added
stringdate 2025-03-12 15:57:16
2025-03-21 13:32:23
| created
timestamp[us]date 2008-09-06 22:17:14
2024-12-31 23:58:17
| id
stringlengths 1
7
| metadata
dict | source
stringclasses 1
value | text
stringlengths 59
10.4M
|
|---|---|---|---|---|---|
2025-03-21T14:48:31.105890
| 2020-05-28T17:04:55 |
361586
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jake B.",
"Noah Snyder",
"William Ballinger",
"https://mathoverflow.net/users/101335",
"https://mathoverflow.net/users/22",
"https://mathoverflow.net/users/33041"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629577",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361586"
}
|
Stack Exchange
|
Unusual skein relation in HOMFLY polynomial
If I take the HOMFLY(PT) polynomial defined by
$$l \,P(L_+) + l^{-1}\,P(L_-) + m\,P(L_0) = 0,$$
I have looked at expressions of the form
(knots that are the same except inside a small disk, where they look like the pictures indicated).
In every case the result had a factor $(l^4 + 2l^2+1-l^2m^2)$.
My question is: why does the expression always have a factor $(l^4 + 2l^2+1-l^2m^2)$?
I understand that this happens when one of the links are disjoint due to the HOMFLY relation relating the disjoint sum (split union) and the connected sum: $$P(L_1 \sqcup L_2)=-\frac{l+l^{-1}}{m} P(L_1 \# L_2),$$
since if you stick this in, you get exactly the factor $(l^4 + 2l^2+1-l^2m^2)$.
Does the relation
perhaps hold in general? According to the proof for the connected sum formula, it shouldn't.
I haven't thought through the details, but HOMFLY describes quantum "GL_t" where t is allowed to vary continuously. When t happens to be an integer this has a quotient to the category of representations of quantum GL_n. The phenomenon you're observing here seems to be related to the quotient to GL_1. That is, for GL_1 the defining representation is invertible and so you expect to get the kind of relation you're looking at.
First, doing a Reidemeister II move in the first diagram in your expression and then subtracting a multiple of the skein relation converts it to the change in the HOMFLY polynomial through a crossing change $P(L_+) - P(L_-)$, so your question is equivalent to asking why this difference is always divisible by $l^4 + 2l^2 + 1 - l^2 m^2$. Alternatively, since any two links with the same number of components differ by some sequence of crossing changes, we want $$P(L) \cong P(L') \mod (l^4 + 2l^2 + 1 - l^2m^2)$$ whenever $L$ and $L'$ have the same number of components.
Now, there are two values of the variable $m$ at which the HOMFLY polynomial is much simpler. If we substitute $m = l + l^{-1}$, then the quantity $(-1)^{c(L) - 1}$ satisfies the skein relation ($c(L)$ is the number of components), and if we substiture $m = -l-l^{-1}$ then the quantity $1$ satisfies the skein relation. Therefore, the residue of $P(L)$ modulo either $m-l-l^{-1}$ or $m+l+l^{-1}$ depends only on $c(L)$, so the same is true modulo $$\operatorname{gcd}(m-l-l^{-1},m+l+l^{-1}) = m^2 - (l+l^{-1})^2$$
At the decategorified level all these spectral sequences should go away and you should just have honest quotients after specializing the variables.
Which conjecture(s) are you refering to? 1.5? Does my statement hold or is it still a conjecture?
It's mostly conjecture 3.1 that's needed for the argument as written, but Noah Snyder's comment correctly points out that there's a much simpler unconditional argument. I'll edit shortly with more details.
|
2025-03-21T14:48:31.106101
| 2020-05-28T17:07:10 |
361587
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ilya Bogdanov",
"Tanya Vladi",
"https://mathoverflow.net/users/158421",
"https://mathoverflow.net/users/17581"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629578",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361587"
}
|
Stack Exchange
|
What is the smallest dimension that allows finding $n$ points at distances $|x_i-x_j|^{\delta/2}$, where $0<\delta<1$, and $x \in \mathbb{R}^n$?
Let $x_1,\cdots,x_n \in \mathbb{R}$, are there $\xi_1,\cdots,\xi_n \in \mathbb{R}^s$, such that
$|x_i-x_j|^{\delta}=||\xi_i-\xi_j||^2$, $0<\delta<1$, what is the smallest $s$ to guarantee the existence of $\xi_i$s for any $x_i\in\mathbb{R}$?
I’m afraid that $s=n$, if it exists. You determine the Gram matrix of the $\xi_i$. If you are sure it will be positive semi-definite, then in most cases it will be positive definite, which yields linear independence of the $\xi_i$.
@IlyaBogdanov I am sure it exists as $|x_i-x_j|^\delta$ is conditionally positive definite. Why is $s=O(n)$? I was thinking it should be $O(n^2)$
If the resulting matrix is positive definite, then you may define the scalar product by that matrix. The same, with some work, holds for the degenerate case. So, if the $\xi_i$ exist, then they exist in $\mathbb R^n$.
|
2025-03-21T14:48:31.106198
| 2020-05-28T17:10:21 |
361588
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris Gerig",
"Gorapada Bera",
"https://mathoverflow.net/users/117723",
"https://mathoverflow.net/users/12310"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629579",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361588"
}
|
Stack Exchange
|
Spectral flow of Dirac operator twisted by instanton
Suppose $E$ is a $SU(2)$-bundle over a closed three manifold $M$ and $S$ is the spinor bundle over $M$. Also assume $D_{A(t)}:\Gamma(S\otimes_{\mathbb c} E)\to \Gamma(S\otimes_{\mathbb c} E)$ is a family of Dirac operators twisted by a family of connections $A(t)$ on $E$, $t\in[0,1]$.
We have also a family of curl-div operators $L_{A(t)}:\Omega^0(E)\oplus\Omega^1(E)\to \Omega^0(E)\oplus\Omega^1(E)$ where
\begin{equation}
L_{A(t)}=
\begin{pmatrix}
0&d^*_{A(t)}\\
d_{A(t)}&*d_{A(t)}\\
\end{pmatrix}
\end{equation}
My question is:
Suppose $A(0)$ and $A(1)$ are two flat connections and $\mathbb A:=\{A(t):t\in[0,1]\}$ is an instanton on $M\times[0,1]$.
Are there any relations between Spectral flow of $\{D_{A(t)}:t\in[0,1]\}$ and Spectral flow of $\{L_{A(t)}:t\in[0,1]\}$?
Note: From the work of Atiyah-Patodi-Singer we could relate them with expressions involving eta invariants and some characteristic classes. Are they getting simplified in our present situations? If the question above was not expressed properly, that's my fault. Please let me know, I would try to modify them.
Just for the record: If $A(0)$ is gauge-equivalent to $A(1)$, i.e. $\mathbb{A}$ is a closed loop in the configuration space modulo gauge, then $SF(L_t)=2SF(D_t)$ and this didn't need the path to be an instanton nor the endpoints to be flat.
Thank you @Chris for pointing out that. Is not Atiyah-Singer index theorem over $M\times S^1$ in that case imply $SF(L_t)=-8SF(D_t)$?
Yup, I hastily messed up the factor (because I'm coincidentally thinking about these operators twisted by other $U(2)$-bundles in my work).
|
2025-03-21T14:48:31.106327
| 2020-05-28T17:16:30 |
361590
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Joseph O'Rourke",
"Nandakumar R",
"https://mathoverflow.net/users/142600",
"https://mathoverflow.net/users/6094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629580",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361590"
}
|
Stack Exchange
|
On some optimal containers of a set of points on the 2D plane
Given a set of N points in general position on the plane, the problem is to give efficient algorithms to find
the smallest semicircular region (semidisk) that contains the points
the smallest circular segment that contains the points (2 variants - 'smallest' could mean either of 'least area' or 'least perimeter').
the smallest sector that contains the points (again, 2 variants)
I am not aware of previous works on these questions.
Note: An O(N^2) algorithm for question 1 has been proposed in https://arxiv.org/abs/2005.10245. No proof of optimality was given, so there could be faster algorithms.
Some thoughts on question 2 are also given there but no algorithm.
Higher dimensional analogs of these questions - hemisphere, spherical segment etc..- arise naturally.
Note: The question of finding the largest semidisk/sector/circular segment contained in a given convex region C too seems unexplored. It is not hard to see that for a given C, the largest semidisk it contains should necessarily have both end points of the diameter on the boundary of C. Based on this property, a basic algorithm (complexity estimate: O(N^4)) has been proposed at https://nandacumar.blogspot.com/2020/07/largest-semidisk-inside-convex-polygon.html
Just to add slightly to your arXiv Lemma 2 concerning how the points hull $P$ touches the minimum semicircle diameter: If $P$ only touches the semicircle arc at exactly one point $x$, then $x$ must be the midpoint (i.e., highest) point of the arc. Otherwise $P$ touches the arc in at least two points.
Yes; that is a property of the hull and the semidisk that had better be stated explicitly - although the subsequent arguments assume it implicitly. Even the 'observation' regarding the 'vertex at the largest minimum distance from an edge' needs tightening - in general, for an edge E on the hull, there appear to be 2 vertices (not 1) which are both at the largest minimum distance from E. But the rotating calipers type of argument still seems to hold. Thank you!
In this reference we have an algorithm to determine the smallest circle containing a convex polygon. Follows a python code which uses this algorithm to find the smallest semi-circle container. The focused example is the same referenced in the OP in the cited paper. I hope the script is self explained. I am a early python programmer...
import math
import numpy as np
from numpy import linalg as LA
import matplotlib.pyplot as plt
from shapely.geometry import Polygon
data0 = [[2.30, 0.15],[0.63, 0.41],[0.37, 0.59],[0.79, 1.47],[2.32, 1.87],[3.6107, 0.72],[2.73, 0.14]]
def sub(p1,p2):
return list(map(lambda i, j: i-j,p1,p2))
def add(p1,p2):
return list(map(lambda i, j: i+j,p1,p2))
def cline(p,v,u):
v = [element * u for element in v]
return list(map(lambda i, j: i+j,p,v))
def max_secant(data):
n = len(data)
dmax = 0
for i in range(n):
for j in range(i):
d = LA.norm(sub(data[i],data[j]))
if d > dmax:
dmax = d
i0 = i
j0 = j
return (i0, j0)
def verify(data, feasible):
internal = True
error = 0.005
for i in range(len(data)):
dif = LA.norm(sub(data[i],feasible[0]))-feasible[1]
if dif > error:
internal = False
return internal
def polar_form(triangle):
(x1,y1) = triangle[0]
(x2,y2) = triangle[1]
(x3,y3) = triangle[2]
M = np.array([[2*(x2-x1),2*(y2-y1)],[2*(x2-x3),2*(y2-y3)]])
b = np.array([-(x1**2-x2**2+y1**2-y2**2),-(x3**2-x2**2+y3**2-y2**2)])
(x0, y0) = list(np.linalg.solve(M,b))
r = LA.norm([x1-x0,y1-y0])
return [[x0,y0], r]
def collect_triangles(data, i0, j0):
triangs = []
for i in range(len(data)):
if i not in [i0, j0]:
triangs.append([data[i0],data[i],data[j0]])
return triangs
def rotate(data):
data0 = []
n = len(data)
dummy = data[0]
for i in range(n-1):
data0.append(data[i+1])
data0.append(dummy)
return data0
def take_extremals(data):
breaks = []
sant = 1
v = sub(data[1],data[0])
n = len(data)
for i in range(1,n-2):
s = np.sign(np.dot(v,sub(data[i+1],data[i])))
if (sant != s):
breaks.append(i)
sant = s
if len(breaks) == 1:
breaks.append(n-1)
return breaks
def mirror(data, p, v):
reflected = []
vn = LA.norm(v)
n = len(data)
v = [v[0]/vn,v[1]/vn]
for i in range(n):
v0 = np.dot(sub(data[i],p),v)
v1 = [v[0]*v0,v[1]*v0]
v2 = add(p, v1)
v2 = [2*v2[0],2*v2[1]]
pr = sub(v2, data[i])
reflected.append(pr)
return reflected
def glue(data1, data2):
sdata = []
n1 = len(data1)
for i in range(n1):
sdata.append(data1[i])
n2 = len(data2)
for i in range(n2):
sdata.append(data2[n2-i-1])
return sdata
def select(data, k1, k2):
datas = []
for i in range(k1, k2+1):
datas.append(data[i])
return datas
def best_circle(data1):
(k1, k2) = take_extremals(data1)
p0b = data1[0]
vb = sub(data1[1],data1[0])
datas = select(data1, k1, k2)
datam = mirror(datas,p0b,vb)
dataf = glue(datam, datas)
(i0, j0) = max_secant(dataf)
v = sub(dataf[i0],dataf[j0])
r = 0.5*LA.norm(v)
p1 = add(dataf[i0],dataf[j0])
p0 = [element*0.5 for element in p1]
triangles = collect_triangles(dataf,i0,j0)
polar = []
polar.append([p0,r])
for i in range(len(triangles)):
polar.append(polar_form(triangles[i]))
feasible = []
for i in range(len(polar)):
if verify(dataf,polar[i]):
feasible.append(polar[i])
bestr = math.inf
for i in range(len(feasible)):
[pc, r] = feasible[i]
if r < bestr:
bestr = r
bestcirc = feasible[i]
return(bestcirc, p0b, vb)
########################
#### main program ####
########################
data1 = data0
circmin = math.inf
for i in range(len(data0)):
(circ, p0x, vx) = best_circle(data1)
if circ[1] < circmin:
circmin = circ[1]
bestcirc = circ
p0b = p0x
vb = vx
data1 = rotate(data1)
print(bestcirc)
#############################
#### plotting the result ####
#############################
(figure, axes) = plt.subplots()
(cx,cy) = bestcirc[0]
r = bestcirc[1]
poly = Polygon(data0)
(x, y) = poly.exterior.xy
xmin = cx - 1.1*r
xmax = cx + 1.1*r
ymin = cy - 1.1*r
ymax = cy + 1.1*r
axes.set_xlim((xmin,xmax))
axes.set_ylim((ymin,ymax))
uncolored_circle = plt.Circle( (cx,cy), r, fill = False)
axes.set_aspect( 1 )
axes.add_artist( uncolored_circle )
plt.plot(x,y)
v12 = vb
nv12 = LA.norm(v12)
v12 = [v12[0]/nv12,v12[1]/nv12]
s1x = cx - v12[0]*r
s1y = cy - v12[1]*r
s2x = cx + v12[0]*r
s2y = cy + v12[1]*r
plt.plot([s1x,s2x],[s1y,s2y])
plt.title( 'Result' )
plt.show()
|
2025-03-21T14:48:31.106679
| 2020-05-28T18:27:34 |
361594
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629581",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361594"
}
|
Stack Exchange
|
Generalized Brunnian links
A Brunnian link of order $n$ is nontrivial link of $n$ rings
that becomes a trivial link of $n-1$ rings if any ring is
removed. They were classified up to link-homotopy by
Milnor in 1954. This suggests the following generalization. Let
$\mathcal{A}$ be an antichain of subsets of $\{1,2,\dots,n\}$. In
other words, $\mathcal{A}$ is a collection of subsets of
$\{1,2,\dots,n\}$ such that if $A,B\in\mathcal{A}$ and $A\subseteq B$,
then $A=B$. An $\mathcal{A}$-link is a link of $n$ rings
$C_1,\dots,C_n$ such that $\{C_{i_1},\dots,C_{i_j}\}$ is a minimal set
of rings whose removal separates all the remaining rings if and
only if $\{i_1,\dots,i_j\} \in\mathcal{A}$. What can be said about
$\mathcal{A}$-links? For what $\mathcal{A}$ do they exist? For
instance, they obviously don't exist if $\mathcal{A}$ consists of the
single set $\{1,2,\dots,n\}$. Can they be classified analogously to
Milnor's result? There is a special case
here.
|
2025-03-21T14:48:31.106780
| 2020-05-28T18:46:44 |
361595
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anton Petrunin",
"Joseph O'Rourke",
"Robert Bryant",
"https://mathoverflow.net/users/13972",
"https://mathoverflow.net/users/1441",
"https://mathoverflow.net/users/6094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629582",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361595"
}
|
Stack Exchange
|
Generalized figures of constant width
Is it known which plane figures $Q$ can rotate touching three given circles $A$, $B$, and $C$?
This question was asked by Lazar Lyusternik in 1946, there is only one reference to this paper that solves the problem in one limit case.
Do you know of any other research on this question?
(I learned this question from Sergei Tabachnikov.)
May I ask: What is the one limit case that solves the problem?
@JosephO'Rourke If A and B are infinitesimally close points and C is a convex region, then in most of the cases Q is a circle. http://mi.mathnet.ru/umn3279
@AntonPetrunin: I suppose you mean "... and $Q$ is a convex region,".
There is another well-known limiting case: When the circles have infinite radii, i.e., when they are lines and form the sides of a triangle. There, one is looking for a (convex) region $Q$ that can be rotated freely (and then translated) so that it remains tangent to all three sides. It is well-known that, unless the angles of the triangle are rational multiples of $\pi$, the only solutions are the 'inscribed' circles, i.e., the circles tangent to all three lines. However, when all the angles are rational multiples of $\pi$, there is an infinite dimensional family of non-congruent solutions.
@RobertBryant Actually, this question was asked by Lyusternik as well; I suppose "Geometric Applications of Fourier..." by Groemer is the right reference.
I have looked at Goldberg's paper referenced by J. J. Castro in his excellent answer. It turns out that there is a simpler (and more general way) to generate Goldberg's non-circular solutions, so I thought that I would just mention that.
The idea is to consider the 'rotor' (Goldberg's term) as fixed and as the envelope of a circle in periodic motion. It's a bit easier if you use complex notation, i.e., think of $\mathbb{R}^2$ as $\mathbb{C}$. Consider a circle of radius $r>0$ and center $z_0=1$, parametrized by $z(s) = 1 + r\mathrm{e}^{is}$, and move it by a circle of rigid motions: $R_t(z) = \mathrm{e}^{it}z + a(t)$, where $a$ is periodic of period $2\pi/n$ for some integer $n\ge 3$. Now take the two envelopes of this $1$-parameter family of circles $R_t\bigl(z(s)\bigr)$. Because of the periodicity of $a$, these will also be envelopes of the circles of radius $r$ centered at the other $n$-th roots of unity, and hence these envelopes will have the property that, when they are moved by the inverse of $R_t$, they will remain tangent to the $n$ circles of radius $r$ centered on the $n$-th roots of unity.
A simple computation shows that the two envelopes of this family are given by
$$
E_\pm(t) = \mathrm{e}^{it}+a(t) \pm \frac{r\bigl(\mathrm{e}^{it}-ia'(t)\bigr)}{\bigl|\mathrm{e}^{it}-ia'(t)\bigr|}.
$$
(Note that this is well-defined as long as $\mathrm{e}^{it}-ia'(t)$ never vanishes, which can be ensured, for example, by choosing $a$ so that $|a'(t)|<1$.)
Setting $a=0$ (or a constant) gives the trivial circle solution. Below, I give an example of $E_-$ (the interior envelope) with $n=3$, $r=1/2$, and $a(t) = \tfrac17\sin(3t)$, or, more precisely, $R_{-s}(E_-)$ as $s$ varies between $0$ and $2\pi$. (Replacing this by $a(t) = \tfrac16\sin(3t)$ yields a nonconvex solution. In general, when $a$ is allowed to be large, the envelopes will have cusps.)
Right --- in some cases we have nontrivial rotors, the question when it happens. Most likely we have only circle for generic choice of $A$, $B$ and $C$. (Actually Lusternik states it as it is known to him --- he says "the answer depends on the choice".)
Lyusternik´s problem was proposed also in the book by Yaglom and Boltyanski "Convex Figures". There is another case which was solved by Goldberg, when the circles (3 or more circles) are of the same radii and its centers are in the vertices of a regular polygon. https://onlinelibrary.wiley.com/doi/abs/10.1002/sapm195837169
|
2025-03-21T14:48:31.107047
| 2020-05-28T18:54:10 |
361597
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Cameron Zwarich",
"https://mathoverflow.net/users/152648",
"https://mathoverflow.net/users/99234",
"kaka Hae"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629583",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361597"
}
|
Stack Exchange
|
If $X$ is separable space then $X^∗$ is separable in all topologies $\tau$ such that $(X^∗,\tau)^∗ =X$?
Let $(X,\|.\|_{X})$ be a separable Banach space and the associated dual space is denoted
by $X^*$. By $w^*$ we shall indicate the weak$-*$ topology on $X^*$.
Let $B_{X^∗}= \{x^∗ \in X^∗ : \|x^∗\|_{X^∗}\leq 1\}$. Since $X$ is separable, the set $B_{X^∗}$ furnished with the relative $w^∗-$topology is compact (by the Alaoglu theorem) and metrizable (see Theorem I.5.85). Note that
$$
X^*=\bigcup_{n}{nB_{X^*}}
$$
hence $X^∗_{w^∗}$ (the space $X^∗$ furnished with the $w^∗-$topology) is separable.
Can we say that : $X^∗$ is separable in all topologies $\tau$ such that $(X^∗,\tau)^∗ =X$?
Yes. Separability of a locally convex space is equivalent to the existence of a dense countable-dimensional subspace. All compatible topologies for a dual pair (i.e. those that give the same continuous linear functionals) agree on the closures of convex sets.
can you give me a book containing this result? please
I had to search for one, but according to Google Books (I don't own the book), it appears in section 2.5 of Barrelled Locally Convex Spaces by Carreras and Bonet.
Thank you very much, @Cameron, for your efforts.
|
2025-03-21T14:48:31.107150
| 2020-05-28T18:58:35 |
361598
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Johann Cigler",
"Pietro Majer",
"https://mathoverflow.net/users/5585",
"https://mathoverflow.net/users/6101"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629584",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361598"
}
|
Stack Exchange
|
A recursion which defines polynomials with integer coefficients?
Let $[n]=1+q+\dots+q^{n-1}$ and $u(n)=\prod_{j=1}^n \gcd([j],[n])$.
Define
$$r(n)=\sum_{d|n,d>1}{(-1)^d \frac{u(n)}{du(\frac{n}{d})^d}r\Big(\frac{n}{d}\Big)^d}+\frac{(1-q)^{n-1}u(n)}{n[n]}$$ with $r(1)=1,$ where $d$ runs over all divisors of $n$ which are greater than $1$.
For $q=1$ the numbers $r(n)$ are integers because in this case $\frac{u(n)}{du(\frac{n}{d})^d}$ is an integer and the right-hand side vanishes for $n>1.$
Computations suggest that in the general case $r(n)$ is a polynomial in $q$ with integer coefficients.
The first elements are $1, 1, -q, 1+q^2+q^3, -q(1-q+q^2), q^2(1+3q+2q^2+2q^3+2q^4+2q^5+q^6), -q(1-q+q^2)^2, \dots.$
For a prime number $p$ we get $r(p)=\sum_{j=1}^{\frac{p-1}{2}}\frac{1}{p}\binom{p}{j}(-q)^j [p-2j]\in \mathbb{Z[q]}$.
Any idea how to prove that $r(n)$ is a polynomial in $q$ with integer coefficients for all $n?$
Is the $\operatorname{gcd}$ in the definition of $u(n)$ a gcd of integer numbers, or polynomials in $q$?
@Pietro Majer: For arbitrary q it is the polynomial gcd.
|
2025-03-21T14:48:31.107240
| 2020-05-28T19:45:20 |
361600
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629585",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361600"
}
|
Stack Exchange
|
A bounded extension operator
Let $\Omega\subset\mathbb{R}^n$ be a bounded domain with smooth boundary $\partial\Omega$. Consider the harmonic extension operator $E :L^2(\partial \Omega) \rightarrow H^{1/2}(\Omega)$ which assigns to a prescribed boundary value $g$ a function $f$ with $f\rvert_{\partial\Omega}=g$ and $\Delta_\Omega f=0$.
I have two questions about this operator:
Can $E$ be bounded from $L^2(\partial \Omega)$ into $L^2(\Omega)$ as a right inverse of the trace operator? (or possibly another modification of $E$ or another extension operator).
Is there any explicit characterization of the range of $E$: $\mathrm{ran}(E):=E\,L^2(\partial \Omega)$?
Finally, any reference on some properties of such operator would be helpful.
|
2025-03-21T14:48:31.107316
| 2020-05-28T20:34:29 |
361604
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/69642",
"user69642"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629586",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361604"
}
|
Stack Exchange
|
Reference request: The transform of a bounded random variable has a zero in the complex plane
Together with coauthors I'm working on a paper where we use the following Proposition:
If a real-valued random variable $X$ has bounded support, then except in the trivial case that $X$ has all its mass in a single point, its moment generating function $$ M(z) = E(e^{zX})$$ has a zero in the complex plane.
Notice that the result is the same whether we are talking about the moment generating function, the characteristic function, the Laplace transform or the Fourier transform. Since the moments of $X$ grow at most exponentially, they are all entire functions and just rotations of each other.
It seems to us that the proposition must be both well-known and important, and we are baffled that we haven't been able to find it stated explicitly with a simple and self-contained proof.
Is there a statement and simple proof of the proposition in the literature?
The proposition is a consequence of the Hadamard factorization theorem: Since $M(z)$ is of order (at most) 1, it can be written as $e^{az+b}$ times a product involving its zeros. If there aren't any, we are left with $M(z) = e^{az}$, and $X$ must have all its mass at $a$.
But it can be proved with much easier complex analysis (see below), and this is why we're asking.
There is a lemma in William Feller's book An Introduction to Probability Theory and Its Applications vol II p 525, that implies our proposition and that seems to have been distilled out of Harald Cramér's proof of the decomposition theorem conjectured by Paul Lévy. It states that if $\exp(c\cdot X^2)$ has finite expectation for some $c>0$ (a weaker condition than boundedness), then either $X$ is normal or its characteristic function has a zero.
But both Cramér's original paper Über eine Eigenschaft der normalen Verteilungsfunktion and Feller's book simply refer to the Hadamard factorization theorem. Feller even says about the lemma that "Unfortunately its proof depends on analytic function theory and is therefore not quite in line with our treatment...".
There is a very reasonable (and useful to us) proof of the same lemma in the recent paper Three remarkable properties of the Normal distribution by Eric Benhamou, Beatrice Guez and Nicolas Paris, so we're certainly not complaining, just wondering if something even simpler has been published.
To establish our Proposition, some easy parts of the proof of the Hadamard theorem will do: If $M(z)$ has no zeros, we can introduce the function $$ K(z) = \int_0^z \frac{M'(t)}{M(t)}\,dt,$$
and we have $M(z) = e^{K(z)}$ throughout the complex plane (actually $K$ is known as the cumulant generating function). Assuming without loss of generality that $X$ is supported on $[-1,1]$, we obtain
$$ \text{Re}(K(z)) \leq \left|z\right|, $$ which leads by the Borel-Carathéodory theorem to
$$ \left| K(z) \right| \leq 4\left| z \right|.$$
This makes $K(z)/z$ a bounded entire function, and Liouville's theorem finishes the proof.
Spelling out the Borel-Carathéodory argument with a couple of more equations will reduce the whole thing to undergraduate complex analysis, and this is what we are thinking of doing.
Regarding this type of results, I would advise you to look in the book "Characteristic Functions" by Eugene Lukacs in which there is a thorough treatment of the interplay between these probability transforms and complex analysis.
It is the content of Theorem 7.2.3 page 202 of Eugene Lukacs book "Characteristic Function".
|
2025-03-21T14:48:31.107551
| 2020-05-28T20:39:08 |
361605
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Nawaf Bou-Rabee",
"https://mathoverflow.net/users/64449",
"https://mathoverflow.net/users/68463",
"sharpe"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629587",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361605"
}
|
Stack Exchange
|
Do Lyapunov functions imply exponential integrability of hitting times?
I have a question of some integrability of hitting times.
Let $X=(\{X_t\}_{t \ge0},\{P_x\}_{x \in E})$ be a diffusion process on a locally compact separable metric space $E$.
We assume that there exist Borel measurable functions $f\colon E \to [1,\infty)$ and $g \colon E \to \mathbb{R}$ such that $\left\{f(X_t)-f(x)-\int_{0}^{t}g(X_s)\,ds\right\}$ is a local martingale in $P_x$ for any $x \in E$. By convention, we will write $\mathcal{L}f$ for $g$. We further assume that $f$ is bounded, and that there exist $\alpha,\beta \in (0,\infty)$ and compact subset $K \subset E$ such that
\begin{align*}
\mathcal{L}f(x) \le -\alpha f(x) +\beta\textbf{1}_{K}(x),\quad x \in E.
\end{align*}
Under the conditions stated above, we can prove that
\begin{align*}
\sup_{x \in E}E_x[\sigma_K]<\infty,
\end{align*}
where $\sigma_K$ is the first hitting time of $K$. Indeed, fix $x \in E \setminus K$ and a localizing sequence $\{\tau_l\}_{l \ge 1}$ . Then, for any $l \ge 1$,
$\left\{ f(X_{t\wedge \sigma_K \wedge \tau_l})-f(x)-\int_{0}^{t\wedge \sigma_K \wedge \tau_l}\mathcal{L}f(X_s)\,ds \right\}_{t \ge 0}$ is a $P_x$-martingale. Therefore, we obtain that for any $t>0$ and $l \ge 1$,
\begin{align*}
-E_{x} \left[\int_{0}^{t \wedge \tau_l \wedge \sigma_K} \mathcal{L}f(X_s)\,ds\right] &= f(x)-E_{x}[f(X_{t \wedge \tau_l \wedge \sigma_K})] \le f(x).
\end{align*}
Because $\mathcal{L}f \le -\alpha f$ on $E \setminus K$, it follows that
\begin{align*}
E_{x}[t \wedge \sigma_K \wedge \tau_l] \le f(x)/\alpha.
\end{align*}
Because $f$ is bounded, Fatou's lemma shows that $\sup_{x \in E \setminus K}E_x[\sigma_K]<\infty$. It clearly holds that $\sup_{x \in K}E_x[\sigma_K]=0.$
My question
We can take increasing compact subsets $\{K_n\}_{n=1}^{\infty}$ of $E$ such that $E=\bigcup_{n=1}^{\infty}K_n$. In this situation, I would want to expect that
\begin{align*}
(1)\quad \lim_{n \to \infty}\sup_{x \in E}E_{x}[\sigma_{K_n}]=0.
\end{align*}
Then, there exists $N \in \mathbb{N}$ such that for any $n \ge N$, $\sup_{x \in E}E_x[e^{\sigma_{K_n}}]<\infty$.
Can we prove $(1)$? If not, please tell me a counterexample.
Do the ${ K_n }$ in (1) have any relation to $K$ in the infinitesimal drift condition? Also, can you expand a bit more on the display before (1) that shows that $E_x( \sigma_K ) \le f(x) / \alpha$?
@NawafBou-Rabee Thank you for your comment. There exists $N \in \mathbb{N}$ such that $K \subset K_N$. Therefore, we can prove that $\mathcal{L}f \le -\alpha f$ on $E \setminus K_n$ for any $n \ge N$ and $\sup_{n \in \mathbb{N}}\sup_{x \in E}E_{x}[\sigma_{K_n}]<\infty$.
Thanks; an expectation seems to be missing in the LHS of the inequality with the upper bound $f(x)$. How is the limit in (1) related to integrability of the exponential moment of $\sigma_{K_n}$?
@NawafBou-Rabee The exponential moment of $\sigma_{K_n}$ is proved by the Khasminskii's lemma. This allows us to conclude that $\sup_{x \in E}E_{x}[e^{\sigma_L}]<\infty$ if $\sup_{x \in E}E_{x}[\sigma_L]<1$, where $L$ is a compact subset of $E$.
It may depend on what exactly you mean by a diffusion on a general metric space.
Here is a counterexample for a discontinuous process. Take $E = \mathbb{R}$, and let $(X_t)$ be the process started at $x \in \mathbb{R}$ and jumping to the origin after a unit exponential time $\tau$. In other words, $(X_t)$ jumps to the origin at rate $1$, and then stays there forever. Take $f(x) = I \{x \ne 0 \}$. Then $g(x) = -f$. The conditions on $f$ are satisfied with $K = \{ 0\}$.
In this situation (1) fails because $\sup_{x \in E}E_{x}[\sigma_{K_n}] = \sup_{x \in E}E_{x}[\sigma_{K}] = E_{x} \tau = 1$.
EDIT. Here is a set-up that should give a counterexample with a continuous process. Take $E = [0, \infty) \times [0,1]$, and consider the system
$$
dX _t = h(X_t, Y_t), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$$
$$
dY _t = g(X_t, Y_t), \ \ \ \ \ \ \ (X_0, Y_0) \in E
$$
with $h$ and $g$ satisfying the following conditions:
$h([0, \infty) \times (0, 1]) = \{0\}$.
$g([0, \infty) \times (0, 1] ) = \{-1\}$
$h(x, 0) = - \varphi (x)$, where $\varphi > 0$ is a function such that the ODE $ z' = \varphi (z), \ z(0) = 0$ escapes to infinity in a finite time (for example, $\varphi (x) = 1 + x ^2$) .
$g(x, 0 ) = 0$.
For $(x,y) \in E$, take now $f(x,y) = (y + t_x) \vee 1$, where $t_x$ is the time when the solution to the ode
$$z' = \varphi (z), \ z(0) = 0$$
reaches $x$. It holds that $t_x\leq t_{expl}$, where $t_{expl}$ is the explosion time for $z$. Hence $f$ is bounded. Furthermore, $f(X_t, Y_t) = (f(X_0, Y_0) - t) \vee 1$, hence $f$ satisfy the inequality
\begin{align*}
\mathcal{L}f \le -\alpha f +\beta\textbf{1}_{K},
\end{align*}
with $\alpha = \frac{1}{\|f \|_{\infty}}$ and $K = \{ (0,0)\}$.
However, (1) is not satisfied because any compact $\mathcal{K} \subset E$ is bounded, and for $x \in (0,\infty)$ such that $\{x\} \times [0,1] \cap \mathcal{K} = \varnothing$ we have $E_{(x,1)}[\sigma_{\mathcal{K}}] \geq 1$ because $\sigma_{\mathcal{K}} \geq 1$ under $P_{(x,1)}$.
Remark.
The example can be modified without much trouble to make $h$ and $g$
continuous. I have a feeling that (1) might actually be true if $E = \mathbb{R}$.
Thank you for your reply. This is a counterexample for a discontinuous process.
Thank you for your very kind reply.
|
2025-03-21T14:48:31.107963
| 2020-05-28T20:44:14 |
361608
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"Michael Greinecker",
"Wojowu",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/35357"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629588",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361608"
}
|
Stack Exchange
|
Is there difference in notion of measurability in classical versus constructive?
Are there notions of measure and examples of sets measurable in that measure in classical logic but not in constructive logic (I think there cannot be counterexamples in other direction)? Are there differences in notion of measurability in each system?
I don't know much about constructive mathematics, but I don't think this question even makes sense. If we have some two models of set theory (for some definition of "model"), one governed by classical logic, one by constructive logic, we can't really compare sets contained in one to sets contained in the other. This same issue comes up in in purely classical setting too though - it doesn't make sense to compare sets between models of ZF and ZFC (say).
@Wojowu There are things one can prove classically but constructively. This does not require a constructivist to accept propositions that are classically wrong (such as all functions being continuous). Then the things one can prove constructively are a subset of what one can prove classically, and the measurability of some set can well be such a thing.
I think that @Wojowu's point was not about the proveability of a certain fixed proposition, but about the fact that even the sets of which we're trying to assert measureability might be different in the two cases—that is, that the very notion of "subset of $\mathbb R$" might differ depending on the logic. (I don't know if this is true when we change logic, but it's certainly true when we change axioms, as the fact that there are models of ZF with all subsets of $\mathbb R$ measureable shows—they're different sets, not just different statements about the same sets!)
|
2025-03-21T14:48:31.108116
| 2020-05-28T21:42:17 |
361610
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bonnevie",
"Zach Teitler",
"https://mathoverflow.net/users/158771",
"https://mathoverflow.net/users/535019",
"https://mathoverflow.net/users/88133",
"jujumumu"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629589",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361610"
}
|
Stack Exchange
|
Characterization of all-orthogonal tensors
In the paper [1], it is proven in Theorem 2 that any $n$-tensor $\mathcal{A}\in\mathbb{R}^{d_1\times...\times d_n}$ can be decomposed as
$$
\mathcal{A}=\mathcal{S} \times_1 U_1 ...\times_n U_n
$$
where $\times_n$ is the mode-$n$ product defined as $\mathcal{T}\times M=\sum_{i_k}\mathcal{T}_{i_1...i_k...i_n}M_{i_kj}$ and $U_n$ are appropriately shaped matrices. More interesting is the condition imposed on $\mathcal{S}$ by the theorem to ensure uniqueness of the decomposition, which the authors label all-orthogonality and which prescribes that
$$
\sum_{i_1,...,i_{k-1},\\i_{k+1},...,i_n}\mathcal{S}_{i_1,...,i_k,...,i_n}\mathcal{S}^{i_1,...,j_k,...,i_n}=\delta_{i_k}^{j_k}a_{k, i_k} \quad \forall k
$$
Where $\delta$ is the kronecker delta and $a_k$ is a set of weights particular to each $k$, so that the right-hand side is a diagonal matrix.
My question is a bit broad, as I am looking for any possible characterization of the family of all-orthogonal tensors. I would largely be happy with results for $3$-tensors, especially if it's easier to say something concrete. A parameterization or further decomposition into simpler matrix/tensor components would be the ideal, but I'd also appreciate some ideas on the space/manifold they inhabit and whether they possess any invariances. Results in $\mathbb{C}$ are also welcome.
I have been trying to attack the problem from different angles, but without much luck. By QR decomposition, it would seem that the QR of an arbitrary matricization $\mathcal{A}_{(k)}\in \mathbb{R}^{D\times d_k}$ separating one mode from the rest (where $D=\prod_{j\neq k}d_j$),
$$
\mathcal{A}_{(k)}=QR
$$
has $R$ being diagonal, since the all-orthogonality condition corresponds to orthogonality of the matricizations. So all-orthogonality means that every matricization of the type above splits into an orthogonal and a diagonal matrix.
Some observations of varying utility:
the all-orthogonality property plays a parallel role to the diagonality property in matrix SVD, which is also what makes me assume there is some underlying simplicity to the condition.
Empirically, there does exist tensors where $R$ in the QR above is the identity for all choices of $k$, i.e. tensors where all matricizations are orthogonal.
The Levi-Civita antisymmetric tensor is all-orthogonal after appropriate scaling, along with any transformations of it where each mode is independently transformed by orthogonal matrices.
[1] "L. De Lathauwer, B. De Moor, and J. Vandewalle, “A Multilinear Singular Value Decomposition,” SIAM J. Matrix Anal. Appl., vol. 21, no. 4, pp. 1253–1278, Jan. 2000."
There is a paper Orthogonal and unitary tensor decomposition from an algebraic perspective by Boralevi, et al. They study a class of tensors they call "odeco" for "orthogonally decomposable". It might not be the same as your "all-orthogonal", but still perhaps something there might be relevant.
@Bonnevie did you ever figure this out?
@jujumumu unfortunately not. Still seems like an interesting problem, I revisit it occasionally, but haven't had the chance to progress much on it.
If you are into graphical tensor network notation, the all-orthogonality condition of an n-tensor implies that if you take two copies of the tensor, pick any n-1 coordinates and connect the matched legs, then the result is a diagonal matrix.
@Bonnevie Have you had any success in characterizing the space of 2x2x2 all orthogonal tensors? I've tried but to no avail. Or found any properties of the dimension of these spaces?
There is another paper by L. De Lathauwer et al. which might be interesting for you: On the Largest Multilinear Singular Values of Higher-Order Tensors
In this paper, they describe when certain all-orthogonal tensors exist, depending on the collection of weights you mentioned. In some cases, this even allows for an explicit construction.
Within the paper A Geometric Description of Feasible Singular Values in the Tensor Train Format, there are further references to papers that have dealt with similar problems - all depending on statements of existence depending on whats similar to above mentioned weights (usually referred to as singular values), though only sometimes explicit constructions. In particular, due to the relation to the quantum marginal problem, there is surprisingly much literature on this.
Given that the literature on even the principle existence of such objects is elaborate, a further decomposition of all-orthogonal tensors might be too much to ask in general, with the exception of some special cases. These cases are a bit similar to decomposition of large matrices into block matrices, but only on a very rough level.
|
2025-03-21T14:48:31.108430
| 2020-05-28T22:00:48 |
361612
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dieter Kadelka",
"Gabe K",
"Gin Pat",
"Iosif Pinelis",
"Pierre PC",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/125275",
"https://mathoverflow.net/users/129074",
"https://mathoverflow.net/users/158772",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629590",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361612"
}
|
Stack Exchange
|
Scalar product of random unit vectors
Let $X,X'$ be two random vectors on the sphere $S^{d-1}$. What is the distribution of their dot product $X\cdot X'$ in the following cases:
$X,X'$ independent with uniform distribution on the sphere $S^{d-1}$
$X\in S^{d-1}$ deterministic, $X'$ uniformly distributed on $S^{d-1}$
?
By conditioning with respect to $X$, the solution to both questions is the same. Now the probability that the dot product belongs to $[-1,x]$ is the ratio of the volume of a certain spherical cap to the volume of the whole sphere, which is a simple integral computation. I got a density $(1-x^2)^{(d-1)/2}\mathrm dx$ up to constant, but I might be wrong. There are formulas on Wikipedia.
In additon we may assume that $X \equiv (1,0,\ldots,0)$.
@DieterKadelka Absolutely, and that's what I actually did. Apologies, a plus sign turned to a negative one; I found $\sqrt{1-x}^{d-3}\mathrm dx$ up to a constant. If $\mathbb S^{d-1}$ is parametrised as $(\sqrt{1-h^2}\cdot\theta,h)$ for $\theta\in\mathbb S^{d-2}$ and $h\in(-1,1)$, then the volume form is $(1-h^2)^{d/2-1}\mathrm d\theta\cdot(1-h^2)^{-1/2}\mathrm dh$.
I think this is worth an answer.
As noted in the comments, by the spherical symmetry, the distribution of the dot product in both parts of your question is the same that of $X\cdot(1,0,\dots,0)$. Moreover, the distribution of $X$ is the same as that of the random vector
$$\frac{Z}{\sqrt{Z_1^2+\dots+Z_d^2}},$$
where $Z=(Z_1,\dots,Z_d)$ is a standard normal random vector.
So, the distribution of the dot product in question is the same that of
$$R:=\frac{Z_1}{\sqrt{Z_1^2+\dots+Z_d^2}}.$$
The distribution of $R$ is obviously symmetric, and the distribution of $R^2$ is the beta distribution with parameters $\frac12,\frac{d-1}2$. It follows that the probability density function (pdf) $f_R$ of $R$ is given by
$$f_R(r)=\frac{\Gamma \left(\frac{d}{2}\right)}{\sqrt{\pi }\, \Gamma
\left(\frac{d-1}{2}\right)}\,\left(1-r^2\right)^{\frac{d-3}{2}}\, 1\{|r|<1\},$$
and the dot product in question has the same pdf.
One beautiful fact here is that when $d=3$, this is a uniform distribution. Actually, this fact is in some sense due to Archimedes, who showed that the sphere and cylinder have the same surface area. More precisely, the height and polar angle are Darboux coordinates for both spaces.
@GabeK : Good point.
Thank you! And if I didn't make mistakes it should be possible to rewrite the distribution $f_R$ as $f(\alpha) = C(d) \sin^{d-2}(\alpha) ,1{0<\alpha<\pi}$ in order ti obtain the distribution of the angle between vectors $X,X'$
|
2025-03-21T14:48:31.108622
| 2020-05-28T23:26:44 |
361613
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/158780",
"ifitnessvn"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629591",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361613"
}
|
Stack Exchange
|
Bandwidth of two-dimensional grid graphs
Suppose that $G$ is a subgraph of the two-dimensional grid and there are $n$ vertices in $G$. What is the maximum possible graph bandwidth of $G$ as a function of $n$?
If $G$ is a square grid graph with dimensions $k$ and $l$ (i.e., the Cartesian product of two paths of length $k$ and $l$), it is known that $G$ has a bandwidth of $\min(k, l) \le \sqrt{kl} = \sqrt{n}$ [1]. Is there an $O(\sqrt{n})$ upper bound for grid graphs in general?
[1] Chvátalová, J., 1975. Optimal labelling of a product of two paths. Discrete Mathematics, 11(3), pp.249-253.
I have once seen the case above, and they (repliers) said that yes there is an O(n−−√). Sorry, I can not remember the source.
|
2025-03-21T14:48:31.108707
| 2020-05-29T00:43:41 |
361617
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lior Silberman",
"Will Sawin",
"https://mathoverflow.net/users/151472",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/327",
"tbg93dk"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629592",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361617"
}
|
Stack Exchange
|
Analytic properties of motivic L-functions twisted by Dirichlet characters
Let $M$ be a pure motive over $\mathbb{Q}$ and consider the (completed) $L$-function $\Lambda(M, s)$ attached to its $\ell$-adic realization. Let us assume that this $L$-function admits analytic continuation and a functional equation relating $\Lambda(M, s)$ and $\Lambda(M^\vee, 1-s)$ as conjectured. Let $\chi$ denote a quadratic Dirichlet character.
Can one deduce the analytic continuation and functional equation of the twisted $L$-function $\Lambda(M, \chi, s)$ from the properties of $L(M, s)$?
Thank you in advance for your help.
No, except for $GL_2$ motives of level 1 (Hecke's theorem) and maybe some other very special examples.
In $\mathrm{GL}_2$ there is Booker's Theorem https://annals.math.princeton.edu/2003/158-3/p11
Thank you both for your replies. Thanks for Brooker's paper, I was unaware of this result. @WillSawin I'm sorry for my lack of knowledge but which Hecke's theorem are you referring to? I guess another example would be an elliptic curve over $\mathbb{Q}$, since then both $M$ and its twist are modular, but it is perhaps wrong to say that the analytic properties of one is deduced from the other...
Searching "Hecke's converse theorem" should bring all the information (on that theorem) you need.
|
2025-03-21T14:48:31.108818
| 2020-05-29T01:07:39 |
361618
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629593",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361618"
}
|
Stack Exchange
|
Correlation of stopping times for integral of Brownian motion increment
Let $\mu(x):=\int_{\epsilon}^{x}\exp\{B_{s+\epsilon}-B_{s-\epsilon}\}ds$, where $(B_{s})_{s\geq 0}$ is a Brownian motion (starting at $B_{0}=0$) and epsilon is small $0<\epsilon\ll 1 $. Consider the stopping time
$$T_{a}:=\inf\{t\geq \epsilon: \mu(t)\geq a \},\text{ for $a\geq 0$.}$$
For "good" deterministic choices $a<b<c<d<$ and $t_{1},t_{2}\geq 0$ we are studying the correlation of
$$\{T_{b}-T_{a}\geq t_{1}\}\text{ and }\{T_{d}-T_{c}\geq t_{2}\}.$$
Q: One approach is by studying a mixing condition: trying to estimate the difference
$$| P[\{T_{d}-T_{c}\geq t_{2}, T_{b}-T_{a}\geq t_{1}\}]-P[\{T_{d}-T_{c}\geq t_{2}\} ]\cdot P[ \{T_{b}-T_{a}\geq t_{1}\}]|$$
when $[a,b]$ and $[c,d]$ are distant enough. In other words, we are trying to find whether there are any constraints on $a<b<c<d$ so that the above difference goes to zero as the distance $c-b$ between the two intervals increases.
If you think this question is ill-posed or too obvious for Mathoverflow, please put your hint in the comments and I will modify/delete the question.
Approach
The process $\mu(x)$ is not Markov because of the $B_{s-\epsilon}$ term and so strong Markov property (SMP) doesn't apply. But in the spirit of (SMP), we introduce the event
$$ E:=\{T_{c}> T_{b}+2\epsilon\}.$$
So just for the sake of understanding (even if we can't get a good control on $E^{c}$), lets focus on the above difference with the first term intersected with $E$:
$$| P[\{T_{d}-T_{c}\geq t_{2}, T_{b}-T_{a}\geq t_{1}\}\cap E]-P[\{T_{d}-T_{c}\geq t_{2}\} ]\cdot P[ \{T_{b}-T_{a}\geq t_{1}]|.$$
Equivalently, by conditioning we are trying to estimate
$$| P[\{T_{d}-T_{c}\geq t_{2}\}\cap E\mid \{T_{b}-T_{a}\geq t_{1}\}]-P[\{T_{d}-T_{c}\geq t_{2}\} ]|.$$
I will update as I find things.
|
2025-03-21T14:48:31.108941
| 2020-05-29T02:07:38 |
361619
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anubhav Mukherjee",
"KSackel",
"https://mathoverflow.net/users/149240",
"https://mathoverflow.net/users/33064",
"https://mathoverflow.net/users/66405",
"no_idea"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629594",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361619"
}
|
Stack Exchange
|
Convex surfaces with transverse boundary (contact geometry)
Suppose I have a compact surface $\Sigma$ in a contact 3-manifold, where the boundary $\partial\Sigma$ is transverse to the contact structure. Am I able to perturb $\Sigma$ rel boundary so that it is convex (in the contact geometry sense)?
If I can't do this rel boundary, what sort of things could I do to perturb $\Sigma$ so that it is convex? For example, I could perturb $\partial\Sigma$ so that it is Legendrian, and then I could pertub $\Sigma$ so that it is convex (assuming $tb(\partial\Sigma)<0$). Would any complications arise if I then perturb this boundary to be transverse after doing this?
I haven't found any explicit mention of these kinds of surfaces, and I am hoping there is just a quick answer to this.
Check etnyres note on convex surfaces. If it is convex surface then you need consider the behavior of the dividing curves.
I thought the statement was the surface $\Sigma$ with Legendrian boundary is convex iff it has dividing curves. So you could have a surface with transverse boundary this theorem doesn't say anything.
You can certainly fix the transverse boundary by modifying the typical proof: a $C^{\infty}$ generic perturbation rel boundary will make the characteristic foliation Morse-Smale, and having a Morse-Smale characteristic foliation implies convexity.
|
2025-03-21T14:48:31.109301
| 2020-05-29T02:25:22 |
361620
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"JCA",
"Nik Weaver",
"https://mathoverflow.net/users/153196",
"https://mathoverflow.net/users/158779",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/64444",
"mathbeginner",
"worldreporter"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629595",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361620"
}
|
Stack Exchange
|
Two isomorphic reduced group $C^*$-algebras
Suppose that $C^*_r(G)\cong C^*_r(H)$, can we conclude that $G\cong H$?
Take two non-isomorphic finite groups $G, H$ with isomorphic group algebras.
In general the conclusion does not hold, as pointed out by JCA. But some classes of groups have the property you are asking for. Such groups are usually called C*-superrigid.
The simplest counterexample is $G = \mathbb{Z}_2\times \mathbb{Z}_2$ and $H = \mathbb{Z}_4$. These groups are not isomorphic but $C^*(G) \cong \mathbb{C}^4 \cong C^*(H)$.
Pro Nik weaver, May I ask you a question: You proved that calkin algebra has outer automorphisms. Does there exist other open questions concerning $C^*$-algebras which have automorphisms?
@mathbeginner: nothing comes to mind.
|
2025-03-21T14:48:31.109395
| 2020-05-29T02:45:16 |
361622
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gautam",
"Kasper Andersen",
"R. van Dobben de Bruyn",
"Richard Stanley",
"https://mathoverflow.net/users/2807",
"https://mathoverflow.net/users/65801",
"https://mathoverflow.net/users/7400",
"https://mathoverflow.net/users/82179"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629596",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361622"
}
|
Stack Exchange
|
SOS polynomials with integer coefficients
A well known theorem of Polya and Szego says that every non-negative univariate polynomial $p(x)$ can be expressed as the sum of exactly two squares: $p(x) = (f(x))^2 + (g(x))^2$ for some $f, g$. Suppose $p$ has integer coefficients. In general, its is too much to hope that $f, g$ also have integer coefficients; consider, for example, $p(x) = x^2 + 5x + 10$. Are there simple conditions we can impose on $p$ that guarantee that $f, g$ have integer coefficients?
Clearly a necessary condition is that $p(n)$ is a sum of two squares for every integer $n$. Could this condition be sufficient? Is there a nice characterization of polynomials with this property?
Not an answer, just a comment: The result that a nonnegative univariate polynomial is a sum of two squares is a direct consequence of the fundamental theorem of algebra. The result by Polya and Szegö is more subtle, it deals with polynomials on $[0,\infty[$.
@KasperAndersen Thanks for the correction. The result I found states that if $p(x) \geq 0$ for all $x \geq 0$ then there exist $f, g, h, k$ so that $p(x) = f(x)^2 + g(x)^2 + x(h(x)^2 + k(x)^2)$. Is that the result you were referring to? If so, I'd be interested in understanding when $f, g, h, k$ have integer coefficients as well.
There is the following result of Davenport, Lewis, and Schinzel [DLS64, Cor to Thm 2]:
Theorem. Let $p \in \mathbf Z[x]$. Then the following are equivalent:
$p$ is a sum of two squares in $\mathbf Z[x]$;
$p(n)$ is a sum of two squares in $\mathbf Z$ for all $n \in \mathbf Z$;
Every arithmetic progression contains an $n$ such that $p(n)$ is a sum of two squares in $\mathbf Z$.
Criterion 3 is really weak! For example, it shows that in 2, we may replace $\mathbf Z$ by $\mathbf N$. Because it's short but takes some time to extract from [DLS64], here is their proof, simplified to this special case.
Proof. Implications 1 $\Rightarrow$ 2 $\Rightarrow$ 3 are obvious. For 3 $\Rightarrow$ 1, factor $p$ as
$$p = c \cdot p_1^{e_1} \cdots p_r^{e_r}$$
with $p_j \in \mathbf Z[x]$ pairwise coprime primitive irreducible and $c \in \mathbf Q$. We only need to treat the odd $e_j$ (and the constant $c$). Let $P = p_1 \cdots p_r$ be the radical of $p/c$, and choose $d \in \mathbf N$ such that $P$ is separable modulo every prime $q \not\mid d$. Suppose $P$ has a root modulo $q > 2d\operatorname{height}(c)$; say
$$P(n) \equiv 0 \pmod q$$
for some $n$. Then $P'(n) \not\equiv 0 \pmod q$, hence $P(n+q) \not\equiv P(n) \pmod{q^2}$. Replacing $n$ by $n+q$ if necessary, we see that $v_q(P(n)) = 1$; i.e. there is a $j$ such that
$$v_q\big(p_i(n)\big) = \begin{cases}1, & i = j, \\ 0, & i \neq j.\end{cases}.$$
If $e_j$ is odd, then so is $v_q(p(n))$, which equals $v_q(p(n'))$ for all $n' \equiv n \pmod{q^2}$. By assumption 3 we can choose $n' \equiv n \pmod{q^2}$ such that $p(n')$ is a sum of squares, so we conclude that $q \equiv 1 \pmod 4$. If $L = \mathbf Q[x]/(p_j)$, then we conclude that all primes $q > 2d\operatorname{height}(c)$ that have a factor $\mathfrak q \subseteq \mathcal O_L$ with $e(\mathfrak q) = f(\mathfrak q) = 1$ (i.e. $p_j$ has a root modulo $q$) are $1$ mod $4$. By Bauer's theorem (see e.g. [Neu99, Prop. VII.13.9]), this forces $\mathbf Q(i) \subseteq L$.
Thus we can write $i = f(\theta_j)$ for some $f \in \mathbf Q[x]$, where $\theta_j$ is a root of $p_j$. Then $p_j$ divides
$$N_{\mathbf Q(i)[x]/\mathbf Q[x]}\big(f(x)-i\big) = \big(f(x)-i\big)\big(f(x)+i\big),$$
since $p_j$ is irreducible and $\theta_j$ is a zero of both. Since $f(x)-i$ and $f(x)+i$ are coprime and $p_j$ is irreducible, there is a factor $g \in \mathbf Q(i)[x]$ of $f(x)+i$ such that
$$p_j = u \cdot N_{\mathbf Q(i)[x]/\mathbf Q[x]}(g) = u \cdot g \cdot \bar g$$
for some $u \in \mathbf Q[x]^\times = \mathbf Q^\times$. Applying this to all $p_j$ for which $e_j$ is odd, we get
$$p = a \cdot N_{\mathbf Q(i)[x]/\mathbf Q[x]}(h)$$
for some $h \in \mathbf Q(i)[x]$ and some $a \in \mathbf Q^\times$. By assumption 3, this forces $a$ to be a norm as well, so we may assume $a = 1$. Write $h = \alpha H$ for $\alpha \in \mathbf Q(i)$ and $H \in \mathbf Z[i][x]$ primitive. Then
$$p(x) = |\alpha|^2 H \bar H,$$
so Gauss's lemma gives $|\alpha|^2 \in \mathbf Z$. Since $|\alpha|^2$ is a sum of rational squares, it is a sum of integer squares; say $|\alpha|^2 = |\beta|^2$ for somce $\beta \in \mathbf Z[i]$. Finally, setting
$$F + iG = \beta H,$$
we get $p = F^2 + G^2$ with $F, G \in \mathbf Z[x]$. $\square$
Footnote: I am certainly surprised by this, given that the version for four squares is clearly false. Indeed, the condition just reads $p(n) \geq 0$ for all $n \in \mathbf Z$. But the OP's example cannot be written as any finite sum of squares in $\mathbf Z[x]$, because exactly one of the terms can have positive degree. (However, it might be different in $\mathbf Q[x]$.)
References.
[DLS64] H. Davenport, D. J. Lewis, and A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964). ZBL0126.27801.
[Neu99] J. Neukirch, Algebraic number theory. Grundlehren der Mathematischen Wissenschaften 322 (1999). ZBL0956.11021.
Thanks for this thoughtful answer. Do you have any thoughts about the situation in Q[x]?
Can't you just clear denominators? Like, multiply with a large square to assume $p \in \mathbf Z[x]$, and apply the criterion of DLS?
|
2025-03-21T14:48:31.109742
| 2020-05-29T03:08:28 |
361625
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629597",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361625"
}
|
Stack Exchange
|
faithful modules over a finite dimensional commutative algebra
Let $A$ be a commutative algebra over a field $k$ which is finite dimensional as a vector space over $k$. Let $M$ be a faithful $A$-module. Does it follow that $dim_k(M)\geq dim_k(A)$?
$A$ is a product of local Artinian rings, so the question is local. The best result I know is in this paper of Gulliksen, who proved that if the socle dimension is at most $3$, then the length of any faithful module is at least the length of the ring. So the answer is yes if $A$ is the product of local Artin rings of socle dimension at most $3$. He also gave counter example when the socle dimension is bigger than $3$.
|
2025-03-21T14:48:31.109940
| 2020-05-29T03:44:37 |
361628
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jochen Wengenroth",
"https://mathoverflow.net/users/152648",
"https://mathoverflow.net/users/21051",
"kaka Hae"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629598",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361628"
}
|
Stack Exchange
|
Can we say that $f$ admits a $m(X,X^*)$-continuous extension to $X$?
Let $X$ be a Banach space equipped with the Mackey topology $m(X,X^*)$. We suppose that $\big(X,m(X,X^*)\big)$ is separable space. Let $H$ be a countable, $m(X,X^*)$-dense subset with $(H=-H)$.
Let $f$ be a sublinear function $m(X,X^*)$-continuous on the subset $H$ of $X$ that goes to $\mathbb{R}$.
Can we say that $f$ admits a $m(X,X^*)$-continuous extension to $X$?
Banach spaces are Mackey, i.e., the Mackey topology is the norm topology.
Then, $f$ admits a continuous extension to $X$?
|
2025-03-21T14:48:31.110005
| 2020-05-29T04:30:46 |
361631
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A beginner mathmatician",
"Andreas Blass",
"Asaf",
"https://mathoverflow.net/users/136860",
"https://mathoverflow.net/users/6794",
"https://mathoverflow.net/users/8857"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629599",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361631"
}
|
Stack Exchange
|
Definition of generic point
I am trying to read a paper named D.S. Ornstein, B. Weiss, Subsequence ergodic theorems for amenable groups, Israel J. Math. 79 (1) (1992) 113–127, doi:10.1007/BF02764805. In this paper the authors have defined generic points.
Let $(X,\mathcal B,\mu)$ be an ergodic finite measure preserving dynamical system. Then a set $B\in\mathcal B$ and a point $x_0\in X$ will be called generic if
$$
\lim\limits_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}1_B(T^ix_0)=\mu(B)
$$
and the same holds for all sets $B^\prime$ in the algebra generated by $B$ namely:
$$
\bigcup_N\bigvee_{i=-N}^NT^i\{B,B^c\}.
$$
I do not understand the meaning of the set $\cup_N\vee_{I=-N}^NT^i\{B,B^c\}$. Can someone help me out?
After defining the generic points the author say "As usual, we denote the common refinement of partitions (or algebras) by $V$." What did they mean by that?
The symbol that you've transcribed as $\lor$ and later as $V$ is actually $\bigvee$ and is being defined as "the common refinement". So $\bigvee_{i=-N}^NT^i{B,B^c}$ means the (coarsest) common refinement of the $2N+1$ partitions $T^i{B,B^c}$, where $i$ ranges from $-N$ to $N$.
@Andreas. Thank you for your answer. I want to ask you if there is some very natural way by which one can describe $B^\prime$? For example if $x_0$ is generic then clearly pointwise ergodic theorem holds true for the particular function $1_B.$ Is there a way by which the function space corresponding to $B^\prime$ can be described?
Take $\chi_{B}$ to be the characteristic function of $B$, so $1-\chi_{B}$ is the characteristic function of the complement. Now by taking products and using complements, one may write all the sets in that algebra by using those functions.
Anyways, that's far from being a research question and should be in mathstackexchange.
|
2025-03-21T14:48:31.110160
| 2020-05-29T04:54:02 |
361633
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629600",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361633"
}
|
Stack Exchange
|
Determine whether a rational function on the codomain of a surjective morphism is regular
Let $X$ be a smooth affine algebraic variety with a (not necessarily free) action by an algebraic torus $T$. Let $Y$ be the quotient stack $X/T$ and let $p:X\rightarrow Y$ be the quotient map. Suppose $f$ is a rational function on $Y$ and its pull-back $p^*f$ is a regular function on $X$. Does it follow that $f$ is also a regular function on $Y$?
Assuming that by a rational function you mean a section of $\mathcal O_Y$ defined on a nonempty open substack, the answer is yes, as long as $X$ is integral (more generally if $X$ is only reduced, we need some density assumption on the open substack).
Indeed, let $V \subseteq Y$ be an open substack with inverse image $U \subseteq X$, let $f \in \mathcal O_Y(V)$, and assume $\rho^* f \in \mathcal O_X(U)$ extends to $g \in \mathcal O_X(X)$. The quotient stack $[X/T]$ is given by the groupoid
$$\begin{array}{ccc} & & T \times X & & \\ & \swarrow & & \searrow & \\ X & & & & X,\! \end{array}$$
where the source and target maps are the projection $s = \pi$ and the action $t = a$; see Tag 0444 for details. In fact, the commutative diagram
$$\begin{array}{ccc}T \times X & \to & X \\ \downarrow & & \downarrow \\ X & \to & Y\end{array}$$
is a $2$-fibre product diagram; see Tag 04M9. Restricting to $V \subseteq Y$ gives
$$\begin{array}{ccc}T \times U & \to & U \\ \downarrow & & \downarrow \\ U & \to & V.\!\end{array}$$
We have a section $g \in \mathcal O_X(X)$, and the pullbacks along $a$ and $\pi$ to $T \times X$ agree on $T \times U$ since they come from $f \in \mathcal O_Y(V)$. Since $X$ is integral, so is $T \times X$, so two sections that agree on $T \times U$ are equal. This gives the cocycle condition to glue to a section of $\mathcal O_Y(Y)$ since $\mathcal O_Y$ is an fppf sheaf; see Tag 06TU (or use Tag 044U for maps to $\mathbf A^1$). $\square$
|
2025-03-21T14:48:31.110300
| 2020-05-29T05:39:56 |
361635
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aaron Bergman",
"Xiao-Gang Wen",
"annie marie cœur",
"https://mathoverflow.net/users/106497",
"https://mathoverflow.net/users/17787",
"https://mathoverflow.net/users/947"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629601",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361635"
}
|
Stack Exchange
|
Derive how the level quantization for 3d quantum Chern-Simons theory path integrals?
Let us consider abelian and non-abelian 3d quantum Chern-Simons theory path integrals:
abelian Chern-Simons theory on non-spin manifolds ---
$$
\int [DA]\exp(i \frac{k}{2\pi} \int_X (A \wedge dA ))
$$
abelian Chern-Simons theory on spin manifolds ---
$$
\int [DA]\exp(i \frac{k}{4\pi} \int_X (A \wedge dA ))
$$
non-abelian Chern-Simons theory ---
$$
\int [DA]\exp(i \frac{k}{4\pi} \int_X \mathrm{Tr}_{} (A \wedge dA + \frac{2}{3} A \wedge A \wedge A))
$$
where $A$ takes values in the Lie algebra valued $\mathcal{G}$ 1-form. So does the Tr take the matrix representations in the Lie algebra $\mathcal{G}$.
What are the correct and rigorous ways to argue the quantization of values of $k$?
I think there are three possible helpful ideas:
extend 3-manifolds $X$ to 4-manifolds $Y$?
large gauge transformation.
Use Wess Zumino Witten like terms.
Could any expert demonstrate these line by line?
None of the actions you write are well-defined. Once you pick a proper definitions, the action will be valued in $\mathbb{R}/\mathbb{Z}$. The quantization follows accordingly. You can find one definition in the appendix to Freed’s https://arxiv.org/abs/0808.2507.
Our recent paper https://arxiv.org/abs/2008.02613 provides an answer. It reveals that how quantization of chiral central charge and Hall conductance depend on the groundstate degeneracy on Riemannian surfaces. Note that groundstate degeneracy is not even defined for an arbitrary spacetime manifold.
Do Riemannian surfaces include all orientable 2d spatial manifolds already? What is the emphasis of the sentence: " groundstate degeneracy is not even defined for an arbitrary spacetime manifold?"
Riemannian surfaces do include all orientable 2d spatial manifolds. Groundstate degeneracy is defined for any Riemannian surfaces, but is not defined for an arbitrary spacetime which is a 3d manifold. In your question, you only have 3d spacetime manifold.
|
2025-03-21T14:48:31.110443
| 2020-05-29T06:28:20 |
361636
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A beginner mathmatician",
"Dabed",
"Lior Silberman",
"R W",
"YCor",
"https://mathoverflow.net/users/136860",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/142708",
"https://mathoverflow.net/users/327",
"https://mathoverflow.net/users/8588"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629602",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361636"
}
|
Stack Exchange
|
Ergodic action on unitary group
Let $G$ be a locally compact Hausdorff group. Assume that $\theta:G\to U_d$ is a group homomorphism where $U_d$ is a finite dimensional unitary group. Consider a action of $G$ on $U_d$ by $g.u:=\theta(g)u,$ $u\in U_d.$ Consider $U=\overline{\theta(G)}.$ Is it true that $G$ acts ergodically on $U$ where $U$ is equipped with the invariant probability measure of $U_d$?
Since $U_d$ is connected, every subgroup proper of it had infinite index, so unless $U=U_d$ it must be a set of measure zero there, and the literal answer is "no".
On the other hand, the action of $G$ on $U$ is ergodic for the Haar measure of $U$. Indeed, let $f\in L^1(U)$ be $G$-invariant (e.g. the characteristic function of a $G$-invariant set). Then by the continuity of the action of $U$ on $L^1(U)$, $f$ is $U$-invariant, hence constant a.e.
($U_d$ being connected, $U$ of finite index means $U=U_d$.)
Yes, you're right. I was thinking "compact Lie group".
Not clear what you mean by saying that the literal answer is "no". Look at the action of $\mathbb Z$ on the unit circle (aka $SO(2)$) determined by an irrational rotation.
In your case the closure of the image is all of $\mathrm{SO}(2)$.
@Daniel. It is better you should make some comment like dumb question" when you are sure not because you feel" something. I am a beginner what is dumb to you may be nontrivial to me..
@A beginner mathmatician Please forgive me I think my use of the English language caused some confusion I meant it was a dumb question on my part and that is why I used the verb feel, I was just trying to confirm my understanding of the problem with the answerer, I also try to learn here so we are in the same boat, my apologies to you I have deleted the comment and maybe I should rephrased it once again here as "Could you explain why this doesn't contradict the Weyl's ergodic theorem on $R/ Z\cong SO(2)$?
@Daniel your question is the same as the one by R W above, and the answer is the same: the action is only ergodic wrt the invariant probability measure of $U$. When $U=U_d$, this happens to be there invariant measure of $U_d$.
Consider the case of a rational circle rotation: the map $x\mapsto x+\alpha$ on $\mathbb{R}/\mathbb{Z}$. When $\alpha$ is irrational, it generates a dense subgroup and hence acts ergodically wrt the invariant measure on the circle. But when $\ alpha=p/q$ is rational, it generates the finite subgroup $U=\frac1q\mathbb{Z}/\mathbb{Z}$. This subgroup has measure zero in the circle, so the action on $U$ is not ergodic for Lebesgue measure on the circle. However, the action is ergodically for the discrete measure on the finite group $U$.
I think I get it now, I needed this concrete example to understand what you proved, this also helped me understand better the weyl's ergodic theorem and what I had read about the rotation number, they make a lot more of sense now.
|
2025-03-21T14:48:31.110672
| 2020-05-29T06:31:12 |
361637
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"Gerry Myerson",
"J.Mayol",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/7410",
"https://mathoverflow.net/users/94414"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629603",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361637"
}
|
Stack Exchange
|
Classification of the behaviours of the logistic map
On this this wikipedia page, it is claimed that the iterative sequence $x_{n+1}=rx_n(1-x_n)$ (the logistic map) starting at a point $[0,1]$ and where $r$ ranges in $[0,4]$ behaves differently according to $r$. For example:
For $r\in[0,1)$, $x_n \to 0$, for all $x_0$;
For $r\in [1,2)$, $x_n \to \frac{1-r}{r}$, for all $x_0$;
For $F<r<4$ (where $F$ is the Feigenbaum constant), almost all of the values of $r$ produce a chaotic behaviour.
As I tried for several hours to find a book where a full characterization (and proof!) of the different behviours of this iterative sequence, I did not find any proof, except for the easiest cases (namely $r \leq 2$, $r=4$ and somes cases with two or four limit points).
Can someone provide me with a reference where this iterative map is studied in its full extent?
Saber N. Elaydi's textbook, Discrete Chaos, goes into some detail about this map. Whether you'll consider it to be the "full extent", I don't know.
@GerryMyerson thank you. In fact I already encountred books where some "parts" of the proof are given, but never the difficult parts (intervals where chaos occurs, or when chaos disappear). It is really strange that there no theorem of the form "for $r \in I_1$ then [...] occurs, for $r \in I_2$ then [...] occurs, etc.".
It's possible that for some of what you want there are no theorems, just numerical evidence.
Try the book "Iterated Maps on the Interval as Dynamical Systems" by Collet and Eckmann. https://www.springer.com/gp/book/9780817649265
If you change the form of your map to $z_{n+1} = z_n^2 + c $ ( conjugation) and take only real values of c ( real slice of Mandelbrot set) then you will find the answers in papers by G. Pastor, M. Romera. Here is for example : Calculation of the Structure of a Shrub in the Mandelbrot Set
The part from 0 to Feigenbaum point in your map is a periodic region where period doubling cascade occurs. In c plane it is from 0.25 to F
The part from F to 4 in your map is a Mandelbrot set antenna. It's structure is described in that paper :
Look also for:
Sharkovskii's theorem
exponential map which transforms plane
"Many questions concerning (discrete) dynamical systems are of a number theoretic or combinatorial nature." Christian Krattenthaler
HTH
Lyubich wrote a nice short survey on this: "The quadratic family as a qualitatively solvable model of chaos", in the October 2000 issue of Notices of the AMS. At just 11 pages long, that survey is fairly terse but it does hit the most important points and cites some of the original papers where things are proved, so you can follow the references to find the proofs. As a general rule the proofs are fairly involved and require one to first build up quite a bit of machinery.
There is also a longer survey by Graczyk and Świa̧tek from around the same time that has more details and a more comprehensive list of references:
Graczyk, Jacek; Świątek, Grzegorz, Smooth unimodal maps in the 1990s. (Survey), Ergodic Theory Dyn. Syst. 19, No. 2, 263-287 (1999). ZBL0941.37024.
|
2025-03-21T14:48:31.110933
| 2020-05-29T07:52:24 |
361641
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Adrien",
"https://mathoverflow.net/users/13552",
"https://mathoverflow.net/users/40804",
"https://mathoverflow.net/users/41291",
"mme",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629604",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361641"
}
|
Stack Exchange
|
1d TQFT minus connection =?
Correct me if I am wrong but I believe at least conceptually (maybe even rigorously) data of a 1-dimensional TQFT and of a vector bundle with connection are equivalent.
Going into more detail (and consequently making more and more mistakes), a vector bundle with connection allows to assign to a point the fibre over that point, and to a path the monodromy (or holonomy? Yes I am that ignorant) along this path.
Three questions, but closely related:
Does there exist equally layman-ish description of going back from a TQFT to a vector bundle?
Clearly this requires the base to be at least a smooth manifold. Is there known any TQFT-like object (and maybe also connection-like object) that would work in the non-smooth context? Say, for topological manifolds, or even arbitrary finite CW-complexes?
What if one leaves manifolds alone but removes the connection? Is there known a version of TQFT that would work for vector bundles with arbitrarily severe restrictions (say, very-very nice algebraic vector bundles on very-very good algebraic varieties) but without any additional structure?
Two remarks:
Sort of minimal version of the question is whether a vector bundle $p:E\to[0,1]$ comes with any kind of map (relation? correspondence?) between $p^{-1}(0)$ and $p^{-1}(1)$. Vague association with motives comes to mind but that's all my mind offers.
Obviously I am tempted to ask the same about 2D-TQFT. But I am (almost) successfully resisting this temptation.
1d TQFT's are in 1-1 correspondence with finite dimensional vector spaces, and the image of the circle is the dimension of that vector space.
I think what you have in mind is instead the notion of $X$-structured TQFT, aka homotopy quantum field theory. Those are defined for pairs of a topological manifold together with a map into some fixed topological manifold $X$.
So in the 1d case, you are looking for symmetric monoidal functors into Vect from the category $Bord_1^X$ which has
objects points together with a map into $X$, so the set of objects is just $X$.
morphisms bordisms between those, i.e. maps between two points of $X$ are intervals equipped with a map into $X$ in a compatible way, i.e. at the end of the day this is just a path in $X$ between your points.
Now you want to look at this up to homotopy, so the bottom line is that $Bord_1^X$ really is just the fundamental groupoid of $X$, and a 1d $X$-HQFT is thus a flat vector bundle on $X$. So not only the connection is important, but it has to be flat.
Now I'm not entirely sure, but I think given a vector bundle on $X$ with a non-necessary flat connection $A$, you get in fact an example of a 2-dimensional HQFT, where very roughly the value on a 2-dimensional surface equipped with a map into $X$ is computed by integrating the pull-back to your surface of the curvature 2-form of $A$. This is basically saying every connection is automatically "2-flat" thanks to the Bianchi identity.
Many thanks for the correction, and the last part is especially interesting! So, what is a flat vector bundle over a non-smooth manifold? And, could you please indicate (say, give a reference for) how to reconstruct it from an $X$-HQFT?
A flat connection on a bundle is a lift of the structure group from $G$ to $G^\delta$, the group made discrete. (This is a topological notion, no smoothness needed.) This is equivalent to a representation $\pi_1(X) \to G$ which gives rise to a $G$-bundle isomorphic to your given bundle.
In general you can always talk about representations of the fundamental groupoid, which are the same as locally constant sheaves on $X$.
As for your last question, rephrasing what Mike said, giving an HQFT, i.e. a functor $\prod_1(X)\rightarrow Vect$, is the same as giving a vector space $V_x$ for each point $x \in X$, and an iso $V_x \cong V_y$ for every path between $x$ and $y$ which depends only on the homotopy class of that path in a way compatible with composition, so this induces a locally constant sheaf (aka locall system) on $X$ in a fairly tautological way (every choice of a contractible $U$ containing $x,y$ identifies $V_x$ and $V_y$ canonically).
Going into more detail (and consequently making more and more mistakes), a vector bundle with connection allows to assign to a point the fibre over that point, and to a path the monodromy (or holonomy? Yes I am that ignorant) along this path.
Yes, this is pretty much correct.
You can indeed assign holonomy to paths.
Furthermore, there is an equivalence between 1-dimensional smooth TFTs
over X and vector bundles with connections over X.
Precise definitions with proofs can be found in the paper
https://arxiv.org/abs/1501.00967.
Does there exist equally layman-ish description of going back from a TQFT to a vector bundle?
Yes.
The underlying vector bundle without a connection can be recovered
by evaluating the smooth TFTs at the smooth family of points given by the manifold itself.
The connection is recovered by differentiating the parallel transport map.
Clearly this requires the base to be at least a smooth manifold. Is there known any TQFT-like object (and maybe also connection-like object) that would work in the non-smooth context? Say, for topological manifolds, or even arbitrary finite CW-complexes?
Yes. Replace the site of smooth manifolds with the site of topological manifolds
or the site of finite CW-complexes.
What if one leaves manifolds alone but removes the connection? Is there known a version of TQFT that would work for vector bundles with arbitrarily severe restrictions (say, very-very nice algebraic vector bundles on very-very good algebraic varieties) but without any additional structure?
Yes, this is the (∞,1)-version of 1-dimensional TFTs.
(“Holonomy” is now not a strict functor, but an (∞,1)-functor,
which no longer produces a connection.)
See, for example, the survey by Lurie
and the more recent work by Chris Schommer-Pries.
|
2025-03-21T14:48:31.111291
| 2020-05-29T08:21:10 |
361642
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"François Brunault",
"https://mathoverflow.net/users/124323",
"https://mathoverflow.net/users/43263",
"https://mathoverflow.net/users/6506",
"tim",
"user43263"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629605",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361642"
}
|
Stack Exchange
|
Existence of solutions of polynomials systems (and their "rough" shape) over $\mathbb{R}$ & friends with positive-dimensional ideals
This is a follow-up (but self-contained) question to my previous one. There I asked about state-of-the-art methods to solve multivariate polynomials systems over non-algebraically closed fields in general.
I learned that the theory is more involved that I thought (I'm not working in algorithmic algebraic geometry, so I'm only familiar with the very basics, like Buchberger's algorithm, or the definition of the dimension of an ideal). Therefore it is necessary to ask a more specific questions that the previous, general one, which is more tailored to my needs.
My setup is the following:
Regarding complexity: I'm interested in solving a large number of polynomial systems (on commodity hardware), on the order of $10^4$. But each of the systems is of relatively small size - my baseline consists of least 6 different variable and 4 equations. If I could tackle this, I'd already be happy. Going further, I don't expect the systems to grow beyond about 20 different variables and 20 equations.
So perhaps I don't actually need the fastest possible algorithm and can make do with simpler, older ones - but I will let you be the judge of that.
Regarding the polynomials: There are no restrictions their coefficients, so, depending on the field I'm working in they can take any number.
Regarding the field: Regarding the field I'm working in, my baseline is $\mathbb{R}$, but I'd also be interested in $\mathbb{Q}$ and $\mathbb{Z}$. If there are methods that are much easier for one field than another, than I will the choice of the field to study be influence by the time I need to invest to learn that method, i.e. the easiest one wins.
Regardin the dimension of the ideal spanned by the polynomials: The ideal has dimension $2$ or $3$ over the complex numbers, in most cases I tested so far with the help of CAS.
What I'm looking for: I'm interested is learning about methods (I'm happy with specific references) that tell me
1) whether the system has a solution at all or not. Working over, e.g., $\mathbb{C}$, this would be easy (e.g. compute a Gröbner basis: If it contain the $1$, if and only if the solution variety is empty). But this doesn't work unfortunately for non-algebraically closed fields. Given the answers from my previous question, I'm inclined to think that answer this question shouldn't be too hard (perhaps even trivial for the expert computational geometer, which I'm not unfortunately).
2) if it has an infinite number of solutions (if the variety is zero-dimensional, things are easy of course), I would like to pick out one single variable, say $n_0\in \{1,\ldots,n\}$, project the solution variety $V(f_1,\ldots,f_s)\subseteq \mathbb{
R}^n$ (supposing we work over the field $\mathbb{R}$) along this variable onto $\mathbb{R}$ to investigate whether there exists an interval $[-\alpha,\alpha]$ around $0$ which is contained in this projected set (I don't need to understand the projected set fully). That is what I menat by "rough shape" in the title.
Are the coefficients exact or only floating-point numbers?
@FrançoisBrunault My investigation has two steps: First, experiments, where I necessarily use floating-point precision. In a second step, based on the experiments, I will want to prove some general statements, where I will allow any real-numbers as coefficients (even if these can't be simulated any more). I'm guessing there seems to be a dramatic difference in terms of the capability of the methods you have in mind between these two cases. While I, for now, would be interested mostly in the floating-point case to do the experiments, it would be good to know if there are [...]
some fundamental issues I might miss when doing experiments, which come from floating points. Thus I would ask you to focus on floating-point, but dt mention where I need to be careful, when passing to exact real coefficients (so that I will know where any pitfalls might lie when generalizing after I have finished experimenting; perhaps; if the exact coefficient case is more difficult I could adjust my investigation accordingly).
If the coefficients are floating point then your question is somewhat ill-posed, because polynomial equation solving is very sensitive to the input. Think of the equation $x^2=\varepsilon$ with $\varepsilon$ near 0, the answer to your question (2) changes completely. You can certainly approximate the coefficients by decimal numbers, and apply the quantifier elimination method I mentioned to get the projection as an explicit semi-algebraic set of $\mathbb{R}$. And it's possible to decide whether a semi-algebraic set is a neighbourhood of $0$. But you have no guarantee that the answer is correct
@FrançoisBrunault Well, by necessity I need to use floating point, since I don't know of a different way to carry out solving the mentioned $10^4$ systems, where the coefficients discretize a high-dimensional cube (though, afterwards, when I want to prove theorems, which is the final goal, of course I will work with coefficient from $\mathbb{R}$).
Regarding floating-point: I can't quite follow you: Why is it not possible to decide if the answer is correct? Would you say, from an algorithmic/simulation standpoint, that it would be better to work with $\mathbb{Z}[x_1,\ldots,x_n]$?
The problem is that, especially with complicated systems of poylnomial equations, there can be losses of precision and they can cumulate. The answer to (1) or (2) depends on finding some sign (think of a quadratic equation) and this is a big problem when your coefficients are not known exactly. So guaranteeing the output is not known (note that even for 1-variable polynomials, determining the roots to some guaranteed precision is a non-trivial task). Of course, you can experiment, but I just wanted to say that the conclusions will not be mathematically proven.
From your earlier post I was worried you had thousands of variables (: Six could very well be in range of CAD methods mentioned by @FrançoisBrunault (assuming you are ok with approximate coefficients, and there are other factors of course.) Have you looked at RAGlib?
@FrançoisBrunault From your last comment, I realized that I think you misunderstood me, resp. I misunderstood you (depending on the viewpoint). When I mentioned I was doing computations in floating-point, I referred to the fact that the input of my coefficients are only with floating point numbers (since that is what I can input in a CAS; and I'm happy with that, that is ok) - and not that my coefficients are other some numbers that I then approximate with floating-point precision before running any algo on them (which is what you had in mind). So in this sense my coefficients are exact! [...]
[...] Now that this is clarified, could you please let me know of some (ideally introductory) references regarding the quantifier elimination method that you mentioned, as well as decidability procedure to decide whether a semi-algebraic set is a neighbourhood of 0?
[...] I guess for (1) you are alluding to something like Khovanskii's fewnomials approach, about which I learned in the previous question, that seems to be a generalization of Descartes' rules of signs?
@tim Thanks a lot, that library would be perfect - if it were open-source. I will need eventually to publish some part of the code and using non open-source software is not so well received in my community. Do you happen to know if there are any open-source packages that offer similar functionality? If I were to use RAGlib, do you now of any publication or article that provides details about the algorithms RAGlib uses internally? I couldn't find much documentation on the website; if I download it, there are only a few files containing examples and explanations how to use the library.
Just expanding my comments to this question and the previous one:
I assume that your polynomials have rational coefficients (which seem to be the case, since you mention they are floating point numbers with fixed precision, in particular they are decimals), and that you are interested in the solutions in $\mathbb{R}^n$.
The assertion that the projection of $V(f_1,\ldots,f_s) \subset \mathbb{R}^n$ to the $x_n$ variable is a neighbourhood of 0 is a first order formula over the reals, namely
\begin{equation*}
\exists a>0, \forall x_n \in [-a, a], \exists x_1,\ldots,x_{n-1} \in \mathbb{R}, \forall i, f_i(x_1,\ldots,x_n) = 0.
\end{equation*}
It is a formula with no free variable, hence is decidable, and CAD softwares like Qepcad or Redlog will output "true" or "false".
Regarding feasibility, my worry is that the semi-algebraic set of $\mathbb{R}$ given by the projection to $x_n$ will probably involve polynomials with gigantic coefficients. You have to experiment to see whether the CAD software can still do it in reasonable time.
Regarding the theory, the heart of the algorithms is the cylindrical algebraic decoposition (CAD), and Alexandre Eremenko's answer to your previous question mentions good references. I know only the basics, but enjoyed reading the book by Bochnak, Coste and Roy. It's good to read them with a particular goal in mind and see how the corresponding algorithm works. You can also look at the documentations of the softwares I mentioned, which give a good idea of what problems they can solve.
Thanks, this really was very helpful!
@user43263 By the way, I realize that this also works (in theory) for polynomials whose coefficients are arbitrary real numbers (which are computable to arbitrary precision, like $\pi$, $e$ or whatever...). Namely, you introduce indeterminates for these coefficients. CAD will output a boolean formula involving (in)equalities between polynomials in these indeterminates (like "$b^2-4ac \geq 0$"), and you just need to evaluate these expressions to sufficient precision to decide whether the boolean formula is true or false.
|
2025-03-21T14:48:31.112052
| 2020-05-29T08:54:13 |
361643
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Display name",
"M. Winter",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/127521"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629606",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361643"
}
|
Stack Exchange
|
Tilings of lattice polytopes by transformations of lattice polytopes
A quasi-lattice polytope is a polytope obtained by reflections, translations, and rotations of lattice polytopes. In a tiling of a lattice polytope by quasi-lattice polytopes, are all quasi-lattice polytopes necessarily lattice polytopes?
If we allow dilations in the general sense to enter the "quasi club", then any figure may just be obtained from dilating a unit hypercube, so the result is obviously false. What if we restrict dilations to transformations induced by matrices with integer entries? This stronger question is true in 1 dimension since $[a,b]$ having integral endpoints implies $[ac,bc]$ has integral endpoints for all $c \in \mathbb{Z}.$
Let's be even more general. Fix a dimension $n,$ let $S$ be a set of affine transformations, and define $S$-lattice polytopes as those in the image of $s: P \to P$ for some $s \in S$ where $P$ is the set of $n$ dimensional lattice polytopes. How large can $S$ be such that the statement "in a tiling of lattice polytopes by $S$-lattice polytopes, all $S$-lattice polytopes are lattice polytopes" is true?
For $n=1, S = \{x \to ax+b | a \in T, b \in \mathbb{R}\}$ where $T = \mathbb{Z}$ works. In fact, $T$ can be replaced by any extension $R \supseteq T$ such that $R$ is linearly independent over $\mathbb{Z}$ (defining an infinite set to be linearly independent iff every finite subset is), and this characterizes all maximal $S$ completely.
You can make $S$ pretty large, e.g. you can let it contain all transformations $T$ so that $\smash{T(\Bbb Z^n)\cap \Bbb Z^n=\varnothing}$. Then no lattice polytope can be decomposed into $S$-lattice polytopes and the statement is trivially satisfied.
The answer to your first question is No.
Consider the 5x5 square as a lattice polytope in two different ways as seen above.
Each tiling of one of these sqares gives a tiling of the other.
But you can choose the tiles in such a way, so that they are lattice polytope in only one of the squares.
So the right square is tiled with quasi-lattice polytopes that are not lattice polytopes.
I considered a 3-4-5 triangle before, but didn't find this counterexample because I only thought about rotating subtiles instead of the entire tile. Your approach should generalize to exclude all rotations with angles $\theta$ such that $\tan \theta = b/a$ where $a, b, \sqrt{a^2+b^2} \in \mathbb{Z}.$ Furthermore, angles whose tangents are irrational will never map a lattice polygon to another lattice polygon, so any counterexamples involving angles not of the aforementioned form, if they exist, will be much harder to find.
@Displayname Irrational angles might be safe. Consider my comment under your post: if you compose $S$ of rotations by irrational angles, then no $S$-lattice polytope can have two vertices on $\Bbb Z^n$ and so no decomposition of a lattice polytope into such is possible.
It's true that a polytope created from a rotation by an angle with irrational tangent cannot have two lattice points. But it may have 1 lattice point. What prevents us from using a different $S$-polytope for every corner and then miraculously filling in the interior?
@Displayname The edges of a lattice polytope must have rational directions (that is, the tangent of the angle with the axes is rational), and so an $S$-lattice polytope (that is, after an irrational rotation) must have irrational edge directions only. But you cannot decompose a lattice polytope only into polytope with irrational edge directions, because some tile must share (a part of) an edge with the lattice polytope, whose edges have rational directions.
|
2025-03-21T14:48:31.112290
| 2020-05-29T08:57:06 |
361644
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629607",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361644"
}
|
Stack Exchange
|
Estimation of parameters through multivariate Taylor expansion?
I do have a function $$f(t) = \prod\limits_{j=1}^{n} \left(1 + \sum\limits_{i=1}^{n} M_{i,j} t_i\right)^{-\alpha_{j}}$$ defined by parameters:
$M_{i,j} \in \mathbb{R}_{+}, \;\forall i \in 1,...,d,\; \forall j \in 1,...,n$
$\alpha_j \in \mathbb{R}_{+}, \; \forall j \in 1,...,n$
over $\mathbb{C}^{d}$ (but i'm mainly interested in the $\mathbb{R}_{+}^{d}$ restriction).
Suppose that I pick a suitable $t^{*}$ (say, inside the radius of convergence), and that I have observations of values of the derivatives up to a certain number of derivations $N$ :
$$g_m = f^{(m)}(t^{*}), \; \forall m \in \mathbb{N}^d, \sum_{i=1}^{d}m_i \le N$$
Can I, through a Taylor expansion around $t^{*}$, estimate parameters $M,\alpha$ from the observations $g_m$? I guess that if I have a lot of them (say $N$ is very big), i could do it. On the over hand, I don't know precisely how much I do need compared to $n,d$. Furthermore, I do not know how to do this estimation.
Could you help me a little?
|
2025-03-21T14:48:31.112392
| 2020-05-29T09:16:25 |
361646
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Martin Rubey",
"Noam Zeilberger",
"Sam Hopkins",
"https://mathoverflow.net/users/1015",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/3032"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629608",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361646"
}
|
Stack Exchange
|
$\mathfrak{sl}_3$ webs without faces having a multiple of 4 sides
In settling the main conjecture of Cyclic action on Kreweras walks, see https://arxiv.org/abs/2005.14031, a rather interesting object popped up.
Recall from
Kuperberg, Greg, Spiders for
rank 2 Lie algebras,
Commun. Math. Phys. 180, No. 1, 109-151
(1996). ZBL0870.17005.
that an $\mathfrak{sl}_3$-web $W$ is a planar graph, embedded in a
disk, with boundary vertices labeled $1,2,\ldots,m$ arranged on
the rim of the disk in counterclockwise order, and any number of
(unlabeled) internal vertices such that
$W$ is trivalent: all the boundary vertices have degree one, while all the internal vertices have degree three;
$W$ is bipartite: the vertices (both boundary and internal) are colored white and black, with edges only between oppositely colored vertices.
An $\mathfrak{sl}_3$-web is non-elliptic or irreducible if all of
its internal faces have at least $6$ sides. Non-elliptic
$\mathfrak{sl}_3$-webs with all boundary vertices white are in
bijection with rectangular standard Young tableaux having $3$ rows:
Khovanov, Mikhail; Kuperberg, Greg, Web bases for $\text{sl}(3)$ are not dual canonical, Pac. J. Math. 188, No. 1, 129-153 (1999). ZBL0929.17012.
Tymoczko, Julianna, A simple bijection between standard $3\times n$ tableaux and irreducible webs for $\mathfrak{sl}_{3}$, J. Algebr. Comb. 35, No. 4, 611-632 (2012). ZBL1242.05277.
We call a non-elliptic $\mathfrak{sl}_3$-web Kreweras, if it has no faces with a multiple of $4$ sides.
One reason why these objects are interesting is the fact that
$$
\sum_W 2^{\kappa(W)}=\frac{4^n}{(n+1)(2n+1)}\binom{3n}{n},
$$
where the sum is over all Kreweras webs $W$ with $3n$ boundary vertices and $\kappa(W)$ connected components. This is the number of Kreweras words, see http://oeis.org/A006335.
The sequence of numbers of connected Kreweras webs, with $3,6,9,12,15,18,\dots$ boundary vertices, is $1,2,12,104,1088,12768,\dots$, unknown to the OEIS. However, Lagrange inversion shows that this equals $2^{n+1}\cdot\frac{(4n+1)!}{(3n+2)!(n+1)!}$. Note that this is $2^n$ times http://oeis.org/A000260.
The sequence of numbers of connected Kreweras webs up to rotation, with $3,6,9,12,15,18,\dots$ boundary vertices, is $1,1,2,10,76,714,\dots$, also unknown to the OEIS.
Question: Can the condition, that a Kreweras web only has faces whose number of sides is not divisible by $4$, be given a representation theoretic meaning?
is there a typo in the question: do you mean "... condition that a $\mathfrak{sl}_3$-web only has faces ... not divisible by 4 [and hence is a Kreweras web]" rather than "... condition that a Kreweras web ..." ?
(In any case, your preprint is interesting.)
Well, it's a language problem. But I mean precisely what you are saying.
@NoamZeilberger: we updated to include some explicit discussion of OEIS sequences you might be interested in- see what is now Section 6.2 of https://arxiv.org/abs/2005.14031.
|
2025-03-21T14:48:31.112596
| 2020-05-29T09:19:05 |
361648
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Deane Yang",
"Guillaume Aubrun",
"S. Dewar",
"erz",
"https://mathoverflow.net/users/122002",
"https://mathoverflow.net/users/53155",
"https://mathoverflow.net/users/613",
"https://mathoverflow.net/users/908"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629609",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361648"
}
|
Stack Exchange
|
Naming convention: looking for better terminology for "centrally symmetric smooth strictly convex bodies"
I have recently found myself researching a certain type of convex body in $\mathbb{R}^2$, namely centrally symmetric smooth strictly convex bodies.
Instead of repeating such a sentence repetitively I have found myself referring to them as "symmetric bodies", but I was hoping there is a more widely used name for them.
I have checked the literature but cannot for the life of me find anything.
Any help is appreciated.
perhaps you could just say "in this article CSSSCB stands for centrally symmetric smooth strictly convex bodies?"
"symmetric convex body" is widely used to mean "centrally symmetric convex body", so I would add an extra adjective here. Maybe "regular symmetric (convex) body" ? (I have seen "regular" to mean "smooth and strictly convex", but it is not standard terminology)
You could try giving them your own name. Call them D-bodies.
@erz I did originally go with this but I found it isn't very nice to read every time.
Centrally symmetric convex body which is absorbing is called "barrel" in the functional analysis. So I suggest you use this term (probably stating explicitly that the barrels you consider are smooth).
This will run into the same issue of me having to call them "strictly convex smooth barrels" which (while better) is still a bit of a mouthful.
|
2025-03-21T14:48:31.112723
| 2020-05-29T09:19:08 |
361649
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629610",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361649"
}
|
Stack Exchange
|
$V$-like actions of $V$
This continues my question about prefix-continuous bijections (since the answer was "yes").
Notation and conventions: Let $A$ be a finite alphabet and $L \subset A^*$ a language. Let $G$ be a group. For words $u, w$, juxtaposition $uw$ denotes word concatenation, and if we have an action $G \curvearrowright L$ and $g \in G, u \in L$ we write $gu$ for the image of $u$ in the action of $g$; concatenation associates first.
An action $G \curvearrowright L$ is veelike (or $G$ acts veelike) if for all $g \in G$ there exists $n \in \mathbb{N}$ such that for all $u \in A^*$ with $|u| = n$ there exists $u' \in A^*$ such that $guv = u'v$ for all $uv \in L$. Now we have the following simple observation.
Thompson's $V$ acts veelike on the regular language defined by the regular expression $\emptyset + (0+1)^* 1$.
Proof. This is the regular language $L$ containing the empty word $\emptyset$ and all words ending in the symbol $1$. To find the action, recall the defining action of $V$ on the boundary of the infinite binary tree. We can see the tree as $\{0,1\}^\omega$, and elements of $V$ can be presented by picking two maximal prefix codes, bijecting them and then rewriting the prefix of the input word according to said bijection. Observe that the set
$$ X = \{x \in \{0,1\}^{\omega} \;|\; \exists m: \forall i \geq m: x_i = 0\} $$
of infinite paths/words that are eventually zero is invariant under its action, and $V$ clearly acts faithfully on it (e.g. because it's a simple group and the action is nontrivial). There is an obvious bijection from $X$ to $L$: just cut out the sequence after the last $1$, if one exists, and map the all zero sequence to the empty word $\emptyset$. The action of $V$ on $L$ is "veelike" in the obvious sense (same formula); that's basically the definition. Conjugating the action of $V$ to $L$, it stays veelike since on all long enough words you do exactly the same rewrites as you would on infinite words beginning that way. Square.
(I have not written a more careful proof than that, it's natural so what could go wrong.)
My question is the following:
On which alphabets $A$ and languages $L \subset A^*$ does Thompson's $V$ admit a veelike action?
In particular:
Does Thompson's $V$ act veelike on the set of binary words $\{0,1\}^*$?
One possible way to prove this would be to find a bijection $\phi : X \to \{0,1\}^*$ such that conjugating the action of $V$ from $X$ to $\{0,1\}^*$ gives you a veelike action.
Does such a bijection exist?
I claim that the bijection of @PierrePC to my previous question does not work (the answer is correct but I did not state all the necessary properties for the bijection). Namely, after conjugation the action indeed rewrites only prefixes, but you need to see the whole word to know how they are rewritten, i.e. the action is not continuous in the right sense.
More concretely, through this bijection the element that just flips the first bit acts
$$<PHONE_NUMBER>0000001... \mapsto<PHONE_NUMBER>0000001... $$
$$<PHONE_NUMBER>0000001... \mapsto<PHONE_NUMBER>0000001... $$
which are conjugated respectively to
$$<PHONE_NUMBER>0000000 \mapsto<PHONE_NUMBER>000000 $$
$$<PHONE_NUMBER>000000 \mapsto<PHONE_NUMBER>000000 $$
which is highly veeunlikely because the length change depends on the $1$ arbitrarily far inside the word.
|
2025-03-21T14:48:31.112960
| 2020-05-29T09:39:29 |
361651
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629611",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361651"
}
|
Stack Exchange
|
Is the set $\operatorname{Unif}(0,\frac{1}{n})$ for odd and even $n$ a 2-alternating capacity?
Let $\Omega$ be a complete metrizable space $\mathscr A$ its Borel $\sigma$-algebra and $\mathscr M$ the set of all probability measures on $\Omega.$ Every non-empty subset $\mathscr P \subset \mathscr M$ defines an upper probability $$v(A)=\sup\{P(A)|P\in\mathscr{P}\},\quad A\in\mathscr A.$$
Then, a $2$-alternating capacity is defined with the following conditions:
$1.$ $v(\emptyset)=0$ and $v(\Omega)=1$
$2.$ $A\subset B\Longrightarrow v(A)\leq v(B)$
$3.$ $A_n\uparrow A\Longrightarrow v(A_n)\uparrow v(A)$
$4.$ If $\mathscr P$ is weakly compact, then we also have $F_n\downarrow F,\,\, F_n\, \text{closed}\Longrightarrow v(F_n)\downarrow v(F)$
$5.$ $v(A\cup B)+v(A\cap B)\leq v(A)+v(B)$
That is also to say that the set $$P_v=\{P\in\mathscr M|P(A)\leq v(A)\,\,\text{for all}\,\, A\in\mathscr A\}$$ is not larger than the closed convex hull of the set $\mathscr P$ determining $v$.
Example $1:$ Let $\Omega=\{1,2,3\}$, $P_0=\{\frac{1}{2},\frac{1}{2},0\}$, $P_1=\{\frac{4}{6},\frac{1}{6},\frac{1}{6}\}$ and let $v$ be the upper probability determined by $\mathscr P=\{P_0,P_1\}$. Then, $$P_v=\Bigg\{\frac{3+t}{6},\frac{3-t-s}{6},\frac{s}{6}\Bigg|0\leq s,t\leq 1\Bigg\}$$ whereas the convex closure of $P$ is the proper subset of $P_v$ determined by $s=t$. In this example, $v$ is $2$-alternating.
Example $2:$ Let $\Omega=\{1,2,3,4\}$, $P_0=\{\frac{5}{10},\frac{2}{10},\frac{2}{10},\frac{1}{10}\}$, $P_1=\{\frac{6}{10},\frac{1}{10},\frac{1}{10},\frac{2}{10}\}$ and let $\mathscr P=\{P_0,P_1\}$. Here $v$ is not $2$-alternating. Let $A=\{1,2\}$ and $B=\{1,3\}$. Then,
$$v(A\cup B)+v(A\cap B)=\frac{15}{10}> v(A)+v(B)=\frac{14}{10}$$
Question: Let $U^n=\operatorname{Unif}(0,\frac{1}{n})$ the uniform measure on unit interval. And $\mathscr{P}_0$ corresponds to those with odd $n$, $\mathscr{P}_1$ contains exactly those with even $n$. Is $\mathscr{P}_0$, $\mathscr{P}_1$ or $\{\mathscr{P}_0,\mathscr{P}_1\}$ a $2$-alternating capacity? If we change the definition with "odd and even $n$ less than $10^{10}$", does $\mathscr{P}_0$, $\mathscr{P}_1$ or $\{\mathscr{P}_0,\mathscr{P}_1\}$ become a $2$-alternating capacity?
The reference of this question is this paper. Examples are from page $254$.
|
2025-03-21T14:48:31.113115
| 2020-05-29T09:46:37 |
361652
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fedor Pakhomov",
"Michael Greinecker",
"arsmath",
"https://mathoverflow.net/users/35357",
"https://mathoverflow.net/users/36385",
"https://mathoverflow.net/users/3711"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629612",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361652"
}
|
Stack Exchange
|
Constructive proof of the approximate Brouwer's Fixed Point Theorem for $\Delta^n$
The approximate Brouwer Fixed Point Theorem (aBFPT) for the standard $n$-simplex is:
Let $f$ be a uniformly continuous function from $\Delta^n$ into itself.
Then for each $\varepsilon>0$ there exists $x\in\Delta^n$ such that
$d(f(x),x)<\varepsilon$.
Does anyone know a reference for a full proof of aBFPT via Sperner's Lemma in Bishop-style constructive mathematics (BISH)?
A constructive proof of aBFPT via Gale's Hex Theorem is in Hendtlass's PhD thesis. A constructive proof for the $2$-simplex is in this paper by van Dalen. That's all I could find.
If you don't know a reference but have a proof handy I'd very much like to see that as well.
Thank you in advance!
Note: The status of BFPT in constructive mathematics has been discussed in this MO post. For quick orientation: aBFPT may be viewed as constituting the constructive core of the full classical BFPT. The latter implies LLPO and is thus inherently nonconstructive (indeed, the two are equivalent over BISH since both are equivalent to WKL in the presence of countable choice). Classically, BFPT can be retrieved from aBFPT by a simple application of the Bolzano-Weierstrass theorem, which is equivalent to LPO over BISH and thus constructively inadmissible.
I don't know the answer, but you might want to look in the computational complexity literature. Approximate Brouwer is an exemplar of a complexity class that has turned out to be interesting, called PPAD.
I am not proficient at all in this topic. But I do know about two papers of Orevkov that might be relevant here ( http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=tm&paperid=1617&option_lang=eng and http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=tm&paperid=2831&option_lang=eng ). As far as I understand, he studied Brower's Fixed Point theorem in the context of Russian constructive mathematics. In particular, Theorem 7.1 from the second paper is aBFPT. Unfortunately, both the papers are in Russian and they seems not to be translated in English.
I just found out that actually there exist an English translation of the second paper. It is in the volume "Problems in the Constructive Trend in Mathematics, IV" that were translated in English.
All van Dalen does is taking the usual argument based on Sperner's lemma and getting rid of having to decide ordering relations by adding some additional slack. The same method works for the usual arguments and does not depend on the dimension being 2.
There are also seems to be an explicit prove for the general case in https://link.springer.com/chapter/10.1007/978-1-4020-8926-8_14
|
2025-03-21T14:48:31.113321
| 2020-05-29T10:32:38 |
361654
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629613",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361654"
}
|
Stack Exchange
|
How does the knot contact homology augmentation polynomial change under a surgery on the base manifold?
So for every knot $K \in S^3$, there is a knot contact homology of $K$, and we can find the augmentation variety for this homology. The defining polynomial is known as the augmentation polynomial $A(K)$, which, as Aganagic et al. conjectured, corresponds to the classical A-polynomial when setting $Q=1$.
Now my question is that, how does this $A(K)$ change when I do surgery on $S^3$. As a concrete example, I want to find $K=\text{unknot}$ and the surgery is the contact surgery manifold obtained on a Legendrian unknots. It has Thurston-Bennequin number $\mathtt{tb} = -2$ and surgery coefficient $-1$.
What I knew was that "contact" surgery with coefficient $-1$ is equivalent to attaching a Legendrian handle on the manifold, so in that case is simply a good old Legendrian surgery.
My thought was that this surgery changes the symplectization from simply $\mathbb R\times S^3$ to $\mathbb R\times M^3$, but now the picture looks like this:
So I think from Ekholm's talk I learned that I need to consider the closed Reeb orbits, but I am stuck. Ekholm in one of his lectures calculated the value of unknot in $\mathbb RP^3$, but I do not understand how he arrived at his answer.
Is my above thinking correct? Or am I completely wrong? Thank you everyone!
|
2025-03-21T14:48:31.113442
| 2020-05-29T12:33:19 |
361661
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Drew Heard",
"Matt",
"https://mathoverflow.net/users/103150",
"https://mathoverflow.net/users/16785"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629614",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361661"
}
|
Stack Exchange
|
A generalization of integral Poincaré duality
In this paper, Felix, Halperin and Thomas define the notion of a Gorenstein space over a field $\mathbb{k}$:
An augmented differential graded algebra $R$ over $\mathbb{k}$ is Gorenstein if $\text{Ext}_R(\mathbb{k},R)$ is concentrated in a single degree and has $\mathbb{k}$-dimension one.
$X$ is Gorenstein over $\mathbb{k}$ if the cochain algebra
$C^*(X,\mathbb{k})$ is Gorenstein.
This definition is motivated by their subsequent results on this being a generalization of the notion of a Poincaré duality space.
Does there exist a parallel notion of a Gorenstein space over $\mathbb{Z}$ which similarly generalizes Poincaré duality over $\mathbb{Z}$?
EDIT: Alternatively, is it thought that no such generalization is available, so that one needs to use the machinery of symmetric spectra to get such a generalization over $\mathbb{Z}$?
A lot of work has been done (starting with Dwyer, Greenlees, and Iyengar) and furthered by Greenlees on when a morphism $f \colon R \to k$ of ring spectra is `Gorenstein', see https://arxiv.org/abs/math/0510247 for example. Taking $R=C^*(X;\mathbb{Q})$ and $f \colon R \to \mathbb{Q}$ the natural map gives you the above definition. Replacing $\mathbb{Q}$ with $\mathbb{Z}$ gives you a reasonable definition. (The theory tends to work best when $k$ is a field however).
Thank you for your comment. Are you aware of whether there is a definition of being Gorenstein over $\mathbb{Z}$ without using spectra, or are you saying that there is no such definition available, and the work of Dywer, Greenless, Iyengar addresses this deficiency?
I'm not aware of anything else, but that could be more to do with my own ignorance! Looking into the references of D-G-I might help you find other sources (e.g., https://link.springer.com/article/10.1007/BF02776063)
Prior to Dwyer-Greenlees-Iyengar, Dwyer and myself (independently) defined Gorenstein conditions for group rings over the sphere $S[G]$, i.e., the suspension spectrum of a topological group.
The definition easily extends to the case of a morphism $R\to k$.
In the orientable case the definition goes like this:
$R\to k$ is said to be Gorenstein of dimension $d$ if:
1) $k$ is finitely dominated as an $R$-module, i.e., $k$ is a retract up to homotopy of a finite $R$-module (this finiteness condition shouldn't be ignored!), and
2) the derived mapping spectrum $\hom_R(k,R)$ is weakly equivalent as an $R$-module to $k[-d] := \Sigma^{-d}k$.
If one wishes to have an unoriented Gorenstein condition, then it seems to me one needs to replace $k[-d]$ by a twisted version of it. In the special case when $R$ is an augmented $k$-algebra spectrum, we can simply require that the $k$-module $\hom_R(k,R)$ is equivalent to $k[-d]$ as a $k$-module (but not necessarily as an $R$-module).
If $R$ is isn't a $k$-algebra, we can fix another $R$-module structure on $k$, call it $k^\xi$, and require that $\hom_R(k,R)$ is
equivalent to $k^\xi[-d]$ as an $R$-module.
Returning the the group ring case $S[G]$, we have:
Theorem. Assume in addition $\pi_0(G)$ is finitely presented. Then
following are equivalent:
1) $S[G] \to S$ is Gorenstein in dimension $d$.
2) $BG$ is a (finitely dominated) Poincaré duality space in dimenson $d$.
And yes, there is a parallel story over $\Bbb Z$, but it is enough to work with dgas instead of ring spectra (for example, the derived $\hom_{R}(\Bbb Z,R)$ is a differential graded module for a discrete ring homomorphism $R\to \Bbb Z$).
|
2025-03-21T14:48:31.113691
| 2020-05-29T13:24:59 |
361664
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Daebeom Choi",
"LSpice",
"Peter Scholze",
"RandomMathUser",
"anon",
"https://mathoverflow.net/users/149169",
"https://mathoverflow.net/users/161405",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/6074",
"https://mathoverflow.net/users/95601"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629615",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361664"
}
|
Stack Exchange
|
Some basic questions on quotient of group schemes
Let $S$ be a fixed base scheme and $G, H$ be group schemes over $S$. Since I am mainly interested in commutative group schemes over fields, we may assume that $G,H$ are commutative and $S$ is a field if this helps.
(1) Let $f:G\to H$ be a morphism of group schemes. To define the cokernel of this map, we need to choose which topology to work with. Some people use the fppf topology (as in van der Geer & Moonen's book) and other people use the fpqc topology (as in Cornell-Silverman). My question is: what is the difference of those two topologies in terms of group schemes? Is fppf quotient and fpqc quotient of group schemes different? Which topology do people prefer when they are working with group schemes?
(2) Let $H$ be a (normal) closed subgroup scheme of $G$. I think there are at least three plausible definitions of the quotient $G/H$:
Categorical quotient: Since $H$ naturally acts on $G$, we can think categorical quotient $G/H$ of the action $H\times G\to G$.
Fppf/fpqc quotient: $G/H$ represents the quotient of $H\to G$ in the category of fppf/fpqc sheaves.
Naive quotient: A group scheme $G/H$ with a surjective (wrt fppf/fpqc topology) map $p:G\to G/H$ such that kernel of $p$ is the inclusion $H\to G$
Are they equivalent in some good situations? In van der Geer & Moonen's book, it is proved that a fppf quotient is also a categorical quotient. But I cannot find proof nor prove other directions.
context of the question (2): Let $f:A\to B$ be an isogeny of abelian varieties with kernel $\ker f$. Then we have the dual exact sequence $0\to \widehat{B}\to \widehat{A}\to \widehat{\ker f}\to 0$. In Milne's book on abelian variety, to prove the dual exact sequence, consider $0\to \ker f\to A\to B\to 0$ as an exact sequence in the category of commutative group schemes over a field and use a long exact sequence with $\text{Hom}(-, \mathbb{G}_m)$. To use the long exact sequence, we need to prove $B$ is $A/\ker f$ as a fppf/fpqc quotient (In fact I don't know which topology to work with. This is why I ask the question (1)...). However, I only know that $B$ is the `naive quotient (3)' $A/\ker f$.
(3) Is the category of commutative group schemes over a field an abelian category? This statement is in Milne's book on abelian variety, but I cannot find proof. The main point is existence of cokernel, i.e. representability of fppf/fpqc quotient. However, I only know the following theorem in Cornell & Silverman,
Theorem. Let $G$ be a finite type $S$-group scheme and let $H$ be a closed subgroup scheme of $G$. If $H$ is proper and flat over $S$ and if $G$ is quasi-projective over $S$, then the quotient sheaf $G/H$ is representable.
and this is too weak to prove our statement.
Also one more quick question: do you know any good reference dealing with sufficiently general group schemes? I know Shatz's paper in Cornell-Silverman, Tate's paper in Cornell-Silvermann-Stevens, and Stix's lecture note, but they focus on finite flat group schemes. Also, I know some other articles & books which mainly focus on affine algebraic groups. Are there some more general references?
Thank you for reading my stupid questions.
Have you looked at Milne's book, Algebraic Groups, CUP 2017. That should answer all your questions over fields.
@anon I take a look at that book, and I think this book contains essentially everything I want. Thank you for your recommendation! Also, I found answers to my question in other books, and I posted it as an answer.
Let me self-respond to my question. First of all, about reference: I found answers to these questions in `Rational Points on Varieties' by Bjorn Poonen. This book contains an excellent summary of essential facts on algebraic groups. Proof of these facts are contained in, of course, SGA 3-1. I still wonder why I cannot think of SGA while searching for reference.
(3) Consider the following theorems:
Theorem. [Theorem 5.2.5 of Poonen's book, Theorem 3.2 of Expose VI$_A$ of SGA 3-1] Let $H$ be a closed normal subgroup scheme of a group scheme of finite type $G$ over an Artinian ring $A$. Then the fppf quotient $G/H$ exists as a group scheme. Also, the quotient map $p:G\to G/H$ is faithfully flat.
Theorem [Theorem 5.2.9 of Poonen's book, Corollary 7.4 of FGA] Let $f:G \to H$ be a homomorphism between algebraic groups over a field. Then $f$ is factored into homomorphism $G\to G/\ker f\to H$, where $G/\ker f\to H$ is a closed immersion.
By combining these two theorems, we can show that the cokernel of a map always exists.
(2) By these two theorems, we know that the fppf quotient always exists. As I mentioned in the question, fppf quotient is also a categorical quotient. Since the categorical quotient is determined by its universal property, the categorical quotient must be the fppf quotient. Hence these two notions of quotient coincide. Equivalence of the fppf quotient and the naive quotient can be shown in a similar way. (I prove the fact in this way because I use the existence of fppf quotient as a Blackbox, but I think this argument is redundant, because in my understanding, what SGA proved is that the categorical quotient is the fppf quotient.)
(1) Since Poonen's book and SGA both use the fppf topology, I think the fppf topology is a better choice. By the above theorems, at least in the commutative algebraic group case, a map between commutative algebraic group schemes is surjective in the category of algebraic group schemes if and only if it is surjective as fppf sheaves. However, I'm not sure whether we can do this with fpqc topology. If we are working with algebraic groups, then everything is finitely presented, so it is hard to imagine that fpqc topology makes any difference. If someone know something about algebraic group schemes over fpqc topology, then please let me know.
As long as the group schemes are all finitely presented, it should make no difference whether one uses the fppf or fpqc topology. However, if you allow non-finitely presented group schemes -- they often occur in the Tannakian formalism -- you should use the fpqc topology; in general, for the correct meaning of $G/H$, the map $G\to G/H$ may fail to be an fppf cover.
So I think for this sort of question (quotients of flat finitely presented group schemes) the best is to use the theory of algebraic stacks and spaces. I am not an expert, so if someone could double check this that would be great.
Let $G$ be an fppf group scheme over a scheme $S$, and $H$ an fppf subgroup scheme of $G$.
Let $\mathcal{X}=[G/H]$ be the stack quotient. Since $G \times H \to G \times G$ is an fppf groupoid, it is algebraic and $G \to \mathcal{X}$ is an fppf presentation of $\mathcal{X}$. Since the inertia is $H$, it is fppf, so $\mathcal{X}$ is a gerbe over the fppf sheaf quotient $G/H$ (which is an algebraic space), and so $\mathcal{X} \to G/H$ is smooth. So $G \to G/H$ is fppf, where $G/H$ is the quotient in algebraic spaces (or in fppf sheafs).
Now if $G/H$ is a nice space, for instance qs (this is always the case in practice, for instance it is if $H \to G$ is qc), then it contains an open subscheme. If the base $S$ is a field, then since $G$ acts transitively in $G/H$ by acting on this subscheme we get that $G/H$ is a scheme (this is the same trick as for proving that a group algebraic space over a field is a group scheme. In fact we also have that an abelian algebraic space over a base $S$ is always an abelian scheme but this is harder to prove).
Remark: if $H \to G$ is proper, then $[G/H]$ is separated.
fppf is familiar, and I guess qc is 'quasi-compact', but what is qs?
fppf='faithfully flat locally of finite presentation', qc is indeed 'quasi-compact' and qs is 'quasi-separated'
|
2025-03-21T14:48:31.114283
| 2020-05-29T14:27:45 |
361668
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"KKD",
"Vít Tuček",
"https://mathoverflow.net/users/135674",
"https://mathoverflow.net/users/6818"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629616",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361668"
}
|
Stack Exchange
|
Checking axiom of Category $\mathcal{O}$
Let $K$ be a finite extension of $\mathbb{Q}_p$ and $G$ be a split connected reductive algebraic group over $K$ with Borel $B$. We have the associated Lie algebras $\mathfrak{g}=$Lie$(G)$ and $\mathfrak{b}=$Lie$(B)$.
Let $M$ be a $U(\mathfrak{g})$-module with $N \subset M$ a finite dimensional $K$-module, which is $B$-invariant and generates $M$ as a $U(\mathfrak{g})$-module.
I read that $M$ is then locally $\mathfrak{b}$-finite, i.e $U(\mathfrak{b}) \cdot m \subset M$ is finite dimensional for all $m \in M$, but I have trouble to see this. As $U(\mathfrak{b})$ seems so big for me, I cannot think of a finite basis for $U(\mathfrak{b}) \cdot m \subset M$ by knowing only $N$.
I am familiar only with complex numbers, but there the result is easy, because being $B$-invariant means that it is a space of highest weight vectors and then your $M$ is covered by Verma modules.
Thanks for your fast answer but I cannot follow you. I'm still learning all these things. By space of highest weight vectors you mean $N$ has a basis of highest weight vectors? Why? And why is $M$ covered by Verma Modules?
The module $U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N$ is locally $U(\mathfrak{b})$-finite and there is a surjective $U(\mathfrak{g})$-homomorphism $\varphi\colon U(\mathfrak{g})\otimes_{U(\mathfrak{b})} N \to M$ given by $u \otimes n \mapsto u\cdot n.$ It is easy to see that every weight space of $M$ has only finitely-many preimages. It follows that $M$ cannot contain infinite-dimensional $U(\mathfrak{b})$-module. In the case $N$ is one-dimensional, the module I just constructed is called Verma module. In case of $N$ not being completely reducible, one would have to do some gymnastics with short exact sequences.
But perhaps for a beginner it's easier to just attack the problem directly. Assume first for simplicity that $m = u \cdot n$ for some $u \in \mathfrak{g}.$ Pick $X \in \mathfrak{b}.$ Then
$$
X \cdot m = X\cdot u \cdot n = Xu \cdot n = [X,u]\cdot n + uX\cdot n = u'\cdot n + u \cdot n',
$$
where $u'$ is some other element of $\mathfrak{g}$ and $n'$ is some other element of $N.$ Now iterate for general $u\in U(\mathfrak{g})$ and repeat for all possible $X \in \mathfrak{b}$. You will see that you pick up at most $\dim \bigotimes^k \mathfrak{g} \otimes N$ possible elements. They might not be all linearly independent, but they surely span $\mathfrak{b} \cdot m.$
Thanks for this detailed answer. Could we argue also in the following way: $U(\mathfrak{g}) \otimes_{U(\mathfrak{b})} N$ lies in Category $\mathcal{O}$, which is closed under submodules and quotients. By the map you have given $M$ is a quotient of $U(\mathfrak{g}) \otimes_{U(\mathfrak{b})} N$ and therefore lies in category $\mathcal{O}$ too. Hence has to be locally $U(\mathfrak{b})$-finite.
Yes. If you already know that $\mathcal{O}$ is closed under quotients then this is a good argument.
|
2025-03-21T14:48:31.114482
| 2020-05-29T15:25:12 |
361672
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lev Soukhanov",
"Mark Grant",
"Moishe Kohan",
"Shijie Gu",
"Willie Wong",
"YCor",
"https://mathoverflow.net/users/114032",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/33286",
"https://mathoverflow.net/users/3948",
"https://mathoverflow.net/users/39654",
"https://mathoverflow.net/users/8103"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629617",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361672"
}
|
Stack Exchange
|
Is the intersection of two distinct sufficiently small metric spheres always empty, a point or a metric sphere of lower dimension?
Let $(X,d)$ be an $n$-dimensional $(n< \infty)$ complete geodesic metric space, where any two points in $X$ are joined by a unique shortest geodesic. Let $S$ be a sufficiently small metric $(n-1)$-sphere in $X$. Let $\epsilon>0$ be the radius of $S$. Pick a point $p \in S$. Find a metric $(n-1)$-sphere $S'$ of radius $0<r < \epsilon/2$ around $p$. A $k$-dimensional metric sphere $S^k$ in $(X,d)$ of radius $r$ centered at $p$ is given by a $k$-dimensional subset of $(X,d)$: $S^k \subset \{x\in X|d(p,x) = r\}$. Here dimension I'm referring to covering dimension.
Is the intersection $S \cap S'$ always an $(n-2)$-dimensional metric sphere?
Just to clarify - what is a k-dimensional metric sphere in a complete geodesic metric space? I believe the answer is "no" regardless of anything, but this clarification is needed.
If you have some notion of k-dimensional sphere in mind such that there is a family with ((n-k)k + n + 1) parameters (say, emitting geodesic segments of length r along some k-dimensional subspace in the tangent space to some point), then the answer is no by dimension counting.
@LevSoukhanov Sorry, I didn't make my question clear. Just modified it.
I still don't understand what you mean by "a metric sphere". Do you mean, the sphere centered at some point of $X$? Also there is no question currently (except in the title, but the post should make sense without reading the tilte).
@Ycor: Yes. I put the question at the end.
You still haven't defined what you mean by sphere, so that it's meaningful for the intersection. A sphere in a metric space (which metric space)? A topological sphere?
I am sorry, but I still do not understand. The subset you are specifying in the definition of k-dimensional sphere is not actually k-dimensional in most cases.
@Lev: I think that it is a typo, the OP probably meant $S^k \subset \ldots$ instead of $S^k =$. // But as stated: obviously $S\cap S'$ is a subset of $S$ and so is a subset of some sphere. So the OP's question essentially boils down to whether the intersection has dimension $(n-2)$....
@Lev: Sorry, like what Willie pointed out it should be a proper subset.
@YCor: Metric sphere I'd like to think of that as the boundary of metric ball. Metric sphere is not necessarily homeomorphic to a topological sphere.
It's not at all obvious (to me at least) why a metric sphere in a geodesically complete metric space of Lebesue dimension $n$ must always have Lebesgue dimension $n-1$.
@Mark Grant: No. But I’m more interested in metric spheres which have Lebesgue dimension n - 1.
Let $C_\alpha$ denote the $n$-dimensional closed Euclidean solid cone with the cone angle $\alpha\in (0, \pi)$. Let $X_\alpha$ be the metric space obtained by gluing two copies $C^\pm_\alpha$ of $C_\alpha$ at their tips, and equipped with the natural path-metric. Let $o\in X_\alpha$ denote the common tip of the cones. I will use it as the center of the first sphere $S=S(o,1)$; I will use $\epsilon=1$. For every point $x\in S$, the sphere $S'=S(x,r)$ is $n-1$-dimensional. However, for each $r\in (0,1)$, for all sufficiently small $\alpha$, for every $x\in S\subset X_\alpha$, the intersection $S'\cap S$ is empty, hence, has dimension $-1$, not $n-2$. By modifying this construction one can build uniquely geodesic spaces where spheres as in your question are $n-1$-dimensional but have arbitrary dimension of their intersection, between $-1$ and $n-1$.
Edit. Here is a generalization, to ensure that the spheres $S, S'$ are connected (and locally path-connected) and the intersection $S\cap S'$ is nonempty.
Let $Y$ be a closed solid cone in the Euclidean space $E^k, 1\le k<n$, with the tip $o$. I will glue $Y$ to the space $X_\alpha$ as above so that $o$ is identified with the common tip $o\in X$ and two boundary rays of $Y$ are identified with geodesic rays in the two cones $C^\pm_\alpha \subset X$. Let $Z$ denote the resulting path-metric space. It is easy to check (say, using Reshetnyak gluing theorem) that $Z$ is a $CAT(0)$ space, hence, is uniquely geodesic.
Moreover, both spheres $S, S'\subset Z$ (defined as before) are connected and locally path-connected. Furthermore,
for every $r$, there is $\alpha$ such that $S\cap S'$ is $(k-2)$-dimensional. (Note that $k-2<n-2$.)
By working a bit harder, one can modify this construction so that spheres are still connected and locally path-connected, while $S\cap S'$ is $(n-1)$-dimensional, where $n$ is the dimension of the ambient space.
What if the metric spheres are connected and locally path-connected?
@Bobbrown: See the edit.
|
2025-03-21T14:48:31.114792
| 2020-05-29T15:40:44 |
361676
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Giorgio Metafune",
"Math604",
"Paul",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/66623",
"https://mathoverflow.net/users/9253"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629618",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361676"
}
|
Stack Exchange
|
Minimal assumption for an elliptic equation
On the disc $\mathbb{D}$ on the disc with a metric $g=e^{2\lambda} \vert dz \vert^2$(let assume $\lambda$ is smooth on $\overline{\mathbb{D}}$) and I consider either
$$\newcommand{\Div}{\operatorname{div}}
\Div_g(X)=e^{-2\lambda}\Div_e(e^{2\lambda} X)=f$$
where $\Div_e$ is the classical divergence, $f\in L^2$ and $e^{2\lambda} f$ with zero average and I want a solution $X\in W_0^{1,2}$ with some control like $\Vert X\Vert_{1,2}\leq C \Vert f\Vert_2$.
Or
$$\Delta_g(\phi)=e^{-2\lambda}\Div_e(\nabla \phi)=f$$
with $f\in L^2$ and I want a solution $\phi\in W_0^{2,2}$ with some control like $\Vert e^{-2\lambda} \nabla \phi\Vert_{1,2}\leq C \Vert f\Vert_2$
Those problem are more or less equivalent, and using Gilbard-Trudinger, I know I got a uniform estimate as soon as I control the modulus of continuity of $\lambda$, i.e. $C$ depends only of this modulus.
But in my case I only get $C_0>0$ such that $e^{-C_0}\leq e^{\lambda}\leq e^{C_0}$,
and I would the estimate above with a $C$ which depends only on $C_0$. Does any one have a reference on the minimal hypothesis I can put on the control of the metric, i.e. $\lambda$, to get a uniform estimate?
P.S.: In fact I have a bit more $\Vert \nabla \lambda \Vert_{L^{2,\infty}} \leq C_0$, but my question is more about the minimal hypothesis to get the uniform estimate rather than can I apply it in my own case of study.
In your second question it is not clear what is $X$ and what is the norm $|\cdot|_{1,2}$ since you are asking for $W^{2,2}$ solutions. Also note that $\Delta_g|$ is not well-defined "pointwise" on $W^{2,2}$ functions, unless $\lambda$ has a gradient.
Indeed, I have edited my question: in fact we can consider $\lambda$ smooth, but I want that the estimate depends only of my Harnack estimate.
you might want to single out the case where $ \lambda$ is harmonic... i have seen lots of work regarding euqations like $-\Delta u(x) + t a(x) \cdot \nabla u(x) =f(x)$ where $ a(x)$ is divergence free and the goal is to obtain estimates independent of $t$. One can normally obtain L^infty bounds on u independent of $t$ (maybe with some assumptions on $a$) but i think Holder estimates are hard to find... try googling 'large advection'
@Math604, Thanks for your answer, but I realize that the second formulation of my problem had a typo. It is corrected, but unfortunately your answer doesn't match anymore.. thx again
so now the equation is just of the form $ \Delta \phi = g(x) f(x)$ so you should be able to apply some standard theory to see what you can get?? or is that not sufficient?
|
2025-03-21T14:48:31.115252
| 2020-05-29T16:52:55 |
361678
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"gradstudent",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/89451"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629619",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361678"
}
|
Stack Exchange
|
About concentration of eigenvalues values of a random symmetric matrix in a specific interval
Given a random symmetric matrix $M$ and two numbers $\lambda_\min$ and $\lambda_\max$ how do we calculate the expected or high probability value of the fraction of its eigenvalues which lie in the interval $[\lambda_{\min},\lambda_{\max}]$?
I would like to know of references which might have achieved such an estimate for even any specific distribution of $M$
In particular I am most curious about the case when $M$ is random PSD, $\lambda_\min =0$ and $\lambda_{\max} \lll \Vert M \Vert_2$
A random PSD matrix $M$ can be constructed by taking $M=WW^T$, with the $n\times n$ matrix elements of $W$ i.i.d. with mean zero and variance $\sigma^2$. For $n\gg 1$ the marginal distribution $\rho(\lambda)$ of the eigenvalues $\lambda$ of $W$ is given by the Marcenko-Pastur distribution
$$\rho(\lambda)=\frac{1}{2\pi\sigma^2 n}\sqrt{\frac{4n\sigma^2 }{\lambda}-1},\;\;0<\lambda<4n\sigma^2 ,$$
and then you can integrate from $0$ to $\lambda_{\rm max}<4n\sigma^2 $ to obtain the desired fraction $f$ of eigenvalues in the interval $(0,\lambda_{\rm max})$,
$$f=\frac{\sqrt{\lambda_{\rm max} \left(4 n {\sigma}^2-\lambda_{\rm max}\right)}}{2 \pi n {\sigma}^2}-\frac{2 \arctan\left(\sqrt{\frac{4 n {\sigma}^2}{\lambda_{\rm max}}-1}\right)}{\pi }+1.$$
For $\lambda_{\rm max}\ll 4n\sigma^2$ this has the asymptotics
$$f\rightarrow\frac{2}{\pi}\sqrt{\lambda_{\rm max}}.$$
$f$ versus $\lambda_{\rm max}/n\sigma^2$.
The $\sqrt\lambda_{\rm max}$ growth is generic for random PSD matrices. It arises because eigenvalue repulsion causes a $1/\sqrt\lambda$ accumulation of the eigenvalues near the hard spectral edge at $\lambda=0$.
Thanks! So this is a very specific distribution for which the computation seems to exist. Would you know of any general results of this kind?
it's not that specific: the matrix elements of $W$ can have any distribution, it does not need to be a Gaussian; the Marcenko-Pastur derivation does assume that the elements of $W$ are i.i.d., but the large-$n$ limit also holds if they are correlated, under similar conditions as the Wigner semicircle law for the GOE. Moreover, the $f\propto\sqrt{\lambda}_{\rm max}$ asymptotics is generic for a hard spectral edge at 0, so it should apply to any random PSD matrix.
|
2025-03-21T14:48:31.115429
| 2020-05-29T17:07:33 |
361680
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andy Sanders",
"https://mathoverflow.net/users/130225",
"https://mathoverflow.net/users/49247",
"mtraube"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629620",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361680"
}
|
Stack Exchange
|
Opers and global differential operators
This is a follow up question to a previous question of mine and my thought of answer to it.
Given a (compact) Riemann surface $\Sigma$, a $SL(n,\mathbb{C})$-oper is a rank $n$ holomorphic vector bundle $E\rightarrow \Sigma$, s.th. $\det(E)\cong \mathcal{O}$ is trivial, together with a holomorphic connection $\nabla:E\rightarrow E\otimes K_\Sigma$ ( $K_\Sigma$ being the canonical bundle of $\Sigma$) satisfying the following list of axioms:
1) There is a flitration by holomorphic subbundles $0=F^0\subset F^1\subset \cdots \subset F^n=E$.
2) The connection is Griffiths transverse: $\nabla(F^i)\subset F^{i+1}\otimes K_\Sigma$ .
3) The connection is non-degenerate in the following sense: The induced map $F^i/F^{i-1}\rightarrow F^{i+1}/F^i\otimes K_\Sigma$ is an isomorphism.
By non-degeneracy $E$ is in fact completely specified by $F^1$ and for $SL(n,\mathbb{C})$ it has to hold $F^1\cong K^{\frac{n-1}{2}}_\Sigma$. Having this, an $SL(n, \mathbb{C})$ is equivalent to an $n$-th order differential operator $P:\Omega^{\frac{-n+1}{2}}\rightarrow \Omega^{\frac{n-1}{2}}$, where $\Omega$ is the sheaf of holomorphic sections of $K_\Sigma$.
Q1 Is the converse true? I.e. given a holomorphic line bundle $L\rightarrow \Sigma$ with sheaf of holomorphic sections $\mathcal{L}$ and an n-th order holomorphic differential operator $P:\mathcal{L}\rightarrow \mathcal{L}$, can I construct from this an $SL(n,\mathbb{C})$ oper?
I think to answer to this is yes, but I cannot track a reference for this.
From the discussion I have in my previous question, it seems that to any rank $n$ vector bundle $H$ of degree zero with a holomorphic connection one gets at least locally a $n$-th order holomorphic differential operator defining its flat sections, due to the existence of a local cyclic section.
Q2 As I think the answer to the first question is yes, the local picture cannot glue in general I suppose. So the question is, do the local cyclic sections glue to a global cyclic section if and only if the bundle $H$ has an oper structure?
I think the answer to your question should be contained, or at least intimated, in these notes of Richard Wentworth: https://arxiv.org/abs/1402.4203.
See also section 2 of the notes https://arxiv.org/pdf/math/0501398.pdf of Beilinson-Drinfeld where they also basically answer your question.
@AndySanders, nice, thanks for the references! The beginning of section 2 in the paper by Beilinson-Drinfeld directly answers the first question with the additional assumption that the differential operator has nowhere vanishing principal symbol. So representing the vector bundle locally on a coordinate open as $\mathcal{D}_1(U)/ \mathcal{D}_1P(U)$ the differential operators $P(U)$ only glue to a differential operator with non vanishing principal symbol if the bundle has in fact an oper structure.
Indeed. I'm glad I could say something useful.
|
2025-03-21T14:48:31.115629
| 2020-05-29T17:11:28 |
361681
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"JustWannaKnow",
"Michael Engelhardt",
"Nate Eldredge",
"https://mathoverflow.net/users/112113",
"https://mathoverflow.net/users/134299",
"https://mathoverflow.net/users/150264",
"https://mathoverflow.net/users/4832",
"lambda"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629621",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361681"
}
|
Stack Exchange
|
Invertibility of discrete Laplacian
In QFT and Statistical Mechanics the discrete Laplacian usually plays a key role when we want to discretize the theory. However, few books (at least to my knowledge) really work the properties of this operator in details, so I'm trying to figure out some of these properties myself.
Let $\Lambda := \epsilon Z^{d}/L\mathbb{Z}^{d}$ be a finite lattice where $\epsilon> 0$ and $L > 1$ are integers such that $L/\epsilon \in \mathbb{N}$ is even. An scalar field over the lattice $\Lambda$ is simply a function $\phi : \Lambda \to \mathbb{C}$, so that the space of all fields is $\mathbb{C}^{\Lambda}$. Because the lattice is a quotient space, we're dealing with periodic boundary conditions. Thus, we can introduce the discrete Laplacian as the linear operator $-\Delta: \mathbb{C}^{\Lambda} \to \mathbb{C}^{\Lambda}$ defined by:
$$(-\Delta \phi)(x) := \frac{1}{\epsilon^{2}}\sum_{k=1}^{d}[2\phi(x)-\phi(x+\epsilon e_{k})-\phi(x-\epsilon e_{k})]$$
with $\{e_{1},...,e_{d}\}$ being the canonical basis for $\mathbb{R}^{d}$.
Now, let $\langle \phi, \varphi \rangle_{\Lambda} := \epsilon^{d}\sum_{x\in \Lambda}\overline{\phi(x)}\varphi(x)$ be an inner product on $\mathbb{C}^{\Lambda}$. If I'm not mistaken, the follwing identity holds:
\begin{eqnarray}
\langle \phi, -\Delta \phi\rangle_{\Lambda} = \sum_{x\in \Lambda}\sum_{y\sim x}|\phi(x)-\phi(y)|^{2} = \sum_{x\in \Lambda}\sum_{y\sim x}(\overline{\phi(x)}-\overline{\phi(y)})(\phi(x)-\phi(y)) \tag{1}\label{1}
\end{eqnarray}
where $y\sim x$ denotes that $|x-y| = 1$, where $|\cdot|$ is the maximum 'norm' on $\mathbb{Z}^{d}$.
My point is the following. We could have assumed $\phi = 0$ outise $\Lambda$ as a boundary condition, instead of our periodic one. In this case, I know that the discrete Laplacian is positive in the sense that:
$$\langle \phi, -\Delta \phi \rangle_{\Lambda} > 0 \quad \mbox{if} \quad \langle \phi, \phi \rangle_{\Lambda} > 0$$
and I'd expect the same property with periodic bondary conditions. However, because of the first equality in relation (\ref{1}), it seems that if we take $\phi$ to be constant everywhere, say $\phi(x) = 1$ for every $x \in \Lambda$, it'd follow that $\langle \phi, -\Delta \phi \rangle_{\Lambda} = 0$ even with $\langle \phi, \phi\rangle_{\Lambda} > 0$. This would lead to a non-invertibility of this operator. I don't know if this is a known fact that I just didn't know yet or if my reasoning is not correct, but I'd appreciate any help here.
Yes, this is well known. The derivative operator with periodic boundary conditions has a zero mode and is not invertible. This remains true for your discretized version. A wealth of information about much more than this is in any textbook on lattice gauge theory.
It shouldn't be too surprising that the injectivity, and the spectral properties in general, are affected by the choice of boundary conditions. You see this in the continuous situation too: the Dirichlet Laplacian on a nice domain is injective, the Neumann Laplacian isn't.
Your reasoning is correct in that the discrete Laplacian for periodic boundary conditions has a zero mode. On the space of fields satisfying $\sum_x\phi(x)=0$, its spectrum is, however, strictly positive, and it can be inverted on that space. This is all well known.
Thanks for the comment! Does this fact depend on the dimension $d$?
This is true for any finite connected graph.
|
2025-03-21T14:48:31.115883
| 2020-05-29T18:55:29 |
361684
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andreas Blass",
"Paul Taylor",
"Peter LeFanu Lumsdaine",
"Simon Henry",
"Taras Banakh",
"Todd Trimble",
"https://mathoverflow.net/users/22131",
"https://mathoverflow.net/users/2273",
"https://mathoverflow.net/users/2733",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/61536",
"https://mathoverflow.net/users/6794"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629622",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361684"
}
|
Stack Exchange
|
Why do elementary topoi have pullbacks?
In the book of Szabo "Algebra of Proofs", Definition 13.1.9 introduces an elementary topos as a cartesian closed category with a subobject classifier. On the other hand, many other sources including Johnstone add to this definition that the category should contain limits of finite diagrams. For the proof that the requirement on limits of finite diagrams can be removed, Szabo refers the reader to the paper "Colimits in topoi" by Robert Pare who writes in the second paragraph of the section "Preliminaries on topoi" that the existence of finite limits follows from the existence of equalizers which can be derived from appropriate application of the subject classifier. But for finding a monomorphism from the subobject classifier we should have the corresponding pullbacks in the category. Why such pulbacks exist? The definition of a subobject classifier works only in one direction: given a monomorphism it yields the characteristic morphism. But for the opposite direction (from a characteristic morphism to a monomorphism) the definition does not say anything on the existence of the corresponding pullbacks.
The Question. Is it true that a cartesian closed category with a subobject classifier indeed has pullbacks?
If yes, could you provide a (desirably simple) proof? Thank you.
Maybe you should clarify how you define subobject classifier in general ? The definition I would consider standard contains the fact that pullback of monomorphisms exists. (And product and pullback of monomorphisms are enough to get all finite limits because fiber product can be constructed as pullback of a diagonal inclusion)
@SimonHenry I take the definition of a subobject classifier from Wikipedia: https://en.wikipedia.org/wiki/Subobject_classifier
The definition of "subobject classifier" on the Wikipedia page indeed omits any requirement that pullbacks of "True" along arbitrary morphisms to $\Omega$ should exist. On the other hand, the first sentence on that page implicitly includes that requirement.
Maybe it's worth mentioning that, if all pullbacks of "true"$:1\to\omega$ exist, then so do all pullbacks of monomorphisms. To pull back a monomorphism $X\to Y$ along a map $f:\to Y$, just compose $f$ with the classifying map of $X\to Y$ and pull "true" back along that composite.
@AndreasBlass So, this requirement must be included in order to have that more simple definition of an elementary topos. Right? By the way, Wikipedia is open source and I encourage specialists to make necessary corrections there.
I don't see a way to deduce that requirement, but there's a difference between "I don't see" and "there does not exist". So I'm inclined to wait for more comments or an answer.
Perhaps one should be aware that Szabo's Algebra of Proofs has a somewhat notorious reputation of having a lot of mistakes in it. (Let me add, for the sake of balance, that I found the book very useful in getting my bearings when I was doing dissertation work; however, I wouldn't take him as an authority on topos theory.)
@ToddTrimble But I have seen this simplified definition of an elementary topos not only in Szabo. For example, it is mentioned on page 84 of Goldblatt's book (https://projecteuclid.org/euclid.bia/1403013939), which I hope has a good reputation. And Goldblatt also gives that standard (incomplete) Wikipedia'a definition of a subobject classifier.
Re Goldblatt: well, unfortunately not entirely. Maybe I'll spend some time thinking about your question.
@ToddTrimble By the way, another question concerning this simplified definition of an elementary topos. Why it does contain an initial object? If we do not assume any coproducts in the definition, how to generate an initial object from terminal, products, and subobject classifier (even with the correct definition)?
There's a theorem that an elementary topos has finite colimits. The usual construction (the theorem is that the power object functor $P: E^{op} \to E$ is monadic) is somewhat complicated though; details are in the book by Mac Lane-Moerdijk. A more elementary approach is to develop enough "internal logic" (conjunction, implication, and universal quantification), and then define the initial object as the internal intersection of all subobjects of 1. More on this in this paper: https://www2.mathematik.tu-darmstadt.de/~streicher/CTCL.pdf
The construction of colimits in a topos was first done by Christian Mikkelsen in his thesis at Aarhus (which is brilliant but difficult to obtain). The monadicity result was then found by Bob Paré.
Start with the observation that a subobject of the minimal subobject $0$ of $1$ is $0$ again. Let $X$ be any object. Then there is a "singleton map" $\sigma: X \to PX$ which is monic, and there is also a map $1 \to PX$ which classifies the subobject $id_X: X \to X$. Take the pullback of the subobject $\sigma$ along $0 \to 1 \to PX$. This gives a subobject of $0$, which is $0$ again, and this gives a map $0 \to X$.
(The last comment was in reply to a comment which has now been deleted.)
@ToddTrimble Thank you for the explanations with the initial object. I have deleted my comments because I understood how to do that but not so elegantly as you wrote.
It should also be noted that the reason why this confusion has arisen is because of ambiguity in the usage of cartesian [closed]. For Freyd and Lawvere, cartesian meant having all finite limits, whereas in computer science it has come to mean just products.
@PaulTaylor: For the record, Mikkelsen’s thesis is no longer difficult to obtain, at least if this is the right one: Christian Juul Mikkelsen, Lattice Theoretic and Logical Aspects of Elementary Topoi, 1976 licentiate thesis at Aarhus University, reprinted in Theory and Applications of Categories in 2022, also available from Mikkelsen’s personal webpage.
@PeterLeFanuLumsdaine. I exchanged emails with Christian Mikkelsen sometime after the above comment and found out about his website, but I didn't know about TAC - that's a useful stable citatation, thanks. (His website is mainly about his own impressive genealogical research, with about 250 of his and his wife's ancestors! It also shows when Denmark switched from patronymics to surnames.)
I'll give a counter-example to the claim that having a subobject classifier and being cartesian closed implies the existence of all finite limits. However, this is based on the definition of sub-object classifier given on wikipedia (linked in the comment above) that I would consider as incorrect:
The wikipedia definition (at the time this is written) only asks that for every monomorphism $U \hookrightarrow X$ there is a unique map $X \to \Omega$ such that $U$ is the pullback of the universal subobject $1 \hookrightarrow \Omega$, but it does not ask that every map $X \to \Omega$ be the classifier of some subobject (i.e. that all pullbacks of the universal subobject exist).
If you add the requirement that every map to $\Omega$ classify something, i.e. that pullback of the map $1 \to \Omega$ exists, then it follows that pullbacks of all monomorphisms exist. Moreover pullbacks of monomorphisms, and the existence of finite products imply (in a $1$-category) the existence of all finite limits: A fiber product $B \times_A C$ can be recovered as the pullback of the monomorphism $A \to A \times A$ along $B \times C \to A \times A$.
Consider the category $C$ of finite sets that are not (isomorphic to) the three element sets, with all functions between them. (feel free to replace three by any odd prime).
$C$ has products: if $|A \times B| = 3$ then $|A|=3$ or $|B|=3$, so $C$ is stable under product in the category of sets. As it is a full subcategory it follows that these are products in $C$ as well.
$C$ has a subobject classifier in the sense of Wikipedia's definition, given by the usual $1=\{\top\} \to \Omega = \{ \bot, \top \}$. Indeed given any mono $A \subset B$ in $C$, its classifying map $B \to \Omega$ in set is also a classyfing map in $C$.
$C$ do not have a subobject classifier in the sense of what I would consider the correct definition: the map $4 \to \Omega$ classying $3 \subset 4$ does not have a pullback, indeed if the pullback $P$ existed there should be exactly three maps $1 \to P$, which is the case for no objects of $C$.
In particular, this is an example of a pullback in $C$ that does not exists.
$C$ is cartesian closed. If $X,Y \in C$ then their exponential $X^Y$ in Set is also in $C$ as $|X^Y|=|X|^{|Y|}=3$ has a unique solution given by $|X|=3$ and $|Y|=1$ hence never happen for $X \in C$. Again as $C$ is a full subcategory stable under product this implies that these are exponential objects in $C$.
Thank you so much for the answer, which has clarified the things.
|
2025-03-21T14:48:31.116598
| 2020-05-29T19:10:56 |
361685
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Loeffler",
"Peter Humphries",
"https://mathoverflow.net/users/2481",
"https://mathoverflow.net/users/3803"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629623",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361685"
}
|
Stack Exchange
|
Non-vanishing of L-values for twists of Hilbert modular forms
Let $F$ be a real quadratic field, and let $f$ be a Hilbert modular form over $F$ of parallel weight 2.
It's known, by a theorem of Rohrlich, that there exist infinitely many Hecke characters of $F$ such that $L(f, \chi, 1) \ne 0$. Does there always exist a Hecke character factoring through the norm map to $\mathbf{Q}$ such that this holds?
(There are some statistics due to Ryan et al, https://math.dartmouth.edu/~jvoight/articles/halfhilbert-022218.pdf, which suggest that this should be true for "almost all" quadratic twists coming from $\mathbf{Q}$; but just one character -- not necessarily quadratic -- would be enough for my purposes!)
I'm sure you already know this; such a character must be a genus character associated to a pair of primitive quadratic Dirichlet characters $\chi_1$, $\chi_2$ modulo fundamental discriminants $D_1, D_2$ such that $D_1 D_2$ is equal to the fundamental discriminant $D$ of $F$.
No, I don't already know that. Why is it true?
(This assertion, if true, would imply that there are only finitely many possible $\chi$. This would contradict the asymptotics of Ryan et al, so I'm a bit sceptical.)
Apologies, this is only the case when you are talking about class group characters. In general, the picture is as follows. Let $\omega : F^{\times} \backslash \mathbb{A}F^{\times} \to \mathbb{C}^{\times}$ be a Hecke character, and suppose that this factors through the norm map $\mathrm{N}{F/\mathbb{Q}}$: $\omega = \chi \circ \mathrm{N}{F/\mathbb{Q}}$ for some Hecke character $\chi : \mathbb{Q}^{\times} \backslash \mathbb{A}{\mathbb{Q}}^{\times} \to \mathbb{C}^{\times}$...
... Then the automorphic induction of $\omega$ is the Eisenstein series $\chi \boxplus \chi \chi_{F/\mathbb{Q}}$, where $\chi_{F/\mathbb{Q}} : \mathbb{Q}^{\times} \backslash \mathbb{A}_{\mathbb{Q}}^{\times} \to \mathbb{C}^{\times}$ is the quadratic Hecke character associated to $F$.
This shows that you can index such characters by Dirichlet characters. An analytic number theorist's approach now would be to consider the average of $|L(f,\chi,1)|^2$ over Dirichlet characters of conductor at most $Q$ and hope to prove an asymptotic as $Q \to \infty$. The size of the indexing family is roughly $Q^2$, whereas the conductor of the square of the $L$-function is roughly $Q^8$, so the ratio of the log of conductors is $4$, which is precisely the barrier at which we can hope to prove an asymptotic. (I.e. if it were $>4$ then we probably wouldn't stand a chance.)
|
2025-03-21T14:48:31.116793
| 2020-05-29T20:04:14 |
361690
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ivan Di Liberti",
"Simon Henry",
"fosco",
"https://mathoverflow.net/users/104432",
"https://mathoverflow.net/users/152679",
"https://mathoverflow.net/users/22131",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/7952",
"varkor",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629624",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361690"
}
|
Stack Exchange
|
Finitary monads on $\operatorname{Set}$ are substitution monoids. Finitary monads on $\operatorname{Set}_*$ are...?
$\DeclareMathOperator\Fin{Fin}\DeclareMathOperator\Lan{Lan}\DeclareMathOperator\Set{Set}$
The present question is intimately related to another question.
It is well known that the category of functors $\Fin \to \Set$ is equivalent to the category of finitary endofunctors $\Set \to \Set$; in this equivalence, finitary monads correspond to what are called substitution monoids on $[\Fin,\Set]$, i.e., to monoids with respect to the monoidal structure
$$
F \diamond G = m\mapsto \int^n Fn \times G^{*n}m \tag{$\star$}
$$ where $G^{*n}$ is the functor
$$
m \mapsto \int^{p_1,\dotsc, p_n} Gp_1 \times \dotsb \times Gp_n \times \Fin(\sum p_i, m).
$$
More precisely, the equivalence $[\Set,\Set]_{\omega} \cong [\Fin,\Set]$ can be promoted to a monoidal equivalence, and composition of endofunctors corresponds to substitution of presheaves in the following sense: let $J : \Fin \to \Set$ be the inclusion functor, then
$$
\Lan_J(F\diamond G) \cong \Lan_JF \circ \Lan_JG \tag{$\heartsuit$}
$$
and
$$
(S\circ T) J \cong SJ \diamond TJ\tag{$\clubsuit$}
$$ for two finitary endofunctors $S,T : \Set \to \Set$. (Kan extending along $J$ and precomposing an endofunctor of $\Set$ with $J$ is what defines the equivalence.)
I would like to prove the same exact theorem, replacing everywhere the cartesian category of sets with the monoidal category of pointed sets and smash product, but I keep failing.
The equivalence of categories
$$
[\Fin_*,\Set_*]\cong [\Set_*,\Set_*]
$$ remains true; and this equivalence must induce an equivalence between the category of finitary monads on pointed sets, and the category of suitable "pointed substitution" monoids, that are obtained from the iterated convolution on $[\Fin_*,\Set_*]$ as
$$
F\diamond' G = m \mapsto \int^n Fn \land G^{*n}m
$$ where $\land$ is the smash product, and $G^{*n}$ iterates the convolution on $[\Fin_*, \Set_*]$ induced by coproduct on domain, and smash on codomain:
$$
G^{*n}m = \int^{p_1,\dotsc,p_n} Gp_1 \land \dotsb\land Gp_n \land \Fin_*(\bigvee p_i, m)
$$ where $\bigvee p_i$ is the coproduct of pointed sets, joining all sets along their basepoint.
This would be the perfect equivalent of $(\star)$.
However, trying to prove the isomorphisms $(\heartsuit)$, $(\clubsuit)$, I find that it is not true that $\Lan_J(F\diamond G) \cong \Lan_JF \circ \Lan_JG$. I am starting to suspect that the generalisation is false as I have stated it, or that it is true in a more fine-tuned sense.
So, I kindly ask for your help:
To what kind of monoids on $[\Fin_*,\Set_*]$ do finitary monads on pointed sets correspond?
Edit: I am led to believe this construction is not a particular instance of an enriched Lawvere theory, because in that framework a theory isn't what it is in mine:
for Power, if $\mathcal V$ is a locally finitely presentable base of enrichment, a theory is an identity on objects functor $\mathcal V_\omega ^{op}\to \mathcal L$ from the subcategory of finitely presentable objects to $\mathcal L$ strictly preserving cotensors; if $\mathcal V = \Set_*$ with smash product, cotensors in $\mathcal V^{op}$ are tensors in $\mathcal V$, thus smash products.
Instead, for me, a theory is a bijective-on-objects functor $\Fin_* \to \mathcal L$ that sends coproduct into smash product (or even a more general monoidal structure on $\mathcal L$).
Or at least, this is what I was led to believe trying to back-engineer the equivalence between finitary monads and Lawvere theories in the case of pointed sets.
Could it be the case that the equivalence works out in the $\mathrm{Set}*$-enriched setting, giving a correspondence between $\mathrm{Set}$-enriched cartesian operads (or substitution monoids) and finitary $\mathrm{Set}_$-monads on $\mathrm{Set}_*$? It looks like your setting is one in which sets are being consistently replaced with pointed sets, which seems suggestive of the setting for enriched Lawvere theories. But perhaps there's an obvious reason this doesn't work out.
Set is Ind(Fin). Is Set$*$ Ind(Fin$*$)?
@მამუკაჯიბლაძე Yes, it is.
@varkor I will address your question tomorrow. Thanks for the attention :) Power's enriched Lawvere theories framework isn't exactly what it's needed here. Because you would have to take the monoidal structure given by smash on Fin_* too
@varkor I edited the question!
Interesting. Your theories are a little unusual, as they no longer look like a class of categories equipped with specified structure and generators, as in the traditional setting. If you haven't checked already, it would be worth comparing your setting to that of Bourke & Garner's, which establishes theory–monad correspondences in a great level of generality (potentially at the cost of some massaging to check your definitions align with theirs).
A distinctive feature of algebraic theories is that they are monoidal functors (in the case of classical Lawvere theories, cartesian). These correspond in turn to promonoidal promonads on the associated arity. Is it the case that if an arity ${\cal A} \hookrightarrow {\cal E}$ is chosen to be monoidal (i.e., $\cal E$ is monoidal, e.g. the base of enrichment, and $\cal A$ is $\otimes$-closed), the $\cal A$-nervous monads must preserve this structure (thus, bestowing theories or the associated promonads with monoidality)?
It's unclear to me whether or not this is true: the definitions are very abstract, so it's difficult to get an intuition without explicitly working through the example. I notice that Bourke & Garner do not seem to give any examples in this vein, but it's not clear whether this is a deliberate omission. I personally would be interested to know the outcome if you do work through it.
It seems to me if you want this to work, as you use the monoidal structure on pointed sets, you everything to be enriched in pointed sets. It means that you should restrict to $Set_*$-enriched functor, which are exactly the functor preserving the $0$-objects. Once you impose this restriction everything should work.
Rather than checking to see if that is a substitution monoid, I think you might have an easier time using Rory Lucyshyn-Wright’s notion of a eleutheric system of arities, seen here. It’s a relatively straightforward condition to check, with several equivalent statements.
Edit: To give a (very brief) description of how this works out: a monoidal subcategory of $\mathcal{V}$ is a symmetric monoidal subcategory $j:\mathcal{J} \to \mathcal{V}$, and a $\mathcal{J}$-ary Lawvere theory is a $\mathcal{V}$-category with a bijective-on-objects, power-preserving functor $\mathcal{J} \to \mathcal{T}$. A system of arities is eleutheric if for every $T: \mathcal{J} \to \mathcal{V}$, the left Kan extension $Lan_jT$ exists and is preserved by $\mathcal{V}(J,-), J \in \mathcal{J}$. This ends up being just enough to ensure that every $\mathcal{J}$-ary theory induces a free $\mathcal{T}$-algebra monad on $\mathcal{V}$.
|
2025-03-21T14:48:31.117291
| 2020-05-29T20:30:40 |
361691
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Clement C.",
"Yuval Peres",
"https://mathoverflow.net/users/37266",
"https://mathoverflow.net/users/7691"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629625",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361691"
}
|
Stack Exchange
|
Distribution of a stopped random sum, with subexponential stopping time
I am trying to find a reference (or, if it's false, a counterexample) for the following sort-of-intuitive fact: if $\tau$ is a stopping time with a subexponential probability distribution, and $(X_n)_{n\geq 1}$ are independent r.v.'s, also subexponential, then $\sum_{n=1}^\tau X_n$ aso has a subexponential distribution.
Specifically, I would like to know if the following statement is known:
Let $(X_n)_{n\geq 1}$ be independent random variables satisfying $\mathbb{E}[e^{X_n}] \leq 1$ for all $n$, and $\tau$ be a stopping time. Suppose $\mathbb{E}[e^{\alpha \tau}] \leq e^\beta$ for some $\alpha >0$ and $\beta<\infty$. Then $\mathbb{E}[e^{\sum_{n=1}^\tau X_n}] \leq 1$.
The hypothesis implies that $M_k=[e^{\sum_{n=1}^k X_n}]$ is a supermartingale, with $M_0=1$. Then the optional stopping theorem for positive supermartingales implies the requested inequality. (see e.g. Williams' book "Probability with martingales").
Note that no moment conditions on the stopping time $\tau$ are needed,
just that it is an almost surely finite stopping time. Alternatively, look up "Wald's third identity"and apply it to the independent variables $e^{X_n}/[\mathbb{E}e^{X_n}]$ and the given stopping time. Note that the case of a general stopping time follows from the case of a bounded stopping time via Fatou's lemma.
Thank you! I can't believe it was so simple -- I got stuck at the almost sure boundedness assumption for $\tau$ in Wald's third identity. If I understand correctly, the assumption on $\tau$ can even be relaxed: all that's needed is almost sure finiteness.
You need some assumption on $\tau$ for Wald's identity to hold. But what you asked is an inequality, so you could apply Wald to $\min{\tau,m}$ and then let $m \to \infty$ using Fatou.
|
2025-03-21T14:48:31.117435
| 2020-05-29T20:38:14 |
361693
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Pietro Majer",
"https://mathoverflow.net/users/6101"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629626",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361693"
}
|
Stack Exchange
|
minimizing weighted length of closed curves
Let $\mathcal{A}$ be the family of closed smooth curves in the right half of the complex plane $\mathbb{C}$ such that any curve in the family must enlose the point $z=1$ and tangent to the $y$-axis at the origin. Then we define the weighted length of curves in the family as
$$L(\gamma):= \int_{\gamma} \frac{2}{1+|z|^2}d|z|,$$where $d|z|$ is the classical length element.
My question is that, is it true that $\inf_{\gamma \in \mathcal{A}}L(\gamma) =\pi$?
I have done some computations for some curves with explicit formulas. For example, if $\gamma$ is a unit circle centered at $z=1$, then $\gamma \in \mathcal{A}$ and $L(\gamma)=4\pi/\sqrt{5}$. Also, it seems that when $\gamma$ is more and more closed to the line segment $[0,1]$ with multiplicity 2, the weighted length $L(\gamma)$ is getting smaller and approaching to $\pi$.
Any ideas or comments are really appreciated. Thank you very much for your time.
You may compare any path $\gamma$ in your minimization class with the path $|\gamma|\wedge1\in[0,1]$. The value of $L$ should decrease.
I just found out two proofs. One is analytic, and the other is geometric.
For the analytic proof, one can use polar coordinates and apply some elemetary inequality. But such proof has limited applications. so I'm presenting a geometric proof.
Note that the metric of the unit sphere is given by $g=\frac{4}{(1+|z|^2)^2}\delta$, where $z$ is the coordinate of point on sphere obtained using stereographic projection map from the north pole, and $\delta$ is the Euclidean metric. Hence $L(\gamma)=l_g(\Pi^{-1}(\gamma))$, where $\Pi$ is the stereographic projection map and $l_g$ is the length function on the unit sphere. For any $\gamma \in \mathcal{A}$, since $\gamma$ enloses the point $z=1$, $\Pi^{-1}(\gamma)$ must be a closed curve on sphere suth that it starts at the south pole, and it must goes to some point in the northern hemisphere and then goes back to the south pole. Therefore, $\Pi^{-1}(\gamma) \ge \pi$, and $"="$ holds if and only if $\gamma$ is a line segment connecting $z=0$ and $z=1$.
|
2025-03-21T14:48:31.117605
| 2020-05-29T21:21:52 |
361694
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Giorgio Metafune",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/41291",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629627",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361694"
}
|
Stack Exchange
|
Set of eigenvalues of the boundary problem
I'm looking for the results about the set of eigenvalues of boundary problem for differential equation
\begin{equation}
\bigl(p(x) u'(x; \lambda) \bigr)' + q(x) u(x; \lambda) = -\lambda w(x) u(x; \lambda), \quad x \in [0, h]
\end{equation}
where $p(x) > 0$, $q(x) > 0$ and $w(x) < 0$, with boundary conditions
\begin{equation}
u'(x; \lambda)|_{0} = u'(x; \lambda)|_{h} = 0.
\end{equation}
It seems that there must be only finitely many positive eigenvalues of the problem. But I haven't found any book where such problem is considered.
Actually, this problem arises in the theory of electromagnetic waves propagation in plane shielded waveguides filled with inhomogeneous medium. In this case, the first equation has the form
\begin{equation}
\bigl(\frac{1}{\varepsilon(x)}u'(x; \lambda)\bigr)' + u(x; \lambda) = \lambda\frac{1}{\varepsilon(x)}u(x;\lambda);
\end{equation}
here $\varepsilon(x) > 0$ is a continuous function that describes permittivity of the medium. So, actually I'm interested particularly of this special case.
Dividing the equation by $-w$ you get a Sturm-Liouville problem which can be studied in the weighted space $L^2_w$. A. Zettl has written a very extensive and detailed book on the topic.
I've managed to find the necessary information in the book "Sturm-Liouville Theory" by Anton Zettl. This book was recommended to me by @Giorgio Metafune in his comment.
Please consider that questions/answers here are not only for the OP to use. In its present form your answer is of very poor quality for any potential reader except yourself.
|
2025-03-21T14:48:31.117739
| 2020-05-29T21:35:55 |
361696
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dirk Werner",
"Willie Wong",
"https://mathoverflow.net/users/127871",
"https://mathoverflow.net/users/213802",
"https://mathoverflow.net/users/33741",
"https://mathoverflow.net/users/3948",
"leo monsaingeon",
"mathishard.butweloveit"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629628",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361696"
}
|
Stack Exchange
|
Dual space of mean-free Sobolev space
I am considering the space $V:=\{v \in H^1(\Omega): \int_\Omega v = 0\}$ of mean free functions. What is the dual space of this space? Is the dual space given by $D:= \{f \in (H^1(\Omega))^*: \langle f, 1\rangle=0\}$? If so, why?
I understand that the dual space $H^1(\Omega)^*$ of $H^1(\Omega)$ consists of all linear functionals, i.e., of mappings from $H^1(\Omega)$ to $\mathbb{R}$. Furthermore, I understand that $\langle f,v\rangle$ denotes the duality pairing between $H^1(\Omega)^*$ and $H^1(\Omega)$ and can be computed by $\langle f,v\rangle = \int_\Omega f v$.
Then, the dual space $V^*$ of $V$ should contain all linear functions from $V$ to $\mathbb{R}$. In my opinion, it should hold, that if $f \in D$ is a constant function, then $\langle f,v \rangle = \int_\Omega fv = f \int_\Omega v = 0 \quad \forall v \in V$. How can I define this property of $f$ in $V^*$ and is this equivalent to $\langle f,1 \rangle=0$?
Frieder
Every element $f\in H^1(\Omega)^$ is a linear functional acting on $V$. However, notice that if $f, g \in H^1(\Omega)^$ such that $f - g = C$, then $\int fv = \int gv$ for every $v \in V$, so $f$ and $g$ should represent the same element in $V^*$.
Are you not happy with $V^$ being just the restriction of $(H^1)^$ to test-functions in $V$? For example in finite dimension we know the dual of $\mathbb R^3$, and if $V=\mathbb R^3\cap{x_3=0}\cong \mathbb R^2$ then the dual is just the restriction of $(\mathbb R^3)^*\cong \mathbb R^3$ to the hyperplane. Is that not enough? As in Willie Wong's comment, the full characterization is then that if two cotangent vectors $f=(f_1,f_2,f_3)$ and $f'=(f'_1,f'_2,f'_3)$ have the same "projection" $(f_1,f_2)=(f_1',f_2')$ then $f=f'$ as elements of $V'$ (which has an obvious geometric interpretation)
In abstract terms, if $V$ is a closed subspace of $W$, then $V^$ is isometric to the quotient space $W^$ mod the annihilator of $V$. In your case, $V$ is the kernel of a continuous linear functional $\ell$, hence the annihilator consists of the span of $\ell$. Please note that it is a little touchy to say that the duality ``can be computed by $\langle f,v \rangle = \int_\Omega fv$''; in dimension 1 the functional $f$ might be a point evaluation.
For anyone looking for a short, compact answer to this question, see the nicely written section 2.1.1 in F. Dubois et al 2003: doi:10.1016/j.matpur.2003.09.002
|
2025-03-21T14:48:31.117943
| 2020-05-29T21:41:59 |
361698
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ali",
"S.Surace",
"https://mathoverflow.net/users/50438",
"https://mathoverflow.net/users/69603"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629629",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361698"
}
|
Stack Exchange
|
conformal changes to Lorentzian curvature
Let $(M,g)$ be a Lorentzian manifold and let $R$ be the curvature tensor. We say $R\leq 0$ if
$$ g(R(X,Y)Y,X) \leq 0\quad \forall \, X,Y \in TM.$$
My question is whether given a Lorentzian manifold $(M,g)$, it is always possible to find a metric $\hat{g}=cg$ such that the curvature of $(M,\hat{g})$ is non-positive.
I suppose you want $X,Y$ to be sections, not elements, of $TM$. What is $c$?
No, this is not possible if the Lorentzian manifold has a null geodesic admitting a pair of conjugate points.
The notion of conjugate points for null geodesics does not depend on the conformal factor.
By the Lorentzian version of the Cartan-Hadamard theorem (see e.g. the book by Beem et al. 1996 Global Lorentzian Geometry. Prop. 11.13) in a spacetime if the inequality you wrote were to hold for any pair of timelike vectors then there would not be causal geodesics with conjugate points.
It makes me wonder if this is the only obstruction, namely if the inequality can be achieved if one removes null conjugate and cut points.
|
2025-03-21T14:48:31.118047
| 2020-05-29T21:56:59 |
361699
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerry Myerson",
"HenrikRüping",
"Per Alexandersson",
"https://mathoverflow.net/users/1056",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/3969"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629630",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361699"
}
|
Stack Exchange
|
Square root of a sequence given by a linear recurrence relation
This question is closely related to this one but in a sense a converse of it. Let us concentrate for simplicity on a third order relation $x_{n+3}=ax_{n+2}+bx_{n+1}+cx_n$ with given intial terms $x_1,x_2,x_3$ and assume $x_n\geq0$ for all $n$. Assume we fix $a,b,c\in\mathbb{R}$ generically and vary $x_1,x_2,x_3\in\mathbb{R}$. We are looking for a second order recurrence $y_{n+2}=ry_{n+1}+sy_n$ such that $y_n^2 = x_n$ for all $n$. For generic $a,b,c$ there is only a finite number of options for the unknown coefficients $(r,s)$. Hence, only for $(x_1,x_2,x_3)$ lying in a certain two-dimensional set can we find a second-order relation satisfied by $\pm\sqrt{x_n}$ (this is because $\pm\sqrt{x_{3}}=\pm r\sqrt{x_{2}}+\pm s\sqrt{x_{1}}$ has to hold). We restrict our attention to $(a,b,c)=(\alpha^2+\beta^2+\alpha\beta,-\alpha^2\beta^2-\alpha^3\beta-\alpha\beta^3,\alpha^3\beta^3)$ for some $\alpha,\beta\in\mathbb{C}$, since in that case we can find a solution $(r,s)=(\alpha+\beta,-\alpha\beta)$.
Now my question is simply: what can we say for general $(x_1,x_2,x_3)$? For a vast majority of $(x_1,x_2,x_3)$ no linear relation of the form $y_{n+2}=ry_{n+1}+sy_n$ will be satisfied for the square roots of $(x_n)$. But can it be that the square roots still have to satisfy a second order recursion of a more general form? Trivially, the roots satisfy a third order non-linear recursion (the initial one), so the question is basically to find out whether the ''order-collapse'' (passing from order 3 to order 2) happens indeed only for special $(x_1,x_2,x_3)$ and not generally. If this is the case, it should also be possible to prove it. Is there a good intuition as to why it should (or should not) hold?
Example 1: Here is a concrete example (see also Problem 6 here). Take $(a,b,c,x_1,x_2,x_3)=(m,m,-1,1,1,4)$. For $m=2,10$ the roots follow a linear second order recurrence with $(r,s)=(\sqrt{m-1},1)$. For other values of $m$ we would have to perturb the initial terms $(x_1,x_2,x_3)=(1,1,4)$ to ensure that the same recurrence still works. But what if we insist on keeping the initial terms as they are and take, say, $m=3$? Is the whole magic lost, or can we still find a relation for the roots that is closely related to $y_{n+2}=ry_{n+1}+sy_n$? Intuitively, I would hope that (e.g., for $m=3$) there is a more general (but still algebraically elegant) second order recursion that holds for any $(x_1,x_2,x_3)$, with coefficients depending on $(x_1,x_2,x_3)$, and that simply degenerates to $y_{n+2}=ry_{n+1}+sy_n$ for $(x_1,x_2,x_3)=(1,1,4)$.
Example 2 (from a comment below): Take $(a,b,c,x_1,x_2,x_3)=(0,0,1,1,4,9)$. We get $(\alpha,\beta)=(\exp(2\pi i/3),\exp(4\pi i/3))$, hence $(r,s)=(-1,-1)$, so the equation $y_{n+2}=-y_{n+1}-y_n$ will drive the square roots, however only for specific choices of the initial terms. E.g., $(x_1,x_2,x_3)=(1,4,9)$ obviously works, but the question is how to generalize the relation to an arbitrary choice of the initial parameters.
Bonus question: Above we have worked with reals. Now we assume additonally $x_1,x_2,x_3,a,b,c\in\mathbb{Z}$. If $\sqrt{x_n}\in\mathbb{Z}$ for all $n$, does it follow that the order-collapse has to occur (i.e., the square roots satisfy a second order (linear?) recurrence)?
Let's look at a simple example, $x_{n+3}=x_n$, $x_1=1$, $x_2=4$, $x_3=9$. Does the sequence $1,2,3,1,2,3,1,2,3,\dots$ satisfy a second-order (linear constant-coefficient) recurrence?
Let try it the other way around. Let start with a generic (say MiPo has distinct roots) second order recurrence $y_n=\lambda_1r_1^n +\lambda_2r_2^n$. If we square this, we get $\lambda_1^2(r_1^2)^n+2\lambda_1\lambda_2(r_1r_2)^n+\lambda_2^2(r_2^2)^n$, and thus the third order recurrence relation must have a minimal Polynomial with roots $r_1^2,r_2^2,r_1r_2$ and the coefficients must looks as above. One should be able to deal with the case where the minimal polynomial hasthe same root twice analogously.
One can check if a sequence satisfies a linear recurrence, by requiring that a certain determinant vanish. In this case, you will have a 3x3-matrix, where the entries are $\sqrt{x_i}$, $\sqrt{x_{i+1}},\dotsc$ (a Toeplitz matrix).
|
2025-03-21T14:48:31.118423
| 2020-05-29T21:59:43 |
361701
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David White",
"https://mathoverflow.net/users/11540"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629631",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361701"
}
|
Stack Exchange
|
Explaining the "free left fibration" functor for infinity categories
This is a cross-post from here
I am reading A. Mazel-Gee's paper "All about the Grothendieck construction". In that paper he explains that the left adjoint ${\mathrm{Cat}_{\infty}}_{/\mathcal{C}}\to \mathrm{coCFib}(\mathcal{C})$ (from $\infty$-categories over $\mathcal{C}$ to cocartesian fibrations over $\mathcal{C}$) to the forgetful functor is the functor that sends $F:\mathcal{D}\to \mathcal{C}$ to the "free cocartesian fibration on F"
$$\mathrm{Fun}([1],\mathcal{C})\times_{\mathcal{C}}\mathcal{D}\to\mathcal{C}$$
I am now wondering if there is a similar explicit description for the left adjoint ${\mathrm{Cat}_{\infty}}_{/\mathcal{C}}\to \mathrm{LFib}(\mathcal{C})$. This would be the composite of the previous functor with the reflexive localization $L:\mathrm{coCFib}(\mathcal{C})\to \mathrm{LFib}(\mathcal{C})$. Now by the results in the paper we have
a commutative diagram of large $\infty$-categories $$\require{AMScd}\begin{CD}\mathrm{Fun}(\mathcal{C},\mathrm{Cat}_\infty) @>{(=)^{gpd}\circ -}>> \mathrm{Fun}(\mathcal{C},\mathcal{S})\\
@V{Gr}V{\simeq}V @V{Gr}V{\simeq}V \\
\mathrm{coCFib}(\mathcal{C}) @>{L}>> \mathrm{LFib}(\mathcal{C}),\end{CD}$$ where $\mathcal{S}$ is the $\infty$-category of spaces, $Gr$ denotes the Grothendieck construction and $(=)^{gpd}$ is the groupoidification functor.
This implies by the naturality of the Grothendieck construction that the fibers of $L(\mathcal{D}\to\mathcal{C})$ over $x$ identify with $(\mathcal{D}_x)^{gpd}$. But it is not straight-up groupoidification as that would take us to $\mathcal{S}_{/\mathcal{C}^{gpd}}$. If I understand correctly the description of the Grothendieck construction as a lax colimit then the functor L should be some kind of "free groupoidification of the fibers". But this is not as explicit as I would like : can we describe this process without referring to the functor by which the coCartesian fibration is classified ?
On the level of model categories, this is presented by the Quillen adjunction $${\mathrm{Set}_{\Delta}^+}_{/\mathcal{C}^\sharp} \leftrightarrows {\mathrm{Set}_\Delta}_{/\mathcal{C}}$$ between the functor forgetting the marked edges and the functor marking all edges ; the model structures are the marked one and the covariant one, respectively. Therefore the functor $L$ is given by a fibrant replacement of a coCartesian fibration $\mathcal{D} \to \mathcal{C}$ in ${\mathrm{Set}_\Delta}_{/\mathcal{C}}$. Do we have explicit such replacements ?
Have you considered emailing Aaron? I think he is also sometimes active in MO Chat. Since you first asked 12 days ago, I think it's safe to email him.
As suggested by David White, I emailed A. Mazel-Gee. Let me paraphrase his answer :
We claim that given a cocartesian fibration $F:\mathcal{D}\to\mathcal{C}$, the free left fibration $LF:\mathcal{E}\to\mathcal{C}$ on $F$ is just given by inverting in $\mathcal{D}$ the morphisms sent to equivalences in $\mathcal{C}$. We will use Corollary 3.11 in this paper by Ayala & Francis. The natural map $\mathcal{D}\to \mathcal{E}$ is a map of coCartesian fibrations so we just have to check that the induced maps on fibers $\mathcal{D}_x\to\mathcal{E}_x$ for $x\in\mathcal{C}$ are localizations. But as I said in my original post, we have $\mathcal{E}_x=(\mathcal{D}_x)^{gpd}$ ; thus $\mathcal{D}\to \mathcal{E}$ is a localization. Now left fibrations reflect equivalences so a morphism in $\mathcal{D}$ get inverted in $\mathcal{E}$ if and only if it gets inverted in $\mathcal{C}$.
It's actually something you already know: It is the fibrewise groupoidification of the free cartesian fibration. The free cartssian fibration functor sends a functor $$p:A\to B\mapsto p': A\downarrow B\to B.$$ This is totally classical and due originally iirc to Ross Street. The thing to look for is the "slice 2-monad".
|
2025-03-21T14:48:31.118685
| 2020-05-29T23:17:40 |
361708
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629632",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361708"
}
|
Stack Exchange
|
Langlands dual group in math vs. Goddard-Nyuts-Olive dual group in physics
Given a group $G$, there is a so-called Langlands dual group $G^{∨}$.
Given a group $G$, there is also a so-called Goddard-Nyuts-Olive dual group $G^{'}$ that relates to the magnetic charge.
Are the two $G^{∨}$ and $G^{'}$ defined in different settings? Or are their definitions related? What are the intuitions and motivations behind their definitions?
Are the two $G^{∨}$ and $G^{'}$ exactly the same?
What are the constraints on $G$ to give such groups: $G^{∨}$ and $G^{'}$? (eg, compact or not, simple, semi-simple, connected, simply-connected, Lie group or not, topological group, etc.)
A Reference on Goddard-Nyuts-Olive dual group:
P. Goddard(Cambridge U.), J. Nuyts(UMH, Mons), David I. Olive(CERN and Bohr Inst.)
Dec 1, 1976, Published in: Nucl.Phys.B 125 (1977) 1-28
DOI: 10.1016/0550-3213(77)90221-8
https://www.sciencedirect.com/science/article/pii/0550321377902218?via%3Dihub
The groups $G^{\vee}$ and $G'$ are the same. It is clear from the description in the paper you refer to.
Their definitions are the same as well, just swapping roots and coroots. You can do it in compact Lie groups or reductive algebraic groups.
There is further definition of $G^{\vee}$ from geometric Satake. Its physical significance has been looked into by Kapustin and Witten. They also revisit the Goddard-Nuyts-Olive work, you refer to, from the hills of Geometric Langlands.
|
2025-03-21T14:48:31.118801
| 2020-05-29T23:19:05 |
361709
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Darth Vader",
"Jon Bannon",
"LSpice",
"https://mathoverflow.net/users/149852",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/6269"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629633",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361709"
}
|
Stack Exchange
|
Is $L(\mathbb{Z}*\mathbb{Z}_{2})$ a free group factor?
This is a reference request for something that is likely to be well-known to operator algebraists. I will not, therefore, include the technical definition of free product of finite von Neumann algebras, but instead refer the reader to Ching - Free products of von Neumann algebras for the definition.
Theorem 3.5 of Dykema - Interpolated free group factors (letting $A=L(\mathbb{Z})$ and $B=\mathbb{C}$) gives that $M_{2}(L(\mathbb{\mathbb{Z}}))*L(\mathbb{Z}_{2})\cong M_{2}(L(\mathbb{F}_{3}))$. Is it known whether or not $L(\mathbb{Z}*\mathbb{Z}_{2})\cong L(\mathbb{Z})*L(\mathbb{Z}_{2})$ is a free group factor, or interpolated free group factor?
I am, of course, interested in related results like the one quoted above, as well, if the original question is still unknown. Please feel free to provide references as answers.
This should just be a comment- but for some reason I couldn't add a comment.
It seems to me that using Corollary 5.3 of this paper by Dykema, we indeed get a positive answer to your question.
Corollary 5.3 states that $L(G \ast H) \cong L(F(2-|G|^{-1}-|H|^{-1}))$, if $G$ and $H$ are nontrivial amenable groups, with $|G|+|H| \geq 5$.( $\infty ^{-1}=0$).
So $L(\mathbb Z \ast \mathbb Z_2)= L(F(1.5))$ according to the above formula (provided I subtracted correctly).
EDIT: I also found Theorem 1.1 in this paper to be very interesting. It relates to reduced $C^{\ast}$-algebras of free product groups.
Perfect! Thanks @Darth Vader!
I had JUST downloaded this paper of Dykema to have a look. Thanks for saving me the work.
@JonBannon you are very welcome. Thank you for accepting the answer :).
It seems I can comment now- even though I couldn't a few minutes ago :P. Nevertheless, I'm happy that the question got resolved.
@DarthVader, there's a reputation cut-off for commenting, so your upvotes are the reason you can now comment: https://mathoverflow.net/help/privileges/comment . (Also, I think that you can always comment on your own posts.)
@LSpice: Thank you- that probably explains it :).
|
2025-03-21T14:48:31.118967
| 2020-05-29T23:41:26 |
361711
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"R. van Dobben de Bruyn",
"YCor",
"https://mathoverflow.net/users/108274",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/82179",
"user267839"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629634",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361711"
}
|
Stack Exchange
|
When $K[X_1,X_2,...,X_n] \to K[Y_1,Y_2,...,Y_m]$ is a flat morphism
Let $K$ be a field and $\varphi: K[X_1,X_2,...,X_n] \to K[Y_1,Y_2,...,Y_m]$ a polynomial $K$-algebra morphism. Assume $n, m \ge 2$. By definition $\varphi$ endows $K[Y_1,Y_2,...,Y_m]$ with a $K[X_1,X_2,...,X_n]$-module structure.
Are there any available criteria to decide when the $K[X_1,X_2,...,X_n]$-module $K[Y_1,Y_2,...,Y_m]$ induced in this way by $\varphi$ a flat $K[X_1,X_2,...,X_n]$-module?
I want also to note that this generalizes this MathSE question.
As far as we consider the case with more than one indeterminantes. The case $n=m=1$ always has a positive answer since $K[X]$ is a PID and in this setting flat = torsion-free. For $n \ge 2$ $K[X_1,X_2,...,X_n]$ is not a PID, so the criterion is not applicable.
An approach is to use a lemma that states that if $R \to R', S \to S'$ are flat modules, then $R \otimes S \to R' \otimes S'$ is a flat $R \otimes S$-module. The point is that obviously not every polynomial map $\varphi: K[X_1,X_2,...,X_n] \to K[Y_1,Y_2,...,Y_m]$ arises from such "atomic" pieces $\varphi_i: K[X_i] \to K[Y_i]$ as tensor product $\bigotimes_i \varphi_i$.
So I'm asking if there exist approaches dealing with this problem or is it too broad?
The geometric way to look at it is that $\phi \colon \mathbf A^m \to \mathbf A^n$ is flat if and only if all fibres have dimension $m-n$. This follows from "miracle flatness"; see e.g. Tag 00R4.
I suspect you mean "$K$-algebra homomorphism" rather than just "ring homomorphism".
@YCor: Yes, thanks.
@R.vanDobbendeBruyn: Thank you, that's exactly what I was looking for!
|
2025-03-21T14:48:31.119095
| 2020-05-30T01:29:52 |
361713
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629635",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361713"
}
|
Stack Exchange
|
Quantiles of a Levy process
Let $X = \{ X_t \in {\bf R}, t \geq 0 \}$ be a 1-dimensional (real) Levy process. Suppose further that the distribution of $X_t$ is not concentrated on a grid. (This forces the distribution of $X_t$ to have a Lebesgue density).
For a fixed $p \in (0,1)$, let $Q_t(p)$ be the quantile function of $X_t$, i.e $$ {\bf P}(X_t \leq Q_t(p)) = p. $$
If $X$ is a symmetric process and $Y_t = X_t + \alpha t$, then $Q_t(1/2)$ is a linear function of $t$. Indeed, $Q_t(1/2) = \alpha t$.
Do Levy process have this property in general? In other words, for a general Levy process $X_t$, can one find a $p$ that makes $Q_t(p)$ a linear function of time?
First of all, $X$ being non-lattice is not enough for $X_t$ being absolutely continuous. A simple counter-example is $$X_t = \sum_{n=1}^\infty \frac{N_t^{(n)}}{n!} \, ,$$ where $N_t^{(n)}$ are independent Poisson processes. If $X$ is non-lattice and not a compound Poisson process, then the distribution of $X_t$ is continuous. However, the definition of the quantile $Q_t(p)$ does not depend on continuity $X_t$; what is needed here is that the support of $Q_t(p)$ is an interval. This is the case when $X$ has a Gaussian component or when the closed semigroup generated by the support of Lévy measure of $X$ is an interval. (A detailed discussion of distributional properties of a Lévy process can be found, for example, in Sato's book Lévy processes and infinitely divisible distributions.)
It is quite simple to see that $Q_t(\tfrac12)$ need not be linear. Consider first the usual Poisson process $X_t$. Then $Q_t(\tfrac12) = 0$ for $t < \log 2$, and $Q_t(\tfrac12) = 1$ when $\log 2 < t < T$, where $T$ satisfies $e^{-T} (1 + T) = \tfrac12$. In particular, $Q_t$ is not linear in this case.
If one insists on a Lévy process $X_t$ with a continuous distribution, then it is enough to add an "epsilon" of Brownian motion to the above example. This is intuitively clear, I hope, but a detailed argument would require a somewhat technical estimate.
|
2025-03-21T14:48:31.119244
| 2020-05-30T02:39:10 |
361714
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D.R.",
"David Roberts",
"GH from MO",
"Sam Hopkins",
"auspicious99",
"https://mathoverflow.net/users/112504",
"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/158836",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/519",
"https://mathoverflow.net/users/80427",
"naf",
"user779120"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629636",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361714"
}
|
Stack Exchange
|
Recent developments in the proof of Fermat's last theorem
I posted on Mathematics Stack Exchange, but was encouraged to post on MathOverFlow instead.
It has been 20 years since Fermat's last theorem was proved by Andrew Wiles.
Has there been any simplification in proof in the last 20 years?
What I do know is only that different proofs of Faltings's theorem were given by Vojta and Bombieri
and Khare-Wintenberger's proof of Serre's Conjecture gives a different proof of FLT (the latter was told to me in a comment on the Mathematics Stack Exchange).
Thanks in advance.
Ken Ribet at the 2020 JMM in Denver, CO, "A 2020 View of Fermat's Last Theorem": https://www.youtube.com/watch?v=mq9BS6S2E2k
I watched that video right away. The video of Ribet explaining it was exciting. About 50 minutes into that video, Ribet was talking about Khare-Wintenberger's proof of Serre's Conjecture.
So maybe based on what you saw in the video, you might like to add an answer to your own question, for the benefit of those who have not watched it?
The proof via Serre's conjecture allows one to avoid difficult results of Ribet and Langlands-Tunnel but requires other non-trivial input so it is not clear that it is really a simplification. But the proof is indeed conceptually simpler.
@auspicious99 Most of that video was a summary of Wiles's proof, not a simplification in proof. However, the answer to the question might be two results an alternate proof of faltings's theorem and a Serre Conjecture(If there's nothing else.).
What I'd like to see is a serious technical discussion about how exactly newer approaches make things simpler. What is no longer needed if Serre's conjecture is used instead? Etc. Ribet's talk was for a general mathematical audience, not for people who want gory details.
@DavidRoberts: Serre's 1987 Duke paper contains a proof that Serre's conjecture implies Fermat's last theorem. So if you accept Serre's conjecture, then you don't need anything beyond Serre's 1987 Duke paper. Of course the proof of Serre's conjecture relies on the original ideas of Taylor and Wiles.
@GH sure, but what goes into the proof of Serre's conjecture is the point. How simpler is Serre's paper+proof of Serre's conjecture vs Wiles+Taylor–Wiles?
@DavidRoberts I would also like to see an answer along the lines of what you suggested. Do you think starting a bounty would help? Or essentially copy-pasting the current question into another post in hopes of getting a better response?
@D.R. I think asking a new, focused question on the form "Is Serre's paper+proof of Serre's conjecture (and all prerequisites) simpler than the proof of FLT via Wiles+Taylor–Wiles (and all prerequisites)?" is worthwhile. Link back to this question if you do.
|
2025-03-21T14:48:31.119455
| 2020-05-30T02:49:10 |
361715
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629637",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361715"
}
|
Stack Exchange
|
Maximum in solution set to a Diophantine equation related to unit fractions
Some time ago, Kellogg communicated to Carmichael a result with an incomplete proof, which was soon after verified as correct. I do not recall the source but recall the result. Define
$$S_n = \{ (x_1,x_2 \dots x_n) \subset \mathbb{N}^n: \sum x_i^{-1} = 1\}.$$
And regard elements of $S_n$ identical up to a permutation as the same element. I am interested in the question:
For each $n$, what is the largest integer $l(n)$ such that it appears as a member of some element of $S_n$?
Define the sequence $u_n$ with $u_1 = 1$ and $u_i = u_{i-1}(u_{i-1}+1).$ Then Kellogg reasoned that $l(n)$ is bounded above by $u_n$.
The question of this post: As this was a very old result, has there since arisen a better description of $l(n)$? Exact values for certain arguments, lower bounds, sharper upper bounds?
$$1={1\over2}+{1\over2}={1\over2}+{1\over3}+{1\over6}={1\over2}+{1\over3}+{1\over7}+{1\over42}={1\over2}+{1\over3}+{1\over7}+{1\over43}+{1\over1806}=\cdots$$ so certainly the sequence $u_n$, which is $1,2,6,42,1806,\dots$, is a lower bound as well for $l(n)$.
$u_n$ is tabulated at the Online Encyclopedia of Integer Sequences.
A proof is attributed to D. R. Curtiss, On Kellogg's Diophantine problem, Amer. Math. Monthly 29 (1922), pp. 380-387.
Also, the OEIS entry says,
Using the methods of Aho and Sloane, Fibonacci Quarterly 11 (1973), 429-437, it is easy to show that $a(n)$ is the integer just a tiny bit below the real number $\theta^{2^n}-1/2$, where $\theta \approx 1.597910218$ is the exponential of the rapidly convergent series $\sum_{n\ge0} \log(1+1/a_n)/2^{n+1}$. For example, $\theta^{32} - 1/2 \approx 3263442.0000000383$.
Others who have written on this problem are
Tanzo Takenouchi, On an indeterminate equation, Proc. Phys.-Math. Soc. Japan(3), 3(1921) 78-92.
David Eppstein, Ten algorithms for Egyptian fractions, Mathematica. in Education and Research, 4(1995) 5-15. This article is also available on the web at http://www.ics.uci.edu/~eppstein/numth/egypt/intro.html – see also https://www.ics.uci.edu/~eppstein/numth/egypt/curtiss.html
O. T. Izhboldin & L. D. Kurlyandchik, Unit fractions, Proc. St. Petersburg Math. Soc., 111(1995) 193-200; MR 99m:ll024.
|
2025-03-21T14:48:31.119636
| 2020-05-30T03:16:20 |
361717
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dmitri Pavlov",
"Robert Furber",
"https://mathoverflow.net/users/402",
"https://mathoverflow.net/users/61785"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629638",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361717"
}
|
Stack Exchange
|
Baire category theorem for uncountable unions
Any compact Hausdorff space $X$ is a Baire space:
if the set $X$ is a meager set (meaning a countable union of nowhere dense subsets,
also known as a set of first category),
then $X$ is empty.
I am interested in analogues of this theorem for uncountable unions.
Specifically, suppose a compact Hausdorff space $X$
is partitioned into a disjoint family $\{Y_i\}_{i∈I}$ of nowhere dense subsets.
To exclude trivial counterexamples such as partitions into singleton subsets,
assume that for any subset $J⊂I$ the union $⋃_{i∈J}U_i$
is a set with the Baire property (meaning it is the symmetric difference
of an open set and a meager set).
If $I$ is countable, then the condition involving the Baire property is trivially satisfied.
Furthermore, any countable collection of nowhere dense subsets
can be easily adjusted to a countable disjoint collection of nowhere dense subsets with the same union
by replacing $Y_i$ with $Y_i∖⋃_{j<i}Y_j$.
Thus, the above assumption is indeed an analogue for uncountable unions
of the assumption of the Baire category theorem.
Under what additional conditions on $X$ (if any)
can we conclude that $X$ is empty?
If additional assumptions are necessary,
I am specifically interested in cases when $X$ is extremally disconnected
or even hyperstonean.
I do not want to impose any countability (or cardinality) assumptions on $X$,
e.g., being metrizable, separable, first countable, etc.,
as done (for example) in a related question about partitioning of Polish spaces.
I also do not want to impose any cardinality assumptions on $I$,
as is done in a related question about Baire spaces for higher cardinalities.
In fact, for hyperstonean spaces the answer is positive
if we assume the nonexistence of real-valued-measurable cardinals
(see Lemma 438B in Fremlin's Measure Theory, which proves a more general result),
which can be seen as evidence in favor of the positive answer to the above question.
The question is then whether the large cardinal hypothesis can be removed
if we assume $X$ to be compact and Hausdorff, and if necessary, extremally disconnected or hyperstonean.
The hyperstonean case can be dealt with using a result from Fremlin's Measure Theory. For every hyperstonean space $X$, we can find a semi-finite measure $\mu$ defined on the sets with the Baire property whose nullsets are exactly the meagre sets and which is inner regular with respect to compact subsets. Therefore $(X, \mathcal{BP}(X), \mu)$ (where $\newcommand{\BP}{\mathcal{BP}}\BP(X)$ is the $\sigma$-algebra of sets with the Baire property) is a compact semi-finite measure space, so we can apply Fremlin's Lemma 451Q. Specialized to this case, this states that if $(E_i)_{i \in I}$ is a pairwise disjoint family of sets in $\BP(X)$ such that for each $J \subseteq I$ we have $\bigcup_{i \in J}E_i \in \BP(X)$, then $\mu\left(\bigcup_{i \in I}E_i\right) = \sum_{i \in I}\mu(E_i)$. In particular, if each $E_i$ is a meagre set, (so $\mu(E_i) = 0$ for all $i \in I$), then $\bigcup_{i \in I}E_i$ is meagre.
[Removed wrong suggestion for a different proof.]
In the absence of the axiom of choice, it is consistent that there is a counterexample to the question for compact Hausdorff spaces. An example is given by the partition of $[0,1]$ into singletons in Shelah's model where all subsets of $\mathbb{R}$ have the Baire property.
Thanks a lot, Fremlin's Lemma 451Q indeed resolves the hyperstonean case! (The numbering appears to be different in the latest version.)
I have a question about your second proof, specifically, about the claim “Integration against a complex finite measure ν:BP(X)→C defines a normal linear functional on L∞(X) iff ν vanishes on meagre sets.” Consider the hyperstonean space H given by the Stone spectrum of the Boolean algebra 2^κ, where κ is a real-valued-measurable cardinal. Consider also a probability measure 2^κ→[0,1] that vanishes on all countable subsets of κ. Passing to H, we obtain a finite measure ν:BP(X)→C that vanishes on all meager subsets of H.
The union H' of all clopen subsets of H corresponding to finite subsets of κ is a dense open subset H'⊂H. The measure ν vanishes on all such clopen subsets. Thus ν cannot be a τ-smooth measure because v(H')=ν(H)≠0. In particular, ν is not a Radon measure and the integration functional cannot be normal, since the ν-integral of the characteristic function of an open subset coincides with its ν-measure.
@DmitriPavlov You are quite right - I went off half-cocked with that second "proof", I will edit the answer to remove it. Of course $\mathcal{BP}(\beta(\kappa)) = \mathcal{P}(\beta(\kappa))$. The measure $\nu$ has to be inner regular with respect to compact sets to define a normal state. (I will also fix the numbering of Fremlin's lemma, he added 451L at some point and bumped up all the later numbers.)
|
2025-03-21T14:48:31.119943
| 2020-05-30T06:52:01 |
361723
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629639",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361723"
}
|
Stack Exchange
|
Graphs with vanishing homology and behaviour of the suspension of that graphs
Let $G$ be a simple graph. The clique complex, $\Delta(G)$ of the graph $G$ is the simplicial complex given by the collection of all complete subgraph of $G.$ Now we define the graph homology $H^{Gr}_\ast(G) : = H_\ast(\Delta(G))$ (the simplicial homology of $\Delta(G)$).
Next, define a contractible graph as follows:
A family $\mathcal{F}$ of graphs $G_1, G_2, \cdots, G_n ,\cdots$ is called contractible if
(1) The trivial graph, $\ast \in \mathcal{F}.$
(2) Any graph of $\mathcal{F}$ can be obtained from the trivial graph by finite series of contractible transformations $\{T_1, T_2, T_3, T_4\}$
where $T_1$: deleting of vertex $v$. A vertex $v$ of a graph $G$ can be deleted, if $N_G(v):= \{ u \in V(G): \text{ the edge }[uv] \in E(G)\} \in \mathcal{F}.$
$T_2:$ Gluing of a vertex $v$. If a subgraph $G_1$ of a graph $G$ is contractible, $G_1 \in \mathcal{F}$ the the vertex $v$ can be glued to the graph $G$ in such a manner that $N_G(v) =G_1,$
$T_3:$ deleting of an edge $[v_1v_2]$. The edge $[v_1v_2]$ of $G$ can be deleted if $N_G(v_1)\cap N_G(v_2)\in \mathcal{F}.$
$T_4:$ Gluing of an edge $[v_1v_2]$. Let two vertices $v_1$ and $v_2$ of a graph $G$ be non-adjacent. The edge $[v_1v_2]$ of $G$ can be glued if $N_G(v_1)\cap N_G(v_2)\in \mathcal{F}.$
Any graph $G \in \mathcal{F}$ is called a contractible graph.
$\mathbf{Question:}$ Let $G$ be a graph such that $H^{Gr}_\ast(G) =0 $ for $\ast>0$ and $\mathbb{Z}$ for $\ast=0.$ Then can we prove that the suspension graph $S(G)$ (suspension is considered as the topological unreduced suspension on the vertices of $G$) is contractible in above sense.
Thank you so much in advance. Any help will be appreciated.
|
2025-03-21T14:48:31.120085
| 2020-05-30T07:17:13 |
361725
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"CNS709",
"Denis Nardin",
"Mark Grant",
"Thomas Rot",
"Tyrone",
"https://mathoverflow.net/users/12156",
"https://mathoverflow.net/users/137622",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/54788",
"https://mathoverflow.net/users/8103"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629640",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361725"
}
|
Stack Exchange
|
Set of all sections of a fiber bundle up to homotopy equivalence
Let $\pi: E \to B$ be a fiber bundle of (topological or differentiable) manifolds. Denote by $[B, E]_{\pi}$ the set of all homotopy classes of sections of the bundle, i.e
\begin{align}
[B, E]_\pi &= \{\sigma: B \to E \ | \ \pi\sigma = \text{id}_B \}/\sim \\
\sigma \sim \sigma' &\iff \exists H: I \times B \to E \ | \ H_0 = \sigma, H_1 = \sigma', \pi H_t = \text{id}_B
\end{align}
Is it known how to calculate such set? With "calculate" I mean to reduce the computation of it to the computation of something more known, as the homology/cohomology/homotopy groups of $E$, $B$ or some combination of them.
Try H. Baues's Obstruction Theory if the more modern texts don't have what you need. In $\S$ 6 he constructs a spectral sequence derived from a Moore-Postnikov decomposition which seems to be what you are asking for.
The keyword is indeed obstruction theory, but in practice this will be very difficult. E.g. if B=S^n\times S^k, with fiber S^k, you are asking for a computation of the homotopy groups of spheres.
@ThomasRot That's indeed my goal: reduce the computation to something well known (or well studied, at least, as the homotopy groups of spheres)
@Tyrone Thank you for the reference, I will look at it (maybe it will be a bit long). Do you know if there is some simplification if $B$ is a sphere?
The machinery is the same, but the computations are obviously going to be simpler. As Thomas points out, the keyword is 'obstruction theory'. If you haven't seen it before then there are better starting places than Baues's book. For instance, Davis-Kirk's 'Lecture Notes in Algebraic Topology' has a good introductory chapter on the subject, as does Hatcher's book. Arkowitz's 'Introduction to Homotopy Theory' constructs Moore-Postnikov factorisations. These are all good starting points you should consult first, but they don't treat the more advanced topics like non-simple fibrations.
A good introduction to obstruction theory is the first few chapters of Mosher and Tangora's book.
@ThomasRot Thank you very much! Where can I find what you state? Thanks everyone
Look at GW Whitehead's "Elements of Homotopy Theory", Section VI.6. In particular, Corollary 6.16 says (once you unpack the notation, and assume $K=B$ and $L=\varnothing$) that if the fibre $F$ is $(n-1)$-connected and $B$ has dimension at most $n$, then homotopy classes of sections (if they exist) are in bijection with $H^n(B;\pi_n(F))$. Here $B$ may not be $1$-connected and so coefficients may be twisted.
@MarkGrant Thank you and thank you all. I didn't know it was so complicated. Unfortunately I am interested in the fibration $RP^n \to PTS^n \to S^n$ of the projectivized tangent bundle, in order to classify its sections, i.e the Lorentzian metric on $S^n$. Unfortunately $RP^n$ is not $(n-1)-$connected. What is the procedure in this case?
The relevant (simple) obstruction theory mentioned in the comments is contained in G. W. Whitehead's Elements of Homotopy Theory, Section VI.6. There an answer to the more general lifting problem is obtained under certain conditions, one of which is that the fibre $F$ is $q$-simple for certain values of $q$ (meaning $\pi_1(F)$ acts trivially on $\pi_q(F)$). The answer is then given in terms of cohomology with local coefficients.
In the non-simple case, you might check the papers of P. Olum (or the work of H. J. Baues citeed in the comments), but the answer is likely to be complicated.
However, you mentioned in the comments that you are interested in the case of the projectivized tangent bundle
$$
\mathbb{R}P^{n-1} \to PTS^n \to S^n
$$
of the $n$-sphere. A section of this bundle is also known as a line field on $S^n$. It is well known that there exists a line field on a closed manifold $N$ if and only if the Euler characteristic $\chi(N)$ is zero (this is proved as Theorem 2.3 in https://arxiv.org/abs/1612.04073, for example). Hence there exist line fields on $S^n$ if and only if $n$ is odd.
In the case $n$ odd, the fibre $\mathbb{R}P^{n-1}$ is not $(n-1)$-simple (the acion of $\pi_1(\mathbb{R}P^{n-1})=\mathbb{Z}/2$ on $\pi_{n-1}(\mathbb{R}P^{n-1})=\mathbb{Z}$ is non-trivial) so the standard obstruction theory does not apply to classify the sections. However, the more general problem of classifying line fields on a closed $n$-manifold $N$ up to homotopy, i.e. classifying sections of the projectivized tangent bundle
$$
\mathbb{R}P^{n-1} \to PTN \to N
$$
up to vertical homotopy, has been attacked by Koschorke in
Koschorke, Ulrich, Homotopy classification of line fields and of Lorentz metrics on closed manifolds, Math. Proc. Camb. Philos. Soc. 132, No. 2, 281-300 (2002). ZBL0994.57024.
There it is claimed (in Example 1.7) that the number of line fields on $S^n$ up to homotopy is given by
$$
\ell(S^n) = \begin{cases} 0 & n\mbox{ even}\\
1 & n\equiv 1(4)\\
2 & n\equiv 3(4), n\ge7\\
\infty & n=3. \end{cases}
$$
I was wondering about the infinitely many line fields on $S^3$, but this is clear: Since $S^3$ is parallelizable, homotopy classes of line fields are just homotopy classes of maps $S^3\to \mathbb{R}P^2$, and $\pi_3(\mathbb{R}P^2)\cong \pi_3(S^2)\cong \mathbb{Z}$.
Yes, this works for $n=1,3,7$ ($S^n$ parallizable) and it gives one example for type (together with $n$ even). $[S^1, RP^0] = e, [S^7, RP^6] = [S^7, S^6] = \pi_1^s = Z_2$. I will definitively look at that paper for the other cases, thank you. I was hoping there'd be more "classical methods"
|
2025-03-21T14:48:31.120566
| 2020-05-30T08:40:03 |
361727
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Francois Ziegler",
"https://mathoverflow.net/users/19276"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629641",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361727"
}
|
Stack Exchange
|
Conjugacy classes in reductive group under adjoint action of parabolic subgroup
Given a reductive group $G$ over a finite field and a parabolic subgroup $P$ , I wonder what are the orbits in $G$ under the adjoint action of $P$. This should be standard, but I can only find results on the orbits inside $P$.
Actually, I would like to know which (say) permutation representations of the parabolic $P\to \operatorname{Sym}(\Omega)$ admit an extension to a $P$-equivariant map of sets $G\to \operatorname{Sym}(\Omega)$. But I do not know in which context this question falls.
I would already be very happy for an answer in the case $G=\operatorname{GL}_3(\mathbb{F}_p)$ and the $(2,1)$- parabolic $P=(\operatorname{GL}_2(\mathbb{F}_p)\times \operatorname{GL}_2(\mathbb{F}_p))\ltimes \mathbb{F}_p^2$. I want to play around with a group-theoretic version of connection and parallel transport on the Grassmanian $\operatorname{Gr}(3,2)$ over $\mathbb{F}_p$.
Thanks and greetings,
Simon
This reminds me of §4 in Wolf, Joseph A., Representations that remain irreducible on parabolic subgroups, Differential geometrical methods in mathematical physics, Proc. Conf. Aix-en-Provence and Salamanca 1979, Lect. Notes Math. 836, 129-144 (1980). ZBL0449.22016...
|
2025-03-21T14:48:31.120932
| 2020-05-30T11:04:06 |
361730
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Clej",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/149786"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629642",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361730"
}
|
Stack Exchange
|
Location of bulk and edges for Gaussian random matrices
I have some trouble to understand the difference between the "bulk" and the "edges" of the spectral density of random matrices (for instance in this question).
From my understanding, all properties of random matrices eigenvalues are actually only valid in the bulk (correlations for instance). But where does the separation begin ?
Let's take the example of Gaussian matrices. For instance, for 8x8 or 100x100 random matrices from the GOE, the spectral density looks like:
I initially thought "the bulk" designated for these matrices the inner part of the Wigner semi-circle, while the edges were the outer parts. Is such a crude guess a valid approximation (for practical applications for instance) ?
It is important to distinguish results that apply only in the limit of a large matrix size $N$, from results that apply for small $N$ as well. The theory of random matrices addresses both large-$N$ properties as well as small-$N$ properties, and there are physical applications in both regimes.
I understand from your question that your interest is in large-$N$ properties of the GOE. Then "bulk" versus edge refers to the support $(-W,W)$ of the Wigner semicircle, rescaled such that $W={\cal O}(1)$. The transition from bulk to edge region is at a separation of order $N^{-2/3}$ from $\pm W$. The deviation of the eigenvalue density from the Wigner semicircle in the edge region is described by the Tracy-Widom law.
But do keep in mind that the theory of random matrices is not restricted to the large-$N$ regime. For example, chaotic scattering from a billiard with an $N\times N$ transmission matrix is described by the circular ensembles for any $N$, even as small as $N=1,2,3,\ldots$. In that case there is no notion of bulk versus edge, but there are universal properties that can be measured in experiments, such as the $T^{-1/2}(1-T)^{-1/2}$ distribution of the transmission probability $T\in(0,1)$ for $N=1$.
Thank you very much for this useful answer! For simulations, is there any investigation which have been conducted to know when the "large N limit" begins? I guess it depends on the applucation.
yes, it depends on the application; for many applications I am familiar with the correction to the large-$N$ result is a factor $1/N$ smaller.
|
2025-03-21T14:48:31.121144
| 2020-05-30T13:35:06 |
361737
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Loïc Teyssier",
"Mathmank",
"Pulcinella",
"https://mathoverflow.net/users/119012",
"https://mathoverflow.net/users/24309",
"https://mathoverflow.net/users/87785"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629643",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361737"
}
|
Stack Exchange
|
Examples of Stokes data
I'm trying to learn Stokes data but can't find an example to get my teeth into it.
Background. It's well-known that on a complex manifold $X$, there is the Riemann Hilbert equivalence
$$\text{regular holonomic D modules}\ \stackrel{\sim}{\longrightarrow} \ \text{perverse sheaves}$$
which for instance sends the regular linear ODE $Pf=0$ to its sheaf of solutions, which forms a local system. As I understand it the point of Stokes data is to give something like
$$\text{holonomic D modules}\ \stackrel{\sim?}{\longrightarrow} \ \text{perverse sheaves + Stokes data}$$
and it should send the linear ODE $Pf=0$ to its sheaf of solutions (plus extra data).
For instance, take $X=\mathbf{P}^1$. Then the above equivalence should send (ignoring shifts)
$$\mathscr{D}_X1 \ \longrightarrow \ \mathbf{C}$$
$$\mathscr{D}_Xe^{1/x} \ \longrightarrow \ \mathbf{C}.$$
These D modules are given by the ODEs $y'=0$ and $y'+y/x^2=0$. So the fact that they are sent to the same local system is not a counterexample of RH, since the second is irregular. I gather that under the $?$ map, $\mathscr{D}_Xe^{1/x}$ is sent to $\mathbf{C}$ along with some extra data at the irregular point $x=0$.
Question. What explicitly is the Stokes data of $\mathscr{D}_Xe^{1/x}$ (and in similar cases)? Is there an obvious relation to Stokes lines of the associated ODE?
It helps to understand irregular singularities as the merging of regular singular points, say $$(x^2-a^2)y'+y=0$$ as $a\to0$.
For nonzero $a$ the data is encoded as monodromy (constant) matrices acting on your local solutions, given by the analytic continuation along loops generating the fundamental group.
Parts of the monodromy data makes it to the limit, namely the loops that are not cut by the merging. In the example that would be a loop encircling once both points $\pm a$, its limit encircling once the singularity $0$ and providing the monodromy part.
Yet part of the data won't make it to the limit, and something is lost if one looks only at the monodromy of the multivalued solutions. In the example that would be the monodromy associated to any one of the loops encircling only one singularity $\pm a$.
In the scalar case, the monodromy $y\mapsto c^\pm y$ around $\pm a$ is given by $$c^\pm=\exp\frac{\pm1}{2a}$$ which gets wild as $a\to 0$. Observe though that $c^+c^-=1$ makes it to the limit as the monodromy of $x\mapsto \exp \frac{1}{x}$.
So, where has the lost data gone? And, what is the link with Stokes lines? In the above example the Stokes data is trivial, but simply considering the modified ODE $$(x^2-a^2)y'+y=x$$ gives a non-trivial example. For $a=0$ the so-called Euler's equation has a unique power-series solution $$\hat y(x) = \sum_n (n!)x^{n+1} $$ which doesn't sum as an analytic object in the usual way. By Borel-Laplace-summing this series you obtain two analytic solutions, each one defined on a sector containing a half-plane, from which you deduce two sectorial systems of solutions. The Stokes data comes from the comparison between these two systems where the sectors overlap.The overlapping location is determined by the bissecting lines of the sectors, i.e. the Stokes lines. In the example you can obtain a Liouvillian representation of the solutions by explicit integration, therefore providing an integral representation for the Stokes data. You end up with formulas with coefficients given by values of the Gamma function (more details are linked at the end).
The discussion above streeses the fact that monodromy data is not a good presentation since it doesn't pass to the limit when a regular system degenerates onto an irregular one. Moreover the distinction monodromy/Stokes data is rather artificial, since Stokes data has also a meaning as gluing of local solutions. I prefer the view where everything is "Stokes data": one can always subsdivide $\mathbb P_1$ into "sectors" attached to the singular points, on which you have a trivial system, and the sectorial systems get compared in the pairwise intersections of said generalized sectors. In the case of a singularity (regular or not) you can form a neighborhood around it by tiling contiguous sectors: the composition of the Stokes operators coming from crossing the corresponding overlaps attached at the singularity gives you the monondromy operator. All the Stokes data passes to the limit in cases of merging.
As the construction shows, the Stokes data is not attached to an element of the fundamental group of $X\setminus sing$, like the monodromy, but rather to the "dual" groupoid of paths linking singular points (with an explicit representation as a path-integral operator).
A rich combinatorics comes from these considerations in the case of higher-Poincaré-rank systems (merging of $>2$ singular points).
To read more about the above topics, look for papers by Christiane Rousseau (Montréal)
(Linear systems) Jacques Hurtubise, Caroline Lambert, and Christiane Rousseau. Complete system of analytic invariants for unfolded differential linear systems with an irregular singularity of Poincaré rank k. Mosc. Math. J., 14(2):309–338, 427, 2014.
(Non-linear, but with a detailed analysis of generalizations of the examples above) a book chapter of mine https://hal-cnrs.archives-ouvertes.fr/hal-01170840
(Non-linear, detailed construction and study of the generalized sectors) with Christiane Rousseau https://hal-cnrs.archives-ouvertes.fr/hal-01890315
(Slightly non-linear, confluence of Stokes data in the Painlevé family) by Martin Klimes https://arxiv.org/abs/1609.05185
Thanks so much for this answer!
To be extremely explicit, you mean that Stokes data is just ``usual monodromy along with the monodromy over the Stokes lines (Stokes factors)''? 2. Is it clear how this definition relates to the Delgine/Sabbah definition of Stokes data as being extra data on the pullback of the local system to the blowup at the singular point?
It depends on how you present it, but yes, this is the amount of information needed. 2. I'm not at all at my ease with a more algebraic-geometrical approach. I understand what happens at an analytic/geometric level, I wouldn't risk saying hopeless mistakes. Have you tried the Singer-Vanderput book on Differential Galois Theory ? You may find useful connections there
Thanks, that book guided me to the reference in my answer.
If the Stokes data is trivial in your first example of $(x^2-a^2)y' + y =0$, then how should we account for the lost monodromy around the points $\pm a$ as the points merge?
Part of it constitutes the exponential torus appearing in the Galois group (see works by e.g. J.-P. Ramis)
Thanks for answering! Can you see the exponential torus appearing in any geometric way in the confluence picture?
Let $\mathcal{L}$ be a local system on $X\setminus x$.
Take $S^1$ (the sphere bundle of $X$ at $x$) and the sheaf $\mathcal{V}$ of ``section germs", i.e.
$$\mathcal{V}(\theta,\theta')\ =\ \lim_{\epsilon\to 0}\mathcal{L}(\text{sector with radius }\epsilon\text{ and angle }\theta\text{ to }\theta').$$
So a section of $\mathcal{V}$ is a flat section of $\mathcal{L}$ defined on a small sector.
TL;DR The Stokes data of $\mathcal{L}$ is the filtration $\mathcal{V}_\alpha\subseteq \mathcal{V}$, consisting of section germs $f$ such that $f e^{-\int \alpha}$ has at worst Laurent poles near zero.
Here $\alpha=\sum_{n\ge -N} a_n z^{n/k}dz$ ranges over all etale germs of meromorphic one forms.
Old, overly wordy answer
To complement Loïc Teyssier's excellent answer, this is the algebro-geometric interpretation of Stokes data, first in the case of $e^{1/x}$.
0. A zeroeth approximation: Stokes data is the information that as $x\to 0$,
$$e^{1/x}\ \longrightarrow \ \begin{cases}
0& \text{if }\text{arg}x\in (-\pi/2,\pi/2)\\
\infty & \text{if }\text{arg}x\in (\pi/2,3\pi/2)
\end{cases}.$$
Here $x\to 0$ along rays (lines to the origin of constant argument). So Stokes data remembers how the limiting behavoir of the solution approaching the singular point depends on the argument.
Let's turn this into sheaf language. Take ODE on the disk $X=\Delta$ with singular point $0$, and local system of solutions $\mathscr{L}$ on $\Delta\setminus 0$. To be able to talk about the limiting behavior of solutions as $x\to 0 $ along rays, take the real oriented blowup at $0$
$$\pi \ :\ \widetilde{X}\ \longrightarrow \ X,$$
then $\pi^{-1}\mathscr{L}$ is a local system containing this information. Identify the fibre above $0$ with $S^1$. Write $\mathscr{V}$ for the restriction of $\mathscr{L}$ to $S^1$; this is where that information is stored.
1. A first approximation: Stokes data is a subsheaf
$$\mathscr{V}^0\ \subseteq \ \mathscr{V}$$
given by the solutions with at worst a finite order pole in the given direction. Thus a germ $f$ lies in $\mathscr{V}^0_\theta$ if the size of $f(re^{i\theta})$ is bounded by $r^{-n}$ for some $n$ (this is not quite true, this needs to hold for a sector containing $\theta\in S^1$). In the $e^{1/x}$ example, this is
$$\mathbf{C}_{(-\pi/2,\pi/2)}e^{1/x} \ \subseteq \ \mathbf{C}_{S^1}e^{1/x}.$$
The actual definition asks for (a little) more information about the limiting behavior.
2. A second approximation: Stokes data is a collection of subsheaves
$$\mathscr{V}^\alpha\ \subseteq\ \mathscr{V}$$
for every $\alpha\in \Omega^1_\Delta(\star 0)$ a meromorphic one form on $\Delta$ with poles only at $0$. A germ $f$ lies in $\mathscr{V}^\alpha_\theta$ iff
$$f(re^{i\theta}) e^{-\int \alpha}$$
is bounded by $r^{-n}$ in a small sector containing $\theta$.
These subsheaves fit together to form a filtration, in that
$$\mathscr{V}^\alpha_\theta\ \subseteq \ \mathscr{V}^\beta_\theta$$
whenever $e^{\int\alpha}e^{-\int \beta}$ has aforementioned boundedness property on a sector containing $\theta$. This gives a partial order on $\Omega^1(\star 0)_\theta$, for which the above is a filtration (a lie: you need to replace $\Omega^1(\star 0)$ by its quotient by the forms with at worst simple poles). Moreover, there's a grading on $\mathscr{V}_\theta$ for which this is the associated filtration.
$\infty$. Stokes data as in $2$ a filtration of $\mathscr{V}$ by a partially ordered sheaf, but using a slightly different indexing poset: you replace the Zariski fibre $\Omega^1(\star 0)_\theta$ with the etale fibre. In practice this means that you consider $\alpha=\sum_{n\ge n_0} a_n x^{n/k}dx$ for all $k\in\mathbf{N}$ and instead of just $k=1$.
So e.g. it contains information that
$$e^{1/x}e^{\int \frac{dx}{\sqrt{x}^5}}\ =\ e^{1/x-2/3\sqrt{x}^3} \ \longrightarrow\ \begin{cases}
0&\text{if }\theta\in \pm(\pi,2\pi/3)\\
\infty&\text{if }\theta\in (-2\pi/3,2\pi/3)
\end{cases}$$
where $\sqrt{x}$ is the positive square root defined off the negative reals.
In this language, Stokes lines are just the phenomenon that $f e^{-\int\alpha}$ flips between satisfying and not satisfying the boundedness condition for only finitely many angles $\theta$, so you can see the Stokes lines directly in the sheaves $\mathscr{V}^\alpha$.
e.g. in the $e^{1/x}$ example, $\theta=\pm \pi/2$ are the two Stokes lines
Everything in this answer comes from
La classification des connexions irrégulières à une variable, by Malgrange. http://www.numdam.org/item/CIF_1982__17__A1_0/
Twisted wild character varieties, by Boalch and Yamakawa. https://arxiv.org/abs/1512.08091
The definition of a Stokes structure on a sheaf is $4.1$ of the first reference (it's the same as I've written above), how to give a Stokes structure in the ODE case is the top of page $7$. A Riemann Hilbert correspondence (which justifies the above definition of Stokes data) is theorem $4.2$.
|
2025-03-21T14:48:31.121911
| 2020-05-30T14:57:01 |
361742
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629644",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361742"
}
|
Stack Exchange
|
Is an integral fusion category pseudo-unitary over a finite field?
Here are two propositions in the book Tensor Categories:
Proposition 9.5.1. A pseudo-unitary fusion category admits a unique
spherical structure.
Proposition 9.6.5. Let $\mathcal{C}$ be a weakly integral fusion
category defined over $\mathbb{C}$. Then $\mathcal{C}$ is
pseudo-unitary.
Question: Is Proposition 9.6.5. true for an integral fusion category defined over a finite field?
The notion of pseudo-unitary (i.e. categorical dimension equals Frobenius-Perron dimension) is defined (in the book) for a fusion category over $\mathbb{C}$, but this notion should exist without problem over a finite field in the integral case.
The combination with Proposition 9.5.1. leads to:
Weaker question: Let $\mathcal{C}$ be an integral fusion category defined over a finite field. Is $\mathcal{C}$ spherical?
|
2025-03-21T14:48:31.121993
| 2020-05-30T15:19:36 |
361743
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Scott Taylor",
"alesia",
"https://mathoverflow.net/users/112954",
"https://mathoverflow.net/users/30679"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629645",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361743"
}
|
Stack Exchange
|
"Basic" loops on standardly embedded surfaces
Take a genus $g$ surface $S$ standardly embedded in $\mathbb{R}^3$, by which I mean it is unknotted. Surface $S$ bounds a volume $V$ that deformation retracts on a standardly embedded planar graph $G$ with $\beta_1 = g$, and that only has degree $3$ vertices.
Among the loops on $S$ that are null homotopic in $V$, there is a subset that are boundaries of embedded disks in $V$ that intersect $G$ exactly once for some choice of $G$ as above.
Do these loops (or perhaps close variants) have a name? Do they have an alternate definition?
They are often called meridians of $G$. Note that there are many graphs $G$ to which $V$ deformation retracts (most nonplanar); if you are not particular about which graph $G$ then they are called meridians of $V$. $V$ is called a handlebody and $G$ is a spine of the handlebody. See, for example, Scharlemann's "Refilling meridians in a genus 2 handlebody complement" arXiv:math/0603705
In a paper I found meridians are defined as essential curves bounding an embedded disk. In particular the condition that it should intersect a (well chosen) spine exactly once is not specified. Does that mean that the condition is always true, or are these meridians more general than what I defined?
They are essentially equivalent ideas. Suppose $D$ is a collection of essential discs in $V$ such that no two are isotopic and such that any other essential disc disjoint from $D$ is isotopic to a disc in $D$. Then cutting $V$ along $D$ produces a collection of 3-balls with 3 scars from $D$ in each boundary. Cone the center of each scar to the center of the 3-ball to get a "tripod". Undoing the cutting along $D$ connects the tripods together (or two edges of a tripod together) to form a graph $G$ to which $V$ deformation retracts.
Also the graph $G$ and the discs $D$ satisfy your requirements, except that $G$ may not be planar, and given any embedded disc in $V$ with boundary essential in $S$ it is contained in such a collection of discs. (That is little harder to prove, but is a standard innermost circle/outermost arc argument. Probably you can find the argument in Jaco's or Hempel's books on 3-manifolds.) Insisting that $G$ be planar is much more restrictive and the corresponding discs are really associated more to $G$ than to $V$. I would still call them meridian discs though.
|
2025-03-21T14:48:31.122197
| 2020-05-30T15:53:57 |
361747
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"JustWannaKnow",
"https://mathoverflow.net/users/150264",
"https://mathoverflow.net/users/69642",
"https://mathoverflow.net/users/7410",
"user69642"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629646",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361747"
}
|
Stack Exchange
|
Connections between two constructions of infinite dimensional Gaussian measures
Let me discuss two possible constructions of Gaussian measures on infinite dimensional spaces. Consider the Hilbert space $l^{2}(\mathbb{Z}^{d}) := \{\psi: \mathbb{Z}^{d}\to \mathbb{R}: \hspace{0.1cm} \sum_{x\in \mathbb{Z}^{d}}|\psi(x)|^{2}<\infty\}$ with inner product $\langle \psi, \varphi\rangle_{l^{2}}:= \sum_{x\in \mathbb{Z}^{d}}\overline{\psi(x)}\varphi(x)$. We can introduce in $l^{2}(\mathbb{Z}^{d})$ the discrete Laplacian as the linear operator:
$$(\Delta \psi)(x) := \sum_{k=1}^{d}[-2\psi(x)+\psi(x+e_{k})+\psi(x-e_{k})]$$
where $\{e_{1},...,e_{d}\}$ is the canonical basis of $\mathbb{R}^{d}$. Because $(-\Delta+m^{2})$ has a resolvent for every $m\in \mathbb{R}$, we can consider its inverse $(-\Delta+m^{2})^{-1}$. It's integral Kernel or Green's function $G(x,y)$ is given by:
\begin{eqnarray}
G(x,y) = \frac{1}{(2\pi)^{d}}\int_{[-\pi,\pi]^{d}}d^{d}p \frac{1}{\lambda_{p}+m^{2}}e^{ip\cdot(x-y)} \tag{1}\label{1}
\end{eqnarray}
where $p\cdot (x-y) = \sum_{i=1}^{d}p_{i}(x_{i}-y_{i})$ and $\lambda_{p} :=2\sum_{i=1}^{d}(1-\cos p_{i})$ is the eigenvalue of $-\Delta$ associated to its eigenvector $e^{ip\cdot x}$.
[First Approach] If $m \in \mathbb{Z}$, let $s_{m} :=\{\phi\in \mathbb{R}^{\mathbb{N}}: \hspace{0.1cm} \sum_{n=1}^{\infty}n^{2m}|\phi_{n}|^{2} \equiv ||\phi||_{m}^{2}<+\infty\}$, $s:=\bigcap_{m\in \mathbb{Z}}s_{m}$ and $s':=\bigcup_{m\in \mathbb{Z}}s_{m}$. Note that $s$ is a Fréchet space when its topology is given by the family of semi-norms $||\cdot||_{m}$ and $s'$ is the dual space of $s$ if $l_{\psi}$ is a continuous linear map on $s$ with $l_{\psi}(\phi) =( \psi,\phi) := \sum_{n=1}^{\infty}\psi_{n}\phi_{n}$. Let $C=(C_{xy})_{x,y \in \mathbb{Z}^{d}}$ be an 'infinite matrix' with entries $C_{xy}:= G(x,y)$. We can consider $C_{xy}$ to be a matrix $C=(C_{ij})_{i,j \in \mathbb{N}}$ by enumerating $\mathbb{Z}^{d}$. Now, let us define the bilinear map:
\begin{eqnarray}
s\times s \ni (\phi, \psi) \mapsto \sum_{n=1}^{\infty}\phi_{i}C_{ij}\psi_{j} \equiv (\phi, C\psi) \tag{2}\label{2}
\end{eqnarray}
Thus, $\phi \mapsto (\phi, C\phi)$ is a quadratic form and we can define:
$$W_{C}:=e^{-\frac{1}{2}(\phi,C\phi)}$$
Using Minlos' Theorem for $s$, there exists a Gaussian measure $d\mu_{C}$ on $s'$ (or $\mathbb{R}^{\mathbb{Z}^{d}})$ satisfying:
\begin{eqnarray}
W_{C}(\psi) = \int_{s'}e^{i(\psi,\phi)}d\mu_{C}(\phi) \tag{3}\label{3}
\end{eqnarray}
[Second Approach] For each finite $\Lambda \subset \mathbb{Z}^{d}$, set $C_{\Lambda}$ to be the matrix $C_{\Lambda} =(C_{xy})_{x,y \in \Lambda}$ where $C_{xy}$ are defined as before. Then, these matrices $C_{\Lambda}$ are all positive-definite, so that they define Gaussian measures $\mu_{\Lambda}$ on $\mathbb{R}^{\Lambda}$. Besides, these are compatible in the sense that if $\Lambda \subset \Lambda'$ are both finite and $E$ is a Borel set in $\mathbb{R}^{\Lambda}$ then $\mu_{\Lambda}(E) = \mu_{\Lambda'}(E\times \mathbb{R}^{\Lambda'\setminus\Lambda})$. By Kolmogorov's Extension Theorem, there exists a Gaussian measure $\nu_{C}$ with covariance $C$ on $l^{2}(\mathbb{Z}^{d})$ which is compatible with $\mu_{\Lambda}$ for every finite $\Lambda$.
Now, It seems that these two constructions occur when the so-called thermodynamics limit is taken in QFT and Statistical Mechanics. Both Gaussian measures $\mu_{C}$ and $\nu_{C}$ are measures on $\mathbb{R}^{\mathbb{Z}^{d}}\cong \mathbb{R}^{\mathbb{N}}$. I don't know if this is true but I'd expect these two constructions to be equivalent in some sense, but it is not obvious to me that they are. For instance, the first construction provides a Gaussian measure on $s'$ and the second one on $l^{2}(\mathbb{Z}^{d})$. Are there any relation between these two measures? Are they equal? Maybe the Fourier transform of $\nu_{C}$ would give $W_{C}$, proving these two are the same. Anyway, I'm very lost here and any help would be appreciated.
The source of the confusion is not saying explicitly what are the sets and $\sigma$-algebras the measures are supposed to be on. For example, a sentence like ''By Kolmogorov's Extension Theorem, there exists a Gaussian measure $\nu_C$ with covariance $C$ on $l^2(\mathbb{Z}^d)$ which is compatible with $\mu_\Lambda$ for every finite $\mu_\Lambda$.'' is asking for trouble because it seems to say the measure $\nu_C$ is on the set $l^2(\mathbb{Z}^d)$, which is false.
Let's go back to basics. A measurable space $(\Omega,\mathcal{F})$ is a set $\Omega$ equipped with a $\sigma$-algebra $\mathcal{F}$. A measure $\mu$ on the measurable space $(\Omega,\mathcal{F})$ is a map from $\mathcal{F}$ to $[0,\infty]$ satisfying the usual axioms. From now on I will only talk about probability measures.
For best behavior, the $\Omega$ should be a (nice) topological space and $\mathcal{F}$ should be the Borel $\sigma$-algebra for that topology. Suppose one has two topological spaces $X,Y$ and a continuous injective map $\tau:X\rightarrow Y$. Then if $\mu$ is a measure on $(X,\mathcal{B}_X)$ where $\mathcal{B}_X$ is the Borel $\sigma$-algebra of $X$, then one can construct the direct image/push forward measure $\tau_{\ast}\mu$ on $(Y,\mathcal{B}_Y)$ by letting
$$
\forall B\in\mathcal{B}_{Y},\
(\tau_{\ast}\mu)(B):=\mu(\tau^{-1}(B))\ .
$$
This is well defined because a continuous map like $\tau$ is also $(\mathcal{B}_X,\mathcal{B}_Y)$-measurable. Technically speaking $\mu$ and $\tau_{\ast}\mu$ are different measures because they are on different spaces. However, one could argue that they are morally the same. For example, one might be given the measure $\tau_{\ast}\mu$ without knwing that it is of that form, and only later realize that it is and thus lives on the smaller set $\tau(X)$ inside $Y$.
The first construction:
Let $s(\mathbb{Z}^d)$ be the subset of $\mathbb{R}^{\mathbb{Z}^d}$ made of multi-sequences of fast decay $f=(f_x)_{x\in\mathbb{Z}^d}$, i.e., the ones for which
$$
\forall k\in\mathbb{N}, ||f||_k:=\sup_{x\in\mathbb{Z}^d}\langle x\rangle^k|f_x|\ <\infty
$$
where $\langle x\rangle=\sqrt{1+x_1^2+\cdots+x_d^2}$.
Put on the vector space $s(\mathbb{Z}^d)$ the locally convex topology defined by the collection of seminorms $||\cdot||_k$, $k\ge 0$.
The strong dual can be concretely realized as the space $s'(\mathbb{Z}^d)$ of multi-sequences of temperate growth. Namely, $s'(\mathbb{Z}^d)$ is the subset of $\mathbb{R}^{\mathbb{Z}^d}$ made of discrete fields $\phi=(\phi_x)_{x\in\mathbb{Z}^d}$ such that
$$
\exists k\in\mathbb{N},\exists K\ge 0,\forall x\in\mathbb{Z}^d,\ |\phi_x|\le K\langle x\rangle^k\ .
$$
The vector space $s'(\mathbb{Z}^d)$ is given the locally convex topology generated by the seminorms
$||\phi||_{\rho}=\sum_{x\in\mathbb{Z}^d}\rho_x\ |\phi_x|$
where $\rho$ ranges over elements of $s(\mathbb{Z}^d)$ with nonnegative values.
The measure $\mu_C$ obtained via the Bochner-Minlos Theorem is a measure on $X=s'(\mathbb{Z}^d)$ with its Borel $\sigma$-algebra $\mathcal{B}_X$.
The second construction:
Let $s_0(\mathbb{Z}^d)$ be the subset of $\mathbb{R}^{\mathbb{Z}^d}$ made of multi-sequences of finite support $f=(f_x)_{x\in\mathbb{Z}^d}$, i.e., the ones for which
$f_x=0$ outside some finite set $\Lambda\subset\mathbb{Z}^d$. Put on the vector space $s_0(\mathbb{Z}^d)$ the finest locally convex topology. Namely, this is the locally convex topology
generated by the collection of all seminorms on $s_0(\mathbb{Z}^d)$ .
Note that $s_0(\mathbb{Z}^d)\simeq \oplus_{x\in\mathbb{Z}^d}\mathbb{R}$.
Let $s'_0(\mathbb{Z}^d)$ be the strong topological dual realized concretely as $\mathbb{R}^{\mathbb{Z}^d}$. One can also define the topology by the seminorms $||\phi||_{\rho}=\sum_{x\in\mathbb{Z}^d}\rho_x\ |\phi_x|$
where $\rho$ ranges over elements of $s_0(\mathbb{Z}^d)$ with nonnegative values.
However, this is the same as the product topology for $s'_0(\mathbb{Z}^d)=\prod_{x\in\mathbb{Z}^d}\mathbb{R}$.
The measure $\nu_C$ constructed via the Daniell-Kolmogorov Extension Theorem is a measure on $Y=s'_0(\mathbb{Z}^d)=\mathbb{R}^{\mathbb{Z}^d}$ with its Borel $\sigma$-algebra for the product topology a.k.a strong dual topology.
The precise relation between the two measures:
We simply have $\nu_C=\tau_{\ast}\mu_C$ where $\tau$ is the continuous canonical injection due to $X=s'(\mathbb{Z}^d)$ being a subset of $Y=\mathbb{R}^{\mathbb{Z}^d}$.
Thank you SO much for this amazing answer! This was exactly what I was searching for!
Let me see if I understand the application of your explanations. Suposse a field is $\phi: \mathbb{Z}^{d}\to \mathbb{R}^{d}$ so that the space of all fields is $\mathbb{R}^{\mathbb{Z}^{d}}$. For each $x \in \mathbb{Z}^{d}$ I can define random variables $f_{x}: \Omega=\mathbb{R}^{\mathbb{Z}^{d}}\to \mathbb{R}$ by $f_{x}(\phi) :=\phi(x)$. Thus, I can calculate, e.g. correlations $\mathbb{E}{\nu{C}}[f_{x}f_{y}] = \mathbb{E}{\mu{C}}[\phi(x)\phi(y)]$ both ways. That's why these are "basically the same"?
You are moving the goalposts again and replying to your question requires making corrections as a preliminary step. I suppose you mean $\phi$ is a function from $\mathbb{Z}^d\rightarrow\mathbb{R}$ and not $\mathbb{Z}^d\rightarrow\mathbb{R}^d$. So $\phi$ means an element of $Y=\mathbb{R}^{\mathbb{Z}^d}$. Then, technically $\mathbb{E}_{\mu_C}[\phi_x \phi_y]$ is not well defined because $\mu_C$ is a measure on $X$ and not $Y$.
I must think more about all this. I'll possibly post another question on applications of these ideas. It might be better.
No need. All you have to do is say $\phi$ designates an element of the space of fields $X$ and not $Y$. Then the equation $\mathbb{E}{\nu_C}[f_x f_y]=\mathbb{E}{\mu_C}[\phi_x \phi_y]$ is correct. This is just a trivial consequence of the abstract change of variable theorem relating expectations for the direct image measure to expectations of the original measure.
I think that what you are looking for is the link between the white noise measure $\mu_C$ and the isonormal process indexed by $\ell^2(\mathbb{Z}^d)$ with covariance structure given by $C$. The white noise measure $\mu_C$ is a Gaussian measure on $s'$ so that for all $\varphi \in s$, $\langle ;\varphi\rangle_{s',s}$ is a centered Gaussian random variable with variance $\langle \varphi ; C \varphi\rangle$. By an approximation argument, you should be able to give some sense to $\langle ; f\rangle$ with $f \in \ell^2(\mathbb{Z}^d)$ so that it is a centered Gaussian random variable under $\mu_C$ with variance $\langle f;C f\rangle$. Now, your second construction gives rise to a Gaussian stochastic process indexed by $\mathbb{Z}^d$ with covariance structure given $C$. By re-indexing, each element $X_j$ of this Gaussian stochastic process admits the representation $\nu_{C}(e_j)$ where $e_j=(0,\dots,0,1,0,\dots)$. Now, again by approximation, you can extend $\nu_C$ to all $\ell^2(\mathbb{Z}^d)$ and it is completely defined, for all $f,g \in \ell^2(\mathbb{Z}^d)$, by
$$ \mathbb{E}\left(\nu_{C}(f)\nu_{C}(g)\right)= \langle f;Cg\rangle ,$$
and $\mathbb{E}(\nu_C(f))=0$. Now, the link is clear and you have the following equality in law under $\mu_C$, for all $f \in \ell^2(\mathbb{Z}^d)$
$$\nu_c(f) = \langle ; f\rangle.$$
This is completely similar to the classical construction of the white noise probability measure on the space of tempered distributions on $\mathbb{R}$ ($S'(\mathbb{R})$) and the classical isonormal Gaussian process indexed by $L^2(\mathbb{R})$.
The underlying probability space is $s'$ then $\langle ;f\rangle$ is the random variable $X$ defined, for all $\omega \in s'$, by $$X(\omega) = \langle \omega; f \rangle$$. The ``difficulty" is that the bilinear pairing is defined for $\omega \in s'$ if $f$ belongs to $s$ and only $\mu_C$ a.e.if $f$ belongs to $\ell^2(\mathbb{Z}^d)$.
I'm still having trouble understanding the origin of such differences. My main interest is statistical mechanics. I've been reading the section Gaussian Free Field (GFF) of two different books and the Hamiltonian on both books is the same. But when the time comes to discuss the thermodynamic limit, one book goes in the $l^{2}(\mathbb{Z}^{d})$ direction and the other goes in the $s'$ direction. But it seems that the target here is the same, but the constructions are differente and the results are different. Do you know why?
I understand your question this way: what is the difference between white noise distribution theory and Malliavin calculus ? In the first theory, the white noise (derivative of Brownian motion) is the central object of interest whereas in the second it is Brownian motion. Both theories have several similarities but they are intrinsically different by nature.
I see. I think this confusion of mine is worth a new post, to put it some context. Anyway, you've been really helpful! Thanks a lot!
|
2025-03-21T14:48:31.123069
| 2020-05-30T16:12:48 |
361748
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Learning math",
"Yuval Peres",
"https://mathoverflow.net/users/35936",
"https://mathoverflow.net/users/7691"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629647",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361748"
}
|
Stack Exchange
|
On the concentration of Lipschitz functions near its expectation, where the vector has identical but not independent, components
Consider the random vector $X:=(X_1\dots X_1) \in \mathbb{R}^n, X_1 \sim \mathcal{N}(0,1).$ Notice the identical components, they're identically distributed but not independent.
Now, I was wondering whether we could expect an inequality like the one we expect in the case where $X$ had iid $\mathcal{N}(0,1)$ components, i.e. in this other extreme case, can we expect that for any Lipschitz function $f: \mathbb{R}^n \to \mathbb{R},$ we have for each $t \ge 0:$
$$ P\left[ |f(X) - Ef(X)| \ge t \right] \le 2 exp \left( - \frac{ct^2}{||f||_{Lip}^2} \right), $$
where $c > 0$ is an absolute constant.
P.S. I might add a similar question on what happens to the above probability when $n \to \infty, t $ is fixed, but I'm still thinking about the question.
P.P.S. if you could cite some reference, that'd be very useful too. I do have access to Michel Lédoux's great book, but I find it a bit difficult for myself to quickly navigate through and get the right inequality I want (but that's just me!).
If $f$ is the sum and all the components are identical then the inequality fails for $t=n$.
Thank you for your answer, but do you mind elaborating on it a bit more? For the sum function, the Lipschitz norm is $\sqrt{n}.$ I wonder if your answer is valid for each $n \ge 1,$ or when $n \to \infty ?$ (The limiting case is also quite important for me) I see that $P[ |\sum_{i=1}^{n} X - \sum_{i=1}^{n} EX| \ge t=n ] = P[nX \ge n] =P[X \ge 1] \nrightarrow 0,$ unlike the RHS term $2 exp(- \frac{cn^2}{n}) = 2 exp (- cn)\to 0$ as $n \to \infty.$ So this does prove that the inequality isn't true in the limiting case. Is that what you had in mind? Thanks again!
For every $c$ the inequality fails if $n$ is large enough.
|
2025-03-21T14:48:31.123243
| 2020-05-30T16:35:09 |
361749
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Vladimir Dotsenko",
"YCor",
"https://mathoverflow.net/users/1306",
"https://mathoverflow.net/users/14094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629648",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361749"
}
|
Stack Exchange
|
Polynomial identities of supercommutative-gradable algebras
All algebras below are associative, and not assumed unital, and, to fix ideas, over the complex numbers.
An algebra $A$ is supercommutative-gradable if it admits a grading $A=A_0\oplus A_1$ in $\mathbf{Z}/2\mathbf{Z}$ ($A_iA_j\subset A_{i+j}$ for $i,j\in\mathbf{Z}/2\mathbf{Z}$) that makes it supercommutative: for $a,b$ homogeneous $ab=ba$ if either $a,b$ has even degree, and $ab=-ba$ for $a,b$ of odd degree.
I insist that by supercommutative-gradable, I assume that such a grading exists, but do not endow $A$ with it: I still view $A$ as a bare algebra, with no fixed grading.
What are the polynomial identities satisfied by supercommutative-gradable algebras? More precisely, in universal algebra terms: what is the variety generated by supercommutative-gradable algebras? [In particular, is it finitely generated? (Edit: Yes!)]
(For readers not familiar with universal algebra or polynomial identities, see addendum below to make the question precise.)
For instance, the class of supercommutative-gradable algebras satisfies the identities $(xy-yx)z-z(xy-yx)$ and $x^2y^2-2xyxy+2yxyx-y^2x^2$, and none of these two follows from the other one. (The identity $(xy-yx)z-z(xy-yx)$ holds because $xy-yx$ always has even degree, hence is central.)
Note: (about the above convention above for the meaning of supercommutative gradable: $\mathbf{Z}$-gradings vs $\mathbf{Z}/2\mathbf{Z}$-gradings)
Let $\mathcal{A}$ be the class of supercommutative-gradable algebras. Some subclasses of $\mathcal{A}$ could compete for being called "supercommutative-gradable algebras", namely the class $\mathcal{A}_{\mathbf{Z}}$ (resp. $\mathcal{A}_{\mathbf{N}}$, resp. $\mathcal{A}_{\mathbf{N}_{>0}}$), those algebras admitting an algebra grading in $\mathbf{Z}$ (resp...) satisfying the supercommutativity rule. Also we have smaller classes $\mathcal{A}^1_{\mathbf{Z}}$, $\mathcal{A}^1_{\mathbf{N}}$ in which we assume the algebra unital with unit of degree $0$. All the obvious inclusions between these classes are strict. However, the question is not sensitive to the choice of class: indeed, if $A\in\mathcal{A}$, then it is quotient of an algebra in $\mathcal{A}_{\mathbf{N}_{>0}}$, which itself (adding a unit) is subalgebra of an algebra in $\mathcal{A}^1_{\mathbf{N}}$. For the former quotient assertion: write $A=A_1\oplus A_2$ (writing $A_2$ rather than $A_0$) and consider the free $\mathbf{Z}$-graded supercommutative algebra $\tilde{A}$ over the vector space $A_1\oplus A_2$ with $A_1,A_2$ of degree $1,2$: then $A$ is canonically quotient of $\tilde{A}$.
Addendum (basic definitions of identities in algebras, varieties)
Fix the associative (non-unital) free $\mathbf{C}$-algebra $\mathbb{F}=\mathbf{C}\langle X_n:n\in\mathbf{N}\rangle$. An element $P\in \mathbb{F}$ is a polynomial identity of a class $\mathcal{C}$ of algebras if $P$ vanishes in every $A\in\mathcal{C}$, that is, if $P$ belongs to the kernel of every homomorphism $\mathbb{F}\to A$ for every $A\in\mathcal{C}$.
The set of polynomial identities of $\mathcal{C}$ forms a 2-sided ideal $I_\mathcal{C}$ of $F$ satisfying strong conditions: it is fully invariant (=stable under all endomorphisms); it is strongly graded, in the sense that it is a graded ideal for the unique algebra grading of $\mathbb{F}$ in the free abelian group $\mathbf{Z}^{(\mathbf{N})}$ (with basis $(e_n)$) for which $X_n$ has degree $e_n$ for every $n$ (for instance $x_1x_2x_1^4x_2-x_2^2x_1^5$ has degree $5e_1+2e_2$, while $x_1^2+x_2^2$ is not strongly homogeneous). Describing polynomial identities of $\mathcal{C}$, in practice, means exhibiting generators of $I_\mathcal{C}$ as a fully invariant 2-sided ideal.
For instance, for $\mathcal{C}$ the class of commutative algebras: the polynomial identities of $\mathcal{C}$ are generated by $X_0X_1-X_1X_0$.
The variety generated by $\mathcal{C}$ is the class of all algebras in which all $P\in I_{\mathcal{C}}$ are polynomial identities. It is also the smallest class of algebras containing $\mathcal{C}\cup\{\{0\}\}$ and stable under taking quotients, subalgebras, and arbitrary (unrestricted) direct products. The mapping $\mathcal{V}\mapsto I_\mathcal{V}$ is a canonical bijection between the "set" of varieties (of associative algebras) and fully invariant 2-sided ideals of $\mathbb{F}$. [It's not properly a set: to make it a set, cheat by fixing a set $X$ of cardinal $2^{\aleph_0}$ and consider $\mathbf{C}$-algebra structures with underlying set $X$.]
A variety of associative algebras $\mathcal{V}$ is finitely based if the ideal $I_\mathcal{V}$ is finitely generated as fully invariant ideal (it's not always the case). To my surprise it's always the case (I expected the contrary, by analogy with groups or Lie algebras in finite characteristic).
I believe that the identity $(xy-yx)z-z(xy-yz)$ generates everything (in characteristic 0, at least). To show that no further identities are needed, it is enough to exhibit one algebra that has no further identities. It follows from an old theorem of Krakowski and Regev that the Grassmann algebra of a countably dimensional vector space works for that purpose.
Actually I was performing computations leading to the opposite conclusion, and precisely that the identity $wxyz+wyxz+zyxw-xwyz-ywxz-zwyz$ does not follow from the identity $[[x,y],z]$. Indeed (with the help of Sagemath) I computed that in the consequences of $[[x,y],z]$ in degree $(1,1,1,1)$ (i.e., in a space of dimension 24) has dimension 14, while the actual set of relations in this degree has dimension 16 (and the above relation is an explicit element testifying this). I'll double check.
I didn't take into account the new relations such as $[[wx,y],z]$ (only those such as $[[x,y],z]w$ or $w[[x,y],z]$). Taking all of them into account, I get all 16 dimensions, and in particular the above relation follows from $[[a,b],c]$. The paper you link seems to yield the full conclusion. Thanks a lot!
PS: free link to Krakowski-Regev's paper at AMS site (rather than JSTOR): https://www.ams.org/journals/tran/1973-181-00/S0002-9947-1973-0325658-5/ D. Krakowski and A. Regev, The polynomial identities of the Grassmann algebra. Trans. Amer. Math. Soc. 181 (1973), 429-438
Happy to help. Is it something that naturally arises in your work?
Just very roughly. I sometimes have to compute cohomology algebras of small-dimensional Lie algebras, hence wondered about the classification of those (quite special) supercommutative associative algebras, and eventually wondered about this. I also have a lot of confusion about this variety: I'm more familiar with varieties of Lie algebras, and the analogy can lead to surprises: for instance, in this very variety, the 2-sided ideal generated by commutators (which are central) is not central.
I see, thanks! By the way, there is a fun geometric interpretation of free algebras in this variety: the free algebra generated by V is isomorphic to the space of even polynomial differential forms on $V^*$ with the "quantised product" $\alpha\cdot \beta=\alpha\wedge \beta+ d\alpha \wedge d\beta$ (see Proposition on p. 782 in the paper https://www.intlpress.com/site/pub/pages/journals/items/mrl/content/vols/0014/0005/a007/).
What OP calls "finitely generated" variety is normally called "finitely based". By Kemer's theorem every variety of algebras over a field of characteristic 0 is finitely based. So if the super-commutative algebras are considered as algebras (ignoring grading) then the answer is "yes". It is still "yes" if the grading is is taken into account.
Thanks, I didn't know or expect this fact! I won't accept the answer because I'm definitely interested in knowing generators of this particular variety, but I'm happy to upvote it.
|
2025-03-21T14:48:31.123792
| 2020-05-30T17:14:56 |
361751
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben Wieland",
"David Roberts",
"Dmitri Pavlov",
"LSpice",
"Moishe Kohan",
"Nicolas Hemelsoet",
"Praphulla Koushik",
"Qfwfq",
"Simon Henry",
"Wojowu",
"https://mathoverflow.net/users/104710",
"https://mathoverflow.net/users/104742",
"https://mathoverflow.net/users/118688",
"https://mathoverflow.net/users/22131",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/39654",
"https://mathoverflow.net/users/402",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/4639",
"https://mathoverflow.net/users/4721",
"xuq01"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629649",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361751"
}
|
Stack Exchange
|
What are the occurrences of stacks outside algebraic geometry, differential geometry, and general topology?
What are the occurrences of the notion of a stack outside algebraic geometry, differential geometry, and general topology?
In most of the references, the introduction of the notion of a stack takes the following steps:
Fix a category $\mathcal{C}$.
Define the notion of category fibered in groupoids/ fibered category over $\mathcal{C}$; which is simply a functor $\mathcal{D}\rightarrow \mathcal{C}$ satisfying certain conditions.
Fix a Grothendieck topology on $\mathcal{C}$; this associates to each object $U$ of $\mathcal{C}$, a collection $\mathcal{J}_U$ (that is a collection of collections of arrows whose target is $U$) that are required to satisfy certain conditions.
To each object $U$ of $\mathcal{C}$ and a cover $\{U_\alpha\rightarrow U\}$, after fixing a cleavage on the fibered category $(\mathcal{D}, \pi, \mathcal{C})$, one associates what is called a descent category of $U$ with respect to the cover $\{U_\alpha\rightarrow U\}$, usually denoted by $\mathcal{D}(\{U_\alpha\rightarrow U\})$. It is then observed that there is an obvious way to produce a functor $\mathcal{D}(U)\rightarrow \mathcal{D}(\{U_\alpha\rightarrow U\})$, where $\mathcal{D}(U)$ is the "fiber category" of $U$.
A category fibered in groupoids $\mathcal{D}\rightarrow \mathcal{C}$ is then called a $\mathcal{J}$-stack (or simply a stack), if, for each object $U$ of $\mathcal{C}$ and for each cover $\{U_\alpha\rightarrow U\}$, the functor $\mathcal{D}(U)\rightarrow \mathcal{D}(\{U_\alpha\rightarrow U\})$ is an equivalence of categories.
None of the above 5 steps has anything to do with the set up of algebraic geometry. But, immediately after defining the notion of a stack, we typically restrict ourselves to one of the following categories, with an appropriate Grothendieck topology:
The category $\text{Sch}/S$ of schemes over a scheme $S$.
The category of manifolds $\text{Man}$.
The category of topological spaces $\text{Top}$.
Frequency of occurrence of stacks over above categories is in the decreasing order of magnitude. Unfortunately, I myself have seen exactly four research articles (Noohi - Foundations of topological stacks I; Carchedi - Categorical properties of topological and differentiable stacks; Noohi - Homotopy types of topological stacks; Metzler - Topological and smooth stacks) talking about stacks over the category of topological spaces.
So, the following question arises:
What are the occurrences of the notion of a stack outside of the three areas listed above?
There would be of course more articles on topological stacks than what I have mentioned above.. Please let me know if you come across any other.. It is just that I have come across only four.. But, I am sure the number would be much less when compared to articles dealing with algebraic stacks..
I did some proofreading, hopefully without changing any meanings. Is it really true that $\mathcal J_U$ is a collection of collections of arrows, or is one of those layers a typo?
@LSpice It is not a typo.. It associates to each object a collection of collection of arrows... :D each collection in that collection is called a cover of $U$.. Thanks for other corrections..
The entire chapter of the book by Bridson and Haefliger is about nonpositively curved complexes of groups which are stacks in the category of simplicial complexes. Orbifolds are widely used in low dimensional topology.
@MoisheKohan thanks for the comment.. can you please make it as an answer adding some more information.. Yes, I also heard of orbifolds as “proper étale Lie groupoids”..
Stack semantics in categorical semantics, no?
A big results of topos theory is the (2,1)-category of Grothendieck toposes identifies with a reflective full subcategory of the category of "localic stacks" (stacks over the category of locales, where the covers are open surjection). But, that's not very far off topological stacks, so I'm not sure this count. This was mostly formulated in the language of localic groupoids however, but this is completely equivalent, and stacks make the theory more natural.
@SimonHenry I am hearing them for the first time.. Any answer with 1 or more references would be welcome :)
Considering that bundle gerbes have already been mentioned twice despite being excluded in the second example, I edited the main post to make it more clear that it asks for examples outside of the three listed areas.
@DmitriPavlov thanks for the edit., it looks better.. twice?? Where?
@PraphullaKoushik: The first time in the answer by Francois Ziegler that has been deleted for this reason. The second time in the answer by Qfwfq.
@PraphullaKoushik https://ncatlab.org/nlab/show/classifying+topos+of+a+localic+groupoid
@DavidRoberts Ok. Thanks.. :)
@DmitriPavlov ok :)
Another application of stacks is in synthetic differential geometry.
Start with the opposite category of germ-determined finitely generated C^∞-rings
and equip it with the appropriately defined Grothendieck topology,
then pass to ∞-stacks.
The resulting category (known as the Dubuc topos)
contains all smooth manifolds, is a Grothendieck ∞-topos
(so in particular, has all homotopy colimits and is cartesian closed),
and allows for a good notion of infinitesimals.
The latter allows to manipulate differential geometric objects
such as vector fields and differential forms
using infinitesimal methods similar to the ones used by Élie Cartan and Sophus Lie,
yet perfectly rigorous.
For instance, the de Rham complex is now precisely the smooth infinitesimal singular cochain complex, and the Stokes theorem is now precisely the definition of the de Rham differential as the singular cochain differential.
Just like for stacks on manifolds, homotopy colimits in this category
have excellent geometric properties.
Even better, if one takes germ-determined finitely generated
differential graded C^∞-rings and takes ∞-stacks on the resulting ∞-site,
then one gets the ∞-stack that has all the excellent properties
listed above, together with excellent geometric properties of homotopy limits (which always exist).
In particular, in this category nontransversal intersections exist and have
desired geometric properties, etc.
This subject is known as derived differential geometry.
I might be able to understand this once I have some understanding of $\infty$-category (I started reading something about infinity categories, so it should not take much time)... Does any other examples comes to your thought that does not have anything to do with infty stacks and only uses just stacks (1-stacks)...
@PraphullaKoushik: You can certainly consider stacks in groupoids over C^∞-rings. These do not have a good theory of all homotopy colimits, but they do have a good theory of ordinary quotients.
Ok. I am hearing about $C^{\infty}$-rings first time... stacks in groupoids over the category of $C^{\infty}$-rings (equipped with some nice Grothendieck topology)... I see there are some references about $C^{\infty}$-rings at https://ncatlab.org/nlab/show/smooth+algebra I will see them..
Stacks are used in complex analysis, for example.
See the papers by Finnur Lárusson, in particular,
Excision for simplicial sheaves on the Stein site and Gromov's Oka principle,
which shows that having the Oka–Grauert property for a complex manifold X is equivalent
to the condition that the presheaf of spaces of holomorphic maps into X
is an ∞-stack in the appropriate Grothendieck topology on the site of Stein manifolds.
A few years ago, Bernstein wrote a note with a new approach to representation theory of algebraic groups using the langage of stacks.
This doesn't sound to be like an occurence "outside algebraic geometry".
@Wojowu : the note is supposed to give another approach to representation theory of algebraic group over local field (using stacks to consider different forms of a group "at once"), which seems outside algebraic geometry.
Thanks. I will see that paper. :)
Stacks over the category locales are very interesting for topos theory:
A big success of topos theory is the fact that the $(2,1)$-categories of Grothendieck toposes and geometric morphisms between them embedded as a reflective full subcategory of the category of localic stacks, that is stack on the category of locales. It is in fact a full subcategory of the category of "Geometric localic stacks", that is those localic stacks comming from localic groupoids.
In my mind this is the results that best convey the idea that Grothendieck toposes are geometric objects. Of course, Grothendieck had the intuition that toposes were geometric object from the very beginning of the theory, but this results is to me really what turn this intuition into something formal.
Note: There are some size issues involved whose discussion will be postponed to the very end.
We will identifies the category of locales with a full subcategory of the category of toposes, by identifying each locales $\mathcal{L}$ with the sheaf topos Sh$(\mathcal{L})$.
The basic idea is fairly simply given $\mathcal{T}$ a topos and $\mathcal{L}$ a locale, you get a category of geometric morphisms Hom$(\mathcal{L},\mathcal{T})$, if you simply drop the non-invertible natural transformations, you get a groupoid Hom$(\mathcal{L},\mathcal{T})$ of geometric morphisms and natural transformations.
This attach to every topos a pre-stack on the category of locales. It can be shown that this pre-stack is a stack for the topology whose coverings are the open surjections between locales (and the coproduct).
This construct a functor from the $(2,1)$-category of toposes to the $(2,1)$-category of localic stacks, which is fully faithful and identifies the category of toposes with a reflective full subcategory of stacks. Stack in the image are called "etale-complete" stack (to be honest one generally talks about étale-complete localic groupoids, but this is a property of the associated stack).
The starting point of this story started with the famous representation theorem of Joyal and Tierney in "An Extension of the Galois Theory of Grothendieck", which can be understood as the construction of the left adjoint, and the proof that it is essentially surjective, though most of the key idea are already present.
The results as presented above appeared in the two paper of Moerdijk:
The classying topos of a continuous groupoids, I & II
As the title suggest the results is mostly stated in terms of groupoids rather than stack, but the theory is really about stacks, and if I remember correctly the connection to stack is explicit mentioned in the paper. I think Bunge's paper "An application of descent to a classification theorem for toposes" is also relevant to the story.
So what I've said above is only correct up to some important size consideration that need to be taken care of.
The category of locales, with the topology of open surjections, do not satisfies the smallness condition needed in order for stackification to be well defined.
Though the point of view we adopt here, is that up to passing to a larger Grothendieck universe stackification is always defined, the question is only wether or not preserve it preserves certain smallness conditions.
In this case stackification do not preserve smallness: there are examples of small pre-stack of locales (in the sense "small colimits of representable") whose stackification is not even "levelwise small", that is $\mathcal{F}(\mathcal{L})$ can fail to be an essentially small groupoids.
But this is actually a good things, because for many Grothendieck topos, the groupoids Hom$(\mathcal{L},\mathcal{T})$ are not essentially small.
Here the appropriate "category of stack" to consider for what I say above to be correct are the large stack that are small colimits (in the category of stack) of representable stacks. This is not a locally small category (but the category of Grothendieck topos isn't either). The fact that the stack attached to a topos is in this category is non trivial, but follows directly from the work of Joyal and Tierney mentioned above.
Thank you :) This is mode detailed than what I can expect.. I will read it carefully and ask if I have any questions :)
Mike Shulman has stack semantics, an application of stacks to logic. This is basically sheaf semantics, a now standard application to logic of sheaves (far from their own origin in geometry), except that sheaf semantics isn't quite powerful enough to capture unbounded quantification in the way that Mike needs in order to do what he wanted to do with set theory (which is what he was doing when he came up with stack semantics).
This is a fairly low-powered application of stacks, as sheaves are almost but not quite sufficient. But simply adopting this approach makes some things easier to talk about, even when one could do them in the old (sheaves-only) way. And if you want to apply this kind of logic to category theory itself instead of to set theory, then the stacks are really necessary.
There are two notions of stack. The one you mention is a sheaf of groupoids. Sometimes these come up on their own. The other notion is a geometric object, often a "bad quotient." This object can be represented as a sheaf of groupoids, but that is only a technical tool. If you had other tools, you might use them instead. For example, if you had a foliation of a manifold, you might want to consider the "space of leaves." You could consider this as a stack on the site of topological spaces, but you could also represent it by the convolution algebra of the equivalence relation. Constructions that are Morita invariant depend only on the stack. So you might say that Connes-style noncommutative geometry is (in part) the study of stacks, or you might say it is a reason that stacks are not more popular.
I do not understand your last sentence "So you might say that Connes-style noncommutative geometry is (in part) the study of stacks, or you might say it is a reason that stacks are not more popular." Can you explain what it means?
Groupoid convolution algebras can indeed be used to study sufficiently nice stacks (e.g., those presented by Lie groupoids), but claiming that convolution algebras are stacks is stretching things too far. For instance, is there an actual theorem (with proof) in the literature that establishes an equivalence of bicategories between sufficiently nice stacks and sufficiently nice algebras? Otherwise one simply cannot refer to algebras as a “notion of stack”.
@DmitriPavlov Yes, but it's an exercise.
@BenWieland: If it is an exercise, can you tell us what algebras (with a precise list of properties) correspond to Lie groupoids and what bimodules between these algebras (again with a precise list of properties) correspond to bibundles between Lie groupoids?
I believe @BenWieland refers to the fact that if you remember the Cartan sub-algebra in the convolution algebra then you can at least recover the initial space and the equivalence relation. It is not quite the same as a stack in the sense of stack of groupoids as you do not quite recover the isotropy, but it fits with the idea that you completely understand the "bad quotient".
@SimonHenry You can recover the isotropy in a complementary way. I'm not alluding to ad hoc constructions, but to the fact, mentioned by Dmitri, that algebras form a 2-category and thus it directly yields a prestack on topological spaces.
?? I agree that there are way to organise algebras in a $2$-category and that this allows to reconstruct at least a differentiable stack from the convolution algebra, but I really don't see a 2-category structure that would get you near to what you claim then ! (i.e. recover the stack attached to the groupoid).
@BenWieland: I know quite a few people who would be extremely interested in seeing a reference for the construction you are alluding to. This would solve a known research problem. For starters, what algebras correspond to Lie groupoids? And how do we characterize bimodules corresponding to bibundles?
@SimonHenry what do you mean by recovering a differentiable stack without recovering a stack? Do you mean smooth functions rather than continuous functions? And if you switch from ad hoc constructions to 2-categories, how can the wrong 2-category yield anything without yielding everything?
@BenWieland : I'm staying vague because you did not gave enough details on the constructions you have in mind. But If you see algebra as a 2-category using bi-module then given two discrete groupe $G$ and $H$, you can consider the "discrete" stack $BG$ and $BH$ (they exists in all the categories of stacks considered here), morphisms from $BG$ tp $BH$ are group morphisms $G \to H$ up to conjugations. The associated convolution algebras are the groups algebra $k[G]$ and $k[H]$. I don't think you can fully recover a group from its group algebra alone...
... (without considering for e.g. the Hopf algebra structure) and anyway morphisms (bimodule) between k[G] and k[H] are quite different from group morphisms...
@SimonHenry Yes, if you want an equivalence of categories, it is going to work everywhere or nowhere, as I said. In particular, it is good to check in easy examples like group algebras. Yes, you need the Hopf structure.
There are "bundle gerbes" (introduced by Murray), which are a particular kind of stacks. People study connections on them, generalizing connections on principal bundles.
How is this an answer to OP's question? Bundle gerbes are stacks on the site Man of smooth manifolds, which is Example 2 in OP's list. The OP specifically requested an example not in his list.
@Dmitri Pavlov: by being an answer to the question in the title, and to the first question in the body of the OP.
@Dmitri Pavlov: but, upon a second reading, I realize the only question was likely intended to be the last one; the first one (along with the title) was probably to be intended as general context to be specified in the body of the OP.
Surely you can see that the two questions are identical, except for the clarifying parenthetical remark? Gerbes are already discussed in the papers cited by the OP in the main post (e.g., in Carchedi's paper), repeating examples from the main post as separate answers is a rather strange thing to do.
Yup: the parenthetical remark in the second question rules out the examples mentioned in the previous paragraphs; which the first question didn't. -- If you, or anyone, thinks my reading of the OP was off-topic (which may be the case), I'm totally fine with my answer being deleted. I'm not going to comment in this thread anymore.
Thanks for your answer.
|
2025-03-21T14:48:31.125144
| 2020-05-30T18:06:46 |
361755
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A beginner mathmatician",
"Anthony Quas",
"Jochen Glueck",
"https://mathoverflow.net/users/102946",
"https://mathoverflow.net/users/11054",
"https://mathoverflow.net/users/136860"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629650",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361755"
}
|
Stack Exchange
|
von Neumann ergodic theorem for $L_p$
Let $\tau:\Omega\to \Omega$ be a measure-preserving transformation with $\mu(\Omega)<\infty$. Define $T:L_p(\Omega)\to L_p(\Omega)$ as $Tf:=f\circ \tau$. I want to prove that for all $1\leq p<\infty$, given $f\in L_p$ there exists $\bar{f}\in L_p$ such that $\|\bar{f}\|_p\leq \|f\|_p,$ $\bar{f}\circ \tau=\bar f$, and $\|\frac{1}{n}\sum_{k=0}^{n-1}T^kf-\bar{f}\|_p\to 0$ as $n\to \infty$.
For $p=2$ this is just von Neumann's mean ergodic theorem. Using this One can easily prove the result for $1\leq p<2$. But how to prove the statement for $2<p<\infty$? Also is it true for $p=\infty$?
Something's odd about the properties of $\overline{f}$: you probably mean $\overline{f} \circ \tau = \overline{f}$ instead of $\overline{f} \circ \tau = f$, right?
@Jochen. Yes. Edited.
False for p infinite. True for finite. See e.g. the book by Krengel, Ergodic theorems. Other sources (that also go further) are [1, Sec. I.2.1] or [2, Theorem 8.8].
[1] T. Eisner, Stability of operators and operator semigroups, Operator Theory:
Advances and Applications, vol. 209, Birkh¨auser Verlag, Basel, 2010.
[2] T. Eisner, B. Farkas, M. Haase, and R. Nagel, Operator theoretic aspects
of ergodic theory, Graduate Texts in Mathematics, vol. 272, Springer, Cham,
2015. http://www.math.uni-leipzig.de/~eisner/book-EFHN.pdf
From Krengel only I am reading this. There is no proof of this fact. He only stated this. As I mentioned I could prove for $1\leq p<2$ but not for $p>2$. Also now I see that he has given only a hint for $p=\infty$. But after lot of try I could not solve that too.
@Abeginnermathmatician: It is indeed in Krengel's book: For $p \in (1,\infty)$ it is an immediate consequence of Theorem 2.1.2 on page 73. Right after the theorem, Krengel explains why it is also true for $p = 1$ (as a consequence of Theorem 2.1.1 on page 72).
@Abeginnermathmatician: For a $p=\infty$ counterexample, consider the shift on 2 symbols with the Bernoulli (1/2,1/2) measure and $f(x)=x_0$. Then the pointwise limit of the averages is 1/2 almost everywhere; for each $n$, there is a set of positive measure (namely the set of $x$’s with $x_0=...=x_{n-1}=0$) where the difference between the average and the limit is 1/2.
|
2025-03-21T14:48:31.125344
| 2020-05-30T18:28:19 |
361757
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Stanley Yao Xiao",
"YCor",
"aleph",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/83189"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629651",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361757"
}
|
Stack Exchange
|
Lattice points in hypercubes
Let $ (\Lambda_n) $ be a family of lattices, $ \Lambda_n \subset \mathbb{Z}^n $, with $ \det\Lambda_n \sim n $ as $ n \to \infty $ (meaning $ \lim_{n\to\infty} n^{-1} \det\Lambda_n = 1$). I am interested in the asymptotics of the number of points of $ \Lambda_n $ in the hypercube $ [0,2)^n $. In particular, is it true that: $$ |\Lambda_n \cap [0,2)^n| = |\Lambda_n \cap \{0,1\}^n| \sim \frac{2^n}{n} $$ as $ n \to \infty $? Does someone know how to prove this type of statement? Would an additional condition on $ \Lambda_n $, such as a lower bound on its minimum, be helpful in establishing such an asymptotic result?
Obviously not: take $\Lambda_n=(n\mathbf{Z})\times\mathbf{Z}^{n-1}$. Then $\Lambda_n\cap [0,2[^n$={0}\times{0,1}^{n-1}$ has cardinal $2^{n-1}=\frac{2^n}{2}\gg \frac{2^n}{n}$. So you need additional conditions; maybe as you suggest with a lower bound on its minimum.
To make the comment by YCor more precise, you need some control on the successive minima on your lattice. If it is too skew then what you're asking for won't work, because there is a lower rank sub-lattice that has too many points in the box.
Thank you both! Are you perhaps aware of a general result that bounds the number of lattice points inside a convex region in terms of successive minima? I know of Davenport's theorem (from the replies to related questions), but it does not seem to give sharp bounds for what I need.
The following answer is from A Reverse Minkowski Theorem. It deals with a sphere, rather than a hypercube. I am unaware of extensions to hypercubes, but the lower bound I will quote from it at least seems non-trivial, and of interest.
Minkowski's first theorem has a "point-counting" version, due to Blichfeldt and van der Corput.
For any lattice $L \subseteq \mathbb{R}^n$ with $\det(L) \leq 1$ and $r > 0$,
$$|L \cap rB_2^n| \geq 2^{-n}\mathsf{vol}(rB_2^n) = \frac{1}{\sqrt{\pi n}}\left(\frac{\pi e r^2}{2n}\right)^{n/2}(1 + o(1)).$$
This result is apparently given for arbitrary norms, so look at the linked paper for the proper citation/statement for $\ell_\infty$.
Of course, to get an asymptotic, one needs an upper bound as well.
As the comments mention, one needs to rule out "skew" lattices (say of the form $t\mathbb{Z}\oplus \frac{1}{t}\mathbb{Z}$) somehow. The authors of the linked paper do this via the notion of stable lattices.
A lattice $L\subseteq \mathbb{R}^n$ is stable if $\det L = 1$, and $\det(L') \geq 1$ for all sublattices $L'\subseteq L$.
They then give upper bounds for a number of point-counting contexts (assuming the underlying lattice is stable) in corollary 1.4. I will not copy them all here (there are 3 cases depending on what the particular value of $r$ one has). None of them appear to obviously (asymptotically) match the above upper bound (and in later work on a similar problem, but without an obvious point-counting application, the authors mention the work I describe above is non-tight).
This is answer is mostly to say that there is recent work towards establishing these kinds of bounds, but I do not believe the hypercube has been examined, nor do I believe tight point-counting asymptotics have been established.
The assumption authors have found useful to get rid of the "skew" lattices described is known as stability, and has been used in other contexts going back a few decades.
Thank you, this is very helpful!
|
2025-03-21T14:48:31.125615
| 2020-05-30T20:36:59 |
361761
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Hajer Bagdady",
"https://mathoverflow.net/users/157500",
"https://mathoverflow.net/users/40804",
"mme"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629652",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361761"
}
|
Stack Exchange
|
Does H-supplmented module have D2?
A module $M$ is called H-supplemented if for every submodule $N$ of $M$ there exists a direct summand $D$ of $M$ such that $M = N + X$ if and only if $M = D + X$ for every submodule $X$ of $M$.
A module $M$ has $D2$ if $A\leq M$ such that $M/A$ is isomorphic to a direct summand of $M$, then $A$ is a
direct summand of $M$.
The question is: Does H-supplemented imply D2?
Isn't the $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$ $H$-supplemented but not $D2$?
Thanks for you reply, I don't see why it's D2. Could you please explain?
@HajerBagdady The claim is that it's not D2. Take $A = (\Bbb Z/2\Bbb Z, 2\Bbb Z/4\Bbb Z)$. If this was a direct summand the module would be isomorphic to $(\Bbb Z/2)^3$.
|
2025-03-21T14:48:31.125702
| 2020-05-30T22:46:52 |
361768
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Michael Renardy",
"Mr. Gentleman",
"https://mathoverflow.net/users/12120",
"https://mathoverflow.net/users/58857"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629653",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361768"
}
|
Stack Exchange
|
Dynamical system described by coupled nonlinear differential equations
Suppose a dynamical system is described by two variables, $x$ and $y$, and they change over time according to the following two coupled nonlinear differential equations:
\begin{equation}
\begin{split}
&\frac{dx}{dt}=-x^\alpha\\
&\frac{dy}{dt}=-x-y^2
\end{split}
\end{equation}
where $\alpha>0$ is a parameter for this system, and the initial condition for $x$ is positive $x(t=0)>0$.
This system is referred to as stable if $|y(t\rightarrow\infty)|<\infty$, and unstable otherwise. For a given $\alpha>0$ and $x(t=0)>0$, under what initial condition of $y$ is the system stable?
The following is some qualitative understanding I have.
First, the first term in the second equation tends to destabilize the system (by pushing $y$ to $-\infty$).
Second, if $x(t=0)=0$, then the system is stable if $y(t=0)\geqslant 0$ and unstable if $y(t=0)<0$, while $x=0$ for all $t>0$. That is, there is at least some (possibly measure-zero) regime where the system is stable. If $x(t=0)>0$ and $y(t=0)=0$, the system is unstable because $y\rightarrow-\infty$ as $t\rightarrow\infty$. So one expects there may be a separatrix between the stable regime and the unstable regime. The goal is to understand this separatrix.
Third, it seems we can focus on the vicinity of $(x, y)=(0, 0)$ and understand the separatrix there. In this regime, it seems if $\alpha$ is sufficiently large, $x$ approaches zero too slowly, so in the second equation it always destabilizes $y$ unless $x(t=0)=0$. That is, it seems the stable regime is really a measure-zero line in the two dimensional space of $x$ and $y$. On the other hand, if $\alpha$ is small, $x$ may approach zero sufficiently fast, and it does not destabilize $y$ if $y(t=0)$ is also large. So there seems to be a value of $\alpha_0$, such that when $\alpha>\alpha_0$, there is only a measure-zero stable regime, and when $\alpha<\alpha_0$, there is an extended stable regime.
I would like to understand (i) what is $\alpha_0$? (ii) when $\alpha<\alpha_0$, what is the separatrix (expressed in terms of $y(t=0)$ as a function of $x(t=0)$ and $\alpha$)? (iii) what happens exactly at $\alpha=\alpha_0$?
The first equation is solvable in closed form, and then the second equation becomes a Riccati equation. For that, you have closed form solutions only for special values of $\alpha$. Some general observations: $y$ remains bounded if and only if it is always nonnegative. A necessary condition for that is that $y(0)$ is positive and $x$ is integrable. Whether $x$ is integrable depends on $\alpha$. If $x$ is not integrable, then $y$ cannot remain nonnegative. It may remain nonnegative if $x$ is integrable and $y(0)$ is large enough.
Thank you. Can you provide more details on what it means to say $x$ is integrable, and for which $\alpha$ it is? For example, if $\alpha=1$, does the system have a stable regime?
By integrable I mean that $\int_0^\infty x(t),dt<\infty$. This is the case if $\alpha<2$.
|
2025-03-21T14:48:31.125939
| 2020-05-30T23:14:16 |
361769
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dmitri Pavlov",
"Reid Barton",
"Sebastian H. Martensen",
"https://mathoverflow.net/users/126667",
"https://mathoverflow.net/users/129677",
"https://mathoverflow.net/users/402"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629654",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361769"
}
|
Stack Exchange
|
How do you prove that the category of weak equivalences of sSet is accessible?
I am trying to prove that the category of simplicial sets is a combinatorial model category by using Proposition A.2.6.15 of Lurie's book, and this requires proving that the weak equivalences are generated by a (small) set of weak equivalences under filtered colimits. It seems so doable, yet I am at a loss. The only references I have found on the matter, state it as a consequence of the fact that sSet is a combinatorial model category so that's no help.
Maybe his alternative characterisation of accessibility (Proposition A.2.6.5) can help?
Any help and references would be much appreciated.
Edit for clarification: I need to prove that sSet is a combinatorial model category via Smith's theorem, meaning that I can only use that sSet is locally presentable and that I have to prove a few conditions for the classes of weak equivalences and cofibrations. One of these conditions is that weak equivalences are generated by filtered colimits from a set. Therefore I cannot use that sSet is a model category, and I have to prove this statement in some more direct way.
Also, weak equivalences are the morphisms that become weak homotopy equivalences in the realisation.
This question is not really meaningful unless you specify which description of the weak equivalences of sSet you take as the definition.
A morphism is a weak equivalence if it is a weak homotopy equivalence in the realisation.
See Corollary 5.1 in The accessibility rank of weak equivalences, which shows that weak equivalences of simplicial sets are finitely accessible.
See also Theorem 4.6 in Model Structures on Ind Categories and the Accessibility Rank of Weak Equivalences, which establishes the same result in a different way.
This proves that sSet is a finitely combinatorial model category,
because it has a set of generating (acyclic) cofibrations
with compact (co)domains (compact simplicial sets are
precisely simplicial sets with finitely many nondegenerate simplices),
and its underlying category is locally finitely presentable
because it admits a set of compact generators given by all representables,
i.e., simplices.
If your goal is to prove that sSet is merely a combinatorial model category (without finiteness), then this follows much more directly and immediately from the fact that sSet is a cofibrantly generated model category (by construction), whose underlying category is a locally presentable category (just like any category of presheaves of sets on a small category). The accessibility of weak equivalences is then an automatic consequence of these facts.
$\def\Exi{{\rm Ex}^\infty}$
Finally, if one merely wants to prove the accessibility of weak equivalences directly, without referring to model structures,
this can be done as follows.
Recall the original definition of simplicial weak equivalences due to Kan:
a simplicial map $f$ is a weak equivalence if $\Exi(f)$ is a simplicial homotopy equivalence.
The functor $\Exi$ is an accessible functor between locally presentable categories because $\Exi$ is the filtered colimit of ${\rm Ex}^n$ and each of these preserves filtered colimits.
Thus, it suffices to show that simplicial homotopy equivalences
between Kan complexes form an accessible subcategory.
Indeed, by the simplicial Whitehead theorem,
simplicial homotopy equivalences between Kan complexes can be characterized by a relative-homotopy-lifting property
with respect to the set of morphisms ∂Δ^n→Δ^n.
The relative-homotopy-lifting property can be reformulated as an ordinary extension
property with respect to a certain morphism in the category of simplicial maps,
see the formula (3.3) in Dugger and Isaksen's paper Weak equivalences of simplicial presheaves.
But such a lifting property singles out an accessible subcategory of the category of simplicial maps between Kan complexes,
see the proof of Corollary A.2.6.8 in Lurie's Higher Topos Theory.
Thanks for the references! But both of those articles seem to conclude that weak equivalences are accessible on the condition that sSet is already a finitely combinatorial model category, and I can't assume that. I think the point in both articles is to show that it is finitely accessible, whereas I am comfortable with any level of accessibility.
Do you know any work around? Or maybe you have another reference?
@SebastianH.Martensen: The “assumption” of being finitely combinatorial is a completely elementary fact: the generating (acyclic) cofibrations of sSet have as their (co)domains compact simplicial sets, i.e., simplicial sets with finitely many nondegenerate simplices.
@SebastianH.Martensen: I amended my answer accordingly.
I am sorry for the communication error but I realise I may not have been clear (or maybe I am just not understanding you correctly); I want to prove that sSet is a combinatorial model category via Smith's theorem, i.e. I only know that sSet is locally presentable, and I have to prove that a bunch of conditions are true for sSet equipped with the chosen cofibrations and weak equivalences. Among these conditions, I need to prove that weak equivalences are generated by filtered colimits from a set. Therefore I cannot use that sSet is a model category directly. Do you know an argument for this?
@SebastianH.Martensen: Yes, I added an answer to this question as the last paragraph to my answer.
Could you elaborate on how the arrow category of simplicial homotopy equivalences presents itself as a limit of accessible categories? Do you mean in the sense that a morphism of simplicial sets X -> Y is a limit over the simplex category for X of morphisms which are each a colimit of morphisms between standard n- and m-simplicies?
@SebastianH.Martensen: I added an argument for the accessibility of simplicial homotopy equivalences with precise references.
|
2025-03-21T14:48:31.126309
| 2020-05-30T23:39:35 |
361770
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Cameron Zwarich",
"https://mathoverflow.net/users/153196",
"https://mathoverflow.net/users/99234",
"mathbeginner"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629655",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361770"
}
|
Stack Exchange
|
Subalgebra of a nuclear $C^*$-algebra
Is every exact $C^*$-algebra a subalgebra of a nuclear $C^*$-algebra?
I know the conclusion is true when the exact $C^*$-algebra is separable, since every separable exact $C^*$-algebra is a subalgebra of $\mathcal{O}_2$.
If a exact $C^*$-algebra is non-separable, does the above conclusion also hold?
Pisier implies that this is still open in his new book "Tensor Products of C*-algebras and Operator Spaces", so unless it's been solved in the past few months it's probably still open.
@Cameron Zwarich, May I ask you a question : in which page does Pisier mention that the problem is still open ?
I'm using the PDF on his website. It's at the beginning of the chapter on Exactness and Nuclearity and in the notes at the end. He doesn't say that it's still open explicitly, but he mentioned it as open in his previous book on operator spaces (right after he states the embedding theorem for separable algebras) and doesn't say anything here.
|
2025-03-21T14:48:31.126408
| 2020-05-31T01:02:23 |
361773
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ville Salo",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/158871",
"solus0684"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629656",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361773"
}
|
Stack Exchange
|
Uniqueness of function with range $\mathbb{S}^2$ under a constraint
Assume $g,f\colon A\subset\mathbb{R}^M\rightarrow\mathbb{S}^2$ are two bijective functions defined on the set $A$. Now assume a constraint $C$: $\forall x,y\in A, \exists R\in SO(3)\colon Rf(x)=f(y)\iff Rg(x)=g(y)$.
Does $C$ imply $f=\pm g$?
[Update]: The answer is "yes" when $C$ is modified to $\forall x,y\in A, \forall R\in SO(3)\colon Rf(x)=f(y)\iff Rg(x)=g(y)$.
I recommend you clarify your formulas a bit. Ignoring the second paragraph, the answer is "no". Constraint $C$ holds for any functions $f, g$: for all $x, y$ we can pick $R$ so that neither equation holds, thus the equivalence holds. And when $A$ is nonempty, it is of course possible to find $f, g : A \to \mathbb{S}^2$ such that $f \neq \pm g$. I also don't get the second paragraph, what is the product on $\mathbb{S}^2$?
My original notation might be wrong. What I meant was that if for any $R\in SO(3)$ for which $Rf(x)=f(y)$ holds $Rg(x)=g(y)$ also holds and vice versa. I think I have to change $\exists R$ to $\forall R$. Right? I removed the second paragraph for the sake of clarity.
This is what I came up with if $C$ is modified to $\forall x,y\in A, \forall R\in SO(3): Rf(x)=f(y)\iff Rg(x)=g(y)$. Let $\theta(R)$ denote the rotation angle corresponding to rotation matrix $R$. We can show that $f(x)\cdot f(y)=\max\limits_{R\in SO(3): Rf(x)=f(y)} \cos{\theta(R)}$. This implies that if $C$ holds $\forall x,y\in A f(x)f(y)=g(x)g(y)$ and consequentially we can show that $\exists R'\in RO(3): g=R'f$. By plugging this into C this implies that $\forall x,y\in A, R\in SO(3): Rf(x)=f(y) \iff RR'f(x)=R'f(y)$ which implies $\forall R\in SO(3): RR'=R'R$ equivalent to $R'=\pm I$.
You are correct, if you change $C$ so $R$ is $\forall$-quantified, then $C$ implies $f = \pm g$.
Pairs $\{v, -v\}$ have distinct stabilizers under $SO(3)$ so setting $x = y$ we see $f = \pm g$ pointwise. But then if $f(x) = g(x)$ and $f(y) = -g(y)$, picking any $R$ such that $Rf(x) = f(y)$, if we were to have $Rg(x) = g(y)$ then we would have
$$ Rf(x) = Rg(x) = g(y) = -f(y) = -Rf(x), $$
a contradiction. Of course neither the choice of domain nor the fact the dimension of the sphere is $2$ matters.
(Your argument is probably correct as well, but I had already written this when I realized $\cdot$ means dot product.)
I suppose the only thing used about the action of $SO(3)$ on $\mathbb{S}^2$ is that the group acts transitively, and whenever $x$ and $y$ have the same stabilizer, we have $gx = y$ for some $g \in Z(G)$.
|
2025-03-21T14:48:31.126639
| 2020-05-31T01:58:40 |
361777
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerhard Paseman",
"Iosif Pinelis",
"Will Sawin",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/51189",
"zeraoulia rafik"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629657",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361777"
}
|
Stack Exchange
|
Is this number theoretic quantity bounded above?
I am considering a combinatorial argument which involves the following quantity. We use the prime counting function $\pi(n)$ and to save on exponents we set $h=\pi(n/2)$. The quantity as a function of integer $n \gt 7$ is
$$(\pi(n)!)^{1/(n-h)}$$
Computations for small $n$ suggest this is always less than $4$, as do rough back-of-the-envelope asymptotic calculations. Is this bounded above for all $n \gt 7$? If so, what is the bound? (I'm hoping it is always less than 3.)
Gerhard "Researching Minds Want To Know" Paseman, 2020.05.30.
nice question ,upvote
I think it is upper bounded by 4 using a rough approximation to the prime counting function. I would appreciate verification. Gerhard "Worth An Acknowledgement To Me" Paseman, 2020.05.30.
Don't you just want to plug an effective form of the prime number theorem into Stirling's formula here?
Your rough back-of-the-envelope asymptotic calculations suggest the limit is $e$, right? I think these calculations, unless you did them in a very strange way, are already a proof.
I often do strange things @Will. I guess I am just looking for confirmation. Gerhard "Anything For These Uncertain Times" Paseman, 2020.05.30.
And the rough calculations suggest something greater than $e$, while the numerical computations suggest something less than $e$. If I can tweak the argument to get a related quantity less than 2, I am within spitting distance of combinatorial proof of. a theorem of Sylvester. Gerhard "More Combinatorial Than Sylvester's Proof" Paseman, 2020.05.30.
Let $k:=\pi(n)$, so that $p_k\le n<p_{k+1}$, where $p_k$ is the $k$th prime. By the last displayed formula in this section of the Wikipedia article,
\begin{equation*}
-1+\ln(k\ln k)<\frac{p_k}k<\ln(k\ln k)
\end{equation*}
if $k\ge6$, whence
\begin{equation*}
n>-k+k\ln(k\ln k),\quad n/2<m_k:=\frac{k+1}2\,\ln((k+1)\ln(k+1)).
\end{equation*}
Therefore, letting
\begin{equation*}
c_1:=1.25506,\quad r(k):=\frac{\ln((k+1)\ln(k+1))}{\ln m_k}
\end{equation*}
and using this result, we get
\begin{equation*}
h=\pi(n/2)\le\pi(m_k)<c_1\frac{m_k}{\ln m_k}=c_1\frac{k+1}2\,r(k).
\end{equation*}
Next,
\begin{multline*}
r(k):=\frac{\ln(k+1)+\ln\ln(k+1)}{\ln(k+1)+\ln[\ln(k+1)+\ln\ln(k+1)]-\ln2} \\
<\frac{\ln(k+1)+\ln\ln(k+1)}{\ln(k+1)+\ln\ln(k+1)-\ln2}<\frac{10}9
\end{multline*}
if
\begin{equation*}
k\ge195,
\end{equation*}
which will be assumed henceforth.
So,\begin{equation*}
h<c_2(k+1),
\end{equation*}
where
\begin{equation*}
c_2:=\frac7{10}>c_1\frac{10}9\Big/2.
\end{equation*}
So, using the trivial inequality $k!\le k^k$, we have
\begin{equation*}
\ln[(\pi(n)!)^{1/(n-h)}]=\frac{\ln(k!)}{n-h}
\le\frac{k\ln k}{-k+k\ln(k\ln k)-c_2(k+1)}
<\frac{k\ln k}{k\ln k}=1
\end{equation*}
and hence
\begin{equation*}
(\pi(n)!)^{1/(n-h)}<e \tag{1}
\end{equation*}
for $k\ge239$, that is, for $n\ge1499$. By direct calculation, (1) holds for $n\le1498$ as well.
It is also easy to see that the upper bound $e$ on $(\pi(n)!)^{1/(n-h)}$ is exact.
Thanks for this! I will go over this and cite it in my write-up. It turns out I have some leeway: h can be a little larger than pi(n/2), and it is enough if I can show a weaker upper bound of three minus epsilon or even four minus epsilon. Do you think we can simplify this given the leeway, but also arrange so that it holds for k greater than 50, so that less computation for small k is needed? (I'm willing to accept verifying k less than 40, so that c1 can drop a bit.) Gerhard "Looking Forward To The Math" Paseman, 2020.05.31.
@GerhardPaseman : If you make $h$ larger, then your expression will get larger. However, if you don't need the exact upper bound, then the increase of $h$ could be OK within certain limits, and even the proof could possibly be simplified. I suspect the proof can be simplified a bit anyway, as many of the first proofs.
I have indeed simplified the proof a bit.
Thank you again. I will upvote and accept this later. It turns out I have a simpler argument and simpler quantity to estimate which achieves my goal. I will post later after I have checked it . Gerhard "Can it Be So Simple?" Paseman, 2020.05.31.
I have further simplified the answer, which now does not use any version of Stirling's formula.
Why the downvote?
So I accept the answer of Iosif Pinelis, but it turns out I don't need the exact result now. I will post some backstory, and then the reason I don't need it now.
Thanks to MathOverflow user Daniel.W, and his question (360323) on strengthening Sylvester's theorem, I've been motivated to read the paper On Arithmetical Series. I approached it after a hint from Emil Jerabek to look at the thesis of Alan Woods. The thesis contained a write-up of Sylvester's method that I finally understood, and this allowed me to try to understand some of the proof in the 1892 paper.
A different version appears in a 1929 paper of Schur (which I have yet to find) and a (mostly) combinatorial one in a paper of Erdos in 1934. However, the arguments are still involved, and the Erdos paper leaves a lot of finitely many exceptions to be explored to yield a full proof.
After looking at the basic relation in Sylvester's paper, I (re-)discovered a result that allowed one to show that there was a number in (m,m+,n] with a prime factor greater than n whenever 4m was at least as big as n^2. This was heartening since previously I could only show it for m bigger than exponential in n. This in turn allowed me to discover a method which involved the quantity in the question above, and simple arguments showed that m only needed to be bigger than a small constant times n. (Iosif's argument and some additional computation shows the small constant is 3.). I then was going to attempt a third method to bridge the remaining gap which is for all m at least n. All of this was then going to be retooled to answer the motivating MathOverflow question.
After seeing Iosif's argument and thinking on simplifying the motivating argument, I found it. Here it is.
Write the product of the integers in (m,m+n] as P=(m+1)...(m+n). Rewrite as W(n!)L, where W are the prime factors of P/(n!) which are at most n gathered together, and L is the product of all the prime factors larger than n.
A key observation of Sylvester, (which I invite the reader to prove) is that W is at most (and for n bigger than 7, strictly less than) (m+n-p+1)...(m+n), where p is $\pi(n)$. This is because W is the product of p distinct prime powers, each one dividing a term of P (and usually different powers divide different terms, we suffer no loss in assuming this).
So if (m,m+n] has only n-smooth numbers, then L=1 and n! Is bigger (not necessarily strictly) than (m+1)...(m+n-p). The literature now expends a lot of effort to show how small m is, and Sylvester himself resorts to existence of primes in (m, 3m/2] to complete his argument. There is an easier way, however.
Write m=jn+i for i non negative. Then rewrite W(n!) = P by dividing out W and dividing out terms in the factorial larger than p. We get p! greater than j^(n-p), if P is n-smooth.
But we can argue with Chebyshev estimates to get j less than 6, and if we are as thorough as Iosif we can get j less than 3 with a small amount of computation needed. I need to perform this step but I believe p less than 50 should be more than sufficient.
So when the dust settles, we have reduced a large portion of Sylvester's argument to showing log(p!) Is less than n-p, using nothing more than grade school arithmetic and Sylvester's observation on W. With care we get that (m,m+n] has a multiple of a prime greater than n when m is at least 3n or greater. If need be, we can turn to the Erdos proof to handle m smaller than 3n.
However, there is more. The motivating question asks for two distinct numbers in the interval which have prime factors bigger than n. We now let L be a product of d many candidates for fixed d of members of (m,m+n]. We now are comparing log(p!) to n-p-d, and we will get the same bound on j, although this bound may start holding for larger m only.
Given the amount of time I spent reading these proofs, I'm surprised not to find this observation (that j is less than 3) is in the literature. We can use this observation, Chebyshev estimates, and work of Nagura or earlier to answer the motivating question affirmatively. That C=18 for two numbers hasn't been proved yet.
Gerhard "Is Confident It Will Be" Paseman, 2020.05.31.
Now, seeing Iosif's replacement, of p^p. the argument has become simpler. Gerhard "Many Thanks Again, Iosif Pinelis" Paseman, 2020.05.31.
It seems my thinking gets organized after posting the question.
The natural logarithm of the quantity $\pi(n)!$ is near $\pi(n)\log(\pi(n)/e) + (\log(\tau\pi(n)))/2$ (a number-theoretic use for $\tau$, the circumference of a unit radius circle). Using an approximation to $\pi(n)$ we get that this is less than $An$ for some $A \lt 2$. But $An/(n-h)$ is bounded above by $2A$, and gets very close to $A$. So with some work the original quantity should be shown to be less than $e^A$.
Verification is still appreciated.
Gerhard "And Still Worth An Acknowledgement" Paseman, 2020.05.30.
It may help to set $k= \pi(n)$ and use upper bounds for the $n$th prime to deduce $n> k \log k+ k \log \log k - k$ because then the numerator and denominator of the logarithm will look quite similar although $\pi(n/2)$ then becomes annoying
I may do that. however, the important thing for me is that $h$ is small relative to $\pi(n)$. Gerhard "Going To Write It Up" Paseman, 2020.05.30.
For the sake of history, I compare $\pi(n)!$ to $k^{n-h}$. If $k=3$ makes that quantity too large, then the conclusion of Sylvester's Theorem follows for his $m \gt 2n$. This also helps toward a generalization I am working on with another MathOverflow user. Gerhard "Not Doing It For Power" Paseman, 2020.05.30.
|
2025-03-21T14:48:31.127787
| 2020-05-31T06:53:27 |
361783
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629658",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361783"
}
|
Stack Exchange
|
The number of perfect squares which can occur in an arithmetic progression of length n
This is a similar question to https://math.stackexchange.com/questions/2023399/the-maximum-number-of-perfect-squares-that-can-be-in-an-arithmetic-progression/3693487#3693487
Let f(n) be the maximum number of squares in an AP (arithmetic progression) of length n. For example, $f(3)=3$, as 1, 25, 49 is a 3-term arithmetic progression with three squares, and $f(4)=3$, as there are no 4 term arithmetic progressions of squares. Also, $f(5)=4$, with the AP 49, 169, 289, 409, 529 as a small example.
Trivially, f is monotone increasing, as adding terms onto an existing AP cannot reduce the number of squares. Also, $f(a+b) \leq f(a)+f(b)$, by concatenation of sequences. It seems to me that the easiest way to find upper bounds on f is to constrain configurations of squares. Let $(0, a, b, c)$ (with $0<a<b<c$ denote a configuration of squares of the form: $M, M+aK, M+bk, M+ck$, where $k>0$. The configuration $(0, 1, 2, 3)$ is a four term arithmetic progression, which we already know is ruled out. Using elliptic curves you can show that $(0, 1, 3, 4)$ and $(0, 1, 4, 5)$ are also impossible (and it looks like many more are impossible as well. On the positive end, there are solutions for any configuration $(0, a, b, c)$ when $c \neq a+b$ (I'm working on a parametric solution).
By eliminating those configurations, I have found that $f(6)=4$, $4 \leq f(7) \leq 5$, and $f(8)=5$ with $1, 25, 49, 73, 97, 121, 145, 169$ as an example.
Up to what $n$ is $f(n)$ known? Specifically, are f(9) and f(10) known?
It's tabulated out to $f(52)=12$ at the Online Encyclopedia of Integer Sequences.
A reference is given to Enrique González-Jiménez and Xavier Xarles, On a conjecture of Rudin on squares in Arithmetic Progressions.
A conjecture is presented which gives a very simple form for $f(n)$ that works for all $n\le52$ except $n=5$.
|
2025-03-21T14:48:31.127963
| 2020-05-31T06:54:21 |
361784
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dominic van der Zypen",
"Johannes Schürz",
"https://mathoverflow.net/users/134910",
"https://mathoverflow.net/users/8628"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629659",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361784"
}
|
Stack Exchange
|
Infinite complete linear hypergraphs with edges of different sizes
Is there an infinite cardinal $\kappa$ with a collection of subsets ${\cal E}$ of $\kappa$ with the following properties?
$\bigcup {\cal E} = \kappa$,
$e \neq f \in {\cal E}$ implies $|e \cap f|=1$
$|e|<\kappa$ for all $e\in {\cal E}$, and
not all members of ${\cal E}$ have the same cardinality.
$\kappa =\omega$?
If you want that all members of $\mathcal{E}$ have different cardinality, then $\kappa$ cannot be a successor cardinal, but any limit cardinal will do the trick. Furthermore, $\vert\mathcal{E}\vert \leq \kappa$, and if $\kappa$ is weakly inaccessible, then you get $\vert \mathcal{E}\vert = \kappa$.
Thanks @JohannesSchürz for your comments! Can you give me a hint on how to construct a set $\cal E$ of finite subsets of $\omega$ that have properties 1-4? If you put it into an answer, I'll accept it and we can close this thread.
Partition $\omega \setminus \{0\}$ into infinitely many finite sets $I_n$ such that $\vert I_n \vert=n$. Define $\mathcal{E}:=\{I_n \cup \{0\} \colon n \in \omega\}$.
|
2025-03-21T14:48:31.128076
| 2020-05-31T07:51:24 |
361788
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629660",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361788"
}
|
Stack Exchange
|
Null preserving transformation
Suppose that $(\Omega,\mu)$ is a measure space. Let $\tau:\Omega\to\Omega$ is a measurable map such that $\mu\circ\tau^{-1}<<\mu$. Then $\tau$ s said to be null preserving. I want to prove the following. If $f:\Omega\to\mathbb R$ is measurable and $\mu(\{f\neq f\circ \tau\})=0,$ then there exists a measurable function $f^\prime$ such that $f^\prime=f^\prime \circ \tau$ and $\mu(f\neq f^\prime)=0.$ If we define $A:=\{x\in\Omega:f(x)=f(\tau(x))\}.$ I can prove that $A$ is $\tau$-invariant mod $\mu.$ A natural way to define $f^\prime$ would be $f^\prime=f1_{B}$ where $B$ is $\tau$-invariant and $\mu(A\Delta B)=0.$ But I can not really see if it works. It will work definitely if we have $B\subseteq A.$ We can have $B$ to be the set $\cup_{k=0}^\infty(A\setminus\ \cup_{k=0}^\infty\tau^{-k}(A\setminus \tau^{-1}A)).$ Can we have that $B\subseteq A$? Also. I want to find some intuitive idea how the construction should be.
I'll rename your $\Omega$ into $X$ to simplify typing.
Let $D=\{x\in X,~\forall n\in\mathbb{N}:~ f(\tau^n(x))=f(x)\}=\bigcap\limits_{n\in\mathbb{N}_0} \{x\in X: f(\tau^{n+1}(x))=f(\tau^n(x))\}$ $=\bigcap\limits_{n\in\mathbb{N}_0}\tau^{-n}(A)=X\backslash \bigcup\limits_{n\in\mathbb{N}_0}\tau^{-n}(X\backslash A)$. Since $\tau$ is null-preserving, $D$ is of full measure. Also, $\tau (D)\subset D$.
Define $g:X\to\mathbb{R}$ as follows. If for $x\in X$ there is $n\in \mathbb{N}_0$ such that $\tau^n(x)\in D$, define $g(x)=f(\tau^n(x))$; otherwise, define $g(x)=0$. Note, that in the first case we get the same value, regardless of $n$ (as long as $\tau^n(x)\in D$).
Since $D$ is of full measure, it is clear that $g=f$ almost everywhere. Let us check $g=g\circ\tau$. If $x\in X$ is such that $\tau^n(x)\in D$, for some $n$, we get $g(x)=f(\tau^n(x))=f(\tau^{n+1}(x))=g(\tau(x))$. Otherwise, $\tau^n(x)\not\in D$, for every $n$, therefore $\tau^{n+1}(x)\not\in D$, for every $n$, and so $g(x)=0=g(\tau(x))$.
|
2025-03-21T14:48:31.128225
| 2020-05-31T08:07:22 |
361791
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Brendan McKay",
"https://mathoverflow.net/users/9025"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629661",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361791"
}
|
Stack Exchange
|
Diophantine bound for homogeneous system under norm conditions for solutions and system
If $b_{ij}\in[0,2^t-1]\cap\mathbb Z$ holds at every $i\in\{1,\dots,k'\}$ and $j\in\{1,\dots,k\}$ where $1\leq k'\leq k$ holds then there are $x_1,\dots,x_k\in\mathbb Z:\sum_{j=1}^kx_jb_{1i}=0\wedge\dots\wedge\sum_{j=1}^kx_jb_{k'j}=0$ with $\sum_{j=1}^{k}|x_j|\neq0$. Further restrict $b_{ij}$ to set $T_t(\beta,q)$ of coprime integers in $[0,2^t-1]\cap\mathbb Z$ with
$b_{\max}-b_{\min}<\beta2^t$ at fixed $\beta\in(0,1)$ and
$\sum_{j=1}^kb_{ij}^2=q$ at every $i\in\{1,\dots,k'\}$.
$B$ is $k'\times k$ matrix of $b_{ij}$.
If $N(B)=\{|(x_1,\dots,x_k)|_\infty:\sum_{i=1}^kx_ib_{1i}=0\wedge\dots\wedge\sum_{i=1}^kx_ib_{k'i}=0\}$ then a good upper bound for $$\alpha(k,t,\beta)=\max_{b_{ij}\in T_t(\beta,q)}\min_{n\in N(B)}n$$ is given by the Bombieri-Vaaler lemma of $O((det(BB'))^{\frac1{2(k-k')}})$ where $B'$ is transpose of $B$.
If $$N_q(B)=\{|(x_1,\dots,x_k)|_\infty:\sum_{i=1}^kx_ib_{1i}=0\wedge\dots\wedge\sum_{i=1}^kx_ib_{k'i}=0\wedge\sum_{i=1}^kx_i^2=q\}$$ where entries $b_{ij}$ of $B$ come from $T_t(\beta,q)$ and $N_q(B)\neq\emptyset$ holds then what is a good upper and lower bound for $q$?
Is there a name for this $q$?
Probably you want $a_i\ne 0$, or at least not all of them 0.
|
2025-03-21T14:48:31.128321
| 2020-05-31T08:26:52 |
361792
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629662",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361792"
}
|
Stack Exchange
|
Comparison of two well-known bases of the integral homology group of based loop group
Let $G$ be a compact simply-connected Lie group. Then one can look at the homology $H_*(\Omega G;\mathbb{Z})$ of the based-loop space $\Omega G$ in at least two different ways:
(1) Via Bott-Samelson's theory. They used Morse theory to show that the Bott-Samelson varieties associated to elements $\omega t_{\lambda}$ of the affine Weyl group lying in the dominant Weyl chamber exhibit an additive basis of $H_*(\Omega G;\mathbb{Z})$.
(Notice the reduced decomposition is obtained by drawing a line segment from the origin to any interior point of the cell represented by $\omega t_{\lambda}$.)
(2) Via affine Grassmannian $Gr_{G_{\mathbb{C}}}$ of $G_{\mathbb{C}}$. It is known that it has the same homotopy type as $\Omega G$ so that
$$ H_*(\Omega G;\mathbb{Z})\simeq H_*(Gr_{G_{\mathbb{C}}};\mathbb{Z})$$
and that the latter is generated by the affine Schubert subvarieties.
Is there any way to compare these two bases? Are they identical?
|
2025-03-21T14:48:31.128415
| 2020-05-31T08:46:12 |
361794
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/93600",
"sleeve chen"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629663",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361794"
}
|
Stack Exchange
|
Suggested papers or reading for PDE (high dimension) reduction to ODE by symmetries
Could anyone please suggest related papers or article about the topic related to my one question below?
Reduce PDE to ODE by dilation symmetry
I also cite a paper in the link above.
We know that if there are number of $n$ states, we need to find $"n-1"$ symmetries to reduce the PDE to ODE. For each iteration (there are $n-1$ iterations), we need to do some tedious but not difficult calculation.
Is there any advanced method based on symmetries reduction (of course, it might based on some special condition of the structure of problems) such that maybe we can take fewer steps?
Thanks in advance.
We really want to find invariant differential relations between the invariant functions. The symmetries might not commute, so the reduction cannot necessarily be carried out in the stages that your problem suggests.
@BenMcKay Yes! But I mean if the equations do have some nice form (please let me know which form is qualified) then we can reduce the PDE to ODE more efficiently.
As you know from your previous question, Peter Olver's book is probably the main source of information on symmetry methods for differential equations. You could also look at the books of Ibrahimov.
|
2025-03-21T14:48:31.128529
| 2020-05-31T09:36:06 |
361797
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/127334",
"zab"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629664",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361797"
}
|
Stack Exchange
|
Oscillatory integral independent of a parameter
Let $h: \mathbb{R} \mapsto \mathbb{C}$ be a positive definite function, continuous at the origin. (In fact, $h$ is the Fourier transform of a finite measure). Define the oscillatory integral
$$Q(t) := \int_0^\infty \frac{\text{Im} \{ h(u)^t\}}{u} du, \quad t > 0.$$
Are there non-trivial examples of $h$ that makes $Q$ independent of $t$?
Is there a known class of functions $\{ h \}$ that have this property?
$h(u)=e^{iu}$ qualifies, doesn't it?
Yes it does. Unfortunately for my purposes, it is a trivial case, i.e it's the Fourier transform of a point mass.
|
2025-03-21T14:48:31.128616
| 2020-05-31T09:45:35 |
361798
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"Libli",
"Robert Bryant",
"Theo Johnson-Freyd",
"https://mathoverflow.net/users/13972",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/26290",
"https://mathoverflow.net/users/78"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629665",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361798"
}
|
Stack Exchange
|
Subgroup $\mathrm{E}_6$ generated by $\mathrm{Spin_7}$ and $\mathrm{SL}_3$
Let $\mathbb{O}$ be the octonion algebra (say over $\mathbb{R}$) and let $J_{3}(\mathbb{O})$ be the set of $3 \times 3$ hermitian matrices with octonion coefficients, that is:
$$ J_3(\mathbb{O}) = \left\{ \begin{pmatrix} \lambda_1 & a & b \\ \overline{a} & \lambda_2 & c \\ \overline{b} & \overline{c} & \lambda_3 \end{pmatrix}, \ \ \lambda_i \in \mathbb{R}, \ a,b,c \in \mathbb{O} \right\}$$
The group $\mathrm{E}_6$ is the group of linear automorphisms of $J_{3}(\mathbb{O})$ which preserve the cubic form:
$$\lambda_1 \lambda_2 \lambda_3 + 2 \mathrm{Re}(a\overline{b}c) - \lambda_2 N(b)^2 - \lambda_3 N(a)^2 - \lambda_1 N(c)^2,$$ where $N$ is the norm over $\mathbb{O}$.
There are many interesting subgroups of $\mathrm{E}_6$ related to this description. $\mathrm{SL}_3(\mathbb{R})$ is one of them. The action of $\mathrm{SL}_3(\mathbb{R})$ on $J_3(\mathbb{O})$ is given by :
$$ \forall g \in \mathrm{SL}_3(\mathbb{R}), \forall A \in J_{3}(\mathbb{O}), \ g\cdot A = g A\,^{t}\! g,$$
where $^{t}\!g$ is the transpose of $g$.
The group $\mathrm{Spin_8}$ can also be seen as a subgroup of $\mathrm{E}_6$ with the action:
\begin{align*} &\forall (g_1,g_2,g_3) \in \mathrm{Spin}_8,\quad \forall A = \begin{pmatrix} \lambda_1 & a & b \\ \overline{a} & \lambda_2 & c \\ \overline{b} & \overline{c} & \lambda_3 \end{pmatrix} \in J_{3}(\mathbb{O}), \\ &g\cdot A = \begin{pmatrix} \lambda_1 & g_1(a) & g_2(b) \\ \overline{g_1(a)} & \lambda_2 & g_3(c) \\ \overline{g_2(b)} & \overline{g_3(c)} & \lambda_3 \end{pmatrix},
\end{align*}
where we identify $\mathrm{Spin_8}$ with $\{(g_1,g_2,g_3) \in \mathrm{SO}_8^3, \ \forall (x,y) \in \mathbb{O}, \ g_3(xy) = g_1(x)g_2(y) \}$.
It is well-known (see Harvey's Spinor and calibrations for instance) that the subgroup of $E_6$ generated by $\mathrm{SO_3}$ and $\mathrm{Spin}_8$ is $\mathrm{F}_4$. I think it is equally well-known (I don't have a reference at hand, but it seems to be an easy corollary of the previous statement) that $\mathrm{E}_6$ itself is generated by $\mathrm{SL}_3$ and $\mathrm{Spin_8}$.
Question : Is there an explicit description the subgroup of $\mathrm{E}_6$ (resp. $\mathrm{F}_4$) generated by $\mathrm{SL}_{3}$ and $\mathrm{Spin}_7$ (resp. $\mathrm{SO}_3$ and $\mathrm{Spin}_7$), where $\mathrm{Spin_7}$ is seen in $\mathrm{Spin}_8$ as $\{(g_1,g_2,g_3) \in \mathrm{Spin}_8, \ g_1(1) = 1\}$?
I think SL(3) contains elements that act on Spin(8) via triality. So I think those elements together with Spin(7) already generate Spin(8).
@TheoJohnson-Freyd : This doesn't look possible to me. If you represent an element of $J_3(\mathbb{O})$ as a $24 \times 24$ matrix, it happens, quite miraculously, that the action of any element $g \in \mathrm{Spin}7$ on $A \in J_3(\mathbb{O})$ can be written as $GAG^{-1}$ for a uniquely defined $G \in \mathrm{SL}{24}$. On the other hand, it is impossible for the action of $F_4$ to be represented by matrix conjugation...
Indeed any element $A \in J_{3}(\mathbb{O})$ can be diagonalized (with at most $3$ different real entries on the diagonal) using a $F_4$ transformation. If the action of $F_4$ could be represented by matrix conjugation, then the dimension of the kernel (in $\mathbb{R}^{24}$) would be the same after diagonalization.
But the dimension of the kernel of a $24 \times 24$ diagonal matrix with at most three different real entries on the diagonal is either 0, 8, 16 or 24. While it is easily shown that there are $24 \times 24$ matrices in $J_3(\mathbb{O})$ which have kernel of dimension $4$.
Huh. I admit I was just guessing. What you say sounds reasonable.
Is this the split $\operatorname E_6$?
@LSpice : it should rather be $\mathrm{E}_{6(-26)}$.
@Libli: Note that your formula for the cubic form is incorrect. Where you have $\mathrm{Re}(abc)$ you should have $\mathrm{Re}(a\bar bc)$. Either that or you should replace $b$ by $\bar b$ in your coordinatization of $J_3(\mathbb{O})$.
I should probably add that Theo Johnson-Freyd was acutally right. I made a decisive mistake when I asked my question
The embedding of $\mathrm{Spin}7$ in $\mathrm{SL}{27}$ I am looking at does not factor through $\mathrm{E}_6$ (though it looks like it does).
Anyway, this is my fault, I should have been more careful. In any case, I won't delete my previous comments (even if they don't apply to the embedding I am referring to in my question) for the sake of this thread's coherence.
@LSpice: If you want the split $\mathrm{E}_6$, you can just replace the octonions by the split octonioins. Everything will work as before, but the stabilizer of that cubic form (same formula) will be $\mathrm{E}^{(6)}_6$, i.e., the split form. (The other three real forms of $\mathrm{E}_6$ don't have a representation on $\mathbb{R}^{27}$.)
N.B.: I am revising my response for clarity. (The actual answer to the question asked by the OP is still the same, but I think that this re-organization, particularly at the end, makes the structure of the argument for the answer more clear. I was inspired to do this because some people had some difficulty following the original.) I should also say that the main idea is essentially the one that Theo Johnson-Freyd proposed in his first comment on the question.
I'll use the more usual notation
$$
J_3(\mathbb{O}) = \left\{\ \left.\begin{pmatrix} \lambda_1 & a_3 & {\overline{a_2}} \\ \overline{a_3} & \lambda_2 & a_1 \\ a_2 & \overline{a_1} & \lambda_3 \end{pmatrix}\ \right| \ \ \lambda_i \in \mathbb{R}, \ a_i \in \mathbb{O} \right\}
\tag 1
$$
and the cubic form given by
$$
C = \lambda_1\lambda_2\lambda_3 + 2\,\mathrm{Re}(a_1a_2a_3)
- \lambda_1\,a_1\overline{a_1} - \lambda_2\,a_2\overline{a_2}
- \lambda_3\,a_3\overline{a_3}\,.
$$
Then $\mathrm{E}_6\subset\mathrm{GL}\bigl(J_3(\mathbb{O})\bigr)\simeq \mathrm{GL}(27,\mathbb{R})$ is the group of linear transformations of $J_3(\mathbb{O})$ that preserve the cubic form $C$ and $\mathrm{F}_4\subset\mathrm{E}_6$ is the subgroup that also fixes $I_3\in J_3(\mathbb{O})$. (Explicitly, $\mathrm{F}_4$ is a maximal compact in this noncompact real form $\mathrm{E}_6^{(-26)}$
of $\mathrm{E}_6$.)
The subgroup $\mathrm{Spin}(8)\subset{\mathrm{SO}(8)}^3$ is defined as the set of triples $g = (g_1,g_2,g_3)$ that satisfy
$$
\mathrm{Re}\bigl(g_1(a_1)g_2(a_2)g_3(a_3)\bigr) = \mathrm{Re}(a_1a_2a_3)
$$
for all $a_i\in\mathbb{O}$. Let $K_i\subset\mathrm{Spin}(8)$ for $1\le i\le 3$ be the subgroup that satisfies $g_i(\mathbf{1}) = \mathbf{1}$ (where $\mathbf{1}\in\mathbb{O}$ is the multiplicative identity). Each of the $K_i$ is isomorphic to $\mathrm{Spin}(7)$, any two of them generate $\mathrm{Spin}(8)$, and the intersection of any two of them is equal to the intersection of all three of them, which is a group isomorphic to $\mathrm{G}_2$, diagonally embedded in ${\mathrm{SO}(8)}^3$ as the automorphism group of the octonions.
As has already been observed, $\mathrm{SL}(3,\mathbb{R})$ acts on $J_3(\mathbb{O})$ preserving $C$ via $a\cdot A = a\,A\,^{t}a$ (usual matrix multiplication), where $a\in\mathrm{SL}(3,\mathbb{R})$ and $A\in J_3(\mathbb{O})$ are arbitrary. This is a faithful action, so, in this way, $\mathrm{SL}(3,\mathbb{R})$ will be regarded as a subgroup of $\mathrm{E}_6$.
Meanwhile, by its very definition, $g = (g_1,g_2,g_3)\in\mathrm{Spin}(8)$ acts on $A\in J_3(\mathbb{O})$ via
$$
g\cdot \begin{pmatrix} \lambda_1 & a_3 & \overline{a_2} \\ \overline{a_3} & \lambda_2 & a_1 \\ a_2 & \overline{a_1} & \lambda_3 \end{pmatrix}
= \begin{pmatrix} \lambda_1 & g_3(a_3) & {\overline{g_2(a_2)}} \\ \overline{g_3(a_3)} & \lambda_2 & g_1(a_1) \\ g_2(a_2) & \overline{g_1(a_1)} & \lambda_3 \end{pmatrix}\tag 2
$$
and this faithful action preserves $C$ as well, so $\mathrm{Spin}(8)$ will also be regarded as a subgroup of $\mathrm{E}_6$.
Now, as mentioned, $\mathrm{SO}(3)$ and $\mathrm{Spin}(8)$ together generate $\mathrm{F}_4\subset \mathrm{E}_6$. Consequently (since there is no connected Lie group that lies properly between $\mathrm{F}_4$ and $\mathrm{E}_6$), it follows easily that $\mathrm{SL}(3,\mathbb{R})$ and $\mathrm{Spin(8)}$ together generate $\mathrm{E}_6$.
We want to show that $\mathrm{SO}(3)$ and $K_1\simeq\mathrm{Spin}(7)$ also suffice to generate $\mathrm{F}_4$ while $\mathrm{SL}(3,\mathbb{R})$ and $K_1\simeq\mathrm{Spin}(7)$ suffice to generate $\mathrm{E}_6$.
To do this, let $h\in\mathrm{SO}(3)\subset\mathrm{SL}(3,\mathbb{R})$ be
$$
h = \begin{pmatrix}0&-1&0\\-1&0&0\\0&0&-1\end{pmatrix} = h^{-1} = {}^th.
$$
Then we have
$$
h\cdot\begin{pmatrix} \lambda_1 & a_3 & \overline{a_2} \\ \overline{a_3} & \lambda_2 & a_1 \\ a_2 & \overline{a_1} & \lambda_3 \end{pmatrix}
= \begin{pmatrix} \lambda_2 & \overline{a_3} & a_1 \\ a_3 & \lambda_1 & \overline{a_2} \\ \overline{a_1} & a_2 & \lambda_3 \end{pmatrix}.
$$
Consequently, for $g = (g_1,g_2,g_3)\in \mathrm{Spin}(8)$, computation yields
$$
h(g_1,g_2,g_3)h = \bigl(\ cg_2c,\ cg_1c,\ cg_3c\ \bigr)\in\mathrm{Spin}(8),
\tag 3
$$
where $c:\mathbb{O}\to\mathbb{O}$ is conjugation, i.e., $c(a) = \overline{a}$.
(Thus, conjugation by $h$ gives an involution of $\mathrm{Spin}(8)$ that, together with the order $3$ homomorphism $k(g_1,g_2,g_3) = (g_2,g_3,g_1)$, generates a group of automorphisms of $\mathrm{Spin}(8)$ isomorphic to $S_3$ that maps isomorphically onto $\mathrm{Out}\bigl(\mathrm{Spin}(8)\bigr)$. I imagine that this is what Theo Johnson-Freyd had in mind with his initial comment on this question.)
Note that $g_i(\mathbf{1}) =\mathbf{1}$ implies that $cg_ic = g_i$. Consequently, from the above formula $(3)$, it follows that if $g\in K_1$, then $hgh\in K_2$.
Thus the subgroup of $\mathrm{E}_6$ generated by $\mathrm{SO}(3)$ and $K_1 \simeq\mathrm{Spin}(7)$ contains $K_2$ and, hence, $\mathrm{Spin}(8)$ (since $K_1$ and $K_2$ generate $\mathrm{Spin}(8)$). Thus, this group is $\mathrm{F}_4$. Similarly, the subgroup of $\mathrm{E}_6$ generated by $\mathrm{SL}(3,\mathbb{R})$ and $K_1 \simeq\mathrm{Spin}(7)$ contains $K_2$
and, hence, $\mathrm{Spin}(8)$. Thus, this group is $\mathrm{E}_6$, as desired.
Similar arguments (using similar choices of $h$) suffice to show that, for any of $i= 1$, $2$, or $3$, the subgroup of $\mathrm{E}_6$ generated by $\mathrm{SO}(3)$ and $K_i$ is $\mathrm{F}_4$, while $\mathrm{SL}(3,\mathbb{R})$ and $K_i$ generate $\mathrm{E}_6$.
|
2025-03-21T14:48:31.129290
| 2020-05-31T10:04:26 |
361799
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629666",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361799"
}
|
Stack Exchange
|
Diffeomorphism for mapping one SDE into another
Let $Y_t,X_t$ be $(\Omega,\mathcal{F},\mathcal{F}_t,\mathbb{P})$-adapted Markov diffusion processes with valued in $\mathbb{R}^n$. (When) does there exist a diffeomorphism $\phi:\mathbb{R}^n\to \mathbb{R}^n$ such that
$$
\phi(X_t)= Y_t \mathbb{P}-a.s?
$$
More generally, since the continuous image (in the above sense) of a diffusion process needs not be a diffusion process since the Ito-formula (and extensions thereof) fail. What conditions do we need on $X,Y$ and $\phi$ such that $Y_t$ is just an $\mathbb{R}^n$-valued Markov process?
This is more of a long comment. As far as I know, the first question is a very difficult (and interesting) one. Here are a few examples of diffusions in $\mathbb R^3$ that are not diffeomorphic in your sense, and may give an idea of the difficulties ahead:
the standard Brownian motion in $\mathbb R^3$;
the Brownian motion along planes $z=c$, i.e. $X_t=(B^1_t,B^2_t,c)$ for $c$ a constant and $B^1,B^2$ independent Brownian motions;
the Brownian motion along planes with orthogonal drift, i.e. $X_t=(B^1_t,B^2_t,z_0+t)$;
the Brownian motion along the (contact) distribution $dz-ydx=0$,¹ i.e.
$$X_t = \left(B^1_t,B^2_t,\int_0^tB^2_s\circ dB^1_s\right);$$
the Brownian motion reflected along the plane $z=0$, i.e. $X_t = (B^1_t,B^2_t,Z_t)$ where $Z_t=B^3_t$ until $B^3$ reaches zero for the first time, then $Z_t=|B^3_t|$.
When one restricts to solution of SDEs of the form $dX_t=b(X_t)dt + \sigma(X_t)dB_t$ with $b,\sigma$ smooth and $\sigma$ of constant rank (points 1 to 4 above), we could imagine that the data of a distribution with a quadratic form on it, together with a vector field, would suffice to characterise the diffusion, but it isn't the case. For $\epsilon_1,\epsilon_2,\epsilon_3$ the standard basis of $\mathbb R^3$, $\epsilon_1 dB^1_t+\epsilon_2dB^2_t$ and
$$ \big( \epsilon_1\cos(x^1)+\epsilon_2\sin(x^1)\big)dB^1_t
+ \big(-\epsilon_1\sin(x^1)+\epsilon_2\cos(x^1)\big)dB^2_t$$
do not define identical diffusions, although the randomness propagates along the same planes $z=c$ and correspond to the same quadratic form. In short, the question you are asking
contains strictly the question of whether two second degree operators are diffeomorphic, which
contains strictly that of whether two Euclidean distributions are diffeomorphic, which
contains strictly the (as far as I know) difficult topological question of whether two distributions are locally diffeomorphic,
so I think the first question is too broad as it stands.
Edit: Now that I think about it more, maybe my argument goes in fact in the other direction: given two unrelated diffusions $X$ and $Y$, there are so many different possibilities that it should be easy to show they are not diffeomorphic. In the examples 1 to 4 above, we can argue by considering iterated brackets, in the somewhat classical description of e.g. Hörmander.
Another example of this: if the diffusions are elliptic with the same quadratic part, for instance $X$ and $Y$ have generators $\frac12\Delta+b_X$ and $\frac12\Delta+b_Y$ for some smooth vector fields $b_X$ and $b_Y$, then a diffeomorphism sending $X$ to $Y$ would have to send the principal symbol of one generator to the other, and hence be an isometry. In this case it should be easy to see whether $b_Y$ is just a version of $b_X$ that got moved around.
¹ See the wikipedia page about contact structures for an illustration. Here I mean distribution in the sense “subbundle of the tangent bundle”.
|
2025-03-21T14:48:31.129533
| 2020-05-31T11:16:38 |
361800
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dieter Kadelka",
"Math_Y",
"S.Surace",
"Yuval Peres",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/2363",
"https://mathoverflow.net/users/68835",
"https://mathoverflow.net/users/69603",
"https://mathoverflow.net/users/7691",
"zeb"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629667",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361800"
}
|
Stack Exchange
|
How to solve this stochastic optimization problem?
How one can solve the following stochastic optimization problem?
\begin{align}
\max\quad& \mathbb{E}[\mathbf{1}^{\mathrm{T}}X]\\
\text{s.t.} \quad& \mathbb{E}[\mathbf{A}X]\leq\mathbf{1}_{m\times 1}\qquad(\ast),
\end{align}
where $\mathbf{A}$ is a random matrix and consequently the solution vector $X$ is also a random vector.
I think about a simpler case. We know that for each realization of $\mathbf{A}$, one can solve the following
\begin{align}
\tilde{X}=\arg\max\quad& \mathbf{1}^{\mathrm{T}}X\\
\text{s.t.} \quad& \mathbf{A}X\leq\mathbf{1}_{m\times 1}\qquad(\ast\ast),
\end{align}
and so we can compute $\mathbb{E}[\mathbf{1}^{\mathrm{T}}\tilde{X}]$ by Solving $(\ast\ast)$ several times (for different realizations of $\mathbf{A}$) and then average. But, what can we do about $(\ast)$?
I don't understand your "consequently the solution vector $X$ is also a random vector". You should present a little bit more detail. What is the background, what is $X_i$, dependent or not or even deterministic? More precise: $X \in ?$.
Is X independent of A?
No, X is not independent of $\mathbf{A}$.
If you optimize for each realization you satisfy (*) so you would need to prove that you cannot do any better than this.
This is a slight addition to RaphaelB4's answer, to explain one approach to extending the discrete case to the general case. This approach begins by thinking about the discrete case in slightly more detail.
Note that RaphaelB4's optimization problem only has one input parameter: the matrix $\hat{A}$ formed by concatenating the finite set of matrices $A_i$ in the (assumed to be discrete) support of the random variable $\mathbf{A}$. Note also that the matrix $\hat{A}$ has the same number of rows as each $A_i$, and that its set of columns is the union of the columns of the $A_i$. So there's really no difference from the case where the matrices $A_i$ are just column vectors. For instance, we could make a new random variable $\mathbf{A}'$ taking values in column vectors, which first randomly picks a matrix $A_i$ and then chooses one of its columns uniformly at random.
Also, note that the optimization problem
$$\max_{\mathbb{E}[\mathbf{A}]X \le 1} 1^TX$$
with a deterministic vector $X$ gives a lower bound on the solution to the optimization problem $(*)$. Since the probabilities are actually irrelevant to the problem, we may set them however we like: so in fact, we only care about the convex hull of the set of matrices $A_i$ that show up in the support of the random variable $\mathbf{A}$.
Putting these two ideas together, we see that the answer to the optimization problem $(*)$ only depends of the convex hull $\mathcal{C}$ of the set of column vectors that show up as columns of matrices in the support of $\mathbf{A}$. If $C \in \mathcal{C}$, then we can easily check that
$$\max_{\mathbb{E}[\mathbf{A}X]\le 1} \mathbb{E}[1^TX] \ge \max_{Cx \le 1} x = \frac{1}{\max(\max_i C_i, 0)}.$$
In fact, a little thought shows that we have
$$\max_{\mathbb{E}[\mathbf{A}X]\le 1} \mathbb{E}[1^TX] = \max_{C \in \mathcal{C}} \frac{1}{\max(\max_i C_i, 0)}.$$
So the problem reduces to minimizing the convex function $\max_i C_i$ over column vectors $C$ in the convex set $\mathcal{C}$. If you have a good description of the support of your random matrix $\mathbf{A}$, then it should be possible to convert this into a good enough description of the convex hull $\mathcal{C}$ to apply a standard convex optimization algorithm (such as the ellipsoid algorithm).
Edit: I seem to have made an implicit assumption. The above is true only if we require that $X \ge 0$ as well. If we don't require the coordinates of $X$ to be positive, then "convex hull" should be replaced by "affine hull" everywhere. This actually makes the problem easier: it should be much simpler to describe the affine hull of the set of columns of matrices in the support of $\mathbf{A}$ than to describe the convex hull.
If we replace $1^\mathrm{T}\mathbf{X}$ with $\min_i X_i$ in the objective function, what could we say?
In that case the exact probabilities still don't matter, but I think we can no longer split each matrix into its columns separately. We can still say that the answer will only depend on the convex hull of the support of the random matrix $\mathbf{A}$, but beyond that the problem seems to have become somewhat harder.
Thank you. Can I have your email?
I consider the case of discrete probability. There exists $n\in \mathbb{N}$, $A_1,\cdots,A_n$ and $p_1,\cdots, p_n>0$ such that $\sum_{i=1}^n p_i = 1$ and $\mathbb{P}(A=A_i)=p_i$. Then we denote $X_1,\cdot,X_n$ such that $X=X_i$ in the event $A=A_i$. The problem ($\star$) becomes $$\max \sum_{i=1}^n p_i 1^T X_i = \hat{1}^T \hat{X},\\ \quad \sum_{i=1}^n p_i A_i X_i = \hat{A} \hat{X}\leq 1_m $$ where $\hat{1}=1_{n p}$, $\hat{A}=(A_1,A_2,\cdots,A_n)\in \mathbb{R}^{m\times np}$ and $\hat{X}=(p_1 X_1,\cdots, p_n X_n)\in \mathbb{R}^{np}$. Therefore it a usual optimisation problem similar as $(\star \star)$ but with $X$ in a larger dimension set.
Remark : In particular for $i_\max$ the event $i$ such that the solution of $(\star,\star)$ with $A_i$ gives the largest value. Then chosing $\hat{X}=(0,\cdots,0,X_{i_{\max}},0,\cdots,0)$ we obtain that the solution of $(\star)$ gives an higher value that the maximum of all possible $(\star,\star)$ and this will also be valid for non discrete probability.
Interestingly, this approach shows that the actual values of the probabilities $p_i$ are irrelevant to the final answer (as long as all the $p_i$ are nonzero).
|
2025-03-21T14:48:31.129916
| 2020-05-31T13:03:17 |
361805
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Nik Weaver",
"https://mathoverflow.net/users/2258",
"https://mathoverflow.net/users/23141",
"santker heboln"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629668",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361805"
}
|
Stack Exchange
|
Extensions of trace class operators
Let $H_i, K_i$ for $i=1,2,3$ be Hilbert spaces with horizontal exact sequences. Assume $T_1, T_2, T_3$ have dense range, that $T_1, T_2, T_3$ are trace class operators and that the squares commute. Assume that $S \colon H_2 \to K_2$ is a compact operator with dense range making the squares commute.
$\require{AMScd}$
\begin{CD}
0 @>>> H_1 @>f_1>> H_2 @>f_2>> H_3 @>>> 0\\
@V VV @V T_1 VV @V T_2 VV @V T_3 VV @V VV \\
0 @>>> K_1 @>>g_1> K_2 @>>g_2> K_3 @>>> 0
\end{CD}
Is it the case that there is a $C > 0$ such that $C \mathrm{Tr} |T_2| \geq \mathrm{Tr} |S|$?
This is related to the question: Are nuclear operators closed under extensions?
What are your morphisms between Hilbert spaces?
Bounded operators
The answer is no. Take $H_2 = K_2$ to be an infinite dimensional Hilbert space and take $H_1 = K_1$ to be an infinite dimensional subspace whose orthocomplement $H_3 = K_3$ is also infinite dimensional. Let $f_1 = g_1$ be the inclusion and $f_2 = g_2$ the orthogonal projection. Let $T_1: H_1 \to K_1$ and $T_3: H_3 \to K_3$ be any trace class operators with dense range such that ${\rm ker}(T_3)$ is infinite dimensional. Let $T_2 = T_1 \oplus T_3$.
Let $S_0: H_2 \to K_2$ be any operator which vanishes on ${\rm ker}(T_3)^\perp$ and takes ${\rm ker}(T_3)$ into $K_1$. Then $S = S_0 + T_2$ will make the diagram (with $T_1$ and $T_3$ unaltered, I am sure you meant) commute. So clearly the trace of $|S|$ can be arbitrarily large. Also, ${\rm ran}(S)$ contains both ${\rm ran}(T_1)$ and ${\rm ran}(T_3|_{{\rm ker}(T_3)^\perp}) = {\rm ran}(T_3)$, so the range of $S$ is dense.
(In the first version of this answer I didn't notice that the ranges had to be dense.)
|
2025-03-21T14:48:31.130043
| 2020-05-31T13:29:12 |
361807
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/9449",
"roy smith"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629669",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361807"
}
|
Stack Exchange
|
The morphisms induced by two Cartier divisors
Let X be a projective variety. We consider two Cartier divisors $D,E$ such that $E\geq D$ and the relative morphisms
$\phi_D: X - - -> \mathbb{P}(H^0(X, O_X(D))^*)$ and $\phi_E: X- - -> \mathbb{P}(H^0(X, O_X(E))^*)$
Is there a relation between them?
I think that the map $\phi_D$ can be factorized via the map $\phi_E$ in the following sense:
It's clear that $L(D)\subseteq L(E)$. If $v$ and $w$ are two rational sections of $O_X(D)$ and $O_X(E)$ respectively, then $H^0(X, O_X(D))\cong L(D)$ via the tensor map $\otimes v$ and $H^0(X, O_X(E))\cong L(E)$ via the tensor map $\otimes w$. So we get a monomorphism
$0\to H^0(X, O_X(D))\to H^0(X, O_X(E))$
which sends a section $s$ to a section $\{(U_j, w_j\cdot \frac{s_j}{v_j})\}$. This means $h^0(X, O_X(D))\leq h^0(X, O_X(E))$.
Now we can consider the dual morphism
$H^0(X, O_X(E))^*\to H^0(X, O_X(D))^*$
and then we would consider the morphism induced to the projective spaces
$F: \mathbb{P}(H^0(X, O_X(E))^*)---> \mathbb{P}(H^0(X, O_X(D))^*)$
that it is not well defined everywhere because in general the dimension of $H^0(X, O_X(E))^*$ is greater or equal than the dimension of $H^0(X, O_X(D))^*$ and so the kernel of our map will be not trivial. In particular $F$ is not defined on $\{[f]: f|_{H^0(X,O_X(D))}=0\}$. Thus we can consider the morphism
$F\circ \phi_E: X---> \mathbb{P}(H^0(X, O_X(D))^*)$
that takes a point $p\in U_j$ and sends it to
$[s\to w_j(p)\frac{s_j}{v_j}(p)]$
This morphism will be not defined on $Bs(E)\cup \phi_E^{-1}\{[f]: f|_{H^0(X,O_X(D))}=0\}$ that corresponds to $Bs(D)\cup supp(E-D)$, right?
Moreover we observe $F\circ \phi_E=\phi_D$ on $X\setminus (Bs(D)\cup supp(E-D))$, right?
Why is it interesting for me this question?
In this way we could study the morphism $\phi_D$ through the morphism $\phi_E$. For example, if $D$ is an effective globally generated divisor, then $mD\geq (m-1)D$ and this permit us to say
$h^0(X, O_X((m-1)D))\leq h^0(X, O_X(mD))$
Moreover, if $D$ is also ample, then we have
$\phi_D=F\circ \phi_{mD}$
where $\phi_{mD}$ is an embedding. If I prove that $F$ has finite fiber, then $\phi_D$ has finite fiber, that is the usual property of ample divisors.
I know this property can be proved by contradiction taking a curve $C$ on a fiber $\phi_D^{-1}(p)$, and so
$0<deg((mD)|C)=(mD).C=mD.C=m\phi_D^*H.C=mH.\phi_*C=0$
but I would understand if my argument is good to prove it in another way.
I want to encourage you in this line of thought, as it seems to me, if I am seeing this correctly, you are using the time honored method of viewing the morphism defined by a divisor D, as the composition of that defined by D+B, and the projection from the subspace spanned by (the image of) B. This has often been useful to me in studying curves.
|
2025-03-21T14:48:31.130229
| 2020-05-31T14:09:11 |
361808
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"https://mathoverflow.net/users/13268"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629670",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361808"
}
|
Stack Exchange
|
Morse theory for vector-valued functions
Let $f:\mathbb{R}^{m+k}\mapsto\mathbb{R}^k$ be a smooth function. I have seen quite a few books for Morse theory for $f$ when $k=1$. Is there a generalization to $k\geq2$? When $k=1$, we can define a Morse function $f$ by checking the eigenvalues of its Hessian at its critical points. What is the corresponding concept of (non-)degeneracy of critical points when $k\geq 2$? Is there a normal (quadratic) form of $f$ near its nondengerate critical points when $k\geq 2$, as given in the classical Morse lemma? Any references will also be appreciated.
I think you might want Cerf theory.
There is a vast literature on singularities, like:
Gibson, Christopher G.; Wirthmüller, Klaus; du Plessis, Andrew A.; Looijenga, Eduard J. N. Topological stability of smooth mappings. Lecture Notes in Mathematics, Vol. 552. Springer-Verlag, Berlin-New York, 1976
Arnold, V. I.; Gusein-Zade, S. M.; Varchenko, A. N. Singularities of differentiable maps. Volume 1. Classification of critical points, caustics and wave fronts. Birkhäuser 1985, 2012
Arnold, V. I.; Gusein-Zade, S. M.; Varchenko, A. N. Singularities of differentiable maps. Volume 2. Monodromy and asymptotics of integrals. Birkhäuser 1988, 2012.
The answer in
Modification of Morse lemma with two functions
shows that this doesn't work in general (depending on what you mean)
|
2025-03-21T14:48:31.130333
| 2020-05-31T14:11:41 |
361809
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Handelman",
"https://mathoverflow.net/users/42278"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629671",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361809"
}
|
Stack Exchange
|
Primitive ideals of inductive limits of $C^*$-algebras
I am trying to understand ideals of direct limits in the category of $C^{\ast}$-algebras.
Let $(A_n,f_n)$ be a direct sequence of $C^{\ast}$-algebras and let $I$ be a primitive (modular) ideal of direct limit $\varinjlim A_n $. Is it true that there exists primitive (modular) ideals $I_i$ of $A_i$ such that $I= \varinjlim I_i$?
Unfortunately I could not find any reference.Any reference or ideas?
No. Let $B$ be a simple unital AF C*-algebra that is not ultrasimplicial (that is, its ordered K$0$ group is not of rank one); we can find examples so that in addition, $B$ is the completion of $A = \lim A_i$ ($A_i$ finite-dimensional C*-algebras) and the induced maps $K_0 (A_i) \to K_0(A{i+1})$ are one to one (not all simple AF C*-algebras can be so realized, but lots can). Then ${0}$ is a primitive ideal, but cannot be a limit of primitive ideals in the $A_i$, since it can only be a limit of the zero ideals, all but finitely many of which are not primitive.
|
2025-03-21T14:48:31.130440
| 2020-05-31T15:46:57 |
361817
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bernhard Boehmler",
"Geoff Robinson",
"Mark Wildon",
"https://mathoverflow.net/users/12826",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/7709"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629672",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361817"
}
|
Stack Exchange
|
What is known about ordinary character values at involutions?
Let $G$ be a finite group and let $\chi$ be the character of an irreducible complex representation $\rho$ of $G$ on $V$.
Let $x$ be an involution in $G$.
I'd like to ask the following
Question 1:
What is known about $\chi(x)?$
1a) Are there criteria when $\chi(x)$ is positive / negative / zero ?
Of course, $1_{\text{Aut}(V)}=\rho(x^2)=\rho(x)^2$, such that the only possible eigenvalues of $\rho(x)$ are $\pm 1$.
Moreover, there is an article written by P.X. Gallagher with the title "Character values at involutions" (DOI: https://doi.org/10.1090/S0002-9939-1994-1185260-1) dealing with the case that $\int_G\chi_1\chi_2\chi_3 \neq 0$, where the integral is in the sense of the Haar measure.
EDIT:
The main parts of Gallagher's results are the following ones:
For an involution $\sigma$ of a finite group $G$ and an irreducible complex representation $R$ of $G$, denote by $q$ the proportion of $-1$'s among the eigenvalues of $R(\sigma)$. Then:
$(*)$ $\frac{1}{h}\leq q \leq 1-\frac{1}{h}$, unless $q = 0$ or $1$,
where $h$ is the index of the centralizer $C$ of $\sigma$.
Moreover, if $\int_G\chi_1\chi_2\chi_3 \neq 0$, then $(*)$ is refined to prove that the proportions of $-1$'s among the eigenvalues of $\rho_1, \rho_2$ and $\rho_3$ (i.e., the corresponding representations) at $\sigma$ are the sides of a triangle on a sphere of circumference 2.
$\ $
1b) When does $\int_G\chi_1\chi_2\chi_3 \neq 0$ happen (necessary / sufficient criteria)?
1c) When does $\int_G\chi_1\chi_2\chi_3 = 0$ happen (necessary / sufficient criteria)?
1d) Are there results apart from Gallagher's result?
1e) Can one deduce additional information, if all considered characters lie in the same 2-block?
Thanks for the help.
Since $\int_G \chi_1\chi_2\chi_3 = \langle \chi_1\chi_2, \overline{\chi}_3 \rangle$, the questions (1b) and (1c) are equivalent to decomposing arbitrary tensor products of representations. This is a tough problem on which there are many interesting results but no general theory. For instance, Saxl has conjectured that for the symmetric group if $\chi = \chi^{(m,m-1,\ldots, 1)}$ is the 'staircase' character then every irreducible character appears in $\chi^2$.
I think your question would be slightly improved if you could briefly indicate why Gallagher's result is related to involutions.
Thank you very much for the comment. I've edited the question.
Just a small remark: Theorem 3.1 on page 456 of Feit's book is also related, see https://books.google.de/books?id=6EvRfXt2mrcC&pg=PA456&lpg=PA456&dq=%22brauer%22%2B%22block%22%2B%22character%22%2B%22involution%22&source=bl&ots=xuJPZ4pBVa&sig=ACfU3U3YDKD89hRrShtbpIxeOwou45556A&hl=de&sa=X&ved=2ahUKEwihsPW41eDpAhURzqQKHUmwDgMQ6AEwAHoECAgQAQ#v=onepage&q=%22brauer%22%2B%22block%22%2B%22character%22%2B%22involution%22&f=false
There are many results about the values of $\chi(t)$ when $t$ is an involution of a finite group $G$ and $\chi$ is an irreducible character: Isaacs' book on Character Theory has many such results collected from the literature, but there are many others scattered around:
For example, if $G = O^{2}(G)$ (equivalently, if $G/G^{\prime}$ has odd order), then $\chi(1) \equiv \chi(t)$ (mod $4$).
(Knörr): We have $\chi(t) = 0$ for every involution $t$ if and only if $|S|$ divides $\chi(1)$, where $S$ is a Sylow $2$-subgroup of $G.$
Regarding block theory, whenever $t $ is an involution of $G$ and $\chi$ is an irreducible character in the principal $2$-block of $G$, we have $\chi(tuv) = \chi(tu)$ whenever $u,v \in C_{G}(t)$ have odd order and $v \in O_{2^{\prime}}(C_{G}(t)),$ which is a consequence of Brauer's Second and Third Main Theorems.
I could probably give several more examples if you gave further clues as to what you are looking for.
Further edit to address a question from comments: if $t$ is an involution of $G$ and $B$ is a $2$-block of $G$, then results of Brauer imply the following facts (among others):
If $B$ has defect group $D$ and $t$ is not $G$-conjugate to an element of $D$, then we have $\chi(t) = 0 $ for every complex irreducible character $\chi \in B$.
If $B$ has dfect group $D$ and some conjugate of $t$ lies in $D$, then there is an irreducible character $\chi \in B$ with $\chi(t) \neq 0$, and we have $\sum_{ \chi \in {\rm Irr}(B)} \chi(1)\chi(t) =0,$ so there are irreducible characters in $B$ taking both positive and negative values at $t$.
Later edit: Another theorem of Brauer is that if $B$ is a $2$-block of defect $d >1$, then the number of irreducible characters in $B$ of degree exactly divisible by $2^{a-d}$ is divisible by $4$. In particular, this implies that if $|G|$ is divisible by $4$ and $t$ is an involution of $G$, then the number of irreducible characters $\chi$ in the principal $2$-block such that $\chi(t)$ is odd is a multiple of $4$.
Thank you very much for the answer. I would be interested in results in the literature concerning the case that all ordinary irreducible characters in question are lying in a non-principal $2$-block which has a non-cyclic defect group...Are there criteria known when they have negative values at involutions / an involution?
I will make a small edit .
Thank you very much.
For the symmetric group, let $\chi^\lambda$ be the symmetric group charater canonically labelled by the partition $\lambda$. Then $\chi^\lambda(x) = 0$ whenever $\lambda = \lambda'$ is a self-conjugate partition and the involution $x$ has an odd number of disjoint transpositions. This isn't very deep: in fact $\chi^\lambda(x) = 0$ for any odd permutation $x$. The converse does not hold: for example $\chi^{(6,3,2,2,2)}(1,2) = 0$.
As a very weak sufficient condition, it follows easily from the Murnaghan–Nakayama rule that if all the $2$-hooks in the partition $\lambda$ are horizontal (i.e. two boxes in the same row) then $\chi^\lambda(1,2) \ge 0$, with strict inequality unless $\lambda$ is a $2$-core (i.e. a staircase partition as in Saxl's Conjecture).
Thank you very much for the answer.
If the involution has no fixed-points, then the Murnaghan-Nakayama rule is cancellation-free. Hence, the character value is (up to a sign) the number of domino tableaux of the shape $\lambda$. The number of such tableaux can be computed via a hook-formula (Fomin-Lulov / James-Kerber).
You can extend this to all involutions, but you need to sum over all possible ways to distribute single boxes in $\lambda$, so that they occupy some skew shape $\lambda/\mu$, then apply the above argument for each shape $\mu$.
Thank you very much for the answer.
Suppose that $\chi$ is an irreducible character of a finite group $G$ and $t$ is an involution in $G$. It is well known that $\chi(1)\equiv\chi(t)$ (mod $2$). Stephen Gagola and Sidney Garrison showed (J. Algebra 74 (1982) 20--51) that if $\chi$ is a faithful orthogonal character of $G$, and $\chi(1)-\chi(t)\equiv4$ (mod $8$), then $G$ has a non-trivial double cover. Moreover if $t$ lies in the commutator subgroup of $G$, then $H^2(G,{\mathbb C}^\times)$ has even order. They also have results related to the restriction of a real-valued character to a Klein-four subgroup of $G$. They used these results to verify that $M_{22}$ has a four-fold cover.
Thank you very much for the answer.
|
2025-03-21T14:48:31.130894
| 2020-05-31T16:34:21 |
361820
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629673",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361820"
}
|
Stack Exchange
|
A computation in a commutative Frobenius algebra
This is already posted here https://math.stackexchange.com/questions/3695584/a-computation-in-a-commutative-frobenius-algebra but I didn't get any answers.
Given a commutative Frobenius algebra (in the category of vector spaces over $\mathbb C$) $A$ with multiplication $m\colon A\otimes A\to A$, comultiplication $c\colon A\to A\otimes A$, identity $1\in A$ and trace $\mathrm{tr}\colon A\to\mathbb C$, does the following expression simplify
$$Z\left[g\right]=\mathrm{tr}\left(f^{\circ g}\left(1\right)\right)$$
where $m\circ c\equiv_\text{def} f\colon A\to A$ and $f^{\circ g}=f\circ f\dots\circ f$ ($g$ times) where $g\in\mathbb N$. Also, why does $Z[1]=\dim A$?
I'm thinking about this a trying to evaluate a 2d TQFT on a closed 2-manifold of genus $g$. Lurie claims that $Z\left[1\right]=\dim A$. Any help in calculating any other specific values would be very helpful. ($Z[0]=\mathrm{tr}\,1$ obviously.)
I don't know what you mean when you ask if the expression simplifies. The generating relations for a commutative Frobenius algebra are equivalent to the fact that the only thing that matters is the diffeomorphism type of the corresponding surface (possibly with boundary) that you draw. (I like to think of this as the idea that the algebraic relations of a Frobenius algebra are precisely those encoding the handlebody theory for surfaces coming from Morse theory.) If you have a genus $1$ surface (so you're computing $Z[1]$), there are many possible expressions, but you need at least one occurrence of each of the four generating elementary operations: the unit, the counit (which you call the trace), multiplication, and comultiplication, and the expression you've written for $Z[1]$ is the simplest expression you could get, in some sense. (There aren't many Morse functions on a genus $1$ surface with only four critical points! This expression corresponds to the only one, in a way you can make precise.) For $Z[g]$, there are other expressions with the same number of each of the elementary terms, which again is minimal; the one you've given is one of the more natural choices, but I could also imagine, for example, writing $$Z[2] = \epsilon \circ m \circ (m \otimes \mathrm{id}_A) \circ (c \otimes \mathrm{id}_A) \circ c \circ \eta.$$
I'm using $\eta \colon \mathbb{C} \rightarrow A$ for the unit (which is more natural) and $\epsilon \colon A \rightarrow \mathbb{C}$ for the counit, for clarity, since I will use the word trace in a more standard manner. It is arguably more complicated, in that you have to include these extra identity morphisms (corresponding to trivial cylinders).
As for verifying $Z[1] = \dim A$, intuitively, you're taking a trivial cylinder, which corresponds to the identity map $\mathrm{id}_A \colon A \rightarrow A$, and gluing the two ends together. This tells you, as a general rule, that you need to take the trace of this map, so that
$$Z[1] = \mathrm{tr}(\mathrm{id}_A) = \dim A.$$
Here are a few more details about how to show this purely algebraically. From drawing surfaces, you'll see that the trivial cylinder can be decomposed in such a way that one finds
$$\mathrm{id}_A = (\mathrm{id}_A \otimes \epsilon) \circ (\mathrm{id}_A \otimes m) \circ (c \otimes \mathrm{id}_A) \circ (\eta(1) \otimes \mathrm{id}_A)$$
If you had a basis $\{e^i\}$ (NB: any Frobenius algebra is automatically finite-dimensional!), and you wrote everything out in this basis accordingly (e.g. $\eta(1) = \eta_i e^i$ and $c(e^i) = c^i_{jk}e^j \otimes e^k$), this tells you that $$\eta_i c^i_{jk}\mu^{k\ell}_{m}\epsilon^m = \delta_j^{\ell}$$ (where I'm assuming Einstein summation). Now we take the input and output and glue them together. What that does, in a basis, is to force the indices $j$ and $\ell$ to be the same. To be concrete, in a basis, we have
$$Z[1] = (\epsilon \circ m \circ c \circ \eta)(1) = \eta_ic^{i}_{jk}\mu^{jk}_{m}\epsilon^{m} = \delta_{j}^{j} = \dim A,$$
where we have used the commutativity, in the form $\mu^{jk}_m = \mu^{kj}_m$.
|
2025-03-21T14:48:31.131282
| 2020-05-31T17:23:42 |
361822
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Deane Yang",
"alvarezpaiva",
"https://mathoverflow.net/users/21123",
"https://mathoverflow.net/users/613"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629674",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361822"
}
|
Stack Exchange
|
History of microlocal analysis
Was the study of pseudo-differential operators the basis for the birth of microlocal analysis? I'm trying to find out how this branch of analysis was developed...
I believe microlocal analysis arose from two different directions. One is introduction of pseudodifferential operators to prove theorems about the regularity of solutions to elliptic PDEs. The other is the introduction of Fourier integral operators to study the propagation of singularities for the solution of a hyperbolic PDE. A seminal work on this is a paper by Peter Lax. I believe it is this one: Asymptotic solutions of oscillatory initial value problems.
Duke Math. J. 24 (1957), 627–646.
I think Calderon-Zygmund singular integrals led to pseudo-differential operators (Kohn and Nirenberg, I think) while WKB asymptotic techniques (as interpreted by Keller, Lax, and Maslov) led to FIO's. Then Hormander wrapped the whole thing up.
|
2025-03-21T14:48:31.131380
| 2020-05-31T17:37:44 |
361824
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"David Galvin",
"Richard Stanley",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/21690",
"https://mathoverflow.net/users/2807"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629675",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361824"
}
|
Stack Exchange
|
Asymptotics of sum involving Stirling numbers
I've encountered the following sum:
$$
s_n = \sum_{j=1}^n {n \brace j}(\alpha n)_j \beta^j.
$$
Here $\alpha$ and $\beta$ are positive constants, $(\alpha n)_j$ is a falling power, and ${n \brace j}$ is the Stirling number of second kind. My computations suggests that $s_n$ grows like $n^n C^n$ (where the constant $C$ depends on $\alpha$ and $\beta$); specifically, $\sqrt[n]{(s_n/n^n)}$ seems to converge to a limit as $n \rightarrow \infty$.
Are there good approaches to figuring out these kinds of limits? A similar situation was discussed in this question, but the sum there had $(\alpha)_j$ rather than $(\alpha n)_j$; the extra dependence on $n$ is causing me headaches!
your earlier question had a rising factorial (Pochhammer) instead of a falling factorial, is that intentional?
Yes, my sum has a falling rather than a rising power. But as observed in a comment in the earlier question, when the argument of the Pochhammer symbol is constant (not depending on n), there's not any difference in difficulty between falling and rising power.
The factor $\beta^n$ seems superfluous.
@RichardStanley I have corrected the sum --- the power of $\beta$ should have been $j$ rather than $n$.
The sum in question can be written (using Latin letters instead of Greek) as
$$ S_n(a,b):= \sum_{k=0}^n {n \atopwithdelims \{ \} k} k! \binom{na}{k} b^k $$
where the round brackets denotes an ordinary binomial. It will be shown as $n \to \infty$
$$ (1)\quad S_n(a,b) \sim \frac{n!}{2}\,\exp{\Big(n\big( h(u_0) + \frac{h''(u_0)}{2}\,u_0^2\big)\Big)} \,
\text{erfc}\big(\sqrt{\frac{n}{2} h''(u_0)} \,\,u_0 \big) $$
where
$$h(u) = \frac{\log(1+b(e^u-1))^a}{u}, \quad u_0=\frac{1}{a} + \text{W}\big(\frac{1-b}{a\,b}\,e^{-1/a}\big) $$
and $W()$ is the Lambert W function.
This follows from an asymptotic analysis of an alternative form,
$$(2) \quad S_n(a,b)=n!\,[u^n] (1+b\,(e^u-1))^{n \,a}$$
where $[u^n]$ is the 'coefficient of' operator. Note that when $b=1,$ the sum can be solved exactly, and gives the form the OP suspects. (This summation is known.) To derive the alternative form, note that sum is the Hadamard product of
$$ \omega_n(z) = \sum_{k=0}^\infty {n \atopwithdelims \{ \} k} k! z^k \quad \text{& } \quad (1+z)^{na}=\sum_{k=0}^\infty \binom{na}{k} z^k $$
The expression for $\omega_n$ can be written in terrms of Eulerian polynomials, but I prefer the polylogarithm at negative argument,
$$ \omega_n(t) = \frac{\text{Li}_{-n}(t/(1+t))}{1+t} $$
Taking the Hadamard product,
$$S_n(a,b) = \frac{1}{2 \pi \,i} \oint \frac{dt}{t(1+t)} \text{Li}_{-n}(\frac{t}{1+t}) \big(1+\frac{b}{t}\big)^{na} \, dt$$
where the integration path surrounds the origin.
Use $\text{Li}_{-n}(t/(1+t))=(-1)^{n+1}\text{Li}_{-n}(1+1/t).$ Let $t \to 1/t$ in the integral and check residue at $\infty.$ Then
$$S_n(a,b) = \frac{(-1)^n}{2\pi i}\oint \frac{ \text{Li}_{-n}(1+t)}{1+t} \big(1+b\,t\big)^{n\,a} \, dt$$
Now use the known generating function
$$\sum_{n=0}^\infty \frac{u^n}{n!} (-1)^n \frac{ \text{Li}_{-n}(1+t)}{1+t} =
\frac{1}{e^u-1-t}$$ Putting this equation in the penultimate and using Cauchy's theorem results in equation (2).
For the asymptotic analysis, write (2) as a countour integral
$$S_n(a,b)=\frac{n!}{2 \pi \,i} \oint \Big( \frac{(1+b(e^u-1))^a}{u} \Big)^n \frac{du}{u} $$
Classic saddle point analysis tells us to write this as
$$S_n(a,b)=\frac{n!}{2 \pi \,i} \oint \exp{\big(n h(u)\big)} \frac{du}{u} $$
where $h$ has been given in (1), then expand $h$ in a power series about $u_0,$ which has also been found explicitly and given in (1). Run the contour through $u_0$ and open the loop so that it becomes a vertical line. Then, so long as certain conditions are satisfied (I have not checked them, other than $h''(u_0)>0$, a necessity) then
$$ S_n(a,b) \sim \exp{(n\ h(u_0))} \frac{n!}{2 \pi} \int_{-\infty}^\infty \frac{dt}{u_0+i\,t} \exp{\big(-\frac{n}{2}\ h''(u_0) t^2 \big)} $$
The integral is solvable in terms of the complementary error function. For $n$ sufficiently large, it will simplify in the asymptotic limit to an exponential, as long as $h''(u_0)$ isn't too small. A potential problem with this analysis is that if $h''(u_0)$ does become small, and much smaller than $h^{(3)}(u_0)$ then a quadratic expansion of $h$ is not sufficient.
For examples of the accuracy of the approximation, a table is shown.
$$
\begin{array}{c|lcr}
n & a & b & \text{true} & \text{asym} & \text{% err}\\
\\
\hline
20 & 1/2 & 1/2 & 2.7717\ 10^{17} & 2.6867\ 10^{17} & 3.07\\
{} & 1/2 & 3/2 & 3.7421\ 10^{21} &3.5745\ 10^{21} & 4.48 \\
{} & 3/2 & 1/2 & 2.1498\ 10^{25} & 2.0862\ 10^{25} &2.96\\
{} & 3/2 & 3/2 & 1.6663\ 10^{23} & 1.5884\ 10^{23} & 4.68\\
200 & 1/2 & 1/2 & 6.634\ 10^{374} & 6.617\ 10^{374} & 0.34\\
{} & 1/2 & 3/2 & 3.336\ 10^{415} & 3.319\ 10^{415} & 0.51\\
{} & 3/2 & 1/2 & 6.158\ 10^{453} & 6.138\ 10^{453} & 0.33\\
{} & 3/2 & 3/2 & 8.720\ 10^{521} & 8.673\ 10^{521}& 0.53
\end{array}
$$
It's really surprising that the approximation for a problem with 2 parameters is so easily characterized.
Added 6/3/2020
There is a much simpler derivation of (2) that bypasses the need for the Hadamard product and the polylogs with negative index. Here it is:
It is known that
$$ {n \atopwithdelims \{ \} k} k! = n^k $$
Then
$$ \sum_{k=0}^n {n \atopwithdelims \{ \} k} k! \binom{na}{k} b^k =$$
$$ n! [u^n] \sum_{k=0}^\infty \binom{na}{k} b^k (e^u-1)^k =
n! [u^n] \big(1+b(e^u-1) \big)^{n \ a} $$
by the binomial theorem.
Thank you, this is really nice!
|
2025-03-21T14:48:31.131742
| 2020-06-03T13:18:20 |
362099
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Math604",
"Simon Pun",
"https://mathoverflow.net/users/66623",
"https://mathoverflow.net/users/80635"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629676",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362099"
}
|
Stack Exchange
|
Continuity of Helmholtz-Hodge projection in $H^1(\Omega)$
Let $\Omega \subset \mathbb{R}^d$ (for simplicity $d = 2$ or $3$) be a bounded Lipschitz domain. For any vector-valued function $\mathbf{f} \in \mathbf{L}^2(\Omega):= \left ( L^2(\Omega) \right )^d$, one can uniquely write
$$ \mathbf{f} = \nabla \phi + \mathbf{w}_{\mathbf{f}},$$
where $\phi \in H_0^1(\Omega)$ and $\mathbf{w}_{\mathbf{f}} \in W:= \left \{ \mathbf{u} \in \mathbf{L}^2(\Omega): \int_\Omega \mathbf{u} \cdot \nabla \phi ~ dx= 0 \text{ for all } \phi \in H_0^1(\Omega) \right \}$. In particular, the function $\phi \in H_0^1(\Omega)$ satisfies the following variational form:
$$
\int_\Omega \nabla \phi \cdot \nabla \varphi ~ dx = \int_\Omega \mathbf{f} \cdot \nabla \varphi \quad \text{for all } \varphi \in H_0^1(\Omega).
$$
Based on this (Helmholtz-Hodge) decomposition, one can define a mapping $\mathbb{P}: \mathbf{f} \mapsto \mathbf{w}_{\mathbf{f}}$. This mapping is obviously linear and bounded in the sense of $\mathbf{L}^2(\Omega)$. In particular, there is a generic constant $C>0$ such that
$$ \|\mathbb{P}(\mathbf{f})\|_{\mathbf{L}^2(\Omega)} \leq \|\mathbf{f} \|_{\mathbf{L}^2(\Omega)}$$
for any $\mathbf{f} \in \mathbf{L}^2(\Omega)$.
My question: if we restrict the domain of the mapping $\mathbb{P}$ to be a subspace of $\mathbf{L}^2(\Omega)$, say $\mathbf{H}_0^1(\Omega) := \left ( H_0^1(\Omega) \right )^d$, can we show the boundedness of the restricted mapping in the sense of $\mathbf{H}_0^1(\Omega)$ (with respect to the $\mathbf{H}_0^1(\Omega)$ semi-norm $\|\mathbf{v} \|_{\mathbf{H}_0^1(\Omega)}^2 := \int_\Omega |\nabla \mathbf{v} |^2~ dx$ for any $\mathbf{v} \in \mathbf{H}_0^1(\Omega)$)? Or under which additional assumptions can we show such a boundedness result of this projection mapping?
do you want to add a boundary condition to $ \phi$? you can make it zero... then if you write out the pde that $ \phi$ satisfies stuff becomes more apparent (or at least for me the pde always helps me)
@Math604 I have added the corresponding variational form for the function $\phi$. It doesn't matter for the boundary condition of the function $\phi$, but it is okay to include it.
i admit i haven't read the question carefully... but $ \phi$ satisfies $ -\Delta \phi= div(f)$ in $ \Omega$ with $ \phi=0$ on $ \partial \Omega$. So from this you can get estimates on $ \phi$ and since we have $ f = \nabla \phi + w$ you can now get estimates on $ w$. I am not sure whether it gives you want you want...but i would try this
@Math604 Yes, you are right. Yet I am interested in the estimate of $\mathbf{w}\mathbf{f}$ in the semi-$H^1$ norm. Under some suitable regularity assumptions, I wonder if we have it hold: $| \mathbf{w}{\mathbf{f}} |{H^1(\Omega)} \leq |\mathbf{f}|{H^1(\Omega)}$?
so it looks like you need to estimate the H^1 norm of $\nabla \phi$.. which should controlled by the $H^2$ norm of $ \phi$. So by elliptic regularity it appears you want the $L^2$ norm of div(f) .... so at least formally it seems you get want you want...but i guess one needs to try and write things out a bit...
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.