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2025-03-21T14:48:31.404683
| 2020-07-02T14:52:20 |
364667
|
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"LSpice",
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|
Stack Exchange
|
An inequality for weights of affine Lie algebras, level, and dual Coxeter number
Suppose $\mathfrak{g}$ is an (untwisted) affine Lie algebra with the normalized invariant form $(\cdot | \cdot)$. Let $\lambda \in \mathfrak{h}^\ast$ be a dominant integral weight such that $\lambda(d)=0$ for $d$ the derivation in the loop algebra construction; that is, $\lambda = c_0\Lambda_0 +c_1\Lambda_1+\cdots+c_n\Lambda_n$, where $c_i \in \mathbb{Z}_{\geq0}$, $\Lambda_i$ the fundamental weights. Let $l$ be the level of $\lambda$.
While doing some (albeit limited) explicit computations for various $\mathfrak{g}$ and $\lambda$, I am encountering an inequality of the form
$$
2l(\lambda|\rho)-h^\vee(\lambda|\lambda) \geq 0
$$
where $h^\vee$ is the dual Coxeter number of $\mathfrak{g}$ and $\rho$ is as usual a choice of weight satisfying $\rho(\alpha_i^\vee)=1$ for all simple coroots $\alpha_i^\vee$.
Question: Should this inequality hold for all such $\lambda$?
I have little intuition for whether this should be expected; this very well could be an artifact of the specific $\lambda$ that appear in my computations. For example, I would be somewhat satisfied if this were to hold under the stronger condition that each $c_i \in \{0,1\}$.
What is $d$?...
@LSpice $d$ is the derivation coming from the loop algebra construction of $\mathfrak{g}$; really, I was just emphasizing that $\lambda$ has no $\delta$ term in the summand. I will add that.
The answer to this is found as theorem 13.11 in Kac, "Infinite dimensional Lie Algebras". To be specific, we have $2k(\Lambda|\rho) \geq h^{\vee} (\Lambda| \Lambda)$ for all $\Lambda \in P^k_+$, with equality if and only if $\Lambda = k \Lambda_j$ mod $\mathbb{C} \delta$. Here $j \in J$, where $J$ is a set depending on the Kac labeling of the Dynkin diagram in question (J should correspond to the set of simple roots with Kac label 1, roughly).
This is actually fantastic! Thank you--I am still woefully unfamiliar with the content of the last two chapters of Kac's book, but I'm not completely surprised that this was "hidden" in there!
|
2025-03-21T14:48:31.404855
| 2020-07-02T15:03:20 |
364669
|
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|
Stack Exchange
|
$AC^p$ curves and pointwise metric speed in abstract metric spaces?
For a fixed "reasonable" metric space $(X,d)$ (say complete, separable, whatever is needed...), a curve $\gamma:[0,1]\to X$ is said to be $AC^p(0,1)$ (absolutely continuous) if
$$
d(\gamma(s),\gamma(t))\leq\int_s^t m(r)dr
\qquad\mbox{for all }0\leq s\leq t\leq 1
$$
for some nonnegative function $m\in L^p(0,1)$ (with an obvious definition for $p=\infty$, corresponding to Lipschitz curves).
Theorem: If $\gamma\in AC^p(0,1)$ for some $p\in [1,\infty]$ then the metric derivative
$$
|\dot\gamma(t)|:=\lim\limits_{h\to 0}\frac{d(\gamma(t+h),\gamma(t))}{h}
$$
exists for a.a. $t\in (0,1)$, it is an $L^p$ function, and it is the smallest admissible function $m$ in the above definition of $AC^p$ curves.
The statement and proof can be found in [Ambrosio, Gigli, Savaré, Gradient Flows in Metric Spaces and in the Space of Probability Measures, thm. 1.1.2 page 24]
I am interested in the following characterization of $AC^p$ curves.
Question: Assume that I have a curve $\gamma:[0,1]\to X$ such that, for some function $m\in L^p(0,1)$, there holds
$$
|\dot\gamma(t)|_+:=\limsup\limits_{h\to 0}\frac{d(\gamma(t+h),\gamma(t))}{h}\leq m(t)
\qquad \mbox{for a.a. }t\in (0,1).
$$
Can I conclude that $\gamma\in AC^p$ with $|\dot\gamma(t)|=|\dot\gamma(t)|_+\leq m(t)$ for a.e. $t$?
Of course this seems very plausible, but so far I cannot prove it by hand and I could not find this statement anywhere in the literature. Is this known? (I suspect that there should be an elementary proof) Can anyone provide a reference?
Quick comment: of course the function $|\dot\gamma(t)|_+$ is some kind of upper metric derivative which presumably should control the metric speed itself, it is exists. The statement would immediately follow if we could prove directly that
$$
d(\gamma(s),\gamma(t))\leq \int _s^t |\dot\gamma(\tau)|_+d\tau,
$$
but so far I am stuck and I don't really see how to proceed from the definition of $|\dot\gamma|_+$.
This is not even true for real-valued functions. The standard counterexample is the Cantor function, which is differentiable a.e. with derivative $0$, but is not constant as any absolutely continuous function with this property would be.
|
2025-03-21T14:48:31.405014
| 2020-07-02T16:23:27 |
364678
|
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"Drew Heard",
"R. van Dobben de Bruyn",
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|
Stack Exchange
|
Is $\mathbb{Q}_p \otimes_{\mathbb{Q}}\mathbb{Q}_p $ coherent?
Let $\mathbb{Q}_p$ denote the field of fractions of $\mathbb{Z}_p$. By the answers to this quesition the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_p$ cannot be a Noetherian ring (alternatively, this follows because the transcendence degree of $\mathbb{Q} \to \mathbb{Q}_p$ is infinite). One could instead hope for the weaker result that this ring is coherent. Is this true?
You can use the following:
Lemma. Let $A = \operatorname{colim}_i A_i$ be a filtered colimit of coherent rings such that $A$ is flat over each $A_i$. Then $A$ is coherent.
For example, this is true if all the transition maps $A_i \to A_j$ are flat.
Proof. Let $I \subseteq A$ be a finitely generated ideal. Then $I = AI_i$ for some finitely generated ideal $I_i \subseteq A_i$ for some $i$. By assumption, $I_i$ is finitely presented as $A_i$-module, i.e. there is an exact sequence
$$A_i^m \to A_i^n \to A_i \to A_i/I_i \to 0.$$
By flatness of $A_i \to A$, the sequence
$$A^m \to A^n \to A \to A/I \to 0$$
is exact as well, i.e. $I$ is finitely presented. $\square$
Example 1. Let $K \subseteq L$ and $K \subseteq M$ be field extensions. Then $A = L \otimes_K M$ is coherent. Indeed, it can be written as a colimit
$$A = \underset{\substack{\longrightarrow \\ K \subseteq L_i \subseteq L \\ K \subseteq M_j \subseteq M}}{\operatorname{colim}} L_i \underset K\otimes M_j,$$
where the colimit runs over all finitely generated subextensions $K \subseteq L_i \subseteq L$ and $K \subseteq M_j \subseteq M$. Each $L_i \otimes_K M_j$ is Noetherian, so in particular coherent, and the transition maps are flat because both $L_i \to L_{i'}$ and $M_j \to M_{j'}$ are.
Example 2. The algebraic integers $\bar{\mathbf Z}$ are coherent as the colimit of all $\mathcal O_K$ for $\mathbf Q \subseteq K$ finite. The transition maps $\mathcal O_K \to \mathcal O_L$ are flat because $\mathcal O_K$ is a Dedekind domain and $\mathcal O_L$ is torsion-free.
Thank you, that is very nice! I have to admit, I expected the answer to be negative, and as such didn't ask what I was most interested in - do you happen to know if $\mathbb{Z}p \otimes{\mathbb{Z}} \mathbb{Z}_p$ is coherent?
At the end, "$\mathcal O_K$ is DVR" should be "is a Dedekind domain", or say that its localizations at primes are DVRs
Dear Remy, thank you for the update. But I don't quite understand the example - it seems you claim that the maps in the filtered colimit factor through the localizations? But this does not appear to be the case in general, already for $\mathbb{Z}_{(p)} \to \mathbb{Z}_p$
Right, that didn't work at all. In fact it's somehow the opposite: the valuations on the intermediate rings (induced by the valuation on $\mathbf Z_p$) are basically never divisorial, because that would give a nontrivial extension of residue fields (for dimension reasons). In particular, $\mathbf Z_p$ cannot be written as a colimit of DVRs that are essentially of finite type. I think it's fair to ask the $\mathbf Z_p$-version as a separate question ― I don't see any way to apply my flatness argument.
|
2025-03-21T14:48:31.405250
| 2020-07-02T17:15:56 |
364681
|
{
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"Chris Gerig",
"Joshua Mundinger",
"Neal",
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"sort": "votes",
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}
|
Stack Exchange
|
Laplacian on a manifold with two boundary components
I am interested in the Laplace equation on knot complements. The full complement of a knot $K$ is in $S^3$, but for compactness, we delete an open tubular neighborhood around $K$. The Laplace PDE on $S^3 - K$ with Dirichlet conditions on the boundary torus has only trivial constant solutions. However, I want to model electrostatic charge of a knot in $\mathbb{R}^3$, and since the potential goes to zero at infinity, we can delete an open ball around the point at infinity. Thus our boundary domain has two boundary components: a torus and an $S^2$. Close to the knot, the potential is large, and far away, the potential is small. So we have the PDE $\Delta \Phi =0$ and the boundary conditions that $\Phi$ is a positive constant $c_1$ on the boundary torus and another positive constant $c_2$ on the boundary sphere with $c_1 > c_2$.
Are there any references for solving PDEs of this form, where the boundary is broken into two disjoint components and the conditions are that the solution must take differing constants on them?
What is it that you wish to know? That a solution exists, is unique, and has certain regularity? Or something else?
I already have existence, but I want to know about uniqueness and about the critical sets of the solutions.
Uniqueness follows immediately from the maximum principle, no?
How so? But I need to know more than just existence and uniqueness.
Uniqueness holds as long as you know the value on all boundary components. This is typically explained in electrodynamics books (such as Griffiths') which is what your question is about.
|
2025-03-21T14:48:31.405738
| 2020-07-02T17:22:30 |
364683
|
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|
Stack Exchange
|
Exponentially suppressed events for bounded difference super-martingales
Let $\{ Z_n \mid n = 0,1,..\}$ be a non-negative super-martingale and assume that it is of bounded difference i.e $\exists ~c_i >0$ s.t $\vert Z_{i+1} - Z_i \vert \leq c_i$. Then we know (Azuma-Hoeffding) that $\mathbb{P} \left [ Z_n - Z_0 \geq t \right ] $ is exponentially suppressed for any $t>0$,
Under what conditions are either or both of the following probabilities also low or even exponentially suppressed ?
$\mathbb{P} \left [ \min_{i=0,\ldots,n} Z_i \geq t \right ] $
$\mathbb{P} \left [ \frac{1}{n+1} \sum_{i=0,\ldots,n} Z_i \geq t \right ] $
The above question might be reminiscent of what happens in the ``Kolmogorov's sub-martingale inequality" which says the following : that if $\{ Z_n \mid n = 1,..\}$ is a non-negative sub-martingale then for any $a >0$ and $m \in \mathbb{Z}^+ $we have, $\mathbb{P} [ \max_{n=1,..m} Z_n > a ] \leq \frac{\mathbb{E} [Z_m ]}{a}$
Though I don't know if there is any direct connection between this and my question.
|
2025-03-21T14:48:31.405849
| 2020-06-29T10:46:41 |
364406
|
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"authors": [
"Ben MacAdam",
"Lucina",
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"https://mathoverflow.net/users/75783"
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|
Stack Exchange
|
Has the covariant Hom-functor of the category of additive categories a left adjoint?
Let $\mathsf{Add}$ denote the (strict) 2-category of small additive categories and additive functors. Because categories of additive functors are itself additive, we have for each additive category $\mathcal{A}$ a covariant Hom-functor
$$\mathsf{Add}(\mathcal{A},-) : \mathsf{Add} \to \mathsf{Add} $$
in a 2-categorical sense, i.e. for each additive functor $F:\mathcal{B}\to\mathcal{C}$ we obtain an additive functor
$$F_*^\mathcal{A} : \mathsf{Add}(\mathcal{A},\mathcal{B}) \to \mathsf{Add}(\mathcal{A},\mathcal{C})$$
Does the functor $\mathsf{Add}(\mathcal{A},-)$ have a left adjoint in a 2-categorical sense? Does there exists for each additive category $\mathcal{B}$ an additive category "$\mathcal{B}\otimes\mathcal{A}$" such that we have for each additive category $\mathcal{C}$ a natural isomorphism (or at least an equivalence) of categories
$$\mathsf{Add}(\mathcal{B} \otimes \mathcal{A},\mathcal{C}) \cong \mathsf{Add}(\mathcal{B},\mathsf{Add}(\mathcal{A},\mathcal{C}))\quad ?$$
Yes, additive categories are enriched in the category of abelian groups (or commutative monoids), which is a (co)complete monoidal closed category. I recommend you look at the construction of the hom-V-category and the tensor V-category in chapter 2 of Kelly’s Basic Concepts of Enriched Category Theory.
Thanks for the reference! Can this be extended to abelian categories such that we can understand e.g. Ext as the derived functor of $Hom(-,-) : \mathcal{A}^\text{op}\otimes \mathcal{A} \to \mathsf{Ab}$?
I'm not a homological algebraist, but based on nlab's page on Ext, maybe?
|
2025-03-21T14:48:31.405995
| 2020-06-29T10:52:41 |
364407
|
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"ABB",
"Mateusz Wasilewski",
"Matthew Daws",
"Nik Weaver",
"Ruy",
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|
Stack Exchange
|
Commutative C*-rings
Let us consider the unital commutative $C^*$-algebra $C[0,1]$. We say $A\subseteq C[0,1]$ forms a C*-subring if it satisfies the following conditions:
1- $A$ is an involutive unital subring (closed under pointwise addtion, multiplication and invoultion operations) whose unit is just the constant function $1$.
2- $A$ is norm-closed.
Q. Any characterization of (the elements of) $A$?
So the important point is: no $*$-operation, and no scalar multiplication?
Dear Matthew, Thanks. Just no scaler multiplication. I made a correction.
A remark: the C*-subring generated by x is formed by all functions that are integer valued at 0 and 1 (see Theorem 5 in "What Can Be Approximated by Polynomials with Integer Coefficients" by Le Baron O. Ferguson).
@ Mateusz Wasilewski. The Theorem that you mentioned to says: For a continuous real-valued function $f$ on the unit interval $[0, 1]$
to be uniformly approximable by polynomials with integer coefficients it is necessary
and sufficient that it be integer-valued at both 0 and 1.
This is not an answer! It is hoped to be a reasonable conjecture, at best!
Based on Mateusz Wasilewski's comment one can build examples of subrings of $C([0,1])$ by choosing a closed subset $K⊆[0,1]$ and taking $A$ to be the set of
functions in $C([0,1])$ taking integer values on $K$. However this is not such a general example for at least two
reasons. It fails to include:
sub-C*-algebras of $C([0, 1])$, and
algebras of the form $R\cdot1$, where $R$ is a closed subring of $\mathbb{C}$.
Algebras of form (1) can be quite big (think of space filling curves) so the choice of $[0,1]$ as base space does not seem
to bring any relevant simplification and indeed it may actually obfuscate the problem. So it is perhaps a good idea to
recast the problem asking instead for the general form of subrings of $C(X)$, where $X$ is a compact topological space.
If $A$ is a subring of $C(X)$, then it is also a subring of the sub-C*-algebra generated by $A$, which in turn
is of the form $C(Y)$, where $Y$ is a quotient of $X$. It is easy to see that $A$ separates points of $Y$ so it might
make sense to require this condition in the original question.
Regarding (2), I haven't given much thought as to what is the general form of a closed subring of $\mathbb{C}$, such as
$\mathbb{Z}+i\mathbb{Z}$ (any ideas?), but at first glance it does not seem to be an entirely trivial question so let's do away with this
difficulty by sticking to real valued functions instead. After all, apart from $\mathbb{R}$ itself, the only closed unital subring of $\mathbb{R}$ is
$\mathbb{Z}$!
So here is my conjecture:
Let $X$ be a compact Hausdorff topological space and let $A$ be a closed unital subring of $C(X, \mathbb{R})$ (continuous real valued functions on $X$), which separates points of $X$. Then there exists a closed subset $K⊆X$
such that $A$ coincides with the set of functions in $C(X)$ taking integer values on $K$.
Hopefully a clever modification of the proof of the Stone-Weierstrass Theorem can prove this as well!
It feels like you're on the right track, but I think the ring generated by $f(x) = x$ on $[0,2]$ is a counterexample. Anything in the ring is integer-valued at $x = 0$, $1$, or $2$, but also must have the same parity at $x = 0$ and $x = 2$ ... maybe your conjecture can be refined to accomodate this?
Yes, you are right! This immediately brings about a bunch of ideals associated to points in the space and a bunch of homomorphisms between them! So far I cannot see how to organize all of this :-(
For instance, I think the discrete unital subrings of $\mathbb{R}^2$ are $\mathbb{Z}^2$ and, for each $n$, the set ${(a,b): a \equiv b , ({\rm mod}, n)}$.
@NikWeaver, what scares me most is the subring of $\mathbb{R}^3$ formed by the integer triples $(x,y,z)$ such that
$x\equiv y (\text{mod } p)$,
$y\equiv z (\text{mod } q)$, and
$z\equiv x (\text{mod } r)$.
Hmm. So maybe for every pair of points in $K$ we have such a condition?
This makes me think of the categorical notion of equalizer: given two homomorphisms $f$ and $g$ from the ring $A$ to the ring $B$, the set ${a\in A: f(a)=g(a)}$ is a subring.
Identity and conjugation are the only continuous automorphisms of $\mathbb{C}$. Nevertheless $\text{Aut}(\mathbb{C})$ is a huge group with cardinality $2^\frak{c}$. I have never heard of a "closed graph Theorem" that applies to automorphisms of $\mathbb{C}$ so, in my ignorance I am allowed to imagine that there is no such Theorem, meaning that there is a discontinuous $\phi\in\text{Aut}(\mathbb{C})$ with closed graph. If this is granted then ${(x, y)\in \mathbb{C}^2: \phi(x)=y}$ is a closed subring (actually a field) of $\mathbb{C}^2$ which is not of any form so far mentioned here!
PS: The above will not be invariant under involution.
Well, a first, simpler question is what are the unital subrings of $\mathbb{Z}^n$? This paper makes it look like this is already pretty complicated. (See Propositon 2.3, for instance.)
Here is an idea that might help separate the analytical aspects of the problem from the sticky algebraic ones. Given a closed subring $A\subseteq C(X, \mathbb {R})$, let $K={x\in X: f(x)\in \mathbb{Z},\ \forall f\in A}$. We then get a subring $B\subseteq C(K, \mathbb {Z})$ by restricting every $f$ in $A$ to $K$, and one might try to prove that $A={f\in C(X, \mathbb{R}): f|_K\in B}$. Of course this says nothing about the algebraic structure of $B$ but, according to @NikWeaver 's comment, people seem to know a lot about this.
Yes, I like this idea.
Let $R$ be a unital subring of $\mathbb{Z}^n$ and choose $n$ points $t_1$, $\ldots$, $t_n$ in $[0,1]$. We can see from Proposition 2.3 of this paper that $R$ can be fairly complicated. Then the set of functions in $C[0,1]$ whose restriction to $\{t_1, \ldots, t_n\}$ belongs to $R + iR$ is an involutive unital subring of $C[0,1]$. I think this shows that it is unreasonable to expect there to be any clean characterization of all involutive unital subrings of $C[0,1]$.
|
2025-03-21T14:48:31.406414
| 2020-06-29T11:10:24 |
364408
|
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|
Stack Exchange
|
Link between characters and isotypic components
I am currently studying finite complex reflection groups using the book written by Lehrer and Taylor called "Unitary Reflection Groups" and I am unsure if I understood isotypic components the right way. Let $G$ be a finite complex reflection group and $I$ an irreducible $G$-module. They define the isotypic component of $I$ in regard to a $G$-module $V$ as the direct sum of all submodules of $V$ isomorphic to $I$. Later on for a $G$-module $M$ they introduce $M^G:=\{m\in M\mid gm=m \forall g\in G\}$ and describe it as the isotypic component corresponding to the trivial representation. My question about this is that $M^G$ doesn't have to be irreducible (since every element of $M^G$ is fixed by the operation of $G$ by definition every subspace of is a submodule, so it is is irreducible if and only if it is onedimensional or $0$). But because every onedimensional subspace is a submodule we can take a basis of $M^G$ and every basis vector spans an irreducible (because onedimensional) subspace. If $\{v_1,\dots,v_k\}$ is a basis for $M^G$ does that imply the isotypic component corresponding to the trivial representation is $M^G=\oplus_{i=1}^k\mathbb{C}v_i$ or did I misunderstand the definition? Because the different submodules $\mathbb{C}v_i\simeq \mathbb{C}$ are clearly isomorphic and irreducible (since they are onedimensional). So $M^G$ is not irreducible itsself, but the sum of irreducible, isomorphic submodules.
What you say about $M^{G}$ is correct.. Likewise, if $I$ is any irreducible $G$-module ,the isotypic component of a $G$-module $V$ corresponding to $I$ is the sum (not necessarily direct) of all submodules of $V$ isomorphic to $I$. As you have noted when $I$ is trivial, the isotypic component associated to a general irreducible $I$ will be a direct sum of the form $I_{1} \oplus I_{2} \ldots \oplus I_{k}$, where each $I_{j}$ is isomorphic to $I$ and the dimension of the isotypic component is kdim(I), but not every submodule of $V$ isomorphic to $I$ need be one of the $I_{j}$ in general.
Thank you for the explanaition! This really helped me. This question can be closed (but I don't know how).
|
2025-03-21T14:48:31.406696
| 2020-06-29T12:27:37 |
364413
|
{
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"Wojowu",
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"https://mathoverflow.net/users/164630",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/41291",
"მამუკა ჯიბლაძე"
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"sort": "votes",
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}
|
Stack Exchange
|
What is the multiplicative order of this number
Let $q, r \in \mathbb{P}$ and $r$ is the next prime to $q$.
What is the multiplicative order of $r$ modulo $\displaystyle\bigg( \prod_{\substack{p \leq q \\\text{p prime}}} p \bigg)$ ?
In other word what is the smallest $k \in \mathbb{N}^*$ verifying :
$$r^k = 1 \pmod{\prod_{\substack{p \leq q \\\text{p prime}}} p}$$
Using Euler theorem we know that $k$ divide $\phi \displaystyle\bigg( \prod_{\substack{p \leq q \\\text{p prime}}} p \bigg) = \prod_{\substack{p \leq q \\\text{p prime}}} (p - 1)$
Many thanks for any help
Edit 20/09/2020 : The sequence is added here : https://oeis.org/A333992
$k$ is the least common multiple of orders of $r$ modulo $p$ for all $p\leq q$, so in particular it divides the least common multiple of all the $p-1,p\leq q$. I doubt you will get a formula any more explicit than that.
@Wojowu Just out of curiosity - dividing that lcm by the number in question gives the sequence $1,1,1,2,3,1,2,5,1,3,3,7,2,22,2,2,1,2,1,1,1,2,2,1,4,12,5,2,6,1,1,5,1,1,2,1,6,82,1,1,...$
Thank you @მამუკაჯიბლაძე and Wojowu for your contributions.
Seems not to be in the OEIS. I encourage you to contribute it.
Thank you Mr Robert, i will add it.
|
2025-03-21T14:48:31.406816
| 2020-06-29T12:59:50 |
364415
|
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"Delmastro",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364415"
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|
Stack Exchange
|
Pell equation and quadratic residues
We say an integer $k$ is Pell if there exist some integers $p,q$ such that
$$
p^2k-q^2=1
$$
In studying a physics system we ended up with two weaker notions of Pell:
We say an integer $k$ is pre-Pell if there exist some integers $p,q$ such that
$$
pk-q^2=1
$$
We say an integer $k$ is weakly Pell if there exists some Pell integer $k'$ such that the product $kk'$ is also Pell.
The physics of the problem led us to conjecture that these two notions are in fact equivalent:
Conjecture: $k$ is pre-Pell if and only if it is weakly Pell.
One of the directions is obvious. We have very superficial knowledge of number theory so the only proof we could come up with was using some very strong conjectures (namely, one of the Hardy-Littlewood conjectures), but I think it is safe to assume that a much simpler, unconditional proof should exist. I would appreciate it if someone could sketch it here, or point me in the right direction.
$k$ is "pre-Pell" if and only if $-1$ is a square modulo $k$, right? Hence, if and only if $k$ has no prime divisor that's one less than a multiple of four, right? "Pell" is a little bit harder, e.g., $34$ has no disqualifying prime divisor, but $q^2-34p^2=-1$ has no integer solutions.
@GerryMyerson Yes, that is correct. (For "pre-Pell" you can also allow a single factor of $2$). Strictly Pell is more complicated, no simple characterisation (in terms of prime factors) is known, to the best of my knowledge.
The answer is yes. As was observed, the condition of being pre-Pell is simply the stipulation that $k$ is a sum of two squares: that is, if $p | k$ and $p \equiv 3 \pmod{4}$ then $p$ must divide $k$ with even multiplicity. If we assume $k$ is square-free, then it is divisible only by $2$ or primes of congruent to $1 \pmod{4}$.
In fact, if $k$ is pre-Pell then all of its odd prime divisors must be congruent to $1 \pmod{4}$. To see this, suppose $x^2 + 1$ is divisible by $p^2$ for some $p \equiv 3 \pmod{4}$. Then in particular it must be divisible by $p$, so the congruence $x^2 + 1 \equiv 0 \pmod{p}$ must be soluble. But this is not the case: $-1$ is a square mod $p$ if and only if $p = 2$ or $p \equiv 1 \pmod{4}$.
We now work on the equivalences. Suppose that $k$ is such that there exists $k^\prime$ such that the negative Pell equation is soluble for both $k^\prime$ and $kk^\prime$. Being a sum of two squares is a necessary condition for negative Pell to be soluble, hence the latter condition implies that $k = (kk^\prime)/k^\prime$ must also be a sum of two squares. To see this, suppose that $kk^\prime$ has a prime divisor $p$ congruent to $3 \pmod{4}$ (otherwise it is obvious that $k$ is a sum of two squares). Suppose that $p^{2m} || kk^\prime$ and $p^{2n} || k^\prime$. Then $p^{2m-2n} || k$, hence $p$ divides $k$ to even multiplicity. Therefore, $k$ is pre-Pell as desired.
The converse is much harder. Suppose that $k$ is a sum of two squares. We can use the fact that for any prime $p \equiv 1 \pmod{4}$, the negative Pell equation $x^2 - py^2 = -1$ is soluble. Thus we may reduce the question to the following: does there exist a prime $p \equiv 1 \pmod{4}$ such that $x^2 - kpy^2 = -1$ is soluble?
Fortunately this can be done via governing fields: that is, for each $k$ there exists a number field $F_k$ such that the 4-rank of the class group of the field $\mathbb{Q}(\sqrt{kp})$ is determined by the splitting behaviour of $p$ in $F_k$. In particular, if the 4-rank of the class group of $\mathbb{Q}(\sqrt{kp})$ is zero, then negative Pell is soluble. We are then done by Chebotarev's density theorem guaranteeing the existence of such a prime (in fact, infinitely many).
Hi, I just came back to this post and got confused again. Pre-Pell is not just the stipulation that $k$ is the sum of two squares: we also require these squares to be coprime, right? Or am I missing something? Thanks!
I will add some comments to the answer to address your question.
Great, thank you very much!
|
2025-03-21T14:48:31.407072
| 2020-06-29T13:02:41 |
364416
|
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"url": "https://mathoverflow.net/questions/364416"
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|
Stack Exchange
|
Why does norm map the $\sigma$-conjugacy classes to the conjugacy classes?
Let $E/F$ be a cyclic extension of order $\ell$ (not assumed prime) of fields of characteristic $0$, and $\Sigma$ its Galois group; we denote by $\sigma$ a generator of $\Sigma$. We denote by $G(E), G(F)$ the groups of $E,F$-rational point on $G={\rm GL}_{n}$. Two elements $g,h\in G(E)$ are called $\sigma$-conjugate if $g=x^{-1}hx^{\sigma}$ for some $x\in G(E)$, and $N(g)=gg^{\sigma}\cdots g^{\sigma^{\ell-1}}$ is called the norm of $x$.
Question. Is the norm map an injection from the set of $\sigma$-conjugacy classes in $G(E)$ into the set of conjugacy classes in $G(F)$ ?
This is the assertion of lemma 1.1 in ''Simple Algebra, Base Change, and the Advanced Theory of the Trace Formula" written by J. Arthur and L. Clozel. Their proof was very hard to understand for me.
What is your question? (The title and body are different.) Are you looking for a yes/no answer? An explanation of their proof? My recollection is that it's just an argument in Galois cohomology.
|
2025-03-21T14:48:31.407177
| 2020-06-29T14:10:46 |
364420
|
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}
|
Stack Exchange
|
Possible limit involving the gamma function
Does $$\lim_{n \to \infty} \int_{0}^{1} \Gamma(x)^{n/(n+1)}dx - n$$ exist?
Here's some background. The integral
$$\int_{0}^{1} \Gamma(x) dx$$
diverges rather slowly. Inserting the exponent $n/(n+1)$ perhaps leads to a nice surprise---that the floor of resulting integral appears to be $n$. For example, for $n = 100$, the integral has a value of $100.759456...$
$\newcommand\Ga\Gamma$
Note that $\Ga(x)=\Ga(1+x)/x$ for $x>0$ and
$-n=1-\int_0^1 x^{-n/(n+1)}\,dx$ for $n>0$.
So, the limit in question is
$$1+\lim_n J_n,$$
where
$$J_n:=\int_0^1 x^{1/(n+1)}f_n(x)\,dx,$$
$$f_n(x):=g(x)-h_n(x),$$
$$g(x):=\frac{\Ga(1+x)-1}x,\quad h_n(x):=\Ga(1+x)\frac{\Ga(1+x)^{-1/(n+1)}-1}x.$$
Letting $c$ stand for any expressions bounded uniformly over all $x\in(0,1)$ and all $n\ge1$, we have $\Ga(1+x)=1+cx$ and $\Ga(1+x)^{-1/(n+1)}=1+cx/n$, so that $h_n(x)=c/n$ and hence $\int_0^1 x^{1/(n+1)}h_n(x)\,dx\to0$. Thus, the limit in question is
$$1+\int_0^1 g(x)\,dx=0.75330\ldots. $$
As seen from the proof, the rate of convergence here is $O(1/n)$. So, the limit value $0.75330\ldots$ is in agreement with the value of the integral you computed for $n=100$.
More generally, if $h(x)$ is any function on $(0,1]$ such that $h(x)-\frac1x$ is continuous at $x=0$, then $\int_0^1 h(x)^{n/(n+1)},dx - n \to \int_0^1 \big( h(x)-\frac1x \big),dx$. The proof of this more general statement along the lines of your solution is even cleaner (it shows that the functional equation for $\Gamma$ is irrelevant, for example).
@GregMartin : The above proof will indeed hold if you write $h(x)$ and $xh(x)$ instead of $\Gamma(x)$ and $\Gamma(1+x)$, respectively, everywhere in the proof, assuming that $h(x)-1/x$ is bounded.
|
2025-03-21T14:48:31.407313
| 2020-06-29T14:49:01 |
364423
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364423"
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|
Stack Exchange
|
Invariant theory for the orthogonal group and Clifford algebras
The first fundamental theorem of invariant theory for
the orthogonal group $O_n(k)$ asserts that the
ring of invariants is generated by the scalar products:
a polynomial function of $m$ vectors $v_1,..., v_m \in k^n$ is necessarily a polynomial in the products $\langle v_i, v_j \rangle$.
In the real case, in dimension 2, that is for $O_2({\bf R})$,
there is a short proof using the complex numbers.
I am wondering if there is also a simple proof in higher dimension, for the $O_n({\bf R})$ case, using Clifford algebras? Any reference welcome.
|
2025-03-21T14:48:31.407399
| 2020-06-29T15:55:31 |
364431
|
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"authors": [
"Asaf Karagila",
"Mohammad Golshani",
"Pedro Sánchez Terraf",
"Timothy Chow",
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"https://mathoverflow.net/users/66044",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364431"
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|
Stack Exchange
|
Which is the more popular approach to forcing in the literature?
There are some interesting questions and answers on the site discussing the different approaches to forcing in set theory, and I understand that the two most important are the ones using countable transitive models (ctm) and Boolean valued ones (bvm), respectively.
My question is primarily about which of those two approaches appears in the literature more often, especially in research articles. I know that it might be difficult to answer, but perhaps an educated guess by the users of the site will be enough for me.
The context of this question is that my team is working on a formal verification of forcing using the ctm approach, and it is important for us to be able (to the extent possible) to represent the actual practice of the subject. It is to be noted that a full formalization of the bvm approach was recently completed by Han and van Doorn.
I know four approaches to forcing: using partial orders, using Boolean valued models, topological approach and categorical approach. For me working with partial orders is the most convenient , but sometimes, it is easier to use the Boolean valued approach.
Thanks @MohammadGolshani for your clarification. Also, I do not know if classical realizability fits in any of those four approaches. In any case, I'd be interested in knowing if my claim that the first two are the main ones is incorrect.
There are two types of "working with forcing":
We can develop the theory of forcing, e.g. iterations, where working with canonical forcing notions is somewhat preferable, so dealing with complete Boolean algebras is somehow the most natural approach, and by extension with Boolean-valued models (well, sometimes).
For example, talking about homogeneity conditions is easy when you have them. But maybe you have a rigid partial order which is forcing equivalent to adding a Cohen real (e.g. construct a tree where each node has a unique number of successors). Or perhaps you want to iterate forcings, but the standard definition of iteration given in terms of partial orders is a pre-ordered set. Being able to forego all of that and just find an invariant is great.
We can use the theory of forcing, e.g. proving various consistency results. In this case it is almost exclusively done with partial orders, and indeed with pre-orders, where we simply ignore all of the forcing theoretic issues that make the "formally correct statements" just a huge pain in the lower-lower-back to state.
I think this is best reflected in Jech's "Set Theory" book (3rd edition, for those keeping track). The basic theory of forcing is developed with Boolean algebras and Boolean-valued models. When forcing is actually used, Jech quickly reverts back to partial orders and pre-orders, instead.
Now you can also talk about forcing with topological spaces, forcing with sheaves (or shivs), etc. This is not very common in set theoretic papers in the last few decades. I won't comment on other subjects, as I'm not an expert.
Finally, a word about the foundations of forcing. When one learns about forcing, it is often confusing. The generic object is seemingly black magic, and what's going on with those Cohen reals encoded in limit steps? And what is this "arbitrarily large, but finite fragment of $\sf ZFC$" that Kunen keeps talking about?
Well, the reality is that we can develop forcing in a lot of different ways:
Just force over countable transitive models of $\sf ZFC$. That's the simplest, most straightforward way to do it. But this requires us to assume more in terms of consistency.
Just force over countable models of $\sf ZFC$. Oh, but then it gets ugly when talking about things like ordinals and whatnot, because these models are not necessarily well-founded. Also, this requires more consistency, although significantly less than before.
Use reflection to argue that we can find countable transitive models of any large enough fragment of $\sf ZFC$, force over those, and use a meta-theoretic argument to conclude the proof.
Use Boolean-valued models to develop forcing as a proper class and argue with Boolean-valued truth that certain statement are consistent. But that's kind of yuck in most cases.
Instead of Boolean-valued models, define an "internal ultrapower" of the universe by extending the filter base that is the dense open sets to a "generic" filter, and use this model, where the forcing theorem and truth lemma still hold, to finish your argument. In some sense this is a neater version of Boolean-valued models, but in another sense it is quite the opposite.
Use Feferman's theory, where we add a constant symbol, postulate that it is a countable transitive elementary submodel of the universe, then force over that model. No additional consistency is needed, as Feferman's theory is finitely consistent (assuming $\sf ZFC$ is), so it is given to us. But it sort of makes this specified model somehow... on a pedestal. Also without further assumptions (which are tantamount to (1) with more power) the models of Feferman's theory are ill-founded, which is yuck to think about from a meta-theoretic point of view.
Use other tricks and machinery to encode forcing and just work syntactically in a theory as weak as $\sf PRA$. (Shudder here.)
But what do people actually end up doing (once they grok forcing)? Well. We force over the universe. We just ignore all of it and force over the universe. Because at the end of the day, the goal is to use forcing, and as all of these approaches lead us to the same way, and we anyway define everything and work internally to whatever set-sized model we may have used, we might as well force over the universe. Simply rub your hands, and a generic appears! Magic!
Thank you very much Asaf.
It was confusing for me, and when first reading Jech I did wonder where the heck did $G$ come from. With Kunen I felt reassured, and by jumping inside a ctm, I was able to understand what it felt to have a generic visit from outer space.
Now, when you force over $V$, do you feel the same :-)? Which of the approaches represents that feeling better?
Read my first paper. See how confused I was. Read my most recent papers. You tell me what approach represents my feelings... :-)
I believe I can get the last one from arXiv. Is the first one on your website?
The first one is also on arXiv, but the published version should be free nowadays anyway.
+1, thanks. I'll take a look. I'll wait a few days before accepting.
Hey, as long as people are reading my papers... :-)
It remains an interesting practical question what approach a proof assistant should take. Requiring additional consistency is probably undesirable and PRA is horrible. People use partial orders in practice so BVMs seem awkward. I don't know how easily proof assistants can implement the meta-theoretic argument of your option 3. So maybe 5 or 6 is the way to go?
@TimothyChow: That's an interesting question. I think that the answer is more arithmetic than you'd expect. Because at the end of the day you care about the statement $p\Vdash\varphi$, which is internal, which means that you can just think about it syntactically from a formalist point of view, which seems to me as how proof assistants are wired.
@AsafKaragila : I don't think proof assistants are necessarily "wired to be syntactic" in this particular sense; on the other hand, from the point of view of, say, reverse mathematics, it's valuable to know that some of these theorems can be proved in very weak systems. Here's a question: How often in practice do set theorists need the fact that some theorem (proved using forcing) is provable in some system that is much weaker than ZF? My impression is that this is rare. One just declares that one is working in ZF and it never matters that some weaker system suffices.
@TimothyChow: You're right. I cannot recall any theorem that I saw was mentioned as "Aha! But we can actually prove this from PRA!", this is best captured in a quote from a thesis by VanLiere about the choice of metatheory. My point is that when we work with forcing, we envision an actual model, but a proof assistant is just a verifying procedure, not a person who needs mental crutches to deal with infinite objects.
@TimothyChow Still on the practical side, in the Isabelle/ZF proof assistant (the one we're using), doing the meta-reasoning (about the argument using Reflection) right now is unfeasible. On the other hand, the other various assistants in which this is most easily set up, have metatheories above than one inaccessible in consistency strength.
And concerning the finite, but arbitrary fragment of ZFC (call it $\Phi$), Kunen says that there is another (fixed) finite $\Delta$ such that if $M\models \Phi + \Delta$, then $M[G]\models \Phi$. Through the formalization we are able to pinpoint $\Delta$ exactly (though, since we have not formalized all the requisites for having $M[G]\models\neg \mathit{CH}$, the $\Delta$ we have today is not yet complete).
@PedroSánchezTerraf: One can argue that meta-reasoning is in fact "the reasoning", because we are proving a theorem about ZFC, rather than proving a theorem in ZFC. Forcing is just useful this way that it lets you work from within a model within the theory, without worrying about the theory itself.
You're right. What we'll be able to do, in our framework, is to prove ZFC theorems (about variously extending ctms).
|
2025-03-21T14:48:31.408047
| 2020-06-29T15:56:37 |
364432
|
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|
Stack Exchange
|
Calabi-Yau structures on dg-categories
A (smooth) dg algebra is called (left) Calabi-Yau if (see for example here)
$$ A^! = A[-n]$$
Here we use the inverse dualizing complex $A^!=\mathbf{R}\operatorname{Hom}_{(A^e)^{op}}(A,A^e)$. In topological string theory (e.g. in the context of matrix factorizations, see here or here), it is interesting to regard $A$ as a dg-category and form a sort of Yondeda-embedding $\hat{A} := \operatorname{Int}(A-{\operatorname{Mod}})$ defined in Toen of it, the full subcategory of fibrant-cofibrant objects in the category of $A$-dg-modules.
9
My questions are the following:
Is the dg-category $\hat{A}$ (where the dg-structure is inherited from $A-{\operatorname{Mod}}$ automatically a Calabi-Yau $A_\infty$-category when regarded as an $A_\infty$-category? So, are both of these definitions of Calabi-Yau compatible? Due to the work of Costello, that would be required for the above construction to actually define a TCFT, and it is often stated it actually does - however I have never actually seen a discussion of this subtle point.
If this is not always the case, is the homotopy category of $\hat{A}$ at least a Calabi-Yau category?
|
2025-03-21T14:48:31.408153
| 2020-06-29T16:15:04 |
364434
|
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"LeechLattice",
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"url": "https://mathoverflow.net/questions/364434"
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|
Stack Exchange
|
$t$-balanced numbers
Disclaimer: throughout this question, we'll assume the truth of Goldbach's conjecture.
For $n$ a large enough composite positive integer, write $r_{0}(n):=\inf\{r>0,(n-r,n+r)\in\mathbb{P}^{2}\}$, $r_{0}^{-}(n):=n-\sup\{p\mid p<n\}$, $r_{0}^{+}(n)=\inf\{p\mid p>n\}-n$, $r_{k+1}^{-}(n)=n-\sup\{p\mid p<n-r_{k}^{-}(n)\}$, $r_{k+1}^{+}(n)=\inf\{p\mid p>n+r_{k}^{+}(n)\}-n$.
Let $t_{\pm}(n):=\inf\{k\mid r_{k}^{\pm}(n)=r_{0}(n)\}$. Write also $k_{0}(n):=\pi(n+r_{0}(n))-\pi(n-r_{0}(n))$. One has $t_{-}(n)+t_{+}(n)=k_{0}(n)-1$.
Say $n$ is $t$-balanced iff $t_{-}(n)=t_{+}(n)=t$. Can one get an asymptotics for the number of $t$-balanced integers not exceeding $x$?
Is $t_{-}(n)$ just the number of primes in the open interval $(n-r_0(n),n)$?
Indeed, that's a way to phrase it.
|
2025-03-21T14:48:31.408226
| 2020-06-29T16:27:22 |
364435
|
{
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"Moishe Kohan",
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"jg1896"
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|
Stack Exchange
|
Pseudoreflection groups in affine varieties
Suppose $\mathsf{k}$ is an algebraically closed field of zero characteristic. Chevalley-Shephard-Todd (C-S-T) Theorem in one of its equivalent versions is the following result:
(C-S-T): Let $G$ be a finite subgroup of $GL(V)$, where $V$ is a finite dimensional vector space. Then the quotient variety $V/G$ is smooth if and only if $G$ is a group generated by pseudoreflections.
My first question is the following:
(Question 1) Let $G$ be a finite group of automorphisms of an irreducible affine smooth variety $X$. When is the quotient variety $X/G$ smooth?
Of course this holds if the action of $G$ is free, but I'm looking for a condition that is similar in nature to that of C-S-T Theorem. I've heard something among these lines: $G$ is generated by elements $g$ such that $X^g = \{ x \in X| g.x=x \}$ is a divisor, etc; but I could not find an exact statement or reference.
I would greatly appreciate an answer, either in form of an adequate reference, or a proof of an exact statement.
My second question is related to the first one as follows. We have a result by Kantor (1977) and Levasseur (1981) that is:
(K-L): Let $V$ be a finite dimensional vector space, $G$ a finite subgroup of $GL(V)$. Then the quotient map $V \mapsto V/G$ induces an isomorphism $\mathcal{D}(V)^G \simeq \mathcal{D}(V/G)$ if and only if $G$ contains no pseudoreflections.
(cf. Levasseur, Thm 5, MR633520)
Here of course $\mathcal{D}(.)$ is the ring of differential operators, and the action of $G$ on this ring is by conjugation.
(Question 2) Let $X$ be an irreducible affine smooth variety, and $G$ a finite group of automorphisms of it. The quotient map $X \mapsto X/G$ induces an injective map $\mathcal{D}(X)^G \rightarrow \mathcal{D}(X/G)$. When is this map an isomorphism?
Again there are sufficient conditions for this to hold, such as the action of $G$ being free (cf. Cannings, Holland, 3.7, MR1305263, 1994). I wonder if there is a condition for this to hold analogue to the one considered by Kantor and Levasseur, i.e., $G$ having no "pseudoreflections", in the sense I suggested in Question (1).
Should $X$ be affine in Question 1 (judging by title looks like that's what you intended)?
Yes, it is, thank you for pointing out. I will edit it.
I do not know about the general case, but over the complex numbers, $X/G$ is smooth if and only if the action on $T_xX$ of the stabilizer $G_x$ of each $x\in X$ satisfies the CST condition. The key is Cartan's theorem on local holomorphic linearization, see the reference I gave here.
This is extremely helpful! I will look Cartan's work. Thank you very much @MoisheKohan
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2025-03-21T14:48:31.408422
| 2020-06-29T16:28:37 |
364436
|
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Stack Exchange
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Counting quadratic curves in $\mathbb P^1 \times \mathbb P^1$ passing through seven points in general position
Let $p_1,\dots,p_7 \in \mathbb P^1 \times \mathbb P^1$ be 7 points in general position.
What is the number of maps $F=(F_0,F_1):\mathbb P^1 \to \mathbb P^1 \times \mathbb P^1$ modulo domain reparametrization for which $F_0$, $F_1$ are degree 2 and such that the image of $F$ passes through $p_1,\dots,p_7$?
Can you reduce to the case of $\mathbb P^2$ by blowing up one of your seven points and then blowing down the two lines through it and adding them to your list of points? I think this will preserve your genericity condition and then you're reduced to cubics on $\mathbb P^2$. If not, adapting the argument from cubics on $\mathbb P^2$ with elliptic curves should work.
By the "generalized Riemann-Hurwitz formula", the number of curves of geometric genus $0$ in a general pencil of $(2,2)$ divisors on $\mathbb{P}^1\times \mathbb{P}^1$ equals the (topological) Euler characteristic of the blowing up of $\mathbb{P}^1\times \mathbb{P}^1$ at the $8$ base points of the pencil, i.e., the number equals $8+2\times 2 = 12$.
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2025-03-21T14:48:31.408527
| 2020-06-29T16:49:07 |
364439
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Stack Exchange
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$X-Y$, where $X$ and $Y$ are sums of Bernoulli random variables
Let $X = x_1 + x_2 + \ldots + x_n$ and $Y = y_1 + y_2 + \ldots + y_n$, where each $x_i$ is an independent Bernoulli random variable with success probability $p_i$ and each $y_i$ is a Bernoulli random variable with success probability $q_i$. In addition, suppose that $\sum_i p_i = c +\sum_j q_j$ for some constant $c>5$. My question is, whether or not the inequality $$ \mathbb{P}(X-Y =2) \geq \mathbb{P}(X-Y=1) $$ always holds?
Do you assume that all random variables $x_1,\ldots,x_n,y_1,\ldots,y_n$ are independent?
Yes, all the variables are independent.
I guess $X$ and $Y$ are what's known as "Poisson binomial" random variables, and the question is basically whether a difference of two independent Poisson binomials is unimodal.
Do you have investigated the limiting case that $X \sim Po(\lambda)$ and $Y \sim Po(\lambda-c)$ with $c > 5$ and $\lambda > 5$ and $X,Y$ independent? What if you replace $c$ by some smaller value?
It seems that it is valid for the limiting cases. While I was searching for the correct answer, I found the following series of results: 1- both $X$ and $Y$ are unimodal and log-concave (https://arxiv.org/pdf/0912.0581.pdf) and 2- when one distribution is unimodal while the other one is log-concave, their difference is also unimodal (statement in the proof of proposition 7 in paper "Precision May Harm: The Comparative Statics of Imprecise Judgement."). Is it correct?
@BjørnKjos-Hanssen: Does unimodality necessarily implies the inequality?
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2025-03-21T14:48:31.408776
| 2020-06-29T17:20:47 |
364440
|
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Stack Exchange
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Relation between variables (vertexes, edges, regions and faces) in three dimensional Voronoi diagram
A Voronoi diagram is a kind of tesselation that divided the medium into polygons in 2D and polyhedrons in 3D. In two dimensions, any Voronoi diagram has vertexes(V), edges(E) and regions(F) that equal the number of sites. Euler’s formula: V - E + F = 2 demonstrates the relationship between these variables. Furthermore, the relation between the vertexes and edges is obtained as: = 3 − 6 with some assumptions. Also in three dimensions, any Voronoi diagram has vertexes, edges, regions and faces. I want to know in three dimensions, is there any relation between these variables like two dimensions? I am new to computational geometry and happy with any kind of regard.
Cross-posted: https://mathoverflow.net/q/364440/37212, https://math.stackexchange.com/q/3739061/14578. Please do not post the same question on multiple sites.
Euler's formula $V-E+F=2$ is not specific to Voronoi diagrams.
Rather, it counts those quantities for a planar graph,
or for a polyhedron in $\mathbb{R}^3$ (whose $1$-skeleton is a planar graph).
See David Eppstein's Twenty Proofs of Euler's Formula.
Euler's formula for polytopes in $\mathbb{R}^4$
(and Voronoi diagrams in $\mathbb{R}^3$) is $$V-E+F-C=0 \;,$$ where
$C$ is the number of three-dimensional cells.
So the hypercube becomes $8 - 16 + 32 - 24 = 0$.
The constants $=2$ and $=0$ in 3D and 4D are the Euler characteristics
of the respective complexes.
Thank you for your regard. Also, I want to know, based on some assumptions and Voronoi diagram, is there any relation like " = 3 − 6" between some variables for three dimensional? or any relationship that determines the number of vertexes?
@ALIN: A "generic" Voronoi diagram has vertices of degree-$3$. That's from where $E=3V-6$ derives. Generic for 3D Voronoi diagrams would mean that no $5$ points are co-spherical, so vertices have degree-$4$.
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2025-03-21T14:48:31.408948
| 2020-06-29T17:45:41 |
364442
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|
Stack Exchange
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How do you find the Cholesky decomposition of the sum of two positive definite matrices without adding the matrices directly?
If you're given two positive definite matrices ($A_1,A_2$) and the Cholesky Decomposition of those two matrices ($L_1,L_2$ such that $A_1=L_1L_1^T, A_2=L_2L_2^T$). Is there a way to find the Cholesky decomposition of $A_1+A_2$ without adding the two matrices directly? i.e. $A_1+A_2 = f(L_1,L_2)f(L_1,L_2)^T$.
No. This question and easier variants get asked regularly on [scicomp.se], and the answer is invariably no. See for instance https://scicomp.stackexchange.com/q/10630/4405 .
TL;DR:
low rank update (e.g., one column added, one element changed...): factorizations can be efficiently updated.
low norm update: you can approximate the inverse using perturbative approaches, or use the pre-update matrix as a preconditioner in an iterative method.
full-rank update (e.g., a diagonal matrix is added): no known solutions apart from recomputing the factorization.
Yes. More generally, for $X \in \mathbb{C}^{m_1 \times n}$, $Y \in \mathbb{C}^{m_2 \times n}$, you can compute the QR decomposition
$$ \begin{bmatrix} X \\ Y \end{bmatrix} = QR . $$
Then, since
$$ X^* X + Y^* Y = \begin{bmatrix} X \\ Y \end{bmatrix}^* \begin{bmatrix} X \\ Y \end{bmatrix} = (QR)^*(QR) = R^* R, $$
it follows that a Cholesky factorization of $X^* X + Y^* Y$ is $R^* R$. This method of computing $R$ will generally be far more accurate than the "direct" method of first computing the sum and then factorizing, essentially because forming $X^*X$ and $Y^*Y$ squares the condition numbers of $X$ and $Y$. Also note that the QR method gracefully handles the rank deficient case, e.g., when $m_1 + m_2 < n$, whereas the direct method may very likely fail.
On the other hand, the work of both methods is comparable in the general case. For real matrices, we can estimate $\sim 2(m_1+m_2)n^2 - \tfrac{2}{3}n^3$ flops for Householder orthogonalization versus $\sim (m_1 + m_2) n^2 + \tfrac{1}{3}n^3$ for the direct method (if we exploit the symmetry in the matrix multiplication). In special cases, the work of the QR decomposition can be reduced. For example, when $X$ is triangular and $Y$ is a single row, the QR decomposition can be computed with $n$ Givens rotations in $O(n^2)$ work. This is exactly the classic rank-1 Cholesky update procedure. It is likely some reduction is possible in the case $X$ and $Y$ are both triangular as in the question, but the flop count will still be $O(n^3)$.
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2025-03-21T14:48:31.409132
| 2020-06-29T18:01:32 |
364444
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Stack Exchange
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A set of objects classically generates the full subcategory of compact objects iff it generates the whole category
Sorry in advance if my question doesn't have the level of this community.
I am studying this paper of Bondal and Van Den Bergh and in particular section 2. Generators and resolutions in triangulated categories.
As long as I was figuring out the definitions of "classically generated", "generated" and "compactly generated", I came up with the Theorem 2.1.2 of this paper which says that in a compactly generated category $\mathcal{C}$, a set of objects $\mathcal{E}\subset \mathcal{C}^{c}$ classically generates $\mathcal{C}^{c}$ if and only if it generates $\mathcal{C}$. For the proof I started to read the reffering paper by Neeman and successfully finished and understood it. But there are some problems of connecting the information of the 2 papers.
First of all, while theorem 2.1.2 uses the notion of "generates" that Bondal and Van Den Bergh define, in Neeman's paper there is nowhere such a notion and maybe instead of this, he uses the notion of the smallest localising triangulated subcategory that contains a set of objects. I think that there is the following connection which I unfortunately can't prove
$$\mathcal{R}\ \text {is the smallest localising triangulated subcategory that contains a set}\ R\subset \mathcal{C}^{c}\ \Leftrightarrow $$$$R^{\perp}=0 $$ where $R^{\perp}$ is defined as in Bondal's paper as a subcategory of $\mathcal{R}$ and $\mathcal{R}$ is the smallest localising triangulated subcategory that contains a set $R\subset \mathcal{C}^{c}$ as in Neeman's paper. I strongly think that this is true under the hypothesis that $\mathcal{R}\subset \mathcal{C}$ where $\mathcal{C}$ is compactly generated.
If the 1. is true then I construct the proof of Theorem 2.1.2 in the following way: by lemma 2.2 in Neeman's Paper(in the main proof of theorem 2.1) we clearly have the implication $\mathcal{E}$ generates $\mathcal{C}$ $\implies$ $\mathcal{E}\subset \mathcal{C}^{c}$ classically generates $\mathcal{C}^{c}$. For the other direction we note that if the smallest localising triangulated subcategory that contains $\mathcal{C}^{c}$ is the whole $C$, then $\mathcal{C}$ consists of coproducts of objects in $\mathcal{C}^{c}$. Then it is easy, since $Hom(M,\bigoplus E_{i})\cong \bigoplus Hom(M,E_{i})$.
As it seems there is a problem of combining these two papers and any help will be accepted.
Edit: To be honest I hadn't found this post that has an answer from Leonid Positselski, so I reform my question: as the proof is for well-generated categories (a generalised notion of compact objects) and uses the Brown representability theorem for triangulated categories I was wondering if there is a simpler proof for compact objects without using it. Maybe the first proof before the introduction of well-generatedness by Krause and others.
Cross-posted on math.stackexchange
Thank you very much for the comment Profesor @JeremyRickard.I also edited the post above.
I agree that the various notions of 'generates' can be confusing. I think the following result may clarify what you are after (this can be found in Lemma 2.2.1 of 'Stable model categories are categories of modules' by Schwede and Shipley).
Let $\mathcal{C}$ be a triangulated category with infinite coproducts and let $\mathcal{P}$ be a set of compact objects. Then the following are equivalent:
(i) The smallest localizing subcategory of $\mathcal{C}$ containing $\mathcal{P}$ is $\mathcal{C}$.
(ii) An object $X \in \mathcal{C}$ is trivial if and only if $[P,X]_* = 0$ for all $P \in \mathcal{P}$ (in the language of the question, $\mathcal{P}^{\perp} = 0$).
Finally, we also have the following, due to Thomason:
Suppose $\mathcal{C}$ is a compactly generated triangulated category, and $\mathcal{A}$ is a set of compact objects, then $\mathop{Loc(A)} \cap \mathcal{C}^{c} = \mathop{Thick(A)}$.
Here $\mathop{Loc}(\mathcal{A})$ denotes the smallest localizing subcategory of $\mathcal{C}$ containing $\mathcal{A}$, and $\mathop{Thick}(\mathcal{A})$ denotes the smallest thick (epaisse in Bondal--Van Den Bergh) subcategory of $\mathcal{C}^c$ containing $\mathcal{A}$.
Thank you very much @Drew Heard for the answer.Actually,after your response and some reading of Lemma 2.2.1 of 'Stable model categories are categories of modules' by Schwede and Shipley i figured out that in particular lemma 1.7 of Neeman's paper gives what i wanted (as Schwede and Shipley explains but it was unclear to me in Neeman's paper)
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2025-03-21T14:48:31.409434
| 2020-06-29T18:35:48 |
364445
|
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Stack Exchange
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Holomorphic tubular neighborhood of divisors at infinity
For the discussion of holomorphic tubular neighborhoods and some criteria for their existence see this question.
Let $X$ be a smooth quasi-projective variety over $\mathbb{C}$. Hironaka tells us that there exists a smooth projective compactification $X \subset \bar{X}$, such that $D =\bar{X}\backslash X$ is a strictly normal crossing divisor. Let $D=\cup_{i}^N D_i$ be its decomposition into irreducible smooth divisors. What is the obstruction to finding such $\bar{X}$, such that each $D_i$ has a holomorphic tubular neighborhood in $\bar{X}$?
Example 1: Let $S$ be a smooth projective variety, $E$ a vector bundle on $S$. The space $X=\text{Tot}(E)$ is compactified to $\bar{X}=\mathbb{P}_S(E\oplus \mathcal{O}_S)$ with the divisor at infinity $D =\mathbb{P}_S(E)\subset\bar{X}$. Its normal bundle should be $\mathcal{O}_{\mathbb{P}(E)}(1)$ and one should be able to find a holomorphic tubular neighborhood by constructing one in each fiber and gluing together.
Example 2: Using the answer by Joey, I have convinced myself that if the divisor $D_i$ is $\mathbb{P}^n$ for some $n\geq 2$ and $n+1=\text{dim}(X)$, then it also has such a neighborhood. If $n=1$, then one needs additionally that $(D_i)^2<0$.
The above examples make me believe that it could be always achievable, as each blow up at a smooth point introduces a new $\mathbb{P}^{n}$ with normal bundle $\mathcal{O}(-1)$.In particular, blowing up seems to only help with the existence of these neighborhoods.
Edit: What I am asking is: Does there exists such a procedure of blowing up the divisor at infinity, that leaves me at the end with $D_i$ having holomorphic tubular neighborhoods? I don't expect to understand all possible compactifications.
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2025-03-21T14:48:31.409566
| 2020-06-29T19:21:42 |
364448
|
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Stack Exchange
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The space of rearrangements of a plane curve
I learned of the paper "Closing curves by rearranging arcs" by L. Alese (arXiv link) by a post on Reddit and was curious about related questions and generalizations. Here a rearrangement of a curve is where you cut the curve into finitely many arcs and then through rigid plane motions glue the arcs back together so that the tangents at the endpoints match.
What does the space of all rearrangements of a curve look like? For straight line segments and circles you get nothing: the curve is the same no matter how you cut and rearrange pieces. For curves with non-constant curvature the possibilities increase. The paper above answers the question "Does the rearrangement space contain a closed curve?"
My guess is the space is determined by the curvature measure along the curve. This is a signed measure whose integral gives the turning number along the curve; it is possibly singular if there are any sharp turns.
It seems like the "boundary" or "completion" of the rearrangement space is determined by the total variation of the curvature measure in some sense, I'm not sure how to make this precise.
Added on:
Say you have a finite signed curvature measure $\kappa$, defined on the Borel sets of $I=[0,1]$. The corresponding angle function is $\theta(t)=\kappa([0,t])$ and the corresponding curve is
$$x(t)=\int_0^t\cos(\theta(s))ds$$
$$y(t)=\int_0^t\sin(\theta(s))ds.$$
A rearrangement of $\kappa$ is determined by a partition of $I$ into sub intervals $I_1=(0,t_1],I_2=(t_1,t_2],\ldots,I_k=(t_{k-1},1]$ together with a permutation $\sigma$ of $\{1,\ldots,k\}$.
This gives a map $P:I\to I$ defined by rearranging the $I_j$ according to $\sigma$ and then stacking them back over the interval $I$, together with saying $P(0)=0$. For example, with the partition $I_1=(0,\frac{3}{4}], I_2=(\frac{3}{4},1]$ and the (cycle notation) permutation $(1 2)$:
$$P(t)=\begin{cases}t+\frac{3}{4}, \text{ if } 0<t\leq\frac{1}{4}\\t-\frac{1}{4}, \text{ if } \frac{1}{4}<t\leq 1\\ 0, \text{ if } t=0\end{cases}.$$
The rearranged curvature measure is defined by
$$\tilde\kappa(A)=\kappa(P(A)),$$
and the space of all rearrangements is given by applying this construction over all partitions and all permutations.
Starting from a given $\kappa$, what does the space of rearrangements look like?
Perhaps it can be useful to link with a question by (supposedly) L. Alese https://mathoverflow.net/questions/363495/surprising-properties-of-closed-planar-curves?noredirect=1&lq=1
That's exactly the content I saw in the Reddit post.
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2025-03-21T14:48:31.409762
| 2020-06-29T19:22:15 |
364449
|
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|
Stack Exchange
|
Upper bounding the sum with hypergeometric and binomial probabilities
Could you please help me upper bound this tricky expression:
$$P(A)=\sum_{i=0}^n{\left( 1 - \dfrac{\binom kq \binom {n-k}{i-q}}{\binom {n}{i}} \right)}^I \binom ni p^i {(1-p)}^{n-i}$$.
So far I only managed to obtain a lower bound by proving $\sum_{i=0}^n{\dfrac{\binom kq \binom {n-k}{i-q}}{\binom {n}{i}}\binom ni p^i {(1-p)}^{n-i}}=\binom kq p^q {(1-p)}^{k-q}$ and applying Jensen's inequality for power function $f(x):=x^I$.
This lower bound is $P(A) \geq {\left(1-\binom kq p^q {(1-p)}^{k-q}\right)}^I$ and works okay for my needs. However, I would like to get a similar upper bound too and this is what causes the main difficulty.
Do you have any suggestions?
What is $I$? Is $p$ in $[0,1]$? What are $n,k,q$?
Oh, yes, sorry. $I, n, k, q$ are natural numbers and $p$ is in $[0,1]$.
Too many variables; not enough constraints.
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2025-03-21T14:48:31.409865
| 2020-06-29T19:54:30 |
364451
|
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Stack Exchange
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Characteristic polynomial of the Gcd matrix
Let $A_n$ be the $n \times n$-matrix with entries $\gcd(i,j)$ and $f_n$ the characteristic polynomial of $A_n$.
Question: Is $f_n$ irreducible over $\mathbb{Q}$ for all $n$ except $n=8$?
This is true for $n \leq 60$.
Any references for this problem?
True for $n \le 180$.
@lhf I do not know whether this problem is stated somewhere else. I came up with it by experimenting with GAP.
Any significance that can be attached to the $n=8$ case? If I'm not wrong $f_8(x)=(x^2-8x+8)(x^6-28x^5+248x^4-900x^3+1324x^2-704x+96)$
Can you clarify, irreducible over which field?
@JukkaKohonen Over $\mathbb{Q}$.
Of course, you must know that $\det A_n=\phi(1)\cdots\phi(n)$, where $\phi$ is Euler's totient function. Actually, it is equivalent over $\mathbb Z$ to ${\rm diag}(\phi(1),\ldots,\phi(n))$.
This question assumes that numbering of rows and columns ($i$ and $j$) goes from $1$ to $n$. If it ranges from $0$ to $n-1$ as seems more logical to me (with gcd of $0$ and $j$ being $j$ for all $j$ of course), then the characteristic polynomial appears to be always irreducible.
Edit. See the calculation modulo $2$ below.
This is not an answer, but it is too long for a comment. The matrix $A_n$ can be decomposed as $A_n=D_n^T\Phi_nD_n$ where the entries of $D_n$ (D for divisors, not for diagonal) are either $1$ if $i|j$, or $0$ otherwise, and $\Phi_n={\rm diag}(\phi(1),\ldots,\phi(n))$, $\phi$ being the Euler's totient function.
Notice that the matrix $D_n$ is the $n\times n$ upper-left block of an infinite upper triangular matrix $D$. Its diagonal is made of $1$s, so that $\det D_n=1$. Thus
$$f_n(X)=\det(XD_n^{-T}D_n^{-1}-\Phi_n).$$
The matrix $D_n^{-T}D_n^{-1}$ is the $n\times n$ upper-left block of $D^{-T}D^{-1}$. The entries of $D^{-1}$ are known to be either $\mu(j/i)$ if $i|j$, or $0$ otherwise. Therefore
$$(D_n^{-T}D_n^{-1})_{ij}=\sum_{k|i,j}\mu\left(\frac ik\right)\mu\left(\frac jk\right).$$
For instance this is zero when $p^2|j$ but $p$ does not divide $i$. Also the diagonal entry corresponding to $j=i$ equals $2^r$, where $r$ is the number of distinct prime factors of $i$. A closed formula is that the $(i,j)$-entry equals
$$\mu\left(\frac{{\rm lcm}(i,j)}{{\rm gcd}(i,j)}\right)2^{P(i,j)},$$
where $P(i,j)$ is the number of distinct prime factors of ${\rm gcd}(i,j)$ for which the valuations agree:
$$v_p(i)=v_p(j)\ge1.$$
Calculation modulo 2. Since $\phi(j)$ is even for every $j\ge3$, the matrix $\Phi_n$ reduces to ${\rm diag}(1,1,0,\ldots,0)$. Thus $f_n(X)=X^n-aX^{n-1}+bX^{n-2}$ where
$$a=(D_n^{-T}D_n^{-1})\binom{\hat1}{\hat1}+(D_n^{-T}D_n^{-1})\binom{\hat2}{\hat2}$$
is the sum of minors obtained by deleting either the first row and column, or the second,
and
$$b=(D_n^{-T}D_n^{-1})\binom{\widehat{12}}{\widehat{12}}$$ is the minor when deleting the two first columns and rows. Because $\det(D_n^{-T}D_n^{-1})=1$, this gives
$$a=(D_nD_n^T)_{11}+(D_nD_n^T)_{22},\qquad b=(D_nD_n^T)\binom{12}{12},$$
the latter being a $2\times2$ minor. These quantities involve only the two first rows $(1,\ldots,1)$ and $(0101\ldots)$ of $D_n$. We obtain
$$a=n^2+\lfloor\frac n2\rfloor^2=n+\lfloor\frac n2\rfloor,\qquad b=n\lfloor\frac n2\rfloor-\lfloor\frac n2\rfloor^2=(n+1)\lfloor\frac n2\rfloor.$$
Hence (mod $2$)
$$f_n(X)=\left\{\begin{array}{lcr} X^n, & & n=0,3, \\
X^{n-1}(X+1), & & n=1 \\ X^{n-2}(X^2+X+1), & & n=2. \end{array}\right.$$
Quite neat decomposition of $A$. Is it used for well known number theoretic results? If so and you easily happen have a reference for that, please post it.
@YaakovBaruch. The decomposition $A_n=D_n^T\Phi_nD_n$ is well known ans easy to verify. I made an exercise (#188) from it in my blog http://perso.ens-lyon.fr/serre/DPF/exobis.pdf (companion to my book on Matrices).
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2025-03-21T14:48:31.410138
| 2020-06-29T20:30:52 |
364458
|
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|
Stack Exchange
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A variant on Wieferich primes
Recall that a Wieferich prime is a prime number $p$ such that
$2^{p-1} \equiv 1 \bmod p^2.$
It is not known whether there are infinitely many Wieferich primes, nor whether there are infinitely many non-Wieferich primes. In fact there are only $2$ known Wieferich primes.
I'm interested in a slightly different condition which I'm hoping is easier to handle. Namely, I can replace the exponent $p-1$ by the order of $2$ in $(\mathbb{Z}/p\mathbb{Z})^\times$. Moreover, I just want this power to hit the identity with odd $p$-adic valuation. Specifically:
Are there infinitely many primes $p$ such that $v_p(2^{\mathrm{ord}_p(2)}-1)$ is odd?
Here $\mathrm{ord}_p(n)$ denotes the order of $n$ in $(\mathbb{Z}/p\mathbb{Z})^\times$ (which divides $p-1$ by FLT), and $v_p$ is the $p$-adic valuation.
Note that the existence of infinitely many non-Wieferich primes would provide a positive answer to my question (since here $v_p(2^{p-1}-1) = 1$).
Ideally I'd also like to know that there collection of such primes has positive density, rather than just being infinite.
Do you possibly mean $v_p(2^{{\rm ord}_p(2)}-1)$ instead of $v_p(2^{{\rm ord}_p(2)})$?
Yes thanks, now fixed.
Since $v_p(2^{{\rm ord}_p(2)}-1)=v_p(2^{p-1}-1)$, you might as well keep the exponent simpler.
@Pace: Nice observation. How do you show this?
@DanielLoughran Let $x={\rm ord}_p(2)$. Write $2^x=1+ap^k$ where $gcd(p,a)=1$. Now, the order of $2$ modulo any larger power of $p$ must be a multiple of $x$, say $xy$. We compute that $2^{xy}=(1+ap^k)^y=1+yap^k+$terms divisible by $p^{k+1}$. We see that this is $1$ modulo $p^{k+1}$ if and only if $p|y$.
Among the first million primes, the only ones with this number not 1 are the Wieferich ones, 1093 and 3511, for them it is 2.
Perhaps you like a discussion of mine http://go.helms-net.de/math/expdioph/fermatquotients.pdf where I look at generalized Wieferich-primes and tried to build some tables with softer requirements in the hope to increase the "database" ( of 2 exemplars) that some pattern occurs and hints to some further idea of analysis. Short overview in indexlist on homepage http://go.helms-net.de/math/index.htm see "Wieferich primes and Fermat quotients..."
Suppose the answer is no and that the finitely many exceptions are all at most $B$. Let $\ell \equiv 1 \pmod{3}$ be prime and consider $n=2^{\ell} -1$. If $p>B$ is a factor of $n$, then $\ell$ is the order of $2$ modulo $p$, so $p$ occurs in $n$ with an even exponent, so $n = x^2c, c \le B!$. Let $y = 2^{(\ell - 1)/3}$. Then $n=2y^3 - 1$ and finally $2y^3 - 1 = cx^2$, so $(x,y)$ is an integral point on one of a finite collection of elliptic curves and there can be only finitely many such. But there are infinitely choices for $\ell$, contradiction. (This is a variant of an old argument of Granville.)
This is a very slick argument! However it doesn't seem to get positive density, which I was originally hoping for. Which paper of Granville does this appear in?
I think one can make the argument a bit simpler, keeping the same basic idea. Since $\ell$ is the order of $2$ mod $p$, we have $p > \ell$. Thus, as long as we choose $\ell > B$, we have $2^{\ell}-1$ coprime to $B!$. So $n$ is actually a square, which is absurd since $n\equiv 3\pmod{4}$. In this variant there's no need to restrict the primes $\ell$ to the progression $\ell\equiv 1\pmod{3}$,
I'm guessing the Granville paper in question is https://mathscinet.ams.org/mathscinet-getitem?mr=841645 . It seems that the state of the art only provides $\log X$ or so non-Weiferich primes up to $X$, even if one assumes the abc conjecture (see e.g. the recent paper https://mathscinet.ams.org/mathscinet-getitem?mr=3983273 ) so I doubt positive density is within reach of current methods.
... though on the other hand, this argument does seem to imply that for every prime $\ell$, there is a prime $p$ with $ord_p(2)=\ell$ and $v_p( 2^{ord_p(2)}-1)$ odd (because $2^\ell-1$ is not a square). Don't know if this is actually helpful for your application.
@Terry Tao: This is a nice observation which could certainly be useful! It seems to give a more elementary proof of infinitude and also gives a stronger conclusion.
Felipe refers to my first ever paper (mathscinet.ams.org/mathscinet-getitem?mr=789713) from 1985 ! However I have a more recent paper that gives a better result along the lines asked for (mathscinet.ams.org/mathscinet-getitem?mr=2997580) which shows that every $2^n-1$, with $n\ne1$ or $6$, has a primitive prime factor that divides it to an odd power.
Thanks Andrew! I'll take a look at your more recent paper.
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2025-03-21T14:48:31.410464
| 2020-06-29T20:57:29 |
364459
|
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Stack Exchange
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How may a largest fixed-point be defined in second order logic?
Adapting from Anil Gupta and & Nuel Belnap, Revision theory of truth, MIT 1993, p. 194, in the context of a second order logic, where $A(x.G)$ is a formula where $G$ only occurs positively, a fixed point
$$\forall x(Gx\leftrightarrow A(x,G))$$
may be isolated. One may also show
$$\forall x(Gx\leftrightarrow\forall H[\forall x(Hx\leftrightarrow A(x,H))\rightarrow Hx])$$
I understand how these together define a least fixed point of $A(x,G)$.
How may one define a largest fixed point of $A(x,G)$ in the context?
Just as the least fixed point is the intersection of all the sets $H$ such that $\forall x\,(A(x,H)\to H(x))$, so (dually) the greatest fixed point is the union of all the sets $K$ such that $\forall x\,(K(x)\to A(x,K))$.
Alternatively, one can use duality to obtain the greatest fixed point of $A(x,G)$ as the complement of the least fixed point of $\neg A(x,\neg G)$.
As in $$Gx\leftrightarrow\exists H(\forall x(Hx\rightarrow A(x,H))\wedge Hx)$$?
Isn't the greatest fixed point just the whole structure in question?
@NoahSchweber No, because the whole structure might not be fixed. A good example would be the Cantor-Bendixson derivative, as a monotone map from closed subsets (of some decent ambient space) to closed subsets. Its greatest fixed point is the perfect piece in the Cantor-Bendixson decomposition of the ambient space into countable + perfect.
@FrodeAlfsonBjørdal That looks correct to me (but it's 11 p.m. here).
@AndreasBlass I see that the truth of the R.S. $$\exists H(\forall x(Hx\rightarrow A(x,H))\wedge Hx)$$ entails $$(Gx\rightarrow A(x,G))\wedge Gx$$. But I do not see that $$Gx$$ (or $$(Gx\rightarrow A(x,G))\wedge Gx$$) entails the R.S.
@AndreasBlass Excuse me, I see how $$(Gx\rightarrow A(x,G))\wedge Gx$$ entails the R.S.
Is $$((Gx\rightarrow A(x,G))\wedge Gx)\leftrightarrow\exists H(\forall x(Hx\rightarrow A(x,H))\wedge Hx)$$ the desired fixed point?
I thought we should also get $$(Gx\leftrightarrow A(x,G))$$.
@FrodeAlfsonBjørdal Now that I'm awake, the formulation in your first comment still looks correct (though I'd prefer not to have the same variable free at some occurrences & bound at others). In particular, the $G$ that it defines is a fixed point, i.e., it does what you asked about in your most recent comment.
@AndreasBlass Thanks! Now I see it.
@AndreasBlass For the possible benefit of others, and for my own notes.
Suppose DF: $$Gx\leftrightarrow\exists H(\forall y(Hy\rightarrow A(y,H))\wedge Hx)$$
Logical consequence (1) of DF:
$$A(x,H)\rightarrow(Gx\leftrightarrow Hx)$$
Logical consequence (2) of DF:
$$Hx\rightarrow(Gx\leftrightarrow A(x,H))$$
Consequence (3):
$$\forall x(Hx\wedge A(x,H))\rightarrow(\forall x(Gx\leftrightarrow Hx)\wedge\forall x (Gx\leftrightarrow A(x,H)))$$
$$\forall x((Hx\rightarrow A(x,H))\wedge Hx)\rightarrow((Hx\wedge A(x,H))$$
$$\exists H(\forall x((Hx\rightarrow A(x,H))\wedge Hx))\rightarrow(\forall x(Gx\leftrightarrow A(x,G)))$$
So if there is a fixed point of A(x,H) then G is such a fixed point.
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2025-03-21T14:48:31.410776
| 2020-06-29T21:32:04 |
364461
|
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|
Stack Exchange
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Fixed points of diffeomorphisms of tori isotopic to identity and their traces under isotopies
Suppose $T^n$ is the $n$-dimensional torus ($n\geq 2$) and $f: T^n\to T^n$ is a diffeomorphism isotopic to the identity and fixing points $x_1,\ldots,x_k\in T^n$. Does there exist an isotopy $\{ f_t: T^n\to T^n\}_{0\leq t\leq 1}$ connecting $f_0=Id$ with $f_1=f$ so that all the loops $\{ f_t (x_i)\}_{0\leq t\leq 1}$, $i=1,\ldots,k$, lie in the same free homotopy class?
I believe this is not always possible: let $d\colon \mathbb R \to [0,1/2]$ send a real number to the distance to the nearest integer. Consider the map
$$F\colon \mathbb R^2, (x,y) \mapsto (x+2d(y),y),$$
which commutes with the $\mathbb Z^2$ action on $\mathbb R^2$ and thus descends to a homeomorphism of $T^2 = \mathbb Z^2 \backslash \mathbb R^2$, denoted $f$. Of course, $f$ is not smooth, but we can clearly make it smooth it by slightly changing the formula above (replace $d$ by a function that is smooth and takes values $0$ on integers and $1/2$ on half-integers).
Geometrically, $f$ can be understood as a Dehn twist on half of the torus and the reverse Dehn twist on the other half.
Clearly, $f$ perseveres $p= [(0,0)]$ and $q = [(0,1/2)]$ and is isotopic to the identity.
We can lift any isotopy $f_t$ with $f_0 = Id$ and $f_1 = f$ to an isotopy $F_t$ of $\mathbb R^2$ from $F_0$, a translation by, say, $(z_1,z_2) \in \mathbb Z^2$, to $F_1 = F$ from above. But then $F_t(p)$ goes from $(z_1,z_2)$ to $(0,0)$, whereas $F_t(q)$ goes from $(z_1,z_2+1/2)$ to $(1,1/2)$, so the loops these curves describe in $T^2$ are not freely homotopic.
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2025-03-21T14:48:31.410900
| 2020-06-29T21:33:59 |
364462
|
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|
Stack Exchange
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Notation question: bigraded direct sum of graded objects
In some work I'm doing I have two graded modules $M$ and $N$ (graded on $\mathbb Z$, say) and need to take, not the usual direct sum, but the bigraded sum consisting of all $M_p \oplus N_q$ (so graded on $\mathbb Z \oplus \mathbb Z$). For the graded sum I would simply write $M\oplus N$. Before I make up a notation for the bigraded sum, is there an already existing notation that anyone has used or seen?
I'm not sure what tags to include on this question, feel free to criticize. I've tried to tag the communities that might have run into this before.
I would denote this $\boxtimes$, the external tensor product. If you treat graded objects as weight decomposition of some $\mathbb{G}_m$ action, then this bigraded sum is indeed the external tensor product of two representations of $\mathbb{G}_m$ as a representation of $\mathbb{G}_m \times \mathbb{G}_m$. I don't know if anyone have used this notation, though.
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2025-03-21T14:48:31.411008
| 2020-06-29T21:49:19 |
364464
|
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"authors": [
"Alexandre Eremenko",
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Stack Exchange
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Under what conditions are two orientation-reversing involutions of a compact surface equivalent?
Let $M$ be a compact, connected, orientable surface and $\varphi_1,\varphi_2$ be two orientation-reversing involutions (i.e., diffeomorphisms for which $\varphi^2=Id$) such that the fixed-point set of both is non-empty. I am trying to understand what conditions guarantee the existence of an equivariant self-diffeomorphism $f$. (I.e., such that $f \circ \varphi_1 = \varphi_2 \circ f$, which is what I call "equivalen[ce]" in the title.)
The fixed-point set is a union of circles. Perhaps conditions can be derived from how those circles from $\varphi_1$ and $\varphi_2$ interact homologically. (More precisely, how the image of their fundamental classes through the inclusion are related to one another. Of course, necessary conditions can arise from this.) Since such a diffeomorphism restricts to a diffeomorphism between the fixed point sets, it is clear that this self-diffeomorphism does not always exist. (E.g., take a $2$-genus surface on $\mathbb{R}^3$ to be lied down, and consider reflection through the middle vertical circle as $\varphi_1$ and through the $xy$ plane as $\varphi_2$. The fixed point sets have a different number of connected components.)
I have two main questions:
Under what conditions on $M$, $\varphi_1$ and $\varphi_2$ (or perhaps their fixed point sets) does this hold?
If the above is a little too broad, then is it known if this always holds for $S^2$? What about $S^1 \times S^1$?
Let me just remark that such a diffeomorphism would especially imply that the orbit spaces are homeomorphic. Now for similar questions, say we look at free actions of $\mathbb{Z}/p$ on S^3, one gets quite a few such actions with non-homeomorphic orbit spaces (these orbit spaces are called lense spaces). Thus this question asks about things which only hold in dimension 2 and only when the order of the diffeomorphism is 2.
Two orientation-reversing involutions of a given closed orientable surface are equivalent if and only they have the same number of fixed point circles and have the same orientation character, in the sense that the quotient surfaces (when there is non-empty fixed point set) are both orientable or both non-orientable.
Here's a sketch.
If there are no fixed points then the orbit map is just the orientation double covering of a non-orientable surface whose Euler characteristic is half that of the given surface, and hence determined by the classification of surfaces. A homeomorphism of quotient surfaces lifts to an equivariant homeomorphism.
More generally, when there are fixed point circles, the orbit map is a surjection onto a connected surface with boundary, with one boundary component for each fixed point circle. Removing annular neighborhoods of the boundary curves we see a double covering of a surface with boundary. The 2-fold covering is connected if and only if the quotient is non-orientable. In each case we know the Euler characteristic of the quotient again is half the original Euler characteristic. Therefore, knowing the number of boundary components we again have determined either the orientable double covering or a trivial double covering, of a specific surface with boundary, which is then completed by equivariantly attaching a set number of invariant annuli joining pairs of boundary components. To do this note that each boundary curve in the target is trivially covered by a pair of boundary curves in the domain. For each such pair of boundary curves one attaches an annulus $S^1\times [-1,1]$ with an involution that interchanges the ends and fixes $S^1\times \{0\}$. The effect in the quotient to to attach $S^1\times [0,1]$ along $S^1\times \{1\}$. Again a homeomorphism of orbit spaces lifts to the required equivariant homeomorphism.
Note that the orientation character condition can be reformulated as saying for each of the two involutions the fixed point set represents a nonzero element of mod 2 homology (in the non-orientable quotient space case) or each represents zero in mod 2 homology (in the orientable quotient space case).
Aside: Involutions of non-orientable surfaces are perhaps more interesting. One has to distinguish fixed point components that are themselves orientation reversing loops. Moreover, a given involution can also have both fixed circles and isolated fixed points, e.g. a standard involution on the projective plane.
Thanks for the answer! The point which is still unclear to me is how to attach these annuli in the last part. Could you please expand a bit?
I added some discussion about attaching the annuli. Hope that's a little clearer. I also added a note about the meaning of the mod 2 homology class represented by the fixed point set.
Thanks for the clarification! The note was also pretty useful. I've accepted the answer, but I wonder if this would also apply for $M$ a compact manifold with boundary. I don't see a point where the argument would fail, but perhaps I'm missing something.
One issue that bothers me is that if $M$ has boundary, then removing the annuli neighbourhoods can leave corners.
If $M$ has boundary, then the fixed point set might contain arcs. One would have to account for the number of such arcs, and whether a fixed point arc connected two different boundary components or had both end points in the same boundary component. I don't think the corners are a problem. They would actually be useful for keeping track of where to re-attach the neighborhoods of the fixed point arcs.
This always holds for anti-conformal involutions of $S^2$. Represent $S^2$ as the Riemann sphere, then the fixed point set of an anti-conformal involution is a circle on the sphere. And this circle
determines the involution uniquely. And the group of automorphisms of the Riemann sphere acts transitively on circles (because it acts transitively on triples of distinct points, and each such triple lies on a unique circle).
But this does not hold for tori. For example, a square torus has two essentially distinct anti-conformal involutions: one fixes two circles and another fixes one circle.
Of course, the square torus is an exception but a generic torus with an involution which fixes two circles, there is another involution which also fixes two circles and they are not conjugate:
represent such a torus by a rectangle in the usual way; then one involution
fixes vertical sides (and a vertical middle line), while another fixes
horizontal sides (and the horizontal middle line). Unless it is a square,
there is no conformal automorphism which interchanges vertical and horizontal sides.
Same applies to a generic torus with an involution fixing one circle (rhombic torus). There is always another involution, and these two involutions are not conjugate to each other.
Remark. Consider the Weierstrass form of the torus:
$$y^2=4(x-e_1)(x-e_2)(x-e_3).$$
When all $e_j$ are real, the unvolution $(x,y)\mapsto (\overline{x},\overline{y})$ has two fixed circles. When $e_1$ is real but $e_2=\overline{e_3}$ are not real,
the same involution has one fixed circle. Just sketch the set of real points of this curve in $x,y$ plane.
Maybe I am blind, but I cant see the involution withone one fixed circle.
Hello, thanks for the answer. Why can we assume that the involutions are anti-conformal? And I also can't see the involution with just one fixed circle on the torus.
Consider a rhombus. Identifying the opposite sides by translations gives you a torus. The involution of this torus with one fixed circle comes from reflection of this rhombus with respect to a diagonal. Since a rhombus has two diagonals, there always two non-conjugate such involutions, unless the rhombus is a square.
@Aloizio Macedo: a conformal involution can only have isolated points fixed. Thus when the set of fixed points is infinite, the involution must be anti-conformal.
@AlexandreEremenko It seems that you are saying that an involution must be either conformal or anti-conformal. If so, I don't understand why. If this and other facts of the first paragraph can be found somewhere, would you have a reference? Thanks again!
@Aloizio Macedo: yes I was only speaking about anti-conformal involutions of Riemann surfaces. The question does not specify what an involution is. I suppose you mean at least homeomorphisms?
@AlexandreEremenko Sorry for not being clear, but the $\varphi$ are assumed to be a priori only diffeomorphisms. To be specific, $\varphi$ is a diffeomorphism such that $\varphi^2=Id$ and reverses orientation. I'll clarify this in the question.
I suppose that for every orientation-reversing involution-diffeo of the sphere, there exists a conformal structure which makes it anti-conformal, but I have to think how to prove this. At lease this is the case when the set of fixed points is a topological circle. But how do you know that the set of fixed points set is a union of circles? Perhaps if you have a reference for this, it can also prove what I just said.
@AlexandreEremenko (In case I was also unclear regarding this, by circles I don't mean "hard" circles as in intersection with a plane, but rather diffeomorphic to circles.) Since $\varphi$ is an involution and orientation-reversing (and the dimension is $2$), it is anti-symplectic. Then the fixed-point set is Lagrangian and therefore $1$-dimensional. Since it is a compact $1$-dimensional manifold, it is a union of circles.
@AlexandreEremenko: any finite group action on a manifold is isometric in some Riemannian metric (obtained by averaging an arbitrary Riemannian metric). On surfaces an isometry is anti-conformal if and only if it reverses orientation, while orientation-preserving isometries are conformal.
@Igor Belgradek: Good point. However if you have TWO involutions, it is not clear why the group they generate is finite.
|
2025-03-21T14:48:31.411860
| 2020-06-29T22:25:37 |
364468
|
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|
Stack Exchange
|
Simultaneous similarity of matrices over finite fields
Suppose $A,B\in SL(3,F_q)$, where $F_q$ is the finite field of order $q$ and $SL(3,F_q)$, the group of matrices with determinant one and entries from $F_q$ , are such that $A$ has eigenvalues in $F_q$ and $B$ has eigenvalues in $\overline{F_q}\setminus F_q$. Also, $A$ is diagonalizable over $F_q$ and $B$ is diagonalizable over $\overline{F_q}$, the closure of $F_q$. I am trying to show that $A$ and $B$ are not simultaneously diagonalizable by a matrix $P\in SL(3,\overline{F_q})$ (i.e., $\nexists P\in SL(3,\overline{F_q})$ such that $(PAP^{-1},PBP^{-1})$ are diagonal). I am considering the approach of looking at the $F_q$ algebra generated by $A$ and $B$ and trying to show it is not isomorphic to the algebra generated by $F_q[PAP^{-1},PBP^{-1}]$. I am looking for some reference which might be helpful in proving the above. Most of the results I saw had been about irreducible representations which is not helpful for my case. I would appreciate it if you could suggest some reference that could be helpful.
Thanks in advance for your time.
What about the case where $A$ is the identity matrix? Then they clearly are simultaneously diagonalizable. If $A$ has distinct eigenvalues then you should be OK.
I think there is a simpler way to see what is going on: first of all, the hypotheses force the characteristic polynomial of $B$ to be irreducible of degree $3$ over $F_{q}.$ On the other hand, if $PAP^{-1}$ and $PBP^{-1}$ are both diagonal, then $PAP^{-1}$ and $PBP^{-1}$ certainly commute. Hence $A$ and $B$ already commute. Since $B$ has irreducible characteristic polynomial, (which is also its minimum polynomial in this situation), and since $B$ leaves every eigenspace of $A$ invariant (as $A$ and $B$ commute), this forces (given the hypotheses) $A$ to have the form $\lambda I$ for some $\lambda \in F_{q}$.
So, if you exclude $A$ being a scalar matrix, what you want to be true indeed is.
Where are utilizing irreducibility of characteristic polynomial?
Because it means that $B$ acts irreducibly on the underlying space, so has no proper non-zero invariant subspace. But each eigenspace of $A$ is $B$-invariant, so any such eigenspace is the whole space (ii $A$ is a scalar matrix).
Your intuition is somewhat correct. The characteristic polynomial of $B$ is cubic over $\mathbf F_q$, and has no roots in $\mathbf F_q$ by assumption, hence is irreducible. Thus, the roots all live in $\mathbf F_{q^3}$, and $\mathbf F_q[B] \cong \mathbf F_{q^3}$. If $A$ and $B$ are simultaneously diagonalisable over $\bar {\mathbf F}_q$, then
$$\mathbf F_q[A,B] \underset{\mathbf F_q}\otimes \bar{\mathbf F}_q \cong \bar{\mathbf F}_q[PAP^{-1},PBP^{-1}]$$
is a subalgebra of the diagonal matrices $\bar{\mathbf F}_q^3$, so a dimension count shows that the inclusion $\mathbf F_q[B] \subseteq \mathbf F_q[A,B]$ must be an equality. Since $A$ has eigenvalues in $\mathbf F_q$ and $\mathbf F_q[B] \cong \mathbf F_{q^3}$, this forces $A$ constant (think about which elements of $\mathbf F_{q^3}$ have characteristic polynomial totally split over $\mathbf F_q$).
(Conversely, if $A$ is constant, as Will Sawin noted, then clearly $A$ and $B$ are simultaneously diagonalisable over $\bar{\mathbf F}_q$).
Thank you so much for answering. It was very helpful. I was considering $A$ to be the matrices that are not conjugate to the central elements. Sorry I forgot to mention that.
|
2025-03-21T14:48:31.412110
| 2020-06-29T23:14:36 |
364469
|
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|
Stack Exchange
|
Abundancy index and non-solvable finite groups
Let $\sigma$ be the sum-of-divisors function. A number $n$ is called abundant if $\sigma(n)>2n$. Note that the natural density of the abundant numbers is about $25 \%$. The abundancy index of $n$ is $\sigma(n)/n$. The following picture displays the abundancy index for the $10000$ first orders of non-solvable groups (see A056866).
Observe that for $G$ non-solvable with $|G| \le 446040$ then $|G|$ is abundant, with minimal abundancy index $\frac{910}{333} \simeq 2.73$.
Question 1: Are the non-solvable groups of abundant order?
Note that the number of integers $n \le 446040$ with $\sigma(n)/n \ge 910/333$ is exactly $19591$, so of density less than $5 \%$ with more than half of them being the order of a non-solvable group. Among those which are not the order of a non-solvable group, the maximal abundancy index is $512/143 \simeq 3.58$, realized by $n=270270$, whereas there are exactly $896$ numbers $n \le 446040$ with $\sigma(n)/n > 512/143$, which then are all the order of a non-solvable group.
Question 2: Is a number of abundancy index greater than $512/143$ the order of a non-solvable group?
Weaker version 1: Is there $\alpha >3$ such that a number of abundancy index greater than $\alpha$ must be the order of a non-solvable group?
Weaker version 2: Is there $\beta < 1$ such that a number $n$ of abundancy index greater than $\beta e^{\gamma} \log \log n$ must be the order of a non-solvable group?
Recall that $\limsup \frac{\sigma(n)}{n \log \log n} = e^{\gamma}$ with $\gamma$ the Euler-Mascheroni constant.
Finally, there are non-solvable finite groups $G$ with $|G| \gg 446040$ and abundancy index less than $\frac{910}{333}$. The non-abelian simple groups $G$ with $|G|=n \le<PHONE_NUMBER>32$ and $\sigma(|H|)/|H|>\sigma(n)/n$ for all non-abelian simple groups $H$ of order less than $n$ are exactly
the 39 the simple groups $\mathrm{PSL}(2,p)$ with $p$ prime in {5, 37, 107, 157, 173, 277, 283, 317, 563, 653, 787, 907, 1237, 1283, 1307, 1523, 1867, 2083, 2693, 2803, 3413, 3643, 3677, 4253, 4363, 4723, 5443, 5717, 6197, 6547, 6653, 8563, 8573, 9067, 9187, 9403, 9643, 10733, 11443}. Let $n_p:=|\mathrm{PSL}(2,p)| = p(p^2-1)/2$.
It follows that for $G$ non-abelian simple with $|G| \le<PHONE_NUMBER>32$ then $|G|$ is abundant, with minimal abundancy index $$\sigma(n_{11443})/n_{11443} = 50966496/21821801 \simeq 2.33.$$
The following picture displays the adundancy index of $n_p$ for $p$ prime and $5 \le p \le 10^6$.
The minimal is $579859520520/248508481289 \simeq 2.3333 \simeq 7/3$, given by $p=997013$.
Question 3: Is it true that $\inf_{p \ge 5, \text{ prime}} \sigma(n_p)/n_p = 7/3$?
Question 4: Is the abundancy index of the order of a non-solvable group greater than $7/3$?
Fun fact: the smallest integer $n$ such that there exists two non-isomorphic simple groups of order $n$ is $20160$, whereas the biggest integer that is not the sum of two abundant numbers is $20161$ (see A048242). Any explanation?
Question 3 seems hard. It has something to do with the factorizations of $p-1$ and $p+1$.
If there are infinitely many primes $p$ such that $p-1$ is 4 times a prime and $p+1$ is 6 times a prime, then question 3 has a positive answer. 997013 is one such prime. We may also rearrange the factors of 2 and 3 to get other sufficient conditions. Any of these follow from well-known conjectures, like https://en.wikipedia.org/wiki/Dickson%27s_conjecture but they are still open.
One question that fits into this theme but you didn't ask is: Among the numbers $n$ such that all groups of order $n$ are solvable, is the abundancy index of $n$ bounded? The answer is no, because the abundancy index of odd numbers is unbounded, and the Odd Order Theorem asserts that any odd natural number is a satisfactory value for $n$.
$29$ is a smaller prime one more than four times a prime and one less than six times a prime.
@GerryMyerson That's great, but I mentioned 997013 because it is briefly considered just before question 3, as the prime $p$ satisfying $5 \leq p < 10^6$ that yields the smallest abundancy index for $|PSL(2,p)|$.
@S.C that's what happens when I read the comments without reading the question.
I can answer Questions 1 and 4.
Make sure you look at S. Carnahan's answer. It deals with Questions 2 and 3.
Questions 1 and 4: If a finite group $G$ is not solvable then $|G|$ is divisible by $|G_0|$ for some finite simple group $|G_0|$.
By the CFSG, either $12\bigm||G_0|$ or $G_0$ is a Suzuki group.
If $12\bigm||G_0|$ then
$$\frac{\sigma(|G|)}{|G|}\geq\frac{\sigma(|G_0|)}{|G_0|}>\frac{\sigma(12)}{12}=\frac{7}{3}.$$
If $G_0$ is a Suzuki group then $320\bigm||G_0|$ and
$$\frac{\sigma(|G|)}{|G|}\geq\frac{\sigma(|G_0|)}{|G_0|}>\frac{\sigma(320)}{320}>\frac{7}{3}.$$
Thus, every non-solvable group has abundancy index larger than $\frac{7}{3}$.
By a theorem of Burnside, dating back to the early 20th (or maybe late 19th) century, if $G$ is a finite non-Abelian simple group, then $|G|$ is divisible either by $12$ or the cube of its smallest prime divisor. Suzuki groups were proved by Thompson to be the only simple groups of order prime to $3$ (actually before the full classification), and no non-Abelian simple group has odd order, by Feit-Thompson- more generally, no non-Abelian simple group has a cyclic Sylow $2$-subgroup. Hence every non-Abelian simple group other than a Suzuki group has order divisible by $12$.
@ThomasBrowning: Question 2 (or its weaker versions) asks about a sufficient condition on an integer to be the order of a non-solvable group. It is not about a necessary condition.
@SebastienPalcoux Then doesn't Question 2 have a negative answer by S. Carnahan's comment? Take n to be odd with large abundancy index.
@ThomasBrowning: Correct! One way for Question 2 (or weaker version 1) to survive would be to exclude the multiple of odd abundant numbers.
As I mentioned in a comment, Question 2 (in its revised form) has a negative answer, because odd natural numbers have unbounded abundancy index, while the Odd Order Theorem implies all groups of odd order are solvable.
Weaker version 2 has a positive answer: If $\beta$ is sufficiently close to 1, then any $n > 1$ whose abundancy index is greater than $\beta e^\gamma \log \log n$ is a multiple of 60, so there is a group of order $n$ that is unsolvable.
As I mentioned in a different comment, Question 3 is true subject to well-known open conjectures, such as [Dickson's conjecture][1]. In particular, it suffices to show that there are infinitely many primes $p$ such that $p-1$ is 4 times a prime and $p+1$ is 6 times a prime.
[1]: https://en.wikipedia.org/wiki/Dickson%27s_conjecture
That’s right! One way for Question 2 (or weaker version 1) to survive would be to exclude the multiple of odd abundant numbers. The least odd number of abundancy index greater than $3$ is $1018976683725$.
|
2025-03-21T14:48:31.412544
| 2020-06-29T23:20:02 |
364470
|
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|
Stack Exchange
|
A set of questions on continuous Gaussian Free Fields (GFF)
As I said in my previous posts, I'm trying to teach myself some rigorous statistical mechanics/statistical field theory and I'm primarily interested in $\varphi^{4}$, but I know that the absense of this term provides important simplifications to the theory and we can give meaning to the theory when this term is not included using functional integrals and Gaussian measures on functional spaces. My intention with this post is to understand the problems involved in the continuous limit of this theory. I know one usually discretizes the theory to define the objects of interest, but I'm trying to understand the origin of these problems starting from the continuous limit. It's very difficult to find such a complete analysis in books or articles, and I usually find myself having to built the whole picture from little pieces of it, so my intention here is to fill the gaps led by this process.
In what follows, I ask 5 questions and try to answer some of them, but I don't know if my answers and my reasoning are correct. I'd appreciate if you could correct me if needed and add more information, if necessary.
First of all, the idea is to give precise meaning to the probability measure:
\begin{eqnarray}
\frac{1}{Z}\exp\bigg{(}-\int_{\mathbb{R}^{d}}\frac{1}{2}\varphi(x)(-\Delta+m^{2})\varphi(x)dx \bigg{)}\mathcal{D}\varphi \tag{1}\label{1}
\end{eqnarray}
Where $\mathcal{D}\varphi$ is a "Lebesgue measure" in the space of fields. Here, the space of fields will be simply $\mathcal{S}'(\mathbb{R}^{d})$. In what follows, $\mathcal{S}'(\mathbb{R}^{d})$ is equipped with the strong topology and its associate Borel $\sigma$-algebra, i.e. the $\sigma$-algebra generated by its open sets.
Question 1: As I said before, I know that it is usual to discretize the theory and define (\ref{1}) by means of thermodynamic + continuous limits. But is it possible to address the problem directly on $\mathbb{R}^{d}$?
My attempted answer: I think that, once you discretized the theory and saw what are the correct limits and objects you need, you can pose the problem directly on $\mathbb{R}^{d}$ at the end of day, but it is by no means obvious, at a first sight, how to properly define (\ref{1}) or even other objects related to it, such as correlations etc.
In what follows, I'll address the problem directly on $\mathbb{R}^{d}$ assuming my answer to the first question is correct and I'm allowed to do it.
Question 2: Is (\ref{1}) a well-defined measure on its own, for all values of $m \ge 0$? How does the ultraviolet divergencies influence the existence of this measure? Does it play any role on its well-definiteness or just on correlation functions?
My attempted answer: I don't think this is well-defined on its own, because I don't think that the "product Lebesgue measure" $\mathcal{D}\varphi$ is well-defined in $\mathcal{S}'(\mathbb{R}^{d})$. However, I know that we can give meaning to (\ref{1}) if we use Minlos-Bochner theorem.
If my answer to question 2 is correct, I must use Minlos-Bochner. Then, (\ref{1}) is the measure $\mu_{G}(\varphi)$ on $\mathcal{S}'(\mathbb{R}^{d})$ induced by $W(f,f):=e^{C(f,f)}$ (using Minlos-Bochner) where:
\begin{eqnarray}
C(f,g):= \frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}\frac{\overline{\hat{f}(\xi)}\hat{g}(\xi)}{|\xi|^{2}+m^{2}}d^{d}\xi \tag{2}\label{2}
\end{eqnarray}
Question 3: Intuitivelly, I know that (\ref{2}) is related to (\ref{1}). This is because $\hat{C}(\xi) = 1/(|\xi|^{2}+m^{2})$ is the Fourier transform of the Green's function $G(x)$ of the massive Laplacian $-\Delta+m^{2}$. Informally: Green's functions are inverse operators and, therefore, the measure induced by Minlos-Bochner theorem is a functional analogue of the usual property that Fourier transform of Gaussians are Gaussians. But, apart from the intuition, how can we related (\ref{1}) to $d\mu_{G}$? In other words, does (\ref{1}) have anything to do with the covariance of $d\mu_{G}$?
My attempted answer: I think the only way to realize $d\mu_{G}$ is the corrected Gaussian measure associated to (\ref{1}) (which was not defined as a Gaussian measure in the first place) is by discretizing the space and recovering the theory with thermodynamic + continuous limits. But starting from Minlos-Bochner's theorem, with covariance (\ref{2}), it doesn't seem obvious to me (apart from intuition) that $d\mu_{G}$ has anything to do with (\ref{1}).
Question 4: As I mentioned before, $d\mu_{G}$ is a Gaussian measure on $\mathcal{S}'(\mathbb{R}^{d})$ while (\ref{1}) seems to be just induced by a bilinear form on $\mathcal{S}(\mathbb{R}^{d})\subset \mathcal{S}'(\mathbb{R}^{d})$. Is (\ref{1}) well-defined only as a subset $\mathcal{S}(\mathbb{R}^{d})$ of $\mathcal{S}'(\mathbb{R}^{d})$? Or is it actually a quadratic form on $\mathcal{S}'(\mathbb{R}^{d})$ (in which case I don't seem to understand it correctly)?
Question 5: If I can, in fact, work the theory directly in the infinite/continuous setup, and all the Gaussian measures are properly defined, is it possible to calculate correlations, say, by using properties of Gaussian measures?
Note: I said, right from the beginning, that the space of fields is $\mathcal{S}'(\mathbb{R}^{d})$ but I know it because I already studied some models before and I knew what was the properly funcional space to consider. However, I believe (not sure) that physicists interpret fields as proper functions e.g. on $\mathcal{S}(\mathbb{R}^{d})$ and (\ref{1}) would be something like a quadratic form $\langle \varphi, (-\Delta+m^{2})\varphi\rangle$ on $\mathcal{S}(\mathbb{R}^{d})$. Then, because of Minlos-Bochner theorem, one notices that $\varphi$ must actually be considered as an element of a larger space $\mathcal{S}'(\mathbb{R}^{d})$ in which (\ref{1}) have no meaning unless $\varphi \in \mathcal{S}(\mathbb{R}^{d})$. This is what I think, but I don't know if I'm completely wrong and fields have physical reasons to be tempered distributions right from the beginning.
Here https://arxiv.org/pdf/2004.04720.pdf there is a description of how to define the continous Gaussian Free Field - altthough it seems that you setup is quite different.
a definition of the continuous GFF along the lines of your Q1-Q5 is Generalized random fields and Lévy's continuity theorem on the space of tempered distributions; from the mathematical side the Euclidean path integral is perfectly well defined, the difficult nontrivial task is to derive this from some physical model.
@CarloBeenakker thanks for the reference! What do you mean by 'derive this from some physical model'? You mean choosing the right covariance so it represents some physical model?
I mean to derive the field theory from a Hamiltonian -- how is $\phi$ related to microscopic degrees of freedom?
Essentially, what is asked is the continuation of my previous MO answer
Reformulation - Construction of thermodynamic limit for GFF
and the solution of the exercise I mentioned at the end of that answer.
There, I explained the construction of Gaussian Borel measures $\mu_m$ on the space $s'(\mathbb{Z}^d)$ of temperate multisequences indexed by the unit lattice in $d$ dimensions.
The measure $\mu_m$ is specified by its characteristic function
$$
p\longmapsto\exp\left(-\frac{1}{2}\sum_{x,y\in\mathbb{Z}^d}p(x)G_m(x,y)p(y)\right)
$$
for $p=(p(x))_{x\in\mathbb{Z}^d}$ in $s(\mathbb{Z}^d)$, the space of multisequences with fast decay.
The discrete Green's function $G_m(x,y)$ is defined on $\mathbb{Z}^d\times\mathbb{Z}^d$
by
$$
G_m(x,y)=\frac{1}{(2\pi)^d}\int_{[0,2\pi]^d}d^d\xi\
\frac{e^{i\xi\cdot(x-y)}}{m^2+2\sum_{j=1}^{d}(1-\cos \xi_j)}\ .
$$
Here we will assume $m\ge 0$ for $d\ge 3$, and $m>0$ if $d$ is $1$ or $2$.
For any integer $N\ge 1$, define the discrete sampling map $\theta_N:\mathscr{S}(\mathbb{R}^d)\rightarrow s(\mathbb{Z}^d)$ which sends a Schwartz function $f$ to the multisequence
$$
\left(f\left(\frac{x}{N}\right)\right)_{x\in\mathbb{Z}^d}\ .
$$
This map is well defined and linear continuous.
Indeed,
$$
\langle Nx\rangle^2=1+\sum_{j=1}^{d} (Nx_j)^2\le N^2\langle x\rangle^2
$$
because $N\ge 1$.
So
$$
||\theta_N(f)||_k:=
\sup_{x\in\mathbb{Z}^d}
\langle x\rangle^k \left|f\left(\frac{x}{N}\right)\right|
\le \sup_{z\in\mathbb{R}^d}\langle Nz\rangle^k|f(z)|\ \le N^k\ ||f||_{0,k}
$$
where we used the standard seminorms
$$
||f||_{\alpha,k}=\sup_{z\in\mathbb{R}^d}\langle z\rangle^k|\partial^{\alpha}f(z)|
$$
for Schwartz functions.
Now consider the transpose map $\Theta_N=\theta_N^{\rm T}$ from $s'(\mathbb{Z}^d)$ to
$\mathscr{S}'(\mathbb{R}^d)$. It is defined by
$$
\langle \Theta_N(\psi),f\rangle=\langle\psi,\theta_N(f)\rangle=\sum_{x\in\mathbb{Z}^d}\psi(x)f\left(\frac{x}{N}\right)
$$
for all discrete temperate fields $\psi$ and continuum test functions $f$.
Essentially,
$$
\Theta_N(\psi)=\sum_{x\in\mathbb{Z}^d}\psi(x)\ \delta_{\frac{x}{N}}
$$
where $\delta_z$ denotes the $d$-dimensional Dirac Delta Function located at the point $z$.
Now $\Theta_N$ is continuous for the strong topologies.
Indeed if $A$ is a bounded subset of Schwartz space
$$
||\Theta_N(\psi)||_A=\sup_{f\in A}|\langle \Theta_N(\psi),f\rangle|=
\sup_{p\in \theta_N(A)}|\langle \psi,p\rangle|
$$
and $\theta_N(A)$ is bounded in $s(\mathbb{Z}^d)$ (because a continuous linear map sends bounded sets to bounded sets).
Suppose we are given sequences $m_N$ and $\alpha_N$ dependent on the UV cutoff $N$.
Define the Borel measure
$$
\nu_N=(\alpha_N\Theta_N)_{\ast}\mu_{m_N}
$$
on $\mathscr{S}'(\mathbb{R}^d)$.
Its characteristic function is
$$
W_N(f)=\int_{\mathscr{S}'(\mathbb{R}^d)}d\nu_N(\phi)\ e^{i\langle\phi,f\rangle}
=\int_{s'(\mathbb{Z}^d)}d\mu_{m_N}(\psi)\ e^{i\langle\psi,\alpha_N\theta_N(f)\rangle}
$$
by the abstract change of variable theorem.
We then get $W_N(f)=\exp\left(-\frac{1}{2}Q_N(f)\right)$
where
$$
Q_N(f)=\frac{\alpha_N^2}{(2\pi)^d}\sum_{x,y\in\mathbb{Z}^d}
f\left(\frac{x}{N}\right)f\left(\frac{y}{N}\right)
\int_{[0,2\pi]^d}d^d\xi\ \frac{e^{i\xi\cdot(x-y)}}{m^2+2\sum_{j=1}^{d}(1-\cos \xi_j)}
$$
$$
=\frac{N^{2-d}\alpha_N^2}{(2\pi)^d}\sum_{x,y\in\mathbb{Z}^d}
f\left(\frac{x}{N}\right)f\left(\frac{y}{N}\right)
\int_{[-N\pi,N\pi]^d}d^d\zeta\ \frac{e^{i\zeta\cdot(\frac{x}{N}-\frac{y}{N})}}{N^2 m_N^2+2N^2\sum_{j=1}^{d}\left(1-\cos \left(\frac{\zeta_j}{N}\right)\right)}
$$
after changing $[0,2\pi]^d$ to $[-\pi,\pi]^d$ by periodicity, then
changing variables to $\zeta=N\xi$, and finally some algebraic rearrangement.
Pointwise in $\zeta\in\mathbb{R}^d$, we have
$$
\lim\limits_{N\rightarrow\infty}
2N^2\sum_{j=1}^{d}\left(1-\cos \left(\frac{\zeta_j}{N}\right)\right)
=\zeta^2
$$
and this is why I put an $N^2$ in the denominator.
Finally, we can pick the right choice for the sequences $m_N$ and $\alpha_N$. For a fixed $m\ge 0$ (or strictly positive if $d=1,2$) we let $m_N=\frac{m}{N}$. Now we pick $\alpha_N$ so that
the prefactor $N^{2-d}\alpha_N^2$ becomes the volume element $N^{-2d}$ for a Riemann sum approximation of a double integral on $\mathbb{R}^d\times\mathbb{R}^d$.
Namely, we pick $\alpha_N=N^{-\frac{d}{2}-1}$.
Equivalently, going back to $\alpha_N\Theta_N(\psi)$, that means choosing
$$
\alpha_N\sum_{x\in\mathbb{Z}^d}\psi(x)\ \delta_{\frac{x}{N}}=\left(\frac{1}{N}\right)^{d-[\phi]}
\sum_{x\in\mathbb{Z}^d}\psi(x)\ \delta_{\frac{x}{N}}
$$
where $[\phi]=\frac{d-2}{2}$ is the (canonical) scaling dimension of the free field. I wrote the last equation in a way to explicitly display the lattice spacing $\frac{1}{N}$.
Now an excellent exercise, for graduate students in analysis, is to show that
$$
\lim\limits_{N\rightarrow \infty}Q_N(f)=\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d}
d^d\zeta\ \frac{|\widehat{f}(\zeta)|^2}{\zeta^2+m^2}
$$
where the Fourier transform is normalized as
$\widehat{f}(\zeta)=\int_{\mathbb{R}^d}d^dx\ e^{-i\zeta\cdot x} f(x)$.
Finally, Fernique's version of the Lévy Continuity Theorem for $\mathscr{S}'(\mathbb{Z}^d)$, shows that the Borel measures $\nu_N$ converge weakly to the one obtained directly in the continuum using the Bochner-Minlos Theorem.
Thanks again for the answer! I'm in a hurry, so I can't work out the details right now, but as usual this is an exellent/very detailed answer so I already accepted it because the bounty was going to expire in a few hours.
Thanks. I don't think accepting the answer does anything with the bounty.
Just a quick question: as I said, I will work the details later but as far as I understood it by reading it quickly, you proceed by taking limits to construct the infinite volume measures. This is something that confuses me a lot: if I want to work directly on $\mathbb{R}^{d}$. I know I can define (by Minlos-Bochner) the Gaussian measure on $\mathcal{S}'(\mathbb{R}^{d})$ with covariance $(-\Delta+m^{2})$, but it is not clear to me that this has to do with my "measure" (\ref{1}) in the first place. My point is: as far as I know, these connections are only possible once you induced (cont)
(...) once you induced these infinite volume measures from the discretized model. Is it true, indeed?
I mean, once you worked out the details and defined an infinite volume measure from the discretized version of the model, you get that the measure on $\mathcal{S}'(\mathbb{R}^{d})$ defined by $(-\Delta+m^{2})$ is the 'appropriate' infinite volume measure, but then these measures can only be 'interpreted' a posteriori.
To take (1) seriously, or what Carlo probably means by starting with a Hamiltonian and a measure with density $e^{-H}$ you need to replace $\mathbb{R}^d$ by a finite set. So you need a lattice with spacing $1/N$ but then the infinite volume lattice must be replaced by a finite box with $L^d$ sites. So you then have two limits to do $L\rightarrow\infty$ with fixed $N$ (the lattice thermodynamic limit of my previous answer), and then the continuum limit already in infinite volume, i.e., $N\rightarrow\infty$ which is what I explained in this answer.
|
2025-03-21T14:48:31.413462
| 2020-06-30T00:36:42 |
364474
|
{
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"Michael Greinecker",
"Pablo Lessa",
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}
|
Stack Exchange
|
Convergence of conditional measures for a convergent sequence of probabilities whose projection is constant
Setting
Suppose $\mu_n$ is a sequence of probability measures on $[0,1]\times [0,1]$ converging to a limit probability $\mu$ meaning that
$$ \lim_{n\to+\infty}\int f(x,y)d\mu_n(x,y) = \int f(x,y)d\mu(x,y)$$
for all continuous $f:[0,1]\times [0,1] \to \mathbb{R}$.
Suppose furthermore that all these probabilities project to the uniform measure on the first coordinate. This implies there are Borel mappings (conditional probabilities) $x \mapsto \mu_{n,x}$ and $x \mapsto \mu_{x}$ from $[0,1]$ to the space of probabilities on $[0,1]$ satisfying
$$\int f(x,y) d\mu_n(x,y) = \int_0^1 \int_0^1 f(x,y)d\mu_{n,x}(y) dx,$$
and
$$\int f(x,y) d\mu(x,y) = \int_0^1 \int_0^1 f(x,y)d\mu_{x}(y) dx.$$
Question
I'm looking for a reference for the fact that
$\lim_{n\to+\infty}\mu_{n,x} = \mu_x$
for almost every $x \in [0,1]$.
More generally, I'm looking for some reference covering the situation when $\mu_n$ are probabilities on some compact space with constant push-forward under some continuous mapping of that space.
Proof
Here's a proof of the claim (I would still love to have a reference).
Take $f(x,y) = h(x)g(y)$ with $h$ and $g$ continuous and notice that
$$\lim_{n \to +\infty}\int_0^1 h(x)(\mu_{n,x}(g) - \mu_x(g)) dx = 0,$$
where we use $m(g)$ for the integral of $g$ with respect to the measure $m$.
Using $h$ to approximate the set $A_{\epsilon} = \lbrace x \in [0,1]: \liminf_{n \to +\infty} \mu_{n,x}(g) - \mu_x(g) > \epsilon\rbrace$ and Fatou's lemma (all functions are bounded)
$$\frac{\epsilon}{2}|A_{\epsilon}| \le \int_0^1 h(x)\liminf_{n \to +\infty}(\mu_{n,x}(g)-\mu_x(g)) dx \le \liminf_{n \to +\infty} \int_0^1 h(x)(\mu_{n,x}(g)-\mu_x(g)) dx = 0,$$
where $|A|$ denotes the Lebesgue measure of $A$. This shows that $A_\epsilon$ has measure $0$.
Since this holds for all $\epsilon > 0$ and also for the function $-g$ we get
$$\lim_{n \to +\infty}\mu_{n,x}(g) = \mu_x(g),$$
for almost every $x$.
Intersecting the full measure sets where this holds, over all $g$ in a countable dense set of continuous functions on $[0,1]$, the claim follows.
What does it mean to approximate a measurable set by a continuous function?
@MichaelGreinecker I was thinking: 1- Take a compact subset of your set which has almost the same measure. 2- The characteristic function of the compact set is a monotone decreasing limit of continuous functions.
This is false. Generally, disintegration behaves poorly with respect to weak convergence. I believe the error in your proof is the first inequality, which I don't see how to justify.
Many counterexamples arise from a well known phenomenon in optimal transport. For any probability measure $\mu$ on $[0,1] \times [0,1]$ with uniform first marginal, there exists a sequence $\mu_n$ of probability measures on $[0,1] \times [0,1]$ with uniform first marginal such that (1) $\mu_n \to \mu$ weakly and (2) each $\mu_n$ is supported on the graph of a continuous function. That is, each $\mu_n$ is of the form $\mu_n(dx,dy)=dx\delta_{f_n(x)}(dy)$ for some continuous $f_n$. See Theorem 9.3 of Ambrosio's lecture notes, for example, and approximate the Borel maps therein in $L^1$ by continuous ones.
Now, for example, if $\mu$ is Lebesgue measure (or more generally if the disintegration $\mu_x$ is nonatomic for a.e. $x$), and $\mu_n$ is supported on the graph of a measurable function for each $n$, then there is no way we can have $\mu_{n,x} \to \mu_x$ weakly, because $\mu_{n,x}$ is a delta for each $n$ whereas $\mu_x$ is not (and the set of delta measures is weakly closed).
Thank you very much! This is a really nice answer. I still don't see how the proof is wrong, but I'll look into it.
I think I found the mistake. Repeating the argument for $-g$ only proves that $\liminf \mu_{n,x}(g) \le \mu_x(g) \le \limsup \mu_{n,x}(g)$ almost surely.
A simple special case of Dan's answer above: Define $f_n:[0,1] \to [0,1]$ by $f_n(x)= nx \mod 1$ and define $g_n:[0,1] \to [0,1]^2$ by $g_n(x)=(x,f_n(x))$. The pushforward $\mu_n=\lambda g_n^{-1}$ of Lebesgue measure $\lambda$ on $[0,1]$ is the uniform measure on the graph of $f_n$. The sequence $\mu_n$ converges weakly to Lebesgue measure $\mu$ on $[0,1]^2$
but for each $x$ in the unit interval, $\mu_{n,x}$ are Dirac measures that cannot converge weakly to $\mu_x=\lambda$.
That is really simple. Thanks! I noticed that restricting $n$ to powers of $2$ this is the mixing property for the iterates of the mapping $f_2$.
|
2025-03-21T14:48:31.413779
| 2020-06-30T03:07:23 |
364477
|
{
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"Dieter Kadelka",
"Eggplant",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364477"
}
|
Stack Exchange
|
Duality problem of an infinite dimensional optimization problem
I am reading the paper "OPTIMAL INEQUALITIES IN PROBABILITY THEORY: A CONVEX OPTIMIZATION APPROACH" by BERTSIMAS and POPESCU. In the paper, the authors derived a duality problem for an infinite-dimensional optimization problem. I am not sure how to derive the following:
The primal is:
$\max_\mu \quad \int_S \textbf{1}d\mu$
subject to $\int_\Omega \bar z^kd\mu=\sigma_\kappa$, $\forall \kappa\in J_k$.
The dual is:
$\min_{y\in \mathcal{R}^{|J_k|}} \quad \sum_{\kappa\in J_k}y_\kappa \sigma_\kappa$
subject to $g(\bar z)=\sum_{\kappa\in J_k}y_\kappa\bar z^\kappa\geq 1$, $\forall \bar z \in S$, and $g(\bar z)=\sum_{\kappa\in J_k}y_{\kappa}\bar z^\kappa\geq 0$, $\forall \bar z\in\Omega$.
In the above, $\mu$ is a probability measure and $S\subseteq \Omega\subseteq\mathcal{R}^n$.
Moreover, $\bar z=(z_1,\ldots,z_n)'$, $\kappa=(k_1,\ldots,k_n)'$, $\bar z^\kappa=z_1^{k_1}\cdots z_n^{k_n}$, and
$$
J_k=\{ \kappa=(k_1,\ldots,k_n)'|k_1+\cdots+k_n\leq k,~k_j\in\mathcal{Z}_+,~j=1,\ldots,n \}.
$$
Could anyone tell me how to deal with this kind of infinite-dimensional optimization problem?
Is the formulation "subject to $\int_\Omega \bar z^k d\mu = \sigma_\kappa, \kappa \in J_k$ correct? This would imply that $\int_\Omega \bar z^k d\mu$ has different values $\sigma_\kappa$, if $|J_k| > 1$.And what is $\bar z \in S$? A function? Not everyone has access to the paper. Please be more specific.
@DieterKadelka Sorry for the confusion. I have added illustrations for the notation.
This is a special case (with $f=1_S$) of the duality
$$s=i,\tag{1}$$
where
$$s:=\sup\Big\{\int f\,d\mu\colon\mu\text{ is a measure, }\int g_j\,d\mu=c_j\ \;\forall j\in J\Big\},$$
$$i:=\inf\Big\{\sum b_j c_j\colon f\le\sum b_jg_j\Big\},$$
$\int:=\int_\Omega$, $\sum:=\sum_{j\in J}$, $f$ and the $g_j$'s are given measurable functions, the $c_j$'s are given real numbers, and $J$ is a finite set such that (say) $0\in J$, $g_0=1$, and $c_0=1$, so that the restriction $\int g_0\,d\mu=c_0$ means that $\mu$ is a probability measure.
In turn, (1) is a special case of the von Neumann-type minimax duality
$$IS=SI,\tag{2}$$
where
$$IS:=\inf_b\sup_\mu L(\mu,b),\quad SI:=\sup_\mu\inf_b L(\mu,b),$$
$\inf_b$ is the infimum over all $b=(b_j)_{j\in J}\in\mathbb R^J$, $\sup_\mu$ is the supremum over all probability measures $\mu$ over $\Omega$, and $L$ is the Lagrangian given by the formula
$$L(\mu,b):=\int f\,d\mu-\sum b_j\Big(\int g_j\,d\mu-c_j\Big)
=\int \Big(f-\sum b_j g_j\Big)\,d\mu+\sum b_j c_j.$$
Indeed, $\inf_b L(\mu,b)=\int f\,d\mu$ if $\int g_j\,d\mu=c_j$ for all $j$, and $\inf_b L(\mu,b)=-\infty$ otherwise. So,
$$SI=s.\tag{3}$$
On the other hand,
$$IS=i.\tag{4}$$
Indeed,
\begin{align}
IS&=\inf_b\Big\{\Big[\sup_\mu \int \Big(f-\sum b_j g_j\Big)\,d\mu\Big]+\sum b_j c_j\Big\} \\
&=\inf_b\Big\{\Big[\sup\Big(f-\sum b_j g_j\Big)\Big]+\sum b_j c_j\Big\},
\end{align}
which is clearly no greater than $i$. On the other hand, if for some $b$ we have $s_b:=\sup\big(f-\sum b_j g_j\big)\in\mathbb R$, then $f\le\sum \tilde b_jg_j$ and
$$\sum\tilde b_jc_j=s_b+\sum b_jc_j=\Big[\sup\Big(f-\sum b_j g_j\Big)\Big]+\sum b_j c_j,$$
where $\tilde b_j:=b_j+s_b\,1_{j=0}$. So, $i$ is no greater than $\inf_b\big\{\big[\sup\big(f-\sum b_j g_j\big)\big]+\sum b_j c_j\big\}=IS$. Thus, (4) is verified as well.
So, by (3) and (4), (1) indeed follows from (2).
In turn, the von Neumann-type minimax duality (2) follows under general conditions when $L(\mu,b)$ is affine in $\mu$ and in $b$ (as it is in our case). A necessary and sufficient condition for the minimax duality
$$\inf_y\sup_x F(x,y)=\sup_x \inf_y F(x,y)$$
whenever $F(x,y)$ is concave in $x$ and convex in $y$ was given in this paper.
Thanks a lot! This perfectly answers my question.
|
2025-03-21T14:48:31.414045
| 2020-06-30T03:22:16 |
364479
|
{
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"authors": [
"Bjørn Kjos-Hanssen",
"Emil Jeřábek",
"Jochen Glueck",
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"url": "https://mathoverflow.net/questions/364479"
}
|
Stack Exchange
|
Is there a name for order-preserving functions $f$ where “$a\le b$ if and only if $f(a) \le f(b)$”?
This is something only slightly stronger than monotonicity. I think that in category theory this would be a fully faithful functor, but I’m not sure if there is a standard name for this in order theory.
For context, I’m working in a theorem prover (Isabelle) and this is a property of the function which converts words to natural numbers (embedding $\{0 \ldots 2^{32} - 1\}$ into $\mathbb{N}$).
The first sentence is really an understatement: in fact, this is much stronger than monotonicity.
For posets this condition forces injectivity so "order embedding" is natural. But for preorders (=set with reflexive transitive relations), it doesn't imply injective, so "order embedding" would be misleading.
The first thing I thought of was
order embedding
and this is confirmed by an article on monotonicity in order theory.
The first thing I thought of was also order embedding (like Bjørn), but one can also say that $f$ preserves and reflects the order. That way, you understand even if you don't know what an order embedding is. Folks sometimes write that $f$ is a strong order homomorphism in this situation.
I rather think that “order embedding” is more clear even if you see the terms for the first time; most people do not know what “reflects” means, whereas embeddings are commonplace. However, the “preserves and reflects” or “strong homomorphism” terminology has one big advantage, namely that it is applicable more generally to any relational structures, whereas the identification of this property with order embedding relies on reflexivity and antisymmetry of the partial order, and it is not valid in general.
@EmilJeřábek Somehow, your comment makes me think of languages without equality.
|
2025-03-21T14:48:31.414200
| 2020-06-30T04:17:34 |
364483
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364483"
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|
Stack Exchange
|
Gronwall estimate with a Fourier transform
Suppose I have the following equality
$$\hat{u}_\epsilon(t,k) = \alpha(t,k) + \int_0^t\int_{\mathbb{R}^n} e^{ik\cdot(x +\epsilon \phi(s,x))}u_\epsilon(s,x)dxds$$
Where $\alpha(t,k) \geq 0$ and $\alpha(t,\cdot)\in L^1( \mathbb{R}^n)$. Moreover $\phi \in C^\infty\cap L^\infty(\mathbb{R}\times\mathbb{R}^n)$ and $\epsilon \in (0,1]$. Furthermore, $\hat{u}_\epsilon$ denotes the Fourier transform of $u_\epsilon$. I would like to deduce a uniform estimate for $\|\hat{u}_\epsilon(t,\cdot)\|_{L^1}$ with respect to $\epsilon$ (as $\epsilon\rightarrow 0^+$) using Gronwall's inequality. I am wondering if it's possible in general due to the non-linear exponent $e^{ik\cdot(x+\epsilon\phi(s,x))}$. The relevant theorem is the Beurling-Helson theorem which makes me think I cannot get such an estimate unless $\phi$ is linear in $x$. But the fact that the map $x\mapsto x + \epsilon\phi(t,x)$ is nearly the identity makes me think otherwise. Any ideas would be appreciated.
I will more comfortable with the notation $v_\epsilon=\hat{u_\epsilon}$; you have then
$$
v_\epsilon(t,x)=\alpha(t,x)+\int_0^t\int e^{2πix(\xi+\epsilon\phi(s,\xi))} \hat{v_\epsilon}(s,\xi) d\xi ds=\alpha(t,x)+\int_0^t \bigl(\textrm{Op}(e^{2πix\epsilon\phi(s,\xi)})
v(s,\cdot)\bigr) (x) ds
$$
where $\textrm{Op}(e^{2πix\epsilon\phi(s,\xi)})=A_{\epsilon, s}$ is the
operator with symbol
$e^{2πix\epsilon\phi(s,\xi)}$. Let me now assume that $A_{\epsilon, s}$ is bounded on $L^2(\mathbb R^n)$ and that $\alpha$ is in $L^2$: you get with $L^2$ norms
(the triple norm is the operator-norm)
$$
\Vert v_\epsilon(t)\Vert\le\Vert\alpha(t)\Vert+\int_0^t\vert\!\Vert A_{\epsilon, s}\Vert\!\vert
\Vert v_\epsilon(s)\Vert ds,
$$
and if $\vert\!\Vert A_{\epsilon, s}\Vert\!\vert$ as a function of $s$ is in $L^1$, you can use Gronwall. The real problem is to get an estimate for the triple norm (by the way $\phi$ is certainly real-valued): that operator is likely to be a Fourier Integral Operator and not a pseudo-differential operator, so to get the sought bound you should consider
$$
A_{\epsilon, s}^* A_{\epsilon, s},
$$
which will be a pseudo-differential operator under some assumptions on $\phi$.
|
2025-03-21T14:48:31.414362
| 2020-06-30T04:51:48 |
364485
|
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"Anonymous",
"Bugs Bunny",
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"Qiaochu Yuan",
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|
Stack Exchange
|
Category of modules over an Azumaya algebra and the Brauer group
Let $k$ be a field, and let $\alpha \in \mathrm{Br}(k)$. Let $A$ be an Azumaya algebra representing $\alpha$. Then the category $A$–$\mathrm{mod}$ depends only on $\alpha$.
I would like to know whether there's a way to describe which $k$-linear categories arise this way. Thus I'd like to know if there's a way to define the Brauer group of a field $k$ as classifying certain kinds of $k$-linear categories. I'd also like to know if there's a good description of the sum of elements of the Brauer group in terms of categories (is it some sort of tensor product of categories?).
The one condition I can come up with is that it should be a semisimple abelian category over $k$ for which the endomorphism algebra of the unit object is $k$.
An even more bold hope is to express the invariant map $\mathrm{Br}(\mathbb{Q}_p) \to \mathbb{Q}/\mathbb{Z}$ in terms of this category.
If $A$ isn't commutative there won't be a tensor product.
Vector spaces over $k$. This category does not depend on $\alpha$
$k$-linear cocomplete categories admit a "tensor product over $\text{Mod}(k)$" (thinking of them as module categories over $\text{Mod}(k)$) and the only thing you need to know about it to answer this question is that the tensor product of $\text{Mod}(A)$ and $\text{Mod}(B)$ is $\text{Mod}(A \otimes_k B)$. Azumaya algebras then give rise to invertible categories with respect to this tensor product and I believe every invertible such category has this form although I don't know how to prove it. More details about the tensor product in this blog post.
A more explicit characterization isn't hard but also isn't particularly enlightening - over a field, Azumaya algebras are central simple algebras, so you are looking for $k$-linear cocomplete categories where
there's exactly one isomorphism class of simple object
every object is a direct sum of copies of this simple object, and
the (categorical) center (endomorphisms of the identity functor) is $k$.
But the invertibility characterization should hold for $k$ any commutative ring, and is in my opinion very conceptually clean; it tells us that the Brauer group is a kind of "higher Picard group" classifying a categorified version of line bundles. Some difficulties here in general with the difference between the Brauer group and the cohomological Brauer group.
Fix a commutative ring $k$. I'd suspect that any element $\alpha \in H^2(k,\mathbf{G}_m)$ would give rise to an invertible category: look at the "degree 1" subcategory of quasicoherent sheaves on the corresponding $\mathbf{G}_m$-gerbe over $k$. For this to come from an Azumaya algebra, I think you want $\alpha$ to be a torsion element. This is automatic if $k$ is a field (or a regular ring), but not in general.
Are you suggesting that even when the ring is not regular, it might be isomorphic to the cohomological Brauer group? (but not the group defined in terms of Azumaya algebras)
What is the first "it" in the above comment?
I wouldn't be surprised if the statement "if $\mathcal{C}$ is invertible over $\mathrm{Mod}(k)$ then $\mathcal{C} \cong \mathrm{Mod}(A)$ for an Azumaya algebra $A$" will depend on "functional analytic" questions about how completed you want your categories to be. These are answered already by what you want "$\mathrm{Mod}$" to mean: all modules, finitely generated modules, finitely generated projective modules, ...?
You probably mean the first option, and so will work with locally (finitely?) presentable categories. But if you will allow me to work with the last option, then the natural tensor product is only completed for direct sums and idempotents; then Tillmann showed that $\mathcal{C}$ is 2-dualizable iff it is $\mathrm{mod}(A)$ for a finite-dimensional separably algebra, and so invertible iff it $A$ is Azumaya.
Actually, these categories do know each other: Tillmann's result kicks in as soon as you demand that $\mathcal{C}$ have enough compact projective objects. Whether dualizability/invertiblity in ${$locally presentable $k$-linear categories$}$ implies enough compact projectives is, so far as I know, still open, but it is expected to hold.
@TheoJohnson-Freyd Is this invertibility the same as the associated 2-dim TQFT being invertible?
@Anonymous to clarify my "it", the "degree 1" subcategory of quasicoherent sheaves on the corresponding Gm-gerbe over k
@DavidCorwin Yes.
To compare my comments with @Anonymous's example about sheaves of categories over a scheme $X$, I think the situation is as follows. An element of $\mathrm{H}^2(X, \mathbb{G}_m)$ should determine a sheaf of locally presentable categories which is invertible over $\mathrm{QCoh}(X)$. Perhaps you can work slightly smaller ($\mathrm{Coh}$?). But I suspect that if it doesn't come from the honest Brauer class, then it won't have enough "$\mathrm{QCoh}(X)$-compact-projectives".
|
2025-03-21T14:48:31.414694
| 2020-06-30T05:10:52 |
364487
|
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"Michael Greinecker",
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|
Stack Exchange
|
BSDE without volatility
Let $(W_t)_{0\leq t\leq 1}$ be a standard Wiener process on $[0,1]$, and let $\mathcal{F}_t$ be the natural filtration. Consider a BSDE
$$
dX_t=f(t,X_t)dt+\sigma(t,X_t) dW_t
$$
with terminal condition $X_1=x$, where $f(t,\cdot)$ and $\sigma(t, \cdot)$ are $\mathcal{F}_t$-adapted square integrable processes.
My question: is it possible for the BSDE to be well defined if $\sigma(t,X)=0$ for all $t\in [0,1]$ and all $X$? Also, it seems unlikely to me that I can treat such a case as an ODE since reversing time would screw with the progressive measurability of $f$. Am I wrong?
You can treat it as an ODE. Since the resulting function is a deterministic function of time, progressive measurability amounts to the function itself being Borel measurable.
Sorry, I was imagining that the drift is not necessarily deterministic. I will edit my question accordingly.
I am not sure if I understand your question correctly. A typical Brownian BSDE has the form
$$dY_t = f(\omega, t, Y_t, Z_t)dt - Z_t dW_t$$
with terminal condition
$$Y_T = \xi \in \mathcal{F}^{W}_T$$
where $Y$ and $Z$ are two parts of the solution and required to be adapted to the Brownian filtration. If your question boils down to if there is a BSDE with deterministic terminal condition $\xi = x$ and (second part of the) solution process constant zero, the answer is yes. E.g., take
$$f(\omega, t, Y_t, Z_t) = W_{\frac{T}{2}}1_{(\frac{T}{2},\frac{3T}{4}]} - W_{\frac{T}{2}}1_{(\frac{3T}{4},T]}$$
and $x=0$.
|
2025-03-21T14:48:31.414923
| 2020-06-30T05:34:29 |
364490
|
{
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|
Stack Exchange
|
Analytic solution of a compact operator eigenvector (integral equation)
In Sewall Wright's Evolution of Mendalian Population, the equation for the nonrecurrent mutation is
$$\frac{\phi(x)}n = \binom n {nx}\int_0^1 q^{nx}(1-q)^{n(1-x)}\phi(q)\,dq,\quad \forall x\in[0,1],$$
where $n>0$ and $\binom a b$ is the binomial coefficient. We are to solve for a nonzero function $\phi\ne0$. It is an eigenvalue problem of a compact operator (Fredholm equation of the second kind). But what is a general method for solving this integral equation? dirichlet Is there a transform, say Mellin transform, that does the trick?
|
2025-03-21T14:48:31.414994
| 2020-06-30T05:36:39 |
364491
|
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|
Stack Exchange
|
P-adic distance between solutions to S-unit equation
Let $p$ be a fixed prime number and $S$ is a finite set of prime numbers which does not contain $p$. A theorem of Siegel asserts that the number of solutions to the $S$-unit equations are finite; that is, there are only finitely many $S$-unit $u$ such that $1-u$ is also an $S$-unit. Therefor for each such $S$ there exist a lower bound on $|u_1-u_2|_p$ where $u_1$ and $u_2$ are solutions to $S$-unit equations.
My question is: does there exist such a lower bound uniformly? More precisely, does there exist a lower bound for the $p$-adic distance between solutions to the $S$-unit equations that only depends on the size of $S$(and perhaps on $p$)? Here we are assuming $S$ does not contain $p$.
I recall that the following paper looked at similar questions: Bombieri, Mueller, Poe, The unit equation and the cluster principle, Acta Arith. 79 (1997) 361–389.
There cannot be such a uniform bound, unless $p=2$ in which case there are
no solutions at all (because $p \notin S$ but $u$ and $1-u$ cannot both be
$2$-adic units).
Fix $p$ and $e$. We shall construct a set $S$ of uniformly bounded size
and $S$-units $u_1,u_2$ that are congruent modulo $p^e$.
By a theorem of Chen (Sci. Sinica 16 (1973), 157-176),
there are infinitely many primes $q$ such that $q-2$ is
either a prime or the product of two primes (and as usual this remains
true if we require that neither $q$ nor $q-2$ is a multiple of $p$).
Hence $q/2$ is an $S(q)$-unit for some set $S(q)$ of at most $4$ primes.
Therefore there exist distinct $q_1,q_2$ satisfying this condition
with $q_1 \equiv q_2 \bmod p^e$.
Then $u_1 := q_1/2$ and $u_2 := q_2/2$ are both $S$-units for some
$S = S(q_1) \cup S(q_2)$ of size at most $8$
(indeed at most $7$ because $S(q_1),S(q_2)$ both contain $2$);
and $|u_1 - u_2|_p \leq p^{-e}$. $\Box$
|
2025-03-21T14:48:31.415137
| 2020-06-30T06:51:31 |
364493
|
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|
Stack Exchange
|
Harmonic functions for subordinate Brownian motions and the Hölder continuity
This question is about harmonic functions of subordinate Brownian motions.
We write $B=(\{B_t\}_{t \ge 0}, \{P_x\}_{x \in \mathbb{R}^d})$ for the $d$-dimensional Brownian motion. Let $\{S_t\}_{t \ge 0}$ be a subordinator, which is an increasing pure-jump Lévy process starting at zero independent of $B$. We set $Y_t=X_{S_t}$, $t \ge 0$. Then, $Y=(\{Y_t\}_{t \ge 0},\{P_x\}_{x \in \mathbb{R}^d})$ is a subordinate Brownian motion.
I could find some sufficient conditions for nonnegative harmonic functions for $Y$ to satisfy the Harnack inequality (perhaps, KM is a standard reference). As a consequence, we can prove that the harmonic functions are Hölder continuous.
However, this method will generally not allow us to obtain the exact value of the index of Hölder continuity. Is there any preceeding research on the quantitative estimates of the index of the Hölder continuity?
Smoothness of harmonic functions for subordinate Brownian motions is proved in my paper with Tomasz Grzywny:
T. Grzywny, M. Kwaśnicki, Potential kernels, probabilities of hitting a ball, harmonic functions and the boundary Harnack inequality for unimodal Lévy processes, Stoch. Proc. Appl. 128(1) (2018): 1–38, DOI:j.spa.2017.04.004
See Theorem 1.7 and Remark 1.8(b) there.
Let me stress that the work of Ante Mimica and Panki Kim (and their co-authors) had a huge impact on the area. It is a great pity that Ante died so young.
Thank you for your very helpful comment. I will read your joint paper immediately. I'm not familiar with the research area and may have asked some basic questions. I'm sorry.
|
2025-03-21T14:48:31.415262
| 2020-06-30T06:56:56 |
364494
|
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|
Stack Exchange
|
Is the collection of primitive recursive functions a lower set in the poset of computable functions?
If $g:\mathbb{N}\to\mathbb{N}$ is primitive recursive and $f:\mathbb{N}\to\mathbb{N}$ is computable such that $f(n) \leq g(n)$ for all $n\in \mathbb{N}$, does this imply that $f$ is primitive recursive?
No. Let $g$ be the constant function 1.
Let $\{h_n\}$ be a computable list of all primitive recursive functions and let $f_n(x)=\min(h_n(x),1)$.
So $\{f_n\}$ is a computable list of all primitive recursive functions bounded by 1.
Now let $F(n)=1-f_n(n)$. Then $F$ is another computable function bounded by 1, distinct from all the $f_n$, so $F$ is not primitive recursive.
No, as Bjørn explained in his answer, but it is a lower set in the class of functions with primitive recursive graphs. Specifically, if $f:\mathbb{N}^k\to\mathbb{N}$ is such that
The function $g:\mathbb{N}^{k+1}\to\{0,1\}$ such that $g(x_1,\ldots,x_k,y) = 1 \iff f(x_1,\ldots,x_k) = y$ is primitive recursive, and
There is a primitive recursive function $h:\mathbb{N}^k\to\mathbb{N}$ such that $f(x_1,\ldots,x_k) \leq h(x_1,\ldots,x_k)$,
then $f$ is primitive recursive.
The reason is that we can recover $f$ by bounded search: $$f(x_1,\ldots,x_k) = \mu y \leq h(x_1,\ldots,x_k)\,[g(x_1,\ldots,x_k,y) = 1].$$
Note that there are many functions that are not primitive recursive but whose graphs are primitive recursive, for example the Ackermann function has a primitive recursive graph. This fact is useful to show that the inverse Ackermann function is primitive recursive, for example.
|
2025-03-21T14:48:31.415375
| 2020-06-30T07:51:29 |
364497
|
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|
Stack Exchange
|
Can we define cardinality that works under weaker grounds than Scott's cardinals?
Its known that within the perspective of $\sf ZF$ related theories, Scott's definition of cardinality can work under weaker grounds than Von Neumann's! Scott's cardinality works as long as the principle "every set is equinumerous to some well founded set" is in action. Now choice implies that, so it works under choice (even in absence of regularity!), and it can also work in absence of choice and regularity but provided that the above principle is axiomatized.
Can we have a definition of cardinality that is weaker than that of Scott's? That is, it works under all grounds that Scott's cardinality works under, but it can also work at some grounds where Scott's cardinality fails!
Scott's cardinality of a set $x$ is the set of all sets equinumerous to $x$ of the lowest possible rank.
You do understand that MO is not a platform for publishing new results in this manner, right?
No. I don't understand that. Because it's written that we can ask questions and answer them at the same time, and the question is mathematical. So where did I break the rules?
While the stackexchange technology allows it, I take it to be a social norm of MathOverflow that it is ok to ask a question before one knows the answer, then if later it turns out the asker gets the answer they should answer their own question. Asking and answering in one go is so uncommon, though, that this is the only example I can recall in a decade of using the site.
The answer is YES!
Let $\mathcal H_\alpha$ stand for the set of all sets hereditarily strictly subnumerous to ordinal $\alpha$.
Now for any set $x$, $\mathcal H^x_{min}$ is meant to be the minimal $\mathcal H_\alpha$ such that there exists an iterative power of it that is supernumerous to $x$. Formally:
Define: $\mathcal H^x_{min} = min \ \mathcal H_\alpha: \exists \beta [ x \rightarrowtail P_\beta(\mathcal H_\alpha)]$
Where $``\rightarrowtail"$ signify "is injective to".
$P_\beta$ is defined recursively as:
$P_\emptyset(x)=x \\ P_{\beta+1}(x)= P(P_\beta(x)) \\ P_\beta(x) = \bigcup (\{P_\alpha(x):\alpha < \beta\}) \text{ if } \beta \text{ is a limit ordinal} $
Now by $P^x_{min}(S) $ its meant the minimal iterative power of $S$ that is supernumerous to $x$. Formally:
Define: $P^x_{min} (S) = min \ P_\beta (S): x \rightarrowtail P_\beta(S) $
Now we come to define cardinality of a set $x$, denoted by $``|x|"$, as the set of all subsets of $P^x_{min} (\mathcal H^x_{min})$ , that are equinumerous to $x$. Formally:
Define: $|x|= \{y| \ y \sim x \land y \subseteq P^x_{min} (\mathcal H^x_{min}) \}$
Where $\sim$ signify "is bijective to"
This definition of cardinality can work under grounds weaker than those of Scott's cardinality? The latter demands the assumption that "every set is equinumerous to some well founded set", and under that assumption the cardinality [defined here] of any set would be exactly its Scott's cardinal. But this definition can work even when the above assumption fails, but it requires the statement: $$\forall x \exists \alpha \exists \beta : x \rightarrowtail P_\beta (\mathcal H_\alpha)$$ which doesn't imply the above assumption! (See this answer, and this).
Note: by well founded set its meant a set whose transitive closure is well founded with respect to $\in$.
|
2025-03-21T14:48:31.415599
| 2020-06-30T07:54:59 |
364499
|
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|
Stack Exchange
|
Nonconstant smooth maps $f:\mathbb{R}^4-\{0\}\to L\left(\mathbb{R}^4,\mathbb{R}^5\right)$
Does there exist a nonconstant smooth map $f:\mathbb{R}^4\setminus\{0\}\to L\left(\mathbb{R}^4,\mathbb{R}^5\right)$ such that each $f(x):\mathbb{R}^4\to\mathbb{R}^5$ is an isometric linear map and $x\in\bigcap_{i=1}^4\ker\frac{\partial f}{\partial x_i}(x)$ for all $x\in\mathbb{R}^4\setminus\{0\}$?
I'm so used to people writing $f(x)$ to refer to a function rather than its values that it took me a few re-readings to parse that the question is meaningful exactly as written.
@MattF. Yes. Take $f:\mathbb{R}^4-{0}\to L\left(\mathbb{R}^4,\mathbb{R}^8\right)$ such that the matrix $\left(F_{jk}(x)\right)$ of the linear map $f(x):\mathbb{R}^4\to\mathbb{R}^8=\mathbb{C}^4$ with respect to the canonical bases of $\mathbb{R}^4$ and $\mathbb{C}^4$ is given by
$$F_{jk}(x)=\left(\frac{x_k}{|x|}\left(\frac{1}{2}-i\frac{x_j}{|x|}\right)+i\delta_{jk}\right)e^{2i\frac{x_j}{|x|}}.$$
Perhaps the OP can tell us why this particular question is of interest. In particular, why we are being asked about the specific case of $n=4$ and $m=1$ in, say, the more general question for $f:\mathbb{R}^n\setminus{0}\to L(\mathbb{R}^n,\mathbb{R}^{n+m})$.
@RobertBryant : It turns out that such a map $f$ gives rise to an isometric immersion of $\mathbb{S}^{n-1}$ into $\mathbb{S}^{n+m-1}$, which is totally geodesic if and only if $f$ is constant. Since every isometric immersion of $\mathbb{S}^3$ into $\mathbb{S}^4$ is totally geodesic, the answer for my question is no. But I was wondering if perhaps there might be an analytical proof of this that could offer some insights into the open problem of whether there exists a nontotally geodesic isometric immersion of $\mathbb{S}^n$ into $\mathbb{S}^{2n}$ for $n\geq3$.
@SubmanifoldsinSpaceForms: I see; I suspected as much. I think that, in general, it's not good practice to pose questions on MO for which you know the answer without at least revealing that you know the answer and explaining why the argument you have is not satisfactory for your purposes.
@RobertBryant: ok
|
2025-03-21T14:48:31.415755
| 2020-06-30T08:26:16 |
364503
|
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|
Stack Exchange
|
On classifying groups of order $p^5$
Can someone suggest me some source where the author has classified all non-isomorphic groups of order $p^5$ ? I need complete classification (not upto isoclinism), and also in finitely presented form . I found that with increase in value of prime $p$, number of groups increases. So, can we completely classify all groups of order $p^5$ for any prime $p$, in finitely presented form or get their structure description ?
Eamonn O'Brien's papers, or Charles Leedham,'s Green's papers may include references to such results. Marshall Hall dealt with $2$-groups of order at most $64$, I think, so that includes $2^{5}$ obviously.
@Geoff Your comment is useful to me. Thanks!
Probably the classification is uniform for $p\ge 5$ (or at least large $p$). (Uniform doesn't mean the number is independent of $p$, but the description ought to be uniform.)
The number of such groups depends on $p \bmod 12$. See the last paragraph and table at https://kconrad.math.uconn.edu/blurbs/grouptheory/groupsp3.pdf.
@KConrad I have found the general formula by G.Bagnera is $$61+2p+2gcd(p-1,3)+gcd(p-1,4)$$. So, for $p=3$, there are 67 groups in number. In your article, it is written 66. Kindly see into it.
Fixed, and I changed a reference to include Dokchitser's site https://people.maths.bris.ac.uk/~matyd/GroupNames/.
See section 6.5 in https://arxiv.org/abs/1806.07462 (in german)
@spin Your suggestion is really helpful. Although I don't know how to translate it into English, still it contains all the stuff I needed in mathematical form.
@spin Is there any easy way except google translator to convert the language into English ?
Can we get isoclinism class $\phi_2$ for any $p^n$ ordered group ?
Bender in 1927 wrote a paper A determination of the groups of order $p^5$ in English giving a proof of the classification of groups of order $p^5$. The list of groups is also available in the GAP Small Groups Library package.
Actually Bender's paper also contains gaps (at least for $p=3$). I recommend reading Besche, Eick, O’Brien, A millennium project: constructing small groups, Int. J. Algebra & Computation (2003) which summarizes the history of the results.
|
2025-03-21T14:48:31.415933
| 2020-06-30T08:37:51 |
364504
|
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|
Stack Exchange
|
Exactness of injective tensor products
For (algebraic) tensor products, it is well-known that the functor $A\otimes_R \cdot:Mod_R\rightarrow Mod_R$ is only (left-) exact when $A$ is a flat $R$-module. In particular, all vector spaces are flat. What happens in the continuous (archimedean) setting?:
Let $B$ be a separable infinite-dimensional Banach space and suppose that $
f:E\rightarrow F,
$
is a continuous linear injective map from a separable nuclear space $E$ to a separable Banach space $F$, both infinite-dimensional (if it matters). Let $\otimes_{\epsilon}$ denote the injective tensor product of LCS and let $\hat{\otimes}_{\epsilon}$ denote its completion.
Is the map
$
1_{B}\hat{\otimes}_{\epsilon} f: B\hat{\otimes}_{\epsilon} E \rightarrow B\hat{\otimes}_{\epsilon} F,
$
a continuous linear 1-1 map also?
Related:
This post is related to this unanswered post.
Do you mean just the injective tensor product or the completed injective tensor product?
@JochenWengenroth (I made the correction) but indeed I'm interested in the completed injective tensor product.
This is always true (without nuclearity): If $T_j:E_j\to F_j$ are continuous linear maps between Hausdorff locally convex spaces and $E_2$ is complete then $$ T_1\hat\otimes_\varepsilon T_2: E_1 \hat\otimes_\varepsilon E_2 \to F_1\hat\otimes_\varepsilon F_2$$ is injective if so are $T_1$ and $T_2$. This is 16.2.2 in Jarchows's book Locally Convex Spaces.
Thank you very much Jochen. Out of curiousity I'm assuming the result fails if the completion is not taken?
No, without the completions it is much easier.
I just wanted to mention that I really enjoyed your book "Derived Functors in Functional Analysis" :) I had read it during my PhD some years ago
Thank you......
|
2025-03-21T14:48:31.416078
| 2020-06-30T09:32:07 |
364508
|
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|
Stack Exchange
|
Almost-geodesics on a Riemannian Hilbert manifold which are still almost geodesics in some submanifold
Let $H$ be a separable infinite dimensional Hilbert space, and consider it as a Hilbert manifold in the usual way (that is, with the single chart with the identity map). It is known that there always exists (under the above hypothesis) a metric (e.g. the one induced by the inner product, as noted by @Thomas Rot) such that $H$ becomes a complete Riemannian Hilbert manifold (see, for instance, Biliotti L., Mercuri F. (2017) Riemannian Hilbert Manifolds. In: Suh Y., Ohnita Y., Zhou J., Kim B., Lee H. (eds) Hermitian–Grassmannian Submanifolds. Springer Proceedings in Mathematics & Statistics, vol 203. Springer, Singapore). Moreover, we know that Hopf-Rinow Theorem does not hold in general under our assumptions, because $H$ is infinite dimensional. However, we do know that almost-geodesics always exist (see I. Ekeland: The Hopf-Rinow Theorem in infinite dimension. J. Diff. Geometry 13, 287-301 (1978)). Now, recall that a submanifold $M$ of $H$ is said to be totally geodesic if every geodesic in $M$ is also a geodesic in $H$. My question is related to a converse of this notion, which involves almost-geodesics instead of geodesics: under what conditions can we say that an almost-geodesic in $H$ is still an almost-geodesic in $M$, for some infinite dimensional submanifold $M$? Is there any known term for such manifolds?
EDIT: an almost-geodesic, following the above paper by Ekeland, is a $C^{\infty}$ path $c(t)$ from some point $c(0)=p \in H$ to some other point $c(1)=q \in H$ such that
$$ \int_{0}^{1} \| \dot{c}(t) \|_{c(t)} dt \leq \epsilon + d(p,q)$$
for some $\epsilon > 0$. Theorem A in the cited paper by Ekeland states that:
Let $N$ be a complete infinite dimensional Riemannian manifold, and let $p,q \in N$. $\forall \epsilon >0$, there exists some $C^{\infty}$ path $c(t)$ from $p$ to $q$ such that
$$ \int_{0}^{1} \| \dot{c}(t) \|_{c(t)} dt \leq \epsilon + d(p,q)$$
that is, every such path is an almost geodesic.
Here, with 'submanifold' I mean an embedded submanifold.
EDIT: as @Thomas Rot pointed out in the comments, the previous question can be trivially answered. As he observed, an interesting question then is: under what conditions is a manifold $\epsilon$-geodesic?
About your introduction: $H$ with the Riemannian metric induced by the inner product is itself is a complete Hilbert manifold right? Do you mean to study other complete metrics?
Yes, you are right, thanks for noting this. Actually, what I'm most interested in now is not the metric used (we just need one for which $H$ is complete, so that almost geodesics certainly exist), but the existence of almost geodesics on $H$ which are still almost geodesics on $M$. Any metric for which this happens will suffice, so we can consider the inner product
can you define the notion of "almost geodesic" for me?
Sure, I will add the definition in the question
Any curve is an almost geodesic then.
Ah, you're right! I was focusing too much on the theorem and didn't notice this. Thanks!
There are mathematical questions here if you change the quantifiers. For example one can define an $\epsilon$-almost geodesic as a curve that satisfies the bound with the given $\epsilon$. The question is then what submanifolds are $\epsilon$-almost geodesic? Maybe you will need to add some extra conditions on how to make the measurment precise for long geodesics. I would not know the answer (even in the finite dimensional case).
This is a great improvement of the question, thanks!
|
2025-03-21T14:48:31.416331
| 2020-06-30T09:34:54 |
364509
|
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|
Stack Exchange
|
Topologically mixing cellular automata on groups
For which group-alphabet pairs $(G, A)$ does $(G, A^G)$ admit a topologically mixing cellular automaton?
Definitions:
Let $G$ be a (discrete) group. An alphabet is a finite set of cardinality at least two. The full shift on the group $G$ and alphabet $A$ is the group action $(G, A^G)$, where $A^G$ has its product topology and $G$ acts by $gx_h = x_{g^{-1}h}$. A cellular automaton or CA is a continuous function $f : A^G \to A^G$ that commutes with the action, i.e. $g \cdot f(x) = f(g \cdot x)$ for all $g \in G$ and $x \in A^G$. Let's say a dynamical $\mathbb{N}$-system is a pair $(f, X)$ where $f : X \to X$ is continuous and $X$ is a compact metrizable space. If $f$ is a CA, then $(f, A^G)$ is a dynamical $\mathbb{N}$-system. A dynamical $\mathbb{N}$-system $(f,X)$ is topologically mixing if
$$ \forall \mbox{ open sets } U, V \subset X: \exists n_0 \in \mathbb{N}: \forall n \geq n_0: f^{-n}(V) \cap U \neq \emptyset. $$
Some things out of the way:
If $G$ contains a finitely-generated infinite group $H$ and $(H, A^H)$ admits a topologically mixing CA $f$, so does $(G, A^G)$.
Proof: By acting as $f$ separately on each left coset of $H$ we obtain a CA $f' : A^G \to A^G$ whose action is isomorphic (as a dynamical $\mathbb{N}$-system) to the (possibly infinite) product system $\prod_{i \in G/H} (f, A^H)$, and a product of topologically mixing systems is clearly topologically mixing.
In particular, if $\mathbb{Z} \leq G$ then $(G,A^G)$ admits a topologically mixing CA.
Proof: On $A^{\mathbb{Z}}$, the shift by $1 \in \mathbb{Z}$ is itself a CA (since $\mathbb{Z}$ is abelian), and obviously it is topologically mixing. Now apply the previous item.
If $G$ is finite, then topological mixing cannot happen for any alphabet $A$.
Proof: Fixed points of the $G$-action stay fixed, and every set is open.
If $G$ is infinite and locally finite, then topological mixing cannot happen for any alphabet $A$.
Proof: A cellular automaton has a finite neighborhood $N \subset G$ such that $x \mapsto f(x)_g$ factors through $x \mapsto x|_{gN}$. The CA $f$ acts separately on the left cosets of $\langle N \rangle$, and we can apply the previous item to conclude it is not topologically mixing when $\langle N \rangle$ is finite.
Now, I am expecting that the answer to the above question is "Exactly the pairs $(G, A)$ such that $G$ is not locally finite." If I am correct in this, in light of the previous items, the following is the gist of the problem.
Let $G$ be a finitely-generated torsion group. Find a topologically mixing CA on the full shift $(G, A^G)$ (for some alphabet $A$, preferably for all).
Of course $f$ has to be surjective, and I would be especially interested in a bijective example (on some groups, e.g. the Grigorchuk group, injective implies bijective). If you can do just topological transitivity, that's also interesting. If you prefer ergodicity or measure-theoretic mixing notions (for the uniform Bernoulli measure on $A^G$), that's also interesting but presumably harder to do.
I think that now after a few years of daydreaming I see how to do this. It seems a little tricky to write down carefully, and maybe not worth the trouble (especially as I have no idea if anything this complicated is needed). So here's a sort of sci-fi novel version, sorry about the length, there's a lot of moving parts.
Theorem(-ish). Let $G$ be a finitely-generated torsion group. Then there exists a topologically mixing CA on the full shift $(G,A^G)$ for any finite non-trivial alphabet $A$.
Proof(-ish). Take $A$ any nontrivial alphabet and $G$ any finitely-generated torsion group. Suppose $0,1 \in A$. Our cellular automaton $f : A^G \to A^G$ will have two important properties: first, it is conserving, meaning it conserves the number of live/populated cells, that is, cells containing a nonzero symbol. In formula form
$\forall x \in A^G: |\{g \in G \;|\; x_g \neq 0\}| = |\{g \in G \;|\; f(x)_g \neq 0\}|$.
Second, it is reversible, and not just that, it is morally conjugate to its inverse by another cellular automaton (i.e. its inverse has a similar description, up to swapping the order of some steps that were arbitrary in the first place).
I will explain how to construct the CA that realizes the topological mixing property, step by step, interspersed between the ideas of how topological mixing is realized. But let's start by discussing some aspects of the behavior of the final CA immediately, assuming only the above properties.
First, observe that by conservation and torsionness of $G$, if we take a finite configuration (meaning $x \in A^G$ such that its support $|\{g \in G \;|\; x_g \neq 0\}|$ is finite), then its $f$-orbit is finite (and by reversibility $x$ is $f$-periodic). This can be proved by induction on the cardinality of the support: If you have only a single symbol in $x$ at $g$, then after at most $|A|$ time steps you see the same symbol again $gh$, and then use the period of $h$. If you have $n$ symbols, observe that if there exists $R$ such that at all times the symbols stay within distance $R$ from each other for all time, then we can think of them as a single symbol in a conserving cellular automaton. If there is no such $R$, then eventually the finitely many symbols split into at least two groups separated by a sufficient distance that by induction they become periodic separately.
The idea is to take a configuration $x \in U$ to have finite support, let's call this the blob, and observe it until the Poincaré recurrence time predicted by the previous paragraph (in the dynamics of the CA we of course have yet to describe). Let $K \subset G$ be the finite set of cells that become populated at some point during this time. Let $k \in K$ be a "corner" cell, meaning such that a geodesic path exists from $k$ to infinity. Such $k$ exists by a compactness argument.
Now, our CA will admit the construction of "worms" that can contract and expand at some "speed" (e.g. measured with word norm w.r.t. some finite generating set), which will be the maximal asymptotic speed at which you can populate empty areas or completely evacuate areas. There will be things that move faster, but they can only move inside already existing material, informally the "speed of sound is significantly greater than the speed of light". Namely, the worms will have nerves carrying signals governing their behavior. The idea is that we have our worms go pick up a symbol out of the blob (specifically at $k$), and then retreat at the speed of light so that the blob is now left on its own, and will have smaller support. We iterate this process to erase the blob, and then we will build another arbitrary blob in its place.
Before I can describe the worms, let me first describe a way to embed arbitrary one-dimensional cellular automata into cellular automata over arbitrary groups. For this I need the notion of an atom. An atom has a nucleus and a cloud. The nucleus simply consists of three $1$s in a specific constellation, more specifically there is a nucleus "at" $g$ and $g$ is specifically the proton of the nucleus (and the resulting atom), if these live cells are at positions $g, gh, gh'$, where $h, h'$ are some fixed elements such that $d(1,h) = 2, d(h,h') = 1, d(1, h') = 3$, and the radius $6$ ball around the proton contains no other live cells. Observe that such a pair $h, h'$ exists. We fix some large $R$ (specified later) and in the $R$-ball around the proton, we can have any other symbols, as long as they are pairwise separated by a distance of at least $3$. These form the cloud, and we call the cells in its support its electrons. (Obviously all this faux physics terminology is just for mnemonic purposes.)
We say an atom is active if there is no other atom at distance at most $2R$. Note that we can uniquely decompose any configuration in $A^G$ into atoms and other stuff. Now, in each cloud, we can code a large amount of information in the constellation of electrons. In particular, we can store two elements of some finite set $B$, and we can store a pointer to a "follower atom" (specifically its proton) and a "predecessor atom" which are at distance, say, at most $10 R$. This needs a little calculation because the group can have a strange growth rate, but note that there must be infinitely many possible choices of $R$ such that a quantity exponential in $R$ (the amount of data we can code in the cloud) is much larger than the size of the $10 R$ ball (the offset vectors we need to code), as otherwise the group's growth rate is a tower function (while it obviously has an exponential upper bound).
Now, we can use a standard conveyor belt trick. We think of atoms that are their follower's predecessor as connected to their follower, and then every configuration splits into polymers, namely atom chains. On each such chain, as each atom stores an element of $B^2$, we can interpret a conveyor belt: think of the two $B$s as being on top of each other, connect the top $B$-elements into a tape, and the bottom ones into an "upside down" tape, and at the ends of the polymer (if they exist), you join the top and bottom of the $B^2$-symbol. This way, every configuration actually splits into non-coding areas and encoded configurations of the form $B^{\mathbb{Z}}$ (polymers that have at most one end) and $B^{2n}$ (finite polymers).
It is straightforward to simulate any one-dimensional cellular automaton on these encoded tapes. Our starting point, the primordial worm, is obtained by simulating a very large power of the shift. The primordial worm does not move, and it does not eat, it only thinks shifty thoughts. We will slowly evolve it into a worm that can move around, and can eat (and... uneat) things, including other worms.
Let us call the cellular automaton corresponding to the primordial worm $f_{\mathrm{shift}}$. All the action is, so far, "inside" the support of the configuration, in the sense that the support does not essentially grow in the action of $f_{\mathrm{shift}}$. This will be what gives the speed of sound / nerve signal we referred to earlier. Observe that by using a suitable coding of $B^2$, this indeed is a conserving cellular automaton. Actually if we are specifically simulating a shift (or a conserving cellular automaton), one possibility is to have a certain subconstellation corresponding to each $b \in B$, and explicitly allocate a part of the atoms for storing this.
Now, we add to our worms the capability of extending themselves. For this, we will have two special symbols $1$ and $2$. We will construct an involutive cellular automaton that looks at the head of each worm, namely the end of the worm where the top track is moving symbols "out of" the track. Specifically, look at the three $B$-symbols coded on the top track at the head. If the symbols are not of the form $12b$ for any $b \in B$, or $b12$ for any $b \in B$, then nothing happens. If they are one of these, then the idea is to keep the $12$ in place, and possibly insert/remove the $b$ after it. Check that the distance from the head the tail end of the worm is vast enough. (Specifically, check the distance of the $12$ to the tail end is vast, so that the distance is not modified by what happens in the next paragraph.) If it is not vast, do nothing.
If it is indeed vast, then if we are in the case $b12$, read off the coding of a new atom on the top track (in the atoms preceding the $1$, starting with the $b$). Basically part of the $B$-symbols being carried have a position where they store symbols (or holes) that can be used for this purpose. Verify that the coded atom's predecessor arrow agrees with the forward arrow of the atom coding the $2$. If so, then we would like to put down the atom at the head of the worm, extending it (you should literally move the symbols from the special positions of the top track atoms to their positions, so as to keep the CA conservative). In the situation with $12b$ on the top track, you should snatch the atom containing the $b$ and store it in holes on the top track, so that this is an involution. Of course nothing happens if there are already atoms where we want to put something, or we are trying to contract an atom that's not there.
Of course, there's the issue that there may be other worms tring to extend themselves. We simply check that no other worms are nearby before we extend/contract. Note that we should make sure that when worms extend themselves they don't come closer to us, so that we don't e.g. extend two worms and then decide they are now too close to contract themselves back. The standard trick (which was already used when considering whether a polymer is "vast enough") is that the $2$ in a worm's head is its anchor, and we measure distance to the anchor primarily (cancel movement if another anchor is at distance at most $R'$, say), and only secondarily to other cells (cancel movement if a non-anchor cell is at distance at most $R'' \leq R'/10$, say). Let us call the CA that performs this extension/contraction $f_{\mathrm{e/c}}$.
At this point a worm that has one infinite end can extend themselves and contract itself quite freely: the speed of sound is very fast so if we apply $f_{\mathrm{shift}}$ and $f_{\mathrm{e/c}}$ alternately, the data at the head of the worm is completely different (locally), so the instructions to successive applications of $f_{\mathrm{e/c}}$ are independent.
At this point, remember that we have this "corner" cell $k$ where at some point in time (again, of course it depends on this CA we are developing, but whatever it will be at the end, there is a corner $k$...), a live cell appears in the evolution of the blob. What we can already do at this point is grow a worm along a geodesic from infinity so that it is near $k$ at a time when $k$ is populated.
We now want to evolve our worms even further, by allowing them to eat, and our plan is to eat up the cell at $k$ when it is populated, and move it onto our track. This is really a can of worms... For the sake of simplicity I prefer to think of the blob as adversarially as possible, so when I send a worm $w$ to remove a symbol from the blob I cannot discount the possibility that the blob becomes very angry, and uses all its might to shoot a bunch of worms at my worm $w$, eating it up and becoming even more powerful than before. And of course, it's also an issue that our cellular automaton rule couldn't possibly know which worm has "priority" when snatching, $w$ or the other worms.
The next evolution step, which helps with this a bit, is that we will have worms of two different colors, red and blue. This can be indicated by e.g. having a second kind of nucleus, or just something static in the cloud. We take our CA to be
$\prod_{c \in \{\mathrm{red}, \mathrm{blue}\}} f_c \circ f_{\mathrm{shift}} \circ (f_{\mathrm{e/c}} \circ f_{\mathrm{shift}})^n$,
where $n$ is absolutely humongous ($n = 1$ should work fine, but for the mental picture I prefer that we can contract/extend a lot between other things).
The action of $f_c$ is be similar to extension/contraction of worms: At the end of a worm (= end of a polymer), $f_c$ reads an instruction on the top track to move the symbol at offset $t$ (relative to the proton of the atom at the head) to some special part of the encoded $B$-symbol, or vice versa; we refer to this as eating or uneating a symbol. We have to be very careful here, as the CA may not be invertible if we eat a symbol on the tape of another worm that is about to eat something elsewhere (and that elsewhere could be far away, since the worms look at many atoms before their head, before making their decisions). So we now describe the many checks we perform before eating (or uneating) anything.
The idea is, when $f_c$ is applied, worms of color $c$ stake claims over some positions, again some this is guided by values that $f_{\mathrm{shift}}$ moves on the top track at their head. One may think of the worms as putting their fingers on these positions. First each worm will ask all worms near it for permission to make a change. More precisely, each worm carries some values on its top track, one primary value and several inhibitor values (these need not be directly at head, we can actually use any fixed amount of atoms of the worm leading up to the head to store them). If a worm find its primary value in the inhibitors of some other worm, then it does nothing. If after the inhibitions, two worms are still interested in the same coordinate, then neither will touch it. In fact, if two worms are interested in coordinates at distance at most $2R$, then both drop their claim. If after this, a worm is still interested in a particular coordinate, it checks that the modification does not affect any other $c$-colored atoms, i.e. we do not introduce any new $c$-colored atoms, and you do not modify the contents of any existing $c$-colored atoms. (Otherwise, again, the worm simply drops its claim.) Finally, if the change touches an atom from a worm of the other color $\bar{c}$ or it introduces such a worm, then in those worms (possibly the worms are different before and after the modification) we check all the inhibitor values seen locally, and check that our primary value is not among them.
Here there's some playing with numbers to see that this gives an involution, which I didn't do, but I think there isn't anything really nontrivial since we never modify the $c$-colored worms so the set of claims made by $c$-worms is exactly the same, and the only things that can both change in the configuration, and can affect cancellation of claims, are cells that are part of $\bar{c}$-colored worms, and we explicitly check for those both before and after the change.
Ok, so now clearly we can have our worm get close enough to the corner $k$ that we can stake a claim for the nonzero symbol in that position. If this position is part of a $\bar{c}$-colored atom at that point in time, our worm should be of color $c$. We choose the primary value so that no inhibitor values prevent this (including a possible new $\bar{c}$-colored atom that appears when we remove an atom), and we should conversely have among our own inhibitor values all the nearby worms' primary values. The only problem that could still prevent the change is that we actually introduce a $c$-colored atom when making this change. But clearly we can pick the encodings so that there is always at least one atom whose removal does not introduce a $c$-colored atom, and we can just eat this nearby cell instead of $k$. Indeed I think this is automatic if the atoms are sufficiently error correcting and unbordered. This is another point where I didn't work out the details, but I do not think there's anything subtle here that could go wrong.
Ok, so now we believe we can snatch a symbol into our worm, at least momentarily. Now we will begin contracting our worm at maximal speed, what I referred to as "light speed" above. As mentioned, things can actually move much faster in theory, namely information moves much faster in the "nerves" of worms. In any case, it should be an easy induction to show that no finite configuration can extend into empty space faster than the expansion/contraction rate of worms, even if it is guided by possible inhibition signals coming from the area into which it is expanding. I do not want to try to even state that carefully, but hopefully the claim is clear, and at least to me it seems obviously true.
Now at this point, for all we know, we are contracting and a bunch of other worms have gotten very angry and are chasing our worm, trying to snatch its head. (I don't see why something like this couldn't happen since we may actually bring a blue worm into existence. Of course we could probably first deactivate it by being careful about what we snatch, but I feel it's easier to think adversarially about the blob.) The other worms will not see very far into our snatcher worm, since we are contracting at the same speed (or faster) than they are extending, so all we need to do is pick our inhibitor signals so that they do not modify us. They may be angry, but they are very polite, and they will leave us alone if we do so.
Now the blob is smaller, and I said we continue by induction. But there are a couple of problems. First, possibly the angry worms chase us forever, this time I don't think it can happen because the only input they get from us is some inhibitor signals, and they should not be able to use this to guide them along the geodesic direction along which we are contracting. But I don't see a clean argument.
If I had infinitely many geodesic rays then I would be fine, as I could just drag those angry worms to infinity and would have made the blob smaller quicker, but it's not immediately clear to me how to get infinitely many geodesic rays in a group that eventually get separated from each other (maybe there is some variant of Halin's theorem that gives this, or even some simpler argument?) So while contracting, we should also keep snatching things from the chasers. That doesn't slow us down since the snatching steps are between, and not alternative to, contraction/extension, but there is the issue that some of the chasers may be the same color as us. We of course also need to snatch things that are red, so we need to send in both a red and a blue worm.
So it'd be great if we could have red and blue snakes travel sort of in the same place. We can in fact easily do precisely this: allow atoms to be offset by some vector, and store that vector in their nucleus, say. The distances are measured w.r.t. the "virtual position" the offset vector points to, so that we can still argue they are moving at the same geodesic rate, but now by growth rate considerations if we allow a sufficient offset vector size, we can easily have two worms travel along the same geodesic. Now, we can use a red and a blue worm in unison to eat up the chasers. And this also shows that we have a blue worm we can use to red cells in the blob.
At this point, the CA construction is finished, and on the way I have explained why it is able to erase a finite blob, i.e. how to eat its entire population into the internal memory of the red and blue snatcher worms. Now, we need to build an arbitrary blob. Well, just observe that the CA is morally conjugate to its inverse: the inverse is
$\prod_{c \in \{\mathrm{blue}, \mathrm{red}\}} (f_{\mathrm{shift}}^{-1} \circ f_{\mathrm{e/c}})^n \circ f_{\mathrm{shift}}^{-1} \circ f_c$
(where we observe that $f_c$, $f_{\mathrm{e/c}}$ are involutions), and we see that the process works the exact same (actually looking at this formula maybe this is literally conjugate by a CA; anyway we don't need that).
We of course need yet another geodesic for the building process. One could use the same geodesic and offsets, and probably even the same worms with some thinking, but the easiest way is probably to take a two-way infinite geodesic (which exist by compactness because one-way geodesics exist and the graph is transitive), and use the geodesic "in the other direction" for building. This way, the building process doesn't even see the forward process in any way, as long as it's performed sufficiently late. And of course we can have any delay as long as it's big enough, so indeed we get topological mixing.
Square.
I believe that an easy modification to this construction shows that every infinite finitely-generated torsion group admits a physically universal cellular automaton in the sense of Janzing, namely instead of $f_{\mathrm{shift}}$, use a more complicated CA that allows the worms to think, and after the initial extraction process send the information about the pattern to the construction process using one-endedenss of torsion groups. (On non-torsion groups, physical universality should be much easier to exhibit, but of course hasn't been done either.)
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2025-03-21T14:48:31.418349
| 2020-06-30T09:35:25 |
364510
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Stack Exchange
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Does my functional satisfy the Palais Smale condition?
Consider the functional
$$ I(u)=\frac{1}{2} \int_\Omega |\nabla u|^2\ dx + \frac{1}{4} \int_\Omega (1-|u|^2)^2 \ dx - \frac{c}{2} \int_\Omega \langle i\partial_1 u , u\rangle ,$$
where $u:\mathbb{R}^2 \to \mathbb{C}$ is a $H^1$-function ($H^1=$ Sobolev space), $\Omega \subset \mathbb{R}^2$ bounded, $c\in (0,\sqrt{2})$ and $\langle , \rangle$ stands for the scalar product of $\mathbb{C} \equiv \mathbb{R}^2$. I have proved that $I$ is coercive and weakly lower semicontinuous and $C^1$.
I wonder if $I$ satisfies the Palais Smale condition at level $c$ $(PS)_c$, that is, assume that $x_n$ is a sequence s.t $I(x_n)\to c$ and $I ' (x_n)\to 0$, then there exists a converging subsequence of $x_n$ in $H^1(\Omega)$.
Any idea is welcome. Thank you in advance!.
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2025-03-21T14:48:31.418446
| 2020-06-30T12:22:16 |
364515
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"Ben McKay",
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Stack Exchange
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Élie Cartan's paper "Les groupes réels simples, finis et continus" of 1914
Question 1.
Does Élie Cartan's paper
Les groupes réels simples, finis et continus,
Ann. Sci. École Norm. Sup. (3) 31 (1914), 263–355
contain a classification of $\Bbb C$-linear involutions of simple complex Lie algebras?
Question 2.
If not, what kind of a similar/related classification does it contain?
Question 3.
In what book/paper is this Cartan's paper discussed?
MathSciNet contains only the title, while zbMATH
contains a review: a translation into German of a few lines from the introduction.
I do read French. However, it is hard to read a paper on Lie groups or Lie algebras written before Dynkin: the language has changed completely!
Why are you asking about $\mathbf{C}$-linear involutions? doesn't the topic of the paper suggest that it classifies $\mathbf{C}$-skew-linear involution? In any case it is never explicit (the paper doesn't mention automorphisms). And all along it makes this classification, which is equivalent to classifying those skew-linear involutions (more than half of the paper is about exceptional cases EFG). I don't think it gives any information on linear involutions.
About Q3: https://scholar.google.fr/scholar?hl=fr&as_sdt=0%2C5&q=Cartan+Les+groupes+r%C3%A9els+simples%2C+finis+et+continus+1914&btnG= can always help.
@YCor: You are right. We need real structures (anti-linear involutions). However, conjugacy classes of of real structures bijectively correspond to conjugacy classes of $\Bbb C$-linear involutions. See page 442 of Helgason's book, or Serre, Cohomologie galoisienne, III.4.5, Theorem 6 and Example (b). Helgason writes that Cartan discovered this later, in 1929.
The paper and its progeny are discussed at length in Helgason (1978, p. 537):
In his paper [2] Cartan classifies the simple Lie algebras over R. His method, which required formidable computations, used the signature of the Killing form although it often happens that two nonisomorphic real forms of the same complex algebra have the same signature. Cartan's statement ([2], p. 263): “Les groupes réels d'ordre $r$ qui correspondent à une même type complexe d’ordre $r$ se classent en général complètement d’après leur caractère,” is therefore not to be taken literally; cf. Lardy ([1], p. 195). After noticing the equivalence of problems B and B' (§1) Cartan (in [12]) simplified his original treatment (see also Lardy [1]). Following his general theory [1] of automorphism of complex simple Lie groups, Gantmacher [2] gave a simplified treatment of the real classification. For further developments of this method see Murakami [3], Wallach [2], and Freudenthal and de Vries [1]. While Gantmacher used a Cartan subalgebra $\mathfrak h$ of $\mathfrak g$ whose “toral part” $\mathfrak h\cap\mathfrak k$ is maximal abelian in $\mathfrak k$, Araki develops in [1] a new method using a Cartan subalgebra $\mathfrak h\subset\mathfrak g$ whose “vector part” $\mathfrak h\cap\mathfrak p$ is maximal abelian in $\mathfrak p$. In addition to a solution to problem B' (§1) this method gives valuable information about the restricted roots and their multiplicities (cf. Exercises F). A modification is given by Sugiura [2]. In the present work we use the method of Kac [1] which at the same time gives a rather explicit description of the automorphism $\sigma$ of finite order.
Excellent! Thank you! I have Helgason's book in my bookcase, but I have not read it....
Just a comment about Helgason's statement "Cartan's statement...is therefore not to be taken literally". I always felt that this was a little unfair. When Cartan wrote "en général" in that sentence, I feel that he intended it to be read as 'usually' or as we would use the phrase 'it is generally the case that'. Cartan obviously knew examples of non-isomorphic real forms with the same character, but usually the character is enough to distinguish two real forms of the same complex simple Lie algebra. In fact, I think he was alerting the reader to the fact that it was not always sufficient.
@RobertBryant Fair statement (on the statement (on the statement))! ;)
The paper of Cartan is discussed in
D. W. Morris, Introduction to arithmetic groups. Deductive Press, [place of publication not identified], 2015.
The free book can be downloaded from arXiv: https://arxiv.org/abs/math/0106063.
Here is a quote from that book:
The classification of real simple Lie algebras (Theorem 18.1.7) was obtained by E. Cartan [4]. (The intervening decades have led to enormous simplifications in the proof.)
The book discusses the involutive isometries of symmetric spaces, and Cartan involutions due to Cartan subgroups, but doesn't say whether Cartan ever mentioned them.
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2025-03-21T14:48:31.418807
| 2020-06-30T13:00:00 |
364518
|
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Stack Exchange
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Sum involving determinants of binomial coefficients, indexed by partitions
I would appreciate some help proving a conjecture related to combinatorics and representation theory.
Given an integer partition $\lambda\vdash n$, define a polynomial in $N$ whose roots are the negatives of the contents of the partition,
$$ [N]_\lambda=\prod_{\square \in \lambda}(N+c(\square)).$$ This polynomial is closely related to the value of a Schur function evaluated at the $N\times N$ identity matrix. On the other hand, given $\nu\vdash m$ and $\rho\vdash k$ contained in $\nu$, Jacobi-Trudi applied to a skew-Schur function leads to a determinant of binomial coefficients
$$s_{\nu/\rho}(1_N)=\det_{1\le,i,j\le m}\left({N+\nu_i-i-\rho_j+j-1 \choose \nu_i-i-\rho_j+j}\right).$$ The final ingredient I need for my question is another determinant of binomials,
$$A_{\lambda\rho}=\det_{1\le,i,j\le k}\left({\rho_i-i \choose \lambda_j-j}\right).$$
Now, in the course of some physics calculation, I arrived at the quantity
$$ E_{\lambda\nu}(N)=\sum_{\lambda\subset\rho\subset\nu} A_{\lambda\rho}s_{\nu/\rho}(1_N).$$
I thought this was as far as I could push it, but experimentation convinced me that, as a function of $N$, this guy satisfies
$$ E_{\lambda\nu}(N)\propto [N]_{\nu/\lambda}.$$
It is very surprising to me that this sum should factor like this.
The question is how to prove the above conjecture.
For example, if $\nu=(2,2,1)$ and $\lambda=(1)$, the six terms in the sum are
$$\{\frac{1}{24}N(N^2-1)(5N-6),-\frac{1}{2}N^2(N-1) ,\frac{1}{3}N(N^2-1) ,\frac{1}{2}N(N-1),-N^2,N\}.$$ When all these are added, the result is proportional to $N(N-2)(N^2-1)=[N]_{(2,2,1)/(1)}$.
Actually, I think I know the proportionality constant when $\nu$ and $\lambda$ are both hooks:
$$E_{\lambda\nu}(N)= \frac{1}{(m-n)!}{m-n \choose m-n-\ell(\nu)+\ell(\lambda)}[N]_{\nu/\lambda}.$$
What is the actual question then in this case? It seems no "formal" question has been stated in the post....
An obvious but probably unhelpful suggestion is to try to put variables $x_i$ back into the equation.
@user61318 Indeed, you are quite right and Lemma 9.1 in the paper you linked shows that my $E$ can be computed using an analogue of Cauchy-Binet. I would be happy to accept an answer with this content, if you want to write one.
@user61318 Yes, the factorization follows. Let me know if you are going to write an answer (otherwise I will write one myself)
See Problem 80 at http://math.mit.edu/~rstan/ec/ch7supp.pdf (version of 4 July 2020) and the solution at http://math.mit.edu/~rstan/ec/ch7suppsol.pdf.
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2025-03-21T14:48:31.419021
| 2020-06-30T13:36:16 |
364520
|
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|
Stack Exchange
|
Maximal number of smallest circuits in a matroid
It is known (see here for example) that, in a simple graph of odd genus $g$ with $n$ vertices and $m$ edges, the number of cycles of lenght $g$ is at most $\frac{n(m-n+1)}{g}$.
Since this can be be phrased only in terms of circuits and matroid properties (when the graph is connected, we can recover $n$ as the rank of the matroid plus one), I was wondering is this result could be extended to any matroid. More precisey, is it true that, for any matroid of size $m$, rank $r$ and odd genus $g$, the number of circuits of size $g$ is at most $\frac{(r+1)(m-r)}{g}$?
If no, are there some weaker similar bounds on the number of circuits of size $g$?
If yes, what can we say about the equality case? In the case of graphs, it is reached for the Moore graphs. Are there new non-graphical matroids reaching the equality?
If you leave the world of graphical matroids, no such small bound can hold. The number of shortest circuits in a uniform matroid $U_{m,r}$ is $\binom{m}{r+1}$.
Oh, you're right... I was hoping that this bound could have given some matroid theoritic version of Moore graphs, but this probably can't work
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2025-03-21T14:48:31.419147
| 2020-06-30T16:04:33 |
364527
|
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"Barinder Banwait",
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|
Stack Exchange
|
Q-curves and twisting
An elliptic curve $E$ over $\overline{\mathbb{Q}}$ is called a $\mathbb{Q}$-curve if it is isogenous (over $\overline{\mathbb{Q}}$) to all its Galois conjugates -- see Are Q-curves now known to be modular? for example.
If I take a finite Galois extension $K / \mathbb{Q}$ and an elliptic curve $E / K$ whose base-extension to $\overline{\mathbb{Q}}$ is a $\mathbb{Q}$-curve, then all the Galois conjugates $E^{\sigma}$ are also defined over $K$, but the isogenies between them might not be. Supposing $E$ to be non-CM for simplicity, then what you get instead is a $K$-isogeny from each conjugate $E^{\sigma}$ to some possibly non-trivial quadratic twist of $E$. Let me say $E$ is a strong $\mathbb{Q}$-curve over $K$ if it's non-CM and it's actually $K$-isogenous to all its Galois conjugates. (Clearly any $\mathbb{Q}$-curve over $K$ becomes a strong $\mathbb{Q}$-curve over some finite extension $L / K$, but I want to keep $K$ fixed here.)
It's easy to produce examples of $\mathbb{Q}$-curves which aren't strong $\mathbb{Q}$-curves, by taking a strong $\mathbb{Q}$-curve and applying a quadratic twist by an element of $K^\times / K^{\times 2}$ that's not stable under $Gal(K / \mathbb{Q})$. However, I can't find any examples which aren't of this form.
Are there $\mathbb{Q}$-curves which are not twists of strong $\mathbb{Q}$-curves?
(I'm chiefly interested in the case when $K$ is a real quadratic field here.)
I think there are examples of $\mathbb{Q}$-curves defined over a quadratic field $K$ which are not strong $\mathbb{Q}$-curves over $K$ in Jordi Quer's paper "$\mathbb{Q}$-Curves and Abelian Varieties of $\mathrm{GL}_2$-Type". He uses the term $\mathbb{Q}$-curves completely defined over $K$ instead of strong $\mathbb{Q}$-curves (I will stick to Quer's terminology). The key result is Corollary 3.3, which then can be applied for instance to the family of $\mathbb{Q}$-curves with an isogeny of degree $3$ that he writes in Section 6:
$$
C^{(a)}\colon Y^2 = x^3 -3\sqrt{a}(4+5\sqrt{a})X+2\sqrt{a}(2+14\sqrt{a}+11a).
$$
Suppose that $C^{(a)}$ is not CM (the curve is CM only for 9 values of $a$). In the terminology of the article, the sets {a} and {3} are dual bases with respect to the degree map, and $K_d = \mathbb{Q}(\sqrt{a})$. The curve $C^{(a)}$ is defined over $K_d$ and by Corollary 3.3 if the quaternion algebra $(a,3)_\mathbb{Q}$ is different in the Brauer group from $(-1,a)_{\mathbb{Q}}^x$ for all $x\in\{0,1\}$ then there is no curve $\overline{\mathbb{Q}}$-isogenous to $C^{(a)}$ completely defined over $K_d$ (when he writes isogenous in Corollary 3.3 he means over $\overline{\mathbb{Q}}$, not just over $K_d$).
It looks like your notion of strong $\mathbb{Q}$-curve over $K$ is what Peter Bruin and Andrea Ferraguti refer to as a $\mathbb{Q}$-curve being completely defined over $K$. Such curves have $L$-function factoring as a product of $L$-series of newforms for $\Gamma_1(N)$. This then seems to coincide with the definition of strongly modular given by Xevi Guitart and Jordi Quer. This latter set of authors provide an explicit example of an elliptic $\mathbb{Q}$-curve (which they call a building block after Elisabeth Pyle's thesis) over $K = \mathbb{Q}(\sqrt{-3})$ which is not strongly modular, and state that no curve isogenous to it over $\overline{\mathbb{Q}}$ and defined over $K$ can be strongly modular:
$$ Y^2 = X^3 + 4aX^2 + 2(a^2 + b\sqrt{-3})X, $$
for $a,b \in \mathbb{Q}$. I haven't checked the details, but this might give you what you're after?
It is known that strongly modular implies completely defined over $K$ (I'm excluding CM to be safe), but the converse is not always true. I think that the example by Guitart and Quer is an instance of a building block which is not strongly modular, but building blocks are completely defined over $K$ if I understood correctly.
(However, the converse is true if $K$ is a quadratic field.)
@FrançoisBrunault I'm sorry, that seems to be a contradiction?
You are saying that (a) strongly modular $\Leftrightarrow$ completely def / K when K is quadratic; (b) the Guitart--Quer example is a building block that is not strongly modular, and (c) building blocks are completely def / K. Since the Guitart-Quer example is over a quadratic field, (a), (b), (c) can't all be true at once.
@DavidLoeffler Right, sorry these things always get me confused. I think that for this particular curve, the field of complete definition is $\mathbb{Q}(\sqrt{-2}, \sqrt{-3})$. This is explained in Section 3 of https://arxiv.org/abs/math/0611663 One would have to write down the isogeny, I haven't done that...
(c) above looks like the culprit to me. In the 1-dimensional case, it's saying "elliptic $\mathbb{Q}$-curves are completely defined over $K$", which is not true in general
In any case, what I wrote about building blocks cannot be true, as David points out.
This answer is really helpful -- thanks! I was hoping for examples over real quadratic fields, but if there are examples over imaginary quadratic fields then probably there will be examples over real fields as well.
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2025-03-21T14:48:31.419585
| 2020-06-30T19:52:41 |
364537
|
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|
Stack Exchange
|
Polynomials of minimum degree that interpolate primes in intervals
Given an interval $[a,b]$ what is the minimum degree of univariate polynomials in $\mathbb Q[x]$ that passes through all primes between $a$ and $b$ (denoted by $\mathbb P[a,b]$ with total number of primes in $[a,b]$ given by $\pi(b-a)=|\mathbb P[a,b]|$)?
$\forall x,y\in\{1,\dots,\pi(b-a)\},f(x),f(y)\in\mathbb P[a,b]$ and $f(x)<f(y)\iff x<y$.
It is at most $O(\pi(b-a))$ and is there a reason to believe it cannot be $o(\pi(b-a))$?
What do you mean by saying a polynomial "passes through" a prime? What points are you interpolating?
Equivalently, you want to interpolate the points $(i, p_i)$, $i = m \ldots n$ where $p_i$ is the $i$'th prime.
The prime $k$-tuples conjecture implies that for each integer $k > 2$ and each $d$ from $1$ to $k-1$, there are infinitely many $m$ such that with $n=m+k$ the minimum degree of the interpolating polynomial is $d$.
The conjecture says that $(p_{m+1}-p_m, p_{m+2}-p_m, \ldots, p_{m+k} - p_m)$ is, for infinitely many $m$, any increasing $k$-tuple of even integers such that there is no prime $q$ for which this covers all residues mod $q$. In particular this implies that for a given
$k>2$, if $A$ is the product of all primes $\le k$ and $d \ge 1$ there are infinitely many $m$ such that
$p_j = p_m + A (j-m)^d$ for $m \le j \le m+k$. Thus in such a case, if $d \le k-1$ the minimum degree is $d$.
EDIT: see also OEIS sequence A335435.
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2025-03-21T14:48:31.419716
| 2020-06-30T21:18:17 |
364539
|
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|
Stack Exchange
|
When is the characteristic polynomial of the character table of a cyclic group irreducible?
Let $G=C_n$ be the cyclic group and $f_n$ the characteristic polynomial of its character table (over $\mathbb{C}$) in the ordering so that the character table is given by the discrete Fourier transform matrix (https://en.wikipedia.org/wiki/DFT_matrix) without the factor.
Question 1: For which $n$ is $f_n$ irreducible (over the smallest field extension of $\mathbb{Q}$ containing its coefficients)?
I did input this into GAP using the command IrrDixonSchneider to obtain the character table (I hope this is does not change much with respect to irreducibility) and tested for irreducibility with GAP. GAP seems to use another numbering for the character table, so lets call $g_n$ the characteristic polynomial according to GAP (maybe someone can clarify the ordering which GAP uses to get the character table of the cyclic group. I try to see what it is but Im not sure at the moment).
For $n \leq 23$ it was true that the $n$ such that $g_n$ is irreducible coincides with the sequence https://oeis.org/A280862 , which are those $n$ such that $a_n \psi_n = \varphi_n$, where $\varphi_n$ is the Euler phi function (https://oeis.org/A000010), $\psi_n$ is the reduced totient function (https://oeis.org/A002322) and $a_n$ is the greatest common divisor of all $(d-1)$'s, where the $d$'s are the positive divisors of $n$ (https://oeis.org/A258409).
Question 2: Is this true?
Question 3: Does being irreducible depend on the ordering used to obtain the character table here (or even for a general group)?
What is the meaning of the characteristic polynomial of the character table? There's no canonical ordering on the elements of $C_n$, nor its character; they are not in natural bijection, so there is no way to view the character table as naturally the matrix of an endomorphism of a vector space.
Isn't the matrix you are looking at more or less the DFT matrix?
@JoshuaMundinger Thanks, I clarified for question 1 that I indeed mean the DFT matrix. For question 2, I used GAP so Im not sure what the ordering is what GAP uses so I try to find that out now. Thanks for the comments.
@GeoffRobinson Can reordering change the property of being irreducible?
Well, you can think of the character table as the matrix $[\omega^{(i-1)(j-1)}]: 1 \leq i,j \leq n,$ where $\omega$ is a complex primitive $n$-th root of unity. Admittedly, reordering the rows and columns will lead to a different characteristic polynomial.
I'm not sure what the relationship between the characteristic polynomials is after permuting rows and columns.
@GeoffRobinson, in this case, choosing a different primitive $n$th root of unity $\omega$ just conjugates your matrix (right?), so it doesn't change the characteristic polynomial.
@LSpice: Yes, changing the root of unity that way and leaving the $(i,j)$-entry of the same form does not affect the characteristic polynomial. But I think that if you reorder the linear characters and the powers of a generator in different ways, you might in general replace the character table $C$ by a matrix of the form $PCQ^{T}$, where $P,Q$ are different permutation matrices. This replaces the characteristic polynomial det($C-xI$) by the polynomial $\pm$det($C - xP^{T}Q$))
@GeoffRobinson, agreed that it's not a general phenomenon; I just meant that, in the very special case of cyclic groups, there is a canonical family of compatible choices that define the matrix up to conjugacy.
@LSpice: Yes, I would agree with that.
I think a strong pointer to the answer (for the cyclic group of order $n>2 $, where the character table $C$ is chosen so that the $(i,j)$-entry is $\omega^{(i-1)(j-1)}$ for a fixed complex primitive $n$-th root of unity $\omega$) is as follows:
We have $C\overline{C}^{T} = nI,$ but (with this labelling) we note that $C$ is also symmetric. Hence we have $C\overline{C} = nI.$ As noted in my answer to MO363691 (and also previously noted by Denis Serre in his answer to his own question MO78050), we have $\overline{C} = PC$ where $P$ is a permutation matrix of order $2$ (the number of $2$-cycles in the associated permutation is $\frac{n-1}{2}$ if $n$ is odd, and $\frac{n-2}{2}$ if $n$ is even- this is the number of complex conjugate pairs of linear characters of the cyclic group of order $n$ which are not real-valued).
Now we have $CPC = nI$, so that $P = nC^{-2}$ and (since $P^{2} = I$), we have $C^{2} = nP$. Now the eigenvalues of $C$ are all in the set $\{ \sqrt{n},-\sqrt{n}, i\sqrt{ n}, -i\sqrt{n} \}.$ Since $P$ has the eigenvalue $-1$ with multiplicity $\lfloor \frac{n-1}{2}\rfloor$, we see that $\lfloor \frac{n-1}{2}\rfloor$ of the eigenvalues of $C$ are pure imaginary of absolute value $\sqrt{n}$ . If $n$ is odd, then $\frac{n+1}{2}$ of the eigenvalues of $C$ are real of absolute value $\sqrt{n}$, while if $n$ is even, $\frac{n+2}{2}$ of the eigenvalues are real of absolute value $\sqrt{n}$.
Furthermore, as in the answers to MO3639691 and MO78050 that det(C) is real if $n \equiv 1$ or $2$ (mod $4$), and det(C) is pure imaginary if $n \equiv 0$ or $3$ (mod $4$).
Let $F$ denote the field generated by the coefficients of the characteristic polynomial of $C.$ Then the minimum polynomial $p(x)$ of $C$ in $F[x]$ is certainly a divisor of $(x^{2}-n)(x^{2}+n)$ from our knowledge of the eigenvalues of $C$.
Hence any irreducible factor of $p(x)$ has degree $1$ or $2$. Since the characteristic polynomial of $C$ has degree $n >2,$ the characteristic polynomial of $C$ is definitely not irreducible in $F[x].$
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2025-03-21T14:48:31.420129
| 2020-06-30T21:25:44 |
364540
|
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|
Stack Exchange
|
Local freeness of $\pi_*F(r)$ from flatness of $F$
In 'Fundamental Algebraic Geometry' by Fantechi there is a lemma in section 5.3.2, page 119:
LEMMA 5.5 Let $S$ be a noetherian scheme and let $F$ be a coherent sheaf on $\mathbb{P}^n_S$. Suppose there exist some integer $N$ such that for all $r \ge N$ the direct image $\pi_*F(r)= \pi_*(F \otimes O_{\mathbb{P}^n_S}(r))$ is locally free. ($\pi$ is the structure morphism $\mathbb{P}^n_S \to S$).
Claim: Then $F$ is flat over $S$.
Now the author leaves as exercise to show that the converse(!) of the above lemma holds as well: if $F$ is flat over $S$ then $\pi_*F(r)$ is locally free for all sufficiently large $r$.
Does anybody know how the proof of the converse works or where I can find it?
About the choice of suitable $N$ I conjecture that Serre's vanishing theorem may play an important role in the proof:
Let $A$ a ring and $F$ coherent on $\mathbb{P}^n_A$. Then for all $i \ge 1$ there exist $m_0$ with $H^i(\mathbb{P}^n_A, F(m))$ for all $m \ge m_0$.
But that's just my suspucion, not more since I otherwise not know how a candidate for $N$ can be found.
See for example Hartshorne, Theorem III.9.9. The idea is to use a Čech resolution and Serre vanishing.
@R.vanDobbendeBruyn: You mean that one which starts with "Let $T$ be an integral noetherian scheme. Let $X \subset \mathbb{P}^n_T$ be a
closed subscheme... And the claim is that $X$ is flat over $T$ iff Hilbert polynomial $P_t$ is independent of $t$. Here is required that $T$ is integral, especially reduced, the book seems to work without it but thats subtile.
Another more important problem is how can we obtain a closed subscheme $X \subset \mathbb{P}^n_T$ in the thereom from coherent sheaf $F$ on $\mathbb{P}^n_S$? $F$ induces a relative spectrum (a candidate for closed subscheme $X$ in the theorem) $X_F:=\operatorname{Spec}_S(F) \to S$ in well known way but is $X_F \subset \mathbb{P}^n_T$?
$F$ is by definition only a $O_{\mathbb{P}^n_S}$-module not a $O_{\mathbb{P}^n_S}$-algebra. that is a priori there is no map $O_{\mathbb{P}^n_S}=O_S[x_0,...x_n] \to F$ corresponding to $X \subset \mathbb{P}^n_T$. Do you know how relate $X$ and $F$?
This can be deduced from semicontinuity theorem . For sufficiently large $N$, higher direct images of $F(N)$ vanish. Then the direct image is locally free.
You're right, the reference doesn't cover the precise statement, but as @Mohan points out Thm III.12.11 does (together with Thm III.8.8).
I also think the (very short) proof of Thm III.9.9 (i)$\Rightarrow$(ii) applies verbatim in this context, because that direction uses neither that $\mathscr F$ is the structure sheaf nor that $T$ is integral. In fact the first is not even stated as an assumption, and the second is only used for (iii)$\Rightarrow$(ii).
yes of course you are right, the proof in 9.9 solves the problem. sorry, as I wrote my last comment I didn't took a glance into theorem's proof of 9.9, only into it's statement.
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2025-03-21T14:48:31.420379
| 2020-06-30T21:41:20 |
364542
|
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"Andreas Weingartner",
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|
Stack Exchange
|
On the density map of the abundancy index
Let $σ$ be the sum-of-divisors function. Let $σ(n)/n$ be the abundancy index of $n$. Consider the density map $$f(x) = \lim_{N \to \infty} f_N(x) \ \ \text{ with } \ \ f_N(x) = \frac{1}{N} \#\{ 1 \le n \le N \ | \ \frac{\sigma(n)}{n} < x \}. $$
In this paper, Deléglise mentioned that Davenport proved that $f$ is continuous, and proved that $0.752 < f(2) < 0.7526$ (bounds improved by Kobayashi in his PhD thesis).
Let $\alpha = f^{-1}(1/2)$ be the median abundancy index, i.e. the number $\alpha$ such that the integers of abundancy index greater than $\alpha$ have natural density exactly $1/2$.
$$\begin{array}{c|c}
N & f_N^{-1}(1/2) \newline \hline
1 &1.00000000000000 \newline \hline
10 &1.50000000000000 \newline \hline
10^2 &1.54838709677419 \newline \hline
10^3 &1.51485148514851 \newline \hline
10^4 &1.52707249923524 \newline \hline
10^5 &1.52501827363944 \newline \hline
10^6 &1.52384533012867 \newline \hline
10^7 &1.52381552194973 \newline \hline
10^8 &1.52381084043829
\end{array}$$
The above table suggests that $\alpha \simeq 1.52381$.
Question 1: What is known about the median abundancy index? Is it even mentioned somewhere? It is true that $|\alpha-1.52381|<10^{-5}$?
Let $(b_n)_{n \ge 1}$ be the sequence of integers such that for all $k<b_n$ then $$|\sigma(k)/k - \alpha| > |\sigma(b_n)/b_n - \alpha|.$$ This is the lexicographically first sequence of integers whose adundancy index strictly converge to the median adundancy index. Let us call this sequence the buddhist sequence in reference to the Middle Way in buddhism philosophy. Assuming that $|\alpha-1.52381|<10^{-5}$, here are the first terms of this sequence together with the distance of their adundancy index from $1.52381$: $$ \begin{array}{c|c}
n & b_n & |\sigma(b_n)/b_n -1.52381| \newline \hline
1 & 1 & 0.52381000000000 \newline \hline
2 & 2 & 0.02381000000000\newline \hline
3 & 21& 0.00000047619048\newline \hline
4? & 22099389? & 0.0000002693327?
\end{array} $$
Observe that $b_3=21$, $\sigma(21)/21 = 32/21$ and $|32/21-1.52381|<10^{-6}$, which is statistically unexpectable, as shown if we consider the variation $(b'_n)$ taking $22$ as initial term:
$$ \begin{array}{c|c}
n & b'_n & |\sigma(b'_n)/b'_n -1.52381| \newline \hline
1&22& 0.112553636363636 \newline \hline
3&26& 0.0915746153846153 \newline \hline
4&27& 0.0423285185185187 \newline \hline
5&46& 0.0414073913043478 \newline \hline
6&58& 0.0279141379310344 \newline \hline
7&62& 0.0245770967741934 \newline \hline
8&74& 0.0167305405405405 \newline \hline
9&82& 0.0127753658536585 \newline \hline
10&86& 0.0110737209302325 \newline \hline
11&94& 0.00810489361702116 \newline \hline
12&106& 0.00449188679245283 \newline \hline
13&118& 0.00161372881355915 \newline \hline
14&122& 0.000780163934426037 \newline \hline
15&3249& 0.000659067405355485 \newline \hline
16&14337& 0.000478759154634911
\end{array} $$
So there is a very good chance that $\alpha = 32/21$. If so the buddhist sequence ends with its third term and $b_3=21$ should be called the Buddha number. If not, then we know that the set of abundancy indices is dense, so that the buddhist sequence must have a next term $b_4$, but $\sigma(b_3)/b_3$ is already too close to $\alpha$ compared to its above conjectured approximation, so we cannot conjecture the next term. A possible candidate for $b_4$ is mentioned in above table.
Question 2: Does the buddhist sequence end with its third term? If not what are the next terms?
Below are some additional computations with 10 samples of 100001 random integers between $10^{20}$ and $10^{21}$ suggesting that $\alpha = 32/21$ should be correct (sage lists are numbered from 0).
sage: import random
sage: for t in range(10):
....: L=[]
....: for i in range(100001):
....: b=random.randint(10**20,10**21)
....: q=sum(divisors(b))/b
....: L.append(q)
....: L.sort()
....: print((32/21-L[50000]).n())
-2.01727393333164e-8
0.00244355476044226
0.00201824866273585
-0.00130445314014877
-0.000322772616778371
0.00102756546533326
-6.74774915307343e-10
-1.48849650772673e-19
-0.0000572173485145812
-6.52303473965081e-20
Observation: One sample provides a median close to 32/21 with 20 digits, one with 19 digits, one with 10, one with 8, one with 5, one with 4 and four with 3.
How to explain such statistical irregularities?
A number with abundancy index greater (resp. less) than $2$ is called an abundant (resp. deficient) number, because the sum of its proper divisors (or aliquot sum) exceeds (resp. subceeds) itself. In the same flavour, a number with abundancy index greater (resp. less) than the median abundancy index $\alpha$ could be called an advantaged (resp. disadvantaged) number.
There is a Collatz-like problem (called Calatan-Dickson conjecture) related to the aliquot sum $s$ asking whether all the aliquot sequences $(s^{\circ r}(n))_{r \ge 0}$ are bounded. A value of $\alpha-1 \simeq 0.52381$ suggests heuristically a positive answer to this problem because $\alpha-1$ is the median for $s(n)/n$, although there are serious counter-example candidates like $n=276$ as $s^{\circ 100}(276)>10^{19}$. There are five such candidates less than $1000$ called the Lehmer Five (see this webpage dedicated to recent advances on the aliquot sequence).
The following picture displays $f_N$ for $N=10^7$ (which should be a good approximation of $f$, according to above table).
Observe that the function $f$ seems to make a jump around $\alpha$, whereas it is continuous, so it should be non-differentiable there; moreover the phenomenon happens around many other points (with a Cantor set or fractal flavour), which leads to:
Question 3: Is $f$ a Weierstrass function? What is the meaning of these jumps?
It may be worth having a look at Andreas Weingartner, The distribution functions of $\sigma(n)/n$ and $n/\phi(n)$, II, Journal of Number Theory 132 (2012), 2907-2921, and part one, Proceedings of the American Mathematical Society, 135 (2007), 2677-2681.
@GerryMyerson: Thanks for pointed out these references about the function $f$ (or more precisely $A=1-f$). I went through them, but unfortunately, none answer (or even mention) the specific questions of this post.
You might try writing to the author to see what he knows about your questions.
@GerryMyerson: Yes I did that just after your first comment. Moreover the author is a MathOverflow user: Andreas Weingartner.
My papers are about the size of $1-f(x)$ as $x\to \infty$. For your problem, you need to approximate $f(x)$ for $1<x<2$. Besides the paper by Deleglise, there is a more recent paper by Kobayashi (A new series for the density of abundant numbers), that shows how to get upper and lower bounds for $f(x)$ for a fixed value of $x$.
@AndreasWeingartner: Good! Then Kobayashi method should be applied to $x=32/21$ to check whether $f(32/21) = 1/2$, or at least to find good bounds for that.
For question 1:
Just for fun I calculated bounds on $\alpha$ in 2018, but have not published them. By using the generalized Deleglise method from my thesis, we find
$$ 1.523812 < \alpha < 1.5238175, $$
so $\alpha\neq 32/21$.
This was found by calculating the density bounds for equally spaced $x$, then narrowing in when we bracket density $1/2$. Here are some relevant bounds:
$$ 0.50003297 \leq f(1.523812) \leq 0.50018578 $$
$$ 0.4999934 \leq f(1.523813) \leq 0.5001300 $$
$$ 0.49995299 \leq f(1.523814) \leq 0.5000895 $$
$$ 0.49991554 \leq f(1.523815) \leq 0.5000560 $$
$$ 0.4998909 \leq f(1.523816) \leq 0.500012223 $$
$$ 0.49986562 \leq f(1.523817) \leq 0.500001975 $$
$$ 0.49985411 \leq f(1.5238175) \leq 0.49981476 $$
For question 3:
The function $f$ is known to be singular, that is, continuous, non-constant, and differentiable almost everywhere with derivative zero. Thus, it is not a Weierstrass function.
As for the meaning of the jumps, one way of thinking of these is in terms of the series described in my paper "A new series for the density of abundant numbers." The series for $f(x)$ suddenly gains large terms at certain values of $x$, causing the jumps.
Bibliographic details for the cited paper: International Journal of Number Theory, Vol. 10, No. 01, pp. 73-84 (2014). https://doi.org/10.1142/S1793042113500814
According to these bounds, here are the first terms of the the buddhist sequence: $$b_1=1, \ b_2=2, \ b_3=21, \ b_4 \in {142174, 142442, 143246 }, \ b_5=?$$
|
2025-03-21T14:48:31.420853
| 2020-06-30T22:05:17 |
364543
|
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|
Stack Exchange
|
Can a generic rope exert the same force as a rope with loops?
This is a spin-off of my previous question. I will take a moment to reintroduce the notation. The problem concerns elastic "ropes" in $\mathbb{R}^3$, which are modeled as sequences of points $\gamma=(x_1,x_2,\dots,x_m)$. A rope is supported on the union of line segments
$$
S(\gamma) := \bigcup_{j=1}^{m-1} \overline{x_jx_{j+1}} \subset\mathbb{R}^3.
$$
The $j$-th segment of the rope has direction
$$
\tau_j := \frac{x_{j+1} - x_j}{|x_{j+1}-x_j|},
$$
and the force $F_\gamma$ associated to the rope $\gamma$ is described by the vector-valued measure
$$
F_\gamma := \sum_{j=1}^{m-1} \tau_j (\delta_{x_{j+1}} - \delta_{x_j}).
$$
This time I am curious about "generic" ropes $\gamma$. We say that $\gamma$ is generic if the points $\{x_1,x_2,\dots,x_m\}$ are in general position. That is, the points are all distinct, no three are collinear, and no four are coplanar. In particular, no two of the line segments $\overline{x_jx_{j+1}}$ and $\overline{x_kx_{k+1}}$ intersect except when $j=k+1$ or vice versa (so there are no loops in $S_\gamma$).
Question: Suppose that $\gamma$ and $\gamma'$ are two ropes with the same force, so
that $F_\gamma = F_{\gamma'}$. If $\gamma$ is generic, does it follow that
$\gamma'$ is also generic?
What I believe I can show is that, if $\gamma'$ is not generic, it must have at least three loops. I would also be curious if there was a reasonable strengthening of the "generic rope" condition that satisfied this conjecture. So what I am really after is something like the question below.
Question' Is there a set $S$ of "good ropes" that can be described explicitly with the following properties:
For two ropes $\gamma,\gamma'$ with $\gamma\in S$ and $F_\gamma=F_{\gamma'}$, $\gamma'$ is guaranteed not to have any self-intersections (that is, $S(\gamma')$ has no loops).
The set $S$ is "generic" or at least "dense".
For example, I am happy to consider only ropes that do not have any corners with angle $2\pi/3$, if that were to help at all.
|
2025-03-21T14:48:31.421011
| 2020-07-02T18:58:12 |
364687
|
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|
Stack Exchange
|
Vector-valued interpolation for sublinear operators
Grafakos in his $\textit{Classical Fourier Analysis}$ formulates (see Exercise 4.5.2 therein) the following vector-valued version of the Riesz-Thorin interpolation theorem.
$\textbf{Theorem}$
Let $1\le p_0, q_0,p_1,q_1, r_0, s_0, r_1, s_1\le\infty$ and $\theta\in(0,1)$ satisfy
\begin{align*}
\frac{1-\theta}{p_0}+\frac{\theta}{p_1}&=\frac{1}{p},\qquad \frac{1-\theta}{q_0}+\frac{\theta}{q_1}=\frac{1}{q},\\
\frac{1-\theta}{r_0}+\frac{\theta}{r_1}&=\frac{1}{r},\qquad \frac{1-\theta}{s_0}+\frac{\theta}{s_1}=\frac{1}{s},
\end{align*}
and let $T$ be a linear operator mapping $L^{p_0}(\mathbb{R}^n, \ell^{r_0})$ to $L^{q_0}(\mathbb{R}^n, \ell^{s_0})$ and $L^{p_1}(\mathbb{R}^n, \ell^{r_1})$ to $L^{q_1}(\mathbb{R}^n, \ell^{s_1})$.
Then $T$ maps $L^{p}(\mathbb{R}^n, \ell^{r})$ to $L^{q}(\mathbb{R}^n, \ell^{s})$.
My $\textbf{question}$ is whether one can weaken the assumption that $T$ is linear? More precisely, does an analogous result hold for an operator of the form
$$
T: \{f_j\}_{j\in\mathbb{N}}\rightarrow\{M f_j\}_{j\in\mathbb{N}},
$$
where $M$ is a sublinear operator i.e. $|M(f+g)(x)|\le |Mf(x)|+|Mg(x)|,\quad x\in\mathbb{R}^d$?
In the scalar-valued there is an interpolation theorem for sublinear operators and it goes by the name of Marcinkiewicz-Zygmund.
I would appreciate any hints or perhaps a reference to suitable literature.
How do you define sublinearity in the vector-valued case?
In general let's say $|T(f+g)|{\ell^s}\le|Tf|{\ell^s}+|Tg|{\ell^s}$, $s\in[s_0,s_1]$.
In the particular application I have in mind $T$ is of the form $T({f_j}{j\in\mathbb{N}})={Mf_j}_{j\in\mathbb{N}}$, for some sublinear operator $M$. Thanks for a good question!
Is your first condition sufficient in the scalar case? It looks as a very weak requirement compared to sublinearity.
I'm not sure if I understand what you mean. In the scalar case the norm $|\cdot|_{\ell^s}$ is replaced by the standard absolute value and then this condition becomes exactly the sublinearity condition for $T$.
|
2025-03-21T14:48:31.421159
| 2020-07-02T19:40:51 |
364689
|
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|
Stack Exchange
|
Characters, centralizers and cosets
I am trying to understand/count the number of solutions of a number of equations in finite groups and came across the following class function:
$$ \theta_\chi(x) = \sum_{y\in G} |C_G(xy)y \cap C_G(x)| \chi([y,x]) $$
where $\chi$ is an irreducible character of $G$.
Calculating its values seems to be a delicate task as one has to understand very well the centralizers of $G$, their cosets and their intersections.
**Two questions:
Has any one seen/studied this formula or a similar one?
Any idea/suggestion about how to simplify this expression?**
The function $\theta_\chi$ can be used to calculate the coefficients of the class function
$$f_3(g) = |\{(x,y,z)\in G^3 : [x,y]=[x,z]=[y,z]=g \}|$$
as a linear combination of the irreducible character of $G$. More precisely, if we let
$$ m_\chi = \sum_{x\in G} \theta_\chi(x) $$
then
$$f_3(g) = \sum_{\chi \in Irr(G)} \frac{m_\chi}{|G|}\chi(g) $$.
Is $C_G$ the centraliser or the conjugacy class?
Also, could you say anything further about how this arose? It looks like an interesting expression, but I can't guess how one would arrive at it.
@LSpice: In this case, $C_{G}(x)$ means the centralizer of $x$.
@LSpice Just added more info about the context.
|
2025-03-21T14:48:31.421281
| 2020-07-02T19:53:22 |
364691
|
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|
Stack Exchange
|
Structural properties of polytopes for mainstream integer or linear programs
Are there any papers/textbooks/monographs that describe distinguishing properties of the polytopes that arise when solving the linear relaxation of well-known integer programs? For example, it is well-known that the feasible polytope for the vertex cover problem has the half-integrality property that all corners are $0$, $1/2$, or $1$. Are there any surveys that describe how the polytopes for e.g. set cover, traveling salesman, facility location, Steiner tree, are different from one another?
(I realize of course that there could be many different polytopes depending on what formulation one chooses, but I hope this gets my idea across)
http://or.stackexchange.com
|
2025-03-21T14:48:31.421491
| 2020-07-02T21:06:32 |
364695
|
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|
Stack Exchange
|
An analogue of Noether's Problem for non-rational varieties
For the sake of simplicity, let $\mathsf{k}$ be algebraically closed and of zero characteristic. Varieties are irreducible.
The (linear) Noether's Problem (which goes back to the early 1910's in Burnside's and Noether's work on invariants) is the following: let $V$ be a finite dimensional vector space, and $G \subset GL(V)$ a finite group. When is the variety $V/G$ rational?.
Now this problem is intrinsically very nice and has a miriad of applications to moduli questions (cf. C. Böhning https://arxiv.org/abs/0904.0899; J-L. Colliot-Thélène, J-J. Sansuc, The rationality problem for fields of invariants under linear algebraic groups MR2348904 );
PI-algebras (E. Formanek, The polynomial identities and invariants of n×n matrices, MR1088481); the inverse Galois problem (D. Saltman, Groups acting on fields: Noether’s problem, MR810657); etc.
It is also interesting that the group actions that give a positive solution to Noether's Problem are very rich in nature.
Some questions closely related to this Problem have arisen in my research, and I wonder if they are adressed in some paper, or anyone knows something about then. I thought about these questions, but I could only think of examples coming from elliptic curves (such as multiplication by $[n]$ map), but these are, to my taste, a little bit too 'simple'.
(Q1): Let $X$ be a smooth projective curve of genus at least 1. What are the finite groups $G$ of $Aut_\mathsf{k} \, X$ such that $X/G$ is a smooth projective curves of the same genus?
(Q2): Let $X$ be a non-rational variety and $G$ a finite group of $Aut_\mathsf{k} \, X$. When $X/G$ is birationally equivalent to $X$ itself?
Edit Both questions were already answered in the comments. The questions were not difficult in the end.
(Q1): If $X$ is a smooth projective curve, and $G$ a finite group acting on it, then $X/G$ is smooth projective curve. Using Riemann-Hurwitz formula one shows that Q1 can only hold when $X$ has genus $0$ (always) or genus $1$.
For Q1, doesn't Riemann-Hurwitz imply that if genus is $>1$, the group has to be trivial? For the second, isn't it possible to take $X=Y\times\mathbb{P}^1$ for a non-stably rational $Y$ and take any finite subgroup of automorphisms of $\mathbb{P}^1$, acting on the second factor?
Thank you both. You properly answer my questions.
Q2 maybe becomes more interesting if you ask for X general type, rather than X non-rational... it's not immediately clear to me whether there should be an easy counterexample.
@dhy to my understanding the first counter-example for rational $X$ (the affine space in particular) is due to Saltman in 1984 or 1985
@jg1896 Sorry, by counter-example here I actually mean example where $X$ is birational to $X/G$. The reason is that in the general type case for most $X$ and $G$ it will be possible to distinguish $X$ and $X/G$ via natural invariants, e.g., the $h^{0,i}$.
@dhy Don’t general type varieties only have finitely many rat’l self-maps? It seems to me a general type example is thus impossible (bc iterating a self-map of positive degree would give infinitely many self-maps).
@DanielLitt good point!
|
2025-03-21T14:48:31.421719
| 2020-07-02T21:16:17 |
364696
|
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|
Stack Exchange
|
Embedded surfaces in pseudo-Anosov mapping tori
Given a genus $g$ (where $g \geq 2$) closed surface $S$, and a pseudo-Anosov map $\varphi: S \to S$, one can construct the mapping torus, which turns out to be a hyperbolic $3$-manifold, which we'll call $M$. Inside $M$, one way to construct other closed embedded surfaces is via the following construction:
Take a (multi-)curve $\gamma$ in $S$ such that $\varphi(\gamma)$ is disjoint from $\gamma$.
(Post comment edit) The curves $\gamma$ and $\varphi(\gamma)$ bound a subsurface $S'$ with two boundary components, namely $\gamma$ and $\varphi(\gamma)$.
Take the union (in $M$) of $S'$ (where we think of $S'$ as the subsurface of the fiber over the point $0 \in S^1$) and $f_{[0,1]}(\gamma)$, where $f$ is the suspension flow of $\varphi$.
Questions:
Is every embedded surface in $M$ homologous to a surface obtained via the above construction?
If the answer to the above question is no, does the answer become yes if we are allowed to change the monodromy map $\varphi$ and the fiber $S$, while keeping the total space $M$ fixed?
P.S. I'm new to $3$-manifold topology, and thus am not sure if the construction I described has a well-known name already. If it does, someone could perhaps point that out in the comments, and I can edit this question accordingly.
EDIT: I realize that it might not be possible to realize surfaces of very high genus via this construction, since there might not be enough room on $S$, so maybe the question should be what are the surfaces that can be realized this way.
That union, which I'll call $K$, is not a surface. Up to homeomorphism, $K$ is the quotient of the union of $S$ with $S^1 \times[0,1]$ where $S^1 \times 0$ is identified with $\gamma$ and $S^1 \times 1$ is identified with $\phi(\gamma)$. The topological space $K$ is not a 2-manifold near any point $x \in \gamma \cup \phi(\gamma)$, because the relative homology group $H_2(K,K-x;\mathbb Z)$ is isomorphic to $\mathbb Z^2$, not to $\mathbb Z$ as is required to be a 2-manifold. Intuitively, instead of just 2 surface sheets meeting near $x$, there are 3 of them.
Yes, of course. I forgot to write down a step in the construction where I look at a subsurface $S'$ whose two boundary components are $\gamma$ and $\varphi(\gamma)$. That leads to the result space being a manifold with two boundary components, with a cylinder glued together to the two boundary components, which definitely is a surface
The surfaces you describe here are sometimes called "cut and cross-join surfaces". I believe they were first introduced in: Cooper, Long & Reid, Bundles and finite foliations. Invent. Math. 118 (1994), no. 2, 255–283, and further developed in Masters, Thick surfaces in hyperbolic 3-manifolds. Geom. Dedicata 119 (2006), 17–33.
|
2025-03-21T14:48:31.421933
| 2020-07-02T22:07:54 |
364699
|
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|
Stack Exchange
|
Oddness of intersection form of surface bundle
Let $\Sigma_g$ be a Riemannian surface of genus $g$. Let $M^4$ be a surface bundle: $\Sigma_g \to M^4 \to \Sigma_h$. When $g=1$, $M^4$ is called a torus bundle.
My question: is there a torus bundle whose intersection form contains an odd diagonal element (if we choose a basis and view the intersection form as a matrix)?
If $M^4=\Sigma_1\times \Sigma_h$, then $M^4$ is spin and its intersection form has only even diagonal elements.
More generally:
For a given fiber $\Sigma_g$, is there a $\Sigma_g$-bundle whose intersection form is odd?
what is the meaning of "odd diagonal element"?
We choose a basis and view the intersection form as a matrix. I modified my question. What is E(1) 4-manifold?
$E(1)= CP^2 # 9 \bar{CP^2}$ which is a torus bundle as you desired. For more details please have a look at the chapter 3 of Gompf and Stipsicz book. In paticular you will be interested in elliptic surfaces.
@AnubhavMukherjee: It seems to me that the elliptic fibration you are talking about has (generically) 12 nodal fibres, so it is not a torus bundle.
In fact, its Euler characteristic is $12$, whereas for a torus bundle it is $0$.
@FrancescoPolizzi maybe you are right. I was thinking about singular fibration though.
Suppose $\pi\colon M \to \Sigma_g$ is an oriented smooth torus bundle. If $w_2(M) = 0$, then also the second Wu class $v_2(M) = 0$ and $M$ has even intersection form (the converse holds if $H_1(M;\mathbb Z)$ has no $2$-torsion, but we do not need this here). I claim that this is always the case in our situation.
Even better, I claim that $M$ is always parallelizable: stably, $TM$ agrees with the vertical tangent bundle $T_{\pi}$, whose classifying map $E \to B\text{SL}_2(\mathbb R)$ can be identified with the map
$$E \xrightarrow{\pi} \Sigma_g \xrightarrow{(1)} B\text{Diff}^+(T^2) \xrightarrow{(2)} B\text{SL}_2(\mathbb Z) \xrightarrow{(3)} B\text{SL}_2(\mathbb R),$$
where (1) is the classifying map of $\pi$, (2) is induced from applying $\pi_1$, and (3) is induced from extending coefficients. Since $H^2(B\text{SL}_2(\mathbb Z);\mathbb Z) = \mathbb Z/12$ is torsion, the map composition of (1), (2) and (3) is trivial on second cohomology and hence nullhomotopic, as $B\text{SL}_2(\mathbb R) = K(\mathbb Z,2)$. Thus, $T_{\pi}$ is trivial and $M$ is stably parallelizable. Since $\chi(M) = \chi(\Sigma_g)\chi(T^2) = 0$, $M$ is parallelizable.
If the base is not a surface, I believe that it is possible for torus bundles to be non-spin, see Johannes Ebert's thesis (the last pages of chapter 5), although no concrete examples are constructed there.
For higher genus, note that there are examples of surface bundles over surfaces whose total space has signature $4$, in particular, its intersection form cannot possibly be even.
Also, the total space of the (unique!) nontrivial $S^2$-bundle over $S^2$ is diffeomorphic to $\mathbb CP^2 \# \overline{\mathbb CP^2}$, which has odd intersection form.
Thank you for the answer. I just like to confirm what you said: All torus bundle with base $\Sigma_g$ are spin and have even intersection form.
For surface bundles over surfaces whose total space has signature 4, the total space is not spin. But can we conclude that the intersection form be odd?
Yes, I think the above argument shows that the total space of every torus bundle with base a surface is spin (even parallelizable), and therefore its intersection form is even.
The signature of any even intersection form over Z is divisible by 8.
You mentioned that ``the total space of the (unique!) nontrivial $S^2$-bundle over $S^2$ is diffeomorphic to $CP^2#CP^2$, which has odd intersection form.'' ie there exists a 2-cocycle $c \in H^4(CP^2#CP^2;Z)$ such that $\int_{CP^2#CP^2} c^2 =k$ and $k$ is odd. I have two questions here: (1) is $k=\pm 1$? (2) what is the evaluation of the 2-cocycle $c$ on the fiber $S^2_{fiber}$: $\int_{S^2_{fiber}} c =?$.
|
2025-03-21T14:48:31.422203
| 2020-07-03T00:33:57 |
364701
|
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"authors": [
"Aleksandar Milivojević",
"Jens Reinhold",
"John Greenwood",
"Tim Campion",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Realizing Stiefel-Whitney classes via vector bundles
Let $X$ be a CW complex. If $E$ is a vector bundle over $X$, then it's well-known that the Stiefel-Whitney classes $w_j(E) \in H^j(X,\mathbb F_2)$ of $E$ are determined from the classes $w_{2^k}(E)$ (for $2^k \leq j$) via the Wu formula, using the cup product and the action of the Steenrod algebra.
Question 1: Does the Wu formula imply any further relations? This is a purely algebraic question which I make more precise in (a) and (b) below.
That is, let $H$ be a nonnegatively-graded $\mathbb F_2$-algebra with an unstable action of the Steenrod algebra satisfying the Cartan formula and $Sq^{|x|}(x) = x^2$ for all homogenenous $x \in H$. Let $W$ be the set of sequences $(w_j \in H^j)_{j \in \mathbb N}$ with $w_0 = 1$ and satisfying the Wu formla.
(a) For any sequence $(v_{2^k} \in H^{2^k})_{k \in \mathbb N}$, does there exist $w \in W$ (necessarily unique) with $w_{2^k} = v_{2^k}$ for all $k \in \mathbb N$?
Presumably (i) the Whitney sum formula and (ii) the universal formula for the Stiefel-Whitney classes of a tensor product of vector bundles are compatible with the Wu formula, so that $W$ is a commutative ring using (i) for addition and (ii) for multiplication.
(b) Is $W$ a polynomial algebra on whichever generators from (a) do exist?
Question 2: What restrictions beyond the Wu formula are there restricting the Stiefel-Whitney classes of a vector bundle $E$ on a CW complex $X$? This is a genuinely topological question.
Over here Mark Grant describes one such restriction, but ideally I'd like a more systematic discussion.
If it simplifies matters to assume that $X$ is finite, or even a compact manifold, then that's fine.
I once asked a related question: https://mathoverflow.net/questions/239482/vector-bundles-with-exactly-one-nonzero-sw-class
Isn't Question 2 answered by the integral cohomology of $BO(n)$?
@JohnGreenwood I'm not sure what you mean? I agree that the integral cohomology of $BO(n)$ will pull back to "integral characteristic classes" which probably contain some information that the Stiefel-Whitney classes don't. Do you have something more specific in in mind?
Answers to https://mathoverflow.net/q/257617/41291 might (or might not) contain relevant information
Here is a start at answering Question 1: there are indeed further relations between the $w_{2^k}$, or at least conditions on the $w_{2^k}$.
For example, consider the case where $H$ has multiplication which is null except for what is implied by the multiplication being unital. (For example, $H$ may be the cohomology of a suspension space.)
In this case, the Wu formula reduces to
$$Sq^i(w_j) = \binom{j-1}{i} w_{i+j}$$
So if the $w_{2^k}$'s are given, we are forced to define $w_{2^k + j'} = Sq^{j'} w_{2^k}$ for $0 \leq j' \leq 2^k-1$, which gives us the definition of each $w_j$. So now in the case where $j = 2^k+1$ and $1 \leq i \leq 2^k - 1$, the Wu formula stipulates that $Sq^i Sq^1 w_{2^k} = 0$. This is always the case for $i = 1$, but for all other $i$, the relation $Sq^i Sq^1 = 0$ does not hold in the Steenrod algebra, so I believe there are examples of $H$'s and $w_{2^k} \in H^{2^k}$ where this equation does not hold. So this is an example of some kind of further condition which $w_{2^k}$ may be required by the Wu formula to satisfy.
This answer reflects a fundamental confusion on my part: The Wu formula is a formula satisfied by the Stiefel-Whitney classes of the tangent bundle of a smooth manifold. Presumably it is not satisifed by an arbitrary vector bundle. So the case considered here, where the multiplication is trivial, is not terribly relevant.
The Wu formula does hold for arbitrary vector bundles. I am having trouble locating Wu's original paper at the moment, but see for example May's Concise p.197. The Wu formula describes the action of the Steenrod algebra on the mod 2 cohomology of BO (generated as a Z/2-algebra by the universal Stiefel-Whitney classes)
What does hold only for closed manifolds, and also goes under the name "Wu's formula", at least in Milnor-Stasheff, is the formula Sq(total Wu class) = total Stiefel-Whitney class. This restriction to closed manifolds is to be expected, since the Wu classes use the (mod 2) Poincare duality structure in their definition. However, the "Wu's formula" linked to in the question is the one concerning the Steenrod algebra action on H*(BO;Z/2).
@AleksandarMilivojevic Ah, thanks, that's helpful!
|
2025-03-21T14:48:31.422509
| 2020-07-03T02:44:18 |
364706
|
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|
Stack Exchange
|
Is the cohomology of Hilbert modular surfaces spanned by special cycles?
We consider the Hilbert modular surface $X$ that parametrize abelian surface with real multiplication by $\mathcal{O}_F$ and $\mathfrak{a}$-polarization, where $F$ is a real quadratic field with discriminant $D$.( see section 2 of this paper). On the surface we can define special cycles $T(Dr)$, which parametrize abelian surfaces with special endomorphism $s$ such that $s \circ s=rNm(\mathfrak{a})\mathrm{Id}$.
Now the question is that do we have a nice way to characterize the subspace of cohomology of $X$ spanned by the special cycles? In particular, is the whole cohomology group generated by special cycles? If not, can we describe the complement?
And how is this related to Zucker's conjecture?
If the cohomology was spanned by cycles, then the etale cohomology (as a Galois representation) would be built from twists of Artin representations. But there are definitely examples where this cannot happen. For instance, if there is an elliptic curve $E/F$ with good reduction everywhere, then the twisted tensor induction of $V_p E$ from $F$ to $\mathbf{Q}$ appears in $H^2(X,\mathbf{Q}_p)$.
|
2025-03-21T14:48:31.422620
| 2020-07-03T02:47:27 |
364707
|
{
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"authors": [
"Pietro Majer",
"Quarto Bendir",
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"url": "https://mathoverflow.net/questions/364707"
}
|
Stack Exchange
|
Federer's questions on the mass and comass norms
In Federer's book "Geometric measure theory", he says in section 1.8.3 (where $\|\cdot\|$ is the comass norm):
Very little appears to be known about the structure of the convex sets $\wedge^m(\mathbb{R}^n)^\ast\cap\{\phi:\|\phi\|\leq 1\}$. What are their extreme points?
In section 1.8.4 he says:
Suppose $S$ and $T$ are mutually orthogonal subspaces of an inner product space $V$, $s:S\to V$ and $t:T\to V$ are the inclusion maps, $\xi\in\operatorname{im}\wedge_ps$ and $\eta\in\operatorname{im}\wedge_qt$. The equation $\|\xi\wedge\eta\|=\|\xi\|\cdot\|\eta\|$ holds if either $\xi$ or $\eta$ is simple. [...] I do not know whether the above equation holds in case neither $\xi$ nor $\eta$ is simple.
Are these matters understood by now?
edit 1, for completeness: given a finite-dimensional real inner product space $V$, the comass of $\phi\in\wedge^mV^\ast$ is defined by
$$\|\phi\|=\sup\Big\{\phi(\xi):\text{decomposable }\xi\in \wedge^mV\text{ with }|\xi|\leq 1\Big\}$$
and the mass of $\xi\in\wedge^mV$ is defined by
$$\|\xi\|=\sup\Big\{\phi(\xi):\phi\in\wedge^mV^\ast\text{ with }\|\phi\|\leq 1\Big\}.$$
Here $|\cdot|$ is the standard norm on $\wedge^mV$, defined by an orthonormal basis $e_{i_1}\wedge\cdots\wedge e_{i_k}$. It seems to me that the mass and comass are essentially analogous to (and perhaps the inspiration for) the Gromov norm on homology.
edit 2 : As pointed out in the answers to the question Pietro Majer links to, the question of 1.8.4 is addressed in Frank Morgan's article "The exterior algebra $\Lambda^k({\mathbb R}^n)$ and area minimization" (Linear Algebra and its Applications Volume 66, April 1985, Pages 1–28) where certain cases (by dimension) are proved.
Some references to recent results on Federer's problems are listed here : https://mathoverflow.net/questions/176544/open-problems-in-federers-geometric-measure-theory?rq=1
Thanks, I'd seen that when posting but hadn't realized that the second question there is equivalent to the second question here
I have to admit it's hard for me to understand, from a high-level perspective, why the comass is hard to understand. As I understand it, it's just the minimization of a linear function with a quadratic constraint. It feels like it should be more tractable!
|
2025-03-21T14:48:31.422807
| 2020-07-03T04:14:59 |
364712
|
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"Gjergji Zaimi",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364712"
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|
Stack Exchange
|
Bound on the chromatic number of square of bipartite graphs
In continuation of the previous question, what is a strict upper bound on the chromatic number of the square of a bipartite graph?
I think the chromatic number number of the square of the bipartite graph with maximum degree $\Delta=2$ and a cycle is at most $4$ and with $\Delta\ge3$ is at most $\Delta+1$. This is because the edge set of a connected bipartite graph consists of the edges of a union of trees and a edge disjoint union of even cycles (with or without chords). Now, the square of a cycle requires at most $4$ colors, and the square of a tree requires at most $\Delta+1$ colors. Thus, the required number of colors is $\Delta+1$ in the latter case. Am I right here? Any counterexamples? Thanks beforehand.
The maximum degree of $G^2$ for general $G$ is at most $\Delta^2$, so we immetiately get an upper bound $\chi(G^2)\le \Delta^2+1$.
An example that is close to optimal is the incidence graph of the points and lines of a finite projective plane of order $q$. Here we have $2(q^2+q+1)$ vertices and the graph is regular of degree $\Delta=q+1$. The square of this graph has maximal cliques of size $q^2+q+1$ and in fact this is also equal to the chromatic number, so $\chi (G^2)=\Delta^2-\Delta+1$. In particular you shouldn't expect a linear bound in $\Delta$ without further conditions.
thanks! well, I thought that the maximum degree of the square of a graph is twice that of the original graph, by looking at cases like the square of cycles and trees. What about the half-square graphs of a bipartite graph? I think at least in that case we can bound the chromatic number in linear way
The finite projective plane gives a quadratic chromatic number even if you just take the half square.
Would it be of any use if the major degree vertices induce a forest, like in $K_{m,n}\ \ m\neq n$?
|
2025-03-21T14:48:31.423201
| 2020-07-03T04:39:38 |
364714
|
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|
Stack Exchange
|
Dimensions of the intersection of 8 quadrics
Suppose $e_i,q_i \in \mathbb{R}^3$, $1\leq i \leq 3$ with $\Vert e_i \Vert=1$ are known.
Define the projection on the plane orthogonal to $e_i$
$P_i= I-e_i e_i^T$ where $I$ is the $\mathbb{R}^{3\times 3}$ identity matrix.
Suppose to have the unknowns $k \in \mathbb{R}$, $u,v \in \mathbb{R}^3$, $e_i^T u \neq 0$, and the following system of equations
$u^T v=0$
$\Vert u \Vert= 1$.
$k e_i \times u + P_i v e_i^T u -q_i e_i^T u =0 $ for $1\leq i \leq 3$.
Each $k e_i \times u + P_i v e_i^T u -q_i e_i^T u =0 $ provides three equations, but only two are linearly independent. The system has therefore 8 equations (8 quadrics).
There are clearly sets of $q_i$ for which the system has no solution (it is sufficient to take any $q_i$ such that $e_i^T q_i \neq 0$). Is there any choice of the $q_i$ such that the system has multiple solutions in $k$,$u$,$v$?
Another way of formulating the question is the following. Given constants $e_i, e_i\in \mathbb{R}^3$ and $P_i= I-e_i e_i^T$, is the following function $(k,u,v) \rightarrow (q1,q2,q3)$ injective?
$\displaystyle q_i = k \frac{e_i \times u}{e_i^T u} + P_i v $
|
2025-03-21T14:48:31.423306
| 2020-07-03T04:43:21 |
364715
|
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|
Stack Exchange
|
Reference request : Convergence of radial basis function interpolation or spline interpolation as points become dense, for a continuous function
Is there any proof for this. Kindly request a reference in case available or any related documents towards this.
PS : I am specifically interested in the case of scattered data (irregularly placed), although any reference on the case of data placed on a uniform grid would also be valuable.
|
2025-03-21T14:48:31.423372
| 2020-07-03T06:41:17 |
364720
|
{
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"Andrew Yuan",
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"https://mathoverflow.net/users/48303",
"popoolmica"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364720"
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|
Stack Exchange
|
Diagonalization of the generalized 1-particle density matrix
Let $\mathscr{H}$ be a complex separable Hilbert space and $\mathscr{F}$ be the corresponding fermionic Fock space generated by $\mathscr{H}$. Let $\rho: \mathscr{L}(\mathscr{F}) \to \mathbb{C}$ be a bounded linear functional on all bounded operators of $\mathscr{F}$ with $\rho(I)=1$ and $\rho(A^*)=\rho(A)^*$, and define the 1-particle density matrix (1-pdm) by the unique bounded self-adjoint $\Gamma: \mathscr{H}\oplus \mathscr{H}^* \to \mathscr{H}\oplus \mathscr{H}^*$ such that
$$
\langle x|\Gamma y\rangle = \rho([c^*(y_1)+c(y_2)][c(x_1)+c^*(x_2)])
$$
where $x=x_1 \oplus \bar{x}_2$ and $y=y_1 \oplus \bar{y}_2$ (I use the notation $\bar{x} (\cdot) = \langle x|\cdot\rangle$) and $c,c^*$ are the annihilation/creation operators.
In references V. Bach (Generalized Hartree-Fock theory and the Hubbard model)[Theorem 2.3] and J.P Solovej (Many Body Quantum Mechanics)[9.6 Lemma and 9.9 Theorem], the authors claim that (under suitable conditions) $\Gamma$ is diagonalizable by a Bogoliubov transform $W:\mathscr{H}\oplus \mathscr{H}^* \to \mathscr{H}\oplus \mathscr{H}^*$ so that $W^* \Gamma W = \operatorname{diag}{(\lambda_1,...,1-\lambda_1,...)}$. The main idea of the proof is that $\Gamma$ is diagonalizable by an orthonormal basis, and that if $x\oplus \bar{y}$ is an eigenvector with eigenvalue $\lambda$, then $y\oplus \bar{x}$ is an eigenvector with eigenvalue $1-\lambda$. The proof is fine when $\lambda\ne 1/2$, since the 2 eigenvectors are orthonormal to each other. However, if $\lambda=1/2$, then things become a little more difficult. J.P Solove solves this in the case where the eigenspace of $\lambda =1/2$ is even-dimensional, but as far as I know, I can't understand why would it be.
Question. Is there something I'm forgetting? If not, is there a way or are there references that complete the proof?
EDIT. It's clear that if $\mathscr{H}$ is finite-dim, we see that the eigenspace of $\lambda=1/2$ is even-dim (due to the fact that the orthogonal complement is even-dim and $\mathscr{H}\oplus \mathscr{H}$ is even-dim). However what happens if $\dim \mathscr{H} =\aleph_0$ (infinitely countable)
Could you point out where Solovej explicitly considers the case $\lambda=1/2$?
@popoolmica He proves this in the 9.6 lemma and applies it in equation (73).
|
2025-03-21T14:48:31.423534
| 2020-07-03T08:26:31 |
364724
|
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|
Stack Exchange
|
Antipodal vertices in spectral graph embeddings
Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.
Under which condistions does the following hold:
If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.
For example, is this true if $G$
is walk-regular, (resp. vertex-transitive)
is 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive),
is distance-regular (resp. distance-transitive), or
has a symmetry $\phi\in\mathrm{Aut}(G)$ mapping $v\mapsto v'$ for all $v\in V$.
Especially in the last case I can imagine that we have $u_v = \pm u_{v'}$ for all $v\in V$ and eigenvectors $u\in \smash{\Bbb R^V}$ of $G$ (not necessarily to $\theta_2$).
But I am specifically interested in the case $\theta_2$ and whether we then always have the negative sign.
Do you have any evidence/heuristics for why this should hold in some cases? Also what is special about $\theta_2$?
@AntoineLabelle My intuition comes from spectral embeddings of $G$, where you assign points in Euclidean space to the vertices of $G$, and you do this by somehow using the eigenvectors of $G$ to some eigenvalue. It seems to be common intuition, that the $\theta_2$-embedding is "as expanded as possible" (imagine the vertices as repelling each other, but the edges keeping them together, and then the $\theta_2$-embedding is an equilibrium configuration under these condition). And to be as expanded as possible, it is plausible that antipodal vertices are embedded opposite to each other.
@AntoineLabelle My evidence is that it holds for all the graphs I have checken (e.g. the skeletons of regular polytopes, but also many others).
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2025-03-21T14:48:31.423807
| 2020-07-03T08:31:32 |
364725
|
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"Jeremy Rickard",
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"bergfalk",
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|
Stack Exchange
|
Acyclic extensions of acyclic simplicial complexes
Say an abstract simplicial complex $X$ is acyclic if its reduced integral simplicial homology groups $\tilde{\mathrm{H}}^{\Delta}_p(X)$ vanish for all $p\geq 0$. Is it the case that, for any $n>0$, any $n$-dimensional acyclic simplicial complex $X$ on a set $V$ may be extended to an $n$-dimensional acyclic simplicial complex $Y$ on $V$ with a complete $(n-1)$-skeleton, i.e., satisfying $Y^{n-1}=[V]^n$? Here the last expression denotes the collection of size-$n$ subsets of $V$, the vertex-set of $X$.
I suspect this to be true, but I'm finding it tricky to rigorously argue so I'm partly wondering if I'm missing anything and/or if there are any theorems about that might ease the job. Ultimately I'm interested in extensions $Y$ which conserve further properties of $X$, but this question seems to me the place to begin.
I don't understand what role $X$ plays here? If $Y^{n-1}=[V]^n$ and $m<n$, then $Y$ contains every $m$-dimensional simplicial complex $X$ on $V$.
@JeremyRickard, fair point, when dim$(X)$<dim$(Y)$ the question's not very interesting. But (as the edit now emphasizes) when dim$(X)$=dim$(Y)$?
Maybe I am getting something wrong, so lets see: suppose we have a simplicial complex consisting of a single solid triangle with an edge attached to some vertex. This 2-dimensional complex has four vertices and is (as far as I understand) acyclic. What would be its acyclic extension?
@M.Winter It seems to me that you would have 3 choices of 2-dimensional acyclic extensions, each taking the form of all of the edges and 3 of the faces of a tetrahedron.
@bergfalk But hasn't the skeleton of the tetrahedron (with a single face added if you like) a non-vanishing first homology group (it is not simply connected)?
@M.Winter Yes, but I'm not sure what this has to do with our discussion, as the acyclic extension proposed in the previous comment has 3 faces, not 1.
@bergfalk Oh sorry, I missed that. I understand the setting now. Interesting question.
|
2025-03-21T14:48:31.423971
| 2020-07-03T09:02:38 |
364727
|
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"Gordon Royle",
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|
Stack Exchange
|
An eigenvalue upper bound for 1-walk-regular graphs
Let $G$ be a graph and suppose that $G$ is 1-walk-regular (or, if you prefer, vertex- and edge-transitive, or distance-regular).
Let $\theta_1>\theta_2>\cdots>\theta_m$ be the distinct eigenvalues of its adjacency matrix.
It is known that $\theta_1=\deg(G)$ (the vertex-degree of $G$).
Now, I believe that the following holds:
$$\theta_2 < \mathrm{deg}(G)\cdot\cos\Big(\frac{\pi}{2\mathrm{diam}(G)}\Big),$$
where $\mathrm{diam}(G)$ denotes the diameter of $G$.
This bound can be quite off.
However, if $G$ is antipodal (that is, for every vertex there is a unique maximally distant vertex), then I believe we even have
$$\theta_2 \le \mathrm{deg}(G)\cdot\cos\Big(\frac{\pi}{\mathrm{diam}(G)}\Big),$$
and this bound is actually attained with equality in many cases.
Question:
Are these bounds known (or are there counterexamples)?
Is known for which graphs the second inequality is satisfied with equality?
Update
As requested in the comments, I provide a list of some graphs that attain the second bound.
Because I am most familiar with polytope theory, all my examples are skeleta of polytopes.
The list includes the skeleton of ...
an even-sided polygon (the edge-graph is the even cycle),
a cross-polytopes (the edge-graph is the completement of a disjoint union of edges; these are the only antipodal graphs of diameter 2),
the cuboctahedron (degree 4, diameter 3, $\theta_2=2$),
the icosidodecahedron (degree 4, diameter 5, $\theta_2=1+\sqrt 5$),
the 24-cell (degree 8, diameter 3, $\theta_2=4$),
the 600-cell (degree 12, diameter 5, $\theta_2=3(1+\sqrt 5)$).
...
I think I have an understanding for why it works with these polytopes, and there are more of these in higher dimensions.
In the light of these examples (and my idea of why they work) I wonder whether there is a graph that attains the bound and is
not the edge-graph of a polytope, or
not vertex/edge-transitive, or
not of even degree.
Note that the bound can also be arbitrarily bad. E.g., numerical experiments suggest that the bound becomes worse for the crown graphs with increasing degree.
Update 2
I will explain how the second inequality is motivated and might be proved.
I consider a spectral embedding of the graph to the eigenvalue $\theta_2$. Because the graphs is 1-walk-regular, all its vertices are embedded on a sphere (of, say, radius $r=1$), and all edges will be embedded with the same length, say $\ell$. Without going into the details, this length can be expressed as
$$(*)\quad \ell=\sqrt{1-\frac{\theta_2}{\mathrm{deg}(G)}}.$$
Now I assumed (but I have not proof for this, see this qestion) that antipodal vertices are embedded "opposite to each other", that is, if $i,j\in V(G)$ are antipodal, then their embeddings satisfy $v_i=-v_j$.
Now, if there is a path of length $\mathrm{diam}(G)$ from $i$ to $j$, all vertices on the sphere, all edges of the same length, one can determine a lower bound on the length of these edges so that this path is possible.
One can imagine how trigonometry enter the picture here.
This lower bound on the edge length translated to an upper bound on the eigenvalue $\theta_2$ via $(*)$.
With this approximate reasoning I can explain how I came to the examples I know of. Compare the image of the cube and the cuboctahedron below:
In the cuboctahedron, the path connecting antipodal vertices is "flat", while in the cube it is not.
That is, in the cuboctahedron the edges are as short as possible for reaching from one end of the circumsphere to the other in only $\mathrm{diam}(G)=3$ steps.
Here are some consequences for graphs that attain the bound:
the spectral embedding of the graph to $\theta_2$ must decompose into embeddings of flat cycles of length $2\mathrm{diam}(G)$.
the degree must be even as every flat cycle entering a vertex must leave the vertex in the opposite direction.
In the case of vertex- and edge-transitive polytopes this might allow the following characterization:
The edge-graph attains the bound, if and only if the vertex-figure is centrally symmetric.
More vague, for vertex- and edge-transitive graphs $G$ a characterization might be the following:
$G$ attains the bound if and only if the stabilizer $\Gamma_i\subseteq\mathrm{Aut}(G)$ at a vertex $i\in V$ induces a centrally symmetric symmetry on the neighborhood $N(i)$, whatever this means exactly.
It amazes me that up to that point all the examples I know of have been polytopal. I see no reason for why this should be the case.
The last bound is tight for an even cycle. Do you have examples for which it is tight and which are not Abelian Cayley graphs?
@smapers I added a list of examples.
Can you outline the proof of your inequality, or give a link?
Usually the proof of an inequality suggests where to look for
cases of equality (by listing all the $\leq$'s that must be $=$).
Here is an infinite family for you - edge-transitive and antipodal, so not really answering the question: $L(K_n \times K_2)$ has valency $2(n-2)$, diameter $3$ and second eigenvalue $n-2$. Here $K_n$ is a complete graph, $\times$ is the tensor product and $L$ is the line graph. Unless I am mistaken, $n=3$ gives the $6$-cycle and $n=4$ the cuboctahedron.
@GordonRoyle Awesome construction. It indeed generalizes the sequence of hexagon and cuboctahedron with skeleta of further higher-dimensional polytopes ($d=5$: runcinated 4-simplex, $d=6$: stericated 5-simplex, ...). I was not aware of this construction for their edge-graphs.
@Noam I included a sketch for my reasoning.
@M.Winter There is a 24-vertex $4$-regular graph that is a double-cover of the Cuboctahedron with diameter 3 and second eigenvalue $2$ - it seems like this might be an obvious geometric construction, but I don't know enough geometry.
@GordonRoyle It does not correspond to anything geometric that I can immediately think of. Can you say something more about the comstruction of this graph?
Your question reminds me of Chris Godsil's characterization of when a distance-regular graph is isomorphic to the 1-skeleton of its \theta_2-eigenpolytope -- which I believe is what you are calling the spectral embedding w.r.t. \theta_2 (see https://doi.org/10.4153/CJM-1998-040-8)
Here is the graph as requested from the comments.
Take the cuboctahedron and then assign colours to the edges as follows:
There are 8 triangles and 6 quadrilaterals in this graph, and it takes a moment to confirm that each triangle has a unique red edge, and each quadrilateral has either one or three red edges. In particular, all faces have an odd number of red edges.
Now form a double cover by replacing each vertex $v$ by a pair of non-adjacent vertices $v$, $v'.$ If $vw$ is a black edge of the cuboctahedron, then add a matching $\{vw, v'w'\}$ (i.e., a "straight" matching) and if it is a red edges of the cuboctahedron, then add a matching $\{vw', v'w\}$ (i.e., a "crossed" matching).
(Of course, this can all be expressed as a "voltage graph" over $\mathbb{Z}_2$, where the black edges correspond to $0$ and the red edges to $1$.)
Anyway, this removes all the triangles and quadrilaterals, and so the resulting graph has diameter 3, girth 5, and characteristic polynomial
$$
(x - 4) \cdot x^{3} \cdot (x + 2)^{5} \cdot (x - 2)^{7} \cdot (x^{2} + 2x - 2)^{4}.
$$
It is not antipodal.
|
2025-03-21T14:48:31.424485
| 2020-07-03T10:22:14 |
364735
|
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|
Stack Exchange
|
Estimate on Mobius function
Let $\mu(n)$ be the Mobius function, how to estimate
$$\sum_{1\le i<j\le x}\mu(i)\mu(j) $$
as $x$ goes to $\infty$? Are there some references on this?
If $S$ is your sum then
$$ \left\lvert \sum_{1\leq n\leq x}\mu(n)\right\rvert^2 = 2S+ \sum_{1\leq n\leq x}\mu(n)^2.$$
The second sum on the right is $(\frac{6}{\pi^2}+o(1))x$, and hence estimating $S$ is equivalent to estimating $\lvert \sum_{n\leq x}\mu(n)\rvert$, a classical problem of analytic number theory.
In particular, assuming the Riemann Hypothesis, the left-hand side is $O(x^{1+o(1)}$), and hence (assuming RH)
$$ S \ll x^{1+o(1)}.$$
Unconditionally we can show that $S=o(x^2)$, but cannot show $S\ll x^{2-\epsilon}$ for any $\epsilon>0$.
"The best we can conclude unconditionally is $S = o(x^2)$". What do you mean by this? I'm no analytic number theorist, but I'd be surprised if $|\sum_{n \le x} \mu(n)| \ll \frac{x}{\log\log x}$ wasn't known unconditionally.
Yes, sorry, I was being imprecise - I meant that in terms of improving the exponent nothing better is known. The standard proof of the prime number theorem can be adapted to show that $S\ll x^2 \exp(-O(\sqrt{\log x}))$, for example, and I believe better quantitative results are known.
|
2025-03-21T14:48:31.424607
| 2020-07-03T11:06:14 |
364739
|
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|
Stack Exchange
|
Kan liftings and projective varieties
Regard the following two bicategories:
$\operatorname{dg-\mathcal{B}imod}$, with objects dg categories, and morphisms categories from $C$ to $D$ being the categories of $C$-$D$-bimodules. Composition is given by the dg tensor product. Note that this might be considered as the bicategory of Chain-complex-enriched profunctors/ relators (see Coend Calculus), so the machinery developed in this paper should (as chain complexes form a Bénabou-cosmos) show that every 1-morphism in this category has both a left and a right adjoint, and Right Kan Liftings exist here
$\mathcal{V}ar$, with objects smooth projective varieties and morphism categories from $X$ to $Y$ the derived categories $D^b (X \times Y)$, as considered here. The 1-morphisms here encode kernels for Fourier-Mukai-Transformations. Using the Serre kernel, it is shown in that paper that every 1-morphism in this bicategory has a left and a right adjoint.
These bicategories are very closely related; it was shown e.g. by Toen that if one takes the dg enhancements of the derived categories $D^b_{dg}(X)$, $D^b_{dg}(Y)$, then the dg functors between those are in one-to-one correspondence with the elements of $D^b(X \times Y)$, to name just one similarity. Therefore, I would suspect that also the second category possesses Right Kan Liftings, but I am not sure how to construct them (as already the proof that the Serre kernel lets us construct adjoints there is very nontrivial).
Is this indeed true? How are they constructed? And most importantly (as from the fact that $\mathcal{V}ar$ has adjoints, a lot can be deduced about Grothendieck duality etc.) do they also have interesting uses in the study of smooth projective varieties, or can their construction even be extended to more general schemes?
Years ago I thought for a long time about the possibility to use profunctors to describe Fourier-Mukai theory: this question looks very natural to me. I would start with this question: dg-functors $D(X)\to D(Y)$ correspond to objects of $D(X\times Y)$. Is this also a functor, and not only a correspondence on objects? An equivalence of categories? What about dg-profunctors?
@Fosco I also think this is the right question to ask. For reference, this is statement 8.15 of this paper, checking functoriality seems to boil down to going through here (referenced in 8.9). Essentially, the latter paper proves that certain stable model categories have a compact generator, and one would have to show that this assignment even works in a functorial way. I'll need a bit more time to check if this actually works out.
|
2025-03-21T14:48:31.424818
| 2020-07-03T12:21:59 |
364741
|
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|
Stack Exchange
|
Example of a representation of a finite group where Weyl's unitary trick is necessary?
Is there an example of a representation $\rho: G \rightarrow GL(V)$ for some finite group $G$ where say $W \subset V$ is a $G$-invariant subspace for $\rho$ but the orthogonal complement (in the standard sense) $W^{\perp}$ is not G-invariant? I understand one could "unitarize" the representation using Weyl's averaging trick (and getting a new inner product) but my question is to find an example where one cannot do away with this averaging (and use the usual inner product)
What is the "usual inner product"?
Also, what if this usual inner product is not G-invariant?
That’s not “Weyl’s unitary trick”. Weyl’s trick is to replace a noncompact group by a compact one to which Hurwitz’s averaging method then applies.
@FrancoisZiegler Thanks for pointing out. I was using the term based on some lecture notes I stumbled upon. For instance see Theorem 3.10 here:https://math.berkeley.edu/~teleman/math/RepThry.pdf
Try $G=\{1,-1\}$ and $\rho\colon G \to GL_2\mathbb{R}$ where $\rho(-1)$ is the matrix
$$\left(\begin{matrix}-1&2\\ 0&1\end{matrix}\right)$$
Take $W$ to be the span of
$$\left(\begin{matrix}1\\ 0\end{matrix}\right)$$
and use the standard inner product on $\mathbb{R}^2$.
|
2025-03-21T14:48:31.424958
| 2020-07-03T13:24:49 |
364743
|
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"authors": [
"Fedor Petrov",
"Gerhard Paseman",
"Gjergji Zaimi",
"H A Helfgott",
"Sam Hopkins",
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|
Stack Exchange
|
Alternating sum over collections of sets
Let $\mathbf{P}$ be a collection of subsets of a finite set $X$. Let $\mathscr{S}$ be the set of all subsets $\mathbf{S}\subset \mathbf{P}$ such that $\bigcup_{S\in \mathbf{S}} S = X$. Can one give a sensible upper bound on the sum
$$\Sigma = \sum_{\mathbf{S}\in \mathscr{S}} (-1)^{|\mathbf{S}|},$$
where $|\mathbf{S}|$ is the number of elements of $\mathbf{S}$? In particular: is the absolute value of the sum bounded by the number of minimal elements of $\mathscr{S}$?
(For a strategy that does not work, see Alternating sum over collections closed under containment).
What if every set $S$ in $\mathbf{P}$ is of cardinality $\leq l$, and $|X|=m\geq l$? Can one give a non-trivial bound in terms of $m$ and $l$?
Is not it the same question?
How? It's a special case. Or can every question of the other kind be put in this framework?
It would not surprise me if the simplicial complex $\mathscr{S}$ you describe has a name, but I'm not sure how one might google for that...
Well, $\mathbf{P}$ need not be the set of all subsets of $X$ of cardinality $\leq l$.
Ahh, I missed that you allow $\mathbf{P}$ to be not necessarily all the subsets.
Why a special case? Any contaitment-closed family has such a structure, does not it?
@FedorPetrov Not obvious to me - am I being dense?
@FedorPetrov: equivalently you are saying any simplicial complex can be realized via this construction. That's plausible but like H.A. I do not see it automatically.
Let me try to prove that the question about minimal elements is equivalent to the previous, namely:
Theorem. Assume that $\mathbf{P}$ is a finite set and $\mathscr{S}$ is a family of subsets of $\mathbf{P}$ which is closed under taking over-sets. Then there exists a finite set $X$ and an injection $\varphi:\mathbf{P}\to 2^X$ such that
$$
\mathscr{S}=\{\mathbf{S}\subset \mathbf{P}:\cup_{j\in \mathbf{S}}\varphi(j)=X\}.
$$
Proof.
For any set $\mathbf{S}\subset \mathbf{P}$ such that $\mathbf{S}\notin \mathscr{S}$ choose an element $x_{\mathbf{S}}$ which does not belong to all sets $\varphi(i),i\in \mathbf{S}$, and does belong to all $\varphi(j),j\notin \mathbf{S}$. Define $X=\sqcup_{\mathbf{S}} \{x_{\mathbf{S}}\}$, $\varphi$ is already defined. If $\mathbf{S}\notin \mathscr{S}$, then $\cup_{j\in \mathbf{S}}\varphi(j)\ne X$, because of the element $x_{\mathbf{S}}$. Now take
$\mathbf{S}\in \mathscr{S}$. Fix any element $x_{\mathbf{T}}\in X$, where $\mathbf{T}\notin \mathscr{S}$. Since all over-sets of $\mathbf{S}$ belong to $\mathscr{S}$, we conclude that $\mathbf{T}$ is not an over-set of $\mathbf{S}$, i.e., there exists $j\in \mathbf{S}\setminus \mathbf{T}$. The set $\varphi(j)$ covers $x_{\mathbf{T}}$. Since the element $x_{\mathbf{T}}\in X$ was arbitrary, we conclude that $\cup_{j\in \mathbf{S}}\varphi(j)=X$.
OK, I think this works, and disproves that the absolute sum is bounded by the number of minimal sets of $\mathscr{S}$. At the same time, the set $X$ is pretty large (it can be of size $2^{|\mathbf{P}|}$ or close to that), so one can still have a useful bound for this problem (see my answer) and not for the other one.
The two versions of the problem are fully equivalent. Suppose $X$ is a finite set and $\mathbf{P}$ is a collection of subsets of $X$. Let's define
$$f(\mathbf{P})\mathrel{\mathop:}=\sum_{S\in \mathbf{P}}(-1)^{|S|} \qquad \text{and} \qquad g(\mathbf{P})\mathrel{\mathop:}= \sum_{S_1,S_2,\dots,S_r\in \mathbf{P}\\ S_1\cup \cdots \cup S_r=X}(-1)^r.$$
Let's also denote by $\widehat {\mathbf{P}}$ the set of all subsets which contain some element of $\mathbf{P}$. The following holds:
$$g(\mathbf{P})=g(\mathbf{\widehat{\mathbf{P}}})=f(\widehat{\mathbf{P}}).$$
To prove the first equality notice that if $A_0\subset A_1$ are subsets such that $A_0\in \mathbf{P}$ and $A_1\notin \mathbf{P}$ then
$$g(\mathbf{P}\cup\{A_1\})-g(\mathbf{P})=\sum_{S_1,S_2,\dots,S_r\in \mathbf{P}\\ A_1\cup S_1\cup \cdots \cup S_r=X}(-1)^{r+1}$$
however the collections that index the sum on the right split into those that contain $A_0$ and those that don't. These two cancel each other out and the sum evaluates to zero. Since we can keep adding subsets to $\mathbf{P}$ one by one, this shows that $g(\mathbf{P})=g(\widehat{\mathbf{P}})$. Finally, the equality $g(\widehat{\mathbf{P}})=f(\widehat{\mathbf{P}})$ was proven by Fedor in the previous question (sidenote: this is referred to as Rota's crosscut theorem).
A third equivalent formulation is to ask for bounds on the Euler characteristic of the simplicial complex obtained by using $X$ as a set of vertices and adding a simplex for $S$ whenever the complement of $S$ is in $\widehat{\mathbf{P}}$. Thus your questions become:
What is the largest Euler characteristic of a simplicial complex with $N$ facets?
What is the largest Euler characteristic of a simplicial complex with $N$ facets and $m$ vertices?
What is the largest Euler characteristic of a simplicial complex on $m$ vertices if all facets have dimension $\geq m-l$
The answer to Q1 is $\binom{N-1}{ \lfloor (N-1)/2 \rfloor}$ by the Sagan-Yeh-Ziegler paper. They construct a simplicial complex with $N$ vertices, $\binom{N}{\lfloor N/2\rfloor}$ facets, with Euler characteristic \binom{N-1}{ \lfloor (N-1)/2 \rfloor}, which also gives a simplicial complex with the same Euler characteristic but $N$ facets and $\binom{N}{\lfloor N/2\rfloor}$ vertices. The answer to Q2 was conjectured to be $e^{O(\log N\log m)}$ by David Speyer here, and I don't know what the status of this is.
For Q3, if $m-l\le \frac{m}{2}$ then we can use the same example in Q1 which gives the answer $\binom{m-1}{\lfloor (m-1)/2\rfloor}$. If $m-l\geq m/2$ then the number of facets is at most $\binom{m}{l}$ and assuming Speyer's conjecture the correct upper bound should be $e^{O(\log m \cdot \log \binom{m}{l})}$.
I'm confused, isn't the SYZ paper bounding the Euler characteristic of a simplicial complex with a given number of vertices, not facets?
@SamHopkins Yes, it is, but it implies the bound for fixed facets, as mentioned in David's answer I linked to.
How do the bound $N\leq \binom{m}{\lfloor m/2\rfloor}$ and the bound $\chi \leq \binom{N-1}{\lfloor (N-1)/2\rfloor}$ imply that $\chi \leq \binom{m-1}{\lfloor (m-1)/2\rfloor}$? I must be missing something.
What I am looking for now is something that is both (a) unconditional and (b) at least as good as the (unconditional) bound I clumsily gave in my own answer below. The bound there is good enough for my purposes, but of course I'd prefer something more literate than what I was able to give!
@HAHelfgott I made a small edit to clarify. I just meant that the case $m-l\le m/2$ can still be optimized with the SYZ construction.
(The above answer is nice, but I'd perhaps be a bit careful with "fully equivalent", in that the dependencies may be different: the number of minimal elements in $\mathbf{P}$ may be much larger than $X$, no? In that case, we are losing something when we reduce the new problem to the old problem and back.)
@HAHelfgott The number of minimal elements of $\mathbf{P}$ is the same as the number of minimal elements of $\widehat{\mathbf{P}}$ which is what i meant by fully equivalent.
Well, sure, but if you are interested in a bound with a good dependency on $m$ and $l$, reduction to the older problem may not necessarily be the way to go, no?
@HAHelfgott I'm not changing the sets in this reduction. The ground set $X$ stays the same, so $m$ doesn't change, and the minimal sets stay the same as well. As far as the comment about an unconditional bound: Once you know that a simplicial complex on $m$ vertices has $\chi \le \binom{m}{m/2}\le 2^m$ this is automatically better than the bound $l^m$ for $l\geq 3$.
Yes, here $X$ doesn't change, but the bound given in the older question has (in effect) the set of minimal elements of $S$ as its ground set.
@HAHelfgott Yes, I was being dense. You're right. I just hope this clarifies that knowing the optimal answer to one question leads to knowing the optimal answer for the other.
On Sagan-Yeh-Ziegler: is the Möbius function of a lower-order ideal $\mathcal{F}$ simply $\sum_{S\in \mathcal{F}} (-1)^{|S|}$? In that case a direct application to $\mathcal{F} = {X\setminus S: S\in \widehat{P}}$ does give the bound $\chi \leq \binom{m-1}{\lfloor (m-1)/2\rfloor}$ immediately.
Well, minus that, really. Yes, that seems to be stated in equation (1), section 2, of Sagan-Yeh-Ziegler.
@HAHelfgott That's correct.
But then $\chi\leq 2^m$ already follows from the trivial bound $f(\widehat{P})\leq |\widehat{P}|\leq 2^m$ and $g(P) = f(\widehat{P})$.
@HAHelfgott Good point :) If we want to get a considerable distance away from that trivial bound we need to restrict the class of simplicial complexes in a way that doesn't include the SYZ example.
Well, $\chi\leq 2^m$ is not in itself obvious, its proof now is elegant, and it's strong enough for me.
Here's a very naive but arguably non-trivial bound. (Please feel free to do better!)
Just choose a set $S_0$ in $\mathbf{P}$. It is clear that, for $\mathbf{S}\subset \mathbf{P}$ not containing $S_0$, if $\mathbf{S}$ is in $\mathscr{S}$, then the contributions of $\mathbf{S}$ and $\mathbf{S}\cup \{S_0\}$ to the sum $\Sigma$ cancel out. Hence $$\Sigma = - \mathop{\sum_{\mathbf{S}\subset \mathbf{P}}}_{\mathbf{S}\not\in \mathscr{S} \wedge (\mathbf{S}\cup \{S_0\}\in \mathscr{S})} (-1)^{|\mathbf{S}|} = - \sum_{T\subset S_0, T\neq \emptyset}\, \sum_{\mathbf{S}\in \mathscr{S}_{X\setminus T}} (-1)^{|\mathbf{S}|},$$
where, for $Y\subset X$, we denote by $\mathscr{S}_{Y}$ the set of all subsets $\mathbf{S}\subset \mathbf{P}$ such that $\bigcup_{S\in \mathbf{S}} S = Y$.
Thus, $|\Sigma|\leq a_{m,l}$, where $a_{m,l}$ is given by the following recurrence relation:
$$a_{m,l} = \sum_{i=1}^{\min(l,m)} \binom{l}{i} a_{m-i,l},$$
with $a_{0,l}=1$.
It is easy to show that $l^m\leq a_{m,l} \leq ((e-1) l)^m$.
Have you looked at Moebius inversion on finite posets? My only reference for it is chapter 5 section 7 in " Algebras, Lattices, Varieties". Gerhard "Looking For A Mu Twist" Paseman, 2020.07.03.
It looks as if it might conceivably be related, but I'm completely unfamiliar with the formalism. Can you see a more precise link?
|
2025-03-21T14:48:31.425591
| 2020-07-03T15:26:00 |
364753
|
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|
Stack Exchange
|
Minimax theorems in nonconvex setting
Let $X$ be a topological space, $Z$ be a compact convex subset of $\mathbb R^m$, and let $f:X \times Z \to \mathbb R$ be a continuous function (w.r.t the product topology on $X \times Z$).
Question. Under what minimal additional assumptions do we have the duality $\sup_{z \in Z}\inf_{x \in X}f(x,z) = \inf_{x \in X}\sup_{z \in Z}f(x,z)$ ?
The case $m=1$ (in which case $Z$ must be a compact interval in $\mathbb R$) was proved in Theorem 1 of Minimax theorems in a fully non-convex setting under additional connectivity conditions on the sets $\arg\min f(\cdot,z)$ and $\arg\max f(x,\cdot)$ for all $(x,z) \in X \times Z$.
shouldn't sup and inf be changed on the RHS?
Good catch. Fixed. Thanks!
|
2025-03-21T14:48:31.425681
| 2020-07-03T15:41:00 |
364754
|
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|
Stack Exchange
|
Is identification of double points of an immersion smooth?
Let $f:M^m\to N^n$ be a generic map between smooth manifolds $n>m$. Depending on the pair $(m,n)$ generic maps will have a singular set of double points $\Sigma_2\subset M$.
Let $\phi:\Sigma_2\to \Sigma_2$ be the map of sets that sends $x\in \Sigma_2$ to the other point $y\in \Sigma_2$ such that $f(x)=f(y)$.
$\phi$ can be thought as a $\mathbb{Z}/2$-action on $\Sigma_2$.
For the sake of concreteness one can think of a generic immersion of a $M^3\to N^4$, then the set of double points is generically of dimension 2.
Is $\Sigma_2$ a smooth submanifold?
Is $\phi$ a smooth or at least a continuous map? In other words how bad is the action?
How does this generalize to triple points and $n$-points? I.e., can we have any kind of group of $n$ elements acting?
Don't you generically get more than just double points? I'm thinking of the case where N is $\mathbb{R}^4$ and M is a union of three hyperplanes in N in general position. Then there is a $1$-dimensional locus of triple intersection, and that should be stable under small perturbations. Are you just limiting yourself to maps where that doesn't happen?
@AchimKrause yes generically you have also triple points, so you will have a non empty $\Sigma_3$.
My definition for $\Sigma_n$ is the set points that belong to a fiber of cardinality exactly $n$. So $\Sigma_2 $ is disjoint from $\Sigma_3$ (triple points are not double points for me, sorry if this is the most used definition).
In other words, just consider the double points for the sake of this question but don't assume the map does not have triple points.
It may be reasonable to generalise $\mathbb Z/2\mathbb Z$ not to $\mathbb Z/n\mathbb Z$ or another $n$-element group, but to $\operatorname{Sym}_n$.
I think the answer to the first 2 questions is yes. Most of the details are in the thesis of Ralph Herbert:
Herbert, Ralph J., Multiple points of immersed manifolds, Mem. Am. Math. Soc. 250, 60 p. (1981). ZBL0493.57012
The important construction here is the $r$-tuple point manifold $\Delta_r(f)$ of the immersion $f:M^m\looparrowright N^n$, which for $r\ge2$ is defined as follows. Let $F(M,r)\subseteq M^{(r)}$ be the ordered configuration space of $r$-tuples $(x_1,\ldots , x_r)$ of points of $M$ such that $x_i\neq x_j$ whenever $i\neq j$; it is an open submanifold of $M^{(r)}$. Now consider the restriction of the $r$-th Cartesian power of $f$ to this configuration space, which will by abuse of notation will be denoted $f^{(r)}:F(M,r) \looparrowright N^{(r)}$. Generically, $f^{(r)}$ is transverse to the thin diagonal $d_r(N)= \{(n,\ldots , n)\}\subseteq N^{(r)}$ (see Herbert; also Gollubitsky and Guillemin, Stable mappings and their singularities, Chapter III, Corollary 3.3). Then $$\Delta_r(f) :=(f^{(r)})^{-1}(d_r(N))= \{(x_1,\ldots , x_r)\in F(M,r) \mid f(x_i)=f(x_j)\mbox{ for all }1\le i,j\le n\} $$
is a submanifold of $F(M,r)$ of codimension $rn-n$, therefore of dimension $rm-(rn-n)=n-r(n-m)$. If $M$ is a closed manifold, then so is $\Delta_r(f)$ (compactness is not obvious, it uses the fact that $f$ is locally an embedding). Note that $\Delta_r(f)$ carries smooth free actions of the symmetric group $\mathfrak{S}_r$ and the symmetric group $\mathfrak{S}_{r-1}$ which permute the last $r$ and $r-1$ coordinates, respectively. Let $M_r(f):=\Delta_r(f)/\mathfrak{S}_{r-1}$.
Now consider the map
$$
\mu_r(f): M_r(f)\to M,\qquad (x_1,[x_2,\ldots , x_r])\mapsto x_1
$$
given by projection onto the first co-ordinate. This $\mu_r(f)$ can be shown to be an immersion. Its image is the set of $x\in M$ such that $|f^{-1}f(x)|\ge r$.
Now restrict to the case $r=2$. The immersion $\mu_2(f):M_2(f)\looparrowright M$ is not an embedding when $f$ has triple points or higher. But if we remove the points of the domain where $\mu_2(f)$ fails to be injective, we get an injective immersion onto you $\Sigma_2$ (the "pure" double points), which in fact is an embedding. (A perhaps more convincing argument using general position which works for all $\Sigma_r$ is given on page 25 of Herbert.)
Thus $\Sigma_2\subseteq M$ is an embedded submanifold, and the involution $\phi:\Sigma_2\to \Sigma_2$ is smooth as it is conjugate to the restriction of invoution on $M_2(f)=\Delta_2(f)$ which permutes the factors.
For triple points and higher, I don't see any group action on $\Sigma_r$, since when $f^{-1}f(x)=\{x,x_2,\ldots , x_r\}$ there is no natural way to order the set $\{x_2,\ldots, x_r\}$.
|
2025-03-21T14:48:31.426131
| 2020-07-03T16:30:10 |
364756
|
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|
Stack Exchange
|
Algebras Morita equivalent with the Weyl Algebra and its smash products with a finite group
My question os motivated, naturally, by the problem of classifying symplectic reflection algebras up to Morita equivalence (a classical reference for rational Cherednik algebras is Y. Berest, P. Etingof, V. Ginzburg, "Morita equivalence of Cherednik algebras", MR2034924; the most up do date work in this subject I know of is I. Losev, Derived equivalences for Symplectic reflection algebras, https://arxiv.org/abs/1704.05144);
and also the problem of understading rings of differential operators on irreducilbe affine complex varieties $X$ up to Morita equivalence (a nice discussion of this lovely problem in the intersection of ring theory and algebraic geometry can be found in Y. Berest, G. Wilson, "Differential isomorphism and equivalence of algebraic varieties", MR2079372)
Given that, my questions are:
(Question 1): What are the more general known conditions on a symplectic reflection algebra $H_{1,c}(V,\Gamma)$ that imples it is Morita equivalent to $\mathcal{D}(V) \rtimes \Gamma$?
(Question 2): What are the recent developments made in the study of equivalence of rings of differential operators up to Morita equivalence (and in particular Morita equivalent to the Weyl algebra) since Berest, Wilson [op. cit.]?
(Question 3): Etingof in "Cherednik and Hecke algebras of varieties with a finite group action", MR3734656, introduces more general versions of rational Cherednik algebras and discuss the possibility of extending the results in Y. Berest, O. Chalykh, Quasi-invariants of complex reflection groups ,MR 2801407, in this setting. So, being optmistic, one hipothetically could obtain results similar as those discussed in Berest, Etingof, Ginzburg [op. cit] regarding Morita equivalence of these generalized rational Cherednik algebras with smash products of rings with differential operatos with a finite groups. Has this line of inquiry lead to results relevant to this discussion so far?
(Question 4): This is totally unrelated to the previous questions. It is more of a very open question in ring theory: are there interesting simple Noetherian algebras, coming from another areas than those above, which are Morita equivalent to a Weyl algebra or a smash product of it with a finite group?
I do not know much on recent developments related to the first three questions asked. However, i know of some old results related mainly to the fourth question:
If $A_1$ is the Weyl algebra over an alg closed field of zero char, with the two generators denoted $p,q$ and $I$ is a non-zero, right ideal, then $M_2(End_{A_1}(I))\cong M_2(A_1)$ and $A_1$ is Morita equivalent to $End_{A_1}(I)$. Furthermore, these algebras are not generally isomorphic: Pick for example $I=p^2A_1+(pq+1)A_1$. Its endomorphism ring is isomorphic to $\{x\in Q|xI\subseteq I\}$, where $Q$ is the quotient division ring of $A_1$. This is not isomorphic to $A_1$ but it is Morita equivalent to it. If you are interested in this example, this is presented in An example of a ring Morita equivalent to the Weyl algebra $A_1$, S.P. Smith, J. of Alg, 73, 552 (1981).
Another result which may be of interest -regarding your fourth question- is that:
If the semigroup $k\Lambda$ has the same quotient field with $k[t]$, then $D(K)$ is Morita equivalent to $A_1$.
Here $K$ stands for certain subalgebras of $k[t]$ and $D(K)$ for the ring of differential operators on $K$. This is shown in: Some rings of differential operators which are Morita equivalent to $A_1$, Ian Musson, Proc. of the Am. Math. Soc., 98, 1, 1986
Finally, if you are interested in examples involving smash products with finite group algebras, i do not have some readily available but i think it is natural to look for such in the graded version of Morita equivalence.
I hope these are of some interest to the OP. Sorry in advance if these are too old and you are already aware of them.
P.S.: One more thing which might be of some interest with respect to the second question: The article Rings graded equivalent to the Weyl algebra, J. of Alg., vol. 321, 2, 2009, generalizes some results of Y. Berest, G. Wilson and Stafford, in the setting of graded module categories. (also i think this article is the first -although i am not sure- which introduces the terminology "graded Morita equivalence")
I already knew these papers, but anyhow thanks for reminding me of them! Hopefully I can check if they lead to new developments related to the fourth question!
|
2025-03-21T14:48:31.426408
| 2020-07-03T17:07:47 |
364758
|
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|
Stack Exchange
|
Difference between semilinear and fully nonlinear
I'm confused why the Hamilton Jacobi Bellman equation:
$$\frac{\partial u}{\partial t}(t,x)+\Delta u(t,x) -\lambda||\nabla u(t,x) ||^{2}=0$$
is considered fully nonlinear, but not semilinear.
By definition, fully nonlinear means the equation is nonlinear in its highest-order terms. But the highest-order terms for this equation are in the $\Delta u(x,t) = \sum_{i=1}^{n} \frac{\partial ^2u}{x_{i}^2}$ , which are linear.
A semi-linear PDE reads
$
\mathcal Lu=F(u),
$
where $\mathcal L$ is a linear operator and $F$ is a function.
A quasi-linear PDE with order $m$ reads
$
\mathcal L\bigl((\partial_x^\alpha u)_{\vert \alpha\vert\le m-1}\bigr) u=F\bigl((\partial_x^\alpha u)_{\vert \alpha\vert\le m-1}\bigr)
$ where
$\mathcal L\bigl((\partial_x^\alpha u)_{\vert \alpha\vert\le m-1}\bigr) $ is a linear operator of order $m$ whose coefficients are functions of $(\partial_x^\alpha u)_{\vert \alpha\vert\le m-1}$ and $F$ is a function.
Your equation is not semi-linear, but quasi-linear.
In my experience, fully nonlinear refers to an equation of the form $F(x,Du,D^2u)=0$ such that the linearization at $u=u_0$ is a linear equation of the required type (elliptic, hyperbolic etc.). In the example you mention I think the standard would be to call it a quasi-linear equation, and I am quite sure some people would call it semilinear since the nonlinear term does not depend on second order derivatives. Let me add that there is no universally accepted terminology for this, and in many cases the description depends on the problem (and its history).
|
2025-03-21T14:48:31.426530
| 2020-07-03T17:07:51 |
364759
|
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|
Stack Exchange
|
Let $T$ be a maximal torus of $SU(k+1)$. Who is the normalizer $N(T)$ of $T$ in $O(2k+2)$?
I'm reading an article which I cannot understand a paragraph very well.
$T$ is a maximal torus of $SU(k+1)$ acting linearly on $\mathbb{C}^{k+1}$. And here is what is written that I cannot fully understand:
Note that the normalizer $N=N(T)$ of $T$ in $O(2k+2)$ normalizes the
centralizer $C(T)$ of $T$ in $O(2k+2)$. Assume $k\geq 2$. Since $C(T)$ is a maximal
torus of $U(k+1)$ which is also a maximal torus of $O(2k+2)$, we see that
$C(T)$ is the identity component of the normalizer $N$. Moreover, $N$ is
generated by $C(T)$ , the complex conjugation $c$ and the symmetric group
$S_{k+1}$ of permutations of complex coordinates.
First of all, I cannot see why $C(T)$ should be a maximal torus of $U(k+1)$. I can see why it should be connect, since it is the union of maximal tori in $U(k+1)$ containing $T$. My supervisor also said that it is because a element in $C(T)$ lies on $N(T)$, and we can assume, using complex conjugation $c$, that every element $n\in N(T)$ is complex-linear on $\mathbb{C}^{k+1}$ (and I cannot see why). And that, up to permutation, $n$ preserves the decomposition $\mathbb{C}^{k+1}=\mathbb{C}\oplus\cdots\oplus \mathbb{C}$ (which I also cannot see why). And furhter he said that such an element must Lie on the maximal torus $S^1\times\cdots\times S^1\subset U(K+1)$; factorization compatible with the decomposition $\mathbb{C}^{k+1}=\mathbb{C}\oplus\cdots\oplus \mathbb{C}$ (and, again, I cannot see why).
Given this, why should $C(T)$ be the indentity component of $N(T)$?
Why don't you calculate? Take $k=2$. Take the diagonal maximal torus in ${\rm SU}(3)$; it is 2-dimensional. Compute its centralizer in $U(3)$; it is the 3-dimensional diagonal torus. Clearly this 3-dimensional torus is a maximal torus in $U(3)$ and a maximal torus in ${\rm SO}(6)$. You continue calculations....
I have calculated it to the usual torus in SU(n), but I would like to understant it more geometrically/generally, for an arbitrary maximal torus of $SU(n)$. I know that all need can be stracted from those calculations, but I would like to understand the techinics used by the author, which are much more general.
It is the same! You take an arbitrary maximal torus $T\subset {\rm SU}(n)$, you consider its natural complex representation in $V:={\Bbb C}^n$, this representation uniquely decomposes in a direct sum of one-dimensional complex representations, you take eigenvectors as basis vectors, and you get the "usual" diagonal torus. Then you should calculate!
|
2025-03-21T14:48:31.426708
| 2020-07-03T17:08:18 |
364760
|
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|
Stack Exchange
|
Cohomology of a family of twisted cubic curves (Hartshorne III, 12.9.2)
I'm trying to understand following Example in Hartshorne (Chapter III, Example 9.8.2
& Example 12.9.2):
Let $X_1 \subset \mathbb{P}^{3}$ be a twisted cubic curve not containing the point
$(0:...:1) $ (will become important when we project $X_0$ to $\mathbb{P}^{2}$)
and consider the automorphisms parametrized by
$\sigma_t, t \in k$ of $\mathbb{P}^{3}$ defined by $(x_0:...:x_3) \mapsto
(x_0:...:t \cdot x_3)$. Let $X_t:= \sigma_a(X_t)$. For $t \in k \neq 0$ all
$X_t$ are isomorphic as abstract schemes to $X_1$. Then the
$X_t$ form a flat family parametrized by $\mathbb{A}^1 - \{0\}$.
Now according to (9.8) this family extends uniquely to a flat family defined
over all of $\mathbb{A}^1$, and with $X_0$ happen crazy things.
Following Hartshorne it can be proved (for details see book pages 259/260 )
that set theoretically $X_0$ coincides with
$X_1$ but as a scheme $X_0$ admits an embedded component $(0:0:0)$, in this case
it is a "double point" and the concern of this question is how does it effect
on cohomology.
Recall that the twisted cubic curve in $\mathbb{P}^{3}$ was defined by the
parametric equations $x_0 = t^3 , x_1 = t^2 u, x_2 = tu^2 , x_3 = u^3$ . In other words,
it is just the $3$-uple embedding of $\mathbb{P}^{1}$ in $\mathbb{P}^{3}$ (Chap. II Example 7.8.5 and Chap I, Ex. 2.12).
Example 12.9.2 (page 289) contains some things about the cohomology of the members of the
family $X_a$ I not pretty understand. It says:
Example 12.9.2. In the flat family of (9.8.4), we have $h^0 (X_t, O_{X_i}) = 1$ if
$t \neq 0$, and $2$ if $t = 0$, because of the nilpotent elements. On the other
hand, $h^1 (X_t, O_{X_t}) = 0$ for $t \neq 0$, since $X$, is rational,
and $h^1 (X_t, O_{X_t}) =h^1 (X_t, O_{X_t})_{red})=1$
since $(X_0)_{red}$ is a plane cubic curve. So in this case the
functions $h^0 ,h^1$ both jump up at $t = 0$.
Several caclulations I not completly understand and would like to discuss:
I) on $h^0 (X_t, O_{X_t}) = 1$
Since $X_1 \cong X_t$ for $t \neq 0$ we can assume $t=1$. $X_1$ is a twisted cubic
curve of degree $3$ embedded by $3$-uple Veronese $\nu: \mathbb{P}^{1} \to \mathbb{P}^{3}$.
The first idea came up to me is to use the overkiller $X_1 \cong \mathbb{P}^{1}$ and to think only
about the cohomology of $\mathbb{P}^{1}$. I think that Hartshorne used at this point
another argument to reach $h^0 (X_t, O_{X_t}) = 1$, because looking at the used argument
for $h^1 (X_t, O_{X_t}) = 0$ where $t \neq 1$ he only used that $X_1$ is rational,
that is birational equivalent to $\mathbb{P}^{1}$, not neccessary isomorphic.
And indeed genus $g_1=h^1 (X_1, O_{X_1})$ is by Theorem 8.19 a birational invariant.
This leads me to suspicion that to establish $h^0 (X_1, O_{X_1}) = 1$
Hartshorne used another argument. Which one?
I'm looking for an argument using cohomology theory that for a rational, non singular irreducible
cubic curve of degree $3$ in $\mathbb{P}^3$ the space of global sections
has dimension one?
II) on $h^0 (X_0, O_{X_0}) = 2$
How the fact that $X_0$ is not reduced anymore rases the the dimension to $2$? More specificaly: the $X_0$ difers from $X_1$ by the observation the point $(0:0:0) \in X_0$ in an embedded point, $O_{X_0, (0:0:0)}$ has nilpotent elements. How to see formally that these contribute exactly an additional one dimension to the global sections:
that is $h^0 (X_0, O_{X_0})= h^0 (X_1, O_{X_1})+1$?
III) $h^1 (X_t, O_{X_0}) =h^1 (X_t, O_{X_t})_{red})$
Why passing from $X_0$ to associated reduced $(O_{X_0})_{red}$ not changes $h^1$?
(recall that the birationality argument not work anymore since $X_0$ isn't smooth.)
A smooth connected projective curve has no nontrivial functions. The question is not appropriate for this site, which is for research level questions.
To be fair, I have known plenty of research-level mathematicians who flinch at questions from Hartshorne's book.
@RP_: Certainly, but probably not at determining $H^0(C,\mathscr{O}_C)$ for a smooth projective curve $C$.
I have to admit that the first question is far from
beeing a research one. I was desperate at that time, sorry. The answer
I meanwhile found is quite simple because $X_1$ is rational
we can find two rational maps $X_1 \dashrightarrow P^1$ and
$P^1 \dashrightarrow X_1$ which can be uniquely extended to usual
morphisms since $X_1$ and $P^1$ are regular. As both maps have dense
images the induced maps on global sections are injective, that is
we obtain a chain of inclusions
$k=H^0(P^1, O_{P^1}) \subset H^0(X_1, O_{X_1}) \subset
H^0(P^1, O_{P^1})$. Thus $h^0(X_1, _{X_1})=1$ and I) is solved. I mean that this should be the shortest argument.
I made also some thoughts about II and III and whould be very
thankful if somebody could look through and tell me if my approach is ok.
We know that $(X_0)_{red}= X_1$ and $X_0$ is everywhere reduced exept in
$(0:0:0)$. That is $X_1$ can be considered as closed subscheme of $X_0$ and
the structure sheaves of $X_1$ and $X_0$ are related to each other
by exact sequence
$$0 \to J \to O_{X_0} \to O_{(X_0){red}}=O{X_1} \to 0$$
The ideal sheaf $J$ has support only at $(0:0:0)$ so it's support is
zero dimensional and applying long cohomology sequence
to the sheaf
sequence above we conclude III) since $h^1(X_0, J)=0$. Only problem is
why $h^0(X_0, J) \cong J_{(0:0:0)}$ is one-dimensional $k$-space?
$O_{X_0,red}$ is not $O_{X_1}$. $H^1(X_0, J) = H^2(X_0, J) = 0$ suffices.
@Functor: wait, I claimed not $O_{X_0,red}= O_{X_1}$ - former would be just a $O_{X_0}$-algebra locally given by $R \to R/(Rad(R))$ - but that $O_{(X_0){red}}=O{X_1}$ and that just follows from definition $X_1:=(X_0){red}$. But with what I agree, is that the above sequence involves abusion of notation, correctly it should be given as $$0 \to J \to O{X_0} \to j_*O_{(X_0){red}}=j*O_{X_1} \to 0$$ where $j:(X_0)_{red} \to X_0$ canon closed embedding. But of course what you wrote about cohomology groups is the correct argument.
|
2025-03-21T14:48:31.427106
| 2020-07-03T17:25:32 |
364762
|
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|
Stack Exchange
|
An explicit negative solution to the Lüroth problem for non-algebraically closed fields
Let $\mathsf{k}$ be a field of characteristic $0$, and consider $\mathsf{k}(x,y)$.
If $\mathsf{k}$ is algebraically closed, then every field $L$ such that the inclusion $\mathsf{k} \subset L \subset \mathsf{K}(x,y)$ holds is a purely transcendental extension of the base field (i.e., Castelnuovo's Theorem implies a positive solution to the Lüroth problem in two dimensions).
Now suppose that $\mathsf{k}$ is not algebraically closed.
Question: can we have a finite group $G$ of field automorphisms of $\mathsf{k}(x,y)$, fixing $\mathsf{k}$, such that $\mathsf{k}(x,y)^G$ is not a purely transcendental extension of $\mathsf{k}$?
I am looking for an explicit example of $G$ such that $\mathsf{k}(x,y)^G$ is not rational.
@PaceNielsen yes I wanted the field fixed. I wil edit the question. Nice argument though!
In positive characteristic if you allow finite group schemes, there are such examples with field of invariants general type.
@Mohan this is very interesting. Do you have an easy example/reference for such an example?
I think they are called Zariski surfaces. May be you could look them up.
This version of the Luroth problem for the fields of invariants is sometimes called the Noether problem: https://encyclopediaofmath.org/wiki/Noether_problem#:~:text=The%20question%20of%20the%20rationality,a%20field%20of%20rational%20functions.&text=In%20general%2C%20the%20answer%20to%20Noether's%20problem%20is%20negative.
According to the first paragraph in Shafarevich's paper "On Luroth's problem" (found here http://www.math.ens.fr/~benoist/refs/Shafarevich.pdf) the field of rational functions on the surface $z^2+y^2=x^3-x$ over $\mathbb{R}$ is an example of a non-rational field $F$, containing $\mathbb{R}$ of transcendence degree $2$, that embeds in $\mathbb{R}(u,v)$ (fixing $\mathbb{R}$). I don't know whether or not this embedding can be chosen so that $\mathbb{R}(u,v)/F$ is Galois.
This gives an example of a non-rational subfield of $k(x,y)$, but does it arise as the fixed field under some group $G$ of automorphisms of $k(x,y)$? (I don't think that is automatic, is it?)
I second that. It does not seem clear to me how this field is of the form $\mathbb{R}(x,y)^G$ for an appropriate finite $G$. But I might be missing something.
Just to amplify: of course it is not automatic for a subfield $L$ of $k(x,y)$ (with the degree of the extension assumed finite) to be the field of invariants under some group $G$. If it were, then it would have to be Galois, but e.g. $k(x^3,y) \subset k(x,y)$ is not a Galois extension if $k$ does not contain a primitive 3rd root of unity.
So the next step would probably be to work out the embedding that Shafarevich claims exists, find its degree, and check if it is Galois.
@PaceNielsen thank you very much for suggestion. I will check if the embedding can be made as a Galois extension. Luckly, yes.
Geometrically, the question translates into degree of unirationality of this surface. Namely, if there is a degree two dominant rational map from $\mathbf{P}^2$ to this cubic, then the corresponding field extension would automatically be Galois. Typically, degree two parametrization of cubic hypersurfaces comes from a line defined over a ground field, as in http://www.math.uni-bonn.de/~huybrech/Notes.pdf, Corollary 1.18. However in this case the only line I see (at infinity) is passing through singular points, and the construction doesn't work.
Higher degree unirational parametrizations of cubics with a rational point are explained in https://arxiv.org/pdf/math/0005146.pdf.
|
2025-03-21T14:48:31.427360
| 2020-07-03T17:26:03 |
364763
|
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|
Stack Exchange
|
Cusp forms with integer Fourier-coefficients
Let $S_k(\Gamma_1(N))(\mathbb{Z})$ be the set of modular forms of weight $k$ and level $N$ with integer Fourier coefficients. Then is true that any cusp form can be written as $\mathbb{Q}$ linear combination of Hecke eigenforms with integer coefficients ?
No, very few eigenforms can be scaled to have integer FCs. However, given any eigenform scaled so its coefficients are algebraic integers, you can take the "trace" by summing up all conjugate forms to get a (typically non-eigen) form with integer coefficients.
Related: https://mathoverflow.net/questions/364787/hecke-eigenform-with-integer-fourier-coefficients
No. Take $k = 24$ and $N = 1$. Then $\Delta^{2} = q^{2} - 48q^{3} + 1080q^{4} + \cdots \in S_{K}(\Gamma_{1}(N),\mathbb{Z})$. However, if we write $\Delta^{2} = c_{1} f_{1} + c_{2} f_{2}$, where $f_{1}$ and $f_{2}$ are the Hecke eigenforms (with coefficients in $\mathbb{Q}(\sqrt{144169})$) then (if we order $f_{1}$ and $f_{2}$ appropriately), $c_{1} = \frac{1}{24 \sqrt{144169}}$ and $c_{2} = -\frac{1}{24 \sqrt{144169}}$.
A charming, decisive, minimalist example! :)
|
2025-03-21T14:48:31.427454
| 2020-07-03T18:13:29 |
364769
|
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|
Stack Exchange
|
Finding a non-negative similar matrix
Let $B$ be some matrix. Is there a way to decide whether there exists an invertible matrix $P$, such that the matrix:
$$
A = P^{-1}BP
$$
Is non-negative? (That is - all the entries in the matrix are non-negative)
Basically - given a linear transformation, I want to find a different base, in which the transformation matrix of this linear transformation is non-negative.
For example, for the matrix:
$$
B=\begin{bmatrix} 0 & -1 \\ 1 & 2
\end{bmatrix}
$$
I have the matrix:
$$
P=\begin{bmatrix} 1&0\\ 1&1
\end{bmatrix}
$$
And
$$
P^{-1}=\begin{bmatrix}1&0\\ -1&1
\end{bmatrix}
$$
And:
$$
A=P^{-1}BP=
\begin{bmatrix}1& 0\\ 1 &1
\end{bmatrix}
$$
Is there a way to efficiently decide if such matrix $P$ exists, and to find it?
I assume that you are talking about real matrices. Well, notice that, if such an $A$ exists, then trace($B^{m}$) is non-negative for all non-negative integers $m$.
a necessary condition is that $B$ has a real nonnegative eigenvalue $\lambda$, and any other eigenvalue $\lambda'$ satisfies $|\lambda'|\leq\lambda$ (Perron-Frobenius theorem)
In the 2 by 2 case, if $tr(B) \geq 0$ and $B$ has real eigenvalues, then it is similar to a nonnegative matrix (symmetric, even).
There are necessary and sufficient conditions so that some power of $B$ is conjugate to a matrix with strictly positive entries, but this is not what you are are looking for. In general, conjugacy to a primitive matrix (a matrix with all entries nonnegative, and some power of the matrix is strictly positive) is a very difficult prolem.
|
2025-03-21T14:48:31.427603
| 2020-07-03T18:44:48 |
364772
|
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|
Stack Exchange
|
Paradox in additive combinatorics
Let $S$ be an infinite set of positive integers. Let us define the following quantities:
$N_S(z)$ is the number of elements of $S$, less or equal to $z$
$r_S(z)$ if the number of positive integer solutions to $x+y\leq z$,
with $x,y\in S$ and $z$ an integer
$t_S(z)$ if the number of positive integer solutions to $x+y= z$,
with $x,y\in S$ and $z$ an integer
We assume here that
$$N_S(z) \sim \frac{a z^b}{(\log z)^c}$$
where $a,b,c$ are positive real numbers with $b\leq 1$. This covers primes, super-primes, squares and more.
We have:
$$r(z)\sim \frac{a^2\Gamma^2(b+1)}{\Gamma(2b+1)} \cdot \frac{z^{2b}}{(\log z)^{2c}}$$
$$r'(z)\sim \frac{a^2\Gamma^2(b+1)}{\Gamma(2b)} \cdot \frac{z^{2b-1}}{(\log z)^{2c}}$$
For details about these results, see my previous MO question, here. For super-prime numbers, see this OEIS entry, and especially this paper. I mentioned earlier, and this seems to be a well known and trivial fact, that $t(z) \sim r'(z)$ on average.
Barring congruence restrictions, a conjecture states that if $r'(z) \rightarrow \infty$ as $z\rightarrow \infty$, then almost all large enough integer $z$ can be written as $z=x+y$ with $x,y\in S$. I will call this conjecture A. Because of congruence restrictions, I worked with pseudo-primes instead of primes. They are generated as follows. A positive integer $k$ belongs to $S$ (set of pseudo-primes) if and only if $R_k < N'_S(k)$ where the $R_k$'s are independent random deviates on $[0, 1]$. Here
$$N'_S(z) \sim \frac{abz^{b-1}}{(\log z)^c}.$$
Note that $N'_S(z)$ is the asymptotic derivative of $N_S(z)$.
Examples:
For pseudo-primes, $a=b=c=1$.
For pseudo-super-primes, $a=b=1, c=2$.
For pseudo-super-super-primes, $a=b=1, c = 3$.
For my test power set, $a=1, b= \frac{2}{3}, c=0$.
Pseudo-super-primes are extremely rare compared to primes, yet all but a finite number of integers can be expressed as the sum of two pseudo-super-primes. This is compatible with results obtained here and intuitively, it makes sense. Pseudo-super-super-primes are even far more rare, and here the conjecture A seems to fail: it looks like not only a large chunk of integers can not be written as the sum of two pseudo-super-super-primes, but these exceptions seem to represent the immense majority of all positive integers. Now the paradox.
Paradox
My test power set (see definition in the example section) consists of integers that are even far more rare than pseudo-super-super-primes, yet for them conjecture A works, as expected. Perhaps this is caused by the fact that these integers are far more abundant than pseudo-super-super-primes among the first million integers, but asymptotically they become far less abundant than pseudo-super-super-primes.
My question
How do you explain my paradox? Is conjecture A wrong? Or is it possible that if you look at extremely, massively large integers (probably well above $10^{5000}$), they can always be expressed as a sum of two pseudo-super-super-primes despite the fact that the opposite is true for smaller integers that only have a few hundreds digits?
Update: I posted a new MO question suggesting that there is no paradox. See here.
What is the definition of $r'(z)$? You don't state what you mean by that.
$r'(z)$ is the derivative of $r(z)$, and $r(z)$ is asymptotically equivalent to the number of solutions in positive integers to $x+y\leq z$ with $x,y\in S$. By asymptotically, I mean when $z\rightarrow\infty$. – Vincent Granville 7 hours ago
I must be missing something here. $r(z)$ is not a continuous function so how can it have a derivative? The right-hand side of the asymptotic does have a derivative. Is there something going on with a discrete analog like the finite difference method?
$r(z)$ is the asymptotic continuous version of the function counting the number of solutions to $x+y\leq z$, and it is assumed here to be equal to $a z^b/(\log z)^c$, and $r'(z)$ is its derivative.
Also, $r′(z)\sim r(z)-r(z−1)\sim dr(z)/dz$. See how I derived all these results, in a previous MO question: https://mathoverflow.net/questions/363055/goldbach-conjecture-and-other-problems-in-additive-combinatorics/
What is your reason for believing conjecture A fails for pseudo-super-super-primes? Is it just computational? If so then the likeliest explanation is that the computational evidence is misleading and conjecture A does hold for large enough integers. The existence of small counterexamples for results that hold for all sufficiently large integers is a well-known phenomenon in additive number theory (e.g. see Waring's problem), and the sparser the sets you're working with, the larger the threshold for the 'eventual' behaviour to kick in would be.
@Thomas. Yes I agree with you, glad that one more time this "conjecture A" is believed to be true. My guess is that my problem is purely computational, dealing with too small numbers. I worked more on this, and I have more and more theoretical evidence that conjecture A is correct.
I wrote:
My test power set (see definition in the example section) consists of integers that are even far more rare than pseudo-super-super-primes, yet for them conjecture A works, as expected. Perhaps this is caused by the fact that these integers are far more abundant than pseudo-super-super-primes among the first million integers, but asymptotically they become far less abundant than pseudo-super-super-primes.
Indeed, that's the explanation. If you check my new MO question here, you have the following result. Let us denote as $w(z)$ the number of positive integers less or equal to $z$ that can not be written as $z=x+y$, with $x,y \in S$. These integers are called exceptions. We have
$$w(z) \approx \int_0^z \exp\Big(-\frac{1}{2} r'(u)\Big)du.$$
The total number of exceptions (unless small, say $<50$) is well approximated by $w(\infty)$ when averaged over a large number of sets $S$ that have the same statistical distribution of elements. And because $b>\frac{1}{2}$ we have $w(\infty)<\infty$. Of course the range varies greatly across multiple sets, but it is correct on average.
In particular,
For pseudo-super-primes, $w(\infty) \approx \int_2^\infty
\exp(-u\cdot(\log u)^{-4})du \approx 26341$.
For pseudo-super-super-primes, $w(\infty) \approx \int_2^\infty
\exp(-u\cdot (\log u)^{-6})du > 10^7$. Still, it is finite.
For my test power set, $w(\infty) \approx 65$ (see here).
Note that I used $\int_2^\infty$ instead of $\int_0^\infty$ due to a singularity at $1$ that should be ignored.
|
2025-03-21T14:48:31.428095
| 2020-07-03T18:45:54 |
364773
|
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|
Stack Exchange
|
Cellular chain complex of $G$-CW-complexes & their differentials
I not completly understand EXAMPLE 2.31 (page 19) dealing with homology of
$G$-CW-complexes. Source: http://www.ltcc.ac.uk/media/london-taught-course-centre/documents/LTCC-notes-Lecture3-2019.pdf
Recall, the cooking recipe for cellular homology
in case classical CW-complexes without any $G$-structure.
$X$ is $k$-dimensional CW-complex if
we have a decomposition $X_0 \subset X_1 \subset ... \subset X_{n-1} \subset X_n=X$
where every $X_{i+j}$ arises as pushforward of $X_j$
after attatching $I_j$ many $j$ cells $\cong D^j$ along $S^{j-1} \subset D^j$ to $X_j$.
The we build a cellular chain complex
$$C_k(X) \to C_{k-1}(X) \xrightarrow{\text{d}} ... \to C_1(X) \xrightarrow{\text{d}} C_0(X) \to 0$$
where $C_j := \bigoplus_{i \in I_j} \mathbb{Z}$ is free $\mathbb{Z}$-module of rank $I_j$ and $d$-degree maps. Then compute homology
of this complex.
My question is if we now would like to compute cellular homology of a $G$-CW-complex
how the $C_j(X)$ would look like. Would they also be free $\mathbb{Z}$-modules or
will they have additional structure? Maybe as $\mathbb{Z}G$-modules?
The reason why I worry about this is following example 2.31:
Consider EXAMPLE 2.31. We deal with finite cyclic group $G:= \mathbb{Z}/n$ that act on circle $S^1$ considered
as $G$-CW-CW-complex with $n$ vertices ($0$-cells) and $n$ $1$-cells.
The obviuos action by $G$ is that the generator $t := \bar{1}$ of
$G$ shift the $1$-cells resp
$0$-cells conterclockwise:
Then in the example following complex is considered:
$$\mathbb{Z}G \to \mathbb{Z}G \to \mathbb{Z} \to 0$$
Is this the analogon of cellular complex $C_k(X) \xrightarrow{\text{d}} ... \xrightarrow{\text{d}} C_0(X) \to 0$ from above? Then the message is $C_1(X)=C_0(X)= \mathbb{Z}G$?
Set theoretically that's fine since the number of $1$- and $0$-cells coinsides
with $\vert G \vert$ but as what are $C_j(X)$ and the cellular complex are considered?
As $\mathbb{Z}$-modules and $\mathbb{Z}G$-structure play no role for homology or as a proper $\mathbb{Z}G$-complex? And what are the differentials $d$ here? Maps of $\mathbb{Z}$ or $\mathbb{Z}G$-modules?
More generally, let $X$ is $k$-dimensional $G$-CW-complex if
we have a decomposition $X_0 \subset X_1 \subset ... \subset X_{n-1} \subset X_n=X$
of $G$-cells and assume that this $G$-CW has $I_j$ many $j$ cells.
What is $C_j(X)$ explicitly and again what are the $d$-maps of the corresoponding cellular chain complex
$$C_k(X) \xrightarrow{\text{d}} C_{k-1}(X) \xrightarrow{\text{d}} ... \to C_1(X) \xrightarrow{\text{d}} C_0(X) \to 0$$
Are they maps of $\mathbb{Z}$ or $\mathbb{Z}G$-modules?
The $G$-action on $X$ makes $C_\ast(X)$ into a chain complex of $\mathbb ZG$-modules. In general I recommend Brown's bible "Cohomology of Groups".
@ChrisGerig: Thank you for the reference. Page 14 provides precisely the answer I was looking for! That is every $C_j(X)$ is a free $\mathbb ZG$ where the number of generator coinsides with number of orbits because $G$-action on cells is free
|
2025-03-21T14:48:31.428284
| 2020-07-03T20:27:29 |
364779
|
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"erz",
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|
Stack Exchange
|
Going from a class of path functions to a topology
Someone asked a version of this 10 years ago, but no satisfactory answer was given, so I want to try again.
I have a class of functions F from the reals to a set S (the specifics are complicated and don't matter). I'm curious to know whether one can find a metric topology or any topology T such that F is exactly the set of continuous functions from the reals to S relative to T. So I'd be interested in either a strong set of necessary conditions (to prove that there isn't any such topology) or a weak set of sufficient conditions (to prove that there is).
Here are some necessary conditions of the sort I'm thinking of:
Constant: Every constant function is in F.
Composition with continuous functions: If f is in F and g:R->R is continuous then f(g(t)) is in F.
Splicing: If f and g are in F and f(0) = g(0), then the function h which is equal to f on negative numbers and equal to g on non-negative numbers is also in F.
Separation: (for Hausdorff topologies). If f is in F, u and v are in S, u!=v and f(0)=u then there is some open interval I around 0 such that v is not in f(I).
Can any one suggest any good theorems of this kind?
how about if $f(t)=g(t)$, for $t\in A$, then $f(t)=g(t)$, for $t\in \overline{A}$?
|
2025-03-21T14:48:31.428401
| 2020-07-03T20:28:16 |
364780
|
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|
Stack Exchange
|
Is there a conjectured uniform Lindelof hypothesis for Hurwitz zeta functions
Consider $\zeta(s, a) = \sum_{n=1}^{\infty} (n+a)^{-s}$ (alternatively, consider its functional equation Dirichlet series $\sum_{n=1}e(a n) n^{-s}$). What is the expected growth-rate of $\zeta(1/2 + i T, a)$ for $a$ away from 0?
When $a=\frac{m}{n}\in\mathbb{Q}$ these are finite linear combinations of Dirichlet L-functions mod $n$. Hence, assuming GRH, $\zeta(1/2 + i T, m/n) \ll_n T^{\epsilon}$.
If the Lindelof hypothesis is a phenomenon of Euler products, arithmetic, zero-free regions, then one could expect that the implied constant grows large with $n$, and that there is no such bound for irrational $a$.
If the Lindelof hypothesis is an analytic phenomenon, one could expect that $\zeta(1/2 + i T, a)\ll T^{\epsilon}$ uniformly for $a$ away from $0$.
Are there any computations or conjectures pointing one way or the other? Or other ways that we do or not have nice control of the behaviour with respect to $a$.
One comment on Lindelof being an analytic phenomenom. As far as I know there is "nothing" that is known or conjectured to fail Lindelof. To make this precise, Dirichlet series $\sum a_n n^{-s}$ with $a_n\ll n^\epsilon$ that extend to an entire function with polynomial growth on vertical lines (this class includes all L functions) all appear to satisfy the Lindelof hypothesis, and have sub-polynomial growth to the right of $1/2$. The bound on the coefficients is meant to exclude counterexamples like $\zeta(2s-1)$.
Indeed, linear combinations of functions satisfying Lindelof will also satisfy Lindelof, but rarely have Euler products. Some can be arranged to retain nice functional equations, but this seems un-necessary.
Right. There provably cannot be non-trivial linear combinations between things with either Euler products, or satisfying a Riemann hypothesis. On the other hand, the exponential twists above form a (continuous) basis for everything with a degree 1 growth rate, so Lindelöf for them gives Lindelöf for all other small Dirichlet series. I don’t know of an analogous continuous family of higher degree functions.
|
2025-03-21T14:48:31.428569
| 2020-07-03T20:52:50 |
364781
|
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|
Stack Exchange
|
Given a chain of commuting matrices over the complex numbers, can we build one over the real numbers?
Suppose we have two $n\times n$ matrices $A$ and $B$ with entries in $\mathbb{R}$, and two non-scalar matrices $X$ and $Y$ with entries in $\mathbb{C}$, such that $AX=XA$, $XY=YX$, and $BY=YB$.
Is it necessarily the case that there exist non-scalar matrices $X'$ and $Y'$ with entries in $\mathbb{R}$ such that $AX'=X'A$, $X'Y'=Y'X'$, and $BY'=Y'B$?
(Here "non-scalar" just means that the matrices aren't scalar multiples of the identity matrix.)
Changing $X$ into $X-(Tr(X)/n)I_n$ and similarly for $Y$, this can be restated as follows: define $V$ as the set of pairs $(X,Y)$ with the given condition and with trace zero. If $V$ contains $(X,Y)$ both nonzero, does it contain a real point with the same condition? I have no idea whether this can be useful, but at worst it's a harmless comment.
Just a comment (hence in the right place): the obvious thing to do is to put $X' = X + \overline X$ and $Y' = Y + \overline Y$, but that might be scalar, so use $X' = i(X - \overline X)$ and/or $Y' = i(Y - \overline Y)$ if necessary; but then of course it's no longer obvious that $X'$ and $Y'$ commute. If $X$ is semisimple, then the centraliser of either choice for $X'$ is both larger than that of $X$, and closed under conjugation, hence contains either choice for $Y'$; and similarly for $Y$; but I don't see how to reduce to this case (since $X_\text s$ or $Y_\text s$ could be scalar).
In fact, this shows that you can reduce to the case where $X$ and $Y$ are both nilpotent, which means, since they commute, they can be simultaneously conjugated to strictly upper triangular elements; but I think that this conjugation cannot also be assumed to keep $A$ and $B$ real.
Update: it turns out the answer is no! See for instance Example 4.14 here.
|
2025-03-21T14:48:31.428717
| 2020-07-03T22:28:58 |
364784
|
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|
Stack Exchange
|
Extension of a theorem of Bisch to cyclotomic integers of fixed degree
Theorem 3.2 in this paper (Bisch, 1994) states that for a finite depth ${\rm II}_1$ subfactor of integral index, every intermediate subfactor are also of integral index. As an application, every such subfactor of prime index must be maximal, moreover, many possible indices of such subfactors can be ruled out as follows: let $\alpha_1, \dots , \alpha_n > 0$ such that $\prod_i \alpha_i \in \mathbb{N}$, then one among the $\alpha_j \not \in \mathbb{N}$ cannot be the index of such a subfactor (proof: if all of them are, you get a contradiction with Bisch's theorem by considering a direct product of subfactors). For example, there is no such subfactor of index $n^{p/q} \not \in \mathbb{N}$ (where $n, p, q \in \mathbb{N}^{*})$, or there is none of index $n(3-\sqrt{5})$, because there are ones of indices $2$ and $\phi =(3+\sqrt{5})/2$, but $2 \cdot \phi \cdot n(3-\sqrt{5}) = 4n \in \mathbb{N}$. So we can a lot retrict the set of possible such indices like that (I don't know how much).
Now there is an other way to make such restrictions coming from fusion categories, using a theorem in this paper (Etingof-Nikshych-Ostrik, 2005), which implies that the index of such subfactors must be a cyclotomic integer, i.e. an element of $\mathbb{Z}[c_d]$ for some integer $d$, with $c_d=2\cos(2\pi/d)$.
Question: Can Bisch's theorem be extended to cyclotomic integers of fixed degree (i.e. if the index of a finite depth ${\rm II}_1$ subfactor is in $\mathbb{Z}[c_d]$, then so is for the index of the intermediate subfactors)?
As an application, we could get even more restrictions on possible indices of such subfactors, moreover, if such a subfactor is of index $\alpha \in \mathbb{Z}[c_d]$ and if $\beta$ is the index of an intermediate subfactor then $\beta\mathbb{Z}[c_d]$ would be a divisor ideal of $\alpha\mathbb{Z}[c_d]$; in particular, if $\alpha\mathbb{Z}[c_d]$ is a prime ideal, then the index of intermediate subfactors would only be units of $\mathbb{Z}[c_d]$ or units multiple of $\alpha$.
At first glance it looks like Bisch's argument applies without modification to the ring of integers in any field. Am I missing something?
@NoahSnyder: He used the following which should not hold for non-integral index subfactor: << We prove a simple criterion for finite depth of subfactors $N$ of a ${\rm II}_1$ factor $M$ with integer index. Such subfactors arise for instance from spin- and vertex model commuting squares >>.
The sentence you quote isn't part of the proof.
@NoahSnyder: Yes. Then a stronger statement should hold (but perhaps more difficult to apply): the index of the intermediates of such a subfactor of index $α$ must be in the ring of integers of $\mathbb{Q}(α)$. Why Dietmar did not write such a fundamental result in his paper if its proof is is exactly the same as for the rational integers?
Because it's a paper about when indices can be rational non-integers?
@NoahSnyder: According to my understanding of Bisch's proof, its result extends to every "integrally closed" ring of algebraic integers (as you wrote), and is false in general in the non-"integrally closed" case, because $3+\sqrt{5} \in R=\mathbb{Z}[\sqrt{5}]$ which is not integrally closed whereas $3+\sqrt{5} =2\frac{3+\sqrt{5}}{2}$ (product of index of two TLJ subfactors) and $\frac{3+\sqrt{5}}{2} \in \mathbb{Z}[\frac{1+\sqrt{5}}{2}]$ the integral closure of $R$.
|
2025-03-21T14:48:31.429319
| 2020-07-03T22:59:11 |
364787
|
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|
Stack Exchange
|
Hecke eigenform with integer Fourier coefficients
Is it true that for any even $k$ and $N,$ there always exist a Hecke eigenform with integer Fourier co-efficient of weight $k$ and level $N$ ?
Please use a high-level tag like "nt.number-theory". I added this tag now.
https://www.lmfdb.org/
Are you asking about holomorphic modular forms? Cusp forms? In either case, no as $M_2(1) = 0$. If you exclude this example and allow Eisenstien series, then yes.
There are lots of Eisenstein series with integer coefficients, as several people have pointed out. CM forms (coming from Hecke characters over im. quad. fields) also often have small coefficient fields. On the other hand, I believe it's expected that if you fix an $N$ and look at cuspidal, non-CM eigenforms of level $N$, trivial character and weight $k$, then the degrees of the coefficient fields go to $\infty$ with $k$. The LMFDB database that @Pasten links to is a great way of getting a feel for this -- you can search by degree of coefficient field.
@DavidLoeffler This is also my feeling, but I don't know how to prove that even in the simplest case. E.g. how can one show that the only level $1$ eigenforms with $\Bbb Q$-coefficients are those of weight $12, 16, 18, 20, 22, 26$? Also it seems that for level $1$ and any weight, the coefficient field always has degree equal to the dimension (which implies the previous result). Is that true/proven? Heuristically I imagine that it's almost impossible for polynomials with such large coefficients to be reducible.
@WhatsUp The statement about level 1 modular forms is Maeda's conjecture. See here: https://orbilu.uni.lu/bitstream/10993/11499/1/%5Bbw%5DPTsaknias.pdf. There is lots of numerical evidence, everyone believes that it's true, and nobody has a clue how to prove it.
@DavidLoeffler Thanks for the reply, I now find lots of information by searching for Maeda's conjecture.
No, eg. $k$ odd and $N=1$. If $k=4m+6n$ then can't you just restrict an Eisenstein series for $N=1$ to a smaller group?
$k$ is even, sorry.
|
2025-03-21T14:48:31.429512
| 2020-07-03T23:53:19 |
364793
|
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"url": "https://mathoverflow.net/questions/364793"
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|
Stack Exchange
|
Integrability condition for flat connections
I am reading Kobayashi's book "Differential geometry of complex vector bundles". More precisely, I am on section 2 of chapter 1, page 5.
Kobayashi is trying to prove that if $E$ is a vector bundle on some manifold $M$, with a flat connection $D$, then it admits a "flat structure" $\{U,s_U\}$ which consists on an open cover of $M$ and a local frame of $E$ such that the transition functions are locally constant.
In order to do this, he starts with some arbitrary local frame $s'$ and looks for functions $a:U \rightarrow GL(r,\mathbb{C})$ such that in the frame $s_U= s' a$ the connection $1$-form is $0$.
Therefore, if $\omega'$ is the connection $1$-form in the frame $s'$, what he is trying to do is solve the following equation for $a$
$$
\omega' a + da = 0.
$$
He claims that solutions exists since the "integrability condition" for this equation is obtained by differentiating it
$$
0=(d\omega') a -\omega' \wedge da = (d\omega')a + (\omega' \wedge \omega')a = \Omega' a,
$$
which is true since we assumed that the connection is flat.
My question is what does he mean by the "integrability condition". Moreover, why is that the integrability condition for that equation? And, also, why can he use the fact $da=-\omega' a$ when computing it?
I think he might be using some form of the Frobenius theorem, since I know that it is what you use from a "global" point of view.
In any way, I want to know precisely in this context what he means by that "integrability condition", maybe it is just something basic or standard that I am missing.
It is the Frobenius theorem. You can see a long discussion of different ways to write the Frobenius theorem in my lecture notes. https://github.com/Ben-McKay/introduction-to-exterior-differential-systems/blob/master/main.pdf
On the manifold $X=U\times \operatorname{GL}_r$, with points written $x=(m,a)$, each tangent space $T_x X$ contains a linear subspace $V_x$ consisting of tangent vectors on which $a^{-1}da=-\omega'$. The problem is to prove that these $V_x$ spaces form a smooth subbundle $V \subset TX$, and that this subbundle is closed under bracket. For this we can take any framing $e_1,\dots,e_n$ of tangent vector fields on $U$, and associate to each the vector field $e'_i$ which projects to $e_i$ and satisfies $da=-\omega'a$. These are a basis for $V_x$, for each $x$, so $V$ is a smooth subbundle. Note that the $e_i'$ project to the $e_i$, so brackets project to the brackets. Bracket closure (i.e. we can apply the Frobenius theorem to $V$) is precisely flatness of the connection, using the equations $L_v \xi=d(i_v \xi)+i_vd\xi$ and $d\omega'(e_i,e_j)=L_{e_i}(\omega'(e_j))-L_{e_j}(\omega'(e_i))-\omega'([e_i,e_j])$.
I think there is one thing that you still need to prove. You have found that there exists some integral submanifold $Y\subset X$ with $T_x Y = V_x$ for any $x$, but not that this $Y$ is (at least locally) the image of a section of $p:X\rightarrow U$. In order to do this I guess you have to prove that at every point $p_*:V_x\rightarrow T_{p(x)} U$ is an isomorphism and then apply the inverse function theorem?
I guess my last statement is true since the form $da+\omega' a$ in $U\times GL_r$ is a pullback of a form in $U$, and $V_x$ is its kernel?
@G.Gallego: sorry I left so many gaps in the explanation; yes, the inverse function theorem, as you say.
I think I can give a more explicit proof of this fact.
Let us take $x^\nu$ coordinates on $U$ and write $\omega'=\sum_\nu A_\nu dx^\nu$ and $df = \sum_\nu \partial_\nu f dx^\nu$. Our differential equation then becomes the PDE
$$
\partial_\nu f(x) + f(x) A_\nu (x) = 0.
$$
Now, if we write $F_\nu (x,y) = -yA_\nu(x)$, we can regard our equation as
$$
\frac{\partial f}{\partial x^\nu} = F_\nu (x,f(x)).
$$
This is the kind of equation that appears in the classical form of the Frobenius theorem (see Spivak, theorem 1 in chapter 6). The integrability condition for this equation is
$$
\partial_\nu F_\mu - \partial_\mu F_\nu + \sum_k \partial_{y^k} F_\mu F_\nu^k - \sum_k \partial_{y^k} F_\nu F_\mu^k=0.
$$
But this precisely means
$$
\partial_\nu A_\mu - \partial_\mu A_\nu - \sum_k A_{k,\mu} A^k_{\nu} + \sum_k A_{k,\nu} A^k_{\mu}=0,
$$
which is the $dx^\nu \wedge dx^\mu$ component of the form $\Omega'=d\omega' + \omega' \wedge \omega'$.
Been trying to do this myself I get confused about something.To apply the frobenius theorem we need the functions $f$ to be defined a priori right ? I just don't see where that $f$ comes from.
@Idon'tknow $f$ is the solution to the PDE, and the Frobenius theorem precisely gives you the existence of such $f$. In Spivak’s, my $f$ is denoted by $alpha$.
Hm okay, but then in your notation what is Spivak's $f$ ? I am just kinda confused since we want to determine $A$, but $A$ and $f$ are not defined a priori, so that $F$ is not defined a priori.@G. Gallego
Spivak’s $f$’s are my $F$’s. You do not want to determine the $A$’s. The $A$’s are the components of the connection $1$-form in the initially chosen arbitrary frame $s’$. That is, they are already determined.
Maybe it can be confusing: what I now call $f$ is what in the original post I called $a$. Those $a$’s (now $f$’s) are the “variables” of the differential equation. These are “frame change functions” from the arbitrary initial frame $s’$ to a frame giving a flat structure (that is, a frame where the connection form vanishes).
Thank you G. Gallego !
|
2025-03-21T14:48:31.429879
| 2020-07-04T03:13:43 |
364802
|
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|
Stack Exchange
|
Relaxing Meyniel graphs: condition for strongly perfect instead of very strongly perfect
A Meyniel graph, $\mathcal{G}$ is a graph in which every cycle of odd length at least 5 has at least 2 chords.
First off, I have a technical question which is very important to me:
what is meant by every cycle of odd length at least 5 has at least 2 chords?
Does this mean every induced subgraph of $\mathcal{G}$ which are cycles of odd length at least 5 has at least 2 chords?
Or does this mean every cycle of odd length at least 5 "contained in $\mathcal{G}"$ has at least 2 chords? Where contained refers to a logical sense of containment in $\mathcal{G}$ (I refrain from using "subgraph" here because this is not necessarily what I mean).
For example, suppose a graph $\mathcal{L}$ has the house X graph as a subgraph (say the other subgraphs are not relevant to the discussion). See:https://mathworld.wolfram.com/HouseGraph.html
Then clearly, $\mathcal{L}$ has $C_{5}$ (chordless cycle of length $5$) also as a subgraph. However, I say here by my definition of containment, which I'm trying to define but it's very hard for me to make precise, that all odd cycles of length at least $5$ contained in $\mathcal{L}$ have 3 chords.
Here is my attempt to make the essence or significance of my definition more precise: **
take all subgraphs which are cycle graphs of odd length.
Case 1: If I can add edges from the original graph to that subgraph, without changing whether that subgraph is an odd cycle graph of length at least 5, then alter the graph by adding such edges. The number of chords of that cycle graph is then the number of chords of this "altered graph". This altered graph (not the original graph) is what is considered as "contained" in $\mathcal{G}$ then.
Case 2: If I cannot add edges from the original graph to that subgraph, without changing whether that subgraph is an odd cycle graph of length at least 5, then doing nothing (do not alter the graph). Then in this case the number of the chords of that cycle graph is the number chords of just the number of chords of that cycle graph with no alteration. That cycle graph is considered as "contained" in $\mathcal{G}$**
Take another example, Graph From Berge Paper
https://www.researchgate.net/publication/267671221_Strongly_perfect_graphs page 6.
By my definition of containment, in this graph all cycles of length $5$ contained in the graph have $0$ chords. However, it is clear that an induced cycle graph of length $5$ will have $1$ chord. Berge and Duchet say this graph is perfect but not strongly perfect (if an induced cycle graph of length $5$ with no chords existed, then the graph would not even be perfect by the strong perfect graph theorem).
This brings us to my main question:
It was proved by Ravindra that a Meyniel graph, a graph in which every odd cycle of length five or more has at least two chords, is strongly perfect.
A strongly perfect graph is a graph in which every induced subgraph $H$ has an independent/stable set meeting all maximal cliques of $H$. Note such a stable set is called a strong stable set.
It was later proved that Meyniel graphs are very strongly perfect graphs, an even stronger condition, which means that every vertex of an induced subgraph, $H$, belongs to an independent set of $H$ meeting all maximal cliques of $H$.
Now, coming to the main point. Here is my conjecture:
Conjecture: A graph $\mathcal{G}$, where every cycle of length of $5$ contained in $\mathcal{G}$ has at least $1$ chord and every cycle of odd length $\geq 7$ contained in $\mathcal{G}$ has at least $2$ chords, is strongly perfect. Here, containment of odd cycles and number of chords of odd cycles means what I attempted to define in bold above.
Some motivation for this is notice that the house graph (house graph not house X graph): https://mathworld.wolfram.com/HouseGraph.html
is strongly perfect. I can find a strong stable set in every possible induced subgraph (I verified for all 31 possible induced subgraphs).
Thus my point is that if I relax it that every cycle of length of $5$ contained in $\mathcal{G}$ has at least $1$ chord, then is the graph strongly perfect?
Is there any known results on this, or any research papers you could point me to?
Do you think the conjecture is true or do you think there exists a counterexample. I am sorry for my lack of familiarity with unified graph theory terminology. This concept of containment always confuses me and I want to be as technical as possible as this result is very important to me. I have highlighted in bold my main points. Thanks.
You are using the term 'induced cycle' inappropriately and it must absolutely be used correctly to be able to discuss these concepts. Also, please include citations for Ravindra, and for the later result that Meyniel graphs were proven strongly perfect.
You ask:
Does this mean every induced subgraph of G which are cycles of odd length at least 5 has at least 2 chords?
No, that doesn't make sense. If an induced subgraph is an induced cycle, then it contains no chords. The defining property of Meyniel graphs does not make any reference to induced subgraphs. Every odd cycle size 5 and larger contains two chords. This is a way of defining a class of graphs without an induced add cycles, since the definition prescribes how the cycles are broken (with 2 chords).
The house graph (i.e. the complement of $P_5$) is not Meyniel since the 5 vertices form a cycle but there is only one chord.
You mention that Meyniel graphs were shown to be strongly perfect and then that they were shown to be very strongly perfect. You seem to have missed the fact that Meyniel graphs were actually shown to be equivalent to the class of very strongly perfect graphs. This is important, as you later claim that the house graph (co-$P_5$) is very strongly perfect, but that graph is not Meyniel. The vertex at the top of the house is the example vertex which is not in an independent set which hits every maximal clique. Since with that top vertex, you form a maximum independent set with one of the vertices at the bottom of the house, and then the side of the house opposite your chosen bottom vertex is a maximal clique which is not hit by the independent set.
As for your conjecture, relaxing Meyniel graphs in a way that allow for 5-cycles with a single chord, I am not aware of this relaxation being studied, but I suspect it has. Meyniel graphs are also called (5,2)-odd chordal graphs, meaning every odd cycle of size 5 or more has 2 chords. In this notation, your conjecture states:
(Conjecture:) a $C_5$-free (7,2)-odd chordal graph is strongly perfect.
GraphClasses.org mentions a number of minimal superclasses of Meyniel graphs:
$(C_5,$house)-free
(antihole, odd hole)-free = no induced $C_5, \overline{C}_6, \overline{C}_7, C_7, \overline{C}_8, \overline{C}_9, C_9$... $\subseteq$ Perfect
bip* graphs $\subseteq$ Perfect
strongly even-signable = building-free and even-signable
locally perfect graphs $\subseteq$ Perfect
slim graphs = probe Meyniel graphs $\subseteq$ Perfect
quasi-Meyniel $\subseteq$ Perfect
strongly perfect $\subseteq$ Perfect
So there are (at least) 5 classes (and their equivalents) which are a superclass of Meyniel (i.e. a Meyniel relaxation) which are still perfect graphs and their relationship to strongly perfect graphs is non-comparable (meaning the class is not known to be a subclass or superclass of strongly perfect graphs). I do not know where your Meyniel relaxation lies among these.
Here is an example of one of your graphs... A 7-cycle with two chords and no induced $C_5$ (i.e. any 5-cycle contains a chord):
This graph is strongly perfect as the independent set {2,5,6} seems to hit every maximal clique. However, I believe the following graph is a counterexample to your conjecture:
Please verify that this satisfies your understanding of 'contains' and that this has the property that every odd cycle size 7 or larger has 2 chords and every 5-cycle has at least one chord. Then note that it is not strongly perfect.
|
2025-03-21T14:48:31.430469
| 2020-07-04T06:00:43 |
364807
|
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|
Stack Exchange
|
Isometric immersions of $\mathbb{S}^n$ into $\mathbb{S}^{2n}$
A major open problem in submanifold geometry is to determine whether there exists a nontotally geodesic isometric immersion of $\mathbb{S}^n$ into $\mathbb{S}^{2n}$. Every isometric immersion of $\mathbb{S}^n$ into $\mathbb{S}^{n+p}$ is totally geodesic if $p<n$, and nontotally geodesic examples are known for $p>n$ and also for $p=n=2$. Case $p=n\geq3$ remains open. It can be shown that the existence of a nontotally geodesic isometric immersion of $\mathbb{S}^n$ into $\mathbb{S}^m$ is equivalent to the existence of a nonconstant solution $Q:\mathbb{R}^{n+1}\setminus\{0\}\to O\left(n+1,m+1\right)$ to the linear first-order PDE system
$$\frac{\partial Q}{\partial x_i}x=0,\ 1\leq i\leq n+1,$$
where $O\left(n+1,m+1\right)=\left\{Q\in\mathbb{R}^{\left(m+1\right)\times\left(n+1\right)}:Q^TQ=I\right\}$. What parametric representation for $O\left(n+1,m+1\right)$ might be useful to addressing this problem?
|
2025-03-21T14:48:31.430554
| 2020-07-04T06:00:50 |
364808
|
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|
Stack Exchange
|
Homeomorphisms and "mod finite"
Suppose $f:C\to C$ is a homeomorphism, where $C=\{0,1\}^{\mathbb N}$ is Cantor space.
Suppose $f$ preserves $=^*$ (equality on all but finitely many coordinates). Does it follow that $f$ also reflects $=^*$?
That is, if
$$x=^* y \implies f(x)=^* f(y)$$
does it follow that:
$$x=^* y \iff f(x)=^* f(y)?$$
Using Axiom of Choice we can show that a bijection of $C$ that preserves $=^*$ need not reflect $=^*$, so the continuity assumptions are not superfluous.
@samerivertwice for $f\in{0,1}^{\mathbf{N}}$ and $n\in\mathbf{N}$ the $n$-th coordinate of $f$ is $f(n)$.
Define $f : C \to C$ by the formula
$$ f(x) = x_0 \cdot (x \oplus \sigma(x)) $$
where $\cdot$ is word concatenation, $\oplus : C \times C \to C$ is coordinatewise xor, and $\sigma(x)_i = x_{i+1}$ is the shift.
Clearly this map is continuous and preserves $=^*$. It is a bijection because you can deduce the preimage one coordinate at a time, which amounts to summing the prefix, $f^{-1}(x)_i = \bigoplus_{j \leq i} x_i$. By compactness $f$ is a homeomorphism. However, $f(0^\omega) = 0^\omega$ and $f(1^\omega) = 10^\omega$, so $f$ does not reflect $=^*$.
The idea is that a surjective non-injective one-dimensional cellular automaton forgets a finite amount of ("global") information on every step. I picked the xor CA $x \mapsto x \oplus \sigma(x)$ and also wrote the single bit that's forgotten (in the form of the first coordinate) to make it a homeomorphism.
This is simple enough that you can randomly stumble upon the formula, but turning xor injective this way is actually a pretty important idea in cellular automata theory. To name just one example, Kari's proof of the undecidability of reversibility of cellular automata on free abelian groups of rank $\geq 2$ (i.e. whether the CA is a homeomorphism on the full shift) uses this trick to turn a cellular automaton injective if a Turing machine halts; you are applying xor on a one-dimensional snake and you cut off its head if the machine halts.
As for the follow-up question from the comment on the other answer, the preservation of $E_0$ is uniform with $m_0 = n_0+1$.
This is great. This $f$ makes $f^{-1}\circ \sigma\circ f$ not preserve $=^*$ too, so it subsumes all my follow up questions
I can confirm that every statement here is correct. Formalized in Lean at https://github.com/bjoernkjoshanssen/bay/blob/main/ville-salo.lean
Every statement? Did you formalize the snake construction? ;)
Every statement in the first paragraph only .
Even with a homeomorphism, preserving $=^*$ does not imply reflection. There might be an easy example, I'm not sure; at any rate it follows from a result in topological dynamics (the "absorption lemma" of Giordano–Putnam–Skau) that such examples exist.
Let me elaborate a bit: by results of Giordano–Putnam–Skau, there exists a minimal homeomorphism of $X=\{0,1\}^{\mathbb N}$ which induces the relation that you denote $=^*$ (and which is often denoted $E_0$); there is an explicit construction (via a Bratteli diagram) in a paper of Clemens. By this, I mean that being in the same $\varphi$-orbit is the same as being $=^*$-equivalent.
Denote such a homeomorphism by $\varphi$, fix some point $x \in X$, and let $E$ be the relation obtained from $=^*$ by splitting the class of $x$ in its positive semi-orbit under $\varphi$, and its negative semiorbit. (So, all classes are the same as $=^*$, except that the class of $x$ has been split in two pieces). Giordano–Putnam–Skau prove that there is a homeomorphism $g$ of $X$ such that for all $x,y \in X$ one has
$$x=^*y \Leftrightarrow g(x) \mathrel E g(y).$$
Since $E$ is contained in $=^*$, yet not equal to it, such a $g$ preserves $=^*$ without reflecting it.
Edited to add: the book "Cantor minimal systems" by Ian Putnam is a good reference for information and bibliographical data about this area. Clemens's paper can be found here: Generating equivalence relations by homeomorphisms.
Surprised to see Bratteli diagrams make an appearance here... amazing!
I'm not sure how needed they are to answer your question. They do play a key part in Giordano--Putnam--Skau's work, via a connection that Vershik made with topological dynamics. Equivalence relations induced by minimal homeomorphisms of the Cantor space are surprisingly complicated...
My follow-up question would be whether it helps if $E_0$ is preserved uniformly in the sense that whenever $x(n)$ and $y(n)$ agree above $n_0$ then there is an $m_0$ depending only on $n_0$ such that $f(x)(m)$ and $f(y)(m)$ agree above $m_0$... but that's for another day perhaps
Offhand, I do not know how to answer that one. The anwser might well be different!
@JulienMelleray: Pardon my orbit-equivalence ignorance, but what's the result you are referring to that implies there exists a homeomorphism $g$ such that $x =^* y \iff g(x) E g(y)$?
@VilleSalo It's the "absorption theorem", theorem 12.1 in Putnam's book mentioned in my answer. It also follows from theorem 2.3 in Giordano--Putnam--Skau's paper in Crelle (which is, I think, the earliest proof). Indeed one can use the fact that both relations are induced by minimal homeomorphisms which have the same (unique) invariant measure, hence are orbit equivalent by Theorem 2.3 of Giordano--Putnam--Skau's Crelle article.
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