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2025-03-21T14:48:30.144053
| 2020-03-26T10:00:59 |
355735
|
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"Martin Hairer",
"UnclePetros",
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"julian"
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|
Stack Exchange
|
Stochastic invariant subset
Let us consider a stochastic differential equation (SDE),
$$
dx_{t}=f\left( x_{t}\right) dt+\sigma\left( x_{t}\right) dW_{t}%
$$
and a compact set $C\subset\mathbb{R}^{n}$.
Given a stochastic Lyapunov function $\Phi\left( x_{t}\right) $ for this SDE
with respect to $C$, i.e.
(i) $\Phi$ is positive definite.
(ii) $L\Phi\left( x\right) $ is not necessary to be nonpositive in $C$ but
$L\Phi\left( x\right) <0$ for all $x\notin C$, where $L$ is the
infinitesimal generator of the SDE.
How can I prove that $C$ is an invariant set with respect to the solutions of
the SDE? In this I work with convergence in probability.
This seems wrong to me. Consider $\sigma(x)=\sqrt{2}$ and $f(x)=-x$. Then $L=\triangle-x\cdot\nabla$. $\Phi(x)=|x|^2$ is a Lyapunov function with $C=\overline{B}_1(0)$. But $C$ is certainly not invariant.
Just after exiting the compact, is the probability of the state to return to C strictly lower than 1 or is always 1?
The probability of return is $1$.
How can I prove that?
This follows by Birkhoff‘s theorem.
I am not a specialist in this theorem, but it seems to mention Cosmology and Theory of relativity. How can it be applied to the stochastic problem that I pose?
https://en.wikipedia.org/wiki/Ergodic_theory#Probabilistic_formulation:_Birkhoff%E2%80%93Khinchin_theorem
@julian The return probability is $1$ in your example, but the assumptions of the OP don't even guarantee that, take BM in high enough dimension with $\Phi(x) = 1/|x|$ outside the unit ball. (Unless "positive definite" is meant to be "proper".)
|
2025-03-21T14:48:30.144189
| 2020-03-26T10:08:30 |
355737
|
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"url": "https://mathoverflow.net/questions/355737"
}
|
Stack Exchange
|
Maximizing quadratic forms
Consider the maximization problem
$$\text{maximize} \quad Q(x)= \sum_{i<j} \Big(\sum_{k} a_{ik}a_{jk}\Big) x_i x_j \quad \text{subject to} \quad \sum_{i}x_i^2=1,$$
and let $M$ be maximum value obtained by $Q$ under such constraint. Suppose that $a_{ij}=a_{ji}$ for all $i,j$ and that
$$ \begin{cases} a_{ij}=0 &\text{if $i+j\equiv 0\mod 2$,} \\
a_{ij}>0 & \text{if $i+j\equiv 1\mod 4$,}
\\
a_{ij}<0 & \text{if $i+j\equiv 3\mod 4$.}
\end{cases} $$
I was wondering if one could get an interesting upper bound on $M$ by using the above information on the signs of the $a_{ij}$'s. Namely, I am interested in a better upper bound than that provided by the Cauchy-Schwarz inequality, but still that does not require for every instance to use Lagrange multipliers (I am not interested in the precise value of $M$, nor in where the maximum is attained).
|
2025-03-21T14:48:30.144277
| 2020-03-26T10:23:34 |
355738
|
{
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"Cristóbal Guzmán",
"https://mathoverflow.net/users/39129"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355738"
}
|
Stack Exchange
|
Implicit function theorem for subdifferentiable convex functions
I am trying to find a method to apply the implicit function theorem for subdifferential convex functions. The original theorem provides an equation for the partial derivative of the implicit function w.r.t $x$ by using the chain rule. There are some variants for non-differentiable functions but these only guarantee that a solution $x=x(y)$ to the equation
$$
F\big(x(y),y\big) = 0
$$
exists and do not give any information about the partial derivative of $x$ respect to $y$. Is there any way to adapt this in terms of subdifferentials instead of partial derivatives?
Thanks
I am not an expert on this, but I suspect this book is useful: Dontchev-Rockafellar. Implicit Functions and Solution Mappings: A View from Variational Analysis. https://link.springer.com/book/10.1007/978-0-387-87821-8
I do not know if your question has an answer, if it has please share it !
With the regular subderivative you might have trouble. But the notion of "conservative field" is a generalisation of the derivative that shows some nice properties. see https://link.springer.com/article/10.1007/s10107-020-01501-5 and https://proceedings.neurips.cc/paper_files/paper/2021/file/70afbf2259b4449d8ae1429e054df1b1-Paper.pdf . For instance the chain rule is rigorously obtained for non differentiable functions that have a conservative field.
In short a concervative field is a set value mapping that is equal almost everywhere to the differential of your function and that verifies some other properties such as closedness of its graph. Only functions that are differentiable almost everywhere can have conservative fields, that is the case for all convex functions. The subdifferential of a convex function is a conservative field for the same function.
This might help you but I do not garantee that it will be sufficient.
I realised there is an answer in "Optimization and nonsmooth analysis" Frank H. Clarke page 255 .
The answer uses the clarke subdifferential (it corresponds to the subdifferential of a convex function).
|
2025-03-21T14:48:30.144421
| 2020-03-26T10:39:24 |
355739
|
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|
Stack Exchange
|
Second derivative estimates
I am in big trouble since I don't see how to proceed (I don't need the exact calculation) with the following estimates.
In one of his papers, Lin proves the following result:
Let's consider a bounded, smooth domain $\Omega$ in $\mathbb{R}^n$ and let $(a_{ij})$ be a symmetric $n$x$n$ matrix-valued function on $\Omega$ wich satisfy $\lambda I \leq a(x) \leq \lambda^{-1}I$. Put
$$L_a=\sum_{ij}a_{ij}(x)\dfrac{\partial^2}{\partial x_i \partial x_j}$$
and let $u\in C^{1,1}(\Omega)$ be the solution of the following Dirichlet problem:
$$
\begin{cases}
L_au=-f &\text{ in }\Omega \\
u=0 &\text{ in }\partial \Omega\end{cases}
$$
where $f \in L^n(\Omega)$.
Theorem. There is a positive constant $p=p(n,\lambda)$ such that $$
\Vert D^2u\Vert_{L^p(\Omega)}\leq c(n,\lambda,p,\Omega)\Vert f\Vert_{L^n(\Omega)}
$$
for any $u\in C^{1,1}(\Omega)$, $u=0$ in $\partial \Omega$, where $f=L_a u$ and $D^2u$ is the Hessian matrix of $u$.
Now, I want to show an other statement that use this last theorem.
N.B. The definition of convex function can be found here.
I am not interested actually in the first derivative estimates (this is what Evans says in [5]) but I want just an estimate of the second derivative. So my questions are:
How can I use Lin's estimate?
Is it an easy application of the Lin theorem or there something deeper in the Lin paper that I have to see?
Please somebody help me.
|
2025-03-21T14:48:30.144534
| 2020-03-26T10:47:11 |
355740
|
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"authors": [
"Benjamin Steinberg",
"Carl-Fredrik Nyberg Brodda",
"Diego Martinez",
"YCor",
"https://mathoverflow.net/users/120914",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/147609",
"https://mathoverflow.net/users/15934"
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}
|
Stack Exchange
|
Ascending sequences of idempotents in inverse semigroups
I've enocuntered the following question in my current research, and I'd appreciate any help you could give me. This is probably well known to experts on the subject.
Let $S = \langle K \rangle$ be a finitely generated inverse semigroup. Recall that the set $E$ of idempotents (i.e. elements $e \in S$ such that $e^2 = e$) is partially ordered via $e \leq f$ when $ef = fe = e$ (idempotents always commute in inverse semigroups).
Question: Can $S$ have an infinite ascending sequence of idempotents? That is, are there $e_n \in E$ such that $e_1 < e_2 < \dots < e_n < \dots$?
By $e < f$ I mean that $e \leq f$ and $e \neq f$. The only instance of the latter behavior I've come accross is the semigroup $S = (\mathbb{N}, \min)$ (and its' relatives), where $n \cdot m := \min\{n, m\}$. In this case we have that $S$ is equal to its' semilattice of idempotents, and $1 < 2 < \dots$, but this semigroup is not finitely generated.
Yes. Indeed, for $X$ a set, let $G_X$ be the group of partial bijections of $X$, that are defined and identity outside a countable subset. I claim that, for $X$ uncountable, every countable subset of $G$ is contained in a (5-generator) finitely generated submonoid (and hence in a finitely generated inverse submonoid).
The claim being granted, and using that the power set of $\omega$ contains a chain isomorphic to $(\mathbf{Q},\le)$, one obtains such a chain of idempotents in a suitable inverse monoid.
Note: the same claim was proved by Sierpinski and Banach in the 1930's for the monoid of all self-maps of every set, and by Galvin (1995) for the group of all permutations of every set.
Now let me prove the claim, inspired by Galvin's proof. Let $(f_n)_{n\in\mathbf{Z}}$ be a sequence in $G_X$. So there exists an infinite countable subset $X_{0,0}$ such that for every $n$, each $f_n$ is defined and identity outside $X_{0,0}$. Choose for all other $(m,n)\in\mathbf{Z}^2$ an infinite countable susbet $X_{m,n}$, pairwise disjoint. Henceforth, all maps are assumed to be defined and identity outside $X'=\bigcup_{m,n}X_{m,n}$. Also fix a bijection $X_{0,0}\to X_{m,n}$ for all $(m,n)\neq (0,0)$, so that we identify $X'$ to $X_{0,0}\times\mathbf{Z}^2$.
Define
$u$ as the permutation $(x,m,n)\mapsto (x,m+1,n)$;
$r$ as the permutation $(x,0,n)\mapsto (x,0,n+1)$, $(x,m,n)\mapsto (x,m,n)$ for $m\neq 0$;
$f$ as the permutation $(x,m,n)\mapsto (f_m(x),m,n)$ for $n\ge 0$ and $(x,m,n)\mapsto (x,m,n)$ for $n\ge 0$.
I claim that for every $m$ we have $f_m\in\langle u,u^{-1},r,r^{-1},f\rangle$, where $\langle\cdots\rangle$ means the submonoid generated (actually, it follows that $f_m\in\langle u,r,f\rangle_{\text{inverse-monoid}}$).
Indeed, write $g_m=u^mfu^{-m}$: then $g_m$ is like $f$, but shifted $m$ times to the right. Then one sees that $g_m(r^{-1}g_mr)^{-1}=f_m$, and the claim is proved.
[Note 1: observe that $f_m$ is written as a word of length $\le 2+2(2m+1)=4m+6$ with respect to the given generators: since this only depends on $m$, this shows that $G_X$ is "strongly distorted" (as monoid, and as inverse monoid) and in particular strongly bounded, a.k.a. Bergman's property.]
[Note 2: Probably it's also true for $X$ countable, with some further preliminary lemmas. Also with only two generators.]
[Note 3: from Vagner-Preston, every countable inverse monoid embeds into $G_{\aleph_1}$. As corollary, every countable inverse monoid embeds into a 3-generated one. This is probably well-known?]
Note 3: It is a result of Chris Ash from 1978 that every countable inverse semigroup embeds into a 2-generated inverse semigroup.
Thank you to both. @Ycor, what do you mean by "strongly distorted"? Would you have a reference where I could read about Bergman's Property?
@Carl-Fredik, would you have the reference at hand? Or, maybe, just a sketch of the proof.
@DiegoMartínez The claim appears in [Byleen, Karl,
"Inverse semigroups with countable universal semilattices". Semigroups and their applications (Chico, Calif., 1986), 37–42, Reidel, Dordrecht, 1987]. There he attributes it to Ash, and says it appears in a survey article; I have not been able to find this survey. I do not know what the proof looks like.
It is shown in proposition 4.2 of https://www.cambridge.org/core/services/aop-cambridge-core/content/view/FD52B0C08506F91AD2747E37AA789A34/S0013091502000974a.pdf/countable_versus_uncountable_ranks_in_infinite_semigroups_of_transformations_and_relations.pdf that every countable subset of the symmetric inverse monoid is contained in a 2-generated inverse subsemigroup.
@DiegoMartínez given a variety of structures (e.g., inverse monoids), say with finite signature (= finitely many finitary laws), an object $G$ is strongly distorted if there exists a numerical sequence $(a_n)$ such that for every sequence $(g_n)$ in $G$ there exists a finite subset $F\subset G$ such that for every $n$, $g_n$ belongs to the substructure generated by $F$ and has length $\le a_n$ with respect to $F$ in some reasonable sense.
|
2025-03-21T14:48:30.144861
| 2020-03-26T12:15:15 |
355750
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A. Gupta",
"Carlo Beenakker",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/42940"
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|
Stack Exchange
|
A system of inequalities involving a skew-symmetric integer matrix
Which skew-symmetric integer matrices $S$ satisfy the following inequalities
$SV_i \ne z_iE_i$ for all $i = 1,\cdots, n$
where
$V_i$ denotes the column with integer entries such that the $i$-th entry is $-1$ and all other entries are non-negative integers,
$z_i \in \mathbb Z$, and $E_i$ denotes $i$-th standard basis vector in $\mathbb Z^n$.
The question asked originally was as follows: Notation as above can we
find an integer matrix $V$ with $-1$ on the main diagonal and all other entries non-negative so that $SV$ has diagonal form.
for $n=2$ this is obviously not possible, and also for $n=3$ it seems excluded (irrespective of the integer constraint)
after the last edit the question has become totally different; please, at least keep the first question visible, don't just delete it and replace it by a different question, that is really demotivating for MO users who make an effort to think about your question.
@Carlo Beenakker Your very kind comment led us to re-think about our problem and reformulating the question in its current form is the what we really wish to ask.
|
2025-03-21T14:48:30.144976
| 2020-03-26T12:37:17 |
355752
|
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"authors": [
"ABIM",
"frafour",
"https://mathoverflow.net/users/145915",
"https://mathoverflow.net/users/36886"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355752"
}
|
Stack Exchange
|
Generalized compact open topology?
Let $X,Y$ be topological spaces. The compact-open topology on $C(X,Y)$ is generated by the sub-basic open sets
$$
\left\{U_{K,O}: \mbox{ K is compact in X and O is open in Y}\right\}\\
U_{K,O}:=\left\{f \in C(X,Y):\, f(K')\subseteq O \right\}
.$$
However, if $X$ is not Hausdorff then sometimes this topology may be oddly-behaved.
Instead, consider this generalization (which clearly is equal to the compact-open topology when $X$ is Hausdorff) of the compact-open topology; generated by the sub-basic sets:
$$
\left\{V_{K',O}:(\exists K \subseteq X \mbox{ compact})\; K'\subseteq K, \mbox{ K is closed in X and O is open in Y}\right\}\\
V_{K',O}:=\left\{f \in C(X,Y):\, f(K')\subseteq O \right\}.
$$
Is this topology studied? If so where?
So, equivalently, $K'$ is compact and has compact closure, right?
Exactly... Is this a studied object?
|
2025-03-21T14:48:30.145073
| 2020-03-26T12:47:23 |
355753
|
{
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"authors": [
"Enrico",
"Jef",
"https://mathoverflow.net/users/110362",
"https://mathoverflow.net/users/52811"
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|
Stack Exchange
|
Rank 3 Lagrangian vector bundles on an elliptic curve
Let $k$ be an algebraically closed field of characteristic zero (feel free to assume $k= \mathbb{C}$) and $E$ an elliptic curve over $k$ with identity $P \in E(k)$.
I am interested in certain morphisms from $E$ to $LG(3,6)$, the Lagrangian Grassmannian of $3$-dimensional Lagrangian subspaces of a $6$-dimensional symplectic vector space over $k$, namely those morphisms $E \rightarrow LG(3,6)$ such that the pullback of $\mathcal{O}(1)$ on $LG(3,6)$ (coming from the Plücker embedding) to $E$ along this morphism is isomorphic to $\mathcal{O}_E(6P)$.
More concretely, I am interested in rank $3$ vector bundles $V$ on $E$ with the following two properties:
There exists a surjection $\mathcal{O}_E^{\oplus 6} \twoheadrightarrow V$ whose kernel is a Lagrangian subvectorbundle of $\mathcal{O}_E^{\oplus 6}$ (where we put the standard symplectic form on $\mathcal{O}_E^{\oplus 6}$).
We have an isomorphism $\text{det}\, V \simeq \mathcal{O}_E(6P)$.
An example of such a $V$ is given by $\mathcal{O}_E(2A)\oplus \mathcal{O}_E(2B) \oplus \mathcal{O}_E(2C)$ where $A,B,C \in E(k)$ sum to zero, i.e. the divisor $A+B+C$ is linearly equivalent to $3P$.
Question: Is every $V$ of this form?
I believe that if $V$ is a direct sum of line bundles then it must necessarily be of the form described above. So we could equivalently ask: is every $V$ satisfying the above two conditions a direct sum of line bundles?
The variety $LG(3,6)$ is a homogenous space for the algebraic group $Sp_6$, but I haven't been able to find results in the literature which treat this specific case.
Thanks in advance!
Reversing the problem (since you are using a Mukai--style construction), you may ask for the existence of an homogeneous bundle $F$ on $SG(3,6)$ of rank 5 and $c_1(F)=4$. The only non--degenerate example I can come up with is $F= \bigwedge^2 R^{\vee} \oplus \mathcal{O}(1) \oplus \mathcal{O}(1)$, where $R$ is the rank 3-tautological. This corresponds to an elliptic curve as double (multi)-linear section of $(\mathbb{P}^1)^3$. Otherwise you can use $R^{\vee}$ (or $Q$) as a bundle, but then the embedding is degenerate in $SG(3,6)$, and you should embed $E$ directly in a 3-dimensional quadric.
Can you explain why the problem you pose is related to the one given here, and what $SG(3,6)$ stands for? I'm not really familiar to Mukai-style constructions.
The Lagrangian Grassmannian $LG(k, 2k)$ is a special case of the symplectic Grassmannian $SG(k,n)$ (basically I typed the comment in a hurry using the notation I am most used to, sorry).
By Mukai-style I mean the vector bundle method (see for example section 3 of http://library.msri.org/books/Book28/files/mukai.pdf or section 5 of "Algebraic Geometry V"). Finding a rank 3 bundle with 6 sections on $E$ gives you a morphism to $Gr(3,6)$. Then your original $E$ can be recovered by taking zero locus of an appropriate number of sections of $\bigwedge^i R^{\vee}$. Note that $LG(3,6)$ is itself the zero locus of $\bigwedge^2 R^{\vee}$ on $Gr(3,6)$.
I think your guess is correct and one can proceed as follows (some details are missing though).
Let $V$ be a six dimensional symplectic vector space and $F$ be a rank three-vector bundle on $E$ wiht an exact sequence
$$ 0 \longrightarrow G \longrightarrow V \otimes \mathcal{O}_E \longrightarrow F \longrightarrow 0,$$
where $G$ is a Lagrangian subbundle of $V \otimes \mathcal{O}_E$.
Assume furthermore that $\det(F) = \det(G^*) = \mathcal{O}_{E}(6P)$.
Let $W \subset V$ be a generic Lagrangian subspace and consider the map:
$$ \phi : G \longrightarrow V/W \otimes \mathcal{O}_{E}.$$
The genericity of $W$ implies that it is generically on E an ismorphism. Furthermore, $\phi$ is (globally) injective as $G$ is torsion free. We denote by $Z \subset E$ the subscheme corresponding to the degeneracy locus of $\phi$. Since $\det(G^*) = \mathcal{O}_{E}(6P)$, we have the linear equivalence $Z \sim 6P$.
We have an exact sequence:
$$ 0 \longrightarrow G \longrightarrow V/W \otimes \mathcal{O}_E \longrightarrow \mathcal{F} \longrightarrow 0,$$
where $\mathcal{F}$ is scheme theoretically supported on $Z$.
The vector space $W \subset V$ is generic and $E$ is a curve, so that the corank of $\phi$ is exactly $1$ on $Z$. As a consequence $\mathcal{F}|_{Z}$ is a line bundle on $Z$.
Let $Z_{red} = \{P_1, \ldots, P_l\}$ with the $P_i$ distincts. We write:
$$ \mathcal{F} = \bigoplus_{i=1}^{l} \mathcal{F}_i,$$
where $ \mathcal{F}_i$ is the restriction of $\mathcal{F}$ to the connected components of $Z$ corresponding to $P_i$.
For any subbundle $F$ of $V \otimes \mathcal{O}_E$ whose quotient is a vector bundle, we denote by $F^{\perp} = (V/F)^*$.
We have han exact sequence:
$$ 0 \longrightarrow G^{\perp} \longrightarrow V^*/(W^{\perp}) \otimes \mathcal{O}_E \longrightarrow \mathcal{H} \longrightarrow 0,$$
where $\mathcal{H}$ is scheme theoretically supported on a subscheme of $E$ linearly equivalent to $6P$.
We similarly split $\mathcal{H}$ as $\bigoplus_{i=1}^q \mathcal{H}_i$, where the $\mathcal{H}_i$ correspond to the various connected component of the support of $\mathcal{H}$.
The bundles $G$ and $W \otimes \mathcal{O}_E$ being Lagrangian, the skew-symmetric form $\sigma : V \longrightarrow V^*$ induces isomorphisms:
$$ \sigma_{G} \ : \ G \stackrel{\sim}\longrightarrow G^{\perp} \ \textrm{and} \ \sigma_{V/W} \ : \ V/W \stackrel{\sim}\longrightarrow V^*/(W^{\perp})$$
which are compatible with the maps:
$$ G \longrightarrow V/W \ \textrm{and} \ G^{\perp} \longrightarrow V^*/(W^{\perp}).$$
We deduce that $\mathcal{H}$ and $\mathcal{F}$ are equal and that up to a reordering of the we have $\mathcal{H}_i = \mathcal{F}_i$, for all $i$.
For all $i \in \{1, \ldots, l\}$, the skew symmetric isomorphism $\sigma$ induces a skew-symmetric isomorphism:
$$\sigma_i : \mathcal{F}_i \stackrel{\sim}\longrightarrow \mathcal{F}_i,$$
which lifts to a skew-symmetric isomorphism:
$$h^0(\sigma_i) \ : \ H^0(E,\mathcal{F}_i) \stackrel{\sim}\longrightarrow H^0(E, \mathcal{F}_i).$$
The skew-symmetry of the isomorphism $h^0(\sigma_i)$ forces the dimension of the vector spaces $H^0(E,\mathcal{F}_i)$ to be even. As a consequence, of the Riemman-Roch formula on $E$, the multiplicity of $P_i$ as a connected component of $Z$ must always be even.
The generic situation (that is when $E \longrightarrow LG(3,6)$ is a generic point in a component of $Hom(E, LG(3,6))$ should correspond to the case:
$Z_{red} = \{A,B,C\}$ with $A,B,C$ distincts and $Z = \{2A,2B,2C\}$ as a subscheme of $E$.
Now I would like to deduce from this that we have a map:
$$ \mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C) \longrightarrow G$$
which is generically an isomorphism (I have a vague idea why this should be true, but I don't have a precise argument to offer, perhaps someone else will find).
If we have such a map which is generically an isomorphism, then it must be an isomorphism, owing to the relation $\det(G) = \det(\mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C))$.
We conclude that $F \simeq \mathcal{O}_E(2A) \oplus \mathcal{O}_E(2B) \oplus \mathcal{O}_E(2C)$ as $G^* \simeq F$.
|
2025-03-21T14:48:30.145608
| 2020-03-26T13:02:14 |
355755
|
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|
Stack Exchange
|
Measure of the convex hull of a ball and a point
I need to prove the following statement:
Let $B_s(z)$ be a ball centered at $z$ of radius $s$ s.t. $0\not\in B_s(z)$. Moreover let $K_s(z)$ the convex hull of $\{0\}\cup B_s(z)$.
Then
$$ \left|K_{\frac{s}{r}}\left(\frac{z}{r}\right)\cap B\right|\geq\alpha>0, \text{for $r$ large enough}, $$
where $|\cdot|$ is the Lebesgue measure, $B=B_1(0)$.
I have no clue on how to prove it...
By picturing $K_s(z)$ it is a cone with vertex in $0$ and the ball as the "top".
As $r$ goes to infinity, the ball $B_{\frac{s}{r}}\!\left(\frac{z}{r}\right)$ goes inside the unit ball centered in the origin, so for $r$ large enough $\big|K_{\frac{s}{r}}\!\left(\frac{z}{r}\right)\cap B\big|=\big|K_{\frac{s}{r}}\left(\frac{z}{r}\right)\big|$. By intuition $|K_r(z)|=C_nr^n$ so to me seems false..
Is my intuition false?
Yes, this statement is obviously false, Where did you get it from?
"Sets of finite perimeter and geometric measure theory" F.Maggi, page 264 in the middle of the page ^^'
Maybe i formulated it wrong.. It is strange that the book made an error like this
|
2025-03-21T14:48:30.145736
| 2020-03-26T13:33:43 |
355758
|
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|
Stack Exchange
|
$u_t=Au+F(u)$ where $A$ is the infinitesimal generator of $C_0$-semigroup
I asked this question on Mathematics Stackexchange, but got no answer.
In Pavel's book: Nonlinear Evolution Operators and Semigroups - Applications to Partial Differential Equations, we have the following definitions and result:
Consider the following semilinear equation:
$$\tag{5.59}\begin{cases}u_t(t)=Au(t)+F(u(t)), ~0 \leq t \leq T,\\u(0)=u_0 \in H. \end{cases} $$
where $A:D(A)\subset H \rightarrow H$ be the infinitesimal generator of $C_0$-semigroup $S(t)$ with $\|S(t)\|\leq 1$.
Definition 5.1. (1)A function $u:[0,T]\rightarrow H$ is said to be a classical solution of $(5.59)$ if $u(0)=u_0$ and $$\tag{5.68}u \in C([0,T];D(A))\cap C^1([0,T];H)$$
and $(5.59)$ is satisfied. (2) The function $u$ is a classical solution on $[0,\infty[$ of $(5.59)$ if $(5.68)$ holds for every $T>0$ and $(5.59)$ is satisfied for every $t>0$.
By a mild solution of $(5.59)$ we mean a continuous function $u:[0,T] \rightarrow H$ satisfying $$\tag{5.69} u(t)=S(t)u_0+\int_{0}^{t} S(t-s)F(u(s))\, ds, ~0 \leq t \leq T.$$
Theorem 5.8 If $F:H\rightarrow H$ is locally Lipschitz then for every $u \in H$, there is a unique mild solution $u:[0,T_{\max}[\rightarrow H$ of $(5.59)$ with either $T_{\max}=+\infty$, or $T_{\max}<\infty$ and $\lim_{t \uparrow T_{\max}}\|u(t)\|=+\infty$. If $u_0 \in D(A)$, the mild solution $u$ is classical. (2) If $F$ is Lispschitz continuous, then $T_{max}=+\infty.$
My question:
I would like to know if Theorem 5.8 is valid for $ A $ being just the infinitesimal generator of a $C_0$-semigroup, that is, without the hypothesis that $\|S(t)\|\leq 1$.
I imagine that the answer is positive in the case that $S(t)$ is uniformly bounded, that is, there is $M>0$ such that $\|S(t)\|\leq M$ for $t \geq 0$.
Assuming I have proven that $A-\lambda I$ is the infinitesimal generator of a $C_0$-semigroup of contractions, for some $\lambda>0$. Then, $A$ is the infinitesimal generator of a $C_0$-semigroup $T(t)$ satisfyng $\|T(t)\|\leq e^{\lambda t}$, for all $t \geq 0$. I don't know if the result is valid in this case.
Can anyone help me? Please.
Could this DOI:10.2307/2001143 be of some help ?
Yes the theorem is true in the general case. The semigroup is assumed to be contractive for the sake of simplicity. Of course the assumption is not restrictive since there is always an equivalent norm which makes the semigroup contractive (this is the same idea in the proof of Hille-Yosida theorem to pass from the uniformly bounded case to the contractive case). You can find the general statement
(for Lipschitz continuous functions) of the theorem in Theorem 1.4 pp 185. (see also Theorem 1.2) of the book
A. Pazy, Semigroups of linear operators and applications to partial differential equations. Springer 1983.
Edit:
As for the second part on the improved regularity, the result is true for locally Lipschitz functions, when $X$ is reflexive (in your case it is a Hilbert space). See Proposition 4.3.9 pp 60 (and Remark 4.3.10) in the book
T. Cazenave and A. Haraux, An Introduction to Semilinear Evolution Equations, Oxford 1998.
PS: One should note the definition of Lipschitz continuous function in pp 55.
By using Theorem 1.4 I agree with the first part of Theorem 5.8, that is, If $F:H\rightarrow H$ is locally Lipschitz then for every $u \in H$, there is a unique mild solution $u:[0,T_{\max}[\rightarrow H$ of $(5.59)$ with either $T_{\max}=+\infty$, or $T_{\max}<\infty$ and $\lim_{t \uparrow T_{\max}}|u(t)|=+\infty$.
I am still in doubt with respect to the second and third part of the theorem, I have not found the respective results in the Pazy's book.
@VictorHugo for the second part see Theorem 1.7 (and theorems before) and for the third part see Theorem 1.2 (for global solution) as I said in my answer.
In Theorem $1.7$ you have the hypothesis that $f$ is Lipschitz continuous from $Y=D(A)$ into $Y$ with the graph norm (which is a strong condition.). After Theorem $1.7$ there is a remark saying that: If in the previous theorem we assume only that $f: [t_ 0, T] \times Y \rightarrow Y$ is locally Lipschitz continuous in $Y$ uniformly in $[t_0, T]$ we obtain, using Theorem 1.4, that for every $u_0 \in D(A) $ the initial value problem possesses a classical solution on a maximal interval $[t_0, t_\max[$. It would be interesting if this comment was valid with $f$ only locally Lipschitz in $X$.
@VictorHugo see my edit
Thank you! @S.Maths
@VictorHugo Welcome!
Just to clarify, does the proof of Proposition 4.3.9 depend on whether or not $F$ is locally Lipschitz? (It seems not)
Your claim is true. Observe that there is no loss of generality to assume $\|T(t)\|\leq M$ for all $t\geq 0$ since you can add a constant to $A$, which you then absorb into $F$. Now the standard fixed point argument works like charm and $[0,T]\ni t\mapsto u(t)$ is seen to be Lipschitz for every $T>0$.
This is a comment to complement @S.Maths response.
Suppose we show that $$\|u(t)-u(t')\|_X\leq C|t-t'| \hbox{ for all } t,t' \in [0,T]$$ and some $C>0$.
Since $F$ is locally Lipschitz, there is $L_M>0$ such that $$\|F(v)-F(w)\|\leq L_M\|v-w\|$$ for all $v,w \in X$ such that $\|v\| \leq M$ and $\|w\| \leq M$.
Taking into account the continuity of the norm, the function $\|u(\cdot)\|:[0,T] \rightarrow \mathbb{R}$ is continuous on the compact set $[0,T]$.
Therefore, there is $M>0$ such that $\|u(t)\| \leq M$ for all $t \in [0,T]$. Thus, $$\|F(u(t))-F(u(t'))\|\leq L_M\|u(t)-u(t')\|\leq CL_M|t-t'|$$ for all $t,t' \in [0,T]$, which proves that $t \in [0,T] \mapsto F(u(t))$ is Lipschitz continuous, no matter if $F$ is locally Lipschitz or globally Lipschitz.
What do you mean by "no matter if $F$ is locally Lipschitz or locally Lipschitz."
Sorry, it should be "no matter if $F$ is locally Lipschitz or globally Lipschitz."
I wanted to say that the same proof applies in the case where$F$ is globally lipschitz.
Yes, this is obvious, because in locally compact metric spaces "locally Lipschitz" is equivalent to "globally Lipschitz on every compact subset". In your case the space is $[0,T]$ and the function is $F(u)$.
|
2025-03-21T14:48:30.146141
| 2020-03-26T13:52:05 |
355761
|
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|
Stack Exchange
|
Completely split primes in non-anticyclotomic $\mathbb{Z}_p$-extensions
In his colloquium paper "The Structure of Selmer Groups" Greenberg writes the following:
If $K$ is an imaginary quadratic field ... it is conjectured that for any [non-anticyclotomic] $\mathbb{Z}_p$-extension of K at most one prime of $K$ can split completely. (One can prove that at most two can.)
(end of the first paragraph on page 3)
I am looking for a reference in which this conjecture is formulated and also where the claim in parentheses is proven. I have searched through most, if not all, of the author's articles and much of the existing literature but have not been able to find anything regarding this question. Perhaps the conjecture is folklore or it follows easily from some other more well-known conjecture that I have not considered.
A further question after this would be if there is anything one conjectures (or can even prove) in the general CM case?
Thanks in advance!
The article Sur les ideaux dont l'image par l'application d'Artin dans une $\mathbb{Z}_p$-extension est triviale of Michel Emsalem provides a satisfactory answer to the general question of how many places can split in a $\mathbb{Z}_p$-extension.
As such, I will mark this question as answered.
|
2025-03-21T14:48:30.146273
| 2020-03-26T13:53:18 |
355762
|
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|
Stack Exchange
|
descent implies hyperdescent
My question concerns the Propsition 5.12 of the paper of Bhatt-Mathew on arc-topology where they claim that the functor $X\mapsto D^b_{\text{cons}}(X,\Lambda)^{[-n,n]}$ which assigns to a qcqs scheme $X$ the subcategory of the full derived category $D(X_{ét},\Lambda)$ of étale sheaves of $\Lambda$-modules of amplitude in $[-n,n]$ spanned by those objects which are bounded with constructible cohomology. Here they claim that this functor satisfies $v$-descent and this automatically implies $v$-hyperdescent. In the proof they use the fact this functor is bounded(in the derived category) and hence takes values in $\text{Cat}_{2n+2}$ the ($\infty$)-category of $2n+2$-categories. It seems to me that we can generalise the situation as follows.
Let functor $\mathcal{F}\colon\text{Sch}_R\to\mathcal{C}$(+ some properties) be a functor where $\mathcal{C}$ is an infinity category(+ some properties) and $\tau$ be a Grothendieck topology on $\text{Sch}_R$. Then I want to study the implication ``$\tau$-descent $\implies$ $\tau$-hyperdescent''.
Here I assume that $\mathcal{F}$ is locally of finite presentation(``finitary'' according to loc. cit.).
In what generality of $\mathcal{C}$ is it is true ? I think I can prove the implication in small cases like when $\mathcal{C}=\text{Cat}_1$.
Definitions
We say that $\mathcal{F}$ satisfies $\tau$-descent if for each $\tau$-covering $X'\to X$ the natural map $$
\mathcal{F}(X) \to\lim(\mathcal{F}(X')\rightrightarrows\mathcal{F}(X'\times_X X')\rightrightarrows \cdots)$$
is an equivalence in $\mathcal{C}$.
The functor is said to satisfy $\tau$-hyperdescent if for every hypercover $$X_{.}=(\cdots \rightrightarrows X_1 \rightrightarrows X_0 \to X_{-1}=X)$$ of $X$ in the $\tau$-topology we have that $$
\mathcal{F}(X)\to \lim(\mathcal{F}(X_0)\rightrightarrows \mathcal{F}(X_1) \rightrightarrows\cdots)$$ is an equivalence in $\mathcal{C}$.
It is certainly true that descent implies hyperdescent whenever $\mathcal C$ is a $n$-category for some $n<\infty$ (it wasn't clear from your question whether you knew this or not). This is because, for any $\infty$-site $\mathcal A$:
A presheaf $F:\mathcal A^{op}\to\mathcal C$ is a sheaf or hypersheaf if and only if $\mathrm{Map}(c, F(-)):\mathcal A^{op}\to\mathcal S$ is for all $c\in\mathcal C$.
Every sheaf $F:\mathcal A^{op} \to \mathcal S_{\leq n}$ is a hypersheaf, because truncated objects in an $\infty$-topos are hypercomplete.
Here is a more general condition. Suppose that $\mathcal C$ is generated under colimits by cotruncated objects, i.e., truncated objects in $\mathcal C^{op}$. Any $n$-category satisfies this, but also the $\infty$-category of coconnective spectra in any presentable $\infty$-category (and this is usually not an $n$-category for any finite $n$). For example, if $R$ is a ring then $D(R)_{\leq 0}$ is compactly generated by cotruncated objects; many sheaves in the paper of Bhatt and Mathew take values in this $\infty$-category.
Under this condition on $\mathcal C$, every sheaf $F:\mathcal A^{op}\to\mathcal C$ is a hypersheaf. Indeed, suppose $X_\bullet\to X$ is a hypercover. We want to show that the map $F(X) \to \mathrm{lim}_{n\in\Delta^{op}} F(X_n)$ is an equivalence in $\mathcal C$. Since $\mathcal C$ is generated under colimits by cotruncated objects, it suffices to check after applying $\mathrm{Map}(c,-)$ for $c$ cotruncated. This means that $\mathrm{Map}(c,-)$ takes values in $n$-truncated spaces for some $n$ (depending on $c$). In particular $\mathrm{Map}(c,F(-))$ is a hypersheaf, whence the result. For related observations see Definition 3.1.4 in https://arxiv.org/pdf/2002.11647.pdf and the following results.
|
2025-03-21T14:48:30.146518
| 2020-03-26T14:12:21 |
355764
|
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|
Stack Exchange
|
O(1) or o(logn) discrepancy for multiples of an irrational for at least one sub interval
O(1) or o(logn) discrepancy for multiples of an irrational for at least one sub interval.
Using $\{x\}$ to denote the fraction part of $x$ we can define for any $I\subset [0,1]$,
$$E(n,\theta, I) ={ \left|\{\,\{\theta\},\{2\theta\},\dots,\{n\theta\} \,\} \cap I \right|}-n|I|$$
$ $
$$\Delta_{sup}(n,\theta)=\sup_I |E(n,\theta,I)|$$.
The Equidistribution Theorem says that the sequence $a_i=(i\theta)$, $i\in\mathbb{Z}_{\geq1}$ is equidistributed modulo 1 when $\theta$ is irrational hence
$$\Delta_{sup}(n,\theta)=o(n)$$ for all irrational $\theta$.
Furthermore if $\theta$ has bounded partial denominators in its continued fraction expansion then we have
$$\Delta_{sup}(n,\theta)\ll \log n$$
(See Theorem 1.B of W Schmidt "Lectures on irregularities of distribution" $^*$
My question asks how small can $E$ be as a function of $n$.
1) Does there exist an irrational number $\theta$ and a fixed interval $I\subsetneq [0,1]$ with $E(n,\theta, I)=O(1)$?
If this false or too difficult:
2) Does there exist an irrational number $\theta$ and a fixed interval $I\subsetneq [0,1]$ with $E(n,\theta, I)=o(\log n)$?
If this false or too difficult:
3) Does there exist an irrational number $\theta$ and an interval $I\subsetneq [0,1]$ of fixed size whose position is allowed to vary with $n$, s.t. $E(n,\theta, I)=o(\log n)$?
$*$ "Schmidt, Wolfang M., Lectures on irregularities of distribution. (Notes by T. N. Shorey), Tata Institute of Fundamental Research, Lectures on Mathematics and Physics: Mathematics, 56. Bombay: Tata Institute of Fundamental Research. vi, 128 p. (1977). ZBL0434.10031.")
Isn't it known that the golden ratio minimizes the irregularities of distribution across all irrationals? (per the discussion about the coefficients of the continued fraction expansion). I don't imagine that $I$ really has any bearing on this, as I would expect all fixed intervals to perform asymptotically the same. (This doesn't cover your (3), admittedly)
@StevenStadnicki Thanks for your comment Steven. Yes I think it's possible that the interval doesn't matter at least for certain ranges of the error. This isn't true for general equidistributed sequences - for the Van der Corput sequence you have the analogous discrepancy to be $\leq1$ for any elementary interval $[w/2^k, (w+1)/2^k]$ so it would be interesting to have a proof.
Maybe van Aardenne-Ehrenfest, T. "Proof of the Impossibility of a Just Distribution of an Infinite Sequence Over an Interval." Proc. Kon. Ned. Akad. Wetensch. 48, 3-8, 1945 is relevant here, or https://mathworld.wolfram.com/DiscrepancyTheorem.html
@GerryMyerson Thank you for the reference. I think this is more related to showing that there is at least one interval with discrepancy $\geq C \log n$ whereas I am interested in seeing if there is at least one interval with $\leq C \log n/\log \log n$ say. However if we could show that for the irrational number sequence that all intervals of the same size have approximately the same discrepancy (as Steven proposed) then this result would show that 1) , 2) and 3) are all false.
I'm not sure how to do this though!
I think you are asking for so-called "bounded remainder sets". Given $\theta$, there exist intervals $I$ having discrepancy $O(1)$, namely those whose length is in $\mathbb{Z} + \theta \mathbb{Z}$. The classical reference is a paper of Kesten:
H. Kesten,On a conjecture of Erdös and Szüsz related to uniform distribution mod 1, Acta Arith. 12(1966), 193–212.
Thank you that is an extremely elegant and satisfactory result! To summarise this shows that for any irrational, 1) and hence 2) and 3) are true for intervals of sizes ${n\theta}$ only. Fantastic!
|
2025-03-21T14:48:30.146776
| 2020-03-26T15:28:57 |
355770
|
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|
Stack Exchange
|
Can every diffeomorphism be rescaled into a volume preserving one?
This is a cross-post. Let $D \subseteq \mathbb{R}^2$ be the closed unit disk, and let $f:D \to D$ be a diffeomorphism.
Does there exist a smooth $h \in C^{\infty}(D)$ such that $h\cdot f$ is an area-preserving diffeomophism of $D$?
(Clarification: by $h\cdot f$ I mean multiplication of a scalar by a vector, i.e. $\big( h\cdot f\big) (x):=h(x)\cdot f(x)$).
Clearly, such an $h$ must send the entire boundary $\partial D$ either to $1$ or to $-1$. Another necessary condition is $\det\big(d (h\cdot f)\big) = 1$. Since
$$
d (h\cdot f)=h df+dh \otimes f=h df+f \cdot (\nabla h)^T,
$$
the matrix determinant lemma implies that
$$
1=\det\big(d (h\cdot f)\big)=h^2\det(df)+(\nabla h)^T \operatorname{adj}( hdf ) f. \tag{1}
$$
In fact, I am not even sure whether the PDE $(1)$ always has a solution for every diffeomorphism $f$. (that is even when omitting the requirement that $h\cdot f$ would be a diffeomorphism, or even a map from $D$ into $D$-when looking only at the PDE with no other restrictions on the result-does there always exist a solution?)
What is meant by $h\cdot f$? (I'd read it a priori as $h\circ f$, but then the question would be trivial taking $h=f^{-1}$). Where does "conformally" intervene in the question? If I were reading the title, I'd interpret it as whether for each $f$ there exists $h$ conformal such that $h\circ f\circ h^{-1}$ is area-preserving.
@YCor I meant for the standard (scalar) multiplication of a vector by a scalar. Your are right that the title was a bit misleading. I have edited the question to address this, and also clarified what do I mean by $h \cdot f$. Thanks for your comment.
Surely yes? Pull the volume form $\omega$ back through $f$. $f^\star \omega$ is another volume form. Then the question is, does there exist $h$ so $\omega=h (f^\star\omega)$ which presumably can be shown to be true. Moser's theorem?
@AlexArvanitakis Hi, I don't follow your reduction. Why is the question equivalent to finding a function $h$ such that $ \omega=h (f^\star\omega)$? According to my calculation, $(hf)^*\omega \neq h (f^\star\omega)$. I also don't see the connection to Moser's theorem. If you could elaborate that would be great.
If it is possible, then
$$\mathrm{area}\,f^{-1}(S)=\mathrm{area}\,S$$
for any sector $S\subset D$.
But this identity might fail --- so, not in general.
|
2025-03-21T14:48:30.147032
| 2020-03-26T15:33:06 |
355773
|
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|
Stack Exchange
|
intuition for lattices in p-adic (or other non-Archimedean) vector spaces?
I could use some help to jumpstart my intuition for lattices in vector spaces over non-Archimedean fields, like $\mathbb{Q}_p$ and $\mathbb{F}_q((t))$.
I have some intuition for $\mathbb{Z}$-lattices in $\mathbb{Q}$- or $\mathbb{R}$-vector spaces, though perhaps it's not sophisticated enough. The question formed while watching some talks about affine Grassmannians and Witt-vector analogs, realizing that my prior intuition was leading me astray.
I'd like to develop some intuition, in particular, for $R$-lattices in $K$-vectors spaces, where $R$ is $\mathbb{Z}_p$ or $\mathbb{F}_p[[t]]$ and $K$ is $\mathbb{Q}_p$ or $\mathbb{F}_p((t))$.
As a starting point, I would like to have a good feel for 2-dimensional lattices $R^2 \hookrightarrow K^2$.
Question: What is a good way to visualize these objects? Or what is a minimal list of properties that I should aim to capture in a mental image, especially to reflect similarities and differences to the case of $\mathbb{Z}$-lattices?
[Edit]: Details of current intuition:
My intuition for $\mathbb{Z}$-lattices mostly stems from the standard grid $\mathbb{Z}^2 \hookrightarrow \mathbb{R}^2$ and its images under $GL_2(\mathbb{R})$ action, moving the basis elements around arbitrarily, and then imagining higher dimensional analogs, to the extent possible.
I don't remember the full list of ways I noticed my intuition was leading me astray in the context of affine Grassmannians, but three insufficiencies with my current intuition are:
1) Unlike $\mathbb{Z} \subset \mathbb{Q}$, my mental image of $\mathbb{Z}_p \subset \mathbb{Q}_p$ already occupies two dimensions, rather than one, where I think about the $p^{n+1}\mathbb{Z}_p$ cosets as $p$ 2D blobs nested inside each $p^n\mathbb{Z}_p$ coset. So visualizing a rank-2 lattice here is akin to the difficulty of visualizing a plane curve over $\mathbb{C}$ using the intuition of plane curves over $\mathbb{R}$ - we've already got an ambient space of 4 dimensions.
2) $\mathbb{Z} \subset \mathbb{R}$ is cocompact, which is not the case for $\mathbb{Z}_p \subset \mathbb{Q}_p$. This is connected to the fact that if we move outward in $\mathbb{R}$, we keep encountering elements of $\mathbb{Z}$, whereas for my image of $\mathbb{Q}_p$, $\mathbb{Z}_p$ is located directly at the center, not occuring periodically.
3) My mental images for $\mathbb{Q}_p$ and $\mathbb{F}_p((t))$ are more or less the same, but I would like a mental image that somehow distinguishes them.
Could you talk about what your intuition is for $\mathbb Z$-lattices, to get some feel for what's intuitive to you? (For example, "just picture a $\operatorname{GL}_2(K)$-translate of the standard embedding" probably isn't what you want, but what is?)
One distinction between the archimedean and non-archimedean situations is that $\mathbf Z$ in $\mathbf R$ is discrete with a compact quotient while $\mathbf Z_p$ inside $\mathbf Q_p$ is compact with a discrete quotient (i.e., $\mathbf Z_p$ is open in $\mathbf Q_p$).
@LSpice in my mind I know that is true, but not yet in my heart. The points I added are my best at putting into words why not.
@KConrad hmm, yes, an interesting reversal, which certainly does suggest the need for a different mental image. Is there a high-level explanation for why this is the case? My initial hunch is that it's connected to the fact that the real and p-adic expansions go off in different directions.
As a topological space you can embed $\Bbb{Q}p$ into the real line, sending $\sum{j=J}^\infty a_j p^j\in \Bbb{Q}p, a_j\in 0 \ldots p-1$ to $\sum{j=J}^\infty a_j p^{-3j}\in \Bbb{R}$. If $U\subset \Bbb{R}^n$ is open then $U\cap \Bbb{Q}_p^n$ is open in $\Bbb{Q}_p^n$, and a lattice in $\Bbb{Q}p$ is a particular kind of open set closed under $p$-adic addition. The $p$-adic addition is obtained from the real addition by identifying $\sum_j (a_j+b_j p) p^{-3j}\sim \sum_j (a_j+b{j-1}) p^{-3j},b_j\in 0,1$
|
2025-03-21T14:48:30.147493
| 2020-03-26T15:49:19 |
355776
|
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|
Stack Exchange
|
Invariant subspaces and isotypic decomposition (reference request)
If $G$ is a (say) compact group and $V=\bigoplus_{i\in I}V_i$ the isotypic (a.k.a. primary) decomposition of a $G$-module, then
any $G$-invariant subspace $W\subset V$ writes $W=\bigoplus_{i\in I}(W\cap V_i)$.
While this isn’t hard to prove (similar to Hoffman-Kunze 1971, §7.5 for a single operator), it seems silly to redo it in a paper. Unfortunately, the only reference I’m familiar with omits the proof (Kirillov 1976, §8.3), and when using it (e.g. Lie groups VIII.3.1) Bourbaki refers to such an abstrusely worded version (Algebra VII.2.2) that unpacking it takes as much work as a direct proof.
Q: What is a good reference to quote for this? Bonus points if the case of non-algebraically closed fields is spelled out.
A better Bourbaki reference is Algèbre VIII, §4, Proposition 4 (unfortunately not yet translated, as far as I know). It works for semi-simple modules, no fields involved.
@abx Would you like to make your comment an answer? It is the one I wish I could accept, so far.
OK, done. $\qquad$
As suggested by the OP, I am turning my comment into an answer: a better Bourbaki reference is Algèbre VIII (new edition), §4, Proposition 4 d) (unfortunately not yet translated, as far as I know). It works for semi-simple modules over an arbitrary ring.
Seems perfect, thanks. The earlier edition has it as VIII, §3, Proposition 9 b): “Tout sous-module $\mathrm N$ de $\mathrm M$ est somme directe des $\mathrm N\cap\mathrm M_\lambda$”.
For compact groups you can quote IV.2.7 in Naimark-Stern. There the $T$-isotypical component is described as the image of an operator $E^T$.
For general semisimple categories it may be better to give a short modern proof. Define the center of the category. It is a product of division rings. Each simple object gives an idempotent in the center. This idempotent gives compatible idempotents in $Hom(V,V)$ and $Hom(W,W)$, which you can split off.
Thank you. Naimark and Stern’s long section 2.7 (pp. 199-205) has much useful material on the primary decomposition of a given $V$ (what they call $S$) and projectors onto a $V_i$ (what they call $\smash{E^T}$), but they don’t seem to state let alone prove results on restriction to arbitrary invariant $W\subset V$, do they?
They don't, you are right. They prove that $E^T$ is the projection $V\rightarrow V_{i}$, killing all other isotypical components. The operator does not depend on a representation. Since $V_i$ (or $W_i$) is the 1-eigenspace of $E^T$ on $V$ (or $W$), it is a basic Linear Algebra exercise to deduce the statement you need.
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2025-03-21T14:48:30.147719
| 2020-03-26T15:53:49 |
355777
|
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|
Stack Exchange
|
Geodesic convexity and the Geometric Hessian
This is an elementary question in differential geometry. We know that for a smooth real-valued function $f$ defined on an open geodesically convex set of a Riemannian manifold $ \mathcal{X} \subset \mathcal{M}$, $f$ is geodesically convex if and only if its Riemannian Hessian is positive semidefinite on $\mathcal{X}$. Here, the Riemannian Hessian is defined as the second covariant derivative of $f$, hence a tensor of type (0,2). This result can be found in [1].
My question is whether the above result applies to the alternative definition of Riemannian Hessian as a linear mapping on the tangent space at a point, hence a tensor of type (1,1). This alternative definition is also called the ''geometric Hessian'' in [2]. Specifically, let $T_x{\mathcal{M}}$ be the tangent space at $x \in \mathcal{X}$. The geometric Hessian at $x$, denoted by $\text{Hess} f(x): T_x{\mathcal{M}} \to T_x{\mathcal{M}}$, is defined as,
$$\text{Hess} f(x) [\eta_x] = \nabla_{\eta_x} \text{grad} f, \;\forall \eta_x \in T_x{\mathcal{M}},$$
where $\nabla$ is the Riemannian connection and $\text{grad} f$ is the gradient vector field.
I am wondering if the following statement is true:
For a smooth function $f: \mathcal{X} \to \mathbb{R}$, $f$ is geodesically convex if and only if $\text{Hess} f(x) \succeq 0$ for all $x \in \mathcal{X}$.
References:
[1] C. Udriste (1994), Convex Functions and Optimization Methods on Riemannian manifolds, Mathematics and its Applications. Dordrecht: Kluwer Academic Publishers. ISBN 0-7923-3002-1.
[2] P-A. Absil , R. Mahony, and R. Sepulchre (2008), Optimization Algorithms on Matrix Manifolds, Princeton University Press.
The Riemannian metric can be used to turn one Hessian into the other. So the assumptions are the same in both versions.
@SebastianGoette Thank you for the comment. I have posted a proof.
Thanks to Prof. Absil for verifying the following proof.
Let $\mathfrak{X}(\mathcal{M})$ and $\mathfrak{F}(\mathcal{M})$ be the set of smooth vector fields and scalar functions on $\mathcal{M}$, respectively. By Theorem 6.2 in [1], $f$ is geodesically convex in $\mathcal{X}$ if and only if its second covariant derivative $\nabla^2 f: \mathfrak{X}(\mathcal{M}) \times \mathfrak{X}(\mathcal{M}) \to \mathfrak{F}(\mathcal{M})$ is positive semidefinite on $\mathcal{X}$, i.e., for all vector fields $\eta \in \mathfrak{X}(\mathcal{M})$, the corresponding function $\nabla^2 f[\eta, \eta]$ is non-negative on $\mathcal{X}$.
Let $\nabla$ be the Riemannian connection.
At a particular point $x \in \mathcal{X}$,
\begin{equation}
\nabla^2 f(x)[\eta, \eta] = \langle \nabla_{\eta_x} \text{grad} f, \eta_x \rangle_x = \langle \text{Hess} f(x) [\eta_x], \eta_x \rangle_x \geq 0,
\end{equation}
where the second equality holds by definition of the geometric Hessian (Definition 5.5.1 in [2]).
References:
[1] C. Udriste (1994), Convex Functions and Optimization Methods on Riemannian manifolds, Mathematics and its Applications. Dordrecht: Kluwer Academic Publishers. ISBN 0-7923-3002-1.
[2] P-A. Absil, R. Mahony, and R. Sepulchre (2008), Optimization Algorithms on Matrix Manifolds, Princeton University Press.
|
2025-03-21T14:48:30.147933
| 2020-03-26T15:53:56 |
355778
|
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|
Stack Exchange
|
Parallel transport of vector along piecewise smooth loop on high-dimensional manifold
In this https://math.stackexchange.com/questions/2568300/gauss-bonnet-like-statement-connecting-parallel-transport-and-curvature question, it was discussed that the rotation of a vector that is parallely transported along a piecewise smooth simple loop $\gamma$ in a $\textbf{2-dimensional}$ manifold is exactly the integral of the gaussian curvature over the interior of the loop. The proof seems to be similar to the one of the $2$-dimensional Gauss-Bonnet theorem.
I am interested in a high-dimensional analogue of this result, let's say in the case of hypersurfaces, i.e. for an $m$-dimensional manifold M embedded in $\mathbb{R}^{m+1}$ (assume also that $m$ is even if necessary). In this case (Chern-)Gauss-Bonnet theorem becomes like theorem 5.4.3 here https://people.math.ethz.ch/~salamon/PREPRINTS/difftop.pdf and a more general statement (not just for hypersurfaces) are theorems 4.2 and 4.4 (for manifolds with boundary) here https://math.berkeley.edu/~alanw/240papers00/zhu.pdf.
I wait for a result of the type $\int_{interior} K dvol_M= rot_{\gamma} \mod something$. I am not sure what "something" should be. I think that this is connected to my lack of understanding about the sum of angles of a geodesic triangle in a high-dimensional manifold. Any idea or reference is highly appreciated.
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2025-03-21T14:48:30.148068
| 2020-03-26T16:15:59 |
355783
|
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|
Stack Exchange
|
Logconcavity of height of Dyck paths
A finite sequence $a_i$ is called logconvace in case $a_i^2 \geq a_{i-1} a_{i+1}$.
Question : For a fixed $n$, is the sequence $a_{n,k}$ giving the number of Dyck paths of semilength $n$ having height $k$ logconcave? (see http://oeis.org/A080936)
A starting point is the product formula $(2k^2+6k+1-3n)(2n)!/((n-k)!(n+k+3)!)$ given in the reference, which is, however, valid only for $(n+1)/2\leq k\leq n$. Computing $T(n,k-1)T(n,k+1)/T(n,k)^2$ gives you three natural factors, which are all less than $1$.
I have an idea for a combinatorial approach that I'll just sketch for now. (1) Given two height $k$ paths, follow the first path until you reach height $k$, then insert the entire second path, and finally complete the first path. (2) Given a height $k-1$ path and a height $k+1$ path, build the same kind of composite path by interrupting the first at height $k-1$. Both of these approaches create height $2k$ paths. Why does (1) generate more? Or starting from the height $2k$ paths, why are there more ways to split them into two height $k$ paths than heights $k-1$ and $k+1$ paths?
A stronger property than log-concavity, is real-rootedness of $\sum_k t^k a_{n,k}$.
However, for $n=4$, this polynomial is $1 + 7 t + 5 t^2 + t^3$ which is not real-rooted.
|
2025-03-21T14:48:30.148204
| 2020-03-26T17:17:02 |
355787
|
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|
Stack Exchange
|
Quasi-isometries and E-unitary inverse semigroups
Let $S = \langle K\rangle$ be a finitely generated inverse semigroup, where $K \subset S$ is a fixed, finite and symmetric set of generators.
Preliminaries: Recall that we say that $s, t \in S$ are $\mathcal{L}$-related if $s^{-1}s = t^{-1}t$. Given an $\mathcal{L}$-class $L \subset S$, we may construct its' Schützenberger graph $\Lambda(L, K)$, whose vertices are the points of $L$ and where $x, y \in L$ are connected by an edge labeled by $k \in K$ if $kx = y$. We consider $L$ equipped with the natural path metric $d_L$ via $\Lambda(L, K)$. Another studied congruence in $S$ is $\sigma$, where $s \sigma t$ if $sx = tx$ for some $x \in S$. The quotient $S/\sigma$ is a group $G$ known as the maximal homomorphic image of $S$. Moreover, we say that $S$ is E-unitary if whenever $s \sigma t$ and $s \mathcal{L} t$ then $s = t$, i.e. the quotient map embeds every $\mathcal{L}$-class into $G$.
Question: Let $S = \langle K \rangle$ be a fin. gen. E-unitary inverse semigroup. Let $L \subset S$ be an $\mathcal{L}$-class. Is the quotient map $L \rightarrow G$ a quasi-isometric embedding? That is, are there constants M, C > 0 such that for all $x, y \in L$
$$ \frac{1}{M} d_L(x, y) - C \leq d_G(x\sigma, y\sigma) \leq M d_L(x, y) + C$$
where $d_G$ is the path metric in the left Cayley graph of $G$ with respect to the generating set $K \sigma$. Observe that the right inequality above is true for any $M \geq 1$, since any geodesic between $x, y$ falls down to a path between $x\sigma, y\sigma$.
Partial results/remark: it's clear that if $L$ has only finitely-many $\mathcal{R}$-classes then the quotient map is going to be a quasi-isometry. Indeed, the $\mathcal{H}$-class of the idempotent of $L$ is a group included in $G$, and that inclusion of groups is a quasi-isometry. Since $L$ has only finitely-many $\mathcal{R}$-classes, then so is the inclusion of $L$ into $G$, i.e., the quotient map.
Motivation: In [1] quasi-isometries in monoids are studied, from the point of view of the Cayley graph. However, little is said about my inquiry, and I haven't been able to find any reference on this in the literature. My guess is the answer should be yes, but any help is greatly appreciated.
[1] Gray and Kambites, Groups acting on semimetric spaces and quasi-isometries of monoids, Trans. Ame. Math. Soc. 365 (2013) 555--578.
What if you take the free E-unitary cover of the free abelian group of rank 2 generated by x,y and add the idempotent relations $xx^{-1}=1=x^{-1}x$. This should give an E-unitary inverse semigroup where maximal group image is the free abelian group where Schutzenberger graphs have finitely many y edges but all horizontal x edges through any vertex. Then the Schutzenberger graph of y does not quasisometrically embed because the distance from (n,0) to (n,1) is 2n in the Schutzenberger graph and is 1 in the group.
That should be 2n+1 not 2n.
The answer is no. Let $G$ be a free abelian group of rank 2 generated by $x,y$. Let $S$ be the Meakin-Margolis expansion of $G$. It consists of all pairs $(X,g)$ with $X$ a finite connected subgraph of the Cayley graph of $G$ containing the origin and $g$. The product is $(X,g)(Y,h)=(X\cup gY,gh)$. The projection to $G$ is an idempotent pure homomorphism, so $S$ is E-unitary, and $S$ is generated by the edge from the origin to (1,0) and the edge from the origin to (0,1). Call these generators $x,y$ respectively. Now let $T$ be the quotient of $S$ by the relations $xx^{-1}=1=x^{-1}x$. It is not hard to see that $T$ is E-unitary since it is sandwiched between $S$ and $G$. Its elements can be viewed as pairs $(X,g)$ with $X$ a connected subgraph of the Cayley graph which contains the origin and $g$ with only finitely many vertical $y$ edges and containing the horizontal line through any vertex of $X$. Such graphs are precisely the Schutzenberger graphs of $T$.
These graphs in general do not quasi-isometrically embed as soon as they have a $y$ edge. For example of you take the Schutzenberger graph of $y$ you have the lines $x=0$ and $x=1$ and the edge from $(0,0)$ to $(0,1)$. So the distance from $(n,0)$ to $(n,1)$ in this graph is $2|n|+1$ while in the Cayley of $G$ the distance is $1$. So the embedding is not a quasi-isometry.
@MarkSapir any quotient of an E-unitary inverse semigroup that is above the maximal group image is E-unitary
@MarkSapir, likely. It is elementary and probably in Petrich's book too.
|
2025-03-21T14:48:30.148789
| 2020-03-26T17:26:33 |
355788
|
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|
Stack Exchange
|
A Jacobian of the Cayley graph
Let $\pi$ be a finitely generated discrete group. Let $F$ be a finitely generated free group with an epimorphism to $\pi$. Let $K$ be the kernel of the epimorphism. Consider the Abelianization of $K$. This is an infinitely generated free Abelian group. Let $T$ be the Pontryagin dual, which is an infinite-dimensional torus (something like a Jacobian of the Cayley graph). The group $\pi$ acts on it. Did anybody study such a thing?
I'm not aware of any literature on its Pontryagin dual, but the abelianization of $K$ is called the relation module of the finitely generated group $\pi$. It is a module over $\mathbb{Z}[\pi]$, and has a large literature (though in general it is hard to work with).
As a random example of something that is known about this, the Lyndon Identity Theorem says that the relation module of a one-relator group is cyclic; see Theorem 6.2 of my notes on one-relator group (see https://www3.nd.edu/~andyp/notes/OneRelator.pdf) for more details.
|
2025-03-21T14:48:30.148885
| 2020-03-26T17:26:34 |
355789
|
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|
Stack Exchange
|
Representation of algebras as bounded nilpotent operators
Let $A$ be a real/complex algebra (just a real/complex vector space with a multiplication; none PI's are required). Let $\mathcal{H}$ be a real/complex Hilbert space. Let $\operatorname{B}(\mathcal{H})$ be the algebra of bounded endomorphisms of $\mathcal{H}$ and let $\operatorname{Nil}(\mathcal{H})\subset\operatorname{B}(\mathcal{H})$ be the vector subspace of nilpotent bounded operators, i.e., such that $T^2=0$.
My question is:
There are known conditions on $A$ and on $\mathcal{H}$ to ensure the existence of representations $\rho:A\rightarrow \operatorname{B}(\mathcal{H})$ which factor trough the inclusion $\operatorname{Nil}(\mathcal{H})\hookrightarrow\operatorname{B}(\mathcal{H})$?
In other words:
Under which conditions there exists an algebra homomorphism $\rho:A\rightarrow \operatorname{B}(\mathcal{H})$ whose image is contained in $\operatorname{Nil}(\mathcal{H})$?
Thank you for any help.
P.S: I'm specially interested when the Hilbert space is the Sobolev space of some nice bounded open set $U\subset \mathbb{R}^n$ and related structures.
All infinite-dimensional separable Hilbert spaces are isometric hence they have the same algebras of bounded operators. As you you look at algebraically defined entities, the question does not depend on the particular incarnation of $H$.
Thank you for this remark.
What are "PI's" in the first sentence, and can you clarify what you mean by "subalgebra of those nilpotent bounded operators which mutually commute"?
PI = Polynomial Identity.
And I tried to clarify the meaning of "subalgebra of those nilpotent bounded operators which mutually commute" editing my question. Actually, it was bad written. Sorry for this and thanks for the comment.
Thank you, but I'm still confused. How do I tell whether $T$ belongs to Nil$(\mathcal{H})$? It has to commute with everything else in Nil$(\mathcal{H})$. So I already need to know what is in Nil$(\mathcal{H})$.
I changed again my question. I forgot the commuting requirement. Now $\operatorname{Nil\mathcal{H}}$ is just the vector subspace of bounded operators such that $T^2=0$. This is enough for my problem.
The problem is still ill-posed, because the set of nilpotent bounded operators is not a vector space. (The sum of two nilpotents need not be nilpotent.) Let's interpret the question as: which algebras admit faithful representations as algebras of nilpotent bounded operators?
Note that "nilpotent" usually means $T^n = 0$ for some $n$, not $T^2 = 0$. But whatever definition you take, it will automatically be shared by $A$ and any faithful representation of $A$. So the question is just: which nilpotent algebras can be faithfully represented as algebras of bounded operators?
In the finite dimensional case, a theorem of Cayley tells us that every algebra can be faithfully represented as an algebra of matrices. So in this case every nilpotent algebra has a faithful matrix representation. I think the same is true of any algebra whose dimension is countable. Sketch of proof: assume $A$ is unital and let $T_1, T_2, \ldots$ be a vector space basis for $A$. We regard $A$ as acting on itself by left multiplication and need to find an inner product with respect to which every left multiplication operator is bounded. We can do this by taking the $T_i$ to be mutually orthogonal, and inductively taking the Hilbert space norm of $T_i$ to be sufficiently large so as to ensure that the compression of $M_{T_j}$ (the left multiplication operator), for any $j < i$, to ${\rm span}(T_1, \ldots, T_i)$ has norm at most $(2-1/i)r_j$ where $r_j$ is the norm of its compression to ${\rm span}(T_1, \ldots, T_j)$.
Uau! This is exactly what I need. Thank you very much!
You are welcome!
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2025-03-21T14:48:30.149131
| 2020-03-26T17:48:34 |
355792
|
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|
Stack Exchange
|
When do objects in the image of a functor $G$ have a unique action as algebras over the codensity monad of $G$?
Let $G:\mathcal{B}\longrightarrow \mathcal{A}$ be some functor which admits a right Kan extension along itself, $(\operatorname{Ran}_G G, \eta:\operatorname{Ran}_G G \circ G \rightarrow G)$.
The functor $\operatorname{Ran}_G G : \mathcal{A}\rightarrow \mathcal{A}$ is canonically a monad, for a unit $u:1_{\mathcal{A}}\rightarrow \operatorname{Ran}_G G$ and multiplication $m: \operatorname{Ran}_G G \circ \operatorname{Ran}_G G \rightarrow\operatorname{Ran}_G G$ obtained from the universal property of Kan extensions: this is called the codensity monad of $G$.
Given any $B$ in $\mathcal{B}$, the object $G(B)$ can be viewed as an algebra over the codensity monad of $G$, namely as the algebra $(G(B),\eta_B)$, where $\eta$ is the natural transformation associated to the Kan extension.
Is it known when $(G(B),\eta_B)$ is in some sense the unique $\operatorname{Ran}_G G$-algebra structure that can be put on $G(B)$?
My question is motivated by the following examples investigated in http://www.tac.mta.ca/tac/volumes/28/13/28-13.pdf:
If G is the inclusion of $\operatorname{FinSet}$ in $\operatorname{Set}$, then the Eilenberg-Moore category of the codensity monad of $G$ is the category of compact Hausdorff spaces. Now, if we pick a finite set $B$ there is only one topology we can put on it so that it becomes $T_1$, namely the discrete topology, hence there is only one "way" of regarding $B$ as a compact Hausdorff space.
Fix some field $k$. If G is the inclusion of $\operatorname{FDVect}$ in $\operatorname{Vect}$, then the Eilenberg-Moore category of the codensity monad of $G$ is the category of linearly compact $k$-vector spaces. Again, if we pick a finite dimensional vector space $B$ there is only one topology we may endow it with so that it becomes linearly topologised and $T_1$, namely the discrete topology. So, as before, there is only one "way" of seeing a finite dimensional vector space as a linearly compact vector space.
Did you see 5.2, 5.3, 5.4 in the reference you mention?
@IvanDiLiberti Thank you for pointing that out. I did not know that when $G$ is fully faithful, then the natural transformation $\operatorname{Ran}_G F \circ G \rightarrow F$ associated to the Kan extension is a isomorphism.
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2025-03-21T14:48:30.149286
| 2020-03-26T18:19:01 |
355793
|
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|
Stack Exchange
|
An analog of the representation dimension for algebras
The representation dimension of a finite dimensional algebra $A$ is defined as
$repdim(A)= \inf \{ gldim(B) | B=End_A(M)$ for a generator-cogenerator $M \}$.
It was shown by Iyama that it is always finite, answering a question of Auslander.
Define $urepdim(A)= \sup \{ gldim(B) | gldim(B) < \infty , B=End_A(M)$ for a generator-cogenerator $M \}$.
Question 1: Is $urepdim(A)$ finite?
Question 2: What is $urepdim(A)$ in case $A$ is hereditary (of Dynkin type)?
Define $frepdim(A)= \sup \{ findim(B) | B=End_A(M)$ for a generator-cogenerator $M \}$, where $findim(B)$ denotes the finitistic dimension of $B$.
Question: Is $frepdim(A)$ finite?
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2025-03-21T14:48:30.149364
| 2020-03-26T18:34:00 |
355795
|
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|
Stack Exchange
|
On a mysterious reference of Grothendieck
These days I found a mysterious page on Google books describing a book entitled On the De Rham cohomology of schemes by Grothendieck, Coates, and Jussila.
At once I thought this was an error and Google books had miss-indexed Exposé IX of Dix Exposés sur la Cohomologie de Schémas. However, the references diverge both in their titles:
Crystals and the De Rham Cohomology of Schemes
vs.
On the De Rham cohomology of schemes,
as well as their numbers of pages (53 in Dix Exposés vs. 106 in this reference).
The book is even cited in Cisinski 's Habilitation thesis as
[Gro66] A. Grothendieck « On the de Rham cohomology of schemes » Publ. Math. IHES 29 (1966), p. 93--103.
Maybe this is "On the de Rham cohomology of algebraic varieties", but again, the references diverge in page number and title.
So, does this book really exists, and, if so, is it available anywhere?
(My reasons for not promptly deciding that this is just an indexing eror is that Google Books usually does a great job of indexing unpublished/unavailable/lost references, such as the second volume of Görtz--Wedhorn's textbook on schemes (see here) or abandoned book projects)
Past issues of Publ. IHES (like most French journals) are freely available at Numdam: Issue 29 (1966) is here: Grothendieck "On the de Rham cohomology of algebraic varieties" 95-103.
It might be generated by a Google bug: frequently for a thesis, PhD or Habilitation, Google refers to names of the thesis committee as authors. Also "I. Coates" doesn't seem to exist and it's probably John Coates. Coates was certainly too young in 1965 to be in such a committee. Anyway, it might be, for any reason, that this "google books" page was automatically generated and messes up several facts (it would sound unlikely that a then published 100-page paper of Grothendieck would be now beyond radars).
Tlön, Uqbar, Orbis Tertius
I think your hunch is correct that Google is in error, and that "Crystals and the de Rham cohomology of schemes" refers to the article in Dix Exposés. I doubt there is a book by the same name and author. There is also "On the de Rham cohomology of algebraic varieties" but that's different. The first is a long proposal for crystalline cohomology (which hadn't been developed at that point), and the second is an extract from a letter to Atiyah giving a proof of the algebraic de Rham theorem. Does that help?
Yes; thanks! My hopes on reading such an article are now shattered, but oh well :P
As an addendum to Donu's answer, I quote from the Table des matieres of Dix Exposés:
IX GROTHENDIECK (Alexander), Crystals and the De Rham Cohomology of schemes (notes by I. Coates and O. Jussila), IHES Decembre 1966, 54 p.
So it is certainly Dix Exposés. And "I. Coates" is most likely John Coates. According to Wikipedia Coates was born in '45 (so he would've been 21 at the time), and he moved to Paris to study at the ENS after obtaining a Bachelor's degree from ANU.
The notes can be found here, page 306 (=p314 of the pdf); it's written "I. Coates" as well at that page.
Thanks for the information! :)
PS: I got the confirmation from Pierre Berthelot (who's quoted in this chapter) that it's John coates, and "I. Coates" is a typo.
|
2025-03-21T14:48:30.149614
| 2020-03-26T18:45:28 |
355796
|
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|
Stack Exchange
|
Equation $p\cdot q\cdot r=a^3-1+43\cdot (b^2-1)$
$p\cdot q\cdot r=a^3-1+43\cdot (b^2-1)$
p, q, r are primes.
a, b integers>0.
Is this equation a Mordell equation?
Has this equation infinitely many solutions?
Any motivation for this question? To a non-expert it looks completely arbitrary.
@abx I have not yet a particular motivation
No motivation?!
Is there any reason you picked 43?
Please don't ask random "unmotivated" questions. Note that this site was created for professional mathematicians and PhD students to help their research efforts. Of course anyone can ask and answer questions, but this is the level expected here.
@Robert, perhaps OP is in the $43$ cabal that has been asking questions involving $43$ here and on m.se for a while now. E.g., https://math.stackexchange.com/questions/3574600/is-a-number-congruent-to-344-pmod-559-always-congruent-to-10m-pmod-41 and https://math.stackexchange.com/questions/3566264/3484-cdot-n-is-always-congruent-to-216-pmod-344 and https://math.stackexchange.com/questions/3564437/integers-of-the-form-215s129 and more.
@GerryMyerson I don't know about a cabal but from the quirks of idiom in certain recent questions, e.g. https://mathoverflow.net/questions/355415/has-this-diophantine-equation-infinitely-many-solutions-for-a-b-not-multiple , my guess is that a single user is creating unregistered accounts, using them to post these rather unmotivated questions, and then just creating new accounts the next time a question occurs to them
Consider e.g. $b=1$ and $a = 2k+1$, so the equation becomes $pqr =2 k (4k^2 + 6k + 3)$.
The generalized Bunyakovsky conjecture implies there are infinitely many primes $k$ such that $4 k^2 + 6 k + 3$ is prime, so that we have a solution with $p=2, q=k, r=4k^2 + 6 k + 3, a=2k+1, b=1$. To date all nontrivial cases of that conjecture, including this one, remain open.
|
2025-03-21T14:48:30.149897
| 2020-03-26T18:57:47 |
355797
|
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|
Stack Exchange
|
Moduli, Teichmüller spaces and mapping class group of a sphere with four punctures
In the complex analytic setting, it is easy to see that the moduli space of a sphere with four punctures is $\mathcal{M}=\mathbb{CP}^1 / { 0,1,\infty }$, since I can use a Moebius transformation to send three of the punctures at those location whilst the fourth puncture is free to move.
On the other hand, the moduli space can be obtained as the quotient of the Teichmüller space by the mapping class group. Teichmüller space can be described using e.g. Fenchel-Nielsen coordinates, for each simple closed curve on the sphere we have two coordinates (a length and an angle). The mapping class group can be described using Dehn twists.
I always assumed that the three boundary points ${0,1,\infty}$ could be interpreted as sending to zero the length of three distinct curves separating the punctures into one of the three sets $(12)(34)$, $(23)(41)$ or $(13)(24)$, but I have recently realized that these curves are related to each other by suitable elements of the mapping class group. So I would naively say that the corresponding limits should be identified as well.
How do I reconcile the above picture of the moduli space with the one coming from Teichmüller modulo mapping class group? How do I see three boundary points in this picture?
The very recent issue of BAMS (57, N3, 2020) contains a nice big paper of Milnor addressing this subj.
Could you tell me what BAMS stands for?
Ok, I could find it but it seems to me that there is very little material relevant for my question. It does not seems to me that this particular moduli space is mentioned in any example. Could you point me at the pag. You think is relevant?
In your description of moduli space you say:
I can use a Moebius transformation to send three of the punctures at
those location whilst the fourth puncture is free to move.
That assumes that the punctures have names. Let's call them $a, b, c, d$ and we will agree to send them to $0, 1, \infty, z$. (I prefer to use letters for the names, as then we can't confuse the punctures with numbers.)
In your discussion of Teichmuller space, you identify three curves. The first curve, call it $\beta$, separates $a$ and $b$ from $c$ and $d$, while the second curve, call it $\gamma$, separates $a$ and $c$ from $b$ and $d$. Now, since these punctures have names, there is no mapping class that sends $b$ to $c$. Thus there is no mapping class that sends $\beta$ to $\gamma$.
Except there obviously is...
The solution lies in understanding which mapping class group you are using. In your definition of moduli space you are using the pure mapping class group (fixing all punctures). In your definition of Teichmuller space you are using the full mapping class group (of orientation preserving homeomorphisms, up to isotopy).
In order to send the two curves into each other I only need Dehn twists, which are in the pure MCG. More precisely a Dehn twist along Beta followed by one along alpha sends alpha into beta.
Perhaps you are using half-twists instead of full twists? (But half-twists are not pure.) Or perhaps you think that $\alpha$ and $\beta$ meet once, and that the product of the twists behaves as it does on a torus? (But $\alpha$ and $\beta$ meet twice...) Hmmm. I suggest you draw some pictures. The easiest way is to draw a round disk $D$ in the plane, add three dots inside (labelled $a$, $b$, and $c$) and draw the curves $\beta$ and $\gamma$ as simply as is possible. They intersect an even number of times (aka twice) - after all, they are curves in the plane!
yes, I am trying to draw some. I was using this software to experiment https://flipper.readthedocs.io/en/latest/
Just to make it clear, you would disagree with me that a sequence of pure mapping class group elements (Dehn full twists) would send those three lines into each other?
I think that using that software I was applying half twists, if i do the drawings by hand I get the same as the the software applying twice its MCG elements...
How can you prove formally that up to pure mappings class group any simple closed curve is equivalent to one of the three described in my question?
There is a "slick" (ie uninformative) proof using the classification of surfaces. Let $\sigma$ be a curve in the four-times punctured sphere $F$. Suppose that $\sigma$ separates the punctures in the same way that $\beta$ does. Consider $F - \sigma$. By the classification of surfaces, this is a disjoint union of pants, say $P$ and $Q$. Similarly $F - \beta$ is a disjoint union of pants, say $R$ and $S$. Up to swapping names, there are homeomorphisms of $P$ with $R$ and of $Q$ with $S$ preserving punctures. Glue these homeomorphisms together. QED. Exercise: find a direct proof.
It seems to me that a direct proof would be equivalent to show that each vector (p,q) with p and q relatively prime can be written as g . e, where g is a matrix in the free group generated by ta={{1,2},{0,1}} and tb={{1,0},{2,1}} and e is one of the three vectors {1,0},{0,1},{1,1}.
A vector (p,q) represents a curve on the sphere, the matrices ta,tb two Dehn twist generating the pure MCG and those three vectors my three curves. Am I right?
Yes, that should work. Exercise: Now do the same for the five-times punctured sphere. (This is harder.)
|
2025-03-21T14:48:30.150265
| 2020-03-26T19:23:45 |
355799
|
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|
Stack Exchange
|
Use $\mathbb{P}(\vert \hat{s}_n-s\vert > x)\leq a(n,x)$ and $\mathbb{P}(\vert \hat{s}_n-s_n\vert > x)\leq b(n,x)$ to bound $\vert s_n - s\vert$
Let $s, s_n\in\mathbb{R}$ and $\hat{s}_n$ be a random variable.
I have two concentration inequalities:
$$\mathbb{P}(\vert \hat{s}_n-s\vert > x)\leq a(n,x)$$ for all $n\geq1$ and $x>0$;
and
$$\mathbb{P}(\vert \hat{s}_n-s_n\vert > x)\leq b(n,x)$$ for all $n\geq1$ and $x>0$.
Is there a way to bound $\vert s_n - s\vert$?
What are $a(n,x)$ and $b(n,x)$?
Yes, there is a way. Let
$$X:=\hat s_n-s,\quad h:=s_n-s,\quad a(x):=a(n,x),\quad b(x):=\limsup_{y\uparrow x}b(n,y).$$
We have
$$P(|X|>x)\le a(x)\quad\text{and}\quad P(|X-h|\ge x)\le b(x)$$
for all $x>0$, and we need to bound $|h|$ in terms of the functions $a$ and $b$.
The simple but key observation is that the event $\{|X-h|<|h|/2\}$ implies the event $\{|X|>|h|/2\}$. So,
$$a(|h|/2)\ge P(|X|>|h|/2)\ge P(|X-h|<|h|/2)\ge1-b(|h|/2),$$
whence $c(|h|/2)\ge1$, where $c:=a+b$. So,
$$|s_n-s|=|h|\le h_*:=2c^{-1}(1),$$
where $c^{-1}$ is the generalized inverse of the function $c\colon[0,\infty)\to\mathbb R$ given by the formula
$$c^{-1}(u):=\sup\{x\ge0\colon c(x)\ge u\}$$
for $u\in(0,1]$;
if $c$ is continuously and strictly decreasing from $c(0)\ge1$ to $c(\infty-)=0$, then $c^{-1}$ is the usual inverse of the function $c$.
If e.g. $a(x)=b(x)=2e^{-x^2/\sigma^2}$ for some real $\sigma>0$, then the upper bound $h_*$ on $|s_n-s|$ is $2\sigma\sqrt{\ln4}$, which is proportional to $\sigma$.
|
2025-03-21T14:48:30.150372
| 2020-03-26T19:48:21 |
355800
|
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|
Stack Exchange
|
Constructing set with maximal independent subset
What is the minimal $m$ such that there exists a set $A = \{a_1,...a_n\}$ of vectors : $a_i \in \{0,1\}^m$ ($n$ is given) such that every subset of vectors of size $k$ is independent, but only with scalars $\alpha_i = \{-1,0,1\}$? What if we also require that $||a_i||_0 = s$?
Is there a way to find such a set?
are you working over $\mathbb{R}$?
Edited, basically an answer over $\mathbb{R}$ would be ok, but I am interested in the case where all the coefficients of the linear combination can only be $-1,1,0$
Take $a_i=e_i$, where $e_i$ is the standard vector basis.
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2025-03-21T14:48:30.150459
| 2020-03-26T20:17:29 |
355805
|
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|
Stack Exchange
|
Coronavirus in NYC: What is happening in the near-term?
Although I am motivated to have some different questions answered (see below), I would like to build or use a model to predict the number of COVID-19 Cases in NYC in the near term (perhaps through the first week of April). Please regard this post as a reference request for models for this statistic.
I am a mathematician at City College of New York who is alarmed at the coronavirus numbers coming out of the city. I am not an epidemiologist. New York City is the current epicenter of the epidemic in the US, with cases growing substantially faster than elsewhere in the country. This is probably due to the density of the population. I think the people of the city deserve to understand what the city is about to experience, and I couldn't find good predictions (preferably aimed at the NYC experience).
I base my understanding of current numbers of infected on data coming from the NY Department of Health. This page only displays the current snapshot, but the data has been compiled in the covid-19-ny github respository. I have posted a Jupyter notebook on my website in which I generate the graphs attached to this post.
Up until very recently, the confirmed cases in NYC was doubling every 1.55 days. This is the graph supporting this observation generated in the jupyter file mentioned above. For reasons indicated below, it is not clear if this pattern has broken in recent days or not.
Question: When will there be a substantial deviation from this exponential growth rate? In particular, will the pattern hold for the next 7 days (until April 2nd)?
Note that the graph decreases in slope in the last 3 or 4 days. The hopeful reason to believe this decrease is the drastic measures enforcing and encouraging social distancing in the city and state. Also of course awareness of the issue has grown in the public.
However, I fear the real reason for the decrease in slope is less pleasant. In the graph below, I show the number of newly confirmed cases each day. We observe that in recent days the number of newly confirmed cases is hovering around 3000. According to covidtracking.com the number of tests performed on Wednesday, March 25th in New York State was 12,209. Presumably about half of those tests were conducted in New York City, and the rate of positive tests statewide was about 42%.
The positive test rate is probably slightly higher in the city. This means that you can't expect the number of confirmed cases to go up much more than 50% of 6000, which is more or less in line with the number of newly confirmed cases in recent days. Thus, I would expect that the number of newly confirmed cases in the city to begin growing linearly (with slope ~3000) unless testing capacity increases dramatically, despite the fact that we would continue to see exponential growth at some rate if there was ample testing.
This belief is inline with recent news articles suggesting that hospitals are being recommended to only test patients who show symptoms of the virus if they require hospitalization. I believe the original advisories are here. In the absence of the ability to test everyone, I expect criteria for a confirmed case will need to be changed as occurred in China.
Taken together, I believe this suggests that the number of confirmed cases is no longer a meaningful statistic in NYC. I suggest the last day of meaningfulness is around March 23rd. If you believe the growth rate of doubling every 1.55 days persisted beyond that date, then given enough testing, the number of confirmed cases in NYC would be have been 29000 yesterday, March 25th, and would be about 46000 today, March 26th.
In the massively over-simplistic model of unabated exponential growth, the number of "confirmed" cases becomes rapidly horrific. By April 2nd, there would be 1 million confirmed cases of COVID-19 in NYC. I am not an epidemiologist, so I do not want to suggest that this behavior will persist to this date or even that it is currently ongoing.
Let me point out some issues.
Certainly continued exponential growth is impossible in a finite population. However NYC has 8.6 million people, so it is also not clear it is an issue in the near term I am discussing.
I don't understand the effect social distancing and the mass school and workplace closings will have. There is a lag between when someone is infected and when they are confirmed. Certainly changes to the trajectories due to societal changes will not manifest in the data for a while after the changes are made.
We don't really understand the number of people who are asymptomatic. (In Iceland it is possibly 50%.) This could cause exponential growth to slow sooner than we might otherwise expect.
I don't know of good comparisons in other places of the world. Are there similar densely populated cities in Italy that have already been through this? Is Wuhan a good model?
Final remark: I believe mathematicians are morally obligated to act in the current epidemic, and to attempt to steer policy. NYC is already largely shut down, and I regret not realizing what was going on before recently. In other parts of the world, there may still be an opportunity to have a substantial impact on policy.
That paper (and subsequent) are worth to consider: Coronavirus: Why You Must Act Now.
Tomas Pueyo: https://medium.com/@tomaspueyo/coronavirus-act-today-or-people-will-die-f4d3d9cd99ca
Washigton Post provided oversimplified but beautiful visualizations what is going on: https://www.washingtonpost.com/graphics/2020/world/corona-simulator/
State of art analysis provided by Imperial College Group of Neil Ferguson: https://www.imperial.ac.uk/mrc-global-infectious-disease-analysis/news--wuhan-coronavirus/ and previous reports
I am sympathetic to your concerns. To make this post a better fit to this forum, I recommend adding a brief preface closer to the forum goals. To me the preface would go something like: "Although I am motivated to have some different questions answered (see below), I would like to build or use a model to predict [[put your main statistic here, e.g. number of actual cases]]. Please regard this post as a reference request for models for this statistic. Other information is also welcome at [[put alternate contact on user page]]." Gerhard "Wants To Make Things Better" Paseman, 2020.03.26.
@GerhardPaseman even with your changes, I suspect this question would be inappropriate for this site, and more appropriate, for example, at the stats.SE.
The number of "confirmed" cases of infection has no signification, nowhere. It depends of how and how much testing is done, which varies from place to place and from day to day. No one here has any idea of the number of real cases. It may be twice, ten times, 100 times, or even 1000 times the number of "confirmed" cases. Can you give a convincing argument that it is impossible that there is now, in the US 80 million people infected, or more? And by the way, this would be very good news.
@Mark careful, that's how Pontius Pilate got into trouble. Matthew 27:24.
|
2025-03-21T14:48:30.150963
| 2020-03-26T20:18:09 |
355806
|
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|
Stack Exchange
|
How many persons pass your 1.5 meter neighbourhood during 1 week ? If the distribution is power law what is the exponent?
Consider a graph with vertices being people (in some region), and make an edge if one person pass another closer than say 1.5 meter during say one week.
(Such a graph might be thought a kind of useful for modelling epidemics on graphs (papers, igraph) ). It is quite a similar to much studied "social networks".
Of course, such a graph would much depend on the region and probably time period, but still we might hope for some universality and graphs for big cities and "normal" time periods, might be similar.
Question 0: Have such or similar graphs been studied ?
Question 1: What is rough estimate of an average vertex degree of such a graph ?
I.e. how many persons pass your 1.5 meter neighbourhood during 1 week ?
Any ideas are welcome.
Pursuing an analogy with other social networks, one may think that degree distribution has power form:
"that is, the fraction P(k) of nodes in the network having k connections to other nodes goes for large values of k as P(k) = C*k^l+... ". I.e. graph is scale-free network, which seems was rather unexpected feature of such graphs uncovered in 90-ies (it is not true for Erdos-Renyi model).
It somehow reflects that there are quite many vertices with extremely large number of edges, which might be true for our situation also, because imagine supermarket cashier - thousands persons pass nearby him, or public person who shake many hands.
For example, for citation graph exponent estimate is 1.7 and 2.1 for social graphs (see MO302559 ).
Question 3: If the power law is true what can be the estimate of the exponent ?
I.e. what is "l" in P(k) = C*k^l+... ?
Question 4: What can be the mathematical random graph model for such graph - Barabasi-Albert, Watts-Strogatz, etc... ?
PS
Bonus Question: If confinement-lockdown-quarantine happens, what happens with the graph above ?
The only thing I was able to google which at least somehow close the question is the following claim:
"According to a new U.K. study, you will shake 15,000 hands in your lifetime. " (link).
PSPS
Technical off-topic question: What are current technical abilities to get such information ? Does "Google/Apple" or whatever have such an information ?
Operators of mobile network clearly have information on clients positions, but the cell-size is about 500-1.5km in cities, so such an information would not be precise enough.
Some people have tracking their positions turned on, for example, my daughter's cell phone provides me information on her position, via google service "family link", so in principle such information might be partially available to Google/Apple. It is was widely discussed that Israel's parliament allowed temporary
mobile phone tracking of infected people, probably similar things has been done on China and South Korea. I vaguely remember there has been some volunteer projects where such information has been collected by volunteers, I cannot a provide a link at the moment.
It seems having such information (made anonimized) would be helpful for modeling epidemics.
PSPSPS
See also:
Suggestions for reducing the transmission rate?
Relevant mathematics to the recent coronavirus outbreak
https://stats.stackexchange.com/search?q=covid-19
• Concerning the first of the two questions in the title, "How many persons pass your 1.5 meter neighbourhood during 1 week?"
Here is a graph from Mixing patterns between age groups in social networks, showing daily average number of contacts per person in each age group. The data is based on the EpiSims social contact network, which forms the basis of most studies in the literature on this topic. A "contact" happens when two persons are in the same "room", where the size of the room was constructed such that the individuals would come close enough to transmit a disease. The 1.5 meter distance seems a reasonable proxy for this. If I multiply the number of contacts by 7 (to convert from daily to weekly average) I would estimate that 140 is a representative number in the 20-50 age range. For the elderly it is about half that number.
• Concerning the second question in the title: For the EpiSims social contact network the power law exponent of the degree distribution is 2.8.
Thank you very much for your answer ! It is unexpected for me. I might a little remark that multiplying by 7 may not always be fully correct since we meet partly same persons everyday - family members, work-colleagues, but, of course, I do not think , it will change answer much.
If you would have some time please take a look on the question: https://cstheory.stackexchange.com/questions/47957/power-law-for-degree-distribution-of-random-knn-graphs may be you know something about it or can suggest something ...
|
2025-03-21T14:48:30.151302
| 2020-03-26T20:44:19 |
355808
|
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|
Stack Exchange
|
Is a symplectic camel actually prohibited from passing through the eye of a needle?
Gromov's symplectic nonsqueezing theorem asserts that in the symplectic space ${\bf R}^{2n}$ with canonical coordinates $p_1,\dots,p_n,q_1,\dots,q_n$, and two radii $0 < r < R$, it is not possible to symplectomorphically map the ball $B(0,R)$ into the cylinder $\{ p_1^2+q_1^2 \leq r^2 \}$. It is often known as the "symplectic camel theorem", with several authors interpreting this theorem as an assertion that a symplectic camel (in the sense of a ball $B(0,R)$ that can only evolve by a continuous one-parameter family of symplectomorphisms cannot pass "through" the eye of a needle of radius $r$ less than $R$.
In the past I have accepted this interpretation uncritically, but on closer inspection it seems that the standard formulation of the nonsqueezing theorem does not actually imply a statement of this form. To make precise what an "eye of a needle" is, one needs a codimension one obstacle to serve as the "needle". I will somewhat arbitrarily choose the hyperplane $\{q_n=0\}$, minus the above-mentioned cylinder, as the needle, in which case one has a precise mathematical question.
Question. Let $n \geq 2$ and $0 < r < R$, and let $N \subset {\bf R}^{2n}$ denote the "needle" $N := \{ q_n = 0; p_1^2+q_1^2 > r^2 \}$. Does there exist a family $S(t): B(0,R) \to {\bf R}^{2n} \backslash N$, $t \in [0,1]$ of symplectomorphisms varying continuously in $t$ (say in the uniform topology), such that $S(0)$ takes values in the left half-space $\{ q_n<0\}$ and $S(1)$ takes values in the right half-space $\{q_n>0\}$?
Certainly a counterexample to Gromov's non-squeezing theorem (using a symplectomorphism that is connected to the identity) would allow one to construct a positive answer to this question, by first moving the ball far away from the needle, transforming it into a subset of the cylinder, sliding that subset through the needle and then far on the other side, then undoing the transformation. However it is not clear to me that the non-squeezing theorem in its standard formulation prevents some more exotic way to slide this "camel" through the "needle" (for instance, if it is possible to symplectically deform the ball into an "L-shape" object that resembles the union of two half-cylinders, which could then be maneuvred through the needle).
Eliashberg & Gromov sketched a proof in their paper "Convex symplectic manifolds" (Section 3.4). Written in the 4-dimensional case it says:
For $r>0$ define the subspace $X(r)\subset\mathbb{R}^4$ to be the union of the half-space $\lbrace q_2<0\rbrace$ and the half-space $\lbrace q_2>0\rbrace$ and the 3-ball $\lbrace(p_1,p_2,q_1,0)\;|\;p_1^2+p_2^2+q_1^2<r^2\rbrace$. For $R>r$ there is no 1-parameter family of symplectic embeddings $S_t:(B(0,R),\omega_\text{std})\to (X(r),\omega_\text{std})$ with the image of $S_{t\le0}$ in $\lbrace q_2<0\rbrace$ and the image of $S_{t\ge1}$ in $\lbrace q_2>0\rbrace$.
McDuff & Traynor ("The 4-dimensional symplectic camel and related results") go on to show that $X(r_1)$ and $X(r_2)$ are symplectomorphic if and only if $r_1=r_2$. Oh they also give Eliashberg--Gromov's proof of the camel theorem (Theorem 5.2), reducing it to the monotonicity lemma as in the usual Gromov nonsqueezing theorem.
Chris' answer is correct, but it should be noted that the camel problem is also resolved in Viterbo's "Symplectic topology as the geometry of generating functions," specifically in Section 5. The proof is truly different, in that there is no invocation of any pseudo-holomorphic curve techniques, so you never need to invoke the monotonicity lemma, for example.
|
2025-03-21T14:48:30.151560
| 2020-03-26T20:51:14 |
355809
|
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"url": "https://mathoverflow.net/questions/355809"
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|
Stack Exchange
|
Classes of algebras where derived equivalence preserves the global dimension
Question: Are there known classes $X$ of finite dimensional algebras in the literature that have the property that in case $A, B \in X$ are derived equivalent, they share the same global dimension?
Note that usually derived equivalences very rarely preserve the exact value of the global dimension, for example take a linear oriented hereditary algebra $kQ$ of Dynkin type $A_n$ which has global dimension one and is derived equivalent to its Koszul dual which has global dimension $n-1$.
A trivial class of examples is taking $X$ the class of semisimple algebras.
A non-trivial recent class is the class of algebras having a simple-preserving anti-duality, see theorem 5.10 in https://arxiv.org/pdf/1607.03513.pdf (and this is the only non-trivial class I am aware of in the literature).
|
2025-03-21T14:48:30.151647
| 2020-03-26T21:11:03 |
355810
|
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|
Stack Exchange
|
Smooth affine curve with no immersion in projective plane
(1) I am trying to find an example of a smooth affine curve $C$ over $k$ with no immersion $C \to \mathbb{P}^2_k$ (for me a curve is an integral separated dimension one scheme of finite type over $k$).
Here I will discuss my attempt so far.
Since the immersion factors, $C \to U \to \mathbb{P}^2_k$ as a closed immersion followed by an open immersion I first tried considering special cases of the open $U$. If we restrict to principal affine opens $U = D(q) \subset \mathbb{A}^2_k \subset \mathbb{P}^2_k$ for some $q \in k[x,y]$ then I can find an example. What I mean is I can exhibit a curve that has no closed immersion into any $D(q) \subset \mathbb{A}^2_k$. The idea is to show that a smooth curve embedded in $D(q)$ must have a trivial canonical bundle and then produce an example of an affine curve with a nontrivial canonical bundle which I have done. This is an easy computation using that $C \cong \mathrm{Spec}{(k[x,y,q^{-1}]/(f))}$ and $(f, \partial_x f, \partial_y f)$ is the unit ideal since $C$ is smooth. Then one can explicitly compute that
$$ \Omega_{C/k} = R \mathrm{d}{x} \oplus R \mathrm{d}{y} / (\partial_x f \mathrm{d}{x} + \partial_y f \mathrm{d}{y}) \cong R $$
where $R = k[x,y,q^{-1}]/(f)$.
This leads me to my second question:
(2) Is the fact that $\Omega_{C/k} = \mathcal{O}_{C}$ for affine plane curves $C \subset D(q)$ a consequence of a more general fact or is this computation the ``correct'' way of seeing why this is true?
One possible example is to take a degree 4 curve in $\mathbb{P}^2_k$ such that the tangent line at a point $P$ intersects the curve in at least one other point. Then the canonical bundle is the pullback of $\mathcal{O}_{\mathbb{P}^2_k}(1)$ but we know that $\{ P \}$ is not a hyperplane section by construction so $K_C \not\sim \ell \cdot [P]$ and thus removing the point $P$ gives an affine curve with a nontrivial canonical bundle. Notice that this example is immersed in $\mathbb{P}^2_k$ since it is an open subset of a complete smooth curve in $\mathbb{P}^2_k$, therefore, the vanishing of the canonical bundle cannot be an obstruction to such an immersion.
To go beyond this, I tried to find an example $C$ with no immersion $C \to \mathbb{A}^2_k$ which may be easier. We can take the closure $C \to \overline{C} \to \mathbb{A}^2_k$ and since $C$ is smooth by assumption $C \subset \overline{C}_{\text{smooth}}$. Furthermore, $\overline{C}_{\text{smooth}} = \overline{C} \cap (D(\partial_x f) \cup D(\partial_y f))$ where $\overline{C} = V(f)$ for some $f \in k[x,y]$. Therefore, I have affine opens $C \cap D(\partial_x f)$ and $C \cap D(\partial_y f)$ on which the canonical bundle must vanish (these are affine since $\overline{C} \to \mathbb{A}^2_k$ is affine (closed immersion) and $C \to \overline{C}$ is affine since $C$ is affine and $\overline{C}$ is separated). But I am not sure how to proceed from here since it seems hard / impossible to come up with a curve which has no size 2 affine open cover trivializing its canonical bundle.
I would be most interested in an example with $k = \mathbb{C}$ but in the unlikely case that a nice example requires some not algebraically closed field that would be very interesting.
Finally, I was wondering about the converse of the partial result I had found:
(3) Is $\Omega_{C/k} = \mathcal{O}_C$ the only obstruction to having a closed immersion $C \to D(q) \subset \mathbb{A}^2_k$ for some $q \in k[x,y]$? That is, for such a curve does there always exist such a $q \in k[x,y]$ and a closed immersion $C \to D(q) \subset \mathbb{A}^2_k$.
Many thanks!
Can you say why $X \to \mathbb{P}^2$ has to be a closed immersion (in particular why an immersion since closed is obvious)? E.g. if you took a curve $C$ in $\mathbb{P}^2$ with one cusp $P$ and removed the cusp point then $C \setminus P$ should admit a smooth projective model $X$ (the normalization of $C$) but the image of $X$ must contain $C$ since it is closed but that is not smooth. What has gone wrong with my reasoning? Thanks for the answer.
For example, I thought that genus 2 curves all have a hyperelliptic affine model $y^2 = f(x)$ giving an immersion of this affine part in $\mathbb{A}^2$ but we know that genus 2 complete curves are not plane curves so their smooth projective model can't embed in $\mathbb{P}^2$.
No worry. I appreciate your response.
This may be relevant: https://mathoverflow.net/questions/9751/is-every-smooth-affine-curve-isomorphic-to-a-smooth-affine-plane-curve
|
2025-03-21T14:48:30.152050
| 2020-03-26T21:16:05 |
355811
|
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|
Stack Exchange
|
Law of large numbers for Harris recurrent Markov chains
I'm trying to familiarize myself with the details of the proof that the Markov chains produce by the Metropolis-Hastings algorithm have a law of large numbers. I've found a half dozen or more references to the proof in Meyn, Tweedie and Glynn's book. I believe that I follow most of it, but I'm not sure on some of the step in one part of the proof where they prove that a certain function is harmonic.
I realize that this may not be research level, but I have had problems finding alternative treatments because everyone cites this!
Here is the relevant section:
Here is my "understanding" of the steps working from the bottom:
To go from 4 to 3 involves a re-expression of $\lim_{n \rightarrow \infty} \sum_{k = 1}^n g(\Phi_k)$.
To go from 3 to 2, one applies the "smoothing property of conditional expectation". In the process the $g(\Phi_1)/n$ term vanishes because the numerator becomes a constant.
To go from 2 to 1 follows from the Markov property. Starting from $x$ and conditioning on $\Phi_1$ is like starting at $\Phi_1$. This is reflected in the conditioning on $\mathcal{F}^{\Phi}_1$ being dropped and $\mathrm{P}_x$ becoming $\mathrm{P}_{\Phi_1}$.
Where I am getting lost is in the indexing changes between steps. I keep thinking perhaps the index in the subscript in step 3 should maybe be $k$ rather than $k+1$ or that the minus sign out to be a plus sign.
If anyone could expand upon or annotated the steps in this derivation I'd be very grateful. Apologies if this is trivial (I sort of hope it is!)
The argument given in the book is actually quite misleading. It is much easier just to notice that the event $\{S_n/n\to \langle g,\pi\rangle \}$ from the definition of the function $g_\infty$ is (time shift) invariant, and therefore this function is harmonic by Theorem 17.1.3.
Oh my god, thank you! I kept thinking exactly what you said, and didn't quite understand what 17.1.6 was adding, but that's the one that everyone cites, so I thought maybe I was missing some technical detail, but then I had trouble tracking their indicies...I thought I was going insane.
|
2025-03-21T14:48:30.152245
| 2020-03-26T22:01:28 |
355818
|
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|
Stack Exchange
|
The last step in Ahlfors' and Sario's proof of the triangulability of surfaces
In their book, Riemann Surfaces, Ahlfors and Sario write, at the bottom of pg. 109 to the top of pg. 110,
"Consider the sequences $\{V_n\}$ and $\{W_n\}$ introduced by Lemma 46B. We will show that there exist closed Jordan regions $J_n$,
such that $V_n \subset J_n \subset W_n$, whose boundaries $\gamma_n$ have only a finite number of common points.
They will then form a covering of finite character."
No proof is given for the last sentence. How does one prove it?
The precise content of Lemma 46B and the definition of "covering of finite character" will be given below,
for convenience here. The essence of the question is then extracted.
Establishing the property of the cited last sentence constitutes the last step in the proof of triangulability of surfaces.
The triangulation of a surface is constructed directly from $\{J_n\}$, based on the property in the cited last sentence. See the Remarks below, for further comments on the importance of the property in the cited last sentence.
Regarding the property (in the second cited sentence), "whose boundaries $\gamma_n$ have only a finite number of common points",
the authors mean that for all $m$, $n$: $\gamma_n \cap \gamma_m$ consists of at most a finite number of points.
Here are the given properties of the open Jordan regions, $V_n$, $W_n$, and the desired and established properties of the closed Jordan regions, $J_n$,
all in the connected surface (2-dim second countable manifold), $F$; the enumeration of the properties follows that of the authors.
(An open Jordan region in $F$ is a subset of $F$ whose closure is homeomorphic to a closed disk in the Euclidean plane, in such a manner that the open region corresponds to the open disk. A closed Jordan region is the closure of an open Jordan region.)
Lemma 46B: There exist sequences $V_n$, $W_n$ ($n = 1, 2, ...$) of open Jordan regions in $F$ satisfying:
(B1) $\overline{V}_n \subset W_n$.
(B2) $\bigcup_n V_n = F$.
(B3) No point of $F$ belongs to infinitely many $\overline{W}_n$.
Definition of 'covering of finite character': Closed Jordan regions $J_n$ ($n = 1, 2, ...$) in $F$, form a covering of finite character if
(A0) $\bigcup_n $Int$(J_n) = F$, where Int$(J_n)$ denotes the interior of $J_n$.
(A1) Each $J_n$ meets at most a finite number of others.
(A2) For all $m$, $n$: $\gamma_n \cap \gamma_m$ consists of at most a finite number of points or arcs (possibly both), where $\gamma_n := \partial J_n$.
The constructed regions ${J_n}$: The closed Jordan regions $J_n$, $n = 1, 2, ...$, in $F$ are constructed recursively in a manner such that for all $n$,
$V_n \subset J_n \subset W_n$ and their boundaries $\gamma_n := \partial J_n$ satisfy:
For all $n$, $\gamma_n$ meets $\gamma_{n-1} \cap \cdots \cap \gamma_1$ in at most a finite number of points.
The essential question: Does (A1) follow from (A0), (B3), and the $\{\gamma_n\}$'s property,
for all $m$, $n$: $\gamma_n \cap \gamma_m$ consists of at most a finite number of points?
Remarks: If the sequences $\{V_n\}$ and $\{W_n\}$ are of finite length, then the whole business is trivial;
so wlog the sequences are of infinite length. The result (A0) follows immediately from (B2) since Int$(J_n) \supset V_n$, for all $n$.
The property (A2) is satisfied by construction. Thus only the verification of (A1) remains. From the purpose of triangulation, it would suffice to show that a subsequence of $\{J_n\}$ satisfies (A1). Thus if $F$ is compact, then we are done. The strength of the approach to triangulation that is adopted by the authors, is that it also handles noncompact surfaces (as well as surfaces with or without boundary).
Once $\{J_n\}$ have been constructed, the sequence $\{V_n\}$
has no further role to play, since (A0) can play the role of (B2). The same might be said of the sequence $\{W_n\}$ since it follows from (B3)
that no point of $F$ belongs to infinitely many $J_n$. This leads to the following modified question.
Modified question: Does (A1) follow from (A0) if, in addition, no point of $F$ belongs to infinitely many $J_n$?
I received the following observation by email; it constitutes an answer to my question.
The last paragraph of the proof of Lemma 46B shows that the following result, which is stronger than (B3), holds: Each $\overline{W}_n$ meets at most a finite number of the other $\overline{W}_m$'s. Therefore, in particular, the $J_n$'s satisfy (A1) because, by construction, each $J_n \subset W_n$.
Whether or not the weaker condition, (B3), is sufficient for (A1) to hold (as in ``The Essential Question''), remains an open question; but an answer to the latter question is not required for the proof of triangulability of surfaces.
|
2025-03-21T14:48:30.152576
| 2020-03-26T23:08:35 |
355820
|
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|
Stack Exchange
|
Sum of $\sum_{1\leq k\leq k'\leq n}\frac{k^{\alpha}}{k'^{\alpha}}$
I want to calculute or estimate of order $O(n^{2-\varepsilon})$, where $\varepsilon>0$, of the following sum for $0<\alpha<1$
$$\sum_{1\leq k\leq k'\leq n}\frac{k^{\alpha}}{k'^{\alpha}}.$$
Bound it by an integral.
Consider all pairs $(k,k')$ with $\frac{n}{4}\leq k\leq\frac{n}{2}\leq k'\leq\frac{3n}{4}$. There are $\left(\frac{n}{4}\right)^2$ such pairs and for them $\frac{k}{k'}\geq\frac{1}{3}$. So the sum is bounded from below by
$\left(\frac{n}{4}\right)^2\left(\frac{1}{3}\right)^\alpha$,
and thus as $n\to\infty$ is not $O(n^{2-\epsilon})$ for any $\epsilon>0$.
What about $O(\frac{n^2}{\log(n)})$.
the factor $3^{-\alpha}$ is $O(1)$
|
2025-03-21T14:48:30.152659
| 2020-03-27T00:13:32 |
355826
|
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|
Stack Exchange
|
A metric $w$ on a Kahler manifold is extremal if and only if the gradient vector field of the scalar curvature is holomorphic
I am trying to understand the calculation in An introduction to Extremal kahler metrics. On the fourth line of page 55 the author calculated that $\int_{M} - 2 S R^{\bar k j} \partial_{j} \partial_{\bar k} \varphi = \int_{M} \varphi \nabla_{j} \nabla_{\bar k}( -2 S R^{\bar k j})$, why is this true?
My thinking is：Consider $-2S R^{\bar k j}\partial_{\bar k} \varphi$, which is a section of $T^{1,0} M$, then we look at the quantity $\nabla_{j} -2 S R^{\bar k j} \partial_{\bar k} \varphi = \nabla_{j} -2S R^{\bar k j} \partial_{\bar k} \varphi - 2S R^{\bar k j}\partial_{j} \partial_{\bar k} \varphi$. This line of reasoning will result in different order of covariant derivatives than then one appearing in the reference. I do not really see how the equality should follow.
My apology if this is too elementary for mathoverflow, but I have been stuck for a while and could not find any thing related to this either online or in math stack exchange.
still looking for an answer...
It would be helpful to know exactly where in that document the calculation is done. Without that it's hard to say exactly. However, it looks like it's using integration by parts twice.
oh I am sorry, and it is in P55. I totally forgot to put it in the description.
wouldn't integration by parts result in $\nabla_{\bar k}\nabla_{j}R^{\bar k j}S$? @GabeK
Initially that seems like it would be the case, but I suspect the order of differentiation doesn't change the answer here. In particular, curvature only appears when you take mixed covariant derivatives of vector fields, so we should expect that $\int_{M} - 2 S R^{\bar k j} \partial_{j} \partial_{\bar k} \varphi = \int_{M} - 2 S R^{\bar k j} \partial_{\bar k} \partial_{j} \varphi $, at which point integration by parts twice gets what you want. I'm not 100% sure of that though, so that should definitely be checked.
That actually makes sense, thank you. One could first change the differentiation order for $\partial_{\bar k} \partial_{j} \varphi$ as you suggested, then it works through. But it is strange that if we do not do this, and get $\nabla_{\bar k} \nabla_{j} R^{\bar k j} S$, then we switch the order of the covariant derivative, it will certainly produce some extra term.
That was confusing to me as well, which is why I wasn't 100% sure. Somehow the extra terms will need to cancel and I didn't see why that is the case.
"it will certainly produce some extra term": in fact, if you just write down the error term that you get when commuting these 2 covariant derivatives, you will get precisely zero
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2025-03-21T14:48:30.152861
| 2020-03-27T02:03:40 |
355828
|
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|
Stack Exchange
|
Embedding 2-complexes null homotopically into 2-complexes
Whitehead's conjecture states that if $L$ is an aspherical 2-complex and $K$ is a subcomplex of $L$, then $K$ is also aspherical. It is known by work of Howie and Luft that if the Whitehead conjecture is false, then there is a counterexample where $L$ is constructed as the limit of a series of nested spaces $K= K_0 \subset K_1 \subset K_2 \subset \cdots$ where the inclusion of $K_i$ in $K_{i+1}$ is nullhomotopic and each $K_i$ is not aspherical (or should I just say spherical?).
I do not think about 2-complexes very often so I apologize if this is too basic, but I can't picture any examples of such inclusions of 2-complexes $A \subset B$ where the inclusion is null homotopic and $\pi_2(A) \neq 0$. What are some examples of this? Ideally, I would like for $A$ and $B$ to both be finite. Any example will probably have $\pi_2(B) \neq 0$, since otherwise you solve Whitehead.
Given a 2-complex $A$, can we always find another 2-complex $B$ containing $A$ as a subcomplex so that $A$ is nullhomotopic in $B$? If you could, then this would yield a solution to the Whitehead conjecture, since you could just do that over and over again, and the resulting limit space would be aspherical by compactness. So there are presumably obstructions, does anyone know some?
Edit: Whoops - In light of Mike's observation I should assume that $H_2(A) = 0$. For the setup of the Whitehead conjecture, we have $H_2(K) = 0$, so I definitely should have included that initially - sorry.
For the final question: if $A \subset B$ is an inclusion of 2-complexes, then $H_2(A)$ injects into $H_2(B)$ (as $H_3(A,B) = 0$ for dimensions reasons). So for instance $S^2$ can't be null as a subcomplex of any 2-complex.
I'm not certain that this question is any easier than the Whitehead conjecture itself!
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2025-03-21T14:48:30.153033
| 2020-03-27T02:28:36 |
355829
|
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|
Stack Exchange
|
Equal products of consecutive integers
Summary: What are the non-trivial solutions to the question: Find two sequences of consecutive integers whose products are the same. There are four known solutions, all of thich consist of small integers, and there is no clear pattern connecting them. Given the small number of known solutions and the lack of any clear pattern among them, I suspect this question contains more number theory than at first appears.
The "Puzzle Corner" of MIT News for March/April 2020 gives a "speed" problem by Sorab Vatcha: "Find seven consecutive integers whose product equals the product of four consecutive integers." The obvious "speed" solution is: 0, 1, 2, 3, 4, 5, and 6; and 0, 1, 2, and 3.
This leads to the general question of whether there are any "nontrivial" solutions to find two distinct sequences of integers whose products are the same. The obvious trivial solutions involve (1) sequences containing 0, (2) replacing all of the integers in a sequence with their negatives, (3) adding or deleting values 1 or -1, and (4) one sequence has length 1 (and hence contains one integer that is the product of the integers in the other sequence). These criteria reduce the problem to finding two distinct sequences of integers $\ge 2$ of length $\ge 2$ whose products are the same.
There is a less obvious criterion of non-triviality: (5) The two sequences must not overlap. If the sequences overlap, then the overlapping part can be deleted from both of them, yielding a soltion with shorter sequences. This interacts with the prohibition (4) against sequences of length 1: Removing the overlapped part may reduce one sequence to length 1, and the shorter solution may be trivial also. And indeed, there is a large family of solutions constructed this way: If the product of $a \cdots b$ is $P$, then $a \cdots (P-1) = (b+1) \cdots P$.
There are non-trivial solutions. The one with the smallest product is $5 \cdot 6 \cdot 7 = 14 \cdot 15 = 210$. Is there an enumeration of all solutions?
The known non-trivial solutions are:
$5 \cdots 7 = 14 \cdot 15 = 210$,
$2 \cdots 6 = 8 \cdots 10 = 720$,
$19 \cdots 22 = 55 \cdots 57 = 175560$, and
$8 \cdots 14 = 63 \cdots 66 = 17297280$.
I have run a number of computer searches that have not revealed any further solutions: (a) all sequences with products less than $10^{30}$, (b) sequences with numbers less than 10,000,000 and length less than 10, (c) sequences with numbers less than 1,000,000 and length less than 100, (d) sequences with numbers less than 100,000 and length less than 1,000, and (e) sequences with numbers less than 10,000 and length less than 10,000.
See also https://math.stackexchange.com/questions/991728/equal-products-of-consecutive-integers/ and https://math.stackexchange.com/questions/3346618/non-trivial-solutions-to-equal-products-of-consecutive-integers.
It appears no further solutions are known: http://oeis.org/A163263
Note the literature references in the above OEIS entry.
A connection linking those numbers in pairs is that $210$ divides $175560$ and $720$ divides $17297280$.
Suppose $m(m+1)\times\dotsb\times(m+j)=n(n+1)\times\dotsb\times(n+k)$, $m<n$. Multiplying by $(n-1)!(m-1)!$ we get $(m+j)!(n-1)!=(n+k)!(m-1)!$ So the problem is equivalent to finding solutions to $a!b!=c!d!$.
The observation is made in many places (see the linked questions and OEIS sequence) that the sequence with larger numbers cannot contain a prime. It should also be noted that sufficiently large solutions cannot contain both three times a prime and two times a different prime. This can be extended using information on maximal gaps to limit how many large primes can appear as factors in a large sequence. An interesting variant is when a product of n consecutive positive integers divides another sequence of n consecutive positive integers. Except for n factorial dividing every such, or a sequence dividing itself, I am not aware of examples beyond the ones listed above and trivial modifications. Listing such pairs of n tuples would go far toward solving this problem.
Gerhard "Unless Erdos Already Did It" Paseman, 2020.03.26.
"Unless Erdos Already Did It" - you made my day!
At least the first two numbers are related to remarkable approximations to $\pi$.
$210\pi \approx 660$, which simplifies to $$\pi \approx \frac{22}{7}$$ (Archimedes)
$720\pi \approx 2262$, or $$\pi \approx \frac{377}{120}$$ (Ptolemy)
$\frac{377}{120}$ is not a convergent but a semiconvergent, actually closer to $\pi$ than the more usual $\frac{333}{106}$.
Both approximations are given by integrals.
$$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx = \frac{22}{7}-\pi$$ (Dalzell)
$$\frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx = \frac{377}{120}-\pi$$
The relationship between $\pi$ approximations and products of consecutive integers is shown in series related to these integrals.
The next two integrals following this pattern are
$$\frac{1}{4}\int_0^1 \frac{x^6(1-x)^8}{1+x^2}dx = \frac{566053}{180180}-\pi$$
and
$$\frac{1}{8}\int_0^1 \frac{x^7(1-x)^{10}}{1+x^2}dx = \frac{18113671}{5765760}-\pi$$
The ratios of these denominators to the third and fourth numbers are notably simple as well.
$$\frac{180180}{175560} =\frac{39}{38}$$
$$\frac{5765760}{17297280} =\frac{1}{3}$$
On an unrelated note, the third number is the order of the sporadic simple group $J_1$ :)
This is because numbers with lots of small factors tend to coincide. The number of seconds in a week is equal to the order of $J_2$, for example.
19202122 is also a good denominator for a rational approximation to $\pi$, but not 63646566.
@DaveBenson Sure, I was not trying to make a serious point here.
The order of $J_2$ does not look like totally unrelated to these numbers. $604800$ is the product of two small factorials, $5!7!$. $720$ is also $3!5!$, besides $6!$. Sequence https://oeis.org/A175430 lists both.
@JoachimKönig No, of course, I knew that.
|
2025-03-21T14:48:30.153414
| 2020-03-27T03:59:56 |
355832
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|
Stack Exchange
|
Descent properties for rational topological cyclic homology
Descent properties can be extremely useful for studying $\operatorname{TC}$ (topological cyclic homology), since it is a sheaf in many well behaved topologies.
I was wondering what is known about $\operatorname{TC}_\mathbb{Q}$ in this regard.
It seems clear to me that any site which has $\operatorname{TC}$ as a sheaf, and is also given in terms of a cd-structure will have $\operatorname{TC}_\mathbb{Q}$ as a sheaf, but I am particularly interested in the étale topology.
What do you call $TC$?
@OlivierBenoist topological cyclic homology
|
2025-03-21T14:48:30.153481
| 2020-03-27T08:05:49 |
355837
|
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|
Stack Exchange
|
Can consistent proxi-finite extensions of ZFC be additive?
The definition of "strongly coming from $HF$" is given in this posting.
Question: if $\varphi, \psi$ each is strongly coming from $HF$, then can the following be true?
$Con (ZFC + \varphi) \land Con (ZFC + \psi) \to Con (ZFC + \varphi + \psi)$
Let $\phi$ state "all recursive theories extending ZFC are inconsistent" and let $\psi$ be $\neg\phi$. Then I think both fit the definition and are consistent with ZFC, assuming Con(ZFC).
@MonroeEskew, I don't see how $\phi$ qualify as strongly coming from $HF$?
Oh I thought you removed quantifiers are replaced them with existentials. So you're essentially asking if ZFC is consistent with all sentences that are true in HF. Well assuming there is a transitive model of ZFC, then yes.
@MonroeEskew, my definition is more precise than just all sentences that are true in HF, its known that ZFC is inconsistent with all of those, I'm asking if ZFC is consistent with all sentences that are STRONGLY COMING form HF. There is a difference. In the first case take $\forall x (x \text{ is infinite} \to \phi)$ this is trivially true in $HF$ and the same sentence but with $\neg \phi$ in the consequent is also true in $HF$, but we cannot add both to ZFC. I hoped for the strongly coming feature to overcome this?
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2025-03-21T14:48:30.153612
| 2020-03-27T08:55:25 |
355840
|
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|
Stack Exchange
|
Restriction of product of automorphic forms
Let $W \subset V$ be quadratic spaces over a number field $F$.
Let $G_n=SO(V)$ and $G_m=SO(W)$ and we consider $G_m$ as a subgoup of $G_n$ via a diagonal embedding.
Let $f$ be an automorphic form of $G_n$ and $g$ an automorphic form of $G_m$.
I am wondering whether the function $h$ on $G_m$ defined by $h(k)=f(k)g(k)$ is automorphic form on $G_m$.
Except for $\mathfrak{z}$-finiteness, I verified that other properties of autumorphic forms does hold. But I am doubtful $\mathfrak{z}$-finiteness hold.
Do you have any idea on this?
Thank you very much!
This is a special case of restricting automorphic forms on a larger group $G$ to a smaller (sub)group $H$, of course. So far as I know, orthogonal groups are not special in this regard, although, yes, there are some obvious natural maps among them.
Certainly if the map $H\to G$ is a $k$-morphism (with groundfield $k$, or, being more careful, over some localization of the ring of integers of $k$), then the restrictions of left $G_k$-invariant functions are left $H_k$-invariant. This is the immediate part.
Under very mild hypotheses and/or normalizations, "right $K$-finiteness" is preserved, as is "moderate growth".
But $\mathfrak z$-finiteness (where $\mathfrak z$ is the center of the universal enveloping algebra) is very rarely preserved. Likewise, and in parallel, if we require automorphic forms to generate irreducibles under right translation, this property will very rarely be preserved under restriction.
The rarity of this is already visible on $\mathbb R$ with the Laplacian: very rarely is the product of two $\Delta$-eigenfunctions $\Delta$-finite...
In the automorphic context, indeed, computing decomposition coefficients of such a restriction (or product) occasionally produces very interesting Euler products. Rankin-Selberg and Langlands-Shahidi et al are instances of this.
EDIT: still, yes, it is true in some rather special situations (see work of Kudla and Rallis on "first occurrence", for example, and "Howe conjectures"), that for suitable $H_1\times H_2\subset G$, for rather degenerate (e.g., "minimal") (automorphic and other) repns on $G$, restriction to $H_1\times H_2$ and projection to certain $\pi_1$-components on $H_1$ produces an irreducible on $H_2$. (Also see Segal-Shale-Weil...)
This is not quite what is happening in the literal "map to Fourier-Jacobi coefficients" story. There we have a two-step-nilpotent abelian radical, and the map is "integrate along the center of that unipotent radical". This is a $G$-hom, so preserves $\mathfrak z_G$ eigenvalues. But it does not immediately promise eigenfunction properties for the Levi component. Yes, holomorphy is preserved, for example. Is this more in the direction of what you're wanting?
Maps to Fourier-Whittaker or Fourier-Jacobi or Fourier-Bessel or whatever components are $G$-homs, so preserve all $G$-related properties. Averaging on the left, etc., is a $G$-hom for the right action of $G$. This is mostly disjoint from the restriction-to-subgroup maps, which are not-so-often given by integration along some sort of "complementary subgroup".
|
2025-03-21T14:48:30.153838
| 2020-03-27T09:17:47 |
355842
|
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|
Stack Exchange
|
A "singular" Tischler theorem
The Tischler theorem says that a compact manifold $M$ admitting a closed nowhere vanishing $1$-form $\alpha$ fibers over the circle. I was wondering if anything could be said about the case where $\alpha$ vanishes on some small (codimension-$1$? $0$-dimensional?) set, $sing(\alpha)$. I guess to hope for a fiber-bundle structure on $M/sing(\alpha)$ is too much, but maybe there are some extra conditions one could impose to get such a result?
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2025-03-21T14:48:30.154014
| 2020-03-27T09:46:05 |
355844
|
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|
Stack Exchange
|
measures on groups without assuming a locally compact group topology
I'm interested in knowing whether there exists any kind of theory for measures on groups without assuming that it's the Haar measure for a locally compact group topology.
Translastion-invariant, locally finite measure on a topological group.
As I recall, there is something like this in Hewitt & Ross Vol. I
Hewitt, E.; Ross, K. A., Abstract harmonic analysis. Vol. I: Structure of topological groups. Integration theory. Group representations, Die Grundlehren der mathematischen Wissenschaften. 115. Berlin-Göttingen-Heidelberg: Springer-Verlag. VIII, 519 p. (1963). ZBL0115.10603.
What the exact conditions were, I do not remember. But the conclusion was that the measure is essentially Haar measure on a locally compact group which is in sense a "completion" of the given group.
What is the question you're answering? The question as asked now is extremely broad (and vague), and hence since your answer is accepted, it might have an interpretation I didn't guess.
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2025-03-21T14:48:30.154115
| 2020-03-27T09:54:38 |
355845
|
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|
Stack Exchange
|
simple Lie groups over C
For an affine algebraic group over $\mathbb{C}$ which is simple, in the sense of no normal subgroups closed in the Zariski topology except for finite central subgroups and the whole thing, I'm trying to refresh my memory of how to show that in that case your Lie algebra over C is going to be simple in the sense of no ideals except for zero and the whole thing.
We have a correspondence between subalgebras and subgroups, and any subalgebra of your Lie algebra will give rise to a Lie group and closed immersion of that (relative to the strong topology) into your original Lie group, but I would presume that there will be no guarantee the range will be Zariski closed. So I'm just trying to remind myself how to fill in the rest of the argument. Is this result stated in my first paragraph actually correct for $\mathbb{C}$?
My main reason for asking this is that I'm wondering whether it can be generalised to non-archimedean local fields, but I thought it might be convenient to start by thinking about the case of the complex numbers.
For a Lie group, "Zariski closed" makes no sense in general. It turns out to make sense in a unique way in a simple complex Lie group, but this takes some energy to prove. Maybe you just want to start with $G(\mathbf{C})$ for some complex algebraic group.
Thank you for reminding me of that point, so I made a few modifications to my question which I hope now make it a more sensible question.
The sketch would be: if $\mathfrak{g}$ has a nontrivial abelian ideal, then $G(\mathbf{C})$ has a nontrivial abelian normal subgroup (taking the exponential), hence its Zariski closure yields a contradiction. So $\mathfrak{g}$ is semisimple (and nontrivial: I assume the group has positive dimension since otherwise the claim is false). If $\mathfrak{g}$ were not simple, it would have a simple factor: then the centralizer of this factor is not trivial, and hence the centralizer of this factor in the group is a nontrivial proper Zariski-closed subgroup.
Thanks, seems as though that would generalise easily enough to non-archimedean local fields of characteristic zero?
Yes: if $G$ is a (finite-dimensional) Lie group over a complete normed field of characteristic zero (a) if its Lie algebra is not semisimple, then $G$ has a nontrivial abelian normal subgroup (b) if its Lie algebra is semisimple and not simple, then $G$ has a positive-dimensional proper normal subgroup, obtained as centralizer of some simple factor.
There are some very good books of Tony Springer, of J. Humphreys, and of A. Borel, all with the title Linear Algebraic Groups, which cover this material in detail.
|
2025-03-21T14:48:30.154648
| 2020-03-27T10:00:37 |
355847
|
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|
Stack Exchange
|
Mapping class group of torus, why is $(ST)^3=S^2$?
In the context of topological quantum field theories, I am interested in the mapping class group of a torus. Here I can consider the torus as a square with identified edges and also decorated with directed curves (see below).
The mapping class group is $\mathsf{SL}(2,\Bbb Z)$ and is generated by $S$ and $T$, which correspond to a 90deg rotation and a Dehn twist respectively. I take the Dehn twist to be along the vertical (meridian). These generators should satisfy $S^2=C$, and $(ST)^3=S^2$, where $C$ is conjugation corresponding to flipping both directions of the torus. The first is easy to see
However, I cannot get the second property, here is my working
Note in this second case I am not mapping back to the original square using the periodicity to make it clear what transformations I am doing.
$(ST)^3=\Bbb 1$ is true for the modular group $\mathsf{PSL}(2,\Bbb Z)$ but then we should also have $S^2=\Bbb 1$.
Can anyone point out where I am going wrong?
Flip the direction of rotation for $S$, or choose the other meridian for $T$.
We can see this at the level of matrices. Define
$$S_1 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \qquad S_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ to be the two choices of our rotation matrix. Similarly define
$$T_1 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \qquad T_2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$ to be the two choices for our Dehn twist. (I suppose there really are four choices - the off-diagonal entries could just as well be negative, depending on the orientation of the meridian you're Dehn twisting around - but I'll ignore that. It's not too hard to deal with that possibility.)
Then you can do the algebra to check that $S_1^2 = S_2^2 = -\mathrm{Id} = C$, and that $(S_iT_j)^3 = -\mathrm{Id}$ if $i = j$ and $+\mathrm{Id}$ otherwise. This tells you that you need to swap one of $S$ or $T$ as described above.
Thanks a lot! So simple. I was just trying use a right-handed convention for my rotations and my Dehn twists. I didn’t realise but you’re right the S is typically done counter-clockwise. Do you think there is much to be learned from this difference?
|
2025-03-21T14:48:30.154837
| 2020-03-27T11:25:22 |
355851
|
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|
Stack Exchange
|
The infinite graph
How to define the infinite (countable) graph which is constructed as follows?
The picture depicts a finite graph.
Where is the infinity!?
If I understand your picture correctly, let $V$ be the set of all dyadic rationals in $[0,1]$, and let $(p,q)$ be an edge if and only if $p = \frac{k}{2^n}$, where $k$ is an odd number (or $0$), and $q = p + \frac{1}{2^m}$ for some $m \geq n$. Is this what you have in mind?
@WillBrian Yes, that's it. Thanks!
This easy question could be community. I had 1: a nicer construction, and 2: an axiomatization.
To do this formally, first formulate a definition by induction, and then take the union.
The basis step is a graph $G_0$ with two vertices and one edge.
For the inductive step, assuming $G_i$ is defined, embed $G_i$ into a graph $G_{i+1}$ as follows: for any edge $a$---$b$ of $G_i$ that is not an edge of $G_{i-1}$, attach a new vertex $v_{a,b}$ and edges $a$---$v_{a,b}$, $v_{a,b}$---$b$.
Now define the final graph to be the union of the nested sequence of graphs $G_0 \subset G_1 \subset G_2 \subset \cdots$.
As said in the comments, this construction can also be encoded with a tiny bit of number theory.
In the construction of $G_{i+1}$, you probably want to add a new $v_{a,b}$ only for those edges $a-b$ of $G_i$ that were not already in $G_{i-1}$. Otherwise, a single edge $a-b$ will result in "new" vertices and edges at all later stages.
This is a right remark. Thanks!
Thanks for the correction @AndreasBlass
|
2025-03-21T14:48:30.154979
| 2020-03-27T12:03:34 |
355854
|
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|
Stack Exchange
|
Is Steiner symmetrization "Turing complete"?
This question stems from intuition so it is a little soft. It concerns performing computation using transformations on sets. The idea is that a rearrangement like Steiner symmetrization might be "computationally strong enough" to be used to execute arbitrary programs.
First we probably should pick an appropriate set $\mathcal{B} \subset \mathcal{P}(\mathbb{R}^d)$, maybe the Lebesgue or Borel measurable sets.
Then consider a set of transformations $(T_i)_{i \in I}$ where $T_i: \mathcal{B} \longrightarrow \mathcal{B}$ for every $i \in I$.
We also need a program function $f: \mathcal{B} \longrightarrow I$.
Lastly we need some condition that decides when the program finishes running, formally $E: \mathcal{B} \longrightarrow \{\mathit{True}, \mathit{False}\}$.
Then computation is done as follows:
Some $M_0 \in \mathcal{B}$ is the input
$M_{n+1} = T_{f(M_n)}(M_n)$
If $E(M_n) = \mathit{True}$ then the program stops running and the output is $M_n$
1. Question: How are such systems of computation formalized? (i.e. does there exist a "name" for this)
2. Question: Can such systems be "Turing complete" in some sense? How would one show this?
I will also give the specific example that made me wonder: $\mathcal{B}$ are the Lebesgue measurable sets and $(T_i)_{i \in I}$ are the Steiner symmetrizations in all directions.
Then $f$ is something I don't know and $E$ could be a condition like the set being convex.
I am sorry for this question being a little vague and would appreciate if someone could improve it.
|
2025-03-21T14:48:30.155109
| 2020-03-27T12:17:02 |
355857
|
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|
Stack Exchange
|
Free groups are CT-groups
A group $G$ is called CT-group if being commutative elements is transitive relation on $G\setminus\{1\}$ i.e. if $ 1 \neq x,y,z\in G $ and $[x,y]=1, [y,z]=1 $ then $[x,z]=1$.
I encountered the fact that free groups are CT-group, but every proof to this is very long and work for much general case (I know its true also for hyperbolic torsion free groups, which is nice but overkill for me).
Any one knows a self contained proof for this? The papers I saw uses arguments from algebraic topology, which are way over my knowledge, I only need the case of free groups.
The centralizer of an element in a free group is cyclic....
There's an easy self-contained proof using basics on tree automorphisms (namely that an automorphism of a nonempty tree without fixed point preserves a unique axis). Now everything depends on what's "way over your knowledge".
OMG @SteveD, this is so simple, I don't know how I missed it, thanks!
@YCor could you send reference to this proof, it sound interesting (I do my thesis is geometric group theory, this is exactly my type of things).
(1) A free group admits a free action on a tree. (Namely its Cayley graph with respect to a free generating subset).
(2) If a group acts freely on a nonempty tree, then it's a CT-group.
(Indeed, if $u,v$ are non-identity elements, they are loxodromic since the action is free, if they commute, they have the same axis, and conversely if they have the same axis then $[u,v]$ is identity on this axis and hence is identity.)
(This argument doesn't take for granted that centralizers of nontrivial elements are cyclic, and actually proves it.)
|
2025-03-21T14:48:30.155272
| 2020-03-27T12:41:47 |
355858
|
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|
Stack Exchange
|
Comparison of product topology and colimit topology in sequence spaces
In Munkres Theorem 20.4 it is shown that the (relative) uniform topology induced by:
$$
d(x,y)\triangleq \sup_{n \in \mathbb{N}} d(x_n,y_n)
$$
is strictly finer than the product topology on $\prod_{n \in \mathbb{N}} \mathbb{R}$ and its relative topology on $\ell^{\infty}(\mathbb{R})$.
Now, for each positive integer $n$, let $K_k\triangleq \left\{x \in \prod_{n \in \mathbb{N}} \mathbb{R}:\,
|x_n|\leq k
\right\}$. Clearly $K_k \subseteq \ell^{\infty}(\mathbb{R})\subseteq \prod_{n \in \mathbb{N}} \mathbb{R}$ and $K_k \subseteq K_{k+1}$ defines an inductive system in the category Top.
How does the inductive limit topology on $\bigcup_{k \in \mathbb{N}} K_k$ compare to (restriction of) the uniform topology thereon? My intuition tells me that they coincide, is this correct?
In the definition of $K_n$, what are $t$ and $|\cdot|$? Does it mean something like $\forall t \in \mathbb N, \lvert x_t\rvert \le n$, where $\lvert\cdot\rvert$ is the usual absolute value? (Also, you currently ask for the inductive-limit topology on $\bigcup_{n \in \mathbb N}$; I guess you mean $\bigcup_{n \in \mathbb N} K_n$?)
Exactly, thanks for pointing that out.
Oh, I didn't know this gives the weak$\star$-topology. Do you have a reference? So in a nutshell its stronger than the product topology but weaker than the $\ell^{\infty}$ topology (strictly in both cases).
It is not clear to me what topology you are considering on $K_k$, the norm or product topology. If the former, your conjecture is correct, if the latter you get the so-called weak $\ast$-topology as the dual of $\ell^1$.(I deleted my previous comment since I then clocked on this ambivalence)
The quickest way is via Google—“weak topology” will give you an article on Wikipedia which starts with the weak topology on a Banach space, then goes on to the weak $\ast$-topology on a dual Banach space. Any text on Banach spaces will discuss this (sorry, I don’t have access to a library to give a more precise reference for the obvious reason)
Sorry, I messed up my first comment when I deleted and repaced it. As was correctly stated there, the topology you get is not the weak $\ast$-topology but the finest one on the whole space which agrees with the latter on bounded sets—commonly known as the bounded weak $\ast$-topology. Sorry for confusing the issue—mea culpa. This is a complete lc topology and its convergent sequences are the norm bounded ones which converge in the product topology. You could check out the Banach-Steinhaus theorem.
As user131781 noted the correct answer depends on the topology of $K_n$. If it is the norm topology the result is a special case of the following more general case:
Let $K$ be a metric space with topology $\cal{O}$ and $K_n$, $n \in \mathbb{N}$ a sequence of subsets with $K_n \subset K^o_{n+1}$ ($X^0$ the interior of $X$) and with $K_n \uparrow K$. Let $\cal{O}_1$ be the inductive topology on $K$ induced by the injections $i_n \colon K_n \to K$. By definition $\cal{O}_1$ is finer than $\cal{O}$. But since $(K,\cal{O})$ is metrizable to show that $\cal{O}$ is finer than $\cal{O}_1$ it suffices to show that each convergent sequence $x_n \to x$ w.r.t. $\cal {O}$ converges w.r.t. $\cal{O}_1$. But this is a consequnce of the assumption $K_n \subset K^o_{n+1}$.
|
2025-03-21T14:48:30.155510
| 2020-03-27T12:50:20 |
355859
|
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|
Stack Exchange
|
Strengthening of Shoenfield's result on the recursive omega-rule
It is trivial to show that Peano artithmetic ($\mathsf{PA}$) supplemented with the $\omega$-rule is complete. Joseph Shoenfield (`On a Restricted $\omega$-Rule', Bull. Acad. Polon. Sci. 7 (1959): 405–7) showed that this is true even if we replace the $\omega$-rule with the recursive $\omega$-rule; i.e., we admit
$$\frac{\phi(\bar{0}), \phi(\bar{1}),\ldots}{\forall x \ \phi(x)}$$
only if there exists a recursive function enumerating the proofs of $\phi(\bar{0}), \phi(\bar{1}),\ldots$.
Is it known whether this result can be strengthened? E.g., is $\mathsf{PA}$ $+$ the primitive recursive $\omega$-rule complete? $\mathsf{PA}$ $+$ the Kalmár-elementary $\omega$-rule?
In short the answer is yes.
Let us consider Schütte-style proof of completeness of $\omega$-logic. This proof works as follows. For any sequent $\Gamma$ we define it's canonical (cut-free) pre-proof (i.e. possibly ill-founded proof tree that locally obeys the rules of $\omega$-logic). The general idea here is to define pre-proof, whose conclusion is $\Gamma$, so that each possible rule is applied at some point. Next we show that if there is an infinite path in the cannonical pre-proof of $\Gamma$ there is an infinite path, then $\Gamma$ is false. Henceforth a sequent $\Gamma$ is true in $\mathbb{N}$ iff the canonical pre-proof of $\Gamma$ is well-founded.
The usual construction of the canonical pre-proof is in fact Kalmar elementary. Moreover, it should be even polynomial.
|
2025-03-21T14:48:30.155638
| 2020-03-27T12:50:42 |
355860
|
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|
Stack Exchange
|
Possible Birkhoff spectra for irrational rotations
Let $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ be the unit circle (think of it as of the interval $[0,1)$ with endpoints identified). Assume that $\alpha$ is irrational and consider the rotation by $\alpha$, denoted $T_\alpha\colon [0,1)\to[0,1)$ where $T_\alpha(x)=x+\alpha\bmod 1$. Let $C\subset [0,1)$ be a fat Canot set, that is a Cantor set with positive Lebesgue measure $\lambda(C)$. Set $\chi_C$ to be the characteristic function of $C$.
What are the possible limit points of the Birkhoff averages of $\chi_C$ (a.k.a. Birkhoff spectrum of $\chi_C$), that is what can we say about the set
$$
B(C)=\bigcup_{x\in[0,1)} S_C(x),
$$
where
$$
S_C(x)=\text{limit points of the sequence }\left(\frac1n\sum_{j=0}^{n-1}\chi_C(T^j_\alpha(x))\right)_{n=1}^\infty.
$$
What can we say about the set $B(C)$?
Remarks. By the Birkhoff ergodic theorem for $\lambda$-a.e. $x\in [0,1)$ we have
$$
\lim_{n\to\infty} \frac1n\sum_{j=0}^{n-1}\chi_C(T^j_\alpha(x))=\lambda(C).
$$
Furthermore, by the unique ergodicity of $T_\alpha$ we have
$$
\limsup_{n\to\infty} \frac1n\sum_{j=0}^{n-1}\chi_C(T^j(x))\le\lambda(C).
$$
But $C$ is nowhere dense, so $U=[0,1)\setminus C$ is nonempty, open and dense, and the same holds true for $T_\alpha^{-k}(U)$ for $k=1,2,\ldots$. Therefore the set
$$
D=\bigcap_{k=0}^\infty T_\alpha^{-k}(U)
$$
is residual.
For $x\in D$ we clearly have $S_C(x)=\{0\}$. It follows that $0,\lambda(C)\in B(C)$ and $B(C)\subseteq [0,\lambda(C)]$. Can it happen that $B(C)=\{0,\lambda(C)\}$? Can it happen that $B(C)= [0,\lambda(C)]$?
|
2025-03-21T14:48:30.155746
| 2020-03-27T13:00:22 |
355861
|
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|
Stack Exchange
|
What is the most general formulation of Gauss's circle problem?
The (in)famous Gauss circle problem asserts that the usual error term for counting integral points inside a circle of radius $R$ centred at the origin can be made much smaller. In particular, by using a simple geometry of numbers argument one gets
$$\displaystyle \# \{(x,y) \in \mathbb{Z}^2 : x^2 + y^2 \leq R^2\} = \pi R^2 + O \left(R \right).(1)$$
Gauss's circle problem is the conjecture that the error term $O(R)$ can be replaced by $O(R^{1/2})$. By the work of Voronoi, one now has the exponent $2/3$ (i.e., an error term of $O(R^{2/3})$) and by work of Huxley one can replace this exponent by $131/208$.
One can formulate Gauss's circle problem in modern language. Let $N(X)$ denote the number of (necessarily principal) ideals in $\mathbb{Z}[i]$ having norm bounded by $X$. Then the conjecture is equivalent to the assertion that
$$\displaystyle N(X) = \frac{\pi X}{4} + O \left(X^{1/4} \right),$$
the factor of $1/4$ introduced because the Gaussian integers $\pm x \pm iy$ generate the same ideal.
If we instead fix a number field $K$ of degree $n \geq 2$ and fix an ideal class $C$ in $\mathcal{O}_K$, denote by $N_K(C;X)$ for the number of ideals belonging to $C$ having norm bounded by $X$. Then the analogous theorem to (1) is
$$\displaystyle N_K(C;X) = \frac{\kappa(K) X}{h(K)} + O_n \left((h(K) R(K))^{1/n}(1 + \log h(K) R(K))^{(n-1)^2/n} X^{1- 1/n} \right),(2)$$
where $h(K)$ is the class number of $K$, $R(K)$ the regulator of $K$, and $\kappa_K$ the residue at $s = 1$ of the Dedekind zeta function $\zeta_K(s)$ (this being a theorem of Korneel Debaene; see https://arxiv.org/abs/1611.10103). In particular, for $n = 2$ (2) exactly recovers (1).
My question is, what is the correct formulation of 'Gauss's circle problem' for (2)? What is the expected optimal exponent in $X$, and what is the expected dependency on the regulator/class number terms?
|
2025-03-21T14:48:30.155884
| 2020-03-27T13:06:41 |
355864
|
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|
Stack Exchange
|
About the cohomology of $BG^\delta$. Making a Lie group discrete
Let $G$ be a connected Lie group. Recall that the topological group $G^\delta$ is $G$ endowed with the discrete topology. The inclusion $G^\delta \to G$ induces a map between the classifying spaces $\eta: BG^\delta\to BG$.
Question 1
Let $\eta^*:H^*(BG,\mathbb{Z})\to H^*(BG^\delta,\mathbb{Z})$ be the induced map in integral cohomology.
By Corollary 1 in Milnor, On the homology of Lie groups made discrete,
we get that $\eta^*$ is injective.
On the other hand, by Lemma 10 in the same paper, we learn that the kernel of
$\eta_{\mathbb{Q}}^*:H^*(BG,\mathbb{Q})\to H^*(BG^\delta,\mathbb{Q})$ (notice the rational coefficients here)
is equal to the kernel of $\eta^*_{\mathbb{Q}}:H^*(BG,\mathbb{Q})\to H^*(B\Gamma,\mathbb{Q})$, where $\Gamma<G$ is a discrete cocompact group.
Consider $G=U(n)$. Then $H^*(BG,\mathbb{Z})= \mathbb{Z}[c_1,c_2,\dots,c_n]$ injects in $H^*(BG^\delta, \mathbb{Z})$, in particular $\eta^*(c_1)\neq 0$.
However, we can take $\Gamma= \{\mathbb{1}\}$, which implies that $H^*(B\Gamma,\mathbb{Q})=H^*(\mathbb{R},\mathbb{Q})$, hence $\eta^*_{\mathbb{Q}}$ is trivial and in particular $\eta^*_{\mathbb{Q}}(c_1) = 0$.
Why doesn't this give a contradiction?
I will only attempt to answer your first question. The reason there is no contradiction is that it is not true for arbitrary spaces that $H^{\ast}(X;\mathbb Q) = H^{\ast}(X;\mathbb Z) \otimes \mathbb Q$.
For instance, take $X = B\mathbb Q$. Then $H^2(X;\mathbb Z) = \text{Ext}(\mathbb Q,\mathbb Z)$ is a $\mathbb Q$ vector space of uncountable dimension. In particular, we can find a map $B\mathbb Q \to K(\mathbb Z,2)$ that is injective on second cohomology with $\mathbb Z$ coefficients. But clearly the map induced on rational cohomology is zero.
Probably it's obvious but why the map on rational cohomology is zero? I tried to compute $H^2(B\mathbb{Q},\mathbb{Q}) = H^2(\mathbb{Q},\mathbb{Q}) = Ext^2_{\mathbb{Z}[\mathbb{Q}]}(\mathbb{Z},\mathbb{Q})$ but I can't see why it should be trivial.
I would compute it differently, $H^2(B\mathbb Q;\mathbb Q) = \text{Hom}(H_2(B\mathbb Q;\mathbb Q),\mathbb Q)$, and $H_2(B\mathbb Q;\mathbb Q) = \text{Tor}_1(\mathbb Q,\mathbb Q) = 0$ since $\mathbb Q$ is flat.
|
2025-03-21T14:48:30.156057
| 2020-03-27T13:56:10 |
355868
|
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|
Stack Exchange
|
Uniquely describing a polytopal complex by prescribing the local structure around its vertices
Let $C$ be a $d$-dimensional (abstract) polytopal complex.
Most of what I say below could be asked in this general setting, but for a start, let's further restrict to simple polytopal spheres, that is, $C$ is homeomorphic to the $d$-sphere, and at each vertex meet exactly $d+1$ facets.
My goal is to formalize and then prove a certain kind of statement.
For a warmup, here are two examples of what I am after:
Assume that each vertex of $C$ is incident to $d+1$ facets, each one combinatorially equivalent to the $d$-cube. From that I can already deduce that $C$ is combinatorially equivalent to the (boundary of the) $(d+1)$-cube.
Assume that $d=2$ and each vertex is incident to a 4-gonal cell and two 6-gonal cells. This already suffices to conclude that $C$ is combinatorially equivalent to the (boundary of the) permutahedron.
In general, I want to prove something like this:
If each vertex of a simple polytopal sphere $C$ looks the same locally (that is, is incident to the same type of facets), then $C$ is already uniquely determined.
So how to prove this?
Intuitively, this is easy.
Take, for example, the second example from above (and see the image below): if you try to draw its edge graph, you will start with a single vertex (the white dot) surrounded by a 4-gon and two 6-gons. You then see that there are several vertices for which only a single face is missing (the black dots), and you add these.
You go on until you realize that now all the vertices satisfy the constraint.
You never made a choice, and so the result must be unique.
My question is the following:
Question: How can this proof be formalized?
There seem to be some subtleties, some are hinting at certain generalizations:
You need to use that $C$ is a polytopal sphere, otherwise the complex might not be unique.
For example, there are infinitely many possibilities to realize a polytopal torus in which every vertex is incident to exactly three 6-gons. Maybe "sphere" can be replaced by simply connected, though.
We do not necessarily need simple, but it is the "simplest" of several conditions that ensure that we never have to make a choice when placing the next facet. Here is a case where above reasoning fails because we dropped simplicity: if we want every vertex of a 2-dimensional complex to be surrounded by a 3-gon and three 4-gons, there are two solutions: this one and this one.
How do I know that there is always a vertex at which the next facet is determined?
How can one be sure that the order in which I add new facets (which are forced on me) does not make a difference. Or does it?
Re: the last question - "How can one be sure that the order in which I add new facets (which are forced on me) does not make a difference. Or does it?" - since a facet can never stop being forced once it is forced, I think a routine application of the 'diamond lemma' proves confluence, i.e., that it does not matter which order you carry out the forced facet additions.
@SamHopkins This was already a very helpful hint. I wasn't aware of this lemma. Just a note: one might want to apply above reasoning to tilings of $\Bbb R^n$ too. In that case, the process of adding further facets isn't finite. If I understood correctly, this finiteness is an important condition in the diamond lemma. Is there away around this?
In general there is no way around some kind of finiteness condition to apply the diamond lemma.
Sorry, but there are counterexamples to your local to global quest:
Consider the uniform small rhombicuboctahedron versus the elongated square gyrobicupola (Miller's solid, one of the Johnson solids). - Both show up the same vertex figures all over, still they are globally different (i.e. non-unique) outcomes (polyhedra).
--- rk
This counterexample is already contained in my post in the third to last bullet point. This is why I need the rule for adding one more tile that the placement must be unique, as it is in the simple case.
Just want to let you know that similar finds to Miller's solid lately were found within 4D, cf. http://www.polytope.net/hedrondude/polychora.htm#pseudouniform
|
2025-03-21T14:48:30.156481
| 2020-03-27T14:09:54 |
355869
|
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|
Stack Exchange
|
Earliest use of deconvolution by Fourier transforms
From a previous discussion here Origin of the convolution theorem, it was shown that the property of convolution $y(t)$=$a$*$b$ becoming a multiplication after Fourier transform: $F$$(y(t))$= $F(a)F(b)$, was well known by early 1900s and clearly mentioned in 1941.
I was searching the earliest known use of deconvolution by Fourier transforms. Surprisingly, the term deconvolution is quite recent as per the unabridged version of the Oxford English Dictionary (OED). In deconvolution, two functions are divided in the Fourier domain to recover the original function, say $a$, if $y(t)$ and $b(t)$ are known. For example, if we wish to recover $a$, we can divide $F(y(t))$ by $F(b)$ and do an inverse transform to get $a$. It may not be a rigorous way, but it is a popular technique in spectroscopy from an empirical perspective.
OED mentions a 1967 paper titled "Posteriori image-correcting "deconvolution" by holographic fourier-transform division" in Physics Letters. The authors show the following:
The authors cite Maréchal and Croce as the first example in Comptes Rendus, however Gallica Original Paper does not have a single equation and the word Fourier is mentioned only in the first two lines! So this reference seems to be incorrect.
I am not interested in image analysis, but rather in the earliest known use of division process in the Fourier domain to recover original functions.
a) I wanted to know if mathematicians were using this approach well before 1960s to recover an original function from given convolution?
b) Spectroscopists call the division in the Fourier domain as deconvolution, what do mathematician call the process of division of two functions in the Fourier domain?
Update (30 Mar 2020)
From the detailed response by Tom Copeland, and the Table 1 shown in History of Convolution one can see another reference from 1943,
G. Doetsch, Theorie und Anwendung der Laplace-Transformation. New York: Dover, 1943
and the note 200 reads:
The reference to Picherle is given as "I. Studi sopra alcune operazioni funzionali. Mem. Accad. Bologna (4) 7 (1886)."
However the Table 1 of the convolution history mentions 1907. No reference is provided.
Thanks.
I would look at the history of Green (Green's) functions, the Heaviside operational calculus and associated Bromwich-Laplace transform and fractional.calculus, and integral transforms as discussed by Titchmarsh and Hardy for correlates of deconvolution, particularly where the Green/influence function or integral kernel is of the form G(x-y).
@M.Farooq --- Pincherle's 1886 paper is online here --- it does not seem to address Fourier transformations or convolutions, at least not in any explicit way (did I overlook it?)
Thanks Dr. Carlo. I think the reference by G. Doetsch, Theorie und Anwendung der Laplace-Transformation is not correct. If we see the Table 1, here https://pulse.embs.org/january-2015/history-convolution-operation/, it mentions a 1907 reference by Pincherle. No citation is provided there.
indeed, I think I have identified this 1907 reference; I copy the relevant text in the answer box.
By the way, I think that there is a common tendency in mathematics history to attribute 'generously', identifying the first place where recognisable precursors of an idea appear, not just the first place it appears in its modern form. (Fourier would probably not recognise what we now call the Fourier transform.) Quite possibly that's the spirit of the reference to Maréchal and Croce, although I am no expert on the physics and so am not confident I'd recognise the idea in the paper even if it were there ….
An early use of division in Fourier space to undo a convolution is Fourier Treatment of Optical Processes (1952), by Peter Elias, David S. Grey, and David Z. Robinson. (This paper precedes the paper by Maréchal and Croce cited in the OP.)
Following the OP's and Copeland's lead to Pincherle suggests this 1907 publication Sull'inversione degli integrali definiti. The convolution theorem for Laplace transforms [called "funzioni generatrici" – generating functions; "funzione determinante" is the inverse transform] is stated and used to invert the convolution by dividing the transformed functions:
From eq. (10), or the equivalent (10'), we immediately find the answer to question number 4. Indeed, equation (d)
$$\frac{1}{2\pi i}\int a(x-t)f(t)dt=g(x)$$
is equivalent [for the Laplace transforms] to $Gg=Ga\cdot Gf$ or $\gamma(u)=\alpha(u)\phi(u)$, and hence the function $f$ is determined by [the inverse transform of] $\gamma(u)/\alpha(u)$.
I am also wondering that mathematicians must be using this a little bit earlier given that convolution was known in 1900s? Why there is a delay of 50 years for undoing convolution? What do you think?
Thank you very much! I think that is it.
Thanks for the link to the article by Pincherle. Have you access to Heaviside's Electromagnetic Theory Vol. 3? Supposedly the section "The solution of definite integrals by differential transformation" has some presentation of the inverse Laplace/Bromwich/Fourier-Mellin transform, which Heaviside discussed with Bromwich.
@TomCopeland --- Heavide's book is online here --- I looked at the section you mention but could not find anything hinting at deconvolution.
Thanks. Note that eqn. 11 on p. 236 is a statement of the inverse Laplace transform of $1/p^{n+1}.$ Heaviside is taking the Laplace transform of $\int_0^{\infty} e^{-px} (x^n/n!)dx$ with $p = d/dx = D$. Then, with $H(x$) the Heaviside step fct., evaluating $D^{-(n+1)}H(x)= H(x)x^{n+1}/(n+1)!$, which is the inverse Laplace trf of $1/p^{n+1}$. Bromwich felt this method generalized by Heaviside was superior to the use of the inverse Laplace transform as a complex contour integral, which Bromwich was the first to present, I believe.
In addition, this is a simple example of the convolution theorem: with the H(x) introduced explicitly, $\int_0^{\infty} e^{-Dt} (t^n/n!)dt H(x) = \int_0^{\infty} H(x-t) (t^n/n!)dt = \int_0^{x} H(x-t)(t^n/n!)dt = H(x) \frac{x^{n++1}}{(n+1)!}$.
Rather, $D^{-(n+1)}H(x)=H(x) \frac{x^{n+1}}{(n+1)!}$, so $$\int_0^{\infty} e^{-tD} \frac{t^n}{n!}dt H(x)= D^{-(n+1)
}H(x)= H(x)\frac{x^{n+1}}{(n+1)!}$$ is equivalent to taking the inverse LPT of $$\frac{1}{p}\int_0^{\infty}e^{-xp} \frac{x^n}{n!}dx = \frac{1}{p^{n+2}}$$ (which explains why I've seen this expression for the LPT in some older literature).
Early uses of deconvolution via integral transforms:
A) Signal processing:
$$ \int_{\infty}^{\infty} K(y-x) h(x)dx
= \int_{-\infty}^{\infty} e^{-i 2 \pi \omega (y-x)} h(x)dx $$
$$= e^{i2 \pi \omega y}\hat{h}(\omega)=H(\omega)$$
is an example of a convolution.
Dephasing $H$ and taking an inverse FT amounts then to deconvolution of the type you designate :
$$\int_{-\infty}^{\infty} \frac{H(\omega)}{e^{i2 \pi \omega y}}e^{i2 \pi \omega x} d \omega= h(x).$$
This must have been done at least by the researchers, such as Schwinger, at the MIT Radiation Lab during WW II in the development of radar.
B) Deconvolution via Fourier transforms of the Wiener-Hopf integral equation published in 1931:
Lawrie and Abrahams present in "A brief historical perspective of the Wiener-Hopf technique", the solution developed by Wiener and Hopf of the convolutional equation
$$ \int_{0}^{\infty} k(x-y) f(y) dy =\left\{\begin{matrix}
g(x), & x > 0\\
h(x), & x<0
\end{matrix}\right.$$
where $f(x)$ and $h(x)$ are unknown. For $h(x)=0$, the solution specializes to the inverse transform of a ratio of Fourier transforms
$$ FT[HV(x)g(x)]/ FT[k(x)] = FT[HV(x)f(x)]. $$
$ HV$ is the Heaviside step function.
(Norbert Wiener had a vast range of interests, and, since signal propagation/processing had recently become important due to the development of telegraphs, power lines, telephones, radar, and x-ray diffraction, it seems plausible that he was one of the earliest to publish on deconvolution through the Fourier transform. The Mellin and Laplace transforms and deconvolutions are better suited for development of the operational/algebraic calculus explored by Lebnitz, Euler, and dozens after them.)
C) Operational calculus, fractional calculus, differential algebra:
For Heaviside's operator calculus (and use by Dirac), see the discussion, references, and comments at Ron Doerfler's post at his website Dead Reckonings. (Synowiec is also cited below, and see this note by Davis on Bromwich's views of the Heaviside calc.)
For differential algebra in general, read "Some highlights in the development of algebraic analysis" by Synowiec in which symbolic methods, the Heaviside calc, and the Laplace transform are stressed, but Norbert Wiener's Fourier transform method is only briefly mentioned with a reference to his 1926 book On the Operational Calculus. Pincherle's contributions are mentioned as well as by Dominguez.
Quoting Dominguez (from his timeline table):
1907 Despite the many occurrences and uses of the CCO, none of the previous authors made a complete study of it. The earliest one is, perhaps, that made by the Austrian-born mathematician Salvatore Pincherle (1853–1936) in connection with the solution of the complex integral equation
$$ \frac{1}{2 \pi i} \int_{|z| = P} k(s-z) f(z) dz = g(s)$$
where $P > 0$ and $k(z)$ and $g(z)$ are given functions, while $f(z)$ is unknown. Pincherle succeeded in the solution of this CCO using as tool the Laplace transform. .... . These results are the basis for the deconvolution method established in [35].
Pincherle also developed an axiomatic approach to the fractional calculus (which can be based on the Mellin convolution). See "The Role of Salvatore Pincherle in the Development of Fractional Calculus" by Mainardi and Pagnini. The solution to the op eqn.
$$ D^r HV(x)f(x) = HV(x)g(x)$$
is $$HV(x)f(x) = D^{-r}HV(x)g(x) = D^{-r}D^rHV(x)f(x),$$
which can be expressed as a deconvolution.
From "Operational method for the solution of fractional differential equations with generalized Riemann-Liouville fractional derivatives" by Hilfer, Luchko, and Tomovski:
In the 1950’s, Jan Mikusinski proposed a new approach to develop an operational calculus for the operator of differentiation .... This algebraic approach was based on the interpretation of the Laplace convolution as a multiplication in the ring of the continuous functions on the real half-axis. The Mikusinski operational calculus was successfully used in ordinary differential equations, integral equations, partial differential equations and in the theory of special functions.
Thanks, Unfortunately, the original paper is not available anywhere. I checked HathiTrust and Internet Archive. Hathi Trust has all copies of Sitzungsberichte (1931) but it is locked due to copyright even from viewing a single page.
A close connection between the symbolic calculus and FT deconvolution methods is illustrated by Bracewell's derivation of Eddigton's formula for correction of optical spectral line broadening due to limited resolution of the spectrometer. See the 1913 article by Eddington (https://academic.oup.com/mnras/article/73/5/359/972797) and the section titled Eddington's Formula in Bracewell's book Fourier Analysis and Imaging. Very likely Bracewell used FT deconvolution when he did R&D on radars and imaging by radiotelescopes before 1954, well before he wrote his books on the FT.
Thanks Tom, I have updated the post. Your pointer to Pincherle as the earliest one is the right version as it can be seen in the early post.
|
2025-03-21T14:48:30.157221
| 2020-03-27T14:27:56 |
355871
|
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|
Stack Exchange
|
Can the Lie derivative of a Riemannian metric be expressed in terms of the Lie derivative of a Lorentzian metric?
On a Lorentzian manifold with metric (M,g) with a vanishing Euler-Poincare characteristic, there exists a line element vector X which has a collinear vector u (Manifold Theory: An introduction for mathematical physicists p183). Let M be endowed with a Riemannian metric $g^{+}$. A Lorentzian metric can then be constructed as $ g_{\alpha\beta}=g^{+}_{\alpha\beta}-2u_{\alpha}u_{\beta}$. X is timelike and globally hyperbolic so we don't have any closed causal curves.
Consider an open subset that contains $g^{+}$ on M with the Lie derivative along an arbitrary vector $\xi$, $L_{\xi}g^{+}_{\alpha\beta} $. The question is then, can one express this Lie derivative along $\xi$ as $ L_{X}(g_{\alpha\beta}+2u_{\alpha}u_{\beta})$? If not, please provide the details to fix this expression.
I don't understand the question. We fix a Riemannian metric $g^+$, and then ask if, for any vector field $\xi$, there is a nonzero vector field $X$ so that $L_{\xi}g^+=L_X g^+$? I am not seeing any Lorentzian geometry in that question, but I am not sure I have the right question.
Ben, we are on a Lorentzian manifold with metric g. It is constructed as shown. So yes, I would like to be able to use $L_X (g+2uu)$ where u exists only on that manifold.
@Kolten Welcome to MO. Please rewrite the question using more standard terminolgoy. Your definition of $X$ and $u$ doesn't make any sense to me. Is $X$ related to $g$? Is the $g_{\alpha\beta}$ the Lorentzian metric you started with? But then $g^+$ can't be arbitrary as your phrasing suggests. Are you really asking if $L_\xi g^+$ is expressible for any $\xi$ with $L_X$ or is it a typo?
If I understood the problem correctly, then $L_{X-\xi}g^+=0$, a nonzero Killing field for the Riemannian metric. Generic Riemannian metrics do not have these. Therefore the answer is no, if I have the right question.
Vit, X exists on a Lorentzian manifold. That means there is a unit vector collinear with X from which $g=g+-2uu$ So u is related to X. There is no typo, but from the other comments it appears that even though $\xi$ is arbitrary, one cannot replace it with X. But that bothers me why that cannot be so. Ben has a comment in that regard.
Ben, do you have any suggestion on how to invoke the line element vector X in the Lie derivative, instead of the vector $\xi$ associated with the Riemannian metric? I assume you agree that $L_{\xi}g^{+}=L_{\xi}(g+2uu)$.
Why is $\xi$ associated with the Riemannina metric? It is just a vector. Also, what is the definition of "line element vector"? I have not see that term before.
Willie, a line element field (X,-X) at a point p on M is a regular timelike vector field unique up to a sign. u is a unit vector collinear with one of the pair in the line element field. If $\xi$ is a vector in the Riemannian open subset of M, it might even vanish on part of M, whereas X does not. But I assume if we work with the non-vanishing part of $\xi$ locally in that subset, there should be no problem with using $ L_{\xi}g^{+}=L_{\xi}(g+2uu)$.
You question has a typo maybe? In the OP you want $L_\xi g^+ = L_X (\ldots)$.
Willie, that is what I would like to say, but Ben points out that I can't. Do you have any insight how to associate the Riemannian vector with the Lorentzian one, or am I wasting my time with that and should use the non-zero part of $\xi$?
If $\xi$ is just some random vector field, there is absolutely nothing you can say. You are wasting your time even if you try to use the non-zero part of $\xi$.
What if I just used the line element vector X on M instead of any particular Riemannian vector? In other words, $ L_{X}g^{+}=L_{X}(g+2uu)$
should hold in the Riemannian open subset on M. Although I have a calculation that involves
$L_{\xi} g^{+}$ and holds only in that set. But that set belongs to M where X exists. That, again, is my problem.
|
2025-03-21T14:48:30.157508
| 2020-03-27T14:41:46 |
355872
|
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|
Stack Exchange
|
Injectivity of an integral operator
Consider the operator
$$K:L^2(0,1)\rightarrow L^2(0,1) \\ u\rightarrow\int_0^1k(s,x)u(s)ds.$$
with $k\in L^2((0,1)\times(0,1)).$
I want to know under what assumption the kernel is reduced to zero. i.e. $ker(K)={0}$.
I can say that if $k$ is a Green function for some differential operator this will be true. But what about the general case? Can we obtain a criteria for the injectivity by some expansion process on the $L^2$ basis?. Thank you.
In my opinion, in this generality it iss impossible to say something non trivial. Nevertheless take a look at "Bounded Integral Operators on L2 spaces" by Halmos and Sunder.
Your operator $K$ is a Hilbert-Schmidt operator since its kernel belongs to $L^2$. As a result this is a compact operator whose spectrum contains a sequence of eigenvalues $\\{\lambda_k\not=0\\}$ with finite multiplicities such that $\lim_k\vert \lambda_k\vert=0$. To deal with the self-adjoint case, you can find an orthonormal set $\\{\mathbf e_k\\}$ such that
$$
K\mathbf e_k=\lambda_k\mathbf e_k \quad \mathbb R\ni\lambda_k, \quad (\lambda_k)\in \ell^2
.$$
As a result, 0 will always belong to the spectrum even if $\\{\mathbf e_k\\}$ is an orthonormal basis, but $K$ will be injective in that case. Setting $$
E=\overline{\text{span}\\{\mathbf e_k\\}},$$
you get
$
\ker K=E^\perp.
$
In the non-self-adjoint case, maybe the equality $ker(K)=ker(K^*K)$ could be useful.
Thank you sir for the answer. In fact, it is quite difficult to compute the eigenvectors in my case. Thank you again.
@Gustave Well, I believe that the most important thing is the Hilbert-Schmidt structure provided by the $L^2$ assumption on the kernel: you get (in the self-adjoint case) an infinite diagonal matrix with a diagonal in $\ell^2$.
|
2025-03-21T14:48:30.157655
| 2020-03-27T14:49:16 |
355873
|
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|
Stack Exchange
|
The valuation of finite extension of an non-archimedean field
Let $(k,|.|)$ be a non-archimedean complete field and $(L,|.|)$ be a finite extension of $(k,|.|)$, $[L:k]=n$, such that $L=k(\xi)$. Let $\phi$ the homomorphism of $k$-Banach algebra
$$\begin{array}{rcl}
\phi: (L,|.|)&\to &(k^n,||.||)\\
f_1+f_2\xi+\cdots+f_n\xi^{n-1}&\mapsto &(f_1,f_2,\cdots,f_n)\\
\end{array}$$
where $||.||$ is the maximum norm.
Since we know that, all $k$-norms over a finite $k$-vector space are equivalent, the operator $\phi$ must be bounded.
Is there any reference for the computation of the operator norm of $\phi$? Or less precise,
an explicit $c$ such that
$$\max |f_i|\leq C|f_1+f_2\xi+\cdots+f_n\xi^{n-1}|.$$
|
2025-03-21T14:48:30.157734
| 2020-03-27T14:56:24 |
355874
|
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|
Stack Exchange
|
Eigenvalues of adjacency matrix of a k-regular graph
If $A_G$ is the adjacency matrix of a k-regular graph, let $B = J+xA_G$, where J is the matrix whose elements are all 1s and $x\in R$ is a scalar. If $\lambda_1\geq\lambda_2\geq \dots \geq \lambda_n$ are eigenvalues of $A_G$, how do we prove that $\underset{x\in R}{\min} \lambda_{max}(J+xA_G) = \frac{n\lambda_n}{\lambda_n-\lambda_1}$?
I know that e (the vector of all 1s) is an eigenvector of B with eigenvalue $n+x\lambda_1$, but what about the other eigenvectors? If we know other eigenvectors of B, then we can compare among them and pick the maximum. Thanks!
If $G$ is regular, then $J$ and $A_G$ are simultaneously diagonalizable (i.e. they have a common set of eigenvectors).
That is, the eigenvalues of $xA_G$ and $J$ (to the same eigenvectors) just add up to the eigenvalues of $B=J+xA_G$.
Note that the spectrum of $J$ is $\{0^{n-1}, n^1\}$, and that the eigenvalue $n$ corresponds to the eigenvector $(1,...,1)$.
The corresponding eigenvalue of $A_G$ is $k=\lambda_1$.
If $x$ is positive or only a little bit negative, then the largest eigenvalue of $B$ is $n+x\lambda_1$.
But if $x$ is more negative, then at some point the largest eigenvalue of $B$ will be $0+x\lambda_n$ (note that $\lambda_n<0$).
So the moment when these two values coincide is when the minimum is attained:
$$n+x\lambda_1 = x\lambda_n \quad\implies\quad x=\frac{n}{\lambda_n-\lambda_1}.$$
If you plug this into $\lambda_n x$ you found the desired value.
Thanks! So the key is J ,in general, has a common set of eigenvectors with any matrix, right?
@RayyyyySun Not with every matrix, just with those for which $(1,...,1)$ is an eigenvector, and all other eigenvectors are perpendicular to $(1,...,1)$.
I see, Thanks! really appreciate :)
|
2025-03-21T14:48:30.157890
| 2020-03-27T15:49:26 |
355879
|
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|
Stack Exchange
|
Open/closed/constructible subsets of locally free sheaves
(Cross-posted from math.SE since I'm not sure what is a suitable platform. Link on https://math.stackexchange.com/questions/3597258/open-closed-constructible-subsets-of-locally-free-sheaves)
Is there a "canonical" way to define open/closed/constructible subsets of a locally free sheaf/vector bundle $\mathcal{F}$ on a scheme $X$? There is a clear way to do this with topological vector bundles, but are things defined the same way in algebraic geometry?
I came across them in a paper (https://arxiv.org/abs/1910.05207), but couldn't find any other sources discussing possibly related concepts specifically in algebraic geomery other than those looking at subsets of global sections of these vector bundles.
My guess was that this has something to do with interpreting a locally free sheaf/vector bundle $\mathcal{F}$ locally as (relative) $\text{Spec} (\text{Sym}^\cdot \mathcal{F}^\vee)$ (which are also discussed in section 4 of the paper above), but I'm not sure how to formulate a precise definition other than what was suggested at the beginning of this question. It may be helpful to list interesting examples/non-examples.
|
2025-03-21T14:48:30.158003
| 2020-03-27T16:13:05 |
355881
|
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|
Stack Exchange
|
Connectedness of the stabilizer in a semisimple group of a semisimple element in the Lie algebra: a reference request
Let $G$ be a (connected) semisimple algebraic group over an algebraically closed field $k$ of characteristic 0.
We consider the adjoint representation
$$ {\rm Ad}\colon G\to {\rm GL}({\mathfrak g}),$$
where ${\mathfrak g}={\rm Lie}\ G$.
I am looking for a reference to a proof of the following assertion:
Proposition. For any semisimple element $X\in{\mathfrak g}$,
its stabilizer ${\rm Stab}_G(X)\subset G$ with respect to the adjoint representation
is connected.
I think that I can prove the proposition; see my proof below.
Proof. Our semisimple element $X$ is contained in a Cartan subalgebra ${\mathfrak t}$ of ${\mathfrak g}$, which is the Lie algebra of a maximal torus $T$ of $G$. Then ${\mathfrak t}$ is an algebraic subalgebra of ${\mathfrak g}$.
Let ${\langle X\rangle_{\rm alg}}\subset {\mathfrak g}$ denote the smallest algebraic subalgebra of ${\mathfrak g}$ containing $X$;
then ${\langle X\rangle_{\rm alg}}\subseteq {\mathfrak t}$. It follows that ${\langle X\rangle_{\rm alg}}={\rm Lie}\ S$ for some subtorus $S\subseteq T$.
Now (in characteristic 0) we have
$${\rm Stab}_G(X)=\bigcap_{Y\in {\langle X\rangle_{\rm alg}}} {\rm Stab}_G(Y)=C_G(S),$$
where $C_G(S)$ denotes the centralizer of the torus $S$ in $G$.
By Theorem 22.3 of Humphreys' book "Linear Algebraic Groups",
$C_G(S)$ is connected, as required.
Edit: A similar argument shows that for any commutative subalgebra ${\mathfrak a}\subset {\mathfrak g}$ consisting of semisimple elements, its centralizer in $G$
$$ C_G({\mathfrak a}):=\bigcap_{X\in {\mathfrak a}} {\rm Stab}_G(X)$$
is connected (because the "algebraic closure" $\langle {\mathfrak a}\rangle_{\rm alg}$ of $\mathfrak a$ is the Lie algebra of some torus $S\subset G$).
A reference: Steinberg, Torsion in reductive groups,
Advances in Math. 15 (1975), 63–92, Corollary 3.11.
In positive characteristic $p$: see loc. cit., Theorem 3.14. It says that if (and only if) $p$ is not a torsion prime for $G$, then $C_G({\mathfrak a})$ is connected for any commutative subalgebra ${\mathfrak a}\subset {\mathfrak g}$ consisting of semisimple elements.
See also:
Connectedness of centralizers of semisimple Lie-algebra elements under the action of a semisimple algebraic group
Steinberg says there is a different proof for $k = \mathbb{C}$, and gives reference to "Lemma 5" in Volume 2 of Seminaire Chevalley. I do not find such a lemma there, and I wonder what is the intended reference.
@spin: I think it is an erroneous reference...
|
2025-03-21T14:48:30.158190
| 2020-03-27T16:50:32 |
355883
|
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|
Stack Exchange
|
Number of components of self-index complementary graphs
Let $G$ be a simple graph. We say this graph is self-index complementary ($SIC$) if $\lambda_1 (G)=\lambda_1 (\overline{G})$, where $\lambda_1(G)$ denotes the index of the adjacency matrix of the graph $G$. Index of a graph is its maximum eigenvalue.
In this case, I am not interested in the regular graphs. I found a method to generate an infinite family of irregular graphs which are $SIC$.
During my examination, I observed that such graphs have special structures. One of this property is as follow:
If $G$ is $SIC$ then it has at most two connected components.
I found all $SIC$ graphs up to $9$ vertices and the data verify the conjecture.
My question is: Is there any proof for this observation or is there any counter example?
Suppose $G$ has complete graphs $K_{p_1},\dots,K_{p_k}$ as its components where $k\geq 3$ and $p_1>p_2\geq\dots\geq p_k$. Then $G$ is not regular, has the complete multipartite graph $K_{p_1,\dots,p_k}$ as its complement, and $\lambda(G)=p_1-1$. The characteristic polynomial of $K_{p_1,\dots,p_k}$ is known: https://www.sciencedirect.com/science/article/pii/S0012365X1100327X.
Can you arrange for $p_2,\dots,p_k\in{0,1,\dots,p_1-1}$ so that $p_1-1$ is a root of the polynomial appeared in that paper?
I do not think this method works, since the number of edges effects on the largest eigenvalue, and I think in the first place we must control the number of edges and triangles in both graphs and its complements.
|
2025-03-21T14:48:30.158321
| 2020-03-27T17:00:51 |
355884
|
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|
Stack Exchange
|
Smoothness of the generic fiber implies smoothness over a dense open of the target?
I can prove the following result without too much trouble:
Let $f: X \to S$ be a proper flat morphism of finite presentation with $S$ irreducible having generic point $\xi \in S$. Then the following are equivalent,
(1) the generic fiber $X_\xi \to \mathrm{Spec}{\kappa(\xi)}$ is smooth
(2) there exists a smooth fiber $X_s \to \mathrm{Spec}(\kappa(s))$ for some $s \in S$
(3) there exists a dense open set $U \subset S$ such that $f : X_U \to U$ is smooth.
A possible reference is [EGA-IV-4-12.2.4] then use properness to conclude that the image of the nonsmooth locus is closed.
However, I am interested in the case that $f$ is not proper. It is easy to show that (2) does not imply (3) without properness. E.g. take
$$ X = \mathrm{Spec}{k[x, y, z]/(y(xz - 1))} \to \mathrm{Spec}{k[z]} = \mathbb{A}^1_k $$
which has only one smooth fiber (over $z = 0$). However, is it true that (1) still implies (3) without properness? Explicitly, is the following true:
Let $f: X \to S$ be a flat morphism of finite presentation with $S$ irreducible having generic point $\xi \in S$. Then if $X_\xi \to \mathrm{Spec}(\kappa(\xi))$ is smooth then there is a dense open set $U \subset S$ such that $f : X_U \to U$ is smooth?
If this is true can somebody point me to a reference? If it is false can somebody provide a counterexample?
Many thanks.
You probably want to use Chevalley's theorem on the constructibility of the pushforward of constructible sets (EGA IV, 1.8.4.) here.
It looks like that does the trick. Thanks
I think you might find it helpful to look at "spreading out" in Bjorn Poonen's book on Rational points. For example, you can replace "smooth" by "proper", "unramified" "etale" "separated" "flat" etc.
Wow, this "spreading out" theorem is much stronger than I expected. Thanks for the reference.
A tiny typo: the reference should be [EGA-IV3, 12.2.4]
As Will Sawin said in his comment, this is true using Chevalley's theorem. A sketch goes as follows.
Suppose that $X_\xi \to \mathrm{Spec}(\kappa(\xi))$ is smooth. Let $N \subset X$ be the nonsmooth locus. For a morphism of finite presentation, the smooth locus is retrocompact open so $N$ is constructible. Thus by Chevalley's theorem $f(N)$ is constructible so $S \setminus f(N)$ is constructible. However, $\xi \in S \setminus f(N)$ because the generic fiber is smooth so $S \setminus f(N)$ contains a dense open $U \subset S \setminus f(N)$. Then $X_U \to U$ is smooth because $f$ is flat + finite presentation and each fiber above $s \in U$ is smooth since $U$ does not intersect $f(N)$ meaning the fiber over $s \in U$ does not intersect $N$.
|
2025-03-21T14:48:30.158628
| 2020-03-27T17:12:13 |
355886
|
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|
Stack Exchange
|
if $\max_{z \in K} |\zeta(z+it)-f(z)|<\epsilon.$ then is this $\lim_{t\to \infty} \inf \frac {|\zeta'(z+it)|}{|f'(z)|} $ a finit limit?
Universality theorem of Riemann zeta function states that :Let $K$ be a compact subset with connected complement lying in the strip $\{1/2 < \operatorname{Re}(z)<1\}$, and let $f : K \rightarrow \mathbb{C}$ be continuous, holomorphic on the interior of $K$, and zero-free on $K$. Then for each $\epsilon>0$, there exists $t>0$ such that
$$\max_{z \in K} |\zeta(z+it)-f(z)|<\epsilon.$$ , which means the approximation of such holomorphic function by Riemann zeta function in the strip now my question here is : if we have $\max_{z \in K} |\zeta(z+it)-f(z)|<\epsilon.$ then is this $\lim_{t\to \infty} \inf \frac {|\zeta'(z+it)|}{|f'(z)|} $ a finit limit?
Note: The motivation of this question is to know more about growth rate of the ratio $\frac {\zeta'(z+it)|}{f'(z)} $
Is the infimum over $z\in K$? By the same universality theorem this infimum will oscilate between arbitrarily close to 0 and arbitrarily large arbitrarily often, so there is no limit.
Thanks for your comment , I have a wrong typo I meant limit of derivative ratio
$\zeta'$ is also universal (there should be a reference here) so the same holds.
|
2025-03-21T14:48:30.158735
| 2020-03-27T17:22:53 |
355887
|
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|
Stack Exchange
|
Hopping geodesics
Is there a complete metric space $X$ with the following property?
For any pair of points $p,q\in X$ there is unique minimizing geodesic $[pq]_X$ that connects $p$ to $q$, but the map $(p,q)\mapsto [pq]_X$ is not continous.
Comments
For sure such space cannot be compact (or proper).
If we fix one end $p$, then there is a classical example --- the wheel; it is a unit circle with continuum of spikes from to the center $p$ to each point on the circle.
An example of noncomplete space with this property can be constructed by starting with a long circle and applying the following construction countably many times: Given a geodesic space $X$ construct a space $X'$ where to each pair of points $p,q\in X$ such that $|p-q|_X>1$ we add a unit segment with ends in $p$ and $q$.
This question was asked at MSE: https://math.stackexchange.com/questions/482311/on-continuously-uniquely-geodesic-space-ii?noredirect=1&lq=1, but the discussion was inconclusive.
Yes, such examples do live. The following answer was given in "Metric spaces of non-positive curvature" by Bridson and Haefliger; thanks to GGT and Moishe Kohan.
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2025-03-21T14:48:30.158848
| 2020-03-27T17:26:26 |
355888
|
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|
Stack Exchange
|
Reference request for certain kinds of FC-groups
This reference request has a general part and a specific part. In both cases, it feels like I could make faster progress in searching the literature or reconstructing arguments, if I knew the correct terminology to search for or the correct papers/books to use as jumping-off points. (I am not a group theorist, geometric or otherwise, and my own undergraduate education in group theory largely confined itself to the finite setting.)$\newcommand{\cF}{{\mathcal F}}$
Some current work my collaborators and I are doing on Fourier algebras of virtually abelian groups has led me, while playing with various toy examples, to consider the family $\cF$ of finitely generated groups in which the centre has finite index. By old results of B. H. Neumann (with precursors in work of Schur and Baer, apparently), the family $\cF$ admits various alternative characterizations:
Theorem (various, but mostly Baer and B. H. Neumann + follklore?)
Let $G$ be an infinite, finitely generated group. The following are equivalent:
1) $Z(G)$ has finite index in $G$;
2) There is a free abelian, finite-rank subgroup $N\leq Z(G)$ which has finite index in $G$;
3) There is an injective homomorphism $G\to {\mathbb Z}^d \times F$ for some $d\in {\mathbb N}$ and some finite group $F$;
4) The derived subgroup $G'$ is finite;
5) Every conjugacy class in $G$ is finite.
(Condition 1 is sometimes known as FZ; Condition 4 is sometimes known as FD; and Condition 5 is usually known as FC. Note that most of these equivalences fail in at least one direction if we drop the condition of being finitely generated.)
It follows from Neumann's proof (Proc LMS 1951) that if $G\in \cF$ then the set $G_{\rm tor}$ of torsion elements forms a finite subgroup of $G$, which contains $G'$.
Q1. Has there been a systematic study of those $G\in\cF$ where $G_{\rm tor}$ is abelian (but $G$ is not)? Is there a name for this class of groups?
Having failed to look up examples of groups satisfying the condition in Q1, some playing around led me to the following example which must surely be known in a more general/systematic context:
Example. Let $S= \{\pm 1\}\times {\mathbb Z}^2$ and consider the following bijections on $S$:
$$ a: ( \pm 1, x, y ) \mapsto ( \pm (-1)^y , x+1, y) $$
$$ b: (\pm 1, x,y) \mapsto (\pm 1, x, y+1) $$
$$ t: (\pm 1, x,y) \mapsto (\mp 1, x,y) $$
Informally $b$ is just vertical translation, while $a$ acts as horizontal translation on even rows and as a "glide reflection" on odd rows.
Clearly $t$ commutes with both $a$ and $b$, while some elementary calculations show that $[a,b]=t$. It follows that if we let $G$ be the group generated by $a$ and $b$, then $G$ is finitely generated with derived subgroup $\langle t \rangle$, and $G/G'$ is free abelian of rank $2$. Some more calculations indicate that $G_{\rm tor}=\langle t \rangle$ and so $G$ has no nonabelian finite subgroups. We also have $Z(G) = \langle a^2, b^2, t\rangle$ and so $G/Z(G)$ has order $4$, hence is also abelian.
Q2. Is this group $G$ a particular case of some well-known family of 2-step nilpotent groups? More specifically, have groups with the presentation
$$
G_n = \langle a, b \mid [a,b]^n= [a,[a,b] ] = [b, [a,b] ] = e\rangle
$$
been studied?
(EDIT: blunder/typo in the original presentation pointed out by Mark Sapir; now fixed.)
(UPDATE: Yves Cornulier has pointed out in comments that the groups in Q2 are $H_3({\mathbb Z})$ modulo the appropriate finite-index subgroup of the centre, so Q2 is answered to all intents and purposes; I'm leaving it up here for the record.)
Q1: it's easy to produce non-abelian f.g. FC-groups whose torsion subgroup is abelian. Maybe the simplest is to take $C_3\rtimes\mathbf{Z}$ where the generator of $\mathbf{Z}$ acts by inversion on the cyclic group $C_3$. In general, the groups $C_n\rtimes_m\mathbf{Z}$ (where the positive generator of $\mathbf{Z}$ acts by multiplication by $m$ coprime to $n$) are well-known: they provide examples of non-isomorphic residually finite groups with isomorphic profinite completion, for instance for $(n,m)=(11,4)$ vs $(n,m)=(11,9)$.
@MarkSapir Quite right; I shall correct the blunder/typo. I meant that the commutator of the generators is central and has order $n$
@YCor Thanks, that's a good point; for some reason I was hung up on trying to get ${\bf Z}^2$ as a quotient. Well the examples you point out will also serve our purposes well. I'd still be interested to know more about the groups in Q2 (now that the typo has been corrected)
This is the Heisenberg group over $\mathbf{Z}$ modulo the index $n$ subgroup of its center.
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2025-03-21T14:48:30.159159
| 2020-03-27T17:51:18 |
355891
|
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|
Stack Exchange
|
Simplicial simple homotopy vs. cellular simple homotopy
I recently started reading up on simple homotopy theory. Here is a question I stumbled upon.
In his 1938 Paper Simplicial Spaces, Nuclei and m-Groups Whitehead introduced the notion of elementary expansions and elementary collapses of simplicial complexes. Essentially a simplicial complex $K'$ is obtained from $K$ through an elementary collapse, by removing two simplices $p$ and $q$ from $K$ such that:
$p$ is a maximal simplex of $K$.
$q < p$ is a (maximal, proper) free face of $p$ (i.e. not contained in any other simplex but itself and $p$).
An elementary expansion is the obvious inverse operation. There are topological realizations of these operations, with the collapse given by "pushing the free face onto the others". Lets call a map of simplicial complexes a simplicial simple homotopy equivalence, if it is homotopic to the composition of such maps.
Later on he decided that CW-complexes were a more convenient setting to work in (Simple Homotopy Types), and this was where he developed the now famous results on whitehead torsion. This is the setting that most of the following material (such as Cohens Book A course in simple homotopy theory) are presented in and probably most familiar to most topologists. By a cellular simple homotopy equivalence, I mean a map of CW-complexes as in 3.
It seems to me, that it is somewhat folklore knowledge, that the second setting is a generalization of the prior in the following sense:
Let $K$,$K'$ be abstract simplicial complexes and $f:|K| \to |K'|$ a map between their realizations. Then $f$ is a simplicial simple homotopy equivalence, if and only if it is a cellular simple homotopy equivalence, with respect to the obvious $CW$-structures on $|K|$ and $|K'|$.
Is this even true?
I imagined this would be a simple consequence of the simplicial approximation theorem, but couldn't figure out an easy proof. I also skimmed most of the papers from the time period I could find on the matter, but didn't really get a satisfying answer.
If yes, I would be really thankful for a reference, or a sketch of a proof.
Kamps-Porter wrote a book that covers the simplicial perspective on simple homotopy theory. I gave it to a friend of mine, so I don't have it at hand, but it might be helpful to look into.
I managed to get my hands on a copy. As far as I can tell, there is no direct answer to my question in there. However I find the abstract homotopy perspective on simple homotopies very enlighting, so this might prove very helpful in the long run. Thanks!
Turns out I should have read the original material Simplicial Spaces, Nuclei and m-groups a little more carefully. It was in there all along. I'm still supprised nobody ever explicitly stated this though. But I guess it was such common knowledge at the time that nobody bothered.
My question can be rephrased in the following way. For a simplicial complex $K$ (all complexes are taken to be finite) denote by $E_S(L)$ the set of inclusions of simplicial complexes $L \to K$ that are homotopy equivalences, modulo the equivalence relation generated by $L\to K \xrightarrow{s} K' = L\to K' \implies L \to K \sim L \to K'$ for $s$ a composition of (simplicial) elementary expansions (this actually has set size). This is the description Siebenmann chooses in Infinite Simple Homotopy Types. Further denote by $E_C(K)$ the obvious analogon in the CW setting. The latter of course is just the underlying set of the geometric description on the Whiteheadgroup $WH(X)$ as it is constructed in Cohens simple Homotopy Theory for example. As both equivalence relations identify homotopic maps and using the standard mapping cylinder arguments, my question rephrases to:
What is the Kernel of the obvious forgetful map $E_S(L) \to E_C(L) = WH(L) \cong WH( \pi_1(L))$ for $L$ connected. Here I mean kernel in the sense of pointed sets, as the lefthandside will only a posteriori be a group (one could of course bother with proving its a group first..).
The way Whitehead proved in Simple Homotopy Types that the last isomorphism is injective, is effectively by transforming each inclusion $L \to K$ to the form $ L \to K' = L \cup \bigcup e_i^n \bigcup e_i^{n+1}$ through cellular simple expansions and collapses $(n \geq 1)$. The corresponding element in $WH( \pi_1(L))$ is then given by $ \pi_{n+1}( K', K'^{n} \cup L) \to \pi_n(K'^n \cup L, L)$. Explicitly one checks using Hurewicz theorem and universal coverings that this is in fact an isomorphism of free $\mathbb Z (\pi_1(L))$ modules, with basis given by the cells $e_i$. One then shows, that all the elementary matrix operations making the corresponding matrix trivial on the $WH(\pi_1(L))$ side have an analogue on the simple homotopy side, proving injectivity.
What was known to me when I asked the question, was that this works in the cellular category, i.e. with cellular simple equivalences. It turns out, before passing to the cellular setting, Whitehead did the analogous proof in the simplicial world, which turns out to require a lot more technical arguments, but it is completely done in 1 and the statement is implicitly proven in the proof of theorem 20 there.
Very roughly, one first proves that subdivisions are simplicial simple equivalences, so that one can work in the p.l. category instead. Here one has the p.l. mapping cylinder see 1). The attachment of a p.l. cell along a p.l. boundary map, is then to be understood as taking the cylinder along the boundary and then gluing the cell on top of it. (I guess this circumvents problems with pushouts in the p.l. category). He then shows that the same simplicifactions as in the cellular setting are valid in the p.l. setting using such cylinders and simplicial approximation, thus, showing that $E_S(L) \to WH(\pi_1(K))$ and hence $E_S(L) \to E_C(L)$ has trivial kernel.
In particular this gives a positive answer to my original question.
|
2025-03-21T14:48:30.159553
| 2020-03-27T21:12:48 |
355895
|
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|
Stack Exchange
|
Normalized $p(x)-\mathrm{laplacian}$ is uniformly elliptic?
The normalized $p(x)-\mathrm{laplacian}$ is defined by
$$-\Delta_{p(x)}^{N} u = -\operatorname{tr}\Big( \big( I + \frac{(p(x)-2)}{|Du|^{2}}Du \otimes Du\big)D^{2}u\Big)=0 ,$$ from now on, two hypotheses are necessary the function $p$ is Lipschitz and $\inf p >1$. Under these conditions, can there be any hope that this operator will be uniformly elliptical? If the answer is negative, there is some condition on $ p $ so that this operator is uniformly elliptical in addition to imposing $p (x) = \mathrm{constant}$?
OBS: We work in a bounded domain in $\mathbb{R}^{n}$.
well, it depends on your definition of "uniformly elliptic" for nonlinear operators...
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2025-03-21T14:48:30.159633
| 2020-03-27T21:26:59 |
355896
|
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|
Stack Exchange
|
Random variable corresponding to sum of density functions
The distribution of functions of random variables is well-studied for various different and general cases, but I didn't find much result for the reverse.
Suppose that $X_1, X_2$ are (probably independent) random variables and we have $X_1\sim f(x_1)$ and $X_2\sim g(x_2)$. Construct a new distribution for the new random variable $X$ as follows:
$$X\sim p(x)=\frac{1}{\kappa}(f(x)+g(x))$$
where $\kappa$ is a normalization factor such that the function is a density function.
Is it possible to express $X$ as a function of $X_1$ and $X_2$ in a way that their PDF is $p(x)$?
More formally, find a function $G$ such that
$$X=G(X_1, X_2)\sim p(x)$$
Note: "$\sim$" denotes the probability density function of a random variable.
Note 2: If we had $p(x)=f(x)*g(x)$, where $*$ is the sign for convolution, then I think we may write $G(X_1, X_2) = X_1+X_2$.
Note 3: I have asked the same question here for product of density functions.
This is not really a research level question; I'm voting to migrate to http://math.stackexchange.com.
I assume $X_1 \sim f(x_1)$ means that that the distribution of $X_1$ has density function $f$.
Note first that your $\kappa$ can only be $2$, otherwise the integral of $p$ will not equal $1$. Then $p$ is the density of a mixture of $X_1, X_2$; it corresponds to flipping a coin to decide whether to take $X_1$ or $X_2$. In notation, if $A$ is an event of probability $1/2$ that is independent of $(X_1, X_2)$, then $p$ is the density of the random variable $X = 1_A X_1 + 1_{A^c} X_2$. You can also think of it as $X = Z X_1+ (1-Z) X_2$ where $Z$ is a Bernoulli(1/2) random variable independent of $(X_1, X_2)$.
In general such a random variable cannot be written as a function of $X_1, X_2$ alone.
Thank you very much for the answer. So, it is not possible to write $1_A$ in terms of $(X_1, X_2)$?
@SMA.D: No, because as I said, $A$ must be independent of $(X_1, X_2)$. If it could be written as a function of $(X_1, X_2)$, it would be independent of itself and therefore could only have probability 0 or 1.
|
2025-03-21T14:48:30.159822
| 2020-03-27T21:35:59 |
355898
|
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"Simon Henry",
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|
Stack Exchange
|
Is the inclusion functor from gaunt strict $n$-categories to weak $(\infty,n)$-categories fully faithful?
I'm now second-guessing an assertion I made here so let me ask it as a question.
Let $Cat_n$ be the 1-category of strict $n$-categories;
Let $\widetilde{Cat_n}$ be the $(\infty,1)$-category obtained from $Cat_n$ by localizing at the weak equivalences (as presented by the folk model structure);
Let $Gaunt_n \subseteq Cat_n$ be the fully faithful inclusion of the gaunt $n$-categories, i.e. those strict $n$-categories where every equivalence is an identity.
Let $\widetilde{Cat_n} \to Cat_{(\infty,n)}$ be the "inclusion" functor (which is probably not fully faithful for $n \geq 3$).
Questions:
Is the composite functor $Gaunt_n \to Cat_n \to \widetilde{Cat_n}$ a fully faithful functor of $(\infty,1)$-categories?
Is the composite "inclusion" functor $Gaunt_n \to Cat_n \to \widetilde{Cat_n} \to Cat_{(\infty,n)}$ a fully faithful functor of $(\infty,1)$-categories?
I'm pretty sure the answer to (1) is yes, and intuitively the answer to (2) should also be yes, but I'm not quite sure.
Barwick and Schommer-Pries do show that if we further restrict to the inclusion $\Theta_n \to Cat_{(\infty,n)}$, or even the slightly larger $\Upsilon_n \to Cat_{(\infty,n)}$, then we get a fully faithful $(\infty,1)$-functor. But I'm not sure about all gaunt $n$-categories.
This is easy to see using Rezk's $\Theta_n$-space model for $(\infty,n)$-categories. The category of gaunt n-categories is equivalent (via the cellular nerve functor) to the full (simplicially enriched) subcategory of Rezk $\Theta_n$-spaces spanned by the "discrete-valued" ones.
@AlexanderCampbell Thanks. I think what confuses me now, as discussed with Simon below, is the fact that the cellular nerve doesn't generally produce a functor $Cat_n \to Cat_{(\infty,n)}$ unless Rezk completion is applied, which makes the situation a bit murky. Do you have any insight on how to produce the "correct" functor $Cat_n \to Cat_{(\infty,n)}$ and check that it agrees with the cellular nerve on $Gaunt_n$?
A reasonable guess (which we know works for $n=2$) of a right Quillen functor from $n$-categories to $n$-quasi-categories is the nerve/singular functor induced by a Reedy cofibrant replacement of the full inclusion $\Theta_n \to \mathbf{Cat}_n$. Compose this with Ara's right Quillen functor from $n$-quasi-categories to Rezk $\Theta_n$-spaces to get a right Quillen "classifying diagram" functor. Since gaunt $n$-categories "see" weak equivalences as isomorphisms, this agrees on gaunt $n$-categories with the full inclusion I mentioned above.
@AlexanderCampbell That does sound like a reasonable guess. I wonder how it compares in general to taking the Rezk completion of the cellular nerve as Simon suggests below... I also wonder if there's a model-independently characterization of the "correct" functor...
I will see an $(\infty,\infty)$-category as a functor $\Theta^{op} \to \text{Space}$ that satisfies the usual Segal condition, i.e., i.e. preserve the pushouts encoding the various type of compositions (the globular sum), and the Rezk completeness condition at all level (the map from the space of $n$-cell to the space of "invertible $n+1$-cells" defined in the appropriate way is an equivalence). These naturally form an $(\infty,1)$-category.
This corresponds to an "inductive" (by opposition to coinductive style) definition of weak $(\infty,\infty)$-category, but if you add the assumption that every cell of dimension $>n$ is invertible, then you recover something equivalent to other classical definition of weak $(\infty,n)$-categories. You can also do everything with $\Theta_n$ directly to avoid this.
I claim that $\text{gaunt}$, as a $1$-category identifies with the full subcategory of these functor $\Theta^{op} \to \text{Space}$ as above that takes values in discrete spaces.
Indeed, a functor $\Theta^{op} \to \text{Set}$ satisfiying the Segal conditions is the same as a strict $\infty$-category by the classical $\Theta$-nerve theorem.
If you unfold what the Rezk completeness condition means in this special case it exactly means that every isomorphism in the category (in the strict sense) is an identity, hence that your category is Gaunt.
$\text{gaunt}_n$ corresponds to these that are further more $n$-categories: if your category is Gaunt and all cell of dimension $>n$ are invertible, then it only has identity cell of dimenion $>n$, so it is a $n$-category.
Thanks! I'm still a bit confused. (1) This gives fully-faithfulness at the 1-categorical level, but how do I know that I still have fully-faithfulness after deriving these functors? (2) You seem to be identifying a strict $n$-category with its cellular nerve as Alex does above -- but in general isn't this the wrong functor? It carries non-gaunt $n$-categories to non-complete $\Theta_n$-spaces. This shouldn't be a problem because we're restricting to gaunt $n$-categories in the end, but it seems to me that we need to set up a model structure on $Gaunt_n$ or something to derive this functor...
@TimCampion : For (1), when I talk about $\infty$-categories as functor $\Theta^{op} \to$ Space, I mean by "Space" the $(\infty,1)$-category of spaces. So there is no $1$-categorical level in what I say.
Regarding (2) I have to admit I didn't think about it. In this point of view, isn't the "right functor" taking the cellular nerve of a strict $\infty$-category (to get a Cellular set seen as a discrete cellular space) and then taking the Rezk completion ? if so then Gaunt object are already Rezk complete, so Rezk completion does nothing.
I'm not 100% sure this is the "right functor", since it's the composite of a right adjoint and a left adjoint, so it's not clear to me that the resulting functor is a right adjoint. I'd be most comfortable constructing the "right functor" as something like Rezk's classifying diagram. Probably this is still equivalent to the cellular nerve for gaunt $n$-categories, but there's something to check.
|
2025-03-21T14:48:30.160490
| 2020-03-27T22:45:18 |
355910
|
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|
Stack Exchange
|
Spectral decomposition of a $4\times4$ real nonsymmetric matrix with unknown elements
I'm trying to eigendecompose the following matrix $A$, i.e. to find $Q$ and $\Lambda$ such that
$$
A = \begin{bmatrix}
-\alpha & \alpha & -\gamma^{-1} & 0\\
\beta & -\beta & 0 & -\gamma^{-1}\\
-1 & 0 & \alpha & -\beta\\
0 & -1 & -\alpha & \beta
\end{bmatrix}=Q\Lambda Q^{-1}
$$
where $\alpha>\beta>0$ and $\gamma>0$.
Notice that
$A$ is a Hamiltonian matrix, i.e. $JA$ is symmetric where $J=\begin{pmatrix}0 & I_2 \\ -I_2 & 0\end{pmatrix}$, $I_2$ is the $2\times2$ identity matrix.
The characteristic polynomial of a real Hamiltonian matrix is even. Thus, if $λ$ is an eigenvalue of $A$, then $−λ$, $\bar λ$ and $−\bar λ$ are also eigenvalues. It follows that $\text{trace} A=0$.
$A$ can be written in block notation: $A=\begin{pmatrix}B & -\gamma^{-1}I_2 \\ -I_2 & -B^T \end{pmatrix}$, with $B=\begin{pmatrix}-\alpha & \alpha \\ \beta & -\beta\end{pmatrix}$.
the elements on the antidiagonal are all $0$.
sum of row $1$, sum of row $2$, sum of column $3$ and sum of column $4$ are all equal to $-\gamma^{-1}$.
sum of row $3$ and sum of column $2$ are equal to $\alpha-\beta-1$.
sum of row $4$ and sum of column $1$ are equal to $\beta-\alpha-1$.
can we use these facts to find $Q$ and $\Lambda$?
Moreover, the characteristic polynomial is
$$
p_A(\lambda) = \lambda^4-[(\alpha+\beta)^2+2\gamma^{-1}]\lambda^2+\gamma^{-1}(2\alpha^2+2\beta^2+\gamma^{-1})
$$
hence
$$
2\lambda^2 = (\alpha+\beta)^2+2\gamma^{-1} \pm \sqrt{(\alpha+\beta)^4-4\gamma^{-1}(\alpha-\beta)^2}
$$
and the eigenvalues are either real or complex depending on the values of the parameters.
I don't know how explicit or simple you want the expressions for $\Lambda$ and $Q$ to be, but here is a description. Using your notations, we write $A$ as
$\begin{bmatrix}
B&-\gamma^{-1}{\rm{I}}_2\\
-{\rm{I}}_2&-B^{\rm{T}}
\end{bmatrix}$
where
$B=\begin{bmatrix}
-\alpha&\alpha\\
\beta&-\beta
\end{bmatrix}$.
Let $\lambda$ be an eigenvalue and
$\begin{bmatrix}
\mathbf{v}_{2\times 1}\\
\mathbf{w}_{2\times 1}
\end{bmatrix}$
an eigenvector for it. We have
$$
\begin{bmatrix}
B&-\gamma^{-1}{\rm{I}}_2\\
-{\rm{I}}_2&-B^{\rm{T}}
\end{bmatrix}
\begin{bmatrix}
\mathbf{v}\\
\mathbf{w}
\end{bmatrix}
=\lambda
\begin{bmatrix}
\mathbf{v}\\
\mathbf{w}
\end{bmatrix}\Rightarrow
\begin{cases}
B\mathbf{v}-\gamma^{-1} \mathbf{w}=\lambda\mathbf{v}\\
-\mathbf{v}-B^{\rm{T}}\mathbf{w}=\lambda\mathbf{w}
\end{cases}
$$
Solving the second equation for $\mathbf{v}$ and substituting in the first equation yields
$$
\mathbf{v}=-\left(B^{\rm{T}}+\lambda{\rm{I}_2}\right)\mathbf{w},\,\quad
\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)\mathbf{w}=\mathbf{0}.
$$
Thus the eigenvalues are the roots of the quartic
$$
{\rm{det}}\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)=0.
$$
Notice that changing $\lambda$ to $-\lambda$ changes the matrix
$BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}$ to its transpose. Hence (as you mentioned) if $\lambda$ is an eigenvalue, so is $-\lambda$. Writing the roots of the quartic as $\lambda_1,-\lambda_1$ and $\lambda_2,-\lambda_2$, for each $\lambda_i$ the $2\times 2$ matrix
$BB^{\rm{T}}+\lambda_i(B-B^{\rm{T}})+(\gamma^{-1}-\lambda_i^2)\,{\rm{I}_2}$
is not full rank and thus there are non-zero column vectors $\mathbf{w}_{ir}$ and $\mathbf{w}_{ic}$ ($i\in\{1,2\}$) with
$$
\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)\mathbf{w}_{ic}=\mathbf{0}\quad
\mathbf{w}_{ir}^{\rm{T}}\left(BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2)\,{\rm{I}_2}\right)=\mathbf{0}
$$
Plugging into previous equations, the corresponding eigenvalues could be computed. The result is
$$
Q=\begin{bmatrix}
-\left(B^{\rm{T}}+\lambda_1{\rm{I}_2}\right)\mathbf{w}_{1c}&
-\left(B^{\rm{T}}-\lambda_1{\rm{I}_2}\right)\mathbf{w}_{1r}&-\left(B^{\rm{T}}+\lambda_2{\rm{I}_2}\right)\mathbf{w}_{2c}&
-\left(B^{\rm{T}}-\lambda_2{\rm{I}_2}\right)\mathbf{w}_{2r}\\
\mathbf{w}_{1c}&\mathbf{w}_{1r}&\mathbf{w}_{2c}&\mathbf{w}_{2r}
\end{bmatrix}
$$
which satisfies
$$
A=Q
\begin{bmatrix}
\lambda_1& & &\\
&-\lambda_1& &\\
& & \lambda_2 & \\
& & & -\lambda_2
\end{bmatrix}
Q^{-1}.
$$
Thank you very much! Could you explain why the matrix $M = BB^{\rm{T}}+\lambda_i(B-B^{\rm{T}})+(\gamma^{-1}-\lambda_i^2),{\rm{I}2}$ is of rank one? And why do you consider the two equations $M w{ic} = 0$ and $w_{ir}^T M = 0$?
@soundwave I edited my answer; I want $BB^{\rm{T}}+\lambda(B-B^{\rm{T}})+(\gamma^{-1}-\lambda^2),{\rm{I}_2}$ to be singular, so the rank is $0$ or $1$. The $\lambda_i$'s are defined to be the roots of the quartic equation given by the vanishing of the determinant. Once you have such a $\lambda_i$, you use my computations of eigenvectors in the beginning of the solution to express the upper part $\mathbf{v}_i$ in terms of the lower part $\mathbf{w}_i$ that should be a non-zero (column) vector in the null space of $BB^{\rm{T}}+\lambda_i(B-B^{\rm{T}})+(\gamma^{-1}-\lambda_i^2),{\rm{I}_2}$.
Finally, notice that $-\lambda_i$ is also among the eigenvalues, and substituting $\lambda_i$ with its opposite in $BB^{\rm{T}}+\lambda_i(B-B^{\rm{T}})+(\gamma^{-1}-\lambda_i^2),{\rm{I}2}$ results in the transpose matrix. A vector in the column null space of the transpose is the transpose of a vector in the row null space of the original matrix. That's where $\mathbf{w}{ic}$ and $\mathbf{w}_{ir}^{\rm{T}}$ come from.
Thank you for the explanation! With "column null space" and "row null space" do you mean, respectively, the cokernel and the kernel? I'm reading that the kernel of a matrix $A$ coincides with the cokernel of $A^\rm T$, and vice versa the cokernel of $A$ coincides with the kernel of $A^\rm T$. But wouldn't this mean that $w_{ic} = w_{ir}$? I'm a bit confused since I don't understand very well.
@soundwave By the "column null space" I mean the kernel as if you think of the matrix as a linear map; i.e. the subspace of vectors $\mathbf{w}$ with $A\mathbf{w}=\mathbf{0}$. The "row null space" is defined by
${\mathbf{w}|\mathbf{w}^{\rm{T}}A=\mathbf{0}}$. What I do is very concrete, I'm picking vectors with these properties. As for the linear algebra fact you mentioned, you must be careful about what you mean by "coincide". Do you mean an isomorphism? In that case you don't get the same spaces, but spaces that may be identified. And again, there is no reason to appeal to such a thing.
|
2025-03-21T14:48:30.160999
| 2020-03-27T22:47:52 |
355911
|
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|
Stack Exchange
|
Random walks on infinite directed regular graphs
Let us consider a directed graph $\Gamma=(V,E,s,t)$ ($V$ set of vertices, $E$ set of edges, $s,t: E \rightarrow V$ are the "source" and "target" maps).
Assume that $\Gamma$ is bi-regular, that is there are two integers $d_1 \geq 1$ and $d_2 \geq 1$ such that $|s^{-1}(v)|=d_1$ and $|t^{-1}(v)|=d_2$ for every $v \in D$. Assume that $\Gamma$ weakly connected (there is an undirected path relying any two vertices).
And finally assume that $\Gamma$ has no forward-closed finite subset (i.e. if $S$ is a finite subset of $V$, there is an edge in $E$ with source in $S$ and target not in $S$). In particular $\Gamma$ is infinite.
Consider the standard discrete-time forward random walk on $\Gamma$, starting at $x \in V$: at time $0$ you are at $x$ with probability $1$; for any $n \geq 0$, at time $n+1$, conditional to being at $y$ at time $n$, your odds of being at any of the $d_1$ forward-neighbors of $y$ (i.e. $t(e)$ for $e \in E$, $s(e)=y$) is $1/d_1$.
Let $p^n_{x,y}$ be the probability of being at $y$ at time $n$.
Is it true that for every $y$, $p_n(x,y) \rightarrow 0$ as $n \rightarrow \infty$?
Here are some remarks. This question is related to Fedja's beautiful answer to this question. Fedja proves the result when $\Gamma$ is an undirected regular graph (seen as an undirected graph by replacing each undirected edge by two directed edges going both way). Unfortunately, I have not been able to extend his argument to my directed case.
The hypothesis that $\Gamma$ has no forward-closed finite subset is certainly necessary: If $S$ was such a subset, and $x \in S$, then you would be sure to stay in the finite set $S$ forever, so $\sum_{y \in S} {p^n_{x,y}} = 1$ and one of these $p^n_{x,y}$ at least can not tend to $0$.
The hypothesis that $|s^{-1}(v)|=d_1$ (or at least $s^{-1}(v)$ finite) for all $v$ is necessary to define the random walk, but that $|t^{-1}(v)|=d_2$ (or at least is finite) for all $v$ is also necessary for the theorem to be true. Without it, consider the graph with $V=\mathbb Z$, and for every $a \in \mathbb Z$, there is one directed edge from $a$ to $a+1$ and one directed edge from $a$ to $0$ (called the "speedy return" edge). Thus $|s^{-1}(a)|=2$ for every $a \in \mathbb Z$, but $|t^{-1}(0)|=\infty$. The probability $p^n_{0,0}$ is $\geq 1/2$ since every path that ends up with the "speedy return edge" goes from $0$ to $0$.
Here is a counterexample:
Let $G_1$ be the digraph with vertex set $\mathbb N$, two loops at $0$, an edge from $0$ to $1$, and for every $i \geq 1$ an edge from $i$ to $(i+1)$ and two parallel edges from $i$ to $(i-1)$. Let $G_2$ be any countable digraph in which every vertex has $2$ outgoing edges and $4$ incoming edges, and let $f \colon V(G_2) \to V(G_1)$ be such that $|f^{-1}(0)| = 0$ and $|f^{-1}(i)| = 1$ for every $i \geq 1$.
Let $G$ be the digraph obtained from $G_1 \uplus G_2$ by adding all edges from $v$ to $f(v)$. Then $G$ is bi-regular with $d_1 = 3$ and $d_2 = 4$. Moreover, $G$ is weakly connected and has no finite forward closed sets since every vertex has a forward edge connecting it to the forward ray in $G_1$.
The simple random walk on $G$ almost surely enters $G_1$ after finitely many steps (and remains in $G_1$ thereafter since there are no edges from $G_1$ to $G_2$). But the simple random walk on $G_1$ is just a biased random walk on $\mathbb N$ with bias towards $0$. This random walk is irreducible, aperiodic (because of the loops at $0$), and positive recurrent. Thus $\lim_{n \to \infty} p_n(x, i) = \mu(i)$ independently of the starting point $x$, where $\mu \neq 0$ is the invariant probability measure on $\mathbb N$.
|
2025-03-21T14:48:30.161247
| 2020-03-27T22:52:17 |
355912
|
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"url": "https://mathoverflow.net/questions/355912"
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|
Stack Exchange
|
Connections between spectral geometry and critical point/Morse theory
I am researching electrostatic knot theory, which is essentially the theory of harmonic functions on knot complements. I want to understand the number of critical points of the electric potential, their indices, and their placement. I also want to understand how these phenomena change under knot isotopy.
In particular, hyperbolic knots have a unique complete metric of constant curvature -1, and thus possess a well-defined Laplacian operator. There are a number of papers discussing the spectral theory of the Laplacian for hyperbolic knots, like Futer, Kalfagianni, Purcell (2009). Harmonic functions (AKA the kernel of the Laplacian) also have special critical point sets, notably critical points of harmonic functions are all saddle nodes. However, after some searching, I couldn't find anything that relates the critical sets of harmonic functions to the spectrum of the Laplacian, even outside the context of knot theory. I suspect this could be a potentially fruitful path to follow, but does anyone here have any suggestions of where to look?
|
2025-03-21T14:48:30.161358
| 2020-03-27T23:21:07 |
355913
|
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"Steven Clark",
"Sylvain JULIEN",
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"joriki"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355913"
}
|
Stack Exchange
|
What can this $\int_{0}^{t} (\pi(x)-Li(x)) dx$ tell us about primes distribution?
Many papers I have read which are related to primes distribution only it discussed sign and refinement Bounds of $\pi(x)-Li (x)$ with $\pi(x)$ is a prime counting function and $Li (x)$ is the logarithm integral $x$ , I computed $\int_{0}^{1} (\pi(x)-Li (x)) $ I have got $\log 2$, this result attracted my attention to the behavior of $\int_{0}^{t} (\pi(x)-Li (x))dx $ , such that i can't juge weither it is divergent or convergent integral , But my question here is:
Question: What can this $\int_{0}^{t} ( \pi(x)-Li (x)) $ tell us about primes distribution ?
Note that you merely computed the integral of $\operatorname{Li}(x)$, as $\pi(x)=0$ on $[0,1]$.
Up to the sign.
Apart from that, I doubt that the integral converges, as Littlewood proved the difference $\pi(x)-Li(x)$ changes sign infinitely often. I think he also proved this difference is not bounded.
What do you mean you can't tell whether the integral $\int_0^t(\pi(x)-\text{li}(x)),dx$ converges or diverges? Do you really mean whether $\underset{t\to\infty}{\text{lim}}\left(\int_0^t(\pi(x)-\text{li}(x)),dx\right)$ converges or diverges?
The integral does not converge. See https://academic.oup.com/blms/article-abstract/31/4/424/277640?redirectedFrom=fulltext.
On the other hand, a proof that this integral is less than $K.t^{\alpha}$ almost everywhere (i.e. on a set of the form $(t_{0},\infty)\setminus J$ where $J$ has Lebesgue measure $0$) for some $\alpha<2$ and an absolute constant $K$ would imply that there exists some $1/2\leq\sigma_{0}<1$ such that $\zeta$ does not vanish in the right half plane $\Re(s)>\sigma_{0}$, which quantifies the "randomness" of the distribution of the primes: the less $\sigma_{0}$, the less random this distribution.
|
2025-03-21T14:48:30.161519
| 2020-03-27T23:43:28 |
355914
|
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"authors": [
"Emil Jeřábek",
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|
Stack Exchange
|
A countable expression expressible in $\mathrm{FO}_3$ with only one binary predicate?
Transitivity on a set is defined as follows
$$\forall x \forall y\forall z( T(x,y) \land T(y,z) \rightarrow T(x,z))$$
Now if we wanted to count total number transitive relations which are defined on an arbitrary set, then we don't have a closed-form expression to do so. The sequence is given here https://oeis.org/A006905
My question is whether you know of any non-trivial expression, which requires at least three variables and one binary predicate and only uses universal quantifiers to be expressed (i.e requires $\mathrm{FO}_3$ with one binary predicate and only universal quantifiers), which has a closed-form counting expression. i.e a generator function that takes the cardinality of domain elements and maps them to the model count of the $\mathrm{FO}_3$ expression.
Through this investigation, I want to understand if there is a link between the countability of a certain process with its expressibility in first-order logic.
Example of what it means that a closed-form counting expression exists.
$$\forall_x\forall_y \mathsf{Smokes}(x) \land \mathsf{Friend}(x,y) \rightarrow S(y)$$
If we have $n$ individuals, then the number of models of this expression is given below. Notice this entire expression is expressed in $\mathrm{FO}_2$
$$a(n) = (3^{n}+4^{n})^{n}$$
Source: https://www.ijcai.org/Proceedings/16/Papers/607.pdf
Also asked at CSThE.
What is FO3/FO2/FOL/F03/F02? doesn't occur in the linked paper.
@YCor $\mathrm{FO}_k$ is the fragment of first-order logic consisting of formulas that use at most $k$ distinct variables. (They may be reused.)
Alright, I found a simple answer. The number of equivalence relations on a set is given as $\sum_{k=0}^{n} S(n, k)$. But I would still appreciate interesting answers. Also, I would be happy to know if there is some theory that tells us whether a closed-form for a sequence would exist or not.
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2025-03-21T14:48:30.161674
| 2020-03-27T23:51:05 |
355916
|
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|
Stack Exchange
|
Chow variety of 1-cycles on abelian surface
It is an easy exercise to show that on a K3 surface, a smooth genus $g$ curve moves in a $g$-dimensional linear system. Nearly the same exercise shows that on an abelian surface, the corresponding linear system is dimension $g-2$. But you now also have a $2$-dimensional space of translations. I'm wondering how these translations combine with the linear system to give the Chow variety of $1$-cycles.
For concreteness, let $A$ be an abelian surface over $\mathbb{C}$. Fix an effective line bundle $L$ and corresponding linear system $|L| \cong \mathbb{P}^{g-2}$, all of whose elements are curves in the class $\beta \in H_{2}(A, \mathbb{Z})$. Let $\text{Chow}_{\beta}(A)$ be the variety parameterizing algebraic 1-cycles with fixed homology class $\beta$. Is it true that
$$\text{Chow}_{\beta}(A) \cong |L| \times A,$$
where a tuple $(C, a)$ represents the curve $t_{a}^{*}C$? I'm thinking it can't quite be that simple because some curves in the linear system might be invariant under certain translations, right? For example, if a curve $C \in |L|$ in invariant under translation $t_{a}$ by $a \in A$, then the points $(C, 0)$ and $(C, a)$ would be identified in the product. Is the right answer a quotient of what I write in the equation above? Is there a concrete answer here?
I've been looking in the literature, and can't find anything on this. I'd appreciate any help, and also sources if possible!
The situation for divisors on abelian surfaces is fully understood. Read Mumford's Abelian varieties for the behaviour of line bundles w.r.t. translations.
For a curve $C$ Mumford introduces the subgroup $H(C)$ of $A$ consisting of all $a \in A$ such that translation by $a$ preserves $C$ (as a curve, not just its class). So is the idea that $\text{Chow}_{\beta}(A) = (|L| \times A)/\sim$ where $(C, a) \sim (C,b)$ if $a-b \in H(C)$?
It is not a product, but a projective bundle over $\operatorname{Pic}^{\beta }(A)\cong \hat{A} $, with fiber $|L|$ over $L\in \operatorname{Pic}^{\beta }(A)$.
|
2025-03-21T14:48:30.161823
| 2020-03-28T01:22:25 |
355920
|
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|
Stack Exchange
|
Square integrable borel probability measures on Euclidean spaces are the law of random variables from an atomless polish space
Could someone provide me with a reference or proof for the following: Let $(\Omega, \mathcal{A}, P)$ be an atomless probability space, with $\Omega$ a Polish space. Given $f$ a random vector on $(\Omega,\mathcal{A},P)$ denote by $L(f)$ its law, i.e. the induced measure on the Euclidean target space. Then, for any $k$, and any square integrable measure $\mu$ on $\mathbb{R}^{k}$, there exists a random vector $f$ on $(\Omega,\mathcal{A},P)$ with $L(f)=\mu$.
In Cardaliaguet's Notes on Mean Field Games, subsection 6.1 pg 43, he mentions the result as 'recall'. Meanwhile Keisler, Sun on Why saturated probability spaces are necessary, Lemma 2.1(ii), they mention something stronger as a well-known result.
The proof I like to remember goes as follows.
Since all uncountable Polish spaces are Borel isomorphic, it suffices to prove this for $\Omega = \mathbb{R}$ and $k=1$, so that $P, \mu$ are both measures on $\mathbb{R}$, with $P$ atomless. Let $F,G : \mathbb{R} \to [0,1]$ be their respective cumulative distribution functions. Then the pushforward $F_* P$ is Lebesgue measure $\lambda$ on $[0,1]$ (easy exercise; show that $(F_* P)([0,a]) = a$). Now let $G^\leftarrow : [0,1] \to \mathbb{R}$ defined by $G^\leftarrow(t) = \sup\{ x : G(x) < t\}$ be the "inverse" of $G$. Show that $G^\leftarrow_* \lambda = \mu$ (also an exercise, or see Theorem 1.2.2 of Durrett, Probability: Theory and Examples). We conclude that $(G^{\leftarrow} \circ F)_* P = \mu$, which is to say that if we consider $G^{\leftarrow} \circ F$ as a random variable on $(\mathbb{R}, P)$, its law is $\mu$.
The square integrability is not needed, and this works equally well if $\Omega$ and $\mathbb{R}^k$ are replaced by any other Polish spaces.
|
2025-03-21T14:48:30.161981
| 2020-03-28T03:18:06 |
355924
|
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|
Stack Exchange
|
How to tell, roughly, which PDE's are interesting to analyse?
How can one tell which PDE's are, roughly speaking, perhaps more interesting to analyse?
Physical motivation is one reason. For example, the KdV $u_t+u_{xxx} - 6uu_x=0$ for a function $u:\mathbb R\times\mathbb R\to\mathbb R$ is a model for many physical systems, such as the propagation of shallow water waves along water bodies with low height. It is also a completely integrable PDE, which makes its analysis more tempting. But putting aside complete integrability (many studied PDE do not have completely integrable variants) why not study the derivative quasilinear equation $u_t + u_{xxx} - 6uu_{xxx}$ or the another derivative semilinear equation $u_t + u_{xxx} - 6uu_{xx}=0$ (I am only putting the KdV here as an example, and am not really asking this particular question for the KdV)?
Aside from physical motivation, how can one pick PDE to analyse?
Edit: Just to be clear, I am asking this from the perspective of PDE research. When I say "analyse," I mean "do PDE research on."
PDEs can come from other sources than physics: geometry , stochastic processes and economics are examples of such fields.
A PDE is interesting to analyze if mathematicians are interested in its analysis. What kinds of PDEs are other folks in your field thinking about, and why do they find them interesting?
“Interesting” PDEs often have more geometrical (e.g. integral) formulations, whose analysis can feel different from “write an arbitrary PDO and crank in the Strichartz estimates”.
I will echo the answer by Denis Serre. In mathematical, and related theoretical work, one often finds two different situations: "problems in search of solutions" and "solutions in search of problems". If you have some independent motivation for studying a PDE, that is your problem and you need to solve it, which is pretty self-explanatory. By your question, you're probably not in this first situation. On the other hand, if you set yourself the task to study a certain method or technique (a "solution"), then some PDEs naturally manifest as relevant "problems".
Dear Igor, actually, what I can see is that PDE research for beginners usually begins with a specific problem being chosen for them, and they learn tools to solve it and then move on to (usually only modestly) more difficult but somewhat similar specific problems. In this way they branch out a little bit, but how far they branch out depends on the individual. I was trying to get at understanding what's beyond this, or maybe how to pick my own problems.
David, in addition to all the answers here, if you have time to spend check out the book: Polyanin, Zaitsev - Handbook of Nonlinear Partial Differential Equations, 2008. You have a few thousand PDEs there and you maybe find some of them interesting to analyze.
My impression is that a PDE, or rather a class of PDEs, is interesting to the analysts when it is relevant for some analytical tool. Let me take a few examples. The list is not exhaustive.
linear constant coefficient PDEs in the whole space ${\mathbb R}^n$ are treated with Fourier analysis.
Elliptic (scalar) PDEs are analyzed with the maximum principle. This is particularly true in the modern treatment.
Linear and semi-linear evolution PDEs are the realm of semi-group theory.
Dispersive linear PDEs obey to Strichartz inequalities.
So-called integrable equations or systems are treated by Lax pairs and spectral theory.
Othere examples in homogenization or in hyperbolic conservation laws are solved by using compensated compactness.
Strichartz inequalities are also useful for semilinear problems (problems where the nonlinearity does not have derivatives), e.g. the nonlinear Schrodinger equation and nonlinear wave equation with power nonlinearities.
|
2025-03-21T14:48:30.162261
| 2020-03-28T04:19:29 |
355926
|
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|
Stack Exchange
|
Examples of invariant measures for Hamiltonian PDE
If $X$ is a symplectic space and $H$ is a Hamiltonian on $X$, then we have the non-normalized Gibbs measure $e^{-\lambda H}dm$ for any $\lambda\in\mathbb R$ with $dm$ being the Haar measure on $X$, which is a family of invariant measures for the Hamiltonian flow.
I am seeking examples of invariant measures for various Hamiltonian partial differential equations (PDE's). If you name a certain PDE and a family of invariant measures for it, please explain whether we know if those are the only invariant measures possible for that PDE.
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2025-03-21T14:48:30.162328
| 2020-03-28T04:27:27 |
355928
|
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|
Stack Exchange
|
What is symplectic rigidity?
What is an explanation for what the theory of symplectic rigidity is and what kind of questions it can answer? I was led to this after reading about the symplectic non-squeezing theorem of Gromov.
Rigidity, as used throughout mathematics and not just symplectic geometry, indicates that some structure attached to an object captures more data than one would "expect" from the underlying object itself. There are a number of helpful examples already in this Wikipedia entry. (I write "expect" in quotes because often times the "expectation" is far from correct, and might not even match your initial intuition.)
Gromov's non-squeezing theorem is typically the first example most people encounter in symplectic geometry. It states if there exists a symplectic embedding of the standard ball $B^{2n}(R)$ into $\mathbb{R}^{2n}$ such that its image lies completely in the cylinder $Z^{2n}(r) = B^2(r) \times \mathbb{R}^{2n-2}$, then $R \leq r$. However, there certainly exist volume-preserving embeddings $B^{2n}(R) \hookrightarrow Z^{2n}(r)$. Hence, this is an example of symplectic rigidity if you take the setting of volume-preserving geometry as your expectation. In other words, symplectic geometry is more rigid, or stronger, than volume-preserving geometry, since it sees more information.
Volume-preserving geometry is not the only setting from which you can build your "expectation." Depending on the situation, you may take:
Almost symplectic geometry: $(M^{2n},\omega)$ with $\omega \in \Omega^2(M)$ such that $\omega^n \neq 0$ (we drop the closedness condition).
The underlying smooth geometry
Obviously, if you have symplectic rigidity when compared with almost symplectic geometry, then you also have symplectic rigidity with respect to volume geometry and smooth geometry, for example. On the other hand, sometimes the problem at hand doesn't have an interpretation in volume-preserving geometry, and so you have to settle for rigidity with respect to smooth geometry.
Here's another example of rigidity: there is no closed exact Lagrangian submanifold embedded in standard $\mathbb{R}^{2n}$. However, in the smooth category, there are many $n$-dimensional closed submanifolds of $\mathbb{R}^{2n}$. So symplectic geometry is more rigid than smooth geometry. (Here's an example where it wouldn't make sense to compare this to volume-preserving or almost symplectic geometry.)
There's a whole industry of rigid invariants nowadays in symplectic geometry; the industrial revolution occurred in 1985 with Gromov's famous paper Pseudo holomorphic curves in symplectic manifolds. In fact, both examples mentioned so far first appeared in that paper. Often, these rigid invariants are tied up intimately with counts of pseudo-holomorphic curves (e.g. Gromov-Witten theory and Quantum Cohomology). Gromov non-squeezing can be seen as a zeroth order approximation to these subtle curve counts: it boils down to the existence of a particular pseudo-holomorphic curve. There's also the Fukaya Category, which is by now a big machine which can be thought of as encoding rigid invariants coming from Lagrangian submanifolds, and is itself intimately tied up with physics via mirror symmetry (this is all a long story in itself).
Finally, I should mention on the flip side that you can ask if symplectic geometry exhibits phenomena in which the weak "expectation" actually matches. This does happen, and is called flexibility. (This is particularly fruitful when combined with contact geometry, in which there are overtwisted discs and loose Legendrians.) As a purely symplectic example of flexibility, for any symplectic manifold $(M^{2n},\omega)$ and any manifold $L$ with $\dim L < n$ there is a forgetful map from the space of isotropic embeddings $\{f \colon L \hookrightarrow M \mid f^*\omega = 0\}$ to the space of formal isotropic embeddings, i.e. embeddings $f \colon L \hookrightarrow M$ and bundle monomorphisms $F \colon TL \rightarrow f^*TM$ (over $L$) with $F^*\omega = 0$. The forgetful map is just given by taking the same underlying embedding and $F = df$. This forgetful map induces a homotopy equivalence (and is hence referred to as an h-principle, where the "h" stands for "homotopy").
Thank you for the very nice answer!
|
2025-03-21T14:48:30.162591
| 2020-03-28T04:43:28 |
355929
|
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|
Stack Exchange
|
Derivation of an uncertainty principle from the symplectic non-squeezing theorem
Is there a derivation of an uncertainty principle or uncertainty-type principle from the symplectic non-squeezing theorem?
Yes. see:The symplectic camel and phase space quantization, by Maurice De Gosson.
Journal of Physics A: Mathematical and General, Volume 34, Number 47
|
2025-03-21T14:48:30.162639
| 2020-03-28T05:00:44 |
355931
|
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|
Stack Exchange
|
Current optimal regularity level for Benjamin-Ono equation
Ionescu and Kenig showed global wellposedness for the Benjamin-Ono equation for all (in particular, low) regularities $s\geq 0$. At $s=0$ they used modified $X^{s,b}$ spaces in order to avoid logarithmic divergences. The techniques they used seem to be classifiable as "wellposedness" techniques. My question is whether there has been more progress below the $s=0$ regularity level using wellposedness techniques.
|
2025-03-21T14:48:30.162813
| 2020-03-28T05:03:39 |
355932
|
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"Aurel",
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|
Stack Exchange
|
Explicit fundamental domain for the action of $\operatorname{O}(n,1)(\mathbb{Z})$ on $\operatorname{O}(n,1)(\mathbb{R})$
Minkowski computed explicit fundamental domains for the action of $\operatorname{SL}_n(\mathbb{Z})$ on $\operatorname{SL}_n(\mathbb{R})/\operatorname{SO}_n(\mathbb{R})$ for each $n \leq 6$. In the case where $n = 2$, one obtains the familiar fundamental domain for the action of $\operatorname{SL}_2(\mathbb{Z})$ on the complex upper half-plane. The case where $n = 3$ is studied in detail in the paper entitled "Hecke Operators and the Fundamental Domain for $\operatorname{SL}(3, \mathbb{Z})$" by Daniel Gordon et al.
Are there analogous computations in the literature of explicit fundamental domains for the action of the orthogonal group $\operatorname{O}(n,1)(\mathbb{Z})$ on $\operatorname{O}(n,1)(\mathbb{R})$, at least for some small values of $n$? I am particularly interested in the case where $n = 2$.
What I know: I understand that computing such fundamental domains is difficult in general. I'm aware of the construction of Borel and Harish-Chandra via Siegel domains, but I'm not sure whether it's possible to make their construction explicit in the way that Minkowski was able to do.
$O(2,1)(\mathbb{Z})$ is commensurable with $\mathrm{PSL}_2(\mathbb{Z})$ and $O(3,1)(\mathbb{Z})$ with $\mathrm{PSL}_2(\mathbb{Z}[i])$, so you should be able to recover fundamental domains for these two from fundamental domains present in the literature.
related: https://mathoverflow.net/questions/143543/algorithm-to-compute-the-integral-orthogonal-group
For information about these groups up through dimension 17, see:
Vinberg, È. B.
The groups of units of certain quadratic forms. (Russian)
Mat. Sb. (N.S.) 87(129) (1972), 18–36.
English translation [Math. USSR-Sb. 87 (1972), 17–35].
Vinberg shows up through dimension 17 that $O(n,1; \mathbb{Z})$ has a finite index subgroup generated by reflections. He gives an explicit polygon for this reflection group, i.e., a fundamental domain for its action. Then $O(n,1; \mathbb{Z})$ is generated by this reflection group along with the symmetry group of the polygon.
He has a later paper with Kaplinskaja that studies 18 and 19:
Vinberg, È. B.; Kaplinskaja, I. M.
The groups O18,1(Z) and O19,1(Z). (Russian)
Dokl. Akad. Nauk SSSR 238 (1978), no. 6, 1273–1275.
English translation: Soviet Math. Dokl. 19 (1978), no. 1, 194–197.
I believe there are some references for (slightly) higher dimensions as well. You might look at some papers of Allcock, for example.
Edit: For the special case $n=2$, it is reflections in the sides of a $(2,4,\infty)$ triangle, i.e., angles $\pi/2$, $\pi/4$, and one ideal vertex. This can be seen directly from the diagram in Table 5 of Vinberg's paper.
Richard Borcherds found descriptions of the groups (and in principle the fundamental domains) up to $n=23$.
https://doi.org/10.1016/0021-8693(87)90245-6
|
2025-03-21T14:48:30.163009
| 2020-03-28T06:52:41 |
355935
|
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|
Stack Exchange
|
$\lambda_2$ of Laplacian of a regular graph
Given a $d$-regular graph $G=(V,E)$ with $|V| =n$. We know that the smallest eigenvalue of the normalized laplacian matrix of $G$ is $0$. I have seen the formulation of the second smallest eigenvalue $\lambda_2$ in terms of the rayleigh quotient. However, the below paper mentions it in a way which I am not aware of.
How to derive the below (or similar) formulation of $\lambda_2$? Or am I horribly wrong in my interpretation?
$\lambda_2 = \min\limits_{\overline{v} \in \mathbb{R}^n, ||\overline{v}||=1} \dfrac{\mathbb{E}_{(u,v) \in E} ||\overline{u}-\overline{v}||^2}{\mathbb{E}_{u,v \in V} ||\overline{u}-\overline{v}||^2}$
Source : Proof of Theorem 7.1 (Section 7) of https://arxiv.org/pdf/1205.2234.pdf
Can you explain the notation? What do the $\Bbb E$ stand for and what is the $\bar u$?
For each vertex $u \in V$, $\overline{u}$ is a vector of unit length in $\mathbb{R}^n$.
And $\mathbb{E}$ denotes the expectation. For eg, the numerator can be written of as, $\sum_{(u, v) \in E} 1/|E| \cdot ||\overline{u}-\overline{v}||^2$ and the denomiator as, $\sum_{u, v \in V} 1/|V|^2 \cdot ||\overline{u}-\overline{v}||^2$.
Then there is still something off. You cannot use $\bar v$ as a bound variable in the $\min$ and the sum (in the expectation value) at the same time. Also, which vector is $\bar u$ exactly? This seems not well defined to me.
I just checked the paper, and the formula there is different from what you wrote. First, the minimum is over all assignments of unit vectors to vertices (if I understand $\bar u$ as you explained). Second, the expectation value in the denominator is not over $V$ but over some other set $\mathcal P_1$ from some "bisection", whatever that means.
They are calling the graph $G_1 = (P_1, E_1)$ as a regular expander which I have relabelled to $G = (V,E)$. The choice for vectors in $V \setminus P_1$ does not matter in the minimization.
|
2025-03-21T14:48:30.163200
| 2020-03-28T06:53:59 |
355936
|
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|
Stack Exchange
|
Contrasting theorems in classical logic and constructivism
Is it possible there are examples of where classical logic proves a theorem that provably is false within constructivism? Is so what are some examples?
What are some examples of most contrasting theorems provable in these two logics that does not fall in 1.?
Relevant, I think, are Brouwerian counterexamples. There are also no doubt many examples such as the following: Osvald Demuth, The differentiability of constructive functions (Russian, 1969). The MR review begins: "An increasing constructive function $[\cdots]$ on the constructive interval $[0,1]$ is given which $[\cdots]$ is nowhere differentiable."
Brouwerian counterexamples are mostly good for convincing classical mathematicians that intuitionism makes no sense.
This is a very natural question, but as it happens one needs some more background to give a natural answer (is my humble opinion).
For clarity let me give a summary first indication:
As to your Question 1. : This is commonly thought of as not being possible in strict terms, because 'constructive mathematics' is usually interpreted as 'that part of alternative constructive mathematical theories which is compatible with classical mathematics.'
Broadly speaking, this interpretation of 'constructive mathematics' is known as BISH (from Bishop-style mathematics). BISH is seen as the common core of classical math (CLASS),
intuitionistic mathematics (INT) and 'Russian' recursive mathematics a la Markov (RUSS).
Now if we include INT and RUSS in your term 'constructivism'. then the answer to question 1. is yes.
This because both INT and RUSS have additional axioms (compared to BISH) that are false classically. You could equally say that CLASS has an additional axiom compared to BISH (namely excluded middle) that is provably false in INT and RUSS.
Famous examples of conflicting theorems are:
a) All total functions from $\mathbb{R}$ to $\mathbb{R}$ are continuous (a theorem both in INT and RUSS)
b) The unit interval is compact in the Heine-Borel sense (a theorem both in CLASS and INT, but false in RUSS).
[Finally, if perhaps you mean only propositional first-order logic, then the answer is no again, because usually the term 'intuitionistic logic' is used to indicate that part of classical propositional logic which is valid constructively, which you can think of as `the part without excluded middle'.]
In the comment above, Brouwerian counterexamples are mentioned. Brouwerian counterexamples are explicit constructive examples where we can demonstrate that the classical 'solution' has insufficient constructive content (no general algorithm possible, or some other constructive deficiency). Often, Brouwerian counterexamples can be generalized to conflicting theorems in INT and/or RUSS. Brouwerian counterexamples are often useful in BISH to indicate that there can be little hope of constructivizing a certain classical theorem.
[Update 29 March | After your elucidation in the comments below, I feel I can give:]
Some remarks for Question 2:
a) An important example concerns axioms of choice. In constructivism (BISH, INT, RUSS) there are a number of choice axioms which may or may not conflict with each other and/or CLASS, depending on the precise phrasing and of course the intended logical strength.
In general, constructivists do not dismiss all forms of the idea that the statement for all $x$ there is $y$ such that $P(x,y)$ implies the existence of a choice function $f$ such that $P(x, f(x))$ holds for all $x$.
But the difference with CLASS (or between the others) can still be quite sharp, because of the very different interpretation of what is a 'meaningful' statement.
For a constructivist, 'meaningful' usually implies that we are talking about mathematical structures that can be constructed like (and therefore from !) the natural numbers $\mathbb{N}$, if we give ourselves enough time to do so. (Strict finitism is also a field of study in constructive mathematics, but since this is quite difficult it has not attracted many researchers).
In classical mathematics, a theory is often considered 'meaningful' already if we have good reason to believe that the theory is consistent. Whether the theory describes structures that we can actually build from scratch (viz. $\mathbb{N}$) is often not a matter of interest.
Strong intuitionistic axioms of choice sometimes expose this huge philosophical difference by leading to theorems which are false in CLASS, but sometimes they also lead to INT having the 'same' theorem as in CLASS (and you can only tell the difference by carefully interpreting the difference in meaning).
b) Another important example (also) concerns 'information' and 'interpretation'. Very very often, contradiction between CLASS and INT or RUSS can be meaningfully avoided by rephrasing. For example, by changing 'for all $x$' (CLASS) into 'for all $x$ for which we can determine whether $x\geq 0$ or $x\leq 0$' (INT), we might obtain the 'same' theorem.
This is imho a most salient point that Bishop added to Brouwer's views: why look for contradiction if you can rephrase for accordance?
However, imho it can also be very clarifying to study these contradictions. Usually one learns better how to rephrase for accordance when one has a clear idea where the conflict arises...
This answer is spot on! I'd like to contribute the realizability topos given by infinite-time Turing machines as a further intriguing environment. Any topos has an "internal logic", but the one of this one is particularly challenges many mathematical intuitions shaped by classical logic. In this topos, there is no surjection $\mathbb{N} \to \mathbb{R}$ (as you would expect from CLASS), but there is an injection $\mathbb{R} \to \mathbb{N}$. This observation is due to Andrej Bauer.
(a) comes from the fact that the equality of two real numbers is not decidable. (a) just means that there are less well-defined total functions in INT and RUSS than in CLASS.
@IngoBlechschmidt yes that is a nice addition. Am I correct in thinking that you mean an injection from Baire space $\mathbb{N}^{\mathbb{N}}$ into $\mathbb{N}$? Perhaps this also leads to an injection from $\mathbb{R}$ into $\mathbb{N}$ but I don't see this immediately, being poorly knowledged on toposes.
@Franka I meant $\mathbb{R}$, but indeed now that you say it I recall that Andrej proved it for $\mathbb{N}^\mathbb{N}$. Slides illustrating Andrej's proof are here (see slide 24/25). The argument easily adapts to $\mathbb{R}$, this is a fun exercise; if you want me to spell out the details, I'll gladly do so!
@IngoBlechschmidt on principle i would of course like to try it myself... but wisdom or (rather more likely) experience tells me i'd be very happy if you would spell it out for me :-) . do you have my email address?
@Franka Thank you for providing motivation! I just added the proof to a paper draft of mine, please have a look at Footnote 9 in Section 2.2 and share any difficulties in following the argument so that I can improve it, for the benefit of all who are interested in this curious injection. :-) Section 2 of this paper aims to give a leisurely introduction to the internal language of the effective topos and is hopefully readable even without extensive knowledge of toposes.
@IngoBlechschmidt thanks Ingo! It will take me some time to read ( 27 pages...:-) ) but I will do my best and get back to you. Cheers Franka
@FrankaWaaldijk For 2 I meant even if you do not get total contradiction like true and false perhaps we can get something that does not imply other under reasonable considerations or even perhaps contradicts other under reasonable considerations.
@VS. After your explanation I added some remarks on question 2. to my answer. The remarks are a bit generic, because specific examples would take too much detail (thus obscuring clarity for non-insiders). Hope you still can get the gist, if not I could try to specify a bit more.
@FrankaWaaldijk For Pi_2^0 statements such contradiction cannot occur (that is true in classical and false in constructive) by cs.nyu.edu/pipermail/fom/2006-March/010113.html. How about for Sigma_2^0 statements which can be shown to be true in classical nevertheless false in constructive? Perhaps it is possible to prove P=NP in classical and P\neq NP in constructive.
@VS. Your remark concerns Peano Arithmetic PA only...outside of PA there are many $\forall\exists$ statements which are true in CLASS but not in INT and/or RUSS, or vice versa. It so happens that P=NP is equivalent to a $\Pi_0^2$ statement so in that particular case INT and CLASS will agree.
The terminology seems quite confusing! If I've understood right: when discussing mathematics, 'intuitionistic' contradicts 'classical' while 'constructive' is a weakening of both; but for logics, 'intuitionistic' is a weakening of 'classical'?
This is partly a common situation in axiomatics... when we drop Euclid's fifth postulate, we get a weakening of Euclidean geometry. This weakening then allows us to add a non-parallelity axiom to obtain a non-Euclidean geometry which is contradictory to what we started with. Similarly intuitionistic first-order logic allows us to add non-classical axioms to obtain either full-blown INT or RUSS, which are contradictory to CLASS. Not adding these axioms gives us BISH, say. Historically Brouwer was by far the first (INT), which is why the terms intuitionistic and constructive get mixed.
@RobinSaunders : see above... (couldn't add your tag because to do so exceeded the text limit)
@FrankaWaaldijk FYI P\neq NP is \Pi^0_2 statement while P=NP is \Sigma^0_2 statement. So I still dont follow if P=NP is possible in Classical while P\neq NP is possible without LEM or LLPO? Also note P\neq NP in Classical implies P\neq NP without LEM or LLPO since P\neq NP is a \Pi^0_2 statement.
No, this is not possible. PA is conservative over Heyting Arithmetic HA (which is PA without excluded middle) with respect to $\Pi_0^2$ statements. PA proving P$\neq$NP implies that already HA proves P$\neq$NP.
@FrankaWaaldijk Ok.
@FrankaWaaldijk If a statement is provable then is it necessarily in Sigma1? I think that will resolve my confusion. Thank you.
There are several ways one could interpret the word "constructivism" here, and the answer depends on what you meant by it.
Bishop-style constructivism is a generalization of Brouwerian intuitionistim, Russian constructivism, and classical mathematics. It is mathematics done without excluded middle (of course, you can still use excluded middle on those instances that you can prove to hold using other means) and general axiom of choice, but you still have countable choice. Thus, anything you prove in this setting is true in classical mathematics as well.
There are other forms of constructivism which are Bishop-style constructivism extended with additional principles and axioms. These additional principles often contradict classical logic, and so you get consequences that are classically false. Here are some examples:
In the internal language of the effective topos (an older name for this is Russian constructivism) the following are valid statements:
There are countably many countable subset of $\mathbb{N}$.
There is an increasing sequence in $[0,1]$ that has no accumulation point.
The Cantor space $2^\mathbb{N}$ and the Baire space $\mathbb{N}^\mathbb{N}$ are homeomorphic.
Every map $f : [0,1] \to \mathbb{R}$ is continuous.
There exists a continuous unbounded map $f : [0,1] \to \mathbb{R}$.
There is a covering of $\mathbb{R}$ by intervals $(a_n, b_n)_n$ with rational endpoints such that $\sum_{k = 1}^n |b_n - a_n| < 1$ for all $n \in \mathbb{N}$.
There is a subset $S \subseteq \mathbb{N}$ which is not finite and not infinite.
There exists an infinite binary rooted tree in which every path is finite.
The ordinals form a set, i.e., they are not a proper class. One has to be careful about how ordinals are defined and how to precisely understand the notions of "class" and "set", but these are technical details.
In the internal language of the realizability topos $\mathsf{RT}(K_2)$ (an older name for this is Brouwerian intuitionisism) the following statements are valid:
Every map $f : X \to Y$ between complete separable metric spaces is continuous.
Every map $f : [0,1] \to \mathbb{R}$ is uniformly continuous.
Every map $f : \mathbb{R} \to \{0,1\}$ is constant, or equivalently, if $\mathbb{R} = A \cup B$ and $A \cap B = \emptyset$ then $A = \mathbb{R}$ or $B = \mathbb{R}$.
There are many other examples. I recommend taking the effort to get used to these amazing new worlds of mathematics.
What do the applied mathematicians say about these outrageous statements?
I don't think they know about them, or at least I never got a reaction. There's a positive side to all of this, too. For example, in the effective topos everything is computable (in a precise sense). I'd imagine at least some applied mathematicians would appreciate that. In general, I find computer scientists will take anything that helps them solve their problems, even if it's outrageous. Mathematicians are more hung up on value judgements and respecting traditions.
Oh, there's another topos (a model of intuitionistic bounded Zermelo set theory) that would interest applied mathematicians. In it all maps are differentiable and nilpotent infinitesimals exist, so you get to do analysis engineering-style, with $dx$'s and $dy$'s that are very small and whose squares are negligible.
What would you say to the typical high school math teacher who says many of these statements can be disproven by drawing a picture? They are so antithetical to geometric and physical intuition.
They are only antithetical to geometric and physical intuition under one understanding of "space" and "continuity" – the one you are aware of. It is entirely possible to develop different kinds of intuition (speaking from first-hand experience and having taught PhD students the alternatives) that let you "internalize" these statements. So, it's really the high-school teachers who determine what kind of intuitions the young minds will develop. In the current educational system there is only one kind. The resulting uniformity of thought gives the illusion of absolute truth.
Perhaps the closest we get in practice to alternative geometric intuitions is when engineers and physicsts, despite having been taught $\epsilon\delta$ proofs, insist on using infinitesimals. They don't have the necessary formal training to use them properly (because they were taught $\epsilon\delta$ proofs), but they rely on their intuitions quite reliably.
And by the way, it's not always about logic and foundations. One can change define "space" (in ZFC) in such a way that Banach-Tarski goes away and all subspaces of $\mathbb{R}$ are measurable, and the definition is arguably simpler than the usual definition of topological space (but of course one needs to get used to it, as is usual with new math). See for instance this paper by Alex Simpson.
Just briefly, here is an alternative high-school explanation of why all maps are continuous: motion is continuous, obviously, since if I want to get from $A$ to $B$ I cannot just teleport (despite many sci-fi movies). Therefore, all functions $\mathsf{Time} \to \mathsf{Space}$ are continuous. We can also have teleportation, but then we need to modify $\mathsf{Time}$ to account for it.
It seems like you have to bite the bullet. A lot of things I can draw on the board that are usually meant to represent geometrical shapes or functions on $\mathbb R$ cannot represent the “real” mathematical things according to constructive math (i.e. discontinuous but piecewise continuous functions, or anything inducing such a function like a Euclidean shape). Pedagogically this sounds like a nightmare. Or maybe if people are interested in studying things corresponding to “classical” intuition, then you have to convince them to study finitary surrogates for those?
@MonroeEskew: a piecewise continuous map, say $f(x) = \mathsf{if}; x < 0 ;\mathsf{then}; 0 ;\mathsf{else}; 1$ has as its domain the set $D = {x \in \mathbb{R} \mid x < 0 \lor x \geq 0}$. The strange thing is, that constructively $D$ is a proper subset of $\mathbb{R}$, even though $\mathbb{R} \setminus D = \emptyset$. It's a phenomenon that is difficult to explain classically. It's not that $D$ has fewer points than $\mathbb{R}$. Rather, the points of $D$ have more information. Each point of $D$ comes with information on whether it is negative or non-negative.
@BenCrowell: I think your criticism is not entirely fair. Of course one can never draw a completely accurate graph of a discontinuous function – but the same holds for any graph, because pencils are not infinitely thin. As Monroe says, it's better to just bite the bullet and try to understand how it's possible to have a proper subset $D \subseteq \mathbb{R}$ such that $\mathbb{R} \setminus D = \emptyset$. Once you get over that hurdle, a lot fo constructive math starts to make more sense.
@BenCrowell: I think physicists tend to use something closer to nilpotent infinitesimals of synthetic differential geometry (SDG), rather than non-standard analysis. They say things like "we may ignore $dx^2$", which directly translates to "$dx$ is a square-nilpotent" in SDG. (Of course, there's still a lot of similarity with non-standard analysis.)
@AndrejBauer Is this the right way to understand it? If you want to treat classical analysis (or just undergrad calculus) in a way that doesn’t diverge too dramatically from common notions, but is still faithful to constructivism, then you do a kind of calculus of partial functions, where the domains tend to shrink as you introduce more information about the definitions of the functions. So classical analysis is about “something” that is just a lot harder to describe from the constructivist standpoint?
Perhaps a reasonable analogy can be take from topology: let $R_d$ be the reals with the discrete topology and $R_e$ the reals with Euclidean topology. The identity map $\mathrm{id} : R_d \to R_e$ is a continuous injection, so $R_d$ is a "subspace" of $R_e$ (not what topologists would say). It's this sort of thing that's going on.
If you want to do classical analysis, then it's best to use classical logic. You can do classical analysis constructively, but it will be clumsy and unnatural. The situation is symmetric: you can do constructive math in classical logic by "simulating" then using models, but that's a lot of overhead.
@AndrejBauer In Riemannian geometry we add a point at infinity to compactify the plane. Is it known that proofs that essentially utilize compacitification are non-constructive and cannot be constructivized? That is there is no proof in the sense that HA cannot prove it but PA can if we cannot avoid compactification?
It is "essentially known" that the vast majority of mathematics is constructive, or can be constructivized. Specifically, there should be no trouble with one-point compactifications. You're a bit off there with HA and PA, as those are not strong enough to talk about analysis. You'd want a richer formalism.
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2025-03-21T14:48:30.164443
| 2020-03-28T07:59:11 |
355937
|
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|
Stack Exchange
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balls and boxes optimization
There are $n$ balls, among which $m$ balls are bad, and hence $n-m$ are good. We are given a number of boxes. We want to put balls into boxes such that all the good balls (or most of them, e.g., $99$%) are in good boxes. A box is called good if containing only good balls. The thing is that we cannot distinguish a good ball with a bad one. We can only know whether a box is good or not once putting balls into it. A trivial solution is to put only one ball into each box. However, this requires $n$ boxes. My objective is to minimize the number of boxes required to achieve that with high probability, all or most of the good balls are in good boxes. Is this a known problem? What is the structure of an optimal strategy?
your description of bad box is unclear. a bad box is one with (at least) one bad ball? so a box is neither good nor bad until a ball is put into it? is that how to understand the "we cannot know...until" part?
group testing
@kodiu yes, a box is bad as long as one bad ball is put in it; we can tell a box is good or bad after putting balls.
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2025-03-21T14:48:30.164547
| 2020-03-28T08:39:45 |
355938
|
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Stack Exchange
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How does the constant bound on Hormander's $L^2$ estimate for non-degenerate phase depend on the cut-off function?
A classical estimate, due to Hormander, assets that the integral operator
$$Tu = \int_{\mathbb{R}^{d}} e^{i\varphi(x,y)/h} a(x,y) u(x) dx, \ \ a \in \mathcal{C}_{c}(\mathbb{R}^{2d}), \ \ \varphi \in \mathcal{C}(\mathbb{R}^{2d})$$ can be extended to a bounded operator on $L^{2}$, if the mixed Hessian of $ \varphi $ is non-vanishing on the support of $a$. This has an sharp estimate
$$ \| Tu \|_{L^2} \leq C h^{d/2} \| u \|_{L^2} , \ \ \text{as} \ \ h \rightarrow 0 $$
for some constant $C$. If the phase $\varphi$ behaves somewhat more badly, then it is difficult to obtain a similar bound as above.
In my case, I have a phase $\varphi$ on $\mathbb{R}^2$ which has vanishing Hessian on the diagonal, and I would like to know what kind of estimate I can get for the above operator. It seems to be a common practice in this case to split the domain of integration into two parts
$$ Tu = T \chi u + T ( 1 - \chi )u $$
where $\chi$ contained the singular part of the phase. If I set $\chi$ to be a cut-off function supported on a ball of radius $r$, then using the cut-off $\chi(|x-y|)$ it is easy to bound to first operator with $ \| T\chi\|_{\mathcal{L}(L^2)} \leq C r$. On the support of $(1-\chi)a$, I should be able to apply Hormander's estimate and conclude that $ \| T(1-\chi) \|_{\mathcal{L^2}} \leq C_{r}h $, the constant here depends on $r$.
In order to get a full intuition of whether this method would work, I wonder if anybody could tell me if there is a way to characterize the precise dependency of $C_r$ on $r$, or more generally, the amplitude $a$?
Many thanks in advance!
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2025-03-21T14:48:30.164702
| 2020-03-28T09:06:18 |
355942
|
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Stack Exchange
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Why not this reflective-size set theory be the foundational theory of sets?
Language: first order logic with equality + primitives of membership $\in$ and a constant $W$ signifying the world of all sets.
Axioms:
Extensionality: $\forall z (z \in x \leftrightarrow z \in y) \to x=y$
Comprehension: $\exists x: x=\{y \in W| \phi\} $; if $x$ doesn't occur in $\phi$
Reflection: $\forall\vec{x} \in W [\exists x \phi \to \exists x \in W \phi]$; if $\phi$ in $L(=,\in)$ with all parameters from $\vec{x}$.
Size: $ x \in W \leftrightarrow x \subset W \land |x|<|W|$
Foundation: $x \neq \emptyset \to \exists y \in x (y \cap x =\emptyset)$
/Theory definition finished.
Now to me the above axiomatic system looks very "natural". All notions encountered in its axioms are pretty much elementary of logic and set conception. However this theory proves to be so strong relative to ZFC, its consistency strength is near to a Mahlo. In my own experience I know of no consistent theory that is presented in alike elementary manner and at the same time enjoying such strength. Possibly "ZFC + CH fails for every multipleton set" is of comparable elementary presentation, and it is much stronger than this, however the semantics of it are much more complicated and its not immediately clear how it conforms with the usual semantics of ZFC, that it cannot be regarded as constituting a direct extension of it. However with the above theory we do have a motivation that directly flow from conceptions about ZFC itself, that in some sense can be seen as a direct extension of it much as ZFC itself can be seen as a natural extension of Zermelo. This line of theories can show a nice spectrum, if we change "size" axiom by "$W$ is transitive", we get a theory of a strength in the vicinity of second order arithmetic, if we instead replace it by the axiom stating that "every subset of an element of $W$ is an element of $W$", then we get a theory in the strength of ZFC. So the theory here comes as a upgrade of that up to size limitation which blows it up to near Mahlo strength. I advocate this set theory to be the maximal of known naturally looking set theories.
Now I see a tendency to coin mathematical-formal concepts in the hereditarily accessible sets of ZFC, since that's what the axioms of ZFC can grant, the underlying rationale is that discourse about those sets is granted by the most natural conception about sets that is embodied by ZFC. However, here I'd suggest that this limit of naturalness that's built after whats granted to exists after ZFC, is to be set up to instead be what's granted to exist by this theory! As this theory can be viewed as an embodiment of natural conceptions about sets, since it extends ZFC from conceptions inherent to ZFC itself and is expressed in a quite elementary manner.
Question: What mathematical arguments can be raised against this stance? I'm not asking for a philosophical opinion here! I'm asking about a mathematical technical workup that can in some sense invalidate this approach, or can show that adopting this theory to organize thought about sets can cause impediments with some advanced techniques that ZFC grants in a flawless manner, that makes it unworthy of being considered as a foundation of natural set theoretic thought?
I'm not a logician, so this may be stupid : isn't this the theory of classes and sets?
@user44191, you mean Muller's. http://philsci-archive.pitt.edu/1372/1/SetClassCat.PDF
@user44191, I don't know which theory of classes and sets you have in your mind, the known theories are NBG, MK, Ackermann's, Muller's, and Quine's ML. In any case this theory is way stronger than all of those.
This theory looks similar to ones considered by Bernays in "Zur Frage der Unendlichkeitsschemata in der axiomatischen Mengenlehre" (there is an English translation, "On the Problem of Schemata of Infinity in Axiomatic Set Theory").
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2025-03-21T14:48:30.165119
| 2020-03-28T10:17:58 |
355944
|
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|
Stack Exchange
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On algebras where indecomposable modules are determined by their dimension vectors
Let $A$ be a finite dimensional quiver algebra such that any two indecomposable modules with the same dimension vector are isomorphic.
Question: In case $A$ has $n$ simple modules and finite global dimension, does $A$ have Loewy length at most $n$?
Yes.
Let $A=kQ/I$ be a quiver algebra with $n$ simple modules and finite global dimension such that $\text{rad}^nA\neq0$. For a vertex $i$ of the quiver $Q$, I'll use $e_i$ to denote the corresponding primitive idempotent of $A$, and $S_i$ to denote the corresponding simple (right) $A$-module.
Since $\text{rad}^nA\neq0$, there is a path of length $n$ that is nonzero in $A$. Since $Q$ has $n$ vertices, this path must pass through one of them twice. So there is a directed cycle $c$ that is nonzero in $A$.
Suppose $c$ starts and ends at vertex $i$, so $c=e_ice_i\neq0$.
Let $K$ be a submodule of the indecomposable projective $e_iA$ that is maximal subject to $c\not\in K$, and let $X=e_iA/K$. Then $c$ spans the socle of $X$, so $X$ is a module whose head and socle are both isomorphic to $S_i$, and so $\text{rad}X$ and $X/\text{soc}X$ have the same dimension vector. They are also both indecomposable, since they have simple socle and head respectively.
Since $A$ has finite global dimension, the No Loops Conjecture implies that $S_i$ is not a summand of the head of $\text{rad}X$, so $\text{rad}X\not\cong X/\text{soc}X$.
What do you mean with $c \neq K$? Probably that c not $\in$ K?
@Mare Yes, sorry. Fixed.
Thanks, a very nice proof.
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2025-03-21T14:48:30.165264
| 2020-03-28T10:54:25 |
355946
|
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|
Stack Exchange
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I found a (probably new) family of real analytic closed Bezier-like curves; is it publishable?
Given $n$ distinct points $\mathbf{x} = (\mathbf{x}_1, \ldots, \mathbf{x}_n)$ in the plane $\mathbb{R}^2$, I associate a real analytic map:
$f_{\mathbf{x}}: S^1 \to \mathbb{R}^2$
with the following properties. If $C_n(\mathbb{R^2})$ is the configuration space of $n$ distinct points in $\mathbb{R}^2$, then each there is a real analytic map
$$f: C_n(\mathbb{R}^2) \times S^1 \to \mathbb{R}^2$$
such that $f_{\mathbf{x}}(-) = f(\mathbf{x},-).$$
The image of $f_{\mathbf{x}}$ is in the convex hull of $\mathbf{x}$. Also, if $\mathbf{x}'$ is a permutation of the points in $\mathbf{x}$, then $f_{\mathbf{x}'} = f_{\mathbf{x}}$. Finally, if you apply a Euclidean transformation to the configuration $\mathbf{x}$, then the image of the curve $f_{\mathbf{x}}$ gets transformed by the same Euclidean transformation. Here are some plots.
The points in $\mathbf{x}$ are shown in red, and the corresponding curve is shown in blue.
My question is, is there any interest in such Bezier-like curves? Can I publish them somewhere? I am not familiar with the area. Could someone possibly suggest some journal(s) by any chance?
Actually, I found originally a related family, where given $n$ distinct points $\mathbf{x} = (\mathbf{x}_1,\ldots,\mathbf{x}_n)$ in $\mathbb{R}^3$, I associate a real analytic map:
$$f_{\mathbf{x}}: S^2 \to \mathbb{R}^3,$$
with similar properties as the first family of maps. My work does not extend further to higher dimension though. Any comments and/or suggestions are welcome.
Edit 1: after discussing with @DanieleTampieri, and looking at the third figure, it does seems like my maps could possibly be used in boundary detection problems possibly. It is interesting that a single formula seems to accomplish what is usually done algorithmically.
Edit 2: following one of @JochenGlueck's comments, I did the following experiment. I started with a configuration of $4$ points and plotted the corresponding curve. Then I added $6$ points at random inside the convex hull of the initial $4$ points, and plotted the corresponding curve. The new curve looks like it passes through the original $4$ points now, interestingly. Here are the corresponding two plots.
Edit 3: I wrote a GUI interface using the Python library Tkinter. Now I can place the points on a canvas, press a button and see the resulting curve. This will enable me to experiment further with these curves. After consulting with a few people, I think this may fit in a journal of Computer Graphics perhaps. It may not be appropriate in a journal of Approximation theory or Numerical Analysis I think, as it does not really contain results, as of now. In any case, I will let things stew a bit more in my brain before writing things up. Thank you all.
Edit 4: I saw how to generalize my maps in two different directions: as a sequence of maps, and in higher dimension. More specifically, I have defined, for each positive integer $m$, a real analytic map:
$$f_m: C_n(\mathbb{R}^d) \times S^{d-1} \to \mathbb{R}^d,$$
such that, given $\mathbf{x} \in C_n(\mathbb{R}^d)$ (where $C_n(\mathbb{R}^d)$ denotes the configuration space of $n$ distinct points in $\mathbb{R}^d$), the map $f_m(\mathbf{x},-)$ maps the $d-1$ dimensional sphere to a good approximation of the boundary of the convex hull of $\mathbf{x}$. In the cases I considered, it seems that small values of $m$ suffice in practice (I mostly experimented in 2d and a little bit in 3d). Here is the plot of the images of some sample points on the sphere, for the case where $\mathbf{x}$ is the configuration of the $4$ vertices of a regular tetrahedron. I used $m=3$. I also include a 2d example, with $m=3$ (my previous map corresponds to $m=1$).
Edit 5: I uploaded a short note on arXiv. In case someone is interested in knowing how my maps are defined: Rational Maps and Boundaries of Convex Hulls, arXiv:2004.04538. I submitted this short note to a journal for review (sorry for all the updates).
Edit 6, UPDATE: Peter Olver found a counterexample to my conjecture in the arXiv article in Edit 5, so I will withdraw it, BUT this led to a fruitful collaboration where, after modifying the definition of the maps, we were actually able to prove convergence in the preprint Continuous Maps from Spheres Converging to Boundaries of Convex Hulls, arXiv:2007.03011. However, the maps are only piecewise rational, yet they are continuous. I am curious to know what people think of our preprint, for the ones who are interested enough to read it!
From my point of view, your algorithm can be interesting in character recognition and tracing a bitmap (transforming a bitmap into a smooth, scalable image). I don't know if your result is new nor can be classified as "prior art", though any journal in approximation theory is potentially interested.
@DanieleTampieri, thank you for your comments. I got a few upvotes then a few downvotes. I wish more people would take the time to write comments. I do not work in approximation theory, so constructive criticism would be welcome too.
You're welcome: and if you write down your work, whether you decide to publish it or not, I'd be glad to have a look at it.
@DanieleTampieri, thank you very much! I would have to read the surrounding literature first. If you are familiar with the area, would you like to collaborate perhaps? I will write you an email.
Bezier curves also have applications in shape optimization problems. See Haslinger, J., & Mäkinen, R. A. (2003). Introduction to shape optimization: theory, approximation, and computation. Society for Industrial and Applied Mathematics. I don't know whether it will be useful for you or not.
A remark about the properties listed in the paragraph before the plots: Due to the plots I'm under the impression that your map has, in addition, some kind of "maxmimality" property which seems to be important from a geometric point of view and which you have not listed: Say, if $\overline{\mathbf{x}}\in\mathbb{R}^2$ denotes the arithmetic mean of $\mathbf{x}$, then e.g. $\tilde f_{\mathbf{x}}(s) = f_{\mathbf{x}}(s)/2+\overline{\mathbf{x}}/2$ also has all the properties listed before the plots, but is "contracted" towards $\overline{\mathbf{x}}$ - which your plots don't seem to be.
@JochenGlueck, it is a very valid remark. I have some idea how to formulate this maximality, but not how to prove it.
@Malkoun: Thank you for your response! Another small observation just occurred to me: If we compare your first plot the other ones, it appears that your curve gets "closer" to the points in the tuple $\mathbf{x}$ if there are more points inside the convex hull of $\mathbf{x}$. So if you start with a tuple $\mathbf{x}$, and then add several (maybe random?) points from the convex hull of $\mathbf{x}$, will this push your curve closer to the boundary of the convex hull?
@JochenGlueck, I did the experiment you suggested. The new curve seems to pass through the original points in $\mathbf{x}$ (consisting of $4$ points) after adding $6$ random points inside the convex hull of the original $\mathbf{x}$.
|
2025-03-21T14:48:30.165761
| 2020-03-28T10:58:21 |
355947
|
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"Alfred",
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|
Stack Exchange
|
Eigenvalues and eigenvectors of Gaussian random matrices
Let us assume we have a square matrix $A$ whose entries are sampled from a standard Gaussian distribution of mean $0$. Do we have any information about the distribution of its eigenvalues?
Particularly, I'm aware that there are different results on symmetric gaussian matrices (or, the Gaussian orthogonal ensemble of $A$):
The eigenvalues follow a semicircle law
Is there an equivalent result for standard, non-symmetric Gaussian matrices?
I have answered this for the ensemble of nonsymmetric matrices, but do note that your statements on the symmetric case (GOE) are not correct: the eigenvalues follow a semicircle law, not a Gaussian distribution.
You are right. Thanks for the answer, I'll edit for future readers!
This is the Ginibre ensemble, see Eigenvalue statistics of the real Ginibre ensemble for the eigenvalue distribution. For an $N\times N$ matrix with $N\gg 1$ there are on average $\sqrt{2N/\pi}$ eigenvalues on the real axis, uniformly in the interval $(-\sqrt N,\sqrt N$). The rest of the eigenvalues fill a disc of radius $\sqrt N$ in the complex plane, uniformly except for a depleted strip along the real axis. Here is a scatter plot of the eigenvalues for $N=100$ (taken from arXiv:1305.2924).
|
2025-03-21T14:48:30.165878
| 2020-03-28T11:10:35 |
355948
|
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|
Stack Exchange
|
Distance function and its approximation
An easy and quick question:
Consider a function $u\in C(\Omega)$, where $\Omega$ is a bounded domain in $\mathbb{R}^n$.
Define a function $Q$ that measures the distance of a point $(x,y) \in\mathbb{R}^{n+1}$ from the graph of $u$. That is, consider the function
$$Q(x,y)=\inf_{z\in \mathrm{graph}(u)} d_z (x,y)$$
where $d_z(\zeta)=|\zeta - z|$.
My question is:
why the function $Q$ should be the limit of smooth approximations $Q_p \geq Q$ given by
$$Q_p(x,y)=\bigg[\int_\Omega \{ d_{(\xi,u(\xi))}(x,y)\}^{-p} \mathrm{d}\xi \bigg]^{-1/p}\quad \textrm{for } (x,y) \notin \mathrm{graph}(u)$$
why the derivative is
$$ DQ_p(x,y)=(Q_p(x,y) )^{p+1} \int_\Omega |(x-\xi, y-u(\xi)|^{-p-2}(x-\xi,y-u(\xi))\,\mathrm{d}\xi$$
I just tried to show the uniformly convergence. Can somebody check this following calculation?
Let $d_\xi (x,y) = d_{(\xi,u(\xi))}(x,y)\$. One have
$$d_\xi (x,y) \leq \sup_{\xi} d_\xi (x,y) \rightarrow \dfrac{1}{d_\xi (x,y) } \geq \dfrac{1}{\sup_{\xi} d_\xi (x,y)}$$
so applying the integral $$\int_\Omega \bigg( \dfrac{1}{d_\xi (x,y) } \bigg)^p \geq \bigg( \dfrac{1}{\sup_{\xi} d_\xi (x,y)} \bigg) ^p |\Omega|$$
and so we have
$$ \bigg[\int_\Omega \{ d_{(\xi,u(\xi))}(x,y)\}^{p} \mathrm{d}\xi \bigg]^{1/p}\geq \dfrac{1}{\sup_{\xi} d_\xi (x,y)} |\Omega|^{1/p}$$
and accordingly
$$\dfrac{1}{\bigg[\int_\Omega \{ d_{(\xi,u(\xi))}(x,y)\}^{p} \mathrm{d}\xi \bigg]^{1/p}} \leq \dfrac{1}{\bigg( \dfrac{1}{\sup_{\xi} d_\xi (x,y)} \bigg) |\Omega|^{1/p}} $$
Now to prove the uniformly convergence, one need
$$\lim_{p\rightarrow \infty } \sup_{(x,y)} | Q_p(x,y) - Q(x,y)| \rightarrow 0$$
By the calculation above, we have
$$\lim_{p\rightarrow \infty } \sup_{(x,y)} | Q_p(x,y) - Q(x,y)|\leq \lim_{p\rightarrow \infty } \sup_{(x,y)} \bigg| \dfrac{1}{\bigg( \dfrac{1}{\sup_{\xi} d_\xi (x,y)} \bigg) |\Omega|^{1/p}} - \inf_\xi d_\xi(x,y)\bigg|$$
and using the fact that, in general, $ \dfrac{1}{\sup_x \dfrac{1}{f(x)}} \geq \inf_x f(x)$
follows
$$\lim_{p\rightarrow \infty } \sup_{(x,y)} | Q_p(x,y) - Q(x,y)|\leq \lim_{p\rightarrow \infty } \sup_{(x,y)} \bigg| \big(\dfrac{1}{|\Omega|^{1/p}} -1\big) \dfrac{1}{\sup_{\xi} d_\xi (x,y)}\bigg|$$
Now one can put $\big(\dfrac{1}{|\Omega|^{1/p}} -1\big)$ out of the supremum and calculete the limit for $p\rightarrow \infty$ the result is $0$.
Is this right?
The second identity appears to follow simply from the chain rule applied to
$Q_p(\eta) = f_p(g_p(\eta))$, where
\begin{align*}
f_p(t) &=t^{-1/p},\\
g_p(x,y) &= \int_\Omega \left(d_{(\xi,u(\xi))}(x,y)\right)^{-p}\,\mathrm{d}\xi,\\
d_{(\xi,u(\xi))}(x,y) &= |(x-\xi,y-u(\xi))|
\end{align*}
by definition.
As for the first identity, it looks like $Q_p$ measures the inverse of the distance function $d$ in an $L^p$ norm, and thus the limit $p\to\infty$ should give the supremum norm for this inverse, hence the infimum of the distance itself.
Thank you @gmvh . I wanted a more formal proof of the first question, for example show that $Q_p(x,y)$ converges uniformly to $ Q(x,y)$...any idea?
While I believe that "can you check this"-type questions are strongly discouraged on MO, your step-by-step derivation of a more formal statement of what I stated informally looks fine to me (but then I don't know what level of rigour you're aiming for).
|
2025-03-21T14:48:30.166085
| 2020-03-28T11:37:28 |
355953
|
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|
Stack Exchange
|
A solution of a system of equations that involve directional derivatives
[Edited on 29-March-2020 to make the question clearer]
Let $f, g : [0,1]^2 \to \mathbb{R}$ be two smooth functions, which are strictly increasing and concave in each coordinate. That is, for every $0 < x,y < 1$ we have $\frac{\partial f}{\partial x}(x,y) > 0$, $\frac{\partial f}{\partial y}(x,y) > 0$, $\frac{\partial^2 f}{\partial x^2}(x,y) < 0$, $\frac{\partial^2 f}{\partial y^2}(x,y) < 0$, and the same holds for $g$.
A point $(x_0,y_0) \in [0,1]^2$ is a solution if
$$\frac{\partial f}{\partial x}(x_0,y_0) = \frac{\partial g}{\partial x}(1-x_0,1-y_0),$$
$$\frac{\partial f}{\partial y}(x_0,y_0) = \frac{\partial g}{\partial y}(1-x_0,1-y_0).$$
I would like to know under which conditions on $f$ and $g$ this system has a unique solution.
Isn't this impossible, since $g(1-x,1-y)$ is now strictly decreasing in each variable rather than increasing like $f(x,y)$? Or am I missing something?
The function $g(1-x,1-y)$ is indeed decreasing in each variable, but the directional derivative of $g$ in each variable at the point $(1-x,1-y)$ is still positive, or do I miss your question?
Sorry, I still see a contradiction. For simplicity freeze $y$ and let $h(x) = g(1-x,1-y)$. According to what you are saying, $\frac{\partial}{\partial x} h(x) > 0$ but $h(x)$ is monotone decreasing. Is that not a contradiction?
I guess the confusion is because of the way we interpret the expression $\frac{\partial g}{\partial x}(1-x)$: for me, $\frac{\partial g}{\partial x}(1-x)$ is the derivative of $g$ evaluated at the point $1-x$. For you, this is the derivative of the function $h(x) = g(1-x)$ evaluated at $x$. And these two derivatives are indeed of different signs. I apologize for the confusion.
OK, now I understand! Thanks for explaining. But then the solution is pretty straightforward. In one variable, using the above notation, you want $f'(x) = g'(1-x) = -h'(x)$. The only possible solution is $h(x)=-f(x)$ or $g(x) = -f(1-x) + C$. Or with two variables, $g(x,y) = -f(1-x,1-y) + C$, for a constant $C$.
Igor, you are a bit quick for me. In the one dimensional case, $x \mapsto f'(x)$ is decreasing and $x \mapsto g'(1-x)$ is increasing, and therefore, by imposing the appropriate inequalities on $f'(0)$, $f'(1`)$, $g'(0)$, and $g'(1)$ we deduce that there is exactly one point $x$ such that $f'(x) = g'(1-x)$. [continues in the next comment]
Since we consider the two-dimensional case, for every $y$ there is at most one $x$ that satisfies $$\frac{\partial f}{\partial x}(x,y) = \frac{\partial g}{\partial x}(1-x,1-y),$$
and for every $x$ there is at most one $y$ that satisfies $$\frac{\partial f}{\partial y}(x,y) = \frac{\partial g}{\partial y}(1-x,1-y).$$ It is not clear to me how we derive that there is a unique pair $(x,y)$ such that both equations hold for it.
Note that $f$ and $g$ are given functions, and we look for a solution $(x,y) \in [0,1]^2$. We do NOT look for a set of functions $f$ and $g$ that satisfies the equations for every $x$ and $y$.
Ah, I see. Sorry, I totally misread your question again!
If you are happy with using the intermediate value theorem in one dimension (in order to put desired hypotheses on $f(x)$ and $g(x)$), then you might also be happy with using its generalization in two or higher dimensions: the Poincaré-Miranda theorem.
Thanks, Igor. The Poincare-Miranda Theorem allows one to derive conditions for the existence of a solution. Are there tools that allow one to derive conditions for uniqueness of solution?
Let $z=z_0$ with $F(z)$, $G(z)$ solve the 1-dimensional version of your problem, as you described in the comments. Then $f(x,y) = F(x+y)$ and $g(x,y) = G(x+y)$ satisfy your hypotheses, while any $(x_0,y_0)$ on the line $x_0+y_0=z_0$ is a solution to your equation, meaning that the solution is not unique.
If you strengthen your hypotheses such that the full Hessian of $f$ is negative definite as a matrix, same for $g$, you can exclude such counterexamples. Then uniqueness does hold: if there are two distinct solutions, draw a straight line between them, restrict $f$ and $g$ to that line, and repeat the 1-dimensional uniqueness argument.
|
2025-03-21T14:48:30.166616
| 2020-03-28T12:04:21 |
355954
|
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|
Stack Exchange
|
Densest safe disk packing
Inspired by current regulations regarding the minimal distance to be kept among people to prevent spreading of the COVID-19 virus and the maximal number of people in a group that is not subjected to that distance-regulation, I'd like to ask the following:
given
a minimal distance $R\ \gt\ 0$ to be kept between individuals in a "group"
a maximal number $k$ of individuals that are not subjected to the minimal-distance rule for groups, i.e. for each individual there are maximally $k-1$ others that are closer than $R$.
Modeling the individuals as unit circles leads to the
Question:
how many unit disks can be maximally packed in a square with sidelength $a=n*R$ under the constraint that for each disk the distance of the $k$-th nearest disk is at least $R$.
Can one ever do better than just taking the best solution for $k=1$ and replace each disk by $k$ copies of that disk?
Can the disks overlap? Or does "packing" imply that they don't?
no, they are not allowed to overlap
Bumped by Community Bot every four months since April of 2020.
This is an initial idea for an approximate solution for finding the densest safe packing of an infinite set of disks in the euclidean plane:
determine the smallest regular hexagon in which $k$ unit circles can be packed, for small $k$ known solutions can be found here and let $\varrho$ denote the radius of that hexagon's incircle
place copies of that hexagon at each vertex of a triangular grid with edge-lengths equal to $2\varrho+R$ and rotated to render the hexagon's sides orthogonal to the grid's edges.
exploit potential for further improvements by convoluting the unit disks inside the hexagons with open disks of radius $\frac{R}{2}$ and checking whether the distance of gridlines can be reduced without generating an overlap of convoluted circles in different hexagons.
I posted this as an answer to make visible that there are ideas for a solution. Please feel free to let me know if you think it isn't appropriate.
|
2025-03-21T14:48:30.166795
| 2020-03-28T13:14:58 |
355960
|
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"Alexandra L.",
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|
Stack Exchange
|
Deciding whether a system of linear integer inequalities has infinitely many solutions
I have a quick question that I am struggling to find a solution to:
Given a system of linear integer inequalities $A\textbf{x} \leq \textbf{b}$, where $A\in \mathbb{Z}^{m\times n}$ and $\textbf{b}\in\mathbb{Z}^{m}$, is it possible to decide whether this system has infinitely many solutions?
Any help or pointers to the relevant literature are appreciated!
EDIT: I am only interested in integer solutions, i.e., the question is whether there are infinitely many integer solutions.
I am not sure. If a polytope determined by the system is bounded then there are definitely finitely many integer solutions, but if this polytope is not bounded can it still be the case that we have finitely many solutions?
Welcome Alexandra! Yes, $4x \leq 6$, $-4x \leq -2$ in $x$ and $y$ with $y$ unconstrained is unbounded but has no integer solutions.
|
2025-03-21T14:48:30.166889
| 2020-03-28T15:56:47 |
355966
|
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|
Stack Exchange
|
Martingale optional stopping before a stopping time
Here’s an easy one, I hope:
Suppose $\tau$ is a stopping time and $(M_t)$ is a martingale which together satisfy the hypotheses of the optional stopping theorem so that $\mathbb{E}[M_\tau]= \mathbb{E}[M_0]$.
Will it also be the case that $\mathbb{E}[M_\sigma]= \mathbb{E}[M_0]$ for an arbitrary random time $\sigma \leq \tau$, for example $\sigma= \tau-1$?
Thank you very much for your help.
No. The problem is that $\tau-1$ is not a "stopping time" for the martingale.
Example. Martingale $(X_n), n=0,1,2$ is the standard random walk on $\mathbb Z$. Our sample space is $[0,1)$ with Lebesgue measure.
\begin{align}
X_0(\omega) &= 0,\qquad \omega \in [0,1).
\\
X_1(\omega) &= \begin{cases}
1,\quad & \omega \in [0,1/2),
\\
-1,\quad & \omega \in [1/2,1).
\end{cases}
\\
X_2(\omega) &= \begin{cases}
2,\quad & \omega \in [0,1/4),\\
0,\quad & \omega \in [1/4,1/2),\\
0,\quad & \omega \in [1/2,3/4),\\
-2,\quad & \omega \in [3/4,1).
\end{cases}
\end{align}
The filtration $\mathcal F_0, \mathcal F_1, \mathcal F_2$ for this martingale:
$\mathcal F_0$ is the sigma-algebra generated by $X_0$, so it is the trivial sigma-algebra with atom $[0,1)$.
$\mathcal F_1$ is the sigma-algebra generated by $X_0, X_1$, so it is the sigma-algebra with atoms $[0,1/2), [1/2,1)$.
$\mathcal F_2$ is the sigma-algebra generated by $X_0, X_1, X_2$, so it is the sigma-algebra with atoms $[0,1/4),[1/4,1/2),[1/2,3/4),[3/4,1)$.
Define stopping time $\tau = 2 \wedge \min\{n : X_n < 0\}.$ Then
\begin{align}
\tau(\omega) &= \begin{cases}
2,\quad & \omega \in [0,1/4),\\
2,\quad & \omega \in [1/4,1/2),\\
1,\quad & \omega \in [1/2,3/4),\\
1,\quad & \omega \in [3/4,1).
\end{cases}
\\
\sigma(\omega) &= \begin{cases}
1,\quad & \omega \in [0,1/4),\\
1,\quad & \omega \in [1/4,1/2),\\
0,\quad & \omega \in [1/2,3/4),\\
0,\quad & \omega \in [3/4,1).
\end{cases}
\end{align}
Now $\tau$ is a stopping time, so of course $\mathbb E[X_\tau] = 0$. But compute $\mathbb E[X_\sigma] = 1/2 \ne 0$, which shows that $\sigma$ is not a stopping time. Computations:
\begin{align}X_\tau(\omega) &= \begin{cases}
2,\quad & \omega \in [0,1/4),\\
0,\quad & \omega \in [1/4,1/2),\\
-1,\quad & \omega \in [1/2,3/4),\\
-1,\quad & \omega \in [3/4,1).
\end{cases}
\\
X_\sigma(\omega) &= \begin{cases}
1,\quad & \omega \in [0,1/4),\\
1,\quad & \omega \in [1/4,1/2),\\
0,\quad & \omega \in [1/2,3/4),\\
0,\quad & \omega \in [3/4,1).
\end{cases}
\end{align}
Calculation that shows $\sigma$ is not a stopping time:
$$
\{\sigma = 0\} = [1/2,1) \notin \mathcal F_0.
$$
For future reference: An "easy one" like this should be posted at https://math.stackexchange.com, not here.
My mistake on SE vs here. Thank you for putting that nicely, and for the clear solution.
Even if $\sigma$ is a stopping time with $\sigma\le\tau$ the answer is No.
Example: Take your martingale to be a standard 1-dimensional Brownian motion $M$ with $M_0=0$, and then define $\sigma:=\inf\{t: M_t=1\}$ and $\tau:=\inf\{t>\sigma: M_t=0\}$. These are both finite (a.s.) stopping times. And clearly $0<\sigma<\tau$ and $M_\sigma=1$ but $M_\tau=0$ (a.s.), so $\Bbb E[M_\tau]=0=\Bbb E[M_0]$ but $\Bbb E[M_\sigma]=1\not=\Bbb E[M_0]$.
This is not quite an example because $M$ and $\tau$ don't satisfy the hypotheses of the optional stopping theorem, even though they do happen to satisfy its conclusion.
Granted. What are the optimal hypotheses for Optional Stopping?
The weakest hypothesis I know is to have ${ M_{t \wedge \tau} : t \ge 0}$ uniformly integrable.
I guess the original question is not quite well posed, as "the hypotheses of the optional stopping theorem" is ambiguous.
|
2025-03-21T14:48:30.167218
| 2020-03-28T16:11:11 |
355968
|
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|
Stack Exchange
|
Understanding traveling waves as critical points of the constrained energy
I am trying to understand a (straightforward I guess) statement of an old (but outstanding) paper on stability of solitary waves. Let us consider the following functionals:
$$
V(u)=\dfrac{1}{2}\int_\mathbb{R} u^2dx \quad \hbox{and}\quad E(u)=\int_\mathbb{R}(\tfrac{1}{2}uMu-\tfrac{1}{2}u^2-F(u))dx.
$$
As you can guess, these functionals are conserved quantities of a certain one-dimensional PDE. This PDE has some special kind of solution called traveling wave solutions of speed $c$ (that is, $u(t,x):=\phi_c(x-ct)$ is a solution of the PDE). Note that the fixed profile also depends on $c$ (I mean, not only the speed of the wave but also the "scale" of the profile depends on $c$). Traveling waves of speed $c$ satisfy the following equation:
$$
\qquad \qquad M\phi_c+c\phi_c-\phi_c-f(\phi_c)=0. \qquad \qquad (*)
$$
Here $M$ denotes a non-specified differential operator and $f$ denotes the non-linearity (both of them quite general, but satisfying some regularity and growth assumptions). From now on we shall denote the profile $\phi_c$ just by $\phi$ and we will assume that $c>0$ is fixed. Then, it is clear from the $(*)$ that $\phi$ satisfies $E'(\phi)+cV'(\phi)=0$. Nevertheless, (here is where I got confused) the authors say that from this latter identity we can see $\phi$ as a critical point of $E$ subject to the constraint $V(u)=V(\phi)$.
I don't really understand this last statement, in particular, it bothers me the fact that on the identity $E'(\phi)+cV'(\phi)=0$ there is a factor $c$ appearing, which I don't know how to connect with the statement "$\phi$ is a critical point of $E$ subject to the constraint $V(u)=V(\phi)$". Could anyone explain me how to see this equivalence? I am particularly confused because my knowledge of optimization on functionals is almost zero, I was thinking in a kind of "Lagrange multipliers" but I don't really understand how to implement this method for functionals.
PS: Sorry, I forgot to say, in the definition of $E(u)$ the function $F$ denotes a primitive of the nonlinearity $f(u)$, that is, $F$ satisfies that $F'(x)=f(x)$ and $F(0)=0$. If it helps, $f$ also satisfies that $f(0)=0$.
Edit: I understand that Lagrange multipliers have a $\lambda$ factor on the constraint, but on our equation we don't have any $\lambda$ but $c$, that is, exactly the speed of the traveling wave (that is specifically my doubt). My problem is that $c$ is not a parameter, it is FIXED (positive, arbitrary, but fixed). So it cannot play the role of $\lambda$ on the Lagrange multipliers method.
The problem is precisely that you should not fix a priori the parameter $c$, because it will be exactly the Lagrange multiplier (or rather, $\lambda=-c$ will).
Think of it like this: Let me introduce a new parameter $R>0$, which I think of as a prescribed $L^2$ energy level.
Then the minimization problem
$$
\min\limits_{u\in(\dots)}\Bigg\{E(u):\qquad V(u)=R\Bigg\}
$$
is a constrained minimization problem. Presumably under structural assumptions on your operator $M$ this problem is well-behaved (convex, coercive, and what have you), hence there should be at least a solution $u$.
If you work out the Lagrange multipliers setting, the outcome is that the (Fréchet) gradient $E'(u)$ should be colinear (in $L^2$) with the gradient $V'(u)$. In other words, there is a real number $\lambda$ such that
$$
E'(u)=\lambda V'(u)
$$
Note carefully that I'm purposedly omitting the dependence of both $c,u$ on $R$ (and in fact they need not be unique, more on this later).
Writing $c:=-\lambda$ and given your explicit expressions of $V,E$ we get the PDE
$$
E'(u)+cV'(u)=0
\qquad
\Leftrightarrow\qquad
Mu-u-f(u)+cu=0.
$$
Usually your PDE is of the form $\partial_tu=-Mu+u+f(u)$, hence you can interpret $-cu$ as a convective derivative $\partial_t(u(x-ct))$ and you get indeed a travelling wave.
For any "$V$-energy" level $R>0$ that you chose you get solutions $(u_c,c)$, the typical case being that a value of $R$ gives rise a finitely many solutions. In your partcular case, varying continuously $R>0$ is what gives the whole family of travelling waves.
NB I am fully aware that this is just a purely heuristic explanation, but given that the OP is not very specific I cannot really "prove" anything here.
Note on the non-uniqueness: very often the orientation $c>0$, corresponding to waves moving to the right $u(x-ct)$, is completely arbitrary. In this case one usually also gets left-moving waves $u(x+ct)$, so for any solution $(c,u_c)$ you get another (reflected) solution $(-c,u_{-c})$, which means that the pairs of (constrained minimizers, Lagrange multipliers) are not unique. But this strongly depends on the operator $M$ (and its invariance under symmetries such as $x\to-x$).
PS: sorry for using $u$ instead of $\phi$, but I guess old habits die hard.
|
2025-03-21T14:48:30.167551
| 2020-03-28T16:16:33 |
355970
|
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"LSpice",
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|
Stack Exchange
|
does the ratio of the count of rational numbers on an $n\times n$ grid to $n^2$, converge as $n$ tends to infinity
Suppose we order the rational numbers using the diagonal method (used to prove they are countable) using an $n\times n$ grid. Now suppose we count the distinct rational numbers (those points on the grid where gcd(numerator, denominator) = 1. Denote this by $S_n$.
Does $\lim_{n\to\infty} n^2/|S_n|$ exist, and if so what is it?
Some quick computations suggest that it converges to approx 1.64...
It may not be immediately apparent, but this is not a current research-level problem. Wikipedia says the answer was proven in 1881 and gives as reference Hardy & Wright 2008, Theorem 332.
thanks for the reference
I think I don't understand the statement of the problem. Are you counting rational numbers in a certain box (which one?), or rational numbers on the lattice points of a certain bounded portion of the lattice (in which case the answer seems to be that there are $n^2$)? I'm sure I'm misunderstanding a clear statement, but maybe I'm not the only one.
@LSpice The question is asking for distinct rational numbers in the bounded section. For example, for $n = 2$ (i.e. on the 4 points), there are 3 distinct rational numbers: $2/1, 1/2, 1/1 = 2/2$.
@user44191, oh, so we are counting the image of the mapping $(m, n) \mapsto m/n$ from lattice points to rational numbers. That's what I was missing.
The limit you are looking for is the reciprocal of the probability that two random positive integers are coprime. This probability (with the proper intepretation) is the density of square-free numbers:
$$\prod_{\text{$p$ is prime}}\left(1-\frac{1}{p^2}\right)=\left(\sum_{n=1}^\infty \frac{1}{n^2}\right)^{-1}=\frac{6}{\pi^2}.$$
Note that
$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}=1.6449340668\dots$$
|
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