added
stringdate 2025-03-12 15:57:16
2025-03-21 13:32:23
| created
timestamp[us]date 2008-09-06 22:17:14
2024-12-31 23:58:17
| id
stringlengths 1
7
| metadata
dict | source
stringclasses 1
value | text
stringlengths 59
10.4M
|
---|---|---|---|---|---|
2025-03-21T14:48:31.234895
| 2020-06-12T16:40:24 |
362885
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Robert Israel",
"Tio Miserias",
"Wojowu",
"dohmatob",
"https://mathoverflow.net/users/13650",
"https://mathoverflow.net/users/1508",
"https://mathoverflow.net/users/159512",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/78539",
"pinaki"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630077",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362885"
}
|
Stack Exchange
|
A simple proof that non negative polynomials have even degree
I am looking for a simple proof that a non-negative polynomial in n variables has always even degree. I have proved it but using Artin's Theorem ( Every non-negative polynomial is the sum of squares of rational functions), but I think that there are simpler proofs.
I know that this question could be off-topic, but I couldn't get answers at MSE.
If you have posted the question on Math.SE before, you should link to it here. Anyway, it is easy to reduce to the case of one variable by restricting yourself to lines, and hopefully you know how to do it in the case of one variable.
Yes, I know the one variable case. How do you reduce it to this case?
If the original polynomial has odd degree, then you can find a line such that the restriction to that line also has odd degree.
@TioMiserias This can be solved via an elementary argument. See my answer below.
This can be proved via an elementary argument.
Univariate case
Let's first look at the univariate case. Suppose $f(x) = x^{2r+1} + \sum_{0 \le j \le 2r}a_jx^j \in \mathbb R[x]$ is a nonnegative polynomial of odd degree $d=2r+1$, assumed to be unitary w.l.o.g. Then, in the limit as $x \longrightarrow -\infty$, we have the following inconsistent diagram
$$
0 \le \frac{f(x)}{x^{2r}} = x + o(1) \rightarrow -\infty,
$$
where the LHS is because $f(x) \ge 0$, and $x^{2r} > 0$ for $x \ne 0$.
Edit: correct solution for multivariate case
For the multivariate case, the story is similar. I've added a modification to address some counterexamples in the comments section. Apart from this small modification, the DNA of my strategy is essentially the same as before. We first prove the following lemma
Lemma. There always exist $s \in \mathbb R^n$ such that $m_s(x) := f(s_1 x,\ldots,s_n x) \in \mathbb R[x]$ is a unitary univariate real polynomial with the same degree $d$ as the $n$-variate polynomial $f(x_1,\ldots,x_n) \in \mathbb R[x_1,\ldots,x_n]$.
For example, in the counterexample of Robert Israel, simply take $(s_1,s_2) = (100, 1)$, to get $f(s_1 x, s_2 x) = 100 x^3 - x^3 + 100 x^2 = 99x^3 + 100 x^2$, a 3rd degree univariate polynomial.
Proof of Lemma.
To see why such a tuple exists, suppose w.l.o.g that $f$ only contains monomials of degree $d$ (everything else deleted), i.e suppose $f(x_1,\ldots,x_n) = \sum_{|\alpha| = d}c_\alpha \Pi_{i=1}^n x_i^{\alpha_i}$ where the $c_\alpha$'s a real numbers which are not all zero. Recall that for $\alpha_1,\ldots,\alpha_n \in \mathbb N$, the notation $|\alpha|$ is shorthand for $\alpha_1 + \dots + \alpha_n$.
Now, for $s \in \mathbb R^n$, consider the unitary univariate polynomial $m_s(x) := f(s_1x,\ldots, s_n x) \in \mathbb [x]$ of degree $\le d$. One computes
$$
m_s(x) = \sum_{|\alpha| = d}c_\alpha \Pi_{i=1}^n s_i^{\alpha_i}x_i^{\alpha_i} = \sum_{|\alpha| = d}c_\alpha x^{|\alpha|}\Pi_{i=1}^n s_i^{\alpha_i} = x^d\sum_{|\alpha| = d}c_\alpha \Pi_{i=1}^n s_i^{\alpha_i}=:x^d f(s_1,\ldots,s_n).
$$
But because $f$ is not an identically zero polynomial, we can always choose $s \in \mathbb R^n$ such that $f(s_1,\ldots,s_n) \ne 0$. For this choice of $s$, $m_s(x)$ has degree exactly equal to $d$, and we're done. $\Box$
Mindful of the above lemma, we can assume w.l.o.g that $f(x,\ldots,x)$ is a univariate polynomial of degree $d=2r+1$ (for there is always a subsitition $x_j \leftarrow s_j x_j$ for all $j=1,\ldots,n$, which lands us here). Thus $f(x,\ldots,x) = x^{2r+1} + \text{ lower order terms in }x$. Thus, in the limit as $x \longrightarrow -\infty$, we have
$$
0 \le \frac{f(x,\ldots,x)}{x^{2r}} = x + o(1)\rightarrow -\infty,
$$
which is again, an inconsistent diagram.
We conclude that any polynomial nonnegative polynomial in any number of variables must have even degree!
It's not necessarily true that $f(x,x,\ldots,x)$ has degree $2r+1$. Consider $f(x,y) = x^3 - y^3 + x^2$. But what you can say is that for generic $s_1, \ldots, s_n$, $f(s_1 x, \ldots, s_n x)$ has degree $2r+1$.
Except the highest order term may vanish on the main diagonal, e.g. if the polynomial is $x^2-2xy+y^2+$lower terms
@RobertIsrael Good catch. Let $d$ be the degree of original $n$-variate polynomial $f(x_1,\ldots,x_n)$. A simple fix to my original argument is to consider a substition of the form $f(s_1 x, \ldots, s_n x)$. There always exists such a subsitition which converts $f$ into a univariate polynomial of degree $d$. Please see updated answer.
|
2025-03-21T14:48:31.235172
| 2020-06-12T17:02:32 |
362889
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Clement C.",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/37266"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630078",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362889"
}
|
Stack Exchange
|
A measure of non-uniformity of a vector/probability distribution?
In the course of a research project about discrete probability distributions, my coauthors and I keep seeing some quantity appear, and I would like to understand whether it has been studied or has a characterization in the literature (I couldn't find any besides the 3 references below).
Let $p$ be a discrete probability distribution over some discrete set $\mathcal{X}$, seen as a vector $(p_i)_{i\in\mathcal{X}}$. The quantity is then
$$
s(p) := \frac{\lVert p\rVert_3^3 - \lVert p\rVert_2^4}{\lVert p\rVert_2^3} \tag{$\dagger$}
$$
(one could also consider $
t(p) := \frac{\lVert p\rVert_3^3 - \lVert p\rVert_2^4}{\lVert p\rVert_2^4}
$ if it's more convenient, both are equivalent for my purposes.) Note that the lack of homogeneity is just apparent, as we enforce $\lVert p\rVert_1=1$ (it's a probability distribution).
This quantity intuitively captures "how non-uniform" $p$ is: for instance, $s(p)=0$ iff $p$ is uniform on its support, and [1,2] give a (loose) connection between $t(p)$ and the total variation distance of $p$ to the closest uniform distribution. The numerator also appears in the analysis of the empirical estimator for the collision probability of a distribution [3].
We also have the view that, looking at the random variable $Y=p(X)$ for $X\sim p$,
$$
\mathbb{E}[Y] = \lVert p\rVert_2^2, \qquad \operatorname{Var}[Y] = \lVert p\rVert_3^3 - \lVert p\rVert_2^4
$$
which maybe may be useful? For instance, by Chebyshev's inequality,
$$
\mathbb{P}\{ |p(i) - \lVert p\rVert_2^2 |> \alpha\lVert p\rVert_2^2 \} \leq \frac{t(p)}{\alpha^2}
$$
thus giving a lower bound on the total mass (under $p$) of elements whose probability is within a factor $\alpha$ of $\lVert p\rVert_2^2$.
Do the quantities $s(p)$ (or $t(p)$) appear in the literature, and are either of them known to characterize (in some quantitative sense) how "non-uniform" a probability vector is?
[1] Batu, Tuğkan; Canonne, Clément L. Generalized uniformity testing. 58th Annual IEEE Symposium on Foundations of Computer Science—FOCS 2017, 880--889, IEEE Computer Soc..
[2] Diakonikolas, Ilias, Daniel M. Kane, and Alistair Stewart. Sharp bounds for generalized uniformity testing. In Advances in Neural Information Processing Systems, pp. 6201-6210. 2018.
[3] Diakonikolas, Ilias; Gouleakis, Themis; Peebles, John; Price, Eric. Collision-based testers are optimal for uniformity and closeness. Chic. J. Theoret. Comput. Sci. 2019.
Why the downvote? Is there something wrong with my question, and if so, how can I improve it?
isn't the $L_2$ norm $||p||_2^2$ the typical way to measure nonuniformity of a distribution?
@CarloBeenakker It is definitely one way. But it has some issues as well, and doesn't quite cut it: for instance, when the support size $k$ is large enough you can have two distributions with same $\ell_2$ norm, but one with all elements within a multiplicative factor say $2$ of $1/k$, and the other where a single (or a constant number of elements) with probabity $\approx 1/\sqrt{k}$ and all the others probability $\approx 1/k$.
|
2025-03-21T14:48:31.235650
| 2020-06-12T18:43:18 |
362899
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Qfwfq",
"Robbie Lyman",
"Wojowu",
"YCor",
"https://mathoverflow.net/users/135175",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/4721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630079",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362899"
}
|
Stack Exchange
|
Can one turn finite-dimensional vector subspaces into a cancellative semigroup?
Let $V$ be a vector space over some field and let ${\rm Fin}\,V$ be the family of all finite-dimensional subspaces of $V$. Is it possible to turn ${\rm Fin}\,V$ into an commutative cancellative semigroup $(S,+)$ such that $F,G\subseteq F+G$ for all $F,G\in S$.
Note that here $+$ is not the familiar Minkowski sum.
depending on your view of set theory, although there are countably many isomorphism classes of finite-dimensional $k$-vector spaces, there doesn't seem to be a good reason to assume there is a set consisting of all finite-dimensional $k$-vector spaces.
@Rylee Lyman: the op is assuming they're all subspaces of a fixed $V$.
If $V$ itself is finite-dimensional, then the answer is no - let $W$ be a codimension one subspace and let $W_1,W_2$ be two distinct complementary subspaces (of dimension one). Then necessarily $W+W_1=W+W_2=V$ (since $V$ is the only subspace containing both $W$ and $W_1$, same for $W_2$)
It would be nice not to denote this law by $+$ since $F+G$ already has a well-established meaning.
If $V$ is finite-dimensional, Wojowu explained that the answer is no. If $V$ is infinite-dimensional, we'll show that the answer is yes, and we can even have it be a monoid with the zero-dimensional space $0$ as the identity.
Let $S$ be the set of finite-dimensional subspaces of $V$, excluding the zero-dimensional space $0$. Let $\prec$ be a well ordering on $S$ whose order type is the initial ordinal of the cardinality of $S$.
Let $T_n$ be the set of finite multisets of at least $n$ elements whose elements lie in $S$. We well-order $T_2$ by 'colexicographical ordering':
If $A \subsetneq B$, then $A < B$;
If neither $A$ nor $B$ are subsets of each other, then let $a := \max(A \setminus B)$ and $b := \max(B \setminus A)$. Then $A < B$ if and only if $a \prec b$.
Then $T_2$ is also well-ordered by the initial ordinal of the cardinality of $S$.
Now, we construct a map $f : T_2 \rightarrow S$ by transfinite induction:
Given a partial function $g : T_2 \rightarrow S$ (initially the empty partial function), let $A$ be the first element of $T_2$ (in its well-ordering) such that:
$A$ is not already in the domain of $g$;
no element of $A$ is in the image of $g$.
Then we let $g'$ be the partial function which extends $g$ by setting $g'(A)$ to be an arbitrary element $W \in S$ such that:
For all $B \subsetneq A$ with $|B| \geq 2$, $g(B)$ is a subspace of $W$ (note that the colexicographical ordering on $T_2$ ensures that we've already defined $g$ on all such subsets $B$);
For all $U \in A$, $U$ is a subspace of $W$;
$W$ occurs later in the lexicographical ordering on $S$ than any element in $A$ or in the image of $g$.
Such an element $W$ necessarily exists because the first two conditions only narrow the set of candidate spaces to a set of equal cardinality to $S$ (this is where we require that $V$ is infinite-dimensional), and the third condition removes strictly fewer elements from the set of candidates (this is where we require the well-ordering to be with the initial ordinal [of that cardinality], rather than an arbitrary ordinal) so the set of candidates is non-empty.
Proceed inductively, and let $f$ be the limit of this process.
$f$ is defined on all finite multisets of $\geq 2$ elements from $S \setminus \textrm{Image}(f)$. It's injective, and when $A \subseteq B$, $f(A)$ is a subspace of $f(B)$.
We now define a function $h : S \rightarrow T_1$ as follows:
If $U$ is in the image of $f$, then $h(U)$ is its preimage;
otherwise, $h(U)$ is the singleton set containing $V$.
We can think of $h$ as 'decomposing' a vector space into its 'prime factor' vector spaces, and $f$ as 'multiplying' a multiset of 'prime factors' back into a vector space. Given that, we can now define:
$U + W := f(h(U) \sqcup h(W))$
where $\sqcup : T_1 \times T_1 \rightarrow T_2$ is the disjoint union operation on multisets. Then, it's easy to see that:
$U + W + X = f(h(U) \sqcup h(W) \sqcup h(X))$
for all spaces $U,W,X$, so we indeed have a commutative semigroup. It's also cancellative, because $f$ is injective, $h$ is injective, and $\sqcup$ is cancellative.
We can then throw in the zero-dimensional space $0$ as an identity element, obtaining a monoid.
|
2025-03-21T14:48:31.235931
| 2020-06-12T19:06:28 |
362901
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jason Starr",
"https://mathoverflow.net/users/103164",
"https://mathoverflow.net/users/13265",
"user347489"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630080",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362901"
}
|
Stack Exchange
|
$\mathscr Coh_{X|S} $ is algebraic and of finite type
Let $S$ be a Noetherian scheme and $X$ a projective $S$-scheme.
Reading Laumon-Moret--Bailly's "Champs Algebriques", Theorem <IP_ADDRESS>:
$ \mathscr Coh_{X|S} $ and $ \mathscr Fib_{X|S,r} $ are algebraic stacks of finite type.
The proof of "finite type" part seems to be done by proving that the $\mathfrak Isom$ sheaf above $U\in \operatorname{Aff}/S$ is a separated $U$-scheme of finite type.
Is there any reference that this property of the $\mathfrak Isom$ sheaf implies that the stack is of finite type?
I know that geometric properties of the algebraic stacks are "transferred" to stacks by a presentation/atlas, but I am not aware of such a thing for the $\mathfrak Isom$ sheaf.
Thank you for any reference on the subject.
It is recommended to always use a top-level tag in your questions, see here: https://meta.mathoverflow.net/questions/1075/frequently-asked-questions-about-tagging-on-mathoverflow for example the [algebraic-geometry] tag.
It is not true that every stack with finite type diagonal is of finite type. If that were true, then every algebraic space over $S$ would be of finite type. The proof of "finite type" for those two stacks uses "limit arguments" and the equivalence of "limit preserving" with "finitely presented".
|
2025-03-21T14:48:31.236170
| 2020-06-12T20:02:11 |
362905
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630081",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362905"
}
|
Stack Exchange
|
The definition of flat sheaves of rings on algebraic stacks
Reading Sheaves on Artin stacks by M. Olsson I find this definition (3.7, (i)):
Let $\mathcal X$ be an algebraic stack on a scheme $S$. A sheaf of rings $\mathcal A$ on $\mathcal X_{\textrm{lis-et}}$ is flat if for any smooth morphism $f \colon U \to V$ in $\operatorname{Lis-Et}(X)$, the natural map of sheaves on $U_{\textrm{et}}$
$$
f^{-1}(\mathcal A_V) \to \mathcal A_U
$$
is faithfully flat.
On the other hand, I read the book Champs algébriques by G. Laumon and L. Moret-Bailly, and in (12.7) there I find the following definition (I translate and adapt a little):
A sheaf of rings $\mathcal A$ on an $S$-algebric stack $\mathcal X$ is flat if for any arrow $\varphi \colon U \to V$ in $\operatorname{Lis-Et}(X)$ with $\varphi$ smooth, the homomorphism of étale rings
$$
\varphi^{-1}\mathcal A_V \to \mathcal A_U
$$
is flat.
Now, I must be missing something, but why does the first definition require faithful flatness whereas the second one just flatness? Are those definitions actually equivalent for some reason I can't figure out?
The book by Laumon and Moret-Bailly goes on to say (Examples 12.7.1) that the structure sheaf $\mathcal O_{\mathcal X}$ on an algebraic stack $\mathcal X$ is indeed flat. Does this hold if we assume the first definition of "flatness"? (I certainly hope so...)
|
2025-03-21T14:48:31.236291
| 2020-06-12T20:03:14 |
362906
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630082",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362906"
}
|
Stack Exchange
|
Binomial transform of Dirichlet series
Let $\Theta(s)$ be a Dirichlet series , and let $\beta$ be its abscissa of convergence:
$$\Theta(s)=\sum_{n=1}^{\infty}\frac{\theta(n)}{n^{s}}\;\;\;\;\;\;\Re(s)>\beta$$
And let $\left\{a_{n}\right\}_{n\in\mathbb{N}}$ be a sequence, whose ordinary generating function is $g(x)$:
$$g(x)=\sum_{n=0}^{\infty}a_{n}x^{n}\;\;\;\;\;\ |x|<x_{0}\leq 1$$
We want to write the related Dirichlet series:
$$\Phi (z)=\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{\theta(k+1)}{(k+1)^{z}}$$
in terms of $\Theta(s)$ and $g(x)$. A practical example of such a construction is the everywhere-convergent series for the Riemann zeta function:
$$\zeta(z)-\frac{1}{z-1}=\sum_{n=0}^{\infty}\left | G_{n+1} \right |\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{1}{(k+1)^{z}}$$
$\left | G_{n+1} \right |$ being the absolute Gregory coefficients.
My attempt :
We use the formula (due to Apostol) :
$$\theta(k+1)=\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}\Theta(\sigma+it)(k+1)^{\sigma+it}dt\;\;\;\;\;\;\sigma>\beta$$
Ignoring issues of convergence, we have:
$$\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{\theta(k+1)}{(k+1)^{z}}$$$$=\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}\Theta(\sigma+it)\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(k+1)^{\sigma+it-z}dt$$
Now, we have that :
$$\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(k+1)^{-\mu}=\frac{1}{\Gamma(\mu)}\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\int_{0}^{\infty}x^{\mu-1}e^{-(k+1)x}dx$$
$$=\frac{1}{\Gamma(\mu)}\sum_{n=0}^{\infty}a_{n}\int_{0}^{\infty}x^{\mu-1}e^{-x}\left(1-e^{-x}\right)^{n}dx=\frac{1}{\Gamma(\mu)}\int_{0}^{\infty}x^{\mu-1}e^{-x}g(1-e^{-x})dx\;\;\;\Re(\mu)>0$$
Assuming this last integral admits analytic continuation, and denoting it by $\Omega (\mu)$, we have:
$$\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{\theta(k+1)}{(k+1)^{z}}=\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}\Theta(\sigma+it)\Omega(\sigma+it-z)dt$$
But i doubt that this kind of integrals could ever be evaluated, even using the residue theorem, due to the limit, and the $T^{-1}$ factor. Is there an alternative to my approach ?
Mellin transform interpolation is related formally at least to Newton series interpolation. One example is its use to interpolate the Bernoulli polynomials from their relation to the Mellin kernel for the Riemann zeta function (times $(s-1)!$) to obtain the Hurwitz zeta function. Maybe you could modify the arguments.
|
2025-03-21T14:48:31.236438
| 2020-06-12T21:02:38 |
362911
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"DCM",
"https://mathoverflow.net/users/61771"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630083",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362911"
}
|
Stack Exchange
|
Isometries and complex differentials
Assume that A is a linear operator of $L^2(D)$ onto istelf, where $D$ is the unit disk. Assume that $f\to \partial A[f](z)$ and $f\to \bar \partial A[f ](z)$ are isometries. Whether it implies that $A$ is a contraction?
Just a suggestion, but I think you might want to think about what you're taking as the codomains of $\partial$ and $\bar \partial$, and in what sense you want to be taking derivatives (assuming that's what $\partial$ and $\bar \partial$ stand for that is).
|
2025-03-21T14:48:31.236506
| 2020-06-12T22:21:16 |
362916
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"François Brunault",
"LSpice",
"Nate Eldredge",
"Theo Johnson-Freyd",
"YCor",
"Zach Hunter",
"https://mathoverflow.net/users/130484",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/6506",
"https://mathoverflow.net/users/78"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630084",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362916"
}
|
Stack Exchange
|
Is it all right to invite a professor to cowrite a paper
I managed to improve the techniques of a year-old paper to get a significantly stronger result. I used original ideas, but a large part relies upon results of the original paper. (to the point that half the paper may be recapping the lemmas of the past paper)
Q1) Is it acceptable to ask the author of the original paper if they would like to collaborate?
I have heard that Paul Erdos would do this, which is something I highly admire. However, I am an undergraduate, so I fear my invitations would not be appreciated.
Q2) If it is acceptable, how do I make sure to make my invitation not pushy or expectant?
I'm thinking of writing something like:
I was wondering if anyone else had already achieved this result, and
if it would be fit for publishing. If yes, while I fear it may be rude
to invite you to collaborate, please let me know if you are at all
interested in getting involved.
If you've already proved the result, what would be left for the collaborator to do?
An option is to start a discussion on the subject, for instance sending a draft, and not evoke a suggestion of collaboration at the first email, but rather insist on mathematical questions. If the person you're asking gets involved in the discussion, at some point you could suggest a join project.
This could also be appropriate for Academia.SE.
I am currently on roughly the other end of this question. I was flattered to be asked, but said that I hadn't contributed anything beyond the original paper, and so didn't merit co-author status. The asker wasn't a student, but a colleague; so, when he insisted, I relented. Had he been a student, I would almost certainly have insisted that he take sole authorship.
@NateEldredge as he would have more experience with the field, and with writing papers, even minimal involvement from him could help me write a higher quality paper in a shorter amount time. meanwhile at least one benefit for him would be getting another publication without as much effort? (I am not yet too familiar with the politics of academia so I'm not sure how valuable this is too him, this is why I'm hesitant as I don't want to be a charity case)
As others have said: you should certainly get in touch with the author and send a draft, if for no other reason than feedback. Also, FWIW, I have had the experience of feeling "I can prove the results, but the writing would be improved if I coauthored with Person X".
If your work raises new questions and you feel that the professor's expertise would be useful to solve them, then it is natural to ask him to collaborate.
I think a preliminary step is to talk to an advisor or someone near you who can read your paper. The intention you state is good, but I think it could be phrased more artfully. Here is what I recommend.
Line up one or two people at your institution (people who taught one of your classes, or someone in your department who might have contact with the professor whose work you have used). Get their opinion on the advisability of reaching out and making your request. Ask them for how they might word your request.
If possible, contact a student or coauthor of this professor and relate your concerns. As I understand it, your first concern is to get your work published and your second is not to step on any toes, especially those of the person whose work you have extended. If this student or coauthor is in the relevant fields, ask them for a sense of how publishable your result is.
Based on this feedback, choose to approach the professor with the intention of announcing your work to him and ask if he (pronoun presumption on my part) is interested in reading it. Then let him make the next move.
In all of this, pick someone to tell of your process. There should be no reason for anything horribly wrong to happen. If and when it does happen, it helps to have someone who already knows your side of the story.
Gerhard "Plans To Backup His Backups" Paseman, 2020.06.12.
|
2025-03-21T14:48:31.236802
| 2020-06-12T23:25:30 |
362919
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/158968",
"https://mathoverflow.net/users/35520",
"ofer zeitouni",
"user158968"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630085",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362919"
}
|
Stack Exchange
|
What is the "mode" of the Girsanov density?
Consider the Girsanov density $$\exp\left(\int_0^Tf(s,B(s))dB(s)-\frac12\int_0^Tf^2(s,B(s))ds\right)$$
Is there a notion of "mode" of this density?
For example, is there a continuous path $z(t)$ with $z(0)=0$ that maximizes the functional
$$z\mapsto \exp\left(\int_0^Tf(s,z(s))dz(s)-\frac12\int_0^Tf^2(s,z(s))ds\right)$$
This is the same as maximizing the functional
$$z\mapsto \int_0^Tf(s,z(s))dz(s)-\frac12\int_0^Tf^2(s,z(s))ds$$
I tried taking a functional derivative but it gets a bit messy. Can you relate the maximum of this functional to the mode of the drift?
One could argue that the minimizer of the Onsager-Machlup functional serves as such a mode.
@oferzeitouni https://projecteuclid.org/download/pdf_1/euclid.cmp/1103904077 found this paper, thanks!
|
2025-03-21T14:48:31.236894
| 2020-06-12T23:56:02 |
362921
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Francois Ziegler",
"Ramiro Hum-Sah",
"https://mathoverflow.net/users/15629",
"https://mathoverflow.net/users/157706",
"https://mathoverflow.net/users/19276",
"paul garrett"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630086",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362921"
}
|
Stack Exchange
|
On how to diagonalize a Casimir element
$\DeclareMathOperator\U{U}\DeclareMathOperator\SU{SU}\DeclareMathOperator\SO{SO}$I'm trying to read the physics paper Two Dimensional QCD as a String Theory. I'm struggling with my ignorance about some computational aspects regarding Lie algebras.
Section 2.3 of the aforementioned paper states:
The representations of $\U(N)$ are labeled by the
Young diagrams, with
$m$ ($m\leq N$) boxes of length $n_{1}\geq n_{2} \geq \dotsb n_{m} \geq 0$. Such a representation $R$ has dimensions $d_{R}$ and Casimirs $C_{2}^{\U(N)}(R)$ given by \begin{gather*}
C_{2}^{\U(N)}(R)=N\sum_{i=1}^{m}n_{i} + \sum_{i=1}^{m}n_{i}(n_{i}+1-2i); \\
d_{R}=\frac{\Delta(h)}{\Delta(h^{0})}, \\
\Delta(h)=\prod_{1\leq j \leq N}(h_{i}-h_{j}), h_{i}=N+n_{i}-1, h_{i}^{0}=N-i.
\end{gather*}
Clarification: The statement "$C_{2}^{\U(N)}(R)$ are Casimirs of the group" does not make sense because of the fact that the quadratic Casimir element $C_{2}$ of $\U(N)$ is by definition a bilinear form on the universal enveloping algebra of the Lie algebra of $\U(N)$ and the $C_{2}$ shown in the paper is a number. I suppose that what the author write as $C_{2}^{\U(N)}(R)$ is the eigenvalue of $C_{2}^{\U(N)}$ associated to the representation $R$ labeled by the partition $(n_{1},\dotsc,n_{m})$.
The question:
I'm asking for your kind help to identify some references where I can learn how to derive the formulas from above and possibly for other cases such as $\SU(N)$, $\SO(N)$ or some symplectic groups if possible.
I'd even be happy if someone could recommend a physical derivation of the formulas.
My own preferred description would be in terms of a "highest weight" of the irreducible. In fact, the same computation would apply to any Verma module, not only their irreducible quotients... Is this the sort of thing that would address your question?
Thanks for your valuable comment paul garrett. I'm familiar with the construction that obtain the highest weight representations of a Lie algebra from its Cartan subalgebra and its set of positive roots. If I understand well, your'e suggesting me that I should compute use the theorem of highest weight representations to compute the irreducible representations of some Lie algebra and then use something like the Weyl dimension formula to obtain ...
... the formulas I need. My problem is basically that I'm not so good in computing such things on concrete examples and I'm not convinced that I were capable to compute the formulas of my question.It would be very beneful for me to see a reference with explicit computations.
See https://arxiv.org/abs/0807.3696 for derivation.
Let $E_i^j$ be the ${\rm U}(N)$ generators and $V$ be the fundamental representation. Decompose $V^{\otimes k}$ by the Schur-Weyl duality and apply the second Casimir $C_2 = E_i^j E_j^i$ to $V_R^{{\rm U}(N)} \otimes V_R^{S_k}$.
After some computation, $C_2$ becomes a sum over the permutations $\sum_{r \neq s} (rs)$. This is a sum of Jucys-Murphy elements whose eigenvalues are the contents of the box; see Wikipedia. By summing the contents over $R$, you can derive the Casimir eigenvalue $C_2^{{\rm U}(N)}(R)$.
That paper of Gross also gives $C_2^{{\rm SU}(N)}(R)$. His result agrees with Freudenthal's formula of $C_2=\langle \mu,\mu+2\rho\rangle$ by recalling that the Dynkin label $(m_1 , m_2 , \dots , m_{N-1})$ is related to the partition $(n_1, n_2 , \dots, n_N)$ by
\begin{equation}
m_i = n_{N-i} - n_{N+1-i} \ge 0.
\end{equation}
First of all, you're right that by "the Casimirs" the authors mean the eigenvalues of the quadratic Casimir operator on the irreps in question — this is a common phrasing in the physics literature.
For $\mathfrak{su}(n)$, the Young diagram with $m$ rows of lengths $n_i$ corresponds to the highest weight $\mu=\sum_i n_i\lambda_i$ (cf. e.g. these lecture notes). From there, the Weyl dimension formula cited in your comment should give the dimensions. For the Casimir eigenvalues, these notes may be of help. Note that there may be factors of 2 (and possibly of the dimension of the irrep) by which conventions used in the physics and mathematics literature may differ in defining the eigenvalues.
The specific formula for Casimir eigenvalues is the one of Casimir-van der Waerden-Freudenthal mentioned in this answer.
|
2025-03-21T14:48:31.237200
| 2020-06-13T02:34:10 |
362926
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aoi Koshigaya",
"Nulhomologous",
"https://mathoverflow.net/users/129738",
"https://mathoverflow.net/users/158462"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630087",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362926"
}
|
Stack Exchange
|
Definitions of formal schemes
In Demazure's Lectures on p-Divisible Groups, an affine formal scheme over a field $k$ is defined as $\mathop{\mathrm{Spf}}(A)$ for some profinite $k$-algebra $A$. However, in EGA I Chap. 10, an affine formal scheme is defined as $\mathop{\mathrm{Spf}}(A)$ for some admissible ring $A$. My question is: Is Demazure's definition a special case of EGA's definition? If not, is there a more general definition which contains both of them as special cases.
Your question is not about formal shemes, but about if a profinite $k$-algebra, for $k$ a field, is an admissible (topological) ring (with the topology given by being profinite). And the answer is clearly affirmative. So Demazure's definition can be seen as a special case of EGA's definition.
@Nulhomologous But it seems that an infinite product of $k$ is a profinite $k$-algebra which is not admissible.
I don't see how an infinite product of k could be a profinite k-algebra.
@Nulhomologous According to Demazure, a profinite $k$-algebra is equivalent to a left exact functor from the cat $\mathcal C$ of finite $k$-algebras to the cat of sets. Let $I$ be any set. Then the constant sheaf on $\mathcal C$ with value in $I$ (i.e. $R \mapsto$ locally constant function from $\mathop{\mathrm{Spec}}(R)$ to $I$) corresponds to the product of copies $k_i$ of $k$ indexed by $I$. In particular, the constant $p$-divisible group $\mathbb Q_p/\mathbb Z_p$ over $k$ should corresponds to an infinite product of $k$.
|
2025-03-21T14:48:31.237311
| 2020-06-13T02:43:14 |
362927
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Tanmoy Paul",
"https://mathoverflow.net/users/76412"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630088",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362927"
}
|
Stack Exchange
|
On examples of subspaces of $C(X)$ for which state spaces are Choquet Simplices
Let $C(X)$ be the Banach space of all Real valued continuous functions on a compact Hausdorff space $X$. What are examples of uniformly closed subspace $\mathcal{A}$ of $C(X)$ such that $\mathcal{A}$ separates points, containing constants and the state space of $\mathcal{A}$ is a Choquet Simplex.
The state space of $\mathcal{A}$ viz. $S_{\mathcal{A}}$ is defined as $\{\Lambda\in\mathcal{A}^*:\|\Lambda\|=1 ~\mbox{and}~ \Lambda (1)=1\}$. By a Choquet Simplex we mean a compact convex subset $K$ of a locally convex topological vector space $E$ such that for each $p\in K$, there exists a unique measure $\lambda$ on $K$ such that $f(p)=\int_K f(t)d \lambda (t)$, $\forall~ f\in E^*$. When $K$ is metrisable then $\lambda$ can be assumed to satisfy $S(\lambda)\subseteq ext (K)$ but for non metrisable case $S(\lambda)\subseteq \overline{ext(K)}$, in some sense these measures are 'maximal'. $ext(K)$ represents the set of all extreme points of $K$.
On page 148 of Convexity Theory and its Applications in Functional Analysis, L. Asimow and A.J. Ellis say that every Dirichlet algebra (an algebra where the real parts of its elements are dense in the space of real-valued continuous functions) gives rise to a simplicial state space. The disc algebra is an example of a Dirichlet algebra.
Yes, but I need an example with Real scalar. The space $C(X)$ I considered with Real scalars.
|
2025-03-21T14:48:31.237440
| 2020-06-13T02:50:14 |
362928
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bill Johnson",
"Dongyang Chen",
"https://mathoverflow.net/users/2554",
"https://mathoverflow.net/users/41619"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630089",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362928"
}
|
Stack Exchange
|
The Calkin representation for Banach spaces
Let $X$ be an infinite dimensional Banach space. Let $\Lambda_{0}$ be the set of all finite dimensional subspaces of $X$ directed by the inclusion $\subseteq$. For each $\alpha\in \Lambda_{0}$, let $I_{\alpha}:=\{\beta\in\Lambda_{0}:\alpha\subseteq \beta\}$. Then $\{I_{\alpha}:\alpha\in \Lambda_{0}\}$ is a filter basis and hence is contained in some ultrafilter $\mathcal{U}$.
For an infinite dimensional Banach space $Y$, let $(Y^{*})_{\mathcal{U}}$ be the ultrapower of $Y^{*}$ with respect to $\mathcal{U}$. Let $\widehat{Y}$ be the subspace of $(Y^{*})_{\mathcal{U}}$ defined by $$\widehat{Y}:=\{(y^{*}_{\alpha})_{\mathcal{U}}\in (Y^{*})_{\mathcal{U}}:w^{*}-\lim_{\mathcal{U}}y^{*}_{\alpha}=0\}.$$ For an operator $T:Y\rightarrow X$, we define $\widehat{T}:\widehat{X}\rightarrow \widehat{Y}$ by $\widehat{T}((x^{*}_{\alpha})_{\mathcal{U}})=(T^{*}x^{*}_{\alpha})_{\mathcal{U}}.$ It is easy to see that $\widehat{T}=0$ if $T$ is compact.
Question 1. Is $T$ compact if $\widehat{T}=0$?
Question 2. Let $K$ be a compact, convex and balanced subset of $B_{X}$ and let $\epsilon>0$. We set $A:=K+\epsilon B_{X}$ and define the gauge of $A$ by $$\|x\|_{A}:=\inf\{t>0:x\in tA\}, \quad x\in X.$$
It is easy to see that $$\epsilon\|x\|_{A}\leq \|x\|\leq (1+\epsilon)\|x\|_{A}, \quad x\in X.$$ We set $Y:=(X,\|\cdot\|_{A})$ and let $j:Y\rightarrow X$ be the formal identity. Is there a constant $C$ such that $\|\widehat{j}\|\leq C\cdot \epsilon$?
Thanks!
Are you asking because you are thinking about following up on a lemma in my paper with March Boedihardjo?
Yes. I want to improve Theorem 2.1 in your paper with March Boedihardjo and characterize the bounded compact approximation property.
The answer to question 1 is yes.
Suppose that $T^*B_{X^*}$ is not compact. Since it is norm closed, there is
$\epsilon >0$
and an infinite subset $S$ of $B_{X^*}$ so that
$\|T^{*}x_1^*-T^{*}x_2^*\| > \epsilon$
for all
$x_1^*\not= x_2^*$
in
$S$. Let $x^*$ be any weak$^*$ limit point of $S$. For $\alpha$ in $\Lambda_0$ pick $x_\alpha^*$ in $S$ with
$T^*x^*_\alpha \not= T^*x^*$
so that
$\|x^* - x_\alpha^*\|_\alpha < 1/\dim \alpha$, where
$\|z^*\|:= \|z^*_{|\alpha}\|$. Then by the choice of $\Lambda_0$,
$x^*_\alpha \to x^*$ weak$^*$
and hence
$\widehat{T}(x^{*}_\alpha -x^{*})_\alpha =0$, which means
$\|T^*x_\alpha^* - T^*x^*\| \to 0$.
Since $S$ is $\epsilon$-separated, this forces $T^*x_\alpha^* = T^*x^*$ eventually, which is a contradiction.
Thanks, Bill. I think that in you answer, $|T^{}x^{}-T^{}x^{}{\alpha}|{\alpha}<1/\textrm{dim} \alpha$ should be $|x^{}-x^{}{\alpha}|{\alpha}<1/\textrm{dim} \alpha$.
Moreover, I think that for each $\alpha$, we can pick $x^{}_{\alpha}\in S$ such that $|T^{}x^{}_{\alpha}-T^{}x^{*}|\geq \frac{\epsilon}{2}$. Can we do it?
Yes; sorry for the typo. I'll fix it.
But I am not sure that we can pick $x^{}_{\alpha}\in S$ such that $|T^{}x^{}_{\alpha}-T^{}x^{*}|\geq \epsilon/2$ for all $\alpha$.
The answer to question 2 is yes.
The proof you already know since you proved that $T$ compact implies $\hat{T}$ is zero. (For someone who has not thought about this, it is immediate from the elementary fact that a bounded net in $X^*$ that converges to zero weak$^*$ must converge uniformly to zero on compact subsets of $X$.) So if
$(x^*_\alpha)_\alpha$ is in $\widehat{X}$
with $\sup \|x_\alpha^\alpha \| \le 1$ and
$x^*_\alpha \to 0$ weak$^*$, then
$x^*_\alpha \to 0$
uniformly on $K$. Since the unit ball of $Y$ is contained in
$K+\epsilon B_{X}$,
it follows that
$\|\hat{j}\| \le \epsilon$,
so $C$ can be one.
|
2025-03-21T14:48:31.237657
| 2020-06-13T04:36:08 |
362933
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Brian",
"GH from MO",
"Thomas Bloom",
"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/385",
"https://mathoverflow.net/users/70508"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630090",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362933"
}
|
Stack Exchange
|
Asymptotics on a double sum over primes
I am attemping to find asymptotics of
$$\sum_{p \leq n}\ln p \left( \sum_{k=1}^\infty \left(\left\{\frac{n}{p^k} \right\} - \left\{\frac{n-1}{(p-1)p^k} \right\} \right) - \left\{\frac{n-1}{p-1} \right\} \right) $$
where $\{ x\}$ denotes the fractional part of $x$ and $p$ denotes prime. On January 2016, Fedor Petrov consequently has shown
$$\sum_{p \leq n}\ln p\sum_{k=0}^\infty \left\{\frac{n-1}{(p-1)p^k} \right\} = (1-\gamma)n+O(n\epsilon_n),$$
where $\gamma$ denotes the Euler-Mascheroni constant and $\epsilon_n \to 0$. This leaves us with determining asymptotics for
$$\sum_{p \leq n}\ln p\sum_{k=1}^\infty \left\{\frac{n}{p^k} \right\},$$
which I suspect is determined somewhere in literature since this is the fractional part discussed in Legendre's Theorem.
I think the asymptotic for your third sum is the same as the second sum (namely $(1-\gamma)n+o(n)$), so to get an asymptotic overall you will at least need to know the second order terms in more detail than you give.
You don't mention the third sum - do you know the asymptotics for $\sum_{p\leq n} \ln p { \frac{n-1}{p-1}} $?
@Thomas Bloom I should have specified where the asymptotics for the second and third sum come from. To be exact, Petrov has shown that $\begin{align} \sum_{p \leq n} \ln p \sum_{k \geq 1} \left{ \frac{n-1}{(p-1)p^k}\right} &= o(n) \ \sum_{p \leq n} \ln p \left{ \frac{n-1}{p-1}\right} &= (1-\gamma)n + o(n). \end{align}$
Beyond the current terms, I'm not sure how I would go about finding higher order terms. If you would like more information about the current estimates, here is the link: https://mathoverflow.net/q/228738/70508
Please always include a top-level tag like "nt.number-theory"
|
2025-03-21T14:48:31.237806
| 2020-06-13T06:50:48 |
362935
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alessandro Codenotti",
"Alex Ravsky",
"Anonymous",
"Henno Brandsma",
"Smolin Vlad",
"Will Brian",
"Wojowu",
"https://mathoverflow.net/users/142738",
"https://mathoverflow.net/users/2060",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/43954",
"https://mathoverflow.net/users/49381",
"https://mathoverflow.net/users/70618",
"https://mathoverflow.net/users/89233"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630091",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362935"
}
|
Stack Exchange
|
Uncountable subspaces of the real line
Definition 1. A topological space $\langle X, \tau \rangle$ is meager if $X = \bigcup_{n \in \omega}A_n$, where each $A_n$ is nowhere dense in $\langle X, \tau \rangle$.
Definition 2. A topological space is of the second category if it is non-meager.
Definition 3. A topological space is perfect if it does not contain isolated points.
I have the following question about the second category property of uncountable subspaces of the real line:
Question. Is it provable in ZFC (is it consistent with ZFC) that every perfect uncountable subspace of the real line contains a perfect uncountable subspace that is of the second category?
@BjørnKjos-Hanssen This space is of the second category because it is complete metric space
@BjørnKjos-Hanssen Cantor set is of first category as a subset of $\mathbb R$, but is second category in itself.
Do you mean second category in $\Bbb R$ or second category in itself (i.e. Baire)? That makes a big difference.
@HennoBrandsma, yes, i mean second category in itself
It is certainly not provable in ZFC, since under Martin's Axiom, every subset of $\mathbb R$ of size less than the continuum is meager.
@Anonymous, is it meager in itself?
Maybe I'm missing something, but doesn't this follow from the fact that every uncountable perfect subsets of the reals contains a Cantor space?
@AlessandroCodenotti, a counterexample to your statement is the Bernstein set
@Smolin A Bernstein set isn't perfect, since it isn't closed (in fact a Bernstein set isn't even Borel because of the same reason: every borel subset of the reals is either countable or it contains a Cantor set)
@AlessandroCodenotti, i define perfect space (Definition 3 in my question) as a topological space without isolated points
Yes, under MA every subset of $\mathbb R$ of size less than the continuum is meager in itself.
If you don't require perfect sets to be closed let $A\subseteq\Bbb R$ be perfect uncountable. If $A$ contains an uncountable closed set, then $A$ contains a Cantor set, otherwise $A$ is a Bernstein set, and Bernstein sets are Baire spaces, so second category in themselves
@Anonymous, do you mean subsets without isolated points? Because there exists a countable subset of the real line with isolated points that is of the second category in itself. Could you give please some link?
@AlessandroCodenotti, it is not clear why is it true that either set contains a Cantor subspace or is a Bernstein set, because by definition in your link, set is a Bernstein set iff it and its complement intersect with every Cantor space, i.e. it is not enough does not contain a cantor subspace in order to be a Bernstein set
Yes, I should have said every subset of size less than the continuum without isolated points. I think that this is in Miller's article in the Handbook of Set-Theoretic Topology.
This is a good question -- not sure why someone voted to close it. (@Anonymous is right about what happens under MA -- so that bit is solved -- but I really like the "is it consistent" part of the question. I wonder what happens under CH?)
@WillBrian But CH implies MA, isn't it?
@AlexRavsky: Correct. And in case CH holds, the statement of Anonymous becomes "under MA every countable subset of $\mathbb R$ is meager in itself." Notice that we're talking about sets of size less than $\mathfrak{c}$, not $\leq!\mathfrak{c}$. (I suppose we should add "without isolated points" to make this true, although in the context of this question that was probably just assumed.)
@AlexRavsky: Oh, but now I understand your point a bit better. The comment of Anonymous solves the problem assuming MA+$\neg$CH, because then there are uncountable subsets of $\mathbb R$ with size $<!\mathfrak{c}$. Under CH, MA technically still holds, but it doesn't apply directly to uncountable sets.
|
2025-03-21T14:48:31.238168
| 2020-06-13T07:36:54 |
362938
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Hermi",
"https://mathoverflow.net/users/168083"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630092",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362938"
}
|
Stack Exchange
|
Conditional distribution for the random walk on $\mathbb{Z}$
Define a random walk on the integers $\mathbb{Z}$ with step distribution $F$ and initial state is zero which is a sequence $S_n$ of random variables and its increments are iid random variables $X_i$ with common distribution $F$, that is,
$$S_n=\sum_{i=1}^n X_i$$
Can we find a distribution $F$ such that for some $0<i<n$, when conditioned to go from 0 to 0 versus from 0 to 1 does not stochastically order, that is, $S_i$ condition on $S_n=1$ versus $S_i$ condition on $S_n=0$ does not stochastically order.
In fact, this is an example of a discrete random walk which when conditioned to form a bridge violates monotone coupling.
Suppose the possible increments are $+3$, $+2$, $-1$, and $-3$. The specific probabilities don't matter, as long as they're positive.
Then conditional on $S_2=1$, we have $S_1\leq 2$ with probability $1$ (the first two steps must be $2$ and $-1$ in some order).
But conditional on $S_2=0$, we have $S_1=3$ with probability $1/2$ (the first two steps must be $3$ and $-3$ in some order).
Thanks for your nice example, James!
|
2025-03-21T14:48:31.238271
| 2020-06-13T08:36:04 |
362941
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"https://mathoverflow.net/users/148980",
"https://mathoverflow.net/users/25510",
"niran90"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630093",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362941"
}
|
Stack Exchange
|
Conformal mapping between two right-angled triangles
I want to derive a conformal mapping $f\!:\!A\!\to\! B$ where $A=\{ (x,y)\ |\ x\!\in\![0,1]\ \text{ and }\ 0 \leq y \leq x \}$ and $B=\{ (x,y)\ |\ x\!\in\![0,1]\ \text{ and }\ 0 \leq y \leq \frac{1}{\sqrt{3}}x \}$. The regions $A$ and $B$ are the right-angled triangles with angles $\{\frac{\pi}{4},\frac{\pi}{2},\frac{\pi}{4}\}$ and $\{\frac{\pi}{6},\frac{\pi}{2},\frac{\pi}{3}\}$, respectively.
Can anyone help me derive an explicit equation for $f(x,y)$? I am vaguely familiar with the Schwartz-Christoffel mapping and the Schwartz triangle mapping, but I do not have a rigorous enough understanding of complex analysis to apply these to the above case. Any advice would be much appreciated!
There are two expressions depending on what you prefer: hypergeometric functions or elliptic functions.
Let $f$ be the Schwarz-Christoffel map of the upper half-plane onto $A$,
and $g$ the Schwarz-Christoffel map onto $B$ (both sending $(0,1)$ to $(0,1)$). Then your map $A\to B$ is $g\circ f^{-1}$. Explicit formulas:
$$f(z)=\frac{\int_0^z\zeta^{-3/4}(\zeta-1)^{-1/2}d\zeta}{\int_0^1\zeta^{-3/4}(\zeta-1)^{-1/2}d\zeta},$$
$$g(z)=\frac{\int_0^z\zeta^{-5/6}(\zeta-1)^{-1/2}d\zeta}{\int_0^1\zeta^{-5/6}(\zeta-1)^{-1/2}d\zeta}.$$
All integrals can be expressed in terms of hypergeometric functions.
Alternative method, using elliptic integrals and their inverses. Reflect both triangles with respect to the long side. In the first case you obtain a square, and the mapping function
$$f^{-1}(z)=1/\wp^2(z,1,0).$$
Similarly, the map $g$ can be expressed in terms of a standard elliptic integral of the first kind corresponding to hexagonal lattice. The second triangle has to be reflected with respect to the side $(0,1)$, to obtain an
equilateral triangle.
Remark on computation. Elliptic integrals can be expressed in terms of theta-functions (see Whittaker-Watson, for example). Theta-series are converging so fast they they are even suitable for computation by hand, without a computer. The little book by N. I. Akhiezer, Elements of the theory of elliptic functions, contains less material but is more user-friendly than Whittaker-Watson.
Thanks so much! With regards to your first solution, unfortunately I have no experience with hyper-geometric functions and evaluating complex integrals. Is the evaluation of these integrals straightforward? If not, can you suggest how I can solve these numerically? (I am unsure how to 'numerically' represent an integral over the entire upper-half plane)
@niran: the integrals are not over the entire half-plane, they are line integrals. If you are not familiar with hypergeometric functions, look to some complex analysis textbook, for example, Whittaker-Watson. Also Maple and other computer systems easily compute these integrals.
Inversion of $f$ in the first approach can be a problem, but in the second approach I wrote $f^{-1}$ for you in the form convenient for computation. $\wp$ function is called WeierstrassP in Maple.
I am not as clear on how to apply your second solution. Would you mind elaborating on how to derive $g(z)$ as well?
The best thing is to use $g$ from the first solution and $f^{-1}$ from second. Then you don't need to invert anything. The second approach does not give you a ready formula for $g$ but rather for $g^{-1}$.
I see that $\wp$ can take different types of arguments. What's the significance of the $1$ and $0$ in $\wp^2(z,1,0)$? Are these the elliptic invariants or the half-periods of $\wp$?
$g_1=1,g_3=0$. Periods are $1$ and $i$
Ok. I am getting $f^{-1}(1)=0.90558...$ when I use your equation on Wolphram Alpha. Shouldn't $f^{-1}(1)=1$ instead?
Since the Weierstrass function, $\wp[z,{g_2,g_3}]$, in Maple/Mathematica/Wolfram take in arguments $g_2$ and $g_3$, should I be using $\wp(z,0,0)$ instead? (I'm assuming $g_2 = 0$ as well?)
@nitan: I wrote in my answer correctly: $\wp(z,1,0)$ this is consistent with Maple notation. For $g_2=g_3=0$ $\wp$ is not defined.
Ok thanks for clarifying - I was just confused because you mentioned $g_1$ and $g_3$ previously, not $g_2$ and $g_3$ (I'm guessing this was a typo?). So back to my previous question - shouldn't $f^{-1}(1) = 1$ hold? I get $f^{-1}(1) = 0.90558$ when I use $g_2 = 1$ and $g_3 = 0$.
|
2025-03-21T14:48:31.238555
| 2020-06-13T09:12:32 |
362944
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Hermi",
"Iosif Pinelis",
"https://mathoverflow.net/users/168083",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630094",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362944"
}
|
Stack Exchange
|
How to prove that a Brownian bridge $\mathbb{P}(M[0, 1/2]\geq s)\leq 2\mathbb{P}(B(1/2)\geq s/2)?$
Consider a Brownian bridge $B: [0,1]\to \mathbb{R}$ with $B(0)=B(1)=0$. Let $M[0, 1/2]=\max_{x\in[0,1/2]}B(x)$. How to prove that
$$\mathbb{P}(M[0, 1/2]\geq s)\leq 2\mathbb{P}(B(1/2)\geq s/2)?$$
Actually, here is a "no big max" argument that could be used in the proof of construction Airy line ensemble. The "no big max" means the top curve between $(a, b)$ cannot get too high.
(Definition of Brownian bridge) If $\{B(t): t\geq 0\}$ is standard Brownian motion, then $\{Z(t): 0\leq t\leq 1\}$ is a Brownian bridge process when $$Z(t)=B(t)-tB(1).$$
Let $B_t:=B(t)$. We have to show that
\begin{equation}
P(M_{1/2}\ge s)\le2P(B_{1/2}\ge s/2) \tag{1}
\end{equation}
for $s\ge0$, where $M_T:=\max_{0\le t\le T}B_t$ for $T\in(0,1)$.
We shall prove (1) by first obtaining an explicit expression for $P(M_T\ge s)$.
Indeed, for $t\in[0,1]$, we can write
$$B_t=W_t-tW_1,$$
where $W$ is a standard Wiener process. For each real $x$, consider
\begin{equation}
C^{-x}_t:=W_t-tx,
\end{equation}
so that $C^{-x}$ is a Wiener process with drift $-x$. Then
\begin{align*}
P(M_T\ge s|W_1=x)&=P(\max_{0\le t\le T}(W_t-tW_1)\ge s|W_1=x) \\
&=P(\max_{0\le t\le T}(W_t-tx)\ge s|W_1=x) \\
&=P(\max_{0\le t\le T}C^{-x}_t\ge s|C^{-x}_1=0) \\
&=e^{-2s^2}G\Big(\frac{(1-2T)s}{\sqrt{(1-T)T}}\Big)+G\Big(\frac{s}{\sqrt{(1-T)T}}\Big)
\end{align*}
by Theorem 3.1 (with $-x,s,1,T,0$ in place of $\mu,\beta,u,t,\eta$, respectively), where $1-G$ is the standard normal cdf.
Note that $P(M_T\ge s|W_1=x)$ does not depend on $x$ (which of course should have been expected, since the Brownian bridge $B$ is independent of $W_1$). We conclude that
\begin{equation}
P(M_T\ge s)
=e^{-2s^2}G\Big(\frac{(1-2T)s}{\sqrt{(1-T)T}}\Big)+G\Big(\frac{s}{\sqrt{(1-T)T}}\Big).
\end{equation}
In particular,
\begin{equation}
P(M_{1/2}\ge s)
=\tfrac12\,e^{-2s^2}+G(2s)=:l(s).
\end{equation}
On the other hand, since $B_{1/2}$ equals $\tfrac12\,W_1$ in distribution, we have $P(B_{1/2}\ge s/2)=P(W_1\ge s)=G(s)$.
So, it remains to show that
\begin{equation}
d(s):=l(s)-2G(s)\le0.
\end{equation}
For real $s>0$, let
\begin{equation}
d_1(s):=d'(s)e^{2 s^2}
=\sqrt{\tfrac{2}{\pi }} e^{3 s^2/2}-2 s-\sqrt{\tfrac{2}{\pi }}.
\end{equation}
Then $d_1'(s)=3 \sqrt{\frac{2}{\pi }} e^{3 s^2/2} s-2$ and hence $d_1$ is clearly $-+$; that is, $d_1$ may change its sign at most once (on $(0,\infty)$) and only from $-$ to $+$.
So, $d_1$ is down-up -- that is, there is some $c\in[0,\infty]$ such that $d_1$ is decreasing on $(0,c]$ and increasing on $[c,\infty)$. Also, $d_1(0)=0$ and $d_1(\infty-)=\infty$. So, $d_1$ is $-+$ and hence $d$ is down-up. Also, $d(0)=0=d(\infty-)$. So, $d<0$ (on $(0,\infty)$), as desired.
You can prove it via standard computations. See (2.1), (2.7), and (2.8) of On the maximum of the generalized Brownian bridge.
Let $B_t:=B(t)$. For $t\in[0,1]$, we can write
$$B_t=W_t-tW_1,$$
where $W$ is a standard Wiener process. We have to show that
$$P(M_{1/2}\ge s)\le2P(B_{1/2}\ge s/2)$$
for $s\ge0$, where $M_{1/2}:=\max_{0\le t\le1/2}B_t$. The Brownian bridge $B$ is independent of $W_1$. Therefore and by the reflection principle,
$$\tfrac12\,P(M_{1/2}\ge s)=P(M_{1/2}\ge s,W_1\ge0)
=P(\max_{0\le t\le1/2}(W_t-tW_1)\ge s,W_1\ge0)
\le P(\max_{0\le t\le1/2}W_t\ge s)
=2P(W_{1/2}\ge s)=2P(W_1\ge s\sqrt2)=2P(B_{1/2}\ge s/\sqrt2);$$
here we also used the fact that $B_{1/2}$ equals $\tfrac12\,W_1$ in distribution.
So,
$$P(M_{1/2}\ge s)\le4P(B_{1/2}\ge s/\sqrt2).$$
The upper bound $4P(B_{1/2}\ge s/\sqrt2)$ on $P(M_{1/2}\ge s)$ is better than your desired upper bound $2P(B_{1/2}\ge s/2)$ for all $s\ge0.992$.
@BobO. : What do you mean by the `first "="'?
@BobO. : This holds because (i) $P(W_1\ge0)=1/2$ and (ii) $B$ is independent of $W_1$ and hence $M_{1/2}$ is independent of $W_1$ (read the equality in your latter comment right-to-left).
Why $B(1/2)\sim 1/2W(1)$? It seems $B(1/2)=W(1/2)-(1/2)W(1)$.
@BobO. : Both $B_{1/2}$ and $\frac12,W_1$ are zero-mean normal random variables. Also, $Var, B_{1/2}=Var(W_{1/2}-\frac12,W_1)=Var,W_{1/2}+(\frac12)^2Var,W_1-Cov,W_{1/2}W_1=\frac12+\frac14-\frac12=\frac14=Var(\frac12,W_1)$. So, $B_{1/2}$ and $\frac12,W_1$ are equal in distribution.
|
2025-03-21T14:48:31.238789
| 2020-06-13T09:51:46 |
362946
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fedor Petrov",
"https://mathoverflow.net/users/35520",
"https://mathoverflow.net/users/4312",
"ofer zeitouni"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630095",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362946"
}
|
Stack Exchange
|
Theorem 5.3 ([Okounkov-01]) in Borodin and Gorin's lecture note
In this lecture note: https://arxiv.org/pdf/1212.3351.pdf, Theorem 5.3(P28):
Suppose that the $\lambda \in \mathbb{Y}$ is distributed according to the Schur measure $\mathbb{S}_{\rho_1; \rho_2}$. Then $X(\lambda)$ is a determinantal point process on $\mathbb{Z}+1/2$ with correlation kernel $K(i,j)$ defined by the generating series
\begin{align}\label{1}
(1)\sum_{i, j\in \mathbb{Z}+1/2} K(i,j)v^iw^{-j}=\frac{H(\rho_1; v)H(\rho_2; w^{-1})}{H(\rho_2; v^{-1})H(\rho_1; w)}\sum_{k=1/2, 3/2, 5/2, \dots}(\frac{w}{v})^k
\end{align}
where $$H(\rho; z)=\sum_{k=0}^{\infty}h_k(\rho)z^k=\exp(\sum_{k=1}^{\infty}p_k(\rho)\frac{z^k}{k})$$
My question is how to rewrite (1) as
$$K(i,j)=\frac{1}{2\pi i}\oint\oint\frac{H(\rho_1; v)H(\rho_2; w^{-1})}{H(\rho_2; v^{-1})H(\rho_1; w)}\frac{\sqrt{vw}}{v-w}\frac{dvdw}{v^{i+1}w^{-j+1}}$$
with integration going over two circles around the origin $|w|=R_1$, $|v|=R_2$ such that $R_1<R_2$ and the functions $H(\rho_1; u), H(\rho_2; u^{-1})$ are holomorphic in the annulus $R_1-\epsilon<|u|<R_2+\epsilon$.
Did you try to ask the authors?
Isn't it Cauchy's theorem?
|
2025-03-21T14:48:31.238890
| 2020-06-13T10:13:45 |
362947
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630096",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362947"
}
|
Stack Exchange
|
Continuity of the involution in Banach *-algebras
My question is concerned with the involution in Banach *-algebras.
1- Should the involution be assumed continuous in every Banach *-algebra?
If the answer is negative,
2- Does there exist any characterization of Banach *-algebras whose involution is continuous (isometry)?!
For 1), this is § 36 (pp. 190) of Bonsall & Duncan: The following are equivalent for a Banach *-algebra $A$, and the set $\text{Sym}(A')$ of all continuous self-adjoint linear functionals on $A$
The linear involution $f \mapsto f^*$ is continuous on $A$
$\text{Sym}(A')$ is separating on $A$
$\text{Sym}(A)$ is closed on $A$
|
2025-03-21T14:48:31.238964
| 2020-06-13T10:16:18 |
362948
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A. Gupta",
"https://mathoverflow.net/users/42940"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630097",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362948"
}
|
Stack Exchange
|
Projective group of Plucker quadric over the reals
A somewhat elementary question but seemingly difficult to find a suitable reference:
Consider the six-dimensional real space $\wedge^2(\mathbb R^4)$ with basis $e_i \wedge e_j \ (i < j)$ where $e_i, \ i = 1, \cdots, 4$ denotes the standard basis of $\mathbb R^4$
(Thus each vector $v \in \wedge^2(\mathbb R^4)$ is given as
$v = \sum_{1 \le i < j \le 4} \xi_{ij}e_i \wedge e_j$).
The Plucker quadric is defined as the zero locus of the quadratic function $$ q(v): = \xi_{12}\xi_{34}
- \xi_{13}\xi_{24} + \xi_{14}\xi_{23}. $$
With respect to the given basis the polarization of $q$ is given by the matrix
$$ P:= \begin{pmatrix} 0 & 0 & 0
& 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}. $$
Let $G$ be the isometry group of the quadratic form $q$, that is,
$$ G = \{ X \in GL(6, \mathbb R) \mid X^TPX = P \}. $$
Consider the Plucker quadric as a projective variety $\mathcal G$ in $\mathbf P(\wedge^2(\mathbb R^4))$. Let $\mathscr G$ be the group of projective transformations in $PGL(6, \mathbb R)$ that preserve
$\mathcal G$ as a set. Is it true that $\mathscr G = G/Z$ where
$Z$ denotes the center of $GL(6,\mathbb R)$. If not then what is $\mathscr G$? Will the answer change if $\mathbb R$ is throughout replaced by $\mathbb Q$.
What is (are) a suitable reference (s) for this topic?
In the case of an algebraically closed field such as $\bar {\mathbb Q}$ (where the main interest lies) this question has an affirmative answer as can be seen in M. Berger : Geometry II.
|
2025-03-21T14:48:31.239099
| 2020-06-13T11:53:50 |
362952
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexander Osipov",
"YCor",
"https://mathoverflow.net/users/112417",
"https://mathoverflow.net/users/14094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630098",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362952"
}
|
Stack Exchange
|
Is there a condensation from $\aleph_1^{\aleph_0}$ onto a metrizable compact space?
Is there a condensation (continuous bijective mapping) from $D^{\aleph_0}$ onto a metrizable compact space ?
$D$ - discrete space of cardinality $\aleph_1$.
CH implies it is a positive answer. In general, I don’t know the answer.
Strengthening the question. Is there a condensation from $D^{\aleph_0}$ onto $N^N$?
What's $N$? also, is it implicit that there exists a condensation of $\aleph_0^{\aleph_0}$ onto a metrizable compact space?
$N$ is a space of natural numbers. We can assume that $N^N$ is a space of irrational numbers. Sierpinski proved that irrational numbers admits a condensation onto [0,1].
If $|D| < \aleph_\omega$, then there is a condensation from $D^\omega$ onto $\omega^\omega$ (the Baire space) if and only if there is a partition of $\omega^\omega$ into exactly $|D|$ Borel sets.
As far as I know, this theorem was first proved by me and Arnie Miller in
"Partitions of $2^\omega$ and completely ultrametrizable spaces," Topology and its Applications 184 (2015), pp. 61-71.
See Theorem 3.9. I still don't know what happens for $|D| \geq \aleph_\omega$.
It is a theorem of Hausdorff that $\omega^\omega$ can be partitioned into $\aleph_1$ Borel sets, regardless of whether $\mathsf{CH}$ holds or not. This, together with the theorem quoted above, provides a positive answer to your question.
It is consistent with any allowed value of $\mathfrak{c}$ that there is a partition of $\omega^\omega$ into $\kappa$ Borel sets for every $\kappa \leq \mathfrak{c}$. (This is Theorem 3.11 in the linked paper.) It is also consistent with any allowed value of $\mathfrak{c}$ that the only partitions of $\omega^\omega$ into Borel sets have size $\leq \aleph_0$, $\aleph_1$, and $\mathfrak{c}$ (see Corollary 3.16 in the linked paper), or size $\leq \aleph_0$, $\aleph_1$, $\kappa$, and $\mathfrak{c}$ for any particular $\kappa$ between $\aleph_1$ and $\mathfrak{c}$ (see Proposition 3.17 and the comments following). Generally, though, it's not known how to get some prescribed list of sizes and not others.
The answer is affirmative and can be derived from
Theorem (Banakh, Plichko). The Hilbert space $\ell_2(\aleph_1)$ condenses onto the Hilbert cube.
By the way, this theorem is related to Problem 1 from the Scottish Book.
In order to answer the original problem of Alexander Osipov, it suffices to construct a condensation of $\aleph_1^\omega$ onto the Hilbert space $\ell_2(\aleph_1)$. This can be done as follows. Using the Torunczyk's characterization of the Hilbert space topology, one can prove that $\ell_2(\aleph_1)$ is homeomorphic to the countable power of the hedgehog $$H(\aleph_1)=\bigcup_{\alpha\in\aleph_1}[0,1]\cdot e_\alpha\subset \ell_2(\aleph_1)$$ where $(e_\alpha)_{\alpha\in\aleph_1}$ is the standard orthonormal basis of the Hilbert space.
Next, using the observation that $[0,1]$ is the union of irrationals and rational, one can show that $H(\aleph_1)$ is a continuous bijective image of $\aleph_1\times (\omega^\omega\sqcup \omega)$ and then $H(\aleph_1)^\omega$ is a continuous bijective image of $(\aleph_1\times(\omega^\omega\sqcup\aleph_1))^\omega\cong \aleph_1^\omega$.
Taras, Thanks!!!
|
2025-03-21T14:48:31.239322
| 2020-06-13T12:00:12 |
362953
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Martin Argerami",
"Math Lover",
"Nik Weaver",
"YCor",
"Yemon Choi",
"https://mathoverflow.net/users/129638",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/3698",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630099",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362953"
}
|
Stack Exchange
|
Weak center is same as center for $C^{\ast}$-Algebra?
Let $A$ be a $C^{\ast}$-algebra. We say $A$ is weakly commutative if $ab^*c=cb^*a$ for all $a,b,c \in A$ and define weak center of $A$ as $$Z_w(A)= \{ v \in A : av^*c=cv^*a \;\forall a,c \in A \}.$$
Are these notions of weak commutativity and weak center the same as the usual notions of commutativity and center in $C^{\ast}$-algebras?
By using approximate identity, I have managed to prove that both notions weakly commutative and commutative are same. It is also clear that center of $A$, i.e., $Z(A)$, contains $Z_w(A)$ but reverse inclusion is not clear. Any ideas?
By the way the star sign on $b$ and $v$ is redundant (by the obvious change of variable $b\mapsto b^*$). In particular, the definition doesn't involve the star map. My opinion in this case is that the question should be amended (despite already having an answer, which could be a comment) to make it non-trivial (currently it's clearly not research-level, namely not seriously thought).
@YCor: I asked this question as I was trying to figure out definition of center of TRO. Please see my comment below.
Yes but asking a question about $C^$-algebras to which $M_2(\mathbf{C})$ is a trivial counterexample is not serious (and actually every non-abelian unital $C^$-algebra is a trivial counterexample, for the same reason). I'd have recommended to delete it, but unfortunately this is not possible now it has an answer.
@YCor: I agree, please accept my sincere apologies.
You don't have to, I'm not offended.
I’m voting to close this question for the reasons mentioned by @YCor
Add to the closing reasons that the question was also asked verbatim in MSE.
Not the same. The center of $M_2$ is $\mathbb{C}\cdot I_2$, but its weak center is $\{0\}$. E.g. $I_2$ is not in the weak center because not all $a,c \in M_2$ commute.
Thank you. Actually I was trying to define Center for TRO and this definition what I called weak center was suggested here: https://mathoverflow.net/questions/362862/definition-of-center-of-ternary-ring-of-operators. Do you know any definition of center for TROs?
I know very little about TROs, but it seems reasonable that there should be a good definition of center for them. My first guess would be ${z: abz^*c = az^*bc$ for all $a,b,c}$?
Oh, you can't do that, never mind. Not sure how a definition could go, then.
I searched in literature, I found the definition of commutativity Of TRO but I didn’t found the definition of center anywhere. In case you come up with something, please do comment on the question I have linked in above comment.
Okay. I'd think that really understanding the definition of commutativity should give some clues.
The definition of center which was suggested in the question I linked above in comment is motivated by definition of commutativity only. At first look, it seems like a good definition but as turned out that it doesn’t coincide with usual definition of commutativity for $C^{\ast}$-algebra. I can’t really think of any other definition for center as of now.
|
2025-03-21T14:48:31.239543
| 2020-06-13T12:02:17 |
362954
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"M. Winter",
"YCor",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/22377",
"verret"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630100",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362954"
}
|
Stack Exchange
|
When is a $k$-distance-transitive graph already distance-transitive?
Call a (finite and connected) graph $k$-distance-transitive if its symmetry group acts transitively on the pairs in each one of the sets
$$D_\delta:=\{(i,j)\in V\times V\mid \mathrm d(i,j)=\delta\},\qquad\text{for all $\delta\in\{0,...,k\}$}.$$
For $k=\mathrm{diam}(G)$ (the diameter of $G$), these are the distance-transitive graphs.
Question: Is it known that if $k$ is just large enough (e.g. relative to $\mathrm{diam}(G)$, or in an absolute sense), that every $k$-distance-transitive graph is already distance-transitive?
Or are there examples of $k$-distance-transitive graphs of arbitrary large diameter, that are not distance-transitive even if $k=\mathrm{diam}(G)-1$?
It seems you're considering connected graphs. You're also assuming graphs are finite, right?
@YCor Yes, I will edit this in. What has this question to do with (metric) geometry?
Metric geometry is, to a large extent, the study of metric spaces, including their isometry groups. Connected graphs provide a very nice and important supply of metric spaces. The notion of distance transitivity (isometry group is transitive on pairs at given distance) is a purely metric notion.
Just a small comment: if we have a $k$-distance-transitive graph, and the girth is also large (I think at least $2k$), then it must also be $k$-arc-transitive. It is known that there are no finite $8$-arc-transitive graphs except cycles. So, if you want examples with large $k$, then you need small(ish) girth.
Actually, here's an improvement on this argument: it is known that there is an absolute constant $c$ such that, in a finite $2$-arc-transitive graph, fixing (pointwise) a ball of radius $c$ around a vertex fixes everything. (This is due to Trofimov and Weiss. I think you can take $c$ to be $5$ or $6$.) Now, if your family is going to be $k$-distance-transitive for large $k$ smaller than the diameter, with $k$ growing, then I think it cannot have the property above. In particular, the large $k$ examples cannot be $2$-arc-transitive. In particular, they must have girth $3$.
Not an answer, but a related result that I wanted to share.
Note that $k$-distance-transitivity is generalized by $k$-walk-regularity. In
Camara, van Dam, Koolen, Park: "Geometric aspects of 2-walk-regular graphs"
the authors prove the following (see Corollary 6.4):
Theorem. Let $G$ be a $k$-walk-regular graph of degree $d$ with an eigenvalue $\theta\not\in\{-d,d\}$ of multiplicity at most $k$. If $k\ge 2$, then $G$ is distance-regular.
So $k$-walk-regular graphs that are not distance-regular must have only eigenvalues of large multiplicity (excluding $\theta=\pm d$).
Maybe something similar is true in the case of distance-transitivity.
|
2025-03-21T14:48:31.239740
| 2020-06-13T12:07:52 |
362955
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/143284",
"https://mathoverflow.net/users/69642",
"trisct",
"user69642"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630101",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362955"
}
|
Stack Exchange
|
Differentiability of characteristic functions and moments of the corresponding measure
Let $f$ be the characteristic function of a real-valued random variable $X$. It is known that if $f$ has a $k$-th order derivative (for some even $k$) then $\mu$ has a finite $k$-th order moment. Is there any reference or discussion the case when $k$ is odd? For example, if we only know that $f$ has its first order derivative, do we know that $\mu$ has finite first moment?
A good reference regarding this type of results is the book by Eugène Lukacs "Characteristic Function". For example, chapter 2.3 "Characteristic functions and moments" provides results in this direction. Theorems 2.3.1, 2.3.2 and 2.3.3 might be what you are looking for.
To complement my post and to counter the down vote: in the same chapter there is an example due to Zygmund of a characteristic function with a continuous first derivative but for which the associated law has an infinite first moment...
Thanks. I saw the example. But it was provided without proof that it is differentiable. Did Zygmund prove it? Can you give a reference to the proof of differentiability?
I think it was stated by Zygmund in this paper https://projecteuclid.org/euclid.aoms/1177730443 but without proof. For a proof, you should have a look at his fundamental work on trigonometric series.
Zygmund's example is the discrete random variable $X$ where
$\mathbb{P}(X=\pm n) = \frac{C}{n^2\log(n)}$ for integer $n \geq 2$. $C$ is a unique constant that makes this a probability distribution. You can calculate that $X$ does not have a finite first moment because
$$\mathbb{E}(|X|)= \sum_{n=2}^\infty \frac{2C}{n \log(n)} = +\infty$$. It diverges because $\int_2^\infty \frac{1}{x\log(x)}dx$ diverges.
The characteristic function is
$$\varphi(t) = C\sum_{n=2}^\infty \frac{\cos(nt)}{n^2 \log(n)}.$$
$\varphi(t)$ is differentiable at zero and the derivative is $\varphi'(0)=0$. Zygmund's argument does not rely on any fancy theory of trigonometric series. You can just check the differentiability directly.
Specifically, look at the quotients
$$\frac{\varphi(h) - \varphi(0)}{h} = C\sum_{n=2}^\infty \frac{\cos(nh) - 1}{hn^2\log(n)}$$.
Split this into two pieces for a $M \in \mathbb{N}$ to be chosen soon.
$$=C\sum_{n=2}^M \frac{\cos(nh) - 1}{hn^2\log(n)} + C\sum_{n=M+1}^\infty \frac{\cos(nh) - 1}{hn^2\log(n)}$$.
For the first sum, where $nh$ is small, we use the Remainder Theorem estimate
$$|\cos(nh) -1|\leq \frac{n^2h^2}{2}$$
and for the second sum, where $nh$ is big, we use the estimate
$$|\cos(nh) -1|\leq 2.$$
This all leads to the estimate, where $K$ is some constant independent of $h$ whose value changes line to line,
$$\left|\frac{\varphi(h) - \varphi(0)}{h}\right| \leq K \sum_{n=2}^M \frac{h}{\log(n)} + K\sum_{n=M+1}^\infty \frac{1}{n^2\log(n) h}$$.
The first sum is trivially less than $KMh$. For the second sum,
$$\sum_{n=M+1}^\infty \frac{1}{n^2 \log(n) h} \leq \frac{1}{h\log(M)} \sum_{n=M+1}^\infty \frac{1}{n^2} \leq \frac{K}{hM\log(M)}.$$
These calculations prove that
$$\left|\frac{\varphi(h) - \varphi(0)}{h}\right| \leq KMh + \frac{1}{M \log(M) h}$$.
Given $h$, choose integer $M=M(h)$ so that $\frac{1}{M \sqrt{\log(M)}}\approx h$. Then both terms go to $0$ as $h \to 0$, proving that $\varphi(t)$ is differentiable at $t=0$.
|
2025-03-21T14:48:31.240070
| 2020-06-13T12:51:57 |
362959
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"TheSimpliFire",
"https://mathoverflow.net/users/113397",
"https://mathoverflow.net/users/35520",
"ofer zeitouni"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630102",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362959"
}
|
Stack Exchange
|
A function with unexpectedly simple Legendre transformation
Let $I(x) = \frac{1}{2\pi} \int_{-2}^2 \sqrt{4-y^2}\ln|x-y|dy$. Then $I(x)$ is a concave function and
\begin{equation}
I(x)=
\begin{cases}
\frac{1}{4}x^2-\frac{1}{2}, &\text{if } |x|\leq2 \\
\frac{1}{4}x^2-\frac{1}{2}-\frac{1}{4}|x|\sqrt{x^2-4}+\ln \frac{|x|+\sqrt{x^2-4}}{2} &\text{if } |x|>2
\end{cases}
\end{equation}
Let $J(x)$ be the Legendre transform of $\frac{1}{2}x^2 - I(x)$, i.e. $J(x): = \sup_{y\in \mathbb{R}}\{xy - \frac{1}{2}y^2 + I(y)\}$, then $J(x)$ has a quite simple form
\begin{equation}
J(x) =
\begin{cases}
x^2-\frac{1}{2} &\text{if } |x|\leq1 \\
\frac{1}{2}x^2 + \ln |x| &\text{if } |x|>1
\end{cases}
\end{equation}
The simple form of $J(x)$ suggests some hidden deeper truth. My question is, can we find $J(x)$ directly without calculating $I(x)$ explicitly?
for $y>2$, the function $y^2/2-I(y)$ is precisely the rate function for the top eigenvalue of GOE/GUE, up to a constant, see section 6 of https://link.springer.com/content/pdf/10.1007/PL00008774.pdf. Thus you are asking about the log-mgf of the top eigenvalue, which in turns can be computed using the Selberg integral. I am however not sure this qualifies as a simpler, or deeper, truth..
Claim. $J(x)=\tfrac12x^2+\ln|x|+c$ for $|x|>1$ with $c$ constant.
Proof: Here, an explicit form for $I(x)$ is not needed but I can't use it to prove $c=0$.
For $|x|>1$, the substitution $y=2\cos t$ followed by
$(x-\cos t)(x-\cos s)=x^2-1$ yields $$I'(2x)=\frac1{2\pi}\int_{-2}^2\frac{\sqrt{4-y^2}}{|2x-y|}\,dy=x-\frac1\pi\int_0^\pi\frac{x^2-1}{|x-\cos t|}dt=x-\sqrt{x^2-1}$$ where $'=d/dx$. Thus $I'(x)+1/I'(x)=x$ for $|x|>2$.
We have $J(x)=xy_*-\tfrac12y_*^2+I(y_*)$ where $x-y_*+I'(y_*)=0$. This is equivalent to $y_*-x+\frac1{y_*-x}=y_*$ using the above identity, which implies $y_*=x+\frac1x$. Thus $J(x)=\frac12x^2-\frac1{2x^2}+I(x+\frac1x)$ so we now evaluate $I(x+\frac1x)$.
Let $K(x)=xI^*(x+\frac1x)$ where $^*=d/d(x+\frac1x)$ so the above identity simplifies to $K(x)^2-(x^2+1)K(x)+x^2=0$. By inspection, we have $K(x)=1$ or $x^2$; however, if $K(x)=x^2$ then $I'$ behaves like $x$ asymptotically, which is a contradiction. Thus $K(x)=1$, so $I^*(x+\frac1x)=\frac1x$ implies $I'(x+\frac1x)=\frac1x(1-\frac1{x^2})$ or that $I(x+\frac1x)=\frac1{2x^2}+\ln|x|+c$. This holds for $|x|>1$ so $J(x)=\tfrac12x^2+\ln|x|+c$ for $|x|>1$.
$I(2)=1/2$ proves $c=0$ but that requires direct evaluation.
|
2025-03-21T14:48:31.240228
| 2020-06-13T13:25:29 |
362962
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Michael Engelhardt",
"YCor",
"https://mathoverflow.net/users/134299",
"https://mathoverflow.net/users/14094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630103",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362962"
}
|
Stack Exchange
|
Fourier Transform of an even function
Let $S^n$ be an $n$-dimentional unit sphere.
Consider $f: S^n \longrightarrow R_+$, where $f$ is an even continuous function.
Denote
$$
F(f):=\int_0^{\infty}\int_{S^n}f(y)g\left(\frac{|xy|}{t}\right)dy\frac{dt}{t^{n+1}},
$$
where $x \in S^n, \, t>0$, and function $g$ is such that
$$
\int_{0}^{\infty}s^jg(s)ds=0, \quad j=0,2,4,\ldots, 2\left[(n-1)/2\right]
$$
$$
\int_1^{\infty}s^{\alpha}|g(s)|ds< \infty, \quad \alpha>n-1.
$$
Find the Fourier Transform of $F$.
What does it mean for a function to be "even" on the unit sphere for $n>1$?
I guess "even" just means $f(x)=f(-x)$ all $x$.
Ah - parity-even as it's called in some quarters. Yes, that must be it.
This is not a full answer, just an outline. It may require additional regularity assumptions on $f$ and $g$. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\SO}{SO}$ For $x\in\bR^{n+1}\setminus 0$ we set $$\bar{x}:=\frac{1}{|x|}x.$$
I assume that $xy$ denotes the inner product. Note that
$$
F[f](x)=\int_0^\infty\left(\int_{S^n} f(y)g(|xy|/t)dy\right) t^{-n-1}dt
$$
$$
= \int_0^\infty\left(\int_{S^n} f(y)g(|x| |\bar{x} y|/t)dy\right) t^{-n-1}dt
$$
($t=s|x|$)
$$
= |x|^{-n}\int_0^\infty\left(\int_{S^n} f(y)g(|\bar{x} y|/s)dy\right) s^{-n-1}ds=|x|^{-n}F(\bar{x}).
$$
For $s>0$ define $\newcommand{\eT}{\mathscr{T}}$
$$
\eT_s:L^2(S^n)\to L^2(S^n),\;\;\eT_s[f](x)=\int_{S^n}f(x)g(|xy|/s) dy,\;\;\forall x\in S^{n}.
$$
(This requires some assumption on $g$.) Observe next that we have a right action of $\SO(n+1)$ on $L^2(S^n)$. For $A\in\SO(n+1)$ define
$$
L^2(S^n)\ni f\mapsto A^*f\in L^2(S^n),\;\;A^*f(x)=f(Ax).
$$
Note that
$$
\eT_s[A^*f](x) = \int_{S^n}f(Ax)g(|Axy|/s) dy= \int_{S^n}f(Ax)g(|AxAy|/s) dy
$$
$$
= \int_{S^n}f(Ax)g(|xy|/s) dy=\eT_s[f](Ax)
$$
so that
$$\eT_s[A^*f]=A^*\eT_s[f].
$$
In other words, the transformation $\eT_s$ is equivariant with respect to the action of $\SO(n+1)$ and thus, according to Schur's Lemma, it acts as multiplication by constants on the irreducible components of this $\SO(n+1)$ representation on $L^2(S^n)$.
These are the spaces of homogeneous harmonic polynomials or, equivalently, the eigenspaces of the Laplacian on the round $n$-dimensional sphere. As such they coincides with the restrictions to the sphere of homogeneous harmonic polynomilas.
Denote by $\newcommand{\bH}{\mathbb{H}}$ $\bH_d$ space of (restrictions to $S^n$) of homogeneous polynomials of degree $d$ on $\bR^{n+1}$. Thus, $\forall s>0$, $d>0$ there exists a constant $c_d(s)$ such that
$$
\eT_s[P]=c_d(s)P,\;\;\forall P\in \bH_d.
$$
Let me explain how to find this constant. Denote by $\newcommand{\bx}{\boldsymbol{x}}$ $\bx^+=(1,0,0,\dotsc,0)\in\bR^{n+1}$ the North Pole in $S^n$ and choose $P\in\bH_d$ such that $P(\bx^+)=1$. Then
$$
\eT_s[P](\bx^+)=c_d(s)P(\bx^+)=c_d(s).
$$
Hence
$$
c_d(s)=\int_{S^n}P(y)g(|\bx^+y|/s) dy.
$$
Fortunately the spaces $\bH_d$ are well understood and the above integral can be explicitly described as a $1$-dimensional integral involving $g$ and Legendre polynomials. This is the so called Funk-Hecke formula ; see Sec. 1.4 of
C. Muller: Analysis of Spherical Symmetries in Euclidean Spaces, Springer Verlag, 1998.
Now observe that if $f\in\bH_d$ and $x\in S^n$ then
$$
F[f](x)=\int_0^\infty \eT_s[P] s^{-n-1} ds=\left(\int_0^\infty c_d(s) s^{-n-1} ds\right)P.
$$
Thus, everything boils down to computing Fourier transforms of homogeneous functions of the form,
$$\frac{1}{|x|^{n+d}}P_d(x), $$
where $P_d$ is a homogenous harmonic polynomial of even degree $d$ in $n+1$ variables.
|
2025-03-21T14:48:31.240455
| 2020-06-13T14:02:02 |
362963
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630104",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362963"
}
|
Stack Exchange
|
Is the Hitchin fibration proper?
By Hitchin fibration I mean the usual morphism from the coarse moduli space of semi-stable Higgs bundles to the Hitchin base (i.e. the direct sum of spaces of global sections of powers of the canonical bundle).
Is this morphism known to be proper? If so what is the standard reference?
Yes, it is proper. A reference for this fact is Theorem 6.11 in
C. Simpson, "Moduli of representations of the fundamental group of a smooth projective variety II", Pub. Mat. IHÉS, Tome 80 (1994), p. 5-79
|
2025-03-21T14:48:31.240522
| 2020-06-13T15:50:08 |
362972
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ceka",
"Jochen Wengenroth",
"Justthisguy",
"https://mathoverflow.net/users/21051",
"https://mathoverflow.net/users/84963",
"https://mathoverflow.net/users/91418"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630105",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362972"
}
|
Stack Exchange
|
Sobolev extension from a discrete set of points
Let $1 > \alpha > 0$ and fix some $C > 0$. Consider $\Omega \subset \mathbb{R}^n$ a bounded domain and $Y \subset \Omega$ a discrete (finite) set of points. For $f: Y \to \mathbb{R}$ define
$$\|f\|_{C^\alpha(Y)} := \sup_{y \in Y} |f(y)| + \sup_{y\neq z \in Y} \frac{|f(y) - f(z)|}{|y - z|^\alpha}.$$
Is there an extension map $E: C^\alpha(Y) \to H^s(\Omega)$, where $s = n/2 + \alpha$, such that $Ef|_Y = f$, and there is a $C_1 = C_1(C, \Omega) > 0$ (independent of $Y$)
$$\|f\|_{C^\alpha(Y)} \leq C \implies \|Ef\|_{H^s(\Omega)} \leq C_1.$$
We require that $C_1$ is independent of $Y$; here $H^s$ denotes the Sobolev space of index $s$. In case of a negative answer, what is the optimal Sobolev exponent $s$ one can get on the right hand side (we always require the extension to be continuous so we may define the restriction)?
Some comments about the problem. Firstly, one can easily check that the extension defined as
$$Ef(x) := \min_{y \in Y} \Big(f(y) + |x - y|^\alpha \|f\|_{C^\alpha(Y)}\Big)$$
satisfies the required properties as a map $E: C^\alpha(Y) \to C^\alpha(\Omega) \subset H^{\alpha - \varepsilon}(\Omega)$, for any $\varepsilon > 0$. So the question has positive answer for $s = \alpha - \varepsilon$.
Also, by Sobolev embedding theorems, we know that $H^{n/2 + \alpha} (\Omega) \subset C^\alpha(\Omega)$. So we expect there should be a better Sobolev exponent for some other choice of extension.
Finally, if the answer to my question is negative, it would be interesting to consider the extension map $E$ into spaces $H^s(\Omega)$, where $s \geq 1$ (say for $n \geq 2$). If $s = 1$ one could expect a ``harmonic extension" to be optimal; also, in this case if one takes a piecewise-linear extension on the triangulation determined by points in $Y$, I believe the answer is positive if $\alpha \geq \frac{1}{2}$. However this is not satisfactory, as in the application I have in mind $\alpha$ is very small.
References to literature are very welcome!
There is a series of articles of Fefferman in the Anals of Math. about such problems.
Thanks. It's indeed relevant, although I haven't been able to extract useful information yet. There are related papers by P. Shvartsman on Whitney-Sobolev extensions; e.g. ``Whitney-type extension theorems for jets generated by Sobolev functions". Unfortunately, these do not consider extension in L^2-based Sobolev spaces.
For $E \subset \mathbb R^n$ finite, $0 < s < 1$ and $sp > n$, let $|f|{L^{s,p}(E)}$ be the trace norm $ |f|{L^{s,p}(E)} = \inf{|F|_{L^{s,p}(\mathbb R^n)} : F|E = f} $. Here $| \cdot |{L^{s,p}(\mathbb R^n)}$ is the homogeneous fractional $L^{s,p}$ seminorm. I know a proof that the standard Whitney extension operator is bounded from $L^{s,p}(E)$ to $L^{s,p}(\mathbb R^n)$ following along the lines of Shvartsman's work, but it's a little long. Would you be interested in this?
|
2025-03-21T14:48:31.240710
| 2020-06-13T16:24:36 |
362974
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gordon Royle",
"M. Winter",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/1492"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630106",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362974"
}
|
Stack Exchange
|
Upper bound for smallest eigenvalue of infinite family of graphs
Let $\left\{G_{n}\right\}_{n=1}^{\infty}$ be a sequence of regular simple connected graphs with at least one edge such that $G_i$ is an induced sub-graph of $G_{i+1}$ and is not equal to $G_{i+1}$.
Is there a constant $c<0$ such that for large enough $i$ the smallest eigenvalue of $G_{i}$ is smaller or equal to $c$? In other words, can the sequance of smallest eigenvalues of $G_{n}$ approach to $0$?
Thanks!
Do you know regular graphs of smallest eigenvalue arbitrarily close to 0?
But a graph with minimum eigenvalue at least $-1$ is a disjoint union of cliques, so you can’t have smallest eigenvalues near $0$.
|
2025-03-21T14:48:31.240785
| 2020-06-13T16:29:31 |
362975
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"F Zaldivar",
"Gro-Tsen",
"https://mathoverflow.net/users/17064",
"https://mathoverflow.net/users/3903"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630107",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362975"
}
|
Stack Exchange
|
Norm on tensor product of fields
Let $F$ be an algebraically closed field of characteristic $p$ equipped with an absolute value $|\cdot|:F \rightarrow \mathbb{R}_{\ge 0}$ with respect to which $F$ is complete.
Define $|\cdot|_{prod}$ on the ring $F\otimes _{\mathbb F_p} F$ in the following way. If $c\in F\otimes _{\mathbb F_p} F$, then
$$|c|_{prod}:=\inf\left(\max_{1\le i\le n}\{|c_{1,i}||c_{2,i}| \}\ : \ c=\sum^{n}_{i=1}c_{1,i}\otimes c_{2,i}\right)$$
where the infimum is taken over all the possible ways to write $c$ as a sum of pure tensors. Does $|\cdot|_{prod}$ define a norm on $F\otimes _{\mathbb F_p} F$?
I am able to show that $|\cdot|_{prod}$ defines a semi-norm, which is submultiplicative and non-archimedean, but I am not able to find whether there does exist some $x\ne 0$ s.t. $|x|_{prod}=0$.
My guess is that such elements don't exist and I am also able to show that no non-zero pure tensor has absolute value zero, but I still can't show this for a general element of $F\otimes _{\mathbb F_p} F$.
The following may be relevant: https://mathoverflow.net/questions/103945/valuations-on-tensor-products
Also, the reference: https://arxiv.org/abs/1302.1381
especially Theorem 9.
Your definition of the norm is evocative of the projective cross-norm of Banach spaces. I wonder if there's something to draw from this.
You can probably find this in most books on non-Archimedean functional analysis, see for instance Proposition 17.4 in Schneider's book.
The rough idea is to reduce to a tensor product of finite-dimensional spaces and then to norms associated to bases. You can now compute directly.
By the way, you probably want to assume your norms to be non-Archimedean.
OK, I found the argument on normed vector spaces in these online notes. I will explain it in the case at hand.
Observe that for $a \in \mathbf{F}_p \subset F$ we have $|a| = 1$ if $a \not = 0$ and $|0| = 0$.
Let $c \in F \otimes_{\mathbf{F}_p} F$. By linear algebra, there are minimal sub $\mathbf{F}_p$-vector spaces $V, W \subset F$ such that $c \in V \otimes_{\mathbf{F}_p} W \subset F \otimes_{\mathbf{F}_p} F$. Then $\dim(V) = \dim(W) < \infty$ and this integer is called the rank of $c$.
Write $c = \sum_{i = 1, \ldots, n} x_i \otimes y_i$ with $x_i \not = 0$ and $y_i \not = 0$ for all $i = 1, \ldots, n$.
If $n$ is minimal, then $n$ is the rank of $c$ and $x_i \in V$ and $y_i \in W$. Since our ground field is $\mathbf{F}_p$ is finite, we have only a finite number of cases here and hence the infimum over these cases is $> 0$.
If $n$ is strictly bigger than the rank of $c$, then $x_1, \ldots, x_n$ must be linearly dependent (otherwise $V$ would be the span of $x_1, \ldots, x_n$ and have bigger dimension). Let $\sum a_i x_i = 0$ be a nontrivial $\mathbf{F}_p$-linear relation. After renumbering we may assume $a_n \not = 0$ and $|x_n| \geq |x_i|$ for all $i$ with $a_i \not = 0$. Thus we may assume $x_n = \sum_{i < n} b_i x_i$ for some $b_i \in \mathbf{F}_p$ not all zero and we may assume $|x_n| \geq |x_i|$ for those $i < n$ with $b_i \not = 0$. It follows that $|x_n| = \max_{i < n} |b_ix_i|$.
Set $y'_i = y_i + b_i y_n$. Then we see that $c = \sum_{i \leq n} x_i \otimes y_i = \sum_{i < n} x_i \otimes y'_i$. Finally,
$$
\max_{i \leq n} |x_i|\cdot |y_i| =
\max_{i < n} \max(|x_i| \cdot |y_i|, |b_i x_i| \cdot |y_n|) \geq
\max_{i < n} |x_i| \cdot |y'_i|
$$
So by induction on $n$ we win.
|
2025-03-21T14:48:31.241359
| 2020-06-13T16:37:56 |
362976
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630108",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362976"
}
|
Stack Exchange
|
Schur triples question
So I've been reading through On the number of monochromatic Schur triples by Datrovsky on finding the minimal number of Schur triples. This means you're trying to 2-colour the set of the smallest n positive integers in such a way as to minimize the number of monochromatic triples (a,b,c) where a+b=c.
In the first paragraph or so he somehow gets a formula for these with something discrete Fourier-esque (called equation (1.1) in the paper). With a lot of algebra I think I've managed to convince myself that this formula holds but I'm not sure how one would think of this or when it would generalize. In particular I don't follow the steps in corollary 1 where he generalizes to Schur k-tuples. It also seems like the same trick is reused in a few places in the paper so I'd really like to get a handle on exactly what tools are being used here.
If anyone could give some insight I'd really appreciate it.
Fourier analysis is a well-established tool in additive combinatorics,
and counting questions are routinely translated into equivalent questions about the Fourier transform.
This survey of Gowers might give you an idea of the breadth and scope: https://arxiv.org/abs/1608.04127.
|
2025-03-21T14:48:31.241480
| 2020-06-13T16:51:01 |
362977
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Benjamin Steinberg",
"DJWilliams",
"LSpice",
"Maxime Ramzi",
"diracdeltafunk",
"https://mathoverflow.net/users/102343",
"https://mathoverflow.net/users/15934",
"https://mathoverflow.net/users/159556",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/38068",
"https://mathoverflow.net/users/54637",
"spin"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630109",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362977"
}
|
Stack Exchange
|
How much does Ext tell me about isomorphisms?
So this was a question I sort of stumbled on and realised I was quite stumped. Suppose we have two finitely generated $R$-modules $M, N$ (I have the group ring $R=\mathbb{Z}[G]$ in mind) which appear in short exact sequences
$$0\to X\to M\to Y\to 0,$$
$$0\to X\to N\to Y\to 0.$$
I know that $M$ and $N$ are indecomposable and I know that ${\rm Ext}^1(Y, X)\cong\mathbb{Z}/2 \Bbb Z$. Are $M$ and $N$ isomorphic?
If I can find a homomorphism between them which fits into a commutative diagram, then obviously this is true. But must such a homomorphism exist?
Don't you mean $Ext^1(Y,X)$ ?
ext(X,Y) classifies sequences 0->Y->M->X->0. If it was the other way I think you are OK.
Sorry, that was a typo. My bad.
Shouldn't your question be how much Ext tells you about isomorphism (as an adjective) of modules, not about isomorphic modules, since the point is to deduce, not to assume, the modules are isomorphic?
@LSpice yes, sorry. I should have thought more on the wording of this question. First time posting, so I didnt think. My own stupid fault.
So $M$ and $N$ are indecomposable. In particular these short exact sequences are nonsplit, so they correspond to non-zero elements of $\operatorname{Ext}^1(Y,X)$. If $\operatorname{Ext}^1(Y,X) \cong \mathbb{Z}/2\mathbb{Z}$, this means that these two sequences are equivalent, and in particular $M$ and $N$ are isomorphic.
@spin that was exactly my logic, however I wasnt sure since I am not wholly comfortable with ext. I need to read more on it. It also seemed to me that they would be isomorphic if they belonged to inverse classes. However once again I was not sure of this.
Well, $x \mapsto -x$ is the identity map on $\mathbb{Z}/2\mathbb{Z}$, so being in inverse classes is the same as being in the same class! Perhaps I'm misunderstanding what you mean? The important point is that $M$ and $N$ being indecomposable implies that neither extension corresponds to the identity element in $\text{Ext}^1$, and there is only one non-identity element in $\mathbb{Z}/2\mathbb{Z}$. Small note: this implication isn't entirely trivial, you need to use the fact that if $X = 0$ or $Y = 0$ then we would have $\text{Ext}^1(Y,X) = 0 \neq \mathbb{Z}/2\mathbb{Z}$.
@diracdeltafunk sorry, I was unclear. I meant if ext was $\mathbb{Z}/3$, for example. Then would that mean that any two indecomposable modules in an exact sequence $0\to X\to ?\to Y\to 0$ are still isomorphic? This just seems too good to be true. I suppose I am still unhappy with exactly how powerful ext is at classifying extensions (and how that relates to the modules). Could you also explain your small note further, please. Which implication do you mean?
Ah, yes, the assumption that $\text{Ext} \cong \mathbb{Z}/2$ is very strong and it is what allows us to conclude that any two non-split extensions are equivalent. If $\text{Ext} \cong \mathbb{Z}/3$ this will not be the case, however, any two non-split extensions will still be isomorphic, as you suspected! This is because if $0 \to X \xrightarrow{f} Z \xrightarrow{g} Y \to 0$ represents an element $a \in \text{Ext}^1(Y,X)$, then $-a$ is represented by $0 \to X \xrightarrow{-f} Z \xrightarrow{g} Y \to 0$ (check this by computing the Baer sum of these two extensions and verifying it splits)
(cont.) However, for $\lvert \text{Ext}^1 \rvert > 2$ we can't use this argument anymore, because even after taking the quotient of $\text{Ext}$ by $a \sim -a$ we will have more than one nontrivial class. In general, though, I agree that $\text{Ext}$ is a spookily powerful tool: that's why it's so popular and deeply studied! Anyway, the implication I was talking about was "if $M$ and $N$ are indecomposable then neither extension corresponds to the identity element of $\text{Ext}^1$". This is true, but only because we had also assumed $\text{Ext}^1 \neq 0$. Still very simple, but not immediate.
@Diracdeltafunk thank you so much. This all seems so much clearer ($Ext$ really is powerfu!l). So just to clarify, if $Ext(Y, X)$ is $\mathbb{Z}/2$ or $mathbb{Z}/3$ then any indecomposable modules that occur in between $X, Y$ must be isomorphic (if the latter, then the exact sequences don't have to be equivalent, though) since they cant belong to the trivial class. Did you mean $|Ext^1|>3$ above, by the way?
Yes, thanks, that was a typo! But that's exactly right.
|
2025-03-21T14:48:31.241774
| 2020-06-13T16:55:40 |
362978
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"KConrad",
"https://mathoverflow.net/users/3272"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630110",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362978"
}
|
Stack Exchange
|
English translation of a Russian paper by Gordin and Lifšic
Unfortunately I can't read Russian, I was wondering if there is an English translation of this paper
“The central limit theorem for stationary Markov processes”, Dokl. Akad. Nauk SSSR, 239:4 (1978), 766–767
I apologize in advance if this is not the suitable place to make this question.
The paper, available in original form at http://www.mathnet.ru/links/c6486f384b0ab5aeb7e0dee994974fad/dan41622.pdf, is quite short. From this .pdf you can copy and paste the text into Google Translate yourself, perhaps a bit at a time, and what comes out (ignoring the notation, which is math rather than Russian) seems reasonable for computer translation. Its first suggested translation for Доказательство is "Evidence" rather than the more standard "Proof", but in context I think you could figure out what was meant.
This journal was transllated into English as
Soviet Mathematics. Doklady Many US libraries subscribed it.
If you have access to a university library, and it does not have it, use ILL.
Here is the exact reference for the translation:
Gordin, M. I.; Lifshits, B. A.
The central limit theorem for stationary Markov processes. (English. Russian original) Zbl 0395.60057
Sov. Math., Dokl. 19, 392-394 (1978); translation from Dokl. Akad. Nauk SSSR 239, 766-767 (1978).
|
2025-03-21T14:48:31.241894
| 2020-06-13T17:24:13 |
362980
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"0xbadf00d",
"LL 3.14",
"https://mathoverflow.net/users/153203",
"https://mathoverflow.net/users/91890"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630111",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362980"
}
|
Stack Exchange
|
How is this bound for a Wasserstein contraction coefficient in this paper obtained?
I'm trying to understand the following conclusion from this paper (see below for the relevant paragraphs):
I'm not sure whether they really mean that it follows from the statements of Lemma 3.2 (since I absolutely don't see how this conclusion is possible) or from its proof (which I could imagine).
Taking a closer look to the proof of Lemma 3.2-ii, we see that the result actually is that for all $r\in[r_0,1)$ and for all $\alpha\in(0,1)$, there is a $K\ge0$ such that $$\operatorname E\left[\rho_r(X^x_1,X^y_1)\right]\le\alpha\rho_r(x,y)+K\;\;\;\text{for all }x,y\in E\tag1.$$ And, as always, this extends to bounds for $\operatorname E\left[\rho_r(X^x_n,X^y_n)\right]$ by the semigroup property.
In the proof the additive $K$ corresponds to a bound on an integral restricted to an open ball. Maybe the assumption $\rho(x,y)\ge K_\ast$ can be used to show this integral is $0$?
No, you just have to take $K_*$ such that $K \leq (\alpha_*-\alpha)\rho(x,y)$, and so for example $K_* = \frac{K}{\alpha_*-\alpha}$.
If I am not missing something, a good way of doing the proof could be:
Since $\alpha<1$, there exists $\alpha_*\in (\alpha,1) \subset (0,1)$. Now take $K_* = \frac{K}{\alpha_*-\alpha}$ and $\rho(x,y) ≥ K_*$. Then
$$
\begin{align*}
\alpha\,\rho(x,y) + K &= \alpha\,\rho(x,y) + (\alpha_*-\alpha)\,K_*
\\
&≤ \alpha\,\rho(x,y) + (\alpha_*-\alpha)\,\rho(x,y) = \alpha_*\,\rho(x,y)
\end{align*}
$$
Thank you for your answer. A few remarks: It should be $K_\ast:=\frac K{\alpha_\ast-\alpha}$, since otherwise $K_\ast$ is negative. And the first inequality in your displaced equation is an equality.
Yes, corrected ;)
They proceed in the proof of Lemma 3.8 by showing a rather simple inequality: https://i.sstatic.net/hJRkW.png. It seems like they are claiming that $$1-\alpha_\ast+\alpha_\ast d(x,y)\le\frac{1+\alpha_\ast\beta K_\ast}{1+\beta K_\ast}d(x,y),$$ but I'm not able to see how they obtain this. They say that it follows from $d(x,y)\ge1+\beta K_\ast$ and so I've tried to add $d(x,y)-(1+\beta K_\ast)\ge0$ to the right-hand side. However, I wasn't able to arrange the terms so that they match the right-hand side. Do you've got an idea?
|
2025-03-21T14:48:31.242067
| 2020-06-13T17:31:40 |
362981
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lennart Meier",
"Victor TC",
"https://mathoverflow.net/users/112348",
"https://mathoverflow.net/users/2039",
"https://mathoverflow.net/users/43326",
"user43326"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630112",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362981"
}
|
Stack Exchange
|
Bousfield $p$-completion on spectra
Bousfield p-completion on spaces is a functor $(-)^{\wedge p}$ whose main property is that a map $f:X\rightarrow Y$ induces an isomorphism $f_{\ast}:H_\ast(X,\mathbb{F}_{p})\rightarrow H_\ast(Y,\mathbb{F}_p)$ if and only if $f^{\wedge p}:X^{\wedge p}\rightarrow Y^{\wedge p}$ is a homotopy equivalence.
Let $X$ be now be a connective CW spectra, it is known that $H\mathbb{F}_p$-localization agrees with $M\mathbb{F}_p$-localization (both on X, the latter is referred as the $p$-completion). Let $f:E\rightarrow F$ a map between two connective CW spectra, is there any connection between $f^{\wedge p}$ and $({H\mathbb{F}_p})_{\ast}(f)$ (or $({H\mathbb{F}_p})^{\ast}(f)$)?, of course, an analogous version of the previous property is desirable, although there may be more restrictive conditions so that this property holds.
It can very well happen that the map is zero in homology, but is not nullhomotopic, e.g. almost any map between shifts of sphere spectra. In some sense, the more detailed relationship is described by the Adams spectral sequence.
@LennartMeier Do you mean a map $f$ with $H_\ast(f,\mathbb{F}_p)=0$ but $f^{\wedge p}\not\simeq\ast$?
Yes. This is fullfilled for any map $\Sigma^n\mathbb{S} \to \mathbb{S}$ with $n$ positive, whose order is a power of $p$, e.g. the Hopf map with $n=1$ and $p=2$.
In view of the way the question is formulated, an obvious answer is $(H\mathbb{F}p)(f)=(H\mathbb{F}p)(f^{\wedge p})$, (and similarly for cohomology), is this what is really asked for?
@user43326 Actually I had the suspicion that (under some restrictions) $(H\mathbb{F}p)\ast(f)$ is an isomorphism if and only if $f^{\wedge p}$ is a stable homotopy equivalence.
@VictorTC As a matter of fact it is more or less the definition of the localization (p-completion is the localization with respect to $HF_p$)
Besides, it is in your first paragraph, or am I missing something?
@user43326 No, I meant this property is well known on topological spaces, but besides the common name 'p-completion', its definition on spectra looks quite different.
OK, now I understood your post. It is actually easier to construct and requited properties for spectra than spaces, but the point is that your "main property" is a part of the definition of categorical localizations. So if you accept the fact that it is a localization, then you also accept that the property holds...
@user43326 Thank you!, I checked the characterization of $E_{\ast}$-localization according to Bousfield's paper, and indeed this property follows from that.
Welcome. It is as a matter of fact the definition of the localization with respect to a class of morphism (here $E_*$-equivalences) in any category.
|
2025-03-21T14:48:31.242394
| 2020-06-13T18:00:38 |
362984
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Akira",
"https://mathoverflow.net/users/99469"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630113",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362984"
}
|
Stack Exchange
|
Besov or Triebel-Lizorkin spaces versus Lorentz spaces
I first asked this question on math.stackexchange here but it seems it is more a research level question ...
At the $0$ order of derivatives of Sobolev spaces and for a fixed integrability order $p$, we find Besov spaces $\dot{B}^0_{p,q}$, Triebel Lizorkin spaces $\dot{F}^0_{p,q}$ and Lorentz spaces $L^{p,q}$, with in particular if $p≥ 2$
$$
\begin{align*}
\dot{B}^0_{p,1} ⊂ \dot{B}^0_{p,2} ⊂ L^p &= \dot{F}^0_{p,2} \subset \dot{F}^0_{p,p} =\dot{B}^0_{p,p} ⊂ \dot{B}^0_{p,\infty}
\\
L^{p,1} \subset L^{p,2} ⊂ L^{p} &= L^{p,p} \subset L^{p,\infty}.
\end{align*}
$$
So, the ordering is well understood when remaining either in the $B,F$ setting, either in the Lorentz setting. so my question is Is there any embedding from one setting into the other one? (except the trivial embeddings on/from $L^p$).
The way of building these spaces is different (in one case one cuts in frequency and in the other one cuts in height). However, Sobolev embeddings tells us that we can trade a bit of local regularity for a bit of local integrability. Moreover the function $|x|^{-a}$ is in both $\dot{B}^{0}_{d/a,\infty}$ and $L^{d/a,\infty}$, but not in $L^{d/a}$.
For interested readers, this question has got an answer at https://math.stackexchange.com/questions/3654495/besov-or-triebel-lizorkin-spaces-versus-lorentz-spaces.
|
2025-03-21T14:48:31.242517
| 2020-06-13T19:07:56 |
362989
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Blake H",
"Pat Devlin",
"https://mathoverflow.net/users/159565",
"https://mathoverflow.net/users/22512"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630114",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362989"
}
|
Stack Exchange
|
Maximum-weight perfect matching in a 3-regular, complete, 3-partite hypergraph
Let $H=(V, E)$ be a weighted hypergraph such that $V=A\cup B \cup C$, where $A,B,C$ are disjoint sets of size $n$, and $E=A\times B\times C$. In my particular case, $\forall e\in E$, $ wt(e)\in\{0,\pm 1, \pm 3\}$.
I'm trying to figure out whether or not there is an efficient algorithm to compute the maximum cardinality perfect matching for this case. I've attempted generalizing max-flow, for this constrained case, but I didn't really get anywhere. I also tried translating it to a classical graph, since we have a lot of tools to deal with those. For example, I let the vertices correspond to the edges of $H$, and having an edge connect two vertices if they map to disjoint edges in $H$. Then the weights correspond to the sum of the weights of the vertices in the new graph (the weights of the edges in hypergraph). The issue with this is that I'm unsure exactly how to characterize the optimal solution in this way. It is clear that I'm trying to find the maximum weight $n$-length path that corresponds to disjoint edges in $H$, but I'm not sure how to start finding an efficient algorithm for that.
If you know the answer then a hint is fine, but I'm mostly concerned if this is already known to be NP-complete.
Yup, NP-hard.
Consider the case where edges only have two weights. Then we have a hypergraph of light edges, and the problem is to pick a matching using as many light edges as possible. This is NP-hard, and in fact I believe it was one of Karp’s original 21 NP-complete problems.
https://en.m.wikipedia.org/wiki/3-dimensional_matching
That said, you might still want to solve this problem, in which case a lot is known. There are analyses of random-type strategies (also if the edge weights are random) as well as Dirac-type results (“if there are lots of light edges, then...”).
There are also results on fractional matchings if that floats your boat. Those are easier of course.
But yea. Matchings in graphs are actually fine, but matchings in hypergraphs are suddenly much harder (even in the $r$-partite case)!
Thanks a lot! After thinking about the problem some more yesterday, I found out that all I really need to do is verify that a matching is maximum-weight perfect. If I can find that one of these problems that is NP-complete, then I should be able to verify whether or not a given the matching is maximum weight, perfect, correct?
Not at all. That’s more or less the same as the original problem. The decision problem would be “is there a matching of weight at least k?” That problem is in NP. If you’re given a matching and asked if it’s best, then that problem is still NP-hard. (If you were good at that, you could use that to make a poly-time algorithm to start with a matching and find a matching that does better. You could then iterate this to find a maximum-weight matching.)
Oh of course. That makes sense now, I appreciate all the help!
|
2025-03-21T14:48:31.242832
| 2020-06-13T19:51:26 |
362991
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ali Enayat",
"Andrés E. Caicedo",
"François G. Dorais",
"Gerald Edgar",
"Gerhard Paseman",
"Joe Shipman",
"Joel David Hamkins",
"Julien Narboux",
"LSpice",
"Paul Taylor",
"Philip Ehrlich",
"Theo Johnson-Freyd",
"Todd Trimble",
"Zvonimir Sikic",
"https://mathoverflow.net/users/18939",
"https://mathoverflow.net/users/1946",
"https://mathoverflow.net/users/2000",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/2733",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/30676",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/454",
"https://mathoverflow.net/users/6085",
"https://mathoverflow.net/users/74325",
"https://mathoverflow.net/users/78",
"https://mathoverflow.net/users/9269",
"https://mathoverflow.net/users/94293"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630115",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362991"
}
|
Stack Exchange
|
Who first characterized the real numbers as the unique complete ordered field?
Nearly every mathematician nowadays is familiar with the fact that
there is up to isomorphism only one complete ordered field, the
real numbers.
Theorem. Any two complete ordered fields are isomorphic.
Proof. $\newcommand\Q{\mathbb{Q}}\newcommand\R{\mathbb{R}}$Let us observe first that every complete ordered field $R$ is
Archimedean, which means that there is no number in $R$ that is
larger than every finite sum $1+1+\cdots+1$. If there were such a
number, then by completeness, there would have to be a least such
upper bound $b$ to these sums; but $b-1$ would also be an upper
bound, which is a contradiction. So every complete ordered field is
Archimedean.
Suppose now that we have two complete ordered fields, $\R_0$ and
$\R_1$. We form their respective prime subfields, that is, their
copies of the rational numbers $\Q_0$ and $\Q_1$, by computing
inside them all the finite quotients
$\pm(1+1+\cdots+1)/(1+\cdots+1)$. This fractional representation
itself provides an isomorphism of $\Q_0$ with $\Q_1$, indicated
below with blue dots and arrows:
Next, by the Archimedean property, every number $x\in\R_0$
is determined by the cut it makes in $\Q_0$, indicated in yellow, and since $\R_1$
is complete, there is a counterpart $\bar x\in\R_1$ filling the
corresponding cut in $\Q_1$, indicated in violet. Thus, we have
defined a map $\pi:x\mapsto\bar x$ from $\R_0$ to $\R_1$. This map
is surjective, since every $y\in\R_1$ determines a cut in $\Q_1$,
and by the completeness of $\R_0$, there is an $x\in\R_0$ filling
the corresponding cut. Finally, the map $\pi$ is a field
isomorphism since it is the continuous extension to $\R_0$ of the
isomorphism of $\Q_0$ with $\Q_1$. $\Box$
My expectation is that this theorem is familiar to almost every
contemporary mathematician, and I furthermore find this theorem
central to contemporary mathematical views on the philosophy of
structuralism in mathematics. The view is that we are entitled to
refer to the real numbers because we have a categorical
characterization of them in the theorem. We needn't point to some
canonical structure, like a canonical meter-bar held in some
special case deep in Paris, but rather, we can describe the
features that make the real numbers what they are: they are a
complete ordered field.
Question. Who first proved or even stated this theorem?
It seems that Hilbert would be a natural candidate, and I would
welcome evidence in favor of that. It seems however that Hilbert
provided axioms for the real field that it was an Archimedean
complete ordered field, which is strangely redundant, and it isn't
clear to me whether he actually had the categoricity result.
Did Dedekind know it? Or someone else? Please provide evidence; it
would be very welcome.
This sounds like something Tarski would have known and published (who first proved it). Have you looked at his literature? Gerhard "Society Seems Disordered And Incomplete" Paseman, 2020.06.13.
I'm fairly sure Tarski must have known it. But I suspect it must have been known much earlier.
See commentary on Twitter at https://twitter.com/JDHamkins/status/1271893175314141185
This is a great question, but, like all questions about history of science and mathematics, it seems like it should go on HSM. (The standard response is that that site is less active, and at least part of that is because there's so much HSM activity here!)
@LSpice The standard response is really that the quality of HSM is awful and many experts and interested scholars avoid it as a result. Sending a question there is usually a disservice.
@AndrésE.Caicedo, right, but it's circular: HSM will never get any better if the most qualified people keep avoiding it, so those who are interested in HSM should ask and answer questions there, not re-purpose MO for it. We are very strict with new users that MO is only for research-level questions in mathematics; we should be equally strict amongst ourselves.
@LSpice If a new user wrote such a well-written and clear question, I would strongly support keeping it here on MO. And when I am working with graduate students, I strongly encourage them to look into the history, and read original papers, as part of their research.
@LSpice I think your understanding of what this site is is perhaps too narrow. I feel the question belongs here.
@JoelDavidHamkins Hilbert's completeness axiom is not the
least upper bound axiom, it is not redundant. Hilbert chose his version of the completeness axiom to be independent of Archimedes axiom.
@JoelDavidHamkins see [Gio13] Eduardo Giovannini. Completitud y continuidad en Fundamentos de la Geometría de Hilbert : acerca
del Vollständigkeitsaxiom. THEORIA. An International Journal for Theory, History and Foundations
of Science, 28(1) :139–163, 2013.
@LSpice One can turn your argument around: it is precisely the excessive efforts to limit interesting questions on MO that has led in recent years to a reduced level of quality and engagement with MO. Have you observed this? To the extent that you are successful in transferring mathematically interesting questions to another site, I would argue that you are working towards the decline of MO. Let's have the interesting questions here! My MO policy has always been: questions are welcome on MO if they are of interest to research-level mathematicians.
@JoelDavidHamkins, I have not noticed any diminution of quality, and I don't know how to measure anyone's engagement but my own. This is one area where SEDE can say for sure, but I suspect that, for a lot of users, the idea of the reduced level of quality and engagement is rather like https://imgs.xkcd.com/comics/the_pace_of_modern_life.png . (I want to say again that this is nothing against your question! I don't want to kill MO; I want HSM to thrive.)
@JoelDavidHamkins It's also hard for me to tell if there's really been diminution of quality, but if you see a question that you feel has been wrongly closed/deleted, please consider making an appeal at https://meta.mathoverflow.net/questions/223/requests-for-reopen-and-undelete-votes-for-on-hold-closed-and-deleted-question With the scope of your influence at this site, I'd bet this could help bend back MO in a direction you'd prefer.
@Julien Narboux Hilbert's completeness axiom is "R is maximal Archimedean ordered field" ("R is maximal ordered field" characterizes Conway's No)
My of-the-duff guess would have been Landau's Grundlagen. But I looked it up, its date is 1930, so the existing answers below show that I was wrong.
Joel, I believe this was first explicitly stated and proved by E.V. Huntington in his classic paper: Complete sets of postulates for the theory of real quantities, Trans. Am. Math. Soc. vol. 4, No. 3 (1903), pp. 358-370. See Theorem II', p. 368.
Edit (June 14, 2020): It is perhaps worth adding that in 1904, the year following the publication of Huntington's paper, O. Veblen published his paper A System of Axioms for Geometry, Trans. Am. Math. Soc. vol. 5, no. 3, pp. 343-384, in which he introduced the idea of a categorical system of axioms. He illustrated his conception with Huntington's above mentioned characterization of the reals (pp. 347-348). No doubt, this is mentioned in the paper referred to below by Ali Enayat.
EDIT (8/27/23)
In light of the doubts raised by Zvonimir Sikic regarding Huntington being the originator of the idea of a categorical system of axioms, it should be noted that while Huntington was well aware that there existed categorical axiom sets in the literature prior to his isolation of the concept, including those axiom sets mentioned by Sikic, he held the view that the earlier writers neither isolated the idea nor expressed any interest in it. He made this point explicitly as follows in 1913 when he provided a novel categorical characterization of Euclidean geometry.
More important than the question of independence is the proof of the sufficiency of the postulates to determine a unique type of system; or, to use a phrase of Veblen’s, the proof that the postulates form a categorical set. Little attention seems to have been paid to this question except by the present writer in connection with the foundations of analysis, and by Veblen in connection with the foundations of geometry; and yet there appears to be no other way of proving that all the propositions of a science are deducible from a given set of postulates, than by showing that the postulates form a ‘sufficient' or ‘categorical' set.” (pp. 524-525)
—Huntington, E. V. 1913. “A Set of Postulates for Abstract Geometry, Expressed in Terms of the Simple
Relation of Inclusion.” Mathematische Annalen 73 (4):522–59. (pdf from archive.org)
For what it's worth, David Hilbert was the managing editor of Mathematische Annalen at the time of the publication of Huntington's paper.
Edit: 8/28/23
I should have mentioned in my previous edit, that contrary to Zvonimir Sikic contention, Huntington's paper of 1903 does indeed contain an axiomatization of the ordered field of real numbers as well as a proof of it categorical nature. In Section 2 he provides an axiomatization of the ordered field of real numbers (consisting of 14 axioms including the Dedekind Completeness axiom) and in Theorem II' on page 368 he proves that the axiomatization is categorical.
Huntington's work was based on previous work by Dedekind and Hilbert, but nevertheless, the following article corroborates Ehrlich's claim that E.V. Huntington can be credited to be the first to have formulated and proved the
categoricity of the second order theory of the real field (in modern parlance). Completeness and Categoricity. Part I: Nineteenth-century Axiomatics to Twentieth-century Metalogic, by S. Awodey & E. H. Reck (a preprint can be found via http://www.andrew.cmu.edu/user/awodey/preprints/cc/ccI.pdf). It appeared in print in History and Philosophy of Logic, 23: 1–30 (2002).
What about Hoelder's paper from 1901: 'Der Quantitat und die Lehre yom Mass', in Berichte uber die Verhandlungen der Koniglich Siichsuschen Gesellschaft der Wissenschaften zu Leipzig, Matematisch-Physische Classe, 1-64.
It is proved there that Archemedean systems of magnitudes with no minimal magnitudes, are isomorphicaly and densely embeddable in R+ (if they have minima they are isomorphic to Z+). If a system of magnitudes is continuous, i.e. does not have empty Dedekind cuts, then it is isomorphic to R+. Systems of magnitudes is a linearly ordered semigroup with restricted difference.
@philipehrlich Thanks a lot. I completely agree that neither Hilbert nor Holder are originators of the idea of a categorical system of axioms (I suppose it was Veblen). But was that the question? Let me give another example. Who is the originator of the categoricity of the axiomatization of N? I would say Dedekind, although he didn't have the idea of a categorical system of axioms and did not even axiomatize N.
@Zvonimir Sikic. Yes, I suspect Dedekind would get credit with respect to N. See, however, my recent comment regarding your claim about Huntington in your answer to Joe's question.
Hilbert's „Über den Zahlbegriff” published in 1900. is the first modern axiomatization of Archimedean fields. To axiomatize R, Hilbert added the maximality axiom (i.e. R is a maximal Archimedean field). He asserted that the existence of the model of the axioms is „nur einer geeigneten Modification bekkanter Schlussmethoden” i.e. „is just a suitable modification of known methods of reasoning”, and uniqueness trivially follows from maximality.
I guess that „bekkanter Schlussmethoden” were the following: If S is an ordered field it contains an isomorphic copy of Q, so let’s call it Q. Every point in S determines a cut in Q. If S is also Archimedean then every cut in Q determines none or one point in S. If S is also maximal then every cut in Q must determine one point in S. Namely, if it doesn’t it could be extended to an Archimedean field with this property, which is by Dedekind construction isomorphic to R. Hence, there is, up to isomorphism, exactly one maximal Archimedean ordered field which is isomorphic to R and every other is, up to isomorphism, contained in R.
This kind of argument has been well known since the time of Dedekind's „Stetigkeit und irrationale Zahlen", i.e. for more than 30 years. Everyone interested in the topic knew the argument, so Hilbert did not consider it necessary to repeat it. This is why I believe that Hilbert proved that maximality uniquely determines R just by stating that the proof is "a suitable modification of known methods of reasoning" (even though he didn't actually formulate the result in terms of isomorphism and didn't give the proof himself).
Hölder published a more general results in „Die Axiome der Quantität und die Lehre vom Mass” from 1901. He proved, in modern terms, that every continous system of magnitudes is isomorphic to R (it is a trivial corollary that every complete ordered field is isomorphic to R). A continous system of magnitudes was axiomatized with trichotomy, associativity, density, positivity (a < a + b), difference (if a < b then there is c such that a+c= b), and Dedekind's axiom (there is no gaps). In his terminology „there is a [real] measure-number for each given magnitude and there is a magnitude for each given [real] measure-number“ and the correspondence preserves the defining properties. It follows taht „if one has two systems of magnitudes, both fulfilling axioms I to VII, then the systems can be explicitly related to each other such that the sums [and order] of corresponding magnitudes also correspond“.
Huntington, in his „A complete set of postulates for the theory of absolute continuous magnitude“ from 1902, wrote: “The set of postulates adopted in the present paper is nearly the same as the second set of Hölder’s, with the exception that Dedekind's postulate of continuity is here replaced by Weierstrass's”. So, Huntington is not concerned with Archimedean fields but with Hölder's systems of magnitudes. He proved in „Complete Sets of Postulates for the Theory of Real Quantities“ from 1903. that „any two assemblages, M (<, +) and M ' (<, +), which satisfy the [Hölder's] postulates 1-10 are equivalent; that is, they can be brought into one-to-one correspondence in such a way that when a and b in M correspond to a' and b' in M', we shall have, a' < b' whenever a < b and a + b will correspond to a' + b' “.
So, Huntington proved the categoricity of (R, <, +) in Section 1 of his 1903 paper, as Hölder did in #15 of his 1901 paper. As Philip Ehrlich pointed out, Huntington also proved the categoricity of (R, <, +, x) in section 2 of his 1903 paper, which Hölder apparently did not. So it may seem that Huntington did something more after all. However, in #16 Hölder defined multiplication in (R,<,+) and proved in #17 that it is unique, so there was no need for special treatment of (R, <, +, x).
Hence, as far as categoricity is concerned, there is nothing in Huntington's work that we do not already find in Hölder's. Therefore, it is not surprising that Ebbinghaus et. al. in references to the chapter on real numbers and their axiomatization (in their book "Zahlen") do not mention Huntington, only Hilbert and Hölder.
How does uniqueness follow trivially from maximality, unless one undertakes the Huntington argument? And does Hilbert claim uniqueness up to isomorphism? Does Hölder? To my way of thinking, realizing that this is an important phenomenon is a significant part of the accomplishment. Huntington does this very clearly.
In particular, saying that the axioms were already known earlier is not the same as saying that the categoricity result was known and proved earlier. Was the categoricity theorem even stated clearly by anyone before Huntington?
Joel, I edited my answer to respond to your comments. I think that Hilbert was aware of the uniqueness of the Archimedean field that satisfies his axioms because he could certainly have considered the argument I offer as his “bekkanter Schlussmethoden” accessible to everyone.
I don't think that Hilbert had a general concept of categoricity (he used the rather vague semantic/syntactic term complete) and it could be that Huntington's concept of sufficiency was a first step towards that. But the idea was well known in special cases (e.g. Dedekind's theorem that every two simply infinite systems are similar), so I think that in the special case of Archimedean fields the priority is Hilbert's.
Thank for the edit. The argument you offer is the same as Huntington's, the same as what I describe in the OP. From a contemporary perspective, to be sure, this is an elementary argument. But I worry that we make a mistake by viewing things this way from a contemporary outlook. The question is who was first to express and prove the categoricity result. You argue in effect that Hilbert could have done so. Probably that is right, since he was a very talented expert individual. But did he actually formulate the result and give a proof? If not, it seems that Huntington is still the right answer.
I've edited it once more to make myself clearer and I would like to see Ehrlich's comment.
Joel, I just read your proof for the first time (before I just skimmed over it, thinking ok I know that) and saw that you can delete the first paragraph (the Archimedean derivation) and the text "by the Archimedean property" in the third paragraph.
I don't think so, because the axioms are merely that it is a complete ordered field, so you need to know that it is Archimedean in order to know that every number is determined by the cut it makes in the copy of $\mathbb{Q}$. There are nonarchimedean ordered fields with elements above every rational, and it seems one must argue that we're not in that case, which is what the first paragraph argument does.
For example, in order to know that two different points can't have the same cut in $\mathbb{Q}$, one should consider their difference, which would be smaller than any positive rational number, and so by the Archimedean property they must be equal.
By the Archimedean property, every number x∈R0 determines a cut in Q0, [don't need Archimedean property for that], and since R1 is complete, there is a counterpart x¯∈R1 filling the corresponding cut in Q1 [completenes of R1 guaranties that a cut in R1 define x¯, not a cut in Q1, you have to prove that]. Thus, we have defined a map π:x↦x¯ from R0 to R1. This map is surjective, since every y∈R1determines a cut in Q1, and by the completeness of R0, there is an x∈R0 filling the corresponding cut [the same problem].
It is easier to use supremums: Let X and Y be complete ordered fields. They both contain Q (because they are ordered fields). For every x∈X define Qx = {q∈Q: q<x} and y=supQx in Y (y exists because Y is complete). Function y=y(x) is injective because x<x' implies Qx< Qx' which implies y<y'. It is surjective because for every y∈Y, y=y(x) for x=supQy in X (x exists because X is complete).
For injectivity, you have to know that distinct numbers in $\mathbb{R}_0$ don't determine the same cut in $\mathbb{Q}_0$, and this amounts to Archimedean. Once that is addressed, your argument is fundamentally the same as mine—you simply fold in the proof of Archimedeanness at each step by using completeness. Meanwhile, I find it insightful to point out that completeness implies Archimedean, since this unifies our understanding of the field, so it is pedagogically better to establish that it is Archimedean. Especially so since Hilbert's axioms include Archimedean and Huntington's do not.
In any case, since it is true that completeness implies Archimedean, there seems to be no mathematical reason to object to the observation. One wouldn't be weakening an assumption or somehow making a stronger result by avoiding this observation.
Yes, you are right, but it is easy to fold in the injectivity, because there is no q between x and y iff y - x < 1/n for every n iff {1/n} has no infimum. But ( I repeat) to define x¯ you should know that cuts in Q1 defines it. In the Archimedean field X, cuts in Q do the same as cuts in X, but it has to be proved. My point is that using completenes you can bypass this proof. Of course, I agree that completeness=Archimedean+Cauchy is "insightful and unifies" but you introduced Archimedean as a guarantee that "every number x∈R0 determines a cut in Q0" which is true in every ordered field.
Yes, you are right. I have edited my post to say "is determined by the cut it makes", which is what I had meant.
@Zvonimir Sikic. Your claim about Huntington's paper of 1903 is mistaken. In Section 2 he provides an axiomatization of the ordered field of real numbers (consisting of 14 axioms including the Dedekind Completeness axiom) and in Theorem II' on page 368 he proves that the axiomatization is categorical.
@Philip Ehrlich, Yes you are right, I ignored section 2. But there is a reason for that which I explain in my edited answer.
@Zvonimir Sikic. Your point about Hölder proving the uniqueness of multiplication (assuming the left and right distributive laws) is a good one. So, while, Hölder may not have proved categoricity, what he proved certainly naturally suggests it.
The key question here is that Dedekind complete implies Archimedean, as in the first paragraph of Joel's Proof.
This was shown by Otto Stolz, in Zur Geometrie der Alten, insbesondere über ein Axiom des Archimedes, Mathematische Annalen, 22(4):504–519, 1883.
Apparently Stolz was persuaded to retract this paper.
I confess that I can't read German with any fluency and only got through the first part of the paper, but it does seem to me that that contains the correct argument that Joel states. Therefore Stolz deserves the credit.
The later part of the paper looks rather strange, but that's in part because I didn't follow the text.
However, Mikhail knows more about the history and will no doubt explain.
Footnote: I conjecture that a version of Conway's "surreal" numbers using computably enumerable subsets instead of general ones might be able to avoid Stolz's argument and yield a Dedekind complete but non-Archimedean field.
Some years ago, I put this conjecture to (the late) John Conway, he scratched his head and said "maybe".
This is very interesting, if it is true that Stolz has a clear statement of the categoricity result, since it is much earlier than Huntington. Can we have any German readers translate the relevant parts? To my way of thinking, however, the categoricity result goes strictly beyond merely observing that completeness implies Archimedean. That is key, of course, but categoricity is an important further idea of its own, of central importance and naturally in contemporary thought. It is precisely the question—when did this change occur? Huntington has essentially the contemporary view.
Or are you claiming only that Stolz has the complete$\to$Archimedean result only? If so, then this seems insufficient for credit for the categoricity result.
@JoelDavidHamkins: As a categorist, I'm not keen on the (model-theoretic) notion of "categoricity". After the complete/Archimedean question, I don't see the difficulty. (There are constructivity issues, on which my view is here, but you asked the question classically.)
Stolz is one of my favorite historical figures and I have extensively read and written about his work (e.g. https://www.researchgate.net/publication/225675816_The_Rise_of_non-Archimedean_Mathematics_and_the_Roots_of_a_Misconception_I_The_Emergence_of_non-Archimedean_Systems_of_Magnitudes). The claim that he had a notion of categoricity strike me as far fetched. He also never wrote about ordered fields. He worked primarily on the positive cones of ordered abelian groups, and in 1891 succeeded in proving that the positive cone of a Dedekind complete ordered abelian group is Archimedean,...
(continued), his earlier attempts having beed criticized by others including Veronese. This is all discussed in the paper referred to above.
@PaulTaylor To my way of thinking, the categoricity results of Dedekind and Huntington and so forth are the origin of the philosophy of structuralism in mathematics, because they enable us to refer to our familiar mathematical structures by identifying the features that characterize those structures up to isomorphism. I would think this is as important in category-theoretic foundations as in any other, whether set theory or type theory or what have you. So I was surprised to hear that you disparage the categoricity results (which are not theorems of model theory but mathematics generally).
I am confused by the failure to recognize the set-theoretic issue. Without requiring Archimedeanity, Conway’s Surreals are ordered, and satisfy the field axioms, and have no gaps! The only reason they can’t be a complete ordered field, if you use a no-gap definition of “completeness” rather than one involving sequences or the Archimedean property or a least upper bound property, is that they are too big to be a set.
You need to give a simple answer to “why aren’t the surreal numbers a second complete ordered field?”
If we were to use a surreals-style "no-gap" account of completeness, then the real field $\mathbb{R}$ itself would fail to be complete, since it has an unfilled gap between $0$ and the positive elements. To be sure, the surreal field is based on the idea of iteratively filling all such gaps, but no set-sized structure can be complete in that no-gaps sense. This is why, when speaking of the complete ordered field, we use the LUB account of completeness, which the surreal numbers lack. For example, there is no LUB in the surreal numbers to the infinitesimal surreal numbers.
Joe, Conway’s No is non-Archimedean ordered field which has Dedekind's gaps. Perhaps, your point is that it satisfies Cantor's axiom? Because it does. If you worry that No is a real class, you may limit it to the ordered field of surreal numbers built up to the "moment" ε0 and it is still non-Archimedean ordered field which has Dedekind's gaps, but satisfies Cantor's axiom. Let's add that No is to ordered fields what R is to Archimedean ordered fields. Just as every Archimedean ordered field is contained in R, so every ordered field is contained in No.
Yes, there are "gaps" in Conway's number system. They are discussed on pages 37--38 of On Numbers and Games.
They’re not really “gaps” that are legal in ZFC! That’s my point. Any “gap” that can be defined by giving two sets of Numbers L and R such that every element of L is < every element of R is already filled by a surreal number. You can only make a true “gap” by using left and right “classes”, not sets!
What in the world does "legal in ZFC" mean? You make ZFC sound like a police state! The fact is that $\mathbb{R}$ has no gaps at all (neither "legal" nor "illegal")!
@FrançoisG.Dorais $\mathbb{R}$ does have unfilled gaps, everywhere, in the sense that Joe is using, since one can find two disjoint sets $L$, $R$ with every element of $L$ strictly below every element of $R$, but no element $x\in\mathbb{R}$ having $L<x<R$. Take $L=(-\infty,0]$ and $R=(0,\infty)$. The legality question is just whether $L$ and $R$ are sets or proper classes. In the surreals, all gaps defined by sets this way are inhabited by the surreal number ${L\mid R}$, which was created specifically to occupy this gap. But there are unfilled surreal gaps specified by proper classes.
If we consider dense linearly ordered sets, without requiring addition or multiplication properties, but still require the LUB property (in the form “if Y is a nonempty subset of X such that there exists x in X with y<=x for all y in Y, then there is a least such x”), what possibilities arise that are not simply intervals of reals? I can think of a few….
|
2025-03-21T14:48:31.244871
| 2020-06-13T20:03:02 |
362992
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex M.",
"AlexArvanitakis",
"InfiniteLooper",
"Konstantinos Kanakoglou",
"Mirco A. Mannucci",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/22757",
"https://mathoverflow.net/users/54780",
"https://mathoverflow.net/users/85967",
"https://mathoverflow.net/users/86526"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630116",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362992"
}
|
Stack Exchange
|
The inner product of a Clifford Algebra
Any Clifford algebra $\operatorname{Cl}(k, p)$ carries an induced inner product, which is the "trace" on its 0-blade: $\langle AB\rangle_0$ for given elements $A, B$ of the algebra.
This inner product, when restricted to the generating vector space $V$, gives back the inner product on $V$.
Now, my question (maybe entirely trivial, but I could not find it in the standard literature, for instance in Lounesto's great book Clifford algebras and spinors):
What is the interplay of the Clifford product and its induced inner product? Are there any formal laws?
PS. I have done some little googling, and I came up with some references on associative algebras with inner product, but unfortunately there seems to be a certain latitude as far as their definition.
PPS In light of the comments below, I realize that the phrasing of my post is not as clear as it should. Please refer to the comments below and my additional replies for the proper context.
Would have said the symmetrised "Clifford product" is this "induced inner product" up to a factor of 2 but perhaps I'm misunderstanding one or both of the definitions
@AlexArvanitakis they are not the same, though they are related: you refer to (ab+ ba) /2 which is the symmetric side of the clifford product. I was referring to _0, which is taking the o-dimensional component of the product. On vectors these two are the same, but not in general
I am afraid your question is not clear (at least the way it is phrased). Do you mean that you are looking for some relation between the $0$-component of the product $A\cdot B$ (for arbitrary elements $A$, $B$) and the bilinear form asssociated to the quadratic form involved in the definition of the Clifford algebra ?
@KonstantinosKanakoglou I agree: my question is not as crystal clear as it should, and I apologize. Let me see if I can make it a bit clearer: you start from a vector space with an inner product (V, Q), you build the clifford algebra. Now, the clifford agebra is an associative algebra PLUS the inner product given by the 0-blade (which of course restrict to the starting Q for V).
Now, forget completely about V, and just look at what you have: you have a unital associative algebra AND a inner product on it. Now, comes the question: forget entirely about V, Q and just look at what you have: you have an associative unital algebra, with an inner product. Now, can I describe how these two products are related in the algebra from an entirely algebraic standpoint?
Why I am asking: I want to clarify the structure of the clifford algebra per se, regardless of the underlying vector space. Does it make a bit more sense?
If this is the case (and if i understand correctly), isn't $xy+yx=B(x,y)1$ the relation you are looking for ?
@KonstantinosKanakoglou this is certainly true if x and y are vectors. But, as I have said, I wish to forget about V. Not sure it holds for arbitrary x and y in the algebra (see above what I said to Alex)
Recalling the construction of the Clifford algebra $C$, if we start from a vector space $V$ then the Clifford algebra is a suitable quotient of the tensor algebra $T(V)$. The canonical projection $\pi : T(V)\to C$, enables one to view $\pi (V)$ as a subspace of $C$, and this subspace generates $C$ (as an algebra). So, in this sense (if by vectors you mean the elements of $\pi(V)$), yes you are right. But their multiplication is in $C$.
So, you have a relation between the multiplication of generators of $C$ and the bilinear form associated to the quadratic form (via which $C$ is defined). What else do you need here?
I see that I am still unable to make myself clear (my fault). What you say is true, but play one second the following game: take your favorite Clifford (mine is Cl(3, 0) and FORGET completely HOW it has been generated. There is no grading, no V, nothing, only the algebra, with its structure as unital associative
algebra. Now, it so happens that the 0-blade trace of the product endows the algebra with a inner product (so the entire algebra is a vector space with an inner product). Now I ask: ok, what is the interplay of the two products? Are they completely separated (remember, no V)? I suspect not. I suspect there are some structural relations between them.....
PS Notice that I have not contradicted myself: the 0 blade exists independently, simply because it is a unital algebra (1 is there as a subspace of the algebra).
Mirco, maybe it is due to my insufficient knowledge, but i do not really understand the expression: "the 0-blade trace of the product endows the algebra with a inner product". What is the 0-blade on the first place ?
take two elements of the algebra, say V and W. Multiply them: V W. Now expand the result in any basis you like: VW = a * 1 + ....................... a is the 0-blade. Fact: a is a inner product (you can amuse yourself with some calculation, for instance try it on Cl(0, 1), aka the complex numbers seen as a clifford . The result will perhaps surprise you....)
Have you tried to figure out what is the signature of the bilinear from on $Cl(p,q)$ ?
oh yes. That is easy: write down the standard basis 1 e1...... e1e2......... etc. Each squared is either + 1 or -1 . That determines uniquely the signature of the form on Cl(p, q). For instance, going back to Cl(0, 1). in that case the signature is + -, which when restricted to the generating one-dim space is simply -. It is kind of funny: C from this angle is 2-dimensional space-time + -.......
if you do not like that basis approach there is some combinatorial formula which tells you the signature of Cl(p, q). But that is perhaps beside the point.
This trace on the 0-blade is presumably the honest matrix trace in the usual faithful matrix representation? In that case you can find very explicit formulae for all values of the trace, in the physics/quantum field theory literature.
When I used the word "trace" I put a quote around it. But now, on a second thought, you may have a point. 1 is certainly represented as the diagonal of ones in any matrix representation, so it should be easy to verify what happens when two elements A and B get multiplied: what is the coefficient of 1? Indeed their inner product ...
@MircoA.Mannucci: Right at the bottom of the Wikipedia entry on the Clifford scalar product: is this what you wanted? The fact that Clifford transposition is the adjoint under the scalar product?
$
\newcommand\lcontr{\,\lrcorner\,}
\newcommand\rcontr{\,\llcorner\,}
\newcommand\lcontrr{{\rfloor}}
\newcommand\rcontrr{{\lfloor}}
\newcommand\form[1]{\langle#1\rangle}
\newcommand\Ext\bigwedge
\newcommand\rev\widetilde
$
I hope that this is in the vein of what you were looking for.
In a review (Marcel Riesz’s Work on Clifford Algebras, 1993)
of Marcel Riesz's lecture notes (Clifford Numbers and Spinors, 1958)
on Clifford algebras, Pertti Lounesto demonstrates a natural linear isomorphism
between $\Ext(V)$ and $Cl(V)$.
Given a symmetric bilinear form $\form{\cdot,\cdot}$ on a vector space $V$
over a field $K$ with characteristic $\not=2$,
this extends to a bilinear form on $\Ext V$.
On $k$-blades (i.e. simple $k$-vectors), we define
$$
\form{x_1\wedge\cdots\wedge x_k, y_1\wedge\cdots\wedge y_k}
= \det\form{x_i, y_j},
$$
which is to say that we take the determinant over the matrix with entries
$a_{ij} = \form{x_i, y_j}$.
For blades $A, B \in \Ext V$ with different grades we define $\form{A, B} = 0$,
and then extend by linearity to the entirety of $\Ext V$.
Though maybe not obvious when described like this,
this extension of the form on $V$ is natural,
see here and here.
With this form on $\Ext V$,
the adjoints of the exterior product
$$
\form{X \wedge Y, Z} = \form{Y, X \lcontr Z},\quad
\form{X \wedge Y, Z} = \form{X, Z \rcontr Y}
$$
(where $X, Y, Z \in \Ext V$ are arbitrary multivectors)
are found to be the left ($\lcontr$) and right ($\rcontr$) contractions
of the Clifford algebra defined by $\form{\cdot,\cdot}$ on $V$.
They are indispensable when studying Clifford algebras for applications.
There are various different conventions that can be used when defining them;
see the appendix of this preprint
(Compendium on Multivector Contractions, 2022)
by André Mandolesi.
It's worth noting that the alternative contractions
$$
X\lcontrr Y = \rev X\lcontr Y,\quad X\rcontrr Y = X\rcontr\rev Y,
$$
where $\rev X$ is the reverse of $X$, are particularly popular;
see Lounesto or this article
(The Inner Products of Geometric Algebra, 2002)
by Leo Dorst for an exposition on their basic properties.
We may note that $a \lcontr b = a \rcontr b = \form{a, b}$ for $a, b \in V$.
Now define the products
$$
aX = a\lcontr X + a\wedge X,\quad
Xa = X\rcontr a + X\wedge a,
$$
which can be seen to be exactly the Clifford algebra products of $a$ and $X$.
By this definition, this product is self-adjoint:
$\form{aX, Y} = \form{X, aY}$ with an analogous formula for $a$ on the right.
But any blade $A$ can be written as $A = a_1a_2\cdots a_k$ for some $k$
and $a_1,\dotsc, a_k \in V$ using Clifford products (which are associative);
hence
$$
\form{AX, Y}
= \form{a_1a_2\cdots a_kX, Y}
= \form{a_2a_3\cdots a_kX, a_1Y}
= \form{a_3a_4\cdots a_kX, a_2a_1Y}
= \form{X, \rev AY}.
$$
By linearity, the same holds for when $A$ is an arbitrary multivector.
Thus,
$$
\form{X, Y} = \form{X\cdot 1, Y} = \form{1, \rev XY} = \form{\rev XY}_0.
$$
It's worth noting that the contractions
have a very direct connection with the Clifford product.
For any $X, Y \in \Ext V$,
$$
X\lcontr Y = \sum_{j=0}^n\sum_{k=0}^n\form{\form{\rev X}_j\form{Y}_k}_{k-j},\quad
X\rcontr Y = \sum_{j=0}^n\sum_{k=0}^n\form{\form{X}_j\form{\rev Y}_k}_{j-k},
$$
where $\form{\cdot}_l = 0$ if $l < 0$.
For $a \in V$ and $X \in \Ext V$ in particular,
$$
a\lcontr X = \frac12(aX - \hat Xa),\quad
X\rcontr a = \frac12(Xa - a\hat X),
$$
where $\hat X$ is the grade involution (i.e. main involution) applied to $X$.
__
|
2025-03-21T14:48:31.245622
| 2020-06-13T20:33:22 |
362996
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Stefan Dawydiak",
"https://mathoverflow.net/users/88921"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630117",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362996"
}
|
Stack Exchange
|
Kazhdan-Lusztig basis elements appearing in product with distinguished involution
My apologies if the below is too malformed to make sense.
Let $(W,S)$ be the affine Weyl group of a reductive group $G$, and let $\{C_w\}$ be the Kazhdan-Lusztig $C$-basis (an answer in terms of the $C'_w$-basis is of course equally useful, though). Let $H$ be the corresponding Hecke algebra over $\mathbb{Z}[q,q^{-1}]$, and write $h_{x,y,z}$ for the structure constants of $H$ in the $C_w$-basis.
It is know that $W$ is partitioned into finitely-many two-sided cells, and finitely-many left cells, each of which contains a distinguished involution $d$ (all these terms understood in the sense of Kazhdan-Lusztig and Lusztig). Let $w\in W$ and let $\Gamma$ be the left cell containing $w$. Let $d$ be the distinguished involution in $\Gamma$.
Consider the structure constants $h_{w,d,z}$ for $z\in\Gamma$. What is known about them? For example, is it known in general for which $z$ they are nonzero?
For example, in the case when $d=s$ is a simple reflection, we have
$$
C_wC_s=-(q^{1/2}+q^{-1/2})C_w,
$$
so that $h_{w,d,w}=-(q^{1/2}+q^{-1/2})$ and $h_{w,d,z}=0$ otherwise.
In the other extreme, let $W_0\subset W$ be the finite Weyl group, with longest element $w_0$. Then the lowest two-sided cell contains $W_0$-many distinguished involutions, all conjugate to $w_0$. In this case for $w$ in the same left cell as $w_0$, $C_wC_{w_0}$ is again proportional to $C_w$, and $h_{w,w_0,w}$ is itself proportional to the number of $\mathbb{F}_q$-points of the finite-dimensional flag variety of $G$.
It seems too much to hope for that $C_wC_d$ will always be proportional to $C_w$, but perhaps something like this is almost true?
Indeed even $C_dC_d$ is not always proporitional to $C_d$. In type $\tilde{B}3$, this happens for the distinguished involution $d=s_2s_0s_1s_3s_2$, one has $h{d,d,w}=-(v+v^{-1})$ for $w=s_2s_3s_1s_0s_2s_3s_1s_0s_2$. It can be checked that $w$ and $d$ are even in the same left cell.
|
2025-03-21T14:48:31.245819
| 2020-06-13T20:41:16 |
362997
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gjergji Zaimi",
"Gutiérrez",
"Joel Kamnitzer",
"Martin Rubey",
"Per Alexandersson",
"Richard Stanley",
"Sam Hopkins",
"https://mathoverflow.net/users/1056",
"https://mathoverflow.net/users/206706",
"https://mathoverflow.net/users/2384",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/2807",
"https://mathoverflow.net/users/3032",
"https://mathoverflow.net/users/438"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630118",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362997"
}
|
Stack Exchange
|
Bender-Knuth involutions for symplectic (King) tableaux
First let me recall the combinatorial theory of the characters of $\mathfrak{gl}_m$, a.k.a., Schur polynomials. For a partition $\lambda$, a semistandard Young tableaux of shape $\lambda$ is a filling of the boxes of (the Young diagram of) $\lambda$ with positive integers such that entries strictly increase down columns and weakly increase along rows. For such a tableau $T$ we define $\mathbf{x}^{T} := \prod_{i} x_i^{a_i(T)}$ where $a_i(T):=\#\textrm{$i$'s in $T$}$. For $\lambda$ a partition with at most $m$ parts, the generating function
$$ s_{\lambda}(x_1,\ldots,x_m) := \sum_{T} \mathbf{x}^{T},$$
where the sum is over all semistandard Young tableaux of shape $\lambda$ with entries in $\{1,\ldots,m\}$, is the character of the irreducible, finite-dimensional representation $V^{\lambda}$ of $\mathfrak{gl}_m$ with highest weight $\lambda$. This is all classical.
Now, the characters of $\mathfrak{gl}_m$ are invariant under the action of the Weyl group of $\mathfrak{gl}_m$, a.k.a. the symmetric group $\mathfrak{S}_m$. Bender and Knuth defined certain operators on the set of semistandard tableaux, now called Bender-Knuth involutions, which allow one to see this symmetry combinatorially (the involutions swap the quantities $a_i(T)$ and $a_{i+1}(T)$).
King (see paper cited below) defined tableaux for the symplectic Lie algebra. Namely, for a partition $\lambda$ with at most $n$ rows, a symplectic tableau of shape $\lambda$ is a filling of the boxes of $\lambda$ with the symbols $\overline{1}<1<\overline{2}<2<\cdots <\overline{n}<n$ (with the symbols totally ordered that way) such that:
the entries strictly increase down columns and weakly increase down rows (semistandard condition);
entries $i$ and $\overline{i}$ do not appear below row $i$ (symplectic condition).
For such a tableau $T$ we define $\mathbf{x}^{T} := \prod_{i} x_i^{a_i(T)}$ where $a_i(T):=\#\textrm{$i$'s in $T$} - \#\textrm{$\overline{i}$'s in $T$}$. Then King showed the generating function
$$ sp_{\lambda}(x_1,\ldots,x_m) := \sum_{T} \mathbf{x}^{T},$$
where the sum is over all symplectic tableaux of shape $\lambda$, is the character of the irreducible, finite-dimensional representation $V^{\lambda}$ of $\mathfrak{sp}_{2n}$ with highest weight $\lambda$.
Now, $sp_{\lambda}(x_1,\ldots,x_m)$ must be invariant under the action of the Weyl group of $\mathfrak{sp}_{2n}$, i.e., the hyperoctahedral group $\mathfrak{S}_2 \wr\mathfrak{S}_n$. In other words, $sp_{\lambda}(x_1,\ldots,x_m)$ is invariant under permuting and negating the exponents of the $x_i$.
Question: Are there Bender-Knuth-like involutions for symplectic tableaux that allow one to see this symmetry combinatorially?
I thought this should be well-known, but googling "symplectic Bender-Knuth" did not seem to turn up anything useful. Note that for negating $a_i(T)$, I believe the usual Bender-Knuth involution should work; but for swapping the values of $a_{i}(T)$ and $a_{i+1}(T)$, the symplectic condition causes problems if one tries to naively apply the usual Bender-Knuth involution.
King, R. C., Weight multiplicities for the classical groups, Group theor. Meth. Phys., 4th int. Colloq., Nijmegen 1975, Lect. Notes Phys. 50, 490-499 (1976). ZBL0369.22018.
EDIT:
In case it's helpful, let me mention another way to think about Bender-Knuth involutions, using Gelfand-Tsetlin patterns. Recall that a Gelfand-Tsetlin pattern of size $n$ is a triangular array
$$\begin{array}{c c c c c}
a_{1,1} & a_{1,2} & a_{1,3} & \cdots & a_{1,n}\\
& a_{2,2} & a_{2,3} & \cdots & a_{2,n} \\
& & \ddots & \cdots & \vdots \\
& & & a_{n-1,n} & a_{n,n} \\
& & & & a_{n,n}
\end{array}$$
of nonnegative integers that is weakly decreasing in rows and columns. There is a well known bijection between semistandard Young tableaux of shape $\lambda = (\lambda_1,\ldots,\lambda_n)$ with entries $\leq n$ and GT patterns with $0$th (i.e., main) diagonal $(a_{1,1},a_{2,2},\ldots,a_{n,n})=(\lambda_1,\ldots,\lambda_n)$. Moreover, as shown in Proposition 2.2 of the paper of Berenstein and Kirillov below, the $i$th Bender-Knuth involution for $i=1,\ldots,n-1$ acting on the set of these tableaux can be realized by toggling (in a piecewise-linear manner) along the $i$th diagonal of the corresponding GT pattern.
For symplectic tableaux, there is also a GT pattern-like model. Namely, the $n$-symplectic tableaux of shape $\lambda=(\lambda_1,\ldots,\lambda_n)$ are in bijection with ``trapezoidal'' arrays
$$\begin{array}{c c c c c c c c}
a_{1,1} & a_{1,2} & a_{1,3} & \cdots & \cdots & a_{1,2n-2} & a_{1,2n-1} & a_{1,2n} \\
& a_{2,2} & a_{2,3} & \cdots & \cdots & a_{2,2n-2} & a_{2,2n-1} \\
& & a_{3,3} & \cdots & \cdots & a_{3,2n-2} \\
& & & \vdots & \vdots \\
& & & a_{n,n} & a_{n,n+1}
\end{array}$$
of nonnegative integers that are weakly decreasing in rows and columns, and where again we have $(a_{1,1},a_{2,2},\ldots,a_{n,n})=(\lambda_1,\ldots,\lambda_n)$; see for instance Lemma 2 of the paper of Proctor cited below. It might be reasonable to try to realize the symplectic Bender-Knuth operations by toggling along diagonals of these trapezoidal arrays; but note that this trapezoid shape has $2n$ diagonals, which is a lot more than the $n$ involutions we expect to generate the relevant hyperoctahedral group.
Kirillov, A. N.; Berenstein, A. D., Groups generated by involutions, Gelfand-Tsetlin patterns, and combinatorics of Young tableaux, St. Petersbg. Math. J. 7, No. 1, 77-127 (1996); translation from Algebra Anal. 7, No. 1, 92-152 (1995). ZBL0848.20007.
Proctor, Robert A., Shifted plane partitions of trapezoidal shape, Proc. Am. Math. Soc. 89, 553-559 (1983). ZBL0525.05007.
A symplectic analogue of the Bender-Knuth involution appears as Theorem 6.12 in the Ph.D. thesis of Sheila Sundaram, available at http://library.mit.edu/F/K4V8G7A637R9YSTMJGEGVT74CNTEATF3887MYEAEXQXCRFJ2RE-00369?func=find-acc&acc_sequence=013157766.
@RichardStanley: I think the argument in Theorem 6.12 is incomplete. (Note also she uses the convention $1<\overline{1}<2<\overline{2}<\cdots<n<\overline{n}$, but this is ok since we can swap $i$ and $\overline{i}$ in the BK way as I mentioned.) Imagine a one-column tableau with first entry $\overline{1}$ and second entry $2$. Sundaram does not really specify what to do here to swap $\alpha_1$ and $\alpha_2$.
@SamHopkins Are you looking to prove CSP on King tableaux? Sounds like a fun project! We should collaborate on some paper regarding CSP - I have a few conjectures waiting to be solved :)
@PerAlexandersson: I did not ask this question with any particular application in mind. It's just that after SSYT for the general linear group, symplectic tableaux are the next 'nicest' tableaux modeling the characters of classical groups, and I thought this should be something known.
@SamHopkins: I don't understand the problem with Sheila Sundaram's proof. The hyperoctahedral group is generated by $(1,\bar 1)$ and the transpositions $(i,i+1)$. To swap $i$ and $i+1$, we can proceed as with $GL_n$ semistandard tableaux. To swap $1$ and $\bar 1$, the same trick works.
When the shape is $(1,1)$, there are five symplectic tableaux. To show that $sp_{1,1}$ is symmetric in $x_1$ and $\bar x_2$, we observe that it is symmetric in $x_1$ and $\bar x_1$ and also in $x_2$ and $\bar x_2$. In this case, the involution simply negates the unique entry $1$ (respectively $2$), except for the special tableau having weight $1$.
@MartinRubey: You cannot directly swap $i$ and $i+1$ like she is trying to do since $\overline{i}$ is in-between.
@MartinRubey: Moreover, if instead you try to swap say $\overline{i}$ and $i+1$, well you will again run into problems, because the character is not symmetric with respect to all $(2n)!$ permutations of ${1,\overline{1},\ldots,n,\overline{n}}$. The swapping of $\alpha_i(T)$ and $\alpha_{i+1}(T)$ has to involve changing entries for all four symbols $i,\overline{i},i+1,\overline{i+1}$.
Oh, I see, Hm, you got me interested in fixing this :-)
A first natural guess is to try and perform evacuation on the skew tableau of boxes labeled ${i,\overline{i}, i+1, \overline{i+1}}$, but this unfortunately breaks the symplectic condition sometimes. Is there a special notion of symplectic evacuation or symplectic jdt on King tableaux?
@GjergjiZaimi: it might be possible to extract such an involution from the crystal structure on King tableaux defined by Lee, mentioned below: https://arxiv.org/abs/1910.04459.
@SamHopkins Are there any updates? I am working on fixing Sundaram's proof but want to know if someone already knows a fix to it.
@Gutiérrez: no updates that I know (beyond what are recorded in comments on this page) and I’m still very interested- so looking forward to seeing what you’ve done!
Let the hyperoctahedral group $\mathbb{S}_n\wr\mathbb{S}_2$ act naturally on the set $\{1, 2, ..., n\} \cup \{1', 2', ..., n'\}$. Then, we can say $\mathbb{S}_n\wr\mathbb{S}_2$ is generated by the transposition $(1\quad 1')$ and the permutations $(i\quad i\!+\!1)(i'\quad i\!+\!1')$ for $i = 1, ..., n-1$. We want to show that the number of King's tableaux (or King's patterns) of a fixed shape $\lambda$ and a given weight $x^\alpha$ is equal to that of weight $s.x^\alpha = x^{s(\alpha)}$ for all generators $s$. In particular, it is enough to define an action of $\mathbb{S}_n\wr\mathbb{S}_2$ on the set of King's tableaux such that $s.(\text{weight}(T)) = \text{weight}(s.T)$ and such that $s^2.T = T$ for all tableaux $T$ and generators $s$.
For $s = (1\quad 1')$, one can use type A Bender--Knuth involutions. To do this, one relabels the tableau from the alphabet $1<1'<2<2'<\cdots<n<n'$ to the alphabet $1<2<\cdots<2n$, applies the 1st Bender--Knuth involution, and relabels back.
For $s = (i\quad i\!+\!1)(i'\quad i\!+\!1')$, we first decompose the generator as a product of simple transpositions, $s = (i'\quad i\!+\!1)(i\!+\!1\quad i\!+\!1')(i\quad i')(i'\quad i\!+\!1)$. One now performs the four associated type A Bender--Knuth involutions. The result might not be a King tableau. It is shown in my MSc thesis that postcomposition with a suitable "rectifying" map gives a King tableau, and that this action of $s$ is involutory.
The text is available in my webpage (Proposition 5.9 and Appendix A).
This is really nice! Thank you for taking the time to work through the details of this in your thesis. I guess the combinatorics are slightly more involved than what you might expect at first blush.
For anyone seeing this answer now, the relevant paper of Álvaro Gutiérrez has been published at https://doi.org/10.37236/12571
For any semisimple Lie algebra $ \mathfrak g $ and any crystal $ B $ of a $\mathfrak g$-representation, we have an action of the cactus group $ C_{\mathfrak g} $ on $ B $. We have a surjective group homomorphism onto the Weyl group $ C_{\mathfrak g} \rightarrow W_{\mathfrak g} $. The action of $ C_{\mathfrak g} $ on $ B$ makes manifest the Weyl group symmetry in the weight multiplicities of the representation attached to $ B $. This is explained in our paper https://arxiv.org/abs/1708.05105.
In Halacheva's paper, https://arxiv.org/abs/2001.02262, she proved that for $ \mathfrak g = \mathfrak{sl}_n $, the Bender-Knuth moves generate this action of the cactus group (see also Chmutov-Glick-Pylyavskyy https://arxiv.org/abs/1609.02046). However, the Bender-Knuth moves are not the actions of the usual generators of the cactus group.
So for $ \mathfrak{sp}_{2n} $, we have the cactus group action (which is generated by involutions) on tableaux which makes manifest the Weyl group symmetry. The only remaining question is whether there are certain elements of the cactus group whose action looks like "Bender-Knuth" on these symplectic tableaux.
Yes, it's the last step of what does the action look like on tableaux that I'm interested in.
My first reaction was that I'd expect a Bender-Knuth / Cactus group action on the De Concini-Procesi / Kashiwara-Nakashima symplectic tableaux, similar to https://arxiv.org/abs/1609.02046.
I don't know if it's useful, but a crystal structure on King tableaux has recently been constructed in: https://arxiv.org/abs/1910.04459
Does this paper answer your question: https://arxiv.org/abs/2104.11799
@JoelKamnitzer: I only just saw this comment now. Interesting paper, thanks for making me aware of it! However, I don't think it quite answers my question, because the model of symplectic tableaux is not the King tableaux I am interested in (look at Definition 2.1).
Another relevant reference I'll leave here for posterity: https://arxiv.org/abs/2207.08446v1 (but again it does not work with King tableaux).
|
2025-03-21T14:48:31.246623
| 2020-06-13T20:57:47 |
362998
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"GH from MO",
"Gerhard Paseman",
"Igor Belegradek",
"LSpice",
"Nik Weaver",
"Stanley Yao Xiao",
"YCor",
"erz",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/152049",
"https://mathoverflow.net/users/1573",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/53155",
"user676464327"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630119",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362998"
}
|
Stack Exchange
|
Math graduate school applications and personal hardships
This question is on mathematics, career, and personal life.
I.) Assumptions.
i.) Suppose that I am a Junior in college majoring in mathematics at a top research institution in the U.S. I have "mostly" good grades in mathematics (say, around 2/3 of my grades are A's or A-'s), but I have some B's and B+'s. I've taken a good number of graduate level courses with at least half of them A or A-'s.
ii.) Earlier in my college career, there was a noticeable dip in my grades (A's becoming B's and B+'s) due to personal issues.
iii.) My goals:
(Short term:) enroll in a top pure math graduate program in the U.S. (ideally a top 6 institution) where there is a strong research community in my field as well as other fields in mathematics which may be helpful to my own research or broadening my perspective.
(Long term:) become a professor in pure mathematics at a top 30 institution (or equivalent) studying geometric topology or some related field.
II.) My Questions:
a.) Is it appropriate to talk about personal matters in one's personal statements or to have such a discussion in letters of recommendation, if it has negatively impacted one's academic performance?
b.) What are graduate school admission committees' views in regards to personal hardships? I know the short answer to this is "it depends," so I would appreciate getting the various viewpoints available on the MathOverflow community. I am sure it also depends on the school; school-specific viewpoints would be much appreciated.
this is not an answer to your two questions, hence a comment: I have found it counterproductive to take "becoming a professor" as a goal in life, in particular at an early stage of one's career as a scientist; I would approach graduate school as an opportunity to learn skills that will serve you well wherever life takes you, and explore a great variety of career options.
I'm not sure whether this question is of interest for an international audience.
Investors (people who are willing to spend resources in some area) are looking for a good chance of return. Having turned your situation around is good, but investors are willing to look at someone who is resilient enough to try turning it around, who knows they are resilient, and who can communicate that. If you can portray a positive and realistic future and attitude, that gives the investor confidence. Gerhard "Explain Turnaround, But Don't Dwell" Paseman, 2020.06.13.
@YCor, I don't think it's for MO at all, but much more for AcademiaSE; but, setting that aside, why should it not be of interest to an international audience? (The only location-specific things that I can see are two mentions of the US, which could probably even be deleted without changing the substance of the question.)
@Carlo Beenakker Thank you for the advice. I guess what I should have said instead was that "I intend to become a mathematician working in academia." Following Thurson's advice (https://mathoverflow.net/posts/44213/revisions), my ultimate goal is to deepen mankind's (and hence my own) understanding of mathematics. (I just didn't think this was worth mentioning since it is )
@LSpice The reason why I wanted to hear from MO is because different fields in academia seems to have different cultures. (e.g. at least to my understanding, experimental physics seems to care less about an applicants grades and put more emphasis on his/her capacity to do research.) I wanted to know hear the math community's answer to my question.
@Gerhard Paseman Your comment seems very plausible. In practice, to what extent do admission committees value such resilience? Based on what I hear from people it's mostly about grades, GRE scores, and how much a well-known faculty member recommends you.
@MattF. Thanks. I just made the post shorter.
My only experience with admission committees was applying to Berkeley. It is possible I got admitted in to graduate school by good fortune, even though I left undergraduate study five years earlier. And stellar recommendation letters are good too. However, I wrote a personal essay where I mustered all my confidence and talked about using my industry experience to help me in my graduate studies. I think I would not have been accepted if I did not display that kind of attitude. Again, acknowledge, but don't dwell. Gerhard "Believe That You Are Worthy" Paseman, 2020.06.13.
I find the focus on "top N" a bit unhealthy
You set your goals too high, which is OK if you don't get unhappy by not achieving them. In order to get into a top graduate school you need to be in the top few percent (1%, 2%, 5% etc., depending on the school). Your grades will not justify this as the majority of applicants are straight A's. If you have interesting research papers (preferably with no co-author), or have showed extraordinary talent to your recommenders, that might counterbalance it. I suggest that you apply to at least 30 graduate schools. Some schools do ask recommenders to mention special circumstances they are aware of.
My two cents: one, I think a brief explanation of why your grades dropped would be helpful to the admissions committee, as long as it doesn't get too self-involved or whiny. Two, many stellar mathematicians didn't go to a top graduate school. Getting into a top school is good for your ego but it isn't vitally important.
@NikWeaver There are surely practical benefits to getting into a top school, right? You'll probably do better on the postdoc market, get more connections, and be exposed to different areas of mathematics than lesser known schools. (For instance, when famous foreign scholars come to give talks, I assume most of them are in the famous schools.) Here, by "famous," I mean people whose research is gaining a lot of attention at that point in time.
To have a successful research career in math there is no need to go to top grad school. The chance of becoming a top mathematician is very low - I would not obsess about it. But if you are, I suggest the following experiment. Write the list of invited ICM speakers for several recent ICMs. Go to math genealogy site, and see where they got the PhD from (compare with the total number of PhDs from that school). If you are focused on US market, run a similar test with a list of junior faculty at top US schools.
This question was asked a while ago, but never got an answer. The comments focused on how a goal of "getting into a top 6 PhD program" followed by a "top 30 professorship" was a potentially unhealthy goal. I agree with that, so this answer is aimed at anyone applying to grad school, even beyond the "top 6." Let's start with question (a), regarding personal statements.
Yes, you can use your personal statement to explain relevant personal hardships, how they affected your performance, what you learned from them, and how you've grown.
Here is some general advice about grad school applications, from Swarthmore. Their advice item (2) is to do as well as possible on outside measurements like the GRE, REUs, study abroad, and Putnam exam. Their (3) is to take as many challenging math courses as possible (it sounds like the OP did this). Challenging courses and good letters of recommendation matter more than perfect grades. Their (5) concerns applying for grants to fund your PhD, (6) is about letters of recommendation, (7) is about your personal statement, and (8) is about emailing professors at places you apply.
The advice from Swarthmore is backed up by web pages of various PhD programs, e.g., this one from UMD, previous MO questions, MSE questions, and advice on academia.SE. The idea that it's ok to have some Bs, especially in hard courses, appears in those links and is confirmed here and here.
Let's turn to question (b) regarding how graduate school admissions committees view personal hardships.
Yes, done right, your history of overcoming a personal hardship can become a strength in your application.
The last link above also includes advice to use your personal statement to present yourself as a "whole package." A candidate can use their personal statement to strengthen their application if they explain (briefly) the personal hardship and how they learned from it and bounced back from adversity. You can find more advice here regarding the personal statement and using it to tie together your record. There have also been previous advice threads about using the personal statement. The general theme is that, if written convincingly, the personal statement can definitely help you, especially if it's related to your motivation, determination, grit, resilience, and current ability to succeed in their program. The story in that thread (and responses to it) show that for sure the personal statement and overcoming hardships, can play in your favor as the applicant.
The advice at all these sources consistently says to keep the story short and to the point. Aim for around two pages for the personal statement. Since letters of recommendation are consistently put among the top criteria (whereas, personal statements are not), it might be wise to talk to at least one letter writer about your concerns that the grade dip will harm your chances of being accepted, and to explain to them what happened. You can't really tell them how to write their letter, but if they agree with you during that conversation, odds are that they will add a sentence or two backing up your story and expressing their confidence regarding you despite this previous hiccup.
Lastly, I want to re-emphasize the point made in some of the comments that setting goals like "being accepted to a top N university" is not wise for mental health, and is not necessary for career success. Certainly it's true that there's a prestige bias in mathematics (happy to share links to published research that proves this if you want). However, the advice on MO, MSE, and academia.SE is consistent on the point that it's better to choose a field of study that you love (instead of whichever you think might best help you get a job), choose an advisor you have a good relationship with (instead of the most famous / influential advisor), and choose a graduate program where you can be fulfilled, productive, and happy (e.g., taking into account the location, culture, other students, your romantic partner, etc.) instead of whichever is the most prestigious. In math, we're so used to optimization problems, but in life, it's important to optimize on a number of different fronts simultaneously (and, honestly, for happiness, it's better to take an attitude of "satisficing," e.g., "this job might not be the best I can get, but it's good enough for me and I'm happy with it"). Constantly climbing and defining your self worth in terms of something as fickle as external rankings of your university, is not a recipe for a happy life. Good luck with your applications, PhD experience, and life!
Just to comment to thank David for his service in writing down answers to these questions, which may help many others, especially young people, in the future.
|
2025-03-21T14:48:31.247761
| 2020-06-13T21:23:10 |
363000
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"KConrad",
"MCS",
"Wojowu",
"efs",
"https://mathoverflow.net/users/109085",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/120369",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/3272"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630120",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363000"
}
|
Stack Exchange
|
A.C.M. van Rooij's *Non-archimedean functional analysis* (1978) is very out-of-print! Anyone know of any good alternatives?
(This is a literature/reference question.)
So... long story short:
(1) In my present research, I needed a theory of continuous functions from the $p$-adic integers to the $q$-adic integers. Unable to find a quick answer online, I ended up writing a paper (un-published) in which I worked out the basic details of this theory, developed an integration theory, translation-invariant measures, a Fourier transform, and so on.
(2) Just recently, after traveling down the rabbit hole of the literature, I've found that much (if not all) of what I did in my paper—along with quite a bit of extra details—were covered in A.C.M Rooij's Non-Archimedean Functional Analysis (1978). Unfortunately, this book is now woefully out-of-print; cheapest available copy I could find is $250 (US).
(3) I was fortunate enough to be able to at least find a copy of the table of contents of Rooij's book; scroll down to the bottom of the linked page to see it. Chapters 5 through 9 are what I need.
My question: What are some non-out-of-print books that cover this material? (Preferably with as little abstract nonsense as possible)
I ask because the books I immediately pull up when searching for "non-archimedean functional analysis" end up being riddled with either rigid analytic spaces and/or modern abstract nonsense, as opposed to the clean and direct analysis that I'm looking & hoping for. Any help would be much appreciated.
The book can be found online, through certain repositories which collect copies of books online. I believe the site policy disallows me from sharing a link directly, but it shouldn't be hard to find if you search for how to access scientific literature for free.
since it's out of print, it's probably fair use to download it for personal use.
I've searched, but haven't been able to find more than the front matter, and Chapters 1 & 3.
https://libgen.is/
This books exists in university libraries (I first saw it in such places when I was a student). If you have an academic affiliation, you could request the book by interlibrary loan. Perhaps current events may slow down the process of getting it that way, but being out of print does not mean the book is genuinely inaccessible if you have borrowing privileges at a university library.
@EFinat-S You're a life-saver! Thanks a million!
|
2025-03-21T14:48:31.247973
| 2020-06-13T22:04:06 |
363002
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Geva Yashfe",
"Joseph O'Rourke",
"Pietro Majer",
"Ville Salo",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/6094",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/75344"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630121",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363002"
}
|
Stack Exchange
|
Monotile that tiles when you apply a rubber band
My (non-mathematician) friend asked me a physics/tilings question that maybe someone here is interested in dissecting, or can point to the literature if this problem has been studied.
Does there exist a tile such that when you put a bunch of copies of it on a table and push from all sides, they always form a tiling?
My friend illustrated with physical (uniform density) lozenge tiles that they do not have this property, by throwing some on the table, and pushing them together. More specifically this suggests the stronger property that a typical initial configuration will get stuck. The tiles this was demonstrated on had positive friction.
The informal question as stated is a bit ambiguous. I am not going to try to formalize the physics of the problem, but I'll at least try to specify how the force is applied in a hopefully unambiguous (but somewhat arbitrary) way. You can suggest a better variant in the same spirit if e.g. it's easier to solve or mine misses the point for a "stupid" reason. (edit: I have added a "physics-free" formalization below.)
Let's say a tile is a nice enough subset $P \subset \mathbb{R}^2$, you can pick what that means. E.g. if going for a negative answer, you can choose something like "simply connected convex polygon". If going for a positive answer, I could imagine something like piecewise smooth being helpful. (For physics considerations it's a zero friction rigid body, and let's say of uniform density.)
Let $G = \mathbb{R}^2 \rtimes S^1$ be the rototranslation group (so no flips), which acts on $\mathbb{R}^2$ from the left. A partial tiling is a subset of $T \subset G$ such that the interiors of $t \cdot P$ for distinct $t \in T$ are disjoint. We say a partial tiling $T$ fills $C \subset \mathbb{R}^2$ if $T \cdot P \supset C$.
A physical jam is a finite partial tiling $T \subset G$ such that, assuming the tiles have zero friction and behave according to physics, if you stretch a rubber band around the convex hull of $T \cdot P = \bigcup_{t \in T} \{t \cdot P\}$, the tiles will not budge. Intuitively, jams always exist aplenty, just put some tiles on the table, stretch the band around them and let go (if there's a third dimension available there's a problem with that strategy, but you see what I mean).
Definition. A tile $P$ is a physical rubber band monotile if all $r > 0$, there exists $R > 0$ such that every jam whose convex hull contains the ball of radius $R$ fills the ball of radius $r$.
In terms of this, the question is:
Is there a physical rubber band monotile?
Observe that any physical rubber band monotile admits a partial tiling that fills the entire plane. In the usual terminology, $P$ tiles the plane under rototranslations, and such $P$ is sometimes called a monotile.
In case this question is non-trivial, here's some starters:
Is the equilateral (or any) triangle a physical rubber band monotile? Is the square (or any other rhombus, e.g. the lozenge)? Is the hexagon? Any of the pentagon monotiles?
I'm also interested in higher dimensions of course (my friend may or may not be). In one dimension I was able to solve the problem myself.
Physics-free formulation
Pick a (compatible) metric for $G$ and topologize the set of closed sets of $G$ with the Hausdorff metric, and the set $\mathcal{T}$ of all finite partial tilings with the induced metric. Let $c : \mathcal{T} \to \mathbb{R}_+$ be the (continuous) map that takes a partial tiling $T$ to the length of the boundary curve of the convex hull of $T \cdot P$. Paths in $\mathcal{T}$ starting from a finite partial tiling amount to moving the tiles in a continuous way (adding or removing a tile would necessarily be a jump because interiors must stay disjoint).
A weak jam is a finite tiling $T \in \mathcal{T}$ such that there does not exist a path $p : [0,1] \to \mathcal{T}$ with $p(0) = T$ and $x \mapsto c(p(x))$ strictly decreasing. A strong jam is a finite tiling $T \in \mathcal{T}$ such that there does not exist a path $p : [0,1] \to \mathcal{T}$ with $p(0) = T$, and $x \mapsto c(p(x))$ nonincreasing and $c(p(1)) < c(p(T))$. Every strong jam is a weak jam, obviously. The difference is whether we allow moving tiles so that the rubber band length stays constant.
Definition. A tile $P$ is a strong (resp. weak) rubber band monotile if all $r > 0$, there exists $R > 0$ such that every weak (resp. strong) jam whose convex hull contains the ball of radius $R$ fills the ball of radius $r$.
Every strong rubber band monotile is a weak rubber band monotile, obviously. In terms of these, the question is:
Do strong/weak rubber band monotiles exist?
It is easy to prove that no rectangle is a strong rubber band monotile, by arranging the rectangles into a bigger rectangle and removing all but the boundary tiles. I'd say that's definitely also a physical jam. Perhaps @GerhardPaseman's answer shows that the square is not even a weak rubber band monotile.
It seems that the definition of "rubber band monotile" merged with the subsequent question...
@PietroMajer: thanks!
I changed an "a" to a "the" in the definition of rubber band monotile, which may drastically change the problem, but which seems more natural after seeing @GerhardPaseman's suggestion.
The existence of "maximally random jammed (MRJ) configurations" seems relevant. For example:Atkinson, Steven, Frank H. Stillinger, and Salvatore Torquato. "Existence of isostatic, maximally random jammed monodisperse hard-disk packings." Proceedings of the National Academy of Sciences 111, no. 52 (2014): 18436-18441. PNAS link.
One more small detail: $c(T)$ is the length of the convex hull of $T\cdot P$, isn't it?
@PietroMajer yes that is what I meant
I suspect not, at least for regular polygons. (For nonregular polygons, there are orientation issues which I believe won't be solved by rubber bands.) Let me illustrate with squares.
Consider 8 squares in a three by three arrangement with a central hole. This configuration by itself does not tile when you apply a rubber band, but if you have a larger configuration with more tiles, some of these eight tiles might get pushed in.
Now push four edge squares out, and place a diamond tilted square in the middle, then push the squares back in. Now you have a nontiling arrangement. I suspect you can expand this to get larger nontiling arrangements which resist rubber bands.
Gerhard "Hope This Doesn't Snap Back" Paseman, 2020.06.13.
This seems to work, and give arbitrarily large configurations with minor modification. However, it's not clear to me that it continues working if force is applied unevenly to the different sides (it looks like it might snap into a 3x3 square of squares if the symmetry
of the configuration is broken?)
Maybe the question should be extended to "typical" configurations in some sense, or to configurations in which some "external shell" is randomly perturbed.
My guess is also that it's impossible, and your suggestion indeed should work for the square with the uniform rubber band. The comment of @GevaYashfe is also relevant for figuring out whether this also proves the square is not a weak rubber band monotile, i.e. whether this jam is strong.
In the 9-square configuration, let's rotate the central diamond, and let the N,S,W,E squares move respectively S,N E,W, always touching the corresponding vertices of the diamond, and let the other square follow staying in touch. Then the length of the boundary of the convex hull (the rubber) decreases, because it is made by 4 unit segments plus 4 hypotenuses of 4 right-angled triangles with a unit cathetus, and a cathetus whose length decreases.
A small check also show that one can do the whole 90 degrees rotation of the diamond, till the tiling reaches the 3x3 minimal configuration (that is, in the process the 4 corner squares never overlap the central square)
So if I understand @PietroMajer correctly, this is not a strong jam. I give a weak jam in the question, and this is presumably a physical jam, so so far I suppose we believe the square is neither a physical nor a weak rubber band monotile, but do not yet know whether it is a weak rubber band monotile, i.e. it might not have a strong jam. To me strong jams seem like the most interesting of these, so would be nice to figure out whether they exist. (Sorry about the choice of terminology, it's maybe not the most convenient.)
I think it is not even a weak jam, for the length of the octagon is strictly decreasing in the process
No, my question (after edit) contains a weak jam.
What I'm saying is just that for the 9 squares tiling T with the diamond in the middle, there exists a path of tilings p(x) with p(0)=T and c(p(x)) strictly decreasing.
And I agree with you.
|
2025-03-21T14:48:31.248541
| 2020-06-14T02:42:44 |
363006
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Math1016",
"Robert Bryant",
"https://mathoverflow.net/users/13972",
"https://mathoverflow.net/users/41200"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630122",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363006"
}
|
Stack Exchange
|
Moment map of $\mathrm{O}(n)$-action on $\mathbb{C}^n$
Let $(\mathbb{C}^n, \omega_0)$ be the complex Euclidean space of dimension $n$ with the standard Kähler structure $\omega_0$. I am looking for a Hamiltonian $\mathrm{O}(n)$-action on $(\mathbb{C}^n, \omega_0)$ such that the preimage $\mu^{-1}(0)$ of $0 \in \mathrm{o}(n)^{*}$ under its moment map $\mu: \mathbb{C}^{n} \rightarrow \mathrm{o}(n)^{*}$ consists of the unique point $0 \in \mathbb{C}^{n}$.
I would appreciate any explanation or references where such an example is explained. Thank you.
I notice that you wrote "Hamiltonian" action and "its moment map" rather than "Poisson" and "the equivariant moment map". (A Hamiltonian action has a moment map (which may not be unique), but it may not be $\mathrm{Ad}^$-equivariant.) Is there any significance to this? Did you mean for us to assume that $\mu$ is $\mathrm{Ad}^$-equivariant? (In your particular situation, there always will exist such an equivariant moment map, and it will be unique.)
Thanks for the comment, Robert. Yes, I meant that $\mu$ is $\mathrm{Ad}^{*}$-equivariant. I can accept uniqueness of such a moment map. Then, what is a Hamiltonian $\mathrm{O}(n)$-action on $(\mathbb{C}^{n}, \omega_{0})$ except for the canonical restriction of the linear $\mathrm{O}(n; \mathbb{C})$-action?
In that case, because you are assuming that $0 = \mu^{-1}(0)$, the action must fix $0$ (since $\mu^{-1}(0)$ must be a union of orbits). Moreover, it must be an isolated fixed point. The induced action on $T_0\mathbb{C}^n\simeq\mathbb{R}^{2n}$ must then be a representation of $\mathrm{O}(n)$ that fixes the symplectic structure on $T_0\mathbb{C}^n$ and that has no trivial summands. By dimension count, it must be two copies of the standard represention of $\mathrm{O}(n)$ on $\mathbb{R}^n$, i.e., the standard rep'n on $\mathbb{C}^n$. But this rep'n doesn't work, so there is not one that does.
Thank you for verifying non-existence of the desired moment map. It is interesting to consider moment maps combined with representations of $\mathrm{O}(n)$. Do you know what book explains their relation?
I don't know a book that specifically addresses this sort of use of representation theory in symplectic geometry, but, since representation theory is a basic tool in geometry when symmetries are present, it's used often in symplectic problems. Maybe a book that discusses the geometry of the moment map?
Thank you so much, Robert!
|
2025-03-21T14:48:31.248729
| 2020-06-14T04:03:21 |
363008
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/122662",
"Đào Thanh Oai"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630123",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363008"
}
|
Stack Exchange
|
A new theorem (discovered in 2013) equivalent to Brianchon theorem (the old theorem) discovered in XIX century?
In 2013, I found a new problem as follows: Let six points $A_1$, $A_2$, ...$A_6$ lie on a circle $(O_1)$, and the six points $B_1$, $B_2$,...,$B_6$ lie on another circle $(O_2)$. If the quadruples $A_i$, $A_{i+1}$, $B_{i+1}$, $B_i$ lie on a circles with centers $C_i$ for $i=1,2,...,5$ then $A_6$, $A_1$, $B_1$, $B_6$ lie on a circle (namely center of the new circle is $C_6$) and $C_1C_4$, $C_2C_5$, $C_3C_6$ are concurrent.
The proof [3][4] show that the hexagon $C_1C_2C_3C_4C_5C_6$ circumscribed around a conic section with two Focus are $O_1$, $O_2$
The Theorem was re-discovered by Szilasis in 2017 (see [5])
Converse of theorem: Let a hexagon $C_1C_2C_3C_4C_5C_6$ which $C_1C_4, C_2C_5, C_3C_6$ are concurrent then exist many configuration of eight circles as figuration above.
If the converse theorem is true, so the theorem is equivalent with an old theorem due to Brianchon
Special case: Let a hexagon $C_1C_2C_3C_4C_5C_6$ circumscribed a conic with two focus $O_1$, $O_2$. Draw six circles $(C_1)$, $(C_2)$,....,$(C_6)$ with centers $C_1$, $C_2$,...$C_6$ and radii $C_1O_1$, $C_2O_1$, ...., $C_6O_1$ respectively. Let circles $(C_i)$ meets circle $(C_{i+1})$ again at $A_i$ then six points $A_1, A_2,..., A_6$ lie on a circle, namely circle $(A)$. Similarly: Draw six circles $(C_1)$,$(C_2)$,....,$(C_6)$ with centers $C_1$, $C_2$,...$C_6$ and radii $C_1O_2$, $C_2O_2$, ...., $C_6O_2$ respectively. Let circles $(C_i)$ meets circle $(C_{i+1})$ again at $B_i$ then six points $B_1, B_2,..., B_6$ lie on a circle, namely the circle $(B)$.
Question 1: Is the converse of theorem true?
Question 2: How can one prove the radius of circle $(A)$ is equal to the radius of circle $(B)$?
The dual of theorem on the ball creat a beautifull basket (see theorem dual in [8])
Some relation problem
Dao's theorem on six circumcenters
Some Problems On Apollonian Gasket
Can generalization of a generalization Pascal theorem, Pappus theorem to Higher Dimensions?
A chain of six circles associated with six points on a circle (in Mobius plane)
$N$-$th$ closed chain of six circles
Reference:
1 - Dao, O.T.: Problem 3845, Crux Mathematicorum, 39, Issue May 2013
2-https://www.geogebra.org/material/show/id/Zk3F5y5X
3 - J. Chris Fisher, Problem 3945, Crux Mathematicorum, Volume 40, Issue May, 2014
[4]-Michel Bataille, Solution to Problem 3945, Crux Mathematicorum, Volume 41, Issue May, 2015
[5]-Gábor Gévay, A remarkable theorem on eight circles, Forum Geométrico rum, Volume 18 (2018), 401--408
[6]-Ákos G.Horváth, A note on the centers of a closed chain of circles
[7]-Dao Thanh Oai, The Nine Circles Problem and the Sixteen Points Circle, International Journal of Computer Discovered Mathematics ISSN 2367-7775, June 2016, Volume 1, No.2, pp. 21-24.
[8]-Dao Thanh Oai, Cherng-Tiao Perng, On The Eight Circles Theorem and Its Dual, International Journal of Geometry, Vol. 8 (2019), no. 2, page 49-53
Thanks Dr @David Roberts
I checked the converse theorem by computer, the converse is true. But I am looking for a proof.
Three points collinear
|
2025-03-21T14:48:31.248949
| 2020-06-14T05:30:22 |
363011
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/142929",
"user142929"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630124",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363011"
}
|
Stack Exchange
|
Irrationality or transcendence of $i^{i\Omega}$ and $2^\Omega$, with $\Omega=W(1)$ and $W(x)$ being the main branch of Lambert $W$ function
In this post we denote the main (or principal) branch of the Lambert $W$ function as $W(x)$, I add as reference that Wikipedia has the article Lambert $W$ function. The particular value $W(1)=\Omega$ is known as the omega constant, see the Wikipedia Omega constant. We denote the imaginary unit as $i=\sqrt{-1}$.
Question. I would like to know if it is possible to deduce what about the irrationality or transcendence of each one of the following real constants $i^{i\Omega}$ and $2^{\Omega}$. I emphasize that I'm asking what work can be done about these, before I'm accepting an answer. Many thanks.
I'm asking what about the state of art, or what work can be done for my question. I don't know if my question is in the literature, if this is the case please refer the literature answering as a reference request and I try to search and read the statements from the literature. For the first question I tried a variant of Gelfond's constant, and for the second question I wondered about it, if I'm remember well and I'm right, in the context of an application of the six exponentials theorem.
I was inspired in my question posted on Mathematics Stack Exchange MSE 3579844 (asked Mar 13) that previously I've also asked on MathOverflow, and in the linked section of the Wikipedia's article with title List of unsolved problems in mathematics. As reference I know also the statement of Gelfond–Schneider theorem
For the persons interested in this kind of questions, there is an excellent preprint on arXiv by Wadim Zudilin as arXiv:2004.11029 and title Diophantine problems related to the Omega constant
Many thanks for the edit (I had not noticed the typo) @J.W.Tanner
|
2025-03-21T14:48:31.249224
| 2020-06-14T06:12:31 |
363012
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerry Myerson",
"Will Sawin",
"https://mathoverflow.net/users/142929",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/18060",
"user142929"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630125",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363012"
}
|
Stack Exchange
|
On a conjecture about the arithmetic function that counts the number of twin primes
This is cross-posted from the question that I've asked with same title on Mathematics Stack Exchange two months ago, which has remained unanswered.
Given a positive real number $x$ we will write positive integer or real number $x$) as $$\pi_2(x)=\#\{\text{ primes }p\leq x\,:\,p+2\text{ is also a prime}\}$$
the arithmetic function that counts the number of primes $p$ less than a given positive real $x$ satisfying that $p+2$ is also a prime. (Such primes are known as twin primes. It is an open problem if there are infinitely many twin primes. Inspired by the Second Hardy–Littlewood conjecture, a few experiments using Pari/GP scripts to state the following two conjectures.
Conjecture 1. One has $$\pi_2(x+y)\leq \pi_2(x)+\pi_2(y)+1$$
for all integer $x\geq 2$ and all integer $y\geq 2$.
Conjecture 2. There exists an integer constant $K\geq 2$ such that $$\pi_2(x+y)\leq \pi_2(x)+\pi_2(y)+1$$
holds, for all integer $x\geq K$ and all integer $y\geq K$.
Question. Are these known, or is it possible to prove or refute any of previous conjectures? Can you find counterexamples for the first conjecture or add heuristics to know what about the veracity of this kind of conjectures?
I've tested the first conjecture for the segments of integers $2\leq x,y\leq 500$. I've added the second conjecture since I don't know if it is possible to find easily a counterexample for Conjecture 1.
Edit: Additionally/alternatively you can investigate what work can be done for/what about the veracity of the conjectures on assumption that the twin prime conjecture holds, or under assumption of other well-known conjectures. If this question is hard, I should to accept an answer that provide the more helpful discussion about what work can be done for our questions currently. See the contributions in comments of the users of the linked MSE post.
References:
[1] G. H. Hardy and J. E. Littlewood, Some Problems of 'Partitio Numerorum.' III. On the Expression of a Number as a Sum of Primes, Acta Math. (44), J. E. (1923) pp. 1–70.
In the unlikely event that the Twin Prime Conjecture is false, then of course both conjectures here are true for sufficiently large $x,y$ since we'll have $\pi_2(x+y)=\pi_2(x)=\pi_2(y)$.
Many thanks for your clarification and help, then I'm going to delete the question in next minures @GerryMyerson
I've undelete the post @GerryMyerson in case that there is more feedback (notice that I know about your excellent knowledges but I believe that I deleted my post abruptly , just as my decission). Thus I'm sorry and thanks for your patience, I hope that there is more feedback in next few days, many thanks again for your attention.
I'm glad you undeleted. My comment did not at all completely resolve your question. Please don't consider my comment as grounds for deleting your question.
The most interesting thing to do is to study this conjecture under the condition of the first Hardy-Littlewood conjecture. One could try to show that conjecture 1 is false under that assumption by the same strategy as the second Hardy-Littlewood conjecture - i.e. by finding an admissible tuple which contains too many twin pairs. This should be more fruitful to perform a computer search on than looking at twin primes directly.
Many thanks @WillSawin for your attention and help, also thanks again for previous professor in comments. I'm not a professional mathematician but I understand many of the details of the answers that provide the professors here for my questions, thus feel free to add your thoughts as an answer since these are according the Question and my previous Edit (that one can to invoke conjectures, heuristics or other reasonings to provide an answer).
Many thanks for the edit professor @JoshuaZ
Many thanks for your edit @kodlu
|
2025-03-21T14:48:31.249508
| 2020-06-14T06:18:07 |
363013
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aleksei Kulikov",
"Arthur B",
"Carlo Beenakker",
"https://mathoverflow.net/users/104330",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/51546",
"https://mathoverflow.net/users/8737",
"student"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630126",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363013"
}
|
Stack Exchange
|
Analyze a function defined in terms of an integral
Here is a question that really has puzzled me for quite a while. I happened to see this function defined in terms of an integral
$$f(x):=\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2}dy.$$I want to analyze the behavior of the function when $x \rightarrow \infty$.
The strange thing is that when I used Mathematica to plot the function, the graph indicates that $\lim_{x\rightarrow \infty} f(x)=0$. However, it is easy to see that $\liminf_{x\rightarrow \infty}f(x) \ge \frac{\pi}{4}$, since
$$\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2} \, dy \ge \int_0^{\pi/2}\frac{2e^{x+e^x\cos y}\sin y}{1+\left(e^{e^x\cos y}\right)^2}\, dy\\
=-\Big(\tan^{-1}\left(e^{e^x \cos y}\right)\Big)\Big|_{0}^{\pi/2}\\=\tan^{-1}\left(e^{e^x}\right)-\pi/4$$
Now I have two questions:
First, why the result from Mathematica is different from what I obtained?
Second, does $\lim_{x\rightarrow \infty} f(x)$ exist?
Maybe this question is not so suitable for mathoverflow, since it is just a calculus problem. However, I just feel so confused about the contradiction of numerical result and math. I want to understand the reason behind this situation. Any comments are really appreciated. Thank you very much.
Below is the code and picure I got from Mathematica....
you probably made a coding error in Mathematica, when I plot it the limit is about 1.57
You missed $2$ in antiderivative, so $\liminf$ is at least $\frac{\pi}{2}$ (which coincides with the number Carlo wrote). I bet that it won't be hard to prove that the the loss in multiplying by $\sin (y)$ is negligible as $x\to \infty$ and so the limit is actually $\frac{\pi}{2}$.
@CarloBeenakker, thank you very much for the comment... As you can see from my updates, my code should be fine, but when I tried to plot the the graph for $x \in [0,20]$, something went wrong...
@AlekseiKulikov, exactly! Thank you very much!
Mathematica seems to be plotting the function just fine...
If we look a bit at the integrand, it's clear that most of the mass is around $y = \pi/2$ as $x$ increases which should let us introduce a $\sin y$ term and use the antiderative you've already found.
We can try to cut the integral at $\pi/2 - 1/x$.
Let $$I = \int_0^{\pi/2 - 1/x} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy + \int_{\pi/2 - 1/x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy = I_0 + I_1$$
Notice that $f(u) = e^u / (1 + e^{2u})$ is a decreasing function of $u$ and thus that
$$I_0 < (\pi/2 - 1/x) 2 e^x \frac{e^{e^x \cos (\pi/2-1/x)}}{1 + e^{2 e^x \cos (\pi/2-1/x)}} $$
When $x \rightarrow \infty$ the $\cos (\pi/2 - 1/x)$ behaves as $1/x$ and the logistic function $1-\sigma(u)$ behaves as $e^{-u}$, so the right term behaves as $\pi e^{x - e^x /x}$ which converges to $0$.
For $I_1$, we note that if $y \in [\pi/2-1/x,\pi/2]$, $1 - \frac{1}{2x^2} < \sin y \leq 1$
$$I_1 \left(1-\frac{1}{2x^2}\right)< \int_{\pi/2-x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}\sin y}{1 + e^{2 e^x \cos y}} \leq I_1 $$
The middle term converges to $\pi /2$ and thus so does $I_1$ and so does $I$.
Thank you very much! By the way, I got the same picture for $x$ up to 10, but when I plotted the graph for $x \in [0,20]$, it gives me the limit going to $0$, as you can see from my updates....
The support of the mass becomes too small and the numerical integration misses it. If you do a change of variable which blow up the region around pi/2 it'll be more stable.
|
2025-03-21T14:48:31.249764
| 2020-06-14T06:40:17 |
363014
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Sergei Akbarov",
"https://mathoverflow.net/users/18943"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630127",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363014"
}
|
Stack Exchange
|
Are linear continuous mappings open on totally bounded sets?
Let $X$ and $Y$ be locally convex spaces, and $\varphi: X\to Y$ a linear continuous mapping. Suppose first that $S$ is a compact set in $X$. Then $\varphi$, being considered as a mapping from $S$ to $\varphi(S)$,
$$
\varphi\Big|_S:S\to \varphi(S)
$$
is open in the sense that for any open set $U$ in $S$ (with respect to the topology induced from $X$) its image $\varphi(U)$ is an open set in $\varphi(S)$ (with respect to the topology induced from $Y$).
This is strange, I was sure that the same must be true if $S$ is not necessarily compact, but just totally bounded (since we can consider the extension of $\varphi$ to the completions of the spaces $X$ and $Y$), but recently I understood unexpectedly that I can't write an accurate proof. Does this mean that there is a counterexample?
So my question:
Is $\varphi\Big|_S:S\to \varphi(S)$ an open mapping for each totally bounded set $S$ in $X$?
Bounded sets, e.g., unit balls in normed spaces, are always totally bounded for weak topologies. So you only have to find such a ball with two incompatible weak topologies, which is easy to do—say the ball of a suitable dual space with the weak and the weak $\ast$ topology.
Thank you! I was stuck in this, you helped me!
|
2025-03-21T14:48:31.249877
| 2020-06-14T07:43:02 |
363017
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"Zach Teitler",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/88133"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630128",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363017"
}
|
Stack Exchange
|
Analytical decomposed form of a specific traceless symmetric tensor
Assume an m-way tensor $\mathcal{Z}$.
$\mathcal{Z}_{p_1 p_2 ... p_m} = 0$ if any different indices match
and $\mathcal{Z}_{p_1 p_2 ... p_m} = 1$ otherwise.
It is a symmetric tensor. Now if it is 2-way tensor, i.e., a matrix, I can decompose it by diagonalization (of a symmetric hollow matrix). For a general tensor, probably I can do a numerical tensor decomposition (e.g., a symmetric tensor decomposition).
But I was wondering, since it is such a simple tensor (elements are either 1 or 0), is there an analytical decomposed expression for this tensor?
I want to avoid storing the full tensor and then decompose it numerically.
I am a not a mathematician so I apologize if my terminologies are not correct.
You could store it as a function which associates to any $m$ indices the expressions you wrote down above. Then you don't need to store any entries; you just check a list of $m$ indices for duplicate indices.
What kind of decomposition are you looking for? 2. In the correspondence between symmetric tensors and polynomials, this corresponds to an elementary symmetric polynomial.
Thanks to Zach Teitler for the comment that this tensor is associated with elementary symmetric polynomial (ESP). I searched for the decomposed form of ESP and found a paper (Power Sum Decompositions of Elementary Symmetric Polynomials by Lee (2015)).
I followed that and all I had to do was to translate the algebraic expressions in the context of tensors. There might be small errors in my part but this is basically what I found.
The $N$-way, $M$-dimensional tensor $\mathcal{Z}$ is equivalent to the ESP $S_{N}^{M}$ -
\begin{align*}
\mathcal{Z}_{p_1 ... p_N}
&= \sum_{\mu (I) = 1}^{R} \: \alpha_{\mu} \: t_{p_1}^{\mu} ... t_{p_N}^{\mu}
\\
S_{N}^{M}
&= \sum_{1 \leq p_1 \leq ... \leq p_N \leq M} X_{p_1} ... X_{p_N}
\quad (2 \leq N \leq M)
\end{align*}
where R is the rank.
The vector $\alpha$ and matrix $t$ are defined as -
\begin{align*}
t_{p}^{\mu (I)}
&= - 1, \quad \mbox{if} \: p \in \{ I \}
\\
&= + 1, \quad \mbox{otherwise}
\end{align*}
\begin{align*}
\alpha_{\mu (I)}
&= (-1)^{n_I} \: \frac{1}{2^{N-1}} \:
\begin{pmatrix}
M - k - n_I - 1 \\
k - n_I
\end{pmatrix}, \quad \mbox{if} \: k = \frac{N-1}{2} \:
\mbox{(odd)}
\\
&= (-1)^{n_I} \: \frac{1}{2^N \: (M-N)} \:
\begin{pmatrix}
M - k - n_I - 1 \\
k - n_I
\end{pmatrix} \: (M - 2 \: n_I), \quad \mbox{if} \: k = \frac{N}{2} \: \mbox{(even)}
\end{align*}
where $ \{ I \} \subset \{ M \} $ and $ n_I = | I | \leq k $. All the subsets are ordered and assigned an integer which belongs to $ \{ \mu (I) \} $. R has an upper bound ($ R \leq \sum_{r=0}^{k} \: \begin{pmatrix}
M \\ r \end{pmatrix}$).
|
2025-03-21T14:48:31.250084
| 2020-06-14T10:20:11 |
363026
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex Ravsky",
"ArtOfProblemSolving",
"LSpice",
"Max Alekseyev",
"RobPratt",
"https://mathoverflow.net/users/141766",
"https://mathoverflow.net/users/159578",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/43954",
"https://mathoverflow.net/users/7076"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630129",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363026"
}
|
Stack Exchange
|
What is the best way to partition the $4$-subsets of $\{1,2,3,\dots,n\}$?
Also asked on MSE: What is the best way to partition the $4$-subsets of $\{1,2,3,\dots,n\}$?.
Consider the set $X = \{1,2,3,\dots,n\}$. Define the collection of all $4$-subsets of $X$ by $$\mathcal A=\{Y\subset X: Y\text{ contains exactly $4$ elements}.\}$$
I want to partition $\mathcal A$ into groups $A_1,A_2,\dots, A_m\subset \mathcal A$ (each of them is a collection of $4$-subsets of $X$) such that $\bigcup_{i=1}^m A_i=\mathcal A$ and such that the intersection of any two distinct $4$-subsets in each $A_k$ has cardinality at most $1$, i.e. such that for all $i\in\{1,\dots,m\}$ and $Y_1, Y_2\in A_i$, we have $$Y_1\neq Y_2 \implies \lvert Y_1\cap Y_2\rvert \le 1.$$
My question: What can be said about the smallest $m$ (depending on $n$) such that such a partition exists?
My thoughts: I was expecting that each $A_i$ can contain "roughly" $\frac n4$ elements, so we would have $$m(n)=\Theta\left(\frac{\binom n4}{\frac n4}\right)=\Theta(n^3).$$ In particular, we would have $m(n)\le c n^3$ for some constant $c\in\mathbb R$.
However, I am neither sure if this is correct, nor how to formalize this.
For every pair of elements of $X$, the 4-subsets containing them must be in distinct $A_i$, implying that $m(n)\geq \frac{(n-2)(n-3)}2$.
We can show that if $n$ is a power of an odd prime then $m(n)\le n^2$.
By the way, $\binom X 4$ is a handy and fairly well understood notation for the set of $4$-subsets of $X$.
This problem can be reformulated in terms of graph coloring:
Let the graph $G=(V,E)$, $V=\mathcal A$, $(x,y)\in E \leftrightarrow x \cap y \geq 2$
Then a partition of $\mathcal{A}$ into groups $A_1,A_2,…,A_m$ corresponds to a $m$-coloring of $G$.
The graph $G$ has degree no more than $6{n\choose 2}$, so $G$ can be colored in no more than $6{n\choose 2}+1$ colors by Brooks' theorem.
A lower bound of $\chi$ is presented in Max Alekseyev's comment, which can be interpreted as a clique of $G$.
The degree of every node is $$\binom{4}{2}\binom{n-4}{2}+\binom{4}{3}\binom{n-4}{1}=(3n-11)(n-4).$$
@RobPratt Oh yes you are right, thanks!
|
2025-03-21T14:48:31.250263
| 2020-06-14T10:34:04 |
363027
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dieter Kadelka",
"Kurt. Z",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/159579"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630130",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363027"
}
|
Stack Exchange
|
How to compare the minimums of two discrete convex functions?
I have a question that troubled me for a long time.
If I have two convex discrete function $f(·)$ and $g(·)$ such that $f(·) \ge g(·)$. (may be not necessary?)
Let $x_1 = \text{argmin } f(·)$, and $x_2 = \text{argmin } g(·)$. How to prove that $x_1 \le x_2 $?
Actually, I want to find the upper bound of the minimum $x^*$ of $f(x)$ in order to reduce the enumeration scope to find a global minimum more efficiently.
One possible sufficient condition: if we prove $\Delta f(x) \ge \Delta g(x)$, then $x_1 \le x_2 $. (I hope it's true.)
Why the sufficient condition is correct? Or, there are other approaches to prove $x_1 \le x_2 $?
First a note: This is not a research level question. My answer is too long for a comment.
I think that by discrete function $h$ you mean that $h$ is defined on some $K := \{\ldots,k,k+1,k+2,\ldots\}$, possibly unbounded and that $\Delta h(x) = h(x+1)-h(x)$. Then $h$ is convex iff $\Delta h$ is not decreasing. To fix notation let $x^*$ be the largest minimum point of $h$. I now restrict to the case $K = \mathbb{N}_0$. Then
$$x^* = \sup \{k \in K \colon \Delta h(k) \leq 0\}$$
if there is any $k \in \mathbb{N}$ with $\Delta h(k) > 0$. Otherwise $h$ may have a minimum point, but no largest. Now if $x_2 \in \mathbb{N}$ exists necessarily $\Delta f(x_2) \geq \Delta g(x_2) > 0$, from which $x_1 \leq x_2$ immediately follows. There are simple examples that this condition is not necessary.
Thank you very much. I find my big mistake that $\Delta h$ is not increasing! Your answer reminds me. One more question, what if $h$ is not convex, are there any approaches to find bounds for minimum point of $h$? (I'm just curious about this. )
@Kurt Z: As far as I know: No. Convexity implies that local minima are global minima. There is a huge literature about finding global minima for the general case.
|
2025-03-21T14:48:31.250411
| 2020-06-14T10:37:44 |
363028
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jana",
"Sebastian",
"https://mathoverflow.net/users/32151",
"https://mathoverflow.net/users/4572"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630131",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363028"
}
|
Stack Exchange
|
Indecomposability of local systems and vector bundles
Let $M$ be a (connected) complex manifold, $L$ be a local system on $M$ and $\mathcal{L}$ the vector bundle associated to $L$. If $L$ is indecomposable, does it imply that $\mathcal{L}$ is also indecomposable?
No, this does not hold in general. For example, there do exist irreducible flat connections on the trivial holomorphic bundle of rank two over a compact Riemann surface of genus $g\geq2.$
On the other hand, if you consider irreducible local systems with unitary monodromy over compact Riemann surfaces then the associated vector bundle will be stable, hence indecomposable.
Are there similar examples in higher dimension as well?
Yes. For example, you can take the product of Riemann surfaces.
|
2025-03-21T14:48:31.250506
| 2020-06-14T10:54:23 |
363030
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"DesmosTutu",
"https://mathoverflow.net/users/147649"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630132",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363030"
}
|
Stack Exchange
|
Bounds for associated Legendre polynomials
I am trying to analyze the behaviour of the Associated Legendre polynomials $P_{n}^{m}$ on $[0,1]$. More specifically, I am trying to get upper bounds for $P_{n}^{m}$ on $[0,1]$. Bernstein's inequality for the Legendre polynomial is a classical result which states the following
\begin{align*}
|P_{n}(x)| \leq \sqrt{\frac{2}{\pi n}}\frac{1}{(1-x^2)^{1/4}}.
\end{align*}
I am interested in bounds of the above type for $P_{n}^{m}$'s. For a fixed $n\in \mathbb{N}$, normalizing $P_{n}^{m}$ appropriately as below, for every $|m|\leq n$ we have
\begin{align*}
\frac{(n-m)!}{(n+m)!} \int_{0}^{1} P_{n}^{m}(x)^2 dx = \frac{C}{2n+1}.
\end{align*}
So specifically, I am interested in upper bounds on $[0,1]$ for the following functions, for a fixed $n$, as $m \in \mathbb{Z}$ varies in $[-n,n]$.
\begin{align*}
L_{n}^{m}(x) := \sqrt{\frac{(n-m)!}{(n+m)!}} ~P_{n}^{m}
\end{align*}
I did a bit of searching and found that bounds for the above collection of normalized Associated Legendre functions are available in this and this and I state them below.
\begin{align}
\sqrt{\frac{(n-m)!}{(n+m)!}}~ |P_{n}^{m} (x)| & \leq \frac{1}{2^m m!} \sqrt{\frac{(n+m)!}{(n-m)!}} (1-x^2)^{m/2} := A_{n}^{m}(x) , \tag{1}\label{1}\\
\sqrt{\frac{(n-m)!}{(n+m)!}}~ |P_{n}^{m} (x)| & \leq \frac{1}{n^{1/4}} \frac{1}{(1-x^2)^{1/8}} =: f_{n}(x).\tag{2} \label{2}
\end{align}
I tried to see how good these bounds are. It seemed that $A_{n}^{m}$ is a good approximate for $L_{n}^{m}$ near 1 (which is expected as it captures the vanishing of $L_{n}^{m}$ at 1), but elsewhere it is not good.
And the bound $f_n$ appears to be just an upper bound which does not necessarily capture any feature of $L_{n}^{m}$.
Hence my question is: Are some other better bounds (than the ones in \ref{1} and \ref{2}) known for $L_{n}^{m}$'s?
Thanks!
More generally for a polynomial with real root see https://math.stackexchange.com/questions/4738155/got-a-factored-version-of-the-taylors-series
Please have a look at the following paper
https://www.sciencedirect.com/science/article/pii/S0021904598932075
In particular, Eq. (6) to get the answer to your question.
Lohöfer, G., Inequalities for the associated Legendre functions, J. Approximation Theory 95, No. 2, 178-193 (1998). ZBL0923.33002.
|
2025-03-21T14:48:31.250668
| 2020-06-14T11:31:27 |
363032
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dieter Kadelka",
"Eric",
"Henry",
"Ilkka Törmä",
"Julian Rosen",
"Sean Eberhard",
"Ville Salo",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/12565",
"https://mathoverflow.net/users/129185",
"https://mathoverflow.net/users/20598",
"https://mathoverflow.net/users/5263",
"https://mathoverflow.net/users/66104",
"https://mathoverflow.net/users/75935",
"mathworker21"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630133",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363032"
}
|
Stack Exchange
|
What is the asymptotic dynamics of the winning position in this game?
$n$ players indexed $1,2,...,n$ play a game of mock duel. The rules are simple: starting from player $1$, each player takes turns to act in the order $1,2,...,n,1,2,...$. In his turn, a player randomly chooses one of the other remaining players as target, and fires one shot at him. If hit, the target is removed from the game. Whoever survives to the last is the winner. All players have the same marksmanship $0\lt p\le 1$, meaning that their shots hit target with probability $p$.
An example: if $p=1$ and $n=3$, player $1$ randomly chooses $2$ or $3$ as his target and fires one shot. The survivor of $1$'s shot will shoot next and fire at the only other remaining player, player $1$. Hence in this case the three players have winning probabilities $0,0.5,0.5$ respectively.
Question: which player has the highest winning probability for large $n$'s?
To be more specific, let's define $B(n)$ to be the player with the highest winning probability for the $n$-players game. Are there any regularity about where $B(n)$ tends to occur in $(1,2,...,n)$? In particular:
Does $\lim_{n \to \infty} B(n)/n$ exist for any $p$?
Is $\lim_{n \to \infty} B(n)=1$ if $p$ is sufficiently small?
Edit: Clarified the rules to address possible ambiguity suggested in the comments.
If my calculation is correct, the winning probability for player $k$ in the $n$ players game, $s(n,k)$, is given by the recursive formula:
$$
\begin{align}
& s(n,1)=(1-p)s(n,n)+ps(n-1,n-1)\\
& s(n,2)=(1-p)s(n,1)+\frac{n-2}{n-1}ps(n-1,1)\\
& s(n,k)=(1-p)s(n,k-1)+\frac{n-k}{n-1}ps(n-1,k-1)+\frac{k-2}{n-1}ps(n-1,k-2)
\end{align}
$$
for $k\gt 2$.
For large $n$'s and $p=1$, plotting $s(n,k)$ against $k$ looks like sinusoid:
If you plot the peaks of these curves (i.e. $B(n)$) against $n$, you get this:
At first glance it seems $B(n)$ just oscillates between $1$ and $n$ in a self-similar fashion. But actually the hill tips wear off further and further away from the diagonal line as $n$ increases and the valleys also grow flatter:
One can wonder whether the hills will eventually fall to small bumps or be flattened relative to the diagonal line, meaning that $\lim_{n \to \infty} B(n)/n=0$.
On the other hand, $B(n)=1$ becomes more often as $p$ gets smaller. Blue curves are $B(n)$ for $p=0.5,0.48,0.46$ respectively, for $n$ up to 5000 (larger $n$'s take exponentially more time to compute):
Can we claim that $\lim_{n \to \infty} B(n)=1$ for small $p$? Or is there always exceptions?
Do the players decide who they fire at, or is that also random? Assuming they decide, then the question isn't necessarily well-formed yet, as there could be multiple Nash equilibria (intuitively the first player has an advantage, but the other players could gang up on that player, etc).
@SeanEberhard There's no communication. The rule states that "each player takes turns to fire one shot randomly at one of the remaining opponents". You can check this is also an equilibrium. It's a very symmetric one. There's absolutely no reason for others to gang up against the 1st one, since they all have the same marksmanship and the situation is symmetric.
Related to https://oeis.org/A071818
@Eric you didn't answer sean's question. they can choose to fire at someone without needing to previously communicate.... Also, can you tell me how the situation is symmetric? The first person goes first
@mathworker21 I answered him by saying that the rule states "each player takes turns to fire one shot randomly at one of the remaining opponents", period. If you insist on looking at the game from a game theory point of view, all I can say is that this rule happens to be consistent with the equilibrium in which everyone fire one shot randomly at one of his opponents in his turn. This equilibrium is symmetric in the sense that each person has the same strategy (randomly firing one shot at his opponents).
@Eric I think there's ambiguity regarding "fire one shot randomly". Does it imply that the target is chosen (uniformly?) at random, that the shot has a chance of missing, or both?
I suggest the OP explain carefully in an appendix to the question what game this strategy is a Nash equilibrium for, since everyone is clearly interested. But it does seem to be just a distraction, and this is really about a zero player game.
The rule is not clear to me either. I gather from the description (I didn't peruse the formulas) that a target is chosen uniformly at random among the living, and you kill them with probability $p$. But it seems $B(n)$ could depend on $p$.
@IlkkaTörmä I see your point. I've edited the rule to address this point. And I also added an example to clarify. Hope it's clear now.
@VilleSalo You understood the rule correctly. I edited the rule to address this and also added an example to clarify. All players adopting the strategy "randomly choose one other player to shoot in my turn" is a Nash equilibrium for a game where the rule is changed to allow players to freely choose their target.
@Eric: Ok, I do not see that without a proof and it seems very doubtful to me (why not shoot $B(n)$ instead?), but it is beside the point.
@VilleSalo Let's consider $n$ players game and it's your turn to shoot. If all other players are adopting the "random shooting" strategy, killing anyone would give you the same winning probability. Because anyone you kill, you always become the last to shoot in a $n-1$ players game.
You are absolutely correct.
I agree, after thinking about it for more than 1 minute, it's clear that this question is not at all about a game. But you made that kind of hard on your reader by saying "game" in the title, first line of the question, etc.
@Eric: I don't understand your recursion formula. For $p = 1$ we get $s(n,1) = s(n-1,n-1)$. Do you have an explanation? Why not $s(n-1,k)$ or some mixture of these values?
@DieterKadelka p=1 means player 1 will remove his target for certain. Whoever his target happens to be, after that target is removed, there're now n-1 players and he becomes the last one of them to shoot.
I think an interesting question is whether, with $p=1$, you have $\lim\limits_{n\to \infty} \frac{\min s(n,k)}{\max s(n,k)} = 1$ or not
|
2025-03-21T14:48:31.251093
| 2020-06-14T11:46:48 |
363033
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/35520",
"ofer zeitouni"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630134",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363033"
}
|
Stack Exchange
|
Solving integral equation with an unknown probability distribution
Considering this system of integral equations, where $\gamma \in \mathbb{R} $ and $\alpha\in \mathbb{C}$ are the unknown to solve :
$$ 1=\int_{-\infty}^{\infty} p(u) \frac{ -1}{\gamma-\left(u-z^{*}+\tau \alpha^*\right)\left(u-z+\tau \alpha\right)}\mathrm{d}u$$
$$\alpha^*=\int_{-\infty}^{\infty}p(u) \frac{ u-z+\tau \alpha}{\gamma-\left(u-z^{*}+\tau \alpha^*\right)\left(u-z+\tau \alpha\right)}\mathrm{d}u$$
$$\alpha=\int_{-\infty}^{\infty}p(u) \frac{ u-z^*+\tau \alpha^*}{\gamma-\left(u-z^{*}+\tau \alpha^*\right)\left(u-z+\tau \alpha\right)}\mathrm{d}u$$
Where $\alpha^*$represents the complex conjugate of $\alpha$ and $p(u)$ is a probability distribution. Is it possible to solve these equations without further information on $p(u)$? Or is it at least possible to change their representation to make them more friendly ?
The solution $\alpha$ would then give an equation representing the boundaries of the eigenvalues of a random matrix :
$$ 1=\int_{-\infty}^{\infty}p(u) \frac{ 1}{\left(u-z^{*}+\tau \alpha^*\right)\left(u-z+\tau \alpha\right)}\mathrm{d}u$$
Therefore I see $\gamma$ and $\alpha$ as intermediate steps, the boundaries only depend directly on $p(u)$ and not $\alpha$ strictly speaking.
Any remark, reference or advice is always very appreciated. Thank you.
it would help if you formulated the random-matrix problem that you are interested in; that might also clarify the role of the parameters $z$ and $\tau$, which you have left unspecified (are they known?)
Isn't p actually also an unknown? aren't you trying to find p,$\alpha$ so that the equations are true for all $z$ complex? and what is $\tau$? another given parameter? If $\tau$ is real, then the third equation is the same as the second.
|
2025-03-21T14:48:31.251351
| 2020-06-14T12:18:15 |
363034
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Denis T",
"Jeremy Rickard",
"MoreauT",
"Theo Johnson-Freyd",
"Todd Trimble",
"https://mathoverflow.net/users/152969",
"https://mathoverflow.net/users/22989",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/78",
"https://mathoverflow.net/users/81055"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630135",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363034"
}
|
Stack Exchange
|
Idempotent complete category which is not abelian
In the following, we work with additive categories.
We say that a category is weakly idempotent complete if all the epimorphisms which admit a section have kernel. It's equivalent to the dual statement: all the monomorphisms which admit a retract have cokernel.
A stronger notion is the notion of idempotent complete. A category is idempotent complete if each morphism in $\{f:A \rightarrow A \;|\; f^2=f\}$ has a kernel or equivalently a cokernel.
As suggested by the name of these notions, an idempotent complete category is weakly idempotent.
We know that an abelian category is idempotent complete.
We have the following examples :
• We consider the category $K_{vect} ^{ ^{\ge n}}$ of vector spaces over a field $K$ with no non-nul vector space of smaller dimension than $n$. Then $K_{vect} ^{\ge n}$ is not weakly idempotent complete since the trivial projection $K^{n+1} \twoheadrightarrow K^n$ has a section but no kernel.
• We consider the category $K_{vect} ^ {^{\equiv 0 [2]}}$ of vector spaces over a field $K$ with pair dimension or infinite dimension. Then $K_{vect} ^ {^{\equiv 0 [2]}}$ is weakly idempotent complete. But it's not idempotent complete since the projector $K^{2} \stackrel{\begin{pmatrix}Id & 0 \\ 0 & 0\end{pmatrix}}{\rightarrow} K^2$ has no kernel or cokernel.
So this is the question :
Is someone knows an example of an idempotent complete category which is not abelian?
(here we don't need to have Grothendieck's axioms)
Thanks you,
Timothée
I think the most interesting example is Deligne-Milne category $Rep(GL_t)$ for integral parameter $t$. https://www.jmilne.org/math/Books/DMOS.pdf
Free abelian groups. Divisible abelian groups. Abelian groups with no elements of order $101$. More generally, if you take an additive subcategory of an abelian category that is closed under taking direct summands, then it will be idempotent complete, and although it may be abelian, it will typically not be.
Could you post your answer as an answer and not as a comment ? If you do this I could accept the answers :)
Very often people post answers in comments when they are hesitant about whether a question is on-topic here: they don't necessarily want to encourage more such questions, but also they don't want to leave the OP empty-handed. Of course I'm not sure that's Jeremy's thinking, but just to let you know this sometimes happens.
The OP asked for responses as an answer. I have made this answer CW, and encourage everyone to make a big list. I start with the answer that comes up in my work, and then compile the answers from the comments. (As this answer is CW, feel free to reorganize, remove this intro, etc.)
Project modules over a ring $R$. Finitely generated projective modules over $R$.
Free abelian groups. (Example 1. for $R = \mathbb{Z}$.)
Vector bundles over a fixed base $M$ (Example 1. for $R = \mathcal{O}(M)$.)
Divisible abelian groups.
Abelian groups with no elements of order $101$.
Deligne-
Milne category $Rep(GL_t)$ for $t$ integral. https://www.jmilne.org/math/Books/DMOS.pdf
Smooth manifolds.
The homotopy category of any idempotent complete additive ∞-category is still idempotent complete, and they are typically not abelian. For example the homotopy category of spectra, the derived category of any Grothendieck abelian category, or the category $DM(S)$ of Voevodsky for $S$ a locally Noetherian scheme... If you want small examples you can take compact objects there instead.
Smooth manifolds? The OP said to work with additive categories.
@ToddTrimble Well, it is CW :) Anyway, I added it just because my favourite example of idempotent completion is to start with the category of open domains in R^n and smooth maps...
Hey, it's one of mine too! https://mathoverflow.net/questions/162552/idempotents-split-in-category-of-smooth-manifolds
|
2025-03-21T14:48:31.251601
| 2020-06-14T12:31:06 |
363035
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"Marouani",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/159582"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630136",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363035"
}
|
Stack Exchange
|
Bounded form in complex complete manifold
If $\alpha$ is a bounded form in a complex complete manifold $X$ (i.e $\sup_X|\alpha (x) |<\infty$, then $d\alpha$ is it also bounded?
Rq: if $d\alpha$ is bounded then \alpha is not necessary bounded, take for exemple $dx_1\wedge dx_2...\wedge dx_{2n} =d\Gamma$ on $\mathbb{C}^n.$
You are assuming that $X$ has a complete Riemannian metric? Or Finsler? Hermitian? Kaehler?
X is a hermitian manifold
On $\mathbb{C}$, try $\omega=\sin(|z|^2) \, dz$.
|
2025-03-21T14:48:31.251667
| 2020-06-14T12:46:39 |
363036
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Michael Greinecker",
"https://mathoverflow.net/users/35357",
"https://mathoverflow.net/users/5656",
"kenneth"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630137",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363036"
}
|
Stack Exchange
|
Can Schauder's fixed point theorem apply to a metric space?
I am currently reading the existence proof of Mean Field Game equation, which is a coupled system of Hamilton-Jacobi-Bellman equation and Fokker-Planck equation, see page 42 of the paper here. The idea is to obtain a fixed point of a continuous mapping $\Psi: S \mapsto S$. The space $S$ is given by $C([0, T], \mathcal P(\mathbb T))$, where $P(\mathbb T))$ is Wasswerstein metric space.
The proof is closed up with Schauder's FPT, which is usually applied for Banach space. I am not sure this generalization can be found in any literature?
I guess $\mathcal{P}(\mathbb T)$ is treated as a closed convex subspace of the space of finite signed measures on $\mathbb{T}$ endowed with the weak*-topology. The set $S$ is only a subset of the space you described. As the Wikipedia entry, you linked to, states, the fixed-point argument works in any local convex Hausdorff topological vector space.
@MichaelGreinecker Thanks for your reply. Yes, I agree with you. It seems to work for much more general setup from Wiki.
|
2025-03-21T14:48:31.251826
| 2020-06-14T12:49:16 |
363037
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Moishe Kohan",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/39654",
"https://mathoverflow.net/users/40804",
"mme"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630138",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363037"
}
|
Stack Exchange
|
Does cohomology ring determine a compact symmetric space?
Suppose that $M_1, M_2$ are compact connected symmetric spaces with isomorphic integer cohomology rings. Does it follow that $M_1$ is diffeomorphic to $M_2$?
The only result I am aware of is this old paper:
Wolf, J. A., Symmetric spaces which are real cohomology spheres, J. Differ. Geom. 3, 59-68 (1969). ZBL0191.20101.
(specifically, the unnumbered corollary in the introduction) where Joe Wolf proves that each symmetric space which is an integer (co)homology $n$-sphere, is diffeomorphic to $S^n$.
Related: If $M_1,M_2$ are assumed simply connected and have isomorphic real cohomology graded algebras, do you know if they are diffeomorphic? Here I added simply connected, otherwise we could take $S^2$ vs $P^2(\mathbf{R})$.
@YCor: Good question. Wolf gives one simply-connected example of a real homology sphere $SU(3)/SO(3)$: I am not sure if it is diffeomorphic to $S^5$.
@Moishe This space is sometimes known as the Wu manifold. Its second homotopy/homology group is $\Bbb Z/2$, and otherwise it has the homology of $S^5$.
@MikeMiller: Thanks! So one has to work over the integers.
|
2025-03-21T14:48:31.251970
| 2020-06-14T13:58:10 |
363043
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Roberts",
"https://mathoverflow.net/users/142929",
"https://mathoverflow.net/users/4177",
"user142929"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630139",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363043"
}
|
Stack Exchange
|
From Zurab's integral representation for the Apéry's constant to almost impossible integrals
I would like to know if the following integrals are known, or in case that aren't in the literature we can calculate these in closed-form (in terms of elementary and standard functions). I wondered about it when I tried to get variants of an integral representation from a paper of Zurab (identity $(22)$ from [1], that I refer as Zurab's integral in the title), in next two contexts. If my post is more suitable for Mathematics Stack Exchange or you've feedback (in particular if you know these integrals) please add your feedback in comments.
Question 1. The first, just as comparison to [1], calculate if possible $$\int_0^{\pi}\frac{J_0(x)}{\binom{1}{x/\pi}}dx,\tag{1}$$
where $J_0(x)$ denotes the Bessel function. Many thanks.
For Question 1 I tried to combine an integral representation from the Wikipedia Bessel function and calculations from Wolfram Alpha online calculator, in particular I did an attempt to exploit the section Bessel's integral that refers previous Wikipedia (reference Nico M. Temme, Special Functions: An introduction to the classical functions of mathematical physics (2nd print ed.). New York: Wiley, (1996) pp. 228–231), and the output int x(pi-x)/sin(x) cos(a x) dx that provides me Wolfram Alpha online calculator.
Added see my comments: If isn't possible to get a simple closed-form in terms of particular values of elementary/special functions that I evoke but the integrals (1) can be expressed as series in some simplified form, I should consider it also as an answer.
Question 2. I would like to know as reference request or in other case if it is possible to compute the integrals $$c_{n,m}:=\int_0^{1/2}\frac{x(x-1/2)}{\sin^2(2\pi x)}\, \sin(2\pi(2m+1)nx)dx\tag{2}$$
for integers $m\geq 0$ and integers $n\geq 1$. Many thanks.
Thus for Question 2 if you know a reduction formula or reference for these integrals $(2)$ from the literature helpful here, answer my question as a reference request and I try to search and read from the literature their closed-forms.
If I remember well I could to find just a choice of integration by parts that was suitable in the computations that I did few months ago, but it doesn't solve the problem, since I didn't know how to finish my (tedious) calculation to get in closed-form $(2)$. This is just a curiosity that I wondered few months ago, again from Zurab's integral as starting point in an attempt to combine his integral and a definition of a certain function (and arithmetical functions that I omit here from page 79 of [2]) I wrote $$-\frac{7\zeta(3)}{8\pi^3}=\sum_{n=1}^\infty\sum_{m=0}^\infty\frac{\mu(n)\chi_1(n)}{n^2}\cdot\frac{(-1)^m c_{n,m}}{(2m+1)^2},$$ where $\zeta(s)$ is the Riemann zeta function (and those other arithmetic functions from the mentioned article [2]).
References:
[1] Zurab Silagadze, Sums of Generalized Harmonic Series. For Kids from Five to Fifteen, RESONANCE, September 2015. arXiv:1003.3602
[2] Manuel Benito, Luis M. Navas and Juan Luis Varona, Möbius inversion from the point of view of arithmetical semigroup flows, Biblioteca de la Revista Matemática Iberoamericana, Proceedings of the "Segundas Jornadas de Teoría de Números" (Madrid, 2007), pp. 63-81. (Semantic Scholar page, pdf)
I emphasize that, in case that this question is interesting for MathOverflow, I'm asking as a reference request since I don't know if these definite or indefinite integrals in $(1)$ and $(2)$ are well-known from the literature, and in other case what work can be done to compute the closed-forms of the integrals in Question 1 and Question 2 (hints or some approach) in terms of particular values of elementary functions or standard functions.
If isn't possible to get simple closed-forms that I evoke but the integrals $(1)$ and/or $(2)$ can be expressed as series in some simplified form, I should consider it also (in the spirit of what work can be done about these questions) as an answer. My motivation is learn from the professors of the site. Many thanks.
Many thanks for your edit @DavidRoberts
No problems! I enjoyed reading the Silagadze paper :-)
$c_{n,m}=0$ if $n$ is even; for $n$ odd some experimentation indicates it has the form
$$c_{2n+1,m}=\frac{1}{\pi^3}\bigl(a_{n,m}-\tfrac{7}{8}(2n+1)(2m+1)\zeta(3)\bigr),\;\;\text{with}\;\;a_{n,m}=a_{m,n}\in\mathbb{Q}\geq 0.$$
I have not found a closed-form expression for the coefficients $a_{n,m}$, they are quite unwieldy.
Some values of $a_{n,m}$ for small $n,m$:
$$\{a_{0,0},a_{0,1},a_{0,2},a_{0,3},a_{0,4},a_{0,5}\}=\left\{0,2,\frac{110}{27},\frac{20804}{3375},\frac{3187348}{385875},\frac{323770282}{31255875}\right\},$$
$$\{a_{1,0},a_{1,1},a_{1,2},a_{1,3},a_{1,4},a_{1,5}\}=\left\{2,\frac{3187348}{385875},\frac{88717098057736}{6093243231075},\frac{3060320225351036566502}{146665526640505747875},\frac{2020487623534710670386246274}{74353310943126393099796875},\frac{13319855208745022209825602301502381584}{397806359157656353958636885223796875}\right\}.$$
Many thanks for your answer.
Thank you very much again for your propositions in first paragraph for the sequence $c_{n,m}$.
To evaluate
\begin{align}
c_{n,m}&=\int_0^{1/2}\frac{x(x-1/2)}{\sin^2(2\pi x)}\, \sin(2\pi(2m+1)nx)\,dx\\
&=-\frac{1}{8}\int_0^{1}\frac{x(1-x)}{\sin^2(\pi x)}\, \sin(\pi(2m+1)nx)\,dx
\end{align}
we first notice by changing $x\to 1-x$, that the integral vanishes if $n$ is even. In the following $c_{2n+1,m}$ is calculated.
We apply twice a result obtained in this answer:
\begin{equation}
\int_0^1\frac{x(1-x)}{\sin \pi x}f(x)\,dx=2\sum_{n=0}^\infty \int_0^1x(1-x)f(x)\sin\left( (2n+1)\pi x \right)\,dx
\end{equation}
With $f(x)=\frac{\sin(\pi(2m+1)(2n+1)x)}{\sin(\pi x)}$, first:
\begin{equation}
c_{2n+1,m}=-\frac{1}{4}\sum_{p=0}^\infty \int_0^1\frac{x(1-x)}{\sin(\pi x)}\sin(\pi(2m+1)(2n+1)x)\sin\left( (2p+1)\pi x \right)\,dx
\end{equation}
then with$f(x)=\sin(\pi(2m+1)(2n+1)x)$:
\begin{equation}
c_{2n+1,m}=-\frac{1}{2}\sum_{p,q=0}^\infty \int_0^1x(1-x)\sin(\pi(2m+1)(2n+1)x)\sin\left( (2p+1)\pi x \right)\sin\left( (2q+1)\pi x \right)\,dx
\end{equation}
A decomposition of the product of the three sines is
\begin{align}
\sin(\pi&(2m+1)(2n+1)x)\sin\left( (2p+1)\pi x \right)\sin\left( (2q+1)\pi x \right)\\
=&\frac{1}{4}\left[
\sin\left( (N-2p+2q)\pi x \right)-\sin\left( (N-2p-2q-2)\pi x \right)\right.\\
&\left.-\sin\left( (N+2p+2q+2)\pi x \right)+\sin\left( (N+2p-2q )\pi x\right)
\right]
\end{align}
where $N=(2m+1)(2n+1)$ is an odd integer. Using the identity
\begin{equation}
\int_0^1 x(1-x)\sin((2s+1)\pi x)\,dx=\frac{4}{\pi^3}\frac{1}{(2s+1)^3}
\end{equation}
if $s$ is an integer, we obtain
\begin{align}
c_{2n+1,m}=-\frac{1}{2\pi^3}\sum_{p,q=0}^\infty&\left( \frac{1}{(N-2p+2q)^3}- \frac{1}{(N-2p-2q-2)^3}\right.\\
&\left.- \frac{1}{(N+2p+2q+2)^3}+ \frac{1}{(N+2p-2q)^3} \right)
\end{align}
\begin{align}
=-\frac{1}{2\pi^3}\sum_{p,q=0}^\infty&\left( \frac{1}{(2p+2q+2-N)^3}- \frac{1}{(2p+2q+2+N)^3}\right.\\
&\left.+ \frac{1}{(2p-2q+N)^3} -\frac{1}{(2p-2q-N)^3} \right)
\end{align}
\begin{align}
=-\frac{1}{32\pi^3}\sum_{q=0}^\infty&\left[\psi^{(2)}\left( \frac{N}{2}+1+q \right)-\psi^{(2)}\left(- \frac{N}{2}+1+q \right)\right.\\
&\left.+\psi^{(2)}\left( -\frac{N}{2}-q \right)-\psi^{(2)}\left( \frac{N}{2}-q \right)
\right]
\end{align}
where $\psi^{(2)}$ is a polygamma function. Using recurrence identities and reflection formula, we find
\begin{align}
c_{2n+1,m}&=-\frac{1}{16\pi^3}\sum_{q=0}^\infty\sum_{k=1}^N\left[\left( \frac{N}{2}+1+q -k\right)^{-3}
-\left(\frac{N}{2}-q -k\right)^{-3}
\right]\\
&=\frac{1}{32\pi^3}\sum_{k=1}^N\left[
\psi^{(2)}\left(\frac{N}{2}+1-k \right)+\psi^{(2)}\left(- \frac{N}{2}+k \right)
\right]\\
&=\frac{1}{16\pi^3}\sum_{k=1}^N
\psi^{(2)}\left(\frac{N}{2}+1-k \right)\\
\end{align}
With the known value $\psi^{(2)}\left(1/2 \right)=-14\zeta(3)$ and recurrence and reflection identities,
\begin{align}
c_{2n+1,m}&=\frac{1}{16\pi^3}\sum_{k=1}^N \psi^{(2)}\left(\frac{1}{2}+\frac{N+1}{2}-k \right)\\
&=\frac{1}{16\pi^3}\sum_{r=1}^{(N-1)/2}\psi^{(2)}\left(\frac{1}{2}+r \right)+\sum_{r=0}^{(N-1)/2}\psi^{(2)}\left(\frac{1}{2}-r \right)\\
&=\frac{1}{16\pi^3}\left[\psi^{(2)}\left(\frac{1}{2}\right)+2\sum_{r=1}^{(N-1)/2}\psi^{(2)}\left(\frac{1}{2}+r \right)\right]\\
&=\frac{1}{8\pi^3}\left[-7N\zeta(3)+2\sum_{r=1}^{(N-1)/2}\sum_{s=0}^{r-1}\frac{1}{\left( s+\frac{1}{2} \right)^3}\right]
\end{align}
which seems to be correct when compared to the values obtained by @CarloBeenakker.
Many thanks for your answer. I'm going to study the propositions that you're shared.
|
2025-03-21T14:48:31.252445
| 2020-06-14T15:09:30 |
363046
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Nate Eldredge",
"Tobsn",
"Yfmvch",
"https://mathoverflow.net/users/159586",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/85194"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630140",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363046"
}
|
Stack Exchange
|
"Universal" properties of Brownian motion
I've been learning some stochastic calculus (mostly through Oksendal) recently and while I understand the definition of Brownian motion given by Oksendal, I am curious if there are more "categorical" reasons we should be especially interested in this particular stochastic process above others.
Now when I say "categorical," I don't necessarily mean actual category-theoretic statements (although those are welcome), but more broadly properties of Brownian motion that intuitively justify its study to the exclusion of other processes.
As an example, when I first read the martingale representation theorem I thought it was a type of universal statement (until I read more carefully and realized that it of course applies only to martingales with respect to the filtration associated to Brownian motion).
Sorry if this is too vague, but with a background in algebra I get fidgety studying objects that feel somewhat arbitrarily selected.
Have you read about Donsker's theorem? It says roughly that Brownian motion is "universal" in the same sense that the central limit theorem says the normal distribution is universal: a large class of random objects converge to it in the scaling limit.
Thank you! This is just the kind of statement I was looking for. I had seen it in the particular case of scaled simple random walk, but not in this generality.
Another reason for Brownian motion on $\mathbb{R}^{n}$ being 'canonic' is that its intrinsic metric is the Euclidean distance. A fact for which analogues exist also on manifolds and infinite dimensional Hilbert spaces, keyword here: Varadhan short-time asymptotics.
The Brownian motion is, up to a scaling factor, the only continuous¹ process in $\mathbb R^d$ whose increments are stationary ($B_{t+dt}-B_t$ has the distribution of $B_{dt}$ for $t,dt\geq 0$), independent ($(B_{t_{i+1}}-B_{t_i})_{0\leq i<k}$ are independent for $t_0<\cdots<t_k$) and rotationally invariant ($RB_t$ has the distribution of $B_t$ for all rotations $R:\mathbb R^d\to\mathbb R^d$).
In fact, a continuous process whose increments are stationary independent has the form $t\mapsto AB_t+tv$, for a well-chosen $v\in\mathbb R^d$ and $A:\mathbb R^d\to\mathbb R^d$ linear. Then it is not difficult to see that the rotational invariance forces $v$ to be zero and $A$ a constant multiple of the identity.
I am not sure this particular result has a name, but it follows directly from the description of the Lévy processes. Similar results in Lie groups are known as Hunt's theorem.
¹ Here, by continuous process, I mean a process whose samples are almost surely continuous.
Donsker's theorem as pointed out in comments is helpful. Here's another helpful characterization.
Brownian motion is the centered continuous Gaussian process satisfying:
-Self similarity. I.e. $B_{ct}=c^H B(t)$ in distribution for some $H$.
-Stationary, independent increments. I.e. $B_t-B_s=B_{t-s}$ in distribution and are independent.
(Also normalized so $E[B_1^2]=1$ but that is for convenience)
You can build the covariance from this characterization.
|
2025-03-21T14:48:31.252692
| 2020-06-14T15:45:15 |
363048
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Manfred Weis",
"Nik Weaver",
"Stanley Yao Xiao",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/31310"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630141",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363048"
}
|
Stack Exchange
|
Advice for contacting journals for the first time
What can be recommended for contacting journals with the purpose of checking whether they would be interested in an publishing an article if one is not affiliated to a research institution and still is unknown to the professional math community?
Specifically I would like to know whether it is acceptable to initially only state the problem whose solution is claimed and give some indication that there is strong evidence for the correctness of that solution, e.g. by providing experimental results, proving acquaintance with the problem and attempts to solve it.
Are such initial contacts tolerated by journals or do they insist on the availability of a mature document when they are initially contacted?
Based on your description, it does not seem like you solved any particular problem that would be of interest to some part of the mathematical community but rather found some numerical evidence that something is true. There are journals who publish experimental results, as long as the computations are interesting (Mathematics of Computation comes to mind). You can try emailing one of the editors of a journal like this and see if they are interested.
@StanleyYaoXiao I am very sure that I have found the solution to an interesting problem, specifically an algorithm to find a solution; the experimental evidence would be to run it on test instances and document its performance.
@StanleyYaoXiao I wouldn't have a problem to provide details on that algorithm on MO, but that would be self advertising, which is deprecated. If however someone asked for it the situation may be a different one.
If you believe you have solved a problem that would be of interest to some mathematical community, then your best bet is to prepare a manuscript carefully and submit to a journal. Your manuscript should 'look like' a math paper: i.e., preferably in LaTeX, demonstrate a good knowledge of existing literature, and has good exposition.
@ManfredWeis it sounds like you are an independent researcher with little involvement in the mathematical "community". In that case you are really to be commended for having the wherewithal to pursue research on your own. Good for you!
However, you might not be familiar with important aspects of this community. For instance, when a mathematician says that they have "solved" a problem this is understood to mean that they have a rigorous proof. If you mean something different (having amassed experimental evidence of some sort) then that greatly limits the possible venues for publication. Stanley's first comment about journals for experimental math is very on point.
@NikWeaver I am convinced that I have a rigorous proof as understood by professional mathematicians that utilizes a new idea for transforming the problem to one that is already solved and I'm sure that every argument in the proof can easily be checked and nodded off. So even if the proof turned out to be flawed its ideas would shed a new light on the problem. The numerical experiments would only be performed before thinking about communicating the proof, primarily to gain confidence in its correctness or that it is worth looking at.
Well, if you have a rigorous proof of a theorem which answers an important open problem then you need to write a "mature document" that presents it, so that people can check your proof. As you can see by Stanley's and my reactions, laying emphasis on numerical experiments is likely to lead people into assuming you don't have a proof.
Is the situation this: you think you've solved a famous problem, but you're afraid no one will take you seriously because you are an unknown? If so, the way to deal with that is to make your paper as clear and correct as possible. Numerical experiments might best be described in an appendix --- that would convey their significance without leading people to think that's all you have.
@NickWeaver that's exactly my situation...
Well, good luck!
Sending an editor an outline like you describe won't have much of an effect. All an editor will do, if they respond at all, is say "why don't you submit the paper, and then we will decide whether to send it for peer review and to accept it". They couldn't possibly be any more committal based on just an email.
Writing mathematical papers is hard! Graduate students and early postdocs struggle a lot with that, and typically rely on considerable input from their advisors or other more senior colleagues to produce a paper that is submittable.
What I would recommend is that you first find a professional mathematician who is interested in your problem, and contact them with the questions that you would ask an editor. If you cannot find a mathematician who is interested in your problem, then that does not bode well for your journal submission (since if the editor cannot find anybody interested in it, they will reject the paper).
If the point is to get published in a journal, you need to do it their way. Their job is to take mature papers, have them vetted to make sure they are mature enough to commit to the literature, and then publish them with the stamp of their reputation. If you are unsure whether your paper merits their approval, you probably should not ask. (This is different from trying to determine publishability. If you contact an editor or two, they might give you their opinion if it does not take much time. I think Alex B. has a correct and charitable reading of the situation here.)
If the goal instead is to gauge interest in the problem, there are better ways. Probably the best is to look up conference proceedings of the area and see how much play the problem is given. Use these to get names and addresses of experts who can probably tell you how and when to approach journals, if you still want to go that route. There is less barrier to self publishing, and you can use your MathOverflow user page to talk up about yourself and your work. The user page can serve as a big business card should you wish to contact these experts.
Gerhard "Don't Need No Stinkin' Journals" Paseman, 2020.06.14.
Gauging interest in the problem wouldn't be hard as it is of some fame; that is also the reason why I am careful not to claim that I have solved it but only that I am very convinced to have solved it. The situation is like riding a tiger and not knowing how to get off graciously.
|
2025-03-21T14:48:31.253441
| 2020-06-14T15:48:37 |
363049
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Denis Nardin",
"Tim Campion",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/43054"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630142",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363049"
}
|
Stack Exchange
|
Examples of commutative ring spectra with graded-artinian coefficients?
Question: What are some ring spectra $E$ satisfying the following conditions?
The coefficients $E_\ast = \pi_\ast(E)$ are graded-commutative;
There is a Kunneth spectral sequence $E_\ast(X) \otimes_{E_\ast} E_\ast(Y) \Rightarrow E_\ast(X \wedge Y)$ for finite spectra $X,Y$;
The coefficients $E_\ast$ form a graded-artinian ring (i.e. satisfies the descending chain condition for homogeneous ideals).
Clearly there is a canonical sequence of examples $E = H\mathbb Q, K(1), K(2), \dots, H \mathbb F_p$. More generally, if $E_\ast$ is a graded field then $E$ is an example -- but this essentially reduces to the examples listed.
I'm particularly curious if the telescope $T(n)$ is (can be chosen to) satisfy (1,2,3).
I'm not sure of what you mean with "Künneth SS", but there's a Künneth SS of the form $\mathrm{Tor}{E\ast}(E_\ast X,E_\ast Y)\Rightarrow E_\ast(X\wedge Y)$ for all spectra $X,Y$ and commutative ring spectrum $E$ (type of convergence may vary).
@DenisNardin I want to be a bit loose with what "commutative ring spectrum" means -- in order to have $Tor_{E_\ast}(E_\ast X, E_\ast Y) \Rightarrow E_\ast(X \wedge Y)$, does it suffice for $E$ to be a spectrum with maps $E \wedge E \to E \leftarrow \mathbb S$ which endow $E_\ast$ with the structure of a graded commutative ring? Does one need to assume the maps are $E_1$? $E_\infty$?
For the version I know you need $E_\infty$. There's another variant due to Adams with different restrictions (you need $E$ to be written as a colimit of finite spectra with special properties) which might be more suited to your goals. Regardless, I don't think you can discard the higher Tor in the signature of the spectral sequence in general.
@DenisNardin Of course you're right. That kind of sinks my original motivation for this question, but maybe it's still of some interest with an amended "Kunneth" requirement...
Perhaps an observation that might be useful is that the Künneth formula holds "on the nose" (i.e. $E_\ast X\otimes_{E_\ast} E_\ast Y\cong E_\ast(X\wedge Y)$) whenever $E_\ast X$ is a flat $E_\ast$-module (this doesn't even require the Künneth spectral sequence to exist!).
|
2025-03-21T14:48:31.253721
| 2020-06-14T16:07:05 |
363053
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gabe Goldberg",
"https://mathoverflow.net/users/102684"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630143",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363053"
}
|
Stack Exchange
|
For which ordinals do we have $V_\alpha = L_\alpha$?
Some elements of $L$ become constructible only in levels higher than its rank level. So I ask:
Let $V$ be such that $V = L$.
For which ordinals $\alpha$ do we have $V_\alpha = L_\alpha$?
Indeed, we have this for $\omega$ and if $\alpha$ is the class of all ordinals. But do we have this for other types of ordinals? Can we, for instance, have $V_\alpha = L_\alpha$ for a singular cardinal?
Yes; in fact the first $\alpha>\omega$ with $V_\alpha=L_\alpha$ has cofinality $\omega$. To obtain this $\alpha$, define $f:Ord\to Ord$ by $f(\xi)=$ the smallest $\eta$ such that $V_\xi\subseteq L_\eta$. Such an $\eta$ exists because of the assumption that $V=L$. Write $f^n$ for the $n$-fold iterate of $f$. Then $\sup_{n\in\omega}f^n(\omega+1)$ is the desired $\alpha$.
More generally, the ordinals you asked about are exactly the fixed points of $f$. Since $f$ is a normal function (increasing and continuous), its fixed points constitute a closed unbounded class of ordinals.
More explicitly, $V_\alpha = L_\alpha$ if and only if $\alpha \leq \omega$ or $\alpha$ is an $\aleph$-fixed point, i.e., $\aleph_\alpha = \alpha$.
To see this, assume $V_\alpha = L_\alpha$.
Obviously $\alpha\leq \aleph_\alpha$ (for all ordinals $\alpha$).
On the other hand, $|L_\alpha| = \omega\cdot |\alpha|$ (by induction) while $|V_\alpha| = \aleph_\alpha$ (by GCH and induction) so if $V_\alpha = L_\alpha$, then $\aleph_\alpha = \omega\cdot |\alpha|$, which implies $\alpha \leq \omega$ or $\aleph_\alpha = \alpha$.
I guess I should also say that if $\aleph_\alpha = \alpha$, then $V_\alpha\subseteq L_\alpha$ since $L_\alpha$ is closed under powersets by the condensation lemma.
|
2025-03-21T14:48:31.253875
| 2020-06-14T16:26:41 |
363055
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex B.",
"Gerry Myerson",
"KConrad",
"Stanley Yao Xiao",
"Vincent Granville",
"dohmatob",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/140356",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/3272",
"https://mathoverflow.net/users/35416",
"https://mathoverflow.net/users/78539"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630144",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363055"
}
|
Stack Exchange
|
Goldbach conjecture and other problems in additive combinatorics
The field is also known as additive number theory. I am interested in sums $z=x + y$ where $x \in S, y\in T$, and both $S, T$ are infinite sets of positive integers. For instance:
$S = T$ is the set of primes (leading to Goldbach's conjecture)
$S$ is the set of squares and $T$ is the set of primes, leading to the deeper Hardy
and Littlewood's Conjecture $H$, see my previous question here
A possible approach to check if $S+T = \{x+y, x\in S, y \in T\}$ covers all sufficiently large integers is as follows.
Define $N_S(x)$ as the number of elements in $S$ that are smaller or equal to $x$, and $N_T(y)$ as the number of elements in $T$ that are smaller or equal to $y$. The $n$-th element of $S$ is $N_S^{-1}(n)$, and $n$-th element of $T$ is $N_T^{-1}(n)$. The number $r(z)$ of solutions to
$$N_S^{-1}(x) + N_T^{-1}(y) \leq z$$ is asymptotically
$$r(z) \sim \int_0^{N_S(z)} N_T(z-N_S^{-1}(x)) dx.$$
The number $t(z)$ of ways that an integer $z$ can be written as $x+y$ with $x\in S, y\in T$ is thus
$$t(z) = r(z) - r(z-1) \sim \frac{dr(z)}{dz}$$
as $z$ becomes larger and larger. So in order to prove that for $z$ large enough, $z$ is the sum of an element of $S$ and an element of $T$, one "only" has to prove that $t(z) > 0$ for $z$ large enough.
Question
Is it possible to solve this problem using extremely precise approximations in all the asymptotic derivations discussed here? For instance, if $S$ is the set of prime numbers, then $N_S(z) \sim z/\log z$ and $N_S^{-1}(z)=z\log z$, but this is not precise enough to prove that every large enough even integer is the sum of two primes. You need far better approximations. Likewise, if $S$ is the set of squares, then $N_S(z) \sim \sqrt{z}$ and $N_S^{-1}(z)=z^2$, but this is not enough to prove that every large enough non-square integer is the sum of a square and a prime.
One issue is with the integral, which is only the first term in an Euler - Maclaurin series expansion to approximate $r(z)$. You need to use more than just the first term. If $S=T$ are the sets of squares, rather precise formulas are available for $r(z)$: see the Gauss-circle problem, here (Wikipedia) and here (MSE).
Another question is whether my method is equivalent to the circle method.
Note
Besides $N_S(x), N_S^{-1}(x), N_T(y), N_T^{-1}(y), r(z), dr(z)/dz$, another quantity of interest is the probability for an integer $z$ to belong to $S$: it is defined as $dN_S(z)/dz$, for instance, equal to $1/\log z$ if $S$ is the set of primes.
Illustration
When $S$ is the set of squares and $T$ the set of primes, I made all the computations in my previous question: see here. I also added a lot of new material recently, for instance: among the first 750,000 integers, $z=78754$ is the last one to admit only one ($r(z) = 1$) decomposition as $z=x^2+y$ with $x$ integer and $y$ prime. That is, if $z>78754$ then $r(z) > 1$. Likewise:
$z=101794$ is the last one with $r(z) =2$
$z=339634$ is the last one with $r(z) =3$
$z=438166$ is the last one with $r(z) =4$
$z=383839$ is the last one with $r(z) =5$
The sequence of $z$'s with $r(z)=1$ is listed at the bottom of my previous question, see here. I searched for this sequence to see if it had been discovered, but could not find any reference.
Conclusion
If my approach (assuming it is new!) ever leads to a proof of some famous conjectures, the proof will be very technical, difficult and long. It is beyond my reach but some mathematicians with experience dealing with extremely precise (second- or third-order approximations) to the asymptotics involved might have an answer about the feasibility of my approach. Just to give an idea of the many problems, it may require excellent asymptotics about a function more complex than the Lambert function (again, this briefly outlined in my previous question).
Maybe the following is true for sums of two primes and sums of a prime and and a square: there only finitely many $z$'s that can be expressed as $z=x+y$ in less than $k$ different ways, with $x\in S, y \in T$, regardless of $k$. This would imply that all but a finite number of $z$'s can be expressed as the sum in question.
What is "dx" ? What do you mean by "$r(z)$ is asymptotically ...." ? What is going to infinity in that statement ?
It seems to me that you have outlined something similar to the Hardy-Littlewood circle method, which was discovered over 100 years ago. It is well-known that if you can approximate certain integrals over certain subsets of the unit circle (the "minor arcs") then you can solve pretty much any problem in additive number theory. The problem is, we do not know how to approximate these integrals well enough in the 'binary' case (i.e., binary Goldbach conjecture). In three or more variables this can be done, and was already done by Vinogradov for the ternary Goldbach problem 80 years ago.
@dohmatob: $z$ going to infinity; and $dx$ the standard notation for an integral, as integration is over $x$.
This reasoning is not complete, asymptotics do not tell you everything. A counterexample is $S=T={\text{even integers}}$.
@Alex: yes of course. There are implicit assumptions of "independence" in my reasoning, and I was well aware of your counter-example. $S$ = multiples of $p$, $T$ = multiples of $q$ have no such issue if $p, q$ are co-prime. Even square + prime has to a much smaller extent, the same issue: you must discard all $z$ that are a square, but their density is zero, so impact is non-existent in some way.
Maybe the following is true for sums of two primes and sums of a prime and and a square: there only finitely many $z$'s what can be expressed as $z=x+y$ in less than $k$ different ways, with $x\in S, y \in T$, regardless of $k$. This would imply that all but a finite number of $z$'s can be expressed as the sum in question.
It seems what you are asking is "If we have a precise asymptotic for the number of elements of a set, can we solve binary additive problems involving that set?"
The answer in general seems to be `no'. Let's consider Goldbach's conjecture that every large integer $n$ is the sum of two primes. It is not hard to see from pigeonholing that the typical $n$ will have at most $O( n / \log^2 n)$ solutions to $n=p+q$ within the primes. In fact, classical sieve theory easily establishes a uniform upper bound of this form unconditionally.
Now pick a rapidly increasing sequences of numbers $n'$ and remove from the set of primes those primes arising in solutions to $n'=p+q$ for that given $n'$. For each $n'$ we have removed at most $O(n' / \log^2 n')$ elements from the full set of primes, and so the asymptotic of the counting function of our set has not changed, however the assertion that every large integer is the sum of two elements from our modified set is now false.
You might object that my modified set of primes will not satisfy the more precise asymptotics (with error terms) that hold for the primes, such as the consequences of the (Generalized) Riemann Hypothesis or the Elliott-Halberstam conjectures. And this is true. However, there has been a lot of effort put into trying to deduce solutions to additive problems conditional on these conjectures, and even assuming these conjectures there is no known proof of either of the two famous additive problems (Goldbach and twin primes). Indeed there is an obstruction related to the "parity problem" in sieve theory which also enters the picture.
This does give rise to the following interesting question, which I do not know the answer to:
Does there exist a set of integers that satisfies the asymptotic behavior of the primes in arithmetic progressions (with the error term implied by GRH), but which fails to satisfy weak Goldbach?
A negative answer to this question would fairly conclusively yield a negative answer to your question.
The formula
$$r(z) \sim \int_0^{N_S(z)} N_T(z-N_S^{-1}(x)) dx$$
can be re-written in a more appealing way. With the change of variable $u=N_S^{-1}(x)$ it becomes
$$r(z) \sim \int_0^{z} N_T(z-u)N'_S(u) du,$$
where $N'_S(u)$ is the derivative of $N_S(u)$ with respect to $u$. With an additional change of variable $u=zv$ it becomes
$$r(z) \sim z\int_0^{1} N_T(z(1-v))N'_S(zv) dv.$$
Likewise
$$t(z) \sim r'(z) = \frac{dr(z)}{dz} =z\int_0^{1} N'_T(z(1-v))N'_S(zv) dv .$$
An interesting case is when $S=T$ and
$$N_S(u) \sim \frac{a u^b}{(\log u)^c}, \mbox{ with } 0<a, 0<b\leq 1, \mbox{ and } c \geq 0.$$
This covers sums of two primes ($a=1, b=1, c=1$) and sums of two squares ($a=1, b=\frac{1}{2}, c=0$). We have:
$$r(z) \sim \frac{a^2b z^{2b}}{(\log z)^{2c}}\cdot \int_0^1 (1-v)^b v^{b-1}dv =
\frac{a^2b z^{2b}}{(\log z)^{2c}}\cdot \frac{\Gamma(b)\Gamma(b+1)}{\Gamma(2b+1)}$$
$$r'(z) \sim \frac{2 a^2 b^2 z^{2b-1}}{(\log z)^{2c}}\cdot \int_0^1 (1-v)^b v^{b-1}dv =
\frac{2a^2 b^2 z^{2b-1}}{(\log z)^{2c}}\cdot \frac{\Gamma(b)\Gamma(b+1)}{\Gamma(2b+1)}$$
Notes
Solutions such as $z=x+y$ and $z=y+x$ count as two solutions: $(x,y)$
and $(y, x)$.
The asymptotic formula for $t(z) \sim r'(z)$, representing the number
of solutions to $z=x+y$ with $x\in S, y\in T$ is true only on
average, as $z$ becomes larger and larger. There may still be
infinitely many integer $z$'s for which $t(z)=0$ even if $r'(z)
\rightarrow\infty$ as $z\rightarrow\infty$.
We assume that the sets $S$ and $T$ are "well balanced", both for
small and large values. For instance, if you remove the first
$10^{5000}$ elements of $S$, the asymptotic formula for $N_S(u)$
remains unchanged, but this is likely to cause many formulas to
fail.
On some tests, I noticed that there are more solutions (on average)
to $z=x+y$ with $x\in S, y\in T$ (here $x, y, z$ are integers), if
$z$ is even.
If $S=T$ is the set of primes, some adjustments must be made because
the primes are not "well balanced", they are less random than they
seem (for instance the sum of two odd primes can not be an odd
number, but there are also more subtle issues). This is best
described in the Wikipedia entry about Goldbach's conjecture
(see section about heuristics).
To generate a set like $S$, one way is as follows. Use a random
number generator function $U$ returning independent uniform deviates on $[0,
1]$. If $U(k) < N'_S(k)$ then add the integer $k$ to the set $S$,
otherwise discard it. Do that for all integers.
For sums involving three terms, say $R+S+T$, you can proceed as follows: first work on $S'=R+S$ and derive all the asymptotics for $S'$ using the methodology proposed here. Then work on $S'+T$.
If there are singularities in the functions $N_S$ or $N_S'$, they
must be handled properly in the integral formulas, unless the
integrals are improper but converging.
Generalization of the formula
It also works if $S\neq T$. Say
$$N_S(u) \sim \frac{a_1 u^{b_1}}{(\log u)^{c_1}}, N_T(u) \sim \frac{a_2 u^{b_2}}{(\log u)^{c_2}}$$
with $0<a_1,a_2, 0<b_1, b_2 \leq 1$, and $c_1, c_2 \geq 0$. Then
$$r(z) \sim
\frac{a_1 a_2 z^{b_1 + b_2}}{(\log z)^{c_1+c_2}}\cdot \frac{\Gamma(b_1 +1)\Gamma(b_2+1)}{\Gamma(b_1 + b_2+1)}$$
$$r'(z) \sim
\frac{a_1 a_2 z^{b_1 + b_2 -1}}{(\log z)^{c_1+c_2}}\cdot \frac{\Gamma(b_1 +1)\Gamma(b_2+1)}{\Gamma(b_1 + b_2)}$$
In particular, it applies to sums of a square and a prime, see here. A generalization to sums of $k$ sets is discussed in my new MO question, here.
Concerning "publish here when double-checked": you posted a long question and two of the three answers were written by you. This kind of "work in progress" writing is more suited to a personal blog or webpage. Even if MO may generate more views than an individual's own page, the site is not meant to be a place for people to record their ideas for trying to solve famous open problems.
I will keep that in mind. I wish you can save a draft before publishing, maybe there is a way to do it that I am unaware of. In the short-term, I only have one more question to ask, it will be posted as a separate question.
If you want to save a post you wrote here, you can click "edit" to see the original .tex code again and then copy the code for the post to your own laptop, tablet, etc. That is what you can do to get copies of questions and answers. Comments you write in a comment box have a very limited editing time of just a few minutes, so if you want a copy of those after the editing period is closed on a comment then just copy/paste the text and fix anything that is copied over poorly (e.g., math code).
Eleven versions of an answer to your own question.
Here is a possible path to prove Golbach's and other conjectures in additive number theory, such as the deeper Hardy and Littlewood's $H$ conjecture (All but 21 non-square integers are the sum of a square and a prime). The idea is to try to prove a far deeper, more generic, and stronger result which is just a pure analytical result, not even connected anymore to number theory, in the same way that the roots of Riemann's function is a purely analytical problem that can solve many number theory problems.
It goes as follows. Note that Golbach can be rewritten as follows: each sufficiently large positive integer $z$ can be written as $z=(p-1)/2 + (q-1)/2$ where $p,q$ are odd primes.
Step #1: remove almost all primes from the set $S$ of primes, but still keep infinitely many of them. Just keep a tiny fraction of them and the conjecture (now much stronger) still remains true. The number of primes less than $z$ is $~ z / \log z$, and we are removing so many of them that the number of elements in $S$ that are less than $z$, after removing all these primes, is of the order $z^{2/3}$. In order to achieve this and keep $S$ "well balanced", keep only the primes closest to $z^{3/2}$, for $z=2, 3, 4, 5$ and so on. Thus the new set $S$ satisfies $N_S(x) \sim x^{2/3}$. Based on my above answer, on average each element of $S$ has still a growing number of solutions to $z= x+y$ with $x\in S, y\in S$, as $z$ is growing.
Step #2: now $S$ is "well balanced" (this concept still needs to be defined, this is the most difficult part of the problem), and any well-balanced set $S$ with $r'(z) \rightarrow \infty$ (this is the case here) satisfies the following conjecture ($w$ is an integer):
$$m(z) = \min_{w\geq z} t(w) \rightarrow \infty \mbox { as } z\rightarrow \infty.$$
That is, not only each integer $z$ can be represented as $z=x+y$ with $x\in S, y\in S$, in at least one way, but it actually can be represented in that form in an growing number of ways as $z$ increases.
Illustration
I created 50 different sets $S$ that satisfy the requirements of steps #2, with $N_S(x)\sim \frac{3}{2} x^{2/3}$. The blue curve is the average value of $t(z)$ on the Y-axis, with $z$ (an integer) between $2$ and $250000$ on the X-axis. The red curve represents the minimum $t(z)$ for each $z$ computed across the 50 sets. Even that minimum seems to be growing indefinitely.
Below is the source code to produce these charts. They come from the last part of the code, producing the text file Prob4.txt. It is written in Perl.
$N=500000;
$Nsamples=50;
$a=1;
$b=1/3;
$seed=50000;
srand($seed);
open(OUT,">prob.txt");
open(OUT1,">prob1.txt");
open(OUT2,">prob2.txt");
open(OUT3,">prob3.txt");
for ($sample=0; $sample<$Nsamples; $sample++) {
# -> use better rand generator?
%hash=();
$c=0;
for ($k=2; $k<$N; $k++) {
$r=rand();
if ($r < $a/($k**$b)) {
$hash{$k}=1;
$c++;
print OUT "$sample\t$c\t$k\n";
}
}
#-----------------
$max_z=-1;
@count=();
foreach $x (keys(%hash)) {
foreach $y (keys(%hash)) {
$z=$x+$y;
if ($z< $N) {
$count[$z]++;
if ($z>$max_z) { $max_z=$z; }
}
}
}
#------------------
$c=0;
@max=();
for ($k=2; $k<$N; $k++) {
$cn=$count[$k];
if ($cn eq "") { $cn=0; $count[$k]=0; }
$max[$cn]=$k; # largest z for which z = x + y has k solutions
$c+=$count[$k]; # cumulative count
print OUT1 "$sample\t$k\t$cn\t$c\n";
}
for ($k=0; $k<40; $k++) {
print OUT2 "$sample\t$k\t$max[$k]\n";
}
#-------------
#compute largest z for which z = x + y has k or fewer solutions
#
print "sample: $sample -- max: $max_z\n";
$min=999999999;
for ($k=$max_z; $k>1; $k--) {
if ($count[$k]< $min) { $min=$count[$k]; }
if ($k < $N/2) { print OUT3 "$sample\t$k\t$min\n"; }
}
}
close(OUT);
close(OUT1);
close(OUT2);
close(OUT3);
--------------------------------------------------------------------
# post analysis
@amin=();
@minmin=();
for ($k=2; $k<$N/2; $k++) { $minmin[$k]=999999999; }
open(IN,"<prob3.txt");
while ($i=<IN>) {
$i=~s/\n//g;
@aux=split(/\t/,$i);
$sample=$aux[0];
$k=$aux[1];
$min=$aux[2];
$amin[$k]+=$min;
if ($min<$minmin[$k]) { $minmin[$k]=$min; }
}
close(IN);
open(OUT,">prob4.txt");
for ($k=2; $k<$N/2; $k++) {
$avg=$amin[$k]/$Nsamples;
print OUT "$k\t$avg\t$minmin[$k]\n";
}
close(OUT);
A very interesting reference
In a paper by Andrew Granville, published in Project Euclid (see here) one can read the following:
This is very similar to what I discuss here. While Andrew comes up with $N_S(x) \propto \sqrt{x \log x}$, a stronger result than my $N_S(x) \propto x^{2/3} $ since he needs even fewer primes than me, it is using the same idea that you only need to work with a tiny subset of all primes to prove Goldbach. His argument is probabilistic thus not a proof, mine is non-probabilistic but I don't end up with a proof either. Note that my $x^{2/3}$ can be reduced to $x^\mu$ for any $\mu> \frac{1}{2}$, based on the results featured in my previous answer: that way, we continue to have $r'(z)\rightarrow\infty$ as $z\rightarrow \infty$, which is what we need. If you use $\mu = \frac{1}{2}$, it won't work: $r'(z)=\frac{\pi}{4}$ is a constant, and the primes left are just as rare as square integers. We all know sums of two squares do not cover all integers, but only a very small set of integers, of density zero.
If Andrew Granville had used the more profound law of the iterated logarithm (rather than the central limit theorem), he would probably have ended up with a formal proof of the following result: the density of even integers $z$ that can not be written as $z=x+y$ with $x, y$ belonging to his very small subset of primes, is zero. I believe this statement may have been formally proved already, if you consider the set of all primes, rather than a tiny subset of primes.
Note: Maybe an example of a well-balanced set $S$ is one where the gaps between successive elements is a monotonic (increasing) function. I guess we can make that happen for the tiny subsets of primes needed to prove Goldback, with $\mu=\frac{2}{3}$. However, well-balanced also requires some congruence features. For instance, if all elements of $S$ are odd, then $S$ can not be well-balanced.
|
2025-03-21T14:48:31.254950
| 2020-06-14T16:37:51 |
363058
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexander Chervov",
"HenrikRüping",
"https://mathoverflow.net/users/10446",
"https://mathoverflow.net/users/3969"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630145",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363058"
}
|
Stack Exchange
|
Eigenvectors of a symmetric sum of tensor products
Let $A$ and $B$ be two (finite-dimensional) Hermitian matrices and $n$ be a positive integer. We define the matrix
$$
L_i = A\otimes \dots\otimes A\otimes B\otimes A\otimes \dots\otimes A~,
$$
where there are $n$ factors and $B$ is the $i$-th factor (with $1 \leq i \leq n$). We then define the "symmetric sum"
$$
L = \sum_{i=1}^n L_i~.
$$
Question: Can we say anything about the eigenvectors of $L$ knowing the eigenvectors of $A$ and $B$?
This question seems simple (and maybe it is) but it has resisted my best attempts. If $A$ and $B$ commute, this is trivial so I am interested in any case where they don't commute. I am not necessarily asking for a general solution of the problem and would be interested to see any (non-commuting) example. Even the case where $A= \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} $ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ was too complicated (the constraint satisfied by the eigenvectors is hard to solve). I feel like some form of this problem should have appeared before and that's why I am asking here!
For some background, this problem arises in quantum information theory, when trying to determine the optimal quantum measurement that distinguishes two states, involving $n$ copies of the system. Note that for this application, we have $\mathrm{Tr}\,B = 0$.
Thanks a lot for your help!
Edit: I don't have enough reputation to comment but wanted to mention that I am interested in the limit where $n$ becomes large.
For $n=2$ I got the eigenvalues $0,0,\pm \sqrt{2(a^2+b^2)}$ in your example.
For n=2, and difference, not sum - see some remarks here: https://math.stackexchange.com/questions/146259/deta-otimes-b-b-otimes-a-0-why-why-rkm-n2-n-why-x-and-x-i may be @David E Speyer argument can be modified..
|
2025-03-21T14:48:31.255110
| 2020-06-14T18:38:35 |
363061
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Claudio Rea",
"Mateusz Kwaśnicki",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/152651",
"https://mathoverflow.net/users/25510"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630146",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363061"
}
|
Stack Exchange
|
Subharmonic functions with reasonably non absolute continuous laplacian
Does it exist a compact supported measure $\mu$ in the plane $\Bbb R^2$ with the following two properties?
1) Points have $\mu$-measure zero,
2) If $\Delta u=\mu$, then the polar set of $u$ has positive $\mu$-measure.
($\Delta$ is the laplacian and the polar set of $u$ is the set of the points where $u=-\infty$).
NOTE
a) Condition 2) is independent of the choice of $u$ because the polar set does not change by addition of a harmonic function.
b) $\mu$ cannot be absolutely continuous with respect to the Lebesgue measure because polar sets of subharmonic functions have zero Lebesgue measure.
If $\mu$ is a uniform measure on the standard (ternary) Cantor set $C$, then the logarithmic potential of $\mu$ is infinite on $C$ — I think this answers your question?
An easier example is the equilibrium measure of the generalized Cantor set of zero capacity (see the book of Nevanlinna, Analytic functions, or this article.
To Mateusz: by uniform measure you mean that the measure of a ball only depends on the radius?
To Alexander. My computer doesn’t open the core article. Could you tell me the reference or, if it is short, just send to<EMAIL_ADDRESS>?
|
2025-03-21T14:48:31.255229
| 2020-06-17T08:01:48 |
363312
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Oscar Randal-Williams",
"https://mathoverflow.net/users/318"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630147",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363312"
}
|
Stack Exchange
|
The homotopy type of the simplicial space obtained by free adding degeneracies to a semi-simplicial space
Let $\text{sTop},\text{ssTop}$ denote the categories of simplicial, semi-simplicial spaces respectively. There is a functor $E:\text{ssTop}\rightarrow \text{sTop}$ that is left adjoint to the forgetful functor $E:\text{sTop}\rightarrow \text{ssTop}$.
Given a semi-simplicial space $X_{\bullet}$, $EX_{\bullet}$ is the simplicial space with $p$-simplices $$EX_{p}:=\coprod_{\beta :[p]\twoheadrightarrow [q]}(X_{q},\beta ),$$ where $[n]$ denotes the linearly-ordered set $\{0<1<\cdots < n\}$.
The simplicial data of $EX_{\bullet}$ is as follows. Let $\gamma :[r]\rightarrow [p]$ be a map of finite linearly ordered sets. For a surjection $\beta: [p]\twoheadrightarrow [q]$, factor $\beta\circ \gamma : [r]\rightarrow [q]$ as the composition $\gamma_{1}\circ \beta_{1}: [r]\rightarrow [q]$, where $\beta_{1}:[r]\twoheadrightarrow [s]$ is a surjection and $\gamma_{1}:[s]\hookrightarrow [q]$ is an injection, and define $\gamma : EX_{p}\rightarrow EX_{r}$ on $(X_{q},\beta )$ to be the map $\gamma_{1}^{*}:(X_{q},\beta )\rightarrow (X_{s}, \beta_{1})\subset EX_{r}$ (more details are given on page 3 of Ebert and Randal-Williams' Semi-simplicial spaces, arXiv:1705.03774).
Is it known if (or under which conditions) the thick geometric realization $\Vert EX_{\bullet}\Vert$ and the thin geometric realization $\vert EX_{\bullet}\vert$ are weakly equivalent? If the $p$-simplices $X_{p}$ are empty for $p>2$, then the thick and thin geometric realizations are weakly equivalent by directly checking that the degeneracy maps $s_{i}:EX_{p}\rightarrow EX_{p+1}$ are cofibrations for all $i$ and $p$ (in this case, $EX_{\bullet}$ is what Segal calls a "good" simplicial space, so its thick and thin geometric realizations are weakly equivalent).
The degeneracy maps are always just the inclusion of some terms in the disjoint union, so are cofibrations. So $EX_\bullet$ is always good.
|
2025-03-21T14:48:31.255365
| 2020-06-17T08:41:19 |
363316
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mikhail Borovoi",
"https://mathoverflow.net/users/4149"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630148",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363316"
}
|
Stack Exchange
|
Books on integration on semisimple Lie groups
Can anyone suggest me some good books where I can find integration theory on semisimple Lie groups (using KAK, KAN and other type of decompositions)?
I have read Knapp's book "Lie groups beyond introduction". He discusses integration formula for Iwasawa decomposition. I want to know all the integration formulas for semisimple Lie group, and I need a book built up from very basic theory of Lie groups and Lie algebras.
Helgason's book "Groups, geometry and Analysis " has this. But I want some book where things start from more basic treatments. It seems to me that in order to read "Groups, geometry and Analysis" I need to read Helgason's first book also. Till now I am quite comfortable with Lie group and Lie algebra theory as Knapp treats it, which do not require any Riemannian geometry.
Maybe you have no choice but to look through Helgason's book "Differential Geometry, Lie Groups, and Symmetric Spaces". I think you can read about the structure of semisimple groups and skip Riemannian geometry, if you don't want to learn it now.
|
2025-03-21T14:48:31.255577
| 2020-06-17T09:38:50 |
363317
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jugendtraum",
"Mikhail Katz",
"Moishe Kohan",
"https://mathoverflow.net/users/159748",
"https://mathoverflow.net/users/28128",
"https://mathoverflow.net/users/39654"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630149",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363317"
}
|
Stack Exchange
|
Systole of Riemann surfaces of genus $g$
In Buser and Sarnak's "On the period matrix of a Riemann surface
of large genus", we get
$$\frac4{3}\le\limsup_{g\rightarrow\infty}\frac{\max\{\operatorname{sys}(S)|S\in\mathcal{M}_g\}}{\log g}\le 2$$
Here $\operatorname{sys}(S)$ is the shortest length of closed noncontractible geodesics in $S$, and $\mathcal{M}_g$ is the moduli space. However, I want to ask if there exists a constant $C>0$ such that
$$\liminf_{g\rightarrow\infty}\frac{\max\{\operatorname{sys}(S)|S\in\mathcal{M}_g\}}{\log g}\ge C\quad ?$$
Any help will be appreciated.
Yes, there is such $C$: It boils down to constructing for each $b$ a connected 3-valent graph of girth $\ge C \log(b)$ and with the 1st Betti number $b$ (the number $b$ is your genus $g$). The existence of such graphs is usually proven by a simple counting argument (a probabilistic argument). I will write a proof when I have more time.
@MoisheKohan Thanks! I known the latter result, but why It boils down to constructing such trivalent graph? Does it related to gluing pair of pants along trivalent graph?
Not quite: Using pairs of pants would give a surface of small systole (cuffs of the pants are uniformly bounded). Instead, one glues together ribbon graphs which leads to a hyperbolic surface with boundary. Then one "doubles" such a surface "with a half-twist" along each boundary component.
@MoisheKohan Sorry, I'm a little confused. Why It boils down to constructing such trivalent graph?
Sorry, currently I just do not have time. I will come back to this in about a week, unless somebody writes an answer which would be just fine as well.
@MoisheKohan Thanks! BTW I think it's better if you give me a reference:)
@MoisheKohan , Note that strictly speaking, Buser-Sarnak are dealing with the homology systole rather than the homotopy systole. But a paper by Parlier saves the day.
@MoisheKohan, we proved the result (see my answer) but it's considerably more complicated than the argument you sketched. It would be nice to have a simpler proof, perhaps with a better constant.
Stephane Sabourau and I proved a lower bound for the liminf over the genus. The lower bound is approximately $\frac17$. The paper is on the arxiv: http://arxiv.org/abs/2407.02041 and is due to appear in the Proceedings of the AMS.
|
2025-03-21T14:48:31.255750
| 2020-06-17T09:56:56 |
363320
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Math Lover",
"Matthew Daws",
"https://mathoverflow.net/users/129638",
"https://mathoverflow.net/users/406"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630150",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363320"
}
|
Stack Exchange
|
Question about a paper on approximate identities
I am currently reading this paper on approximate identities of ternary Banach algebras. Assume that $(A, [.,.,.])$ is a ternary Banach algebra. A bounded net $(e_{\alpha}, f_{\alpha})$ is said to be left-bounded approximate identity for $A$ if $\lim_{\alpha}[e_{\alpha}, f_{\alpha},a]=a$ for all $a \in A$. Can someone clarify my following questions on the same paper:
What is the definition of bounded net? Also at many places in the same paper for instance in Theorem $2.2$, the product $e_{\alpha}f_{\alpha}$ is used. What is the meaning of product in ternary algebra?
All these terms are not explained in paper. Thank you very much in advance!
P.S: Copy of the same question On MSE can be found here.
Having look briefly at the paper you link to, I also cannot make sense of it, because as you say, they repeatedly seem to use a normal "product" on a ternary Banach algebra, which is of course not defined. For example, I cannot find a meaning for equation (2.1)
@MatthewDaws, exactly it’s not clear what they really meant.
Cross-posted on Math.StackExchange and now has a very interesting answer...
|
2025-03-21T14:48:31.255865
| 2020-06-17T10:13:04 |
363322
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Derek Holt",
"Geoff Robinson",
"Louis ",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/158239",
"https://mathoverflow.net/users/35840"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630151",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363322"
}
|
Stack Exchange
|
Split extension of finite group and Sylow subgroup by abelian $p$-group
Let $p$ be a prime and let $A$ be an abelian normal $p$-subgroup of a finite group $G$. Hence, for any Sylow $p$-subgroup $H$ of $G$, it holds that $A$ is contained in $H$ and so $A$ is a normal subgroup of any Sylow $p$-subgroup of $G$. So now fix a Sylow $p$-subgroup $P$ of $G$.
How to show that the following two conditions are equivalent:
(1) There is a split exact sequence $1\to A \to G \to G/A \to 1$ .
(2) There is a split exact sequence $1\to A \to P \to P/A \to 1$ .
?
My try: I think I can prove (1) implies (2). Indeed, if (1) holds then there is a subgroup $B$ of $G$ such that $B\cong G/A$ and $G=BA$ and $B\cap A=1$. So then $P=P\cap (BA)=(P \cap B)A$ , and we moreover have $(P\cap B)\cap A \subseteq B \cap A=1$. Thus $P$ is a semidirect product of $A$ and $P\cap B\cong P/A$ . Thus the extension $1\to A \to P \to P/A \to 1$ splits. Is this correct ? And I have no idea how to prove (2) implies (1).
Please help.
What you said so far was OK. You are trying to prove a known theorem of Gaschutz. I don't know a way to do the other implication without some use of group cohomology.
@Geoff Robinson: Thank you very much for your comment , and I'm okay to use Group Cohomology (as I have also given the tag ...) ... could you please detail out a proof or give any reference ? Thanks again
I think the right reference is at MR0051226. The original paper is in German, J Reine Angew Math 190 (1952) 93-107
@Geoff Robinson: unfortunately, I don't know any German ... could you please provide a proof as an answer ? Thanks
This is an old and well-known result, and it also follows easily from basic results on the cohomology of finite groups, which you can find in the classic text by Cartan and Eilenberg on Homological Algebra. The restriction map $H^k(G/A,A) \to H^k(P/A,A)$ is injective for all $k > 0$. So I don't think it is really a suitable question for MO.
There are proofs that do not use the language of cohomology explicitly. The extension splits if and only if the identity map on $A$ extends to a crossed homomorphism $G \to A$, the kernel of which is a complement. There is a short slick proof in which the existence of such a crossed homomorphism $P \to A$ is used to prove that it exists for $G \to A$ (although it is not true that every such map $P \to A$ extends to $G \to A$). I don't know a reference for that immediately.
I have found a reference: (10.4) in Aschbacher's book, "Finite Group Theory".
@Derek Holt: thanks for the awesome reference ... could you slightly elaborate the point why the restriction map $H^2(G/A, A)\to H^2(P/A, A)$ is injective ? Is there any references for this ?
It's Theorem 10.1 in Chapter XII of Cartan and Eilenberg, Homological Algebra.
|
2025-03-21T14:48:31.256074
| 2020-06-17T10:26:45 |
363325
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630152",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363325"
}
|
Stack Exchange
|
Finding a r-atomic solution to the univariate truncated Hausdorff moment problem
Suppose i have a certain $t>0$, and observations of moments of a random variable $X$ given by $\mu_0=1,\mu_1,...,\mu_n$.
How can i:
Check that a measure with support $[0, \frac{1}{t}]$ with thoses $n$ first moments indeed exists
Construct an atomic (discrete) solution $\nu(x_i) = \omega_i$ for some atoms $x_i$ and associated probabilities $\omega_i$, such that: $$\forall i \in 1,...,r, \;x_i \in [0,\frac{1}{t}] \text{ and }{\omega_i >0}, \; \text{ and } \sum_{i=1}^{r} \omega_i = 1$$
I found a lot of things on moment problems in general, and on the Hausdorff moment problem, but i have trouble finding the right solution to this quick problem.
By rescaling, without loss of generality, $t=1$.
Let $c:=(c_0,\dots,c_n)$, where $c_p:=\mu_p$ (so that $c_0=1$). Let $M(c)$ denote the set of all (probability) measures $\nu$ on $[0,1]$ with moments $\int x^p\,\nu(dx)=c_p$ for all $p\in\{0,\dots,n\}$. A necessary and sufficient condition for $M(c)\ne\emptyset$ is well known; see e.g. Karlin--Studden, Ch. IV, Theorem 1.1.
The set $M(c)$ of measures is compact and convex, and hence $M(c)$ has an extreme point if $M(c)\ne\emptyset$. Also, by elementary linear algebra, any extreme point of $M(c)$ is a measure supported on at most $n+1$ points. Then a solution $(x,\omega)=(x_1,\dots,x_r,\omega_1,\dots,\omega_r)$ to your problem exists for some natural $r\le n+1$.
Finding such a solution can be reduced to semidefinite optimization; see e.g. part (c) of Proposition 3.1.
|
2025-03-21T14:48:31.256202
| 2020-06-17T11:07:53 |
363327
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Astro",
"Iosif Pinelis",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/75659"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630153",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363327"
}
|
Stack Exchange
|
Relating maximum/minimum value and variance of data points
Given the expression
$T = \max(|\beta_{\max}-1|,|\beta_{\min}-1|)$
Can we relate $T$ to variance $\sigma^2$ of values $\beta_i^2$, assuming we know $\sigma^2$, $\beta_{\max}$ and $\beta_{\min}$.
PS: What if values $\beta_i^2$ are distributed with mean $\mu=1$.
Let $b_i:=\beta_i$. Then $1-T\le b_i\le1+T$ for all $i$, and these inequalities provide the best bounds on the $b_i$'s given the information we have. So,
$$A\le b_i^2\le B \tag{1}$$
for all $i$, where
$$A:=\max[0,1-T]^2,\quad B:=(1+T)^2,$$
and inequalities (1) provide the best bounds on the $b_i^2$'s given the information we have.
So, for the variance $\sigma^2$ of the $b_i^2$'s we have
$$0\le\sigma^2\le(B-\mu)(\mu-A), \tag{2}$$
where $\mu$ is the mean of the $b_i^2$'s. Moreover, inequalities (2) provide the best bounds on $\sigma^2$ given the information we have.
The latter two statements follow from
Lemma: Let $X$ be any random variable such that $EX=\mu$ and $A\le X\le B$ for some real $A$ and $B$. Then for the variance $\sigma^2$ of $X$ we have
$$0\le\sigma^2\le(B-\mu)(\mu-A). \tag{3}$$
Moreover, inequalities (3) provide the best bounds on $\sigma^2$ in the conditions of this lemma.
Proof. Let $Y:=X-\mu$. Then $EY=0$ and $-a\le Y\le b$, where $b:=B-\mu$ and $a:=\mu-A$. So,
$$0\le E(b-Y)(Y+a)=ba-EY^2=ba-\sigma^2.$$
So,
$$\sigma^2\le ba=(B-\mu)(\mu-A),$$
so that (3) holds. Moreover, $\sigma^2=ba=(B-\mu)(\mu-A)$ if $Y$ takes values $-a,b$ with probabilities $\frac b{a+b},\frac a{a+b}$, respectively, that is, if $X$ takes values $A,B$ with probabilities $\frac b{a+b},\frac a{a+b}$, respectively.
@losif Thanks! Result in (2) is nice! But solving it for T, so as to express T in terms of $\sigma^2$ looks tricky because of max() involved. Any tips on that?
@Astro : If you want to solve the equation $(B-\mu)(\mu-A)=C$ for $T$ (for a given $C$), consider the two cases, $T\le1$ and $T>1$, separately. In the second case, the equation $(B-\mu)(\mu-A)=C$ is quadratic, and in the first case it is of degree $4$. This is what you have, anyway.
If you know $\beta_{\max}$ and $\beta_{\min}$ then I would have thought you know $T$.
If you do not know $\beta_{\max}$ and $\beta_{\min}$ then you can say $T \ge \sigma$, with equality only when you have a distribution with half the probability at $1-\sigma$ and half the probability at $1+\sigma$ (and a mean of $1$)
|
2025-03-21T14:48:31.256364
| 2020-06-17T13:00:42 |
363330
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Karagila",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/7206"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630154",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363330"
}
|
Stack Exchange
|
Historical origin of the empty set
The question is in the title:
Who first claimed the existence / necessity of the empty set ? When did this happen ?
Of course I know that the notation $\emptyset$ goes back to André Weil, and that the first axiom of ZF is that there exists an empty set, but I ask whether this is an older concept.
The first step towards the invention of the empty set is the invention of zero, which is not yet universally used among mathematicians :)
"It can be justifiably argued that Boole had inventented the empty set" [in The Mathematical Analysis of Logic (1847)].
The Empty Set, the Singleton, and the Ordered Pair, Akihiro Kanamori (2003).
And on Aki's website: http://math.bu.edu/people/aki/8.pdf
|
2025-03-21T14:48:31.256463
| 2020-06-17T13:27:27 |
363334
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris H",
"R. van Dobben de Bruyn",
"Will Sawin",
"YCor",
"https://mathoverflow.net/users/128502",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/82179"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630155",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363334"
}
|
Stack Exchange
|
Is a group determined by the number of ways its elements multiply to the identity under some ordering?
Let $G$ be a group, and for each ordered $n$ tuple $(g_1,...g_n)$ of elements of $G$, consider the function $f_n$ that outputs the number of permutations $\sigma\in S_n$ for which $g_{\sigma(1)}g_{\sigma(2)}...g_{\sigma(n)}=e$.
Is the group determined up to isomorphism by these functions $f_n$?
More precisely, if we have a set $G$, and two collections of functions $f^1_n,f^2_n$ on $n$ tuples of elements of $G$, such that there exists group structures $\mu_1,\mu_2:G\times G\rightarrow G$, $\epsilon_1,\epsilon_2:\ast\rightarrow G$ on $G$ giving rise to $f^i_n$ in the manner described above, is there a bijection $X\rightarrow X$ which identifies the groups $(\mu_1,\epsilon_1)$, and $(\mu_2,\epsilon_2)$?
Note that we can recover identity from $f_1$, inverses from $f_2$, and by looking at $3$ tuples, we can partially deduce the group law, in that we can say when $gh=k$ or $hg=k$ holds for any $g,h,k\in G$. So in particular, this has an affirmative answer when $G$ is abelian. For the general case, I wasn't able to find a similarly elementary argument.
In the title "multiset" sounds weird, as for a multiset in a non-abelian group, to "multiply to 1" doesn't make sense. The first question is a bit unclear: is $n$ fixed? it seems not, after reading the sequel. Also in the second question you mean "unique up to isomorphism", or maybe up to permutation of $X$ preserving $f$.
Edited for clarity, I wasn't quite sure how to phrase the title, if you can think of better wording that would be very helpful.
Actually what you count is more precise, since you retain the number of permutations.
It's equivalent to know, for each $g$ and $h$, the multiset ${gh, hg}$. (And knowing that multiset is equivalent to knowing the set.) You'e shown that counting gives us this information. Conversely, if we have $n$ elements, and count all the ways to choose two of them, multiply them in any order, then choose another, and multiply it in any order, and so on, until all are chosen, we will count each ordering the same number $2^{n-2}$ of times. So to recover your information, we can sum over the possible pairwise products and then divide by $2^{n-2}$.
Some more things you can recover from the $f_n$: orders of elements, subgroups, when two elements commute, centralisers, centre, conjugacy classes, normal subgroups, .... I don't know enough group theory to know if a subset of these properties recovers the group (maybe assuming $G$ is finite), but this is all pretty strong evidence that at least $(G,\mu_1)$ and $(G,\mu_2)$ are abstractly isomorphic.
An example of what can happen is $(G,\mu_1) = H_+ \times H_-$ and $(G,\mu_2) = H_+ \times H_-^{\operatorname{op}}$ (which is abstractly isomorphic using $(g_+,g_-) \mapsto (g_+,g_-^{-1})$). A possible stronger question could be if it is always of this form. A more moderate strengthening is whether there exists an isomorphism that respects the $f_n$.
|
2025-03-21T14:48:31.256690
| 2020-06-17T13:40:53 |
363336
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben Smith",
"Gerry Myerson",
"KConrad",
"Will Sawin",
"https://mathoverflow.net/users/156215",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/3272"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630156",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363336"
}
|
Stack Exchange
|
What is the significance of Friedlander-Iwaniec and related theorems?
On p.177 of Number Theory Revealed: A Masterclass by Andrew Granville, the author states that "One can ask for prime values of polynomials in two or more variables." (though he later mentions Landau's $n^2+1$ conjecture so I'm not sure why single-variable polynomials are omitted). For example, there are some classical results, like Fermat's Christmas theorem and Euler's $6n+1$ theorem. More recent results include the Friedlander-Iwaniec theorem, which states that there are infinitely many prime numbers of the form $a^2+b^4$ for integers $a$ an $b,$ and the result of Heath-Brown on the infinitude of primes of the form $x^3+2y^3$ for integers $x$ and $y.$ Granville's book mentions other recent advances by Maynard and collaborators of Heath-Brown and Iwaniec.
Initially, I wondered why the Friedlander-Iwaniec theorem is considered to be a breakthrough, so I asked Why does the infinitude of primes of a certain form matter? on math.stackexchange, but it was closed for being "opinion-based". Granville's book cleared up a part of this question for me because it states that the Friedlander-Iwaniec theorem was the first example in which the polynomial is "sparse" in taking on integer values, and that later results of the same type based on norm forms were inspired by Friedlander-Iwaniec (although I don't know to what extent the proofs of the former are related to the proof of the latter).
My question is this: is the area of exploration quoted at the beginning of this post an island in that it is considered to be an inherently interesting question requiring no further justification, or are there external motivations for pursuing it? Some satisfactory motivations (this list is by no means exhaustive) might be that other problems reduce to it, that the techniques developed in the course of solving these problems are widely applicable elsewhere, real world applications (e.g., cryptography), or maybe these are instances of much more general problems for which "Mathematics is not yet ripe" in the words of Erdős.
I think questions about prime values of polynomials are considered inherently interesting.
All these questions are special cases of Bunyakovsky's conjecture, or, if you want, the Bateman-Horn conjecture. Certainly mathematics is not yet ripe for such problems.
You can also view them as special cases of the general problem of when special sequences of numbers are prime. One generally expects to see transfers of techniques from one special case of this problem to another, but not as much with other areas of number theory and mathematics. So to a large extent I think the results on primes-of-a-special-type problems that are considered most interesting are those that point to way to further solution of primes-of-a-special-type problems.
One could also note that primes of the form $a^2+b^4$ is a special case (in some sense) of primes of the form $a^2 +1$, since infinitely many primes of the form $a^2+1$ implies infinitely many primes of the form $a^2+b^4$. (Of course this is the reverse of the special case direction between the numbers themselves.)
I don't think many who work on these problems are motivated by applications to cryptography. As Ben Smith points out in the comments, there are some applications of Bateman-Horn to cryptography, where we would like to find primes represented by special polynomials for use in certain cryptographic algorithms. Conditional on not just the existence of such primes but the Bateman-Horn conjecture predicting how many there are in a large interval, we can generate such primes in polynomial time by randomly sampling values of the polynomial and primality testing.
However, since these conjectures are very very hard to prove, but we have very strong reasons to believe that they are true, working on them might not help with cryptography very much.
We might want to avoid these special primes for RSA, but primes represented by polynomials do have cryptographic applications, especially in pairing-based cryptography. See e.g. https://eprint.iacr.org/2006/372 for a survey of constructions: you'll see (univariate) polynomial-value primes appearing in several places to get efficient cryptographic parameters, and Bateman-Horn used to (conditionally) prove heuristic algorithmic running times.
@BenSmith Thanks! I edited my answer to account for this.
Suppose you suspect that among some subset $\mathcal{S}$ of the natural numbers, there ought to be infinitely many primes. How would you confirm that this is indeed the case?
Clearly, the "bigger" $\mathcal{S}$ is (measured in terms of natural density), the easier the question should be. The biggest set is of course $\mathbb{N}$ itself, and it was a highly non-trivial theorem of Euclid that confirmed this. What about the next most obvious "big" sets, the arithmetic progressions? Indeed, consider the set $\mathcal{S}(a; q)$ by
$$\displaystyle \mathcal{S}(a; q) = \{qx + a : x \in \mathbb{N}\}.$$
In the special case $q = 4, a = 3$ one can use Euclid's argument again to show that there are infinitely many primes in the set (but far from the correct density). A much more difficult argument shows that the case $q = 4, a = 1$ also gives infinitely many primes. But we expect that there are infinitely many primes in $\mathcal{S}(a; q)$ whenever $\gcd(a,q) = 1$. This is not known until the work of Dirichlet, almost 2000 years after Euclid!
The most extensive generalization of the theorem of Dirichlet known is the Chebotarev density theorem, which shows that in any number field there are infinitely many primes which split completely (over simplifying here).
So far, all of the sets considered have log density one. If we put $\mathcal{S}(X) = \# \{n \leq X : n \in \mathcal{S} \}$ then the log density is the infimum among all non-negative real numbers $\delta$ such that $\mathcal{S}(X) < X^{\delta}$ for all $X$ sufficiently large. Log density one sets can still be very very hard to count primes in: indeed, we expect the set of primes $p$ such that $p + 2$ is also prime to be infinite, and this set (if infinite) is expected to have log density one.
Thus, producing an infinite set of primes which have log density less than one is a formidable task. Both the Friedlander-Iwaniec theorem and Heath-Brown's theorem are examples where this can be achieved. These should be considered highly exceptional in the sense that similar, perhaps even easier looking polynomials are not amenable to their methods! Perhaps the most infamous is the polynomial $F(x,y) = 4x^3 - y^2$, which counts discriminants of elliptic curves in short Weierstrass form.
Describing Euclid's proof of the infinitude of the primes as "highly nontrivial" is rather excessive. By that measure, all proofs that involve a noncomputational idea are highly nontrivial.
@KConrad would you accept "highly nontrivial at the time"?
@Gerry Myerson I think it would be better not to use such a label in the answer. I’d call the work by Archimedes nontrivial (especially considering the nature of his results achieved without a good algebraic notation). Once I looked at the proof of infinitude of the primes in a copy of Euclid and was disappointed: it treats three primes as the “general case”. I understand their lack of good notation made it hard to describe the general case with general notation, but it was still disappointing.
|
2025-03-21T14:48:31.257197
| 2020-06-17T13:51:47 |
363337
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aurélien Djament",
"Leonid Positselski",
"Victor Ostrik",
"https://mathoverflow.net/users/2106",
"https://mathoverflow.net/users/4158",
"https://mathoverflow.net/users/76506"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630157",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363337"
}
|
Stack Exchange
|
Semisimple Abelian categories with infinite sums
A semisimple category is an abelian category in which every object is a finite direct sum of simple objects.
A) Why does one impose the finiteness condition here?
B) If one condsiders infinite direct sums does something go wrong?
C) If B) works with no problems, then is this equivalent to an abelian category where exact sequences split?
See https://mathoverflow.net/questions/327899/name-for-abelian-category-in-which-every-short-exact-sequence-splits/327944#327944
A) It depends on what you are interested in. If you do not impose the finiteness condition, then it means that you are describing a different class of abelian categories. Which class is that, depends on additional conditions which you may want to impose instead of finiteness of the direct sums.
B) Nothing goes wrong, but you have to make some decisions. Firstly, if you want every object to be a (possibly infinite) direct sum of simple objects, then it is natural to impose the condition that all (set-indexed infinite) direct sums exist in your category. Further, you may want the condition that isomorphism classes of simple objects form a set. Under the previous assumptions, this is equivalent to the condition that your category has a generator, or a set of generators.
Imposing these conditions allows you to describe precisely what the objects of your category are. To describe the morphisms in a natural way, you may want to impose a further condition that, for any simple object $S$ in your category $\mathcal A$, the functor $\operatorname{Hom}_{\mathcal A}(S,{-})\colon\mathcal A\to \mathcal Ab$ preserves infinite direct sums. Under the previous assumptions, this is equivalent to the condition that $\mathcal A$ satisfies the axiom Ab5, or in other words, that $\mathcal A$ is a Grothendieck abelian category (as we've already assumed that $\mathcal A$ has a generator).
Then your category $\mathcal A$ is equivalent to the Cartesian product, taken over some set $X$, of the categories $D_x{-}Mod$ of (possibly infinite-dimensional) modules/vector spaces over some division rings (skew-fields) $D_x$, $\,x\in X$.
It seems to be an open question whether the condition that $\mathcal A$ is Ab5 can be dropped (i.e., whether it follows from the conditions that $\mathcal A$ has infinite direct sums, every object is a direct sum of simple objects, and there is only a set of isomorphism classes of simple objects).
C) No, it is not equivalent. In the classical terminology going back to 1960's, a Grothendieck abelian category in which every short exact sequence splits is called "spectral". The term comes from functional analysis and suggests an analogy with the distinction between the discrete and continuous spectrum in the spectral theory of operators in a functional space.
A spectral category in which all objects are direct sums of simple objects is called discrete. A spectral category having no simple objects is called continuous. It is known that there are many nonzero continuous spectral categories.
On the other hand, I am not aware of any example of a category with a generator, with infinite direct sums, in which all short exact sequences split, but which is not Grothendieck.
References:
A related question was discussed on MO in Name for abelian category in which every short exact sequence splits
P. Gabriel, U. Oberst. Spektralkategorien und reguläre Ringe im von-Neumannschen Sinn. Math. Zeitschrift 92, #5, p.389-395, 1966.
B. Stenström. Rings of quotients. An introduction to methods of ring theory. Springer, 1975. Sections V.6-7 and XII.1-3.
L. Positselski, J. Šťovíček. Topologically semisimple and topologically perfect topological rings. Electronic preprint https://arxiv.org/abs/1909.12203, Section 2.
"On the other hand, I am not aware of any example of a category with a generator, with infinite direct sums, in which all short exact sequences split, but which is not Grothendieck.": taking the opposite category of any non-zero spectral Grothendieck category gives an example.
@AurélienDjament Yes, you are right. Thank you.
|
2025-03-21T14:48:31.257456
| 2020-06-17T14:45:41 |
363341
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Friedrich Knop",
"Nick Gill",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/801",
"https://mathoverflow.net/users/89948",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630158",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363341"
}
|
Stack Exchange
|
A group action on another group action quotient: how to best describe the resulting structure and does it have a name?
Suppose I have an action $\alpha:G\times X\to X$ of a group $G$ on a set $X$ and, on top of that, an action $\beta:H\times(X/G)\to(X/G)$ of another group $H$ on the set of $G$-orbits.
Is there a nice way to turn this into a single group-action-like thing on $X$?
Maybe more rigorously: is there some group $\Xi(G,H)$ and an action of it on $X$, maybe with some additional data, which would enable me to reconstruct $G$, $H$, $\alpha$ and $\beta$? Does such $\Xi(G,H)$ have a name? Are its properties studied?
I made some moves but I am not sure if they are in right direction. From $\alpha$ and $\beta$ I can make (using axiom of choice) a map $\gamma:H\times X\to X$ and maps $\delta:X\to G$ and $\varepsilon:H\times H\times X\to G$ such that $\gamma(1,x)=\alpha(\delta(x),x)$ and $\gamma(h_1,\gamma(h_2,x))=\alpha(\varepsilon(h_1,h_2,x),\gamma(h_1h_2,x))$. Clearly one may write down appropriate identities, but all this looks like overkill.
Later I realized that one piece of data is missing: a map $\beta:H\times G\times X\to G$ satisfying $\gamma(h,\alpha(g,x))=\alpha(\beta(h,g,x),\gamma(h,x))$.
It's a bit of a torture to my brain to understand "is there some group ?(G,H)" not as "is there some group? $(G,H)$" along with a typo on spacing.
@YCor Is this better?
I'd have liked just another letter (an unknown in math is usually denoted by some random letter, not a question mark, especially since the question mark is also used in math sentences for its usual meaning) but at least it debugs a bit (even if "[?]" sounds like missing ref in a laTeX draft...). One criterion for a good notation would be something you could read to an audience.
@YCor To an audience I would tell "something(G,H)". But do you have a suggestion? Say, $\Xi(G,H)$?
$\Xi(G,H)$ looks/sounds perfect!
Isn't that what the wreath product is for?
@FriedrichKnop How exactly? Do you mean $H^X\rtimes G$ acts on $X$ in such a way as to in particular recover the action of $H$ on $X/G$? Or you mean something entirely different?
Perhaps the "permutation direct product" would work? It's described in the opening paragraph of a 1982 paper of Bailey, Praeger, Rowley and Speed. On first glance it seems stronger than your set-up because, as I understand it, you aren't specifying exactly what $H$ does to the elements of $X$... But certainly such an action would allow you to recover the set-up you describe. I think!
@NickGill Although it certainly is a wreath product of sorts, I - sorry for my stupidity - still do not see how exactly to use it. Do you mean something like the action of $H^X\rtimes G$ on $(X/G)\times X$ given by$$(\eta,g)\cdot(\chi,x)=(\eta(x)\cdot\chi,g\cdot x)?$$
Sorry, I'm not sure that I understand the question: I'm suggesting you use a direct product rather than a wreath product. I, and some co-authors used the permutation direct product in this paper -- https://arxiv.org/pdf/2005.13869.pdf . It's described briefly in the middle of p.4. The explanation isn't as good as the one in the other paper I mentioned, but I give the link in case you can't access the other one.
I guess my suggestion will work provided the action of $H$ on $X/G$ preserves orbit sizes, i.e. $|\gamma|=|\gamma^h|$ for all $h\in H$ and $\gamma \in X/G$. Is this the case in the situation that you are considering? (If it isn't, then my instinct is that there is no very satisfying answer to the question... But I could be wrong.)
Well, thinking some more, if your $H$-action doesn't preserve orbit sizes, then I guess you could "pad out" sets in $X/G$ in the same orbit of $H$ with dummy elements, and then the underlying set in $X$ corresponding to each such $H$-orbit $\Lambda$ would admit a bijection to a direct product $\Lambda\times \lambda$ (where $\lambda \in \Lambda$) and you could apply the permutation direct product to that? Is that too convoluted though?
@NickGill Thanks for additional explanations. Could I create confusion with my notation $H^X$? I meant the set of all maps from $X$ to $H$ rather than fixed points (which do not make sense in this context).
As for your last comment - do you suggest to extend the action to the disjoint union $$\coprod_{\lambda\in X/G,\Lambda\in X/H,\lambda\in\Lambda}\Lambda\times\lambda?$$
|
2025-03-21T14:48:31.257870
| 2020-06-17T15:30:44 |
363345
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"YCor",
"https://mathoverflow.net/users/14094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630159",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363345"
}
|
Stack Exchange
|
Tangent cone of metric graph
I am starting to study some lecture notes about metric geometry and I would appreciate it if someone could some questions regarding the notion of the tangent cone.
Consider 3 half lines joined by their point of origin. You get a network of 3 roads and a junction point. This is an Aleksandrov space of non-positive curvature. It is even geodesically complete. Now, I am having trouble defining the tangent cone at this junction point:
Is it isometric or BiLipschitz to $R^k$ for some $k>0$ ?
What is the Hausdorff dimension of the space at this point ? is it 1 ?
is the junction point the boundary of the Aleksandrov space ?
These questions might be trivial so I apologize in advance but I just didn't find enough examples that talk about this.
It looks like an exercise. If $X$ is your space and $o$ the joining point (I assume the metric is the geodesic one), it's obvious that $X$ is isometric to its tangent cone at $0$. More generally, if $X$ is a proper metric space and and has a 1-parameter subgroup of non-isometric dilations fixing a point $o$, then every tangent cone of $(X,o)$ is isometric to $(X,o)$ as pointed metric space.
Loosely speaking, if you are standing at the origin then there are only three directions you can travel. So the space of directions $S_o$ is only 3 points. Taking the product of the space of directions with $[0,\infty)$ and identifying $S_o \times \{0\}$ to a point gives you the tangent cone, which in this case is isometric to your starting space.
If you are looking for resources on tangent cones, I would see Nikolaev's paper 'The tangent cone of an Aleksandrov space of curvature $\leq k$' or Halbeisen's 'On tangent cones of Alexandrov spaces with curvature bounded below'
|
2025-03-21T14:48:31.258033
| 2020-06-17T15:35:30 |
363346
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"JSch",
"LSpice",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/159768",
"https://mathoverflow.net/users/2383"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630160",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363346"
}
|
Stack Exchange
|
Algebraic geometry additionally equipped with field automorphism operation
I am looking for some facts on theory, which is essentially algebraic geometry but with field automorphisms added as 'basic' operations. (Precisely, I mean universal algebraic geometry for (universal) algebra being a field $\mathbb{F}$ and operations being polynomials and field automorphisms) I am also interested in computational aspects.
I am mainly interested in case of field of multivariate rational functions, i.e. $\mathbb{F} = K(x,y)$.
An algebraic set in such setting could be $$\{(f,g) \in (K(x,y))^2 \mid x^2-y^2-f \cdot g(x=\frac{x+y}{2}, y = \frac{x-y}{2}) = 0\}.$$ and ideals are additionally closed under field automorphisms.
I would be grateful for reference to any source that considers theory like this, also the computational aspects.
Do you get one distinguished automorphism, or the full automorphism group (say, over the prime field)?
A finite number of automorphisms (and their compositions), but one automorphism is already an interesting subcase.
You should look into difference algebra, which is exactly this setting. (It's called difference algebra because the most classical version looked at the automorphism on polynomials that sends $f(x)$ to $f(x+1)$, which can be used to state the theory of difference equations.)
The hottest area in this is the study of the model theory of fields with automorphisms. Zoe Chatzidakis has some lecture notes on the subject.
And conversely it seems to me that this is one of the most studied settings in model theory.
|
2025-03-21T14:48:31.258169
| 2020-06-17T16:48:24 |
363352
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Kashif",
"Piyush Grover",
"https://mathoverflow.net/users/30684",
"https://mathoverflow.net/users/62012"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630161",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363352"
}
|
Stack Exchange
|
Are there alternative regularizations for optimal transport problems besides entropic regularization?
I see that most of the regularization done involves an entropy term.
Has there been any work done on other regularization methods? In particular, I'm wondering if anyone has done a regularization involving a deviation of the induced OT map from some other fixed map. For example, the term may be $\varphi(\nabla\phi,I)$ for the case where the fixed map is the identity. Here, $\nabla\phi$ is the transport map (gradient of a convex function $\phi$) and $\varphi$ is some divergence measure.
Perhaps not exactly what you have in mind: F. Santambrogio, J. Louet, and L. De Pascale have considered a gradient regularization for the Monge problem, i-e
\begin{equation}
\min\limits_{T\#\mu=\nu} \int |T(x)-x| d\mu(x) +\varepsilon \int |\nabla T(x)|^2
\tag{$P_{\varepsilon}$}
\end{equation}
(I'm not writing any measure in the gradient penalization since the choice of such a reference measure matters quite a lot, as one can imagin and as in entropic optimal transport.) One expects that $(P_{\varepsilon})$ converges to the classical transport problem as $\varepsilon \to 0$, but the result is more subtle than meets the eye. See their paper.
One type of deviation from another map goes under name of "optimal transport over linear/nonlinear dynamics'. Described in the continuous form of OT (Brenier-Benamou style), here the baseline dynamics are given either by a linear or nonlinear dynamical system, and the job is to find the deviation to baseline that leads to desired transport in optimal way. This idea generalizes optimal control to optional transport of measures. Google scholar with these terms will help you.
Are you talking about this paper?
https://arxiv.org/pdf/1502.01265.pdf
Yes, thats one if the first ones to tackle this topic
|
2025-03-21T14:48:31.258321
| 2020-06-17T18:04:26 |
363358
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"6666",
"Donu Arapura",
"https://mathoverflow.net/users/4144",
"https://mathoverflow.net/users/88180"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630162",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363358"
}
|
Stack Exchange
|
Non-isotrival fiber bundle over compact Riemann surface
In this paper, Kodaira constructed a fiber bundle $\Phi:M_{m,n}\to S$ from a compact complex surface $M_{m,n}$ to a compact Rieman surface $S$ of genus $>0$. In particular, (on p.212) for any point $u\in S$, the fibre $C_u =\Phi^{-1}(u)$ is a compact Riemann surface
which is an $m$-sheeted cyclic (branched) covering surface of $R$ with two branch points, where $R$ is a compact Riemann surface of genus $\not=0$. I want to know why this fiber bundle is not isotrivial.
Many thanks.
Kodaira's examples have index $\tau>0$. If $M\to S$ were isotrivial, then it is not hard to see that after pulling back to a finite unramified cover of $S$, the surface becomes a product. But this would force $\tau(M)=0$ [See added note below].
You can look at the book Compact Complex Surfaces by Barth, (Hulek), Peters, and Van de Venn for further explanation.
There are examples of what are sometimes called Kodaira surfaces, where nonisotriviallity is essentially immediate. Namely, find a compact curve $S$ in $M_g$ (which exists once $g>2$), and pull back the "universal" curve.
Added Explanation The index is the signature of the intersection form. By a theorem of Hirzebruch, it can also be computed as
$$\tau(M)= \frac{1}{3}(c_1^2(M)-2c_2(M))$$
It follows that if $M'\to M$ is a finite unramified cover, then $\tau(M')=0$ if and only if $\tau(M)=0$. In particular, if $M'$ can be chosen as a product of curves, then it can be checked that $\tau(M')=0$, so $\tau(M)=0$.
Thanks, but how to see that Kodaira's examples have index $\tau>0$?
Kodaira's paper that you linked says "The purpose of this note is to exhibit a series of compact connected complex 4-manfolds ... with positive indices." You should read it.
Sorry, I checked the book you mentioned but still one some questions, I can see the pullback as the product of curves has index $0$, but why would that imply $\tau(M)=0$? Is index invariant under base change?
OK, I'll expand my answer.
|
2025-03-21T14:48:31.258491
| 2020-06-17T18:47:24 |
363360
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex M.",
"Ben McKay",
"Michael Albanese",
"QGravity",
"anonymous67",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/21564",
"https://mathoverflow.net/users/54774",
"https://mathoverflow.net/users/54780",
"https://mathoverflow.net/users/64606"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630163",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363360"
}
|
Stack Exchange
|
To what extent is a vector bundle on a manifold with boundary determined by its restriction to the interior?
Let $M$ be a manifold with boundary $\partial M$ and interior $M_0$. Let $E\rightarrow M_0$ be a fixed vector bundle. How many extensions of $E$ to a vector bundle $E'\rightarrow M$ are there, up to isomorphism? In terms of the bundles monoid: the restriction of $E'$ to $M_0$ gives a monoid morphism $\mathrm{Vec}_k(M)\rightarrow \mathrm{Vec}_k(M_0)$. Is it surjective/injective?
Intuitively, the bundle $E'|_{\partial M}$ is "the limit" of $E$ at $\partial M$, and therefore should be fixed up to isomorphism.
And perhaps in the same vein, the inclusion $\iota : M_0 \rightarrow M$ induces $\iota_*:\pi_1(M_0)\rightarrow \pi_1(M)$. Is this map surjective/injective? Can the bijectivity be deduced from a tubular neighborhood of $\partial M$ in $M$?
Counter-examples are appreciated.
The inclusion $M_0 \hookrightarrow M$ is a homotopy equivalence, so a vector bundle on $M$ is determined by its restriction to $M_0$
@MichaelAlbanese: I don't get this one, maybe you can point out my mistake, please? If $X$ is a smooth manifold, and if $D$ is a relatively compact open subset with smooth boundary $\partial D$, then we have two bundles over $\overline D$: the intrinsic tangent bundle $T \overline D$ and the restriction of $TX$ to $\overline D$. These bundles coincide on $D$, but have different ranks on $\partial D$, apparently contradicting your comment above. Where is my mistake? Maybe the naive restriction of $TX$ to $\overline D$ is not a bundle over $\overline D$?
@AlexM. They have the same rank. Maybe you're thinking of $T(\partial\overline{D})$ which has rank one less than that of $T\overline{D}$ and $T\overline{X}$. The relationship between these bundles is that $T\overline{D} \cong TX|{\overline{D}}$ and $T\overline{D}|{\partial\overline{D}} \cong TX|_{\partial\overline{D}} \cong T(\partial\overline{D})\oplus\varepsilon^1$.
As I indicated in my comment, the inclusion $\iota : M_0 \to M$ is a homotopy equivalence. This can be shown using the fact that the boundary $\partial M$ has a collar neighbourhood; it then boils down to showing the inclusion $(0, 1) \hookrightarrow [0, 1)$ is a homotopy equivalence. Actually, one needs to show that there is a homotopy inverse $j : [0, 1) \to (0, 1)$ to $i$ such that $i\circ j$ and $j\circ i$ are homotopic to identity maps relative to $[\frac{1}{2}, 1)$. This is not difficult, see this answer for some details.
On any paracompact space $X$, there is a natural bijection between isomorphism classes of real vector bundles on $X$ of rank $r$ and $[X, BO(r)]$, the set of homotopy classes of maps $X \to BO(r)$; see section $1.2$ of Hatcher's Vector Bundles and K-Theory for example. In particular, given a map $f : X \to Y$, we get an induced map $f^* : [Y, BO(r)] \to [X, BO(r)]$ which corresponds to pulling back a vector bundle by $f$. The analogous statement is true for complex vector bundles too, one just replaces $BO(r)$ with $BU(r)$.
In the case that $f$ is a homotopy equivalence, then $f^*$ is a bijection: if $g$ is the homotopy inverse of $f$, then $g^*$ is the inverse of $f^*$. In particular, for the homotopy equivalence $\iota : M_0 \to M$, we see that there is a bijection between isomorphism classes of real/complex rank $r$ bundles on $M$ and $M_0$ given by $E \mapsto \iota^*E = E|_{M_0}$.
Finally, as $\iota : M_0 \to M$ is a homotopy equivalence, the induced map $\iota_* : \pi_1(M_0) \to \pi_1(M)$ is an isomorphism.
As Ben McKay indicates in the comment below, the above does not deal with smooth bundles but topological bundles. The statement for smooth bundles is also true, but requires a bit more work. The key is that every real rank $r$ vector bundle on a smooth manifold $M$ has a classifying map $M \to \operatorname{Gr}_r(\mathbb{R}^N)$ which is unique up to homotopy where $N = r + \dim M + 1$; this is Theorem 3.3.4 of Hirsch's Differential Topology. It follows that isomorphism classes of topological real rank $r$ vector bundles on $M$ are in bijection with $[M, \operatorname{Gr}_r(\mathbb{R}^N)]$; that is, the inclusion $\operatorname{Gr}_r(\mathbb{R}^N) \hookrightarrow \operatorname{Gr}_r(\mathbb{R}^{\infty})$ induces a bijection $[M, \operatorname{Gr}_r(\mathbb{R}^N)] \to [M, \operatorname{Gr}_r(\mathbb{R}^{\infty})] = [M, BO(r)]$.
If the classifying map of a bundle is smooth, then the bundle itself is smooth (the pullback of a smooth bundle by a smooth map is smooth). As every continuous map between smooth manifolds is homotopic to a smooth one, every topological vector bundle on $M$ is isomorphic to a smooth one. Moreover, two smooth maps are homotopic if and only if they are smoothly homotopic which implies that every topological vector bundle is isomorphic to a unique smooth vector bundle up to smooth isomorphism. It follows that isomorphism classes of smooth real rank $r$ vector bundles on $M$ are in bijection with $[M, \operatorname{Gr}_r(\mathbb{R}^N)]$.
Now we can argue as before to deduce that $\iota^*$ induces a bijection between the set of isomorphism classes of smooth real rank $r$ bundles on $M$ and $M_0$. Again, the statement is also true for smooth complex bundles.
Thanks. Can you explain what the space $BO(k)$ (and $BU(k)$) is?
For a topological group $G$, there is a space $BG$ called its classifying space; here the two groups of interest are the orthogonal group and the unitary group. In this case, we have explicit models for $BO(r)$ and $BU(r)$, they are the grassmannians $\operatorname{Gr}_r(\mathbb{R}^{\infty})$ and $\operatorname{Gr}_r(\mathbb{C}^{\infty})$ respectively.
There is still the problem of smooth structure of the vector bundle, once the topological structure is determined.
@BenMcKay: I believe this can also be dealt with. I will edit my answer later today.
@MichaelAlbanese is this true that any (topological or smooth) principal $G$-bundle on a manifold with boundary $M$ is determined by its restriction to its interior?
I think there is a one-to-one correspondence between topological and smooth principal $G$-bundles, as has been proved in the following paper
https://edoc.hu-berlin.de/bitstream/handle/18452/11495/214.pdf?sequence=1
@QGravity: Yes, by the same argument: principal $G$-bundles have a classifying space $BG$, and a manifold with boundary is homotopy equivalent to its interior.
@MichaelAlbanese Thanks ... Now if I have a connection on a principal $G$-bundle on a manifold with boundary is determined by its restriction to the interior? This seems strange if it is true since we consider connections up to gauge transformations since in the presence of boundary, we need to consider gauge transformations that become identity at the boundary while if we restrict to the interior, then it seems that all gauge transformations are allowed since there is no boundary.
@QGravity: The difference of two connections is a smooth tensor, so if two connections agree on the interior, they are in fact the same.
|
2025-03-21T14:48:31.259174
| 2020-06-17T19:10:23 |
363363
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"EGME",
"KConrad",
"Milo Moses",
"https://mathoverflow.net/users/130113",
"https://mathoverflow.net/users/159298",
"https://mathoverflow.net/users/3272"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630164",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363363"
}
|
Stack Exchange
|
Analytic continuation over boundaries
In D.J Newman's paper
A simple analytic proof of the prime number theorem
there is the following theorem:
Suppose $|a_n|<1$ and form the Dirichlet series $F(s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$ which clearly converges to an analytic function for $\Re(s)>1$. If, in fact, $F(s)$ is analytic throughout $\Re(s)\geq1$, then $\sum_{n=1}^{\infty}\frac{a_n}{n^s}$ converges throughout $\Re(s)\geq1$
I do not understand what Newman means by "Analytic Throughout" $\Re(s)\geq1$.
He clearly is not supposing that the function must converge for $\Re(s)\geq1$ for it to be analytic since otherwise, the theorem would be useless, and so I can only assume that he means that the function has an analytic continuation over the line $\Re(s)\geq1$. Since analytic functions are defined on open sets I can only assume this must mean that $F(s)$ has an analytic continuation to some set that contains the real numbers. Am I correct?
If I am, what conditions would one need to show that a function has an analytic continuation throughout $\Re(s)\geq1$? Is it sufficient to show that
$$\lim_{x\to1^+}F(x+ti)$$
exists for all $t\in\mathbb{R}$?
Any insights are appreciated.
He means the function has an analytic continuation from the open half-plane ${\rm Re}(s) > 1$ to the closed half-plane ${\rm Re}(s) \geq 1$. By definition, to say a function is analytic on a closed set means it is analytic on an open set containing that closed set. It is convenient to be able to talk about a function on an open set having an analytic continuation to its boundary without having to always throw in "larger open set containing the closure of the original open set".
Newman is not saying $F(s)$ has an analytic continuation to the "real numbers" but to the vertical line ${\rm Re}(s) = 1$, meaning to an open set that contains that line (and, in the setting of the theorem, it means analytic continuation to an open set containing ${\rm Re}(s) \geq 1$).
There are no simple general conditions that let you check whether a Dirichlet series has an analytic continuation from a half-plane of known convergence to a point on the boundary line. Each class of important examples can require new ideas. In some sense it is like dealing with analytic continuation of a power series from an open disc to its boundary. There is no simple method to check whether a generic power series with radius of convergence 1 has an analytic continuation to a point on the unit circle (assuming the coefficient tend to $0$, a necessary condition for the power series to converge at some point on the unit circle).
To emphasize the subtlety of analytic continuation of Dirichlet series on the boundary of where they are known to converge, one of the consequences of the work by Wiles on Fermat's Last Theorem is that the Dirichlet series defining the $L$-function of an elliptic curve over $\mathbf Q$ has an analytic continuation to all of $\mathbf C$ from its initial "easy" half-plane of absolute convergence ${\rm Re}(s) > 3/2$. Even the analytic continuation of all such Dirichlet series to the line ${\rm Re}(s) = 3/2$ was unknown before his work. (Of course some special cases were known previously.) The upshot is that the hypothesis of analytic continuation in Newman's theorem is a very serious one, and you don't verify it without knowing something significant about the actual example you want to apply it to.
In Newman's proof of the Prime Number Theorem, he wants to apply his theorem to the function $1/\zeta(s)$, which for ${\rm Re}(s) > 1$ has the Dirichlet series representation $\sum \mu(n)/n^s$ with coefficients $\mu(n)$ that are the Moebius function, which is bounded (values are $0$, $1$, and $-1$). Proving $1/\zeta(s)$ has an analytic continuation from ${\rm Re}(s) > 1$ to ${\rm Re}(s) \geq 1$ basically involves showing $\zeta(s)$ has an analytic continuation from ${\rm Re}(s) > 1$ to ${\rm Re}(s) > 0$ except for a simple pole at $s = 1$ (this is done in nearly any analytic number theory book) and then proving $\zeta(s) \not= 0$ for ${\rm Re}(s) = 1$. The nonvanishing of $\zeta(s)$ on the line ${\rm Re}(s) = 1$ (this is "automatic" at $s = 1$ from the pole, which turns into $1/\zeta(s) = 0$ at $s = 1$) is often regarded as the key analytic property of the zeta-function in the proof of the Prime Number Theorem. The proof is not really hard, but it does require a clever idea. It is not something anyone is going to figure out just by starting at the definition of the zeta-function when ${\rm Re}(s) > 1$ or staring at a formula that analytically continues the zeta-function to ${\rm Re}(s) > 0$.
Convergence of limits from inside a domain of convergence of a series to a point on the boundary is not sufficient to imply convergence of the series at the boundary point, e.g., consider $\sum_{n \geq 0} (-1)^nz^n$ as $z \rightarrow 1^{-}$ or $\sum_{{\rm odd} \, n \geq 1} (-1)^{(n-1)/2}/n^s = 1 - 1/3^s + 1/5^s - 1/7^s + \cdots$ as $s \to 0^+$. The power series does not converge at $z = 1$, the Dirichlet series does not converge at $s = 0$, but both series have limit $1/2$ as $z \to 1^-$ or as $s \to 0^+$. The value of the power series limit is easier to see since the power series equals $1/(1+z)$ for $|z| < 1$ and that simple formula gives you an analytic continuation for all of $\mathbf C - \{1\}$, which at $z = 1$ is $1/2$. The Dirichlet series has an analytic continuation to $s = 0$ since, well, there's a more complicated formula you can write down that matches the series for ${\rm Re}(s) > 0$ and makes sense and is analytic on a bigger half-plane than ${\rm Re}(s) > 0$. Without seeing such a formula, which I won't write down here, I don't think it is obvious that the limit of that Dirichlet series (it is the $L$-function of the nontrivial character mod $4$) as $s \to 0^+$ is $1/2$. Read an analytic number theory book that discusses analytic continuation of Dirichlet $L$-functions and you'll see how such analytic contiuation is proved. It's not as easy as the case of a geometric series.
A power series converging on the open unit disc is analytic there, but if it has radius of convergence 1 then it need not be analytic at each point on the unit circle to which it converges. (If the radius of convergence of the power series is bigger than $1$ then the situation is different!) In fact, if a series converging on the open unit circle has an analytic continuation to each point on the unit circle, then by compactness of the closed unit disc the power series has radius of convergence greater than 1. Therefore a power series with radius of convergence 1 that converges on the closed unit disc, such as $\sum z^n/n^2$, is not analytic somewhere on the unit circle even though it converges on the whole unit circle. The series $\sum z^n/n^2$ has a name, the "dilogarithm", and is denoted ${\rm Li}_2(z)$ (you can replace the exponent $2$ in the denominator with $k$ and get ${\rm Li}_k(z)$, hence the notation). It has an analytic continuation from the open unit disc to all of $\mathbf C$ except the point $z = 1$, and on the closed unit disc (including $z = 1$) it is continuous.
The situation with Dirichlet series is more subtle: $\sum_{{\rm odd} \, n \geq 1} (-1)^{(n-1)/2}/n^s$ converges if and only if ${\rm Re}(s) > 0$, but it has no analytic singularity on the imaginary axis. In fact, this series extends analytically to all of $\mathbf C$ (an entire function). Nothing strange happens anywhere on the imaginary axis as far as analytic behavior is concerned. The proof that a power series has an analytiic singularity somewhere on the boundary of its disc of convergence does not carry over to a Dirichlet series and the boundary of its half-plane of convergence because the boundary of a half-plane is not compact, unlike a circle.
I can carry over the dilogarithm example to the setting of Dirichlet series, since every power series $\sum c_kz^k$ could be interpreted as a vertically periodic Dirichlet series by a change of variarbles $z = 1/2^s$, so $|z| = 1/2^{{\rm Re}(s)}$. Then $|z| < 1$ corresponds to ${\rm Re}(s) > 0$ and $\sum c_kz^k = \sum c_k/2^{ks}$. This is a Dirichlet series supported on the powers of $2$. (I could have used $z = 1/3^s$ or other options, but picked $z = 1/2^s$ for concreteness.) Since $2^s$ has period $2\pi i/\log 2$, the function $\sum c_k/2^{ks}$ is unchanged when we add an integral multiple of $2\pi i/\log 2$ to $s$. Let's consider
$$
f(s) = \sum_{k \geq 1} \frac{1/k^2}{2^{ks}} = {\rm Li}_2(1/2^s).
$$
This series converges for ${\rm Re}(s) \geq 0$, just as ${\rm Li}_2(z)$ converges for $|z| \leq 1$. Since ${\rm Li}_2(z)$ extends analytically from $|z| < 1$ to $\mathbf C - \{1\}$, $f(s)$ extends analytically from ${\rm Re}(s) > 0$ to $\mathbf C - A$ where $A = \{s : 1/2^s = 1\} = (2\pi i/\log 2)\mathbf Z$. Therefore $f(s)$ has half-plane of convergence ${\rm Re}(s) \geq 0$, but is not analytic at the points in $A$ on the imaginary axis.
Thank you so much! One thing still left unanswered about my question of the case that the limit does exist everywhere on the boundary. You state that convergence at one point on the boundary does not imply that it converges everyone on said boundary, but if it does converge everywhere on the boundary does that mean that it can be analytically continued over that boundary?
The reason I'm being so particular is that I have a specific example in mind, and if this were true it would save me a huge headache.
I added a paragraph at the end showing that what you are asking for is not true. The analogue for power series is already not true, so it would not be realistic to expect the version for Dirichlet series is true either, and in any case I have given a concrete example (or counterexample, depending on your point of view).
Ok, this makes sense. Thanks again for the clarification. I really appreciate the work that you're doing here on MathOverflow.
This is a great answer to a question about a detail that is often not explained or glossed over in texts, and which is important. This detail makes a “singular” appearance in Landau’s theorem. To complete the question, I would ask: if a D. series converges at some point $s_0$ on its line of convergence then does that means that there will necessarily be some analytic continuation of the series to this point?
@EGME The answer is no, and my answer has an example, which I'll summarize here. The power series ${\rm Li}2(z) := \sum{k\geq 1} z^k/k^2$ converges uniformly on $|z| \leq 1$, so it's continuous on the closed unit disc, but turns out not to be analytic at $z = 1$ (but it's analytic elsewhere on $|z| \leq 1$). Replacing $z$ with $1/2^s$, the Dirichlet series $\sum_{k\geq 1} (1/k^2)/2^{ks}$ converges uniformly on ${\rm Re}(s) \leq 0$, so it's continuous on that closed half-plane, but turns out not to be analytic at $s = 0$, or at any $s$ where $2^s = 1$ (i.e., $s\in (2\pi i/\log 2)\mathbf Z$
|
2025-03-21T14:48:31.259962
| 2020-06-17T19:16:25 |
363365
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"L.F. Cavenaghi",
"Michael Renardy",
"https://mathoverflow.net/users/12120",
"https://mathoverflow.net/users/94097"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630165",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363365"
}
|
Stack Exchange
|
When linear strongly elliptic operators are invertible?
I am studying Pazy's book "Semigroups of Linear Operators and Applications to Partial Differential Equations" and when considering an operator like:
A linear differential operator, $$A : W^{2m,p}(\Omega)\cap W^{m,p}_0(\Omega) \subset L^p(\Omega) \to L^p(\Omega),$$
where $\Omega$ is a bounded open set on $\mathbb{R}^n$ with smooth boundary, $1 < p < \infty$. If one assumes that $A$ is strongly elliptic, in order to show (via his argument) that $-A$ generates an analytic semigroup one must assume that $0\in \rho(A)$ (the resolvent set) (i.e, $A$ is invertible). So, my question is: is this operator necessarily invertible? He does not mention it anywhere.
Well, $0\in\rho(A)$ means $A$ is invertible, by definition. But this condition is certainly not necessary for $A$ to generate an analytic semigroup.
@MichaelRenardy, yeah, I am sorry, I think I wrote it badly, the point is, is it true that $0\in\rho(A)$? I didn't want to write it as a claim, this is indeed my question. I totally agree that it is not necessary to show that $A$ generates an analytic semigroup, but it is definitely necessary on his argument.
|
2025-03-21T14:48:31.260072
| 2020-06-17T19:26:20 |
363367
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Neal",
"https://mathoverflow.net/users/20796",
"https://mathoverflow.net/users/25510"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630166",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363367"
}
|
Stack Exchange
|
Can we attain the maximum and minimum of a Rayleigh quotient over any subspace?
Let $M\in\mathbb{C}^{n\times n}$ be a Hermitian matrix and let $E$ be a subspace of $\mathbb{C}^n$.
$$\mbox{Are } \sup_{x\in E\\
x\neq0}\dfrac{x^*Mx}{x^*x}\mbox{ and }\inf_{x\in E\\
x\neq0}\dfrac{x^*Mx}{x^*x} \mbox{ always attained by some vector }x\in E, x\neq0 ?$$
If $\dim(E)=1$, then the answer would be yes, as the quotient will be constant for any $x\in E, x\neq0$.
If $\dim(E)=n$, then $E=\mathbb{C}^n$ and the max and min values of the Rayleigh quotient correspond to the largest and smallest eigenvalues of $M$.
Is the answer positive for any subspace $E$?
Yes, for example when all eigenvalues are equal, that is when M is proportional to I.
As the Rayleigh quotient is scale invariant, you can restrict to unit vectors in $E$. This subspace of $E$ is compact and the Rayleigh quotient is continuous, so the Rayleigh quotient attains its max and min.
You can restrict the inf/sup to $\{x\in E : \|x\|=1\}$, which is compact, so this follows from the extreme value theorem.
|
2025-03-21T14:48:31.260196
| 2020-06-17T19:51:33 |
363372
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dominic van der Zypen",
"Geoffrey Irving",
"Gerhard Paseman",
"Steven Stadnicki",
"https://mathoverflow.net/users/22930",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/7092",
"https://mathoverflow.net/users/8628"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630167",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363372"
}
|
Stack Exchange
|
"Gray code" for building teams
Motivation. In a team of $n$ people, we had the task to build subteams of a fixed size $k<n$ such that every day, $1$ person of the subteam is replaced by another person in the team, but not in the subteam (so that the new subteam consisted of $k$ again). The question arose if and for what choices of $k$ and $n$ the subteam schedule can be built to contain each choice of $k$ people out of the team exactly once.
Formal version. For any set $X$ and positive integer $k$, let $[X]^k$ be the collection of subset of $X$ having $k$ elements. For $n\in\mathbb{N}$ let $[n] =\{1,\ldots,n\}$. For integers $1< k < n$ we define a graph $G(n, k)$ by $V(G(n,k)) = [[n]]^k$ and $$E(G(n,k)) = \big\{\{A, B\} : A, B \in [[n]]^k \land |A\cap B| = k-1\big\}.$$
For what choices of $1< k < n$ does $G(n,k)$ have a Hamiltonian path? Bonus question: Replace "path" by "cycle" in the original question. (The bonus question need not be answered for acceptance.)
I'll have to go digging to find the specific answer, but this is well-studied and you can find good discussion of it in vol. 4 of Knuth's Art Of Computer Programming, on combinatorial algorithms.
The k=2 version (and variant for n odd) is used in cribbage tournaments with seating arrangements, often where one player is a fixed point and the other 2k-1 form a cycle. This gives a Gray code of the partitions and not just of the teams. Gerhard "For Two Player Games, Naturally" Paseman, 2020.06.17.
Beautiful answer, thanks @bof (and everybody else)!
This seems to be possible for all choices of $k$ and $n$. I found a page here by Dr. Ronald D. BAKER describing a more than sixty year old 'revolving door algorithm'.
When enumerating the k-element subsets of and n-set we are implicitly enumerating the partitions of the n-set into parts, one of size s=k and the other of size t=n-k. Hence the problem is often described as the enumeration of (s,t)-combinations. Suppose the think of the set as a collection of people and imagine them being divided into two adjacent rooms, k people in one room with the remaining n-k people in the other room. Now further imagine that there is a revolving door connecting the two rooms, and a change consists of an individual from each room entering that revolving door and exchanging sides. This analogy is the source of the moniker revolving door algorithm.
W. H. Payne created the following algorithm in 1959. The call to visit might, for example, output the k-subset or it might do the computations of an algorithm which needs all k-element subsets. Each k-subset is referenced by an index-list $c_k \dots c_2c_1$, the indices of the elements belonging to the subset sorted in order. Notice the code makes extensive use of conditionals, the branching command goto and line labels†.
algorithm RevDoorSubset(n, k)
Array C[1..k+1]
R1: for i ← 1 to k do // initialize C
C[i] ← i-1
end for
C[k+1] ← n
R2: visit(C[ ], k) // Do whatever is needed w/ subset (just print?)
R3: if (k is odd) // the easy cases
if ( C[1]+1 < C[2] )
C[1] ← C[1]+1
goto R2
else
j ← 2
goto R4
fi
else
if ( C[1] > 0 )
C[1] ← C[1]-1
goto R2
else
j ← 2
goto R5
fi
fi
R4: if ( C[j] ≥ j ) // try to decrease C[j]
C[j] ← C[j-1]
C[j-1] ← j-2
goto R2
else
j ← j+1
fi
R5: if ( C[j] + 1 < C[j+1] ) // try to increase C[j]
C[j-1] ← C[j]
C[j] ← C[j] + 1
goto R2
else
j ← j + 1
if ( j ≤ k)
goto R4
fi
fi
return
end
I've converted the algorithm to JavaScript; I don't think MathOverflow supports Stack Snippets but I've managed to host a working demo here in the Sandbox on Meta Stack Exchange. Click the 'Run code snippet' below the code, change the values of $n$ and $k$ and click 'Generate'.
You can also try it online here in case the Sandbox fails.
It seems this algorithm produces Hamiltonian cycles, but to prove it I'll need some sleep first.
The following recursive description of a revolving door sequence is taken from here, where it is also proved that it generates a Hamilton cycle. The $k$-subsets of $\{1,\dots,n\}$ are identified with the bitstrings of length $n$ with exactly $k$ entries equal to $1$. Let $R(k,n)$, denote the binary $\binom{n}{k}\times n$-matrix whose rows correspond to the required sequence of the $k$-sets. Then
$$R(0,n)=\begin{pmatrix}0&0&\dots&0\end{pmatrix},\qquad R(n,n)=\begin{pmatrix}1&1&\dots&1\end{pmatrix},$$
and for $1\leqslant k\leqslant n-1$, we obtain $R(k,n)$ by writing down the $\binom{n-1}{k}$ rows of $R(k,n-1)$, putting an additional $0$ in front, and below this writing the $\binom{n-1}{k-1}$ rows of $R(k-1,n-1)$ in reverse order, putting an additional $1$ in front.
The precise Knuth reference mentioned in a comment above is Section <IP_ADDRESS>. Generating all combinations in TAOCP, Volume 4A - Combinatorial Algorithms, Part 1.
Here's an interesting related result: If $n=2k-1$ then $\binom{n}{k}=\binom{n}{k-1}$, and we can hope that in the process every $(k-1)$-set is visited exactly once. That this is possible used to be the middle level conjecture which is now a theorem due to Torsten Mütze: Proof of the middle levels conjecture. Proceedings of the London Mathematical Society 112.4 (2016): 677-713.
Theorem. The graph $G(n,k)$ is Hamiltonian if $n\ge3$ and $0\lt k\lt n$.
Proof. If $k=1$ or $k=n-1$ it's obvious, because $G(n,k)\cong K_n$ in those cases. Now consider the graph $G=G(n,k)$ where $2\le k\le n-2$. Let
$$S=\{A\in[[n]]^k:1\in A\},\quad S'=\{A\in[[n]]^k:1\notin A\}.$$
Since the induced subgraphs $G[S]$ and $G[S']$ are isomorphic to $G(n-1,k-1)$ and $G(n-1,k)$ respectively, they are Hamiltonian by the inductive hypothesis. Moreover, since the graphs are edge transitive, we can choose a Hamiltonian cycle $C$ in $G[S]$ containing the edge
$$\{A,B\}=\{\{1,\dots,k\},\ \{1,\dots,k-1,k+1\}\}$$
and a Hamiltonian cycle $C'$ in $G[S']$ containing the edge
$$\{A',B'\}=\{\{2,\dots,k,k+2\},\ \{2,\dots,k+1\}\}.$$
Now we get a Hamiltonian cycle in $G$ by removing the edges $\{A,B\}$ and $\{A',B'\}$ from $C\cup C'$, and replacing them with $\{A,A'\}$ and $\{B,B'\}$.
Note that the graph is also Hamiltonian for $k = 0, n$.
“By convention, the singleton graph $K_1$ is considered to be Hamiltonian even though it does not posses a Hamiltonian cycle” - https://mathworld.wolfram.com/HamiltonianCycle.html
I would say both the empty and singleton graphs are $n$-connected for any $n$, since deleting any number of edges/vertices leaves them connected, but will admit that I can’t find a reference that agrees with me. :)
|
2025-03-21T14:48:31.260684
| 2020-06-17T19:56:52 |
363373
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Martin Seysen",
"https://mathoverflow.net/users/105705"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630168",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363373"
}
|
Stack Exchange
|
A lattice with Monster group symmetries
The book Mathematical Evolutions contains the following excerpt:
A last, famous, example is the following. It is known that in the space
of one hundred and ninety six thousand eight hundred and eighty three
dimensions there is a wonderful lattice whose symmetry group has order
808 017 424 794 512 875 886 459 904 961 710 757 005 754 368 000 000 000
= $2^{46} \cdot 3^{20} \cdot 5^9 \cdot 11^2 \cdot 13^3 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 41 \cdot 47 \cdot 59 \cdot 71$
Of course, no-one has really seen this lattice: we know that it exists,
but an explicit construction is lacking. Nevertheless, one can construct
the symmetry group of the lattice, the so-called monster group $M$ of
R. Griess and B. Fischer.
The existence of such a lattice is not immediate: for example, the simple group $A_5$ has a faithful three-dimensional representation, but there is no lattice in $\mathbb{R}^3$ with an (orientation-preserving) point symmetry group isomorphic to $A_5$.
However, a remark in a paper by Conway reveals the existence of a (probably unimodular) integral lattice in 196884 dimensions which is closed under the Griess algebra multiplication:
Norton has shown that the lattice $L$ spanned by vectors of the form $1, t, t \ast t'$, where $t$ and $t'$ are transposition vectors, is closed under the algebra multiplication and integral with respect to the doubled inner product $2(u, v)$. The dual quotient $L^{\star} / L$ is cyclic of order some power of 4, and we believe that in fact $L$ is unimodular. The vectors $t, u, v, w$ of Table 3 all lie in $L$.
Now, if we take the intersection of the 196884-dimensional lattice $L$ with the orthogonal complement of the fixed axis $1$, we obtain a 196883-dimensional integral lattice $L'$ with Gram determinant $|L'| = 3|L|$. This is an explicit construction of a lattice with the properties alluded to in the Mathematical Evolutions excerpt.
The orientation-preserving point symmetry group of this lattice $L'$ is isomorphic to the Monster group $M$, and the full point symmetry group is isomorphic to the direct product $M \times C_2$ (since the ambient dimension is odd, so the negated identity matrix $-I$ is an orientation-reversing symmetry which commutes with everything).
I'd like to understand this 'wonderful lattice' $L'$ in greater detail. So, now for a pair of concrete questions:
Question 1: what are the shortest nonzero vectors in the lattice $L'$?
The shortest vectors I can find are those of the form $x := t - t'$ (where $t$ and $t'$ are the axes of two type-2A involutions whose product is also a type-2A involution), which each have doubled inner product $2(x, x) = 448$. The number of such vectors is<PHONE_NUMBER>687194563957260000000, one for each arc (oriented edge) in the Monster graph. If these are indeed the shortest vectors, then the sphere packing corresponding to this lattice would be moderately dense (but not record-breakingly so, and well below Minkowski's lower bound on lattice density).
Question 2: does the set of shortest nonzero vectors generate the lattice?
The fact that the Monster graph is connected implies that any vector of the form $t - t'$ (where $t, t'$ are any pair of axes of type-2A involutions) is generated by the aforementioned norm-448 vectors. But I'm not immediately convinced that these generate the lattice $L'$, because Norton's lattice $L$ was described as being generated by vectors of the form $1, t, t \ast t'$, rather than just vectors of the form $1, t$.
I don't know if this helps, but I have uploaded a python package (with optimized C subroutines) that can do some calulations in the moster and also generate random 2A transposition axes $t$. See:
https://github.com/Martin-Seysen/mmgroup
|
2025-03-21T14:48:31.260963
| 2020-06-17T19:58:19 |
363374
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chilote",
"Neal",
"https://mathoverflow.net/users/17773",
"https://mathoverflow.net/users/20796",
"https://mathoverflow.net/users/47542",
"kodlu"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630169",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363374"
}
|
Stack Exchange
|
Is there a method to find a vector that optimizes a Rayleigh quotient over a subspace?
Let $M\in\mathbb{C}^{n\times n}$ be an arbitary Hermitian matrix and let $E$ be a subspace of $\mathbb{C}^n$.
Is there a method to find vectors $y,z\in E$ such that
$$\dfrac{y^*My}{y^*y}=\sup_{x\in E\\
x\neq0}\dfrac{x^*Mx}{x^*x}\mbox{ and }\dfrac{z^*Mz}{z^*z}=\inf_{x\in E\\
x\neq0}\dfrac{x^*Mx}{x^*x}\hspace{3mm}?$$
If $E=\mathbb{C}^n$ then $x$ and $y$ will be any eigenvector of $M$ corresponding to the largest and smallest eigenvalues of $M$ respectively, so the method to find them is standard. How to find $y$ and $z$ when $E$ is a proper subspace of $\mathbb{C}^n$ with $\dim(E)>1$?
Is there any reason you can't choose a basis of $E$, write $M$ in that basis, and proceed according to standard methods?
@Neal could you please be a little more specific? I must be blind because I do not see a natural approach to this problem.
If you have a subspace then you can find a basis. What's the problem?
Pick a basis $e_i$ of $E$, let $N$ be the $\dim(E)\times\dim(E)$ matrix with $N_{ij} = M(e_i,e_j)$. Then extremize the Rayleigh quotient of $N$ over $E$. By construction this will yield the extremal values of the Rayleigh quotient of $M$ over $E$.
|
2025-03-21T14:48:31.261070
| 2020-06-17T20:06:08 |
363375
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Arkadij",
"Moishe Kohan",
"https://mathoverflow.net/users/109370",
"https://mathoverflow.net/users/39654"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630170",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363375"
}
|
Stack Exchange
|
Smoothness of distance at an end of a manifold
Suppose $(X,g)$ is Riemannian manifold. Let $p\in X$ be a point. The distance with respect to g
\begin{align*}
& \text{dist}_g(p,-):X\to [0,\infty)\\
& \text{dist}_g(p,q) = \text{inf}\{\text{length}_g(\gamma) \,|\,\gamma:[0,1]\to X, \gamma(0) = p,\gamma(1)=q,\gamma \text{ piecewise smooth}\}\,.
\end{align*}
is a continuous function, but it is in general not smooth. However, assume that $X$ is not compact. For simplicity let's assume that it has a single end. Can one choose a compact $K$, s.t. $p\in K\subset X$, and $\text{dist}_g(p,-)|_{X\backslash K}$ is smooth?
My thinking was that if I could assume that $(X,g)$ is ALE and $K$ is the set, such that $X\backslash K$ has an ALE chart on it, it should be true. But this seems rather restrictive.
Oh no, an example would be a 1-ended complete hyperbolic surface of infinite genus.
Thank you. Could you explain a bit more? I don't know much about hyperbolic geometry.
Every complete, connected hyperbolic surface $S$ is isometric to the quotient of the hyperbolic plane ${\mathbb H}^2$ by a discrete subgroup $\Gamma$ of isometries of ${\mathbb H}^2$ acting freely on ${\mathbb H}^2$. Given a point $z\in {\mathbb H}^2$ and $\Gamma$ as above, one defines the Dirichlet domain $D=D_{\Gamma,z}$ by
$$
\{w\in D: d(w,z)\le d(w,\gamma z) \forall \gamma\in \Gamma\}.
$$
This domain is convex (as intersection of half-planes) and is polygonal. Topologically, one obtains $S$ by identifying boundary edges of $D$ via some elements of $\Gamma$. Thus, $D$ is noncompact if and only if $S$ is.
Now, to relate $D$ and the non-smoothness locus of the distance function on $S$, let
let $\pi: {\mathbb H}^2\to S$ be the covering map (the quotient via the $\Gamma$-action). Let
$p:=\pi(z)$. Then (if you look at the definition of the boundary of $D$), $\pi(\partial D)$ is exactly the non-smoothness locus of the function $d^2(p, \cdot)$ on $S$ (I prefer to square to avoid the nonsmoothness at $p$, otherwise, it is the same as the nonsmoothness locus of the distance function $d(p, \cdot)$): A point $q$ belongs to
$\pi(\partial D)$ precisely when there is more than one minimizing (unit speed) geodesic from $p$ to $q$. The loop formed by such distinct geodesics corresponds to an element $\gamma\in \Gamma$ such that a preimage of $q$ in ${\mathbb H}^2$ lies on the bisector of the pair $p, \gamma(p)$.
It remains to observe that $\pi(\partial D)$ is noncompact (equivalently, is unbounded in $S$) if and only if $D$ is noncompact.
Now, since you want a 1-ended example, just take a hyperbolic surface with one cusp or a 1-ended hyperbolic surface of infinite genus...
|
2025-03-21T14:48:31.261266
| 2020-06-17T21:18:45 |
363382
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Gerry Myerson",
"Gordon Royle",
"Hollis Williams",
"Timothy Chow",
"YCor",
"Yemon Choi",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/119114",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/1492",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630171",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363382"
}
|
Stack Exchange
|
Papers with a large number of coauthors
I recently submitted a paper to the preprint arXiv, which was rejected because we didn't list all of the authors on the first page. We chose to follow the polymath model, using a generic name for our group, with a footnote linking to a place with all the actual contributors.
I was surprised that our paper was rejected, as there are many arXiv papers by "D.H.J. Polymath". Does anyone know if this is a recent change?
More importantly, what should be the practice for papers involving a huge number of coauthors in a collaborative setting? Is there a way to petition the arXiv for exceptions in such cases?
"For named collaborations, it is acceptable to only use the collaboration name within the metadata, however a complete list of all authors and their affiliations must be contained in the full printed text." source
The tag "publishing" was made a synonym of "journals", so "journals" is supposed to encompass all that concerns "publications".
@CarloBeenakker : That's interesting. As far as I can tell, D.H.J. Polymath flagrantly violates this rule.
@PaceNielsen : The sheer number of authors doesn't seem to me to be an automatic reason to suppress the names. The physics papers with 5000+ coauthors do explicitly list all the names, even when the number of pages required for the names far exceeds the number of remaining pages in the article. On the other hand, it does seem that there could be situations where using a pseudonym has a strong justification. Anyone read The Bourbaki Gambit by Carl Djerassi?
As someone who has tried and failed to post a "paper by John Rainwater" to the arXiv, not to mention years ago being knocked back when I tried to post my typeset version of Johnson's Cohomology in Banach algebras, I am not particularly surprised that the arXiv rejected your submission while smiling on the DHJ Polymath submissions. Dismayed, yes, but surprised, no.
@TimothyChow If memory serves right, Euler has some papers on the arXiv, though that wasn't a case of a pseudonym
Not exactly Maths, but the paper announcing the discovery of gravity waves, published in Physical Review Letters (https://physics.aps.org/featured-article-pdf/10.1103/PhysRevLett.116.061102) has over 1000 authors, ranging from Abbott to Zweizig.
https://www.nature.com/news/physics-paper-sets-record-with-more-than-5-000-authors-1.17567
@Gordon Royle: 1000 co-authors is not actually that many for a particle physics paper. In experimental particle physics, it's common to have papers with more than 5000 authors (I won't give examples as there are so many by this point).
This is a matter of opinion, but whatever the policy is, it should be consistently applied. If D.H.J. Polymath is allowed to post papers without listing the members, then other collaborations should be allowed to do so as well. D.H.J. Polymath shouldn't get special treatment just because it is famous.
I'm gradually becoming convinced that the mathematical community as a whole needs to pay more attention to arXiv policies. Seemingly small questions of policy can have a huge impact on mathematical research, simply because such a huge fraction of all mathematical research gets posted to the arXiv—far more than is submitted to any single mathematics journal. There is a myth that there are no barriers to publishing on the arXiv, but this is certainly not true. Figuring out the right policies is not a trivial problem; the founders of viXra stated some legitimate concerns about the arXiv, but I believe that viXra has failed to solve the problems it was intended to solve. Perhaps the American Mathematical Society and other professional societies should investigate some of the concerns and complaints about the arXiv that are being raised by mathematical researchers, and try to change some things about the arXiv if doing so would serve the community better.
|
2025-03-21T14:48:31.261584
| 2020-06-18T00:21:09 |
363388
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630172",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363388"
}
|
Stack Exchange
|
McKean-Singer formula in Heat Kernels and Dirac Operators book
I'm reading "Heat Kernels and Dirac Operators" by Berline, Getzler and Vergne. The setting is: $E \to M$ is a $\mathbb{Z}_2$-graded vector bundle on a compact Riemannian manifold $M$
and $D : \Gamma(M, E) \to \Gamma(M, E) $ is a self-adjoint Dirac operator (especially $D^2$ is Laplace). We denote
by $D^{\pm}$ the restrictions of $D$ to $\Gamma(M, E^{\pm})$. I have some troubles to understand the last part in the second variant of the proof of Theorem 3.50 McKean-Singer:
It is clamed that the operator $d(e^{-tD^2})/dt$ has a smooth kernel equal to $-D^2e^{-tD^2}$. This don't make any sense to me. Recall if $P: E\to E$ is an integral operator and $s \in E_x, x \in M $ then it's kernel $K: M \times M \to \mathbb{R}$ is defined by equation
$$ Ps_{\vert x}= \int_{\{x \} \times M} K(x,y) s_{\vert y} dy$$
Especially, $K$ maps from $M \times M$, while $-D^2e^{-tD^2}$ (for fixed $t$) maps from $E$. Any idea how $-D^2e^{-tD^2}$ can be interpreted as an integral kernel?
The assertion is supposed to be that $d(e^{-tD^2})/dt$ has the same smooth kernel as $-D^2 e^{-tD^2}$, i.e. they are the same operator. This is because $e^{-tD^2}$ is the solution operator to the heat equation: $\frac{du}{dt} = -D^2 u$. I'd be comfortable saying this is the definition of $e^{-tD^2}$, but you might prefer to construct $e^{-tD^2}$ using the pseudodifferential calculus in which case this fact would follow from the spectral theorem.
|
2025-03-21T14:48:31.261718
| 2020-06-18T00:22:58 |
363390
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"MMM",
"Todd Trimble",
"https://mathoverflow.net/users/153407",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630173",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363390"
}
|
Stack Exchange
|
Asymptotic behavior of a random geometric sum
Let $S_n$ denote a simple random walk with i.i.d. increments $X_i$ such that $P(X_1 = 0) = P(X_1=1) = 1/2$, i.e. $$S_0 = 0, \ S_n = X_1 + \dots + X_n.$$
The behavior of $S_n$ as $n \to \infty$ is clear, namely $S_n /n \to 1/2$ a.s. Now, let $q < 1$ and consider the random sum
$$ \sum_{k=0}^{n-1} q^{S_k}. $$
What can we deduce about the asymptotic behavior of this sum as $n$ tends to infinity? Any ideas how to approach this or references to work that have examined such sums?
Edit: From the answers below, we have $$ \sum_{k=0}^{n-1} q^{S_k} \to \sum_{k=0}^\infty q^{S_k}.$$
Now, for instance $$ \frac{q^{S_n}}{q^{n/2}} \to 1 \quad a.s.$$
My question is, whether we can find a similar sequence for $\sum_{k=0}^{n-1} q^{S_k}$ that describes its growth behavior, i.e. a function $f(n)$ (ideally a deterministic function) such that $$\frac{ \sum_{k=0}^{n-1} q^{S_k}}{f(n)} \to 1 $$
almost surely as $n \to \infty$?
$\newcommand\D{\overset D=}$
If $|q|<1$, then, by the strong law of large numbers, there is a positive integer-valued random variable (r.v.) $N$ such that $S_k>k/4$ a.s. and hence $|q|^{S_k}<|q|^{k/4}$ a.s. for all $k\ge N$; so, the sum $\sum_{k=0}^{n-1} q^{S_k}$ converges to a real-valued r.v. $\sum_{k=0}^\infty q^{S_k}$ a.s.
If $|q|\ge1$, then $|q|^{S_k}\ge1$ for all $k$, and hence the sum $\sum_{k=0}^{n-1} q^{S_k}$ diverges.
Concerning the edits to your question (where you apparently assume that $0<q<1$): It is of course incorrect that $\frac{q^{S_n}}{q^{n/2}}\to1$ a.s. Indeed, $\frac{q^{S_n}}{q^{n/2}}\to1$ a.s. can be rewritten as $S_n-n/2\to0$ a.s., which is clearly false -- because, say by the law of the iterated logarithm, $\limsup_n|S_n-n/2|=\infty$.
Next, it was shown above in this answer that the sum $\sum_{k=0}^{n-1} q^{S_k}$ converges to a real-valued r.v. $L:=\sum_{k=0}^\infty q^{S_k}$ a.s. Note that
\begin{aligned}
\sum_{k=0}^{n-1} q^{S_k}&=1+\sum_{k=1}^{n-1} q^{S_k} \\
&=1+q^{X_1}\sum_{k=1}^{n-1} q^{S_k-X_1} \\
&=1+q^{X_1}\sum_{k=1}^{n-1} q^{T_{k-1}} \\
&=1+q^{X_1}\sum_{j=0}^{n-2} q^{T_j},
\end{aligned}
where $T_j:=S_{j+1}-X_1=X_2+\dots+X_{j+1}\D S_j$ (with $T_0:=0$) and $\D$ denotes the equality in distribution. Note also that the $T_j$'s are independent of $X_1$. Letting now $n\to\infty$, we get the key identity for the limit r.v. $L$:
$$L\D1+q^X L, \tag{1}$$
where $X\D X_1$ and $X$ is independent of $L$.
From here, it follows that the r.v. $L$ is non-degenerate, that is, $P(L=c)\ne1$ for any real $c$. Indeed, otherwise (1) would imply that $c=1+q^X c$ a.s., which is of course false.
Thus, the condition
$$\frac{\sum_{k=0}^{n-1} q^{S_k}}{f(n)}\to1\quad\text{a.s.}$$
holds with $f(n)\equiv L$, but it cannot hold for any deterministic $f$.
Thank you very much. My main question was still something more precise (I did not make it clear enough before). See my Edit.
Thanks. The part was with $q^{S_n} / q^{n/2} \to 1 $ was stupid. It seems that it is difficult to match the growth behavior of $q^{S_n}$. Would you have any idea for what (random) function $f(n)$, we have $q^{S_n} / f(n) \to 1$ a.s.?
@BenC. : The obvious random $f(n)$ such that $q^{S_n}/f(n)\to1$ is $q^{S_n}$, and I don't think you can get anything more transparent than that.
I thought that this might be possible, because at least in expectation $E (q^{S_n}) / ((1+ q)/2)^n \to 1$. So I thought that maybe the same holds without the expectation...
@BenC. : The convergence of the expectations by itself may mean little about the convergence of the corresponding random variables, as it is the case here. Other than this, I have nothing to add to my answer and comments.
The function $\lim \limits_{n \to \infty} f(x)=\sum_{k=0}^{n-1} q^{s_k}x^k$ becomes,
$f(x)=\sum_{k=0}^{\infty} q^{s_k}x^k$.
From root test we get that radius of convergence of the series is
$\lim \limits_{n \to \infty} |q|^{\frac{s_n}{n}}=\sqrt{|q|}$.
Hence, $f(1)$ converges for $q<1$.
But for $q \geq1$ and $n \to \infty, \sum_{k=0}^{n-1} q^{s_k}$ is asymptotic to $nq^{\frac{n}{2}}$
( as $\lim \limits_{n \to \infty} \frac{s_n}{n}=\frac{1}{2}$)
Could you please explain why the asymptotic claim before the last limit statement holds?
|
2025-03-21T14:48:31.262097
| 2020-06-18T00:26:11 |
363391
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630174",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363391"
}
|
Stack Exchange
|
Classification of finite simple groups with abelian Sylow 2-subgroups
In this MathSE question,
classification of finite simple groups with Abelian Sylow 2-subgroups,
credit is rightly given to John Walter. But in the introduction to his paper, Walter explicitly states that "It seems to be a very difficult problem to show that these are the only examples." Is there a later reference, perhaps earlier than the complete classification theorem, that states that Walter, et al. found them all?
Thanks for your help.
The remark of Walter in his paper is referring specifically to the groups of Type (3) in his classification, that is, simple groups $S$ such that, for each involution $\tau \in S$, we have $C_S(\tau) \cong \langle \tau \rangle \times {\rm PSL}(2,q)$ with $q \equiv \pm 3 \bmod 8$.
These include the first Janko group $J_1$ (with $q=5$) and the groups of Ree type $^2G_2(q)$ with $q=3^k$ and $k$ odd.
It was quickly proved that any unknown simple group of this type must have similar properties to the groups of Ree type. John Thompson devoted a lot of time trying to prove that there were no further groups of this type, and he eventually reduced it to a problem in algebraic geometry, which was finally settled by Bombieri in 1980 in the paper:
Bombieri, Enrico (1980), appendices by Andrew Odlyzko and D. Hunt, "Thompson's problem ($\sigma^2=3$)", Inventiones Mathematicae, 58 (1): 77–100, doi:10.1007/BF01402275, ISSN 0020-9910, MR 0570875.
Of course "Thompson's Problem ($\sigma^2=3$)" is a strange title for a mathematical paper, but it was solving an important problem! I think Bombieri proved it for sufficiently large $q$, and the appendices of the paper describe computer calculations to settle the remaining small values.
So yes, this was resolved before the complete classification, but not so long before. I remember at the time that people were speculating that this problem might turn out to be the last one to be resolved.
It is described in Gorenstein's book on finite simple groups.
|
2025-03-21T14:48:31.262269
| 2020-06-18T01:44:19 |
363397
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fedor Petrov",
"Sam Hopkins",
"Will Jagy",
"darij grinberg",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/3324",
"https://mathoverflow.net/users/4312"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630175",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363397"
}
|
Stack Exchange
|
Canon in algebraic combinatorics and how to study
1) In subjects such as algebraic geometry, algebraic topology there is a very basic standard canonical syllabus of things one learns in order to get to reading research papers.
Is there a similar canon in algebraic combinatorics? (e.g., does someone working in matroids have knowledge of symmetric functions and vice versa?)
2) I want to know how much of EC 1,2 does a regular algebraic combinatorics researcher know? Do I try to solve the vast breadth of problems (at least the ones with difficulty level less equal 3- let's say) in those two books? How about attempting at reading and solving Bourbaki's Lie Groups and Lie Algebras chapters 4-6? This seems like the most read book of Bourbaki, and a treasure trove of Coxeter Group-Root System material. How do I go about studying Macdonald's Symmetric Functions and Hall Polynomials? I mention these books because they appear to be listed as a reference in many of the papers I see. But these are enormous, and reading and solving problems from cover to cover is probably impractical (Is it?).
I want to know how people tackle these kind of classic references. As well as how to practically study algebraic combinatorics.
3) Can someone point out if there is a list of topics-books/notes/videos (similar to https://math.stackexchange.com/questions/1454339/undergrad-level-combinatorics-texts-easier-than-stanleys-enumerative-combinator but with topics such as matroid theory, Coxeter groups, crystal bases included)?
Can the criminality be defeated? No. Should we fight against the criminality? Yes. The same with exercises in EC 1,2.
There is no established concept of what algebraic combinatorics is (though this isn't much different from other subjects: e.g., do Gröbner bases belong to algebraic geometry?). At best you can try to cluster mathematicians according to their joint knowledge. EC1-2 form a major cluster in the sense that someone who knows the material of one chapter is rather likely to know that of another; still, very few are really deeply familiar with the whole territory.
I think algebraic topologists and algebraic geometers might dispute your first sentence.
The "canonical" reference on crystal bases is now the eponymous book by Bump and Schilling. On Coxeter groups, most tend to recommend Björner/Brenti for a first combinatorial introduction (there are also notes by Heckman geared more towards geometers).
Matroid theory now has several long monographs devoted to it: Oxley, Welsh, White, White again...
It's a very strange though not unusual idea that one must study the subject before starting to work in it. No, you really don't, at least not in combinatorics.
Let me clarify how the process works. You study the subject to be smart, to learn basic ideas, tools and techniques. For that you take a class and do the homeworks, with exercises carefully chosen by the instructor. The EC1, EC2 is an opus magnum — part monograph, part textbook, and part reference source. I don't think it's meant to be read cover-to-cover with an attempt to solve all exercises (note aside - there are also supplementary exercises).
When you are mature and confident enough, you start working on a problem. The problem will guide you to tools, ideas and concepts that you don't yet know. If they are from a chapter of EC that you have not studied, you read it and go through some exercises which seem relevant. If they are from a different area altogether (say, from commutative algebra), you read up on that. Then go back to the problem and hope your newly acquired tools will prove helpful.
It can happen that once you learn the true nature of the new tools you realize that they are inapplicable or too weak/general to be used for your purposes. You are then back to square one, enriched with some new knowledge which you might find useful later in your work. But it doesn't mean you have to study the whole area before starting to work on a problem.
Let me leave you with a quote from Béla Bollobás on a related point, see here.
For me, the difference between combinatorics and the rest of mathematics is that in combinatorics we are terribly keen to solve one particular problem by whatever means we can find. So if you can point us in the direction of a tool that may be used to attack a problem, we shall be delighted and grateful, and we’ll try to use your tool. However, if there are no tools in sight then we don’t give up but we’ll try to use whatever we have access to: bare hands, ingenuity, and even the kitchen sink. Nevertheless, it is a big mistake to believe that in combinatorics we are against using tools — not at all. We much prefer to get help from “mainstream” mathematics rather than use “combinatorial” methods only, but this help is rarely forthcoming. However, I am happy to say that the landscape is changing.
Dear Igor, I noticed you were a faculty name on a rectangular tiling thesis...You might enjoy (and know references) for a three-dimensional tiling problem (not mine...) https://math.stackexchange.com/questions/3772410/building-a-cube-from-small-bricks-such-that-no-lines-can-be-pushed-through-betwe that generalizes a Martin Gardner problem
I would suggest that much of what I said in my answer to another MO question applies here, in spades. To a first approximation, the canon is the empty set. Start with a problem, and learn what you need as you go along.
See also How to escape the inclination to be a universalist or: How to learn to stop worrying and do some research.
|
2025-03-21T14:48:31.262682
| 2020-06-18T05:52:14 |
363404
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Francois Ziegler",
"Robert Bryant",
"YCor",
"https://mathoverflow.net/users/13972",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/19276"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:630176",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363404"
}
|
Stack Exchange
|
Emergence of the orthogonal group
Do we know what mathematician first considered, and perhaps named, what we call the group $\mathrm O(n)$, or $\mathrm{SO}(n)$, for some $n>3$?
I mean it specifically as group (not Lie algebra) acting on Euclidean $n$-space. For $n=3$ Jordan (1868) seems a definite upper bound, but for higher $n$ it seems not clear to me that even Cartan (1894) thought in those terms, describing as he does $\mathsf B_l$ and $\mathsf D_l$ as “projective groups of a nondegenerate surface of second order in spaces of $2l$ and $2l-1$ dimensions.” Also please disregard any implicit occurrence of $\mathrm{SO}(4)$ in quaternion theory.
In Jordan's 1868 paper all groups are finite, so how could the real group $\mathrm{(S)O}(3)$ be defined?
@YCor §§14-15 “Il nous reste à examiner le cas où parmi les rotations du groupe, il en existe une $A_\varrho$ dont l’amplitude $\varrho$ soit infiniment petite (...) Le groupe contenant une rotation quelconque autour d’un quelconque de ces axes, contiendra toutes les rotations possibles (...) 6ème type. Il est formé par l’ensemble de tous les mouvements de rotation possibles.”
Oh, thanks. My knowledge on that part of history was mainly based on Wüssing's book, and I actually see that this 1868 memoir is indeed mentioned therein (p196-197), which I had overseen or forgotten.
Your quote about Cartan thinking of $B_n$ and $D_n$ as 'projective groups..." is actually Cartan describing the lowest dimensional homogeneous space of these groups (except, of course, for a few exceptional cases such as $D_2$, which is not simple, and therefore should be left out of the description).
If you go just a little bit further in Cartan's 1894 Thesis, to Chapitre VIII, Section 9, you'll see that Cartan describes linear representations as well. For example, of $B_\ell$, he writes "C'est le plus grand groupe linéare et homogéne de l'espace à $2\ell{+}1$ dimensions qui laisse invariante la forme quadratique
$$
{x_0}^2 + 2x_1x_{1'} +2x_2x_{2'} + \cdots + 2x_\ell x_{\ell'}"
$$
with a similar description for $D_\ell$.
In fact, he gives the lowest dimensional representation of each of the simple groups over $\mathbb{C}$, including the exceptional ones and, except for $\mathrm{E}_8$, he explicitly describes the equations that define the representation. For example, he writes down an explicit homogeneous cubic in 27 variables and states that $\mathrm{E}_6$ is the the subgroup of $\mathrm{GL}(27,\mathbb{C})$ that preserves this cubic form.
For the summary theorem on the linear representations, see Chapitre VIII, Section 10, where he lists each of the lowest representations and notes the various low dimensional exceptional isomorphisms as well.
Remark 1: Cartan continues to refer to groups of type $B$ and $D$ merely as "the largest groups preserving a quadratic form in $n$ variables" or similar terms for a long time. Even in his papers of 1913–1915 classifying the real forms of the complex simple Lie groups, he uses such terminology, though he clearly finds the special case of the compact real forms of special interest.
The first place that Cartan actually refers to 'orthogonal groups' that I can recall are in his 1926–27 papers on the classification of Riemannian symmetric spaces. There, he begins referring to any subgroup of $\mathrm{GL}(n,\mathbb{R})$ that preserves a quadratic form as 'an orthogonal group' and then, later, finally refers to the maximal group that preserves a positive definite quadratic form as 'the orthogonal group'. I don't recall when or whether he used any notation such as $\mathrm{O}(n)$ or $\mathrm{SO}(n)$.
Whether the term 'orthogonal group' was original to him, I can't say. By that time, of course, Weyl had already started his research on compact Lie groups, and it may be that Weyl had already used the term 'orthogonal group' well before Cartan.
Remark 2: Euler's article (Problema algebraicum ob affectiones prorus singulares memorabile, Novi commentarii academiae scientiarum Petropolitanae 15 (1770) 1771, 75–106) discusses the problem of parametrizing the solutions of the equation $A^TA = I_n$ where $A$ is an $n$-by-$n$ matrix for $n=3$, $4$, and $5$, particularly the rational solutions. He does not use the terminology 'orthogonal' or 'group'. Nevertheless his article does contain some remarkable formulae that clearly anticipate the development of the algebra of quaternions.
For example, identifying $\mathbb{R}^4$ with the quaternions $\mathbb{H}$ in the usual way, it is a now-standard fact that every special orthogonal linear transformation $M$ of $\mathbb{R}^4=\mathbb{H}$ can be written, using quaternion multiplication, in the form $M(X) = A\,X\,\bar B$ where $A$ and $B$ are unit quaternions and $X\in\mathbb{H}$. (This is now the usual way that the double cover $\mathrm{Spin}(3)\times\mathrm{Spin}(3)\to\mathrm{SO}(4)$ is introduced.) Meanwhile conjugation $c:\mathbb{H}\to\mathbb{H}$ is orthogonal but has determinant $-1$, so every element of the non-identity component of $\mathrm{O}(4)$ can be written as $$M'(X) = Ac(X)\bar B = A\,\bar X\, \bar B = A\overline{BX} = Ac(BX).$$ Remarkably, Euler gives this formula for parametrizing $\mathrm{O}(4)$ in the form of the product of matrices $L_A\,c\,L_B$ (where $L_P$ denotes left multiplication by the quaternion $P$), many years before the 'official' discovery of quaternions.
Thanks! I should definitely have read further. Would you say that Cartan has a chance of being first to “consider” the group, though not name it other than by one subscripted letter? Ironically Frobenius (1878) seems to do almost the opposite...
@FrancoisZiegler: I wouldn't venture to say that. From what Carlo wrote in his answer, it would appear that Hurwitz already was dealing with $\mathrm{SO}(n)$ and $\mathrm{U}(n)$ as matrix groups in 1897, which seems pretty advanced for the time. Are you sure that Lie and/or Killing didn't recognize that the stabilizer groups of nondegenerate quadratic forms were of Lie's type B or D? Since the relation between 'linear fractional transformations' and $\mathrm{SL}(2)$ was well understood by Lie, you would expect that he'd have realized that 'projective groups' were matrix groups in disguise.
I don’t know! The problem with Lie and Killing is that they are hard to catch talking about anything but Lie algebras. E.g. Lie (1893, p. 317) describing $\mathfrak{so}(n)$ as “the group $x_\nu p_k-x_k p_\nu\quad (\nu,k=1\dots n)$”.
@FrancoisZiegler: Hmmm. I remember that Lie used $p_k$ for what we call the vector field $\frac{\partial}{\partial x_k}$, so he's discussing the group acting on $\mathbb{R}^n$ with 'infinitesimal generators' $x_\nu,\frac{\partial}{\partial x_k} -x_k,\frac{\partial}{\partial x_\nu}$, which is, of course, what we now call the (special) orthogonal group acting on $\mathbb{R}^n$. He surely knew that this was a linear action on $\mathbb{R}^n$. Lie's whole point was that the infinitesimal generators determine the group, so specifying the group by its Lie algebra would have been natural for him.
Right, that’s how we recognize what group Lie is talking about. (I’d say he used $p_k$ for the hamiltonian generating infinitesimal translations in the $k$th direction.) I also recall now that Euler could be said to have “considered” — but certainly not named — $\mathrm O(4)$ in (1771, §20). But that was rather isolated and ahead of his time.
@FrancoisZiegler: Thanks for pointing out Euler's 1771 paper. I had a look at it. Doesn't he consider $\mathrm{O}(5)$ as well, starting in $\S28$? (Of couse, he didn't name it as a group.) Also, he has the formula for the double cover $S^3\to\mathrm{SO}(3)$ in $\S33$ and quaternion multiplication in $\S34$! I had not known this before.
About $\mathrm O(5)$: absolutely and again I hadn’t read that far. Euler seemed only limited by the length of the alphabet :-) If you would include that in the answer I’d like to accept it. About quaternion multiplication, I think he almost had it but inserted a $$\mathrm{diag}(1,-1,-1,-1)$$ for which I’d love to understand the reason — see formula (5) of this answer which goes into more details.
@FrancoisZiegler: I think that the map $c = \mathrm{diag}(1,-1,-1,-1)$ (which is conjugation in the quaternions) is there so that the product will parametrize $\mathrm{O}(4)$, i.e., the product is the matrix for $$X\mapsto Ac(BX) = A\overline{BX} = A\overline {X},\overline{B}.$$ This shows how Euler parametrizes the non-identity component of $\mathrm{O}(4)$ via left multiplication by two (unit) quaternions $A$ and $B$ and conjugation as $AcB$. To get the identity component, i.e., $\mathrm{SO}(4)$, you'd use the product $AcBc$, for the linear map $X\mapsto Ac(B(cX)) = AX\overline{B}$.
When you say “the product” you mean Euler’s in (5), right? That’s brilliant and would show that he was not 1 but 2 steps ahead of everyone... One has to wonder if Gauss had read this paper; maybe even Cayley / Hamilton absorbed more of it than they let on.
@FrancoisZiegler: Yes, of course, the formula $(5)$ what I meant. I assumed from the beginning that Euler was doing that because he essentially says in the text that he wants to show how to parametrize the rational elements of $\mathrm{O}(4)$, i.e., solve the equations defining a matrix in $\mathrm{O}(4)$ in rational numbers.. (Although I confess that my Latin is very bad, I am sure that this is essentially what he is saying.)
Perhaps the next person to consider (and famously parametrize, again without naming them) “les coëfficients propres à effectuer la transformation de deux systèmes de coordonnées rectangulaires” is Cayley (1846, equation 16 on p. 120).
There may be an earlier source, but Adolf Hurwitz 1897 is one upper bound:
A. Hurwitz, Über die Erzeugung der Invarianten durch Integration, Nachr. Ges. Wiss. Göttingen (1897), 71–90.
Hurwitz’s paper introduced and
developed the notion of an invariant measure for the matrix groups
SO(N) and U(N). He also specified a calculus from which the explicit
form of these measures could be computed in terms of an appropriate
parametrisation — Hurwitz chose to use Euler angles. This enabled him
to define and compute invariant group integrals over SO(N) and U(N).
source: A. Hurwitz and the origins of random matrix theory in mathematics
Thanks. Indeed Hurwitz definitely speaks of “der orthogonalen Gruppe” (pp. 72, 75). He also cites e.g. Kronecker (1890) who had “die Gesammtheit der orthogonalen Transformationen” (p. 376), “die Mannigfaltigkeit der orthogonalen Transformationen” (p. 455).
Perhaps the first to speak of orthogonal transformations (though not orthogonal group) is J. J. Sylvester in A Demonstration of the Theorem that every Homogeneous Quadratic Polynomial is reducible by real orthogonal substitutions to the form of a sum of Positive and Negative Squares, Phil. Mag. (4) 4 (1852) 138-142. He refers to Boole who earlier called them rectangular transformations in On the Theory of Linear Transformations (1851, p. 88).
I guess the switch may have been because “orthogonal matrix” was preferable to “rectangular matrix”. As to groupe orthogonal, the name seems to have been first used by Jordan in Traité des substitutions et des équations algébriques (1870), p. 155.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.