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https://mathoverflow.net/questions/406308
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1
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Let $W$ be a one dimensional standard Brownian motion, and $\sigma: [0, \infty) \to \mathbb R$ a Borel function with $c < \sigma < C$ for some constants $c, C > 0$.
Does there exist some $M > 0$ such that, conditional on $W\_t > -1$ for all $0 \leq t \leq 1$ and $W\_1 > M$,
$\int\_0^1 \sigma(t) \, dW\_t \geq 0$, almost surely?
**Remark:** The statement would be true if the integrator had uniformly bounded variation, but I suspect it does not hold for Brownian motion which has unbounded variation.
|
https://mathoverflow.net/users/173490
|
Conditioning the stochastic integral on the driving Brownian motion being large
|
No such $M$ exists for the following $\sigma$. Partition $[0,1]$ into countably many intervals, with endpoints $t\_0=0,t\_1,t\_2,...$ and let $\sigma$ take value 1 on the odd ones and value 2 on the even ones. Given any $M$, the following event $A\_M$ has positive probability:
$A\_M$ requires that the increments $W\_{t\_k}-W\_{t\_{k-1}}$ are in $(9,10)$ for the first $M$ odd values of $k$, and in $(-6,-7)$ for the first $M$ even values of $k$, with $W\_t-W\_{t\_{k-1}}>-1$ for $t \in [t\_{k-1},t\_k]$ when $0<k<2M$.
Then given $A\_M$, we have $W\_{t\_{2M}}>2M$, so the event $W\_1>M$ holds with high probability, yet on $A\_M$, the stochastic integral considered is $<-2M$ if you integrate up to $t\_{2M}$, and the integral over $[t\_{2M},1]$ is likely to be $<2M$.
|
1
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https://mathoverflow.net/users/7691
|
406377
| 166,562 |
https://mathoverflow.net/questions/406373
|
0
|
I fairly new in the field of Stochastic Processes and Markov Chains so excuse my ignorance.
My question is: If we have a sequence of Markov chains such that each one has a stationary distribution $\pi^{(n)}$ and the chains converge in some way to another Markov chain that has stationary distribution $\pi$, can we say that the $\pi^{(n)}$'s converge to $\pi$ (in some way)?
More precisely:
Let $G$ be a simple (ie no loops or multiple edges), finite, connected graph. Suppose that we have a sequence of Markov chains over $G$. Let $\boldsymbol{P}\_1, \boldsymbol{P}\_2, \dots$ denote the corresponding transition matrices. Assume that all chains have a stationary distribution (for example, this can be guaranteed when the weights on each edge are positive since $G$ is connected), call them $\pi^{(n)}$. Now say that $\boldsymbol{P}\_n\to\boldsymbol{P}$ in some way (for example, let's say that we have entry-wise almost sure convergence, or $\|\boldsymbol{P}\_n-\boldsymbol{P}\|\to 0$ for some matrix norm). Suppose that $\boldsymbol{P}$ is a stochastic matrix with stationary distribution $\pi$. Then can we say that $\pi^{(n)}\to\pi$ in some way (similar to the way that the matrices converge)?
My feeling is that there should exist such theorems (maybe with some stronger assumptions). I tried to find such results but I was not successful. Can someone give a reference about such results?
|
https://mathoverflow.net/users/416334
|
Convergence of stationary distributions of a sequence of Markov Chains
|
We assume that the Markov chains are on a finite state space, that $P\_n \to P$ pointwise, and the limit matrix $P$ is irreducible, so its stationary measure $\pi$ is unique. Let $\pi^{(n\_k)} \to \mu$ be a convergent subsequence of $\pi^{(n)}$. Then $\pi^{(n\_k)}P\_{n\_k}=\pi^{(n\_k)}$, so continuity of multiplication implies that $\mu P=\mu$. Thus $\mu=\pi$. Since this holds for every convergent subsequence and the simplex of probability vectors is compact, we conclude that $\pi^{(n)} \to \pi$.
|
5
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https://mathoverflow.net/users/7691
|
406382
| 166,565 |
https://mathoverflow.net/questions/406381
|
3
|
A **normed division algebra over $\mathbb{R}$** is a pair $(A,\lVert{-}\rVert)$ with
* $A$ an $\mathbb{R}$-algebra with a unit $1\_A$;
* $\lVert{-}\rVert\colon A\to\mathbb{R}\_{\geq0}$ a norm on $A$;
such that:
* For each $a\in A$, there exists a unique $a^{-1}\in A$ such that $a^{-1}a=1\_A=aa^{-1}$;
* For each $x,y\in A$, we have
$$\lVert xy\rVert=\lVert x\rVert\lVert y\rVert.$$
**Hurwitz's theorem** then states that the only associative normed division algebras over $\mathbb{R}$ are the real numbers $\mathbb{R}$, the complex numbers $\mathbb{C}$, and the quaternions $\mathbb{H}$.
---
The isomorphisms classes of associative division algebras over $\mathbb{Q}\_p$ [are classified](https://math.stackexchange.com/a/321519) by $\mathrm{Br}(\mathbb{Q}\_p)\cong\mathbb{Q}/\mathbb{Z}$. Is there a classification of the associative *normed* division algebras over $\mathbb{Q}\_p$?
(E.g. is $\mathbb{C}\_p$ such an algebra? Is there a normed division algebra over $\mathbb{Q}\_p$ which we could call the "$p$-adic quaternions $\mathbb{H}\_p$"?)
|
https://mathoverflow.net/users/130058
|
Is there a classification of the $p$-adic normed division algebras?
|
Clearly you are assuming some kind of finite-dimensionality over the center.
To classify the finite-dimensional associative division algebras over $\mathbf Q\_p$, or more generally over a local field, it's standard to fix the center. A *$K$-central algebra* here will mean a $K$-algebra whose center is $K$, so $\mathbf C$ and ${\rm M}\_2(\mathbf C)$ are each $\mathbf C$-central algebras, not $\mathbf R$-central algebras. And rather than say each finite-dimensional associative $\mathbf R$-division algebra (meaning $\mathbf R$ is *in* the center but possibly not the whole center) has to be $\mathbf R$, $\mathbf C$, or $\mathbf H$, it would be better to say the only finite-dimensional associative $\mathbf R$-central division algebras are $\mathbf R$ and $\mathbf H$, while the only finite-dimensional associative $\mathbf C$-central division algebra is $\mathbf C$.
Associative division algebras having center equal to a local field and being finite-dimensional over the center are discussed in Pierce's book *Associative Algebras*. Chapter 15 is on cyclic division algebras and chapter 17 is on division algebras over local fields. Don't expect an easy method to determine which cyclic algebras are division algebras in general, but it's possible to give an easy method "in principle" in special cases. For instance, if $A$ is an $F$-central simple algebra with $\dim\_F(A) = p^2$ for a prime number $p$, then either $A \cong {\rm M}\_2(F)$ or $A$ is a division ring. In practice you may need to some work to figure out if such a central simple algebra given to you in an abstract form is or is not the matrix ring.
For a field $F$, a *quaternion algebra over* $F$ is defined to be an $F$-central simple algebra of dimension $4$. An example is ${\rm M}\_2(F)$, and sometimes it is the only example ($F = \mathbf C$ and $F$ finite). We call ${\rm M}\_2(F)$ the "split" or "trivial" quaternion algebra over $F$. All other quaternion algebras over $F$ are division rings, and when $F$ is $\mathbf Q\_p$ or any other local field there is one nontrivial quaternion algebra over $F$. Over $\mathbf Q\_2$ this algebra is $\mathbf H(\mathbf Q\_2)$, but for $p > 2$ we have $\mathbf H(\mathbf Q\_p) \cong {\rm M}\_2(\mathbf Q\_p)$. A uniform description of the nontrivial quaternion algebra over $\mathbf Q\_p$ for all $p$ uses a cyclic algebra construction based on the quadratic unramified extension of $\mathbf Q\_p$. (**Note**: there are *infinitely many* nonisomorphic quaternion algebras over $\mathbf Q$. The contrast between that and finiteness of the number of quaternion algebras over $\mathbf R$ and $\mathbf Q\_p$ is analogous to the contrast with quadratic extension fields: $\mathbf R$ and each $\mathbf Q\_p$ have only finitely many quadratic extension fields up to isomorphism, while $\mathbf Q$ has infinitely many.)
You could call the unique nontrivial quaternion algebra over $\mathbf Q\_p$ "the $p$-adic quaternions" but that label is not standard. It's more often called the nonsplit or nontrivial quaternion algebra over $\mathbf Q\_p$.
The recent [book](https://math.dartmouth.edu/%7Ejvoight/quat-book.pdf) by John Voight on quaternion algebras has an account on quaternion algebras over local fields in Chapter 13.
A brief account on the history of the quaternion algebra construction over fields other than $\mathbf R$ is described [here](https://mathoverflow.net/questions/156338/history-of-quaternion-algebras).
|
9
|
https://mathoverflow.net/users/3272
|
406384
| 166,566 |
https://mathoverflow.net/questions/406293
|
12
|
Let $\mathbb{C}P^n$ be the $n$-dimensional complex projective space and denote $[z\_0:\dots:z\_n]$ its points. If we glue $[z\_0:\dots:z\_n]$ and $[\overline{z\_0}:\dots:\overline{z\_n}]$ for any $[z\_0:\dots:z\_n]\in\mathbb{C}P^n$, where $\overline{z}$ denotes the complex conjugation of $z$, then we obtain a quotient space $\overline{\mathbb{C}P^n}:=\mathbb{C}P^n/[z\_0:\dots:z\_n]\sim[\overline{z\_0}:\dots:\overline{z\_n}]$.
Now I am interested in these $\overline{\mathbb{C}P^n}$'s, but I know neither if these $\overline{\mathbb{C}P^n}$'s have a formal name nor where I can find their properties.
I'd appretiate if you can provide some relavant references.
|
https://mathoverflow.net/users/392187
|
A quotient space of complex projective space
|
This answer gives information about the cohomology of $\overline{\mathbb CP^n}$. Perhaps someone will recognize this as the cohomology of a familiar space.
The conjugation is an action of $\Sigma\_2$ on $\mathbb CP^n$. We are interested in the orbit space of this action. Recall that the fixed point of this action is homeomorphic to $\mathbb RP^\infty$.
The quotient space $\mathbb CP^n/\mathbb RP^n$ has a free action of $\Sigma\_2$ (away from the basepoint). The fact that for free action strict quotient is equivalent to homotopy quotient implies that there is a homotopy pushout square
$$
\begin{array}{ccc}
\mathbb RP^n \times \mathbb RP^\infty & \to & \mathbb RP^n \\
\downarrow & & \downarrow \\
\mathbb CP^n \times\_{\Sigma\_2} E\Sigma\_2 & \to & \overline{\mathbb CP^n}
\end{array}
$$
It is perhaps instructive to consider what happens when $n=\infty$. One can realize the limit diagram as a diagram of classifying spaces:
$$
\begin{array}{ccc}
B\Sigma\_2\times B\Sigma\_2 & \to & B\Sigma\_2 \\
\downarrow & & \downarrow \\
BO(2) & \to & \overline{\mathbb CP^\infty}
\end{array}
$$
From this square one can, in principle, calculate the cohomology of $\overline{\mathbb CP^n}$ using the Meier-Vietoris exact sequence. If one localizes away from the prime $2$, the answer is pretty simple. In this case $\mathbb RP^\infty\simeq \*$, and we obtain an equivalence away from the prime $2$:
$$
\mathbb CP^n \times\_{\Sigma\_2} E\Sigma\_2 \xrightarrow{\simeq} \overline{\mathbb CP^n}.
$$
If $\Lambda$ is a ring where $2$ is invertible, then $H^\*(BO(2);\Lambda)\cong \Lambda[p\_1]$, where $p\_1$ is a class in dimension $4$ (the first Pontryagin class). The cohomology of $\mathbb CP^n \times\_{\Sigma\_2} E\Sigma\_2$ is isomorphic to the truncation $\Lambda[p\_1]/\_{(p\_1^{\left\lfloor \frac{n}{2}\right\rfloor+1})}$. Notice that for $n=1$ you get the cohomology of a point, and for $n=2$ you get the cohomology of a sphere, as expected.
Cohomology with mod 2 coefficients is more complicated/interesting. In the limit when $n=\infty$ you get the following diagram in cohomology
$$
\begin{array}{ccc}
H^\*(\overline{\mathbb CP^\infty};\mathbb Z/2) & \to & \mathbb Z/2[w\_1, w\_2] \\
\downarrow & & \downarrow \\
\mathbb Z/2[x] & \to &\mathbb Z/2[x, y]
\end{array}
$$
Where the homomorphism on the right side sends $w\_1$ to $x+y$ and $w\_2$ to $xy$. The Poincare series of $\tilde H^\*(\overline{\mathbb CP^\infty};\mathbb Z/2)$ comes out to be $\frac{t^4}{(1-t)(1-t^2)}$.
For finite $n$, we have to take a truncation of this diagram. If I am not mistaken, the diagram in cohomology comes out to be the following
$$
\begin{array}{ccc}
H^\*(\overline{\mathbb CP^n};\mathbb Z/2) & \to & \mathbb Z/2[w\_1, w\_2]/\_{(w\_2^{n+1})} \\
\downarrow & & \downarrow \\
\mathbb Z/2[x]/\_{(x^{n+1})} & \to &\mathbb Z/2[x, y]/\_{(x^{n+1})}
\end{array}
$$
According to my calculations, the Poincare series of $\tilde H^\*(\overline{\mathbb CP^n};\mathbb Z/2)$
turns out to be $$t^4\frac{(1-t^{n-1})(1-t^n)}{(1-t)(1-t^2)}.$$
More explicitly, this equals to
$$t^4(1+t+t^2+ \cdots +t^{n-2})(1+t^2+t^4+\cdots +t^{n-2})$$ if $n$ is even, and $$t^4(1+t^2+t^4+\cdots + t^{n-3})(1+t+t^2+\cdots +t^{n-1})$$ if $n$ is odd. Once again, this seems to give the right answer for $n=1, 2$, so I hope this is a good sign.
|
12
|
https://mathoverflow.net/users/6668
|
406389
| 166,569 |
https://mathoverflow.net/questions/406288
|
1
|
\begin{gather\*}
M\_g=(x\_1\times x\_2\times\dotsb\times x\_n)^{1/n} \\
M\_a=\frac1 n\times (x\_1+x\_2+\dotsb+x\_n).
\end{gather\*}
Is it true that $$\lvert M\_g-M\_a\rvert \leq (\max(x\_i) /\min(x\_i)) \times(\max(x\_i) - \min(x\_i))?\label{1}\tag{1}$$
And is it true that
$$\lvert M\_g-M\_a\rvert\leq (\max(x\_i) /\min(x\_i)) \times\left( \frac1{n^2}\sum \limits \_{(i, j) \in \{1,\dotsc,n\}^2} \lvert x\_i-x\_j\rvert\right)?\label{2}\tag{2}$$
Ps.: $x\_i\in\mathbb R\_+^\*$.
|
https://mathoverflow.net/users/110301
|
Compare AM and GM
|
The inequality $(2)$ (even with factor $\frac12$ in the r.h.s.) follows from the inequality quoted in [this answer](https://mathoverflow.net/q/215094):
$$M\_a - M\_g \leq \frac1{2n\min\_k x\_k} \sum\_{i=1}^n (x\_i - M\_a)^2.$$
First, we notice that
\begin{split}
(x\_i - M\_a)^2 &\leq \max\_k x\_k\cdot |x\_i-M\_a|\\
&= \max\_k x\_k\cdot \left|\frac1n \sum\_{j=1}^n (x\_i - x\_j)\right| \\
& \leq \frac{\max\_k x\_k}n \sum\_{j=1}^n |x\_i - x\_j|.
\end{split}
Then
$$\frac1{2n\min\_k x\_k} \sum\_{i=1}^n (x\_i - M\_a)^2 \leq \frac12 \frac{\max\_k x\_k}{\min\_k x\_k} \frac1{n^2} \sum\_{i,j=1}^n |x\_i - x\_j|.$$
QED
|
3
|
https://mathoverflow.net/users/7076
|
406390
| 166,570 |
https://mathoverflow.net/questions/406368
|
9
|
Let $G$ be a finitely presented group and let $L(G)$ be the Magnus Lie algebra associated to the lower central series of $G$. This $L(G)$ is a graded Lie ring generated by its degree 1 piece $L\_1(G) = G^{ab}$. Must $L(G)$ be a finitely presented Lie ring? If it makes it easier, I would be happy to tensor with the rationals (though I would prefer not to).
|
https://mathoverflow.net/users/416222
|
Is the Magnus Lie algebra of a finitely presented group finitely presented
|
$\newcommand{\Z}{\mathbf{Z}}$No.
Take the Baumslag-Solitar group $$G=\mathrm{BS}(1,3)=\langle t,x\mid txt^{-1}=x^3\rangle=\mathbf{Z}\ltimes\_3\mathbf{Z}[1/3]$$
then the lower central series satisfies $G^1=G$, $G^i=2^{i-1}\mathbf{Z}[1/3]$ for all $i\ge 2$, which has index $2^{i-1}$ in $\mathbf{Z}[1/3]$. So $G^1/G^2\simeq\Z$ and $G^i/G^{i+1}\simeq\Z/2\Z$ for all $i\ge 2$. For $i\ge 2$ define $x\_i$ as the nontrivial element of $G^i/G^{i+1}$, which has degree $2$ in the Magnus Lie algebra. Then in the Magnus Lie algebra all $x\_i$ commute, and $[t,x\_i]=x\_{i+1}$.
This Lie $\Z$-algebra is not finitely presented (even after tensoring by $\Z/2\Z$). Indeed for every odd $i\ge 5$, we can define a central extension of this Lie algebra with a central element $z\_i$ of degree $i$, with nontrivial additional brackets $[x\_j,x\_k]=z\_i$ if $j+k=i$ (this was essentially defined by Vergne a while ago). So $H\_2$ is infinite-dimensional, since it's nonzero in each odd degree $\ge 5$.
(I don't have, for the moment, an example for which it would remain infinitely presented after tensoring with $\mathbf{Q}$.)
|
11
|
https://mathoverflow.net/users/14094
|
406397
| 166,574 |
https://mathoverflow.net/questions/406385
|
3
|
In Grunbaum's paper, [A result on graph coloring](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-15/issue-3/A-result-on-graph-coloring/10.1307/mmj/1029000043.full?tab=ArticleLinkCited), the following conjecture was posed:
>
> Let $G$ be a graph with $n$ nodes with $\Delta(G) < k$. There exists a proper $k$-coloring $c:V(G)\to [k]$ such that for each color $i\in [k]$, we have $\lfloor n/k\rfloor \le |c^{-1}(i)| \le \lfloor n/k\rfloor+1$.
>
>
>
The aforementioned paper demonstrated it was true for $k \le 4$, and mentions an upcoming paper proving $k = \lfloor k/3\rfloor,\lfloor k/2\rfloor$. I was wondering if this problem has received any further attention.
I feel that recently there has been a lot of results about applying Lovász Local Lemma to get specialized colorings of bounded degree graphs (such as frugal colorings and DP-colorings). I would assume that these methods would be well-suited to at least getting asymptotic results for the above conjecture. Has this been attempted/considered recently?
|
https://mathoverflow.net/users/130484
|
Proper graph colorings with similar sized color classes
|
As mentioned in Grunbaum's paper, this is a conjecture of Erdős from 1964. It was solved completely by Hajnal and Szemerédi in 1970 and is now known as the Hajnal-Szemerédi theorem. A simpler proof was given by Kierstead and Kostochka in 2008. See the wikipedia page on [equitable coloring](https://en.wikipedia.org/wiki/Equitable_coloring) for more information and links.
|
5
|
https://mathoverflow.net/users/2233
|
406398
| 166,575 |
https://mathoverflow.net/questions/406414
|
1
|
In the question ([$C(X)$ as finitely generated $C^\*$-algebra](https://mathoverflow.net/questions/233512/cx-as-finitely-generated-c-algebra)), the answer show that spectrum of an abelian unital finitely generated C\*-algebra is homeomorphic to compact subset of $\mathbb{C}^{n}$. I would like to know more about the details of the proof of this fact. Also I would like to know, if the number of generators is related to the $n$ in anyway?
Thank you in advance.
|
https://mathoverflow.net/users/172458
|
finitely generated C*-algebra as $C(X)$
|
Let $X$ be a compact Hausdorff space. Let $C(X)$ be the set of all continuous functions $f:X\rightarrow\mathbb{C}$. I claim that $X$ embeds into $\mathbb{C}^{n}$ if and only if the $C^{\*}$-algebra $C(X)$ is generated by $n$ functions. To prove this fact, we need to the complex Stone Weierstrass theorem.
We say that a subset $\mathcal{A}\subseteq C(X)$ separates points if whenever $x,y\in X,x\neq y$, there is some $f\in\mathcal{A}$ with $f(x)\neq f(y)$.
Theorem: (Complex Stone Weierstrass theorem) Suppose that $\mathcal{A}$ is a closed dense subalgebra of $C(X)$ (here $C(X)$ is given the topology induced by the supremum metric) that separates points and where if $f\in\mathcal{A}$, then $\overline{f}\in\mathcal{A}$. Then $\mathcal{A}=C(X)$.
Suppose that $C(X)$ is generated as a $C^{\*}$-algebra by $f\_{1},\dots,f\_{n}:X\rightarrow\mathbb{C}$. Then whenever $x,y\in X,x\neq y$, there is some $i$ with $f\_{i}(x)\neq f\_{i}(y)$. Therefore, the mapping
$f\_{1}\times\dots\times f\_{n}:X\rightarrow\mathbb{C}^{n}$ is continuous and injective, and since $X$ is compact, the mapping $f\_{1}\times\dots\times f\_{n}$ is an embedding.
Suppose to the contrary that $f\_{1}\times\dots\times f\_{n}:X\rightarrow\mathbb{C}^{n}$ is an embedding. Then whenever $x,y\in X$, there is some $i$ with $f\_{i}(x)\neq f\_{i}(y)$, so $f\_{1},\dots,f\_{n}$ satisfies the conditions of the complex Stone Weierstrass theorem. We conclude that $f\_{1},\dots,f\_{n}$ generates the $C^{\*}$-algebra $C(X)$.
More generally, if $\kappa$ is a cardinal, then $C(X)$ is generated by at most $\kappa$ many functions if and only if $X$ embeds into $\mathbb{C}^{I}$ for some set $I$ with $|I|\leq\kappa$, but the the smallest infinite $|I|$ where $X$ embeds into $\mathbb{C}^{I}$ is actually a well-known cardinal invariant.
Proposition: Suppose $X$ is an infinite compact Hausdorff space. Then the least cardinal $\kappa$ such that $C(X)$ is generated by at most $\kappa$ many functions is precisely $w(X)$ ($w(X)$ is the size of the smallest basis of $X$).
|
3
|
https://mathoverflow.net/users/22277
|
406425
| 166,584 |
https://mathoverflow.net/questions/406427
|
13
|
Is the consistency of classical third-order arithmetic provable in the logic of a topos with natural numbers?
(My guess would be yes, but I haven't seen this anywhere.)
Edit: in the original version I used the name PA$\_3$ as an abbreviation for classical third-order arithmetic, and comments have followed suit, but I've since learnt that this name refers to a different theory (the classical theory of third order functions).
|
https://mathoverflow.net/users/170446
|
Consistency proof in topos logic
|
Let $\Omega\_{\neg\neg} = \{p \in \Omega \mid \neg\neg p \Rightarrow p\}$ be the object of $\neg\neg$-stable truth values, and let us write $P\_{\neg\neg}(A) = {\Omega\_{\neg\neg}}^A$ for the object of $\neg\neg$-stable subobjects of $A$. Observe that $\Omega\_{\neg\neg}$ is a complete Boolean algebra, and this fact can be shown in the internal logic of a topos.
Now, it seems to me that one can build a model of $\mathsf{PA}\_n$ for each $n$, by interpreting its logic in $\Omega\_{\neg\neg}$, and the (higher-order) predicates as elements of the iterates of the $\neg\neg$-powersets ${P\_{\neg\neg}}^k(\mathbb{N})$.
**Supplemental:** I initially claimed we can have a model of $\mathsf{PA}\_\infty$, but Paul pointed out I was overstating the case. Indeed, to have a model of $\mathsf{PA}\_\infty$ we would need a single object that encompasses all finite iterates $\neg\neg$-powersets ${P\_{\neg\neg}}^k(\mathbb{N})$ at once, says something like $\coprod\_{n \in \mathbb{N}} {P\_{\neg\neg}}^k(\mathbb{N})$. However, in an elementary topos such an object need not exist. A counter-model is the set $V\_{\omega + \omega}$ of sets of rank below $\omega + \omega$.
|
17
|
https://mathoverflow.net/users/1176
|
406445
| 166,595 |
https://mathoverflow.net/questions/406367
|
25
|
Is there a result showing that something along the lines of the three body problem is undecidable? Or are they known to be decidable or neither?
I mean problems along the lines of the following formulated in some suitable system:
Given masses, velocities and positions in 3 dimensions and a distance d (assume all expresses in rational multiples of G...ie G=1 so no using G as a non-computable oracle) can one decide whether, acting under the influence of Newtonian gravity only,
1. Any of the point masses get within d units of another.
2. Whether any of the masses ever get beyond d units from one of the other masses.
3. Any of the bodies escapes to infinity relative to one of the others.
---
If necessary to make the problem well-defined one could stipulate that the initial positions are choosen to avoid ever allowing an exact collision of the point particles (also wonder if that is decidable).
---
More generally is their some result letting one embed arbitrary computations into a system of bodies acting only under gravity?
|
https://mathoverflow.net/users/23648
|
Decidability of 3 body problem
|
The paper [Undecidability in $\mathbb{R}^n$: Riddled Basins,
the KAM Tori, and the Stability of the Solar System](http://philsci-archive.pitt.edu/13175/1/parker2003.pdf) by Matthew W. Parker (*Philosophy of Science* **70** (April 2003), 359–382) comes close to answering your question. A classical problem in the same spirit as your question is the [stability of the solar system](https://en.wikipedia.org/wiki/Stability_of_the_Solar_System). As Parker notes, there are quite a few informal claims in the literature that this type of problem is uncomputable, but for the most part, they gloss over the crucial question of how to define computability in the context of real numbers (sometimes referred to as *[real computation](https://en.wikipedia.org/wiki/Real_computation)* or *[computable analysis](https://en.wikipedia.org/wiki/Computable_analysis)*).
Parker offers his own approach to real computation and analyzes the question accordingly. However, as far as I know, neither Parker nor anyone else has analyzed the setup you suggested, which is to restrict the set of initial conditions to the rational numbers, and ask if the subset of (say) "stable configurations" is computable in the traditional sense.
|
14
|
https://mathoverflow.net/users/3106
|
406449
| 166,596 |
https://mathoverflow.net/questions/406424
|
1
|
Is there a solution to this integral?
$$\int\_{0}^{y} x^{-a} \exp \left[- \frac{(b - cx^{-d})^2}{2} \right] dx,$$
where $a > 0$ and $d > 0$.
|
https://mathoverflow.net/users/103291
|
Solution to $\int_{0}^{y} x^{-a} \exp \left[- \frac{(b - cx^{-d})^2}{2} \right] dx$
|
By using series expansion, change of variable and Eq. (3.381.9) of the book: "I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, 8th ed. Burlington, MA, USA: Academic Press, 2015", I was able to find this solution:
$$\int\_{0}^{y} x^{-a} \exp \left[- \frac{(b - cx^{-d})^2}{2} \right] dx
\\= \exp(-b^2/2) \sum\_{k=0}^{\infty} \frac{(bc)^k}{k!} \Gamma\left(\frac{dk+a-1}{2d}, \frac{c^2 y^{-2d}}{2} \right)\frac{1}{2d(c^2/2)^{\frac{dk+a-1}{2d}}}.$$
I've run some simulations and it works.However, there might be some convergence problem for some values, I think so.
|
3
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https://mathoverflow.net/users/103291
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406452
| 166,597 |
https://mathoverflow.net/questions/406451
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11
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$\newcommand{\Grp}{\mathrm{Grp}}$Consider the category of groups $\Grp$, and within it we have the solvable groups $S$. Then any group $G$ determines the functor from solvable groups: $$h\_G:=\text{hom}\_{\Grp}(\\_,G):S^{\mathrm{op}}\rightarrow \text{Set}$$
Does this functor determine the group $G$? More concretely, if we have a natural bijection $$\text{hom}\_{\Grp}(A,G)\cong \text{hom}\_{\Grp}(A,G')$$ for all solvable groups $A$, must this be induced by an isomorphism $G\rightarrow G'$?
Are there any circumstances/more restrictive hypotheses where the answer to this is known to be affirmative, for conceptual reasons?
By other circumstance, I mean imposing finiteness conditions, or other "well behavior conditions", or looking at group objects in a more exotic category, etc. I mean conceptual reasons in the sense of being independent of a classification result of all the objects involved, it wouldn't surprise me if this result were true for finite groups, but only verifiable by induction/case checking for the simple groups. Though interesting, I am more interested in any setting where we have a well understood reason for this to hold, or counterexamples/obstacles to its potential truth.
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https://mathoverflow.net/users/128502
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Are groups determined by their morphisms from solvable groups?
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Let $G, G'$ be two non-isomorphic Tarski monsters of prime exponent $p$ or two non-isomorphic torsion-free Tarski monsters. Then for every solvable group $A$, $\mathbb{hom}(A,G)\cong \mathbb{hom}(A,G')$.
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24
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https://mathoverflow.net/users/157261
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406454
| 166,598 |
https://mathoverflow.net/questions/406458
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2
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Consider [*General Set Theory*](https://en.wikipedia.org/wiki/General_set_theory) ($ \mathsf { GST } $) axiomatized by the following.
1. **Axiom of Extensionality:** The sets $ x $ and $ y $ are the same set if they have the same members:
$$ \forall x \forall y \bigl ( \forall z ( z \in x \leftrightarrow z \in y ) \rightarrow x = y \bigr ) \text . $$
2. **Axiom of Adjunction:** If $ x $ and $ y $ are sets, then there exists a set whose members are just $ y $ and the members of $ x $:
$$ \forall x \forall y \exists z \forall w ( w \in z \leftrightarrow w \in x \lor w = y ) $$
3. **Axiom Schema of Separation:** If $ x $ is a set and $ \phi $ is any property, then there exists a set $ y $ containing just those elements $ z $ in $ x $ which satisfy the property $ \phi $:
$$ \forall x \exists y \forall z \bigl ( z \in y \leftrightarrow z \in x \land \phi ( z ) \bigr ) \text . $$
It's rather straightforward to verify that $ \mathsf { GST } $ proves the existence of every hereditarily finite set; i.e. $ \mathsf { GST } \vdash \exists ! x \, \phi ( x ) $ for some formula $ \phi $ defining the intended set. Since the family of hereditarily finite sets gives a model $ \mathfrak M $ of $ \mathsf { GST } $, those are the only sets that provably exist. My question is the validity of the following statement:
>
> If $ \mathsf { GST } \vdash \exists ! x \, \phi ( x ) $, $ \mathsf { GST } \vdash \exists ! x \, \psi ( x ) $ and $ \mathfrak M \models \forall x \forall y \bigl ( \phi ( x ) \land \psi ( y ) \rightarrow x = y \bigr ) $, then $ \mathsf { GST } \vdash \forall x \forall y \bigl ( \phi ( x ) \land \psi ( y ) \rightarrow x = y \bigr ) $.
>
>
>
My motivation is the fact that $ \mathsf { GST } $ is mutually interpretable with $ \mathsf { PA } $. For $ \mathsf { PA } $ (and even much weaker arithmetical theories like Robinson's) we know the following:
1. For every $ n \in \mathbb N $ there is a numeral $ \overline n $ in the language, which is interpreted as $ n $ in the standard model.
2. As $ \mathbb N $ gives a model of $ \mathsf { PA } $, natural numbers are the only objects that $ \mathsf { PA } $ proves to exist.
3. The equality relation between natural numbers is representable in $ \mathsf { PA } $ in the sense that for any $ m , n \in \mathbb N $, $ m = n $ implies $ \mathsf { PA } \vdash \overline m = \overline n $, and $ m \ne n $ impiles $ \mathsf { PA } \vdash \neg \, \overline m = \overline n $.
My question stated above is about representability of equality of hereditarily finite sets in $ \mathsf { GST } $. The similarities explained above make me feel that it must be true.
I understand that $ \mathsf { GST } $ might not be strong enough for proving the desired representability, and one might need to strengthen it (as the family of hereditarily finite sets gives a model for $ \mathsf { ZFC } $ without infinity, we can see that much stronger candidates are available). But since very weak arithmetical theories are enough for proving representability of equality for natural numbers, I suspect that no additional strength would be necessary for set theoris either.
---
**P.S.** I've only seen the claim that $ \mathsf { GST } $ and $ \mathsf { PA } $ are mutually interpretable in a couple of sources. I don't know of any sources where a proof can be found. I'd really appreciate it if someone could give a reference in the comments section.
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https://mathoverflow.net/users/76416
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Representation of the equality relation between hereditarily finite sets in weak set theories
|
This question is based on a misunderstanding of the situation already in $\mathsf{PA}$: equality is representable in $\mathsf{PA}$ only for *terms*, or at the very least relatively simple formulas. To see this consider, given an arbitrary sentence $\varphi$, the formula (modulo obvious abbreviations) $$\psi(x)\equiv(x=0\wedge\varphi)\vee(x=1\wedge\neg\varphi).$$ Then regardless of what $\varphi$ is we have $\mathsf{PA}\vdash\exists!x\psi(x)$, but if $\varphi$ is independent of $\mathsf{PA}$ then $\mathsf{PA}$ cannot decide whether or not (say) $\forall x(\psi(x)\leftrightarrow x=0)$ holds.
Similarly, only a weak form of "representability of equality" will hold in $\mathsf{GST}$. In fact, your guess must fail for any incomplete theory (with at least "two incompatible element-defining formulas" in the obvious sense).
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4
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https://mathoverflow.net/users/8133
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406459
| 166,601 |
https://mathoverflow.net/questions/406407
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1
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I recently saw a question here on mathoverflow: «For what n and t can a square be partitioned into n similar rectangles in t congruence classes?», where Joseph Gordon gave a proof that, indeed, a square can be partitioned into n non-congruent similar rectangles for any $n\ge3$. His method involves the use of Fibonacci-numbers. For n=3, the aspect ratio,r, of the similar rectangles is the square of the plastic number (also known as the Padovan constant).
Using his method I did some calclulations for a few n (stopped at n=16, where I found r approximately 1.6180355) and the results seem to suggest that as n grows larger, the aspect ratio of the similar rectangles tends towards the golden ratio. Can this be proved/disproved in any way?
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https://mathoverflow.net/users/415477
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Tiling a square with similar non-congruent rectangles. What is the aspect ratio of the rectangles as n grows large?
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You are correct.
Let $\varphi$ be the golden ratio and $r\_n$ be the ratio you want (long side to short).
According to the answer you mention, for a certain positive constant $A\_n:$ $$r\_n=\frac{F\_{n-1} A\_n + F\_{n-2}}{F\_{n-2} A\_n + F\_{n-3}}. \tag{\*}$$ It follows that (for odd $n$) $$\frac{F\_{n-1}}{F\_{n-2}} <r\_n < \frac{F\_{n-2}}{F\_{n-3}}. $$ The same is true for even $n$ except the inequalities are reversed. Furthermore, the two rationals differ by exactly $\frac{1}{F\_{n-2}F\_{n-3}}.$
It is also well known that (again, up to reversing the inequalities) $$\frac{F\_{n-1}}{F\_{n-2}} < \varphi < \frac{F\_{n-2}}{F\_{n-3}}. \tag{\*\*}$$ This establishes the result and shows that the convergence is rapid.
---
The exact values of $A\_n$ were not needed here, but in fact they converge quickly to $\frac{\sqrt 5}{\varphi}$ the root of the linear equation $$(A-1)\varphi^2-1.$$
$A\_n$ is the real root of the cubic polynomial $$f\_n(A)=(A-1)(F\_{n-1} A + F\_{n-2})^2 - (F\_{n-2} A + F\_{n-3})^2.$$ We see from $(\*\*)$ that $f\_n$ is almost exactly $$(A-1)(F\_{n-2}\varphi A + F\_{n-3}\varphi)^2 - (F\_{n-2} A + F\_{n-3})^2\\ =\left((A-1)\varphi^2-1\right)(F\_{n-2} A + F\_{n-3})^2.$$
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1
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https://mathoverflow.net/users/8008
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406466
| 166,604 |
https://mathoverflow.net/questions/406485
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2
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A famous theorem by Sumihiro states that, given a normal quasi-projective variety $X$ with a regular $G$-action (where $G$ is a connected linear algebraic group), there is a G-equivariant
projective embedding $$X\hookrightarrow \mathbb{P}^n,$$ where $G$ acts on $\mathbb{P}^n$ linearly.
The closure of the image of this map thus gives a $G$-equivariant completion of $X,$ i.e. a projective variety with a $G$-action which has a $G$-invariant open subvariety isomorphic to $X.$
Now, in the case that the quasiprojective variety $X$ over $\mathbb{C}$ is also nonsingular, and the group $G$ is equal to $\mathbb{C}^\*,$ is it true that one can always find a **nonsingular** $\mathbb{C}^\*$-equivariant completion of $X$? I have seen some authors using this result without explanation/references, so I am wondering where it follows from.
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https://mathoverflow.net/users/114985
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$\mathbb{C}^*$-equivariant smooth completion of a quasiprojective variety
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Since you seem to work in characteristic zero (even over $\mathbb{C}$), you can first take any $\mathbb{C}^\times$-equivariant completion, and then take its **canonical** resolution of singularities (see, e.g., Bierstone, E. and P. Milman, “Functoriality in resolution of singularities”, Research Institute
for Mathematical Sciences Publications, 44, no. 2 (2008): 609–39); it is automatically $\mathbb{C}^\times$-equivariant.
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3
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https://mathoverflow.net/users/4428
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406486
| 166,608 |
https://mathoverflow.net/questions/406343
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9
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We say a complex manifold $X$ has the *resolution property* if every coherent sheaf $\mathcal{M}$ on $X$ admits a surjection $\mathcal{E}\twoheadrightarrow \mathcal{M}$ by some finite rank locally free sheaf $\mathcal{E}$.
It is well-known that any projective manifold has the resolution property ([SGA 6, Expose II, 2.2]). In addition, any smooth compact complex surface has the resolution property too. (Schuster, *Locally free resolutions of coherent sheaves on surfaces*, 1982)
>
>
> >
> > My question is: if $X$ and $Y$ have the resolution property, is it true that $X\times Y$ must have the resolution property? In particular does the product of two smooth compact complex surfaces always have the resolution property?
> >
> >
> >
>
>
>
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https://mathoverflow.net/users/24965
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Does $X\times Y$ have the resolution property if both $X$ and $Y$ have?
|
Please find below a short argument in the case of schemes. The answer is positive for algebraic spaces too; in that case it can be proven using the characterization: $X$ has the resolution property $\Leftrightarrow$ $X = X'/GL\_n$ with $X'$ quasi-affine $+$ products of quasi-affines are quasi-affine $+$ a bit more work.
Lemma. Let $S$ be a separated scheme. Let $X$ and $Y$ be quasi-compact and quasi-separated schemes over $S$. Then if $X$ and $Y$ have the resolution property, so does $X \times\_S Y$.
Proof. Let $X = U\_1 \cup \ldots \cup U\_n$ and $Y = V\_1 \cup \ldots \cup V\_m$ be affine open coverings. Choose finite type quasi-coherent ideals $\mathcal{I}\_i \subset \mathcal{O}\_X$ such that $U\_i = X \setminus V(\mathcal{I}\_i)$. Choose finite type quasi-coherent ideals $\mathcal{J}\_j \subset \mathcal{O}\_Y$ such that $V\_j = Y \setminus V(\mathcal{J}\_j)$. Denote $\mathcal{K}\_{ij} = (\text{pr}\_1^{-1}\mathcal{I}\_i)(\text{pr}\_2^{-1}\mathcal{J}\_j) \subset \mathcal{O}\_{X \times\_S Y}$ the finite type quasi-coherent ideal generated by the product of the pullback of $\mathcal{I}\_i$ and the pullback of $\mathcal{J}\_j$. Then the affine open $U\_i \times\_S V\_j$ is equal to $X \times\_S Y \setminus V(\mathcal{K}\_{ij})$. As $X$ and $Y$ have the resloution property there exist surjections $\mathcal{E}\_i \to \mathcal{I}\_i$ and $\mathcal{F}\_j \to \mathcal{J}\_j$ with $\mathcal{E}\_i$ finite locally free on $X$ and $\mathcal{F}\_j$ finite locally free on $Y$. Then we see that there is a surjection $\text{pr}\_1^\*\mathcal{E}\_i \otimes \text{pr}\_2^\*\mathcal{F}\_j \to \mathcal{K}\_{ij}$. Since $X \times\_S Y = \bigcup\_{i = 1, \ldots, n, j = 1, \ldots, m} U\_i \times\_S V\_j$ we conclude $X \times\_S Y$ has the resolution property by [Lemma Tag 0F89](https://stacks.math.columbia.edu/tag/0F89).
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7
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https://mathoverflow.net/users/152991
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406487
| 166,609 |
https://mathoverflow.net/questions/406426
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7
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Let $K$ be a $p$-adic field (a finite extension of the field of $p$-adic numbers ${\mathbb Q}\_p$).
Let $T$ be a $K$-torus with character group $X={\sf X}^\*(T)$ and cocharacter group $Y={\sf X}\_\*(T)=X^\vee$.
I would like to have *explicit cocycles* for all Galois cohomology classes in $H^1(K,T)$.
We can compute $H^1(K,T)$ via double duality.
The cup product pairing defines an isomorphism
$$ H^1(K,T)\overset\sim\longrightarrow {\rm Hom}\big(H^1(K,X),{\Bbb Q}/{\Bbb Z}\big),$$
where I write $H^1(K,X)$ for $H^1({\rm Gal}(\overline K/K),X)$. See Milne, Arithmetic Duality Theorems, Corollary I.2.3.
Let $L/K$ be a finite Galois extension of degree $d$ splitting $T$.
We have
$$H^1(K,T)=H^1(\Gamma\_{L/K}, Y\otimes\_{\Bbb Z} L^\times),$$
where $\Gamma\_{L/K}={\rm Gal}(L/K)$.
The finite group $\Gamma\_{L/K}$ acts on $X$ and $Y$, and we have
$$H^1(K,X)=H^1(\Gamma\_{L/K}, X).$$
The canonical pairing of $\Gamma\_{L/K}$-modules
$$ Y\times X\to {\Bbb Z}$$
induces an isomorphism
\begin{align\*}
H^1(\Gamma\_{L/K}, X)&\overset\sim\longrightarrow
{\rm Hom}\bigg(H^{-2}\big(\Gamma\_{L/K},\, Y\otimes\_{\Bbb Z} ({\Bbb Q}/{\Bbb Z})\,\big),
\ {\Bbb Q}/{\Bbb Z}\bigg)\\
&\overset\sim\longrightarrow {\rm Hom}\big(H^{-1}(\Gamma\_{L/K}, Y),{\Bbb Q}/{\Bbb Z}\big);
\end{align\*}
see Brown, Cohomology of Groups, Corollary VI.7.3.
Thus we obtain an isomorphism
\begin{equation\*}
\lambda\colon\,(Y\_\Gamma)\_{\rm tors}= H^{-1}(\Gamma\_{L/K}, Y)\overset\sim\longrightarrow H^1(K,T)=H^1(\Gamma\_{L/K}, Y\otimes\_{\Bbb Z} L^\times),
\end{equation\*}
where $\Gamma={\rm Gal}(\overline K/K)$, $\ Y\_\Gamma$ denotes the group of coinvariants of $\Gamma$ in $Y$,
and $(\ )\_{\rm tors}$ denotes the torsion subgroup of the group in the parentheses.
>
> **Question.** Let $y\in Y$ be a cocharacter whose image in $Y\_\Gamma$ is of finite order (say, of order dividing $d$).
> Write explicitly a cocycle
> in $Z^1(\Gamma\_{L/K}, Y\otimes\_{\Bbb Z} L^\times)$ representing $\lambda[y]\in H^1(\Gamma\_{L/K}, Y\otimes\_{\Bbb Z} L^\times)$.
>
>
>
I think that this is possible, because I know the corresponding formula for $K={\Bbb R}$.
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https://mathoverflow.net/users/4149
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Explicit cocycles for the first Galois cohomology of a $p$-adic torus
|
Answer of James S. Milne: Most probably, this homomorphism
$$\lambda\colon\, H^{-1}(\Gamma\_{L/K}, Y)\overset\sim\longrightarrow H^1(\Gamma\_{L/K}, Y\otimes\_{\Bbb Z} L^\times)$$
is just the cup-product with the fundamental class in $H^2(\Gamma\_{L/K}, L^\times)$.
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1
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https://mathoverflow.net/users/4149
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406509
| 166,614 |
https://mathoverflow.net/questions/406417
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5
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The $\mathbf{A}^1$-invariance of vector bundles have been discussed in, for example, [this paper](https://projecteuclid.org/journals/duke-mathematical-journal/volume-166/issue-10/Affine-representability-results-in-A1-homotopy-theory-I--Vector/10.1215/00127094-0000014X.short) by Asok, Hoyois and Wendt. This of course implies storng $\mathbf{A}^1$-invariance results for the first etale cohomology group $H^1\_{\mathrm{et}}(-,\mathbb{G}\_m)$. Are there any similar results for $H\_{\mathrm{et}}^2(-,\mathbb{G}\_m)$ or the Brauer groups?
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https://mathoverflow.net/users/100553
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$\mathbf{A}^1$-invariance of Brauer groups and $H^2_{\mathrm{et}}(-;\mathbb{G}_m)$
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(For $i=0$, the map $H\_{\mathrm{et}}^{0}(\operatorname{Spec} A,\mathbb{G}\_{m}) \to H\_{\mathrm{et}}^{0}(\operatorname{Spec} A[t],\mathbb{G}\_{m})$ is an isomorphism if and only if $A$ is reduced.)
For $i=1$, it is a theorem of Traverso that $H\_{\mathrm{et}}^{1}(\operatorname{Spec} A,\mathbb{G}\_{m}) \to H\_{\mathrm{et}}^{1}(\operatorname{Spec} A[t],\mathbb{G}\_{m})$ is an isomorphism if and only if $A$ is seminormal.
For $i=2$, at least when $A$ is a field, the map $H\_{\mathrm{et}}^{2}(\operatorname{Spec} A,\mathbb{G}\_{m}) \to H\_{\mathrm{et}}^{2}(\operatorname{Spec} A[t],\mathbb{G}\_{m})$ is an isomorphism if and only if $A$ is perfect; thus if $A$ is regular with perfect fraction field we have a similar positive result (see Auslander, Goldman, *The Brauer group of a commutative ring*, ([link](https://www.ams.org/journals/tran/1960-097-03/S0002-9947-1960-0121392-6/S0002-9947-1960-0121392-6.pdf)) 7.5, 7.7 respectively). In general (for the torsion at least) we only have to worry about the $p$-torsion for primes $p$ that are not invertible in $A$. There are some additional positive results in Knus, Ojanguren, *A Mayer-Vietoris sequence for the Brauer group* ([link](https://core.ac.uk/download/pdf/82736669.pdf)) Theorem 3.6 and I'd be interested in a more complete characterization.
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6
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https://mathoverflow.net/users/15505
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406516
| 166,616 |
https://mathoverflow.net/questions/406475
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10
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Let $R$ be a ring (possibly noncommutative with zero-divisors). A non-unit and non-zero-divisor element $r \in R$ will be called *irreducible* if for all $a,b \in R$ such that $r=ab$, then $a$ or $b$ is a unit.
Many extensions $R$ of $\mathbb{Z}$ do not keep its set of irreducible elements, for example $2 = (1+i)(1-i)$ in $\mathbb{Z}[i]$. More generally ([Chebotarev's density theorem](https://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem)) if $R$ is the ring of integers of a Galois extension of $\mathbb{Q}$ of degree $n$ then the prime numbers that completely split in $R$ have density $1/n$ (so that there are infinitely many ones).
Here are examples of extensions of $\mathbb{Z}$ keeping its irreducible elements:
* $\mathbb{Z}[X]$: every prime number is an irreducible polynomial, but $\mathbb{Z}[X]$ contains also infinitely many other irreducible polynomials (up to units),
* $\mathbb{Z} + \mathbb{Q}\epsilon$ (with $\epsilon^2=0$): it keeps the irreducibles of $\mathbb{Z}$ but there is no new ones up to units (see [this comment](https://math.stackexchange.com/questions/4271735/is-there-an-integral-extension-of-mathbbz-keeping-its-prime-elements#comment8888446_4271735)).
Observe that the extensions of $\mathbb{Z}$ mentioned above (keeping its irreducible elements) either have infinitely many new irreducible elements (up to units) or have none (up to units).
In [this answer](https://math.stackexchange.com/a/4277145/84284) Keith Conrad provides an example with finitely many new irreducible elements (up to unit):
>
> Extend $\mathbf Z[x]$ by inverting all nonconstant irreducible
> elements except for $x$. That gives you a UFD whose primes elements
> are the ordinary prime numbers and $x$, up to units.
>
>
>
A *finite extension* of $\mathbb{Z}$ is a ring which is an extension of $\mathbb{Z}$ and a $\mathbb{Z}$-module of finite type. The example above is not.
**Question**: Is there a *finite* extension $R$ of $\mathbb{Z}$ keeping its irreducible elements, and also with new irreducible elements (up to units), but only finitely many (up to units)?
As Keith Conrad pointed out in [this comment](https://math.stackexchange.com/questions/4275346/existence-of-a-finite-extension-of-the-primes#comment8904743_4277145), if $R$ is an integral domain, then we are in the situation of Chebotarev's density theorem mentioned above. So $R$ must be noncommutative and/or with zero-divisors.
---
"up to units" means that if $r$ is an irreducible element and $u,v$ are units, then $r$ and $urv$ count for one.
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https://mathoverflow.net/users/34538
|
Existence of a finite extension of ℤ providing a finite extension of the primes
|
This post started as some minor observations, but I believe it now contains a full proof that there is no such ring. Throughout, we let $R$ be an example of the kind wanted.
**Observation 1**: $R$ is indecomposable.
*Proof of observation 1*: Write $R=S\_1\times S\_2$, with $S\_1,S\_2\neq 0$. Suppose, by way of contradiction, that $S\_1$ has finite characteristic $m>1$. Let $p$ be any integer prime divisor of $m$. Then $p=(p,p)$ is not irreducible (since $(1,0)(p,p)=0$), contradicting our assumption that the primes of $\mathbb{Z}$ stay irreducible in $R$ (and using your definition of irreducible, which includes not being a zero-divisor).
Thus, each $S\_i$ has characteristic $0$. Moreover, the argument above shows that each integer prime cannot be a zero-divisor in any $S\_i$.
If an integer prime $p$ is a unit in $S\_i$, then $S\_i$ contains a copy of $\mathbb{Z}[1/p]$, which is not finitely generated as a $\mathbb{Z}$-module, contradicting the fact that submodules of finitely generated $\mathbb{Z}$-modules are finitely generated.
Also, if $p$ is not irreducible in $S\_1$ (or, by symmetry $S\_2$), say $p=ab$ where $a,b$ are nonunits, then $(p,p)=(a,p)(b,1)$ is a product into nonunits, contradicting the irreducibility of the primes of $\mathbb{Z}$.
The elements $\{(p,1)\, :\, p>1\text{ prime}\}$ are not units, not zero-divisors, and are easily shown to be irreducible in $R$. They are not associate to each other, nor to the integer primes $(p,p)\in R$. Thus, this would contradict our assumption of having only finitely many new primes (up to associates).
**Observation 2**: If $r\in R$, then the subring $\mathbb{Z}[r]\subseteq R$ is isomorphic to $\mathbb{Z}[x]/I$ where $I$ is a principal ideal generated by a monic polynomial $f$.
*Proof of observation 2*: Since $\mathbb{Z}[r]$ is a $\mathbb{Z}$-submodule of $R$, it must be finitely generated. Hence, $r$ satisfies some monic polynomial. Let $f\in \mathbb{Z}[x]$ be the monic polynomial of smallest degree satisfied by $r$.
Now, $\mathbb{Z}[r]\cong \mathbb{Z}[x]/I$ where $I$ is the ideal of polynomials satisfied by $r$. It thus suffices to show that $I=f\mathbb{Z}[x]$. Supposing otherwise, let $g\in I-f\mathbb{Z}[x]$.
Working over $\mathbb{Q}[x]$ for a moment, take $d=\gcd(f,g)$, a monic polynomial. Since $f$ is monic, we know that $d\in \mathbb{Z}[x]$. By the extended Euclidean algorithm, we can write $d$ as a $\mathbb{Q}[x]$-linear combination of $f$ and $g$. Hence $cd\in I$ for some minimal $c\in \mathbb{Z}\_{>0}$. We know $c\neq 1$ by the minimality of the degree of $f$ (since $d$ is monic).
Let $p$ be any prime dividing $c$. Then $p\cdot (c/p)d(r)=0$, and hence $p$ is a zero-divisor in $R$, which is a contradiction.
**Observation 3**: The monic polynomial $f$ is a power of an irreducible polynomial $q\in\mathbb{Z}[x]$.
*Proof of observation 3*: Assume $f$ is not a power of an irreducible, so $f=gh$ over $\mathbb{Z}[x]$, where $\gcd(g,h)=1$ and $\deg(g),\deg(h)\geq 1$.
By an argument used previously, we can find some constant $c\in \mathbb{Z}\_{>0}$ that is a $\mathbb{Z}[x]$-linear combination of $g$ and $h$, say $c=gg'+hh'$.
Let $J=g\mathbb{Z}[x]$ and $K=h\mathbb{Z}[x]$. Let $S\_1=\mathbb{Z}[x]/J$ and $S\_2=\mathbb{Z}[x]/K$. Consider the map $\varphi\colon \mathbb{Z}[x]/I\to S\_1\times S\_2$ where $a+I\mapsto (a+J,a+K)$. This is an injective ring homomorphism. Thus, we can view $\mathbb{Z}[r]$ as a subring of this direct product. Moreover the image of $\mathbb{Z}[r]$ contains the elements $(c+J,0+K)$ and $(0+J,c+K)$. (Indeed, $h(r)h'(r)\mapsto (hh'+J,hh'+K) = (c+J,0+K)$.) Since the image of $\mathbb{Z}[r]$ contains $(1+J,1+K)$, we see (by a trivial use of Dirichlet's theorem) that the image of $\mathbb{Z}[r]$ contains elements of the form $(p+J,1+K)$ and $(1+J,p+K)$, with $p$ an integer prime. Thus (suppressing the coset notation) we have $(p,p)=(p,1)(1,p)$ in this image. Back inside $\mathbb{Z}[r]$, write this factorization as $p=ab$. Without loss of generality, $a$ is a unit in $R$, with inverse $a^{-1}$.
The ring $\mathbb{Z}[r,a^{-1}]$ is a finitely generated $\mathbb{Z}$-module (being a $\mathbb{Z}$-submodule of $R$), and it naturally maps onto $(\mathbb{Z}[x]/J)[1/p]$, which is not a finitely generated $\mathbb{Z}$-module, which is a contradiction.
**Observation 4**: $\deg(q)=1$.
*Proof of observation 4*: If $\deg(q)>1$, then there are infinitely many primes $p$ that factor nontrivially in $\mathbb{Z}[x]/(q)$ (by an argument supplied by John Voight) and such a factorization lifts to $\mathbb{Z}[x]/(q^n)\cong \mathbb{Z}[r]$; see [this link](https://mathoverflow.net/questions/406752/hensels-lemma-bezouts-identity-and-the-integers) for the full argument.
Now, if such a prime $p$ is to be irreducible in $R$, one of those factors must become a unit $u$ in $R$. But then $\mathbb{Z}[r,u^{-1}]$ is a commutative ring, mapping to $(\mathbb{Z}[x]/(q))[a^{-1}]$ (where $a$ is that corresponding nontrivial factor, but modulo $q$ rather than $q^n$), which is not a finitely generated $\mathbb{Z}$-module, giving us a contradiction, as before.
**Observation 5**: If $J$ is the set of nilpotent elements in $R$, then $J$ is a (nilpotent) ideal.
*Proof of observation 5*: An arbitrary element $r\in R$ satisfies $q^n$, with $q(x)=x-k\in \mathbb{Z}[x]$, and hence $r=k-t$ where $t$ is nilpotent of index $n$. Therefore, $r(k^{n-1}+k^{n-2}t + \cdots + t^{n-1})=k^n$. If $k\neq 0$, then $k^n$ is not a zero-divisor (since all the primes of $\mathbb{Z}$ are not zero divisors), and hence $r$ is not a zero-divisor. Thus, the zero-divisors are exactly the nilpotent elements.
Next, if $t\in R$ is nilpotent, say of index $n\geq 1$, and $r\in R$ is arbitrary then $t^{n-1}(tr)=0$ with $t^{n-1}\neq 0$. Hence $tr$ is a zero-divisor, hence nilpotent. Thus $J$ is closed by right multiplication from $R$; and by left multiplication by a symmetric argument.
In particular $J$ is closed under multiplication. By the paper [Rings in which nilpotents form a subring](https://www.jstor.org/stable/44000112) by Janez Ster ([arXiv version here](https://arxiv.org/abs/1510.07523)), we know that $J$ is also closed under addition, since $R$ satisfies Koethe's conjecture. (Quick argument: The Jacobson radical $J(R)$ is nilpotent, since tensoring up to $\mathbb{Q}$ keeps it in a finite-dimensional $\mathbb{Q}$-algebra.) Thus, $J$ is an ideal.
**Observation 6**: The type of ring we want cannot exist.
*Proof of observation 6*: Fix a $\mathbb{Z}$-basis for $J$, say $t\_1,\ldots, t\_k$. Then a $\mathbb{Z}$-basis for $R$ is $1,t\_1,\ldots, t\_k$ (since every element of $R$ is an integer shift from a nilpotent, by observation 4).
The units of $R$ are exactly $\pm 1+t$ for some $t\in J$. Thus, the associates of a prime $p\in \mathbb{Z}$ are of the form $\pm p+t'$ where $t'\in J$ is divisible by $p$. Thus $p+t\_1$ is not associate to any of the integer primes. A similar argument shows that $p+t\_1$ and $p'+t\_1$ are not associate, for distinct primes $p,p'$. Further, $p+t\_1$ is not a unit, and not a nilpotent (hence not a zero-divisor). It thus suffices to show that $p+t\_1$ is irreducible, and then we'll have infinitely many "new" nonassociate irreducibles.
If $p+t\_1$ factors as $(a+t)(b+t')$ with $a,b\in \mathbb{Z}$ and $t,t'\in J$, then $ab=p$. Hence, without loss of generality, $a=\pm 1$. Thus, $a+t$ is a unit, so every factorization is trivial.
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5
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https://mathoverflow.net/users/3199
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406520
| 166,617 |
https://mathoverflow.net/questions/406436
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1
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Let $f(n)$ be [A007814](https://oeis.org/A007814), the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Then we have an integer sequence given by
\begin{align}
a(0)=a(1)&=1\\
a(2n)& = a(n)+a(n-2^{f(n)})+a(2n-2^{f(n)})\\
a(2n+1) &= a(n)
\end{align}
which can also be expressed as
$$a(2^{m}(2n+1))=\sum\limits\_{k=0}^{m}\binom{m+1}{k}a(2^{k}n)$$
I conjecture that
$$a(n)=\sum\limits\_{j=0}^{2^{wt(n)}-1}(-1)^{wt(n)-wt(j)}\prod\limits\_{k=0}^{wt(n)-1}(1+wt(\left\lfloor\frac{j}{2^k}\right\rfloor))^{t\_{k+1}+1}$$
where
$wt(n)$ is [A000120](https://oeis.org/A000120), number of $1$'s in binary expansion of $n$ (or the binary weight of $n$)
and
$$n=2^{t\_1}(1+2^{t\_2+1}(1+\dots(1+2^{t\_{wt(n)}+1}))\dots)$$
Maybe it might be helpful, that it is a special case of
$$a(n)=\sum\limits\_{j=0}^{2^{wt(n)}-1}m^{wt(n)-wt(j)}\prod\limits\_{k=0}^{wt(n)-1}(1+wt(\left\lfloor\frac{j}{2^k}\right\rfloor))^{t\_{k+1}+1}$$
where
\begin{align}
a(0)& = 1\\
a(2n)& = a(n)-ma(n-2^{f(n)})+a(2n-2^{f(n)})\\
a(2n+1) &= a(n)+(m+1)a(2n)
\end{align}
Is there a way to prove it?
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https://mathoverflow.net/users/231922
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Formula from the recurrence relation
|
The conjectured formula can be proved by induction on $\mathrm{wt}(n)$.
For $\mathrm{wt}(n)=0$, we have $n=0$ and the conjectured formula trivially holds.
Now, for a positive integer $\ell$, suppose that the conjectured formula holds for all $\mathrm{wt}(n)<\ell$. Let us prove it for $\mathrm{wt}(n)=\ell$. So, let $n=2^{t\_1}(1+2^{t\_2+1}(1+\dots(1+2^{t\_\ell+1}))\dots)$ with $t\_j\geq 0$ and $n'=2^{t\_2}(1+\dots(1+2^{t\_\ell+1}))\dots)$, so that
\begin{split}
a(n) &= a(2^{t\_1}(2n'+1)) = \sum\_{m=0}^{t\_1} \binom{t\_1+1}{m} a(2^kn') \\
&=\sum\_{m=0}^{t\_1} \binom{t\_1+1}{m} \sum\_{j=0}^{2^{\ell-1}-1} (-1)^{\ell-1-\mathrm{wt}(j)} (1+\mathrm{wt}(j))^{m+t\_{2}+1} \prod\_{k=2}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{j}{2^{k-1}}\rfloor))^{t\_{k+1}+1} \\
&=\sum\_{j=0}^{2^{\ell-1}-1} (-1)^{\ell-1-\mathrm{wt}(j)} \left[(2+\mathrm{wt}(j))^{t\_{1}+1} - (1+\mathrm{wt}(j))^{t\_{1}+1}\right] \prod\_{k=1}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{j}{2^{k-1}}\rfloor))^{t\_{k+1}+1} \\
&=\sum\_{j=0}^{2^{\ell-1}-1} \left[(-1)^{\ell-\mathrm{wt}(2j+1)}\prod\_{k=0}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{2j+1}{2^k}\rfloor))^{t\_{k+1}+1} + (-1)^{\ell-\mathrm{wt}(2j)} \prod\_{k=0}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{2j}{2^k}\rfloor))^{t\_{k+1}+1}\right] \\
&=\sum\_{j'=0}^{2^\ell-1} (-1)^{\ell-\mathrm{wt}(j')} \prod\_{k=0}^{\ell} (1+\mathrm{wt}(\lfloor \tfrac{j'}{2^k}\rfloor))^{t\_{k+1}+1}, \\
\end{split}
where $j'$ corresponds to values $2j$ and $2j+1$ (in other words, $j=\lfloor j'/2\rfloor$). QED
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2
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https://mathoverflow.net/users/7076
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406527
| 166,618 |
https://mathoverflow.net/questions/406531
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1
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Let $G$ be a Lie group and let $M \rightarrow B$ be a $G$-equivariant vector bundle with typical fiber $E$.
Suppose that $B$ and $E$ are both symplectic manifolds with a $G$-Hamiltonian action. Can we conclude from this that $M$ is also a symplectic manifold with a Hamiltonian action?
I have been told that given a $G$-equivariant vector bundle of the form $G\times\_K V \rightarrow G/K$, where $V$ is a $K$-vector space of even dimension and $K$ is the stabilizer of some point $x \in V$, then since $G/K$ (which is identified with a coadjoint orbit) and $V$ have $G$-Hamiltonian actions then $G\times\_K V$ also has a symplectic structure with $G$-Hamiltonian action. Could you please explain why this is true?
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https://mathoverflow.net/users/172459
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Symplectic structure on a vector bundle
|
Let $f:M\rightarrow B$ be the symplectic fibration. A fibration whose typical fibre $F$ is a symplectic manifold, and the coordinate change are elements of the group of symplectomorphisms of $(F,\omega\_F)$. For the first question, a natural idea is to have the symplectic form $\omega\_M$ of $M$ compatible with the symplectic form of the fibres, that is the fibers are symplectic submanifolds of $M$. Thurston has shown that if there exists a form $[c]\in H^2(M,\mathbb{R})$ whose restriction to $F$ is $[\omega\_F]\in H^2(F,\mathbb{R})$, for every $n>>>0$, there exists a compatible form in the class of $c+nf^\*[\omega\_B]$.
Since The bundle is a vector bundle, $H^2(F,\mathbb{R})=0$, thus the form $c$ exists.
W. Thurston, Some simple examples of symplectic manifolds, Proc. Amer.
math. Soc. 55 (1976), 467-468.
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2
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https://mathoverflow.net/users/80891
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406536
| 166,619 |
https://mathoverflow.net/questions/406539
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5
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Let $k\in\mathbb N$. Given a finite graph with two subsets of vertices $X$ and $Y$, Menger's Theorem gives a criterion for when there are $k$ pairwise disjoint paths starting in $X$ and ending in $Y$.
Now let $X\_1$, $Y\_1$, $\dots$,$X\_k$, $Y\_k$ be subsets of vertices. Is there a "similar" condition for when there are $k$ pairwise disjoint paths, one path starting in $X\_1$ and ending in $Y\_1$, a second path starting in $X\_2$ and ending in $Y\_2$, $\dots$, and a $k^{\text {th}}$ path starting in $X\_k$ and ending in $Y\_k$?
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https://mathoverflow.net/users/51389
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Menger's theorem with restrictions on where the paths can begin and end
|
There is no known necessary and sufficient condition like in Menger's theorem.
However, there is a polynomial-time algorithm that decides if the paths exist. This is one of the main results of the Graph Minors Project of Robertson and Seymour (see [Graph Minors XIII](https://www.sciencedirect.com/science/article/pii/S0095895685710064)). The algorithm is quite involved and uses many of the results from the graph minors series. Robertson and Seymour solve the problem when each $X\_i$ and $Y\_i$ have size $1$, but your problem can be reduced to this case by shrinking each $X\_i$ and each $Y\_i$ to a single vertex.
The $k=2$ case (with $|X\_1|=|X\_2|=|Y\_1|=|Y\_2|=1$) has a nice necessary and sufficient condition for $4$-connected graphs. The required paths do not exist if and only if $G$ is planar and $X\_1, X\_2, Y\_1, Y\_2$ occur in this cyclic order on the outerface of $G$. This was proved independently by Thomassen and Seymour.
No necessary and sufficient conditions are known for $k \geq 3$.
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10
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https://mathoverflow.net/users/2233
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406540
| 166,620 |
https://mathoverflow.net/questions/404004
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8
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Given a continuum $X$ (compact metrizable connected $X$) let $K(X)$ denote the hyperspace of nonempty compact subspaces of $X$ with the Vietoris topology and let $C(X)$ denote the (closed) subspace of $K(X)$ consisting of connected sets.
It is well known (and has been discussed on [MO before](https://mathoverflow.net/questions/354092/which-compact-metrizable-spaces-have-continuous-choice-functions-for-non-empty-c)) that if there is a continuous choice function $K(X)\to X$ then $X\simeq[0,1]$, indeed any other continua doesn’t even have a continuous choice function for the subspace of $K(X)$ of unordered pairs.
The situation is different when looking at $C(X)$: for example the “star” $X$ obtained by joining $n$ arcs at their endpoint has a continuous selection for $C(X)$ by looking at the tree partial order induced by choosing the common point as root and then mapping $C\in C(X)$ to its $\min$ with respect to this partial order. In a similar way we get a continuous selection for $C(X)$ whenever $X$ is a dendrite.
It is also easy to see that $S^1$ doesn’t have a continuous selection for $C(S^1)$ by using a combination of Borsuk-Ulam, $C(S^1)\simeq D^2$ and $\partial C(S^1)\cong F\_1(S^1)$, where $D^2$ is the unit disk in $\Bbb R^2$ and $F\_1(X)$ is the space of singletons of $X$.
This suggests the following: suppose $X$ is a continuum with a continuous selection $C(X)\to X$. Must $X$ be a subcontinuum of a dendrite? If not is there a characterization of the continua $X$ with a continuous selection $C(X)\to X$?
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https://mathoverflow.net/users/49381
|
When does $C(X)$, $X$ a continuum, admit a continuous choice function?
|
I will post a CW answer to my own question to remove it from the unanswered list since Benjamin doesn't seem interested in turning his comment into an answer, but this is all informations taken from the section of *Hyperspaces* that he suggested.
To begin with let's fix the terminology *selectible* to refer to a continuum $X$ for which there exist a continuous selection $C(X)\to X$.
The simple answer to the question at the end, namely "is there a characterization of the selectible continua?" is no.
What is known (according to the book by Illanes and Nadler, I'm not aware of any recent developement however) is more or less the following:
* If $X$ is a selectible continuum, then $X$ is a dendroid.
* Say that a selection $s\colon C(X)\to X$ is rigid if whenever $A,B\in C(X)$ are such that $s(B)\in A\subseteq B$, then $s(A)=s(B)$ (note that the definition of rigid selection in *Hyperspaces* has a typo, this is the correct definition as given in the original reference *Rigid Selections and Smooth Dendroids* by L.E.Ward). Then there is a characterization of the rigidly selectable continua: those are exactly the smooth dendroids by the paper of Ward mentioned above. The idea here is that if $X$ is a smooth dendroid then the "tree" partial order (with root a point witnessing smoothness) is closed in $X\times X$, so the map sending each subcontinuum to its minimum is a rigid selection. Conversely if $s\colon C(X)\to X$ is a rigid selection, then $\{\{s(A)\}\times A\mid A\in C(X)\}$ is a closed partial order on $X$ which agrees with the "tree" order with root $s(X)$, so that $s(X)$ is a point witnessing the smoothness of $X$.
* Not all dendroids are selectible. Given two points $x,y$ in a dendroid let $[x,y]$ denote the unique arc joining them and let $(x,y)$ denote $[x,y]\setminus\{x,y\}$. Following definition 75.10 in *Hyperspaces* say that a dendroid $X$ is of type $N$ (a drawing justifies this name and clarifies the definition) if there are two points $p,q\in X$ and two sequences of arcs $[p\_np\_n'],[q\_n,q\_n']$ with points $p\_n''\in(q\_nq\_n')$ and $q\_n''\in(p\_n,p\_n')$ such that:
$$\begin{align} [p,q] &=\lim [p\_n,p\_n']=\lim [q\_nq\_n'] \\
p &= \lim p\_n = \lim p\_n' = \lim p\_n'' \\
q &= \lim q\_n = \lim q\_n' = \lim q\_n''\end{align}.$$
Then any dendroid of type $N$ is nonselectible.
A simple example of a dendroid of type $N$ is the doubly infinite broom: start with the segment between $(0,0)$ and $(1,0)$ in $\Bbb R^2$ and add segments between $(0,0)$ and $(1,1/n)$ and between $(1,0)$ and $(0,-1/n)$ for all $n\in\Bbb N^+$.
* More surprisingly there are even contractible dendroids that are not selectible, this is example 75.9 in *Hyperspaces* and the construction is originally due to Maćkowiak.
* selectibility is not preserved by neither monotone nor open maps. Nonselectibility isn't either, this is exercise 75.28 in *Hyperspaces*.
* There are various open questions about selectibility apart from a classification of selectible continua, some of them are in chapter 76 of *Hyperspaces*.
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2
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https://mathoverflow.net/users/49381
|
406551
| 166,623 |
https://mathoverflow.net/questions/406548
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4
|
Let $C\_p$, $1<p<\infty$, be the Schatten-$p$-class.
Let $x\in C\_p$ be positive and $\|x\|\_p =1$.
Let $E$ be the conditional expectation onto the diagonal part.
If $\|E(x)\|\_p \ge 1-\delta $ for some small $\delta>0$, then can we get an estimate for
$\|x - E(x)\|\_p$ in terms of $\delta$? In particular, if $\delta\to 0$, then do we have $\|x - E(x)\|\_p\to 0$?
PS. Of course, is it not true for $p=1$.
|
https://mathoverflow.net/users/91769
|
Norm estimate for the difference between a positive operator and its expectation
|
Yes. This follows from the uniform convexity of the Schatten $p$ classes.
Indeed, for every unitary diagonal operator $u$, we have $\| (x + u x u^\*)/2\|\_p \geq \| E( x+uxu^\*)/2\|\_p = \|E(x)\|p \geq 1-\delta$ ($E$ is a contraction). Therefore, we have $\|x - uxu^\*\|\_p\leq\varepsilon$ (where $\varepsilon = \varepsilon(p,\delta)$ is given by uniform convexity, and goes to $0$ with $\delta$). This implies that $\|x-y\|\_p \leq\varepsilon$ for every $y$ in the closed convex hull of $\{uxu^\* \mid u \textrm{ unitary diagonal}\}$. But $E(x)$ belongs to this closure, so we have $\|x - E(x)\|\_p \leq \varepsilon$.
The fact that $E(x)$ belongs to the closure of the unitary orbit is a very general fact. It is completely elementary in finite dimension, and it therefore holds by approximation for Schatten classes. It is also true in arbitrary $II\_1$ factors, see [Examples of non-proper conditional expectations onto von Neumann subalgebras of $II\_{1}$ factors](https://mathoverflow.net/questions/273934/examples-of-non-proper-conditional-expectations-onto-von-neumann-subalgebras-of).
**Edit on october 20 2021 :** It was asked in the comments whether the statement holds for general semifinite von Neumann algebras. The answer is yes. That is, if $1<p<\infty$ (and $q$ is the dual exponent), if $\mathcal{M}$ is an arbitrary semifinite von Neumann algebra (with trace $tr$), $\mathcal{N} \subset \mathcal M$ is a sub-von Neumann algebra with trace-preserving conditional expectation $E: \mathcal M \to \mathcal N$, then for every $x \in L\_p(\mathcal M)$ of norm $1$, $\|E(x)\|\_p \geq 1-\delta$ implies $\|x-E(x)\|\_p \leq \varepsilon(p,\delta) = O(\delta^{\min(1/p,1/q)})$.
It might be possible to adapt the preceding proof, at least for factors, but here is a simpler proof (arguably less enjoyable). By duality, there is $y \in L\_q(\mathcal N)$ of norm $1$ such that $tr(y E(x)) = 1-\delta$. $E$ being a trace-preserving conditional expectation, we have $tr(yx) = tr(y E(x))$, so by Hölder
$$ 1-\delta = tr(y(x+E(x))/2) \leq \| (x+ E(x))/2\|\_p.$$
By uniform convexity, $\|x-E(x)\|\_p \leq \varepsilon$.
Observe that, mutatis mutandis (working with Haagerup's $L\_p$ spaces), the same proof applies even when $\mathcal N$ is not semifinite, we only need the existence of a conditional expectation.
|
3
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https://mathoverflow.net/users/10265
|
406560
| 166,625 |
https://mathoverflow.net/questions/406544
|
2
|
It is known that if a real continuous function $f(x)$ satisfies a local $\alpha$-Hölder condition on a closed interval $[a,b]$, the box dimension of the graph of $f(x)$ on $[a,b]$ will be not greater than $2-\alpha$.
But if the proposition above is taken inversely, that is, if the box dimension of the graph of a real continuous function $f(x)$ on $[a,b]$ will be not greater than $2-\alpha$, how much can we guarantee that $f(x)$ satisfies a local $\alpha-$Hölder condition on a closed interval $[a,b]$?
For instance, the box dimension of the graph of $\sqrt{x}$ is one on $[0,1]$, and $\sqrt{x}$ satisfies a local $1$-Hölder condition on $[a,1]$ given any small positive $a$. Here we may say that $\sqrt{x}$ does not satisfy a local $1$-Hölder condition only at $0$, and one point of exception may not affect the value of the box dimension. So how much is such exception allowed that it does not affect the value of the box dimension?
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https://mathoverflow.net/users/152618
|
A question about box dimension and Hölder condition
|
The box dimension is not countably stable, so even one exceptional point could drive it up. For example, given $\epsilon>0$, the function $f(x)=x^{1/2} \sin(1/x)$ on $[0,1]$, with $f(0)=0$, is locally Lipschitz on $(0,1]$, yet the graph of $f$ has box dimension at least $1.25$. Indeed, for $j<k$, to cover the graph above $[1/(2\pi(k+j)), 1/(2\pi(k+j+1))]$, at least $ck^{3/2}$ squares of side length $1/(4k^2)$ are needed. Summing over $j<k$ we get $c\_1 k^{5/2}$ squares of side length $1/(4k^2)$.
You can get a result of the type you want by working with the Hausdorff or packing dimension of the graph, as these notions are stable under countable union.
In order to prove the packing dimension of the graph of $f$ is at most $2-\alpha$, it suffices to assume that the exceptional set (where the local $\alpha$-Hölder condition fails to hold) has packing dimension at most $1-\alpha$. [This condition is sharp, unless you make further assumptions on the behavior of $f$ on the exceptional set.]
To see this, partition the graph of $f$ to the part over the exceptional set (that clearly has packing dimension at most $2-\alpha$, and for each integer $m$, the part $A\_m$ where the $\alpha$-Hölder condition holds with constant $m$. The usual argument you referred to shows that for each $m$, the box dimension of $A\_m$ is at most $2-\alpha$. Then use the connection between packing dimension and Box dimension (see, e.g. [1] [2] or [3]) to conclude.
[1] Falconer, Kenneth. Fractal geometry: mathematical foundations and applications. John Wiley & Sons, 2004.
[2] Mattila, Pertti. Geometry of sets and measures in Euclidean spaces: fractals and rectifiability. No. 44. Cambridge university press, 1999.
[3] Bishop, Christopher J., and Yuval Peres. Fractals in probability and analysis. Vol. 162. Cambridge University Press, 2017.
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2
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https://mathoverflow.net/users/7691
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406576
| 166,632 |
https://mathoverflow.net/questions/406557
|
11
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A few months ago, I read a nice elementary proof of Lang's theorem:
**Theorem:** Let $G$ be a connected linear algebraic group over $\overline{\mathbb{F}}\_p$ and let $F : G \to G$ be a Frobenius map. Then the Lang map $L : G \to G$ defined by $L(g) = g^{-1}F(g)$ is surjective.
The proof showed that $L$ was both finite and dominant, which I believe is quite a standard approach. The unique feature of this proof was that it showed finiteness by exhibiting a finite generating set for $\overline{\mathbb{F}}\_p[G]$ as an $\overline{\mathbb{F}}\_p[G]$-module (with the module structure induced by $L$). This was done by just playing around with the functions so it required no deep background. Unfortunately, I didn't realise at the time that I would need this later, so I didn't write down the reference and now I can't find it.
**Question:** Does anyone know where I can find this proof of Lang's theorem?
To clarify, I know that there are other proofs (e.g. in Digne-Michel) of Lang's theorem, but this elementary proof would fit perfectly into the context where I need it.
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https://mathoverflow.net/users/175051
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Reference request: Elementary proof of Lang's theorem
|
The following is a rewrite of the proof of Lemma G from Steinberg's [On theorems of Lie-Kolchin, Borel, and Lang](https://doi.org/10.1016/B978-0-12-080550-1.50032-9). Let $G$ be an algebraic group over a finite field $k=\mathbb{F}\_q$ and let $A$ be the coordinate ring of $G$.
**Background from algebraic groups:** We write $\Delta$ for the map $A \to A \otimes\_k A$ induced by pull back along the multiplication map $G \times G \to G$. Let $A^{\vee}$ be the dual vector space $\text{Hom}\_k(A,k)$. The map $\Delta : A \to A \otimes A$ induces a dual map $\Delta^{\vee}: A^{\vee} \otimes A^{\vee} \to A^{\vee}$. Conceptually, $A^{\vee}$ can be thought of as measures on $G$ and $\Delta^{\vee}$ is convolution; in particular, if $\lambda\_i$ is evaluation at a point $g\_i \in G(k)$, then $\Delta^{\vee}(\lambda\_1 \otimes \lambda\_2)$ is evaluation at $g\_1 g\_2$.
For $\lambda \in A^{\vee}$ and $\theta \in A$, set $R\_{\lambda}(\theta) = (\text{Id}\otimes \lambda) (\Delta(\theta)) \in A$. If $\lambda$ is evaluation at a point $g \in G(k)$, then $R\_{\lambda}(\theta)$ is the right translation of $\theta$ by $g$. For $\theta \in A$, let $R(\theta) = \{ R\_{\lambda}(\theta) : \lambda \in A^{\vee} \}$. If $G(k)$ were Zariski dense in $G$, then $R(\theta)$ would be the span of the right translates of $\theta$; in our setting, $G(k)$ is finite, but I still think of this as the motivation for $R(\theta)$.
**Lemma 1** Let $\theta \in A$ and let $\phi \in R(\theta)$. Then $R(\phi) \subseteq R(\theta)$.
**Proof** We have $R\_{\lambda\_1}(R\_{\lambda\_2}(\theta)) = R\_{\Delta^{\vee}(\lambda\_2 \otimes \lambda\_1)}(\theta)$. $\square$
**Lemma 2** For any element $\theta$ of $A$, the vector space $R(\theta)$ is finite dimensional.
**Proof** Let $\Delta(\theta) = \sum \alpha\_i \otimes \beta\_i$, by the definition of the tensor product, the sum is finite. I claim that $R(\theta)$ is contained in the span of the $\alpha\_i$. Indeed, $R\_{\lambda}(\theta) = \sum \alpha\_i \lambda(\beta\_i)$. $\square$
**Proof of Lang's theorem**: Let $F$ be the Frobenius map $G \to G$, so $F^{\ast}(\theta) = \theta^q$ for $\theta \in A$. Let $L$ be the Lang map $g \mapsto g^{-1} F(g)$. We want to show that $A$ is finite over $L^{\ast} A$.
Let $\theta$ be an arbitrary element of $A$; we will show that $\theta$ is integral over $L^{\ast} A$. Since $A$ is finitely generated as a $k$-algebra, this will imply that $A$ is finite over $L^{\ast} A$.
Using Lemma 2, choose a basis $\theta\_1$, $\theta\_2$, ..., $\theta\_r$ of $R(\theta)$. Using Lemma\_1, we have
$$\Delta(\theta\_i) = \sum\_j \theta\_j \otimes \beta\_{ij}$$
for some $\beta\_{ij}$ in $A$. In other words, for $x$ and $y \in G(S)$, for any $k$-algebra $S$, we have
$$\theta\_i(xy) = \sum\_j \theta\_j(x) \beta\_{ij}(y).$$
Now, put $x = g$ and $y = L(g)$. So we have
$$\theta\_i(g)^q = \theta\_i(F(g)) = \sum\_j \theta\_j(g) \beta\_{ij}(L(g))$$
or, in other words,
$$\theta\_i^q = \sum\_j \theta\_j L^{\ast}(\beta\_{ij}).$$
We have now shown that $\theta\_i^q$ is in the $L^{\ast}(A)$-module generated by $\theta\_1$, $\theta\_2$, ..., $\theta\_n$. Thus, the ring $L^{\ast}(A) \langle \theta\_1, \theta\_2, \ldots, \theta\_n \rangle$ is finitely generated as an $L^{\ast}(A)$ module and, in particular, $\theta$ is integral over $L^{\ast}(A)$. $\square$.
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10
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https://mathoverflow.net/users/297
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406580
| 166,634 |
https://mathoverflow.net/questions/406588
|
4
|
I have precedently posted the same question on Math.Stackexchange (<https://math.stackexchange.com/questions/4277856/quasi-compact-surjective-morphism-of-smooth-k-schemes-is-flat>), but to no avail; I hope this is not too low-level for this site.
In the article "The Greenberg functor revisited'' (<https://doi.org/10.1007/s40879-017-0210-0>) by Bertapelle and González-Avilés, one can read, before the statement of Theorem A.12, that a given morphism, call it $f\colon X\to Y$, "is a quasi-compact [even affine] and surjective morphism of smooth $k$-schemes [$k$ is a field] and therefore faithfully flat and locally of finite presentation.'' This might be trivial, but I have no idea how to prove flatness: the most general result of this kind which I have been able to find in the literature is EGA IV.15.4.2, which however requires one of the following conditions (I am keeping the same notation as in EGA), none of which I am able to deduce from the information above:
* $f$ is universally open;
* for any $x\in X$, $f$ is open at the generic points of the irreducible components of $X\_{f(x)}$ containing $x$;
* for any $x\in X$, $\dim\mathcal{O}\_{X,x}=\dim\mathcal{O}\_{f(x)}+\dim(\mathcal{O}\_{X,x}\otimes\_{\mathcal{O}\_{f(x)}}\kappa(f(x)))$
I would be grateful for an explanation of why $f$ is flat.
|
https://mathoverflow.net/users/171983
|
Quasi-compact surjective morphism of smooth k-schemes is flat
|
The simplest blowup morphism $\mathrm{Bl}\_0(\mathbb{A}^2) \to \mathbb{A}^2$ (with center at a point) is not flat.
EDIT. Here is an example with affine morphism. Let
$$
X = \{ x\_1y\_1 + x\_2y\_2 + x\_3y\_3 = 0 \} \subset \mathbb{A}^4\_{x\_1,x\_2,x\_3,x\_4} \times \mathbb{A}^4\_{y\_1,y\_2,y\_3}
$$
and let $f \colon X \to \mathbb{A}^3$ be the projection to the second factor. This example, however, is **singular** at the point $(0,0)$.
EDIT 2. Consider the variety
$$
\bar{X} = \{x\_1y\_1 + x\_2y\_2 + x\_3y\_3 = 0\} \subset
\mathbb{P}^2\_{x\_1:x\_2:x\_3} \times
\mathbb{A}^3\_{y\_1,y\_2,y\_3}.
$$
It is smooth, because the projection to $\mathbb{P}^2$ is a fibration with fiber $\mathbb{A}^2$. On the other hand, the projection $\bar{f} \colon \bar{X} \to \mathbb{A}^3$ is not flat, because the dimension of the fiber jumps at $0$.
Now let
$$
X = \bar{X} \cap
((\mathbb{P}^2 \setminus C) \times \mathbb{A}^3),
$$
where $C$ is a smooth conic. Then
1. $X$ is smooth, because it is open in $\bar{X}$;
2. $X$ is affine over $\mathbb{A}^3$ because $\mathbb{P}^2 \setminus C$ is affine,
3. the map $f \colon X \to \mathbb{A}^3$ is surjective, because the smooth conic $C$ cannot contain a fiber of $\bar{f}$ (a line or the plane),
4. the map $f$ is not flat, because the dimension of the fiber still jumps at $0$.
|
6
|
https://mathoverflow.net/users/4428
|
406591
| 166,639 |
https://mathoverflow.net/questions/406592
|
0
|
Consider the following block matrix:
$$
A = \begin{bmatrix}
0 & I\\
-M & -I
\end{bmatrix}
$$
Suppose matrix $M$ is positive definite and satisfies $M\succeq \alpha I$, where $\alpha>0$ is a constant. When will matrix $A$ be Hurwitz stable, i.e., all of the eigenvalues have negative real parts?
|
https://mathoverflow.net/users/172027
|
Conditions for a block matrix to be Hurwitz stable
|
The eigenvalues of $A$ are $(-1 \pm \sqrt{1-4s})/2$ where $s$ is an eigenvalue of $M$. If $s \ge 1/4$, these have real part $-1/2$, while if $0 < s < 1/4$, they are both real and negative. So it's always true when $M$ is positive definite.
|
1
|
https://mathoverflow.net/users/13650
|
406595
| 166,641 |
https://mathoverflow.net/questions/406559
|
0
|
For each $n\ge 1$, consider $X^i\_t=1-\beta t + W^i\_t$ for $i=1,\ldots n$ and $t\ge 0$, where $\beta>0$ and $(W^i\_t)\_{t\ge 0}$ are independent Brownian motions. $\phi\equiv \big((\phi^1\_t)\_{t\ge 0},\ldots, (\phi^N\_t)\_{t\ge 0}\big)$ is said to be an allocation strategy if every $(\phi^i\_t)\_{t\ge 0}$ is progressively measurable w.r.t. the Brownian filtration $\big(\mathcal F\_t:=\sigma(W^1\_s,\ldots, W^N\_s, s\le t)\big)\_{t\ge 0}$,
$$\phi^i\_t\ge 0 \quad\mbox{ and }\quad \sum\_{i=1}^n\phi^i\_t\le 1,\quad \forall t\ge 0.$$
Denote
$$X^{\phi,i}\_t:=X^i\_t+\int\_0^t \phi^i\_sds \quad \mbox{and} \quad \tau^{\phi}\_i:=\inf\{t\ge 0: X^{\phi,i}\_t\le 0\}.$$
Let $S^{\phi}\_n:=\sum\_{1\le i\le n}{\bf 1}\_{\{\tau^{\phi}\_i=\infty\}}$ be the number of $X^{\phi,i}$ that never hits zero. I am interested in the asymptotic order of
$$\sup\_{\phi} S^{\phi}\_n,$$
where the the supremum is taken over all allocation strategies. My question is whether one has $0<\alpha<1$ and $C>0$ s.t.
$$0<\liminf\_{n\to\infty}\frac{\sup\_{\phi} S^{\phi}\_n}{n^{\alpha}} \le \limsup\_{n\to\infty}\frac{\sup\_{\phi} S^{\phi}\_n}{n^{\alpha}}\le C\quad \left( \mbox{or}\quad 0<\liminf\_{n\to\infty}\frac{\mathbb E[\sup\_{\phi} S^{\phi}\_n]}{n^{\alpha}} \le \limsup\_{n\to\infty}\frac{\mathbb E[\sup\_{\phi} S^{\phi}\_n]}{n^{\alpha}}\le C\right)?$$
Any answers, comments or references are highly appreciated!
|
https://mathoverflow.net/users/261243
|
Number of drifted Brownian motions that never hit zero under allocation
|
I think its O(1).
$$\lbrace \tau^{\phi}\_i > t \rbrace = \lbrace \tau^{\phi}\_i > t , \int\_0^t \phi^i\_sds > \frac {\beta t} 2 \rbrace \cup \lbrace \tau^{\phi}\_i > t,\int\_0^t \phi^i\_sds < \frac {\beta t} 2 \rbrace = $$,
so $$1\_{\lbrace \tau^{\phi}\_i > t \rbrace} \le 1\_{ \lbrace \tau^{\phi}\_i > t , \int\_0^t \phi^i\_sds > \frac {\beta t} 2 \rbrace } + 1\_ {\lbrace \tau^{\phi}\_i > t,\int\_0^t \phi^i\_sds < \frac {\beta t} 2 \rbrace} $$ $$$$
$$\Sigma 1\_{ \lbrace \tau^{\phi}\_i > t , \int\_0^t \phi^i\_sds > \frac {\beta t} 2 \rbrace } \le \Sigma 1\_{ \lbrace \int\_0^t \phi^i\_sds > \frac {\beta t} 2 \rbrace } \le \frac 2 { \beta }$$ by $\Sigma \phi\_i < 1$ while $$\lbrace \tau^{\phi}\_i > t,\int\_0^t \phi^i\_sds < \frac {\beta t} 2 \rbrace \subset \lbrace W\_t > \frac {\beta t } 2 \rbrace$$ which has probability, say $e^{\frac {-\beta t } 2}$, maybe a little different but not materially so. Therefore the expected number of paths for which $\tau^{\phi}\_i > t $ is $\le \frac 2 { \beta } + n e^{\frac {-\beta t } 2}$ and t is at your disposal.
|
1
|
https://mathoverflow.net/users/143907
|
406607
| 166,643 |
https://mathoverflow.net/questions/406601
|
5
|
I am specifically referencing the property that, given a braided monoidal category with a braiding $c$ and left and right unitors $\lambda, \rho$,
$$
\lambda\_A \circ c\_{A,I}=\rho\_{A},
$$
for any object $A$. This equation is stated in almost every reference (first done so by [Joyal and Street](https://www.sciencedirect.com/science/article/pii/S0001870883710558)) defining a braided category, yet an explicit or reproducible proof is, to my knowledge, not given anywhere. The closest thing to a proof I can find is a [series of hints on Mathematics SE](https://math.stackexchange.com/a/920526/982236) asked in 2014, however this answer (by Turion) has several inconsistencies/typos, making it hard to understand. Is there some reference where this proof is actually 'spelled out'? How can one prove this relation given the braided monoidal category axioms?
|
https://mathoverflow.net/users/419447
|
What is the proof of the compatibility of a braiding with the unitors?
|
Using the notation of [Joyal and Street](https://www.sciencedirect.com/science/article/pii/S0001870883710558) §2, here’s a proof of $\newcommand{\r}{\rho}\newcommand{\l}{\lambda}\newcommand{\x}{\otimes}\newcommand{\comp}{\!\!\cdot\!}\r\_A = \l\_A \comp c\_{A,I}$. Since $\l$ and $c$ are invertible, it suffices to prove $\r\_A \comp (\l\_A \comp c\_{A,I} \x I) = \l\_A \comp c\_{A,I} \comp(\l\_A \comp c\_{A,I} \x I)$. Notation: I will write composition as $\comp$, and it binds tighter than $\x$; and will write all associators just as $\alpha$, to reduce clutter. I’ll black-box details of steps that are purely about symm. mon. cats., not involving the braiding — these are most easily seen using coherence for SMC’s.
$$\begin{align}
\r\_A \comp (\l\_A \comp c\_{A,I} \x I)
&= (\l\_A \comp c\_{A,I}) \comp \r\_{A \x I} & \text{naturality of $\r$} \\
&= \l\_A \comp c\_{A,I} \comp (A \x \r\_I) \comp \alpha & \text{SMC facts}\\
&= \l\_A \comp (\r\_I \x A) \comp c\_{A,I \x I} \comp \alpha & \text{naturality of $c$}\\
&= \l\_A \comp (\r\_I \x A) \comp \alpha \comp (I \x c\_{A,I}) \comp \alpha \comp (c\_{A,I} \x I) & \text{axiom (B1)} \\
&= \l\_A \comp (I \x \l\_A) \comp (I \x c\_{A,I}) \comp \alpha \comp (c\_{A,I} \x I) & \text{SMC facts}\\
&= \l\_A \comp (I \x \l\_A \comp c\_{A,I}) \comp \alpha \comp (c\_{A,I} \x I) & \text{functoriality of $\x$}\\
&= (\l\_A \comp c\_{A,I}) \comp \l\_{A \x I} \comp \alpha \comp (c\_{A,I} \x I) & \text{naturality of $\l$}\\
&= (\lambda\_A \comp c\_{A,I}) \comp (\lambda\_A \x I) \comp (c\_{A,I} \x I) & \text{SMC facts}\\
&= (\lambda\_A \comp c\_{A,I}) \comp (\lambda\_A \comp c\_{A,I} \x I) & \text{functoriality of $\x$}
\end{align}
$$
I found this proof using string diagrams, writing down (B1) for $A,I,I$ (as Joyal and Street suggest), and then contemplating how to connect that to the desired equation.
|
5
|
https://mathoverflow.net/users/2273
|
406623
| 166,646 |
https://mathoverflow.net/questions/406608
|
4
|
Let $U(n)$ be the compact manifold of unitary $(n \times n)$-matrices and let $\mu\_n$ denote the Haar-probability measure on $U(n)$. For $m < n$ does there exists a measurable (maybe even continuous or smooth) map
$$
F: \ U(n) \rightarrow U(m)
$$
with the property, that
$$
\mu\_m(A) = \mu\_n(F^{-1}(A))
$$
for every $A \in \mathcal{B}(U(n))$?
Are there maybe certain necessary conditions on $(m,n)$?
|
https://mathoverflow.net/users/409412
|
A 'projective' property of the Haar U(n) measure
|
For any $m<n$ the $n\times n$ unitary matrix $\Omega$ has the block decomposition
$$\Omega=\begin{pmatrix} A&B\\ C&D\end{pmatrix},$$
where $A$ has dimensions $m\times m$, $D$ has dimensions $(n-m)\times(n-m)$, $B$ has dimensions $m\times(n-m)$ and $C$ has dimensions $(n-m)\times m$. Up to a set of measure zero, the matrix $D$ will not have a unit eigenvalue, so $I-D$ is invertible. We then define the continuous map $F$ from $U(n)$ to $U(m)$ by
$$F(\Omega)=A+B(I-D)^{-1}C.$$
One readily checks that $F(\Omega)$ is unitary$^\ast$ and as David Speyer points out $F(\Omega)$ inherits$^{\ast\ast}$ the Haar measure from $\Omega$.
---
$^\ast$ More generally, for any $\Omega\in U(n)$ and $V\in U(n-m)$ the matrix $U=A+B(I-VD)^{-1}VC$ is unitary. One can think of $V$ as the reflection matrix of a barrier that closes off $n-m$ scattering channels. Then the remaining $m$ channels have scattering matrix $U=A+\sum\_{k=0}^\infty B(VD)^kVC$, where $k+1$ counts the reflections off the barrier. As a further check for the unitarity of $U$, I can offer a [Mathematica Notebook.](https://ilorentz.org/beenakker/MO/unitarity_check.nb)
$^{\ast\ast}$ Since for any $g\in U(m)$, $G={{g\;\;0}\choose{0\;\;I}}\in U(n)$ one has $F(G\Omega)=gF(\Omega )$, the measure remains left-invariant.
|
3
|
https://mathoverflow.net/users/11260
|
406640
| 166,649 |
https://mathoverflow.net/questions/406631
|
0
|
Let $p$ be a point in the $2$-dimensional hyperbolic space $H\_2$. Consider a normal coordinate system $(x,y)$ at centered at $p$. Let $o\_x\in H\_2$ be the point of coordinate $(x,0)$ and let $C\_x$ be the hyperbolic circle centered in $o$ passing through $p$.
I have difficulties with the $2$ following questions:
* Given a point $q$ which has coordinates $(x\_q>0,y\_q)$, is there always a $x$ such that $q$ is in the interior of $C\_x$? It is true in Euclidean geometry, I am having doubts for hyperbolic geometry...
* In case it is not true, what it the curve $f(y)$ such that points $(x\leq f(y),y)$ are not in the interior of any $C\_{x'}$ while $(x>f(y),y)$ are for some $x'$?
Thank you for your help.
|
https://mathoverflow.net/users/176470
|
Half space vs growing balls in the hyperbolic plane
|
Your question is actually best reformulated in terms of horospheres and horoballs. Indeed, the set of points described in normal coordinates as $\{(x,0), x\ge 0\}$ is precisely a geodesic ray $\gamma$ in the hyperbolic plane issued from the point $p$. Then the union of the interiors of all your $C\_x$ is precisely the horoball in the hyperbolic plane centered at the limit point $\gamma\_\infty\in\partial \mathbf H^2$ of the ray $\gamma$ and passing through the point $p$.
Now it is more convenient (at least for me) to use the upper hlaf space model, and take $p=i$ and $\gamma$ to be the vertical geodesic ray directed upwards, so that $\gamma\_\infty=\infty$. Then the corresponding horosphere (or rather the horocycle, as its dimension is 1) is just the horizontal line $y=1$, and it is clear that even if the directing vector of a geodesic issued from $p$ has a positive vertical component, then the resulting endpoint still may very well end up to be below the horosphere $y=1$. Therefore, the answer to your first question is "no".
As for your second question, by applying some elementary hyperbolic geometry one can explicitly describe all the tangent vectors at the point $p=i$ (I continue talking about the upper half space model) such that the endpoints of the corresponding geodesic are within the horoball centered at $\infty$ and passing through $p$. Let $(x,y)$ be the coordinates of the tangent vector (since I am using the upper half space model, $x$ and $y$ are swapped here compared to your original formulation), and let $\gamma$ be the geodesic issued from $p$ in the direction $(x,y)$. Then the condition is that $\gamma(d)$ is inside the horoball (i.e., its vertical coordinate is $>1$), where $d=\sqrt{x^2+y^2}$ is the length of the tangent vector. In other words, $d$ should be less than the hyperbolic distance between $p$ and the point $p'$ where $\gamma$ intersects the horosphere $\{y=1\}$. The geodesic $\gamma$ is an arc of the circle centered at $(\frac{y}x, 0)$, whence $p'=(2\frac{y}x,1)$, so that
$$
d(p,p') = arcosh \bigl( 1+ 2 (\tfrac{y}x)^2 \bigr)
$$
(e.g., [see](https://math.stackexchange.com/questions/160338/expression-of-the-hyperbolic-distance-in-the-upper-half-plane)). Thus, the condition is
$$
\sqrt{x^2+y^2} > arcosh \bigl( 1+ 2 (\tfrac{y}x)^2 \bigr)
$$
or
$$
\cosh \sqrt{x^2+y^2} > \bigl( 1+ 2 (\tfrac{y}x)^2 \;.
$$
You are asking whether, for a given $x$ there is a threshold value $y\_0=y\_0(x)$ separating "good" and "bad" point (once again, I remind, that $x$ and $y$ are swapped compared to your original question). I am not sure this is true though. For instance, for $x=1$ the corresponding equation seems to have two roots.
|
1
|
https://mathoverflow.net/users/8588
|
406642
| 166,650 |
https://mathoverflow.net/questions/406618
|
7
|
I'm confused by a subtle point in the definition of analytic sets. Suppose I have a Polish space $X$. Now I start with the collection of Borel sets in $X$ and take all their continuous images in $X$. Do I get the entire family of analytic sets in this way? In other words, can I say in good conscience that the **analytic sets in $X$** are the continuous images of the **Borel sets in $X$**?
Let me state the question another way. By definition a set $A\subseteq X$ is analytic if it is the image $A=f(B)$ of some Borel set $B\subset P$ in some Polish space $P$ using some continuous mapping $f:P\to X$. I don't like referring to an external space $P$. What happens if I try to simplify the definition by requiring $P=X$; will I still get all the analytic sets?
|
https://mathoverflow.net/users/106622
|
In a Polish space, is every analytic set the continuous image of a Borel set from the same Polish space?
|
The answer is positive.
If $X$ is countable all subsets of $X$ are Borel, so they are their own continuous image through the identity function.
If $X$ is uncountable then it contains a copy of the Cantor space, but the Cantor space contains a copy of the Baire space $\mathcal N$ (which is necessarily $G\_\delta$ in $X$, unrelated to your question but interesting nonetheless is the fact that a Polish space contains $\mathcal N$ as a *closed* subspace iff $X$ is not $\sigma$-compact) and being the continuous image of $\mathcal N$ is one of the many characterizations of analytic sets.
Edit: Following the comment by Samuel I realized that the definition used in the original question is different from the one I had in mind and the continuous function is required to have domain the whole of $X$. In that case the answer is negative: the [Cook continuum](https://eudml.org/doc/213964) is a compact metric space $X$ with the property that every continuous function $X\to X$ is either constant or the identity. Clearly no analytic but not Borel subset of $X$ is the image of a Borel set through a continuous function $X\to X$.
|
6
|
https://mathoverflow.net/users/49381
|
406643
| 166,651 |
https://mathoverflow.net/questions/406546
|
4
|
Consider Motzkin paths with the following weight:
All up-steps and the horizontal steps on height $0$ have weight $1$, all down-steps have weight $t$ and the horizontal steps on even heights have weight $1-t$ while those on odd heights have weight $t-1.$
Computations suggest that the weight $c\_n(t)$ of all paths of length $n$ is
$$c\_n(t)=\sum\_j\binom{\lfloor{\frac{n-1}{2}}\rfloor}{\lfloor{\frac{j-1}{2}}\rfloor}\binom{\lfloor{\frac{n}{2}}\rfloor}{\lfloor{\frac{j}{2}}\rfloor}t^{j-1}.$$
Is there a simple way to prove this? The use of continued fractions seems to be rather hard.
Could it help that $c\_n(t)$ is the special case for $q=-1$ of the q-Narayana polynomials
$\sum{\binom{n}{j}\_q \binom{n-1}{j}\_q \frac{q^{j^2+j}}{[j+1]\_q} t^j}$?
|
https://mathoverflow.net/users/5585
|
A special class of weighted Motzkin paths
|
Let $F\_k(x)$ be a generating function for the total weight of Motzkin paths of length $n$ starting and ending at height $k$ and not going below that height. Let $G\_k(x)$ be a similar generating function with an additional restriction that paths do not come to height $k$, except at the beginning and at the end. Then $c\_n(t)$ is the coefficient of $x^n$ in $F\_0(x)$.
Clearly, we have $F\_{k+2}(x)=F\_k(x)$ for any $k\geq 1$, and $G\_{k+2}(x)=G\_k(x)$ for any $k\geq 0$. Taking into account up and down steps, we also have
$$G\_k(x) = 1 + tx^2F\_{k+1}(x),$$
which we will use for $k\in\{0,1\}$.
Finally, taking into account the weight of horizontal steps, we have
$$\begin{cases}
F\_0(x) = \frac{1}{1-(G\_0(x)-1+x)},\\
F\_1(x) = \frac{1}{1-(G\_1(x)-1+(t-1)x)},\\
F\_2(x) = \frac{1}{1-(G\_0(x)-1+(1-t)x)}.
\end{cases}
$$
This gives us a system of 5 equations w.r.t. $F\_0(x), F\_1(x), F\_2(x)$ and $G\_0(x), G\_1(x)$. Solving it, we get
$$F\_0(x) = \frac{1-2tx+\left(t^{2}-1\right) x^{2}-\sqrt{1-2\left(1+ t^{2}\right) x^{2}+\left(1-t^2\right)^2 x^{4}}}{2 x \left(-1+\left(1+t \right) x \right) t}.$$
Extracting the coefficient of $x^n$ is now a technical task.
---
**ADDED.** Proving the required identity can be done in three straightforward steps:
1. Notice that the sum in question equals the coefficient of $z^n$ in
$$\big((1+z^2)(1+z^2t^2)\big)^{n/2} (1+z^2t) \big(z ((1+z^2)(1+z^2t^2))^{-1/2} + (1+z^2t^2)^{-1}\big).$$
2. Apply Lagrange–Bürmann formula to turn the above expression into a generating function.
3. Verify that the obtained generating function is the same as $F\_0(x)$.
|
3
|
https://mathoverflow.net/users/7076
|
406646
| 166,652 |
https://mathoverflow.net/questions/406653
|
1
|
Assume that $X\sim \mathcal N(\sigma\_1,\mu\_1)$ and $Y\sim \mathcal N(\sigma\_2,\mu\_2)$.
I want to estimate $\frac{\mu\_1+\mu\_2}{2}$ after observing $X,Y$.
In my setting, $\sigma\_1,\sigma\_2$ are known and we want to estimate the average of the means (e.g., what is the MLE of it?).
---
In the special case where we know that $\mu=\mu\_1=\mu\_2$ (i.e., we estimate the same quantity from observations with different variances), [it is known](https://arxiv.org/pdf/2010.11537.pdf) that the MLE is:
$$
X\cdot \frac{\sigma\_2^2}{\sigma\_1^2+\sigma\_2^2} + Y\cdot \frac{\sigma\_1^2}{\sigma\_1^2+\sigma\_2^2}.
$$
How does this generalize to arbitrary $\mu\_1,\mu\_2$?
|
https://mathoverflow.net/users/47499
|
Estimating the average of two gaussians' mean
|
The maximum likelihood estimator (MLE) for $(\mu\_1,\mu\_2)$ is $(X,Y)$. So, by the [functional invariance of the MLE](https://en.wikipedia.org/wiki/Maximum_likelihood_estimation#Functional_invariance) (that is, simply by definition), the MLE of $g(\mu\_1,\mu\_2):=(\mu\_1+\mu\_2)/2$ is $g(X,Y):=(X+Y)/2$, which also, obviously, maximizes the [profile likelihood](https://en.wikipedia.org/wiki/Likelihood_function#Profile_likelihood)
$$L\_{X,Y}(\mu):=\sup\{L\_{X,Y}(\mu\_1,\mu\_2)\colon(\mu\_1+\mu\_2)/2=\mu\}$$
in $\mu$, where $L\_{X,Y}(\mu\_1,\mu\_2)$ is the likelihood.
---
One may note that the MLE $(X+Y)/2$ of $(\mu\_1+\mu\_2)/2$ does not depend on $(\sigma\_1,\sigma\_2)$.
|
3
|
https://mathoverflow.net/users/36721
|
406655
| 166,656 |
https://mathoverflow.net/questions/406651
|
5
|
Let $G$ be a simple graph with vertex set $V$, such that for any two vertices $u,v\in V$, we have at least $k$ edge-disjoint paths of length $2$ (i.e., formed by $2$ edges) connecting $u$ with $v$. Let $n=|V|$ be the total number of vertices of $G$.
---
**Question:** What is the minimum value of $k$, expressed as a function of $n$, to ensure that $G$ must be a complete graph?
|
https://mathoverflow.net/users/115803
|
Graph combinatorial optimization problem
|
The answer is $k=n-2$. To see this, first note that $k \geq n-2$, since the complete graph on $n$ vertices minus an edge has the desired property for $k=n-3$. For the other inequality suppose that $G$ is an $n$-vertex graph such that for all distinct $u, v \in V(G)$, there are at least $n-2$ edge-disjoint paths between $u$ and $v$. We claim that $G$ must be a complete graph. Suppose not. Then $ab \notin E(G)$ for some $a,b$. Let $c \notin \{a,b\}$. Then there are at most $n-3$ edge-disjoint paths of length $2$ between $a$ and $c$, which is a contradiction.
|
6
|
https://mathoverflow.net/users/2233
|
406656
| 166,657 |
https://mathoverflow.net/questions/406614
|
2
|
Let $M$ be an almost surely continuous martingale that is not almost surely constant in time - that is, it is not the case that almost surely, $M\_t = M\_0$ for all $t$.
Assume further that $M$ is a time homogeneous Markov process, and that it is transitive, in the sense that for any measurable subset $U$ of $\mathbb R$ with nonzero Lebesgue measure, we have
$$\mathbb E\big[\int\_0^\infty \mathbf 1\_U (M\_t) \, dt \big] > 0.$$
Suppose $f: \mathbb R \to \mathbb R$ is a continuous function such that $f(M\_t)$ is a martingale.
**Question:** Does if follow that $f$ is necessarily an affine linear function? That is, $f(x) = a + bx$ for some $a, b \in \mathbb R$.
|
https://mathoverflow.net/users/173490
|
If a continuous function of a Markov martingale is a martingale, does the function have to be affine linear?
|
If you allow for an arbitrary starting point, then just use the optional stopping theorem for $f(M\_t)$: $$\begin{aligned} f(x) & = \mathbb E^x f(M(\tau\_{(a,b)})) \\ & = \mathbb P^x(M(\tau\_{(a,b)}) = b) f(b) + P^x(M(\tau\_{(a,b)}) = a) f(a) \\ & = \frac{x-a}{b-a} f(b) + \frac{b-x}{b-a} f(a) , \end{aligned}$$
as desired. Here $\tau\_{(a,b)}$ is the first hitting time of $a$ or $b$; it is finite almost surely because $M$ is transitive.
If the starting point (or distribution) is fixed, follow the same argument after the first hitting time of any given $x$ (the latter is almost surely finite because the process is transitive) and use the strong Markov property.
Note: one-dimensional diffusions have been studied by E.B. Dynkin and others, and an essentially complete description is available in terms of the speed measure and the scale function. I think the above argument belongs to this theory (although it is very simple, of course).
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4
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https://mathoverflow.net/users/108637
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406657
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https://mathoverflow.net/questions/403397
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25
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Consider an integer of the form $$N = 1 + \sum\_{i=1}^r d\_i^2$$ where $d\_i \in \mathbb{N}\_{\ge 3}$ and $d\_i^2$ divides $N$.
**Question**: Must $r$ be greater than or equal to $9$?
*Checking* (with SageMath): It is true for $N \le 500000$.
*Remark*: If it is true in general then it is *optimal* because $$144 = 1 + 3^2 + 3^2 + 3^2+ 4^2 + 4^2 + 4^2 + 4^2 + 4^2 + 6^2$$
---
**Motivation** (from tensor category theory): If above question has a positive answer then by [1, Proposition 8.14.6] and [2, Theorem 3.4] a simple integral modular fusion category (over $\mathbb{C}$) would be of rank $ \ge 10 $. And for so, if required, we can also assume $d\_i$ not a prime-power by [3, Corollary 6.16]. Then the (SageMath) checking is $N \le 10^6$; and above optimality would be unchanged because $$ 116964 = 1 + 6^2 + 18^2 + 38^2 + 38^2 + 114^2 + 114^2 + 171^2 + 171^2 + 171^2$$
We did not add this additional assumption at the beginning because it seems to be true without it (experimentally), and without it the question is (number theoretically) more interesting.
With this additional assumption, the next example with $r=9$ is $$ 396900 = 1 + 18^2 + 30^2 + 70^2 + 70^2 + 210^2 + 210^2 + 315^2 + 315^2 + 315^2 $$
Above two examples with $r=9$ can be excluded from coming from a fusion ring because:
* First one: consider the simple object of FPdim $6$, multiply it with its dual (which must be itself here) and apply FPdim (ring homomorphism). Then you get $6^2=1+\cdots$, but there is no FPdim $\le 35$ (of a non-trivial simple object) which is not a multiple of $6$.
* Second one: as above, $18^2 - 1 = 323$ is odd and the only odd FPdim for a non-trivial simple object is $315$ , but $323-315 = 8$ and there is no non-trivial simple object of FPdim $\le 8$.
*Conclusion*: if above two examples are the only ones for $r=9$ (I checked that there is no other one for $N \le 10^6$), and if above question has a positive answer (we can put the additional assumption if required), then a simple integral modular fusion category (over $\mathbb{C}$) would be of rank $ \ge 11 $.
*References*
[1]: P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik; Tensor categories. Mathematical Surveys and Monographs (2015) 205.
[2]: J. Dong, S. Natale, L. Vendramin; Frobenius property for fusion categories of small integral dimension. J. Algebra Appl. 14 (2015), no. 2, 1550011.
[3]: D. Nikshych, Morita equivalence methods in classification of fusion categories, Hopf algebras and tensor categories, 289-325,
Contemp. Math., 585, Amer. Math. Soc. (2013).
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https://mathoverflow.net/users/34538
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Sum of squares and divisibility
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I hope so. But please double check (or, better, simplify) the argument below.
Denote $N=qs^2$ for $q$ squarefree. Then each $d\_i$ divides $s$, say $d\_i=s/m\_i$ and we get $$q=1/s^2+\sum\_{i=1}^r 1/m\_i^2, \quad\quad\quad (\heartsuit)$$
and the sum of $r+1$ square reciprocals is integer. Since $d\_i\geqslant 3$, we get $m\_i\leqslant s/3$. Subtracting the summands with $m\_i=1$ we get $(\heartsuit)$ with smaller $r$ and the same condition $m\_i\leqslant s/3$ (possibly $q$ is no longer squarefree, but we do not care anymore.) So, now each $m\_i$ is at least 2. So, we assume that $(\heartsuit)$ holds with $r\leqslant 8$, some integer $q$ and $s/3\geqslant m\_i\geqslant 2$.
If $r=8$ and all $m\_i$ are equal to 2, then $1/s^2$ must be an integer which is absurd. Otherwise $\sum 1/m\_i^2\leqslant 7\cdot 1/4+1/9$ and $1/s^2+\sum 1/m\_i^2\leqslant 1/36+7/4+1/9<2$, thus $q=1$.
Further we denote $s=m\_{r+1}$, and $(\heartsuit)$ reads as $$1=\sum\_{i=1}^{r+1} 1/m\_i^2. \quad\quad\quad(\clubsuit)$$
And we have a
**Special Condition.** $m\_{r+1}\geqslant 3\max(m\_1,\ldots,m\_r)$ and $m\_i$ divides $m\_{r+1}$ for all $i$.
What we do below is finding all solutions of $(\clubsuit)$ with $r\leqslant 8$. They all fail Special Condition. Let me start with listing them:
(a) $r=3$, $(2,2,2,2)$;
(b) $r=8$, $(2,2,2,3,6,6,6,6,6)$;
(c) $r=8$, $(2,2,3,3,4,4,4,4,6)$;
(d) $r=8$, $(2,2,2,4,4,4,6,6,12)$;
(e) $r=6$, $(2,2,2,4,4,4,4)$;
(f) $r=8$, $(2,2,2,3,3,12,12,12,12)$;
(g) $r=7$, $(2,2,2,3,4,4,12,12)$;
(h) $r=8$, $(2,2,2,3,9,4,4,36,36)$;
(i) $r=8$, $(2,2,2,3,4,4,12,15,20)$;
(j) $r=7$, $(2,2,3,3,3,3,6,6)$;
(k) $r=5$, $(2,2,2,3,3,6)$;
(l) $r=7$, $(2,2,2,3,3,7,14,21)$;
(m) $r=7$, $(2,2,2,3,3,9,9,18)$;
(n) $r=8$, $(3,3,3,3,3,3,3,3,3)$
(o) $r=0$, $(1)$.
Now goes a rather lengthy and boring proof that there are no other solutions of $(\clubsuit)$.
Denote by $2^\beta$ the maximal power of 2 which divides one of the numbers $m\_i$, $i=1,\ldots,r+1$. We have $\beta\geqslant 1$: otherwise all $1/m\_i^2$ are at most $1/9$, and not all equal to $1/9$, and RHS of $(\clubsuit)$ is less than 1. If we have exactly $h$ indices $i$ for which $2^\beta$ divides $m\_i$, then multiplying $(\clubsuit)$ by $2^{2\beta}$ and considering the expression modulo 4 we get $4|h$. Thus $h=8$ or $h=4$ or $h=0$.
Consider several cases.
0. $h=0$, all $m\_i$'s are odd. Then either $r=0$ and we get solution (0), or they all are not less than 3, and $\sum 1/m\_i^2\leqslant 9\cdot 1/9=1$, with equality corresponding to solution (n).
1. $h=8$ and $\beta \geqslant 2$. Then RHS of $(\clubsuit)$ does not exceed $8\cdot \frac1{16}+1\cdot \frac14<1$.
2. $h=8$ and $\beta=1$. Then eight even $m\_i$'s may be denoted by $2s\_i$, $s\_i$ are odd, $i=1,\ldots,8$, and $(\clubsuit)$ reads as
$$
\frac14\sum\_{i=1}^8\frac1{s\_i^2}+\frac{\varepsilon}{\ell^2}=1,\quad\quad\quad(\smile)
$$
where $\varepsilon=0$ if $r=7$ and $\varepsilon=1$ if $r=8$ (and $\ell$ is odd). At most three $s\_i$'s may be equal to 1, others are at least 3, also $\ell\geqslant 3$, thus LHS of $(\smile)$ is at most
$\frac14(3\cdot1+5\cdot \frac19)+\frac19=1$, the equality case is unique, it is the solution (b).
3. $h=4$ and $\beta\geqslant 3$. Analogously, $(\clubsuit)$ reads as
$$
\frac1{4^\beta}\sum\_{i=1}^4 \frac1{s\_i^2}+\sum\_{i=1}^{r-3}\frac1{\ell\_i^2}=1,\quad\quad\quad(\ast)
$$
where $s\_i$ are odd. At most 3 $\ell\_i$'s are equal to 2. Assume that at least one of $\ell\_i$'s is at least 4. Then LHS of $(\ast)$ is at most
$\frac1{64}\cdot 4+3\cdot\frac14+1\cdot\frac19+1\cdot\frac1{16}<1$, a contradiction. So, all $\ell\_i$'s are equal to 2 or 3, and we get contradiction modulo 8 after multiplying by $4^\beta$.
4. $h=4$, $\beta=2$. Now we get
$$
\frac1{16}\sum\_{i=1}^4 \frac1{s\_i^2}+\sum\_{i=1}^{r-3}\frac1{\ell\_i^2}=1,\quad\quad\quad(\diamond)
$$
where $s\_i$ are odd and $\ell\_i$ not divisible by 4. Assume that at most one $\ell\_i$'s is equal to 2, then LHS of $(\diamond)$ does not exceed
$\frac1{16}\cdot 4+1\cdot \frac14+4\cdot \frac19<1$, a contradiction. Thus we get $r\geqslant 5$ and we may suppose $\ell\_{r-3}=\ell\_{r-4}=2$, and $(\diamond)$ reads as
$$
\frac1{16}\sum\_{i=1}^4 \frac1{s\_i^2}+\sum\_{i=1}^{r-5}\frac1{\ell\_i^2}=\frac12.\quad\quad\quad(\diamond\diamond)
$$
If all $\ell\_i$'s ($i=1,\ldots,r-5$) are odd, we get a contradiction modulo 8 after multiplying by $16$. If none of $\ell\_i$'s ($i=1,\ldots,r-5$) equals to 2, then one of them is at least 6, and LHS of $(\diamond\diamond)$ does not exceed $\frac1{16}\cdot 4+2\cdot \frac19+\frac1{36}=1/2$. There exists unique equality case, it corresponds to solution (c). Thus, we may suppose that $r\geqslant 6$ and $\ell\_{r-5}=2$, and $(\diamond\diamond)$ reads as
$$
\frac1{16}\sum\_{i=1}^4 \frac1{s\_i^2}+\sum\_{i=1}^{r-6}\frac1{\ell\_i^2}=\frac14.\quad\quad\quad(\diamond\diamond\diamond)
$$
If exactly one of $\ell\_i$'s ($i=1,\ldots,r-6$) is even (recall that it is not however divisible by 4), we get a contradiction modulo 8 after multiplying by 16. If two of $\ell\_i$'s ($i=1,\ldots,r-6$) are even, then they are at least 6, and LHS of $(\diamond\diamond\diamond)$ is at most
$\frac1{16}(3+\frac19)+2\cdot \frac1{36}=\frac14$. Again, the equality case is unique, it provides a solution (d). Thus, all $\ell\_i$'s ($i=1,\ldots,r-6$) are odd. If $r=6$, we immediately get $s\_1=\ldots=s\_4=1$, this gives solution (e). So, $r\geqslant 7$, at most 3 of $s\_i$'s are equal to 1 and $$\sum\_{i=1}^{r-6}\frac1{\ell\_i^2}\geqslant \frac14-\frac1{16}\left(3+\frac19\right)=\frac1{18}.$$
Thus there is 3 or 5 between $\ell\_i$'s, $1\leqslant i\leqslant r-6$.
Well, proceed with cases. Assume that $\ell\_{r-6}=5$, but $\ell\_{r-7}\ne 3$ (or $r=7$ and $\ell\_1=5$). Then $\sum \frac1{s\_i^2}=4-16\sum\_{i=1}^{r-6}\frac1{\ell\_i^2}\geqslant 4-\frac{32}{25}$. Therefore at least three $s\_i$'s are equal to 1, say $s\_2=s\_3=s\_4=1$, and $(\diamond\diamond\diamond)$ reads as $\frac1{16s\_1^2}+\frac{\varepsilon}{\ell\_{r-7}^2}=\frac1{16}-\frac1{25}=\frac9{16\cdot 25}$ (here $\varepsilon=0$ if $r=7$ and $\varepsilon=1$ if $r=8$). We see that $r=8$, and we get an equation $\frac1{\ell\_1^2}=\frac{9}{16\cdot 25}-\frac1{16s\_1^2}=\frac{(3s\_1-5)(3s\_1+5)}{16\cdot 25\cdot s\_1^2}$. If 5 does not divide $s\_1$, then $3s\_1\pm 5$ is coprime to $25s\_1^2$, and the numerator of $\frac{(3s\_1-5)(3s\_1+5)}{16\cdot 25\cdot s\_1^2}$ may be equal to 1 only if both $3s\_1-5$ and $3s\_1+5$ are powers of 2, which is impossible (powers of 2 never differ by 10). So, $s\_1$ is divisible by 5 and $\frac9{16\cdot 25}\geqslant \frac1{\ell\_1^2}\geqslant \frac8{16\cdot 25}$, that is, $\ell\_1=7$, but this does not provide a solution.
Assume then that $\ell\_{r-6}=3$. Analogously, we get $\sum \frac1{s\_i^2}+\frac{16\varepsilon}{\ell\_{r-7}^2}=16(\frac14-\frac19)=\frac{20}{9}$. If all $s\_i$'s are at least 3, then LHS is at most $4/9+16/9=20/9$, and equality holds in the unique case, corresponding to solution $(f)$.
So let us suppose that $s\_4=1$, we get $\sum\_{i=1}^3 \frac1{s\_i^2}+\frac{16\varepsilon}{\ell\_{r-7}^2}=\frac{11}9$. If all $s\_1,s\_2,s\_3$ are at least 3, we get $r=8$ and $\frac{11}9\geqslant \frac{16}{\ell\_1^2}\geqslant \frac{11}9-3\cdot \frac19=\frac{8}9$, that is impossible (recall that $\ell\_1$ is odd). So, without loss of generality $s\_3=1$ and $\frac1{s\_1^2}+\frac1{s\_2^2}+\frac{16\varepsilon}{\ell\_1^2}=\frac29$. Clearly $s\_1,s\_2\geqslant 3$. If $r=7$, i.e., $\varepsilon=0$, this yields a solution with $s\_1=s\_2=3$, that is (g). So assume that $r=8$. Since $\frac{16}{\ell\_1^2}<\frac2{9}$, we get $\ell\_1\geqslant 9$. Thus $\frac1{s\_1^2}+\frac1{s\_2^2}\geqslant \frac29-\frac{16}{81}=\frac2{81}$. It yields $\min(s\_1,s\_2)\leqslant 9$.
If now $s\_1,s\_2$ are at least 5, then $\frac{16}{\ell\_1^2}\geqslant \frac29-\frac2{25 }=\frac{32}{225}$, or $\frac{225}2\geqslant \ell\_1^2$, this yields $\ell\_1=9$. Thus $\frac1{s\_1^2}+\frac1{s\_2^2}=\frac2{81}$. If $s\_1=s\_2=9$, we get solution (h). Otherwise, say, $s\_2=7$,
this does not lead to a solution.
Finally, if $s\_2=3$, we get an equation $\frac1{s\_1^2}+\frac{16}{\ell\_1^2}=\frac19$, or
$\frac{16}{\ell\_1^2}=\frac19-\frac1{s\_1^2}=\frac{(s\_1-3)(s\_1+3)}{9s\_1^2}$. If 3 does not divide $s\_1$, then $s\_1\pm 3$ are coprime to $9s\_1^2$, so both $s\_1-3$ and $s\_1+3$ must be powers of 2. This happens only for $s\_1=5$, and we get solution
(i). If $s\_1=3k$, we get $\frac{(k-1)(k+1)}{9k^2}=\frac{16}{s\_1^2}$, thus $k^2-1=(k-1)(k+1)$ is a positive perfect square, which is impossible.
5. Uff, now let $h=4$ and $\beta=1$. We get
$$
\frac14\sum\_{i=1}^4\frac1{s\_i^2}+\sum\_{i=1}^{r-3}\frac1{\ell\_i^2}=1
$$
for odd $s\_i$ and $\ell\_i$. Multiplying by 4 and considering modulo 8 we see that $r-3$ must be even, thus $r$ is odd, so $r\leqslant 7$. Therefore $\sum\frac1{\ell\_i^2}\leqslant \frac49$ and $\sum\frac1{s\_i^2}\geqslant 4(1-\frac49)=\frac{20}9$. This implies that at least two of $s\_i$'s are equal to 1, say $s\_3=s\_4=1$. We get from above $\frac1{s\_1^2}+\frac1{s\_2^2}\geqslant \frac29$, so either $r=7$, $\ell\_1=\ldots=\ell\_4=3$, $s\_1=s\_2=3$,
that's solution (j), or, say, $s\_2=1$, and we get $\frac1{4s\_1^2}+\sum \frac1{\ell\_i^2}=\frac14$. There is a solution (a) with $s\_1=1$ and $r=3$. Otherwise $s\_1\geqslant 3$ and we get $\sum \frac1{\ell\_i^2}\in (\frac14-\frac1{36},\frac14)$. Thus, one of $\ell\_i$'s is equal to 3, say $\ell\_{r-3}=1$, and $\frac1{4s\_1^2}+\sum\_{i=1}^{r-4}\frac1{\ell\_i^2}=\frac5{36}$. Assume that all remaining in LHS $\ell\_i$'s are greater than 3, and that $s\_1$ is greater than 3. Then $\frac1{4s\_1^2}+\sum\_{i=1}^{r-4}\frac1{\ell\_i^2}\leqslant \frac1{100}+3\cdot \frac1{25}=\frac{13}{100}<\frac5{36}$.
Next case. $s\_1=3$. Then $\sum\_{i=1}^{r-4}\frac1{\ell\_i^2}=\frac19$. If $r=5$, we get solution (k). Otherwise $r\geqslant 6$, then remaining $\ell\_i$'s are at least 5, not all equal to 5, thus $\sum\_{i=1}^{r-4}\frac1{\ell\_i^2}\leqslant \frac1{25}+\frac1{25}+\frac1{49}<\frac19$.
Finally, let $s\_1>3$, but $\ell\_{r-4}=3$. Then $\frac1{4s\_1^2}+\sum\_{i=1}^{r-5}\frac1{\ell\_i^2}=\frac1{36}$. If all $\ell\_i$'s, $i\leqslant r-5$, are at least 11, then LHS is at most $1/100+2/121<1/36$. So, one of them equals to 7 or 9.
This leads to equations $\frac1{4s\_1^2}+\frac1{\ell\_1^2}=\frac{13}{36\cdot 49}$ and $\frac1{4s\_1^2}+\frac1{\ell\_1^2}=\frac{5}{4\cdot 81}$, respectively.
For $\frac1{4s\_1^2}+\frac1{\ell\_1^2}=\frac{13}{36\cdot 49}$ note that 7 must divide both $s\_1$ and $\ell\_1$ (otherwise we get a contradiction modulo 7 multiplying by 49, due to 13 being quadratic non-residue modulo 7). If, say, $s\_1=7z$, $\ell\_1=7w$, we get an equation $\frac1{4z^2}+\frac1{w^2}=\frac{13}{36}=\frac14+\frac19$. This is only possible for $z=1$, $w=3$ (for $w=1$ LHS is too large, otherwise too small). Surprisingly, this solution does not enjoy SC again.
For $\frac1{4s\_1^2}+\frac1{\ell\_1^2}=\frac{5}{4\cdot 81}$, we analogously get that $s\_1$, $\ell\_1$ must be divisible by 9 (since 5 is not a quadratic residue modulo 3). So, we get an equation $\frac1{4z^2}+\frac1{w^2}=\frac{5}4$ which has the unique solution $z=w=1$, which leads to solution (m).
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8
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https://mathoverflow.net/users/4312
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406674
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https://mathoverflow.net/questions/406638
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6
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$\newcommand{\U}{\mathcal{U}}$
$\newcommand{\P}{\mathbb{P}}$
$\newcommand{\Q}{\mathbb{Q}}$
$\newcommand{\F}{\mathcal{F}}$
Recall the following equivalent definitions of a Ramsey ultrafilter over $\omega$:
>
> **Theorem (Ramsey Ultrafilter).** Let $\U$ be a non-principal ultrafilter over $\omega$. TFAE:
>
>
> 1. For any partition $F : [\omega]^n \to k$, there is a homogeneous set $H \in \U$.
> 2. For any partition $F : [\omega]^2 \to 2$, there is a homogeneous set $H \in \U$.
> 3. For every partition of $\omega$, $\{A\_n : n \in \omega\}$ such that $A\_n \notin \U$ for all $n$, there is a sequence $\langle{x\_n : n \in \omega}\rangle$ such that $x\_n \in A\_n$ and $\{x\_n : n \in \omega\} \in U$.
> 4. Given a decreasing sequence of sets $A\_0 \supseteq A\_1 \supseteq \cdots$ in $\U$, there exists a set $\{x\_i\}\_{i \in \omega} \in \U$ such that for all $n \in \omega$, $x\_{n+1} \in A\_{x\_n}$.
>
>
>
I wish to generalise this notion to countable posets, and am in particular more interested in criterion (4). More specifically, let $(\P,\leq)$ be a countable poset, and let $\U$ be an ultrafilter over $\P$. Let's suppose we say that $\U$ is a *Ramsey ultrafilter* over $\P$ if for any partition $F : [\P]^n \to k$, there is a homogeneous set $H \in \U$.
I wish to attain an equivalence that is of something like the following:
>
> **Conjecture.** Let $\U$ be a non-principal ultrafilter over a countable poset $(\P,\leq)$ (with possibly more assumptions required on $(\P,\leq)$). TFAE:
>
>
> 1. $\U$ is Ramsey. That is, for any partition $F : [\P]^n \to k$, there is a homogeneous set $H \in \U$.
> 2. (?) Let $\F = \{A\_p : p \in \P\}$ be a family of sets in $\U$ such that $p < q \implies A\_p \supseteq A\_q$. Then there exists a set $\Q \subseteq \P$, $\Q \in \U$, such that for all $p,q \in \Q$, if $p < q$ then $q \in A\_p$.
>
>
>
If this is false in full generality, what additional assumptions should we impose on $(\P,\leq)$? Several assumptions I'm more than willing to impose are:
1. Every chain in $\P$ is well-ordered.
2. For all $p,q \in \P$, there are only finitely many $r$ such that $p < r < q$.
But I wish to stay away from the linear order case.
---
**EDIT**: Allow me to explain the main obstacle I face in generalising the equivalence: The most accessible proof of (3) $\implies$ (4) appears to be Jech's book, Lemma 9.2. His proof is as follows:
>
> [Let] $X\_0 \supseteq X\_1 \supseteq \cdots$ be sets in $D$ [where $D$ is a Ramsey ultrafilter]. Since $D$ is a p-point, there exists $Y \in D$ such that each $Y - X\_n$ is finite. Let us define a sequence $y\_0 < y\_1 < \cdots$ in $Y$ as follows:
>
>
> * $y\_0 = $ the least $y\_0 \in Y$ such that $\{y \in Y : y > y\_0\} \subseteq X\_0$.
> * $y\_1 = $ the least $y\_1 \in Y$ such that $\{y \in Y : y > y\_2\} \subseteq X\_{y\_0}$.
> * $\dots$
> * $y\_n = $ the least $y\_n \in Y$ such that $\{y \in Y : y > y\_{n-1}\} \subseteq X\_{y\_{n-1}}$.
>
>
> For each $n$, let $A\_n = \{y \in Y : y\_n < y \leq y\_{n+1}\}$. Since $D$ is Ramsey, there exists a set $\{z\_n\}\_{n=0}^\infty$ such that $z\_n \in A\_n$ for all $n$.
>
>
> We observe that for each $n$, $z\_{n+2} \in X\_{z\_n}$: Since $z\_{n+2} > y\_{n+2}$, we have $z\_{n+2} \in X\_{y\_{n+2}}$, and since $y\_{n+1} \geq z\_n$, we have $X\_{y\_{n+1}} \subseteq X\_{z\_n}$ and hence $z\_{n+2} \in X\_{z\_n}$.
>
>
> Thus if we let $a\_n = z\_{2n}$ and $b\_n = z\_{2n+1}$, for all $n$, then either $\{a\_n\}\_{n=0}^\infty \in D$ or $\{b\_n\}\_{n=0}^\infty \in D$; and in either case we get a sequence that satisfies [the property (4)].
>
>
>
For general posets, we cannot so simply split to two subsets $a\_n = z\_{2n}$ and $b\_n = z\_{2n+1}$. If there are infinitely many branches, then we cannot guarantee one such subset will be in $D$. It also makes no sense to consider $\sigma$-complete Ramsey ultrafilters.
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https://mathoverflow.net/users/146831
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Ramsey ultrafilters on partial order
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Presumably in (2) you meant to assume the sets $A\_p$ belong to $\mathcal U$.
The nontrivial thing is to show that (1) implies (2). The main point is that the Ramsey property of $\mathcal U$ implies that $\mathbb P$ has a $\mathcal U$-large suborder isomorphic either to $\omega$, $\omega^\*$, or an infinite discrete order. Since (2) only depends on the $\mathcal U$-almost everywhere structure of $\mathbb P$, one only needs to check that (2) holds for these three orders. The case $\mathbb P = \omega$ is standard and the other two cases are basically trivial. Here are the details.
Enumerate $\mathbb P$ as $(p\_n)\_{n < \omega}$ and for $n < m$, set $f(p\_n,p\_m) = 0$ if $p\_n < p\_m$, $f(p\_n,p\_m) = 1$ if $p\_n > p\_m$, and $f(p\_n,p\_m) = 2$ if $p\_n$ and $p\_m$ are incomparable.
If $f$ has a $\mathcal U$-large 0-homogeneous set $H$, then you are essentially in the standard situation since $H \cong \omega$, so you finish using the standard argument.
If $f$ has a $\mathcal U$-large 1-homogeneous set $H$, then letting $n = \min\{k : p\_k\in H\}$, $p\_n$ is the maximum element of $H$. So if $(A\_p)\_{p\in \mathbb P}$ is a sequence with $A\_p\in \mathcal U$ and $p < q$ implies $A\_p\supseteq A\_q$, then for $p < q$ in $H\cap A\_{p\_n}$, $q\in A\_p$ since $p \leq p\_n$ and hence $A\_{p\_n}\subseteq A\_p$.
If $f$ has a $\mathcal U$-large 2-homogenous set, then there is a $\mathcal U$-large set of incomparables, and so (2) holds vacuously.
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https://mathoverflow.net/users/102684
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406676
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https://mathoverflow.net/questions/406663
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|
$\newcommand{\Cats}{\mathsf{Cats}}\newcommand{\MonCats}{\mathsf{MonCats}}\newcommand{\BrMonCats}{\mathsf{BrMonCats}}\newcommand{\SymMonCats}{\mathsf{SymMonCats}}\newcommand{\CMon}{\mathsf{CMon}}\newcommand{\Mon}{\mathsf{Mon}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\Ab}{\mathsf{Ab}}\newcommand{\Grp}{\mathsf{Grp}}\newcommand{\Sets}{\mathsf{Sets}}\newcommand{\eHom}{\mathbf{Hom}}\newcommand{\C}{\mathcal{C}}\newcommand{\V}{\mathcal{V}}\newcommand{\Obj}{\mathrm{Obj}}$Let $\C$ be a $\V$-enriched category, let $V\in\Obj(\V)$, and let $A\in\Obj(\C)$.
* The **tensor of $V$ with $A$** (also called the **copower of $V$ with $A$**) is, if it exists, the object $V\odot A$ of $\C$ such that we have a $\V$-natural isomorphism
$$\eHom\_\C(V\odot A,-)\cong\eHom\_\V(V,\eHom\_\C(A,-)).$$
* Dually, the **cotensor of $V$ with $A$** (also called the **power of $V$ with $A$**) is, if it exists, the object $V\pitchfork A$ of $\C$ such that we have a $\V$-natural isomorphism
$$\eHom\_\C(-,V\pitchfork A)\cong\eHom\_\V(V,\eHom\_\C(-,A)).$$
Moreover, $\C$ is called **$\V$-co/tensored** if it has all co/tensors.
An example of these is given by any co/complete category $\mathcal{C}$, whose $\Sets$-co/tensors are given by
\begin{align\*}
X\odot A &\cong \coprod\_{x\in X}A,\\
X\pitchfork A &\cong \prod\_{x\in X}A.
\end{align\*}
Another example is given by the category $\Ab$:
* $\Ab$ is enriched over itself: given $A,B\in\Obj(\Ab)$, we have an abelian group $\eHom\_\Ab(A,B)$ whose product $(f,g)\mapsto f\*g$ is obtained via pointwise multiplication, i.e. by defining $[f\*g](a)=f(a)g(a)$. This relies crucially on the commutativity of $B$, which ensures that $f\*g$ is again a morphism of groups:
\begin{align\*}
[f\*g](ab) &= f(ab)g(ab)\\
&= f(a)\color{red}{f(b)}\color{blue}{g(a)}g(b)\\
&= f(a)\color{blue}{g(a)}\color{red}{f(b)}g(b)\\
&= [f\*g](a)[f\*g](b).
\end{align\*}
* $\Ab$ is tensored over itself via the tensor product of abelian groups $(A,B)\mapsto A\otimes\_\mathbb{Z}B$;
* $\Ab$ is cotensored over itself via the internal $\eHom$ given above, $(A,B)\mapsto \eHom\_\Ab(A,B)$.
Now, $\Ab$ is not enriched over $\Grp$, as there is no sensible tensor product for the latter; however it is "faux co/tensored" over it, as we have isomorphisms
\begin{align\*}
\eHom\_\Ab(G^\mathrm{ab}\otimes\_\Z A,B) &\cong \eHom\_\Grp(G,\eHom\_\Ab(A,B)),\\
\eHom\_\Ab(A,\eHom\_\Grp(G,B)) &\cong \eHom\_\Grp(G,\eHom\_\Ab(A,B)),
\end{align\*}
so $G“\odot”A=G^\mathrm{ab}\otimes\_\Z A$ and $G“\pitchfork”A=\eHom\_\Grp(G,B)$.
This situation is not exclusive to $\Grp$ and $\Ab$: it also occurs with $\Mon$ and $\CMon$, with $\BrMonCats$ and $\SymMonCats$, and seems more generally to occur with pairs of the form $(\Mon(\C),\CMon(\C))$ for $\C$ a symmetric monoidal category.
---
A second related point is that one may use the $\Sets$-co/tensoring of $\Grp$ together with the forgetful functor $|{-}|\colon\Grp\to\Sets$ to $\Sets$ to define "half-tensor products" $\triangleleft$ and $\triangleright$, given by
\begin{align\*}
G\triangleleft H &= |H|\odot G,\\
&\cong \coprod\_{h\in H}G,\\
G\triangleright H &= |G|\odot H,\\
&\cong \coprod\_{g\in G}H.
\end{align\*}
As noted [here](https://math.stackexchange.com/questions/616582), $G\triangleleft H$ is the free group on symbols $a\otimes b$ quotiented by the left distributivity relations $(a+b)\otimes c\sim a\otimes c+b\otimes c$, and similarly for $\triangleright$.
Because of this last point, while monoids in $(\Ab,\otimes\_\Z,\Z)$ are rings, the "monoids" in $(\Grp,\triangleleft,?)$ should be [near-rings](https://en.wikipedia.org/wiki/Near-ring)―rings with not necessarily commutative addition and only the left distributive law. The problem, however, is that $\triangleleft$ doesn't give $\Grp$ a monoidal category structure: it seems to form at best some variant of the notion of a "[lax monoidal category](https://ncatlab.org/nlab/show/lax+monoidal+category)" on it.
As with the faux co/tensors above, this kind of tensor product seems to occur also in many other contexts, including $\MonCats$ with $\Cats$-tensors or perhaps $\BrMonCats$ with "faux $\MonCats$-tensors".
---
This is a rather strange situation: we have these very natural "faux co/tensors", but the usual category-theoretic notions don't quite capture them. [Over Zulip](https://categorytheory.zulipchat.com/#narrow/stream/229199-learning.3A-questions/topic/is.20Ab.20tensored.20over.20Grp.3F), Reid Barton suggested grouping the categories $\Sets$, $\Grp$, $\Ab$, $\Ab$, $\ldots$ ($=(\Grp\_{\mathbb{E}\_n}(\Sets))\_{n\in\mathbb{N}}$) into an "$\mathbb{N}$-graded monoidal category", but again the units and associators seem to be problematic...
So---shortly---what exactly is going on here?
* How should we view these "faux co/tensors" of $\Mon\_{\mathbb{E}\_{n}}(\C)$ by $\Mon\_{\mathbb{E}\_{\leq n-1}}(\C)$?
* What exactly are the "faux monoidal structures" $\triangleleft$ and $\triangleright$ on $\Grp$, whose monoids are supposed to recover near-rings?
|
https://mathoverflow.net/users/130058
|
This is not a tensor: tensoring abelian groups over groups
|
The one line answer is that the category $\mathsf{Ab}$ of abelian groups is enriched over the **skew-monoidal category** $\mathsf{Gp}$ of groups, and that this "faux-tensor" defines a **skew-action** of the skew-monoidal category $\mathsf{Gp}$ on $\mathsf{Ab}$.
A *skew-monoidal structure* on a category $\mathcal{C}$ consists of a "tensor product" functor $\boxtimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}$, a "unit" object $I \in \mathcal{C}$, and "associativity and unit constraint" natural transformations $\alpha \colon (X \boxtimes Y) \boxtimes Z \to X \boxtimes (Y \boxtimes Z)$, $\lambda \colon I \boxtimes X \to X$, and $\rho \colon X \to X \boxtimes I$, satisfying the original five coherence axioms of Mac Lane. The important point is that these associativity and unit constraints are *not required to be invertible*. This notion was introduced by Szlachányi in his paper
>
> Kornél Szlachányi. Skew-monoidal categories and bialgebroids. Adv. Math. 231 (2012), no. 3-4, 1694--1730. <https://doi.org/10.1016/j.aim.2012.06.027>
>
>
>
and has been much studied since, especially by the Australian school of category theory.
The "half-tensor products" of groups that you describe are part of a skew-monoidal structure on the category $\mathsf{Gp}$ of groups. This skew-monoidal structure is an instance of the family of examples described in Example 2.7 of my paper:
>
> Alexander Campbell. Skew-enriched categories. Applied Categorical Structures 26 (2018), no. 3, 597--615. <https://doi.org/10.1007/s10485-017-9504-0>
>
>
>
The tensor product $G \boxtimes H$ of two groups $G$ and $H$ is the group you denote by $G \triangleleft H$, i.e. the copower of $G$ by the underlying set of $H$. Note that group homomorphisms $G \boxtimes H \to K$ correspond to functions $G \times H \to K$ that are group homomorphisms in the first variable. The unit object is the free group on one generator, i.e. $\mathbb{Z}$. The associativity and unit constraints are a little more complicated to describe, but suffice it to say that they are not invertible.
This skew-monoidal structure on $\mathsf{Gp}$ is closed: the functor $- \boxtimes H$ has a right adjoint which sends a group $K$ to the group $[H,K]$ of all functions from $H$ to $K$ with the pointwise group structure; this group $[H,K]$ is the *internal hom* for this skew-monoidal structure on $\mathsf{Gp}$. Thus $\mathsf{Gp}$ is also a **skew-closed category** in the sense introduced by Ross Street in his paper:
>
> Ross Street. Skew-closed categories. J. Pure Appl. Algebra 217 (2013), no. 6, 973--988. <https://doi.org/10.1016/j.jpaa.2012.09.020>
>
>
>
Now, just as one can define categories enriched over monoidal categories, one can also define categories enriched over skew-monoidal categories. (In the terminology of my paper cited above, this is the same thing as a "left normal skew-enrichment" over the skew-monoidal category. Enrichment over skew-closed categories is defined in Street's paper cited above.)
We can define an enrichment of $\mathsf{Ab}$ over the above skew-monoidal structure on $\mathsf{Gp}$ as the change of base of the usual self-enrichment of $\mathsf{Ab}$ along the inclusion functor $\mathsf{Ab} \to \mathsf{Gp}$ equipped with the lax monoidal structure whose tensor constraint $A \boxtimes B \to A \otimes B$ is the homomorphism $U(B) \odot A \to A \otimes B$ whose component at an element $b \in B$ is $-\otimes b \colon A \to A \otimes B$.
Unpacking this, we have that, for each pair of abelian groups $A$ and $B$, the hom-group $\underline{\operatorname{Hom}}(A,B)$ is the usual group of group homomorphisms from $A$ to $B$, with its pointwise group structure, but where we have forgotten that it's abelian. For each triple of abelian groups $A$, $B$, and $C$, the composition homomorphism $\underline{\operatorname{Hom}}(B,C) \boxtimes \underline{\operatorname{Hom}}(A,B) \to \underline{\operatorname{Hom}}(A,C)$ corresponds to the usual composition function $\operatorname{Hom}(B,C) \times \operatorname{Hom}(A,B) \to \operatorname{Hom}(A,C)$, but where we have forgetten that it's a group homomorphism in the second variable. Similarly, the unit homomorphisms $\mathbb{Z} \to \underline{\operatorname{Hom}}(A,A)$ simply pick out the identity homomomorphisms.
(Note that this enrichment of $\mathsf{Ab}$ over $\mathsf{Gp}$ can also be seen an instance of Example 2.7 of my paper cited above, since the category of abelian groups is equivalent to the category of group objects in $\mathsf{Gp}$.)
As you've spelled out in your question, the hom-functor $\underline{\operatorname{Hom}} \colon \mathsf{Ab}^\mathrm{op} \times \mathsf{Ab} \to \mathsf{Gp}$ is part of a two-variable adjunction, and so there are defined tensoring and cotensoring operations of an abelian group by a group. In particular, the tensoring operation defines a **skew-action** of the skew-monoidal category $\mathsf{Gp}$ on the category $\mathsf{Ab}$, in the sense of the paper:
>
> Stephen Lack and Ross Street. Skew-monoidal reflection and lifting theorems. Theory Appl. Categ. 30 (2015), Paper No. 28, 985--1000. <http://tac.mta.ca/tac/volumes/30/28/30-28abs.html>
>
>
>
Note that a skew-action of a skew-monoidal category $\mathcal{V}$ on a category $\mathcal{C}$ is simply an oplax monoidal functor $\mathcal{V} \to \operatorname{Fun}(\mathcal{C},\mathcal{C})$.
|
8
|
https://mathoverflow.net/users/57405
|
406678
| 166,664 |
https://mathoverflow.net/questions/406680
|
1
|
I'm trying to follow an argument in C. Giraud's "**High Dimensional Statistics**" (2nd Ed, p. 11 / $\S$ 1.2.3). The specific page is accessible via Google Books [here](https://www.google.com/books/edition/Introduction_to_High_Dimensional_Statist/YiM_EAAAQBAJ?hl=en&gbpv=1&dq=high%20dimensional%20statistics&pg=PR19&printsec=frontcover) but the formatting is awful.
Suppose that $\epsilon\_j \sim {\cal N}(0,1), j = 1, \ldots, p$ and these are all independent.
Then $\mathbb{P}[ \max\_j |\epsilon\_j| \geq x] = 1 - (1 - \mathbb{P}[|\epsilon\_1| > x])^p$. I understand the steps to get the expression on the right from the one on the left.
What I don't understand is the next part, where a limit is taken (?) in $p$:
$\mathbb{P}[ \max\_j |\epsilon\_j| \geq x] = 1 - (1 - \mathbb{P}[|\epsilon\_1| \geq x])^p \sim^{p \rightarrow \infty } p \mathbb{P}[ |\epsilon\_1| \geq x]$
I guess the result is approximate but I cannot grasp the necessary steps. Any help is appreciated!
|
https://mathoverflow.net/users/423486
|
tail probability of max of Gaussians
|
The correct version of this formula is
$$P(\max\_j |\epsilon\_j| \ge x) = 1-(1-P(|\epsilon\_1| \ge x))^p
\underset{x \to\infty}\sim p \,P(|\epsilon\_1| \ge x)$$
for each real $p>0$,
which follows because $(1-u)^p=1-(p+o(1))u$ as $u\to0$.
(The reproduction quality of the preview of the book is indeed terrible. It is also clear that $1-(1-P(|\epsilon\_1| \ge x))^p
\underset{p\to\infty}\to1$ for each real $x$.)
|
1
|
https://mathoverflow.net/users/36721
|
406681
| 166,665 |
https://mathoverflow.net/questions/406689
|
1
|
I posed [this question on math.stackexchange.com](https://math.stackexchange.com/q/4279365/64809) but have gotten no answer. So I post the question here in order to obtain an answer.
---
$\forall x\in \mathbf R^{n+1}$, let $x\_{(0)}\le x\_{(1)}\le\,\cdots\le x\_{(n)}$ denote the non-decreasing rearrangement of $x$. Suppose $x,y\in\mathbf R^{n+1}$,
$$\sum\_{i=0}^k x\_{i}\le \sum\_{i=0}^k y\_{i},\quad \forall k\in\{0,1,2,\cdots,n\} \tag1$$
and
$$x\_{i-1}\le y\_i,\quad\forall i\in\{1,2,\cdots,n\}. \tag2$$
Show
$$\sum\_{i=0}^k x\_{(i)}\le \sum\_{i=0}^k y\_{(i)},\quad \forall k\in\{0,1,2,\cdots,n\}.$$
---
Here is my attempt at the proof with one obstacle I fail to overcome.
Let $l\_i$ be the permutation of $\{0,1,2,\cdots,n\}$ such that $y\_{(i)}=y\_{l\_i},\, \forall i\in\{0,1,2,\cdots,n\}$. For any $k\in\{1,2,\cdots,n\}$,
$$\sum\_{i=0}^k (y\_{(i)}-x\_{(i)})=I\big(0\in\{l\_i|i\in\{0,1,\cdots,k\}\}\big)(y\_0-x\_{(k)})+\sum\_{\substack{i=0\\l\_i\neq0}}^k (y\_{l\_i}-x\_{l\_i-1})+\Big(\sum\_{\substack{i=0\\l\_i\neq0}}^k x\_{l\_i-1}-\sum\_{i=0}^{k-1}x\_{(i)}\Big)$$
where $I$ stands for the characteristic function. Every term in the first summation term on the right hand side of the above equation is nonnegative by inequality $(2)$. The term in the second parenthesis is nonnegative by virtue of the definition of $x\_{(i)}$. Presumably I should use Inequality $(1)$ to show the nonnegativity of the first term of the above equation. But I do not see how.
|
https://mathoverflow.net/users/32660
|
A sufficient condition for weak majorization from below
|
We should prove that the sum of any $k$ $y$'s is not less than the sum of certain $k$ $x$'s (indeed, this property is equivalent to the condition that the sum of $k$ smallest $y$'s is not less than the sum of $k$ smallest $x$'s.) Let our $k$ $y$'s contain $y\_i$'s for $i=0,1,\ldots,p$, but do not contain $y\_{p+1}$ (there exists unique such $p$). Bound the sum $y\_0+\ldots+y\_p$ by $x\_0+\ldots+x\_p$ and other $y\_i$'s by corresponding $x\_{i-1}$'s.
|
1
|
https://mathoverflow.net/users/4312
|
406691
| 166,667 |
https://mathoverflow.net/questions/406604
|
3
|
**Let $\Omega$ be domain in $\mathbb{C}^n$. Suppose we have taken two distinct points from $\Omega$. Does there exist a domain $U$ in $\mathbb{C}$ such that there is a holomorphic function from $U$ to $\Omega$ whose range contains these two points?**
I tried to prove the identity theorem in several complex variables. Then my mind gave me that above question.
|
https://mathoverflow.net/users/422133
|
Holomorphic connectedness in several complex variables
|
Let $\Omega$ be a domain in $\mathbb C^n$. Fix two points $z\_0$, $z\_1$ in $\Omega$. Then
there exists a curve $\alpha : [0, 1] \to \Omega$ connecting these points. Using the Weierstrass approximation theorem there is a polynomial map $P : [0, 1] \to\Omega$ with
$P(0) = z\_0$ and $P(1) = z\_1$. Then it is easy to choose a simply connected domain
$D\subset\mathbb C$, $[0, 1]\subset D$, such that $P(D)\subset\Omega$. By the Riemann mapping theorem we can conclude that $z\_0$, $z\_1$
lie on an analytic disc $\phi : \mathbb D \to \Omega$.
Even more, this is also true for complex manifolds according to a result by Winkelmann.
|
2
|
https://mathoverflow.net/users/343739
|
406697
| 166,669 |
https://mathoverflow.net/questions/406699
|
6
|
Is there an example of a birational morphism of smooth complex projective varieties $f\colon X\to Y$, that cannot be factored into a chain $X\to X\_1\to\cdots\to X\_n\to Y$ of blow-down along smooth centers?
(By weak factorization theorem, we know in general that $f$ can be factorized into a zig-zag of blow-ups and blow-downs along smooth centers.)
|
https://mathoverflow.net/users/nan
|
Birational morphism that is not successive blow-down along smooth centers?
|
Let $X \subset \mathbb{P}^2\_{x\_i} \times \mathbb{P}^6\_{y\_j}$ be given by the equations
$$
x\_1y\_1 + x\_2y\_2 + x\_3y\_3 = x\_1y\_4 + x\_2y\_5 + x\_3y\_6 = 0.
$$
It is smooth because its projection to $\mathbb{P}^2$ is a $\mathbb{P}^4$-fibration. This also implies that the rank of the Picard group of $X$ is 2. Now let
$$
f \colon X \to \mathbb{P}^6
$$
be the projection. It is a birational morphism, and if it is a sequence of smooth blowups, it is itself a smooth blowup (because the difference of the Picard ranks is 1). But it is not a smooth blowup, because $f$ has 1-dimensional fibers over a codimension 2 subvariety of $\mathbb{P}^6$ and 2-dimensional fiber over a point.
|
11
|
https://mathoverflow.net/users/4428
|
406702
| 166,670 |
https://mathoverflow.net/questions/406705
|
7
|
Suppose that $T \in TC(l^2( \mathbb{Z}))$ is trace class.
Consider its kernel $ T(i,j) = \langle e\_i, T e\_j \rangle $ where $ \{e\_i\}\_{i \in \mathbb{Z}}$ is an ONB for $l^2( \mathbb{Z})$. Now, consider the operator given by the kernel $T(i,j) K(i,j) $ for some numbers $K(i,j)$ such that $\sup\_{i,j} \vert K(i,j)\vert < \infty$.
Is this operator still trace class?
My thoughts:
The new operator is the Hadamard/Schur product of $K \circ T$ for some operator $K$ which we do not know is bounded. If $K$ is bounded in operator norm then $\vert \vert K \circ T \vert \vert\_1 \leq \vert \vert K \vert \vert\_\infty \vert \vert T \vert \vert\_1 $. But our condition on $K$ is not enough to ensure that.
We can split $K$ and $T$ up into real and imaginary parts and then their real and imaginary parts into positive and negative parts. By a triangle inequality we can estimate each of the 16 terms $K\_i \circ T\_j$ where $K\_i$ and $T\_j$ are positive. By the Schur product theorem then $K\_i \circ T\_j$ is also positive. Hence we can compute its trace norm by
\begin{align\*}
\vert \vert K\_i \circ T\_j \vert \vert\_1 &= Tr( K\_i \circ T\_j ) = \sum\_n K\_i(n,n) T\_j(n,n)\\
&\leq \sup\_{n} K\_i(n,n) \sum\_n T\_j(n,n) \leq \sup\_{n} K\_i(n,n) \vert \vert T\_j \vert \vert\_1 \leq \sup\_{n} K\_i(n,n) \vert \vert T \vert \vert\_1
\end{align\*}
where $K\_i(n,n), T\_j(n,n) \geq 0$ since $K\_i$ and $T\_j$ are positive operators.
We can bound the matrix elements of the real and imaginary parts of $K$, but unfortunately we can't bound the the matrix elements of the positive and negative parts. I suspect one can use this insight to construct a counterexample.
|
https://mathoverflow.net/users/143779
|
If I multiply the coefficients of a trace-class operator with bounded complex numbers is it still trace class?
|
It's a little more complicated than I thought! Frederik Ravn Klausen pointed out an error. Still, I maintain that the product needn't even be bounded.
As the answer to [this question](https://math.stackexchange.com/questions/2907991/hadamard-product-optimal-bound-on-operator-norm) shows, in $M\_n$ you can find a unitary $U$ and an matrix $A$ whose entries all have modulus 1, whose Hadamard product $A\bullet U$ has operator norm $\sqrt{n}$. Using the duality between operator norm and trace class norm, find a trace class matrix $B$ with trace norm 1 such that ${\rm tr}(B^\*(A\bullet U)) = \sqrt{n}$.
We have an identity ${\rm tr}(B^\*(A\bullet U)) = {\rm tr}((A\bullet B)^\*U)$. Therefore the trace norm of $A\bullet B$ is at least (in fact, exactly) $\sqrt{n}$. Now $\bigoplus 2^{-n} B\_{5^n}$ is trace class, and its Hadamard product with $\bigoplus A\_{5^n}$ is unbounded.
|
10
|
https://mathoverflow.net/users/23141
|
406706
| 166,672 |
https://mathoverflow.net/questions/406713
|
1
|
Let $G=(V\_1,E\_1)$ be a simple graph with vertex set $\{v\_1,v\_2,\ldots,v\_n\}$ and let $G'=(V\_2,E\_2)$ be another copy of $G$ with vertex set $\{u\_1,u\_2,\ldots,u\_n\}$. Assume $V\_1\cap V\_2= \emptyset$.
Let $H=(V,E)$ be a graph with $V=V\_1 \cup V\_2$ and $E=E\_1\cup E\_2\cup \{u\_1v\_1，u\_2v\_2, \ldots, v\_nv\_n\}$. It is obtained from $G$ by a graph operation as above. So, is there any name of this operation?
|
https://mathoverflow.net/users/375270
|
What's the name of the graph operation of connecting two copies of a graph with a perfect matching?
|
I dont't know of a standard name, but it is $G \square K\_2$, where $\square$ denotes the [Cartesian product](https://en.wikipedia.org/wiki/Cartesian_product_of_graphs).
|
3
|
https://mathoverflow.net/users/2233
|
406714
| 166,674 |
https://mathoverflow.net/questions/406619
|
2
|
Let $H(q)$ be the set of reduced residues $mod(q)$ and $\Phi(a)$ Euler totient function. How can I evaluate
$\min\_{q\leq x}\frac{1}{q}\sum\_{a\epsilon\ H(q)}\frac{\Phi (a)}{a}$
|
https://mathoverflow.net/users/169583
|
Minimal value for the specific summatory Euler Phi function
|
Let us find a good approximation for your sum for a given large $q$. I will use the notation $\varphi(a)$ for the Euler function. First of all, by Möbius inversion,
$$
\sum\_{a\in H(q)}\frac{\varphi(a)}{a}=\sum\_{a\leq q}\left(\sum\_{d\mid (a,q)}\mu(d)\right)\frac{\varphi(a)}{a}=\sum\_{d\mid q}\mu(d)S\_d(q),
$$
where
$$
S\_d(q)=\sum\_{bd\leq q}\frac{\varphi(bd)}{bd}.
$$
To evaluate this sum, notice that
$$
\frac{\varphi(m)}{m}=\sum\_{d\mid m}\frac{\mu(d)}{d},
$$
hence
$$
S\_d(q)=\sum\_{bd\leq q}\sum\_{d\_1\mid bd}\frac{\mu(d\_1)}{d\_1}=\sum\_{d\_1\leq q}\frac{\mu(d\_1)}{d\_1}\#\{bd\leq q: d\_1\mid bd\}.
$$
Now, $d\_1\mid bd$ iff $\frac{d\_1}{(d,d\_1)}\mid b$, so the weight inside our sum is
$$
\left[\frac{q/d}{d\_1/(d,d\_1)}\right]=\frac{q(d,d\_1)}{dd\_1}+O(1).
$$
Therefore,
$$
S\_d(q)=\frac{q}{d}\sum\_{d\_1\leq q}\frac{\mu(d\_1)(d,d\_1)}{d\_1^2}+O(\ln q)
$$
(the constant in O is absolute)
The sum in the main term is
$$
\sum\_{d\_1\leq q}\frac{\mu(d\_1)(d,d\_1)}{d\_1^2}=\sum\_{d\_1}\frac{\mu(d\_1)(d,d\_1)}{d\_1^2}+O\left(\sum\_{d\_1>q}\frac{d}{d\_1^2}\right)=\prod\_{p}\left(1-\frac{(p,d)}{p}\right)+O(dq^{-1}),
$$
because $(d,d\_1)$ is multiplicative as a function of $d\_1$. Thus,
$$
S\_d(q)=\frac{q}{d}\prod\_{p}\left(1-\frac{(p,d)}{p}\right)+O(\ln q)=\frac{q}{d}\prod\_{p\mid d}\left(\frac{p}{p+1}\right)\prod\_p\left(1-\frac{1}{p^2}\right)+O(\ln q)=
$$
$$
=\frac{6q}{\pi^2 d}\prod\_{p\mid d}\frac{p}{p+1}+O(\ln q).
$$
Plugging this into the initial formula, we obtain
$$
\sum\_{a\in H(q)}\frac{\varphi(a)}{a}=\frac{6q}{\pi^2}\sum\_{d\mid q}\frac{\mu(d)}{d}\prod\_p \frac{p}{p+1}+O(\sigma\_0(q)\ln q).
$$
Using multiplicativity once more, we get
$$
\sum\_{a\in H(q)}\frac{\varphi(a)}{a}=\frac{6q}{\pi^2}\prod\_{p\mid q}\left(1-\frac{1}{p+1}\right)+O(\sigma\_0(q)\ln q).
$$
So, up to an error of order $q^{-1+o(1)}$ the quantity you want to minimise is
$$
\frac{6}{\pi^2}\prod\_{p\mid q}\left(1-\frac{1}{p+1}\right),
$$
therefore your guess about primorials above is essentially correct and the minimum is asymptotic to
$$
\frac{6}{\pi^2}\prod\_{p\leq \ln x}\left(1-\frac{1}{p+1}\right)=\frac{6}{\pi^2}\prod\_{p\leq \ln x}\left(1-\frac{1}{p}\right)\left(1-\frac{1}{p^2}\right)^{-1}\sim \frac{e^{-\gamma}}{\ln\ln x}.
$$
|
2
|
https://mathoverflow.net/users/101078
|
406715
| 166,675 |
https://mathoverflow.net/questions/406683
|
8
|
$\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\GL{\operatorname{GL}}$This question is about an assertion in [Mixed Hodge polynomials of character varieties](https://arxiv.org/abs/math/0612668), by Hausel and Rodriguez-Villegas. Fix positive integers $g$ and $n$ and let $\zeta$ be a primitive $n$-th root of unity.
Let
$$H = \left\{ (A\_1, A\_2, \ldots, A\_g, B\_1, B\_2, \ldots, B\_g) \in \GL\_n^{2g}(\CC) : \prod\_{i=1}^g (A\_i B\_i A\_i^{-1} B\_i^{-1}) = \zeta \cdot\mathrm{Id}\_n \right\}.$$
Let $M = H/{\GL\_n(\CC)}$, acting by conjugation, and let $\tilde{M} = (\CC^{\ast})^{2g} \backslash M$ acting by rescaling the $A\_i$ and $B\_i$.
At the top of page 4, Hausel and Rodriguez-Villegas say that $M$ is "cohomologically a product" of $(\CC^{\ast})^{2g}$ and $\tilde{M}$ which I interpret to mean that, although the $(\CC^{\ast})^{2g}$-bundle $M \to \tilde{M}$ may not be trivial, we nonetheless have
$$H^{\ast}(M, \ZZ) \cong H^{\ast}(\tilde{M}, \ZZ) \otimes\_{\ZZ} H^{\ast}((\CC^{\ast})^{2g}, \ZZ).$$
Conveniently, $H^{\ast}((\CC^{\ast})^{2g}, \ZZ)$ is torsion free, so there are no $\operatorname{Tor}\_1$ terms in Künneth.
This is not obvious to me. Why is it true? For what more general versions of character varieties is it true?
|
https://mathoverflow.net/users/297
|
Why is the $\operatorname{GL}_n$ character variety "cohomologically" the product of the $\operatorname{PGL}_n$ character variety and a torus?
|
You need to keep reading to the proof of Theorem 2.2.12.
The main point is that the $PGL\_n$-character variety $\tilde{\mathcal{M}}\_n/\mathbb{C}$ is both a quotient by a torus $\mathcal{M}\_1/\mathbb{C}\cong (\mathbb{C}^\*)^{2g}$ of the $GL\_n$-character $\mathcal{M}\_n/\mathbb{C}$ and also a quotient of a finite group $\mu\_n^{2g}$ of the $SL\_n$-character variety $\mathcal{M}^\prime\_n/\mathbb{C}$ and the action of a finite group gives the invariant part in cohomology in one factor while trivial in the other (since it is a torus).
Thus, quoting the proof of Theorem 2.2.12, we have: \begin{eqnarray\*}H^∗(\mathcal{M}\_n/\mathbb{C}) &=&(H^∗(\mathcal{M}^\prime\_n/\mathbb{C}\times \mathcal{M}\_1/\mathbb{C}))^{\mu\_n^{2g}}\\ &=& H^∗(\mathcal{M}^{\prime}\_n
/\mathbb{C})^{\mu\_n^{2g}}\otimes H^∗(\mathcal{M}\_1/\mathbb{C})\\ &=& H^∗(\tilde{\mathcal{M}}\_n/\mathbb{C})\otimes H^∗(\mathcal{M}\_1/\mathbb{C}).\end{eqnarray\*}
This phenomenon occurs for other character varieties too.
For example, for free groups, see Theorem 2.4 in my paper with Florentino *[Singularities of free group character varieties](https://arxiv.org/abs/0907.4720)*. Note that the theorem is stated for $GL\_n$ and $SL\_n$ but it generalizes to connected reductive $G$ and its derived subgroup $DG$.
More generally, for $G\cong DG\times\_F T$ (central isogeny theorem for connected complex reductive $G$), we have:
$$
\begin{eqnarray\*}
\mathrm{Hom}^0(\Gamma,G)&\cong&\mathrm{Hom}^0(\Gamma,DG \times T)/\mathrm{Hom}(\Gamma,F)\\
&\cong&(\mathrm{Hom}^0(\Gamma,DG) \times \mathrm{Hom}^0(\Gamma,T))/\mathrm{Hom}'(\Gamma,F),
\end{eqnarray\*}$$
by Prop. 5(1) in *[G-Character varieties for G=SO(n,C) and other not simply connected groups](https://arxiv.org/abs/1303.7181)* by Sikora. Note that the superscript $0$ denotes the connected component of the trivial representation and $\mathrm{Hom}'(\Gamma,F)$ denotes the subgroup of $\mathrm{Hom}(\Gamma,F)$ mapping $\mathrm{Hom}^0(\Gamma,DG \times T)$ to itself. This then gives a similar decomposition for corresponding components of the character varieties by quotienting by $PG$ (since $F$ is central and so commutes with the conjugation action).
The above paragraph is quoted from my paper with Sikora *[Varieties of Characters](https://arxiv.org/abs/1604.02164)*.
See also my paper with Ramras *[Covering spaces of character varieties](https://arxiv.org/abs/1402.0781)* for related theorems. In particular, for "exponent canceling" groups we can remove the superscript $0$ above (for some $G$). Examples of exponent-canceling groups are free groups, surface groups, free abelian groups, and RAAGs.
|
7
|
https://mathoverflow.net/users/12218
|
406720
| 166,678 |
https://mathoverflow.net/questions/406671
|
2
|
Consider the compact Lie groups $U(l)$ (the unitary group) and $U(1) \times SU(l)$ for some natural number $l$. Both the groups have the same Lie algebra $\frak{gl}\_l$. Which means that they both have the "same" dense Peter-Weyl subalgebra $PW$ of their $C^\*$-algebra of continuous functions.
The Gelfand-Naimark theorem says that the $C^\*$-algebras of $U(l)$ and $U(1) \times SU(l)$ are non-isomorphic.
How to resolve the fact that both $C^\*$-algebras are completions of the same algebra PW? Is it the case that both spaces have different norms, and or algebra structure, on PW? I guess this must be what is happening?
|
https://mathoverflow.net/users/153228
|
Gelfand-Naimark and Peter-Weyl for the unitary group
|
There were silly errors in my comments, so let me start again. You have a canonical surjection $U(1) \times SU(k) \to U(k)$ given by scalar multiplication, so that any representation of $U(k)$ lifts to a representation of $U(1) \times SU(k)$; the question is whether or not every representation of $U(1) \times SU(k)$ descends to a $U(k)$.
Now, the kernel of this canonical surjection is $\langle (\omega\_k,\omega\_k^{-1}I\_k)\rangle \cong \mathbb{Z}\_k$, where $\omega\_k$ is a primitive $k$th root of unity. Then, for any integer $n$ not divisible by $k$, the map $\pi\_n : U(1) \times SU(k) \to U(1)$ given by $\pi\_n(z,u) := u^n$ is a representation of $U(1) \times SU(k)$ that doesn’t descend to $U(k)$, since $\pi\_n(\omega\_k,\omega\_k^{-1}I\_k) = \omega\_k^n \neq 1$.
In terms of Peter–Weyl algebras, what’s going on is that the Peter–Weyl algebra $\mathcal{O}(U(k))$ of $U(k)$ can be identified with the proper $\ast$-subalgebra of $\mathbb{Z}\_k$-invariants in the Peter–Weyl algebra $\mathcal{O}(U(1)\times SU(k))$ of $U(1) \times SU(k)$, where $\mathbb{Z}\_k$ acts through translation by the central element $(\omega\_k,\omega\_k^{-1}I\_k)$.
|
1
|
https://mathoverflow.net/users/6999
|
406722
| 166,679 |
https://mathoverflow.net/questions/405736
|
4
|
(This is a spin-off of [Determine the minimal elements of a Dynkin system generated by a finite set of finite sets](https://mathoverflow.net/q/405305))
Let $\Omega$ be a finite set. A [Dynkin system](https://en.wikipedia.org/wiki/Dynkin_system) on $\Omega$ is a subset of the power set of $\Omega$ containing $\Omega$, which is closed under complements in $\Omega$ and which is closed under union of disjoint sets.
Given any subset $\mathcal E$ of the power set of $\Omega$, there is a unique minimal Dynkin system $\delta(\mathcal E)$, the Dynkin system generated by $\mathcal E$.
Any Dynkin system is characterised by its minimal non-empty sets. If $\Omega$ has at most three elements, the Dynkin systems are in bijection with the set partitions. If $\Omega$ has four elements, there are four Dynkin systems whose minimal non-empty sets do not form a set partition:
* $\{\{0, 1\}, \{0, 3\}, \{1, 2\}, \{2, 3\}\}$
* $\{\{0, 1\}, \{0, 2\}, \{1, 3\}, \{2, 3\}\}$
* $\{\{0, 2\}, \{0, 3\}, \{1, 2\}, \{1, 3\}\}$
* $\{\{0, 1\}, \{0, 2\}, \{0, 3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}$
**Questions:**
* Can we compute the number of Dynkin systems?
* Can we compute the number of Dynkin systems up to relabelling?
|
https://mathoverflow.net/users/3032
|
What is the number of finite Dynkin systems?
|
We can compute the number of Dynkin systems for small $n$ using an almost-brute-force method. For efficient computation we represent a set ${a\_i}$ as a binary number $\sum\_i 2^{a\_i}$; then $\Omega = \{0, 1, 2, \ldots, n-1\}$ is represented by $2^n - 1$ and we can compute the complement with bitwise exclusive or by $2^n - 1$. This numeric representation also gives a straightforward total ordering to the subsets of $\Omega$.
The computational state consists of three sets: a subset of $\Omega$ which is included in the system; a subset which is excluded from the system; and a subset which is as yet undecided. Initially the included set is $\{ \emptyset, \Omega \}$, the excluded set is empty, and the undecided set contains the rest of the powerset of $\Omega$. At each stage we case split on the condition of the least (in the total ordering) undecided subset: either it is in the included set, in which case we apply the closure properties and verify that they don't require an excluded subset; or it isn't, in which case we add it (and its complement) to the excluded set.
I did have some ideas for approaches which should be more efficient asymptotically, but at the sizes which are practical to calculate they slow the computation down.
Python code, considerably optimised over the version I posted earlier in a comment:
```
def extend_closure(omega, included, extension, excluded):
# We start with a closure which we want to extend
closure = set(included)
# Use a queue
q = set([extension])
while q:
x = next(iter(q))
q.remove(x)
closure.add(x)
# Ensure the complement closure too
if (omega ^ x) not in closure:
# Shortcut contradictions
if (omega ^ x) in excluded:
return None
q.add(omega ^ x)
for y in closure:
if (x & y) == 0 and (x | y) not in closure:
# Shortcut contradictions
if (x | y) in excluded:
return None
q.add(x | y)
return closure
def enumerate_dynkin(n):
omega = (1 << n) - 1
def inner(included, lb, excluded):
yield included
for min in range(lb, (omega + 1) >> 1):
if min in included or min in excluded:
continue
# Consider the implications of adding min
ninc = extend_closure(omega, included, min, excluded)
if ninc:
for sys in inner(ninc, min + 1, set(excluded)):
yield sys
# Consider systems without min
excluded.add(min)
excluded.add(omega ^ min)
for sys in inner(set([0, omega]), 1, set()): yield sys
for n in range(9):
print(n, sum(1 for _ in enumerate_dynkin(n)))
```
On my desktop machine, with the pypy3 implementation of Python, this computes up to $n=7$ inclusive in under ten seconds, and $n=8$ in about 36 days.
Results:
```
n systems
0 1
1 1
2 2
3 5
4 19
5 137
6 3708
7 1506404
8 230328505024
```
In principle the same approach could be extended to handle relabelling: in the branch of the case split where we exclude a subset, we could also exclude all subsets which are equivalent to it under relabelling. But actually calculating that gets a bit tricky. It might actually be easier to generate, group by basic statistics like the number of subsets of each cardinality, and then test for relabelling equivalence within each group. I haven't implemented either approach.
|
4
|
https://mathoverflow.net/users/46140
|
406723
| 166,680 |
https://mathoverflow.net/questions/406712
|
3
|
Let $(f\_n)\_n:X \to \mathbb R$ be a sequence of measurable functions on a measurable space $X$ converging pointwise to a function $f:X \to \mathbb R$, and let $(\mu\_n)\_n$ be a sequence of finite measures (e.g probability measures) on $X$ such that each $f\_n$ is integrable w.r.t $\mu\_n$.
>
> **Question.** *Under what additional conditions do we have $\int\_X (f\_n-f)\,d\mu\_n \to 0$ ?*
>
>
>
**Note.** In case it helps, it may be assumed that $\mu\_n$ converges (in some sense) to a measure $\mu$ on $X$.
|
https://mathoverflow.net/users/78539
|
Dominated convergence theorem when the measure space also varies with $n$
|
$\newcommand\ep\varepsilon$The conjunction of the following conditions is enough:
1. The $f\_n$'s are uniformly bounded: $|f\_n|\le M$ for some real $M>0$ and all $n$;
2. $X$ is Polish;
3. $f\_n\to f$ uniformly on every compact $K\subseteq X$;
4. $\mu\_n\to\mu$ weakly for some $\mu$.
Indeed, take any real $\ep>0$. By [Prokhorov's theorem](https://en.wikipedia.org/wiki/Prokhorov%27s_theorem#:%7E:text=In%20measure%20theory%20Prokhorov%27s%20theorem,on%20complete%20separable%20metric%20spaces.) and in view of conditions 2 and 4, $\mu\_n(X\setminus K)<\ep$ for some compact $K\subseteq X$ and all $n$. Write
$$I\_n:=\int\_X(f\_n-f)\,d\mu\_n=I\_{1,n}+I\_{2,n},$$
where
$$I\_{1,n}:=\int\_K(f\_n-f)\,d\mu\_n,\quad
I\_{2,n}:=\int\_{X\setminus K}(f\_n-f)\,d\mu\_n.$$
By condition 3, $I\_{1,n}\to0$. By conditions 1 and 3, $|f|\le M$ and hence $|f\_n-f|\le2M$. So, $|I\_{2,n}|\le2M\ep$. So, $\limsup\_n|I\_n|\le2M\ep$, for every real $\ep>0$. So, $I\_n\to0$, as desired.
|
6
|
https://mathoverflow.net/users/36721
|
406728
| 166,682 |
https://mathoverflow.net/questions/406727
|
1
|
This is a follow-up to my [previous question](https://mathoverflow.net/questions/406653/estimating-the-average-of-two-gaussians-mean).
Assume that $X\sim \mathcal N(\mu\_1,\sigma\_1^2)$ and $Y\sim \mathcal N(\mu\_2,\sigma\_2^2)$.
I want to estimate $\frac{\mu\_1+\mu\_2}{2}$ after observing $X,Y$.
In my setting, $\sigma\_1,\sigma\_2$ are known and we want to estimate the average of the means with the goal of minimizing the expected squared error.
---
Intuitively, if $\sigma\_1<\sigma\_2$, we should have an estimate that's closer to $X$ than to $Y$. How can we formalize this?
|
https://mathoverflow.net/users/47499
|
Estimating the average of two gaussians' mean with minimal squared error
|
$\newcommand\si\sigma$Clearly, the best estimator of $\mu\_1$ is $X$, no matter what $\si\_1$ and $\si\_2$ are. Similarly, the best estimator of $\mu\_2$ is $Y$, no matter what $\si\_1$ and $\si\_2$ are. So, one may argue, a good estimator of $(\mu\_1+\mu\_2)/2$ is the substitution estimator $(X+Y)/2$.
As was shown in the [previous answer](https://mathoverflow.net/a/406655/36721), $(X+Y)/2$ is indeed the maximum likelihood estimator of $(\mu\_1+\mu\_2)/2$. It is also easy to see that $(X+Y)/2$ is the only unbiased linear estimator (of the form $aX+bY$) of $(\mu\_1+\mu\_2)/2$, as well as the minimax linear estimator of $(\mu\_1+\mu\_2)/2$ (with respect to the quadratic loss function).
On the other hand, one may argue that, if $\si\_1<\si\_2$, then the uncertainly about $\mu\_1$ is less than that about $\mu\_2$, and so $X$ has to be given a greater weight than $Y$. However, if there are no restrictions on $\mu\_1,\mu\_2$, then $|\mu\_1|$ and $|\mu\_2|$ can be much greater than both $\si\_1$ and $\si\_2$, so that $\si\_1$ and $\si\_2$ will matter little, if at all.
Thus, as it was said in a comment, without restrictions on $\mu\_1,\mu\_2$, you will hardly get the desired result.
One way to impose (soft/fuzzy) restrictions on $\mu\_1,\mu\_2$ is to suppose that $(\mu\_1,\mu\_2)$ has the prior bivariate normal distribution with means $0,0$, variances $b^2,b^2$, and correlation $0$. Then the Bayes estimator of $(\mu\_1,\mu\_2)$ (assuming the quadratic loss function) is
$$(\hat\mu\_1,\hat\mu\_2):=\Big(\frac X{1+\sigma\_1^2/b^2},\frac Y{1+\sigma\_2^2/b^2}\Big).$$
Then the corresponding estimator of $(\mu\_1+\mu\_2)/2$ is
$$\frac{\hat\mu\_1+\hat\mu\_2}2=\frac12\,\Big(\frac X{1+\sigma\_1^2/b^2}+\frac Y{1+\sigma\_2^2/b^2}\Big).$$
Here, as desired, the weight/coefficient of $X$ is greater than that of $Y$ if $\sigma\_1<\sigma\_2$.
If you now let $b\to\infty$, thus making the prior knowledge (i.e., the restrictions on $\mu\_1,\mu\_2$) insignificant, then you get back the nice old estimator $(X+Y)/2$ from the [previous answer](https://mathoverflow.net/a/406655/36721).
|
1
|
https://mathoverflow.net/users/36721
|
406732
| 166,685 |
https://mathoverflow.net/questions/406735
|
2
|
Is there a name for an adjunction between two categories such that
i) the unit of the adjunction is a natural isomorphism,
ii) the counit of the adjunction is a natural isomorphism?
|
https://mathoverflow.net/users/153228
|
Name for a categorical adjunction that is a "semi-equivalence"
|
An adjunction for which the unit is a natural isomorphism is called a [coreflective adjunction](https://ncatlab.org/nlab/show/coreflective+subcategory). An adjunction for which the counit is a natural isomorphism is called a [reflective adjunction](https://ncatlab.org/nlab/show/reflective+subcategory).
|
8
|
https://mathoverflow.net/users/152679
|
406739
| 166,687 |
https://mathoverflow.net/questions/406644
|
1
|
We are given a $n\times n$ symmetric matrix $M$ whose entries are positive integers.
Let $z\_{i,j}:=\frac{M\_{i,j}}{M\_{i,j}+\sum\_{k\neq i,j}\min(M\_{i,k},M\_{k,j})}$ for all $1\le i<j \le n$, and
$z:=\sum\_{i<j}z\_{i,j}$.
---
**Question:** Can we prove that the maximum value $z^\*$ of $z$ over *all* matrices $M$ defined above is upper bounded by $\alpha n$ for some absolute constant $\alpha$?
---
---
***Note:*** *$M$ can be viewed as the adjacency matrix of a given undirected multi-edge graph. Alternatively, for all $1\le i<j\le n$, $M\_{i,j}$ can be viewed as an integer capacity in a symmetric flow network. For all $1\le i<j\le n$, $z\_{i,j}$ can be viewed as a function of the amount of "flow" between the two nodes $i$ and $j$ passing through paths of length at most $2$ (i.e., containing $2$ edges).*
|
https://mathoverflow.net/users/115803
|
Combinatorial graph optimization problem on integer adjacency matrices
|
Consider $$M\_{i,j} = \begin{cases} 1 & \textrm{if } i \equiv j \pmod 2 \\ N & \textrm{if } i \not\equiv j \pmod 2\end{cases}$$ where $N > 1$. Then
$$\min(M\_{i,k}, M\_{k,j}) = \begin{cases} 1 & \textrm{if } i \equiv k \pmod 2 \;\vee\; k \equiv j \pmod 2 \\ N & \textrm{if } i \not\equiv k \pmod 2 \;\wedge\; k \not\equiv j \pmod 2 \end{cases}$$
In particular, if $i \not\equiv j \pmod 2$ then $\forall k: \min(M\_{i,k}, M\_{k,j}) = 1$. We get $$z \ge \sum\_{i < j, i \not\equiv j \pmod 2} \frac{N}{N+n-2} = \left\lfloor \frac{n^2}{4} \right\rfloor \frac{N}{N+n-2}$$ and by choosing sufficiently large $N$ we can exceed $\frac{n^2 - 4}{4}$.
|
2
|
https://mathoverflow.net/users/46140
|
406744
| 166,688 |
https://mathoverflow.net/questions/406626
|
4
|
I am looking for a triangulation of an $n$-dimensional simplex such that all sub-simplices are of comparable size, and are "as close as possible" to a regular simplex : the latter property could be formalized as "the minimum $n$-dimensional angle is bounded away from $0$, uniformly in the size of the triangulation".
I have read that the Delaunay triangulation had this kind of properties, but I did not find quantitative results about it, like lower bounds on the minimum angle.
Thank you!
|
https://mathoverflow.net/users/422680
|
Triangulation of a simplex
|
You are looking for the edgewise subdivision:
*Edelsbrunner, H.; Grayson, D. R.*, [**Edgewise subdivision of a simplex**](http://dx.doi.org/10.1007/s004540010063), Discrete Comput. Geom. 24, No. 4, 707-719 (2000). [ZBL0968.51016](https://zbmath.org/?q=an:0968.51016).
The basic idea is to slice your simplex k times along each coordinate hyperplane. Then you get some small uniform shapes, which are not simplices except at the corners (or in dimension 2). So then you slice each one of those shapes up in the same ad hoc way. If you do the details right, IIRC you can even get it so that every edge length is either $1/k$ or $\sqrt{2}/k$. Probably you can use that to figure out the minimal angle.
Another way, with worse constants but which is simpler to describe completely, is this: Split your $n$-simplex into $n+1$ (distorted) cubes with vertices at the barycenter, then subdivide the cubes cubically, and then split the little cubes back into simplices.
|
6
|
https://mathoverflow.net/users/59302
|
406748
| 166,690 |
https://mathoverflow.net/questions/370967
|
1
|
Let $H$ be a Hilbert space. A vector subspace $W\subset B(H)$ is called a Fredholm subspace if there is an upper bound for the absolute value of Fredholm index of all Fredholm operators $T$ in $W$.
>
> Is there a classification of all $C^\*$-algebras $A$ which admit an irreducible representation $\phi:A \to B(H)$ in some Hilbert space $H$ such that $\phi(A)$ is a Fredholm subspace of $B(H)$?
>
>
>
>
> Is there a classification of all $C^\*$-algebras $A$ which admit a faithful representation $\phi:A \to B(H)$ in some Hilbert space $H$ such that $\phi(A)$ is a Fredholm subspace of $B(H)$?
>
>
>
One can consider the terminology "Fredholm algebra" for any such $C^\*$-algebras.
**Edit:** We add an example according to comment by Yemon Choi.
Put $H=\ell^2$ let $S$ be the shift operator on $\ell^2$ and $n$ be a fixed integer. Then this is a finite dimensional Fredholm subspace of $B(\ell^2)$:
$$\{P(S)\mid \text{P is a polynomial of degree at most n}\}.$$
|
https://mathoverflow.net/users/36688
|
Fredholm $C^*$-algebras
|
Any unital $C^\*$-algebra $A$ has an irreducible representation $\phi$ such that every Fredholm operator in $\phi(A)$ has index 0.
To see it, let me first repeat something from Nik Weaver's previous answer: if $\pi$ is a representation of $A$ such that $\pi(A)$ intersects the compact operators trivially, then any Fredholm operator in $\pi(A)$ is actually invertible.
Now to prove my claim, observe that we may assume that $A$ is simple (up to replacing $A$ by $A/I$ where $I$ is a maximal, proper, closed two-sided ideal). Now let $\phi$ be any irreducible representation of $A$. If $\phi$ is finite-dimensional, then the result is clear and was already mentioned. If $\phi$ is infinite-dimensional then $\phi(A)$ intersects the compacts trivially because of simplicity, so Nik's observation above applies.
|
1
|
https://mathoverflow.net/users/14497
|
406749
| 166,691 |
https://mathoverflow.net/questions/406760
|
4
|
Let $P\_i=V\_{i}V\_{i}^{\top}\in\mathbb{R}^{m\times m}$ where $\forall i\in[T]: V\_{i}\in\mathbb{R}^{m\times n}$ is a “tall” matrix (i.e., $m \ge n$) with orthonormal columns. Note that these matrices are symmetric PSD.
Is the product of all these matrices, i.e., $P\_T P\_{T-1}\cdots P\_1$, necessarily diagonalizable?
|
https://mathoverflow.net/users/100796
|
Is the product of projection matrices diagonalizable?
|
No. Take
$$v\_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \qquad v\_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \qquad v\_3 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.$$
Then
$$P\_1=v\_1 v\_1^T = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \qquad P\_2=v\_2 v\_2^T = \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix} \qquad P\_3=v\_3 v\_3^T = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$$
so
$$P\_1 P\_2 P\_3 = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$$
which is not diagonalizable. (Replace $v\_2$ by $\tfrac{1}{\sqrt{2}} \left[ \begin{smallmatrix} 1 \\ 1 \end{smallmatrix} \right]$ if you wanted orthonormal columns instead of just orthogonal ones.)
---
Regarding the random eigenvalues: For $(x,y,z)$ in $\mathbb{R}^3$, let $$f(x,y,z) = (x^2+y^2+z^2) \mathrm{Id}\_3 - \begin{bmatrix} x\\y\\z \end{bmatrix} \begin{bmatrix} x&y&z \end{bmatrix}.$$
So $f(x,y,z)/(x^2+y^2+z^2)$ is orthogonal projection onto $\left[ \begin{smallmatrix} x\\y\\z \end{smallmatrix}\right]^{\perp}$.
I computed $f(1,0,0) f(u,v,w) f(x,y,z) f(1,0,0)$ for randomly chosen $(u,v,w,x,y,z)$ and, on my third try, I found the following example:
$$f(1,0,0) f(2,-3,-2) f(2,-3,1) f(1,0,0) = \begin{bmatrix}
0 & 0 & 0 \\
0 & 58 & -66 \\
0 & 33 & 143 \\
\end{bmatrix}.$$
The characteristic polynomial is
$$\lambda (\lambda^2 - 201 \lambda + 10472),$$
which has complex roots.
|
7
|
https://mathoverflow.net/users/297
|
406761
| 166,693 |
https://mathoverflow.net/questions/406773
|
6
|
I was looking at two sequences of integers, both with prominent place is combinatorics. The first one appears, for instance, in [Stieltjes moment sequences for pattern-avoiding permutations](https://arxiv.org/pdf/2001.00393.pdf) (see page 23)
$$a\_n=\sum\_{k=0}^n\frac{\binom{2k}k\binom{n+1}{k+1}\binom{n+2}{k+1}}{(n+1)^2(n+2)}.$$
The second appears in lattice path enumerations as [Motzkin numbers](https://en.wikipedia.org/wiki/Motzkin_number) (a close cousin of Catalan numbers)
$$b\_n=\sum\_{k=0}^{\lfloor\frac{n}2\rfloor}\frac{\binom{n}{2k}\binom{2k}k}{k+1}.$$
In modulo $2$ arithmetic, I run into what seems to be a (happy) coincidence. Let me ask:
>
> **QUESTION.** Is this true?
> $$a\_n\equiv b\_n \mod 2.$$
>
>
>
**ADDED.** $a\_n$ or $b\_n$ is even iff $n$ is part of [this sequence](https://oeis.org/A081706) listed on OEIS. In fact, whenever that happens, $\nu\_2(a\_n)=1$ and $\nu\_2(b\_n)\in\{1,2\}$.
|
https://mathoverflow.net/users/66131
|
A moment sequence and Motzkin numbers. Modular coincidence?
|
Let's start with $b\_n$. Since Catalan number $C\_k$ is odd iff $k=2^m-1$, from [Lucas theorem](https://en.wikipedia.org/wiki/Lucas%27s_theorem) it follows that
$$b\_n=\sum\_{k=0}^n \binom{n}{2k}C\_k \equiv\sum\_{m\geq 0}\binom{n}{2(2^m-1)}\equiv 1+\nu\_2(\lfloor n/2\rfloor+1)\pmod2,$$
where $\nu\_2(\cdot)$ is the 2-adic valuation.
---
Now consider $a\_n$. From the recurrence given in [OEIS A005802](http://oeis.org/A005802),
$$(n^2 + 8n + 16)a\_{n+2}=(10n^2 + 42n + 41)a\_{n+1}-(9n^2 + 18n + 9)a\_n,$$
we have $a\_{2k+1}\equiv a\_{2k}\pmod2$ for all $k$. It's therefore sufficient to consider even $n$, when
$$a\_n = \frac{1}{(n+1)^2}\sum\_{k=0}^n C\_k \binom{n+1}{k+1} \binom{n+1}{k} \equiv \sum\_{m\geq 0} \binom{n+1}{2^m} \binom{n+1}{2^m-1}\equiv \nu\_2(n+2)\pmod2.$$
---
Hence, for all $n$ we have
$$a\_n\equiv b\_n\pmod{2}.$$
|
8
|
https://mathoverflow.net/users/7076
|
406786
| 166,699 |
https://mathoverflow.net/questions/406785
|
10
|
Let $A$ and $B$ be skew lines in $\mathbb{R}^3$. Choose four points $a\_1, a\_2, a\_3, a\_4$ on $A$ and four points $b\_1, b\_2, b\_3, b\_4$. For all $i,j \in [4]$ draw a line segment from $a\_i$ to $b\_j$. Since $A$ and $B$ are skew, none of these line segments intersect each other. Thus, this is a straight line drawing of $K\_{4,4}$ in $\mathbb{R}^3$ without crossings.
>
> Is this drawing of $K\_{4,4}$ **knotted**?
>
>
>
Recall that a drawing of a graph $G$ is *knotted* if some cycle of $G$ is drawn as a non-trivial knot. I suspect that the answer is no, but could not prove it. The motivation for this problem comes from a [paper](https://arxiv.org/abs/2009.12989) of David Wood and myself, where we determine the maximum number of copies of a fixed tree in various sparse graph classes. A negative answer to the above question would essentially solve the problem exactly for the class of *knotless* graphs. These are the graphs that have a knotless embedding in $\mathbb{R}^3$.
One approach would be to use the classification of knots with small [stick number](https://en.wikipedia.org/wiki/Stick_number), but I am hoping there is a more elegant solution. The answer might depend on the positions of the points on $A$ and $B$. I would be happy if at least one choice of points yields a knotless drawing.
|
https://mathoverflow.net/users/2233
|
Is this drawing of $K_{4,4}$ knotted?
|
Here is a proof that all such embeddings are knotless. Consider the four half-planes bounded by $A$ which each contain one of the points $b\_i$. Then these planes give an open-book decomposition which contains $K\_{4,4}$. Each plane contains four edges, but if $L$ is a cycle of $K\_{4,4}$, then it can only contain at most two edges from each plane, since it can contain at most two of the edges which meet each $b\_i$. This gives an arc presentation of $L$ which has arc-index at most $4$. But it is [known](https://doi.org/10.1017/S0305004100074181) that every non-trivial knot has arc index at least $5$. Therefore the embedding of $K\_{4,4}$ is knotless.
---
Below is my original answer, giving one explicit knotless embedding.
Let the points on $A$ be $\{(-4,0,0), (-1,0,0), (1,0,0), (4,0,0)\}$. Let the points on $B$ be $\{(0,-3,1), (0,-2,1), (0,2,1), (0,3,1)\}$. Look at the projection $\pi$ of $K\_{4,4}$ onto the $xy$ plane, which has exactly four crossings. A non-trivial knot has at least three crossings in any diagram.
Suppose $L$ is a non-trivial knot or link embedded in this $K\_{4,4}$. Then $L$ has at most 8 edges. If $\pi(L)$ contains all four crossings, then $L$ is a pair of unlinked circles. If $\pi(L)$ contains three crossings, then by symmetry, it doesn't matter which three are chosen, and $L$ is an unknot. Therefore this embedding is knotless. There is however a Hopf link in this $K\_{4,4}$, which uses two opposite crossings, so this embedding is not linkless.
Different choices of points could give many more crossings in the projection, and I wouldn't like to try to extend this argument to anything much more complicated.
|
5
|
https://mathoverflow.net/users/126206
|
406788
| 166,701 |
https://mathoverflow.net/questions/406782
|
1
|
Suppose it is given that there exists a ‘strictly positive’ vector $\vec x \in (0,1)^k$, which lies on the probability simplex $\sum\_i x\_i = 1$. What is the least number of inequalities of the form $\vec g\_i^T \vec x \geq 0$ required to ensure that the solution set is a singleton vector? Note that there is no restriction on $\{\vec g\_i \}$ so one can pick the ‘best’ possible vectors (it should be consistent in the sense that there must exist one solution on the simplex with positive components).
I believe the answer would be $2(k-1)$ — I know it definitely is sufficient but can't seem to be able to show that it is necessary as well. Construction for sufficiency — take pairs of $\vec g\_i, - \vec g\_i$ so essentially you would have $k-1$ equations of form $\vec x^T \vec g\_i = 0$ and $\sum\_i x\_i = 1$. We have $k$ equations so this has a unique solution and it's not hard to see that by suitable choice of $\vec g\_i$, we can make the solution have positive components.
|
https://mathoverflow.net/users/424786
|
How many inequalities do I need to ensure a unique solution?
|
$\newcommand\v\vec\newcommand\R{\mathbb R}$For any natural $k\ge2$, $k$ inequalities (but not fewer than $k$) will suffice.
Indeed, take any vector $\v a\in(0,\infty)^k$ such that $\v1\cdot\v a=1$, where $\v1:=(1,\dots,1)\in\R^k$ and $\cdot$ denotes the dot product. Take any linearly independent vectors $\v g\_1,\dots,\v g\_{k-1}$ in $\R^k$ that are orthogonal to $\v a$, and then let $\v g\_k:=-\v g\_1-\dots-\v g\_{k-1}$. Then, for each $\v x\in\R^k$, the inequalities
\begin{equation}
\v g\_1\cdot\v x\ge0,\ \dots,\ \v g\_{k-1}\cdot\v x\ge0,\ \v g\_k\cdot\v x\ge0 \tag{0}
\end{equation}
imply the equalities
\begin{equation}
\v g\_1\cdot\v x=\dots=\v g\_{k-1}\cdot\v x=0. \tag{1}
\end{equation}
Also, the vector $\v1$ is linearly independent of $\v g\_1,\dots,\v g\_{k-1}$ -- otherwise, we would have $1=\v1\cdot\v a=0$, since (1) holds for $\v x=\v a$. So, the system of equalities (1) together with the equality
\begin{equation}
\v1\cdot\v x=1 \tag{2}
\end{equation}
has a unique solution.
Moreover, equalities (1) and (2) do hold for $\v x=\v a$.
Thus, $\v x=\v a$ is the unique solution of the system of inequalities (0) and equality (2),
as desired.
---
Also, for any natural $k\ge2$, any $k-1$ inequalities of the form
\begin{equation}
\v g\_1\cdot\v x\ge0,\dots,\v g\_{k-1}\cdot\v x\ge0 \tag{0a}
\end{equation}
will not suffice to identify $\v a$, even together with the equality (2).
Indeed, suppose the contrary, so that $\v a$ is the only solution in $(0,\infty)^k$ of the system (0a)--(2).
Consider first the case when $\v a$ is not orthogonal to
\begin{equation\*}
V:=\text{span}(\v g\_1,\dots,\v g\_{k-1}).
\end{equation\*}
Take any nonzero vector $\v h\in\R^k$ orthogonal to $V$ and such that $\v1\cdot\v h\ge0$, so that $\v1\cdot(\v a+\v h)\ge1>0$, and also short enough so that $\v a+\v h\in(0,\infty)^k$. Let
\begin{equation}
\v b:=\frac{\v a+\v h}{\v1\cdot(\v a+\v h)}\in(0,\infty)^k. \tag{3}
\end{equation}
Then $\v b\ne \v a$, while (0a) and (2) hold with $\v x=\v b$. This contradicts $\v a$ being the only solution in $(0,\infty)^k$ of the system (0a)--(2).
Next, consider the case when $\v a$ is orthogonal to $V$ and $\v g\_1,\dots,\v g\_{k-1}$ are linearly dependent, so that there is a nonzero vector $\v h\in\R^k$ orthogonal to $V$ and to $\v a$ and such that $\v1\cdot\v h\ge0$, which also is short enough so that $\v a+\v h\in(0,\infty)^k$. Then again, for $\v b$ as in (3), we have $\v b\ne \v a$, while (0a) and (2) hold with $\v x=\v b$. This contradicts $\v a$ being the only solution in $(0,\infty)^k$ of the system (0a)--(2). Again, a contradiction.
Finally, consider the case when $\v a$ is orthogonal to $V$ and $\v g\_1,\dots,\v g\_{k-1}$ are linearly independent. Then there is a vector $\v h\_1\in V$ such that $\v g\_1\cdot\v h\_1=1$ and $\v g\_j\cdot\v h\_1=0$ for $j=2,\dots,k-1$. Letting now $\v h:=t\,\v h\_1$ for a small enough $t>0$ so that $\v1\cdot\v h\ge-1/2$ and $\v a+\v h\in(0,\infty)^k$, we get $\v1\cdot(\v a+\v h)\ge1/2>0$.
So, again for $\v b$ as in (3), we have $\v b\ne \v a$, while (0a) and (2) hold with $\v x=\v b$. This contradicts $\v a$ being the only solution in $(0,\infty)^k$ of the system (0a)--(2). Once again, a contradiction.
Thus, $k$ is the smallest number of inequalities of the form $\v g\_j\cdot\v x\ge0$ that will suffice.
|
4
|
https://mathoverflow.net/users/36721
|
406792
| 166,703 |
https://mathoverflow.net/questions/406468
|
1
|
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\GL{GL}
$Let $F$ be local field of characteristic zero and $(W,\langle,\rangle)$ be a $2n$-dimensional symplectic space over $F$.
Let $X,X^\*$ be maximal totally isotropic subspaces of $W$, which are dual with respect to $\langle,\rangle$.
Let $e\_1,\dotsc,e\_n$ be a basis of $X$ and $f\_1,\dotsc,f\_n$ be a basis of $X^\*$. Let $B\_{n}=M\_nU\_n$ be the parabolic subgroup of $\Sp(W)$ stabilizing the flag
$$\langle e\_1\rangle \subset \langle e\_1, e\_2\rangle \subset \dotsb \subset\langle e\_1, e\_2, \dotsc, e\_{n}\rangle.$$
Let $\{ \chi\_i \}$ be characters of $F^{\times}$ and $U\_n$ the unipotent radical of $B\_n$.
Let $\pi$ be the irreducible generic unramified representation of $\Sp(W)$ that is a subquotient of $\operatorname{Ind}\_{B\_{n}}^{\Sp(W)} (\chi\_1 \otimes \dotsb \chi\_{n} )$. (Here, the induction is normalized.)
Then for any $1 \le i \le n$, let $W\_i=\left<e\_i,f\_i\right>$ and $B\_{i}$ be the parabolic subgroup of $\Sp(W\_i)$ stabilizing $\left<e\_i\right>$. Let $\pi\_i=\operatorname{Ind}\_{B\_{i}}^{\Sp(W\_i)} \chi\_i$.
Then I am wondering whether the irreducible unramified constituent of $\pi\_i$ is also generic. Is it true? And I also guess $\pi\_i$ is irreducible.
Any comments are welcome!
|
https://mathoverflow.net/users/35898
|
Part of some generic representation is also generic?
|
>
> Let $\pi$ be the irreducible generic unramified representation of $Sp(W) $ that is a subquotient of $Ind(\chi\_1, \dots, \chi\_n)$.
>
>
>
I think the key here is to realise that this *does not exist* for all values of $\chi$. If the induction is irreducible (which is true for "sufficiently general" tuples of unramified characters $(\chi\_1, \dots, \chi\_n)$), then it is generic and unramified. However, when the induction is reducible, there is usually *no* subquotient which is unramified and generic at the same time; usually there will be a unique "largest" factor which is the generic one, and a unique "smallest" factor which is the unramified one. For instance, if $n = 1$ and $\chi\_1 = |\cdot|^{1/2}$, then the induction has 2 factors, the trivial rep and the Steinberg; the former is unramified but not generic, and the latter is generic but not unramified!
If you take $n = 1$ and $\chi\_1 = 1$, then the representation is again reducible and the two factors are "the same size" in some sense -- they are representations of $SL\_2$ which are conjugate in $GL\_2$. In this case, exactly one of the two factors is generic, and exactly one of the two is spherical. However, which one has which property depends on how you interpret the definitions! There are two conjugacy classes of hyperspecial max compacts in $SL\_2(F)$ [hence two meanings of "unramified"] and two conj classes of Whittaker characters [hence two meanings of "generic"]. So the question is still not well-defined enough to be answerable.
If you work with "sufficiently general" characters that the induced rep of $G$ is irreducible, then the $\pi\_i$ are all irreducible too. Maybe that's not a very interesting statement, but I'm struggling to find an interpretation of the question which makes sense in the reducible cases.
|
2
|
https://mathoverflow.net/users/2481
|
406797
| 166,706 |
https://mathoverflow.net/questions/406803
|
3
|
Let $\Lambda$ be a simply laced root lattice and $w$ a Coxeter element of the Weyl group of $\Lambda$.
**Question:** Is it true that the action of $w$ on the $\mathbb{F}\_2$-vector space $\Lambda/2\Lambda$ is semisimple?
Motivation: I want to show that, if $F$ denotes the subset of $w$-fixed points on $\Lambda/2\Lambda$, then $w$ has no nonzero fixed points on $\left(\Lambda/2\Lambda\right)/F$. This is more or less equivalent to showing that $w$ acts semisimply on its $1$-eigenspace. A proof of the latter claim would already be great!
|
https://mathoverflow.net/users/110362
|
Action of Coxeter element on mod $2$ root lattice is semisimple
|
The answer seems to be negative.
According to [this](https://mathoverflow.net/q/52100) answer, an operator $X\in\operatorname{M}(n,\mathbb{F}\_q)$ is semisimple if and only if $X^{q^m}=X$, where $m=\operatorname{lcm}(2,\ldots,n)$.
Now consider the standard realization of the root system of type $\mathsf{A}\_n$ inside $\mathbb{Z}^{n+1}$, where the simple roots are $\alpha\_i = e\_i-e\_{i+1}$ for $i=1,\ldots,n$. Then
$$\Lambda = \{ (x\_1,\ldots,x\_{n+1})\in\mathbb{Z}^{n+1} \mid x\_1+\ldots+x\_{n+1}=0 \},$$
and the action of $W(\mathsf{A}\_n)\cong S\_{n+1}$ is by permutations of the entries.
One possible choice of the Coxeter element $w$ is the long cycle
$$w = (1\ \ 2\ \ \ldots\ \ n+1).$$
For $n=3$ one gets $\dim(\Lambda/2\Lambda)=3$ and $m=\operatorname{lcm}(2,3)=6$. Since $\operatorname{ord}(w)=n+1=4$ and $2^6 = 64 \equiv 0 \pmod{4}$, one has $w^{2^m} = \operatorname{id}$, while $w$ act non-trivially on $\Lambda/2\Lambda$.
The same considerations work for $\mathsf{A}\_n$, $n$ odd, and for $\mathsf{D}\_n$ (where the Coxeter number equals $2n-2$ and $2^m$ always gives an even remainder modulo $2n-2$).
The conjecture holds for $\mathsf{A}\_n$, $n$ even, because in this case $2^m\equiv 1\pmod{n+1}$.
|
2
|
https://mathoverflow.net/users/5018
|
406810
| 166,709 |
https://mathoverflow.net/questions/406738
|
3
|
Let $wt(n)$ be [A000120](https://oeis.org/A000120), number of $1$'s in binary expansion of $n$ (or the binary weight of $n$)
and
$$n=2^{t\_1}(1+2^{t\_2+1}(1+\dots(1+2^{t\_{wt(n)}+1}))\dots)$$
Then we have an integer sequence given by
$$a(n)=\sum\limits\_{j=0}^{2^{wt(n)}-1}m^{wt(n)-wt(j)}\prod\limits\_{k=0}^{wt(n)-1}(1+wt(\left\lfloor\frac{j}{2^k}\right\rfloor))^{t\_{k+1}+1}, a(0)=1$$
Let
$$s(n,m)=\sum\limits\_{k=0}^{2^n-1}a(k)$$
then I conjecture that for any $m$
$$s(n,m)=\sum\limits\_{k=0}^{n+1}k!{n+1\brace k}(m+1)^{n-k+1}$$
Is there a way to prove it?
Similar questions:
* [Recurrence for the sum](https://mathoverflow.net/questions/405174/recurrence-for-the-sum)
* [Pair of recurrence relations with $a(2n+1)=a(2f(n))$](https://mathoverflow.net/questions/406902/pair-of-recurrence-relations-with-a2n1-a2fn)
* [Sequence that sums up to INVERTi transform applied to the ordered Bell numbers](https://mathoverflow.net/questions/407290/sequence-that-sums-up-to-inverti-transform-applied-to-the-ordered-bell-numbers)
* [Sequences that sums up to second differences of Bell and Catalan numbers](https://mathoverflow.net/questions/407758/sequences-that-sums-up-to-second-differences-of-bell-and-catalan-numbers)
|
https://mathoverflow.net/users/231922
|
Sum with Stirling numbers of the second kind
|
The [same idea](https://mathoverflow.net/q/405210) of grouping terms by the number of unit bits, as well as grouping by the value of $\mathrm{wt}(j)$ (representing $j$ via individual bits) works here:
\begin{split}
s(n,m) &= \sum\_{\ell=0}^n \sum\_{t\_1 + \dots + t\_\ell \leq n-\ell} \sum\_{j\_1,\dots,j\_\ell\in\{0,1\}} m^{\ell-\sum\_{i=1}^\ell j\_i} \prod\_{k=1}^{\ell} (1+\sum\_{i=1}^k j\_i)^{t\_k+1} \\
&=[x^{n+1}]\ \sum\_{\ell=0}^n \sum\_{j\_1,\dots,j\_\ell\in\{0,1\}} m^{\ell-\sum\_{i=1}^\ell j\_i} \prod\_{k=0}^{\ell} \frac{(1+\sum\_{i=1}^k j\_i)x}{1-(1+\sum\_{i=1}^k j\_i)x} \\
&=[x^{n+1}]\ \sum\_{\ell=0}^n \sum\_{J=0}^\ell \sum\_{j\_1,\dots,j\_\ell\in\{0,1\}\atop j\_1+\dots+j\_\ell=J} m^{\ell-J} \prod\_{k=0}^{\ell} \frac{(1+\sum\_{i=1}^k j\_i)x}{1-(1+\sum\_{i=1}^k j\_i)x} \\
&=[x^{n+1}]\ \sum\_{\ell=0}^n \sum\_{J=0}^\ell m^{\ell-J} \sum\_{d\_0,d\_1,\dots,d\_J\geq 1\atop d\_0+d\_1+\dots+d\_J=\ell+1}\prod\_{k=1}^{J+1} \left(\frac{kx}{1-kx}\right)^{d\_k} \\
&=[x^{n+1}]\ \sum\_{\ell=0}^n m^{\ell+1}[y^{\ell+1}]\ \sum\_{J=0}^\ell \prod\_{k=1}^{J+1} \frac{kxy}{(1-kx-kxy)m} \\
&=[x^{n+1}]\ \sum\_{J=0}^n \prod\_{k=1}^{J+1} \frac{kx}{1-kx-kmx} \\
&=\sum\_{J=0}^n (J+1)!\ [x^{n-J}]\ \prod\_{k=1}^{J+1} \frac{1}{1-k(m+1)x} \\
&=\sum\_{J=0}^n (J+1)!\ (m+1)^{n-J} \left\{n+1\atop J+1\right\}.
\end{split}
It also shows that the summands in the last formula corresponds to fixed values of $\mathrm{wt}(j)=J$.
|
2
|
https://mathoverflow.net/users/7076
|
406817
| 166,713 |
https://mathoverflow.net/questions/406769
|
4
|
A (real) normed space $(V, \lVert \cdot \rVert\_V)$ is called strictly convex if for all $x, y \in V \setminus \{ 0 \}$ we have
\begin{equation}
\lVert x + y \rVert\_V = \lVert x \rVert\_V + \lVert y \rVert\_V \implies \exists c >0 \enspace x = cy.
\end{equation}
A completion of a normed space $(V, \lVert \cdot \rVert\_V)$ is a Banach space $( W, \lVert \cdot \rVert\_W)$ such that there exists an isometric embedding $i \colon V \to W$ such that $i[V]$ is dense in $W$. It is unique up to a (surjective) isometry.
---
Question
--------
If $(V, \lVert \cdot \rVert\_V)$ is a strictly convex (real) normed space, does it follow that its completion is, again, strictly convex?
---
I'm not sure whether it (regardless of whether the answer is positive or negative) is a well-known fact, but my initial tries to find a counter-example have failed. It possibly is due to the fact that I tried some modifications of rather well-behaved spaces, like $\ell^p$ for $p \in (1,\infty)$, so they might have some other properties, which are preserved by completion.
|
https://mathoverflow.net/users/170491
|
Is a completion of strictly convex normed space strictly convex?
|
No, the completion of a strictly convex normed space can fail to be strictly convex.
To put it differently, there are non strictly convex Banach spaces with a dense strictly convex subspace.
Here is a possible construction. To make things easier, it is tempting to start with a space where there is a good control on the non strictly convex part. For example, let $E,F$ be you favorite separable infinite dimensional strictly convex Banach spaces (for example $\ell\_2$), and consider $E\oplus\_{\ell\_1} F$, that is the direct sum of $E$ and $F$ for the norm $\|(v\_1,v\_2)\| = \|v\_1\|\_E + \|v\_2\|\_F$. In that way, for $v=(v\_1,v\_2)$ and $w=(w\_1,w\_2)$, $\|v+w\| = \|v\|+\|w\|$ if and only if for each $i \in \{1,2\}$, $v\_i$ and $w\_i$ are positively proportional. But if $v$ and $w$ are not proportional, this implies that there is a nonzero linear combination of $v$ and $w$ that is in $0\oplus F$ or $E\oplus 0$. In other words, any subspace $V \subset E\oplus\_{\ell\_1} F$ which interesects trivially $E \oplus 0$ and $0 \oplus F$ is strictly convex.
But it is a small exercise that, given a finite number of infinite codimensional subspaces $E\_i$ of a separable Banach space, there is a dense subspace which does intersects them all trivially.
(solution: if $(x\_n)$ is a dense sequence, construct by induction a sequence $(y\_n)$ such that $\|x\_n-y\_n\| \leq \frac 1 n$ and $\operatorname{span}(y\_1,\dots,y\_n) \cap E\_i = \{0\}$ for all $i$; the space spanned does the job).
|
6
|
https://mathoverflow.net/users/10265
|
406818
| 166,714 |
https://mathoverflow.net/questions/204724
|
12
|
(Split off from [Does every CAT(0) space embed in a measurable integral of $\mathbb{R}$-trees?](https://mathoverflow.net/questions/204710/does-every-cat0-space-embed-in-a-product-of-trees/204723) )
Fix an integer $k \ge 2$, and let
$MC0\_k \subset \mathbb{R}^{\binom{k}{2}}$ be the set of possible squared-distances between $k$ points (not necessarily distinct) in any CAT(0) space. This is clearly closed under scaling, and the fact that a product of CAT(0) spaces is a CAT(0) space implies that $MC0\_k$ is a convex set. What are its extreme rays?
(Recall that an *extreme* *point* in a convex set $C$ is a point that is not in the interior of any line segment in $C$. In the context of convex cones, as here, an *extreme ray* consists of points that aren't in the interior of a line segment that is not contained in a ray through the origin.)
One guess is that the extreme rays in $MC0\_k$ are trees with $k$ marked points, where there is at least one marked point at each vertex of valence $>3$. For $k=4$, this includes embeddings in $\mathbb{R}$ as well as tripods, with the central vertex marked.
Note that this is related to the (known-difficult) question of characterizing which lengths can appear as distances in a CAT(0) space, i.e., characterizing $MC0\_k$ in the notation above.
|
https://mathoverflow.net/users/5010
|
What are the extremal CAT(0) metrics?
|
Let me describe a 6-point counterexample.
Let $K$ be a 2-dimensional cone with total angle $\theta=2{\cdot}\pi+\varepsilon$, where $\varepsilon$ is small and positive (any $0<\varepsilon<\tfrac\pi2$ will do).
Note that $K$ is CAT(0).
Consider the following 6 points in $K$: the tip $p$ + and an orbit $\{x\_1,x\_2,x\_3,x\_4,x\_5\}$ of the rotation by angle $\tfrac\theta5$.
Suppose that these 6 points admit an embedding into a product of thees, say $L$.
Let $q$ be the image of $p$; denote by $\Sigma\_q$ its space of directions.
Note that any closed geodesic in $\Sigma\_q$ has length either $2{\cdot}\pi$ or at least $3{\cdot}\pi$. (The latter statement can be prove along the same lines as Gromov's flag condition.)
Denote by $y\_i$ the image of $x\_i$. For any $i$ (mod 5) there is a flat geodesic quadrilateral $qy\_{i-1}y\_iy\_{i+1}$ in $L$ that is isometric to the quadrilateral $px\_{i-1}x\_ix\_{i+1}$.
It follows that there is a closed geodesic in $\Sigma\_q$ of length $\theta$ --- a contradiction.
**Comments:**
* There is a similar example with 5 points --- take a plane convex quadrilateral with line segment attached at one vertex. By Reshetnyak theorem, it is a CAT(0) space. The vertices of the quadraliteral + the end of the segment form a 5-point subspace that cannot be embedded in a product of trees.
* Another related observation: [octahedron comparison holds for products of trees](https://arxiv.org/abs/2212.06445).
That is, if you choose six points in the product of trees and label them by vertices of an octahedron, then there is a configuration in the Euclidean space such that edges are not getting larger and the diagonals are not getting smaller.
It is unknow if it holds in general CAT(0) space.
|
6
|
https://mathoverflow.net/users/1441
|
406833
| 166,720 |
https://mathoverflow.net/questions/406789
|
3
|
Let $u$ be an harmonic function in a cylindrical domain $B\_2^{n-1}\times(-1,1)\subset\mathbb{R}^n$, and suppose its level sets $\Gamma\_t=\{u=t\}$ are graphs of functions on $B\_2^{n-1}$.
Consider a linear parametrization of $u$:
$$u\_t:=u-t.$$
Then the nodal set of $u\_t$ is $t$-level set of $u$:
$$\{u\_t=0\}=\{u=t\}.$$
We can regard this $u\_t$ to be the level set function of evolving surface $\Gamma\_t=\{u=t\}$. If we denote $v$ as the vector field indicating the speed of the flow of $\Gamma\_t$ follows, then by level set equation
$$|v|=\frac{1}{|\nabla u|}.$$
Since we know that $v$ is normal to $\Gamma\_t$, $v$ is parallel to $\nabla u$, hence
$$
v=\frac{\nabla u}{|\nabla u|^2}.
$$
On the other hand, since $u$ is constant on its level surfaces $\Gamma\_t$, its Laplace-Beltrami operators are identically zero:
$$0=\Delta\_{\Gamma\_t}u=\Delta u+H\_{\Gamma\_t}\partial\_{\nu}u-\partial\_{\nu\nu}u=H\_{\Gamma\_t}\nabla u-\partial\_{\nu\nu}u.$$
Therefore with the above expression, we obtain
$$
v=\frac{|H\_{\Gamma\_t}|}{|\partial\_{\nu\nu}u|}\nu,
$$
where $\nu:=\nabla u/|\nabla u|$ is a unit normal vector to $\Gamma\_t$.
So we have that, the surface $\Gamma\_t$ follows some geometric flow somehow related to mean curvature flow (if $|\partial\_{\nu\nu} u|=1$ then it is a mean curvature flow). My questions are:
1. Is there any name or related category of this flow?
2. I am interested in the $C^{2,\alpha}$ regularity of $\Gamma\_t$, if $\Gamma\_t$ was a graph of $C^{2,\alpha}$ function, then will $\Gamma\_s$ be $C^{2,\alpha}$ in some neighborhood of $t$? If so, can we control $C^{2,\alpha}$ norm of $\Gamma\_s$ by $\Gamma\_t$?
Just to note, different formulation of the curvature equation above ($0=H\_{\Gamma\_t}\nabla u-\partial\_{\nu\nu}u$) is that
$$|\nabla u|\Delta\_1 u=\Delta\_\infty u,$$
where $\Delta\_p$ is $p$-Laplacian. It is interesting for me that two operators with duality have some relation. Unfortunately I don't have much knowledge in this direction ($p$-laplacian) so I could not have any meaningful result.
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https://mathoverflow.net/users/151368
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Geometric flow by the level sets of a harmonic function
|
There is a lot of current study of the level sets of harmonic functions in this exact way.
See these papers: <https://arxiv.org/abs/1209.4669> <https://arxiv.org/abs/2108.08402> and <https://arxiv.org/abs/1911.06754> .
Some may not look directly related to what you're asking, but note that anytime you are using the co-area formula, it's like integrating $\frac{d}{dt} \int\_{u=t} F$ with respect to $t$, so it ends up being related to what you're asking.
In terms of the $p$-Laplacian, there is some relationship between your flow and inverse mean curvature flow. See <https://arxiv.org/abs/1812.05022> .
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3
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https://mathoverflow.net/users/1540
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406836
| 166,721 |
https://mathoverflow.net/questions/406609
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0
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This is a continuation of [Number of drifted Brownian motions that never hit zero under allocation](https://mathoverflow.net/questions/406559/number-of-drifted-brownian-motions-that-never-hit-zero-under-allocation)
For each $n\ge 1$, consider $X^i\_t=1+\beta t + W^i\_t$ for $i=1,\ldots n$ and $t\ge 0$, where $\beta>0$ and $(W^i\_t)\_{t\ge 0}$ are independent Brownian motions. $\phi\equiv \big((\phi^1\_t)\_{t\ge 0},\ldots, (\phi^n\_t)\_{t\ge 0}\big)$ is called an allocation strategy if every $(\phi^i\_t)\_{t\ge 0}$ is progressively measurable w.r.t. the Brownian filtration $\big(\mathcal F\_t:=\sigma(W^1\_s,\ldots, W^n\_s, s\le t)\big)\_{t\ge 0}$,
$$\phi^i\_t\ge 0 \quad\mbox{ and }\quad \sum\_{i=1}^n\phi^i\_t\le 1,\quad \forall t\ge 0.$$
Denote
$$X^{\phi,i}\_t:=X^i\_t+\int\_0^t \phi^i\_sds \quad \mbox{and} \quad \tau^{\phi}\_i:=\inf\{t\ge 0: X^{\phi,i}\_t\le 0\}.$$
Let $S^{\phi}\_n:=\sum\_{1\le i\le n}{\bf 1}\_{\{\tau^{\phi}\_i=\infty\}}$ be the number of $X^{\phi,i}$ that never hits zero. Clearly,
$$\frac{\mathbb E[S^{\bf 0}]}{n}~=~\mathbb P[X^1\_t>0, \forall t\ge 0]~=~1-e^{-\beta},$$
where $\bf 0$ stands for the strategy with $\phi^i\equiv 0$ for $i=1,\ldots, n$. Can we can show
$$\lim\_{n\to\infty}\frac{\mathbb E[S^{\phi}]}{n}~~=~~1-e^{-\beta}$$
for all the strategies $\phi$? Any answers, comments or references are highly appreciated!
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https://mathoverflow.net/users/261243
|
Does fixed allocation increase the proportion of positively drifted Brownian motions surviving forever?
|
This is not an answer to your question, but a similar result related to the case $\beta=0$ can be found in [Optimal surviving strategy for drifted Brownian motions with absorption](https://projecteuclid.org/journals/annals-of-probability/volume-46/issue-3/Optimal-surviving-strategy-for-drifted-Brownian-motions-with-absorption/10.1214/17-AOP1211.full) (with arxiv version <https://arxiv.org/pdf/1512.04493.pdf>).
More precisely, the (asymptotic) optimal strategy is given by
$$\phi^i\_t:=\begin{cases}
1 & \mbox{if } \tau^{\phi}\_i>t \mbox{ and } X^{\phi,i}\_{t}=Z\_t \\
0 & \mbox{otherwise}
\end{cases}
,\quad \mbox{where } Z\_t:=\min\left\{X\_{t}^{\phi,i}:~ \tau^{\phi}\_i>t\right\}.$$
Adopting this strategy, $\mathbb E[S^{\phi}]=O(\sqrt{n})$. Namely, for the any strategy the number of processes never hitting zero is of order $\sqrt{n}$. Therefore, I believe your claim is true and it can be shown using the same arguments of Proposition 2.6 (see also Lemmas 2.1, 2.4 and 2.5).
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0
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https://mathoverflow.net/users/nan
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406837
| 166,722 |
https://mathoverflow.net/questions/406752
|
9
|
Factorization in the ring $\mathbb{Z}[x]/(x^2+1)\mathbb{Z}[x]\cong \mathbb{Z}[i]$ is well known. For instance, $5$ and $13$ (and any prime $\equiv 1\pmod{4}$) are no longer prime.
The factorization of $5$ lifts to $\mathbb{Z}[x]/((x^2+1)^2)\mathbb{Z}[x]$, but it isn't a simple consequence of Hensel's lemma. The trouble with using Hensel's lemma is that it requires a form of Bezout's identity, which does not generally hold over $\mathbb{Z}[x]$. Instead, for any two coprime polynomials $a,b\in \mathbb{Z}[x]$, all we know in $\mathbb{Z}[x]$ is that there exists some polynomials $c,d\in \mathbb{Z}[x]$ such that $ac+bd\in \mathbb{Z}-\{0\}$.
For example, $1+2x$ and $1-2x$ are relatively prime, but the "smallest" positive integer that is a linear combination is
$$
(1+2x)+(1-2x) = 2.
$$
Note that $5=(1+2x)(1-2x) + 4(1+x^2)$, and the coefficient on $(1+x^2)$ is divisible by $2$. This lucky coincidence is what allows us to lift the factorization of $5$.
The factorization of $13$ as $(2+3x)(2-3x)\pmod{(1+x^2)}$ also lifts (as pointed out in the comments below by Johan). Here, the smallest positive integer $\mathbb{Z}[x]$-linear combination of $2+3x$ and $2-3x$ is $4$, and that's not enough. But the smallest $\mathbb{Z}[x]$-linear combination of $2+3x$, $2-3x$, and $1+x^2$ is $1$, which is enough.
**Question**: Given a monic irreducible polynomial $q(x)\in \mathbb{Z}[x]$, of degree at least $2$, and an integer $n\geq 2$, is there some integer prime $p$ that factors in $\mathbb{Z}[x]/q(x)^n\mathbb{Z}[x]$?
This question arose thinking about [this problem](https://mathoverflow.net/questions/406475/existence-of-a-finite-extension-of-%E2%84%A4-providing-a-finite-extension-of-the-primes).
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https://mathoverflow.net/users/3199
|
Hensel's lemma, Bezout's identity, and the integers
|
Assume that $\mathcal{O} = \mathbb{Z}[x]/(q(x))$ is the ring of integers of a number field $K$. One can search for a prime $p$ which is not inert in $\mathcal{O}$ and such that there is a principal prime ideal above $p$. I think such primes exist because the density of principal prime ideals is $1/h$, where $h$ is the class number of $\mathcal{O}$ (see [this question](https://mathoverflow.net/questions/375142/how-many-non-principal-prime-ideals-does-a-number-field-contain)), and because the prime ideals of $\mathcal{O}$ lying above an inert prime have density 0 (because their norm is $p^{[K:\mathbb{Q}]}$).
Write $p \mathcal{O} = \mathcal{P} \cdot I$ where $\mathcal{P} = (a)$ is a prime ideal, $I = (b) \neq \mathcal{O}$ is coprime to $\mathcal{P}$ and $p=ab$ in $\mathcal{O}$. Then you will have $p=a(x)b(x)+q(x)r(x)$ for some $a(x), b(x), r(x) \in \mathbb{Z}[x]$, and Hensel's lemma will work because $\langle a(x), b(x), q(x) \rangle = \mathbb{Z}[x]$. It will give a factorisation of $p$ in $\mathbb{Z}[x]/(q(x))^n$ for every $n \geq 2$.
|
6
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https://mathoverflow.net/users/6506
|
406841
| 166,725 |
https://mathoverflow.net/questions/405634
|
3
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Define
$$RT(n,K\_l,f(n))=ex\_l(n,f(n))=\max\_G\{e(G): K\_l \not\subset G, v(G)=n, \alpha(G)\leq f(n)\}$$
and the Ramsey-Turán density function $f\_l:(0,1] \to \mathbb{R}$ as
$$f\_l(\alpha)=\lim\_{n\to \infty}\frac{ex\_l(n,\alpha n)}{{n\choose 2}}.$$
By Turán Theorem we have that $f\_l(\alpha)=1-\frac{1}{l-1}$ for every $\alpha \geq \frac{1}{l-1}$.
How to prove that this limit exists for every $\alpha \in (0,1]$?
I was trying to prove that the sequence $(ex\_l(n,\alpha n)/{n\choose 2})\_n$ is monotone, but I could only prove that
$$ex\_l(n,s)\leq\frac{n \ ex\_l(n-1,s)}{n-2},$$
which is not enough since $ex\_l(n-1,\alpha n) \geq ex\_l(n-1,\alpha(n-1)).$
I also thought of using that $\alpha(G) \leq \alpha n$ and $\alpha(G)\leq \alpha(n-1)$ only give different restrictions once every $\approx 1/\alpha$ values of $n$ in a row. But it would also be necessary to show that every jump is in the same direction, which I couldn't do.
**Edit 1:** In [Simonovits and Sós](https://www.semanticscholar.org/paper/Ramsey-Tur%C3%A1n-theory-Simonovits-S%C3%B3s/b6acab1b394d5096f2cd7221022553cc2cad2f7d) survey on Ramsey-Turán theory, when defining $RT(n,L^{(r)},s)$ for $r$-uniform hypergraphs, they say that the limit
$$\lim\_{n \to \infty}\frac{RT(n,L^{(r)},\varepsilon n)}{n^r}$$
exists and follows relatively easily from vertex-multiplication. I don't quite know how vertex-multiplication works but maybe it helps.
|
https://mathoverflow.net/users/225950
|
Ramsey-Turán density function is well defined
|
I answered the question in detail for $RT(n, K\_l, \alpha n)$ [here](https://math.stackexchange.com/a/4282372/383078) on Math StackExchange.
|
1
|
https://mathoverflow.net/users/106323
|
406842
| 166,726 |
https://mathoverflow.net/questions/406775
|
9
|
It seems to me that a considerably simpler proof [see below] of Birkhoff's ergodic theorem can be obtained for bounded observables than for more general $L^1$ observables. Therefore, I feel like it would be nicer to prove Birkhoff's ergodic theorem by starting off with the bounded case and then extending to the general case, than by "directly" tackling the general case. But can this extension from $L^\infty$ to $L^1$ be done in any "straightforward" elementary manner?
Let me give one possible version of how to make this question precise:
Let $(X,\mathcal{X},\mu)$ be a probability space. A *Markov operator* on $L^1(\mu)$ is a linear, monotone, unity-preserving, integral-preserving function $P \colon L^1(\mu) \to L^1(\mu)$.
Suppose we have a sequence $(P\_n)\_{n \geq 1}$ of Markov operators on $L^1(\mu)$ such that **(1)** for every $f \in L^\infty(\mu)$ the following statements hold:
* $P\_n(f) \overset{\mu\textrm{-a.s.}}{\to} \int\_X f \, d\mu\ $ as $n \to \infty$;
* for all $m,n \geq 1$, $\|nP\_n(f) - mP\_m(f)\|\_{L^\infty(\mu)} \leq |m-n|\|f\|\_{L^\infty(\mu)}$;
and **(2)** for every $f \in L^1(\mu)$ the following statements hold:
* there exists a value $P\_\infty[f] \in \overline{\mathbb{R}}$ such that $\,\liminf\_{n \to \infty} P\_n(f) \overset{\mu\textrm{-a.s.}}{=} P\_\infty[f]$;
* for all $m,n \geq 1$, $\|nP\_n(f) - mP\_m(f)\|\_{L^1(\mu)} \leq |m-n|\|f\|\_{L^1(\mu)}$.
(Obviously, if $f \in L^\infty(\mu)$ then $P\_\infty[f]=\int\_X f \, d\mu$. We will also see from the proof of the result in the Note below that the same holds for all $f \in L^1(\mu)$ with $f \geq 0$.)
>
> Do we necessarily have that for all $f \in L^1(\mu)$, $P\_n(f) \overset{\mu\textrm{-a.s.}}{\to} \int\_X f \, d\mu\ $ as $n \to \infty$?
>
>
>
(Since $P\_n(-f)=-P\_n(f)$, this is equivalent to saying that for all $f \in L^1(\mu)$, $P\_\infty[f]=\int\_X f \, d\mu$.)
**Remark.** I expect that if the answer is *yes*, then not all the conditions given above will be necessary to prove it.
---
**Note.** As below, it is not hard to show [under considerably weaker conditions than those given above] that for each $f \in L^1(\mu)$, $\ P\_n(f) \overset{L^1(\mu)}{\to} \int\_X f \, d\mu\ $ as $n \to \infty$.
*Proof*: Without loss of generality take $f \overset{\mu\textrm{-a.s.}}{\geq} 0$. For each $k>0$ we have $P\_n(f \wedge k) \overset{\mu\textrm{-a.s.}}{\to} \int\_X f \wedge k \; d\mu\ $ as $n \to \infty$. Hence, by monotonicity of Markov operators it is clear that
$$ \liminf\_{n \to \infty} P\_n(f) \overset{\mu\textrm{-a.s.}}{\geq} \int\_X f \, d\mu. $$
But due to the integral-preservation of Markov operators, we also have that $\int\_X P\_n(f) \, d\mu = \int\_X f \, d\mu$ for all $n$; and so, since $P\_n(f) \overset{\mu\textrm{-a.s.}}{\geq} 0$ (by monotonicity of Markov operators), Fatou's lemma gives that
$$ \int\_X \liminf\_{n \to \infty} P\_n(f) \, d\mu \leq \int\_X f \, d\mu. $$
Hence it follows that
$$ \liminf\_{n \to \infty} P\_n(f) \overset{\mu\textrm{-a.s.}}{=} \int\_X f \, d\mu. $$
The result is then an immediate consequence of the following Scheffé-like lemma.
**Lemma.** *Given a sequence $(Y\_n)\_{n \in \mathbb{N}}$ of integrable random variables $Y\_n$ that is uniformly bounded below, if $\ Y \!:=\! \underset{n \to \infty}{\liminf} Y\_n\ $ is integrable and $\mathbb{E}[Y\_n] \to \mathbb{E}[Y]$ as $n \to \infty$, then $Y\_n \overset{L^1}{\to} Y$.*
*Proof of Lemma.* Without loss of generality take $Y=0$. We have that $Y\_n \wedge 0$ converges pointwise to $0$ as $n \to \infty$, and so since $(Y\_n)$ is uniformly bounded below, the dominated convergence theorem can be applied to give that $\mathbb{E}[Y\_n \wedge 0] \to 0$ as $n \to \infty$. Since $|Y\_n|=Y\_n-2(Y\_n \wedge 0)$ and $\mathbb{E}[Y\_n] \to 0$ as $n \to \infty$, it follows that $\mathbb{E}[|Y\_n|] \to 0$ as $n \to \infty$. So we are done.
---
**Short proof of BET for bounded observables.**
I constructed the following proof after doing a bit of Googling of proofs of the ergodic theorem. (In particular, it combines ideas from the paper <https://doi.org/10.1214/074921706000000266> of Keane and Peterson - which, to be fair, is already a pretty impressively short proof even for general $L^1$ observables - with proofs I've seen in online lectures notes on ergodic theory; but, of course, I also use boundedness to help get a short proof.)
I will prove the theorem for ergodic transformations, but it is not at all difficult to extend (in the appropriate manner) to more general measure-preserving transformations. *I have tried to write out the proof sufficiently explicitly to be smoothly readable in about 2-5 minutes.*
**Theorem.** *Let $(X,\mathcal{X},T,\mu)$ be an ergodic measure-preserving dynamical system and let $f \colon X \to \mathbb{R}$ be a bounded measurable function. Then $P\_N(f):=\frac{1}{N} \sum\_{i=0}^{N-1} f \circ T^i \, \overset{\mu\textrm{-a.s.}}{\to} \, \mathbb{E}\_\mu[f]\,$ as $N \to \infty$.*
**Proof.** Since $\overline{\lim}\_{N \to \infty} P\_N(f)$ is $T$-invariant, it is $\mu$-a.e. equal to a constant $\overline{f}$. We will show $\overline{f} \leq \mathbb{E}\_\mu[f]$; then, applying this to $-f$ as well as $f$ gives the result. Fixing arbitrary $\varepsilon>0$, let $g=f+\varepsilon-\overline{f}$. Define the monotone sequences
\begin{align\*}
m\_N^{[g]} := & \, \max\big\{ \, g \ , \ g \!+\! (g \circ T) \ , \ \ldots\ldots \ , \ g \!+\! (g \circ T) \!+\! \ldots \!+\! (g \circ T^{N-1}) \, \big\} \\
E\_N^{[g]} := & \, \{ x \in X : m\_N^{[g]}(x)>0 \}.
\end{align\*}
For $N \geq 2$,
\begin{align\*}
(m\_N^{[g]} - g)(x) &= \max\big\{ \, 0 \ , \ g(T(x)) \ , \ \ldots\ldots \ , \ g(T(x)) \!+\! \ldots \!+\! g(T^{N-1}(x)) \, \big\} \\
&= (m\_{N-1}^{[g]})^+(T(x)).
\end{align\*}
Integrating over $E\_N^{[g]}$ gives
\begin{align\*}
\int\_{E\_N^{[g]}} g \, d\mu \ &= \ \int\_{E\_N^{[g]}} m\_N^{[g]} \, d\mu - \int\_{E\_N^{[g]}} (m\_{N-1}^{[g]})^+ \circ T \, d\mu \\
&= \int\_X (m\_N^{[g]})^+ \, d\mu - \int\_{E\_N^{[g]}} (m\_{N-1}^{[g]})^+ \circ T \, d\mu \\
&\geq \int\_X (m\_N^{[g]})^+ \, d\mu - \int\_X (m\_N^{[g]})^+ \circ T \, d\mu \ = \ 0.
\end{align\*}
Now
\begin{align\*}
\bigcup\_{N=1}^\infty \! E\_N^{[g]} &= \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P\_n(g)(x) > 0 \} \\
&= \{x \, : \, \exists n \in \mathbb{N} \text{ s.t. } P\_n(f)(x) > \overline{f} - \varepsilon \},
\end{align\*}
which is a $\mu$-full measure set by definition of $\overline{f}$. Hence by the dominated convergence theorem, $\int\_X g \, d\mu \geq 0$, i.e. $\mathbb{E}\_\mu[f] \geq \overline{f}-\varepsilon$. But $\varepsilon$ was arbitrary. **QED.**
**Remark.** Contained within the above is a proof of a "maximal ergodic theorem" that does not rely on $g$ being bounded but only on $g$ being $\mu$-integrable; in other words, the above proof goes through without modification for any $f \in L^1(\mu)$ for which it is known that $\overline{f}$ is finite. But I see no trivial way of showing that $\overline{f}$ is finite except when $f$ is bounded.
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https://mathoverflow.net/users/15570
|
Can Birkhoff's ergodic theorem for integrable functions easily be deduced from Birkhoff's ergodic theorem for bounded functions?
|
There is a simple reduction of the Birkhoff ergodic Theorem for $L^1$ functions to the bounded case using Kakutani-Rokhlin towers, that I learned from H. Furstenberg and B. Weiss decades ago. We use the notation of the post, and assume that $T$ is ergodic. Given $f \in L^1(X)$ we may assume it is nonnegative, and then (by subtracting the fractional part and adding 1) that it takes values in the positive integers. Consider the subset
$$Y=\{(x,k): 1 \le k \le f(x) \}$$
of $X \times {\mathbb N}$, endowed with the product $\sigma$-algebra, the probability measure $\nu$ determined by $$\nu(A \times{k})=\frac{\mu(A)}{\int\_X f \, d\mu}$$
for measurable sets $A \subset X$ where $\min\_A f \ge k$, and the transformation $S$ defined by $S(x,k)=(x,k+1)$ if $k<f(x)$ and $S(x,k)=(T(x),1)$ if $k=f(x)$.
Then $S$ is ergodic on $(Y,\nu)$, since if a measurable set $E \subset Y$ is invariant under $S$, then $E\_1:=\{x \in X \,: (x,1) \in E\}$ is invariant under $T$.
**Claim:** The ergodic theorem in $(Y,\nu,S)$ for the indicator function
$$h(x,k):={\mathbf 1}\_{\{k=1\}}$$ (along the sequence of return times to the base $\{k=1\}$ of the tower) implies the ergodic theorem for $f$ in $(X,\mu,T)$.
**Proof:** consider the Birkhoff sums
$$R\_n(x):=\sum\_{j=0}^{n-1}f(T^jx) \,.$$
For each $x \in X$, we have
$$\{\ell \ge 0 : h(S^\ell(x,1))=1 \}= \{R\_j(x) \; : \: j \ge 0\} \,$$
whence we conclude that for $\mu$-almost every $x \in X$,
$$\lim\_{n \to \infty} \frac{n}{R\_n(x)}=
\lim\_{n \to \infty} \frac{ \sum\_{\ell=0}^{R\_n(x)-1}h(S^\ell(x,a)) }{R\_n(x)} =\int\_Y h \, d\nu =\frac{\mu(X)}{\int\_X f \, d\mu} =\frac{1}{\int\_X f \, d\mu}\,.$$
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9
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https://mathoverflow.net/users/7691
|
406846
| 166,727 |
https://mathoverflow.net/questions/406849
|
0
|
Let $X\_1, \ldots, X\_n$ be independent and identically distributed random variables. Let $f:\mathbb{R}^n \to \mathbb{R}$ be a bounded difference function, i.e., for any $x,y \in \mathbb{R}^n$ that differ only in coordinate $i$, we have $|f(x) - f(y)| \leq c\_i$. Under this setting, [McDiarmid’s inequality](https://en.wikipedia.org/wiki/Doob_martingale#McDiarmid%27s_inequality) provides a well-known concentration result.
Under the same setting, do we also have a central limit theorem for an appropriately scaled version of $f(X\_1, \ldots, X\_n) - E[f(X\_1, \ldots, X\_n)]$ ? I am ok with making additional assumptions on $(X\_i)\_i$ sequence.
|
https://mathoverflow.net/users/16976
|
CLT for bounded difference functions
|
In the general setting you propose, the answer is negative. Suppose that $X\_i$ take the values $\pm 1$ with equal probability. Let $h: {\mathbb R} \to {\mathbb R}$ be a Lipschitz function with constant 1, e.g., $$ (\*) \quad h(x)=\max\{x,0\} \, ,$$ and define
$$f\_n(x\_1,\dots,x\_n):=\sqrt{n} \cdot h \left(\frac{x\_1+\dots +x\_n}{\sqrt{n}}\right) \,.$$
(In the special case (\*), one can skip dividing and multiplying by $\sqrt{n}$.)
Then $f\_n$ satisfy the bounded difference condition with constants $c\_i=2$ on the support of the distribution.
(If you want the bounded difference condition as stated to hold everywhere, consider the functions
$\widetilde{f}\_n(x\_1,\dots,x\_n):=f\_n({\rm sign}(x\_1),\dots,{\rm sign}(x\_n))$ instead.)
Let $Z$ be a standard normal variable. Then we have the convergence in law
$$\frac{f(X\_1, \ldots, X\_n) - E[f(X\_1, \ldots, X\_n)]}{\sqrt{{\rm Var} [f(X\_1, \ldots, X\_n)]}}
\Rightarrow \frac{h(Z)-E[h(Z)]}{\sqrt{{\rm Var}[h(Z)] }}\,. $$
The distribution of the right-hand side is not Gaussian for most functions $h$.
In the special case (\*), this distribution will have a density which is the right half of the Gaussian density, shifted and scaled to have mean zero and variance 1.
**Postscript:** Perhaps you can post another question with the particular bounded difference functions $f\_n$ you have in mind.
One proof of Mcdiarmid's inequality uses Martingales, so perhaps in your application a Martingale CLT will be useful. See <https://en.wikipedia.org/wiki/Martingale_central_limit_theorem> and the references therein.
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2
|
https://mathoverflow.net/users/7691
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406851
| 166,730 |
https://mathoverflow.net/questions/406859
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1
|
In [Tka] the author writes:
>
> "Every topological space $X$ can be represented as an open continuous image of a completely regular submetrizable space $Y$ (in other words, $Y$ admits a continuous one-to-one mapping onto a metrizable space) — the corresponding construction is given on p. 331 of [Eng]".
>
>
>
But I can not find this statement on this page. Is there a source in which this statement is explicitly formulated and proved?
[Tka] M. G. Tkachenko. Topological groups for topologists: part I, Bol. Soc. Mat. Mexicana (3), 5, 1999, 237-279.
[Eng] R. Engelking, General Topology, Heldermann Verlag, Berlin 1989.
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https://mathoverflow.net/users/142738
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Open images of submetrizable spaces
|
It seems quite likely that it was intended to be page 331 of the Engelking's book published in 1977 in PWN, Warszava1 - rather then the later edition by Heldermann (revised and completed edition, 1989).
I will quote in full exercise 4.2.D (page 331 in the older edition, page 264 in the newer edition):
>
> 4.2.D. (a) (Ponomarev [1960]) Prove that a $T\_0$-space $X$ is first countable if and only if $X$ is a continuous image of a metrizable space under an open mapping.
> Hint. Let $X$ be a first-countable space and let $\{U\_s\}\_{s\in S}$ be a base for $X$.
> Consider the Baire space $B(\mathfrak m) = \prod\_{i=1}^\infty X\_i$ where $X\_i = S$ with the discrete topology, and the subset $T \subseteq B(\mathfrak m)$ consisting of all points $\{s\_i\}$ such that $\{U\_{s\_i}\}\_{i=1}^\infty$ is a base at a point $x\in X$; assign the point $x \in X$ to the point $\{s\_i\}\in T$.
>
> (b) (Michael [1971a]) Show that every first-countable space $X$ is a continuous image of a first-countable Hausdorff space under an open mapping, and deduce that in (a) the assumption that $X$ is a $T\_0$-space can be omitted.
>
> Hint (Shimrat [1956]). Define in $B(\mathfrak m)$, where $m = |X|$, a family $\{A\_x\}\_{x\in X}$ of pairwise disjoint dense subsets and consider the subset $\bigcup\_{x\in X}(\{x\}\times A\_x)$ of the Cartesian product $X \times B(\mathfrak m)$.
>
> Remark. The construction in the above hint shows that every topological space is a continuous image of a Hausdorff space under an open mapping. Isbell proved more in [1969]: every topological space is a continuous image of a hereditarily paracompact and hereditarily strongly zero-dimensional space under an open mapping. Further information related to parts (a) and (b) can be found in Junnila [1978] and R. Pol [1981].
>
> (c) (Arhangelskii [1963], Franklin [1965]) Observe that the construction outlined in Exercise 2.4.G(b) shows that sequential spaces can be characterized as the images of metrizable spaces under quotient mappings and Frechet spaces can be characterized as the images of metrizable spaces under hereditarily quotient mappings.
>
>
>
* *Arkhangel’skij, A. V.*, Bicompact sets and the topology of spaces, Trans. Mosc. Math. Soc. 13, 1-62 (1965); translation from Tr. Mosk. Mat. O.-va 13, 3-55 (1965). [ZBL0162.26602](https://zbmath.org/?q=an:0162.26602), [MR0150733](https://mathscinet.ams.org/mathscinet-getitem?mr=0150733).
* *Franklin, S. P.*, [**Spaces in which sequences suffice**](http://dx.doi.org/10.4064/fm-57-1-107-115), Fundam. Math. 57, 107-115 (1965). [ZBL0132.17802](https://zbmath.org/?q=an:0132.17802), [MR180954](https://mathscinet.ams.org/mathscinet-getitem?mr=180954).
* *Isbell, J. R.*, [**A note on complete closure algebras**](http://dx.doi.org/10.1007/BF01691060), Math. Syst. Theory 3, 310-312 (1969). [ZBL0182.34303](https://zbmath.org/?q=an:0182.34303), [MR252296](https://mathscinet.ams.org/mathscinet-getitem?mr=252296).
* *Junnila, H. J. K.*, Stratifiable pre-images of topological spaces, Topology, Vol. II, 4th Colloq. Budapest 1978, Colloq. Math. Soc. Janos Bolyai 23, 689-703 (1980). [ZBL0437.54019](https://zbmath.org/?q=an:0437.54019), [MR588817](https://mathscinet.ams.org/mathscinet-getitem?mr=588817).
* *Michael, Ernest A.*, [**On representing spaces as images of metrizable and related spaces**](http://dx.doi.org/10.1016/0016-660X(71)90005-5), General Topology Appl. 1, 329-343 (1971). [ZBL0227.54009](https://zbmath.org/?q=an:0227.54009), [MR293604](https://mathscinet.ams.org/mathscinet-getitem?mr=293604).
* *Ponomarev, V.*, Axioms of countability and continuous mapping, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 8, 127-134 (1960). [ZBL0095.16301](https://zbmath.org/?q=an:0095.16301), [MR0116314](https://mathscinet.ams.org/mathscinet-getitem?mr=0116314); in Russian.
* *Pol, Roman*, [**On category-raising and dimension-raising open mappings with discrete fibers**](http://dx.doi.org/10.4064/cm-44-1-65-76), Colloq. Math. 44, 65-76 (1981). [ZBL0479.54008](https://zbmath.org/?q=an:0479.54008), [MR633099](https://mathscinet.ams.org/mathscinet-getitem?mr=633099).
I will add that the same exercises is mentioned in the answer to this question on Mathematics Stack Exchange: [Every Tychonoff space is an image of a Moscow space under a continuous open mapping.](https://math.stackexchange.com/q/353593) The references in that answer might be worth looking at, too.
1This older edition used to be available online - I do not know whether it is still accessible somewhere. Here is a [Wayback Machine snapshot](https://web.archive.org/web/20120419192949/http://matwbn.icm.edu.pl/kstresc.php?tom=60&wyd=10&jez=pl) of the website where it used to be.
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https://mathoverflow.net/users/8250
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406862
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https://mathoverflow.net/questions/395720
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11
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**Remark 1:** On p.384 of volume 3 of Gauss's *Werke*, which is a part of an unpublished treatise on the arithmetic geometric mean, Gauss makes the following remark:
>
> **On the theory of the division of numbers into four squares**:
> The theorem that the product of two sums of four squares is itself a sum of four squares, is most simply represented as follows: let $l,m,\lambda,\mu,\lambda',\mu'$ be six complex numbers such that $\lambda,\lambda'$ and $\mu,\mu'$ are conjugate. Let $N$ denote the norm, than $$(Nl+Nm)(N\lambda + N\mu)=N(l\lambda+m\mu)+N(l\mu'-m\lambda')$$ and also $$(N(n+in')+N(n''+in'''))(N(1-i)+N(1+i))=N((n+n'+n''-n''')+i(-n+n'+n''+n'''))+N((n+n'-n''+n''')-i(n-n'+n''+n'''))$$
> From this it is easy to derive the following two propositions, in which different representation of a number by a sum of four squares refer to the different value systems of the four roots, taking into account both the signs and the sequence of the roots. **1.** If the fourfold of a number of the form $4k+1$ can be represented by four odd squares, then it can be represented half as often by one odd and three even squares, and vice versa, if a number can be represented in this way, that it can be represented twice as often in the first way. **2.** If the fourfold of a number of the form $4k+3$ can be represented by four odd squares, then it can be represented half as often by one even and three odd squares, and vice versa, if a number can be represented in this way, than its quadruple can be represented twice as often in the first way.
>
>
>
Gauss than says that a certain identity of theta functions can be derived by these two theorems, namely the assertion (written here in modern notation):
$$\vartheta\_{00}(0;\tau)^4 = \vartheta\_{01}(0;\tau)^4 + \vartheta\_{10}(0;\tau)^4$$
(Gauss denotes the three theta functions by $p(y),q(y),r(y)$).
***Notes on remark 1:***
* The first identity in this passage is intimately connected with quaternions - actually it is a kind of pre [Cayley-Dickson construction](https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction), since Gauss essentially says here that the quaternions multiplication rules arise from the rules of complex arithmetic and can be constructed by them. To see this more clearly, lets make the following steps:
$$Nl+Nm=N(l+mj)=N(l-mj)$$
$$N\lambda+N\mu = N(\lambda+\mu j)=N(\lambda+j\mu)$$
Since the quaternions algebra is a composition algebra, the norm is multiplicative, so:
$$N(l-mj)\cdot N(\lambda+j\mu) = N((l-mj)\cdot(\lambda+j\mu)) = N((l\lambda -(mj)(j\mu))+l(j\mu)-(mj)\lambda) = N((l\lambda+m\mu))+l(\bar{\mu}j)-m(\bar{\lambda}j)) = N((l\lambda+m\mu))+(l\bar{\mu}-m\bar{\lambda})j) = N(l\lambda+m\mu)+N(l\bar{\mu}-m\bar{\lambda})$$
Here the associativity of quaternions is used, as well as the fact that (for example) $j\mu = \bar{\mu}j$. For more on Gauss's anticipation of the quaternions algebra, look at the (partially answered) post [Motivating unpublished statements of Gauss about congruences and quaternions](https://mathoverflow.net/questions/426535/motivating-unpublished-statements-of-gauss-about-congruences-and-quaternions).
* The second identity follows directly from the first by substituting $\lambda = 1-i$ and $\mu = 1+i$. Since $N(1-i)+N(1+i)=4$, this identity enables one to generate new representations of an integer $4s$ as sum of four squares by simply changing the signs and order of the different numbers in the representation of $s$ as sum of four squares. For example, if $s = 13 = 2^2+2^2+2^2+1^2$ than this identity implies $52=4s = 5^2+3^2+3^2+3^2$.
**Remark 2**: On p. 1-2 of volume 8 of Gauss's *Werke* there is an additional note on the representation of numbers as sums of squares. According to Dickson's "history of the theory of numbers":
>
> Gauss noted that every decomposition of a multiple of a prime $p$ into $a^2+b^2+c^2+d^2$ corresponds to a solution of $x^2+y^2+z^2\equiv 0 \pmod{p}$ proportional to $a^2+b^2,ac+bd,ad-bc$ or to the sets derived by interchanging $b$ and $c$ or $b$ and $d$. For $p\equiv 3 \pmod{4}$, the solutions of $1+x^2+y^2\equiv 0 \pmod{p}$ coincide with those of $1+(x+iy)^{p+1}\equiv 0 \pmod{p}$. From one value of $x+iy$ we get all by using: $$(x+iy)\frac{(u+i)}{(u - i)}$$ (where $u = 0,1,\cdots, p-1$). For $p\equiv 1 \pmod{4}, p = a^2+b^2$; then $b\frac{(u+i)}{a(u-i)}$ give all values of $x+iy$ if we exclude the values $a/b$ and $b/a$ of $u$.
>
>
>
***Notes on remark 2:***
* The result of Gauss on the correspondence between the representation of a multiple of a prime number $p$ as sum of four squares and the solution to the congruence $x^2+y^2+z^2\equiv 0 \pmod{p}$ is straitforward to prove: $$x^2+y^2+z^2 = (a^2+b^2)^2+(ac+bd)^2+(ad-bc)^2=(a^2+b^2)^2+(a^2+b^2)(c^2+d^2)= (a^2+b^2)(a^2+b^2+c^2+d^2)\equiv 0 \pmod{p}$$. Therefore, what remains to be settled is the other results mentioned in remark 2.
* The result on the correspondence between the solutions $(x,y)$ of the congruence $1+x^2+y^2\equiv 0 \pmod{p}$ and the solution of a certain imaginary congruence of degree $p+1$ was checked by me by taking specific examples: for example, if $p=7$, than $x = 3, y = 2$ is a solution, and $1+(3+2i)^8 = -238-28560i = 7\cdot(-34-4080i)$ is a Gaussian integer multiple of $7$.
* The papers "*On the Computation of Representations of Primes as Sums of Four Squares*" and "*The circle equation over finite fields*" mention results equivalent to Gauss's results in this remark. In particular, Gauss's method of generating new solutions to the congruence $1+x^2+y^2\equiv 0 \pmod{p}$ by $x+iy = (x\_0+iy\_0)(\frac{u+i}{u-i})$ is refered to as "the method of diophantus" in section 2.2 of the first paper i mentioned. Since both papers appear not to be very advanced, i believe that explaning the results in remark 2 is an easy task in the standards of mathoverflow.
**Questions**
* The two propositions which Gauss mentions in remark 1 are unclear to me, and I mean that the propositions themself are unclear, not their derivation. It is simply not formulated clearly. Therefore, I'd like to get an explanation of the two propositions which Gauss mentions, as well as an explanation of its proof.
* I'd like to understand how the theta function identity in remark 1 follows from the two propositions.
* What is the expalanation of the results in remark 2? they are complicated and i don't have a clue of understanding its proof.
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https://mathoverflow.net/users/118562
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Explanation of several unpublished remarks of Gauss on representations of a given number as sums of two, three and four squares
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Let me add a few remarks concerning 2. If $p \equiv 3 \bmod 4$, then ${\mathbb F}\_p(i) = {\mathbb F}\_{p^2}$. The relative norm of $x+iy$ is the product of $x+iy$ and its conjugate $x-iy$, but the latter is the image of the Frobenius automorphism, i.e., $x-iy = (x+iy)^p$. This shows that $x^2 + y^2 = (x+iy)(x-iy) = (x+iy)^{p+1}$.
If $x+iy$ is an element with norm $-1$, i.e., with $x^2 + y^2 = -1$ in ${\mathbb F}\_p$, then you get all other elements with norm $-1$ by multiplying $x+iy$ by an element with norm $1$. By Hilbert's Theorem 90, such elements have the form $\frac{c+di}{c-di}$; if $d = 0$, this quotient is $1$, so you may assume $d \ne 0$ and cancel $d$; writing $u = c/d$ then proves the second claim.
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3
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https://mathoverflow.net/users/3503
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406866
| 166,732 |
https://mathoverflow.net/questions/406864
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2
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I'm working with the notion of direct integrals as in Dixmier. Briefly: Given a measurable space $X$ and a family of separable Hilbert spaces $(H\_x)\_{x\in X}$, a measurable structure is a subspace $Y$ of the sections $\Pi\_x H\_x$ which satisfies the following axioms:
(1) for all $u\in Y$, $x\mapsto \|u(x)\|\_{H\_x}$ is measurable,
(2) given a section $u$, if for all $v\in Y$ one has that $x\mapsto (u(x),v(x))\_{H\_x}$ is measurable, then already $u\in Y$,
(3) there is a sequence $(u\_n)\_n$ in $Y$ such that for each $x\in X$, the sequence $(u\_n(x))\_n$ is dense in $H\_x$.
If the family of spaces is constant, that is to say $H\_x = H$ for some fixed separable Hilbert space $H$, then the measurable functions from $X$ to $H$ are a measurable structure.
**Question**: Are there other examples of measurable structures for such a constant family of spaces?
The maximality condition (2) makes it hard for me to come up with further examples. For instance, I tried to use the measurable functions in a closed subspace of $H$, but then condition (2) would force that all functions with values in the orthogonal complement were measurable as well.
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https://mathoverflow.net/users/70434
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Measurable structures for direct integrals
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Up to isomorphism, there are no other examples. But literally speaking, there are other examples. You could choose a non-measurable map $U$ from $X$ to the unitary group $\mathcal{U}(H)$, for instance by fixing a nontrivial unitary $U\_0 \in \mathcal{U}(H)$ and putting $U(x) = U\_0$ when $x$ belongs to a nonmeasurable set, while $U(x) = 1$ when $x$ belongs to the complement. Then you define $Y$ as the space of sections of the form $x \mapsto U(x) u(x)$, where $u$ is a measurable function from $X$ to $H$.
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4
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https://mathoverflow.net/users/159170
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406874
| 166,735 |
https://mathoverflow.net/questions/406877
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4
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Let $f(x),g(x)$ be polynomials in $\mathbb{Q}[x]$. If $\mathrm{deg}(f)\geq2$ and $f$ irreducible, is the composition $f(g(x))$ always reduced (has no repeated irreducible factors)?
(If we do not ask $\mathrm{deg}(f)\geq2$ we can take $f(x)=x-1, g(x)=x^2+1$; if we do not ask $f$ be reducible, we can take $f(x)=(x-1)x$ and $g=x^2+1$.)
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https://mathoverflow.net/users/nan
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Does irreducible polynomial remain reduced by pre-composition?
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It may have repeated irreducible factor. Take $f(x)=x^2+1$ and $g(x)=x+f(x) h(x)$ so that $g'(i)=0$. Then $f^2$ divides $f(g) $.
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12
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https://mathoverflow.net/users/4312
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406883
| 166,738 |
https://mathoverflow.net/questions/406884
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2
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I am trying to find smooth functions $f : \mathbb{R}\_+ \to \mathbb{R}\_+$ such that the quantity $$\Delta\_f(x) := 2f(x)-f(2x)$$
is positive for $x$ large enough and has the **greatest asymptotic growth**.
---
It seems clear that one cannot go beyond a linear growth, since $\Delta\_f$ vanishes for linear functions and will likely be negative for super-linear ones. However, one can construct many examples of almost linear asymptotic growths.
For example, plugging $f(x) := x^a$ for some $a\in(0,1)$ yields $\Delta\_f(x) = (2-2^a) x^a$.
Choosing $f(x) := \frac{x}{\ln x}$ yields $\Delta\_f(x) \sim 2 \ln 2 \frac{x}{(\ln x)^2}$ which has a larger asymptotic growth.
---
What would be your candidates for even larger asymptotic growths? And is there a way to prove a (sublinear) a priori upper bound on $\Delta\_f$?
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https://mathoverflow.net/users/50777
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Largest asymptotic growth for $2f(x)-f(2x)$
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Let us discretise the problem by setting $a\_n=2^{-n}f(2^n)$, $b\_n=2^{-n-1}\Delta\_f(2^n)$. Then your relation becomes,
$$b\_n=a\_n-a\_{n+1}.$$
since $a\_n,b\_n$ are non-negative, we conclude that
$$\sum\_{n=1}^\infty b\_n<\infty.$$
This is a necessary and sufficient condition. Indeed, take any summable sequence $b\_n$ of positive numbers, then we can define
$$a\_n=\sum\_{k=n}^\infty b\_k$$ as the sequence of partial sums, and obtain your equation on the sequence $x\_n=2^n$. Then you can interpolate by choosing $f(x)$ arbitrarily on the interval $(1,2)$.
Returning to your original notation, the growth condition becomes
$$\int\frac{\Delta\_f(x)}{x^2}dx<\infty.$$
This is a necessary and sufficient condition.
For example, we cannot have $\Delta\_f(x)\sim x/\log x$, but can have
$$\Delta\_f(x)\sim\frac{x}{\log x(\log\log x)^{1+\epsilon}},$$
and so on.
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3
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https://mathoverflow.net/users/25510
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406891
| 166,741 |
https://mathoverflow.net/questions/406886
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4
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$\DeclareMathOperator{\Inv}{Inv}\DeclareMathOperator{\Erg}{Erg}$This is mostly curiosity on my part and I hope that the MO community might be able to help.
For $c\in (0,4]$ consider the logistic map
$$
T\_c:[0,1]\to[0,1],\;\;T\_c(x)=cx(1-x).
$$
Denote by $\Inv\_c$ the collection of Borel probability measures on $[0,1]$ that are $T\_c$-invariant and by $\Erg\_c\subset \Inv\_c$ the subset of consisting of ergodic ones.
**Question 1.** There are simple examples such ergodic measure
measure, e.g., measures concentrated on a periodic orbit. Do there exist ergodic measures not concentrated on a periodic orbit?
**Question 2.** This is a bit more vague. Is it known how the set $\Erg\_c$ evolves with changing $c $ inside the set Borel probability measures on $[0,1]$?
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https://mathoverflow.net/users/20302
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Ergodic measures for the logistic map
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When $c=4$, the map $T\_4(x)=4x(1-x)$ on the unit interval is semi-conjugate to the transformation $z\mapsto z^2$ of the unit circle via $z\mapsto\frac{1}{2}-\frac{1}{4}\left(z+\frac{1}{z}\right)$:
$$
T\_4\left(\frac{1}{2}-\frac{1}{4}\left(z+\frac{1}{z}\right)\right)=\frac{1}{2}-\frac{1}{4}\left(z^2+\frac{1}{z^2}\right).
$$
Thus the pushforward of the Lebesgue measure on the unit circle (which is an ergodic measure for $z\mapsto z^2$) by $z\mapsto\frac{1}{2}-\frac{1}{4}\left(z+\frac{1}{z}\right)$ provides an an absolutely continuous ergodic measure for $T\_4:[0,1]\rightarrow [0,1]$.
In general, there is a very deep [theorem](https://annals.math.princeton.edu/2002/156-1/p01) of Lyubich stating that, aside from a measure zero subset of parameters, for every $c$ the map $T\_c$ is either *hyperbolic* (i.e. has an attracting periodic orbit) or *stochastic* (i.e. admits an absolutely continuous invariant measure).
---
**Added:** Another well-studied example is $c\approx 3.57$ that happens at the end of period-doubling cascade $-$ the largest parameter for which the topological entropy is zero. For this parameter, the map $T\_c$ is infinitely renormalizable (the aforementioned paper of Lyubich shows that the set of such parameters is of measure zero). For this $c$, there exists a *Feigenbaum attractor*. This is an invariant Cantor set on which the dynamics of $T\_c$ is conjugate to a "2-adic adding machine". Ergodic probability measures for such an interval map are classified [here](https://www.researchgate.net/publication/228926349_An_ergodic_adding_machine_on_the_Cantor_set); these are supported either on periodic orbits of period $2^n$ or on the Feigenbaum attractor.
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10
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https://mathoverflow.net/users/128556
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406892
| 166,742 |
https://mathoverflow.net/questions/406872
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2
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Say $\ell^2-\ell-\delta\equiv0\bmod p$ is a polynomial $x^2-x-\delta$ with root $\ell$ and we lift to $\ell'^2-\ell'-\delta\equiv0\bmod p^2$ by Hensel lifting my question is following: if $g^y\equiv \ell^2\bmod p$ held at a generator $g$ then is there an explicit generator $g'$, which can be found in polynomial time, for which $$g'^y\equiv\ell'^2\bmod p^2$$ holds true?
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https://mathoverflow.net/users/10035
|
A problem on generators and Hensel lifting
|
Write $g'=g(1+ap)$ so that
$$g'^y\equiv g^y(1 + pya)\pmod{p^2}.$$
It follows that we can take
$$a = \frac{(\ell^2/g^{y}\bmod p^2)-1}{py},$$
which can be computed in polynomial time.
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2
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https://mathoverflow.net/users/7076
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406898
| 166,744 |
https://mathoverflow.net/questions/406857
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5
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Let $K$ be a finite extension of $\mathbb{Q}\_p$. Let $F$ be a Lubin–Tate formal group law defined over $K$ with endomorphism $f(T)$ corresponding to $\pi$ (a uniformizer of $K$). Then one can define the logarithm of $F$ to be $\lambda\_F(T) = \lim\_{n\rightarrow\infty}\pi^{-n}f^n(T)$. Then $\lambda\_F(T) = T + \text{higher-degree terms}$, so the logarithm is invertible under function composition. We define $\operatorname{exp}\_F(T)$ to be the inverse of $\lambda\_F$ under composition. Then $\operatorname{exp}\_F(\lambda\_F(T)) = T$. However I don't understand how this is possible if $\lambda\_F$ is not one to one. In particular $\lambda\_F$ sends all torsion points of $F$ to 0. Please let me know what is wrong here.
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https://mathoverflow.net/users/378621
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Question about log and exp of a formal group law
|
The radius of convergence of any formal group $F$ (one-dimensional, finite height) is $1$, in other words $L\_F$ will converge at all $z\in\Bbb C\_p$ with $v(z)>0$.
In particular, the logarithm is convergent at all torsion points of the formal group. What goes wrong, goes wrong with the exponential, whose series is not convergent far enough from the origin to reach any of the torsion points.
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13
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https://mathoverflow.net/users/11417
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406912
| 166,748 |
https://mathoverflow.net/questions/406905
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9
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Suppose $A\_1,\dots,A\_n$ are measurable subsets of the plane that are all related by rigid motions such that $|(A\_1 \cup \dots \cup A\_n)^c| = 0$ and $|A\_i \cap A\_j| = 0$ for all $1 \leq i < j \leq n$, where $|S|$ denotes the Lebesgue measure of $S$.
Must each $A\_i$ have the property that $|A\_i \cap B(r)|/|B(r)| \rightarrow 1/n$ as $r \rightarrow \infty$, where $B(r)$ is the disk of radius $r$ centered at 0?
The answer is clearly “yes” if the $A\_i$’s are all obtained from one another by rotation about a point (e.g., consider the Fatou sets for Newton’s algorithm applied to the polynomial $z^3 - 1$). Maybe this is the only way to tile the plane by $n$ congruent pieces, but I don’t see why it should be true.
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https://mathoverflow.net/users/3621
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Tiling the plane with finitely many congruent pieces
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Write $A\_i=T\_i(A)$ for $A=A\_1$, where each $T\_i$ is a rigid motion. For each $i$, we have $|A\_i\cap B(r)|=|T\_i(A\cap T\_i^{-1}(B(r))|=|A\cap T\_i^{-1}(B(r))|$. The symmetric difference between this set and $A\cap B(r)$ is contained in the symmetric difference of $T\_i^{-1}(B(r))$ and $B(r)$, whose measure is $O(r)$. Hence the measures of $A\_i\cap B(r)$ are all equal up to $O(r)$. Since the sum of their measures is $B(r)$, this implies $|A\_i\cap B(r)|=B(r)/n+O(r)$ for each $i$.
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11
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https://mathoverflow.net/users/30186
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406913
| 166,749 |
https://mathoverflow.net/questions/406921
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1
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Let $X = G/\Gamma$ denote the Iwasawa threefold, where
$$G = \left\{\begin{pmatrix} 1 & z\_1 & z\_3\\ 0 & 1 & z\_2\\ 0 & 0 & 1\end{pmatrix} : z\_1, z\_2, z\_3 \in \mathbb{C} \right\},$$
and $\Gamma$ is the discrete subgroup $$\Gamma=\left\{\begin{pmatrix} 1 & z\_1 & z\_3\\ 0 & 1 & z\_2\\ 0 & 0 & 1\end{pmatrix} : z\_1, z\_2, z\_3 \in \mathbb{Z}[i] \right\}.$$
>
> Can anyone point to a reference for the Hodge diamond of $X$?
>
>
>
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https://mathoverflow.net/users/105103
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Reference for the Hodge diamond of the Iwasawa threefold
|
Should have waited a few minutes, it's on page 49 of Danielle Angela's *Cohomological aspects of non-Kähler manifolds*: <https://arxiv.org/pdf/1302.0524.pdf>
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2
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https://mathoverflow.net/users/105103
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406922
| 166,751 |
https://mathoverflow.net/questions/406920
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1
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Let $A\_{n}(F) $ denote the $n \times n$ skew symmetric matrices over a finite field $F$. Suppose $n$ be even and $N$ be a subspace of $A\_{n}(F) $. Now if all the non-zero matrices in $N$ are invertible, then the maximum the dimension of $N$ will be $n/2$. The upper bound of this maximum dimension follows from Chevalley warning theorem .
Also I know that there is a such type of subspace of dim $n/2$. But the way to get this result is tricky. Does there any simplest process to find this type subspace $N$ of $\dim n/2$? Provide me some example of such type subspace.
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https://mathoverflow.net/users/215016
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Problem concerning about an $n$-subspace of $ A_{n}(F) $
|
If you take $n/2 \times n/2$ matrix $A$, with an irreducible minimal polynomial (as can be constructed with a [companion matrix](https://en.wikipedia.org/wiki/Companion_matrix)), the subspace $W = \text{span}\left(I, A, \ldots, A^{n/2-1}\right)$ will be a subspace of matrices of dimension $n/2$ where all non-zero matrices are invertible.
You can create the subspace of $n \times n$ skew symmetric matrices by taking matrices of the form
$$\left[\begin{array}{cc}0 & X \\ -X & 0 \end{array}\right]$$
for $X \in W$.
|
2
|
https://mathoverflow.net/users/7838
|
406925
| 166,753 |
https://mathoverflow.net/questions/406863
|
1
|
Let $f: \mathbb R \to \mathbb R$ be a Lipschitz strictly monotone (so, in particular, invertible) function. Let $u: \mathbb R \to \mathbb R$. If $f \circ u \in BV$ can we conclude that $u \in BV$?
|
https://mathoverflow.net/users/157076
|
If $f \circ u \in BV$ and $f$ is strictly monotone, then is $u \in BV$?
|
The answer is no.
E.g., let $f(x):=\min(1,|x|)x$ for real $x$. Then $f$ is Lipschitz and strictly monotone.
Let then $g$ be any function in $BV$ such that $g(1/n)=(-1)^n/n^2$ for all natural $n$; it is easy to see that such a function exists. (For instance, let $g=0$ on $(-\infty,0]\cup(1,\infty)$ and let $g$ be monotonic on $[\frac1{n+1},\frac1n]$ for each natural $n$. Then the total variation of $g$ is $2\sum\_{n=1}^\infty1/n^2<\infty$.)
Finally, let $u:=f^{-1}\circ g$. Then $f\circ u=g\in BV$, whereas $u(1/n)=(-1)^n/n$ for all natural $n$. So, $u$ is not in $BV$, because the total variation of $u$ is no less than $\sum\_{n=1}^\infty|u(\frac1n)-u(\frac1{n+1})|=\infty$.
|
6
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https://mathoverflow.net/users/36721
|
406930
| 166,755 |
https://mathoverflow.net/questions/406901
|
1
|
Let $(X(t))\_{t \in [-1,1]}$ be a centered non-stationary smooth gaussian process with covariation function $\rho(t,s) = \mathbb E[X(t)X(s)]$. For $t\_0 \in (-1,1)$ and $\epsilon \in (-1-t\_0,1-t\_0)$, define
$$
p\_X(t\_0,\epsilon) : = \mathbb P(X(t) = 0\,\text{ for some } t \in [t\_0-\epsilon,t\_0+\epsilon])
$$
>
> **Question.** *What is a good **upper-bound** for $p\_X(t\_0,\epsilon)$ which is valid for **small** $\epsilon$ (i.e for $\epsilon \to 0^+$) ?*
>
>
>
**A concrete example.** The GP I have in mind is $X(t) := tU + (1-t^2)^{1/2}V$, where $(U,V) \sim N(0,I\_2)$, for which the covariation function is
$$
\rho(t,s) = ts +(1-t^2)^{1/2}(1-s^2)^{1/2}.
$$
|
https://mathoverflow.net/users/78539
|
Lower-bound on zero-crossing probability of the nonstationary gaussian process $X(t) = tU+(1-t^2)^{1/2}V$, with $(U,V) \sim N(0,I_2)$
|
$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example,
$$p\_X(t\_0,\ep)=P\Big(m\_1<\frac VU<m\_2\Big),$$
where
$$m\_1:=\min\_{t\in[t\_0-\ep,t\_0+\ep]}r(t)
=r(t\_0+\ep),\quad
m\_2:=\max\_{t\in[t\_0-\ep,t\_0+\ep]}r(t)
=r(t\_0-\ep),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$
this follows because $r$ is a continuous decreasing function on the interval $(-1,1)$.
Letting now
$$\th\_j:=\arctan m\_j$$
and using the rotational symmetry of the distribution of $(U,V)$, we get
$$p\_X(t\_0,\ep)=\frac{\th\_2-\th\_1}{\pi}.
$$
|
1
|
https://mathoverflow.net/users/36721
|
406932
| 166,756 |
https://mathoverflow.net/questions/406543
|
3
|
The following theorem and proof are in **Applications of the proper forcing axiom**, the Baumgartner's paper in the book *Handbook of Set-theoretic topology*.
$3.6$
***THEOREM***. Assume PFA. Suppose that for each $\alpha < \omega\_1$ a set $S\_\alpha \subseteq \omega\_1$ is given such that, for every limit ordinal $\beta < \omega\_1$, $S\_{\alpha} \cap \beta$ has ordertype $< \beta$. Then there is a closed unbounded set $C$ such that $ \forall \alpha < \omega\_1\ C \cap S\_{\alpha}$ is finite.
***PROOF***. Let $P$ consist of all $p$ for which there is a closed unbounded set $C \subseteq \omega\_1$ containing only limit ordinals so that $p$ is a finite subset of the enumerating function of $C$. Let $Q$ be the set of all pairs $(p, x)$, where $p \in P$ and $x \in [\omega\_1]^{<\omega}$. Let $(p\_1, x\_1) \leq (p\_2, x\_2)$ iff $ p\_1 \supseteq p\_2, \ x\_1 \supseteq x\_2$ and $\forall \alpha \in x\_2 \ \text{range}(p\_1 - p\_2) \cap S\_\alpha =0$. Now force with $Q$.
As usual we have to start by proving that $\forall \alpha < \omega\_1$ the set $D\_\alpha =\{ (p, x) \in Q : \alpha \in dom(p)\}$ is dense in $Q$.
For $D\_0$ let we choose $(p, x) \in Q$ such that $\alpha\_0 \in x,\ \omega \in S\_{\alpha\_0},\ p=\{(1, \omega . 2)\}$. For each $p' \supseteq p$ such that $0 \in dom(p')$ and every $x'\supseteq x,\ \omega \in \text{range}(p' - p) \cap S\_{\alpha\_0}$. Thus $(p', x') \nleq (p,x)$.
This counterexample shows $D\_0$ isn't dense. Do we not need density? Or should we add something to the terms and definitions?
|
https://mathoverflow.net/users/169636
|
Adding a closed unbounded set containing of only limit ordinals with a special property
|
Maybe the following idea works: Given a condition as above, also require the following:
(1) for each $\alpha, f(\alpha)$ is indecomposable,
(2) suppose $dom(p)=\{\beta\_0 < \beta\_1 < \cdots < \beta\_n\}$. Then there exists a closed subset $C$ of $f(\beta\_n)$ of order type $\beta\_n$ such that $C \cap \bigcup\_{\gamma \in x}S\_\gamma \subseteq range(p)$, for each $i$, $C \cap f(\beta\_i)$ has order type $\beta\_i.$ Furthermore, if $\beta\_i$ is limit, then $C \cap f(\beta\_i)$ is unbounded in $ f(\beta\_i)$.
|
3
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https://mathoverflow.net/users/11115
|
406945
| 166,764 |
https://mathoverflow.net/questions/406946
|
1
|
This statement is proved by [Vizing](https://books.google.co.kr/books?id=leL0Y5N0bFoC&pg=PA30&lpg=PA30&dq=V.%20G.%20Vizing,%20Vertex%20colorings%20with%20given%20colors%20(Russian).%20Diskret.%20Analiz.%2029%20(1976),%203%E2%80%9310.&source=bl&ots=zKFBHSs3nD&sig=ACfU3U0UAZFYC02b2_YFOK1NUSg8KP91BA&hl=ko&sa=X&ved=2ahUKEwif1YyHlODzAhVVeXAKHUviAYQQ6AF6BAgCEAM#v=onepage&q=V.%20G.%20Vizing%2C%20Vertex%20colorings%20with%20given%20colors%20(Russian).%20Diskret.%20Analiz.%2029%20(1976)%2C%203%E2%80%9310.&f=false) and [Erdos & Rubin](https://old.renyi.hu/%7Ep_erdos/1980-07.pdf) (page 30) independently.
But I cannot find Vizing's paper (It's too old) and Erdos & Rubin's paper only says 'It is easily proved'.
I think it is related to the proof of the following statement
>
> $K\_{2,r}$ is $r$-choosable.
>
>
>
as Erdos & Rubin's paper mentioned.
But I have no idea with how to prove the generalized one.
Would you help me?
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https://mathoverflow.net/users/384338
|
$K_{k,m}$ is $k$-choosable if and only if $m<k^k$
|
Assume that $K\_{k,m}$ is not $k$-choosable with some lists of admissible colors. Let $A\_1$, $A\_2$, $\ldots$, $A\_k$ be sets of admissible colors in the small part (that with $k$ vertices). Choose arbitrarily colors $a\_i\in A\_i$ for all $i=1,\ldots,k$. The large part must contain a vertex with admissible colors $a\_1,\ldots,a\_k$, otherwise we may color each vertex in the large part. This yields that all $a\_i$'s are distinct, in other words all $A\_i$'s must be disjoint. Then all sequences $(a\_1,\ldots,a\_k)$ are different, and the large part must contain $k^k$ distinct vertices. On the other hand, if $m=k^k$, all $A\_i$'s are indeed disjoint, and the large part contain all such $k^k$ vertices, we can not color it.
|
1
|
https://mathoverflow.net/users/4312
|
406947
| 166,765 |
https://mathoverflow.net/questions/406734
|
4
|
Recall the construction of the reduced crossed product:
>
> Let $\Gamma$ be a discrete group and $A$ be a $C^\*$-algebra with an action $\alpha: \Gamma\to \operatorname{Aut}(A)$. Consider the $\*$-algebra $C\_c(\Gamma,A)$ of finitely supported functions $\Gamma \to A$ with the $\alpha$-twisted multiplication and involution. Then we can build a canonical faithful representation of $C\_c(\Gamma,A)$ as follows: start with a faithful representation $A \subseteq B(H)$. This induces a new faithful representation $\pi: A \to B(H \otimes \ell^2(\Gamma))$ by $\pi(a)(\xi \otimes \delta\_g) = \alpha\_{g}^{-1}(a)\xi \otimes \delta\_g$. Considering the left regular representation $\lambda: \Gamma \to U(\ell^2(H)): g \mapsto (\delta\_h \mapsto \delta\_{gh})$, we obtain an induced faithful representation
> $$C\_c(\Gamma,A) \to B(H \otimes \ell^2(\Gamma)): \sum\_{s \in \Gamma} a\_s s \mapsto \sum\_{s \in \Gamma} \pi(a\_s)(1 \otimes \lambda\_s)$$
> which induces a $C^\*$-norm on $C\_c(\Gamma,A)$. The reduced crossed product $A \rtimes\_r \Gamma$ is the $C^\*$-completion of $C\_c(\Gamma,A)$ with respect to this norm, and does not depend on the choice of faithful representation $A\subseteq B(H)$.
>
>
>
Let $\Gamma$ be a discrete group and let $\varphi: A \to B$ be a $\Gamma$-equivariant completely positive contraction between the $\Gamma$-$C^\*$-algebras $A$ and $B$. I want to show the following:
>
> The induced map $C\_c(\Gamma,A) \to C\_c(\Gamma,B): \sum\_{s \in \Gamma} a\_s s \mapsto \sum\_{s \in \Gamma}\varphi(a\_s)s$ is bounded, hence extends uniquely to a map $\varphi \rtimes\_r \Gamma: A \rtimes\_r \Gamma \to B \rtimes\_r \Gamma.$
>
>
>
**Attempt:** Let $\pi\_A: A \to B(H\_A \otimes \ell^2(\Gamma))$ and $\pi\_B: B \to B(H\_B \otimes \ell^2(\Gamma))$ be faithful representations as above. Then by the $C^\*$-identity.
$$\|\sum\_s \varphi(a\_s)s\|^2 = \|\sum\_s \pi\_B(\varphi(a\_s))(1 \otimes \lambda\_s)\|^2$$
$$=\|\sum\_{s,t} (1 \otimes \lambda\_{s^{-1}}) \pi\_B(\varphi(a\_s^\*)\varphi(a\_t)) (1\otimes \lambda\_t)\|.$$
This looks like something we could apply Cauchy-Schwarz for completely positive maps on, but the surrounding factors $1 \otimes \lambda\_{s^{-1}}$ and $1 \otimes \lambda\_t$ complicate this. Maybe I need to apply Cauchy-Schwarz on some carefully crafted matrix. Does anybody see how I can continue?
Of course, other approaches are also welcome! I am also interested in the following: if $\varphi$ is injective, then is the extension $A \rtimes\_r \Gamma \to B \rtimes\_r \Gamma$ also injective?
|
https://mathoverflow.net/users/nan
|
A completely positive equivariant map $\varphi: A \to B$ induces a map $A \rtimes_r \Gamma \to B \rtimes_r \Gamma.$
|
This is Exercise 4.1.4 in the book by Brown+Ozawa (is that where this question ultimately comes from?) The "Hint" is "The reduced case is easy." Hmm.
Well, it might perhaps help to look at this section of the book. Indeed, Proposition 4.1.5, *and its proof*, shows that if $F\subseteq\Gamma$ is a finite-set, and $P:\ell^2(\Gamma)\rightarrow \ell^2(F)$ the projection, then
$$ (1\otimes P)x(1\otimes P) = \sum\_{s\in\Gamma} \sum\_{p\in F\cap sF} \alpha\_{p^{-1}}(a\_s)\otimes e\_{p, s^{-1}p} \in A\otimes M\_F(\mathbb C)
\cong M\_F(A). $$
Here $x = \sum a\_s\lambda\_s \in C\_c(\Gamma, A) \subseteq A \rtimes\_r \Gamma \subseteq \mathcal B(H\otimes\ell^2(\Gamma))$ and $e\_{p,q}$ are the matrix units of $M\_F(\mathbb C)$. The point of this is to show that the norm of $x$, acting on $H\otimes\ell^2(\Gamma)$, does not depend upon the particular representation $A\subseteq\mathcal B(H)$ (because $M\_F(A)$ has a unique norm).
However, this formula also gives us a convenient way to compute the norm of $\varphi\rtimes 1$. For, if $y=(\varphi\rtimes 1)x$, then $y=\sum\_s \varphi(a\_s)\lambda\_s$, and so
$$ (1\otimes P)y(1\otimes P) = \sum\_{s\in\Gamma} \sum\_{p\in F\cap sF} \beta\_{p^{-1}}(\varphi(a\_s))\otimes e\_{p, s^{-1}p}
= \sum\_{s\in\Gamma} \sum\_{p\in F\cap sF} \varphi(\alpha\_{p^{-1}}(a\_s))\otimes e\_{p, s^{-1}p}, $$
the first equality by applying to above to $B\rtimes\_r\Gamma$, and the second equality using that $\varphi$ is equivariant. However, this then equals
$$ (\varphi\otimes 1\_F)\big( (1\otimes P)x(1\otimes P) \big), $$
where $(\varphi\otimes 1\_F)$ is the dilated map $M\_F(A)\rightarrow M\_F(B)$. As $\varphi$ is CCP by assumption, $\varphi\otimes 1\_F$ is contractive, and so
$$ \| (1\otimes P)y (1\otimes P) \| \leq \|(1\otimes P)x(1\otimes P)\|
\leq \|x\|. $$
Taking the SOT limit as $F$ increases, so $P\rightarrow 1$, shows that $\|y\| \leq \|x\|$ as required.
|
7
|
https://mathoverflow.net/users/406
|
406976
| 166,774 |
https://mathoverflow.net/questions/406959
|
0
|
When I first saw the definition of general Sobolev spaces with real exponent I immediately got interested in the following problem: pick several of your favourite irregular functions/distributions and now ask how singular are they, i.e. what is the largest exponent $s$ such that this function/distribution belongs to $H^s$. For simplicity let me restrict to the interval $[-\pi,\pi]$ where the critical $s$ such that $f \in H^{-s}$ (note the minus sign) can be found by investigating the convergence of the series $\sum\_{n} \frac{|c\_n|^2}{n^{2s}}$ where $c\_n$ are Fourier coefficients.
**Example 1** Let us take $f=\chi\_{[0,1]}$. Then after some computations this boils down to the convergence of $\sum\_{n} \frac{\sin^2{n}}{n^{2+2s}}$ and thus we get that $f \in H^s$ for $s<\frac12$ but $f \notin H^\frac{1}{2}$.
**Example 2** If we take the Dirac delta distribution $\delta$ then the sequence of Fourier coefficients is constant and one gets that $\delta \in H^s$ for $s<-\frac12$ but $\delta \notin H^{\frac12}$.
**Example 3** Let us now take $f(x)=\frac{1}{\sqrt{|x|}}$. Then $f \notin L^2=H^0$ but $f \in H^s$ for $s<0$
So for these examples the *critical exponents* are $s\_0=\frac12,-\frac12,0$ respectively. In all these cases $f \notin H^{s\_0}$ but $f \in H^s$ for $s<s\_0$.
>
> Question Is it possible that for a critical exponent as above in fact $f \in H^{s\_0}$ (but $f \notin H^s$ for $s>s\_0$?)?
>
>
>
This should follow from general facts concerning Dirichlet series but I'm not an expert in this field and I will be happy if someone could give me an answer
|
https://mathoverflow.net/users/24078
|
What is the critical exponent for irregular function in the Sobolev scale?
|
Consider the distribution with Fourier series $$\sum\_{n \ge 1} \frac{\cos(nt)}{n^{1/2+\alpha} \log n} \,.$$
This will be in the Sobolev space $H^s=H^{s,2}$ for $s \le \alpha$ but not for $s>\alpha$.
|
1
|
https://mathoverflow.net/users/7691
|
406980
| 166,775 |
https://mathoverflow.net/questions/406985
|
-2
|
Is there a exact formula for number of decompositions of $2n-1$ into a difference of two squares?
Examples:
```
3: 1 | 21: 1
4: 1 | 22: 1
5: 2 | 23: 3
6: 1 | 24: 1
7: 1 | 25: 2
8: 2 | 26: 2
9: 1 | 27: 1
10: 1 | 28: 2
11: 2 | 29: 2
12: 1 | 30: 1
13: 2 | 31: 1
14: 2 | 32: 3
15: 1 | 33: 2
16: 1 | 34: 1
17: 2 | 35: 2
18: 2 | 36: 1
19: 1 | 37: 1
20: 2 | 38: 3
```
|
https://mathoverflow.net/users/147145
|
The number of decompositions of $2n-1$ into a difference of two squares?
|
Any odd composite integer $m$ can be written as $m=pq$. We can switch it to a difference of squares:
>
> $$m=pq$$
> $$4m=4pq$$
> $$4m=2pq+2pq$$
> $$4m=2pq+2pq+q^{2}-q^{2}+p^{2}-p^{2}$$
> $$4m=\left(p+q\right)^{2}-\left(q-p\right)^{2}$$
> $$m=\frac{\left(p+q\right)^{2}}{4}-\frac{\left(q-p\right)}{4}^{2}$$
> $$m=\left(\frac{p+q}{2}\right)^{2}-\left(\frac{q-p}{2}\right)^{2}$$
>
>
>
Since $m$ is odd, $p$ and $q$ are also odd. Therefore, every possible pairs $p,q$ such that $m=pq$ will be solutions. We can now use the number of divisors function $\sigma(m)$ to conclude that:
>
> The number of decompositions of $2n-1$ into a difference of two squares is defined by: $$\left\lceil \frac{\sigma\left(2n-1\right)}{2}\right\rceil$$
>
>
>
|
0
|
https://mathoverflow.net/users/147145
|
406986
| 166,779 |
https://mathoverflow.net/questions/406978
|
7
|
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\trace{trace}$Let $A \in \SL(2,\mathbb{R})$ and $\trace(A)>2$. Is it true that $$\lVert A\rVert \leq \lVert A^2\rVert,$$
where $\lVert \rVert$ is the operator norm that is the first singular value?
$$\lVert A \rVert =\sqrt{\lambda\_{\text{max}}(A^\*A)}=\sigma\_{\text{max}}(A).$$
Let me mention that if the condition $\trace(A)>2$ is removed, then the above statement is not true; see Jeppe Stig Nielsen's [answer](https://math.stackexchange.com/a/4282670) to [Is it true that $\lVert A\rVert \leq \lVert A^2\rVert$ for $A\in \operatorname{SL}(2, \mathbb{R})$?](https://math.stackexchange.com/questions/4282612/is-it-true-that-a-leq-a2-for-a-in-sl2-mathbbr) on MSE.
My attempt:
I think it is true: The operator norm satisfied
$$\lVert A\rVert=\sup\left\{\lVert Ax\rVert \,\middle\vert\, \text{$x\in\mathbb{R}^2$ and $\lVert x\rVert=1$}\right\}$$
where the symbols $\lVert \rVert$ inside the brackets on the right-hand side denote the standard (Euclidean) length of a vector in $\mathbb{R}^2$. So $\lVert A\rVert$ is the maximal length of the image of a unit vector. On the other hand, $\trace(A)>2$, so one of the eigenvalues is greater than the other one.
|
https://mathoverflow.net/users/127839
|
Is it true that $\lVert A\rVert \leq \lVert A^2\rVert$ for $A\in \operatorname{SL}(2, \mathbb{R})$ when $\operatorname{trace}(A)>2$?
|
We can do this by a calculation. The assumptions on the determinant and trace are equivalent to having eigenvalues $\lambda,1/\lambda$, with $\lambda>1$. We can rotate the first eigenvector to the $e\_1$ position, and then
$$
A=\begin{pmatrix} \lambda & b \\ 0 & 1/\lambda \end{pmatrix} ,
$$
so
$$
A^\*A=\begin{pmatrix} \lambda^2 & \lambda b \\ \lambda b & b^2 +1/\lambda^2
\end{pmatrix}
$$
In general, the eigenvalues of a $B\in\textrm{SL}(2,\mathbb R)$ are $T/2 \pm \sqrt{T^2/4-1}$, with $T=\textrm{tr}\; B$.
In the case of $A^\*A$, we are interested in the large eigenvalue (obtained for $+$). Clearly, this is increasing in $T$. So now the question is if $(A^2)^\*A^2$ has a larger trace $T\_2$, and the same calculation now gives
$$
T\_2 = \lambda^4 + \frac{1}{\lambda^4} + b^2 \left( \lambda+\frac{1}{\lambda}\right)^2 .
$$
This indeed satisfies $T\_2\ge \lambda^2+1/\lambda^2 + b$. In fact, we have $\lambda^4+1/\lambda^4\ge \lambda^2+1/\lambda^2$ for $\lambda\ge 1$, or, equivalently, $f(x)\equiv x^4-x^3-x+1\ge 0$ for $x\ge 1$, since $f'(x)\ge 0$ in this range and $f(1)=0$.
|
11
|
https://mathoverflow.net/users/48839
|
406987
| 166,780 |
https://mathoverflow.net/questions/407002
|
4
|
Let $E$ be an elliptic curve over $\mathbf{Q}$. Then we can base-change $E$ to $\mathbf{C}$ and apply the uniformization theorem to obtain:
$$E(\mathbf{C}) \cong \mathbf{C}/(\mathbf{Z} + \mathbf{Z} \tau ) $$
for some complex number $\tau$ in the upper half plane. I've done a few numerical tests on Sage, and I've found that the *real part* of $\tau$ seems to be a *rational number*. So I wanted to ask: is it known to be true that if $E$ is defined over $\mathbf{Q}$ and $\tau$ is given as above, then $\rm{Re }\,\,\tau$ is a rational number? And if it is, would anyone be able to provide a reference for a proof?
---
*Important note*: an earlier version of the question said that $E$ was an elliptic curve defined over $\mathbf{C}$, *not* over $\mathbf{Q}$, which explains some of the comments. I've just edited it to say that $E$ must be defined over $\mathbf{Q}$.
|
https://mathoverflow.net/users/394740
|
The real part of the period of an elliptic curve
|
However, if $E$ is defined over $\mathbb R$, then it's always possible to find a $\tau$ of the form either $\tau=ti$ or $\tau=\frac12+ti$ so that $E(\mathbb C)$ is analytically isomorphic (over $\mathbb R$, even) to $\mathbb C/(\mathbb Z+\mathbb Z\tau)$. So possibly the examples you were looking at are defined over $\mathbb R$, which you can check by seeing if $j(E)\in\mathbb R$.
**Addendum** If $E$ has complex multiplication, then $\mathbb Q(\tau)$ is an imaginary quadratic field. If $E$ does not have CM, then my recollection is that $\tau$ is transcendental over $\mathbb Q$. There is further information in the answer to [When is the period of elliptic curve over the rationals transcendental?](https://mathoverflow.net/questions/28385/when-is-the-period-of-elliptic-curve-over-the-rationals-transcendental)
|
8
|
https://mathoverflow.net/users/11926
|
407005
| 166,787 |
https://mathoverflow.net/questions/406902
|
2
|
Let $f(n)$ be [A053645](https://oeis.org/A053645), distance to largest power of $2$ less than or equal to $n$; write $n$ in binary, change the first digit to zero, and convert back to decimal.
Let $g(n)$ be [A007814](https://oeis.org/A007814), the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Then we have a pair of an integer sequences given by
\begin{align}
a\_1(0)=a\_1(1)&=1\\
a\_1(2n)& = a\_1(n)+a\_1(2n-2^{g(n)})\\
a\_1(2n+1) &= a\_1(2f(n))
\end{align}
and
\begin{align}
a\_2(0)=a\_2(1)&=1\\
a\_2(2n)& = a\_2(n)+a\_2(n-2^{g(n)})+a\_2(2n-2^{g(n)})\\
a\_2(2n+1) &= a\_2(2f(n))
\end{align}
Let
$$s\_k(n)=\sum\limits\_{j=0}^{2^n-1}a\_k(j)$$
then I conjecture that
$$s\_1(n)=1 + \sum\limits\_{i=0}^{n} i(i+1)^{n-i}$$
and
$$s\_2(n)=(n+1)s\_2(n-1)-(n-2)s\_2(n-2), s\_2(0)=1, s\_2(1)=2$$
where $s\_1(n)$ is [A047970](https://oeis.org/A047970) and $s\_2(n)$ is [A006183](https://oeis.org/A006183).
Is there a way to prove it?
Similar questions:
* [Sum with Stirling numbers of the second kind](https://mathoverflow.net/questions/406738/sum-with-stirling-numbers-of-the-second-kind)
* [Recurrence for the sum](https://mathoverflow.net/questions/405174/recurrence-for-the-sum)
* [Sequence that sums up to INVERTi transform applied to the ordered Bell numbers](https://mathoverflow.net/questions/407290/sequence-that-sums-up-to-inverti-transform-applied-to-the-ordered-bell-numbers)
* [Sequences that sums up to second differences of Bell and Catalan numbers](https://mathoverflow.net/questions/407758/sequences-that-sums-up-to-second-differences-of-bell-and-catalan-numbers)
|
https://mathoverflow.net/users/231922
|
Pair of recurrence relations with $a(2n+1)=a(2f(n))$
|
As proved in [this answer](https://mathoverflow.net/q/405996), for $n=2^tk$ with $2\nmid k$, we have
$$a\_1(n)=\sum\_{i=0}^t \binom{t}{i} a\_1(2^i(k-1)+1).$$
Then for $n=2^{t\_1}(1+2^{t\_2}(1+\dots(1+2^{t\_m}))\dots)$ with $t\_1\geq 0$ and $t\_j\geq 1$ for $j\geq 2$, we have
\begin{split}
a\_1(n) &= \sum\_{i\_1=0}^{t\_1} \binom{t\_1}{i\_1} \sum\_{i\_2=0}^{t\_2+i\_1} \binom{t\_2+i\_1}{i\_2} \sum\_{i\_3=0}^{t\_3+i\_2} \dots \sum\_{i\_\ell=0}^{t\_\ell+i\_{\ell-1}} \binom{t\_\ell+i\_{\ell-1}}{i\_\ell} \\
&=\prod\_{j=1}^\ell (\ell+2-j)^{t\_j},
\end{split}
where $\ell := \lfloor (m+1)/2\rfloor$.
Correspondingly,
\begin{split}
s\_1(n) &= \sum\_{m=0}^{n} \sum\_{t\_1+t\_2+\dots+t\_{m}\leq n-1}\ \prod\_{j=1}^\ell (\ell+2-j)^{t\_j}\\
&= \sum\_{m=0}^n\ \sum\_{t\_1+t\_2+\dots+t\_{m}+t\_{m+1} = n}\ \prod\_{j=1}^{\ell+1} (\ell+2-j)^{t\_j}\\
&=\sum\_{m=0}^n [x^{n-m}]\ \ell! \left(\frac{1}{1-x}\right)^{m-\ell}\prod\_{j=1}^{\ell+1} \frac1{1-jx} \\
&=\sum\_{m=0}^n [x^{n-m}]\ \ell! \left(\frac{1}{1-x}\right)^{m-\ell}\sum\_{q\geq \ell+1} S(q,\ell+1) x^{q-\ell-1}\\
&=\sum\_{m=0}^n \ell! \sum\_{q\geq l+1} S(q,\ell+1) \binom{n-q}{n-q-m+\ell+1}.
\end{split}
Then grouping terms $m=2\ell-1$ and $2\ell$ we have
\begin{split}
s\_1(n)&=\sum\_{\ell\geq 0} \ell! \sum\_{q\geq l+1} S(q,\ell+1) \binom{n-q+1}{n-q-\ell+2} \\
&=1 + \sum\_{\ell\geq 0} \ell \sum\_{q= l+1}^{n+1} S(q,\ell+1) (n-q+1)\_{l-1}\\
&=1 + \sum\_{q=1}^{n+1} \sum\_{\ell\geq 0} \ell S(q,\ell+1) (n-q+1)\_{l-1}\\
&=1 + \sum\_{q=1}^{n+1} \sum\_{\ell\geq 0} (S(q+1,\ell+1) - S(q,\ell+1) - S(q,\ell)) (n-q+1)\_{l-1} \\
&=1 + \sum\_{q=1}^{n+1} \left(\frac{(n-q+3)^q}{n-q+2} - \frac{(n-q+3)^{q-1}}{n-q+2} - (n-q+2)^{q-1}\right) \\
&=1 + \sum\_{q=1}^{n+1} \left((n-q+3)^{q-1} - (n-q+2)^{q-1}\right) \\
&=1 + \sum\_{i=0}^{n} i(i+1)^{n-i}.
\end{split}
$s\_2(n)$ can be treated similarly.
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1
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https://mathoverflow.net/users/7076
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407010
| 166,788 |
https://mathoverflow.net/questions/406990
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4
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Can a Vopenka cardinal be supercompact?
I asked a [weaker question](https://mathoverflow.net/questions/400732/is-vopenkas-principle-ord-has-the-tree-property-consistent) on here before. Unfortunately, I don't know enough set theory to see whether the positive answer there generalizes to a positive answer here.
|
https://mathoverflow.net/users/2362
|
Can a Vopenka cardinal be supercompact?
|
If $\kappa$ is almost huge with target $\lambda,$ then $V\_{\lambda}$ thinks that $\kappa$ is a supercompact Vopenka cardinal.
I'll take for granted the standard facts about almost huge cardinals listed here: <https://neugierde.github.io/cantors-attic/Huge>
In particular, $\kappa$ is Vopenka (in $V$), and this is just an assertion about $V\_{\kappa+1}$ so $V\_{\lambda}$ also sees that $\kappa$ is Vopenka. Also $\kappa$ is $\theta$-supercompact for every $\theta<\lambda,$ i.e. every $\mathcal{P}\_{\kappa}(\theta)$ has a normal fine measure. Again these facts are all absolute down to $V\_{\lambda},$ so $V\_{\lambda}$ thinks that $\kappa$ is supercompact.
Also by a standard elementary embedding argument, these facts about $\kappa$ and $\lambda$ imply that there are stationarily many $\alpha<\kappa$ such that $V\_{\kappa}$ believes $\alpha$ to be a supercompact Vopenka cardinal.
|
9
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https://mathoverflow.net/users/109573
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407024
| 166,792 |
https://mathoverflow.net/questions/406795
|
2
|
Let $\Delta\_n(x\_1, \ldots, x\_n)$ denote the Vandermonde determinant $\displaystyle \prod\_{1 \leq i < j \leq n}(x\_j - x\_i)$. Let $c\_1, \ldots, c\_n$ and $K$ be nonnegative integers satisfying
$$c\_1 + \cdots +c\_n = K + m\frac{n(n-1)}{2},$$
where $m$ is a positive integer. Then the coefficient of $x\_1^{c\_1}\cdots x\_n^{c\_n}$ in the polynomial $(x\_1 + \cdots + x\_n)^{K}\Delta\_n(x\_1^m, \ldots, x\_n^m)$ is
\begin{equation}
\sum\_{\sigma \in S\_n}\mathrm{sgn}(\sigma)\frac{K!}{\prod\_{i=1}^{n}(c\_i - \sigma(i)m + m)!}.
\end{equation}
My questions are the following:
1. Is it possible to express this coefficient as product of terms? We know that this is possible for $m = 1$ (see Lemma $9.8$ in Tao and Vu: Additive Combinatorics on page $336$) and in the case of $c\_i = k-i$ for $i = 1, \ldots, n$ for any positive integer $m$ (see Lemma $9.9$ in Tao and Vu: Additive Combinatorics on page $336$).
2. If that is not possible in general for $m \geq 2$, for which choices of $c\_i$, is it possible (in particular for $m = 2$)?
Any help in this regard will be greatly appreciated. Thanks.
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https://mathoverflow.net/users/93753
|
Coefficient of a term in a several variable polynomial multipled with Vandermonde determinant
|
Note that $\Delta\_n(x\_1^m,\ldots,x\_n^m)=\det(x\_i^{k\_j})$, where $k\_j:=(j-1)m$. Two key observations are the following (both work for arbitrary non-negative integers $k\_1<k\_2<\ldots<k\_n$, if $\sum c\_j=\sum k\_j+K$):
1. (algebraic) $$\left[x\_1^{c\_1}x\_2^{c\_2}\ldots x\_n^{c\_n}\right](x\_1+\ldots+x\_n)^K\det(x\_i^{k\_j})=\frac{(K-\sum k\_j)!}{\prod c\_j!}\det(c\_i^{\underline{k\_j}}),\quad\quad\quad(\heartsuit)$$
where we use Knuth's notation $x^{\underline{n}}=x(x-1)\ldots(x-n+1)$. To prove $(\heartsuit)$, replace the polynomial $F:=(x\_1+\ldots+x\_n)^K\det(x\_i^{k\_j})$ to $$\tilde{F}=
\left(x\_1+\ldots+x\_n-\sum k\_j\right)^{\underline{K}}\det(x\_i^{\underline{k\_j}})$$
(which has of course the same highest degree coefficients as $F$)
and apply [this](https://mathoverflow.net/a/214930/4312) formula for the sets $A\_i=\{0,1,\ldots,c\_i\}$. Note that if $a\_i\in A\_i$, and $\tilde{F}(a\_1,\ldots,a\_n)\ne 0$, expanding the determinant we see that there must exist a permutation $\pi$ for which $a\_i(a\_i-1)\ldots (a\_i-k\_{\pi\_i}+1)\ne 0$, i.e., $a\_i\geqslant k\_{\pi\_i}$, in particular $\sum a\_j\geqslant \sum k\_j$. Further, $\sum a\_j-\sum k\_j$ is a non-negative integer which can not be equal to $0,1,\ldots,K-1$, thus $\sum a\_j\geqslant \sum k\_j+K=\sum c\_j$. But $a\_j\leqslant c\_j$ for all $j$, thus this all is possible only if $a\_j=c\_j$ for all $j$. Therefore, in the RHS of the cited formula there is (at most) unique non-zero term, and this immediately gives $(\heartsuit)$.
2. (combinatorial) If $c\_1<c\_2<\ldots<c\_n$ are non-negative integers and $\sum c\_j=\sum k\_j+K$, then both parts of $(\heartsuit)$, which we denote by $\theta(c\_1,\ldots,c\_n)$ are equal to the number of paths from the point $(k\_1,\ldots,k\_n)$ to the point $(c\_1,\ldots,c\_n)$, for which each of $K$ paths constitutes of increasing exactly one coordinate by 1, and all intermediate points belong to a domain $\{(t\_1,\ldots,t\_n):t\_1<t\_2<\ldots<t\_n\}$. This may be proved by induction on $K$ for fixed $k\_1,\ldots,k\_n$. The base case $K=0$ is clear. For the induction step, note that $\Theta$ satisfies a recurrence
$$
\Theta(c\_1,\ldots,c\_n)=\Theta(c\_1-1,c\_2,\ldots,c\_n)+
\Theta(c\_1,c\_2-1,\ldots,c\_n)+\ldots+\Theta(c\_1,c\_2,\ldots,c\_n-1),
$$
which corresponds to considering the summand taken from the last bracket $x\_1+\ldots+x\_n$. And in RHS we may remove all summands in which $\Theta$ has two equal arguments (say, if $c\_5-1=c\_4$, remove the fifth summand), because they are anyway equal to 0, as is seen from RHS of $(\heartsuit)$ (or from general properties of antisymmetric polynomials, if you prefer.) Now exactly the same recurrence is enjoyed by the number of paths (consider the last step). This proves the induction step.
So, your question reduces either to computing the determinants, or computing the number of such paths. In combinatorics, such paths are traditionally viewed as *skew standard Young tableaux*: consider the Young diagrams $\lambda$ with row lengths $c\_1\leqslant c\_2-1\leqslant c\_3-2\leqslant \ldots \leqslant c\_n-(n-1)$ and $\mu$ with row lengths $k\_1\leqslant k\_2-1\leqslant k\_3-2\leqslant \ldots \leqslant k\_n-(n-1)$. Then the path corresponds to getting $\mu$ from $\lambda$ be consecutively adding the boxes, with a condition that on each step we get a Young diagram again. If you write the numbers $1,2,\ldots,K$ consequently when adding the boxes, you get what is called a skew standard Young tableaux.
Now $(\heartsuit)$ reads as Feit's determinantal formula for the number of skew standard Young tableaux, case $m=1$, that is, $\lambda=\emptyset$, corresponds to some variant of a hook length formula, and the case $c\_j=\ell+j$ (up to permutation it is your second case) corresponds to counting skew standard Young tableaux of a shape which is a reverse Young diagram ($\mu$ is a rectangle), that is, again to hook length formula.
An interesting and rather enigmatic story is that there are other shapes for which the number of skew standard Young tableaux may be computed as a product formula. Some such cases were found in a series of papers by Igor Pak, Greta Panova and Alejandro Morales which you can find on Igor's [homepage](https://www.math.ucla.edu/%7Epak/papers/research.htm#s).
|
4
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https://mathoverflow.net/users/4312
|
407039
| 166,797 |
https://mathoverflow.net/questions/406941
|
11
|
Have there been any successful mathematicians that also happen to be mathematical fictionalists? Let's say success is defined by at least one article published in a non-pay journal.
I ask because this seems like a very extreme position for a working mathematician to have. Also, fictionalism is a very recent position.
I think it would be interesting to hear their point of view, if they exist.
|
https://mathoverflow.net/users/38066
|
Mathematical fictionalism
|
As I suggested in response to a [related MO question](https://mathoverflow.net/q/298749), one difficulty with answering this type of question is that most mathematicians outside of logic and set theory lack well-developed "positions" on the types of questions that occupy much of the attention of philosophers of mathematics.
As a preamble, let me ask this: For a mathematician X to be a fictionalist, is it necessary for X to know the meaning of the word "fictionalist" in, say, Hartry Field's sense?
Maybe the answer is no; maybe X just needs to espouse certain beliefs about mathematics to be a fictionalist, just like M. Jourdain spoke prose all his life without knowing it. But fictionalism is far more specific than prose, and it seems unlikely that X's beliefs would line up neatly with fictionalism unless X had studied fictionalism explicitly. More likely, X's beliefs would agree with fictionalism in some ways and would disagree in other ways. But if you insist that X know what the word "fictionalist" means, then you narrow the pool of candidates hugely. In any case, the only plausible way to find out is to conduct a formal survey. You might have to do this yourself if the surveys you have encountered don't already answer your question.
---
Having said that, I have noticed some aspects of fictionalism being espoused implicitly or explicitly by some mathematicians, but I have also noticed other aspects that seem to be almost universally rejected.
For MO readers who haven't heard of fictionalism, here's a caricature. Hartry Field draws an analogy with Oliver Twist. Did Oliver Twist travel to London? Answer: Yes, *according to a certain story*, but no, not *literally*, since Oliver Twist did not really exist. Analogously, mathematics, if we take its discourse at face value, makes assertions about abstract objects. But abstract objects are not real (Field is a *nominalist*), so mathematical theorems are true only *according to a certain story*. But wait, you say, isn't mathematics essential for doing science, and science surely deals with the real world? Field's response is to try to develop "science without numbers" by re-developing the scientifically applicable parts of mathematics, not on the basis of *abstract* objects, but on the basis of *concrete* objects, such as "regions of space."
If we accept this caricature, then I have certainly encountered mathematicians who, in one way or another, reject the "reality" of certain mathematical objects. Most commonly, I find this happening with regard to infinite set theory. You'll probably be able to find plenty of readers right here on MO who would agree with something like this: "Does the cardinality of the natural numbers differ from the cardinality of the real numbers? Yes, *according to a certain story*; but no, not *literally*, because infinite sets aren't real." But said readers are more likely to call themselves *formalists* than *fictionalists*, if they admit to being any kind of -ists at all.
On the other hand, the nominalist preoccupation with abstract versus concrete isn't something that you'll find resonating with many mathematicians. My reaction to Field's "science without numbers" is that his allegedly "concrete" objects seem just as abstract as standard mathematical objects. I think that this reaction is typical among mathematicians. Even the aforementioned "formalists" will generally agree, if you put the question to them, that symbols and finite strings are abstractions, while at the same time are "real" in a sense that (say) infinite sets are not. Replacing an abstract theory of numbers with an abstract theory of regions of space strikes mathematicians as being a pointless exercise.
|
26
|
https://mathoverflow.net/users/3106
|
407040
| 166,798 |
https://mathoverflow.net/questions/406511
|
4
|
I have been doing research on the Niemeier lattices with root systems of type, $A\_{k}^n$ and I am particularly interested in the finite groups permuting the constituent root systems. These groups seemed random at first, and they appeared to have a connection to certain projective linear groups over Galois fields with a prime number of elements. These root systems correspond to the divisors of $24$. I looked into the structures of the groups that permute the separate root systems and I was intrigued:
We will use the notation $K\_{n}$ for the group that induces permutations on the corresponding Niemeier lattice $A\_{k}^n$ (where $n·k = 24$).
Cases $K\_{1}$ and $K\_{2}$
---------------------------
$K\_{1}$ can only be the trivial group as there is only $1$ vertex to act on and the only group that acts faithfully over any singleton set is trivial.
$K\_{2}$ acts faithfully on $2$ vertices, corresponding to the exchange between the two root systems.
Case $K\_{3}$
-------------
This group is isomorphic to the symmetric group of degree $3$ (order $3!$ or $6$). It also happens(?) to be isomorphic to the projective special linear group, $L\_{2}(2)$
Case $K\_{4}$
-------------
This group is isomorphic to the alternating group of degree $4$ (order $4!/2$ or $12$). This one is also isomorphic to another linear group, namely $L\_{2}(3)$.
Case $K\_{6}$
-------------
This group is isomorphic to the symmetric group of degree $5$ (order $5!$ or $120$), but the striking thing about this is that it’s acting primitively on $6$ vertices, not $5$ vertices. This group is related to the projective special linear group $L\_{2}(5)$. It has one representation that acts primitively on the $6$ vertices of $PG\_{1}(5)$, and another representation on $5$ vertices, (the alternating group, $A\_{5}$), and both are contained within the symmetric group over $6$ vertices, $S\_{6}$.
Case $K\_{8}$
-------------
This group is isomorphic to $AGL\_{3}(2)$ an affine group of rank $3$, over the Galois field of $2$ elements (order $8·7·6·4$ or $1344$). like all the other $K\_{n}$ This group acts primitively on $8$ vertices. This one contains two distinct representations of the projective special linear group, $L\_{2}(7)$ one of which is the action of the group $L\_{2}(7)$ primitive over the $8$ vertices of $PG\_{1}(7)$ the other is a group isomorphic to the first, another linear group $L\_{3}(2)$. this one acts primitively over the $7$ vertices of $PG\_{2}(2)$
Case $K\_{12}$
--------------
This group is isomorphic to one of the sporadic groups, namely $M\_{12}$ the Mathieu group of degree $12$ (order $8·9·10·11·12$ or $95040$). this group contains two separate representations of the linear group $L\_{2}(11)$, but unlike the previous two cases, only one of these groups is maximal within $K\_{12}$ (namely the group $L\_{2}(11)$ and its primitive action over the $12$ vertices of $PG\_{1}(11)$). The other one is more subtle, as it’s not maximal in $M\_{12}$, however it *is* maximal in another sporadic group $M\_{11}$ (the stabiliser of either a point or a total within $M\_{12}$). This representation of $L\_{2}(11)$ acts primitively over $11$ vertices.
Case $K\_{24}$
--------------
This group is the most notable one, as it contains all the groups mentioned previously. It is isomorphic to a very interesting sporadic group called $M\_{24}$ the Mathieu group of degree $24$ (order $3·16·20·21·22·23·24$ or $244823040$). It contains a linear subgroup of type $L\_{2}(23)$, but unlike the previous cases there is only a single instance of this group, and there is no additional representation over $n - 1$ vertices. There is a lot of structure happening here, and frankly I haven’t enough space for that.
A recapitulation of the groups $K\_{n}$
---------------------------------------
**Orders:**
$|K\_{1}| = 1 $
$|K\_{2}| = 2 $
$|K\_{3}| = 6 $
$|K\_{4}| = 12 $
$|K\_{6}| = 120 $
$|K\_{8}| = 1344 $
$|K\_{12}| = 95040 $
$|K\_{24}| = 244823040 $
**Other facts:**
$K\_{3} \cong S\_{3} \cong L\_{2}(2) $
$K\_{4} \cong A\_{4} \cong L\_{2}(3) $
$K\_{6} \cong S\_{5} \supset L\_{2}(5) \cong A\_{5} $
$K\_{8} \cong AGL\_{3}(2) \supset (L\_{2}(7) \cong L\_{3}(2)) $
$K\_{12} \cong M\_{12} \supset M\_{11} \supset L\_{2}(11) $
$K\_{12} \cong M\_{12} \supset L\_{2}(11) $
$K\_{24} \cong M\_{24} \supset L\_{2}(23) $
$M\_{24} \supset \text{ (all previous ones)} $
I have only two questions:
1. Why do these $K\_{n}$ have their respective structures instead of just $S\_{n}$? What do they have in common?
2. What sort of structure must these $K\_{n}$ maintain under their respective group actions?
I will also remind us of what we defined the groups $K\_{n}$ to be: the group of permutations of the separate root systems of the Niemeier lattices of type $A\_{k}^n$ (with $n·k = 24$)
Edit: I forgot to mention: As of writing this, I have no proper experience in Niemeier lattices. I’d prefer it if the explanation were limited to group theory and a basic understanding of lattice theory.
Update: I had just found something interesting. These $8$ groups seem closely related to certain Umbral Groups (particularly, the ones of lambdacy $2$, $3$, $4$, $5$, $7$, $9$, $13$, and $25$) [7th November 2021, 21:05]
Proof of the Umbral Moonshine Conjecture:
<https://arxiv.org/pdf/1503.01472.pdf>
Umbral Moonshine and the Neiemeir Lattices:
<https://arxiv.org/pdf/1307.5793.pdf>
|
https://mathoverflow.net/users/178390
|
Structure of the permutation groups acting on the root systems of Niemeier lattices of type $A_{k}^n$
|
The information required for answering this question is contained in [1]. There it is scattered over many chapters; so I will give a summary here.
Let $N$ be a Niemeier lattice, $R$ be the sublattice of $N$ generated by
its roots and and $R^\*$ be the dual lattice of $R$. Then $G = R^\*/R$ is
an Abelian group, the *glue group*. The elements of $G$
are called the *glue vectors*
of $G$. Since $N$ is unimodular, the subgroup $H = N/R$ of $G$ has
order $|G|^{1/2}$ and also index $|G|^{1/2}$ in $G$. See [1],
Chapter 4.3 for an introduction to gluing theory.
If $R = R\_1 \oplus \ldots \oplus R\_n$ then $G$ has structure
$G= G\_1 \oplus \ldots \oplus G\_n$ with $G\_i = R\_i^\* / R\_i$.
For each glue vector $v \in G$ let $l(v)$ be the norm of its shortest
representative in $R^\*$. Then we have a decomposition
$v = v\_1 + \ldots + v\_n$ with $v\_i \in G\_i$; and we have
$l(v) = l(v\_1) + \ldots l(v\_n)$, where $l(v\_i)$ is the norm of the
shortest representative of $v\_i$ in $R\_i^\*$.
The glue groups of the root lattices and their length functions $l$
are known; for the case $A\_n$ see [1], Chapter 4.6. The glue group
of $A\_n$ is cyclic of order $n+1$. If $g$ is a generator of that
group then we have $l(k\cdot g) = \frac{k \cdot (n+1-k)}{n+1}$
for $0 \leq k \leq n$.
Since $N$ is an even lattice, $l(v)$ must be an even integer for
all $v \in H$. All roots of $N$ lie in $G$; so $l(v) > 2$ must hold
for all nonzero glue vectors $v\in H$.
The conditions listed above impose some rather severe restrictions
on the subgroup $H$ of $G$, which are discussed in [1], Chapter 18.4.
Obviously, a permutation of the root lattices preserving a Niemeier
lattice must also preserve the subgroup $H$ and the norm function $l$
on $H$.
Thus the subgroup $H$ of $G = R^\* / R$ and the norm function $l$ on
$H$ is the structure preserved by the permutation groups $K\_n$ of
the Niemeier lattices $A\_k^{24/k}$.
An answer to the question why a certain group $K\_n$ has a certain
structure is to some extent opinion based. For the *large* cases
$K\_{24/(n-1)}, n = 2,3,$ the Abelian group $G$ happens to be a vector space
over $\mbox{GF}\_n$, and the norm function turns out to be proportional to
the Hamming weight of a vector. So we have to look for linear codes over
$\mbox{GF}\_n^k$, where $(n-1)\cdot k = 24$, with Hamming distance
at least 8 in case $n=2$ and 6 in case $n=3$.
Such codes are known as Golay codes and also discussed in [1].
For the *small* cases $K\_k, k \leq 8$ we have to look for
permutation groups on $k$ letters. Here the linear groups
$\mbox{SL}\_2(k-1)$ have such a permutation representation.
[1] Conway, Sloane, Sphere packings, Lattices and Groups.
|
2
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https://mathoverflow.net/users/105705
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407047
| 166,799 |
https://mathoverflow.net/questions/407046
|
5
|
Fix a (countable) set $\mathcal{P}$ of atomic propositional variables. Recall a *Kripke model $\mathcal{K}$ for intuitionistic propositional logic (IPL)* consists of:
* A preorder $(W,\leq)$
* For each $w \in W$, a (classical) valuation $\varphi\_w\colon \mathcal{P} \to 2$
such that for all $w \leq v$ and $x \in \mathcal{P}$, having $\varphi\_w(x) = 1$ implies $\varphi\_v(x)=1$.
We can extend the valuations $\varphi\_w$ to a forcing relation $w \Vdash F$ between states $w \in W$ and arbitrary formulae $F$, using the schema [here](https://en.wikipedia.org/wiki/Kripke_semantics#Semantics_of_intuitionistic_logic). We then say $W \Vdash F$ if $w \Vdash F$ for all $w \in W$.
These semantics are sound and complete for IPL, so we can show a formula $F$ is not a tautology of IPL by exhibiting a Kripke model where it doesn't hold. For example, let $\mathcal{P} = \{ P \}$ and let $\mathcal{K}$ be:
$$
(P = \mathsf{true}) \\
\uparrow \\
(P = \mathsf{false})
$$
Then $\mathcal{K} \nVdash P \lor \neg P$ and $\mathcal{K} \nVdash \lnot\lnot P \to P$, showing both are not theorems of IPL.
---
I find it interesting that perhaps the two most famous classical tautologies which fail in IPL ($P \lor \neg P$ and $\lnot\lnot P \to P$) are both disprovable in such a small Kripke model. This leads to the following definition.
Given a formula $F$ which is **not** a theorem of IPL, let the *Kripke rank* of $F$, $\mathrm{krk}(F)$ be the size of the smallest Kripke model $\mathcal{K} \nVdash F$. Some simple observations:
* $\mathrm{krk}(F) = 1$ iff $F$ is not a theorem of classical logic.
* $\mathrm{krk}(P \lor \neg P) = \mathrm{krk}(\lnot\lnot P \to P) = 2$.
My (very broad) question is:
**What are the possible values for $\mathrm{krk}(F)$?**
I know a complete answer to this question is probably too much to expect, but here are some particularly relevant subquestions:
* Are finite values $> 2$ possible?
* Are arbitrarily large finite values possible?
* Is every finite value possible?
* Is there any possible infinite value (e.g. $\omega$)?
* What about any/arbitrarily large cardinal values?
---
**Aside:** you can also do this all for intuitionistic first-order logic (IFOL), by defining a Kripke model for IFOL as a functor from a preorder $(W,\leq)$ to the category of $\mathcal{L}$-structures and model-theoretic embeddings. I'm also interested in answers to the general question in this setting: is it any different from the IPL case?
|
https://mathoverflow.net/users/136473
|
Possible values of "Kripke rank" for formulae in IPL
|
The finite model property of intuitionistic logic implies that every unprovable formula has finite rank. On the other hand, all positive integers are ranks of some formulas; there are many families of formulas one could use to show this, but for example, the formulas
$$\bigvee\_{i=0}^n\Bigl(\bigwedge\_{j<i}p\_j\to p\_i\Bigr)$$
have rank $n+1$ (any countermodel has size at least $n+1$ as nodes witnessing the failure of each of the disjuncts have to be pairwise distinct).
For first-order logic, infinite rank is also possible (e.g., take the double-negation shift formula $\forall x\,\neg\neg P(x)\to\neg\neg\forall x\,P(x)$). Uncountable ranks are impossible, as any unprovable formula has a countable countermodel.
|
8
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https://mathoverflow.net/users/12705
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407056
| 166,802 |
https://mathoverflow.net/questions/407062
|
3
|
Let $X$ be a Polish space and let $(\mu\_i)\_{i=1}^{\infty}$ be a sequence of probability measures in the Wasserstein space $\mathcal{P}(X)$ on $X$. Let $(\beta\_i)\_{i=1}^{\infty}$ be a summable sequence in $(0,\infty)$. For every positive integer $k$, define the probability measures
$$
\nu\_k = (\sum\_{1\leq i\leq k}\beta\_i)^{-1} \sum\_{1\leq i\leq k} \beta\_i \mu\_i
$$
and define the probability measure
$$
\nu\_{\infty} = (\sum\_{i=1}^{\infty}\beta\_i)^{-1} \sum\_{i=1}^{\infty} \beta\_i \mu\_i.
$$
Is it true/clear that
$
\lim\limits\_{k\to\infty} \mathcal{W}(\nu\_k,\nu\_{\infty}) = 0,
$ quantitatively (here $\mathcal{W}$ denotes the Wasserstein distance on $\mathcal{P}(X)$).
|
https://mathoverflow.net/users/36886
|
Wasserstein convergence of "series expansion'' of probability measure
|
It is true and clear if the metric space $X$ has a finite diameter, but false in general: Take $\beta\_i=2^{-i}$ and $\mu\_i$ the point mass at $3^i$.
Details: In the case $D=$diam$(X)<\infty$, write $s\_k=\sum\_{i \le k} \beta\_i$
and $s=\sum\_{i<\infty} \beta\_i$. Then $$\nu\_\infty=\frac{s\_k}{s} \nu\_k+\frac{s-s\_k}{s} \gamma\_k$$
for some probability measure $$\gamma\_k=\frac{\sum\_{i > k} \beta\_i \mu\_i}{s-s\_k} \,.$$
Therefore $$\nu-\nu\_k=\frac{s-s\_k}{s} (\gamma\_k-\nu\_k)$$
whence $$\mathcal{W}(\nu\_k,\nu\_{\infty}) \le \frac{s-s\_k}{s} D$$
because the diameter of the Wasserstein metric on $\mathcal{P}(X)$ is $D$.
|
6
|
https://mathoverflow.net/users/7691
|
407063
| 166,804 |
https://mathoverflow.net/questions/407065
|
1
|
Following
[About uniform continuity](https://mathoverflow.net/questions/402661/about-uniform-continuity)
Let $E$ be a topological space, for all $a \in E$, we associate an open set of $E$, $U(a)$ containing $a$.
We will say that $\{U (a), a \in E\}$ is a uniform covering of $E$ if
For any dense $B$ in $E$ the covering $\{U (b), b \in B\}$ covers $E$.
Let $(E, d) $ and $(E', d') $ 2 metrics spaces, and $f$ continuous function of $E$ to $E'$.
Is it true that :
$f$ uniformly continuous
iff
for any uniform cover of $E'$ $U$, for all $b$ in $E$, $V (b) = f^{-1}(U (f (b))$ $V$ is a uniform cover of $E$?
|
https://mathoverflow.net/users/110301
|
Uniform covering and uniform continuity
|
Let $(E',d')$ be the Euclidean plane $\mathbb{R}^2$ with the usual metric.
Let $(E,d)$ be the interval $[-1,1]$ with the usual metric.
Let $f$ be the inclusion map of $[-1,1]$ to the interval $[-1,1]\times \{0\}$; it is an isometry to its image and is uniformly continuous.
Consider the following cover of $E'$:
$$ U(x) = \begin{cases}
B\_1(0) & x\in (-1,1)\times \{0\} \\
E' & \text{otherwise} \end{cases}. $$
Notice that any dense subset $B'\subseteq E'$ must contain at least one point that is not on $(-1,1)\times \{0\}$ and hence $U$ is a uniform cover by your definition.
But let $B\subsetneq E$ be the set of all irrational numbers in that interval. If $x\in B$ then $U(f(x)) = B\_1(0)$ and $f^{-1}(U(f(x)) = (-1,1)$, and hence we see that $\{V(b): b\in B\} = \{(-1,1)\}$ and does not cover $E$.
|
1
|
https://mathoverflow.net/users/3948
|
407074
| 166,808 |
https://mathoverflow.net/questions/407057
|
7
|
Let $k$ be a field, $X$ an algebraic variety, and $G$ a smooth algebraic group, acting on $X$ via $(g,x)\mapsto g\cdot x$.
Fixing $x$ in $X$ a $k$-point, there is a map $f\_x:G\rightarrow X$ sending $g\mapsto g\cdot x$.
Now, assuming that $df\_x:T\_eG\rightarrow T\_xX$ is surjective, how can one show that the orbit $G\cdot x$ is open in $X$?
NB: I expect that some restrictions on the field are necessary, but I'd like to keep them as minimal as possible!
|
https://mathoverflow.net/users/166993
|
Open orbits under the action of an algebraic group
|
$\DeclareMathOperator\Im{Im}$I don't think that smoothness of $X$ is necessary, just the fact that it is geometrically integral (I assume your varieties are geometrically integral). (As noted by Jason Starr, smoothness of $G$ is necessary.)
Here is a possible argument, please let me know if there are any mistakes.
As [noted](https://mathoverflow.net/a/407070) by AlexIvanov, it suffices to prove openness after passing to the algebraic closure of $k$, so we can assume that $k$ is algebraically closed.
Let $Z$ be the scheme theoretic closure of $G$ in $X$. The set theoretic image $\Im(G)$ of $G$ is open inside $Z$ by Chevalley's theorem and translating using the group action. After replacing $X$ with the $G$-stable open subvariety $X \setminus (Z \setminus \Im(G))$, we can assume that the natural map $G \to Z$ is surjective, in other words the orbit is closed. What we need to show at this point is that $Z = X$. We will argue by contradiction. Assume $Z\subset X$ is a proper closed subset.
We know that the tangent space of $Z$ at the base point $x$ agrees with that of $X$ (at $x$) by assumption. Using the fact that $k$ is algebraically closed, we can translate to see that the tangent space of $Z$ at any of its closed points coincides with that of $X$. But note that since $G$ is reduced (because $G$ is smooth), $Z$ is reduced. Since $k$ is algebraically closed, $Z$ is smooth over an open subset. In other words, the dimension of the tangent space of $Z$ for all points in an open dense agrees with the Krull dimension of $Z$. If $Z \subset X$ is a proper closed subset, the dimension of $Z$ is strictly smaller than that of $X$. This is a contradiction, because the dimension of the tangent space of $X$ at any closed point is at least the Krull dimension of $X$.
|
5
|
https://mathoverflow.net/users/339730
|
407081
| 166,810 |
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