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https://mathoverflow.net/questions/412416
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0
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$\DeclareMathOperator\Orb{Orb}\newcommand\abs[1]{\lvert#1\rvert}$The Collatz or the $3n+1$ conjecture is open.
1. Are there non-trivial polynomials $f(x)\in\mathbb Z[x]$ and $g(x)\in\mathbb R[x]$ having unbounded $\abs{f(n)}$, $\abs{g(n)}$ as $n$ grows in $\mathbb N$ and an integer $x\_0\in\mathbb N$ satisfying $\abs{g(x)}<2\abs{f(x)}+1$ for every $x>x\_0$ such that for every $m\in\mathbb N\_{>x\_0}$ there is a $k\in \Orb(2\abs{f(m)}+1)$ such that $k<\abs{g(m)}$?
Here $\Orb(t)$ is the set of numbers $t$ traverses through as we apply Collatz transformations.
2. If $f(x)$ is the *non-polynomial* $(3^x-1)/2$ is there a **polynomial** $g(x)$ as in *1.*?
Note $1$ need not be in $\Orb(\abs{g(x)})$ for every $x\in\mathbb N\_{x\_0}$ for both *1.* and *2.*.
|
https://mathoverflow.net/users/10035
|
Polynomials, $3^x$ and the Collatz conjecture
|
Point 1. is easy to satisfy even with nontrivial polynomials, since if you know $t$ modulo $2^n$ then you know for sure what the first n transformations will be.
For example. I could take $f=2^{20}x+174762$ and $g=6x+2$. Indeed I choose $f$ such that $3(2f(n)+1)+1)$ is always divisible by $2^{20}$, meaning that the first 21 transformations applied to $2f(n)+1$ are
1. $3x+1$ because $2f(n)+1$ is odd;
2. $x/2$ repeated 20 times because $3(2f(n)+1)+1)=2^{20}(6n+1)$ is divisible by $2^{20}$.
After this we will have reached $k = 6n+1$ which is smaller than $g(n)=6n+2$.
Point 2. is likely very hard and I have no idea how to tackle that. Indeed answering 1. with polynomials $f$ and $g$ with $\deg g < \deg f$ will already be very difficult. Although when taking $f$ to be the similar looking non polynomial $f=2(4^x-1)/3$ one can take $g=2$ for silly reasons.
|
1
|
https://mathoverflow.net/users/23501
|
412461
| 168,292 |
https://mathoverflow.net/questions/412463
|
1
|
Let $\{a\_k\}(k\ge 0)$ be a sequence of nonzero real numbers which changes signs infinitely often. Suppose $|a\_k|\to 0 $ and $|a\_k|$ decreases fast. Let $n$ be a positive integer. What's the relation between
$$\sum\_{k\ge 0}\binom {n+k}{k}a\_k $$ and $$\sum\_{k\ge 0}\binom {n+k}{k}\frac{a\_k}{k}. $$ For the asymptotic behavior, as $n\to \infty$, are $$\frac{\sum\_{k\ge 0}\binom {n+k}{k}a\_k}{\sum\_{k\ge 0}\binom {n+k}{k}\frac{a\_k}{k}}=O(1)$$ and
$$\frac{\sum\_{k\ge 0}\binom {n+k}{k}a\_k}{\sum\_{k\ge 0}\binom {n+k}{k}\frac{a\_k}{k}}=O(\frac{n}{\log n})$$ true or not? I tried to use Abel's summation formula, but it seems that Abel's summation formula is not applicable and I don't find relevant references.
|
https://mathoverflow.net/users/159935
|
Relation between $\sum_{k\ge 0}\binom {n+k}{k}a_k $ and $\sum_{k\ge 0}\binom {n+k}{k}\frac{a_k}{k}$
|
Since the second sum cannot start at $k=0$, I assume that both sums start at $k=1$.
Consider a particular example: $a\_k = k\alpha^k$ with $|\alpha|<1$. Then
$$\sum\_{k\geq 1} \binom{n+k}k \frac{a\_k}k = (1-\alpha)^{-(n+1)}-1$$
and
$$\sum\_{k\geq 1} \binom{n+k}k a\_k = (n+1)(1-\alpha)^{-(n+2)}\alpha.$$
Then the ratio of the two $\approx \frac{(n+1)\alpha}{1-\alpha}$, which is not $O(\frac{n}{\log(n)})$.
|
3
|
https://mathoverflow.net/users/7076
|
412466
| 168,295 |
https://mathoverflow.net/questions/412384
|
5
|
It is a well-known fact that given a first-order sentence $\psi$ in prenex normal form $\forall x\_1 \exists y\_1 \forall x\_2 \exists y\_2 \dots \forall x\_n \exists y\_n \theta(x\_1,\dots,x\_n,y\_1,\dots,y\_n)$ (i.e., $\theta$ is quantifier free or at least $\Delta^0\_0$ in the appropriate sense), the truth of $\psi$ in a given structure $M$ can be understood in terms of a game in which two players ($\forall$ and $\exists$) alternatively choose elements of $M$ corresponding to the bound variables of $\psi$, with $\forall$ choosing for the variables $x\_i$ and $\exists$ choosing for the variables $y\_i$. $\exists$ wins if and only if at the end the resulting variable assignment makes $\theta$ true in $M$. The characterization is of course that $M \models \psi$ if and only if $\exists$ has a winning strategy for this game.
Such a strategy can be understood as a series of Skolem functions $f\_1(x\_1),f\_2(x\_1,x\_2),\dots,f\_n(x\_1,x\_2,\dots,x\_n)$ on $M$ with the property that (after expanding $M$ with the $f\_i$'s), $M \models \forall x\_1\dots x\_n \theta(x\_1,\dots,x\_n,f\_1(x\_1),\dots,f\_n(x\_1,\dots,x\_n))$. In the specific case of arithmetic (i.e., the structure $(\mathbb{N},0,+,\cdot)$), these functions are closely related to the idea of realizability. If $\psi$ is provable in Heyting arithmetic, for instance, then one can extract computable functions $f\_1,f\_2,\dots,f\_n$ giving a winning strategy for $\exists$.
Kreisel showed that an analogous weaker statement is true of Peano arithmetic. A proof of $\psi$ in Peano arithmetic gives $\exists$ a computable procedure for beating any particular strategy for $\forall$, provided that $\exists$ is given $\forall$'s strategy in advance. More formally, given a proof of $\psi$ in Peano arithmetic, we can computably extract Turing indices $e\_1,\dots,e\_n$ such that for any functions $g\_1 : \mathbb{N}^0 \to \mathbb{N}$, $g\_2 : \mathbb{N}^1 \to \mathbb{N}$, $\dots$, $g\_n : \mathbb{N}^{n-1} \to \mathbb{N}$, we have that
$ \mathbb{N} \models \theta(a\_0,\dots,a\_n,b\_0,\dots,b\_n), $
where $$a\_i = g\_i(b\_1,\dots,b\_{i-1})$$ and $$b\_i = \Phi\_{e\_i}^{g\_1\dots g\_n}(a\_1,\dots,a\_{i})$$ for each $i \in [1,n]$. ($\Phi\_e^X(n)$ is of course the $e$th Turing functional applied to $n$ given $X$ as an oracle. Assume that we're using some fixed computable tupling function.) This fact is often called Kreisel's 'no-counterexample interpretation' of Peano arithmetic.
Let's say that $\bar{e} = (e\_1,\dots,e\_n)$ is a *weak classical realizer* of $\psi$ if the above condition holds. (I don't know if this concept has an established name.) For the sake of this question, we'll write $\bar{e} \Vdash \psi$ to mean that $\bar{e}$ is a weak classical realizer of $\psi$. (Note that $(\exists \bar{e})\bar{e}\Vdash \psi$ is a lightface $\Pi^1\_1$ condition on $\psi$.) I believe that the commonly studied notions of classical realizers for arithmetic (e.g., realizability in the sense of Krivine) are a priori stronger than this in that while they allow a certain amount of probing of $\forall$'s stragey, they do not necessarily allow arbitrary computation from it.
We can now cosmetically restate Kreisel's result as saying that if $\mathsf{PA} \vdash \psi$, then $(\exists \bar{e})\bar{e} \Vdash \psi$. It's also not too hard to show that if $(\exists \bar{e})\bar{e} \Vdash \psi$, then $\mathbb{N} \models \psi$. (No amount of prep work can beat an unbeatable strategy.)
I'm curious about the extent to which the conditions $\mathsf{PA} \vdash \psi$, $(\exists \bar{e})\bar{e} \Vdash \psi$, and $\mathbb{N} \models \psi$ differ. One immediate observation is that if $\psi$ is a true $\Pi^0\_1$-sentence, then $(\exists \bar{e})\bar{e} \Vdash \psi$ (somewhat trivially), which means that there are many examples of $\mathsf{PA}$-unprovable sentences with weak classical realizers (a Gödel sentence for instance). This leads to a somewhat more nuanced question.
>
> **Question 1.** Is there a sentence $\psi$ such that $(\exists \bar{e})\bar{e} \Vdash \psi$, yet $\mathsf{PA} \cup \{\chi: \mathbb{N}\models\chi,~\chi\text{ a }\Pi^0\_1\text{ sentence}\} \not \vdash \psi$?
>
>
>
In the other direction, there is basically no way that $\mathbb{N} \models \psi$ implies $(\exists \bar{e})\bar{e} \Vdash \psi$, but finding an explicit example of this seems difficult. The theory of $\mathbb{N}$ is lightface $\Delta^1\_1$, which is simpler than the apparent complexity of the set of sentences with weak classical realizers (lightface $\Pi^1\_1$).
>
> **Question 2.** Is there a sentence $\psi$ such that $\mathbb{N} \models \psi$, yet $\neg (\exists \bar{e}) \bar{e} \Vdash \psi$?
>
>
>
|
https://mathoverflow.net/users/83901
|
Which arithmetical sentences have no counterexamples in the sense of Kreisel?
|
In fact all true arithmetical sentences have weak classical realizers. Namely, a true arithmetical sentence $\psi$ $$\forall x\_1\exists y\_1 \ldots \forall x\_n\exists y\_n \theta(x\_1,\ldots,x\_n,y\_1,\ldots,y\_n)$$
is weakly realized by $(e\_1,\ldots,e\_n)$, where $\Phi\_{e\_i}^{\vec{g}}(y\_1,\ldots,y\_{i-1})$ performs an unbounded search for a tuple $(y\_i,\ldots,y\_n)$ such that $\theta(g\_0(),\ldots,g\_n(y\_1,\ldots,y\_n),y\_1,\ldots,y\_n)$ is true and then outputs $y\_i$ from the first found tuple. This works since for any $g\_1\colon\mathbb{N}^0\to\mathbb{N},\ldots,g\_n\colon \mathbb{N}^{n-1}\to\mathbb{N}$ the sentence $\exists y\_1,\ldots, y\_n \theta(g\_0(),\ldots,g\_n(y\_1,\ldots,y\_n),y\_1,\ldots,y\_n)$ is true.
|
3
|
https://mathoverflow.net/users/36385
|
412470
| 168,296 |
https://mathoverflow.net/questions/412433
|
8
|
I would like to know what motivated or led Thom to think that the (un)oriented cobordism groups would correspond with the homotopy groups of some structure (Thom spectum), or with the coefficient groups of a cohomology theory.
|
https://mathoverflow.net/users/167503
|
What motivated Thom to relate the cobordism groups with some homotopy groups?
|
I did not know Thom so I can't speak to all his personal motivations. But I can speak as someone that has read much of his work carefully and I think I have some insights into this.
The primary motivation for his theorem boils down to thinking carefully about the implicit function theorem, and asking when a subset of Euclidean space is the pre-image of a regular value of a smooth function.
These are the problems most students grapple with when learning about smooth manifolds for the first time, and Thom's considerations flowed out of these kinds of naive considerations. Thom just went a little further than most.
The first step is realizing a submanifold of $S^n$ is the pre-image of a regular value of a smooth function
$$f : S^n \to S^m$$
if and only if it has no boundary, is compact and has a trivial tubular neighbourhood. And this is the key insight into the Pontriagin construction.
The next step is asking about general manifolds, and this is where you take the step up to maps to Thom spaces. Specifically the structure of the normal bundle is key in the Pontriagin construction. So if you have a non-trivial normal bundle you need some feature of your map to take that into account. Grassman manifolds are the key object related to maps out of vector bundles. But once you see that Grassman manifolds are the key analogue to the Pontriagin construction that gives you your map defined in a tubular neighbourhood of your submanifold. Coning off as in the Pontriagin construction gives you the Thom space.
From this point of view you can see his theorem as a direct extrapolation from the trivial tubular neighbourhood case. I think for people that like to think categorically it could maybe feel unintuitive since the Thom space is a departure from manifolds.
As has been mentioned, spectra came after the fact, as the idea was germinating around that time. In Thom's paper he largely phrased things in the language of stable homotopy groups.
|
11
|
https://mathoverflow.net/users/1465
|
412475
| 168,297 |
https://mathoverflow.net/questions/412474
|
7
|
Let $G \subseteq \mathbb F\_p^\*$ be a subgroup. Then $G$ is called almost trivial if $G \cap (2-G)$ consists of the element 1.
Then I am wondering how big $G$ can be in terms of $p$. If $G$ is a random set containing 1 then according to the birthday paradox one expects that $G$ and $2-G$ have nontrivial intersection as soon as $\#G \gg \sqrt p$.
>
> I am wondering if maybe one could prove that all almost trivial subgroups of $\mathbb
> F\_p^\*$ have size at most $\tilde O(\sqrt p)$.
>
>
>
Or if a $\sqrt p$ up to log factors bound is out of reach, I was wondering if there is $1/2 < \epsilon < 1$ and some constant $c$ such that for all almost trivial groups we have $\#G < cp^\epsilon$.
I did some numeric experimentation by finding the largest almost trivial subgroup of $\mathbb F\_p^\*$ for all p < 80000. And the largest ratio that I could find for $\frac {\#G}{\sqrt p}$ was $\frac {884}{\sqrt {41549}} = 4.3368\ldots$ because $\mathbb F\_{41549}$ contains an almost trivial subgroup of order 884, so a square root (possibly up to log factors) bound seams reasonable.
P.S. The notion of being almost trivial is based on the notion of almost rational in the paper [Almost rational torsion points on semistable elliptic curves by
Frank Calegari](https://doi.org/10.1155/S1073792801000253).
|
https://mathoverflow.net/users/23501
|
Subgroups of the multiplicative group of a finite field satisfying a certain additive property
|
If we let $S$ be the set of characters of $\mathbb F\_p^\times$ trivial on $G$ then $$\sum\_{\chi \in S} \chi(g) = \begin{cases} \frac{p-1}{|G|} & g\in G \\ 0 & g\notin G \end{cases}$$
so $$\sum\_{\chi\_1,\chi\_2\in S} \sum\_{ g \in \mathbb F\_p \setminus \{0,1,2\} } \chi\_1(g) \chi\_2(2-g) = 0$$ if $G$ is almost trivial.
Now if $\chi\_1 =\chi\_2=1$ then $\sum\_{ g \in \mathbb F\_p \setminus \{0,1,2\} } \chi\_1(g) \chi\_2(2-g)=p-3$ and otherwise, by the bound for Jacobi sums, $\left| \sum\_{ g \in \mathbb F\_p \setminus \{0,1,2\} } \chi\_1(g) \chi\_2(2-g) \right| \leq \sqrt{p}+1 $.
Thus, if $G$ is almost trivial then $p-3 \leq ( ((p-1)/|G|)^2-1) (\sqrt{p}+1)$, or $p \leq c^2 p^{5/2} / |G|^2$ for a constant $c$, meaning $|G| \leq c p^{3/4}$, answering the weaker form of your question.
|
13
|
https://mathoverflow.net/users/18060
|
412477
| 168,299 |
https://mathoverflow.net/questions/412241
|
7
|
Let $G\_{k,n}$ be the grassmannian of $k$-dimensional vector spaces of $\mathbb R^n$. By the Courant–Fisher characterization, the $k$th largest eigenvalue of an $n \times n$ psd matrix $A$ is given by
$$
\tag{1}\label{1}
\lambda\_k = \min\_{V \in G\_{n-k+1,n}} R(A,V),
$$
where $R(A,V):= \max\_{x \in V,\,\lVert x\rVert = 1} x^\top A x$.
Now, let $V$ be drawn according to the Haar distribution on $G\_{k,n}$, and replace the min in \eqref{1} by expectation over $V$.
>
> **Question.** What does $\alpha\_k(A) := \mathbb E\_V [R(A,V)]$ correspond / evaluate to?
>
>
>
**Note.** I'm really only interested in (good) lower-bounds.
Examples
--------
Let $P\_V$ be the orthogonal projector for $V$.
* If $k=1$, then $R(A,V) = v^\top A v$, where $v$ is uniform on the unit-sphere in $\mathbb R^n$, and so $\alpha\_1(A) = \operatorname{trace}(A)/n$.
* If $k=n$, then $P\_V = I\_n$ with probability $1$ and so $\alpha\_n(A) = \lambda\_\text{max}(A)$.
* If $A = I\_n$, then obviously $\alpha\_k(A) = 1$ for all $k \in [n]$.
* If $A = uu^\top$, a rank 1 matrix, then $\alpha\_k(A) = \mathbb E\_V R(A,V) = \mathbb E\_V\lVert P\_V u\rVert^2 = (k/n)\lVert u\rVert^2$.
|
https://mathoverflow.net/users/78539
|
What does $\mathbb E_V \max_{x \in V,\,\|x\|=1} x^T Ax$ evaluate to when $V$ is random $k$-dim suspace of $\mathbb R^n$ and $A$ is fixed psd matrix?
|
In another answer, @dohmatob establishes the lower bound
$$
\alpha\_k(A)\geq\frac{1}{n}\max\Big\{\operatorname{tr}(A),k\lambda\_{\max}(A) \Big\}.
$$
In what follows, we show that this bound is near-optimal for general $A$ in the regime where $k$ is at most a fraction of $n$.
Let $\Lambda$ denote the diagonal matrix of eigenvalues of $A$, and let $G$ denote an $n\times k$ matrix with iid $N(0,1)$ entries. Submultiplicativity gives
\begin{align\*}
\alpha\_k(A)
&=\alpha\_k(\Lambda)
=\mathbb{E}\|(G^\top G)^{-1/2}G^\top\Lambda G(G^\top G)^{-1/2}\|\_{2\to2}\\
&\leq\mathbb{E}\Big[\|(G^\top G)^{-1/2}\|\_{2\to2}\cdot\|G^\top\Lambda G\|\_{2\to2}\cdot\|(G^\top G)^{-1/2}\|\_{2\to2}\Big]
=\mathbb{E}\bigg[\frac{\|\Lambda^{1/2}G\|\_{2\to2}^2}{\sigma\_{\min}(G)^2}\bigg].
\end{align\*}
It is known that $\sigma\_{\min}(G)\sim\sqrt{n}-\sqrt{k}$ when $n$ and $k$ grow proportionately, and there are nonasymptotic results of this flavor, too; see [Vershynin's survey](https://arxiv.org/abs/1011.3027). Next, equation (4.6) from Tropp's paper on [User-friendly tail bounds](https://arxiv.org/abs/1004.4389) implies that (with $B := \lambda^{1/2} \otimes 1\_n$, where $\lambda \in \mathbb R^n$ is the vector of eigenvalues of $A$)
$$
\begin{split}
\|\Lambda^{1/2}G\|\_{2\to2}^2 = \|B \circ G\|\_{2 \to 2}^2
&\leq 2\log(\tfrac{n+k}{\epsilon})\cdot\max(\|\lambda^{1/2}\|^2\_2,k\|\lambda^{1/2}\|\_\infty^2)\\
&= 2\log(\tfrac{n+k}{\epsilon})\cdot\max(\operatorname{trace}(A),k\lambda\_\max(A))
\end{split}
$$
with probability $\geq1-\epsilon$. One may combine these estimates to show that @dohmatob's lower bound is tight up to log factors.
|
2
|
https://mathoverflow.net/users/29873
|
412481
| 168,302 |
https://mathoverflow.net/questions/412480
|
-3
|
Suppose that $f\_{4}(x)$ is a polynomial of degree 4 with no multiple roots, $C$ is the curve defined by $y^{2}=f\_{4}(x)$, I want to show that there is a polynomial $f\_{3}(x)$ of degree 3 with no multiple root such that $C$ is birational equivalent to the curve defined by $y^{2}=f\_{3}(x)$, I completely don't know how to do it, can anyone help me
|
https://mathoverflow.net/users/206621
|
Show that a polynomial of degree 4 is birational equivalent to a polynomial of degree 3
|
As Alexandre pointed it out, just use the linear-fractional transformation to send one point to infinity. Explicitly, denote the four roots of $f\_4$ to be $a,b,c,d$, then there is a unique linear-fractional transformation
$$\sigma:\mathbb P^1\to \mathbb P^1,$$
sending $a,b,c$ to $0,1,\infty$. Denote $\lambda$ to be $\sigma(d)$. Let $C'$ be the curve defined by $y^2=x(x-1)(x-\lambda)$. Then its direct to show that $C$ and $C'$ are birational. Indeed, $C$ with $\sigma^{-1}(\infty)$ removed is isomorphic to $C'$ with $\sigma(\infty)$ removed.
In general, a hyper-elliptic curve of genus $g$ can be represented by the equation $y^2=f(x)$ for some polynomial $f(x)$ with all roots being simple and the degree of $f(x)$ is either $2g+1$ or $2g+2$, depending on whether the infinity point is ramified when considered as a double cover to the $x$-line.
|
5
|
https://mathoverflow.net/users/74322
|
412484
| 168,304 |
https://mathoverflow.net/questions/412485
|
1
|
Let $M \in \mathbb{N}$ and let $\pi \in S\_{M}$ be an involution with at least one fixed point. I'm interested in finding a latin square $A$ of order $M$ such that $A\_{i,j} = \pi(A\_{j,i})$ for each $i,j \in \{1,\ldots, N^2\}$.
A $B\_1$-type latin square is a latin square $A$ indexed by $\mathbb{Z}\_{n,1} = \mathbb{Z}\_n \cup \{\infty\}$ such that $A\_{i+1, j+1} = A\_{i,j} + 1$ for every $i,j \in \mathbb{Z}\_{n,1}$. Theorem 11 [in this survey](https://www.semanticscholar.org/paper/Diagonally-cyclic-latin-squares-Wanless/22cf4ee8bf0412592fa984173a520200359f8a4f) shows that the latin squares I'm looking for exist when $M \geq 2$ is odd, $\pi$ has a unique fixed point and there exists a symmetric $B\_1$-type latin square ($B\_1$-type latin squares of order $M$ exist for every $M \geq 2$ by Theorem 6 in the same document).
I'm particularly interested in the case when $M = N^2$ for some $N \in \mathbb{N}$ and $\pi \in S\_{N^2}$ has exactly $N$ fixed points. By permuting rows and columns we can assume that $\pi$ fixes the last $N$ elements in $\{1,\ldots, N^2\}$ and $\pi(i) = (N^2 - N)/2 + i$ for $i \in \{1,\ldots, (N^2 - N)/2\}$.
My original motivation comes from constructing symmetric finite-state codes.
|
https://mathoverflow.net/users/409086
|
Existence of latin squares with an involutory symmetry
|
To achieve your goal, we can use the "tensor product construction" for Latin squares. It is carried out as follows: Let $X\_a$ be a latin square indexed by $a$, i.e. $X\_a$ is a function $a \times a \rightarrow a$ and $X\_b$ similarly. Then $X\_a \otimes X\_b : (a \times b) \times(a \times b) \rightarrow (a \times b) $ is defined as $((a\_1,b\_1),(a\_2,b\_2)) \rightarrow (X\_a(a\_1,a\_2),X\_b(b\_1,b\_2))$.
The point is that, if there exists permutations $\pi$ and $\rho$ such that $X\_a(i,j) = \pi(X\_a(j,i))$ and $X\_b(i,j) = \rho(X\_b(j,i))$ for all $i$, $j$, then $X\_a \otimes X\_b$ satisfies $X\_a \otimes X\_b (k,l)=(\pi \otimes \rho) (X\_a \otimes X\_b (l,k))$, where $\pi \otimes \rho$ is a permutation on $a \times b$ defined by $(\pi \otimes \rho)(a,b)=(\pi(a),\rho(b))$. The number of fixed points of $(\pi \otimes \rho)$ is the product of that of $\pi$ and that of $\rho$.
For $p$ an odd prime, it's possible to construct latin squares $A\_p(p)$ and $A\_1(p)$ such that the $A\_p(p)$ has all classes invariant under involution, and $A\_1(p)$ has exactly one class invariant under involution.
Index $A\_p(p)$ and $A\_1(p)$ by $\mathbb{Z}\_p$. Then $A\_p(p)$ and $A\_1(p)$ can be defined as $A\_p(x,y)=x+y$ and $A\_1(x,y)=x-y$.
Thus $A\_p(p) \otimes A\_1(p)$ has order $p^2$ and $p$ classes invariant under involution. Call it $B\_p$.
We also need a special latin square $B\_2$ that has two classes invariant under involution:
```
1 2 3 4
4 3 2 1
3 4 1 2
2 1 4 3
```
By the factorization of $N$ into primes $N=2^{n\_0}p\_1^{n\_1}...p\_k^{n\_k}$ (the primes $p\_1,...,p\_k$ are odd) we can construct a latin square $(\otimes^{n\_0} B\_2) \otimes (\otimes^{n\_1} B\_{p\_1}) \otimes ... \otimes (\otimes^{n\_k} B\_{p\_k})$ that has order $N^2$ and $N$ classes in involution.
|
2
|
https://mathoverflow.net/users/125498
|
412486
| 168,305 |
https://mathoverflow.net/questions/412490
|
2
|
We got experimental evidence against the semi strong perfect graph theorem
and would like to learn what is wrong with it.
From [Recognizing the P4-structure of bipartite graph](https://reader.elsevier.com/reader/sd/pii/S0166218X99001043?token=95BAE4CC0839E24CC218B933184ED467786AF8799730EAA190CC2E2E1412A9F12C57EA719ED0FA8FD2E4ABC2A7973546&originRegion=eu-west-1&originCreation=20211225051901)
>
> The P4-structure of a graph $G=(V; E)$ is a hypergraph $H=(V;E)$ such that the hyperedges from $H$ correspond to the vertex sets of the induced P4s in $G$.
> The **semi strong perfect graph theorem** states that a graph is perfect if and only if it has the P4-structure of a perfect graph.
>
>
>
Evidence against this.
Let $G$ be the perfect graph:
```
G=Graph([(0, 1), (0, 2), (0, 3), (1, 2), (1, 6), (2, 9), (3, 4), (3, 5), (4, 5), (4, 7), (5, 10), (6, 7), (6, 8), (7, 8), (8, 11), (9, 10), (9, 11), (10, 11)])
```
Let H be the P4-structure of G with 492 edges given [here](https://pastebin.com/raw/i3NWGFhZ).
$H$ is not perfect according to Sagemath.
>
> Q1 What is wrong with this counterexample?
>
>
>
Code can be tested on [Sagemath online](https://sagecell.sagemath.org/).
|
https://mathoverflow.net/users/12481
|
What is wrong with the experimental evidence against the semi strong perfect graph theorem?
|
This is not a counterexample. $H$ is the $P\_4$-structure of $G$, a certain 4-uniform hypergraph, and $G$ is perfect. All you can derive from the theorem is that any graph with the same $P\_4$-structure $H$ must be perfect.
Besides, $H$ is a hypergraph. While one may try to somehow create a graph out of it, and check whether it's perfect, it has no relation to the theorem in question.
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2
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https://mathoverflow.net/users/11100
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412493
| 168,308 |
https://mathoverflow.net/questions/412471
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3
|
I am working on elliptic curves with torsion group $\mathbb{Z}/14\mathbb{Z}$ over quadratic fields. The curves are constructed using the model $E\_1=[0,a,0,b,0]$ following the formulas on p. 13 of
>
> L. Halbeisen, N. Hungerbuehler, M. Voznyy, A. S. Zargar, *A geometric approach to elliptic curves with torsion groups ℤ/10ℤ, ℤ/12ℤ, ℤ/14ℤ, and ℤ/16ℤ*, arXiv: [2106.06861](https://arxiv.org/pdf/2106.06861.pdf).
>
>
>
Magma [code snippet](https://mega.nz/file/yp5CjTYb#2Q5eMHxQykiDLBWl9xZH5jYOFRyTAsey_D1JvAUTwtA) (on my MEGA) for online [Magma Calculator](http://magma.maths.usyd.edu.au/calc/) is available.
Using $u=-3$ produces a $\mathbb{Z}/14\mathbb{Z}$ curve of rank $1$.
```
u = -3
K: Quadratic Field with defining polynomial $.1^2 + 23 over the Rational Field
ClassNumber(K) = 3
E1 = [ 0, 10656*K.1 + 13600, 0, 37748736*K.1 - 448790528, 0 ]
jInvariant = 1/248278597632*(-64448133136412*K.1 + 699764618102559)
Abelian Group isomorphic to Z/14
Defined on 1 generator
Relations:
14*ts.1 = 0
```
I want to rationalize the coefficient $a=10656\sqrt{-23} + 13600$, or to minimize the size (sum of absolute values of rational and irrational parts) of it as much as possible. It proves to be difficult, as to keep $\mathbb{Z}/14\mathbb{Z}$ torsion, the transformation from $E\_1$ to $E\_2=[0,af^2,0,bf^4,0]$ is required, and the choice of $f$ is unclear so far.
A simple loop in the code produces a curve with a coefficient $a=-28\sqrt{-23} + 605$ of smaller size $|-28|+|605|\lt|10656|+|13600|$, where $f=\sqrt{-23}-5$ and $f^2=-10\sqrt{-23}+2$.
```
jmin = 1 imin = -5 f2 = -10*K.1 + 2 f2/ymin^2 = 1/2048*(-5*K.1 + 1)
E2 = [ 0, -28*K.1 + 605, 0, -4096*K.1 + 63488, 0 ]
jInvariant = 1/248278597632*(-64448133136412*K.1 + 699764618102559)
Abelian Group isomorphic to Z/14
Defined on 1 generator
Relations:
14*ts.1 = 0
```
An attempt to match the $j$-invariants of $E\_1$ and $E\_3=[0,1,0,a\_4,0]$ produces a curve with torsion $\mathbb{Z}/2\mathbb{Z}$.
```
E3 = [ 0, 1, 0, 1/1820984929*(8957952*K.1 + 312162304), 0 ]
jInvariant = 1/248278597632*(-64448133136412*K.1 + 699764618102559)
Abelian Group isomorphic to Z/2
Defined on 1 generator
Relations:
2*ts.1 = 0
```
>
> **Question 1:** Is it possible to rationalize the coefficient $a$? If so, what would be the choice of $f$?
>
>
>
>
> **Question 2:** Is it possible to further minimize the size of $a=-28\sqrt{-23} + 605$? If so, what would be the choice of $f$?
>
>
>
>
> **Question 3:** Does class number of the quadratic field influence the possibility to rationalize $a$? If so, class number $1$ is achieved for $u=2,3,5,6,7,8,$ etc.
>
>
>
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https://mathoverflow.net/users/95511
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Rationalizing and minimizing elliptic curve coefficients
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If I understand correctly: you are starting with $a = 10656\sqrt{-23} +13600$, and you want to choose $f$ in the same number field such that $af^2$ is "as nice as possible", ideally lying in $\mathbf{Z}$; and you have observed that you can get it down to $a' = -28\sqrt{-23} + 605$.
Let's consider the prime factorisation of $a'$. It has the shape $P\_1^2 P\_2 P\_3$ where $P\_i$ are primes, and the prime $P\_2 = \langle 139, (33 + \sqrt{-23})/2\rangle$ is not equal to its Galois conjugate $P\_2^\sigma$. So you can see straight away that no element of the form $a' f^2$ can be in $\mathbf{Z}$, because it will always have even valuation at $P\_2^\sigma$ and odd valuation at $P\_2$, while any element of $\mathbf{Z}$ has to have equal valuations at $P\_2$ and $P\_2^\sigma$.
So the answer to question 1 is "no"; and, insofar as I can understand what question 3 is asking, the answer is "it doesn't matter" (the class number is not the obstruction here). For question 2, I suspect the answer is also "no", but it sounds like a mighty tedious check.
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4
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https://mathoverflow.net/users/2481
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412494
| 168,309 |
https://mathoverflow.net/questions/412495
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1
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Is $\sum\_{\rho \text{ irred. }} \deg(\rho) \chi\_{\rho}(g)=0$ for every froup element $1\neq g \in G$ of the finite group $G$?
I have searched for but not found a proof to this. Probably it is not so difficult, but has as application that:
$$\det(T\_G) = 1$$
where $T\_G = (t\_{gh^{-1}})\_{g,h \in G}$ is the group matrix defined for the functions defined in this [answer:](https://mathoverflow.net/a/412397/165920)
$$\widehat{t\_{x}}(\rho) := \mathbf{1}\_{d\_{\rho}} \exp( \frac{1}{d\_{\rho}} \sum\_{s \in S}\chi\_{\rho}(s) x\_s )$$
where $S$ (with $1 \notin S$) generates the finite group $\left< S \right > = G$.
From this we get, since we know by Frobenius, the factorization of the group determinant :
$$\det(T\_G) = \prod\_{\rho \text{ irred.}} \det( \sum\_{g \in G} t\_g(x) \rho(g) )^{d\_{\rho}} = \prod\_{\rho \text{ irred.}} \det( \widehat{t\_x}(\rho))^{d\_{\rho}} = \prod\_{\rho \text{ irred.}} \det( \mathbf{1}\_{\rho} \exp \left ( \frac{1}{d\_{\rho}} \sum\_{s \in S} \chi\_{\rho}(s) x\_s \right ) )^{d\_{\rho}} $$
$$= \prod\_{\rho \text{ irred. }} \exp( \sum\_{s \in S} \chi\_{\rho}(s) x\_s)^{\deg(\rho)}$$
$$ =\exp\left( \sum\_{\rho \text{ irred.}} \deg(\rho) \sum\_{s\in S} \chi\_{\rho}(s) x\_s \right)$$
and which is equal to:
$$=\exp(\sum\_{s \in S} x\_s \cdot \left ( \sum\_{\rho} \deg(\rho) \chi\_{\rho}(s) \right ) ) =^? \exp(0)=1$$
So if the question answers positive, then the determinant should be equal to $1$ .
Thanks for your help.
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https://mathoverflow.net/users/165920
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Is $\sum_{\rho \text{ irred. }} \deg(\rho) \chi_{\rho}(g)=0$ for every group element $1 \neq g \in G$ of the finite group $G$?
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The assertion holds with deg$(\rho)^2$ replaced with deg$(\rho)$.
You also need the extra requirement that the group element $g$ is non-trivial.
Then the Plancherel Theorem implies that the right hand side equals the trace of the left translation $L\_g$ on $\ell^2(G)$. Then you compute this trace as
$$
\mathrm{tr}(L\_g)=\sum\_{y\in G}\langle L\_g\delta\_y,\delta\_y\rangle=\sum\_{y\in G}\langle \delta\_{gy},\delta\_y\rangle=0.
$$
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3
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https://mathoverflow.net/users/nan
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412496
| 168,310 |
https://mathoverflow.net/questions/412483
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1
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Let $A$ be a non atomic measure on $\mathbb R$. Consider the product measure $\mu := A \times \dots \times A$ on $\mathbb R^n$.
**Question:** Let $M$ be a $n-1$ dimensional smooth submanifold of $\mathbb R^n$. Is it true that $M$ has measure $0$ under $\mu$?
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https://mathoverflow.net/users/173490
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Does a submanifold of nonzero codimension have measure zero under the product of non atomic measures?
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A submanifold $M$ of ${\mathbb R}^{n}$ (with codimension 1) can always (after permuting the coordinates) be locally represented in the form $G\_f=\{(x,y) : y=f(x)\}$ where $x$ runs over $V \subset {\mathbb R}^{n-1}$ and $f:V \to {\mathbb R}$. That implies that $M$ is contained in a countable union of such graphs. Each graph has measure zero under the given product measure by Fubini since $A$ is nonatomic. Thus $M$ indeed has measure zero under the product measure by countable additivity.
A submanifold $M$ of ${\mathbb R}^{n}$ with positive codimension is always a subset of a manifold of codimension 1.
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2
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https://mathoverflow.net/users/7691
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412502
| 168,311 |
https://mathoverflow.net/questions/412504
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22
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This question is about Joyal and Tierney's famous [*An extension of the Galois theory of Grothendieck*](https://doi.org/10.1090/memo/0309). One of the main results states (see the [MathSciNet review](https://mathscinet.ams.org/mathscinet-getitem?mr=756176) by Peter Johnstone):
**Joyal and Tierney's theorem.** *Each Grothendieck topos is equivalent to the topos of equivariant sheaves on a groupoid in the category of locales.*
**Question 1.** What does this statement have to do with Grothendieck's Galois theory?
I have to admit, I don't know much about Grothendieck's Galois theory and which kind of theorems it includes. In my mind, I usually identify the term "Grothendieck's Galois theory" with the following theorem (see [Chua - The étale fundamental group](https://dec41.user.srcf.net/exp/etale_fundamental_group/etale_fundamental_group.pdf) for a concise introduction and SGA 1 for a complete treatment):
**Grothendieck's theorem.** *Let $X$ be a scheme and $x\in X$. Then the category of finite étale covers of $X$ is equivalent to the category of finite continuous $\pi\_1(X,x)$-sets, where $\pi\_1(X,x)$ is the étale fundamental group of $X$ at the base point $x$.*
This is related to Galois theory because whenever $k$ is a field, then the étale fundamental group of $\operatorname{Spec}(k)$ is the absolute Galois group of $k$.
A more precise version of Question 1 would be: does Joyal and Tierney's theorem imply Grothendieck's theorem? (But an answer showing that Joyal and Tierney's theorem implies a modified version of Grothendieck's theorem would satisfy me as well.)
**Question 2.** In [an answer by Zhen Lin to "What does a proof in an internal logic actually look like?"](https://math.stackexchange.com/questions/394194/what-does-a-proof-in-an-internal-logic-actually-look-like/394370#394370) it is mentioned that the proof of Joyal and Tierney's theorem uses the [internal language](https://ncatlab.org/nlab/show/internal+logic) of toposes. In which way does it use the internal language? Which kind of theorems are proved internally, how are they interpreted externally, and why are the external interpretations of these theorems helpful in proving Joyal and Tierney's theorem? Skimming the text I don't see any logic at all.
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https://mathoverflow.net/users/471475
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An extension of the Galois theory of Grothendieck
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The point of view where this title comes from is that Grothendieck's theorem can be seen as a characterization of toposes of the form $BG$ for $G$ a profinite group. It shows that some toposes can be represented as $BG$.
I think before Joyal–Tierney's paper it was also known how to generalize from profinite group to general localic groups.
Joyal–Tierney's theorem shows that if you replace "pro-finite group" by "localic groupoid" then you actually get all Grothendieck toposes this way.
You can't directly recover Grothendieck's theorem from Joyal–Tierney's theorem in the sense that the theorem as stated above doesn't tell you for which toposes the localic groupoid can be chosen to be a profinite group. But if you are familiar with the method used in the paper and how the groupoid is obtained (which in my opinion are even more important than the theorem itself) then it is fairly easy to recover Grothendieck's theorem. For example, it immediately follows from Joyal–Tierney's paper that a topos is of the form $BG$ for $G$ a localic groups if and only if it admits a point $\* \to \mathcal{T}$ which is an open surjection (which does feel similar to Grothendieck's theorem in terms of a fiber functor).
Regarding the use of internal logic, it is definitely not essential, it just makes everything simpler (at least if you are ok with its use) but one could do without it.
The main way they use internal logic is that in the first sections they prove some results about sup-lattice and frames locales, that are later applied not to sup-lattices and frames, but to sup-lattices and frames in a topos $\mathcal{T}$.
I believe there are also a few places where they make a claim about a morphism of locales $f:X \to Y$ and then only prove it when $Y$ is the point (sorry I don't have the paper with me to give precise reference).
Another place where one can consider they use a bit of internal logic — though this one might be only at the level of intuition — is when they show that every topos admits an open cover by a locale. They do this by considering the topos as a classifying topos of some theory $T$ and considering the propositional theory $T'$ of "enumerated $T$-models", that is, $T$-models that are explicitly given as a subquotient of the natural numbers. Though if I remember correctly, they present the argument in a way that doesn't directly involve any logic… (and in any case there are other proofs of this results that are purely in terms of sites, for example the one in MacLane and Moerdijk's book "[Sheaves in geometry and logic](https://link.springer.com/book/10.1007/978-1-4612-0927-0)").
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18
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https://mathoverflow.net/users/22131
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412510
| 168,314 |
https://mathoverflow.net/questions/412381
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10
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Let $p, q$ be two distinct prime number. I'm trying to provide a non-trivial upper bound for the sum
$$S(p, q) = \sum\_{1 \leq x < p} \sum\_{1 \leq y < q} \frac{1}{\|x / p\| \, \|y / q\| \, \|x/p + y/q\|},$$
where $\|t\|$ denotes the distance of $t \in \mathbb{R}$ from the nearest integer.
Precisely, I'm interested in having $S(p, q) = o((pq)^2)$ as $p, q$ go to infinity in some way.
I know that $\min(p, q) \to +\infty$ doesn't suffices, since $S(p, q) \geq (pq)^2 / (q - p)$ (considering $x = 1$ and $y = q - 1$), and we can take a sequence of primes $p\_k < q\_k$ such that $p\_k \to +\infty$ and $q\_k - p\_k$ is bounded.
Maybe $S(p, q) = o((pq)^2)$ as $p \to +\infty$ and $q/p \to +\infty$ ?
The motivation comes from the fact that
$$\sum\_{\substack{1 \leq z < pq \\ (pq, z) = 1}} \frac1{|p^{-1}z \bmod q|\,|q^{-1}z \bmod p| \, z} = \frac{S(p, q)}{(pq)^2} ,$$
where $|p^{-1}z \bmod q| = \min\{|r| : r \in \mathbb{Z}, pr \equiv z \pmod q\}$, and similarly for $|q^{-1}z \bmod p|$. Therefore, $S(p, q) = o((pq)^2)$ means that, on average, $|p^{-1}z \bmod q|$, $|q^{-1}z \bmod p|$, and $z$ cannot be all small.
Thanks for any help
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https://mathoverflow.net/users/357523
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Non-trivial upper bound for a sum related to $p^{-1}z \pmod q$ and $q^{-1}z \pmod p$
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If you just want $o()$, the story is rather simple. Let $a\_{xy}$ be the remainder of $qx+py\mod pq$ where all remainders modulo $P$ are assumed to be between $-P/2$ and $P/2$. Note that all $a\_{xy}$ are distinct, so if we have any set $Z$ of pairs $(x,y)$, then $\sum\_{(x,y)\in Z}\frac 1{a\_{xy}}\le 2(1+\log|Z|)$. What we want to show is just
$$
\sum\_{0<|x|<p/2, 0<|y|<q/2}\frac 1{|xya\_{xy}|}=o(1)\,.
$$
Now for $k=0,1,2\dots$ consider $Z\_k=\{(x,y): 2^k\le |xy|<2^{k+1}\}$ and note that $|Z\_k|\le C(k+1)2^{k}$. Thus the sum over $Z\_k$ is at most
$2^{-k}(1+\log|Z\_k|)\le C(k+1)2^{-k}$ regardless of $p,q$. Thus the only danger is that the sum over $Z\_k$ for some fixed $k$ does not tend to $0$, i.e., that there exists $C>0$ such that $ap+bq+c=0$ for some $a,b,c$ with $0<|a|+|b|+|c|<C$ along a subsequence of pairs $(p,q)$ you are considering (in which case the corresponding term alone gives a positive constant). If you eliminate this possibility in any way ($q/p\to+\infty$ is more than enough), then you are in good shape.
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8
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https://mathoverflow.net/users/1131
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412520
| 168,319 |
https://mathoverflow.net/questions/407352
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3
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It is known that an algebraic function with non-solvable monodromy group can not be represented by radicals. Where can we find a detailed proof about the nonrepresentability by radicals and entire (or meromorphic) functions of algebraic functions in some English references?
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https://mathoverflow.net/users/11966
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Nonrepresentability by radicals and entire (or meromorphic) functions of algebraic functions
|
See e.g.:
* Vladimir Arnold: Collected Works. Springer
* [Khovanskii, A.: Topological Galois Theory - Solvability and Unsolvability of Equations in Finite Terms. Springer 2014](https://doi.org/10.1007/978-3-642-38871-2)
* [Khovanskii, A.: One dimensional topological Galois theory. 2019](https://arxiv.org/abs/1904.03341)
* [Askold Khovanskii - Publications](https://www.math.toronto.edu/askold/)
For the topological proofs without connection to representability in meromorphic functions, see e.g.: [Is there an alternative proof of the Abel–Ruffini theorem?](https://math.stackexchange.com/questions/1550990/is-there-an-alternative-proof-of-the-abel-ruffini-theorem).
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3
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https://mathoverflow.net/users/94085
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412521
| 168,320 |
https://mathoverflow.net/questions/412527
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2
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Easton's theorem can give a very weak nontrivial constraint on continuum function, but it does not hold for singular cardinals. So:
1. What are the non-trivial constraints on continuum function in singular cardinals?
2. Is it possible to well-define a concept of "**Continuum function maximum of singular cardinals**" by these constraints?
3. How strong is the consistency strength of this definition?
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https://mathoverflow.net/users/143814
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Continuum function maximum
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The intricacies of arithmetic at singular cardinals notwithstanding, I think you're looking for something which doesn't exist.
Given any cardinals $\kappa,\lambda$, singular or regular, there is a (set) forcing extension preserving cardinals and cofinalities in which $2^\kappa>\lambda$. Upper bounds only appear in a meaningful sense when we try to control the value of the continuum function on *many* inputs simultaneously. For example, famously Shelah proved that $2^{\aleph\_\omega}<\aleph\_{\omega\_4}$ **provided that** $2^{\aleph\_n}<\aleph\_\omega$ for all $n<\omega$, but it's perfectly consistent that $2^{\aleph\_\omega}\ge\aleph\_{\omega\_4}$ since for example we could have $2^{\aleph\_0}=\aleph\_{\omega\_{17}+42}$ already.
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2
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https://mathoverflow.net/users/8133
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412528
| 168,324 |
https://mathoverflow.net/questions/412534
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1
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$Define: X \text { is cardinal definable} \iff \\\exists \text { cardinal } \kappa \, \exists \text { cardinals } \lambda\_1,.., \lambda\_n <^\rho \kappa \ \exists \phi : \\ X=\{ y \in V\_{\rho(\kappa)} \mid \phi^{V\_{\rho (\kappa)}} (y,\lambda\_1,..,\lambda\_n)\}$
Where: $\lambda\_i <^\rho \kappa \iff \rho(\lambda\_i) < \rho(\kappa)$, and $\rho$ is the rank function; and "*cardinal*" is defined after Scott's.
>
> Now, is the principle that every set is cardinal definable consistent with ZF?
>
>
>
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https://mathoverflow.net/users/95347
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Is every set being cardinal definable consistent with ZF?
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Yes, assuming ZF is consistent it is, because it follows from ZFC + V=HOD. This is because the ordertype of the class of (ordinal-)cardinals is just that of the ordinals. That is, if $X$ is definable over $V\_\alpha$ from ordinal parameter $\beta<\alpha$, then $X$ is definable over $V\_{\aleph\_\alpha}$ from ordinal parameter $\aleph\_\beta$, since $\xi\mapsto\aleph\_\xi$ restricted to $\alpha$ is also definable over $V\_{\aleph\_\alpha}$.
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5
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https://mathoverflow.net/users/160347
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412536
| 168,327 |
https://mathoverflow.net/questions/412551
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0
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I would like to know if some widely believed conjecture, be it GRH, Hardy-Littlewood conjecture, or any other would imply the following statement for some $l>1$:
$l$-th power radioprimal growth conjecture ($l$-PRG conjecture for short)
Let $r\_{0}(n):=\inf\{r>0,(n-r,n+r)\in\mathbb{P}^{2}\}$, $r\_{i+1}(n):=\inf\{r>r\_{i}(n),(n-r,n+r)\in\mathbb{P}^{2}\}$ and $k\_{i}(n):=\pi(n+r\_{i}(n))-\pi(n-r\_{i}(n))$.
There exists $C\_{l}>1$ such that there are infinitely many $n$ such that $i\leq C\_{l}$ implies $k\_{i-1}(n)=i^{l}$.
I would also like to know if $C\_{l}$ can be taken arbitrarily large, and if the hypothetical existence of a maximal $l$ such that the $l$-PRG conjecture holds implies $r\_{0}(n)<K(\varepsilon)\log^{l+\varepsilon} n$ for all sufficiently large $n$ and all $\varepsilon>0$.
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https://mathoverflow.net/users/13625
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$l$-th power radioprimal conjecture
|
Assuming the strengthened version of Hardy-Littlewood conjecture I discuss [here](https://mathoverflow.net/a/395878/30186) (which follows from Dickson's conjecture), the following much stronger result holds: let $a\_0,\dots,a\_m$ be any sequence of natural numbers with $a\_0\geq 1$ and $a\_{i+1}\geq a\_i+2$ for $i<m$. Then there are infinitely many integers $n$ such that $k\_i(n)=a\_i$ for $i\leq m$.
Let $N$ be some fixed natural number to be chosen later. Consider the set $T=\{N,2N,\dots,(a\_m-m)N\}\cup\{-(a\_i-i)N\mid 0\leq i\leq m\}$, and $S$ the set of other numbers in the interval $[-(a\_m-m)N,(a\_m-m)N]$. If $N$ is divisible by all small enough primes, then $T$ is an admissible tuple. Let $n$ be any integer given by the strengthened Hardy-Littlewood conjecture above for these $T,S$. Then clearly its first $m+1$ primality radii are $(a\_i-i)N$ for $0\leq i\leq m$, and you can count there are $a\_i$ primes between $n-(a\_i-i)N$ and $n+(a\_i-i)N$, so $k\_i(n)=a\_i$.
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2
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https://mathoverflow.net/users/30186
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412554
| 168,333 |
https://mathoverflow.net/questions/412201
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5
|
I am reading the paper "Mori Dream Spaces and GIT" by Hu and Keel.
<https://arxiv.org/abs/math/0004017>
I cannot understand the proof of Lemma 1.6 in it.
Let $X$ be a normal projective variety.
Assume that $D$ is a divisor on $X$ such that $R(X,D) = \bigoplus\_{m \in \mathbb{Z}\_{\ge0} } H^0(X,\mathcal{O}\_X(mD))$ is finitely generated.
Lemma 1.6 states that the natural rational map $\varphi\_D \colon X \dashrightarrow Y := Proj(R(X,D))$ is a rational contraction, i.e, there exits a resolution $\Phi \colon \tilde{X} \to Y$ of $\varphi\_D$ such that $\Phi\_\*(\mathcal{O}\_{\tilde{X}} (E)) \simeq \mathcal{O}\_Y$ for any effective exceptional divisor $E$.
Moreover, $D$ can be written as $ D = \varphi\_D^\*(A) + F $ for an ample divisor $A$ on $Y$ and a $\varphi\_D$-fixed divisor $F$.
Outline of the proof of the paper is as follows:
(1) The finite generation of $R(X,D)$ implies that $D$ can be written as $D = M+F$ for a movable divisor $M$ and an effective divisor $F$ such that
$Sym\_k(H^0(X,M)) \to H^0(X,kM)$ is surjective and $H^0(X,kM) \to H^0(X,kM +rF)$ is an isomorphism for any $k>0$ and $r>0$ after replacing $D$ by its multiple.
(2) After taking appropriate resolution, we may assume that $\varphi\_D$ is a regular
and $M = \varphi\_D^\* A $. Now we can see $F$ is $\varphi\_D$-fixed by the isoromorphism $H^0(X,kM) \simeq H^0(X,kM +rF)$.
Here is my question :
[1] How can wee see that $\varphi\_D$ is a rational contraction? Is it a well-known result?
[2] Why does the isomorphism $H^0(X,kM) \simeq H^0(X,kM +rF)$ implies that
$F$ is $\varphi\_D$-fixed?
Can anyone who understands these results answer my questions?
|
https://mathoverflow.net/users/472874
|
Rational contraction and Proj of section ring
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Let's assume that $|kD|=|kM|+kF$ where $|M|$ is base point free and moreover $Sym ^kH^0(M)\to H^0(kM)$ is surjective for any $k>0$. This can be achieved replacing $X$ by an appropriate resolution and $k$ by a multiple. Let $f:X\to Y$ be the morphism induced by $|M|$ so that $f^\*A=M$.
For [1], we wish to show that $f\_\* \mathcal O \_X=\mathcal O \_Y$. The Stein factorization $X\to Z\to Y$ is defined by ${\rm Spec} f\_\* \mathcal O \_X$. We have $H^0(X,kM)=H^0(f\_\*(kM))=H^0(f\_\* \mathcal O \_X\otimes \mathcal O \_Y(kA))$ and so $|kM|$ in fact defines the morphism $X\to Y$. Since $Sym ^kH^0(M)\to H^0(kM)$ is surjective, then $|M|$ defines the same morphism.
For [2], Since $H^0(X,kM)=H^0(X,kM+rF)$, then
$H^0(\mathcal O \_Y(kA))\cong H^0(\mathcal O \_Y(kA)\otimes f\_\* \mathcal O \_X(rF))$. For $k\gg r$, the sheaf $\mathcal O \_Y(kA)\otimes f\_\* \mathcal O \_X(rF)$ is globally generated and so $f\_\* \mathcal O \_X(rF)\cong \mathcal O \_Y$. (See also Lemma 3.2 <https://arxiv.org/pdf/1104.4981.pdf>).
Hope this helps.
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4
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https://mathoverflow.net/users/19369
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412560
| 168,334 |
https://mathoverflow.net/questions/412532
|
3
|
I am looking for a paper "**Linear spaces with disjoint elements and their conversion into vector lattices**" by A. I. **Veksler**.
It was published in 1967 in Research Notes of Leningrad State Pedagogical University.
The paper is in Russian, and is called "**Линейные пространства с дизъюнктными элементами и преврщение их в векторные структуры**".
The name of the journal volume is Ученые записки Ленинградского государственного педагогического института им. А. И. Герцена. Т. 328 : Вопросы современной математики / редкол.: И. Я. Бакельман, К. А. Бохан, А. Л. Вернер [и др.]. – Ленинград : Изд-во Ленинград. госуд. ин-та, 1967. – 282 с.
>
> Can anybody help me finding this paper? Either original, or a translation into English will do.
>
>
>
The topic of the paper is an axiomatic approach to the notion of orthogonality (or disjointness) in vector spaces, where a binary relation is introduced, which satisfies certain properties, with an eye on the notion of disjointness in vector lattices (but I don't know the details, since I cannot access the paper). Note that orthogonality in vector spaces appears in several contexts:
* In the presence of a bilinear form (the usual orthogonality);
* In the presence of multiplication (declare $e\bot f$ if $ef=0$);
* In the presence of an order (if $E$ is a Vector lattice, then $e\bot f$ if $|e|\wedge |f|=0$);
* If the space consists of functions into a vector space (say that $f,g:X\to F$ are disjoint if $f^{-1}(0)\cup g^{-1}(0)=X$).
There is also a lot of research done on the maps preserving disjointness, in various contexts, but i don't know to what extend it is generalizable. Hence, I wonder what is done on this topic from the axiomatic standpoint. So far the aforementioned paper is the only one I came across, and even that is not accessible.
>
> Does anybody know any other literature on this subject?
>
>
>
|
https://mathoverflow.net/users/53155
|
Looking for a paper on axiomatic orthogonality in a vector space
|
This journal published by the [Herzen University](https://en.wikipedia.org/wiki/Herzen_University) is not yet available in electronic form.
A paper version can be found in multiple libraries, including the [National Library of Russia](https://webservices.nlr.ru/util/?method=recordFormat&vid=07NLR_VU1&sysid=002516234&format=037&base=NLR01).
They will [scan](http://nlr.ru/shop/pages/edd/edd.php?ss=2d20b9b25f672960b04e17a61ec1f84a) papers for a [nominal fee of about 0.27 USD per page](http://nlr.ru/shop/pages/srv/srv.php?p=5&ss=2d20b9b25f672960b04e17a61ec1f84a).
|
2
|
https://mathoverflow.net/users/402
|
412562
| 168,335 |
https://mathoverflow.net/questions/412566
|
5
|
Let $G$ be a locally compact group, $C\_0(G)$ the $C^\*$-algebra of continuous functions on $G$ that vanish at infinity, $C\_b(G)$ the $C^\*$-algebra of bounded continuous functions on $G$. We know that $C\_b(G)$ is the multiplier algebra of $C\_0(G)$, and we denote the strict topology on $C\_b(G) = \mathcal{M}\bigl( C\_0(G) \bigr)$ by $\beta$.
Now for a strongly continuous unitary representation $(\pi, H)$ of $G$, functions of the form $\omega\_{\pi,\eta,\xi} : g \in G \to (\pi(g)\eta \mid \xi) \in \mathbb{C}$ are in $C\_b(G)$, and we call them matrix coefficients of the representation $\pi$. Since we can form direct sum and tensor product of two strongly continuous unitary representations, as well as the contragredient representation, we see that (the linear span of) matrix coefficients of all strongly continuous unitary representations of $G$ form a $\*$-subalgebra $A\_0(G)$ of $C\_b(G)$.
**Question.** Is $A\_0(G)$ strictly dense in $C\_b(G)$, i.e. with respect to the $\beta$-topology?
Note that in the compact case, $C\_b(G) = C\_0(G) = C(G)$ and the $\beta$-topology is the same as the norm topology on the $C^\*$-algebra $C(G)$ of continuous functions on $G$. In this case, The answer to the question is affirmative by Peter-Weyl. In the case where $G$ is discrete, then one can check easily that all finitely supported functions on $G$ are already matrix coefficients of the left (or right) regular representation, so the answer to the question is again affirmative. Based on these considerations, here are some sub-questions with some bias on their possible answers.
**Q1.** Does the question have an affirmative answer for unimodular $G$?
**Q2.** Can we construct some counter-example for non-unimodular $G$?
**Q3.** Does the question have an affirmative answer if $G$ is a real Lie group? What if the Lie group $G$ is nilpotent, or solvable, or semisimple/reductive?
|
https://mathoverflow.net/users/128540
|
Density of matrix coefficients of unitary representations of a locally compact group
|
First, some remarks that may help with literature-searching.$\newcommand{\fsnorm}[1]{{\Vert#1\Vert}\_{\rm B}}$
$\newcommand{\supnorm}[1]{{\Vert#1\Vert}\_\infty}$
The algebra you have denoted by $A\_0(G)$ is known as the *Fourier--Stieltjes algebra* of $G$, and is usually denoted by $B(G)$, so I will do that from now on. $B(G)$ has been much studied: it turns out to be complete in a natural submultiplicative norm $\fsnorm{\cdot}$, so it is a commutative Banach algebra of functions on $G$; in fact it is also a dual Banach space for this norm, and multiplication is separately weak-star continuous, so it is an example of a *dual Banach algebra*.
This norm dominates the supremum norm, so $\fsnorm{\cdot}$-convergence implies $\supnorm{\cdot}$-convergence. If one takes the $\fsnorm{\cdot}$-closure inside $B(G)$ of $B(G)\cap C\_c(G)$, the resulting algebra is an ideal in $B(G)$ called the *Fourier algebra* of $G$, usually denoted by $A(G)$. One can show that $A(G)$ is a $\supnorm{\cdot}$-dense subalgebra of $C\_0(G)$ that is closed under conjugation and separates points of $G$.
(Aside: in general $B(G)\cap C\_0(G)$ is larger than $A(G)$.)
---
Given some $f\in C\_b(G)$, an $\varepsilon>0$, and a finite set $E\subset C\_0(G)$, it suffices to find $h\in B(G)$ such that $\supnorm{fg-hg}\leq \varepsilon$ for all $g\in E$.
Since $C\_0(G)$ is $\beta$-dense in $C\_b(G)$ we can find $k \in C\_0(G)$ such that $\supnorm{ fg-kg } \leq \varepsilon/2$ for all $g\in E$. Since $B(G)\cap C\_c(G)$ is $\supnorm{\cdot}$-dense in $C\_0(G)$ we can find $h\in B(G)\cap C\_c(G)$ such that $\supnorm{k-h} \max\_{g\in E} \supnorm{g} \leq\varepsilon/2$. Then
$$
\supnorm{ fg -hg} \leq \supnorm{fg-kg} + \supnorm{k-h}\supnorm{g} \leq \varepsilon \quad\hbox{for all $g\in E$,}
$$
as required.
|
8
|
https://mathoverflow.net/users/763
|
412574
| 168,338 |
https://mathoverflow.net/questions/412575
|
1
|
I would like example of measures which shows that the following propositions are false:
>
> **Proposition 1:** Let $\mathfrak{B}$ be the Borel $\sigma$-algebra of a topological space $X$ and $\mu:\mathfrak{B}\to\overline{\mathbb{R}}$ be a measure. Then for all
> $B\in\mathfrak{B}$ with $\mu (B)<\infty$ and $\varepsilon \in(0,\infty)$ there's a closed set $F\subseteq B$ such that $\mu(B\setminus F)<\varepsilon $.
>
>
>
>
> **Proposition 2:** Let $\mathfrak{B}$ be the Borel $\sigma$-algebra of a topological space $X$ and $\mu:\mathfrak{B}\to\overline{\mathbb{R}}$
> be a measure. Suppose that there's an open cover
> $\{O\_n\}\_{n\in\mathbb{N}}$ such that $\mu (O\_n)<\infty$ for all
> $n\in\mathbb{N}$. Then for all $B\in\mathfrak{B}$ and $\varepsilon\in(0,\infty)$ there's an open set $O\supseteq B$ such that
> $\mu(O\setminus B)<\varepsilon $.
>
>
>
---
I tried to find some counterexamples in some books but could not find any. I even looked for one in the book "Measure Theory" written by V.I. Bogachev.
In the Theorem 3.3 (which in the page 74 of the "[Measure and Integration](https://digitalcommons.wayne.edu/cgi/viewcontent.cgi?article=1050&context=mathfrp)" written by John L. Menaldi) the author "proves" the two propositions above. I believe that his proof is flawed since the Theorem 3.3 says that every Borel measure is outer regular which [isn't true](https://en.wikipedia.org/wiki/Regular_measure#Measures_that_are_neither_inner_nor_outer_regular).
I tried to find the error in that proof but I failed.
---
I'm asking this question here since a similar question (see [here](https://math.stackexchange.com/questions/4340259/are-there-a-compact-k-and-an-open-o-such-that-k-subseteq-b-subseteq-o-and)) that I asked in math.stackexchange didn't receive any answer.
Thank you for your attention!
|
https://mathoverflow.net/users/143671
|
Find a Borel measure such that the closed sets aren't arbitrarily close to the Borel sets with finite measure
|
Let $X$ be the Sierpinski space $\{∅,\{b\},\{a,b\}\}$. Then the point $b$ is Borel but not closed. Let $\mu(a)=\mu(b)=1$. Take $B=\{b\}$, and it's a counterexample to proposition 1. Take $B=\{a\}$, and it's a counterexample to proposition 2.
|
3
|
https://mathoverflow.net/users/125498
|
412579
| 168,339 |
https://mathoverflow.net/questions/412587
|
2
|
Where can I find any more or less explicit semi-orthogonal decompositions of derived categories of perfect complexes or of bounded derived categories for singular schemes that are proper over a ring R?
I guess that one can take a "Beilinson-type" decomposition of $D^{perf}(\mathbb{P}^n(\operatorname{Spec}\,R))$ where $R$ is not regular; one can probably take the "inverse image" of this decomposition to something like $D^{perf}(\mathbb{P}^n\times V(\operatorname{Spec}\,R))$. Yet I would like to have some references for this argument. Moreover, I would like to find some "more interesting" examples. This question is certainly related to examples of explicit descriptions of derived categories of singular varieties.
|
https://mathoverflow.net/users/2191
|
Semi-orthogonal decompositions over singular schemes
|
There is a notion of base change for semiorthogonal decompositions, applying which you can induce semiorthogonal decompositions over rings from those over fields, see Kuznetsov, Alexander. Base change for semiorthogonal decompositions. Compos. Math. 147 (2011), no. 3, 852--876. You can also find nontrivial examples of semiorthofonal decompositions for singular toric surfaces in <https://arxiv.org/abs/1809.10628>.
|
3
|
https://mathoverflow.net/users/4428
|
412589
| 168,341 |
https://mathoverflow.net/questions/412585
|
2
|
Let $P$ be the polytope obtained as the convex hull of $n$ points in $\mathbb R^d$. This is the $V$-representation of $P$. Note that $P$ can also be represented as an intersection of closed half-spaces in $\mathbb R^d$. This is the $H$-representation of $P$.
>
> **Question.** *In terms of $n$ and $d$, what is a good upper-bound on the number of half-spaces in the smallest $H$-representation of $P$ ?*
>
>
>
|
https://mathoverflow.net/users/78539
|
Optimal number of half-spaces in the $H$-representation of the convex hull of $n$ points in $\mathbb R^d$
|
By the [Upper Bound Theorem](https://en.wikipedia.org/wiki/Upper_bound_theorem), the maximum number of $(d-1)$-dimensional faces of an $n$-vertex polytope is achieved by the cyclic polytopes. This number can be explicitly written via the [Dehn-Sommerville equations](https://en.wikipedia.org/wiki/Dehn%E2%80%93Sommerville_equations).
|
3
|
https://mathoverflow.net/users/2233
|
412590
| 168,342 |
https://mathoverflow.net/questions/412541
|
11
|
Recall the definition of *[cardinal definable](https://mathoverflow.net/questions/412534/is-every-set-being-cardinal-definable-consistent-with-zf)*, where every set being cardinal definable is [proved](https://mathoverflow.net/questions/412534/is-every-set-being-cardinal-definable-consistent-with-zf/412536#412536) consistent relative to ZF + V=HOD. To re-iterate it:
$Define: X \text { is cardinal definable} \iff \\\exists \text { cardinal } \kappa \, \exists \text { cardinals } \lambda\_1,.., \lambda\_n <^\rho \kappa \ \exists \phi : \\ X=\{ y \in V\_{\rho(\kappa)} \mid \phi^{V\_{\rho (\kappa)}} (y,\lambda\_1,..,\lambda\_n)\}$
Where: $\lambda\_i <^\rho \kappa \iff \rho(\lambda\_i) < \rho(\kappa)$, and $\rho$ is the rank function; and "*cardinal*" is defined after Scott's as an equivalence class under *bijection* of sets of the lowest possible rank.
>
> Now, is the principle stating that *every set is cardinal definable* consistent with $\sf ZF + \neg AC$?
>
>
>
A related question replacing *cardinal* by *ordinal* in the above question would lead to $\sf V=HOD$, which is known to prove $\sf AC$ and so would be inconsistent with $\sf ZF + \neg AC$.
|
https://mathoverflow.net/users/95347
|
Is every set being cardinal definable consistent with ZF + negation of Choice?
|
This is consistent. Kanovei constructed a model $M$ with an infinite Dedekind finite set of reals which is lightface projectively definable. By descending to $L(R),$ we can further assume it satisfies $V=L(R).$
Clearly choice fails in this model. Since it satisfies $V=L(R),$ every set is definable from an ordinal and a real. Of course, any ordinal is definable from an $\aleph,$ so we just need to check that every real is cardinal definable.
Let $A$ be the definable Dedekind finite set of reals, and fix a canonical surjection $f: A \rightarrow \omega$ (as exists for any infinite set of reals). Let $\langle q\_n \rangle$ be an enumeration of the rationals. Define $A\_r=f^{-1}(\{n: q\_n<r\}).$ For $r<s,$ $|A\_r|<|A\_s|$ since these are Dedekind finite sets. Clearly $r$ is definable from the cardinality of $A\_r,$ so we're done.
Reference:
*Kanovej, V. G.*, [**On the nonemptiness of classes in axiomatic set theory**](http://dx.doi.org/10.1070/IM1978v012n03ABEH001997), Math. USSR, Izv. 12, 507-535 (1978). [ZBL0427.03044](https://zbmath.org/?q=an:0427.03044).
|
12
|
https://mathoverflow.net/users/109573
|
412591
| 168,343 |
https://mathoverflow.net/questions/412594
|
1
|
(I've tried [Math SE](https://math.stackexchange.com/questions/4342665/intuition-behind-using-random-variables-in-monte-carlo-methods-localization), but have so far come up empty handed, so I'm trying my luck here.)
I would like to get a better intuitive understanding of why [Monte Carlo](https://en.wikipedia.org/wiki/Monte_Carlo_method) works so well in approximating a solution to complex problems, such as calculating irrational numbers or the Particle Filter / Monte Carlo Localization to interpret noisy data from a sensor for instance. I have restated my question in two parts.
**Part 1:** One of [the key's to random sampling](https://youtu.be/OgO1gpXSUzU?t=215) is that the random sample will reflect the characteristics of the population from which it is drawn. So when we approximate Pi [by defining an area](https://en.wikipedia.org/wiki/Monte_Carlo_method#Overview) with a circle-quadrant within a square box and let random samples inside vs. outside the quadrant approach the ratio of Pi, are we in essence re-sampling our "deterministicly" defined circle-quadrant and box, *using random variables as a mechanism to be able to re-sample the pre-defined areas over and over again*?
**Part 2:** My question stated differently with Monte Carlo Localization (MCL); if I understand correctly, we first [scatter random particles](https://en.wikipedia.org/wiki/Monte_Carlo_localization#Example_for_1D_robot) (our random sampling/"hypotheses") across our map and then attach larger weight to the particles that overlap/match with our noisy sensor readings as more important. We then repeat to do a new re-sampling. Are the random variables here used as *a "fair" measurement mechanism to allow the obscured truth hidden in our noisy sensor reading to be re-sampled so that we approach a cleaner approximate sensor-reading/localization*?
[Edit] ***So in summary**, are the random variables here acting as a tool/mechanism for us to be able to create multiple observations of an otherwise deterministic experiment? And if so, at each iteration which we are observing and measuring, how does this avoid simply **adding random noise** to the data?*
**Or is there a better way to view the work the random variables do here?**
Also, if appropriate, any recommended further reading (including books/papers) that you think can help with the intuition in this mechanism is much appreciated!
|
https://mathoverflow.net/users/102097
|
Intuition of the "work" done by random variables in Monte Carlo methods (incl. MCL)
|
**Q:** *Are random variables acting as a tool for us to be able to create multiple iterations (to measure and average) of an otherwise deterministic experiment?*
**A:** An answer has two ingredients:
* firstly, many deterministic questions can be formulated as asking for the expectation of a random variable. For example, the integral $\int\_a^b f(x)dx$ is the expectation of $(b-a)f(x)$ for a random variable $x$ which is uniformly distributed in the interval $(a,b)$.
* secondly, the law of large numbers allows us to approximate the expectation by random sampling.
This is the essence of the Monte Carlo method. A further refinement could then be to reduce the variance of the estimator ([importance sampling](https://en.wikipedia.org/wiki/Importance_sampling)).
---
**Q:** Follow-up question: *Why doesn't each iteration simply add random noise to the data?*
**A:** After $N$ iterations, the signal has increased by a factor $N$, while the uncertainty due to the noise has increased by a factor $\sqrt N$, hence the *relative* uncertainty, which is what matters for the expectation, decays as $1/\sqrt N$.
|
4
|
https://mathoverflow.net/users/11260
|
412596
| 168,345 |
https://mathoverflow.net/questions/412581
|
6
|
$\DeclareMathOperator\GL{GL}$Let $K$ be a number field and $R$ its ring of integers. Let $G$ be a connected reductive closed subgroup of $\GL\_{n,K}$. On p55 of Brian Conrad's notes [Reductive group schemes](http://math.stanford.edu/%7Econrad/papers/luminysga3.pdf), he claims that the schematic closure $\overline{G}$ of $G$ inside $\GL\_{n,R}$ is a flat closed $R$-subgroup of $\GL\_{n,R}$.
First, I assume by schematic closure, he means the reduced induced structure on the topological closure. I can see that this closure $\overline{G}$ is an $R$-subgroup of $\GL\_{n,R}$, and that it satisfies $\overline{G}(R) = \{g\in \GL\_n(R) : g\cap \GL\_{n,K}\in G\}$, where in the brackets we view $g$ as a closed subscheme of $\GL\_{n,R}$.
Why is $\overline{G}$ necessarily flat? (Is this obvious?) Is this also true if $R$ is an arbitrary domain?
|
https://mathoverflow.net/users/88840
|
Flatness of the closure of a closed subgroup of the generic fiber of an algebraic group inside an integral model of the ambient group
|
$\DeclareMathOperator\GL{GL}$A colleague asked me this question some time ago and here is the answer I sent him.
Let $R$ be a domain with fraction field $K$. Let $A$ be the $R$-algebra underlying the group scheme $\GL\_{n,R}$ over $R$.
Suppose given a closed subgroup scheme $H$ of $\GL\_{n,K}$. Let $B$ be the underlying $K$-algebra of $H$.
We are then given by assumption a surjection $\phi:A\_K\to B$ and an injection $\lambda:A\to A\_K$ (given by $a \mapsto a\otimes\_K 1$). Let $\mu = \phi\circ\lambda$ and consider the $R$-module $\mu(A)$,
which is a sub-$R$-module of $B$. The following fact can be checked from the definitions: $\mu(A)$ has a natural $R$-algebra structure, compatible with the $R$-algebra structure of B via $\lambda$.
In fact the R-algebra $\mu(A)$ is (by definition) the underlying $R$-algebra of the scheme-theoretic image of $H$ in $\GL\_{n,R}$. Furthermore, we have $\mu(A)\_K\simeq B$ and
$\mu(A)$ is (clearly) torsion free.
Now consider the Hopf algebra structure of B, which is given by a morphism of $K$-modules $c:B\to B\otimes\_K B$. It is easy to see (diagram chasing) that
if $x\in \mu(A)\subseteq B$, then $c(x)$ lies in the image of the natural map $\mu(A)\otimes\_R\mu(A)\to B\otimes\_K B$. So if the natural
map $\mu(A)\otimes\_R\mu(A)\to B\otimes\_K B$ is an injection, we obtain a natural map $\mu(A)\to \mu(A)\otimes\_R\mu(A)$ and this then
(checking this is elementary) defines a Hopf algebra structure on $\mu(A)$. In particular, if the natural
map $\mu(A)\otimes\_R\mu(A)\to B\otimes\_K B$ is an injection, then the scheme-theoretic closure of $H$ in $\GL\_{n,R}$ is indeed a subgroup scheme, which is reduced if $H$ is reduced
(which is what would happen if $K$ is of characteristic 0).
So the basic condition, which must be satisfied, is that the natural map $\mu(A)\otimes\_R\mu(A)\to B\otimes\_K B$ is injective. This will be true iff
$\mu(A)\otimes\_R\mu(A)$ is torsion free. For example, if $\mu(A)$ happens to be flat over $R$ then $\mu(A)\otimes\_R\mu(A)$ will also be flat and thus torsion free.
This is what will happen if $R$ is a Dedekind domain (where torsion free is equivalent to flat) or if $R$ is a valuation ring (where again torsion free is equivalent to flat).
So if we work over either of these two rings, then the scheme-theoretic closure of $H$ in $\GL\_{n,R}$ is indeed a subgroup scheme and it is even flat over $R$.
|
9
|
https://mathoverflow.net/users/17308
|
412597
| 168,346 |
https://mathoverflow.net/questions/412441
|
1
|
Let $G$ be a discrete group. Let $(A,\alpha)$ and $(B,\beta)$ be $G$-$C^\*$-algebras and $\varphi: A \to B$ be $G$-equivariant and completely positive. All crossed products in this post are full (= universal).
I want to prove that there is a completely positive map
$$\varphi \rtimes G: A \rtimes\_\alpha G \to B \rtimes\_\beta G$$
such that $(\varphi\rtimes G)(\sum\_s a\_s s) = \sum\_s \varphi(a\_s)s$.
I managed to prove the following using the $G$-equivariant Stinespring theorem:
>
> Assume that $u: G\to B(H)$ is a unitary representation and $\sigma: A \to B(H)$ a completely positive map satisfying $\sigma(\alpha\_g(a)) = u\_g \sigma(a)u\_g^\*$ (i.e. $\sigma$ is $G$-equivariant where $B(H)$ has the $G$-action induced by $u$). Then there is a unique completely positive map
> $$\sigma \rtimes G: A \rtimes G \to B(H)$$
> satisfying $\sigma \rtimes G(\sum\_s a\_s s) = \sum\_s \sigma(a\_s)u\_s.$
>
>
>
I think I might be able to use this result to prove the result I want: Maybe the following works:
Let $(u,\pi)$ be a covariant representation of $B \rtimes G$ on $H$ where $\pi$ is chosen faithful. Then consider the composition $\sigma: A \to B(H)$ defined by
$$A \stackrel{\varphi}\to B \stackrel{i}\to B \rtimes G \stackrel{\pi}\to B(H)$$
which is completely positive and satisfies $\sigma(\alpha\_g(a)) = u\_g \sigma(a)u\_g^\*$. Hence, by the above result, we obtain an induced map
$$\sigma \rtimes G: A \rtimes G \to B(H).$$
If we can check that $(\sigma \rtimes G)(A \rtimes G) \subseteq \operatorname{Im}(\pi)$ then we can define
$$\varphi \rtimes G(x) = \pi^{-1}(\sigma \rtimes G)(x) \in B \rtimes G$$
which would yield the desired extension. However, I don't think the above inclusion holds.
|
https://mathoverflow.net/users/216007
|
A completely positive equivariant map $\varphi: A \to B$ induces a map on the full crossed products
|
(Too long for a comment). Your notation is confusing you, I think.
A *covariant* representation is not really of $B\rtimes G$, but is of the pair $(B,\beta)$. In particular, $\pi:B\rightarrow B(H)$ (notice the domain!) and not a representation of $B\rtimes G$. The covariant condition, that $\pi(\beta\_t(b)) = u\_g \pi(b) u\_g^\*$, ensures that $\pi$ *extends* to a $\*$-representation, say $\tilde\pi: B\rtimes G\rightarrow B(H)$. (Of course, the point is that *any* $\*$-representation of $B\rtimes G$ comes from some pair $(\pi,u)$, and so we can confuse them!)
Thus, in your argument, we *do not expect* that $(\sigma \rtimes G)(A\rtimes G) \subseteq \operatorname{Im}(\pi)$ but rather in the image of $\tilde\pi$.
Your argument now seems to be to pick $(\pi,u)$ "universal", so that $\tilde\pi$ will be an isometry, and then show that for this $\pi$ we get that $\sigma\rtimes G$ maps into the image of $\tilde\pi$. I think this is essentially obvious. First notice that
$$ \sigma = \tilde\pi \circ i \circ \varphi \implies
\sigma(a) = \pi(\varphi(a)) \quad (a\in A), $$
because by definition $\tilde\pi \circ i = \pi$. By density, we can consider a finite sum $\sum a\_s s \in A\rtimes G$, which has
$$ (\sigma \rtimes G) \Big(\sum a\_s s\Big) = \sum \sigma(a\_s) s
= \sum \pi(\varphi(a\_s)) s, $$
which is clearly a member of the image of $\tilde\pi$.
|
2
|
https://mathoverflow.net/users/406
|
412598
| 168,347 |
https://mathoverflow.net/questions/412599
|
3
|
Let $\phi: S^{k-1}\to E\setminus Q$ be a continuous map, here
* $E$ is an infinite dimensional Hilbert space
* $Q$ is a compact set in $E$
I need extend $\phi$ to $B^{k}$ such that
$\phi: B^{k}\to E\setminus Q$ is still continuous.
If $Q := \{u\in E:\|u\|\_E=1\}\setminus E\_{k+1}~ \text{where}~ E\_{k+1} ~\text{is any $k+1$-dimension subspace of} ~E$ (not compact), that's Ok, how about other $Q$?
|
https://mathoverflow.net/users/166368
|
extend a continuous map on sphere to ball such that the image is out of a compact set
|
Extending $\phi$ to the ball is equivalent to proving that $\phi:S^{k-1}\to H\setminus Q$ is nulhomotopic.
To prove this, you can consider the map $F:\phi(S^{k-1})\times Q\to E;(x,y)\to\frac{y-x}{|y-x|}$, which has compact image, so there is some vector $v$ with $|v|=1$ outside the image of $F$. So you can homotope $\phi(S^{k-1})$ to $\phi(S^{k-1})+kv$ with $k$ as big as you want without intersecting $Q$ at any point. If you take $k$ big enough, $\phi(S^{k-1})+kv$ and $Q$ will be separated by a hyperplane (perpendicular to $v$, for example), so you can just linearly homotope $\phi(S^{k-1})+kv$ to a point in its side of the hyperplane without intersecting $Q$.
|
2
|
https://mathoverflow.net/users/172802
|
412602
| 168,348 |
https://mathoverflow.net/questions/412605
|
7
|
The OEIS sequence [A000085](http://oeis.org/A000085) is defined by
$$ a\_n \!=\! (n-1)a\_{n-2} + a\_{n-1} \;\text{with }\; a\_0\!=\!1, a\_1\!=\!1.$$
If $n$ of the form $b^2-b+1, b \in \mathbb{N}, b > 2, \;\text{then: }\;$ $$ \left\lfloor \frac{a\_n}{a\_{n-1}} \right\rfloor > \left\lfloor \frac{a\_{n-1}}{a\_{n-2}} \right\rfloor$$
How to prove this?
|
https://mathoverflow.net/users/33646
|
Question on OEIS A000085
|
Your statement is (almost) proved in the paper [On solutions of $x^d=1$ in symmetric groups](https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/on-solutions-of-xd-1-in-symmetric-groups/8F90642D9472FA7326164E54BE3BE57B). The $a\_n$ in your post corresponds to $T\_n$ in the paper, and the author defines $R\_n = T\_n/T\_{n-1}$. The author then set out to prove $n-1 \leq R\_n^2-R\_n \leq n$\*. The proof is on p.161 and can be modified to make the inequalities strict, simply by changing all $\leq$s into $<$ and all $\geq$s into $>$.
If $n=b^2-b+1$, then $b^2-b<R\_n^2-R\_n<b^2-b+1$, and $b^2-b-1<R\_{n-1}^2-R\_{n-1}<b^2-b$. Thus, $R\_n>b$ and $R\_{n-1}<b$, and your statement follows.
\*This is paraphrased from p.161 of the paper; formula 2.8 in the paper is wrong, so I circumvented it.
|
18
|
https://mathoverflow.net/users/125498
|
412612
| 168,351 |
https://mathoverflow.net/questions/412617
|
0
|
Let $G$ be the heat kernal, i.e. for $0\le t<s$ and $x,y\in\mathbb R$
$$G(t,x;s,y):=\frac{1}{\sqrt{4\pi(s-t)}}\exp\left(-\frac{(y-x)^2}{4(s-t)}\right).$$
For $T>0$, let $\mathcal H\_T:=\{h:[0,T]\to [0,1]:~ h,~h' \mbox{ are both continuous on } [0,T]\}$ be endowed with the norm
$$\|h\|\_T := \max\_{0\le t\le T}|h(t)| + \max\_{0\le t\le T}|h'(t)|,\quad \forall h\in\mathcal H\_T.$$
Define the map $F$ on $\mathcal H\_T$ as follows: $F[h]=\big(F[h](s): 0\le s\le T\big)$ with
$$F[h](s):=\int\_{-s}^{\infty}\left(\int\_0^s G\big(A(u), -u;A(s),y\big)h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A(u)\big)^2} du\right)dy + \int\_{-s}^{\infty}\left(\int\_0^{\infty} G\big(0,x;A(s),y\big)\rho(x)dx\right)dy,$$
where $A: \mathbb R\_+\to\mathbb R\_+$ denotes the inverse of the function
$$\mathbb R\_+\ni t\mapsto \int\_0^t (1+h(r))^2dr\in \mathbb R\_+$$
and $\rho: \mathbb R\_+\to \mathbb R\_+$ is a probability density, i.e.
$$\int\_0^{\infty}\rho(x)dx =1.$$
Can we prove the existence of $T>0$ s.t. $F$ is a contraction w.r.t. $\|\cdot\|\_T$? Any answer, comments or references are appreciated.
PS : I strongly believe the answer is yes, while I am not familiar with the inequalities related to the heat kernal, especially when estimating the derivative $F[h]'$. Bty, $\circ$ stands for the composition of functions.
PS 2 : A straightforward computation yields for all $h\_1, h\_2 \in\mathcal H\_T$,
$$|A\_1(t)-A\_2(t)|\le 4t\max\_{0\le r\le t}|h\_1(r)-h\_2(r)|,\quad \forall 0\le t\le T.$$
|
https://mathoverflow.net/users/261243
|
Is this a contraction mapping for small $T$?
|
$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\De}{\Delta}\newcommand\R{\mathbb R}$**Edit:** This answer is insufficient, even though (almost) all the reasoning appears relevant to the problem. I will try to come back and fix it.
---
Suppose that $T\in(0,1]$.
Integrating in $y$, simplify the expression for $F$ as follows:
\begin{equation\*}
F[h](s)=I\_1(h)+I\_2(h), \tag{1}
\end{equation\*}
where
\begin{equation\*}
I\_1(h):=\int\_0^s \frac{1}{2} \Big(\text{erf}\Big(\frac{s-u}{2 \sqrt{A\_h(s)-A\_h(u)}}\Big)+1\Big) h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A\_h(u)\big)^2}\, du,
\end{equation\*}
\begin{equation\*}
I\_2(h):=\int\_0^{\infty} \frac{1}{2} \Big(\text{erf}\Big(\frac{s+x}{2 \sqrt{A\_h(s)}}\Big)+1\Big)\rho(x)\,dx.
\end{equation\*}
Here the notation $A\_h$ reminds us that $A$ depends on $h$.
Let $H\_T:=\mathcal H\_T$. Let $B\_T$ be the set of all bounded real-valued functions on $[0,T]$, endowed with the sup-norm.
Since $0\le h\le1$, the operator that maps any $h\in H\_T$ to the function $[0,T]\ni u\mapsto\big(1+h(u)\big)^2$ in $L\_T$ is Lipschitz with a universal Lipschitz constant.
Take any $h\_0,h\_1$ in $H\_T$ with
\begin{equation\*}
\de:=\|h\_1-h\_0\|<1/16. \tag{2}
\end{equation\*}
Letting
\begin{equation\*}
J\_h(t):=\int\_0^t (1+h(r))^2\,dr,
\end{equation\*}
we see that $J\_h$ is $4$-Lipschitz, $t\le J\_h(t)\le4t$, $A\_h$ is $1$-Lipschitz, $|J\_{h\_1}(t) -J\_{h\_0}(t)|\le8t\de$, and hence
\begin{equation\*}
\frac u4\le A\_{h\_j}(u)\le\frac{A\_{h\_{1-j}(u)}}{1-8\de}\le\frac u{1-8\de}, \tag{3}
\end{equation\*}
\begin{equation\*}
|A\_{h\_1}(u)-A\_{h\_0}(u)|\le\frac{8\de u}{1-8\de}\le16\de u; \tag{4}
\end{equation\*}
here in what follows, $0\le t\le T$, $0<u<s\le T$, and $j\in\{0,1\}$, unless specified otherwise.
For brevity, let us refer to the Lipschitz maps with Lipschitz constants depending only on $\de$ as good-Lipschitz. In particular, the map $H\_T\ni h\mapsto A\_h$ is good-Lipschitz. Therefore, the maps $H\_T\ni h\mapsto\big(1+h\circ A\_h(u)\big)^2$ and
\begin{equation\*}
H\_T\ni h\mapsto h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A\_h(u)\big)^2} \tag{5}
\end{equation\*}
are good-Lipschitz, for each $u\in[0,T]$. The latter map is also bounded by a constant depending only on $\de$; let us refer to such maps as well-bounded.
For $h\_t:=h\_0+t(h\_1-h\_0)$, $\tau\_t:=A\_{h\_t}(s)-A\_{h\_t}(u)$, and $\De\tau:=\tau\_1-\tau\_0$, we have
$0<\frac{s-u}4\le\tau\_t\le s-u$ (since $J\_h$ is $4$-Lipschitz and $A\_h$ is $1$-Lipschitz), $|\De\tau|\le32\de s$ (by (4)), and hence
the absolute value of the derivative of the map
\begin{equation\*}
[0,1]\ni t\mapsto
\dfrac{1}{2} \Big(\text{erf}\Big(\dfrac{s-u}{2 \sqrt{A\_{h\_t}(s)-A\_{h\_t}(u)}}\Big)+1\Big)
\end{equation\*}
is
\begin{equation\*}
\begin{aligned}
& \frac{ s-u }{4 \sqrt{\pi } \tau\_t^{3/2}}\ \exp\Big\{-\frac{(s-u)^2}{4\tau\_t}\Big\}\,|\De\tau| \\
& \le\frac1{4^{-1/2} \sqrt{\pi } (s-u)^{1/2}}\ \exp\Big\{-\frac{s-u}4\Big\}\,32\de s.
\end{aligned}
\end{equation\*}
The integral in $u\in(0,s)$ of the latter expression is $\le CT$; here and in what follows, $C$ will denote various well-bounded expressions.
Therefore and because $|\text{erf}|\le1$ and the map (5) is good-Lipschitz and well-bounded, we conclude that $I\_1$ is $CT$-Lipschitz.
It is similar but easier to show that $I\_2$ is $CT$-Lipschitz as well.
Indeed, still with $h\_t:=h\_0+t(h\_1-h\_0)$, let $\rho\_t:=A\_{h\_t}(s)$, and $\De\rho:=\rho\_1-\rho\_0$.
Then
$0<\frac s4\le\rho\_t\le s$ (since $J\_h$ is $4$-Lipschitz and $A\_h$ is $1$-Lipschitz), $|\De\rho|\le16\de s$ (by (4)), and hence
the absolute value of the derivative of the map
\begin{equation\*}
[0,1]\ni t\mapsto
\frac{1}{2} \Big(\text{erf}\Big(\frac{s+x}{2 \sqrt{A\_{h\_t}(s)}}\Big)+1\Big)
\end{equation\*}
is
\begin{equation\*}
\begin{aligned}
&\frac{|s+x|}{4 \sqrt{\pi } \rho\_t^{3/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho\_t}\Big\}\,|\De\rho| \\
&=\frac{|s+x|}{4 \sqrt{\pi } \rho\_t^{1/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho\_t}\Big\}\,
\frac{|\De\rho|}{\rho\_t} \\
&\le\frac{|s+x|}{4 \sqrt{\pi } \rho\_t^{1/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho\_t}\Big\}\,
64\de\le C;
\end{aligned}
\end{equation\*}
so, the integral in $u\in(0,s)$ of the latter expression is $\le CT$. So, $I\_2$ is $CT$-Lipschitz as well.
We conclude that $F$ is $CT$-Lipschitz and hence a contraction if $T$ is small enough.
|
1
|
https://mathoverflow.net/users/36721
|
412632
| 168,355 |
https://mathoverflow.net/questions/412636
|
0
|
How can I visualise PDF of distribution defined by quantiles, that I predict with my neural network? Now I'm passing quantiles to the histogram, but I don't think it is the correct way for visualising. I don't know it's simple/effectively to find the derivation of quantile function defined by the neural network. Exist any way how can comfortable be quantile function predicted by neural network converted to PDF for visualisation of distribution?
Chart: <https://wandb.ai/markub/rl-toolkit/runs/2sgcmjr8/overview?workspace=user-markub>
Thanks.
Have a nice day.
|
https://mathoverflow.net/users/473380
|
Visualization PDF of distribution defined by quantiles
|
Smoothed histograms should be a way to go. See e.g. [this paper](https://www.cs.cmu.edu/%7Ejgc/publication/PublicationPDF/Analysis_Of_Uncertain_Data_Smoothing_Of_Histograms.pdf) and, more generally, these [search results](https://www.google.com/search?q=smoothed%20histogram&oq=smoothed%20histogram&aqs=chrome..69i57.7416j0j1&sourceid=chrome&ie=UTF-8).
|
0
|
https://mathoverflow.net/users/36721
|
412638
| 168,356 |
https://mathoverflow.net/questions/412630
|
6
|
I have recently encountered a triangular array $(a\_{i,j})\_{0\le i\le j}$, each line of which might (should?) have a combinatorial interpretation in terms of $S\_{2n+1}$. Here it is (the first entry of each line is the label $2n+1$, and the first $1$ on the right is $a\_{0,0}$):
```
1 | 1
3 | 2 4
5 | 24 80 16
7 | 720 3136 1120 64
9 | 40320 209408 102144 10752 256
11 | 3628800 21441024 12869120 1892352 84480 1024
13 | 479001600 3130103808 2188865536 402980864 25479168 585728 4096
15 | 87178291200 618377527296 487356047360 105995681792 8484270080 278806528 3727360 16384
```
Since the entries in each line sum to $(2n+1)!$, I would expect these entries to correspond to certain subsets of $S\_{2n+1}$. In particular $a\_{n,0}=(2n)!$, which should naturally correspond to a subgroup $S\_{2n}$. Further we have for example
$$a\_{n,n}=2^{2n};\ a\_{n,n-1}=2^{2n-1}\binom {2n+1}3;\ a\_{n,n-2}=2^{2n-1}\Bigl[2n\binom {2n+1}5-\binom {2n+2}6\Bigr].$$
If ever there is a combinatorial interpretation, it must be independent of the conjugacy classes, as e.g. in $S\_7$, no classes could constitute $a\_{3,3}=64$.
My goal is of course to find a direct formula (or a recursion formula) for these numbers. Note that $a\_{n,1}$ can have big prime factors, e.g. $ a\_{13,1}=3130103808=2^{11}×3^2×13×13063$.
An intriguing feature is that we can "interpolate" between the lines, more precisely between the diagonals, i.e. try to decompose $(2n)!$ similarly as $b\_{n,0}+\cdots+b\_{n,n}$ with the $b\_{i,j}$ obeying the same formulas as the $a\_{i,j}$, just with $2n$ instead of $2n+1$. But then it turns out that the row sums would be $(2n)!+(2n-1)!$, meaning that, if we look again for an interpretation in terms of $S\_{2n}$, we'd have to disconsider the column $b\_{n,0}$, as this gives just the 'superfluous' $(2n-1)!$ contribution. But this being said: why not? (And who knows, it might even give a hint towards coming up with a common interpretation combining even *and* odd rows.)
The first even rows would be logically
```
2 | (1) 2
4 | (6) 16 8
6 | (120) 368 320 32
8 | (5040) 16896 19712 3584 128
```
|
https://mathoverflow.net/users/29783
|
Is there a combinatorial interpretation of this array in terms of $S_{2n+1}$?
|
You can use <https://www.findstat.org> to obtain a candidate for a conjectural solution, see <https://www.findstat.org/StatisticsDatabase/St000389oMp00093oMp00127oMp00066oMp00090> for details:
```
after adding 1 to every value
and applying
Mp00090: cycle-as-one-line notation: Permutations -> Permutations
Mp00066: inverse: Permutations -> Permutations
Mp00127: left-to-right-maxima to Dyck path: Permutations -> Dyck paths
Mp00093: to binary word: Dyck paths -> Binary words
to the objects (see `.compound_map()` for details)
your input matches
St000389: The number of runs of ones of odd length in a binary word.
```
I did check the first few generating polynomials. From the definition of the map, it is easy to see that the leading term is $2^{n-1}$, the number of compositions of $n$. Since the first map sorts the cycles by their minimal elements, its image starts with $1$, which implies that the first ascent of the Dyck path has length one. Thus, the constant terms of the generating polynomials vanish. The linear terms vanish for even $n$, because there is no Dyck path of semilength $n$ with a single odd ascent.
```
1
q
2*q^2
4*q^3 + 2*q
8*q^4 + 16*q^2
16*q^5 + 80*q^3 + 24*q
32*q^6 + 320*q^4 + 368*q^2
64*q^7 + 1120*q^5 + 3136*q^3 + 720*q
128*q^8 + 3584*q^6 + 19712*q^4 + 16896*q^2
256*q^9 + 10752*q^7 + 102144*q^5 + 209408*q^3 + 40320*q
512*q^10 + 30720*q^8 + 462336*q^6 + 1838080*q^4 + 1297152*q^2
1024*q^11 + 84480*q^9 + 1892352*q^7 + 12869120*q^5 + 21441024*q^3 + 3628800*q
```
|
9
|
https://mathoverflow.net/users/3032
|
412639
| 168,357 |
https://mathoverflow.net/questions/412648
|
6
|
Let $G$ be a finite group. Let $p$ be a prime number such that $p \mid |G|$.
Let Irr$(G)$ denote the set of ordinary irreducible characters of $G$.
For $\chi\in$ Irr$(G)$ define $e\_{\chi} := \frac{\chi(1)}{|G|}\sum\_{g\in G} {\chi(g^{-1})\cdot g}$.
Let $B$ denote a $p$-block of $G$. Let Irr$(B)$ be the set of those ordinary irreducible characters of $G$ which lie in $B$.
Define $e\_B:=\sum\_{\chi\in \text{Irr}(B)} {e\_{\chi}}$. This can be rewritten as
$$e\_B = \sum\_{g\in G} {\bigg(\underbrace{\frac{1}{|G|}\sum\_{\chi\in \text{Irr}(B)} {\chi(1)\chi(g^{-1})}}\_{=:a\_g}\bigg)} \cdot g.$$
**Question:**
>
> Is it always true that $a\_g$ lies in $\mathbb{Q}\_p[\zeta]$ for $\zeta = \exp(\frac{2\pi i}{p^n-1})$ for some positive integer $n$?
>
>
>
Example:
>
> Set $G:=A\_5$ and $p:=5$. Then there are exactly two $p$-blocks of $G$.
>
>
> In the coefficients of some of the $e\_{\chi}$ there is the number $z:=\exp(\frac{2\pi i}{5})$ involved. The number $z$ cannot be expressed via the $\zeta$ above and $e\_{\chi}$ does not lie in $\mathbb{Q}\_p[\zeta]$ for $\zeta$ as above.
>
>
> But the two block idempotents $e\_{B\_0}$ and $e\_{B\_1}$ do not involve $z$ anymore.
>
>
>
In other examples, the ${e\_{\chi}}'s$ add up nicely to ${e\_{B\_j}}'s$, or such $z$'s don't appear at all.
If this is always the case, I would be fond of a reference.
Thank you very much for the help.
|
https://mathoverflow.net/users/12826
|
Question concerning the coefficients of block idempotents
|
Yes, this is true. By block orthogonality relations due to R. Brauer, it is true that $a\_{g}=0$ whenever the element $g$ has order divisible by $p$. But when $g$ has order prime to $p$, it is clear that $a\_{g}$ lies in $\mathbb{Q}[\omega]$ for some $t$-th root of unity $\omega$, where $t$ is not divisible by $p$.
Furthermore, the same block orthogonality relations imply that when the order of $g$ is prime to $p$, then the class function $\theta = \sum\_{\chi \in B} \chi(g^{-1} )\chi$ vanishes of all non-identity elements of $P$, a Sylow $p$-subgroup of $G$. Taking the inner product of $\theta$
(restricted to $P$) with the trivial character of $P$ then shows that $[G:P] a\_{g}$ is an algebraic integer.
These two facts give what you want. The necessary block orthogonality relations can be found in almost any text which deals with blocks and characters, eg those by G.Navarro or by M. Isaacs.
|
9
|
https://mathoverflow.net/users/14450
|
412659
| 168,361 |
https://mathoverflow.net/questions/412670
|
1
|
I accidently found a class of rank-deficient square matrices that are their own Moore-Penrose pseudo inverse:
$$\boldsymbol{A}\in\mathbb{R}^{n\times n}: a\_{(i,i)}=n-1,\, a\_{\lbrace i,j\rbrace}=-1\implies\Big(\frac{1}{n}\boldsymbol{A}\Big)^+ =\Big(\frac{1}{n}\boldsymbol{A}\Big)$$
as an exemplary $4\times 4$ calculation done with WA demonstrates
>
> **Questions:**
>
>
> * have all Moore-Penrose "pseudo unit-matrices" been characterized, what about other kinds of pseudo inverses?
> * do these "pseudo unit-matrices" play a special role in the theory of pseudo inverses, resp., in linear algebra?
>
>
>
|
https://mathoverflow.net/users/31310
|
Characterization of pseudo unit-matrices
|
All symmetric idempotent matrices are their own Moore-Penrose pseudo-inverse; these are matrices with eigenvalues equal to either 0 or 1.
Your matrices are in this class.
|
1
|
https://mathoverflow.net/users/11260
|
412674
| 168,367 |
https://mathoverflow.net/questions/412663
|
1
|
Let $\mu$ be a probability measure on $\mathbb R^n$ and let $P$ be a compact polytope in $\mathbb R^n$. For any $x \in \mathbb R^n \setminus P$, let $p(x) \in P$ be (unique!) point in $P$ which is closest to $x$ and let $d(x) := \|x-p(x)\|$ be the distance from $x$ to $p$ and $u(x) := (x-p(x))/d(x) \in S\_{n-1}$, where $S\_{n-1}$ is the unit-sphere in $\mathbb R^n$.
>
> **Question.** *Is it true that that for every $\epsilon>0$, there exists a measurable set $P^\epsilon$ such that $P \subseteq P^\epsilon$, $\mu(P^\epsilon) \le \mu(P) + \epsilon$, and the mapping $x \mapsto u(x)$ is Lipschiitz-continuous on $\mathbb R^n \setminus P^\epsilon$ ?*
>
>
>
For example, if $P$ is just a half-space, $u$ is a constant, and thus Lipschitz-continuous on $\mathbb R^n\setminus P$.
**Related.** [Conditions for Lipschitz continuity of boundary normal vector to closed set with "unique closest-point" poperty (e.g closed convex sets)](https://mathoverflow.net/q/412656/78539)
|
https://mathoverflow.net/users/78539
|
On the Lipschitz continuity of the unit-normal vector field of a polytope
|
**EDIT:** The OP changed the question while I was writing this answer. In this answer, we take the domain of $u$ to be $\{x\in\mathbb{R}^n:d(x)\geq\epsilon\}$ for some fixed $\epsilon>0$. This is a different use of $\epsilon$ than in OP's current problem formulation.
Let $B\_\epsilon$ denote the open ball centered at the origin of radius $\epsilon$. Below, we show how to factor $u$ as the composition of a $2$-Lipschitz map $g\colon\mathbb{R}^n\to\mathbb{R}^n$ with a $(1/\epsilon)$-Lipschitz map $f\colon(\mathbb{R}^n\setminus B\_\epsilon)\to S^{n-1}$.
To analyze each map, we will use the fact that [projection onto a nonempty closed convex set is contractive](https://math.stackexchange.com/questions/3809431/geometric-proof-that-projections-on-convex-sets-are-contractive).
First, define $g$ by $g(x):=x-p(x)$, where $p$ denotes projection onto $P$. Then
$$
\|g(x)-g(y)\|=\|(x-p(x))-(y-p(y))\|\leq\|x-y\|+\|p(x)-p(y)\|\leq2\|x-y\|.
$$
Next, define $f$ by $f(x):=x/\|x\|$. Since the domain of $f$ is the complement of $B\_\epsilon$, we can instead write $f(x)=q(x)/\epsilon$, where $q$ denotes projection onto the closure of $B\_\epsilon$. Then
$$
\|f(x)-f(y)\|=\|q(x)/\epsilon-q(y)/\epsilon\|=(1/\epsilon)\|q(x)-q(y)\|\leq(1/\epsilon)\|x-y\|.
$$
Combining these estimates, we have that $u=f\circ g$ is $(2/\epsilon)$-Lipschitz.
|
1
|
https://mathoverflow.net/users/29873
|
412676
| 168,368 |
https://mathoverflow.net/questions/412592
|
2
|
Let $K$ be a convex body in $\mathbb R^d$ which contains the origin and let $\theta \in (0,1)$.
>
> **Question.** *Is it always possible to find $n$ points $x\_1,\dotsc,x\_n \in \mathbb R^d$ such that
> $$
> \theta K \subseteq \operatorname{conv}\{x\_1,\dotsc,x\_n\} \subseteq K
> \tag{1}\label{1}
> $$
> and $n \le C\_\theta d$ for some constant $C\_\theta < \infty$ which only depends on $\theta$?*
>
>
>
I'm interested in the case where $\theta \to 1$.
**Note.** I don't care about constructive / algorithmic solutions. I'm only interested in existence.
Some known results:
-------------------
* From **Theorem 1.1** of "[Approximating a convex body by a polytope using the epsilon-net theorem](https://arxiv.org/abs/1705.07754)", we know that if $\theta=1/d$, then $n = 500d$ points drawn uniformly at random of from $K$ satisfy \eqref{1} with probability $1-e^{-d}$. The issue with this result is that it only works for $\theta \ll 1$.
* From **Theorem 1.2** of the same paper, we know that if $n= \mathcal O(dc\_\theta^d)$ (with $c\_\theta := 1 / (1 - \theta)>1$) points are drawn uniformly at random from $K$, then \eqref{1} is satisfied with probability $1-e^{-d}$. The issue with this result is that the dependence of $n$ on $d$ is exponential. We need linear.
>
> On second thought it seems the bounds implied by the above results are an artifact of the complete randomness in the points $x\_1,\dotsc,x\_n$.
>
>
>
|
https://mathoverflow.net/users/78539
|
Existence of fine approximate of a convex body in $\mathbb R^d$ with convex hull of $\mathcal O(d)$ points
|
This is false for any fixed $\theta\in(\frac{1}{2},1)$, we will use the balls of radius $1$ in $\mathbb{R}^d$ as a counterexample. Given $\theta$, if we want $
\theta K \subseteq \operatorname{conv}\{x\_1,\dotsc,x\_n\} \subseteq K
$, we will need at least one of the $x\_i$ to be in the convex closure of each [hyperspherical cap](https://en.wikipedia.org/wiki/Spherical_cap#Hyperspherical_cap) of height $1-\theta$. This implies each point of the sphere needs to have some $x\_i$ at distance $<\sqrt{2(1-\theta)}$ from it (that´s the radius of the hyperspherical cap of height $1-\theta$). Call $R=\sqrt{2(1-\theta)}$, so that $R<1$ if $\theta>\frac{1}{2}$.
Then, the union of the $n$ balls $B\_i$ of radius $R<1$ and centers the points $x\_i$ has to cover the sphere. But each $B\_i$ will cover a spherical cap of the sphere with height $\leq h$, for some $h<1$ (concretely, $h=1-\sqrt{1-R}$). So, calling $V\_n$ the $n^{th}$ dimensional volume and $C^n\_h$ the hyperspherical cap of $S^n$ and height $h$, we just have to prove that $\frac{V\_{n}(S^n)}{V\_n(C^n\_h)}$ grows more than linearly for any $h<1$.
To do that we just use the volume formulas [here](https://en.wikipedia.org/wiki/Spherical_cap#Hyperspherical_cap) and we have
$$\frac{V\_{n}(S^n)}{V\_n(C^n\_h)}=\frac{\int\_0^\pi \sin^n(t)dt}{\int\_0^{\arccos(1-h)}\sin^n(t)dt}.$$
In this quotient, the term below decreases exponentially, as $\sin(t)$ is bounded in $[0,\arccos(1-h)]$ by some constant $<1$. But the term above decreases very slowly, as it is in fact $\frac{\sqrt{\pi}\cdot\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(1+\frac{n}{2}\right)}$, which decreases more or less like $\frac{1}{\sqrt{n}}$.
This means that the quotient above increases exponentially, so you´ll need $n$ to grow at least exponencially on $d$.
|
2
|
https://mathoverflow.net/users/172802
|
412685
| 168,371 |
https://mathoverflow.net/questions/412683
|
1
|
Reading a proof of the Schwarz lemma for the Kähler-Ricci flow from p22 of [these lecture notes](https://arxiv.org/pdf/1212.3653.pdf). I am confused as to what they mean by taking $$\inf \_{x \in M} \{\hat{R}\_{i \bar i j \bar j}(x) \mid \{\partial\_{z^{1}}, \ldots, \partial\_{z^{n}} \} \text{ is orthornormal w.r.t. } \hat{g} \text{ at } x, \ i, j = 1, \ldots, n\} \ .$$ Is this lower bound even invariantly defined on the manifold?
|
https://mathoverflow.net/users/473447
|
Schwarz lemma and bisectional curvature lower bound
|
Their idea is correct, but its formulation (and the chosen notation) is indeed a bit sloppy. What they seem to do, in reality, is to define
$$ - \hat C = \inf \_{x \in M} \inf \{ \hat R (e\_i, \bar {e\_i}, e\_j, \bar {e\_j}) \mid \{e\_1, \bar {e\_1}, \ldots, e\_n, \bar {e\_n} \} \text{ is orthornormal w.r.t. } \hat{g} \text{ at } x, \ i, j = 1, \ldots, n\} \ .$$
The inner infimum exists, because $i,j$ take a finite number of values, and the orthonormal frames $\{ e\_1, \bar {e\_1}, \dots, e\_n, \bar {e\_n} \}$ at $x$ form a space homeomorphic to $O(2n)$ which is compact.
They tacitly accept that this inner infimum will be a continuous function of $x$ (up to the reader to prove this!), therefore the outer infimum will be finite ($M$ is assumed compact).
Finally, in equation (2.22) they just work at some $x$, taking $\{ e\_1, \bar {e\_1}, \dots, e\_n, \bar {e\_n} \}$ to be precisely the frame $\{ \partial\_1, \bar \partial\_1, \dots, \partial\_n, \bar \partial\_n \}$ corresponding to the normal coordinates at $x$ with respect to $\hat g$.
|
2
|
https://mathoverflow.net/users/54780
|
412688
| 168,372 |
https://mathoverflow.net/questions/412323
|
1
|
The well-known theorem of [Abhyankar–Moh–Suzuki](https://en.wikipedia.org/wiki/Abhyankar%E2%80%93Moh_theorem) says the following:
Let $f=f(t), g=g(t) \in k[t]$, $k$ is a field of characteristic zero.
If $k[f,g]=k[t]$, then $\deg(f) \mid \deg(g)$ or $\deg(g) \mid \deg (f)$.
Let us concentrate on the case $k=\mathbb{C}$ (I wonder if there is a difference in my following question between $k=\mathbb{C}$ and $k=\mathbb{R}$).
Let $f=f(t), g=g(t) \in \mathbb{C}[t]$ satisfy the following conditions:
**(1)** $f=hf\_1$, $g=hg\_1$, where $\gcd(f\_1,g\_1)=1$
(= $h$ is the gcd of $f$ and $g$) and each of
$\deg(h)$, $\deg(f\_1)$, $\deg(g\_1)$ is $\geq 1$.
**(2)** $\deg(f\_1)=\deg(g\_1)$, hence $\deg(f)=\deg(g)$.
**(3)** $\mathbb{C}(f,g)=\mathbb{C}(t)$.
>
> Should such $f$ and $g$ generate $\mathbb{C}[t]$, namely, $\mathbb{C}[f,g]=\mathbb{C}[t]$?
>
>
>
I have not found an example of such $f$ and $g$ not generating $\mathbb{C}[t]$, but maybe I am missing something easy.
Maybe one (or more) of the following ideas could help:
**(i)** The following known criterion:
$\mathbb{C}[f(t),g(t)]=\mathbb{C}[t]$ iff $(f'(t),g'(t))\neq 0$ and $t\mapsto (f(t),g(t))$ is injective, see [this](https://math.stackexchange.com/questions/2733573/why-mathbbcf-1t-f-2t-mathbbct-iff-f-1t-f-2t-neq0-and?noredirect=1&lq=1).
**(ii)** The theory of SAGBI bases (I will try to find relevant papers, I remember there are ones. I think there is a result saying that if $f', g' \in \mathbb{C}[f,g]$, then $\mathbb{C}[f,g]=\mathbb{C}[t]$).
**(iii)** The theory of sub-resultants.
(**(iv)** Maybe if $\deg(h)=1$, then I can show that $\mathbb{C}[f,g]=\mathbb{C}[t]$. However, I prefer not restricting $\deg(h)$).
Somewhat relevant questions: [Is there a converse of Abhyankar-Moh-Suzuki theorem?](https://mathoverflow.net/questions/304189/is-there-a-converse-of-abhyankar-moh-suzuki-theorem) and [Generalizations of Abhyankar-Moh theorem (embeddings of the line in the plane)](https://mathoverflow.net/questions/295158/generalizations-of-abhyankar-moh-theorem-embeddings-of-the-line-in-the-plane).
Any comments are welcome!
**Edit:** I have now found a counterexample for the case $\deg(g) \geq 2$:
$f=t^{15}+t^2$, $g=t^{15}$. Conditions **(1),(2),(3)** are being satisfied, but $\mathbb{C}[f,g]=\mathbb{C}[t^2] \subsetneq \mathbb{C}[t]$.
So only when $\deg(h)=1$ perhaps there is a positive answer.
Therefore, and in view of idea **(i)**, we will add a fourth condition:
**(4)** $f',g'$ are not simultaneously zero.
Then, the above counterexample is not a counterexample anymore, since $f'=15t^{14}+2t$, $g'=15t^{14}$ have a common zero at $0$.
|
https://mathoverflow.net/users/72288
|
A variation on Abhyankar–Moh–Suzuki theorem
|
The answer is no. You can simply choose $$f=t(t^2-t+1)$$
$$g=t(t^2+1)$$
These satisfy all your conditions, but $\mathbb{C}[f,g]\subsetneq \mathbb{C}[t]$.
Let us check this:
$(1)-(2)$: $f=tf\_1$ and $g=tg\_1$ where $f\_1=t^2-t+1$, $g\_1= t^2+1$ are both of degree $2$ and without common factor.
$(3)$: $\mathbb{C}(f,g)=\mathbb{C}(t)$ because $g-f=t^2$, so $\frac{g}{g-f+1}=t$.
Finally, we observe that
$\mathbb{C}[f,g]=\mathbb{C}[f-g,g]=\mathbb{C}[t^2,t^3+1]=\mathbb{C}[t^2,t^3]\subsetneq \mathbb{C}[t]$.
|
3
|
https://mathoverflow.net/users/23758
|
412696
| 168,377 |
https://mathoverflow.net/questions/412656
|
1
|
Let $C$ be a nonempty closed subset of $\mathbb R^n$. It is known that any such set satisfies the following condition
>
> **(Unique CPP a.e).** *For **almost every** $x \in \mathbb R^n$, there exists a unique point in $c(x) \in C$ such that $\|x-c(x)\| = \mbox{dist}(x,C) := \inf\_{c \in C}\|x-c\|$.*
>
>
>
For example, see **Theorem 1** of <http://tzamfirescu.tricube.de/TZamfirescu-110.pdf>.
In particular, if $C$ is a closed convex subset of $\mathbb R^n$, then it is a folklore fact that $C$ verifies he unique **CPP** **everywhere**.
For any $\epsilon > 0$, let $C^\epsilon := \{x \in \mathbb R^n \mid d(x,A) \le \epsilon\}$ be the $\epsilon$-expansion of $C$.
>
> **Question.** *Under what minimalistic conditions on $C$ does there exist a function $\epsilon:(0,\infty) \to [0,\infty)$, such that $\limsup\_{L \to \infty} \epsilon(L) = 0$ and the mapping $u\_C:x \mapsto (x-c(x))/\|x-c(x)\|$ is $L$-Lipschitz a.e on $\mathbb R^n \setminus C^{\epsilon(L)}$ for any sufficiently large $L>0$ ?*
>
>
>
* I'm particularly interested in the case of sets of the form $C := \cap\_{i \in I} C\_i$ with $C\_i := \{x \in \mathbb R^n \mid f\_i(x) \le 0\}$, where $f\_i:\mathbb R^n \to \mathbb R$ are sufficiently smooth functions. Even the case where $I$ is a singleton $C$ is the sublevel set of a sufficiently smooth function is already interesting.
* The case where $C$ is a closed convex subset of $\mathbb R^n$ has been solved here <https://mathoverflow.net/a/412676/78539>: it suffices to take $\varepsilon(L) = 2/L$.
|
https://mathoverflow.net/users/78539
|
Conditions for Lipschitzness of boundary normal vector, almost everywhere
|
Given a nonempty closed set $C\subseteq\mathbb{R}^n$, let $S\_C\subseteq\mathbb{R}^n$ denote the set of points for which the closest point in $C$ is not unique. Suppose $C$ satisfies each of the following:
* $S\_C$ is nonempty (i.e., $C$ is not convex),
* there exists $\rho>0$ such that $C^\rho\cap S\_C=\emptyset$, and
* the closest point mapping $c\colon(\mathbb{R}^n\setminus S\_C)\to C$ is continuous.
(Every nonempty closed nonconvex subset $C\subseteq\mathbb{R}^n$ with smooth boundary that I can think of satisfies these property.)
Let $r\_C<\infty$ denote the supremum of such $\rho$. Below, we show that for every $\epsilon\in(0,r\_C)$, it holds that $u\_C$ is **not** Lipschitz on $\mathbb{R}^n\setminus (C^\epsilon\cup Z)$ for any null set $Z$.
Select $p\in\mathbb{R}^n$ of distance $r\_C$ from $C$ for which there exist $x,y\in C$ with $x\neq y$ such that
$$
\|x-p\|=\|y-p\|=r.
$$
Then for each $t\in(0,1)$, the points $x(t):=tx+(1-t)p$ and $y(t):=ty+(1-t)p$ are of distance less than $r\_C$ from $C$, and so $c(x(t))=x$ and $c(y(t))=y$. Furthermore,
$$
u\_C(x(t))=\frac{x(t)-c(x(t))}{\|x(t)-c(x(t))\|}=\frac{x(t)-x}{\|x(t)-x\|}=\frac{p-x}{\|p-x\|},
$$
and similarly
$$
u\_C(y(t))=\frac{p-y}{\|p-y\|}.
$$
Notably, for small $t$, $x(t)$ and $y(t)$ are arbitrarily close to each other, while $u\_C(x(t))$ and $u\_C(y(t))$ remain a fixed distance apart.
Given $\epsilon\in(0,r\_C)$, a null set $Z$, and a constant $L>0$, we now show that $u\_C$ is not $L$-Lipschitz on $\mathbb{R}^n\setminus (C^\epsilon\cup Z)$. Fix $\delta>0$ to be selected later. Select $t\_0\in(0,1/2)$ small enough so that $x(t\_0)$ and $y(t\_0)$ avoid $C^\epsilon$ and $\|x(t\_0)-y(t\_0)\|<\delta$. Select a continuous path $q\colon[0,1]\to \mathbb{R}^n\setminus S\_C$ such that $q(0)=x(t\_0)$, $q(1)=y(t\_0)$, and $q(s)\not\in Z$ for almost every $s\in(0,1)$. By the continuity of $c$, there exist $s\_0,s\_1\in[0,1]$ such that $q(s\_0),q(s\_1)\not\in Z$ and
$$
\max\Big\{\|q(s\_0)-x(t\_0)\|,\|q(s\_1)-y(t\_0)\|,\|c(q(s\_0))-x\|,\|c(q(s\_1))-y\|\Big\}<\delta.
$$
As we will see, $q(s\_0)$ and $q(s\_1)$ witness that $u\_C$ is not $L$-Lipschitz on $\mathbb{R}^n\setminus (C^\epsilon\cup Z)$.
After some straightforward manipulations, we have
\begin{align\*}
\|u\_C(q(s\_0))-u\_C(x(t\_0))\|
&=\bigg\|\frac{q(s\_0)-c(q(s\_0))}{\|q(s\_0)-c(q(s\_0))\|}-\frac{x(t\_0)-x}{\|x(t\_0)-x\|}\bigg\|\\
&\leq\frac{4\delta}{\|x(t\_0)-x\|}=\frac{4\delta}{(1-t\_0)r\_C}\leq\frac{8\delta}{r\_C},
\end{align\*}
and similarly
$$
\|u\_C(q(s\_1))-u\_C(y(t\_0))\|\leq\frac{8\delta}{r\_C}.
$$
Then
$$
\|u\_C(q(s\_0))-u\_C(q(s\_1))\|
\geq\|u\_C(x(t\_0))-u\_C(y(t\_0))\|-\frac{16\delta}{r\_C}
=\frac{\|x-y\|-16\delta}{r\_C},
$$
but
$$
\|q(s\_0)-q(s\_1)\|
\leq\|x(t\_0)-y(t\_0)\|+2\delta
\leq 3\delta.
$$
Selecting $\delta<\min\{\frac{1}{32},\frac{1}{6Lr\_c}\}\cdot\|x-y\|$ then gives
$$
\|u\_C(q(s\_0))-u\_C(q(s\_1))\|>L\|q(s\_0)-q(s\_1)\|,
$$
as claimed.
|
1
|
https://mathoverflow.net/users/29873
|
412698
| 168,378 |
https://mathoverflow.net/questions/412697
|
5
|
Let $M$ be a topological manifold. We know that $M$ is stably smoothable if and only its tangent microbundle, up to stabilization, admits a reduction to vector bundle.
Now I wonder if there is a relative version of this fact. Suppose $M$ and $N$ are topological manifolds and $f: M \to N$ is a locally flat topological embedding and assume it admits a normal microbundle. We also assume that the normal microbundle is equivalent to a vector bundle. Suppose $M$ and $N$ are both stably smoothable. I wonder if one can prove that there are smoothable stablizations $M = M \times {\mathbb R}^m$ and $N' = N \times {\mathbb R}^{m + n}$ such that the natural extension $f': M' \to N'$ (extending $f$ by the inclusion ${\mathbb R}^m \to {\mathbb R}^{m+n}$) is a smooth embedding.
If the above speculation is true, I also wonder if such stable smoothings are in any sense unique.
|
https://mathoverflow.net/users/46433
|
Stable smoothing of topological manifolds relative to an embedding
|
I think the answer is yes, it follows from this: if $M$ is a triangulable manifold of dimension $n$ greater than 4, and if the total space $E$ of a vector bundle over $M$ is smoothable, then the smooth structure is concordant to one where the zero section is a smooth submanifold.
To see why it follows, stabilize $M$ so that it has dimension at least $5$. Then the normal bundle of the embedding satisfies the requirements of our theorem since the tubular neighborhood is assumed to be a vector bundle and is an open subset of a smoothable space. By the above proposition and the concordance extension theorem, we can find a concordant smooth structure on $N$ which agrees with the smooth structure on the tubular neighborhood of $M$ which we have manufactured to have $M$ as a submanifold.
So why is the first proposition true? Pick a triangulation of $M$. Over an n-simplex $\Delta$ we inductively apply the fact that any smooth structure on $\Delta \times \mathbb{R}^k$ which is a product structure when restricted to some set of faces $I$ is concordant relative $I \times \mathbb{R}^k$ to a product structure. This is a consequence of the product structure theorem. After this process is completed, the smooth structure on our vector bundle has the property that the restriction to every $n$-simplex in $M$ is a smooth submanifold. Hence, $M$ is a smooth submanifold. The vector bundle requirement ensures that any two trivializations over a shared face $\sigma$ induce the same smooth structure on $\sigma \times \mathbb{R}^k$, meaning the notion of a local smooth product structure on a vector bundle is well defined.
|
4
|
https://mathoverflow.net/users/134512
|
412707
| 168,381 |
https://mathoverflow.net/questions/412039
|
12
|
Let $u\in\mathrm{GL}\_n$ be a unipotent element, let $\mathcal{B}\_u$ be the variety of Borel subgroups containing $u$, and let $d=\dim \mathcal{B}\_u$. Then Springer theory tells us that $H^{2d}(\mathcal{B}\_u,\overline{\mathbb{Q}}\_{\ell})$ is an irreducible representation of $S\_n$ in a natural way, and that all irreducible representations of $S\_n$ come this way.
Now, combining with the branching rule of symmetric groups, viewing as representations of $S\_{n-1}$ one has that
\begin{equation}
H^{2d}(\mathcal{B}\_u,\overline{\mathbb{Q}}\_{\ell})\cong\bigoplus\_{u'}H^{2d'}(\mathcal{B}\_{u'},\overline{\mathbb{Q}}\_{\ell}),
\end{equation}
where $u'$ runs over some unipotent elements of $\mathrm{GL}\_{n-1}$ whoese corresponding Young diagrams are obtained from that of $u$ by removing a corner box.
**Question.** Is there a purely geometric proof of the above displayed formula?
|
https://mathoverflow.net/users/56217
|
Branching rule of $S_n$ and Springer theory
|
This is a nice question. I have never seen this before.
Let us write
$$\mathcal B\_u = \{ V\_0 \subset V\_1 \subset \cdots \subset V\_{n-1} \subset V\_n = \mathbb C^n : u V\_i \subset V\_i \}. $$
Let $ \lambda $ be the Jordan type of $ u $. Then we can partition $ B\_u $ into locally closed subsets depending on the Jordan type of $ u \rvert\_{V\_{n-1}} $; such a Jordan type $ \mu$ must necessarily be obtained by removing one box from $ \lambda$. So we have
$$ \mathcal B\_u = \bigsqcup\_\mu B\_u^\mu \quad B\_u^\mu = \{ V\_\bullet : u\rvert\_{V\_{n-1}} \text{ has Jordan type $ \mu$ } \}.$$
Let $ G(n-1, n)\_u $ denote the set of all $n-1$-dimensional $u$-invariant subspaces of $ \mathbb C^n $ and partition $ G(n-1,n)\_u $ into locally closed subsets $ G(n-1,n)\_u^\mu $ according to the Jordan type of the restriction of $ u $ to the subspace.
We have $ \mathcal B\_u \rightarrow G(n-1,n)\_u $ and $ \mathcal B^\mu\_u $ is the preimage of $ G(n-1,n)^\mu\_u$. Note that the fibre of $ \mathcal B\_u \rightarrow G(n-1,n)\_u $ is $ \mathcal B\_{u\rvert\_{ V\_{n-1}}}$.
Now, here are two facts that I don't see right away, but which must be
true:
1. Each piece $ \mathcal B\_u^\mu $ has the same dimension.
2. $ G(n-1,n)^\mu\_u $ is irreducible and simply-connected (or at least that the bundle $\mathcal B^\mu\_u \rightarrow G(n-1,n)\mu\_u$ is topologically trivial on components).
**Edit:** I think these facts should be contained in the classic paper by Spaltenstein <https://www.sciencedirect.com/science/article/pii/S138572587680008X>
Assuming these facts, we get
$$H(\mathcal B\_u) = \bigoplus\_\mu H(\mathcal B\_u^\mu) = \bigoplus\_\mu H(\mathcal B\_{u(\mu)})$$
where again the direct sum ranges over all partitions made by deleting one box from $\lambda$ and where $ u(\mu)$ denotes a unipotent element of Jordan type $ \mu$, and $ H(X) $ denotes top (co)homology. This gives the desired decomposition.
|
5
|
https://mathoverflow.net/users/438
|
412709
| 168,382 |
https://mathoverflow.net/questions/412733
|
2
|
Given a graph $G$ and $k\le \delta(G)$, the $k$-out model, $\mathcal{G}\_{k-out}(G)$ is a distribution on subgraphs $H$ of $G$ determined by the following: for $v\in G$, we choose a uniformly random subset $S\_v\subset N(v)$ of size $k$, we conclude by taking $H$ to have the edges $\{(v,u):v \in G, u \in S\_v\}$.
Bohman and Frieze showed that whp, $\mathcal{G}\_{3-out}(K\_n)$ contains a hamiltonian cycle. My intuition believes that the situation should change if we instead look for the square of a hamiltonian cycle (for an $n$-vertex graph $G$, we say it contains the square of a hamiltonian cycle if there are distinct $v\_1,...,v\_n \in G$ such that $(v\_i,v\_{i+1}),(v\_i,v\_{i+2})\in E(G)$ for all $i$ (here addition is done modulo $n$)).
More precisely, I believe that for any finite $k$, we have that asymptotically almost surely $\mathcal{G}\_{k-out}(K\_n)$ does not contain the square of a hamiltonian cycle. Is this correct?
|
https://mathoverflow.net/users/130484
|
$k$-out model containing the square of a hamiltonian cycle
|
Your statement is correct, because there are too few triangles in $\mathcal G\_{k-out}(K\_n)$.
For three vertices $a,b,c \in \mathcal G\_{k-out}(K\_n)$ to form a triangle, the edges $ab$, $bc$ and $ca$ should all be present, and the probability is $O(n^{-3})$ after $k$ is fixed. Thus the expected number of triangles is $O(1)$ for fixed $k$. Because a square of hamiltonian cycle contains $n$ triangles (for $n\geq7$), the probability for a square of hamiltonian cycle to appear is $O(1/n)$ by the Markov inequality.
|
3
|
https://mathoverflow.net/users/125498
|
412735
| 168,392 |
https://mathoverflow.net/questions/412684
|
1
|
I'm looking for a solution to the following integral. However, it seems it doesn't have a solution.
$$\int\limits\_{\theta-1}^{x} \left(\frac{y+1-\theta}{\theta} \right)^{-\frac{\displaystyle1}{\displaystyle \epsilon+3}} y^{-\frac{\displaystyle\epsilon+4}{\displaystyle\epsilon+3}}dy,$$
where $\theta \ge 1$, $\epsilon \ge 0$, $x > \theta -1$ and it depends on a number of other parameters.
This equation appears in the context of Physical Layer Security, which is an area of study in digital communications (telecom).
|
https://mathoverflow.net/users/103291
|
Is there a solution to $\int_{\theta-1}^{x} \left(\frac{y+1-\theta}{\theta} \right)^{-\frac{1}{\epsilon+3}} y^{-\frac{\epsilon+4}{\epsilon+3}}dy$?
|
With some effort (the lower integration limit requires care) I found this answer for the definite integral:
$$I(x)=\int\limits\_{\delta}^{x} \left(y-\delta\right)^{-\frac{1}{ \epsilon+3}} y^{-\frac{\epsilon+4}{\epsilon+3}}dy=\frac{ (-2/\delta)^{\frac{2}{{\varepsilon}+3}} \pi ^{3/2} }{\sin \left(\pi\frac{{\varepsilon}+2}{{\varepsilon}+3}\right)\Gamma \left(\frac{{\varepsilon}+1}{2 {\varepsilon}+6}\right) \Gamma \left(\frac{\varepsilon+4}{{\varepsilon}+3}\right)}$$
$$\qquad\qquad+\;\delta^{-1}({\varepsilon}+3) x^{-\frac{1}{{\varepsilon}+3}} (x-{\delta})^{\frac{{\varepsilon}+2}{{\varepsilon}+3}} \, \_2F\_1\left(1,\frac{{\varepsilon}+1}{{\varepsilon}+3};\frac{{\varepsilon}+2}{{\varepsilon}+3};\frac{x}{{\delta}}\right),$$
for $\delta\equiv \theta-1>0$ and $x>\delta$, $\varepsilon>0$. Each of the two terms on the right-hand-side is complex, but the imaginary parts cancel.
---
Here is a Mathematica command to test this result against a numerical evaluation of the integral:
>
> Plot[{((-4)^(1/(3 + eps)) delta^(-(2/(3 + eps))) [Pi]^(3/2)
> Csc[((2 + eps) [Pi])/(3 + eps)])/( Gamma[(1 + eps)/(6 + 2 eps)] Gamma[
> 1 + 1/(3 + eps)]) + ((3 + eps) x^(-(1/(3 + eps))) (-delta + x)^((
> 2 + eps)/(3 + eps))
> Hypergeometric2F1[1, (1 + eps)/(3 + eps), (2 + eps)/(3 + eps), x/
> delta])/delta,
>
> NIntegrate[ y^(-(eps + 4)/(eps + 3))\*(y - delta)^(-1/(eps + 3)), {y, delta,
> x}]}, {x, delta, 2}]
>
>
>
with this output for $\varepsilon=0.2,\delta=0.3$ (two indistinguishable curves)
|
5
|
https://mathoverflow.net/users/11260
|
412737
| 168,393 |
https://mathoverflow.net/questions/412722
|
3
|
I am looking for some metric for distribution with support on the interval $[0,1-\epsilon]$, that will be based on the ratio of their moments.
That is, if $X\sim f(x)$, $Y\sim g(y)$, I'm looking for a metric $d(f,g)$ such that
$\frac{\lvert\mathbb{E}X^k-\mathbb{E}Y^k\rvert}{\mathbb{E}X^k}$ is small for all $0<k$ $\iff$ $d(f,g)$ is small.
Of course, I could just define the distance to be the sum of these ratios, but I am not sure what this means. Can two distributions be very different but have a very close moment-ratio?
I was able to show that the 1-Wasserstein distance is small if $\lvert\mathbb{E}X^k-\mathbb{E}Y^k\rvert$ is small, and vice-versa, but this is not strong enough, I want the ratio of the moments.
Is there some natural metric to look at?
|
https://mathoverflow.net/users/103133
|
Comparing two distributions based of the ratio of their moments
|
$\newcommand\ep\epsilon$There is no such metric, because for any $\ep\in(0,1)$ and any real $k>0$ there are random variables $X$ and $Y$ with different pdf's $f$ and $g$ supported on $[0,1-\ep]$ such that $EX^k=EY^k$.
However, the [pseudometric](https://en.wikipedia.org/wiki/Pseudometric_space) $d\_k$ defined by the formula
$$d\_k(f,g):=|\ln EX^k-\ln EY^k|=\Big|\ln\frac{EY^k}{EX^k}\Big|$$
will have the desired property:
$$d\_k(f,g)\to0\iff\frac{EY^k}{EX^k}\to1\iff\frac{|EY^k-EX^k|}{EX^k}\to0. \tag{1}$$
---
The OP has now said that all moments, for all $k>0$, were meant to be close.
Letting then
$$d(f,g):=\sum\_{k=1}^\infty 2^{-k}\frac{d\_k(f,g)}{1+d\_k(f,g)}, \tag{2}$$
we get a metric $d$ such that
$$d(f,g)\to0\iff\frac{|EY^k-EX^k|}{EX^k}\to0\text{ for each natural }k. \tag{3}$$
Indeed, if $d(f,g)\to0$ then, by (2), for each natural $k$ we have $\dfrac{d\_k(f,g)}{1+d\_k(f,g)}\to0$ and hence $d\_k(f,g)\to0$. Vice versa, if for each natural $k$ we have $d\_k(f,g)\to0$, then $\dfrac{d\_k(f,g)}{1+d\_k(f,g)}\to0$ and hence, by (2) and [dominated convergence](http://dominated%20convergence,), $d(f,g)\to0$. So, $d(f,g)\to0$ if and only if for each natural $k$ we have $d\_k(f,g)\to0$. Thus, (3) follows from (1).
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1
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https://mathoverflow.net/users/36721
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412740
| 168,394 |
https://mathoverflow.net/questions/412743
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-1
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For sets $A,B$ we write $A\approx B$ if there is a bijection between $S$ sand $B$.
If $\kappa$ is a cardinal, let $\kappa^{<\kappa}$ denote the collection of subsets of $\kappa$ having cardinality $<\kappa$.
If $\kappa,\lambda$ are cardinals, does $\kappa^{<\kappa} \approx \lambda^{<\lambda}$ imply $\kappa=\lambda$?
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https://mathoverflow.net/users/8628
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Does $|\kappa^{<\kappa}|=|\lambda^{<\lambda}|$ imply $\kappa=\lambda$?
|
No.
Consider when $2^{\aleph\_0}=2^{\aleph\_1}=2^{\aleph\_2}=\aleph\_3$, with $\kappa=\aleph\_1$ and $\lambda=\aleph\_2$.
---
If you allow for one of these to be singular, then consider $\kappa=\beth\_\omega$ and $\lambda=\kappa^+$. Then $\lambda^{<\lambda}=\lambda^\kappa=2^\kappa\cdot\lambda$, and on the other hand since $\kappa$ is a strong limit cardinal, $\kappa^{<\kappa}=2^\kappa$ as well.
If, on the other hand, you require that both are regular, then there is no provable counterexample, since $\sf GCH$ implies that $\kappa^{<\kappa}=\kappa$ for all regular cardinals.
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7
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https://mathoverflow.net/users/7206
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412744
| 168,395 |
https://mathoverflow.net/questions/412742
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1
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This is followup to the following question: [On the Lipschitz continuity of the unit-normal vector field of a polytope](https://mathoverflow.net/q/412663/78539)
---
Let $C$ be a (nonempty) closed convex subset of $\mathbb R^n$. Note that to every $x \in \mathbb R^n$ corresponds a point $c(x) \in C$ which is closest to $x$. Let $C' := \mathbb R^n \setminus C$ and define $u:C' \to \mathbb R^n$ by $u(x) := (x-c(x))/\|x-c(x)\|$.
>
> **Question.** *What are generic conditions on $C$ under which $u\_C$ is Lipschitz-continuous on $C'$ ?*
>
>
>
Are there any such conditions linked with a "classical" notion of smoothness or curvature of (the boundary of) a closed convex set ?
Examples
--------
* If $C := \{0\}$, then $u$ is the function $x \mapsto x/\|x\|$, which is definitely not Lipschitz on its domain $C'=\mathbb R^n\setminus\{0\}$.
* If $C = B\_n$, the unit-ball in $\mathbb R^n$, then $u$ is the map $x \mapsto x/\|x\|$, which is $1$-Lipschitz on $C'$.
* If $A$ is a positive-definite matrix and $C = \{x \in \mathbb R^n \mid x^\top A x \le 1\}$, then $u$ is the map $x \mapsto Ax/\|Ax\|$, which is $\mbox{cond}(A)$-Lipschitz on $C'$, where $\mbox{cond}(A) := \|A\|\_{op}\|A^{-1}\|\_{op}$ is the condition number of $A$. The previous example is a particular case with $\mbox{cond}(A) = 1$.
|
https://mathoverflow.net/users/78539
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Sufficient conditions for the boundary unit-normal vector field of a closed convex set to be Lipschitz continuous
|
(1) A necessary condition: For every $x\in\partial C$, it holds that the [polar cone](https://en.wikipedia.org/wiki/Dual_cone_and_polar_cone) of $C-x$ is one-dimensional.
(The polar cone is always at least one-dimensional by part 4 of Prop 4 of [these lecture notes](https://people.orie.cornell.edu/dsd95/teaching/orie6300/lec05.pdf).)
Suppose there exists $x\in\partial C$ such that the polar cone of $C-x$ has dimension at least $2$, and select distinct unit vectors $v$ and $w$ from this polar cone. Then $u(x+\epsilon v)=v$ and $u(x+\epsilon w)=w$ for every $\epsilon>0$. (This is part 2 of the same Prop 4.) Sending $\epsilon\to0$ gives arbitrarily close inputs whose outputs are a fixed distance apart, thereby breaking Lipschitz.
(2) A sufficient condition: $C$ is a compact sublevel set $\{x\in\mathbb{R}^n:f(x)\leq y\_0\}$ of some twice continuously differentiable convex function $f\colon\mathbb{R}^n\to\mathbb{R}$ and $y\_0>\min(f)$.
Indeed, since $y\_0>\min(f)$, it holds that $\nabla f(x)\neq0$ for every $x\in\partial C$, and by compactness, there exists $\delta>0$ such that $\|\nabla f(x)\|>\delta$ for all $x\in\partial C$. Since $x\mapsto \nabla f(x)$ is Lipschitz on $\partial C$ (again, by compactness), it follows that the mapping $n\colon\partial C\to S^{n-1}$ defined by $n(x):=\nabla f(x)/\|\nabla f(x)\|$ is also Lipschitz. Finally, the nearest point mapping $c\colon(\mathbb{R}^n\setminus C)\to\partial C$ is $1$-Lipschitz, and so the composition $u=n\circ c$ is Lipschitz.
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1
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https://mathoverflow.net/users/29873
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412746
| 168,396 |
https://mathoverflow.net/questions/412729
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1
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Perhaps stupid question.
>
> **Question**: Can "almost supermartingale" theorem be equally applicable to prove the convergence of some algorithms solving non-random optimization problems?
>
>
>
Attempt for a non-random version of "almost supermartingale" theorem (without any proof):
>
> Let the non-negative sequences be $\{V^k\}$, $\{S^k\}$, and $\{ U^k \}$ for $k=0,1,2,\ldots$. Let $\beta\_0, \beta\_1$ be non-negative scalars satisfying $\sum\_{k=0}^\infty \beta\_k < \infty$. Assume
> $$V^{k+1} \leq \left( 1 + \beta\_k \right) V^k - S^k + U^k$$
> and
> $$\sum\_{i={\color{red}{0 \text{ or } 1?}}}^\infty U^i < \infty.$$
> Then,
>
>
> 1. $V^k \rightarrow V^\infty$
> 2. $\sum\_{k=0}^\infty S^k < \infty$.
>
>
>
---
EDIT:
Additionally, let
$V^k := \| x^k - x^\star \|\_P^2$ where $P$ is a positive semidefinite matrix. Then, using the above theorem, can one say for a subsequence $n\_k$ $\lim\_{k \rightarrow \infty} x^{n\_k} = x^\star$? or do we need more information to prove this?
---
For the completeness, I have also copied the "almost supermartingale" theorem from [[Theorem 30, page 300]](https://large-scale-book.mathopt.com/LSCOMO.pdf).
>
> **Theorem**: Let $V^k$, $S^k$, and $ U^k$ be $\mathcal{F}\_k$-measurable random variables satisfying $V^k \geq 0$, $S^k \geq 0$, and $ U^k\geq 0$ for $k=0,1,2,\ldots$. Let $\beta\_0, \beta\_1$ be non-negative scalars satisfying $\sum\_{k=0}^\infty \beta\_k < \infty$.
> Assume
> $$\mathbb{E}\left[V^{k+1} \mid \mathcal{F}\_k \right] \leq \left( 1 + \beta\_k \right) V^k - S^k + U^k$$
> and
> $$\sum\_{i={\color{red}{0 \text{ or } 1?}}}^\infty U^i < \infty$$
> almost surely. Then,
>
>
> 1. $V^k \rightarrow V^\infty$
> 2. $\sum\_{k=0}^\infty S^k < \infty$.
>
>
>
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https://mathoverflow.net/users/123235
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Can we invoke "almost supermartingale" Theorem for deterministic sequences?
|
For $k=0,1,\dots$, let $v\_k:=V^k$, $s\_k:=S^k$, $u\_k:=U^k$, and $b\_k:=\beta\_k$, so that the $v\_k$'s, $s\_k$'s, $u\_k$'s, and $b\_k$'s are nonnegative real numbers such that $\sum\_{k=0}^\infty b\_k<\infty$,
\begin{equation\*}
\sum\_{k=0}^\infty u\_k<\infty, \tag{1}
\end{equation\*}
and
\begin{equation\*}
v\_{k+1}\le c\_k v\_k-s\_k+u\_k \tag{2}
\end{equation\*}
for all $k$, where
\begin{equation\*}
c\_k:=1+b\_k\ge1,
\end{equation\*}
so that
\begin{equation\*}
\prod\_{k=0}^\infty c\_k \text{ converges (to a number in $[1,\infty)$).} \tag{3}
\end{equation\*}
It follows by (2) that $v\_{k+1}\le c\_k v\_k+u\_k$ for all $k$ and hence, by induction on $k$,
\begin{equation\*}
v\_k\le\Big(v\_0+\sum\_{j=0}^{k-1}u\_j\Big)\prod\_{j=0}^{k-1}c\_j \tag{4}
\end{equation\*}
and hence
\begin{equation\*}
0\le v\_k\le M:=\Big(v\_0+\sum\_{j=0}^\infty u\_j\Big)\prod\_{j=0}^\infty c\_j<\infty. \tag{5}
\end{equation\*}
Similarly to (4),
\begin{equation\*}
v\_k\le\Big(v\_n+\sum\_{j=n}^{k-1}u\_j\Big)\prod\_{j=n}^{k-1}c\_j \tag{6}
\end{equation\*}
for any natural $k$ and $n$ such that $k>n$. Therefore,
\begin{equation\*}
\limsup\_{k\to\infty}v\_k\le\Big(v\_n+\sum\_{j=n}^\infty u\_j\Big)\prod\_{j=n}^\infty c\_j
\end{equation\*}
and hence, in view of (1) and (3),
\begin{equation\*}
\limsup\_{k\to\infty}v\_k\le\big(\liminf\_{n\to\infty}v\_n+0\big)\times1
= \liminf\_{n\to\infty}v\_n.
\end{equation\*}
So, in view of (5), there exists
\begin{equation\*}
v\_\infty:=\lim\_{k\to\infty}v\_k\in[0,M|\subset[0,\infty). \tag{7}
\end{equation\*}
Similarly to (4), for all $k$
\begin{equation\*}
v\_k\le\Big(v\_0+\sum\_{j=0}^{k-1}u\_j\Big)\prod\_{j=0}^{k-1}c\_j - \sum\_{j=0}^{k-1}s\_j,
\end{equation\*}
whence
\begin{equation\*}
\sum\_{j=0}^{k-1}s\_j\le\Big(v\_0+\sum\_{j=0}^{k-1}u\_j\Big)\prod\_{j=0}^{k-1}c\_j.
\end{equation\*}
Letting now $k\to\infty$ and recalling (1) and (3), we see that
\begin{equation\*}
\sum\_{j=0}^\infty s\_j<\infty. \tag{8}
\end{equation\*}
Relations (7) and (8) are what was desired.
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1
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https://mathoverflow.net/users/36721
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412747
| 168,397 |
https://mathoverflow.net/questions/412409
|
5
|
$\newcommand{\C}{\mathbb C}$A question [asked recently](https://mathoverflow.net/questions/412398/how-to-construct-non-abelian-functions) was as follows:
>
> For the symmetric group $G:=S\_3$, is it possible to construct functions $t\_g\colon\C\to\C$ that satisfy the convolution identity
> \begin{equation}
> t\_g(x+y)=\sum\_{h\in G} t\_{gh^{-1}}(x)t\_h(y) \tag{\*}\label{\*}
> \end{equation}
> for all $g\in G$ and all complex $x$, $y$?
>
>
>
This question was then deleted by the OP.
Of course, identity \eqref{\*} trivially holds if $t\_g(x)=0$ for all $g$ and $x$.
Otherwise, I think the question may be of interest to other users and therefore will provide an answer to it on this page.
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https://mathoverflow.net/users/36721
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A functional equation for a family of functions indexed by the symmetric group $S_3$
|
I think there is another "degenerate" way to deal with the question as posed, for a general finite group $G$ (later note: which may be developed into more interesting solutions, as we have done in later edits). The key is the connection with idempotents (not necessarily central) of the group algebra $\mathbb{C}G.$
When $G$ is Abelian, the group algebra $\mathbb{C}G$ is clearly commutative, so all idempotents are central. Hence the case of non-central idempotents only arises when $G$ is non-Abelian).
Notice that if such functions $t\_{g}(x)$ exist for $ g \in G$ and $x \in \mathbb{C}$, then we must have $t\_{g}(0) = t\_{g}(0+0) = \sum\_{h \in G} t\_{gh^{-1}}(0)t\_{h}(0)$.
This means that the element $T$ of $\mathbb{C}G$, defined by $T = \sum\_{g \in G} t\_{g}(0)g$, satisfies $T^{2} = T$, so is either idempotent or zero.
Going in the other direction (ignoring the trivial case $t\_{g}(x) = 0$ for all $g,x$), suppose that we are given an idempotent $E = \sum \_{g \in G} \lambda\_{g}g$
of $\mathbb{C}G.$ Then we may define $t\_{g}(x) = e^{x}t\_{g}(0) = e^{x}\lambda\_{g}$ for all $g \in G$ and all $x \in \mathbb{C}$.
Then for any $g \in G$ and $x,y \in \mathbb{C}$, we have
\begin{eqnarray\*}
\sum\_{h \in G} t\_{gh^{-1}}(x)t\_{h}(y) & = & e^{x+y} \sum\_{h \in G} \lambda\_{gh^{-1}} \lambda\_{h} = e^{x+y} \lambda\_{g} \\
& = & e^{x+y}t\_{g}(0) = t\_{g}(x+y),
\end{eqnarray\*}
using the fact that $E^{2} = E.$
One obvious case is when $E = \frac{1}{|G|} \left( \sum\_{g \in G} g \right)$, which gives rise to the case that $t\_{g}(x) = \frac{e^{x}}{|G|}$ for all $g \in G, x \in \mathbb{C}$ .
Another easy case is when $E = 1\_{G}$, which gives rise to the case $t\_{g}(x) = e^{x}$ if $g = 1\_{G}, 0$ otherwise (for $g \in G, x \in \mathbb{C}$).
A more interesting example when $G = S\_{3}$ and $E$ is the non-central idempotent
$ \frac{1+(12)}{2} - \frac{1}{6} \left(\sum\_{ \sigma \in S\_{3}} \sigma \right)$,
in which case ( for any $x \in \mathbb{C}$), we have $e^{-x}t\_{g}(x) = \frac{1}{3}$ for $g = 1\_{G}$ or $g = (12)$,
and $e^{-x}t\_{g}(x) = \frac{-1}{6}$ otherwise.
This method gives existence of some solutions to the convolution equations, but there are other types of solution, as you already showed. In the solutions above we can also replace $e^{x}$ by $e^{\alpha x}$ for a fixed constant $\alpha$ (independent of $x$).
Later edit: Let us try to develop these arguments further and obtain more information about general solutions to the convolution equations.
Suppose that we solutions $\{t\_{g}(x) : g \in $G$ \}$ for the convolution equations. For each $x \in \mathbb{C}$, let $E(x) = \sum\_{g \in G} t\_{g}(x) g$ in the group algebra $\mathbb{C}G$. We have seen that $E(0)$ is an idempotent of $\mathbb{C}G$.
The convolution equations imply that $E(x) = E(0 + x) = E(0)E(x)$ and $E(x) = E(x+0) = E(x)E(0)$ for each $x \in \mathbb{C}$. Hence $E(x)$ always lies in the subalgebra $E(0)\mathbb{C}G E(0)$ of the group algebra $\mathbb{C}G$, and this subalgebra is isomorphic to the endomorphism algebra of the right $\mathbb{C}G$-module $E(0)\mathbb{C}G$. Also, $E(0)$ is the identity element of this algebra. Also, the convolution equations imply that
$E(x)E(y) = E(x+y) = E(y+x) = E(y)E(x)$ for all $x,y \in \mathbb{C}$.
We also remark that $E(x)$ is a unit in the algebra $E(0)\mathbb{C}G E(0)$, and that its inverse is $E(-x)$.
Let us consider the case that $E(0)$ is a primitive idempotent of $\mathbb{C}G$. In that case, Schur's Lemma tells us that $E(0)\mathbb{C}GE(0)$ consists of scalar multiples of $E(0)$.
In that case, for each $x \in \mathbb{C}$, we have $E(x) = \mu\_{x}E(0)$ for some complex number $\mu\_{x}$, and by the remarks above, we have $\mu\_{x}\mu\_{y} = \mu\_{x+y}$ for all $x,y \in \mathbb{C}$. This forces $\mu\_{x} = e^{\alpha x}$ for each $x \in \mathbb{C}$, for some constant $\alpha.$
Hence the only solutions for which $E(0)$ is a primitive idempotent of $\mathbb{C}G$ are the "degenerate" ones described earlier.
This indicates how to deal with the case that $E(0)$ is a general idempotent of $\mathbb{C}G$. We may decompose $E(0)$ as a sum of mutually orthogonal primitive idempotents of $\mathbb{C}G$, say
$E(0) = \sum\_{j= 1}^{t} E\_{j}(0)$, where $E\_{i}(0) E\_{j}(0) = \delta\_{ij} E\_{i}(0)$ for each $i,j.$
In the case that no two of the $E\_{i}(0)$ are conjugate via a unit of $\mathbb{C}G$, the situation is reasonably clear. For then we see that
$E(0) \mathbb{C}G E(0)$ is a commutative algebra which is the direct sum of the $1$-dimensional algebras $E\_{j}(0)\mathbb{C}G E\_{j}(0)$. Then we may see that there are complex constants $a\_{j} : 1 \leq j \leq t$ such that
$ E(x) = \sum\_{j=1}^{t} e^{a\_{j}x} E\_{j}(0).$
Even later edit: In comments, it is asked what is necessary if we insist the determinant ${\rm det}[t\_{gh^{-1}}(x)]$ is equal to $1$ for all $x$. We illustrate with one example. When $G = \langle u \rangle$ is cyclic of order $2$, there are just two mutually orthogonal idempotents in the group algebra $\mathbb{C}G$. These are $ E\_{1} = \frac{1\_{G} + u}{2} $ and $E\_{2} = \frac{1\_{G} - u}{2}. $
Then the above methods give a general solution to the convolution equations of the form $ \sum\_{g \in G}t\_{g}(x)g = e^{c\_{1}x} E\_{1} + e^{c\_{2}x}E\_{2}.$
This gives the determinant ${\rm det}[(t\_{gh^{-1}}(x))])$ to be $e^{c\_{1}x+c\_{2}x}$. Hence this determinant is $1$ for all $x$, if and only if $c\_{2} = -c\_{1}.$
Incidentally, to partially address one question of the OP in comments, this example for the cyclic group of order two may be ``lifted" to a solution of the convolution equations for $G = S\_{n}$ for any $n$. I don't give all details, but we may choose constants $c\_{1}$ and $c\_{2}$ and set
$t\_{g}(x) = \frac{1}{n!}\left( e^{c\_{1}x} + e^{c\_{2}x} \right)$ whenever $g$ is an even permutation and $t\_{g}(x) = \frac{1}{n!} \left( e^{c\_{1}x} - e^{c\_{2}x} \right)$ if $g$ is an odd permutation. However, we then find that the matrix $[t\_{gh^{-1}}(x)]$ is singular for all $x$ when $n > 2.$
Returning to the general solutions, in the case that two or more of the $E\_{j}(0)$ are conjugate via a unit of the group algebra $\mathbb{C}G$, the analysis is more subtle.
Latest edit (also addressing some issues in comments): Let us recall that for any finite group $G$, the group algebra $\mathbb{C}G$ is isomorphic to a direct sum of (mutually annihilating) full matrix algebras $ \bigoplus\_{\chi} M\_{\chi(1)}(\mathbb{C})$, where $\chi$ runs over the complex irreducible characters of $G$. This gives a decomposition of $1\_{G}$ as a sum of mutually orthogonal primitive idempotents
that is $1\_{G} = \sum\_{\chi} \sum\_{j = 1}^{\chi(1)} E\_{j}(\chi)$. Here, $E\_{j}(\chi)$ is represented by an idempotent matrix of trace $1$ in any irreducible representation of $G$ which affords $\chi$, and is represented by $0$ in any irreducible representation of $G$ affording an irreducible character $\mu \neq \chi.$ Also, for each $\chi$, we have $\sum\_{j=1}^{\chi(1)} E\_{j}(\chi)$ is a central idempotent of $\mathbb{C}G$, and acts as the identity in any irreducible representation of $G$ affording character $\chi$.
This decomposition of $1\_{G}$ into a sum of mutually orthogonal primitive idempotents is not quite unique in general ( but it is when $G$ is Abelian). In general, it is unique up to conjugation by a unit of $\mathbb{C}G$ (and reordering the idempotents).
For non-commutative $G$, we now outline how to construct some solutions to
the convolution equations which have ${\rm det}([t\_{gh^{-1}}(x)]) = 1$ for all $x$, but the list is not exhaustive in general.
We take a set of pairwise distinct (this is not really necessary, but is useful for purposes of exposition) non-zero complex constants $\{ c\_{j}(\chi) : \chi \in {\rm Irr}(G), 1 \leq j \leq \chi(1) \}$ with the property that
$\sum\_{\chi,j} \chi(1) c\_{j}(\chi) = 0.$ We further insist that $\{ e^{c\_{j}(\chi)} : \chi \in {\rm Irr}(G), 1 \leq j \leq \chi(1) \}$ still consists of $\sum\_{\chi} \chi(1)$ distinct elements. This is always possible to achieve for non-trivial $G$. For example, we may take all but one of the $c\_{j}(\chi)$ to be real, positive and pairwise distinct ( with distinct exponentials, all real and greater than one). The zero sum condition then forces the choice of the remaining $c\_{j}(\chi)$, and it clearly must be negative with negative exponential.
It is convenient to introduce exponentials of elements of the group algebra $\mathbb{C}G.$ To do this, we take an element $A$ of $\mathbb{C}G$, and take its component $A\_{\chi}$ in the matrix algebra summand $M\_{\chi}(\mathbb{C})$. Then we define $\exp(A)$ to be the unique element of $\mathbb{C}G$ which has component $\exp(A\_{\chi})$ ( the usual matrix exponential) in $M\_{\chi}(\mathbb{C}).$ As usual, if $A$ and $B$ are commuting elements of $\mathbb{C}G$, then we have $\exp(A+B) = \exp(A)\exp(B).$
Now for any complex number $x$, we define the element $E(x)$ to be
$\exp{\left(\sum\_{\chi} \sum\_{j = 1}^{\chi(1)}c\_{j}(\chi)x E\_{j}(\chi) \right)}.$
Then we see that whenever $\sigma$ is an irreducible representation of $G$ affording character $\mu$, $\sigma(E(x))$ is a diagonalizable matrix with
distinct eigenvalues $e^{c\_{j}(\chi)x} : 1 \leq j \leq \chi(1)$ (strictly speaking, we need to avoid the case that $x$ has the form $\frac{2\pi i}{c\_{r}-c\_{s} }$ for any $r \neq s$).
It follows (with the noted exceptions) that in the regular representation of $G$, $E(x)$ has the eigenvalue $e^{c\_{j}(\chi)x}$ with multiplicity $\chi(1)$ for $1 \leq j \leq \chi(1)$, for each irreducible $\chi$.
Since we assumed that $\sum\_{\chi} \chi(1) \left( \sum\_{j=1}^{\chi(1)} c\_{j}(\chi) \right) = 0$, we see that the determinant of the image of
$E(x)$ in the regular representation is $\prod\_{\chi}\left( \prod\_{j = 1}^{\chi(1)} e^{c\_{j}(\chi)x} \right)^{\chi(1)} = 1$ for all $x$.
But writing $E(x) = \sum\_{g \in G} t\_{g}(x) g$ in the group algebra $\mathbb{C}G$, we see that $E(x)$ is represented by the matrix $[t\_{gh^{-1}}(x)]$ in the regular representation.
Before treating the explicit case $G = S\_{3}$, we remark that in general it is not straightforward to determine explicit formulae for the primitive idempotents of the group algebra $\mathbb{C}G$ when $G$ is a non-Abelian group. There is an explicit general procedure due to A. Young when $G$ is the symmetric group $S\_{n}.$ However, for the case $G = S\_{3}$, we may proceed directly.
We know from the theory described above that there should be four mutually orthogonal primitive idempotents of the group algebra $\mathbb{C}G$ when $G$ is the symmetric group $S\_{3}.$ Two of these are easy, but we also need to decompose the central idempotent corresponding to the degree $2$ irreducible character. This idempotent is $ X = \frac{3(1)- [(1) +(123) +(132)]}{3}.$
We have written the idempotent in this fashion to make it clear that it commutes with the idempotent $Y = \frac{(1) + (12)}{2}.$ Then we see easily that $XY = YX = \frac{(1) + (12)}{2} - \frac{ \sum\_{g \in G} g}{6}.$
This is a primitive idempotent of $\mathbb{C}G$, and may serve as our idempotent $E\_{1}(\chi)$, where $\chi$ is the degree $2$ irreducible character. The primitive idempotent $E\_{2}(\chi)$ is then easily seen to be
$\frac{(1)-(12)}{2} - \frac{ \sum\_{g \in G} {\rm sign}(g) g}{6}.$
Hence we may set $E\_{1}( \lambda\_{0}) = \frac{\sum\_{g \in G} g }{6}$,
$E\_{1}( \lambda\_{1}) = \frac{\sum\_{g \in G} {\rm sign}(g)g }{6}$,
where $\lambda\_{0}$ is the trivial character, $\lambda\_{1}$ is the sign character.
Then we may choose any four non-zero complex constants $c\_{0},c\_{1},c\_{2},c\_{3}$ with $c\_{0} + c\_{1} + 2(c\_{2}+c\_{3}) = 0$
and set $E(x) = e^{c\_{0}x} E\_{1}(\lambda\_{0}) + e^{c\_{1}x}E\_{1}(\lambda\_{1}) + e^{c\_{2}x}E\_{1}(\chi) + e^{c\_{3}x}E\_{2}(\chi)$.
This leads to $t\_{1}(x) = \frac{e^{c\_{0}x}}{6} + \frac{e^{c\_{1}x}}{6} +
\frac{e^{c\_{2}x}}{3} + \frac{e^{c\_{3}x}}{3}$, $t\_{(12)}(x) = \frac{e^{c\_{0}x}}{6} - \frac{e^{c\_{1}x}}{6} +
\frac{e^{c\_{2}x}}{3} - \frac{e^{c\_{3}x}}{3}$, etc. , producing a solution to the convolution equations with ${\rm det}([t\_{gh^{-1}}(x)]) = 1$ for all $x$.
|
4
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https://mathoverflow.net/users/14450
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412750
| 168,398 |
https://mathoverflow.net/questions/412736
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6
|
Let $\mathfrak{g}$ be a (finite dimensional) semi-simple Lie algebra over a field $k$ and let $x \in \mathfrak{g}$. By definition, we have the equivalence:
$$ \mathrm{rk}(\mathrm{ad}\_x) = 0 \iff x = 0,$$
where $\mathrm{rk}(\mathrm{ad}\_x)$ is the rank of $\mathrm{ad}\_x$ seen as an element of $\mathrm{End}(\mathfrak{g})$. I would like to know if there is a classification of elements $x \in \mathfrak{g}$ such that $\mathrm{rk}(\mathrm{ad}\_x) \leq 1$? I am primarily interested in the case where $k = \mathbb{C}$ and $\mathfrak{g}$ is simple of classical type.
|
https://mathoverflow.net/users/37214
|
Rank one adjoint operators on a Lie algebra
|
Another approach.
To show it's impossible (the rank can't be 1), it is enough to show this when the field (assumed of char 0) is algebraically closed, and in turn it's enough to show the result in case $\mathfrak{g}$ is simple. If $x$ has $\mathrm{ad}(x)$ of rank 1, $x$ has centralizer of codimension 1. It is known (see e.g. [this MathSE answer](https://math.stackexchange.com/a/3004226/35400)) that $\mathfrak{g}$ has no subalgebra of codimension 1, unless $\mathfrak{g}$ is 3-dimensional. But for $\mathfrak{sl}\_2$, the operator $\mathrm{ad}\_x$ has rank 2 for every nonzero $x$ (alternatively, all 2-dimensional subalgebras have a trivial centralizer, so can't be centralizer of an element).
A similar approach can be used (with a little further work) to classify which $\mathfrak{g}$ admit an $x$ with $\mathrm{ad}(x)$ of rank 2, and classify such elements $x$.
|
6
|
https://mathoverflow.net/users/14094
|
412759
| 168,400 |
https://mathoverflow.net/questions/412758
|
4
|
Let $G$ be a finite group. Does there exist a finite group $\Gamma$ and a surjection $f\colon \Gamma \rightarrow G$ such that the center $Z(\Gamma)$ lies in the kernel of $f$?
Of course, this is only an interesting question if $Z(G) \neq 1$. It is also pretty trivial if we don't insist that $\Gamma$ is finite since we can then just take $\Gamma$ to be a free group.
I don't have a strong feeling as to whether or not such a $\Gamma$ exists.
|
https://mathoverflow.net/users/473591
|
Extending a finite group to avoid the center
|
Every finite group $G$ is quotient of a finite group with trivial center. Namely choose $p$ odd prime not dividing the order of $G$. Let $V\_p$ be the abelian group of functions $G\to\mathbf{Z}/p\mathbf{Z}$ with sum zero. The action of $G$ on $V\_p$ by translation $g\cdot f(h)=f(g^{-1}h)$ is faithful and fixes no nonzero element. So the semidirect product $G\ltimes V\_p$ has trivial center and admits $G$ as quotient.
|
12
|
https://mathoverflow.net/users/14094
|
412760
| 168,401 |
https://mathoverflow.net/questions/412751
|
2
|
Let $G\subseteq \mathbb{C}^n$ be a bounded domain. Consider the Carathéodory metric $C\_G$ on $G$. If $G=\mathbb{D}^n$ (unit polydisc), then $C\_G(a,z)=\max\_{1\leq j\leq n}p(a\_j,z\_j)$, where $p$ denotes the Poincaré metric on $\mathbb{D}$.
My question: Is there a similar formula in case $D=\mathbb{B}^2\times \mathbb{D}\subseteq \mathbb{C}^3$, where $\mathbb{B}^2$ denotes the unit ball in $\mathbb{C}^2$? More generally, is there a formula (or some estimate) for the Carathéodory metric of the product domain, in terms of the Carathéodory metric on the components?
|
https://mathoverflow.net/users/159066
|
Carathéodory metric on product domain
|
For a general formula, you should check out Kobayashi's book *Hyperbolic Complex Spaces*. In Proposition 3.1.11 of that text, he proves the following:
>
> Let $X$ and $Y$ be complex spaces. For $(x,y), (x',y') \in X\times Y$, we have
> $$
> \text{max}\{ c\_X(x,x'),c\_Y(y,y')\} \leq c\_{X\times Y}((x,y),(x',y')) \leq c\_X(x,x') + c\_Y(y,y').
> $$
>
>
>
|
2
|
https://mathoverflow.net/users/56667
|
412765
| 168,402 |
https://mathoverflow.net/questions/412767
|
3
|
Recall the Coifman–Meyer Theorem as stated in [Grafakos and Oh - The Kato-Ponce Inequality](https://arxiv.org/abs/1303.5144).
**Theorem**: Let $m \in L^{\infty}\left(\mathbf{R}^{2 n}\right)$ be smooth away from the origin and satisfy
$$
\left|\partial\_{\xi}^{\alpha} \partial\_{\eta}^{\beta} m\right|(\xi, \eta) \leq C(\alpha, \beta)(|\xi|+|\eta|)^{-|\alpha|-|\beta|}
$$
for all $\xi, \eta \in \mathbf{R} \setminus\{0\}$ and $\alpha, \beta \in \mathbf{Z}^{n}$ multi-indices with $|\alpha|,|\beta| \leq 2 n+1$. Then for all $f, g \in \mathcal{S}\left(\mathbf{R}^{n}\right)$,
$$
\left\|\int\_{\mathbf{R}^{2 n}} m(\xi, \eta) \widehat{f}(\xi) \widehat{g}(\eta) e^{i\langle\xi+\eta,\rangle} d \xi d \eta\right\|\_{L^{r}\left(\mathbf{R}^{n}\right)} \leq C(p, q, r, m)\|f\|\_{L^{p}\left(\mathbf{R}^{n}\right)}\|g\|\_{L^{q}\left(\mathbf{R}^{n}\right)}
$$
where $\frac{1}{2}<r<\infty, 1<p, q \leq \infty$ satisfy $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$.
I was wondering if it makes sense to expect the following estimate to hold,
$$
\left\|\int\_{\mathbf{R}^{2 n}} m(\xi, \eta) \widehat{f}(\xi) \widehat{g}(\eta) e^{i\langle\xi+\eta,\rangle} d \xi d \eta\right\|\_{L^{p}\left(\mathbf{R}^{n}\right)} \leq C(p, m)\|fg\|\_{L^{p}\left(\mathbf{R}^{n}\right)}
$$
for some Schwartz functions $f,g$ and exponent $p\geq 1?$
**Edit:** Thanks to Terry's remark, the estimate does not necessarily hold for any multiplier. Actually, I had the following estimate which I wanted to show,
$$\|D^\alpha(fg)\|\_p\leq C(\|gD^\alpha f\|+\|fD^\alpha g\|\_p).$$
Here $\alpha\in (0,1)$ and $D^\alpha = (-\Delta)^\alpha.$
Naturally, one would write $D^\alpha(fg)$ as a double integral involving the Fourier transforms of $f$ and $g$. I was hoping that if the modified Coifman-Meyer estimate was true then this estimate would hold after applying Littlewood decomposition.
|
https://mathoverflow.net/users/68232
|
Does the following version of the Coifman–Meyer Theorem exist?
|
No. For instance, if $f,g$ have disjoint supports, the right-hand side vanishes, but there is no reason to expect the left-hand side to vanish; and one can easily construct examples where it does not, for instance by first selecting a pair $f,g$ of bump functions with disjoint support and then localising $m$ to a small region of pairs $(\xi,\eta)$ where $\hat f(\xi)$ and $\hat g(\eta)$ are non-zero and don't vary too much.
More generally, one of the first sanity checks to see if a given estimate $X \leq CY$ is plausible (for some non-negative quantities $X,Y$ and some unspecified constant $C$) is to ask "If $Y$ vanishes, it is then obvious that $X$ must also vanish?" If the answer is "no", it is highly unlikely that the estimate is going to be true. (A "yes" answer, on the other hand, is inconclusive. A good followup question is "If we know $Y$ is very small, does this look like it could force $X$ to be small as well?". Depending on the type of estimate being considered, "bounded" or "finite" can be a more suitable adjective here than "small".)
|
11
|
https://mathoverflow.net/users/766
|
412770
| 168,404 |
https://mathoverflow.net/questions/412773
|
1
|
Let $A$ be a $C^\*$-algebra. Endow $A$ with the strict topology for which a net $\{a\_i\}\_{i \in I}$ converges to $a \in A$ if $$\|a\_i b-ab\| + \|ba\_i-ba\| \to 0$$
for all $b \in A$. Is it true that if $\{a\_i\}\_{i \in I}$ is a bounded net of positive elements that converges strictly to $a \in A$, then $a$ is also positive?
Remarks: (1) If $A$ is unital, the strict topology coincides with the norm-topology and the result is well-known then.
(2) If $A= B\_0(H)$, then the strict topology on bounded subsets of $A$ is the strong$^\*$- topology and then the statement is obviously true.
|
https://mathoverflow.net/users/216007
|
Convergent bounded net of positive operators converges to a positive operator
|
Yes, $a$ is positive, too.
*Proof.*
For every $b \in A$ we have $b^\* a\_i b \to b^\* a b$, so $b^\*a b$ is positive.
Now, let $(e\_j)$ be an approximate identity in $A$. Then it follows that $(e\_j a e\_j)$ is a net of positive elements that converges in norm to $a$; hence, $a$ is positive.
|
4
|
https://mathoverflow.net/users/102946
|
412779
| 168,409 |
https://mathoverflow.net/questions/412784
|
1
|
Sorry in advance if this is not sufficiently research-level, it is really more of a reference request since the proof is not difficult. Let $\mathcal{Y}$ be a compact set, let $\{X\_n\}$ denote a sequence of random variables, and let $f(x,y)$ and $g(y)$ be "nice" functions. Suppose that for each fixed $y\in\mathcal{Y}$, we have $$\liminf\_{n\to\infty} f(X\_n,y)\geq g(y)$$ almost surely. What is the appropriate theorem to cite that says that (for sufficiently nice functions), we have $$\liminf\_{n\to\infty} \min\_{y\in\mathcal{Y}} f(X\_n,y)\geq \min\_{y\in\mathcal{Y}} g(y)$$ almost surely?
|
https://mathoverflow.net/users/70190
|
Pointwise almost sure convergence implies global convergence
|
First here, without loss of generality (wlog) $g=0$ (otherwise, replace $f(x,y)$ and $g(y)$ by $f(x,y)-g(y)$ and $0$, respectively). Second, in view of the almost-sure (a.s.) condition and the a.s. desired conclusion, wlog each random variable $X\_n$ takes only one value, say $x\_n$ -- which let us assume for simplicity to be a real number.
So, the desired result takes this simplified form:
>
> If
>
> $$\liminf\_{n\to\infty} f(x\_n,y)\ge0 \tag{1}$$
> for some "nice" function $f$ and all $y$ in a compact set $Y:=\mathcal Y$, then
> $$\liminf\_{n\to\infty}\min\_{y\in Y} f(x\_n,y)\ge0. \tag{2}$$
>
>
>
Consider now the following example: Let $Y:=[-1,1]$ and let $f(x,y):=-xy^2\,e^{-xy^2}$ for real $x$ and $y\in Y$. Probably everyone will agree that this function $f$ is nice. Let $x\_n\to\infty$ (as $n\to\infty$). Then $f(x\_n,y)\to0$ for each $y\in Y$, so that (1) holds. However, $\min\_{y\in Y} f(x\_n,y)=-1/e$ for each large enough $n$ (namely, for each $n$ such that $x\_n\ge1$), so that (2) fails to hold.
The obvious reason for the just described phenomenon is that the sequence $(x\_n)$ is unbounded.
So, it is natural to assume that $|x\_n|\le M$ for some real $M>0$ and all $n$. It is also natural to assume that the "nice" function $f$ is continuous. So, $f$ is uniformly continuous on the compact set $[-M,M]\times Y$. Also, passing to a subsequence, wlog assume that $x\_n\to x\_\infty$ for some real $x\_\infty$.
So, $f(x\_n,y)\to f(x\_\infty,y)$ uniformly in $y\in Y$ and hence
$$\lim\_{n\to\infty}\min\_{y\in Y} f(x\_n,y)
=\min\_{y\in Y} f(x\_\infty,y)
=\min\_{y\in Y}\lim\_{n\to\infty}f(x\_n,y)\ge0,
$$
by (1). So, we have deduced, very simply, (2) from (1).
Essentially, all we used here is that any function continuous on a compact set $K$ is uniformly continuous on $K$. So, one can hardly expect a reference involving something more substantial than this.
|
3
|
https://mathoverflow.net/users/36721
|
412788
| 168,413 |
https://mathoverflow.net/questions/396004
|
1
|
I'm currently reading this [paper](https://www.ams.org/journals/proc/1996-124-02/S0002-9939-96-03154-1/S0002-9939-96-03154-1.pdf) (and working on a similar one). The main goal is to study the Hammerstein integral equation (in $\mathcal{C}(I,E))$:
$$x(t) = \int\_{0}^{t} K(t,s)f\big(s,x(s)\big)ds,\quad t\in I;$$
where $I=[0,1]$, $K $ is a scalar kernel, $E$ a Banach space and $f:I \times E \rightarrow E$ is a given function.
Let's say that in my hypothesis, I need:
>
> $H\_1$. There exist $a:I \rightarrow (0,+\infty)$ with $a(.)\in L^{\infty}(I)$ and $b:[0,+\infty) \rightarrow(0,+\infty)$ a nondecreasing function such that $\|f(s, x)\| \leq a(s) b\big(\|x\|\big)$ for a.e. $s \in I$ and $x\in E$,
>
>
> $H\_2$. there exists at least one solution $r \in \mathcal{C}\big(I,(0, \infty)\big)$ to the inequality
> $$ b\left(\|r\|\_{\infty}\right) \int\_{0}^{t} \left |K(t, s) \right |a(s) d s \leq r(t), \quad t \in I$$
> where $\|.\|\_{\infty}$ is the sup norm in $\mathcal{C}\big(I,(0, \infty)\big)$.
>
>
>
But I'm wondering if those are reasonable assumptions. I'm looking for an example of the Hammerstein integral equation with the hypothesis $(H\_1)$ and $(H\_2)$, take $\color{blue}{E=\mathbb{R}}$ for simplification.
|
https://mathoverflow.net/users/102228
|
Simple example of Hammerstein integral equation
|
It is a question what is reasonable and whether you can conclude something interesting from it: If, for instance, the integral operator with kernel $\lvert K\rvert$ acts in $L\_\infty$ and $f$ grows sublinear in its second argument (with a bound independent of its first argument), the hypothesis will be satisfied with a sufficiently large constant $r$.
It is unlikely to find a real-world problem in which the former fails but your hypotheses are satisfied (though it is of course easily possible to construct **artifical** such examples - for instance, just modify $f$ arbitrarily for very large arguments after you calculated a possible constant $r$).
As mentioned, the main question is whether you can conclude something interesting from the hypotheses: If it is just another existence result, it is unlikely that your result is not already covered by the hundreds or thousands of existence results for Hammerstein equations with very similar hypotheses which are all essentially only such small generalizations of the sublinear growth condition.
|
1
|
https://mathoverflow.net/users/165275
|
412799
| 168,418 |
https://mathoverflow.net/questions/412525
|
21
|
We have two well known definitions of the [semidirect product](https://en.wikipedia.org/wiki/Semidirect_product) $N \rtimes H$ of groups:
1. (Internal semidirect product) We write $G = N \rtimes H$ if $N$ is a normal subgroup of $G$, $H$ is another subgroup of $G$, $N \cap H = \{1\}$, and $G = NH$.
2. (External semidirect product) Given a homomorphism $\phi: H \to \mathrm{Aut}(N)$ from $H$ to the automorphism group of $N$, denoted $h \mapsto \phi\_h$, we write $N \rtimes H = N \rtimes\_\phi H$ to be the set of pairs $(n,h)$ with $n \in N, h \in H$ and group law $(n\_1,h\_1) (n\_2,h\_2) = (n\_1 \phi\_{h\_1}(n\_2), h\_1 h\_2)$.
The two are equivalent in the sense that every external semidirect product is an internal semidirect product (identifying $N$ with $\{ (n,1): n \in N\}$ and $H$ with $\{(1,h): h \in H\}$) and conversely every internal semidirect product is canonically isomorphic to an external semidirect product.
I recently had occasion to use the version of this concept when the two groups $N,H$ are allowed to intersect [EDIT: in a group normal in $H$], which then imposes two additional compatibility conditions on the homomorphism $\phi$, but I was unable to locate the standard name for the concept, which I will for the sake of this question call the "semidirect product $N \rtimes\_K H$ of $N,H$ relative to $K$". It can be defined internally or externally:
1. (Internal relative semidirect product) We write $G = N \rtimes\_K H$ if $N$ is a normal subgroup of $G$, $H$ is another subgroup of $G$, $N \cap H = K$ is a normal subgroup of $H$, and $G = NH$.
2. (External relative semidirect product). Given a commuting square
$\require{AMScd}$
\begin{CD}
K @>> \iota > N\\
@V \iota V V @VV V\\
H @>>\phi> \mathrm{Aut}(N)
\end{CD}
of homomorphisms using the conjugation action of $N$ on itself [EDIT: with $\iota$ denoting inclusion maps, and obeying the additional compatibility condition
$$ \phi\_h(k) = h k h^{-1}$$
for all $h \in H$ and $k \in K$, which among other things makes $K$ a normal subgroup of $H$] we write $N \rtimes\_{K,\phi} H = N \rtimes\_{K,\phi} H = N \rtimes\_K H$ to be the quotient of $N \rtimes\_\phi H$ by $\{ (k, k^{-1}): k \in K \}$ (which one can check to be a normal subgroup of $N \rtimes\_\phi H$).
One can verify that every external relative semidirect product is an internal relative semidirect product, and conversely that every internal relative semidirect product is isomorphic to an external relative semidirect product.
This is such a basic operation that it must already be well known, so I ask
***Question 1:*** What is the term for this operation in the literature?
Also,
***Question 2:*** Does this operation have a category-theoretic interpretation (e.g., as the universal object for some diagram)?
EDIT: One can also think of $N \rtimes\_{K,\phi} H$ as the amalgamated free product $N \ast\_K H$ quotiented by the relations $h n h^{-1} = \phi\_h(n)$ for $h \in H$, $n \in N$. So maybe "amalgamated semidirect product" could be a better name than "relative semidirect product".
|
https://mathoverflow.net/users/766
|
What is the name of this relative semidirect product of groups?
|
This construction appears in the literature as Remark 1.4.5 in "Pseudo-reductive groups" by Conrad, Gabber and Prasad.
There, it is called a 'non-commutative pushout'. It is a generalization of their 'standard construction' which produces most of the pseudo-reductive groups.
|
7
|
https://mathoverflow.net/users/44668
|
412800
| 168,419 |
https://mathoverflow.net/questions/412798
|
0
|
Let $A$ be a $C^\*$-algebra acting on a Hilbert space that admits a cyclic unit vector $\xi \in H$. Pose $S\_\xi = \{\eta \in A \xi: \| \eta \| = 1\}$, and for each $\eta \in S\_\xi$, pose $A\_\eta = \{a \in A : a \xi = \eta\}$, so $A\_\eta \ne \emptyset$ by definition; and let $c\_\eta = \inf\{\| a \| : a \in A\_\eta\}$. Clearly, $c\_\eta \ge 1$ for all $\eta \in S\_\xi$. I am interested in the following questions.
**Question 1**. Is it true that $c\_\eta = 1$ for all $\eta \in S\_\xi$ ?
**Question 1 (weak form)**. Is it true that $S(A)\xi$ is dense in $S(H)$, where $S(X)$ denotes the closed unit ball of a normed space $X$?
**Question 2**. If the answer of Question 1 turns out to be negative, can there be any examples for which $\sup\{c\_\eta : \eta \in S\_\xi\} = +\infty$ ?
**Question 3**. What if we drop the assumption of $\xi$ being cyclic but merely asks $A$ acts non-degenerately on $H$ ?
|
https://mathoverflow.net/users/128540
|
Controlling norm of operators sending a fixed vector to another
|
Set $A = C([0,1])$ and $H = L^2([0,1])$ (with respect to Lebesgue measure) with $\xi=1$ the constant function. Then $A\xi$ is the image of $C([0,1])$ in $L^2([0,1])$, which is a norm-decreasing injective but not bounded below inclusion. Then $S\_\xi$ is the intersection of the unit ball of $L^2$ with the continuous functions, and for each $\eta$ we see that $A\_\eta$ is the singleton $\{\eta\}$.
Thus, Q1 (both versions) have a negative answer, this example gives a positive answer to Q2, and Q3 can be reduced to the main question by restricting the action of $A$ to the cyclic subspace generated by $\xi$.
However, if $A$ were the compact operators on $H$ then we'd have a positive answer to Q1.
|
2
|
https://mathoverflow.net/users/406
|
412807
| 168,422 |
https://mathoverflow.net/questions/412713
|
4
|
Consider a smooth projective threefold $\overline W$, constructed in section 4 of [this paper](https://arxiv.org/pdf/0810.0957.pdf).
This threefold is a resolution of singularities of the quotient of a product of a K3 surface and $\mathbb CP^1$ by $(\rho, \psi)$, where $\rho$ is a non-symplectic, non-fixed-point-free involution on the K3 surface and $\psi$ is an involution on $\mathbb CP^1$ (the paper (sec 4) contains more details about the construction).
Then $\overline W$ is simply-connected and it has a smooth K3 surface in its anticanonical system.
Here is my question:
>
> Is $\overline W$ irrational? How can one show it?
>
>
>
|
https://mathoverflow.net/users/69559
|
Irrationality of some threefolds
|
I am just posting my comment as an answer. All such threefolds are rational.
By the hypotheses on the involution of the K3 surface, the quotient surface is a rational surface. The projection from the threefold to the rational surface has geometric generic fiber isomorphic to $\mathbb{P}^1$, the other factor in the product. Every fixed point of the involution on $\mathbb{P}^1$ gives a section of this projection.
A normal projective variety that admits a surjective morphism to a target variety is birational (over that target variety) to a product of the target variety and $\mathbb{P}^1$ if and only if the geometric generic fiber is isomorphic to $\mathbb{P}^1$ and there exists a section. Therefore, the threefold is birational to a product of the quotient rational surface and $\mathbb{P}^1$, i.e., it is rational.
|
3
|
https://mathoverflow.net/users/13265
|
412810
| 168,425 |
https://mathoverflow.net/questions/412815
|
2
|
Let $X,Y$ be complex manifolds of $\dim X=n$, $\dim Y=m>1$, $U\subset X$ open and $g\colon U\to Y$ holomorphic embedding. Then $g(U)$ is a submanifold of codimension $m-n\ge1$. It seems clear that $Y\setminus g(U)$ is connected, think for example at $\Bbb C\setminus\{0\}$. I'm searching for a reference, as it seems a well known fact.
|
https://mathoverflow.net/users/70148
|
Complement of complex submanifolds of codimension $\ge1$ is connected
|
Every (real) codimension $2^+$ embedding has path connected complement.
The proof is by using the tubular neighborhood theorem to reduce to showing that the total space of the normal bundle of $g(U)$ in $Y$ minus the zero section is connected, and noticing that the normal bundle is a rank $2^+$ vector bundle.
See [this MO answer](https://mathoverflow.net/questions/71895/on-a-special-case-of-alexander-duality).
|
3
|
https://mathoverflow.net/users/125498
|
412819
| 168,429 |
https://mathoverflow.net/questions/412657
|
4
|
Let $F$ be a field which has a positive characteristic $p \ge 2$ and $(\mathfrak{g},[p])$ be a restricted Lie algebras over a field $F$ where $[p]$ is a $p$-th power map on $\mathfrak{g}$. $(\mathfrak{g},[p])$ is called **simple restricted** if $(\mathfrak{g},[p])$ has no non-trivial $p$-ideals (i.e. ideals $\mathfrak{h}$ of $(\mathfrak{g},[p])$ satisfies $x^{[p]} \in \mathfrak{h}$ for all $x \in \mathfrak{h}$), Similarly $(\mathfrak{g},[p])$ is called **restricted simple** if $\mathfrak{g}$ is simple as ordinary Lie algebra (more detailed definitions are found in [R. Block and R. Wilson - Classification of the restricted simple Lie algebras (1988), pp. 116](https://doi.org/10.1016/0021-8693(88)90216-5)).
According to that paper restricted simple Lie algebras are simple restricted but its inverse is fails in general. I want to see an example that simple restricted Lie algebras but it is not restricted simple as possible as easy (e.g. low-dimension), moreover an explicit description of its non-trivial ideals. Thus my question is the following:
*Question.1-(1): Find an example of a simple restricted Lie algebra $(\mathfrak{g},[p])$ that it is not restricted simple and satisfies the above conditions.*
*Question.1-(2): What is a non-trivial ideal of $(\mathfrak{g},[p])$ in Question.1-(1)?*
|
https://mathoverflow.net/users/167704
|
Simple restricted but not restricted simple Lie algebras
|
The simple restricted Lie algebras are exactly the minimal $p$-envelopes of the simple Lie algebras. In fact, if $(\mathfrak{g}, [p])$ is a simple restricted Lie algebra over a field $\mathbb{F}$ of characteristic $p>0$, then $[\mathfrak{g}, \mathfrak{g}]$ is simple as an ordinary Lie algebra and its minimal $p$-envelope is isomorphic to $\mathfrak{g}$. Conversely, if $L$ is a simple Lie algebra over $\mathbb{F}$, then its minimal $p$-envelope $\mathfrak{g}$ is a simple restricted Lie algebra. This gives rise to a one-to-one correspondence between simple restricted Lie algebras and simple (ordinary) Lie algebras. There are many references for this fact. For instance, this is nicely explained in Section 4 of [Viviani - Simple finite groups and their infinitesimal deformations](http://ricerca.matfis.uniroma3.it//users/viviani/PUBLISHED-VERSIONS/Survey-Liealg(PUBLISHED).pdf).
The simplest (and smallest) explicit example answering both the questions is given by the 3-dimensional Lie algebra $L=\mathbb{F}a+\mathbb{F}b+\mathbb{F}c$ with $[a,b]=c$, $[b,c]=a$ and $[c,a]=b$ over a field $\mathbb{F}$ of characteristic 2. In this case, $L$ is simple as an ordinary Lie algebra, and its minimal $p$-envelope $\mathfrak{g}=\mathbb{F}a+\mathbb{F}b+\mathbb{F}c+\mathbb{F}a^{[2]}+\mathbb{F}b^{[2]}$ is a simple restricted Lie algebra. Of course, $\mathfrak{g}$ is not simple as an ordinary Lie algebra, $[\mathfrak{g},\mathfrak{g}]$ being a non-trivial ideal.
|
6
|
https://mathoverflow.net/users/14653
|
412830
| 168,433 |
https://mathoverflow.net/questions/412705
|
2
|
I am aware that the implicit and inverse function theorems can be generalized to infinite dimensional cases, but I am having difficulty in applying it to a specific calculation.
Let $C^1\_{\mathbb{R}}[0,1]$ be the space of real-valued $C^1$ functions on the interval $[0,1]$. If we impose the following norm:
$$
\begin{equation}
\lVert f \rVert := \lVert f \rVert\_{\sup}+\lVert f' \rVert\_{\sup}
\end{equation}
$$
for $f \in C^1\_{\mathbb{R}}[0,1]$, it is clear that $C^1\_{\mathbb{R}}[0,1]$ is a Banach space over $\mathbb{R}$.
Similarly, $C\_{\mathbb{R}}[0,1]$ is the Banach space of real-valued continuous functions on $[0,1]$ with the supremum norm.
Then, we can think of the operator $F:C\_{\mathbb{R}}^1[0,1] \to C\_{\mathbb{R}}[0,1]$ defined by
$$
\begin{equation}
F(f):=\sinh(f)+(f')^2
\end{equation}
$$
How do I show that (or is it indeed true that) $F$ is strongly $C^1$ and the derivative at each point is a linear isomorphism? More generally, how about the case $f \to G(f,f')$ for any smooth function $G : \mathbb{R}^2 \to \mathbb{R}$ whose Jacobian determinant never vanishes?
I appreciate any help.
|
https://mathoverflow.net/users/56524
|
Showing that a nonlinear operator over function spaces is differentiable and locally invertible?
|
The derivative of $F$ at $f \in C^1[0,1])$ is by definition the linear operator $F'(f): C^1[0,1] \rightarrow C[0,1]$ given by
$$ F'(f)\dot{f} = \lim\_{t\rightarrow 0} \frac{F(f+t\dot{f}) - F(f)}{t}. $$
Here, you get
$$
F'(f) = (\cosh f)\dot{f} + 2f'\dot{f}'.
$$
To apply the inverse function theorem, you need a linear operator $G(f): C[0,1] \rightarrow C^1[0,1]$ such that $F'(f)G(f)g = g$. In other words, given any $g in C[0,1]$, you need to be able to solve for $\dot{f} \in C^1[0,1]$ so that
$$
(\cosh f)\dot{f} + 2f'\dot{f}' = g.
$$
It's easy to check that if $f = 0$, then this is not possible. $F'(0)$ does have a right inverse, but it is not a bounded linear map from $C[0,1]$ to $C^1[0,1]$. So you cannot invert $F$ near $f = 0$ using the inverse function theorem for Banach spaces. In order to use the inverse function theorem, you need to "regain" the derivative you lost because $F'(f)$ is a first order differential operator.
So you want to restrict to $f$ such that the linear ODE above is a nondegenerate one, i.e., you can write it in the form
$$
\dot{f}' = \dots,
$$
where the right side is continuous on $[0,1]$. Next, you show that, for a given $f$, this ODE always has a solution $\dot{f} \in C^1[0,1]$. You also need to show that the map from $g$ to $\dot{f}$ can be chosen to be a bounded linear map from $C[0,1]$ to $C^1[0,1]$. This is all straightforward using the existence and uniqueness theorem for ODEs.
Offhand, I don't see how to show that $F$ is locally invertible on a neighborhood of $0$ in $C^1[0,1]$. However, you can show that it is locally invertible in a neighborhood of $0$ in $C^k[0,1]$ for $k$ sufficiently large (including $k = \infty$) by using the Nash-Moser inverse function theorem, as mentioned by @DCM.
|
2
|
https://mathoverflow.net/users/613
|
412837
| 168,436 |
https://mathoverflow.net/questions/412701
|
2
|
Let $C$ be a (nonempty) compact subset of euclidean $\mathbb R^n$, and consider the set-valued map $p\_C:\mathbb R^n \to 2^C$ defined by
$$
p\_C(x) = \{c \in C \mid \|x-c\| = \mbox{dist}(x,C)\},
$$
where $\mbox{dist}(x,C) := \inf\_{c \in C}\|x-c\|$ is the distance of $x$ from $C$.
>
> **Question.** *Under what minimal conditions on $C$ is $p\_C$ Lipschitz-continuous w.r.t Hausdorff distance ?*
>
>
>
**Note.** The case where $C$ is closed and convex is fully solved. Indeed, in such a case, $p\_C$ is single-valued and non-expansive (and therefore $1$-Lipschitz).
|
https://mathoverflow.net/users/78539
|
On the Lipschitz continuity of $x \mapsto \arg\min_{c \in C}d(x,c)$ w.r.t Hausdorff distance
|
Seems like in fact convex sets are the only ones for which $p\_C$ is continuous. To prove this, we can begin by noticing that for a set $C$ with the property, $p\_C$ can only take one point sets as values. If not, let $x,a\_1,a\_2$ be different points with $a\_1,a\_2\in p\_C(x)$. Then any point $y$ in the open segment from $x$ to $a\_i$ has $p\_C(y)=\{a\_i\}$, so $p\_C$ cannot be continuous at $x$.
So, if for a compact $C$, $p\_C$ is continuous, then it is single valued. So [this](https://math.stackexchange.com/questions/4344363/if-a-compact-set-is-not-convex-some-ball-is-tangent-to-it-at-several-points) shows that $C$ has to be convex.
Edit: From Dustin G. Mixon's comment above, it seems like the nearest point mapping is well studied. A reference that if a compact set $C$ is not convex, then the nearest point mapping is not single valued would be welcome.
|
1
|
https://mathoverflow.net/users/172802
|
412839
| 168,438 |
https://mathoverflow.net/questions/412764
|
3
|
In their classical paper on fluctuations in coin tossing [On Fluctuations in Coin-Tossing](https://www.pnas.org/content/35/10/605), Chung and Feller give a precise formula for the conditional probability of the number of positive “sides” of a random walk with an even number of steps, given a particular outcome for the endpoint.
$$
\mathbf{P}\left(\sum\_{j=1}^{2n} \mathbf{1}\_{(0,\infty)}\left(\frac{\mathsf{X}\_{j-1}+\mathsf{X}\_j}{2}\right)=2k\, \Bigg|\, \mathsf{X}\_{2n}=2\ell\right)
$$
is equal to
$$
\frac{\ell}{\binom{2n}{n-\ell}}\, \sum\_{i=\ell}^{k} \binom{2i}{i-\ell} \binom{2n-2i}{n-i} \cdot \frac{1}{i(n-i+1)}
$$
for a simple random walk $(\mathsf{X}\_0,\mathsf{X}\_1,\dotsc,\mathsf{X}\_{2n})$.
For their Theorem 1
$$
\mathbf{P}\left(\sum\_{j=1}^{2n} \mathbf{1}\_{(0,\infty)}\left(\frac{\mathsf{X}\_{j-1}+\mathsf{X}\_j}{2}\right)=2k\right)\,
=\, u\_{2k} u\_{2n-2k}\, ,
$$
for
$$
u\_0=1\, ,\ \text{ and }\ u\_{2k}\, =\, \mathbf{P}(\mathsf{X}\_{2k}=0)\, =\, \frac{1}{2^{2k}} \binom{2k}{k}\, ,\ \text{ for $k=1,2,\dots$,}
$$
the generalization to odd times was performed somewhat recently by Gessel in slides
[Chung–Feller Theorems](https://people.brandeis.edu/%7Egessel/homepage/slides/chung-feller-slides.pdf) last few slides,
and by Grünbaum in an article
[A Feynman–Kac approach to a paper of Chung and Feller on fluctuations in the coin-tossing game](https://arxiv.org/abs/1810.06092v1)
published in [Proceedings of the American Mathematical Society](https://doi.org/10.1090/proc/14758). The answer for that generalization is
$$
\mathbf{P}\left(\sum\_{j=1}^{2n+1} \mathbf{1}\_{(0,\infty)}\left(\frac{\mathsf{X}\_{j-1}+\mathsf{X}\_j}{2}\right)=2k+1\right)\,
=\, u\_{2k} u\_{2n-2k} \cdot \frac{2k+1}{2(n+1)}\, .
$$
Incidentally, the problem was apparently also stated as an exercise in McKean's textbook:
[Probability: The classical limit theorems](https://www.cambridge.org/us/academic/subjects/mathematics/abstract-analysis/probability-classical-limit-theorems?format=HB&isbn=9781107053212) Exercise 3.4.2 on p139.
But to the best of my knowledge nobody has stated a generalization of Chung and Feller's conditional distribution results in Theorem 2 to odd numbers of steps. (I might have missed it somewhere.)
*Note the analog of their Theorem 2a was the limit/specialization of their Theorem 2 to $\ell=0$ giving a uniform distribution. It makes sense from their Theorem 2a formula if you cancel the $\ell$ in the numerator outside their sum with the $i$ (which is necessarily $0$ in the single-summand sum) in the denominator inside the sum.*
As a secondary question, I wonder why combinatorialists sometimes derive useful formulas and advertise them as Gessel (and possibly McKean?) did, but then do not publish them. I have asked that question specifically over at AcademiaSE with more references for an example related to this question:
[Is it common for combinatorialists to not publish all their results? If so, why?](https://academia.stackexchange.com/questions/180689/is-it-common-for-combinatorialists-to-not-publish-all-their-results-if-so-why)
|
https://mathoverflow.net/users/471081
|
What is the generalization of the formula for Chung and Feller's Theorem 2 to odd numbers of steps?
|
Speaking only for myself, the reason that I don't publish all my results is that I write very slowly. (It has nothing to do with the nature of combinatorics.) It takes a long time for me to arrange my work into something publishable; it's much quicker to put some results into a slide presentation for a talk.
But I'd like to thank you for the references. I do intend to publish this result some day and it's good to know that there are other references to it.
For the benefit of readers who, like myself, think in terms of counting lattice paths rather than sums of random variables associating with random walks, it may be hepful to restate these formulas in terms of lattice paths. We consider paths made up of up steps $(1,1)$ and down steps $(1,-1)$, starting at the origin. (The horizontal components are irrelevant and are included only for convenience in visualization.) The result of Chung and Feller is that among the $4^n$ paths of length $2n$, the number with $2k$ steps above the $x$-axis is $\binom{2k}{k}\binom{2n-2k}{n-k}$. The analogous result for paths of odd length is that the number of paths of length $2j + 2k − 1$ with $2j − 1$ steps above the $x$-axis and $2k$ steps below is
$\frac{j}{2(j+k)}\binom{2j}{j}\binom{2k}{k}$. (A path of odd length cannot end on the $x$-axis; if it ends above the $x$-axis it must have an odd number of steps above the $x$-axis. Switching $j$ and $k$ allows us to count paths with an even number of steps above the $x$-axis.)
|
5
|
https://mathoverflow.net/users/10744
|
412841
| 168,439 |
https://mathoverflow.net/questions/412818
|
1
|
I'm looking for a solution to the following integral.
$$\int\_{\lambda}^{y}(x-a)^{-b}x^{-c}\exp\left( -d x^{-e} \right)dx,$$
where $b,c,d,e> 0$ and $0< a < \lambda < y$.
This equation appears in the context of Physical Layer Security, which is an area of study in digital communications (telecom).
|
https://mathoverflow.net/users/103291
|
Is there a solution to $\int_{\lambda}^{y}(x-a)^{-b}x^{-c}\exp\left( -d x^{-e} \right)dx$?
|
There is no closed-form expression in terms of known functions, but if $a$ is small, you could use the power series in terms of the [exponential integral](https://en.wikipedia.org/wiki/Exponential_integral#Generalization),
$$\int\_{\lambda}^{y}(x-a)^{-b}x^{-c}\exp\left( -d x^{-e} \right)\,dx=\sum\_{p=0}^\infty \frac{b\Gamma(p+b)}{p!\Gamma(1+b)}\frac{a^p}{e}
\left[y^{-k}E\_{1-k/e}\left(d y^{-e}\right)-\lambda^{-k}E\_{1-k/e}\left(d \lambda^{-e}\right)\right].
$$
where $k=b+c+p-1$.
|
3
|
https://mathoverflow.net/users/11260
|
412846
| 168,442 |
https://mathoverflow.net/questions/412826
|
2
|
Let $X,Y$ be real vector spaces, $T: X\to Y$ be a linear map, and fix a nonempty $S\subseteq X$ (we do not assume that $S$ is neither convex nor compact (indeed, right now we do not assume any topological structure on $X$ and $Y$)). Everything below is just basic linear algebra, hence you may prefer to go directly to the final question.
A point $x \in S$ is called an *extreme point* if $x=\alpha y+(1-\alpha)z$, for some $\alpha \in (0,1)$ and $y,z \in S$, implies $y=z$. Denote the set of extreme points with $\mathrm{Ext}\_X(S)$.
The following question is related to the commutation of $T$ and the $\mathrm{Ext}\_X$ operator, cf. [this](https://math.stackexchange.com/questions/1922019/linear-map-of-extrem-points-and-its-injectivity-surjectivity?noredirect=1&lq=1) and [this one](https://math.stackexchange.com/questions/1591561/preservation-of-extreme-points-under-linear-transformation).
If $T$ is an injection then the condition $Tx=\alpha Ty+(1-\alpha)Tz$, with $x,y,z \in S$ and $\alpha \in (0,1)$, is equivalent to $x=\alpha y+(1-\alpha)z$, and similarly $Ty=Tz$ is equivalent to $y=z$. Hence $Tx$ is an extreme point of $T[S]$ if and only if $x$ an extreme point of $S$. Hence:
>
> **Fact**. If $T \in \mathcal{L}(X,Y)$ is injective, then $\mathrm{Ext}\_Y(T[S])=T[\mathrm{Ext}\_X(S)]$.
>
>
>
Now, let us suppose that $T$ is not necessarily injective.
Define $\hat{X}:=X/\mathrm{Ker}(T)$ be the quotient vector space and $\hat{T}: \hat{X}\to Y$ by $\hat{T}(x+\mathrm{Ker}(T))=Tx$ for all $x \in X$.
Then $\hat{T}$ is a well-defined linear injection.
We obtain by the Fact above (applied to $\hat{T}$ and $S+\mathrm{Ker}(T)$) that:
$$
\mathrm{Ext}\_{Y}(\hat{T}[S+\mathrm{Ker}(T)])=\hat{T}[\mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))],
$$
so that
$$
\mathrm{Ext}\_{Y}(T[S])=\hat{T}[\mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))].
$$
(As it has been observed in the linked thread, the linear map $T:\mathbf{R}^2\to \mathbf{R}$ defined by $T(x,y)=x$ is such that $\mathrm{Ext}\_Y(T[S])\subsetneq T[\mathrm{Ext}\_X(S)]$ if $S$ is the closed unit ball of $\mathbf{R}^2$ with the $1$-norm. However, the above identity is correct since $\mathrm{Ext}\_{Y}(T[S])=\mathrm{Ext}\_T([-1,1])=\{-1,1\}$ and $\hat{T}[\mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))]=\hat{T}(\mathrm{Ext}\_{\hat{X}}([-1,1]\times \mathbf{R}))=\{-1,1\}$.)
>
> **Preliminary Question 1.** Is it true that, if $T \in \mathcal{L}(X,Y)$, then $\mathrm{Ext}\_Y(T[S])\subseteq T[\mathrm{Ext}\_X(S)]$?
>
>
>
Considering the above premises, this can be rewritten equivalently as follows:
>
> **Preliminary Question 2.** Is it true that, if $T \in \mathcal{L}(X,Y)$, then $\hat{T}[\mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))]\subseteq T[\mathrm{Ext}\_X(S)]$?
>
>
>
Suppose that the left hand side is nonempty (otherwise there is nothing to prove) and pick $x+\mathrm{Ker}(T)\in \mathrm{Ext}\_{\hat{X}}(S+\mathrm{Ker}(T))$, i.e., if $\hat{T}(x+\ker{T})=\alpha \hat{T}(y+\mathrm{ker}(T))+(1-\alpha) \hat{T}(z+\mathrm{ker}(T))$, for some $\alpha \in (0,1)$, then $y-z \in \mathrm{Ker}(T)$. Equivalently, if $Tx=\alpha Ty+(1-\alpha)Tz$ then $Ty=Tz$.
We would like to show, *if* the Question has a positive answer, then $Tx \in T[\mathrm{Ext}\_X(S)]$, i.e., there exists $h \in \mathrm{Ext}\_X(S)$ such that $Tx=Th$.
However, this suggested the counterexample: pick $T:\mathbf{R}^2\to \mathbf{R}$ as the example above and $S:=\{0\}\times (0,1)$, so that
$$
\mathrm{Ext}\_Y(T[S])=\{0\}\not\subseteq \emptyset= T[\mathrm{Ext}\_X(S)]
$$
Hence, the Preliminary Questions have a negative answer. To avoid similar conterexamples with empty sets, we assume that $X,Y$ are Banach spaces, $T$ is continuous, $S$ (and so also $T(S)$) is compact convex. Hence we come to our final question:
>
> **Question** Suppose that $X,Y$ are Banach spaces, $T$ is continuous, and $S\subseteq X$ is compact convex. Is it true that $\mathrm{Ext}\_Y(T[S])\subseteq T[\mathrm{Ext}\_X(S)]$?
>
>
>
For each $x \in S$, the coset $x+\mathrm{Ker}(T)$ is closed and convex, hence $H:=(x+\mathrm{Ker}(T)) \cap S$ is nonempty compact convex. Continuing the reasoning above, the answer should be positive in Hilbert spaces, picking the unique $h=h(x) \in H$ which minimizes the distance of $H$ from the origin (and, more generally, in Banach spaces with a strictly convex norm). Is it correct also in general Banach spaces?
|
https://mathoverflow.net/users/32898
|
Commutation of linear maps and extreme points
|
The answer to the question is **yes**.
*Proof.* Let $y$ be an extreme point of $T[S]$. Then
$$
F :=T^{-1}(\{y\}) \cap S
$$
is non-empty, compact and a face of $S$. By the Krein-Milman theorem, $F$ has an extreme point $x$, and since $F$ is a face of $S$, it follows that $x$ is also an etreme point of $S$. Since $Tx = y$, this proves the claim.
|
4
|
https://mathoverflow.net/users/102946
|
412849
| 168,443 |
https://mathoverflow.net/questions/410447
|
2
|
It is known that Brownian motion is almost surely locally Holder continuous, on a range that is random, i.e. depends on the particular path. This question explores the maximal range on which Brownian motion is Holder continuous.
Let $W$ be a standard Brownian motion, and let $C > 0$ and $0 < \alpha < \frac{1}{2}$ be constants.
Define the parametrized family of random variables $H\_{C, \alpha}$ by
$$H\_{C, \alpha} := \sup \big \{T > 0 \, \big | \, |W\_t - W\_s| \leq C |t - s|^\alpha \text{ for all } t, s \in [0, T] \big \}$$
**Question:** What is known about the probability distribution of $H\_{C, \alpha}$? Does it admit a density, or an expression in terms of known distributions?
|
https://mathoverflow.net/users/173490
|
On the range of Holder continuity of Brownian motion
|
**Claim:** Let $0<\alpha<1/2$ and $0<C<\infty$. The random variable $H\_{C,\alpha}$ has an absolutely continuous distribution.
**Proof:** Let $\{{\cal F}\_t\}\_{t \ge 0}$ be the standard Brownian Filtration. Given a non-negative rational $q$, define the ${\cal F}\_q$-measurable random boundary functions $\psi\_q^+:[q,\infty) \to (-\infty,\infty)$ and $\psi\_q^-:[q,\infty) \to (-\infty,\infty)$ by
$$\psi\_q^+(t)=\inf\_{s \le q} \; [W\_s+C(t-s)^\alpha]$$
and
$$\psi\_q^-(t)=\sup\_{s \le q} \; [W\_s-C(t-s)^\alpha]\,.$$
Observe that both these functions are a.s. continuous, and $\psi\_q^+$ is nondecreasing.
For $$q \in G^+=\{q \in {\mathbb Q}: \psi\_q^+(q)>W\_q\} \,,$$ consider the stopping time
$$H\_{C, \alpha}^+(q) := \inf \big \{t \ge q \,: W\_t = \psi\_q^+(t)\} \,.$$
Similarly, for $$q \in G^-:=\{q \in {\mathbb Q}: \psi\_q^-(q)<W\_q\} \,,$$ consider the stopping time
$$H\_{C, \alpha}^-(q) := \inf \big \{t \ge q \,: W\_t = \psi\_q^-(t)\} \,.$$
Since $W$ is locally $\beta$-Holder continuous for $\beta \in (\alpha,1/2)$, we infer that with probability 1, we have
$$H\_{C,\alpha} \in \{H\_{C, \alpha}^+(q) : q \in G^+\} \cup \{H\_{C, \alpha}^-(q) : q \in G^-\} \,,$$
so by symmetry, it suffices to prove that for fixed $q \in G^+$, the stopping time
$H\_{C, \alpha}^+(q)$ has an absolutely continuous distribution. We deduce this from the next Lemma using the Markov property of $W$ at time $q$. QED
**Lemma:** Let $\psi:[0,\infty) \to [r,\infty)$ be a nondecreasing function, with $\psi(0)=r>0$. Denote $H:=\inf \{t \ge 0 : W\_t=\psi(t)\},$ where the infimum of the empty set is $\infty$.
Then for all $t,\epsilon>0$ and some absolute constant $A$, we have
$$P\Bigl(H \in [t,t+\epsilon]\Bigr) \le Ar^{-2}\epsilon \,.$$
**Proof:** For $b>0$, let $\tau\_b:=\inf \{t \ge 0 : W\_t=b\}.$
Then $\tau\_1$ has a Levy distribution, with density bounded above by some absolute constant $A>0$. See e.g. <https://en.wikipedia.org/wiki/First-hitting-time_model>
Since $\tau\_b$ has the same law as $b^2 \tau\_1$, the density of $\tau\_b$ is bounded above by $Ab^{-2}$. Take $b=\psi(t) \ge r$. Then
$$P\Bigl(H \in [t,t+\epsilon]\Bigr) \le P\Bigl(\tau\_b \in [t,t+\epsilon]\Bigr) \le Ab^{-2}\epsilon \,,$$
and the lemma follows.
|
2
|
https://mathoverflow.net/users/7691
|
412850
| 168,444 |
https://mathoverflow.net/questions/412854
|
1
|
In [this paper](https://www.researchgate.net/publication/254211850_Non-archimedean_function_spaces_and_the_Lebesgue_dominated_convergence_theorem), the authors explain that the full generality of the Lebesgue dominated convergence theorem holds for functions on a compact zero-dimensional space $X$ taking values in a metrically complete non-archimedean field $K$ if and only if $X$ is finite. However, they remark that "On the other hand, in [11, Theorem 4.13] Katsaras showed a Lebesgue dominated convergence theorem for a certain class of measures $μ$ on $X$"
The paper in question is:
A.K. Katsaras, *On $p$-adic vector measure spaces*, J. Math. Anal. Appl., 365 (2010)
Unfortunately, this paper does not seem to be available. Also, be warned, there is a paper by the same author published in 2009 with a nearly identical title which *is* available, but which does not contain what I need.
I would very much like to know what the details are about this version of Katsaras' Lebesgue Dominated Convergence Theorem.
More generally, being a late-stage PhD student whose research has taken him into this subject, and who has no one to talk to —none of the mathematicians at my university know the slightest bit about non-Archimedean (functional) analysis— a book or paper dealing with convergence theorems and the like for non-Archimedean valued integrals/measures would be much appreciated. Ideally, I'd like an expert to talk to; most of the literature takes such a general approach to the subject that I can barely make heads or tails of what I'm reading.
|
https://mathoverflow.net/users/120369
|
Non-Archimedean Lebesgue dominated convergence theorem
|
Here's the correct DOI link for the paper referenced: <https://doi.org/10.1016/j.jmaa.2009.10.059>
Paper is in Open Archive so I am pretty sure you can view it yourself.
AFAICT, the reference is correct (Theorem 4.13 in the above-linked paper is a Dominated Convergence Theorem). Here is the full bibliographic reference:
>
> Journal of Mathematical Analysis and Applications
> Volume 365, Issue 1, 1 May 2010, Pages 342-357
>
>
>
|
2
|
https://mathoverflow.net/users/3948
|
412856
| 168,446 |
https://mathoverflow.net/questions/412844
|
1
|
Consider
$$X\_t=X\_0 + \int\_0^t b(s)ds+ \int\_0^t \sigma(s)dW\_s,\quad \forall t\ge 0,$$
where $X\_0\ge 0$ is a random variable of density $\rho$, $(W\_t)\_{t\ge 0}$ is an independent Brownian motion and $b,\sigma$ are measurable function "as nice as possible". Set $\tau:=\inf\{t\ge 0: X\_t\le 0\}$ and $Y\_t:=X\_{t\wedge \tau}$ for $t\ge 0$. Denote by $p(t,\cdot)$ be the density of $Y\_t$ restricted on $(0,\infty)$, i.e.
$$\mathbb E[f(Y\_t){\bf 1}\_{\{Y\_t>0\}}] = \int\_0^{\infty}f(x)p(t,x)dx,\quad \mbox{for all continuous and bounded function } f: \mathbb R\to\mathbb R.$$
It is known that $p$ satisfies the Fokker–Planck equation as below :
\begin{eqnarray}
\partial\_t p &=& \frac{\sigma(t)^2}{2}\partial^2\_{xx} p - b(t)\partial\_x p,\quad \forall t>0, x>0 \\
p(0,x) &=& \rho(x),\quad \forall x\ge 0 \\
p(t,0) &=& 0,\quad \forall t>0.
\end{eqnarray}
From the probabilistic point of view, it is straightforward to see $p\ge 0$ and the function
$$m(t):=\int\_0^{\infty}p(t,x)dx \in [0,1]$$
is non-increasing. From the PDE point of view, why the above Fokker–Planck equation together with the initial and boundary conditions implies $p\ge 0$ and $m$ is non-increasing?
Any answer, comments or references are appreciated.
PS : When $b\equiv 0$ and $\sigma\equiv 1$, one has in view of the reflection principle,
$$p(t,x)=\int\_0^{\infty}\frac{1}{\sqrt{2\pi t}}\left(\exp\left(-\frac{(x-y)^2}{2t}\right)-\exp\left(-\frac{(x+y)^2}{2t}\right)\right)\rho(y)dy$$
which clearly satisfies the desired properties. For general $b,\sigma$, do we still have the closed-form expression of $p$?
|
https://mathoverflow.net/users/261243
|
PDE interpretation of some properties of the solution to Fokker–Planck equations
|
Differentiating $m(t)$ and using the equation leads to
$$
m'(t):=\int\_0^{\infty}\left(
\frac{\sigma^2(t)}{2}\partial^2\_{xx} p(x,t) - b(t)\partial\_x p(x,t)\right)\,dx=
-\frac{\sigma(t)^2}{2}\partial\_{x} p(0,t)<0.
$$
The last inequality follows from the Zaremba-Giraud theorem for the sign of the solution's normal derivative at the boundary. See, for example, Nazarov A.I., [*A Centennial of the Zaremba–Hopf–Oleinik Lemma*](https://arxiv.org/abs/1101.0164).
For $b\equiv0$
$$
p(t,x)=\int\_0^{\infty}\frac{1}{\sqrt{2\pi \int\_0^t{\sigma^2(\tau)}\,d\tau}}\left(\exp\left(-\frac{(x-y)^2}{2\int\_0^t{\sigma^2(\tau)}\,d\tau}\right)-\exp\left(-\frac{(x+y)^2}{2\int\_0^t{\sigma^2(\tau)}\,d\tau}\right)\right)\rho(y)dy.
$$
For the general case the Green's function of the first BVP cannot be expressed via elementary functions.
|
2
|
https://mathoverflow.net/users/14551
|
412869
| 168,450 |
https://mathoverflow.net/questions/412855
|
4
|
As the title suggests, I have the following question:
>
> Is there a compact complex manifold with $b\_1(X)=b\_2(X)=b\_3(X)=b\_4(X)=0$?
>
>
>
Clarification:
Denote by $b\_k$ the $k$th Betti number of a compact complex manifold of positive dimension.
|
https://mathoverflow.net/users/105103
|
Is there a compact complex manifold with $b_1(X)=b_2(X)=b_3(X)=b_4(X)=0$?
|
There are complex manifolds with reduced cohomology vanishing in arbitrarily high degrees. Namely, product of two odd-dimensional spheres admits complex structure coming from representing it as a quotient of $(\Bbb C^n \setminus {0})\times (\Bbb C^m \setminus {0})$ by diagonal action of $\Bbb C$. They are known as Calabi-Eckmann manifolds.
Easiest example is $(\Bbb C^n \setminus {0})\times (\Bbb C^m \setminus {0}) / (e^At, e^At)$, where $A$ is a scalar matrix with non-real value $a$. It maps to a product of complex projective spaces by further quotiening, giving it structure of principal bundle with fiber being elliptic curve $\Bbb C / (\Bbb Z\cdot 1, \Bbb Z \cdot a)$. This map is the algebraic reduction of the manifold.
This easy example admits deformations of several types; for example, you can take different contracting flow instead of linear diagonal one. Most deformations do not have complex submanifolds at all, and have algebraic dimension zero as "most" generic complex manifolds.
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10
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https://mathoverflow.net/users/81055
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412874
| 168,452 |
https://mathoverflow.net/questions/412892
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8
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Let $\mathbb{S}$ be the Sierpiński space, the two pointed space $\{ 0, 1 \}$ with open sets $\{0 \}$, $\emptyset$, $\{ 0, 1 \}$. We give $\{ 0, 1 \}$ a partial order where $0 < 1$.
Let $X$ be a topological space. Consider the space $Y = \prod\_{f : X \rightarrow \mathbb{S}} \mathbb{S}$. This space is compact by Tychonoff's theorem.
There is a natural map $f : X \rightarrow Y$ and the closure of the image of $f$ in $Y$, $X\_0$. This is a closed subspace of a compact space and so it is compact.
Is $X\_0$ the Stone–Čech compactification of $X$ under certain conditions? I suspect that it always is. However, some have told me that this construction only works for $T\_{ 3 \frac{1}{2}}$ spaces. I would find it quite helpful if someone could link me to a source explaining the limitations of this construction if there are some. Alternatively, I think there is an adjoint functor theorem in terms of cogenerators? See the special adjoint functor theorem at [nLab](http://ncatlab.org/nlab/show/adjoint+functor+theorem#statement). It seems like every space is a subspace of a product $\prod\_{i \in I} \mathbb{S}$.
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https://mathoverflow.net/users/30211
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Is this space the Stone–Čech compactification?
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No, the closure of the image of $f$ in $Y$ is never the Stone-Čech compactification of $X$ unless $X$ is empty. In particular, consider the element $a\in Y$ which is $1$ on every coordinate. Note that the only open subset of $Y$ that contains $a$ is $Y$ itself. So, if $X$ is nonempty, then $a$ will be in the closure $X\_0$ of the image of $f$. This means that $X\_0$ contains a point whose only neighborhood is the whole space $X\_0$. Also, $a$ is not in the image of $f$ (no element of $X$ is in every closed subset of $X$), so $a$ is not the only point of $X\_0$. This means $X\_0$ is not $T\_1$, so it cannot be the Stone-Čech compactification of $X$.
The main moral here is that the Stone-Čech compactification is not about making a space compact: it is about making a space compact *Hausdorff*. So if you have some sort of universal construction that does not enforce the Hausdorffness condition, there is no reason to expect to get the Stone-Čech compactification.
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16
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https://mathoverflow.net/users/75
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412894
| 168,463 |
https://mathoverflow.net/questions/412891
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2
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Is the maximum nearest neighbor distance between points of the process, over all the infinitely many points of a stationary Poisson point process of intensity $\lambda$ in $\mathbb{R}^d$, almost surely finite? What is its distribution?
I am interested in the case $d=1$ and $d=2$. Also, if it is any easier, you can replace *between points of the process* by *between an arbitrary location in $\mathbb{R}^d$ and the closest point of the process*.
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https://mathoverflow.net/users/140356
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Maximum nearest neighbor distance for a Poisson point process
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$\newcommand\la\lambda\newcommand\de\delta\newcommand\R{\mathbb R}$This maximum (or, rather, supremum) distance, say $M$, is $\infty$ almost surely (a.s.).
Indeed, recall that a (simple) Poisson point process of intensity $\la\in(0,\infty)$ on $\R^d$ is a random Borel measure $m$ over $\R^d$ such that, for any pairwise disjoint bounded Borel subsets $A\_1,\dots,A\_k$ of $\R^d$, the random variables (r.v.'s) $m(A\_1),\dots,m(A\_k)$ are independent Poisson r.v.'s with respective parameters $\la|A\_1|,\dots,\la|A\_k|$, where $|\cdot|$ is the Lebesgue measure. It is not hard to show (see e.g. [Proposition 9.1.III (ii, iii), p. 4](https://link.springer.com/book/10.1007/978-0-387-49835-5)) that $m=\sum\_{i=1}^\infty\de\_{X\_i}$, where $\de\_x$ is the Dirac delta measure supported on $\{x\}$, for $x\in\R^d$, and
the $X\_i$'s are random points in $\R^d$ that are a.s. pairwise distinct. So,
$$M=\sup\_i\min\{\|X\_k-X\_i\|\colon k\ne i\}$$
a.s., where $\|\cdot\|$ is the Euclidean norm.
Now take any real $a>0$ and any natural $n$. Take the hypercube $C\_{a,n}:=[0,3na)^d$ and partition it naturally into $n^d$ congruent smaller hypercubes each with edgelengths $3a$. In each of these $n^d$ hypercubes $C\_j$ ($j=1,\dots,n^d$) each with edgelengths $3a$, take the central sub-hypercube, say $B\_j$, with edgelengths $a$.
Note that
$$p:=P\big(m(B\_j)=1,m(C\_j\setminus B\_j)=0\big) \\
=P\big(m(B\_1)=1\big)\,P\big(m(C\_1\setminus B\_1)=0\big)>0$$
for each $j=1,\dots,n^d$.
Then the probability $P(M\ge a)$ will be no less than the probability that at least one of the $n^d$ congruent smaller hypercubes $C\_j$ contains exactly one point of the random points $X\_i$ and that one point is in $B\_j$. The latter probability is
$$1-(1-p)^{n^d}\to1$$
as $n\to\infty$. So, $P(M\ge a)\ge1$ for all real $a>0$, and hence $P(M=\infty)=1$.
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4
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https://mathoverflow.net/users/36721
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412895
| 168,464 |
https://mathoverflow.net/questions/412899
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5
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I am investigating whether or not there exist $\epsilon > 0$ such that $\zeta(s) \neq 0$ on the strip $1-\epsilon < \Re(s) \leq 1$.
Suppose not. Then given $\delta > 0$ there exists a zero of zeta $\rho$ such that $1 -\delta < \Re(\rho) < 1 $. Hence, there exists a sequence of zeros $\{ z\_n \}\_{n=1}^\infty$ with increasing imaginary parts that have real part getting closer and closer to the line $\Re(z)=1$.
Can we somehow estimate the density of this infinite set of "fake zeros." The idea is to compare the density of this set with the best known estimate for the density of zeros that are on the critical line.
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https://mathoverflow.net/users/8435
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Density of fake zeros of Zeta
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There are, provably, very few zeros with real part close to $1$ (or bigger than $0.51$ for that matter). These theorems go under the name of "zero density estimates", and they have a vast literature. See Chapter 11 in Ivić: The Riemann zeta function (1985). The book was reprinted in 2013 by Dover Publications.
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19
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https://mathoverflow.net/users/11919
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412901
| 168,467 |
https://mathoverflow.net/questions/326116
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22
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I noticed that the only official reason given for awarding Edward Witten the Fields medal was his 1981 proof of the positive mass theorem with spinors, so I was assuming that the proof was fully rigorous.
However, I came across this paper <https://projecteuclid.org/download/pdf_1/euclid.cmp/1103921154> by Taubes and Parker which claims to make Witten's proof 'mathematically rigorous' and to justify assumptions which Witten made about Dirac operators. Does this mean that the Witten proof is not rigorous, or is it just the case that there were some unjustified lemmas to clear up which do not affect the validity or rigour of the argument (similar to the case of Perelman's proof of the Poincaré conjecture, where some lemmas and slight gaps had to be filled in)?
I am just curious as I have never really heard of the Taubes-Parker paper so I was assuming that the Witten paper was fully rigorous.
Later: I realise now that this is a bit of a misleading question, as 'proofs' can in reality come in many different flavours and/or levels of rigour ie. all the i's dotted and t's crossed, or some things left unsaid or which still remain to be proved.
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https://mathoverflow.net/users/119114
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Is Witten's proof of the positive mass theorem rigorous?
|
The positive mass theorem is more or less to do with the geometry of a type of positive scalar curvature condition.
Witten's work considers harmonic spinors, which are solutions to a certain linear elliptic system of partial differential equations. In his paper he presents a calculation which proves a rigidity theorem for harmonic spinors under a type of positive scalar curvature, directly comparable to Bochner's famous rigidity theorem for harmonic 1-forms in positive Ricci curvature. It is a little more complicated only since spinors are more complicated than differential forms.
Given the existence of a harmonic spinor with certain asymptotics, an integrated version of Witten's Bochner-type formula proves that the relevant positive scalar curvature condition implies the nonnegativity of the mass, now being expressed as the integral of the sum of squares of expressions built out of the harmonic spinor.
Witten's proof of the existence of such harmonic spinors is openly incomplete; he says "We have shown that the Dirac operator has no zero eigenvalue. Using this fact, we presume that standard methods can be used to yield" a key analytical step. The problem is existence of a Green's function for the relevant elliptic operator over noncompact spaces. Parker and Taubes gave a complete proof. I think it is not completely accurate to say that their proof only consists of "standard methods," since care is needed about weighted Sobolev spaces which can be somewhat delicate.
So I think it is inaccurate/misleading to put Witten's proof of positive energy theorem with some of his other works in terms of "inspiration and insight" for mathematics, or to just cite "origin in supergravity" (as Atiyah's laudatio does). His work here is a pretty direct and rigorous mathematical argument, in the vein of standard differential geometry. The gap is only due to his not being an expert in PDE methods. Even so he makes plausible that the relevant harmonic spinors exist. I think that a PDE expert reading his paper would likely even find it informally convincing.
As far as the Fields medal goes, I think some of his other work must have been more relevant. It's not hard to imagine someone else having discovered the main parts of Witten's proof, having to do with differential identities for spinors, with much less fanfare. The calculation relating the mass to the spinor is maybe the most striking part but I think many people (whether reasonably or not) would not regard it as a high point of mathematics (or whatever Fields medal is supposed to be about).
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9
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https://mathoverflow.net/users/156492
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412909
| 168,471 |
https://mathoverflow.net/questions/412917
|
2
|
**I. General Question**
Consider a one-parameter family of vector bundles $E\_t$ on a smooth projective variety $X$ with fixed Chern character $v$. Suppose $E\_t$ is Gieseker stable when $t\neq 0$ and $E\_0$ is *not* Gieseker semi-stable. **Is there a way to find the limit $\lim\_{t\to 0}E\_t$ in the Gieseker semi-stable moduli space $\overline{\mathcal{M}}\_v(X)$?**
Typically, the limit is a semi-stable coherent sheaf and not necessarily locally free. I'm wondering if there is a way that allows us "modify" the family and produce the semi-stable sheaf at special fiber after a suitable base change?
A counterpart that I have in mind is the Stable Reduction theorem [[Harris-Morrison](https://link.springer.com/book/10.1007/b98867), Prop. 3.17] for curves, which provides us an algorithm to find the limiting stable curve for a flat family of curves with bad singularities at the special fiber. I'm not sure if "stable reduction" is too much to hope for in the realm of the vector bundles.
**II. An Example**
Here is an example I'm working on. On a smooth cubic threefold $X$, there is a rank-two vector bundle $E$ associated to an elliptic quintic curve $C$ on $X$ via Serre construction of the ideal sheaf $I\_C$. One can show $c\_1(E)=0$ and $c\_2(E)=2$. When $C$ is projectively normal, $E$ is Gieseker stable, while when $C$ is not projectively normal (equivalently, $C$ is contained in a hyperplane), $E$ is *not* Gieseker semi-stable (in fact, it is slope semi-stable, though). See [[Markushevich and Tikhomirov](https://arxiv.org/pdf/math/9809140.pdf), Prop. 2.6]. Therefore, a smooth family $C\_t$ of elliptic quintics on $X$ with $C\_0$ contained in a hyperplane will produce a family of vector bundles $E\_t$ in question.
So how to find the limit $\lim\_{t\to 0}E\_t$ in $\overline{\mathcal{M}}\_{2,0,2}(X)$?
[[Markushevich and Tikhomirov](https://arxiv.org/pdf/math/9809140.pdf), Prop. 2.5] indicates there is a unique pair of lines $L\_1$ and $L\_2$ on $X$ associated to the non projectively normal curve $C\_0$. On the other hand, [[Druel](https://arxiv.org/abs/math/0002058), Theorem 3.5] and [[Beauville](https://arxiv.org/abs/math/0005017), Prop. 6.2(a)] indicate the strict semi-stable sheaves have the form $I\_{M\_1}\oplus I\_{M\_2}$ for a pair of lines $M\_1,M\_2$ on $X$.
These results seem to suggest to us what the limit sheaf is. However, I still cannot find a geometric family of sheaves in $\overline{\mathcal{M}}\_{2,0,2}(X)$ with $I\_{L\_1}\oplus I\_{L\_2}$ appears at time 0.
I appreciate it if anyone could help. Happy new year!
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https://mathoverflow.net/users/74322
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Is there a stable reduction for a family of vector bundles?
|
Consider the Harder-Narasimhan filtration of $E\_0$. Assume for simplicity it has length 2:
$$
0 \to F\_0 \to E\_0 \to E\_0/F\_0 \to 0,
$$
where $F\_0$ and $E\_0/F\_0$ are semistable and the slope of $F\_0$ is bigger than the slope of $E\_0/F\_0$. If $B$ is the parameter space of your family and $i \colon X \to X \times B$ is the embedding of the central fiber, consider the sheaf $E'$ on $X \times B$ defined by the exact sequence
$$
0 \to E' \to E \to i\_\*(E\_0/F\_0) \to 0.
$$
Note that the new family $E'$ has $E'\_t = E\_t$ for $t \ne 0$. On the other hand, if you restrict this sequence to the central fiber and take into account the isomorphism $L\_1i^\*(i\_\*G) \cong G$ (because the normal bundle of the central fiber is trivial of rank 1), you get an exact sequence
$$
0 \to E\_0/F\_0 \to i^\*E' \to E\_0 \to E\_0/F\_0 \to 0,
$$
which implies that $E'$ is flat over $B$ and $E'\_0$ fits into an exact sequence
$$
0 \to E\_0/F\_0 \to E'\_0 \to F\_0 \to 0.
$$
Now the sheaf $E'\_0$ is "more semistable" than $E\_0$, so iterating this procedure you can eventually modify your family to the one with semistable central fiber.
|
5
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https://mathoverflow.net/users/4428
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412919
| 168,472 |
https://mathoverflow.net/questions/412924
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5
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In the book R. M. Young, An introduction to non-harmonic Fourier series, I came across the following problem (page 18):
**Problem.** Show that the sequence $\left \{ \frac{1}{x+1},\frac{1}{x+2},\frac{1}{x+3}, \dots \right \}$ is complete in $L^2[0,1]$.
I tried to apply the Müntz-Szasz theorem but it didn't work out. Any ideas or approaches how to show completeness here? Thanks!
|
https://mathoverflow.net/users/106425
|
Completeness of the sequence $\left \{ \frac{1}{x+1},\frac{1}{x+2},\frac{1}{x+3}, \dots \right \}$ in $L^2[0,1]$
|
Let $S$ be the Hilbert subspace of $L^2[0,1]$ spanned by the sequence. For every positive integer $n$, the function $\frac{n}{x+n}=\frac{1}{1+x/n}$ lies in $S$. For $n$ large, this function is very close to the constant $1$ function in sup-norm, hence $1\in S$. Now we prove by induction that, for every nonnegative integer $m$, the functions $x^m$ and $\frac{x^m}{x+n}$ ($n=1,2,\dotsc$) lie in $S$. We have already established the base case $m=0$. So let us assume that $m>0$ and the functions $x^{m-1}$ and $\frac{x^{m-1}}{x+n}$ ($n=1,2,\dotsc$) lie in $S$. Then $x^{m-1}-\frac{nx^{m-1}}{x+n}=\frac{x^m}{x+n}$ lies in $S$. Moreover, $\frac{nx^m}{x+n}=\frac{x^m}{1+x/n}$ lies in $S$ as well, so by the same approximation argument as before, $x^m\in S$.
It follows that every polynomial lies in $S$, therefore $S=L^2[0,1]$ by the Weierstrass approximation theorem.
|
18
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https://mathoverflow.net/users/11919
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412929
| 168,475 |
https://mathoverflow.net/questions/412923
|
33
|
I recently came across this problem from [USAMO 2005](https://artofproblemsolving.com/wiki/index.php/1995_USAMO_Problems):
"A calculator is broken so that the only keys that still
work are the $\sin$, $\cos$, $\tan$, $\arcsin$, $\arccos$ and $\arctan$ buttons. The display initially shows $0$. Given any positive rational number $q$, show that
pressing some finite sequence of buttons will yield $q$. Assume that the
calculator does real number calculations with infinite precision. All
functions are in terms of radians."
A surprising question whose ingenious solution actually shows how to generate the square root of any rational number.
I'd like to pose the following questions related to this problem:
>
> **What is the smallest set of real functions, continuous at all points of $\mathbb{R}$, which can be applied to $0$ to yield a sequence containing all the rational numbers?**
>
>
>
It's also interesting perhaps weaken this to allowing finite numbers of discontinuities so you can use the rational functions for example:
>
> **What is the smallest set of real functions, continuous except at a finite set of points, which can be applied to $0$ to yield a sequence containing all the rational numbers?**
>
>
>
Note that these are slightly different questions to the one above in that we are asking not only to be able to produce any rational from $0$ but to produce all of them at some point after starting at $0$. In the case of the USAMO question you can generate a complete sequence of rationals as well as any given rational but this may not always be true. (See solution for details)
For the second question note that from the theory of continued fractions of rational numbers the functions $f(x)=1/x$, $g(x)=x+1$ will generate any given rational starting from $0$. For example since
$$\frac{355}{113} = 3+\cfrac{1}{7+\cfrac{1}{16}}$$
we have $\frac{355}{113}=g^{[3]}(f(g^{[7]}(f(g^{[16]}(0)))))$.
If we also throw in $h(x)=x-1$ we again have every inverse included hence this set of three functions will generate all rationals.
So we know that the smallest set must contain either $1$, $2$ or $3$ functions.
In fact as [pregunton](https://mathoverflow.net/users/115044/pregunton) noted in this [related question](https://mathoverflow.net/q/413087/7113) the functions $f(x)=x+1$ and $g(x)= -1/x$ generate the modular group which acts transitively on $\mathbb{Q}$ and this gives an elegant example with only two functions.
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https://mathoverflow.net/users/7113
|
What is the smallest set of real continuous functions generating all rational numbers by iteration?
|
It is enough with one continuous function. First, I'll give a simple example with one function which is discontinuous at one point. To do it, consider the function $$f:(0,\pi+1)\to(0,\pi+1)$$ with
$$
f(x) = \begin{cases}
x+1 &\text{if $x<\pi$,} \\
x-\pi &\text{if $x>\pi$,} \\
1 &\text{if $x=\pi$.}\\
\end{cases}
$$
**Claim**:
The sequence
$$1,f(1),f^2(1),\dots \tag{$\*$}$$ is dense in $(0,\pi+1)$.
To verify the claim, it is enough to see that the image is dense in the interval $(0,1)$, and that is true because for every $n$, the number $\lceil n\pi\rceil-n\pi$ is in the image, and the sequence of multiples of $\pi$ modulo 1 is dense in $(0,1)$ due to $\pi$ being irrational.
Let $A$ denote the image of the sequence $(\*)$.
Since $A$ is dense in $(0,\pi+1)$, we can find an homeomorphism $h:(0,\pi+1)\to\mathbb{R}$ with $h(A)=\mathbb{Q}$ (using that $\mathbb{R}$ is countable dense homogeneous, see for example [this reference](http://matwbn.icm.edu.pl/ksiazki/fm/fm74/fm74118.pdf)). We can also suppose $h(1)=0$ changing $h$ by $h-h(1)$ if necessary.
Then the function $F=hfh^{-1}$ does the trick, because $$F^n(0)=hf^nh^{-1}(0)=h(f^n(1)),$$ so $h(A)$, which is $\mathbb{Q}$, is the image of the sequence $0,F(0),F^2(0),\dots$
To prove that the problem can be solved with one continuous function, we can apply the same argument but taking instead of $f$ a continuous function $g:\mathbb{R}\to\mathbb{R}$ such that $0,g(0),g^2(0),\dots$ is dense in $\mathbb{R}$. As Martin M. W. noticed in his answer, those functions are known to exist (they are called transitive maps), [this paper](https://www.researchgate.net/publication/236026813_Some_classes_of_transitive_maps_on_R) gives examples of them.
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33
|
https://mathoverflow.net/users/172802
|
412931
| 168,476 |
https://mathoverflow.net/questions/412933
|
5
|
Consider the following statements in $\sf ZF$:
>
> (S) If $A, B$ are nonempty sets, then there is a surjection $s:A \to B$, or there is a surjection $t:B\to A$.
>
>
>
>
> (I) If $A, B$ are sets, then there is an injection $i:A\to B$, or there is an injection $j:B\to A$.
>
>
>
Note that (I) implies (S). Assuming $\sf AC$, both statements are true.
**Question.** Is there a model of $\sf ZF$ in which (S) holds, but not (I)?
|
https://mathoverflow.net/users/8628
|
Existence of surjection vs injection over $\sf ZF$
|
The answer is no. Both (I) and (S) are equivalent to AC over ZF. Indeed, for any set $S$ the class of ordinals $\alpha$ such that $\alpha$ injects into $S$ (resp. $S$ surjects onto $\alpha$) is a set, so there is some ordinal $\beta$ outside this set. Assuming (I) (resp. (S)), there is an injection $S\to\beta$ (resp. surjection $\beta\to S$). In the latter case, we also get an injection $S\to\beta$, by taking any element $S$ to the least element in its preimage. In either case, we see $S$ is well-orderable. Thus both (I) and (S) imply well-ordering theorem and hence AC.
Since both (I) and (S) are equivalent to AC, they are also equivalent to each other.
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16
|
https://mathoverflow.net/users/30186
|
412934
| 168,478 |
https://mathoverflow.net/questions/412893
|
4
|
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$Let $ \Sigma\_g $ be a compact orientable surface of genus $ g $. Let the subgroup $ \pi\_1(\Sigma) $ of $ \SL\_2(\mathbb{R}) $ be a Fuchsian representation of the fundamental group. Then $ \Sigma\_g $ admits a hyperbolic structure
$$
\pi\_1(\Sigma\_g)\backslash \SL\_2(\mathbb{R})/\SO\_2(\mathbb{R})
$$
Moreover, the 3-dimensional coset space
$$
\pi\_1(\Sigma\_g)\backslash SL\_2(\mathbb{R})
$$
is a locally Riemannian homogeneous space admitting $ \widetilde{\SL\_2} $ geometry and it is isometric to the unit tangent bundle of $ \Sigma\_g $.
Consider a similar situation for 3-manifolds. Let $ M $ be a hyperbolic 3-manifold
$$
M \cong \pi\_1(M) \backslash \SL\_2(\mathbb{C})/\SU\_2
$$
where $ \pi\_1(M) $ a Kleinian representation of the fundamental group of $ M $.
Then what can we say about the six-dimensional manifold
$$
\pi\_1(M) \backslash \SL\_2(\mathbb{C})
$$
My first thought was that maybe that we could identify $ \pi\_1(M) \backslash \SL\_2(\mathbb{C}) $ with the space given by compactifying each fiber of the tangent space of $ M $ into a 3-sphere. Upon further reflection, though, that doesn't seem promising. $ M $ is isometrically covered by hyperbolic 3 space, which is contractible. And the tangent bundle over a contractible space is always trivial. So by covering the tangent bundle of $ M $ is also trivial. So I'm worried that taking one point compactifications would just give me a trivial bundle (cartesian product of $ M $ with the three sphere) which doesn't seem quite right. On the other hand the tangent bundle to $ \Sigma\_g $ is trivial but the unit tangent bundle isn't trivial. So maybe stuff made from trivial bundles isn't always trivial? Anyway, not sure if I'm missing something obvious but that's my question.
Basically, **For a hyperbolic 3-fold $ M $ can we say something interesting about the relationship between $ \pi\_1(M) \backslash \SL\_2(\mathbb{C}) $ and the tangent bundle of $ M $? (As in the two dimensional case where for a hyperbolic 2-fold $ \Sigma\_g $ we have that $ \pi\_1(\Sigma\_g) \backslash \SL\_2(\mathbb{R}) $ is isometric to the unit tangent bundle of $ \Sigma\_g $.)**
|
https://mathoverflow.net/users/387190
|
Generalizing a result about hyperbolic 2-folds to hyperbolic 3-folds
|
The (orientation-preserving) isometry group $G=PSL(2,{\bf C})$ acts on the bundle of (positively oriented) orthonormal frames on ${\bf H}^3$.
An ad hoc argument using the Iwasawa decomposition $G=KAN$ shows that this action is transitive. Indeed, considering the upper half space model and fixing the standard frame in $(0,0,1)$ you can first use the action of $K=PSU(2)$ to move that frame to any other frame at $(0,0,1)$. Then you can use the action of $A$ (the diagonal matrices) to move it to any frame at any $(0,0,z)$. Finally you can use the action of $N$ (the upper triangular matrices acting as translations) to move that frame to any other frame at any $(x,y,z)$.
This shows transitivity of the action and hence that $G$ bijects to the bundle of positively oriented, orthonormal frames on ${\bf H}^3$. (Injectivity is a standard argument.)
Dividing out $\Gamma=\pi\_1M$ you get that $\Gamma\backslash G$ bijects to the bundle of positively oriented, orthonormal frames on $M$.
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5
|
https://mathoverflow.net/users/39082
|
412937
| 168,481 |
https://mathoverflow.net/questions/412927
|
4
|
Let $G(t)$ be a probability generating function of some integer and non-negative r. v. $X$. Suppose that
$$\lim\_{t\to1}G'(t)=+\infty.$$
That is
$$
\mathbb{E}X=+\infty.
$$
Can you show that
$$
\lim\_{t\to1}\frac{(G'(t))^2}{G''(t)}=0\,?
$$
There is an example satisfying the conjecture:
$$
G(t)=\frac{6}{\pi^2}\text{PolyLog}(2,t).
$$
|
https://mathoverflow.net/users/43017
|
Ratio of the first squared and the second moment
|
This is correct.
Denote $G(t)=\sum\_{k=0}^\infty p\_k t^k$, where $p\_i\geqslant 0$ and $\sum p\_i=1$. Then we are given that $\sum kp\_k=\infty$ and should prove that
$$
\lim\_{t\to 1-0} \frac{(\sum\_{k=1}^\infty kp\_k t^{k-1})^2}{\sum\_{k=2}^\infty k(k-1)p\_kt^{k-2}}=0.
$$
Since both numerator and denominator tend to infinity when $t$ goes to $1-0$, we have for arbitrary positive integer $N>1$:
$$C:=\limsup\_{t\to 1-0} \frac{(\sum\_{k=1}^\infty kp\_k t^{k-1})^2}{\sum\_{k=2}^\infty k(k-1)p\_kt^{k-2}}=\limsup\_{t\to 1-0} \frac{(\sum\_{k=N}^\infty kp\_k t^{k-1})^2}{\sum\_{k=N}^\infty k(k-1)p\_kt^k}\\
=\left(\sum\_{k=N}^\infty p\_k\right)\cdot \limsup\_{t\to 1-0} \frac{(\sum\_{k=N}^\infty kp\_k t^{k-1})^2}{(\sum\_{k=N}^\infty k(k-1)p\_kt^{k-2})(\sum\_{k=N}^\infty p\_k)}\leqslant 2 \sum\_{k=N}^\infty p\_k,
$$
since by Cauchy–Bunyakovsky–Schwarz we have
$$
\left(\sum\_{k=N}^\infty k(k-1)p\_kt^{k-2}\right)\left(\sum\_{k=N}^\infty p\_k\right)\geqslant \left(\sum\_{k=N}^\infty \sqrt{k(k-1)}p\_k t^{k-1}\right)^2
\geqslant \frac12
\left(\sum\_{k=N}^\infty kp\_k t^{k-1}\right)^2.
$$
It remains to observe that $\sum\_{k=N}^\infty p\_k$ may be arbitrarily small, thus $C=0$.
|
4
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https://mathoverflow.net/users/4312
|
412941
| 168,484 |
https://mathoverflow.net/questions/412896
|
3
|
Suppose I have a measurable space $(\Omega, \Sigma)$ and a function $f: \mathbb{R} \times \mathbb{\Sigma} \rightarrow [0,1]$ such that for any $x \in \mathbb{R}$ the tuple $(\Omega, \Sigma, f(x, \\_))$ is a measure space.
Is it the case that
$$\int\_{\mathbb{R}} \left[ \int\_{\Omega} df(x, \\_) \right] dx = \int\_{\Omega} d \left[ \int\_{\mathbb{R}} f(x, \\_) dx \right] \ ?$$
I suspect that this comes down to the question of whether we can interchange the Radon-Nikodym derivative and integration operations, but I am not sure. If this question doesn't make sense because of a missing assumption please let me know :)
|
https://mathoverflow.net/users/265390
|
Does the Radon-Nikodym derivative commute with integration?
|
Provided $x\mapsto f(x,B)$ is Borel measurable for each $B\in\Sigma$, you can define
$$
\mu(B):=\int\_{\Bbb R} f(x,B) dx,
$$ and check (using Tonelli's theorem) that $\mu$ is a measure.
If $H\ge 0$ is a measurable function on $\Omega$,
then
$$
\int\_\Omega H d\mu=\int\_{\Bbb R}\left[\int\_\Omega H(\omega) d\_\omega f(x,\omega)\right] dx,
$$
by the usual simple function approximation.
|
3
|
https://mathoverflow.net/users/42851
|
412942
| 168,485 |
https://mathoverflow.net/questions/412906
|
3
|
I was reading this question [The connected component of the idele class group](https://mathoverflow.net/questions/350676/the-kernel-of-the-global-class-field-theory-homomorphism) but I am very confused about the structure of the *solenoids* $(\widehat{\mathbb{Z}}\times\mathbb{R})/\mathbb{Z}$, (where $\mathbb{Z}$ acts diagonally), more specifically why it is connected. More generally, if we replace $\widehat{\mathbb{Z}}$ by $\mathbb{Z}\_p$, is the resulting quotient group $(\mathbb{Z}\_p\times \mathbb{R})/\mathbb{Z}$ still connected?
Hoping if someone can give me any hints or answers!!
|
https://mathoverflow.net/users/177957
|
Is $(\mathbb{Z}_p\times \mathbb{R})/\mathbb{Z}$ connected?
|
No originality here, but I would tell the story as follows. Consider the subgroup $(\mathbb{Z}\times\mathbb{R})/\mathbb{Z}$ of $(\mathbb{Z}\_p\times\mathbb{R})/\mathbb{Z}$. It is dense, because $\mathbb{Z}$ is dense in $\mathbb{Z}\_p$. It is also connected, because it is isomorphic to $\mathbb{R}$ (with a coarser topology than the standard one). Therefore, in the group $(\mathbb{Z}\_p\times\mathbb{R})/\mathbb{Z}$, the connected component of the identity is dense, whence it equals $(\mathbb{Z}\_p\times\mathbb{R})/\mathbb{Z}$.
The same proof works for $(\widehat{\mathbb{Z}}\times\mathbb{R})/\mathbb{Z}$.
|
6
|
https://mathoverflow.net/users/11919
|
412943
| 168,486 |
https://mathoverflow.net/questions/412951
|
8
|
It looks like there is some literature out there on what might be called 'non-Archimedean complex analysis' e.g. [Benedetto - An Ahlfors Islands Theorem for non-archimedean meromorphic functions](https://arxiv.org/abs/math/0407142) and [Cherry - Lectures on Non-Archimedean Function Theory](https://arxiv.org/abs/0909.4509). I am mainly working on non-Archimedean functional analysis right now and need to become better acquainted with the non-Archimedean analogues of basic results that might be encountered in a first course on complex analysis, up to and including Liouville's Theorem e.g. Cauchy integral formulas, holomorphic functions etc. for a few spectral theory proofs.
Of course I am well aware that with many results there will be no such analogue. What I would like to know is if there exists a good introduction to this area that I could look at, that starts with the fundamentals. For example, is there an analogue of 'holomorphic iff analytic'? Any advice much appreciated.
This relates to my earlier question: [Non-emptiness of spectrum σ(a) in non-Archimedean Banach algebras](https://mathoverflow.net/questions/412631/non-emptiness-of-spectrum-sigmaa-in-non-archimedean-banach-algebras)
|
https://mathoverflow.net/users/197447
|
Literature on non-Archimedean analogues of basic complex analysis results
|
Benedetto has a textbook that discusses basic $p$-adic analysis, although his aim is to study $p$-adic dynamics. And it's for a single variable. But might be a good place to get some information.
[Dynamics in One Non-Archimedean Variable, Robert L. Benedetto, Graduate Studies in Mathematics, Volume 198, 2019, American Mathematical Society](https://bookstore.ams.org/gsm-198/)
I'll also mention that these days a lot of $p$-adic analysis is done on Berkovich space, rather than on $\mathbb Q\_p$ or $\mathbb C\_p$. An introduction to analysis on the Berkovich line can be found in the book of Baker and Rumely.
[Potential Theory and Dynamics on the Berkovich Projective Line,
Matthew Baker, Robert Rumely, Mathematical Surveys and Monographs
Volume 159, 2010, American Mathematical Society.](https://bookstore.ams.org/surv-159)
|
7
|
https://mathoverflow.net/users/11926
|
412954
| 168,490 |
https://mathoverflow.net/questions/412965
|
8
|
Let $f$ be a weight $2$ cusp form for the group $\Gamma\_0(N)$. I was experimenting with integrals of the form
$$ \int\_r^s f(z) \, dz$$
where $r, s \in \mathbf{P}^1(\mathbf{Q})$ and the integral above is over the geodesic in the upper-half plane connecting $r$ and $s$. When I plotted these period integrals, I noticed that the resulting values formed a *rank $2$ lattice* in $\mathbf{C}$. I have two questions about this:
1. **Why would the above period lattice have rank $2$?** Since we are integrating $f(z) \, dz$ over paths in the relative homology group $H^1(X\_0(N), \{\text{cusps}\}, \mathbf{Z})$, I would expect the rank of the period lattice to grow with the rank of this homology group. Why is the rank of the period lattice always $2$, even if the rank of the homology group is bigger?
2. Inside the period lattice obtained from integrating over $H^1(X\_0(N), \{\text{cusps}\}, \mathbf{Z})$, we can consider the sublattice from integrating over just the ordinary homology group $H^1(X\_0(N), \mathbf{Z})$, omitting paths that are not closed loops in $X\_0(N)$. (Equivalently, the sublattice is formed by only plotting $\int\_r^s f(z) \, dz$ when $r$ and $s$ are $\Gamma\_0(N)$-equivalent.) **What is the index of this sublattice inside the full period lattice?** Can one answer this in general?
|
https://mathoverflow.net/users/394740
|
What is the rank of the period lattice of modular forms?
|
For Q1: I'm assuming your $f$ is a normalized Hecke *newform* of level $N$ with coefficients in $\mathbf{Z}$. Then the choice of $f$ determines a splitting of the homology as
$$H\_1(X\_0(N), \{cusps\}, \mathbf{Q}) = (f\text{-generalised eigenspace}) \oplus \text{(other stuff)}.$$
By the Eichler--Shimura isomorphism, we see that the $f$-generalized eigenspace is semisimple (i.e. it's a genuine eigenspace) and has dimension 2. When we integrate against $f$, it kills the "other stuff"; so the map $H\_1(X\_0(N), \{cusps\}, \mathbf{Z}) \to \mathbf{C}$ given by integration against $f$ factors through the image of $H\_1(X\_0(N), \{cusps\}, \mathbf{Z})$ in the $f$-eigenspace of $H\_1(X\_0(N), \{cusps\}, \mathbf{Q})$, and a little elementary linear algebra shows that this eigenspace quotient has to be free of rank 2.
So the image in $\mathbf{C}$ has rank *at most* 2. Since this lattice has non-trivial intersection with both $\mathbf{R}$ and $i \mathbf{R}$, its rank must be exactly 2.
For Q2, the comparison of lattices arising from "usual" and "relative" homology: one can show that the cokernel of the natural map from $H\_1(X\_0(N)) \to H\_1(X\_0(N), \{cusps\})$ is a free $\mathbf{Z}$-module of finite rank which is *Eisenstein* as a Hecke module, i.e. every Hecke eigenvalue system that shows up looks like an Eisenstein series (this is essentially the Manin--Drinfeld theorem). It follows that, for a given cuspidal eigenform $f$ as above, the only primes $p$ that can divide the index of the two period lattices are those such that $f$ is congruent mod $p$ to an Eisenstein series. If I remember correctly, you see this concretely with the unique weight 2 newform of level 11: it is congruent to an Eisenstein series mod 5, and the quotient between the two lattices is a cyclic group of order 5.
If you want to learn more, then you should read about *modular symbols*; the books by [John Cremona](https://johncremona.github.io/book/fulltext/index.html) and [William Stein](https://wstein.org/books/modform/modform/) are both excellent references.
|
14
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https://mathoverflow.net/users/2481
|
412968
| 168,493 |
https://mathoverflow.net/questions/412969
|
4
|
This is essentially a reference request. According to the classification of the finite simple groups, there are 26 (or arguably 27) sporadic groups. Let us denote these by G\_1 , ... , G\_{26} (resp. G\_{27}).
>
> **Question**: *Which G\_i can be embedded into which G\_j?*
>
>
>
I was unable to find the answer to this question in the literature, but I suspect that the experts will know where to look. Note that it is not too difficult to [find](https://en.wikipedia.org/wiki/Sporadic_group) a diagram describing the weaker property of "which G\_i is a *subquotient* of which G\_j."
|
https://mathoverflow.net/users/203598
|
Embeddings of the sporadic simple groups
|
See Table 2, p. 362 in:
* Wilson, Robert A.
Is the Suzuki group Sz(8) a subgroup of the Monster? Bull. Lond. Math. Soc. 48 (2016), no. 2, 355-364. <https://doi.org/10.1112/blms/bdw012>
Which lists all simple groups that are contained in a sporadic group.
It seems the question you ask about (which sporadics can be embedded into another sporadic) was settled in these papers:
* Griess, Robert L., Jr. The friendly giant. Invent. Math. 69 (1982), no. 1, 1-102. <https://doi.org/10.1007/BF01389186>
* Wilson, Robert A. Is J1 a subgroup of the Monster? Bull. London Math. Soc. 18 (1986), no. 4, 349-350. <https://doi.org/10.1112/blms/18.4.349>
|
10
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https://mathoverflow.net/users/38068
|
412974
| 168,494 |
https://mathoverflow.net/questions/412960
|
14
|
In Quantum theory, groups and representations, Peter Woit writes:
>
> A fundamental principle of modern mathematics is that the way to
> understand a space $M$, given as some set of points, is to look at $F(M)$,
> the set of functions on this space.
>
>
>
I was wondering what some examples of this "fundamental principle" is across different fields in mathematics.
*Woit, Peter*, [**Quantum theory, groups and representations. An introduction**](https://dx.doi.org/10.1007/978-3-319-64612-1), Cham: Springer (ISBN 978-3-319-64610-7/hbk; 978-3-319-64612-1/ebook). xxii, 668 p. (2017). [ZBL1454.81004](https://zbmath.org/?q=an:1454.81004).
|
https://mathoverflow.net/users/473920
|
What are some examples of understanding a space by studying the functions on this space?
|
The idea of studying the relationship between structured spaces and appropriate spaces of functions thereon could be described as one of the basic principles of functional analysis, perhaps even the defining one.
Examples:
* completely regular spaces and continuous functions—general, bounded or of compact support (in the locally compact case);
* $\sigma$-algebras and (bounded) measurable functions;
* measure spaces and $L^p$-spaces (strictly speaking, equivalence classes of functions);
* smooth manifolds, including open subsets of euclidean space, and spaces of smooth functions, sometimes combined with growth conditions;
and finally, but the list could go on,
* complex manifolds and holomorphic functions, again often combined with growth conditions.
The next link in the chain is a consideration of the duals of these function spaces. Here there are two main streams:
* representation theorems—the cases where these duals have explicit descriptions, either as spaces of functions themselves or of measures (duality for $L^p$-spaces, Riesz representation theorem);
* the cases where they are used to define new types of objects (Schwartzian distributions, the Bourbakian approach to measure theory).
|
9
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https://mathoverflow.net/users/159073
|
412976
| 168,495 |
https://mathoverflow.net/questions/412944
|
6
|
Let $F$ be a number field. For an irreducible cuspidal automorphic representation $\pi$ of $\operatorname{GL}\_n(\mathbb{A}\_F)$, we say that $\pi$ is symplectic (or orthogonal) if $L(s,\pi,\bigwedge^{2})$ (or $L(s,\pi,\operatorname{Sym}^2)$) has a pole at $s=1$.
I am wondering whether if $\pi=\bigotimes \pi\_v$ is symplectic (or orthogonal), then $\pi\_v$’s are also symplectic (or orthogonal) for all places $v$?
(Here, $\pi\_v$ is symplectic (or orthogonal) means that its corresponding Weil–Deligne group representation by local Langlands correspondence is of such type.)
Any comments are welcome!
|
https://mathoverflow.net/users/35898
|
Global symplectic (orthogonal) type of automorphic representation compels its type to all its local components?
|
Here is a proof of the claim using results from Arthur's monograph [The Endoscopic Classification of
Representations: Orthogonal and
Symplectic Groups](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.294.6966&rep=rep1&type=pdf).
Let $N = 2n$ be an even integer, and $\pi$ a cuspidal automorphic representation satisfying $\pi^\vee \cong \pi$.
* Since $\pi \cong \pi^\vee$, the Rankin-Selberg $L$-function $L(\pi \times \pi, s)$ has a simple pole at $s = 1$. We have the factorisation $L(\pi \times \pi, s) = L(\pi, S^2, s) L(\pi, \wedge^2, s)$ and both factors are non-vanishing at $s = 1$, so exactly one of them must have a pole; thus $\pi$ is orthogonal or symplectic, in the sense of the question, but never both.
* The discussion preceding Theorem 1.5.3 of Arthur shows that $\pi$ defines an element of his set $\tilde{\Phi}\_{\mathrm{sim}}(N)$ of global parameters, and this lies in the subset $\tilde{\Phi}\_{\mathrm{sim}}(G)$ for a uniquely determined quasi-split group $G$ whose Langlands dual $\widehat{G}$ is either $SO\_{2n}$ or $Sp\_{2n}$.
* Theorem 1.5.3 of Arthur shows that $\widehat{G}$ is symplectic if $\pi$ is of symplectic type, and $\widehat{G}$ is orthogonal if $\pi$ is of orthogonal type. (This is a very deep theorem, despite sounding like a tautology!)
* Theorem 1.4.2 of op.cit. now shows that there is a cuspidal automorphic representation $\sigma$ of $G$ such that, for every $v$, the Weil–Deligne representation associated to $\pi\_v$ is the image in $GL\_N$ of the ${}^L G$-valued parameter associated to $\sigma\_v$. So, in particular, it is symplectic (resp. orthogonal) if $\pi$ is.
|
4
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https://mathoverflow.net/users/2481
|
412977
| 168,496 |
https://mathoverflow.net/questions/412958
|
3
|
I use the convention of the Weinberg QFT textbooks, that is, $(-,+,+,+)$.
According to Weinberg QFT vol 2 p. 369, he says the Euclidean Dirac operator
\begin{equation}
{D}:=[i\partial\_i +t\_\alpha A\_{i \alpha}]\gamma\_i
\end{equation}
is Hermitian. Here, $\partial\_4:=-i\partial\_0$, $A\_{4\alpha}:= iA^0\_\alpha$ and $\gamma\_4:=i \gamma^0$. $\alpha$ is the gauge index.
Let us simplify and focus on the case of $U(1)$ gauge theory so that there is no $\alpha$. Also, assume that $A\_0, A\_1, A\_2, A\_3$ are real-valued Schwartz functions on $\mathbb{R}^4$.
Then, the above $D$ must be a densely defined unbounded operator on the Hilbert space $[L^2(\mathbb{R}^4)]^4$.
My question is, what would be the maximal domain of $D$ that makes it self-adjoint? Physicists would not care about this kind of subtlety but, I feel uncomfortable about not distinguishing between Hermitian and self-adjoint.
Could anyone please clarify?
|
https://mathoverflow.net/users/56524
|
The exact domain on which the Euclidean Dirac operator is self-adjoint
|
Recall that a densely defined operator $T$ on a Hilbert space $H$ is *essentially self-adjoint* if and only if its minimal closure $\overline{T}$ is self-adjoint, if and only if $T$ is symmetric and has a unique extension to a self-adjoint operator (i.e., the minimal closure $\overline{T}$).
In general, you can consider the spin Dirac operator $D$ on a complete Riemannian spin manifold $(M,g)$ with spinor bundle $S$ (e.g., the Dirac operator you consider on Euclidean $\mathbb{R}^4$). By [a theorem of Wolf](https://math.berkeley.edu/%7Ejawolf/publications.pdf/paper_050.pdf), $D$ (as well as its square $D^2$) is essentially self-adjoint on the subspace $C\_c^\infty(M,S)$ of compactly supported smooth spinor fields. More generally, it follows from [a theorem of Chernoff](https://www.sciencedirect.com/science/article/pii/0022123673900037) that $D^k$ is essentially self-adjoint on $C\_c^\infty(M,S)$ for every $k \in \mathbb{N}$.
What about the domain of the minimal closure $\overline{D}$? [By a theorem of Gromov and Lawson](http://www.numdam.org/item/PMIHES_1983__58__83_0.pdf), if the curvature of the spinor Levi-Civita connection is uniformly bounded in the appropriate sense, then the domain of $\overline{D}$ is the first $L^2$ Sobolev space $H^1(M,S) = L^{1,2}(M,S)$ of square-integrable spinor fields with square-integrable weak covariant derivatives (w.r.t. the spinor Levi-Civita connection). In your case, this means that the domain of the unique self-adjoint extension of the Dirac operator on Euclidean $\mathbb{R}^4$ is precisely the space $H^1(\mathbb{R}^4) \otimes \mathbb{C}^4$, where $H^1(\mathbb{R}^4)$ is the Sobolev space of square integrable functions on $\mathbb{R}^4$ with square-integrable weak first-order partial derivatives (or equivalently square-integrable weak gradient).
|
4
|
https://mathoverflow.net/users/6999
|
412979
| 168,498 |
https://mathoverflow.net/questions/410524
|
4
|
Let $S\_{g,b}$ denote the orientable connected compact surface of genus $g$ with $b$ boundary components. A group homomorphism $\varphi\colon G\to \text{Homeo}^+(S\_{g,b})$ is said to be free $G$-action if $\varphi(a)$ has no fixed point for all non-trivial $a\in G$. Two free group actions $\varphi\_1,\varphi\_2\colon G\to \text{Homeo}^+(S\_{g,b})$ are said to be equivalent if there is $\mathscr H\in \text{Homeo}^+(S\_{g,b})$ such that $\varphi\_2(a)=\mathscr H^{-1}\circ \varphi\_1(a)\circ \mathscr H$ for all $a\in G$.
A theorem of Nielsen says that any two free $\Bbb Z/n\Bbb Z$-actions on a closed orientable connected surface are equivalent.
>
> Does there exist a classification theory of inequivalent free $\Bbb
> Z/n\Bbb Z$-actions on every $S\_{g,b}\ (b\neq 0)$?
>
>
>
Any reference/idea will be helpful.
|
https://mathoverflow.net/users/363264
|
Inequivalent free $\Bbb Z/n\Bbb Z$-actions on orientable compact bordered surface
|
Did you try to use the double $T$ of the surface $S$? Any fixed point-free action of $\mathbf Z/n$ on $S$ induces a fixed point-free action on the closed orientable surface $T$. Moreover, the induced action commutes with the natural orientation-reversing involution on $T$. By Nielsen's Theorem that you mentioned, you've reduced to study orientation-reversing involutions on $T$ that commute with a given action of $\mathbf Z/n$ on $T$ and whose set of fixed points is homeomorphic to a disjoint union of $b$ circles. Such involutions induce orientation-reversing involutions on the quotient surface $U=T/(\mathbf Z/n)$. Now, orientation-reversing involutions on a closed orientable surface $U$ whose set of fixed points is a disjoint union of circles are easily classified. It should not be too difficult to decide which ones lift to orientation-reversing involutions on $T$ having a set of fixed points composed of $b$ circles.
With this approach you already get conditions on the mere existence of fixed point-free actions of $\mathbf Z/n$ on $S$. Indeed, the double $T$ of $S$ has genus $2g+b-1$. So that $n$ has to divide
$$
-\chi(T)=2(2g+b-1)-2=4g+2b-4
$$
in order for a fixed point-free action of $\mathbf Z/n$ on $S$ to exist. For example, if $S$ is a pair of pants, i.e., $g=0$ and $b=3$, the natural number $n$ has to divide $2$. Adopting the above strategy, I think you only get $1$ fixed point free action of $\mathbf Z/2$ on the pair of pants $S$: the one where the nontrivial element of $\mathbf Z/2$ acts on $S$ by exchanging $2$ boundary circles, and by acting antipodally on the third. That case probably was clear anyway.
|
5
|
https://mathoverflow.net/users/85592
|
412992
| 168,502 |
https://mathoverflow.net/questions/412989
|
10
|
Consider the below $q$-series identity. One of the things I like about this expansion is how nicely the difference on the left hand side factors to the right hand side of the equation.
$$\prod\_{k\geq1}(1+q^k)^3-\prod\_{k\geq1}(1+q^{3k})
=3q\prod\_{n\geq1}(1+q^n)(1+q^{9n})^2(1+q^n+q^{2n}+\cdots+q^{8n}).$$
I have a "not-so-neat" proof, so
>
> **QUESTION.** Can you provide a "nifty" justification? *Caveat: you decide what is "nifty"*.
>
>
>
|
https://mathoverflow.net/users/66131
|
Looking for a "clever" argument for a $q$-series identity
|
Here's a proof that indicates a systematic method for proving such identities. Let $\eta(z) = q^{1/24} \prod\_{n=1}^{\infty} (1-q^{n})$, with $q = e^{2 \pi i z}$. The identity you state in the equation is the same as
$$
\frac{\eta^{3}(2z)}{\eta^{3}(z)} - \frac{\eta(6z)}{\eta(3z)} = 3 \frac{\eta(2z) \eta^{2}(18z)}{\eta^{2}(z) \eta(9z)}.
$$
Replacing $z$ with $2z$ and multiplying by $\eta(2z) \eta(4z) \eta^{2}(6z)$ gives the equivalent
$$
\frac{\eta^{4}(4z) \eta^{2}(6z)}{\eta^{2}(2z)} - \eta(2z) \eta(4z) \eta(6z) \eta(12z) = 3 \frac{\eta^{2}(4z) \eta^{2}(6z) \eta^{2}(36z)}{\eta(2z) \eta(18z)}.$$
Results about $\eta(z)$ in [Gordon and Hughes - Multiplicative properties of $\eta$-products. II](https://doi.org/10.1090/conm/143/01008), and Ligozat - Courbes modulaires de genre 1 ([MR](https://mathscinet.ams.org/mathscinet-getitem?mr=422158)) guarantee that each of the three terms in the above equation are in the space of weight $2$ cusp forms for the group $\Gamma\_{0}(72)$. This space has dimension $5$, and if $f = \sum\_{n=1}^{\infty} a\_{n} q^{n}$ is a function in the space with $a\_{1} = a\_{2} = \dotsb = a\_{7} = 0$, it follows that $f = 0$. It suffices to verify that the coefficients of $q^{1}$ through $q^{7}$ are the same for the two sides of the equivalent identity above.
|
16
|
https://mathoverflow.net/users/48142
|
412994
| 168,503 |
https://mathoverflow.net/questions/412988
|
64
|
I am an enthusiastic but ever-so-slightly naive PhD student and have been 'following my nose' a lot recently, seeing whether topics that I have studied can be generalised or translated in various ways into unfamiliar settings; exploring where the theory breaks down etc.
When doing this, I have found it very difficult to assess whether it is going to 'work' in the more general sense of whether it could lead to a viable project for a PhD thesis or perhaps a short research paper. I guess it becomes easier to get a feel for these things as one gains experience and a better sense of perspective. Of course, one added complication over the past year has been that due to various lockdowns it been difficult to get to know other mathematicians and run ideas past them in the natural way that would have occurred in previous years.
Suppose that a wise and experienced pure mathematician wishes to generalise a particular theory or shed some light on an open problem and will devote, say, at least 6 months to it. What reasonable steps should be taken to maximise the likelihood of this being a fruitful endeavour? My main concern personally would be (is?) a previously unforeseen obstacle rearing its ugly head only after a significant amount of time and energy has been invested that brings the whole thing crashing down. How can this scenario be avoided when exploring something brand new?
EDIT: Although I have referred to my own circumstances above, my question relates primarily to the more general issue.
|
https://mathoverflow.net/users/197447
|
How do pure mathematicians assess whether their research ambitions can be realistically achieved?
|
Over decades, and across multiple research fields, I've noticed a way to predict I'm on track to make progress. **I discover something interesting, only to learn it is already known**.
As a student, this was incredibly discouraging, and in fact I stopped some lines of research for this very reason. But by now I'm used to it: I start looking at a new area, and have an insight. Arg, it turns out people knew it 10 years ago. I read some more, think some more, and have a new insight. Careful searching reveals a paper with that result from three years ago. Too bad—but that paper is fascinating, and I can deeply appreciate it and feel kinship with the author. Thinking about it leads to another insight, which I start writing up. Oof, then I see a preprint from a month ago which says the same thing.
What I've learned over time is that this pattern of rediscovery, particularly if the dates of things I've been rediscovering get more and more recent, is a reliable sign I'm on a good path, and that I'm building my intuition in an area other people care about.
So keep following your nose, check back with the literature regularly, and take any rediscoveries as a green light, not a red light.
|
99
|
https://mathoverflow.net/users/1227
|
413001
| 168,506 |
https://mathoverflow.net/questions/413011
|
14
|
I preface my question by admitting I know no algebraic geometry nor algebraic number theory. I do know some algebraic topology. I'm a student.
Recently I learned about sheaf cohomology. Then a little bit of etale cohomology, as much as I could stomach having never studied algebraic geometry. Then I came across [Artin-Verdier duality](https://en.wikipedia.org/wiki/Artin%E2%80%93Verdier_duality), in particular the notion that $\mathcal{O}\_K$ is 'like a 3-manifold.' This led me to the interesting area of *arithmetic topology* that wants to understand some larger 'arithmetic $\leftrightarrow$ topology dictionary.'
Now I've done some reading to try to grasp the big picture of arithmetic topology. But I'm unclear on one point:
>
> Is the analogy pursued by this arithmetic $\leftrightarrow$ topology dictionary formal, in any sense?
>
>
>
So far I've seen it said how this analogy gives a nice way of thinking about number-theoretic things with topology (e.g. prime ideals are like links, and their factors are like the constituent knots.) The words 'inspire' and 'motivate' are used a lot. And there are precise comparisons to be made between the objects on either side (e.g. the algebraic fundamental group of $\mathrm{Spec} ~\mathbb{Z}$ is isomorphic to the classical fundamental group of $S^3$.)
But I'd like to know whether there is some larger framework that rigorously explains why this analogy exists.
|
https://mathoverflow.net/users/472967
|
Is Mazur's analogy between arithmetic and topology formal, in any sense?
|
For this analogy, like most analogies in mathematics, and indeed like most philosophical principles in mathematics, one can certainly make *a part of it* formal and rigorous, but I don't think any true formal statement could ever capture *all* of what we mean by the analogy.
In particular, by well-chosen definitions, one can write down statements of the form "If X is either a 3-manifold or the ring of integers of a number field, then something is true about X", where "something" is expressed the same way in each case. But there's no reason to expect that there is a single statement that implies all true such statements.
In particular, one can certainly not get an equivalence of categories between some category of 3-manifolds and some category of number fields (as [Wojowu suggests in the comments](https://mathoverflow.net/questions/413011/is-mazurs-analogy-between-arithmetic-and-topology-formal-in-any-sense#comment1058517_413011)), or any kind of correspondence between one 3-manifold and one number field, that respects the interesting structure like Artin-Verdier duality. (Thus I think the equivalence of fundamental groups between $S^3$ and $\mathbb Z$ is a red herring.)
Note that in Verdier duality, a pretty fundamental concept is an orientation. Any 3-manifold is either oriented or has a double cover to be oriented. But from the form of Artin-Verdier duality, for a number field to be oriented, it would have to contain the $n$th roots of unity for all $n$, which is impossible. So the "oriented double cover" in this setting is actually a cover of infinite degree! Covering spaces and dualizing sheaves are some of the concepts we absolutely do want to match up, so I don't think there's any way to wriggle out of this.
|
14
|
https://mathoverflow.net/users/18060
|
413014
| 168,514 |
https://mathoverflow.net/questions/336755
|
13
|
Padé approximants are often better than Taylor series at representing a function. Given a Taylor series, one can use Wynn's epsilon algorithm to easily produce the Padé approximants to it.
[Volterra series](https://en.wikipedia.org/wiki/Volterra_series) are a generalization of Taylor series that can also model "memory" phenomena. Does there exist a similar generalization of Padé approximants that can model these phenomena, or an algorithm like Wynn's to compute them from the Volterra series?
I would be at least happy to know the answer for the discrete Volterra series, which (I think) would be equivalent to something like a multivariate Padé approximant.
[Originally asked at MSE](https://math.stackexchange.com/questions/3297364/taylor-series-is-to-volterra-series-as-pad%C3%A9-approximant-is-to), but seems too advanced for that site.
|
https://mathoverflow.net/users/24611
|
“Taylor series” is to “Volterra series” as “Padé approximant” is to _________?
|
>
> "I would be at least happy to know the answer for the discrete Volterra series, which (I think) would be equivalent to something like a multivariate Padé approximant."
>
>
>
Multi-variate versions of Padé approximants got attention in the 1970s, for example a generalization to functions of two variables was published by J.S.R. Chisolm in 1973 (see ["**Rational Approximates Defined from
Double Power Series**"](https://www.ams.org/journals/mcom/1973-27-124/S0025-5718-1973-0382928-6/)) and a generalization to functions of ***countably*** more variables was studied at around the same time by P.R. Graves-Morris and D.E. Roberts who were both at the University of Kent in Canterbury, leading to this generalization being called the "Canterbury approximant" (see ["**Calculation of Canterbury approximants**"](https://www.sciencedirect.com/science/article/abs/pii/0010465575900685?via%3Dihub)).
The original papers about Canterbury approximants never seemed to get many citations, so it's easy to look through the entire list of papers that cited them to find ***further*** generalizations (for example, to functions of an uncountably infinite number of variables). Some of the most significant post-Canterbury papers on multivariate versions of Padé approximants that can be found in this list are:
* [Annie Cuyt "**How well can the concept of Padé approximant be generalized to the multivariate case?**" (May 1999)](https://www.sciencedirect.com/science/article/pii/S037704279900028X)
* [Philippe Guillaume and Alain Huard. "**Multivariate Padé approximation**" (September 2000)](https://www.sciencedirect.com/science/article/pii/S037704270000337X)
Finally, one of the original authors of the Canterbury approximant co-wrote a book with George Baker Jr. called "[**Padé Approximants**](https://books.google.ca/books?hl=en&lr=&id=Vkk4JNLKbLoC&oi=fnd&pg=PP1&ots=kUWgJtTqM0&sig=rFjYf82DbW2YwlNPUzs4pdfpKUQ#v=onepage&q&f=false)" (first edition: 1982, second edition: 1996) which provides you with any developments which they considered significant between the aforementioned papers from the 1970s and about the time of the aforementioned papers from 1999-2000.
Finally, some steps towards a generalization to functions of ***complex*** variables can be found in my analogous question: [“Taylor series” is to “Volterra series” as “Laurent series” is to \_\_\_\_\_\_\_\_\_?](https://mathoverflow.net/q/413016/66334)
|
11
|
https://mathoverflow.net/users/66334
|
413017
| 168,515 |
https://mathoverflow.net/questions/413015
|
3
|
This is a continuation of [my previous question](https://mathoverflow.net/questions/412137/a-baire-subset-of-reals-that-is-not-suslin-measurable). Recall that a subset $A \subseteq {}^\omega\omega$ is *analytic* if it is the continuous image of the Baire space. I would like to know if there exist two models $N \subseteq M$ such that:
1. $M \models \mathsf{ZFC}$ (or just $M \models \mathsf{ZF} + \text{There exists a non-principal ultrafilter}$).
2. $N \models \mathsf{ZF} + \text{Every subset of reals is analytic}$.
3. $({}^\omega\omega)^M = ({}^\omega\omega)^N$.
From the comments of previous posts, it's worth noting that:
1. Under $\mathsf{ZF} + \mathsf{DC}$, one can construct the universal analytic set and prove that it is not analytic. Thus, $N \not\models \mathsf{DC}$.
2. A (somewhat trivial) example of a model in which every subset of reals is analytic is the Feferman-Levy model, where the reals is a countable union of countable sets.
|
https://mathoverflow.net/users/146831
|
A submodel of set theory with all reals which every set is analytic
|
In fact, the principle "Every set is analytic" is not consistent with $\mathsf{ZF}$ in the first place. We don't need choice to get a surjection $h$ from Baire space to the set of continuous maps on Baire space. But once we have such an $h$, the "diagonalizing" set $\{x: x\not\in h(x)\}$ can't be analytic.
---
Let me show how to get such an $h$ in $\mathsf{ZF}$ alone. Working in $\mathsf{ZF}$, let $C$ be the set of continuous maps $\omega^\omega\rightarrow\omega^\omega$. To each $\gamma\in C$ we assign the set $$S(\gamma)=\{(\sigma,\tau)\in\omega^{<\omega}\times\omega^{<\omega}: \forall f\succ \sigma(\gamma(f)\succ\tau)\}.$$ By your favorite coding mechanism we can identify $S(\gamma)$ with some $\hat{S}(\gamma)\in\omega^\omega$. But $\hat{S}(\gamma)=\hat{S}(\gamma')$ implies $\gamma=\gamma'$ for continuous $\gamma,\gamma'$, so in fact $\hat{S}$ gives an injection from $C$ to $\omega^\omega$.
Now turning a surjection into an injection without choice is hard, but the converse is trivial: for $x\in\omega^\omega$, let $h(r)=\hat{S}^{-1}(r)$ if $r\in ran(\hat{S})$, and let $h(r)$ be the always-zero map otherwise.
|
8
|
https://mathoverflow.net/users/8133
|
413018
| 168,516 |
https://mathoverflow.net/questions/413002
|
1
|
We know that various [Markov categories have deterministic morphisms](https://scholar.google.com/citations?view_op=view_citation&hl=en&user=ofP7CgYAAAAJ&alert_preview_top_rm=2&citation_for_view=ofP7CgYAAAAJ:W7OEmFMy1HYC). This suggests that they support faithful functors into them from categories of categories. One interesting Markov category is the Kleisli Category of the [Distribution Monad](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/distribution+monad), KlDM. I need to know if KlDM supports a faithful functor into it from one or both of the following categories:
1. The category of small categories
2. The category of finitely presented categories
3. The category of concrete categories
|
https://mathoverflow.net/users/417530
|
What category of categories have faithful functors into Markov categories?
|
Following N. Virgo's point (but even more simply), any concrete category $\mathcal C$ admits (by definition) a faithful functor $F: \mathcal C \to Set$. The free functor $G$ from $Set$ to the Kleisli category of the distribution monad is faithful as well. So $GF: \mathcal C \to Kl(Dist)$ is a faithful functor. Conversely, if a category $\mathcal C$ admits a faithful functor $H: \mathcal C \to Kl(Dist)$, then since the forgetful functor $U: Kl(Dist) \to Set$ is also faithful, it follows that $UH$ is faithful, i.e. $\mathcal C$ is concrete. That is, $\mathcal C$ admits a faithful functor to $Kl(Dist)$ iff $\mathcal C$ is concrete.
For example, the category of small categories is concrete (the functor $Mor : Cat \to Set$ which carries a category to its set of morphisms is faithful). Likewise, so is the category of small finitely-presented categories. So these categories are concrete / admit faithful functors to $Kl(Dist)$.
The category of concrete categories is not locally small for any reasonable notion of morphism I can think of, so it is not concrete / it does not admit a faithful functor to $Kl(Dist)$.
|
2
|
https://mathoverflow.net/users/2362
|
413021
| 168,517 |
https://mathoverflow.net/questions/413007
|
5
|
For integer-valued sequences $(x\_n)\_{n=0}^\infty$, consider recurrences of the form
$$x\_n=ax\_{n-1}+(bn+c)x\_{n-2} \tag{$\*$}\label{star}$$
for $n\ge2$, where $a,b,c$ are integers.
There seem to be many such sequences in the OEIS — see e.g. [A000085](http://oeis.org/A000085), [A001475](https://oeis.org/A001475), [A005425](https://oeis.org/A005425), and [A000898](https://oeis.org/A000898) — with various combinatorial descriptions.
>
> **Question:** Is there a common combinatorial description for all integer-valued sequences $(x\_n)\_{n=0}^\infty$ satisfying recurrences of the form \eqref{star} — at least for natural $a$, $b$ ,$c$ and for some initial conditions (on $x\_0$, $x\_1$, depending on $a$, $b$, $c$)?
>
>
>
|
https://mathoverflow.net/users/36721
|
A common combinatorial description for a certain type of recurrences
|
So, the Fibonacci numbers can be constructed from this recursion, so it is natural to look for generalizations of those (and combinatorial models for the Fibonacci numbers).
The Fibonacci number $f\_n$ (up to some index shift), can be seen as the number of integer compositions of $n$ (list of numbers summing to $n$), using only the numbers $1$ and $2$.
Now, the recursion above, $x\_n = a x\_{n-1} + (bn+c)x\_{n-2}$
hints that we also look at integer compositions using $1$ and $2$, but the $1$ comes in $a$ different colors.
Moreover, the 2s have two types, either, one of $c$ colors,
or it is a 'special' 2, with one of $b$ colors and a label.
The label is between 1 and the total sum of entries up to that point.
For example, $a=3$, $b=3$, $c=2$ $n=8$ has the object
$$
1\_3, \; 2\_{1, 1}, \; 2\_{3, 5}, \; 2\_{2}, \; 1\_{1}.
$$
The numbers sum to $n$, and subscripts of the $1$ is between $1$ and $a$. The $2's$ with only one subscript has
the subscript between $1$ and $c$. Finally, each $2$
with two substripts, has the first subscript (color) between $1$ and $b$. The second subscript must be between $1$
and the sum of all numbers in the list up to that point.
For example, for $2\_{3, 5}$, the 5 is the maximum allowed since $1+2+2=5$.
Below is Mathematica code for generating such compositions:
```
a = 3;
b = 3;
c = 2;
Clear[f];
f[0] := {{}};
f[1] := Table[{{1, j}}, {j, a}];
f[n_] := f[n] = Join[
Join @@ Table[
Append[f, {1, aa}],
{f, f[n - 1]}, {aa, a}]
,
Join @@ Table[
Append[f, {2, bb}],
{f, f[n - 2]}, {bb, b}]
,
Join @@ (Join @@ Table[
Append[f, {2, cc, j}],
{f, f[n - 2]}, {cc, c}, {j, n}])
];
```
The natural initial condition is $x\_0 =0$ and $x\_1 = a$. (In the code above, I use $f$ instead of $x$)
|
5
|
https://mathoverflow.net/users/1056
|
413023
| 168,519 |
https://mathoverflow.net/questions/412940
|
54
|
Define the function $$S(N, n) = \sum\_{k=0}^n \binom{N}{k}.$$
For what values of $N$ and $n$ does this function equal a power of 2?
There are three classes of solutions:
* $n = 0$ or $n = N$,
* $N$ is odd and $n = (N-1)/2$, or
* $n = 1$ and $N$ is one less than a power of two.
There are only two solutions $(N, n)$ outside of these three classes as far as I know: (23, 3) and (90, 2). These were discovered by Marcel Golay in 1949. There are no more solutions with $N < 30{,}000$.
I've written more about this problem [here](https://www.johndcook.com/blog/2022/01/01/turning-the-golay-problem-sideways/).
By the way, I looked in Concrete Mathematics hoping to find a nice closed form for $S(N, n)$ but the book specifically says there isn't a closed form for this sum. There is a sum in terms of the hypergeometric function $\_2F\_1$ but there's no nice closed form.
|
https://mathoverflow.net/users/136
|
When do binomial coefficients sum to a power of 2?
|
The case $n=2$ was settled by Nagell in 1948 and suspected (?) by Ramanujan in 1913, but in an equivalent form.
As John points out in his growing [blog post](https://www.johndcook.com/blog/2022/01/01/turning-the-golay-problem-sideways/), the $n = 2$ case is a quadratic equation which, via the quadratic formula, requires that $2^n - 7 = x^2$ for some integer $x$.
Motivated by who-knows-what, Ramanujan posted the following in 1913 (J. Indian Math.).
>
> Question 464. $2^n - 7$ is a perfect square for the values $3, 4, 5, 7, 15$ of $n$. Find other values.
>
>
>
A posted "solution" just verified the values of $n$ he gave and did not address whether there are other solutions. The same problem was proposed by Ljunggren in a Norwegian journal in 1943; in 1948 Nagell proved that there are no other solutions, using a quadratic field with $\sqrt{-7}$ and focusing on values of $x$ rather than $n$.
Skolem, Chowla, and Lewis (referencing Ramanujan but not aware of Nagell's solution) solved the problem using $p$-adic techniques in 1959, prompting Nagell to republish his easier 1948 proof in English.
Meanwhile, *in another part of the forest*, error-correcting codes arose. With that motivation, Shapiro and Slotnick essentially reconstructed Nagell's approach in 1959. Their subsequent results make use of other error-correcting code structures; techniques in coding veer away from the binomial sum question. As van Lint explained in a 1975 survey,
>
> Although as far as perfect codes are concerned the problem has been settled, the purely number-theoretic problem of finding all solutions of (5.2) remains open.
>
>
>
where (5.2) is the more general $\sum\_{i=0}^e \binom{n}{i} (q-1)^i = q^k$ where $q$ is a power of a prime.
Bringing Nagell into the error-correcting code literature occurred by 1964 (Cohen). The OEIS entries [A215797](https://oeis.org/A215797), [A060728](https://oeis.org/A060728), and [A038198](https://oeis.org/A038198) address the problem from different viewpoints.
---
There's one reference to another solution that I have not been able to track down. In a 1998 textbook on error correcting codes, John Baylis writes (p109)
>
> ...so $2+n+n^2$ must be a power of 2. It was shown in 1930 that $n = 1, 2, 5$ and 90 are the only positive integers for which this is true.
>
>
>
Any idea what 1930 result he has in mind?
---
References:
Baylis, Error-Correcting Codes, Chapman & Hall, 1998.
Berndt, Choi, Kang, The problems submitted by Ramanujan to the Journal of the Indian Mathematical Society, Contemporary Mathematics 236, 1999.
Cohen, A note on double perfect error-correcting codes on $q$ symbols, Information and Control 7, 1964.
Nagell, The diophantine equation $x^2 + 7 = 2^n$, Arkiv Math. 4, 1961 (English version of his 1948 article published in Norwegian).
Shapiro, Slotnick, On the mathematical theory of error correcting codes, IBM Journal, January 1959 (available through IEEE).
Skolem, Chowla, Lewis, The diophantine equation $2^{n+2} - 7 = x^2$ and related problems, Proc. AMS 10, 1959.
van Lint, A survey of perfect codes, Rocky Mountain J. Math. 5, 1975.
|
24
|
https://mathoverflow.net/users/14807
|
413024
| 168,520 |
https://mathoverflow.net/questions/413020
|
5
|
1. Consider the co-monad $M:=\Sigma^\infty \Omega^\infty$ on the category of spectra.
It is clear that given a pointed space $X$, $M\Sigma^\infty X=\Sigma^\infty E(X)$,where $E(X)$ is the free unital $E\_\infty$-algebra on $X$.
As $\Sigma^\infty$ commutes with colimit, $M\Sigma^\infty X$ is equivalent to
the free $E\_\infty$-algebra on the spectrum $\Sigma^\infty X$, that is,
to: $$
\lor\_n (\wedge^n (\Sigma^\infty X)\_{hS\_n}).\tag{1}
$$
Is this right?
2. It looks like given an arbitrary spectrum $A$, there is no equivalence between
$\Sigma^\infty\Omega^\infty A$ and $
\lor\_n (\wedge^n (A)\_{hS\_n})
$
Is there any general condition for having such an equivalence?
|
https://mathoverflow.net/users/170573
|
On the endofunctor $\Sigma^\infty\Omega^\infty$
|
For connected spectra $A$, there is an equivalence
$$\Sigma^\infty \Omega^\infty A \simeq \bigvee\_{n=1}^\infty A^{\wedge n}\_{h\Sigma\_n}$$
if and only if $A$ is a wedge summand of a suspension spectrum.
This is Theorem 1.2 in N. Kuhn's paper *Suspension Spectra and Homology Equivalences* (TAMS, 1983). Kuhn calls spectra that are wedge summands of suspension spectra *spacelike*.
I am guessing that if $A$ is not connected then such an equivalence can not exist. It is easy to see that there can not be an equivalence of ring spectra.
|
4
|
https://mathoverflow.net/users/6668
|
413030
| 168,524 |
https://mathoverflow.net/questions/412984
|
2
|
A uniformly random $r$-regular bipartite graph on $n$ vertices is known to be $r$-edge connected. That is, with high probability as $n$ grows large, the minimum size of a cut in a random $r$-regular bipartite graph is $r$.
I was wondering if there is a similar statement for the following small extension: Is a uniformly random $(r+1,r)$-biregular bipartite graph almost surely $r$-edge connected?
An $(r+1,r)$-biregular bipartite graph is a bipartite graph where the left side vertices all have degree $r+1$, and the right side vertices have degree $r$.
It seems very reasonable that this extension is true since the graph has a higher density of edges than the $r$-regular case.
Thank you
|
https://mathoverflow.net/users/256325
|
Is a random $(r+1,r)$-biregular bipartite graph $r$-edge connected w.h.p?
|
The problem can be solved for $r\geq7$ by the following 2nd-eigenvalue results.
The first is from the paper [Edge-Disjoint Spanning Trees, Edge Connectivity, and Eigenvalues in Graphs](https://onlinelibrary.wiley.com/doi/abs/10.1002/jgt.21857).
>
> Theorem 1.6. Let $k$ be an integer with $k \geq 2$ and $G$ be a graph with minimum degree $\delta \geq k$. If $\lambda\_2(G)<δ − 2(k−1)/(δ+1)$, then the edge connectivity of $G$ is at least $k$.
>
>
>
In our case we have $\delta=k=r$, so we need $\lambda\_2(G) < r-2(r-1)/(r+1)$.
The second is from the paper [Spectral gap in random bipartite biregular graphs and applications](https://arxiv.org/abs/1804.07808).
>
> **Theorem 4** (Spectral gap). Let $A$ be the adjacency matrix of a bipartite, biregular random graph uniformly sampled from all biregular graphs with $n$ and $m$ vertices for each part and degrees $d\_1$, $d\_2$. Without loss of generality, assume $d\_1 \geq d\_2$ or, equivalently, $n \leq m$.
> Then:
> (i) Its second largest eigenvalue $\lambda\_2(A)$ satisfies $\lambda\_2(A) \leq \sqrt{d\_1-1}+\sqrt{d\_2-1} + \epsilon\_n'$ a.a.s with $\epsilon\_n' \rightarrow 0$ as $n \rightarrow \infty$.
>
>
>
Here we have $d\_1=r+1$ and $d\_2=r$, so we have $\lambda\_2(A) \leq \sqrt{r}+\sqrt{r-1}+\epsilon\_n'$ a.a.s. This bound guarantees $r$-edge-connectivity when $r \geq 7$.
|
1
|
https://mathoverflow.net/users/125498
|
413035
| 168,526 |
https://mathoverflow.net/questions/413040
|
5
|
Does the 2-category of Grothendieck topoi have exponential objects?
There are size issues: Since Grothendieck topoi are supposed to have a small set of generators, the collection of objects of a Grothendieck topos has to be a *class* and cannot be a [conglomerate](https://en.wikipedia.org/wiki/Conglomerate_(mathematics)). However, classes (in contrast to conglomerates) don't have nice closure properties: if $A$ and $B$ are classes, then the conglomerate $B^A$ is, in general, *not* a class.
So for the purposes of my question, drop the small set of generators from the Giraud axioms. :-)
|
https://mathoverflow.net/users/473904
|
Does the 2-category of topoi have exponential objects?
|
No. [Some, but not all topoi are exponentiable](https://ncatlab.org/nlab/show/exponentiable+topos).
|
11
|
https://mathoverflow.net/users/2362
|
413041
| 168,528 |
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