $((X\_1\oplus\cdots\oplus X\_n)\_\infty,\|\,\|\_{2,s})$ by the nullset $\|\,\|\_{2,s}=0$; the result follows.
For your second question: I'm not sure, but I suspect that the author intends to say that it suffices to show that it suffices to show that $(Y\_1\oplus\cdots\oplus Y\_n)\_\infty$ has property $(S)$, where $Y\_j=(X\_j)\_{\mathcal{U}}$ and the author is swapping out notation; I suppose this works as long as the condition $\varepsilon\_0(X\_i)<1$ is stable under ultrapowers. Indeed, it appears that the author makes exactly that claim midway through page 439: "On the other hand, $\varepsilon\_0(X)=\varepsilon\_0((X)\_{\mathcal{U}})$..."
| 3 | https://mathoverflow.net/users/349327 | 442258 | 178,456 |
https://mathoverflow.net/questions/442250 | 4 | Suppose that $f \colon [0, 1] \to \mathbb{R}$ is a $1$-Lipschitz function.
Define the uniform norm $\|f\|\_\infty = \sup\_{x} |f(x)|$.
Given $\{U\_j\}\_{j=1}^\infty$ independent and identically distributed uniform random variables on the unit interval $[0, 1]$, I am curious to know about the behavior of the random variable
$$
\Delta\_n(f) := \Big|\max\_{j \leq n} |f(U\_j)| - \|f\|\_\infty\Big|
$$
I would roughly expect that $\Delta\_n \sim \frac{1}{n}$, with high probability simply because on average one of the $n$ samples $U\_1, \dots, U\_n$ should lie within $1/n$ of the maximum, and therefore, the error should be of this order. I am also interested to what degree this can be made uniform over Lipschitz functions $f$, i.e., by studying the maximal quantity $\sup\_{f \in \mathcal{F}} \Delta\_n(f)$ where $\mathcal{F}$ is a suitable family of Lipschitz functions (say with $f(0) = 0$). Is this question studied? If so, are there any references available?
| https://mathoverflow.net/users/121486 | Rate of convergence of sample maximum, $\Big|\max_{j \leq n} |f(U_j)| - \|f\|_\infty\Big|$ | $\newcommand\De\Delta\newcommand{\Ga}{\Gamma}\newcommand\ga{\gamma}$Let $X\_j:=g(U\_j)$, where $g:=|f|$. Then the $X\_i$'s are i.i.d. random variables and $g$ is a $1$-Lipschitz function. We have
$$M:=\max\_{[0,1]}g=g(u)$$
for some $u\in[0,1]$.
For any real $h>0$,
\begin{align\*}
P(\De\_n(f)>h)&=P(M-\max\_1^n X\_j>h) \\
&=P(\max\_1^n X\_j<M-h) \\
&=P(X\_1<M-h)^n \\
&=P(g(U\_1)<g(u)-h)^n \\
&=P(g(u)-g(U\_1)>h)^n \\
&\le P(|U\_1-u|>h)^n \tag{1}\label{1} \\
&\le(1-h)^n\le e^{-nh};
\end{align\*}
the $\le$ inequality in \eqref{1} holds because $g$ is $1$-Lipschitz.
So, $\De\_n(f)$ is stochastically dominated by an exponentially distributed random variable with mean $1/n$, which confirms your expectation. In particular, $E\De\_n(f)\le1/n$.
---
Consider now
\begin{equation\*}
\De\_n(F):=\sup\_{f\in F}\De\_n(f),
\end{equation\*}
where $F$ is the set of all $1$-Lipschitz functions on $[0,1]$. Let us show that
\begin{equation\*}
E\De\_n(F)\sim\frac{\ln n}{2n} \tag{2}\label{2}
\end{equation\*}
and
\begin{equation\*}
\De\_n(F)\Big/\frac{\ln n}{2n}\to1 \tag{3}\label{3}
\end{equation\*}
in probability (as $n\to\infty$).
Indeed, it is easy to see that
\begin{equation\*}
\De\_n(F)=\frac12\,\max(2G\_1,G\_2,\dots,G\_n,2G\_{n+1})
=\frac{M\_n}2+L\_n, \tag{4}\label{4}
\end{equation\*}
where
\begin{equation\*}
M\_n:=\max(G\_1,G\_2,\dots,G\_n,G\_{n+1}),
\end{equation\*}
\begin{equation\*}
0\le L\_n\le\frac12\,\max(G\_1,G\_{n+1}), \tag{5}\label{5}
\end{equation\*}
\begin{equation\*}
G\_i:=U\_{n:i}-U\_{n:i-1}
\end{equation\*}
for $i=1,\dots,n+1$; $U\_{n:1}\le\dots\le U\_{n:n}$ are the order statistics for the $U\_j$'s; $U\_{n:0}:=0$; and $U\_{n:n+1}:=1$.
A result by Fisher 1929 (see e.g. [formula (1.7)](https://arxiv.org/abs/1909.06406v1)) is that for $x\in[0,1]$ we have
\begin{equation\*}\label{eq:fisher}
P(M\_n>x)=\sum\_{j=1}^{n+1}(-1)^{j-1}\binom{n+1}j(1-jx)\_+^n,
\end{equation\*}
where $u\_+:=\max(0,u)$. So,
\begin{equation\*}
EM\_n=\int\_0^1dx\,P(M\_n>x)
=\sum\_{j=1}^{n+1}(-1)^{j-1}\binom{n+1}j\int\_0^1dx\,(1-jx)\_+^n
=\frac{\psi(n+2)+\ga}{n+1}\sim\frac{\ln n}n, \tag{6}\label{6}
\end{equation\*}
where $\psi:=\Ga'/\Ga$ and $\ga=0.577\dots$ is Euler's gamma; these results for $EM\_n$ were also given at the end of Section 1 of the linked paper.
Writing $EM\_n^2=\int\_0^1dx\,2x\,P(M\_n>x)$, we similarly get
\begin{equation\*}
Var\,M\_n\sim\frac{\pi^2}{6n^2}.
\end{equation\*}
So, by \eqref{6} and Chebyshev's inequality,
\begin{equation\*}
M\_n\Big/\frac{\ln n}n\to1 \tag{7}\label{7}
\end{equation\*}
in probability.
In view of \eqref{5}, it is easy to see that $EL\_n=O(1/n)$ and hence $L\_n/\frac{\ln n}n\to0$ in probability.
Now \eqref{2} and \eqref{3} follow from \eqref{4}, \eqref{6}, and \eqref{7}. $\quad\Box$
---
From the first equality in \eqref{4} and formula (2.2) of the linked paper, one can get an exact expression for the cdf of $\De\_n(F)$: for all real $x\ge0$,
\begin{equation\*}
P(2\De\_n(F)\le x)\\
=\sum\_{k=0}^{n-1}(-1)^k\binom{n-1}k
[(1-kx)\_+^n-2(1-(k+\tfrac12)x)\_+^n+(1-(k+1)x)\_+^n].
\end{equation\*}
This yields an exact expression for $E\De\_n(F)$:
\begin{equation\*}
E\De\_n(F)=\frac1{2(n+1)}\Big(\psi(n)+\ga+\frac{2\sqrt{\pi}\,\Ga (n)}{\Ga(n+1/2)} -\frac1n\Big),
\end{equation\*}
which, in turn, again yields \eqref{2}.
| 4 | https://mathoverflow.net/users/36721 | 442259 | 178,457 |
https://mathoverflow.net/questions/442262 | -1 | Is it possible to build a 1-priodic smooth function from a rapidly decreasing sequence such that the sequence be the Fourier coefficients of the function?
More precisely:
Let $\lbrace c\_k\rbrace \_{k \in \mathbb{Z}}$ be rapidly decreasing sequence. Is it possible to show that there is a 1-priodic smooth function $F: \mathbb{R}\rightarrow \mathbb{C}$ such that $\lbrace c\_k\rbrace$ is the sequence of the Fourier coefficients of $f = F|\_{[0,1]}$
| https://mathoverflow.net/users/137242 | Building a smooth function from a rapidly decreasing sequence | **Yes**. (Assuming that that **rapidly decreasing** means that $\sup\_{k\in \mathbb Z}\vert k^m c\_k \vert <\infty$ for all $m=0,1,2,\dots$.)
The function must be $f(t)=\sum\_{k\in \mathbb Z} c\_k \exp(2\pi i k t)$.
Just write $f\_N(t)=\sum\_{\vert k \vert \le N} c\_k \exp(2\pi i k t)$. As $\vert c\_k\vert = O(k^{-2})$, this the sequence $(f\_N)$ is Cauchy in $L^2([0,1])$ and thus $f=\lim\_{N\rightarrow \infty}f\_N$ exists in $L^2([0,1])
$. Clearly $f$ is $1$-periodic and has the correct Fourier coefficients. I claim that it is also smooth.
Taking derivatives of $f\_N$ produces powers of $k$, but they do not affect summability, as $c\_k$ is rapidly decreasing. Hence the $k$th derivative $f\_N^{(k)}$ is also Cauchy in $L^2$. Thus $(f\_N)$ is Cauchy in the Sobolev space $H^k([0,1])$. This implies that $f\in H^k$ for every $k\in \mathbb N$ and hence it must be smooth.
| 1 | https://mathoverflow.net/users/126651 | 442264 | 178,459 |
https://mathoverflow.net/questions/441742 | 1 | $\newcommand{\EE}{\mathbb{E}}$
Let $A\in\mathbb{R}^{m\times n}$ and $X$ be an isotropic random vector in $\mathbb{R}^n$, i.e. it holds that $\EE(XX^T) = I\_n$.
How to calculate $$M = \EE\left(\tfrac{XX^T}{\|AX\|^2}\right)$$
and/or $$AMA^T = \EE\left(\tfrac{AXX^TA^T}{\|AX\|^2}\right) = \EE\left(\tfrac{AXX^TA^T}{X^TA^TAX}\right)?$$
For $X$ being a (scaled) random unit coordinate vector $\sqrt{n}e\_k$ where $k$ is uniformly distributed on $\{1,\dots,n\}$ that's simple to do, but would be interested in other isotropic distributions, e.g.
* normally distributed $X$,
* Rademacher vectors $X$ ($X\_i = \pm 1$ with equal probability and independently)
* $X$ uniformly
distributed on the sphere of radius $\sqrt{n}$.
I would also be happy with (bounds on) the smallest and largest eigenvalues of $M$ and $AMA^T$.
| https://mathoverflow.net/users/9652 | Calculating $\mathbb{E}\left(\tfrac{XX^T}{\|AX\|^2}\right)$ for isotropic random vectors $X$ | Assuming $X\sim N(0,I)$, writing $A^TA=\sum\_i L\_i u\_i u\_i^T$, since $sign(u\_i^TX)$ and $sign(u\_k^TX)$ are independent and mean-zero, $u\_i^TMu\_k=0$ if $i\le k$ because the denominator
$\|AX\|^2=\sum\_i L\_i Z\_i^2$ is independent of the signs, where $Z\_i=u\_i^TX$ are iid $N(0,1)$.
This shows that $M$ is diagonalizable in the same basis as $A^TA$. It remains to compute the diagonal terms in the basis $(u\_i)$, and
$$
u\_i^TMu\_i = E\Big[ \frac{Z\_i^2}{\sum\_k L\_k Z\_k^2}\Big]
$$
where $Z\_i=u\_i^TX$ are iid $N(0,1)$. Unfortunately, I do not think explicit expressions are available for general $L\_k$ for the expectation in the right-hand side.
| 1 | https://mathoverflow.net/users/141760 | 442268 | 178,460 |
https://mathoverflow.net/questions/442271 | 8 | I think this is a silly question, but I'm quite confused. In Lurie's "Rotation Invariance In Algebraic K-Theory" Notation 3.2.4. he defines a filtered spectrum $\mathbb{A}$ given by
$$\mathbb{A}\_n = \begin{cases}
\mathbb{S} &\text{ if } n =0 \\
0 &\text{ otherwise}\end{cases} $$
He then notes (example 3.2.9) that $$\mathbb A \otimes \mathbb A$$ takes the form
$$
\cdots \to 0 \to 0 \to \mathbb S \to \Sigma \mathbb S \to 0 \to \cdots
$$
It seems to me like this should (as a filtered spectrum) be the same as $\mathbb A + \Sigma \mathbb A(1)$, since there are no nontrivial maps $\mathbb S \to \Sigma \mathbb S$. But later (Construction 3.4.4) he writes a fiber sequence
$$
\mathbb A \otimes \mathbb A \to \mathbb A \to\_\beta \Sigma^2 \mathbb A (1)
$$
and gives $\beta$ a name ("the anchor map"). But $\beta$ corresponds to a square
$\require{AMScd}$
\begin{CD}
\mathbb{S} @>>> 0\\
@VVV @VVV \\
0 @>>> \Sigma^2\mathbb S
\end{CD}
and therefore a map of spectra $\mathbb S \to \Sigma \mathbb S$, and therefore (to my eyes) should equal zero.
What am I missing here? If this map is zero, why does it get a name? And if it's nonzero, why doesn't this produce a nontrivial element of $\pi\_{-1}\mathbb{S}$?
| https://mathoverflow.net/users/131360 | Why isn't the anchor map in Lurie's "Rotation Invariance in Algebraic K-Theory" zero? | It's zero as a map of filtered spectra, but it is considered as a map of $\mathbb{A}$-bimodules.
| 12 | https://mathoverflow.net/users/39747 | 442273 | 178,461 |
https://mathoverflow.net/questions/441651 | 3 | Let $\Omega$ be an open subset of $\mathbb{C}^n$ and let $f$ be analytic in $\Omega$. Assume $P\in\mathbb{C}[z\_1,\ldots,z\_n]$ is a polynomial whose irreducible factors are all of multiplicity one.
If $f$ vanishes on the zero set $Z(P)$ of $P$, how to justify that $f/P$ is analytic in $\Omega$ ?
(asked on [MSE](https://math.stackexchange.com/questions/3164508), but didn't get a convincing answer)
| https://mathoverflow.net/users/89429 | Factorization of an analytic function in $\mathbb{C}^n$ | *This is a copy of [my answer](https://math.stackexchange.com/a/4654372/312406) from math.SE*
I think you could argue stalk wise. Take a point $x \in \Omega$. By general theory, the ring $\mathcal O\_{\mathbb C^n, x}$ of convergent power series at $x$ is a UFD. Since $P$ is square-free, and $f$ vanishes at $\{P =0\}$, $P$ is a factor of $f$ in $\mathcal O\_{\mathbb C^n,x}$ by the analytic Nullstellensatz. So there is a quotient $\overline{g\_x} = f/P$ in $\mathcal O\_{\mathbb C^n,x}$. This is represented by an actual holomorphic function $g\_x \in \mathcal O(U\_x)$ in some neighbourhood $U\_x$ of $x$, and $g\_x \cdot P = f$ holds on all of $U\_x$. By the identity theorem, the $g\_x$ glue to a holomorphic function $g$ on $\Omega$.
| 1 | https://mathoverflow.net/users/111897 | 442278 | 178,463 |
https://mathoverflow.net/questions/442253 | 4 | Let $\mathcal{A}$ be an abelian category, it is known that any complex $A^{\bullet}$ admits a distinguished triangle
$$B^{\bullet}\rightarrow A^{\bullet}\rightarrow C^{\bullet}\rightarrow B^{\bullet}[1]$$
in the unbounded derived category $D(\mathcal{A})$ for some $B^{\bullet}\in D^-(\mathcal{A})$ and $C^{\bullet}\in D^+(\mathcal{A})$.
In other words, any complex can be decomposed into a complex bounded above and a complex bounded below.
Can we do the same thing in the unbounded homotopy category $K(\mathcal{A})$?
| https://mathoverflow.net/users/nan | Decompose an unbounded (cochain) complex in the homotopy category | Yes. Let
$$\tau^{\leq0}A^\bullet:= \cdots\to A^{-2}\to A^{-1}\to\ker(d^0)\to0\to\cdots$$
be the usual truncation of $A^\bullet$. Then the mapping cone of the inclusion map $\tau^{\leq0}A^\bullet\to A^\bullet$ is homotopy equivalent to
$$\rho^{>0}A^\bullet:=\cdots\to0\to\ker(d^0)\to A^0\to A^1\to\cdots,$$
so there is a distinguished triangle
$$\tau^{\leq0}A^\bullet\to A^\bullet\to\rho^{>0}A^\bullet\to(\tau^{\leq0}A^\bullet)[1]$$
in $K(\mathcal{A})$.
Or another way to see this triangle: If the contractible complex
$$K^\bullet:=\cdots\to0\to\ker(d^0)\xrightarrow{\sim}\ker(d^0)\to0\to\cdots$$
is concentrated in degrees $-1$ and $0$, then there is a degreewise split short exact sequence
$$0\to\tau^{\leq0}A^\bullet\to A^\bullet\oplus K^\bullet\to\rho^{>0}A^\bullet\to0.$$
| 6 | https://mathoverflow.net/users/22989 | 442279 | 178,464 |
https://mathoverflow.net/questions/442277 | 3 | The following theorem is well-known in the ordinary analysis textbook:
**Theorem**: Assume the function $f:U\to\Bbb R^n$ is Lipschitz continuous on an open set $U\subset\Bbb R^m$, then $f$ is almost everywhere differentiable on $U$.
My question:
**Question**: Assume the function $f:U\to\Bbb{R}^n$ is Lipschitz continuous on an open set $U\subset\Bbb R^m$. Prove or disprove that $f$ is almost everywhere $C^1$ on $U$.
Basically, this means that the point where $f$ is not differentiable would be a zero-measure closed set and on the open set where $f$ is differentiable, the gradient should be continuous. Is there any counterexample?
---
Thank for all the comments. This problem is solved now. The key point is that there exists a function $f$ which is differentiable everywhere with bounded derivative but the set consisting of discontinuous point of $f'$ can have positive measure. See [Differentiable function with discontinuous derivative](https://math.stackexchange.com/questions/2664789/characterizing-discontinuous-derivatives)
| https://mathoverflow.net/users/113353 | Is there any strengthened version of Rademacher's Theorem or any counterexample? | The best thing you can do is the following: for every $\epsilon > 0$ there exists a $C^1$ function $g: U \to \mathbf{R}^n$ so that
\begin{equation}
\mathcal{H}^m(\{ f \neq g \} \cup \{ Df \neq Dg \} ) < \epsilon.
\end{equation}
This can be deduced from the Whitney extension theorem; you can find a proof in Leon Simon's lecture notes on GMT (Theorem 5.3, pp. 32-33).
*Edit.* It looks like I misread your question, sorry about that. Here is how to adapt the counterexample.
Let $C \subset [0,1]$ be a fat Cantor set. This is closed, has empty interior, and measure $\lambda := \mathcal{H}^1(C) \in (0,1)$. Define $f: x \mapsto \int\_0^x \mathbf{1}\_C$. This is $1$-Lipschitz and has $f(1) = \lambda$; moreover $f' = 0$ on the open complement $C^c$.
Now, if there were an open set $U \subset [0,1]$ of full measure, and along which $f$ is $C^1$, then $f' = 0$ on $U$, because $C$ has empty interior. But then on the one hand $\int\_{[0,1]} f' = \int\_{[0,1] \cap U} f' = 0$, and on the other hand $\int\_{[0,1]} f' = f(1) - f(0) > 0$; this is absurd.
| 7 | https://mathoverflow.net/users/103792 | 442281 | 178,465 |
https://mathoverflow.net/questions/442270 | 9 | nlab presents a [proof](https://ncatlab.org/nlab/show/locally+compact+topological+space#categorytheoretic_properties) that the category of locally compact Hausdorff spaces does not admit infinite products in general. In particular it shows that there is no infinite product of $\mathbb{R}$, since such a product would be a topological vector space, by the universal property of the projection maps, and it would also be a product in $\mathbb{R}\text{Vect}$. It is known that every locally compact Hausdorff topological vector space is finite dimensional, see [here](https://terrytao.wordpress.com/2011/05/24/locally-compact-topological-vector-spaces/) for example.
However since linear maps in $\mathbb{R}\text{Vect}$ are not in general proper, such a proof doesn't work in the case that you restrict to proper continuous maps between locally compact Hausdorff spaces.
I wanted to know if there is still a known counterexample that means that this category will not have products?
One reason that suggests to me they might is that topological spaces with local homeomorphisms (étale maps) has products, and they are very weird; for example $\mathbb{R}\times\mathbb{R}^2=∅$ in that category. It also massively restricts what maps can be onto compact spaces (they must come from a compact space, which is another 'intuition' for why it might have products.
| https://mathoverflow.net/users/163483 | Does the category of locally compact Hausdorff spaces with proper maps have products? | Suppose $X,Y$ are locally compact Hausdorff spaces admitting a product $X\otimes Y$ in the category $\mathcal{H}$ of all such spaces. If one of $X,Y$ is empty, then $X\otimes Y$ exists, so I'll assume that neither is.
Then the points of $X\otimes Y$ are exactly the maps $\ast\rightarrow X\otimes Y$. Since $X\otimes Y$ is Hausdorff, any such map is proper. Because of this, both of the categorical projections
$$X\xleftarrow{\pi\_X} X\otimes Y\xrightarrow{\pi\_Y}Y$$
are surjective. They are continuous and proper by assumption.
Working now in $Top$, the category of all spaces, we have the Tychonoff product $X\times Y$ and an induced map
$$\theta:X\otimes Y\rightarrow X\times Y$$
factoring the projections. By the above observation, $\theta$ is surjective.
>
> **Lemma**: Suppose that $f:A\rightarrow B$ and $g:B\rightarrow C$ are maps of spaces $A,B,C$. If $f$ is surjective and $g\circ f$ is proper, then $g$ is proper. $\quad\blacksquare$
>
>
>
Apply the lemma to the composite
$$pr\_X\circ \theta=\pi\_X$$
to conclude that $pr\_X$ is proper. This is only possible if $Y$ is compact. Similarly, $X$ must also be compact.
Of course, if $X,Y$ are compact, then the product $X\otimes Y$ exists in $\mathcal{H}$ and is given by the Tychonoff product $X\times Y$.
>
> If $X,Y$ are locally compact Hausdorff and nonempty, then the product $X\otimes Y$ exists in $\mathcal{H}$ if and only if both $X,Y$ are compact.
>
>
>
| 11 | https://mathoverflow.net/users/54788 | 442282 | 178,466 |
https://mathoverflow.net/questions/442111 | 11 | Consider the algebra $B=P(\omega)/\_{\mathrm{fin}}$ (the quotient of the power set of natural numbers modulo the ideal of finite sets). Is there an infinite strictly descending chain $\{A\_i\mid i\in I\}$ of subalgebras of $B$, such that there is an embedding of $A\_{i+1}$ into $A\_i$, but there is no embedding of $A\_i$ into $A\_{i+1}$.
2nd EDIT: Klaas Pieter Hart expanded his answer into a paper “Many subalgebras of $P(\omega)/\_{\mathrm{fin}}$” that can be found at [arXiv](https://arxiv.org/abs/2303.08491).
1st EDIT: Corrected question according to the remark in the comments.
| https://mathoverflow.net/users/22019 | A strictly descending chain of subalgebras of $P(\omega)/_{\mathrm{fin}}$ | There is a family $\{K\_X:X\subseteq\mathfrak{c}\}$ of separable compact
zero-dimensional spaces such that there is a continuous surjection of $K\_X$
onto $K\_Y$ if and only if $X\subseteq Y$.
These spaces are continuous images of $\omega^\*$, the Stone space of
$\mathcal{P}(\omega)/\mathit{fin}$.
So Stone duality then yields a family
$\{B\_X:X\subseteq\mathfrak{c}\}$ of subalgebras of
$\mathcal{P}(\omega)/\mathit{fin}$ such that $B\_Y$ embeds into $B\_X$ if and
only if $X\subseteq Y$. Thus one can even get chains of length $\mathfrak{c}$ or a chain of the same order type as the real line.
We can make such spaces using variations on Alexandroff's double-arrow space,
which is the set
$D=([0,1]\times\{0,1\})\setminus\{\langle0,0\rangle,\langle0,1\rangle\}$
ordered lexicographically and endowed with the order topology;
The resulting space will be denoted $\mathbb{A}$.
(We drop the points $\langle0,0\rangle$ and $\langle0,1\rangle$ because they
would be (the only) isolated points of $\mathbb{A}$.)
Pictorially we have taken the unit interval $[0,1]$ and split all points
$x$ of the open interval $(0,1)$ into two copies:
The space $\mathbb{A}$ is compact and separable, hence a continuous image
of $\omega^\*$.
The variations will be obtained by specifying a subset $X$ of $(0,1)$ and taking
$\mathbb{A}\_X=\{\langle x,i\rangle\in D: x\in X \to i=0\}$; that is,
by splitting the points of $(0,1)\setminus X$ only.
Thus we can write $\mathbb{A}=\mathbb{A}\_\emptyset$,
and $[0,1]=\mathbb{A}\_{(0,1)}$.
One direction in the if and only if will be immediate by the following
observation: if $X\subseteq Y$ then there is a natural continuous map $h\_{X,Y}$ from
$\mathbb{A}\_X$ onto $\mathbb{A}\_Y$; it sends
the points $\langle x,i\rangle$ to $\langle x,i\rangle$ if $x\notin Y$;
the points $\langle x,i\rangle$ to $\langle x,0\rangle$ if $x\in Y\setminus X$;
and $\langle x,0\rangle$ to $\langle x,0\rangle$ if $x\in X$.
We shall construct a family $\{A\_\alpha:\alpha\in\mathfrak{c}\}$ of subsets
of $(0,1)$ (all disjoint from $\mathbb{Q}$)
and put
$S\_X=\mathbb{Q}\cup\bigcup\_{\alpha\in X}A\_\alpha$ for $X\subseteq\mathfrak{c}$.
Then $K\_X$ will be $\mathbb{A}\_{S\_X}$.
It is readily seen that $h\_{S\_X,S\_Z}=h\_{S\_Y,S\_Z}\circ h\_{S\_X,S\_Y}$
whenever $X\subseteq Y\subseteq Z$, so the maps dual to the surjections
$h\_{S\_\emptyset,S\_X}$ yields an embedding of $\mathcal{P}(\mathfrak{c})$
into the family of subalgebras of the clopen algebra of $K\_\emptyset$
ordered by the subset relation and so also into the family of
subalgebras of $\mathcal{P}(\omega)/\mathit{fin}$.
To this end we let $\mathcal{F}$ be the set of all maps $f$ that satisfy:
$\operatorname{dom}f$ is a co-countable subset of $[0,1]$ and
$f:\operatorname{dom}f\to[0,1]$ is continuous.
For every $f\in\mathcal{F}$ we let
$S(f)=\{x\in\operatorname{dom}f:f(x)\neq x\}$ and
$E(f)=\operatorname{dom}f\setminus S(f)$.
We choose a subset $C(f)$ of $\operatorname{dom}f$ such that the
restriction $f:C(f)\to f[S(f)]$ is a bijection.
We apply Theorem 2.0 of [*A method for constructing ordered continua*,
Topology and its Applications, 21 (1985), 35--49](http://dx.doi.org/10.1016/0166-8641(85)90056-2)
to this family and obtain a pairwise disjoint family
$\{V\}\cup\{A\_\alpha:\alpha\in\mathfrak{c}\}$
of Bernstein sets in $(0,1)$ with the following properties:
all are disjoint from $\mathbb{Q}$ and
for every $f\in\mathcal{F}$ such that $f[S(f)]$ (and hence $C(f)$) has
cardinality $\mathfrak{c}$ the intersections
$C(f)\cap A\_\alpha$ and $f[C(f)\cap A\_\alpha]\cap V$ have
cardinality $\mathfrak{c}$, for all $\alpha$.
For later use we note that for a given function $f$ in $\mathcal{F}$ the
sets $E(f)$ and $S(f)$ are closed and open in $\operatorname{dom}f$ respectively and
hence completely metrizable, and the set $f[Sf)]$ is analytic.
Therefore for each of the three sets we know that either it is countable
or it contains a copy of the Cantor set.
Thus, to show that one is countable it suffices to show that its intersection
with some Bernstein set is countable.
The following claim establishes the other direction in our if and only if.
Let $\alpha\in\mathfrak{c}$ and let $X$ and $Y$ be subsets of $(0,1)$
such that $A\_\alpha\subseteq X$ and $(A\_\alpha\cup V)\cap Y=\emptyset$.
We claim that there is no continuous surjection from $\mathbb{A}\_X$
onto $\mathbb{A}\_Y$.
To why this claim suffices assume that $U$ and $V$ are subsets
of $\mathfrak{c}$ and that $U\not\subseteq V$.
Take $\alpha\in U\setminus V$ and note that $A\_\alpha\subseteq S\_U$
and that $(A\_\alpha\cup V)\cap S\_V=\emptyset$, so that
there is no continuous surjection from $\mathbb{A}\_{S\_U}$
onto $\mathbb{A}\_{S\_V}$.
For brevity we write $A=A\_\alpha$.
We first observe that the points of $A$ are not split in $\mathbb{A}\_X$,
so its subspace topology is the same as its subspace topology in $[0,1]$.
Now let $s:A\_X\to A\_Y$ be continuous and let $t:A\_Y\to[0,1]$ be the natural
map that identifies $\langle y,0\rangle$ and $\langle y,1\rangle$ for all
$y$ outside $Y$.
The composition $t\circ s$ is continuous and we let $g$ denote its restriction
to $A$.
Since $A$ is dense in $\mathbb{A}\_X$ we can recover $t\circ s$ from $g$.
If $x\in X$ then $(t\circ s)(x,0)=\lim\_{y\to x}g(y)$, and
if $x\notin X$ then $(t\circ s)(x,0)=\lim\_{y\uparrow x}g(y)$ and
$(t\circ s)(x,0)=\lim\_{y\downarrow x}g(y)$, where in each case $y$ runs
through $A$.
By one half of Lavrentieff's theorem
(Theorem 4.3.20 in Engelking's *General Topology*)
we can find a $G\_\delta$-set $G$ that contains~$A$ and a continuous
map $f:G\to[0,1]$ that extends $g$.
The complement of $G$ in $[0,1]$ is a countable union of closed sets,
each of which is countable because they are closed and disjoint from
the Bernstein set $A$.
It follows that $f$ belongs to the family $\mathcal{F}$.
The arguments given above yield information about the locations of
$s(x,0)$ and $s(x,1)$ whenever $x\in G$.
Indeed, if $x\in G\setminus X$ then $(t\circ s)(x,0)=f(x)=(t\circ s)(x,1)$
and so in all cases $s(x,i)\in\{\langle f(x),0\rangle,\langle f(x),1\rangle\}$.
To summarize: divide $\mathbb{A}\_X$ into two sets $P$, the points with
first coordinate in $G$, and $Q$, the points with first coordinate outside $G$.
The set $Q$ is countable and once we show that $f[G]$ is countable we can
conclude that $s[\mathbb{A}\_X]$ is countable and that $s$ is definitely not
surjective.
Our first claim is that $E(f)$ is countable.
We show this by showing that its intersection with the Bernstein set $A$
is countable.
Indeed, let $x\in E\cap A$, then $f(x)=x$ and hence
$s(x)=\langle x,0\rangle$ or $s(x)=\langle x,1\rangle$;
remember: the points of $A$ are split in $\mathbb{A}\_Y$.
So $E\cap A$ is the union of two sets
$E\_0=\{x\in E\cap A:s(x)=\langle x,0\rangle\}$
and $E\_1=\{x\in E\cap A:s(x)=\langle x,1\rangle\}$.
By continuity of $s$ we have for every $x\in E\_0$ an interval $(p\_x,q\_x)$
with rational end points that is mapped by $s$ into the interval
$[0,\langle x,0\rangle]$ in $\mathbb{A}\_Y$.
It follows that if $x<y$ in $E\_0$ then $y\in(p\_y,q\_y)\setminus (p\_x,q\_x)$.
We see that $x\mapsto(p\_x,q\_x)$ is injective and $E\_0$ is countable.
Likewise $E\_1$ is countable and we conclude that $E(f)\cap A$ is countable
and hence so is $E(f)$ itself.
The second claim is that $f[S(f)]$ is countable.
We show that $f[S(f)]$ is countable by showing that
$f[C(f)\cap A]\cap V$ is countable.
The contrapositive of the properties of our family of Bernstein sets
then shows that $C(f)$ and hence $f[S(f)]$ is countable.
As above: let $x\in C(f)\cap A$ and assume $f(x)\in V$.
Then $s(x)=\langle f(x),0\rangle$ or $s(x)=\langle f(x),1\rangle$.
As above we split $C(f)\cap A$ into two pieces:
$C\_0=\{x:s(x)=\langle f(x),0\rangle\}$ and $C\_1=\{x:s(x)=\langle f(x),1\rangle\}$.
If $x\in C\_0$ then we take a rational interval $(p\_x,q\_x)$ that gets mapped
into $[0,\langle f(x),0\rangle]$ by $s$.
If $x,y\in C\_0$ and $f(x)<f(y)$ then $y\in(p\_y,q\_y)\setminus (p\_x,q\_x)$,
and we conclude that $x\mapsto(p\_x,q\_x)$ is injective and $C\_0$ is countable.
Likewise $C\_1$ is countable.
It follows that $f[G]=E(f)\cup f[S(f)]$ is countable.
16-03-2023: I put a more leisurely explanation on [ArXiV.org](https://arxiv.org/abs/2303.08491)
| 9 | https://mathoverflow.net/users/5903 | 442284 | 178,467 |
https://mathoverflow.net/questions/442272 | 3 | The motivation for this question comes from the study of lifts of an orbifold chart, which is simplified as the following:
Suppose that $ U $ is an open connected subset of $ \mathbb{R}^n $ and $ G $ is a finite group of diffeomorphisms of $ U $, which naturally acts on $ U $. Let $ \pi:U\rightarrow U/G $ denote the quotient map.
Assume we are given two smooth maps $ f,g:V\rightarrow U $ and a point $ x\in V $ such that $ \pi\circ f=\pi\circ g $ and $ f(x)=g(x) $. Obviously, $ f $ and $ g$ play the role of lifts of the map $ \pi\circ f=\pi\circ g $ along $ \pi $. There are examples that show that $ f $ and $ g $ cannot coincide, even on any open neighborhood of $ x $. So the question arises what is the difference between $ f $ and $ g$?
Is it possible to find a smooth map $ h:U'\rightarrow U $ defined on an open neighborhood $U'\subseteq U$ of $ f(x)=g(x) $ such that $ f|\_{V'}=h\circ g|\_{V'} $ for some open neighborhood $V'\subseteq V$ of $ x $?
An example of a situation where there is no such an $h$ would be welcome.
| https://mathoverflow.net/users/131015 | The difference between two lifts along an orbifold chart | Let $\mathbb Z/2$ act on $U=\mathbb R$ by reflection through the origin $x=0$. Let $V=\mathbb R$ as well.
Consider $f,g:\mathbb R\rightarrow \mathbb R$ given by
$$
f(x)=\begin{cases} e^{-1/x^2} \qquad \text{if }\qquad & x>0 \\
0 & x=0 \\
-e^{-1/x^2} & x<0
\end{cases}
$$
and $g(x)=e^{-1/x^2}$ for $x\not=0$ and $g(0)=0$.
So $g$ "bounces" at the origin, and $f$ continuous through. In the quotient both $f$ and $g$ "bounce" at the singularity. You cannot find an $h$ as you want, as $f$ is injective, but $g$ is not.
The situation is much better if $f$ and $g$ are assumed to be diffeomorphisms. Then $h$ exists, and in fact is given by applying an element of $G$ to $U$.
| 3 | https://mathoverflow.net/users/12156 | 442286 | 178,468 |
https://mathoverflow.net/questions/442296 | 6 | If G is an almost simple group, then Aut(G) is complete?
Apologies - I meant to post this on Stack Exchange
Just wondering if anyone has a reference to the above - it's quoted on Wikipedia (so supposing that means it's true...) and seems a plausible result.
I've tried generalising the proof of the version when G is simple. The proof I've seen for a simple group G relies on $Inn(G)$ char $Aut(G)$, which I'm not sure holds in the almost simple case?
I'm presuming there is a CFSG free proof.
| https://mathoverflow.net/users/499057 | If G is an almost simple group, then Aut(G) is complete? | This isn't true. Let $S$ be a simple group such as ${\rm PSL}(4,5)$ with outer automorphism group isomorphic to $D\_8$ (dihedral of order $8$), and let $G = S.2$ be an extension of $S$ by a non-central involution in $D\_8$.
Then ${\rm Aut}(G) = S.2^2$ and ${\rm Aut}({\rm Aut}(G)) = S.D\_8 = {\rm Aut}(S)$.
| 14 | https://mathoverflow.net/users/35840 | 442302 | 178,471 |
https://mathoverflow.net/questions/441891 | 4 | This question was first asked [here](https://math.stackexchange.com/questions/4648121/a-set-with-a-topology-and-a-partial-ordering-question), on math stack exchange, but wasn't able to attract any attention. Now that I am thinking more, it feels like the most suitable place for this question is here.
Suppose I have a topological space whose underlying set $X$ has a partial ordering on it. A prior this partial ordering has no relation to the topology. Suppose there is a [meet semi-lattice](https://en.wikipedia.org/wiki/Semilattice) $I \subseteq X,$ and an arbitrary set of open sets $\{A\_i\}\_{i\in I}$ satisfying $i\in A\_i$ for all $i\in I$.
>
> What conditions (or relations between topology and the partial ordering) would guarantee the existence of another family of open sets $\{B\_i\}\_{i\in I}$ satisfying $i\in B\_i\subseteq A\_i$ and $$B\_i\cap B\_j\subseteq B\_{i\wedge j}$$ for all $i\in I$ ?
>
>
>
| https://mathoverflow.net/users/54507 | When is this topology compatible with the partial ordering? | One of the standard topologies to consider would be the lower-cone topology, whose basic open sets are the lower cones $i{\downarrow}=\{j\mid j\leq i\}$. In this topology, the open sets are exactly the downsets. If $i\in A\_i$ is open, then indeed $i{\downarrow}\subseteq A\_i$, and furthermore $(i{\downarrow})\cap(j{\downarrow})=(i\wedge j){\downarrow}$, so these sets $B\_i=i{\downarrow}$ fulfill exactly your desired criterion.
But since you require only inclusion $B\_i\cap B\_j\subseteq B\_{i\wedge j}$ rather than equality, the upper cone topology also has your feature. That is, the open sets are the upsets, then we may take $B\_i=i{\uparrow}\subseteq A\_i$, and simply observe that $(i{\uparrow})\cap (j\uparrow)\subseteq (i\wedge j){\uparrow}$ because anything above $i$ and $j$ is also above $i\wedge j$. Indeed, for this topology we would have $(i{\uparrow})\cup(j{\uparrow})\subseteq(i\wedge j){\uparrow}$.
But of course, there are other topologies that also have your feature. The indiscrete topology, in which the only open sets are $X$ and $\emptyset$, we would have $A\_i=X$, and taking $B\_i=X$ fulfills your property. Also, in the discrete topology, in which every set is open, we can take $B\_i=\{i\}$ and fulfill your property.
Finally, let me point out that if one wants to insist on the identity $B\_i\cap B\_j=B\_{i\wedge j}$, then this implies that the topology must be included in the lower cone topology as in the first example above. The reason is that if $j\leq i$, then $i\wedge j=j$ and so $j\in B\_j=B\_{i\wedge j}= B\_i\cap B\_j$ and so in particular $j\in A\_i$ and thus $A\_i$ must contain every element of $i{\downarrow}$. In particular, in this case every open neighborhood $A\_i$ of any $i$ must contain the lower cone below $i$, and consequently, every open set must be downward closed. Thus, with the stronger $B\_i\cap B\_j=B\_{i\wedge j}$ requirement, the topology is contained in the lower cone topology.
| 3 | https://mathoverflow.net/users/1946 | 442305 | 178,472 |
https://mathoverflow.net/questions/442303 | 0 | Let $f(x)$ be a real-valued function defined in $(0, \infty)$. I am curious what kind of $f(x)$ has the following representations:
$$
f(x) = \sum\_{j=0}^\infty a\_j e^{-jx}, \quad \forall x \in (0, \infty).
$$
My initial thought was that consider the transformation $s = e^{-x}, s \in (0, 1)$, then we want to find $\{a\_j\}\_{j=1}^\infty$ such that
$$
f(-\log s) = \sum\_{j=0}^\infty a\_j s^j, \quad \forall s \in (0, 1)
$$
That means we may consider the Taylor expansion for $f(-\log s)$ at $s=0$, but then $-\log s = \infty$, for which the Taylor series is not well defined (at infinity).
So is there a method to determine $a\_j$ if $f(x)$ can be represented as an exponential sum?
| https://mathoverflow.net/users/165697 | What kind of functions can be represented as infinite linear combinations of exponential functions? | $\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$You were almost there.
Indeed, suppose that
$$f(x)=\sum\_{j=0}^\infty a\_j e^{-jx}\in\R \quad \forall x\in(0,\infty). \tag{1}\label{1}$$
Then
$$g\_f(s):=f(-\ln s)=\sum\_{j=0}^\infty a\_j s^j\in\R \quad \forall s\in(0,1).$$
So, the radius of convergence of the power series $\sum\_{j=0}^\infty a\_j z^j$ is $1$. So, $g\_f$ can be extended to the analytic function $G\_f$ on the open unit disc
$D$ in $\C$ given by the formula $G\_f(z)=\sum\_{j=0}^\infty a\_j z^j$ for $z\in D$.
Vice versa, if the function $g\_f$ can be extended to an analytic function $G\_f$ on
$D$, then $G\_f(z)=\sum\_{j=0}^\infty a\_j z^j$ with $a\_j=G\_f^{(j)}(0)/j!$ and hence \eqref{1} holds.
Thus, \eqref{1} holds for some $a\_j$'s if and only if the function $g\_f$ can be extended to an analytic function on $D$.
Similarly, \eqref{1} holds for some $a\_j$'s if and only if the function $g\_f$ can be extended to a real-analytic function on the interval $(-1,1)$.
---
For an illustration, suppose that $f(x)=\dfrac1{1-e^{-x}}$ for $x\in(0,\infty)$. Then $g\_f(s)=f(-\ln s)=\dfrac1{1-s}$ for $s\in(0,1)$. Clearly, the function $g\_f$ can be extended to the analytic function $G\_f$ on $D$ given by the formula
$$G\_f(z):=\dfrac1{1-z}=\sum\_{j=0}^\infty a\_j z^j=\sum\_{j=0}^\infty z^j$$
with $a\_j=G\_f^{(j)}(0)/j!=1$. So,
$$f(x)=\sum\_{j=0}^\infty a\_j e^{-jx}=\sum\_{j=0}^\infty e^{-jx}$$
for $x\in(0,\infty)$.
| 1 | https://mathoverflow.net/users/36721 | 442307 | 178,473 |
https://mathoverflow.net/questions/442310 | 3 | [This has been edited in response to comments from Fedor Petrov]
Suppose that $n=dm$ with $d,m>1$. Consider an $n\times n$ matrix $M$ such that
1. All diagonal entries are equal to one
2. Each row has $d$ ones, and all other entries are zero.
This can be seen as the adjacency matrix of a certain kind of directed graph; I am not sure whether that perspective is useful or not. Anyway, I will consider $M$ as a matrix over a finite field $\mathbb{Z}/p$, and I would like to prove that the rank is at least $m$ (which is achieved in the case where $M=J\_d\oplus\dotsb\oplus J\_d$ and $J\_d$ is the all one matrix ). Random experiments (with $m,d\leq 4$ and $p\in\{3,5,7\}$ and $\sim 10^5$ samples) suggests that the rank is always at least $m(d-1)$ with very high probability, but the above example shows that this is not always true. Does anyone know a proof that the rank is always at least $m$?
| https://mathoverflow.net/users/10366 | Lower bound on the rank of a graph | $m$ is a tight bound.
If the matrix be composed of $m$ blocks $d\times d$ with all 1's, its rank equals $m$.
For any $M$ satisfying your conditions, you may color $\{1, 2,\ldots,n\}$ in $d$ colors so that $M\_{i, j}=0$ for all $i<j$ of the same color (if $1, 2,\ldots,k-1$ are already colored, you may choose an appropriate color for $k$, since there exist at most $d-1$ forbidden colors). Then there exist $m$ numbers of the same color, and corresponding rows and columns form a unitriangular minor, thus rank of $M$ is not less than $m$.
| 2 | https://mathoverflow.net/users/4312 | 442313 | 178,475 |
https://mathoverflow.net/questions/442309 | 2 | I have the following $(n+1)\times (n+1)$ matrix
$$P = \begin{bmatrix}
f(0) & g(0) & 0 & 0 & 0 & \dots & 0\\
f(1) & 0 & g(1) & 0 & 0 & \dots & 0\\
f(2) & 0 & 0 & g(2) & 0 & \dots & 0\\
f(3) & 0 & 0 & 0 & g(3) & \dots & 0\\
\vdots & \vdots & & & & \ddots & 0\\
f(n-1) & 0 & 0 & 0 & 0 &\dots & g(n-1)\\
f(n) & 0 & 0 & 0 & 0 &\dots & 0
\end{bmatrix}$$
where $0< f(x)< 1$, $0< g(x)< 1$, and $f(x)+g(x)<1$, $f(x)$ is increasing, $g(x)$ is decreasing. Also, $g(x)$ decreases faster than $f(x)$ is increasing, so their sum $f(x)+g(x)$ is decreasing.
I want to show that by raising $P$ to any power $k$, the summation of the elements of the $i$-th row, i.e. $\sum\_{j=1}^{n+1} P^{k}\_{i,j}$ is decreasing with $i$.
I have taken the following inductive approach:
* I know that the summation of the *rows* of $P^t$ is decreasing when $t=1$, since $f(x)+g(x)$ is decreasing.
* Suppose that the summation of the rows of $P^t$ is decreasing up to $t=k-1$. I want to show that it holds for $t=k$.
* The $i$-th row of $P^k$ can be written as follows:
\begin{equation} P^k\_{i,:} = P\_{i,:} P^{k-1} \end{equation}
from which the summation is derived as
\begin{equation} \sum\_{j=1}^{n+1} P^k\_{i,j} = f(i)\sum\_{j=1}^{n+1} P^{k-1}\_{1,j} + g(i)\sum\_{j=1}^{n+1} P^{k-1}\_{i+1,j}. \end{equation}
I cannot get past this point.
Also, I haven't found any way to compute a closed-form expression of $P^k$, although it has a very specific structure: it is the concatenation of a non-zero column-row pair (the first column and the last row) and a diagonal matrix.
Edit: I do not care about the summation of the last row, only for all the $n$ first rows.
| https://mathoverflow.net/users/477568 | Summation of rows of a matrix P^k is decreasing with the power k | A counterexample is given by
$$P=\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & \frac{1}{2} \\
\frac{1}{2} & 0 & 0 \\
\end{array}
\right),$$
so that then
$$P^2=\left(
\begin{array}{ccc}
0 & 0 & \frac{1}{2} \\
\frac{1}{4} & 0 & 0 \\
0 & \frac{1}{2} & 0 \\
\end{array}
\right).$$
---
Here the conditions on $P$ are satisfied only non-strictly. However, by approximation, this does not matter. For instance, one may replace $P$ by the close matrix
$$P\_h=\left(
\begin{array}{ccc}
h & 1-2h & 0 \\
2 h & 0 & \frac{1}{2} \\
\frac{1}{2} & 0 & 0 \\
\end{array}
\right)$$
for small enough $h>0$.
Then $P\_h$ will strictly satisfy all the conditions on $f$ and $g$, and yet $P\_h^2$ will be arbitrarily close to $P^2$, and hence your desired conclusion will fail to hold.
---
The OP has changed the question, now saying " I do not care about the summation of the last row, only for all the n first rows."
Then a counterexample is
$$P=\frac1{100}\left(
\begin{array}{ccccc}
1 & 98 & 0 & 0 & 0 \\
2 & 0 & 50 & 0 & 0 \\
25 & 0 & 0 & 25 & 0 \\
48 & 0 & 0 & 0 & 1 \\
49 & 0 & 0 & 0 & 0 \\
\end{array}
\right),$$
so that $P^2$ is bad: the sum of the 2nd row of $P^2$ ($\approx0.27$) is less than the sum of the 3rd row of $P^2$ ($\approx0.37$).
| 3 | https://mathoverflow.net/users/36721 | 442321 | 178,477 |
https://mathoverflow.net/questions/442324 | 11 | Gödel's incompleteness theorem shows that there are undecidable statements, i.e., formal logical claims which neither have proofs nor disproofs.
In doing so, Gödel famously enumerated all well-formed formulas with his so-called "Gödel numbering". Thus, the following question is well posed:
>
> What proportion of well-formed formulas in ZFC are undecidable?
>
>
>
I seems clear to me that this proportion should converge. Moreover, there are really only two reasonable options - 0% or 100%. My money is on 100%, since that feels in line with the sort of logical weirdness inherent in this area, though I do not have any rigorous heuristic.
| https://mathoverflow.net/users/159298 | Are 100% of statements undecidable, in Gödel's numbering? | This is going to depend sensitively on your exact choice of Gödel number, and the limit will often not be defined. Pick your favorite very short undecidable statement S. Then for any statement A, you have the nearly the same length statements like "A or S", "A and S", "A and True", "A or False". However, one can construct Gödel numberings where any use of an "or" is much longer than any use of an "and" or the reverse, which will change how common these statements are.
So, the answer can be, with a little work, $100\%$, $0\%$, or undefined (which will probably be the case for most natural numbering choices). It is likely that for any computable $x$ where $0 \leq x \leq 1$, there is a numbering which gives an answer of $x$ to your question.
| 16 | https://mathoverflow.net/users/127690 | 442325 | 178,479 |
https://mathoverflow.net/questions/442291 | 5 | Let $I=(0,1)$ and let $C>1$ be a constant. Let $L^2(I)$ and $H^1(I)$ be the standard Sobolev spaces on $I$. Suppose that $U$ is a **subspace** of $H^1(I)$ with the additional property that:
$$ \| u\|\_{H^1(I)} \leq C \|u\|\_{L^2(I)}\quad \forall\, u\in U.$$
By compact embedding of $H^1(I)\subset L^2(I)$, it is trivial to see that $U$ must be finite dimensional (note here that the fact that $U$ is a subspace is vital).
My question is whether one can put an explicit bound on the dimension of $U$ in terms of the constant $C$.
Thanks,
| https://mathoverflow.net/users/50438 | Bounds on dimension of a subspace | Just use the Wirtinger's inequality that says that if $u=0$ at the center of an interval $I\subset \mathbb R$, then $\int\_I|u|^2\le (|I|/\pi)^2\int\_I|u'|^2$. Thus, if the dimension is $>n$, we can create a function that vanishes at the centers of $n$ subintervals of the interval $(0,1)$ of length $1/n$ and have $\|u\|\_{L^2}\le \frac 1{\pi n}\|u'\|\_{L^2}$, so we must have $\pi n\le C$. On the other hand, we can take trigonometric polynomials $f(x)=\sum\_{|k|\le N}c\_ke^{2\pi i k x}$ and observe that for them $\|f'\|\_{L^2}\le 2\pi N\|f\|\_{L^2}$ and the dimension is $2N+1$, so the bound is tight if we use the norm of the derivative alone. Including the norm of the function into the definition of $H^1$ norm changes the answer by $O(1)$ (and brings it down), so the inequality $\rm{dim\,}U\le \frac{C}\pi+1$ still holds and is nearly the best possible one. That is, of course, in a perfect alignment with Giorgio's comment, but this case is simple enough to do it without formally invoking the spectral theory for the Laplacian.
| 6 | https://mathoverflow.net/users/1131 | 442326 | 178,480 |
https://mathoverflow.net/questions/442280 | 0 | I am dealing with a problem of the form ($a<b$)
$$
\displaystyle \max\_{v \in C^1([a, b])} \int\_a^b v(x)~\mathrm{d}x, \quad \mathrm{s.t.} \int^b\_a \big(-o'(x)v(x)-v'(x)o(x)\big)f(x)~\mathrm{d}x \leq 0,~ \frac{o(b)}{o(a)} \leq v \leq 1
$$
where $o \in C^\infty(\mathbb{R})$ with $o>0$ and $o'< 0$ on $(a, b)$ and $f \in C([a, b])$ with $f>0$ on $(a, b)$. I would already be very happy if someone could provide me with input for the case $f(x)=1$.
Problem is, I can find little to no publications on such problems. Moreover, the usual optimality conditions are useless because we lose $v$ in the derivative of both objective and constraint. My suspicion is that
$$
v(x) = \frac{o(b)}{o(x)}
$$
for all $x \in [a, b]$ is optimal, since it solves the ODE $-o'v-v'o$ and is $1$ on the boundary $b$.
Furthermore, using Gronwall, I was able to show that if some optimal $\tilde{v}$ exists and
$$
-o'(x)\tilde{v}(x) -\tilde{v}'(x)o(x) \leq 0
$$
on some interval $(c, d) \subseteq (a, b)$ holds, then we get the upper bound
$$
\tilde{v}(x) \leq \frac{o(d)}{o(x)}
$$
on $\tilde{v}$ for all $x \in (c, d)$, which is at least somewhat consistent with my hypothesis. Any input would really be appreciated.
| https://mathoverflow.net/users/500621 | Constrained linear optimization problem on $C^1$ | As you suggest, let me consider the case $f \equiv 1$. Without loss of generality, assume also that $a = 0$ and $b = 1$. Let $\sigma := o(1)/o(0) \in (0,1)$. The problem becomes
$$
\sup\_{v \in C^1([0,1])} \int\_0^1 v(x) \mathrm{d}x
\quad \text{s.t.} \quad
v(0)\leq \sigma v(1) \quad \text{and} \quad \sigma \leq v(x) \leq 1.
$$
First, note that I changed the maximum to a supremum, since it is not clear *a priori* that the supremum will be achieved (it will turn out not to be the case). Second, note that the function $o(x)$ has disappeared from the problem, only remaining through the single real value $\sigma \in (0,1)$.
Since $v(x) \leq 1$ on $[0,1]$, one has $\int\_0^1 v \leq 1$ for any admissible $v \in C^1([0,1])$. Let us check that $1$ is the value of the supremum, but that it is not achieved.
* If $v \in C^1([0,1])$ is such that $v(x) \leq 1$ on $[0,1]$ and $\int\_0^1 v = 1$, then $v \equiv 1$. But the constraint $v(0) \leq \sigma v(1)$ becomes $1 \leq \sigma$, which fails because $\sigma \in (0,1)$. So the supremum cannot be achieved.
* For $n \in \mathbb{N}^\*$, let $v\_n(x) := 1$ for $x \in [1/n,1]$ and $v\_n(x) := 1 + (\sigma-1)n^2(x-1/n)^2$ for $x \in [0,1/n]$. Then $v\_n \in C^1([0,1])$ because the junction at $x=1/n$ is $C^1$, $\sigma \leq v\_n \leq 1$ on $[0,1]$, and $\sigma = v(0) \leq \sigma v(1) = \sigma$ because $v(1) = 1$. Moreover, one checks that $\int\_0^1 v\_n \geq 1 - 1/n$. So the supremum is indeed equal to $1$.
| 1 | https://mathoverflow.net/users/50777 | 442328 | 178,481 |
https://mathoverflow.net/questions/442327 | 1 | Let $\{W(t),\,t \in [0,1]\}$ be a standard Brownian motion in $\mathbb{R}^d$, starting from $0$. Let $U$ be a non-empty open set such that $0 \in \partial U$.
Which conditions on $U$ are necessary and sufficient for
$
\mathbb{P}\{\exists t\in[0,1]:W(t) \in U\}=1
$?
Most likely, there are no simple conditions on $U$, but maybe it can be somehow reformulated in terms of potential theory?
For example, is it true that $\mathbb{P}\{\exists t\in[0,1]:W(t) \in U\}=1$ if and only if $0$ is a regular point for $U$? The definition of a regular point is as in <https://encyclopediaofmath.org/wiki/Potential_theory> : we can assume for simplicity that $U$ is bounded, then regularity of $0$ means for each continuous function $\phi$ on $\partial U$ $\lim\_{x \to 0} H\_{\phi}(x)=\phi(0),\,x\in U$, where $H\_{\phi}(x)$ is the generalized solution for the Dirichlet problem on $U$ with boundary values $\phi$.
| https://mathoverflow.net/users/499882 | Brownian motion hitting open set starting from its boundary |
>
> is it true that $\mathbb{P}\{\exists t\in[0,1]:W(t) \in U\}=1$ if and only if $0$ is a regular point for $U$?
>
>
>
Yes, some books even take the definition of regular points to be that eg. see online notes ["Classical potential theory and Brownian motion"](https://digitalassets.lib.berkeley.edu/math/ucb/text/math_s6_v3_article-09.pdf). In terms of conditions besides the [Port-Stone book](https://www.sciencedirect.com/book/9780125618502/brownian-motion-and-classical-potential-theory), I would look at the Garnett-Marshall book "Harmonic measure" eg. the section on the Wiener-series for regular points in terms of capacities.
| 1 | https://mathoverflow.net/users/99863 | 442329 | 178,482 |
https://mathoverflow.net/questions/442315 | 1 | $\newcommand\SS{P}\newcommand\TT{Q}$I will call a Gaussian probability measure $\SS$ on $\mathbb{R}^d$ *isotropic* if its covariance matrix is diagonal with non-vanishing determinant; i.e. $\Sigma\_{i,i}>0$ for $i=1,\dots,d$ and $\Sigma\_{i,j}=0$ whenever $i\neq j$ for each $i,j=1,\dots,d$.
*Note: My definition of "isotropic" includes "the usual isotropic Gaussian measures," which, from my limited understanding, are assumed to have a covariance of the form $\sigma I\_d$ for some $\sigma>0$.*
Let $\mathcal{P}$ the set of *isotropic* Gaussian probability measures on $\mathbb{R}^d$ and let $\mathcal{Q}$ be the set of probability measures on $\mathbb{R}^d$ with Lebesgue density equipped with TV distance.
Consider the *information projection (of I-projection)* defined by
\begin{align}
\pi:\mathcal{Q} &\rightarrow \mathcal{P}
\\
\pi(\TT) &:= \operatorname\*{argmin}\_{\SS\in \mathcal{P}}\, D(\SS\parallel\TT)
\end{align}
I'm looking for references on the following "elementary properties" of the I-projection:
* Is the I-projection $\pi$ Lipschitz, at-least locally?
* Are there error bounds on $D(\pi(\TT),\TT)$ when $\TT$ is a Gaussian measure on $\mathbb{R}^d$ with non-singular covariance...
| https://mathoverflow.net/users/491352 | References: error and stability estimates for information projection | Assuming you want to minimize the Kullback–Leibler divergence
$$D(P\parallel Q)=\int dP\,\ln\frac{dP}{dQ}$$
over all isotropic Gaussian $P$, "the" minimizer is in general not unique and, accordingly, not Lipschitz even on the set of measures $Q$ where it is unique.
The idea of a counterexample is quite simple: Suppose that $d=1$. Let $Q\_h$ be the probability measure with pdf $q\_h$ given by the formula
$$q\_h(x)=c\_h\big((1+h)\,f\_{1,a}(x)\,1(x>0)
+(1-h)\,f\_{-1,a}(x)\,1(x<0)\big)$$
for real $x$, where $f\_{t,a}$ is the pdf of the normal distribution $N(t,a^2)$, $a>0$ is small enough (the condition $0<a<\sqrt{2/\pi}$ should do), $h$ is a real number very close to $0$, and $c\_h(\approx1/2)$ is the normalizing factor.
Since $a$ is rather small, $Q\_h$ is somewhat close to the mixture of the rather narrow normal distributions $N(1,a^2)$ and $N(-1,a^2)$ with slightly unequal weights, $c\_h\,(1+h)$ and $c\_h\,(1-h)$ respectively. So, a minimizer $P\_h$ of the Kullback–Leibler divergence $D(P\parallel Q\_h)$ in $P$ should be sufficiently close to $N(1,a^2)$ or $N(-1,a^2)$ depending on whether the small perturbation $h$ is $>0$ or $<0$, respectively. Thus, an infinitesimally small change from, say, $h>0$ to $-h<0$ will result in quite a nonnegligible change from $P\_h\approx N(1,a^2)$ to $P\_{-h}\approx N(-1,a^2)$. (If $h=0$, then there will be two minimizers.)
I can write down details later, if you want them.
---
Responding to the comment of the OP about a possible relation of your question to a result by Csiszár: Your question concerns the existence and uniqueness, for any **given probability measure (PM) $Q$**, of a PM $P\_{\mathcal S,Q}\in\mathcal S$ such that $D(P\_{\mathcal S,Q}\parallel Q)\le D(P\parallel Q)$ for all **$P\in\mathcal S:=\mathcal P$**, which latter is the set of all isotropic Gaussian PM's.
In contrast, [Csiszár's result](https://www.researchgate.net/publication/3084724_Information_projections_revisited) is that for any given PM $P$ **(rather than $Q$)** there is a unique PM $\tilde P^{\mathcal S,P}$ such that
$D(\tilde P^{\mathcal S,P}\parallel Q)+D(P\parallel\mathcal S)\le D(P\parallel Q)$ for all $Q\in\mathcal S$ **(rather than $P\in\mathcal S$)**, where $D(P\parallel\mathcal S):=\inf\_{Q\in\mathcal S}D(P\parallel Q)$. (So, this looks like some kind of Pythagorean inequality.)
A corollary to this result by Csiszár is that $D(\tilde P^{\mathcal S,P}\parallel Q)\le D(P\parallel Q)$ for all $Q\in\mathcal S$ **(rather than $P\in\mathcal S$)**.
So, if $D$ were a metric, your question would be about the existence and uniqueness of a PM in $\mathcal S$ closest to $Q$. On the other hand, again if $D$ were a metric, the mentioned corollary from Csiszár's result would say that the length of the projection of the segment $\tilde P^{\mathcal S,P}Q$ of the segment $PQ$ onto $\mathcal S$ is no greater than the length of $PQ$, for any $Q$. The latter property would be equivalent to your "closest" property if $D$ were a Euclidean metric. But $D$ is not a metric at all. So, Csiszár's result says something different from what your question is about.
(The comparison of your question to Csiszár's result got more complicated than necessary because you interchanged the usual order of the arguments $P$ and $Q$ of $D(P\parallel Q)$, which was also used by Csiszár. So, I have edited your post, and mine, accordingly.)
| 3 | https://mathoverflow.net/users/36721 | 442343 | 178,485 |
https://mathoverflow.net/questions/442317 | 3 | Let $A$ be a a two-sided noetherian semiperfect ring and assume that the injective dimension of the left and right regular modules are equal to $n \geq 1$.
Let $\Omega^n(mod A)$ be the category of $n$-th syzygy modules consisting of modules that are projective or a direct summand of a module of the form $\Omega^n(M)$ for some $M$.
For $A$ as above those are the maximal Cohen-Macaulay $A$-modules $CM A$. Let $\underline{CM} A$ denote the stable category.
>
> Question: Is there a good reference that the functor $\Omega^1: \underline{CM} A \rightarrow \underline{CM} A$ has the property that $\Omega^1(X) \cong \Omega^1(Y)$ implies $X \cong Y$? $\Omega^1$ should even be an equivalence.
>
>
>
I only know references for the case when $A$ is an Artin algebra or a local commutative Gorenstein ring.
| https://mathoverflow.net/users/61949 | $\Omega$ for noetherian semiperfect rings | This seems to be answered in Proposition 4.2 of Kameyama, Kimura and Nishida, "On stable equivalences of module subcategories over a semiperfect Noetherian ring".
| 2 | https://mathoverflow.net/users/460592 | 442350 | 178,490 |
https://mathoverflow.net/questions/442298 | 4 | It is well known that complex projective three space $\mathbb{C}\mathbf{P}^3$ is a complex manifold. However it also possess a non-integrable almost-complex structure (as discussed in this [article](https://www.sciencedirect.com/science/article/pii/S0393044020301728#b12) for instance). Can somebody give an explicit constructuion of this almost-complex structure. I also wonder how, of if, it interacts with the homogeneous action of the unitary group $U(3)$ (or indeed $SU(3)$) on $\mathbb{C}\mathbf{P}^3$.
| https://mathoverflow.net/users/491434 | Non-integrable almost complex structure for complex projective $3$-space | There are lots of non-integrable almost-complex structures on $\mathbb{CP}^3$, but the one you are looking for is, I suspect, the following.
First I will explain the twistor fibration, which is a submersion $t \colon \mathbb{CP}^3 \to S^4$ with fibres $S^2$. To do this, fix an identification $\mathbb{C}^4 \cong \mathbb{H}^2$ with the quaternionic plane. Now each complex line in $\mathbb{C}^4$ generates a quaternionic line in $\mathbb{H}^2$. This gives a map $t \colon \mathbb{CP}^3 \to \mathbb{HP}^1$. The fibre of $t$ over a point $p \in \mathbb{HP}^1$ is the set of all complex lines in $\mathbb{C}^4$ that lie in the quaternionic line corresponding to $p$. This quaternionic line is a copy of $\mathbb{C}^2$ inside $\mathbb{C}^4$ and so $t^{-1}(p) \cong \mathbb{CP}^1 \cong S^2$. Meanwhile, $\mathbb{HP}^1$ is the 1-point compactification of $\mathbb{H}$ (just as $\mathbb{CP}^1$ is the 1-point compactification $\mathbb{C}$) and so the range of $t$ is diffeomorphic to $S^4$. This gives the fibration $t \colon \mathbb{CP}^3 \to S^4$.
Now I will explain how to use this to give a *non*-integrable almost-complex structure on $\mathbb{CP}^3$. The tangent spaces to the fibres of $t$ determine the so-called vertical tangent bundle, which is a complex line-subbundle $V \leq T\mathbb{CP}^3$. Meanwhile, the Fubini-Study metric in $\mathbb{CP}^3$ gives a complement $H$ to $V$ and so we have a decomposition into complex summands $T\mathbb{CP}^3 = V \oplus H$. With this in hand, we can define $J \colon T\mathbb{CP}^3 \to T\mathbb{CP}^3$ by $J = - i \oplus i$. In other words, $J$ agrees with the standard complex structure on $H$ and *minus* the standard complex structure on $V$.
In the language of twistor geometry, $\mathbb{CP}^3$ is the twistor space of $S^4$. The Aityah-Hitchin-Singer twistor almost complex structure is the standard one on $\mathbb{CP}^3$ (which is integrable when the base Riemannian 4-manifold is self-dual) whilst the one I described here is the Eells-Salamon twistor almost complex structure and this is *never* integrable.
You also asked about how to describe this by thinking of $\mathbb{CP}^3$ as a homogeneous space, so now I'll try and explain something in that direction. The isometries of $S^4$, namely $\mathop{SO}(5)$, lift to $\mathbb{CP}^3$ preserving the twistor fibration. It follows that they also preserve both the integrable and non-integrable complex structures. The action of $\mathop{SO}(5)$ on $\mathbb{CP}^3$ is transitive with stabiliser of a point isomorphic to $\mathop{U}(2)$. From this perspective, $\mathbb{CP}^3 \cong \mathop{SO}(5)/\mathop{U}(2)$. Write
$$
\mathfrak{so}(5) = \mathfrak{u}(2) \oplus \mathfrak{p}
$$
(orthongonal with respect to the Killing form). Then an $\mathop{SO}(5)$-invariant almost-complex structure on $\mathbb{CP}^3$ corresponds to a choice of almost complex structure on $\mathfrak{p}$ which is invariant under the action of $\mathop{U}(2)$. As a $\mathop{U}(2)$-representation,
$$
\mathfrak{p} \cong \mathbb{C}^2 \oplus \Lambda^2 \mathbb{C}^2
$$
(where $\mathbb{C}^2$ is the fundamental representation of $\mathop{U}(2)$). This corresponds to the splitting $T\mathbb{CP}^3 = H \oplus V$ desribed above. It follows that there are precisely two $\mathop{SO}(5)$-invariant almost-complex structures on $\mathbb{CP}^3$ (up to overall sign), and they are the standard one and the one in which $J$ has been reversed on $\Lambda^2\mathbb{C}^2 \cong V$.
Finally, as a remark, you can check that the non-integrable almost-complex structure given here actually has $c\_1=0$ (e.g. by using the $\mathop{SO}(5)$-invariant description above). This means it is not even isotopic to the integrable complex structure.
| 8 | https://mathoverflow.net/users/380 | 442352 | 178,491 |
https://mathoverflow.net/questions/442336 | 0 | Let $X,Y$ be real separable Hilbert space, and let $HS(X,Y)$ be the space of Hilbert-Schmidt operators from $X$ to $Y$, endowed with the [Hilbert-Schmidt norm](https://en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_operator). Let $x\in X$ and $y\in Y$. I am interested in finding the derivative of the map
$$
HS(X,Y)\ni A\mapsto \|y-Ax\|\_Y^2\in \mathbb{R}.
$$
As $HS(X,Y)$ is a Hilbert space, the derivative (as a bounded linear operator from $HS(X,Y)$ to $\mathbb{R}$) can be identified as an element in $HS(X,Y)$. Is there an easy expression of it?
| https://mathoverflow.net/users/91196 | Derivative with respect to a Hilbert-Schmidt operator | Let me denote by $F : HS(X,Y) \to \mathbb{R}$ the map which you defined as $F(A) := \|y-Ax\|\_Y^2$. By standard arguments, you indeed have that $F$ is $C^1$ and, for any $A \in HS(X,Y)$,
$$
\mathcal{L}(HS(X,Y),\mathbb{R}) \ni DF\_{\rvert A} = \begin{cases}
HS(X,Y) \to \mathbb{R}, \\
B \mapsto 2 \langle Bx , Ax - y \rangle\_Y.
\end{cases}
$$
As you recall, $HS(X,Y)$ is a Hilbert space for the scalar product
$$
\langle A, B \rangle\_{HS} := \sum\_{i \in I} \langle A e\_i, B e\_i \rangle\_Y
$$
for any orthonormal basis $(e\_i)\_{i \in I}$ of $X$.
Given $A \in HS(X,Y)$, by the Riesz representation theorem, we look for some $C \in HS(X,Y)$ such that $DF\_{\rvert A} = \langle C, \cdot \rangle\_{HS}$. Thus, we must have, for all $B \in HS(X,Y)$,
$$
\begin{split}
\sum\_{i \in I} \langle C e\_i, B e\_i \rangle\_Y
& = 2 \langle Bx, Ax-y \rangle\_Y \\
& = 2 \left\langle B \left(\sum\_{i\in I} \langle x, e\_i \rangle\_X e\_i \right), Ax-y \right\rangle\_Y \\
& = \sum\_{i\in I} \left \langle Be\_i, 2 \langle x, e\_i \rangle\_X (Ax-y) \right \rangle\_Y.
\end{split}
$$
Hence, one can define $C$ by $C e\_i := 2 \langle x, e\_i \rangle\_X (Ax-y)$.
Then $C \in HS(X,Y)$ and $\|C\|\_{HS} = 2 \|x\|\_X \| Ax-y\|\_Y$.
Eventually, one can thus identify the derivative as
$$
HS(X,Y) \ni DF\_{\rvert A} : \begin{cases}
X \to Y, \\
z \mapsto 2 \langle x, z \rangle\_X (Ax-y).
\end{cases}
$$
| 2 | https://mathoverflow.net/users/50777 | 442358 | 178,494 |
https://mathoverflow.net/questions/442018 | 2 | Let $M\_1$ and $M\_2$ be von Neumann algebras acting on Hilbert spaces $H\_1,H\_2$ and consider $M=M\_1\overline\otimes M\_2$ acting on $H\_1\otimes H\_2$. Let $K$ be an $M$-invariant subspace (so that $P\_K\in M\_1'\overline\otimes M\_2'$).
Assume that $PMP = B(K)$.
Consider the von Neumann algebras $R\_1 = P M\_1\otimes 1 P $ and $R\_2 = P1\otimes M\_2 P$.
Then $R\_1\vee R\_2=B(K)$ and $R\_1\subset R\_2'$, so both $R\_i$, $i=1,2$, are factors.
Therefore $R\_1\subset R\_2'$ is an irreducible subfactor inclusion.
My question is: Do I overlook something that already implies that $R\_1 = R\_2'$? If not, which irreducible subfactor inclusions can arise this way?
A more general question would be if normal \*homomorphisms preserve relative commutants. This has surely been answered but I could not find it.
| https://mathoverflow.net/users/485160 | Question on tensor product of von Neumann algebras and subfactors | I would say you get $R\_1 = R\_2'$, and also $R\_1 = B(K')$ for some $K'$.
1. The projection $P$ is minimal in $M\_1' \mathbin{\bar\otimes} M\_2'$:
In general, if $M \subset B(H)$ is a von Neumann algebra and $P \in M'$ with range $K$, you have $(M P)' = P M' P$ on $K$.
(You can find this in "Takesaki I", Proposition II.3.10.)
From your assumption $M P = B(K)$, you get $P M' P = \mathbb C P$.
2. The algebras $M\_i$ can be assumed to be type I factors:
First look at the type (I, II, III) decomposition of $M\_i'$.
Use the fact that algebras of type II and III cannot contain minimal projections, and that the type I component of the tensor product $M'$ only comes from the type I components of $M\_i'$.
This reduces the situation to the type I case.
Then you can further reduce it to the factor case by looking at the central support of $P$.
3. Now we know that $H\_i = H\_{i 1} \otimes H\_{i 2}$ for $i = 1, 2$, with $M\_i = B(H\_{i 1}) \mathbin{\bar\otimes} \mathbb C$, $M\_i' = \mathbb C \mathbin{\bar\otimes} B(H\_{i 2})$, and $P$ is a rank $1$ projection on $H\_{1 2} \otimes H\_{2 2}$.
Again from point 1 you get
$$
R\_2' = P B(H\_{1 1} \otimes H\_{1 2} \otimes H\_{2 2}) P = B(H\_{1 1}) P = R\_1.
$$
| 2 | https://mathoverflow.net/users/9942 | 442360 | 178,496 |
https://mathoverflow.net/questions/442361 | 2 | Let $x\_1,\ldots,x\_n$ be $n$ complex numbers, and define $x\_I:=\sum\_{i\in I}x\_i$ for any set $I\subseteq[n]$. Finally, declare the family $(x\_1,\ldots,x\_n)$ to be "sumset-distinct" if the $2^n$ numbers $(x\_I)\_{I\subseteq [n]}$ are pairwise distinct. My questions are:
1. Has this notion been studied and if yes under what name?
2. Is there a simple characterization of sumset-distinct families?
3. Are there simple methods to determine whether a given family is sumset-distinct. For example, are the $n^{th}$ roots of unity $\left(e^{\frac{2ik\pi}{n}}\right)\_{1\le k \le n}$ sumset-distinct?
| https://mathoverflow.net/users/477827 | Sumset-distinct numbers | The set $\{x\_i\}$ is called in the literature a sum-distinct set.
There is a big open problem in the area, due to Erdős and Moser: determining $f(n)=\min\{ \max(S): |S|=n, S\subseteq \mathbb{N}, S \text{ sum-distinct}\}$. The current lower bounds on $f$ are of the form $f(n)\ge C 2^n/\sqrt{n}$. A trivial upper bound is $f(n)\le 2^{n-1}$ by taking $x\_i=2^{i-1}$.
There are some useful characterizations which were used in the literature to attack this problem. One characterization used by [Dubroff--Fox--Xu](https://arxiv.org/abs/2006.12988) is that if $\varepsilon\_i$ are i.i.d random variables taking the values $\pm 1$ with equal probabilities then $\sum \varepsilon\_i x\_i$ takes each of its $2^n$ possible values with probability $2^{-n}$ iff $S$ is a sum-distinct set.
[A related (earlier) observation of Elkies](https://www.sciencedirect.com/science/article/pii/0097316586901160) is that if $x\_i$ are integers valued then the Laurent polynomial $A(z)=\prod\_i(1+(z^{x\_i}+z^{-x\_i})/2)$ has a very particular coefficient structure if $S$ is sum-distinct (and in particular the constant coefficient is $1$).
| 7 | https://mathoverflow.net/users/31469 | 442367 | 178,500 |
https://mathoverflow.net/questions/442304 | 4 | I obtained the very strange formula above and at begining I was just wanted know how to interpretate it. But now when I know what is this with help of @Carlo Beenakker, I am leaving it as a proof. BTW I found some [article](https://www.google.com/url?q=https://www.peertechzpublications.com/articles/AMP-5-141.pdf&sa=U&ved=2ahUKEwjDor-tpNb9AhXX_yoKHYGHAC4QFnoECAQQAg&usg=AOvVaw2L5byXDO356phO20CxHXRS) about non formal take of that theorem (I hope my take on this could be colled formal). I made some changes because I am ritarded and I realised that half of calculation was unnecesary.
**proof** Let $f(x)$ be some an analytic function with a given Taylor series. By using the theorem on divergent summation derived by Ramanujan, we can associate to it the convergent integral $\int\_x^{\infty}f (t) dt $ so we can write the summation in terms of Euler–Maclaurin formula as follows
$$
\sum\_{k=x}^{\infty}f (k)=\sum\_{n=-1}^{\infty} \frac {f^{(n)}(x)\zeta (-n)}{n!}.
$$
We can use Riemann's zeta functional equation to derived some transformation of the above expression:
$$
\begin{split}
\sum\_{k=x}^{\infty} f (k) & = -F (x)+\frac {f (x)}{2}+ \sum\_{k=1}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\
& = -F (x)+\frac {f (x)}{2} +\sum\_{n=1}^{\infty}\sum\_{k=1}^{\infty} \frac { f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!} \\
& = -F (x)+\frac {f (x)}{2} +\sum\_{n=1}^{\infty}\sum\_{k=0}^{\infty} \frac { f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!} \\
& = -F (x)+\frac {f (x)}{2} +\sum\_{n=1}^{\infty}\int\_{0}^{\infty}\sum\_{k=0}^{\infty}\frac{f^{(k)}(x) t^{k}(e^{-2 \pi i nt}+e^{2 \pi i nt})}{k!}dt \\
& = -F (x)+\frac {f (x)}{2} + \sum\_{n=1}^{\infty}\int\_{0}^{\infty}f(t+x)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt
\end{split}
$$
And now, for x=0, without any strange equations it is possible to write down formula as
$$ \frac {f (0)}{2}+\sum\_{k=1}^{\infty} f (k)=\sum\_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)$$
PS: I feel like. Is it all Euler-Maclaurin formula? Always has been. Watch this trivial proof of Abel-Plana formula
$$
\begin {split}
\sum\_{k=x}^{\infty} f (k) & = -F (x)+\frac {f (x)}{2}+ \sum\_{k=1}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\
& =-F (x)+\frac {f (x)}{2}+ \sum\_{k=1}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \int\_0^{\infty}\frac {t^k}{e^t-1}dt}{k !}\\
& =-F (x)+\frac {f (x)}{2}+ \sum\_{k=0}^{\infty}\frac{f^{(k)}(x)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \int\_0^{\infty}\frac {t^k}{e^t-1}dt}{k !}\\
& =-F (x)+\frac {f (x)}{2}+ \int\_0^{\infty}\frac {\frac {f (x+\frac {t}{2\pi i})}{2\pi i}+\frac {f (x+\frac {t}{-2\pi i})}{-2\pi i}}{e^t-1} dt \\
&=-F (x)+\frac {f (x)}{2}+i\int\_{0}^{\infty} \frac{f (x+it)-f (x-it)}{e^{2\pi t}-1} dt \\
\end {split}
$$
| https://mathoverflow.net/users/500629 | $\frac {f (0)}{2}+ \sum_{k=1}^{\infty}f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)$ | *This is an answer to the question as it was originally formulated, it has now been heavily edited.*
---
I consider this formula in the OP,
$$\sum\_{k=1}^{\infty} f (k) = \int\_{0}^{\infty}f(t)dt+ \sum\_{n=1}^{\infty}\int\_{0}^{\infty}f(t)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt.\qquad\qquad(\ast)
$$
This is *almost* the [Extension of the Poisson Summation Formula](https://doi.org/10.1006/jmaa.1997.5767) using [Cesàro summation](https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation), see Eq. 3.4 of the cited paper:
$$\sum\_{k=-\infty}^\infty f(k)-\int\_{-\infty}^\infty f(t)\,dt=\sum\_{n=1}^\infty \int\_{-\infty}^\infty f(t)\left(e^{2\pi int}+e^{-2\pi int}\right)\,dt.\qquad\qquad(\ast\ast)$$
The expression on the left-hand side is to be evaluated in the Cesàro sense, when the sum over $f(k)$ diverges (for example, when $f(k)$ is a polynomial). The formula $(\ast\ast)$ follows from the identification of the distributions
$$\sum\_{k=-\infty}^\infty \delta(x-k)-1=\sum\_{n=1}^\infty\left(e^{2\pi inx}+e^{-2\pi inx}\right).$$
Now I can take an even function $f(t)=f(-t)$, and then I almost find from $(\ast\ast)$ the expression $(\ast)$, up to an extra term $f(0)$ on the left-hand-side.
---
Here is a check that a term $\tfrac{1}{2}f(0)$ is missing from the left-hand-side of equation $(\ast)$ from the OP: take the case $f(t)=e^{-t}$; then the left-hand-side of equation $(\ast)$ evaluates to $\frac{1}{e-1}-1$, while the right-hand-side evaluates to $\frac{1}{e-1}-\tfrac{1}{2}$, so there is a $1/2$ difference.
| 3 | https://mathoverflow.net/users/11260 | 442373 | 178,502 |
https://mathoverflow.net/questions/442376 | 5 | Given an infinite family $\{\mathcal{F}\_{\lambda}$, $\lambda <\kappa\}$, $\kappa \geq \omega\_0$, of (ultra)filters of a set $X$, how it is defined the infinite tensor/Fubini product $$\bigotimes\_{\lambda <\kappa}\mathcal{F}\_\lambda$$ as a family of subsets of $X^{\kappa}$?
Recall that given two ultrafilers $\mathcal{F}$ and $\mathcal{G}$ of a set $X$, the tensor/Fubini product $\mathcal{F}\otimes \mathcal{G}$ of them is defined as the collection of sets $$\{A\subset X\times X:\{\xi \in X:\{\nu \in X:(\xi,\nu)\in A\}\in \mathcal{G}\}\in \mathcal{F}\}.$$
Is there any example in the bibliography of their use? More precisely, is there any basic bibliography in this topic?
For instance, for $\kappa=\omega\_0$, if I take an ultrafilter $\mathcal{F}$ in a discrete space $X$ and I do the iterated tensor product $$\mathcal{F}\bigotimes \mathcal{F}=\mathcal{F}^2 \subset \mathcal{P}(X\times X),$$ $$\mathcal{F}^n\bigotimes \mathcal{F} =\mathcal{F}^{n+1}\subset \mathcal{P}(X^{n+1}),\, n<\omega\_0,$$
how I sould define $\mathcal{F}^{\omega\_0}$ in $X^{\omega\_0}$?
| https://mathoverflow.net/users/117312 | Infinite tensor/Fubini product of ultrafilters | The product of ultafilters $F\_\lambda$ for $\lambda<\kappa$ is defined on $\kappa\times X$, not $X^\kappa$, and it is defined relative to a fixed ultrafilter $\mu$ on the index set $\kappa$. Namely, for $Y\subseteq\kappa\times X$, we denote $Y\_\lambda=\{x\in X\mid (\lambda,x)\in Y\}$ for the various sections of $Y$ and then say
$$Y\in\int \mathcal{F}\_\lambda d\mu\qquad\text{if and only if}\qquad \{\lambda\in\kappa\mid Y\_\lambda\in \mathcal{F}\_\lambda\}\in\mu.$$
This directly generalizes the product measure case you defined, in the event that all $F\_\lambda$ are the same, taking $X=\kappa$ and looking at subsets of $X\times X$.
This kind of product ultrafilter occurs routinely in the set-theoretic large cardinal literature, since many large cardinals involve this kind of measure, particularly in the case where these ultrafilters are countably complete.
In your comment below you seek to generalize to infinite exponents. One common way to do this in the large cardinal context is with *extenders*, which are in effect the natural limit of the finite products. If we have measures $\mathcal{F}\_i$ defined for $i\in I$, then we can define the natural measure on $X^I$ concentrating on sets with finite support as follows. For each finite $s\subseteq I$, we have a natural measure $\mathcal{F}\_s=\Pi\_{i\in s}\mathcal{F}\_i$ defined by the Fubini product. And now we say that a set $Y\subseteq X^I$ has support $s$ if there is a set $Y\_0\subseteq X^s$ such that $Y$ consists of all $I$-tuples extending a sequence in $Y\_0$. And we define that $Y$ has measure one for the extender if $Y\_0\in\mathcal{F}\_s$. That is, a set in the full product is large, if it contains a set with finite support that is large for the finite Fubini product on that finite set of indices.
The main point is that one can define ultrapowers by an extender using the finite-support functions on $I$, and then the extender ultrapower is the direct limit of the system of finite Fubini products. This conception of limits of ultrapowers is central in the large cardinal inner model theory, particularly in the case of strong cardinals, whose embeddings are realized by such extender measures.
| 4 | https://mathoverflow.net/users/1946 | 442378 | 178,504 |
https://mathoverflow.net/questions/442384 | 1 | I have a sequence of $p$-dimensional infinitely divisible random vectors $S\_n'$, such that $S\_n' \Longrightarrow X$, as $n \to \infty$.
Suppose the following assumptions
1. The characteristic functions are given by:
$$\varphi\_{S\_n'}(u)=\exp\left\{ \int\_{\mathbb R^p} \left[e^{iu'x} - 1 - i u'x \right] \, d\nu\_n \right\}, \quad \varphi\_{X}(u) = \exp\left\{ \frac{- u'\sigma u}{2} + \int\_{\mathbb R^p} \left[e^{iu'x} - 1 - iu'x \right] d\nu \right\} $$
($\nu\_n$ and $\nu$ are Lévy measures)
2. $E[S\_n']=\int\_{\mathbb R^p}x \,d\nu\_n=0$ and
\begin{equation}\label{0}\tag{0}
\sup\_n \int\_{\mathbb R^p} |x|^2 \, d\nu\_n(x) \leq C< \infty
\end{equation}
3. Let $\mathcal C\_\#$ be the class of continuous and bounded functions vanishing on a neighborhood of $0$. Then:
\begin{equation}\label{I}\tag{I}
\int f \, d\nu\_n \to \int f \, d\nu \quad (n \to \infty),\quad \forall f \in \mathcal C\_\#
\end{equation}
4. First, for any $\epsilon>0$, define the symmetric non-neg-definite matrix $\sigma\_{n,\epsilon}$ as:
\begin{equation}\label{II}\tag{II}
\langle u, \sigma\_{n,\epsilon}u \rangle := \int\_{|x|\leq \epsilon} \langle u ,x\rangle^2 \, d\nu\_n(x), \quad u \in \mathbb R^p
\end{equation}
Then:
\begin{equation}\label{III}\tag{III}
\lim\_{\epsilon \downarrow 0} \limsup\_{n \to \infty} \left| \langle u, \sigma\_{n,\epsilon}u \rangle - \langle u, \sigma u \rangle \right|=0
\end{equation}
(Where $\sigma$ appears in the characteristic function $\varphi\_X$, see hypothesis 1)
**Question:**
Since $\int\_{\mathbb R^p} x \, d\nu\_n =0$ for all $n$, I suspect that
\begin{equation}\label{IV}\tag{IV}
\int\_{\mathbb R^p} x \, d\nu = 0
\end{equation}
How to show this?
*Remarks:*
* Note that (\ref{I}) is almost like a weak convergence of measures. If I could show that (\ref{I}) holds for every continuous and bounded function, I would actually have that $\nu\_n \Longrightarrow \nu$ ($n \to \infty$). In this case, together with (\ref{0}), I could conclude (\ref{IV}) via a uniform integrability argument. Thus, defining $\mathcal C$ be the class of continuous and bounded functions, I think my question boils down to showing that:
\begin{equation}\label{V}\tag{V}
\int f \, d\nu\_n \to \int f \, d\nu \quad (n \to \infty), \quad \forall f \in \mathcal C
\end{equation}
I spent a few hours trying to solve this using hypothesis 1-4, but I couldn't. It may also be that there is another way of showing (\ref{IV}).
I appreciate any help.
**Update**
So far, Iosif Pinelis gave a counterexample. I apologize for the omission, but I forgot to write some assumptions related to $S\_n'$. I will put the updates here.
The $S\_n'$ arise as follows: let $(X\_{jn})\_{1\leq j \leq n}$, $X\_{jn} \sim \mu\_{jn}$, be a triangular array of $p$-dimensional random vectors (row independent) such that:
* $E X\_{jn}= \int\_{\mathbb R^p} x \, d \mu\_{jn}=0$
* $\lim\_{n \to \infty} \max\_{1\leq j \leq n} P(|X\_{jn}|> \epsilon)=0$, for all $\epsilon > 0$
* Defining $S\_n := \sum\_{j=1}^n X\_{jn}$, we have $var(S\_n):=\sum\_{j=1}^n \int\_{\mathbb R^p} |x|^2 \, d\mu\_{jn} \leq C < \infty$, for all $n \in \mathbb N$.
* $S\_n \Longrightarrow X$, as $n \to \infty$. (Here, $X$ is the same limit of the $S\_n'$ given above)
* $\nu\_n(E):= \sum\_{j=1}^n \int\_E d\mu\_{jn}, \quad E\, \,\hbox{ borelian set.}$
Definig first $Y\_{jn}:= [X\_{jn}]^{[1]}$ (This the compound Poisson radom variable i.e $Y\_{jn} \sim CP(\mu\_{jn}, 1)$ ), we define:
$$S\_n' := \sum\_{j=1}^n Y\_{jn}$$
It is easy to show that $E[S\_n']=E[S\_n]=0$ and $var[S\_n']=var[S\_n]$. By an argument of Accompanying Law (section 3.7 from the [Varadhan'lecture notes](https://www.math.nyu.edu/%7Evaradhan/course/PROB.ch3.pdf)), we have that $S\_n'$ is such that
$$S\_n' \Longrightarrow X, \quad (n \to \infty)$$
The assumptions 1-4 given above are true using the theorem 8.7, page 41, from the [Sato's book](https://zbmath.org/?q=an:0973.60001):
| https://mathoverflow.net/users/479236 | How to show that $\int x \,d\nu = 0$ using a pseudo-weak convergence of measures? | $\newcommand\de\delta$A counterexample is given by $p=1$, $\nu(dx):=|x|^{-5/2}\,1(0<|x|<1)\,dx$, and $\nu\_n(dx):=|x|^{-5/2}\,1(1/n<|x|<1)\,dx$.
---
The OP has added certain conditions. The only consequence of those additional conditions that matters in this context is that $\int d\nu\_n=n$ for all $n$ (so that $\mu\_{jn}:=\nu\_n/n$ be a probability measure for any $n$ and $j$). This extra condition on $\nu\_n$ is easy to satisfy by the following slight modification of the definition of the measure $\nu\_n$:
$$\nu\_n(dx):=|x|^{-5/2}\,1(\de\_n<|x|<1)\,dx,$$
where $\de\_n:=(1+\frac34\,n)^{-2/3}$ (so that $\de\_n\to0$ as $n\to\infty$).
| 2 | https://mathoverflow.net/users/36721 | 442388 | 178,507 |
https://mathoverflow.net/questions/442389 | 2 | I have a question about the p.d.f. of exponential family. Suppose $(X,\mathcal{F})$ is a measurable space and $\{F\_{\theta},\theta\in \Theta\}$ is a distribution family on $(X,\mathcal{F})$. When $\{F\_{\theta},\theta\in \Theta\}$ is dominated by a $\sigma$ finite measure $\mu$, $\{F\_{\theta},\theta\in \Theta\}$ is a exponential family if and only if
$$
\frac{d F\_\theta}{d\mu}=\exp\{\eta(\theta)^\top T(x)-\xi(\theta)\}h(x),\forall \theta\in\Theta,
$$
where $h(x)$ is non-negative measurable which is stated in many references like Lehman and Casella, Theory of Estimation.
It is also said that $h(x)$ is not important because it can be absorbed in $d\mu$(so we can even let $h(x)=1$). Let $\lambda(A)=\int\_Ah(x)d\mu,\forall A\in\mathcal{F}$,then
$$
\frac{d F\_\theta}{d\lambda}=\exp\{\eta(\theta)^\top T(x)-\xi(\theta)\}.
$$
**I am confused with the formula $\frac{d F\_\theta}{d\lambda}$ above**. I know that since $F\_\theta\ll \lambda\ll\mu$, we can derive that ($\mu$-a.e.)
$$
\begin{aligned}
\frac{d F\_\theta}{d\mu}&=\frac{d F\_\theta}{d\lambda}\frac{d \lambda}{d\mu},
\\
\exp\{\eta(\theta)^\top T(x)-\xi(\theta)\}h(x)&=\frac{d F\_\theta}{d\lambda}h(x).
\end{aligned}
$$
But we don't know wheather $h(x)>0$ $\mu-$ a.e. or not, how can we get
$$
\frac{d F\_\theta}{d\lambda}=\exp\{\eta(\theta)^\top T(x)-\xi(\theta)\}?
$$
Should we emphasize $h(x)>0$ in the definition? Or maybe we just consider the set $\{x\in X:h(x)>0\}$? If so, how to interpret that $h(x)$ can be set $1$? Does this problem correspond with the support of a distribution?
Any help would be appreciated.
| https://mathoverflow.net/users/500703 | p.d.f. of exponential family | $\newcommand\la\lambda$Let $F:=F\_\theta$ and $R:=\exp\{\eta(\theta)^\top T-\xi(\theta)\}$, so that
$$\frac{dF}{d\mu}=Rh,$$
which means that
$$\int g\,dF=\int gRh\,d\mu \tag{1}\label{1}$$
for all measurable functions $g$ such that either one (or, equivalently, both) of the latter two integrals exist (say in $[-\infty,\infty]$).
In a standard manner, we can show that the condition that $\la(A)=\int\_A h\,d\mu$ for all measurable $A$ is equivalent to the condition that
$$\int f\,d\la=\int fh\,d\mu \tag{2}\label{2}$$
for all measurable functions $f$ such that either one (or, equivalently, both) of the latter two integrals exist in $[-\infty,\infty]$.
Letting now $f=gR$ in \eqref{2}, from \eqref{1} we get
$$\int gR\,d\la=\int g\,dF$$
for all measurable functions $g$ such that either one (or, equivalently, both) of the latter two integrals exist in $[-\infty,\infty]$. This means
$$\frac{dF}{d\la}=R,$$
as desired.
---
(The set $\{x\colon h(x)>0\}$ plays no role in this consideration -- since we only multiply here, and never divide.)
| 3 | https://mathoverflow.net/users/36721 | 442390 | 178,508 |
https://mathoverflow.net/questions/442383 | 10 | [In a recent course in Bonn](https://people.mpim-bonn.mpg.de/scholze/SixFunctors.pdf), P. Scholze explains a formalization of a six-functor formalism due to L. Mann. In this axiomatization, three of the functors $f\_!,f^\*,\otimes$ are "constructed" (in the form of a lax monoidal functor) and then the other three are defined as right adjoints.
Now, in the derived category of holonomic D-modules over an algebraic variety in characteristic zero, we have four functors which are simple to construct:
1. $f\_+$: the usual direct image of D-modules, which is the analog of the functor $f\_\*$;
2. $f^!$: the exceptional inverse image, which plays the same role as the homonymous functor in other six-functor formalisms;
3. $\otimes^\mathsf{L}\_{\mathcal{O}}$: the left derived functor of the O-module tensor product;
4. $\mathrm{D}$: the "Verdier duality", which also works as in other six-functor formalisms.
Let me be precise: the functor $\otimes^\mathsf{L}\_{\mathcal{O}}$ is **not** one of the six functors. (For example, in a six-functor formalism one would expect that the inverse image functor is monoidal with respect to the tensor product. This does not happen here.) But we can make it work!
Since $f^!$ is defined as a shift of $\mathsf{L}f^\*$ (the O-module functor, which is also **not** one of the six-functors), and $\mathsf{L}f^\*$ is monoidal with respect to $\otimes^\mathsf{L}\_{\mathcal{O}}$, we can put $\otimes^!:=\otimes^\mathsf{L}\_{\mathcal{O}}[-\dim]$. Then $f^!$ is monoidal with respect to $\otimes^!$. Finally, as the usual inverse image of D-modules $f^+$ is defined as the Verdier dual of $f^!$, we can define $\otimes$ as the Verdier dual of $\otimes^!$.
This tensor product $\otimes$ has a right adjoint $\underline{\operatorname{Hom}}(-,-)=\mathrm{D}(-\otimes\mathrm{D}(-))$ and the functors $f\_+,f^!$ have left adjoints $f\_!,f^+$; constituting a full six-functor formalism. (Which satisfies every nice property imaginable. And makes the analogy between holonomic D-modules and wildly ramified perverse sheaves in positive characteristic very clear.)
My question is: **how could we use (or modify) L. Mann's foundations to establish this six-functor formalism?** If we already know everything about this six-functor formalism, I guess it should not be too bad to prove that it enters in Mann's axiomatization. But those foundations should give simpler proofs to known (or not) statements; not repose on them, right?
| https://mathoverflow.net/users/131975 | Can we use Mann's six-functor formalism with D-modules? | The six functor formalism applies to $D$-modules, but you need to extend the theory to possibly non-smooth schemes. For this, we see that hypersheaves on the site of pairs $(X,Z)$, with $Z$ a closed subscheme of a smooth scheme $X$, with maps the obvious commutative squares, with coverings those maps $(X,Z)\to (X',Z')$ inducing an étale covering $Z\to Z'$ form an $\infty$-category equivalent to the $\infty$-category of hypersheaves on the usual big étale site of our ground field of characteristic zero (but we could work with arithmetic $D$-modules à la Berthelot for fields of positive characteristic, using the work of Caro). Sending $(X,Z)$ to $D$-modules over $X$ that are supported on $Z$, we get a hypersheaf of symmetric monioidal stable $\infty$-categories. What precedes says that there is a unique (hyper)sheaf of symmetric monoidal stable $\infty$-categories on the big étale that coincides with usual $D$-modules on smooth schemes (the proof is an interpretation of Kashiwara's theorem). This says how to define $\otimes$ and $f^\*$ for any map of schemes of finite type over the ground field $f:Z\to Z'$. To define $f\_!$, we reduce by descent as above to the case where $f$ is induced by a map of pairs $(X,Z)\to (X',Z')$, in which case we use the $g\_!$ induced by $g:X\to X'$ restricted to $D$-modules supported on $Z$ and $Z'$ respectively.
| 3 | https://mathoverflow.net/users/1017 | 442419 | 178,519 |
https://mathoverflow.net/questions/442397 | 2 | I am trying to lower bound the following function, $n \ge 3$ is a natural number:
$$l(n):=\frac{\log(n)}{\log(n)-\frac{1}{n}(\tau(n)\log(\tau(n))+(n-\tau(n))\log(n-\tau(n)))}$$
where $\tau(n)$ counts the number of divisors. Is this function exponential or polynomial in terms of $\log(n)$?
Thanks for your help.
| https://mathoverflow.net/users/165920 | Lower bound on a function of the number of divisors? | Tried to write a comment but got too long, not sure if correct or what you were looking for but hopefully is useful otherwise let me know and I will delete it.
What I did was to rewrite the denominator as follows
$\ln(n)-\frac{1}{n}(\tau(n)\ln(\tau(n))+(n-\tau(n))\ln(n-\tau(n)))\\
=\ln(2)+\ln(n/2)-((\tau(n)/n)\ln(\tau(n))+(1-\tau(n)/n)\ln(n-\tau(n)))\\
=\ln(2)+(\tau(n)/n)\ln(n/2)+(1-\tau(n)/n)\ln(n/2)+(\tau(n)/n)\ln(\frac{1}{\tau(n)})+(1-\tau(n)/n)\ln(\frac{1}{n-\tau(n)})\\
=\ln(2)+(\tau(n)/n)\ln(\frac{1/2}{\tau(n)/n})+(1-\tau(n)/n)\ln(\frac{1/2}{1-\tau(n)/n})\\
=\ln(2)+\sum\_{i=1}^2 p\_i\ln(q\_i/p\_i)\\
\leq\ln(2)
$
where the last inequality is because we have that $p\_1=\tau(n)/n, p\_2=1-\tau(n)/n,q\_1=q\_2=1/2$ then $\sum\_{i=1}^2 p\_i=1$ and $\sum\_{i=1}^2 q\_i=1$ and we can apply [Gibbs inequality](https://en.wikipedia.org/wiki/Gibbs%27_inequality) $\sum\_{i=1}^n p\_i \log \frac{q\_i}{p\_i} \leq 0\qquad (\sum\_{i=1}^n p\_i \log \frac{p\_i}{q\_i} \geq 0)$
Therefore we get that
$l(n):=\frac{\log(n)}{\log(n)-\frac{1}{n}(\tau(n)\log(\tau(n))+(n-\tau(n))\log(n-\tau(n)))}\geq\frac{\log(n)}{\log(2)}$
| 3 | https://mathoverflow.net/users/142708 | 442425 | 178,521 |
https://mathoverflow.net/questions/435315 | 1 | Theorem 1 of [LS] [Liebeck and Seitz - On the subgroup structure of exceptional groups](https://doi.org/10.1090/S0002-9947-98-02121-7) says:
>
> Let $X = X(q)$ be a quasisimple group of Lie type in characteristic $p$, and suppose that $X < G$, where $G$ is a simple adjoint algebraic group of exceptional type, also in characteristic $p$. Assume that [$q$ is large]. Then … there is a closed connected subgroup $\overline X$ of $G$ containing $X$, such that every $X$-invariant subspace of the Lie algebra $L(G)$ is also $\overline X$-invariant ….
>
>
>
[LS] emphasises that the condition is only on $q$, not on $p$; that is, it is OK for the characteristic $p$ to be small, as long as we take a large power of it (usually $q > 9$). Subgroups $X$ as described are called generic. The much more recent [Litterick - On non-generic finite subgroups of exceptional algebraic groups](https://doi.org/10.1090/memo/1207) justifies the restriction to non-generic groups by saying that the generic case is well understood, and points to [LS], which I take indirectly to be indicating that [LS] remains the state of the art.
What, if anything, is known about generic subgroups, in the above sense, for small $q$, especially in the spirit of the quoted Theorem 1?
| https://mathoverflow.net/users/2383 | Generic finite subgroups, associated to small finite fields, of reductive algebraic groups | The state of the art for subgroups $\operatorname{PSL}\_2(q)$, which constitute most of the cases where $q > 9$ in [LS], is now given by David Craven's Memoir [Maximal $\operatorname{PSL}\_2$ Subgroups of Exceptional Groups of Lie Type](https://doi.org/10.1090/memo/1355) ([arxiv version](https://arxiv.org/abs/1610.07469)). For maximal subgroups, complete information for groups of type $F\_4$ and $E\_6$ is given in another preprint [The Maximal Subgroups of the Exceptional Groups $F\_4(q)$, $E\_6(q)$ and $^2E\_6(q)$ and Related Almost Simple Groups](https://arxiv.org/abs/2103.04869) of Craven. The case where $X(q)$ has untwisted rank more than $\frac{1}{2}\operatorname{rank}(G)$ (but arbitrary $q$) is given by Liebeck and Seitz in [Maximal Subgroups of Large Rank in Exceptional Groups of Lie Type](https://doi.org/10.1112/S0024610704006179).
In all, this leaves a very small number of subgroup types; I'm not sure anyone has worried about the remaining cases explicitly.
| 3 | https://mathoverflow.net/users/3935 | 442434 | 178,525 |
https://mathoverflow.net/questions/442222 | 5 | Let $G=(V, E)$ be an acyclic digraph (DAG) with all in- and out-degrees at most $k$. Is it true that the edges of $G$ may be always colored properly in $2k$ colors?
In the discussion of [this](https://mathoverflow.net/a/442170/4312) question it is proved (for $k=2$, but the proofs work verbatim for other values of $k$) that $G$ has $|E|/(2k)$ disjoint edges.
| https://mathoverflow.net/users/4312 | Chromatic index of an acyclic digraph | Here is a solution (obtained while discussing the question with my colleague András Sebő).
Consider the undirected graph $H$ obtained from $G$ as follows: the partite sets of $H$ are two copies $V\_1$ and $V\_2$ of $V(G)$, and for each arc $(u,v)$ in $G$ we add an edge in $H$ between the copy of $u$ in $V\_1$ and the copy of $v$ in $V\_2$.
The graph $H$ is bipartite and has maximum degree $k$. By [Kőnig's theorem](https://en.wikipedia.org/wiki/K%C5%91nig%27s_theorem_(graph_theory)) $H$ has a proper $k$-edge-coloring, that is a partition of its edges into $k$ matchings $M\_1,\ldots,M\_k$. The edges of $H$ are in bijection with the arcs of $G$, and if we consider the image of each matching $M\_i$ in $G$ we obtain a subgraph of $G$ with in-degree and out-degree at most 1 (that is, a disjoint union of directed cycles and directed paths). As $G$ is acyclic, the image of each matching $M\_i$ in $G$ is a union of paths, so the edges of $G$ can be partitioned into $k$ subgraphs, each being a union of paths. As a union of paths is 2-edge-colorable, $G$ has a partition of its edges into $k$ 2-edge-colorable subgraphs, and is therefore $2k$-edge-colorable, as desired.
| 3 | https://mathoverflow.net/users/45855 | 442440 | 178,527 |
https://mathoverflow.net/questions/442416 | 1 | For a long time, I had a false belief that the space/stack $\text{Coh}^{tf}\_{c\_1,c\_2}S$ of torsion-free sheaves $\mathcal{E}$ on a smooth algebraic surface $S$ was not connected, since if you take its map into its double dual
$$0\ \to \ \mathcal{E}\ \to\ \mathcal{E}^{\vee\vee}\ \to \ \mathcal{Q}\ \to\ 0$$
the length of the torsion sheaf $\mathcal{Q}$ gives another invariant labelling connected components.
---
What are some "typical" examples of (say, flat one parameter) families of torsion-free sheaves where the torsion quotient jumps in length?
| https://mathoverflow.net/users/119012 | Families of torsion-free sheaves whose length jumps | Consider the sheaf $F$ on $\mathbb{P}^2 \times \mathbb{A}^1$ defined from the exact sequence
$$
0 \to
F \longrightarrow
\mathcal{O}(-1)^{\oplus 3} \stackrel{(x,y,tz)}\longrightarrow
\mathcal{O} \longrightarrow
\mathcal{O}\_P \to
0,
$$
where $(x,y,z)$ are the homogeneous coordinates on $\mathbb{P}^2$, $t$ is the coordinate on $\mathbb{A}^1$, and $P$ is the point $((0,0,1),0)$.
On the one hand, if one restricts to the fiber over $0 \ne t \in \mathbb{A}^1$, one obtains
$$
F\_t \cong \Omega\_{\mathbb{P}^2},
$$
in particular, this sheaf is locally free.
On the other hand, over $0 \in \mathbb{A}^1$, one obtains an exact sequence
$$
0 \to F\_0 \to \mathcal{O}(-2) \oplus \mathcal{O}(-1) \to \mathcal{O}\_P \to 0,
$$
hence this sheaf is not locally free.
| 1 | https://mathoverflow.net/users/4428 | 442442 | 178,528 |
https://mathoverflow.net/questions/442456 | 4 | Let $X$ be an irreducible smooth projective variety of pure dimension $d$ over the complex numbers and $Z\subset X\times X$ a codimension $d$ irreducible smooth closed subvariety.
>
> Is there a smooth hyperplane section $H$ on $X$ such that $H\times X$ contains a cycle in the rational equivalence class of $Z$?
>
>
>
I expect the answer to be "no". Is it "yes" if one drops smoothness of $H$?
| https://mathoverflow.net/users/497064 | Cycles contained in ample enough hypersurfaces | Let $X:=\Sigma\_g$ be a smooth curve of genus $g\geq 2$ and $Z=\Delta$, where $\Delta \subset \Sigma\_g \times \Sigma\_g$ is the diagonal.
Then $\Delta$ is not rationally equivalent to a union of fibres of $p\_1 \colon \Sigma\_g \times \Sigma\_g \to \Sigma\_g$, for instance because $\Delta^2=2-2g <0$.
| 8 | https://mathoverflow.net/users/7460 | 442458 | 178,533 |
https://mathoverflow.net/questions/442412 | 1 | Let $L\_1$, $L\_2$ be two subspaces in $\mathbb{R}^n$ and $\dim(L\_1) = \dim(L\_2) = s<n$. Let $$0 \leq \theta\_1 \leq \dotsb\leq \theta\_{s} \leq \pi/2$$ be the principal angles between $L\_1, L\_2$. Let there be two unit vectors $u\_1\in L\_1$ and $u\_2 \in L\_2$ and the principal angle between $\operatorname{span}\{u\_1\}$ and $\operatorname{span}\{u\_2\}$ is $\alpha \in [0,\pi/2]$. Is there any known upper bound on $\alpha$ in terms of $\theta\_{s}$?
Here is my try: Let $U\_1, U\_2 \in \mathbb{R}^{n \times s}$ be two orthogonal matrices such that their columns are orthonormal bases of $L\_1$ and $L\_2$ respectively. Now any unit vector in $L\_j$ is essentially of the form $U\_j \alpha\_j$ where $\lVert \alpha\_j\rVert\_2 =1 $ for $j \in \{1,2\}$. Also, the singular values of $U\_1^\top U\_2$ are $\{\cos \theta\_1, \dotsc, \cos \theta\_s\}$. Now, I don't have any idea how to proceed any further. Any help will be appreciated.
| https://mathoverflow.net/users/151115 | Principal angles between subspaces and angle between unit vectors | $\newcommand\si\sigma\newcommand\al\alpha\newcommand\be\beta$Let $U:=L\_1$ and $V:=L\_2$. Without loss of generality, $s=\dim U=\dim V\ge1$. By the section [Angles between subspaces](https://en.wikipedia.org/wiki/Angles_between_flats#Angles_between_subspaces) of the Wikipedia article, there are orthonormal bases $(a\_1,\dots,a\_s)$ and $(b\_1,\dots,b\_s)$ of $U$ and $V$, respectively; nonnegative integers $\si$ and $\al$; and $\be\_1,\dots,\be\_\al$ in the interval $(0,\pi/2)$ such that for any $u\in U$ and $v\in V$ we have
$$u\cdot v=\sum\_{i=1}^\si u^i v^i+\sum\_{i=\si+1}^{\si+\al} u^i v^i\cos\be\_i,$$
where $u\cdot v$ is the dot product of $u$ and $v$; the $u^i$'s are the coordinates of $u$ in the basis $(a\_1,\dots,a\_s)$; and the $v^i$'s are the coordinates of $v$ in the basis $(b\_1,\dots,b\_s)$.
So, if $\si+\al<s$ or $\si+\al>1$, then $u\cdot v=0$ and hence $\angle(u,v)=\pi/2$ for some nonzero $u\in U$ and $v\in V$.
Otherwise, if $\si+\al\ge s$ and $\si+\al\le1$, then $s=1$.
Thus, excluding the trivial case $s=1$, the maximal angle between $\text{span}\{u\}$ and $\text{span}\{v\}$ for nonzero vectors $u\in U$ and $v\in V$ is always $\pi/2$.
---
This conclusion can also be obtained more elementarily, as follows. Suppose that $s\ge2$ and let $(a\_1,\dots,a\_s)$ be any basis of $U$. Take any nonzero $v\in V$. Then for for some nonzero $(u^1,\dots,u^s)\in\mathbb R^s$ and $u:=\sum\_{i=1}^s u^i a\_i$ we have $u\ne0$ and
$u\cdot v=\sum\_{i=1}^s u^i(a\_i\cdot v)=0$ and hence $\angle(u,v)=\pi/2$. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 442460 | 178,534 |
https://mathoverflow.net/questions/442449 | 1 | Random variable $X\geq 0$ and its variance exists. How to prove
$$\mathbb{P}(X\geq(1-t)\mathbb{E}(X))\geq \frac{t^2\mathbb{E}(X)^2}{\mathbb{E}(X^2)}\enspace\text{for}\enspace t\in(0,1]$$
$$\mathbb{E}(\exp(-\lambda(X-\mathbb{E}(X))))\geq\exp(\lambda^2\mathbb{E}(X^2)/2)\enspace\text{for}\enspace \lambda\geq0$$
$$\mathbb{P}(X\leq(1-t)\mathbb{E}(X))\leq \exp(-\frac{t^2\mathbb{E}(X)^2}{2\mathbb{E}(X^2)})\enspace\text{for}\enspace t\in(0,1]$$
Thanks!
---
Update: Thanks to [Alapan Das](https://mathoverflow.net/users/156029/alapan-das) and [Iosif Pinelis](https://mathoverflow.net/users/36721/iosif-pinelis) I found that the first inequality is Paley-Zygumund Inequality. It can be transformed to the shape similar to the third equation i.e.
$$\mathbb{P}(X\leq(1-t)\mathbb{E}(X))\leq 1-\frac{t^2\mathbb{E}(X)^2}{\mathbb{E}(X^2)}\enspace\text{for}\enspace t\in(0,1]$$
but the third one is a tighter bound. I wonder how to improve it.
| https://mathoverflow.net/users/500967 | Prove inequality on expectation | As Alapan Das noted, the first inequality for $t\in[0,1]$ is the [Paley–Zygmund inequality](https://en.m.wikipedia.org/wiki/Paley%E2%80%93Zygmund_inequality). For $t<0$ or $t>1$, the first inequality is false if e.g. $X=1$.
The second inequality is false for all real $\lambda\ne0$ if e.g. $X=1$.
The third inequality is false for all real $t>0$ if e.g. $X=1$. It is also false for all real $t\le-2$ if e.g. $P(X=1)=\frac1{1-t}=1-P(X=0)$.
---
The OP has changed the 3rd inequality from
$$P(X\ge(1-t)EX)\ge\exp-\frac{t^2\,(EX)^2}{2EX^2}$$
for an unspecified $t$ to
$$p\_3:=P(X\le(1-t)EX)
\le\exp-\frac{t^2\,(EX)^2}{2EX^2}\tag{3a}\label{3a}$$
for $t\in(0,1]$,
at the same time asserting that the upper bound in \eqref{3a} on $p\_3$ is "tighter" than the Paley–Zygmund bound
$$1-\frac{t^2\,(EX)^2}{EX^2}$$
on $p\_3$. In fact, it is vice versa:
$$1-\frac{t^2\,(EX)^2}{EX^2}<\exp-\frac{t^2\,(EX)^2}{EX^2}
<\exp-\frac{t^2\,(EX)^2}{2EX^2}$$
(if $EX^2\ne0$).
| 2 | https://mathoverflow.net/users/36721 | 442462 | 178,535 |
https://mathoverflow.net/questions/442451 | 2 | The following is a heuristic for the situation where a decision algorithm or a human, might solve a problem by reducing the entropy of the "search space" at every computation step $i$:
Let $X\_{i+1} \subset X\_i$ for $i = 1,\cdots,t-1$ be finite subsets contained all in a finite set $X\_1$.
We assume that there are given functions $f\_i: X\_i \rightarrow X\_{i+1}$ which are surjective and also we assume that $|X\_{i}| > |X\_{i+1}|$.
In each step $i=1,\cdots,t-1$ we might define an entropy ([the idea is borrowed from David Ellerman](http://philsci-archive.pitt.edu/13213/1/Logic-to-information-theory3.pdf)):
$$H\_{f\_i}:=H\_{i} := - \sum\_{x \in X\_{i+1}} \frac{|f\_i^{-1}(x)|}{|X\_i|} \log\_2\left[\frac{|f\_i^{-1}(x)|}{|X\_i|}\right].$$
We can also consider the function $\hat{f} = f\_t \circ f\_{t-1} \circ \cdots f\_2 \circ f\_1$ which maps from $X\_1$ to $X\_t$.
and consider the corresponding entropy:
$$\hat{H} = H\_{\hat{f}}$$
My question is, if it can be proved that: $H\_i \ge \hat{H}$ and $H\_i \ge H\_{i+1}$?
I did a few small [experiments in Sagemath](https://sagecell.sagemath.org/?z=eJzdU8tugzAQvCPxD5ZysSmkIX0qau703AsR4kCDHbkFg2yIkr_vrjENNPmC-mDZs7O7s2OQddvojqi-bs-kMES1vldyQaT56PUX33fyyKkI03DHNr5HYFUp2ZKKK5qyARDv6ghQKfcdzXIHLsi-qdu-40SqI9eGE9ErqNYoV0Y0mpwgSFIH4DpDHUFP7IJIASCwsMnym58NZRP-2D4758uiLGeZvDL8NhWaGA5aT6NYzbteK1JUFb3qZoVacMfGOpeqg1mJ-PcWJYBEpq9phnM5Fruv0kC1y6o50D_w1DcAHGk9NzyZ9_a9wU5R0nIcwTGrov4sC3LakMMeZghLhuSFMgR0ZevgIXgKXoI4Du0R93Cd-x7EbTh4BSh4xg2DwSNua0sBEnCUyVa5e9DhGsUQFGg2qvG91L7fUZpGG6oA2E0BZCTo0e-n4HutlqqjSWKlJhKCK6gIrkh0RRfqwK1ryrAoHud1YmQ-XJ0YeRc7YKIIr1eqELxShuCg5sZvPT6wbOO5_klWiNEQOBEe3rZ2qCELcwD7ASWMFSw=&lang=sage&interacts=eJyLjgUAARUAuQ==) with divisor sets, but although it seems that the answer is yes, I have no idea if it is true or how it can be proved.
Thanks for your help.
| https://mathoverflow.net/users/165920 | Entropy reduction? | Suppose $X\_1=\{1,\dots,m+n\}$, where $m,n\to\infty$ and $\frac mn\,\log\_2 n\to0$. Suppose that $f\_1(x)=x$ for $x\in\{1,\dots,m\}$ and $f\_1(x)=m+1$ for $x\in\{m+1,\dots,m+n\}$, so that $X\_2=\{1,\dots,m+1\}$. Suppose also that $f\_2(x)=x$ for $x\in\{1,\dots,m-1\}$ and $f\_2(x)=m$ for $x\in\{m,m+1\}$. Then
$$H\_1=\frac n{n+m}\,\log\_2\frac{n+m}n
+m\frac 1{n+m}\,\log\_2\frac{n+m}1\to0,$$
whereas
$$H\_2=\frac2{m+1}\,\log\_2\frac{m+1}2
+(m-1)\frac 1{m+1}\,\log\_2\frac{m+1}1\to\infty,$$
so that $H\_i\not\ge H\_{i+1}$ for $i=1$.
---
However, letting $\hat f\_s:=f\_s\circ \cdots\circ f\_1$, we of course will have
$$H\_{\hat f\_s}\ge H\_{\hat f\_{s+1}} $$
for all $s=1,2,\dots$.
This follows because $l(p):=-p\log\_2 p$ (with $l(0):=0$) is concave in $p\in[0,1]$, so that $l(p+q)\le l(p)+l(q)$ if $0\le q\le1-p\le1$.
| 3 | https://mathoverflow.net/users/36721 | 442463 | 178,536 |
https://mathoverflow.net/questions/425318 | 5 | It is classical that Euclidean normal currents are dense in the space of all currents.
This can be achieved through mollification.
What I want to know if this is still true for metric currents.
In particular I am interested if the space of Normal 1-currents is always dense in the space of all 1-currents for any separable metric space.
Even more specifically I want to know if it is possible for a metric space to have metric currents while having no normal metric currents.
It would make sense that a positive density result could be achieved for nicely supported measures in Banach spaces, but I cannot be sure how to proceed for badly behaved metric spaces since mollifications would lead outside of the space.
| https://mathoverflow.net/users/56713 | Are normal metric currents dense in the space of all metric currents? | The answer is no, and it is pretty trivial apparently.
Let us look at $[0,1]\subseteq\mathbb R$ and define on it the simplest current
$$
T(fd\pi)=\int\_0^1f(t)\pi'(t)dt.
$$
Now we enumerate the rational numbers on it $\{q\_n\}\_{n\in\mathbb N}\subseteq[0,1]$ and given any positive $1>\varepsilon>0$ and any positive element $r\in\ell^1$ such that
$$
\|r\|\_1\leq\varepsilon/2
$$
so we can define the compact set
$$
K=[0,1]\setminus\bigcup\_n B(q\_n,r\_n).
$$
Since we have that $K$ is completely disconnected and $\mathcal L^1(K)>1-\varepsilon$ we have that
$$
T\_{|K}(fd\pi)=\int\_K f(t)\pi'(t)dt
$$
is a current but it is not normal anymore.
We can conclude that $K$ as a metric space has non trivial currents but cannot have normal currents since if it had any they would be supported on curves, and since $K$ is completely disconnected, that cannot happen.
| 0 | https://mathoverflow.net/users/56713 | 442470 | 178,538 |
https://mathoverflow.net/questions/442472 | 6 | Given a set $S$, a tight apartness relation on $S$ is a relation $\#$ which is tight, irreflexive, symmetric, and a comparison, or more specifically, a relation $\#$ such that
* for all elements $a \in S$ and $b \in S$, $a = b$ if and only if $\neg a \# b$
* for all elements $a \in S$, $\neg a \# a$
* for all elements $a \in S$ and $b \in S$, $a \# b$ implies that $b \# a$
* for all elements $a \in S$, $b \in S$, and $c \in S$, $a \# c$ implies that $a \# b$ or $b \# c$.
Assume that the category of sets is an elementary topos. This means that it contains a subobject classifier $\Omega$.
If the category of sets is Boolean, the subobject classifier $\Omega$ automatically has a tight apartness relation given by negation of equality, because in such a case, $\Omega$ is just the set with two elements and thus has decidable equality.
Is the converse of the above statement true? Namely, if $\Omega$ has a tight apartness relation $\#$, then is the category of sets Boolean?
| https://mathoverflow.net/users/483446 | Does a tight apartness relation on a subobject classifier imply the elementary topos is Boolean? | Yes. The tightness axiom, $(a=b)\iff \neg (a\#b)$, implies
$$\neg\neg(a=b) \iff \neg\neg\neg(a\#b) \iff \neg(a\#b) \iff (a=b).$$
Taking $b=\top$, we find that $\neg\neg a \iff a$ for all $a: \Omega$. This is the law of double negation, which is equivalent to excluded middle.
| 9 | https://mathoverflow.net/users/49 | 442476 | 178,539 |
https://mathoverflow.net/questions/442468 | 6 | An elliptic curve is (for the purpose of this question) a cubic algebraic curve defined by an equation (short Weierstrass equation) of the form
$$\displaystyle E\_{a,b} : y^2 = x^3 + ax + b, a, b \in \mathbb{Z}, 4a^3 + 27b^2 \ne 0.$$
The elliptic curves with $a = 0$ are called *Mordell curves*.
Do we have examples of Mordell curves with large algebraic rank, say exceeding 10?
| https://mathoverflow.net/users/10898 | Mordell curves with large rank | As far as I know, this record is still held by a 3-isogenous pair of
Mordell curves of rank $\bf 17$ that I found and announced in February 2016.
(This superseded a rank-16 pair from earlier that month, and curves of ranks
13, 14, 15 from October 2009.)
The curves $y^2 = x^3 + b$ and $y^2 = x^3 - 27 b$
are always related by an isogeny of degree $3$,
and in particular have the same rank.
The record $b,-27b$ have
$$ b = -908800736629952526116772283648363 $$
(in factored form, $-2195745961 \cdot 413891567044514092637683$).
The curve $y^2 = x^3 - 27 b$ has $17$ independent points
```
[-110315760690, 152299457785937151],
[-218829008658, 118569576333381183],
[194693247690, 178654854781822599],
[-12083686365, 156639252691623474],
[179588218407, 174154202398188288],
[660796972800, 559532270810391651],
[481938369495, 369425010854453724],
[532637728899, 419104420151289750],
[891937317975, 856808203106532276],
[1556910033324, 1948958451538253955],
[1369152212199, 1609695603071293320],
[-249954149276, 94452185380426435],
[527526224524, 413931980240076925],
[2095375244992, 3037184017947911267],
[3020920353232, 5252935870900542563],
[45908680009155, 311058636438867847974],
[209109621212430, 3023855428577131273599].
```
| 13 | https://mathoverflow.net/users/14830 | 442487 | 178,544 |
https://mathoverflow.net/questions/375991 | 2 | In David Mumford's book Algebraic Geometry I, Complex Projective Varieties the
proof of (3.25) **Specialization principle** on page 53 contains an argument
I not understand.
General assumptions: all our varieties are over $\mathbb{C}$. The statement is:
>
> (3.25) **Specialization principle.** Let $Z \subset \mathbb{P}^n \times \mathbb{P}^m $
> be $r$-dimensionally subvariety and $X =p\_1(Z) \subset \mathbb{P}^n$ for
> $p\_1: \mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^n $. Suppose
> $\operatorname{dim} X=r $ and let
>
>
> $$\phi= \text{ res } p\_1: Z \to X $$
>
>
> is almost everywhere (= on an open set) finite to one. [... ]
>
>
> Then the map
>
>
> $$ F: X\_1 \to \mathbb{N}, x \mapsto \# \phi^{-1}(x)$$
>
>
> where
> $X\_1 := \{x \in X \ \vert \ X \text{ smooth at } x \text{ and } \phi^{-1}(x)
> \text{ finite } \}$ is *lower semi-continuous* in the *Zariski* topology on $X\_1$.
>
>
>
Lower semi-continuous means that for every $n \in \mathbb{N}$ the set
$\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is closed.
The first part of the proof shows that $ F: X\_1 \to \mathbb{N}$ is
*lower semi-continuous* in the *classical* topology, recall a smooth complex
variety can be canonically endowed with classical analytical topology
considering it as a complex manifold.
Mumford remarks that we can observe that for every $n \in \mathbb{N}$ the set
$\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is *constructible*, ie a union of finite
number of locally closed sets in Zariski topology.
Then Mumford claims that this implies that $F$ is
*lower semi-continuous* in the *Zariski* topology.
That's the point I not understand. Does anybody see
why the implication $F$ lower semi-continuous with resp classical topology
and that $\{x \in X \ \vert \ \# \phi^{-1}(x) \ge n \}$ is *constructible* imply
$F$ lower semi-continuous with resp Zariski topology?
| https://mathoverflow.net/users/108274 | Comparison of classical and Zariski topologies with constructible sets | First, for lower semi-continuity, we should consider sublevel set, i.e.
$$\{x\in X:\ \#\phi^{-1}(x)\leq n\}$$
Now, since $\#\phi^{-1}(x)$ is a constructible function, the above subset, denoted as A, satisfies that
\begin{equation}
A=\bigcup\_{i=1}^N (Z\_{i,1}-Z\_{i,2})
\end{equation}
where $Z\_{i,j}$'s are all Zariski closed subsets. Also, from the lower semi-continuity of $\#\phi^{-1}(x)$ in the classical topology, A is closed in the classical topology. Taking the classical closure of the above equation on $A$, we get
$$A=\bigcup\_{i=1}^N Z\_{i,1}$$
which is Zariski closed. This shows that $\#\phi^{-1}(x)$ is also lower semi-continuous in the Zariski topology.
| 1 | https://mathoverflow.net/users/134247 | 442489 | 178,545 |
https://mathoverflow.net/questions/442249 | 3 | Let $A$ be any set and $S$ be a subset of $A^2$ such that sets $\{a\in A\mid (x,a)\in S\}\mathbin\triangle\{a\in A\mid (y,a)\in S\}$ and $\{a\in A\mid (a,x)\in S\}\mathbin\triangle\{a\in A\mid (a,y)\in S\}$ are finite for any $x,y\in A$. What can be the set $S$?
There are two simple examples of the set $S$, when sets $\{a\in A\mid (x,a)\in S\}$ and $\{a\in A\mid (a,y)\in S\}$ are finite (cofinite) for any $x,y\in A$. If $A=\mathbb{Z}$, then $S$ can be $\{(a,b)\in\mathbb{Z}^2\mid a < b\}$. Are there any other examples?
| https://mathoverflow.net/users/144883 | An almost uniform subset of the Cartesian square | At first, let us introduce some relevant definitions.
**Definition.** A $S\subseteq X\times Y$ is called
$\bullet$ *horizontally finite* in $X\times Y$ if for every $y\in Y$ the set $\{x\in X:(x,y)\in S\}$ is finite;
$\bullet$ *horizontally cofinite* in $X\times Y$ if for every $y\in Y$ the set $\{x\in X:(x,y)\notin S\}$ is finite;
$\bullet$ *vertically finite* in $X\times Y$ if for every $x\in X$ the set $\{y\in Y:(x,y)\in S\}$ is finite;
$\bullet$ *vertically cofinite* in $X\times Y$ if for every $x\in X$ the set $\{y\in X:(x,y)\notin S\}$ is finite.
**Proposition 1.** Is a set $S\subseteq X\times Y$ is horizontally cofinite and vertically finite, then either $X$ is finite or $Y$ is finite or else $X$ and $Y$ are countable.
*Proof.* Assume that the sets $X$ and $Y$ are infinite. Choose any countably infinite set $C\subseteq Y$. Since $S$ is horizontally cofinite, for every $y\in C$ the set $S^{-1}(y)=\{x\in X:(x,y)\in S\}$ is cofinite in $X$. Assuming that $X$ is uncountable, we can find an element $x\in \bigcap\_{y\in C}S^{-1}(y)$. For this element the set $S(x)=\{y\in Y:(x,y)\in S\}$ contains $C$ and hence is infinite, which contradicts the vertical finiteness of $S$. This contradiction shows that the set $X$ is countable. Then the set $\bigcup\_{x\in X}S(x)$ is countable too. Assuming that $Y$ is uncountable, we can find a point $y\in Y\setminus\bigcup\_{x\in X}S(x)$, for which the set $S^{-1}(y)=\emptyset$ is not cofinite in $X$. This contradiction shows that the set $Y$ is countable. $\square$
Now take any set $S\subseteq A\times A$ such that all horizontal sections of $S$ are almost equal and all vertical sections of $S$ are almost equal. The almost equality $X=^\* Y$ of two sets $X,Y$ means that $X\Delta Y$ is finite. This implies that there exist some sets $X,Y\subseteq A$ such that
$\{a\in A:(x,a)\in S\}=^\*Y$ and $\{a\in A:(a,y)\in S\}=^\*X$ for every $x,y\in A$. Moreover, we can assume that
$\bullet$ $X=\emptyset$ if $X$ is finite;
$\bullet$ $X=A$ if $X\setminus A$ is finite;
$\bullet$ $Y=\emptyset$ if $Y$ is finite;
$\bullet$ $Y=A$ if $X\setminus A$ is finite.
Now to understand a possible structure of $S$ we should analyze the structure of the intersections $S\_{11}=S\cap (X\times Y)$, $S\_{01}=S\cap ((A\setminus X)\times Y)$, $S\_{10}=S\cap (X\times (A\setminus Y))$ and $S\_{00}=S\cap ((A\setminus X)\times(A\setminus Y))$.
The set $S$ satisfies the condition from the problem if and only if it satisfies the following four conditions:
$00)$ The set $S\_{00}$ is horizontally finite and vertically finite in $(A\setminus X)\times (A\setminus Y)$;
$01)$ The set $S\_{01}$ is horizontally finite and vertically cofinite in $(A\setminus X)\times Y$;
$10)$ The set $S\_{10}$ is horizontally cofinite and vertically finite in $X\times (A\setminus Y)$;
$11)$ The set $S\_{11}$ is horizontally cofinite and vertically cofinite in $X\times Y$.
**Proposition 2.** If the set $A$ is uncountable, then $X=Y\in\{\emptyset,A\}$.
*Proof.* First we show that $X\in\{\emptyset,A\}$. Assuming that $X\notin\{\emptyset,A\}$, we conclude that the sets $X$ and $A\setminus X$ are infinite, by the choice of the set $X$. Since $A$ is uncountable, either $X$ or $A\setminus X$ is uncountable. If $X$ is uncountable, then the condition (10) and Proposition 1 ensure that the set $A\setminus Y$ is finite and hence $Y$ is uncountable. Applying Proposition 1 to the condition (01), we conclude that the set $A\setminus X$ is finite, which contradicts our assumption. If $A\setminus X$ is uncountable, then we can apply Proposition to the condition (01) and conclude that $Y$ is finite and hence $A\setminus Y$ is uncountable. In this case we can apply Proposition 1 to the condition (10) and conclude that the set $X$ is finite, which contradicts our assumption. This contradiction shows that $X\in\{\emptyset,A\}$.
By analogy we can show that $Y\in\{\emptyset,A\}$.
Assuming that $X\ne Y$, we conclude that either $X=\emptyset$ and $Y=A$ or else $X=A$ and $Y=\emptyset$.
First we consider case $X=A$ and $Y=\emptyset$. In this case the set $S$ is
horizontally cofinite and vertically finite in $A\times A$. By Proposition 1, $A$ is finite or countable, which is not the case. By analogy we can derive a constradiction assuming that $X=\emptyset$ and $Y=A$. This contradiction shows that $X=Y\in\{\emptyset,A\}$ and completes the proof. $\square$
It remains to analyze the structure of the set $S$ in case of countably infinite set $A$, which can be identified with $\omega$.
First observe that every set $S\subseteq X\times Y$ can be thought as a multivalued function assigning to each $x\in X$ the set $S(x)=\{y\in Y:(x,y)\in S\}$.
Under such description, sets $S\subseteq X\times Y$ which are horizontally finite and vertically finite correspond to finite-valued functions $S:X\multimap Y$ whose inverse $S^{-1}:Y\multimap X$ are also finite-valued.
In particular, $S\subseteq\omega\times\omega$ is horizontally and vertically finite if and only if there exist two non-decreasing unbounded functions $f,F:\omega\to\omega$ such that $S(x)\subseteq[f(x),F(x)]$ for all $x\in\omega$.
On the other hand, a set $S\subseteq\omega\times\omega$ is vertically finite and horizontally cofinite if and only if there exists two nondecreasing unbounded functions $f,F:\omega\to\omega$ such that $\{y\in\omega:y<f(x)\}\subseteq S(x)\subseteq\{y\in\omega:y<F(x)\}$ for all $x\in\omega$.
| 4 | https://mathoverflow.net/users/61536 | 442502 | 178,548 |
https://mathoverflow.net/questions/442495 | 5 | Let $X$ be a topological space and $x \in X$ a point. Let $\Omega$ be the set of open sets (viꝫ. the topology) of $X$, and $\Omega\_x$ the set of *germs* around $x$ of open sets, that is, $\Omega\_x = \Omega/{\sim}$ where $U\sim V$ means there exists a neighborhood $W$ of $x$ such that $U\cap W = V\cap W$. (Note that $U$ and $V$ need not be neighborhoods of $x$.)
Clearly, $\Omega$, partially ordered by inclusion, is a distributive lattice, with join and meet operations being given by $\cup$ and $\cap$. This structure passes to the quotient in the obvious way, so $\Omega\_x$ is a distributive lattice.
But $\Omega$ is more than a distributive lattice: it is a **frame** meaning that it has arbitrary joins (given by $\bigcup$), and that finite meets distribute over arbitrary joins. (A frame also has arbitrary meets, but they are not considered part of the structure and need not be preserved by homomorphisms.) Now this structure *does not* pass to the quotient $\Omega\_x$ in the sense that the canonical surjection $\Omega \to \Omega\_x$ does not, in general, define a frame structure on its target so that it is a frame homomorphism. (Indeed, it is easy to construct families $U\_i$ and $V\_i$ of open sets such that $U\_i \sim V\_i$ for every $i$ but $\bigcup\_i U\_i$ and $\bigcup\_i V\_i$ do not have the same germ, e.g., take $U\_i = \varnothing$ in $X = \mathbb{R}$ and $V\_i$ the complement of the closed interval $[-\frac{1}{i}, \frac{1}{i}]$ for $i\geq 1$ around $x=0$. But note that in this example, the join of the classes $[U\_i] = [V\_i]$ under $\sim$ still exists in $\Omega\_x$, and is $[\varnothing]$. So this counterexample does not answer the following question:)
**Question:** Is $\Omega\_x$ itself a frame (albeit not a quotient frame of $\Omega$)? In other words:
* Does the sup of an arbitrary family of germs of open sets exist?
* If so, do finite intersections (=meets) distribute over these arbitrary joins?
(In case the answer is negative, a counterexample with $X=\mathbb{R}$ would be most appreciated.)
| https://mathoverflow.net/users/17064 | Do germs of open sets around a point form a frame? | I just cooked up a counterexample in my head. Joseph Van Name beat me, but still here it is - to make things completely explicit. I hope I did not make any mistakes.
I will reason with closed sets and intersections; to get the answer with open sets and unions, just take the complement of everything.
So let $x = 0$, and let $V\_i = \{\frac{1}{2^i n} \;|\; n \in \mathbb{N}^\*\} \cup \{0\}$. Clearly the intersection of all $V\_i$ is $\{0\}$.
However, it is easy to find a family of closed sets $V'\_i$ that locally coincide with $V\_i$, but whose intersection is larger. Namely set $V'\_i = \{1, \frac{1}{2}, \ldots, \frac{1}{f(i)}\} \cup V\_i$, where $f: \mathbb{N}^\* \to \mathbb{N}^\*$ is some (let's say increasing) function. Then one easily sees that
$$\bigcap\_{i \in \mathbb{N}} V'\_i = \left\{ \frac{1}{k} \;|\; k \in \mathbb{N}^\* \text{ with } f(v\_2(k)) \geq k \right\} \cup \{0\}.$$
This set can be made arbitrarily "thick" by choosing a function $f$ that grows fast enough. So if there were a largest such set, it would have to be at least all of $V\_0$. But it is clear that the intersection of sets locally equal to $V\_i$ can never be all of $V\_0$.
| 5 | https://mathoverflow.net/users/39348 | 442507 | 178,550 |
https://mathoverflow.net/questions/436049 | 4 | Arinkin-Gaitsgory have defined the notion of singular support for any quasismooth $Y$
$$\text{SS}(\mathcal{F})\ \subseteq\ \text{Sing}(Y)$$
and $\mathcal{F}$ any ind-coherent sheaf, where $\text{Sing}(Y)$ is the category of singularities of $Y$ (the classical truncation of the minus one shifted cotangent complex). Thus when $Y$ is a smooth classical scheme, $\text{Sing}(Y)=Y$.
On the other hand, there is a classical notion of singular support of a D module/$\ell$ adic sheaf/constructible sheaf... on any (at least smooth) $Y$
$$\text{SS}(\mathcal{F})\ \subseteq\ T^\*Y.$$
See e.g. section 2.2 of [D-Modules, Perverse Sheaves, and Representation Theory](https://personal.math.ubc.ca/%7Ecautis/dmodules/hottaetal.pdf).
What precisely is the relation between these two notions of singular support?
| https://mathoverflow.net/users/119012 | Two notions of singular support? | When $X$ is a smooth scheme, the derived loop space (i.e. odd tangent bundle) $\mathcal{L}X\simeq\mathbb{T}\_X[-1]$ has $\mathrm{Sing}(\mathbb{T}\_X[-1]) = T^\*X$. Furthermore, there is a Koszul duality:
$$\mathrm{Coh}(\mathbb{T}\_X[-1])^{B\mathbb{G}\_a \rtimes \mathbb{G}\_m} \simeq F\mathcal{D}(X)$$
where the right-hand side is filtered $\mathcal{D}$-modules. Forgetting the $B\mathbb{G}\_a$-action corresponds to taking the associated graded, and this Koszul duality becomes the linear Koszul duality of [Mirkovic--Riche](https://arxiv.org/abs/0804.0923):
$$\mathrm{Coh}(\mathbb{T}\_X[-1])^{\mathbb{G}\_m} \simeq \mathrm{Coh}(\mathbb{T}^\*\_X)^{\mathbb{G}\_m}$$
and the notion of singular support on the left corresponds to the classical support on the right (i.e. singular support of the corresponding $\mathcal{D}$-module).
This was first written up by [Ben-Zvi--Nadler](https://arxiv.org/abs/1002.3636). I have a follow-up [paper](https://arxiv.org/abs/2301.06949) for stacks and there are some references in the intro. I should say that I think none of us write up the compatibility of singular support, but one can use for example the point-wise characterization in [Section 6 of Arinkin--Gaitsgory](https://arxiv.org/abs/1201.6343).
| 2 | https://mathoverflow.net/users/6059 | 442510 | 178,551 |
https://mathoverflow.net/questions/258159 | 26 | I'm trying to understand how the six functor philosophy applies to representation theory. Consider the category of classifying stacks $BG$ (assume $G$ discrete for simplicity). To every stack we can assign a triangulated category $D(BG)=D(BG,k):= D(Rep(G))$ the (perhaps bounded) derived category of the abelian category of representations of $G$ (in vector spaces over $k$ algebraically closed of characteristic 0). As far as I understand the formalism of six functors should translate (if at all) as follows:
If $\pi : BG \to pt$ then:
* $\pi\_{!}(-)$ is a derived version of $(-)^G$ (invariants) = group cohomology.
* $\pi\_\*(-)$ is a derived version of $(-)\_G$ (coinvariants) = group homology .
Generally if $\pi: BH \to BG$ then:
* $\pi\_!(-) = \mathbb{L}Ind^G\_H(-)$,
* $\pi\_\*(-) = \mathbb{R}Coind^G\_H(-)$
* $\pi^\*(-) = \mathbb{L}Res^H\_G(-)$.
I have several questions questions:
>
> 1. Are there any mistakes/wrong intuitions in the above?
> 2. I'm missing the upper shrieks $\pi^!$ and as such also the duality
> functor. How do these look in general? What is the "dualizing
> representation" of a group? Is it something familiar?
> 3. How much of this carries over to the category of algebraic stacks (on the big etale site of schemes) $BG$ for $G$ linear algebraic group? (suppose all this happens over some fixed field for simplicity).
>
>
>
| https://mathoverflow.net/users/22810 | Yoga of six functors for group representations? | The accepted answer here is on a rather negative note -- I don't think that's fair! In fact, I think the correct answer is that all of this works, except that it is $\pi\_\ast$ that gives group cohomology, and $\pi\_!$ that gives group homology.
More precisely, consider the category $C$ of locally compact Hausdorff spaces, say, with the Grothendieck topology of open covers. (One could consider many variants, but this will do.) For any stack $X$ on $C$ (in fact, any sheaf of $\infty$-groupoids), one can define a category $D(X)$, defined via descent. (If $X$ is represented by a locally compact Hausdorff space, this is the category of sheaves with values in $D(k)$, which agrees with the derived category of sheaves of $k$-vector spaces when $X$ is finite-dimensional.) Moreover, $D(X)$ is a closed symmetric monoidal category, and for a morphism $f: X\to Y$ between such stacks, one can define functors $f^\ast$ (with right adjoint $f\_\ast$) and (for a large class of $f$'s) a functor $f\_!$ (with right adjoint $f^!$).
In particular, this applies to $f: BH\to BG$ for discrete $H$ and $G$. In that case, $D(BG)$ is just the derived category of $G$-representations. Moreover any such map $f$ is "etale" and hence $f^\ast=f^!$, both of which agree with restriction. Then $f\_\ast$ is the right adjoint, which is coinduction; while $f\_!$ is the left adjoint, which is induction.
There's probably a better reference for this kind of $6$-functor formalism on topological stacks, but in any case I wrote up something in my notes on [$6$ functors](https://people.mpim-bonn.mpg.de/scholze/SixFunctors.pdf), see in particular Lecture 7 (for the construction of the six functors on locally compact Hausdorff spaces) and the appendix to Lecture 4 (for passing to stacks).
Regarding your last question, yes, everything also works with etale sheaves, and general stacks (as above). But one difference is that in general $D(BG)$ may not be the derived category of its heart.
[Let me note that I strongly disagree with Matthias Wendt here that $B(\mathbb Z/2\mathbb Z)=\mathbb{RP}^\infty$. They are homotopy equivalent, but other than that they are completely different objects, and in particular their $D(X)$'s are rather different, and many of the problems mentioned in that answer are just problems with that approach. It is indeed true that for $f: \mathbb{RP}^\infty\to \ast$ the functors $f\_!$ and $f^!$ are not defined, but for $B(\mathbb Z/2\mathbb Z)\to \ast$ they are.]
| 12 | https://mathoverflow.net/users/6074 | 442524 | 178,556 |
https://mathoverflow.net/questions/442500 | 3 | The similar and more general question is asked [here](https://mathoverflow.net/questions/334335/lipschitz-function-of-independent-subgaussian-random-variables?noredirect=1&lq=1), whose setting is random vectors.
Let $X$ be $\sigma$-sub-Gaussian and $f$ is a Lipschitz function w.r.t. constant $L$. How to prove can the subgaussian moment of $f(X)$ be bounded in terms of $\sigma$ and $L$?
It might not work on random vectors but I think it works on random variable.
| https://mathoverflow.net/users/500967 | Lipschitz function of subgaussian random variable | $\newcommand{\si}{\sigma}\newcommand\R{\mathbb R}$One of mutually equivalent definitions of a ($\si$)-sub-Gaussian random variable (r.v.) $X$ is as follows:
\begin{equation\*}
Ee^{X^2/\si^2}\le2. \tag{1}\label{1}
\end{equation\*}
It is now clear that the answer to the question is no. Indeed, suppose that $X\sim N(0,1)$, so that $X$ is $\si$-sub-Gaussian for some universal positive real constant $\si$. For any real $a$ and $x$, let now $f\_a(x):=x+a$, so that $f\_a$ is $L$-Lipschitz for $L=1$. Then
for any real $b>0$ we have $Ee^{f\_a(X)^2/b^2}\to\infty$ as $a\to\infty$, so that the sub-Gaussian norm of $f\_a(X)$ cannot be bounded in terms of $\si$ and $L$.
However, the sub-Gaussian norm of $f(X)-Ef(X)$ **can** be bounded in terms of $\si$ and $L$, as follows. Let $Y$ be an independent copy of $X$. Then, for any real $c>0$, by Jensen's inequality for the convex function $u\mapsto\exp\frac{(t-u)^2}{c^2}$,
\begin{equation\*}
\begin{aligned}
E\exp\frac{(f(X)-Ef(X))^2}{c^2}&=E\exp\frac{(f(X)-Ef(Y))^2}{c^2} \\
&=\int\_\R P(X\in dx)\exp\frac{(f(x)-Ef(Y))^2}{c^2} \\
&\le\int\_\R P(X\in dx)E\exp\frac{(f(x)-f(Y))^2}{c^2} \\
&=E\exp\frac{(f(X)-f(Y))^2}{c^2} \\
&\le E\exp\frac{L^2(X-Y)^2}{c^2} \\
&\le E\exp\frac{2L^2X^2+2L^2Y^2}{c^2} \\
&=\Big(E\exp\frac{2L^2X^2}{c^2}\Big)^2 \\
&\le E\exp\frac{4L^2X^2}{c^2}\le2
\end{aligned}
\end{equation\*}
by \eqref{1} if
\begin{equation\*}
c=2L\si.
\end{equation\*}
So, the sub-Gaussian norm of $f(X)-Ef(X)$ is $\le2L\si$, if $\si$ is the sub-Gaussian norm of $X$. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 442530 | 178,558 |
https://mathoverflow.net/questions/442533 | 0 | One of my research problem can be reduced to a question of the following form
>
> Given a set family $\mathcal{F}$ of $[n]$ , such that every element of $[n]$ lies in exactly $K$ sets in $\mathcal{F}$, can w partition $\mathcal{F}$ into $K$ subfamilies $\mathcal{F}\_i$ such that each subfamily is a partition of $[n]$?
>
>
>
Edit: It looks like the problem as stated admits an easy counterexample. I have posted a more interesting version [here](https://mathoverflow.net/questions/442538/jigsaw-puzzle-on-set-family-ii).
| https://mathoverflow.net/users/475875 | "JigSaw Puzzle" on Set Family | No. E.g., let $n=3$ and $\mathcal F=\{\{1,2\},\{1,3\},\{2,3\}\}$, so that $K=2$. However, no subset of $\mathcal F$ is a partition of $[n]$.
| 3 | https://mathoverflow.net/users/36721 | 442534 | 178,559 |
https://mathoverflow.net/questions/442539 | 4 | Suppose we have an elementary embedding $j: M\to N$, a forcing notion $\mathbb{P}\in M$, and a strong master condition $q\in j(\mathbb{P})$. A strong master condition for $j$ and $\mathbb{P}$ is a condition $q\in j(\mathbb{P})$ such that for every dense set $D\subseteq \mathbb{P}$ with $D\in M$, there is a condition $p\in D$ such that $q\leq j(p)$. In his chapter of the handbook on page 814, James Cummings claims that the set $G=\{p\in\mathbb{P} : q\leq j(p)\}$ is an $M$-generic filter. In some cases I find this totally believable, but putting some mild conditions on $j$, a problem arises that I have a question about.
My concern is that you should not be able to build a generic over $M$ in $M$. Now, I understand that in general, $q$ will not be a member of $M$, but sometimes I think this should be the case. For example, if $j$ is merely definable over $M$ (even though I know this is not a hypothesis given) and $M$ and $N$ are class transitive models, then I see issues. In particular, $k$ restricted to any $V\_{\alpha}$ will be a member of $M$, i.e. every initial segment of $N$ is a member of $M$, so work in a big enough $V\_{\alpha}$ to contain $k(\mathbb{P}).$ For example, if $\kappa$ is measurable, go up to $\aleph\_{\kappa+\omega}$, and take the image of that under the ultrafilter. This will be enough.
So, now that I've given some particulars, I can state some specific questions.
1. If $j$ is definable and $M,N$ are class models, will $q$ never be definable over $M$ or in $M$? When is the strong master condition in $M$ and when is it not?
2. Maybe just in general some more context on this matter?
| https://mathoverflow.net/users/498641 | Given an elementary embedding $j: M\to N$ and a strong master condition $q\in N$, how are we able to construct a generic over $M$ from $M$? | You’re right; this kind of strong master condition cannot exist in the usual context where $j : M \to N$ is a class of $M$. The relevance is when we’re in an intermediate stage of lifting an embedding. Suppose $j : V \to N$ is definable from parameters in $V$ and $\mathbb P$ is $\mathrm{crit}(j)=\kappa$-c.c. and has the property that for some $\kappa$-directed-closed $\mathbb Q$ in $V^{\mathbb P}$, $\mathbb P \* \dot{\mathbb Q}$ completely embeds into $j(\mathbb P)$ in $N$, where the complete embedding follows $j \restriction \mathbb P$. If we force with $j(\mathbb P)$, then we get a lifted embedding $j : V[G] \to N[G’]$, and $G’$ also generates a $\mathbb Q$-generic $H$. If $N$ is closed enough, then there is a strong master condition $q$ for $j$ and $\mathbb Q$ in $N[G’] \subseteq V[G’]$. Notice that $V[G’]$ is **not** a subclass of $V[G]$.
| 3 | https://mathoverflow.net/users/11145 | 442541 | 178,561 |
https://mathoverflow.net/questions/442551 | 0 | Are there infinitely many ones in the simple continued fraction for pi? I know that there’s a probability distribution given through Gauss-Kuzmin, but is there a proof that there’s infinitely many ones? How about proofs that there are infinitely many of any positive integer in pi’s simple continued fraction?
| https://mathoverflow.net/users/500847 | Infinitely many ones in continued fraction of pi? | Nothing like that is known. There is likewise no proof that the decimal expansion of $\pi$ has infinitely many 1's (or any other specific digit).
The continued fraction expansions of algebraic numbers of degree greater than $2$ are equally opaque: none are known to have any specific positive integer appearing in them infinitely often. And the decimal expansion of no specific irrational algebraic number is proved to contain a specific digit infinitely often even though of course we expect each digit appears equally often.
These kinds of things are just hopeless to expect to be provable at present.
| 6 | https://mathoverflow.net/users/3272 | 442554 | 178,563 |
https://mathoverflow.net/questions/442092 | 3 | Let $(X, d)$ be a compact metric space.
* We say that $\{x\_1, \cdots, x\_n\} \subseteq X$ is an $\varepsilon$-**covering** of $X$ if for any $x \in X$, there exists $i \in \{1, \ldots, n\}$ such that $d(x, x\_i) \leq \varepsilon$. Let
$$
\operatorname{Cov} (X, \varepsilon) := \min \{n: \exists \varepsilon \text {-covering of } X \text { with size } n\}
$$
be the $\varepsilon$-covering number of $X$.
* We say that $\{x\_1, \cdots, x\_n\} \subseteq X$ is an $\varepsilon$-**packing** of $X$ if $d(x\_i, x\_j)\ge\varepsilon$ for all distinct $i, j$. Let
$$
\operatorname{Pack} (X, \varepsilon) := \max \{n: \exists \varepsilon \text {-packing of } A \text { with size } n\}
$$
be the $\varepsilon$-packing number of $A$.
The metric spaces $(X, d)$ and $(X', d')$ are said to be isometric (denoted by $X \cong X'$) if there is a bijective isometry between them. Then we have a theorem.
>
> **Theorem** Let $(X, d)$ be a compact metric space and $f:X \to X$ be $1$-Lipschitz. Then $f$ is an isometry if and only if $f$ is surjective.
>
>
>
I have found the proof of one direction from [here](https://math.stackexchange.com/q/495648/368425) and the other one from [here](https://math.stackexchange.com/a/2672454/368425), i.e.,
>
> * $\implies$ Let $f$ be an isometry. Assume the contrary that there is $y \in X$ such that $y \notin Y:= f(X)$. Then there is $\varepsilon>0$ such that $d(y, Y)\ge \varepsilon$. Let $n:= \operatorname{Cov} (X, \varepsilon/2)$. Because $X \cong Y$, we get $n = \operatorname{Cov} (Y, \varepsilon/2)$. Let $C:=\{x\_1, \ldots, x\_n\}$ be an $\varepsilon/2$-covering of $X$. It follows that $y \in C$. Then $C \setminus \{y\}$ is an $\varepsilon/2$-covering of $Y$. Hence $\operatorname{Cov} (Y, \varepsilon/2) \le n-1$. This is a contradiction.
> * $\impliedby$ Let $f$ be surjective. Fix $x, y\in X$ such that $x \neq y$. Fix $\varepsilon>0$ such that $\delta := d(x,y) - \varepsilon/2 > 0$. Let $S$ be an $\varepsilon/4$-covering of $X$ that minimizes the quantity
> $$
> \mathcal N(S) := \operatorname{card} (\{ (s\_1, s\_2) \in S^2 : d(s\_1, s\_2) \ge \delta\}).
> $$
> Because $f$ is $1$-Lipschitz and surjective, $f(S)$ is also an $\varepsilon/4$-covering of $X$. Hence $\mathcal N(f(S)) \ge \mathcal N(S)$. This implies if $s\_1, s\_2 \in S$ with $d(s\_1, s\_2) \ge \delta$, then $d(f(s\_1), f(s\_2)) \ge \delta$. Now we pick $s\_1, s\_2 \in S$ with $d(s\_1,x)\le \varepsilon/4$ and $d(s\_2,y)\le \varepsilon/4$. Then
> $$
> d(s\_1, s\_2) \ge d(x, y)-d(s\_1, x)-d(s\_3, y) \ge d(x, y)- \varepsilon/2= \delta.
> $$
> So $d(f(s\_1), f(s\_2)) \ge \delta= d(x,y) - \varepsilon/2$. The claim then follows by taking the limit $\varepsilon \to 0^+$.
>
>
>
Now let $(X, d)$ and $(X', d')$ be compact metric spaces and $f:X \to X'$ be $1$-Lipschitz.
* Just as in the proof of direction $\implies$ above, if $f$ is an isometry and $\operatorname{Cov} (X, \varepsilon)= \operatorname{Cov} (X', \varepsilon)$ for all $\varepsilon>0$, then $f$ is surjective.
* The proof of direction $\impliedby$ above uses the quantity $\mathcal N(S)$ which looks related to $\operatorname{Pack} (S, \delta)$.
I would like to ask if below statement is true, i.e.,
>
> If $f$ is surjective and $\operatorname{Pack} (X, \varepsilon)= \operatorname{Pack} (X', \varepsilon)$ for all $\varepsilon>0$, then $f$ is an isometry.
>
>
>
Thank you so much for your elaboration!
| https://mathoverflow.net/users/99469 | If $X,X'$ have the same $\varepsilon$-packing numbers and $f:X \to X'$ surjective $1$-Lipschitz, then $f$ is an isometry | Consider two metrics on $\{x,x',y\}$ defined by
$$|x-y|\_1=|x'-y|\_1=|x-y|\_2=3, \quad |x-x'|\_1=|x-x'|\_2=1, \quad |x'-y|\_2=2.$$
Denote by $X\_1$ and $X\_2$ the corresponding metric spaces.
Note that $\mathrm{pack}\_\varepsilon X\_1\equiv \mathrm{pack}\_\varepsilon X\_2$.
Indeed, for both spaces we have
* $\mathrm{pack}\_\varepsilon=1$ if $\varepsilon>3$,
* $\mathrm{pack}\_\varepsilon=2$ if $3\geqslant \varepsilon>1$, and
* $\mathrm{pack}\_\varepsilon=3$ if $1\geqslant \varepsilon>0$.
The identity map on the set $\{x,x',y\}$ defines an onto short map $X\_1\to X\_2$ which is not an isometry.
| 2 | https://mathoverflow.net/users/1441 | 442567 | 178,568 |
https://mathoverflow.net/questions/442571 | 2 | Consider the series
$$f(z)=\sum\_{n\ge1}\dfrac{z^n}{n^2\binom{2n}{n}}$$
which converges for $|z|\le4$.
One has $f(0)=0$, $f(1)=\pi^2/18$, $f(2)=\pi^2/8$, $f(3)=2\pi^2/9$, and $f(4)=\pi^2/2$.
Naive question: are there any other reasonably "explicit" evaluations (for instance for
$z=-1$, $-2$, $1/2$...) ? Note that the series for $z=1$ is related to Ap'ery's proof of the irrationality of $\zeta(2)$.
| https://mathoverflow.net/users/81776 | Special values of $\sum_{n\ge1}z^n/(n^2\binom{2n}{n})$ | I interpret the request of the OP for an "explicit" evaluation of the series as a request for a "closed-form" expression, which exists (it seems to go back to Euler, [here](https://math.stackexchange.com/a/383181/87355) are several proofs):
$$f(z)=\sum\_{n\ge1}\dfrac{z^n}{n^2\binom{2n}{n}}=2 \arcsin^2\,(\tfrac{1}{2}\sqrt{z}),\;\;|z|\leq 4.$$
| 8 | https://mathoverflow.net/users/11260 | 442573 | 178,570 |
https://mathoverflow.net/questions/442578 | 0 | Let $u(x) \in H^1\_0$ is a complex function in sobolev space extension by zero.
How to find $J'(u)$ for
$$
J(u)= \int\limits\_0^l |u(x)|^2\operatorname{d\!}x\;??
$$
In $L\_2$ it's easy:
$$
J'(u) = \left(\int\limits\_0^l|u(x)|^2 \operatorname{d\!}x\right)'=\big(\|u(x)\|^2\big)'= 2 u(x),
$$
but it does not work with $H^1\_0$, where
$$
\|u(x)\|^2\_{H^1\_0} = \int\limits\_0^l |u'(x)|^2 \operatorname{d\!}x
$$
| https://mathoverflow.net/users/492767 | Derivative in Sobolev space extended by zero | We shall assume that $l\in(0,\infty)$, so that for any $h\in H\_0^1$ we have
$$\int\_0^l|h|^2=\int\_0^l dx\,\Big|\int\_0^x h'\Big|^2
\le\int\_0^l dx\,\Big(\int\_0^l|h'|\Big)^2 \\
\le\int\_0^l dx\,l\,\int\_0^l|h'|^2=l^2\|h\|^2\_{H\_0^1}. \tag{1}\label{1}$$
Next, for any $u$ and $h$ in $H\_0^1$, in view of \eqref{1},
$$J(u+h)-J(u)-2\int\_0^l\Re(\bar uh)=\int\_0^l|h|^2=o(\|h\|\_{H\_0^1})$$
as $\|h\|\_{H\_0^1}\to0$. So, the ([Fréchet](https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative#Definition)) derivative $J'(u)$ of $J$ at $u\in H\_0^1$ is the linear functional on $H\_0^1$ given by the formula
$$J'(u)(h)=2\int\_0^l\Re(\bar uh)$$
for $h\in H\_0^1$.
| 3 | https://mathoverflow.net/users/36721 | 442580 | 178,571 |
https://mathoverflow.net/questions/442399 | 5 | The completeness game $G\_{\gamma}(P)$ for a partial order $P$ has players COM and INC play alternatingly and descendingly elements of $P$ with player INC playing first and player COM playing at limit steps. INC wins the game if at some point $\alpha<\gamma$ (necessarily a limit ordinal) there are no legal moves left. Otherwise, COM wins.
By a theorem of Foreman, for a successor cardinal $\kappa=\lambda^+$, $P$ is $<\kappa$-distributive (i.e. every sequence of $<\kappa$ ordinals is in $V$) iff INC does not have a winning strategy in $G\_{\lambda+1}(P)$.
Let $\kappa=\lambda^+$ be a successor cardinal.
* Assume INC has a winning strategy in $G\_{\kappa}(P)$. Does INC have a winning strategy in $G\_{\lambda+1}(P)$? I.e. if INC always wins eventually, can they uniformly bound the time it takes for them to win?
* If $P$ is $<\kappa$-closed (every sequence of $<\kappa$ conditions in $P$ has a lower bound) and $Q$ is $\kappa$-cc., $P$ is $<\kappa$-distributive in $V[G]$, where $G$ is $Q$-generic. Can INC have a winning strategy in $G\_{\kappa}(P)$ in $V[G]$? What if $Q$ is $\kappa$-Knaster?
---
**Edit:** The answer to the first question is "no". If $(T,\leq)$ is an $\omega\_1$-Suslin-tree, $(T,\geq)$ is $<\omega\_1$-distributive (see e.g. Lemma 15.28 in Jech, noting the different definition of distributivity). However, INC has an easy winning strategy in $G\_{\omega\_1}((T,\leq))$, by simply playing any node strictly above COMs last play, since a play in which COM wins corresponds to a cofinal chain in $(T,\leq)$.
It still remains if such an object can be added by a tame forcing notion.
| https://mathoverflow.net/users/138274 | Are the completeness Games $G_{\lambda+1}(P)$ and $G_{\lambda^+}(P)$ equivalent for INC? | The answer to the second question is no as well.
Suppose $\dot\tau$ is a $Q$-name for a strategy for the player INC in the game $G\_\kappa(P)$. Let us pretend that COM opens the game instead of INC and note that this is unproblematic. We will define by induction a descending sequence $\langle p\_\alpha\mid\alpha<\kappa\rangle$ of conditions in $P$ and $Q$-names $\langle \dot p\_\alpha\mid\alpha<\kappa\rangle$ so that
* $\Vdash\_Q\check p\_\beta\leq \dot p\_\alpha\leq\check p\_\alpha$ for all $\beta\leq\alpha$
* $\Vdash\_Q$"The sequence $\langle \dot p\_\beta\mid\beta\leq\alpha\rangle$ make up the fist $\alpha+1$-many moves of player COM in a game of $G\_{\check\kappa}(\check P)$ in which INC plays according to $\dot\tau$ and the final response of $\dot \tau$ to this is a condition $\leq\check p\_\alpha$."
for all $\alpha<\kappa$. It follows that if $G$ is $Q$-generic then COM wins against $\dot\tau^G$ by playing the conditions $\langle \dot p\_\alpha^G\mid\alpha<\kappa\rangle$, so $\dot\tau$ is not a winning strategy for INC.
Let us turn to the construction. Assume $\alpha<\kappa$ and that $\dot p\_\beta, p\_\beta$ are defined for all $\beta<\alpha$. Let $s\_0$ be some lower bound of $\langle p\_\beta\mid\beta<\alpha\rangle$ which exists as $P$ is ${<}\kappa$-closed. Build descending sequences of maximal possible length
$$\langle s\_\gamma\mid\gamma\leq\xi\rangle,\ \langle t\_\gamma\mid\gamma<\xi\rangle$$
and an antichain $A\_\xi:=\{q\_\gamma\mid \gamma<\xi\}\subseteq Q$ so that
* $s\_\gamma\leq t\_\gamma\leq s\_{\gamma+1}$ for all $\gamma<\xi$ and
* $q\_\gamma\Vdash\_Q$"if COM plays $\langle \dot p\_\beta\mid\beta<\alpha\rangle^\frown\check s\_\gamma$ and INC follows $\dot\tau$, INC's final play is $t\_\gamma$".
The construction is straightforward and can only break down for two reasons: Either there is no lower bound of $\langle t\_\gamma\mid\gamma<\xi\rangle$ or $A\_\xi$ is a maximal antichain in $Q$. The latter must happen first since $P$ is ${<}\kappa$-closed and $Q$ is $\kappa$-cc. We set $p\_\alpha$ to be the final $s\_\xi$ and by mixing of names, we can find a $Q$-name $\dot p\_\alpha$ so that
$$q\_\gamma\Vdash\_Q\dot p\_\alpha=\check s\_\gamma$$
for all $\gamma<\xi$. This completes the construction.
Finally, let me remark that this is sensitive to the order of play in $G\_\kappa(P)$: If $G\_\kappa'(P)$ is the game where INC plays at limit steps instead of COM then it is possible that INC has a winning strategy for $G\_\kappa'(P)$ in $V^Q$:
Let $Q$ be the poset of finite partial functions $q:\omega\_1\rightarrow 2$ ordered by inclusion and let $P=\mathrm{Add}(\omega\_1, 1)$. $Q$ is ccc and $P$ is ${<}\omega\_1$-closed. Let $G$ be $Q$-generic and consider the strategy for INC for the game $G\_\kappa'(P)$ where INC adds a bit of information which agrees with the generic function $\bigcup G$ at the least ordinal which is not yet in the domain of any condition played so far. Assume toward a contradiction that some game in which INC follows this strategy lasts for $\omega\_1$-many rounds. The result is a function $f\colon\omega\_1\rightarrow 2$ which agrees with $\bigcup G$ on a club $C$. Now $Q$ is ccc, so there is a club $C'\subseteq C$ with $C'\in V$. It follows that $f\notin V$, but we must have $f\upharpoonright\alpha\in V$ for all $\alpha<\omega\_1$. This is impossible by a result of Spencer Unger: As $Q\times Q$ is ccc, $Q$ has the $\omega\_1$-approximation property, see Lemma 1.2 in [Fragility and indestructibility II](https://www.math.toronto.edu/sunger/Indestructible2.pdf), Annals of Pure and Applied Logic 166 (2015) 1110-1122.
| 4 | https://mathoverflow.net/users/125703 | 442582 | 178,572 |
https://mathoverflow.net/questions/442557 | 6 | Let $\mathsf{C}$ be a *possibly large* category with a Grothendieck topology satisfying the [*Weakly Initial Set of Covers*](https://ncatlab.org/nlab/show/WISC) condition: there is for each $X$ a *set* (not a proper class) of covering families of $X$ such that if $\{U\_i \to X\}$ is an arbitrary covering family, it is refined by one in the set. In the case where $\mathsf{C}$ is small, Mac Lane and Moerdijk construct a *subobject classifier* $\Omega$ in the category of sheaves on the site $\mathsf{C}$. I want to know: if $\mathsf{C}$ is allowed to be large (but satisfying the weakly initial set of covers condition), can we still construct $\Omega$?
---
For example, $\mathsf{Top}$ is a large category, and we may take the covering families on $X$ to be sets of local homeomorphisms $f\_i \colon V\_i \to X$ such that the union of the images $\bigcup\_i f\_i(V\_i)$ is all of $X$. Every such covering family is, one can check, refined by an *open cover* of $X$, and there are clearly only a set of open covers of $X$.
If one has a weakly initial set of covers for each object of $\mathsf{C}$,
then the [plus construction](https://ncatlab.org/nlab/show/plus+construction+on+presheaves) shows that every presheaf on $\mathsf{C}$ may be sheafified, great.
If $\mathsf{C}$ is *small* (so that there are only a set of arrows in the entire category), then there is for each $X \in \mathsf{C}$ a *set* of sieves on $X$, where I remind you that a *sieve* is a collection $S$ of arrows $f\colon U \to X$ that forms a "right ideal" in the sense that if $h\colon V \to U$ is an arbitrary arrow and $f$ belongs to $S$, then $fh$ belongs to $S$. Say a sieve *covers* $X$ if it contains a covering family. Mac Lane and Moerdijk define what it means for a sieve to be *closed,* show that every covering sieve has a *closure* (which is again covering) and define
$$\Omega(X) = \{ \bar S : S \text{ covers } X \}.$$
This is clearly a set when $\mathsf{C}$ is small, and the assignment $X \mapsto \Omega(X)$ is functorial in this case as well, since covering sieves pull back along any arrow $f\colon Y \to X$ to covering sieves on $Y$, and the closure of a pullback sieve is the pullback of the closure. Anyway, this presheaf turns out to be a *sheaf,* and it is the subobject classifier above.
If $\mathsf{C}$ is large, one only gets a set $\Omega(X)$ if one restricts to those covering sieves on $X$ generated by, for example, a weakly initial set of covers for $X$. The problem, then, is that given such a sieve $S$ and an arrow $f\colon Y \to X$, we know that the pullback sieve $f^\* S$ covers $Y$ so is *refined by* a sieve generated by an element of the weakly initial set of covers for $Y$, but may not be *equal* to such a sieve, so $\overline{f^\*S}$ need not belong to $\Omega(Y)$, and—to me, at least—there doesn't appear to be a natural choice of an element of $\Omega(Y)$ to associate to $S$.
| https://mathoverflow.net/users/135175 | Subobject classifier for sheaves on large sites with WISC | To answer your question directly, WISC does not imply the existence of subobject classifiers.
Notice that when there are only trivial covers, WISC is trivially satisfied, so it suffices to find a category with a proper class of sieves.
Consider the ordered class of ordinals with a terminal object adjoined.
Then each ordinal generates a distinct sieve, so the category of presheaves cannot have a subobject classifier.
---
The crux of the matter is that *closed* sieves are orthogonal to *dense* (= covering) sieves.
As in the example above, not having "too many" dense sieves implies nothing about not having "too many" closed sieves.
To avoid having to make pedantic distinctions between classes and conglomerates and so on, let me just talk about large and small sets.
Consider the following:
**Axiom G.**
There is a small set $\mathcal{C}'$ of objects in $\mathcal{C}$ such that every dense sieve on every object in $\mathcal{C}$ contains a dense sieve generated by morphisms with domains in $\mathcal{C}'$.
It is not hard to see that axiom G implies WISC.
In fact, axiom G implies the category of sheaves is a Grothendieck topos.
(I think the converse is also true, at least under some additional assumptions about the category of sheaves.)
So perhaps axiom G is a bit too strong.
We could weaken it as follows:
**Axiom LG.**
For every object $X$, the slice category $\mathcal{C}\_{/ X}$ satisfies axiom G.
The point is that to ensure that there are not "too many" closed sieves (resp. dense sieves) on $X$, we only need to know axiom G is satisfied for $\mathcal{C}\_{/ X}$.
Thus axiom LG is enough to guarantee WISC and also the existence of a subobject classifier.
On the other hand, axiom LG does *not* imply even local-smallness of the category of sheaves!
For example, consider the category $\textbf{LH}$ of all topological spaces and local homeomorphisms between them.
Say a sieve on an object in $\textbf{LH}$ is a cover if it is jointly surjective.
This site satisfies axiom LG but not axiom G.
Thus, the category of sheaves on $\textbf{LH}$ is a pretopos and has a subobject classifier.
I claim the category of sheaves is neither well-powered nor locally small.
Observe that $\textbf{LH}$ has the unpleasant property that it has a *full* subcategory that is totally disconnected (i.e. all morphisms are endomorphisms) but not small.
We can use this to find a non-small set of distinct closed sieves of $\textbf{LH}$ (= subsheaves of the terminal sheaf).
This demonstrates the failure of the category of sheaves to be well-powered.
But in a category with a subobject classifier, local-smallness implies well-poweredness, so we also deduce that the category of sheaves is not locally small.
| 6 | https://mathoverflow.net/users/11640 | 442587 | 178,574 |
https://mathoverflow.net/questions/442550 | 3 | Currently I am working on studying stochastic integrals of the form: $$Z\_\infty = \int\_0^\infty e^{-f(t)}\mathop{d}S\_t$$
where $S\_t$ is a Compound-Poisson process with Exponentially-distributed increments. We know that the integral converges so long as $f$ is continuous on $[0,\infty)$ and $\lim\_{t\to\infty}f(t) \to \infty$ holds true. $Z\_\infty$ effectively models a perpetuity where $f$ is our discount rate function, and $S\_t$ is the cashflow aggregation process.
As far as I am aware, in the case where $f$ is linear, $Z\_\infty$ is Gamma distributed. I am interested in knowing if there are further results for when the discounting is a power-law rate; $f(t) = \lambda t^\alpha $.
| https://mathoverflow.net/users/500844 | Are there any known results on the probability distributions of perpetuities with power law discount rates? | This is more of an extended comment trying to study this integral.
In the similar spirit as [here](https://math.stackexchange.com/questions/2586336/stochastic-integral-of-poisson-process/2587999#2587999) we study the Laplace transform. As explained [here](https://personal.ntu.edu.sg/nprivault/MA5182/stochastic-calculus-jump-processes.pdf) we have the Lévy-Khintchine formula for compound Poisson processes for a process whose increments have distribution $\nu$:
$$E(exp(r\int\_{0}^{T}f(t)dY\_{t}))=exp(\lambda\int\_{0}^{T}\int\_{-\infty}^{\infty}(e^{yrf(t)}-1) \nu(dy) dt )$$
and so in the above case of $\nu(y)=e^{-\mu y}1\_{y\geq 0}$ and $f(t)=exp(-t^{\alpha})$ we have
$$exp(\lambda\int\_{0}^{T}\int\_{0}^{\infty}(e^{yr exp(-t^{\alpha})}-1) e^{-\mu y} dt ).$$
I tried in Mathematica to possibly get exact formulas for the above for general $a>1$ and $T=\infty$ but didn't get anything back. It might require (keyhole) contour methods to study. You could also try to directly invert its Laplace transform but that seems tricky.
| 1 | https://mathoverflow.net/users/99863 | 442588 | 178,575 |
https://mathoverflow.net/questions/442480 | 4 | Suppose $M = K\backslash G/\Gamma$ is a quotient of a symmetric space by a lattice. I don't know all of the proper adjectives to apply here (e.g. what should be said about $G$ and so on), but I wouldn't be upset if we just took $G=\operatorname{O}(1,n)$ and $K = \operatorname{O}(n)$ so that $M$ is a finite volume hyperbolic orbifold.
Edit: as pointed out by @MoisheKohan in the comments, I really shouldn't be asking about the real hyperbolic setting. Let's instead take $G=\operatorname{U}(1,n)$ and $K = \operatorname{U}(n)$ so that $M$ is a finite volume *complex* hyperbolic orbifold.
Now suppose that $M' \subset M$ is a connected component of an orbifold locus. I believe it follows that $M'$ is also a lattice quotient (of smaller dimension), and feel free to correct me if I'm wrong about that. Now what can be said about the relationship between the arithmeticity of $M$ and $M'$? E.g. does the (non)arithmeticity of one imply the (non)arithmeticity of the other?
| https://mathoverflow.net/users/151664 | Inheritance of arithmeticity properties in orbifold strata | Here is what I think is the correct setup:
Let $X$ be a symmetric space of noncompact type, $\Gamma$ is a lattice in the isometry group of $X$. Then $\Gamma$ has finitely many $\Gamma$-conjugacy classes of finite subgroups. Let $\Phi$ be one of these finite subgroups (I will pick one from each conjugacy class). Then $\Gamma$ fixes a symmetric subspace $Y\subset X$. The (set-wise) $\Gamma$-stabilizer $\Gamma\_Y$ of $Y$ is a sublattice in $\Gamma$: It acts on $Y$ as a lattice. The quotient-orbifold $Y/\Gamma\_Y$ has a natural projection $Q\_Y$ to the orbifold $X/\Gamma$, the projection map $Y/\Gamma\_Y\to X/\Gamma$ is an isometric immersion in the sense of orbifolds. This defines a (finite) stratification of the orbifold locus of $X/\Gamma$: The strata are $Q\_Y$'s.
This picture serves a correction to the erroneous claim in the post "I believe it follows that..."
Now, one can ask meaningful questions relating arithmeticity of $\Gamma$ and the sublattices $\Gamma\_Y$. If memory serves me well, arithmeticity of $\Gamma$ implies arithmeticity of each $\Gamma\_Y$. However, the converse is false: It is surely false for $X={\mathbb H}^n$: One can build examples in all dimensions using the Gromov--Piatetsky-Shapiro construction. I did not check Deligne-Mostov nonarithmetic examples, but, likely, you will find some arithmetic complex one-dimensional immersed totally geodesic suborbifolds. Since you like DM-examples, you should try finding such.
| 2 | https://mathoverflow.net/users/39654 | 442589 | 178,576 |
https://mathoverflow.net/questions/437141 | 4 | Let $G=GL(n, F)$, $B$ be a Borel subgroup and let $B=AN$ be the Langlands decomposition. Let $\nu \in \mathfrak{a}^\*\_{\mathbb{C}}$ be in the positive Weyl chamber. Consider the normalized induced representation $I = I^G\_B(\nu)$. Let $F=\mathbb{R}$ and $J$ be the unique generic irreducible subrepresentation of $I$ which exists thanks to Theorem 6.2 for $F=\mathbb{R}$ of the article by Casselman-Shahidi titled "On irreducibility of standard modules for generic representations". It is known that the intertwining operator
$$
M(w\_0) : I^G\_B(\nu) \rightarrow I^G\_{\bar{B}}(w\_0(\nu)),
$$
corresponding to the longest Weyl group element annihilates every nonzero subrepresentation of $I^G\_B(\nu)$. I was wondering if something more can be said here. For example
1. Is $J$ annihilated by all intertwining operators $M(w)$ where $w\in W$? I am mostly interested in the case of $\nu = \rho$ where $\rho$ is the half sum of positive roots.
2. In the same article the authors have proved analogue of Theorem 6.2 for nonarchimedean field and can be found as Theorem 1. What can we say about the analogue of the above theorem when $F$ is nonarchimedean field?
3. If more general results are known or expected in this direction, a reference would be much appreciated.
Any comments, suggestions or references are welcome. Please feel free to edit if you find something erroneous in the question. Thank you.
| https://mathoverflow.net/users/58056 | Intertwining operators and induced representation | I will answer question 1 for when $\nu=\rho$ and $F$ is $p$-adic (since I am not too familiar with $F=\mathbb R$; I assume a similar argument works for $F=\mathbb R$ as well). It suffices to check that $I\_B^G(\rho)$ is not isomorphic to $I\_B^G(w\rho)$ for $w\ne1\in W$. But this is clear, since
$$\begin{align\*}
\hom\_G(I\_B^G(\rho),1\_G)&\cong\hom\_G(1\_G,I\_B^G(-\rho))\\
&\cong\hom\_G(1\_G,\mathrm{Ind}\_B^G(1\_B))\\
&\cong\hom\_B(1\_B,1\_B)\\
&\cong\mathbb C,
\end{align\*}$$
where $\mathrm{Ind}\_B^G$ denotes naive induction, but
$$\begin{align\*}
\hom\_G(I\_B^G(w\rho),1\_G)&\cong\hom\_G(1\_G,I\_B^G(-w\rho))\\
&\cong\hom\_G(1\_G,\mathrm{Ind}\_B^G(\rho-w\rho))\\
&\cong\hom\_B(1\_B,\rho-w\rho)\\
&=0,
\end{align\*}$$
since $\rho\ne w\rho$.
| 1 | https://mathoverflow.net/users/123673 | 442596 | 178,578 |
https://mathoverflow.net/questions/442555 | 1 | $\newcommand{\fg}{\mathfrak g}\newcommand{\ee}{\varepsilon}$Let $\fg$ be the complex simple Lie algebra of type G$\_2$.
We consider its root system as follows (though it is probably not necessary to state the question):
\begin{equation\*}
\Delta^+(\mathfrak g,\mathfrak h) = \left\{
\begin{array}{ll}
\ee\_2-\ee\_3,&\ee\_1-2\ee\_2+\ee\_3,\\
\ee\_1-\ee\_2,&\ee\_1+\ee\_2-2\ee\_3,\\
\ee\_1-\ee\_3,&2\ee\_1-\ee\_2-\ee\_3
\end{array}
\right\},
\end{equation\*}
where $\mathfrak h$ is a Cartan subalgebra of $\mathfrak g$ with $\mathfrak h^\*=\{\sum\_{i=1}^3 a\_i\ee\_i: a\_i\in\mathbb C,\; \sum\_{i=1}^3a\_i=0\}$.
We will use its root decomposition
\begin{equation\*}
\mathfrak g=\mathfrak h\oplus \bigoplus\_{\alpha\in\Delta(\mathfrak g,\mathfrak h)} \mathfrak g\_\alpha.
\end{equation\*}
Let $(\pi,V\_\pi)$ be a non-trivial irreducible representation of $\mathfrak g$.
Suppose $v\in V\_\pi$ is in the weight space of weight zero (i.e. $v\in V\_\pi(0)$) and satisfies
\begin{equation\*}
\pi(X\_{\ee\_1-\ee\_2})(v)=0
\quad\text{ and }\quad
\pi(X\_{\ee\_1+\ee\_2-2\ee\_3})(v)=0,
\end{equation\*}
where $X\_{\ee\_1-\ee\_2}\in\mathfrak g\_{\ee\_1-\ee\_2}$ and $X\_{\ee\_1+\ee\_2-2\ee\_3}\in \mathfrak g\_{\ee\_1+\ee\_2-2\ee\_3}$ are non-trivial.
>
> Can we ensure that $v=0$?
>
>
>
Note that the story would be very different if we replace the roots $\ee\_1-\ee\_2$ and $\ee\_1+\ee\_2-2\ee\_3$ by any pair $\alpha,\beta\in\Delta(\mathfrak g,\mathfrak h)$ with $\alpha$ short, $\beta$ long and non-orthogonal.
In that case, one easily obtains that $\pi(X\_{a\alpha+b\beta})(v)=0$ for all $a,b\in\mathbb Z$, which implies that $\mathbb C v$ is $\mathfrak g$-invariant since $\mathbb Z\alpha+\mathbb Z\beta=\operatorname{Span}\_\mathbb Z\Delta(\mathfrak g,\mathfrak h)$, contradicting the assumptions on $\pi$.
| https://mathoverflow.net/users/20052 | About certain elements in the zero weight space of an irreducible representation of the complex simple Lie algebra of type G$_2$ | The answer is no. Here is some SAGE code:
```
sage: G2, A1xA1 = [WeylCharacterRing(x,style="coroots") for x in ["G2","A1xA1"]]
sage: b = G2.maximal_subgroup("A1xA1")
sage: G2(2,2).branch(A1xA1,rule=b)
A1xA1(0,0) + A1xA1(1,1) + A1xA1(1,3) + A1xA1(1,5) + 2*A1xA1(2,2) + A1xA1(2,4) + A1xA1(2,6) + 2*A1xA1(3,1) + 2*A1xA1(3,3) + A1xA1(3,5) + 2*A1xA1(4,0) + 2*A1xA1(4,2) + 2*A1xA1(4,4) + 2*A1xA1(5,1) + 2*A1xA1(5,3) + A1xA1(5,5) + A1xA1(6,0) + 3*A1xA1(6,2) + A1xA1(6,4) + 2*A1xA1(7,1) + 2*A1xA1(7,3) + A1xA1(8,0) + A1xA1(8,2) + A1xA1(8,4) + A1xA1(9,1) + A1xA1(9,3) + A1xA1(10,2) + A1xA1(0,4)
```
The presence of A1xA1(0,0) shows that the trivial module appears when this representation of G2 is restricted to A1xA1, giving a nonzero vector satisfying your given conditions.
| 3 | https://mathoverflow.net/users/425 | 442598 | 178,579 |
https://mathoverflow.net/questions/442056 | 2 | My question is on Brascamp-Lieb-inequality on the Euclidean sphere (which can be viewed as an analogue of Young's inequality on the sphere) obtained in [1].
(See also this question:
[Brascamp-Lieb inequalities on the sphere](https://mathoverflow.net/questions/325211/brascamp-lieb-inequalities-on-the-sphere) )
The inequality reads:
Let $e\_{i},i=1,\dots,N$, be orthogonal vectors in
$\mathbb{R}^{N}$. If $f\_j$ are positive functions on $[-1,1]$, $N \geq 3$, and $p\geq 2$ then
$$
\int\_{S^{N-1}} \prod\_{j=1}^N f\_j(x\cdot e\_j) d\mu(x)
\leq \prod\_{j=1}^N \left(\int\_{S^{N-1}} f\_j(x\cdot e\_j)^p d\mu(x)\right)^{1/p},
$$
where $\mu$ is the uniform Borel probability measure on the sphere $S^{N-1}$.
Fix **$(x\_{1},\dots,x\_{N})\in \mathbb{R}^{N}\setminus S^{N-1}$**. Let $0<\alpha\_{j}<1$. Suppose that
$$f\_{j}(t)=|t-x\_{j}|^{-\alpha\_{j}},\quad x\_{i}\neq x\_{j},\;\forall i\neq j.$$
Does the inequality $(1)$ hold true for these particular functions $f\_{j}$ **for some $1\leq p<2$** ?
That is, is it true that
$$
\int\_{S^{N-1}} \prod\_{j=1}^N
|\theta\cdot e\_j-x\_{j}|^{-\alpha\_{j}}
d\mu(\theta)
\leq C\prod\_{j=1}^N \left(\int\_{S^{N-1}} |\theta\cdot e\_j-x\_{j}|^{-p\alpha\_{j}}
d\mu(\theta)\right)^{\frac{1}{p}}\qquad (\*)$$
for some $1\leq p<2$ ?
[1] EA Carlen, EH Lieb, and M Loss. "A Sharp Analogue of Young's Inequality on $S^N$ and Related Entropy Inequalities." <https://arxiv.org/abs/math/0408030>
Edit Summary:
As @fedja pointed out, the estimate is immediately false when
$(x\_{1},\dots, x\_{N})$ lies on $\mathbb{S}^{N-1}$.
Latest Edit:
Instead of asking whether the inequality $(\*)$ holds with $p=1$ I tried to simplify the question to whether $(\*)$ is true for some $p\in [1,2[$.
**Most recent Edit**:
Simplified the functions $f\_{j}$ to $f\_{j}(t):=|t-x\_{j}|^{-\alpha\_{j}}$ instead of $f\_{j}(t)=|t^k-x\_{j}|^{-\alpha\_{j}}$ where $k$ is an odd integer.
| https://mathoverflow.net/users/116555 | (simplified further) Is Brascamp-Lieb inequality on the sphere applicable for these functions for some $1\leq p<2$ | Unfortunately, the answer to the question as stated is "No" even if $N=2$. Let $k=1,\alpha\_1=\alpha\_2=\frac 12$. Choose $x=(x\_1,x\_2)$ on the unit circle (say $(3/5,4/5)$). Then the integrals on the right are finite but the integral on the left diverges (you basically integrate inverse distance to $x$). Perhaps, you meant something else?
| 1 | https://mathoverflow.net/users/1131 | 442601 | 178,581 |
https://mathoverflow.net/questions/442503 | 20 | The six functor formalism in a given cohomology theory consists of for each space a derived category of sheaves and six different ways to construct functors between those categories (four involving a morphism and two only a single space). It then consists of many *coherences* - these are isomorphisms between certain compositions of these functors and, in modern formulations, homotopies between certain combinations of these isomorphisms, 2-homotopies between certain compositions of these homotopies, and so on.
In [Peter Scholze's notes on six functor formalisms](https://people.mpim-bonn.mpg.de/scholze/SixFunctors.pdf) he gives a precise definition, attributed to Lukas Mann, and closes it by saying:
>
> We note that no further coherences are necessary here: Adjoints automatically acquire all
> relevant coherences.
>
>
>
What *mathematical* claim is being made here? How do we know the coherences acquired are all the relevant ones? Does that knowledge give us an *algorithm* to prove a desired coherence results?
In the case of a three-functor formalism, including only $\otimes, f^\*, f\_!$ (i.e. ignoring the adjoints mentioned in the quoted passage), I know basically how to answer all these questions. The three-functor formalism is a functor from a certain infinity-category of correspondences to the infinity-category of all infinity-categories. Each functor arises from a correspondence, so a composition of functors arises from a composition of correspondences. Checking two functors are isomorphic means computing the relevant correspondences and checking they're isomorphic, and this works for all the classical isomorphisms of the six functors formalism that involve only those functors (Leray spectral sequence with compact supports, symmetry and associativity of tensor product, functoriality of pullback, Künneth formula, proper base change, projection formula, tensor products are compatible with pullbacks). Checking two such isomorphisms are the same means evaluating two isomorphisms of correspondences, which ultimately give isomorphisms of schemes, and checking they're the same.
But when adjoints appear I no longer no how to do this. Am I supposed to draw some diagrams in which correspondences are connected with strings?
| https://mathoverflow.net/users/18060 | When (or why) is a six-functor formalism enough? | When defining a homotopy-coherent structure, you have to strike the correct balance between supplying enough data (so that all isomorphisms (between isomorphisms, ...) that you need later are actually defined), and not supplying "too much" data (because if you include two isomorphisms between $A$ and $B$, you might later have to say that after all these two should be same, coherently...).
Generally, if you have some ($\infty$-)category $I$ and a functor towards $\mathrm{Cat}\_\infty$, such that all arrows go to left adjoint functors, then you also get a second functor from $I^{\mathrm{op}}$ towards $\mathrm{Cat}\_\infty$ giving the diagram of the right adjoints functors; and this procedure can be reversed, giving equivalences of $\infty$-category of such data. Informally, the formation of adjoints is unique (up to contractible choice) and functorial. Thus, one can hope that all the relevant coherence isomorphisms involving adjoint functors can be deduced from the data that is already present.
In the case of $6$-functor formalisms, it seems to me to indeed be the case that all the expected maps and isomorphisms that involve the right adjoint functors (internal Hom, $f\_\ast$, $f^!$) do indeed follow automatically. As an example, the base change formula involving $\ast$-pushforward and $!$-pullback is just the adjoint of the base change for $\ast$-pullback and $!$-pushforward. Or the formula $\mathrm{Hom}(f^\ast A,f^! B)=f^!\mathrm{Hom}(A,B)$ follows by passing to taking a partial right adjoint in the projection formula. In general, however, this is a heuristic statement; it is slightly difficult to justify by a theorem because the classical encodings of $6$-functor formalisms consists of some (slightly random) collection of functors, maps, and isomorphisms.
Let me note that there are some maps where I thought for a while that they would not be captured by this abstract notion, but only later realized that actually they are. This the equivalence between $f^\ast$ and $f^!$ for "etale" maps $f$, and between $f\_\ast$ and $f\_!$ for "proper" maps $f$. More generally, for "separated" maps $f$, one expects a natural transformation $f\_!\to f\_\ast$ that should be an isomorphism for proper $f$. In Gaitsgory-Rozenblyum's approach to $6$ functors, they make such transformations part of the datum by working with a more subtle $(\infty,2)$-categorical version of correspondences that allows proper maps of correspondences. There seemed to be a general sentiment that this is really necessary: In fact, Gaitsgory-Rozenblyum explicitly say in their work that they think that $(\infty,2)$-categories are critical to define and construct $6$-functor formalisms. This was, to me, a major psychological roadblock, as my knowledge of $(\infty,2)$-categories lacks far behind that of $(\infty,1)$-categories.
But in fact $(\infty,1)$-categories are enough! And you can automatically deduce the expected isomorphisms $f^\ast=f^!$ for etale $f$ and $f\_!=f\_\ast$ for proper $f$ (and $f\_!\to f\_\ast$ for separated $f$), as in Lecture 6 of my notes. (Really what happens is that there are inductively defined comparison maps, and they may or may not be isomorphisms; but in any case one does not have to supply further data.) So if one would have incorporated such maps $f\_!\to f\_\ast$ as part of the data of $6$-functor formalism, one should also ask that these maps are in fact the same as the automatic maps (together with all coherence isomorphisms involving them...). This is probably true in the Gaitsgory-Rozenblyum encoding, but I really haven't tried to check this.
As a final note, it would be really really nice if one could find a nice algorithm or graphical calculus or such that would help one verify expected commutative diagrams involving all $6$ functors. I'm not aware of any work in this direction. But let me note that (as I was made aware of by my student Adam Dauser) the passage from a $6$-functor formalism towards a symmetric monoidal $(\infty,2)$-category where morphisms are given by "Fourier-Mukai kernels" is an instance of something that Lurie has written down in his notes on the cobordism hypothesis (see for instance Corollary 3.3.35 for $n=2$), which is an area that very much uses such graphical calculus...
| 19 | https://mathoverflow.net/users/6074 | 442604 | 178,582 |
https://mathoverflow.net/questions/442147 | 4 | *This is a follow-up to [a recent question of mine](https://mathoverflow.net/questions/442104/example-of-trickiness-of-finite-lattice-representation-problem):*
For $n\in\mathbb{N}$ let $C(n)$ be the smallest $k$ such that every bounded lattice with cardinality $\le n$ which is isomorphic to the congruence lattice of some finite algebra, is isomorphic to the congruence lattice of some algebra with cardinality $\le k$.
>
> What do we know about the growth rate of $C(n)$?
>
>
>
I'm especially interested in *lower bounds*, since right now my main interest is pedagogical (explaining why the finite congruence representation problem is difficult). Keith Kearnes' answer to my above-linked question uses a theorem of Bunn/Grow/Insall/Thiem which implies that for infinitely many $n$ we have $C(n)\ge 2(n-3)$. However, I don't know whether $C$ is known to grow faster than linear.
| https://mathoverflow.net/users/8133 | How large must algebras with a given congruence lattice be? | **What do we know about the growth rate of $C(n)$?**
We know the exact value of $C(n)$ if, in its definition,
we restrict to the class of distributive lattices.
Otherwise we only have partial results. Let me say a few words.
**Claim.**
If $L$ is a finite distributive that has a maximal chain
$\ell\_1<\ell\_2<\cdots < \ell\_n$, then
1. $L$ is representable as the
congruence lattice of an algebra of size $n$, and
2. $L$ is not representable as the congruence
lattice of any algebra of size $< n$.
Reasons:
For for the first item, observe that
if $L$ is a finite distributive lattice
with a specified maximal chain
$\gamma: \ell\_1<\ell\_2<\cdots < \ell\_n$,
then it has a $0,1$-embedding into a Boolean lattice $B$
in such a way that the chain $\gamma$ remains a maximal chain.
Next, any Boolean lattice $B$ with an $n$-element
maximal chain is isomorphic to a power-set
lattice $\mathscr{P}(S)$ where
$S=\{s\_1,\ldots,s\_{n-1}\}$ is a set of size $n-1$.
Assume that $0\notin S$.
Next,
$\mathscr{P}(S)$ can be embedded into the lattice
$\Pi(\{0\}\cup S)$ of partitions of $\{0\}\cup S$
by mapping $U\subseteq S$ to the partition where $\{0\}\cup U$
is one cell and the other cells are singletons.
This is a $0,1$-embedding.
Altogether we have $0,1$-embeddings
$L\leq B\cong \mathscr{P}(S)\leq \Pi(\{0\}\cup S)$,
which compose to a $0,1$-embedding of
$L$ into a partition lattice
on a set of size $|\{0\}\cup S|=n$.
From here we cite
Quackenbush, R.; Wolk, B.
Strong representation of congruence lattices.
Algebra Universalis 1 (1971/72), 165-166.
which shows that any $0,1$-distributive sublattice $L$
of the lattice of the lattice of equivalence relations
on an $n$-element set is a representation of $L$
as a congruence lattice.
For the second item of the claim, it is impossible
to represent any lattice $L$ with a maximal chain
of $n$-elements as a congruence lattice
on a set of size $k<n$, since the full
lattice of partitions on a $k$-element set
does not have a chain of $n$ elements.
In particular, if we relativize the definition
of $C(n)$ to the class of distributive
lattices, we get the following.
Let $C\_{\textrm{Dist}}(n)$ be
the least $k$ such that every bounded *distributive*
lattice with cardinality $\leq n$ which is isomorphic
to the congruence lattice of some finite algebra,
is isomorphic to the congruence lattice of some
algebra with cardinality $k$.
With this definition and the preceding remarks,
we get $C\_{\textrm{Dist}}(n)=n$. (If you look
at all bounded distributive
lattices with cardinality $\leq n$,
they are all representable, and the hardest
to represent is the $n$-element chain.
This can be represented as the congruence lattice
of an $n$-element algebra, but of no smaller algebra.)
What I said above about distributive lattices
can be extended a little bit.
A finite distributive lattice is a sublattice
of a power of the $2$-element lattice, $\mathbf{2}$.
The same kinds of results mentioned above can be proved for
the smallest nondistributive lattices
$M\_3$ and $N\_5$. Namely, $M\_3$ can be represented as a congruence
lattice on a $3$-element set, $N\_5$ can be represented
as a congruence lattice on a $4$-element set, and any sublattice
of a power of either representation is a congruence representation.
For this see
Snow, John W.
Every finite lattice in $\mathscr{V}(M\_3)$
is representable.
Algebra Universalis 50 (2003), no. 1, 75-81.
and
Snow, John W.
Subdirect products of hereditary congruence lattices.
Algebra Universalis 54 (2005), no. 1, 65-71.
When you move to $M\_4$, it starts to get complicated, as is
explained in [an answer](https://mathoverflow.net/q/442105) to an earlier question.
In fact, it is not hard to produce a $0,1$-embedding of
$M\_q$ into the lattice of partitions of a $q$-element
set whenever $q$ is an odd prime, but the least size
congruence representation of $M\_{p+1}$ has
size $2p$ when $p$ is an odd prime. Thus, if $p$ and $q$
are odd primes satisfying $p<q<2p$, then there
is a $0,1$-embedding of $M\_{p+1}$ into $M\_q$
and then into the lattice
of partitions of a $q$-element set, but there is
no congruence representation of $M\_{p+1}$ on a $q$-element set.
Let me jump ahead to identify what seems to be the hardest
part of this problem. In
Pálfy, Péter Pál; Pudlák, Pavel
Congruence lattices of finite algebras and
intervals in subgroup lattices of finite groups.
Algebra Universalis 11 (1980), no. 1, 22-27.
Palfy and Pudlak isolate three properties of a
finite bounded lattice $L$ of size strictly
greater than $2$:
(A) $L$ is simple.
(B) For any nonzero $x\in L$ there exist
$y\_1, y\_2\in L$ such that $x\vee y\_1=x\vee y\_2=1$
and $y\_1\wedge y\_2=0$.
(C) Any $x\in L$ that is not zero and not an atom
dominates at least $4$ atoms.
Palfy and Pudlak show two things about these properties.
They show that if $L$ is any finite lattice,
then it can embedded as an upper
interval $[u,1]$ in a lattice $L'$ which satisfies
(A), (B), and (C) and has size $|L'|=5|L|+1$.
If one can represent $L'$ as a congruence
lattice of a finite algebra $\mathbf{A}$, then one
can represent $L$ as the congruence lattice
of $\mathbf{A}/u$. Thus, $L'$
(satisfying (A), (B), (C)) is only slightly larger
than $L$ (not necessarily satisfying (A), (B), (C)),
yet any upper bound on the size
of a minimal congruence lattice representation for $L'$
is also an upper bound on the size
of a minimal congruence lattice representation for $L$.
Second, they show that if $L$
satisfies (A), (B), and (C)
and $L$ is representable as the congruence
lattice of a finite algebra $\mathbf{A}$, then there
is representation of $L$ as the congruence lattice
of a transitive $G$-set $\mathbf{B}$ satisfying $|B|\leq |A|$.
This says that, if $L$ is sufficiently complicated
and representable as a congruence lattice, then it
is representable as the congruence lattice of
a transitive $G$-set for some finite group $G$.
Such $G$-sets are isomorphic to $G/H$
under the action of left multiplication by elements of $G$
for some subgroup $H\leq G$.
Hence, if $L$ satisfies the conditions (A), (B), and (C),
and $L$ is representable as the congruence
lattice of a finite algebra, then a minimal
congruence representation may be assumed
to be of the form of a transitive $G$-set $G/H$
of size $|B|=[G:H]$, and the lattice $L$
will be isomorphic to interval in $\textrm{Sub}(G)$
of subgroups containing $H$.
If we try to estimate $C(n)$
for some $n$ equal to the size of some
Palfy-Pudlack lattice (defined as satisfying
their conditions (A), (B), (C)), we
find that we are asking for the least
index $[G:H]$ in a finite group
if the interval in $\textrm{Sub}(G)$
of subgroups containing $H$ has size $n$.
The work of Palfy and Pudlak suggests that the hard
lattices to represent are those satisfying (A), (B), and (C).
If $L$ is such a lattice, which is representable
as a congruence lattice and it has size $n$,
then to compute $C(n)$ we must be able to
determine $[G:H]$ where $G/H$
is the smallest algebra affording a congruence
representation of $L$.
This has been attempted for the sequence
of lattices $L=M\_n$, for which see
Baddeley, Robert; Lucchini, Andrea
On representing finite lattices as intervals
in subgroup lattices of finite groups.
J. Algebra 196 (1997), no. 1, 1-100.
They give $G$-set congruence
lattice representations for $M\_n$ for
$n$ of the form
$1, 2, q+1, q+2, ((q^t+1)/(q+1))+1$,
$q$ a prime power. I believe that it is still
unknown whether $M\_n$ is representable for other
values of $n$.
| 3 | https://mathoverflow.net/users/75735 | 442610 | 178,584 |
https://mathoverflow.net/questions/442572 | 1 | Let $G=\operatorname{SO}^{+}\_{2n}(2)$. I did some Magma computation and found there were $3$ orbits on the natural $G$-set when $n=2,3,4$. The orbit sizes are $1$-$9$-$6$, $1$-$35$-$28$, $1$-$135$-$120$.
Is there a formula for the orbit sizes in general? Thank you.
| https://mathoverflow.net/users/488802 | Orbit sizes of $G=\operatorname{SO}^{+}_{2n}(2)$ | As Mikko Korhonen said in the comments, the (non-zero) singular and non-singular vectors form single orbits by Witt's lemma.
So it remains to count the singular vectors, which are (row) vectors ${\mathbf v}$ over ${\mathbb F}\_2$ such that ${\mathbf v}M {\mathbf v}^{\mathsf T} = 0$, where $M$ is the degree $2n$ matrix
$$\left(\begin{array}{cc}{\mathbf 0}&{\mathbf E}\\{\mathbf 0}&{\mathbf 0}\end{array}\right)\quad{\rm with}\quad {\mathbf E}=\left(\begin{array}{ccccc}0&0&\cdots&0&1\\0&0&\cdots&1&0\\&&\cdots&&\\0&1&\cdots&0&0\\1&0&\cdots&0&0\end{array}\right).$$
So we have to count the number of vectors ${\mathbf v}=(x\_1,x\_2,\ldots,x\_{2n})$ with
$$x\_nx\_{n+1}+x\_{n-1}x\_{n+2}+\cdots x\_1x\_{2n}=0.$$
This is a straightforward induction. For the base case $m=1$ there are three pairs $x\_1,x\_2$ with $x\_1x\_2=0$. For the inductive step, we have $2^{2n-3} + 2^{n-2}$ choices for $x\_1,\ldots,x\_{n-1}x\_{n+2},\ldots,x\_{2n}$
with $x\_{n-1}x\_{n+2}+\cdots x\_1x\_{2n}=0$ and the number of of choices for $x\_1,x\_2,\ldots,x\_{2n}$ is
$$3(2^{2n-3} + 2^{n-2}) + (2^{2n-3} - 2^{n-2}) = 2^{2n-1} + 2^{n-1}.$$
| 4 | https://mathoverflow.net/users/35840 | 442615 | 178,586 |
https://mathoverflow.net/questions/442297 | 5 | Reading [Chapter V](https://drive.google.com/file/d/1x9JpMslhI87jr6Utl4gX7Qf-etGbs6To/view?usp=sharing), pages (73-97) in *Proof Theory* (Springer, 1977), by Kurt Schütte, I have encountered a peculiar problem which puzzles me.
On page 96, a map $\rm{Nr}:\overline{\rm{OT}}\rightarrow \mathbb{N}$ is defined using the Cantor pairing function $\pi(m,n)=\frac{1}{2}(m+n)(m+n+1)+m$, where $\rm{OT}$ is the set of *ordinal terms* defined on page 86 and $\overline{\rm{OT}} = \rm{OT}\cup\{\Gamma\_{0}\}$. Namely (p.97),
1. $\rm{Nr}(0) := 1$, $\rm{Nr}(\Gamma\_{0})=0$.
2. $\rm{Nr}(\alpha) = \displaystyle\prod\_{i=1}^{n}P\_{\pi\left(\rm{Nr}(\alpha\_{i})-1,\rm{Nr}(\beta\_{i})-1\right)}$, for $\alpha = (\alpha\_{1},\beta\_{1})\ldots(\alpha\_{n},\beta\_{n})\in\rm{OT}$ (where $P\_{0}=2$ and for $k\ge 1$, $P\_{k}$ is the $k$-th odd prime number.
On the same page he provides a definition of the inverse $\tau$ of $\rm{Nr}$, thus proving Theorem 14.17 that $\rm{Nr}$ is a bijection.
All very nice. However, what maps to $31=P\_{10}$? Well, $10=\pi(0,4)$, and a simple calculation shows that it should be the term $\alpha = (0,((0,0),0))$. But this is not a principal term, i.e. $\alpha\not\in\rm{OT}$ because it represents $\varepsilon\_{0}$, and the unique ordinal term which represents this ordinal is $\beta=((0,0),0)\in\rm{OT}$. But $\rm{Nr}(\beta) = 5$. Then again, what element of $\rm{OT}$ maps to $31$?
What am I not understanding of Schütte's argument?
| https://mathoverflow.net/users/23204 | A possible flaw in Theorem 14.17 in Kurt Schütte's -Proof Theory- | Fortunately, there appears to be no flaw in Schütte's construction. Only in my understanding of it.
Ordinal terms are defined on page 86. The term $\alpha=(0,((0,0),0))$ does NOT represent $\varepsilon\_{0}$. After a more careful look at the definitions on page 86 and the definition of $\psi$ on page 84 I see now that $\alpha$ represents
$$\psi0(\psi(\psi00)0) = \psi0\varepsilon\_{0} = \phi0\varepsilon\_{0}^{'} = \omega^{\varepsilon\_{0}^{'}} = \omega^{\varepsilon\_{0}+1} = \varepsilon\_{0}\cdot\omega,$$
where $\alpha'$ denotes the successor of $\alpha$, and $\varepsilon\_{0} = \psi(\psi00)0$, which is the ordinal represented by the term $((0,0),0)$.
Therefore, $\alpha$ is indeed a principal term, properly answering my question "what maps to 31?" The map $\rm{Nr}$ is indeed a bijection.
| 4 | https://mathoverflow.net/users/23204 | 442631 | 178,591 |
https://mathoverflow.net/questions/442540 | 2 | If $A\subset\mathbb{R}^2$ is a Borel measurable set and $p\_\theta$ is projection onto the line spanned by $(\cos\theta,\sin\theta)$, then it is well known that for almost every $\theta\in[0,2\pi]$, $p\_\theta(A)$ has Hausdroff dimension the min of 1 and the Hausdorff dimension of $A$.
My question is: let's say $A$ has Hausdorff dimension 1, what would be an example where $p\_\theta(A)$ has Hausdorff dimension <1 for at least 3 different $\theta\in[0,2\pi]$? What if the number 3 is replaced by countable infinity?
| https://mathoverflow.net/users/147078 | Exceptional set for Marstrand's projection theorem | Theorem 1.1 of Orponen's paper "On the packing dimension and category of exceptional sets of orthogonal projections" contains a (relatively short) construction of a compact set $K\subseteq\mathbb{R}^2$ with $\infty>\mathcal{H}^1(K)>0$ and a dense $G\_\delta$ set $\Omega\subseteq[0,2\pi)$ such that $p\_\theta(K)$ has zero Hausdorff dimension, for all $\theta\in\Omega$. I would be interested if an expert here had an easy example for the simpler version of the question that we're asking here, however.
| 1 | https://mathoverflow.net/users/349327 | 442635 | 178,592 |
https://mathoverflow.net/questions/442638 | 2 | $\DeclareMathOperator\Mod{Mod}$I would like to compute the mapping class group (homeomorphism preserving orientation modulo those isotopic to the identity) of the sphere $S^2$ minus $n$ points $p\_1,\dots, p\_n$: $\Mod(S^2\setminus \lbrace p\_1, \dots, p\_n\rbrace)$. I already know that the mapping class group of the disk $D^2$ minus $n$ points is the braid group $B\_n$, whose presentation is well known.
I already know a presentation of $\Mod(S^2\setminus \lbrace p\_1, \dots, p\_n\rbrace)$ and it can be found for example in the book
Benson Farb, Dan Margalit - A Primer on Mapping Class Groups (Princeton Mathematical)-Princeton University Press (2011), page 258, Chapter 9.
However, I don't understand from there how to get the usual presentation if this group, which is
$$\langle \sigma\_1, \dots, \sigma\_{n-1} | rel. of B\_n, (\sigma\_1,\dots, \sigma\_{n-1})^n, \sigma\_1\dots\sigma\_{n-2}\sigma\_{n-1}^2\sigma\_{n-2}\dots \sigma\_1 \rangle. $$
Thus the question is:
How can use the mapping class group $\Mod(D^2)\cong B\_n$ to get the above presentation of $S^2\setminus \{p\_1,\dots, p\_n\}$?
An idea could be also sufficient :).
| https://mathoverflow.net/users/158821 | How to get a presentation of the mapping class group of the $n$-punctured sphere | $\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\Emb{Emb}\DeclareMathOperator\fix{fix}$There's a variety of ways to do this. If you take "mapping class group" to mean "isotopy classes of diffeomorphisms" then a fairly natural approach is to consider the braid group on $n$ strands to be $\pi\_0$ of diffeomorphisms of $S^2$ that preserve $n$ points and fix a disc pointwise. The mapping class group of the sphere with $n$ marked points removes this disc constraint. You have a fibre sequence
$$ \Diff(S^2, n, \fix D^2) \to \Diff(S^2, n) \to \Emb(D^2, S^2 \setminus n) $$
You get the presentation you seek by looking at the homotopy long exact sequence, and assembling the associated short exact sequence for $\pi\_0 \Diff(S^2, n)$. This notation again means isotopy classes of diffeomorphisms of $S^2$ that preserves $n$ points, as a set.
The space of embeddings of $D^2$ in the $n$-times punctured $S^2$ has the homotopy-type of $SO\_2 \times (S^2 \setminus n)$, i.e. a circle cross a wedge of circles.
| 5 | https://mathoverflow.net/users/1465 | 442640 | 178,594 |
https://mathoverflow.net/questions/442622 | 3 | Let $M$ be a connected and simply connected compact Riemannian manifold (without boundaries). Fix a point $p\in M$.
The diameter set $D\_p$ of $p$ is the set of points that maximize the distance from $p$, i.e., $D\_p=\lbrace q : d(p,q)=\max\_r d(p,r)\rbrace$.
The cut locus $C\_p$ is the set of points at which geodesics (starting at $p$) stop to be length minimizing.
It is clear that $D\_p \subset C\_p$. What are conditions under which equality holds? Does it always hold or is it, for instance, true for symmetric spaces?
| https://mathoverflow.net/users/485160 | Cut locus for simply connected manifolds | The equality $D\_p = C\_p$ holds for compact symmetric spaces (CROSSes) of rank $1$, but not in general for higher rank symmetric spaces.
A rank $1$ CROSS is isometric to a round sphere or projective space with Fubini-Study metric. For a sphere, we of course have $D\_p = C\_p = \{-p\}$. On $\mathbb{K}P^n$ with $\mathbb{K}\in\{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\}$, we have $D\_p = C\_p \cong \mathbb{K}P^{n-1}$.
For higher rank, consider $Sp(n)$ (the $n\times n$ quaternionic unitary matrices) with bi-invariant metric and set $p = I$, the identity matrix.
We claim that $D\_p = \{-I\}$. Indeed, given $q\in D\_p$, we may select a minimizing geodesic $\gamma$ from $p$ to $q$. This geodesic is a $1$-parameter subgroup, so can be conjugated to lie within the standard maximal torus $T^n\subseteq Sp(n)$. On the torus (which inherits its standard flat metric) clearly $-I$ is the farthest point from $I$. So, $q$ is conjugate to $-I$, which is central.
On the other hand $C\_p$ contains infinitely many points. To see this, recall that $Sp(n)\setminus \{p\}$ deformation retracts onto $C\_p$. This implies that in cohomology we have $H^3(C\_p)\neq 0$ for $n\geq 2$
| 5 | https://mathoverflow.net/users/1708 | 442641 | 178,595 |
https://mathoverflow.net/questions/442642 | 3 | Given an abstract simplicial complex $K$ on a set of vertices $V$, we can form a semi-simplicial set by $F(K)$ sending $F(K)\_n$ to be the set of ordered $(n+1)$-tuples of vertices in $V$ forming an $n$-simplex in $K$; face maps are defined in the obvious way. One can add in degeneracies by hand to make this a simplicial set.
In general, $|F(K)|$ is not homotopy equivalent to $K$; for example, $|F(\Delta^n)|$ is a wedge of $n$-spheres. (By contrast, the "singular simplicial set" where we don't require the $(n+1)$-tuple to form a $n$-simplex but merely any simplex, is homotopy equivalent and even homeomorphic in the geometric realization to $|K|$: see [this question](https://mathoverflow.net/questions/326431/turning-simplicial-complexes-into-simplicial-sets-without-ordering-the-vertices)) This I learned from the discussion [here](https://golem.ph.utexas.edu/category/2017/08/simplicial_sets_vs_simplicial.html). However, I wanted to know what information in the literature exists about the relationship between $|K|$ and $|F(K)|$ in general: for example, are there some restrictions on the homology or homotopy of $|F(K)|$ based on the homology or homotopy of $|K|$?
| https://mathoverflow.net/users/120548 | Simplicial set from all orderings of simplicial complex | If I understand this correctly, this construction shows up in various places in the study of homological stability. The one main result I know connecting the topology of $F(K)$ and $K$ is as follows. Say that $K$ is **weakly Cohen-Macaulay** of dimension $n$ if $K$ is $(n-1)$-connected and for all simplices $\sigma$ of $K$, the link of $\sigma$ is $(n-2-dim(\sigma))$-connected. For instance, it is easy to see that this holds if $K$ is a combinatorial triangulation of an $(n-1)$-connected manifold of dimension at least $n$ (the "dimension at least $n$" is here because I'm silently using the convention that only nonempty spaces can be $p$-connected for $p \geq -1$). We then have the following theorem:
**Theorem**: Let $K$ be a simplicial complex that is weakly Cohen-Macaulay of dimension $n$. Then $F(K)$ is $(n-1)$-connected.
I think this should be attributed to Randal-Williams--Wahl; see Theorem 2.14 of their paper [here](https://arxiv.org/abs/1409.3541). An alternate proof by Hatcher--Vogtmann can be found as Proposition 2.10 in their paper [here](https://arxiv.org/abs/1508.04334).
| 4 | https://mathoverflow.net/users/317 | 442644 | 178,596 |
https://mathoverflow.net/questions/442600 | 12 |
>
> For each real $A>0$, let $x\_A$ denote the positive root $x$ of the polynomial $x^5-3x-A$. Is the function $(0,\infty)\ni A\mapsto x\_A$ elementary?
>
>
>
~~[I am using this [definition of elementary functions](https://en.wikipedia.org/wiki/Elementary_function#Basic_examples), except for the algebraic functions mentioned there.]~~
*Edit:* After the edifying answer by [Timothy Chow](https://mathoverflow.net/a/442656/36721), I would like to clarify the use of the term *elementary functions* in this context just as Timothy Chow suggested: "the functions that can be expressed using some finite combination of constant functions, the identity function, $\exp$, $\log$, composition, and arithmetic operations".
| https://mathoverflow.net/users/36721 | Can the positive root of this polynomial be expressed elementarily? | I believe that the class of functions that Iosif Pinelis is interested in is what I would call *exponential-logarithmic functions* or *EL functions*; that is, they are the functions that can be expressed using some finite combination of constant functions, the identity function, $\exp$, $\log$, composition, and arithmetic operations (${+}{-}{\times}{\div}$). Some authors call this class of functions *elementary functions*, but that term is now more commonly used in a different sense, which includes algebraic functions. But Iosif Pinelis explicitly says he does not want to include all algebraic functions.
[Alexandre Eremenko's answer](https://mathoverflow.net/a/442617) cites Ritt, which shows that the function in question is not expressible using radicals only. However, as [Kevin Casto comments](https://mathoverflow.net/questions/442600/can-the-positive-root-of-this-polynomial-be-expressed-elementarily#comment1142413_442617), this result does not immediately answer the question of whether the function is an EL function.
[Kevin Casto's answer](https://mathoverflow.net/a/442650) cites a result of mine to show that the answer to [a version of] Iosif Pinelis's question is negative, but conditional on Schanuel's conjecture. So it is natural to ask if we can avoid assuming Schanuel's conjecture.
The answer is yes. The reference I usually cite is *Abel's Theorem in Problems and Solutions*, by V. B. Alekseev, Kluwer, 2004. Specifically, see the Appendix by Khovanskii. There are various theorems in this appendix which yield the desired negative answer, but perhaps the most explicit statement can be found in Section A.8:
>
> If an algebraic equation is not solvable by radicals then it remains non-solvable using the logarithm, the exponentials, and the other meromorphic functions in the complex plane.
>
>
>
| 19 | https://mathoverflow.net/users/3106 | 442656 | 178,601 |
https://mathoverflow.net/questions/442436 | 2 | We are given a vector $n$-dimensional random vector $\mathbf{X}$ whose components are the Bernoulli random variables $X\_1, X\_2, \ldots X\_n$, such that the probability $\mathbb{P}(X\_1=X\_2=\ldots=X\_n=0)$ is null.
---
**Question:** For $n>1$, is there any joint distribution of $\mathbf{X}$’s components and a vector $\mathbf{v}\in [0,1]^n$ such that
$$\frac{\mathbb{E}\langle\mathbf{X},\mathbf{v}\rangle}{\mathbb{E}\,{\sum\_{\!i=1}^{\!n}\!X\_i}}>\frac{1}{n}\mathbb{E}\left[\frac{\langle\mathbf{X},\mathbf{v}\rangle}{{\sum\_{\!i=1}^{\!n}\!X\_i}}\right]\ ?$$
| https://mathoverflow.net/users/115803 | On the mean value taken by Bernoulli random variables with joint distribution constraints | $\newcommand\X{\mathbf X}\newcommand\v{\mathbf v}$
Indeed, $1\le\sum\_{i=1}^n X\_i\le n$ and hence
$$n\sum\_{i=1}^n X\_i\ge n=En\ge E\sum\_{i=1}^n X\_i,$$
so that
$$\frac1n\,E\frac{\langle\X,\v\rangle}{\sum\_{i=1}^n X\_i}
=E\frac{\langle\X,\v\rangle}{n\sum\_{i=1}^n X\_i}\le
E\frac{\langle\X,\v\rangle}{E\sum\_{i=1}^n X\_i}
=\frac{E\langle\X,\v\rangle}{E\sum\_{i=1}^n X\_i},$$
as claimed. $\quad\Box$
| 4 | https://mathoverflow.net/users/36721 | 442664 | 178,605 |
https://mathoverflow.net/questions/438809 | 2 | I recently asked in [this thread](https://mathoverflow.net/questions/434320/lower-bounds-for-pattern-complexity-of-aperiodic-subshifts) about lower bounds on the complexity in the case where we have an aperiodic subshift. If I denote $c\_n(\Omega)$ as the number of possible patterns on $Q\_n=\{0,...,n−1\}^d$ and had hoped for estimates of the form
$$ \liminf\_{n\to \infty} \frac{c\_n(\Omega)}{n^d}\geq C\_d. $$
In the same thread, Ville Salo pointed to a work by Julien Cassaigne showing this is not the case when $d=3$. However, this question was motivated by a more specific class of **linearly repetitive** subshifts. Since the counterexample was motivated by a conjecture by Lagarias and Pleasents, [Repetitive Delone sets and quasicrystals](https://arxiv.org/abs/math/9909033), I noticed that the same paper references that this result is true when $\Omega$ is linearly repetitive. This is stated in Conjecture 1.2b and is said to be proven by Lenz in [Aperiodic linearly repetitive Delone sets are densely repetitive](https://arxiv.org/abs/math/0208132).
However I wished to verify whether this is me misinterpreting the result stated, as this is stated in the language of Delone sets and I am interested in such a result for subshifts. It seems like the difference in setting isn't important for Cassaigne's result in the [counterexample paper](https://oleron.sciencesconf.org/conference/oleron/pages/pleasants.pdf), but I'm not sure why that is necessarily so? I'm wondering whether the transitions between the two settings, might not mean that the result by Lenz is valid in subshifts?
| https://mathoverflow.net/users/143153 | Lower bounds for pattern complexity of linearly repetitive aperiodic subshifts | Just to give closure to this question, I think an earlier paper by Boris Solomyak, [Nonperiodicity implies unique composition for self-similar translationally finite Tilings](https://link.springer.com/article/10.1007/PL00009386), actually explicitly solves my question in the desired setting. In the paper by David Lenz, he shows the property of the complexity function in *Lemma 2.2* . This lemma follows easily from Lenz's *Lemma 2.1*, which states that a return vector that is 'too small' must be a period of the minimal subshift. Lenz mentions that *Lemma 2.1* is proven similarly to *Lemma 2.4* in Solomyak's paper.
In Solomyak's paper, dealing with self similar tilings with finite local complexity, the aforementioned lemma is phrased as follows.
>
> Let $\mathcal{U}$ be a self similar tiling with translation symmetry group $\mathcal{K}$. There exists a constant $N\geq 1$ such that, for any $x,y\in \mathbb{R}^d$, if a patch $P$ occurs in $\mathcal{U}$ while $x+P$ also occurs in $\mathcal{U}$ and $B\_r(y)\subseteq \text{supp}(P)$, then $\Vert x\Vert \leq \frac{r}{N}$ implies that $x\in \mathcal{K}$.
>
>
>
| 0 | https://mathoverflow.net/users/143153 | 442685 | 178,613 |
https://mathoverflow.net/questions/442686 | 11 | I am a new learner of Iwasawa theory and currently reading the famous [paper](https://link.springer.com/article/10.1007/s00222-013-0448-1) by Skinner-Urban in 2014, and the following-up works by many other people. When reading these papers, I found that some people tend to work over the *general* symplectic group $\mathrm{GSp}\_{2n}$ and *general* unitary group $\mathrm{GU}\_{r,s}$ (i.e. the ones with similitude factor) rather than the groups $\mathrm{Sp}\_{2n}$ and $\mathrm{U}\_{r,s}$ without similitudes. Yet meanwhile, some people often work with the ones without similitude, especially when doing quite explicit calculations (e.g. the pullback formula in Skinner-Urban's paper). Still, at least from the perspective of readers as naive and stupid as me, some authors just "randomly" switch the groups they are working with, without any hint.
So **I wonder** that what are the essential differences between the ones with similitude and those without similitudes? Why do people prefer one over the other?
I am so sorry if this post does not fit this advanced site and thank you all for commenting and answering!
| https://mathoverflow.net/users/161208 | Do people prefer working on $\mathrm{GSp}$ and $\mathrm{GU}$ rather than $\mathrm{Sp}$ and $\mathrm{U}$, and why? | *Symplectic case*: Here are two reasons (not necessarily the only ones) why $\operatorname{GSp}\_{2n}$ is more convenient to work with than $\operatorname{Sp}\_{2n}$.
* Firstly: there is **no Shimura datum with underlying group $\operatorname{Sp}\_{2n}$**; you need the "extra room" provided by $\operatorname{GSp}\_{2n}$. In practice, this means that although you can define Siegel Shimura varieties as varieties over $\overline{\mathbb{Q}}$ working purely with $\operatorname{Sp}$, there is no sensible way of describing canonical models over number fields until you bring in the larger group
* Secondly: The general symplectic group has a **richer theory of Hecke operators**, because you can consider the action of matrices like $\operatorname{diag}(1, ..., 1, p, ..., p)$, which normalise $\operatorname{Sp}\_{2n}$ but aren't contained in it. Having this larger Hecke algebra means that, although strong multiplicity one doesn't hold for either group (except for $n = 1$), it is "closer to being true" for $\operatorname{GSp}$ than $\operatorname{Sp}$ (i.e. the $L$-packets are smaller).
*Unitary case*: Here the issue is a bit more nuanced because there are various ways of choosing the Shimura datum: roughly, you can choose a Shimura datum for $U(a, b)$ which looks like $z \mapsto \operatorname{diag}(z / \bar{z}, \dots, z / \bar{z}, 1, \dots, 1)$, or you can choose one for $\operatorname{GU}(a, b)$ which looks like $\operatorname{diag}(z, \dots, z, \bar{z}, \dots, \bar{z})$. Each of these has its merits for different things: working with **$\operatorname{GU}(a, b)$ gives you a nicer moduli-space interpretation** for your Shimura variety (it is of PEL type), so its canonical models are easier to describe; but it is the **$U(a, b)$ one which has the nicer Hecke theory**.
I have the impression that there is a shift under way in the recent literature. 10 years ago, most of the arithmetic theory of automorphic forms on unitary groups (Shimura varities, Galois representations etc) was developed using $GU(a, b)$, as in the Skinner-Urban paper, or the Harris–Taylor book on local Langlands. However, more recently, people seem to have got better at working with non-PEL Shimura data; thus the recent work on unitary special cycles (arithmetic Gan–Gross–Prasad conjectures, etc) is mostly being done in the context of actual unitary groups. [This paper of Rapoport–Smithling–Zhang](https://arxiv.org/pdf/1906.12346.pdf) and [this paper of Jetchev](https://arxiv.org/pdf/1410.6692.pdf) have some useful discussion of the relations between the $U$ and $GU$ Shimura varieties.
| 13 | https://mathoverflow.net/users/2481 | 442695 | 178,616 |
https://mathoverflow.net/questions/442674 | 0 | From the definition of sub-Gaussian distribution $X$ w.r.t. $\sigma$ i.e.
$$\mathbb{P}(|X-\mathbb{E}(X)|\geq t) \leq 2 \exp(-\frac{t^2}{2\sigma^2}).$$
It's natural that when $X \sim \mathcal{N}(\mu, \sigma^2)$,
$$\mathbb{P}(X-\mu\geq t) \leq \exp(-\frac{t^2}{2\sigma^2}).$$
But this bound is too loose since when $t=0$, $\mathbb{P}(X-\mu\geq t) = \frac{1}{2} << 1$.
Thus I guessed somehow
$$\mathbb{P}(X-\mu\geq t) \leq \frac{1}{2}\exp(-\frac{t^2}{2\sigma^2}) \tag{1}$$
when $t\geq0$.
It can be verfied using the computer that $(1)$ is true. But how to prove it?
---
Mill's inequality $$\mathbb{P}(X-\mu\geq t) \leq \sqrt{\frac{2}{\pi}}\frac{\exp(-\frac{t^2}{2\sigma^2})}{\frac{t}{\sigma}}$$ helps solve my guess when $t$ is big. But when $t$ is near zero, how to solve it?
---
I have previously mistyped $\mathbb{P}(X-\mu\geq t)$ by $\mathbb{P}(X-\mu\leq t)$. Great thanks to [Brendan McKay](https://mathoverflow.net/users/9025/brendan-mckay) to point this out!
| https://mathoverflow.net/users/500967 | Upper bound about Gaussian tail bound | Letting
$Z:=\frac{X-\mu}\sigma$ and $z:=\frac t\sigma$, rewrite the inequality in question as
$$d(z):=P(Z\ge z)-e^{-z^2/2}/2\le0 \tag{1}\label{1}$$
for real $z\ge0$.
Note that
$$d'(z)=(z-\sqrt{2/\pi})e^{-z^2/2}/2.$$
So, $d$ is decreasing on the interval $[0,\sqrt{2/\pi}]$ and increasing on the interval $[\sqrt{2/\pi},\infty)$. Also,$d(0)=0$ and $d(z)\to0$ as $z\to\infty$. So, inequality \eqref{1} follows. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 442699 | 178,619 |
https://mathoverflow.net/questions/442603 | 2 | Let $X$ be a smooth intersection of two quadrics in $\mathbb{P}^{2n+1}$.
Question: Does there exist a smooth projective variety $Y$ (different from $X$) of dimension $2n-1$ such that $Y$ is birational to $X$ and if $U$ is the maximal open subset of $Y$ which is isomophic to an open subset of $X$, then $Y \setminus U$ has co-dimension at least $2$ ?
| https://mathoverflow.net/users/174161 | Examples of projective manifold birational to intersection of two quadrics | When n=1 they are elliptic curves, so the answer is no.
When n>1, then $Y$ and $X$ have dimension at least 3. By the Lefschetz hyperplane theorem for $X$, we have $\mathrm{Pic}(X) \cong \mathbb{Z}$. As the codimension of $Y\setminus U$ is 2 we have $\mathrm{Pic}(Y) \cong \mathrm{Pic}(U)$ and the latter is a quotient of $\mathrm{Pic}(X)$ (by excision for the inclusion $U\subset X$). As $Y$ is projective, $\mathrm{Pic}(Y) \cong \mathbb{Z}$. As $\mathrm{Pic}(U) \cong \mathbb{Z}$, then again by excision the codimension of $X\setminus U$ in $X$ is at least 2.
The pullback of the ample generator ($\mathcal{O}\_X(1)$) of $\mathrm{Pic}(X)$ to $U$ extends to the ample generator $L$ of $\mathrm{Pic}(Y)$. By Hartog's extension type results: $H^0(Y,L^{\otimes m})\cong H^0(U,L^{\otimes m}|\_U)
\cong H^0(X,\mathcal{O}\_X(m))$. As some power is very ample, it follows that $Y \cong X$.
So for any $n$, the answer is no.
| 5 | https://mathoverflow.net/users/17630 | 442700 | 178,620 |
https://mathoverflow.net/questions/442676 | 5 | How many monotone mappings $[n] \times [n] \to [n]$ exist? Here:
* $[n]$ denotes $\{1, 2, \ldots, n\}$,
* Monotone means that if $x\_1 \le x\_2$ and $y\_1 \le y\_2$, then $f(x\_1, y\_1) \le f(x\_2, y\_2)$.
I'm interested in the answer up to $2^{\Theta(\cdot)}$-notation. To give an example, I would be absolutely happy if the answer is $2^{\Theta(n \log n)}$, while $2^{\Theta(n^2)}$ would kill my idea.
Anything strictly less than $2^{\Theta(n^2)}$ would be an improvement for me, but I would like an upper bound with a quasi-linear exponent (if it exists, of course).
If we instead consider monotone $[n] \to [n]$, then the number of mappings is around $\binom{2n-1}{n}$ (we can think of it as $2n - 1$ balls, where $n$ balls represent the numbers $1, \ldots, n$, and remaining $n-1$ balls represent positions where the function value increases by $1$).
I've tried to apply this idea to the $[n]^2 \to n$, but I failed to utilize monotonicity on *both* arguments, and could only get $n^{O(n^2)}$.
The answer for $[n]^2 \to [n]$ is clearly between $n^{\Theta(n)}$ (we get this much even for $[n] \to [n]$) and $n^{n^2}$ (it's the number of all possible functions $[n]^2 \to [n]$).
I'm also interested in the same question for mapping $[n]^d \to [n]$, where $d \in \mathbb N$ (and, in computer science terms, $d$ is *not* a fixed parameter).
[Dedekind numbers](https://en.wikipedia.org/wiki/Dedekind_number) answer the question for monotone boolean functions, which can be thought of as $[2]^d \to [2]$.
P.S.: No idea which tags to use.
| https://mathoverflow.net/users/151271 | Number of monotone functions on a square grid | Thanks to [Alex Lazar](https://mathoverflow.net/users/128452/alex-lazar) for referring me to the right keywords.
This is called something like "the number of [plane partitions](https://en.wikipedia.org/wiki/Plane_partition) that fit inside the $n \times n \times n$ box", and [can be computed as](https://mathworld.wolfram.com/PlanePartition.html) :
$$
PL(n,n,n)=\frac{G^3(n+1) \cdot G(3n+1)}{G^3(2n+1)},
$$
where $G$ is the [Barnes G-function](https://en.wikipedia.org/wiki/Barnes_G-function). Mathematica tells me that the highest-order term in the Taylor expansion of its natural logarithm is
$$\frac 32 (3 \ln 3 - 4 \ln 2) n^2 \approx 0.785 n^2.$$
So the answer to my question is $2^{\Theta(n^2)}$.
| 3 | https://mathoverflow.net/users/151271 | 442711 | 178,623 |
https://mathoverflow.net/questions/442385 | 1 | Let $q = p^s$ and $r = q^m$, where $p$ is a prime, $s$ and $m$ are positive integers. Let $N>1$ be an integer dividing $r - 1$, and put $n = (r - 1)/N$.
Let $\alpha$ be a primitive element of $\mathbb{F}\_r$, $\theta = \alpha^N$, and $Tr\_{r/q}$ be the trace function from $\mathbb{F}\_r$ to $\mathbb{F}\_q$. The set
$$C(r,N) = \{ (Tr\_{r/q}(\beta), Tr\_{r/q} (\beta \theta),\dots, Tr\_{r/q}(\beta \theta^{n-1})) : \beta \in \mathbb{F}\_r\}$$
is an irreducible cyclice code of length $n$ over $\mathbb{F}\_q$.
Let's define the set
$$Z(r,a) = \#\{x \in \mathbb{F}\_r: \text{Tr}\_{r/q}(x) = 0 \}.$$
The paper "The weight distribution of some irreducible cyclic codes" by Cunsheng Ding makes the following statement:
>
> Hence, for any $\beta \in \mathbb{F}\_r^\*$ the Hamming weight of the codeword
>
>
> $$c(\beta) = (Tr\_{r/q}(\beta), Tr\_{r/q} (\beta \theta),\dots, Tr\_{r/q}(\beta \theta^{n-1}))$$
>
>
> of the code $C(r,N)$ is equal to
>
>
> $$n - \frac{Z(r, \beta) - 1}{N}.$$
>
>
>
Can anyone provide any proof for this statement?
| https://mathoverflow.net/users/269936 | Weight of a codeword in a cyclic code as a function of the number of solutions of an equation involving the trace function | The codeword $c(\beta)$ in your question is defined on a set via the trace map that does not include the zero element. Thus its number of nonzero elements is simply the length *minus* the number of $k\in \{0,1,\ldots,n-1\}$ which give $Tr(\beta \theta^k)=0.$
From the equidistribution properties of the trace function and the fact that we are considering a subfield of index $N$ the number of zeroes of the trace function *minus* 1, in the extension field, needs to be divided by $N$ when projected into the subfield.
| 1 | https://mathoverflow.net/users/17773 | 442715 | 178,624 |
https://mathoverflow.net/questions/442714 | 7 | Recently, I have been discussing **inverses** with a tenth grade class and **integrals** with an eleventh/twelfth grade class, and this has led me to the following wonder:
>
> **Wonder.** Is there a "reasonable" way to quantify which of finding an inverse and finding an antiderivative is more "difficult"? Meaning, which of these processes is "easier" or which of these processes is more likely to lead to a "pleasant" inverse or antiderivative?
>
>
>
For example, one can use composition to create functions that are easy to invert but difficult to find an antiderivative for; e.g., $x \mapsto e^{x^3}$. (More generally, I am thinking about [**Liouville's Theorem**](https://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra)) on when it's possible to express an antiderivative using elementary functions.) On the other hand, if one simply restricted to the set of polynomials in $\mathbb{R}[x]$, then they would all be straightforward to integrate, but finding a pleasant inverse for those of degree five or higher would be almost always impossible. (In this polynomial context, I am thinking about the [**Abel-Ruffini Theorem**](https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem).)
Ways to refine the language of this question or pointers to pre-existing explorations of related wonders are all welcome. (Efforts to re-tag will be welcomed, too.)
| https://mathoverflow.net/users/22971 | Quantifying difficulty of integrals versus inverses | Finding integrals is easier, and finding inverses is more difficult, in many ways.
We can consider both integrability and invertibility in two ways:
* *Global integrability*: Is there a bound on $\int\_a^b f(x)\,dx$?
* *Elementary integrability*: Is there an elementary $F$ with $F'=f$?
* *Global invertibility*: Is $f$ a bijection from its domain to its range?
* *Elementary invertibility*: Is there an elementary $g$ with $f(g(x))=g(f(x))=x$ on some domain?
**Comparing the Elementary Concepts**
* It is easy to integrate a cubic $ax^3+bx^2+cx+d$, but awkward to find the elementary inverse. I have never found it worthwhile to memorize the old formulas and doubt many others have.
* A variety of probability distributions have elementary cdfs (which are the integrals of the pdfs), but non-elementary quantiles (which are the inverses of those cdfs). E.g. this occurs for beta distributions $B(a,b)$ with integer $a,b$; for the continuous Bernoulli distribution $\mathcal{CB}(\lambda)$; for the Wigner semicircle distribution; and for Erlang distributions, i.e. gamma distributions with integer $k$. Also, many probability distributions have nice-ish cdfs but not nice-ish quantiles; e.g. for the normal distribution, the first good asymptotic for the cdf is
$$F(x)\sim 1-\exp(-x^2/2))/\sqrt{2\pi x^2}$$ but the first good asymptotic for the quantile function is $$Q(1-p)\sim\sqrt{-\log(4\pi p^2)-\log(-\log(2\pi p^2)/2)}$$
* In the universe of 100 simple elementary expressions of the form $f\_1(x)\star f\_2(x)$, where each $\star$ is one of $+,-,\times,/$ and each $f$ is one of $x,\log (x),e^x,\sqrt{x},x^2$:
+ 87 have elementary integrals. (The remaining non-elementary integrals can be expressed using Erf, Erfi, and ExpIntegralEi, except for $\int e^x/\log(x)\, dx$, for which Mathematica provides no other expression.)
+ Only 44 have elementary inverses on appropriate domains. (This includes the four like $x^2\pm\sqrt{x}$, which have elementary inverses so messy that probably few people on this site could find them without assistance.)
**Comparing the Global Concepts**
* The globally integrable functions have nice closure properties: they are a ring. By contrast, the globally invertible functions are not closed under either subtraction or multiplication, and not even closed under monotonic functions like $\log$ or $\tan$ because $\log(\log(x))$ has a smaller domain than $\log$, and $\tan(\tan(x))$ has singularities. (There is one set of invertible functions with nice closure properties: the unbounded monotonic functions on $\mathbb{R}^{\ge 0}$ with $f(0)=0$. That set is closed under addition, multiplication, composition, and functional inverses, but it excludes the very example in the post of $x\to e^{x^3}$, favoring things like $x\to e^{x^3}-1$.)
* The map $f\to f’$ takes any bounded differentiable function to a globally integrable function. The closest analog for globally invertible functions is that the maps $f\to \int f^2$ and $f\to \int e^f$ take almost any function to globally invertible functions (in the first case one has to exclude functions which vanish on an interval), but those maps are not as nice; eg, for reasonable domains, their images may not include the identity.
* For rational functions $p(x)/q(x)$, it is easy to specify which are globally integrable: those where $q$ has no real roots and $\deg(q)\ge \deg(p)+2$. It is awkward to specify which rational functions are globally invertible, e.g. $ax^3+bx^2+cx+d$ is invertible iff (try it without looking first!)
>
> $b^2\le 3ac$ and either $a$ or $c$ is non-zero.
>
>
>
| 8 | https://mathoverflow.net/users/nan | 442717 | 178,625 |
https://mathoverflow.net/questions/442709 | 2 | I would appreciate it if a reference could be given for the following claim.
Let $g$ be a Riemannian metric on $\mathbb R^n$, $n\geq 2$ that is equal to the Euclidean metric outside some compact set. Let us define for each $\alpha\in (0,1)$ the fractional Laplace--Beltrami operator on $\mathbb R^n$ via
$$ (-\Delta\_g)^\alpha u = \int\_0^\infty t^{-1-\alpha} (e^{\Delta\_g}-1)u \,dt.$$
Prove that the operator $(-\Delta\_g)^\alpha$ maps the **Schwartz space** $S(\mathbb R^n)$ continuously into itself.
I suppose that the following estimate on the heat kernel
$$ |e^{t\Delta\_g}(x,y)| \leq C t^{-\frac{n}{2}} e^{-c\frac{|x-y|^2}{t}} \quad \forall \, t>0 x,y \in \mathbb R^n,$$
is used to prove this but would anyone have a reference for this?
| https://mathoverflow.net/users/50438 | On the Fractional Laplace-Beltrami operator | This is false even for the euclidean fractional Laplacian. For a cheap proof, the Fourier transform of $(-\Delta)^{a/2}u$ is $|\xi|^a\widehat u(\xi)$ which is not smooth at 0. For a more solid proof, it is not difficult to prove that $(-\Delta)^{a/2} u$ for $u$ a compactly supported function decays like $\sim|x|^{-n-a}$.
| 2 | https://mathoverflow.net/users/7294 | 442721 | 178,626 |
https://mathoverflow.net/questions/442694 | 2 | My understanding is that convex hull of n points in 4D could have O(n²) edges in the worst case. Source: [https://sites.cs.ucsb.edu/~suri/cs235/ConvexHull.pdf](https://sites.cs.ucsb.edu/%7Esuri/cs235/ConvexHull.pdf)
This same source writes
>
> In 4D, there are n points in general
> position so that the edge joining every
> pair of points is on the convex hull!
>
>
>
But I can't seem to demonstrate this in practice. I tried the follow distributions of points:
* Gaussian distribution (e.g., `V = randn(n,4)`)
* Uniform distribution in box (e.g., `V = rand(n,4)`)
* Uniform distribution on 3-sphere (e.g., `V=normalize_each(randn(n,4));`)
The 3-sphere lead to the worst number of edges but it still empirically looks linear as I increase n.
Is there a known pathological distribution of points that really hits this quadratic behavior as n increases?
| https://mathoverflow.net/users/23064 | Example of worst case distributions for 4D convex hull | I'm making my comment an answer so this doesn't appear on unanswered lists.
The convex hull of points on the [moment curve](https://en.wikipedia.org/wiki/Moment_curve)
$$
(t, t^2, t^3, t^4)
$$
form what is known as a [cyclic polytope](https://en.wikipedia.org/wiki/Cyclic_polytope).
This is a "neighborly polytope" with the property
(in this case with $d=4$) that every pair of vertices
is connected by an edge.
| 3 | https://mathoverflow.net/users/6094 | 442728 | 178,629 |
https://mathoverflow.net/questions/442726 | 9 | Let $F$ by a finite field, and $R(x\_1,x\_2,\ldots,x\_n) := r\_1(x\_1,x\_2,\ldots,x\_n)/r\_2(x\_1,x\_2,\ldots,x\_n)$ a rational function in $n$ variables, frequently in analytic number theory or harmonic analysis one comes across the need to estimate the exponential sum$$ S:= \sum\_{x \in F^n} = e( R(x\_1,x\_2,\ldots,x\_n) ). $$
In the one variable case, a result of Bombieri (using Weil's proof of RH for curves) gives a nearly complete understanding in particular:
$$|S| \lesssim |F|^{1/2} $$if the degree of $deg(r) = deg(r\_1) + deg(r\_2)$ is greater than one, where the constant only depends on (things that depend on) the degree of $f$.
In the multivariate case one still hopes for estimates of the form $|S| \lesssim |F|^{n/2}$ unless something pathological happens (with again perhaps a reasonable dependence of the implied constant on $r$).
There is a whole literature of such estimates stemming from Deligne's work on the Riemann Hypothesis for varieties by authors such as Katz, Hooley, Adolphson-Sperber, Kowalski, Fouvry, Michel, etc. Unfortunately, the hypotheses of these theorems often involve complicated concepts from algebraic geometry that make it difficult for the non-algebraic geometer to understand (much less apply). To take one example, a theorem of Hooley gives a rather general approach to the bivariate case, but then you need to say things about the geometry of the curves $r\_i(x,y) - t = 0$ in the closure $\overline{F}$ which do not seem easy to analyze. It seems that things only get more complicated in more variables, with theorem statements involving even more exotic concepts (e.g. sheafs, mondromy, etc).
My (vague) question is: Why aren't simpler theorem hypotheses possible?
I welcome any attempts to explain this. But for concreteness, let me ask two better posed variants of this:
[1] My sense is that two sources of complications arise from (1) trying to have an explicit dependence of the implied constant on $r$, and (2) exceptional conspiracies between $r$ and the field. If we only want some very crude constant (depending on the degree of $r$) and an estimate for sufficiently large characteristic for a fixed function do things get simpler?
[2] In what ways can square root cancellation really fail asymptotically for rational (or just polynomial) sums? We know the answer is basically never in one variable thanks to Bomberi's theorem. In the multivariable case, the only sums I'm aware of have dummy variables that can be factored out. By which I mean something like$$\sum\_{x,y} e(x^2 +2xy + y^2) = \sum\_{x,y} e( (x+y)^2 ) = |F| \sum e(x^2) \sim |F|^{3/2}.$$ Are there functions where asymptotic square root cancelation fails for a `deep' reason that can't be explained elementarily by properties of the sum (obviously one can always save at least a factor of $|F|^{1/2}$ by fixing all but one variables)?
| https://mathoverflow.net/users/630 | Why are Deligne-type exponential sum estimates so hard to use? | There are a lot of subtle reasons such exponential sums can fail to exhibit square-root cancellation. First let me comment on two reasons suggested in your answer:
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> (1) trying to have an explicit dependence of the implied constant on $r$
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This should never be a problem. Katz proved a while ago simple explicit Betti bounds that suffice for problems of this type, so in any kind of sophisticated cohomological argument the explicit constant should be no worse than one arising from Katz (which will depend only on the degree and number of variables, but not be incredibly far from the optimum).
>
> (2) exceptional conspiracies between
> and the field.
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This can indeed pose a problem, and removing it can simplify things - basically, it should almost always let us use only the classical, characteristic zero topological versions of sheaves and monodromy and such instead of their exotic, étale cohomological incarnations, but not remove them entirely.
Why is there so much difficulty in the statements of these results, and a seeming need to invoke difficult geometric concepts?
One reason, or at least one way of shedding light on the difficulty, is that the same geometric tools (sheaves, monodromy, and more elementary geometric concepts) that let us prove upper bounds for exponential sums in some cases let us prove lower bounds in other cases. So we can rig up exponential sums that fail to exhibit square-root cancellation for more or less subtle geometric reasons. Thus any bound requires, one way or another, avoiding these examples.
Here's a simple one. The sum
$$ \sum\_{a \in F} \left(\sum\_{x\_1,x\_2\in F} e( x\_1 +x\_2 +a /(x\_1x\_2))\right)^n,$$ which can be expanded into a $2n+1$-variable sum, admits full square-root cancellation if and only if $n$ is divisible by 3.
Why? The monodromy computed by Katz of the hyper-Kloosterman sum is $SL\_3$. Katz's computation shows that the sum over $x\_1,x\_2$ always has full square-root cancellation, but the sum over $a$ cancels *if and only if* the $n$th tensor power of the standard representation of the monodromy group admits an invariant vector, which happens if and only if $n$ is divisible by $3$.
One, more elementary, example, arose in a conversation with a couple mathematicians who were specifically interested in character sums of products of linear factors. I therefore considered:
$$ \sum\_{x,y \in F} \chi \left( xy \prod\_{\zeta\_1,\zeta\_2\in \mu\_3} (1 + \zeta\_1 x+\zeta\_2 y) \right) $$
where $F$ is a field of size congruent to $1$ mod $6$, $\chi$ is a character of the multiplicative group of $F$ of order $2$ (i.e. a Legendre symbol - certainly something that appears often in analytic number theory - if $|F|$ is prime) and $\mu\_3$ are the third roots of unity in $F$.
This product of linear factors can actually be expressed as a rational function of $\frac{xy}{ x^3+y^3+1}$ by an algebraic manipulation related to the Dwork family and expressed in that variable one can see that it does not admit square-root cancellation.
So for any bound on multiplicative character sums one somehow needs to rule out examples like this one.
A third example concerns sums of the special form $$\sum\_{x\_1,\dots,x\_n\in F, y\in F^\*} e( y f(x\_1,\dots, x\_n))$$ where one can trivially eliminate the variable $y$ but this just reduces one to the not-obviously-simpler problem of counting zeroes of $f$. Counting zeroes of $f$ can be as hard as counting points on basically any algebraic variety and sum - like abelian varieties - do not exhibit square-root cancellations for reasons that are geometrically meaningful (in this case, related to the topology of a complex torus) but not at all apparent from staring at the equations.
| 15 | https://mathoverflow.net/users/18060 | 442729 | 178,630 |
https://mathoverflow.net/questions/442740 | -2 | Let $f$ be a polynomial with real coefficients in several indeterminates $x\_1, \dots, x\_n$. Suppose that
$$ f = g^2 $$
for some polynomial $g$.
Is it true that we can find polynomials $h\_1, \dots, h\_m$ which only involve monomials of even degree such that $f = {h\_1}^2 + \dots + {h\_m} ^2$?
| https://mathoverflow.net/users/136356 | can the square of a polynomial be written as a sum of squares of polynomials with only even degree terms? | There do not necessarily exist polynomials $h\_1,\ldots,h\_m$ which only involve monomials of even degree such that $f = h\_1^2 + \cdots + h\_m^2$, even in the case when $f$ is a univariate polynomial.
A polynomial $h$ that only involves monomials of even degree is an even polynomial: it satisfies $h(x) = h(-x)$ for all $x$. A sum of squares of even polynomials must also be an even polynomial. However, the square of a polynomial need not be even: for example, if $g(x) = x+1$ then $f = g^2$ is not an even polynomial, and hence it can't be expressed as $f = h\_1^2 + \cdots + h\_m^2$ for even polynomials $h\_1,\ldots,h\_m$.
| 2 | https://mathoverflow.net/users/8049 | 442741 | 178,634 |
https://mathoverflow.net/questions/442727 | 6 | **Question:** Let $k$ be an algebraically closed field, and $A,B$ abelian varieties over $k$. Suppose there exists an isogeny $A\to B$. Does this imply existence of a *separable* isogeny $A\to B$?
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**Known Cases:** The answer is yes when $k$ has characteristic zero (because every isogeny is separable). More interestingly, the answer is also yes for elliptic curves over $\overline{\mathbb{F}\_p}$: Since every isogeny factors into a Frobenius map and a separable map, it suffices to prove the result for $B=A^{(p)}$.
* If $A$ is ordinary, then the Verschiebung map (dual to Frobenius) is separable, and this is true for all isogenous curves as well. So if the field of definition of $A$ is $\mathbb{F}\_{p^k}$, then the composition of Verschiebung maps
$$A\simeq A^{(p^k)}\to A^{(p^{k-1})}\to\cdots\to A^{(p^2)}\to A^{(p)} $$
is separable. (This argument applies more generally to any ordinary abelian variety over $\overline{\mathbb{F}\_p}$.)
* If $A$ is supersingular, we can take any prime $\ell\neq p$ and use the fact that the $\ell$-isogeny graph of supersingular curves is connected.
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**Further thoughts:**
I don't think the second bullet point above is the right place to look for a generalization; connectedness of $\ell$-isogeny graphs of abelian varieties is a very difficult problem in general, and it's also much stronger than we need. I also don't know how to approach the problem when $k\neq\overline{\mathbb{F}\_p}$ has characteristic $p$.
| https://mathoverflow.net/users/404359 | Is there a separable isogeny between any two isogenous abelian varieties? | The answer is no. I think Asvin's example in the comments is correct but there may be a few things to check. I will give a different example that is easier to check, using Moret-Bailly's famous example of a nonisotrivial supersingular abelian surface.
Let $E$ be a supersingular elliptic curve over $F$, the algebraic closure of a finite field of characteristic $p$ and $A = E \times E$. The kernel of Frobenius on $A$ is isomorphic to $\alpha\_p \times \alpha\_p$ and we take $B = A/G$ where $G$ is the subgroup scheme of kernel of Frobenius given by $\{y=tx\} \subset \alpha\_p \times \alpha\_p$ where $t$ is transcendental over $F$. Then $k$ will be the algebraic closure of $F(t)$, $B$ is defined over $k$ but does not descend to $F$.
Now, any separable isogeny with source $A$ is $A \to A/H$, where $H$ is a finite etale subgroup of $A$, but such a subgroup is defined over $F$ so $A/H$ is defined over $F$ and can't be isomorphic to $B$.
| 9 | https://mathoverflow.net/users/2290 | 442749 | 178,636 |
https://mathoverflow.net/questions/442747 | 4 | Summary
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The probability distribution (pdf) of a random walk in 1 dimension is represented by a Bessel function. On the other hand, the pdf of a Brownian motion in free space is represented by a Gaussian distribution. While it is possible to derive the diffusion equation from the master equation of a random walk by taking the limit of small jumps, **is it possible to directly derive the Gaussian distribution as some asymptotic form of the Bessel function?** In other words, can I derive the pdf of a Brownian motion directly from the pdf of a random walk, rather than from the equation of the random walk?
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Consider a continuous time random walk on a neareast-neighbour 1d lattice, where each jump is of size $d$. The probability of being at site $n$, i.e. being at a distance $nd$ of the origin is described by:
$$\dot{P}\_{n}(t)=\frac{1}{2}\left(P\_{n-1}(t)+P\_{n+1}(t)\right)-P\_{n}(t)$$
The solution is $P\_n(t)=I\_n(t)e^{-t}$, where $I\_n(t)$ is the Bessel function.
When we extend the random walk on the real line, i.e $d\to 0$, the probability is described by the diffusion equation:
$$\frac{\partial p(x,t)}{\partial t}=D\frac{\partial^2p(x,t)}{\partial x^2}$$
The solution is $p(x,t)=\frac{1}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}$.
Is it possible to derive the expression of $p(x,t)$ from our expression of $P\_n(t)=I\_n(t)e^{-t}$, instead of starting from the master equation? I.e. how does the Bessel function relate to the Gaussian distribution in the limit $d\to0$?
If the question is not clear, or you would like more details, please let me know and I will edit the question.
| https://mathoverflow.net/users/420641 | Derive the solution of the diffusion equation from the solution of a random walk | To carry out the limit, it helps to start from an integral representation of the Bessel function,
$$P\_n(T)=e^{-T}I\_n(T)=\frac{1}{2\pi}\int\_{-\pi}^\pi \exp [i k n+T \cos k-T]\,dk.$$
For $T\gg 1$ this may be approximated by expansion of the exponent to second order in $k$,
$$P\_n(T)\approx\frac{1}{2\pi}\int\_{-\infty}^\infty \exp[ikn-\tfrac{1}{2}Tk^2]\,dk=(2\pi T)^{-1/2}e^{-n^2/2T}.$$
Define $x=nd$, $t=\tau T$, $D=\tfrac{1}{2}d^2/\tau$ to arrive at
$$d^{-1}P\_n(T)\approx \frac{1}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}\equiv p(x,t). $$
Here is a plot for $d=0.1$, $D=1$, $t=1$; blue is $p(x,t)$, red is $d^{-1}P\_n(T)$, the two curves are indistinguishable.
| 6 | https://mathoverflow.net/users/11260 | 442750 | 178,637 |
https://mathoverflow.net/questions/442696 | 3 | My question concerns differences between the spectral *radius* $\rho$ and *norm* $\| \cdot \|$ of Markov operators in infinite-dimensional Banach spaces. This is far from my area of expertise, that is more "mixing for finite, typically reversible, Markov chains", where spectral properties are often simpler.
I have a Markov chain on a state space $\Omega := \{ x \in \mathbb R^n \mid \sum\_i x\_i = 0 \}$..
* The support of the jumps is discrete: let $P(x,y) := \mathbb P\_x(X\_1 = y)$; then, $\sup\_{x \in \mathbb R^n} | \{ y \in \mathbb R^n \mid P(x,y) > 0 \} | < \infty$.
* Non-zero jump probabilities $P(x,y) \in [0,1]$ are uniformly bounded away form $0$ and $1$.
* It is irreducible: given $x \in \Omega$ a positive-Lebesgue-volume set $A \subseteq \Omega$, there is a finite path $x = y\_0, y\_1, ..., y\_k \in A$ such that $P(y\_{\ell-1}, y\_\ell) > 0$ for all $\ell$.
* It has a unique invariant distribution, which I denote $\pi$.
* Importantly, $P$ *is not* reversible wrt $\pi$—ie, $P$ is not self-adjoint: $P \ne P^\star$.
I also view $P$ as an operator on the set of functions: $(Pf)(x) := \int\_\Omega f(y) P(x, dy)$. (If $\Omega$ were discrete, then $(Pf)(x) = \sum\_{y \in \Omega} P(x,y) f(y)$ in the usual matrix–vector way.)
Constant functions are always eigenfunctions of $P$ with eigenvalue $1$. So, the spectral radius and norm of $P$ on $L^2(\Omega, \pi)$ is always $1$. I thus restrict to the subspace of $L^2(\Omega, \pi)$ orthogonal to the constant functions—as I believe is standard?—which I denote $B$.
>
> I have a 'spectral-gap' bound $\rho(P) \le 1 - \kappa$ of $P$ on $B$, for some $\kappa > 0$. I want to show *something along the lines of*
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> > $\rho\bigl( \tfrac12 (P + P^\star) \bigr) \le \tfrac12 \bigl( 1 + \rho(P) \bigr) \le 1 - \kappa/2$.
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> Recall that $P$ is a Markov operator, so $\rho(P) \le \| P \| \le 1$. I'm not bothered about constants on $\kappa$.
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Highly relevant is [this MO question](https://mathoverflow.net/questions/100688/convexity-of-spectral-radius-of-markov-operators-random-walks-on-non-amenable-g), but it doesn't address quite what I want here.
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I know the standard facts.
* $\rho(T) = \inf\_{k\ge1} \| T^k \|^{1/k} \le \| T \|$ for all bounded, linear operators $T$.
* $\rho(T) = \| T \|$ if $T$ is *normal* ($T T^\star = T^\star T$) of which *self-adjoint* ($T = T^\star$) is a special case.
One approach would be to prove that $\rho(P) = \| P \|$, even though $P$ is not normal in my case. Then,
>
> $\rho\bigl( \tfrac12 (P + P^\star) \bigr) = \| \tfrac12 (P + P^\star) \| \le \tfrac12( \| P \| + \| P^\star \| ) = \| P \| = \rho(P)$,
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since $P + P^\star$ *is* self-adjoint, $\| P \| = \| P^\star \|$ and $\rho(T) = \| T \|$ if $T = T^\star$.
It seems very likely that $\rho(P) = \| P \|$, in my (poorly-informed) view; how much 'nicer' do you want $P$ to be? The only concrete way I know to prove this is normality: $P P^\star = P^\star P$. But, this does not hold.
Another approach would be to relate the two spectral radii quantities more generally, at least for Markov operators. Note that $\rho$ is neither subadditive nor submultiplicative, and $\rho(T) \ne \rho(T^\star)$ for general $T$, in general.
| https://mathoverflow.net/users/59264 | Spectral Radius and Spectral Norm for Markov Operators | If I understood correctly, you are asking whether the spectral gap $\gamma=1-\rho$ of a non-reversible Markov chain $P$ provides any universal control on the Poincaré constant (which is the spectral gap of the additive reversibilization of $P$, or in your notation, $1-\|P\|$). The answer is no, even on finite state spaces: consider the Markov chain on $\{0,1\}^n$ which, at each step, replaces the current state $x=(x\_1,\ldots,x\_n)$ with either $(x\_2,\ldots,x\_{n},0)$ or $(x\_2,\ldots,x\_{n},1)$, each with probability a half. This chain has the maximum possible spectral gap, namely $\gamma=1$. Yet, its Poincaré constant tends to $0$ as $n\to\infty$.
| 2 | https://mathoverflow.net/users/477827 | 442753 | 178,638 |
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