parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/410044
|
8
|
I am wondering if there is any literature on general formula of the Moebius function of subgroup lattices of any finite abelian group $G$? What I know is
When $G$ is cyclic, the Moebius function is simply the classical number theoretic one.
When $G=(\mathbb Z/p\mathbb Z)^r$, the formula involves the number of $k$-dimensional linear subspace of $G$.
But is there some formula for any finite abelian group?
|
https://mathoverflow.net/users/175127
|
Moebius function of finite abelian groups
|
Since every interval of the subgroup lattice $\mathcal{L}(G)$ of a
finite abelian group $G$ is isomorphic to the subgroup lattice of some
finite abelian group, we can restrict ourselves to
$\mu(\hat{0},\hat{1})$, where $\hat{0}$ is the bottom element (the
trivial subgroup of $G$) and $\hat{1}$ is the top element (the group
$G$ itself) of $\mathcal{L}(G)$. If $G$ and $H$ have relatively prime
orders, then $\mathcal{L}(G\times H)= \mathcal{L}(G) \times
\mathcal{L}(H)$. Thus (EC1, second ed., Proposition 3.8.2)
$\mu\_{G\times H}(\hat{0},\hat{1}) =
\mu\_G(\hat{0},\hat{1})\mu\_H(\hat{0},\hat{1})$, so we can assume that $G$
has prime power order $p^n$. Then $\mathcal{L}(G)$ is atomic (i.e.,
$\hat{1}$ is a join of atoms, or $G$ is generated by the subgroups of
order $p$) if and only if $G$ is elementary abelian (a product of
groups of order $p$). Since for any finite lattice $L$,
$\mu(\hat{0},\hat{1})\neq 0$ implies that $L$ is atomic (EC1,
Corollary 3.9.5), we have $\mu(\hat{0},\hat{1})=0$ unless $G$ is
elementary abelian. Finally, if $G$ is elementary abelian of order
$p^n$, then it is well-known (e.g., EC1, equation (3.34)) that
$\mu(\hat{0},\hat{1})=(-1)^n p^{{n\choose 2}}$.
|
20
|
https://mathoverflow.net/users/2807
|
410062
| 167,799 |
https://mathoverflow.net/questions/409898
|
6
|
Apologies if this is not quite at the level of MathOverflow, but it has already been asked at [MSE](https://math.stackexchange.com/questions/1400501/question-about-hartshornes-proof-of-halphens-theorem) and gone unresolved for several years despite a bounty.
Hartshorne states the theorem as follows:
>
> **Proposition IV.6.1**. A curve $X$ of genus $g\geq 2$ has a nonspecial very ample divisor $D$ of degree $d$ if and only if $d\geq g+3$.
>
>
>
The necessity is shown, and then sufficiency. The idea is to show that the set $S$ of divisors $D \in X^d$ such that there exists $D' \sim D$ and points $P,Q \in X$ with $E = D'-P-Q$ an effective special divisor has dimension $\leq g+2$. Because $d\geq g+3$ this means there is some $D\notin S$ that is nonspecial and very ample of degree $d$.
Hartshorne shows that the set of divisors of the form $E+P+Q$ in $X^d$ that are nonspecial with $E$ a special effective divisor has dimension $\leq g+1$.
The part that confuses me comes afterwards. Namely, as $E$ is special the Riemann-Roch tells us that $\dim |E| \geq d-1-g$, and similarly that $\dim |E+Q+P| = d-g$. Because the difference between these two dimensions is at most 1, this somehow implies that the set of divisors $S$ as above has dimension $\leq g+2$.
I don't understand this implication. We have divisors of the form $E+Q+P$, each of which gives a linear system of dimension $d-g$, and all of which form a set of divisors of dimension $\leq g+1$. How does the difference in dimension of $E$ and $E+P+Q$ tell us the dimension of $S$?
|
https://mathoverflow.net/users/169571
|
Hartshorne's proof of Halphen's theorem
|
I think this Hartshorne's explanation is very rough. By using Picard variety, this step can be proved more clearly as follows.
Write $\mathrm{Pic}^d(X)$ for the scheme which parametrized all line bundles of degree $d$ on $X$, $\mathrm{Div}^d(X)$ for the scheme which parametrized all effective divisors of degree $d$ on $X$, and $D\_{\mathrm{univ}}\subset X\times \mathrm{Div}^d(X)$ for the universal effective divisor of degree $d$.
Then, by the universality of $\mathrm{Pic}^d(X)$, the line bundle $\mathcal{O}\_{X\times \mathrm{Div}^d(X)}(D\_{\mathrm{univ}})$ induces a morphism
$$
\varphi\_d:\mathrm{Div}^d(X) \to \mathrm{Pic}^d(X).
$$
This morphism can be written as $\varphi\_d(D) = \mathcal{O}\_X(D)$.
Hence each fiber of $\varphi\_d$ is linearly equivalent class.
In particular, for any $L\in \mathrm{Pic}^d(X)$, it holds that $\dim(\varphi\_d^{-1}(L)) = \dim H^0(X,L) - 1$.
Write $\mathrm{SpDiv}^d\subset \mathrm{Div}^d(X)$ for the closed subscheme determied by all special effective divisors.
Then, by Riemann-Roch, for any $D\in \mathrm{SpDiv}^d$, it holds that
$$
\dim(\varphi\_d^{-1}(\mathcal{O}\_X(D))) = 1+d-g+l(K-D)-1 \geq 1+d-g.
$$
Since $\dim(\mathrm{SpDiv}^d) = g-1$, it holds that
$$
\dim(\varphi\_d(\mathrm{SpDiv}^d)) \leq (g-1)-(1+d-g) = 2g-2-d.
$$
Now, let us consider the following two morphisms:
\begin{align\*}
f:X\times X\times \mathrm{Div}^{d-2}(X)\to \mathrm{Div}^d(X), (P,Q,D)\mapsto P+Q+D, \\
g:X\times X \times \mathrm{Pic}^{d-2}(X) \to \mathrm{Pic}^d(X), (P,Q,L)\mapsto L\otimes \mathcal{O}(P+Q).
\end{align\*}
Then we obtain the following commutative diagram:
$$
\require{AMScd}
\begin{CD}
X\times X\times \mathrm{SpDiv}^{d-2} @>{\subset}>> X\times X\times \mathrm{Div}^{d-2}(X) @>{f}>> \mathrm{Div}^d(X) \\
@. @V{\psi}VV @VV{\varphi\_d}V \\
@. X\times X\times \mathrm{Pic}^{d-2}(X) @>{g}>> \mathrm{Pic}^d(X),
\end{CD}
$$
where $\psi := \mathrm{id}\_X\times \mathrm{id}\_X \times \varphi\_{d-2}$.
Write $T:= g(\psi(X\times X\times \mathrm{SpDiv}^{d-2}))\subset \mathrm{Pic}^d(X)$.
Then
$$\dim(T) \leq 2g-2-(d-2)+2 = 2g-d+2.$$
Since $\dim(\mathrm{Div}^d)) = d$, and $\dim(\mathrm{Pic}^d) = g$,
the dimension of general fibers of $\varphi\_d$ are $d-g$.
Hence
$$\dim(\varphi\_d^{-1}(T)) \leq (2g-d+2)+(d-g) = g+2,$$
(where we note that if $T \subset \varphi\_d(\mathrm{SpDiv}^d)$, then the dimension of general fibers of $\varphi\_d^{-1}(T) \to T$ are $>d-g$, however, in this case, since $\varphi\_d^{-1}(T)\subset \mathrm{SpDiv}^d$, it holds that $\dim(\varphi\_d^{-1}(T)) \leq \dim(\mathrm{SpDiv}^d) = g-1 < g+2$).
Moreover, by our construction, the scheme $\varphi\_d^{-1}(T)$ parametrizes all effective divisors $D\subset X$ which are linearly equivalent to $E+P+Q$, the sum of a special effective divisor $E\subset X$ and points $P,Q\in X$.
This is desired conclusion.
|
3
|
https://mathoverflow.net/users/442153
|
410071
| 167,800 |
https://mathoverflow.net/questions/410032
|
4
|
Who was the first to consider that categories were semi-simplicial sets (and in particular groupoids were simplicial sets)?
I think there was a concept of nerve of a covering in algebraic topology before (maybe Alexandroff).
|
https://mathoverflow.net/users/429204
|
Who introduced nerves in category theory?
|
In Peter Johnstone's 1977 "Topos theory" (p.48) the simplicial description of categories is attributed to Grothendieck and he cites the "Technique de la descente"-series of Bourbaki seminars 1959-62 for it. I guess what he has in mind is in particular prop.4.1 on page 108 of the third installment [Préschémas quotients](http://archive.numdam.org/article/SB_1960-1961__6__99_0.pdf) from 1961.
|
11
|
https://mathoverflow.net/users/470544
|
410076
| 167,803 |
https://mathoverflow.net/questions/409912
|
15
|
Let $S$ be a closed orientable surface of genus at least $2$. I'm interested in the connected components of $\operatorname{Hom}(\pi\_1(S),\operatorname{SL}\_n(\mathbb{R}))$ for $n$ at least $3$.
I know that for $n$ odd there is only $3$ connected components. My first question is:
1. For $n$ even, do we still have $3$ connected components?
Suppose $n$ is odd. There is the connected component containing the trivial representation, the Hitchin component and another one. My second question is:
2. What do we know about the connected component which doen't contains the trivial or Hitchin representations? Can we construct an explicit representation in it? Can any representation in it be deformed to have image in a compact group?
Thanks
|
https://mathoverflow.net/users/197544
|
What do the components of $\operatorname{Hom}(\pi_1(S),\operatorname{SL}_n(\mathbb{R}))$ look like?
|
$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\R{\mathbb{R}}$Just to set the notation, as in the OP's question, let $S$ be a closed orientable surface of genus $g$ at least 2, and $G$ is a Lie group. The question concerns connected components of $\Hom(\pi\_1(S),G)$ with respect to the subspace topology where we identify $\Hom(\pi\_1(S),G)$ with a subset of $G^{2g}$ by evaluation. The question in particular concerns the case $G=\SL(n,\mathbb{R})$, however this case is very much related to three other cases, so I will summarize all four cases. We assume $n\geq 2$ everywhere below.
The first paper to mention is:
Goldman, W., *Topological components of spaces of representations.*
Invent. Math. 93 (1988), no. 3, 557–607.
In that paper, among other foundational results, it is shown that for $G=\SL(2,\mathbb{R})$ there are $2^{2g+1}+2g-3$ components, and for $G=\PSL(2,\mathbb{R})$ there are $4g-3$ components.
Then next foundational paper to mention (as already noted by others) is:
Hitchin, N., *Lie groups and Teichmüller space*. Topology 31 (1992), no. 3, 449–473.
In that paper, for $G=\PSL(n,\mathbb{R})$ and $n\geq 3$, there are 3 components if $n$ is odd, and 6 if $n$ is even.
Remark: Since $\PSL(n,\mathbb{R})=\SL(n,\mathbb{R})$ when $n$ is odd, this also gives 3 components for $G=\SL(n,\mathbb{R})$ for all odd $n$.
Next, let's mention the paper in Eugene Xia:
Xia, E., *Components of $\Hom(\pi\_1,\PGL(2,\mathbb{R}))$*. Topology 36 (1997), no. 2, 481–499.
In that paper, it is shown there are $2^{2g+1}+4g-5$ components for $G=\PGL(2,\mathbb{R})$.
Generalizing the above result, we have:
Oliveira, A., *Representations of surface groups in the projective general linear group*. Internat. J. Math. 22 (2011), no. 2, 223–279.
In this paper, it is shown there are $2^{2g+1}+2$ components for $G=\PGL(n,\mathbb{R})$ when $n\geq 4$ and even (which completes the picture for this group given prior results listed above).
Lastly, we mention (out of order):
Bradlow, S.; García-Prada, O.; Gothen, P., *Representations of surface groups in the general linear group*. Proceedings of the XII Fall Workshop on Geometry and Physics, 83–94, Publ. R. Soc. Mat. Esp., 7, R. Soc. Mat. Esp., Madrid, 2004.
In this paper, it is shown there are $3(2^{2g})$ components when $G=\GL(n,\mathbb{R})$ and $n\geq 3$.
Finally to answer your question. As explained to me by a much more knowledgeable friend (André Oliveira), implicit in the latter two papers is that for $G=\SL(n,\mathbb{R})$ there are $2^{2g+1}+2$ components when $n\geq 4$ and even.
Here what I understood (any errors are my own):
There is only one topological type, determined by the second Stiefel-Whitney class (which takes 2 values). For one value the moduli space is connected. This component contains only representations homotopic to those which factor through $\SO(n)$. For the other value, there is one component like the previous one, but also $2(2^{2g})$ components coming from a choice of a square root of the canonical line bundle for each of the two Teichmüller components in the $\PSL(n,\R)$ case.
Remarks:
1. For the $\GL(2,\R)$ case see:
Baraglia, D.; Schaposnik, L., *Monodromy of rank 2 twisted Hitchin systems and real character varieties*. Trans. Amer. Math. Soc. 370 (2018), no. 8, 5491–5534.
2. For $g=1$ and any real reductive $G$, or $g\geq 2$ and $G$ compact or complex reductive, the number of components is equal to the order of $\pi\_1([G,G])$. For the $g=1$ case, see Proposition 5.1 (and the references in its proof) in Sikora, A., *Character varieties of abelian groups*. Math. Z. 277 (2014), no. 1-2, 241–256. For the $g\geq 2$ and $G$ complex reductive (with references for the compact case), see the Appendix of Lawton, S.; Ramras, D., *Covering spaces of character varieties*. With an appendix by N. Ho and C. Liu. New York J. Math. 21 (2015), 383–416.
3. For non-orientable $S$ and $G=\PSL(2,\R)$ or $\PGL(2,\R)$ see Palesi, F., *Connected components of spaces of representations of non-orientable surfaces*. Comm. Anal. Geom. 18 (2010), no. 1, 195–217. and also Palesi, F., *Connected components of PGL(2,R)-representation spaces of non-orientable surfaces*. Geometry, topology and dynamics of character varieties, 281–295, Lect. Notes Ser. Inst. Math. Sci. Natl. Univ. Singap., 23, World Sci. Publ., Hackensack, NJ, 2012.
|
5
|
https://mathoverflow.net/users/12218
|
410077
| 167,804 |
https://mathoverflow.net/questions/353599
|
7
|
Recently, I proved the following Lipschitz-continuity like result for convex polytopes:
>
> Let $A\in\mathbb R^{m\times n}$ and $b,b'\in\mathbb R^m$ be given such that $\{x\,:\,Ax\leq 0\}=\{0\}$ (which is equivalent to boundedness of all induced polytopes) and that $\{x\in\mathbb R^n\,:\,Ax\leq b\},\{x\in\mathbb R^n\,:\,Ax\leq b'\}$ are non-empty. Then
> $$
> \delta\big(\{x\in\mathbb R^n\,:\,Ax\leq b\},\{x\in\mathbb R^n\,:\,Ax\leq b'\}\big)\leq \Big( \max\_{A\_0\in\operatorname{GL}(n,\mathbb R),A\_0\subset A}\|A\_0^{-1}\| \Big)\|b-b'\|\_1
> $$
> where the operator norm $\|\cdot\|$ as well as the Hausdorff metric $\delta$ are taken with respect to $(\mathbb R^n,\|\cdot\|\_1)$. Also $A\_0\subset A$ is short for "every row of $A\_0$ is also a row of $A$" so the above maximum is taken over all invertible submatrices of $A$.
>
>
>
What this intuitively means is that if two polytopes have parallel faces (i.e. they are both described by the same $A$ matrix), but the *location* of these faces differs ($b\neq b'$), then the distance between the polytopes is upper bounded by the distance between the vectors $b,b'$ times a "geometrical" constant coming from $A$.
Hence the function $b\mapsto\{x\in\mathbb R^n:Ax\leq b\}$ (with suitable domain such that the co-domain equals all non-empty subsets of $\mathbb R^n$) is Lipschitz continuous with a constant determined by $A$.
This came up as a lemma to something only vaguely related which is why I don't have a problem with posting it publicly. Actually if this was a known result then my manuscript could be shortened by 3 pages. Thus my quesiton is:
>
> Is the above result known and, if so, where can in be found in the (convex polytope-)literature?
>
>
>
I would be surprised if nobody has thought about this until now. While I haven't seen this result in the books of Grünbaum and Schrijver or the few papers on convex polytopes I am aware of, this is not the field I usually work in; hence this might very well be known but beyond my mathematical horizon. Thanks in advance for any answer or comment!
|
https://mathoverflow.net/users/116991
|
Lipschitz-continuity of convex polytopes under the Hausdorff metric
|
This is a classic question in the literature on linear programming, since it is related to the stability of the feasible set (and hence the solutions) under perturbation of the parameters.
The classic work in this field is:
* A. J. Hoffman, Approximate solutions of systems of linear inequalities, *J. Res. Nat. Bur. Standards*, 1952.
A more recent work, which essentially states your result as a special case (see Theorem 2.4 and the Lipschitz constant on p.19), is:
* W. Li, The sharp Lifshitz constants for feasible and optimal solutions of a perturbed linear program, *Linear Algebra and Its Applications*, 1993.
|
3
|
https://mathoverflow.net/users/76565
|
410080
| 167,805 |
https://mathoverflow.net/questions/410115
|
5
|
**Problem**: Given three positive integers $0 < n\_1 < n\_2 < n\_3$. Is there always a real number $x$ such that
$$\cos n\_1 x + \cos n\_2 x + \cos n\_3 x < -2?$$
|
https://mathoverflow.net/users/141801
|
Is there always a real $x$ such that $\cos n_1 x + \cos n_2 x + \cos n_3 x < -2$?
|
The answer is negative. If $n\_3=n\_1+n\_2$, then
$$\cos n\_1 x + \cos n\_2 x + \cos n\_3 x\geq -2.$$
Indeed, the left-hand side equals
$$(1+\cos n\_1 x)(1+\cos n\_2 x)-1-\sin n\_1 x\sin n\_2 x,$$
where $1+\cos n\_j x\geq 0$, and $\sin n\_1 x\sin n\_2 x\leq 1$.
Furthermore, if $n\_2=2 n\_1$ and $n\_3=3 n\_1$, then
$$\min\_{x\in\mathbb{R}}\,(\cos n\_1 x + \cos n\_2 x + \cos n\_3 x)=-\frac{17+7\sqrt{7}}{27}\approx-1.31557.$$
More generally, the minimum of the [Dirichlet kernel](https://en.wikipedia.org/wiki/Dirichlet_kernel) is known to great precision, see e.g. [here](http://www.idmercer.com/dirichletkernel.pdf).
|
17
|
https://mathoverflow.net/users/11919
|
410117
| 167,819 |
https://mathoverflow.net/questions/410123
|
11
|
By a result of Godefroy and Kalton if $X,Y$ are separable Banach spaces and $X$ embeds isometrically into $Y$, then $X$ embeds with a linear isometry into $Y$.
Is this result known to fail for nonseparable spaces? That is, is there a known example of two (necessarily nonseparable) Banach spaces $X,Y$ such that $X$ embeds isometrically into $Y$, but such that there is no linear isometric embedding of $X$ into $Y$?
This question was previously asked on [MSE](https://math.stackexchange.com/questions/4192857/what-is-an-example-of-two-banach-spaces-x-y-such-that-x-embeds-isometrically) but received no answer there.
|
https://mathoverflow.net/users/470606
|
What is an example of two Banach spaces $X,Y$ such that $X$ embeds isometrically but not linearly into $Y$?
|
Yes, if $H$ is a nonseparable Hilbert space then it embeds isometrically into the Arens-Eells space ${\rm AE}(H)$, but not linearly isometrically, or even linearly homeomorphically. See Theorem 5.21 of my book *Lipschitz Algebras* (second edition).
As I explain in the notes to that chapter, a more general version of this statement was claimed in a paper of Godefroy and Kalton, but their proof is erroneous and, as far as I can tell, not fixable. However, some of the ideas of my argument are based on theirs.
|
13
|
https://mathoverflow.net/users/23141
|
410137
| 167,823 |
https://mathoverflow.net/questions/410135
|
0
|
This part is from page 107 in Michael E. Taylor's book Partial Differential Equations III.
In this part, we want a proof for the existence of smooth solution of the PDE
$\Delta u=f(x, u)$ on $U$ with boudary condition $\left.u\right|\_{\partial U}=g$ where $g$ is smooth
under the assumption that $\frac{\partial f}{\partial u} \geq 0$.
After making the temporary assumption that for $|u| \geq K, \partial\_{u} f(x, u)=0$ we have that
there exists soomth solution of the PDE
$\Delta u=f(x, u)$ on $U$ with boudary condition $\left.u\right|\_{\partial U}=g$ when $g$ is smooth.
we construct a sequence $f\_{j}(x, u)=f(x, u)$, for $|u| \leq j，f\_{j}(x, u)=f(x, u)$ and when $|u| \geq K\_{j}, \partial\_{u} f\_{j}(x, u)=0$
so we have solutions $u\_{j} \in C^{\infty}(\bar{U})$ to
$\Delta u\_{j}=f\_{j}\left(x, u\_{j}\right),\left.\quad u\_{j}\right|\_{\partial U}=g$
And we have $\sup \_{U}\left|u\_{j}\right| \leq \sup \_{U} 2|\Phi|$
where $f(x, 0)=\varphi(x) \in C^{\infty}(\bar{U})$, take $g \in C^{\infty}(\partial U)$, and let $\Phi \in C^{\infty}(\bar{U})$ be the solution to
$\Delta \Phi=\varphi$ on $U, \quad \Phi=g$ on $\partial U$
Then we can see that the sequence $\left(u\_{j}\right)$ stabilizes for large $j$, then we finish the proof.
How does this proof finish the proof, by Arzela-Ascoli? But this does not meet the conditions of Arzela-Ascoli theorem, actually I don't even know how to use Arzela-Ascoli here.
|
https://mathoverflow.net/users/469129
|
A proof for the existence of smooth solution of PDE in form $\Delta u=f(x, u)$ in Michael E. Taylor's book Partial Differential Equations III
|
It is much simpler than Arzela-Ascoli.
Since you have the uniform bound
$$ \sup\_U |u\_j| \leq \sup\_U 2|\Phi| $$
For simplicity I will assume your $K\_j = j$. Take $J = \sup\_U 2|\Phi|$, then for every $j,k \geq J$ you have that
$$ f\_j(x,u\_j) = f\_k(x,u\_j) = f(x,u\_j).$$
This tells you that for every $j,k \geq J$ you have
$$ \Delta u\_j = f\_k(x,u\_j). $$
By the uniqueness of the solution to the problem with the temporary truncation, you find therefore that $u\_k = u\_j$ for every $j,k\geq J$.
(This is what the author means by "stabilizes for large $j$"; that after some $J$ the sequence $\{u\_j\}$ becomes the constant sequence.)
|
4
|
https://mathoverflow.net/users/3948
|
410144
| 167,826 |
https://mathoverflow.net/questions/410089
|
5
|
I have been reading ['Improved bounds for the sunflower lemma'](https://annals.math.princeton.edu/2021/194-3/p05) by Alweiss, Lovett, Wu and Zhang (Ann. of Math., Vol. 194(3), 2021), and have some gaps in my understanding of the paper. They are as follows:
1. In Lemma: 2.8 (pg. 802), the authors bound the number of bad pairs $(W,S\_{i})$ by a product of four bounds obtained earlier- (i) the bound on the number of sets of the form $W \cup S\_{i}$, (ii) the number of sets of the form $A = S\_{i} \cap S\_{j}$ (once $S\_{j}$ is chosen and fixed in a certain manner), (iii) the number of sets in $\mathcal{F}^{'}$ that contain the set $A$, and (iv) the number of sets of the form $S\_{i} \cap W$. Each of the bounds in i)-iv) are clear to me, but I am unable to understand how the product is indeed a bound for the number of bad pairs, both heuristically and in terms of constructing an injection from the set of bad pairs into the set encoded as above.
2. In Lemma 2.10 (pg. 803), a new weighting $\sigma^{'}$ is defined on $\mathcal{F}^{'}$. Am I right in thinking that $\forall \, S \, \in F^{'}: \sigma^{'}(S) = 1$? In this case, the spreadness criteria seems to go through. If not, how is $\sigma^{'}$ defined explicitly?
3. On pg. 804, the authors construct a recursively-defined, strictly-decreasing sequence of integers, $(w\_{i})\_{i =1}^{r}$, with $w\_{i+1} = \lfloor (1 - \varepsilon)w\_{i} \rfloor + 1$ for some small, fixed $\varepsilon$, uniformly bounded below by $w^{\*}.$ It can be shown that $w^{\*} \geq \frac{1}{\varepsilon}$. The authors claim that $\exists K > 0: r \leq \frac{K \, \log(w)}{\varepsilon}.$ I am unable to prove this remark, and would prefer, if possible, a constructive proof.
My Attempt: Here are some observations, which don't seem to lead to a proof of what is required. We have that $w\_{i} \geq (1 - \varepsilon)^{i}w.$ Moreover, we have that $w\_{i} - w\_{i+1} \geq \varepsilon w\_{i} - 1,$ and hence, we also have that $$w\_{i} - w\_{i + 1} \geq \varepsilon w\_{r} - 1 \geq \varepsilon w^{\*} - 1.$$ Thus, the distance between each pair of successive terms is at least $\varepsilon w^{\*} - 1$, and so, assuming that this is positive, the number of integers in the sequence, i.e., $r$ is bounded above by $$\frac{w - w^{\*} + 1}{\varepsilon w^{\*} - 1}.$$ We also observe that $$1 \leq w\_{i} - w\_{i + 1} \leq \varepsilon w\_{i} \leq \varepsilon w\_{0} = \varepsilon w,$$ and hence, a lower bound for $r$ is $$\frac{w - w^{\*} + 1}{\varepsilon w}.$$
4. On pg. 808, the proof of Theorem: 4.2 concludes by stating that $\mathcal{F}$ cannot be $C \, \log(w)$ spread for large enough $C$. This apparently follows from the improved version of Theorem 2.5 from Rao's paper (also proved independently by Tao). Below is my argument- I am not certain that it is correct.
Any help on any or all of these questions is appreciated.
Edit: I would especially appreciate any insights regarding the first question, even if you don't have the time to answer the others.
|
https://mathoverflow.net/users/nan
|
Questions on 'Improved bounds for the sunflower lemma'
|
Let me try to answer your questions.
1. We can recover the pair $(W, S\_i)$ from the four quantities you mentioned. So the number of their combinations upper bounds the number of pairs $(W, S\_{i})$, which is why we multiply the number of options for each one.
2. Indeed, we switch from a set system with not necessarily uniform weights to a multi set system with uniform weights to simplify the proof.
3. You are asking for a constructive proof. Assume that $w\_{i} \geq \frac{2}{\varepsilon}$. Then $w\_{i+1} \leq (1 - \varepsilon) w\_{i} + 1 \leq (1-\frac{\varepsilon}{2}) w\_{i}$. So you reach $w\_{i} < \frac{2}{\varepsilon}$ after at most $O\left(\frac{\log w}{\varepsilon}\right)$ steps.
4. I don't understand your question. If you replace our Theorem 2.5 with Rao's result it removes all the $\log\log$ terms we have, which is the point we are making.
|
6
|
https://mathoverflow.net/users/20709
|
410160
| 167,833 |
https://mathoverflow.net/questions/410125
|
2
|
Say we have the line segment $L(t) = [0,t]$, and randomly remove open intervals of length $1$ from $L(t)$ until no more open intervals of length $1$ remain. Define $u(t)$ as the expected measure of what remains.
Note $u(t) = t$ for $0 \le t < 1$ and $u(t) = t-1$ for $1 \le t < 2$. In general, we have the integral equation $$u(t) = \frac{2}{t-1} \int\_0^{t-1} u(s)\ ds.$$ Rearranging and differentiating, we get the delay differential equation $$u'(t) = \frac{-1}{t-1} u(t) + \frac{2}{t-1} u(t-1).$$ Note the equilibrium solution is $u(t) = c(t+1)$ for some constant $c$.
Main question - what's the value of $c$? My guess is $c = \frac{1}{4}$, but I have no proof. More generally, what are the asymptotics?
One additional observation. Using the transformation $u(t) = \frac{v(t)}{t-1}$, we can write our delay differential equation in the simplified form $$v'(t) = \frac{2}{t-2} v(t-1).$$
Here's the motivation for this question. Say we have a very large auditorium, and pairs of people fill it up, who sit together in adjacent seats. When no more pairs can be seated, about what fraction of the seats remain unoccupied? The answer is about $\frac{1}{e^2}$. Now instead of pairs, we seat groups of $n$ people, all of whom sit adjacently. As $n$ gets very large, about what fraction of seats are expected to remain unoccupied when the auditorium is filled? The answer should be the same as for my question.
|
https://mathoverflow.net/users/795
|
Asymptotics of a delay differential equation
|
Let us show that
\begin{equation\*}
u(t)\ge c(1+t)\text{ for some real $c>1/4$ and all $t\ge4$. }\tag{1}
\end{equation\*}
Let
\begin{equation\*}
u\_n(z):=u(n+z),\quad w\_n(z):=u\_n(z)-\tfrac14\,(1+n+z);
\end{equation\*}
here and in what follows, $z\in[0,1]$ and $n=0,1,\dots$.
Then for $n\ge1$
\begin{equation\*}
w\_n(z)=\frac2{n-1+z}\,\Big(\sum\_{k=0}^{n-2}J\_k+\int\_0^z w\_{n-1}(x)\,dx\Big),
\end{equation\*}
where
\begin{equation\*}
J\_k:=\int\_0^1 w\_k(x)\,dx.
\end{equation\*}
Next, $J\_0=1/8=-J\_1$, and hence
\begin{equation\*}
w\_n(z)=\frac2{n-1+z}\,\Big(\sum\_{k=2}^{n-2}J\_k+\int\_0^z w\_{n-1}(x)\,dx\Big) \tag{2}
\end{equation\*}
for $n\ge4$. Moreover, $J\_2=0.01\ldots>0$ and
\begin{equation\*}
w\_3(x)=\frac{x (3 x-14)+16 \ln (x+1)}{4 (x+2)}>0
\end{equation\*}
for $x\in(0,1]$, so that $J\_3>0$. It follows now from (2) by induction that $w\_n>0$ on $(0,1]$ and $J\_n>0$ for all $n\ge3$.
Now it follows from (2) that
\begin{equation\*}
J\_n\ge\frac2n\,\sum\_{k=2}^{n-2}J\_k \tag{3}
\end{equation\*}
for all $n\ge4$. Recall that $J\_n>0$ for all $n\ge2$. So,
\begin{equation\*}
b:=\min\_{2\le k\le10}\frac{J\_k}{6+k}>0,
\end{equation\*}
and
\begin{equation\*}
J\_n\ge b(6+n) \tag{4}
\end{equation\*}
for $n=2,\dots,10$.
Also,
\begin{equation\*}
\frac2n\,\sum\_{k=2}^{n-2}(6+k)=9 - \frac{36}n + n\ge6+n
\end{equation\*}
for $n\ge12$. So, by induction, it follows from (3) that (4) holds for all $n\ge2$. So, in view of (2),
\begin{equation\*}
w\_n(z)\ge\frac2n\,\sum\_{k=2}^{n-2}J\_k
\ge b(6+n) \tag{5}
\end{equation\*}
for $n\ge4$. So,
\begin{equation\*}
u(n+z)=u\_n(z)=w\_n(z)+\tfrac14\,(1+n+z)\ge b(6+n)+\tfrac14\,(1+n+z)
\ge c(1+n+z)
\end{equation\*}
for $n\ge4$, where $c:=b+\tfrac14>\tfrac14$. Thus, (1) is proved.
---
It is much easier to show that
\begin{equation\*}
u(t)\le C(1+t)\text{ for some real $C>0$ and all $t\ge0$. }
\end{equation\*}
|
2
|
https://mathoverflow.net/users/36721
|
410162
| 167,834 |
https://mathoverflow.net/questions/410168
|
2
|
Lets consider two views of zeta functions of curves.
For the following, let $\mathbb{F}\_p$ be the field with $p$ elements where $p$ is prime, and let $\overline{\mathbb{F}\_p}$ be the algebraic closure of $\mathbb{F}\_p$. Let $X$ be a smooth curve over $\text{Spec}(\mathbb{F}\_p)$.
**View 1:** Using a Weil cohomology theory, obtain a chain complex $C^\*$. Applying the cohomology construction to the frobenius map $F : X \rightarrow X$, we get a map $f : C^\* \rightarrow C^\*$. Chain complexes do not have determinant in the ordinary sense, but there is a unique generalization, which is $f\_\* \mapsto \text{Det}(f\_\*) := \prod\_{i = 0}^n \text{det}(f\_n)^{(-1)^{i+1}}$ (this only works for certain complexes). We consider the **generalized characteristic polynomial**.
$$ \text{Det}(1 - t f\_\*) = \prod\_{i = 1}^n \text{det}(1 - ft)^{-1^{i+1}}$$
A Poincare duality for the étale cohomology leads quickly to a functional equation. Viewing the zeta function this way seems to accord well with the proofs of the weil conjectures, which are higher analogues of a simple theorem on finite dimensional vector spaces: the lefshetz fixed point theorem.
**View 2:** The global zeta function $Z(f, \psi)$ is
$$ \int\_{\mathbb{I}} f(x) \psi(x) d \mu^\times (x)$$
This view is accompanied by several variants, each expressing the L-function as the Fourier transform of some character.
In this view (the view of class field theory) there is a correspondence between characters on the absolute Galois group $\pi\_1(\mathbb{Q})$ and characters on $\mathbb{A}^\times / \mathbb{Q}^\times$. This view is explained [here](https://server.mcm.ac.cn/%7Ezheng/Lafforgue.pdf) and [here](https://mathoverflow.net/questions/7656/why-does-the-gamma-function-complete-the-riemann-zeta-function).
Both of these views have toy versions in finite dimensional vector spaces using sets instead of schemes. But it is not clear from the second view that zeta functions should arise as characteristic polynomials.
Can anyone give an explanation of why we should expect the (higher analogue of the) characteristic polynomial to be a character? Is there a toy-model where the two views can be unified here? I am searching for some relatively simple observations about the relationships between characteristic polynomials and characters to the effect that these views would seem like the higher analogue of a natural observation about char polys on finite dimensional vector spaces.
|
https://mathoverflow.net/users/30211
|
Connecting two pictures of the Zeta function
|
The quickest answer I can give is that the Lefschetz formula gives the identity $$\prod\_{i }\det (1 - ft, H^i(X))^{ (-1)^{i+1}} = \prod\_{v \in |X| } \frac{1}{1 - t^{\deg v}} $$
and then the product on the right side can be expanded out into a sum over the divisors $D$ on $X$ of the function $t^{\deg D}$ (i.e. a character), which can be further represented as an integral over the ideles by replacing each divisor with a corresponding set of ideles.
The first step, the product identity, can be proved as follows: Using the identity $(1- u)^{-1} = e^{ \sum\_{m=1}^{\infty} u^m /m } $ and the closely related $ \det (1-ft)^{-1} = e^{ \sum\_{n=1}^{\infty} \operatorname{tr}(f^n) t^n /n }$, it suffices to prove
$$\sum\_i (-1)^i \sum\_{n=1}^{\infty} \operatorname{tr}(f^n, H^i(X)) t^n/n = \sum\_{v \in |X|} \sum\_{m=1}^{\infty} t^{ m \deg v} /m$$
which, extracting the coefficients of $t^n$, follows from the Lefschetz formula by
$$ \sum\_i (-1)^i \operatorname{tr}(f^n, H^i(X)) = \# X^{ f^n} = \#X( \mathbb F\_{q^n}) = \sum\_{ v \in |X| , \deg v \mid n} \deg v$$ since each closed point of degree $\deg v$ corresponds to $\deg v$ points over $\mathbb F\_{q^n}$ whenever $n$ is a multiple of $v$.
The second step, $$\prod\_{v \in |X|} \frac{1}{ 1- t^{\deg v}} = \sum\_{D \geq 0} t^{ \deg D}$$ follows from the definition of effective divisors as formal sums of closed points.
The third step,
$$ \sum\_{D \geq 0} t^{ \deg D} = \int\_{ a \in \mathbb I\_F} 1\_{a \textrm{ integral}} t^{ \deg a} d\mu $$
follows since each idele has a divisor, the divisor is effective if and only if the idele is integral (locally at each place), the degree of that divisor equals the degree of the idele, and the measure of the adeles with a given divisor is independent of the divisor (since the measure is multiplicatively invariant.
---
But I'm pretty sure for the question that you don't just want this, the standard proof of the formula, but something more intuitive. Why should we expect something like this is the case?
I think that the right answer is not to seek some totally separate argument but to concentrate on each step of this and try to figure out why it is intuitive and natural from the right perspective. For example, in the case of finite schemes, the Lefschetz formula, involving only $H^0$, as trivial. We can thus think of this formula as the natural generalization of that simple identity to the higher-dimensional case.
But if you only want to know why the higher analogue of the characteristic polynomial is the sum of a fixed function against a varying character, there is another approach. I kind of think the best way to present it is to answer your question with a question, i.e.
>
> Why is the sum over all tilings of $1 \times n$ rectangles with $1 \times 1$ and $1 \times 2$ blocks of the *character* $t^n$ equal to $\frac{1}{ \det (1- Mt)}$, where $M$ is the $2 \times 2$ matrix $ \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$?
>
>
>
to which one answer could be
>
> Grouping terms, this sum is the generating function $\sum\_n F\_n t^n$ of the Fibonacci series $F\_n$. The Fibonacci series satisfy the recurrence relation $F\_{n+2} = F\_{n+1}+ F\_n$, which can be expressed with the matrix $M$ as $\begin{pmatrix} F\_{n+1} \\ F\_{n+2} \end{pmatrix} = M \begin{pmatrix} F\_n \\ F\_{n+1} \end{pmatrix}$, and the identity $\sum\_n F\_n t^n = \frac{1}{1-t-t^2} = \frac{1}{ \det (1- Mt)}$ can be easily proven by multiplying both sides by $\det (1- Mt)$ and applying the recurrence relation.
>
>
>
but to which another answer could be
>
> Generating functions tend to be related to characteristic polynomials of matrices (and other nice mathematical functions) by a number of identities. The generating function $\sum\_n F(n) t^n$, where $F(n)$ counts objects of a particular type, can be expressed as a sum over objects of that type of $t^n$, and when objects of that type have a natural notion of combination, composition, or multiplication, the function $t^n$ is frequently a character.
>
>
>
|
4
|
https://mathoverflow.net/users/18060
|
410172
| 167,836 |
https://mathoverflow.net/questions/410167
|
1
|
Suppose $X \in \{0,1\}^{n \times m}$ is a matrix generated according to the following generative process:
$$Z\_{ij} \sim \text{Bernoulli}(p) \implies X\_{ij} = \frac{Z\_{ij}}{\sum\_{k=1}^m Z\_{ik}}.$$
Is there a name for the distribution of $X$? Is there a closed-form for $E[XX^\top]$? I am struggling to find any information on the behavior of this random matrix.
|
https://mathoverflow.net/users/128729
|
Moments of rescaled Bernoulli random matrix
|
It is apparently assumed that the $Z\_{ij}$'s are independent, as we will do here -- since otherwise hardly anything can be said. Suppose also that $m\ge2$ and $0<p<1$.
The $ab$-entry of the matrix $Y:=XX^\top$ is
\begin{equation}
Y\_{ab}=\sum\_{r\in[m]}X\_{ar}X\_{br}
=\sum\_{r\in[m]}\frac{Z\_{ar}Z\_{br}}{\sum\_{k\in[m]}Z\_{ak}\sum\_{k\in[m]}Z\_{bl}},
\end{equation}
where $[m]:=\{1,\dots,m\}$ and $a,b$ are in $[n]$; here we assume the convention $\frac00=0$.
So,
\begin{equation}
EY\_{ab}=m \,E\frac{Z\_{a1}Z\_{b1}}{\sum\_{k\in[m]}Z\_{ak}\sum\_{k\in[m]}Z\_{bl}}.
\end{equation}
If $a\ne b$ then
\begin{equation}
\begin{aligned}
EY\_{ab}&=m\,E\frac{Z\_{a1}}{\sum\_{k\in[m]}Z\_{ak}} \, E\frac{Z\_{b1}}{\sum\_{k\in[m]}Z\_{bl}} \\
&=m \Big(E\frac{Z\_{a1}}{\sum\_{k\in[m]}Z\_{ak}}\Big)^2 \\
&=m p^2\Big(E\frac1{1+\sum\_{k\in[m]\setminus\{1\}}Z\_{ak}}\Big)^2 \\
&=m p^2\Big(E\frac1{1+B\_{m-1,p}}\Big)^2 \\
&=\frac{\left(1-q^m\right)^2}{m},
\end{aligned}
\end{equation}
where $q:=1-p$ and $B\_{m-1,p}$ is a binomial random variable with parameters $m-1,p$.
Similarly, the diagonal entries of the matrix $EY$ are
\begin{equation}
\begin{aligned}
EY\_{aa}&=m E\Big(\frac{Z\_{a1}}{\sum\_{k\in[m]}Z\_{ak}}\Big)^2 \\
&=m p\,E\Big(\frac1{1+\sum\_{k\in[m]\setminus\{1\}}Z\_{ak}}\Big)^2 \\
&=m p\,E\Big(\frac1{1+B\_{m-1,p}}\Big)^2 \\
&=mp q^{m-1} \, \_3F\_2\left(1,1,1-m;2,2;-\frac{p}{q}\right),
\end{aligned}
\end{equation}
where $\_3F\_2$ is the hypergeometric function.
|
2
|
https://mathoverflow.net/users/36721
|
410178
| 167,838 |
https://mathoverflow.net/questions/368988
|
3
|
Let $X\_m$ denote a set of $m\geq 3$ lines in $\mathbb{R}^2$ that are not all parallel. Consider the problem of determining a closed path of $kn$ points in $X\_m$ $k, n \in \mathbb{Z}^+$, such that the "forward orbit" of the path has angles of incidence with the lines in $X\_m$ which are strictly contained in a given set $\{\theta\_1, \theta\_2, ..., \theta\_n\}$, and further each angle must occur in a particular order when traversing the sequence of points.
In other words, if $p\_1, p\_2, ..., p\_n, p\_{n+1}$ (with $p\_{n+1} = p\_1$) are a set of such points in $X\_m$, join points $p\_i, p\_{i+1}$ with a line segment, so that that line segment makes angle $\theta\_i$ with line $L\_j \subset X\_m$, where $p\_{i+1} \in L\_j$. The question then is whether such a closed path can be found for a given $X\_m$, set of incidence angles $\{\theta\_1, \theta\_2, ..., \theta\_n\}$, and ordering at which the angles must appear.
This appears to be similar to a weakening of the problem of finding periodic billiard paths in polygons, however it still appears to be hard, and I am wondering if anyone is aware of a solution, or an analogous problem that is discussed in literature.
|
https://mathoverflow.net/users/129192
|
Finding particular closed paths in geometric plane regions
|
The answer, given in this paper: <https://arxiv.org/abs/2112.02207>, turns out to be "yes", there always exist such closed curves. The theorem stating the answer to this question is given in the introduction to the linked paper, and I restate it here for reference:
***Theorem:*** *For any space $X\_m$ with labeled lines, let $\theta\_1, \theta\_2,...,\theta\_n$, $n\geq m \geq 3$, be any sequence of acute or right angles, and let $L\_{a\_1}, L\_{a\_2}, ..., L\_{a\_n}$ be a sequence of line labels such that no two consecutive labels are the same, including $L\_{a\_n}$ and $L\_{a\_1}$, and each of the $m$ possible labels occur at least once in the sequence. Then there exists a closed curve $\Gamma$ over $X\_m$ that admits an incidence angle sequence $\theta\_1, \theta\_2, ..., \theta\_n$ with respect to the line sequence $L\_{a\_1}, L\_{a\_2}, ..., L\_{a\_n}$ when traversed in a fixed direction. Moreover, there are either uncountably many such curves, or no more than $2^n$.*
|
0
|
https://mathoverflow.net/users/129192
|
410189
| 167,842 |
https://mathoverflow.net/questions/410140
|
2
|
If $H= (V,E)$ is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph), a *matching* is a set $M\subseteq E$ such that $e\_1\cap e\_2 = \emptyset$ whenever $e\_1\neq e\_2 \in M$. The *matching number* $\mu(H)$ of a hypergraph $H=(V,E)$ with $V$ finite is the maximum number of elements a matching can have.
For infinite hypergraphs $H=(V,E)$, we let $$\mu(H) = \sup\{|M|:M\subseteq E\text{ is a matching}\}.$$ This definition agrees with the above definition for finite hypergraphs.
**Question.** If $H=(V,E)$ is a hypergraph with $V$ infinite, is there necessarily a matching $M\_0\subseteq E$ such that $|M\_0|=\mu(H)$?
|
https://mathoverflow.net/users/8628
|
Matching number in infinite hypergraphs
|
Let $(P,\leq)$ be a poset, let $H$ be the set of maximal chains in $P$ —- so $H$ will be the set of vertices. For $a\in P$, let $E\_a$ denote the set of all chains in $H$ containing $a$; these sets are the edges.
Now, $E\_a$ and $E\_b$ are disjoint iff $a$ and $b$ are incomparable. So the edges are pairwise disjoint iff the corresponding elements form an antichain.
It remains to find a poset containing arbitrarily large antichains but having no infinite ones. One such example is $\mathbb Z\_{\geq0}^2$ with a componentwise relation.
This provides a counterexample, if infinite edges are allowed.
|
3
|
https://mathoverflow.net/users/17581
|
410197
| 167,844 |
https://mathoverflow.net/questions/410122
|
5
|
In a Riemannian symmetric space $Q$, it is well known that the existence of a totally geodesic submanifold at a point $p \in Q$ is equivalent to the existence of a Lie triple system at $p$, i.e., a subspace $V$ of $T\_{p}Q$ such that $[u, [v, w]] \in V$ for all $u,v,w \in V$; see for example Theorem 7.2 in Helgason, *Differential geometry, Lie groups, and symmetric spaces*.
As far as I understand, the proof carries over directly to the case where the ambient manifold is a *semi-Riemannian* symmetric space.
**Question**. Does anyone know where in the literature I can find a statement/proof of this more general result?
|
https://mathoverflow.net/users/74033
|
Proof of equivalence between Lie triple systems and totally geodesic submanifolds
|
Try pages 71 and 72 of Cartan's "La géométrie des groupes de transformations" <https://eudml.org/doc/235668>
|
1
|
https://mathoverflow.net/users/21123
|
410199
| 167,845 |
https://mathoverflow.net/questions/410186
|
14
|
I apologize in advance if this is well-known, but I can't seem to find the answer in the literature. Let me be precise about my question. I am looking for concrete examples of locally compact Hausdorff groups $G$ such that that there exists $a, b \in G$ with $a \ne b$, but for any continuous representation $\pi : G \to \operatorname{GL}\_n(k)$ with $k$ a valuated field and $n$ a positive integer, we have $\pi(a) = \pi(b)$.
I am also interested in a weaker form: the case where $k$ is the field $\mathbb{C}$ of complex numbers.
If it turns out that there are no such groups, please indicate a reference or sketch a proof.
Also a relevant statement which my gut tells me should be true, but I don't know enough Lie theory yet to give a complete proof: for every real Lie group $G$, there are enough finite dimensional, continuous complex representations of $G$ to distinguish points of $G$. I was thinking maybe this can be proved by exploiting the close relationship between representations of a Lie group and representations of its Lie algebra. It would be nice if someone could give a sketch if this can indeed be achieved or describe why this "hand-waving" actually does not work.
Edit: Sorry for the poor choice of terminology. To truly get the "weaker form" as YCor pointed out, I should replace "valuated field" by "a field $k$ with an absolute value" $|\cdot|: k \to \mathbb{R}\_{\ge 0}$, such that (1) $|ab| = |a| |b|$; (2) $|a| = 0$ if and only if $a=0$; (3) $|1 + a| \le 2$ for all $a$ with $|a| \le 1$ (or equivalently by a well-known argument, the triangle inequality). To exclude the discrete topology induced by this absolute value, we exclude the trivial absolute value that $|a| = 1$ for all $a \in k^\times$.
|
https://mathoverflow.net/users/128540
|
Examples of locally compact groups that do not admit enough finite dimensional representations
|
There is an example which satisfies something much stronger: there exist nontrivial groups $G$ such that *any* homomorphism (not even necessarily continuous) $\pi:G\to GL\_n(k)$ for *any* field $k$ (and, in fact, any commutative integral domain) is trivial, so in particular for *any* $a,b\in G$ we have $\pi(a)=\pi(b)$. Since any group can be given a locally compact Hausdorff topology, namely the discrete topology, these will in particular answer your question.
As for examples of such groups $G$, we can take any finitely generated which has no finite quotients, e.g. the [Higman group](https://en.wikipedia.org/wiki/Higman_group) mentioned in the comment by Terry Tao.
This result is apparently due to Mal'cev, but I don't have a reference at hand so here is a sketch of an argument. It is enough to show that if you have a finitely generated subgroup $\Gamma$ of a group $GL\_n(R)$ for some integral domain $R$ (e.g. the image of $G$ under a representation), then $\Gamma$ is [residually finite](https://en.wikipedia.org/wiki/Residually_finite_group) (hence trivial, if $\Gamma$ is a quotient of $G$ which has no finite quotients). The main idea of the proof is to replace $R$ by some ring which is finitely generated over $\mathbb Z$, meaning it is a quotient of a polynomial algebra over $\mathbb Z$. Then $R$ is a [Jacobson ring](https://en.wikipedia.org/wiki/Jacobson_ring), so in particular the intersection of all maximal ideals of $R$ is trivial, and moreover for all maximal ideals $m$ of $R$ we have $R/m$ finite. Now if $\gamma\in\Gamma$ is any nontrivial element, then there is a maximal ideal $m$ not containing all coefficients of $\gamma$, so $\gamma$ has nontrivial image in the finite group $GL\_n(R/m)$.
|
12
|
https://mathoverflow.net/users/30186
|
410205
| 167,847 |
https://mathoverflow.net/questions/410204
|
5
|
I have a stupid question about a topology on $C\_c(X)$. Here $X$ is locally compact Hausdorff. Can assume $\sigma$-compact if it helps.
**Definition (topology on $C\_c(X)$):** For each compact $K \subset X$, $C\_K(X)$ is the set of functions in $C\_c(X)$ with support in $K$. $C\_K(X)$ is given a Banach space structure with the sup norm. Call this topology $\tau\_K$.
Let $\beta$ be the set of $V \subset C\_c(X)$ which are convex, balanced, and which have $V\cap C\_K(X) \in \tau\_K$ for each compact $K$.
Define $\tau$ to be the collection of sets in $C\_c(X)$ given by $\bigcup (\phi\_i + W\_i)$ where $W\_i \in \beta$. This gives a topology on $C\_c(X)$ making it a locally convex topological vector space with $\beta$ as a local basis. Moreover, the topology $\tau\_K$ coincides with the subspace topology $\tau|\_{C\_K(X)}$.
For all the above facts see Rudin's *functional analysis* $\S 6.3$ onwards.
**Question:** Take the product topology $\tau \times \tau$ on $C\_c(X) \times C\_c(X)$. Say $V\subset C\_c(X)\times C\_c(X)$ is convex and balanced.
Suppose for every pair of compact sets $K, F \subset X$, we have that $V \cap C\_K(X)\times C\_F(X) \in \tau\_K \times \tau\_F$. Is $V$ open in the product topology $\tau\times \tau$?
I came across this question while studying pseudomeasures and induced representations from Folland's *Harmonic analysis*.
**Attempt:** Say $(\phi\_1, \phi\_2) \in V$. It suffices to find $W\_1, W\_2 \in \beta$ such that $(\phi\_1,\phi\_2) + W\_1\times W\_2 \subset V$...Pretty lost at this point.
Also posted [here](https://math.stackexchange.com/questions/4326256/product-of-inductive-limit-topologies-on-c-cx-times-c-cx).
|
https://mathoverflow.net/users/105628
|
Product of inductive limit topologies on $C_c(X)\times C_c(X)$
|
The topology $\tau$ you describe makes $(C\_c(X),\tau)$ the *colimit (or inductive or direct limit) in the category LCS of locally convex spaces* of the system $C\_K(X)$ with inclusions $i\_{K,L}:C\_K(X) \hookrightarrow C\_L(X)$ for compact subsets $K\subseteq L$ of $X$, i.e., the inclusions $i\_K:C\_K(X)\to C\_c(X)$ have the *universal property* that, for every family of continuous linear maps (the morphisms of the category) $f\_K:C\_K(X)\to Y$ with $f\_L\circ i\_{K,L}=f\_K$, there is a unique morphism $f:C\_c(X)\to Y$ with $f\_K=f\circ i\_K$. A big advantage of the categorical viewpoint is that you easily get permanence properties like *colimits commute with colimits*.
Since the category LCS is *additive* (you can add two morphisms $(f+g)(x)=f(x)+g(x)$ and this distributes over the composition -- sometimes the term *pre-additive category* is used) general abstract nonsense tells you that the product is also a coproduct and hence a colimit which commutes with colimits.
Therefore and without any calculations, the product topology $\tau\times \tau$ is the colimit topology of the system $C\_K(X)\times C\_L(X)$.
**EDIT.** I hope that the following helps. Whenever you have colimits $E=$colim$\_I E\_i$ and $F=$colim$\_J F\_j$ in a category having codropucts (denoted by $\oplus$) it is always true that $E\oplus F =$colim$\_{I\times J} E\_i\oplus F\_j$. Since LCS is additive, coproducts and products are "the same" so that you can replace $\oplus$ by $\times$. (You see that I am quite reluctant to replace the abstract argument by concrete calculations.)
|
6
|
https://mathoverflow.net/users/21051
|
410207
| 167,848 |
https://mathoverflow.net/questions/410176
|
5
|
*Previously [asked and bountied](https://math.stackexchange.com/questions/4318032/does-the-absolute-fragment-of-second-order-logic-satisfy-a-strong-lowenheim-skol) at MSE:*
Let $\mathsf{SOL\_{abs}}$ be the "forcing-absolute" fragment of second-order logic - that is, the set of second-order formulas $\varphi$ such that for every (set) forcing $\mathbb{P}$ and every (set-sized) structure $\mathfrak{A}$ (in the ground model $V$) we have $$\mathfrak{A}\models\varphi\quad\iff\quad\Vdash\_\mathbb{P}(\mathfrak{A}\models\varphi).$$ Note that this is in fact definable since the second-order theory of a structure in $V\_\kappa$ is calculated, uniformly, in $V\_{\kappa+1}$; by contrast, the set of forcing-absolute second-order properties of *the universe* is not so nice (to put it mildly).
Under a "mild" large cardinal assumption, $\mathsf{SOL\_{abs}}$ has a **weak downward Lowenheim-Skolem property for countable languages**, namely every satisfiable countable $\mathsf{SOL\_{abs}}$-theory $T$ has a countable model. This is just a consequence of the appropriate amount of large-cardinal-given absoluteness applied to the $\Sigma^1\_{\omega+1}$ sentence "$T$ has a countable model," with $L(\mathbb{R})$-absoluteness being more than enough.
However, I don't see how to get the full downward Lowenheim-Skolem property for countable languages:
>
> Does $\mathsf{ZFC}$ + [large cardinals] prove that every structure in a countable language has a countable $\mathsf{SOL\_{abs}}$-elementary substructure?
>
>
>
(To clarify since there is a stronger notion of second-order elementary equivalence in use as well, I mean the "weak" version of elementarity here: $\mathfrak{A}\preccurlyeq\_{\mathsf{SOL\_{abs}}}\mathfrak{B}$ iff for every $\mathsf{SOL\_{abs}}$-formula $\varphi(x\_1,...,x\_n)$ with only **object** variables we have $\varphi^\mathfrak{A}=\varphi^\mathfrak{B}\cap\mathfrak{A}^n$.)
The problem is that for a given structure $\mathfrak{A}$, the property "is a countable substructure of $\mathfrak{A}$" might be too wild for our large cardinal assumption to get useful purchase.
|
https://mathoverflow.net/users/8133
|
Can the forcing-absolute fragment of SOL have a strong Lowenheim-Skolem property?
|
Yes, and a proper class of Woodin cardinals suffices. For $n<\omega$ and $X$ a set of ordinals, $M\_n(X)$ denotes the minimal iterable proper class model $M$ of ZFC with $n$ Woodin cardinals above $\mathrm{rank}(X)$ and $X\in M$.
Because we have a proper class of Woodin cardinals, $M\_n(X)$ exists for every set $X$ of ordinals. If $x$ is a real, then for $n$ even, $M\_n(x)$ is boldface-$\Sigma^1\_{n+2}$-correct,
and for $n$ odd, is boldface-$\Sigma^1\_{n+1}$-correct. All forcing below is set forcing.
Claim 1: Given a structure $\mathfrak{B}$ and a set $B$ of ordinals coding $\mathfrak{B}$, and given an element $x\in\mathfrak{B}$ and a second order formula $\varphi$, if "$\mathfrak{B}\models\varphi(x)$" is forcing absolute then for all sufficiently large $n<\omega$, $M\_n(B)\models$"It is forced by $\mathrm{Coll}(\omega,\mathrm{rank}(\mathfrak{B}))$ that $\mathfrak{B}\models\varphi(x)$".
(Here '"$\mathfrak{B}\models\varphi(x)$" is forcing absolute' just means that
the truth value of "$\mathfrak{B}\models\varphi(x)$" is independent of which generic extension of $V$ we are working in, but here $\mathfrak{B}$ and $x$ are fixed.) (Note I'm not saying that the converse of the claim holds.)
Proof: Let $G$ be $\mathrm{Coll}(\omega,\mathrm{rank}(\mathfrak{B}))$-generic over $V$, hence also over $M\_n(B)$. Then $V[G]\models$"$\mathfrak{B}\models\varphi(x)$".
Since $\mathfrak{B}$ is countable in $V[G]$, this is just a projective statement there, about some reals in $M\_n(B)[G]$. But one can show that (with our large cardinal assumption) $M\_n(y)$ is not changed by forcing, so $(M\_n(B))^V=(M\_n(B))^{V[G]}$, and $V[G]\models$"$M\_n(B)[G]=M\_n(B,G)$", so $M\_n(B)[G]$ is at least boldface-$\Sigma^1\_{n+1}$-correct in $V[G]$. So for sufficiently large $n$,
$M\_n(B)[G]\models$"$\mathfrak{B}\models\varphi(x)$", as desired.
Now working in $V$, let $\pi:H\to V\_\gamma$ be elementary, with $\gamma$ a sufficiently large ordinal, and $H$ transitive and countable, and $\mathrm{rg}(\pi)$ containing $B,\mathfrak{B}$. The proper class mice $M\_n(B)$ are determined by set-sized mice $M\_n^\#(B)$, and these are definable from $B$ over $V\_\gamma$,
so are also in $\mathrm{rg}(\pi)$. Let $\pi(\bar{B},\bar{M}\_n^\#(\bar{B}))=(B,M\_n^\#(B))$. Then because $\pi$ is elementary, in fact $\bar{M}\_n^\#(\bar{B})=M\_n^\#(\bar{B})$, i.e. the collapse of $M\_n^\#(B)$ is the true $M\_n^\#(\bar{B})$.
Claim 2: $\pi\upharpoonright\bar{\mathfrak{B}}:\bar{\mathfrak{B}}\to\mathfrak{B}$ is elementary with respect to forcing absolute second order formulas.
Proof: Suppose $\bar{\mathfrak{B}}\models\varphi(\bar{x})$ where $\varphi$ is forcing absolute, but $\mathfrak{B}\models\neg\varphi(x)$ where $x=\pi(\bar{x})$. Clearly $\neg\varphi$ is also forcing absolute, so in particular, "$\mathfrak{B}\models\neg\varphi(x)$" is forcing absolute, so by Claim 1, for sufficiently large $n$, $M\_n(B)$ models 'it is forced by $\mathrm{Coll}(\omega,\mathrm{rank}(\mathfrak{B}))$ that $\mathfrak{B}\models\neg\varphi(x)$'. So $M\_n(\bar{B})$ models the same about $\bar{\mathfrak{B}}$ and $\bar{x}$ (as the theory gets encoded into the $\#$ versions). We have an $(M\_n(\bar{B}),\mathrm{Coll}(\omega,\mathrm{rank}(\bar{B})))$-generic $g\in V$ (as $M\_n^\#(\bar{B})$ is countable). So for large $n<\omega$, $M\_n(\bar{B})[g]$ models "$\bar{\mathfrak{B}}\models\neg\varphi(\bar{x})$". But $M\_n(\bar{B})[g]=M\_n(\bar{B},g)$, so this model is boldface-$\Sigma^1\_{n+1}$ correct, so with $n$ large enough, this gives that $\bar{\mathfrak{B}}\models\neg\varphi(\bar{x})$, a contradiction.
|
8
|
https://mathoverflow.net/users/160347
|
410213
| 167,850 |
https://mathoverflow.net/questions/408965
|
1
|
Let $Q$ be a quiver of type $ADE$, $I$ is the set of vertices of $Q$. Let $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ be a Nakajima quiver variety for such quiver (here ${\mathbf{v}}=(v\_i)\_{i \in I}$ is the dimension vector and ${\mathbf{w}}=(w\_i)\_{i \in I}$ is framing). We can associate to $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ a pair of weights $\lambda:=\sum\_i w\_i\omega\_i$, $\mu:=\lambda-\sum\_i v\_i\alpha\_i$.
We assume that $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ is nonempty (so $\mu$ is a weight of the irreducible representation $L\_{\lambda}$). Assume also that $w\_i=\delta\_{i,k}$ for some $k \in I$ (i.e. all $w\_i$ are zero except $w\_k$ which is one). In other words we assume that $\lambda=\omega\_k$ for some $k$.
If $\mu$ is conjugate to $\omega\_k$ via the action of the Weyl group (this for example always holds if $\omega\_k$ is minuscule) then the quiver variety $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ is one point.
If $\mu$ is not conjugate to $\omega\_k$ then $\mathfrak{M}({\mathbf{v}},\mathbf{w})$ has dimension at least two.
Question: do we have a good (explicit) description of the variety $\mathfrak{M}({\mathbf{v}},{\mathbf{w}})$ when $\lambda=\omega\_k$ (i.e. $w\_{i}=\delta\_{i,k}$)?
|
https://mathoverflow.net/users/144206
|
Nakajima quiver varieties for ADE quiver with one dimensional framing
|
We, at least, have the explicit answer for their Betti numbers. For an example, let $Q$ be type $E\_8$, and $k$ be the triplet vertex. Take $\mu = 0$. The normalized Poincare polynomial of $\mathfrak M(\mathbf v,\mathbf w)$ was computed in <https://arxiv.org/pdf/math/0606637.pdf> as $$1357104 + 2232771t^2 + 2002423t^4 + 1317308t^6 + 716312t^8 + 342421t^{10} + 148512t^{12}
+ 59490t^{14} + 22162t^{16} + 7687t^{18} + 2463t^{20} + 726t^{22} + 192t^{24} + 44t^{26} + 8t^{28} + t^{30}.$$
Here the constant term corresponds to the middle degree, while the highest term corresponds to the degree 0.
|
4
|
https://mathoverflow.net/users/3837
|
410215
| 167,851 |
https://mathoverflow.net/questions/410047
|
3
|
In number theoretical estimations, often we take the logarithms of a natural number to express it properly. A perfect example of this is the von-Mangoldt function. I am looking for an analogous arithmetic function of the logarithm function which is of the form $$\beta(n)=\sum\_{p\_{i}^{\alpha\_{i}}\mid n}\alpha\_{i}\cdot \log{\log{p\_{i}}}$$ where $n=p\_{1}^{\alpha\_{1}}\cdots p\_{s}^{\alpha\_{s}}.$ Is this function well known in literature?
**P.S.** I am sorry if this post doesn't suit for MathOverflow and would delete it if suggested.
|
https://mathoverflow.net/users/164499
|
Literature on analogous arithmetic function of logarithm function
|
Your function $\beta$ is a completely additive function, in the sense that $\beta(nm)=\beta(n)+\beta(m)$ for all $n,m$.
There is a vast literature on the statistical behavior of additive functions, e.g. Erdös and Kac's 1940 paper and various textbooks: Kubilius' book "Probabilistic methods in the theory of numbers", Elliott's books and (more recently) Tenenbaum's book. And this is just to name a few...
These paper and books focus on the *general* theory rather than specific examples, and I do not know if your specific example got any attention (is there a reason it should have?). A large portion of the classical literature (e.g. the Erdös-Kac paper and Kubilius' book) seems to focus on strongly additive functions (those that satisfy $\alpha(p^k)=\alpha(p)$), so not capturing your $\beta$.
A good source for the treatment of completely additive functions is Billingsley's 1974 paper "The Probability Theory of Additive Arithmetic Functions".
His Theorem 3.1 applies to your $\beta$ and guarantees that $\beta(n)$ suitably normalized has a gaussian limiting distribution, just as in the classical Erdös-Kac theorem.
In more detail, given a completely additive (real) function $\beta$ let
$$A\_n = \sum\_{p \le n} \frac{\beta(p)}{p}, \qquad B\_n = \sum\_{p \le n} \frac{\beta^2(p)}{p}.$$
A sufficient condition for
$$ \frac{\beta(m) - A\_n}{B\_n}$$
to tend in distribution to standard gaussian distribution, where $m$ is a number chosen uniformly at random from $\{1,2,\ldots,n\}$, is that
$$ \frac{\max\_{p \le n}|\beta(p)|}{B\_n} \to 0.$$
Verifying this for $\beta(p)=\log \log p$ (or any power of $\log \log p$) is an exercise in Mertens' theorems and partial summation.
|
3
|
https://mathoverflow.net/users/31469
|
410221
| 167,853 |
https://mathoverflow.net/questions/410209
|
20
|
>
> **Problem**: Given three positive integers $0 < n\_1 < n\_2 < n\_3$ such that
> $$n\_1 + n\_2 \ne n\_3, \quad n\_2 \ne 2n\_1, \quad n\_3 \ne 2n\_1, \quad n\_3 \ne 2n\_2,$$
> is there always a real number $x$ such that
> $$\cos n\_1 x + \cos n\_2 x + \cos n\_3 x < -2?$$
>
>
>
This is a follow-up of the question in MO: [Is there always a real $x$ such that $\cos n\_1 x + \cos n\_2 x + \cos n\_3 x < -2$?](https://mathoverflow.net/questions/410115/is-there-always-a-real-x-such-that-cos-n-1-x-cos-n-2-x-cos-n-3-x-2)
I checked that (brute force), the statement is true for all positive integers $1\le n\_1 < n\_2 < n\_3 \le 1000$
with $n\_1 + n\_2 \ne n\_3, \, n\_2 \ne 2n\_1, \, n\_3 \ne 2n\_1, \, n\_3 \ne 2n\_2$.
For $n\_3 > 1000$, I randomly generate some $n\_1, n\_2, n\_3$ without finding a counterexample.
I also found a related problem in MO:
[The maximum of a real trigonometric polynomial](https://mathoverflow.net/questions/35538/the-maximum-of-a-real-trigonometric-polynomial).
Perhaps the problem here is easier to deal with.
My idea is to consider the case $\cos n\_1 x = -1$ or
$\cos n\_2 x = -1$ or $\cos n\_3 x = -1$.
For example, $\cos n\_3 x = -1$ leads to $x = \frac{(2k\_3 + 1)\pi}{n\_3}$ where $0\le k\_3 \le n\_3; k\_3\in \mathbb{Z}$;
We consider $f(k\_3) = \cos \frac{(2k\_3 + 1)n\_1\pi}{n\_3} + \cos\frac{(2k\_3 + 1)n\_2\pi}{n\_3} - 1$. However, I failed to go proceed.
I also tried non-negative trigonometric polynomials
and found this article: "Extremal Positive Trigonometric Polynomials", [https://www.dcce.ibilce.unesp.br/~dimitrov/papers/main.pdf](https://www.dcce.ibilce.unesp.br/%7Edimitrov/papers/main.pdf)
Theorem 4 gives the necessary and sufficient condition for
$f(\theta) = a\_0/2 + \sum\_{k=1}^n a\_k \cos k \theta$ to be non-negative on $\mathbb{R}$.
|
https://mathoverflow.net/users/141801
|
(update) Is there always a real $x$ such that $\cos n_1 x + \cos n_2 x + \cos n_3 x < -2$?
|
In principle this problem can be resolved numerically in finite time, by exploiting the dichotomy between structure (small linear relations between the frequencies $n\_1,n\_2,n\_3$) and randomness (equidistribution), though I do not know if the approach below can actually be implemented in a feasible amount of time. (One could in fact use quantitative equidistribution theorems on tori for this, such as Proposition 1.1.17 of [this book of mine](https://terrytao.wordpress.com/books/higher-order-fourier-analysis/), but I will take a more explicit approach here as it will likely give better numerical constants.)
We first claim that if $\cos(n\_1 x) + \cos(n\_2 x) + \cos(n\_3 x) \geq -2$ for all $x$ then there must be a non-trivial linear relation $a\_1 n\_1 + a\_2 n\_2 + a\_3 n\_3 = 0$ between the $n\_1,n\_2,n\_3$ with integers $a\_1,a\_2,a\_3$ with $|a\_1| + |a\_2| + |a\_3| \leq C$ for some effectively computable absolute constant $C$. Indeed, use the Weierstrass approximation theorem to find a polynomial $P: {\bf R} \to {\bf R}$ such that
$$ \int\_0^1 \int\_0^1 \int\_0^1 P( \cos(2\pi \theta\_1) + \cos(2\pi \theta\_2) + \cos(2\pi \theta\_3) )\ d\theta\_1 d\theta\_2 d\theta\_3 > \sup\_{-2 \leq x \leq 3} P(x)$$
(this can be done by choosing $P$ to be large and positive on $[-3,-2]$ and small on $[-2,3]$). Then we have
$$ \int\_0^1 \int\_0^1 \int\_0^1 P( \cos(2\pi \theta\_1) + \cos(2\pi \theta\_2) + \cos(2\pi \theta\_3) )\ d\theta\_1 d\theta\_2 d\theta\_3
\neq \int\_0^1 P( \cos(2\pi n\_1 \theta) + \cos(2\pi n\_2 \theta) + \cos(2\pi n\_3 \theta) )\ d\theta.$$
But expanding out the polynomial and extracting the Fourier coefficients we see that the LHS and RHS in fact agree unless there is a non-trivial collision $a\_1 n\_1 + a\_2 n\_2 + a\_3 n\_3 = 0$ with $|a\_1| + |a\_2| + |a\_3| \leq \mathrm{deg} P$. In order to make this algorithm run in as feasible a time as possible it is desirable to get $\mathrm{deg} P$ as small as one can; presumably this can be done numerically since for each fixed choice of degree, one has to solve a linear program in the coefficients of $P$.
Next, once one has restricted to the case $a\_1 n\_1 + a\_2 n\_2 + a\_3 n\_3=0$ for some fixed $a\_1,a\_2,a\_3$, one can perform a similar argument to see that either one has $\cos(2\pi \theta\_1) + \cos(2\pi \theta\_2) + \cos(2\pi \theta\_3) \geq -2$ on the entire hyperplane $\{ (\theta\_1,\theta\_2,\theta\_3): a\_1 \theta\_1 + a\_2 \theta\_2 + a\_3 \theta\_3 =0\}$ (in which the answer to your question is negative), or else there must be a second constraint $b\_1 n\_1 + b\_2 n\_2 + b\_3 n\_3 = 0$ with $(b\_1,b\_2,b\_3)$ independent of $(a\_1,a\_2,a\_3)$ and with $|b\_1|+|b\_2|+|b\_3|$ also bounded. After reducing $n\_1,n\_2,n\_3$ to lowest terms, this leaves one with an explicit finite list of candidate triples $(n\_1,n\_2,n\_3)$ for which the question can be decided by case-by-case check. (To reduce the number of cases, one can normalise $0 \leq a\_1 \leq a\_2 \leq a\_3$ (with $n\_1,n\_2,n\_3$ now arbitrary integers) rather than $0 < n\_1 < n\_2 < n\_3$.)
|
26
|
https://mathoverflow.net/users/766
|
410227
| 167,854 |
https://mathoverflow.net/questions/410222
|
8
|
The classification of oriented compact smooth manifolds up to oriented cobordism is one
of the landmarks of 20th century topology. The techniques used there form the part of the foundations of differential topology and stable homotopy theory.
It is a popular knowledge to find the oriented bordism groups $\Omega\_d^{SO}$ in dimensions lower or equal to 8, <http://www.map.mpim-bonn.mpg.de/Oriented_bordism>,
which we have
$$\Omega\_0^{SO}=\mathbb{Z}$$
$$\Omega\_1^{SO}=0$$
$$\Omega\_2^{SO}=0$$
$$\Omega\_3^{SO}=0$$
$$\Omega\_4^{SO}=\mathbb{Z}$$
$$\Omega\_5^{SO}=\mathbb{Z}/2$$
$$\Omega\_6^{SO}=0$$
$$\Omega\_7^{SO}=0$$
$$\Omega\_8^{SO}=\mathbb{Z} \oplus \mathbb{Z}$$
$$\Omega\_9^{SO}=\mathbb{Z}/2 \oplus \mathbb{Z}/2$$
$$\Omega\_{10}^{SO}=\mathbb{Z}/2$$
$$\Omega\_{11}^{SO}=\mathbb{Z}/2$$
do we happen to know other dimensions in the literature? For any $d \leq 28$? Also are the manifold generators already known for those $d \leq 28$? Are manifold generators systematically constructible? References (precisely which pages) are surely welcome! Thank you in advance.
|
https://mathoverflow.net/users/27004
|
Oriented bordism in higher dimensions (e.g. $12 \leq d \leq 28$)
|
Most of the main results needed for this calculation can be found in Wall's paper "[Determination of the oriented cobordism ring](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/cobord.pdf)", but [this note](http://math.uchicago.edu/%7Emay/REU2016/REUPapers/Gwynne.pdf) by Gwynne might be helpful to express this in more modern language. Here is the exective summary:
* All of the homotopy groups are a direct sum $\Bbb Z^r \oplus \Bbb (Z/2)^s$. Bordism classes of oriented manifolds are completely determined by their Pontrjagin and Stiefel-Whitney numbers.
* The mod-2 cohomology of $MSO$ is the same as the mod-2 cohomology of $BSO$, a polynomial ring on $w\_2, w\_3, \dots$ whose Poincare series is
$$
\prod\_{i \geq 2} \tfrac{1}{1-t^i}.
$$
* Rationally, the ring is a polynomial algebra $\Bbb Q[x\_4, x\_8, x\_{12}, \dots]$ on generators in degrees that are a power of $4$. This tells us the rank $r$ of each group. The Poincare series for the free part of $\Omega^{SO}\_\*$ is thus
$$
p\_{free}(t) = \prod\_{j \geq 1} \tfrac{1}{1-t^{4i}}.
$$
* $2$-locally, the bordism spectrum $MSO$ is a wedge of suspensions of Eilenberg--Mac Lane spectra $H\Bbb Z/2$ and $H\Bbb Z$. This allows us to write
$$
H^\*(MSO) \cong \bigoplus\_\text{free summands}H^\*(H\Bbb Z) \oplus \bigoplus\_\text{torsion summands} H^\*(H\Bbb Z/2).
$$
Turning this into a Poincare series expression using the Poincare series for the cohomology of Eilenberg--Mac Lane spectra, we can solve for the Poincare series of the torsion part in $\Omega^{SO}\_\*$.
$$
p\_{tors}(t) = \left[(1-t) \prod\_{k \geq 2, k \neq 2^i-1} \left(\tfrac{1}{1-t^k}\right)\right] - \left[\frac{1}{1+t}\prod\_{k \geq 1}\left(\tfrac{1}{1-t^{4k}}\right)\right]
$$
I asked Mathematica for a calculation of these groups out to degree 28. Assuming I didn't make a typo, here they are.
$$
\begin{array}{c|l}
n & \Omega^{SO}\_n \\
\hline
0 & \Bbb Z\\
1 & 0\\
2 & 0\\
3 & 0\\
4 & \Bbb Z\\
5 & \Bbb Z/2\\
6 & 0\\
7 & 0\\
8 & \Bbb Z^2\\
9 & (\Bbb Z/2)^2\\
10 & \Bbb Z/2\\
11 & \Bbb Z/2\\
12 & \Bbb Z^3\\
13 & (\Bbb Z/2)^4\\
14 & (\Bbb Z/2)^2\\
15 & (\Bbb Z/2)^3\\
16 & \Bbb Z^5 \oplus \Bbb Z/2\\
17 & (\Bbb Z/2)^8\\
18 & (\Bbb Z/2)^5\\
19 & (\Bbb Z/2)^7\\
20 & \Bbb Z^7 \oplus (\Bbb Z/2)^{20}\\
21 & (\Bbb Z/2)^{15}\\
22 & (\Bbb Z/2)^{11}\\
23 & (\Bbb Z/2)^{15}\\
24 & \Bbb Z^{11} \oplus (\Bbb Z/2)^{10}\\
25 & (\Bbb Z/2)^{28}\\
26 & (\Bbb Z/2)^{22}\\
27 & (\Bbb Z/2)^{31}\\
28 & \Bbb Z^{15} \oplus (\Bbb Z/2)^{23}\\
\end{array}
$$
(The OEIS doesn't seem like anybody interested in bordism theory has invested the effort into adding this type of information.)
|
15
|
https://mathoverflow.net/users/360
|
410228
| 167,855 |
https://mathoverflow.net/questions/410225
|
1
|
I have heard about the Schottky problem and the related Novikov's conjecture about the characterization of matrices in the Siegel upper half-space which are indeed the Riemann matrix of a compact Riemann surface.
Instead of such a global statement, I was wondering about the following: given $\Omega$ a Riemann matrix of an existing compact Riemann surface, what kind of operations on $\Omega$ could guarantee me that the result is again the Riemann matrix of a surface?
For example, is it true that $2\Omega$ is the Riemann matrix of a surface if $\Omega$ is?
The last specific question comes from the desire for a geometric interpretation with surfaces of the theta functions appearing in the formula expressing the product of two theta functions with a matrix $\Omega$ as a sum of product of theta functions with a matrix $2\Omega$.
Thank you in advance for your insights
|
https://mathoverflow.net/users/201087
|
operations on matrices preserving the property of being the Riemann matrix of a surface
|
No, this is not true. If $C$ is a general curve of genus $\geq 4$ with period matrix $\Omega$ and $p$ is a prime, $p\Omega$ is not the period matrix of a curve. This is proved (in an equivalent, more geometric, form) by Donagi and Livné, Ann. Sc. Norm. Sup. Pisa 28, no. 2 (1999), p. 323-339 — see §7.
|
1
|
https://mathoverflow.net/users/40297
|
410231
| 167,856 |
https://mathoverflow.net/questions/410223
|
4
|
Let $m$ be a positive integer divisible by $6$, and let $q$ be one of $8,9$,
or a prime $\gt 3$.
**Question :** Is there always an $x\in [1,m]$, coprime to $m$, such
that $x\not\equiv\pm 1 \ \mod{q}$ ?
The main difficulty in that problem I think is that it combines two very different requirements, the congruence (and coprimality) conditions and the
bound $1\leq x \leq m$. Using the Chinese remainder theorem, one can find a $y\_q$ satisfying the congruence requirement and coprime to $m$, but the problem is that $y\_q$ might not be in $[1,m]$, and if we take the remainder of $y\_q$ when divided by $m$ we might lose the congruence property.
On the other hand, for a fixed $q$ it is easy to show that a counterexample, if it exists, must be astronomically large.
For example, for $q=8$, we can argue as follows : If $5\not\mid m$ we can take $x=5$, so we can assume $2\times 3\times 5 \mid m$. Next, if $11\not\mid m$ we can take $x=11$, so we can assume $2\times 3\times 5 \times 11 \mid m$, and
so on ...
This is a cross-post (with a few elements removed) from an older [MSE post](https://math.stackexchange.com/questions/4310651/compute-gcd-of-values-of-a-quadratic-polynomial).
|
https://mathoverflow.net/users/10341
|
System of congruences with bound condition
|
Yes. Denote $m=2^ab$ where $b$ is odd. One of numbers $b\pm 2$ is not congruent to $\pm 1$ modulo $q$.
|
5
|
https://mathoverflow.net/users/4312
|
410232
| 167,857 |
https://mathoverflow.net/questions/409470
|
2
|
In Aczel's Constructive Set Theory (CZF), no non-degenerate complete lattice can be proved to be a set. There are hallmark examples of complete lattices that are proper classes in CZF, including the Dedekind–MacNeille completion of a lattice/poset.
Is there a way to talk about the collection of Complete Lattices in CZF without introducing a hierarchy? This may seem like a splitting of hairs but I’m concerned about making statements about “all complete heyting algebras”. Is this concern unfounded or is there a way to avoid any potential issues with such statements?
|
https://mathoverflow.net/users/312621
|
Collection of proper classes with in CZF
|
Unfortunately the only answer I can give is that there is not a good way to do this in general, and since no one else has answered yet, that is probably the only real answer.
Having said that, there are a few techniques that work in some special situations, and can still be useful.
The most relevant one here is that you can still talk about all lattices that arise as Dedekind-MacNeille completions of set sized posets by proving statements of the form "for every poset, the Dedekind-MacNeille completion has X property." If I recall correctly you can get a more general version of this by talking about lattices arising from set presented formal topologies.
It is also sometimes possible to give a proof in the meta theory along the lines "if $\mathbf{CZF}$ proves a certain binary relation is the ordering of a (class sized) lattice, then it also proves the lattice has X property."
Finally, adding (suitably formulated) inaccessible sets to $\mathbf{CZF}$ is not so bad. It is still constructive in some sense, and the consistency strength is still a long way below many commonly studied theories with classical logic.
|
4
|
https://mathoverflow.net/users/30790
|
410233
| 167,858 |
https://mathoverflow.net/questions/410226
|
4
|
$X = (x\_1,...x\_n) \in \mathbb{R}^n, X \sim \mathcal{N}(O, \Sigma\_X)$ and $Y = (x\_1,...x\_n) \in \mathbb{R}^n, Y \sim \mathcal{N}(O, \Sigma\_Y)$ are two independent gaussian vectors.
If $\Sigma\_Y - \Sigma\_X$ is positive semidefinite, then $\forall \alpha \in \mathbb{R}^n$, $\alpha\Sigma\_Y\alpha^T \ge \alpha\Sigma\_X\alpha^T $. Because $X \alpha^T \sim \mathcal{N}(O, \alpha\Sigma\_X\alpha^T)$ and $Y \alpha^T \sim \mathcal{N}(O, \alpha\Sigma\_Y\alpha^T)$, we have that $\forall \varepsilon \in \mathbb{R}\_+^\*$, $ P(|X \alpha^T| \le \varepsilon) \ge P(|Y \alpha^T| \le \varepsilon)$
I am trying to show that it is also true for the infinity norm, ie that
$\forall \varepsilon \in \mathbb{R}\_+^\*$, $ P(||X||\_\infty \le \varepsilon) \ge P(||Y||\_\infty \le \varepsilon)$, where $ ||X||\_\infty = \max\_{i = 1..n}|x\_i|$. Do you think this is true, and do you have clues for how to show it? Otherwise, is there a counter example?
|
https://mathoverflow.net/users/470763
|
L_infinity norm of two gaussian vector
|
$\newcommand\R{\mathbb R}$Let $\|\cdot\|$ be any norm on $\R^n$. Take any real $t$. Let $Z$ be a random vector in $\R^n$ such that (i) $Z$ is independent of $X$ and (ii) $Z\sim N(0,\Sigma\_Y-\Sigma\_X)$. Then $X+Z$ equals $Y$ in distribution.
So, it suffices to show that
\begin{equation\*}
P(\|X\|\le t)\ge P(\|X+Z\|\le t). \tag{1}
\end{equation\*}
Note that
\begin{equation\*}
P(\|X+Z\|\le t)=Eg(Z), \tag{2}
\end{equation\*}
where
\begin{equation\*}
g(z):=P(\|X+z\|\le t)=\int\_{\R^n}dx\, f(x)1(\|x+z\|\le t)
\end{equation\*}
and $f$ is the pdf of $X$. The functions $f$ and $x\mapsto1(\|x+z\|\le t)$ are log concave, and hence the function $x\mapsto f(x)1(\|x+z\|\le t)$ is log concave.
So, by the [Prékopa–Leindler theorem](https://en.wikipedia.org/wiki/Pr%C3%A9kopa%E2%80%93Leindler_inequality#Applications_in_probability_and_statistics), $g$ is a log-concave function. Also, the function $g$ is even. So,
$g(z)\le g(0)=P(\|X\|\le t)$ for all $z\in\R^n$, and hence (1) follows from (2).
|
2
|
https://mathoverflow.net/users/36721
|
410242
| 167,861 |
https://mathoverflow.net/questions/410230
|
5
|
Let $S\_p(X)$ be the $p$-th singular chain group and $\mathcal S(X)$ be the singular chain complex of a topological space $X$. There is a barycentric subdivision operator (which is also a chain map) $\operatorname{sd}: S\_p(X) \to S\_p(X)$ as is defined in standard topology textbooks such as Munkres. There is a chain homotopy from $\operatorname{sd}$ to $\operatorname{id}\_{\mathcal S(X)}$.
Let $D\_p$ be the subgroup of $S\_p(X)$ generated by $T - \operatorname{sd}(T)$ such that $T \in S\_p(X)$. Then $\mathcal D:= (D\_p)$ is a chain subcomplex of $\mathcal S(X)$. **I wonder whether the homology of the quotient chain complex $\mathcal S(X)/\mathcal D$ is isomorphic to the homology of $\mathcal S(X)$.**
In fact, this is how I used to understand the proof of the excision theorem of singular homology "intuitively". However, now I begin to doubt this intuition.
There are two possible ways I can think of to prove the statement above. One is to show $\mathcal D$ is acyclic and use zig-zag lemma, the other is to construct a chain homotopy between $\mathcal S(X)/\mathcal D$ and $\mathcal S(X)$.
This question has been posted in MSE ([link](https://math.stackexchange.com/questions/4323525/singular-chain-complex-modulo-subdivision)) for two days, but there is no answer.
|
https://mathoverflow.net/users/166613
|
Homology of singular chain complex modulo subdivision
|
I believe $\mathcal D$ is not acyclic, unless $X$ is a discrete space, and therefore $S(X)/\mathcal D$ is not quasi-isomorphic to $S(X)$.
In fact, assuming that I did not make a mistake in the claim below, we have the following description of $\mathcal D$: it is zero in degree zero, and is isomorphic to $S(X)$ in positive degrees. It follows that $H\_0(\mathcal D)=0$, $H\_1(\mathcal D)$ is the huge group $S\_1(X)/\partial S\_2(X)$, and for $i>1$ $H\_i(\mathcal D)\cong H\_i(X)$. However, the inclusion $\mathcal D\hookrightarrow S(X)$ induces zero in homology.
It is obvious that $D\_0$ is the zero group. For higher degrees, we have the following claim
**Claim**: for all $n>0$, the operator $T\mapsto T-sd(T)$ defines an injective homomorphism $S\_n(X)\to S\_n(X)$. It follows that $D\_n\cong S\_n(X)$ for all $n>0$.
Sketch of proof (hope it is correct, I did not check every detail): Define a partial ordering on the set of singular $n$-simplices of $X$. Given two singular simplices $\sigma, \tau\colon \Delta^n \to X$, we say that $\sigma\le \tau$ if $\sigma=\tau$, or there exists a contracting linear map $h\colon \Delta^n\to \Delta^n$ such that $\sigma=\tau \circ h$. By contracting we mean that there exists a $0\le r<1$ such that $\mbox{dist}(h(x),h(y))\le r\cdot \mbox{dist}(x, y)$. It is not hard to check that this is a partial order. Moreover $\sigma=\sigma\circ h$ for some contracting map only if $\sigma$ is constant.
Now let $0\ne T\in S\_n(X)$. We can write $T=\Sigma\_{i=1}^m n\_i \sigma\_i$, where $\sigma\_1, \ldots, \sigma\_m$ are pairwise distinct singular simplices, and $n\_i$ are non-zero integers. If all $\sigma\_i$ are constant, then $T-sd(T)=T\ne 0$. Otherwise, among the $\sigma\_i$s there exists a non-constant simplex $\sigma\_{i\_0}$ that is maximal with respect to the order that we defined. The singular chain $T-sd(T)$ is a linear combination of $\sigma\_i$s and simplices that are smaller than $\sigma\_i$. It is clear that there is no summand that can cancel $\sigma\_{i\_0}$, so $T-sd(T)\ne 0$.
|
7
|
https://mathoverflow.net/users/6668
|
410248
| 167,863 |
https://mathoverflow.net/questions/409871
|
6
|
This problem was posted on another forum and was given at the 1992 Miklós Schweitzer Competition. This competition is known for its very difficult problems and this one seems no exception. I also can't find a solution anywhere online. Here is the problem:
Let $E \subset [0,1]$ be Lebesgue measurable with measure $\lvert E\rvert < 1/2$. Define
$$h (s) = \int \_ {E^c} \frac{dt}{{(s-t)}^2}$$
where $E^c = [0, 1]\setminus E$. Prove that there exists $t \in E^c$ such that
$$\int\_E \frac {ds} {h (s) {(s-t)} ^ 2} \leq c {\lvert E \rvert} ^ 2$$
with some constant $c$ that is independent of $E$.
The tricky part of the problem is the quadratic convergence to zero with the measure of $E$. It is easy to see that the left hand side of the inequality is linearly bound with $\lvert E\rvert$. Any ideas (or links to sites with solutions to Miklós Schweitzer problems?)
|
https://mathoverflow.net/users/19673
|
A problem concerning a divergent function on $[0, 1]$
|
I asked a colleague in Hungary, and he found the solution here (on page 170):
<http://real-j.mtak.hu/9393/1/MTA_MatematikaiLapok_1992.pdf>
It is in Hungarian, but with some effort and Google translator, finally I understood.
EDIT This is the translation of the argument above. I keep the same notation but write $E^c$ for $[0,1] \setminus E$ insetad of $\overline E$.
a) Let $$F=\{t \in E^c: |I \cap E| \leq K|E||I|\}$$ for every interval $t \in I$. In other words, $F$ consists of all points for which the maximal function of $\chi\_E$ is less than $K|E|$. Using the maximal inequality, one selects $K$ (independent of $E$) such that $|F| \approx 1$, hence $|E\_1|:=|E^c \cap F| \geq 1/2$.
b) Given $s \in E$ let $I\_s$ be an interval containing $s$ such that $|I\_s \cap E|=(1/2)|I\_s|$. Such a $I\_s$ exists in all Lebesgue points of $E$, again using that $|E| <1/2$ and a continuity argument.
c) If $s \in E$ and $t \in E\_1$, then applying a) to $J=I\_s \cup [s,t]$ (or the other way around) we get $(1/2)|I\_s|=|I\_s \cap E| \leq |J\cap E| \leq K|E|(|t-s|+|I\_s|)$. Then $|t-s| \geq \frac{|I\_s|}{4K|E|}$ if $K|E| \leq 1/4$.
d) Suppose that $K|E| \leq 1/4$. Then
$$
\int\_{E\_1} \frac{dt}{(s-t)^2} \leq 2 \int\_{\frac{|I\_s|}{4K|E|}}^\infty \frac{du}{u^2}=\frac{8K|E|}{|I\_s|}$$ but also, using b),
$$h(s)=\int\_{E^c} \frac{dt}{(s-t)^2}\geq \int\_{I\_s \cap \geq E^c} \frac{dt} {(s-t)^2} \geq \frac{1}{|I\_s|^2} \frac 12 |I\_s|=\frac{1}{2|I\_s|}.$$ Summing up,
$$
\int\_{E\_1} \frac{dt}{(s-t)^2} \leq 16 h(s) K|E|.$$
e) If $K|E| \geq 1/4$ then
$$
\int\_{E\_1} \frac{dt}{(s-t)^2} \leq
\int\_{E^c} \frac{dt}{(s-t)^2} =h(s) \leq 4h(s) K|E| \leq 16 h(s) K|E|.$$
f) Finally,
$$
\int\_{E\_1}dt\int\_{E}\frac{ds }{h(s)(s-t)^2}=\int\_{E}\frac{ds}{h(s)}\int\_{E\_1}\frac{ dt}{(s-t)^2} \leq 16K|E|^2 $$ and there is a point $t \in E\_1$ for which the statement holds with $c=32 K$ since $|E\_1| \geq 1/2$.
PS This does not deserve badges for maths...maybe for translating from hungarian!
|
8
|
https://mathoverflow.net/users/150653
|
410251
| 167,864 |
https://mathoverflow.net/questions/410255
|
5
|
I am interested in a morphism of $S$-schemes $f : X \to Y$ such that $X$ and $Y$ are proper over $S$ and there is some $s\_0 \in S$ such that $f : X\_{s\_0} \to Y\_{s\_0}$ is a closed immersion. Is it true that there is an open neighborhood $U \subset S$ of $s\_0$ so that $X\_U \to Y\_U$ is a closed immersion?
Because $f$ is automatically proper, I hoped to use the characterization of closed immersions as proper monomorphisms. Using an argument on formal neighborhoods of the fibers, I think I can show that $f : X \to Y$ is unramified at each point of the fiber and hence should be unramified on some neighborhood of the fiber. Maybe I can also show that $f$ is radicial in a similar fashion?
I am okay with assuming that $X, Y$ are noetherian and flat over $S$ if this helps.
|
https://mathoverflow.net/users/154157
|
Proper morphisms that are closed immersion on a fiber
|
This is proved in [EGA III, tome 1, Proposition 4.6.7(i)](http://www.numdam.org/item/PMIHES_1961__11__5_0/).
|
9
|
https://mathoverflow.net/users/70322
|
410259
| 167,866 |
https://mathoverflow.net/questions/382285
|
5
|
I posted this question on [Math Stack Exchange](https://math.stackexchange.com/questions/3988744/find-the-maximum-trigonometric-polynomial-coefficient-a-k) but did not get any answer. I am trying my luck here.
>
> Let $n,k$ be given positive integers and $n>k$. If for all real numbers $x$ we have $$A\_{1}\cos{x}+A\_{2}\cos{(2x)}+\cdots+A\_{n}\cos{(nx)}\le 1$$
> Find the maximum value of $A\_{k}$.
>
>
>
I don't know if this question has been studied
If $n=2$ it is easy to solve it.
|
https://mathoverflow.net/users/38620
|
Find the maximum trigonometric polynomial coefficient $A_{k}$
|
It is a known result:
$$\max A\_k = 2 \cos \frac{\pi}{\lfloor \frac{n}{k} \rfloor + 2}, \quad 1\le k \le n.$$
See:
1. Theorem 6 and the references therein, "Extremal Positive Trigonometric Polynomials", [https://www.dcce.ibilce.unesp.br/~dimitrov/papers/main.pdf](https://www.dcce.ibilce.unesp.br/%7Edimitrov/papers/main.pdf)
2. Theorem 16.2.4 in: Qazi Ibadur Rahman and Gerhard Schmeisser, “Analytic Theory of Polynomials”, 2002.
|
0
|
https://mathoverflow.net/users/141801
|
410270
| 167,870 |
https://mathoverflow.net/questions/410269
|
5
|
Let $(X\_1,X\_2,\ldots)$ be a stationary ergodic process with each $X\_n$ a real random variable taking values in $[-1,+1]$. Suppose that $\mathbb{E}[X\_n]=0$. Let $S\_n = \sum\_{k=1}^n X\_k$. Is the process $(S\_1,S\_2,\ldots)$ necessarily recurrent, in the sense that there exists some $M$ such that almost surely $|S\_n| \leq M$ infinitely often?
|
https://mathoverflow.net/users/23661
|
Recurrence of ergodic processes
|
Yes. I will use some different notation, but the idea is the same.
Let $(\Omega,\mu)$ be a probability space, and let $\sigma \colon \Omega \to \Omega$ be an ergodic measure preserving transformation. Let $f$ be a measurable function taking values in $[-1,1]$ such that $\int f\,d\mu=0$. Write $S\_nf(\omega)=f(\omega)+\ldots+f(\sigma^{n-1}\omega)$.
I claim that for $\mu$-a.e. $\omega$, $|S\_nf(\omega)|\le c$ infinitely often for any $c>\frac 12$. Here is a proof by contradiction. Suppose that $c>\frac 12$ and there is a set of positive measure $A$ such that $|S\_nf(\omega)|\le c$ only finitely many times for $\omega\in A$. Then since $c-(-c)>1$, then for $\omega\in A$, either $S\_nf(\omega)>c$ for all sufficiently large $n$ or $S\_nf(\omega)<-c$ for all sufficiently large $n$.
Negating $f$ if necessary, we may pick an $N>0$, and a subset $B$ of $A$ of positive measure such that $S\_nf(\omega)>c$ for all $n\ge N$. Now we consider the induced (first return) dynamical system on $B$, which is also ergodic. The $k$th return of $\omega$ to $B$ satisfies $t\_k(\omega)/k\to 1/\mu(B)$.
But by definition of $B$, we see (by splitting up the summation) that if $\omega\in B$,
$$
S\_{t\_{kN}}f(\omega)=
\sum\_{j=0}^{k-1}
S\_{t\_{(j+1)N}(\omega)-t\_{jN}(\omega)}
f(\sigma^{t\_{jN}(\omega)})>ck.
$$
This implies $S\_{t\_{kN}(\omega)}f(\omega)/t\_{kN}(\omega)=
[S\_{t\_{kN}(\omega)}f(\omega)/k] / [t\_{kN}(\omega)/k ]$ has a positive
limit superior. That contradicts the Birkhoff ergodic theorem.
|
4
|
https://mathoverflow.net/users/11054
|
410276
| 167,873 |
https://mathoverflow.net/questions/410191
|
0
|
This is a variant on the question posed [here](https://mathoverflow.net/questions/409899/is-every-matrix-involution-over-a-ufd-diagonalisable), in which the OP asks for a characterisation of the diagonalisable involutions in $\operatorname{GL}\_n(A)$, where $A$ is a $k$-algebra for some field $k$ of characteristic $\neq 2$. The accepted [answer](https://mathoverflow.net/questions/409899/is-every-matrix-involution-over-a-ufd-diagonalisable/410012#410012) shows that, if every finitely-generated projective module is free, then every involution is in fact diagonalisable.
Let us now restrict ourselves to the case where $A = k[x\_1,...,x\_l]$ is a polynomial ring over $k$. The Quillen-Suslin theorem says that all algebraic vector bundles over affine space are trivial, which translates to the condition that all finitely-generated projective modules over $A$ are free, so the solution above does apply. However, this feels to me like "killing a bee with a hand cannon" since the Quillen-Suslin theorem is quite a nontrivial result. Is there a more direct proof in this restricted setting?
As an aside, I'm hoping for a more explicit answer than just "use Quillen patching" as that will of course solve the problem, but I understand if that is the approach experts would take.
|
https://mathoverflow.net/users/175051
|
Order 2 matrices with entries in the polynomial ring over a field are diagonalisable
|
As abx pointed out in the comments, I misread the fact that this is an equivalence, so it is unlikely that there is a more elementary proof.
|
1
|
https://mathoverflow.net/users/175051
|
410280
| 167,875 |
https://mathoverflow.net/questions/410096
|
9
|
This identity came up in my research:
$$
\sum\_{m=1}^n m^2 \frac{(\frac{xy}n + m-1)\_{2m-1} (n+m-1)\_{2m-1}}{(x+m)\_{2m+1} (y+m)\_{2m+1}} = \frac{n^2}{(x^2-n^2) (y^2 - n^2)}.
$$
Here $n$ is a fixed positive integer and $x,y$ are variables. So for each $n$ this is an identity of rational functions in $x,y$. I denote by $(x)\_n$ the falling factorial (Pochhammer symbol)
$$
(x)\_n = x(x-1)\cdots (x-n+1).
$$
It may be worth pointing out that all the falling factorials that appear are of this form:
$$
(x+m-1)\_{2m-1} = x \prod\_{i=1}^{m-1} (x^2-i^2)
$$
This identity looked very foreign to me because of the quadratic argument $\frac{xy}n$ inside the falling factorial. I managed to prove it eventually, but I wonder if there is a simple/standard approach or if this is known and/or related to something interesting.
EDIT: If you prefer binomial identities, by taking the partial fraction expansions this reduces to
$$
\sum\_{m=\max(a,b)}^n \binom{2m}{m-a} \binom{2m}{m-b} \binom{\frac{ab}{n}+m-1}{2m-1} \binom{n+m-1}{2m-1} = 0
$$
for positive integers $a,b,n$ satisfying $a<n, b<n$.
|
https://mathoverflow.net/users/89514
|
Identity involving a quadratic term inside the Pochhammer symbol
|
Your identity is not just a curiosity. It is a special case of a result that has been used to obtain many quadratic and cubic identities for hypergeometric series. Perhaps the most general formulation is given by Warnaar (Summation and transformation formulas for elliptic hypergeometric series, Constr. Approx. 18 (2002), 479–502, Lemma 3.1). There it is formulated as a theta function identity. In the rational limit case, when one replaces a Jacobi theta function $\theta(x|\tau)$ by $x$, it becomes
\begin{multline\*}\sum\_{k=0}^n (a\_k^2-b\_k^2)(c\_k^2-d\_k^2)\prod\_{j=0}^{k-1}(a\_j^2-c\_j^2)(b\_j^2-d\_j^2)
\prod\_{j=k+1}^n(a\_j^2-d\_j^2)(b\_j^2-c\_j^2)\\=
\prod\_{j=0}^n(a\_j^2-c\_j^2)(b\_j^2-d\_j^2)-
\prod\_{j=0}^n(a\_j^2-d\_j^2)(b\_j^2-c\_j^2).
\end{multline\*}
Here, $a\_j$, $b\_j$, $c\_j$ and $d\_j$ are arbitrary sequences. Your identity seems to be the case $a\_j=x$, $b\_j=n+1$, $c\_j=(n+1)(j+1)/y$, $d\_j=j+1$. Since $d\_n=b\_j$, the first term on the right-hand side vanishes. Of course you also need to replace $n$ by $n-1$ and $k$ by $m-1$.
As Ira Gessel points out, this is an indefinite summation, so once you guess the identity the proof is trivial by induction on $n$.
(I removed my claim that there are typos in your formula. I missed that you use falling rather than rising factorials.)
|
6
|
https://mathoverflow.net/users/10846
|
410284
| 167,876 |
https://mathoverflow.net/questions/410234
|
3
|
I am reading K.D.Bierstedt's paper [Gewichtete Räume stetiger vektorwertiger Funktionen und das injektive Tensorprodukt. I. Journal für die reine und angewandte Mathematik 259 (1973): 186-210](https://www.degruyter.com/document/doi/10.1515/crll.1973.259.186/html). It is written in German, and, perhaps, because of that I don't understand several details, and I hope that it will be proper to ask people here to clarify this to me.
As far as I understand, Bierstedt defines the $\varepsilon$-product $X\varepsilon Y$ (up to permutation $X$ and $Y$ which will be important for the comparison with Jarchow's definition below) of locally convex spaces $X$ and $Y$ as the space of linear continuous maps
$$
\varphi:X\_p'\to Y
$$
where $X\_p'$ is the space of linear continuous functionals $f:X\to\mathbb{C}$ equipped with the *topology $p$ of uniform convergence on precompact sets $T\subseteq X$*, by which I suppose he means [totally bounded sets](https://en.wikipedia.org/wiki/Totally_bounded_space) (i.e. such $T$ that for each neighbourhood of zero $U$ there is a finite set $A$ such that $T\subseteq U+A$).
And Bierstedt endows $X\varepsilon Y$ with the topology of uniform convergence on polars $U^\circ$ of neighbourhoods $U$ of zero in $X$.
And on page 197 he states the proposition ("Satz 9(3)") which I guess sounds in English like this:
>
> If $X$ and $Y$ are complete and one of them has the approximation property, then $X\varepsilon Y$ coincides with the usual injective tensor product $X\check{\otimes}\_\varepsilon Y$:
> $$X\varepsilon Y=X\check{\otimes}\_\varepsilon Y$$
>
>
>
My first question is:
**Q1. Do I understand this correctly?**
I did not find this paper in English, and I don't speak German, that is why I have doubts.
If everything is correct, then a problem for me is that I don't understand how this is proved. Bierstedt gives a very meager explanation, which I don't understand, maybe because German is a problem for me. Everything would be more or less simple, but I stucked in the following
>
> Lemma. If $X$ and $Y$ are complete, then $X\varepsilon Y$ is complete as well.
>
>
>
And my second question is
**Q2. Is this lemma true? (And if yes, how is it proved?)**
I have H.Jarchow's book, where he proves a similar fact (Theorem 16.1.5), but he defines $X\varepsilon Y$ differently, and what he proves seems to be not equivalent to Bierstedt's statament. In the case which is interesting for me, i.e. when $X$ and $Y$ are complete, the difference, as far as I understand, is that Jarchow endows $X'$ with another topology, which he denotes by $\gamma$, and this is the *finest locally convex topology which coincides with the $X$-weak topology on polars $U^\circ$ of neighbourhoods of zero in $X$*. And according to Jarchow, the space $X\varepsilon Y$ consists of other operators, the linear continuous mappings
$$
\varphi:X\_\gamma'\to Y.
$$
So the impression is that Bierstedt's and Jarchow's $\varepsilon$-products are related to each other like this:
$$
X\varepsilon\_{\text{Bierstedt}} Y\subseteq X\varepsilon\_{\text{Jarchow}} Y
$$
I don't understand why there must be an equality here and why these two topologies $p$ and $\gamma$ on $X'$ must coincide.
Or, perhaps I miss something in this picture...
Can anybody explain to me how this problem is resolved? Why is Bierstedt's proposition true?
|
https://mathoverflow.net/users/18943
|
$\varepsilon$-product in Bierstedt's paper
|
I don't have access to Bierstedt's article right now. Nevertheless, let me try to answer your question.
0. Yes, precompact sets are also called totally bounded.
1. The lemma is true and probably this is what Bierstedt says.
2. Yes, $X\varepsilon Y$ is complete whenever so are $X$ and $Y$. The proof should be quite standard (for a Cauchy net $T\_i$ you get a pointwise limit because of the completeness of $Y$ and then you show that this limit is in the space and that the convergence is with respect to the topology of $X\varepsilon Y$).
3. This strange topology $\gamma$ on $X'$ (the finest locally convex topology which coincides with the weak$^\*$-topology on all equicontinuous sets) has a another description (this is probably due to Grothendieck, at least, it is closely related to Grothendieck's construction of the completion) which you can see, e.g., in Köthe's *Topolgical Vector Spaces I*, §21.9(7) (in my edition, page 271): It is the topology of uniform convergence on all compact subsets of the completion of $X$ (which makes sense since the completion has the same dual). This explains the difference between Bierstedt's definition and the one in Jarchow: They coincide, if every compact subset of the completion is contained in the closure of a precompact subset of $X$. This is not always the case but, trivially, it holds for complete spaces which is, of course, the most important case.
I prefer Bierstedt's definition because the topology of precompact convergence seems more natural to me than $\gamma$.
**Edit.** Both definitions differ from the one of Laurent Schwartz (in his *Théorie des Distribution à Valeurs Vectorielles*) who endowed $X'$ with the topology of uniform convergence on *absolutely convex compact sets* -- Bierstedt denotes this original $\varepsilon$-product as $X\tilde\varepsilon Y$. Two advantages are the symmetry $X\tilde\varepsilon Y\cong Y\tilde\varepsilon X$ (the isomorphism is given by the transposition of operators) and $X\tilde\varepsilon \mathbb C\cong X$. Neither of the modifications of Bierstedt and Jarchow satisfies this because $X\varepsilon \mathbb C \cong \tilde X$ (the completion of $X$) but $\mathbb C\varepsilon X\cong X$. For complete spaces $X$ all three definitions coincide.
|
3
|
https://mathoverflow.net/users/21051
|
410289
| 167,878 |
https://mathoverflow.net/questions/410212
|
4
|
Let $X$ be a manifold and $E\to X$ a complex vector bundle and let's work in $H^\bullet(X,\mathbb{Z})$. Given the total Chern class of $E$, $c(E)=1+c\_1(E)+\cdots+c\_n(E)$, we can define the total Segre class of $E$ to be $s(E)=1+s\_1(E)+\cdots+s\_n(E)$ to be the inverse of the total Chern class.
Equivalently, recalling that $c(E)=\prod\_{i=1}^n1+\alpha\_i(E)$, with $\alpha\_i(E)$ the $i$-th Chern root of $E$, we can define $$s(E)=\prod\_{i=1}^n\frac{1}{1+\alpha\_i(E)}=\prod\_{i=1}^n\sum\_{k=0}^\infty (-\alpha\_i(E))^k.$$
I am interested in the real analogue, now working in $H^\bullet(X,\mathbb{F}\_2)$, that is to say the following:
given $V\to X$ a real vector bundle, the total Stiefel-Whitney class of $V$ is $$w(V)=1+w\_1(V)+\cdots+w\_n(V)=\prod\_{i=1}^n1+\sigma\_i(V)$$ (where the $\sigma\_i(V)$ are the Stiefel-Whitney roots of $V$) and we can analogously define its inverse as $$p(V)=\prod\_{i=1}^n\frac{1}{1+\sigma\_i(V)}=\prod\_{i=1}^n\sum\_{k=0}^\infty\sigma\_i(V)^k=1+p\_1(V)+\cdots+p\_n(V).$$
This all seems quite natural but I have never seen it anywhere, the only sources I could find are [1] Stiefel-Whitney Homology Classes and Riemann-Roch Formula of Matsui and Sato and [2] Axioms for Stiefel-Whitney Homology Classes of Some Singular Spaces of Veljan, but they work in a different, more general context.
To conclude, I'm asking if these classes have been studied and what else is known about them.
My main concern is the integral of $p\_n(V)$ with [1] and [2] suggesting that $\int\_Xp\_n(V)=\chi(V)\operatorname{mod} 2$, but I would appreciate more sources.
|
https://mathoverflow.net/users/148223
|
Real analogue of Segre classes
|
Unfortunately $\int\_Xp\_n(V)\neq\chi(V) \operatorname{mod} 2$ in general. In particular $\int\_Xp\_n(TX)=0$ and it can be shown as follows.
By the multiplicativity of the Stiefel-Whitney classes, given two bundles $V, W$ such that $V\oplus W=\mathbb{R}^d$, then $p\_k(V)=w\_k(W)$. It is now easy to show that $p\_n(TX)=0$ since any $n$-manifold can be embedded into $\mathbb{R}^{2n}$. Calling $\nu$ the normal bundle we have $p\_n(TX)=w\_n(\nu)=0$, with the last equality given by Corollary 11.4 in Characteristic Classes of Milnor and Stasheff.
I still don't know about the general case.
|
2
|
https://mathoverflow.net/users/148223
|
410294
| 167,879 |
https://mathoverflow.net/questions/410293
|
4
|
In Girard's $\Pi^1\_2$-logic, a *dilator* $D$ is a endofunctor which commutes with pull-back and direct limit on $\mathrm{ON}$, the category whose objects are ordinals and morphisms are strictly increasing functions. For dilator $D\_0,D\_1$, an *embedding* from $D\_0$ to $D\_1$ is a natural transformation from $D\_0$ to $D\_1$.
**My question**
Is a following statement true?
>
> If dilators $D\_0,D\_1$ are bi-embedable, that is there is embeddings $T\_0\colon D\_0\Rightarrow D\_1, T\_1\colon D\_1\Rightarrow D\_0$, then $D\_0=D\_1$.
>
>
>
I think this statement is true because the fact embeddings of dilators is equal to injective homomorphism when dilators are considered as structures. However I can't check this fact because the paper is written in French….
*Girard, Jean-Yves; Ressayre, Jean Pierre*, Elements de logique $\Pi^1\_n$, Recursion theory, Proc. AMS-ASL Summer Inst., Ithaca/N.Y. 1982, Proc. Symp. Pure Math. 42, 389-445 (1985). [ZBL0573.03029](https://zbmath.org/?q=an:0573.03029).
|
https://mathoverflow.net/users/149565
|
Are bi-embeddable dilators equal?
|
No. For example consider the dilator $D$ that maps a well-order $A$ to the well-order consisting of denotations
1. $(2n;x,y)$, where $y<\_Ax$ are elements of $A$;
2. $(2n+1;x)$, where $x\in A$.
The denotations are compared by lexicographical order. For a morphism $f
\colon A\to B$ we as usual put $D(f)((m;x\_1,\ldots,x\_k))=(m;f(x\_1),\ldots,f(x\_k))$. Now consider the dilator $D'$ that omits from $D$ all the denotations of the form $(0;x,y)$. Clearly, $D$ and $D'$ are bi-embeddable, but not equal.
|
5
|
https://mathoverflow.net/users/36385
|
410296
| 167,880 |
https://mathoverflow.net/questions/410295
|
2
|
For any real random variable $X$, define
$$\|X\|\_{2,1}=\int\_0^\infty \sqrt{\Pr(|X|>t)}dt.$$
This quantity (it is not a norm) appears in various problems, e.g. the multiplier central limit theorem (see, e.g., Section 2.9 in [this book](https://link.springer.com/book/10.1007/978-1-4757-2545-2)) or in L-statistics (see, e.g., [this paper](https://arxiv.org/abs/1910.07572)). Problem 2.9.1 of the book cited above mentions the inequality $\|X\|\_{2,1}^2\ge E(X^2)/4$. I think we have actually better. For all $x\ge 0$,
$$\|X\|\_{2,1} \ge \int\_0^x \sqrt{\Pr(|X|>t)}dt\ge x \sqrt{\Pr(|X|>x)},$$
which implies that
$$\begin{array}{rcl}
\|X\|\_{2,1}^2 & = & \int\_0^\infty \|X\|\_{2,1} \sqrt{\Pr(|X|>x)}dx \\
& \ge & \int\_0^\infty x \Pr(|X|>x)dx \\
& =& E[X^2]/2.
\end{array}$$
My question is: is this bound sharp (I don't think it is)? If not, what is the best constant in the inequality?
|
https://mathoverflow.net/users/174236
|
Comparison between $\|X\|_2$ and $\|X\|_{2,1}$
|
$$\begin{aligned}
\|X\|\_{2,1}^2&=\int\_0^\infty\int\_0^\infty ds\,dt\,\sqrt{P(|X|>s)}\sqrt{P(|X|>t)} \\
&\ge\int\_0^\infty\int\_0^\infty ds\,dt\,P(|X|>\max(s,t)) \\
&=E\int\_0^\infty\int\_0^\infty ds\,dt\,1(|X|>\max(s,t)) \\
&=E|X|^2=EX^2.
\end{aligned}$$
So, we have an improvement of your bound. Moreover, the lower bound $EX^2$ on $\|X\|\_{2,1}^2$ is exact: it is attained when $P(|X|=c)=1$ for some real $c$.
|
7
|
https://mathoverflow.net/users/36721
|
410303
| 167,882 |
https://mathoverflow.net/questions/410302
|
2
|
If $G=(V,E)$ is a simple, undirected graph, $C\subseteq V$ is said to be a *vertex cover* if for every $e\in E$ we have $C\cap e \neq \emptyset.$
If $G=(V,E) $ is infinite, is there necessarily a vertex cover $C\_0\subseteq V$ of $G$ such that for every $v\in C\_0$ we have that $C\_0\setminus\{v\}$ is no longer a vertex cover of $G$?
|
https://mathoverflow.net/users/8628
|
Minimal vertex-covering set
|
Yes: take an inclusion-maximal independent set (exists by Zorn lemma) and pass to a complement.
|
1
|
https://mathoverflow.net/users/4312
|
410306
| 167,883 |
https://mathoverflow.net/questions/409941
|
2
|
Let $ M $ be a smooth manifold.
Recall that a manifold $ M $ is smooth homogeneous if there exists a Lie group acting transitively on $ M $.
Recall that a manifold $ M $ is Riemannian homogeneous if it admits a metric with respect to which the isometry group is transitive, moreover this metric can always be chosen to have nonnegative curvature.
And recall that a manifold $ M $ is a linear group orbit if there exists a representation $\pi:G \to GL(V) $ and a vector $v \in V$ such that the orbit of $ v$
$$
\mathcal{O}\_v:=\{ \pi(g)v:g \in G\}
$$
is diffeomorphic to $ M $.
The fundamental group of a linear group orbit is always finite by abelian (i.e. has finite commutator subgroup):
[How bad can $\pi\_1$ of a linear group orbit be?](https://mathoverflow.net/questions/206618/how-bad-can-pi-1-of-a-linear-group-orbit-be)
And the fundamental group of a Riemannian homogeneous space is also always finite by abelian (i.e. has finite commutator subgroup):
<https://www.uni-muenster.de/imperia/md/content/theoretische_mathematik/diffgeo/mr1783960.pdf>
This condition on the fundamental group holds in both cases for essentially the same reason. In both cases the manifold $ M $ is the total space of a vector bundle (the vector bundle is trivial if $ M $ is Riemannian homogeneous see [noncompact Riemannian homogeneous is trivial vector bundle over compact homogeneous](https://mathoverflow.net/questions/410334/riemannian-homogeneous-always-trivial-vector-bundle-over-compact-homogeneous) but possibly nontrivial if $ M $ is a linear group orbit) over a compact Riemannian homogeneous base $ B $. And thus $ M $ deformation retracts onto $ B $. And a quotient of compact groups always has $ \pi\_1 $ with finite commutator subgroup (<https://math.stackexchange.com/questions/4321106/transitive-action-by-compact-lie-group-implies-almost-abelian-fundamental-group/4359177#4359177>).
Are the following three properties equivalent?
* $ \textbf{(1)} $ $ M $ is Riemannian homogeneous
* $ \textbf{(2)} $ $ M $ is a linear group orbit
* $ \textbf{(3)} $ $ M $ is smooth homogeneous and $ \pi\_1(M) $ has finite commutator subgroup
How about if we assume $ M $ compact? In other words, are the following three (actually four I added one) properties equivalent:
* $ \textbf{(cc)} $ $ M $ admits a transitive action by a compact Lie group (so a fortiori is compact)
* $ \textbf{(1c)} $ $ M $ is compact and Riemannian homogeneous
* $ \textbf{(2c)} $ $ M $ is compact and a linear group orbit
* $ \textbf{(3c)} $ $ M $ is compact and smooth homogeneous and $ \pi\_1(M) $ has finite commutator subgroup
|
https://mathoverflow.net/users/387190
|
Riemannian homogeneous equivalent to linear group orbit
|
$ \textbf{(1)} \implies \textbf{(2)} $
Up to diffeomorphism a Riemannian homogeneous space is just a trivial vector bundles over a compact Riemannian homogeneous space ([noncompact Riemannian homogeneous is trivial vector bundle over compact homogeneous](https://mathoverflow.net/questions/410334/riemannian-homogeneous-always-trivial-vector-bundle-over-compact-homogeneous)). Compact Riemannian homogeneous spaces are linear group orbits ([Compact linear group orbit equivalent to linear compact group orbit](https://mathoverflow.net/questions/409511/compact-linear-group-orbit-equivalent-to-linear-compact-group-orbit/409532#409532)) and vector spaces are linear group orbits and cartesian products of linear group orbits are linear group orbits so $ M $ Riemannian homogeneous implies $ M $ a linear group orbit
$ \textbf{(2)} $ is strictly weaker. The tangent bundle of $ S^2 $ is nontrivial so cannot be Riemannian homogeneous. However it arises as a linear group orbit by consider the orbit of the vector $ (1,0,0) \in \mathbb{C}^3 $ with respect to the standard representation of $ SO\_3(\mathbb{C}) $.
$ \textbf{(2)} \implies \textbf{(3)} $
See [How bad can $\pi\_1$ of a linear group orbit be?](https://mathoverflow.net/questions/206618/how-bad-can-pi-1-of-a-linear-group-orbit-be)
$ \textbf{(3)} $ is strictly weaker. The Moebius band has abelian $ \pi\_1 $ and a transitive action by $ SE\_2 $ but is not a linear group orbit (I think, actually that claim is open here <https://math.stackexchange.com/questions/4359525/is-it-possible-to-realize-the-moebius-strip-as-a-linear-group-orbit> but at least we can be sure that the Moebius strip is not Riemannian homogeneous since it is a nontrivial vector bundle)
$ \textbf{(cc)} \implies \textbf{(1c)} $
Since $ M $ admits a transitive action by a compact group $ K $ then $ M $ is compact and $ K $ can always be taken to act by isometries ( by pushing forward the biinvariant metric on $ K $ to a left invariant metric on $ M $).
$ \textbf{(1c)} \implies \textbf{(2c)} $
Since $ M $ is compact then $ K=Iso(M) $ is compact so $ M $ is a quotient of the compact group $ K $ and thus by Mostow-Palais $ M $ is a linear group orbit.
$ \textbf{(2c)} \implies \textbf{(cc)} $
If $ M $ is assumed compact and $ M $ is a linear group orbit of some group $ G $ then the maximal compact subgroup of $ G $ acts transitively see [Compact linear group orbit equivalent to linear compact group orbit](https://mathoverflow.net/questions/409511/compact-linear-group-orbit-equivalent-to-linear-compact-group-orbit/409532#409532) . Thus we have
$$
\textbf{(cc)}=\textbf{(1c)}=\textbf{(2c)}
$$
Now for
$ \textbf{(cc)}=\textbf{(1c)}=\textbf{(2c)} \implies \textbf{(3c)} $
The three equivalent conditions above imply that $ M $ is compact, admits a transitive action by a Lie group, and has finite by abelian fundamental group, see either of the first two links in the question or see <https://math.stackexchange.com/questions/4321106/transitive-action-by-compact-lie-group-implies-almost-abelian-fundamental-group>
$ \textbf{(3c)} \implies \textbf{(cc)}=\textbf{(1c)}=\textbf{(2c)} $
True for dimension 2. True for dimension 3. Not sure in general!
|
0
|
https://mathoverflow.net/users/387190
|
410309
| 167,884 |
https://mathoverflow.net/questions/410275
|
3
|
Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?
My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).
However thanks to Ben McKay for the argument that $ M $ does not have a transitive isometry group. Also this fact is stated in
"Isometries of 2-Dimensional Riemannian Manifolds into Themselves" by
Sumner Byron Myers.
|
https://mathoverflow.net/users/387190
|
Is the Moebius strip Riemannian homogeneous?
|
Suppose by contradiction it is. Write it as $G/K$ where $G$ is the identity component of the isometry group and $K$ is compact, and $G$ acts faithfully on $G/K$. Since $G$ is connected, maximal compact subgroups are connected. Since $G/K$ is not contractible, $K$ is not maximal compact and it follows that maximal compact subgroups have codimension $1$. Since in noncompact simple Lie groups, maximal compact subgroups have codimension $\ge 2$, it follows that $G$ has no noncompact simple factor.
Suppose by contradiction that $G$ is not solvable. Then it has a simple compact connected subgroup $S$. Let $K'$ be a maximal compact subgroup. Then $K'\cap S$ has codimension $\le 1$ in $S$. Since a simple compact group has no subgroup of codimension $1$, it follows that $S\subset K'$. In turn, since $K$ has codimension $1$ in $K'$, by the same argument we deduce that $S\subset K$. Hence $S$ is contained in the intersection of all conjugates of $K$, which is trivial. Contradiction.
So $G$ is a connected solvable Lie group. Let $M$ be a closed connected normal subgroup of codimension $1$. If $MK=G$ then $M$ has smaller dimension and acts transitively, so we can argue by induction. Hence, assuming that $G$ has minimal dimension, we have $MK\neq G$, so $K\subset M$. Since in an abelian connected Lie group the intersection of all codimension 1 closed connected subgroups is trivial, we deduce that $K\subset \overline{[G,G]}$. The latter is nilpotent, and hence $K$ is a central torus in $\overline{[G,G]}$. Since the action of $G$ on the maximal torus of $\overline{[G,G]}$ is trivial, we deduce that $K$ is central, hence trivial. So $G$ is a 2-dimensional Lie group, and in particular is orientable.
(We have proved that if a noncompact connected surface can be endowed with a homogeneous Riemannian metric, then it is diffeomorphic to a Lie group, and hence to the plane, the cylinder. Of course various approaches to this result exist and some have already been mentioned in the comments.)
|
2
|
https://mathoverflow.net/users/14094
|
410312
| 167,885 |
https://mathoverflow.net/questions/410315
|
1
|
Let $(X, d)$ be a uniformly discrete metric space of bounded geometry, that is, $\sup\_{x \in X} |B\_r(x)| < \infty$ for every $r \geq 0$ and there is a uniform $\delta > 0$ such that $d(x, y) \geq \delta$ for all $x \neq y \in X$.
Let $G$ be the so-called *wobbling group of $X$*, that is, the set of all bijections $f \colon X \rightarrow X$ such that
$$ \sup\_{x \in X} d(x, f(x)) < \infty. $$
**Question:** is $G$ countable?
|
https://mathoverflow.net/users/147609
|
Countability of the wobbling group of a bounded geometry metric space
|
Say that a metric space $X$ is *hyperdiscrete* if the distance map $X\times X\to\mathbf{R}\_{\ge 0}$ is a proper map, i.e. for every $R$, the set of $(x,y)$ such that $d(x,y)\le R$ is finite. For instance, with the Euclidean distance, $\mathbf{N}$ is not hyperdiscrete, but the set $\{0,1,4,\dots\}$ of squares in $\mathbf{N}$ is hyperdiscrete. The terminology is from my book with P. de la Harpe, Example 3.A.14(7) ([ArXiv link](https://arxiv.org/abs/1403.3796), [EMS link](https://www.ems-ph.org/books/book.php?proj_nr=209)).
Let $X$ be a proper discrete metric space (i.e., every bounded subset is finite— note this forces $X$ to be countable) and $W(X)$ its wobbling group (= permutations of $X$ at bounded distance to identity).
>
> **Proposition.** Equivalent statements:
>
>
> 1. $W(X)$ is countable
> 2. $W(X)$ has cardinal $<$ continuum
> 3. $W(X)$ is reduced to finitely supported permutations
> 4. $X$ is hyperdiscrete.
>
>
>
Suppose (4) fails. Then there exists an injective sequence $(x\_n)\_{n\ge 0}$ with $d(x\_n,x\_0)$ tending to infinity, such that $d(x\_{2n},x\_{2n+1})$ bounded. Then there is an injective homomorphism from the group $(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$ into $W(X)$, mapping $(a\_n)$ the permutation exchanging $x\_{2n}\leftrightarrow x\_{2n+1}$ whenever $a\_n=1$, and identity elsewhere. So (2) fails. Hence (2) implies (4).
The implications (3)$\Rightarrow$(1)$\Rightarrow$(2) are trivial.
Finally, if $X$ is hyperdiscrete and $f$ is a permutation at distance $\le R$ to identity, then since $\{(x,y):d(x,y)\le R\}$ is finite, its projection to $X$ is finite, and $f$ has to be identity elsewhere, so $f$ is finitely supported. Hence (4) implies (3).
|
5
|
https://mathoverflow.net/users/14094
|
410319
| 167,887 |
https://mathoverflow.net/questions/410305
|
5
|
The basic question is:
Does vanishing of homology with trivial coefficients imply triviality of an infinite-dimensional Lie algebra?
My question is motivated by acylic groups in group theory. In particular, there is a an acylic group $\textrm{Aut}\_f(\mathbb{R})$ of autohomeomorphisms of $\mathbb{R}$ with finite support.
In the same vein is it true that Lie algebra of finite vector fields on $\mathbb{R}$ is acyclic? I know that the cohomology of all vector fields is described by Goncharova's theorem, but I don't know the answer for finite vector fields (although I have seen that the answer is mentioned as known somewhere).
This is an extension of a [post on MSE](https://math.stackexchange.com/questions/4325519/finite-dimensional-lie-algebras-with-trivial-homology).
|
https://mathoverflow.net/users/143549
|
Infinite dimensional Lie algebras with trivial homology
|
There exist acyclic infinite-dimensional Lie algebras, i.e., for which the trivial homology vanishes in all nonzero degree (recall that $H\_0$ of every Lie algebra is always 1-dimensional over the ground field $K$).
>
> **Lemma.** Let $\mathfrak{g}$ be the increasing union of Lie subalgebras $\mathfrak{g}\_n$. Suppose that for fixed $k$, we have $H\_k(\mathfrak{g}\_n)=0$ for all $n$ large enough (say $n\ge N$). Then $H\_k(\mathfrak{g})=0$.
>
>
>
Proof: this is essentially formal "diagram chasing". If $b\in Z\_k(\mathfrak{g})\subset\bigwedge^k\mathfrak{g}$ (spaces of $k$-cycles and $k$-chains), there exists $n$, which we can choose $\ge N$, such that $b\in \bigwedge^k\mathfrak{g}\_n$. Then $d\_k(b)=0$, so $b\in Z\_k(\mathfrak{g}\_n)$. Since $H\_k(\mathfrak{g}\_n)=0$, there exists $c\in\Lambda^{k-1}\mathfrak{g}\_n$ such that $d\_{k-1}(c)=b$. Hence $b\in B\_k(\mathfrak{g})$ (space of $k$-boundaries). This proves the lemma.
>
> **Corollary.** Under the same assumptions, if for some sequence $(N\_k)$, we have $H\_k(\mathfrak{g}\_n)=0$ for all $n\ge N\_k$ and all $k\ge 1$, then $H\_k(\mathfrak{g})=0$ for all $k$. $\Box$
>
>
>
So it is enough to have such an "increasing sequence of Lie algebras" $(\mathfrak{g}\_n)$ at disposal. We can't choose them finite-dimensional (since $H\_1$ and $H\_3$ [can't be both zero then](https://math.stackexchange.com/a/4327464/35400), at least in char. zero). However there are examples in the literature:
For instance choose $\mathfrak{g}\_n$ as the Lie algebra of formal power series vector fields $$\sum\_{i=1}^nf\_i\frac{\partial}{\partial x\_i},\quad f\_i\in K[\![x\_1,\dots,x\_n]\!].$$
Gelfand and Fuks proved in [1] (see [2], Corollary 1) that $H\_k(\mathfrak{g}\_n)=0$ for all $1\le k\le n$.
References:
[1] I. M. Gel′fand and D. B. Fuks, Cohomologies of the Lie algebra of formal vector fields, Izv. Akad. Nauk SSSR Ser. Mat. 34 (1970), 322–337 (Russian). MR 0266195
[2] Victor Guillemin and Steven Shnider, Some stable results on the cohomology of the classical infinite-dimensional Lie algebras, Trans. Amer. Math. Soc. 179 (1973), 275-280.
[Link at AMS site, unrestricted access](https://www.ams.org/journals/tran/1973-179-00/S0002-9947-1973-0327856-3/)
|
8
|
https://mathoverflow.net/users/14094
|
410338
| 167,892 |
https://mathoverflow.net/questions/410339
|
0
|
Let $d$ be a positive integer, and suppose $c$ is an integer such that $\gcd(c,d) = 1$. Then the following identity holds:
$$\displaystyle \left \lvert \{b \pmod{d} : b^2 \equiv c \pmod{d} \}\right \rvert = \sum\_{\substack{\chi \pmod{d} \\ \chi^2 = \chi\_0}} \chi(c).$$
What is the correct analogue for the right hand side when we replace the left hand side with
$$\displaystyle \left \lvert \{b \pmod{d} : b^k \equiv c \pmod{d}\}\right \rvert$$
with $k \geq 3$?
|
https://mathoverflow.net/users/10898
|
Analogues of an identity involving quadratic characters
|
If $\gcd(c,d) = 1$ we have
$$\displaystyle \left \lvert \{b \pmod{d} : b^k \equiv c \pmod{d} \}\right \rvert = \sum\_{\substack{\chi \pmod{d} \\ \chi^k = \chi\_0}} \chi(c).$$
This is not really anything to do with numbers - it works in any finite abelian group, and here we are applying it to the multiplicative group of invertible residue classes mod $c$.
It's a consequence of orthogonality of characters for the group $G/ G^k$.
|
7
|
https://mathoverflow.net/users/18060
|
410341
| 167,893 |
https://mathoverflow.net/questions/410323
|
4
|
I have the following exact sum for the expectation of an event
$$\sum\_{m=0}^{nk} \sum\_{j=1}^n (-1)^{j-1}\binom{n}{j} \binom{(n-j)k}{m} / \binom{nk}{m}$$
which is exactly correct but I want to give an upper bounding approximation that is easier to interpret. In particular, I want to see the expectations dependence on the parameter $k$. In simulation, it appears that as $k \rightarrow \infty$ we have that the sum tends towards $n \log n$. However, I am unsure how to show such a result analytically. My attempt has been to use some form of a stirling approximation on the binomial coefficients like
$$\binom{n}{k} \leq \frac{n^k}{k!}$$
which can reduce the summation in the following way
$$\sum\_{m=0}^{nk} \sum\_{j=1}^n (-1)^{j-1}\binom{n}{j} \frac{((n-j)k)^m}{m!} \cdot \frac{m!}{(nk)^m} = \sum\_{m=0}^{nk} \sum\_{j=1}^n (-1)^{j-1}\binom{n}{j} \left(1-\frac{j}{n}\right)^m$$
and from here maybe we can upper bound the $(1-j/n)^m$ to simplify further to a sum which depends only on $j$? Again comparing to simulation, it seems that when $k \ll \log n$ the sum will be something like $nk$ and for $k \gg \log n$ we will have the $n \log n$ (thus removing the dependence on $k$) but I can't seem to derive such a result.
Any help or advice would be *greatly* appreciated!
|
https://mathoverflow.net/users/214997
|
Approximating binomial coefficient sum
|
Actually we may simply compute this sum (and I guess that your expectation may be computed differently to give the answer in the below simplified form).
We start with $1/{nk\choose m}=(nk+1)\int\_0^1 x^m(1-x)^{nk-m}dx$ by the Beta function $B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$ formula. Next, we sum up over $m$ with fixed $j$ to get
$$
\sum\_{m=0}^{nk}\frac{(n-j)k\choose m}{nk\choose m}=(nk+1)\int\_0^1 (1-x)^{jk}\sum{(n-j)k\choose m}x^m(1-x)^{(n-j)k-m}dx\\=(nk+1)\int\_0^1(1-x)^{jk}\left(x+(1-x)\right)^{(n-j)k}dx=(nk+1)\int\_0^1(1-x)^{jk}dx=
(nk+1)\int\_0^1 t^{jk}dt.
$$
Then sum up over $j$ to get
$$
\sum\_{m=0}^{nk} \sum\_{j=1}^n (-1)^{j-1}\binom{n}{j} \binom{(n-j)k}{m} / \binom{nk}{m}=(nk+1)\int\_0^1\sum\_{j=1}^n(-1)^{j-1}{n\choose j}(t^k)^jdt\\=(nk+1)\int\_0^1(1-(1-t^k)^n)dt=(nk+1)\left(1-\frac1k\int\_0^1(1-s)^ns^{1/k-1}ds\right)\\=(nk+1)\left(1-\frac{\Gamma(n+1)\Gamma(1/k)}{k\Gamma(n+1/k+1)}\right)=(nk+1)\left(1-\frac{1\cdot 2 \cdot 3\cdot \ldots n}{(1+1/k)(2+1/k)\ldots (n+1/k)}\right).
$$
Now your observation follows, for example, from
$$
\frac{\ell}{\ell+1/k}=\left(\frac{\ell}{\ell+1}\right)^{1/k}\cdot e^{O(1/(\ell^2 k))},
$$
thus multiplying against $\ell=1,\ldots,n$ we get
$$
\frac{1\cdot 2 \cdot 3\cdot \ldots n}{(1+1/k)(2+1/k)\ldots (n+1/k)}=\frac{e^{O(1/k)}}{(n+1)^{1/k}}=\exp\left(-\frac{\log n+O(1)}k\right),
$$
therefore in the $\log n=o(k)$ regime using $1-\exp(-t)\sim t$ for small $t=\frac{\log n+O(1)}k$ we indeed get $n\log n$ asymptotics for your sum.
|
8
|
https://mathoverflow.net/users/4312
|
410344
| 167,896 |
https://mathoverflow.net/questions/410334
|
2
|
Is it true that a manifold $ E $ admits a metric with respect to which the isometry group is transitive **($ E $ is Riemannian homogeneous) if and only if $ E $ is the total space of a $ K $ equivariant vector bundle** where $ K $ is a compact group acting transitively on the base $ B $ of the bundle?
If the vector bundle $ E $ is not assumed equivariant then this fails. The Moebius strip is the total space of a vector bundle but it is not Riemannian homogeneous.
This question has a slightly similar flavor to:
[Homogeneity of a projective vector bundle](https://mathoverflow.net/questions/322350/homogeneity-of-a-projective-vector-bundle)
|
https://mathoverflow.net/users/387190
|
noncompact Riemannian homogeneous is trivial vector bundle over compact homogeneous
|
The only if direction fails. That is, there are $K$-equivariant vector bundles which are not homogeneous. For example, the Mobius band has the form $O(2)\times\_{O(1)} \mathbb{R}$, and is not Riemannian homogeneous as you mention.
On the other hand, it seems the if direction is true. In fact, I think I can prove that $M$ must be diffeomorphic to the trivial bundle over a compact homogeneous space, at least if $M$ is connected.
Let $G$ denote the identity component of the isometry group (which still acts transitively). Fix a point $p\in M$ and let $G\_p$ denote the isotropy group. Then $G\_p$ is compact. This follows because the action is proper (see, e.g, [this paper](https://arxiv.org/pdf/0811.0547.pdf)), and then $G\_p\times \{p\}\subseteq G\times M$ is the inverse image of the compact set $\{(p,p)\}\subseteq M\times M$ under the map $G\times M\rightarrow M\times M$ given by $(g,m)\mapsto(gm,m)$. Moreover, because the action is proper, we have a diffeomorphism $M\cong G/G\_p$.
Writing $H$ for the maximal compact subgroup of $G$, we therefore have (up to conjugacy) inclusions $G\_p\subseteq H\subseteq G$. From this, one can form the fiber bundle $H/G\_p\rightarrow G/G\_p\rightarrow G/H$.
Now, $G/H$ is diffeomorphic to Euclidean space, which is contractible. Thus, this bundle is trivial, so $G/G\_p$ is diffeomorphic to $(H/G\_p)\times (G/H)$.
|
1
|
https://mathoverflow.net/users/1708
|
410347
| 167,898 |
https://mathoverflow.net/questions/410333
|
10
|
It is known that a random series
$$
\sum\_{n\geq 1} X\_n
$$
whose terms $X\_n$ are independent converges a.s. if and only if it converges in probability.
Is it true that a martingale $(Y\_n)$ converges a.s. if and only if it converges in probability? If not, are there any counter-examples? Thanks.
|
https://mathoverflow.net/users/20302
|
Martingales converging in probability but not a.s
|
$\newcommand{\N}{\mathbb N}\newcommand{\si}{\sigma}\newcommand{\F}{\mathcal{F}}\newcommand{\Om}{\Omega}\newcommand{\Z}{\mathbb{Z}}$A counterexample can be obtained as follows. Let $T\_1,T\_2,\dots$ be independent (say) geometrically distributed random variables with fast growing means, say with $ET\_i=2^i$.
Let $Y\_n=0$ for $0\le n\le T\_1$. For time moments $n$ after that, let the values of $Y\_n$ coincide with the positions of a simple random walk $W^{(1)}\_\cdot$ starting from $0$ at time $T\_1$ -- but only till the time, say $\nu\_1$, of the first return of $W^{(1)}\_\cdot$ to $0$. This is the first "step".
After this, let $Y\_\cdot$ stay at $0$ for time $T\_2$. After that, let the values of $Y\_n$ coincide with the positions of a simple random walk $W^{(2)}\_\cdot$ starting in state $0$ at time $\nu\_1+T\_2$ -- but only till the time, say $\nu\_2$, of the first return of $W^{(2)}\_\cdot$ to $0$, where $W^{(2)}\_\cdot$ is independent of $W^{(1)}\_\cdot$. This is the second "step".
Continue doing such "steps" indefinitely (assuming that the walks are independent of the $T\_j$'s). Thus, we get a martingale $(Y\_n)$, with respect to a certain filtration of $\sigma$-algebras. The walks occur increasingly rarely; they start at very uncertain times (with standard deviations $\asymp2^i$ and corresponding very flat distributions); and the walks last comparatively short times. Therefore, for any given large time moment $n$, $P(Y\_n\ne0)$ is small. So, $Y\_n\to0$ in probability.
However, because the walks occur infinitely many times, clearly $Y\_n\not\to0$ almost surely.
---
Here are formal details.
**Construction:** Let $T\_1,T\_2,\dots$ be independent geometrically distributed random variables (r.v.'s) (defined on some probability space $(\Om,\F,P)$) with means $ET\_i=1/p\_i$, where $0<p\_i<1/2$ and
\begin{equation\*}
\sum\_{i\in\N}p\_i^{1/2}<\infty. \tag{1}
\end{equation\*}
So,
\begin{equation\*}
P(T\_i=t)=p\_i q\_i^{t-1}\,1(t\in\N) \tag{2}
\end{equation\*}
for real $t$, where $q\_i:=1-p\_i$.
Let $R\_1,R\_2,\dots$ be independent Rademacher r.v.'s (defined on the same probability space $(\Om,\F,P)$) that are independent of the $T\_i$'s; so, $P(R\_i=1|(T\_j))=1/2=P(R\_i=-1|(T\_j))$ for all $i\in\N$.
Define (the starting times of the walks) $S\_1,S\_2,\dots$ and (the finishing times of the walks) $F\_0,F\_1,F\_2,\dots$ by the following recursion:
\begin{equation\*}
F\_0:=0
\end{equation\*}
and, for $k\in\N$,
\begin{equation\*}
S\_k:=F\_{k-1}+T\_k,\quad F\_k:=\inf\{n\in\N\colon n>S\_k,W^{(k)}\_n=0\}, \tag{3}
\end{equation\*}
where
\begin{equation\*}
W^{(k)}\_n:=\sum\_{i\in\N}R\_i\,1(S\_k<i\le n). \tag{3a}
\end{equation\*}
Note that $F\_0=0<\infty$ and $F\_k=\min\{n\in\N\colon n>S\_k,W^{(k)}\_n=0\}<\infty$ for all $k\in\N$ almost surely (a.s.), because the simple random walk is recurrent. So, $S\_k<\infty$ for all $k\in\N$ a.s.
Now, for all $n\in\N$ let
\begin{equation\*}
Y\_n:=\sum\_{k\in\N}1(S\_k<n\le F\_k)W^{(k)}\_n. \tag{4}
\end{equation\*}
Note that a.s. at most one summand in (4) is nonzero, since $S\_k=F\_{k-1}+T\_k>F\_{k-1}$ a.s. for all $k\in\N$.
To complete the construction, for each $n\in\N$ let
\begin{equation\*}
\F\_n:=\si(R\_1,\dots,R\_n,\{1(T\_k\le j)\colon k\in\N,j\in\N,j\le n\}),
\end{equation\*}
the $\si$-algebra generated by $R\_1,\dots,R\_n,\{1(T\_k\le j)\colon k\in\N,j\in\N,j\le n\}$. Clearly, $(\F\_n)\_{n\in\N}$ is a filtration.
**Showing that $((Y\_n,\F\_n))\_{n\in\N}$ is a martingale:**
Abusing notation as is commonly done, for a r.v. $X$ let us write $X\in\F\_n$ to mean that $X$ is $\F\_n$-measurable. Then obviously $1(F\_0\le n)=1(0\le n)\in\F\_0:=\{\Om,\emptyset\}\subseteq\F\_1$ for all $n\in\N$. Using now the relations
\begin{equation\*}
1(S\_k\le n)=\sum\_{j=1}^{n-1}1(T\_k=j)1(F\_{k-1}\le n-j),
\end{equation\*}
$1(T\_k=j)=1(T\_k\le j)-1(T\_k\le j-1)$, and
\begin{equation\*}
\{F\_k\not\le n\}=
\bigcap\_{m=1}^n\Big(\{S\_k\not\le m-1\}\cup\Big\{\sum\_{i=1}^m R\_i\,1(S\_k\le i-1)\ne0\Big\}\Big)
\end{equation\*}
for natural $k,j$ (which follow by (3) and (3a)), we conclude by induction on $k$ that
\begin{equation\*}
\{S\_k\le n\}\in\F\_n,\quad \{F\_k\le n\}\in\F\_n \tag{5}
\end{equation\*}
for all natural $k,n$.
Hence, by (4) and (3a), the sequence $(Y\_n)\_{n\in\N}$ is adapted to the filtration $(\F\_n)\_{n\in\N}$. Moreover, by (4), for all $n\in\N$ we have
\begin{equation\*}
Y\_{n+1}-Y\_n=\sum\_{k\in\N}R\_{n+1}\,1(S\_k\le n,F\_k\not\le n);
\end{equation\*}
also, $R\_{n+1}$ is independent of $\F\_n$. So, in view of (5), $((Y\_n,\F\_n))\_{n\in\N}$ is a martingale.
**Showing that $Y\_n\to0$ in probability (as $n\to\infty$):** By (4),
\begin{equation\*}
1(Y\_n\ne0\}=\sum\_{k\in\N}1(S\_k<n\le F\_k) \tag{6}
\end{equation\*}
and hence
\begin{equation\*}
P(Y\_n\ne0\}=\sum\_{k\in\N}P(S\_k<n\le F\_k). \tag{7}
\end{equation\*}
Here and in what follows, $n$ and $k$ are natural numbers.
Next, consider the duration
\begin{equation\*}
D\_k:=F\_k-S\_k
\end{equation\*}
of the $k$th walk. Since the $R\_i$'s are independent and independent of the $T\_j$'s,
the r.v.'s $D\_1,D\_2,\dots$ are independent and independent of the $T\_j$'s, and the $T\_j$'s are also independent. So, the r.v.'s $T\_1,D\_1,T\_2,D\_2,\dots$ are independent.
Also,
\begin{equation\*}
S\_k=T\_1+\sum\_{j=2}^k(D\_{j-1}+T\_j). \tag{8}
\end{equation\*}
So, $S\_k$ and $D\_k$ are independent and hence
\begin{equation\*}
P(S\_k<n\le F\_k)=\sum\_{d\in\N}P(D\_k=d)P(S\_k<n\le S\_k+d).
\end{equation\*}
Using (8) and the independence of $T\_1,D\_1,T\_2,D\_2,\dots$, for all $d\in\N$ we have
\begin{equation\*}
P(S\_k<n\le S\_k+d)=
P(n-d\le S\_k<n)\le\sup\_{m\in\Z} P(m-d\le T\_k<m)
=\sum\_{t=1}^d p\_kq\_k^{t-1}.
\end{equation\*}
So,
\begin{equation\*}
\begin{aligned}
P(S\_k<n\le F\_k)&\le\sum\_{d\in\N}P(D\_k=d)\sum\_{t=1}^d p\_kq\_k^{t-1} \\
&=\sum\_{t\in\N}^d p\_kq\_k^{t-1}\sum\_{d=t}^\infty P(D\_k=d) \\
&=\sum\_{t\in\N}^d p\_kq\_k^{t-1}P(D\_k\ge t) \\
&=\sum\_{t\in\N}^d p\_kq\_k^{t-1}P(D\_1\ge t) \\
&\le\sum\_{t\in\N} p\_kq\_k^{t-1}ct^{-1/2},
\end{aligned}
\end{equation\*}
since, by (say) the reflection principle, $P(D\_1\ge t)\le ct^{-1/2}$ for $t\in\N$. Here and in what follows, $c$ denotes various universal positive real constants. Therefore and because $0<p\_k<1/2$,
\begin{equation\*}
\begin{aligned}
P(S\_k<n\le F\_k)&\le 2c\sum\_{t\in\N} p\_k(1-p\_k)^t t^{-1/2} \\
&\le 2cp\_k \sum\_{t\in\N} e^{-p\_k t}t^{-1/2} \\
&\le 2cp\_k \int\_0^\infty e^{-p\_k t}t^{-1/2}\,dt =cp\_k^{1/2}.
\end{aligned}
\tag{9}
\end{equation\*}
Also, for each $k\in\N$, $P(S\_k<n\le F\_k)\le P(F\_k\ge n)\to0$ (as $n\to\infty$), since $F\_k<\infty$ a.s. So, by (7), for each $K\in\N$,
\begin{equation\*}
\limsup\_n P(Y\_n\ne0\}\le \limsup\_n \sum\_{k=K}^\infty cp\_k^{1/2}.
\end{equation\*}
Letting now $K\to\infty$ and recalling (1), we see that $\limsup\_n P(Y\_n\ne0\}=0$ and hence $Y\_n\to0$ in probability (as $n\to\infty$).
**Noting that $Y\_n\not\to0$ a.s. (as $n\to\infty$):** It follows (say by (6)) that $|Y\_{S\_k+1}|=1$ for all $k\in\N$. Also, $S\_k\to\infty$ a.s. as $k\to\infty$. So, $Y\_n\not\to0$ a.s. (as $n\to\infty$).
|
11
|
https://mathoverflow.net/users/36721
|
410350
| 167,899 |
https://mathoverflow.net/questions/410346
|
13
|
It is well-known that every closed, connected and orientable 3-manifold $\mathcal{M}$ can uniquely be decomposed as
$$\mathcal{M}=P\_{1}\#\dots\# P\_{n}$$
where $P\_{i}$ are prime manifolds, i.e. manifolds which can not be written as a non-trivial connected sum of two 3-manifolds.
>
> **My main question is the following:** Is there are similar result for
> 3-dimensional compact, connected and orientable manifolds with
> connected boundary?
>
>
>
I know that in this case there are two types of connected sums, namely:
* the *internal* connected sum, which is defined as the connected sum of manifolds without boundary by choosing balls, which are purely in the interior. This also includes the hybrid case, i.e. the connected sum of a manifold with boundary with a manifold without boundary and
* the *boundary* connected sum, where one choses balls on the boundaries and glues them together.
In any case, can any 3-manifold with boundary (with the properties written above) be decomposed in some kind of connected sum?
>
> **As a second, very related question:** If I fix the boundary to be some compact, oriented surface of arbitrary genus, is there some kind
> of classification/decomposition result classifying all manifolds
> having a boundary homeomorphic to our chosen boundary surface?
>
>
>
(This question was previously posted to MathStackExchange)
|
https://mathoverflow.net/users/259525
|
Classification of 3-dimensional manifolds with boundary
|
With regard to the first question: There are a couple of versions of unique decompositions for 3-manifolds with boundary, with respect to boundary connected sum. See Gross, Jonathan L. [The decomposition of 3-manifolds with several boundary components](https://www.ams.org/journals/tran/1970-147-02/S0002-9947-1970-0258047-X/S0002-9947-1970-0258047-X.pdf).
Trans. Amer. Math. Soc. 147 (1970), 561–572, and Swarup, G. Ananda
[Some properties of 3-manifolds with boundary](https://watermark.silverchair.com/21-1-1.pdf?token=AQECAHi208BE49Ooan9kkhW_Ercy7Dm3ZL_9Cf3qfKAc485ysgAAAtgwggLUBgkqhkiG9w0BBwagggLFMIICwQIBADCCAroGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMZEUGhHgvLLIZivQuAgEQgIICi50hLPr_v6maxuVF689cH4ib3d-7moKJLPNO_Q6FN6U4asTbGSylvz_VWiaa2ua4yF9naozOsg4WoczKWpbf5tcUZ1-iwEUWYp95NAsrNpEEyxNdpSlRokkzbP6dXTH-DXZE86WDEmmnXmxNHKg7H_85gbxhXg3vQkIzgkaMQ7-VtuLCtBJX6EusXpuT-astW0snaicJbmZSWKZkf2kC4iGcOmcU9Xrdw-8N2jjaZ1DSY-QkUpC7HaoQQXftH0ttFCdblg2BUt31RTfWP31Z56iftIo0RSVmDaQcIO7OkNZXsWModQihLClPIyojYNeipouXUtL9FCNmbrwYq9XE376SiKGiGpXwhiWr6VYbKp_VRHidIehY4TcaMljNrTuRsMKr69ddnqBkBMQzmBDnjOnPqUvAULJ2w5Q5-zbfenE24erE7cagTo_4tKduSAkcr5-hrwmDwOKGh5osANaLLEjS-K1TFHwO8gKi-bCi6nGRISN58qqwFPz1m-7kmizrFEcVFOumSRDLvqWOTQaInvyz83jbqj2P3P8jJKqodbZJUHbzRg80IdwYVMr0eVWiQKgIQiRp08wKSeEDlwNFc8CSNMSTEUQ9ssPBslZ10yDJG9WsCvoK00fZi8OMLyDevEfEQQYmj0hlaVALPdA6ZIszQc0_h0attOMD-e2r8SWBCE-V2YHEU_dUs6dR7YuD9ycOEvFB9SQWVvJHiywLRPe50lRksGK_odEYGLArWInDN-KZdZCiifzGsfQ-eT6hN_0u93ZKc5LXNwaOA_R-OnS9u32Vucxm9Lf751eeXza9T8G0eKraBQuJSSOz0EUasoTlN2Kp3PM8hfn9mnO5RDtHde5BklZPCQygqA). Quart. J. Math. Oxford Ser. (2) 21 (1970), 1–23.
|
9
|
https://mathoverflow.net/users/3460
|
410361
| 167,900 |
https://mathoverflow.net/questions/410357
|
7
|
A divisor $d$ of a natural integer $N$ defines a permutation
of $\{0,\ldots,N-1\}$ by considering
$$x\longmapsto \pi\_{d\vert N}(x)=\left\lfloor \frac{x}{d}\right\rfloor+\frac{N}{d}
\left( x\pmod d\right)$$
for $x \pmod d$ in $\{0,\ldots,d-1\}$.
One can show that the group $A(N)$ generated by all permutations $\pi\_{d\vert N}$
associated to divisors of $N$ is abelian. (The group
$A(N)$ is in fact generated by $\pi\_{p\vert N}$ for $p$ running through
all prime divisors of $N$.)
Example: $A(p^k)$ is cyclic of order $k$ if $p$ is prime.
It seems that $A(N)$ is fairly small:
I have no example of $N\geq 3$ such that $A(N)$ has more than $N-2$ elements.
(It is however not cyclic
in general.)
*What does the group $A(N)$ measure?*
|
https://mathoverflow.net/users/4556
|
An abelian group associated to divisors of an integer $N$
|
All the $\tau$'s fix $0$ so we can look at how they behave on the set $\{1,2,\dots,N-1\}$. We have
$$\tau\_{d|N}(kd+r)=k+\frac{rN}{d}=\left[\frac{N}{d}(kd+r)\right]\_{\pmod{N-1}}$$
where $[a]\_{\pmod{N-1}}$ is the unique integer in $\{1,2,\dots, N-1\}$ congruent to $a\pmod{N-1}$. So $\tau\_{d|N}$ is the same as multiplication by $\frac{N}{d}$ modulo $N-1$.
Therefore we see that the group $A(N)$ is the subgroup of $(\mathbb Z/(N-1)\mathbb Z)^{\times}$ generated by the divisors of $N$, therefore
$$|A(N)|\le \varphi(N-1)\le N-2$$
as predicted by the OP.
|
3
|
https://mathoverflow.net/users/2384
|
410363
| 167,901 |
https://mathoverflow.net/questions/410365
|
7
|
Let $f : X\rightarrow S$ be a flat finite type morphism of schemes with $S$ integral and Noetherian. Let $\eta\in S$ be the generic point.
Let $\{\sigma\_i\}$ be a collection of sections of $f$ (possibly infinite), which are Zariski dense in $X\_\eta$. I'm interested in additional conditions on $f,\{\sigma\_i\}$ under which one or both of the following properties are satisfied:
**P1: There exists a point $s\in S - \{\eta\}$ such that $\{\sigma\_i\}$ is dense in the fiber $X\_s$.**
**P2: There exists a nonempty open $U\subset S$ such that P1 holds for every $s\in U$.**
Clearly a necessary condition is that $S - \{\eta\}$ needs to be ``large'' (e.g., P1 will often fail if $S$ is the spec of a discrete valuation ring). From now on lets assume $S - \{\eta\}$ is Zariski dense in $S$.
My intuition is that under mild additional assumptions there should be some kind of semicontinuity result for the dimensions of the Zariski closures inside fibers. In particular, the set of $s\in S$ such that $\{\sigma\_i\}$ is not Zariski dense in $X\_s$ should be closed in $S$. However I'm not aware of any results in this direction.
Here are some specific questions:
(1) Do P1, P2 hold under the above assumptions?
(2) What if we also assume $f$ has geometrically irreducible fibers?
(3) What if $X$ is a semisimple affine algebraic $S$-group scheme?
|
https://mathoverflow.net/users/88840
|
When must a set of sections which is Zariski dense in the generic fiber also be dense in some special fiber?
|
Building on Yosemite Stan's example, for any scheme over any number ring, even if you take all the sections, it will still not be Zariski dense in any special fiber, because all sections go through the rational points of the special fiber, which are not dense as the base field is finite. Of course, there are many examples of schemes over number rings with Zariski dense global sections.
Similar examples work for $\mathbb A^1$ over any curve of dimension $1$ over a countable field. There are countably many closed points, order them, and choose the $n$th section to agree with at least one of the previous $n-1$ sections at all of the first $n-1$ closed points. Then in the $n$'th point there will be at most $n$ distinct sections, so they won't be Zariski dense.
Over an uncountable field, the same construction proves that any countable set of points can be the set where the sections aren't dense, so there will not be an open set where the sections are dense. However, one can prove there is at least one closed point where the sections are dense.
First, pick a countable subset of the sections that is Zariski dense at the generic point. Then, for every $d$, by Zariski density we can find some finite set of sections that don't satisfy any nontrivial degree $d$ equation over the generic point, and then the set where those sections do satisfy some nontrivial degree $d$ equation is contained in a proper closed subset. Over un uncountable field, the complement of the union of countably many proper closed subsets will be nonempty, and any point in that nonempty closed set does the trick.
|
7
|
https://mathoverflow.net/users/18060
|
410369
| 167,902 |
https://mathoverflow.net/questions/403877
|
1
|
I have been looking at constructions satisfying the Johnson-Lindenstrauss Lemma (e.g., projections onto random subspaces, random Gaussian matrices, random Rademacher matrices, etc.). It seems that other than the "random subspace" construction, none of the other projections are true projections in the sense that the vectors are not orthonormalized. Is this purely out of convenience/efficiency?
More formally, suppose a matrix $X$ with $X\_{ij} \sim \text{Rademacher}(1/2)$ preserves distances with high probability. Is it also true that $P\_X := X^\top(XX^\top)^{-1}X$ will also preserve distances with the same probability?
The most general version of the JL lemma that I have been able to find is [here](http://cs.brown.edu/people/mriondat/augustseminar/papers/Matousek-VariantsJohnsonLindenstrauss.pdf), which shows that any matrix with mean-zero, unit-variance, uniformly subgaussian, and independent entries satisfies the Lemma, but that clearly does not hold for $P\_X$.
|
https://mathoverflow.net/users/128729
|
Johnson-Lindenstrauss with Orthogonalization
|
Suppose $v$ has unit norm and $\|Xv\|\in[1-\epsilon,1+\epsilon]$. Then
$$\|X^\top(XX^\top)^{-1}Xv\|=\|(XX^\top)^{-1/2}Xv\|\in\bigg[\frac{1-\epsilon}{\sigma\_\max(X)},\frac{1+\epsilon}{\sigma\_\min(X)}\bigg].$$
If $X$ has subgaussian entries, then one may control its extreme singular values with standard techniques. (See [Vershynin's treatment](https://arxiv.org/abs/1011.3027), for example.) Then a union bound gives that the random projection $X^\top(XX^\top)^{-1}X$ is JL with slight loss in the distortion parameter $\epsilon$.
|
1
|
https://mathoverflow.net/users/29873
|
410371
| 167,903 |
https://mathoverflow.net/questions/410049
|
7
|
I was thinking of a way to prove [this](https://math.stackexchange.com/questions/586862/non-vanishing-vector-fields-on-non-compact-manifolds) and I realised that for my approach the lemma from the title would be useful, and it´s an interesting question on its own. Obviously it is true if the manifold is compact or $\mathbb{R}^n$, but I don´t see it in general. This is the precise question:
Let $M$ be a connected manifold and $X$ a compact inside it, then is the union of $X$ and all the relatively compact components of $M\setminus X$ compact?
P.S. I already asked this [question](https://math.stackexchange.com/questions/4322564/is-the-union-of-a-compact-and-the-relatively-compact-components-of-its-complemen) in MSE a few days ago, but after asking a few people I thought it could be more appropiate for MO.
|
https://mathoverflow.net/users/172802
|
Is the union of a compact and the relatively compact components of its complementary in a manifold compact?
|
Actually, there is an elementary proof of this. I will imitate the one given in O.Forster Lectures on Riemann Surfaces. We assume $M$ is connected. Let $Y$ be equal to the union of $X$ with all the relatively compact components of $M \setminus X$
Let $U$ be a relatively compact, open subset containing $X$ and let $C\_j$, $j \in J$ be the connected components of $M \smallsetminus X$. Let $bU$ be the boundary of $U$, which is compact and disjoint from $X$.
* Claim 1. Every $C\_j$ meets $U$: If $C\_j \subset M \setminus U$, its closure in $M$ is also contained in $M \setminus U$, but $C\_j$ is a connected component of $M \setminus U$ so $C\_j = \overline{C\_j}$ which conttradicts connectedness of $M$.
* Claim 2. Only finitely many $C\_j$ intersects $bU$: This is because $bU$ is compact and the $C\_j$ are open and disjoint, anc cover $bU$.
* Claim 3. $Y$ is closed: Let $J\_0$ consist of the indices corresponding to relatively compact components. Then $M \setminus Y = \bigcup\_{j \not \in J\_0} C\_j$, which is open.
By Claim 2, we can find $j\_1, \ldots , j\_k \in J\_0$ be such that $C\_{j\_i}$ intersects $bU$. Then , by Claim 1 again, any other $C\_j$ is contained in $U$. Therefore,
$$ Y \subset U \cup C\_{j\_1} \cup \ldots \cup C\_{j\_k}$$
RHS is relatively compact by the choice of $U$ and the $j\_i$, and LHS is closed by Claim 3, so LHS is compact too.
|
3
|
https://mathoverflow.net/users/176397
|
410372
| 167,904 |
https://mathoverflow.net/questions/410377
|
0
|
I'm a newbie in the field of mathematical research and I'm not able to find the following paper:
>
> " M. Radić, A definition of the determinant of a rectangular matrix,
> (Serbo-Croatian summary) Glasnik Mat. Ser. III 1(21) (1966), 17–22 "
>
>
>
Please could you provide me a link to this paper.
|
https://mathoverflow.net/users/470930
|
Where can I find the following math paper?
|
You can find the paper at this [link](https://books.google.it/books?id=5z0GkLQ6LfgC&pg=PA17&redir_esc=y&hl=it#v=onepage&q&f=false). The table of contents is at the [journal site](https://web.math.pmf.unizg.hr/glasnik/).
|
4
|
https://mathoverflow.net/users/19520
|
410381
| 167,905 |
https://mathoverflow.net/questions/410330
|
7
|
Consider $f\_n(x) = \min\_{|z|=x} \Re \sum\_{j=1}^{n} \frac{z^j}{j}$, a real function of positive variable $x>0$.
I am interested in lower bounds on $f\_n(x)$. Specifically, I ask: what lower bounds can be given on $f\_n(x)$ in the regime where $x=1+O(1/n)$ and $n$ tends to $\infty$?
Trivially, $f\_n(x) \ge - \sum\_{j=1}^{n} \frac{x^j}{j}$ which has order of magnitude $-\log n$ (times a constant) in the aforementioned regime. However, equality cannot be achieved here, since $z^j$ cannot equal to $-x^j$ for both $j=1$ and $j=2$. I don't even know if $f\_n(x)$ is eventually negative.
|
https://mathoverflow.net/users/31469
|
Asymptotics of truncated logarithm on a cricle
|
So, we have a function $u\_n(z) = \Re \sum\_{j = 1}^n \frac{z^j}{j}$. As a real part of an analytic function, it is a harmonic function. We are interested in its behaviour on the circle $|z| = x = 1 + \frac{c}{n}$, where $c > 0$ is some constant and we want to estimate the minimal value of $u\_n$ on it. Let me start with the upper bound from my comment:
$u\_n$ is a harmonic function, thus it satisfies the minimum principle in the form $$\min\_{|z| \le x} u\_n(x) = \min\_{|z| = x} u\_n(x)$$ (if you only heard of the maximum principle, then it is it applied to $-u\_n$).
Thus, we have $f\_n(x) \le u\_n(-1)$. On the other hand $u\_n(-1)\to -\log(2)$, so $\limsup f\_n(x) \le -\log(2)$ (in reality, $u\_n(-1) \le -\log(2) + \frac{1}{n}$, so actually $f\_n(x) \le -\log(2) + \frac{1}{n}$).
Now for the lower bound. According to the Lemma 2 from the [link](https://math.dartmouth.edu/%7Ecarlp/pvrev.pdf) provided by Lucia, we have $f\_n(1) \ge -\log(2) - \frac{2}{n}$. For $z$ with $|z| = x$ let us denote $w = \frac{z}{|z|}$. We have
$$u\_n(z) \ge u\_n(w) - |u\_n(z) - u\_n(w)| \ge -\log(2) - \frac{2}{n} - |u\_n(z)-u\_n(w)|.$$
So, if we can give a uniform upper bound on $|u\_n(z)-u\_n(w)|$ then we get a uniform lower bound on $f\_n(x)$. We have
$$u\_n(z) - u\_n(w) = \sum\_{j = 1}^n \frac{z^j-w^j}{j} = \sum\_{j=1}^n w^j \frac{x^j-1}{j}.$$
Since $x = 1 + \frac{c}{n}$ for $j\le n$ we have $1\le x^j\le 1 + \frac{Cj}{n}$ for some $C$ depending on $c$ (something like $C = e^c$ should work). Plugging this in and taking the absolute values (remember that $|w| = 1$) we get
$$|u\_n(z) - u\_n(w)| \le \sum\_{j = 1}^n \frac{C}{n} = C,$$
which proves the desired bound. Note though that the final constant $D$ in $f\_n(x) \ge -D$ depends on $c$ even if we are concerned with only the big values of $n$. I don't know if it can be made approaching $\log(2)$ (it definitely can't be smaller by the minimum principle argument from above).
|
2
|
https://mathoverflow.net/users/104330
|
410384
| 167,908 |
https://mathoverflow.net/questions/410337
|
5
|
I’ve considered the diffusion equation $$\frac{\partial f(x,t)}{\partial t}=\frac12 \frac{\partial^2 f(x,t)}{\partial x^2}$$ with the conditions $f(x,0)=\delta(x)$ and $f(-1,t)=f(1,t)=0\ \forall t>0$ and I’ve found the solution $$f(x,t)=\sum\_{n=0}^\infty \cos\left[ \left( n+\frac12 \right)\pi x \right] e^{-\frac12 (n+1/2)^2\pi^2 t}$$then I’ve considered the function$$\Lambda(t)=-\frac d{dt}\int\_{-1}^1f(x,t)dx=\sum\_{n=0}^\infty (-1)^n \left(n+\frac12\right)\pi\ e^{-(n+1/2)^2\pi^2 t/2}=\\=\frac\pi2\left(e^{-\frac{\pi^2} 8t}-3\,e^{-\frac{9\pi^2} 8t}+5\,e^{-\frac{25\pi^2} 8t}-\ldots\right)=\frac \pi 4 \vartheta\_1'(0,e^{-\pi^2 t/2})$$where $\vartheta\_1'(u,q)=\dfrac{\partial}{\partial u}\vartheta\_1(u,q)$ and $$\vartheta\_1(u,q)=2\,q^{1/4}\sum\_{n=0}^\infty (-1)^n q^{n(n+1)}\sin\left[\left(2n+1\right)u \right]$$is a Jacobi theta function.
It turns out that $$\lim\_{t\to0^+}\frac{\Lambda(t)}{g(t)}=1$$ where $g(t)=\sqrt{\dfrac2{\pi t^3}}e^{-\frac1{2t}}$ (so *Mathematica* says and the approximation is very good as the values of $\Lambda$ and $g$ differ for less than $1\%$ for $0<t<0.7$), but I don’t know how to prove it.
Thanks in advance for your help.
|
https://mathoverflow.net/users/470349
|
Approximation for a series involving the derivative of a Jacobi theta function
|
Write your $\Lambda$ as
$$\Lambda(t)=\frac{1}{2}\sum\_{-\infty}^\infty f(n),$$
where
$$f(x)=\pi(x+1/2)\sin\pi(x+1/2)\exp\left(-\pi^2(x+1/2)^2t/2\right)=y\sin y\,e^{-ty^2/2},$$
where $y=\pi(x+1/2),$ and $t>0$.
Then use [Poisson's summation formula](https://en.wikipedia.org/wiki/Poisson_summation_formula)
$$\sum\_{-\infty}^\infty f(n)=\sum\_{-\infty}^\infty \hat{f}(2\pi n),$$
where
$$\hat{f}(s)=\int\_{-\infty}^\infty f(x)e^{-isx}dx$$
is the Fourier transform. This Fourier transform can be explicitly computed:
$$\hat{f}(s)=\frac{1}{\sqrt{2\pi}}t^{-3/2}e^{is/2}\left((s/\pi+1)e^{-(s/\pi+1)^2/(2t)}-(s/\pi-1)e^{-(s/\pi-1)^2/(2t)}\right).$$
The trick is that in the Fourier transform your parameter $t$ will stand in the denominator of the exponent, so the series $\sum\hat{f}(n)$ will be asymptotic to the sum of $3$ terms (with $n=0,\pm1$), z when $t\to 0$,
$$\sum\_{-\infty}^\infty\hat{f}(2\pi n)\sim \hat{f}(0)+\hat{f(2\pi)}+\hat{f}(-2\pi)\sim 2\hat{f}(0)=2\sqrt{\frac{2}{\pi t^3}}e^{-1/(2t)},\quad t\to 0+,$$
and since $\Lambda(t)$ is $1/2$ of this sum, we obtain
the answer that wrote.
(In the computation of Fourier transform, I started with the $e^{-y^2/(2t)}$, and then used the transformation rules
of Fourier transform: multiplied my function on $y$, then on $\sin y$, and then scaled by $\pi$ and added $1/2$ to the argument).
Remark. This approximation is related to the famous computation of the [age of Earth](https://www.math.purdue.edu/~eremenko/dvi/ageofearth.pdf) by Lord Kelvin. Roughly speaking, the series corresponds to the exact solution for the heat equation inside the spherical Earth, while the asymptotics corresponds to the
[flat Earth approximation](https://www.math.purdue.edu/~eremenko/dvi/ageofearth2.pdf). It is unclear from his papers on the subject, whether Kelvin knew the exact solution and Poisson's formula.
|
5
|
https://mathoverflow.net/users/25510
|
410411
| 167,919 |
https://mathoverflow.net/questions/410164
|
7
|
Let $M$ be a strongly causal Lorentzian manifold. If $M$ has dimension 4, a theorem of [Hawking, King, and McCarthy](https://aip.scitation.org/doi/abs/10.1063/1.522874) (see Thm 5) says that $M$ is determined up to conformal isomorphism by its class of null geodesic curves (where the parameterization of the null geodesic is forgotten) and thence by the causal relation $J^+$ on $M$.
**Question:** Does this theorem also hold in dimensions other than 4?
My hunch would be that it continues to hold in higher dimensions, but in lower dimensions I'm not so sure. At any rate, I don't see how to adapt Hawking, King, and McCarthy's argument (which involves choosing coordinates using a certain configuration of null geodesics) to dimension 2.
**EDIT:** As came up in the comments, I should clarify that a priori I don't want to assume anything about the smooth structure on $M$. In fact, I would prefer not to assume anything about the topology either -- just take $M$ as a set of points, equipped with the relation $J^+$. From this data, when can one recover the topological, smooth, and conformal structures on $M$?
|
https://mathoverflow.net/users/2362
|
In which dimensions is a strongly causal Lorentzian manifold determined conformally by its causal structure?
|
Trying to recover as much of the topology/geometry from the causal order as possible has been studied quit a bit since the early paper of Hawking et al that you cite. A quick summary of my understanding of the situation is that there are some purely order-theoretic topologies (like Alexander or Scott topologies) that reproduce the topology of the original Lorentzian manifold when the causal relation is sufficiently non-pathological. However, it is very difficult to find a set of conditions on partially ordered sets (posets) that single out precisely those posets that come from the causal order of smooth (or even topological) Lorentzian manifolds. AFAIK, that is an open problem . (Please correct me if I'm wrong!)
You can find a relatively recent collection of results in that direction here:
>
> *Martin, Keye; Panangaden, Prakash*, [**A domain of spacetime intervals in general relativity**](http://dx.doi.org/10.1007/s00220-006-0066-5), Commun. Math. Phys. 267, No. 3, 563-586 (2006). [ZBL1188.83071](https://zbmath.org/?q=an:1188.83071) [arXiv:gr-qc/0407094](https://arxiv.org/abs/gr-qc/0407094)
>
>
>
One can start walking up and down the citation tree from here to find more recent progress.
**Added:** If one restricts attention to generalizing just Thm 5 of Hawking-King-McCarthy to other dimensions, as requested by Tim, then this was done in Thm 1.2 here
>
> *Peleska, Jan*, [**A characterization for isometries and conformal mappings of pseudo- Riemannian manifolds**](http://dx.doi.org/10.1007/BF02192656), Aequationes Math. 27, 20-31 (1984). [ZBL0539.53017](https://zbmath.org/?q=an:0539.53017).
>
>
>
All dimension higher than 2 are covered, which agrees with Willie's counter example.
|
3
|
https://mathoverflow.net/users/2622
|
410412
| 167,920 |
https://mathoverflow.net/questions/410410
|
1
|
A rooted tree is a tree with a distinguished root node. When a rooted tree is embedded in a plane, a cyclic ordering is induced on the subtrees of the root. Such trees are called rooted plane trees.
Given a tree $T$, is there an algorithm that can output a rooted plane tree $T\_r$ isormorphic to $T$ such that every node $v$ except the root is labelled with a number $f(v)$ so that $v$ is the $f(v)$-th child of its pararent?
What's the complexity?
|
https://mathoverflow.net/users/148974
|
Construct a rooted plane tree with nodes labelled
|
Easy answer: Denote the root by $(0)$. Denote its $a\geq 0$ children by $(0,1),(0,2),\ldots,(0,a)$.
Denote recursively the $b$ children of a vertex $(0,a\_1,\ldots,a\_{k-1})$ at height $k-1$ by $(0,a\_1,\ldots,a\_{k-1},1),(0,a\_1,\ldots,a\_{k-1},2),\ldots,(0,a\_1,\ldots,a\_{k-1},b)$.
Draw the vertex $(0,a\_1,\ldots,a\_{k-1},a\_k)$ at the point $(a\_k,k)$ of $\mathbb N^2$ and join it to its ancestor at $(a\_{k-1},k-1)$ by an edge.
The answer to your question is then the function $(0,\ldots,a\_k)\longmapsto a\_k$.
The tree constructed above yields a tree which is immerged in the plane but not embedded (there is an overlay of all first descendants of vertices of given height). If you want to embed
the tree, you have to replace the first coordinate $a\_k$ of $(a\_k,k)$ by the number of
vertices $(0,b\_1,\ldots,b\_k)$ which are
lexicographically smaller than $(0,\ldots,a\_k)$.
|
1
|
https://mathoverflow.net/users/4556
|
410417
| 167,923 |
https://mathoverflow.net/questions/410421
|
3
|
Let $\mathbb{G}$ be a $C^\*$-algebraic compact quantum group. Consider the associated dense Hopf$^\*$-subalgebra $\mathcal{O}(\mathbb{G})$ and let $S: \mathcal{O}(\mathbb{G})\to \mathcal{O}(\mathbb{G})$ denote its antipode. In the book "Compact quantum groups and their representation categories", it is shown that $S$ is unbounded when $\mathbb{G}$ is not of Kac-type. The converse is also true, namely if $\mathbb{G}$ is of Kac-type, then $S$ is bounded. To see this, note that the assumption implies that $S$ is $\*$-preserving, and thus $S$ is a positive map because
$$S(x^\*x) = S(x)S(x^\*) = S(x)S(x)^\*\ge 0$$
for any $x \in \mathcal{O}(\mathbb{G})$. Note that $\mathcal{O}(\mathbb{G})$ is an operator system, and since a positive unital map on an operator system is necessarily bounded (See Paulsen's book "Completely bounded maps and operator algebras", proposition 2.1), we conclude that $S$ is bounded. Moreover, it also follows that $\|S\| \le 2\|S(1)\| = 2.$
**Question**: Can we say something more about the norm $\|S\|$? Is it possible that $\|S\| = 2?$
|
https://mathoverflow.net/users/216007
|
Norm antipode on a Kac-type compact quantum group
|
$S$ is an anti-$\*$-homomorphism, and extends by continuity to $A = C(\mathbb G)$ (the closure of $\mathcal O(\mathbb G)$ acting on the GNS space for the Haar state). Let $A^{\operatorname{op}}$ be the opposite $C^\*$-algebra to $A$. Then we can consider $S$ as a map $A\rightarrow A^{\operatorname{op}}$, which is now a $\*$-homomorphism, and hence contractive.
The relevant result in the book of Neshveyev and Tuset is Proposition 1.7.9. This also shows that (equivalently) $S^2=\operatorname{id}$. So in fact $S$ is an isometry.
|
3
|
https://mathoverflow.net/users/406
|
410426
| 167,926 |
https://mathoverflow.net/questions/410387
|
7
|
Let $\mathcal{M}$ be a compact, connected, oriented 3-manifolds with non-empty connected boundary $\partial\mathcal{M}$. Then, following [this article](https://www.ams.org/journals/tran/1970-147-02/S0002-9947-1970-0258047-X/S0002-9947-1970-0258047-X.pdf), it is stated that $\mathcal{M}$ can be written as
$$\mathcal{M}=P\_{1}\#\_{\partial}\dots\#\_{\partial}P\_{n}$$
where $\#\_{\partial}$ denotes the boundary connected sum and where $P\_{i}$ are $\partial$-prime manifolds, i.e. 3-manifolds with non-empty connected boundary, which are not homeomorphic to a 3-ball and for which a decomposition like $P=Q\_{1} \#\_{\partial}Q\_{1}$ implies that either $Q\_{1}$ or $Q\_{2}$ is a closed 3-ball. So this is basically a generalization of the famous prime decomposition ("Kneser-Milnor theorem") to the case of manifolds with boundary.
Now, lets say I only consider manifolds $\mathcal{M}$ with the property that $\partial\mathcal{M}$ is homeomorphic to the 2-torus $T^{2}=S^{1}\times S^{1}$. Then, if I understand the theorem above correctly, $\mathcal{M}$ has to be a prime-manifold: Suppose that $\mathcal{M}$ can be decomposed as $\mathcal{M}=P\_{1}\#\_{\partial}P\_{2}$ for two prime manifolds. However, the boundary connected sum has the property that $\partial\mathcal{M}=(\partial P\_{1})\# (\partial P\_{2})$, which is not possible in our case, since $\partial\mathcal{M}=T^{2}$ and the 2-torus cannot be obtained as the connected sum of two other manifolds.
What I am wondering is the following:
>
> Is every compact, connected, oriented 3-manifold $\mathcal{M}$ with non-empty connected boundary $\partial\mathcal{M}\cong\_{\mathrm{homeo.}} T^{2}$ of the form $$\mathcal{M}\cong\_{\mathrm{homeo.}}\overline{T^{2}}\#\mathcal{N},$$ where $\overline{T^{2}}=S^{1}\times D^{2}$ denotes the solid torus and where $\mathcal{N}$ is a closed, orientable and connected 3-manifold. The connected sum here is the internal one.
>
>
>
(This is a follow-up question to this [MathOverflow post.](https://mathoverflow.net/questions/410346/classification-of-3-dimensional-manifolds-with-boundary) )
|
https://mathoverflow.net/users/259525
|
Decomposition of manifolds with toroidal boundary
|
The situation is similar to the one of closed manifolds. One defines "boundary-prime" manifolds as those that cannot be decomposed nontrivially in a boundary-connected sum.
Note that if $M$ is connected, has nonempty boundary and is not prime, then $M$ is never boundary-prime. Namely, take a 2-sphere $S\subset M$ separating $M$ into two components none of which is a ball. Connect $S$ to $\partial M$ by a 1-handle $D^2\times [0,1]$ ($D^2\times \{0\}\subset S$, $D^2\times \{1\}\subset \partial M$). Now, remove from $S$ the open disk equal to $int(D^2)\times \{0\}$ and add to $S$ the annulus $\partial D^2\times [0,1]$. The resulting surface $S'$ is a 2-dimensional disk with $\partial S'\subset \partial M$. This disk will cut $M$ in two components, none of which is a ball.
This works no matter what $\partial M$ is, in particular, if $\partial M=T^2$.
Thus, **in what follows (until the concluding paragraph), I will assume that $M$ is prime.**
Lemma. If $\partial M$ is a torus, then $M$ is necessarily boundary-prime.
Proof. Let $D\subset M$ be a properly embedded disk splitting $M$ in two components, none of which is a 3-ball. (Splitting means that you remove from $M$ an open tubular neighborhood of $D$.) Since $D$ separates $M$, $\partial D$ separates $T^2=\partial M$, hence, bounds a disk $D'\subset T^2$. Taking the union $D\cup D'$ we obtain a non-properly embedded 2-sphere in $M$. Pushing the disk $D'$ slightly into $M$, we obtain a 2-sphere $S\subset M$ disjoint from the boundary. Since $D$ was splitting $M$ into two submanifolds none of which is a 3-ball, the same holds for $S$. Hence, $M$ is not prime, contradicting the standing assumption. qed
We continue the discussion of prime manifolds $M$ with toral boundary.
Such a manifold still can have compressible boundary. However, if this is the case, a boundary-compressing $D$ disk in $M$ is necessarily nonseparating. (Since every separating loop in the torus $T^2$ bounds a disk in $T^2$.)
Cutting $M$ open along $D$ results in a manifold $M'$ with spherical boundary. If $M'$ is homeomorphic to the 3-ball, then $M$ itself is a solid torus, $\hat{T}=D^2\times S^1$. Otherwise, attaching $B^3$ along $\partial M'$ results in a closed 3-manifold $N$ which is not $S^3$. Then $M= N\# \hat{T}$. But there is one more possibility, namely, $\partial M$ is incompressible. There are many manifolds like that, for instance, the exterior of any nontrivial knot in $S^3$.
Hence, we obtain the trichotomy for 3-manifolds $M$ which need not be $\partial$-prime.
To conclude: Suppose that $M$ is a connected (not necessarily prime) 3-manifold and $\partial M$ is homeomorphic to $T^2$. Then one of the following mutually exclusive properties holds:
1. $M$ is not prime, equivalently, is not $\partial$-prime.
2. $M=\hat{T}$.
3. $M$ is prime and $\partial M$ is incompressible.
|
5
|
https://mathoverflow.net/users/39654
|
410427
| 167,927 |
https://mathoverflow.net/questions/410358
|
3
|
>
> Theorem(Guillemin Sternberg Marle) Let $(M, \omega, \mu) $ be a symplectic manifold together with a Hamiltonian group action. Let $p$ be a point in $M$ such that $O\_p $ is contained in the zero level set of the moment map. Denote $G\_p$ the stabilizer and $O\_p$ the orbit of $p$. There is a $G$-equivariant symplectomorphism from a neighborhood of the zero section of the bundle $T^\*G ×\_{G\_p} V\_p$ equiped with a symplectic model to a neighborhood of the orbit $O\_p$.
>
>
>
I'm looking for a reference of this theorem. ( The references that people recommend are the articles of Marle which is written in french and the article The normal form for the moment map by Guillemin and Sternberg which I can't download from internet). Could you please give some references where I can find the proof of this theorem.
|
https://mathoverflow.net/users/172459
|
Proof of the Hamiltonian slice theorem
|
The following textbooks contain a proof of the symplectic slice theorem:
* [Juan-Pablo Ortega and Tudor S. Ratiu: Momentum Maps and Hamiltonian Reduction](https://link.springer.com/book/10.1007/978-1-4757-3811-7)
* [Gerd Rudolph and Matthias Schmidt: Differential Geometry and Mathematical Physics](https://link.springer.com/book/10.1007/978-94-007-5345-7)
You might also have a look at my [PhD thesis](https://arxiv.org/abs/1909.00744). One of the main results is a symplectic slice theorem in infinite dimensions. Thus there also a lot of analytical questions that need to be answered, so maybe not the best point to start learning this topic.
|
5
|
https://mathoverflow.net/users/17047
|
410429
| 167,928 |
https://mathoverflow.net/questions/410438
|
12
|
For any $a = [a\_0; \dots; a\_n]\in \mathbb{P}^n(\mathbb{Q})$, the corresponding Galois group $G\_a$ of $f(X) = a\_n X^n + \cdots + a\_1 X + a\_0\in \mathbb{Q}[X]$ is a subgroup of $S\_n$. (I'm interested primarily in the case where $f$ is irreducible here, but we can take $G\_a = \text{Aut}(k/\mathbb{Q})$ for the splitting field $k$ of $f$ regardless.) For a given group $G \subset S\_n$, what exactly can be said about the size of the set of points $a$ with $G\_a = G$? The group $G = S\_n$, for example, occurs exactly when the resolvent of $f$ is irreducible, and this happens for $a$ in a Zariski-dense set by Hilbert's irreducibility theorem. Generalizing from that, is there a specific sense in which $A(X) = \{a\in \mathbb{P}^n(\mathbb{Q}):\, G\_a\subset X\}$ has $A(H)\subset A(G)$ "small" in $A(G)$ for $H < G$? That is, is there a result invoking some notion like the dimension of a variety (although we're necessarily working over $\mathbb{Q}$ here, or at least some other ground field that isn't algebraically closed), a thin set in the sense of Serre, etc. that makes this vague idea of "smallness" more precise?
|
https://mathoverflow.net/users/61829
|
Structure of coefficients of polynomials giving a specified Galois group
|
Every Galois group is Zariski dense, so that's not a very exciting measurement. In fact, for any degree n field, minimal polynomials of generators of that field are Zariski dense. This follows from the fact that the map taking an element to its minimal polynomial is finite-to-one, thus sends Zariski dense sets to Zariski dense sets. (Or, more generally, for any rank n étale algebra, characteristic polynomials of nondegenerate elements are Zariski dense.)
Typically the more precise question one asks is the number of such polynomials of height up to $H$, or really the asymptotics as a function of $H$. Taking Weil height, this is equivalent to asking the number of polynomials $\sum\_{i=0}^n a\_i X^i \in \mathbb Z[x]$ with a given Galois group, where the $a\_i$ are relatively prime and $|a\_i|<H$. (In other words, the Weil height is the maximum coefficient, after clearing denominators and removing common factors.)
This is almost an active research question. The almost is because researchers usually set $a\_n=1$. For this problem, the best results known are [in the recent work of Bhargava](https://arxiv.org/abs/2111.06507), and according to that paper the same methods work for the varying $a\_n$ case, but you will not find exact estimates for that version in the paper.
The basic shape is that, for each Galois group, the number of such polynomials should be proportional to $H$, with the actual power varying greatly with the choice of Galois group, and only upper and lower bounds known for most groups rather than the exact figure.
|
16
|
https://mathoverflow.net/users/18060
|
410439
| 167,931 |
https://mathoverflow.net/questions/410436
|
2
|
In Iwasawa theory, one of the fundamental results is the following structure theorem for finitely generated modules over the ring $\Lambda = \mathbf{Z}\_p[[T]]$.
>
> If $M$ is a finitely generated torsion $\Lambda$-module, then there is a pseudo-isomorphism
> $$M \to \left( \bigoplus\_{i=1}^s \Lambda/(p^{n\_i}) \right) \oplus \left( \bigoplus\_{j=1}^t \Lambda/(f\_i) \right) $$
> where the $f\_i \in \Lambda$ are distinguished polynomials.
>
>
>
**My question is the following:** are the numbers $s$ and $t$ in the above theorem uniquely determined by $M$? That is, is the number of direct summands in the above decomposition uniquely determined by $M$?
As to *why* I'm asking this, I'm in a situation where I have a finitely generated torsion $\Lambda$ module $M$ and I happen to know that $M$ is generated by $n$ elements $x\_1, \dots, x\_n$. (**EDIT**: I also know that the elements $\{x\_1,...,x\_n\}$ form a *minimal* generating set, so no strict subset of $\{x\_1,...,x\_n\}$ generates all of $M$.) I'd then like to use this to conclude that there are $n$ summands in the direct-sum composition above (i.e: that $s+t=n$). To do this, however, one needs to know that $s$ and $t$ are uniquely determined by $M$, which is why I'm asking this question.
Any tips would be appreciated. Thanks!
|
https://mathoverflow.net/users/394740
|
Structure theorem for finitely generated $\Lambda$-modules - uniqueness part
|
By Nakayam's lemma, since $M$ is finitely generated, $x\_1,\dots, x\_n$ generate $M$ if and only if they generate $M/ (p, T)M$ (where $(p,T)$ is the maximal ideal of the local ring $\mathbb Z\_p[[T]]$.)
So $x\_1,\dots, x\_n$ are a minimal generating set if and only if they are a basis of $M/ (p,T)M$, and thus in this case $n$ is the rank of $M/(p,T)M$.
If your pseudo-isomorphism were an isomorphism, you'd be done at this point, as that rank would be $s+t$. However, for a general torsion Iwasawa module, we can have, for example $s=t=0$ with $M$ and thus its minimal generating set nonempty, as happens when $M$ is finite.
So you need some additional condition on $M$.
|
1
|
https://mathoverflow.net/users/18060
|
410446
| 167,934 |
https://mathoverflow.net/questions/410393
|
9
|
Let $A$ be a $C^\*$ algebra of operators acting on some Hilbert space $H$, and $A\_0$ is a norm dense $\*$-subalgebra of $A$. Suppose there exists some unit vector $\xi \in H$, such that (i) $A\_0 \xi$ is dense in $H$; (ii) $a\xi = 0$ if and only if $a=0$ for all $a \in A\_0$, i.e. $\xi$ is separating for $A\_0$.
**Question**: is $\xi$ also separating for $A$, i.e. is it true that $a \xi = 0$ implies $a=0$ for all $a \in A$?
It is clear that if the vector state $\omega\_\xi$ of $\xi$ is a trace on $A$, then the answer is affirmative by (i). More generally, if $\omega\_\xi$ is a KMS state (so that we have sufficient control of $\omega\_\xi$ from being away from a trace using suitable automorphisms of $A$), then the answer is still affirmative. But in general, I don't know the answer, and suspect that there is a counter-example.
|
https://mathoverflow.net/users/128540
|
Can one detect a cyclic and separating vector for a concrete $C^*$-algebra using a dense subalgebra?
|
Here's a counter-example. Take $A\_0:=\mathbb{C}[F(s,t)]\subset A:=\mathrm{C}^\*\_{\mathrm{r}}(F(s,t))$, where $F(s,t)$ is the free group on $\{s,t\}$, $E\colon A\to \mathrm{C}^\*\_{\mathrm{r}}(F(s))$ the canonical conditional expectation, and $\psi\colon \mathrm{C}^\*\_{\mathrm{r}}(F(s)) \cong C(\mathbb{R}/\mathbb{Z})\ni f\mapsto 2\int\_0^{1/2} f(r)\,dr$.
Note that the conditional expectation $E$ is faithful and the state $\psi$ is faithful on the algebra $\mathbb{C}[F(s)]$ of trigonometric polynomials.
Put $\varphi=\psi\circ E$ and let $(\pi,H,\xi)$ denote the GNS-triplet.
Since $\varphi$ is faithful on $A\_0$, the vector $\xi$ is separating on $\pi(A\_0)$.
It is also cyclic because it is a GNS vector.
However $\xi$ is not separating for $\pi(A)$, since $\psi$ is supported on $[0,1/2]$ and $\pi$ is faithful.
|
14
|
https://mathoverflow.net/users/7591
|
410452
| 167,936 |
https://mathoverflow.net/questions/410285
|
9
|
Consider matrices $M$ of size $L\times L$ over a finite field $\mathbb{Z}\_p$, for simplicity focus on $p$ prime. The size $L$ is even. We want to find the order of a specific class of matrices, namely we want to find the smallest non-zero integer $n$ such that $$M^n=1,$$ where 1 is here the identity matrix of size $L\times L$. The matrix $M$ has a specific block structure, which originates from a specific linear cellular automaton. We have $$M=AB,$$ where
$$A=
\begin{pmatrix}
1 & 1 & & && & \\
1 & -1 && & & & \\
& & 1 & 1 && &\\
& & 1 & -1 && & \\
&&&& \ddots & & \\
& & & &&1 & 1\\
& & & &&1 & -1\\
\end{pmatrix}$$
and
$$
B=
\begin{pmatrix}
-1 && & & && 1&\\
& 1 & 1 && && \\
& 1 & -1 && && \\
&&& \ddots & & & \\
& & &&1 & 1&\\
& & &&1 & -1&\\
1 && & & && 1\\
\end{pmatrix}$$
You can see $B$ as $CAC^{-1}$ where $C$ is a cyclic shift over $L$ variables. Separately $A$ and $B$ have simple properties, but their alternating product becomes complicated.
I did some numerical testing, choosing first $\mathbb{Z}\_3$. The interesting thing is (which also motivates me to look deeper into this) that the order $n$ seems to have a complicated behaviour as a function of $L$ and it is not clear to me what is the precise source of this. Sometimes $n$ is very big, seemingly exponentially growing with $L$. For example $n(L=46)=354292=2^2\cdot 23\cdot 3851$. Or $n(L=58)=9565940=2^2\cdot 5\cdot 29\cdot 16493$. But if $L$ is divisible by 6 then we get much lower numbers, $n(60)=120$ for example. I understand that it is natural to see the prime $p$ somehow reflected in the function $n(L)$, but what I don't understand is how the big primes mentioned above can enter the game.
**Update:** Further numerical experiments showed that the smallest orbit happens when $L=2p^m$, in which case $n=2L$, we have simple linear growth. It gets complicated and quickly growing for most other $L$.
The origin of the problem is the following cellular automaton. Consider $L$ variables $s\_j\in \mathbb{Z}\_p$, with $j=1,2,\dots,L$. We define a dynamics on the model, such that we perform two operations cyclically. In each operation we group two neighbouring variables into a pair, and we alternate the ways how we make the pairing. For each pairing we perform the update
$$(s\_j,s\_{j+1})\to (s\_j+s\_{j+1},s\_j-s\_{j+1})$$
What changes is that at every second update $j$ is even or odd. If you represent this linear operation with matrices, you get $A$ and $B$ and we want to iterate the product $M=AB$.
|
https://mathoverflow.net/users/150048
|
Find the order of a class of finite matrices over finite fields
|
Big primes enter the picture as follows: Compute the characteristic polynomial $P\_L$ of your square matrix $M$ of size $L$.
Factor $P\_L$ over your working prime $p$. This gives you the eigenvalues of the involved Jordan-blocs over $\mathbb F\_p$. These eigenvalues are elements of field extensions of degree at most $L$ of your groundfield $\mathbb F\_p$. You can thus get large primes dividing the order of $M$ if they are divisors of $p^k-1$ for $k$
the degree of (at least) one irreducible polynomial
(over $\mathbb F\_p$) dividing $P\_L$.
A fast way for computing the order of $M$ is thus to compute the characteristic polynomial $P\_L$ of $M$,
factor it over $\mathbb F\_p$ and check then if prime-divisors of $p^k-1$ (for $k$ the degree of an involved irreducible polynomial) divide the order.
The order of $M$ over $\mathbb F\_p$ is necessarily a divisor of $U=p^{L-1}\prod\_{k=1}^{L}(p^k-1)$. One can of course remove a factor $p^k-1$ from this product if $P\_L$ mod $p$ has no irreducible divisor modulo $p$ of degree $k$. One can further reduce the size of $U$ by taking greatest common divisors of all involved factors. The exponent $L-1$ of $p$ in $U$ can of course be replaced by $\mu-1$ where $\mu$ is largest occuring multiplicity among the irreducible divisors of $P\_L \pmod p$. (There seem indeed to be often multiplicities in $P\_L$: This leads to possibly non-trivial Jordan-blocks in $M$ which contribute a factor $p^d$ to the order of $M$ where $d+1$ is the dimension of the largest involved Jordan bloc.)
The computational bottleneck of this approach is the factorization of $p^k-1$. The rest is computationally easy using fast exponentiation. Concretetely,
You have to check if $M^{U/q}$ is still the identity for $q$ any prime divisor of $U$. If this is the case, replace $U$ by $U/q$ (and recheck if you can remove one more power of $q$ from $U$). Going through all prime divisors of $U$ gives you at the end the exact order of $M$ over $\mathbb F\_p$.
**Experimental observation** The characteristic polynomial $P\_L$ of the square
matrix $M=AB$ of even size $L$ seems to be given by
$$P\_L=(x^2+4)Q^2\_{L/2}$$
where $Q\_1=1,\quad Q\_2=x+2$ and
$Q\_n=(x+2)Q\_{n-1}-xQ\_{n-2}$ for $n\geq 3$.
(The polynomials $Q\_n$ are 'almost' orthogonal polynomials and seem to satisfy
the divisibility property $P\_a\vert P\_b$ if $a\vert b$.) This formula holds for $L\leq 60$ even.
|
4
|
https://mathoverflow.net/users/4556
|
410455
| 167,938 |
https://mathoverflow.net/questions/410444
|
4
|
We denote by $C[0, 1]$ the space of continuous functions on $[0, 1]$ under the supremum norm, equipped with the Borel sigma algebra.
A *covering* of $C[0, 1]$ is a (possibly countably infinite) collection of Borel sets $E\_i$ such that $\bigcup\_i E\_i = C[0, 1]$.
A covering is said to have *finite eccentricity* if there exist constants $0 < c \leq C$ such that each $E\_i$ contains a open ball of radius $c$ and is contained within an open ball of radius $C$. We call $c$ and $C$ the *eccentricity constants* of the cover.
A covering is said to be *point finite* if every $f \in C[0, 1]$ lies in finitely many of the $E\_i$.
**Question:** Let $\epsilon > 0$ and $0< \delta < 1$ be arbitrary. Does there exist a point finite covering of $C[0, 1]$ of finite eccentricity with eccentricity constants $(1-\delta)\epsilon$ and $(1+\delta)\epsilon$?
|
https://mathoverflow.net/users/173490
|
Does $C[0, 1]$ admit a covering by sets of arbitrarily small eccentricity?
|
To give a positive answer to the question it is enough to, for a fixed $\varepsilon$, give a collection of disjoint balls in $C[0,1]$ of radius $\varepsilon$ which is dense in $C[0,1]$. Indeed, then for $\delta$ as small as you want you can take the sets $E\_i$ to be the balls and adjoin every point $f$ outside the balls to one ball at distance $<\delta$ of $f$ (there are countably many balls so you can do this in such a way that the resulting sets are Borel), obtaining a covering such that each point is in just one set, and with eccentricity constants $\varepsilon,\varepsilon+\delta$.
To build this collection of balls we first consider a sequence of distinct points $0,1,a\_3,a\_4,\dots$ dense in $[0,1]$. Now for $n\geq2$, let $A\_n$ be the set of functions which take some arbitrary values in the points $a\_1,a\_2,\dots,a\_n$ and are defined by linear interpolation in the rest of the interval. Then $A\_n$ is isometric to $\mathbb{R}^n$ in the square metric, with the coordinates given by the values of the function in $a\_1,\dots,a\_n$. Moreover, in this coordinates, the inclusion of $A\_n$ in $A\_{n+1}$ is given by a linear inclusion of $\mathbb{R}^n$ into $\mathbb{R}^{n+1}$ which preserves the first $n$ coordinates.
Now we can consider a tiling of $A\_2$ by balls of radius $\varepsilon$ (in the square metric of $\mathbb{R}^2$, squares of side $2\varepsilon$), and we call $C\_2$ the collection of their centers. As the inclusion $A\_2\to A\_3$ is linear and preserves the first $2$ coordinates, we can extend this tiling to a tiling of $A\_3$ by cubes of radius $\varepsilon$. More precisely, the set $C\_3$ of centers of cubes in $A\_3$ will be given in coordinates by $\{p+k(0,0,\dots,0,2\varepsilon);p\in C\_2,k\in\mathbb{Z}\}$. Concretely, that implies $C\_2\subseteq C\_3$. This way we can extend this tilings of $A\_n$ to tilings of $A\_{n+1}$ by balls (hypercubes) of radius $\varepsilon$, with centers in a set $C\_n$, with $C\_n\subseteq C\_{n+1}\forall n$. Note that by construction, the points of the $C\_n$ are at distance $\geq 2\varepsilon$ from each other.
So, the collection of balls of radius $\varepsilon$ and center in $\cup\_{n=2}^\infty C\_n$ will be disjoint, and it is dense in $\cup\_{n=2}^\infty A\_n$. As $\cup\_{n=2}^\infty A\_n$ is dense in $C[0,1]$, the collection of balls is dense in $C[0,1]$ too. So now we can just use the construction of the first paragraph to get the cover we want.
|
5
|
https://mathoverflow.net/users/172802
|
410458
| 167,940 |
https://mathoverflow.net/questions/410462
|
30
|
Consider the set of ultrafilters $\beta(\mathbb N)$ on $\mathbb N$.
Any function $f\colon\mathbb N\to\mathbb N$ extends to a function $\beta f\colon \beta \mathbb N \to \beta\mathbb N$. We say that two ultrafilters $\mathcal U$ and $\mathcal V$ are *isomorphic* if there is some bijection $f$ with $f(\mathcal U) = f(\mathcal V)$. Since there are only $2^{\aleph\_0}$ many bijections of $\mathbb N$, but $2^{2^{\aleph\_0}}$ many ultrafilters on $\mathbb N$, we know that there are many isomorphism classes of free ultrafilters.
On the other hand, in any proof that I have seen using ultrafilters, it does not seem to matter which ultrafilter is chosen. This leads me to the following **Question: is there some way in which all free ultrafilters are the 'same'?**
I have thought of some possibilities what it could mean for ultrafilters to be the 'same'. We can see any ultrafilter $\mathcal U$ as an ordered set, using the partial order $\subseteq$. I can imagine that if $\mathcal U$ and $\mathcal V$ are free ultrafilters, they are isomorphic as partial orderings. This seems pretty weak though.
Another possibility would be to consider the action of $\operatorname{Homeo}(\beta\mathbb N)$ on $\beta\mathbb N$. Does it act transitively?
It might be interesting to consider the Rudin–Keisler ordering $\leq\_{\text{RK}}$ on $\beta\mathbb N$. It is defined by $\mathcal U\leq\_{\text{RK}} \mathcal V$ iff there is a function $f\colon\mathbb N\to\mathbb N$ with $\beta f(\mathcal V) = \mathcal U$. It is known that there exist free ultrafilters that are not minimal for the Rudin–Keisler ordering, while it is independent of ZFC whether there exists free ultrafilters that are not minimal. Presumably, a minimal ultrafilter is not the 'same' as a not-minimal ultrafilter. However, even then it might be consistent with ZFC that all free ultrafilters are the 'same'.
|
https://mathoverflow.net/users/470870
|
Are all free ultrafilters 'the same' in some sense?
|
Certain important properties are shared by all free ultrafilters. In many applications of ultrafilters, especially more elementary applications, only these properties are used. In such a situation, it does not matter which free ultrafilter is chosen -- any one will do.
But there are some proofs that require (or seem to require) special kinds of ultrafilters.
One important example, mentioned in the comments by Benjamin Steinberg, are algebraically special ultrafilters, such as the idempotent ultrafilters used to prove Hindman's Theorem, or the minimal idempotents used in the ultrafilters proof of the Hales-Jewett Theorem. These ultrafilters are "algebraically special" in the sense that they have special properties defined using the algebraic-and-topological structure $(\beta \mathbb N,+)$. However, if we don't want to look at all this extra structure on $\beta \mathbb N$, an idempotent ultrafilter may not be too different from any other. In fact, every idempotent is equivalent to many non-idempotents (in the sense of being isomorphic to them, as described in your first paragraph).
Another special kind of ultrafilter is a $P$-point. These are definable in the topological space $\beta \mathbb N$, without considering any other dynamic or algebraic structure, and a $P$-point cannot be isomorphic to a non-$P$-point. One important characterization of $P$-points is: an ultrafilter $\mathcal U$ is a $P$-point if and only if for any sequence $\langle x\_n \rangle$ of real numbers, $r = \mathcal U$-$\lim x\_n$ if and only if there is a subsequence of $\langle x\_{n\_k} \rangle$ converging to $r$ with $\{n\_k :\, k \in \mathbb N\} \in \mathcal U$. I have seen this property of $P$-points used in proofs before, although no particularly famous examples spring to mind. I remember a theorem of mine, in a paper with Piotr Oprocha, where we need to take $\mathcal U$-limits that do *not* have this property. So this is an application of ultrafilters where it is important to use a *non*-$P$-point.
From an order-theoretic point of view, Ramsey ultrafilters are quite special. One can write a short proof of Ramsey's Theorem (the infinitary version) using a Ramsey ultrafilter. (I like this proof for pedagogical purposes, and have given it to graduate students in the past, since it nicely parallels the proof that measurable cardinals are Ramsey.)
So some applications of ultrafilters really do use special ultrafilters. To address some of your other questions:
It is a theorem of ZFC, not just a consistency result, that there are RK-incomparable ultrafilters. This is due to Kunen and Frolik.
It is also a theorem of ZFC that the action of homeomorphisms on $\beta \mathbb N \setminus \mathbb N$ is not transitive. In fact, Kunen proved (from ZFC only) that $\beta \mathbb N \setminus \mathbb N$ contains *weak $P$-points*. These are defined as points of $\beta \mathbb N \setminus \mathbb N$ that are not contained in the closure of any countable subset of $\beta \mathbb N \setminus \mathbb N$ not already containing the point. It isn't too hard to see that, because $\beta \mathbb N \setminus \mathbb N$ is compact, not every point has this property. And no self-homeomorphism of $\beta \mathbb N \setminus \mathbb N$ can map a weak $P$-point onto a non-weak $P$-point.
Finally, is there a meaningful sense in which all free ultrafilters are the same? Maybe. It is an open question whether all free ultrafilters can (consistently) have the same *Tukey type*. Roughly, there is a means of comparing partial orders via maps called Tukey reductions, much coarser than the notion of isomorphism described in your post. It is course enough that -- maybe -- all ultrafilters compare to all others. But this is consistently not the case. For example, a $P$-point is never Tukey-equivalent to a non-$P$-point.
|
38
|
https://mathoverflow.net/users/70618
|
410463
| 167,941 |
https://mathoverflow.net/questions/410375
|
7
|
I have two discrete groups $G\_1$ and $G\_2$ sitting in the following exact sequences:
$1\to H\_1\to G\_1\to K\_1\to 1$ and $1\to H\_2\to G\_2\to K\_2\to 1$.
$H\_1$, $K\_1$, $H\_2$ and $K\_2$ are all non-abelian free groups of ranks $k+n$, $k$, $k+l+n$ and $k+l$ respectively. Also $k,l>1$ and $n\geq 1$. Somehow I feel that $G\_1$ and $G\_2$ are not isomorphic! May be there is some easy way to see if it is true or not.
Edit: The Hochschild-Serre spectral sequence gives the following.
$0\to H\_1(H\_1, {\Bbb Z})\_{K\_1}\to H\_1(G\_1, {\Bbb Z})\to H\_1(K\_1, {\Bbb Z})\to 0$
$0\to H\_1(H\_2, {\Bbb Z})\_{K\_2}\to H\_1(G\_2, {\Bbb Z})\to H\_1(K\_2, {\Bbb Z})\to 0$
with action of $G\_i$, $i=1,2$, is trivial on $\Bbb Z$. The action of $K\_i$ on $H\_1(H\_i, {\Bbb Z})$, $i=1,2$, giving the co-invariant $ H\_1(H\_i, {\Bbb Z})\_{K\_i}$ is mysterious!
Edit: Let $S\_1$ and $S\_2$ be two $2$-manifolds with non-abelian free fundamental groups of ranks $k$ and $k+l$ respectively. Consider the configuration space $C(S\_i)$ of $2$-tuple of ordered different points in $S\_i$, $i=1,2$. Then taking the projection to one coordinate gives a fibration $C(S\_i)\to S\_i$ with fiber $S\_i$ minus a point (Fadell-Neuwirth fibration theorem). Hence we get two exact sequences as above with $n=1$: $G\_i=\pi\_1(C(S\_i))$, $K\_i=\pi\_1(S\_i)$ and $H\_i=\pi\_1(S\_i- \{\mbox{point}\})$ for $i=1,2$.
Furthermore, consider the configuration spaces of $m$-tuple of ordered different points of $S\_i$. Then the claim is that the fundamental groups are not isomorphic. These groups are poly-free and hence the title of this thread.
|
https://mathoverflow.net/users/126243
|
Distinguishing poly-free groups
|
Euler characteristic is multiplicative in the setting of your exact sequences
$1\to H\_i\to G\_i\to K\_i\to 1$,
i.e. $\chi(G\_i)=\chi(H\_i)\chi(K\_i)$.
(You can see this directly by building a model for each $G\_i$ as a graph of graphs, or by more sophisticated arguments.)
In your case, this gives
$\chi(G\_1)=\chi(H\_1)\chi(K\_1)=(k-1)(k+n-1)$
while
$\chi(G\_2)=\chi(H\_2)\chi(K\_2)=(k+l-1)(k+n+l-1)$.
In particular, if $l>0$ we can see that $\chi(G\_2)>\chi(G\_1)$, which distinguishes the two groups.
|
5
|
https://mathoverflow.net/users/1463
|
410465
| 167,942 |
https://mathoverflow.net/questions/410443
|
7
|
Let $ G $ be the real points of a linear algebraic group and $ G' $ a Zariski closed subgroup. **Then is $ G/G' $ a Cartesian product
$$
(K/K') \times F
$$
where $ F $ is contractible?** Here $ K,K' $ are maximal compacts of $ G,G' $.
Some relevant information:
Let $ G,G' $ be Lie groups with finitely many connected components (for example the real points of linear algebraic groups). They can be expressed as cartesian products
$$
G= K \times E\, , \, G'=K' \times E'
$$
where $ K,K' $ are maximal compacts and $ E,E' $ are contractible. According to [Mostow, G. D., Covariant fiberings of Klein spaces II, Amer. J. Math. 84 (1962), 466–474] $ G/G' $ is of the form
$$
G/G' \cong K \times\_{K'} F
$$
where $ F \times E' \cong E $ and the $ \times\_{K'} $ means taking the Cartesian product but then identifying
$$
(k\_0,f\_0) \sim (k\_0k,kf\_0k^{-1})
$$
for all $ k \in K' $. We can essentially summarize this by saying that $ G/G' $ is a $ K' $ homogeneous vector bundle over $ K/K' $.
Follow-up: Great answer. Exactly what I wanted. $ SO\_3(\mathbb{C})/SO\_2(\mathbb{C}) $ is the normal bundle over the 2 sphere (so 4 dimensional) and is nontrivial. Upon further reflection I found a less beautiful but more minimal counterexample. $ SE\_2(\mathbb{R}) $ is a linear algebraic group. There is a Zariski closed subgroup given by taking $ x\_{1,1}x\_{2,2}=1 $ and $ x\_{1,3}=0 $ and the quotient is the Moebius strip.
|
https://mathoverflow.net/users/387190
|
Is a quotient of real linear algebraic groups always a Cartesian product of compact and contractible factors?
|
The answer is no.
At least when $G$ and $H$ are semisimple, the quotient $G/H$ is diffeomorphic to the normal bundle of $K\_G/K\_H$ inside $G/H$ (where $K\_G$ and $K\_H$ denote respectively maximal compact subgroups of $G/H$), but this normal bundle might not be trivial.
For a concrete example take $G= SO(n+1,\mathbb C)$ and $H= SO(n,\mathbb C)$ (seen as real algebraic groups). Then $G/H$ is the smooth complex affine quadric of dimension $n$ and $K\_G/K\_H = SO(n+1,\mathbb R)/SO(\mathbb R)$ is the real sphere of dimension $n$ inside it. In particular, it is a real form of $G/H$, so its normal bundle is isomorphic to the tangent bundle of the sphere, which is trivial if and only if $n= 1,3$ or $7$ !
|
7
|
https://mathoverflow.net/users/173096
|
410467
| 167,944 |
https://mathoverflow.net/questions/410099
|
3
|
Preliminaries
=============
Let $ n $ be an integer such that $ n \geq3 $. Denote $ \left[ n \right] \equiv \{1,2, \ldots ,n \} $. Let $ P $ be a non-empty subset of $ \left[ n \right] $ such that $ \left|P \right| \geq 3 $. Denote $ X\_{P} \equiv \left( x\_{i} \colon i \in P \right) $ as an ordered alphabet. Let $ \mathbb{F} $ be a field such that $ \operatorname{char} \left( \mathbb{F} \right) \neq 2$. Denote $ \mathbb{F} \left[ X\_{P} \right] $ as the polynomial ring over $ \mathbb{F} $ in the (commuting) letters of $ X\_{P} $.
Polynomials as functions
------------------------
Let $ P\_{1},P\_{2} $ be non-empty subsets of $ P $ such that $ \left|P\_{1} \right| = \left|P\_{2} \right|$, then the polynomial $ f $ in the letters of $ X\_{P\_{1}} $, denoted by $ f \left( X\_{P\_{1}} \right) \in \mathbb{F} \left[ X\_{P\_{1}} \right]$, and the polynomial $ f $ in the letters of $ X\_{P\_{2}} $, denoted by $ f \left( X\_{P\_{2}} \right) \in \mathbb{F} \left[ X\_{P\_{2}} \right]$, will be considered equivalent as *functions*.
Multivariate Polynomial to univariate polynomial and back
---------------------------------------------------------
A (multivariate) polynomial in $ \mathbb{F} \left[ X\_{P} \right] $ can be uniquely written as a (univariate) polynomial in the letter/variable $ x\_j $ for some $ j \in P $, or in other words as a polynomial in $ \left( \mathbb{F} \left[ X\_{P \setminus \{ j \}} \right] \right) \left[ x\_{j} \right] $. Conversely, a polynomial in $ \left( \mathbb{F} \left[ X\_{P \setminus \{ j \}} \right] \right) \left[ x\_{j} \right] $ can be uniquely written as a polynomial in $ \mathbb{F} \left[ X\_{P} \right] $. Denote $ f \left( X\_{P \setminus \{ j \}} ;x\_{j} \right) $ as *the* polynomial in $ \left( \mathbb{F} \left[ X\_{P \setminus \{ j \}} \right] \right) \left[ x\_{j} \right] $ which is equivalent to the polynomial $ f \left( X\_{P} \right) $, for any $ f \left( X\_{P} \right) \in \mathbb{F} \left[ X\_{P} \right]$.
Univariate resultants and multivariate polynomials
--------------------------------------------------
For any $ j \in P $ and $ k \in P \setminus \{ j \} $, consider the polynomials $ f \left( X\_{P} \right) \in \mathbb{F} \left[ X\_{P} \right]$ and $ g \left( X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \} \cup \{ k \}} \right) \in \mathbb{F} \left[ X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \} \cup \{ k \} }\right]$ and denote $ y \equiv x\_{j} \equiv x\_{k} $. Then the (univariate) resultant of the polynomials $ f \left( X\_{P \setminus \{ j \}} ;y \right) \in \left( \mathbb{F} \left[ X\_{P \setminus \{ j \}} \right] \right) \left[ y \right]$ and $ g \left( X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \in \left( \mathbb{F} \left[ X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \} }\right] \right) \left[ y \right]$, with respect to the variable $ y $, is well-defined, and shall be denoted by
$$ \operatorname{res}\_{y} \left( f \left( X\_{P \setminus \{ j \}} ;y \right) ,g \left( X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \right) $$
Recursive definition
====================
The polynomial function $ q \left( X\_{ \left[ n \right]} \right) \in \mathbb{F} \left[ X\_{\left[ n \right]} \right] $ is defined recursively by the condition $$ q \left( X\_{ \left[ 3 \right]} \right) \equiv -x\_{1}^2-x\_{2}^2-x\_{3}^2+2x\_{1}x\_{2}+2x\_{1}x\_{3}+2x\_{2}x\_{3} $$ and the rule $$ q \left( X\_{ \left[ n \right]} \right) = \operatorname{res}\_{y} \left( q \left( X\_{P \setminus \{ j \}} ;y \right) ,q \left( X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \right)$$ for $ n \geq 4 $, any non-empty $ P \subseteq \left[ n \right] $ such that $ \left|P \right| \geq 3 $ and any $ j \in P $.
Example
-------
Suppose that $ n = 4 $, $ P = \{1,2,3\} $ and $ j = 3 $. Then $ \left[ n \right] = \{1,2,3,4\} $, $ \left| P \right| = 3 \geq 3 $ and $ j \in P $; furthermore
$$
\begin{align}
q \left( X\_{P \setminus \{ j \}} ;y \right) & = q \left( X\_{\{1,2,3\} \setminus \{ 3 \}} ;y \right) \\
& = q \left( X\_{\{1,2\}} ;y \right) \\
& = -y^2+2 \left(x\_{1}+x\_{2} \right)y- \left( x\_{1}-x\_{2} \right)^2
\end{align}
$$
and
$$
\begin{align}
q \left( X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) & = q \left( X\_{\left( \{1,2,3,4\} \setminus \{1,2,3\} \right) \cup \{ 3 \}} ;y \right) \\
& = q \left( X\_{\{3,4\}} ;y \right) \\
& = -y^2+2 \left(x\_{3}+x\_{4} \right)y- \left( x\_{3}-x\_{4} \right)^2
\end{align}
$$
Therefore
$$
q \left( X\_{ \left[ 4 \right] } \right) = \operatorname{res}\_{y} \left( q \left( X\_{ \{1,2 \}} ;y \right) ,q \left( X\_{ \{3,4 \}} ; y \right) \right)
$$
is unambiguous.
Is this well-defined?
=====================
>
> Is the polynomial function defined above well-defined, in the sense that it is invariant (not even up to sign) of the choice of a subset $ P $ and its element $ j $?
>
>
>
Small examples are evidence for the positive answer.
Some ideas
----------
It is not hard to show by induction that the degree of $ q \left( X\_{ \left[ n \right] };y \right) \in \left( \mathbb{F} \left[ X\_{\left[ n \right] \setminus \{ i \}} \right] \right) \left[ y \right]$ for any $ i \in \left[ n \right] $ is $ 2^{n-2} $, so to obtain that, in general,
$$
\begin{align}
\operatorname{res}\_{y} \left( q \left( X\_{P \setminus \{ j \}} ;y \right) ,q \left( X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \right)
& =
\operatorname{res}\_{y} \left(q \left( X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right)
, q \left( X\_{P \setminus \{ j \}} ;y \right) \right)
\end{align}
$$
But this is obviously not enough. Small examples are evidence to $ q \left( X\_{ \left[ n \right] } \right) $ being a symmetric function in $ X\_{ \left[ n \right] } $, so $ q \left( X\_{ \left[ n \right] } \right) $ must be a multi-symmetric function in $ X\_{P \setminus \{ j \}},X\_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} $, but I don't know how to prove the former claim.
Sidenotes
---------
I have in the past posted another question regarding this object, but that discussion is irrelevant to the current question, so I will not link it here.
|
https://mathoverflow.net/users/91155
|
Polynomial function defined recursively by a resultant - is it well defined?
|
Notice that
$$
\let\eps\varepsilon
q(x\_1,x\_2,x\_3)
=-\bigl(\sqrt{x\_1}+\sqrt{x\_2}+\sqrt{x\_3}\bigr) \bigl(-\sqrt{x\_1}+\sqrt{x\_2}+\sqrt{x\_3}\bigr) \bigl(\sqrt{x\_1}-\sqrt{x\_2}+\sqrt{x\_3}\bigr) \bigl(-\sqrt{x\_1}-\sqrt{x\_2}+ \sqrt{x\_3}\bigr)\\
=-\prod\_{\eps\_1,\eps\_2\in\{\pm1\}}\left(
\sqrt x\_3+\sum\_{I<3}\eps\_i\sqrt{x\_i}\right).
$$
So the roots of this polynomial (as a one in $x\_3$) are exactly
$$
\left(\sum\_{i <3}\eps\_i\sqrt{x\_i}\right)^2
$$
(each root appears in the above list twice, but it is a simple root in $q$).
Now a straightforward induction shows that
$$
q(x\_1,\dots,x\_n)
=\prod\_{\eps\_1,\dots,\eps\_{n-1}\in\{\pm1\}}\left(
\sqrt x\_n+\sum\_{I<n}\eps\_i\sqrt{x\_i}\right),
$$
and the (simple) roots of this polynomial (as a one in $x\_n$) are exactly
$$
\left(\sum\_{i <n}\eps\_i\sqrt{x\_i}\right)^2.
$$
The inductive step here is just an application of the standard formula
$$
\mathop{\mathrm{res}}(A,B)
=a\_0^kb\_0^n\prod\_{i,j}(\lambda\_i-\mu\_j),
$$
where the $\lambda\_i$ and the $\mu\_j$ list all the roots of
$$
A(x)=a\_0x^n+\dots \quad\text{and}\quad
B(x)=b\_0x^k+\dots,
$$
respectively. (Here we use that the degrees are powers of two!)
**Remark.** Surely, the polynomial defined by the above formulas, is symmetric in all variables, since its degree (in $\sqrt{x\_i}$) is divisible by 4.
|
2
|
https://mathoverflow.net/users/17581
|
410470
| 167,946 |
https://mathoverflow.net/questions/410437
|
3
|
In *"On numbers and games"*, Conway writes that the surreal Numbers form a *universally embedding totally ordered Field.* Later Jacob Lurie proved that (the equivalence classes of) the partizan games form a *universally embedding partially ordered abelian group.*
As far as I can tell from Lurie's paper, the latter means:
Let $\mathbb{U}$ denote the class of equivalence classes of partizan games. Furthermore let $S \subseteq S'$ be partially ordered abelian groups. Suppose that $\phi: S \rightarrow \mathbb{U}$ is an order-preserving homomorphism. Then there exists an order-preserving homomorphism $\phi': S' \rightarrow \mathbb{U}$ such that $\phi' \mid S = \phi$.
Now the thing is: I had before gotten to the (informal) conception that "universally embedding" in this context means that every (set-sized) partially ordered abelian group is isomorphic to some subgroup of $\mathbb{U}$, and vice versa that every totally ordered field is isomorphic to some subfield of $\mathbf{No}$. My question now is: is that even true? And what does this have to do with the statement of Lurie's paper?
|
https://mathoverflow.net/users/470978
|
What does it mean for the surreal numbers/partizan games to be "universally embedding"?
|
Besides the two types of structures being considered, there are two types of universality here. They are indeed related, but the stronger one (a la Lurie and I believe also Conway) is in my opinion the "right" one.
The story I like to imagine is this. We have some putatively-"universal" object $\mathcal{U}$ that we understand well, and some mysterious object $\mathcal{A}$ of the same type. Certainly at a minimum we should be able to find an embedding $\varphi$ of $\mathcal{A}$ into $\mathcal{U}$. However, what if we *subsequently* find out that $\mathcal{A}$ itself was really just a tiny fragment of some larger structure $\mathcal{B}$? We'd not only like to be able to embed $\mathcal{B}$ into $\mathcal{U}$, we'd also like to not have to undo the work we've already done: we want to find an embedding $\psi:\mathcal{B}\rightarrow\mathcal{U}$ which **extends** the already-thought-up embedding $\varphi$. (One way this story can become literally true is if we're trying to show that the "$\forall\exists$-theory" of $\mathcal{U}$ is reasonably simple, which happens e.g. in computability theory for various partial orders.)
To see an example of how this can play out, consider $\mathbb{Z}$ and $\mathbb{Q}$ as linear orders. Obviously each is "weakly universal" for all finite linear orders: any finite linear order embeds into both $\mathbb{Z}$ and $\mathbb{Q}$. However, only $\mathbb{Q}$ is "strongly universal" for finite linear orders: letting $\mathbf k$ denote the usual linear order with underlying set $\{1,\dotsc,k\}$, given any embedding $\varphi:\mathbf 2\rightarrow\mathbb{Z}$ there is some finite linear order $\mathbf n\supseteq\mathbf 2$ such that no extension of $\varphi$ embeds $\mathbf n$ into $\mathbb{Z}$ (basically just take $n$ much greater than $\varphi(2)-\varphi(1)$). By contrast, the usual "back-and-forth" argument for $\mathbb{Q}$ shows that $\mathbb{Q}$ is not subject to this sort of silliness. Note that the stronger type of universality brings with it a sort of "homogeneity" — a relevant notion here is **Fraïssé limit**.
|
5
|
https://mathoverflow.net/users/8133
|
410471
| 167,947 |
https://mathoverflow.net/questions/410459
|
9
|
Let $f \in S\_2(\Gamma\_0(N))$ be a newform with trivial character. I want to compute the Petersson norm $\lVert f\rVert^2$ of $f$, *not* normalized by $1/[\operatorname{SL}\_2(\mathbf{Z}):\Gamma\_0(N)]$, as in Gross–Zagier.
From [Numerical evaluation of the Petersson product of elliptic modular forms](https://mathoverflow.net/questions/115100/numerical-evaluation-of-the-petersson-product-of-elliptic-modular-forms), I came across the formula $$\lVert f\rVert^2 = \frac{(k-1)!}{2^{2k - 1}\pi^{k + 1}}L(\operatorname{Sym}^2(f),2)$$ with $k = 2$. I implemented this in Magma. If $N$ is not square-free, I guess the correct Euler factor for the symmetric square at $p^2 \mid N$ by testing if the functional equation for the symmetric square is satisfied with $1 \pm x$ or $1 \pm px$.
However, comparing with the result of PARI/gp (which *is* normalized by $1/[\operatorname{SL}\_2(\mathbf{Z}):\Gamma\_0(N)]$, so removing that normalization), it seems that I have to multiply my result by $N$ if $N$ is square-free, the reason for which I don't understand (maybe it's a convention of the implementation of the symmetric square $L$-function in Magma?). It is even worse for $N$ not square-free, e.g. $f \in S\_2(\Gamma\_0(125))^+$ or $S\_2(\Gamma\_0(147))^{w\_3,w\_{49}}$, where the normalization factors seem to be $125$ and $147 \cdot 7/8$, respectively.
PARI's code is hard to read, and Petersson scalar products are not implemented for $N \neq 1$ in Sage.
Can someone please shed light on this?
|
https://mathoverflow.net/users/471019
|
Computing the Petersson norm of newforms of weight 2 from the symmetric square $L$-function
|
My guess is that the formula you are trying to use is only valid for $N=1$, and thus needs correction in general.
Maybe Shimura's paper can help sort this out. <https://doi.org/10.1002/cpa.3160290618>
In (2.1), which Shimura writes for $\Gamma\_1(N)$ but that doesn't matter when the character is trivial, his definition is
$$\langle f,f\rangle={3/\pi\over [SL\_2(Z):\Gamma\_0(N)]}\int\_\Phi |f|^2 dx dy.$$
Then in (2.5) you have
$$\langle f,f\rangle={\Gamma(2)\over (4\pi)^2}\cdot\mathop{\rm res}\limits\_{s=2} D(s,f,f)$$
where by the last display of Section 1 he defines
$$D(s,f,f)=\sum\_{n=1}^\infty {a\_n^2\over n^s}.$$
Now a comparison of Euler products gives that the local factors
of $D(s,f,f)\zeta(2s-2)$ and $L(s,Sym^2 f)\zeta(s-1)$ match, at least away from $p$ that divide $N$ (this discrepancy is the issue that David Loeffler raises). This Euler product comparison is mentioned in another paper of Shimura, see (0.4) of <https://doi.org/10.1112/plms/s3-31.1.79>
Anyway, this gives the answer up to the bad factors,
namely
$$\int\_\Phi |f|^2 dx dy={[SL\_2(Z):\Gamma\_0(N)]\over 3/\pi}\langle f,f\rangle$$
$$=[SL\_2(Z):\Gamma\_0(N)]{\pi\over 3}{1\over (4\pi)^2}\mathop{\rm res}\limits\_{s=2} D(s,f,f)$$
$$=[SL\_2(Z):\Gamma\_0(N)]{\pi\over 48\pi^2}{1\over\zeta(2)}L(2,Sym^2 f)\prod\_{p|N} C\_p$$
$$=N\prod\_{p|N}(1+1/p)\cdot{1\over 8\pi^3}L(2,Sym^2 f)\prod\_{p|N}C\_p$$
Note that this matches your asserted formula when $N=1$ and $k=2$.
In the more general case, considering a bad prime $p|N$, the Euler factor from $\zeta(s-1)/\zeta(2s-2)$ evaluated at $s=2$ exactly cancels out factor of $(1+1/p)$ in the index formula. Meanwhile, the Euler factor of $L(s,Sym^2f)$ when $p$ exactly divides $N$ is $(1-1/p^s)^{-1}$, as is the Euler factor of $D(s,f,f)$ in this case (since $a\_p^2=1$).
Finally, when $p^2|N$, the Euler factor of $D(s,f,f)$ is trivial since $a\_p^2=0$, while that of $L(s,Sym^2f)$ can be known either by theory or trial-and-error computation. For the theoretical side, one can presuably work with the $p$-minimal twist of $f$ where this minimality allows twists with nontrivial Nebentypus - see 2.1 of Coates and Schmidt, particularly (2.12). <https://doi.org/10.1515/crll.1987.375-376.104>
I think one aspect is that if $v\_p(N)$ is odd then the Euler factor of $L(s,Sym^2f)$ is trivial; while if $v\_p(N)$ is even and $f$ is itself $p$-minimal then the factor is $(1+p/p^s)^{-1}$; and otherwise the Euler factor comes from that of the $p$-minimal twist (though perhaps not completely transparently, again with this $(1+p/p^s)^{-1}$ possibly appearing).
|
2
|
https://mathoverflow.net/users/334725
|
410483
| 167,949 |
https://mathoverflow.net/questions/410489
|
5
|
The May Recognition Theorem establishes an equivalence between the $\infty$-categories
* The $\infty$-category of grouplike $E\_n$ monoids
* The $\infty$-category of pointed $(n-1)$-connected spaces
There is also an equivalence between the $\infty$-categories
* The $\infty$-category of $E\_1$ monoids
* The $\infty$-category of pointed $(\infty,1)$-categories whose core is a connected space
Is there analogous generalization to higher dimension? E.g. something like an equivalence between
* The $\infty$-category of $E\_n$ monoids
* The $\infty$-category of pointed $(\infty,n)$-categories that are sufficiently trivial in dimensions below $n$
?
|
https://mathoverflow.net/users/445753
|
Are $E_k$ monoids higher categories?
|
This is closely related to the Baez–Dolan [stabilization hypothesis](https://ncatlab.org/nlab/show/stabilization+hypothesis).
There are numerous proofs of this statement.
One line of reasoning is to establish
a general 1-category statement first: given a symmetric monoidal presentable (∞,1)-category $C$,
the (∞,1)-category of $C$-enriched categories with one object
is equivalent to the (∞,1)-category of ∞-monoids in $C$,
i.e., algebras over the operad $\def\E{{\rm E}} \E\_1$ in $C$.
Iterating this result $n$ times, we obtain that
the (∞,1)-category of categories enriched in the (∞,1)-category of categories enriched in … (repeat $n$ times) … in $C$, with a single $k$-morphism for all $0≤k<n$ is equivalent
to the (∞,1)-category of $\E\_1$-algebras in $\E\_1$-algebras in … (repeat $n$ times) … in $\E\_1$-algebras in $C$.
By Dunn's additivity theorem, the latter (∞,1)-category is equivalent to $\E\_n$-algebras in $C$.
Depending on your preferences, this sketch can be formalized
using model categories, quasicategories, etc., see the nLab article cited above for some references.
|
6
|
https://mathoverflow.net/users/402
|
410493
| 167,953 |
https://mathoverflow.net/questions/410450
|
4
|
$\DeclareMathOperator\Bl{Bl}\DeclareMathOperator\Hilb{Hilb}\DeclareMathOperator\Sym{Sym}\newcommand\Sch{\mathit{Sch}}\newcommand\Sets{\mathit{Sets}}$Let $X$ be a smooth variety. The Hilbert scheme of two points $X^{[2]}$ on $X$ can be obtained by blowup the diagonal of the symmetric product $\Sym^2X:=(X\times X)/\mathbb Z\_2$:
$$X^{[2]}\cong\Bl\_{\Delta}(\Sym^2X)\to \Sym^2X.$$
By definition, this space represents the Hilbert functor \begin{gather\*}
\Hilb^2\_X:\Sch^\text{op}\to \Sets \\
T\mapsto \{{\text{flat families of subschemes of $X$ of length 2 over $T$}}\}/{\sim}.
\end{gather\*}
I'm interested in the space $\Bl\_{\Delta}(X\times X)$, which is a double cover of $X^{[2]}$ branched along the exceptional divisor. In fact it is obtained as the fiber product square:
$\require{AMScd}$
\begin{CD}
\Bl\_{\Delta}(X\times X) @>>> \Bl\_{\Delta}(\Sym^2X)\\
@VVV @VVV\\
X\times X @>>> \Sym^2X.
\end{CD}
I'd like to know if there is a modular interpretation of the double cover $\Bl\_{\Delta}(X\times X)$. To me, it should be something like the "Hilbert scheme" of ordered two points: Its general point parameterizes two points with an order; When the two points collide and become a fat point, it forget the order. However, I don't know how to formulate the modularity more precisely.
So perhaps here is what I want to ask: Is there a moduli functor $\mathcal{M}:\Sch^\text{op}\to \Sets$
representable by the variety $\Bl\_{\Delta}(X\times X)$?
|
https://mathoverflow.net/users/74322
|
$\text{Bl}_{\Delta}(X\times X)$ as the "Hilbert scheme" of ordered two points on $X$
|
Let $X$ be a smooth variety, and $X[n]$ the variety obtained from $X^n = X\times\dots \times X$ by blowing-up the diagonals in order of increasing dimension.
The variety $X[n]$ is a wonderful compactification (the added boundary divisor is simple normal crossing) of the configuration space of $n$ ordered points on $X$. It is known as the Fulton-MacPherson configuration space and it has been constructed in various ways here:
W. Fulton, R. MacPherson, *A Compactification of Configuration Spaces*, Annals of Mathematics, Second Series, Vol. 139, No. 1 (Jan., 1994), pp. 183-225.
In particular, in Theorem $4$ of the paper above you can find the precise definition of the moduli functor you are looking for.
When $n = 2$ the space $X[2]$ is just the blow-up of $X\times X$ along the diagonal. In particular, when $X = C$ is a curve then $C[2] = C\times C$ and $C[3]$ is $C\times C\times C$ blown-up along the small diagonal since the three bigger diagonals are divisors.
|
4
|
https://mathoverflow.net/users/14514
|
410500
| 167,955 |
https://mathoverflow.net/questions/410507
|
1
|
I'm trying to show the curvature of a one dimensional vector bundle with a Riemannian metric vanishes, no matter what the connection is. I found this can be done for orientable bundles, because an orientable Riemannian line bundle is trivial.
My questions are,
1. Are Riemannian line bundles trivial?
2. If not, how to prove their curvature vanish?
|
https://mathoverflow.net/users/471097
|
Riemannian vector bundle
|
Given a connection $A\_{i\alpha}^\beta$ the curvature is
$$F\_{ij\alpha}^\beta=\frac{\partial A\_{j\alpha}^\beta}{\partial x^i}-\frac{\partial A\_{i\alpha}^\beta}{\partial x^j}+A\_{i\gamma}^\beta A\_{j\alpha}^\gamma-A\_{j\gamma}^\beta A\_{i\alpha}^\gamma.$$
In the case of a line bundle one can only have $\alpha=\beta=1$ so denoting $A\_i=A\_{i1}^1$ you get
$$F\_{ij}=\frac{\partial A\_j}{\partial x^i}-\frac{\partial A\_i}{\partial x^j}.$$
In general this is nonzero.
Now, given a bundle metric $g\_{\alpha\beta}$, metric-compatibility of the connection means that $A\_{i\alpha}^\gamma g\_{\beta\gamma}+A\_{i\beta}^\gamma g\_{\alpha\gamma}=\partial\_i g\_{\alpha\beta}.$ Differentiating relative to $x^j$, and subtracting off the same equation with $i$ and $j$ swapped, you get the skew-symmetry
$$F\_{ij\alpha}^\gamma g\_{\beta\gamma}+F\_{ij\beta}^\gamma g\_{\alpha\gamma}=0.$$
Returning again to the case of a line bundle, this says $2F\_{ij}=0$ i.e. $F=0.$
|
2
|
https://mathoverflow.net/users/156492
|
410513
| 167,959 |
https://mathoverflow.net/questions/410505
|
2
|
Suppose $K$ is a number field and $E$ is an elliptic curve defined over $K$. *My question is:* how do you compute the local cohomology group $H^1(K\_v, \, E[p^{\infty}])$?
As to why I'm asking this, it comes up in [Iwasawa theory for elliptic curves](https://arxiv.org/abs/math/9809206) by Greenberg in page 74. In these notes, Greenberg is looking at the elliptic curve $E = 11a3$ defined over $\mathbf{Q}$. He claims that
$$H^1(\mathbf{Q}\_{11}, E[5^{\infty}]) \cong \mathbf{Z}/5\mathbf{Z}. $$
I'd like to understand how one would prove this result. Greenberg's notes say that this follows from the local duality theorems, but I'm not exactly sure how it follows from them. In terms of what I tried to figure this out, I tried to use the fact that $E[5^{\infty}]$ sits in an exact sequence of $G\_{\mathbf{Q}\_{11}}$-modules:
$$0 \to \mu\_{5^{\infty}} \to E[5^{\infty}] \to \mathbf{Q}\_5/\mathbf{Z}\_5 \to 0.$$
I tried applying the long-exact sequence in cohomology to this sequence, but I still couldn't show how to derive the result above. Does anyone know how to derive the fact above that $H^1(\mathbf{Q}\_{11}, E[5^{\infty}]) \cong \mathbf{Z}/5\mathbf{Z}$? Is there some other fact that I need to complete the calculation?
|
https://mathoverflow.net/users/394740
|
Calculating the Galois cohomology group $H^1(K_v, \, E[p^{\infty}])$
|
The corollary 3.4 of chapter 1 Arithmetic Duality theorems by Milne says that:
If $K$ has char 0 local field, there is a canonical pairing $$H^r(K,A^t)\times H^{1-r}(K,A)\to \mathbb{Q}/{\mathbb{Z}}$$ so because eliptic curves are self-dual for computing $H^1(K,E)$ it is enough to compute $E(K)$, It also implies $H^2(K,E)=0$.
Now just look at the exact sequence $$0\to E(5^n)\to E\to E\to 0$$.
|
3
|
https://mathoverflow.net/users/65846
|
410525
| 167,962 |
https://mathoverflow.net/questions/410535
|
1
|
Suppose that $f(t)$ is a square-integrable, band-limited function, i.e. the Fourier transform $\hat f$ has compact support.
**Problem:** Under which assumptions on a function $g(t)$ is the map $h(t) := e^{ig(t)}f(t)$ band-limited, i.e. $\hat h$ has compact support.
My idea would be to use the Paley-Wiener theorem which says $f$ extends to an entire function of exponential type (in particular, it has order less or equal than 1). Now one has to find assumptions on $g$ so that $h$ is again of exponential type (maybe $g$ must be a linear function then?).
Thanks in advance for any help!
|
https://mathoverflow.net/users/170539
|
Under which assumptions is $e^{ig(t)}f(t)$ of exponential type if $f$ is of exponential type?
|
A function is entire and of exponential type if and only if its Fourier transform has bounded support (This is Paley-Wiener theorem). Since your
$f$ belongs to this class, then, assuming that $g$ is also entire, $h=e^gf$ will belong to this class if and only if $g$ is linear.
If you do not want to assume that $g$ is entire, since $h$ and $f$ are entire, $e^g$ is meromorphic, with poles contained in the zero set of $f$,
and of exponential type in the sense that $T(r,e^g)=O(r),$ where $T$ is the Nevanlinna characteristic. This is a complete characterization: take any meromorphic function $F$ of exponential type whose poles belong to the zero set of $f$, and take $g=\log F$.
|
2
|
https://mathoverflow.net/users/25510
|
410539
| 167,967 |
https://mathoverflow.net/questions/410512
|
3
|
Let $\mathcal I$, $\mathcal C$ be $2$-categories (or $(\infty, 2)$-categories, I'm interested in both cases) and assume that $\mathcal I$ is small, $\mathcal C$ has enough weighted (co)limits as you need. We can define the $2$-category of (co)lax functors $\operatorname{Fun}^\text{(co)lax}(\mathcal I, \mathcal C)$. (**Edit**: I wrote an incorrect description for this which is now deleted)
Is there a nice way to describe (co)limits in (co)lax functor categories? I don't expect it's too simple like the non-lax case (which is computed pointwise) because lax functors specialize to monads, and colimits of algebras are usually complicated. But more specifically:
1. Is there a known formula for them? By this, I mean an expression for enough data (value on cells under the diagram) in terms of, say, some weighted (co)limits in $\mathcal C$.
2. Are they simple to compute in some cases, e.g., filtered colimits?
3. In particular, if $\mathcal C$ is (co)complete, can we say that $\operatorname{Fun}^\text{(co)lax}(\mathcal I, \mathcal C)$ is (co)complete?
|
https://mathoverflow.net/users/109318
|
(Co)limits in lax functor categories
|
It depends on what kind of morphisms between your lax functors you're interested in. (As Tim says, lax functors are not the objects of the Gray internal hom.)
For any 2-category $I$, there is a 2-category $I'$ such that lax functors out of $I$ are the same as strict functors out of $I'$. Thus, the 2-category of lax functors and *strict* natural transformations is equivalent to an ordinary $\rm Cat$-enriched functor category, and thus inherits all limits and colimits from its codomain 2-category $C$ computed pointwise. In addition, it is strictly 2-monadic and 2-comonadic over a set-indexed power of $C$.
The 2-category of lax functors and *pseudo* natural transformations between them is equivalent to the category of algebras and pseudomorphisms for this 2-monad or 2-comonad. Therefore, by the classical work of Blackwell-Kelly-Power on 2-monads (*Two-dimensional monad theory*), it has PIE-limits (using the monad) and PIE-colimits (using the comonad), again computed pointwise, and in particular all pseudolimits and pseudocolimits.
Similarly, the 2-categories of lax functors and lax or colax transformations are the categories of algebras and lax or colax morphisms. Such categories don't in general have even all pseudolimits or pseudocolimits, but there are significant classes of limits and colimits that they do have -- again computed pointwise. Steve Lack initiated this study in his paper [Limits for lax morphisms](http://web.science.mq.edu.au/%7Eslack/papers/talgl.pdf), and later he and I characterized precisely the class of limits that such 2-categories admit in our paper [Enhanced 2-categories and limits for lax morphisms](https://arxiv.org/abs/1104.2111).
Note that all of these limits and colimits *are* computed pointwise. It's true that colimits of monads *on a fixed category* are complicated, but here we are talking about (co)limits of monads *in a fixed 2-category* which are a different beast, and in particular involve changing the object *on* which the monad lives.
|
5
|
https://mathoverflow.net/users/49
|
410544
| 167,970 |
https://mathoverflow.net/questions/410360
|
1
|
Let $n$ and $d$ be positive integers with
$$
n,d \to \infty, \quad n/d \to \rho \in (0,\infty).
$$
Let $\Sigma\_d$ be a psd matrix such that
* $\mbox{trace}(\Sigma\_d) = 1$.
* $\|\Sigma\_d\|\_{op} = \mathcal O(1/d)$.
* The empirical eigenvalue distribution of $d \cdot \Sigma\_d$ converges weakly to some distribution $D$ on $\mathbb R$.
Let $W$ be a random $n \times d$ matrix with iid rows from $N(0,\Sigma\_d)$. Finally, let $a,b \ge 0$ be fixed constants, and define random $A$ and $B$ by
$$
\begin{split}
A &= WW^\top + a I\_n,\\
B &= WW^\top + b I\_n.
\end{split}
$$
>
> **Question.** *What is an analytic formula for the limiting value of the (random) scalar $\dfrac{1\_n^\top B^{-1}AB^{-1} 1\_n}{d}$, where $1\_n := (1,1,\ldots,1) \in \mathbb R^n$ ?*
>
>
>
Observations
------------
If we replace $1\_n$ by $z$, where $z \sim N(0,I\_n)$ is independent of $W$, then
$$
\begin{split}
\mathbb E\_z[z^\top B^{-1} A B^{-1} z/d] &= \mbox{trace}(B^{-1} A B^{-1}/d)\\
&= \frac{1}{d}\sum\_{i=1}^n \frac{\lambda\_i + a}{(\lambda\_i + b)^2}\\
& \overset{a.s}{\to} \int\_0^\infty \frac{\lambda + a}{(\lambda + b)^2}\,\mu(\mathrm{d}\lambda)\\
&= m\_\mu(-b) - (a-b) m\_\mu'(-b),
\end{split}
$$
where $\mu$ is the LSD of $WW^\top$, and $m\_\mu$ is its Stieltjes transform (which can be evaluated via $D$ and Silverstein's fixed-point equation).
>
> *Maybe this computation is somehow linked to the an answer to my main question ?*
>
>
>
|
https://mathoverflow.net/users/78539
|
Limiting value of $\dfrac{1_n^\top B^{-1} A B^{-1} 1_n}{d}$, where $A=WW^\top + a I_n$, $B = WW^\top + b I_n$, and $W \sim N(0,\Sigma_d)$
|
The distribution of $v^TB^{-1}AB^{-1}v$ is the same for every vector $v$ in the unit sphere either deterministic or independent of $W$. Once this is established, you are allowed to take $v=z/\|z\|$ independent of $W$; you can then apply the argument given at the end of the question together with concentration of the quadratic form in $z$ (Hanson Wright inequality) and $\|z\|^2/n\to^{as}1$.
The fact that the distribution of $v^TB^{-1}AB^{-1}v$ is the same for all $v$ in the unit sphere can be seen as follows. Let $u=R^Tv$ for a rotation $R\in O(n)$. Then
$$u^TB^{-1}AB^{-1}u
=
v^T R B^{-1} R^T R A R^T R B^{-1} R^T v
= v^T \tilde B^{-1} \tilde A \tilde B^{-1} v
$$
where $\tilde A = RAR^T = \tilde W\tilde W^T + aI$ where $\tilde W=RW$
and $\tilde B = RBR^T = \tilde W\tilde W^T + bI$.
But $W=^d\tilde W$ by rotational invariance of the Gaussian distribution.
|
1
|
https://mathoverflow.net/users/141760
|
410551
| 167,971 |
https://mathoverflow.net/questions/410519
|
3
|
Let $E\rightarrow M$ be a vector bundle.
Kirill Mackenzie in the book General theory of Lie groupoids and Lie algebroids associates a Lie algebroid to $E\rightarrow M$ in the following steps:
1. talk about zero-th and first order differential operators on $E\rightarrow M$, which are some nice maps of sections $\Gamma(M,E)\rightarrow \Gamma(M,E)$.
2. realise these first order differential operators as sections of a vector bundle $\text{Diff}^1(M)\rightarrow M$.
3. do some pullback along some morphism of vector bundles. Call the pullback as the Lie algebroid of derivations on $E\rightarrow M$.
The book also talks about Linear vector fields and says how these are related to the Lie algebroid of derivations.
I do not fully understand the idea of linear vector fields and first/zeroth order differential operators. But, that is not what I want to ask.
Given a principal bundle $P\rightarrow M$, one can consider the morphism of tangent bundles $TP\rightarrow TM$. As the action of $G$ on $TP$ is nice, this would induce a morphism of vector bundles $(TP)/G\rightarrow TM$. The Lie bracket on $\mathfrak{X}(P)=\Gamma(P,TP)$ is nice enough to induce a Lie bracket on $\Gamma(M,(TP)/G)$. Thus, we have a vector bundle over $M$, a morphism of vector bundles to the tangent bundle of $M$, a Lie bracket on sections of this vector bundle, and some more extra nice properties. This is a Lie algebroids. This vector bundle $(TP)/G\rightarrow M$ is called the Atiyah vector bundle and the Lie algebroid is called the Atiyah Lie algebroid.
Given a vector bundle $E\rightarrow M$, one can consider the associated principal bundle $GL(E)\rightarrow M$ and consider the construction mentioned above which would result in a Lie algebroid over $M$.
1. Are these two procedures giving same Lie algebroid? I think the answer to this is positive, but I am not sure.
2. Why would one want to involve differential operators and linear vector fields to associate a Lie algebroid when there is an easier way of considering the associated Atiyah Lie algebroid?
Even though I mentioned that I am not asking about the idea behind linear vector fields/differential operators, please see if you can say a few lines which might make my understand better.
|
https://mathoverflow.net/users/118688
|
Lie algebroid associated to a vector bundle
|
**Question 1.** The two procedures indeed give the same Lie algebroid. One possible way of seeing this is by considering the flows of vector fields: a section of $T(GL(E))/GL(n)$ is a vector field on the frame bundle that is $GL(n)$-invariant, and this means that its flow is by $GL(n)$-equivariant diffeomorphisms. But this is, in turn, equivalent to a family of vector bundle isomorphisms of $E$. When we differentiate this family, we obtain a linear vector field on $E$.
More conceptually, the categories of $GL(n)$-principal bundles and rank $n$-vector bundles (where we only consider automorphisms) are equivalent. Therefore the symmetries are the same, and in particular, the infinitesimal symmetries are the same. But these are precisely the $GL(n)$-invariant vector fields on the one hand, and linear vector fields on the other.
The link to differential operators is also not so difficult to see: a section of $E^{\*}$ is the same as a linear function on $E$. So given a section of $E^\*$, we may differentiate it by a vector field on $E$. In general this will not return a section of $E^\*$, rather it will just be some function on $E$. Linear vector fields are precisely the vector fields on $E$ that send linear functions to linear functions. So we can think of linear vector fields as differential operators for $E^\*$ (and by dualizing, for $E$ as well).
**Question 2.** The question of whether principal bundles or differential operators are 'easier' is probably a matter of one's background. But the Atiyah algebroid is useful for studying both.
I believe the reason Atiyah originally introduced the Atiyah algebroid was to study the question of the existence of holomorphic connections. Note that connections $\nabla$ are a type of differential operator: given a vector field $X$, then we get a first-order differential operator $\nabla\_{X} : \Gamma(E) \to \Gamma(E)$, whose *symbol* is the vector field $X$.
The Atiyah algebroid sits in a nice short exact sequence of Lie algebroids:
$0 \to End(E) \to At(E) \to TM \to 0. $
The right-hand map is the anchor. From the perspective of principal bundles, this is the map you constructed in your question, and the kernel is the 'vertical bundle'. From the perspective of differential operators, where we view sections of the Atiyah algebroid as differential operators on $E$ of order $1$, the anchor map becomes the symbol. Hence, the kernel $End(E)$, corresponds to the differential operators of order $0$.
Looking back at the definition of a connection, you can see that it is precisely giving you a section of the symbol (i.e. anchor) map. So the existence of a connection on $E$ is exactly the same thing as an isomorphism $At(E) \cong End(E) \oplus TM$ (respecting the extension structure). In the world of smooth manifolds, short exact sequences of bundles always split: every bundle admits a connection. In the holomorphic world this is no longer the case: the existence of holomorphic connections is measured by the extension class of $At(E)$, which is a cohomology class in $H^{1}(T^{\*}M \otimes End(E))$, called the Atiyah class of $E$.
|
2
|
https://mathoverflow.net/users/69713
|
410554
| 167,974 |
https://mathoverflow.net/questions/410336
|
0
|
In Kanamori's [Bernays and Set Theory](https://www.researchgate.net/publication/220366286_Bernays_and_Set_Theory) pages 20-21, a first order reflection principle due to Bernays is mentioned, that of:
>
> $$\sf \varphi \to \exists y \, (\text {Trans}(y) \land \varphi^y)$$ for formulas $\varphi$ without $\sf y$ or any class variables, where $\sf Trans(y) $ asserts that $\sf y$ is transitive; $\sf \varphi ^ y$ denotes the relativization of the formula $\varphi$ to the set $\sf y$, i.e., $\exists x$ is replaced by $\exists x \in \sf y$ and $\forall x$ by $\forall x \in \sf y$. Starting with the observation that set parameters $a\_1,...,a\_n$ can
> appear in $\varphi$ and $\sf y$ can be required to contain them by introducing clauses $\exists x(a\_i \in x)$ into $\varphi$. Bernays just with his schema established Pair, Union, Infinity, and Replacement (schema)—in effect achieving a remarkably economical presentation of ZF.
>
>
>
My question: what is the proof of replacement from reflection?
It's easy to see that Reflection can prove Pairing, Union, and Infinity. But the proof of Replacement is what's escaping me.
I tried to reflect the formula: $\exists x (a \in x) \land \forall m \in a \exists ! z: \phi(m,z)$ on the transitive set $\sf y$, but what we'll get is a set (which is $\sf y$) that contains the $z$'s of $\phi^\sf y$ among its elements, and I don't know how those are the same $z$'s of $\phi$?
|
https://mathoverflow.net/users/95347
|
What is the proof of replacement from Bernays first order reflection?
|
The above reflection scheme does not imply the existence of an empty set, nor does it imply the existence of more than one element. In the domain with one element which is an element of itself, the reflection scheme holds.
Let T0 be the theory whose axioms are extensionality, foundation and the above reflection scheme. If ZF is consistent, then pairing, union, power set and some instances of separation are not provable from T0.
Proof:Let A=ωU{{1},{{1}}}. Let U(y) be a formula which holds iff y∈A or (y∉Vω and for every x∉Vω which is in the transitive closure of {y}, the intersection of Vω with the transitive closure of x is A).
Then relativized to U, the above reflection scheme holds, but pairing, union, some instances of separation, and power set do not.
Let T1 be the theory whose axioms are the axioms of T0, power set, and all instances of separation. Then if ZF is consistent, there are instances of replacement not provable from T1.
Proof: Define a sequence s by s0=0, and s(n+1)=U{b|there is a "formula with parameters from V(sn) such that b is the least ordinal for which holds in Vb"}
Let t=U{sn| n∈ω}. Then all the axioms of T1 hold in Vt, but there are instances of replacement that do not. For example,
let ψ(n,x) be a formula which holds iff n∈ω and x=sn.
|
1
|
https://mathoverflow.net/users/133981
|
410568
| 167,978 |
https://mathoverflow.net/questions/410565
|
3
|
Suppose that $ M $ is non-orientable with transitive action by a Lie group $ G $. **Does that imply that some Lie group $ G' $ acts transitively on the orientable double cover $M'$?**
This is true for compact dimension 2: the Klein bottle is an $ \operatorname{SE}\_2 $ manifold and so is the torus. The projective plane is an $ \operatorname{SO}\_3 $ manifold and so is the sphere.
Morally I think the answer should be yes in general since an orientable double cover should be better behaved than/at least as well behaved as the original non orientable manifold.
|
https://mathoverflow.net/users/387190
|
Transitive action on non-orientable $ M $ lifts to orientable double cover
|
There is a general theory for lifting Lie group actions to covering spaces, see Bredon's monograph "Introduction to compact transformation groups", chapter 1, section 9. In the case of orientation covers one gets that any (effective, continuous) action of a Lie group $G$ on a non-orientable manifold lifts to a $G^\prime$ action on the orientation cover, where $G^\prime$ consists of all lifts of elements of $G$. Thus $G^\prime$ surjects onto $G$ with kernel of order two, and in particular, $G^\prime$ is a Lie group.
If the $G$-action is transitive, then so is the $G^\prime$-action. (Indeed, if $\tilde x, \tilde y$ are points in the covering space, which project to $x, y$, respectively, and $g\in G$ with $g(x)=y$, then one of the two lifts of $g$ takes $\tilde x$ to $\tilde y$).
|
7
|
https://mathoverflow.net/users/1573
|
410571
| 167,981 |
https://mathoverflow.net/questions/409916
|
6
|
Working in *mono-sorted* first order logic with equality $``="$ and membership $``\in"$:
* *Define:* $set(x) \equiv\_{df} \exists y \, (x \in y)$
* Axiomatize:
1. **Extensionality:** $( a \subseteq b \land b \subseteq a \to a=b)$
2. **Separation:** $(set(a) \to \exists \ set \ x : \forall y \, (y \in x \leftrightarrow y \in a \land \phi ))$
3. **Reflection:** $ (\varphi \to \exists \ set \ x : \text { trs}(x) \land \varphi^x)$
where formulas $\phi, \varphi$ do not use $``x"$; $\varphi$ do not use $``="$; $\varphi^x$ is obtained from $\varphi$ by merely bounding all of its quantifiers by $``\subseteq x"$; and $``\text { trs}" $ stands for *is transitive*.
This theory can prove: Set existence, Empty, Pairing, set Union, Power, Infinity, over the set world of it, also it proves [Substitution](https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory#The_axioms_in_Kelley%27s_General_Topology). It can also prove class comprehension scheme. So it appears to prove all axioms of [$\sf MK$](https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory#The_axioms_in_Kelley%27s_General_Topology)-$\sf Reg.$-$\sf Choice$.
>
> Is this theory consistent?
>
>
>
>
> If so, is it equivalent to [Bernay's reflection](https://en.wikipedia.org/wiki/Reflection_principle#As_new_axioms)?
>
>
>
|
https://mathoverflow.net/users/95347
|
What's the consistency strength of this form of reflection?
|
This theory is inconsistent.
We note that by 1 and 2 that if set(x) and y⊆x, then set(y).
(a) There is a v such that ∀x(set(x)-->x∈v).
Proof:Suppose not. Then ∀v∃s∃t(s∈t∧s∉v). By 3 there is transitive x such that
```
set(x) and ∀v(v⊆x-->∃s∃t(s⊆x∧t⊆x∧s∈t∧s∉v). In particular
(x⊆x-->∃s∃t(s⊆x∧t⊆x∧s∈t∧s∉x). But this is impossible.
```
Suppose that ∀x(set(x)-->x∈V). Then ∃w∀t(t∈V-->t∈w). By 3 there is transitive x
such that set(x) and ∃w(w⊆x∧∀t(t⊆x∧t∈V-->t∈w)). Since t⊆x implies set(t),
t⊆x implies t∈x. By 2, there is a c such that t∈c<-->(t∈x∧t∉t). Since set(c),
c∈c<-->c∉c.
|
1
|
https://mathoverflow.net/users/133981
|
410575
| 167,983 |
https://mathoverflow.net/questions/410580
|
2
|
A Banach space $X$ has the *weak Phillips property* if the canonical projection $X^{\*\*\*}\to X^{\*}$ is sequentially weak$^{\*}$-weak continuous [[FreedmanÜlger2000](https://doi.org/10.1090/S0002-9939-00-05703-8), [Ülger2001](https://doi.org/10.4064/cm87-2-1)].
Let $1<p<\infty$ and $E = (\oplus\_{n=1}^{\infty}\ell^1\_n)\_{\ell^p}$ be the $\ell^p$ direct sum of $(\ell^1\_n)\_{n=1}^{\infty}$.
**Question:** Does the space of compact operators $K(E)$ have the weak Phillips property?
|
https://mathoverflow.net/users/164350
|
Does $K( (\oplus_{n=1}^{\infty}\ell^1_n)_{\ell^p})$ have the weak Phillips property?
|
Yes, it does. This is essentially an unpublished result due to Hermann Pfitzner, see III.3.6 and III.3.7 in *$M$-ideals in Banach spaces and Banach algebras* by P. Harmand, W. Werner and myself ([Zbl 0789.46011](https://zbmath.org/?q=an%3A0789.46011)) along with the fact that $K(E)$ is an $M$-ideal in its bidual $L(E)$.
|
4
|
https://mathoverflow.net/users/127871
|
410588
| 167,986 |
https://mathoverflow.net/questions/410601
|
3
|
Let ${\cal U}$ be a free ultrafilter on $\omega$. Is the linearly ordered set $(\omega+1)^\omega/{\cal U}$ complete?
|
https://mathoverflow.net/users/8628
|
Is $(\omega+1)^\omega/{\cal U}$ complete for ${\cal U}$ free ultrafilter?
|
No. Every ultraproduct by a free ultrafilter on $\omega$ is $\aleph\_1$-saturated. And infinite $\aleph\_1$-saturated linear orders cannot be complete!
*Proof:* Let $L$ be an infinite $\aleph\_1$-saturated linear order. Since $L$ is infinite, it contains an infinite increasing sequence or an infinite decreasing sequence. Without loss of generality, let's say we have an increasing sequence $(a\_n)\_{n\in\omega}$.
By $\aleph\_1$-saturation, the partial type $\{x>a\_n\mid n\in\omega\}$ is realized in $L$, so the set $\{a\_n\mid n\in\omega\}$ is bounded above. Suppose $b$ is an upper bound. Then the partial type $\{x>a\_n\mid n\in\omega\}\cup \{x<b\}$ is realized in $L$, so $b$ is not a least upper bound. Thus $L$ is not complete.
|
6
|
https://mathoverflow.net/users/2126
|
410606
| 167,991 |
https://mathoverflow.net/questions/385019
|
4
|
This is a question in stable homotopy theory which I will boil down to a pure combinatorics question. If you're not interested in the homotopy theory, feel free to skip to the end for the combinatorial formulation.
**Homotopy theory:**
The question is basically whether the stable version of Serre's method of killing homotopy groups leads directly to a subexponential upper bound on the homotopy groups of a finite spectrum. I learned [here](https://mathoverflow.net/questions/291507/growth-of-stable-homotopy-groups-of-spheres/) that the size of the stable stems (measured as $\log |\pi\_k \mathbb S|$) is conjectured to grow roughly linearly, but that no subexponential bound seems to be known.
The motivating observation for the following approach is the simple fact that the dimension $\operatorname{dim} \mathcal A^k$ of the Steenrod algebra grows subexponentially in $k$ (Proof: by Milnor's description of the dual $\mathcal A\_\ast$, we have that $\dim \mathcal A^k$ counts certain partitions of $k$, and the number of partitions grows subexponentially). Since killing homotopy groups just keeps peeling off Eilenberg-MacLane spectra, I have some hope that when one adds up all of these subexponential contributions, the result might still be subexponential.
I know very little about the mod $p^n$-Steenrod algebras for $n \geq 2$, but there are recent results by Mathew and by Burklund giving good bounds on the exponents of the stable stems, so for the purposes of this post I'm going to ignore this issue and blithely pretend that all homotopy groups I see have exponent $p^1$.
So let $X = X\_{\geq 0}$ be a connective $p$-local spectrum, and let $X\_{\geq k}$ denote the $k$-connective cover of $X$. Assume that $X$ has finite homotopy groups in each degree. Consider the fiber sequence $\Sigma^{k-2} H\_{k-1} X\_{\geq k-1} \to X\_{\geq k} \to X\_{\geq k-1}$ (obtained by using Hurewicz and rotating the most obvious fiber sequence). This gives us the bound
$$\operatorname{dim} H\_n(X\_{\geq k}) \leq \operatorname{dim} H\_n(X\_{\geq k-1}) + \operatorname{dim} (H\_{k-1} X\_{\geq k-1}) \operatorname{dim}(\mathcal A^{n-k+2})$$
So let us set $h\_{n,k} = \operatorname{dim} H\_n(X\_{\geq k})$ and $a\_{n} = \operatorname{dim}(\mathcal A^{n}$). The goal is to get a subexponential bound on $h\_{k,k}$, say when $X = M(p)$ is the mod $p$ Moore spectrum so that $h\_{n,0} = \delta\_{n,0}$ is just the Kronecker delta.
**Combinatorics:**
Here's the **Question:**
>
> Let $h\_{n,k}$ be natural numbers defined for $n,k \in \mathbb N$, where $h\_{n,k} = 0$ for $n < k$. Let $a\_n$ be natural numbers defined for $n \in \mathbb N$ satisfying an inequality $a\_n \leq \exp(c \log(n)^d)$ for some $c,d>0$ (by convention, $a\_n = 0$ for $n < 0$). Suppose that we have the inequality
>
>
> $$h\_{n,k} \leq h\_{n,k-1} + h\_{k-1,k-1}a\_{n-k+2}$$
>
>
> for all $n \in \mathbb N$ and $k \geq 1$. As a boundary condition, suppose that $h\_{n,0} = \delta\_{n,0}$ is just the Kronecker delta. Does there follow an upper bound for $h\_{k,k}$ which is subexponential in $k$?
>
>
>
**Remarks:**
Because of the simplifying assumption made about the exponents of the groups involved, I'm not certain that a positive answer to the combinatorial question would give a subexponential bound on the stable stems, but I suspect the simplifying assumption can only make things worse for us, so it probably would.
|
https://mathoverflow.net/users/2362
|
Is there an elementary subexponential upper bound on the size of the stable stems?
|
There is indeed a subexponential bound on the size of the stable stems, which can be seen already at the $E\_2$ page of the May spectral sequence. See [John Palmieri's comment](https://mathoverflow.net/questions/385019/is-there-an-elementary-subexponential-upper-bound-on-the-size-of-the-stable-stem#comment981054_385019). I've explained this in a bit more detail [here](https://mathoverflow.net/a/385081/2362).
|
0
|
https://mathoverflow.net/users/2362
|
410608
| 167,992 |
https://mathoverflow.net/questions/410498
|
2
|
I'm wondering if anyone has shown Fatou's lemma for sums when the limits are taken over nets.
That is, has anyone proved the following?
>
> Let $\{x\_{k,\alpha}\}\_{k\in \mathbb N,\alpha \in \mathcal A}\subseteq \mathbb R\_+$ with
> $\mathcal A$ a directed set. Then
>
>
> $$\sum\_{k=1}^{\infty} \sup\_{\bar \alpha}\inf\_{\alpha\geq \bar \alpha}
> x\_{k,\alpha}\leq \sup\_{\bar \alpha}\inf\_{\alpha\geq \bar \alpha}
> \sum\_{k=1}^{\infty} x\_{k,\alpha}.$$
>
>
>
This result is not true if the sum were replaced by a general measure.
|
https://mathoverflow.net/users/470906
|
Fatou's lemma and dominated convergence for nets and the counting measure
|
If $S \mathrel{:=} \sup\_{\overline\alpha} \inf\_{\alpha \ge \overline\alpha} \sum\_{k = 1}^\infty x\_{k, \alpha}$ were strictly less than $\sum\_{k = 1}^\infty \sup\_{\overline\alpha} \inf\_{\alpha \ge \overline\alpha} x\_{k, \alpha}$, then there would be some $N$ such that $S$ was strictly less than $\sum\_{k = 1}^N \sup\_{\overline\alpha} \inf\_{\alpha \ge \overline\alpha} x\_{k, \alpha}$, and then some $\overline\alpha\_0$ such that $S$ was strictly less than
$$\sum\_{k = 1}^N \inf\_{\alpha \ge \overline\alpha\_0} x\_{k, \alpha} \le \inf\_{\alpha \ge \overline\alpha\_0} \sum\_{k = 1}^\infty x\_{k, \alpha} \le \sup\_{\overline\alpha} \inf\_{\alpha \ge \overline\alpha} \sum\_{k = 1}^\infty x\_{k, \alpha} = S.$$
|
2
|
https://mathoverflow.net/users/2383
|
410609
| 167,993 |
https://mathoverflow.net/questions/410617
|
6
|
I'm dealing with the expression $x = \frac{1}{3}y(y+1)(2y+1)^2(2y^2+2y+1)$. What is this approximately, if one is explicitly writing y in terms of x? There's no general formula for sixth powers unfortunately.
Also can one given an approximation of this so that the difference between the true y and the approximation go to zero? (Not just the ratio).
|
https://mathoverflow.net/users/265714
|
Solution to sixth order equation
|
For $x,y>0$ there is a unique solution $y(x)$ to $x = \frac{1}{3}y(y+1)(2y+1)^2(2y^2+2y+1)$ given by
$$y=\tfrac{1}{2} 3^{-1/3} \sqrt{\frac{\left(\sqrt{11664 x^2-3}+108 x\right)^{2/3}+3^{1/3}}{\bigl(\sqrt{11664 x^2-3}+108 x\bigr)^{1/3}}}- \tfrac{1}{2}.$$
Here is a plot of $y$ versus $x$.
|
14
|
https://mathoverflow.net/users/11260
|
410621
| 167,998 |
https://mathoverflow.net/questions/410665
|
4
|
For which fields $K$ and integers $n>1$ does the group ring $K(\mathbb{Z}/n\mathbb{Z})$ have idempotents distinct from $0$ and $1$?
|
https://mathoverflow.net/users/104638
|
Idempotents in group rings of finite cyclic groups
|
More generally, when $G$ is a finite group and $\mathbb{K}$ is a field, the group algebra $\mathbb{K}G$ has no non-trivial idempotent if and only if the regular $\mathbb{K}G$-module is indecomposable. If $G$ is non-trivial and the characteristic of $\mathbb{K}$ is either zero or is coprime to $|G|$, then this is never the case, since the augmentation ideal is a direct summand of the regular module in that situation.
If $|G|$ is divisible by the characteristic of $\mathbb{K}$, say $p$, then either $G$ is a $p$-group, in which case the regular $\mathbb{K}G$-module is indecomposable (eg by a theorem of J.A. Green), or else $|G|$ is divisible by some prime $q \neq p,$ in which case $G$ has a conjugacy class of elements of order $q$, and the projective cover of the trivial $\mathbb{K}G$-module has dimension at most $\frac{|G|}{q},$ so that $\mathbb{K}G$ contains an idempotent which is neither zero nor $1$.
Hence when $\mathbb{K}$ has prime characteristic $p$ and $G$ is a finite group, the group algebra $\mathbb{K}G$ has an idempotent other than $0$ or $1$ if and only if $G$ is not a $p$-group, while if $\mathbb{K}$ has characteristic zero and $G$ is a non-trivial finite group, then $\mathbb{K}G$ always has idempotents other than $0$ or $1$.
|
5
|
https://mathoverflow.net/users/14450
|
410670
| 168,010 |
https://mathoverflow.net/questions/410581
|
0
|
This posting comes as a possible salvage to [this](https://mathoverflow.net/questions/409916/whats-the-consistency-strength-of-this-form-of-reflection) earlier presented reflective theory which was [proved inconsistent](https://mathoverflow.net/questions/409916/whats-the-consistency-strength-of-this-form-of-reflection/410575#410575). Attesting the particulars of the underlying language in reflection axioms.
Working in *mono-sorted* first order logic with equality $``="$ and membership $``\in"$:
* *Define:* $set(x) \equiv\_{df} \exists y \, (x \in y)$
* Axiomatize:
1. **Extensionality:** $( a \subseteq b \land b \subseteq a \to a=b)$
2. **Separation:** $(set(a) \to \exists \ set \ x : \forall y \, (y \in x \leftrightarrow y \in a \land \phi ))$
3. **Reflection:** $ (\varphi \to \exists \ set \ x : \text { trs}(x) \land \varphi^x)$
where formulas $\phi, \varphi$ do not use $``x"$; $\varphi$ do not use $``="$; $\varphi^x$ is obtained from $\varphi$ by merely bounding its quantifiers by $``\subseteq x"$ if the quantified variable appears only on the right of symbol $\in$, and by $``\in x"$ whenever it appears on the left. $``\text { trs}" $ stands for *is transitive*.
>
> Can this syntactic restriction on $\varphi^x$ manage to escape inconsistency?
>
>
>
I conjecture that this reflection scheme would be equivalent to second order [Bernays reflection](https://www.researchgate.net/publication/220366286_Bernays_and_Set_Theory) schema (page 21) over the rest of axioms of this theory.
|
https://mathoverflow.net/users/95347
|
Is this reflection schema equivalent to second order Bernays reflection?
|
This theory is inconsistent.
We note that by 1 and 2 that if set(x) and y⊆x, then set(y).
(a) There is a v such that ∀x(set(x)-->x∈v).
Proof:Suppose not. Then ∀v∃s∃t(s∈t∧s∉v). By 3 there is transitive x such that
```
set(x) and ∀v(v⊆x-->∃s∃t(s∈x∧t∈x∧s∈t∧s∉v). In particular
(x⊆x-->∃s∃t(s∈x∧t∈x∧s∈t∧s∉x). But this is impossible.
```
Suppose that ∀x(set(x)-->x∈V).
(b)V∉V
Proof:Suppose V∈V. By 2, there is an x such that set(x) and ∀y(y∈x<-->y∈V∧y∉y).
```
Then x∈x<-->x∉x.
```
Then ∃w(w∉w∧∀t(t∈V-->t∈w)). By 3, there is a transitive x such that
∃w(w∈x∧(w∉w∧∀t(t∈x∧t∈V-->t∈w))). But then x=w, and thus w∈w.
|
2
|
https://mathoverflow.net/users/133981
|
410682
| 168,016 |
https://mathoverflow.net/questions/410577
|
17
|
Let $p$ be a prime and $n \in \mathbb N$. Does there exist a closed manifold which is of type $n$ after $p$-localization?
When $n= 0$ the answer is yes. When $p = 2$ and $n = 1$ we can take $\mathbb R \mathbb P ^2$.
Other than that, I'm not sure.
**Notes:**
* Recall that a finite CW complex $X$ is said to be of type $n$ if $\widetilde{K(n)}\_\ast X \neq 0$ but $\widetilde{K(m)}\_\ast X = 0$ for $m < n$, where $K(n)$ is the $n$th Morava $K$-theory at the prime $p$.
* Recall the [thick subcategory theorem](https://ncatlab.org/nlab/show/thick+subcategory+theorem) tells us that for every $n$ (and every $p$) there exists a finite CW complex which is of type $n$ after $p$-localization, and conversely that every finite CW complex is of type $n$ for some $n$ after $p$-localization. The question is whether "finite CW complex" can be upgraded to "closed manifold".
* When $n \geq 1$, a closed manifold of type $n$ can't be orientable (since if it's orientable, then by Poincare duality its reduced rational homology is nonvanishing, and $K(0) = H\mathbb Q$).
|
https://mathoverflow.net/users/2362
|
For which $n$ does there exist a closed manifold of (chromatic) type $n$?
|
After discussing this with Tim we came up with the following answer:
The first steifel whiteny class $\omega\_1$ of $M$ can be written as the following composition:
$$M \to BO(n) \to BO \to BAut(\mathbb{S}) \to BAut(\mathbb{Z}) \simeq B\mathbb{Z}/2$$
But if $M$ is of type $\ge 2$ then $[M,BO]\simeq [\Sigma^\infty M, bo] \simeq 0$ since $bo$ is of height $\le 1$. So $M$ must be orientable in cotradiction with the third point.
**Conclusion:** All closed smooth manifolds are of type $\le 1$.
~~Oh and I believe that at odd primes, type $1$ complexes can be realized by Lens manifolds.~~ Here I was uncareful. This is wrong as it conflicts with the Tim's third point as was pointed out by Gregory Arone in the comments.
|
10
|
https://mathoverflow.net/users/22810
|
410684
| 168,017 |
https://mathoverflow.net/questions/410655
|
3
|
Let $G$ be a reductive group over a number field $\mathbb{Q}$ and $K$ be a maximal compact subgroup of $G$.
Let $\Gamma$ be an arithmetic subgroup of $G(\mathbb{Q})$.
Let $\mathfrak{g}$ be the complexfied Lie alogebra of $G$ and $Z(\mathfrak{g})$ the center of the universal enveloping algebra $U(\mathfrak{g})$ of $\mathfrak{g}$.
Then one can define the automorphic form on $\Gamma \backslash G(\mathbb{R})$ as function on $\Gamma \backslash G(\mathbb{R})$ satisfying 'smooth', 'right $K$-finite', 'moderate growth', '$Z(\mathfrak{g})$-finite' conditions.
Let $X$ be an element in $\mathfrak{g}$ and $\phi$ is an automorphic form $\Gamma \backslash G(\mathbb{R})$.
Then some book says that $X\phi$ satisfies also $K$-finite conditions. But there is no proof on this and I can't check it well.
Would you let me know why $X\phi$ is also $K$-finite?
It seems that it would use some compatibility property of $\mathfrak{g}$-actions and $K$-actions, (i.e., $kX\phi=((Ad(k)X))k\phi$ for $k\in K$, $X\in \mathfrak{g}$.)
Any comments are appreciated!
|
https://mathoverflow.net/users/35898
|
Why differential operator preserves $K$-finiteness of automorphic form?
|
Consider the map $\mathfrak{g}\otimes C^\infty(\Gamma\backslash G)\to C^\infty(\Gamma\backslash G)$, $X\otimes f\mapsto Xf$.
The group $K$ acts on both sides and the map is equivariant. If $f$ lies in a finite-dimensional $K$-module $M$, then $Xf$ lies in the finite-dimensional $K$-module that is the image of $\mathfrak{g}\otimes M$.
|
3
|
https://mathoverflow.net/users/nan
|
410685
| 168,018 |
https://mathoverflow.net/questions/410640
|
3
|
Consider the following nonlocal ODEs on $[1,\infty)$.
#1)
$$\begin{align}
r^2 f''(r) + 2rf'(r)-l(l+1) f(r) &= -\frac{(f'(1) + f(1))}{r^2}\\
f(1) &= \alpha \\
\lim\_{r\to \infty} f(r) &= 0
\end{align}$$
#2
$$\begin{align}
r^2 f''(r) + 2rf'(r)-l(l+1) f(r) &= -\frac{(f'(1) + f(1))}{r^2}\\
f'(1) &= \beta \\
\lim\_{r\to \infty} f(r) &= 0
\end{align}$$
where $l$ is a positive integer, $\alpha, \beta \in \mathbb{R}$.
Both ODEs are uniquely solvable, and so we can define the Dirichlet to Neumann operator $T: \alpha \mapsto \beta$. I am trying to prove some properties of $T$.
Define the following norm $\lVert \cdot \lVert$:
$$\lVert f \lVert^2 := \int\_1^{\infty} r^2 f'(r)^2 dr + l(l+1) \int\_1^{\infty} f(r)^2 dr$$
It can be shown that $\lVert f \lVert \leq C |f'(1)|$ for any $f$ solving the above ODE, where $C$ is independent of $f$ and $l$. It then follows that $|f(1)| \leq |\int\_1^{\infty} f'(r) dr| \leq C' \sqrt{\int\_1^{\infty}r^2 f'^2} \leq C' \lVert f \lVert \leq C'C|f'(1)|$
And so we have for any $f$ solving the above ode,
$$|f(1)| \leq \bar C |f'(1)|$$
for some $\bar C$ that is independent of $f$ and $l$.
However, the other way around is not true. In fact, it holds that
$$|f'(1)| \leq C \sqrt{l(l+1)}|f(1)|$$
for any $f$ solving the above ODE, where $C$ is independent of $f$ and $l$. I don't know how to prove this inequality. A weaker inequality is the following,
$$\lVert f \lVert \leq C \sqrt{l(l+1)} |f(1)|$$
which I also was not able to prove.
Any help is appreciated.
|
https://mathoverflow.net/users/138705
|
Dirichlet to Neumann operator for a nonlocal ODE
|
You have a linear ODE with explicit coefficients: the solutions can actually be written down explicitly via [variation of constants](https://en.wikipedia.org/wiki/Variation_of_parameters). To summarize the result1 when $\ell > 1$, let $g\_{a,b}(r)$ be given by
$$ g\_{a,b}(r) = \frac{a}{r^{\ell+1}} + \frac{b}{r^2} $$
you find that
$$ r^2 g\_{a,b}'' + 2r g\_{a,b}' - \ell(\ell+1) g\_{a,b} = -\frac{(\ell+2)(\ell-1) b}{r^2} $$
As
$$ g\_{a,b}(1) = a + b , \qquad g'\_{a,b}(1) = -(\ell+1) a - 2b $$
for $g\_{a,b}$ to solve the equation you indicated requires $(\ell+2)\ell-1)b = g\_{a,b}(1) + g'\_{a,b}(1)$. So we need (when $\ell > 1$)
$$ - \ell a - b = (\ell+2)(\ell - 1)b \implies a = \frac{1 - \ell - \ell^2}{\ell} b $$
which yields
$$ \alpha = \frac{1-\ell^2}{\ell} b, \qquad \beta = - \frac{b}{\ell}(1 + 2\ell - 2\ell^2 - \ell^3) $$
so the Dirichlet to Neumann map has norm exactly
$$ \frac{\ell^3 - 2\ell^2 + 2\ell + 1}{\ell^2 - 1} = O(\ell) = O(\sqrt{\ell(\ell+1)})$$
---
1 The cases where $\ell = 0$ and $1$ requires more care.
When $\ell = 0$, the equation you wrote is only solvable when $b = 0$, and the Dirichlet-to-Neumann map has norm 1.
When $\ell = 1$, variation of constants gives
$$ g\_{a,b} = \frac{1}{r^2}(a + b\ln(r)) $$
for which
$$ r^2 g\_{a,b}'' + 2r g'\_{a,b} - 2 g\_{a,b} = -\frac{3b}{r^2} $$
which then requires $a = -2 b$, and hence $\alpha = -2b$ and $\beta = 5b$, and the D2N map has norm $5/2$.
|
3
|
https://mathoverflow.net/users/3948
|
410686
| 168,019 |
https://mathoverflow.net/questions/410678
|
2
|
Given two SDEs $X^1$, $X^2$ :
$$X^i\_t=1+t+\int\_0^t\sigma\_i(s)dW\_s,\quad \forall t\ge 0,$$
where $\sigma\_i:\mathbb R\_+\to [1/2,1]$ are non-decreasing s.t. $\sigma\_1(t)\le \sigma\_2(t)$ for all $t\ge 0$. Can we prove $\mathbb P[\inf\_{0\le s\le t}X^1\_s>0]\ge \mathbb P[\inf\_{0\le s\le t}X^2\_s>0]$ for all $t\ge 0$?
|
https://mathoverflow.net/users/261243
|
Does higher volatility of SDE imply lower probability of staying positive?
|
Yes. This follows by time change: The process $(X^i\_t)$ equals the process $(1+t+W\_{\tau\_i(t)})$ in distribution, where
$$\tau\_i(t):=\int\_0^t\sigma\_i(s)^2\,ds,$$
so that $\tau\_2\ge\tau\_1$ and hence for the corresponding inverse functions we have $\tau\_2^{-1}\le\tau\_1^{-1}$.
It follows that
$$\begin{aligned}
&\inf\_{s\in[0,t]}(1+s+W\_{\tau\_2(s)}) \\
&=\inf\_{u\in[0,\tau\_2(t)]}(1+\tau\_2^{-1}(u)+W\_u) \\
&\le\inf\_{u\in[0,\tau\_1(t)]}(1+\tau\_1^{-1}(u)+W\_u) \\
&=\inf\_{s\in[0,t]}(1+s+W\_{\tau\_1(s)}).
\end{aligned}$$
So, for all real $x$,
$$
\begin{aligned}
&P(\inf\limits\_{s\in[0,t]}X^1\_s>x) \\
&=P(\inf\_{s\in[0,t]}(1+s+W\_{\tau\_1(s)})>x) \\
&\ge
P(\inf\_{s\in[0,t]}(1+s+W\_{\tau\_2(s)})>x) \\
&=P(\inf\limits\_{s\in[0,t]}X^2\_s>x).
\end{aligned}$$
Thus, the desired result follows.
|
3
|
https://mathoverflow.net/users/36721
|
410689
| 168,021 |
https://mathoverflow.net/questions/410693
|
3
|
Let $R$ be a ring and $I$ an ideal. I am interested under which conditions the following holds:
>
> **Claim.** Suppose that any two elements in $I$ have a non-trivial $\operatorname{gcd}$. Then $I$ is contained in some non-trivial principal ideal.
>
>
>
For simplicity you may assume that $R$ is a UFD so that $\operatorname{gcd}$s exist and are well-defined (a GCD domain probably suffices for this too but I have no experience with them). Although this question looks quite simple, I only know of one non-trivial example where this holds. Namely $R=\mathbb Z\_p[[T]]$ and by extension $R=\mathcal O[[T]]$ for $\mathcal O$ the ring of integers of some local field extension.
I asked this question some while ago on Math.SE ([here](https://math.stackexchange.com/questions/4317163/a-criterion-for-whether-an-ideal-is-contained-in-a-principal-ideal)) but have received no comment/answer. There you can also find a proof of the above claim for $R=\mathbb Z\_p[[T]]$. As far as I can tell, this proof does not generalize notably.
>
> Are there some reasonable conditions one can put on $R$ to make the above claim work? By reasonable I mean something not making the assertion trivial (e.g. not taking a PID but maybe UFD+Noetherian suffices).
>
>
>
Thanks in advance!
|
https://mathoverflow.net/users/127269
|
A criterion for whether an ideal is contained in a principal ideal
|
This is true in UFDs. Let $f$ be any element of $I$ and let $\pi\_1,\dots, \pi\_n$ be its irreducible factors (choosing one from each class of irreducible factors that differ by a unit). If, for some $i$, $I \subseteq (\pi\_i)$ then we are done. Otherwise, for each $i$ let $g\_i$ be an element of $I$ that is nonzero mod $\pi\_i$.
Then $$ h= \sum\_{i=1}^n g\_i \prod\_{ 1\leq j \leq n, j \neq i} \pi\_j$$ is an element of $I$. Because $R$ is a UFD, $\pi\_i$ is a prime ideal, so $R / \pi\_i$ is an integral domain. For $j\neq i$, since $\pi\_j$ is irreducible and not a unit times $\pi\_i$, $\pi\_j$ is not a multiple of $\pi\_i$, so $\pi\_j$ is not a zero-divisor mod $\pi\_i$. Thus
$$ h \equiv g\_i \prod\_{ 1\leq j \leq n, j \neq i} \pi\_j \not\equiv 0 \mod \pi\_i$$
Since $h$ is not a multiple of $\pi\_i$ for any irreducible factor $\pi\_i$ of $f$, $f$ and $h$ do not have a gcd, contradicting the assumption.
|
2
|
https://mathoverflow.net/users/18060
|
410694
| 168,023 |
https://mathoverflow.net/questions/410677
|
8
|
What is the étale fundamental group of the circle $X({\bf R})$, where
$$
X(k) = \{(x,y) \in k^2 \mid x^2+y^2 = 1\}?
$$
I know that there is a sequence
$$
1 \rightarrow \pi\_1^{et}(X({\bf C})) \rightarrow \pi\_1^{et}(X({\bf R})) \rightarrow Gal({\bf C}/{\bf R}) \rightarrow 1
$$
with $Gal({\bf C}/{\bf R})= \{z\mapsto z, z\mapsto \bar{z}\}$. The first group is the profinite completion of $\bf Z$ since $X({\bf C})$ is the projective complex line with two points removed, but I don't know if the sequence splits or not.
More generally, what can be said about that sequence for general smooth varieties over ${\bf R}$?
|
https://mathoverflow.net/users/6129
|
Etale fundamental group of the circle
|
As Donu explained, the sequence splits by choosing an $\mathbf R$-point of $X$. So the only question remaining is what the $\operatorname{Gal}(\mathbf C/\mathbf R)$-action on $\pi\_1^{\text{ét}}(X\_{\mathbf C})$ is. I claim that the action is trivial, because the two points at infinity $V(x^2+y^2)$ are not defined over $\mathbf R$. (By contrast, for $\mathbf G\_{m,\mathbf R}$ we get the nontrivial action where a generator $\sigma \in \operatorname{Gal}(\mathbf C/\mathbf R)$ acts by $-1$, using the same argument as below.)
Indeed, consider the tower $Y\_n \to X\_{\mathbf C} \to X$ where $Y\_n \to X\_{\mathbf C}$ is the unique cover of degree $n$. The composite $Y\_n \to X$ is Galois as $\pi\_1^{\text{ét}}(X\_{\mathbf C}) \trianglelefteq \pi\_1^{\text{ét}}(X)$ is normal and $n\mathbf Z \subseteq \mathbf Z$ is characteristic (in other words, any conjugate $Y'\_n$ would contain $X\_{\mathbf C}$ as $X\_{\mathbf C} \to X$ is normal, hence $Y'\_n \cong Y\_n$ as $X\_{\mathbf C}$-covers since $Y\_n$ is the unique degree $n$ cover).
We need to compute $\operatorname{Gal}(Y\_n/X)$. This depends on some choices of isomorphisms: we choose $X\_{\mathbf C} \stackrel\sim\to \mathbf G\_{m,\mathbf C}$ via $(x,y) \mapsto x+yi$, and we write $t = x+yi$. Under this identification, the generator $\sigma$ of $\operatorname{Gal}(X\_{\mathbf C}/X) = \operatorname{Gal}(\mathbf C/\mathbf R)$ acts on $\mathbf G\_{m,\mathbf C}$ via $x+yi \mapsto x-yi$, i.e. the $\mathbf C$-semilinear map $\sum\_j c\_jt^j \mapsto \sum\_j \bar c\_jt^{-j}$.
Then $Y\_n$ can be identified with $\operatorname{Spec} \mathbf C[t^{\pm1/n}]$. The Galois group of $Y\_n$ over $Y\_1$ is $\mu\_n(\mathbf C)$, where $\zeta \in \mu\_n(\mathbf C)$ acts as the $\mathbf C$-linear map $t \mapsto \zeta t$. The conjugation action of $\sigma \in \operatorname{Gal}(X\_{\mathbf C}/X)$ on $\operatorname{Gal}(Y\_n/X\_{\mathbf C})$ is therefore trivial, as
$$(\sigma \circ \zeta \circ \sigma^{-1})(t) = (\sigma \circ \zeta)(t^{-1}) = \sigma (\zeta^{-1}t^{-1}) = \overline{\zeta^{-1}}t = \zeta t.$$
Thus, we see that the conjugation action of $\operatorname{Gal}(\mathbf C/\mathbf R)$ on $\pi\_1^{\text{ét}}(X\_{\mathbf C})$ is trivial. $\square$
|
10
|
https://mathoverflow.net/users/82179
|
410695
| 168,024 |
https://mathoverflow.net/questions/410578
|
1
|
This question is an example in the book Introduction to Probability Models 11th edition (Sheldon M.Ross). 3.6.2 A random graph:
A graph has $V$ nodes and a set $A$ of pairs of nodes in $V$ called arcs.
$V = \{1,2,...n\}$ and $A = \{(i,X(i), i=1,...,n\}$. The probability that node $i$ is connected to node $j$ is:
$P\{X(i)=j\}=\frac{1}{n}, j=1,2,...,n$
If from each node $i$, we select at random ONE of the $n$ nodes (including $i$ itself), what is the probability the graph is connected (there is a path between each pair of the $\binom{n}{2}$ nodes)?
I understand the proof of this part that this probability is: $P\{graph\ is\ connected\} \approx\sqrt{\frac{\pi}{2n(-1)}}.$
In the second part, we want to calculate the probability that there are 2 components in the random graph. Say nodes $1,2,3,...,k$ is connected, and nodes $k+1,...,n$ is connected. Let $C$ denotes the number of connected components. $P\_n(i)=P\{C=i\}$.
To calculate $P\_n(2)$, we need to use:
$P\{X(i)\in\{1,2,...,k\}, for\ all\ i=1,...,k\}=\left(\frac{k}{n}\right)^k$
$P\{X(i)\in\{k+1,...,n\}, for\ all\ i=k+1,...,n\}=\left(\frac{n-k}{n}\right)^{n-k}$
$P\{nodes\ 1,2,3,...,k\ form\ a\ connected\ subgraph\}=P\_k(1)$
$P\{nodes\ k+1,...,n\ form\ a\ connected\ subgraph\}=P\_{n-k}(1)$
So the point I DON'T understand is: because there are $\binom{n-1}{k-1}$ ways of choosing a set of $k-1$ nodes from the nodes $2$ through $n$, we have:
$P\_n(2)=\sum\limits\_{k=1}^{n-1}\binom{n-1}{k-1}\left(\frac{k}{n}\right)^k\left(\frac{n-k}{n}\right)^{n-k}P\_k(1)P\_{n-k}(1)$
Why we choose $k-1$ from $n-1$ rather than choose $k$ from $n$?
|
https://mathoverflow.net/users/471176
|
how to compute the probability that a random graph has two components?
|
There are several alternative derivations. It seems that in the derivation you are following, the first component is defined as the component of the node 1, so if this component has size $k$, then only $k-1$ additional nodes from $\{2,\ldots,n\}$ must be chosen for this component.
|
1
|
https://mathoverflow.net/users/7691
|
410696
| 168,025 |
https://mathoverflow.net/questions/410664
|
1
|
Actually, I have asked this question in <https://math.stackexchange.com/questions/4330127/orthogonal-transformation-of-multivariate-bernoulli-gaussian-distribution>, but I think mathoverflow might be more appropriate for it.
Recently, I studied multivariate Bernoulli-Gaussian distribution which is very useful for sparse signal processing. Suppose $X = (X\_{1}, \cdots, X\_{n})$ are i.i.d BG($p, \sigma^{2}$), we can know that $\mathbb{E}(X) = \mathbf{0}$ and Cov($X$) = diag($p \sigma^{2}$). Suppose $A$ is an orthogonal matrix with proper size and $Y = AX$, we can also know that $\mathbb{E}(Y) = \mathbf{0}$ and Cov($Y$) = diag($p \sigma^{2}$). Is it possible for us to know the distribution of $Y$ and is it possible for us to know $Y = (Y\_{1}, \cdots, Y\_{n})$ are i.i.d? Thank you.
|
https://mathoverflow.net/users/471275
|
Orthogonal transformation of multivariate Bernoulli-Gaussian distribution
|
Each $X\_j$ has the Bernoulli-Gaussian distribution,
$$P(x\_j)=(1-p)\delta(x\_j)+pN(x\_j;0,\sigma^2).$$
To characterize the distribution of the variables $Y\_i=\sum\_{j}A\_{ij}X\_j$, for an orthogonal $n\times n$ matrix $A$, I calculate the moment generating function:
$$F(z\_1,z\_2,\ldots z\_n)=\mathbb{E}\left[e^{\sum\_{i}z\_iy\_i}\right]=\mathbb{E}\left[e^{\sum\_{ij} z\_i A\_{ij} x\_j}\right]=\prod\_j \mathbb{E}\left[e^{\sum\_i z\_i A\_{ij} x\_j}\right]$$
$$\qquad
= \prod\_j \left(1-p + p \exp\left[\tfrac{1}{2} \sigma^2 \sum\_{k,l} z\_k z\_l A\_{kj}A\_{lj}\right]\right).$$
For $p\neq 0,1$ this does not factorize in terms dependent only one single $z\_j$, so the $Y\_i$ variables are not independent.
|
1
|
https://mathoverflow.net/users/11260
|
410697
| 168,026 |
https://mathoverflow.net/questions/410533
|
7
|
$\DeclareMathOperator\PSL{PSL}$ Let $ \mathbb{Z} $ be the ring of integers and $ \mathbb{R} $ the field of real numbers. Let $ \Sigma\_g $ be a surface of genus $ g \geq 2 $. Let $ \pi\_1(\Sigma\_g) $ be the fundamental group of the surface. There are many way to embed $ \pi\_1(\Sigma\_g) $ into $\PSL\_2(\mathbb{R}) $. There are, however, no ways to embed $ \pi\_1(\Sigma\_g) $ into $ \PSL\_2(\mathbb{Z}) $. Given some $ g \geq 2 $, is there a good way (an algorithm) to find a real algebraic integer $ \alpha $ such that $ \pi\_1(\Sigma\_g) $ embeds in $ \PSL\_2(R) $? Here $ R $ is the ring
$$
R:=\mathbb{Z}[\alpha]
$$
(Preferably $ \alpha $ is just a (real) quadratic extension. That is, $ \alpha $ is the root of some polynomial $ x^2+bx+c $ where $ b^2-4c \geq 0 $ and $ b,c \in \mathbb{Z} $.)
History of the question: The original question claimed that surface groups do not embed in $ \PSL\_2(\mathbb{Q}) $ and asked for embeddings into $ \PSL\_2(\mathbb{F}) $ where $ \mathbb{F} $ is a finite degree field extension of $ \mathbb{Q} $. The claim that surface groups do not embed in $ \PSL\_2(\mathbb{Q}) $ is false. In fact there are many such embeddings.
The first edit of the question fixed this and asked instead for an embedding into $ \PSL\_2(R) $ for $ R $ a finite rank extension of $ \mathbb{Z} $ by algebraic integers. The current version of the question is the second edit.
|
https://mathoverflow.net/users/387190
|
Do surface groups embed into PSL_2 over a real quadratic integer ring?
|
For the new question an answer is given by arithmetic Fuchsian groups. For example it is well-known that the reflection group associated with the regular right-angled pentagon in $\mathbb H^2$ contains every surface group as it contains the fundamental group of the non-orientable surface of Euler characteristic -1 as an index-4 subgroup. On the other hand it is an index-10 subgroup in the (2,4,5)-triangle group $\Delta$ (by dividing the pentagon into triangles from the center with vertices on the pentagon's and on the middles of its edges). The latter is known to be arithmetic by a result of Takeuchi (see for example item 6 in the table in section 13.3 of Maclachlan--Reid <https://zbmath.org/?q=an%3A1025.57001>). Its trace field is $\mathbb Q(\sqrt 5)$, which means that $\Delta$ will be realisable as a subgroup of $\mathrm{PSL}\_2$ over a quadratic extension $F$ of $\mathbb Q(\sqrt 5)$, and contained in the ring of integers $\mathbb Z\_F$. There are many choices for such an extension, for example one can take $F = \mathbb Q(\sqrt 2, \sqrt 5)$ since the quaternion algebra ramifies only at primes dividing 2. So in principle you get an embedding of $\Delta$, and hence of any surface group, into $\mathrm{PSL}\_2(\mathbb Z\_F)$. You can ask Sage to compute an integral basis for the latter to get a complete answer to your question.
Another possibility to avoid the arithmetic machinery would be to use hyperbolic geometry to compute generators for the triangle group directly and see where they lie, this has probably been done but I don't know a reference (and I'm not sure how to choose the walls as to get something which would lie where it should, i.e. $\mathbb Z\_F$).
This may be not optimal, maybe all surface groups embed into $\mathrm{PSL}\_2$ over a real quadratic field. For example the (2, 4, 6)-triangle group is arithmetic with trace field $\mathbb Q$, it embeds into $\mathrm{PSL}\_2(\mathbb Z[\sqrt 2])$ if I'm not mistaken (see the table in Maclachlan--Reid) so its surface subgroups do as well. However I don't know which genera will be realised this way.
---
**EDIT** For the new question a positive general answer is given by computations of John Voight which he recorded in <https://arxiv.org/pdf/0802.0911.pdf>. The paper gives a complete list of all "Shimura curves" (a particular family of hyperbolic 2-orbifolds) whose underlying surface has genus at most 2. In particular his table 4.1 indicates that the curve associated to the full unit group of the unique maximal order in the $\mathbb Q$-quaternion algebra of discriminant 26 has genus 2 and no singularities, so the image of this group in $\mathrm{PSL}\_2(\mathbb R)$ is isomorphic to the surface group of genus 2 and it is contained in $\mathrm{PSL}\_2(\mathbb Z[\sqrt 26])$.
This representation should be computable explicitely using software developed by Voight and others, that i'm not too familiar with (i don't think it's available in Sage currently).
|
4
|
https://mathoverflow.net/users/32210
|
410715
| 168,030 |
https://mathoverflow.net/questions/410716
|
0
|
I'm stuck on a passage I do not understand, which reads:
>
> $$\int\_{r<|y|<1} \bigg| \frac{1}{(|y|^2 - r^2)^s |y|^n} - \frac{1}{|y|^{n+2s}}\bigg|\ \text{d}y$$
> $$\int\_1^r \bigg| \frac{1}{(t^2 - r^2)^s} - \frac{1}{t^{2s}}\bigg|\ \frac{\text{d}t}{t}$$
> $$r^{s}\int\_1^{1/r} \bigg| \frac{1}{(\tau^2 - 1)^s} - \frac{1}{\tau^{2s}}\bigg|\ \frac{\text{d}\tau}{\tau}$$
>
>
>
I understood there is a sort of rescaling between the second and the third line, but I cannot find a way to get the passage between the first and the second line.
I thought about a change of variable like $|y|^n = t$ but there is something that I miss, and it doesn't work.
This is an integral in $\mathbb{R}^n$, where $0<s<1$.
|
https://mathoverflow.net/users/471329
|
Understanding the performed change of variable in this integration
|
Note that the integrand only depends on the absolute value of $y$, and not on the direction of the vector.
That means it is a good idea to substitute $t = |y|$.
Let's denote $$f(t) = \bigg| \frac{1}{(t^2 - r^2)^s t^n} - \frac{1}{t^{n+2s}}\bigg|$$ for positive real $t$.
Denote $t\mathbb S^n$ to be the sphere of radius $t$.
Then we have
$$\int\_{r<|y|<1}f(|y|) dy = \int\_r^1\int\_{t\mathbb S^n}f(t)dudt = \int\_r^1\text{Vol}(t\mathbb S^n)f(t)dt = \text{Vol}(\mathbb S^n)\int\_r^1f(t)t^{n-1}dt.$$
This reduces to almost the same formula that you want. It seems to me that the volume of the unit sphere has been neglected in your text, and the integration goes from $1$ to $r$ which is strange as $r<1$.
For the third line you can substitute $\tau = \frac tr$, but it seems you already knew how to do that.
|
2
|
https://mathoverflow.net/users/470870
|
410720
| 168,032 |
https://mathoverflow.net/questions/410735
|
1
|
A finite, simple, undirected graph $G=(V,E)$ is said to be *(vertex)-critical* if removing any vertex decreases its [chromatic number](https://en.wikipedia.org/wiki/Graph_coloring). By $\delta(G)$ we denote the *minimum degree* of $G$. The *degeneracy* of $G$ is defined by $$\text{degen}(G) = \max\{\delta(H): H \text{ is an induced subgraph of }G\}.$$
Obviously, for any graph $G$ we have $\delta(G) \leq \text{degen}(G)$.
**Question.** Is there a vertex-critical graph $G$ such that $\delta(G) < \text{degen}(G)$?
|
https://mathoverflow.net/users/8628
|
Degeneracy vs minimal degree in critical graphs
|
There is such a graph, provided by the graph6 string
```
U???????wE?kAgWoAdGpQQ`GdA^?ldgsRW[C~w??
```
and is [uploaded to the House of Graphs](https://hog.grinvin.org/ViewGraphInfo.action?id=48171).
The graph is critical, as verified by the SageMath code:
```
a=Graph('U???????wE?kAgWoAdGpQQ`GdA^?ldgsRW[C~w??');
a.chromatic_number() #returns 5
for p in a.vertices():
b=a.copy()
b.delete_vertex(p)
if b.chromatic_number()==5:
print('The graph is not critical');
break
```
(You should not see the message "The graph is not critical".)
The graph has minimum degree $4$ and degeneracy at least $5$:
```
a.degree(0) #returns 4
a.delete_vertex(0);min(a.degree()) #returns 5
```
|
3
|
https://mathoverflow.net/users/125498
|
410739
| 168,036 |
https://mathoverflow.net/questions/410368
|
10
|
I am looking for a simple reference to the following fact:
>
> If $f:\Omega\to\mathbb{R}$ is continuous, where $\Omega\subset H$ is an open subset of a separable Hilbert space $H$, then for any $\varepsilon$ we can find a $C^\infty$ smooth function $f\_\varepsilon:\Omega\to\mathbb{R}$ such that $|f(x)-f\_\varepsilon(x)|<\varepsilon$ for $x\in\Omega$.
>
>
>
This result is stated on page 758 in [2] and it follows from results of [1] and the existence of a smooth partition of unity in $H$.
Is there any more direct and more recent reference? It is quite hard to read the original paper of Bonic and Frampton since they discuss much more general results and I believe there should be a much more straightforward reference to this fact.
**[1] R. Bonic, J. Frampton,** Smooth functions on Banach manifolds. *J. Math. Mech.* 15 (1966), 877–898.
**[2] J. Eells,** A setting for global analysis. *Bull. Amer. Math. Soc.* 72 (1966), 751–807.
|
https://mathoverflow.net/users/121665
|
Density of smooth function in Hilbert spaces
|
I think the paper of Bonic and Frampton is a good reference, but it is true that for a specific result like the one you are after, there is some amount of lingo that one can bypass. Here is my take on their proof, very heavily inspired by their Theorem 1. I must add that there might very well be a more readable proof of the fact in a classical book, I did not go through an extensive search.
Let $H$ be a separable Hilbert space,¹ and fix $\phi:\mathbb R\to\mathbb R\_+$ a smooth function with $\{\phi>0\}=(0,\infty)$, for instance $t\mapsto\mathbf 1\_{t>0}\exp(-1/t)$.
>
> **Lemma 1 (bump function).**
> There exists a smooth function $f:H\to\mathbb R$ with bounded support.
>
>
>
In particular, we immediately get (using translations and dilations, and considering squares) that the collection of open sets $\{f>0\}$, for $f:H\to\mathbb R\_+$ smooth, is a basis.
*Proof.*
The function $Q:x\mapsto|x|\_H^2/2$ is smooth (its derivatives are $D\_xQ=\langle x,\cdot\rangle\_H$, $D\_x^2Q = \langle\cdot,\cdot\rangle\_H$, $D\_x^kQ=0$ for $k\geq3$), so $x\mapsto\phi(1-Q(x))$ is such a function by smoothness of composite functions.
>
> **Lemma 2 (Partition of unity).**
> Let $\mathscr U$ be a collection of open sets in $H$ with union $\Omega$. There exists sequences of open sets $U\_n\in\mathscr U$ and smooth functions $f\_n:H\to\mathbb R\_+$ such that
>
>
> 1. $\operatorname{supp}(f\_n)\subset U\_n$;
> 2. every point $x\in\Omega$ admits a neighbourhood $V\_x$ such that $\{f\_n>0\}$ intersects $V\_x$ only for finitely many $n$.
> 3. the function $\sum\_{n\geq0}(f\_n)\_{|\Omega}$ (well-defined and smooth according to the previous point) is constant equal to 1.
>
>
>
*Proof.*
It suffices to find sequences that satisfy 1. and 2., together with the fact that for every $x\in\Omega$, $f\_n(x)>0$ for at least one $n$. Indeed, if $\mathscr V$ is the collection of every open ball whose closure belongs to some $U\in\mathscr V$, and $\tilde f\_n$ is given by the modified lemma, then the collection of $f\_n:=\tilde f\_n/\sum\_{m\geq0}\tilde f\_m$ satisfies 1., 2. and 3. (the sum might not be well-defined and smooth out of $\Omega$, but it is fine in a neighbourhood of $\operatorname{supp}(\tilde f\_n)$ by definition of $\mathscr V$).
We know from the bump function lemma that open sets of the form $\{g>0\}$ for $g:H\to\mathbb R\_+$ and $\operatorname{supp}(g)\subset U\in\mathscr U$ cover $\Omega$. Passing to a countable subcover,² we find a countable collection of smooth non-negative functions $g\_n$ such that for every $x\in\Omega$, we have $g\_n(x)>0$ for at least one $n$.
Define
$$f\_n:x\mapsto g\_n(x)\prod\_{m<n}\phi\big(2^{-n}-g\_m(x)\big).$$
Let us show that for an appropriate choice of $U\_n$, it satisfies 1., 2. and the modified 3. It is clearly non-negative and smooth. We have
$$\{f\_n>0\} = \{g\_n>0\}\cap\bigcap\_{m<n}\{g\_m<2^{-n}\},$$
so $\operatorname{supp}(f\_n)\subset\operatorname{supp}(g\_n)\subset U\_n$ for a well-chosen $U\_n\in\mathscr U$; this is point 1. For fixed $x\in\Omega$, if $n(x)$ is the first $n$ such that $g\_n(x)>0$ (which exists by the above), then clearly $x\in\{f\_{n(x)}>0\}$; this is point 3. Now setting
$$V\_x = \{g\_{n(x)}>g\_{n(x)}(x)/2\},$$
clearly $V\_x$ is an open neighbourhood of $x$. Moreover, for all $n>n(x)$ large enough, $2^{-n}<g\_{n(x)}(x)/2$ and $\{f\_n>0\}\subset\{g\_{n(x)}<2^{-n}\}$ cannot intersect $V\_x$. This is point 2.
Using the lemma, the result you state is classical.
>
> **Proposition.**
> For $\Omega\subset H$ open, $u:\Omega\to\mathbb R$ continuous and $\varepsilon>0$, there exists $u\_\varepsilon:\Omega\to\mathbb R$ smooth such that $|u-u\_\varepsilon|\_\infty<\varepsilon$.
>
>
>
*Proof.*
Let $\mathscr U$ be the collection of open balls in $\Omega$ such that $|u-u(\text{centre})|\_\infty<\varepsilon/2$ over said ball. Since $u$ is continuous, it is an open cover of $\Omega$. By the lemma, there exists sequences $(f\_n)$, $(x\_n)$, $(r\_n)$ satisfying points 1., 2. and 3. with $U\_n=B\_{x\_n}(r\_n)$ the open ball with centre $x\_n$ and radius $r\_n$. The sum
$$u\_\varepsilon:=\sum\_{n\geq0}u(x\_n)f\_n$$
satisfies $|u-u\_\varepsilon|\_\infty<\infty$, since
$$|(u-u\_\varepsilon)(x)| = \left|\sum\_{n\geq0}\big(u(x)-u(x\_n)\big)f\_n(x)\right|\leq\sum\_{n\geq0}\varepsilon f\_n(x)=\varepsilon.$$
---
¹ It is possible that the result is true for $H$ larger (not separable), using things like: from every cover $\mathscr U$ of $H$, we can extract a (possibly uncountable) subcover $\mathscr V$ with a map $\mathscr V\mapsto\mathbb N$ such that all $U,V\in\mathscr V$ with the same label are disjoint. This is true in every paracompact space, but I am not entirely sure we can modify Lemma 2 to get this general case.
² $H$ is separable, hence it admits a countable basis, and the same holds for $\Omega$. For every pair $U\subset V$ of elements in the countable basis, take one $W$ in the support-of-smooth-functions basis such that $U\subset W\subset V$, if it exists; the result is a countable subcover. This relies on choice as is, but if the countable dense subset is given one can find a (big) explicit such countable cover.
|
2
|
https://mathoverflow.net/users/129074
|
410740
| 168,037 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.