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2025-03-21T14:48:31.293877
| 2020-06-16T21:09:05 |
363275
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"Jianrong Li",
"https://mathoverflow.net/users/11877",
"https://mathoverflow.net/users/36721"
],
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}
|
Stack Exchange
|
Estimate an expression about probability about Bernoulli random variables
Given $v_{ij} \in \{0,1\}$, $i \in \{1,2\}$, $j \in \{1,2,\ldots,n\}$. Let $X_1, X_2, \ldots, X_n$ be random variables, $P[X_i=1]=P[X_i=0]=1/2$, $i \in \{1,\ldots, n\}$. By checking many examples, I think that the following is true: when $|v_1|, |v_2|$ are large enough,
\begin{align}
\frac{1}{2} \sum_{x_1=0}^n \sum_{x_2=0}^n | P[ \sum_{r=1}^n v_{1r}X_r = x_1, \sum_{r=1}^n v_{2r}X_r = x_2 ] - P[ \sum_{r=1}^n v_{1r}X_r = x_1+1, \sum_{r=1}^n v_{2r}X_r = x_2+1 ] | < 1-\epsilon.
\end{align}
Is there some method to prove this? Thank you very much.
For example, let $n=4$, and $v_{1}=(v_{11},\ldots,v_{14})=(1,1,1,0), v_2=(v_{21},\ldots,v_{24})=(0,0,1,1)$, then the following Python codes gives the result $0.3446349999999999$.
import numpy as np
N = 10**5
XX = [np.random.randint(2, size=4) for n in np.arange(N)]
r=0
for x1 in range(0,5):
for x2 in range(0,5):
P1 = list(map(lambda X: (X[0]+X[1]+X[2]==x1)&(X[2]+X[3]==x2), XX))
P2 = list(map(lambda X: (X[0]+X[1]+X[2]==x1+1)&(X[2]+X[3]==x2+1), XX))
r=r+abs(np.mean(P1)-np.mean(P2))
r/2
Let $n=4$, and $v_{1}=(v_{11},\ldots,v_{14})=(0,1,1,1), v_2=(v_{21},\ldots,v_{24})=(1,1,1,1)$, then the following Python codes gives the result $0.31340999999999997$.
import numpy as np
N = 10**5
XX = [np.random.randint(2, size=4) for n in np.arange(N)]
r=0
for x1 in range(0,5):
for x2 in range(0,5):
P1 = list(map(lambda X: (X[1]+X[2]+X[3]==x1)&(X[0]+X[1]+X[2]+X[3]==x2), XX))
P2 = list(map(lambda X: (X[1]+X[2]+X[3]==x1+1)&(X[0]+X[1]+X[2]+X[3]==x2+1), XX))
r=r+abs(np.mean(P1)-np.mean(P2))
r/2
Are the $X_i$'s independent?
@Iosif Pinelis, thank you very much! Yes, the $X_i$'s are independent.
We shall assume that the $X_i$'s are independent. The problem can be restated as follows: show that for some $h\in(0,1)$, all natural $n$, and all subsets $J$ and $K$ of the set $[n]:=\{1,\dots,n\}$
we have
\begin{equation*}
S:=\sum_{x,y}|P(X_J=x,X_K=y)-P(X_J=x+1,X_K=y+1)|\le2-h, \tag{1}
\end{equation*}
where $X_J:=\sum_{i\in J}X_i$ and the sum in (1) is over all all integers $x,y$. Write
\begin{equation*}
S\le T+U, \tag{2}
\end{equation*}
where
\begin{equation*}
T:=\sum_{x,y}|P(X_J=x,X_K=y)-P(X_J=x+1,X_K=y)|,
\end{equation*}
\begin{align*}
U&:=\sum_{x,y}|P(X_J=x+1,X_K=y)-P(X_J=x+1,X_K=y+1)| \\
&=\sum_{x,y}|P(X_J=x,X_K=y)-P(X_J=x,X_K=y+1)|.
\end{align*}
By the independence of the $X_i$'s,
\begin{equation*}
P(X_J=x,X_K=y)=\sum_z P(X_{J\cap K}=z)P(X_{J\setminus K}=x-z)P(X_{K\setminus J}=y-z).
\end{equation*}
Hence,
\begin{align*}
T&\le\sum_zP(X_{J\cap K}=z)\,\sum_y P(X_{K\setminus J}=y-z) \\
&\times\sum_x|P(X_{J\setminus K}=x-z)-P(X_{J\setminus K}=x+1-z)| \\
&=\sum_x|P(X_{J\setminus K}=x)-P(X_{J\setminus K}=x+1)|=:D_{|J\setminus K|},
\end{align*}
where $|\cdot|$ denotes the cardinality.
Similarly, $U\le D_{|K\setminus J|}$, so that, by (2)
\begin{equation*}
S\le D_{|J\setminus K|}+D_{|K\setminus J|}. \tag{3}
\end{equation*}
Note that $D_0=1$ and, by this answer, for $k\ge1$ we have
\begin{align*}
D_k=\frac1{2^k}\,\Big(2\binom k{m+1}-1\Big)\le\frac34,
\end{align*}
where $m:=\lfloor (k-1)/2\rfloor$. So, by (3),
\begin{equation*}
S\le 1+3/4=7/4
\end{equation*}
if $J\ne K$.
In the remaining case when $J=K$,
\begin{equation*}
S=\sum_x|P(X_J=x)-P(X_J=x+1)|=D_{|J|}\le1.
\end{equation*}
Thus, in all cases (1) holds with $h=2-\max(7/4,1)=1/4>0$, as desired.
thank you very much for your great answer! I am trying to understand every step of your proof. In the formula before (3), it is said that $\sum_z P(X_{J \cap K}=z) \sum_y P(X_{K\backslash J}=y-z)=1$. I am trying to understand this step. I think that $$\sum_z P(X_{J \cap K}=z) \sum_y P(X_{K\backslash J}=y-z) \=\sum_z P(X_{J \cap K}=z) \sum_y P(X_{K\backslash J}=y) = \sum_{y,z} P(X_{J \cap K}=z) P(X_{K\backslash J}=y) \=\sum_{y,z}P(X_{K}=y+z)=\sum_y P(X_{K}=y)=1$$. Is my understanding correct? I checked that when $k=2$, $D_k=3/4$. So maybe for $k \ge 1$, $D_k \ge 3/4$?
sorry, I mean when $k \ge 1$, $D_k \le 3/4$.
@JianrongLi : Oops! I did indeed miss $3/4$; this is now corrected -- thank you for pointing this out. As for your detalization of the multi-line display before (3), what I had in mind was something like this, but a bit simpler: just using the fact that $\sum\limits_t P(X_M=t)=1$.
thank you very much!
|
2025-03-21T14:48:31.294357
| 2020-06-16T21:14:07 |
363277
|
{
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"M. Winter",
"Steven Stadnicki",
"https://mathoverflow.net/users/108884",
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],
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"sort": "votes",
"url": "https://mathoverflow.net/questions/363277"
}
|
Stack Exchange
|
How dense can a transitive sets of points be?
How dense can a finite set of points on the $d$-dimensional unit sphere be if I require that the symmetry group of that arrangement is still transitive on the points?
As a measure for density I use the radius of the largest spherical cap not containing any point in its interior.
For $d=1$ we can get arbitrarily dense. For $d=2$ I suppose the densest set is some orbit of the icosahedral group. Is there something known about general $d$? E.g. is there any $d>1$ for which one can get arbitrarily dense again?
The Archimedean polyhedra are the place to start. I'm not actually sure that the convex hull of such a point set has to be Archimedean, but it's certainly plausible that it would.
@StevenStadnicki I expected something like this. I wonder if there is anything denser than these point sets.
|
2025-03-21T14:48:31.294472
| 2020-06-16T21:17:34 |
363279
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlos Esparza",
"Emil Jeřábek",
"Ian Agol",
"Jeremy Rickard",
"LSpice",
"Peter LeFanu Lumsdaine",
"YCor",
"https://mathoverflow.net/users/123448",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/1345",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/2273",
"https://mathoverflow.net/users/22989",
"https://mathoverflow.net/users/2383"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363279"
}
|
Stack Exchange
|
If $A$, $B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for all abelian groups $G$, must $A$ and $B$ be isomorphic?
$\DeclareMathOperator\Hom{Hom}$The question is in the title. If the isomorphism $\Hom(A, G) \cong \Hom(B, G)$ is natural in $G$ then this is just the Yoneda Lemma. If $A$ and $B$ are finitely generated this is also true by the structure theorem.
However this sounds like it should be false in general, else it would imply by Yoneda that if $\Hom(A, -)$ and $\Hom(B, -)$ are (a priori not naturally) isomorphic, then they are also isomorphic in a natural way (though possibly by a different set of isomorphisms).
The question of course immediately generalizes to $R$-modules.
Edit: Some context (that isn't really relevant for the question)
I'm interested in this question in light of the universal coefficient theorem for Cohomology. A positive answer would imply that knowing all Cohomology groups of a space, with arbitrary coefficients, would already determine its Homology (although I think it is conceivable that this topological statement can be proven in a different way).
I 'migrated' this question from math.stackexchange (I deleted the question there) since it had already been asked
If $A=C_2^{(\aleph_0)}$ and $B=C_2^{(\aleph_1)}$, the these are isomorphic for every $G$ iff $\alpha^{\aleph_0}=\alpha^{\aleph_1}$ for all cardinal $\alpha$. This is false under CH (taking $\alpha=2$) but I don't know if it's consistent in general (=in ZFC). In case that yields a counterexample, one might restrict to $A,B$ countable, or assume GCH.
@YCor I think Eric Wofsey commented on mathSE (on a deleted answer to the linked question) that there always is a cardinal $\alpha$ such that $\alpha^\kappa \neq \alpha^\lambda$ for $\kappa \neq \lambda$
Ah, thanks indeed, I should have checked before. So the question has a positive answer when $A$ is an elementary abelian $p$-group for some prime $p$ (i.e., if for every $B$ the question has a positive answer for this given $A$); we can say that $A$ is "recognizable".
If $G=U(1)$ is a topological group, then this follows from Pontryagin duality (with $G=S^1=U(1)$, and the discrete topology on $A$ and $B$), where one remembers the topology on $Hom(A,G)$.
https://en.wikipedia.org/wiki/Pontryagin_duality#The_Pontryagin_duality_theorem
@IanAgol very interesting... although I suppose that since $S^1$ has noncontinuous group automorphisms it is not possible to get the topology of $\mathrm{Hom}(A, G)$ just from the group structure.
@CarlosEsparza that’s right, the topology on Hom(A,G) is not unique. Somehow this is dual to the fact that Hom(Z,A) determines A.
I just stumbled across the answer to this in Fuchs' 2015 book on Abelian Groups.
The papers
Hill, Paul, Two problems of Fuchs concerning tor and hom, J. Algebra 19, 379-383 (1971). ZBL0228.20027.
and a 1974 paper of Sebel'din that I've not been able to find, and seems to be in Russian, which means that I probably wouldn't be able to understand it even if I did:
Homomorphism groups of complete direct sums of torsion-free abelian
groups of rank 1 [Russian]. Tomsk. Gos. Univ., Tomsk 1, 121–122 (1974)
give counterexamples where the groups $A$ and $B$ are $p$-groups and torsion-free groups respectively.
But from Fuchs' account, Sebel'din's example is fairly simple.
Let $S=\mathbb{Z}^{(\omega)}$ be the direct sum of countably many copies of $\mathbb{Z}$, and let $A=S\oplus\mathbb{Q}$ and $B=S\oplus\mathbb{Q}\oplus\mathbb{Q}$.
Then $A\not\cong B$, but $\operatorname{Hom}(A,G)\cong\operatorname{Hom}(B,G)$ for all abelian groups $G$.
It's a fun exercise to verify this, and for me it was an ISHTOT moment when I did, so I won't spoil it for you.
Google tells me Ishtot is some sort of Indian spirit. It does not know it as an acronym. Could you clarify?
@EmilJeřábek If it's an Indian spirit, then that makes it much better! I meant "I Should Have Thought Of That".
Oh, I see. Thank you.
@EmilJeřábek, you missed a golden opportunity to say "Oh, ISHTOT".
Is this (spoiler alert) the right idea?
@CarlosEsparza I don’t think Hom(Q,G) is necessarily Div(G) if G is not torsion-free. Consider that e.g., every automorphism of $G=\mathbb Z_{p^\infty}$ lifts to a homomorphism $\mathbb Q\to\mathbb Z_{p^\infty}$.
@CarlosEsparza As Emil says, $\operatorname{Hom}(\mathbb{Q},G)\not\cong\operatorname{Div}(G)$ if $\operatorname{Div}(G)$ has torsion, since $\operatorname{Hom}(\mathbb{Q},G)$ is always a vector space over $\mathbb{Q}$. But the important point here is that if $\operatorname{Div}(G)$ has torsion then $\operatorname{Hom}(\mathbb{Q},G)$ is an infinite dimensional vector space.
I should have also pointed out that if $G$ has torsion, then $\mathrm{Div}(G)$ may not be divisible. E.g., if $G$ is a $p$-group, then $\mathrm{Div}(G)$ is the Ulm subgroup $U^1(G)$, which is not divisible if $G$ has Ulm length at least $2$.
@EmilJeřábek True. Although that can easily be fixed by taking $\operatorname{Div}(G)$ to be the largest divisible subgroup of $G$ rather than the group of divisible elements.
The paper of Sebel’din is available in full text here: А. М. Сбельдин, Группы гомоморфизмов вполне разложимых абелевых групп без кручения, Известия высших учебных заведений Математика, 1973, номер 7, 77–84
|
2025-03-21T14:48:31.294868
| 2020-06-16T21:36:05 |
363280
|
{
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"LSpice",
"Marouani",
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"url": "https://mathoverflow.net/questions/363280"
}
|
Stack Exchange
|
Relationship between volume and area
Let $\mu(z) dV_n$ be a measure in $\mathbb{C} ^n$.
Let $B_n(r) := \{z \mid \|z\| < r\}$ be the ball of radius $r$ in $\mathbb C^n$, and $\partial B_n(r) $ be the corresponding sphere.
In $\mathbb{C} $ how can we find the following inequality?
$$
\operatorname{Vol}_{\mu}(B_1(r))=\int_{B_1(r)} \mu(z) dV_1(z)=
\int_0^r\left(\int_{\partial B_1(t)} \mu dz\right)dt\geq \int_0^r \left[\int_{\partial B_1(t)}(\mu)^{ \frac{1}{2}} \right]^2\frac{1}{2\pi t} dt
$$
And can we generalize this inequality in $\mathbb {C}^n$?
I tried to proofread this (for example, don't mix text and math in an equation; write $\operatorname{Vol}(B)$ $\operatorname{Vol}(B)$, not Vol$(B)$ Vol$(B)$), but I'm not sure I got everything right. For example, you asked whether you could generate the inequality in $\mathbb C^n$, and I changed that to 'generalize'. Please feel free to revert or re-edit if I got it wrong.
Please don't self-vandalize your own post. You can delete it if you like, as long as it has no answer (and also undelete it later).
You say "Let $\mu(z)dV$ is a measure on $\mathbb C^n$". So I guess $\mu$ is a function (namely, the density of your measure w.r.t Lebesgue). Q: What does it then mean to say "Let $B_\mu(r)$ be a ball in $\mathbb C^n$ ?". More precisely, $B_\mu(r) = ???$
$Vol(B_\mu(r)) $ is the volume of the ball with respect to the measure $\mu dV$
$B(r) ={z\in\mathbb {C^n} ;|z|<r} $
This still doesn't make sense to me. What is subscripted set $B_{\mathbf{\mu}}(r) = ???$ which appears in your integrals ?
You can forget the index but if we measure it is with respect to $\mu$
OK, I've updated you question to more tightly reflect what you seem to intend.
See my post below for a complete solution to the general problem. It's based on the co-area formula + Jensen's inequality. The particular case of $\mathbb C$ (i.e $n=1$) is a consequence of the elementary fact that a circle of radius $t$ has length (i.e Hausdorff measure) equal to $2\pi t$.
The problem can be solved via co-area formula and Jensen's inequality. We will do it Bourbaki style, i.e from $n$-dimensional case to particular case $n=1$.
Instead of $\mathbb C^n$, we can equivalently see the problem as a problem in $\mathbb R^m$, where $m=2n$ (i.e we isomorphically map real dimensions for each complex dimension). So, let $dV_m$ denote volume measure in $\mathbb R^m$ and $dS_{m-1}$ denote the corresponding surface area measure, i.e $(m-1)$-dimensional Hausdorff measure. The mapping $F: z \mapsto \|z\|$ on $\mathbb R^m$ has jacobian determinant $1$ (except at $z=0$, where it is undefined). Also note that for all $t \ge 0$, we have
$$
F^{-1}(\{t\})=\{z \in \mathbb R^m \mid F(z) = t\} = \{z \in \mathbb R^m \mid \|z\| = t\} = \partial B_m(t).
$$
By the coarea-formula (see Corollary 1.4, for example), we have
$$
\begin{split}
\int_{B_m(r)}\mu(z)dV_m(z) &= \int_{0}^r\left(\int_{F^{-1}(\{t\})}\frac{\mu(z)}{|Jac_F(z)|}dS_{m-1}(z)\right)dt\\
&= \int_{0}^r\left(\int_{\partial B_m(t)}\mu(z)dS_{n-1}(z)\right)dt\\
&= \int_{0}^r\left(\int_{\partial B_m(t)}\frac{\mu(z)}{S_{m-1}(\partial B_m(t))}dS_{m-1}(z)\right)S_{n-1}(\partial B_m(t))dt\\
&\ge\int_{0}^r\left(\int_{\partial B_m(t)}\frac{\mu(z)^{1/2}}{S_{m-1}(\partial B_m(t))}dS_{m-1}(z)\right)^2S_{m-1}(\partial B_m(t))dt\\
&= \int_{0}^r\left(\int_{\partial B_m(t)}\mu(z)^{1/2}dS_{m-1}(z)\right)^2\frac{1}{S_{m-1}(\partial B_m(t))}dt,
\end{split}
$$
where the inequality is an applicaiton of Jensen's inequality on the convex function $x \mapsto x^2$ and the probability measure $A \mapsto S_{m-1}(A \cap \partial B(t))/S_{m-1}(\partial B(t))$.
In particular, if $n=2$, we have $m=2\cdot 1 = 2$, $dS_{m-1} = dS_1$ which is the arc-length measure, and so $S_1(\partial B(t)) =$ length or circle of radius $t$, which equals $2\pi t$.
So there is no square!
The data in your question are still not clear. See my comments in the question above.
Ok, the situation is clearer now. My answer above solves both your particular (case $n=1$) and general problems (cases $n > 1$).
Thank you for the answer
OK, cool. If it anwsers your question, don't forgot to upvote and accept it. The is a button to the left for that :)
Let us continue this discussion in chat.
I'm sorry, I don't know how I can upvote and accept the answer
At the top left side of the the answer, click on the check mark to accept. There is also an up arrow (to upvote the answer), and a down arrow (to downvote).
|
2025-03-21T14:48:31.295148
| 2020-06-16T21:40:43 |
363281
|
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"authors": [
"Nils Matthes",
"https://mathoverflow.net/users/14349"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363281"
}
|
Stack Exchange
|
Does fibrewise vanishing of cohomology imply its vanishing (non-proper case)?
Let $f: X\rightarrow S$ be a quasi-compact, quasi-separated, flat morphism of schemes, with $S$ locally Noetherian. Also, fix an integer $i\geq 1$.
Question 1: Does $H^i(X_s,\mathcal{O}_{X_s})=0$, for all $s\in S$, imply that $R^if_\ast\mathcal{O}_X=0$?
The answer is yes, if $f$ is also assumed to be proper (flat base change+theorem of formal functions+easy direct proof for $S$ Artinian local). In general, I would expect that there are some simple counterexamples but, to my surprise, I didn't find anything in the literature - probably I simply looked in the wrong places! I should also add that I would be particularly interested in counterexamples in characteristic zero, i.e. where $f$ is a morphism of schemes over a field of characteristic zero.
Here is a variant of question 1:
Question 2: Assume in addition that $f$ has a section $e: S\rightarrow X$. What is the answer to question 1 then?
The "raison d'être" for asking question 2 in addition is that a colleague explained to me how, granting existence of a section, the assumptions above ensure that $H^0(X_s,\mathcal{O}_{X_s})=\kappa(s)$, for all $s\in S$, implies that $f_\ast\mathcal{O}_X=\mathcal{O}_S$. So, at least in this very special case, the fibres of $f$ "remember" what the global sections of $f$ are.
Here is one simple counterexample:
Example. Let $X = \mathbf A^2 \setminus 0$ and $S = \mathbf A^1$, where $f \colon X \to S$ is the first coordinate projection. The fibres are affine, so have no higher coherent cohomology.
But if $V \subseteq S$ is affine open containing $0$, then $f^{-1}(V)$ has an affine open cover by $U_1 = V \times \mathbf G_m$ and $U_2 = (V \setminus 0) \times \mathbf A^1$. Then the function $\tfrac{1}{xy}$ on $U_{12}$ is not a sum of functions on $U_1$ and $U_2$, so it gives a nonzero class in $(R^1f_* \mathcal O_X)_0$. $\square$
(Note that this map has many sections.)
Great, that's as simple as I was hoping for! Many thanks.
|
2025-03-21T14:48:31.295312
| 2020-06-16T22:53:15 |
363285
|
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"Christian Remling",
"Giorgio Metafune",
"Jochen Glueck",
"Paul",
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|
Stack Exchange
|
Integrability of a function under a condition on its Fourier transform
Let $f\in L^1(\mathbb{R})$ and continuous on $\mathbb{R}$ such that its Fourier transform $\hat f$ equals zero in a neighborhood of zero.
Let $F$ be function such that $\hat F$ exists and
$$\hat f(x) =x\hat F(x),\quad \forall x\in \mathbb{R}$$
Prove that $F\in L^1(\mathbb{R})$.
Any hints on how to prove that?
I already asked this question on MSE, I hope it will have a chance here
Could you say more about where this question/problem arose, and why it is supposed to have a positive answer? That is: does this come from a paper you are reading? A course you are taking? etc
it is a question of a professor without indications. If F is a primitive of f (F '= f) we see that $ \hat f (x) = i x \hat F (x), \forall x \in \mathbb {R} $
What precisely does "such that $\hat F$ exists" mean?
As for your previous question I changed the title to make it a bit more informative, and I changed the format of the link. May I kindly suggest for future questions that you try to (i) choose titles that contain a bit more concrete information about the question (phrases such as "Prove that" don't really add much information to the title) and (ii) use urls within links instead of writing down the entire url in plain text?
@Jochen Glueck thanks for the formatting (title and link)
For your question, I am a little embarrassed because the question does not specify the meaning. I believe in the sence of moderate distributions. If the problem is badly posed I will wait seven september to see the professor.
The assumption about $\widehat{F}$ existing is unnecessary. We could define $F(x)=\int_{-\infty}^x f(t), dt \in C_0$ and take its distributional FT. In fact, it's also immediately (from $\widehat{f}=x\widehat{F}$) that $F\in L^2$, so we don't really need distributions.
We have $\widehat{f}=0$ near zero, so we can write $\widehat{F}=\widehat{g}\widehat{f}$ with a function $\widehat{g}\in C^{\infty}$, $\widehat{g}(x)=1/x$ for $|x|\ge a>0$.
Since $\widehat{g}'' = -(t^2 g)\,\widehat{}\in L^1$, we have $|g(t)|\lesssim 1/t^2$. Moreover, $g\in L^2\subseteq L^1_{\textrm{loc}}$. So $g\in L^1$, and thus also $F=g*f\in L^1$, since $f\in L^1$ by assumption.
We wrote the same proof!
@ Christian Remling
Thank you very very very much
Assume that $\hat f$ vanishes in $[-2a,2a]$ and take $\phi$ odd, vanishing in $[-a,a]$ and equal to $1/x$ if $|x| \ge 2a$. Then $\hat F=\phi \hat f$ and $F=\psi*f$ where $\psi$ is the inverse Fourier transform of $\phi$
$$\psi (\xi)=2 \int_a^\infty \phi(x)\sin (\xi x)\, dx.$$ Then $\psi$ is bounded for $|\xi| \le 1$ and, integrating by parts twice, decays at least as $1/\xi^2$ at infinity. This shows that $\psi \in L^1$ and $F \in L^1$, too.
@ Giorgio Metafune Thank's so much
@ Giorgio Metafune According to my teacher, we can only express a fourier transform or its inverse by an integral if the function is in L ^ 1, here the function is in L ^ 2, the inversion formula $\psi (\xi)=\int_0^\infty \phi(x)e^{(i\xi x)}, dx.=2 \int_a^\infty \phi(x)\sin (\xi x), dx.$ is not justified
If $\phi_R(x)=\phi \chi_{(0,R)}$, then $\phi_R \to \phi$ and $\hat{\phi_R} \to \hat{\phi}$ in $L^2$. But $\hat{\phi_R}=\int_0^R \phi (x) \sin(\xi x)dx \to \psi$ pointwise.
thank's so much
|
2025-03-21T14:48:31.295673
| 2020-06-16T23:07:38 |
363286
|
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|
Stack Exchange
|
Bass-Serre theory textbook
I am a PhD freshman working on topological graph theory and geometric group theory. I would like to learn some Bass-Serre theory. What do you think is the best introductory textbook in this topic? Thank you in advance for your recommendations.
Serre’s book and Scott-Wall’s survey remain by far the best sources.
I like the beginning of Dunwoody and Dicks book. They might be a little too brief for a first introduction but they have a number of nice applications and I love they prove the universal covering tree is a tree using homology rather than normal forms.
As mentioned by Andy Putman in the comments, the classical (and probably the best) references are Serre's book Trees and Scott and Wall's paper Topological methods in group theory.
Serre's approach is elementary and essentially self-contained, based on combinatorial arguments. Scott and Wall's approach is based on algebraic topology (in particular, they use covering theory). If you are already familiar with some algebraic topology, then you may find Scott and Wall's approach more intuitive, more visual. Personally, I began by reading Serre's book, but I prefer the point of view of graphs of spaces.
You can also take a look at Henry Wilton's blog Geometric group theory. There are several blogposts dedicated to the Bass-Serre theory. In complement to Serre's book, an interesting complement is Higgins' article The fundamental groupoid of a graph of groups. Looking at the fundamental groupoid instead of the fundamental group makes some arguments more natural.
Depending on how far you want to go, there is a very short and very well written note by Wenyuan Yang here : http://bicmr.pku.edu.cn/~wyang/ggt/BassSerre.pdf
You can find the construction of the Bass-Serre tree for Amalgamated sums and HNN extensions and some related theorems.
"Trees" by JP Serre.
As it's been quoted several times, let me mention that his book "Arbres amalgames $\mathrm{SL}_2$", in French, was published in 1977 by Astérisque, and later (1980) translated into English and published as "Trees" at Springer.
|
2025-03-21T14:48:31.295854
| 2020-06-16T23:49:58 |
363288
|
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|
Stack Exchange
|
Integral Milnor-Moore theorem
Given a field K of char. zero the theorem of Milnor Moore
states that taking the enveloping hopf algebra defines an embedding
$\mathcal{U} $ from Lie algebras over K into hopf algebras over K.
Taking the enveloping hopf algebra is left adjoint to taking the primitive elements
$\mathcal{P}$ and so the Milnor-Moore theorem says that for any Lie algebra $L$ over K the unit $L \to \mathcal{P}\mathcal{U}(L)$ is an equivalence.
If a field K is of positive char., the theorem of Milnor-Moore remains valid if
one replaces Lie algebras by restricted Lie algebras over K, and the enveloping hopf algebra by the
enveloping restricted hopf algebra:
Taking the enveloping restricted hopf algebra defines an embedding
$\mathcal{U}_\mathrm{res} $ from restricted Lie algebras over K into hopf algebras over K.
$\mathcal{U}_\mathrm{res} $ is left adjoint to taking the primitive elements
$\mathcal{P}$ (that carry a canonical restricted Lie algebra structure).
So similarly the Milnor-Moore theorem says that for any restricted Lie algebra $L$ over K the unit $L \to \mathcal{P}\mathcal{U}_\mathrm{res}(L)$ is an equivalence.
By a theorem of Fresse equipping a Lie algebra over K with the structure of a restricted Lie algebra over K is the same as equipping it with divided powers.
So one could interprete the Milnor-Moore theorem as saying that there is an embedding
from Lie algebras with divided powers over K into hopf algebras over K
left adjoint to taking primitive elements (with their canonical structure of a Lie algebra with divided powers).
Divided power Lie algebras make sense over any commutative ring R
and one can ask if there is an embedding from Lie algebras with divided powers over R into hopf algebras over R left adjoint to taking primitive elements. Especially I would like to know this for R the integers. Is this true and does anyone know a reference?
Hadrian, did you ever work this out? I've been wondering lately if there's a Milnor-Moore over the sphere spectrum or some localization.
@Jonathan Beardslay. I never worked this out, mainly since I could not understand the divided power version of Koszul duality good enough to control the unit of the envelope-primitives adjunction. I heard there is an analogous Milnor-Moore theorem in chromatic homotopy theory. Here one can of course omit divided powers by Tate vanishing.
|
2025-03-21T14:48:31.296010
| 2020-06-17T01:49:25 |
363292
|
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"Andrej Bauer",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
What is the groupoid cardinality of the category of vector spaces over a finite field?
For any groupoid, it's groupoid cardinality is the sum of the reciprocals of the automorphism groups over the isomorphism classes. Let us consider the category of vector spaces over a finite field $\mathbb F_q$ with only invertible morphisms allowed.
Then, the size of the automorphism groups are $g_n= \prod_{i=1}^n(q^n-q^{i-1}) = q^{n\choose 2}\prod_{i=1}^n(q^i-1)$ so it's groupoid cardinality is the following infinite sum:
$$\sum_{n \geq 1}\frac{1}{g_n}.$$
Is there a closed form expression for it? Note that as $q\to 1$ in an appropriate sense, the general linear group is supposed to converge to the symmetric group and the groupoid of vector spaces should converge to the groupoid of finite sets.
In this sense, the above infinite sum is an analog of the usual one for $e$ and can perhaps reasonably be called a q-analog of $e$.
Alternatively, in the case of abelian, (even semi-simple) categories, is there a better replacement for the groupoid cardinality that I should be using?
Is the sum $q$-hypergeometric, i.e., the ratio $g_{n+1}/g_n$ is a rational function of $q^n$? If it is then you can apply the $q$-Gosper algorithm.
Upon substituting $x=\frac{1}{q}$ we obtain
$$\sum_{n\geq 0}\frac{1}{|\mathrm{GL}_n(\mathbb F_q)|}=\sum_{n\geq 0}\frac{x^{n^2}}{(1-x)(1-x^2)\cdots(1-x^n)}$$
and this evaluates to the product $\prod_{i\geq 1}\frac{1}{(1-x^{5i-4})(1-x^{5i-1})}$ by the first Rogers-Ramanujan identity.
You can also interpret the second Rogers-Ramanujan identity as giving a groupoid cardinality for the category of affine spaces over $\mathbb F_q$. where the sum becomes $\sum_{n\geq 0}\frac{1}{|\mathrm{AGL}_n(\mathbb F_q)|}$. (Note that I wrote these sums to contain a term for $n=0$, corresponding to the zero dimensional vector/affine space.)
I don't think that there is a good explanation for this evaluation that avoids proving the R-R identities first. In particular it is unclear how to directly relate the exponents of the infinite product side with the category of vector/affine spaces.
Sorry, let me repost that comment with correct links. I wonder if there's a way to do it using cylindric plane partitions as in https://arxiv.org/abs/1209.1807 and https://doi.org/10.1090/proc/13373. Cylindric plane partitions are known to be related to affine crystals (https://arxiv.org/abs/math/0702062).
@SamHopkins That sounds very cool. I would love to see if there is a connection there.
WOW! I know, such comments are discouraged, but - WOW!
I can't help seconding @მამუკაჯიბლაძე 's sentiments. I suppose the following is a standard question about the Rogers-Ramanujan identities (which are new to me), but: Why 5?
|
2025-03-21T14:48:31.296208
| 2020-06-17T02:16:42 |
363293
|
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|
Stack Exchange
|
Is the bound of Spencer on discrepancy tight?
Suppose $\mathcal{S} = \{S_1,\ldots,S_m\}$ is a set system in $[n]=\{1,\ldots,n\}$, which means that for each $i$, $S_i\subset [n].$ Define the discrepancy of $\mathcal{S}$ by
$$disc(\mathcal{S})=\min_{\chi}\max_{i=1,\ldots,m}|\chi(S_i)|,$$
where $\chi:[n]\to \{-1,+1\}$ is a coloring and for a subset $S$ of $[n$], denote $\chi(S)=\sum_{j\in S}\chi(j)$. Spencer proved that for all set system $\mathcal{S} = \{S_1,\ldots,S_m\}$ with $m\geq n$,
$$disc(\mathcal{S})=O(\sqrt{n\ln(2m/n})$$
in "Six Standard Deviations Suffice".
Is the result tight? In other words, whether there exsits a set system $\mathcal{S} = \{S_1,\ldots,S_m\}$ such that $disc(\mathcal{S})=\Omega(\sqrt{n\ln(2m/n})$?
Some books mentioned there exists such a set system only for $m=n$. Like "The Probabilistic Method" by Alon and Spencer, "Geometric Discrepancy" by Matousek. I cannot find a proof for general $m>n$.
Yes, the bound is tight, and here is a proof sketch. In fact, we will show that a random sequence of subsets $\mathcal{S} = \{S_1, S_2, \dots, S_m\}$ provides the lower bound up to constants. Precisely, sample each $S_i$ just by including each element of $[n]$ in the set with probability $1/2$.
We assume $n, m$ are large. Fix a coloring $\chi: [n] \to \{-1, 1\}$. Note that the distribution of $\chi(S_i)$ is approximately normal with variance $\Theta(\sqrt{n})$. Therefore, $\mathrm{Pr}_{S_i}[\chi(S_i) \le c\sqrt{n\log(2m/n)}] \le 1 - \frac{n}{m}$ for a small constant $c$. Therefore, $$\Pr_\mathcal{S}[\mathrm{disc}(\mathcal{S}) \le c\sqrt{n\log(2m/n)}] \le \left(1-\frac{n}{m}\right)^m < \exp(-n).$$ This suffices to union bound over all $2^n$ colorings $\chi$.
|
2025-03-21T14:48:31.296353
| 2020-06-17T02:36:34 |
363294
|
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|
Stack Exchange
|
Bounding the probability that two binomials are equal
Note: This question was migrated from this earlier post, where it initially appeared. Following suggestions, I moved this into its own question.
Let $B_{n,p}$ denote the usual binomial random variable (i.e., the probability that it equals $k$ is given by ${n \choose k} p^k (1-p)^{n-k}$). I would like some reference (or proof) for the following:
For all integers $0 \leq t < n$ and all $0 < p < 1$, we have $t \cdot \mathbb{P}(B_{n+t,p} = B_{n-t,p}) \leq \dfrac{100}{p}$, where the variables $B_{n+t,p}$ and $B_{n-t,p}$ are independent binomials.
I believe this could be done as follows, but I didn’t work it out fully...
Let $f(t)$ denote the quantity in question (thinking of $n$ and $p$ as fixed). I imagine that $f(t)$ is unimodal with a maximum taken at some relatively small $t$. If we replace the difference of Binomials with their normal approximations, this suggests the maximum should occur when $t = C \sqrt{np(1-p)} /p$. And values of $t$ this small, it is easy to see the desired bound on $f(t)$ simply because that probability is always at most $C/\sqrt{np(1-p)}$.
So we’d just need to argue that if $t \geq C \sqrt{np(1-p)}/p$, then $f(t)$ is decreasing [note that if convenient, we can safely throw in an extra constant here without any concern].
I can also prove a weaker bound involving some extra $\log(np(1-p))$ factor (there about) via a naive approach that wastefully bounds the probability in question using some concentration results such as Bernstein’s inequality. But that’s not the way to go about it, and it gives us the wrong answer.
Hoping for a nice argument or (perhaps better?) a reference.
Thanks!
Added remark: I’m really just asking about the probability that two independent binomials $B_{m,p}$ and $B_{k,p}$ are equal, so one might reasonably hope this is already known.
You are dividing by $t$ to obtain the upper bound on probability but $100/p$ is already huge, so something is wrong in your right hand side. Is it 100 p/t?
No, this is what I want. It’s immediate unless $t$ is sort of just the right size. Consider for instance $t=f(n) \sqrt{np(1-p)}/p$. If $f(n)$ is bounded, then we’re fine simply because the probability is always at most $C/\sqrt{np(1-p)}$ (anticoncentration of the binomial). On the other hand, if $f(n)$ is too big (e.g., if $f(n) \geq C *\sqrt{\log(np)}$ [and assuming $np(1-p) \geq 10$ or something]), then by Bernstein’s inequality the probability is at most $C / np$, so we’re also fine. So! We really only care about $100 \leq t p/\sqrt{np(1-p)} \leq 100 \sqrt{\log(np)}$.
Isn't it just the central limit theorem?
A local limit theorem is all fine and good, but as I recall the error terms would be bigger than I can tolerate. I could be wrong though, and I’m certainly not an expert in the existing local limit theorems.
$$\mathbb{P}(B_{n+t,p}=B_{n-t,p})=\sum_{k=0}^{n-t}\frac{(n+t)!(n-t)!}{(n+t-k)!(n-t-k)!(k!)^2}p^{2k}(1-p)^{2n-2k}. $$ We write $a_k$ the terms in this sum. We have $$ \frac{a_{k+1}}{a_k}=\frac{p^2(n+t-k)(n-t-k)}{(1-p)^2(k+1)^2}$$ The formal function $f(k)=\frac{a_{k+1}}{a_k}$ is decreasing in $k$. And there is a $k^*$ such that $f(k^*)\approx 1$. It corresponds to a $a_{k^*}=\max_k a_k$ .There we have $f'(k^*)< - \frac{1}{k^*}$. Therefore for $k$ not too far from $k^*$ $$ a_k = a_{k^*}\prod_{k^*\leq i < k}f(i)\approx a_{k^*}\prod_{k^*\leq i < k} (1+(i-k^{*})f'(k^*)) \leq a_{k^*} \exp(-\sum_{l=0}^{k-k^*}\frac{l}{k^*})\approx a_{k^*}e^{-\frac{(k-k^*)^2}{2k^*}}$$ So one should get $$\sum_{k}a_k \leq C\sqrt{k^*} a_{k^*}.$$Moreover using the TCL we have
$$a_{k^*}= \mathbb{P}(B_{n+t,p}=B_{n-t,p}=k^*) \approx \frac{1}{\sqrt{\sigma^2}}e^{-\frac{1}{\sigma^2}(k^*-(n+t)p)^2)}\times \frac{1}{\sqrt{\sigma^2}}e^{-\frac{1}{\sigma^2}(k^*-(n-t)p)^2}\leq \frac{1}{\sigma^2} e^{-\frac{2(tp)^2}{\sigma^2}}$$ where $\sigma^2 = Cnp(1-p)\approx k^*$. Finally $$pt\sqrt{k^*}a_{k^*}\leq \frac{pt}{\sigma} e^{-\frac{2(tp)^2}{\sigma^2}}\leq \sup_{x\geq0} x e^{-2x^2}\leq C $$
Do you have a reference for the local limit theorem you’re using? In general, I’d be concerned about the error terms involved.
This is probably not helpful, but for $p=1/2$ the sum can be evaluated to give $2^{-2n}\binom{2n}{n-t}$.
|
2025-03-21T14:48:31.296642
| 2020-06-17T03:45:54 |
363297
|
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|
Stack Exchange
|
How to pronounce "⊗¬-category"?
In his PhD thesis, Categorical Structure of Continuation Passing Style, Thielecke studies the ⊗¬-categories, which are premonoidal categories with structure (namely, a functor ¬ which is adjoint to itself), in order to model the semantics of continuation-passing style languages.
I've been studying his work for some time now, but I'm not really familiar to literature about category theory. So my question is, and it might sound silly: how should I pronounce "⊗¬-category"? In fact, I could rephrase this: if I were to write about those categories in a paper, how should I refer to them? Should one write it in full, or simply just refer to them as ⊗¬-categories all the time and leave the pronunciation as an open problem?
A version of this problem was solved by Victor Borge. https://www.youtube.com/watch?v=Qf_TDuhk3No
@AndreasBlass, I would like to write a comment but, I cannot stop laughing...
When $\lnot$ is an identity-on-objects functor, I've seen something called (monoidal) categories with converse; so perhaps you could refer to these as monoidal categories with negatiion? Also, @AndreasBlass: bahahahahahaah
If you're planning to write a paper, there's no advantage in trying to write out the pronunciation: you may just use the symbols. If I was trying to pronounce the name, I would say "tensor-not category" or similar, analogously to "dagger category" for †-category or "star algebra" for *-algebra.
@varkor I actually find it incredibly distracting to have symbols that I cannot pronounce in my head while reading…
How about "TN-category" (for "tensor-negation")?
I shared an office with Hayo Thielecke in the late 1990s.
The pronunciation he used was "Tensor-NOT-category".
I'm still working on my paper on the formalization of the CPS-calculus, I'll take note of that. Thank you very much!
|
2025-03-21T14:48:31.296799
| 2020-06-17T03:55:54 |
363298
|
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|
Stack Exchange
|
Some questions about the contruction of center stable manifolds for cubic NLKG by Lyapunov-Perron method
In Nakanish & Schlag: Invariant manifolds and Dispersive Hamiltonian Evolution Equations,, on theorem 3.22, they use Lyapunov-Perron methods to conctruct center stable manifolds for focusing cubic nonlinear Klein-Gordon equation
\begin{equation}
\partial_{tt} u -\Delta u+ u -u^3=0, \quad (x,t) \in \mathbb{R}^3 \times \mathbb{R}
\end{equation}
here there are some details of proof I cannot figure out. More precisely, the theorem includes the following result (and we are now considering only in radial case, i.e. the following functions are all radial):
If the solution $(u,\dot u)$ remains into $B_\nu(Q,0)$ for all time $t\ge0$, where $B_\nu (Q,0) \subset \mathbb{H}= H^1 \times L^2$ is a $\nu$ ($\nu \ll 1$) ball with center $(Q,0)$, and $Q$ is the correponding ground state, then they claim that for any forward time $t$, $u(t)$ lies entirely in center stable manifolds (here we do not present the concrete construction of the center stable manifolds), and it suffices to check that
\begin{align*}
&\quad \; \lVert (\lambda,\gamma) \rVert_{X_T}
\\
&= \lVert \lambda\rVert_{L^1(0,T)} + \lVert \lambda \rVert_{L^\infty(T,\infty)} + \lVert \gamma \rVert_{S_t(0,T)} + \lVert (\gamma,\dot \gamma) \rVert_{L^\infty((T,\infty);H^1 \times L^2)} \lesssim \nu,
\tag{1}
\end{align*}
where $u=Q+v=Q+ \lambda(t) \rho + \gamma(t,x)$, and $(\gamma,\rho)_{L^2}=0$, $S_t(0,T)= L_t^2 L_x^6 \cap L_t^2 H_x^1$, and $\rho$ is the eigenvector of unique negative eigenvaule of linearized operator $L_+=-\Delta+1 -3Q^2$. And from the construction of center stable manifolds, we can see that
\begin{equation}
\lambda(t)
=e^{-kt} \left( \lambda(0) + \frac{1}{2k} \int_0^\infty e^{-ks} N_\rho(v)(s) ds \right) + \frac{1}{2k}\int_0^\infty e^{-k|t-s|} N_\rho (v) (s) ds ,
\end{equation}
\begin{equation}
\gamma(t)= \cos (wt) \gamma(0) + \frac{\sin (wt)} {w} \dot{\gamma}(0 )+ \int_0^t \frac{\sin(t-s) w}{w} N_c(v)(s) ds,
\end{equation}
where $w^2=P^+ \circ L_+$, $N_\rho(v)= (\rho,N(v))_{L^2}$, $N_c=P_+ (N(v))$, here $N(v)=3Qv^2+v^3$ and $P^+$ is the spectral projection of $L^+$ onto $(1,\infty)$ (since we can see that $\sigma(L_+)=(1,\infty) \cup \{-k^2\}$ ). And with the properties of $\gamma$, for the second equation above, we have Strichartz estimates
\begin{equation}
\lVert \gamma \rVert_{S_t} \lesssim \lVert \gamma(0) \rVert_{H^1} + \lVert \dot \gamma (0) \rVert _{L^2} + \lVert N_c(v) \rVert_{L_t^1 L_x^2}.
\end{equation}
My question is simple, and is involved in the following two estimates: with the preliminaries before, we want to check $(1)$. And they claim the following two inequlities:
\begin{align*}
\lVert \lambda \rVert _{L^1(0,T) \cap L^\infty(T,\infty)}
&\lesssim \lVert N_\rho\rVert_{L^1(0,T) \cap L^\infty(T,\infty)} \\
&\lesssim \nu+\nu(\lVert \lambda \rVert _{L^1(0,T) \cap L^\infty(T,\infty)}+\lVert \gamma \rVert_{S_t(0,T)} + \lVert \gamma \rVert_{L_t^\infty((0,\infty);H^1 \times L^2)}),
\end{align*}
and
\begin{equation}
\lVert \gamma \rVert_{S_t(0,T)} \lesssim \nu + \nu(\lVert \lambda \rVert_{L^1(0,T)+ \lVert \gamma \rVert_{S_t(0,T)} }),
\end{equation}
uniformly in $T >0$. But how can we check the estimates above, since I always get the higher order of $\lVert \gamma (t)\rVert_{S_t}$ and $\lVert \lambda \rVert_{L^1(0,T)}$ on the right hand side and I can not thrrow them away, which makes me confused.
Some more questions: By orthogonality between $\rho$ and $\gamma$ in $L^2$, it is obvious that $\lVert (\lambda,\dot \lambda) \rVert_{L^\infty} \lesssim \nu$ and $\lVert (\gamma, \dot \gamma) \rVert _{H^1 \times L^2} \lesssim \nu$, I also wonder why they add $L^\infty(T,\infty)$ of $\lambda$ and $L_t^\infty H_x^1(T,\infty)$ of $\gamma$ in $\lVert \cdot \rVert_{X_T}$?
|
2025-03-21T14:48:31.297024
| 2020-06-17T05:08:10 |
363300
|
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|
Stack Exchange
|
Explanations simple enough that non-mathematical audiences can understand
The following (debatable) quote is attributed to Einstein:
"You do not really understand something unless you can explain it to your grandmother."
I feel very enlightened when there is a simple explanation of an important idea in mathematics. Below are some of my favorite ones.
My Question: Are there other explanations like this?
(Credit to my analysis professor, many years ago): The geometric series $\sum_{n=1}^{\infty} \frac{1}{2^n} = 1$ can be explained as follows: take a disc. Cut it in half. Now take half of the disc, and cut that in half. Repeat this process. Then we have $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} +...$ disc. But we started out with a whole disc, so the total is a single disc!
Definitions, etc.
By "explanation," I am asking for a proof or a heuristic argument which is simple (in the sense defined below).
Ideally, the fact is a central part of some subfield of mathematics. (For instance, the geometric series is certainly an important example in calculus since it is the main idea behind many comparison tests. It is also the most basic example of a series that can be computed explicitly (besides perhaps telescoping series). ) The facts themselves must also be simple.
By "simple", I mean you can explain it to someone without any mathematical background (say a child under 10 years old). In particular, words like "derivative," "group," and "Riemann curvature tensor," are considered to be "too hard," but expressions like "speed/velocity," "symmetry" and "how much a surface/curve curves" are acceptable. (In this regard, words from elementary physics (e.g. Newtonian mechanics, electromagnetism, wave mechanics) are great, but quantum mechanics and relativity are too hard. Notions from middle/high school (e.g. Euclidean geometry) are great too.)
Simple pictures are okay too, although the picture cannot be too complicated. (Again, the main criteria here is that your average non-mathematical audience can understand it.)
i don't understand your examples. The first one is a proof that the sum of $2^{-n}$ is $1$, while the other two are interpretations (which give no mathematical insight)
@erz Thank you. I edited my question.
Related Math.SE question: Visually stunning math concepts which are easy to explain. I think the only simple explanations (in your definition) are those who have a visual representation. Most 10 year olds don't even know what a square root is.
From the Gale-Shapley paper on stable marriages: https://gilkalai.files.wordpress.com/2020/03/tyi.png
The first "proof" seems to put a lot of faith in that "dot dot dot". What if I started off by taking a square, keeping half of it, keeping 2/3 of what's left, keeping 3/4 of what's left, and so on?
I’m voting to close this question because I think it is based on a false premises. "Proofs that anyone can understand" are more often than not "handwaving that could equally well justify claims that are false"
Pythagoras theorem.
Albert Einstein wrote about two pivotal moments in his childhood. The first involved a compass that his father showed him when he was four or five. The second involved his early exposure to Euclidean plane geometry. He was impressed by the idea that a mathematical assertion could “be proved with such certainty that any doubt appeared to be out of the question”.
Steven Strogatz discusses a breathtakingly simple proof of the Pythagorean theorem whose provenance is traced to Einstein as a child. "Though we cannot be sure the following proof is Einstein’s, anyone who knows his work will recognize the lion by his claw."
Einstein's first proof.
The proof relies on the insight that a right triangle can be decomposed into two smaller copies of itself. That’s a peculiarity of right triangles. If you try instead, for example, to decompose an equilateral triangle into two smaller equilateral triangles, you’ll find that you can’t. So Einstein’s proof reveals why the Pythagorean theorem applies only to right triangles: they’re the only kind made up of smaller copies of themselves.
The New Yorker article on cutting a right-angled triangle into two smaller triangles says, "Their similarity becomes clear if you imagine picking them up, rotating them, and arranging them like so, with their hypotenuses on the top and their right angles on the lower left". However, the two little triangles need to be reflected to directly correspond in similarity to the larger triangle (unless picking up and putting down does this).
|
2025-03-21T14:48:31.297355
| 2020-06-17T05:20:17 |
363302
|
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|
Stack Exchange
|
Reductive Lie groups and existence of maximal compact subgroup
I am reading Knapp's book "Lie groups beyond an introduction" (2nd edition). I am struggling to understand the following point. Recall that $G$ is a reductive Lie group. If the Lie algebra $\mathfrak g$ of $G$ is reductive and equipped with a involution and a nondegenerate symmetric bilinear form $B$ on $\mathfrak g$ which is $\theta$-invariant and $Ad$-invariant and the following hold.
We have a decomposition $\mathfrak g=\mathfrak k\oplus \mathfrak p$ with respect to the eigenspaces of $\theta.$
$G$ has a compact subgroup with Lie algebra $\mathfrak k.$
The map $(k,\exp X)\mapsto k\exp X$ from $K\times \exp\mathfrak p$ to $G$ is a diffeomorphism.
The bilinear form $B_\theta (X,Y):=-B(X,\theta Y)$ is positive definite on $\mathfrak g.$
Every automorphism $Adg$ for $g\in G$ is inner.
From this Knapp concludes that $K$ has to be a maximal compact subgroup. he argues as the following (Page 446). Let $K$ is properly contained in a compact subgroup $K_1$. Take $k_1\in K_1\setminus K.$ Then for some $k\in K$ and $X\in\mathfrak p$, we have $k\exp X=k_1$ which implies that $\exp X\in K_1.$ Since $(\exp X)^n=\exp (nX)$ is in $K_1$ the sequence $\exp(nX)$ must have a convergent subsequence. I get easily to this point. Now Knapp says that this contradicts 3.! I do not understand this. Can someone help me out?
In the topological space $K\times \exp {\mathfrak p}$ the sequence $a_n=(1,nX)$ has no convergent subsequence.
@MikhailBorovoi. Can you please explain a little more? $\mathfrak p$ has the topology of an Euclidean space and thus $nX$ must be a divergent sequence for $X\neq 0.$ But the diffeomorphism is given from $K\times \exp p$. I do not know if the exponential map is a homeomorphism on $\mathfrak p.$
Actually it is a diffeomorphism $K\times {\frak p}\to G$ (this is what Knapp means). And then you should use the definition of the product topology in $K\times {\frak p}$.
"I do not know if the exponential map is a homeomorphism on $\frak p$". Yes, it is a diffeomorphism. You should use the definition of $\frak p$.
@MikhailBorovoi. Thank you very much.
@MikhailBorovoi. One more thing. Can you suggest me some god books where I can find integration theory on semisimple Lie groups using KAK and KAN type decompositions? Helgason does not have this. Knapp has this for KAN only.
Unfortunately, no, I cannot. I do algebra only and know nothing about integration. But you can ask this in a separate question on MO.
|
2025-03-21T14:48:31.297662
| 2020-06-18T14:25:57 |
363437
|
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|
Stack Exchange
|
What is the Borel complexity of this set?
Problem. What is the Borel complexity of the set
$$c(\mathbb Q)=\{(x_n)_{n\in\omega}\in\mathbb R^\omega:\exists\lim_{n\to\infty}x_n\in\mathbb Q\}$$
in the countable product of lines $\mathbb R^\omega$?
Remark. It is easy to see that $c(\mathbb Q)$ is an $F_{\sigma\delta\sigma}$-set in $\mathbb R^\omega$.
Question. Is $c(\mathbb Q)$ of type $F_{\sigma\delta}$ in $\mathbb R^\omega$?
Maybe a naive point, but: It'd be enough to continuously reduce an arbitrary $\Sigma^0_4$ set in $\omega^\omega$ to $c(\mathbb{Q})$. With such an argument you can show the toy version that $c({0})$ is $F_{\sigma \delta}$ but "no simpler" than that, so far I wasn't able to do yours though. (I won't say complete because I'm not familiar with how good a Polish space this is, though I suspect very.)
Your set is an intersection of $G_{\delta \sigma}$ set and an $F_{\sigma \delta}$ set, by stating that you get arbitrarily close to some rational number arbitrarily late, and that a limit exists. I'm guessing that means my approach will not work.
@VilleSalo Good point! Thank you. By the way, my idea of proving that $c(\mathbb Q)$ is an $F_{\sigma\delta}$ does not work, unfortunately. So, the problem remains open.
I think I can reduce any $\Sigma^0_3$ and thus the answer is "no", but not 100% sure and it's complicated. I'll try to find some time if no one answers before.
Ok, I am curious to see this reduction.
The argument is some standard recursion theoretic trickery (I'm educated enough to recognize that, but not educated enough to know what this is called).
Theorem. Every $\Sigma^0_3$ subset of Cantor space continuously reduces to your set.
It follows from this that if your set were $F_{\sigma \delta}$ then every $\Sigma^0_3$ subset of Cantor space would be $\Pi^0_3$, a contradiction with the properness of the Borel hierarchy. I think.
Proof. In front of you stand $\omega$ many creatures. Every second, some subset of these creatures blinks. You have set up video surveillance and have every blink of every creature on tape. By watching these tapes in a dovetailing fashion, you eventually find out about every blink. The event that one of these creatures blinks infinitely many times models (= is an intuitive interpretation of) a general $\Sigma^0_3$ subset of Cantor space: such a set is of the form $\bigcup_n \bigcap_m \bigcup_k C_{n,m,k}$ where $C_{n,m,k}$ is clopen, and a point being in $C_{n,m,k}$ means the time between the $(m-1)$th and the $m$th blink of creature $n$ is $k$ seconds.
Finding a continuous function means we keep watching the tapes, and keep outputting real numbers (actually it's enough to output approximations to the real numbers, but I'll just output exact numbers directly).
We'll keep track of the following data: An irrational number $r \in (0,1)$, and rational numbers $1 = q_0 > q_1 > q_2 > q_3 > \cdots > r$, such that $[r, q_i]$ is disjoint from $\frac{1}{i} \mathbb{Z}$. Note that for any $r \in (0,1)$ we can find such numbers. These should also be indexed by the current time, but I'll keep that implicit.
Now, keep watching the tapes, and do as follows
every second, output $r$ (i.e. $x_t$ will be the value $r$ output at time $t$)
if creature $i$ blinks, increase $r$ to some irrational number between $(r + q_i)/2$ and $q_i$, and pick new values for the numbers $q_j$ for $j > i$ if needed
I claim that the resulting sequence $(x_t)_t$ is in $c(\mathbb{Q})$ if and only if some creature blinks infinitely many times. Namely, suppose that the $i$th creature blinks infinitely many times, but no creature with smaller index does. Then after some time $t$, no creature $j < i$ blinks. At that point, we have picked some value $q_i \in \mathbb{Q}$, and since the $i$th creature keeps blinking, we eventually increase $r$ to $q_i$ by the second rule, thus $(x_t)_t \in c(\mathbb{Q})$.
Suppose then that after some time, no creature ever blinks again. Then from that point on, we keep outputting $r$, which is irrational, so $(x_t)_t \notin c(\mathbb{Q})$.
Suppose then that no creature blinks infinitely many times, but there are infinitely many blinks altogether. Since the sequence is monotone increasing and bounded from above (by $q_0 = 1$), there is some limit $\lim_t x_t = r'$. Suppose it is rational, say $r' = n/i$. Consider the last time the $i$th creature blinks. At this time, the value $r$ satisfies $r < q_i$, and $[r, q_i]$ is disjoint from $\frac{1}{i} \mathbb{Z}$. But the way the process is defined, $r$ never increases past $q_i$ unless the $i$th creature blinks, so $r' \in [r, q_i]$ cannot be equal to $n/i \in \frac{1}{i} \mathbb{Z}$. Square.
I think the actual complexity of your set is hinted by the fact it is an intersection of a $F_{\sigma \delta}$ and a $G_{\delta \sigma}$ set, but that's more fine-grained than I'm willing to touch right now, since I'm not exactly a boldface expert.
edit
Scratch that, I will touch it a little.
Theorem. Every set that is an intersection of a $\Sigma^0_3$ and a $\Pi^0_3$ subset of Cantor space continuously reduces to your set.
I imagine this pinpoints the exact complexity, since as mentioned your set is an intersection of a $F_{\sigma \delta}$ and a $G_{\delta \sigma}$ set.
Proof. We now have a $\Sigma^0_3$ set $A$ and a $\Pi^0_3$ set $B$, and are interested in reducing $A \cap B$ to your set $c(\mathbb{Q})$. For $A$ we have the set of $\omega$ many magnificent creatures from the previous proof, which occasionally blink. For $B$ we have another $\omega$ many cute creatures, which occationally wag their tails. We interpret $A \cap B$ as the event that some magnificent creature blinks infinitely many times, and no cute creature wags its tail infinitely many times.
In the even coordinates of the output, keep writing (the current value of) $r$ according to the protocol of the previous proof. In the odd coordinates, by default also write $r$, but write $r+\frac{1}{2^i}$ whenever the $i$th creature wags its tail.
If the point of Cantor space is in $B$, then eventually the first $n$ cute creatures stop wagging their tails, and thus the value eventually stays in $[r-\frac{1}{2^n}, r+\frac{1}{2^n}]$ where $r$ is where the monotone sequence converges. If it is additionally in $A$, then $r \in \mathbb{Q}$ so $(x_t)_t \in c(\mathbb{Q})$.
If on the other hand the point is not in $B$ then there is no convergence because we keep seeing consecutive values $(r, r+\frac{1}{2^i})$ for any $i$ that never stops wagging its tail, so $(x_t)_t \notin c(\mathbb{Q})$. If the point is in $B$ but not in $A$, then we have convergence to some element of $\mathbb{R} \setminus \mathbb{Q}$, and again $(x_t)_t \notin c(\mathbb{Q})$. Square.$\;\;\;\;\;$
Cute argument (with all these blinking creatures). Thank you.
|
2025-03-21T14:48:31.298082
| 2020-06-18T14:38:31 |
363439
|
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|
Stack Exchange
|
Conditions on continuity under Lebesgue measure
Let $h : X \times I \rightarrow \mathbb{R}$ be a continuous function, where $X$ is a compact set of $\mathbb{R}^k$, for some $k$.
Set $\hat{h}(x,t) = 1$ if $h(x,t) \neq 0$, $0$ otherwise.
Define $g : I \rightarrow \mathbb{R}$ by
$g(t) = \int_X \hat{h}(x,t) d\mu$, where $\mu$ is Lebesgue measure on $X$.
Under what conditions can we assert that $g$ is continuous?
Leo, thank you for your effort in helping me find a solution. I think
you missed part of the problem, though. The function $h$ definitely
depends on both variables: $h(x,t)$. My question pertains to
$\hat{h}$; not to $h$. $\hat{h}$ is defined in terms of $h$ as:
$\hat{h}(x,t) = 1$ if $h(x,t) \neq 0$, $0$ otherwise.
If it helps clarity, fix $t$ at any value and set
$ A_t = \{x \in X \mid h(x, t) \neq 0\} $.
Then an alternate definition of $g$ is, $g(t) = \mu(A_t)$.
The question then is, "Is $g$ continuous as a function of $t$?"
Choose any $t_0 \in I$. It seems intuitive, since $h$ is continuous,
that in some geometric sense we have
(I) $ A_t \rightarrow A_{t_0}$, and consequently that
(II) $\mu(A_t) \rightarrow \mu(A_{t_0})$,
which thus means that $g(t) \rightarrow g(t_0)$, in turn implying that
$g$ is continuous.
The difficulty lies in capturing the geometric notion, (I), "measure-theoretically" so that we can assert (II).
if $h(t,x)=h(t)$ does not depend on $x$ and takes on the valued $0$ and is non trivial then $g(t)=1$ if $h(t)\neq 0$ and $g(t)=0$ otherwise, so is is a discontinuous function. I really don't see how the dependence on $x$ would change anything, so essentially $g$ must be identically zero or never vanish (well, of course you can construct ad-hoc counterexamples to this rough idea, but if you're going after a general statement then it seems hopeless to me...)
You want the continuity of an integral depending on a parameter $g(s):=\int_S f(x,s)d\mu(s)$. The standard hypotheses on $f$ (“Carathéodory”) are those that allow to apply the Dominated Convergence Thm. This is no research topic, so the question should be better placed on math.stackexchange.com.
We can equivalently define $\hat h$ as the indicator function of an open set $U\subset X\times I$, so $g$ is lower semi-continuous.
Set $A$ the boundary of $U$ and $V=(X\times I)\setminus U$. $U,A,V$ is a partition of $X\times I$, with $U,V$ open and $A$ closed. Set $g_V(t)=\int_X1_V(x,t)dx$, and $g_A=\int_X1_A(x,t)dx$. We have $g+g_A+g_V=1$, $g_U$ and $g_V$ are lower semi-continuous and $g_A$ is upper semi-continuous.
If $g_A$ is continuous, then $g+g_V=1-g_A$ is continuous, so $g$ and $g_V$ are continuous.
If $g_A$ is not continuous, then at least one of the other functions is discontinuous.
A case in which we can easily prove that $g_A$ is continuous is be the case $g_A=0$, that is "the trace of the boundary of $U$ on $X\times \{t\}$ is always negligible for the 1-dimensional Lesbesgue measure".
|
2025-03-21T14:48:31.298311
| 2020-06-18T14:38:59 |
363440
|
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|
Stack Exchange
|
On a transcendental number defined as a variation involving the Lambert $W$ function in the nested square root representation of the golden ratio
Define the real number $\xi$ satisfying
$$\xi=\sqrt{1+W\left(1+\sqrt{1+W(1+\ldots)}\right)}\tag{1}$$
where $W(x)$ denotes the main branch of the Lambert $W$ function, as reference I add that Wikipedia has the article with title Lambert $W$ function. Then $(1)$ is similar to the well-known representation for the golden ratio $(1+\sqrt{5})/2$, where were the Lambert $W$ function appears in an alternating way instead of the corresponding nested square root, see the subsection 4.4 Alternative forms from the Wikipedia Golden ratio, or [1] in Spanish). From this we conclude the following easy statement.
Claim. One has that $\xi\approx 1.3918$ satisfies the identity $$e^{\xi^2-1}=\frac{1}{\xi-1},\tag{2}$$ and thus $\xi$ is transcendental.
Proof sketch. As usual we deduce $\xi^2-1=W(1+\xi)$ from $(1)$, and calculating with the inverse $W^{-1}(x)$ we conclude $(2)$. By contradiction we prove the transcendence of $\xi$, as an application of Lindemann-Wierstrass theorem (see reference [2]).$\square$
I don't know if my claim is in the literature or if next questions are in the literature, if this post have a good mathematical content and is on topic, please feel free to refer the literature answering my questions as a reference request that I'm going to search and read those statements from the literature.
Question 1. The encyclopedia Wolfram MathWorld have an article with title e Continued Fraction. My purpose to write the post was to know how a mathematician tries to take advantage from a simple claim (previous) to deduce more related and advanced statements. Can you show any continued fraction representation related to our real number $\xi$ (as you see the examples of Wolfram MathWorld for $e$ are diverse/varied)? Many thanks
I wanted to ask previous question as an invitation and with the purpose to learn what can be a good/interesting continued fraction or issue concerning continued fractions, associated to our real number $\xi$. Next question also is in this spirit, I don't know if it can be made some work about it.
Question 2. I would like to know/determine some (wider) class of functions $f(x)$ for which I can to repeat my experiment/claim and state that
the real number $\xi_f$ defined as $\xi_f=\sqrt{1+f\left(1+\sqrt{1+f(1+\ldots)}\right)}$ will be a real transcendental number. Many thanks.
I mean with this second question a case study of what conditions are required for the real functions $f(x)$ with the purpose to get an extension of my Claim (if this is in the literature feel free to answer this second question as a reference request). I'm asking what work can be done for an extension of my claim (thus an example of a wider and suitable class of functions $f(x)$). Isn't required to find nice-closed forms, just decide what is a suitable class of functions for which I can to prove transcendence.
References:
[1] Samuel G. Moreno and Esther M. García Caballero, Uno, dos y $\ldots\,$¡$\phi$!, Miniaturas matemáticas de La Gaceta de la RSME, La Gaceta de la Real Sociedad Matemática Española, Vol. 20 (2017), Núm. 1, Pág. 170.
[2] The section Numbers proven to be transcendental from the Wikipedia Transcendental number.
Feel free to add in comments your feedback about the post, many thanks and good day.
Many thanks for your attention and edit @Glorfindel
Just for fun, note that it also works for:
$$\xi_i=\sqrt{-1+W\left(-1+\sqrt{-1+W(-1+\ldots)}\right)}\tag{1}$$
which yields the complex number $\xi_i =0.430438...+ 0.96479388...i$,
that satisfies the identity:
$$e^{\xi_i^2+1}=\frac{\xi_i-1}{\xi_i^2+1}$$
Many thanks your expressions/formulas always have a great mathematical beauty @Agno
A quick thought for your Question 2.
If the limit for $\xi_f$ exists, then as you noted we'll get $\xi_f ^2 - 1 = f(\xi_f +1)$. If we define $g(x) = f(x+1) - x^2 +1 = f(x+1) - (x+1)^2 + 2(x+1)$, then we have $x^2 - 1 = f(x+1)$ iff $g(x) = 0$.
So your second question more or less reduces to saying ``we have a function $g$, and we would like to know when the equation $g(x) = 0$ forces $x$ to be transcendental."
Or perhaps a bit more to the point would be to define $h(x) = \sqrt{1+f(x)}$. Then you are wondering about fixed points of $h$, and you're hoping there's only one.
I'm afraid that question at the moment is too broad to really sink your teeth into (or to make much progress) since it seems to be the same as asking "which fixed points are transcendental." But I certainly like the creativity of the idea!
You are incredible, many thanks! I'm going to study your answer for Question 2. About your last paragraph I'm not a professional mathematician and I'm asking the post with the idea about prospecting if it is possible to state something interesting for Question 1 and Question 2. Many thanks again for your answer and words!
Glad to help, and thanks for the cool question. Even if you're "not a professional mathematician," don't let that stop you from thinking about cool stuff like this! :-)
|
2025-03-21T14:48:31.298641
| 2020-06-18T15:57:58 |
363446
|
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|
Stack Exchange
|
Is the symplectomorphism group of a compact manifold a tame Fréchet Lie subgroup of $\operatorname{Diff}(X)$?
In the famous paper Hamilton, Richard S. The inverse function theorem of Nash and Moser.
Bull. Amer. Math. Soc. (N.S.) 7 (1982), no. 1, 65–222, Hamilton introduced the category of tame Fréchet Lie groups and showed that for a compact symplectic manifold $(X,\omega)$, the identity component $\operatorname{Diff}_0(X)$ is a tame Fréchet Lie group whose action on the space of symplectic forms isotopic to $\omega$ is tame. Then he asked (Problem 2.5.5) whether the stabilizer $\operatorname{Diff}_\omega(X)\cap\operatorname{Diff}_0(X)$ is a tame Lie subgroup. I have two questions:
Since the symplectomorphism group $\operatorname{Diff}_\omega(X)$ is modelled on the subspace of closed 1-forms (using Weinstein's trick), is it true that it is itself a tame Fréchet Lie group? This may be well known, but I was not able to find references.
Is the answer to Hamilton's Problem 2.5.5 known? Again, any reference would be appreciated.
|
2025-03-21T14:48:31.298749
| 2020-06-18T16:48:55 |
363455
|
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"Benjamin Steinberg",
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|
Stack Exchange
|
Geometric content of area of a word in geometric group theory?
Where does the idea of 'area' come from in Geometric Group Theory? The wikipedia article states that this definition was 'inspired' from Riemannian geometry:
Gromov's proof was in large part informed by analogy with filling area functions for compact Riemannian manifolds where the area of a minimal surface bounding a null-homotopic closed curve is bounded in terms of the length of that curve
Could someone explain what the relationship is between a geometric notion of area, and the formal definition [written down for completeness]:
Let $G = \langle S | R \rangle$ be a finitely presented group. Let $w$ be a word in the free group $F(S)$. Then if $w =_G 1$, we can write:
$$
w = \prod_{i=1}^n u_i r_i^{\pm 1} u_i^{-1} \quad u_i \in F(S); r_i \in R
$$
The area of $w$ is defined as $\min \{ n : w = \prod_{i=1}^n u_i r_i^{\pm 1}u_i^{-1} \}$
It's intuitively obvious how the length of the perimeter can be related to $|w|$: if we think of the path walked by $w$ on the Cayley graph starting from (say) the identity, the path will be a loop (as $w =_G 1$). The perimeter is the number of edges we need to traverse, which is the length of the word. On the other hand, this definition of area given above is not transparent. It seems to be saying something like:
count the minimum number of 'irreducible' components needed to write $w$ down.
I am unable to see the geometric content of this definition. I would greatly appreciate one, either by analogy, or a direct explanation on the Cayley graph of $G$.
You might find this introduction by Martin Bridson useful: The geometry of the word problem.
Let $X$ be the presentation complex of $G=\langle S\mid R\rangle$. Any element $g\in G$ can be realised as a (based) loop $w:S^1\to X$, which we can take to be a cellular map.
A van Kampen diagram is a simply connected, planar 2-complex $D$ with a cellular map $D\to X$. An embedding of $D$ into the plane defines a natural boundary $\partial D$, namely the loop round the "outside" of $D$; $D$ is said to be a van Kampen diagram for $w$ if the map $w:S^1\to X$ factors through the boundary of $D$.
The term area can now be very naturally motivated: the area of $w$ is just the minimum number of 2-cells in a van Kampen diagram for $w$. You can easily illustrate this definition by working out the areas of words in the presentation
$\langle a,b\mid [a,b]\rangle$
for $\mathbb{Z}^2$. For instance, the area of $[a^n,b^n]$ is $n^2$, but the area of $[a,b]a[a,b]a[a,b]\ldots a[a,b]a^{-n}$ is $n$, and both of these can be proved by drawing pictures in the plane.
As @YCor says in comments, this is all explained in Bridson's survey The geometry of the word problem.
+++++++++++++
As requested, here's the definition of the presentation complex $X$ associated to the presentation $\langle S\mid R\rangle$.
Take a graph (ie a 1-dimensional CW-complex) $X^{(1)}$ with one vertex, and with edges in bijection with the elements of the generating set $S$. The fundamental group of this graph is naturally isomorphic to the free group $F(S)$, and so each relator $r\in R$ can be realised as a loop $\rho_r$ in $X^{(1)}$. We now use the loops $\rho_r$ as attaching maps for the 2-cells of $X$, and the Seifert--van Kampen theorem tells us that the fundamental group of $X$ is isomorphic to $G$. Note also that the Cayley graph $\mathrm{Cay}_S(G)$ is naturally the 1-skeleton of the universal cover of $X$.
Thank you. I am unaware of this notion of 'presentation complex'. I presume the survey The geometry of the word problem. spells it out in full detail. It would still perhaps be nice if you could append its definition to the answer so the answer is self-contained.
I've added the definition to the answer.
Don’t you want your planar complex simply connected?
I certainly do! Thanks, @BenjaminSteinberg!
If you want a concrete example of a presentation complex, start with the standard presentation $\pi_1(S) = \langle a_1, b_1 ,a_2, b_2 \mid a_1 b_1 a_1^{-1} b_1^{-1} a_2 b_2 a_2 b_2^{-1}\rangle$ of the fundamental group of a closed oriented surface $S$ of genus $2$.
Use this presentation to define a gluing diagram on a regular octagon with angles $2\pi/8$ in the hyperblic plane $\mathbb H^2$. That octagon is a fundamental domain for the deck transformation action of $\pi_1(S)$ on $\mathbb H^2$, and the translates of that octagon under the action define a tiling of $\mathbb H^2$. That tiling is the presentation complex for the standard presentation of $\pi_1(S)$.
This also gives a concrete example of the relation between the geometric group theory notion of area and true area in the hyperbolic plane, where the latter is scaled so that the octagon has area 1.
|
2025-03-21T14:48:31.299117
| 2020-06-18T17:10:26 |
363457
|
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|
Stack Exchange
|
Identifying a subset with as few tests as possible
Informal description: You are given a set of $n$ blood samples, each having probability $p$ of being infected with a disease. Your goal is to determine the set $P$ of infected samples with as few tests as possible (on average). Each test is applied to a subset $S$ (of your choice) of the samples and returns positive if at least one of the samples is infected ($P \cap S \neq \varnothing$). What's the optimal way to choose the subsets to test to determine $P$ as efficiently as possible?
Formal description:
Let $n\in\mathbb{N}$. A test protocol $\mathscr{T}$ for subsets of $\{1,\ldots,n\}$ is a finite binary tree in which each non-leaf node $x$ is labeled by a subset $S_x$ of $\{1,\ldots,n\}$ and the two edges descending from the node $x$ are labeled “positive” and “negative”. For a test protocol $\mathscr{T}$ and a subset $P \subseteq \{1,\ldots,n\}$, we define a branch $\mathscr{B}_P = (x_0,\ldots,x_r)$ in the tree (= path from the root $x_0$ to a leaf $x_r$) as follows: $x_0$ is the root and, so long as $x_i$ is not a leaf, we let $x_{i+1}$ be the node attained by following the edge $(x_i, x_{i+1})$ labeled “positive” resp. “negative” according as $P \cap S_{x_i} \neq \varnothing$ resp. $P \cap S_{x_i} = \varnothing$. (In other words, the test tells us to test $S_{x_0}$ where $x_0$ is the root of $\mathscr{T}$, then test $S_{x_1}$ where $x_1$ is the node reached from $x_0$ by following the positive or negative branch according as $P \cap S_{x_0}$ is inhabited or empty, and so on until we reach a leaf $x_r$.) Calling $x_P$ the leaf (previously denoted $x_r$) where the branch $\mathscr{B}_P$ associated to $P$ terminates, we say that the test protocol $\mathscr{T}$ is decisive when $P \mapsto x_P$ is a bijection between subsets of $\{1,\ldots,n\}$ and leaves of $\mathscr{T}$, i.e., $P \mapsto \mathscr{B}_P$ is a bijection between subsets of $\{1,\ldots,n\}$ and branches of $\mathscr{T}$. The length $r$ of the branch $\mathscr{B}_P$ is then called the testing length $\ell(P)$ of the subset $P$ for the decisive protocol $\mathscr{T}$.
Now let $0<p<1$ be given: what is $\ell_{\mathrm{min}}$ (in function of $n$ and $p$) the smallest possible expected value $\sum_{P\subseteq\{1,\ldots,n\}} p^{\#P}\,(1-p)^{(n-\#P)}\,\ell(P)$, for a decisive protocol $\mathscr{T}$, of the testing length $\ell(P)$ of a subset $P$ that is drawn by choosing whether $i \in P$ using a Bernoulli distribution with probability $p$ independently for each $i$?
Examples:
The simplest decisive test protocol consists of testing each sample on its own, i.e., create a balanced binary tree with depth $n$ and $S_{x_i} = \{i+1\}$ for $x_i$ a node at depth $i$. This has $\ell(P) = n$ for every subset $P$ and provides a trivial upper bound on $\ell_{\mathrm{min}}$.
If $p$ is very small, we can create a test protocol which starts by testing whether any sample is infected, i.e., $S_{x_0} = \{1,\ldots,n\}$, so the negative branch can conclude immediately that $P = \varnothing$, whereas in the positive branch we use, say, the trivial test described above (pruning the cases where $n-1$ samples have tested negative and we know there is a positive). This provides an upper bound of $(1-p)^n + (n+1)(1-(1-p)^n) = 1 + n(1-(1-p)^n)$ on $\ell_{\mathrm{min}}$.
A lower bound on $\ell_{\mathrm{min}}$ comes from information theory: the subset $P$ has $n(-p\,\log_2 p - (1-p)\,\log_2(1-p))$ bits of information, so $\ell_{\mathrm{min}}$ must be at least this value. (But clearly this lower bound is not optimal since when $p\to 0$ this tends to $0$ whereas we can't do less than $1$ test.)
However, when $p=\frac{1}{2}$, the lower bound just given coincides with the trivial upper bound of $n$, so $\ell_{\mathrm{min}} = n$.
A few quick thoughts.
This is called the group testing problem. If folks wanted to learn more, I suppose they could look it up, and here is a substantial survey on the question (it likely answers whatever you want to know). But that might ruin the fun.
I would really like to say that if you increase $p$, then the best algorithm only gets slower...
The following algorithm works in at most $1 + 2np \log(n)$ steps on average, so for $p \leq n^{-c}$, this matches the information theory lower bound within a multiplicative constant.
(i) Initially test the entire set.
(ii) If you test a set, and it contains at least one infected element, then cut the set into two almost equal-sized pieces, and recursively test each piece.
[To analyze that algorithm, perhaps consider the problem where we know that exactly $k$ elements are infected. Then the above algorithm tests at most $1+2k \lceil \lg(n) \rceil$ sets, where $\lg$ is the log base $2$ and $\lceil x \rceil$ denotes the ceiling function (to prove this bound, draw the binary tree of what is tested in this algorithm. Note that each infected element has at most $\lceil \lg(n) \rceil$ sets above it, and each of those contributes at most $2$ tests to the total count). Then just take the expected value of both sides, and we're done since the expected value of $k$ is $np$.]
For larger values of $p$ (e.g., $p = 1 / \log(n)$), I'm not sure what should be the truth. For all $p \geq 1/2$, I would like the answer to be $n$ (see point (1) above).
Added afterwards: I looked at that survey, and I'm kicking myself for not getting the full answer to this. See their Theorem 1.3 which answers your question fully. I suppose I'll post this in a comment to this answer. If you don't want to know, don't look.
Call the algorithm I described above Algorithm 1. To improve it, first partition $[n]$ into $np$ roughly equal sets. Then perform Algorithm 1 on each of those $np$ sets. This runs in time $k \log(1/p) + \mathcal{O}(k)$, which matches the information theory lower bound if $\log(1/p) \to \infty$.
This isn't intended to be a complete answer, just a formalization of the most "obvious" idea.
A natural approach is to try to maximize the information gained with each test. If we are currently at the node $x$ of the test protocol $\mathscr{T}$, then for every subset $B \subseteq \{1, ..., n\}$ we can (in principle) compute the conditional probability $\mathbb{P}[B\mid x]$ that testing the subset $B$ will give a positive result, and then try to select the subset $B$ such that $\mathbb{P}[B\mid x]$ is as close to $\frac{1}{2}$ as possible, since this choice of $B$ will then maximize the conditional entropy $H(B \mid x)$. I will call this protocol the greedy strategy, and will use the symbol $\mathscr{G}$ to refer to this protocol.
It is unclear if the greedy strategy can be implemented in practice for large values of $n$. Even computing one of the conditional probabilities $\mathbb{P}[B \mid x]$ seems like it could be difficult, if we arrive at the node $x$ after making a sufficiently complex sequence of choices. However, for $p > 0.245...$, it is possible to work out what the greedy strategy will do.
The simplest case is the case $p > \frac{3-\sqrt{5}}{2} \approx 0.382$. In this case, the greedy strategy recommends that we always choose $B$ of size $1$, and the expected length of the greedy strategy in this case is $\ell(\mathscr{G}) = n$.
More generally, we can at least predict what the greedy strategy will do in its first step. The greedy strategy will pick a set $B$ of size $m$, where $m$ maximizes $H(1-(1-p)^m)$ among all choices $m \le n$. So the cutoff where we go from picking a set of size $m$ to picking a set of size $m+1$ occurs when $1-(1-p)^m = (1-p)^{m+1}$.
In particular, for $0.382... > p > 0.245...$, the first thing the greedy strategy will do is to examine a set $B_1$ of size $2$. If none of the samples in $B_1$ is infected, then we end up recursively applying the greedy strategy on a set of size $n-2$.
What if at least one of the samples in our first set $B_1$ of size $2$ is infected? Now there are several different choices that we could make in the next step: choose a set $B_2$ (of size $2$) which is disjoint from $B_1$, or choose a set $B_2$ (with the size of $B_2$ to be determined) such that $|B_2 \cap B_1| = 1$. The conditional probability that an element of $B_1$ is infected is $\frac{p}{1 - (1-p)^2} = \frac{1}{2-p} > \frac{1}{2}$, so if we choose $B_2$ to intersect $B_1$, then we may as well take $B_2$ to be a subset of $B_1$ of size $1$. Oddly enough, the greedy strategy always prefers to try taking $B_2$ to be another disjoint subset of size $2$.
So in the range $0.382... > p > 0.245...$, the greedy strategy will always begin by breaking up the set $\{1,...,n\}$ into groups of size $2$ (with one element left over if $n$ is odd), and testing each group. Then it will try testing the first element from one of the groups of two that contains an infected sample. If that first element is uninfected, then we know the second element of the group of two is infected and can ignore it. If that first element is infected, then we have no information about the second element of the group of two, so the greedy strategy will try to pair this element up with the leftover element if $n$ is odd, or will hold onto it for later if $n$ is even. This process then continues in an obvious way.
Thus in the range $0.382... > p > 0.245...$, the greedy strategy is equivalent to the following strategy: as long as there are at least $2$ unknown samples, we test the first two unknown samples together, throw them both away if the group tests negative, and otherwise immediately test the first of the two samples if the group tests positive, throwing both away if the first tests negative, and throwing just the first away if it tests positive. This gives us the recurrence
$$\ell(\mathscr{G}_n) = 2-(1-p)^2 + p\ell(\mathscr{G}_{n-1}) + (1-p)\ell(\mathscr{G}_{n-2}),$$
which has the solution
$$\ell(\mathscr{G}_n) = \frac{2-(1-p)^2}{2-p}n + \frac{(1-p)^2-p}{(2-p)^2}(1 - (p-1)^n).$$
Can anyone continue the analysis of the greedy strategy? Does it end up doing something simple in the end?
The following paper by Price and Scarlett appeared today on arXiv. It considers $k$ defective items, but given a fixed $p,$ one can choose $k=c p n,$ for example to have control over the probability of failure of this algorithm, via, say the Chernoff bound.
Picking $c=2,$ for example would give probability of failure $P_{err}$ upper bounded by
$$P_{err}\leq (e/4)^t\approx \frac{1}{1.47^t}$$
by the multiplicative Chernoff bound.
A Fast Binary Splitting Approach to Non-Adaptive Group Testing
From the abstract:
In this paper, we consider the problem of noiseless non-adaptive group testing under the for-each recovery guarantee, also known as probabilistic group testing. In the case of $n$ items and $k$ defectives, we provide an algorithm attaining high-probability recovery with $O(k \log n)$ scaling in both the number of tests and runtime, improving on the best known $O(k^2 \log k · \log n)$ runtime previously available for any algorithm that only uses $O(k \log n)$ tests.
We recursively work with groups of items of geometrically vanishing sizes, while maintaining a list of “possibly defective” groups and circumventing the
need for adaptivity. While the most basic form of our algorithm requires $\Omega(n)$ storage, we also provide
a low-storage variant based on hashing, with similar recovery guarantees.
|
2025-03-21T14:48:31.300230
| 2020-06-18T17:28:44 |
363462
|
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|
Stack Exchange
|
Slice category over Set
I would like to define a category as $\bf Cat \downarrow Set$ (which would be the slice category of $\bf Cat$ over the object $\bf Set$. However, since $\bf Set$ is not an object of $\bf Cat$, I cannot do that. However, using the slice definition still gives us some category where the objects are $\bf Set$-valued functors and arrows are functors in $\bf Cat$ along with some 2-isomorphism.
So I don't think this is the result of some slice but it seems so close ! Moreover I could also extend this definition to all the categories, but again, since their is no category of all categories, I don't think it is the result of any slicing.
My question is : is there a standard notation for these kind of construction ? Am I completely lost with those size issues?
Thanks a lot :)
What you're describing is the (2-)comma category $(\mathbf{Cat} \hookrightarrow \mathbf{CAT}) \downarrow (\mathbf{Set} : \mathbf{1} \to \mathbf{CAT})$, where $\mathbf{Cat}$ is the (2-)category of small categories, $\mathbf{1}$ is the terminal category and $\mathbf{CAT}$ is the (2-)category of locally small categories. The objects of this comma category are small categories with functors into $\mathbf{Set}$ and morphisms are functors between small categories commuting with the functors into $\mathbf{Set}$. Comma categories are a generalisation of (co)slice categories that often allow you to describe structure which is "slice-like", but isn't quite.
Alternatively, if you're happy with the objects being locally small categories, you can simply take the (2-)slice category $\mathbf{CAT}/\mathbf{Set}$.
Thanks for your answer, that's exactly what I was looking for. Do you have any references on $\bf CAT$ ? I was not able to find something about it. I looked into enriched categories to see that it is equivalent to $\bf Cat$ enriched with $\bf Set$.
I'm not sure what might be considered a canonical reference, but see for instance Section 1.7 of M-Completeness Is Seldom Monadic Over Graphs, which describes foundational issues regarding treating $\mathbf{CAT}$ as a category versus a 2-category.
@141592653 for category related things it is always nice to check https://ncatlab.org/nlab/show/2-category - references are on the bottom of the page.
Thank you very much, you were very helpful
Here is how you get rid of the "size" issues in constructions like this:
An object of the slice (2-)category that you want is a functor
$F:{\mathbf C}\to{\mathbf{Set}}$, which is otherwise known as a
presheaf on ${\mathbf C}^{\mathsf{op}}$.
The Grothendieck construction transforms this a functor $P:{\mathbf E}\to{\mathbf C}^{\mathsf{op}}$ that is a discrete fibration.
It's an exercise to work out what the 1- and 2-cells of this 2-category look like in terms of discrete fibrations.
|
2025-03-21T14:48:31.300457
| 2020-06-18T17:29:42 |
363463
|
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|
Stack Exchange
|
Exit time estimate for a simple continuous-time random walk
Let $S = (S_t, t \geq 0)$ be a simple one-dimensional continuous-time random walk with total jump rate one, $S_0 = 0$. Denote by $T_k$ the time when $S$ exits the interval $I_k = [-k,k] \cap \mathbb{Z}$. Let also for an interval of integers $I$, $\lambda (I)$ be the principle Dirichlet eigenvalue of the normalized discrete Laplacian on $I$ defined as $\Delta f(y) = \frac 12 [f(x+1) + f(x-1) - 2f(x) ]$ for $f$ vanishing outside $I$; that is, $\lambda(I) = 1 - \cos (\frac{\pi}{\ell + 1})$, where $\ell$ is the length of $I$.
In [1] in Section 5 the following inequality is used
$$
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
P (T_k > q) \leq c (\lambda (I_k) q + 1 ) ^{1/2} e^{- \lambda (I_k) q},
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
$$
where $c >0 $ is a constant.
The authors say that (1) can be obtained "from standard estimates" and give two references. One of the references is not accessible to me, while I have not found the other very helpful.
I would thus appreciate any reference where inequalities of type (1) involving the principle eigenvalue of the Laplacian are discussed.
[1] Ramírez, A. F.; Sidoravicius, V., Asymptotic behavior of a stochastic combustion growth process, J. Eur. Math. Soc. (JEMS) 6, No. 3, 293-334 (2004). ZBL1049.60089.
Let $\tau_k$ denote the number of steps for discrete time simple RW on the integers (started at 0) to exit the interval $[-k,k]$. Let $\gamma:=\cos(\frac{\pi}{2k+2})$ so that $\lambda=1-\gamma$.
The formula in [1] page 243, line -5, gives (bounding the alternating series there by twice the first term and using
$\cot(x) \le 1/\sin(x) \le \pi/(2x)$ for $x\in[0,\pi/2]$ ) that
$$P[\tau_k>n] \le 8\gamma^n \,.$$
The estimate for continuous time RW follows:
$$P[T_k>q] \le \sum_n P[{\rm Poisson}(q)=n] \cdot P[\tau_k>n] \,$$
whence
$$P[T_k>q] \le \sum_n \frac{q^n e^{-q}}{n!}\cdot 8\gamma^n =8e^{q\gamma-q}=8e^{-q\lambda}\,.$$
[1] Spitzer, Frank. Principles of random walk. GTM Vol. 34. Second edition, Springer.
@Sinusx: I found a detailed discussion in Spitzer’s book, see revised answer.
I have deleted my first comment as it is no longer pertinent to the answer.
|
2025-03-21T14:48:31.300639
| 2020-06-18T18:06:52 |
363467
|
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|
Stack Exchange
|
Any reflexive space has the property of Banach-Saks?
We say that a Banach space $(X,\|.\|)$ has the Banach-Saks property if every bounded sequence $(x_m)_m$ in $X$ admits a subsequence $(x_{m_n})_n$ which converges in the sense of Cesàro, that is, there is a vector $x$ in $X$ such that
$$
{\displaystyle \left \| {\frac {\sum_{n = 1}^{N}x_{m_{n}}}{N}} - x \right \| \underset{N}{\longrightarrow} 0.}
$$
Remark
Any Banach space satisfying the Banach-Saks property is reflexive.
Question
IS the converse implication true, i.e. does every reflexive space satisfy the Banach-Saks property?
The answer appears to be no, see https://math.stackexchange.com/a/93881/822
|
2025-03-21T14:48:31.300719
| 2020-06-18T18:30:04 |
363469
|
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|
Stack Exchange
|
is constant sheaf perverse on a Cohen-Macaulay variety?
Let $X$ be a connected Cohen-Macaulay algebraic variety of dimension $d$, say, over $\mathbb C$. Is it true that $\underline{{\mathbb C}_X}[d]$ is a perverse sheaf, where $\underline{{\mathbb C}_X}$ is the constant sheaf on $X({\mathbb C})$? This is true if $X$ is a set-theoretic complete intersection by a theorem of Le Dung Trang.
I think this fails in general. Take the cone $C$ over a smooth projective pair $(X,L)$ with $H^i(X,L^j) = 0$ for $0 < i < d = \dim(X)$ and all $j$, so $C$ is CM. Deligne's formula for the IC-sheaf says $IC_C = \tau^{\leq -1} (j_* \mathbf{C}[d+1])$ is perverse, where $j:C-{0} \subset C$. The cone of the map $\mathbf{C}[d+1] \to IC_C$ is then a bunch of skyscraper sheaves supported in a whole range of degrees if $X$ is interesting (as $C-0$ is homotopic to an $S^1$-bundle over $X$ attached to $L$). So I think $\mathbf{C}[d+1]$ is not perverse. I can try to write an explicit example later.
I agree, I think $X$ being an abelian variety provides a counterexample: the cone is the skyscraper tensored with the primitive part of $H^*(X)$ which in the abelian variety category will be nonzero in every possible degree.
Unless I'm mistaken, that does not quite work because it's not CM: the $\mathcal{O}$-cohomology of the abelian variety causes the cone to not be CM. But you can replace the abelian variety with the blowup of projective space along an abelian variety to reproduce the same phenomenon.
ah, that's right, we need $X$ with $H^i(X,O)=0$ for $i>0$ but with some primitive cohomology, your example appears to do the job. Thanks!
|
2025-03-21T14:48:31.300873
| 2020-06-18T18:30:20 |
363470
|
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|
Stack Exchange
|
Reference for the algorithm to find the intersection between a subspace and positive orthant
I came across this algorithm, in this question Algorithm for the intersection of a vector subspace with a cone of non-negative vectors ;
Is there any reference for the algorithm described in the question?
|
2025-03-21T14:48:31.300927
| 2020-06-18T19:59:01 |
363473
|
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|
Stack Exchange
|
Is the existence of undecidable propositions decidable?
In proving his first incompleteness theorem Godel constructed a proposition that is undecidable, i.e. neither provable nor disprovable within a consistent formal system $F$ that contains elementary arithmetic.
My question is:
Can the (true) statement $D$="There exist propositions in $F$ that are undecidable" be proved within some formal system $F$ to which the first incompleteness theorem applies?
Note that $D$ is expressible as $$\neg (\forall _n \exists_m [(\Pi_m \text{ proves } P_n) \lor (\Pi_m \text{ proves }\neg P_n)])$$
where $P_n$ is the $n$-th proposition and $\Pi_m$ is the $m$-th proof. Hence this should be a valid arithmetical proposition in the Godel encoding.
No. If $F$ proves that some statements are undecidable in $F$, then $F$ would prove $F$ is consistent (as otherwise it proves everything), which is impossible by Godel's second incompleteness theorem.
To build off of Wojowu's statement, it is possible for a system F to prove "If F is consistent, then there are undecidable statements in F." But what this really is in some sense a theorem of F + Con F being imported into F.
@Wojowu Thank you for your answer, much appreciated.
@IvanMeir In principle there are very weak theories that violate second incompleteness theorem (although, in comparision to say $\mathsf{PA}$, for this theories it is far less clear what should count as "natural" formalization of provability). The most well-known examples of theories violating G2 are due to D. Willard, see for example his paper "Self-verifying axiom systems, the incompleteness theorem and related reflection principles". There is also my recent example https://arxiv.org/abs/1907.00877 . By slight modifications of those examples it is easy to get provably undecidable sentences.
@FedorPakhomov Thank you for your response which is very interesting. So if I understand correctly, there exist theories that the first incompleteness theory applies to but not the second in which case we should be able to prove undecidability for certain sentences? I was wondering how much of standard arithmetic do such theories contain?
@IvanMeir The key limitation of this kind of theories is that they couldn't prove totality of the successor function. It isn't immediately clear that those theories directly provide examples of provably undecidable sentences. However, it is easy to modify those theories so that this examples will be present. If you add an additional unary predicate $P(x)$ to the theory $H_{<\omega}$ from my paper, then the same argument used to show that $H_{<\omega}\vdash \mathsf{Con}(H_{<\omega})$ will show that $\exists x;P(x)$ is provably undecidable for $H_{<\omega}$.
|
2025-03-21T14:48:31.301154
| 2020-06-18T21:26:52 |
363477
|
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|
Stack Exchange
|
Has it already been shown that a digraph is $k$-connected if and only if there is a set of $k$ disjoint paths between every pair of vertices?
My question concerns graph theory.
There is a conjecture I know to be true, but I am not sure whether it has been proven before. It is a fairly simple result. It may be well-known; I am not sure.
Begin Informal Conjecture
I do give a more formal statement of the conjecture later in this post, but first, let us at something which is more intuitive:
Suppose $v$ and $w$ are nodes.
There are paths from $v$ to $w$.
However, there are no really short paths from $v$ to $w$.
By this, I mean that is no edge from $v$ to $w$
Every path from $v$ to $w$ contains at least one vertex other than $v$ and $w$
If you must remove $k$ nodes to make there are no more paths from vertex $v$ to vertex $w$, then there is a set of at least $k$ non-overlapping/separate paths from node $v$ to vertex $w$.
Note that when I say that we remove $k$ nodes to break all paths from $v$ to $w$, we are not allowed to remove nodes $v$ and $w$. We have to remove nodes from the graph other than $v$ or $w$.
By "overlapping/separate" paths, I do allow that both paths begin at the same starting node and end at the same final node. However, but the paths never meet, or cross each-other, after leaving their shared origin and before arriving at their mutual destination. The two path interiors have no nodes in common.
End Informal Conjecture
Begin formal definition of $k$-vertex-connected
Let $k \in \mathbb{N}$
Let $D$ be an arbitrary directed graph of at least 2 vertices.
Let $v$ and $w$ be arbitrary vertices in $D$.
Suppose that $V^{\prime}$ is an set of $(k-1)$ or fewer vertices in digraph $D$ such that that $v, w \notin V^{\prime}$
Let $D^{\prime}$ denote $D^{\prime}$ denote the graph formed by $D^{\prime}$ by removing the vertices in set $V^{\prime}$
$D$ is said to be $k$-vertex-connected if and only if there exists a path from vertex $v$ to vertex $w$ in digraph $D^{\prime}$
End formal definition of $k$-vertex-connected
Begin formal definition of $k$-pathy
For any $k \in \mathbb{N}$
For any directed graph $D$
$D$ is $k$-pathy if and only if for every vertices $v$ and $w$ in $D$, if there exists a path in $D$ from $v$ to $w$ and every path from $v$ to $w$ contains at least $2$ edges, then there exists a set of paths $PS$ from $v$ to $w$ such that for every pair of paths $P, Q \in PS$ if $P \neq Q$, then for all vertices $x$ in $D$, if $x$ is in path $P$ and $x$ in path $Q$, then $x \in \{v, w\}$
End formal definition of $k$-pathy
Begin Formal Conjecture
For any $k \in \mathbb{N}$
For any directed graph $D$
$D$ is $k$-pathy if and only if $D$ is $k$-vertex-connected.
End Formal Conjecture
Begin Remark
Being able to say that there are $k$ non-overlapping paths between two nodes is extremely useful in proofs.
It is also very concrete You can draw a picture of $k$ non-overlapping paths very easily.
As an example of a silly application (student-level homework), is if there are at least three paths between two nodes, then there are at least two paths of odd length or two paths of even length. If you glue the two odd paths together or glue the two even paths together, you will get a cycle. The cycle will contain an even number of edges.
Thus, if there exists two nodes in a graph, such that there exists three paths between the two nodes, then there exists an even cycle in the graph.
End Remark
I did not read this but it sounds like https://en.wikipedia.org/wiki/Menger%27s_theorem
@VilleSalo You are right. Menger's Theorem says that if there is no directed edge from $x$ to $y$, then minimum number of vertices, distinct from $x$ and $y$, whose removal disconnects $x$ and $y$) is equal to the maximum number of pairwise internally vertex-disjoint paths from $x$ to $y$.
This is one of the most fundamental/often used theorems in graph theory. More than 90 years ago, Karl Menger proved the Theorem: Menger's Theorem states that:
if there is no directed edge from $x$ to $y$, then the minimum number of vertices, distinct from $x$ and $y$, whose removal disconnects $x$ and $y$, is equal to the maximum number of pairwise internally vertex-disjoint paths from $x$ to $y$.
|
2025-03-21T14:48:31.301454
| 2020-06-18T23:21:38 |
363482
|
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|
Stack Exchange
|
is there a ‘nice’ lattice on the set of unlabelled graphs with $n$ vertices?
It is easy to endow the set of vertex-labelled graphs with $n$ vertices with a lattice structure: take the union and the intersection of the edge set as meet and join respectively.
However, I wonder whether there is a ‘nice’ lattice structure on unlabelled graphs, i.e., graphs up to isomorphism.
I'd also be happy with a lattice structure on a (reasonably) large subset, if this makes things any easier, e.g., connected or planar graphs on $n$ vertices.
To clarify: a very non-‘nice’ way to make the set of unlabelled graphs on $n$ vertices into a lattice is to pick an arbitrary total order.
An idea for a slightly nicer way might be to set $G < H$ if and only if $H$ has more edges than $G$, however this does not produce a lattice.
A "natural" poset structure on the set of $n$-vertex unlabelled graphs is the subgraph ordering. However, this is not a lattice for $n\geq 5$. This poset has some nice properties: it is a quotient of a boolean algebra and hence Sperner, for instance. See pp. 49-50 of my book Algebraic Combinatorics, second ed.
@Martin: there is, probably, no poset structure on the set of unlabeled graphs with $n$ vertices with the property that the ordering $\le$ can be computed in polynomial time (in $n$). This is because such an ordering could be used to solve graph isomorphism, by checking whether $G \le H$ and $H \le G$. But maybe this is too obvious to merit an answer. E.g. you might consider the subgraph ordering to be nice.
@QiaochuYuan, I find that this is actually a rather important observation. But indeed, for me, "nice" does not imply "explicit".
@QiaochuYuan, it occurred to me that we could have an ordering where $G<H$ or even just the covering relation is easy to compute. At least, I don't see right now how this would yield an easy isomorphism test.
I guess this class is way too small, but since the questions is still unanswered, I will give you an example I like.
An EFG (Edge Firing Game) is defined from an undirected graph $G$ with a distinguished vertex called the sink. It starts from an orientation of $G$ and generates a set $L$ of orientations as follows. The initial orientation is in $L$, and if an orientation is in $L$ and has a vertex (other than the sink) with only incoming edges, then the orientation obtained by reversing all these edges also is in $L$.
This operation defines a relation between the orientations in $L$, see above example (the sink is the black square); it is the covering relation of a distributive lattice. This was shown by James Propp in 1993 together with many other interesting results on lattices and orientations.
Interestingly, this game actually encodes all distributive lattices: each of them may be obtained from an EFG, which we have shown in 2001.
The question asks for a lattice whose elements are the unlabeled graphs on $n$ vertices. What you're describing is a method for constructing a lattice given some graph.
Well, one may view it as a lattice on a set of directed graphs (with the same underlying undirected graph), and since the question says that a lattice "on a (reasonably) large subset" is interesting, I gave it a try. But I do agree that, certainly, this class is not "reasonably" large, and that this does not make a very relevant answer to the question :/
|
2025-03-21T14:48:31.301720
| 2020-06-19T00:01:31 |
363483
|
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|
Stack Exchange
|
First order asymptotics for maxima of stationary Gaussians with vanishing covariance
Let $G$ be a centered stationary Gaussian process indexed by the integer lattice $\mathbb Z^d$. A straightforward Borel-Cantelli argument shows that
$$\limsup_{m\to\infty}\frac{1}{\sqrt{\log m}}\left(\max_{\|x\|\leq m}G(x)\right)\leq\sqrt{2d\mathrm{Var}[G(0)^2]}.$$
In general, there need not be matching lower bound (e.g., if $G(x)=G(0)$ for all $x\in\mathbb Z^d$). However, in some cases, I know that we do have a matching lower bound, for example, in the case where the $G(x)$ are all independent.
That said, I'm trying to find a reference for the following result:
Question. Denote the covariance function $C(x):=\mathrm{E}[G(0)G(x)]$. Suppose that $C(x)\to0$ as $\|x\|\to\infty$. Do we then have that
$$\liminf_{m\to\infty}\frac{1}{\sqrt{\log m}}\left(\max_{\|x\|\leq m}G(x)\right)\geq\sqrt{2d\mathrm{Var}[G(0)^2]}?$$
If so, does anyone know of a reference for such a result?
I strongly suspect the result is true, in part because I do have a reference for the same statement when $G$ is a continuous stationary Gaussian process on $\mathbb R^d$.
For $d=1$, see the paper "Maxima of stationary Gaussian processes", by Pickands
(ZW 1967), Theorem 3.4; I think (but have not checked) that a similar method works for $d>1$. Maybe the book of Adler and Taylor has further references.
Thanks for that! This is exactly what I was looking for in 1d. I'll look into the higher dimensional case.
|
2025-03-21T14:48:31.301843
| 2020-06-19T01:59:17 |
363484
|
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|
Stack Exchange
|
is this the correct universal property of free Lie superalgebras?
Consider a $\Bbb Z_2$ graded set $A$.
Universal property of free Lie superalgebra $FLS(A)$: Let $\mathfrak g$ be a Lie superalgebra and let $\Phi: A \to \mathfrak g$ be a set map which preserves the $\mathbb Z_2$ grading. Then $\Phi$ can be extended to a Lie superalgebra homomorphism $\Phi: FLS(A) \to \mathfrak g$.
My questions.
Is this the correct universal property of free Lie superalgebras?
Why do we need the set map to preserve the $\Bbb Z_2$ grading?
What is the definition of Lie sub-superalgebra generated by a set $S$ where $S$ is a subset of $FLS(A)$?
More generally, if $\mathfrak g = \frac{FLS(A)}{R}$ then the Lie sub-superalgebra generated by $S$ in $\mathfrak g$ is equal to $\mathfrak g = \frac{FLS(S)}{R}$ where $s$ is a subset of $\mathfrak g$?
Thank you.
|
2025-03-21T14:48:31.301931
| 2020-06-19T03:03:29 |
363486
|
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|
Stack Exchange
|
Why aren't $B_n$ and $C_n$ the other way around?
In the classification of complex simple Lie algebras/groups, I have always been vaguely puzzled why $B_n$ and $C_n$ are labeled the way they are. I always instinctively want to put the special orthogonal groups together, and so I am tempted to use the letter $B$ for what is standardly called $C$, and vice versa. Looking at the Dynkin diagrams of affine Weyl groups reinforces this instinct of mine, because the vertex of degree 3 makes $\tilde D_n$ look more like $\tilde B_n$ than $\tilde C_n$, at least in my eyes.
Is there some intuitive reason for the standard notation? Or is just a historical accident with no particular rhyme or reason behind it?
$B$, $C$, and $D$ all rhyme, so sadly there is no appeal to rhyme to order them differently. :-)
"Historical convention" (going back to Lie?) is probably the correct explanation, but note that under what I would call the "standard combinatorial folding procedure" as described by Stembridge in Folding by automorphisms, we produce the Type $B_n$ root system from the Type $A_{2n-1}$ root system, and the Type $C_{n}$ root system from the Type $D_{n+1}$ root system. Though note, as discussed in this prior MO question, that there are two "dual" folding procedures which both arise in Lie theory.
In particular, when classifying simple Lie algebras it is nice to first construct all the simply laced Dynkin diagrams/root systems/Lie algebras; and then show that the non-simply laced ones can all be obtained from the simply laced ones via folding along diagram automorphisms. In this second step, B comes from A, and C comes from D.
Interesting. Do we have any indication that whoever first introduced the labeling was influenced by this fact?
@TimothyChow: some comments in this recent question https://mathoverflow.net/questions/363404/emergence-of-the-orthogonal-group/363416 suggest the names of the types of the classical groups might go back to Lie. I guess one would have to read his papers to see why these letters are chosen.
In fact, various sources are suggesting to me that the names go back to the 1885 paper of Lie, "Allgemeine Untersuchungen uber Differentialgleichungen die eine continuirliche endliche Gruppe gestatten": https://link.springer.com/article/10.1007/BF01446421.
|
2025-03-21T14:48:31.302224
| 2020-06-19T04:10:13 |
363490
|
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|
Stack Exchange
|
Galois action of Weil restriction
Let $K/\mathbb{Q}$ be a quadratic field. Let $E$ be an elliptic curve defined over $K$ but not over $\mathbb{Q}$, and let $\bar{E}$ be the Galois conjugate of $E$. Then by the descent theory (for instance, see section h), we can construct an abelian surface $A:=Res_{K/\mathbb{Q}}(E)$, which is isomorphic to $E\times \bar{E}$ over $K$ (but not like to be over $\mathbb{Q}$).
With this setup, my question is: Knowing the $Gal(\overline{K}/K)$-action attached to the Tate module $T_{\ell}(E)$, how to compute the $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$-action attached to the Etale cohomology $H^2_{et}(A, \mathbb{Q}_{\ell})$? For instance, given a good prime $p$ of $A$ (i.e. the reduction of $A$ at $p$ is still smooth), then what is the trace of Frobenius $p$ at $H^2_{et}(A, \mathbb{Q}_{\ell})$? In particular, if there is an isogeny from $E$ to $\bar{E}$, can we understand the transcendental part of $H^2_{et}(A,\mathbb{Q}_{\ell})$ as symmetric square of $T_{\ell}(E)$?
I apolgies if the question is too complicated. Any suggestion or reference is welcome. And please feel free to add conditions if you think that is necessary for get the answer.
The Tate module of $A$ is the induced module of the Tate module for $E$.. The $H^1$ for $A$ is the dual of its Tate module, and $H^2(A)=\wedge^2 H^1(A)$.
Thanks @anon, I think you are right. It's kind of embarring that I didn't realize that the induced representation is the answer.
|
2025-03-21T14:48:31.302363
| 2020-06-19T04:26:08 |
363491
|
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|
Stack Exchange
|
Genus of the graph complement
Suppose a simple undirected graph $G$ with $n$ vertices has (minimum) genus $g$. What is the genus of its complement?
My intuitive guess is that the answer is something like
$$\text{genus of }K_n - g$$
but I wasn't able to find any information on this topic online. Is there a known solution to this problem?
This is my first time on MathOverflow, so I apologize if my question is missing something -- thanks for the help :)
Can you add the definition of genus of a graph (or a link)?
If $G$ is any graph on five vertices, other than $K_5$ or its complement, then both $G$ and its complement are planar, while $K_5$ isn't.
@M.Winter I take it to be the minimum $g$ such that the graph has a 2-cell embedding into a surface of genus $g$.
Your intuitive guess can be off by something linear in $n$: clearly $K_{n-2,2}$ is planar, its complement $K_{n-2}$ (plus an isolated edge) has genus $(n-5)(n-6)/12$, which is $\approx n/3$ less than the genus of $K_n$.
To get a feeling and some (counter-)examples one could play around with known coloring results and the coloring-genus-connection given by the Hadwiger conjecture (now a theorem).
@M.Winter I added a definition for graph genus (which concurs with that given by Gerry Myerson). To clarify, is your suggestion to search for results on the chromatic number of the graph complement, then apply the theorem to generate a bound on the genus?
|
2025-03-21T14:48:31.302499
| 2020-06-19T05:40:38 |
363493
|
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|
Stack Exchange
|
An algebraic property that makes a per-C*-algebra complete
Let $A$ be a normed *-algebra with $\|x^*x\|=\|x\|^2$. Suppose that for every subset S of A, the left annihilator
${\displaystyle \mathrm {Ann} _{L}(S)=\{a\in A\mid \forall s\in S,as=0\}\,}$
is generated as a left ideal by some projection p of A.
Q. Is $A$ norm complete (i.e., an AW*-algebra)?
How about the subalgebra of those functions in $\ell_\infty({\bf N})$ that take only finitely many values?
Yes, it is a nice countra-example. Thanks.
|
2025-03-21T14:48:31.302583
| 2020-06-19T06:27:00 |
363495
|
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|
Stack Exchange
|
Surprising properties of closed planar curves
In https://arxiv.org/abs/2002.05422 I proved with elementary topological methods that a smooth planar curve with total turning number a non-zero integer multiple of $2\pi$ (the tangent fully turns a non-zero number of times) can always be split into 3 arcs rearrangeable to a closed smooth curve. Here comes one example of what I called the 2-cut theorem.
I am now writing the introduction to my thesis and I would like to cite more examples of counterintuitive properties of planar closed curves like, I think, the one above. The inscribed square problem, asking whether every Jordan curve admits an inscribed square (pic from Jordan curves admitting only acyclic inscriptions of squares), alredy came to my mind.
Such a property is still a conjecture for the general case, but proofs have been provided for several special cases and the easier inscribed rectangle problem can be solved with a beautiful topological argument (3Blue1Brown made a very nice video about that https://www.youtube.com/watch?v=AmgkSdhK4K8&t=169s).
My question: what are other surprising/counterintuitive properties of closed planar curves you are aware of?
Note: this question was originally posted on MathStackExchange since I thought it was suitable for a more general audience. Apparently, it requires a slightly broader knowledge of literature.
Post at m.se is https://math.stackexchange.com/questions/3723039/surprising-properties-of-closed-planar-curves
See this answer https://mathoverflow.net/a/269261/53155
I'd like to post links to some MO questions related to Jordan curves (I wish I've done this right away). Perhaps this will attract additional attention to them. https://mathoverflow.net/questions/357508/chord-of-fixed-length-traveling-around-a-jordan-curve and https://mathoverflow.net/questions/296585/enveloping-a-jordan-curve-with-a-trace-of-another-one?noredirect=1&lq=1
maybe this also qualifies: "Can a Jordan arc be included into a Jordan loop?"
https://mathoverflow.net/questions/75213/including-a-jordan-arc-into-a-jordan-loop-can-the-magi-go-home-by-another-way
The Four Vertex Theorem, due to Mukhopadhyaya in 1909, states that a plane closed simple smooth curve with positive curvature has at least four vertices, where a vertex is a local maximum or minimum of the curvature. There's a detailed discussion in a chapter of Fuchs and Tabachnikov, Mathematical Omnibus.
The assumption of positive total curvature is not enough, you need a simple curve. For some Pascal's Limaçons admit only two vertices.
Chakerian's Theorem (proved in this paper) that a closed curve of length L in the unit ball in $\mathbb{R}^n$ has total curvature at least L.
(In this later paper Chakerian gave a simpler proof and noted that equality holds iff the curve is of length $2\pi n$ and winds round the unit circle $n$ times.)
This is fun! Does there exist something similar of a sphere of surface $S$ instead of a curve?
One result that I initially found a bit surprising is Grayson's theorem. It's a little bit of a different flavor than the other examples but I think it's interesting and worth a mention.
Given a closed planar curve $\gamma$ which is smooth enough ($C^2$ is sufficient but it's possible to be deal with less regular curves), there is a process known as curve shortening flow, which deforms the curve using the flow
$$ \frac{\partial \gamma}{\partial t} = \kappa N. $$
Here, $\kappa$ is the unsigned curvature and $N$ is the unit normal vector. This is called curve shortening flow because it tends to shrink curves.
Grayson's theorem states that if you start with a closed curve that does not cross itself, then under the flow it will remain embedded and eventually become convex. Gage and Hamilton had earlier shown that a convex curve shrinks to a round point, so this shows that any embedded curve shrinks to a round point.
Grayson's original proof is pretty involved. Now there are more conceptual proofs of this result (Huisken has a particularly nice one), so it's a bit less surprising. Nonetheless, it's possible to draw some really crazy curves and somehow the flow avoids itself and makes the curve convex. A good resource for this is Anthony Carapetis' website, which has an applet demonstrating the flow. Note that if you allow the curve to cross itself, the theorem fails and you can get local kinks to appear.
Thank you! I awarded the bounty for the concise and clear presentation, and for the link to the interactive resource.
I post as an answer to not overload the original entry and to contribute myself to the nice flow of results.
Browsing the book "Unsolved problems in geometry: Unsolved Problems in Intuitive Mathematics" by Croft, Falconer and Guy (after a good hint from the Reddit user Giovanni_Resta), the following properties of closed convex planar curves came out.
Given any closed convex planar curve, there always exist three concurrent chords bisecting each other and cutting at prescribed angle. The reference given is "On chords of convex curves", H. Steinhaus - Bull. Acad. Polon. Sci. Cl. III, 1957, which I was not able to find online. The general problem for simple closed curve is apparently still open and very much related to the general inscribed square problem I already mentioned in the question.
Given any closed convex planar curve, there exists a point in its interior which lies on 4 normals through 4 distinct boundary points. The theorem was proven by Heil also for 3D convex bodies and 6 normals and it is conjectured for convex bodies of $\mathbb{R}^d$ and $2d$ normals. The reference given is "Concurrent normals and critical points under weak smoothness assumptions", E. Heil - NYASA, 1985.
To those, I add a characterization for $C^3$ closed planar curves (not necessarily simple), which I found in "Robust fairing via conformal curvature" - K. Crane, U. Pinkall, and P. Schröder, ACM Trans. Graphics 32, 2013.
A $C^3$ planar curve $\gamma$ parametrized on $[a,b]$ and with curvature $k$ is closed iff $\int_a^b k'\gamma=0 $.
If it is not surprising, it is at least a weird relation between dependent descriptors.
How about the ``Oval's Problem" of Benguria and Loss. This is a (somewhat surprisingly) open problem connecting spectral theory with plane geometry. The conjecture is that for any simple closed curve $\sigma$ of length $2\pi$ and any periodic function $f:[0,2\pi]\to \mathbb{R}$ one has
$$
\int_0^{2\pi} |f'(s)|^2+\kappa(s)^2 |f(s)|^2 ds \geq \gamma \int_{0}^{2\pi} |f(s)|^2 ds
$$
for $\gamma=1$. Here $s$ is the arclength parameter and $\kappa(s)$ is the geodesic curvature. In other words, the lowest eigenvalue of $-\frac{d^2}{ds^2}+\kappa(s)^2$ is bounded from below by $1$ on a closed curve. This lowest bound is achieved by the circle, but is actually achieved by a whole family of "ovals" that deteriorate to a multiplicity two line segment. Interestingly, this problem is related the sharp Lieb-Thirring inequality which is a purely spectral theoretic problem (this was Benguria and Loss's motivation). It is also related to minimal surface theory.
In their paper, Benguria and Loss show that this inequality holds with $\gamma=\frac{1}{2}$. This is actually, sharp if one expands the set of curves to include those that total turning number 1 so one has to use the curve is closed in some way (and not just that $\int_0^{2\pi} k(s)ds=2\pi$. Various other authors have worked on parts of this problem: Burchard and Thomas showed the ovals where local minima of the lowest eigenvalue (so the problem is solved near curves in the family, Linde showed closed convex curves have $\gamma\geq 0.6085$, Denzler showed there is a closed convex curve that minimized the value of $\gamma$ and Bernstein and Mettler discussed the symmetry of the problem (related to projective geometry) and showed some weaker geometric inequalities held for closed convex curves (but not for curves of turning number 1).
|
2025-03-21T14:48:31.303130
| 2020-06-19T06:41:41 |
363496
|
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|
Stack Exchange
|
Contraction elements in unital *-rings
Let $A$ be a unital *-ring. Let us put, $A^+=\{\sum_1^N x^*_ix_i: x_i\in A , N\in \mathbb{N} \}$. We write $x\geq0$ if $x\in A^+$.
Suppose that for every $n\in \mathbb{N}$ and $a\in A$, there exsits $x\in A$ with $a=nx$.
Let us assume for natural numbers $m,n$ we have $mx^*x-n\geq0$. Can we conclude that $mxx^*-n\geq0$ as well?
Assume WLOG that $n,m>0 $ and note that $mx^*x-n\geq0$ iff $x^*x\geq n/m$ iff the spectrum $\sigma(x^*x)\subseteq[n/m,\infty)$ and since $\sigma(x^*x)$\{0}$=\sigma(xx^*)$\{0} it follows that $\sigma(xx^*)\subseteq [n/m,\infty)$ so $mxx^*-n\geq 0$.
Note that, $A$ is not a Banach algebra. How do you define the spectrum? Why is it non-empty?!
|
2025-03-21T14:48:31.303213
| 2020-06-19T08:21:19 |
363503
|
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|
Stack Exchange
|
Are Chow groups invariant under universal homeomorphisms?
Let $f:X\to Y$ be a universal homeomorphism of schemes, $R$ a coefficient ring. Which assumptions on $f$ and $R$ are suffient to ensure that the pullback map $f^*$ of $R$-linear Chow groups is bijective (clearly, it suffices to verify injectivity)? Actually, I am mostly interested in the case where $Y$ is a smooth variety over a field $k$ and $X=Y\times_{Spec\,k}Spec\,k^{perf}$; yet I would be deeply grateful for any references on this subject!
I am afraid this will happen only in trivial cases. For instance if $Y$ is a smooth projective curve over $\mathbb{Q}$, with Jacobian variety $J$, $CH^1(Y)=\operatorname{Pic}(Y) $ is finitely generated, while $\operatorname{Pic}(X) =\mathbb{Z}\oplus J(\bar{\mathbb{Q}})$ is not.
Well, in characteristic $0$ only "trivial" universal homeomorphisms exist, yes.:)
@MikhailBondarko how about the resolution of a cusp? (Tag 0BRC)
This is easy to answer in equicharacteristic under the additional assumption that $f$ is finite flat and $X$ and $Y$ are irreducible and generically reduced, using $\mathbf Z[\tfrac{1}{p}]$-coefficients, where $p$ is the exponential characteristic (i.e. $1$ if we're in characteristic $0$).
Indeed, in this case push and pull shows that the composition
$$\operatorname{CH}_*(Y) \stackrel{f^*}\to \operatorname{CH}_*(X) \stackrel{f_*}\to \operatorname{CH}_*(Y)$$
equals the degree of $f$ (see e.g. Fulton, Example 1.7.4). Since $f$ is radicial, the extension $K(Y) \to K(X)$ has degree $p^n$ for some $n \in \mathbf N$, hence the same goes for $f$. Thus $f^*$ is injective with $\mathbf Z[\tfrac{1}{p}]$-coefficients.
Similarly, if $V \subseteq Y$ is an integral subscheme with image $W$, then $f_*[V] = p^r[W]$ for some $r \in \mathbf N$, and $f^*[W] = m[V]$ for some $m \in \mathbf Z_{>0}$. Pushing forward gives
$$p^n[W] = f_*f^*[W] = mp^r[W],$$
so $m$ is a power of $p$ as well (these equalities hold in $Z_*(Y)$, so there is no torsion), showing surjectivity with $\mathbf Z[\tfrac{1}{p}]$-coefficients. $\square$
This is enough to go from $k$ to $k^{\operatorname{perf}}$ with $\mathbf Z[\tfrac{1}{p}]$-coefficients by a limit argument (see e.g. Tag 0FH6).
Example. Injectivity is not true integrally, even for divisors on smooth projective $\bar{\mathbf F}_p$-varieties. For example, consider a $\pmb\mu_p$-quotient $X \to Y$ of smooth projective varieties with $X$ a complete intersection of dimension $\geq 2$. Then $\mathbf{Pic}^\tau_X = 0$ but $\mathbf{Pic}^\tau_Y = (\pmb\mu_p)^\vee = \mathbf Z/p$. See for example Cor. 1.2 of this preprint of mine for a modern account (but the result is much older).
I'm not sure if the same phenomenon occurs for a base change along a purely inseparable field extension.
For surjectivity with $\mathbf Z$-coefficients, by the argument above, it suffices to move a subscheme to a linearly equivalent one for which $[K(V):K(W)] = [K(X):K(Y)]$. This seems plausible at least in the quasi-projective case.
Remark. If you don't assume $f$ is flat or l.c.i. or $Y$ is regular, I'm not sure what the definition of $f^*$ is. (At least Fulton does not define this, nor does the Stacks project as far as I can tell.)
If you work in equal characteristic, you have a chance of survival. In characteristic zero, this amounts to check that Chow groups do see nilpotent extensions. In characteristic $p>0$, if $p$ is invertible in your coeffcient Ring $R$, what you want is true as well (in fact this is true for higher Chow groups). I do not know any explicit reference though (but this does not mean such thing does not exist). However, this is an easy consequence of combinations of results which are documented.
To see it first observe that higher Chow groups are representable in $DM^{eff}(k,R)$ at least for equidimensional affine schemes: this is Theorem 4.2 in Bivariant cycle cohomology, by Friedlander and Voevodsky (in the book "Triangulated categories of motives"). This extends to non affine schemes as follows: using Suslin's beautiful paper Motivic complexes over nonperfect fields, we may replace $k$ by its perfection without changing $DM^{eff}$ with $R$ coefficients, and then apply the main result from Kelly's book which allows to replace arguments of resolution singularities as in the paper of Friedlander and Voevodsky above by resolution by $\ell$-alterations with $\ell\neq p$, which exist by a well known result of Gabber.
Finally, note that all motivic invariants tend to be invariant under universal homeomorphisms after you invert the exponential characteristic: this may be seen a joint paper of mine with Déglise for $DM$ (a way to summarize what I wrote above is that, using Suslin's paper, alle the representability results in section 8 of my joint paper with Déglise are true over arbitrary fields), and, more generally, in a paper of Khan and Elmanto for $SH$.
If you are happy to tensor with $\mathbf{Q}$, then all this extend to arbitrary schemes, since $DM$ satisfies $h$-descent with rational coefficients.
Thank you! Yet does there exist an easier proof? Is it really necessary to invert p? Since $X$ is universally homeomorphic to $Y$, any cycle in it is a pullback of a cycle in $Y$, and any cycle in $\mathbb{A}^1\times X$ is a pullback of a cycle in $\mathbb{A}^1\times Y$; doesn't this help?
I am pretty sure that the Frobenius is not invertible with integral coefficients: this would contradict the $p$-adic version of Bloch-Kato (obtained by replacing $\ell$-adic cohomology with synthomic cohomology). As for the existence of an elementary proof away from $p$-torsion, I do not see why not (not that you may use Suslin's paper as a basis to reach such a thing, since most of it is about cycles themselves).
|
2025-03-21T14:48:31.303585
| 2020-06-19T08:40:31 |
363506
|
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|
Stack Exchange
|
Embedding Calabi-Yau manifolds in projective space
When studying homological mirror symmetry, a lot of work is done not in the setting of complex manifolds, but of smooth (quasi-)projective varieties, see e.g. a paper from Orlov. However, the actual case one wants to consider is usually that of Calabi-Yau-manifolds.
It is usually not explained why the former case includes the latter, although it seems to be used very often that it does. My question is therefore: Can every (possibly non-compact) Calabi-Yau be embedded into projective space, giving a smooth quasiprojective variety (I read that it sometimes can't be included as a projective one)? I assume that to show this one would have to (like for Fano manifolds) construct an ample line bundle over it, as that would already imply the result. If it is possible, how can it be shown, and more importantly if not, why is Orlov's approach even justified?
Maybe a few further remarks: It seems like to embed a complex manifold $X$ into projective space, it must neccessarily be Kähler, as the Kähler structure of $\mathbb{C} P^n$ restricts to one on $X$, but this is obviously fulfilled here. Also, I read the article on ncatlab about Calabi-Yau-varieties, and to me it seems like it also supposes that these are equivalent to Calabi-Yau manifolds, although it only describes the analytification direction.
A quick remark, for the compact case: if your definition of Calabi-Yau manifold is the strongest one, namely holonomy equal to SU(n) or, equivalently, simples connected, trivial canonical bundle and no holomorphic p-forms except for bottom and top degree, then it is automatically projective, by Kodaira's criterion.
Thanks a lot, this seems like a very elegant criterion! However I definitely need this result for non-compact Calabi-Yaus, as the LG-models considered by Orlov become a lot more easy to handle in the compact case.
So, which is your definition of Calabi-Yau? I am curious! P.S. The Kodaira criterion I am speaking about in this situation is "compact Kähler manifold with $H^2(X,\mathcal O_X)=0$ is projective".
To define superstring theories on it, the manifold $X$ needs to be Kähler, and vanishing of conformal anomalies (or of R-symmetry anomalies, as one wants to perform a topological twist) further constrains the first Chern class to be trivial. That should be everyting (I'm not really an expert so I don't know how this compares to the other definitions, but it seems pretty weak). However, partial or more general results are still very much of interest; but the theory here depends crucially on a holomorphic function $W$ with target $X$, and this function becomes trivial in the compact case.
Your definition allows every open subset of a Calabi-Yau manifold to be also a Calabi-Yau manifold. For example, the unit ball in complex Euclidean space is a Calabi-Yau manifold in your sense, but not a quasi projective variety, as its closure in projective space is not a projective variety. So you need a better definition, with some sort of completeness or inextensibility, at the least.
K3 surfaces are not necessarily embeddable in projective space, but they are Calabi Yau in most senses.
@BenMcKay, indeed, I was tacitly thinking about dimension greater than or equal to three!
@BenMcKay Thanks! I can't say I completely understand your arguments, but yes, that makes sense. I wonder if it helps to restrict to Calabi-Yau manifolds of dimension 3, as they are the most important cases in HMS, or as diverietti says to the dimension being $\geq 3$.
@MarkusZetto, you can still take an open ball of dimension three...
My understanding is that the question is now: is every complete Kaehler manifold with trivial canonical bundle and of complex dimension 3 or more biholomorphic to a quasi projective variety? I don't know the answer.
@BenMcKay, stated in this form unfortunately the answer is no for trivial reasons: take a non projective complex torus...
For what it’s worth, these non-compact CYs are generally considered as local models of singularities in compact CYs, which in dimension three are algebraic (see here: https://mathoverflow.net/questions/30629/are-calabi-yau-manifolds-in-dimension-3-algebraic/30634#30634) I don’t know if that means that all noncompact CYs arise this way.
Also, there’s no reason from the physics point of view to restrict to the quasi projective case. To the extent you can make sense of topological string on the manifold, all the mirror symmetry stuff should hold, even if it’s only been proven for algebraic manifolds.
|
2025-03-21T14:48:31.303910
| 2020-06-19T09:41:34 |
363509
|
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|
Stack Exchange
|
The derivative of a $C_0$-semigroup with respect to a perturbation parameter
Let $H$ be a Hilbert space, and $A : H \to H$ be the (semi-bounded) generator of the $1$-parameter $C_0$-semigroup $[0, \infty) \ni t \mapsto \mathrm e ^{-t A}$. Let $B : H \to H$ be a bounded operator, and consider the "perturbation" $[0,1] \ni \varepsilon \mapsto A + \varepsilon B$. I would like to use the formula
$$\mathrm e ^{-t (A + B)} - \mathrm e ^{-t A} = \int _0 ^1 \frac {\mathrm d} {\mathrm d \varepsilon} \ \mathrm e ^{-t (A + \varepsilon B)} \ \mathrm d \varepsilon$$
which I believe is true, but I do not know where to find. (In order to be true, it might be necessary to consider the formula in a strong sense, i.e. applied on some arbitrary $v \in H$.)
Question: Could you please help me with a bibliographic reference for the above?
Davies' "One-parameter semigroups" doesn't have it. Please also notice that I am not interested in a proof, but only in a citable reference.
Furthermore, is it true that $\frac {\mathrm d} {\mathrm d \varepsilon} \ \mathrm e ^{-t (A + \varepsilon B)} = -t \ \mathrm e ^{-t (A + \varepsilon B)} B$ ?
A standard reference is Semigroups of Linear Operators and Applications to Partial Differential Equations by Pazy.
The last statement is not true, unless $A$ and $B$ commute.
@MichaelRenardy: If it is not true, do you know the correct statement? Is it like the Duhamel formula: $\frac {\mathrm d} {\mathrm d \varepsilon} \ \mathrm e ^{-t (A + \varepsilon B)} = \int_0^1 \mathrm e ^{-s t (A + \varepsilon B)} (-t B) \mathrm e ^{-(1-s) t (A + \varepsilon B)} \ \mathrm d s$?
@AlexM. have a look at the Kato reference I gave. You can extract the answer from there.
The answer can be found in chap. IX of T. Kato's "Perturbation Theory for Linear Operators" and is, essentially, an elementary consequence of Duhamel's formula for inhomogeneous linear differential equations of order 1 in Hilbert spaces:
$$\mathrm e ^{-t(A+B)} v - \mathrm e ^{-tA} v = - \int _0 ^t \mathrm e ^{-(t-s) (A+B)} B \ \mathrm e ^{-s A} v \ \mathrm d s$$
for all $v \in H$.
I think Theorem 2.20 in Chapter IX of Kato,
Kato, Tosio, Perturbation theory for linear operators., Classics in Mathematics. Berlin: Springer-Verlag. xxi, 619 p. (1995). ZBL0836.47009.
has something to do with your question. A link to the section can be foud for example here.
|
2025-03-21T14:48:31.304189
| 2020-06-19T09:45:20 |
363510
|
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|
Stack Exchange
|
Maximum nonintersecting interval pick
This surely has been solved in the context of scheduling already! (Shall I ask on some computer SE instead?)
Assume we have a set of closed "intervals" on $\mathbb Z$ ($\mathbb R$ isn't different, I guess) and like to pick the maximum cardinality of nonintersecting intervals from it. I bet the obvious greedy algorithm (first throw out all intervals that completely enclose another, then pick leftmost, repeat) already gives the solution.
A reference would suffice. (If the proof is trivial, please leave it to me :-)
Is this it? https://en.wikipedia.org/wiki/Interval_scheduling
Bingo and THX! I admit I could have found that myself if I wanted to and weren't too lazy (those summer temperatures :-). Question can be closed, I can find the original literature with a formal proof of the greedy algorithm myself.
|
2025-03-21T14:48:31.304282
| 2020-06-19T10:12:11 |
363511
|
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|
Stack Exchange
|
A strange convergence for a semigroup of operators
I am reading B. Simon's "Kato's inequality and the comparison of semigroups", and I am having troubles understanding a part of the proof of Theorem 1 therein, that goes as follows:
Let $A,B$ be semibounded self-adjoint operators on some $L^2(M, \mu)$, such that $\mathrm e ^{-t A} f \ge 0$ for all $f \ge 0$. Denote by $D(A)$ the domain of $A$, and by $Q(A)$ the domain of the quadratic form associated to $A$. If $| \mathrm e ^{-t B} u | \le \mathrm e ^{-t A} |u|$ for all $u$, then $u \in D(B)$ implies $|u| \in Q(A)$.
The proof should be elementary and straightforward, but I am having difficulties with a certain passing to the limit. Specifically, if $u \in D(B)$, then it is clear from the general theory of $1$-parameter semigroups that
$$\lim _{t \to 0} \ \left( u, \frac {1 - \mathrm e ^{-t B}} t u \right) = (u, Bu) \ .$$
(Parantheses denote scalar product.)
What is not clear to me, though, is why
$$\lim _{t \to 0} \ \left( |u|, \frac {1 - \mathrm e ^{-t A}} t |u| \right) = (|u|, A |u|) \ .$$
More specifically, why does the limit even exist? And why does $|u| \in D(A)$, in order to be able to write $A |u|$? Or am I misreading Simon's ideas?
Simon has the unfortunate habit of sometimes writing something like $(u,A u)$, when he means $q_A(u,u)$, where $q_A$ is the quadratic form associated with $A$. In any case, Ouhabaz' article "Invariance of Closed Convex Sets and Domination Criteria for Semigroups" seems to be the better source for domination of semigroups - for example, he gives characterization purely in terms of the quadratic forms where Simon has form and operator domains.
Simon refers to this explicitly in "An abstract Kato inequality...", on page 1068 just after the Definition.
|
2025-03-21T14:48:31.304431
| 2020-06-19T11:38:14 |
363517
|
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|
Stack Exchange
|
Existence of a bounded right inverse to a linear closed surjective operator
Let $A:D_A \subseteq H \to K$ a linear closed surjctive operator between two Hilbert spaces $H$ and $K$.
One would expect that in such a situation there must exist a bounded right inverse of $A$, namely an operator $R:K \to H$ such that $AR=Id_K$. In fact this is certainly true if $A$ is bijective but the proof doesn't seem to go through with the hypothesis of surjectivity.
Any ideas what is going on in this situation ?
EDIT: Although the answers cover my original question, i think it is quite natural at this point to ask wether this is true if $H$ and $K$ are more generally Banach space instead of Hilbert spaces.
Does something go wrong if you simply quotient out the kernel? Then the induced operator is bijective.
But then how do you know that the quotient operator is still closed?
Ok now I see even if the operator is not bounded the kernel is closed, right?
The kernel is closed but I don't think it has to be complemented in $D_A$. I smell a counterexample (at least if you want $R$ to be bounded) but I don't have time to think about it now ...
Maybe differentiation on $L^2(0,\infty)$ would work as a counterexample? The right inverse would essentially have to be the antiderivative and unboundedness of the domain should show that it is unbounded. I don't have time to check the details now, sorry.
Differentiation on $L^2(0,\infty)$ is not surjective.
If the quotient map $q:X \to X/Y$ has a right inverse $j:X/Y \to X$ then $jq:X \to X$ is a projection whose kernel is $Y$. Such a $j$ does not exist if $Y$ is uncomplemented in $X$.
@GiorgioMetafune that depends on what you mean by "uncomplemented". The image of this projection need not be a closed subspace, but every subspace of a pre-Hilbert space has an unclosed algebraic complement.
@Nik Weaver Yes, I am only dealing with closed subspaces in general Banach spaces. In my situation $jq(X)$ is closed since coincides with the kernel of $I-jq$.
It's still true in the unbounded case, and you can see this using polar decomposition. Write $A = BU$ where $B$ is some positive unbounded operator on $K$ and $U$ is the orthogonal projection from $H$ onto a closed subspace $H_0$ followed by some isometry from $H_0$ onto $K$. We can take $B$ to be a multiplication operator, $B = M_f$, on $K = L^2(X)$, and then the fact that $BU$ is surjective implies that $f$ must be bounded away from zero. So $1/f$ is a bounded function and thus $B^{-1} = M_{1/f}$ is a bounded operator. Finally, $U^*B^{-1}$ is the desired bounded right inverse.
You are assuming H is a Hilbert space. The notation may suggest it, but this was never specified explicitly.
Yes, I thought it was obvious from the notation that we're talking about Hilbert spaces.
Yes I forgot to mentioned that the spaces are assumed to be Hilbert.
It seems to me that one can prove the existence of a right inverse without using the polar composition: If we endow $D(A)$ with the graph norm, $A$ becomes a continuous linear surjection between Hilbert spaces and thus has a continuous linear right inverse into $D(A)$ which is also continuous as an operator with values in $H$.
This applies also to real Hilbert spaces and in some situations even to Banach spaces, if the range $K$ is projective, e.g., $\ell^1$.
|
2025-03-21T14:48:31.304680
| 2020-06-19T11:38:39 |
363518
|
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|
Stack Exchange
|
What is a bipartite hypergraph?
Bipartite graphs are very useful, and I am looking for a generalization of this concept to hypergraphs. I found two different definitions of bipartite hypergraphs:
In the Wikipedia page Hypergraph, a bipartite hypergraph is defined as a hypergraph whose vertices can be partitioned into two parts, $X$ and $Y$, such that each hyperedge of cardinality at least 2 contains at least one vertex from $X$ and one vertex from $Y$. It is equivalent to Property B or to 2-colorability.
Aharoni and Kessler (1990), and later Annamalai (2016), use the term bipartite hypergraph in a stronger sense: vertices can be partitioned into two parts, $X$ and $Y$, such that each hyperedge of cardinality at least 2 contains exactly one vertex from $X$, and all its other vertices are from $Y$.
What is the more standard meaning of this term today? And, is there an alternative term to the second (stronger) meaning, that I can use to differentiate it from the first (weaker) meaning?
Both (and more...) notions are used.
Coloring hypergraphs is annoying, and so we come up with silly names like “property B” and “rainbow” (to be fair, rainbow is a good name).
This also comes up for independent sets. Is it a set containing no edges, a set that each edge meets at most once, or other?
The reason there’s no fixed standard is that there are a ton of naturally useful definitions, which unfortunately don’t coincide when edges have more than two elements. (I’ve heard it said, “the problem in coloring $k$-uniform hypergraphs is that $k-1$ doesn’t usually equal 1.”)
This gets worse. Even notions like “cycle” have oodles of possible meanings, and (again) all of them are studied.
In writing or reading a paper, you pick a notion that fits your situation, and go with that. You also (of course) explicitly say what notion you’re going with by telling the reader exactly what you mean by “bipartite” (or whatever).
#hypergraphsAreHard
Yeah, I've noticed that hypergraphs are hard :) Do you know of an alternative term to the second meaning? (for the first meaning there is "2-colorable" and "property B").
I don’t know a special name for that, but that doesn’t say a lot.
Good answer, though this gave me flashbacks to arguing with my advisor about how many vertices should be in common between consecutive links of a hypercycle. Fortunately it turned out that the answer was “it doesn’t matter, they all give the same answer” for the problem I was working on!
As Pat Devin wrote there are many notions of "bipartite hypergraphs".
There is a chapter about this topic in the Hypergraph book by Berge (it can be found online as a pdf, look at chapter 5)
Berge, Claude, Hypergraphs. Combinatorics of finite sets. Transl. from the French, North-Holland Mathematical Library, 43. Amsterdam etc.: North-Holland. x, 255 p. {$} 79.00; Dfl. 150.00 (1989). ZBL0674.05001.
In my dissertation I studied different concepts of "bipartite hypergraphs" with a focus on matching and flow problems. It is also available online: Matchings and Flows in Hypergraphs.
One can look at the various equivalent definitions of bipartite graphs and than try to generalize them for hypergraphs. One property is the 2-colorability, but here, one demands that every subhypergraph is 2-colorable. Another possibility is to forbid certain kind of odd cycles (called strong odd cycles). Hypergraphs without strong odd cycles are than called balanced. It depends on the situation or application how one defines a "bipartite" hypergraph.
|
2025-03-21T14:48:31.304950
| 2020-06-19T11:59:35 |
363523
|
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|
Stack Exchange
|
Iwasawa decompostion and simply connected subgroups
Let $G$ be a semisimple Lie group, i.e. $G$ is connected and Lie algebra of $G$ is semisimple. We know by Iwasawa decomposition, there are connected subgroups $K$, $A$ and $N$ of $G$ such that the multiplication map from $K\times A\times N$ to $G$ is a smooth diffeomorphism. It is well known that $A$ and $N$ are simply connected and Lie algebra of $A$ is abelian. Since Lie algebra of $A$ is abelian and $A$ is connected and simply connected, clearly $A$ is isomorphic to $\mathbb R^n.$ This means that $G$ contains a copy of $\mathbb R^n.$ Moreover, $\{1\}\times A\times\{1\}=A$ is a closed subgroup of $G.$ This shows that any semisimple Lie group is noncompact if $A\neq \{1\}$ !! I am new in this business. Is my claim right? What can be necessary and sufficient condition for compactness of a semisimple Lie group in this direction of thoughts?
$K$ is a maximal compact subgroup, so $G$ is compact just when $G=K$.
@McKay. $K$ is compact if and only if $G$ has finite center!!
A semisimple group has finite centre. (A non-trivial identity component in the centre would give, upon passing to the Lie algebra, a non-trivial Abelian ideal in the Lie algebra.)
@LSpice: I think there is a unique counterexample, if $G$ is simple: $G=\widetilde{SL}_2 \mathbb{R}$ has $K$ noncompact, because in $SL_2\mathbb{R}$, the associated $K$ is a circle. So I think that in general, for any semisimple $G$, $K$ is compact unless $G$ has a factor which is an infinite cover of $SL_2\mathbb{R}$.
@BenMcKay, aha. I am used to thinking only of algebraic groups. I guess that $\widetilde{\operatorname{SL}}_2(\mathbb R)$ is the universal cover?
@LSpice: yes, the universal cover.
As you say, $K$ is compact just when $G$ has finite center. We can see this in the case of the universal covering group of the isometry group of the hyperbolic plane. Connected and simply connected solvables are diffeomorphic to Euclidean space. So for any semisimple Lie group $G=KAN$, we see that $A=1$ and $N=1$ and $K$ has finite center just when $G$ is compact, in which case $G=K$.
|
2025-03-21T14:48:31.305111
| 2020-06-19T12:20:12 |
363524
|
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|
Stack Exchange
|
Riemannian manifolds on which bounded sets are totally bounded
A Riemannian manifold is known to be complete if and only if any of its closed bounded subset is compact.
Is there a similar criterion for Riemannian manifolds for which every bounded subset is totally bounded with respect to the geodesic distance, that is, every bounded set for the geodesic distance can be covered by finitely many geodesic balls of arbitrarily small radius?
Obviously this will happen in any complete Riemannian manifold.
There are manifolds that are bounded without being totally bounded, see https://math.stackexchange.com/q/749891/105127.
What do you actually mean by similar criterion? You started with $P \iff Q$ where $P$ is complete and $Q$ is closed bounded subset is compact. Is your goal to replace $Q$ by "every bounded subset is totally bounded" and ask what is the corresponding generalization of $P$?
(Minor comment: doesn't Hopf Rinow require finite dimensions?)
Convexity would be enough.
Possibly a not very useful observation: for metric spaces the property "every bounded subset is totally bounded" is equivalent to "the Cauchy-completion having the Heine-Borel property.". Hopf Rinow says that finite dimensional Riemannian manifolds is geodesically complete IFF Cauchy-complete; unfortunately the Cauchy completion of an arbitrary Riemannian manifold need not be equipped with a good Riemannian structure.
@WillieWong I am wondering about a necessary and sufficient condition as fruitful as the equivalence for completeness.
|
2025-03-21T14:48:31.305240
| 2020-06-19T12:50:01 |
363530
|
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|
Stack Exchange
|
Invertible matrix and 0-1 vector
Let $n\ge 1$ and $A$ be any $n\times n$ real invertible matrix. Can we always find a 0-1 vector $b\in\{0,1\}^n$ such that each entry of $Ab$ is nonzero? For example, if $A$ is the identity matrix, then we can (only) take $b$ to be the all-one vector.
Edit: The answer given by Pat Devlin is beautiful. From his construction, we know that there exists a parallelepipedon whose each vertex lies on some coordinate plane. Indeed, for $n=3$ from the matrix
$$A=\left(\begin{matrix}1&0&0\\
0&1&-1\\
1&-1&-1
\end{matrix}\right)$$
we can obtain the desired parallelepipedon using the columns of $A$. His solution suggests that the invertibility assumption on $A$ seems inappropriate.
Can we put a rectangular solid (particularly, a cube) in $R^3$ such that each vertex lies on some some coordinate plane. In higher dimension, if the column vectors of $A$ are orthogonal with each other (i.e.,$A^T A$ is a nonsingular diagonal matrix), can we always obtain a desired 0-1 vector $b$ such that each entry of $Ab$ is nonzero?
What is a appropriate assumption on $A$ for a positive solution? Does orthogonal matrices suffice?
@JohnMachacek Perhaps you are misunderstanding the question, because the answer is definitely yes for $n=2$: either one of the columns has all nonzero entries (in which case the corresponding basis vector is the desired $b$) or both have a single zero entry, in which case (to be invertible) each row has a single nonzero entry, hence the all-one vector suffices for $b$.
@SeanClark yes my mistake. I guess all this time off has left my especially slow in the mornings :)
Who and why vote to close ?
Can the vertices of the hypercube be covered by $n$ hyperplanes (the preimages of the coordinate hyperplanes)?
@ZachTeitler More specifically, $n$ hyperplanes (passing through the origin) in general position.
Yes. See answer. :-)
The answer for $n \leq 2$ is yes, and for $n \geq 3$, it's no. (The $n=1$ case is left to the reader)
The case $n=2$: If $v_i$ are the rows of the matrix, then each equation $v_1 \cdot x = 0$ and $v_2 \cdot x = 0$ describes a co-dimension $1$ space. And the four corners of the square $\{0,1\}^2$ cannot be covered by two lines passing through the origin.
The case $n \geq 3$: For $n \geq 3$, we can cover all the vertices of the cube with three independent hyperplanes through the origin. Namely...
$$x_1 = 0$$
$$x_2 -x_3 = 0$$
$$x_1 - x_2 - x_3=0$$
It's easy to see these three conditions describe a co-dimension three space [i.e., the conditions are independent]. To see they cover the vertices of the cube, note that failure of the first and second equations would imply $x_1 =1$ and $x_2 \neq x_3$. Since each variable is $0,1$, this would force $x_2 + x_3 = 1$, so the third equation will be satisfied.
The $n\geq 3$ case phrased as a matrix:
Make the first row $[1, 0, 0, 0, 0, \ldots]$. Then the second row $[0, 1, -1, 0, 0, \ldots]$. Then make the third row $[1, -1, -1, 0, 0, 0, \ldots]$. These three rows are independent (yay!), so just fill up the rest of the matrix with more linearly independent rows until you have a basis (i.e., an invertible matrix). Then see above argument why $A \vec{x}$ will have at least one of the first three coordinates equal to $0$.
Literature:
Also, there are some really cool results about covering the cube with hyperplanes. See for instance this article about covering all but one of the vertices and also this article about covering the vertices in a way where each equation is needed, and every variable actually shows up.
Edit 1: I changed the equations to handle the $n=3$ as well.
Edit 2 (this part added to address orthogonal case): As per comment, perhaps you’d like the matrix to be orthogonal as well. That’s also fine (at least for $n \geq 4$. I haven’t thought about $n=3$ in this case).
Take the first three rows to be $[1, -1, 0, 0, 0, 0, \ldots]$ and $[0,0,1,-1,0,0,\ldots]$ and $[1,1,-1,-1,0,0,\ldots]$ (and renormalize these so that they each have length 1). Then extend this to an orthonormal basis and fill the rest of the matrix with those rows. This works in the same ways as the above example. (This was actually the first construction I gave, but then I swapped it out for the above, which only needed $n=3$.) ((I notice that I don’t enjoy thinking about the orthogonal $n=3$ case. It might be that you can’t make any orthogonal matrix to do it. On the other hand, it could be that I’m bad at visualizing rotations of the cube that put all its corners onto three orthogonal planes...))
Edit 3: You seem to want this to be true for really special matrices. I don’t think that’s really going to happen. The cube fits inside a pretty darn small space (union of three planes), so this seems to give us a lot of wiggle room for all sorts of stuff. :-/
Thank you for your cool construction. What happens if $A$ is an orthogonal matrix?
That’s ok too. I’ll add a note
@W.Wang let me know if there’s anything else you’d like to be answered with this.
|
2025-03-21T14:48:31.305602
| 2020-06-19T14:09:22 |
363537
|
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|
Stack Exchange
|
Brachistochrone for a rolling sphere with slippage
I was recently looking into generalisations of the brachistochrone problem: for example, in this article the authors study the brachistochrone with Amontons-Coulomb friction where a bead slides along a wire from one fixed point to another under the influence of gravity and friction.
In this one the author considers the brachistochrone for a sphere which rolls without friction. It seems like in the literature no-one has studied the brachistochrone/tautochrone curves for a homogeneous sphere with slippage between sphere and surface ie. a sphere which slides and rolls with friction (I am happy to be corrected if I am wrong on this). Is there any reason why this solution would be impossible to obtain or something to do with the effect of slip being too difficult to analyze?
I note that in the first article it is possible to express the solution in terms of elementary functions but perhaps that is no longer possible for the rolling sphere problem.
Let me suggest these generalizations of the brachistochrone, which generate numerous unsolved problems: https://aapt.scitation.org/doi/abs/10.1119/1.14839 and https://aapt.scitation.org/doi/abs/10.1119/1.15423 and https://aapt.scitation.org/doi/abs/10.1119/1.15001
Hi David, I had seen this article as well but this generalization is for a frictionless track, I was interested in a problem with the presence of friction which dissipates energy. I will consider that article again though.
|
2025-03-21T14:48:31.305987
| 2020-06-19T15:02:39 |
363539
|
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|
Stack Exchange
|
Working with truncation of inverse of integers (number of necessary digits)
Ideally I would like to find exact value of $Mr'$ where $r'=\frac{1}{M'}$ where $M'$ ranges from $1$ to $T+1$ and $1\leq M\leq T$ holds. However in real world approximations have finite precision. If I use truncation of $r'$ I get an estimate of $y\leq Mr'$. I just want to get a reasonable approximation of $r'$ such that if $y\leq1$ or $>1$ then $M\leq M'$ or $M'<M$ holds respectively.
The period of rationals can be as large as denominator. So in this case the number of digits should be at least $T$ in worst case since $M$ is upper bound by $T$. Is the lower bound $T$ on number of digits sufficient up to constant multiplicative factors or perhaps we may need $poly(T)$ number of digits or perhaps we need superpolynomial in $T$? What is the upper bound?
Thus my question is following: How many digits in base $b$ should $r'$ be truncated to as a function of $T$ to ensure that $M<M'\leq T+1$ holds iff $0<y<1$ holds (that is it works even at $M'=M+1$ at every $M\in\{1,2,3\dots,T-1,T\}$)?
Well, of these, $1/(T + 1)$ is the closest to $0$ and $T/(T + 1)$ is the closest to $1$, so just however many digits you need to distinguish $1/(T + 1)$ from $0$ and $T/(T + 1)$ from $1$, right?
Yes the resolution is $\frac1{T+1}$ which gives an estimate of possibly at most $\frac{c\log T}{\log b}$ for the number of $b$ digits where $c$ is some constant with $c>0$.
|
2025-03-21T14:48:31.306115
| 2020-06-19T15:17:49 |
363540
|
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|
Stack Exchange
|
Interpretation of the word Random
I have previous knowledge of what a random experiment is, but sometimes I get confused by the use of the word Random.
I can express my doubts as the following questions: if something is random them it is aleatory? if something has a defined distribution like in a random experiment then is not true that it is not random? If someone say something is random it is referring to a specific distribution like in a uniform distribution?
for example in the following paragraph:
Extinction is a process that can depend on a variety of ecological, geographical, and physiological variables. These variables affect different species of organisms in different ways, and should, therefore, yield a random pattern of extinctions.
It says I should be able to describe extinction with a distribution frequency?
This is a better question for http://math.stackexchange.com. At the research level, the closest thing to a precise meaning for the word "random" is something like "is a function on a probability space", but it sounds like that is not what you are looking for.
This is a great question. For instance, what does it mean to say "there is a 90% chance that it rained yesterday?" But I agree with the comments that this isn't the site where you'll get an awesome discussion going. (Maybe try math stackexchange or perhaps even a philosophy forum)
That said! Speaking as a mathematician who happens to specialize in probability, I have absolutely no idea what "random" means.
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2025-03-21T14:48:31.306255
| 2020-06-19T15:20:48 |
363542
|
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|
Stack Exchange
|
Gelfand-Kirillov dimension of the first Weyl algebra
How can we compute the Gelfand-Kirillov dimension (GK for short) of the first Weyl algebra?
As we know we can look at the Weyl algebra as a generalized Weyl algebra in the following way:
Let $A=\mathbb F[h]$ be a polynomial algebra in the variable $h$. Consider its automorphism $\sigma$ defined by $\sigma(h)=h-1$ and the element $t=h$. Then the generalized Weyl algebra $A(\sigma, t)$ is isomorphic to the Weyl algebra $A_1(\mathbb F)$. And the Gelfand-Kirillov dimension (GK for short) of the generalized Weyl algebras are known in Zhao, Mo, and Zhang - Gelfand–Kirillov dimension of generalized Weyl algebras
I understood that GK($\mathbb F[h]$) is 1 and GK($A_1(\mathbb F)$) is 2, which means that $GK(A_1(\mathbb F))=GK(\mathbb F[h])+1$.
Now the question is that why $GK(\mathbb F[h])=1$? And why $GK(\mathbb F[h_1,\ldots,h_m])=m$?
Let $R=\mathbb{F}[h]$. That $GK(R)=1$ comes straight from the definition of Gelfand-Kirillov dimension. Let $R_n$ be the $\mathbb{F}$-subspace of $R$ spanned by monomials of degree at most $n$. Set $f(n)=\dim R_n = n+1$. Then
$$GK(R) = \limsup(\log f(n)/\log n) = 1.$$
A similar argument applies if $R=\mathbb{F}[h_1,\dots,h_m]$. With $R_n$ as above, $f(n)=\dim R_n = \binom{n+m}{m}$ and
$$GK(R) = \limsup(\log f(n)/\log n) = m.$$
Regarding the Weyl algebra $A=A_1(\mathbb{F})$, it helps to know that GK dimension is stable under taking associated graded rings. We have $A=\mathbb{F}\langle x,y : xy-yx-1\rangle$. Take the standard filtration on $A$ with $\deg(x)=\deg(y)=1$, so $gr(A)=\mathbb{F}[x,y]$. Thus, $GK(A)=GK(\mathbb{F}[x,y])=2$.
Good references on GK dimension are "Growth of algebras and Gelʹfand-Kirillov dimension" by Krause & Lenagan, as well as "Noncommutative noetherian rings" by McConnell & Robson.
Thank you very much for your answer! Sorry, I cannot understand why $lim sup (log(n+1)/log n)=1$ and $lim sup(log {n+m\choose m}/log n)=m$?
Try using l'hospital's rule.
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2025-03-21T14:48:31.306393
| 2020-06-19T15:46:19 |
363543
|
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|
Stack Exchange
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Globally hyperbolic spacetimes and future Cauchy developement
Let $(M,g)$ be a globally hyperbolic Lorentzian spacetime with a non-compact Cauchy hypersurface $S \subset M$. Let $ \Omega \Subset S$ be an open subset. Is it true that the future Cauchy development of $\Omega$, denoted by $D^+(\Omega)$, is a compact subset in $M$?
No.
Let $M$ be the maximally extended Schwarzschild solution. Let $S$ be (for example) the $t = 0$ hypersurface that goes through the bifurcate sphere. If you take $\Omega$ sufficiently large, then $D^+(\Omega)$ touches the inner singularity, and hence is not compact.
For a more trivial example: let $(M,g)$ be the portion of Minkowski space with $t\in (-1,1)$ and let $S = \{t = 0\}$. Take $\Omega = B(0,10)$.
On the other hand, the above example shows that it is not enough to ask that $M$ is Cauchy inextendible.
The statement is true if you add the condition that there exists a sequence of events $p_i\in M$ with $p_i \prec p_{i+1}$ such that $\cup (J^-(p_i)\cap S)$ covers $S$. But this is quite strong.
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2025-03-21T14:48:31.306612
| 2020-06-19T15:58:50 |
363546
|
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|
Stack Exchange
|
To what extent is a vector bundle on a smooth manifold determined by its restriction to the complement of a closed smooth submanifold?
The question is a follow-up to this one.
Let $N\subset M$ be a closed smooth submanifold of codimension $k\geq 1$ of a smooth manifold $M$. Let $E_1\rightarrow M$ and $E_2\rightarrow M$ be two vector bundles over $M$. Suppose $E_1|_{M-N}\sim E_2|_{M-N}$. Is it true that $E_1|_N\sim E_2|_N$?
Can we impose some conditions on $k$ and $n=\dim M$ such that under the condition $E_1|_{M-N}\sim E_2|_{M-N}$, we can conclude $E_1\sim E_2$? (a counter example for this statement in general is provided by $E_1=T\mathbb{S}^1\rightarrow\mathbb{S}^1,E_2=\text{mobius}\rightarrow\mathbb{S}^1$ and $N=\{\cdot\}$) (the difference with the previous question is that $M$ and $M-N$ are generally not homotopic).
The answer to the first question is No. Just think of the inclusion of the circle in the 2-dimensional sphere.
@user43326: That doesn't work. Any vector bundle on $S^2$ is trivial on $S^2\setminus S^1 = D^2\sqcup D^2$ and on $S^1$.
@MichaelAlbanese Thank you, yes, we need an example where the inclusion of N in M isn't null homotopic.
Here's an example to show that the answer to the first question is no.
Let $M = \mathbb{CP}^2$ and $N = \mathbb{CP}^1$. Note that $M - N$ is diffeomorphic to $\mathbb{C}^2$ and is therefore contractible. So for any two vector bundles $E_1, E_2 \to \mathbb{CP}^2$ of the same rank, we have $E_1|_{M - N} \cong E_2|_{M-N}$. However, we need not have $E_1|_N \cong E_2|_N$. For example, if $E_1 = \mathcal{O}_{\mathbb{CP}^2}(1)$ and $E_2 = \mathcal{O}_{\mathbb{CP}^2}$, then $E_1|_N \cong \mathcal{O}_{\mathbb{CP}^1}(1)$ and $E_2|_N \cong \mathcal{O}_{\mathbb{CP}^1}$ which are distinguished by the second Stiefel-Whitney class.
Am I right to say that the real version with $\mathbb{RP}^2$ and $\mathbb{RP}^1$ also provides a counter-example in the case of real vector bundles?
@quangtu123: Yes. Although note that the counterexample I mention in my answer works for real and complex bundles: $E_1|N$ and $E_2|N$ are not isomorphic as real bundles and hence cannot be isomorphic as complex bundles either, although $E_1|{M-N}$ and $E_2|{M-N}$ are isomorphic as real bundles and as complex bundles.
|
2025-03-21T14:48:31.306773
| 2020-06-19T16:13:35 |
363547
|
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|
Stack Exchange
|
GCH implies acceptability
I have been studying the concept of acceptability, particularly in its relation to GCH.
There are many versions of it in the sources I have found, with some slight variations, and some of them are claimed to be equivalent to GCH, some of them are claimed not to be equivalent to GCH. For example, Welch, in A Condensed History of Condensation, claims that GCH is equivalent to acceptability but not to weak acceptability. Friedman and Holy, in A Quasi-Lower Bound on the Consistency Strength of PFA, define a slightly different form of weak acceptability and prove that it is equivalent to GCH. Schindler and Zeman, in Fine Structure, define a fine-structural version of acceptability and say that it can be seen as a stronger form of GCH.
I am interested in any further references providing details and clarifying the relation between acceptability and GCH.
I am particularly interested in knowing whether the fine-structural version given in Schindler and Zeman is implied by GCH in some interesting class of models. More specifically, in the models $\textbf{L}[A]$, does GCH imply that for some appropriate choice of $A$ the $J$-structures $J^A_{\alpha}$ are acceptable for some unbounded class of $\alpha$'s in the sense given in Fine Structure Theory?
EDIT
A more basic question may be helpful. In which precise sense, if any, is fine-structural acceptability (defined for the $J$-hierarchy) equivalent to acceptability defined directly for the $L$-hierarchy?
Given GCH, for any infinite cardinal $\kappa$, $H_\kappa = J_\kappa^A$ for some $A\subseteq \kappa$ such that for any limit ordinals $\tau\leq \alpha < \kappa$, if $P(\tau)\cap J_{\alpha+\omega}^A\nsubseteq J_{\alpha}^A$, then there is a surjection from $\tau $ to $J_{\alpha}^A$ in $J_{\alpha+\omega}^A$. With Global Choice, one gets a class $A$ with this property such that $V = L[A]$. It seems harder to design $A = (A_\alpha){\alpha\in \text{Lim}}$ so that $A\alpha$ is amenable to $J_\alpha^A$. Schindler's Set Theory, Ex 11.1 suggests this is possible, so maybe Ralf knows the answer.
Nice comment, thanks.
Another related reference, is the following “An Abstract Condensation Property
” by David Richard Law.
Here is the abstract of it:
Let $A = (A, \dotsc)$ be a relational structure. Say that $A$ has condensation if there is an
$F : A^{< \omega} → A$ such that for every partial order $P$, it is forced by $P$ that substructures of $A$ which are closed under $F$ are isomorphic to elements of the ground model. Condensation holds if every structure in $V$, the universe of sets, has condensation. This property, isolated by Woodin, captures part of the content of the condensation lemmas for $L$, $K$ and other "$L$-like" models. We present a variety of results having to do with condensation in this abstract sense. Section 1 establishes the absoluteness of condensation and some of its consequences. In particular, we show that if condensation holds in $M$, then $M \models \mathrm{GCH}$ and there are no measurable cardinals or precipitous ideals in $M$. The results of this section are due to Woodin. Section 2 contains a proof that condensation implies $\Diamond_κ(E)$ for $\kappa$ regular and $E \subseteq \kappa$ stationary. This is the main result of this thesis. The argument provides a new proof of the key lemma giving GCH. Section 2 also contains some information about the relationship between condensation and strengthenings of diamond. Section 3 contains partial results having to do with forcing "$\operatorname{Cond}(A)$", some further discussion of the relation between condensation and combinatorial principles which hold in $L$, and an argument that $\operatorname{Cond}(G)$ fails in $V[G]$, where $G$ is generic for the partial order adding $\omega_2$ cohen subsets of $\omega_1$.
In the definition of the condensation I think that you have too many $P$s and not enough $A$s.
@AsafKaragila Fixed it.
Oddly enough, the thesis' abstract has that typo, but the definition on page 1 is correct.
Thanks for the reference.
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2025-03-21T14:48:31.307071
| 2020-06-19T16:27:41 |
363550
|
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|
Stack Exchange
|
Integral operator with Bessel kernel
For $x,y\ge 0$, let
$$
k(x,y)= \frac {J_1(2\sqrt{xy})}{\sqrt{xy}},
$$
where $J_1$ is the the Bessel function of the first kind
$$
J_{1}(z)=\sum_{k=0}^{\infty}(-1)^{k} \frac{\left(\frac{z}{2}\right)^{2 k+1}}{k !(k+1) !}.
$$
Does this kernel give a bounded linear operator on $L^2(0,+\infty)$:
$$
Af(x) =\int_{0}^\infty k(x,y) f(y) dy.
$$
Note that the kernel $k(x,y)$ is not square-integrable since
$$
\int_0^\infty \left( \frac {J_1(2\sqrt{x})}{\sqrt{x}}\right)^2 dx= 1.
$$
Cross post math.stackexchange
Yes it is. A sufficient conditions for the boundedness of an integral operator in $L^2(0,\infty)$ of the form
$$Tf(x)=\int_0^\infty K(xy)f(y)\, dy$$ is
$$\int_0^\infty \frac{|K(x)|}{\sqrt x}\, dx =C <\infty.$$ If $K$ is positive the above condition is also necessary and $C$ yields the norm. In our situation $K(x)=\frac{J_1(2 \sqrt x)}{\sqrt x}$ and
$$C=\int_0^\infty \frac{|J_1(2\sqrt x)|}{x}<\infty $$ since $J_1(x) \approx x, x \to 0$ and $|J_1(x)| \le \frac {C}{\sqrt x}, x \to \infty$.
Wonderful! Thank you very much. Do you by any chance know standard/textbook reference for this sufficient condition? Is there a special name for operators with the kernels K(x,y)=K(xy)?
It is easier if I explain how to do: write $$Tf(x)=\int_0^\infty K(z) \frac{f(z/x)}{x} dz$$ and use Minkowski for integrals that is $$|Tf|_2 \le \int_0^\infty |K(z) |f(z/x)1/x|_2, dz$$ (the last $L^2$-norm is with respect to $x$). After changing variable in the $x$-integral you get the formula above. Another way is go to the line writing $x=e^t, y=e^s$ and, after some manipulation one sees that the above formula is equivalent to Young estimate for convolutions with a L^1 function.
|
2025-03-21T14:48:31.307203
| 2020-06-19T17:32:13 |
363557
|
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|
Stack Exchange
|
Composing equal characteristic and mixed characteristic deformations
Suppose $k$ is an algebraically closed field of characteristic $p>0$ and $A$ is a $k$-point on a moduli space of certain objects over $k$. Suppose, moreover, that $A$ deforms to a $k[[t]]$-point $B$ and a $W(k)$-point $C$ (in characteristic $0$). I wonder if there is any literature discuss whether one can "compose" $B$ and $C$ to form a $W(k)[[t]]$-point $D$ that is a $p$-deformation of $B$ and a $t$-deformation of $C$?
I am interested in this question because I am trying to show a $W(k)$-point $C$ as above always has a property $P$ relative to $A$. We know that holds when $A$ verifies a condition $E$. Furthermore, a lift of $B$ to characteristic zero always satisfies $E$, hence having property $P$. Therefore, if we can find a composition of $B$ and $C$, then, as reduction preserves $P$, we are done.
Murphy's law(s) tell you that there exist many kinds of objects with quite arbitrary formal deformation spaces, in particular with ${\rm Def}_A \simeq \operatorname{Spf} W(k)[[t]]/(pt)$, which is the "wedge sum" of $\operatorname{Spf} W(k)$ and $\operatorname{Spf} k[[t]]$...
...even for "easy" objects such as surfaces of general type. What you want will be satisfied if you know that $\operatorname{Def}_A$ is smooth over $W(k)$, but that's a very strong assumption to make.
@PiotrAchinge Thank you! Could you tell me why $\text{Def}_A$ being smooth over $W(k)$ will make that works? I am not very familiar with deformation theory.
(Formal) smoothness basically means that for a surjection of complete $W(k)$-algebras $R'\to R$, the induced $\operatorname{Def}_A(R')\to \operatorname{Def}_A(R)$ is surjective. Your $B\in \operatorname{Def}_A(k[[t]])$ and $C\in \operatorname{Def}_A(W(k))$ together define a $D_0 \in \operatorname{Def}_A(R)$ where $R = k[[t]]\times_k W(k) = W(k)[[t]]/(pt)$ (look up Schlessinger's criteria). Now use the surjection $R' = W(k)[[t]]\to R$ to get your $D\in \operatorname{Def}_A(W(k)[[t]])$.
@PiotrAchinger I have a naive question. Does "$B \in \text{Def}_A k[[t]]$ and $C \in \text{Def}_A W(k)$ together define a $D_0 \in \text{Def}_A(R)$" come from the surjectivity of $\text{Def}_A (U \times_V W) \xrightarrow{} \text{Def}A (U) \times{\text{Def}_A(V)} \text{Def}_A(W)$? I think the criterion only apply to the case when $U,V, W$ are Artinian and $W$ is a small extension of $V$.
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2025-03-21T14:48:31.307358
| 2020-06-19T17:59:17 |
363558
|
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|
Stack Exchange
|
The scalar convergence in $\mathcal{C}(X)$ is topologizable?
Let $(X,\|.\|)$ be a separable Banach space and $\mathcal{C}(X)$ be the collection of all nonempty, closed and convex subsets of $X$. For any $C$ in $\mathcal{C}(X)$ we set
$$
s(x^*, C) := \sup_{x\in C}{\langle x^*, x \rangle}\qquad;\qquad \|C\| := \sup_{x\in C} \|x\|
$$
to define for the set $C$ respectively its support function $s(., C)$ and radius $\|C\|$.
On $\mathcal{C}(X)$ we use the following convergence : a net $\{C_\alpha\}$ convergent scalarly to $C_\infty$ in $\mathcal{C}(X)$ if
$$
\lim_\alpha s(x^*, C_\alpha)=s(x^*, C_\infty)\qquad \forall x^*\in X^*
$$
Can we say that the scalar convergence in $\mathcal{C}(X)$ is topologizable ?
Isn't it just the initial topology with respect to the functions $s(x^,-)$? So a prebase would be given by sets of the form ${C\text{ such that }s(x^,C)\in U}$ for $x\in X^*$ and $U$ a open set of $\overline{\mathbb R}$.
Do you know the Wijsman topology?
I just read the definition on Wikipedia. Is my comment not enough to answer your question?
|
2025-03-21T14:48:31.307459
| 2020-06-19T18:49:41 |
363564
|
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"Ville Salo",
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|
Stack Exchange
|
Bound on the cardinality of the union of a chain
I was trying to go through the proof that submodules of free modules over a principal ideal domain are free and I ran into a set theoretic cardinality problem that I can't find answered anywhere: Suppose I have an index set $I$, and two chains {$U_{i}$}$_{i\in I}$ and {$V_{i}$}$_{i\in I}$ such that for each $i$, $\mid U_{i}\mid\leq\mid V_{i}\mid$. Is it necessarily true that $\mid\bigcup _{i\in I} U_{i}\mid\leq\mid\bigcup _{i\in I} V_{i}\mid$? I can see that this is not true if one doesn't assume the collections are directed families. I've tried well-ordering $I$ and trying to make an injection using transfinite induction by constructing the collection {$C_{i}$}$_{i\in I}$, where $C_{i}=U_{i}\setminus\bigcup _{j<i}U_{j}$ but it's getting me nowhere. I feel like this should be obvious.
For a counterexample take $\omega_1$ as index set, $U_i = i$, and $V_i = \omega$ for all $i$.
I think it's true if you assume that $(I,\lt)$ is totally ordered; $|U_i|\le|V_i|$ for each $i$; $U_i\subseteq U_j$ whenever $i\lt j$; and $V_i\subsetneqq V_j$ whenever $i\lt j$. Would that be good enough?
Proof of @bof's claim: Let $V = \bigcup_i V_i$. Clearly $|I \times V| \geq |\bigcup_i U_i|$. The additional assumption implies $|V| \geq |I|$, so $|I \times V| = \max(|I|, |V|) = |V|$ (for infinite sets that is; the finite case is easy).
@VilleSalo $|V|\ge|I|$ does not necessarily follow, e.g. if $I=\mathbb R$ and $V_i={x\in\mathbb Q:x\lt i}$, but we can replace $I$ with a cofinal well-ordered subset $J$ and then everything is fine, $|V|\ge|J|$ etc.
Good point. Easy to trip up on these things (if you are not a set theorist).
|
2025-03-21T14:48:31.307589
| 2020-06-19T19:30:28 |
363566
|
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|
Stack Exchange
|
Conv A = Dual B
I have two cones $A$ and $B$ in a Euclidean space.
I want to show that $\mathrm{Conv}\,A=\mathrm{Dual}\,B$; that is, $a\in\mathrm{Conv}\,A$ if and only if $\langle a,b\rangle\ge 0$ for any $b\in B$.
The inclusion $\mathrm{Conv}\,A\subset \mathrm{Dual}\, B$ (or equivalently $\mathrm{Dual}\,A\supset \mathrm{Conv}\,B$) is easy --- one has to check $\langle a,b\rangle\ge 0$ for pairs $a\in A$ and $b\in B$.
It remains to show the opposite inclusion $\mathrm{Conv}\,A\supset \mathrm{Dual}\,B$. I see that in principle it can be done by calculations.
Do you know tricks that help to prove the opposite inclusion $\mathrm{Conv}\,A\supset \mathrm{Dual}\,B$? Is there software that can help?
The cones $A$ and $B$ are very concrete; they are given by few algebraic identities and inequalities in $\mathbb R^{10}$.
(In principle I could describe the sets of vectors here, but I do not think it might help.)
I assume that $\operatorname{Conv} A$ denotes the convex hull of $A$? But why do you need to take the convex hull of a cone, given that a cone is already convex?
@JochenGlueck I did not say that cones are convex (in fact they are not).
Thank you very much for your response! Hmm, may I suggest to include a definition of your usage of the word cone in the question? (I just browsed half a dozen books about ordered vector spaces; many of them define a cone indeed to be always convex, while some others do not - and there also seem to be some further differences between the various definitions of a cone in the literature).
@JochenGlueck a set $X$ is a cone if $v\in X$ implies that $\lambda{\cdot}v\in X$ for any $\lambda\ge0$.
Thank you very much.
|
2025-03-21T14:48:31.307732
| 2020-06-19T20:16:32 |
363570
|
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|
Stack Exchange
|
Can all proper sublattices of $\mathbb{Z}^n$ be generated cyclically?
Let $\Lambda \subset \mathbb{Z}^n$ be a proper sublattice (so that $\Lambda \ne \mathbb{Z}^n$). We say that $\Lambda$ is cyclically generated if there exists a matrix $M \in \text{GL}_n(\mathbb{Z})$ and an element $\mathbf{u} \in \mathbb{Z}^n$ such that $\Lambda$ is equal the $\mathbb{Z}$-span of $\{\mathbf{u}, M \mathbf{u}, M^2 \mathbf{u}, \cdots \}$. Is it true that all proper sublattices are cyclically generated? If so, how would one prove this, and if not, what's a counterexample?
@SamHopkins I don't think what you claim is true. For example, take $M = \left(\begin{smallmatrix} 6 & 5 \ 1 & 1 \end{smallmatrix} \right)$ and $\mathbf{u} = (1,1)^T$. Then $M \mathbf{u} = (11, 2)^T$ which plainly does not have the same length as $\mathbf{u}$.
I'm pretty sure the answer is yes, but my solution seems inelegant. Specifically: we can assume WLOG that the sublattice is generated by elements that are multiples $d_i \vec{e}i$ of the standard basis vectors, and such that $d_1 | d_2 | \dots | d_n$. Then we can choose $M$ to be bidiagonal with $1$ on the diagonal and $\frac{d{i + 1}}{d_i}$ below the diagonal, and $d_1 \vec{e}_1$ will be cyclic.
@user44191 why can you make such an assumption? In general one can assume that $\Lambda$ has a basis which forms an upper triangular (or lower triangular) matrix, but not necessarily diagonal. We can only perform row or column operations, not both.
@StanleyYaoXiao You can make such an assumption because $\Lambda = A \Lambda'$ for some $A \in GL_n(\mathbb{Z})$ and $\Lambda'$ of the form specified; if $M', u'$ are a cyclic pair for $\Lambda'$, then $u := Au', M := A M' A^{-1}$ is a cyclic pair for $\Lambda$. And such an $A$ and $\Lambda'$ are guaranteed by Smith normal form.
What user44191 sounds right to me. In particular, why can't we perform both row and column operations?
@WillSawin A lattice can be seen as the column space (or row space, by taking transposes) of some generator matrix. Row operations can change the column space, see this for a simple example.
@WillSawin I am thinking that if you have $n$ column vectors $\mathbf{v}_1, \cdots, \mathbf{v}_n$ say, then taking column operations preserves the lattice spanned by the $\mathbf{v}_j$'s, but row operations do not.
@StanleyYaoXiao It's worth mentioning that the simple case of $M$ being a suitable permutation matrix ("cyclic rotation operator") leads to the case of cyclic lattices, which are known to not include all lattices. One can generalize the notation of cyclic lattices to ideal lattices, where I believe $M$ is the companion matrix of some polynomial, and again gets a strict subset of all lattices.
@Mark fascinating! I didn't know the notion of 'ideal lattices' have been developed. If you have more references on them please let me know, perhaps via email if you don't mind.
If you send $M$ to $g Mg^{-1}$ and $u$ to $gu$ then the matrix $N$ whose columns are $u, Mu, M^2u,\dots, M^{n-1} u$ a becomes $g N$. That gives you the row operations.
@user44191 I think you should post your comment as an answer...
Some standard latticework allows us to write $\Lambda = \text{im}(A)$ for some $n$ by $n$ matrix $A$ (with nonzero determinant). Using Smith normal form, there are some $U, V \in GL_n(\mathbb{Z})$ such that $UAV = D$, where $D$ is a diagonal matrix with diagonal elements $d_1 | d_2 | \dots | d_n$.
Write $\vec{e}_i$ for each of the standard basis vectors. Then let $\vec{u} := U^{-1} d_1 \vec{e}_1$, and let $M := U^{-1}BU$, where $B$ is the bidiagonal matrix with all $1$s on the diagonal, and $\frac{d_{i + 1}}{d_i}$ on the subdiagonal. Note that $det(B) = 1$, and so $B \in GL_n(\mathbb{Z})$; correspondingly, so is $M$. I claim that $(\vec{u}, M)$ is a cyclic pair for $\Lambda$.
Proof: Write $\vec{v}_i := U^{-1} d_i \vec{e}_i$ for each $i$. Note that $\vec{u} = \vec{v}_1$. Then each $\vec{v}_i \in \text{im}(U^{-1}D) = \text{im}(U^{-1}DV^{-1}) = \text{im}(M) = \Lambda$; further, it's not hard to see that they generate $\Lambda$ (by the same reasoning taken backwards).
We can check that $M \vec{v}_i = U^{-1}BU U^{-1} d_i \vec{e}_i = U^{-1} B d_i \vec{e}_i = U^{-1} (d_i \vec{e}_i + d_{i + 1} \vec{e}_{i + 1}) = \vec{v}_i + \vec{v}_{i + 1}$ for $1 \leq i < n$. Noting the base case that $\vec{v}_1 = \vec{u}$, we can use induction to see that $\vec{v}_{i + 1} = M \vec{v}_i - \vec{v}_i \in \text{span}(\{M^j \vec{u}\}_{j = 0}^i)$. As the $\{\vec{v}_i\}$ are a generating set, $\text{span}(\{M^j \vec{u}\}_{j = 0}^{n - 1}) = \Lambda$, and we are done.
|
2025-03-21T14:48:31.308034
| 2020-06-21T14:42:38 |
363712
|
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|
Stack Exchange
|
Existence of a special function
Consider a $C^2$ bounded domain $D$ of $\mathbb{R}^d$. Let $b \subset \partial D$ a non-empty part of the boundary. Let $n(x)$ be the unit outward vector on $\partial D$.
Is there any smooth function $f \in C^\infty(\overline{D})$ such that $f=\partial_n f=0$ on $b$.
PS: It was further required the following condition $f=1$ on $\partial D \setminus b$, but it seems to be impossible in this case.
If there is no direct answer, it would be more helpful to mention a reference where the existence of such a function is studied.
Such a $b$ must be both open and closed in the boundary, so be a union of components. If the boundary is connected, that means that you ask $f$ to be $0$ or $1$ on all of $\partial D$, so you may just take it to have that same value on all of $\overline D$.
@Lspice I don't see why it must be open and closed. I can assume it to be open w.r.t $\partial D$. The same value on whole domain is not what I need.
According to your requirements, unless I am misreading, $b = f^{-1}(0) \cap \partial D$ and $\partial D \setminus b = f^{-1}(1) \cap \partial D$, both of which are closed in $\partial D$.
You're right. I think I should at least modify the second condition. Now, I consider just the first one.
Does $f=0$ satisfy your conditions?
No. Note that I removed the second condition.
You ask for a smooth function that vanishes in a certain range, with (as you say) no other condition. Why doesn't $f = 0$ satisfy that?
From your assumptions, you have a $C^2$ function $\rho:\mathbb R^d\rightarrow \mathbb R$, such that
$$
D=\{x\in \mathbb R^d, \rho(x)<0\}, \quad \partial D=\{x\in \mathbb R^d, \rho(x)=0\},
$$
and
$
x\in \partial D\Longrightarrow d\rho(x)\not=0.
$
As a result, locally the set $D$ is given by an inequality $x_d<\phi(x')$ and the boundary by the equality $x_d=\phi(x')$ where $\phi$ is a $C^2$ function on $\mathbb R^{d-1}$. Now the function $f$ given by
$$
f(x) =\bigl(x_d-\phi(x')\bigr)^2
$$
satisfies your requirements, but is only $C^2$. Going back to the function $\rho$,
you can use a theorem of H.Whitney saying that given the closed set $\partial D$, you can find a $C^\infty$ function $f$ positive on the complement of $\partial D$ and vanishing on $\partial D$. Then of course $df$ must vanish on $\partial D$ since $f$ is non-negative.
Thank you. Could you give your opinion about the condition I removed, If we can get something close?
That removed condition would make the smoothness impossible: the most caricatural case would be the situation where $b$ is reduced to a single point of the boundary.
Why can your locally defined function be pieced together to a global function?
Partition of unity.
@Bazin could you provide a reference to a proof of the existence of such $\rho$? Thanks.
Well, near every point of the boundary, you have a local function $\rho$, so that you can cover the boundary by a finite collection of open sets ${U_k}_{1\le k\le N}$ to which you add $U_0=\Omega$. Using a partition of unity (see e.g. Section 1.4 in Hörmander's ALPDO I) you can construct a global $\rho$.
|
2025-03-21T14:48:31.308270
| 2020-06-21T15:14:15 |
363713
|
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|
Stack Exchange
|
De Rham's theorem for top-forms in manifolds with boundary
In page 79 of Bott-Tu, "Differential Forms in Algebraic Topology", they define the relative de Rham theory as follows:
Let $f:S\to M$ be a smooth map. Define the complex $\Omega^*(f)$ by
$$\Omega^k(f):=\Omega^k(M)\oplus\Omega^{k-1}(S)$$ $$\underline{\mathrm{d}}(\alpha,\beta)=(\mathrm{d}\alpha,f^*\alpha-\mathrm{d}\beta)$$
It is easy to prove that $\underline{\mathrm{d}}^2=0$ which allows us to define the cohomology $H^*(f)$. As a particular case, one can consider a submanifold $\imath:N\hookrightarrow M$ and define
$$\Omega^*(M,N):=\Omega^*(\imath)$$
My interest lies in the case when $N=\partial M$ and $M$ compact, where one can also define the integral of top forms as
$$\int_{(M,\partial M)}(\alpha,\beta):=\int_M\alpha-\int_{\partial M}\beta$$
It is easy to check, using Stoke's theorem, that
$$\int_{(M,\partial M)}\underline{\mathrm{d}}(\alpha,\beta)=0$$
Thus, we have a well defined map
$$\tag{1}\label{one}\int_{(M,\partial M)}:H^n(M,\partial M)\to\mathbb{R}$$
If $\partial M=\varnothing$, then $H^n(M,\partial M)=H^n(M)$ and the previous integral is the standard one. The de Rham's theorem for top-forms then tells us that if $M$ has no boundary
$$\tag{2}\label{two}\int_M:H^n(M)\to\mathbb{R}\quad \text{ is an isomorphism}$$
However, with boundary we have:
1. It is surjective (applying \eqref{two} over the boundary and using elements of the form $(0,\beta)$).
2. Its kernel is isomorphic to $H^n(M)$. Sketch of the proof: for every $[\alpha]\in H^n(M)$, build an element $[(\alpha,\beta)]$ such that $\int_{(M,\partial M)}(\alpha,\beta)=0$ using de Rham's theorem over the boundary. This map is well defined.
I have a heuristic argument to show that $H^n(M)$ is always zero: given $\alpha\in\Omega^n(M)$, take the double of $M$ along the boundary $\partial M$ and extend to some $\widetilde{\alpha}\in\Omega^n(M\sqcup_{\partial M}M)$ such that its integral is zero (using a tubular neighborhood over $\partial M$). Then using \eqref{two} (the double has no boundary) shows that $\widetilde{\alpha}$ is exact and, therefore, its pullback to $M$, which is $\alpha$, is also exact.
This seems a very strong result that I haven't found anywhere, while the proof seems very simple, thus I doubt if there are obstructions to the extension that invalidate the proof.
So the questions I have in mind (all of them are almost the same question) are:
Is $H^n(M)=0$ if $M$ is compact with boundary?
Is there a useful characterization of $H^n(M,\partial M)$ that can be used in this context?
Is there a de Rham's theorem like \eqref{two} for manifolds with boundary (with no prescribed boundary conditions)?
Is there a de Rham's theorem like \eqref{two} for relative cohomology?
If $H^n(M)\neq 0$, is there another a map $G:H^n(M,\partial M)\to \mathbb{R}$ such that $(\int_{(M,\partial M)},G)\to\mathbb{R}^2$ is an isomorphism?
Check out Poincaré-Lefschetz duality (and for example https://mathoverflow.net/questions/19936/pairing-used-in-lefschetz-duality). These are more appropriate for https://math.stackexchange.com/
It is indeed true that $H^n(M)=0$ if $M$ is a compact manifold with boundary. In particular, $H^n(M,\partial M)\cong\mathbb{R}$ by Lefschetz duality (as Chris Gerig mentioned) and the integral (1) is an isomorphism.
The only reference I have found that states this results is:
Differential forms: theory and practice. Steven Weintraub. Academic Press (Elsevier) 2014.
Theorem 8.3.10 for compact manifolds with boundary.
Theorem 8.4.8 for non-compact manifold with boundary using compact support forms.
|
2025-03-21T14:48:31.308618
| 2020-06-21T15:26:04 |
363715
|
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|
Stack Exchange
|
Solution to the matrix equation AX=XB in the form of generalized inverses
It is known that the solution (suppose it exists) of Sylvester matrix equation $AX-YB=C$ can be given in the form of generalized inverses as
$$X=A^{-}C + A^{-}ZB + (I-A^{-}A)W,$$
and
$$Y = - (I-AA^{-})CB^{-} +Z - (I-AA^{-})ZBB^{-}$$
with $W$ and $Z$ being arbitrary.
See J. K. Baksalary and R. Kala [Linear Algebra Appl. 25, 41-43 (1979; Zbl 0403.15010)].
My question is how about the matrix equation $AX=XB$?
Can the solution the matrix equation $AX=XB$ be given in the form of generalized inverses as above?
P.S. Note the when $C=0$ the solution the matrix equation $AX=YB$ has the form
$$X=A^{-}ZB + (I-A^{-}A)W,$$
and
$$Y=Z-(I-AA^{-})ZBB^{-}.$$
But if we demand
$$X=A^{-}ZB + (I-A^{-}A)W=Z-(I-AA^{-})ZBB^{-}=Y,$$
we have to solve a more complicated equation which is not needed.
Thus it seems that the solution of matrix equation $AX=XB$ can not be seen as a special case of matrix equation $AX-YB=C$ in the form of generalized inverses.
isn't $X=0$ a solution?
Yes, X=0 certainly is a solution. But I want to give a formula for all the other ones.
In fact, I can find nonzero solution like this $X= (I-A^{-}A)Z (I-BB^{-})$ with Z being arbitrary. But this is not a general formula for all solutions.
|
2025-03-21T14:48:31.308718
| 2020-06-21T16:14:42 |
363718
|
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|
Stack Exchange
|
Prove that infimum of a metric outer measure is a metric outer measure
Let $X$ be a compact measure space and $\mathcal{O}(X)=\{U\subset X, \textrm{U - open set}\}$
and let $\lambda$: $\mathcal{O}(X) \rightarrow \left[0,+\infty\right]$ be a metric outer measure.
Prove that $\mu^*(A)=\inf_{A\subset B\in\mathcal{O}(X)} \lambda(B)$ is a metric outer measure.
I have just troubles with the last condition:
$A, ~B\in \mathcal{O}(X)~\textrm{and}~\rho(A,B)>0 \Rightarrow \mu^*(A\cup B)=\mu^*(A)+\mu^*(B)$,
where $\rho(A,B):=\inf{\{d(x,y):~x\in A,~y \in B\}}.$
Do you know how to show it or find any sources containing the proof?
you can consider the infinum only among the open sets that are positively separated from the other set
I don't get it, can you explain more?
Ok, first, i think you wat $\mu^$ to be defined for all sets, not just open, right? So let $A,B\subset X$ be such that $\rho(A,B)=r>0$. Let $A'=\bigcup_{x\in A} B(x,\frac{r}{3})$ and $B'=\bigcup_{x\in B} B(x,\frac{r}{3})$, which are open and $\rho(A,B)\ge \frac{r}{3}>0$. Then $\mu^(A)$ can be calculated as infimum of $\lambda(U)$, where $A\subset U\subset A'$, and same for $B$ and $B'$, as well as $A\cup B$ and $A'\cup B'$.
I understand but I still can't see why the condition is satisfied according to what you have written.
Okay, never mind. I drew it, thanks for help!
|
2025-03-21T14:48:31.308826
| 2020-06-21T16:16:26 |
363719
|
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|
Stack Exchange
|
Can we say that $\{f_n\}\text{ is uniformly integrable over }E\setminus (\cap_p B_p)$?
Let $(E,\mathcal{A},\mu)$ be probability space and $\{f_n\}$ be sequence of functions such that
$$
\sup_n\int_{E}|f_n|d\mu<+\infty.
$$
Let $\{B_p\}$ be a sequence non-increasing in $\mathcal{A}$ such that $\mu(\cap_p B_p) =0$ and for every $p$
$$
\{f_n\}\text{ is uniformly integrable over }E\setminus B_p
$$
Can we say that $\{f_n\}\text{ is uniformly integrable over }E\setminus (\cap_p B_p)$?
This would have been a better question for Math.SE and I've flagged it to be migrated there.
No.
Take $E=[0,1]$ with Lebesgue measure. Let $f_n = n 1_{[0, 1/n]}$, so that $\int_E |f_n|\,d\mu = 1$ for every $n$, and $B_p = [0, 1/p]$. Note that $|f_n| \le p$ on $E \setminus B_p = (1/p, 1]$ for every $n$, so that $\{f_n\}$ is indeed uniformly integrable over $E \setminus B_p$. But clearly $\{f_n\}$ is not uniformly integrable over $E \setminus \bigcap_p B_p = (0,1]$.
|
2025-03-21T14:48:31.309117
| 2020-06-21T16:27:01 |
363720
|
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"Ville Salo",
"Vivek Shende",
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"Zach Teitler",
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|
Stack Exchange
|
Short exact sequences every mathematician should know
I'd like to have a big-list of "great" short exact sequences that capture some vital phenomena. I'm learning module theory, so I'd like to get a good stock of examples to think about. An elementary example I have in mind is the SES:
$$
0 \rightarrow I \cap J \rightarrow I \oplus J \rightarrow I + J \rightarrow 0
$$
from which one can recover the rank-nullity theorem for vector spaces and the Chinese remainder theorem.
I'm wondering what other 'bang-for-buck' short exact sequences exist which satisfy one of the criteria:
They portray some deep relationship between the objects in the sequence that is non-obvious, or
They describe an interesting relationship that is obvious, but is of important consequence.
Should every mathematician know what a short exact sequence is?
@VilleSalo I was under the impression every mathematician does know what a short exact sequence is :) I'm a computer science student though, so I wouldn't know :D
@SiddharthBhat I definitely do not think every mathematician knows what a short exact sequence is. While this topic probably appears in most first-year PhD courses, if a mathematician goes into, say PDEs, there is no real reason they would remember this topic. Just like in computer science, if someone goes into systems they might forget the Ford-Fulkerson algorithm, and a theoretical computer scientist might forget the soldering they learned in computer architecture.
I once met a mathematician who had never heard of a homeomorphism. Maybe they were not a true Scotsman though.
@VilleSalo : I once heard the following (probably apocryphal) story about an oral exam. Professor: "Are $M_1$ and $M_2$ homeomorphic?" Student: "$M_1$ is, but $M_2$ isn't."
@DavidWhite: first year PhD courses? I learned them in a (required) first year undergrad course... (But the department I went (Leiden) is pretty strong on algebra ― other universities in the country wait until second year...)
I don't know if it's apocryphal, but it's not a realistic clueless answer (since it would mean something if homeomorphism had a slightly different meaning).
(That said, my education certainly has serious gaps in other places, so it's easy to imagine how some curricula can get you by without exposure to this concept.)
@TimothyChow See https://mathoverflow.net/a/53905/2926, especially Gerry Myerson's first comment.
@DavidWhite Case in point, I'm into PDE and while thanks to my own curiosity I know what a short exact sequence is, I never had to depend on that knowledge neither for research nor for any compulsory class and I'd guess the same holds for many people I know. That is not to say that we don't use them, in fact there is one that is used all the time, but naming a thing of which you only know a single example isn't really helpful.
I did four years maths undergraduate at Cambridge (specialising in probability in 4th). I did 1st and 2nd year algebra courses (Groups and Groups, Rings and Modules). I'm pretty sure I never encountered 'exact sequences'. Just finishing 4yrs of PhD at Cambridge (probability), and still haven't heard of them until this question! -- not that, looking them up now, they're a complicated concept.
@DavidWhite What is a "first-year PhD course"? My PhD is research, not courses... (I've attended some lecture series that I'm interested in, but not in any official sense; I just rock up to a lecture I'm interested in and learn!) I'm aware that the US system works somewhat differently to the European system, and the UK system in particular, but I'm not very clued up on exactly how. Are these PhD courses a US thing?
If I may, I'd suggest that adding a link to Wikipedia would tremendously clarify the question for many readers. Those readers may or may not be mathematicians, but either way, it seems like the question would be improved, unless your goal is actually to turn away anyone who is not a "true enough mathematician" to already know what a short exact sequence is.
Do split short exact sequences count? If so, there are many direct sum decompositions that "every mathematician should know"...
@DavidZ I added a link to Wikipedia. I was under the impression that all pure math students see this idea, hence the tongue-in-cheek title. Clearly, I'm mistaken! I shall write more straightforward titles next time: "Most spectacular exact sequences" for example.
Although the answers so far probably constitute a nice list of examples for short exact sequences, I cannot refrain from noting that this list is quite far away from what the title of the question suggests...
@SamT: Yes, it also varies in the US, but it's common to have 0-2 years of classes in addition to research requirements. (I'm a physicist, but I audited a few "PhD math classes". For example, topology I/II following Hatcher, and topology III following Milnor)
A proof by contradiction: I have not heard of "short exact sequences", I recognize only five of the given examples from my MSc algebra two decades ago (but would have only been able to write two with luck if you'd give me a rough description of what I'm trying to capture); I'd argue I'm a mathematician as I have an Erdös number of four via three distinct coauthors (who have their "3" via three distinct others). I've just not used much number theory. Thus, the OP asks for something that doesn't exist.
Perhaps someone can edit the title. None of the answer give an example of something every mathematician should know.
(Referring to the comment by Durinx) Perhaps this is an answer to the question "Short exact sequences every mathematician knows".
@TimothyChow: If you replace "homemorphic" with "isomorphic" and talk about vector spaces, I have personally seen it (in writing, no less).
@TerryLoring : I think the title is fine as long as one takes it as a figure of speech rather than a literal statement.
Am I right in saying that the majority of people with mathematics degrees do not know what a Hilbert space is? In which case, I highly doubt that every mathematician knows what a short exact sequence is.
@HollisWilliams: I think it depends very much on whether you mean Bachelor's degree, Master's degree or PhD, in which country/institution and possibly whether one has studied pure or applied mathematics, even though the latter two terms are not well defined. Certainly, a course in functional analysis is not required fare for a Bachelor's degree everywhere in the world and it might be possible to sidestep it in some postgraduate degrees, although how sensible that is is another matter.
I first learnt about short exact sequences during my masters, but not in any depth. I'm now doing a PhD focusing mostly on homological algebra, so short exact sequences are of course a very important aspect of my work these days. Also @HollisWilliams I've never studied Hilbert spaces. I don't know if that makes me an ill-informed mathematician, but I've simply had no need for them in my own work. Mathematics is a vast subject, you can't learn it all, so you learn what you need and what interests you. Which will often mean central concepts of certain areas will pass you by.
I think it's interesting that many of the answers to the question don't seem to satisfy the requirements of the question at all, they are of the form "here's an abelian group X, here's a subgroup Y, and now 0 -> Y -> X -> X/Y -> 0 is an exact sequence". These answers should be on another thread called "important subgroups of abelian groups", not here!
I made it out of a PhD program at a US university without knowing what a short exact sequence was.
Maybe this discussion raises the following question: What are notions that everyone in their respective fields would find basic (and would expect mathematicians from other fields to know) - while in reality people from other fields have (almost) never heard of those.
We're going to do all this work compiling this list, and then BuzzFeed is just going to steal all of our content for its next clickbait compilation.
There is one obvious sequence that underlies all vector analysis and a lot that builds up on it, no matter if its applied analysis, PDE, physics or the original foundations of algebraic topology. Yet it is rarely written out, as the people in the applied fields prefer to split it into its constituent statements and the people in pure mathematics are inclined to immediately write down some generalization instead. What I am talking about is of course the relationship between the classic differential operators on 3D vector fields:
$$0 \to \mathbb R\to C^\infty(\mathbb{R}^3;\mathbb{R}) \stackrel{\operatorname{grad}}{\to} C^\infty(\mathbb{R}^3;\mathbb{R}^3) \stackrel{\operatorname{curl}}{\to} C^\infty(\mathbb{R}^3;\mathbb{R}^3) \stackrel{\operatorname{div}}{\to} C^\infty(\mathbb{R}^3;\mathbb{R}) \to 0 $$
FWIW, isn't this clearer when written in terms of cohomology with the $d$ operator + duality? It's unclear at first glance why the objects involved are scalar field -> vector field -> perpendicular vector field -> scalar field. It makes a lot more sense (to me) to think in terms of forms, where we get scalar field[0 form] -> vector field [1 form] -> 2 forms [dual to vector field] -> 3-forms [dual to scalar field].
@SiddharthBhat "[T]he people in pure mathematics are inclined to immediately write down some generalization instead."
@SiddharthBhat It strongly depends on who you are talking to, that's why I deliberately chose to write it this way, because we tend to almost forget about it. I love differential forms, but I had to teach this stuff to electrical engineers for a while. Telling them about differential forms only gets you confusion and the eternal question "will it be on the exam?". But writing it out this way after having done all those "div curl = 0" and "find the potential" calculations for weeks will get you a lot of appreciation for the underlying structure.
There is also the single variable version that every freshman is implicitly exposed to: $0 \to {\mathbb R} \to C^\infty({\mathbb R}; {\mathbb R}) \stackrel{\frac{d}{dx}}{\to} C^\infty({\mathbb R}; {\mathbb R}) \to 0$. That is to say, the fundamental theorem of calculus, complete with "+C". Incidentally the ${\mathbb R}$ is also missing on the left of your long exact sequence.
...well, strictly speaking the full fundamental theorem of calculus also comes with an important splitting of this short exact sequence (the definite integral). Anyway, I submit this sequence as a literal answer to the question title of a "short exact sequence every mathematician should know".
@mlk This sequence may be short and exact, but it is not a short exact sequence.
@J.vanDobbendeBruyn Can you clarify why it's not short exact? $\mathbb R^3$ has trivial De-Rham cohomology, right? Hence the sequence ought to be short-exact?
@SiddharthBhat To clarify my comment, a short exact sequence is an exact sequence that is of the form $0 \to A \to B \to C \to 0$. Sequences with slightly more (or fewer) terms do not quality, even if they are still short in a colloquial sense.
This might be very basic, but the short exact sequence
$$
0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0
$$
is both an injective resolution of $\mathbb{Z}$, and a flat resolution of $\mathbb{Q}/\mathbb{Z}$, making it a very useful exact sequence in many homological computations.
The exponential sheaf sequence:
$$0\to 2\pi i\,\mathbb Z \to \mathcal O_M {\buildrel\exp\over\to}\mathcal O_M^*\to 0$$
where $\mathcal O_M$ is the sheaf of holomorphic functions on the complex manifold $M$, $\mathcal O_M^*$ is the sheaf of non-vanishing functions and
$$\exp : \mathcal O_M \to \mathcal O_M^*$$
is induced by the usual exponential function.
Restricting to sections over any open set $U$ we get a long exact sequence including the map
$$\cdots \to H^0(\mathcal O_U^*)\to H^1(2\pi i\,\mathbb Z|_U) \to \cdots.$$
The cohomology group $H^0(\mathcal O_U^*)$ is the set of non-vanishing holomorphic functions on $U$. Roughly speaking, the cohomology group $H^1(2\pi i\,\mathbb Z|_U)$ gives us ($2π i$ times) an integer for each closed loop in U. This map essentially tells us the winding number, around zero, of each non-vanishing holomorphic function, as we go around such a loop. I say all mathematicians should know this because the winding number is so fundamental. But this short exact sequence is also the starting point for a long path through all kinds of interesting places like the Riemann-Roch theorem.
Also, the image of a line bundle $L\in H^1(M,\mathcal{O}^\times)$ in the next cohomology group corresponds to the first Chern class $c_1(L)\in H^2(M,\mathbb{Z})$.
I've been a mathematician for 30 years and one of my most cited papers introduced a new exact sequence. Yet I have no idea what is a sheaf. So I fail to see how this is an exact sequence every mathematician should know.
@TerryLoring I'm envious. I remember learning this stuff for the first time with great fondness (30 years ago almost exactly). I can never have that feeling again, but you can!
@DanPiponi What is a good place to learn this? This seems like a beautiful way to look at winding.
I learnt it from Gunning's "Lectures on Riemann Surfaces" (1966). Confusingly the sequel to this book is also called "Lectures on Riemann Surfaces (Jacobi Varieties)". You want the one that gives this exact sequence (written as $0\to\mathbb Z\to\mathcal O\to\mathcal O^*\to 0$) at the bottom of page 26.
I just learnt of another use of this exact sequence: On the sheaf of complex functions, this can be used to show that the cokernel pre-sheaf is not sheafy. The cokernel is locally trivial, but globally not. (https://math.mit.edu/~mckernan/Teaching/07-08/Spring/18.726/model2.pdf)
I barely know what a sheaf is, but I guess this is a generalisation of a short exact sequence that literally every mathematician does know: the exactness of $0\to 2\pi i \mathbb{Z}\to \mathbb{C}\stackrel{\exp}{\to}\mathbb{C}_*\to 1$ is essentially Euler’s identity; $\exp(\pi i)=-1$.
@HJRW Exactly right. Now generalise to holomorphic functions from some domain U to the 3 spaces you mention. Now it gets interesting. For example on $\mathbb{C}-{0}$ you can compute log continuously on small subdomains but not globally because log is "multivalued". Sheaves give a nice way to talk about the fact that local logs exist (away from 0) and $H^1$ measures the extent to which you fail to be able to extend these to global functions.
I find it hard to believe that three days have gone by and no one has explicitly mentioned
$$
0 \to \Bbb Z \to \Bbb R \to \Bbb S^1 \to 0
$$
I think a short exact sequence that every teacher should know is
$$ 0 \to \mathbb R^d \to \mathrm{Isom}(\mathbb R^d) \to \mathrm{O}(\mathbb R^d) \to 0, $$
maybe for $d=2$ or $d=3$. Better still, forget about the origin and see $E=\mathbb R^d$ as an affine space.
It is a great visual helper, in the sense that many elementary properties take a concrete sense. It makes it clear that there is a copy of the group on the left hand side (the translations) in the isometry group, and that there is some other component described by the group on the right hand side, although how to perform the decomposition is not obivous (which is of course expected for working mathematicians, but a nice way to introduce it to others). Once we describe how the right hand side sits in the isometry group, it is very visual also that is not canonical, and that in fact there is a choice to be made to define such a copy, introducing what a section is and why it is interesting/important.
More generally: semidirect products for groups.
(These are split, by the way.)
The short exact sequence
$$ 0 \to \mathrm{rad}({\mathfrak g}) \to {\mathfrak g} \to {\mathfrak g}/\mathrm{rad}({\mathfrak g}) \to 0$$
separates a Lie algebra ${\mathfrak g}$ into its solvable radical $\mathrm{rad}({\mathfrak g})$ and its semisimple projection ${\mathfrak g}/\mathrm{rad}({\mathfrak g})$, and is absolutely fundamental in the classification theory of Lie algebras, particularly in the characteristic zero, finite dimensional setting in which Levi's theorem is available to split the above sequence. Given the ubiquity of Lie algebras and Lie groups in mathematics, as well as the wider philosophy of separating algebraic objects into their "solvable" and "simple" components, I would submit that this sequence should be known to any mathematician.
In a somewhat similar spirit, the short exact sequence
$$ 0 \to [G,G] \to G \to G/[G,G] \to 0$$
that separates a group $G$ into its commutator subgroup $[G,G]$ and its abelianisation $G/[G,G]$ is generally the first step towards understanding solvable groups $G$ (because if $G$ is solvable then $[G,G]$ is also solvable with the derived length decremented by one), while the analogous short exact sequence
$$ 0 \to Z(G) \to G \to G/Z(G) \to 0$$
separating a group $G$ into its centre $Z(G)$ and the quotient $G/Z(G)$ is similarly often the first step in understanding nilpotent groups (because if $G$ is nilpotent then $G/Z(G)$ is nilpotent with the nilpotency degree decremented by one). There are of course analogues of these sequences for Lie algebras also.
In the category of topological (or algebraic) groups, one also has the short exact sequence
$$ 0 \to G^0 \to G \to G/G^0 \to 0,$$
where $G^0$ is the identity component and $G/G^0$ is the totally disconnected projection. In principle, this separates the study of such groups into the connected and totally disconnected cases.
One might add to this list$$0\to H_2(G)\to[F,F]/[R,F]\to G\to0$$for perfect $G=F/R$ with free $F$, and similar extensions for Lie algebras, etc.
"Every mathematician should know" is too much to ask, but I do think the following is a great short exact sequence that captures a vital phenomenon:
$$0 \to K(H) \to B(H) \to Q(H) \to 0.$$
$K(H)$ is the compact operators on a Hilbert space $H$, $B(H)$ is the bounded operators, and $Q(H)$ is the Calkin algebra. The "vital phenomenon" is that being invertible modulo the compacts, i.e., being Fredholm, is the same as being invertible in the Calkin algebra.
It won't help you learn module theory but IMHO it deserves to be on a big list ...
I voted up, even if this is not much more than a definition (of Calkin algebra).
@DenisSerre it's a fair comment. I guess the substantive fact is that the Calkin algebra is a C*-algebra --- easy enough for us now that we have the GNS construction, but it wasn't easy for Calkin ...
Would have gone for the Toeplitz extension myself :) (haven't checked to see if it's already here)
Yeah, that's another good one.
@NikWeaver but many textbooks nowadays prove that quotients of $C^$-algebras (by closed two-sided ideals) are once again $C^$-algebras before they get into the GNS construction... (For instance, see Murphy, or Takesaki, or Blackadar.) Is it really needed? Or is this considered to be part of the GNS construction?
@J.vanDobbendeBruyn I guess the modern approach is to define C*-algebras abstractly, and you're right, we don't need GNS to see that the Calkin algebra is an abstract C*-algebra. I should have said "concrete C*-algebra".
@NikWeaver ah yes, the historical distinction between $B^$-algebras and $C^$-algebras... Indeed, looking at Calkin's original paper, it seems that he does not attempt to prove the $B^$-identity, and immediately aims for a faithful representation. (In fact, according to Wikipedia, the $B^$-identity was only introduced a few years later, in 1946.)
@J.vanDobbendeBruyn yes, if I remember right he uses something like an ultraproduct construction to do this. Anyway your point is well-taken.
I strongly doubt there is any short exact sequence that every mathematician should know, but I certainly wish that those of them who know that for a (co)chain complex $(C,d)$
$$
0\to\operatorname{Im}(d)\to\operatorname{Ker}(d)\to H(C,d)\to0
$$
is short exact, would also know that
$$
0\to H(C,d)\to\operatorname{Coker}(d)\to\operatorname{Im}(d)\to0
$$
is short exact too.
It's sad this is so far down the list, because I think it's the only short exact sequence I actually know!
The $\text{Im}(d)$ complex in the second sequence needs to be shifted by one degree, no?
@Feryll Well, $\operatorname{Im}(d)$ can be naturally graded in two ways, shifted wrt each other: the second one comes from the isomorphism $\operatorname{Im}(d)\cong C/\operatorname{Ker}(d)$
An example of a short exact sequence satisfying your first desiderata, but one which you probably won't fully understand till you are further along in homological algebra, is the Universal Coefficient Theorem. The homology version says: if $R$ is a PID, $A$ is an $R$-module, and $C$ is a flat chain complex over $R$, then there is a natural short exact sequence
$$
0 \rightarrow H_n(C) \otimes_R A \rightarrow H_n(C\otimes_R A) \rightarrow Tor_1^R(H_{n-1}(C),A) \rightarrow 0
$$
Moreover, this sequence splits, and the splitting is natural in $A$ but not in $C$.
A related result is the Künneth Theorem: if $R$ is a PID and $X,Y$ topological spaces then there is a natural short exact sequence
$$
0 \rightarrow \bigoplus_{i+j=k} H_i(X;R) \otimes_R H_j(Y;R) \rightarrow H_k(X\times Y; R) \rightarrow \bigoplus_{i+j=k-1} Tor_1^R(H_{i}(X;R),H_j(Y;R)) \rightarrow 0
$$
Furthermore, this sequence splits, but not canonically.
Of course, there is also a dual version, with Ext and product instead of Tor and direct sum. And a universal coefficient theorem in cohomology. Look them up in Hilton and Stammbach's book (Theorems 3.1, 3.3).
Within the category of Banach spaces and bounded linear maps,
$$0\to c_0 \to \ell_\infty \to \ell_\infty / c_0 \to 0$$
is a paradigm example of a short exact sequence that does not split, contrary to any short exact sequence
$$0\to c_0 \to X \to Z \to 0,$$
where $X$ (or $Z$) is separable. Here $c_0$ is the space of sequences convergent to 0 and $\ell_\infty$ is the space of bounded sequences, both endowed with the supremum norm.
The relevant results are Sobczyk's and Phillips-Sobczyk's theorems. See also the paper Sobczyk's Theorems from A to B by Félix Cabello Sánchez, Jesus M. F. Castillo, and David Yost.
The "commutative analog" of my example! Might be worth mentioning that $l_\infty/c_0$ is isomorphic to $C(\beta\mathbb{N}\setminus \mathbb{N})$.
Evaluating at a single element of $\beta {\mathbb N}/{\mathbb N}$ one obtains the short exact sequence $0 \to o({\mathbb R}) \to O({\mathbb R}) \to {\mathbb R} \to 0$, where $O({\mathbb R})$ is the ring of bounded nonstandard reals, $o({\mathbb R})$ is the ideal of infinitesimal nonstandard reals, and ${\mathbb R}$ is the standard reals. Also can't resist mentioning the variant $0 \to o({\mathbb Q}) \to O({\mathbb Q}) \to {\mathbb R} \to 0$, which one way to construct the real numbers in nonstandard analysis. More generally one has the nonstandard hull construction in metric or Banach spaces.
How about the short exact sequence that expresses that every group can be expressed in terms of generators and relators? For any group $G$, there is a short exact sequence (in fact many) of the form $$0\to R\to F\to G\to 0,$$ with $F$ and $R$ being free groups. This expresses $G$ as a free group of generators modulo the relations encoded in $R$.
Of course there are analogous statements in other categories, such as those of modules.
"Analogous" should be taken with a grain of salt, since $R$ might not be free in other categories (including many module categories).
Good point. I was thinking about particularly nice rings. In general projective resolutions might be much longer.
I think this was explicitly part of my first answer to this question (and, indeed, is very worth highlighting to a newcomer like the OP) https://mathoverflow.net/a/363723/11540
For any abelian group $A$, there is a short-exact sequence $$0 \to T(A) \to A \to A/T(A) \to 0,$$
where $T(A)$ is the torsion subgroup of $A$, and $A/T(A)$ is torsion-free.
The kernel-cokernel exact sequence: in an abelian category, given $A \xrightarrow{f} B \xrightarrow{g} C$, the following sequence is exact
$$ 0 \to \ker f \to \ker gf \to \ker g \to \text{coker } f \to \text{coker }gf \to \text{coker} g \to 0$$
The maps are the obvious ones. The map $\ker g \to \text{coker } f$ is the one which factors through $B$.
I don't know if this fits, because it's not short and maybe it is too trivial, but I really think that every mathematician should know. For example, at a very low level, this tells the following basic facts
$gf$ is injective iff $f$ is injective and $\ker g \hookrightarrow{} \text{coker }f$
$gf$ is surjective iff $g$ is surjective and $\ker g \twoheadrightarrow \text{coker }f$
$gf$ is an isomorphism iff $f$ is injective, $g$ surjective and $\ker g \xrightarrow{\cong} \text{coker } f$
If $f$ and $g$ are injective/surjective, so is $gf$.
I think that more cools applications are covered in the following paper by Xiong, which I found just now.
A nice picture of this sequence from Nakaoka's website is the following
Do you remember on exactly which website you found this image? I don't find it at https://www.sci.kagoshima-u.ac.jp/nakaoka .
Given a finitely generated module $M$ over a commutative Noetherian ring $R$, there is a short exact sequence $$0\to M_1 \to R^n \to M\to 0$$
where you map $1$ in each $R$ to a generator of $M$ and $M_1$ (also finitely generated) is called a module of syzygy of $M$. Understanding this sequence (and it's repetitions) is a fundamental problem in commutative algebra and algebraic geometry and has generated countless beautiful results as well as been widely used a versatile tool on it's own. For instance, Hilbert's Syzygy Theorem says that if $R$ is a polynomial ring over a field, and you repeat this with $M_1$, etc, eventually you will get a free module.
Not sure there is one thing that "everyone should know", but for instance Stanley's proof of the Upper Bound Conjecture in combinatorics used a refinement of the Hilbert Theorem above.
Since the OP is just getting started, I'll add that this story is told in section 1.8 of Zimmermann's book on representation theory, where he also explains that (by definition) the $n$-th syzygy $\Omega_M^n$ fits in a short exact sequence $0\to \Omega_M^n \to P_{\Omega_M^{n-1}} \to \Omega_M^{n-1}\to 0$. This is an important reason why the stable module category in representation theory is a triangulated category. Here $P_{\Omega_M^{n-1}}$ is as in my answer, a projective module that maps onto $\Omega_M^{n-1}$
There is also a story of cosyzygies, i.e., cokernels of injective envelopes, which also fits into the framework of short exact sequences.
I suppose many algebraic topologists would agree that the short exact sequence
$$0\longrightarrow \mathbb Z/p \longrightarrow \mathbb Z/p^2 \longrightarrow \mathbb Z/p\longrightarrow 0$$
giving rise to the Bockstein operator in (co)homology and the exact sequence
$$ 0\longrightarrow C \stackrel{\cdot p}\longrightarrow C \longrightarrow C/pC\longrightarrow 0$$
giving rise to the Bockstein spectral sequence of the form
$$H(C/pC) \Longrightarrow H(C)/pH(C)$$ may fit the bill. They are quite simple and lead to remarkably interesting mathematics.
In addition to the Bockstein, this is the key example to see that $\mathrm{Ext}(\mathbb Z/p, \mathbb Z/p)\ne 0$, which is nice to know. I also like $0\to\mathbb Z\to\mathbb Z\to\mathbb Z/n\to 0$, as a good first example of a non-split short exact sequence but also just a general insight into how abelian groups behave differently than vector spaces.
In fact every first grader has to know this short exact sequence: it is how you do addition by “carrying”.
@VivekShende Sounds a bit dramatic but OK. :)
(I think there was a nice thread here about how carrying gives a cocycle, too...)
I saw $0\to \mathbb Z_p\to \mathbb Z_{p^2}\to \mathbb Z_p\to 0$ as an answer, with $p$ prime, but I will add with $p$ not prime and the particular choice $p=10$,
$$0\to \mathbb Z_{10}\to \mathbb Z_{100}\to \mathbb Z_{10}\to 0 $$
for the following reasons:
We all know that the group in the middle is given by $\mathbb Z_{10}\times \mathbb Z_{10}$ twisted by a 2-cocycle.
We know (if we don't, we should!) that the 2-cocyle is given by $f(n,m)=0$ if $n+m<10$ and $f(n,m)=1$ if $n+m\geq 10$, where $n,m\in\{0,\dots,9\}$.
Every child knows (or should know) (in particular, every mathematician -even a non algebraist one-) how to sum 2-digit numbers.
Since the question is what exact sequence we "should" know, I think 3. and 2. are two good reasons, even for the two different meanings of "should" know.
What is the 2-cocycle? Is that the homology group that's generated by $Ker(g)/Im(f)$?
If the Babylonians knew about exact sequences, they would certainly know $0 \rightarrow\mathbb{Z}6 \rightarrow \mathbb{Z}{60} \rightarrow \mathbb{Z}{10} \rightarrow 0$. So why not also $0 \rightarrow \mathbb{Z}p \rightarrow\mathbb{Z}{pq} \rightarrow \mathbb{Z}{q} \rightarrow 0$ for any natural numbers $p, q$?
The Tate extension. Let $k$ be a field, and let $V$ be the space $k((t))$ be the space of Laurent series with coefficients in $k$, considered as a topological vector space. If we write $\operatorname{GL}(V)$ for the group of (topological) automorphisms of $V$, then there is a canonical central extension
$$0\rightarrow k^{\times}\rightarrow\operatorname{GL}^{\flat}(V)\rightarrow\operatorname{GL}(V)\rightarrow0.$$
For a detailed construction, see http://www.its.caltech.edu/~justcamp/notes/sila.pdf. A brief summary: let $L$ be the space $k[[t]]\subset V$. Then for any element $g\in\operatorname{GL}(V)$, $gL$ and $L$ are commensurable (their intersection is of finite codimension in both) and so one can associate to them a one-dimensional vector space, the relative determinant. $\operatorname{GL}^{\flat}(V)$ can be defined as pairs of an element $g$ and a trivialization of the relative determinant.
Here the deep fact is the existence of the object $\operatorname{GL}^{\flat}(V)$. This leads to a host of other central extensions that are central (haha....) to the study of e.g. affine Lie algebras and friends (and thus to conformal field theory.) In a different direction, this short exact sequence is equivalent to the existence of the local residue symbol; in fact, I think Tate's name got attached to this because of his use of the local residue symbol to give a new proof of Riemann-Roch.
I guess the quintessential example, satisfying your second desiderata, is
$$
0 \rightarrow A \stackrel{f}{\rightarrow} B \rightarrow B/f(A) \rightarrow 0.
$$
For example, if $f = \mu_n: \mathbb{Z} \to \mathbb{Z}$ is multiplication by $n$, this means the following is exact
$$
0 \rightarrow \mathbb{Z} \stackrel{\mu_n}{\rightarrow} \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z} \rightarrow 0.
$$
Another example of the same general result is that, if $C$ is finitely presented, then it fits in a short exact sequence
$$
0 \rightarrow N \rightarrow P \rightarrow C \rightarrow 0
$$
where $N$ and $P$ are finitely generated, and $P$ is projective. Think of $P$ as the generators, and $N$ as the relations you quotient out by to get $C\cong P/N$.
Since you asked for a big list, I'll try to restrict myself to one example per answer.
Even though the $\mathbb{Z}/n\mathbb{Z}$ example is very basic, I often find it extremely helpful to think about (e.g. lets you understand the difference between a SES and a direct sum).
@SamHopkins Is it different from a direct sum because it does not split?
@SiddharthBhat: yes, exactly, that's the same thing.
Despite it being frequently used implicitly in papers (a classical example being Milnor's '56 paper about exotic spheres), I have never seen the following spelled out anywhere, so this might be a good place:
Let $\xi^n : E \to B$ be a real, smooth vector bundle over a manifold $B$. There is a short exact sequence of vector bundles over $E$,
$$0 \to \xi^*(\xi) \stackrel{i}{\to} \tau(E) \stackrel{d\xi}{\to} \xi^*(\tau(B)) \to 0,$$
where $i$ sends $(e_1, e_2) \in E \times_B E$ to the tangent vector starting at $e_1$ and pointing in the direction specified by $e_2$. Here, $\tau(M)$ denotes the tangent bundle of $M$. (As always, short exact sequences of vector bundles split.)
It follows from an easy dimension count. Alternatively, one can write out the gluing of the charts, giving something like
$$0 \to \text{colim } \mathbb R^n_\text{d} \times \mathbb R^n_\text{o} \times U_i \to \text{colim } \mathbb R^n_\text{d} \times \mathbb R^n_\text{o} \times TU_i \to \text{colim } \mathbb R^n_\text{o} \times TU_i \to 0,$$
where subscript $d$ and $o$ are just labels to distinguish between the coordinate which tracks the direction and that which tracks the origin, and $\{U_i\}$ is an atlas for $\xi$. This also gives a more rigorous description of $i$.
This must be closely related to another very important short exact sequence - the Atiyah class extension$$0\to\Omega^1\otimes E\to J^1(E)\to E\to0$$
I find it particularly illuminating when talking about connections. A right splitting ($H:\xi^(\tau(B))\to\tau(E)$ such that $d\xi\circ H=\mathrm{id}$) is a choice of horizontal lift, while a left splitting ($K:\tau(E)\to\xi^(\xi)$ such that $K\circ i=\mathrm{id}$), sometimes known as the connector, is the projection on the vertical part (hence $K(v)=0$ if $v$ is horizontal). The introduction of connections is a direct (human) result of the fact that this sequence does not canonically split.
This one is just too much fun to leave out.
Write the braid group on $n$ strands as $B_n$.
By following the strands of a braid $\sigma\in B_n$ we construct a permutation of $n$ items, which we write as $\eta(\sigma)$. This $\eta$ is an epimorphism whose kernel is the pure braid group $P_n$. The pure braids are the braids whose strands end up where they started:
$$
P_n \to B_n \overset{\eta} \to S_n
$$
Short exact sequences form a bridge of sorts between homological algebra and representation theory. For example, Maschke's theorem is the statement that, if $G$ is a finite group and $k$ is a field whose characteristic does not divide the order of $G$, then the $k$-representations of $G$ are completely reducible. This is equivalent to the statement that every short exact sequence of $k[G]$-modules
$$
0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0
$$
splits. This is, in turn, equivalent to the statement that $k[G]$ is a semisimple ring (there are many equivalent forms of what this means; my favorite is that every module is both injective and projective). You can then hit it with the Artin–Wedderburn theorem and write $k[G]$ as a product of matrix algebras. Strictly speaking, you don't need the language of short exact sequences, but many find it clarifying.
Another fundamental (half) short exact sequence is the Jacobi--Zariski sequence. For algebras over operads, for example, it takes the following form: for a triple $C\to B\to A$ of maps of $P$-algebras, there is a half exact sequence of functors
$$0\longrightarrow \mathrm{Der}_B(A,-) \longrightarrow \mathrm{Der}_C(A,-) \longrightarrow\mathrm{Der}_C(B,-) $$
coming from the half exact sequences in Kahler differentials
$$A\otimes_B\Omega_{B\mid C}^1\longrightarrow \Omega_{A\mid C}^1 \longrightarrow \Omega_{A\mid B}^1 \longrightarrow 0 $$
that is exact if $B\to A$ is a cofibration.
Decided to turn into an answer my comment to another answer here.
The Atiyah class $\alpha_E\in\operatorname{Ext}^1(E,\Omega^1\otimes E)$ of a holomorphic vector bundle $E$ is the class of the short exact sequence
$$
0\to\Omega^1\otimes E\to J^1(E)\to E\to0,
$$
where $\Omega^1$ is the cotangent bundle (corresponding to the sheaf of holomorphic 1-forms) and $J^1(E)$ is the sheaf of first order jets of sections of $E$. A good reference is "Rozansky-Witten invariants via Atiyah classes" by Kapranov (Compositio Math. 115 (1999) 71-113). Kapranov notes that there is a dual way to represent this class, using another remarkable short exact sequence
$$
0\to E\to{\mathcal D}^{\leqslant1}\otimes_{\mathcal O}E\to T\otimes E\to0.
$$
Here $T$ is the tangent bundle and ${\mathcal D}^{\leqslant1}$ is the sheaf of differential operators of order $\leqslant1$. (More precisely, this gives the class corresponding to $-\alpha_E$ in view of the canonical isomorphism $\operatorname{Hom}(-,\Omega^1\otimes-)\cong\operatorname{Hom}(T\otimes-,-)$.)
For a free product $A*B$ of groups $A$ and $B$, there is the exact sequence
$1 \to [A,B] \to A*B \to A \times B \to 1$
where $[A,B]$ is the subgroup generated by all elements $[a,b]=aba^{-1}b^{-1}$ and $A \times B$ is the direct product group. The first map is the inclusion and the second one is the intuitive one. This sequence is important for combinatorial and geometric group theory.
A starting point in anabelian geometry (a "thème central de la géométrie algébrique anabélienne", as Grothendieck writes in his Esquisse d'un Programme) can be considered to be the following:
Let $k$ be a field with separable closure $\overline{k}$ and $X$ a quasi-compact, quasi-separated $k$-scheme. If $\overline{x}$ is a geometric point of $X$ and the base change $X_{\overline{k}}$ is connected, then there is a short exact sequence of profinite topological groups:
$$ 1 \to \pi^{ét}_1(X_{\overline{k}}, \overline{x}) \to \pi^{ét}_1(X, \overline{x}) \to \pi^{ét}_1(Speck, \overline{x}) \simeq Gal(\overline{k}/k) \to 1$$
In fact, for $X=\mathbb{P}^1_{\mathbb{Q}}\backslash{\{0,1,\infty\}}$, J. S Milne on p. 30 of his LEC course notes, calls $\pi^{ét}_1(X, \overline{x})$ "arguably, the most interesting object in mathematics" due to the deep motivic ideas and profound connections that surround it, in relation with the already mysterious absolute Galois group of the rationals.
Let $M$ be a smooth manifold and $x:M\rightarrow \mathbb{R}$ a smooth function with $0$ as regular value, such that $X=\{x=0\}\subset M$ is a smooth submanifold. Then
$$
0\rightarrow x C^\infty(M)\hookrightarrow C^\infty(M) \xrightarrow{f\mapsto f\vert_X} C^\infty(X)\rightarrow 0
$$
is a short exact sequence and a right split corresponds to an extension map.
Where does this show up:
For $M=\mathbb{R}$ this is the statement that the remainder in a Taylor series can be written as $R(x)=x^kr(x)$ for a smooth function $r(x)$.
For $M=\mathbb{R}^d\times \bar {\mathbb{R}}^d$ (where $\bar{\mathbb{R}}^d$ is the radial compactification) and $x$ a boundary defining function of $\partial M=\mathbb{R}^d\times S^{d-1} = S^*\mathbb{R}^d$ (co-sphere bundle), this yields
$$
0\rightarrow \Psi_{\mathrm{cl}}^{m-1}(\mathbb{R}^d)\hookrightarrow \Psi_{\mathrm{cl}}^{m}(\mathbb{R}^d) \xrightarrow{\sigma_m} C^\infty(S^*\mathbb{R}^d)\rightarrow 0,
$$
the shorth exact symbol sequence of pseudo-differential operators. Here $\Psi^m(\mathbb{R}^d)=\mathrm{Op}(x^{-m}C^\infty(M))$ with $\mathrm{Op}$ denoting the standard quantisation of symbols $a:\mathbb{R}^d_z\times \mathbb{R}^d_\xi\rightarrow \mathbb{C}$. One can take $x=\langle \xi \rangle^{-1}$ as bdf. of fibre-infinity. A right split is then a quantisation map. The symbol sequence (together with the multiplicativity of $\sigma_m$) allows to construct parametrices of elliptic operators and is thus the starting point of elliptic regularity theory.
For $M=\bar {\mathbb{R}}^d\times \bar {\mathbb{R}}^d$, which is a manifold with corners, the constructions from the previous point yields Melrose's scattering (classical) scattering pseudo-differential operators.
Can you please expand a bit on how the $M=\mathbb{R}$ case corresponds to remainders in Taylor series?
On $M=\mathbb{R}$ we take $x(t)=t$, such that $X={0}$. Now, if a smooth function $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfies $f(0)=0$, then by exactness of the SES we can write $f(t)=t g(t)$ for another smooth function $g:\mathbb{R}\rightarrow \mathbb{R}$. Inductively, if $f(t)=O(t^k)$, then $f(t) = t^k g(t)$ for a smooth function $g:\mathbb{R}\rightarrow\mathbb{R}$. This applies to the remainder in the Taylor series, where $g$ might a priori only be bounded.
Take a bundle $E \rightarrow M$ in $Diff$ and then apply the tangent functor. We get $TE\rightarrow TM$. The kernel of this is the vertical bundle, $VE$ and this all assembles into the short exact sequence:
$0 \rightarrow VE \rightarrow TE \rightarrow TM \rightarrow 0$
A splitting of this gives an Ehresmann connection, aka a horizontal bundle $HE$ such that $TE = VE \oplus HE$
The concept then descends to connections over vector bundles and principal bundles.
All this is in Michor, Kolar & Slovaks Natural Operations in Differential Geometry. It's possible to introduce curvature in this generality and which describes the local integrability of the horizontal bundle. In fact, they introduce it in greater generality over just a manifold, rather than a bundle and there they have cocurvature as well as curvature where the former describes the integrability of the vertical bundle. In the example above, the cocurvature vanishes as the vertical bundle is always integrable.
The following is both elementary and simple, and carries a useful content, so it may qualify at least as something most mathematicians may not regret to know. In the context of formal Laurent series,
$$0 \rightarrow \mathbb{C} \rightarrow \mathbb{C}((z)) \xrightarrow{D} \mathbb{C}((z)) \;\xrightarrow{ \mathrm{Res} }\; \mathbb{C} \rightarrow 0,$$
describes the main features of the operation of formal residue, taking a formal Laurent series $\sum_k c_kz^k$ to its coefficient $c_{-1}$. Among other results these allow to give an easy proof of the Lagrange inversion formula (check e.g. Genealogy of the Lagrange inversion theorem and its answers for other remarks).
A useful short exact sequence of sheaves in algebraic geometry is
$0 \rightarrow \mathcal{O}_C(K_C) \rightarrow \mathcal{O}_C(K_C + P) \rightarrow Q \rightarrow 0,$
where $Q$ is the quotient (a skyscraper sheaf with support in $P$).
Considering the long exact cohomology sequence associated to this short exact sequence leads to a standard proof of the Riemann-Roch formula for algebraic curves, since $L(C,K_C)$ is the space of global sections of the sheaf $\mathcal{O}_C(K_C)$ (see the textbook on Riemann surfaces by Otto Forster).
Up to a twist, this is a special case of the ideal sheaf sequence $I_Z \rightarrow O_X \rightarrow O_Z$ for a closed subvariety $Z$ in a variety $X$. That sequence and its twists are absolutely fundamental tools in projective algebraic geometry.
This probably isn’t a short exact sequence that everyone needs to know, but it’s one of some significance for mathematical physics and differential geometry.
Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Any smooth principal $G$-bundle $\pi : P \to B$ gives rise to a canonical short exact sequence
$$
0 \to \mathfrak{g} \times P \to TP \to \pi^\ast TB \to 0
$$
of $G$-equivariant vector bundles on the total space $P$, where $\mathfrak{g}$ carries the adjoint representation of $G$. Here, the map $\mathfrak{g} \times P \to TP$ is induced by the map that sends $X \in \mathfrak{g}$ to the corresponding fundamental vector field $X_P$ on $P$, while the map $TP \to \pi^\ast TB$ is given by $\pi_\ast$.
As Atiyah first observed, a principal connection can be identified with a splitting of this short exact sequence. Moreover, the gauge action of global gauge transformations on principal connections is compatible with this identification: if $f : P \to P$ is a gauge transformation and $\rho : \pi^\ast TB \to TP$ is a right splitting, then $f$ acts on $\rho$ to yield the right splitting $f_\ast \circ \rho$.
From the (somewhat idiosyncratic) perspective of noncommutative geometry, this short exact sequence can be viewed as relating the $G$-equivariant differential calculus on the total space $P$ to differential calculus on the base $B$ and the differential calculus along the orbits (modelled on that of the structure group $G$); a principal connection, then, tells you how to decompose the total differential calculus into a “direct sum” of basic and orbitwise calculi. Indeed, in noncommutative geometry itself, this short exact sequence—more precisely, a noncommutative analogue of its dual—actually becomes part of the definition of (algebraic) quantum principal bundle and principal connection.
Can you suggest some references(non-physics) which define connection as splitting of the Atiyah sequence... One that I know is the Appendix A of the book "Lie groupoids and Lie algebroids in differential geometry" by K. Mackenzie..
I wish I did! It's all been scattered articles and the odd MO post (including some of yours, in fact)—thanks for the reference to Mackenzie!
Very good answer, very close to this one as well.
On obvious reference is Atiyah's 1957 paper Complex Analytic Connections in Fiber Bundles, where Atiyah writes "Definition: A connection in the principal bundle P is a splitting of the exact sequence A(P)."
That is exactly where Atiyah first observed all this.
I don't know whether this is a SES that every mathematician should know but it does satisfy the first sentence of the body of your question, since one could say it captures triangulability:
$$ 0 \to \text{ker}f \to \Theta_{3}^{H} \overset{f}{\to} \mathbb{Z}/2 \to 0 $$
where:
the abelian group $\Theta_3^H$ is the cobordism group of oriented homology three spheres modulo binding an acylic PL/smooth 4-manifold.
f is the Rokhlin homomorphism, which is 1/8th the signature of a compact, smooth spin(4) manifold that the integral homology sphere bounds.
Galewski, Stern and Matumoto showed in the 1980s that the non-splitting of this SES is equivalent to there being non-triangulable manifolds in every dimension 5 and above. Whereas, Manolescu recently showed that the SES does not in fact split.
Given an abelian category $\mathcal{A}$, and a diagram
$$\require{AMScd}\begin{CD}
W @>f>> Y \\
@VgVV @VVhV \\
X @>>k> Z
\end{CD}$$
we can form an associated sequence
$$0 \to W \xrightarrow{(g,f)} X \oplus Y \xrightarrow{(k,-h)} Z \to 0.$$
Then the diagram is:
Commutative iff the sequence is a complex,
Cartesian iff the sequence is a left exact complex,
co-Cartesian iff the sequence is right exact complex,
semi-Cartesian iff the sequence is a middle exact complex.
$\hphantom{0}$
Standard disclaimer: maybe not every mathematician, etc. etc., but it would behoove anyone working with commutative & homological algebra to be aware of this. It would be quite hoove indeed.
From tensor of vector spaces, I find this exact sequence useful:
$$0\rightarrow U\otimes V'+U'\otimes V\rightarrow U\otimes V \rightarrow (U/U')\otimes(V/V')\rightarrow 0.
$$
For a commutative ring R and an element $x\in R$, denote $(0:x)=\{y\in R\colon yx=0\}$, I find this really good:
$$0\rightarrow (0:x) \rightarrow R \rightarrow Rx\rightarrow 0
$$
It can be used to show that, in a local ring, we have:
$$\mathrm{pd}_R(Rx)\leqslant 1 \iff Rx \text{ is projective } \iff x \text{ is not a zero-divisor}
$$
More generally, your second exact sequence is a special case of $1 \to G_x \to G \twoheadrightarrow G\cdot x$, for a group $G$ acting on a space $X$, and $x$ a point of $X$ with stabilizer subgroup $G_x$.
@Lspice: Is that just an "exact sequence in spirit"? More precisely: in what category does your more general exact sequence live? If $G_x$ is not normal in $G$, then the orbit $G \cdot x$ is not naturally a group...
@JoshuaGrochow, re, it depends on how strict you are about where exact sequences are allowed to live. I regard it as an exact sequence in the category of pointed sets, which is good enough for making a, not long, but less short sequence in cohomology $1 \to G_x^\Gamma \to G^\Gamma \to (G\cdot x)^\Gamma \to \operatorname H^1(\Gamma, G_x) \to \operatorname H^1(\Gamma, G)$, and that's extremely useful in applications even if it doesn't fit ito the most general framework.
@LSpice: Nice! Thanks for sharing. Looks not far from things I know. Is there a good place to read about it? Is it at least an "exact sequence" of pointed $G_x$-sets? And when $G_x$ is normal in $G$, is it actually an exact sequence of groups?
@JoshuaGrochow, re, I learned about it from Serre - Galois cohomology. Specifically, the result I quoted is I.§5.4, p. 50, Prop. 36. It is not an exact sequence of pointed $G_x$-sets, because the common base point of $G_x$ and $G$ is usually not fixed by $G_x$. When $G_x$ is normal, it is the usual exact sequence of groups, and we get a slightly longer exact sequence: Serre I.§5.5, p. 51, Prop. 38.
Note that the first exact sequence here holds for arbitrary modules over any commutative ring, as long as the first nonzero term is understood appropriately (as a sum of canonical images rather than of the tensor products themselves).
The defining short exact sequence for Milnor's $K_2(R)$ ($R$ any ring) is
$$0\rightarrow K_2(R)\rightarrow St(R)\rightarrow E(R)\rightarrow 0$$
where $St(R)$ and $E(R)$ are the Steinberg and elementary groups. If we cheat a little on the definition of "short" this extends to
$$0\rightarrow K_2(R)\rightarrow St(R)\rightarrow GL(R)\rightarrow K_1(R)\rightarrow 0$$
An explicit example that is my favourite is below which is in Humphreys book on Linear Algebraic Groups.
For a field $k$, let $T(n,k)$ denote the group of $n\times n$ non-singular upper triangular group and $D(n,k)$ the non-singular diagonal matrices and $U(n,k)$ upper triangular matrices with 1's in the diagonal.
$1\to U(n,k)\to T(n,k)\to D(n,k)\to 1$
is actually a split-sequence. This can be generalized to any connected solvable group leading to its structure theorem as semi-direct product of maximal unipotent normal subgroup and maximal torus subgroup.
Any split connected solvable group! For example, if $k$ is imperfect of characteristic $2$ and $\theta \in k^\times$ is a non-square, then the group of matrices of the form $\begin{pmatrix} a & b \ b\theta & a \end{pmatrix}$ is solvable, an extension of a so called wound unipotent group by the multiplicative group, but the maximal unipotent subgroup is trivial.
An example that might be useful in virtually all branches of mathematics: If $V$ and $W$ are vector spaces (over the same field $\mathbb{F}$) and if $U \subseteq V$ is a subspace, then the obvious exact sequence
$$ 0 \longrightarrow U \longrightarrow V \longrightarrow V/U \longrightarrow 0 $$
turns into the non-trivial exact sequence
$$ 0 \longrightarrow U \mathbin{\otimes} W \longrightarrow V \mathbin{\otimes} W \longrightarrow (V/U) \mathbin{\otimes} W \longrightarrow 0. \tag*{$(*)$} $$
(I rarely need exact sequences in my work, but simple manipulations like this make quotients and subspaces of tensor products much easier to deal with.)
Can you please describe why $(\star)$ is a very useful exact sequence?
@SiddharthBhat hm, yeah, very useful is a little too strong (so I changed it). I guess my main point is that manipulations of exact sequences make life a lot easier when dealing with subspaces and/or quotients of tensor products.
$\DeclareMathOperator\Aut{Aut}$Wells's exact sequence: if $G$ is a group with normal subgroup $N$, it relates $\Aut(G)$ to $\Aut(N)$ and $\Aut(G/N)$. I'll state it, then explain it; for those who haven't seen it, the notation may make it seem scary, but hopefully my explanation below makes it less scary. It starts with:
$$1 \to Z^1_\alpha(G/N, Z(N)) \to \Aut(G; N) \xrightarrow{f_1} \Aut(N) \times \Aut(G/N).$$
The idea is this: first, we want to restrict our attention to automorphisms of $G$ that send $N$ to itself — this is precisely what $\Aut(G; N)$ is. If we don't do that, things get…complicated. (When $N$ is characteristic, $\Aut(G; N) = \Aut(G)$, so this is a particularly nice case. And if you apply this inductively in a finite group, your base case is characteristically simple groups, whose automorphism groups are easily described.)
Once we have that, we find that any automorphism $\varphi$ also induces an automorphism of the quotient $G/N$, so we certainly get a map from $\Aut(G; N) \to \Aut(N) \times \Aut(G/N)$. This is $f_1$.
Now, what is the kernel of $f_1$? It's exactly the automorphisms of $G$ that fix $N$ pointwise and that induce the trivial automorphism of $G/N$; in other words, all they do is they move around the elements within their respective $N$-cosets. Thus, any such automorphism $\varphi$ is given by a function $c \colon G/N \to N$ such that $\varphi(g) = g \cdot c(\pi(g))$, where $\pi \colon G \to G/N$ is the natural quotient map, and $c(1) = 1$. For $\varphi$ to be a homomorphism (hence isomorphism), it is necessary and sufficient that
$$c(\pi(g) \pi(h)) = h^{-1} c(\pi(g)) h c(\pi(h))$$
for all $g,h \in G$. Applying this equation to arbitrary $h \in N$, we find that we must have $\operatorname{Image}(c) \subseteq Z(N)$, and the equation above is precisely the equation of a "Derivation of $G/N$ into $Z(N)$ consistent with the given action of $G/N$ on $Z(N)$ (by conjugation)". In the notation above, $\alpha$ is the "outer" action of $G/N$ on $N$, which induces an actual action on $Z(N)$, and the notation $Z^1_\alpha(G/N, Z(N))$ is precisely this set of derivations.
What Wells actually did was to show that we can replace $\Aut(N) \times \Aut(G/N)$ with its subgroup $C$ of "compatible pairs" (the notion of compatibility just depends on the action $\alpha$), and then extend this exact sequence with an additional map $C \to H^2_\alpha(G/N, Z(N))$. Malfait further expanded this into a 27-term commutative cube diagram.
There's a really great summary of work on this exact sequence and its extensions by Jill Dietz, that also highlights the role of Buckley's group action in this (and related) sequences.
The sequence
$$0 \rightarrow \Psi^{\mu-1}(\Omega) \overset{\iota}{\rightarrow} \Psi^\mu(\Omega) \overset{\sigma_\mu}{\rightarrow}S_{h}^{\mu}(\Omega \times (\mathbb{R}^{n} \setminus 0)) \rightarrow 0$$
is exact, where $S_{h}^{\mu}(\Omega \times (\mathbb{R}^{n} \setminus 0))$ is the space of the functions $b:\Omega \times (\mathbb{R}^{n}\setminus 0) \rightarrow \mathbb{C}$ positively homogeneous of degree $\mu$, $\iota$ is the inclusion operator and $\sigma_\mu$ is the principal symbol of a pseudodifferential operador $B=op(b)$ with $b \in S^{\mu}(\Omega \times \mathbb{R}^{N})$.
|
2025-03-21T14:48:31.313276
| 2020-06-21T17:00:54 |
363722
|
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"https://mathoverflow.net/users/150150",
"https://mathoverflow.net/users/36721",
"sidi mohamd deval"
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"url": "https://mathoverflow.net/questions/363722"
}
|
Stack Exchange
|
Two inequalities in $\mathbb{R}$
How to prove that for real numbers $a$ and $b$, the following inequalities hold?
$(a|a|^{p-2}-b|b|^{p-2})(a-b)\geq 2^{2-p}|a-b|^p$,if $p\geq 2$
$(a|a|^{p-2}-b|b|^{p-2})(a-b)\geq (p-1)\frac{|a-b|^2}{(|a|+|b|)^{2-p}}$,if $ 1<p< 2 $
Let $l(a,b)$ denote the left-hand side of both inequalities. Then $l(b,a)=l(a,b)=l(-a,-b)$. The right-hand sides of both inequalities have the same properties.
So, without loss of generality (wlog) $a\ge b$ and $a\ge0$. Also, by homogeneity, wlog $a=1$. So, we have these two cases to consider:
$$(i)\ 0\le b\le a=1\quad\text{and}\quad (ii)\ b<0<a=1.$$
In case (i), your first inequality can rewritten as
$$r_{11}(b):=\frac{1-b^{p-1}}{(1-b)^{p-1}}\ge2^{2-p} \tag{1}$$
for $p\ge2$ and $b\in[0,1)$.
For $p\ge2$ and $b\in(0,1)$, we have
$$r_{11}'(b)=(p-1) (1-b)^{-p} \left(1-b^{p-2}\right)>0$$
and $r_{11}(0)=0$, whence $r_{11}(b)\ge1\ge2^{2-p}$, so that (1) holds.
In case (ii), your first inequality can rewritten as
$$r_{12}(c):=\frac{1+c^{p-1}}{(1+c)^{p-1}}\ge2^{2-p} \tag{2}$$
for $p\ge2$ and $c:=-b>0$.
For such $p$ and $c$, we have
$$r_{12}'(c)=(p-1) (1+c)^{-p} \left(c^{p-2}-1\right),$$
so that $r_{12}$ is decreasing on $(0,1]$ and increasing on $[1,\infty)$, with the minimum value $r_{12}(1)=2^{2-p}$, so that (1) holds.
As for your second inequality, in case (i) it can rewritten as
$$r_{21}(b):=\frac{(1+b)^{2-p} \left(1-b^{p-1}\right)}{(1-b) (p-1)}\ge1 \tag{3}$$
for $p\in(1,2)$. Let
$$(Dr_{21})(b):=r_{21}'(b)(p-1)(1-b)^2 (1+b)^{p-1} \\
= -(p-1) b^{p-2}+(p-3) b^{p-1}+b (p-1)-p+3.$$
Then
$$(Dr_{21})''(b)=-(1-b) (3-p) (2-p) (p-1) b^{p-4}<0$$
for ($p\in(1,2)$ and) $b\in(0,1)$, so that $(Dr_{21})(b)$ is concave in $b$. Also, $(Dr_{21})(1)=0=(Dr_{21})'(1)$. So, $Dr_{21}<0$ and hence $r_{21}$ is decreasing on $[0,1)$. Also, $r_{21}(1-)=2^{2-p}$. So, $r_{21}\ge2^{2-p}\ge1$, so that (3) holds.
In case (ii), your second inequality can rewritten as
$$r_{22}(c):=\frac{1+c^{p-1}}{(1+c)^{p-1}}\ge1 \tag{4}$$
for $p\in(1,2)$ and $c:=-b>0$.
For such $p$ and $c$, we have
$$r_{22}'(c)=(1 + c)^{-p} (c^{p-2}-1),$$
so that $r_{22}$ is increasing on $(0,1]$ and decreasing on $[1,\infty)$. Also, $r_{22}(0)=r_{22}(\infty-)=\frac1{p-1}$. So, $r_{22}\ge\frac1{p-1}>1$,
so that (4) holds.
@ losif Pinelis thank you so much but
For $0<b<a=1$,second inequality can rewritten as $r_2(b)=\frac{(1+b)^{2-p}(1-b^{p-1})}{(p-1)(1-b)}$
$r_3(b)=\frac{1-b^{p-1}}{(p-1)(1-b)}$
$r_3'(b)=\frac{-(p-1)b^{p-2}+(p-2)b^{p-1}+1}{(p-1)(1-b)^2}\geq 0$
$r_3(0)=\frac{1}{p-1} >1$
then second inequality is true
@sidimohamddeval : I did copy the exponent $2-p$ incorrectly. Now this is fixed.
For $0<b<a=1$,second inequality can rewritten as $r_2(b)=\frac{(1+b)^{2-p}(1-b^{p-1})}{(p-1)(1-b)}>1$\
$r_3(b)=\frac{1-b^{p-1}}{(p-1)(1-b)}$\
$r_3'(b)=\frac{-(p-1)b^{p-2}+(p-2)b^{p-1}+1}{(p-1)(1-b)^2}\geq 0$
$r_3(0)=\frac{1}{p-1} >1$
then second inequality is true
@ losif Pinelis thank you so much
|
2025-03-21T14:48:31.313453
| 2020-06-21T18:09:33 |
363736
|
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|
Stack Exchange
|
Three integral expressions for integer values of $\zeta(s)$. Could these be further reduced to known integrals?
In this MSE-question I've asked about three, similarly shaped, integrals for integer vales of $\zeta(s)$ that I found numerically:
$$\zeta \left( 3 \right) =\frac12{\int_{0}^{1} \frac{1}{x}\big(\zeta(2)-{\it Li_2} \left(1-x\right)\big) \,{\rm d}x} \tag{1}$$
$$\zeta \left( 4 \right) =\frac{4}{5}{\int_{0}^{1} \frac{1}{x}\big(\zeta(3)-{\it Li_3} \left(1-x\right)\big) \,{\rm d}x} \tag{2}$$
$$\zeta \left( 5 \right) =\frac12{\int_{0}^{1} \frac{1}{x}\big(\zeta(2)-{\it Li_2} \left(1-x\right)\big)^2 \,{\rm d}x} \tag{3}$$
ADDED: found one more:
$$\zeta \left( 3 \right) = \frac32 - \frac14{\int_{0}^{1} \big(\zeta(2)-{\it Li_2} \left(1-x\right)\big)^2 \,{\rm d}x} \tag{4}$$
where ${\it Li_n}(z)$ is the PolyLogarithm.
I have not found any similar expressions at other integer values.
The answer to the MSE-question helped reducing the integral for $\zeta(3)$ to a known integral, however still curious whether the other two could be reduced to something known as well.
Again indicate the series corresponding to the integral
"a known integral" --- if Mathematica knows the integral, does that count?
@Carlo Beenhakker, yes, that counts since Mathematica also yields the indefinite integral for the $\zeta(4)$-case. One down; how about the integral for $\zeta(5)$?
These identities, and many more, follow from a theorem in Integrals of polylogarithmic functions, recurrence relations, and associated Euler sums,
For example,
There are also variations with $\log x$ factor, such as
Thanks Carlo. This will solve the $\zeta(5)$ case in equation (3)! I had actually just found (numerically) this simple equation for integers $n \ge 2$:
$${\int_{0}^{1} \zeta(n)-{\it Li_n} \left(1-x\right) ,{\rm d}x} = (-1)^n \left(1-\sum_{k=2}^{n-1} (-1)^k,\zeta(k) \right)$$
|
2025-03-21T14:48:31.313599
| 2020-06-21T19:06:08 |
363741
|
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|
Stack Exchange
|
Traces and closed walks that do not close before their time
Let $A$ be the adjacency matrix of a graph. Then, as is well-known and trivial to show, $\mathrm{Tr} A^k$ equals the number of closed walks of length $k$.
Is there a similar way to express (a) the number of closed walks of length $k$ that do not return to their origin before $k$ steps? (b) the number of closed paths or trails of length $k$ (paths being walks that do not repeat vertices, and trails being walks that do not repeat edges)?
Let me narrow my question, in part because a closed expression may be hopeless. Say one shows that there are few closed paths of length $\leq 2 k$. Can one give an upper bound on $\mathrm{Tr} A^{2k}$, or on any related quantity, as a result?
I summarized a lot of linked posts on non-backtracking walks in this question on a connection between Chebyshev, Ramanujan, Ihara and Riemann. maybe it helps...
I don't think so. Let $a_k$ be the number of paths of length $k$ starting and ending at a particular vertex, and let $b_k$ be the number of such paths who return to their origin for the first time at step $k$. For convenience set $a_0=1$ and $b_0=0$. Then for $n\geq 1$ we have $a_n = \sum_{k=0}^n b_k a_{n-k}$.
Let $A_v(x),B_v(x)$ be the associated generating functions, depending on the chosen vertex. The identity above is $A_v-1=A_vB_v$ and hence $B_v = 1-1/A_v$.
Now $\sum_v A_v(x)=\sum_{k\geq 0} \mathrm{Tr}(A^k)x^k = \mathrm{Tr}\left(\sum_{k\geq 0} A^k x^k\right)$ and we conclude that
$$\sum_v A_v(x) = \mathrm{Tr} \left((\mathrm{Id}-Ax)^{-1}\right)\,,$$
but this can't lead to a simple formula for $\sum_v B_v(x)$ in general (you will get a formula when the graph is vertex-transitive).
[Editing to add: counting self-avoiding walks is much harder. Even the asymptotic number of self-avoiding walks in $\mathbb{Z}^d$ is not known precisely, see https://www.math.ubc.ca/~slade/spa_proceedings.pdf]
|
2025-03-21T14:48:31.313747
| 2020-06-21T19:27:17 |
363744
|
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|
Stack Exchange
|
Non-Noetherian local ring with nilpotent maximal ideal
Browsing through my notes on Artin rings, I have realized that I don't know an example for this and I wasn't able to google anything relevant.
What is an example for a commutative non-Noetherian local ring $(R,\mathfrak{m})$ with nilpotent maximal ideal, i.e. $\mathfrak{m}^n = 0$ for some $n \in \mathbb{N}$?
Note that in this setting $R$ Noetherian $\Leftrightarrow$ $R$ Artinian.
In this line of thoughts:
Is there a standard name for local rings with a nilpotent maximal ideal?
Take $R = k[x_1,\ldots]$ with $\mathfrak m = (x_1,\ldots)$ and look at $R/\mathfrak m^2$. (However, this does not work for any maximal ideal that is not finitely generated, because it is possible that $\mathfrak m = \mathfrak m^2$, in which case $R/\mathfrak m^2 = R/\mathfrak m$ is a field. An example of such an $\mathfrak m$ is the maximal ideal in a valuation ring with divisible value group.)
In fact in the case $\mathfrak m^2 = 0$ where $R$ contains $k = R/\mathfrak m$, we must have $$R = k \oplus M = \operatorname{Sym}^*(M)/\operatorname{Sym}^{\geq 2}(M)$$ for some $k$-module $M$. This is an example of the type you're looking for if and only if $M$ is not finitely generated.
@R.vanDobbendeBruyn: thanks for making it obvious what to look at! If you post it as an answer, I will accept it.
|
2025-03-21T14:48:31.313865
| 2020-06-21T19:38:16 |
363746
|
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|
Stack Exchange
|
When do algebraic closures exist constructively?
The field of algebraic numbers exists constructively, since we can represent a number by an irreducible polynomial plus an estimate in rational coordinates that separates it from any other root.
More generally, if we have a countably enumerated field with decidable arithmetic, it seems like we can construct the algebraic closure by picking a countable ordering of the irreducible polynomials, then defining an ordering of the roots of each polynomial that respects the orderings chosen for all previous polynomials.
Questions:
Is it correct that something like this construction works for any constructive countable field?
Is there a natural broader class of field for which the algebraic closure constructively exists?
This is proved in Theorem VI.3.5 of "A Course in Constructive Algebra" by Ray Mines, Fred Richman, and Wim Ruitenburg.
I'm not aware of any generalizations of this sort.
Would it generalize to any well ordered field?
It seems so, but I'm not sure.
|
2025-03-21T14:48:31.313984
| 2020-06-21T20:10:38 |
363750
|
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|
Stack Exchange
|
How to describe the compact real forms of the exceptional Lie groups as matrix groups?
I know that $G_2$ can be described as the subgroup of $SO(7)$ preserving a specific element of $\Lambda^3(\mathbb{R}^7)^*$. It can thus be realized as a matrix group. Prof. Robert Bryant did describe in his answer to the post A question on complex semisimple Lie groups and $(\mathbb{C}^2, \omega)$ a nice description of the complex $E_6$ as the group of symmetries of $V = \Lambda^2_0 (\mathbb{C}^8)^*$ endowed with a cubic form on that space. In that description, what is the real structure which gives the compact real form $E_6$? I conjecture it would be induced by the real structure $j \wedge j$ on $V$, where $j$ is a quaternionic structure on $\mathbb{C}^8$ such that $\omega$ (see Prof. @RobertBryant's answer in the above link) is real. It is just a guess. Is it correct please?
I do not really know how to realize the compact real forms of $F_4$, $E_7$ and $E_8$ as matrix groups. Your help is kindly appreciated. References are more than welcome (especially if they can be found online, and I hope "my" library has access to them!). If someone feels like writing a whole answer, then that would be great too. It is time for me to learn more about the exceptional Lie groups.
One last thing. I know that what I am looking for can be found in the relevant E. Cartan's papers. However, while I would definitely learn a lot by going back to the source, yet I don't have as much free time nowadays as I would like to (not to mention that reading Cartan is known to be difficult, and it is not the language barrier, in my case). So is there a simplified and modernized version of that part of Cartan's work please, that would also discuss compact real forms?
Do you mean the simply connected group? the adjoint group? all of them? this strongly affects the question.
@YCor, I understand. I do not require the compact real form to be simply connected. I just want some compact real form for each type. Actually, Prof. Robert Bryant's answer is what I really want to know/learn. I will try to read Cartan.
Cartan describes all of the compact real forms of the simple Lie groups over $\mathbb{C}$ in his first paper that classifies the real forms. In fact, he describes them exactly in the terms that you ask for: A representation of the complex Lie group together with an auxilliary structure, either a real structure on the complex representation space or a Hermitian quadratic form.
For $\mathrm{G}_2$ (resp, $\mathrm{F}_4$, $\mathrm{E}_8$), the compact forms are represented as special orthogonal real matrices of rank 7 (resp., 26, 248). For $\mathrm{E}_6$ (resp. $\mathrm{E}_7$), the compact real forms are represented as special unitary matrices of rank $27$ (resp. $56$).
Explicitly, here are the defining structures in the lowest dimensional representations of the compact real forms of the exceptional groups:
$\mathrm{G}_2$ is the stabilizer of a $3$-form on a real vector space of dimension $7$.
$\mathrm{F}_4$ is the stabilizer of a quadratic form and a cubic form on a real vector space of dimension $26$. (I believe that the cubic form alone is enough to define $\mathrm{F}_4$.)
$\mathrm{E}_6$ is the stabilizer of a cubic form and a positive definite Hermitian form on a complex vector space of dimension $27$. (The cubic form alone only defines the complex $\mathrm{E}_6$.)
$\mathrm{E}_7$ is the stabilizer of a symplectic form, a quartic form, and a positive definite Hermitian form on a complex vector space of dimension $56$. (The quartic form and the Hermitian form by themselves are almost enough to define $\mathrm{E}_7$; they define a group with two connected components, the identity component of which is the compact $\mathrm{E}_7$.)
$\mathrm{E}_8$ is the stabilizer of a $3$-form on a real vector space of dimension $248$.
Ok, I will read Cartan. (French) Après tout, cela fait assez longtemps que je n'ai pas lu un article en français!
Do you think that for the case of $F_4$, some even power of the cubic form may be a power of the quadratic form, or something like that?
Also trying to guess, is the $3$-form nothing but the Cartan $3$-form on the Lie algebra, meaning $(x,y,z) \mapsto \operatorname{tr}(ad_x,[ad_y,ad_z])$?
@Malkoun: In the $\mathrm{F}_4$ case, no power of the cubic form is a power of the quadratic form. They are independent functions on $\mathbb{R}^{26}$. In the $\mathrm{E}_8$ case, yes, the $3$-form is the Cartan $3$-form. (It seems to be sort of a "folk-theorem" that, except for the $3$-dimensional cases (where the result is false), the stabilizer of the Cartan $3$-form on a real simple Lie algebra is the automorphism group of the Lie algebra. This seems to have been observed by many different people at different times, but it's hard to find a first reference.)
What is the name of the Cartan article please? I am at numdam.org, but there seem to be a number of articles by Elie Cartan with similar titles.
Is it "Les groupes réels simples finis et continus"?
I guess when Cartan writes "groupes finis", he means finite-dimensional groups, rather than finite groups.
@Maloun: Yes, there are a couple of different papers, but "Les groupes réels simples finis et continus" is the main one (though Cartan re-proves the classification using symmetric spaces 10 years later). Yes, 'groupes finis' means finite-dimensional Lie groups. (Examples of 'groupes infinis' are things like the diffeomorphism group of a manifold, the symplectomorphisms of a symplectic manifold, the gauge group of a fiber bundle, etc.) "Continus" essentially means 'of positive dimension' (as opposed to 'discrete groups').
Thank you for the dictionary! I am reading the article now.
It should be complemented with E. Cartan's thesis. Ok, Prof. Bryant, I have a question. Do you know of some other reference where the invariant forms are written in a nicer way please? For example, for $E_6$, you gave a really nice description of the complex $E_6$ using $\Lambda^2 (\mathbb{C}^8)^*$, and a non-degenerate element $\omega$ in that space. You wrote the cubic form in a very compact way using wedge products of forms. You claim it is due to Cartan. However, Cartan, if I am not mistaken, writes the forms the way one would write polynomials. It is not clear they are equivalent.
Also the cubic form of $G_2$ can be defined using the octonionic cross-product. So far so good. Now, for $F_4$, $E_7$ and $E_8$, is there a nice way to write the invariant forms (whether symmetric or skew-symmetric), the way you wrote the invariant cubic form in the case of $E_6$ please? Finally, is this compact way of writing the cubic form for $E_6$ due to you please?
@Malkoun: You are right that it is not obvious that the cubic form on $\mathbb{C}^{27}$ that I gave is equivalent to Cartan's cubic form. The equivalence comes from a result of Dynkin, who observed that the restriction of the $27$-dimensional representation of $\mathrm{E}_6$ to its maximal subgroup $\mathrm{Sp}(4,\mathbb{C})$ is an irreducible representation of that subgroup. By dimension count, it must be $\Lambda^2_0(\mathbb{C}^8)$. Now, in this representation, $\mathrm{Sp}(4,\mathbb{C})$ preserves only one cubic form, the one I wrote down. Hence, the desired equivalence is demonstrated.
Let us continue this discussion in chat.
@RobertBryant Is there a good exposition in some reference on Cartan's classification of compact real forms? I do not know French, unfortunately.
@QGravity: The best reference I know is Helgason's "Differential Geometry, Lie groups, and Symmetric Spaces". See Table V of Chapter X, Section 6. If all you want is the classification of the compact real forms of the simple Lie groups, that's much easier and follows directly from the classification of the complex simple Lie algebras over $\mathbb{C}$, but, if you want all the real forms, you need to know the finite order automorphisms of the complex simple Lie algebras, which is more delicate.
Thank you for the reference. What I would like to know is how to express simple roots and fundamental weights of exceptional Lie algebras in terms of the standard basis $e_i$ of $\mathbb{R}^n$ for some $n$ in each of the five cases of exceptional Lie algebras. I would appreciate it if you could introduce a reference.
@QGravity: This information about roots and fundamental weights can also be found in Helgason's book.
@RobertBryant Very much appreciate your help.
There is an abstract way of integrating Lie algebras but I guess you are asking for a more hands on approach. I suggest browsing Exceptional Lie groups by Ichiro Yokota. Usually, it's the compact (or perhaps complex) Lie groups which are treated in the literature, so maybe you should be more specific in what do you think is lacking there.
As far as $F_4$ goes the description can be made rather succinct using so called Jordan algebra:
Take the real vector space of octonionic-Hermitian three by three matrices and endow it with commutative product defined by $A \circ B = \frac{1}{2}(AB+BA).$ The automorphism group of this product is the compact Lie group of type $F_4.$ One can actually define some kind of octonionic determinant for this Jordan algebra and then it can be proved that $F_4$ is the group stabilizing this determinant and trace. If you take just the group stabilizing the determinant, you will obtain noncompact real form of $E_6.$
Thank you. Well, I don't really know if it is lacking in the literature, since I don't know the literature very well, to be honest. Most of the books I have focus on Lie algebras rather than Lie groups, and when they do talk about Lie groups, they usually do something general, rather than something specific. These are great books though.
However, as Weyl wrote in his "Classical Groups" book, though I cannot remember his exact words, even if one knows the general theory, that does not mean one should not work out the examples (he was talking about invariant theory, and Hilbert's theorems).
Ideally, I would like to see the descriptions made by Cartan of the $5$ exceptional Lie groups but in a more modern way. I thank you for the octonionic description of $F_4$ and for the reference. I will check it out.
I apologize for changing the accepted answer. I really thank you for the reference, which seems to take on an octonionic approach. For my purposes though, it is Cartan's work which is closer to what I am currently interested in. I apologize again.
No worries. This site is more about math than about egos. ;)
|
2025-03-21T14:48:31.314618
| 2020-06-21T21:02:35 |
363751
|
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|
Stack Exchange
|
Is the parity of $\omega(n)$ equally distributed?
I recently learned that the prime omega function $\Omega(n)=\Omega\left(p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}\right)=\alpha_1+\alpha_2...+\alpha_k$ is very well studied. In particular, we know that $\Omega(n)$ is equally often even and odd. This statement is, in fact, equivalent to the prime number theorem.
My question is, do we know anything about the distribution of parities of $\omega(n)=\omega\left(p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}\right)=k$?
It is natural to assume that $\omega(n)$ is equally often even and odd, but perhaps it is much harder to show. From what I understand the reason that the distribution of $\Omega(n)$ is so much easier to analyze is that the Liouville lambda function $\lambda(n)=(-1)^{\Omega(n)}$ is very well understood and it's summary function $L(x)=\sum_{n<x}\lambda(n)$ can be related to the Mobius/Mertens function by
$$L(x)=\sum_{d^2<x}M\left(\frac{x}{d^2}\right)$$
The Mertens function is obviously very well studied, but no such inversion formulas are possible for $\omega(n)$ so we cannot use methods like this. I am curious about not only whether or not the result I ask for is known but whether or not the result is easier/harder to prove than the equivalent result for $\Omega(n)$.
https://arxiv.org/pdf/1906.02847.pdf
Perhaps the conjectured relationship $H(x)=\sum\limits_{n\le x}(-1)^{\omega(n)}=\sum\limits_{n\le x}a(n)\ M\left(\frac{x}{n}\right)$ where $a(n)=\sum\limits_{d|n}\mu(rad(d))$ and $rad(d)$ is the square-free kernel of $d$ might be of some use.
@StevenClark, thanks for the input but Peter Humphries link gives us our full expected answer.
Perhaps, Milo, you could summarize that link and post your summary as an answer.
That's a great idea! I'll do that.
In Peter Humphries link he answers the question very well, but by looking at the results cited I learned that this is in fact a special case of a more general phenomenon.
If $f(n)$ is a (real valued) multiplicative function with $\left|f(n)\right|\leq1$, then it's mean value $M=\lim_{x\to\infty}\frac{1}{x}\sum_{n<x}f(n)$ exists. Moreover, if the series
$$\sum_{p}\frac{1-f(p)}{p}$$
diverges then $M=0$. This is theorem 6.4 In Elliot's "Probabilistic Number Theory", attributed to Wirsing. Both $(-1)^{\Omega(n)}$ and $(-1)^{\omega(n)}$ are multiplicative since $\Omega(n)$ and $\omega(n)$ are additive. They both only take values in $\pm1$ and so their mean values must exist. By definition of $\omega$ and $\Omega$ we have
$$\sum_{p}\frac{1-(-1)^{\Omega(p)}}{p}=\sum_{p}\frac{1-(-1)^{\omega(p)}}{p}=\sum_{p}\frac{1-(-1)}{p}=+\infty$$
and thus they both must have average order $0$, meaning equidistribution of parities.
It is true though that the investigation of the parity of $\omega(n)$ is more complicated though. As I mentioned in the question, the equidistribution of parities of $\Omega(n)$ was known before the proof of the PNT to be equivalent to it, and so when the PNT was proved in 1896 the equidistribution of parities of $\Omega(n)$ was settled. The equidistribution of parities of $\omega(n)$, however, was only settled in 1975 by van de Lune and Dressler.
The "general result" of the mean values of multiplicative functions that can be used to settle the equidistribution of $\omega(n)$ is new, namely, Elliot's book was only published in 1979. It is interesting to think that this is so close to the result of van de Lune and Dressler.
should there be an upper bound on $|f(n)|$ in your second paragraph?
@kodlu yes! Sorry, I forgot to but it.
|
2025-03-21T14:48:31.314877
| 2020-06-21T21:07:39 |
363752
|
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|
Stack Exchange
|
Calculation in prismatic cohomology
In the standard references for prismatic cohomology, most theorems are proved in a local context (i.e. with completeness assumptions), and the devissage to the global case (i.e. smooth proper varieties) is often left to the reader. My first (vague) question is: how do these devisagges work? Are there references where these arguments have been written up (possibly in other contexts, e.g. crystalline cohomology)? For example, in Bhatt's Columbia notes he says that he will work in the local $p$-adically complete setting, because standard arguments then yield the general case: I would like to know what these arguments are.
To make things concrete, let me ask a more specific question. How to compute the mod $p$ prismatic cohomology of $\mathbb{P}^n_k$ (I would expect the answer to be $\mathbb{F}_p[[T]][h]/h^{n+1}$)? Etale comparison gives us the torsion-free part, but how to show that there is no torsion?
The theory seems to be formulated for formal schemes (see e.g. the original paper, Def. 4.1). By formal GAGA, in the proper case you can think about it as varieties, but at least that explains the ubiquity of completions. In §4.3 ("generalities on computing prismatic cohomology") they say some words on the (formal) affine line.
I'm not sure I understand your concrete question at the end. What is $k$? And what is the prism structure?
Thank you for the answer to the first question. As to the second, for some reason I was thinking of crystalline cohomology. Would $k=\mathbb{Z}p[[t]]$ work? I don't know how to answer regarding the prism structure. I mean cohomology of the structure sheaf $\mathcal{O}{\Delta}$. Is there extra information that I have to give? Do I have comparison with crystalline and etale for all the choices of this extra information? I did not notice that while reading the notes.
I'll go back to studying.
|
2025-03-21T14:48:31.315040
| 2020-06-21T22:30:15 |
363757
|
{
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"Alexander Chervov",
"Gerry Myerson",
"https://mathoverflow.net/users/10446",
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|
Stack Exchange
|
Classification of space-filling (by identical copies) convex polyhedra in R^3
Is the classification of space-filling (by identical copies) convex polyhedra in R^3 is known ?
There are only 5 "parallelohedra" - filling by translation.
But if relax that property to identical (but not necessary by parallel translation),
there are more as
Wikipedia suggests. But it is not clear is the list described there is final or not.
To be precise I am asking about honeycombs filling the space by copies of single polyhedron,
not several of them, as it happens, for example in famous Weaire–Phelan structure.
I remember working, unsuccessfully, under John Conway on space-filling tetrahedra in 1975. Michael Goldberg, Three infinite families of tetrahedral space-fillers, Journal of Combinatorial Theory, Series A Volume 16, Issue 3, May 1974, Pages 348-354 https://www.sciencedirect.com/science/article/pii/0097316574900582 was probably state-of-the-art at the time.
See also Marjorie Senechal, Which Tetrahedra Fill Space? Mathematics Magazine Vol. 54, No. 5 (Nov., 1981), pp. 227-243. https://www.jstor.org/stable/2689983?seq=1
@GerryMyerson Thank you very much !
|
2025-03-21T14:48:31.315262
| 2020-06-21T23:22:43 |
363759
|
{
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"Alex Ravsky",
"Penelope Benenati",
"Rodrigo de Azevedo",
"Sam Zbarsky",
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"https://mathoverflow.net/users/47135",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/363759"
}
|
Stack Exchange
|
Combinatorial optimization problem on sums of differences between real numbers
We are given an increasing sequence $S$ of positive real numbers $x_1, x_2, \ldots, x_n$, such that $$x_{i+2}-x_{i+1} \ge c\,(x_{i+1}-x_i)$$ for all $i=1, 2, \ldots n-2$, where $c\ge 1$ is constant. Each number $x_i\in S$ is associated with a positive integer weight $w_i$ for all $i=1, 2, \ldots, n$. Let $W$ be the sequence formed by these weights.
Finally, let $$A=\sum_{1\le i < j < k\le n} w_i w_j w_k(x_j-x_i)~,$$ and $$B=\sum_{1\le i < j < k\le n} w_i w_j w_k\min(x_k-x_j, x_j-x_i)~.$$
Question: What is the minimum value for $c$ such that, for all $S$ and $W$, we have $A \le 2B~$? (I am also interested in tight upper bounds).
Conjectures: I believe that $c\ge \frac{3}{2}$ is a sufficient condition to obtain $A \le 2B$ as required (I do not know whether it is also necessary). Furthermore, I think that the worst case w.r.t. $W$ occurs when $w_{n}\gg n$ and $w_{n-1} \gg n$, while all the other weights $w_1, w_2, \ldots, w_{n-2}$ are equal to $1$. Finally, I also conjecture that, for any $\alpha\ge 2$, $c\ge 1+\frac{1}{\alpha}$ is a sufficient condition to obtain $A \le \alpha B$.
For fixed $i,j,k$, we have $\sum_{i < j < k} w_i w_j w_k\min(x_k-x_j, x_j-x_i)=w_i w_j w_k\min(x_k-x_j, x_j-x_i)?$
Now, as I see, given $i < j < k$ we have
$$s_{i,j,k}=w_i w_j w_k\min(x_k-x_j, x_j-x_i)\le w_i w_j w_k(x_j-x_i)\le \sum_{i < j <\ell} w_i w_j w_\ell(x_j-x_i)=
s_{i,j}.$$
Yes, it's correct. I can modify the notation if it's misleading. Thanks.
The notation is confusing. If I understand correctly, the sums in the definitions of $s_{i,j}$ and $s_{i,j,k}$ are both over all tuples $1<i<j<k<n$. If this is correct, you should call the sums $A$ and $B$ or something. If I'm misunderstanding, and you want to be able to take, say $s_{2,4}$, then which variables are you varying as you take the sum?
You are right, the notation is confusing: Both sums are over all $i, j, k \in {1, 2, \ldots, n}$ such that $i < j < k$, that is over all tuples $1 \le i < j < k \le n$. Please note that we can have $i=1$ and $k=n$, which is different from writing "[...] over all tuples $1 < i < j < k < n$." Thank you very much for your answer, I will read it as soon as possible. Thereafter I will also change the notation according to your suggestions.
How is this a combinatorial optimization problem? How do you know it's convex? How can a combinatorial optimization problem be convex?
In the beginning, I used other tags. After that, someone suggested a tag change, and I approved the amendment, thinking s/he was a "moderator." I trusted this suggestion. Referring to your question about the title, I wanted to emphasize that the weights are integers. Anyway, if you have any idea about better tags and/or title, I will be glad to follow your advice. Thank you in advance for your help.
I will assume $c\le 2$, since that seems to be the case you are interested in. The argument below is formulated in terms of taking a given $c$ and computing the optimal $\alpha$, but this is equivalent to your question and the result agrees with your conjecture.
Step 1: We can change the formulation of the question so that the weights can be any nonnegative real numbers. Say we have such weights. We can always convert them to positive integer weights while changing
$$\frac{A}{B}$$
by arbitrarily small $\epsilon$. To do this, we approximate the real numbers by positive rational numbers, then scale all weights up by a large constant factor to make them all positive integers.
Step 2: We now pick some specific $1\le i<j<k\le n$ and set $w_i=w_j=w_k=1$ and set all other weights to $0$. Then
$$\frac{A}{B}=\max\left(1,\frac{x_j-x_i}{x_k-x_j}\right)$$.
Note that for $\ell<j$, we have
$$
x_{\ell+1}-x_\ell\le \frac{x_{j+1}-x_j}{c^{j-\ell}}\le\frac{x_k-x_j}{c^{j-\ell}}
$$
with equality achieved when $k=j+1$ and all the ratios of successive differences are exactly $c$.
Then
$$
x_j-x_i=\sum_{\ell=i}^{j-1}x_{\ell+1}-x_\ell\le \sum_{\ell=i}^{j-1}\frac{x_k-x_j}{c^{j-\ell}}=(x_k-x_j)\sum_{a=1}^{j-1}\frac{1}{c^a}\le \frac{x_k-x_j}{c-1}
$$
where equality is never achieved, but can be achieved to within an error of $\epsilon$ by taking $k=j+1$, taking $j-i$ large, and making all the ratios of successive differences be exactly $c$.
Thus for the special case of $w_i=w_j=w_k=1$ and all other weights being $0$, we have
$$\frac{A}{B}\le \frac{1}{c-1}$$ and the bound is tight.
Step 3: Now if we allow arbitrary weights, we have that termwise the same inequality holds, that is
$$
w_iw_jw_k(x_j-x_i)\le \frac{1}{c-1}w_iw_jw_k\min(x_j-x_i,x_k-x_j)
$$
and summing, we get
$$
A\le \frac{1}{c-1}B
$$
|
2025-03-21T14:48:31.315575
| 2020-06-21T23:28:25 |
363760
|
{
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"Gerry Myerson",
"Vincent Granville",
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"url": "https://mathoverflow.net/questions/363760"
}
|
Stack Exchange
|
Is there always a prime between $\beta z^\alpha$ and $\beta (z+1)^\alpha$ for $z$ large enough, and a fixed $\alpha < 2$?
I am interested if possible in $\beta = \frac{2}{3}, \alpha=\frac{3}{2}$ and $z$ is a positive integer or real number. My choice here is related to some progress I make in additive combinatorics (see my last answer to this MO question). It would make my life easier if this was true.
However, I can be less picky: all I really need it seems, is $\alpha < 2$. The closest $\alpha$ is to $2$, of course the more likely the answer to my question could be positive, but it also makes some arguments in my previous MO question less likely to work out. For $\beta$, you can pick up any positive value that would give a positive answer to my question.
Of course, the answer is "no" for $\alpha\leq 1$. For $\alpha>1$ this is conjecturally true, but not proven. This is a question about maximal prime gaps, specifically whether they are $O(p_n^{\alpha-1})$. See here for known results.
Indeed, it's not even known whether there's always a prime between consecutive squares, so the question here seems hopeless.
Correction from my previous comment: the exponent should be, I think, $\frac{\alpha-1}{\alpha}$, not $\alpha-1$.
Thank you. I guess I will stop wasting my time trying to prove my problem. That was supposed to be the easiest of the two big challenges I am facing, and it turns out to be unproven yet.
|
2025-03-21T14:48:31.315713
| 2020-06-21T23:29:13 |
363761
|
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"Carlo Beenakker",
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"kodlu"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/363761"
}
|
Stack Exchange
|
A probability question from sociology
We know that $\frac{1}{2} \leq a \leq p \leq 1$. And, $n \geq 3$ is a positive odd number, and $t$ is an integer. $a$ satisfies the equation below.
\begin{equation} \small
\begin{aligned}
&\sum_{t=0}^{n-1} \left( {n-1 \choose t} [p a^{t} (1-a)^{n-t-1}+(1-p) (1-a)^{t} a^{n-{t}-1}] \cdot [\frac{a^{t+1} (1-a)^{n-t-1}}{a^{t+1} (1-a)^{n-t-1}+(1-a)^{t+1} a^{n-t-1}}-\frac{a^{t} (1-a)^{n-t}}{a^{t} (1-a)^{n-t}+(1-a)^{t} a^{n-t}}] \right) \\[15pt]
&={n-1 \choose {\frac{n-1}{2}}} a^{\frac{n-1}{2}}(1-a)^{\frac{n-1}{2}} (2p-1)
\end{aligned}
\end{equation}
We want to prove that $a$ converges to $\frac{1}{2}$.
Any references or remarks are appreciated!
This site is full of questions that are not easy, yet we typically do not offer money for solutions. Can you give some motivation for this question that is mathematical rather than monetary? That is, how did this question come up for you?
Is $t$ fixed as $n\rightarrow \infty$?
It seems the left-hand-side is $p$-independent.
If I expand the left-hand-side of your equation around $a=1/2$ I find
$$
\sum_{t=0}^{n-1} {n-1 \choose t} [p a^{t} (1-a)^{n-t-1}+(1-p) (1-a)^{t} a^{n-{t}-1}] $$
$$\times\left[\frac{a^{t+1} (1-a)^{n-t-1}}{a^{t+1} (1-a)^{n-t-1}+(1-a)^{t+1} a^{n-t-1}}-\frac{a^{t} (1-a)^{n-t}}{a^{t} (1-a)^{n-t}+(1-a)^{t} a^{n-t}}\right] $$
$$=\sum_{t=0}^{n-1}\left(a-\frac{1}{2}\right) 2^{2-n} \binom{n-1}{t}+{\cal O}(a-1/2)^3=2a-1+{\cal O}(a-1/2)^3.$$
I similarly expand the right-hand-side,
$${n-1 \choose {\frac{n-1}{2}}} a^{\frac{n-1}{2}}(1-a)^{\frac{n-1}{2}} (2p-1)=2^{1-n} (2 p-1) \binom{n-1}{\frac{n-1}{2}}-\left(a-\tfrac{1}{2}\right)^2 2^{2-n} (n-1) (2 p-1) \binom{n-1}{\frac{n-1}{2}}+{\cal O}(a-1/2)^4.$$
Equating left-hand-side and right-hand-side I solve for $a$,
$$a=\frac{1}{2}+\frac{2^n \left(\sqrt{2^{3-2 n} (n-1) (1-2 p)^2 q^2+1}-1\right)}{4(n-1) (2 p-1) q},\;\;q=\binom{n-1}{\frac{n-1}{2}}.$$
For $n\gg 1$ this solution tends to
$$a\rightarrow\frac{1}{2}+\frac{1}{2\sqrt{2n}}\frac{\sqrt{16 (p-1) p+\pi +4}-\sqrt{\pi }}{2 p-1}.$$
the term $(a-\frac{1}{2}) 2^{2-n} \binom{n-1}{t}$ is the first nonzero term in a Taylor series around $a=1/2$; I then used the identity $\sum _{t=0}^{n-1} \binom{n-1}{t}=2^{n-1}$ to obtain $2a-1$ for the sum.
the term of order $(a-1/2)^2$ sums to zero when summed from $t=0$ to $t=n-1$.
both l.h.s. and r.h.s. are correct up to and including terms of the same order $(a-1/2)^2$. I would hope that makes for a consistent calculation, but I am not certain.
|
2025-03-21T14:48:31.315878
| 2020-06-22T00:39:50 |
363765
|
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"Pedro",
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"url": "https://mathoverflow.net/questions/363765"
}
|
Stack Exchange
|
Wall's presentation of the Steenrod algebra
In the paper "Generators and Relations for the Steenrod Algebra" (C. T. C. Wall, Annals of Mathematics, Second Series, Vol. 72, No. 3 (Nov., 1960), pp. 429-444) Wall shows that there is a presentation of the mod 2 Steenrod algebra $A$ with generators $s_i$ corresponding to $\mathrm{Sq}^{2^i}$ and relations of the form
$$ [s_i,s_j] = T_{ij}\quad 0\leqslant j\leqslant i-2$$
$$s_i^2 +s_is_{i-1}s_i+s_{i-1}^2s_i = T_i $$
where $T_{ij}$ and $T_i$ belong to the subalgebra generated by $s_0,\ldots,s_{i-1}$. Using this presentation Wall was able to solve a conjectured posed earlier by Toda.
Question 1 Has anyone worked out the terms $T_{ij}$ and $T_i$ for low values of $i$ and $j$? I presume that the terms in $T_{ij}$ and $T_i$ are at least cubic in the $s_i$? Is this true? (I'll re-read the paper with a bit more care and probably figure this out, but perhaps someone can confirm this quickly).
Of course this gives presentations of the subalgebras $A(n)$ generated by $s_0,\ldots,s_n$ in $A$. It is easy to see, for example, that
$A(0) = T(x\mid x^2)$
$A(1) = T(x,y\mid \text{add } y^2 + xyx)$.
$A(2) = T(x,y,z \mid \text{add } [x,z]+yxy, z^2+y[z,y])$
In the lovely User's guide to spectral sequence J. McCleary uses the first two presentations above to do some computations in the Adams spectral sequence: it turns out $A(0)$ admits a $(4,12)$ periodic resolution as an $A(1)$-module. A particular case of my question is:
Question 2 Has anyone worked out a presentation of say, $A(3)$? Is it known if each $A(n)$ is finite dimensional? In this case, does one know how quickly $\dim A(n)$ grows like, for example?
I am definitely not well-versed in the story of this algebra, so if the question is answered by either a paper, or book, or if there is literature that points in the right direction, I am happy to keep reading that in lieu of a full answer.
Disclaimer: I presume that with some spare time I can compute the presentation of $A(2)$ suggested by Wall. My question is not "can this be done?" but rather "has this been done systematically enough for low values of $n$?"
Add: I added the presentation of $A(2)$.
Question 1: "Closed forms" for the relations are known, due to Grant Walker in what seems to be unpublished work. The relations are described in Wood's paper "Problems in the Steenrod algebra," (PDF) Theorem 4.18. For example,
$$
s_i^2 = s_{i-1} \chi(s_i) s_{i-1} + s_{i-1}^2 \chi(s_i),
$$
where $\chi$ is the antipode. I put "closed" in quotes because this reduces the problem to having a closed form for $\chi(s_i)$. Wood gives some information about that, too, and in any case, it's easy to compute for small values of $i$ either by hand or using SageMath.
Wood's paper is a nice place to find lots of information about the Steenrod algebra; I recommend it.
Question 2: Certainly each $A(n)$ is finite-dimensional: by Milnor's famous theorem about the dual of the Steenrod algebra, $A(n)$ is dual to a quotient of a polynomial ring, namely
$$
\mathbb{F}_2 [\xi_1, \xi_2, \xi_3, ...] / (\xi_1^{2^{n+1}}, \xi_2^{2^n}, ..., \xi_i^{2^{n+2-i}}, ..., \xi_{n+1}^2, \xi_n, \xi_{n+1}, ...).
$$
So its vector-space dimension is $2^\binom{n+2}{2}$.
Dear John: thank you, this is really useful. I think I almost opened that paper a few days ago!
I was just trying to access this link http://www.idealibrary.com/links/citation/0021%2d8693/188/203 in your website but for some weird reason it redirects to https://www.oladieta.com.br/. I thought you should know.
|
2025-03-21T14:48:31.316120
| 2020-06-22T02:00:40 |
363767
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/363767"
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|
Stack Exchange
|
Lower bound for diagonal Ramsey numbers —- reference request
Using the first moment method, in 1947 Erd\H{o}s gave a lower bound on the diagonal Ramsey numbers $R(k,k)$:
$$
R(k,k) \geq (1+o(1))\frac{k}{e\sqrt{2}} 2^{k/2}.
$$
In 1975 Spenser used the Lov\’asz Local Lemma to improve this by a factor of $2$, to $(1+o(1))(k\sqrt{2}/e)2^{k/2}$.
In between these two lower bounds, there is the one you get via the alteration or deletion method:
$$
R(k,k) \geq (1+o(1))\frac{k}{e} 2^{k/2}.
$$
I’m trying to discover who first noticed this last bound, and when? Was it before or after Spencer’s improvement? I’ve seen the bound in numerous sets of notes online (and learned it as a graduate student), but with no attribution. Maybe it’s just ``folklore’’.
(I’m leading a summer reading group in Ramsey Theory, and I plan on Tuesday to tell them the history of upper and lower bounds on diagonal Ramsey numbers.)
On p.54 of "Recent Developments in Graphy Ramsey Theory" by Conlon, Fox, Sudakov which appeared in the monograph Surveys in Combinatorics 2015, Czumaj et. al. (eds), is the footnote below:
Though we do not know of an explicit reference, a simple application of the deletion method which improves Erdos' bound by a factor of $\sqrt{2}$ was surely known before Spencer's work.
Sounds like it may well be folklore.
|
2025-03-21T14:48:31.316237
| 2020-06-22T03:39:06 |
363769
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/363769"
}
|
Stack Exchange
|
Complex Monge Ampere equal with zero right hand side
Let $(X, \omega)$ be a compact Kaehler manifold, with $K_{X}$ numerically effective. Suppose that $[\zeta] = c_1(K_X)$ and $\int_{X}\zeta^n = 0$. I am interested in solving
\begin{equation}
(\zeta + i \partial \bar \partial f)^n = 0
\end{equation}
for some $f$ smooth. This is immediate if we assume $\zeta \geq 0$. So a solution should be expected to exist by the positivity conjecture in algebraic geometry. But that is certainly a quite hard problem. I wonder whether there has been any results/progresses on solving such an equation.
|
2025-03-21T14:48:31.316300
| 2020-06-22T06:49:12 |
363782
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363782"
}
|
Stack Exchange
|
Functoriality of Ext-functor
Let $X$ be a normal, integral variety and $U \subset X$ an open subset such that the complement of $U$ is of codimension at least $2$. Let $F$ be a coherent sheaf on $X$ such that $\mathcal{E}xt^1_U(F|_U,\mathcal{O}_U)=0$, where $\mathcal{E}xt^1$ denotes sheaf Ext. Does this imply that $\mathcal{E}xt^1_X(F,\mathcal{O}_X)=0$? I know that (locally) $Ext$ commutes with flat base change. But, since the natural inclusion map from $U$ to $X$ is not affine, I do not know if I can use this statement to answer my question (if I could then using the fact the $i_*\mathcal{O}_U=\mathcal{O}_X$ and then the base change statement, would give a positive answer to my question). Is there an analogous sheaf theoretic functoriality statement of Ext?
To summarize, can I say that:
$$\mathcal{E}xt^1_X(F,i_*\mathcal{O}_U) \cong i_*\mathcal{E}xt^1_U(i^*F,\mathcal{O}_U),$$
where $i$ is the natural inclusion of $U$ in $X$?
This is definitely false, even in very simple situations: take $X$ smooth, $Z$ a smooth subvariety of $X$ of codimension 2, and $\mathscr{F}=\mathscr{I}_Z$. Then obviously $\mathscr{E}xt^1_U(\mathscr{I}_{Z}{}_{|U}, \mathscr{O}_{U})=0$, but because of the exact sequence $0\rightarrow \mathscr{I}_Z\rightarrow \mathscr{O}_X\rightarrow \mathscr{O}_Z\rightarrow 0$, $\mathscr{E}xt^1_X(\mathscr{I}_Z,\mathscr{O}_X)$ is isomorphic to $\mathscr{E}xt^2_X(\mathscr{O}_Z,\mathscr{O}_X)$ which is $\neq 0$.
|
2025-03-21T14:48:31.316404
| 2020-06-22T06:55:51 |
363783
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Federico Poloni",
"Gerry Myerson",
"MRm",
"Pat Devlin",
"https://mathoverflow.net/users/145894",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/1898",
"https://mathoverflow.net/users/22512"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/363783"
}
|
Stack Exchange
|
Linear system of equations with prior knowledge on linear dependency of some unknowns
Assume I have a system of linear equations $Au = b$ where A is some $M\times N$ matrix, u is $N\times 1$ unknown variables vector, and b is $M\times 1$. Now, further assume that I know that some elements of $u$ are linearly dependent on the other elements of $u$. I.e., we can write $u$ as $u=[u_1, u_2]$ where $u_1$ is $N_1\times 1$, and $u_2$ is $N_2 \times 1$ and $N_1+N_2=N$ and it is known that $u_1 = Cu_2$, where $C$ is some unknown matrix.
Is there any way that this knowledge (that we know that some elements of u are dependent and we know which elements, but we don't know the dependency matrix C) can help me improve the complexity of solving the system of equations $Au=b$? I.e., is there any way it can simplify the complexity of the gaussian elimination (or any other algorithm for solving a system of linear equations)?
Thanks!
If $Au=b$, $A$ is $M\times N$, and $u$ is $N\times1$, then $b$ had better be $M\times1$, not $N\times1$.
correct - I've updated the question
What do you mean by "solving" a system of linear equations when $M\neq N$? There may be zero solutions, or infinite solutions.
I don't think so. Given any two vectors $u_1$ and $u_2 \neq 0$, there is always a matrix $C$ such that $u_1 = Cu_2$. So you have no additional information, essentially.
Yes, yes, yes! The OP is saying that some of the coordinates are linearly dependent (in that one is a multiple of the other). But that’s true for literally any two real numbers. So the OP is saying literally nothing.
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2025-03-21T14:48:31.316552
| 2020-06-22T07:25:47 |
363786
|
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|
Stack Exchange
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Closed paths, traces and spectra
Let $\Gamma$ be a graph. Write $A$ for its adjacency matrix. It is trivial to show that $\mathrm{Tr} A^k$ equals the number of closed walks of length k, that is, the number of $k$-step walks that return to where they started.
Suppose that we show that $\Gamma$ has few closed paths of length $\leq k$. (A path is a walk without repeated vertices; obviously we are allowing (and requiring) the origin vertex to be the ending vertex.) Can we prove an upper bound on $\mathrm{Tr} A^k$ as a result? If not, can we prove something else about the spectrum of $A$, possibly with additional conditions? (We may, for instance, assume that the degree of $\Gamma$ is very small compared to its number of vertices.)
The same questions can be posed if we show that $\Gamma$ has few closed trails of length $\leq k$. A trail is a walk without repeated edges.
Since a closed walk can be broken down to a number of closed paths (for instance a closed walk of length 6 can be 2 closed paths of length 3), don't you get an upper bound for number of walks from the number of paths?
That would be the naïve expectation, but I don't see how that happens.
But then 6 can be broken in a finite number of ways $6=4+1+1=3+3=2+2+2=...$ and each decomposition can be bounded by $N^a$ where $N$ is the number of paths at most $6$ and $a$ is the number of summands, can't we get at least some function of $N$ as upper bound?
I'm not sure of quite what you have in mind.
Did you try the decomposition into Lyndon words used for understanding the graph zeta function. See for example the article "A Combinatorial Proof of Bass's Evaluations
of the Ihara-Selberg Zeta Function for Graphs" by Foata and Zeilberger https://www.ams.org/journals/tran/1999-351-06/S0002-9947-99-02234-5/
I would put this as a comment but don't have enough (any) reputation...
Closed paths of length $\leq k$ show up when you take the trace of the alternating $k$th power of $A$, which is also equal to
$\sum_{ i_1 < i_2 < \cdots < i_k : } \lambda_{i_1} \cdots \lambda_{i_k}$
where $i_j \in |V(\Gamma)|$ and
$\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_{|V(\Gamma)|}$
are the eigenvalues of the adjacency matrix.
There are always going to be paths of length 2: their contribution to the trace can be expressed in terms of the number of perfect matchings in the graph of different sizes.
You might ask that there are no closed paths of length $2 < L \leq k$. I suspect having few closed paths expresses that the eigenvalues have some kind of `symmetry' expressed by a symmetric polynomial of the eigenvalues being small.
A variant of trails show up if you trace the alternating $k$th power of the non-backtracking operator of the graph as follows.
Namely, say that a closed walk is a weak trail if no edge is traversed more than once in each direction that can be given to the edge (so the edge might be traversed once in each direction).
The non-backtracking operator acts on functions on directed edges of $\Gamma$. A directed edge $e_1$ is `connected' to $e_2$ if the terminus of $e_1$ is the source of $e_2$ but $e_1$ is not $e_2$ with the opposite orientation; note that this is not symmetric. The non-backtracking operator is the adjacency operator of the directed graph $\Gamma^*$ whose vertices are directed edges of $\Gamma$ and directed edges in $\Gamma^*$ are as above.
Note that closed walks in $\Gamma^*$ are in length-preserving one-to-one correspondence with closed non-backtracking walks in $\Gamma$. Closed paths in $\Gamma^*$ are in length-preserving one-to-one correspondence with closed non-backtracking weak trails in $\Gamma$.
Unfortunately, the non-backtracking operator is not self adjoint in general, but if $\Gamma$ is $d$-regular then one can write a Jordan form of the operator in terms of the eigenvalues of the adjacency operator of $\Gamma$ (see e.g. Section 3.1 of the article link of Lubetzky and Peres).
Because the non-backtracking operator is the adjacency operator of $\Gamma^*$, the previous remarks relating the eigenvalues of $\Gamma$ to closed paths also give a relation between some more complicated spectral information of $\Gamma$ and closed non-backtracking weak trails.
Sorry this doesn't exactly answer the question, it was originally intended as a comment.
Thank you! Can you go into a bit more detail on your last comment - on trails?
I added some more information: after thinking about it I realized it is not exactly trails that show up but a weaker variant: I added the details anyway.
@Michael_Magee What happens if, instead of the alternating $k$th power of $A$, we take the symmetric $k$th power of $A$? I am asking because, on the spectral side, I am working with real orthonormal vectors $v_i$ that satisfy $\alpha_i:=\langle v_i, A v_i\rangle$ large (at least for many $i$) without necessarily being eigenvectors. When I consider $\mathrm{Tr} A^{2k}$, all is fine, since $\mathrm{Tr} A^{2k} \geq \sum_i \alpha_i$ gives me a useful lower bound on $\mathrm{Tr} A^{2k}$. (continued in next comment)
@Michael_Magee [cont'd] However, even if we work with $\bigwedge^k A^2$ instead of $\bigwedge^{2 k} A$, we do not necessarily have $\mathrm{Tr} \bigwedge^k A^2 \geq \sum_{i_1<i_2<\dotsc<i_k} \alpha_{i_1}\dotsc \alpha_{i_k}$ (do we?), so I'm in a bit of a pickle.
Also, what exactly do you mean when you say that "closed paths of length ≤k show up when you take the trace of the alternating kth power of A"? It's not the case (as I naïvely understood at first) that the trace is simply the number of closed paths of length $k$ (times $-1$), though that's certainly a term.
(I obviously meant $\mathrm{Tr} A^{2k}\geq \sum_i \alpha_i^{2 k}$ in the above.)
can you actually write down $\mathrm{Tr} \bigwedge A^{k}$ in terms of the number of closed paths in the sense of closed paths given above (as opposed to the weaker sense used in the definition of an Ihara zeta function)? I don't doubt it can be done - I simply want to make sure there are no grounds for confusion.
More precisely, $\mathrm{Tr}\bigwedge^k A$ has a term for each vertex disjoint collection of closed (oriented) paths (as above) whose lengths sum to $k$ (every term comes from a permutation in $\mathrm{Sym}_k$ and the cycle type of this permutation dictates the lengths of the paths). The term also contains the sign of the permutation it comes from: this probably makes it useless for analytic purposes unless you know there are no closed paths of certain lengths.
The intuitive difference between this and Ihara zeta is that Ihara involves reciprocal of a characteristic polynomial and hence by Cauchy identity is related to $\mathrm{Sym}^k A$ whereas $\mathrm{Tr} \bigwedge^k A$ shows up in the characteristic polynomial. So closed paths are telling you something about the coefficients of the characteristic polynomial. But there is also a formula for log of characteristic polynomial (in some region of parameter) in terms of a power series involving all the $\mathrm{Tr} (A^k)$. I don't know if any of this is helpful... of course trying...
to bound roots of a polynomial in terms of its coefficients is in general extremely difficult.
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2025-03-21T14:48:31.316992
| 2020-06-22T08:11:15 |
363790
|
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|
Stack Exchange
|
Necessity of conditions $N$ odd, square-free and $\chi$ quadratic in Kohnen's plus space - modular forms of half-integral weight
Kohnen introduced the "plus" space as a subspace of the space of modular forms of half integral weight, first in his 1980 paper and then generalized the work in a later 1982 paper. Why is the condition $N$ odd, square-free and $\chi$ quadratic necessary?
To elaborate: Let $S_{k+1/2}\left(4N,\left(\frac{4\chi(-1)}{.}\right)\chi\right)$ denote the space of half-integral weight modular forms of level $4N$ and character $\left(\frac{4\chi(-1)}{.}\right)\chi$, where $\chi$ is a Dirichlet character of modulus $N$ and $\left(\frac{a}{b}\right)$ is the Kronecker symbol. Kohnen defines the plus sub-space $S_{k+1/2}^+\left(4N,\left(\frac{4\chi(-1)}{.}\right)\chi\right)$ by attaching certain conditions on the Fourier coefficients of the modular forms, and then develops nice theory analogous to the Atkin-Lehner-Li theory of newforms in the case of integral weight modular forms.
Question: Why are the conditions $N$ odd, sqaure-free and $\chi$ quadratic necessary in the second paper? Kohnen remarks that these are not necessary for few of the stated results, but I cannot figure out where exactly these are required.
The question is motivated by the fact that several authors (like Ueda, Yamana, Manickam-Ramakrishnan-Vasudevan) have generalized this work to other levels (like $8N,16N,32N$) and non-quadratic characters, and also to the full-space (a nice history can be found (but not restricted to) in the introduction of this paper and this paper); but this condition $N$ odd and square free is still there, and in some places $\chi$ is quadratic. I could not find the reason for these conditions.
This is not an answer and I cannot comment due to lack of reputation, but it seems to first pop up in his Lemma 4 on p. 50. This occurs after a long, very technical discussion about double coset operators. The goal is to show that two specific representations of the Hecke algebra are equivalent (one whose image lies in $End_{\mathbb{C}}(S_{k+1/2}(N, \chi))$ and the other in $End_{\mathbb{C}}(S_{2k}(N))$). The Lemma:
Lemma 4 (p. 50). Every elliptic or hyperbolic conjugacy class in $\Gamma_{0}(4N)C_{n}\Gamma_{0}(4N)$ contains an element $\begin{pmatrix} a
& b \\ c & d \end{pmatrix}$ with $d > 0$, $(b, d) = 1$, and $\left( \frac{b}{f}, \frac{t^{2} - 64n^{2}}{f^{2}} \right) = 1$. Here, $t := a + d$ and $f := (d-a, b, c)$.
Now, for squarefree $M$ set
$$\mu(t, f, n, M) := \prod_{p \mid (M, f)} (1+p)\cdot \#\{x \in \mathbb{Z} : 1 \leq x \leq M, (x, M) = 1, x^{2} - tx + n \equiv 0 \mod (fM, M^{2})\},$$
Lemma 5. Let $A \in GL_{2}(\mathbb{Z})$ be an elliptic or hyperbolic matrix with $\det A = 16n^{2}$, $t \equiv 0 \mod{4}$ and $f$ odd. Then for $N$ odd and squarefree there are $\mu(\frac{t}{4}, f, n^{2}, N)$ matrices $B \in \Gamma(1)/\Gamma_{0}(4N)$ with $B^{-1}AB \in \Gamma_{0}(4N)C_{n}\Gamma_{0}(4N)$.
So it seems that the definition of $\mu$ depends on $M$ being squarefree, and in turn this allows Kohnen the counting argument in Lemma 5. I don't know why squarefree-ness is needed from this, though. But maybe this information can help you or someone else determine the answer.
EDIT:
Answer: If $N$ is not squarefree, there exists a case such that all common eigen-subspaces of $S_{k+1/2}^{+}(4N, \chi)$ for Hecke operators have dimension at least 2, whence a strong multiplicity one theorem does not hold (see Ueda's On twisting operators and newforms of half-integral weight).
We may try to decompose our cusp space into eigen-subspaces of twisting operators (why??). Let $$M := \prod_{\substack{p \mid N \\ ord_{p}(N) \geq 2}} p^{ord_{p}(N)}, \quad \Pi := \{p \text{ prime} : p \mid M\}.$$ One can decompose the Kohnen space as
$$S_{k+1/2}^{+}(4N, \chi) = \left( \bigoplus_{\kappa: \Pi \rightarrow \{\pm 1\}} S^{0, \kappa} \right) \oplus Ker\left(R_{\Pi}; S_{k+1/2}^{+}(4N, \chi) \right),$$
where $S^{0, \kappa} := \{f \in S_{k+1/2}^{+}(4N, \chi) : f \mid R_{l} = \kappa(l)f, \forall l \in \Pi\}$ and $R_{\Pi}, R_{l}$ are the twisting operators of the characters $\prod_{l \in \Pi} \left( \frac{*}{l} \right)$ resp. $\left( \frac{*}{l} \right)$. The 'kernel' part consists of oldforms, and each $S^{0, \kappa}$ is stable under the Hecke operators $T(n^{2})$ for $(n, 4N) = 1$.
Why we need squarefree: There exists a case such that $S^{0, \kappa} \cong S^{0, \kappa'}$ as Hecke modules for distinct $\kappa, \kappa': \Pi \rightarrow \{\pm 1\}$. Hence when $M \neq 1$ Kohnen's theory does not work.
|
2025-03-21T14:48:31.317715
| 2020-06-22T08:52:32 |
363792
|
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|
Stack Exchange
|
Examples of metric spaces with measurable midpoints
Given a (separable complete) metric space $X=(X,d)$, let us say $X$ has the measurable (resp. continuous) midpoint property if there exists a measurable (resp. continuous) mapping $m:X \times X \to X$ such that $d(x,m(x,x')) = d(x',m(x,x')) = d(x,x') / 2$ for all $x,x' \in X$.
It seems to be known (e.g see section 6 of this paper) that continuous midpoint spaces (i.e Polish spaces with the continuous midpoint property) include:
Hilbert spaces.
Closed convex subsets of Banach spaces.
Hyperconvex spaces.
CAT(0) spaces.
Hopefully, the collection of measurable midpoint spaces contains much more general examples (for the above list is quite restrictive).
Question. What are some examples of measurable midpoint spaces ?
What you call "Polish space" is usually called "complete metric space". In a Polish space the distance is not fixed: it's just assumed it exists. The distinction is very useful, I think.
Good point. Fixed.
@coudy Sorry, I meant "separable complete metric space", as opposed to "Polish space". Of course separability is essential (maybe not in OP's question who mentions possibly non-separable CAT(0) spaces), it was just not the point of my comment and I forgot it.
Do you know an example where the midpoint is always unique but the map is not continuous?
We will use the Kuratowski–Ryll-Nardzweski selection theorem:
Let $(\Omega, \mathscr{F})$ be a measurable space. Let $E$ be a Polish space. Let $\Gamma$ be a set-valued function from $\Omega$ to $E$; that is, for each $\omega \in \Omega$, let a set $\Gamma(\omega) \subseteq E$ be given. Assume that, for all $\omega \in \Omega$, the set $\Gamma(\omega)$ is nonempyty and closed in $E$. Assume that $\Gamma$ is $\mathscr F$-measurable in the sense:
$$
\text{for every open set }U\subseteq E,\qquad
\{\omega\,:\,\Gamma(\omega) \cap U \ne \varnothing\} \in \mathscr F .
$$
Then there is a measurable selection $\gamma$ for $\Gamma$: that is, a
function $\gamma : \Omega \to E$ with
$\bullet $ $\gamma(\omega) \in \Gamma(\omega)$
$\bullet $ for every open set $U \subseteq E,\quad \gamma^{-1}(U) \in \mathscr F$.
Let $X$ be a locally compact complete separable metric space with the midpoint property. For $a,b \in X$, let $\Gamma(a,b)$ be the midpoint set,
$$
\Gamma(a,b) = \left\{m : d(a,m)=d(b,m) = \frac{d(a,b)}{2}\right\} .
$$
Then $\Gamma$ is a set-valued function from $X \times X$ to $X$. Note $\Gamma(a,b)$ is nonempty and closed. Let $\mathscr F$ be the sigma-algebra of Borel sets in $X \times X$. We will prove (see below) that $\Gamma$ is $\mathscr F$-measurable. An application of the Kuratowski–Ryll-Nardzweski selection theorem then establishes the existence of an $\mathscr F$-measurable $\gamma : X\times X \to X$ with $\gamma(a,b) \in \Gamma(a,b)$.
Proof that $\Gamma$ is $\mathscr F$-measurable:
Let $U \subseteq X$ be open. We have to show $T_U \in \mathscr F$, where
$$
T_U := \{(a,b) \in X \times X\,:\,\Gamma(a,b) \cap U \ne \varnothing\} .
$$
The set
$$
Q := \{(a,b,u) \in X \times X \times X \,:\,
d(a,u) = \textstyle\frac{1}{2}d(a,b)\text{ and }
d(b,u) = \textstyle\frac{1}{2}d(a,b)\}
$$
is a closed set. Write $\pi$ for the continuous "projection" function $(x,y,u) \mapsto (x,y)$.
Then $T_U$ is the projection
$$
T_U = \pi(Q\cap(X \times X \times U)) =
\bigcup_{u \in U}\{(a,b) \in X \times X \,:\,
d(a,u) = \textstyle\frac{1}{2}d(a,b)\text{ and }
d(b,u) = \textstyle\frac{1}{2}d(a,b)\} .
$$
Now in our case, any open set $U$ is a countable union of compact sets,
so the projection is sigma-compact, and therefore Borel.
added
Without assuming locally compact, we do know that the projection of
a Borel set is analytic and thus universally measurable. So if we are given a Borel measure $\mu$ on $X \times X$, we get a $\mu$-measurable midpoint function.
Ah, this one is looking like the "book proof". Thanks! Anecdotally, a few hours ago I stumbled on the "selection theorems" page on wikipedia, and eventually stumbled upon KRN selection theorem, I felt this could eventually give a solution after some very hard work (and learning after learning some brand new stuff too).
Nice argument! Although I must say I was curious how you would do without the local compactness... I think there should be a different argument in the case where the space is not locally compact and the midpoint is unique, but I cannot quite think it through.
@GeraldEdgar Thanks for the nice answer. I think everyone will agree that this is the "book proof".
@GeraldEdgar: You mean Borel measure $\mu$ on $X^2$, right ?
@dohmatob ... yes, fixed.
@GeraldEdgar Your comment about weakening the local-compacity assumption at the price of only having an almost everywhere (w.r.t to some fixed Borel measure $\mu$ on $X\times X$) is on point, because my problem actually originated from probability theory (optimal transport). I was really only interested in showing that a certain map $T:X\times X \to X \times X$ defined by $T(x,y)=(x,m(x,y))$ would be measurable, so that it would make sense to talk about the pushforward measure $T_# \mu$.
If $(X,d)$ is a complete metric space with the algebraic midpoint property (i.e. for all $x$ and $y$ in $X$, there exists $z\in X$ such that $d(x,z)=d(y,z)=d(x,y)/2$) then $X$ is a path metric space. Indeed, for all $x,y\in X$ one can iteratively construct a map $\gamma$ from $[0,1]\cap\mathbb D$ to $X$ such that $d(\gamma(s),\gamma(t))=|t-s|$, and extend it using completeness of $X$. Here $\mathbb D$ is the set of dyadic numbers. This result is Theorem 1.8 in [G].
Suppose that $(X,d)$ is complete, locally compact, and has the algebraic midpoint property. Then $X$ has the measurable midpoint property.
Of course, this does not include general closed convex subsets of Banach spaces, but it covers for instance any complete manifold.
Let us $(U_k)_{k\geq0}$ construct a countable basis of $X$; in particular, it will show that $X$ is second countable. I want the diameter of $U_k$ to tend to zero as $k$ goes to infinity, and every fixed $x$ to be contained in an infinite number of $U_k$. By [G, Theorem 1.10 (Hopf-Rinow)], the closed metric balls of $X$ are compact; then one can take a finite open cover of $B(x_0,1)$ by open balls of radius $1/1$, then a finite open cover of $B(x_0,2)$ by open balls of radius $1/2$, etc. Let also $(z_k)_{k\geq0}$ be a sequence such that $z_k\in U_k$.
Note that for any closed set $F$, the set of pairs $(x,y)$ such that $F$ contains at least one midpoint of $\lbrace x,y\rbrace$ is closed, using the compacity of closed bounded sets. Let $k_0(x,y)$ be the first $k$ such that the closure $\overline U_k$ contains at least one midpoint of $\lbrace x,y\rbrace$, and iteratively $k_{n+1}(x,y)$ is the first $k>k_n(x,y)$ such that the closed intersection
$$ \overline U_k\cap\bigcap_{0\leq m\leq n}\overline U_{k_m(x,y)} $$
contains at least one midpoint of $\lbrace x,y\rbrace$.
Note that $k_n(x,y)$ is measurable, since the set of $(x,y)$ such that $k_n(x,y)\leq K$ is a finite union of closed sets. Then obviously $f_n:(x,y)\mapsto z_{k_n(x,y)}$ is measurable as well. Since the diameter of $U_k$ tends to zero, $(f_n(x,y))_{n\geq0}$ is a Cauchy sequence for all $(x,y)$, and $f(x,y):=\lim_{n\to\infty}f_n(x,y)$ is a well-defined midpoint of $\lbrace x,y\rbrace$. As a limit of measurable functions, it is measurable as well.
[G] M. Gromov, Metric structures for Riemannian and non-Riemannian spaces. 3rd printing (2007).
Thanks. You need separability of $(X,d)$, in order to ensure there existence of a countable basis $(U_k)_k$. It might be nice to make this more explicit in your write-up (btw separability is part of my assumptions).
@dohmatob The separability is a consequence of completeness, local compacity, and the algebraic midpoint property, as mentioned. Indeed, it makes $(X,d)$ into a complete, locally compact path space, hence the closed balls are compact. So $(X,d)$ is a $\sigma$-compact metric space, and it is second countable using the argument I gave in the answer.
Maybe the unit circle embedded in the euclidean plane is an example of a space that has several measurable midpoints structures but no continuous such structure?
Let us choose as the middle point of two points which are not on a diameter
the point in the middle of the shortest arc connecting the two points.
When two points are on a diameter, we may choose one of the middle points
on the two arcs as our middle point, but this cannot be made continuously.
Nice example, thanks. Interestingly, your example seems to be continuous almost everywhere on $X^2$.
In any complete Riemannian manifold $X$, if a measurable midpoint exists, it will be continuous almost everywhere on $X^2$. Indeed, it will be continuous at all $(x,y)$ such that $y$ is not in the cut locus of $x$ (there is no choice for any such pair). The cut locus of $x$ has measure zero in $X$ (and in fact it has Hausdorff dimension at most $\operatorname{dim}M-1$), so the set of singularities $(x,y)$ has measure zero in $X^2$ using Fubini's theorem.
Let us construct an example of a measurable midpoint space that is not a continuous midpoint space. The idea is to create a "jump" of midpoints somewhere. One way to do that is to consider a slit rectangle, e.g.
$$ \tilde E = [0,1]\times[0,1]\setminus \{\frac12\}\times(0,1) $$
endowed with the length metric induced by the canonical euclidean scalar product and $E$ is the completion of $\tilde E$ (obtained by adding two copies of the slit interval, one on the right of the slit and one on its left). Any two points in the same half (e.g. $[0,\frac12]\times [0,1]$ have an obvious midpoint, the Euclidean one; two points in different halves are connected by one or two shortest paths, hence have a midpoint. The midpoints are explicit enough to be easily seen to depend measurably on the endpoints. However, the midpoint between $x=(\frac12,1)$ and $y_t=(t,0)$ jumps at $t=\frac12$, from the center of the left copy of the slit interval to the center of its right copy, which are at distance $1$ one from the other. Hence $E$ does not have the continuous midpoint property.
(I must admit have trouble imagining a polish space with the midpoint property, but without the measurable midpoint property.)
Nice example, thanks. Concerning your last point, maybe you could make it a theorem (possibly with an additional contion on $X$) ? ;)
We are ultimately looking at the complexity of the multivalued function $\mathrm{MidPoint}_\mathbf{X} : \mathbf{X} \times \mathbf{X} \rightrightarrows \mathbf{X}$ assigning some midpoint to the points here. This includes what choice functions there are, but need not be limited to it. The framework to study the complexity of such operations is Weihrauch reducibility.
Just by definition of the midpoint, it follows that the map from a pair of points to the closed set of midpoints (equipped with the lower Vietoris topology) is continuous. This tells us that $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_\mathbf{X}$, where $\mathrm{C}_\mathbf{X}$ is closed choice on $\mathbf{X}$, which maps non-empty closed sets to some element. With $\mathrm{UC}_\mathbf{X}$ I denote the restriction of $\mathrm{C}_\mathbf{X}$ to singletons.
Everything we need about closed choice for this is found here.
Since we are assuming $\mathbf{X}$ to be Polish1, we immediately get the following:
If $\mathbf{X}$ is sigma-compact and midpoints are unique, then $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_\mathbb{N}$. This implies that the midpoint map is piecewise continuous, ie that there is a countable cover of $\mathbf{X} \times \mathbf{X}$ by closed sets, such that on each piece the map is continuous.
If $\mathbf{X}$ is sigma-compact, then $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_\mathbb{R}$. This guarantees that there is a Baire class 1 selection function for midpoints, but we get more [Baire class 1 is equivalent to "preimages of opens are $\Sigma^0_2$, so this is much simpler than Borel measurable]. For example, there always is a midpoint which is low (in the computability-theoretic sense) relative to the space.
If midpoints are unique, then $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{UC}_{\mathbb{N}^\mathbb{N}}$. Since the domain of $\mathrm{MidPoint}_\mathbf{X}$ is a Polish space, this already implies that $\mathrm{MidPoint}_\mathbf{X}$ is Borel measurable.
Without any restrictions, we just get that $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_{\mathbb{N}^\mathbb{N}}$. This doesnt rule out that we could avoid Borel measurable selection functions for the midpoint, but any construction would need to be very weird. The best starting point I can think of is using diagonally non-arithmetic functions.
1 We don't need that the metric defining our midpoints is complete, we just need some equivalent complete metric around.
Thanks. Upvoted. There are many things which seem to be missing. You seem to be using the fact that Polishness of the space implies many things, but it's not clear by what mechanism these conclusions are drawn. Because we're in a metric space, the assumption in case 1 is equivalent to saying that $X$ is locally compact (at this point, I'm fine with such an assumption, since the other posted solutions require it too). Finally, is every 'Baire class 1" function also Borel-measurable ?
Also, my 2 cents is that a version of the arguments you use here can be in principle used to prove the Kuratowski-Ryll-Nardzewski measurable
selection theorem used in Gerald Edgar's answer.
Indeed, this is done in Theorem 4.1 of http://cca-net.de/vasco/publications/borel.pdf.
I've clarified that Baire class 1 is much simpler than Borel measurable. The Polish-requirement is what makes sure that Baire space occurs as subscript where it does.
I don't understand what you mean by "Baire is simpler than Borel measurable". When all the dust settles, do we have a measurable function ? (Note that I don't care about computation, construction, or simplicity; only existence)
Yes, every Baire class 1 function is Borel measurable. The set of Borel measurable functions is the union of the Baire class $\alpha$ functions where $\alpha$ ranges over the countable ordinals. Hence, the "much simpler" statement.
Great. Thanks for the clarification. So, it might be possible to get the existence of a Borel measurable function with requirements less than Polish, right ?
|
2025-03-21T14:48:31.318619
| 2020-06-22T09:13:39 |
363796
|
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|
Stack Exchange
|
The zero entries in the character table of a finite group
When you browse the character tables of the small finite groups (for example here), you can observe that every zero entry corresponds to the value of an irreducible character $\chi$ on a non-central element $g$ such that the degree $\deg(\chi)$ of $\chi$ and the order $|C_G(g)|$ of the centralizer of $g$ in $G$ are not coprime (i.e. $\gcd(\deg(\chi) , |C_G(g)| ) \neq 1$).
Question 1: Is it true in general?
The reciprocal is false, one counter-example is given by $S_4$ which admits an irreducible character $\chi$ and a non-central element $g$ with $\deg(\chi) = 2$, $|C_G(g)| = 8$ but $\chi(g) = 2$. Moreover, the vanishing of $\chi(g)$ for $\chi$ irreducible ang $g$ non-central is not completely determined by $\deg(\chi)$ and $|C_G(g)|$ because for $G = M_{11}$ there is $\chi$ irreducible, $g_1, g_2$ non-central with $\deg(\chi) = 10$, $|C_G(g_1)| = |C_G(g_2)| = 8$, but $\chi(g_1) = 0$ and $\chi(g_2) = 2$.
Question 2: Why for $\chi$ irreducible, $g$ non-central and $\gcd(\deg(\chi) , |C_G(g)| ) \neq 1$ then "often" $\chi(g)$ vanishes? Is it always true in some specific cases, for example when $\deg(\chi) = |C_G(g)| $?
Just to parse the logic, you make the following conjecture: If $G$ is a finite group, $\chi$ a character of $G$, and $g \in G$ such that $|C_G(g)|$ and $\deg(\chi)$ are coprime, then $\chi(g) \neq 0$. Did I get it right?
Dear Sebastien, here are two facts which partially explain your observations:
(1) The central character $\frac{|G|\chi(g)}{|C_G(g)|\chi(1)}$ of $\chi$ at the conjugacy class of $g$ is an algebraic integer. When gcd$(\chi(1),|C_G(g)|)\ne1$, this makes it more likely that $\chi(g)=0$.
(2) Let $p$ be a prime and suppose that $|G|/\chi(1)$ is coprime to $p$. Then Richard Brauer proved that $\chi(g)=0$ if $p$ divides the order of $g$. Note that in this situation $p$ divides gcd$(\chi(1),|C_G(g)|)$.
@TheoJohnson-Freyd: This is the statement that Question 1 asks for, but for now it is just an observation on (very) small groups, not a general conjecture (yet).
@JohnMurray: your comment is very useful and deserves to be posted in answer. What is the reference for (1)?
See (3.7) in Isaacs, Character Theory of Finite Groups, Academic Press or G. James, M. Liebeck, Representations and Characters of Groups, Academic Press.
Maybe I misunderstand something, but it seems to me that if $\chi$ is the irreducible character of $S_6$ corresponding to the partition $(3,3)$ and $g \in S_6$ is a $6$-cycle, then $\chi(g)=0$ as $(3,3)$ does not have hook shape, $\chi(1)=5$ by the hook-length formula or counting standard Young tableaux or a well-know fact about Catalan numbers, and $C_{S_n}(g)=\langle g \rangle$ has order six.
@JohnShareshian: Yes, you got a counter-example for Question 1, perhaps the smallest one.
Here are infinitely many examples showing that the answer to question 1 is negative. Take $n \equiv 1 \bmod 4$ with $n>5$ and let $\chi$ be the character of the symmetric group $S_n$ associated to the partition $(n-2,2)$. Then, as is well-known, for each $w \in S_n$, $\chi(w)$ is obtained by subtracting the number of fixed points of $w$ from the number of $2$-sets fixed (setwise) by $w$. In particular, $\chi(1)={{n} \choose {2}}-n=\frac{n(n-3)}{2}$, and if $w$ has cycle type $(n-4,4)$ then $\chi(w)=0$. Now $w$ generates its own centralizer, which thus has order $4(n-4)$, which is manifestly coprime with $\frac{n(n-3)}{2}$ under the given conditions.
@JohnShareshian Do you expect the existence of an example with $G$ an alternating group (or more generally a simple group)?
I think that when $n=17$, you can use the same $\chi$ as in the answer (restricted to $A_n)$ and $w$ of shape $(9,4,4)$ to get an example in a simple group. There are likely infinitely many examples like this.
A partial answer to Question 2: the following is a theorem of Burnside (see e.g. Isaacs, Theorem 3.8).
Theorem. Let $\chi$ be an irreducible character, let $K$ be a conjugacy class of $G$, and let $g\in K$. Suppose that $\gcd(\chi(1),\#K)=1$. Then either $\chi(g)=0$ or the irreducible representation with character $\chi$ sends $g$ to a scalar (i.e. $g \in {\rm Z}(\chi)$).
The proof uses the integrality property of central characters that John Murray mentions.
Since $\#K=\#G/\#{\rm C}_G(g)$, the hypothesis translates to the assumption that for every prime $p$ that divides $\chi(1)$, the centraliser ${\rm C}_G(g)$ contain a Sylow $p$-subgroup of $G$, so it is a stronger variant of the hypothesis in your Question 2. If for all $p|\chi(1)$ the Sylow $p$-subgroups of $G$ have order $p$, then the hypothesis of Burnside and of your question coincide.
Here is another beautiful little lemma that relates centralizers to zeros of characters. This can be easily read off the character table. It is due to Feit and Thompson and appeared in their Odd-Order Theorem paper.
Lemma Let $N \unlhd G$ and $\chi \in Irr(G)$ such that $N \not \subseteq ker(\chi)$. If $g \in G$ with $N \cap C_G(g)=1$, then $\chi(g)=0$.
Proof $Irr(G)$ can be split up into two disjoint subsets:
$Irr(G)=S \mathop{\dot{\cup}} T$, where $S=\{\chi \in Irr(G): N \subseteq ker(\chi)\}$ and $T=\{\chi \in Irr(G): N \not\subseteq ker(\chi)\}$. Observe that $S$ can be identified with $Irr(G/N)$. If $g \in G$ with $N \cap C_G(g)=1$, then $C_G(g)$ embeds isomorphically into $C_{\overline{G}}(\overline{g})$ (where $\overline{.}$ denotes modding out by $N$), whence $|C_G(g)| \leq |C_{\overline{G}}(\overline{g})|$. By applying the Second Orthogonality Relation twice we get
$$|C_{\overline{G}}(\overline{g})|=\sum_{{\chi \in S}}|\chi(\overline{g})|^2=\sum_{{\chi \in S}}|\chi(g)|^2 \geq |C_G(g)|=\sum_{{\chi \in Irr(G)}}|\chi(g)|^2=\sum_{{\chi \in S}}|\chi(g)|^2 + \sum_{{\chi \in T}}|\chi(g)|^2$$
yielding $\chi(g)=0$ whenever $N \not\subseteq ker(\chi)$.
|
2025-03-21T14:48:31.319043
| 2020-06-22T09:33:38 |
363799
|
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|
Stack Exchange
|
On $\sum_{\substack{1\leq d\leq x\\(d,r)=1}}\frac{\mu(d)}{d}W(\frac{x}{d})$, with $\mu(n)$ the Möbius function and $W(x)$ the Lambert $W$ function
I wondered if it is possible to posed a similar question than Question 2 by Olivier Ramaré from [1] (page 231), although the computational evidence that I have for my conjecture is very small.
Conjecture. For all $x\geq 1$ and for all integers $r\geq 1$ one has that
$$0\leq \sum_{\substack{1\leq d\leq x\\(d,r)=1}}\frac{\mu(d)}{d}W(\frac{x}{d})\tag{1}$$
where here $\mu(n)$ denotes the Möbius function and $W(z)$ denotes the main branch of the Lambert $W$ function.
As reference I know that Wikipedia has the articles with titles Möbius function and Lambert $W$ function.
Question. I would like to know if you can find a counterexample* (an integer) $x$ and its corresponding $r$, or show what work can be done about the veracity of previous conjecture. Many thanks.
*If you are able to find a counterexample easily please try to state/prove an inequality with good mathematical content of the type $C\leq \sum_{\substack{1\leq d\leq x\\(d,r)=1}}\frac{\mu(d)}{d}W(\frac{x}{d})$, being this your inequality true for a suitable constant $C$ and $\forall x\geq 1$ and $\forall r\geq 1$.
If my problem/question is very difficult, after there are some contributions I should consider accepting an available answer.
References:
[1] Olivier Ramaré, Some elementary explicit bounds for two mollifications of the Moebius function, Funct. Approx. Comment. Math. Volume 49, Number 2 (2013), pp 229-240.
I know the reference [1] from Project Euclid. I hope that my Question is clear, interesting and at research level, please feel free to add your feedback in comments. On the other hand as was said my computational evidence is very small, thus please if you can to get easily a counterexample I am looking for an answer for the asterisk question in penultimate paragraph as an alternative question, with the purpose that the post is more interesting.
I think that my problem is interesting as a comparison to Ramaré's Question 2 that I refer from the first paragraph in page 231 of [1], since the main/principal branch of the Lambert $W$ function is an important function with an asymptotic behaviour related to the logarithm, see the section Asymptotic expansions from the linked article dedicated to the function Lambert $W$. Also are arising, in recent years, articles from the literature about analytic number theory and some branchs of the Lambert $W$.
|
2025-03-21T14:48:31.319225
| 2020-06-22T09:33:59 |
363800
|
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|
Stack Exchange
|
Modelling rational spaces of finite $\mathbb{Q}$-type with spaces with finite simplices in every simplicial dimension
EDIT 2
Original question below. I will award the outstanding bounty for an answer to the following question (question (2) in the OP).
Let $X$ be a Kan complex which is connected, nilpotent, and of finite $\mathbb{Q}$-type. Does there always exist a (connected, nilpotent if possible) simplicial set $Y$ with finitely many non-degenerate simplices in each simplicial dimension together with a morphism of simplicial sets $f:X\to Y$ or $f:Y\to X$ which is such that $A_\mathrm{PL}(f)$ is a quasi-isomoprhism?
If yes, a reference or an easy argument for this fact would be fantastic.
If not, are there any additional conditions that would make it true? Are there known counterexamples?
I am currently trying to relax some finiteness conditions for a result I am trying to prove in rational homotopy theory (to avoid having the result holding only for the space given by a single point) and I find myself confronted with the following two closely related questions, which I guess have already been considered many times before. A reference to an answer to (one or both of) those questions - in the positive or the negative - would be extremely welcome.
Let $X$ be a simplicial set. If necessary, assume that $X$ is connected, nilpotent, and of finite $\mathbb{Q}$-type. Is there a "rational" fibrant replacement for $X$, i.e. a map $X\to Y$ that induces a quasi-isomorphism $A_{\mathrm{PL}}(Y)\to A_{\mathrm{PL}}(X)$ and such that $Y$ is Kan? it would be even better if this map were a weak equivalence of simplicial sets. Of special interest to me is the case where $X$ has finitely many non-degenerate simplices in every simplicial degree ($Y$ is of course allowed to have infinitely many).
Let $X$ be a Kan complex which is connected, nilpotent, and of finite $\mathbb{Q}$-type. Is $X$ always "rationally the same" as a connected, nilpotent simplicial set of finite $\mathbb{Q}$-type with finitely many non-degenerate simplices in every simplicial dimension? More precisely, what I think I would like is the existence of a simplicial set $Y$ with finitely many non-degenerate simplices in every simplicial dimension and a morphism $f:X\to Y$ (or the other way around, or even a zig-zag of morphisms) such that $A_{\mathrm{PL}}(f)$ is a quasi-isomorphism of unital commutative algebras (resp. the same for all maps in the zig-zag). I am not asking for $f$ to be a weak equivalence (it would be very nice if it were, but I don't think it is possible in general).
Question (1) might be easy, maybe just taking any fibrant resolution or maybe taking the chains over the geometric realization of the space, but I'm not aware of the fact that $A_{PL}$ sends weak equivalences to quasi-isomorphisms (indeed, Bousfield-Gugenheim say that they suspect that it is not the case just after Prop. 8.3 in their article). I took a look in a couple of standard sources (Bousfield-Gugenheim, the book by Félix-Halperin-Thomas) but without success.
EDIT
Regarding question (1), prompted by the comment by Geoffroy Horel, we should have the following.
The functor $A_{\mathrm{PL}}$ is a right Quillen adjoint when seen as a functor
$$A_{\mathrm{PL}}:sSet^{op}\longrightarrow Com\text{-}alg_{\ge0}\ ,$$
indeed, Bousfield-Gugenheim prove (Lemma 8.2 and Prop. 8.3 in their article) that its left adjoint sends cofibrations of algebras to fibrations of simplicial sets ($=$ cofibrations in the opposite category) and preserves weak equivalences, making it into a left Quillen functor.
By the Ken Brown lemma, it follows that $A_\mathrm{PL}$ preserves weak equivalences between fibrant objects in $sSet^{op}$, i.e. cofibrant objects in $sSet$. But all simplicial sets are cofibrant in the Kan model structure, so that we get that $A_{\mathrm{PL}}$ preserves weak equivalences in general.
Therefore, we can take any fibrant replacement of a simplicial set for question (1) without any issue.
Please let me know if there is any shortcoming in this argument. My doubts mainly arise from the fact that Bousfield-Gugenheim state that they suspect $A_\mathrm{PL}$ not to preserve weak equivalences, so such an easy argument looks a bit suspicious, especially considering that I think that the model categorical argument that are used were already known at the time (they should be in Quillen, but I haven't checked).
I think there is a problem with the direction of the arrow in Question 1. A fibrant replacement of $X$ would be $X\to Y$. Then yes, any way of producing functorially a Kan complex replacement will work. I'm sure that $A_{PL}$ preserves weak equivalences but in that case you don't even need that as the map $X\to Y$ will be a homotopy equivalence (cofibrant source, fibrant target) and $A_{PL}$ will preserve those.
@GeoffroyHorel Salut Geoffroy :) Thanks a lot, this makes sense! We're using the fact that it is a contravariant adjunction, right? Feel free to post it as an answer.
Hi Daniel. Good to see you doing maths:) Yes I completely agree with what you wrote, I have no idea why Bousfield and Gugenheim got confused by this.
@GeoffroyHorel I never really stopped ;) We're hoping to have a really nice article out this summer with Bruno, those questions are related to one of the last results. I can explain what we do to you via skype one of these days if you want, just drop me an email :) Do you know who could have an idea about question 2?
|
2025-03-21T14:48:31.319667
| 2020-06-22T10:02:26 |
363801
|
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|
Stack Exchange
|
Factorizing vector fields near manifolds of singularities
Let $V: \mathbb{R}^n \to \mathbb{R}^n$ be a smooth vector field containing a smooth $k$-dimensional manifold $M$ (with $1\leq k < n$) of singularities: $V(M)=0$.
Suppose furthermore that at every point $x\in M$, the derivative $D_x V$ has $(n-k)$ nonzero eigenvalues, and thus $T_x\mathbb{R}^n = T_x M \oplus F_x$, where $\ker D_x V$ coincides with $T_x S$ and $F_x$ denotes some realization of the quotient space $T_x \mathbb{R}^n / T_x M$.
I am interested in knowing when this geometric splitting induces a corresponding factorization of the vector field. Specifically, I would like to know when I can write $V = W_1 X_1 + \cdots + W_{n-k} X_{n-k} := WX$, so that
$W = (W_1 \,\cdots\, W_{n-k})$, considered as an $n \times (n-k)$ matrix, has full rank and $\text{Im}\,W(x)=F_x$ for each $x \in M$, and
$X(M) = 0$ and $\ker D_xX = T_x M$ for each $x \in M$.
In other words, $V$ is factored so that $X$ encodes the information about the manifold $M$ and $W$ encodes the information about the transverse fiber bundle.
What are the results in this direction? I imagine that it's impossible to ask for such a factorization for general smooth vector fields, especially globally, but I am interested to hear about obstructions or sub-cases where problems of this type are studied. I am aware of some (local) positive results in the case when $V$ is polynomial in its entries; here, one can explicitly construct such factorizations using division algorithms for local rings. I've also heard folklore that the case $k=n-1$ (i.e. when $X$ is a scalar function) should be somehow easier or more obvious.
For example, if the normal bundle is not trivializable this thing is not possible. In particular, if $M$ is non-orientable then this is not possible.
|
2025-03-21T14:48:31.319833
| 2020-06-22T10:03:52 |
363802
|
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|
Stack Exchange
|
A generalization of metric spaces
Let $(L,<,+)$ be a structure such that (1) $<$ is a linear order of $L$, (2) $L$ has a least element 0, (3) $+$ is a binary function on $L$ that behaves like addition of positive real numbers, i.e. commutativity, associativity, $a+0 = a$, and $a< b$ iff there is $c>0$ such that $b =a+c$.
Say a pair $(X,d)$ is an “$L$-metric space” if $d : X^2 \to L$ is a function satisfying the metric space axioms (only the codomain is not necessarily $\mathbb R$).
Question 1: Has this or something similar been studied?
Question 2: Say a topological space is “weakly metrizable” when there exists $L$ and $d$ as above inducing the topology. When is a space weakly metrizable?
I don't know anything about this, but if you don't know where to begin and no one else has a good suggestion, then maybe try looking through some of Leonard M. Blumenthal's work, and the topics of metric lattices and metric semilattices.
What's an easy example of such an $(X, L)$ that can't be 'reduced' to $[0, \infty)$, in the sense that there is (well, isn't) an order-preserving (and, say, sub-additive?) map $L \to [0, \infty)$ such that the resulting $X^2 \to [0, \infty)$ gives the same topology?
@LSpice Good question. Consider an ultrapower of $\mathbb R$ by a nonprincipal ultrafilter on $\mathbb N$, call it $\mathbb R^$. The interval topology on $\mathbb R^$ is not first-countable. But the absolute value of differences gives a weak metric.
It follows from your assumptions that for $a<b\in L$ there is a unique $c$ such that $a+c=b$ and that $L$ is a cancellative monoid: $a+c=b+c$ implies $a=b$. Also addition preserves the order. A cancellative commutative monoid embeds in an abelian group. If I'm not mistaken $L$ embeds in an ordered abelian group $\Lambda$ such that
the positive elements correspond to the non-zero elements of $L$.
Metric spaces where the codomain of the distance function is an ordered abelian group - that is, $\Lambda$-metric spaces - have been defined as a first step to defining $\Lambda$-trees. Ian Chiswell has written an introduction to this subject.
As for the topology that such a metric gives rise to, note that if $\Lambda$ has a minimum non-zero convex subgroup, then this subgroup is archimedean, and thus the topology is $\mathbb{R}$-metrisable. If there is no such convex subgroup, then the balls of radius $\Lambda_0$ - that is, the sets $B(x,\Lambda_0)=\{y\in X:d(x,y)\in\Lambda_0\}$ - where $\Lambda_0$ ranges through the non-zero convex subgroups, gives a base for the topology which is therefore $0$-dimensional. And in fact replacing $\Lambda$ by the (linearly ordered) set of its convex subgroups and putting $d'(x,y)$ equal to the convex subgroup spanned by $d(x,y)$ gives a generalised ultrametric equivalent to $d$.
Generalised ultrametric spaces have been studied for example by Priess-Crampe and Ribenboim. (Of course if $L\subseteq\mathbb{R}_{\geq 0}$, then these are ultrametric spaces.)
My PhD thesis was on this topic, focusing mostly on a combinatorial approach (rather than topological). So what I write below is directed toward Question 1.
I call a structure $\mathcal{R}=(R,+,\leq,0)$ a distance magma if $(R,\leq,0)$ is a total order with least element $0$, and $+$ is a commutative binary operation on $R$ which preserves the order ($a\leq b\Rightarrow a+c\leq b+c$). I call $\mathcal{R}$ a distance monoid if the operation is associative. The last axiom you mention about "solving inequalities" didn't play a special role. (Also, I use $\mathcal{R}$ instead of $L$ since eventually there will be a first order language.)
My own interest started with the special case of fixing a countable set $R\subseteq\mathbb{R}_{\geq 0}$ such that $0\in R$ and $R$ is closed under $x+_R y:=\sup\{z\in R:x\leq y+z\}$, which determines a distance magma $\mathcal{R}=(R,+_R,\leq,0)$. The first main question was:
Question. When is there a countable, universal and ultrahomogeneous $\mathcal{R}$-metric space (so, in other words, a Urysohn-like space relative to the distance set $R$)?
For example if $R=\mathbb{Q}_{\geq0}$ then then the answer is yes, witnessed by the classical rational Urysohn space. Another example is $R=\{0,1,2\}$ which yields the random graph (where $1$ designates an edge and $2$ designates no edge).
It turned out that this question was answered by Delhomme, LaFlamme, Pouzet, and Sauer. There is a more technical general characterization, but, in the special case above, there is a really nice fact:
Fact. There is an $\mathcal{R}$-Urysohn space if and only if the operation $+_R$ is associative.
The idea is that associativity of $+_R$ is combinatorially capturing the ability to amalgamate three-point $\mathcal{R}$-metric spaces, which leads to amalgamation for the class of all finite $\mathcal{R}$-metric spaces.
The "smallest" example of a set $R$ as above for which there is no $\mathcal{R}$-metric spaces is $\{0,1,2,4\}$.
In any case, it became more natural to work in a more general axiomatic framework. Given a countable distance magma $\mathcal{R}$, there is a natural notion of an $\mathcal{R}$-metric space, and a countable $\mathcal{R}$-Urysohn space exists if and only if $\mathcal{R}$ is a distance monoid. So assume $\mathcal{R}$ is a distance monoid and let $\mathcal{U}_{\mathcal{R}}$ denote the $\mathcal{R}$-Ursyosn space. In other words, $\mathcal{U}_{\mathcal{R}}$ is the unique (up to isometry) countable $\mathcal{R}$-metric space such that every finite $\mathcal{R}$-metric space embeds as a subspace of $\mathcal{U}_{\mathcal{R}}$, and any partial isometry between two finite subspaces of $\mathcal{U}_{\mathcal{R}}$ extends to a (total) isometry $\mathcal{U}_{\mathcal{R}}$.
I consider $\mathcal{U}_{\mathcal{R}}$ as a first-order structure in a relational language $L_{\mathcal{R}}$ containing binary relations $d_r(x,y)$ for $r\in R$, which are interpreted as "$d(x,y)\leq r$". Let $T_{\mathcal{R}}$ be the complete theory of $\mathcal{U}_{\mathcal{R}}$ in this language.
If $\mathcal{R}$ is finite then $T_{\mathcal{R}}$ is $\aleph_0$-categorical, but for infinite distance monoids this fails. For example, if $\mathcal{R}=(\mathbb{Q}_{\geq 0},+,\leq,0)$ (so $\mathcal{U}_{\mathcal{R}}$ is the rational Urysohn space), then there are plenty of non-isolated $2$-types over $\emptyset$, which describe "new distances" from irrationals and infinitesimal cuts. For example, $\{\neg d_0(x,y)\}\cup\{d_r(x,y)\leq r:r>0\}$ is a finitely satisfiable $2$-type, which describes two elements at nonzero infinitesimal distance.
In general, any distance $r\in R$ can be viewed as a quantifier-free $2$-type: $\{\neg d_s(x,y):s<r\}\cup\{d_s(x,y):r\leq s\}$. This produces a canonical embedding of $R$ in the set of quantifier-free $2$-types over $\emptyset$, and the ordering on $R$ extends naturally to this space (in a way very similar to a Dedekind-MacNeille completion). A more interesting fact is that the monoid operation also extends to the (quantifier-free) $2$-types. Specifically, given $2$-types $p,q$, define $p+q$ to be the supremum (which exists) of the set of $2$-types $r$ such that the $3$-type $p(x,y)\cup p(y,z)\cup r(x,z)$ is finitely satisfiable. (So $p+q$ is the largest distance that you can consistently put on a triangle with distances $p$ and $q$.)
So altogether, we have a new distance monoid $\mathcal{R}^*$, built from the quantifier-free $2$-types, which extends $\mathcal{R}$. $\mathcal{R}^*$ behaves a little like a "nonstandard" or saturated extension of $\mathcal{R}$ (but it is not exactly this).
In my thesis I proved a characterization of quantifier elimination for $T_{\mathcal{R}}$ in terms of the behavior of $\mathcal{R}^*$.
Theorem. $T_{\mathcal{R}}$ has QE if and only if for any standard element $r\in R$, the operation $x\mapsto x+r$ is continuous on $\mathcal{R}^*$ (where here continuity is with respect to the order topology on $\mathcal{R}^*$).
So QE does happen in most "nice" situations, for example if $\mathcal{R}$ is finite (in which case $\mathcal{R}^*=\mathcal{R}$), or if $\mathcal{R}=(\mathbb{Q}_{\geq 0},+,\leq,0)$. But it can fail by "poking holes" in nice monoid. An example is $\mathcal{R}=(R,+_R,\leq,0)$ where $R=\mathbb{Q}\cap (\{0\}\cup[2,3)\cup (3,\infty))$.
The QE result was really only the starting point, since I was mainly interested in model-theoretic neostability properties of $T_{\mathcal{R}}$ under the assumption of QE. The rest of the thesis was about characterizing such properties (stability, simplicity, SOP$_n$, forking, elimination of imaginaries, etc.) via algebraic/combinatorial properties of $\mathcal{R}$. It turned out to be quite successful: most model-theoretic properties of $T_{\mathcal{R}}$ are controlled by $\mathcal{R}$.
The model-theoretic results became two papers: arXiv 1502.05002 and arXiv 1504.02427. I wrote another paper (arXiv 1509.04950) about extending partial isometries (i.e., the "Hrushovksi property"), which built on results of Solecki about rational-valued metric spaces. What I did was generalized further by Hubička, Konečný, and Nešetřil arXiv 1902.03855. They also proved the Ramsey property for these classes (arXiv 1710.04690).
Question #2 is definitely interesting, but nothing I did was in that direction since I effectively viewed my spaces as discrete. But there might be something in these two older articles:
Alsina & Trillas, On natural metrics, Stochastica 2 (1977).
Narens, Field embeddings of generalized metric spaces, Victoria Symposium on Nonstandard Analysis, Springer, Berlin (1974).
There is also some more recent work by Etedadialiabadi, Gao, Le Maître, and Melleray, which I haven't read closely, but looks related.
There has been a lot of work around a related idea:
F. W. Lawvere,
Metric spaces, generalized logic, and closed categories,
Rendiconti del seminario matématico e fisico di Milano, 1973 - Springer
The analogy between $dist (a, b)+ dist (b, c)≥ dist (a, c)$ and $hom (A, B)⊗ hom (B, C)→ hom(A, C)$ is rigorously developed to display many general results about metric spaces as consequences of a «generalized pure logic» whose «truth-values» are taken in an arbitrary closed category.
It might be useful to look at this to get an idea of where such a theory can go. You would need to follow up citations. The paper was reprinted in TAC reprints see
http://www.tac.mta.ca/tac/reprints/index.html.
For completion's sake and regarding Question 1, let me add a slightly more general concept: scaled spaces.
Let $M$ be a set, let $X$ be a totally ordered set and let $0$ be a symbol such
that $0<x$ for all $x\in X$. An $X$-valued scale on $M$ is a map
$d:M\times M\to X\cup\{0\}$ such that for all $x,y,z\in X$:
$d(x,y)=0\Leftrightarrow x=y$;
$d(x,y)=d(y,x)$; and
$d(x,z)\leq\max\{d(x,y),d(y,z)\}$.
The space $(M,X,d)$ is called a scaled space (ultrametric space if $X\subset(0,\infty)$).
This concept was studied by H. Ochsenius and W. H. Schikhof and then applied to the study of Banach spaces over fields with an infinite rank valuation. To initiate in this topic I recommend three articles:
H. Ochsenius and W. H. Schikhof, “Banach spaces over fields with an infinite rank valuation,” p-Adic Functional Analysis, Lecture Notes in Pure and Appl. Math. 207, 233–293 (Marcel Dekker, 1999).
H. Ochsenius and W. H. Schikhof, “Norm Hilbert spaces over Krull valued fields,” Indagat. Math. 17 (1), 65–84 (2006).
A. Barria Comicheo, Generalized Open Mapping Theorem for $X$-normed spaces,
p-Adic Numbers, Ultrametric Analysis and Applications, vol. 11, (2), 2019, pp. 135--150.
Regarding Question 2, in the context of scaled spaces, it was proved in [1] that a scaled space $(M,X,d)$ is ultrametrizable if and only if $M$ is discrete or there exist $s_1>s_2>\dots$ in $X$ such that $\lim_n s_n=0$.
Is this a strictly more general concept? The original question covers all metric spaces, whereas yours (as you point out) only obviously covers ultrametric spaces.
It is more general in the sense that no algebraic structure is required in the domain nor codomain of the "metric". The price to pay is that it only generalizes ultrametric spaces.
As long as one drops the subtraction axiom ($a<b\Leftrightarrow \exists c, a+c=b$) then $X\cup{0}$ is an "algebraic structure" as in the question, but just with the binary operation of $\max$. (Btw, OP does not impose extra structure on the domain of the metric, just the codomain.) One can go further and relax totality of the order to obtain "lattice-valued" ultrametrics (e.g., as in this paper, though surely the concept is older).
There are several results in topology about metrics taking values in (the positive cone of) a partially ordered Abelian group $\mathbb{G}=(G, <, +, 0)$. The first occurrence of this spaces I'm aware of is dated 1950 by Sikorski (MR0040643 - R. Sikorski, Remarks on some topological spaces of high power). He literally started a new branch of topology that investigated this kind of objects, called $\omega_\mu$-metrizability.
There are also several results about the topological nature of spaces with metrics taking values in other kind of structures. The wider framework I'm aware of has been studied by Reichel in 1977 (MR0458373 - H. C. Reichel, Some results on distance functions). I'll resume here some of its results:
He consider totally ordered semigroups with minimum $(S,<, 0, +)$, i.e. structures such that $(S,<, 0)$ is a total order with minimum $0$, $(S, +)$ is an Abelian semigroup and $<$ is translation invariant, i.e. $a<b$ implies $a+c<b+c$ for every $c$ (I believe you also wanted to assume this kind of invariance?).
We say that a $S$ is continuous if whenever a sequence $\langle x_i\mid i<\gamma\rangle$ is coinitial in $(S\setminus\{0\},<)$, then $\langle x_i+x_i\mid i<\gamma\rangle$ is coinitial as well in $(S\setminus\{0\},<)$.
Notice that every structure of the form you described is a continuous totally ordered semigroup with minimum in this sense.
Let us call $S$-metrizable a space that has a metric in the structure $S$.
It turns out that the behavior of such class of spaces is influenced mostly by the coinitiality of $(S, <)$, i.e. the size of the smallest sequence of elements of $S\setminus \{0\}$ converging to $0$. We call such cardinal number the degree of $S$.
Reichel proved the following:
Theorem 2, countable (Reichel): For a given topological space $(X,\tau)$, the following are equivalent:
$X$ is $S$-metrizable for some continuous totally ordered semigroup $S$ of degree $\omega$.
$X$ is metrizable (with standard metric over real numbers).
(He also characterized spaces that are metrizable over non-continuous totally ordered semigroup $S$ (Theorem 1 of the paper cited above)).
When the degree of $S$ is uncountable, the situation is easier.
Theorem 2, uncountable (Reichel): For a given topological space $(X,\tau)$, the following are equivalent:
$X$ is $S$-metrizable for some continuous totally ordered semigroup $S$ of degree $\kappa>\omega$.
$X$ is $S$-metrizable for $S$ the set of positive elements of some totally ordered Abelian group of degree $\kappa>\omega$.
Combining this with other results in literature, in second case one get the following:
Theorem: For a given topological space $(X,\tau)$, the following are equivalent:
$X$ is $S$-metrizable for some continuous totally ordered semigroup $S$ of degree $\kappa>\omega$.
$X$ is $S$-metrizable for every continuous totally ordered semigroup $S$ of degree $\kappa>\omega$.
$X$ is $S$-metrizable for $S$ the set of positive elements of some totally ordered Abelian group of degree $\kappa>\omega$.
$X$ is ultrametrizable over some total order of coinitiality $\kappa>\omega$ (i.e. the $+$ operation of $S$ coincide with the maximum between elements).
$X$ is a subset of $\lambda^\kappa$ with bounded topology for some $\lambda$.
...(there are several other topological characterizations).
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2025-03-21T14:48:31.320819
| 2020-06-22T10:05:53 |
363803
|
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|
Stack Exchange
|
$f^!=f^*[d]$ for quasismooth maps?
Given a smooth map of schemes $f:X\to Y$ of relative dimension $d$, then there is a natural isomorphism $f^!\simeq f^*[d](2d)$ (in any context where the six operations are defined; see Cesinski-Deglise).
If $f$ is a smooth map of Artin stacks I imagine the same is true (there is a notion of the six operations for Artin stacks: https://arxiv.org/abs/1211.5948).
Question: Let $f:X\to Y$ be a quasismooth map of Artin stacks (say also flat to rule out things like closed embeddings). Is it still true that $f^!\simeq f^*[d](2d)$?
In general, I think this should fail for most non-smooth local complete intersections.
For a specific example, let $X = \mathbb V(xy) \subset \mathbb A^2$ and $Y = *$. Then the stalk at the dualizing complex of $X$ at the origin is $\mathbb Q_l[1] \oplus \mathbb Q_l[2]^{\oplus 2}$, contradicting the local constancy of $f^! \mathbb Q_l = \omega_X$.
Thanks. Should that be $\mathbf{Q}\ell\oplus\mathbf{Q}\ell[1]$?
@Meow I just did the calculation, and I think we were both wrong. It's $H_(X,X-0) = \mathbb Q_l[1] \oplus \mathbb Q_l [2]^{\oplus 2}$. Comes from $\mathbf D i^ \omega_X = i^! \mathbb Z_X.$ How are you getting your answer?
Sorry, that was a guess not a computation on my part. Thanks!
|
2025-03-21T14:48:31.320936
| 2020-06-22T10:13:36 |
363806
|
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Subfunctor of representable is continuous
Given an object $A$ in a category $\mathcal{C}$, the functor $Hom_{\mathcal{C}}(A,-)$ is tautologically continuous i.e. limit-preserving. Let $F$ be a subfunctor of $Hom_{\mathcal{C}}(A,-)$, is $F$ in general continuous? We might assume $F$ is fully-faithful if necessary.
Take $\mathcal C$ a meet-semilattice. Not every sub-upper-set of a principal filter is a filter.
It might be actually true for fully faithful $F$, since there are hardly any. What would be your example of a fully faithful subfunctor of a representable functor?
@მამუკაჯიბლაძე I do have an example for the fully faithful $F$ but it seems not easy to describe in a few lines... Do you have a proof / reference that such $F$ is necessarily continuous? Thanks!
|
2025-03-21T14:48:31.321021
| 2020-06-22T10:56:33 |
363809
|
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"Gerhard Paseman",
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|
Stack Exchange
|
A modern reference to the Zsigmondy Theorem
I need to cite the classical Zsigmondy Theorem, which was proved in 1892.
Is there any modern reference to this theorem?
I mean some standard textbook in Number Theory containing this theorem together with the proof.
If you just need to cite the theorem, what's wrong with citing the original source?
In Linear Forms in Logarithms and Applications by Yann Bugeaud, European Mathematical Society, 2018, there are mentions and references on pages 95 and 106
I plan to include both the original course and the modern one (for those that do not read German, for example).
M. Teleuca, Zsigmondy's theorem and its applications in contest problems, International Journal of Mathematical Education in Science and Technology Volume 44, 2013 - Issue 3, Pages 443-451, https://www.tandfonline.com/doi/abs/10.1080/0020739X.2012.714493?mobileUi=0&journalCode=tmes20
The abstract begins, "In this article, we present a detailed proof of Zsigmondy's theorem."
See also https://math.stackexchange.com/questions/660585/elementary-proof-of-zsigmondys-theorem and Moshe Roitman, On Zsigmondy primes, Proceedings of the
American Mathematical Society, Volume 125, Number 7, July 1997, Pages 1913–1919, https://www.ams.org/journals/proc/1997-125-07/S0002-9939-97-03981-6/S0002-9939-97-03981-6.pdf
There are also notes of Jameson on cyclotomic polynomials. A MathOverflow search should reveal the link. It has a proof following Roitman. Gerhard "My Goto For Cyclotomic Polynomials" Paseman, 2020.06.22.
Indeed, it is very difficult to find Zsigmondy's theorem with a proof in a book. However, it is proved in Appendix B to Chapter 30 in
Berkovich, Ya. G.; Zhmudʹ, E. M. Characters of finite groups. Part 2. Translated from the Russian manuscript by P. Shumyatsky [P. V. Shumyatskiĭ], V. Zobina and Berkovich. Translations of Mathematical Monographs, 181. American Mathematical Society, Providence, RI, 1999.
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2025-03-21T14:48:31.321178
| 2020-06-22T12:06:05 |
363812
|
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Stack Exchange
|
Why does this cluster tilting object form a local slice?
I am reading the paper "Cluster automrphisms", here is the link: http://prospero.dmat.usherbrooke.ca/ibrahim/publications/Cluster_Automorphisms.pdf
In the proof of lemma 3.1 I am stuck: For an acyclic quiver $Q$ and a seed $(X,Q)$ of a cluster algebra $\mathcal{A}$, we can get a cluster tilting object $M= \oplus_{x \in X} M_x$ in the cluster category $\mathcal{C}_Q$ correpsonding to $X$. If $EndM \cong kQ$ or $EndM \cong kQ^{op}$, then how to get that the set $\{ M_x \mid x \in X\}$ forms a local slice in $\mathcal{C}_Q$?
The definition of local slice is not short, so you can see section 2.4 of this paper: https://arxiv.org/pdf/1606.05161.pdf
(If $Q$ is of type ADE, then I can get the result by the equivalence: $ind \mathcal{C}_Q/ add \tau M \cong mod EndM$ and the projective component of the AR quiver of $mod End M$ forms a local slice. For the general case, I find it does not work. Thank you for any help.)
|
2025-03-21T14:48:31.321275
| 2020-06-22T13:33:10 |
363820
|
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Stack Exchange
|
Question concerning Brauer's second main theorem, Brauer correspondent blocks and blocks covered by nilpotent blocks
A version of Brauer's second main theorem is as follows:
Let $G$ be a finite group, $x$ be a $p$-element of $G$, $B\in\mathcal{Bl}(G)$, and $\chi\in$ Irr$(B)$.
If $d_{\chi\mu}^x\neq 0$ and $\mu$ belongs to a block $b$ of $C_G(x)$, then $b^G=B$. Hence we have
$$\chi(xy)=\sum\limits_{b^G=B} \sum\limits_{\mu\in\text{IBr}(b)} d_{\chi\mu}^x\cdot \mu(y)$$
for all $y\in C_G(x)_{p'}$.
Now, let $B$ be a nilpotent 2-block of $G$ having $D:=C_2\times C_2$ as defect group, $x$ an involution in $G$, $y:=1$ and $b^G=B$.
Questions:
i) What can be said about $b$?
ii) Is $l(b)$ or $k(b)$ (or bounds for these numbers) known?
iii) Does $b$ (under certain circumstances) have to be nilpotent as well?
iv) I think that the defect group $\tilde{D}$ of $b$ is contained in $D$. Do we have $D\cong\tilde{D}$?
I would be thankful for references in the literature.
Thanks for the help.
A Brauer correspondent of a nilpotent block is always nilpotent, so $b$ is indeed nilpotent and in this case we $\ell(b) = 1$, $k(b) = 4$. Also, $b$ does indeed have defect group $D$.
Probably the best reference is to fully understand the Brou'e Puig Inventiones(?) paper on nilpotent blocks in this special case, although Brauer's paper on blocks with Abelian defect group and inertial index one (which is essentially equivalent to being nilpotent in the Abelian defect group case), which is an earlier paper, already provides answers to your question in the Klein $4$ defect group case (in the case of question iii), you obtain that $b$ has Abelian defect group $D$ with inertial index one).
Thank you very much for your answer and the references.
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2025-03-21T14:48:31.321404
| 2020-06-22T13:37:52 |
363821
|
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Stack Exchange
|
Can we say that $f_\infty(t)=g_\infty(t) \text{ a.e}$?
Let $(E,\mathcal{A},\mu)$ be a finite measure space. Let $\{f_n\}$ and $\{g_n\}$ two uniformly integrable sequences such that:
$$
\qquad\forall n\geq 1~:~ f_n(t)=g_n(t)~~ \text{ a.e.}\tag1
$$
$$
\qquad f_n\underset{n}{\to} f_\infty\text{ weakly in } L^1 \tag2
$$
and for all subsequence $\{g_{n_i}\}$ of $\{g_n\}$ we have
$$
\qquad\frac{1}{m}\sum_{i=1}^{m}{g_{n_i}(t)}\underset{m}{\to} g_\infty(t) \text{ a.e.} \tag3
$$
Can we say that $f_\infty(t)=g_\infty(t) \text{ a.e}$?
Yes: by uniform integrability and Vitali's convergence theorem, one concludes that the pointwise a.e. convergence (3) can be upgraded to strong $L^1$ convergence, hence also weak convergence.
Moreover, (2) implies that the Cesáro mean
$$
\frac{1}{m}\sum\limits_{i=1}^mf_{n_i}\rightharpoonup f_\infty
\qquad\mbox{ weakly in }L^1
$$
(Apply the usual fact that if some real sequence converges to a finite limit then its Cesáro mean also converges to the same limit, here the real sequence is $\int f_{n_i} \phi\to \int f_\infty\phi$ for any fixed $\phi\in L^\infty$ as $i\to\infty$.)
Since $f_n=g_n$ as elements of $L^1$ the uniqueness of weak limits guarantees indeed $f_\infty=g_\infty$
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2025-03-21T14:48:31.321621
| 2020-06-22T14:01:30 |
363823
|
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Stack Exchange
|
Estimation of a sum involving Stirling's number of second kind and binomial coefficient
Let $S(n, j)$ be Stirling's number of second kind. Let $p\in [0,1]$ and $m \in N$.
Bound from above the following sum:
$$
\sum_{j=0}^m S(n,j) {m \choose j}\, j! \, p^j
$$
Can we assume that $n>m$ ?
This is not yet a full answer, but more an expanded comment, asking firstly whether I've your formula correctly translated.
If I get your formula right, then, for $p=1$ , the results written in form of a matrix-problem looks like
$$ X(1) = S2 \cdot \,^dF \cdot P = \small \begin{bmatrix}
1 & 1 & 1 & 1 & 1 & 1 & \cdots\\
0 & 1 & 2 & 3 & 4 & 5 \\
0 & 1 & 4 & 9 & 16 & 25 \\
0 & 1 & 8 & 27 & 64 & 125 \\
0 & 1 & 16 & 81 & 256 & 625 \\
0 & 1 & 32 & 243 & 1024 & 3125 \\
\vdots & & & & & & \ddots \end{bmatrix}
$$
where $S2$ is the matrix of Stirlingnumbers 2nd kind, $F=[0,1,2!,3!,...]$ the factorials and $\,^dF$ taken as diagonal, and $P$ the upper triangular Pascalmatrix.
If this is correct, then your problem is, with a vector $V(p)=[1,p,p^2,...]$ and its diagonal $\,^dV(p)$ we would ask for estimates in $X(p)$:
$$ X(p) = S2 \cdot \,^dF \cdot \,^dV(p) \cdot P $$
The occuring pattern in $X(p)$ ist not easily to decode, but we can rewrite the matrixformula a bit
$$\begin{array} {} X(p) &= S2 \cdot \,^dF \cdot \,^dV(p) \cdot P \\
&= S2 \cdot \,^dF \cdot (P \cdot P^{-1} \cdot) \,^dV(p) \cdot P \\
&= X(1) \cdot ( P^{-1} \cdot \,^dV(p) \cdot P )\\
&= X(1) \cdot P \cdot \,^dV(p) \\
\end{array}$$
and the constant matrix from the two beginning factors looks like
$$ X(1) \cdot P =\small \begin{bmatrix}
1 & 2 & 4 & 8 & 16 & 32 \\
0 & 1 & 4 & 12 & 32 & 80 \\
0 & 1 & 6 & 24 & 80 & 240 \\
0 & 1 & 10 & 54 & 224 & 800 \\
0 & 1 & 18 & 132 & 680 & 2880 \\
0 & 1 & 34 & 342 & 2192 & 11000
\end{bmatrix}
$$
where we only need to multiply the columns by the parameter $p^c$ where $c$ is the column-number starting from $0$ to get your sought values.
I think a formula for the entries of $ X(1) \cdot P$ is not so difficult to derive and thus to have the upper bounds.
|
2025-03-21T14:48:31.321775
| 2020-06-22T14:04:10 |
363824
|
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|
Stack Exchange
|
Spectral gap of a Markov chain on the nonnegative integers
Let $\lambda_k,\mu_k\in\mathbb R_{\ge0}$ $(k\ge1)$ be nonnegative real numbers such that $\sum_{k=1}^\infty k\lambda_k<\infty,$ let $S=\mathbb Z_{\ge0}$ be the nonnegative integers, let $T=\mathbb R_{\ge0}$ be the nonnegative real numbers and consider the continuous-time Markov chain $X=(X_t)_{t\in T}$ on $S$ with rates
$$Q(n,n+k)=(n+1)\lambda_k\quad(k\ge1),\qquad Q(n,n-k)=(n+1-k)\mu_k\quad(1\le k\le n).$$
(This Markov chain appears in biology as a model of the length of an evolving DNA sequence (Miklós et. al. 2004). I have also examined some properties of this process in a previous Math.StackExchange post.)
In addition, assume $X$ is irreducible and reversible. Next, let $A=\{k\in\mathbb Z_{\ge1}:\mu_k\ne0\}.$ Then it can be shown that the following conditions hold: (1) $\forall k\in\mathbb Z_{\ge1}\quad\lambda_k=0\leftrightarrow\mu_k=0;$ (2) $\gcd A=1;$ and (3) $\forall j,k\in A\quad r_j^{1/j}=r_k^{1/k}<1.$ The converse can also be shown; if (1), (2) and (3) are true, then $X$ must be irreducible and reversible. (The proof of this if-and-only-if condition involves a lot of manipulating detailed-balance equations and a little bit of number theory, in the form of Bezout's lemma and Schur's lemma.)
For example, if $0<\lambda_1=\mu_1$ and if $0=\lambda_k=u_k$ for all integers $k\ge2,$ then we recover the linear birth-death process with immigration with birth rate $\lambda_1,$ death rate $\mu_1$ and immigration rate $\lambda_1,$ whose nonzero rates are
$$Q(n,n+1)=(n+1)\lambda_1\quad(n\ge0,k\ge1),\qquad Q(n,n-1)=n\mu_1\quad(n\ge1).$$
Or, for example, given parameters $\mu\in\mathbb R_{>0},\gamma,r\in(0,1),$ we can let $\mu_k=\mu(1-r)^2r^{k-1}$ and $\lambda_k=\mu(1-r)^2\gamma^kr^{k-1}$ for all $k\ge1.$ Both these examples have been used in, and are of interest in, computational biology.
Now, my question is, what is the spectral gap, $\text{gap}(X),$ of $X$?
For the linear birth-death process with immigration mentioned in the above paragraph, we can show the spectral gap is $\mu_1-\lambda_1$ via the "dual variational formula" for reversible irreducible birth-death processes, which reads: Let $X$ with birth rates $b_i$ ($i\in S$) and death rates $d_i$ ($i\in\mathbb Z_{\ge1}$); then
$$\text{gap}(X)=\sup_{v\in{\cal V}}\inf_{i\in S}R_i(v),$$
where ${\cal V}=(\mathbb R_{>0})^S$ is the set of strictly-positive-real-valued nonnegative-integer-indexed sequences and where
$$R_i(v)=d_{i+1}+b_i-d_i/v_{i-1}-b_{i+1}v_i=\Delta d(i)-\Delta b(i)+d_i(1-v_{i-1}^{-1})+b_{i+1}(1-v_i),$$
where $\Delta d(i)=d_{i+1}-d_i,\Delta b(i)=b_{i+1}-b_i,a_0=0,v_{-1}=1,$ for all $i\in S.$ (This result is Theorem 1.1 in Chen (1996).) Coming back to the linear birth-death process with immigration, note that taking $v=1=(1,1,1,\dots)$ to be the constant sequence with all 1's yields $R_i(v)=\mu_1-\lambda_1$ for all $i\in S,$ and thus $\text{gap}(X)\ge\mu_1-\lambda_1.$ In addition, it turns out that $\mu_1-\lambda_1$ is an eigenvalue, so the bound is sharp, i.e., $\text{gap}(X)=\mu_1-\lambda_1.$
I conjecture that "in general", if $\sum_{k=1}^\infty k\mu_k<\infty$ (a reasonable assumption), then $\text{gap}(X)=\mu-\lambda,$ where $\mu=\sum_{k=1}^\infty k\mu_k$ is "the total deletion rate per site" and $\lambda=\sum_{k=1}^\infty k\lambda_k$ is "the total insertion rate per site". I have no idea how to prove it, though; the coupling is no longer monotone in general, since $X$ can make jumps of size $>\!\!1.$ In addition, I don't know of any generalizations of the Chen's theorem above, so I don't know how to get a lower bound.
Chen, Mufa, Estimation of spectral gap for Markov chains, Acta Math. Sin., New Ser. 12, No. 4, 337-360 (1996). ZBL0867.60038.
Miklós, I., Lunter, G. A., & Holmes, I. (2004). A “long indel” model for evolutionary sequence alignment. Molecular Biology and Evolution, 21(3), 529-540.
crossposted from Math.StackExchange
Spectral gaps of birth-death processes are discussed in Landim, Sethuraman, and Varadhan, Ann. Prob. 24, 1871 (1996). See https://projecteuclid.org/euclid.aop/1041903209
As an aside, you may be interested in Karlin-McGregor theory for such chains.
This paper seems not directly relevant. It doesn't give any equalities, only inequalities that may not be sharp. And my model doesn't seem to fit exactly their class of models. For instance, the process I'm describing is "often" not a birth-death process (if $\exists k\in\mathbb Z_{\ge2}\enspace\mu_k\ne0$ then what I'm describing is not a birth-death process). And I don't see how to generalize their results to this case.
|
2025-03-21T14:48:31.322094
| 2020-06-22T14:28:40 |
363828
|
{
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|
Stack Exchange
|
Finding non-zero elements with $x^*x\leq\frac{1}{n}$
For a given unital Banach *-algebra $A$, let us put $A_+=\{\sum_1^n x^*_ix_i : x_i\in A, n\in \mathbb{N}\}$. We write $x\geq0$ if $x\in A_+$.
What types of (non semi-simple) unital Banach *-algebras contain some non-zero elements $x$ such that
$$ (\forall n\in \mathbb{N})~~~ nx^*x\leq1$$
There are many Banach -algebras with infinitesimal elements. Perhaps, the simplest is the 3-dimensional unital commutative algebra ${\bf C}[X]/\langle X^3\rangle$, where $X^=X$. The hermitian element $X^2$ is infinitesimal as $0 \le (\alpha-(2\alpha)^{-1}X^2)^2 = \alpha^2-X^2$ for every $\alpha>0$.
Sorry, I see, now we have the edit, I understand the correct order of the quantifiers. (This was my failure to read, btw...)
@ Narutaka OZAWA It is an amusing example. Thanks a lot for your nice answer.
|
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