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2025-03-21T14:48:31.430937
| 2020-07-04T06:33:20 |
364809
|
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|
Stack Exchange
|
RKHS norm of Lipschitz functions
Given a set $\mathcal{X}$ and RKHS $\mathcal{H}$ of functions on $\mathcal{X}$, we can recover a (pseudo)metric on $\mathcal{X}$ by $d(x,y)=||\phi_x-\phi_y||_{\mathcal{H}}$, where $\phi_x=k(x,\cdot)$.
It is straightforward to see that any function $f \in \mathcal{H}$ which has RKHS norm less than $L$ is Lipschitz (with respect to our metric above) with constant $L$:
$$|f(x)-f(y)|=|\langle f,\phi_x-\phi_y\rangle|\leq ||f||_{\mathcal{H}}d(x,y), $$
for any $x,y \in \mathcal{X}$.
I am very interested in the following question: if we have a function $f\in \mathcal{H}$ which is Lipschitz with constant $L$, is there anything we can say about it's norm $||f||_{\mathcal{H}}$?
by "anything we can say", do you mean bounding it from above? If yes, then you want your Hilbert space to be isomorphic to a subspace of a Lipschitz space. I would expect a subspace of the Lipschitz space isomorphic to a Hilbert space necessarily be finitely dimensional, but I am not sure.
What happens when $\mathcal{X}=\mathbb{Z}$ and $\mathcal{H}=l^2$?
@erz yes, bounding it above was precisely what I was looking for
I think that in general $L(.)$ and $\Vert.\Vert_\mathcal{H}$ measure quite different things.
Writing $L(f)$ for
$$
\inf\{ M>0:|f(x)-f(x')| \leq Md(x,y) \;\forall \;x,x'\in \mathcal{X}\}
$$
let $\mathcal{X}=\mathbb{Z}$ and $\mathcal{H}=l^2$. Then (unless I've made an embarrassing mistake...) setting $f_n=1_{[-n,n]}$ gives you a sequence in $\mathcal{H}$ with $L(f_n)=1/2$ and $\Vert f_n\Vert_\mathcal{H}=2n$.
Hmm, correct me if I’m wrong but I thought $l^2$ wasn’t an RKHS? Either way it’s an interesting counterexample
What's your definition of a RKHS? I think the usual one is just "a Hilbert space of functions on a set whose evaluation functionals are norm continuous", no?
I think another example is $\mathcal{H} = H^1_0((0,1])$, i.e. the absolutely continuous functions $f : [0,1] \to \mathbb{R}$ with $f' \in L^2([0,1])$, under the norm $|f|_{\mathcal{H}}^2 = \int_0^1 |f'|^2$. Then the "Lipschitz constant" is the Holder norm of exponent 1/2, $L = \sup |f(s)-f(t)|/\sqrt{|s-t|}$. So now let $f_n$ approach $f(x) = \sqrt{x}$ which has infinite $\mathcal{H}$-norm.
|
2025-03-21T14:48:31.431102
| 2020-07-04T07:06:50 |
364810
|
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|
Stack Exchange
|
Image of function contains identity elements
Let $U$ be the set of all nonempty subsets of $[0,1]$ that are a union of finitely many closed intervals (where an "interval" that is a single point does not count as an interval). Let $f:U\times U\rightarrow U$ be such that for any $A,B\in U$:
(a) $f(A,B)\neq [0,1]$.
(b) $f(A,B)\cap A\neq\emptyset$ and $f(A,B)\cap B\neq\emptyset$.
(c) The length (i.e. Lebesgue measure) of $f(X,B)\cap A$ is maximized at $X=A$, and the length of $f(A,X)\cap B$ is maximized at $X=B$.
Is it true that the image of $f$ must be equal to the set $\{A\mid f(A,A)=A\}$?
Note: The original unsolved question asked whether such $f$ exists. The claim of this question is made in the comments of that question, but the commenter could not recall its proof or whether the proof ever existed.
(My deleted answer below is foolishly wrong ... should have proofread it before posting.)
Does condition (c) mean that those are unique maximisers, or could the maxima also be achieved elsewhere?
@LSpice The maxima could also be achieved elsewhere.
|
2025-03-21T14:48:31.431209
| 2020-07-04T07:12:22 |
364811
|
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"Philippe Gaucher",
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|
Stack Exchange
|
Almost combinatorial accessible model categories
Theorem: Assume VP. Let $\mathcal{M}$ be an accessible model category
such that there exists a set of generating cofibrations $I$ and such
that all objects are fibrant. Then it is combinatorial.
Proof: Consider the left determined model structure with respect to $I$: the minimal $I$-localizer is denoted by $\mathcal{W}_I$. Denote by $\mathcal{W}$ the class of weak equivalences of $\mathcal{M}$. Then $\mathrm{cof}(I) \cap \mathcal{W}_I \subset \mathrm{cof}(I) \cap \mathcal{W}$. Therefore all objects of the left determined model structure are fibrant. Since a model structure is characterized by its class of cofibrations and its class of fibrant objects, $\mathcal{M}$ is left determined, and therefore is combinatorial.
Question: Can we remove VP in the statement of the theorem ? I mean, is it as difficult as Smith's conjecture about left determined
model structures ? Note that in the proof, I don't use the fact that
the model category $\mathcal{M}$ is accessible, I only use the fact
that the underlying category is locally presentable. The question is:
is it easier when one already knows that $\mathcal{M}$ is accessible as a
model category ?
Actually, you can, and you don't need accessibility (local presentability of the underlying category is enough).
Under your assumption, for each $i:A \to B$ a generating cofibration, take $B \coprod_A B \hookrightarrow I_A B \to B$ a cylinder object, and let $j_i : B \hookrightarrow I_A B$ be the first leg inclusion. Then the $j_i$ form a generating set of trivial cofibrations.
The proof essentially use that all objects are fibrant, and that you can consider the weak factorization system generated by the $j_i$ (so some smallness or local presentability assumption).
The key step in the proof is to observe that a map that has the RLP against the $j_i$ and is a weak equivalence is a trivial fibration. This easily follows from the fact that weak equivalences between fibrant objects have the "up to homotopy" lifting property against all cofibration and that lifting property against the $j_i$ is enough to rectify this in an actual lifting property. One then check by hand the cofibration, weak equivalence and the map with the lifting property against the $j_i$ forms a model structure, with the same cofirbation and weak equivalences as the one you started from, hence is the the same model structure.
I found it explicitly written out as corollary 3.2 of:
Valery Isaev, On fibrant objects in model categories, Theory and Applications of Categories, Vol. 33, 2018, No. 3, pp 43-66, journal page, arXiv:1312.4327
I knew this paper but I had not paid attention to corollary 3.2. Thanks.
|
2025-03-21T14:48:31.431406
| 2020-07-04T08:16:07 |
364813
|
{
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"authors": [
"Cheng-Chiang Tsai",
"LSpice",
"Monty",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Question on external tensor product
Let $G_1,G_2$ be $p$-adic groups. Let $\rho_1,\pi_1$ be smooth representations of $G_1$ and $\rho_2,\pi_2$ be smooth representations of $G_2$. Assume that $\pi_2$ is admissible.
I am wondering that if $\text{Hom}_{G_1}(\rho_1,\pi_1)=0$, then $\text{Hom}_{G_1 \times G_2}(\rho_1 \boxtimes \rho_2,\pi_1 \boxtimes \pi_2 )=0$?
If it does not true in general, then is it true when $\pi_1$ is 1-dimensional representation?
Maybe I am being very stupid here: under the assumption, wouldn't $\mathrm{Hom}_{G_1}(\rho_1\otimes V_1,\pi_1\otimes V_2)=0$ for any vector spaces $V_1,V_2$ (in particular possibly infinite-dimensional) as the image of $v\otimes w$ is necessarily zero for any $v\in\rho_1$, $w\in V_1$, without even using the smoothness?
@Cheng, Thank you for your comment. Why do u think the image of $v \otimes w$ is necessarily zero?
I edited the question. How about when $\pi_1$ is one-dimensional?
$\DeclareMathOperator\Hom{Hom}$Assuming, as your title suggests, $\boxtimes$ stands for the external tensor product: Pick two different 1D representations $\rho_1$ and $\pi_1$ of $G$. Then $\Hom_{G_1}(\rho_1, \pi_1)$ is $0$ but $\Hom_{G_1 \times G_1}(\rho_1 \boxtimes \pi_1, \pi_1 \boxtimes \rho_1)$ is non-$0$.
@LSpice I believe the map you suggest is only $G_1$-equivariant (diagonally), not $G_1\times G_1$-equivariant: one way goes $(x,y)\mapsto(gx,g'y)\mapsto(g'y,gx)$ and another $(x,y)\mapsto(y,x)\mapsto(gy,g'x)$. Or do you have in mind some other map?
@მამუკაჯიბლაძე you are right.
@LSpice, I think it is true when $\pi_1$ is one-dimensional.
Following my original comment, suppose $\varphi\in\mathrm{Hom}{G_1}(\rho_1\otimes V_1,\pi_1\otimes V_2)$ is such that $\varphi(v\otimes w)\in\pi_1\otimes V_2$ is non-zero. We can always pick a functional $\phi$ on $V_2$ such that $\phi(\varphi(v\otimes w))\not=0$. But then $\phi\circ\varphi|{\rho_1\otimes\langle w\rangle}\in\mathrm{Hom}_{G_1}(\rho_1,\pi_1)\not=0$. Or is it?
@Cheng, Oh you are right! Thank you very much for your enlightening!
|
2025-03-21T14:48:31.431550
| 2020-07-04T08:35:44 |
364814
|
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|
Stack Exchange
|
Question about derivations of $R[X]$
$\DeclareMathOperator\Der{Der}$Consider the polynomial algebra $R[X] = R[x_1,\ldots, x_n]$ over the ring $R = \mathbb{C}[t]$ and let $\Der_{R}(R[X])$ be the Lie algebra of derivations of $R$-algebra $R[X]$.
Assume that for $n$ derivations $D_i= f_{i1}\partial_{x_1}+\dotsb+f_{in}\partial_{x_n}\in\Der_R(R[X])$, $f_{ij}\in R[X]$ we have that the polynomial $$F=\det(f_{ij})_{i, j=1,\ldots, n}\in R\setminus\{0\}.$$
Is it true that there exists an $R$-Lie algebra $\mathcal{C}\subset\operatorname{LND}(R[X])$ such that the set $\mathcal{C}\cap(\bigoplus_{1\leq i\leq n}R[X]\cdot D_i)$ generates $\bigoplus_{1\leq i\leq n}R[X]D_i$ as an $R[X]$-module?
What is LND?...
It is a set of all locally nilpotent derivations.
|
2025-03-21T14:48:31.431626
| 2020-07-04T09:55:05 |
364816
|
{
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"authors": [
"Gerhard Paseman",
"Martin Rubey",
"Nathan Reading",
"https://mathoverflow.net/users/3032",
"https://mathoverflow.net/users/3402",
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|
Stack Exchange
|
How to construct a lattice having a subset of a given relations?
I am given a (smallish, say $n=14$ element) set $X$, and a set $R$ of (a few hundred) quadruples of elements $(a, b, c, d)$ with $a,b,c, d\in X$.
I want to construct lattices on $X$, such that for all incomparable $a, b\in X$ the quadruple $(a,b,a\vee b, a\wedge b)$ is in $R$.
A trivial solution is to use an $n$-element chain as the lattice, which already gives me $14!=$a huge number of lattices. Thus, in a first step I'd like to find all non-isomorphic lattices. (Aside: is there a lattice structure on the set of isomorphism types of lattices on $n$ elements?)
Currently, I do not have a clue how many lattices I could build with my set, but I am certainly more interested in lattices with many incomparable elements.
As a first step, I determined which pairs $a,b$ do not occur as first two elements in one of the quadruples in $R$, because these must be comparable. In the first case I'm interested in, there are a dozen of these.
How could I do an exhaustive search?
Note that a and a join b are comparable. I suggest the next step is to look at those pairs formed from the first and third elements of each quadruple. In particular, if a and c are identical, b must be below a. Gerhard "See About The Pecking Order" Paseman, 2020.07.04.
Sorry, I should have mentioned that the quadruples are really four-element-subsets, that is, a,b,c, and d are all distinct.
Just so I am clear, can you explain the meaning of a meet c? Since you have only a and b as input, I want to understand the fourth coordinate of the output. Gerhard "For A Meeting Of Minds" Paseman, 2020.07.04.
Sorry, that was a typo!
You have an interesting kind of partitioning problem here. (Maybe that's why you are using lattices.)
One thing that should be noted. If d is a join b for some given a and b which are incomparable, then d is NOT a meet c for any c in the lattice. So when you focus on the first coordinate a, you immediately divide X minus a into three sets: those above a, those below a, and those that are incomparable to a..
This suggests an approach. Fix a, and look at the quadruples which have a in the first two coordinates. If two of the quadruples place d as being both above a and incomparable to a, then those two are incompatible conditions; one of them has to be thrown out, which means one of the b's that might lead to the conflict has to be comparable to b.
I recommend fixing a (or a small subset A which you assume to be an antivirus in the lattice), and see what inconsistencies you can generate. These inconsistencies will then inform you of comparability relations among the elements. Even if you have just one pair of incomparable elements, that will determine two other elements, and now you have less than 12! lattices to deal with. If nothing else, you can learn which three element subsets can be antichains in the lattice.
Gerhard "Let's Join Our Heads Together" Paseman, 2020.07.04.
When you say "antivirus" here, do you mean "antichain"? Either this is a lattice term I don't know, or we all have the word "virus" entirely too much on our minds, or autocomplete strikes again...
Oh no! I forgot my mask when posting that! Gerhard "Let's Blame This On Spellcheck" Paseman, 2020.07.06.
I support not editing out the typo. It's such a timely typo.
Nathan "Wish I Could Generate Cool Middle Names Like Gerhard" Reading, 2020.07.06.
Practice makes tolerable. You are invited to copy the style, but not the content. Gerhard "Don't You Steal My Signatures!" Paseman, 2020.07.06.
No, I would never dream of copying the style (outside of this little exchange). Nathan "There Can Only Be One Gerhard '...' Paseman" Reading
|
2025-03-21T14:48:31.431896
| 2020-07-04T10:07:52 |
364817
|
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|
Stack Exchange
|
Questions related to compact complex curves, symmetric products and linear independence
Let $X$ be a compact complex curve and let $L$ be a very ample line bundle over $X$. Denote by $C_n( X )$ the configuration space of $n$ (ordered) distinct points on $X$.
Given distinct points $z_1$, ..., $z_n$ on $X$, so that they form an element in $C_n( X )$, let
$$p_1 = [ (z_2, ... ,z_n) ],$$
$$p_2 = [ (z_1, z_3, ..., z_n) ],$$
$$\vdots$$
$$p_n = [ (z_1, z_2, ..., z_{n-1}) ]$$ in the symmetric product $\operatorname{Sym}^{n-1}X$ of n-1 copies of X.
Note that on the product $X^{n-1}$ of $n-1$ copies of $X$, one can pullback $L$ using the first projection map, and so on, then pullback $L$ using the ($n-1$)-st projection map, and then tensor all these line bundles together. This defines a line bundle on $X^{n-1}$, which descends to a line bundle on $\operatorname{Sym}^{n-1}X$. Let us denote the resulting line bundle, say, by $\operatorname{Sym}^{n-1}( L )$.
Question 1: if $L$ is very ample, is $\operatorname{Sym}^{n-1}( L )$ also very ample? (I would think so)
If this is indeed the case, then $\operatorname{Sym}^{n-1}( L )$ induces an embedding from $\operatorname{Sym}^{n-1}X$ into a complex projective space, namely
$$PH^0(\operatorname{Sym}^{n-1}X, \operatorname{Sym}^{n-1}( L ))^*,$$
the projectivization of the dual of the space of holomorphic sections of $\operatorname{Sym}^{n-1}( L )$). Let us denote this embedding by $f$.
Question 2: Is it true that $f(p_1)$, ..., $f(p_n)$ are linearly independent, as points in complex projective space?
I have a feeling this is true. I also ask similar questions if $p_1,\ldots,p_n$ are instead defined by:
$$p_1 = [(z_1,\ldots,z_1)],$$
$$p_2 = [(z_2,\ldots,z_2)],$$
$$\vdots$$
$$p_n = [(z_n,\ldots,z_n)]$$
in $\operatorname{Sym}^{n-1}X$.
I know that the answers to my questions are yes in the case where $X = P^1(\mathbb{C})$ and $L = \mathcal{O}(1)$. Indeed, the linear independence statement amounts to the fact that the Lagrange basis (related to Lagrange interpolation) is linearly independent (so indeed a basis). And for the modified $p_i$, this amounts to the well known fact that $n$ distinct points on a rational normal curve of degree $n-1$ in $P^{n-1}(\mathbb{C})$ are linearly independent, itself equivalent to the non-vanishing of a Vandermonde determinant, from an algebraic point of view.
I wonder if my statements still hold if $X$ is any compact complex curve of genus $g$ and if $L$ is any very ample line bundle on $X$.
|
2025-03-21T14:48:31.432177
| 2020-07-04T11:17:06 |
364818
|
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|
Stack Exchange
|
Homotopy type of the affine Grassmannian and of the Beilinson-Drinfeld Grassmannian
The affine Grassmannian of a complex reductive group $G$ (for simplicity one can assume $G=GL_n$) admits the structure of a complex topological space. More precisely, the functor $$X\mapsto |X^{an}|$$ that associates to a scheme locally of finite type the underlying topological space of its analytification ([SGA1, XII, 1]) can be left-Kan-extended to a functor $PSh(Sch)\to Top$, and therefore applied to the ind-scheme $Gr_G$.
Some topological properties of this space are known: its connected components are in bijection with the topological fundamental group $\pi_1(G)$ ([Beilinson-Drinfel'd, Quantization of Hitchin's integrable system and Hecke eigensheaves, Proposition 4.5.4]). The étale fundamental group (which is the profinite completion of the topological fundamental group) of each orbit in $Gr_G$ under the left action of $L^+G$ is trivial (Richarz, Affine Grassmannians and the Geometric Satake Equivalences, proof of Proposition 3.1).
Also, $|Gr_G^{an}|$ is known to be homotopy equivalent to a polynomial loop space ([Zhu, An introduction to the affine Grassmannian and the Geometric Satake Equivalence] and [Pressley/Segal, Loop Groups, Sec. 8.3]).
However, my knowledge essentially stops here. For example, the homotopy groups of this polynomial loop space are unknown to me.
Has anyone ever written or thought about the topology of the affine Grassmannian in further detail? Relevant questions to me are the following:
What are the fundamental groups of the connected components of $|Gr_G^{an}|$?
The cohomology of the affine Grassmannian is much more studied and used than its homotopic properties. Is there any chance that the connected components of $|Gr_G^{an}|$ satisfy conditions like "being a simple space", which often allows homotopical issues to be translated into cohomological issues, by means of the homological Whitehead theorem?
Needless to say, the answer to such questions for the Beilinson-Drinfel'd Grassmannian will probably be more difficult. However, any ideas or references in this more sophisticated context would be highly appreciated.
Thank you in advance.
|
2025-03-21T14:48:31.432338
| 2020-07-04T13:43:06 |
364825
|
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|
Stack Exchange
|
Does $\pi_k(M)\neq 0$ implies $\operatorname{ind}(\gamma) < k$?
Cross post from MSE. and sorry if this is an obvious question.
Here is a line of proof of Theorem 1.15 from
Brendle, Simon, Ricci flow and the sphere theorem, Graduate Studies in Mathematics 111. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4938-5/hbk). vii, 176 p. (2010). ZBL1196.53001.
Let us fix two
points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > \frac{\pi}{2}$. Since $\pi_k(M)\neq 0$, there
exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$.
Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact?
Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$.
Very roughly speaking: Geodesics $\gamma:[0,1]\to M$ with $\gamma(0)=\gamma(1)=p$ and $\operatorname{ind}(\gamma)<k$ correspond to critical points of Morse index less than $k$ of the energy functional $E:\Omega_p(M)\to \mathbb{R}$, where $\Omega_p(M)$ is the space of loops based at $p$.
The assumption that $\pi_k(M)\cong\pi_{k-1}(\Omega_p(M))\neq 0$ implies that $H_i(\Omega_p(M))\neq 0$ for some $i\leq k-1$, by the Hurewicz Theorem. Then Morse homology implies there must be a critical point of index less than $k$.
|
2025-03-21T14:48:31.432436
| 2020-07-04T14:23:06 |
364828
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364828"
}
|
Stack Exchange
|
Special fiber of a reflexive sheaf over DVR
Let $f:X \to \mbox{Spec}(R)$ be a flat, projective morphism with $R$ a discrete valuation ring and the special and generic fibers of $f$ are normal and integral. I am looking for examples of rank $1$, reflexive sheaves on $X$ such that its restriction to the generic fiber is reflexive but its restriction to the special fiber $X_k$ is not a reflexive sheaf on $X_k$. Any hint/reference will be most welcome.
|
2025-03-21T14:48:31.432490
| 2020-07-04T15:05:32 |
364830
|
{
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"KReiser",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364830"
}
|
Stack Exchange
|
Slope stability using the Riemann-Roch theorem
I am studying about the (Stability on a curve). Suppose $C$ is a smooth curve of genus g. The Riemann-Roch
theorem asserts that if $E$ is a coherent sheaf on $C$ then the Euler characteristic of $E $ is $\chi(E)= deg(E)+ r(E)(1-g)$
Polarizing $C$ with $H = p$ a point, we find
Then The Hilbert polynomial $P(E,m)= \chi(E(m))$ $=deg(E(m))+r(E)(1-g)$.
My question is how can we rewrite these equation to become $r(E)m +(deg(E)+r(E)(1-g))$?
Where $E(m)$ Is the twisted sheaf
Crossposting from MSE after such a short period of time is not within community norms. One should also link cross-posts together to prevent the duplication of work.
|
2025-03-21T14:48:31.432570
| 2020-07-04T16:12:15 |
364834
|
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"authors": [
"Asvin",
"Bazara",
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"https://mathoverflow.net/users/58001"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364834"
}
|
Stack Exchange
|
When every localization of the polynomial ring over a ring has finitely many idempotents
Let $R$ be a commutative ring such that every localization ring $R_r$ has finitely many idempotents for each non nilpotent element $r\in R$. Why dose every localization ring $R[x]_{f(x)}$ have finitely many idempotents for each non nilpotent polynomial $f(x)\in R[x]$?
By quotient ring $R_r$, do you mean $R/(r)$ or the localization?
@Asvin: thank you for your comment. It was corrected.
|
2025-03-21T14:48:31.432628
| 2020-07-04T17:05:44 |
364837
|
{
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],
"authors": [
"Michael Albanese",
"https://mathoverflow.net/users/21564",
"https://mathoverflow.net/users/27004",
"wonderich"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364837"
}
|
Stack Exchange
|
Instanton numbers for diverse gauge bundles on diverse manifolds --- their relations to characteristic classes
It is standard (?) that the $SU(N)$ gauge theory has the instanton number $n$ quantized as $n \in \mathbb{Z}$
$$
n = { 1 \over 8\pi^2} \int_{\mathcal{M}_{4}} \text{tr} \left(F \wedge F\right) = {1 \over 64 \pi^2} \int_{\mathcal{M}_{4}} d^4 x \, \epsilon^{\mu\nu\rho\lambda} F_{\mu\nu}^\alpha F_{\rho\lambda}^\alpha
\in \mathbb{Z}
$$
Here $F$ is the curvature 2-form $F=d A + A \wedge A$ and $A$ is the gauge 1-connection of the gauge bundle of gauge group $G$. Also $F=F^\alpha T^\alpha$ where $\alpha$ is the Lie algebra indices, with repeated indices summed over.
We know that:
If the gauge group is $G=SU(2)$, the instanton number $n \in \mathbb{Z}.$ I think this can be understood as
$$
n = c_2(V_{SU(2)}) \in \mathbb{Z}?
$$
If the gauge group is $G=SO(3)$ on the spin 4-manifold $\mathcal{M}_{4}$, then $n \in \frac{1}{2}\mathbb{Z}$. (I use the notation to mean that $n \in \frac{1}{2}\mathbb{Z}$ as $n$ takes the half integer values.)
$$
n = p_1(V_{SO(3)})/4 \in \frac{1}{2}\mathbb{Z}?
$$
If the gauge group is $G=SO(3)$ on the non-spin 4-manifold $\mathcal{M}_{4}$, then $n \in \mathbb{Z}/4$.
$$
n = p_1(V_{SO(3)})/4 \in \frac{1}{4}\mathbb{Z}?
$$
What are the general statements we can make for other general $G$ and other manifolds?
Questions:
If the gauge group is $G=U(1)$, the instanton number $n \in 2 \mathbb{Z}.$ True or False? We can express this $n$ as the first Chern class $c_1$ square of associated vector bundle $V_{U(1)}$ as
$$
n = c_1(V_{U(1)})^2 \in 2 \mathbb{Z}?
$$
Is this correct?
If the gauge group is $G=SU(N)$, the instanton number $n \in \mathbb{Z}.$ True or False? We can express this $n$ as the second Chern class $c_2$ of associated vector bundle $V_{SU(N)}$ as
$$
n = c_2(V_{SU(N)}) \in \mathbb{Z}?
$$
If the gauge group is $G=PSU(N)$ on the spin 4-manifold $\mathcal{M}_{4}$, the instanton number can be $1/N$ fractional of $\mathbb{Z}$ values:
$$n \in \frac{1}{N} \mathbb{Z},$$
True or False? What is the precise mathematical invariant to characterize this $n \in \frac{1}{N} \mathbb{Z}$? Is that Pontryagin class $p_1$ when $G=PSU(N)$ is real $\mathbb{R}$? Or some $c_2(V_{PSU(N)})$ when $PSU(N)$ is complex $\mathbb{C}$?
If the gauge group is $G=PSU(N)$ on the non-spin 4-manifold $\mathcal{M}_{4}$, the instanton number can be $1/N^2$ fractional of $\mathbb{Z}$ values: $$n \in \frac{1}{N^2} \mathbb{Z},$$
True or False? What is the precise mathematical invariant to characterize this $n \in \frac{1}{N^2} \mathbb{Z}$? Is that Pontryagin class $p_1$ when $G=PSU(N)$ is real $\mathbb{R}$? Or some $c_2(V_{PSU(N)})$ when $PSU(N)$ is complex $\mathbb{C}$?
If the gauge group is $G=U(N)$ on the spin or non-spin 4-manifold $\mathcal{M}_{4}$, the instanton numbers carry both the $U(1)$ and $PSU(N)$ part with constraints. So there are two instanton numbers
$n_{U(1)}$ and $n_{PSU(N)}$.
What can be their constraints?
$$n_{U(1)} \in \text{what characteristic class or invariant} ?$$
$$n_{PSU(N)} \in \text{what characteristic class or invariant} ?$$
For $G = O(N)$ or $SO(N)$, we have $n = -\frac{1}{2}\int_Mp_1(E)$ where $E$ is a real (respectively orientable) vector bundle of rank $N$. For $G = U(N)$ we have $n = -\frac{1}{2}\int_Mp_1(E) = -\frac{1}{2}\int_Mc_1(E)^2 - 2c_2(E)$ where $E$ is a complex vector bundle of rank $N$, and for $G = SU(N)$ we have $n = -\frac{1}{2}\int_Mp_1(E) = \int_Mc_2(E)$ where $E$ is a complex vector bundle of rank $N$ with $c_1(E) = 0$.
For $G = PSU(N)$, one has natural projective bundles, but I don't know if you can get vector bundles in any natural way. You need a real vector bundle to talk about $p_1$ or a complex vector bundle to talk about $c_1$ or $c_2$. In particular, if you have a group homomorphism $PSU(N) \to O(M)$ or $U(M)$, then we can construct the associated vector bundle and talk about its Pontryagin or Chern classes.
many thanks @Michael Albanese! a systematic answer is still welcome
What would you consider a systematic answer to contain? Do you have references for the quantization in the $PSU(N)$ case?
No, Refs are welcome. I knew the Book by Freed and Uhlenbeck
I don't think they consider $PSU(N)$ in that book. Do you know that $n \in \frac{1}{N}\mathbb{Z}$ for $PSU(N)$ bundles over a spin manifold, or is that a guess? I only ask, because after stating it, you wrote "True or false?".
I guessed "∈(1/)ℤ for () bundles over a spin manifold" - I am not sure how to prove it formally. Refs for proving them are welcome
|
2025-03-21T14:48:31.432882
| 2020-07-04T20:04:00 |
364845
|
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|
Stack Exchange
|
Show that $\sup_{\|g\|\leq \delta_n}\left| \frac{1}{\sqrt{n}}\sum_{i=1}^n g(Z_i)\right|\rightarrow_{\text{a.s.}}0.$ when $\delta_n\rightarrow 0$?
UPDATE: The result below can be understood as an almost sure stochastic equicontinuity condition. I don't know of any result establishing primitives of almost sure stochastic equicontinuity. If you know, this would very likely help me answer the question.
I need the following almost sure convergence result:
Let $\|\cdot\|$ denote a norm on a functional space (could be $\|\cdot\|_{\infty}$ or $\|\cdot \|_2$ in $L^2$, for example). $\mathcal{G}=\{g:\mathcal{Z}\rightarrow \mathbb{R}\}$ of measurable functions. Let $Z_1,Z_2,\dots$ be $i.i.d$ random variables with $E[g(Z_i)]=0$ for any $g\in \mathcal{G}$. Let $\{\delta_n\}$ be a sequence of positive numbers such that $\delta_n\rightarrow 0$.
Then, under ADDITIONAL ASSUMPTIONS,
\begin{equation*}
\sup_{g\in\mathcal{G} \text{ s.t. }\|g\|\leq \delta_n}\left| \frac{1}{\sqrt{n}}\sum_{i=1}^n g(Z_i)\right|\rightarrow_{\text{a.s.}}0.
\end{equation*}
Note that all random variables are defined in the same probability space and are independent draws of the same distribution. The number $n$ here denotes the size of the sample, and the almost sure convergence is over the distribution of the $Z_i$ as $n\rightarrow \infty$. The space $\mathcal{G}$ does not depend on $n$, and $g$ is not random.
I need to find which reasonable assumptions could be made to establish this result. $\mathcal{G}$ is $P$-Donsker? Lipschitz functions? Bounds on moments? etc.
Attending requests for clarification of the background: this result is needed to establish a stochastic equicontinuity condition on a linear regression estimator with a generated regressor (it means that one of the regressors is estimated). This is a high level result, in the sense that it is a result for a generic estimator for the regressor, we want to give conditions that this estimator must satisfy, but not specify the estimator. I am using results in a paper by Chen, Linton and Keilegom (2003) which establish the consistency of the bootstrap for estimators which are based on the optimization of a function of the data, the parameter of interest and a nuisance infinite dimension parameter. So, suppose the model is $E[Y|X,W]=\beta X+\gamma h_0(W)$, but $h_0$ is not known. The term is thus estimated and the regression is done on $\hat{h}(W)$ instead. If it helps, you can think of $h_0(W)=E[V|V\leq 0, W]$, for another variable $V$ which is observed in the data, for example. (FIY: this is a huge simplification of the true situation, so don't give me suggestions about how to estimate the model I just described. In the actual model $V$ is not observed and must be predicted out of sample with machine learning and optimization methods.)
In order to apply one of the results in the paper I mentioned, I need to prove that some quantities such as the one above are $o_{\text{a.s.}}(1)$. For example, I give you a simplified version of one of them:
\begin{equation*}
\sup_{h,h_0\in\mathcal{H} \text{ s.t. }\|h-h_0\|\leq \delta_n}\left| \frac{1}{\sqrt{n}}\sum_{i=1}^n \left[X_i(h(W_i)-h_0(W_i))-E[X_i(h(W_i)-h_0(W_i))\right]\right|
\end{equation*}
I need to show that several objects that look more or less like the one above are $o_{\text{a.s.}}(1)$ as $n\rightarrow \infty$ for any sequence of positive $\delta_n\rightarrow 0$. Note that I can make assumptions about $\mathcal{H}$, $X$ and $W$. For example, I can say that $\mathcal{H}$ is the space of measurable negative Lipschitz functions, and I can say that $E[|X|^4]$ and $E[|W|^4]$ are finite. Basically I need conditions that would allow me to establish this result.
At the essence this is an empirical process. Define $\mathbb{G}_nf=\frac{1}{\sqrt{n}}\sum_{i=1}^n f(Z_i)-E[f(Z_i)]$ and let $\mathbb{G}$ be the Brownian Bridge, then if $\mathcal{G}$ is $P$-Donsker, $\mathbb{G}_ng\rightsquigarrow\mathbb{G}g$ for every fixed $g$. This is a convergence in distribution in a functional space. I need to show that $\sup_{\|g\|_{\mathcal{G}}\leq \delta_n} |\mathbb{G}_n g|\rightarrow_{\text{a.s.}} 0$.
I hope this is clearer and someone can help me. I ran out of ideas of things to try and need a fresh take.
I think you should wait a little bit more than 9 hours on Math Stack Overflow.
To me this seems to some sort of "Rundumschlag" (german). Can you make your question more concrete, specifying the function space (what is the $\sup$-norm) and clarify your problem. For example is your $\sup$ measurable as a supremum over uncountable many functions?
As stated the claim does not hold, as $Z_i$ could be uniform on $[-1,1]$ and for each $n$, the function $g$ could be an odd function from $[-1,1]$ to $[-\delta_n,\delta_n]$, chosen to satisfy $g(Z_i)=\delta_n$ for all $i \le n$. More care is needed in the supremum.
@DieterKadelka Define any normed space, suppose that the norm exists. I even took out the $\sup$ from the description (I simply meant $||g||=\sup_{z\in\mathcal{Z}}|g(z)|$ and you can suppose that the functions $g$ are defined in $\mathcal{Z}$ with image in $\mathbb{R}$ and are bounded.) I think it is clear in the question that I have plenty of freedom to decide the functional space. If you know of a theorem that may achieve this, I can probably assume its requirements. I added explanations to the question. See new paragraph.
@YuvalPeres Yes, it should be obvious that as stated it does not hold, as I mentioned that even to prove this convergence in probability one needs bounded second moments. Your example forgets the requirement in the statement that $E[g(Z_i)]=0$. Please see updated comments.
Note that in my example g is an odd function so the requirement does hold.
@YuvalPeres The expectation is taken over the distribution of $Z_i$. The function $g$ cannot vary in $n$.
@DieterKadelka Sorry, googled "Rundumschlag" and cannot figure out what you mean. If you want more information: I am writing a theorem to prove the consistency of a bootstrap procedure. I can make assumptions in that theorem, I decide which ones they are within reason. However, to prove said theorem I need to use an existing result whose requirements make it necessary for me to establish the result above.
@Caetano: I don't know the correct translation, its something like "sweeping statement" or "sweeping attack". Further, your comment that $g$ cannot vary in $n$ looks strange, since you work with the supremum over all $g$ such that $|g| \leq \delta_n$. In your answer to my comment the norm seems to be $|g| = |g|_\infty$. Please make your question more concrete, otherwise the discussion may be endless.
@DieterKadelka Thanks for explaining. Indeed the norm can be $||g||_{\infty}$. That is what I have in mind, but I could use a different norm, if someone knows of a result which uses a different norm. I don't understand the confusion with the function $g$, but I will try to clarify. Say $g\in\mathcal{G}$ a fixed space which does not vary with $n$. The function is not random, in any way. The only randomness comes from the $Z$. I will update the statement.
@MattF. Thanks! I need to step out for a minute now, but as soon as I return I will add that as well.
@MattF. Done! Hope this is clear now.
@MattF. Glad to adapt. Any suggestion? I'm a bit lost as to what is needed. My background is Statistics and this notation is so standard there that I have a hard time figuring out what is the issue with clarity here.
The function $g$ must depend on $n$ because of the assumption on its norm. I understand you do not want it to depend on the values of $Z_i$ but the notation you use is not consistent with that wish. Please try to reformulate.
@YuvalPeres Perhaps you can explain what you mean? It is really not clear to me. The sup is taken over all the functions in the space $\mathcal{G}$ with image between $-\delta_n$ and $\delta_n$. This means that the group of functions over which the sup is taken changes with $n$. However, a specific function $g$ is fixed. For example, think that one of the functions could be $g(Z)=E[V|Z]-E[V]$, where the expectation is taken over the joint distribution of $V$ and $Z$.
@Caetano If you write out in detail the definitions of the a.s. convergence and supremum, you will see the problem. Assume the norm is $L^\infty$ norm, that the $Z_i$ are uniform on $[−1,1]$ . Then for each $n$ the sup is taken over such a large class of functions $g$, that it includes odd functions from $[−1,1]$ to $[-\delta_n,\delta_n]$ chosen to satisfy $g(Z_i)=\delta_n $ for all $i \le n$.
@Caetano I think it will help clear up the misunderstanding if you write exactly what statement you have proved for convergence in probability; as far as I can tell, the supremum you wrote does not converge to zero even in probability in the example mentioned in my last comment, assuming $\delta_n=1/(1+\log n)$ for instance. If, however, the supremum is outside the probability that the normalized sum is large, then all is well, but that is not what convergence in probability of the supremum means.
@YuvalPeres I think I understand what you mean, thanks! You are saying that for all events, among odd functions from $[-1,1]$ to $[-\delta_n,\delta_n]$ there is always one such that $g(Z_i)=\delta_n$ for that event. Yes, this makes sense, and there may be a problem with the result in probability as well then. I will be thinking about this. On the background I have been working on just relaxing that assumption from the theorem I wanted to use anyway, because I think this requirement need only hold in probability and for sequences $\delta_n=\delta/n^{1/4}$.
@Caetano: From the discussion its not clear to me, what you want to show: $\lim_{n \to \infty} \sup_{g∈G ~s.t. ~|g|≤\delta_n} \left|\frac{1}{\sqrt{n}}\sum_{i=1}^n g(Z_i) \right| = 0$ (here Yuval Peres has a counterexample) or something like $\sup_{g∈G} \lim_{n \to \infty} \left|\frac{1}{\sqrt{n}}\sum_{i=1}^n g(Z_i) \right| = 0$, which seems to be true (I have not checked it). In the second formula the introduction of $\delta_n$ does not make sense.
@DieterKadelka The first is exactly what I want to show, but remember that the question is about what types of conditions would have to be imposed for this to hold. I am sure this would not hold in general. Yuval's counter example is very helpful though. I think it would be ruled out with a requirement that $\mathcal{G}$ has only increasing functions. Also, I am reading up on stochastic equicontinuity and I'm finding several instances where Lipschitz style conditions show up.
@DieterKadelka Also, just FIY, I did not make up the $\delta_n$ in the sup, it is literally the requirement of the theorem I want to use. I am also finding several similar requirements to that whenever stochastic equicontinuity conditions show up, and it is always the sup over the space of the parameters in a shrinking neighborhood of the right value. Requirements are usually that the expectation of the sup goes to zero, or some convergence in probability though, I only saw a.s. convergence in this instance.
@YuvalPeres Ok, I have found a general result for this convergence in probability in Pakes and Pollard Lemma 2.17 (http://www.stat.yale.edu/~pollard/Papers/PakesPollard89Econometrica.pdf) In fact, it's a more general result. $g$ here is their parameter $\theta$. It's in probability though, not almost surely. I will be studying it though, see if I can figure this out.
@Caetano Lemma 2.17 in Pakes and Pollard is not as general as the statement you were making- that statement does not hold for convergence in probability, as the example I indicated earlier shows. The supremum needs to be over a more restricted class.
@YuvalPeres Again, my question has never been whether that statement holds in general. I know it does not hold in general. The question is whether there are any known sufficiently respectable conditions under which it may hold. Lemma 2.17 in Pakes and Pollard does give such conditions for the convergence in probability: Euclidean class with envelope function with a bounded second moment.
@YuvalPeres Overall I am finding that I am having a communication problem here. I think that mathematicians think: "under such and such conditions, does this conclusion hold?" While applied math people think: "which reasonable conditions are sufficient for this conclusion to hold?"
A sufficient condition to establish the result above for the convergence in probability is if $\mathcal{G}$ has finite entropy with an envelope function $M$ which satisfies $E[M(Z)^2]<\infty$. This allows us to apply Theorem 1 in https://cowles.yale.edu/sites/default/files/files/pub/d10/d1059.pdf (Wayback Machine, DOI: 10.1016/S1573-4412(05)80006-6).
Unfortunately I could not find any result for almost sure stochastic equicontinuity, which is probably the answer to this problem. I will look it up further. If you know anything about this, please let me know.
|
2025-03-21T14:48:31.433670
| 2020-07-04T21:16:30 |
364851
|
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|
Stack Exchange
|
Density of integral points on affine cubic surfaces of a certain type
Let $Q(x,y,z)$ be a cubic polynomial with integer coefficients, such that the terms $x^3, y^3, z^3$ do not appear. That is, it is at most quadratic in each of the variables $x,y,z$.
Is there a general method to count integral points $(a,b,c)$ with $\max\{|a|, |b|, |c|\} \leq T$ on the affine cubic surface defined by $Q(x,y,z) = 0$?
The prototypical example is the Markoff surface defined by $Q(x,y,z) = x^2 + y^2 + z^2 - xyz$. Here Zagier showed that the density of integral points is asymptotic to $C (\log T)^2$ for some explicit constant $C > 0$. This is generalized by Baragar and Umeda in this paper.
Their method depends on an explicit descent on the Markoff surface, namely that for a given point $(a,b,c)$ the point $(bc - a, b, c)$ is also a point. Zagier then showed that all integral solutions are generated from the fundamental solution $(3,3,3)$ and by permuting the variables and applying the above map. This means that the set of solutions grows exponentially in size, hence giving the bound $O((\log T)^2)$.
Is there something similar that can be done, using only the fact that all of the monomials in $Q$ are at most quadratic in each variable?
Not in general.
The involutions of the Markov surface such as
$(a,b,c) \leftrightarrow (bc-a,b,c)$ preserve integral points because
$x^2 + y^2 + z^2 - xyz$ is a monic quadratic polynomial in each variable.
That works more generally for any polynomial of the form
$Q(x,y,z) = x^2 + y^2 + z^2 - L(x,y,z)$ with $L$ linear in each variable,
though already in this special case $Q=0$ can have infinitely many solutions
(it doesn't even have to be irreducible, e.g. $(x \pm y \pm z)^2$ works).
For the general cubic polynomial with no $x^3,y^3,z^3$ terms,
there are still rational involutions that fix three of the variables,
but they have denominators and thus need not send integral points to
integral points (and occasionally will send a rational or even integral point
to infinity, when the quadratic coefficient w.r.t. one of the variables
vanishes).
Thank you for your answer. Are there ways to detect when there are "many" fundamental solutions to $Q = k$? Baragar and Umeda gave some examples, but I don't see how to think about this in the general case.
You're welcome. For $Q=k$, do you mean thie for the specific case of
$Q = x^2+y^2+z^2-xyz$, or any $Q$ satisfying the conditions of your question?
I would like as general a statement as possible; if that's too much to ask then just understanding the case $Q = x^2 + y^2 + z^2 - xyz$ would be great since that will likely help me get a "feeling" for this type of phenomenon.
|
2025-03-21T14:48:31.433867
| 2020-07-04T21:52:41 |
364853
|
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|
Stack Exchange
|
Origin and variations of problem on $4xy-x-y$ being square
One of the forms in which the Diophantine equation in question can be found in the literature is this:
Solve the equation \begin{eqnarray}z^{2} = 4xy-x-y \qquad \qquad (\ast)\end{eqnarray} in positive integers $x, y$, and $z$.
There are some other variants of it here and there. It is usually given once the instructor has touched upon the Jacobi symbol.
My questions about the problem are the following two:
Do you happen to know where it was that the problem first appeared? If I understand correctly, the variation in which one is asked to establish that $z^{2} = 4xyt^{u}-t^{v}-y$ does not admit solutions in positive integers might have first appeared in a Russian compilation of problems of olympiad caliber...
What other notable variations of (*) do you know?
Please let me thank you in advance for your learned replies.
https://artofproblemsolving.com/community/c3046h1833823_representation_of_a_square_of_a_certain_shape
Looking at the two answers, I wonder (1) how accessible the Euler--Goldbach correspondence was in London around 1900 and (2) whether Euler or Goldbach considered the triangular number result that Crofton included.
@BrianHopkins Euler's writings and correspondence were published separately.
@individ Yes, Fuss published some of the correspondence in 1845, including the 9 September 1741 letter that Carlo detailed (now indexed as OO754). I still wonder how accessible that was to Crofton.
@BrianHopkins I found an interesting property of these equations. $aX^2+bX+cY^2+dY=jZ^2+pZ$ If the coefficients are $b,d,p- $ At least one is not zero. There will always be solutions. The truth is not always positive and mutually simple, but there will always be solutions. https://math.stackexchange.com/questions/794510/curves-triangular-numbers
I don't know why, but the "here and there" phrase with "here" pointing somewhere really plays with my head. I'm emotionally unable to concentrate on the real content of the post.
A twentieth century source for this problem is the Canadian entry to the 1984 International Mathematics Olympiad. The original handwritten document of the full problem set can be downloaded here.
For variations, in addition to $z^{2} = 4xyt^{u}-t^{v}-y$ [link], there is:
$x,y,m,n,k$ are natural; prove that $4kxy-1 \nmid x^m+y^n$ [link]
Addendum: following up on individ's pointer, here is a two-and-a-half century older attribution to Euler, in a letter to Goldbach dated 9 September 1741, see page 107 of this source.
I have also found similar theorems a long time ago, such as that $4mn-m-1$ can never be a square. Likewise $4mn-m-n$ can never be a square, assuming that $m$ and $n$ are integer numbers.
A discussion of this result of Euler follows on page 45 of this source.
Thanks for digging this up, Carlo. And that's a wonderful link!
@BrianHopkins, Carlo Beenakker: Thank you very much for taking the time to leave your replies... I did not know (or recall) that the "original" problem could be found in the Goldbach-Euler Briefwechsel.
I found an 1899 reference in Google Books: Mathematical Questions and Solutions from the "Educational Times" edited by D. Biddle, vol. 70, Hodgson, London, 1899.
On page 73:
(Professor Crofton, F.R.S.) Show that $4mn-m-n$ cannot be (1) a square; nor (2) a triangular number.
Brief solutions by H. W. Curjel and Allan Cunningham are given. It looks like the question was posed in the February 1, 1898 issue of The Educational Times and Journal of the College of Preceptors (vol. 51, page 87).
Here are the solutions, which could help with making variant problems.
Solutions (1) by H. W. Curjel, M.A.; (2) by Lt.-Col. Allan Cunningham, R.E.
(1) If
$$4mn - m - n = x^2 \quad \text{or} \quad \frac{1}{2}\{y(y+1)\},$$
$$(4m-1)(4n-1) = 4x^2 + 1 \quad \text{or} \quad \frac{1}{2}\{(2y+1)^2+1\},$$
which is impossible, since all the prime factors of $X^2+1$ other than 2 are of the form $4N+1$.
(2) Here
$$4mn - m - n = \frac{1}{4}(4m-1)(4n-1) - \frac{1}{4}.$$
(i) Let this $ = z^2$ (if possible) [I think that should be $x^2$]. Then
$$(4m-1)(4n-1) = 1+(2x)^2,$$
which gives a sum of squares $\{1^2+(2x)^2\} = $ product of two numbers of form $(4m-1)$, which is impossible.
(ii) Let $4mn-m-n = \frac{1}{2}(x^2+x)$, a triangular number (if possible). Therefore
$$ (4m-1)(4n-1) = 2x^2 + 2x + 1 = (x+1)^2 + x^2,$$
a sum of squares, which is impossible, as before.
No. The first mention of this equation was in the works of Euler. The given articles are a rewritten result of it. https://artofproblemsolving.com/community/c3046h1049059__
A triangular number can't be? Write the equation as you imagine it. Its shape can be different.
@individ It would be great to have a specific citation for where Euler discussed this. Regarding your question about the triangular number result, I've included the 1899 published solutions.
@individ --- great find, a 1741 letter by Euler to Goldbach indeed discusses this result!
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2025-03-21T14:48:31.434329
| 2020-07-04T22:03:35 |
364854
|
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Stack Exchange
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Understanding an identity for dyadic sums
I am reading a paper on PDEs and I has been struggling trying to understand some specific identities (that should be very easy). Let me introduce the main notation of the book, which says that each time a sum like $\sum_\lambda f(\lambda)$ appears, then the $\lambda$-variable must to be understood as a dyadic variable, that is, the previous sum has to be understood as $$
\sum_{k=0}^{\infty} f(2^k).
$$
This will also apply to $\mu$ below. Now, just after that, in the proof of a basic lemma the author writes the following sequence of equalities: $$
\sum_\lambda \lambda^s\sum_{\mu\geq \lambda/8} a_{\mu}d_{\lambda}=\sum_\lambda \lambda^s\sum_{\substack{k=-3,\\ 2^k\lambda\geq1}}^{\infty}a_{2^k\lambda}d_{\lambda}=\sum_{k=-3}^\infty\sum_{\lambda\geq 2^{-k}} a_{2^k\lambda}d_\lambda
$$
where $\{a_\lambda\}$ and $\{d_\mu\}$ are two sequences of real numbers indexed by dyadic integers and $s>0$. I don't really understand these identities for two reasons. First, I don't see how to change the $\mu$-variable to the index $2^k\lambda$ appearing in $a$ (I would just put $a_{2^k}$ withouth the $\lambda$). But then, it seems confusing to me that the inequality in the inner sum becomes $2^k\lambda\geq 1$, particularly because in the left-hand side $\lambda$ is bounded from above, and in the other sums $\lambda$ seems to be bounded from below. Does anyone has any explanation?
Edit: The idea of the lemma is actually to prove the following inequality: $$
\sum_\lambda \lambda^s\sum_{\mu\geq \lambda/8} a_{\mu}d_{\lambda} \leq C \left(\sum_\lambda \lambda^{2s}a_\lambda^2\right)^{1/2}\left(\sum_\lambda d_\lambda^2\right)^{1/2},
$$
for some constant $C>0$ (whenever the right-hand side makes sense). Now I am feeling that the proof might be wrong, so if it is the case, I am wondering if anyone has a correct proof for it.
We write $\mu=2^k \lambda$ with $k \ge -3$. The standing assumption $\mu \ge 1$ becomes $2^k \lambda \ge 1$. The RHS of the sequence of equalities you reproduce is missing a factor $\lambda^s$; is this typo in the original paper?
The sequence of equalities you included should thus be written (adding $L$ on the left) as
$$
L:= \sum_\lambda \lambda^s\sum_{\mu\geq \lambda/8} a_{\mu}d_{\lambda}=\sum_\lambda \lambda^s\sum_{\substack{k=-3,\\ 2^k\lambda\geq1}}^{\infty}a_{2^k\lambda}d_{\lambda}=\sum_{k=-3}^\infty \sum_{\lambda\geq 2^{-k}} \lambda^s a_{2^k\lambda}d_\lambda \,.
$$
Thus
$$
L= \sum_{k=-3}^\infty 2^{-ks} \sum_{\lambda\geq 2^{-k}} (2^k\lambda)^s a_{2^k\lambda}\, d_\lambda \,.
$$
Apply Cauchy-Schwarz to the inner sum:
$$
L \leq \sum_{k=-3}^\infty 2^{-ks}\left(\sum_{\lambda\geq 2^{-k}} (2^k \lambda)^{2s} a_{2^k\lambda}^2\right)^{1/2}\left(\sum_\lambda d_\lambda^2\right)^{1/2}
$$
Then substitute $\mu=2^k \lambda$:
$$
L \leq
\sum_{k=-3}^\infty 2^{-ks}\left(\sum_{\mu} \mu^{2s} a_{\mu}^2\right)^{1/2}\left(\sum_\lambda d_\lambda^2\right)^{1/2} \, ,
$$
which yields the inequality you asked about
$$
\sum_\lambda \lambda^s\sum_{\mu\geq \lambda/8} a_{\mu}d_{\lambda} \leq C \left(\sum_\lambda \lambda^{2s}a_\lambda^2\right)^{1/2}\left(\sum_\lambda d_\lambda^2\right)^{1/2} \, ,
$$
with $C:=\sum_{k=-3}^\infty 2^{-ks}$.
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2025-03-21T14:48:31.434529
| 2020-07-04T22:42:29 |
364857
|
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Stack Exchange
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Reference on classifying real subspaces of complex vector spaces (based on restricted complex structure)
Every complex vector space can also been seen as real vector space. If we now choose a real subspace, it may not be a complex subspace (in particular, if it is of odd real dimension).
If the complex vector space was equipped with an inner product (for example, a Hilbert space), we can restrict the imaginary unit (also known as linear complex structure) to any real subspace using the orthogonal projection. We can then classify the types of real subspaces based on the spectrum of this ``restricted complex structure''. In particular, if the restricted complex structure squares to minus identity, i.e., is itself a complex structure, the real subspace is also a complex subspace. In general, the spectrum encodes how being a complex subspace is violated.
I worked this out for myself, but I'm confident that this is standard material in linear algebra of complex vector spaces. However, the standard introductory text books that I checked do not discuss real subspaces of complex vector spaces and their classification in the above way.
Do you know of a standard reference that I could cite when discussing this (in particular, the above mentioned classification based on the spectrum of the restricted complex structure)?
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2025-03-21T14:48:31.434643
| 2020-07-05T00:03:12 |
364859
|
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Stack Exchange
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Manifolds with $w_1(TM)\cup w_1(TM)=0$ and $w_2(TM)=0$ but $w_1(TM)\neq 0$
For a generic dimension $d$, is there an nonorientable manifold $M$ (i.e. $w_1(TM)\neq 0$) with vanishing $w_1(TM)\cup w_1(TM)$ and $w_2(TM)$, i.e.,
$$w_1(TM)\cup w_1(TM)=0, ~~~~~ w_2(TM)=0, ~~~~~w_1(TM)\neq 0?$$
Here $w_i(TM)$ is the $i^{\text{th}}$ Stiefel-Whitney class of the tangent bundle of the manifold $M$. For $d=2$, the Klein bottle is an example.
If such manifolds exist, what kind of structure do they carry? For example, if $w_1(TM)=0$, and $w_2(TM)=0$, then the manifold can be equipped with a spin structure, and we say it is a spin manifold. I'd like to see what is the corresponding structure in the above more complicated case.
A smooth manifold $M$ admits a pin$^+$ structure if and only if $w_2(M) = 0$, and a pin$^-$ structure if and only if $w_1(M)^2 + w_2(M) = 0$; see this page for some information on pin structures. The manifolds you are enquiring about satisfy both conditions and hence admit pin$^+$ and pin$^-$ structures. However, as $w_1(M) \neq 0$, they are non-orientable, so they do not admit spin structures.
There exist manifolds of arbitrary dimension which satisfy your requirements. For example, $K\times S^n$ where $K$ is the Klein bottle.
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2025-03-21T14:48:31.434746
| 2020-07-05T01:40:25 |
364861
|
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Stack Exchange
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$SU(n)$ character variety of integral homology spheres
Suppose that $Y$ is an integral homology $3$-sphere. Is it true that, for any $n \in \mathbb{N}$, there is always an irreducible representation $\pi_{1}(Y) \rightarrow SU(n)$?
I'm also happy to see the answer when the Lie group is replaced by $SL(n, \mathbb{C})$ or $PSL(n, \mathbb{C})$, but keeping irreducible unchanged.
Edit: In this paper, Zentner proved the statement positively for $SL(2, \mathbb{C})$. It might be true that there is a threshold $n$ such that there is no more $SU(N)$ irreducible representations for $N \geq n$. The question was stated in this way because constructing counter-examples to the vanishing statement might be easier.
The Poincaré homology sphere has fundamental group the binary icosahedral group, which is finite and therefore admits only finitely many irreducible representations.
Yes, my motivation of asking this question is to see if this failure holds in much more generality.
So you are asking: for all $n \in \mathbb N$, is there an integral homology 3-sphere $Y$ with an irreducible $\pi_1(Y) \to SU(n)$?
No, I am asking if $Y$ is fixed, can you find such representations. As you pointed out this is not true if Y is the Poincar'e homology sphere, the question is about whether this failure holds for all $Y$.
It should be true that for a given closed 3-manifold with infinite fundamental group, there is an irreducible $SU(N)$ representation for $N$ sufficiently large. But the case of homology spheres and $n=2$ is a well-known open question, so I think your question as stated is probably difficult.
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2025-03-21T14:48:31.435142
| 2020-07-05T02:41:37 |
364863
|
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|
Stack Exchange
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Relationship between cycle length, number of chords, and number of induced $P_{4}$ subgraphs of the cycle
I was wondering if there was a known relationship between the length of cycle, the number of chords of the cycle, and the number of induced $P_{4}$ subgraphs of the cycle.
Here, I am referring to cycles of length greater than or equal to $5$.
Here $P_{4}$ refers to: Path graph on $4$ vertices (there will be $3$ edges then).
I am primarily interested at the point in which the number of induced $P_{4}$ becomes $0$. I was wondering if there is any closed form expression for this. I mean is there some expression involving the length of the cycle and the number of chords (and probably other quantities) that determines when the number of induced $P_{4}$ is 0
Or is there a closed form expression involving the length of the cycle and the number of chords (and probably other quantities) for the exact number of induced $P_{4}$?
Two things are clear to me:
When the number of chords in the cycle increases, the number of induced $P_{4}$ subgraphs decreases. Also when the length of the cycle increases, typically the number of induced $P_{4}$ subgraphs increases.
I did the following examples by hand:
Cycle of length $5$ with $0$ chords: Number of $P_{4}$ induced subgraphs: $5$
Cycle of length $5$ with $1$ chord: Number of $P_{4}$ induced subgraphs: $2$
Cycle of length $5$ with $2$ chords: Number of $P_{4}$ induced subgraphs: $0$
Cycle of length $5$ with $3$ chords: Number of $P_{4}$ induced subgraphs: $0$
Cycle of length $5$ with $4$ chords: Number of $P_{4}$ induced subgraphs: $0$
Cycle of length $5$ with $5$ chords: Number of $P_{4}$ induced subgraphs: $0$
Cycle of length $6$ with $0$ chords: Number of $P_{4}$ induced subgraphs: $6$
Cycle of length $6$ with $1$ chord: Number of $P_{4}$ induced subgraphs: $5$
Cycle of length $6$ with $2$ chords: Number of $P_{4}$ induced subgraphs: $4$
Cycle of length $6$ with $3$ chords: Number of $P_{4}$ induced subgraphs: $3$
Cycle of length $6$ with $4$ chords: Number of $P_{4}$ induced subgraphs: $1$
Cycle of length $6$ with $5$ chords: Number of $P_{4}$ induced subgraphs: $0$
Cycle of length $6$ with $6$ chords: Number of $P_{4}$ induced subgraphs: $0$.
As seen, here for a cycle of length $5$, the number of $P_{4}$ induced subgraphs becomes $0$ when we have a chord of length $2$.
For a cycle of length $6$, the number of $P_{4}$ induced subgraphs becomes $0$ when we have $5$ chords.
Hopefully, I have not made a mistake in the above calculations.
These results would be useful for me for something I'm working on relating to perfect graphs.
If there does not exist a result, relating to what I'm asking, can you to relevant research papers or other sources? Thanks.
Any suggestions or guidance on this?
I agree with
Cycle of length 5 with 0 chords: Number of P4 induced subgraphs: 5
Cycle of length 5 with 1 chord: Number of P4 induced subgraphs: 2
But I'm not sure how to interpret your statement:
Cycle of length 5 with 2 chords: Number of P4 induced subgraphs: 0
There are two ways to have 2 chords in a 5-cycle... crossing chords (in which case the number of $P_4$s is 0, as you wrote) or the two chords could have a common endpoint, making the gem graph, which still has one $P_4$ left in it.
So your question of asking for a closed form for the number of necessary chords needs further clarification: are you asking how many chords are necessary so that no matter how they are added to the cycle, there are no $P_4$s left? Or are you asking for the fewest number of chords one has to add to a cycle (possibly in a clever way) in order to destroy all $P_4$s.
Since $P_4$s are self complementary, you can think about what the complement of the resulting graph is once you reach one of your graphs with zero $P_4$s. The complement will also have no $P_4$s and will be nearly empty of edges... so if we take the interpretation of your question as to ask "What is the fewest number of edges required to add to a cycle strategically in order to destroy all $P_4$s?" we take away edges in the complement until it would be down to a disjoint union of a combination of $K_3+K_1$s (if strategically possible), or $P_3+K_1$s and $2K_2$s (on our chosen partition of $4$-vertex sets).
So the approach in deleting from $\overline{C_n}$ would be to turn every $4$-set of vertices to a triangle ($3$ edges) and an isolated vertex, and when not possible, to reduce a $4$-set down to having $2$ edges (as either a $P_3+K_1$ or a $2K_2$. So on $n=4k$ vertices, you are looking at adjusting (adding in the original, or deleting in the complement) enough edges to leave behind between $2k$ and $3k$ edges.
The complement of $C_n$ has $\left(\frac{n(n-1)}{2}-n\right)$ edges, which is $= \frac{n^2-3n}{2}$ edges, and so you must delete (add) between $\left(\frac{n^2-3n}{2} - \frac{3n}{4}\right)$ and $\left(\frac{n^2-3n}{2} - \frac{n}{2}\right)$ edges, or between $\frac{2n^2-9n}{4}$ and $\frac{2n^2-8n}{4}$ edges.
Checking for $n=5$, this is between $\frac{50-45}{4}$ and $\frac{50-40}{4}$, which is the range [1.25,2.5] = 2, which checks out with your strategic choice of 2 edges.
Check for $n=6$, this is $\frac{72-54}{4}$ and $\frac{72-48}{4}$, which is [4.5,6], the most 'strategic' option would be 5 edges, which corresponds to your hand-computation.
If you computed $C_7$ after your original post, I would think you found a solution of CEIL$\left(\frac{98-63}{4}\right) = 9$ edges.
If you are interested in restricting $P_4$s, you should familiarize yourself with $P_4$-sparse and $P_4$-lite and their related graph classes.
If you are interested in optimal ways of deleting (or adding in the complement) edges from a graph in order to make the graph $P_4$-free, you might want to see this couple of papers which develop and use exact (exponential-time, but FPT) algorithms for deleting graphs until they are $P_4$-free:
Bounded Search Tree Algorithms for Parameterized Cograph Deletion
Defining and identifying cograph communities in complex networks
|
2025-03-21T14:48:31.435490
| 2020-07-05T02:46:18 |
364864
|
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|
Stack Exchange
|
Can a knotted sphere isometrically embed into $\mathbb R^3$?
All smooth simple closed curves in $\mathbb R^3$ (knotted or not) can be isometrically embedded into $\mathbb R^2$ as a circle of equal arclength.
The situation for knotted spheres seems more complicated because metric surfaces are more rigid. Here I am unsure that if given a smooth knotted sphere in $\mathbb R^4$ if there is a way to embed into $\mathbb R^3$ while preserving its metric structure. So my question is:
Does there exist a nontrivial smooth knotted sphere in $\mathbb R^4$ which can
be smoothly isometrically embedded into $\mathbb R^3$? And if so,
are there any concrete examples?
In all of the above, the Riemannian metric structures are assumed to be inherited from the ambient space.
How can a knot be embedded into the plane??
There's a slight inconsistency in terminology, but I believe the author at that point in the question is thinking of "circle" as being a synonym for "knot".
Can you clarify the metrics you're using in the isometries, what exactly you mean by a round circle, and be specific about what the analogous situation is when you move up in dimension?
So here I am thinking of a knot K in R^3 as a simple smooth closed curve and the isometry to a circle (of equal perimeter) does not arise from an ambient isotopy (unless K is the unknot). Assuming K is of unit length, the isometry can be seen by considering an ant walking along the knot at unit speed from a starting point p. A point the ant reaches at time t gets maped to (cos 2 pi t, sin 2 pi t). For x, y on K the distance along K is the same as the associated distance along the unit circle in R^2.
I've made some edits that hopefully help. I removed the adjective "round" since it doesn't really affect my main question (but i just meant a circle as in elementary school as opposed to a topological circle). As for the Riemannian metric structures, they should all be assumed to be induced by the ambient space. And to elaborate on the situation in higher dimension: is there a smooth Riemannian submanifold of R^4 which is a nontrivially knotted topological sphere and can be smoothly isometrically embedded into R^3?
It can be done $C^1$.
Thank you for mentioning it can be done C^1 (via the Nash-Kuiper theorem I presume). However, I am mainly interested in maps with more regularity. Smooth would be ideal, but C^2 would be interesting too.
Maybe you are aware of this, but this paper gives an example of a metric on $S^2$ which does not admit a $C^2$ isometric embedding into $\mathbb R^3$: https://projecteuclid.org/download/pdf_1/euclid.jdg/1214429999.
A spun knot will give a 2-sphere embedded in $\mathbb{R}^4$ whose intrinsic metric embeds into $\mathbb{R}^3$ isometrically as a surface of revolution.
Take a tangle $T$ in $\mathbb{R}^3_+$ with two ends on $\mathbb{R}^2$, that is an embedding $e: ([0,1],\{ 0, 1\}) \to (\mathbb{R}^3_+,\mathbb{R}^2)$. We may assume that $e$ is an isometry. Spin $T$ about $\mathbb{R}^2$ in $\mathbb{R}^4$ to get a surface $S$.
If $T$ is knotted then $S$ will be knotted. Then there exists an isometric embedding $e’:([0,1],\{0,1\})\to (\mathbb{R}^2_+,\mathbb{R})$ so that $d(e’(t),\mathbb{R})=d(e(t),\mathbb{R}^2)$ for all $t,\in [0,1]$ and thus the surface of revolution $S’$ of $e’([0,1])$ about $\mathbb{R}$ in $\mathbb{R}^3$ is isometric to $S$.
|
2025-03-21T14:48:31.435756
| 2020-07-05T03:23:53 |
364866
|
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|
Stack Exchange
|
Equivariant cohomology of a semisimple Lie algebra
Suppose $\mathfrak{g}$ is a real Lie algebra integrating to the connected Lie group $G$. One may consider the $G$-equivariant cohomology of $\mathfrak{g}$ ($\mathfrak{g}^*$) where the $G$-action is the adjoint (coadjoint) representation. That is, the cohomology induced by the cochain complex of $G$-invariant differential forms on $\mathfrak{g}$. (Or, more generally, $G$-invariant $V$-valued forms where $V$ is the target of a linear representation of $G$). When $G$ is compact, it follows from a theorem of Chevalley and Eilenberg that this cohomology vanishes in all degrees. I am however interested in the semisimple case. What is known in general about this cohomology and its vanishing? What is known for specific semisimple Lie algebras? Can this cohomology be tied to the standard Lie algebra cohomology?
Whitehead‘s Lemmas answer this at least in degree 1 and 2: https://en.wikipedia.org/wiki/Whitehead%27s_lemma_(Lie_algebra)
It vanishes as well. The complex of differential forms on $\mathfrak{g}$ is null homotopic through the standard null-homotopy coming from the linear homotopy to the origin. Since the $G$-action preserve this null-homotopy, it is $G$-equivariantly null and so its $G$-invariants subcomplex is null as well.
Could you please also explain where is semisimplicity used?
it is not used anywhere. Its just the fact that as a $G$-space the representation is a contractible space.
So it is also true when $G$ is, say, solvable??
I might be confusing with something else of course, but it seem unconditional to me, its just a contractible complex. Maybe you have another complex then I in mind?
Please get me right - I am not sceptical, I am surprised :)
in this case I think its just true unconditionally and the proof is as I written. the operation of integration of a form along the path from the origin is invariant to the group action, for every group and every representation.
Maybe it would help to note that the terms in this complex are not acyclic objects in the category of representations of the group?
|
2025-03-21T14:48:31.435934
| 2020-07-05T03:38:48 |
364867
|
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|
Stack Exchange
|
Totally geodesic submanifolds of complex Grassmannians
It is known that the connected totally geodesic complex submanifolds of a projective space ${\rm P}V$ equipped with a Fubini-Study metric are precisely the projective subspaces ${\rm P}Z$, where $Z \subseteq V$ is a complex subspace. Direct implication is obvious, and if for the converse we assume that $N\subseteq {\rm P}V$ satisfies the assumptions, we fix $L \in N$ and a basis $H_1,\ldots, H_k$ for $T_LN \subseteq T_L({\rm P}V)\cong {\rm Hom}(L,L^\perp)$, and let $Z = L \oplus \bigoplus_{i=1}^k H_i[L]$.
For Grassmannians, it is again easy to see that every ${\rm Gr}_k(Z)\subseteq {\rm Gr}_k(V)$ is connected totally geodesic and complex, but the same argument for the converse seems to fail. One reason being that if $N\subseteq {\rm Gr}_k(V)$ is a Grassmannian, then $k$ must divide the dimension of $N$.
Looking around I have found papers here and there discussing particular cases where $\dim V = 4$ and $k=2$, but I'd like to know if there's any reference discussing this more general situation.
Thanks.
There is an extensive literature on totally geodesic submanifolds of symmetric spaces. A good place to start to read about this (and references to the preceeding literature) would be, for example,
Bang-yen Chen and Tadashi Nagano, Totally geodesic submanifolds of symmetric spaces, I, Duke Math Journal 44 (1977), 745–755.
They discuss various complex Grassmannians and their totally geodesic submanifolds in particular examples.
|
2025-03-21T14:48:31.436074
| 2020-07-05T06:29:45 |
364871
|
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|
Stack Exchange
|
How to learn a continuous function?
Let $\Omega \subset \mathbb{R}^m$ be an open subset bounded with a smooth boundary.
Problem : Given any bounded continuous function $f:\Omega\to\mathbb{R}$, can we learn it to a given accuracy $\epsilon$? ($\epsilon>0$).
Definition : What do you mean by learning a function to a given accuracy $\epsilon$?
Using samples of $f$, at sufficiently large but finite number of data points that are drawn randomly(iid) from the set $\Omega$ (under a uniform probability distribution), and using a sufficiently large but finite number of registers whose precision (arithmetic) is sufficiently large but finite (this finite precision is an important condition), should be able to compute a function $F$ with only a finite number of computations (they could be additions, multiplications, and divisions but performed using registers of finite precision) such that $\|f-F\|_{L^\infty(\Omega)} \le \epsilon$.
compute a function $F$ : Given any query point $x$, one should give out $F(x)$.
Conjecture: There exists a method of learning such that one can derive a bound on required precision $p$ that depends only on $\Omega$ and $\epsilon$ and is independent of $f$.
Question: Has anyone formulated this problem before (any reference). Has anyone solved it? If I solve it, what is its market value? (mathematics market)
PS: solving means coming up with a method to learn such functions in the defined way.
(please feel free to tag appropriately)
I'm not sure what is the point of limiting the precision of registers, since it's possible to concatenate small registers to a larger one. Limiting the precision only makes sense if the numebr of registers is also limited. Should the number of registers also be limited?
Yes. number of registers is finite.
How are you supposed to catch e.g. all bump functions? Also, if you are working with general continuous functions, infinite precision information is completely local so you can safely allow it and still the problem is impossible to solve.
@VilleSalo : looks like a misnomer. My intention is all the internal computations performed are of finite precision.
You can assume that if you like. I'm just saying it affects nothing.
The randomly part is confusing: there is no uniform distribution on a countably infinite set.
@Benoit Yes. We can just say uniform distribution over the domain...no need of the intermediate countable dense set D.
Ok, then you clearly need to restrict the modulus of continuity uniformly on the family of functions you want to consider, as is explained in other answers and comments to them.
No more rollbacks, please. Let's stabilize this question.
I wonder whatever gave you the idea that this could be possible? It's not that continuity is some a priori limitation of the amount of fluctuations, just that for any given function you can a posteriori find such limitations. I think the property you actually mean here is Lipschitz continuous. But anyway such a “learning theorem” would only make sense if you also include some finite (but arbitrarily small) probability that the process will fail to give something that obeys the $\epsilon$-accuracy.
"They got us in the first half, not gonna lie."
Suppose you were given $N$ samples of $f$ (samples are irregularly placed and being drawn from a countable dense set $D$), If you can construct $F_N$ such that $\limsup\limits_{N\to\infty} |f-F_N|_{L^{\infty}(\Omega)} = \epsilon$ you have learnt the function. This is what I should have got into but I ended up with this!
and I shouldn't have even mentioned the word precision.
The answer is no by a Cantor diagonal argument:
Let $\Omega=(0,1)$.
Let $G$ be all functions that can be computed by a finite number of registers with finite precision. It does not matter where $G$ is learnt from.
The number of states of $n$ registers with precision $m$ is finite, thus the number of functions computable on $n$ registers with precision $m$ is finite. Let the set of such functions be $G_{mn}$.
Thus, $G=\bigcup_{m\in \mathbb{N}} \bigcup_{n\in \mathbb{N}} G_{mn}$ is countable. Label the elements of $G$ by $G_1,G_2,...$
Since there're an infinite number of disjoint intervals contained in $\Omega$, it's possible to avoid each $G_i$ on some interval on $\Omega$.
Let $H_n=[1-10^{-n}+\frac1310^{-n},1-10^{-n}+\frac2310^{-n}]$ be a closed interval on $\Omega$. Since $G_n$ is measurable, we can find a continuous function $f_n$ that agrees with $G_n+1$ on at least half of $H_n$ (i.e. the measure of $\{f_n=G_n+1\}$ is at least half of that of $H_n$).
Let $f$ be a continuous function on $\Omega$ that agrees with $f_n$ on $H_n$ for every $n$.
Then, for each $n$, $||f-G_n||_{L^\infty(\Omega)} \geq ||f-G_n||_{L^\infty(H_n)} \geq ||f_n-G_n||_{L^\infty(\{f_n=G_n+1\})}=1$.
See my question : "compute a function F : Given any query point x, one should give out F(x)"....only once...only for the given single $x$. Need not compute for all $x$ in the domain. Thats what I mean by compute a function.
"Compute a function" is an algorithm or a formula that gives out $F(x)$ for a given $x$. And to do this, for a single given $x$, need to be done with finite registers, computations and ...blah blah blah...
See this to know what I mean by compute a function : https://annals.math.princeton.edu/wp-content/uploads/annals-v169-n1-p08.pdf
Just for illustration the solution function we got is $F(x) = x^2 + 5$. Then the computations required to compute the function $F$ is 1 multiplication and one addition. Just two computations and a couple of registers.
But $x$ could be any element in the domain. If the number of registers and their precision need to compute $F(x)$ is bounded independently of $x$, then my argument still applies.
Agree with you. This finite arithmetic precision to represent data points $x_i$ is problematic.
The answer is NO from general no-free-lunch principles. In particular, the collection of all continuous functions has infinite fat-shattering dimension, and hence is not learnable in your sense. See Alon, Ben-David, Cesa-Bianchi, and Haussler - Scale-sensitive dimensions, uniform convergence, and learnability.
Counterexample: sin(1/x) over (0,1)
Learning the function near 0 requires infinitely many samples.
both the number of samples and there by the arithmetic precision, both should go to infinity and only then we can expect $\lim\limits_{N\to \infty}|f-F_N|_{L^{\infty}(\Omega)} \to \epsilon$ or ofcourse it can better. but this is the best thing that can be said. I should probably use limit supremum.
Technically, a single function can never be a counterexample, since an unrestricted learner can just have it hard-coded inside. You want a family of functions as a counter-example.
You can do this if $f$ is uniformly continuous on $\Omega$.
Then given any $\epsilon>0$ you can find $\delta >0$ s.t if you sample $\Omega$ within $\delta >0$ you can reconstruct $f$ to within $\epsilon$ by using nearest neighbour interpolation, define $F(x)$ to have value $f(x_{\Omega})$ where $x_{\Omega}$ is the nearest sample point to $x$ in $\Omega$. Since $\Omega$ is bounded this only needs a finite number of sampling points which are reasonably evenly distributed. i.e. a random sample of sufficient size would do. You can approximate any real number to a fixed precision using a rationals of bounded height so the precision is also finite.
Clearly you cannot do this uniformly. Different functions $f$ will require a different $\delta$ for a given $\epsilon$.
So you could do this for an equicontinuous family of functions on a compact domain.
If you read my question, " at sufficiently large but finite number of data points", so the number of data points required need not be bounded by a bound independent of $f$. They just have to be finite for a given $f$. The bound I am conjecturing is on required precision, and that be independent of $f$. So your answer is valid for all continuous functions, but the problem I am worried is that you may need to draw sample points infinite times to get an $delta$-even distribution? I hope I am wrong.
Your conjecture specifically states "one can derive a bound on required precision p that depends only on Ω and epsilon and is independent of f".
@AryehKontorovich : Yes bound on precision. Not the number of data points. Precision means arithmetic precision of floating point number. Suppose the function value at a data point $x$ is $f(x)$, the number of bits used to represent the real number $f(x)$ is what I meant by arithmetic precision. Like we use double precison floating point numbers and single precision floating point numbers in matlab.
Yes the precision required in general will depend on $f$ as clearly if you required a large number of data points to get your sampling of $\Omega$ within $\delta$ then you need enough precision to prevent sample data points merging which would not then allow you to pick up the fine details of the function - it could oscillate with arbitrarily small period. If you take only functions with variation bounded by a constant value you can do it I think.
However if you are allowed to sample at any scale then you can compensate for the limited precision of each sample by making more samples over different scales. In general precision and data size are intimately connected and make up to the total number of bits required to store your sample data. This is what you generally want to limit in the real world.
Agree, the finite arithmetic precision to represent data points $x_i$ is problematic. Even if you have a large number of data points, it won't make up for it.
|
2025-03-21T14:48:31.436805
| 2020-07-05T08:39:28 |
364874
|
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|
Stack Exchange
|
Is TA (true arithmetic) interpretable in a recursively axiomatizable theory?
Can there exist a consistent, recursively axiomatizable theory $T$, such that $\forall \phi, TA\vdash \phi \Rightarrow$ $T\vdash \tau(\phi)$, where $\tau$ is some suitable translation from the language of $TA$ to that of $T$?
Edit: By "suitable translation" I was thinking about a translation that preserves the intended meaning of $\phi$ in $TA$. For example, for any theorem $\phi$ of $PA$, you can also prove it ($\tau(\phi)$) in the language of $ZFC$, and $\phi$ and $\tau(\phi)$ share the same intended meaning in the standard model $\mathcal{N}$ of arithmetic. Only now instead of $PA$ and $ZFC$, we're thinking about $TA$ and any recursively axiomatizable theory.
(I apologize if this sounds too sloppy. I had some trouble getting this part straight in my head, but I hope the point is understood)
What’s “suitable”?
As long as the translation $\tau$ is computable (or even arithmetical) this isn't possible. For any arithmetical $\tau$ and recursively axiomatizable $T$ the set ${\varphi\mid T\vdash \tau(\varphi)}$ will be arithmetical and hence couldn't coincide with the set of theorems of $\mathsf{TA}$.
@FedorPakhomov Does not the question ask about inclusion rather than coincidence? Maybe it is trivial to encompass this in your comment, but...
@მამუკაჯიბლაძე Right, I missed it. So we would need to require a bit more from $\tau$. For example, we could only consider $\tau$ that preserve negation, i.e. $T\vdash \lnot \tau(\varphi)\leftrightarrow\tau(\lnot\varphi)$. Then we should have ${\varphi\mid \mathsf{TA}\vdash\varphi}={\varphi\mid T\vdash \tau(\varphi)}$. Otherwise due to completeness of $\mathsf{TA}$ there would be $\psi$ such that $T\vdash\tau(\psi)$ and $T\vdash\tau(\lnot\psi)$, which would imply inconsistency of $T$.
@FedorPakhomov I understand coincidence is impossible. But why is inclusion also impossible? Can you explain how "$\tau$ must preserve negation" could imply there's no such $T$?
@Eric Assume for a contradiction that ${\varphi\mid \mathsf{TA}\vdash\varphi}\subsetneq {\varphi\mid T\vdash \tau(\varphi)}$. Then there is $\psi$ s.t. $\mathsf{TA}\nvdash \psi$ and $T\vdash \tau(\psi)$. However, since $\mathsf{TA}$ is complete, we would have $\mathsf{TA}\vdash \lnot\psi$. Thus we would have $T\vdash \tau(\lnot \psi)$ and further $T\vdash \lnot \tau(\psi)$ (by preservation of negation). Contradiction, since $T$ was consistent and $T\vdash \tau(\psi)$.
Without further requirements on $\tau$, this is trivial. Let $T$ be any first-order theory and let $\top$ be any tautology. Define $\tau(\phi)= \top$.
If you want $\tau$ to be injective, then let $T$ be any first-order theory in the language of $TA$ and define $\tau(\phi)=\top\vee\phi$.
If you also want equivalence, then take $T$ to be any first-order theory in the language of $TA$ (such that $T\subseteq TA$) and define $\tau$ as follows:
-$\tau(\phi)=\phi$ if $\phi$ is not a theorem of $TA$
-$\tau(\phi)=\top\vee\phi$ if $\phi$ is a theorem of $TA$
This $\tau$ is injective and gives you the equivalence:
$\forall \phi(TA\vdash\phi\leftrightarrow T\vdash\tau(\phi))$.
If you want $\tau$ to be an interpretation in the usual sense, as the title of the question suggests, then this is not possible as Fedor Pakhomov has explained in two comments (intepretations preserve negation).
So, you must specify what do you mean by a suitable $\tau$, as Monroe Eskew has already asked in a comment.
EDIT:
The edited version of the question suggests that the OP wants $\tau$ to be an interpretation in the usual sense. There is no such $\tau$ (this was already clear from the comments and the previous version of the answer, but I will unify the argument here for convenience).
Indeed, assume without loss of generality that the signature of $TA$ is finite, and let $\tau$ be an interpretation of $TA$ in $T$. In this case, $\tau$ is computable. Therefore, the set of sentences $\phi$ such that $T\vdash \tau(\phi)$ is RE. However, the set of those sentences is just the set of true sentences of arithmetic, for if $TA\vdash \phi$ then $T\vdash\tau(\phi)$, and if $TA\not\vdash\phi$, then $TA\vdash\neg\phi$, $T\vdash\neg\tau(\phi)$, and $T\not\vdash\tau(\phi)$. This is a contradiction with Tarski's theorem.
Entirely true, and answers the question in its present form perfectly. Still, it is formulated in such way that it is a comment rather than answer.
I agree, I have expanded the answer with more information.
I edited the question to address the indeterminacy. But suppose $\tau$ preserves negation, why does this make $T$ impossible?
Since the language of arithmetic is finite, interpretations in the usual sense are automatically computable (hence arithmetical).
Sure Emil, but the OP may have a different presentation of TA in mind. For example, one with infinitely many primitive constants.
This does not matter. You can’t interpret any extension of TA to any language you want, because you cannot even interpret its finite-language fragment in the usual language of arithmetic.
Sure, there is no interpretation, but my comment was on the issue of the computability of the interpretation. I will edit the answer to eliminate the reference to this issue, thanks.
|
2025-03-21T14:48:31.437185
| 2020-07-05T08:58:52 |
364876
|
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|
Stack Exchange
|
Replacing the usual mean by the weak one
The mean of a non-negative random variable is just its $L_1$ norm. As such, it has a "weak" version:
$$ ||X||_{1,w} = \sup_{t>0} tP(X\ge t).$$
An immediate consequence of Markov's inequality is that the "strong" (i.e., usual) mean always upper-bounds the weak one, and the latter could be finite while the former infinite (such as $X=1/U$, where $U$ is uniform on $[0,1]$).
Question: is there a theory of convergence to the weak mean? Analogues of the CLT, strong law of large numbers, large deviations, measure concentration?
Edit. Let me elaborate. Obviously, the sample mean will not converge to the weak mean. What about the weak sample mean?
The model example here is the (extremely well studied) St. Petersburg paradox: https://en.wikipedia.org/wiki/St._Petersburg_paradox
I think the weak sample mean will converge a.s. to the weak mean iff $EX^2<\infty$; cf. e.g. Theorem A in shorturl.at/lLY59. However, a proof of this (which I don't have) will likely be too long for a usual MO answer.
@IosifPinelis Maybe we should write another paper! :)
Sorry, I forgot to mention the trivial case, $|X|{1,w}=\infty$, when the convergence will obviously hold. So, the conjectured necessary and sufficient condition is "$|X|{1,w}=\infty$ or $|X|_2<\infty$", with a gap between the two terms of the disjunction "or".
|
2025-03-21T14:48:31.437318
| 2020-07-05T09:45:53 |
364880
|
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|
Stack Exchange
|
Convergence of Radon-Nikodým derivative
Imagine we have a sequence of finite measures $\nu_n << \mu_n$ (on the torus $\mathbb{T}^2\subseteq \mathbb{R}^2$) converging weakly to some measures $\nu << \mu$. Do we automatically have that the Radon-Nikodým derivatives $h_n$ of $\nu_n$ wrt. $\mu_n$ converge to the Radon-Nikodým derivative $h$ of $\nu$ wrt. $\mu$?
Does the situation becomes easier if we assume that $0\leq h_n \leq 1$ and $0\leq h\leq 1$?
Does the results carry over if we consider finite random measures instead of deterministic measures? And is the result still available if the convergence is only in the distributional sense, i.e. for compactly supported smooth functions?
Thanks a lot for your help!
The standard example to look at here is the system of Haar functions where things go wrong—these are generally defined on the unit interval but this has no relevance for your query. By the way, your objects, as measures, can be regarded as Schwartzian distributions and then everything is fine except that the derivatives have to be taken in the distributional sense which might not correspond to your requirements.
@user131781 What do mean when you say that everything is fine if I consider them as Schwartzian distributions? I have never heard of taking Radon-Nikodým derivatives in the distributional sense... Do you have a reference? This is perhaps exactly what I need :)
The answer is negative for a.e. pointwise convergence as well as $L^1$ convergence with respect to $\mu$. Let $\mu_n=\mu$ be half the Lebesgue measure, color the torus as a checkerboard with $2^n$ cells, and assume $h_n$ takes value $1$ on black cells and $0$ on white cells. Then $\nu_n=h_n\mu \to \mu$ $*$-weakly, but for all $n$ the function $h_n-h$ is everywhere $\pm\frac12$.
To control densities, you want convergence in total variation (or variations on this).
Thank you very much!
|
2025-03-21T14:48:31.437475
| 2020-07-05T12:37:27 |
364889
|
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|
Stack Exchange
|
Interpretation of a short note of Gauss on the resolution of a special system of inhomogeneous linear equations by roots of unity
My question refers to a 2-pages fragment of Gauss, entitled: "Note on the resolution of a special system of linear equations", which is found on pages 30-31 of volume 8 of his works. Here is a link to Gauss's fragment, https://gdz.sub.uni-goettingen.de/id/PPN236010751?tify={%22pages%22:[36],%22panX%22:0.368,%22panY%22:0.464,%22view%22:%22export%22,%22zoom%22:0.498} . In this fragment, Gauss expresses the solution vector $(P,Q,R,S,T)$ of the following inhomogenous system of $5$ linear equations:
$$ aP + bQ + cR + dS + eT = A $$
$$ bP + cQ + dR + eS + aT = B $$
$$ cP + dQ + eR + aS + bT = C $$
$$ dP + eQ + aR + bS + cT = D $$
$$ eP + aQ + bR + cS + dT = E $$
(note that the coefficents of each row are a cyclic shift of the previous row) by five auxillary parameters $p,q,r,s,t$ which solve another system of $5$ linear equations:
$$ ap + bq + cr + ds + et = 1 $$
$$ bp + cq + dr + es + at = 0 $$
$$ cp + dq + er + as + bt = 0 $$
$$ dp + eq + ar + bs + ct = 0 $$
$$ ep + aq + br + cs + dt = 0 $$
Gauss then expresses $P,Q,R,S,T$ through a bilinear expression in $p,q,r,s,t, A,B,C,D,E$. To find $p,q,r,s,t$ Gauss employs a certain algebraic trick involving the fifth root of unity $\epsilon$ ($\epsilon^5 = 1$).According to the short comment of Fricke on Gauss's result, the final form of Gauss's formulas for $p,q,r,s,t$ involves an expression for the norm of the cyclotomic number $a +b\epsilon + c\epsilon^2+d\epsilon^3+e\epsilon^4$ in the quintic (cyclotomic) field $Q[\epsilon]$ (see also this post: https://math.stackexchange.com/questions/24840/finding-the-norm-in-the-cyclotomic-field-mathbbqe2-pi-i-5 ). The whole Gaussian method looks like an hybrid between techniques of linear algebra and techniques fron alegraic number theory (roots of unity), and therefore it aroused my curiosity (this complicated procedure suggests that Gauss had an idea here...). Here i omitted the final formulas, because they are long and the best way to gain insight into them is simply to look into Gauss's note.
I understand that trying to give a sound interpretation of such a complicated note (and without supporting context) is a very difficult task, but maybe someone who is familiar with advanced algebraic techniques will find familiar patterns in Gauss's fragment (honestly i have no clue or idea about how to dechiper this note). I already posted this question on HSM stack exchange, and i'm aware my question is a little bit historical, but i think that the mathematical nature of this question makes it appropriate to post it here.
Therefore, my question is about giving a rough outline and interpretation of the meaning of Gauss's statements in this note.
Despite not understanding a lot of German, I think I get the general idea. Do you know how a real matrix of the form:
$$A = \left( \begin{array}{cc} a & -b \\
b & a \end{array} \right) $$
can be thought of as the complex number $a+ib$? It is a famous trick. Essentially, multiplying by $a+ib$ amounts to left multiplying by the real matrix $A$, when you think of each complex number as a column vector consisting of its real and imaginary part.
Well, I think Gauss is using a similar trick in that note, but using the field obtained from $\mathbb{Q}$ (or is he using $\mathbb{R}$?) by adjoining a $5$-th root of unity $\epsilon$.
Then each element in $\mathbb{Q}[\epsilon]$ can be written as $a+b\epsilon + c \epsilon^2 + d \epsilon^3 + e \epsilon^4$, or can be thought of as a $5$-dimensional column vector containing the components $a,\ldots,e$. Multiplying by such an element in that field can be then written as multiplication by a $5$ by $5$ rational (or real if he is using $\mathbb{R}$ as the ground field) matrix satisfying the cyclic conditions you have mentioned. I think this is the basic idea.
To solve the equation $ax = b$ in $\mathbb{Q}[\epsilon]$, where $b \neq 0$, all one has to do is multiply by $a^{-1}$. The unique solution is then $x = a^{-1} b$. The auxiliary equation $ay = 1$ has the unique solution $y = a^{-1}$. And so by solving the auxiliary equation, you are actually getting $a^{-1}$. Then the unique solution of the original equation is then just $x = a^{-1} b$, but you have to represent, either $a^{-1}$ or $b$ (since multiplication here is commutative) as a $5$ by $5$ matrix and the other one by a $5$-dimensional column vector. I think this is what Gauss is doing in that note.
Towards the end of that note, I think that Gauss is calculating $a^{-1}$ (using my notation) using the norm, if I am not mistaken. Let us consider a simpler case. If $z = x + i y \in \mathbb{C}$, then its norm $N(z) = z \bar{z} = x^2 + y^2$. So if $z \neq 0$, then $z^{-1} = \bar{z}/(x^2+y^2)$.
I think what Gauss is doing is something similar. One can write $N(a)$, if $a \in \mathbb{Q}[\epsilon]$, as the product of $a$ and its "conjugates" (over the ground field $\mathbb{Q}$), in the Galois theory sense (by definition of the norm). Please look up the resulting formula somewhere (I do not know the formula of the norm of an element in this algebraic number field). My guess is that, if $a \neq 0$, then $a^{-1}$ should be the product of the Galois conjugates of $a$, other than $a$ itself, divided by $N(a)$, which should be given by a formula, somewhere in the literature. I think this is what Gauss is doing. I hope this is enough details.
Edit: Regarding the last formulas in the article, it seems that Gauss did a "translation" too by what he denotes by $n$. So it is a little bit more complex than what I wrote.
See also circulant matrices, for instance on wikipedia, or elsewhere. After rearranging the equations, the $5$ by $5$ matrix becomes a circulant matrix.
Thanks Malkoun! clearly your answer helps a lot and gives the basic idea of the note (i voted your answer). I didn't accept the answer yet because i still need to process it in my mind and see how the pieces of the puzzle combine. Right now i'm still not sure it's a definitive answer (and obviously it might be a definitive answer, but i need to check i understand all aspects of Gauss's note). Thanks a lot! i'll accept the answer once the subject will be clear to me.
No worries. If you need more help, let me know.
To be honest, the note is clear to me, except how he got the solution of the auxiliary system at the end. I mean, one needs to probably compute to check it (sorry, this would take time, so I will not do it).
this is exactly but i wanted to ask - i followed your answer and clearly saw your interpretation fits exactly to the systems of linear equations that descibe the arithmetic in the quintic field $Q[\epsilon]$, but what i don't understand is the formulas right after the systems (immediately after the systems he states several formulas which connect $p,q,r,s,t$ with $a,b,c,d,e$ ) and at the very end (the formulas that involve the norm of the cyclotomic number). If you won't do it, can you at least give a hint about how to arive the solution at the very end of te note?
If my intuition is correct, this is what he is doing. You know how the norm is often used to calculate inverses, right? So for instance, if $z\in \mathbb{C}$, then $N(z) = z\bar{z} = x^2+y^2$ allows us to deduce that if $z \neq 0$, then $z^{-1} = \bar{z}/(x^2+y^2)$. I think Gauss is doing something similar, but in the field $\mathbb{Q}[\epsilon]$. More precisely, if $x \in \mathbb{Q}[\epsilon]$, then write $N(x)$ as the product of $x$ and all its "conjugates", and deduce what $x^{-1}$ is, assuming $x \neq 0$. Does it make sense?
That being said, I am not an algebraic number theorist, but I am sure you can find references for the norm of an element in a cyclotomic field somewhere, and check whether what I wrote is true.
Yes. it makes sense (i just need to understand the concept of "conjugate" in general cyclotomic field, not just $Q[i]$; i'll need to make some googling for this). Thanks.
Let us continue this discussion in chat.
what did you mean when you said "Gauss did a "translation" too by what he denotes by n. So it's a little bit more comlex than what i wrote."? By the way, it should be added to your answer that all congugates to a number $f(\epsilon)$ (in $Q[\epsilon]$) are given by $f(\epsilon^2)...f(\epsilon^4)$ (that is Ernst Kummer's definition of conjugates in cyclotomic field). That explains some of the formulas in Gauss's note.
@user2554, well, I think the best way to understand the last formulas is to do the calculations! I did not do the calculations (I am sorry, but I don't have much free time). However, I have a feeling Gauss did things this way, so as to avoid writing quantities like $\epsilon + \epsilon^4$ and $\epsilon^2 + \epsilon^3$.
|
2025-03-21T14:48:31.438042
| 2020-07-05T12:46:27 |
364890
|
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|
Stack Exchange
|
Moments of complex random variables
My question is that how much information we can get form integer moments of a complex random variable?
Let $\mathcal{Z}$ be a complex value random variable, and assume that we can compute $$\int \mathcal{Z}^k d\mu,$$
For $k \in \mathbb{N}$ and $\mu$ be a measure.
I also am looking for an example of a complex random variable, that its moments positively approaches zero with
$$ 0< \int \mathcal{Z}^k d\mu \asymp \frac{1}{k!}.$$
If moments satisfy the above, can we occlude that $|\mathcal{Z}|$ is bounded almost surely?
One simple example: let $$ \mathcal{Z}= \sum_{m \in A} e^{imt} + \tfrac{1}{2},$$
where $A \subset \mathbb{Z}$ and $t$ is uniformly distributed in $(0, 2\pi].$ Then k-th moment of $\mathcal{Z}$ is 2^{-k}. Therefore moments decay but they fall short of decaying like $1/k!$.
This is to rewrite the excellent example by Mateusz Wasilewski in a more conventional form.
Let $Z:=XU$, where $X$ and $U$ are independent random variables (r.v.'s); $P(X>0)=1$; $X$ is unbounded; $EX^k<\infty$ for all natural $k$; $U=e^{iT}$; $T$ is a r.v. with values in the interval $[0,2\pi)$ and pdf $p$ given by the formula
$$p(t)=\frac1{2\pi}\,\Big(1+2\sum_{n=1}^\infty a_n\cos nt\Big)$$
for $t\in[0,2\pi)$;
$$0<a_n\sim\frac1{n!\,EX^n};$$
and $2\sum_{n=1}^\infty a_n<1$ (so that $p>0$).
Then for all natural $k$
$$EU^k=Ee^{ikT}=\int_0^{2\pi}e^{ikt}p(t)\,dt=a_k$$
and hence
$$EZ^k=EX^k\,EU^k=EX^k\,a_k\sim\frac1{k!}$$
and $EZ^k>0$, whereas the r.v. $|Z|=X$ is unbounded.
Thanks, this is very clear.
Let $u$ be uniformly distributed on the unit circle and let $X$ be a positive random variable with all moments finite. If $u$ and $X$ are independent then $Z:=uX$ has all moment equal to zero, because $\mathbb{E} u^{k} = 0$ for any $k\neq 0$. If you want an example with strictly positive moments then just add to $Z$ and independent positive random variable with moments decreasing very fast. You can easily cook up in this way examples that are not bounded.
EDIT: Here comes an actual example. Our random variable $Z$ will be of the form $Z=uX$, where $u$ is some distribution on the unit circle and $X$ is positive; we assume that $u$ and $X$ are independent. Then the moments are $\mathbb{E}Z^k = \mathbb{E} u^{k} \cdot \mathbb{E} X^{k}$. We want $X$ to be unbounded, so the moments of $X$ will grow to infinity at some rate, but it is not so important. We will use the oscillatory nature of $u$ to produce examples with an arbitrarily fast decay of $\mathbb{E} u^{k}$. The distribution of $u$, being supported in the unit circle, can be represented by the Fourier series $u \sim \sum_{n\in\mathbb{Z}} a_n e^{int}$. Note that $a_n= \mathbb{E} u^{-n}$. We have $a_0=1$ and we want the coefficients $(a_n)$ to decay very fast, while ensuring that the Fourier series above represents a probability measure. The moments are supposed to be real, so $a_n = a_{-n} = \overline{a_n}$. We can therefore rewrite the Fourier series as $1 + 2\sum_{n\geqslant 1} a_n \cos(nt)$. If we take $a_n$ small enough so that $|2\sum_{n\geqslant 1} a_n \cos(nt)|\leqslant 1$, then we have a probability measure. Now just pick $(a_n)$ in such a way that $\mathbb{E} u^{k} \simeq \frac{1}{k! \mathbb{E}X^{k}}$ and you get a counterexample.
Thanks, so basically, to put more in analysis form, an exponential sum like the one I mentioned plus an independent function which has moments like $1/k!$ would do the job.
I am wondering if that is the only possibility, meaning that if we pull that oscillatory part out of $\mathcal{Z}$ we always end up with positive RV with same moments as $\mathcal{Z}.$
Could you give me an example of a RV with moments like $1/k!$?
I actually realised that it's not possible to have such a fast decay of moments, even for not necessarily positive random variables. Indeed, if $X$ is a non-zero random variable with all moments finite then for some $\varepsilon>0$ the probability of the event ${X^2>\varepsilon}$ is positive and we have a moment bound $\mathbb{E} X^{2k} \geqslant C \varepsilon^{2k}$ which decays more slowly than the factorial. So I still don't know if there are examples satisfying your original assumptions.
On the edit: You no longer assume that $t \sim U(0, 2\pi)?$ cause that way all the moments of $u$ are zero. Also, could you please explain a bit more on how to choose $a_n$ such that $\mathbb{E} u^{k} \simeq \frac{1}{k! \mathbb{E}X^{k}}$.
Yes, I said that now $u$ is some distribution on the unit circle, not the uniform one. It is basically a small perturbation of the uniform distribution so that the moments are very small but positive. Could you ask more specifically, which part of the construction is still unclear?
|
2025-03-21T14:48:31.438480
| 2020-07-05T14:12:47 |
364896
|
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|
Stack Exchange
|
Primes which do not divide certain homogeneous polynomials
It is known that if $x^2 + y^2 = z^2$ is a primitive Pythagorean triplet then $z$ is not divisible by any prime of the form $4k-1$. The following is a generalization of this classical result which shows that the source of this property is not the hypotenuse $z$, but the two orthogonal side $x$ and $y$:
Conjecture: Let $f(x,y) = a_0x^n + a_1 x^{n-1}y + a_2x^{n-2}y^2 + \cdots + a_ny^n$, $a_0a_n \ne 0$. Then there are infinitely many primes of the form $8k+3$ which do not divide $f(x,y)$ for any primitive Pythagorean triplet $x^2 + y^2 = z^2$.
Example: Taking $f(x,y) = x^n + y^n$, the conjecture says that for every there are infinitely many primes which do not divide the sum of the two orthogonal sides of any Pythagorean triangle. The conjecture has been proved for $x+y$ (in MSE) and is already known to be true for $x^2 + y^2$. Experimental data shows that $x^3 + y^3$ is not divisible by infinitely many primes of the form $8k+3$ while $x^4 + y^4$ is not divisible by infinitely many primes of the form $8k+3, 8k+5$ and $8k+7$.
Question: Is the conjecture known? I am looking for a proof or disproof of the conjecture.
The conjecture was proved for special case $f(x) = x+y$ in MSE but the general case remains open. Hence it is posted in MO.
Some of the binomials were studied by Legendre who had divisibility results that depended on the exponent, likely motivated by the (then) Fermat conjecture. Gerhard "Which You May Already Know" Paseman, 2020.07.05.
Also, what about x^2 + 3y^2? Gerhard "A Reason For Being Unknown?" Paseman, 2020.07.05.
@GerhardPaseman Since $x^2 + 3y^2 = (x^2 + y^2) + 2y^2$ hence trivially the conjecture is true for $x^2 + 3y^2$ since it is true for $x^2 + y^2$
Perhaps I misunderstand the question, but doesn't $f(x,y) := 4x-3y$ give a counterexample just because of the Pythagorean triple $(x,y,z)=(3,4,5)$? Or if one wants a nonnegative example, take $f(x,y) = 4x+3y$ and use the triples $(x,y,z) = ((p-2)^2-1, 2(p-2), (p-2)^2+1)$ (which is congruent mod $p$ to $(3,-4,5)$) for a given prime $p = 8k+3$.
|
2025-03-21T14:48:31.438636
| 2020-07-05T14:45:27 |
364898
|
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|
Stack Exchange
|
Divisibility between values of polynomials
In this paper Corvaja and Zannier considered the following problem. Let $K$ be a number field, and let $S$ be a finite set of places of $K$ containing the infinite places and let $\mathcal{O}_S$ be its ring of $S$-integers. One may enlarge $S$ so that $\mathcal{O}_S$ is a PID. Suppose that $f_i, g_i$ are polynomials in $\mathcal{O}_S[x,y]$ for $i = 1,2,3$. Then they showed (see Theorem 2) that under certain genericness assumptions that the set of points $(x,y) \in \mathcal{O}_S^2$ satisfying
$$\displaystyle f_i(x,y) | g_i(x,y), i = 1,2,3$$
is not Zariski dense. Moreover, they showed that none of the assumptions they made can be dropped. In particular, there are easy counter-examples when $i = 1,2$.
In my situation, I wish to consider a variant of this problem in the case $i = 2$. Take single variable polynomials $f(x), g(x) \in \mathbb{Z}[x]$ and a polynomial $q(x,y) \in \mathbb{Z}[x,y]$. What can we say about the number of integral points $(x,y) \in \mathbb{Z}^2$ with $\max\{|x|, |y|\} \leq T$ such that
$$\displaystyle q(x,y) | \gcd(f(x), g(y))?$$
Can any aspect of Corjava-Zannier be applied in this setting, and if not, are there any other methods to count such points?
|
2025-03-21T14:48:31.438756
| 2020-07-05T16:33:35 |
364902
|
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"authors": [
"Dmitri Pavlov",
"Nik Weaver",
"Robert Furber",
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|
Stack Exchange
|
Which complete orthomodular lattices arise from von Neumann algebras?
Let $A$ be a von Neumann algebra. Then a classic observation is that the set of projections $\Pi(A)$ is naturally a complete orthomodular lattice.
Question 1: Is the construction $A \mapsto \Pi(A)$ a functor from von Neumann algebras to complete orthomodular lattices?
For this to make sense, I should say what a morphism of von Neumann algebras is -- but I'm not sure what the appropriate choice of morphism is. I should also say what a morphism of complete orthomodular lattices is, and here there is a natural guess -- a morphism $f: L \to M$ should be a function which preserves $(-)^\perp$ and sups (equivalently, infs).
The construction $A \mapsto \Pi(A)$ makes sense even when $A$ is just a $C^\ast$-algebra, except that we only know that $\Pi(A)$ is an orthoposet (maybe an ortholattice?).
Question 2: Let $A$ be a $C^\ast$-algebra, and suppose that $\Pi(A)$ is a complete orthomodular lattice. Does it follow that $A$ is a von Neumann algebra?
Question 3: Assuming the answer to Question 1 is "yes", let $\Pi: vNA \to COML$ be the above functor from von Neumann algebras to complete orthomodular lattices.
(a.) Is the functor $\Pi$ fully faithful?
(b.) Does the functor $\Pi$ have a left or right adjoint?
(c.) What is the essential image of the functor $\Pi$?
(3c) is the title question, of course. It seems there was substantial interest in this question in the '60's and '70's. I've come across work by Holland, Topping, and Fillmore identifying various properties of orthomodular lattices of the form $\Pi(A)$ not shared by all complete orthomodular lattices. I believe there is a characterization those lattices of the form $\Pi(A)$ where $A = B(H)$ is the algebra of all bounded operators on a Hilbert space $H$ (though I'm having trouble tracking down where I read this). But the trail seems to go cold after this period. Did the question just go out of fashion?
I do have the sense that there might be a characterization which has something to do with "having a full set of states". For instance, suppose we define a state on a complete orthomodular lattice $L$ to be a function $f: L \to [0,1]$ which preserves order and directed sups, is additive ($f(0) = 0$, $f(a \vee b) = f(a) + f(b)$ if $a \perp b$), and is normalized so that $f(1) = 1$ (I think maybe this should really be called a "normal state" or something like that?). Say that $L$ has a full set of states if for $a, b \in L$ we have $a \leq b \Leftrightarrow f(a) \leq f(b)$ for all states $f$. Then I believe that if $L = \Pi(A)$, then $L$ has a full set of states. Is the converse true?
In the commutative case (https://mathoverflow.net/questions/71259/which-complete-boolean-algebras-arise-as-the-algebras-of-projections-of-commutat) the answer to (3c) is not known either. In fact, this is a famous open research problem, originally formulated by von Neumann.
@DmitriPavlov Although, as Nik Weaver points out (and you point out in that question), in the abelian case, some form of answer is known (admitting a full set of normal states) -- it's just not entirely satisfactory from some perspectives.
Counterexample for question 2: $\Pi(C([0,1]))$ has two elements, the constant $0$ and $1$ functions. This is a complete Boolean algebra. But $C([0,1])$ is not a von Neumann algebra.
Question 1: Yes, if you take the von Neumann algebra morphisms to be normal $*$-homomorphisms. Restricting any such map to the projections will preserve sups and orthocomplements.
Question 2: No, this already fails in the commutative case. Look up "abelian AW*-algebra" or see the discussion of Stonean spaces in volume one of Kadison-Ringrose.
Question 3: (a) Any von Neumann algebra is generated by its projections, so the functor is faithful. It is not full, just look at $M_2(\mathbb{C})$ where the lattice of projections contains a $0$ and a $1$ and $2^{\aleph_0}$ incomparable elements between those two. There are all kinds of morphisms from this lattice to itself that don't extend linearly.
Question 3: (b) I'm weak on category theory, but I'd imagine the $M_2(\mathbb{C})$ example falsifies both possibilities.
Question 3: (c) I don't think there's any good answer to this question, and that's the reason research in this direction petered out. The idea of asking for a full set of normal states is good, and this does answer the question in the abelian case (again, see Kadison-Ringrose), but I doubt it will work in general. There's just no way to linearize these states on lattices. But counterexamples will take some work.
Thanks! How essential is it to use $M_2(\mathbb C)$ here? After all, in light of Gleason's theorem, this is a very "exceptional" von Neumann algebra. Could one just as well use $M_3(\mathbb C)$ to show that $\Pi$ is not full?
You make a good point. $M_3(\mathbb{C})$ does seem much more rigid ... I'll have to think about this.
I'm actually starting to think that the situation is not as bad as indicated in Nik Weaver's answer. Some notation:
If $\mathcal A$ is a $C^\ast$-algebra or JB-algebra (resp. von Neumann algebra or JBW-algebra), let $\Pi(\mathcal A)$ be its ortholattice of projections. Let $State(\mathcal A)$ (resp. $State_n(\mathcal A)$) be its convex space of states (resp. normal states). If $L$ is a complete orthomodular lattice, let $State_n(L)$ be the convex space of normal states on $L$. If $K$ is a convex space, let $Aff(K)$ be the space of bounded ($\mathbb R$-valued) affine functionals on $K$.
We are led to the following considerations:
By Gleason's Theorem, if $\mathcal A$ is a von Neumann algebra without a direct summand of $M_2(\mathbb C)$, the natural restriction map $State_n(\mathcal A) \to State_n(\Pi(\mathcal A))$ is an isomorphism.
So in order to recover $\mathcal A$ from $\Pi(\mathcal A)$, it typically suffices to recover it from $State_n(\mathcal A)$.
Let $\mathcal A$ be a JB-algebra (resp. JBW-algebra). Then as Banach spaces, there is a natural isomorphism $Aff(State(\mathcal A)) \cong \mathcal A$ (resp. $Aff(State_n(\mathcal A)) \cong \mathcal A$). Moreover, Alfsen and Schultz have shown (see Geometry of State Spaces of Operator Algebras and predecessor works) how to recover the Jordan multiplication on $\mathcal A$ purely from the convex structure on $State(\mathcal A)$ (resp. $State_n(\mathcal A)$), including giving a characterization of those convex spaces for which this can be done. (Briefly, they identify conditions under which a functional calculus can be developed. Since a Jordan multiplication is determined by its squaring operation, this yields a candidate Jordan multiplication, and they identify further conditions ensuring that the candidate multiplication is bilinear and hence actually is a Jordan multiplication.)
Thus if $\mathcal A$ is a $C^\ast$-algebra (resp. von Neumann algebra), then the self-adjoint part of $\mathcal A$ and its Jordan multiplication can be reconstructed from $State_n(\mathcal A)$. But because $\mathcal A$ is not typically isomorphic to its opposite algebra (a fact which taking $State_n(\mathcal A)$ forgets), there is no hope in general to recover the algebra multiplication on $\mathcal A$. However,
Alfsen and Schultz have shown that any JB-algebra (resp. JBW-algebra) $\mathcal A$ is "locally" the self-adjoint part of at most two different $C^\ast$-algebras (resp. von Neumann algebras), which are opposite to each other. Moreover, they have characterized in terms of "orientation" data on the convex space $State(\mathcal A)$ (resp. $State_n(\mathcal A)$) exactly when a global algebra multiplication can be found, and how to distinguish between the different possible choices.
Putting this together, we see that the functor $\mathcal A \mapsto State_n(\mathcal A)$ is "almost injective" on isomorphism classes, its essential image can be explicitly characterized, and the additional data to make the functor 1-to-1 on isomorphism classes has been described. This leaves me suspecting that with the right notion of morphism of oriented convex space, the functor becomes fully faithful.
This is not quite the same as characterizing things in terms of $\Pi(\mathcal A)$, but if we exclude von Neumann algebras with summands of $M_2(\mathbb C)$, then $State_n$ factors through $\Pi$. In particular, we have a description of the essential image of $\Pi$, in terms of the normal state space of a complete orthomodular lattice $L$. It might be nice to have a more direct description, but the state space of $L$ is a pretty natural invariant to consider, so one might also be satisfied with this.
This is very nice, but I don't think the situation is quite as happy as you say in point 3. Any direct sum of counterexample will be a worse counterexample. Maybe Alfsen and Schultz are assuming $\mathcal{A}$ is a factor?
Or am I confused?
Ah, I think I was extrapolating too much from what they said. What does seem to be the case, though, is that "locally" there are at most two ways to turn a JB algebra into a C* algebra, but I don't actually understand how to globalize this statement.
@TimCampion The normal state space of $L^\infty([0,1])$ has no extreme points, so is not compact in any locally convex topology. Normal state spaces are only closed bounded convex sets.
@RobertFurber Thanks, I've edited to fix this. As goes without saying, all errors are my own and not Alfsen and Schultz's.
@TimCampion For some reason I forgot to mention this earlier (maybe because it doesn't relate to the official title of the post) but it is possible to use the Alfsen--Shultz-Hanche-Olsen--Størmer results to turn the state functor on C$^$-algebras and normal state functor on W$^$-algebras into categorical equivalences. I worked out the details of this during my PhD. Essentially all you need is already in the two volumes on state spaces by Alfsen and Shultz.
@TimCampion The maps of state spaces that are images of Jordan homomorphisms are those preserving complementary projective faces (Lemma 5.15 of Geometry of State Spaces...). The simplest way of ensuring that you get $$-homomorphisms was suggested to me by Bram Westerbaan - use Choi's theorem that a 2-positive Jordan homomorphism is a $$-homomorphism, so either keep the state space as an "operator convex set", or reconstruct $M_2(\mathcal{S}(A))$ from an orientation on $\mathcal{S}(A)$. Even in the case when $A \cong A^{\mathrm{op}}$ the orientation is needed to distinguish *-homomophisms
from Jordan homomorphism that don't preserve products, so even for cases as simple as $M_n(\mathbb{C})$. One thing I'll add -- as far as I know, it is an unsolved problem to find a structure defined solely in terms of projections that acts like an orientation of the state space, and so allows us to distinguish *-homomorphisms from Jordan homomorphisms. It needs to be a structure because of course the transpose map induces an isomorphism of projection lattices. Coincidentally, I recently asked an expert about it at a conference who agreed it is a very difficult problem.
@TimCampion Regarding state spaces, there are some references in this old answer and a sentence about the nonunital case.
@TimCampion Although probably you should look into this if you want to know an example of a structure you can add to the projection lattice to get an equivalence: https://arxiv.org/abs/1212.5778
@RobertFurber Thanks so much! I think all the information you've provided would constitute a pretty good answer!
|
2025-03-21T14:48:31.439523
| 2020-07-05T18:38:05 |
364907
|
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|
Stack Exchange
|
How many squares can be formed by using n points?
How many squares can be formed by using n points on a 3 dimensional space?
Like using 4 points, there is 1 square be formed
Using 5 points, still 1 square
Using 6 points, 3 squares can be formed
Let me know if I get this right. Given a set $X$ of points in the plane we denote by $Sq(X)$ the number of squares whose vertices are in $X$. Define $$ Sq_n=\max\big{S(X);;;|X|=n\big}.$$ Find $Sq_n$. For example $Sq_4=Sq_5=1$, $Sq_6=2$. Is this what you are asking?
OEIS A051602
Geoff, you might see four squares. I see four triangles. Gerhard "Is Not Seeing Dead Points" Paseman, 2020.07.05.
I can form three squares with six points. I think this is of interest to combinatorial geometers, and should stay open, especially because of its vagueness. Gerhard "Ambiguity Begets Search For Resolution" Paseman, 2020.07.05.
What is a 3D plane? Do you mean a 2D plane? By the way, the squares don't all have to be the same size, do they?
I mean a 3-dimensional plane and yes, they don't have to be the same size. I pick these tags because there aren't a tag "counting"
I see you changed the answer for $6$ points: originally it was $2$ squares, now it's $3$ squares. Was this originally intended as a question about points in a $2$-dimensional plane, and now it's about points in $3$-dimensional space? Are we trying to hit a moving target?
@GerhardPaseman, "let's leave this vague question open because there are lots of interesting questions similar to it" seems like a bad precedent to set that will discourage askers from posing precise questions.
In some cases I agree @Lspice. In this case I don't, because it reminds readers that there is more mathematics just lying below the surface. The original wording admitted a very specific interpretation which would occur to many. It also admitted other interpretations of interest, which would apply to the naive one (squares on the real two dimensional plane). Gerhard "There Are Other Bad Precedents" Paseman, 2020.07.05.
No, I change 2 square to 3 square is because I made a mistake and I change 2D to 3 D was because of bad finger typing(I see the problem wrong and I typed it wrong...).
Now that we understand the question, can we calculate it for a few small values and check the OEIS again? The link given by @Rob doesn't apply, as it's for ${\bf R}^2$.
Note that for six points, there are at least two configurations that give three squares: a prism, and a regular octahedron. How many squares are determined by the eight vertices of a cube?
@Gerry Interesting observation about the octahedron. One can take a k simplex and reflect it to get a similar construction for k dimensions, again k choose 2 squares on 2k points. Gerhard "Looking For Up-Down Switching Mirror" Paseman, 2020.07.06.
@GerryMyerson I think the vertices of the cube determine only the obvious squares. The story is different for the vertices of the 4-cube which contain many more squares.
In the plane $n$ points can determine at most $O(n^2)$ squares. This is because any two distinct points can determine up to three squares.
In $R^3$ this argument no longer holds, since two points can form the corners of arbitrarily many squares. As Gerhard points out, $O(n^3)$ is an upper bound (in any dimension) given that three points determine at most one square.
One can do a bit better than this. Using the Szemerédi–Trotter theorem one can show that a set of $n$ points in $R^3$ determines at most $O(n^{7/3})$ right triangles. It follows that $n$ points determine at most $O(n^{7/3})$ squares (since a square will contain a right triangle). On the other hand, it is certainly easy to see that there exists point sets with at least $\Omega(n^2)$ squares. The bound of $O(n^{7/3})$ is known to be sharp for the less constrained problem of counting right triangles.
Update: A result of Sharir, Shefer and Zahl shows that the number of mutually similar triangles in a point set in $R^3$ is at most $O(n^{\frac{15}{7}})$, where $15/7 = 2.142\ldots$, which implies the same bound for the number of squares.
Closing the gap for squares, however, seems an interesting and non-trivial problem.
" it is certainly easy to see that there exists point sets with at least $Ω(^2)$ squares." Yes $\mathbb{Z}^2\cap [0.\sqrt{n}]^2$ works.
In k dimensions, take a regular unit k simplex on k points, and copy it an orthogonal distance of 1. This results in k choose 2 unit squares on 2k points. I invite others to count square arrangements in a hypercube.
Combinatorially, there can be no more squares than three sets of an n set. Indeed, since three points of a square determine the fourth, there are at most a fourth as many squares possible as three sets. I imagine Erdos may have an upper bound for planar arrangements, which should be on the order of n^2, since any two points in the plane determine one of three squares containing those two points.
Gerhard "Dots And Spots And Knots..." Paseman, 2020.07.05.
I quickly checked and at least in the case $n=16$, the 4-cube gives more squares than this prism (which gives 28). The 4-cube has 24 square faces alone, and there are others. But in the limit, the prisms are probably better.
By the way, the $2k$ points $\pm e_i\in\Bbb R^k, i\in{1,...,k}$ (where the $e_i$ form the canonical basis) also give rise to $k\choose 2$ squares. So the pattern noted by Gerry (about the regular octahedron) extends with the regular cross-polytope.
|
2025-03-21T14:48:31.439935
| 2020-07-05T18:53:09 |
364908
|
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|
Stack Exchange
|
On integrals of products of Laguerre polynomials
Let $L_k$ be the classical Laguerre polynomial with degree $k$ defined by
$$
L_k(x)=\left(\frac {d}{dx}-1\right)^{k}\bigl\{\frac{x^{k}}{k!}\bigr\}.
$$
Let me define for $a_1, a_2$ positive, $\alpha_1, \alpha_2\in \mathbb N$,
$$
K_{\alpha_1, \alpha_2}(a_1, a_2)=\iint_{t_1\ge 0, t_2\ge 0\\\frac{t_1}{a_1}+\frac{t_2}{a_2}\ge 1}
e^{-t_1-t_2}(-1)^{\alpha_1}L_{\alpha_1}(2t_1)(-1)^{\alpha_2}L_{\alpha_2}(2t_2)dt_1 dt_2.
$$
I would like to prove that $K_{\alpha_1, \alpha_2}(a_1, a_2)\ge 0$.
I know a (not-so-simple) proof of that inequality when $\min(\alpha_1, \alpha_2)=0$, but other cases remain open. Obviously there is a $n$-dimensional question by defining $K_{\alpha_1,\dots, \alpha_n}(a_1,\dots, a_n)$ and I know a proof for $n=1$, but not for $n\ge 2$ for general $\alpha$.
|
2025-03-21T14:48:31.440018
| 2020-07-05T19:50:24 |
364909
|
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|
Stack Exchange
|
Curious inversion formula in additive combinatorics
Let $S$ be an infinite set of positive integers, and $T=S+S=\{x+y, \mbox{ with } x,y\in S\}$.We definte the following functions:
$N_S(z)$ is asymptotic continuous version of the function counting the number of elements in $S$ less or equal to $z$.
$N'_S(z)$, the derivative of $N_S(z)$, is the "probability" for $z$
(an integer) to belong to $S$
$r(z)$ is the asymptotic continuous version of the function counting
the number of solutions $x+y \leq z$ with $x,y\in S$.
$r'(z)$ is the derivative of $r(z)$.
We will work with
$$N_S(z) \sim \frac{az^b}{(\log z)^c}.$$
Here $\frac{1}{2}< b \leq 1$ and $a>0, c\geq 0$. The case $b=1, c=0$ should be excluded. This covers a vast array of sets: sums of primes, sums of super-primes etc. The following is a known result (see here):
$$r(z) \sim \frac{a^2 z^{2b}}{(\log z)^{2c}}\cdot \frac{\Gamma^2(b+1)}{\Gamma(2b+1)}$$
$$r'(z) \sim \frac{a^2 z^{2b-1}}{(\log z)^{2c}}\cdot \frac{\Gamma^2(b+1)}{\Gamma(2b)}$$
More generally (see here):
$$r(z) \sim z\int_0^{1} N_S(z(1-v))N'_S(zv) dv.$$
$$r'(z) \sim z\int_0^{1} N'_S(z(1-v))N'_S(zv) dv .$$
Since $b>\frac{1}{2}$, we have $r'(z) \rightarrow \infty$ as $z\rightarrow \infty$. This guarantees (it's a conjecture) that barring congruence restrictions, $T = S + S$ contains all the positive integers except a finite number of them. The inversion formula is as follows:
Inversion formula
$$N_T(z) = z-w(z), \mbox{ with } w(z) \sim \int_0^z \exp\Big(-\frac{1}{2} r'(u)\Big)du.$$
Since $r'$ is a function of $N'$ and thus, a function of $N$, we have a formula linking $N_T$ to $N_S$. So if you know $N_T$, by inversion (it involves solving an integral equation, though we are only interested in the asymptotic value of the solution) technically, you can retrieve $N_S$, assuming the solution is unique (chances are that the solution is far from unique.)
Note that $w(z)$ represents the number of positive integers, less or equal to $z$, that do not belong to $T=S+S$. These integers are called exceptions; $w(\infty)$ is finite and represents an estimate of the total number of exceptions. I tried to assess the validity of the inversion formula using some test sets $S$, and empirical evidence suggests that it is correct. Essentially, it is based on the following simple probabilistic argument (see proof in my answer to this post). Let $u(z)$ be the probability that $z$ (an integer) is an exception. Then, if $r'(z)\rightarrow\infty$ as $z\rightarrow\infty$ and $S$ is free of congruence restrictions and other sources of non-randomness, then
$$u(z) \sim \exp\Big(-\frac{1}{2}r'(z)\Big).$$
Testing the formula on an example
I created 100 test sets $S$, with $a=1, b=\frac{2}{3}, c=0$, as follows: an integer $k$ belongs to $S$ if and only if $U_k<N'_S(k)$, where the $U_k$'s are independent uniform deviates on $[0, 1]$. I computed various statistics, but I will mention only one here. The theoretical value for $w(\infty)$ is
$$w(\infty) \approx \int_0^\infty \exp\Big(-\frac{\lambda}{2}u^{1/3}\Big)du \approx 63.76, \mbox{ with } \lambda = \frac{\Gamma^2(\frac{5}{3})}{\Gamma(\frac{4}{3})}.$$
Note that the above integral can be computed explicitly. I then conjectured the value $w(\infty)$ for each of the 100 test sets. It ranged from $13$ to $199$, with an average value of $65.88$. Again, $w(\infty)$ is an estimate of the number of exceptions, that is, positive integers that can not be represented as $x+y$ with $x, y \in S$. So the approximate theoretical value is in agreement with the average value inferred from my experiment.
My question
Is this inversion formula well known? Can it be of any practical use? Can it be further refined, maybe generalized to sums of three sets or made more accurate with bounds on the error term?
This is not an answer to the question, but an explanation as to how I came up with the formula for $w(z)$. We assume here that $S$ is a random set. That is, let us consider $X_z$ as a Bernouilli random variable of parameter $N'_S(z)$. A positive integer $z$ belongs to $S$ if and only if $X_z = 1$. Thus $P(z\in S) = N'_S(z)$.
Now we compute $u(z)=P(z\notin S + S)$ when $z$ is an odd positive integer. We have:
$$u(z) = \Big[\prod_{k=0}^z P(k\notin S \mbox{ or } z-k \notin S)\Big]^{1/2}.$$
The exponent $\frac{1}{2}$ is because we count $k + (z-k)$ and $(z-k) + k$ as two solutions when it should appear only once in the product. The following is easy to derive from the above product:
$$u(z) = \Big[\prod_{k=0}^z \Big(1-N'_S(k)N'_S(z-k) \Big)\Big]^{1/2}.$$
Thus $$\log u(z)=\frac{1}{2}\sum_{k=0}^z \log\Big(1-N'_S(k)N'_S(z-k) \Big).$$
Note that since $b\geq \frac{1}{2}$, either $N'_S(k)$ or $N'_S(z-k)$ tends to zero in the product when $z\rightarrow\infty$, so we have the approximation
$$\log\Big(1-N'_S(k)N'_S(z-k) \Big) \approx -N'_S(k)N'_S(z-k).$$
Also,
$$\sum_{k=0}^z N'_S(k)N'_S(z-k) \sim \int_0^z N'_S(v)N'_S(z-v)dv
= z\cdot\int_0^1 N'_S(z(1-v))N'_S(zv)dv\sim r'(z).$$
It follows immediately that
$$u(z) \approx \exp\Big[-\frac{1}{2} r'(z)\Big].$$
Similar reasoning for the case "$z$ even" leads to the same result. To complete the proof, note that
$$w(z)=\sum_{k=0}^z u(k) \sim \int_0^z u(v) dv \approx \int_0^z \exp\Big[-\frac{1}{2} r'(v)\Big] dv.$$
Some open problems
Let $R$ be the set of exceptions, that is, the finite set of positive integers that can not be written as $z=x+y$ with $x,y \in S$. Let
$Y_z$ be a bernouilli random variable of parameter $u(z)$, such that $z\in R$ if and only if $Y_z=1$
$N_R(z)=Y_0 + Y_1 +\cdots + Y_z$ be the number of exceptions less or
equal to $z$
$N_R(\infty)<\infty$ be the total number of exceptions
$M_R$ be the largest exception ($M_R=k$ if and only if $Y_k=1$ and $Y_{k+n}=0$ for $n>0$).
All these variables are random variables. We have established (see above) that $$w(z) = E[N_R(z)] \approx \int_0^z \exp(-r'(v)/2) dv.$$
The big challenge here to make further progress is that the $Y_z$'s are not independent. Problems to investigate are
What are the variances of $N_R(z), N_R(\infty)$ and $M_R$? (they are
finite)
What is the expectation of $M_R$? (it is finite)
Are $N_R(\infty)$ and $M_R$ bounded random variables? If yes, what are the upper bounds?
The fact that $E[N_R(\infty)]$ is bounded does not imply that $N_R(\infty)$ is bounded. It may or may not be the case depending on $a, b, c$. If we had a positive answer to that question, we could make a tiny bit of progress towards proving Golbach's conjecture ($a=1,b=0,c=1$), though there are other big hurdles to overcome (the fact that primes are not random enough, for instance the sum of two odd primes is never an odd number).
Note that the Central Limit Theorem does not apply to $N_R(z)=Y_0+\cdots +Y_z$ because of auto-correlations in the $Y_k$'s, and especially the fact that $\mbox{Var}[N_R(z)]$ is bounded no matter how large $z$ is. It is very obvious based on empirical evidence that $(N_R(z)-E[N_R(z)])/\sqrt{\mbox{Var}(N_R(z))}$ does not tend to a Gaussian variable as $z\rightarrow\infty$, even if $S$ is a very thin set (e.g. $a=\frac{1}{100}, b=0.50001, c=200$) resulting in a very, very, very large yet finite $E[N_R(\infty)]$. Similar examples of lack of convergence to a Gaussian distribution are illustrated in one of my articles, see here.
A possible approach is to use the Borel-Cantelli lemma, or a sharper version of it to the set $R$, see here. Then, since $\sum_{z=1}^\infty u(z) < \infty$, it follows that with probability 1 (that is, almost surely), $Y_z=1$ only for finitely many integer $z$'s, and thus both $N_R(\infty)$ and $M_R$ are almost surely finite.
|
2025-03-21T14:48:31.440632
| 2020-07-05T20:57:58 |
364912
|
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|
Stack Exchange
|
Could prime factorization of n!+1 using the general number field sieve be said to take subfactorial time?
I am interested in the prime factorization using the general number field sieve. This method is said to take subexponential time relative to the number of bits in a number. (Other algorithms are exponential.) I have two questions about this:
Could this operation be said to take subfactorial time?
Does this operation on average take greater than exponential time? If so, then it is between exponential and factorial.
For a number $n$ that has $b=1+\log_2 n=O(\ln n)$ bits
NFS has complexity of the order
$$
\exp\left\{(c+o(1)) (\ln n)^{1/3} (\ln \ln n)^{2/3}\right\}
$$
which is subexponential when compared to the input size in bits. An exponential algorithm would have
complexity of the order
$$
\exp(c b)=\exp(c \ln n).
$$
If the unusual term factorial time means an algorithm takes time asymptotic to $\ln b!$ for the input size $b$ the answers to your questions are both yes.
|
2025-03-21T14:48:31.440725
| 2020-07-05T21:18:14 |
364913
|
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"Francesco Polizzi",
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|
Stack Exchange
|
density of singular K3 surfaces
By singular K3 I mean a smooth complex K3 with Néron-Severi rank equal to 20.
Are singular K3 surfaces dense in the moduli space of polarized K3 surfaces?
In this paper by Hulek and Kloosterman (see page 2) it is said that the answer is positive: https://arxiv.org/pdf/math/0601207.pdf
However, no reference is provided.
This is a standard argument and there probably exists a reference but it's not hard once you rephrase it in terms of the period domain.
The moduli space of K3 surfaces is locally isomorphic to its period domain.
The period domain is an open subset of the vanishing locus of a quadratic polynomial in $\mathbb P^{21}(\mathbb C)$. See, for instance, Huybrechts' lectures on K3 surfaces.
So it suffices to show the singular points are dense in the period domain.
The singular points in the period domain correspond to points whose coordinates generate a lattice of rank two in the complex numbers. Examples include points defined over an imaginary quadratic field. (One can check that these are the only examples but that is not necessary for this argument.)
So it suffices to check that any tuple of complex numbers satisfying this quadratic polynomial can be approximated by a tuple of numbers in $\mathbb Q(i)$ satisfying this quadratic polynomial. To do this, express the polynomial as $xy+$ a bunch of terms not involving $x$ and $y$, approximate each complex number by an element of $\mathbb Q(i)$, then adjust $x$ or $y$ by a little bit to make the quadratic polynomial vanish again.
A reference for the density of singular K3 surfaces in the period domain (see Will Sawin's answer) is
Piatetski-Shapiro, Ilya I.; Shafarevich, I. R., Arithmetic of K3 surfaces, Trudy Mezhdunarod. Konf. Teor. Chisel, Moskva 1971, Tr. Mat. Inst. Steklova 132, 44-54 (1973) ZBL0293.14010.
In particular, this is a screenshot of page 54:
|
2025-03-21T14:48:31.441116
| 2020-07-05T21:22:48 |
364914
|
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|
Stack Exchange
|
Let $G'\triangleleft G<\operatorname{Iso}(M)$ be a normal subgroup. A $G'$-stratum is the union of $G$-strata of lesser dimension
Let $G$ be a group of isometries acting effectively by isometries on a connected Riemannian manifold. And let $G'\triangleleft G$ be a normal subgroup. I am trying to prove that $\dim \operatorname{St}_G(p)\leq \dim \operatorname{St}_{G'}(p)$ for every $p\in M$; where $\operatorname{St}_X(p)$ stands for the stratum of the action of $X$ on $M$ through $p$.
We have the formula
$$ \dim\operatorname{St}_G(p)=\dim G+\dim M_0^{G_p}-\dim N_G(G_p); $$
where $M_0^{G_p}$ is the connected component of the fixed-point set $M^{G_p}$ through $p$.
It is obvious that $\dim M_0^{G_p}\geq\dim M_0^{G'_p}$, since $G'_p\triangleleft G_p$. Thus, the result is true if we show that $\dim N_G(G_p)\geq \dim N_G'(G'_p)$, which I'm not sure if is true. If it is not true, is there any other way to prove my claim?
I'm pretty sure $\dim M_0^{G_p} > \dim M_0^{G'_p}$ should not be strict inequality.
We have to assume that the index $[G':G]$ is finite. In this case:
Let $G'$ be a normal subgroup of $G$ such that the quotient $\Gamma=G/G'$ is finite and acts by isometries in $X'=M/G'$, and $X=X'/\Gamma$. Thus, $(G')^0=G^0$ and, therefore the orbits $G'(p)$ and $G(p)$ have same connected components through $p$. This way, $\nu_pG(p)=\nu_pG'(p)$. Also, as $G_p^0=(G_p')^0$, the orbits associated to their respective slice representations, i.e., their infinitesimal actions on the normal space to the tangent space to their respective orbits, have same dimension. And $(\nu_pG(p))^{G_p}\subset (\nu_pG(p))^{G'_p}$. As the orbits of their associated slice representations have same dimension, the cohomogeneity of the action $(G'_p,(\nu_pG(p))^\dagger)$ can be no greater then the cohomogeneity then the action of $(G_p,(\nu_pG(p))^\dagger)$. Thus, $\dim\operatorname{St}_X(\pi(p))\leq \dim\operatorname{St}_{X'}(\pi'(p))$.
Here, $(\nu_pG(p)^\dagger)$ is the orthogonal complement of $(\nu_pG(p))^{G_p}$ in $\nu_pG(p)$.
|
2025-03-21T14:48:31.441280
| 2020-07-05T23:58:45 |
364919
|
{
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"Gerhard Paseman",
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"Robert Israel",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364919"
}
|
Stack Exchange
|
How smooth can this be?
If $a$ is an even integer then how smooth can $a^2-1$ be?
Approximately how many integers in $a\in[0,t]$ are there such that $a^2-1$ is $k$-smooth?
The numbers of $prime(i)$-smooth $4n^2-1$'s for $n \le 10^6$ start
$0, 1, 3, 6, 9, 15, 22, 35, 56, 76$ (i.e. there is one $3$-smooth, namely $n=1$, there are three $5$-smooth (including $n=2$ and $13$), etc). This sequence does not seem to be in OEIS.
Since $a^2-1$ is $k$-smooth if and only if $a-1,a+1$ are $k$-smooth, the latter question is almost answered by Stormer's theorem https://en.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem , which gives a bound exponential in $\pi(k)$. Presumably the proof of that theorem adapts to numbers separated by 2 rather than 1.
There is a PDF of Klazar from 2010 on this topic, which covers the case of smooth numbers differing by 2. It should come up in a search on Stormers Theorem. Gerhard "Is Typographically Challenged At Present" Paseman, 2020.07.05.
Referring theorem $1$ in http://kam.mff.cuni.cz/~klazar/stormer.pdf?
$26^2-1$ is $5$-smooth, $244^2-1$ is $7$-smooth, $1574^2-1$ is $13$-smooth.
|
2025-03-21T14:48:31.441401
| 2020-07-06T01:32:08 |
364924
|
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"Bjørn Kjos-Hanssen",
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"喻 良"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364924"
}
|
Stack Exchange
|
Finding 1-generic paths through a tree $T \subseteq 2^{<\omega}$
Consider Cantor space $2^\omega$ with the standard topology generated by open sets $[\sigma] = \{ \sigma^\frown x: x \in 2^\omega \}$. If $A \subseteq 2^{<\omega}$ and $x \in 2^\omega$, we say $A$ is dense along $x$ if for every prefix $\sigma \prec x$, there is $\tau \succ \sigma$ such that all finite extensions of $\tau$ are in $A$.
An element $x \in 2^\omega$ is 1-generic if, for every $\Sigma^0_1$ (computably enumerable) set $A \subseteq 2^{<\omega}$ which is dense along $x$, we have $x \in [A]$ ($x$ is a path through $A$). I think this is the standard definition (from here).
Now, suppose $T \subseteq 2^{<\omega}$ is a tree. What conditions can we impose on $T$ that guarantee $[T]$ contains a 1-generic member? Effectively, I'm looking for some type of "generic basis theorem". In particular, if $T$ is infinite and $\Sigma^0_2$, can we guarantee it contains a 1-generic path?
What conditions can we impose on $T$ that guarantee $[T]$ contains a 1-generic member?
An element that is 1-generic relative to $T$ will not be on $[T]$ unless $[T]$ contains a whole clopen cone $[\sigma]$.
Since "most" 1-generics are 1-generic relative to $T$, I suppose this means the condition to impose is basically that $[\sigma]\subseteq [T]$ for some $\sigma\in 2^{<\omega}$.
Effectively, I'm looking for some type of "generic basis theorem". In particular, if $T$ is infinite and $\Sigma^0_2$, can we guarantee it contains a 1-generic path?
No, if we let $T$ consist of all diagonally non-recursive $\{0,1\}$-valued functions then $T$ contains no 1-generic path. This is because one can show that no 1-generic computes a DNR function.
What do you mean by ' "most" 1-generics are 1-generic relative to $T$'?
The set of 1-generics relative to $T$ is comeager in the sense of Baire category.
Bjorn's answer is true even for weak-1-genericicty. More than that, if a recursive tree contains a non-recursive real which is recursive in a sufficiently generic real, then it has a recursive perfect subtree.
I don't think you were asking for 1-genericity relative to $T$ but just plain old normal 1-genericity. I’m going to assume $T$ has no terminal nodes since if it doesn't things get more messy (though I did deal with that way back in my thesis).
The difficulty with any useful basis result here is that you lose if $T$ is too definable. Obviously if T contains a full cone $[\sigma]$ it contains a
generic so let's suppose that $\sim T$ is dense (every string can be extended to meet it). But now if $T$ is $\Pi^0_1$ (and hence also if it is computable) it fails to have any generic paths since T complement itself is the witnessing $\Sigma^0_1$ set. But a really complex T need nor help either.
The best I think you can do for a general answer is the obvious thing you would start with: if $\sigma \in T$ and W is a $\Sigma^0_1$ set then you need an extension of $\sigma$ in $T$ that either meets With or strongly avoids that extension. But that's just another way of stating the genericity requirement. You can probably hide that a bit better but I don't think there are any useful basis type results here.
|
2025-03-21T14:48:31.441632
| 2020-07-06T01:44:36 |
364926
|
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"Milo Moses",
"Sylvain JULIEN",
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"url": "https://mathoverflow.net/questions/364926"
}
|
Stack Exchange
|
Asymptotic for the probability that a number has $k$ prime factors less than $Q$
If we let $\omega_Q(n)$ denote the number of distinct prime factors of $n$ less than a bound $Q$, then what asymptotic formulas exist for $\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]$ as $Q\to\infty$ if $k$ remains fixed (or perhaps very small with respect to n)?
I am asking this question since my study led me to want to bound the quantity
$$\mathbf{E}_{n\in\mathbb{N}}\left[\frac{2^{\omega_Q(n)}}{\sqrt{\omega_Q(n)}}\right]$$
as $Q\to\infty$. Since
$$\mathbf{E}_{n\in\mathbb{N}}\left[\frac{2^{\omega_Q(n)}}{\sqrt{\omega_Q(n)}}\right]=\sum_{n=1}^{\pi(Q)}\left(\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]\right)\left(\frac{2^{\omega_Q(n)}}{\sqrt{\omega_Q(n)}}\right)$$
and
$$\sum_{n=1}^{\pi(Q)}\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]2^{\omega_Q(n)}\sim_{Q\to\infty} c\log(Q)$$
is well understood, good (upper) bounds on $\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]$ could help me in my effort.
For small values of $k$ computations can be done directly, like
$$\Pr_{n\in\mathbb{N}}[\omega_Q(n)=0]\sim\frac{c}{\log(Q)}$$
and
$$\Pr_{n\in\mathbb{N}}[\omega_Q(n)=1]\sim c\frac{\log(\log(Q))}{\log(Q)}$$
The main approach I have been using is noting that $\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]$ is exactly the coefficient of $x^k$ in the polynomial
$$\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)$$
Asymptotics of this full polynomial are easy to come by, for instance as $Q\to\infty$ we have that
$$\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\sim c \log^{x-1}(Q)$$
Heuristically this would suggest that
\begin{align*}
\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]&=\frac{1}{k!}\left.\frac{d^k}{dx^k}\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|_{x=0}\\
&\sim \frac{c}{k!}\left.\frac{d^k}{dx^k}\log^{x-1}(Q)\right|_{x=0}\\
&=\frac{c}{k!}\frac{\log^k(\log(Q))}{\log(Q)}
\end{align*}
This argument is however by no means rigorous so I would appreciate true asymptotics.
If I'm not mistaken, one can deduce from Mertens' theorem that $c=e^{-\gamma}$, right?
@SylvainJULIEN Yes, the way I derived the asymptotics was with Merten's theorems. I decided to use $c$ instead of $e^{-\gamma}$ since the exact value of the constant doesn't really matter to me.
As pointed out in the question, we have that
$$\prod_{p<Q}\left(\frac{x-1}{p}+1\right)=\sum_{k=0}^{\pi(Q)}\Pr_{n\in\mathbb{N}}[\omega_Q(n)=k]x^k$$
which can be derived by showing that on both the RHS and the LHS the coefficient of $x^k$ is equal to
$$\sum_{\substack{S\subseteq \{p<Q\} \\ |S|=k}} \left(\prod_{p\in S}\frac{1}{p}\right)\left(\prod_{p\not\in S}\left(1-\frac{1}{p}\right) \right)$$
Treating the LHS with simple manipulation we get that
\begin{align*}
\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)&=\exp\left(\log\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right)\\
&=\exp\left(\sum_{p<Q}\log\left(\frac{x-1}{p}+1\right)\right)\\
&=\exp\left(\sum_{p<Q}\frac{x-1}{p}+\sum_{p<Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)\right)\tag{1}
\end{align*}
We now note that
$$\sum_{p<Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)=\sum_{p}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)-\sum_{p\geq Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)$$
and thus we can set
$$f_1(x)=\sum_{p}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)$$
and
$$g(x)=\sum_{p\geq Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)$$
Morally, we can think of $g(x)$ as the "error" as $Q\to\infty$ which we must show is inconsequential. We thus get that
\begin{equation}
\sum_{p<Q}\left(\log\left(\frac{x-1}{p}+1\right)-\frac{x-1}{p}\right)=f_1(x)-g(x)\tag{2}
\end{equation}
By Merten's theorem, we have that
\begin{align*}
\sum_{p<Q}\frac{x-1}{p}&=(x-1)\left(\log(\log(Q))+M+\epsilon_Q\right)\\
&=(x-1)\log(\log(Q))+Mx+\epsilon_Q x-M-\epsilon_Q\tag{3}
\end{align*}
where $\epsilon_Q\to 0$ and $M$ is the Meissel-Mertens constant. Substituting (2) and (3) into (1) we get that
\begin{align*}
\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)&=\exp\left((x-1)\log(\log(Q))+Mx+\epsilon_Q x-M-\epsilon_Q+f_1(x)-g(x)\right)\\
&=e^{-M-\epsilon_Q}\log^{x-1}(Q)e^{Mx}e^{f_1(x)}e^{\epsilon_Qx}e^{-g(x)}
\end{align*}
For simplicity's sake, we now define
$$f_2(x)=e^{Mx}e^{f_1(x)}$$
and thus
\begin{equation}
\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)=e^{-M-\epsilon_Q}\log^{x-1}(Q)f_2(x)e^{\epsilon_Qx}e^{-g(x)}\tag{4}
\end{equation}
Taking the derivative $k$ times yields
$$\frac{d^k}{dx^k}e^{-M-\epsilon_Q}\log^{x-1}(Q)f_2(x)e^{\epsilon_Qx}e^{-g(x)}$$
As $Q\to\infty$, the only term that will matter in a product rule decomposition of this equation is the one that grows the fastest. It is easy to show that
$$g^{(n)}(x)=O\left(\frac{1}{x}\right)$$
for any order derivative $(n)$, and so the fastest growing term is the one where $\log^{(x-1)}(Q)$ is differentiated the full $k$ times. Since there are finitely many terms the others are inconsequential in terms of growth and so
\begin{align*}
\frac{d^k}{dx^k}e^{-M-\epsilon_Q}\log^{x-1}(Q)f_2(x)e^{\epsilon_Qx}e^{-g(x)}&\sim_{Q\to\infty}e^{-M-\epsilon_Q}f_2(x)e^{\epsilon_Qx}e^{-g(x)}\frac{d^k}{dx^k}\log^{x-1}(Q)\\
&=e^{-M-\epsilon_Q}f_2(x)e^{\epsilon_Qx}e^{-g(x)}\log^k(\log(Q))\log^{x-1}(Q)
\end{align*}
evaluating at $x=0$ and substituting into (4) yields that
\begin{equation}
\left.\frac{d^k}{dx^k}\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|_{x=0}\sim e^{-M-\epsilon_Q}f_2(0)e^{-g(0)}\frac{\log^k(\log(Q))}{\log(Q)}
\end{equation}
As $Q\to\infty$ we have that $g(0)\to0$ and $f_2(0)=e^{M-\gamma}$ and so
\begin{equation}
\left.\frac{d^k}{dx^k}\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|_{x=0}\sim e^{-\gamma}\frac{\log^k(\log(Q))}{\log(Q)}\tag{5}
\end{equation}
We also see that
\begin{align*}
\left.\frac{d^k}{dx^k}\prod_{p<Q}\left(\frac{x}{p}+1-\frac{1}{p}\right)\right|_{x=0}&=\left.\frac{d^k}{dx^k}\sum_{j=0}^{\pi(Q)} \mathrm{Pr}_{n\in\mathbb{N}}[\omega_Q(n)=j]x^j\right|_{x=0}\\
&=k!\mathrm{Pr}_{n\in\mathbb{N}}[\omega_Q(n)=k]
\end{align*}
and thus we conclude from (5) that
$$\mathrm{Pr}_{n\in\mathbb{N}}[\omega_Q(n)=k]\sim e^{-\gamma}\frac{\log^k(\log(Q))}{\log(Q) k!}$$
which is the desired result
|
2025-03-21T14:48:31.441939
| 2020-07-06T02:49:15 |
364927
|
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"Alex Wertheim",
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|
Stack Exchange
|
Characterization of algebraic integers providing a prime ideal
Let $\alpha$ be an algebraic integer and let $\mathcal{O}_{\mathbb{Q}(\alpha)}$ be the ring of integers of $\mathbb{Q}(\alpha)$.
Question: How to characterize an algebraic integer $\alpha$ such that $\alpha \mathcal{O}_{\mathbb{Q}(\alpha)}$ is a prime ideal of $\mathcal{O}_{\mathbb{Q}(\alpha)}$?
Example: if $\alpha \in \mathbb{Z}$, then $\mathcal{O}_{\mathbb{Q}(\alpha)} = \mathbb{Z}$, so that $\alpha \mathcal{O}_{\mathbb{Q}(\alpha)}$ is a prime ideal iff $\alpha$ is a prime number (up to a sign).
If I'm not mistaken, a necessary condition is that the norm of $\alpha$ must be a prime power, as if $\alpha\mathcal{O}{\mathbb{Q}(\alpha)}$ is a prime ideal, then $\mathcal{O}{\mathbb{Q}(\alpha)}/\alpha\mathcal{O}{\mathbb{Q}(\alpha)}$ is a finite field with cardinality $|N{\mathbb{Q}(\alpha)/\mathbb{Q}}(\alpha)|$. This condition is clearly not sufficient, as taking a prime power in $\mathbb{Z}$ shows.
@AlexWertheim: If I am not mistaken, $α$ is a unit in $\mathcal{O}_{\mathbb{Q}(\alpha)}$ when (by definition) its norm is one.
Trivially, $\alpha=0$ qualifies. As remarked, for $\alpha\ne0$ we must have $|N_{K/\mathbb{Q}}(\alpha)|=p^m$ with $p$ prime and $1\leq m\leq n=[K:\mathbb{Q}]$, where $K=\mathbb{Q}(\alpha)$. If $m=1$ then $\alpha\mathcal{O}{K}$ is a prime ideal. If $m=n$, the ideal is prime iff $p$ is inert in $K$; in that case, $\alpha$ simply equals $p\beta$ for some unit $\beta\in\mathcal{O}{K}$. Don't know whether much can be said in case $1<m<n$.
If $m=n-1$, for $\alpha\mathcal{O}{K}$ to be prime one would have $p\in\alpha\mathcal{O}{K}$, hence $\beta=\alpha/p$ is integral (which can be verified easily enough) of norm $p$, hence $\beta\mathcal{O}{K}$ is prime, and $p\mathcal{O}{K}$ $=$ $\alpha\mathcal{O}{K}\cdot\beta\mathcal{O}{K}$, So $p$ would need to split as the product of a degree $n-1$ and a degree $1$ prime in $K$. But this is not a criterion that can be readily checked from the minimum polynomial of $\alpha$ alone - even if it factorizes as $x^{n-1}(x-a)$ in $\mathbb{F}_{p}[x]$.
$\beta=p/\alpha$, of course.
If $x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_0$ is the minimum polynomial of $\alpha$ over $\mathbb{Q}$, with $|a_{0}|=p^{m}$, where $m\le n$, for $\beta=p/\alpha$ to be an integer, a condition necessary for $\alpha\mathcal{O}{K}$ to be prime, one needs $a{i}\equiv0$ mod $p^{m-i}$ for $0<i<m$.
@MatthévanderLee: you should replace all your comments by one answer (that you could improve at your convenience).
|
2025-03-21T14:48:31.442117
| 2020-07-06T07:08:55 |
364931
|
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"Ben McKay",
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|
Stack Exchange
|
Equivalence generated by Jacobian minors
Let $f,g:\mathbb{R}^m \to \mathbb{R}^n$ be two smooth functions and let $k$ be a strictly positive integer. Write $f \sim_k g$ if at each point in the domain, the determinants of all $k \times k$ minors of the Jacobians of $f$ and $g$ coincide. This is clearly an equivalence relation; and more generally, one could let $f$ and $g$ be smooth maps of smooth manifolds $X \to Y$, and ask that for each $k$-form $\omega$ on $Y$ the pullbacks $f^*\omega$ and $g^*\omega$ agree as $k$-forms on $X$.
Have such equivalences been studied and given an alternate characterization? Here's the sort of thing I'm looking for: if $k=1$ and we are mapping between Euclidean spaces, $f$ and $g$ must lie in the same orbit of the translation group acting on the codomain. Perhaps there is a similar Lie-theoretic reformulation of $\sim_k$ for maps $X \to Y$ even when $k > 1$?
If $k$-forms have the same pullbacks, then we can make $k$-forms with very small support, take a limit, and find that the maps $f$ and $g$ agree.
First, let me point out that the OP's suggested generalization to arbitrary target manifolds $Y$ of asking that $f^*\omega = g^*\omega$ for all $k$-forms on $Y$, is not equivalent to the question about equality of $k$-by-$k$ minors.
To see this, consider the simplest case, $X = Y=\mathbb{R}^1$ and $k=1$. The general $1$-form is of the form $\omega = h(x)\,\mathrm{d}x$ (where $x:\mathbb{R}\to\mathbb{R}$ is the identity function), and $f^*\omega = g^*\omega$ for all $1$-forms $\omega$ is equivalent to
$$
h\bigl(f(x)\bigr)f'(x) = h\bigl(g(x)\bigr)g'(x)
$$
for all functions $h:\mathbb{R}\to\mathbb{R}$, which clearly implies that $f(x) = g(x)$ for all $x\in\mathbb{R}$. Meanwhile, $f^*\mathrm{d}x = g^*\mathrm{d}x$ only implies $f'(x) = g'(x)$, which is the same as requiring that all the $1$-by-$1$ Jacobian minors of $f$ and $g$ are equal, which does imply that they differ by an additive constant.
Instead, I think that the natural generalization is that one endows the $n$-manifold $Y$ with a coframing, i.e., a basis $\omega = (\omega^1,\ldots,\omega^n)$ of the $1$-forms on $Y$ (which, of course, requires that $Y$ be parallelizable, in fact, $\omega:TY\to\mathbb{R}^n$ defines a linear isomorphism $\omega_y:T_yY\to\mathbb{R}^n$ for each $y\in Y$) and then require that
$$
f^*(\omega^{i_1}\wedge\cdots\wedge\omega^{i_k})
= g^*(\omega^{i_1}\wedge\cdots\wedge\omega^{i_k})
$$
for all $i_1<i_2<\cdots <i_k$.
Even this will generally force some 'finite' equations on $f$ and $g$ if $\omega$ is chosen generally (and the dimension of $X$ is greater than $k$). For a multi-index $I = (i_1,\ldots,i_p)$ with $1\le i_1<i_2<\cdots<i_p\le n$, write $|I|=p$ and $\omega^I$ for $\omega^{i_1}\wedge\cdots\wedge\omega^{i_p}$. Then there will be functions $h^I_J$ on $Y$ such that
$$
\mathrm{d}\omega^I = \sum_{|J|=|I|{+}1} h^I_J\,\omega^J.
$$
If the functions $h^I_J$ are not constant, and $f$ and $g$ both have differential rank at least $k{+}1$, then applying the exterior derivative to the equations $f^*\omega^I = g^*\omega^I$ for $|I|=k$ will generally force some relations on the functions $f^*h^I_J$ and $g^*h^I_J$. One can avoid this 'problem' by assuming that the functions $h^I_J$ be constants. For example, when one takes the standard coordinate coframing on $Y = \mathbb{R}^n$, one has $h^I_J=0$. More generally, if $Y$ is a Lie group and the $\omega^i$ are a basis for the left-invariant forms on $Y$, then the $h^I_J$ are constants. In this latter case, when $X$ is connected, one will have $f^*\omega^i=g^*\omega^i$ for all $i$ if and only if $g = \lambda_y\circ f$ where $\lambda_y:Y\to Y$ is left multiplication by $y\in Y$ (regarded as a Lie group). So I think that this is the natural generalization of the OP's case of $\mathbb{R}^n$.
Second, let me point out that, if the differential ranks of $f$ and $g$ are both less than $k$ at every point, then, of course, $f^*\omega = g^*\omega=0$ for all all $k$-forms $\omega$ on $Y$, so there is no further condition implied than the rank condition. Thus, to get an interesting problem, one must assume that the differential ranks of $f$ and $g$ are at least $k$ in order to get an interesting theory.
Once one assumes this, there are some reasonable things to say. For example, if one assumes that the differential ranks of $f$ and $g$ are both at least $k$, then the condition $f^*\omega^I = g^*\omega^I$ for all $|I|=k$ implies that $\mathrm{ker}(f'(x)) = \mathrm{ker}(g'(x))\subset T_xX$ and $\omega_x\bigl(f'(x)(T_x)\bigr) = \omega_x\bigl(g'(x)(T_x)\bigr) \subset\mathbb{R}^n$ for all $x\in X$. Moreover, if one sets $K_x = \mathrm{ker}(f'(x)) = \mathrm{ker}(g'(x))\subset T_xX$ and $Q_x = \omega_x\bigl(f'(x)(T_x)\bigr) = \omega_x\bigl(g'(x)(T_x)\bigr) \subset\mathbb{R}^n$, then the induced isomorphisms
$$
[f'(x)],[g'(x)]:T_xX/K_x\to Q_x
$$
satisfy, for all $x\in X$,
$$
E_k\bigl([f'(x)]\bigr) = E_k\bigl([g'(x)]\bigr):E_k(T_xX/K_x)\to E_k(Q_x),
$$
where $E_k$ is the '$k$th-exterior power' functor on the category of vector spaces and linear maps.
When the differential ranks of $f$ and $g$ are equal to $k$, this is not a very strong condition, so $f$ and $g$ need not be closely related. For example when $m=n=k$, and $X = Y = \mathbb{R}^k$, then the OP's condition on $f$ and $g$ reduces to the assumption that the $f$- and $g$- pullbacks of the standard volume form on $Y$ are equal, and, of course, there are many such pairs $f$ and $g$ besides the translations.
When the differential ranks of $f$ and $g$ are both greater than $k$, though, this is a much stronger condition. In fact, one finds that, when $k$ is odd, this is equivalent to $[f'(x)] = [g'(x)]$ while, when $k$ is even, this is equivalent to $[f'(x)] = \pm[g'(x)]$. Once one is in this situation, at least when one assumes that the differential ranks of $f$ and $g$ are constant, the Cartan equivalence method can be applied. For example, one has the following result:
Proposition: Suppose that $f,g:\mathbb{R}^m\to\mathbb{R}^n$ are smooth maps of constant differential rank greater than $k\ge1$ and suppose that all their corresponding $k$-by-$k$ Jacobian minors are equal. If $k$ is odd, then $g = c + f$ where $c\in\mathbb{R}^n$ is a constant. If $k$ is even, then $g = c \pm f $ where $c\in\mathbb{R}^n$ is a constant.
Every time I read this answer, I learn something more. Thank you for taking the time to set everything down so carefully, and moreover, for answering the question that I should have asked!
|
2025-03-21T14:48:31.442501
| 2020-07-06T07:09:44 |
364932
|
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|
Stack Exchange
|
Online study groups for individual mathematical texts
This has been one of my earlier academic dreams. I have pitched the idea to Prof Ravi Vakil, among others. Recently, because of participation in the mathematical competition hosted by Alibaba, I have decided to revive my decade-old enthusiasm. The basic idea is as follows. Many mathematical texts are difficult to penetrate on one's own, be it due to typos, vagueness, or prerequisite assumptions. Thus it is helpful to study with friends or mentors. People outside of the academic circle, or even graduate students, may find it difficult to meet people with similar interests, let alone reading the same chapter of the same graduate text in mathematics (GTM) book.
Given the success of numerous online Q & A fora, including this one, and the ridiculous level of specialization, it seems plausible that with the right timing, we can now dedicate one group/site to each textbook. Even though it is often advised that one should not read a technical treatise sequentially, there is something satisfying about being able to finish a single math text from cover to cover, that a cursory scan or occasional reference lookup does not conjure. To be honest I have not managed to finish a single book this way during grad school. Note taking on a single text is also a great mnemonic tool in the long term, as I am sure many people still practice.
So my question is simple, are there existing platforms that offer such a service already? I am aware of something tangentially related, such as Fermat's Library. But I don't see a way to expand it into a full-blown book-based service. My recent encounter with Slack suggests that the latter may be a more suitable style than stack exchange, since it encourages multi-threaded conversations around a single topic. The level of specialization there is naturally more fine-grained than MO, down to a specific piece of software for instance. Not all books need to get the same amount of attention, but just having a few books being annotated and Q&A'ed frequently can really facilitate self-paced learning.
Any thoughts or suggestions? Would this be helpful to many students/researchers in math?
Somewhat related post on [math.se]: How to organize math study groups online for long-distance collaboration? As a side note, I'll mention that there were several attempts to use Stack Exchange chatroom as a way to organize a few users of Math.SE together to created a study group or a reading group. (In one case I remember, reddit was also used for communication - in addition to chat.)
And maybe you remember that there is a failed proposal for a Stack Exchange site which is not exactly what you want, but at least related. It was discussed on MathOverflow Meta: area 51 proposal - Math review. And here is one more discussion on meta: Wouldn't it be nice to have subforums on MathOverflow dedicated to learning certain books/papers/etc.?. In any case, the conclusion from those meta discussion seems to be that Stack Exchange is probably not the right place for such project.
These are great references that I wasn't able to locate myself. I may explore Reddit more next.
I used the MO chatroom once and quickly lost focus; people in math/physics do not have similar level of urgency as software engineers:D Something like Slack would be much more useful. Unfortunately they mostly cater towards engineers.
Some of the study groups in Stack Exchange chat that I was able to find are mentioned in this chat conversation. (Although I am not really sure whether some useful insight can be gained from looking through them.)
This question was asked a long time ago but remains on the unanswered queue, so I will try to answer.
The OP correctly identifies that the biggest obstacle to a community working through a book together is timing. Historically, the best way to generate such a community and keep it running on the same timetable is to organize a seminar. Nowadays that can also be an electronic seminar, and there are plenty of examples of successful electronic seminars:
https://s.wayne.edu/echt/echt-reading-seminars/echt-kan-seminar-fall-2023/
https://www.math.uwo.ca/faculty/kapulkin/seminars/hottest.html
https://researchseminars.org/seminar/eCHT
The Kan seminar is famous: homotopy theorists in Boston would work through the classic results the field builds on, each person reading in depth, and one person (rotating) presenting the results of the week and answering questions on those results from the group. I've also organized a number of seminars and reading groups at my university where a bunch of professors got together to work through a book and internalize its content.
Slack is a relatively new innovation, and is not really necessary for this functionality. The point of slack is to get someone's immediate attention. Many academics prefer not to allow their attention to be grabbed so easily. I had a co-author once who communicated with me via slack and I feel somewhat traumatized by that (I have a physiological reaction to the Slack three-tap "knock brush" sound, as it meant I had to drop everything to work on something urgent in the moment whenever I heard the sound). Much better than Slack, for academics, is a Trello board. With Trello, whenever you have something you don't understand, you can put a post-it note about that up on the board. And others can answer it in their own time, or can prepare an answer to give you at the seminar meeting. In theory a MathOverflow-esque site could be used for such a Q & A interface, or MO chat, but the comments make it clear that in practice this hasn't worked very well. It's important to keep the size of the group manageable and make sure everyone is invested.
|
2025-03-21T14:48:31.443007
| 2020-07-06T07:34:53 |
364934
|
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|
Stack Exchange
|
Ultralimit of metric spaces vs. inductive limits of underlying topological spaces
Let $\{(X_n,d_n)\}_{n =1}^{\infty}$ be a sequence of bounded metric spaces such that:
$X_n \subseteq X_{n+1}$ is a metric subspace of $X_n$.
Let $\omega$ denote a non-principal ultrafilter (i.e.: complements of finite subsets of $\mathbb{N}$ are always in $\omega$) and let $X_{\omega}$ denote the corresponding ultralimit metric space.
Let $F:\mathit{Met} \rightarrow \mathit{Top}$ be the forgetful functor. Then the inclusions of $F(X_n)$ into $F(X_{n+1})$ are continuous and thus $(F(X_n),(\mathbb{N},\leq))$ form an inductive system; hence, we can take the inductive limit in $\mathit{Top}$. Denote this inductive limit by $\varinjlim F(X_n)$.
How are $X_{\omega}$ and $\varinjlim F(X_n)$ related? I expect that $\varinjlim F(X_n)$ can be identified with a dense subset of $X_{\omega}$ but maybe I'm mistaken...
You get an isometric embedding. Its image is not always dense. For instance if each $X_n$ is a set of $n$ points pairwise at distance one: both are discrete spaces, so the embedding is closed, and not an equality since the union is countable while the ultralimit isn't.
@YCor your comments is a complete answer --- why not to make an anwer from it (so the question will not show up in unanswered).
There is no need to use $F$ --- you can define direct limit in the category of metric spaces.
|
2025-03-21T14:48:31.443122
| 2020-07-06T10:21:44 |
364938
|
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|
Stack Exchange
|
Probability of factor of particular size
What is the probability that an integer $a$ picked uniformly in $[t/2,t]$ has a factor in $[t^{\alpha},2t^{\alpha}]$ where $\alpha\in(0,1)$? I am interested when $\alpha=1/3$ but if there is a general result it would be useful. Is there a good asymptotic in terms of $t$ and $\alpha$ and is it $1/\mathrm{polylog}(t^\alpha)$?
|
2025-03-21T14:48:31.443180
| 2020-07-06T10:24:03 |
364939
|
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"Badam Baplan",
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|
Stack Exchange
|
Is $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_p$ coherent?
The question is as in the title:
Is the ring $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_p = \mathbb{Z}_p \otimes_{\mathbb{Z}_{(p)}} \mathbb{Z}_p$ coherent?
As shown in the related question, the ring $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_p$ is coherent, and so one cannot reduce to this case. (On the other hand $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_p$ is not Noetherian, and so neither is $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_p$). The flatness lemma used in the answer to the rational case does also not appear to be applicable in this case.
I am tempted to consider a more general question: Let $R$ be a DVR, and let $S$ be a flat $R$-algebra whose special and generic fibers are both coherent rings. Is $S$ a coherent ring?
Never mind, there are counterexamples to my more general question! More precisely, taking $R=\mathbf{R}[x]_{(x)}$ and $S=$germs of $C^\infty$ functions at $0 \in \mathbf{R}$ gives a counterexample. In this case, the special and generic fibers of $S$ are fields, but $S$ is not coherent.
Do we know the global dimension of $\mathbb{Z}p \otimes\mathbb{Z} \mathbb{Z}_p$? Is it $2$?
Dear Badam, I'm not sure, but it seems to be difficult/unknown. There is a paper of Auslander which considers the case of global dimension of algebras over a field, but this is as much as a I could find.
@DrewHeard right, I'm not even sure what gl. dim. of $\mathbb{Q}p \otimes\mathbb{Q} \mathbb{Q}_p$ is! Just thinking out loud because if we knew the gl. dim. to be small we might have coherence for free or at least an easier route to checking it.
|
2025-03-21T14:48:31.443319
| 2020-07-06T10:27:26 |
364941
|
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|
Stack Exchange
|
Pseudodifferential Operators and Functional Calculus
I hope this is not too naive a question for MO. I've been taking a mathematical physics course, and was shown how operators like $\sqrt{1-\Delta}$ could be defined by taking multiplication operators in Fourier space. Later it emerged that functional calculus methods also allow one to construct similarly general operators, and it seems that they are considered equivalent, as some texts do not even specify how they interpreted such formulas when they write them. Why are we allowed to identify operators constructed by these two techniques (if we are)?
For me these two methods are the same: the spectral theorem asserts that self adjoint operators are unitarily equivalent to multiplication operators, and this allows us to define an action of a suitable algebra of functions i.e. a functional calculus. Did you have something else in mind?
"Why are we allowed to identify operators constructed by these two techniques?" The short answer is: the spectral theorem.
The route we took to functional calculus was via Helffer-Sjostrand, not the 'traditional' spectral theorem. But I see now what to do, thank you!
|
2025-03-21T14:48:31.443580
| 2020-07-06T10:51:44 |
364944
|
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|
Stack Exchange
|
Structure/description of a finitely presented group
I am unable to see the structure of the following finitely presented group.
$$\langle a,b,c,d : [a,b]=c=a^p,\ [c,b]=c^p=d^p,\ b^{p^2}=c^{p^2}=1 \rangle$$
I have tried in GAP, but it is not showing any result. Kindly help me out. This presentation is given in the following article by the name of the group $\phi_8(321)b$ on the page 324 under the class of groups of order $p^6$. How to present it in simple way ? ( Here $[a,c]=[b,d]=[a,d]=[c,d]=1$ )
Reference : James, Rodney, The groups of order (p^6) ((p) an odd prime)., Math. Comput. 34, 613-637 (1980). ZBL0428.20013.
It is easy to see that the above presentation defines an infinite group. I am guessing that there is a convention (which is often used in presentations of finite $p$-groups) that a pair of generators commutes unless otherwise specified. So you need to add the relations $[a,c]=[a,d]=[b,d]=[c,d]=1$ to the presentation.
With the extra relations you can define it in GAP for small primes, and it does indeed define a group of order $p^6$. I am not sure exactly what you are asking.
@Derek In GAP, it is not loading, I have tried for $p=3$. Here, in the presentation $$a^{p^3}=b^{p^2}=d^{p^2}=1$$. Since, $a,b$ and $d$ are generators. My question is do the structure involve some kind of non split extension ??
One immediate observation is that this group is the central product of its subgroups $\langle a,b,c \rangle$ and $\langle d \rangle$ of order $p^5$ and $p^2$, so this effectively reduces your question to a question about the smaller group $\langle a,b,c \rangle$, which has nilpotency class 3.
Well every nonabelian finite $p$-group $P$ is a nonsplit extension of $Z(P)$ by $P/Z(P)$ for example, so you need to ask more precise questions.
I am sorry for that. Actually, I have not studied non-split extensions and I am very less familiar with central products. Therefore I was surprised to see that why the product of powers of generators is coming more than $p^6$. So, it must involve something other than direct or semidirect product is all I thought. Thank you for the answer. It really is helping me out.
@DerekHolt Can somehow I change $a^p$ to $a^{p^2}$ by keeping the structure preserved or I will say by keeping the group unaffected ?
|
2025-03-21T14:48:31.443756
| 2020-07-06T11:00:11 |
364945
|
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|
Stack Exchange
|
Combinatorial condition for orientability on simplicial complexes
Let $K$ be a simplicial complex whose geometric realization is a topological or smooth manifold.
Is it possible to restate the condition of orientability of $M$ exclusively in (combinatorial) terms of $K$? Or at least, is there a sufficient condition on $K$ which guarantees that $M$ is orientable?
*By combinatorial terms I mean a condition that can be established on the abstract simplicial complex $K$. Hopefully there exists one apart from looking at the top dimensional simplicial homology (as Oliver suggested). If no other condition is proposed, I will accept that as an answer.
You will see orientability on top level simplicial homology of $K$.
Sure @OlivierBégassat but I ask for some combinatorial condition on the face poset which does not involve homology. Thanks anyway for your comment. I mean, if $M$ was a smooth manifold, then we would have at least the definition of orientability with local homology and the differential geometric one. Is there one relying only on the combinatorics of the simplicial complexes which does not require homology?
What do you mean by the underlying space of a simplicial complex $K$? The only thing I can think of is the geometric realization $|K|$, whose singular homology is isomorphic to the simplicial homology of $K$, so Olivier's comment seems to be a valid answer.
@MarkGrant you are absolutely right, I should have said geometric realization instead of "underlying space". Oliver's comment is a valid answer and I will accept it. I just was looking for other combinatorial condition which did not involve homology. It may not exist!!
I edit the question to clarify it.
I find the question a little confusing. Could you define what you mean by "combinatorial"? as the homological definition of orientation appears (to me) to be combinatorial when talking about simplicial complexes.
You can choose an orientation of some facet $F$ of $K$. This induces an orientation of a neighboring facet $F'$. (By "neighboring" I mean they differ in only one vertex.) If every path from $F$ back to $F$ preserves the orientation of $F$, then $K$ is orientable. Of course it's not a very practical condition to check directly, but it is purely combinatorial (no linear algebra).
Let me phrase @RichardStanley 's comment in a different way. Let $n$ be the dimension. It is orientable iff you can orient each $n$-simplex in such a way that two of them with a common $(n-1)$-face induce opposite orientation on the common $(n-1)$-face. An orientation is an equivalence class of total orderings on the vertices. Two orderings are equivalent if the permutation that relates them is even.
|
2025-03-21T14:48:31.443961
| 2020-07-06T11:29:49 |
364947
|
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|
Stack Exchange
|
Semicharacters on groups
I think, this must be simple, but apparently, I don't have enough intuition, so excuse me. The following construction is useful in the holomorphic duality theory for complex Lie Groups:
Let $G$ be an arbitrary group. A function $s:G\to [1,+\infty)$ is called a semicharacter on $G$, if
$$
s(x\cdot y)\le s(x)\cdot s(y),\qquad x,y\in G.
$$
Question:
is it true that for any group $G$, for any its subgroup $H\subseteq G$ and for any semicharacter $c:H\to [1,+\infty)$ there is a semicharacter $s:G\to [1,+\infty)$ that majorates $c$ on $H$:
$$
c(x)\le s(x),\qquad x\in H
$$
?
I guess this is usually called "submultiplicative".
Yes, this can be called "a submultiplicative function".
No it's not true in general. On $\mathbf{Z}$, consider the function $n\mapsto 2^{|n|}$. If $\mathbf{Z}$ is embedded in the center of the Heisenberg group (or in any group where it's distorted), every subadditive function on the Heisenberg is $\le C\sqrt{|n|}$ on $\mathbf{Z}$, hence every submultiplciative function on $\mathbf{Z}$ extending to a submultiplicative function is $\le\exp( C\sqrt{|n|})$ for some $C>0$. The usual keyword is "distortion".
YCor, and why is each subadditive function on the Heisenberg group $\le C\sqrt{|n|}$ on $\mathbb{Z}$?
Hint: $[e_{12}^n,e_{23}^n]=e_{13}^{n^2}$, where $e_{ij}=I_3+E_{ij}$.
@YCor I want to mention this example in my text. How can I refer to you?
This is classical in geometric group theory, so I don't need attribution. Just happy to help here!
OK, thank you, YCor!
|
2025-03-21T14:48:31.444096
| 2020-07-06T11:39:50 |
364948
|
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|
Stack Exchange
|
Relation between Kirchhoff's Circuital law and Matrix tree Theorem
I'm not a professional mathematician, just an undergraduate student. I was reading Introduction to Graph Theory by West, I came over the topic which discuses the methods to find the spanning trees in a general graph. Kirchhoff's Matrix tree theorem was there. Out of curiosity, I thought, could there be any relation between Kirchhoff's circuital laws and matrix tree theorem. I found this https://cp-algorithms.com/graph/kirchhoff-theorem.html. But I can't relate both. Can someone explain this, or give some reference to study.
Chapter II (pages 12 and following) of Combinatorics of Electrical Networks gives a linear algebra derivation of Kirchhoff's theorem from the circuit laws of Ohm and Kirchhoff.
|
2025-03-21T14:48:31.444189
| 2020-07-06T11:53:17 |
364949
|
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|
Stack Exchange
|
compatification of $\mathbb{R}^2$ under a conformal metric
I want to understand the following question:
Let $\delta$ be the Euclidean metric in $\mathbb{R}^2$. Is there any criteria for smooth function $u$ such that $(\mathbb{R}^2, e^{2u}\delta)$ can be compactified to a compact closed Riemannian surface?
For example, if $u(x,y)=\ln(\frac{2}{1+x^2+y^2})$, then $(\mathbb{R}^2, e^{2u}\delta)$ is the unit sphere minus the north pole. Hence $\mathbb{R}^2$ can be compactified under such metric.
I'm wondering is there a general criteria?
The meaning of "can be compactified" is not clear. The surface with a metric of your type is always conformally equivalent to the plane. The plane, as a Riemann surface can be compactified to the sphere (and to nothing else). But probably you mean compactification not of the Riemann surface but of the metric space. Could you explain the precise meaning of "a metric space can be compactified"?
For exmple, the plane with original, Euclidean metric "can be compactified" or cannot?
As a Riemann surface, the plane can have only one compactification, the sphere.
Introducing a factor to the metric does not change the conformal structure. So whatever factor you put there, the resulting Riemann surface remains conformally equivalent to the plane and thus have one and only one compactification, namely the sphere. In other words, possible compactifications of a Riemann surface do not depend on the metric in the given conformal class.
Perhaps you are thinking of a metric completion, instead of compactification: metric completion means that you add to your metric space the infinite elements corresponding to equivalence classes of Cauchy sequences. If you are asking for which metrics this completion will add exactly one point at infinity, then the criterion is the following: for every $\epsilon>0$ there exists a compact such that for $z,w$ outside this
compact, the distance between $z$ and $w$ is less than $\epsilon$.
Sorry I didn't clarify in my question. I know that if the criteria is satisfied, then $\mathbb{R}\cup {\infty}$ is a complete metric space, under the metric $g=e^{2u}\delta$.
Also, I know that if $\int_{\mathbb{R}^2}e^{2u}dx<\infty$, then the criteria is satisfied. My true question is that, how to make sure such metric is still a Riemannian metric at $\infty$? Or how to choose a local chart in a neighborhood of $\infty$ such that it is a smooth extension of the original metric defined only on $\mathbb{R}^2$?
Choose the chart $1/z$ and the condition is that your $e^u$ written in this chart is smooth at $0$.
I see. Thank you very much!
I answer the following question (which I think is the intention): Under which condition does there exist a compact Riemannian manifold $(M,g)$ and a chart $\phi=(x,y)\colon U\subset M\to \mathbb R^2$ such that the metric $$g_{\mid U}=\phi^*{e^{2u}\delta}?$$
Because every Riemannian metric induces a Riemann surface structure, $M$ must be diffeomorphic to the 2-sphere, see the answer of Alexandre Eremenko.
Then, you can use (conformal) coordiantes $(\tilde x,\tilde y)=\tfrac{1}{x^2+y^2}(x,-y)$ centered at infinity $\infty$ to compute wether for the given function $u$ the $(0,2)$-tensor extends smoothly as a Riemannian metric to $\infty.$ This is a standard exercise in a first course on differential geometry.
Thank you very much for your post. Yes, the question you posted is really what I want to understand. I have trouble in understanding why the metric completion under the metric $g=e^{2u}\delta$ implies that such completed metric is also Riemannian, or why it is also a smooth extension of the original metric defined only on $\mathbb{R}^2$? For example, given $e^{2u}=1/(x^2+4y^2)$, what are such $M$, $U$ and $\phi$? Sorry I have a very poor basis of Riemannian geometry.
Sorry for the typo in the previous comment. It should be $e^{2u}=1/(1+x^2+4y^2)$. Could you give me a reference on how to use the conformal coordinates you gave to proceed computation?
You have $\phi^*\delta=\tfrac{1}{(\tilde x^2+\tilde y^2)^2} \delta$. Therefore you need that $e^{2u\circ\phi(\tilde x,\tilde y)} \tfrac{1}{(\tilde x^2+\tilde y^2)^2}$ extends smoothly to $0$ and that the value of this function at $0$ is bigger than 0.
Thank you very much! This makes a lot of sense.
|
2025-03-21T14:48:31.444483
| 2020-07-06T11:53:22 |
364950
|
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|
Stack Exchange
|
Which line graphs of Cayley graphs are Cayley
When are the line graphs of Cayley graphs Cayley?
From this link we can know when the line graphs of complete graphs are cayley. But, my question pertains to the larger class of Cayley graphs. Are there new results in this direction? Any hints? Thanks beforehand.
Why do you say "The question necessary includes the question of when a cayley graph is edge transitive. "?
@verret Is it just that in order for the line graph to be Cayley, it must at least be vertex transitive? And the vertices of the line graph are the edges of the graph. Or is there something more subtle that I am missing?
@GordonRoyle yes, it is just as you said that the vertices of line graph correspond to edges of the graph itself, which made me think that my question includes the question of whether the graph is edge transitive
Sure, these graphs are a subclass of the class of edge-transitive Cayley graphs, but why is it necessary to solve the question for the larger class to understand the smaller class? Unless you have an actual reason for this, I suggest to remove it.
@verret edited. I thought it was somewhat easier to determine when a graph is edge transitive, so I added it
why the downvote?
One would expect few Cayley graphs to have this property, as their automorphism group is just not big enough. Either way, a complete classification is almost certainly out of reach. Even the example you give of complete graphs is far from trivial (the published proof requires on quite a few group-theoretic results, including the Feit-Thompson Theorem).
|
2025-03-21T14:48:31.444734
| 2020-07-06T13:14:15 |
364956
|
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|
Stack Exchange
|
Minimal value of a function involving the closeness of $\arctan$ to $\pi/2$
It appears that function
$$f(x)=\int_{0}^\infty(\pi/2-\arctan(t))^x dt$$
has a minimum for $x$ near $e$. Can someone prove this?
the minimum seems to be at $x$ bit larger than $e$, near 2.7456.
One only has to note that (i) $f$ is strictly convex where it is finite (because the integrand is strictly convex in $x$) and (ii) the values of $f$ at (say) $2740/1000, 2745/1000, 2750/1000$ are $1.96046\ldots>1.96045\ldots< 1.96046\ldots$, respectively. So, the only minimum of $f$ is in the interval $(2740/1000, 2750/1000)$. Of course, this way one can compute the minimum with any degree of accuracy.
Here is the graph of the function $f$ over the interval $(2740/1000, 2750/1000)$:
Mathematica 12.0 says
f[x_?NumericQ] := NIntegrate[(Pi/2 - ArcTan[t])^x, {t, 0, Infinity}];
NMinimize[{f[x], x > 0}, x]
$$ \{1.96046,\{x\to 2.74563\}\}$$
|
2025-03-21T14:48:31.444839
| 2020-07-06T13:30:37 |
364957
|
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|
Stack Exchange
|
Name for a matrices having a specific property
is there an established name for the property that a square matrix can be made symmetric by permutation of its columns?
Is it possible to recognize those kind of matrices efficiently?
If the permutation matrix $P$ is an exchange matrix, then a square matrix $A$ such that $AP=(AP)^T$ is called persymmetric (see https://arxiv.org/abs/1007.3239)
Here is a suggestion (not an answer) for the second question, at least for real matrices $A$: Suppose that there is a permutation matrix $P$ such that $(AP)^{T} = AP.$
Then $(AP)^{2} = (AP)(AP)^{T} = AA^{T}$, so that $AP$ is a symmetric square root of the positive semidefinite (symmetric) matrix $AA^{T}$. If the non-zero eigenvalues of $AA^{T}$ are all of algebraic multiplicity one, then there are $2^{r}$ real symmetric square roots of $AA^{T}$, where $r$ is the rank of $A$. This is basically a matter of finding an orthonormal basis of (real) eigenvectors for $AA^{T}$.
In view of comments, let me explain further: For exposition's sake, consider the case where $A$ has full rank $n$ and $AA^{T}$ has no non-zero eigenvalue of multiplicity greater than one. After finding an orthonormal basis of (real) eigenvectors for $AA^{T}$, we have an orthogonal real matrix $U$ such that $UAA^{T}U^{T}$ is diagonal. Then $UAA^{T}U^{T}$ has $2^{n}$ symmmetric square roots, all of which are diagonal. If $Q$ is one of these, then $Q^{\prime} = U^{T}QU$ is a symmetric square root of $AA^{T}$, and each symmetric square root of $AA^{T}$ arises in this way. Hence there is such a permutation matrix $P$ with $AP$ symmetric if and only if one (or more) of the $Q^{\prime}$ as above is such that $A^{-1}Q^{\prime}$ is a permutation matrix.
If $A$ has rank $r <n$, but $AA^{T}$ does not have any non-zero eigenvalue of algebraic multiplicity greater than one, then $AA^{T}$ has $2^{r}$ real symmetric square roots $Q^{\prime}$ , and we need to inspect whether any of these $2^{r}$ choices of $Q^{\prime}$ has the same columns as $A$, simply permuted around.
If $AA^{T}$ has a non-zero eigenvalue of algebraic multiplicity greater than one, then this strategy will not work as it stands.
Nice. I understand that $AA^T=UDQQ^TDU^T$ where $U$ is a real unitary matrix, $D$ is a real diagonal matrix with distinct diagonal entries and $Q$ is an arbitrary real unitary matrix. So the square roots of $AA^T$ are represented as $B = UDQ$.
I think your conclusion $2^r$ comes from the sign of the diagonal entries of $D$. But does $Q$ not play the role of supplying distinct square roots as well?
How do you find $P$ for which $AP=UDQ$ for some $Q$ and $D$ with the right signs of the diagonal entries?
Given $A$, it seems you are presuming there exists a known permutation $P$ such that $(AP)^T=AP$. I thought the question was to determine 1) whether there is such a $P$; 2) if it does exist, how to find it. It seems your answer does not answer either question. What am I missing?
@Hans: I have written a more detailed explanation in the answer I posted. This should address your questions.
+1. Nice. This argument holds for any orthonormal $P$ not just a permutation matrix, right?
@Hans : Yes, that's right, but note that $A^{-1}Q^{\prime}$ is always an orthogonal matrix, so you'd need to be looking for a particular type of orthogonal matrix.
|
2025-03-21T14:48:31.445054
| 2020-07-06T14:07:41 |
364960
|
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|
Stack Exchange
|
Independent inner functions on the unit disk
This problem cropped up in a paper that I am writing and I have thought about it for too long to no avail: let $\mathbb{D}$ be the open unit disk in the complex plane and suppose $\varphi:\mathbb{D}\to\mathbb{D}$ is inner ($|\varphi(z)|=1$ a.e. for $z\in\partial\mathbb{D}$) with $\varphi(0)=0$. Can there exist another inner function $\psi:\mathbb{D}\to\mathbb{D}$ such that $\psi(0)=0$ and $\langle \varphi^n,\psi^m\rangle=0$ for all $n,m\in\mathbb{N}$?
Here, $\langle \cdot,\cdot\rangle$ is the inner product on the Hardy space $H^2(\mathbb{D})$,
$$
\langle f,g\rangle=\frac 1 {2\pi}\sup_{0\leq r< 1}\int_0^{2\pi}f(re^{i\theta})\overline{g(re^{i\theta})} \ d\theta.
$$
I am aware that independent holomorphic random variables is an area of interest, but I am nowhere near expert enough to see whether the question is a trivial corollary of some result therein.
|
2025-03-21T14:48:31.445140
| 2020-07-06T14:28:49 |
364961
|
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|
Stack Exchange
|
The provability logic of $I\Delta_0+\Omega_1 $ and complexity theory
Almost 30 years ago, a number of folks in provability logic tried to show that GL (see for instance the excellent survey by Rineke Verbrugge here) is indeed the logic of $I\Delta_0+\Omega_1$ (in the following I refer to it as bounded arithmetics, though technically speaking there are various sub-systems of PA which are also entitled to this label).
If Rineke's revised article (2017) has not missed some very latest breakthrough, looks like whether GL is the provability logic of bounded arithmetic is still quite open.
Now, there was already a sense from the beginning that the reason why lifting the classical proof by Bob Solovay for PA's provability logic to bounded arithmetic is not viable (at least not by brute force), is that this seemingly sedate question is somehow related to deep complexity problems, perhaps even with the infamous P vs NP.
Here is then my question:
has there been any work on the above geared at making fully explicit
the felt connection with complexity theory? By fully explicit I mean a
result such as " if GL is the provability logic of bounded arithmetic
then $\ldots$" (where the dots stand for some complexity theory's
open question). Or perhaps the converse.
Any reference and/or comments will be deeply appreciated.
As far as I am aware, no such connection has been spelled out, except for the one mentioned in the original Berarducci-Verbrugge paper: if bounded arithmetic proves NP=coNP, then its provability logic is GL (this a rather tenuous connection as the premise is widely assumed to be false, while the conclusion may well be true). There was really nothing published about this problem at all besides that single paper.
|
2025-03-21T14:48:31.445289
| 2020-07-06T15:56:40 |
364967
|
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|
Stack Exchange
|
Motivation for the Jacobian Variety
I've been to many talks in Number Theory and for some reason I've yet to fully grasp, we all seem to like Jacobian Varieties a lot. I know that they are Abelian varieties, which give information about their respective curve, but I'm not sure what information exactly. I know of the analytic description of the Jacobian, but I'm still not exactly sure why the Jacobian is so studied.
In his AMS article, What is a motive Barry Mazur seems to suggest that Jacobians encapsulate all cohomology theories. Is this true? How can I see this?
Kleiman's chapter in FGA explained (available separately on arXiv) contains a great and very detailed historical account (albeit possibly a bit tangential to your main question) of the development of the Picard scheme (as well as details for the construction in a very general setting).
If you are a number theorist, you presumably like class groups? Let $X$ be a curve defined over $\mathbb{F}_p$, let $J$ be its Jacobian and let $x$ be an $\mathbb{F}_p$ point of $X$. Let $A$ be the coordinate ring of the affine curve $X \setminus \{ x \}$. Then the class group of $A$ is $J(\mathbb{F}_p)$. (And similar statements can be made for deleting more than one point, or deleting points defined over extensions of $\mathbb{F}_p$.)
Why do you need to delete a point?
Answer 1 to get an affine variety. If someone is truly coming from classical number theory, they may only know class groups of rings, not Pic. Answer 2 if you take $Pic(X)$, you get $\mathbb{Z} \times J(\mathbb{F}_p)$, not $J(\mathbb{F}_p)$.
@Ang I don't have a reference off the top of my head. We have $\mathrm{Pic}(X \setminus { x }) \cong \mathrm{Pic}^0(X)$, since every divisor on $X \setminus { x }$ can be extended to a degree $0$ divisor on $X$ in a unique way. (Here it matters that $X$ is an $\mathbb{F}_p$ point.) The fact that $J(\mathbb{F}_p) \cong \mathrm{Pic}^0(X)$ is the defining property of the Picard functor.
Suppose $X/\mathbb{Q}$ is a (smooth, projective, geometrically integral) curve of genus $g\geq 2$ and $J/\mathbb{Q}$ its Jacobian variety.
If one is interested in determining the (finite, by Faltings) set of rational points $X(\mathbb{Q})$, then it can be useful to compute $J(\mathbb{Q})$ first. The latter is easier because $J(\mathbb{Q})$ is a finitely generated abelian group, and descent theory analogous to elliptic curves allows us to often do this in practice.
If we pick a point $P\in X(\mathbb{Q})$ then we have an associated embedding $i_P: X \hookrightarrow J$. In favorable situations studying this embedding allows us to determine $X(\mathbb{Q})$ from $J(\mathbb{Q})$.
For example, the method of Chabauty-Coleman gives a very concrete instance of this when the rank of $J(\mathbb{Q})$ is less than $g$ (for a friendly introduction to this method see the nice survey of McCallum-Poonen).
The moral is: by replacing $X$ by $J$, we somehow have made the geometry harder but the arithmetic easier.
The relation with motives can be explained in relatively concrete terms. The $\ell$-adic cohomology groups $H^i(X_{\bar{\mathbb{Q}}},\mathbb{Q}_l)$ are zero if $i\neq 0,1$, $2$ and isomorphic to $\mathbb{Q}_l, \mathbb{Q}_l(-1)$ if $i=0, 2$ respectively. (The minus $-1$ denotes the Tate twist.)
So the only interesting degree is $i=1$, and pulling back via $i_P$ will induce an isomorphism $H^1(X_{\bar{\mathbb{Q}}},\mathbb{Q}_l) \simeq H^1(J_{\bar{\mathbb{Q}}},\mathbb{Q}_l)$. This last group (with its Galois action) is isomorphic to the dual of the $\ell$-adic Tate module of $J$.
So $J$ and its torsion points encapsulate all the cohomological information of $X$.
Similar statements will hold for other Weil cohomology theories: the only interesting degree is $1$ and $i_P$ will induce an isomorphism on $H^1$.
Edit: as pointed out in the comments, the geometry of $J$ is arguably easier than that of $X$. A better moral is thus maybe that we have made the space we're considering larger but richer in structure.
I must respectfully disagree that replacing $C$ with $J$ "makes the geometry harder." It does increase the dimension, which one could argues makes the geometry harder, but it introduces a group structure, and I'd sugestt that the geometry of a high dimensional group variety (especially one that's compact) is much less difficult than the geometry of lower dimensional varieties having less structure. Or even ignoring the group structure, $J$ has Kodaira dimension 0, while $C$ has Kodaira dimension 1, again suggesting that $J$'s geometry is simpler than $C$'s.
Thanks for the comment, I'll edit my vague moral to make it more accurate.
What are the favourable situations that allow us to determine $J(\mathbb{Q})$ from $X(\mathbb{Q})$? Also, why does $i_P$ become an isomorphism on $H^1$?
As outlined by the other answers, the Jacobian $J_X$ of a curve $X$ defined over $\mathbb{F}_q$ indeed encapsulates all cohomology information of $X$. In particular one can read the zeta function $\zeta_X$ directly on $J_X$: the numerator of $\zeta_X$ is simply the (reciprocal) polynomial of the Frobenius $\pi_q$ acting on $J_X$.
In particular André Weil's original proof of the Hasse-Weil bound for curves used Jacobians (implicitely). That was a big motivation in his Foundations of algebraic geometry: the algebraic construction of Jacobians over any field.
By the way over $\mathbb{C}$ the Abel-Jacobi map shows that the Jacobian of $X$ is intimately related to the study of abelian integrals. I think historically that was the prime motivation to study Jacobians. A fun fact is that modular functions coming from hyperelliptic integrals can be used to solve algebraic equations. Cf the appendix of Mumford's TATA2.
|
2025-03-21T14:48:31.445699
| 2020-07-06T15:56:57 |
364968
|
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|
Stack Exchange
|
Sieving by composite moduli
A traditional sieve gives a bound on the number of integers $n$ in an interval (say $I=[0,N]$) such that $$n\not\in S_p \mod p$$ for every prime $p$ in a set $\mathcal{P}$, where $S_p\subset \mathbb{Z}/p\mathbb{Z}$.
What happens if instead we are given a set $\mathcal{M}$ of composite moduli, and asked to give an upper bound on the number of integers $n$ in $I$ such that
$$n\not\in S_m \mod m\;\;\; \forall m\in \mathcal{M},$$ where $S_m\subset \mathbb{Z}/m\mathbb{Z}$? Is there any literature on the matter? Note the elements of $\mathcal{M}$ need not be pairwise coprime.
If the set is finite and not redundant (for each m there is number sieved only by m), the standard elementary arguments should work on a large scale just like with primes. Remember to redefine Eulers phi so that phi(m)=m-1. Were you thinking about short intervals? Gerhard "Jacobsthal May Play A Role" Paseman, 2020.07.06.
Oops. There is a twist. One does have to consider common divisors of the m_i. I will think further on this. I have no references at present for this problem for composite moduli. Gerhard "Found More To Throw Out" Paseman, 2020.07.06.
Is $S_p$ the set of integers not divisible by $p$?
No, it's just the set of excluded classes. Call it $\Omega_p$ if you prefer.
A. Sarközy, A note on the arithmetic form of the large sieve,
Sudia Sci Math Hungarica 27, 1992, 83--95
covers the literature until then (there are some later results, adapting this as needed.)
I. Ruzsa, On the small sieve. II. Sifting by composite numbers
Journal of Number Theory 14 (2), 1982, 260--268
|
2025-03-21T14:48:31.445847
| 2020-07-08T11:35:11 |
365113
|
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|
Stack Exchange
|
Cheeger constant of truncated hypercube
Look at the $d$-dimensional hypercube and truncate it. This means one replaces each vertex by a cycle (of length $d$) in such a way the the new graph is 3-regular.
Question 1: What is the asymptotic as $d \to \infty$ of the Cheeger (or isoperimetric or conductance) constant of $TQ_d$?
By asymptotic I mean $\Theta$ (actually $\Omega$ is probably enough; see below).
I am using the convention:
$$
h(G) = \displaystyle \min_{|X|\leq |G|/2} \frac{|\partial X|}{|X|}
$$
where $|\partial X|$ is the set of edges between $X$ and $X^\mathsf{c}$. But since the graph is 3-regular, the various definitions are equivalent up to multiplicative constants (i.e. they are $\Theta$ of each other).
If you cut the hypercube along an hyperplane (parallel to the coordinate planes), then one gets that $h(TQ_d) \leq \frac{1}{d}$. And I am relatively sure there should be a lower bound (i.e. $h(TQ_d)= \Omega(\frac{1}{d})$ but could not find it in the literature.
Also, there is a lower bound on $h$ using the spectral gap $\lambda$ ($h \geq K\lambda$ for some $K$). But $\lambda$ is $O(\frac{1}{d^2})$. Note that the upper bound $h \leq K' \sqrt{\lambda}$ implies $h = O(\frac{1}{d})$ and is coherent with the above.
Question 2: does it matter what kind graph one uses to replace the vertices (instead of a cycle)?
There are two extreme cases (where the replacement graph has either the best or the worst Cheeger constant):
one could either replace a vertex in $Q_d$ by a sequence of expanders $X_d$ (with $|X_d| = d$ and degree bounded by, say, 10)
one could replace a vertex by a line (remove one edge from each cycle)
In the expander case, it's (hopefully) easy to see that the best "cuts" are those from the hypercube.
As an addendum: I did manage to get a lower bound $\Omega(\tfrac{1}{d})$ for question 1 and question 2 (in the case of the line). The proof is very clumsy, so I would be thankful if someone could point me to a reference.
|
2025-03-21T14:48:31.446021
| 2020-07-08T11:48:32 |
365114
|
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|
Stack Exchange
|
How to understand adjoint functors?
I asked this same question on MathUnderflow two weeks ago but didn't receive any answer. Now that I am thinking more, it feels like the most suitable place for this question is here.
I have a good grasp of all different definitions/interpretations of adjoint functors, but still do not know have to interpret the left or right adjoint of a give functor, when it exist. It would be a easy to explain my question through an example.
For a particular example consider the inclusion of groupoids into small categories: $$\mathcal{F}:\mathcal{Grpd}\hookrightarrow\mathcal{Cat}.$$ This functor has left adjoint $\mathcal{L}$ which freely invert all existing morphisms of a category. Also it has a right adjoint $\mathcal{R}$ which extract the (maximal) subcategory of all isomorphisms, called "core groupoid" of a category. In some scene this bi-adjunction align with the free-forgetful philosophy. Also we can compute these adjoint functors explicitly as pointwise Kan extensions, then $$\mathcal{L}(\mathcal{C})=\text{Ran}_{\mathcal{F}}(\text{id}_{\mathcal{Grpd}})(\mathcal{C})=\lim(\mathcal{C}\downarrow\mathcal{F}\xrightarrow{\Pi_{\mathcal{C}}}\mathcal{Grpd}\xrightarrow{\mathcal{F}}\mathcal{Cat}).$$ But I don't understand how to interpret this "limit" as a localization process and same for the right adjoint $\mathcal{R}.$ Can we derive it from this formula? If not, how can we see it?
In general, If we know some adjoint of a given functor exist, what is the process to understand it's effect/outcome?
If I have a functor $G: A\to B$, a left adjoint $F$ is a functor $B\to A$ that, for each $b\in B$ picks out the best approximation of $b$ by $G$(someone), via the unit $b\to GFb$. "best approximation" here is in the sense of "mapping out of $b$, no one will do a better job" (that's why left adjoints are often given by colimits, and right adjoints, where the story is dual, by limits). You already know that probably (hence a comment, not an answer), but that's one of the usual ways to think about adjoints
Nice question Bumblebee. So, let us start with some "metaphysics of adjointness":
THE LEFT AND RIGHT ADJOINTS TO A FUNCTOR
$ \mathcal{F}:\mathcal{C}\hookrightarrow\mathcal{D}$
ARE THE FREE (LEFT) AND CO-FREE (RIGHT) WAYS TO GO BACK FROM $D$ TO
$C$.
If you choose some easy examples, for instance $C=Top$ and $D=Set$ and the functor is simply the forgetful functor which "forgets" the topological structure, Left and Right start from a given set and endow it with a topology, in the most economic way (trivial topology) or in the most rigid one (discrete) . Same happens if you replace $Top$ with $Groups$ (or any other algebraic category).
Now, not all functors which have adjoints are forgetful functors, so matters are slightly more subtle sometimes, but the "general metaphysics" of adjointness still holds true.
Now the second part of your question, the scary formula for your example: rather that filling this page with calculations, I want to give you the heuristics (so far I have told you what adjoints are, not whether they exist or how they are calculated).
Here, I use the second "metaphysical principle of adjointness", namely this:
THINK OF CATEGORIES AS GENERALIZED ORDERS, AND OF ADJOINTNESS AS
GENERALIZED GALOIS CONNECTIONS.
If you look up Galois connections (see here), how they are defined and calculated, you will also understand cats, by generalizing. Same exact story.....
Thank you for this nice explanation. Adjoints were a huge dark spot in my knowledge. I will get back to you later, as needed for more details.
Please do. But not before you have digested galois connections. If you follow that tip, you will "get it" also for the more general notion of adjointness
If you start with a category and only consider what you can see by looking at functors from groupoids, well you’ll only see the invertible morphisms. So the right adjoint is the core.
If you start with a category and only consider what you see by looking at functors to groupoids, well that’s a little harder because a non-invertible morphism can be sent to an invertible one. So to get the left adjoint you have to try to add formal inverses.
Thank you for the answer. How do you think of left or right adjoint functors in general?
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2025-03-21T14:48:31.446311
| 2020-07-08T13:08:34 |
365121
|
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|
Stack Exchange
|
$3$-variable Jacobi style identity linked to generalised Frobenius partitions
I was fiddling around with a family of probabilistic models and came across two "identities", which appear to be linked to generalized Frobenius partitions (more on this below). I would be grateful if anyone with expertise in this area can answer the following questions:
Are these known "identities"? If so where I find them?
Are these equivalent to known "identities" by some non-trivial means? If so then how?
Can the sums $S_{\text{even}}(q,t)$ and $S_{\text{odd}}(q,t)$ be written as products?
The identities:
$2\sum_{m\in \mathbb{Z}}S_{\text{even}}(q,t) q^{m^2+m}z^{2m} = \prod_{n\ge 1}(1+tzq^n+z^2q^{2n})(1+tz^{-1}q^{l-1}+z^{-2}q^{2(l-1)}) + \prod_{n\ge 1}(1-tzq^n+z^2q^{2n})(1-tz^{-1}q^{l-1}+z^{-2}q^{2(l-1)})$
$2t\sum_{m\in \mathbb{Z}}S_{\text{odd}}(q,t) q^{(m+1)^2}z^{2m+1} = \prod_{n\ge 1}(1+tzq^n+z^2q^{2n})(1+tz^{-1}q^{l-1}+z^{-2}q^{2(l-1)}) - \prod_{n\ge 1}(1-tzq^n+z^2q^{2n})(1-tz^{-1}q^{l-1}+z^{-2}q^{2(l-1)})$
Where:
$S_{\text{even}}(q,t) = \sum_{w \in E}q^{(\sum_{i \,\text{odd}}iw_i + \sum_{i \,\text{even}}i(w_i-1))}t^{2(\sum_{i \,\text{odd}}\textbf{1}(w_i \ge 1) + \sum_{i \,\text{even}}\textbf{1}(w_i = 0))}$
$S_{\text{odd}}(q,t) = \sum_{w \in O}q^{(\sum_{i \,\text{even}}iw_i + \sum_{i \,\text{odd}}i(w_i-1))}t^{2(\sum_{i \,\text{even}}\textbf{1}(w_i \ge 1) + \sum_{i \,\text{odd}}\textbf{1}(w_i = 0))}$
$E$ is the set of sequences of non-negative integers which agree with $(0,1,0,1,0,1,...)$ far enough to the right and have no two consecutive $0$'s.
$O$ is the set of sequences of non-negative integers which agree with $(1,0,1,0,1,0,...)$ far enough to the right and have no two consecutive $0$'s.
What I know so far:
In a similar vein to the "general principle" for generalised Frobenius partitions (found in Andrews' book) the product $\prod_{n\ge 1}(1+tzq^n+z^2q^{2n})(1+tz^{-1}q^{l-1}+z^{-2}q^{2(l-1)})$ has a Laurent series expansion in $z$ and the constant term should be counting generalised Frobenius partitions of $n$ (the $q$ exponent) allowing at most 1 repetition in each row and a fixed total number of non-repeats (the $t$ exponent). However it is not obvious (at least not to me) that there is any relationship between this and the sums $S_{\text{even}}(q,t)$ and $S_{\text{odd}}(q,t)$.
Indeed if we specialise by letting $t=1$ (which makes perfect sense probabilistically) then the first few terms can be computed and we get $S_{\text{even}}(q,t) = 1+ q + 3q^2 + 5q^3 + 9q^4 + 14q^5 + 24q^6 + ...$, which appears to agree with Andrews' $\phi_2(q)$, measuring GFP's that have at most $1$ repetition in each row (https://oeis.org/A053993). This is expressible as a product, as in Andrews' book. The other sum, after specialisation, doesn't seem to have a similar interpretation but seems to be expressible as an infinite product too (https://oeis.org/A201077). So, assuming this interpretation of the sum can be proved, we then find that the specialised identities are equivalent to two $2$-variable Jacobi-style identities (which I also haven't been able to identify).
The two $3$-variable identities above seem to be isolating the even/odd $z$ terms of the Laurent expansion in $z$. This is the only purpose of the $\pm$ signs on the RHS. Because of this I expect that these identities will follow by taking the "sign transform" of a simpler identity.
Another observation is that if we let $t=x+x^{-1}$ and $q = x^4$ (another natural thing to do probabilistically) then the sum $S_{\text{even}}(q,t)$ appears to collapse to the usual partition function $\prod_{n \ge 1}(1-x^{2n})^{-1}$ and $S_{\text{odd}}(q,t) = \frac{S_{\text{even}}(q,t)}{(1+x^2)}$. Assuming that this can be proved, we find that the corresponding specialised identities are equivalent to two $2$-variable Jacobi-style identities...and in fact are the even/odd parts of the Jacobi triple product.
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2025-03-21T14:48:31.446526
| 2020-07-08T13:57:44 |
365125
|
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|
Stack Exchange
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Analytic vs Zariski neighbourhood of a fibre
Let $f\colon X\to \mathbb P^1$ be a proper morphism of smooth complex algebraic varieties and let $p\in\mathbb P^1$. Are there a complex disk $\Delta\subseteq\mathbb P^1$ and a Zariski open subset $U\subseteq \mathbb P^1$, with $p\in\Delta\subseteq U$, such that $H^1(f^{-1}(U),{{\mathcal O}^{\rm an}}^*)\to
H^1(f^{-1}(\Delta),{{\mathcal O}^{\rm an}}^*)$ is an injection?
I believe that fails when $X$ is a general hypersurface in $\mathbb{P}^3\times \mathbb{P}^1$ of bidegree $(2,d)$, and $p$ is any point of $\mathbb{P}^1$ over which the projection morphism is smooth.
Could you please expand your comment and explaining why it is not an isomorphism?
Thank you, I think I have understood your answer now. I have modified the question.
I don't think the modified question works either. Let $E$ be a general elliptic curve. Take $X$ to be the quotient of $E\times \mathbb{P}^{1}$ by an involution which is a translation by a point of order 2 on $E$ and is $z \to 1/z$ on $\mathbb{P}^{1}$. Take the map $f : X \to \mathbb{P}^{1}$ that corresponds to the projection $E \times \mathbb{P}^{1} \to \mathbb{P}^{1}$. Let $p \in \mathbb{P}^{1}$ be a point over which $f$ is smooth, then for any disk $\Delta$ centered at $p$ the analytic Picard group of $f^{-1}(\Delta)$ is just $\text{Pic}(E)$. On the other hand for any Zariski open set $p \in U \subset \mathbb{P}^{1}$ the analytic Picard group of $f^{-1}(U)$ is $\text{Pic}(E')$ where $E'$ is the quotient of $E$ by the point of order 2. Under these identifications the restriction map from the analytic Picard group of $f^{-1}(U)$ to the analytic Picard group of $f^{-1}(\Delta)$ becomes the pullback map
$\text{Pic}(E') \to \text{Pic}(E)$ which has a kernel $\mathbb{Z}/2$.
Thank you very much!
Just to be sure that I have understood completely your answer: you are using the fact that the analytic Picard group of $f^{-1}(U)$ is the invariant Picard group in $E\times \tilde U$ by the involution, right?
Yes, this what I had in mind. And the point is that for a Zariski open set $U$, the cover $\widetilde{U}$ is connected while for a small disk $\Delta$, the cover $\widetilde{\Delta}$ has two components.
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2025-03-21T14:48:31.447060
| 2020-07-08T14:09:16 |
365126
|
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|
Stack Exchange
|
Algebraic groups as functors of points vs maximal points over algebraically closed field
So I only started learning about group schemes this summer, and I found two approaches. For the record I am interested in affine group schemes $G$ of finite type over a field $k$ (algebraic groups).
Initially I saw their description as group objects in the category of affine schemes over $k$, which is also equivalent to a group object in the category of functors $\mathit{Alg}_k \to Set$ which is representable, which is also equivalent to a Hopf algebra.
When getting deeper into the theory, namely learning about parabolic and Borel subgroups, many texts started working with what I believe is $|G|$, the closed points of $G$ over the algebraic closure $\overline{k}$, sometimes denoted $\mathrm{spm}(G_{\overline{k}})$. This has the structure of a semi-topological group which seems useful in proving facts about $|G|$. My question is, are there general ways to lift results about $|G|$ to results about $G$?
For example, there is a neat result that for a group homomorphism $\phi: G \to H$ the corresponding map $|G| \to |H|$ has closed image, and I vaguely remember that $\mathrm{Im}(\phi)$ itself is closed in $H$ (correct me if I am wrong), and I wonder if you can prove the latter from the first.
What I tried
I became more interested in the question "how much is the functor of points of a scheme determined by its values on (algebraically closed) fields" but I could not find anything online even after a thorough search, so I produced the following result, which may still be wrong.
Proposition: Let $F,G: X \to Y$ be distinct morphisms of schemes with $X$ reduced with finitely many irreducible components. Then there is a field $L$ so that the functor of points morphisms $F(L),G(L): X(L) \to Y(L)$ are distinct.
Here we can obviously replace $L$ with an algebraically closed field. This would be helpful in lifting any result which can be stated in terms of a comparison between two morphisms.
Disclaimer: This may have been better to post on stackexchange, and I haven't posted any questions in years on either site. However I wanted the opinion of someone who has worked with algebraic groups for a while.
There is a comparison between finite type (sometimes assumed separated and/or integral) $k$-schemes and 'varieties' (in the classical sense, i.e. using only the closed points). See for example Hartshorne, Prop. II.4.10. This means that in finite type situations, considering $\bar k$-points is enough in some sense. However, you need more than just a topological language to describe the objects, e.g. $(X\times Y)(\bar k) = X(\bar k) \times Y(\bar k)$ does not have the product topology. Is this the kind of thing you're looking for?
See Milne Algebraic Groups, CUP, 2017, which uses the modern approach and covers the deeper theory (parabolic, Borel subgroups...).
No one should have to fuss with archaic language that, for example, Borel's Linear Algebraic Groups is written. Milne's book contains some hints for translating the archaic language into the language of modern algebraic geometry.
|
2025-03-21T14:48:31.447279
| 2020-07-08T14:15:09 |
365129
|
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|
Stack Exchange
|
coset of affine Lie algebra
In many books about conformal field theory, when we talk about a coset $\mathfrak{g}_k/\mathfrak{h}_{k'}$, we would talk about how the modules of $\mathfrak{g}_k$ are decomposed into those of $\mathfrak{h}_{k'}$ tensoring those of $\mathfrak{g}_k/\mathfrak{h}_{k'}$, for example, for the vacuum module
$$
\mathcal{R}_\text{vac}[\mathfrak{g}_k] = \mathcal{R}_\text{vac}[\mathfrak{g}_k/\mathfrak{h}_{k'}] \otimes \mathcal{R}_\text{vac}[\mathfrak{h}_{k'}] \oplus ...
$$
I wonder if there is an "extra procedure" where we actually single out the vacuum representation $\mathcal{R}_\text{vac}[\mathfrak{g}_k/\mathfrak{h}_{k'}]$ from the sum, namely, a procedure where we actually get the vertex operator algebra $\mathfrak{g}_k/\mathfrak{h}_{k'}$ alone? (maybe it's as simple as focusing on states annihilated by some $\mathfrak{h}_{k'}$ generators, however I'm not entirely sure)
This is typically given by the commutant, or coset construction. You take the vector subspace of $\mathcal{R}_\text{vac}[\mathfrak{g}_k]$ spanned by vectors $v$ satisfying $Y(u,z)v \in \mathcal{R}_\text{vac}[\mathfrak{g}_k][[z]]$ for all $u \in \mathcal{R}_\text{vac}[\mathfrak{h}_{k'}]$. Equivalently, you take fields that have no singularities when applied to vectors in $\mathcal{R}_\text{vac}[\mathfrak{h}_{k'}]$. This subspace has a vertex operator algebra structure.
One possible reference is section 5.7.2 of Frenkel and Ben-Zvi's "Vertex algebras and algebraic curves".
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2025-03-21T14:48:31.447391
| 2020-07-08T14:25:19 |
365130
|
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|
Stack Exchange
|
Relation between random graph models $G^{(B)}_{n,m}$ and $G_{n,m}$
In Frieze, Alan; Karoński, Michał, Introduction to random graphs, in Section 1.3 Pseudo-Graphs, there is a model of random multi-graphs, which is denoted as $\mathbb{G}^{(B)}_{n,m}$.
Def. A random multi-graph $\mathbb{G}^{(B)}_{n,m}$ contains $m$ edges, where every edge is chosen uniformly at random and independently from ${[n] \choose 2}$.
It is also shown in the same section that $\mathbb{G}^{(B)}_{n,m}$ and $\mathbb{G}_{n,m}$ are equivalent, when $m = O(n)$. ($\mathbb{G}_{n,m}$ is the simple random graph with $m$ edges.) The equivalence is in the sense that for a graph property $\mathcal{P}$, if $\mathbb{P}(\mathbb{G}^{(B)}_{n,m} \in \mathcal{P}) = o(1)$, then $\mathbb{P}(\mathbb{G}_{n,m} \in \mathcal{P}) = o(1)$, for $m = O(n)$.
My question is what happens in the case when $m = \omega(n)$? For example, $\mathbb{G}_{n,m}$ is connected a.a.s. when $m = \frac{1}{2}(n\log{n} + \omega(1))$. What about $\mathbb{G}^{(B)}_{n,m}$?
The relation between $\mathbb{G}_{n,m}$ and $\mathbb{G}^{(B)}_{n,m}$ is not quite an equivalence. All we have is that if the random multigraph does not have a property almost surely, then the random simple graph also does not have this property, within the range $m=O(n)$. The reason essentially is that is within this range, $m$ is so small that there is a constant probability that the random multigraph turns out to be a simple graph. However, the converse implication need not hold, for example for the graph property "The graph is not simple" will hold with probability $0$ for the simple random graph, but not necessarily for the random multigraph.
For larger values of $m$, $\mathbb{G}^{(B)}_{n,m}$ will become non-simple, so there wouldn't be any natural implications anymore. For the specific question you ask, I think Theorem 4.2 from the same book ($\mathbb{G}_{n,m}$ becomes connected precisely when no isolated vertices remain) can give you an answer, provided that the same result holds for $\mathbb{G}_{n,m}^{(B)}$. I haven't read through the entire proof carefully to check this, but I would be very surprised if it did not generalize for some reason.
|
2025-03-21T14:48:31.447552
| 2020-07-08T15:37:22 |
365133
|
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"Dieter Kadelka",
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|
Stack Exchange
|
Existence of solution
Assume that $T$ is an compact integral operator on $L^2(D)$ and let $M$ be its Hilbert norm. here $D$ is the unit conplex disk
I need a reference for the following assertion. There is a function $f$ so that $T^* T f = M^ 2 f$.
What is the Hilbert norm of an operator? Is it the operator norm?
Yes, sorry. That is.
Presumably you want $f$ which is not identically zero :)
In fact it is true by very elementary facts. $T^*T$ is a symmetric positive compact operator on a Hilbert space. Eigenvalues/eigenvectors of a symmetric compact operator $L$ have a variational characterization: they are critical values/critical points of the Raileigh quotient $(Lx\cdot x)/(x\cdot x)$. In particular $|L|$ is an eigenvalue. This is elementary material that you can find in any textbook of Fun.Anal.
Please also note that generic titles such as Existence of a solution should be avoided. Check the guidelines about "how to ask".
Since an answer has already be given, let me leave the following longer but more elementary approach as a complete answer rather than a sketch in a comment. I think this material is actually fairly standard and probably in a book such as Bollobas's Linear Analysis or Kadison+Ringrose's Introduction to the Theory of Operator Algebras (although I am going from very hazy memories which may be mistaken).
As Dieter Kadelka, points out, this is true in greater generality/abstraction.
Proposition 1. Let $T$ be a bounded operator on a Hilbert space $H$, and let $M=\Vert T\Vert$. Then $M^2$ is an approximate eigenvalue of $T^*T$. (See e.g. Wikipedia for the definition.)
$\newcommand{\hnorm}[1]{{\Vert#1\Vert}_H}\newcommand{\ip}[2]{\langle#1,#2\rangle}$
Proof. By definition of $M$ there exists a sequence $(x_n)$ in $H$ with $\hnorm{x_n}=1$ for all $n$ and $\hnorm{Tx_n}\nearrow M$. Now observe:
$$
\begin{aligned}
\hnorm{T^*Tx_n -M^2x_n}^2
& = \ip{T^*Tx_n- M^2x_n}{T^*Tx_n- M^2x_n} \\
& = \hnorm{T^*Tx_n}^2 - 2M^2 \ip{T^*Tx_n}{x_n} + M^4 \\
& = \hnorm{T^*Tx_n}^2 - 2M^2 \hnorm{Tx_n}^2 + M^4 \\
& \leq 2\Vert T^*T\Vert^2 -2 M^2\hnorm{Tx_n}^2
\end{aligned}
$$
We have $\Vert T^*T\Vert \leq M^2$ (actually, equality holds but we don't need this) and $\hnorm{Tx_n} \to M$ as $n\to\infty$. It follows that
$\hnorm{T^*Tx_n-M^2x_n}^2$ is bounded above by something which converges to zero as $n\to\infty$.
Q.E.D.
Lemma 2. Let $K$ be a compact operator on a Banach space $X$ and suppose $\lambda$ is an approximate eigenvalue of $K$. Then $\lambda$ is actually an eigenvalue of $K$.
Proof. By assumption there is a sequence of norm-one vectors in $X$, say $(x_n)_{n\geq 1}$, such that $\Vert Kx_n- \lambda x_n\Vert \to 0$. Since $K$ is compact, there is a subsequence $(x_{n(j)})_{j\geq 1}$ which converges to some $y\in X$; since each $x_{n(j)}$ has norm one, so does $y$. Then $Ky-\lambda y = \lim_j Kx_{n(j)} - \lambda x_{n(j)} = 0$.
Q.E.D.
Corollary 3. Let $T$ be a compact operator on a Hilbert space $H$ and let $M=\Vert T\Vert$. Then $M^2$ is an eigenvalue of $T^*T$.
Proof. Since $T$ is compact so is $T^*T$. Apply Proposition 1 and Lemma 2.
Q.E.D.
(Note that Corollary 3 is one possible starting point for the standard proof of the spectral theorem which Dieter Kadelka uses for his solution.)
A direct reference for this result may be difficult to find. Actually its a direct consequence of the spectral decomposition theorem found in most functional analysis books, f.i. Dunford/Schwartz: Linear Operators II.
You can replace $L^2(D)$ by any Hilbert space $H$. With $T$ also $U := T^* T$ is a compact operator, which is self-adjoint and with $\lambda \geq 0$ for any Eigenvalue of $U$ (since $0 \leq \langle Tx,Tx \rangle = \langle x,Ux \rangle = \langle x,\lambda x \rangle = \lambda \langle x,x \rangle$). Let $\sigma(U) \subset [0,\infty)$ be the (compact) spectrum of $U$. Since $U$ is in addition self-adjoint there is a spectral decomposition of $U$ in the form
$$Ux = \sum_{\lambda \in \sigma(U)} \lambda P_\lambda x$$
with $P_\lambda$ the (finite dimensional, if $\lambda > 0$) projection on the orthogonal Eigenspaces of $U$ corresponding to Eigenvalue $\lambda$. Let $M^2$ be the largest Eigenvalue of $U$ and $f$ be an Eigenvector corresponding to Eigenvalue $M^2$, Then $Uf = M^2 f$ and further $\|T\|^2 = \sup_{x \in H \colon \|x\| \leq 1} \langle Tx,Tx \rangle = \sup_{x \in H \colon \|x\| \leq 1} \langle x,Ux \rangle \leq M^2$, thus replacing $x$ by $f$ even $\|T\|^2 = M^2$.
The spectral theorem is overkill. One of the usual approaches to proving the spectral theorem for compact self-adjoint operators on Hilbert space, is to start with the observation that for $T=T^\ast \in B(H)$ at least one of $\Vert T\Vert$ or $-\Vert T\Vert$ is an approximate eigenvalue of $T$, and for compact operators approximate eigenvalues are eigenvalues
|
2025-03-21T14:48:31.447861
| 2020-07-08T16:03:40 |
365135
|
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|
Stack Exchange
|
Continuous version of conditional probability distributions $( \mathcal{L}(X_t | \mathcal{G}) )_{t \geq 0}$ if $(X_t)_{t \geq 0}$ is continuous?
Let me first explain the setup:
Let $(X_t)_{t \geq 0}$ be a stochastic process on some probability space $(\Omega,\mathcal{F},P)$ with values in a complete and separable metric space $E$ (e.g. $E = \mathbb{R}$) and let $\mathcal{G}$ be a sub-$\sigma$-field of $\mathcal{F}$. The conditional probability distribution $\mathcal{L}(X_t | \mathcal{G})$ can then be seen a as a random variable on $(\Omega,\mathcal{G})$ with values in $\mathcal{P}(E)$, which denotes the set of Borel probability measures on $E$. Then $\mathcal{P}(E)$ can be endowed with a metric that metrizes the weak convergence (also often called narrow convergence) of Borel probability measures on E, turning $\mathcal{P}(E)$ into a complete and separable metric space.
The question I have is about the regularization of the regular conditional probability distribution is the following:
If we now suppose that the paths $[0,\infty) \ni t \mapsto X_t(\omega) \in E$ are continuous for each $\omega \in \Omega$, can we choose versions/modifications of the collection of regular conditional probability distributions $(\mathcal{L}(X_t | \mathcal{G}))_{t \geq 0}$, such that the mappings $[0, \infty) \ni t \mapsto \mathcal{L}(X_t | \mathcal{G})(\omega) \in \mathcal{P}(E)$ become continuous for each $\omega \in \Omega$ ? Does this hold in this generality or do we need additional assumptions ?
If $(X_t)_{t \geq 0}$ is for example a Brownian motion, then the above desired regularisation would follow from Kolmogorov's continuity theorem. But I have not found any results for general continuous processes.
Thanks a lot in advance!
For simplicity take $E=\Bbb R$ and the time interval to be $[0,1]$, and think of $X=(X_t)_{0\le t\le 1}$ as a random element of $C=C([0,1]\to\Bbb R)$, a Polish space. We then have a regular conditional distribution of $X$ given $\mathcal G$, call it $Q=Q(\omega,B)$, $\omega\in\Omega, B\in\mathcal B(C)$. And the induced "marginal conditional distribution" $A\mapsto Q(\omega, \{\pi_t\in A\})$ will be weakly continuous in $t$. Here $\pi_t$, $0\le t\le 1$ are the coordinate projections from $C$ to $\Bbb R$.
Wow, I did not think about it in this way. Thank you so much!!
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2025-03-21T14:48:31.448026
| 2020-07-08T16:04:27 |
365136
|
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|
Stack Exchange
|
"Non-group" ${\rm II}_1$ factors
Do we have existence/examples/criteria for a ${\rm II}_1$ factor that is not isomorphic to $L(G)$ for any group $G$?
Yes. This was established by Connes in 1975. Every L(G) is antiisomorphic to itself, but there exist II_1 factors without this property. See Connes, Alain "Sur la classification des facteurs de type II", C.R. Acad. Sci. Paris Ser A-B 281 (1975), 13-15.
Thanks. Naturally, if the "antiisomorphic to itself" is only a necessary property of group factors, it leads to the question, are there antiisomorphic to itself II_1 factors that are not group factors? And since we are on a roll, are there non-amenable groups for which L(G) is hyperfinite? Or should I pose these as separate questions on MO here?
For your first question: Yes. See V.F.R. Jones "A II_1 factor anti-isomorphic to itself but without involutory antiautomorphisms", Math. Scand. 46 (1990) 103-117. In fact, this question is probably a duplicate (https://mathoverflow.net/questions/345844/abstract-characterization-of-group-von-neumann-algebra-ii1-factor)
For your second question, the answer is no. See Proposition 10.1.3 and Proposition 10.2.2 here: https://www.idpoisson.fr/anantharaman/publications/IIun.pdf (Here I have assumed that your G is i.c.c. and discrete. The situation is more complicated in general.) As you can see, your questions are pretty good ones, even though the answers are known. Keep on trucking!
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2025-03-21T14:48:31.448144
| 2020-07-08T16:15:39 |
365137
|
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|
Stack Exchange
|
Variational formulation of abstract Cauchy problem, possible?
Recently I have come across a method known as "variational method" in which we try to establish weak solutions of various boundary value problems involving ordinary derivatives, partial derivatives, and fractional derivatives. The use of Sobolev spaces is the key to finding such solutions in which the Lax Milgram theorem is applied to show the existence and uniqueness of solutions. The general approach can be seen in the 8th chapter of the book "Functional analysis, Sobolev spaces, and partial differential equations" (publisher link).
Now I was wondering if a similar approach can be applied to check the weak solutions of an abstract evolution equation whose solutions are Banach space-valued functions, i.e, bounded linear operators. The abstract Cauchy problem is:
$$u'(t)=Au(t), \text{for}\hspace{3pt} t\geq 0$$
$$u(0)=x$$
where $A:D(A)\subset X \rightarrow X$ is the linear operator and the generator of a semigroup and $X$ is a Banach space.
So far I am aware of two types of solutions namely classical solution in which $u(0)=x \in D(A)$ and a mild/integral solution in which $u(0)=x\in X$.
Have a look at this Q&A and the references cited therein. Tonti's method is applicable not only to elliptic boundary value problems, but also to general evolution problems.
Another possible reference is Theorem 3.1.7 in
Curtain, Ruth F.; Zwart, Hans, An introduction to infinite-dimensional linear systems theory, Texts in Applied Mathematics. 21. New York, NY: Springer-Verlag. xviii, 698 p. (1995). ZBL0839.93001.
where the equivalence of mild and weak solutions is formulated.
See the book of A. Bensoussane et al. for different notions of solutions in section: 3 Nonhomogeneous linear evolution equations, especially Proposition 3.1 and 3.2 where it is proved that some of these notions coincide.
A. Bensoussan, G. Da Prato, M. C. Delfour and S. K. Mitter, Representation and Control of Infinite Dimensional Systems, Birkhäuser, 2007.
|
2025-03-21T14:48:31.448300
| 2020-07-08T16:59:34 |
365139
|
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|
Stack Exchange
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Probability of sequence of coin flips palindrome
I am flipping a coin at least 3 times, and then until the sequence of coin flips is a palindrome. What is the mean number of flips I will perform? What is the probability I won't stop?
11 or 00 are no palindromes, but 101, 111, 000, 010 are palindromes? I'm irritated.
@DieterKadelka Well, you have to require some minimum length, else you'll never get past one flip.
The unrestricted question was posed on Quora in 2016 and answers there suggest why the three flip minimum makes a more interesting question.
The last section of Problems and Snapshots from the World of Probability could also be relevant; $\S$17.6 "Palindromes" starting on p244. The book is by Blom, Holst, and Sandell, published by Springer in 1994.
Also posted to (but closed on) m.se, https://math.stackexchange.com/questions/3749357/mean-of-a-coin-flipping
Let's distinguish some cases. If the first two flips result in HT or TH then with probability 1 you will flip a palindrome. Indeed, this will happen at the time $n$ when the $n$-th flip coincides with the first one. You can easily calculate the expected value of $n$ to be 4.
If the first two flips result in HH or TT then we distinguish two more cases. Either the third flips coincides with the first two or it doesn't. In the former case then you get a palindrome and it happens with probability 1/4. All the intricacy of the problem is in the last case. Let's prove that with positive probability you won't stop (so in the end the expected value is infinity, as already discussed by mike).
WLOG, assume that the first three flips are TTH. The possible palindromes starting with TTH are of the form TTHTT or TTH$w$HTT where $w$ is a palindrome word (possibly empty). If $n=2k$ or $2k+1$ then the probability of flipping a palindrome word of length $n$ is $\frac{1}{2^k}$. Therefore, by a union bound, the probability of flipping a palindrome (given that the sequence starts with TTH) is less than
$$\frac{1}{4} + \left(2\sum_{k=0}^\infty \frac{1}{2^k}\right)\cdot\frac{1}{8}=\frac{3}{4}$$
All in all, with probability $>\frac{1}{16}$ you will not flip a palindrome (this lower bound is obviously not optimal).
Long comment: I think the expected number of tosses in $\infty$ and you can show it along the following lines: The probability of a palindrome of even length is the the same as the probablility that the sequence of flips exactly reverses the second half of the palindrome, i.e., HHTH HTHH is a palindrome of length 8 and HHTH THTT is the sequence that exactly reverses the second half , however, the reversed sequence will cause a random walk based on it to return to 0, and so occurs later than the first return time to 0, which has infinite expectation. Of course, you have to do something about odd ones.
|
2025-03-21T14:48:31.448521
| 2020-07-08T17:27:30 |
365140
|
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|
Stack Exchange
|
Is operadic desuspension inverse to operadic suspension?
Given a graded vector space $V$ over a field $k$, consider it's suspension $\Sigma V$ such that $(\Sigma V)^i=V^{i-1}$. For an operad of graded vector spaces over a field $\mathcal{O}$, the operadic suspension $\mathfrak{s}\mathcal{O}$ is defined in several different ways depending on the author. Some standard references might be An Alpine Expedition through Algebraic Topology and Operads in Algebra, Topology and Physics. All the definitions I've seen yield isomorphic graded vector spaces, but the operadic structures differ slightly. In the reference above, the operadic structure is not explicitely defined, it is just said to be induced by the one on $\mathcal{O}$, but it seems to be obvious that $\mathfrak{s}^{-1}\mathfrak{s}\mathcal{O}\cong \mathcal{O}\cong \mathfrak{s}\mathfrak{s}^{-1}\mathcal{O}$ as operads (not only as collections of graded vector spaces).
Here I'm interested in the definition given by Benjamin C. Ward in his Thesis (Section 2.1.2), for which I think that property does not hold.
Background definitions
He defines the operadic suspension as
$$\mathfrak{s}\mathcal{O}(n)=\mathcal{O}(n)\otimes\Sigma^{n-1}sign_n$$
where $sign_n$ is the sign representation of the symmetric group on $n$ letters. The symmetric group action on the graded vector spaces is the obvious diagonal action, and a diagonal operadic composition is given by the following operadic insertion on $\{\Sigma^{n-1}sign_n\}$. We may identify $\Sigma^{n-1}sign_n$ with the exterior power $\bigwedge^n k$, so it is spanned by the element $e_1\wedge\cdots\wedge e_n$. Therefore, define the $i$-th insertion map
$$\circ_i:\Sigma^{n-1}sign_n\otimes\Sigma^{m-1}sign_m\to \Sigma^{n+m-2}sign_{n+m-1}$$
as the map
$$(e_1\wedge\cdots\wedge e_n)\otimes (e_1\wedge\cdots\wedge e_m)\mapsto (-1)^{(i-1)(m-1)}(e_1\wedge\cdots\wedge e_{n+m-1}).$$
We may identify the elements of $\mathcal{O}$ with elements of its operadic suspension, so for $a,b\in\mathcal{O}$ we may write $a\tilde{\circ}_i b$ for the insertion in the suspension. We can compute it in terms of $a\circ_i b$ (the insertion in $\mathcal{O}$) in the following way:
$$\tilde{\circ}_i=(\mathcal{O}(n)\otimes\Sigma^{n-1}sign_n)\otimes (\mathcal{O}(m)\otimes\Sigma^{m-1}sign_m)\cong (\mathcal{O}(m)\otimes \mathcal{O}(m))\otimes (\Sigma^{n-1}sign_n\otimes \Sigma^{m-1}sign_m)\to \mathcal{O}(n+m-1)\otimes \Sigma^{n+m-2}sign_{n+m-1}$$
The Koszul sign rule on the isomorphism produces a sign with exponent $(n-1)\deg(b)$ and then the insertions are performed diagonaly, so after the identification we get
$$a\tilde{\circ}_i b=(-1)^{(n-1)\deg(b)+(i-1)(m-1)}a\circ_i b.$$
The operadic desuspension $\mathfrak{s}^{-1}\mathcal{O}$ is defined similarly using $\Sigma^{1-n}sign_n$, so the signs are the same.
Problem
I expected $\mathfrak{s}^{-1}\mathfrak{s}\mathcal{O}\cong \mathcal{O}$ as operads, but I think that the insertions are different. If I compute the insertion induced on $\mathfrak{s}^{-1}\mathfrak{s}\mathcal{O}$ in a similar way as above using the isomorphism
$$(\mathcal{O}(n)\otimes\Sigma^{n-1}sign_n\otimes \Sigma^{1-n}sign_n)\otimes (\mathcal{O}(m)\otimes\Sigma^{m-1}sign_m\otimes \Sigma^{1-m}sign_m)\cong (\mathcal{O}(m)\otimes \mathcal{O}(m))\otimes (\Sigma^{n-1}sign_n\otimes \Sigma^{m-1}sign_m)\otimes (\Sigma^{1-n}sign_n\otimes \Sigma^{1-m}sign_m)$$
Then, the insertion induced on this product is identified with
$$(-1)^{(1-n)(m-1)}a\circ_i b$$
which is of course not the same as $a\circ_i b$. So, for this new operad created by the suspension and desuspension to be isomorphic to the original one, we must have an automorphism $f$ on $\mathcal{O}$ such that $f(a\circ_i b)=f(a)\circ_i f(b)=(-1)^{(1-n)(m-1)}a\circ_i b$. I think this automorphism must be then of the form $f(a)=(-1)^{\varepsilon(a)}a$, with $\varepsilon(a)=\pm 1$. But this implies that $(-1)^{(n-1)^2}f(a\circ_i a)=(-1)^{2\varepsilon(a)}a\circ_i a=a\circ_i a$, which is not true for all $n$.
Question
Is my conclusion about this suspension true or am I mistaken? I'm not so sure that $f$ really needs to be of that form, but I can't really find an morphism that makes the two structures isomorphic. Is this definition of operadic suspension used by any other author?
What you really need to show is that
$$f(a\circ_ib)=(-1)^{(n-1)(m-1)}f(a)\circ_if(b).$$
Here, $n$ is the arity of $a$, $m$ is the arity of $b$, and $\circ_i$ is the infinitesimal composition in $\mathcal{O}$ (once you twist the definition of the infinitesimal composition by your sign, you get the usual equation for operad morphisms). You achieve this with
$$f(a)=-(-1)^{\frac{n(n+1)}{2}}a.$$
Thank you, Fernando, it was very silly after all!
@javi you're welcome. I've faced this kind of sign confusion several times, I know the formulas by heart.
Me too, and sometimes I get stuck in the combinatorics. Is there any idea of how someone would come up with that sign for $f$? Or any kind of tcommon tricks?
You can set $f(a)=(-1)^{\phi(n)}a$ impose the formula you want to get to and solve the recursion to get a definition of $\phi$. I proceeded like this the first time, but now I just try the usual formulas and slightly modify them if necessary.
As a complement to Fernando's answer: the suspension of an operad can be seen as the arity-wise tensor product with the endomorphism operad of a 1-dimensional vector space, spanned by an element of degree 1. See this other question and in particular Pedro's answer.
Also, as to the origin of the sign in Fernando's answer, I found Section 4.2 from Part I of this paper by Thibaut Mazuir, as well as Section 4.2 in The Twisting Procedure by Dotsenko-Shadrin-Vallette useful.
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2025-03-21T14:48:31.448861
| 2020-07-08T17:52:03 |
365141
|
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|
Stack Exchange
|
Consistent of the two definitions of L-moments
I don't know how to prove that the definition
\begin{equation}
\lambda_r = \frac{1}{r} \sum_{j=0}^{r-1} (-1)^j {r - 1 \choose j} E[X_{r-j:r}]
\end{equation}
where
\begin{equation}
E[X_{r:n}] = \frac{n!}{(r - 1)! \: (n - r)!} \int_{0}^{1} x(u) \: u^{r-1} \:(1-u)^{n-r} \: du
\end{equation}
is consistent with
\begin{equation}
\lambda_r = \int_{0}^{1} x(u) P^*_{r-1}(u) du
\end{equation}
where
\begin{equation}
P^*_r(u) = \sum_{k=0}^{r-1} p^*_{r-1,k} u^k,
\end{equation}
and
\begin{equation}
p^*_{r,k} = (-1)^{r-k} {r \choose k} {r + k \choose k} = \frac{(-1)^{r-k} (r + k)!}{(k!)^2 (r - k)!}
\end{equation}
The indication that gives me is "Substituting the definition of the expectation of an order statistic in the first formula, expanding the binomials in $u$ and summing the coefficients of each power of $u$". The indication is from Hosking (1990) - https://doi.org/10.1111/j.2517-6161.1990.tb01775.x
Thanks
Although you have an answer, "The indication that gives me" suggests this is a homework problem or other exercise, in which case it doesn't belong here.
I agree with LSpice. If it's from a paper then it is maybe appropriate for MathOverflow but please add citation.
I'm new here and I don't really know how this works. it is an indication given in an article to go from one formula to another
@rma then it would appreciated and helpful if you could also mention or cite the article.
We have
\begin{equation}
EX_{r-j:r}=\frac{r!}{(r-j-1)!\,j!} \int_0^1 du\,x(u)\,u^{r-j-1}\,(1-u)^j \\
=r\binom{r-1}j \int_0^1 du\,x(u)\,u^{r-j-1}\,(1-u)^j
\end{equation}
and
\begin{equation}
(1-u)^j=\sum_{i=0}^j(-1)^i \binom ji u^i.
\end{equation}
So,
\begin{equation}
\lambda_r=\int_0^1 du\,x(u)\,p_r(u),
\end{equation}
where
\begin{equation}
p_r(u):=\sum_{j=0}^{r-1} (-1)^j \binom{r-1}j^2 u^{r-j-1}\sum_{i=0}^j(-1)^i \binom ji u^i
=\sum_{k=0}^{r-1} p_{r,k}u^k,
\end{equation}
\begin{equation}
p_{r,k}:=(-1)^{r-1-k}\sum_{j=r-1-k}^{r-1} \binom{r-1}j^2 \binom j{r-k-1}
=\frac{(-1)^{k+r-1} (k+r-1)!}{(k!)^2 (r-k-1)!},
\end{equation}
as desired.
The latter equality (which I obtained with Mathematica's help) can be obtained by using Vol. 1 (in Russian), Formula <IP_ADDRESS>:
\begin{equation}
\sum_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m \tag{1}
\end{equation}
with $r-1,k$ in place of $n,m$, respectively. (This book is available online.)
Actually, it is easy to see why (1) is true. Indeed,
\begin{equation}
\binom nk^2 \binom k{n-m}=\binom nm \binom nk \binom m{n-k}
\end{equation}
and hence
\begin{align}
\sum_{k=0}^n\binom nk^2 \binom k{n-m}
& =\binom nm \sum_{k=0}^n\binom nk \binom m{n-k}\\
&=\binom nm \binom{n+m}n
=\binom nm \binom{n+m}m.
\end{align}
Hello, first, thanks for answering.
I just edited my question to fix the two bugs that were there. The indices were not good, they were only up to r-1. I have the same as you, but I don't see how you build $p_ {r, k}$
Thanks and best regards.
@rma : I am not sure what you mean by "build". However, I have added details on $p_{r,k}$.
I meant the last equality in your definition of $P_r (u)$. I don't understand how you got the index k and put $u^k$.
I had come to the same as you, well, except for the final equalities of the combinatorial numbers. I wouldn't have seen that in a thousand years.
(And finally, forgive my English, it is very bad and I use the google translator)
@rma : I just use $u^{r-j-1} u^i=u^k$ for $k:=r-j-1+i$, noting that, for each $j$, the condition $0\le i\le j$ is equivalent to $r-j-1\le k\le r-1$.
Thansk for all!!!
|
2025-03-21T14:48:31.449211
| 2020-07-08T17:56:51 |
365142
|
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|
Stack Exchange
|
Integral involving Bessel function
I have to work out the integral:
$$
\int_0^{\infty} dq \frac{J_0(q \xi)}{q+1}
$$
where $ J_0(z) $ is the Bessel function of the first type and order zero, $\xi \in \mathbb{R}$, $\xi \ge 0 $. So far, I only collected a good amount of trials and errors.
This problem actually comes from physics. I am Fourier transforming the 2-dimensional Coulomb potential $W(q)$ to real space coordinates, when screening effects in the the electron gas are accounted for as in the RPA, in the static and long wavelength limit so that:
$$
W(q) = \frac{2\pi e^2}{A} \frac{1}{q + \Lambda}
$$
where $e$ is the electron charge, $A$ is the area, $q$ the modulus of wavevector $\mathbf{q}$ and $\Lambda$ (real positive constant) the screening wavevector. When transforming to real space coordinates $\mathbf{x}$, one arrives at the integral above (unless I made some mistakes), with $\xi = \Lambda x$, $x$ being the modulus of $\mathbf{x}$. So it would be physically interesting to have the large $x$ limit and small $x$ behaviour.
Hi Manuel, welcome to MO. (i) What is RPA? (ii) What do you mean by "work out the integral"? what do you expect to get in the end - a bound? an explicit expression (hard to believe)?
I mean to evaluate the integral, as it will be a function of $\xi$. It would be nice to find an analytical expression for it, I imagine involving some other special functions, maybe the Bessel of second type, or Struve functions. If there is no expression for that, at least understanding how to have the asymptotic limit for large $\xi$ and for small $\xi$.
About the RPA: it is an approximation used in condensed matter physics. Do not worry for it, if it happens the reader is familiar with this approximation, he might give me some other useful insights in what I am doing. Anyway, thank you very much! :)
The integral can be expressed in terms of Bessel and Struve functions,
$$I(\xi)=\int_0^{\infty} dq\, \frac{J_0(q \xi)}{q+1}=\tfrac{1}{2} \pi \bigl(\pmb{H}_0(\xi)-Y_0(\xi)\bigr).$$
The small-$\xi$ behavior is
$$I(\xi)=\xi-\ln \xi-\gamma_{\rm Euler} +\ln 2 +{\cal O}(\xi^2).$$
For large $\xi$ one has
$$I(\xi)=1/\xi-\tfrac{1}{16}\sqrt\pi\xi^{-3/2}(\cos\xi+\sin\xi)+{\cal O}(\xi^{-2}).$$
According to wikipedia, this is in fact $\frac12\pi \mathbf{K}_0(\xi)$
your $\mathbf{K}_0$ is again a Struve function, right, not a Bessel function?
|
2025-03-21T14:48:31.449395
| 2020-07-08T18:06:10 |
365145
|
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"url": "https://mathoverflow.net/questions/365145"
}
|
Stack Exchange
|
The chromatic polynomial of a line graph
Is there a way to obtain the chromatic polynomial of the line graph of a regular simple graph, having known the chromatic polynomial of the graph?
There already exist characterizations of line graph of a simple graph, like it consists of several cliques of order equal to the maximum degree of the graph, any two being joined at at most one unique vertex. In addition, the skeleton graph of the cliques, that is the graph formed by taking each clique as a vertex, is nothing but the graph itself. Can we use these characterizations along with knowing the chromatic polynomial of the graph beforehand, to determine the required polynomial? What can be said about the complexity of this problem? Can there be a polynomial time algorithm to compute it, starting from the chromatic polynomial of the graph? Thanks beforehand.
If anything were known I might expect it to be discussed under the term "edge chromatic polynomial".
@BenBarber yes, a similar question exists here, but I also ask about complexity and a method starting from the chromatic polynomial of the graph
An observation: there are graphs with the same chromatic polynomial but different edge chromatic polynomials (various trees, say), so any such algorithm cannot depend solely on the chromatic polynomial.
|
2025-03-21T14:48:31.449508
| 2020-07-08T18:36:14 |
365150
|
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"sort": "votes",
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|
Stack Exchange
|
Examples of semisimple Hopf algebras where the category of representations has certain properties
I wish to find an example of a semisimple (hence finite dimensional) Hopf algebra $H$ with the following properties:
$H$ is nontrivial (i.e. not a group algebra or the dual of one);
The category of representations of $H$ is a monoidal category which is not symmetric (i.e. the Grothendieck ring is noncommutative);
There exists a two-dimensional representation $V$ which is inner-faithful (i.e. the only Hopf ideal which annihilates $V$ is the zero ideal).
I've checked various examples of Hopf algebras of low dimension and haven't been able to find an example with these properties. Unless I'm mistaken, the nontrivial examples in dimensions 8 and 12 don't have the desired properties, nor do the sixteen examples in dimension 16 which were classified by Kashina.
Seeking to construct an example with the desired properties, I have also considered abelian extensions of the form $(kG)^* \# \hspace{1pt} kL$ as in this paper. With a judicious choice of $G$ and $L$, one can ensure that the Grothendieck ring of such a Hopf algebra is noncommutative, but none of the examples I have checked have a two-dimensional inner-faithful representation.
Of course, it may also be the case that a Hopf algebra with these properties cannot exist, but I don't see why this should be the case.
|
2025-03-21T14:48:31.449615
| 2020-07-08T19:37:19 |
365159
|
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"RBega2",
"Yuhan",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365159"
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|
Stack Exchange
|
Index bounded Riemannian metrics
Let $L$ be a closed simply-connected smooth manifold with a Riemannian metric $g$. We say $g$ is index bounded if the energy functional (which is assumed to be Morse/Morse-Bott)
$$
E: C^k(L,g) \rightarrow \mathbb{R}
$$
is a bounded function for each $k$. Here $C^k(L,g)$ is the set of closed geodesics with Morse index $k$.
Question 1: When does $L$ admit an index bounded metric? (Examples I know are spheres.)
Question 2: If $L$ admits an index bounded metric, then under what kind of perturbations the index bounded property will be preserved?
My motivation: Morse theory of the energy functional computes the homology of the loop space of $L$. Suppose we know that on the homology level it is finite dimensional in each degree, then when is there a chain level analogue? Question 2 maybe related to the homotopy theory between different minimal model DGA's for the loop space, which I am not very familiar with. There are also parallel questions in the contact/symplectic setting, replacing closed geodesics by Reeb orbits.
Do you have an example of a metric that is not index bounded?
@RBega2 Not yet. I tried for some time to show any simply-connected manifold admits such a metric, but there is not much progress.
|
2025-03-21T14:48:31.449733
| 2020-07-08T19:43:57 |
365160
|
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"authors": [
"D.S. Lipham",
"Taras Banakh",
"Ville Salo",
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|
Stack Exchange
|
The Borel class of a subset of $\mathbb Z^\omega$
Define $F(t)=\ln(t+1)$ for $t\geq 0$.
For each sequence of integers $ s=s_0s_1s_2...\in \mathbb Z^\omega$ define $$t^*_{ s}=\sup_{n\geq 0}F^{n}(|s_n|)$$ where $F^{n}$ is the $n$-fold composition of $F$.
Let $\sigma$ be the shift mapping on $\mathbb Z^\omega$; so $\sigma(s_0s_1s_2...)=s_1s_2s_3...$., and let $\sigma^k$ be the $k$-fold composition of $\sigma$.
Is the set $$\mathbb S:=\{s\in \mathbb Z^\omega:t^*_{\sigma^k(s)}\to\infty \text{ as
}k\to\infty\}$$
an $F_{\sigma\delta}$-subset of $\mathbb Z ^\omega$? Assume $\mathbb Z$ is given the discrete topology and $\mathbb Z ^\omega$ has the product topology.
A positive answer to this question would imply that a certain set in complex dynamics is an Erdős space factor. See this paper, Remark 5.3 in particular, for more about this problem. Essentially a space $E$ is an Erdős space factor if $E\times \mathfrak E\simeq \mathfrak E$ where $\mathfrak E$ is the rational Hilbert space.
EDIT 10/26/20: I proved that the space from complex dynamics (mentioned above) is not an Erdős space factor, answering a question by Dijkstra and van Mill: link to paper. This result implies in particular that the set $\mathbb S$ is not $F_{\sigma\delta}$.
It seems that you obtain $F_{\sigma\delta}$ writing down the definition of your set: $\bigcap_{n\in\mathbb N}\bigcup_{k\in\mathbb N}\bigcap_{m\ge k}{s\in\mathbb Z^\omega:s(m)\ge n}$.
@TarasBanakh Your set is a proper subset of the one I wrote. For instance I believe the sequence $0,1,0,2,0,3,0,4...$ belongs to my set but not yours.
The sequence of $t^*$'s for $0,1,0,2,...$ would be above the sequence $F^{1}(1), 1, F^{1}(2), 2, F^{1}(3),3, …$ which goes to $\infty$
@TarasBanakh I wonder if $t^*\to\infty$ is the same as saying that a subsequence of s goes to infinity? In this case, my set is probably not $F_{\sigma\delta}$. Assuming these sets are equal, how would you prove that? By the way, I have shown that the original set is both $F_{\sigma\delta\sigma}$ and $G_{\delta\sigma\delta}$.
Oh, sorry. I have not noticed that you have iterations of the function $F$.
Definitely this is not the same as being unbounded, as suggested in the penultimate comment, just insert sufficiently long zero runs in the sequence $\mathbb{N}$.
I think your set is $G_{\delta \sigma}$ complete in the appropriate sense, i.e. any $\Sigma^0_3$ subset of Cantor space reduces to it continuously.
Similar to https://mathoverflow.net/questions/363437/what-is-the-borel-complexity-of-this-set/363475#363475 but I have to sleep first.
@VilleSalo I am also very interested in knowing if it is $G_{\delta\sigma}$.
Sorry, I meant to say $G_{\delta \sigma}$ hard instead of complete. I did not actually prove it's $G_{\delta \sigma}$! (Well, I haven't really "proved" anything yet.)
Here's what I had in mind. Consider a $\Sigma^0_3$-set
$$ T = \bigcup_{n \in \mathbb{N}} \bigcap_{m \in \mathbb{N}} \bigcup_{k \in \mathbb{N}} C_{n,m,k}. $$
where each $C_{n,m,k} \subset 2^\mathbb{N}$ is clopen. We show that there is a continuous map $f : 2^{\mathbb{N}} \to \mathbb{Z}^\omega$ such that $f^{-1}(S) = T$ where
$$ S = \{s \in \mathbb{Z}^\omega \;:\; t^*_{\sigma^k(s)} \rightarrow \infty \text{ as } k \rightarrow \infty \} $$
is the set from the question. This proves that $S$ is not $F_{\sigma \delta}$, since that would imply all $\Sigma^0_3$ sets $T$ in $2^\mathbb{N}$ are $\Pi^0_3$.
You are at the concert. On the stage, there is a conductor and $\omega$ many cellists. The conductor is reading a point $x \in 2^\mathbb{N}$. Whenever she notices $x \in C_{n,m,k}$, the conductor cues the $n$th cellist to play the note $m$, assuming it hasn't been played before, and $n$ has played all the notes before $m$. Only one cellist plays at a time, there is a rest when all the events $x \in C_{n,m,k}$ visible so far are exhausted, and if $m$ cannot be played yet because previous notes have not been played, the conductor makes a note of it and it is played once they have. As you listen to these rising scales, you note that $x \in T$ if and only if one of the cellists plays the entire scale $\mathbb{N}$.
From this continuously revealed information you will construct the continuous function $f : 2^\mathbb{N} \to \mathbb{Z}^\omega$. The construction for $f(t) = s$ is thusly. We go through $\ell = 0, 1, ...$ and by default we just set $s_\ell = 100$ for all $\ell$. Whenever the $n$th cellist plays, we do as follows:
if one of the cellists $n' < n$ has played between the last time the $n$th cellist played (or the beginning of time if $n$ hasn't played anything) and the present time, then we set $s_{\ell} = 100$. As long as no cellist $n' < n$ plays again we ensure that also $t^*_{\sigma^\ell(s)} = 100$.
otherwise (if no cellist $n' < n$ has played between), then if the last time $n$ played we set $s_{\ell'} = 100+h$ then we now set such a high value at $s_\ell$ that we have $\lfloor t^*_{\sigma^{\ell'+1}(s)} \rfloor \geq 100+h+1$. Namely, set $s_\ell = \lceil \text{pexp}^{\ell-\ell'-1}(100+h+1) \rceil$ where $\text{pexp}(x) = \exp(x) - 1$. Note that in fact we get precisely $\lfloor t^*_{\sigma^{\ell'+1}(s)} \rfloor = 100+h+1$ because of basic properties of $\text{pexp}$. It also follows, because $\log (103 + h) < 100 + h$ and by induction, that we do not disturb any of the ensured values $t^*_{\sigma^\ell(s)} = 100$ for any $n' \leq n$: those were ensured before setting the value of $s_{\ell'}$.
Now suppose indeed $t \in T$, and some cellist plays infinitely many times. Then if the $n$th cellist is the first cellist that does, then the first item applies only finitely many times for $n$, and after that whenever we set $s_{\ell} = \text{pexp}^{\ell-\ell'-1}(100+h+1)$ we actually set $t^*_{\sigma^{\ell''}(s)} \geq 100+h+1$ for all $\ell'' \in [\ell'+1, \ell]$. So since $n$ plays infinitely many times, actually $t^*_{\sigma^{\ell''}(s)} \rightarrow \infty$ as $\ell'' \rightarrow \infty$.
Suppose then $t \notin T$. If the song is finite, obviously $\lim_\ell t^*_{\sigma^{\ell}(s)} = 100$. Otherwise, whenever $n$ plays for the last time, we have a new ensured value at which $t^*_{\sigma^{\ell}(s)} = 100$, thus $\liminf_\ell t^*_{\sigma^{\ell}(s)} \leq 100$.
An observation:
As far as I can tell all we are using is that $F$ is monotone, $F(h + 2) < h$ for $h \geq 100$ and that $\lfloor F^n(h) \rfloor \rightarrow \infty$ as $h \rightarrow \infty$ for any $n$, and the values range over $[100, \infty) \cap \mathbb{N}$. And I guess $100$ can be replaced by some other number (probably $1$ or $2$ for your function). Maybe I missed some axioms.
edit
Your set is not $G_{\delta \sigma}$ either. Namely, any $\Pi^0_3$ set
$$ T' = \bigcap_{n \in \mathbb{N}} \bigcup_{m \in \mathbb{N}} \bigcap_{k \in \mathbb{N}} D_{n,m,k} $$
clearly continuously reduces to
$$ S' = \{s \in \mathbb{Z}^\omega \;:\; \lim_i s_i = \infty\}. $$
To see this, for each $n$ separately go through $(m,k)$ in lexicographical order, advancing to the next $m$ when you observe the point is not in $D_{n,m,k}$. On step $\ell$, output $n$ if $m$ is updated for $n$, otherwise output $\ell$. This way you construct $g(t) \in \mathbb{Z}^\omega$ for $t \in 2^\mathbb{N}$.
Clearly $t \in T'$ if and only if $m$ is updated for each $n$ only finitely many times. If $m$ is updated infinitely many times for $n$, then the limit of $g(t)$ is at most $n$, while if $m$ is updated finitely many times for each $n = 0, 1, ..., N$ then $g(t)$ stays above $N$ from some point on.
Now it's easy to further reduce to your set, observing that if a sequence satisfies $|s_{i+1} - s_i| \in \{-1,0,1\}$ and $s_i \geq 100$ for all $i$, then $s \in S' \iff s \in S$. Just replace all jumps by arithmetic progressions.
By taking the coordinatewise minimum of this process and the above, I suppose we have
Every set of the form $A \cap B$ for $A \in \Sigma^0_3$ and $B \in \Pi^0_3$ continuously reduces to your set.
But I don't know if your set can be written as $C \cap D$ for $C \in F_{\sigma \delta}$ and $D \in G_{\delta \sigma}$.
Thank you for your answers! I will try to understand soon.
|
2025-03-21T14:48:31.450240
| 2020-07-08T19:49:07 |
365161
|
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"Carlo Beenakker",
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|
Stack Exchange
|
Does anyone know a journal like Expositiones Mathematicae, except not by Elsevier?
Meaning primarily expository articles aimed at a wide-ranging mathematical audience, at the same technical level. But not by such a predatory publisher.
I recomond u this journal called Sarajevo
"The journal does not consider papers of expository nature "
See https://mathoverflow.net/questions/15366/which-journals-publish-expository-work
for the sake of fairness and precision, the term predatory publisher has the specific meaning of a publisher that "charges publication fees to authors without checking articles for quality and legitimacy and without providing the other editorial and publishing services that legitimate academic journals provide". This does not apply to Expositiones Mathematicae, does it?
The MO question Which journals publish expository work? is probably the most useful resource on the web for mathematicians seeking an answer to that question. Can I suggest that people add their answers to that question rather than here? Then all the information is kept in one place.
Carlo, many people use "predatory" for Elsevier, with some justification: they prey on academia and would die without us. The Wikipedia definition is arguably too narrow. (Plus, see the section "Bohannon's experiment" of the Wikipedia page, which shows Elsevier behaving exactly according to that defn of predatory publisher.) I think for Elsevier, parasitic is even more apt: they suck out as much of universities' blood as they can while still leaving us alive.
@CarloBeenakker I agree with Tom Leinster; that Wikipedia finds this page suitable doesn't mean that there consensus on this meaning, which indeed in this case is somewhat restrictive (and we can't be sure that these publishers, do not influence in some way the Wikipedia information to appear on the "good" side).
You can try l'Enseignement Mathématique, published by the European Mathematical Society.
Pete Clark's experience with EM may be worth bearing in mind: see point 2 in his question https://mathoverflow.net/questions/15366/which-journals-publish-expository-work
...and Felipe Voloch's answer to Pete Clark's question: https://mathoverflow.net/a/15404
Some journals publishing only surveys: Bulletin of the American Mathematical Society, Sugaku Expositions, Russian Mathematical Surveys. Some journals also publish surveys besides research articles.
|
2025-03-21T14:48:31.450441
| 2020-07-08T20:26:05 |
365163
|
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"Jonah765",
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"url": "https://mathoverflow.net/questions/365163"
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|
Stack Exchange
|
Dice roll expectation question
Suppose you roll a dice 100 times, How many times would you expect the most common number to show up.
I.e. roll a dice 100 times and document the frequency of each value, then repeat this process infinitely many times and take the mean of the highest frequency from each trial.
Is there a way to derive a formula or approach to calculate such a value? thanks.
@MattF. interesting, thanks. 100 rolls is quite arbitrary but I wonder of there is any significance to this number.
@MattF. seems as though a moderator closed it, not much I can do. It was my error posting in the wrong place originally. If you decide to look further into it perhaps let me know
According to the multinomial probability mass function formula, the expected maximum frequency in $n$ rolls of a fair die is
$$e_n=\frac1{6^n}\sum\frac{n!}{x_1!\cdots x_6!}\,\max(x_1,\dots,x_6),$$
where the sum is taken over all $n$-tuples $(x_1,\dots,x_6)$ of nonnegative integers such that $x_1+\dots+x_6=n$. There seems to be no reason for the existence of a simpler expression for $e_n$.
Mathematica computes
$$e_{30}=\frac{3063261583291047469655}{383808888404050968576}$$
in about 7 sec, and
$$e_{40}=\frac{936567872552422596737147305735}{92829823186414819915547541504}$$
in about 33 sec.
It will likely take too long to compute $e_{100}$.
thanks so much for the response, this was really informative.
|
2025-03-21T14:48:31.450569
| 2020-07-08T20:31:03 |
365164
|
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"Abdelmalek Abdesselam",
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|
Stack Exchange
|
Proving the exponential decay of Green's function for the lattice $-\Delta+p$
The Green function $G(x,y) =G(x-y)$ of the discrete Klein-Gordon operator $-\Delta+p$ on $\mathbb{Z}^{d}$ is given by:
\begin{eqnarray}
G(x-y) = \int_{[-\pi,\pi]^{d}}\frac{d^{d}k}{(2\pi)^{d}}\frac{e^{ik\cdot (x-y)}}{\gamma(k)+p} \quad (x,y \in \mathbb{Z}^{d}) \label{1}\tag{1}
\end{eqnarray}
where $p>0$, $k\cdot (x-y) := k_{1}(x_{1}-y_{1}) +\cdots+k_{d}(x_{d}-y_{d})$ and:
$$\gamma(k) := 2\sum_{j=1}^{d}(1-\cos(k_{j})). $$
Now, I think I've read somewhere (I don't seem to recall what was the article, sadly) that this function has exponential decay, i.e:
\begin{eqnarray}
G(x-y) \le Ke^{\alpha ||x-y||_{\infty}} \tag{2}\label{2}
\end{eqnarray}
where $\alpha$ and $K$ are positive constants and $||x-y||_{\infty} := \text{max}\{|x_{1}-y_{1}|,...,|x_{d}-y_{d}|\}$. I'm trying by myself to prove this result (and I don't think it is very difficult) but my first couple attempts were unsuccessfull. I've tried math stack and nobody helped me either, so I'm starting to doubt my memory and I don't know if this estimate holds. So my question is: does (\ref{2}) hold and, if so, how can I prove it?
you need $p>0$ for the decay.
Oh, right! So the decay really holds. Thanks! It's edited already!
The quickest way to get the decay is the representation as a Neumann series/simple random walk with positive killing rate. See, e.g., Lemma 3 page 10 of https://arxiv.org/abs/0901.4756
Another idea is complex contour deformation. Try it for $d=1$ first assuming WLOG $x-y>0$, and try to create a positive imaginary part in $k$. For higher $d$ you can iterate by adding the cosine expressions of the previous components of $k$ to the mass $p$. BTW finding exact long distance asymptotics is nontrivial.
Do you know any reference on this second approach?
Sounds more or less what I tried to do
|
2025-03-21T14:48:31.450713
| 2020-07-08T20:58:28 |
365166
|
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|
Stack Exchange
|
Expected size of binomial coefficient with Poisson arrivals?
I have a Poisson process where new elements arrive to a set with Poisson intensity $\lambda$. Initially, there are $N_0$ elements in the set. The probability that there are $N_0 + M$ elements in the set at time $t$ is $Pr[N(t) = N_0+M] = \frac{(\lambda t)^M}{M!} e^{-\lambda t}$.
I'm interested in the expected number of subsets of size $k$. That is, I want to compute
$$f(k,t) = \sum_{M=0}^{\infty} \frac{(\lambda t)^M}{M!} e^{-\lambda t} \binom{N_0+M}{k}$$
Is there some formula or approximation for this expectation when $N_0$ is large?
Suppose that a real $c:=\lambda t>0$ and a natural $k$ are fixed, whereas $n:=N_0\to\infty$.
Take any real $m>0$. Then
$$f(k,t)=g_m(k,t)+h_m(k,t),$$
where
$$g_m(k,t):=e^{-c}\sum_{0\le j<m}\frac{c^j}{j!}\,\binom{n+j}k,$$
$$h_m(k,t):=e^{-c}\sum_{j\ge m}\frac{c^j}{j!}\,\binom{n+j}k.$$
For each $j$, $\binom{n+j}k\sim n^k/k!$, whence
$$g_m(k,t)\sim P_m\frac{n^k}{k!},$$
where
$$P_m:=e^{-c}\sum_{0\le j<m}\frac{c^j}{j!}.$$
Also, $\binom{n+j}k\le(n+j)^k/k!=O(n^k+j^k)$ for all $j$, whence
$$h_m(k,t)=O\Big((1-P_m)n^k+\sum_{j\ge0}\frac{c^j}{j!}\,j^k\Big)
=O\Big((1-P_m)n^k+1\Big)
=O\big((1-P_m)n^k\big).$$
Letting now $m\to\infty$ and noting that then $P_m\to1$, we conclude that
$$f(k,t)\sim\frac{n^k}{k!}.$$
|
2025-03-21T14:48:31.450835
| 2020-07-08T21:25:23 |
365170
|
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|
Stack Exchange
|
Ron L. Graham’s lesser known significant contributions
Ron L. Graham is sadly no longer with us.
He was very prolific and his work spanned many areas of mathematics including graph theory, computational geometry, Ramsey theory, and quasi-randomness. His long association with Paul Erdős is of course very well known. Graham’s number, and Graham-Rothschild theorem, and the wonderful book Concrete Mathematics are other well known contributions.
However, some of his contributions may not be as widely known, but deserve to be so. This question is to encourage people to comment on such contributions. I am not familiar with his work on scheduling theory, for example.
He was into magic tricks and the mathematics behind them and co-authored a book on this with Persi Diaconis. And he was into juggling, like Claude Shannon.
Edit: Thanks to @LSpice for pointing out the Meta MathOverflow thread here on personal anecdotes.
MMO thread for anecdotes: https://meta.mathoverflow.net/questions/4619/ron-graham-1935-2020 .
Another wonderful book of his, co-authored with Rothschild and Spencer, is Ramsey Theory. It's been very influential for me, and although I don't think it qualifies as "lesser known", it is worth a mention here.
Seems to be Bruce Lee Rothschild.
The largest small hexagon determines the largest area a plane hexagon of unit diameter can have. (No, it's not the regular hexagon!) I love the title. For further results in this direction, search for Mossinghoff's work on isodiametric polygons.
R. L. Graham, A Fibonacci-like sequence of composite numbers, Mathematics Magazine, Vol. 37, No. 5 (Nov., 1964), pp. 322-324.
Graham defined $S(L_0,L_1)=(L_0,L_1,L_2,\dots)$ to be a sequence of numbers satisfying $L_{n+2}=L_{n+1}+L_n$ for $n=0,1,\dots$. He found relatively prime numbers $M,N$ such that every term of $S(M,N)$ is composite.
Along with Chung, Diaconis, and Holmes, he determined that there are 17,152 different arrangements of the "Stomachion" tangram puzzle into a square. See, here for a nice summary.
The puzzle is attributed to Archimedes. There is some evidence that Archimedes might have been doing similar counting; after all, Hough taught us that Hellenistic Greeks had some relatively sophisticated combinatorics.
I vaguely recall seeing a video back in the early 2000's, wherein Diaconis and Holmes flew down to visit Chung and Graham; the four of them did the counting one weekend.
I have a soft spot for The cover polynomial of a digraph (co-authored with Fan Chung) because it was one of two papers that motivated my Ph.D. thesis problem. The cover polynomial is a kind of digraph analogue of the Tutte polynomial. They later generalized it to the matrix cover polynomial.
One nice thing about the cover polynomial is that it satisfies a rather unexpected combinatorial reciprocity theorem. It was previously known that the rook polynomial of a board determines the rook polynomial of its complement, but the definition of the cover polynomial allows this relationship to be expressed in a particularly nice combinatorial way. I've continued to reap the benefits of Chung and Graham's insight; just a few years ago, a generalization of this reciprocity theorem provided a key step in a joint paper of mine with Patrick Brosnan, proving a conjecture of Shareshian and Wachs on regular semisimple Hessenberg varieties.
The following was posted by Stuart Margolis on my Facebook. I hope he doesn't mind me including it here lightly edited.
Ron Graham wrote a few papers in finite semigroups in the late 1960s that were known only to the Rhodes school of semigroups for many years. They have been rediscovered over the last few years and are as fresh and important today as they were half a century ago.
The paper:
"On Finite O-Simple Semigroups and Graph Theory", R. Graham, MATHEMATICAL SYSTEMS THEORY, Vol, 2, NO. 4, 325-339, 1968,
was the first paper to explicitly look at finite 0-simple semigroups as bipartite group labeled graphs (also called gain graphs, voltage graphs and other names). Among many results, it has the beautiful theorem that classifies finite 0-simple semigroups whose idempotent generated subsemigroup has only trivial subgroups and idempotent generated subsemigroups of finite 0-simple semigroups in general. By a result of Des FitzGerald this work can be extended to study idempotent generated subsemigroups for all finite semigroups.
The results were rediscovered later and given a more topological flavor by C.H. Houghton in the early 1970s. The so called Graham-Houghton graph of a 0-simple semigroup has been a tool of great import in a burgeoning literature on idempotent generated semigroups that has appeared over the last years.
A treatment of this work appears in in section 4.13 of the Rhodes-Steinberg book, "The Q-Theory of Finite Semigroups".
The paper:
Maximal Subsemigroups of Finite Semigroups* N. GRAHAM, R. GRAHAM, AND J. RHODES, JOURNAL OF COMBINATORIAL THEORY 4, 203-209 (1968) does just what its title says- describes maximal subsemigroups of finite semigroups.
The paper remained largely unknown for many years and gets rediscovered every so often. In the last few years the paper "Chains of subsemigroups" by Cameron, Gadouleau, Mitchell and Peresse uses these results to study the longest chain of subsemigroups of a finite semigroup.
My favorite result in combinatorics is Graham-Pollak theorem stating that the minimum number of bicliques (complete bipartite subgraphs) that partition the edge set of the complete graph K_n on n vertices is n-1. There are many such constructions with n-1 bicliques (exercise in Babai-Frankl notes), but the harder part is obtaining the lower bound for which Graham and Pollak used linear algebra.
Graham and Pollak studied this problem in the context of addressing graphs, representing the vertices of a graph by words/addresses of the same length k over the alphabet {0,1,*} such that the distance between any two vertices equals the number of positions in their addresses where one has a 0 and the other has a 1. Graham and Pollak proved that you can always address a graph on n vertices and diameter d with addresses of length at most d(n-1) and conjectured an upper bound of n-1. This is known as the Squashed Cube Conjecture and was proved by Winkler in 1980s (also a chapter in Van Lint-Wilson "A Course in Combinatorics" book). As far as I know, determining the minimum value of k is not known to be NP-hard or not.
There are several variations of the biclique decomposition problem that are still open. For r>3, while asymptotically solved by Alon in 1986, the exact value of the minimum number of complete r-partite r-uniform hypergraphs whose edges partition the complete r-uniform hypergraph on n vertices is not known. Also, for t>1, one can define the minimum number of bicliques whose edges cover K_n such that each edge is covered once and at most t times. This parameter has connection to geometric problems as shown by Zaks in 1979 and Alon in 1997. It is known that this parameter is of order of magnitude n^{1/t}, but the exact value is not known for t>1. For example, when t=2, it is between \sqrt{n-1} and 2\sqrt{n} (lower bound due to Huang and Sudakov in 2012 and upper bound due to Alon 1997). See these slides for more details: https://www.ima.umn.edu/materials/2014-2015/W9.8-12.14/21263/ima-1.pdf
Corollary (Graham). A rational number $p/q$ can be written as a sum of finitely many distinct reciprocals of integer squares iff $p/q \in [0,-1+\pi^2/6)~ \cup ~[1,\pi^2/6)$.
For the statement of the full theorem from which this follows, see On Finite Sums of Unit Fractions, with Graham as the sole author. Link.
This result is not very significant, but in literature surround unit/Egyptian fraction-esque problems, this result is cited often; if not for its use than for its novelty.
Very nice and interesting. Thanks for your answer.
With various collaborators he developed an elegant and deep mathematical theory of juggling, involving for instance the affine Weyl group $\tilde{A}_n$. A seminal early paper is here.
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2025-03-21T14:48:31.451510
| 2020-07-08T21:43:31 |
365171
|
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|
Stack Exchange
|
Does big and nef imply projectivity?
Suppose that we have a compact Kaehler manifold $X$ with big and nef canonical class $c_1(K_{X})$, does it imply that $X$ is projective? By the base point free theorem, big and nef implies semi ample but it is for projective algebraic manifolds. So it seems to suggest that big and nef does not necessarily imply projectivity. But I have seen in literature that people claim that big and nef does imply projectivity.
I guess if $X$ has a big line bundle then it is bimeromorphic to a projective variety, hence it is Moishezon. But Moishezon plus Kaehler equals projective.
@Pop: IMHO, this comment should be expanded into an answer
If $X$ has a big line bundle $L$ then for an appropriate natural number $m$, sections of $L^m$ define a meromorphic map $\varphi: X \dashrightarrow \mathbf P^N$ which is bimeromorphic onto its image. Therefore $X$ is bimeromorphic to the projective variety $\overline{\varphi(X)}$, hence it is Moishezon. But Moishezon plus Kähler equals projective.
It seems to me that the OP was only assuming that $c_1(K_X)$ is big (and nef), in the sense that it contains a Kahler current, not that $K_X$ itself is big. So you need to first use a result of Ji-Shiffman in https://doi.org/10.1007/BF02921329 to get that $K_X$ is a big line bundle.
Another comment: by another result of Demailly-Paun (Thm 2.12 in https://arxiv.org/pdf/math/0105176.pdf) it suffices to assume that $c_1(K_X)$ is nef and has strictly positive self-intersection, since this implies that $c_1(K_X)$ contains a Kahler current. Nefness of course here is in the analytic sense, i.e. it lies in the closure of the Kahler cone.
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2025-03-21T14:48:31.451659
| 2020-07-08T21:59:22 |
365173
|
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|
Stack Exchange
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Can one associate a "nice" topos to a von Neumann algebra?
The question here inspires my present question.
Reyes proves here that the contravariant functor Spec from the category of commutative rings to the category of sets cannot be extended to the category of noncommutative rings in such a way that every noncommutative ring is assigned to a nonempty set. Reyes also proves that it is impossible to suitably extend the Gelfand spectrum functor to the category of noncommutative C* algebras.
If one loosens the demand a Set-valued functor, then there are nice analogues of Gelfand duality. Please allow me to philosophize for a moment (I do so in order for someone to correct my perhaps inaccurate viewpoint). Even in the commutative case the need for "more open sets" in the Zariski topology led to the development of topos theory by Grothendieck in order to support étale cohomology. Toposes extend the notion of locale, which has the noncommutative relative quantale. It is possible to associate quantales to étale groupoids, to which there have been associated homology theories. Simon Henry's work on Boolean toposes (focusing on the von Neumann algebra/measure theoretic setting...his work goes beyond this) uncovers deeper important connections between von Neumann algebras and toposes.
I have begun to wonder if to find a good homology/cohomology theory for von Neumann algebras will require extracting a topos-like geometric object from the projection lattice of the von Neumann algebra and computing some sort of homology/cohomology of that object. I've read in Henry's papers that the kinds of object coming from projection lattices of von Neumann algebras are substantially different from Grothendieck toposes (in some way that I don't know enough about to ask for). The following question is a bit pie-in-the sky, and most likely completely hopeless, but I wonder if there is an "orienting answer":
Question: Is there hope of associating a "nice" topos to a von Neumann algebra?
This question is laughable, but I ask it nevertheless. What I mean by "nice" here is something like "has a computable cohomology of some kind". The philosophy being that toposes may be the right "noncommutative spaces" that may stand in counterpoint to von Neumann algebras.
(I'm going to be a bit informal to be able to go to the point relatively directly, but if you want more details on some specific aspect. I can try to add them)
Toposes are closely related to topological groupoids, in fact, they can be seen as a special type of localic groupoids or localics stack, the "étale-complete localic groupoids". (see the other answer)
So because we know very well how to attach a C* algebra or Von Neuman algebra to a groupoid it is very natural to expect that one can attach C* or Von Neuman algebra to a topos. Maybe not in full generality as topos corresponds to very general topological spaces and C*-algebras are attached to locally compact topological groupoids, but at least for 'nice topos' it should be possible. And also topos corresponds to Groupoid up to morita equivalence only, so the algebra we produce in general is only well defined up to Morita equivalence.
In some sense my work on this topic at the time was an attempt to give a direct description of the C* algebra or Von Neuman algebra one can attach to a topos (without going through groupoids) or to describe some properties of the Von Neuman algebra directly in terme of the topos (for example its modular time evolution).
And in fact it is possible:
To get a von Neuman algebra you should start with a Boolean topos that satisfies some 'measurability' condition, consider an 'internal Hilbert space object' in the topos and look at its algebra of endomorphisms. The construction works better if one assume that the topos $T$ is in addition 'locally separated' and take an Hilbert space of the form $L^2(X)$ for $X$ such that $T/X$ is separated. In this case you get a close connection between what I call measure theory over $T$ and the modular time evolution of the corresponding Von Neuman Algebra. This is essentially what I study in the paper you linked.
For C* algebra things are a bit more complicated, the best construction I could get to is described here.
Now, to go back to your question: can we go the other way and attach a topos to a von Neuman algebra or C algebra ?*
Essentially, no. At least not in a very interesting way if we do not have some additional structures. Of course it is not possible to give a definitive negative answer to this kind of question, so I'll say "probably not".
The problem is better understood in terms of groupoids than in terms of topos: the convolution algebra of a groupoids contains a lot of information on the groupoids, but if you consider it as a mere C*-algebras clearly a lot of information is lost.
For example, let's consider a groupoid $BG$ with only one object $*$ and $Hom(*,*)=G$ a group (Corresponding to the topos $BG$ of sets with a $G$-action). The kind of Von Neuman algebra or C* algebra you will attach to this topos is a Groupe algebra of $G$. Now if $G$ is abelian you will obtain an abelian Von Neuman algebra. But Abelian Von Neuman algebras corresponds to ordinary measurable spaces, so in this case you get two very different types of toposes that corresponds to exactly the same von Neuman algebra (a BG, and a topos of sheaves over a Boolean locale). The isomorphisms between the two Von Neuman algebra you get is induced by a kind of "Fourier transform" whose origin is purely analytic and non-geometric (at least in this picture).
What I read on this type of example is that if you want to construct a topos (or groupoid) out of an algebra you need something more. What this "something more" is can vary a lot, to give two example:
For C*-algebra the notion of Cartan subalgebra sometime allow to reconstruct a groupoids, I don't know the literature on this topic but these slides will give you an idea. I assume a similar theory for Von Neuman algebra might be possible.
One expects there will some connection between module for the algebra that one obtains a some kind of bundle of vector space/hilbert space on the topos. These bundle of vector spaces on the topos generally have a "pointwise tensor product". So one expect the Algebra we obtain to have an additional structure that corresponds to this tensor product, i.e. some sort of "generalized bi-algebra structure". One also expect that this tensor product is enough to recover the geometric object (this is very similar to Tanaka theory). I have a draft that I never finished on this topics if want to see a precise statement.
There are probably other similar story that can be told.
So in some sense I see that C*-algebra/Von Neuman algebra attached to a topos as some kind of invariant, like a homology theory. That reveals a lot of important and sometime hidden informations, but definitely not all the informations.
To finish, I would like to comment on the Bohr topos mentioned in the other answer, as it is is the only such construction present in the literature. I want to emphasize that it does not really answer the question in a satisfying way because the "Bohr topos" is not reall a topos, it is only an ordinary topological space. Indeed, because it is a topos of sheaves on a poset, it actually is a topos of sheaves on a locale, and in fact on a topological space due to a compactness argument, so it will never exhibit any "non-commutative" phenomenon. The construction has been formulated in the language of toposes because many people hope it might be possible to modify the construction to actually produce a topos, and maybe it is, but at the present time what is constructed is really just an ordinary topological space.
Thank you for the answer, Simon, and for the clarification about the Bohr topos construct.
You may be able to accelerate my understanding of why the Bohr topos/topological space is not a good candidate for an invariant of von Neumann algebras. The hope is not so much that the object exhibit noncommutative phenomena, but that it is a "classical" object sensitive enough to distinguish noncommutative objects. Crudely, one might hope that the poset structure (the way abelian subalgebras sit inside the von Neumann/C* algebra) would remember enough about the von Neumann algebra structure to distinguish certain von Neumann algebras from one another.
Said more briefly, are you saying that this "Bohr topos" topological space cannot even detect noncommutative phenomena? According the the above link, it remembers the Jordan algebra structure, and JBW algebras have some connection to cohomology https://www.math.uci.edu/~brusso/ChuRus111114final.pdf
Of course, this is a distinct question to the one I asked above...
By the "above link" in the other comment, I mean https://ncatlab.org/nlab/show/Bohr+topos
What I mean is just that by Gelfand duality, ordinary topological space essentially corresponds to commutative algebras. So when you construct the Bohr topos, maybe it will encode some information about your non-commutative algebra (obviously it does) but it encodes it in a way that is "purely commutative" (in the sense that it is an ordinary space). The Bohr topos itself is not a "non-commutative object". The sort of construction that I think your question ask for should somehow encode the "generalized spaces" aspect of VN algebra (i.e. these related to non-commutativity) in the...
"Generalized space" aspect of toposes (which are encoded in the categorical aspect beyond the poset of subterminal objects).
Ah, I see. Thanks again, Simon!
I think I will accept this answer, since the Fourier transform remark on the abelian groupoid case shows that hoping for the identification to be geometric requires more data. I think this illustrates the problem clearly. Thank you for the other links, as well! Your second bullet (for extra data for a candidate topos construction) is very interesting.
You may want to read about the so called BOHR TOPOS, ie a topos built on a C*-algebra. Here is a reference on nLab
https://ncatlab.org/nlab/show/Bohr+topos
and here is a great discussion on the n-Category Cafe:
https://golem.ph.utexas.edu/category/2011/07/bohr_toposes.html
I may be wrong, but once you are into the Bohr topos, the original algebra appears as * algebra object of the ambient category. At that point you have all the topos-related artillery, and you can do your cohomology there.
PS Another related thing you may want to look into are QUANTALES. Basically, a quantale is the "quantum version" of a locale, and the prototype is built out of subspaces of an algebra. The guy who invented them was Mulvey: see Mulvey and Pellettier. Perhaps they are useful for your endeavor
Thanks for this, Mirco! I'll have a look.
I really like the Bohr topos direction. Thanks again! I am not yet sure to accept this as the definitive answer (since I am naive), but feel like doing so as it is certainly an "orienting answer" and points to something quite nice. In a while I will accept this if nothing even more striking springs up. Thanks again!
In fact, some of this stuff appears in the dissertation linked to in the question...
@JonBannon I have added the link to Mulvey's article. As for my "answer", it was more of an enlarged comment, but glad to help
Imho, the Bohr topos is better viewed as an invariant of a Jordan algebra than as an invariant of a Von Neumann algebra. The poset of commutative subalgebras ov the Von Neumann algebra is best viewed as the poset of associative subalgebras of the Jordan algebra. The Bohr topos then describes all the Jordan algebra structure but not all the Von Neumann algebra structure
|
2025-03-21T14:48:31.452776
| 2020-07-08T22:04:34 |
365175
|
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|
Stack Exchange
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Cohomology theories for spaces vs cohomology theories for spectra
It is a standard consequence of the Brown Representability Theorem for $\operatorname{Ho}(\operatorname{Top}_*)$ that the category of generalized cohomology theories for spaces (pointed CW complexes, more specifically) is equivalent to the stable homotopy category $\operatorname{SHC}$ (defined as the homotopy category of the stable model structure on sequential spectra) via representability. On the other hand, we can define a cohomology theory for spectra (as in Barnes and Roitzheim's Foundations of Stable Homotopy Theory) to be a contravariant functor $E^*:\operatorname{SHC}^{op}\to\operatorname{Ab}_*$ such that
Each exact triangle in $\operatorname{SHC}$ gives rise to a long exact sequence in $\operatorname{Ab}$
$E^n$ preserves products for each $n$ (that is, it sends wedge sums to products of abelian groups)
$E^*$ preserves suspension up to a specified natural isomorphism $E^{n+1}(\Sigma X)\cong E^n(X)$.
Then it is easy to check that $[,E]$ is always a cohomology theory for spectra. But is every cohomology theory for spectra thus represented? Given such a cohomology theory $E^*$, we obtain an associated cohomology theory for spaces by restricting to suspension spectra. Thus, by the result for spaces mentioned above, the question becomes whether a cohomology theory for spectra is determined by its restriction to spaces.
This is certainly true if we require $E^*$ to preserve sequential homotopy colimits, because any spectrum is weakly equivalent to a CW spectrum with basepoint at its unique $0$-cell, which is a sequential homotopy colimit of homotopy cofibers of coproducts of shifted sphere spectra. But without this requirement, does the result still hold? If so, why? If not, is there a standard counterexample?
Regarding the non-equivalence of cohomology theories and the stable homotopy category see also this mathoverflow question: https://mathoverflow.net/questions/117684/are-spectra-really-the-same-as-cohomology-theories -- For representability results for cohomology theories see also the very nice book 'Axiomatic Stable Homotopy Theory' by Hovey, Palmieri and Strickland.
The category of cohomology theories on pointed CW-complexes is not equivalent to the stable homotopy category. The latter projects onto the former, and this projection induces a bijection on isomorphism classes, but there is a kernel, containing superphantom maps, see [Christensen, J.Daniel. “Ideals in Triangulated Categories: Phantoms, Ghosts and Skeleta.” Advances in Mathematics 136, no. 2 (June 1998): 284–339. https://doi.org/10.1006/aima.1998.1735.].
The category of cohomology theories on spectra is equivalent to the stable homotopy category. This is also called Brown representability. You essentially have to show that all cohomology theories are representable, and the rest follows from Yoneda.
By the aforementioned theorems, any cohomology theory for spectra is determined by the induced cohomology theory for spaces up to a non-canonical isomorphism.
I see. But why is it that cohomology theories for spectra are representable?
@DoronGrossman-Naples as I said, that's Brown's representability theorem. You can check it in numerous references, like [Margolis, H. R. Spectra and the Steenrod Algebra. Vol. 29. North-Holland Mathematical Library. Amsterdam: North-Holland Publishing Co., 1983.]
@DoronGrossman-Naples : it's possible that you get sequential colimits back from asking $E^n$ to respect all products, and not just finite ones
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2025-03-21T14:48:31.453033
| 2020-07-09T00:26:45 |
365186
|
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|
Stack Exchange
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A generalization to Bott‘s theorem (from Milnor’s “Morse theory”)
This is Theorem 22.1 of Milnor‘s Morse theory:
Let $M$ be a complete Riemannian manifold, let $p,q\in M$ be so that the space $\Omega’$ of minimal geodesics joining $p$ to $q$ is a topological manifold, and every such geodesic has Morse index $\geq c$. Then $\pi_i(\Omega’)$ is isomorphic to $\pi_i(\Omega)$ for $i\leq c-2$, where $\Omega$ is the space of curves joining $p$ to $q$.
For example, let $M$ be the round $n$-sphere, $p,q$ be the north and south pole, then $\Omega’$ is the $n-1$ sphere.
Is there a generalisation to this theorem, in which we replace $\{p,q\}$ by some closed curve $\gamma$ in $M$, $\Omega’$ by the space (which should assumed to be a manifold) of area-minimising disks in $M$ with boundary $\gamma$, and $\Omega$ by the space of disk in $M$ with boundary $\gamma$?
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2025-03-21T14:48:31.453135
| 2020-07-09T00:47:03 |
365188
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|
Stack Exchange
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Index and length of closed geodesics
Consider the round metric on $S^n$. The geodesics are (multiples of) great circles, and one can verify that this metric is of Morse-Bott type. The Morse indices of the n-covered great circles are (if I recall correctly) $(n-1), 3(n-1), 5(n-1), \dots, (2k-1)(n-1), \dots$.
Observe in particular that the Morse index of geodesics grows linearly with the length. i.e. there is a constant $C>0$ (depending only on the metric) such that for any geodesic $\gamma$, we have $$\operatorname{ind}(\gamma) > C \operatorname{length}(\gamma).$$
Question: let $g$ be a small, suitably generic perturbation of the round metric with the property that all geodesics are non-degenerate (such a metric is called "bumpy", I believe). Is it still the case that the Morse index of the geodesics grows linearly with the length?
I am also interested in the weaker question: does there exist a bumpy metric on $S^n$ with the property that the Morse index of the geodesics grows linearly with the length?
This follows from Bonnet-Myers. For a metric near the round metric (in the $C^\infty$ topology), the sectional curvature will be pinched below by $k > 0$, where $k\thickapprox 1$. Hence a segment of length $>\pi/\sqrt{k}$ of any geodesic will be unstable by Myers' theorem. Hence the index of a geodesic of length $L$ should be at least $L\sqrt{k}/\pi$: divide it up into segments of length $\geq \pi/\sqrt{k}$ which each have index at least 1. In fact, using a comparison theorem, one could probably do better and get an extra constant factor depending on the dimension (Myers theorem is using the weaker input of a lower bound on Ricci curvature, so a lower bound on sectional curvature will give a higher index estimate). You may be familiar with it, but Milnor's book on Morse theory is a good introduction to these topics (see Part III). For comparison theorems I like Gallot-Hulin-Lafontaine.
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2025-03-21T14:48:31.453299
| 2020-07-09T00:55:03 |
365190
|
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|
Stack Exchange
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Fast double exponentiation in finite fields
Let $p$ be a prime, and let $\mathbb{F}_p$ be the finite field with $p$ elements. Let $a$ be a non-zero element of $\mathbb{F}_p$. Can we quickly evaluate $a^{2^r} \mod{p}$? Using repeated squaring, this can be computed in time roughly $O(r\log{p})$. I would like to know if this can be done in time sublinear in $r$. I suspect this is difficult in a general ring, but finite fields are quite nicely behaved from an algorithmic standpoint, so I am hopeful this is possible.
is $r$ large with respect to $p$? If it's sufficiently large we can first calculate $2^r$ mod $p-1$ by repeated squaring using the binary expansion of $r$, then take the binary expansion of that, and do the same thing.
In addition, if you know the factorization of $p-1$ (and hence the exponent $e$ of $\mathbb{Z}/(p-1)$) then you can reduce $r$ mod $e$ before computing $2^r \bmod{p-1}$.
Obviously that should be $(\mathbb{Z}/(p-1)\mathbb{Z})^\times$ in my comment above. If you replace $p$ with an RSA modulus $N = pq$, then the supposed slowness of computing $a^{2^r} \pmod{N}$ is the basis of the Rivest-Shamir-Wagner time lock.
|
2025-03-21T14:48:31.453523
| 2020-07-09T00:59:34 |
365192
|
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|
Stack Exchange
|
Preserves naively $\mathbb{A}^{1}$-homotopic maps
I've been studying $\mathbb{A}^{1}$-homotopy recently and would like some guidance with the question below. Thank you so much.
Setup
Fix $k$ a field of characteristic zero. Let $Sm_{k}$ denote the category of smooth $k$-varieies. Let $\tau$ be a Grothendieck topology on the category of simplicial presheaves of smooth $k$-varieties. Consider this site with the $\tau$-local injective model structure and then Bausfield localize at the family $\{\mathbb{A}^{1}\times U\rightarrow U\}$ for all $U \in Sm_{k}$.
Given a presheaf of groups $\mathrm{G}$, we can define a simplicial presheaf $B\mathrm{G}:Sm_{k}^{op}\rightarrow sSet$ as follows: $B\mathrm{G}(U)=B(\mathrm{G}(U)) \in sSet$ for all $U \in Sm_{k}$.
Question
Let $\mathrm{G}$ and $\mathrm{H}$ be presheaves of groups. Let $f_{1},f_{2}:\mathrm{G}\rightarrow \mathrm{H}$ be considered as morphisms of simplicial presheaves.
I would like to know if the following is true. If $f_{1}$ and $f_{2}$ are naively $\mathbb{A}^{1}$-homotopic, then so are $Bf_{1}$ and $Bf_{2}$.
Although not explicitly stated in the help centre (which seems to be a generic page for all SE websites), it is considered good practice to add at least one top level tag.
Thank you for letting me know.
An $\mathbb{A}^1$-homotopy between $f_0$ and $f_1$ is a morphism $h:G\to H^{\mathbb{A}^1}$. Since the assignment $G\mapsto BG$ is a functor and since we obviously have $B(H^{\mathbb{A}^1})=(BH)^{\mathbb{A}^1}$, the answer is yes (assuming that $h$ is morphism of groups, of course).
So, there is a natural isomorphism between $B\mathbb{A}^{1}$ and $\mathbb{A}^{1}$?
Not as (pre)sheaves. If $X$ is a sheaf of sets and $F$ a (pre)sheaf, then $F^X$ is the (pre)sheaf $U\mapsto F(X\times U)$. For $F$ a sheaf of groups, $F^X$ is a sheaf of groups and one checks rather directly that $B(F^X)=B(F)^X$. There is no structure on $X$. Take $X=\mathbb{A}^1$.
|
2025-03-21T14:48:31.453726
| 2020-07-09T04:00:39 |
365199
|
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"url": "https://mathoverflow.net/questions/365199"
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|
Stack Exchange
|
($1$-)pullbacks of Kan complexes
Ultimately, I'm trying to figure out whether or not the full subcategory in $\mathbf{sSet}$ spanned by Kan complexes is finitely complete (as a $1$-category).
Since fibrations are stable under pullback in general, I know that Kan complexes are closed under finite products, so the question boils down to whether the pullback in the square
$\require{AMScd}$
\begin{CD}
K\times_LK' @>>> K\\
@VVV @VVV\\
K' @>>> L
\end{CD}
where $K$, $K'$, and $L$ are all Kan complexes must have $K\times_LK'$ as a Kan complex also.
In my limited experience, I feel like this isn't true since it's not true in a general model category, but I can't construct a counterexample.
I'm honestly pretty bad at creating Kan complexes in general, and my usual go-to's (simplicial groups and nerves of groupoids) are actually preserved under taking pullbacks (the former because $\mathbf{Grp}$ is complete and limits of simplicial sets/groups are computed levelwise; the latter because the nerve is fully faithful from $\mathbf{Cat}$ to $\mathbf{sSet}$ and pullbacks of groupoids are groupoids). Maybe my intuition is wrong?
It's probably better to try to prove the more general result: Given a commutative cube in $sSet$ where the vertical edges are Kan fibrations except for the 'top left corner' one, and the top and bottom squares are pullbacks, can you prove the last vertical edge is a Kan fibration? That said, I don't know if this is actually true, but it seems that if the special case were true, the more general would be, and likewise if they are false.
Take any simplicial set $X$ which is not a Kan complex. Let $K$ be a Kan replacement of $X$, and let $L$ be a Kan replacement of the pushout $K\amalg_X K$. Then the two maps $K\to L$ are levelwise injective, and the pullback $K\times_L K$ is precisely $X$.
|
2025-03-21T14:48:31.453927
| 2020-07-09T05:04:35 |
365200
|
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"Gerry Myerson",
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"url": "https://mathoverflow.net/questions/365200"
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|
Stack Exchange
|
generalizations of the delone nagell equation
Are there any references that study integer solutions to cubic Diophantine equations of the form $x^3 + 2y^3 = 2^a 3^b$ for $\{a,b\}\subset \{0,1,2,3\}$? I am aware that Nagell solved $x^3 + 2y^3 = 1$ $(x,y)=(1,0)$ is the only solution. I read (in the textbook Topics in Number Theory by Leveque, but no reference was provided) that Nagell proved that $x^3 + 2y^3 =3$ only has $(x,y) = (1,1)$ as its solution. Thanks in advance!
Note that $a=2$ is impossible, by elementary considerations. $a=3$ reduces to $x^3+2y^3=3^b$.
The equation $x^3 + 2y^3 = 1$ has a second integral solution besides $(1,0)$, namely $(-1,1)$. Those are all the integral solutions.
First of all, it is not true that $x^3 + 2y^3 =3$ only has $(1,1)$ as solution. Indeed, Nagell proved that the only solutions are $(1,1)$ and $(-5,4)$. The comments answer your question in some other cases. Actually, there is a more general case of your equation which is of interest and which has been studied, namely Thue equations. I will refer to the book:
Chaohua Jia, Matsumoto Kohji. Analytic Number Theory. Springer, Developments in Mathematics, 2002
In 1909, Thue proved that every equation of the form:
$$ p(x,y)=M $$
with $p$ irreducible polynomial in $\mathbb{Z}$ and $M \in \mathbb{Z}$ different from $0$, only has a finite number of integer solutions. In 1989, Tzanakis and de Weger gave an algorithm for finding bounds on $|x|$ and $|y|$ (see Tzanakis, N.; de Weger, B. M. M. On the Practical Solution of the Thue Equation. J. Number Th. 31, 99-132, 1989). There are some methods (discussed for instance in the book cited above) to find the maximal number of solutions in some particular cases. Your equation is generalised by:
$$ ax^3+by^3=c $$
Nagell and Delaunay proved that, if $b=c=1$, this equation has at most one solution with $xy \neq 0$. Then, the case with $c=1$ or $c=3$ was settled by Nagell, who proved that the equation has again at most one solution with $xy \neq 0 $ except when $a=1$, $b=2$ or $a=2$, $b=1$. More generally, if $p$ is a cubic homegeneous irreducible polynomial with negative discriminant $\Delta \neq -23, -31, -44$, then the equation with $M=1$ has at most three solutions (this was proved by Nagell and Delaunay). An explicit example is the following one ($a \geq 2$):
$$ x^3 + axy^2 + y^3 = 1 $$
It is easy to see that $(0,1)$, $(1,0)$ and $(1,-a)$ are solutions. By the above result, these are the unique ones.
|
2025-03-21T14:48:31.454144
| 2020-07-09T06:38:31 |
365206
|
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|
Stack Exchange
|
Reduction from a continuous tree to a discrete tree
I have a continuous and finite tree graph $C$ with a finite set of points $P$ ($P$ contains all the vertices in $C$ and additional points). I wish to construct a discrete tree $D$ which preserves the distances in $C$, i.e., each point $p_i \in C$ is represented by a vertex $d_i \in D$, and for every $p_1,p_2,p_3 \in C$, s.t. $dist(p_1,p_2) > dist(p_1,p_3)$, it follows that $dist(d_1,d_2) > dist(d_1,d_3)$. How can I construct the tree assuming the distances between points in $P$ can be irrational?
Thanks
|
2025-03-21T14:48:31.454222
| 2020-07-09T08:11:03 |
365210
|
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"Alexandre Eremenko",
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"url": "https://mathoverflow.net/questions/365210"
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|
Stack Exchange
|
Curvature of complete conformal metrics on the open unit disk
Let $D$ be the unit disk in the complex plane, and assume that $g$ is a Riemannian metric on $D$ which is complete and conformal to the standard Euclidean metric. Can it be the case that the Gaussian curvature of $g$ approaches zero as we approach $\partial D$?
Yes. Take the metric with length element $\rho(z)|dz|$ where $\rho(z)=(1-|z|)^{-2}$.
It is complete since $\int^1\rho(t)dt=\infty$, and the curvature
$$-\rho^{-2}\Delta\log\rho=\rho^{-4}({\rho'}^2-\rho\rho'')=-2(1-r)^2\to 0,$$
where $r=|z|$ and the primes indicate differentiation with respect to $r$.
Great example! Is there a limit on the rate at which the curvature can approach zero?
@user160856: since there is no flat complete metric, there must be some such limit on the rate, but I do not see how to determine it at this moment, even for the metrics depending only on $r$.
Why does the fact that there is no flat complete metric on the unit disk imply that there must be some limit on the rate?
@user160856: If I knew I would answer your question in the comment.
|
2025-03-21T14:48:31.454330
| 2020-07-09T09:19:48 |
365213
|
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|
Stack Exchange
|
Oriented cobordism classes represented by rational homology spheres
Any homology sphere is stably parallelizable, hence nullcobordant. However, rational homology spheres need not be nullcobordant, as the example of the Wu manifold shows, which generates $\text{torsion}({\Omega^{\text{SO}}_{5}}) \cong \mathbb Z/2\mathbb Z$. This motivates the following question.
Which classes in $\Omega^{\text{SO}}_{\ast}$ can be represented by rational homology spheres?
Of course, any such class is torsion, as all its composite Pontryagin numbers, as well as its signature, vanish.
The necessary condition pointed out by Jens Reinhold is also sufficient: any torsion class $x = [M] \in \Omega^{SO}_d$ admits a representative where $M$ is a rational homology sphere.
EDIT: This is Theorem 8.3 in $\Lambda$-spheres by Barge, Lannes, Latour, and Vogel. They also calculate the group of rational homology spheres up to rational h-cobordism, and more. I'll leave my argument below:
To prove this, we first dispense with low-dimensional cases: in any dimension $d < 5$ the only torsion class is $0 = [S^d]$. The high dimensional case follows from Claims 1 and 2 below.
I'll write $MX$ for the Thom spectrum of a map $X \to BO$ and $\Omega^X_d \cong \pi_d(MX)$ for the bordism group of smooth $d$-manifolds equipped with $X$-structure. Representatives are smooth closed $d$-manifolds $M$ with some extra structure, which includes a continuous map $f: M \to X$.
Claim 1: if $d \geq 5$ and $X$ is simply connected and rationally $\lfloor d/2 \rfloor$-connected, then any class in $\Omega^X_d$ admits a representative where $M$ is a rational homology sphere.
Claim 2: There exists a simply connected space $X$ such that $\widetilde{H}_*(X;\mathbb{Z}[\frac12]) = 0$, and map $X \to BSO$ such that the image of the induced map $\Omega^X_d = \pi_d(MX) \to \pi_d(MSO) = \Omega_d^{SO}$ is precisely the torsion subgroup, for $d > 0$.
Proof of Claim 1: Starting from an arbitrary class in $\Omega^X_d$ we can use surgery to improve the representative. Since $X$ is simply connected and $d > 3$ we can use connected sum and then surgery on embeddings $S^1 \times D^{d-1} \hookrightarrow M$ to make $M$ simply connected. Slightly better, such surgeries can be used to make the map $M \to X$ be 2-connected, meaning that its homotopy fibers are simply connected. From now on we need not worry about basepoints and will write $\pi_{k+1}(X,M) = \pi_k(\mathrm{hofib}(M \to X))$. These are abelian groups for all $k$.
If there exists a $k < \lfloor d/2\rfloor$ with $\widetilde{H}_k(M;\mathbb{Q}) \neq 0$ we can choose $\lambda \in H_k(M;\mathbb{Q})$ and $\mu \in H_{d-k}(M;\mathbb{Q})$ with intersection number $\lambda \cdot \mu \neq 0$. If $d = 2k$ for even $k$ we can additionally assume $\lambda \cdot \lambda = 0$, since the signature of $M$ vanishes. The rational Hurewicz theorem implies that $\pi_k(M) \otimes \mathbb{Q} \to H_k(M;\mathbb{Q})$ is an isomorphism, and the long exact sequence implies that $\pi_{k+1}(X,M) \otimes \mathbb{Q} \to \pi_k(M)\otimes\mathbb{Q}$ is surjective. After replacing $\lambda$ by a non-zero multiple, we may therefore assume that it admits a lift to $\pi_{k+1}(X,M)$. Such an element can be represented by an embedding $j: S^k \times D^{d-k} \hookrightarrow M$, together with a null homotopy of the composition of $j$ with $M \to X$. In the case $k < d/2$ this follows from Smale-Hirsh theory, in the case $d = 2k$ we must also use $\lambda \cdot \lambda = 0$ to cancel any self-intersections. (Actually there could also be obstructions to this in the case $d=2k$ for odd $k$, but those obstructions vanish after multiplying $\lambda$ by 2.) The embedding and the nullhomotopy gives the necessary data to perform surgery on $M$ and to promote the surgered manifold to a representative for the same class in $\Omega^X_d$.
Performing the surgery gives a new manifold $M'$ where $H_k(M';\mathbb{Q})$ has strictly smaller dimension than $H_k(M;\mathbb{Q})$ and $\widetilde{H}_*(M';\mathbb{Q}) = 0$ for $* < k$. This is seen in the same way as in Kervaire-Milnor. The case $d > 2k+1$ is easy, similar to their Lemma 5.2. In the case $d = 2k+1$ the diagram on page 515 shows that we can kill the homology class $j[S^k]$ and at worst create some new torsion in $H_k(M')$. In the case $d = 2k$ the diagram on page 527 shows that we can kill the homology class $j[S^k]$ and at worst create some new torsion in $H_{k-1}(M')$.
In finitely many steps we arrive at a representative where $\widetilde{H}_k(M;\mathbb{Q}) = 0$ for all $k \leq \lfloor d/2\rfloor$. Poincaré duality then implies that $H_*(M;\mathbb{Q}) \cong H_*(S^d;\mathbb{Q})$. $\Box$.
Proof of Claim 2: Finiteness of the stable homotopy groups of spheres implies that $\pi_d(MX)$ is a torsion group for $d > 0$ for any such $X$. Therefore we can never hit more than the torsion in $\pi_d(MSO)$, all of which is exponent 2 by Wall's theorem. The difficult part is to construct an $X$ where all torsion is hit.
The non-trivial based map $S^1 \to BO$ factors through $\mathbb{R} P^\infty \to BO$, whose image in mod 2 homology generates the Pontryagin ring $H_*(BO;\mathbb{F}_2)$. We can freely extend to double loop maps
$$\Omega^2 S^3 \to \Omega^2 \Sigma^2 \mathbb{R}P^\infty \to BO$$
where the second map then induces a surjection on mod 2 homology. Both $\Omega^2 \Sigma^2 \mathbb{R}P^\infty$ and $BO$ split as $\mathbb{R} P^\infty$ times their 1-connected cover, so the induced map of 1-connected covers $\tau_{\geq 2}(\Omega^2 \Sigma^2 \mathbb{R}P^\infty) \to \tau_{\geq 2}(BO) = BSO$ also induces a surjection on mod 2 homology.
Now let $X = \tau_{\geq 2}(\Omega^2 \Sigma^2 \mathbb{R}P^\infty)$ with the map to $BSO$ constructed above. Take 1-connected covers of the double loop maps above, Thomify, 2-localize, and use the Hopkins-Mahowald theorem to get maps of $E_2$ ring spectra
$$H \mathbb{Z} _{(2)} \to MX_{(2)} \to MSO_{(2)}.$$
(See e.g. section 3 of this paper.)
We can view $MX_{(2)} \to MSO_{(2)}$ as a map of $H\mathbb{Z}_{(2)}$-module spectra, and hence $MX/2 \to MSO/2$ as a map of $H\mathbb{F}_2$-module spectra. The induced map $H_*(MX/2;\mathbb{F}_2) \to H_*(MSO/2;\mathbb{F}_2)$ is still surjective (it looks like two copies of $H_*(X;\mathbb{F}_2) \to H_*(BSO;\mathbb{F}_2))$, and inherits the structure of a module map over the mod 2 dual Steenrod algebra $\mathcal{A}^\vee = H_*(H\mathbb{F}_2;\mathbb{F}_2)$. Both modules are free, because any $H\mathbb{F}_2$-module spectrum splits as a wedge of suspensions of $H\mathbb{F}_2$. In fact the Hurewicz homomorphism $\pi_*(MX/2) \to H_*(MX/2;\mathbb{F}_2)$ induces an isomorphism
$$\mathcal{A}^\vee \otimes \pi_*(MX/2) \to H_*(MX/2;\mathbb{F}_2),$$
and similarly for $MSO$. Therefore the map $\pi_*(MX/2) \to \pi_*(MSO/2)$ may be identified with the map obtained by applying $\mathbb{F}_2 \otimes_{\mathcal{A}^\vee} (-)$ to the map on homology, showing that the induced map $\pi_*(MX/2) \to \pi_*(MSO/2)$ is also surjective. Now any 2-torsion class $x \in \pi_d(MSO)$ comes from $\pi_{d+1}(MSO/2)$, hence from $\pi_{d+1}(MX/2)$ and in particular from $\pi_d(MX)$. $\Box$
|
2025-03-21T14:48:31.454803
| 2020-07-09T10:39:11 |
365218
|
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|
Stack Exchange
|
Probability that random high dimensional vectors are all on the convex hull
Say I pick $n$ i.i.d. random standard normal points in $\mathbb{R}^d$. Roughly, as long as $n$ is much smaller than exponential in $d$, with high probability all points will be on the convex hull. This is because with high probability they will all be near the radius $\sqrt{d}$ sphere and all almost orthogonal, and thus each point is the furthest in its own direction from the origin. Let $p(n,d)$ be the failure probability that at least one point is in the interior of the convex hull.
Question: What's the best upper bound on $p(n,d)$ as a function of $n$ and $d$? I care most about the regime $d \gg 1$, $n \in O(\operatorname{poly}(d))$.
I think that by "all points will be on the convex hull" you mean, that no $X_i \in co(X_1,\ldots,X_n)^0$, when $X_1,\ldots,X_n$ are the generated random $X_i$, It seems that the condition $X_i \not\in ext(co(X_1,\ldots,X_n))$ gives the same probabilities.
Yes, I was glossing over the boundary because it is measure 0.
It's not too bad to see that the probability is at most $2n^2 e^{-d/2e}$. Let $x_1,\ldots,x_n$ be the points. We will use a union bound, so it is sufficient to examine the probability that $x_1$ is in the convex hull of $x_2,\ldots,x_n$. This happens if and only if there are $\lambda_j \in [0,1]$ with $\sum \lambda_j = 1$ and $$x_1 = \sum_{j = 2}^n \lambda_j x_j\,.$$
Take an inner product with $x_1$ to see that this implies $$\| x_1 \|_2^2 = \sum_{j = 2}^n \lambda_j \langle x_1, x_j \rangle.$$
Thus $$P(x_1 \in \mathrm{conv}(x_2,\ldots,x_n)) \leq P( \|x_1\|^2 \leq \max_{j \geq 2} |\langle x_1, x_j \rangle|).$$
If we divide by $\|x_1\|$, the RHS probability bound becomes
$$P\left(\|x_1\| \le \max_{j \ge 2} \left|\left<x_1/\|x_1\|, x_j \right>\right|\right).$$
$\|x_1\|^2 \sim \chi^2_d$ and $\left<x_1/\|x_1\|, x_j\right> \sim N(0,1)$, so from $\chi^2_d$ and $N(0,1)$ tail bounds we have
\begin{align*}
P(\|x_1\| \le t\sqrt{d}) &\le \left(t e^{(1-t^2)/2}\right)^d \\
P\left(\left<x_1/\|x_1\|, x_j\right> \ge t\sqrt{d}\right) &\le \frac{\left(e^{-t^2/2}\right)^d}{t\sqrt{2\pi d}}
\end{align*}
for any $t \in (0,1)$. Matching the base of the exponents gives
\begin{align*}
t e^{(1-t^2)/2} &= e^{-t^2/2} \\
t &= e^{-1/2} \approx 0.606531
\end{align*}
whence union bounding shows
\begin{align*}P(x_1 \in \mathrm{conv}(x_2,\ldots,x_n))
&\le (n-1) \left(1 + \frac{1}{\sqrt{2\pi d/e}}\right) e^{-d/2e} \\
&< 2ne^{-d/2e}
\end{align*}
and so
$$P(\exists~j \text{ s.t. }x_j \in \mathrm{conv}(x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_n)) < 2n^2 e^{-d/2e}.$$
I do not know if this is optimal, but it's worth noting that it's basically the strategy you suggested. When $n$ is exponentially large in $d$ the probability does not tend to $0$ provided the exponent is big enough, which is where this bound breaks.
Thanks, yes this the argument I had in mind, but nicely expressed. I’m interesting in knowing $c$; if you don’t feel like writing that out okay if I edit to fill in later?
Yeah absolutely feel free to edit it in.
I think the exponent can be tightened from $e^{-d/2e}$ to $2^{-d/2}$, but don't have time to complete the proof.
|
2025-03-21T14:48:31.455055
| 2020-07-09T11:04:01 |
365219
|
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|
Stack Exchange
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Is $C(\mathbb{R}^n)$ is a DF-Space?
I recently have begun reading about DF-spaces and its clear to me that $C(K)$ is a DF-space for any compact subset (non-empty) $K$ of some $\mathbb{R}^D$ for finite D, since $C(K)$ is Banach. However, what if we relace $K$ by all of $\mathbb{R}^D$ itself? Is $C(\mathbb{R}^D)$ a DF-space when it is equipped with the usual local-uniform convergence topology?
No. With the topology of uniform convergence on compact sets, $C(\mathbb R^d)$ is a Fréchet space (for the sequence of semi-norms $p_n(f)=\sup\{|f(x)|: |x|\le n\}$) and the intersection of the classes of Fréchet and DF-spaces consists precisely of all Banach spaces.
Here is a proof of this well-known fact: By definition, a DF-space has a fundamental sequence of bounded sets $B_n\subseteq B_{n+1}$, i.e., every bounded subset is contained in some $B_n$. On the other hand, if $p_n$ is an increasing sequence of semi-norms describing the topology of the Fréchet space and $1/c_n=\sup\{p_n(b): b\in B_n\}+1$, then you can check that $B=\bigcup_{n\in\mathbb N} c_n B_n$ is a bounded set such that each bounded set is contained in a multiple of $B$. Then the closed absolutely convex hull of $B$ is a bounded $0$-neighbourhood and its Minkowski functional is a norm giving the topology of the space.
Do you know of a reference where they show that Fréchet + DF = Banach?
I'm far away from my book shelf, but this should be, e.g., in Jarchow's book or Köthe's and certainly also in Bonet and Perez Carreras.
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2025-03-21T14:48:31.455191
| 2020-07-09T11:47:26 |
365222
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Stack Exchange
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Looking for a family of random variables such that only the second clause is fulfilled
Working with the epsilon-delta-criterium, a family $(X_i)_{i \in I}$ on $(\Omega,A,P)$ is uniformly integrable if
i) $sup_{i \in I} E(X_i) <\infty$
ii) $\forall \epsilon>0$ ex. $\delta>0$ s.t. $P(A)<\delta \Rightarrow \int_{A} |X_i|dP < \epsilon $
I easily came up with an example where i) is fulfilled, but ii) isn't ($X_n=1_{[0,\frac{1}{n}]} \cdot n$).
But I am looking for an example where ii) is fulfilled but i) isn't.
I would be really happy if someone had an answer because it would help me understand why we need (i).
What are you taking the supremum of in (i)? Anyway, any example that satisfies (i) can easily be modified not to do so without breaking (ii). This is not research level.
Let $\Omega$ be a single point having probability 1. Then (ii) is trivially satisfied because for any $\delta < 1$, the only $A$ with $P(A) < \delta$ is $A=\emptyset$. So now you can choose any family which violates (i), e.g. $X_n = n$.
I believe that if $\Omega$ is atomless, then (ii) implies (i).
Thank you @NateEldredge for the good example!
Thanks @NateEldredge: Let Ω be a single point having probability 1. Then (ii) is trivially satisfied because for any δ<1, the only A with P(A)<δ is A=∅. So now you can choose any family which violates (i), e.g. Xn=n.
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