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2025-03-21T14:48:31.380207
| 2020-06-27T14:31:33 |
364274
|
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|
Stack Exchange
|
If a Markov semigroup is eventually contractive, can we conclude that it admits a unique invariant measure?
Let $E$ be a separable $\mathbb R$-Banach space, $\rho$ be a complete separable metric on $E$, $\operatorname W_\rho$ denote the Wasserstein metric of order $1$ associated to $\rho$, $\mathcal M_1(E)$ denote the set of probability measures on $(E,\mathcal B(E))$ and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$ with $$\operatorname W_\rho(\mu\kappa_t,\nu\kappa_t)\le ce^{-\lambda t}\operatorname W_\rho(\mu,\nu)\;\;\;\text{for all }\mu,\nu\in\mathcal M_1(\mu,\nu)\tag1$$ for some $c\ge0$ and $\lambda>0$.
Are we able to conclude that $(\kappa_t)_{t\ge0}$ has a unique invariant measure $\mu_\ast\in\mathcal M_1(E)$?
By $(1)$, the adjoint semigroup $(\kappa_t^\ast)_{t\ge0}$ is eventually contractive: Let $t_0\ge0$ with $$ce^{-\lambda t}<1\;\;\;\text{for all }t\ge t_0$$ and $t\ge t_0$. Since the Wasserstein space $$\mathcal S^1(E,\rho):=\left\{\mu\in\mathcal M_1(E):(\mu\otimes\delta_0)\rho<\infty\right\}$$ equipped with $\operatorname W_\rho$ is complete and hence we can apply Banach's fixed-point theorem yielding that there is a unique $\mu_\ast\in\mathcal S^1(E,\rho)$ with $$\mu_\ast\kappa_t=\mu_\ast\tag2.$$ Moreover, for any $\mu_0\in\mathcal S^1(E,\rho)$ and $$\mu_n:=\mu_{n-1}\kappa_t\;\;\;\text{for }n\in\mathbb N,$$ it holds $$\operatorname W_\rho(\mu_n,\mu_\ast)\xrightarrow{n\to\infty}0\tag3.$$
So, all what's left to prove is that $\mu_\ast$ does not depend on $t$, i.e. $\mu_\ast$ is invariant with respect to $\kappa_t$ for all $t\ge t_0$.
BTW: Is this all we can hope for or can we even conclude that $\mu_ast$ must be invariant with respect to $\kappa_t$ for all $t\ge\color{red}0$?
That's fine, but it only proves uniqueness in the space of measures with a finite first moment.
@RW Yes, this space is $\mathcal S^1(E,\rho)$. I would like to show that there is a $\mu_\ast\in\mathcal S^1(E,\rho)$ with $\mu_\ast\kappa_t=\mu_\ast$ for all $t\ge t_0$.
Note that your argument contains an implicit assumption that $\kappa_t \mu \in \mathcal{S}^1$ for every $\mu \in \mathcal{S}^1$ (otherwise the Banach fixed point theorem does not apply). I will also make that assumption. Also, I realized that I have written $\kappa_t \mu$ with $\mu$ on the right; sorry about that.
You have shown that for some fixed $t^* \ge t_0$, that $\kappa_{t^*}$ has an invariant measure $\mu_*$ which is unique in $\mathcal{S}^1$.
Let $t > 0$ be arbitrary. Then we have by the semigroup property that
$$\kappa_{t^*} \kappa_t \mu_* = \kappa_{t+ t^*} \mu_* = \kappa_t \kappa_{t^*} \mu_* = \kappa_t \mu_*$$
which proves that $\kappa_t \mu_*$ is invariant for $\kappa_{t^*}$. By uniqueness, $\kappa_t \mu_* = \mu_*$. This proves that $\mu_*$ is invariant for $\kappa_t$.
If $t \ge t_0$, then your argument shows that $\mu_*$ is in fact the unique invariant measure in $\mathcal{S}^1$ for $\kappa_t$. Otherwise, for $t < t_0$, suppose $\mu' \in \mathcal{S}^1$ is another invariant measure for $\kappa_t$. Let $n$ a large enough integer so that $n t \ge t_0$; then $\mu' = \kappa_t^n \mu' = \kappa_{nt} \mu'$. Since $\kappa_{nt}$ has $\mu_*$ as its unique invariant measure, we have $\mu' = \mu_*$.
We have thus shown that for every $t$, $\mu_*$ is invariant for $\kappa_t$, and is the unique such measure in $\mathcal{S}^1$.
Thank you for your answer. (a) For your uniqueness argument in your second argument to work, we need that $\kappa_t^\ast\mu_\ast\in\mathcal S^1$. It's not clear to me why this is the case. Is it a general fact that a Markov kernel is $\mathcal S^1$-preserving? (b) Just to be sure: The result we've obtained is that $(\kappa_t)_{t\ge0}$ has a unique invariant measure in $\mathcal S^1$, but there might be other invariant measures in $\mathcal M_1\setminus\mathcal S^1$, right?
@0xbadf00d: Good point. In fact, consider the following counterexample: let $\nu$ be a probability measure not in $\mathcal{S}^1$ and take $\kappa_t f(x) = \int f,d\nu$ for all $t$. Then $\kappa_t \mu = \nu$ for every $\mu$ and every $t>0$, so (1) is trivially satisfied (the LHS is always zero). Indeed, this is also a gap in your argument, because $\kappa_t$ has a unique invariant measure and it is not in $\mathcal{S}^1$. I think we need to assume that $\kappa_t$ takes $\mathcal{S}^1$ into $\mathcal{S}^1$ for any of this to work.
@0xbadf00d: I agree that even with this assumption, nothing appears to rule out the possibility of another invariant measure which is not in $\mathcal{S}^1$. I don't know of an example where this happens, however.
|
2025-03-21T14:48:31.380485
| 2020-06-27T15:30:13 |
364277
|
{
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"Jeremy Rickard",
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|
Stack Exchange
|
Dense generator whose closure under finite colimits takes several steps to form?
Let $\mathcal C$ be a locally finitely presentable category, and let $\mathcal C_0 \subseteq \mathcal C$ be a dense generator of finitely-presentable objects. Then
Every object $C \in \mathcal C$ is a colimit of objects of $\mathcal C_0$, and
The closure $\overline{\mathcal C_0}$ of $\mathcal C_0$ under finite colimits comprises precisely the finitely-presentable objects of $\mathcal C$ [1].
I'm interested in cases where the closure process in (2) takes more than one step to form. So inductively define $\mathcal C_{n+1}$ to comprise the finite colimits of objects of $\mathcal C_n$. Then $\overline{\mathcal C_0} = \cup_{n \in \mathbb N} \mathcal C_n$.
Questions:
What is an example of a locally finitely presentable category $\mathcal C$, and a dense generator $\mathcal C_0 \subseteq \mathcal C$ of finitely-presentable objects, such that $\overline{\mathcal C_0} \neq \mathcal C_1$?
We might opt to treat retracts specially -- so what happens if we instead define $\mathcal C_{n+1}$ to comprise the retracts of finite colimits of objects of $\mathcal C_n$?
More generally, we can ask this for higher degrees of presentability (though the iterative construction of the closure of $\mathcal C_0$ under $\kappa$-small colimits may now in principle take transfinitely many steps). I'd be interested in such examples too.
Notes:
This doesn't happen if $\mathcal C$ is a presheaf category and $\mathcal C_0$ is contained in the representables: any finitely-presentable preseheaf is a finite colimit of representables.
EDIT: The following examples are all at least potentially mistaken; see Jeremy Rickard's comments.
My favorite example of a finite-colimit-closure which takes several steps to form is the closure of $\{R\}$ under finite colimits in $Mod_R$, for appropriate rings $R$, e.g. $R = \mathbb Z$. But in this case, although $\{R\}$ is a strong generator of finitely-presentable objects, it is not a dense generator. And I think the finite-colimit closure of the dense generator $\{R \oplus R\}$ takes only one step to form.
Similarly, the finite colimit closure of $\{\mathbb Z\} \subseteq Grp$ takes at least two steps to form, but $\{\mathbb Z\}$ is not dense, and on the other hand, the finite colimit closure of the dense generator $\{F_2\}$ occurs in one step.
For another similar example, in the final paragraph of Section 5.9 of Basic Concepts of Enriched Category Theory, Kelly claims that the walking idempotent is not a colimit (in $Cat$) of copies of the walking arrow. I don't follow his proof sketch, but perhaps if it could be understood, then the argument might show that the walking idempotent is also not the finite colimit of copies of the "composable pair" category $\bullet \to \bullet \to \bullet$, which is dense in $Cat$.
[1] This is not true $\infty$-categorically, where we need to additionally close under retracts (the indexing category for an idempotent not being finite in the $\infty$-categorical sense). For example, not every retract of a finite CW complex is homotopy equivalent to finite CW complex, by the Wall finiteness obstruction.
You’ve linked to a proof that not every abelian group is a colimit of copies of $\mathbb{Z}$. But every finitely generated abelian group is a finite colimit of copies of $\mathbb{Z}$, so the finite colimit closure of ${\mathbb{Z}}$ is reached in one step.
Isn't it always true that every finitely presentable object is a retract of a finite colimit of the generators ? I haven't thought very much about it, but I'm under the impression that looking at the 'finite subdiagram' of the canonical diagram gives that every every object is filtered colimit of finite colimits of the generators ? Is there something going wrong with this ?
I think the "walking idempotent" is a finite colimit of copies of the "composable pair" category. Take the diagram with two copies of $1\to 2\to 3$, with three maps $E,F,G$ from the first copy to the second copy, where $E(1)=1$, $E(2)=E(3)=2$, $F(1)=2$, $F(2)=F(3)=3$, and $G(1)=1$, $G(2)=G(3)=3$.
@SimonHenry I think you're suggesting to look at the subdiagram $\mathcal C_1 / C \subseteq \mathcal C / C$ -- but there's no reason to think this diagram is filtered unless $\mathcal C_1$ is known to be closed under finite colimits!
@JeremyRickard I think you're right on both counts. So the best example above is ${\mathbb Z} \subset Grp$ (if it does indeed work!), which is still not dense. I wonder if a more complicated ring like $k[x,y]$ might work, though...
I'm probably wrong, but what I had in mind was given $C_0/X$ the indexing category of the canonical diagram whose colimits is $X$, look at the class of all finitely generated categories $F \to C_0 /X$. This is a filtered diagram, and each colimits along finitely generated categories can be rewritten as a finite colimit.
I think Simon Henry's comment works to show that there are no examples of (2). That is
Theorem: Let $\mathcal C$ be a locally finitely-presentable category, let $\mathcal C_0 \subseteq \mathcal C$ be a dense generator of finitely-presentable objects. Let $X \in \mathcal C$ be finitely-presentable. Then $X$ is a retract of a finite colimit of objects of $\mathcal C_0$.
So the only question is whether the retract can be eliminated, both in the finitary and the infinitary case.
Remark: I've always been a little hazy on when a colimit can be decomposed using a colimit decomposition of the indexing diagram. But let $K$ be a simplicial set, and let $\{K_I \mid I \in J\}$ be a directed sub-poset of the collection of simplicial subsets of $K$ with $\cup_{I \in J} K_I = K$. Then, according to HTT Rmk <IP_ADDRESS>, for any diagram $F: K \to \mathcal C$ in a cocomplete quasicategory, we have $\varinjlim_{k \in K} F(k) = \varinjlim_{I \in J} \varinjlim_{k \in K_I} F(k)$.
Proof of Thm: Using the Remark, with $K = \mathcal C_0 / X$ and $J$ the (directed) collection of finitely-generated subcategories of $K$, we obtain a $J$-indexed diagram with colimit $X$. Since $X$ is finitely-presentable, we get that $X$ is a retract of the colimit of a finitely-generated subdiagram of $\mathcal C_0 / X$. According to a theorem of Pare, any finitely-generated category admits a final functor from a finite category. So $X$ is equally a retract of a finite colimit of objects of $\mathcal C_0$.
Thanks for clarifying this !
Note to self: it's actually possible to decompose a colimit in terms of a colimit decomposition of the indexing category in complete generality.
Note to self: Still have to understand Reid Barton's argument to see if it generalizes to more general dense generating sets.
|
2025-03-21T14:48:31.380896
| 2020-06-27T15:30:14 |
364278
|
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"S. carmeli",
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|
Stack Exchange
|
Are "large enough" finite etale covers arithmetic?
Let $X$ be a variety over a number field $K$. Then it is known that for any topological covering $X' \to X(\mathbb{C})$, the topological space $X'$ can be given the structure of a $\overline{K}$-variety in such a way so that the morphism $f: X' \to X$ inducing the topological map is a finite etale morphism over $\overline{K}$. However, the variety $X'$ and the morphism $f$ may not descend to $K$.
My question is as follows: does there always exist a further finite etale covering $f' : X'' \to X'$ such that the composition $X'' \to X$ may be defined over $K$?
EDIT: Just to be clear, I'd like all the covers involved to be geometrically connected to avoid trivial solutions.
Let's assume that $X$ admits a $K$-point $x$ and use the corresponding geometric point as the base point. The existence of a rational point is in fact necessary for a positive answer, as explained by S. carmeli.
In terms of etale fundamental groups the question can be paraphrased as follows: given an open subgroup $H\subset \pi_1(X_{\overline{K}},x)$ does there exist an open subgroup $H'\subset H$ such that the action of the Galois group $G_K$ on $\pi_1(X_{\overline{K}},x)$ preserves $H'$.
This is true and follows from $\pi_1(X_{\overline{K}},x)$ being topologically finitely generated. Consider the subgroup $\Gamma_H\subset G_K$ consisting of elements $\gamma\in G_K$ such that $\gamma(H)=H$. Let $h_1,\dots, h_n$ be a set of topological generators of $H$ ($H$ is topologically finitely generated because it has finite index in $\pi_1(X_{\overline{K}})$). Then $\Gamma_H$ can be expressed as $\{\gamma\in G_K|\gamma(h_i)\in H\}$ so $\Gamma_H$ is an intersection of finitely many open subset, hence is an open subgroup. In particular, $\Gamma_H$ has finite index in $G_K$. Take $\Gamma\subset \Gamma_H$ to be an open subgroup which is moreover normal in $G_K$.
Let $g_1,\dots, g_m$ be a set of representatives of cosets of $\Gamma$ in $G_K$. Then $H'=\bigcap g_i(H)$ is an open subgroup with the desired property. Indeed, suppose that $x\in H'$ and $\gamma g_i\in G_K$ are arbitrary elements where $\gamma\in \Gamma$ and $i\in\{1,\dots, m\}$. The result of the action $\gamma \circ g_i(x)$ lies in $H'$ because for each $k=1,\dots, m$ we have $g_k^{-1}\gamma g_i=\gamma'g_j^{-1}$ for some $\gamma'\in \Gamma$ and $j\in\{1,\dots, m\}$ so $\gamma g_i(x)\in \gamma g_ig_j(H)=g_k\gamma'(H)=g_k(H)$.
We can think of this argument as of a generalization of the proof that a compact group acting on a finite-dimensional $\mathbb{Q}_p$-vector space always preserves some $\mathbb{Z}_p$-lattice.
Hah! beat me by a few seconds! When the homotopy exact sequence isn't split, how are you arguing that a normal finite index subgroups correspond to something defined over $K$ if and only if it's $G_K$-stable? (Since you only have an outer action to work with). I imagine there must be some group cohomology argument but I'm blanking at the moment.
@WillChen I had in mind getting an honest action by choosing the algebraic closure of the residue field of the generic point as the base point (though I'm not claiming that the exact sequence splits). Then a cover $X'\to X_{\overline{K}}$ corresponding to the action of the fundamental group on a finite set such that the kernel is Galois-stable gets equipped with the descent datum for $G_K$ acting on $X_{\overline{K}}$, hence the result.
Can you say a few more words on why your particular choice of base point gives rise to an honest action?
@WillChen The action of $G_K$ on $X_{\overline{K}}$ preserves the generic point so it induces an action on the fundamental group with respect to the base point supported at the generic point.
The $G_K$ action certainly preserves the generic point, but are you sure it preserves the geometric generic point? Doesn't it act nontrivially on the geometric point by acting on coefficients? E.g., for the affine line $\overline{K}[t]$, the geometric generic point is $\overline{K}[t]\hookrightarrow \overline{K(t)}$. I don't think the $G_K$-action on the source preserves this injection.
@WillChen Oh, I guess I see my mistake. You're saying that to get an honest action using geometric generic point we would have to pick isomorphisms between $\sigma\circ x$ and $x$ for all $\sigma\in G_K$ in a way compatible with the composition, but that would amount to exhibiting a section of $G_{K(X)}\to G_K$ which might not be possible in the absence of a rational point. Sorry for trying to mislead you!
Here's a simple argument assuming $X$ admits a $K$-rational point, and that $X$ has a finitely generated geometric fundamental group. In fact the "further" covering $X''$ can be chosen to be geometrically Galois over $X$.
Let $\Pi := \pi_1(X_K)$, let $\overline{\Pi} := \pi_1(X_{\overline{K}})$ (assumed to be topologically finitely generated). Let $G_K := \text{Gal}(\overline{K}/K)$.
Since we're working over a field, there's a homotopy exact sequence
$$1\rightarrow \overline{\Pi}\rightarrow\Pi\rightarrow G_K\rightarrow 1$$
from which we get a canonical outer action $G_K\rightarrow\text{Out}(\overline{\Pi})$.
The covering $X'$ (over $\overline{K})$ corresponds to a finite index subgroup $H \le \overline{\Pi}$. It would suffice to find a finite index normal subgroup $\Gamma\lhd \overline{\Pi}$ which is stabilized by $G_K$. Indeed, using the $K$-rational point of $X$, the homotopy exact sequence is split, so the outer action of $G_K$ comes from an honest action, and $\Pi = \overline{\Pi}\rtimes G_K$ relative to this action. If $\Gamma\lhd\overline{\Pi}$ is stabilized by $G_K$, then the subgroup $\Gamma\rtimes G_K\le \Pi$ visibly corresponds to a geometrically connected finite cover of $X_K$ (though it may not be normal inside $\Pi$).
To find this $\Gamma$, let $N\le H$ be the intersection of all the $\overline{\Pi}$-conjugates of $H$, so $N$ is normal and of finite index inside $\overline{\Pi}$. Let $\Gamma$ be the intersection of the kernels of all the surjective homomorphisms $\overline{\Pi}\rightarrow\overline{\Pi}/N$. Since $\overline{\Pi}$ is finitely generated, there are only finitely many such homomorphisms, so $\Gamma$ is also finite index inside $\overline{\Pi}$. Moreover, it's easy to check that $\Gamma$ is characteristic inside $\overline{\Pi}$. Thus, $G_K$ must stabilize $\Gamma$, and hence $\Gamma\rtimes G_K$ will correspond to the desired covering $X_K''\rightarrow X_K$, which is moreover geometrically Galois.
Adding on Will's and Sasha's answers, the condition of having a rational point, or at least a "1-truncated homotopy fixed point" for the action is necessary. For example, let $C_2$ act on the circle $S^1$ by half rotation. The covers of $S^1$ are the standard n-fold ones, and we can ask what it takes to lift the action of $C_2$ to the cover, so that it is "defined over $BC_2$". In particular, we need to lift that half-circle rotation to the n-fold cover, for which the options are $1/2n + k/n$ rounds rotation. For this to be an involution, we need that applying it twice gives the identity, i.e. that $1/n +2k/n$ is an integer. If $n$ is even, this is impossible, and so the double cover of this action on $S^1$ has no cover definable over $BC_2$. To turn this topological picture to arithmetic, take $K=\mathbb{R}$ and let complex conjugation act on $\mathbb{C}^\times$ by $z\mapsto -1/\bar{z}$ (which is a form of the multiplicative group with no rational points). The action on the unit circle is then half rotation, so the Galois story realized to the topological one up to profinite completion.
I would add that what happens topologically is that if we have a fixed point, we can use it to define a "connected" compositum of pointed covers, by taking the component of the tuple of base points lifts. This is what missing in this example esssentially, even though up to isomorphism all covers are actually "the same".
Great, thanks for the correction! So you're explaining that for the non-split form of $\mathbb{G}_m$ over $\mathbb{R}$ the etale fundamental group is the non-split extension of $\mathbb{Z}/2$ by $\widehat{\mathbb{Z}}$ so that there is no subgroup surjecting onto $\mathbb{Z}/2$ whose intersection with the geometric fundamental group is contained in $2\widehat{\mathbb{Z}}$
@SashaP yes that's a much simpler and equivalent way to say this :-)
|
2025-03-21T14:48:31.381508
| 2020-06-27T15:55:25 |
364279
|
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|
Stack Exchange
|
Maximum number of edges in "square" hypergraph
For any set $X$ and any cardinal $\kappa$, let $[X]^\kappa$ denote the subsets of $X$ having cardinality $\kappa$.
A linear hypergraph is a hypergraph such that for all $e\neq e_1 \in E$ we have $|e\cap e_1|\leq 1$. For any positive integer $n$ let $[n] = \{1,\ldots,n\}$. We say that $([n^2], E)$ is a square hypergraph if it is linear and every element of $E$ has $n$ elements.
For any integer $n>1$ let $m(n)$ denote the maximum cardinality of $E$ where $([n^2], E)$ is a square hypergraph. For instance we have $m(2) = 6$. What is the value of $$\lim (\sup)_{n\geq 2}\frac{m(n)}{n^2}\;?$$ (Note: I added $(\sup)$ in case the limit doesn't exist.)
In a linear hypergraph, any pair of vertices is contained in at most one hyperedge. Since any hyperedge contains $n$ vertices, it contains ${n \choose 2}$ pairs. Double counting gives
$$m(n) \leq \frac{n^2 \choose 2}{n \choose 2} = n (n+1).$$
This bound is sharp when $n$ is a prime power (take an affine plane of order $n$), so
$$ \limsup \frac {m(n)}{n^2} = 1. $$
|
2025-03-21T14:48:31.381611
| 2020-06-27T16:49:35 |
364284
|
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"Haldot",
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|
Stack Exchange
|
"Well-known fact" that every irreducible 3-manifold with non-empty boundary has an incompressible surface
I have seen in several sources that this results holds, however none of them included the proof. Does anyone know where I can find one?
Also, it would be great if someone could provide me with a counterexample, where irreducibility of the manifold matters.
Thank you.
The proof might be too long for this fact. However, here is one reference
Algorithmic Topology and Classification of 3-Manifolds by Sergei Matveev in the series Algorithms and computations in Mathematics, Volume 9, 2003, Springer-Verlag.
You may start reading from page 167.
Counterexample: the 3-ball $B^3$. It is irreducible because it is a compact submanifold of $\mathbb{R}^3$ with connected boundary. It has non-empty boundary and $\partial B$ is simply connected (and incompressible surfaces with boundary must send their boundary to curves which do not bound disks in $\partial B$). Hence if there is an incompressible surface it must be closed. Non orientable closed surfaces do not embed in $B^3$. Moreover, $B^3$ is simply connected and
recall that by Dehn's lemma an orientable, closed surface with $\chi(S)\leq 0$ is incompressible if the inclusion is injective at the level of fundamental group.
However the following is true:
If $M$ is compact with non-empty boundary, oriented , irreducible, and $\partial$-irreducible* then either $M=B^3$ or $M$ contains an incompressible
and $\partial$-incompressible surface.
This result relies on the fact that (under the above assumptions), given a class in $H_2(M,\partial M;\mathbb{Z})$ you can represent it by disjoint union of incompressible and and $\partial$-incompressible surfaces.
*
You can find all the details for example in Bruno Martelli. An Introduction to Geometric Topology. https://arxiv.org/pdf/1610.02592.pdf
Proposition 9.4.3 and Corollary 9.4.5.
*Meaning that there are no essential disks, in other words $M$ is not obtained by joining two 3-manifolds with a 1-handle.
@Haldot The definition of incompressible surface I use (also used in the book mentioned above) concerns surfaces with non-positive Euler characteristic. This rules out 2-disks. In the case of 2-disks the interesting notion is the one of essential disk, which are disks that are not boundary parallel (i.e. are not obtained by pushing the interior or a disk in the boundary $\partial M$ into the interior of $M$).
This is the "correct" notion of essential disk (as an essential sphere is one that does not bound a B^3 and an essential surface with $\chi\leq 0$ in necessarily incompressible) because as in the case of spheres finding an essential sphere allows you to decompose the manifold wrt connected sum, fiinding an essential 2-disk allows you to decompose the 3-manifold with respect to boundary connected sum)
Oh, sorry, I deleted my comment before I saw your answer. Thanks!
|
2025-03-21T14:48:31.381831
| 2020-06-27T17:11:43 |
364286
|
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|
Stack Exchange
|
Conjugate point to spacelike hypersurface
Suppose you have a smooth spacelike hypersurface $\Sigma$ in some spacetime (four-dimensional Lorentzian manifold). Let $\gamma$ be a timelike geodesic meeting $\Sigma$ orthogonally and let $p$ be a point on $\gamma$. Then $p$ is said to be conjugate to $\Sigma$ iff there exists a Jacobi field $J$ orthogonal to the tangent vector $\gamma'$, which is zero at $p$, nonzero on $\Sigma$, and comes from a variation of $\gamma$ by geodesics intersecting $\Sigma$ orthogonally.
Why is this the definition? Why not say that $p$ is conjugate to $\Sigma$ iff there exists a Jacobi field $J$ which is zero at $p$ and zero on $\Sigma$ (and thus $J$ is orthogonal everywhere to $\gamma$), the same definition as conjugacy along a geodesic? If you do that, you cannot ask of the variation to come from geodesics intersecting $\Sigma$ (if $q$ is the intersection of $\gamma$ and $\Sigma$, then the orthogonal space to the tangent space at $q$ of $\Sigma$, i.e. $(T_q\Sigma)^{\perp}$, is one-dimensional, so any geodesic intersecting $\Sigma$ orthogonally at $q$ must be a reparametrization of $\gamma$), but why does one care about intersecting orthogonally?
EDIT: If you take $\theta$ to be the expansion of the timelike geodesics through $p$, then $p$ is conjugate to $q$ along $\gamma$ iff $\theta$ goes to infinity at $q$.
I would like to see a proof for that in the hypersurface case: $p$ is conjugate to $\Sigma$ along $\gamma$ iff $\theta$, the expansion of the timelike geodesics orthogonal to $\Sigma$, goes to infinity at $p$. In particular, I do not know how to impose the condition that the Jacobi field $J$ arises from a variation of geodesics orthogonal to $\Sigma$.
Your second definition defines a pair of conjugate points --- the hypersuface is not needed. (Why --- because you can --- why not.)
@AntonPetrunin I know it defines a pair of conjugate points as I said- "same definition as conjugacy along a geodesic". My question stands: why does one care about intersecting orthogonally? I am assuming that the reason is the one I have written in the edit, but I don't know how to prove so I don't know the underlying motivation.
There are two cases: Jacobi fields defined in terms of a geodesic spray from a point and a geodesic spray from a surface. In both cases the differential equation that defines the Jacobi tensor is the same. It is only the initial conditions that are different. Because of this conjugate points or focal points both occur with the expansion diverges.
The reason for the similarity in the differential equation can be seen from either a mathematical point of view or a physical one. I'll start with the maths.
A causal geodesic is length maximal up to its first conjugate point. Similarly the normal geodesic to a surface is length maximal (the distance from the surface to some point) up to its first focal point. In both cases variation of arc length arguments can be used to show this (via the Morse index theory) and thus in both cases the same differential equation results.
The physical point of view comes from analysing the infinitesimal acceleration of a congruence of geodesics. It turns out that the differential of the exponential map defines a Jacobi tensor. Thus the study of the infinitesimal acceleration of a congruence of geodesics is equivalent to the study of the variation of arc length.
Beem, Erhlich and Easley (BEE) have much to say about both points of view (see Chapter 10 and 12). In particular they carefully describe why it is sufficient to study congruences of geodesics (rather than arbitrary causal curves). If you'd like to see the formula for arbitrary congruences of causal curves they are in Hawking and Ellis.
As to why we want congruences that are normal to the surface in the case of focal points; From the math point of view the components of the Jacobi tensor that are not orthogonal to the geodesic aren't interesting (this in Lemma 10.9 in BEE). By this I mean the non-orthongal components don't provide information about variation of arc-length. So we should only consider Jacobi tensors with compoents normal to the geodesic. The question then becomes do we want the Jacobi tensor to initially have zero volume or non-zero volume. This corresponds to the point and surface cases.
From the physical point of view, in the surface case we wish to describe the behaviour of a volume element of the surface. Thus the initial conditions must ensure non-zero volume. Since the normal to the surface is... well normal to the surface the correct initial conditions correspond to a congruence of geodesics each geodesic of which is normal to the surface.
TL;DR: Read Beem, Erhlich and Easley chapter 10 and 12 and season with some Hawking and Ellis. Make sure to read up about variation of arc length and Morse index theory.
|
2025-03-21T14:48:31.382519
| 2020-06-27T17:14:17 |
364288
|
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"Cihan",
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|
Stack Exchange
|
When does an open manifold admit two linearly independent vector fields?
$\DeclareMathOperator{\span}{span}$
$\DeclareMathOperator{\co}{H}$
$\newcommand{\kk}{\mathbb{F}}$
$\newcommand{\qq}{\mathbb{Q}}$
$\newcommand{\zz}{\mathbb{Z}}$
$\newcommand{\rr}{\mathbb{R}}$
$\newcommand{\semi}{\hat{\chi}_2}$
$\newcommand{\ori}[1]{\textbf{(O$_{\pmb{#1}}$)}}$
$\newcommand{\nori}[1]{\textbf{(NO$_{\pmb{#1}}$)}}$
$\newcommand{\rarr}{\rightarrow}$Let $M$ be a smooth connected $d$-manifold (without boundary), and write $\span(M)$ for the maximum number of linearly independent vector fields on $M$. By Poincaré-Hopf, we know that if $M$ is closed, then $\span(M) \geq 1$ if and only if $\chi(M) = 0$. It is also known that every open $M$ satisfies $\span(M) \geq 1$. I am interested in the characterization of the condition $\span(M) \geq 2$. This has been achieved for closed manifolds with works of several people (over the years ~1965-90), as I will explain.
Notation for closed $M$: Given a field $\kk$, write $b_j(M;\kk) := \dim_\kk{\co_j(M;\kk)}$. Write $$\semi(M;\kk) := \left( \sum_{j \geq 0} b_{2j}(M;\kk) \!\!\!\mod{\!2}\right) \in \kk_2$$
for the Kervaire semi-characteristic over $\kk$ (it is always a mod-2 number). For each $0 \leq j \leq d$, write $w_j(M) \in \co^j(M;\kk_2)$ for the $j$-th Stiefel-Whitney class. Write $[M] \in \co_d(M;\kk_2)$ for the $\kk_2$-fundamental class and $$\langle-,[M] \rangle \colon \co^k(M;\kk_2) \rarr \co_{d-k}(M;\kk_2)$$ the associated PD isomorphism. If $M$ is oriented and $d \equiv 0 \!\pmod{4}$, let $\sigma(M)$ denote its signature.
For the characterization of closed $M$ with $\span(M) \geq 2$, first note that when $d=2$ we only have the $2$-torus. Assuming $d \geq 3$, then $M$ has $\span(M) \geq 2$ if and only if in addition to $\chi(M) = 0$, it satisfies one of the following conditions:
$\ori{0}$ : $M$ is orientable, $d \equiv 0\!\pmod{4}$, and $\sigma(M) \equiv 0\!\pmod{4}$.
$\ori{1}$ : $M$ is orientable, $d \equiv 1\!\pmod{4}$, $w_{d-1}(M) = 0$, and $\semi(M;\rr) = 0$
$\ori{2,3}$ : $M$ is orientable and $d \equiv 2,3 \!\pmod{4}$.
$\nori{0,2}$ : $M$ is non-orientable, $d$ is even, and writing $\zz_{w_1(M)}$ for the orientation sheaf, the twisted Bockstein $$\beta^{*}\colon \co^{d-2}(M;\kk_2) \rightarrow \co^{d-1}(M;\zz_{w_1(M)})$$ sends $w_{d-2}(M)$ to $0$.
$\nori{1}$ : $M$ is non-orientable, $d \equiv 1\!\pmod{4}$, $w_1(M)^2 = 0 = w_{d-1}(M)$, and $$\semi(M;\kk_2) = \langle w_2(M)w_{d-2}(M), [M] \rangle \in \kk_2 \, .$$
$\nori{3}$ : $M$ is non-orientable, $d \equiv 3\!\pmod{4}$, and $w_1(M)^2 = 0$.
$\nori{1,3}$ : $M$ is non-orientable, $d$ is odd, $w_1(M)^2 \neq 0$, and $w_{d-1}(M) = 0$.
[I can give precise references with a bit more work, but for now I will note that $\ori{0}$ is due to Frank and independently Atiyah for $d > 4$ and due to Randall when $d=4$, $\ori{1}$ is due to Atiyah, $\ori{2,3}$ is due to E. Thomas, $\nori{0,2}$ is due to Pollina, $\nori{1}$ and $\nori{3}$ are due to Randall, $\nori{1,3}$ is due to Mello.]
For open manifolds, I could only find the following: Non-orientable surfaces necessarily have $\span = 1$, and open orientable manifolds of dimension 2,3 are parallelizable.
Can we similarly characterize open $d$-manifolds $M$ with $\span(M) \geq 2$? For instance, can the method in this answer be adapted to show that $\span(M) \geq 2$ whenever $M$ is orientable and $d \equiv 3\!\pmod{4}$?
Throughout we assume $d>4$ and $d$ odd. Denote by $V_{d,2}$ the Stiefel-manifold of orthonormal $2$-frames in $\mathbb R^d$. Since $V_{d,2}$ is $(d-3)$-connected there is a $2$-field over the $(d-2)$-skeleton of $M$.
The first obstruction to extend this $2$-field over the $(d-1)$-skeleton lies in $H^{d-1}(M;\pi_{d-2}V_{d,2})
=H^{d-1}(M;\mathbb Z_2)$ and is given by $w_{d-1}(M)$. Suppose this class vanishes and consider an extension of the $2$-field over the $(d-1)$-skeleton. But since $M$ is open, there are no $n$-cells for $n>d-1$. Hence the only obstruction to extend a $2$-field from the $(d-2)$-skeleton to the whole open manifold is the Stiefel-Whitney class $w_{d-1}(M)$.
All other obstructions in the theorems you mentioned, are coming from the existence of a $d$-cell of a $d$-dimensional closed manifold.
I see. It is a theorem of Massey that $w_{d-1}(M) = 0$ whenever $d \equiv 3 \pmod{4}$ with $M$ is orientable and closed. I think we can drop the closed assumption by arguing that $w_{d-1}$ still vanishes because it vanishes on compact submanifolds. Thus there is always a $2$-field in this case.
|
2025-03-21T14:48:31.382812
| 2020-06-27T18:48:43 |
364291
|
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"A beginner mathmatician",
"Aleksei Kulikov",
"Christian Remling",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364291"
}
|
Stack Exchange
|
Integrability of Fourier transform of truncated fractional power
Is the Fourier transform of the function $f$ which agrees with $1_{[-1.1]}|x|^\alpha$ on $[-1,1]$ and then decays very fast to zero to become a compactly supported continuous function, is in $L^1(\mathbb R)$, where $\alpha\in(1,2)$? My guess is that the answer is true. If I can show that the identity $\widehat{D^\alpha f}(\zeta)=(2\pi i\zeta)^\alpha\hat{f}(\zeta)$ holds in some sense and we have $D^\alpha(1_{[-1,1]}|x|^\alpha)=\alpha!|x|1_{[-1,1]}$ in some sense. Then $\hat{f}$ would have enough decay at infinity to make it in $L^1(\mathbb R)?$ May be fractional derivatives and their interaction with the Fourier transform might be helpful. But I could not find any good reference (easily readable) which can make the above argument rigorous? For my purpose if one such $f$ exists for which the Fourier transform is integrable is enough?
If $\hat{f}\in L^1(\mathbb{R})$ then $f\in C(\mathbb{R})$ so the answer is no. But this question is more appropriate for math.stackexchange.
@Aleksei. I have edited the question. The edited question is what I meant originally.
The smoothness of this is better than that of $|x|$, which already has Fourier coefficients $\simeq 1/n^2$, so the answer is yes.
@Christian. Ok. I missed that observation. Clearly, for $|x|$ we have decay $|\zeta|^{-2}$ at $\infty.$ But how to prove your statement rigorously?
|
2025-03-21T14:48:31.382946
| 2020-06-27T20:15:52 |
364296
|
{
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"Christian Remling",
"Juan H. Arredondo",
"https://mathoverflow.net/users/48839",
"https://mathoverflow.net/users/520457"
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|
Stack Exchange
|
Fourier transform of a function of bounded variation
I know if $f\in L^2(\mathbb R)$ is two times continuously differentiable, then we must have that the Fourier transform is integrable. Is there any more relaxed condition than this? For example if $f$ is continuously differentiable and $f^\prime $ is of bounded variation, will it imply that $f^\prime$ is integrable. Assume that all the functions are compactly supported.
No, for example the FT of step function has decay $1/x$. (I assume that a hat is missing and you're asking if a BV function has integrable FT.)
When f is of bounded variation the Fourier sine transform F^s(f) and the Fourier cosine transform F^c(f) -which form the Fourier transform of a function f := F^f(f)= F^c(f) -i F^s(f) have different behavior. Respectively, F^s(f) is not (HK-)integrable while F^c(f) is.
If $f'$ is of bounded variation, then $\hat{f}$ will be integrable. To see this, note that by your assumptions, $f''$ (in the sense of tempered distributions) is a finite complex-valued measure, so that $\widehat{f''}$ is bounded, meaning that $|\hat{f}(\xi)|\lesssim 1/(1+|\xi|^2)$.
|
2025-03-21T14:48:31.383047
| 2020-06-27T21:32:26 |
364303
|
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|
Stack Exchange
|
Invertible bimodule for hereditary algebras
Let $A=kQ$ be a path algebra over a field $k$ for a finite acyclic quiver with enveloping algebra $A^e$.
Question: When is it true that $\tau_{A^e}(A) \cong A$ as a left and as a right $A$-modules? (which should mean that $\tau_{A^e}(A)$ is an invertible bimodule)
This holds when $Q$ is the quiver with 3 vertices 1,2,3 and arrows 1->2 and 3->2. Curiously, I found no other examples.
I think this is related to the fact that $\tau_{A^e}^2(A) \cong A$ as $A$-bimodules for this example, and it feels like this is the unique example where this (or even $\tau_{A^e}^k(A) \cong A$ for some $k \geq 1$) holds.
|
2025-03-21T14:48:31.383124
| 2020-06-27T22:01:21 |
364304
|
{
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"Dan Piponi",
"Stefano Gogioso",
"https://mathoverflow.net/users/1233",
"https://mathoverflow.net/users/128347",
"https://mathoverflow.net/users/76468",
"shoosh"
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"sort": "votes",
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}
|
Stack Exchange
|
Fast way to generate random points in 2D according to a density function
I'm looking for a fast way to generate random points in 2D according to a given 2D density function.
For instance something like this:
Right now I'm using a modified version of "Poisson disc" method I found here:
https://www.jasondavies.com/poisson-disc/
To make this fast, it relies on a grid on fixed interval.
In my case, I don't know the right size of the grid cells since the density changes over the plane.
(To generate this image I used no grid at all and it's pretty slow)
Is there a "right way" to generalize this algorithm for this need?
Maybe a different approach altogether?
How is the function specified? Is it blackbox or do you know it in advance?
Do you want the samples to be independent? The fact that you're using a Poisson disc method suggests you're not. It can make a big difference.
@StefanoGogioso function is just code I can calculate at any point
@DanPiponi not independent. If the density is constant, they need to be more or less constant distance from each other
Assuming you can easily compute the infimum and supremum of the density function $f$ on a given square with axis-aligned sides, you can use a quadtree to perform an approximate sampling. Let's assume that $f$ is bounded.
Work in 2D Cartesian coordinates and write $S_{l, b, \delta}$ for the axis-aligned square containing all points $(x,y)$ with $l \leq x < l+\delta$ and $b \leq y < b+\delta$, where $\delta > 0$. Assume that you start with a square domain $S_{l_0, b_0, \delta_0}$, within which you want to sample points. Fix an $\epsilon > 0$ and proceed to build a quadtree as follows, starting from the quadtree with a single node $S_{l_0, b_0, \delta_0}$ (a leaf).
For every node $S_{l, b, \delta}$ of the quadtree, compute the infimum $m_{l, b, \delta} := \inf\{f(x,y) | (x,y) \in S_{l, b, \delta} \}$ and supremum $M_{l, b, \delta} := \sup\{f(x,y) | (x,y) \in S_{l, b, \delta} \}$ of the density function $f$ on the node. These two quantities give bounds for the integral of $f$ over the node:
$$
m_{l, b, \delta}\delta^2
\leq
\int_{S_{l, b, \delta}} f(x,y) \,d(x,y)
\leq
M_{l, b, \delta} \delta^2
$$
If $(M_{l, b, \delta} - m_{l, b, \delta})\delta^2 > \epsilon$ for any leaf $S_{l, b, \delta}$ of the quadtree, proceed to subdivide the leaf into four new leaves, each with side $\delta/2$:
$$
S_{l+\lambda\frac{\delta}{2}, b+\beta\frac{\delta}{2}, \frac{\delta}{2}}
\hspace{5mm} \text{for} \hspace{5mm}
\lambda, \beta \in \{0, 1\}
$$
Because $f$ is bounded, this process is guaranteed to terminate and the maximum depth for the quadtree is $O\left(\log(\epsilon^{-1})\right)$, or more specifically it is bounded above by the smallest $d \geq 0$ such that:
$$
(M_{l_0, b_0, \delta_0} - m_{l_0, b_0, \delta_0}) \delta_0^2 2^{-2d} \leq \epsilon
$$
The number of nodes in the quadtree can be $O\left(\epsilon^{-1}\right)$ in specific ill-behaved cases, but it can be guaranteed to be $O\left(\log(\epsilon^{-1})\right)$ if some additional regularity requirements are imposed on $f$ (I seem to recall that $f$ being Lipschitz continuous with a fixed upper bound on the Lipschitz constant is enough).
When subdivision is terminated, define an approximate integral $I_{l, b, \delta}$ for each node as follows:
if $S_{l, b, \delta}$ is a leaf, set the approximate integral to $I_{l, b, \delta} := M_{l, b, \delta} \delta^2$;
if $S_{l, b, \delta}$ is an internal node, set $I_{l, b, \delta}$ to be the sum of the approximate integrals over its children:
$$
I_{l, b, \delta}
:=
\sum_{\lambda=0}^1 \sum_{\beta=0}^1
I_{l+\lambda\frac{\delta}{2}, b+\beta\frac{\delta}{2}, \frac{\delta}{2}}
$$
Because of the termination condition for quadtree subdivision, we now have the following approximation for every node $S_{l, b, \delta}$ in the quadtree:
$$
\left| I_{l, b, \delta} - \int_{S_{l, b, \delta}} f(x,y) \,d(x,y)\right| \leq \epsilon \delta^2
$$
Once the quadtree is built, $\epsilon$-approximate sampling of a point according to the density function $f$ can be performed efficiently by descending the quadtree. Starting from the root, at every internal node $S_{l, b, \delta}$, randomly choose one of the four children $S_{l+\lambda\frac{\delta}{2}, b+\beta\frac{\delta}{2}, \frac{\delta}{2}}$ with probability given by:
$$
\mathbb{P}(\lambda, \beta) := \frac{I_{l+\lambda\frac{\delta}{2}, b+\beta\frac{\delta}{2}, \frac{\delta}{2}}}{I_{l, b, \delta}}
$$
Once a leaf is reached, sample a uniformly distributed point within it.
This produces independent and identically distributed samples for an $\epsilon$-approximation to $f$, each sample taking $O(\log(\epsilon^{-1}))$ time and using $O(\log(\epsilon^{-1}))$ samples from a uniform probability distribution on $[0,1]$ (one sample for each step of the descent from root to leaf plus two samples for the leaf).
|
2025-03-21T14:48:31.383410
| 2020-06-27T22:30:18 |
364306
|
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"Eduardo Longa",
"Michael Albanese",
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|
Stack Exchange
|
Positive scalar curvature on the total space of a circle bundle
Let $(\Sigma_\gamma,g)$ be a closed and orientable Riemannian surface of genus $\gamma \geq 1$, $(M^3,\tilde{g})$ be a closed, connected and orientable Riemannian $3$-manifold, and $\pi : M \to \Sigma_\gamma$ be a Riemannian fibre bundle whose fibers are minimal circles. Is it known whether the scalar curvature of $(M,\tilde{g})$ can be strictly positive?
It is a theorem of Gromov and Lawson, also Schoen and Yau, that no closed orientable three-manifold which contains an aspherical factor in its prime decomposition can admit a metric of positive scalar curvature, see Theorem IV.6.18 of Spin Geometry by Lawson and Michelsohn. In particular, as the three-manifold you're interested in is aspherical, it does not admit a metric of positive scalar curvature (irrespective of the nature of the fibers).
In fact, thanks to the solution of the elliptisation conjecture, we now know that a closed orientable three-manifold admits a metric of positive scalar curvature if and only if its prime decomposition contains no aspherical factors.
A natural question to ask is whether your manifold can admit a metric of non-negative scalar curvature. In the absence of positive scalar curvature metrics, such a metric is Ricci-flat, and on a three-manifold, a Ricci-flat metric is flat. If $M$ admits a flat metric, it is finitely covered by $T^3$ and hence $b_1(M) \leq 3$. As $M$ is an orientable circle-bundle over $\Sigma_{\gamma}$, the Gysin sequence tells us that $\pi^* : H^1(\Sigma_{\gamma}) \to H^1(M)$ is injective, and hence $\gamma = 1$. If the Euler class of the bundle is non-zero, then $M$ is finitely covered by the Heisenberg manifold $H(3, \mathbb{R})/H(3, \mathbb{Z})$ which does not admit a flat metric. On the other hand, if the Euler class is zero, then $M = T^3$ which certainly does.
Thank you for your answer. Why is $M$ aspherical?
It is a circle bundle over an aspherical manifold. In general, if you have a fiber bundle with two of the three spaces aspherical, then it follows for the long exact sequence in homotopy that the third is also aspherical. Alternatively, one can show directly that a circle bundle over $\Sigma_{\gamma}$ for $\gamma \geq 1$ has universal cover $\mathbb{R}^3$.
|
2025-03-21T14:48:31.383690
| 2020-06-28T00:11:10 |
364310
|
{
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"Marcus M",
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"url": "https://mathoverflow.net/questions/364310"
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|
Stack Exchange
|
Static Widom-Rowlinson model
In Elena Pulvirenti's slides she introduced a $\textbf{static Widom-Rowlinson model of one species}$. Consider $\Lambda\subset R^2$ with periodic boundary conditions, $\Lambda$ set of particle configurations with
$$\Gamma=\{\gamma\subset \Lambda: N(\gamma)\in \mathbb{N}\}, \, $$
where $N(\gamma)$ is the cardinality of $\gamma$. The halo of a configuration is $h(\gamma)=\cup_{x\in \gamma}B_2(x)$ with radius $2$. Let $H(\gamma)=|h(\gamma)|-N(\gamma)|B_2(0)|$ be the Hamiltonian.
Define the grand-canonical Gibbs measure,
$$(1) \mu(d\gamma)=\frac{z^{N(\gamma)}}{\Xi}e^{-\beta H(\gamma)}\mathbb{Q}(d\gamma) $$
where $\mathbb{Q}$ is Poisson point process with intensity 1 and $\Xi$ is the partition function.
Her result is that the 2-species Widom-Rowlinson model is equivalent to 1-species. $\textbf{The 2-species WR model}$ is two types of particles(blue and red) with configurations $\gamma^B, \gamma^R$. The grand-canonical Gibbs measure:
$$(2) \hat{\mu}(d\gamma^R, d\gamma^B)=\frac{1}{\hat{\Xi}}1_{\{\text{red-blue hard-core}\}}z_R^{N(\gamma^R)}z_B^{N(\gamma^B)}\mathbb{Q}(d\gamma^R)\mathbb{Q}(d\gamma^B)$$
where $1_{\{\text{red-blue hard-core}\}}$ means it is $1$ if $d(\gamma^R, \gamma^B)\geq 1$, otherwise is 0, and $z_R=e^{\beta\lambda_R}$ and $z_B=e^{\beta\lambda_B}$.
$\textbf{My question is why 1-species and 2-species are equivalence?}$ I am confused about that fix the centers of the red discs and integrate over the centers of the blue disc, then:
$$\frac{1}{\hat{\Xi}}\int_{\Gamma} 1_{\{\text{red-blue hard-core}\}}z_R^{N(\gamma^R)}z_B^{N(\gamma^B)}\mathbb{Q}(d\gamma^B)=C \frac{z^{N(\gamma^R)}}{\Xi}e^{-\beta H(\gamma^R)}$$
where $(z_B, z_R)\to (\beta, ze^{\beta V_0})$ and $V_0:=|B_2(0)|$.
So let's think about it this way: if $A$ is an event in the two-type model depending only on red then
\begin{align*}
\mathbb{P}(A) &= \frac{1}{\tilde{\Xi}} \sum_{j,k\geq 0} \frac{z_R^k}{k!} \frac{z_B^j}{j!} \int_{S^k} \int_{S^j} 1_{A} \cdot 1_{RBHC} \,dy\, dx \\
&=\frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}\int_{S^j} 1_{RBHC} \,dy \right)\,dx \\
&= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}\int_{(S \setminus h(\gamma_R))^j} \,dy \right)\,dx \\
&= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}(|S| - |h(\gamma_R)|)^j \right)\,dx \\
&= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \exp(z_B |S| - z_B |h(\gamma_R)|)\,dx \\
&= \frac{C}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A e^{- z_B |h(\gamma_R)|)}\,dx \\
&= \frac{C}{\tilde{\Xi}} \sum_{k\geq0}\frac{(z_R e^{-z_BV_0})^k}{k!} \int_{S^k} 1_A e^{- z_B (|h(\gamma_R)| - k V_0))}\,dx \,.
\end{align*}
Doing the change of variables listed completes it.
So actually neither of those matters in the above computation, they're both just the normalizing constant that makes it a probability measure. My $S$ is the underlying space on which this model is defined (so some finite volume subset of R^d).
If you'd like, you can use the identity $1/\Xi = P(\gamma_R = \emptyset)$, and then the above shows that $\Xi = \tilde{\Xi}/C$. This is equivalent to what I described in my previous comment.
|
2025-03-21T14:48:31.383890
| 2020-06-28T02:07:40 |
364311
|
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|
Stack Exchange
|
On a revised quantum Riemann hypothesis
This post provides a revision of the disproved quantum Riemann hypothesis proposed 2 years ago in this post, where you can refer to have more details about the motivations, the notations and the justification of the term quantum.
Let first recall that for $n \in \mathbb{N}$, $\sigma$ the divisor function, and $\gamma$ the Euler–Mascheroni constant, then $$\limsup_{n \to \infty} \frac{\sigma(n)}{n \log \log n} = e^{\gamma}.$$
Next Robin's theorem states that the Riemann hypothesis is true if and only if for $n$ large enough $$\sigma(n) < e^\gamma n \log \log n.$$
Let $\mathfrak{F}$ be the set of irreducible finite index subfactor planar algebras (up to equivalence), and let $\mathfrak{F}_r$ be those of depth $r \ge 2$.
Some reminders:
Watatani's theorem states that for $\mathcal{P} \in \mathfrak{F}$, its biprojection lattice $[e_1, id]$ is finite.
The index of a biprojection $b$ is noted $|b : e_1|$, and $|id:e_1|$ is the index of $\mathcal{P}$, also noted $|\mathcal{P}|$.
The index is multiplicative, i.e. for $b,b'$ biprojections then $$b \le b' \Rightarrow |b':e_1| = |b':b| \cdot |b:e_1|.$$
The index $|b:e_1|$ is a cyclotomic integer, and restricted by Jones index theorem.
Consider the divisor set
$$D(\mathcal{P}) = \{ |b : e_1| \ \text{ with } \ b \in [e_1,id] \},$$
which then is a finite subset of the (generically infinite) number theoretic divisor set $D(|\mathcal{P}|)$ in the ring of cyclotomic integers (of appropriate degree), see this post.
Consider the divisor function
$$\sigma(\mathcal{P}) = \sum_{d \in D(\mathcal{P})} d.$$
Question 1: Is there a (finite) constant $\gamma_r$ such that the following equality holds?
$$\limsup_{\mathcal{P} \in \mathfrak{F}_r, \\ |\mathcal{P}| \to \infty} \frac{\sigma(\mathcal{P})}{|\mathcal{P}| \log \log |\mathcal{P}|} = e^{\gamma_r}$$
Note that $\gamma_2 = \gamma$ because the index of a biprojection of $\mathcal{P} \in \mathfrak{F}_2$ is a rational integer, and for $G$ a finite group (in particular $C_n$), the subfactor planar algebras $\mathcal{P}(G)$ is in $\mathfrak{F}_2$, and there is a 1-1 (Galois) correspondance between $H\le G$ and the biprojection of $\mathcal{P}(G)$, whose indices are precisely given by $|H|$.
FYI, $\mathfrak{F}_2$ contains exactly the finite dimensional Hopf C*-algebra subfactor planar algebras.
Assume that Question 1 admits a positive answer (otherwise the asymptotic should be improved), we can then formulate a (revised) depth $r$ quantum Riemann hypothesis (QRH$_r$) as follows:
(QRH$_r$) For $\mathcal{P} \in \mathfrak{F}_r$ and $|\mathcal{P}|$
large enough then $\sigma(\mathcal{P}) < e^{\gamma_r} |\mathcal{P}| \log \log |\mathcal{P}|$.
Note that QRH$_2$ is (obviously) equivalent to RH.
Question 2: What is about QRH$_r$ when $r>2$?
We can provide a variation, replacing $\mathfrak{F}_r$ by the set of $\mathcal{P} \in \mathfrak{F}$ for which the indices of the biprojections are in $\mathbb{Z}[\zeta_r]$ with $\zeta_r = e^{2\pi i/r}$ and $r$ minimal. Let us call this variation QRH$^r$. If relevant, we can combine both variation as QRH$_{r'}^{r}$.
Note that it would be inappropriate to replace $\sigma(\mathcal{P})$ by $$\tilde{\sigma}(\mathcal{P})=\sum_{b \in [e_1,id]} |b :e_1|,$$
because then above asymptotic would be incorrect, but mostly because there would be no more relation with RH, because if $\mathcal{P}_G$ is the group subfactor planar algebra for the finite group $G$ then $|\mathcal{P}_G| = |G|$ and $\tilde{\sigma}(\mathcal{P}_G)$ is the sum of the order of the subgroups of $G$ (let us call it $\tilde{\sigma}(G)$), which explodes for the abelian $p$-groups as $G = C_2^n$:
$$ \begin{array}{c|c} n&1&2&3&4&5&6&7 \newline \hline
|C_2^n|&2&4&8&16&32&64&128 \newline \hline
\tilde{\sigma}(C_2^n)&3&11&51&307&2451&26387&387987 \end{array} $$
To learn more, you can watch the series Quantum Symmetries and Quantum Arithmetic.
The last video finishes on QRH.
|
2025-03-21T14:48:31.384160
| 2020-06-28T02:14:35 |
364313
|
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"Bjørn Kjos-Hanssen",
"Heisenberg",
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|
Stack Exchange
|
Stochastic integral with respect to a random field
I came across a generalized Black-Scholes equation formulation in this paper.
Let me highlight the basic idea below. Consider a random field $W(t,T)$ where for a fixed $T$, $W$ is a Brownian motion and for a fixed $t$, $W$ is a continuous function. $W$ satisfies the following:
$dW(t,u)dW(t,v)=c(u,v)dt$
$dW(t,u)dW(t,u)=dt$
The author's define the dynamics of some asset as follows:
$$\frac{dS(t)}{S(t)}=\mu(t)dt+\int_{T_1}^{T_2}\sigma(t,u)dW(t,u)du$$
The second term on the right side intrigues me. The paper doesn't really get into explaining the doing calculus with such processes so I just want to ask the following.
If I define $dg(t)=\int_{T_1}^{T_2}\sigma(t,u)dW(t,u)du$ then does it follow that
$$dg(t)dg(s)=\int_{T_1}^{T_2}\sigma(t,u)dW(t,u)du \int_{T_1}^{T_2}\sigma(t,v)dW(t,v)dv\\=\int_{T_1}^{T_2}\int_{T_1}^{T_2}\sigma(t,u)\sigma(t,v)c(u,v)dtdudv$$
Do we have the Ito isometry?
$$E\left[\left(\int_0^t\int_{T_1}^{T_2}\sigma(s,u)dW(s,u)du\right)^2\right]=E\left[\int_0^t\int_{T_1}^{T_2}\int_{T_1}^{T_2}\sigma(s,u)\sigma(s,v)c(u,v)dudvds\right]$$
I was not able to find any literature on such integrals. If anyone can suggest any references that would also help a lot. Thanks!
Let's see what happens if we set some things constant that you haven't ruled out could be constant.
If $\sigma\equiv 1$ and $W(s,u)=W_s$ (so $W$ does not depend on $u$) then $c\equiv 1$ and it looks like your Ito isometry does not hold:
$$t^2 (T_2-T_1)=E\left(\left(t(W_{T_2}-W_{T_1})\right)^2\right)=E\left(\left(\int_0^t\int_{T_1}^{T_2} dW_s du\right)^2\right) \ne E\left(\int\int\int du,dv,ds\right)= t(T_2-T_1)^2$$
@BjørnKjos-Hanssen Thank you
The notations in the question are ambiguous (Bjørn Kjos-Hanssen showed that the other interpretation cannot be correct). I assume that the expression of interest is given by
$$
g(t) = \int_0^t \Sigma(s)\,dW(s)\;,
$$
where $W$ is a $C(\mathbb{R})$-valued Wiener process with covariance $\hat c$ (at time $1$) and $\Sigma(s) \in C(\mathbb{R})^*$ is the finite measure given by $\Sigma(s) f = \int_{T_1}^{T_2}\sigma(s,u)f(u)\,du$. Here, the covariance $\hat c$ is the bilinear map on $C(\mathbb{R})^*$ such that, for measures $\mu$ and $\nu$,
$$
\hat c(\mu,\nu) = \int c(u,v)\, \mu(du)\,\nu(dv)\;.
$$
Itô isometry then indeed reads
$$
\mathbb{E} g(t)^2 = \int_0^t \mathbb{E} \hat c(\Sigma(s),\Sigma(s))\,ds \;,
$$
assuming of course that $\Sigma$ is adapted and square integrable.
Regarding references, any book on SPDEs would do, for example "Stochastic Equations in Infinite Dimensions" by Da Prato & Zabczyk or Section 3 of my lecture notes.
Thank you this helps a lot. This is probably a very basic questions but I don't fully understand how you interpret $c$. On the left side of the second equation you have $c$ evaluated at two measures and on the right side $c$ is evaluated at to real numbers. Why is this possible?
Edited for clarity.
@MartinHairer Usually there is some linear operator $Q$ with finite trace in $C(\mathbb{R})$ so that $W_t-W_s\sim N(0,(t-s)Q)$. What is $Q$ in terms of $c$?
Maybe @MartinHairer can correct me but I think if we were to think of a Hilbert space valued Weiner process, $Qf(\cdot)=\int c(s,\cdot) f(s)ds$, $f\in H$ and $Q\in L(H).$
|
2025-03-21T14:48:31.384389
| 2020-06-28T06:57:20 |
364318
|
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|
Stack Exchange
|
Flat conical hypersurfaces in $\mathbb{R}^4$
What can be said about an isometric immersion $f:\mathbb{R}^3-\left\{0\right\}\to\mathbb{R}^4$ such that $\left\|f(x)\right\|=\left\|x\right\|$ for all $x\in\mathbb{R}^3-\left\{0\right\}$?
That depends on what kind of solutions one is willing to allow. It's not hard to show that such a map is of the form $f(x) = |x|g(x/|x|)$ for some isometric immersion $g:S^2\to S^3$, and there are many such isometric immersions $g$, especially if one only requires that $g$ be $C^1$. It is easy to write down $C^1$ examples that are smooth except at two points in $S^2$, but it is not obvious to me whether there is an example that is $C^2$ everywhere on $S^2$ other than the obvious totally geodesic isometric immersions. I imagine that this is known, but I don't recall a reference.
Interesting. I guess the argument in the paper 'Isometric immersions of constant curvature manifolds' by Ferus uses only $C^2$.
I don't know Ferus' paper. I guess you mean that Ferus assumes that the mapping is at least $C^2$. What conclusion does he draw from this?
@RobertBryant: Exactly. He shows that every isometric immersion of $\mathbb{S}^n$ into $\mathbb{S}^{2n-1}$ is totally geodesic.
|
2025-03-21T14:48:31.384503
| 2020-06-28T07:00:34 |
364319
|
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"LSpice",
"Qiaochu Yuan",
"Steven Landsburg",
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"skbmoore"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364319"
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|
Stack Exchange
|
Approximating power series coefficients --- Why does a clearly illegitimate method (sometimes) work so well?
For reasons that don't matter here,
I want to estimate the power series coefficients
$t_{ij}$ for the rational function
$$T(x,y)= {(1+x)(1+y)\over 1- x y(2+x+y+x y)}=\sum_{i,j} t_{ij}x^iy^j$$
Using a method that I cannot justify, I get
highly accurate estimates when $i=j$ and highly inaccurate estimates when
$|i-j|$ strays at all far from zero.
My questions are:
Q1) Why does my apparently illegitimate method work so well when $i=j$?
Q2) Why does the answer to Q1) not apply when $i\neq j$ ?
(Of course, once the answer to Q1) is known, the answer to Q2) might be
self-evident.)
I'll first present the method, then explain why I think it shouldn't work,
then present the evidence that it works anyway when $i=j$, and then present
the evidence that it rapidly goes haywire when $i\neq j$.
The Apparently Illegitimate Method:
Note that $t_{ij}=t_{ji}$, so we can limit ourselves to estimating
$t_{j+k,j}$ for $k\ge 0$.
I) Define
$$T_k(y)=\sum_jt_{k+j,j}y^j$$
For example, a residue calculation gives
$$ T_0(y)= {1-y-\sqrt{1-4y+2y^2+y^4}\over y\sqrt{1-4y+2y^2+y^4}} $$
It turns out that all of the $T_k$ share a branch point at $\zeta\approx .2956$ and are analytic in the disc $r<\zeta$.
II) Write
$$L_k=\lim_{y\mapsto \zeta} T_k(y)\sqrt{y-\zeta}$$.
Discover that $L_0\approx 1.44641$ and $L_k=L_0/\zeta^{k/2}$.
III) Approximate
$$T_k(y)\approx L_k/\sqrt{y-\zeta}$$
IV) Expand the right hand side in a power series around $y=0$ and equate
coefficients to get
$$t_{ij}\approx \pm{L_0\over\sqrt{\zeta}}\pmatrix{-1/2\cr j\cr}\zeta^{-(i+j)/2}
\approx \pm 2.66036 \pmatrix{-1/2\cr j\cr}\zeta^{-(i+j)/2}\qquad(E1)$$
Remarks:
Obviously one could try to improve this approximation
at Step III by using more terms in the power series for $T_k$ at $y=\zeta$.
This doesn't seem to help, except when $k=0$, in which case the original approximation is already quite good.
For $k\ge 2$, $T_k(y)$ has a zero of
order $k-1$ at the origin. Thus one could modify this method by approximating
$T_k(y)/(y^{k-1})$ instead of $T_k(y)$
This yields
$$t_{ij}\approx \pm{2.66036}\pmatrix{-1/2\cr 1-i+2j}\zeta^{-(i+j)/2}\qquad(E2)$$
(E2) is (much) better than (E1) in the range $i\ge 2j+1$, where it gets
exactly the correct value, namely zero. Otherwise, it seems neither systematically better nor worse.
Why Nothing Like This Should Work: The expansion of $T_k(y)$ at
$\zeta$ contains nonzero terms of the form
$A_{i,j}(\zeta-y)^j$ for all positive integers $j$. (I'm writing $i=j+k$ to
match up with the earlier indexing.) The truncation at Step III throws all
these terms away. Therefore the expansion around the origin in Step IV
ignores (among other things) the contribution of $A_{ij}$ to the estimate
for $t_{ij}$. So unless we can control the sizes of the $A_{ij}$, we
have absolutely no control over the quality of the estimate.
And in fact, even when $k=0$, the $A_{j,j}$ are not small.
For example, $t_{8,8}=8323$ and my estimate for $t_{8,8}$ is a
respectable $8962.52$. But $A_{8,8}$, which should have contributed to that
estimate and got truncated away, is equal to $58035$. It seems remarkable
that I can throw away multiple terms of that size and have the effects nearly cancel.
I'd like a conceptual explanation for this.
But When $i=j$, It Works Anyway:
and these get even better if you truncate just slightly farther out.
Why any explanation can't be too general:
Incidentally, for your original problem you might try the techniques in Pemantle and Wilson’s Analytic Combinatorics in Several Variables: https://www.math.upenn.edu/~pemantle/papers/ACSV.pdf
The paper 'A New Method for Computing Asymptotics of Diagonal Coefficients of Multivariate Generating Functions' by A. Raichev and M. Wilson has the precise machinery that can solve this problem. Get a copy and these brief notes correspond to their symbols, for the diagonal case
$$ f_{n,n} = [x^n \, y^n ] \frac{I(x)}{J(x)} = [x^n \, y^n ] \frac{(1+x)(1+y)}{1-xy(2+x+y+xy)}.$$
Solve the simultaneous system for zeros of the denominator $J$; in Mathematica,
Solve[ { x D[J,x] == y D[J,y], J==0 },{x,y} ].
The proper solution must have both $x$ and $y$ positive. That set is
$$ \mathbf c=(\rho,\rho),\,\rho=(\tau-2/\tau-1)/3, \, \tau=(17+3\sqrt{33})^{1/3} \approx 0.543689.$$
This solution set, with identical $c_1 = c_2$, falls under the purview of a simplified calculation, in which it can be shown
$$ f_{n,n} \sim \rho^{-2n} \frac{1.5009481}{\sqrt{n}}.$$
(The true amplitude can be written in terms of $\rho$, and as many decimal places as wanted are possible, but I'm not going to bother to typeset it.)
For comparison:
$n=40$, $\text{true}=3.4601\times 10^{20}$, $\text{asym} = 3.5261\times 10^{20}$ , $\text{absolute % err} = 1.91\%$.
$n=200$, $\text{true}=7.6554\times 10^{104}$, $\text{asym} = 7.6847\times 10^{104}$ , $\text{absolute % err} = 0.38\%$.
For the non-diagonal case, you will be looking at $f_{an,bn}$. The machinery should work, though it is more complicated. You'll get roots that depend on $(a,b)$ and have to solve a complicated determinant to get the amplitude, also dependent on $(a,b)$. The question is, why does the non-diagonal case deviate rapidly from the diagonal? Is there a way to understand this qualitatively? I believe the answer lies in the modified form
$$ f_{an,bn} \sim c_1(a,b)^{-a n} c_2(a,b)^{-b n} \cdot \operatorname{amp}(a,b)/\sqrt{n}.$$
The amplitude will vary only like a polynomial upon changing $(a,b)$, but the first two factors have an exponential dependence.
Glad it helped! I like to edit by hand, but the GUI also has some nice tools. (The link button looks vaguely like two links of chain, if you squint.) There's also some editing help, which is accessible by clicking on the '?' icon when you post. (Personally I disagree with the advice that [link][1] followed by [1]: url is easier to read than [link](url), but to each their own. Posts aren't meant to be read in source form anyway.)
For n=40 and n=200, my method gives better approximations than yours does. My first question is why my method does so well. I don't think you've addressed that.
@StevenLandsburg True, I haven't answered 'why my method does so well,' or that for certain cases yours might be better. I can guess that for certain $n$ a perturbed $\rho$ could give a better result? By the way, my $\rho,$ given as an explicit algebraic number, appears to be the square of your $\zeta,$ given only numerically. You also have some binomials, which may reduce to that factor $1/\sqrt{n}.$ The answer I presentedgotten in my answer.
@skbmoore: I'm finally getting around to writing some of this up and I'd like to thank you for this helpful answer. Would you prefer to be acknowledged as skbmoore or under your real full name, and if the latter, what is that name? If you'd rather not post here you can write me at<EMAIL_ADDRESS>.
@StevenLandsburg Hello Steven. I'm glad you found my remarks helpful. skbmoore is fine. Since I like to keep up with applications of asymptotics, please send me a notice when you post it (most likely) on the ArXiv.
|
2025-03-21T14:48:31.384971
| 2020-06-28T08:46:31 |
364321
|
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|
Stack Exchange
|
On periods of symmetric algebras
Let $A$ be a symmetric finite dimensional algebra over a field of characteristic two (or even over the field with two elements) such that every simple $A$-module has the same period equal to $n$.
Question: Is the period of the algebra $A$ also equal to $n$?
This is not true in other characteristics, but I am not aware of an example in characteristic two, but there probably is one out there.
|
2025-03-21T14:48:31.385029
| 2020-06-28T08:46:43 |
364322
|
{
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"authors": [
"Vidit Nanda",
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"url": "https://mathoverflow.net/questions/364322"
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|
Stack Exchange
|
Is there discrete Morse theory on acyclic categories?
Forman introduced discrete Morse theory on finite regular cell complexes. Minian introduced a version of discrete Morse theory for posets which generalizes Forman's original Morse theory https://arxiv.org/abs/1007.1930. So I thought the following problem:
Is there a version of discrete Morse theory for acyclic categories which generalizes Minian's Morse theory?
A desire theorem (It is imaginary and is not logical and currect.) is that there is a "function (functor)" from an acyclic category such that the classifying space of the given acyclic category is homotopy equivalent to a CW complex such that the number of cells = the number of "critical objects".
An acyclic category is a small category in which only the identity morphisms have inverses and any morphism from an object to itself is an identity morphism. A poset $P$ is an acyclic category.
You might be interested in this paper, which describes morse theory on a certain class of acyclic poset-enriched categories via localization: https://arxiv.org/abs/1510.01907 If this is the sort of thing you are looking for, let me know and I can describe it in an answer.
Recently, it appeared in ArXiv (https://arxiv.org/abs/2107.06202) a paper which may be related to this question. The title is:
Morse theory for loop-free categories
Note that loop-free categories is the same as acyclic categories. The abstract is:
We extend discrete Morse-Bott theory to the setting of loop-free (or
acyclic) categories. First of all, we state a homological version of
Quillen's Theorem A in this context and introduce the notion of
cellular categories. Second, we present a notion of vector field for
loop-free categories. Third, we prove a homological collapsing theorem
in the absence of critical objects in order to obtain the Morse
inequalities. Examples are provided through the exposition. This
answers partially a question by T. John: whether there is a Morse
theory for loop-free (or acyclic) categories?
|
2025-03-21T14:48:31.385175
| 2020-06-28T09:26:33 |
364324
|
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|
Stack Exchange
|
Infinitely many $n$ such that $\gcd(\lfloor n\sqrt{2}\rfloor, \lfloor n\sqrt{3}\rfloor)=m$
Is it true that for any positive integer $m$ there are infinitely many positive integers $n$ such that $\gcd(\lfloor n\sqrt{2}\rfloor, \lfloor n\sqrt{3}\rfloor)=m$?
$\lfloor x \rfloor$ is the floor function of $x$ and $\gcd$ is the greatest common divisor.
In fact we can prove that this result is true for $\gcd(\lfloor n\sqrt{2}\rfloor, n)=m$ (consider a solution $(x_k,y_k)$ to the Pell equation $x^2-2y^2=-1$, such that $x_k>m$ and take $n=my_k$).
Does this result hold if instead of $\sqrt{3}$ we have $\sqrt{k}$, where $k$ is not a perfect square, $k>2$?
Can this result be generalized in any way?
If $a$ and $b$ are positive real numbers such that $\gcd(\lfloor na \rfloor$, $\lfloor nb \rfloor)=m$ holds for infinitely many positive integers $n$ for each $m$, what can be said about $a$ and $b$?
By the multidimensional equidistribution theorem, for any $m$, the probability of $m$ dividing both $\lfloor{n\sqrt 2}\rfloor$ and $\lfloor{n\sqrt 3}\rfloor$ is $1/m^2$. $gcd(\lfloor{n\sqrt 2}\rfloor, \lfloor{n\sqrt 3}\rfloor)=m$ is equivalent to $m$ dividing both $\lfloor{n\sqrt 2}\rfloor$ and $\lfloor{n\sqrt 3}\rfloor$, but neither of $2m,3m,5m,7m,11m,...$ divides both $\lfloor{n\sqrt 2}\rfloor$ and $\lfloor{n\sqrt 3}\rfloor$. (One should use an effective estimate here, in order to deal with infinite number of primes.)
The answers to your last two questions are both "no". The correct generalization is requiring $a,b$ nonzero and $a/b$ irrational. If $a/b$ is rational, let $a/b=p/q$ ($p,q\in \mathbb Z$), and $q\lfloor na \rfloor-p \lfloor nb \rfloor$ is bounded independent of $n$, so $\gcd(\lfloor na \rfloor,\lfloor nb \rfloor)$ is either bounded or of the order of $n$. If $a/b$ is irrational, the equidistribution theorem works (use one-dimensional equidistribution if only one of $a$ and $b$ is irrational).
@LeechLattice Thanks, I edited the question.
|
2025-03-21T14:48:31.385315
| 2020-06-28T09:32:33 |
364325
|
{
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"Dieter Kadelka",
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"leo monsaingeon"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364325"
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|
Stack Exchange
|
Properties of the total variation norm on space of totally finite measure (from Bogachev)
Let $(X,d)$ be a metric space, $\mathcal{B}$ the Borel $\sigma$-algebra on $X$, and $\mathcal{M}(X)$ the space of totally finite measures on $\mathcal{B}$. Let $\|\mu\|_{TV}$ be the total variation norm on $\mathcal{M}(X)$ defined by $$\|\mu\|_{TV} = \mu^+(X) + \mu^-(X) \label{0}\tag{0}$$ where $\mu^+$, $\mu^-$ is the Jordan-Hanh decomposition of $\mu$. Do we have the following properties (like in the case of probability measures):
$$
\|\mu\|_{TV} = \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace .\label{1}\tag{1}$$
According to Bogachev p.177 vol. 1, \eqref{1} is not true if both measures $\mu^+$ and $\mu^-$ are nonzero, and only the following is valid:
$$
\|\mu\|_{TV} \leq 2 \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 \|\mu\|_{TV}. \label{2}\tag{2}
$$
But I have seen (here for example) the total variation norm used as a metric (written $d_{TV}$) on the space of probability measure, with the following definition
$$ d_{TV}(P,Q):=\sup\limits_{A\in \mathcal B}|P(A) - Q(A)|,\label{3}\tag{3}$$
which seems to me to contradict Bogachev. Is there something I am misunderstanding ?
$$
\|\mu\|_{TV} = \frac{1}{2}\sup \limits_{f \text{msb},\|f\|_{\infty} \leq 1} \int f\ d\mu.\label{4}\tag{4}
$$
I have the same question here, does property \eqref{4} holds for $\mu \in \mathcal{M}(X)$ ?
In addition, do you know of any reference treating these questions for totally finite measure (apart from Bogachev) ?
Thanks !
There's no contradiction, since for probability measures $\mu^-=0$. Also, $|P|{TV}=d{TV}(P,0)$. The "problem" arises when BOTH $\mu^\pm$ are nonzero, as clearly illustrated by Stefan's counterexample.
Thanks, if I understand: for $P$ and $Q$ probability measures, $d_{TV}(P,Q) \neq ||P-Q||{TV}$ where $d{TV}$ is defined in \eqref{3} and $||\cdot||_{TV}$ is defined in \eqref{0}. This is where my mistake comes from I think.
I have difficulties to understand the problem. Let $|\mu| := \sup_{A \in \cal{B}} |\mu(A)$. Then as noted in (2) $|,|$ and $|,|_{TV}$ are equivalent norms. Further $|\mu| = \max{\mu^+(X),\mu^-(X)}$. Where is the problem? In (3) there is no problem, both sides define equivalent metrics.
(1) is certainly not true for general signed measures $\mu$. However, if we restrict to signed measures with $\mu(X)=0$, then it is true with a factor of $2$, i.e.
$$\|\mu\|_{TV} = 2 \sup_{A \in \mathcal{B}} |\mu(A)| \tag{*}.$$
That is, in this special case, the leftmost inequality in (2) is attained.
For one inequality, let $X = B^+ \cup B^-$ be the Hahn decomposition for $\mu$. Note that $\|\mu\|_{TV} = \mu(B^+) - \mu(B^-)$, while $\mu(X) = \mu(B^+) + \mu(B^-) = 0$ so that $\mu(B^+) = -\mu(B^-) = \frac{1}{2} \|\mu\|_{TV}$. Hence taking $A = B^+$ shows the $\le$ inequality in (*).
Conversely, for any $A \in \mathcal{B}$, the defining property of the Hahn decomposition implies $\mu(A \cap B^-) \le 0$ and $\mu(A^c \cap B^+) \ge 0$, and therefore we have $$\mu(A) = \mu(A \cap B^+) + \mu(A \cap B^-) \le \mu(A \cap B^+) \le \mu(B^+) = \frac{1}{2} \|\mu\|_{TV}.$$
A similar argument shows $\mu(A) \ge -\frac{1}{2} \|\mu\|_{TV}$, so that $|\mu(A)| \le \frac{1}{2} \|\mu\|_{TV}$. This shows the $\ge$ inequality.
In particular, taking $\mu = P-Q$ where $P,Q$ are both probability measures, we see that $d_{TV}(P,Q)$ as defined by (3) is exactly half of $\|P-Q\|_{TV}$. So the definitions are the same, up to a constant factor of 2.
Your equation (4) is also off by a factor of 1/2. The identity
$$\|\mu\|_{TV} = \sup_{\|f\|_\infty \le 1} \int f\,d\mu$$
is true for every signed measure. To see one direction, write
$$\int f\,d\mu = \int f\,d\mu^+ - \int f\,d\mu^- \le \mu^+(X) + \mu^-(X) = \|\mu\|_{TV}.$$
For the opposite inequality, take $f = 1_{B^+} - 1_{B^-}$.
Let me just give a simple counter-example to your equation (1): take the real line as space (any metric space with at least two points will do) and use the delta measures $\delta_1$ and $\delta_2$ at two distinct points. Then the real measure $\mu = \delta_1 - \delta_2$ will have total variation norm $2$ but $|\mu(A)| \le 1$ for all measurable subsets $A \subseteq \mathbb{R}$.
As a reliable reference, you can take a look at Rudin's book on Real and Complex Analysis.
For the second question: take $P$ and $Q$ be the two delta measures at different points. Then again $\mu = P - Q$ as above has total variation norm $2$. However, the supremum in your equation (3) is $1$, strictly smaller that $2$: if $A$ contains only one of the points (no matter which), then $|P(A) - Q(A)| = 1$. If $A$ contains none, we have $|P(A) - Q(A)| = 0$. If $A$ contains both points, both measures give $P(A) = 1 = Q(A)$, hence also no contribution to the sup.
Thanks. I have the same question as for @leo monsaingeon comment. To be sure: does this mean that $d_{TV}(P,Q) \neq ||P−Q||{TV}$ where $d{TV}$ is defined in \eqref{3} and $||\cdot||_{TV}$ is defined in \eqref{0} ?
This seems to be the case, I added a few lines.
|
2025-03-21T14:48:31.385761
| 2020-06-28T09:46:46 |
364327
|
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"Geoff Robinson",
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|
Stack Exchange
|
Faithful representation of group of order $p^4$
In the (xi) group of the classification of groups of order $p^4$ given by W.Burnside in his book, "Theory of groups of finite order". The group ($\mathbb{Z}_{p^{2}}\rtimes \mathbb{Z}_{p^{}}) \rtimes_{\phi}\mathbb{Z}_{p^{}} $, have presentation
$$\langle a,b,c : a^{p^{2}}=b^p=c^p=e,\,ab=ba^{1+p},\,ac=cab,\,bc=cb\rangle$$
I was trying to explore about the representation (must be faithful) of the above group in $\mathrm{GL}(4,F_p)$, i.e. group of $4\times 4$ invertible matrices, taken over $F_p$. Please guide me in the right direction, from where to start or if my statement mathematically correct or not ?
One small remark: if $p > 4$ then any element $x$ of ${\rm GL}(4,F_{p})$ of order a power of $p$ satisfies $(x-I)^{4} = 0$, so that we certainly have $(x-I)^{p} = 0$ and $x^{p} = I$.
Hence ${\rm GL}(4,F_{p})$ contains no element of order $p^{2}$ when the prime $p$ is greater than $3$.
Kindly explain me how to get this result ? and the above result is true for $p>3$ or $4$
The explanation is given by linear algebra, as shown, except I just notcied a small typo, which I will correct now.. Also, since $p$ is a prime, the requirement that $p>3$ certainly forces $p \geq 5$, so $p > 4.$
I am sorry but I am not able to see the fact that if $x$ is any $p$-element of $GL(4,F_p)$, then it will satisfy ${(x-I)}^4=0$. Can you please elaborate a bit more ?
Let us suppose that $x$ has order $p^{k}$. Then $x^{p^{k}} = I$, so we have $(x-I)^{p^{k}} = 0$ because $F_{p}$ has characteristic $p$. Hence $1$ is the only root of the characteristic polynomial of $x$. By the Cayley-Hamilton Theorem, we must have $(x-I)^{4} = 0.$
|
2025-03-21T14:48:31.385892
| 2020-06-28T09:49:11 |
364328
|
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|
Stack Exchange
|
Braided category inside braided 2-category
Let $\mathcal{C}$ be a semistrict braided monoidal $2$-category in the sense of [BN] (so in particular a strict $2$-category). Let $\mathcal{C}_1$ be the category of $1$-morphisms (objects) and $2$-morphisms (morphisms) in $\mathcal{C}$. (I hope that this is allowed. I think of $\mathcal{C}_1$ as a coproduct of all the Hom-categories in $\mathcal{C}$)
The semistrict monoidal structure ($\boxtimes$) on $\mathcal{C}$ induces a monoidal structure ($\otimes$) on $\mathcal{C}_1$. Let $R: \boxtimes \Rightarrow \boxtimes^{op}$ be the pseudo-natural equivalence corresponding to the braiding in $\mathcal{C}$. For $1$-morphisms $f:a \to c$ and $g:b \to d$ we can define $f\widehat{\otimes}g:=R_{c,d}\circ (f \otimes g)$ and $g\check{\otimes}f:=(g\otimes f)\circ R_{a,b}$ (analogously on morphisms in $\mathcal{C}_1$). By pseudo-naturality of $R$ we have a natural transformation $c:\widehat{\otimes} \to \check{\otimes}^{op}$ in $\mathcal{C}_1$ satisfying the hexagon equations of a braiding.
So $\mathcal{C}_1$ looks very similar to a braided monoidal category, with "the only difference" that $\widehat{\otimes} \neq \check{\otimes}$. Now, assume that $\mathcal{C}$ had enough structure, so that $\mathcal{C}_1$ is an abelian category. Is there a way of "modding out" the difference of $\widehat{\otimes}$ and $\check{\otimes}$ in $\mathcal{C}_1$? Maybe by considering some kind of coequalizer? If so, what are the conditions on $c$ for being well-defined on the new category?
[BN] Baez, Neuchl - Higher-Dimensional Algebra I: Braided Monoidal 2-Categories
|
2025-03-21T14:48:31.386029
| 2020-06-28T10:53:48 |
364334
|
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"Andrej Bauer",
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|
Stack Exchange
|
Why can't we embed Tarski's truth in PA?
I recently learned that ZFC can prove $Con(PA)$ because it can give a model of PA, but I'm not given the technical details. (My teacher thinks it is too obvious to even mention.)
What plagues me is that my naïve intuition tells me that the modeling procedures can be imitated in PA, exactly in the same way.
Here is my attempt:
Let $eval_F$ and $eval_T$ be evaluation functions for Formulas and Terms. Let $e$ denote any variable assignment.
We can define these functions recursively, exploiting Tarski's lemma. Explicitly,
\begin{align}
eval_T(\ulcorner v_i\urcorner,e) &= e[i] \\
eval_T(\ulcorner o\urcorner,e) &= 0 \\
eval_T(\ulcorner s\tau \urcorner,e) &=eval_T(\ulcorner\tau\urcorner,e)+1 \\
eval_T(\ulcorner \tau_1 + \tau_2 \urcorner,e) &=eval_T(\ulcorner\tau_1\urcorner,e)+eval_T(\ulcorner\tau_2\urcorner,e) \\
eval_T(\ulcorner \tau_1 \cdot \tau_2 \urcorner,e) &=eval_T(\ulcorner\tau_1\urcorner,e) \cdot eval_T(\ulcorner\tau_2\urcorner,e)
\end{align}
and
\begin{align}
eval_F(\ulcorner \bot \urcorner,e) &=0 \\
eval_F(\ulcorner \tau_1 = \tau_2 \urcorner,e) &= \chi_=(eval_T(\ulcorner\tau_1\urcorner,e),eval_T(\ulcorner\tau_2\urcorner,e)) \\
eval_F(\ulcorner \Phi\to\Psi \urcorner,e) &= \mathrm{sgn}((1-eval_F(\ulcorner \Phi \urcorner,e))+eval_F(\ulcorner \Psi \urcorner,e)) \\
eval_F(\ulcorner \forall v_i.\Phi \urcorner,e) &=\begin{cases}
1 & (\forall n.eval_F (\ulcorner\Phi\urcorner,e[i\mapsto n]) = 1) \\
0 & (\mathrm{otherwise})\\
\end{cases}
\end{align}
Then $eval_T$ and $eval_F$ are $\Sigma_1^0$- and $\Sigma_2^0$-defined function respectively. Although $eval_F$ is not decidable, at least we know that $eval_F$ is total over coded PA-formulas and the value is either $0$ or $1$.
If we show that every axiom in PA evaluates to $1$ and the inference rules are truth-preserving, then we can show the soundness of the model:
$$
\forall \phi:\mathrm{Form}. (Provable(\phi) \to \forall e. (eval_F(\phi,e)=1))
$$
If so, we can conclude $Con(PA)$, which is $\neg Provable(\ulcorner \bot \urcorner)$, because $\bot$ evaluates to $0$.
Of course, this violates Gödel's second incompleteness theorem, so I must be wrong somewhere - but I couldn't find where.
I am now suspecting three possibilities:
We cannot in fact well-define $eval_T$ and $eval_F$ in PA.
We cannot prove that $eval_F$ models the axioms of PA.
The inference rules does not preserve truth generated by $eval_F$.
I want to know where my argument fails. Thanks in advance.
P.S.
The most suspicious one for me is the second one, especially induction scheme.
Nonetheless I am convinced that induction scheme is provably evaluated to 1, since it reduces to
\begin{align}
\forall \phi:\mathrm{Form}.\forall e. &\forall i. \bigl(
eval_F(\phi,e[i\mapsto 0])=1 \to \\
&\forall n. (eval_F(\phi,e[i\mapsto n])=1 \to eval_F(\phi,e[i\mapsto n+1])=1) \to \\
&\forall n. (eval_F(\phi,e[i\mapsto n])=1 )\bigr)
\end{align}
which is an instance of induction scheme of outer PA.
The thing is, in formal logic you can't define formulas by recursion like you do like that. If you want to define a formula $A$, then you have to define using symbols which do not involve $A$ itself. The way this can be done in ZFC is by constructing not a single formula $eval_F$, but rather a set, call it $e_F$, which contains all the formulas. Sets can be defined iteratively like that, and we can say that $e_F$ is the set which satsifies the inductive properties you list.
Why is $\mathit{eval}_F$ total? What is the PA formula that expresses $\mathit{eval}_F(n) = k$? If you use recursion to define a formula, you also have to explain how to eliminate the recursions, because in first-order logic there is no way to define formulas by recursion the way you do.
Perhaps the following comment will be helpful: you claim that the logical complexity of $\mathit{eval}_F(n) = k$ is $\Sigma^0_2$. If that were the case, then $\mathit{eval}(\lceil \forall v_i . \Phi\rceil) = k$ would have complexity $\Pi^0_3$, at least the way it's written in your definition.
@Wojowu See this Math SE answer. The fact that you can actually explicitize recursively defined predicates or functions even in PA, without any notions about sets, is the significant result of metamathematics from Gödel's famous paper.
@AndrejBauer Now I got the point. I was bugged in the explicitization step. To evaluatie universal quantification, I need to make infinite sequence to prove the evaluation, which is impossible in PA.
Saying my complaints, "You can't define formulas like that" is a bit confusing, because the fact is you can actually recursively define formulas in PA, namely by β-encoding. The entire issue belongs to me by attempting to use recursion where β-encoding is inapplicable.
First-order logic does not provide for definitions of functions by recursion. For example, the transitive closure of a binary relation $R$, though definable from $R$ by recursion, is not in general first-order definable from $R$.
Peano Arithmetic, though formulated in first-order logic, does have enough axioms to support some definitions by recursion. Specifically, thanks to the availability in PA of sequence coding, a recursive definition can be reformulated as an explicit definition provided each step in the recursion depends on only finitely many previous results. For example, the natural definition of the factorial function ($0!=1$ and $(n+1)!=n!\cdot (n+1)$) can be rewritten as "$x!=y$ iff there is a sequence $a_0,a_1,\dots,a_x$ of length $x+1$ with $a_0=1$, $a_{n+1}=a_n\cdot(n+1)$ for all $n<x$, and $a_x=y$." This reformulation is expressible in PA because coding allows us to replace "there is a sequence" with "there is a number". But it's crucial here that all the "predecessors" of "$x!=y$" in this recursion can be coded into a single number, which requires that there be only finitely many of these predecessors. There's no way to code an infinite sequence of natural numbers into a single natural number.
In your recursive definition of $eval$, the clause for a universally quantified formula unfortunately involves infinitely many predecessors, namely the values for all the alternative assignments $e[i\mapsto n]$. This one clause prevents your recursion from being expressible in PA.
(Any extension of PA that would allow expressing this recursion --- whether by adding more axioms or by strengthening the logic --- and allow proofs by induction of formulas containing such an expression would also prove the consistency of PA along the lines you indicated. So it's a good thing that PA can't express this recursion; if it could, then it would prove its own consistency and therefore would be inconsistent by Gödel's second incompleteness theorem.)
That's why I assumed PA, not arbitrary first-order theory. Anyway, I overlooked that universal quantifier clause involves infinite predecessors, which requires sequences of countable ordinal lengths to encode it. Now I see how second-order arithmetic (sufficiently strong ones) proves Con(PA) and PA doesn't. Thanks!
|
2025-03-21T14:48:31.386466
| 2020-06-28T12:51:49 |
364337
|
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
How can construct three circles in a given triangle such that three internal tangent form an equilateral triangle
How can construct three circles in a given triangle such that three internal tangent form an equilateral triangle?
See also:
Malfatti circles
Heuristically, there ought to be a one-parameter family of such circles by counting degrees of freedom, Since we have two degrees of freedom (radii of three circles) and only two conditions (two angles are $60^\circ$, the third is automatic)
Why is this tagged as an open problem? Is it stated somewhere that it is unsolved? It is not an "open problem" merely because the answer is not known to the OP.
@SamZbarsky I tried from your sugesstion but no. One given triangle, maybe have only one equilateral triangle respectively.
There seems to be even a one-parameter family of such circles with these three tangents intersecting at a single point.
But this does not count, right? Your requirement is $60^\circ$ angles rather than equilateral triangle - which in the degenerate case of a single point is not equivalent, right?
@მამუკაჯიბლაძე Equilateral triangle degenerate to a single point is equivalent
I am in doubt because at this point three tangents might intersect at angles different from $60^\circ$, is it still OK?
Say you decrease the radius of each circle by the same amount, and move all lines (including triangle edges) to maintain tangency with the circles. As this operation would merely replace each line with a parallel line, the new triangle will be similar to the original one, and the "equilateral triangle" property will be preserved.
By taking the radius decrease to be the radius of the smallest circle, it seems that if there's a solution, then there's always a solution in which one of the circles has radius 0. Trying to find such a solution may be simpler than the original problem.
@user1020406 Yes, your comment is right, I checked. But how can find original three circle?
So you confirm that the comment by @SamZbarsky is correct, and there is a whole family of solutions?
@მამუკაჯიბლაძე Yes, there are family of solution. I am sorry to You.
No no please don't be sorry. You are asking question, you want to know the answer, this is nothing to be sorry about - on the contrary, you contribute to this site and you are doing good thing, I believe.
Draw three congruent equilateral triangles of an arbitrary size, with bases on the three sides centered at the incircle tangency points. Then the desired circles are inscribed in the kites delimited by the extended sides of these equilateral triangles as shown. Their common tangents form an equilateral triangle congruent to the other three.
This construction is most easily justified by working backwards. Find a solution among the one-parameter family with an equilateral triangle of the given size. Reflecting it over the common diameter of any two of the circles produces equilateral triangles with bases on the three sides. The quadrilaterals delimited by their extended sides are circumscribed about the circles, and so are kites by the equal opposite angles. That means any two of them are equidistant from the vertex between them, which means their bases are centered at the incircle tangency points.
|
2025-03-21T14:48:31.386703
| 2020-06-28T13:15:05 |
364339
|
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|
Stack Exchange
|
Uniform position for multiple components
(Modified from https://math.stackexchange.com/questions/3730261/uniform-position-theorem-for-reducible-varieties/3730457#3730457)
The uniform position theorem states (roughly) that a general hyperplane section meets an irreducible degree $d$ curve $C \subset \mathbb{P}^r$ at $d$ points with any $r$ of them being independent (e.g. see general/uniform position theorems around section III.1 in vol. I of "Geometry of Algebraic Curves").
Is there a version of this when we remove the "irreducible" condition? For example, what happens if we consider points coming from some collection of disjoint irreducible curves? Note that this excludes cases such as some large number of curves lying in some hyperplane (or others similar to what was mentioned in the MSE link).
This seems plausible for many cases, but I'm not sure how to put this into a coherent statement and irreducibility seems to be an important part of proving this theorem. In principle, one could imagine applying the theorem to each irreducible component and taking an intersection of some finite number of open sets (see link on top). However, it seems like this intersection may be empty (e.g. in the "degenerate" examples in the second paragraph). Also, are the counterexamples of the kind mentioned above the only ones possible?
|
2025-03-21T14:48:31.386810
| 2020-06-28T13:56:23 |
364341
|
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|
Stack Exchange
|
n-dimensional polyhedron with special properties
I'd like to know if there exists a convex face transitive n-dimensional polyhedron with all dihedral angles equal to $\frac{2\pi}{3}$.
For n = 2,3,4 an example can be a regular hexagon, a rhombic dodecahedron and a 24-cell respectively. But I still got no idea what to do for n > 4.
Any help or ideas are appreciated, thank you!
In four dimensions a second such polytope exists, which is the orthoplex/16-cell/dual of the hypercube.
|
2025-03-21T14:48:31.386873
| 2020-06-28T14:25:32 |
364346
|
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"John Klein",
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|
Stack Exchange
|
Rational homotopy invariance of algebraic $K$-theory
Suppose that $R\to S$ is a 1-connected morphism of connective structured ring spectra that induces an isomorphism on rational homotopy groups. Is the induced map of (Waldhausen) K-theory spectra
$$
K(R) \to K(S)
$$
also an equivalence on rational homotopy?
(The case of the map of group rings $S^0[G] \to \Bbb Z[G]$ was answered in the affirmative in one of Waldhausen's early papers.)
I am looking for a solid reference (assuming the result to be true; I believe it is).
For the record: The proof Waldhausen (Algebraic K-theory of topological spaces, I, Prop. 2.2, 1978) gave for S^0[G] --> Z[G] explicitly only uses that the map is a rational homotopy equivalence and 1-connected, hence also applies to any other such map of connective structured ring spectra.
The theorem can be found in more general form in Land, Tamme On the K-theory of pullbacks, Lemma 2.4.
That is exactly what I am looking for! Thanks.
I saw the you mentioned that lemma in your talk today. Great talk by the way. I learned something...
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2025-03-21T14:48:31.386984
| 2020-06-28T14:53:20 |
364348
|
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|
Stack Exchange
|
Asymptotics for solution of transport equation and characteristics
Consider the transport equation $$u_t(t,x) + v(t,x) \cdot \nabla u(t,x) = 0.$$
Suppose that the solution of the characteristic equation
$$\dot X(t) = v(t,X(t)) $$
decays to zero as $t \to \infty$. What happens to the solution $u$ of the PDE as $t \to \infty$? Does it also decay to zero or to the Dirac delta as the weak solution formula
$$\int_{\mathbb R^N} \phi u(t,x)dx = \int_{\mathbb R^N} \phi(X(t,x))u_0(x)dx \qquad \phi \in C^\infty_c$$
suggests?
V is invompressible if its divergence is 0 everywhere. Such flows cannot have sinks as required by your 2nd eq.
@PiyushGrover Thanks. Can you show why if $\mathrm{div} v \neq 0$ then the solution of the ODE cannot decay to zero?
I am saying the opposite, that is if div v=0, then ODE cannot decay to 0 for all initial conditions.. Just take tiny circle around origin and apply divergence thm. Since all traj. are going into that circle, the line integral will be non-zero, but the area integral is 0 if div.v=0.
@PiyushGrover This counterexample is not clear to me: where are you applying the divergence theorem? Let's start over: if div v = 0, is it possible to prove that $X(t) > c \ge 0$ for every $t>0$?
Take a 2D example with 0 divergence. $\dot{x}=x$,$\dot{y}=-y$. See what you get.
@PiyushGrover Ok, in this example the characteristic does not decay: $(e^t, e^{-t})$. How can you deduce in general the lower bound on $X$ from the bound on the divergence? Is it still true if the div is bounded instead of zero?
Divergence controls the rate of expansion/contraction of volumes (or Lebesgue measures), think about case where initial data is uniform small ball around origin. If div=0, this ball will deform but volume will be conserved (e.g. it will become ellipse in the case I mentioned above). If div>0, volume will grow. Hence, a neccesary condition for all traj. to go to 0 is div<0, but certainly it is not sufficient. E.g. $\dot{x}=x$, $\dot{y}=-2y$.
@PiyushGrover Thanks, but can you show a general estimate of the kind $X(t) >c \ge 0$ under this assumption div = 0?
see this: https://mathoverflow.net/questions/327863/role-of-the-divergence-of-the-vector-field-in-transport-equations-mass-concentr
Let me change slightly your notations with the flow $\psi (t,y)$ defined by
$$
\dot \psi(t,y)=v(t, \psi(t,y)),\quad \psi(0,y)=y.
$$
The solution $u$ is constant along the integral curves of the vector field, i.e.
$$
u(t,\psi(t,y))=u(0, y).
$$
Using the inverse function theorem you can introduce $\phi(t,x)$ to be a first integral defined by
$$
x=\psi(t,y)\Longleftrightarrow y=\phi(t,x).
$$
It is possible locally and let us assume that we can do that globally. Then we have
$$
u(t,x)=u(t=0, \phi(t,x))=u_{0}(\phi(t,x)).
$$
Assuming for instance that the initial datum $u_{0}$ is compactly supported or decays at infinity, you will get decay for the solution $u$ whenever $\phi(t,x)$ goes to infinity when $t\rightarrow+\infty$.
The natural condition for decay of $u$ whenever the Cauchy datum
$u_{0}$
is say compactly supported is that the first integral (which is the inverse function of the flow) goes to infinity with $t$.
Thanks! I have some additional questions: 1. Does $\psi \to 0$ imply $\phi \to \infty$ as $t \to \infty$? 2. What happens in general, for example for measure initial data? Do we have convergence to the Dirac delta?
@Zac Since $x=\psi(t,y)$ is equivalent to $y=\phi(t,x)$, we have $x=\psi(t,\phi(t,x))$. Assuming $x\not=0$ we must have that $\phi(t,x)$ goes to infinity, otherwise under a mild continuity assumption, $x=\psi (+\infty,0)=0$.
Thank you! What about the second question on the convergence to the Dirac delta in case of measure initial data?
@Zac Well, under some conditions of regularity and behavior at infinity, the solution of the transport equation is given via the first integral above with the formula $u=u_0(\phi(t,x))$, where $u_0$ is the Cauchy datum. To prove convergence to the Dirac mass at $x=0$, you take $u_0=\delta_0$ which is indeed well localized; I guess that the arguments sketched above show that $\phi(t,x)$ goes to infinity for $x\not=0$, so that $u=u_0(\phi(t,x))=0$, proving that the limit distribution is supported at 0.
It seems that the weak formulation implies that the limit distribution is a measure, thus proportional to the Dirac mass at 0.
But if you take the weak formulation of the solution $\int_{\mathbb R^N} \phi u(t,x)dx = \int_{\mathbb R^N} \phi(X(t,x))u_0(x)dx$, then as $\psi(t) \to 0$ you get the concentration to the Dirac delta (which is also more consistent with the conservation of mass)
|
2025-03-21T14:48:31.387289
| 2020-06-28T15:02:19 |
364349
|
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|
Stack Exchange
|
Directed graph minor theorems
In proving the graph minor theorem, Robertson and Seymour proved a stronger statement, namely that the directed graph minor theorem is true, using the definition
A directed graph is a minor of another if the first can be obtained from a subgraph of the second by contracting edges.
Since then, many more notions of a “directed graph minor” have arisen in various context (see here and here for examples and references). One of the goals in defining these notions is that the definition Robertson and Seymour gave doesn’t obviously capture the notion that if $G$ is a minor of $H$, then $G$ should be “simpler” than $H$.
I’m looking for references to proofs of the graph minor theorem for these more restrictive definitions of “graph minor.” I’m particularly interested in the following definition (Johnson et al., 2001):
A graph $G’$ is a butterfly minor of a directed graph $G$ if $G’$ can be obtained from $G$ by a sequence of the following local operations:
Deleting an edge (a, b);
Contracting an edge (a, b) where b has indegree 1;
Contracting an edge (a, b) where a has outdegree 1.
But other definitions would also interest me, especially if they’re amenable to analyzing computational graphs of circuits or ML devices (neural networks, Bayesian networks)
So directed graphs are not well-quasi-ordered by butterfly minors; see the intro of [BPP]. Furthermore, there are reasons to think that many of the FPT results for graph minors may not hold in the directed setting (ie [PW]).
Yet, perhaps surprisingly, it may still be possible to get a structure theorem for butterfly minors! There is an ongoing project to do this; I believe the most recent paper is "The Directed Flat Wall Theorem" by Giannopoulou, Kawarabayashi, Kreutzer, and Kwon. If you Google the authors you should be able to find more information about the project. This is a big task and the results so far are exciting.
Also notable, there is a relationship between "matching-minors" of (undirected) bipartite graphs and butterfly minors; see [HRW]. Furthermore, Johnson's thesis "Eulerian digraph immersion" is on a different kind of directed minor for 4-regular, directed, Eulerian graphs (same Johnson as mentioned in the question).
I'm not really sure why any of these definitions would be useful for studying graphs of circuits or graphs in machine learning, but I sure hope they are! It's nice to see interest from this direction, and I hope something in the post is helpful.
Thank you for the references! For anyone curious, the counterexample to the graph minor theorem for butterfly minors is $C_{2k}$, where the arcs alternate directions. My motivation for using these in machine learning is that I'm looking at the structure of neural networks. It turns out that if $G\triangleleft H$ and $H$ cannot be trained to compute $f$ then neither can $G$.
|
2025-03-21T14:48:31.387616
| 2020-06-28T15:51:42 |
364354
|
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|
Stack Exchange
|
Natural candidates for super-half-exponential which limit to half-exponential function from above
There are no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.
However super-half-exponentials (functions whose composition grows faster than exponential) have lot of candidates.
Eg: $2^n$ itself.
A little thought gives $f(0,a,n)=2^{n^a}$, $f(1,a,n)=2^{2^{(\log n)^a}}$, $f(2,a,n)=2^{2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)=2^{2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(0,1)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.
The functions grow slower as $k$ increases.
Define $g_{upper}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.
Is $g_{upper}(a,n)$ almost the half-exponential function? That is, does $g_{upper}(a,g_{upper}(a,n))=2^{O(n^{1+\epsilon})}$ hold at every $a\in(0,1)$ and at every $\epsilon>0$?
I do not think inductive arguments depending on finite $k$ work when $k\rightarrow\infty$ without knowing speed of $f(k,a,n)$'s decrease as $k$ increases. It is possible $g_{upper}(a,n)$ is the half-exponential function.
It is also possible $g_{upper}(a,n)$ and $g_{lower}(a,n)$ approach each other where $g_{lower}(a,n)$ is defined in Natural candidates for sub-half-exponential which limit to half-exponential function from below.
That is $g_{upper}(a,g_{upper}(a,n))=2^{O(n)}$ holds which means $g_{upper}(a,n)$ is the half-exponential function.
Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k+1,a',n)$ where $a'\in(a,1)$) with
$$f(k+1,a,n)\ll h(k,n)\ll f(k,a,n)$$
$$h(k,h(k,n))=2^{O(n)}$$
at every fixed $k\in\mathbb Z\cap[1,\infty]$?
+1 interesting question. The definition of $g$ needs explanation. For fixed $n$, the value $f(k,a,n)$ becomes complex for large $k$.
$n$ is unbounded.
|
2025-03-21T14:48:31.387752
| 2020-06-28T16:47:22 |
364356
|
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|
Stack Exchange
|
Minimum cardinality of a cofinal collection of countable subsets of a set
Setup
Let $X$ be a set of cardinality $\kappa\geq \aleph_0$.
Edit:
Based on Todd Eisworth's suggestion:
What is the minimum cardinality of a collection $\hat{X}$ of countable subsets of $X$ such that every countable subset $A \subseteq X$ is contained in some element of $\hat{X}$?
I guess you mean "infinite countable". In ZFC the picture is clear: for $\kappa>\aleph_0$ there're a single orbit. For $\kappa=\aleph_0$, contrary to your claim the orbits are listed in terms of the cardinal of its complement, which is a non-negative integer or $\aleph_0$. I guess this even holds in ZF+DC. If you're really interested in the ZF picture, would you be more explicit?
Yes I mean "infinite countable". However, what do you mean by an orbit in this case (the action of the symmetric group on $\mathbb{N}$ on the set of sequences on X)? In that case, I don't see why there would be a single orbit if $\kappa>\aleph_0$?
Oh, you're just using a misleading terminology. You're actually counting "infinite countable subsets of $X$" modulo nothing (countable subsets modulo permutation suggest modulo permutations of $X$). That means injective maps from $\mathbf{N}\to X$ modulo precomposition by permutations of $\mathbf{N}$.
If I understand correctly that you counting infinite countable subsets of $X$, its cardinal is $0$ for $X$ finite and $\kappa^{\aleph_0}$ for $X$ infinite (=$2^{\aleph_0}$ for $X=\aleph_0$, $=\kappa$ for infinite $X=2^\alpha$).
"non-trivial" means??
The relation you've got in the question is trivial, as you can always take the union of two countable sets to prove they are equivalent. It IS sensible to ask about the minimum number of countable subsets of $\kappa$ needed so that any countable subset of $\kappa$ is a subset of the given family.
@ToddEisworth I incoporated your suggested interpretation of my question. I think you explained the correct thing I'm looking for..
You're writing "is in $\hat{X}$" but seem to mean "is a subset of some element of $\hat{X}$", which is drastically different.
@YCor Yes you're right, but I was worried that such a collection may not exist. Indeed, I modified the question now.
It now makes sense. Write $f:\kappa\mapsto f((\kappa)$ this function, so $f(\aleph_0)=1$, and $f(\aleph_1)=\aleph_1)$. In general, for uncountable $\kappa$ one has $\kappa\le f(\kappa)\le\kappa^{\aleph_0}$. I'm guess it's been studied and hope set theory people will provide some feedback.
If $\kappa\geq 2^{\aleph_0}$ then $f(\kappa)=\kappa^{\aleph_0}$. My guess is that for $\kappa<2^{\aleph_0}$ things could be weird.
We have $f(\aleph_{\alpha+1})=\max(f(\aleph_{\alpha}),\aleph_{\alpha+1})$. In particular, $f(\aleph_n)=\aleph_n$ for all $n$. Also, for $\lambda$ limit of uncountable cofinality, $f(\aleph_\lambda)=\sup_{\alpha<\lambda}f(\aleph_\alpha)$. I have no idea about $f(\aleph_\omega)$.
Clearly $f(\aleph_n)=\aleph_n$ for $n>0$, and for limit cardinals $\kappa$ of uncountable cofinality one has $f(\kappa) = \sup(f(i):i<\kappa)$. Also, in analogy with Hausdorff's successor formula, $f(\kappa^+)=\max(f(\kappa), \kappa^+)$. The interesting question is when $\kappa$ has countable cofinality, and Shelah's pcf theory has something to say about this. It is clear that $\kappa^{\aleph_0} = f(\kappa) \cdot 2^ {\aleph_0}$, so the number $f(\kappa)$ is sometimes seen as "more interesting that $\kappa^{\aleph_0}$, as it helps to separate the "noise" (continuum) from the information.
I see now that my comment repeats most of @YCor 's comment.
Both these comments are very helpful, but since my set-theory is not my strongest field, would it be possible to post some references/ more details on the computation $f(\aleph_n)=\aleph_n, n>0$ etc..
I don't see where it's been mentioned yet that $f(\kappa)$ always has uncountable cofinality. So $f(\aleph_\omega)\gt\aleph_\omega$. Which must be like $2+2=4$ to the set theorists but may be worth pointing out to the general public.
@bof Would it be possible to mention some details since this is far from obvious for a non-set-theorist such as myself
Suppose $f(\kappa)=\lambda=\sum_{n\in\omega}\lambda_n$ with $\lambda_n\lt\lambda$. Let $\hat X$ be a cofinal family of countable subsets of $\kappa$ with $|\hat X|=\lambda$. Write $\hat X=\bigcup_{n\in\omega}\hat X_n$ where $|\hat X_n|=\lambda_n\lt\lambda$. For each $n$ choose a countable set $A_n$ contained in no element of $\hat X_n$; then $\bigcup_nA_n$ is a countable set contained in no element of $\hat X$. Contradiction.
@bof I would hapilly accept this as an answer, hapilly, if you could add a couple references for a non-expert such as myself.
@Rahman.M By $\operatorname{cof}([\lambda^+],\subseteq)=\lambda^{\aleph_0}$ did you mean $\operatorname{cof}([\lambda^+]^\omega,\subseteq)=\lambda^{\aleph_0}$? If so, how do you prove that? We know that $\operatorname{cof}([\lambda^+]^\omega,\subseteq)=\operatorname{cof}([\lambda]^\omega,\subseteq)$ when $\lambda$ is singular and of countable cofinality..
@Rahman.M Of course we are interested in the case where $2^{\aleph_0}\gt\lambda^+$.
https://mathoverflow.net/q/287888/1783
As it is stated in the comments, the question is about the cofinality of $([\lambda]^{\aleph_0}, \subseteq)$.
The following definition is due to Shelah:
$
cov(\lambda, \mu, \theta, \sigma)=min\{|P|: P$ is a family of subsets of $\lambda$ each of size $< \mu$ such that for every $a \subseteq \lambda, |a|<\theta$, for some $\alpha < \sigma$ and $A_i \in P,$ for $i<\alpha,$ we have $a \subseteq \bigcup_{i<\alpha}A_i \}$.
It is easily sean that
1-$cf([\lambda]^{\aleph_0}, \subseteq)=cov(\lambda, \aleph_1, \aleph_1, 2)$.
By a theorem of Shelah (see [Sh: 355], $\aleph _{\omega +1}$ has a Jonsson Algebra, see also Analytic guides and updates for cardinal arithmetic page 23)
2-$\lambda^{\aleph_0}=cov(\lambda, \aleph_1, \aleph_1, 2)+ 2^{\aleph_0}$.
Note that in particular, if $\lambda^{\aleph_0} > 2^{\aleph_0}$, then
$cf([\lambda]^{\aleph_0}, \subseteq)=\lambda^{\aleph_0}$.
(1) is just the definition of "cov" with those particular parameters, so this is just standard cardinal arithmetic, i.e., no need to quote Shelah here!
The notation you are using is probably impenetrable for someone asking the question. Would you mind adding a reference or some definitions?
(2) is more or less obvious (see the comment by Goldstern below the question). Shelah lists this as an "observation" in Sh355 (p. 85). Also a minor point: $\lambda\geq 2^{\aleph_0}$ suffices for the final conclusion--the point being that we can include $\lambda=2^{\aleph_0}$. (not a big deal).
Actually "$\lambda^+\geq 2^{\aleph_0}$" is enough for the last bit (as in bof's comment).
Actually yes...since I'm not set-theorist the "it is easily seen" is prettttttttty far from even seeable... Could you add some details please?
|
2025-03-21T14:48:31.388449
| 2020-06-28T16:51:26 |
364357
|
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|
Stack Exchange
|
Good overviews on $\phi^{4}$-field theory?
I'm looking for nice overviews on $\phi^{4}$-field theory from the mathematical-physics point of view. To be a little more specific, here are some topics I'd like to read about:
(1) What are the motivations, both from the physics and mathematical point of view, to study $\phi^{4}$-theories?
(2) What has been (mathematically) accomplished so far? What are the most important open questions nowadays?
(3) What are the tools used to (rigorously) study these class of models? Renormalization group, cluster expansions, etc.
My motivation for this question is a rather simple one: I'm teaching myself some field theory but it is really hard to find these discussions on books. In general, books are more interested in solving problems and sometimes I find myself studying some models that I don't know anything about. Also, I'm primarily interested in statistical mechanics, so this helps to narrow the question a bit more. Connections with statistical mechanics and QFT are welcome too, but I don't want a purely QFT reference (I'm sufficiently lost in my own area of interest, after all). Thanks in advance!
This reference is a bit older, but it should be a good starting point for items 2 and 3: $\phi^4$ field theory in dimension 4: a modern introduction to its unsolved problems.
Concerning item 1, you might find it instructive to motivate the $\phi^4$ field theory from the perspective of its limitation: when does it apply and when are higher order terms needed? For that perspective I would recommend Higher-order field theories: $\phi^6$, $\phi^8$ and beyond.
This last reference is a chapter from a recent book, A Dynamical Perspective on the $\phi^4$ Model (2019) which has an interesting introductory chapter on the history of the $\phi^4$ model, as well as overviews of more specialized topics.
Note that in the case of purely ferromagnetic interactions, the triviality conjecture discussed in the first reference has recently been proven by Aizenman and Duminil-Copin.
@MartinHairer --- thanks, it's at https://arxiv.org/abs/1912.07973
In terms of physical motivation of $\phi^4$, if I remember rightly the references in the introduction to this paper might have some pointers.
Dashen R and Neuberger H 1983 How to get an Upper Bound on the Higgs Mass, Physical Review Letters 50, 24
Reference [7] of that article points to rigorous proofs of results which almost show the triviality of $\phi^4$ in four dimensions.
|
2025-03-21T14:48:31.388643
| 2020-06-28T17:20:30 |
364359
|
{
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"authors": [
"Richard Stanley",
"https://mathoverflow.net/users/2807",
"https://mathoverflow.net/users/41291",
"მამუკა ჯიბლაძე"
],
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364359"
}
|
Stack Exchange
|
A polynomial formed from the roots of another polynomial ad infinitum
Let $P(x)$ be a monic polynomial of degree $d$ with complex coefficients. Let $r_1(P),r_2(P),\dots, r_d(P)$ denote the set of roots, ordered so that $|r_1(P)| \leq |r_2(P)|\leq\dots\leq |r_d(P)|$. Define the map $T$ by:
$$(TP)(x)=x^d+r_1(P)x^{d-1}+r_2(P)x^{d-2}+\dots+r_d(P),$$
i.e. $TP$ is the monic polynomial whose coefficients are the roots of $P$.
Let us call a monic polynomial periodic if $T^KP=P$ for some $K>0$.
The question is: for any $d>0$, does there exist a periodic polynomial of degree $d$, other than the trivial solution $x^d$?
Remark on the definition of T
The above definition of $TP$ is ambiguous if there are two roots $r_i(P)$ and $r_j(P)$ such that $|r_i(P)|=|r_j(P)|$ and $r_i(P)\neq r_j(P)$. If the roots of $P$ have this property, then you may break the ties however you please. For example, if $P(x)=x^3-x$, then it is up to you whether to set $r_2(P)=1$ and $r_3(P)=-1$ or $r_2(P)=-1$ and $r_3(P)=1$. However, either ordering still must have $r_1(P)=0$, since there is no ambiguity there.
Note that the set of polynomials that have this ambiguity has measure zero, so I suspect such considerations will not influence the solution of the problem anyway.
Empirical Evidence
If $d=1$ then the answer is clearly yes (any $P(x)=x-a$ will do the job, with $a\ne 0$). If $d=2$ then $P(x)=x^2+x-2$ is a fixed point of $T$, so in particular is periodic with period 1.
I examined other low degrees by numerical simulation. Note that this requires relaxing the definition of a cycle, since testing for exact equality of floating point numbers is impossible. Thus, for these simulations, the condition $T^KP=P$ was replaced with $\|T^KP-P\|_\infty<\varepsilon$, with $\varepsilon=10^{-10}$. In particular, these simulations can only find polynomials $P$ that are periodic up to some fixed error tolerance.
The simulation was done by first initializing the coefficients of $P$ using values drawn from a standard normal distribution, and then iteratively applying $T$ 1000 times and checking whether the obtained sequence was eventually periodic (up to error $<\varepsilon$). Note that this method might not find all cycles.
The periods found thusly for low degrees are:
$$
\begin{array}{rc}
d=3 & \text{possible periods}= 1 ; 11 \\
4 & 21 \\
5 & 4 ; 56 \\
6 & 34 ; 44 \\
7 & 10 ; 15 ; 26 ; 234 \\
8 & 3 ; 38 ; 83 ; 292 \\
9 & 256 ; 311 ; 466 \\
10 & 275 ; 336
\end{array}
$$
Furthermore, for degrees $\leq 8$, all of the simulated sequences eventually became periodic, however this was not true for $d=9$ or $10$ (of course, this does not imply that these sequences never become periodic, just that they did not before the simulation ended).
Originally posted on math stackexchange, where bounty period expired without an answer: https://math.stackexchange.com/questions/3724155/a-polynomial-formed-from-the-roots-of-another-polynomial-ad-infinitum
Reverse operator might be more tractable - assigning to $r_1,...,r_d$ their symmetric functions $\sigma_1(r_1,...,r_d),...,\sigma_d(r_1,...,r_d)$
This seems to be equivalent to https://mathoverflow.net/questions/216346.
|
2025-03-21T14:48:31.388854
| 2020-06-28T18:14:16 |
364363
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364363"
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|
Stack Exchange
|
Field theory, Abel-Ruffini theorem, technical question
Let me put the question first.
Let $F,K$ be subfields of $\mathbb{C}$. Suppose that $t,\rho\in \mathbb{C}$ are algebraic over $F$ and $\rho \in K$. If $F(t)\cap K\subset F$, is it true that $F(t,\rho)\cap K\subset F(\rho)$? Any counterexample or any extra conditions to make it true?
This question rises up in an elementary (i.e. without Galois theory) proof of Abel-Ruffini theorem given by A. Skopenkov (See arXiv:1508.03317v4). I believe that there is a gap in the proof of theorem 4 in the article and we actually need to answer the question above, which seems not trivial.
I notice that $\sum f_i(t)\rho^i\in K$ does not imply $f_i(t)\in K$, so I feel that the answer is negative. But I can not provide any counterexample.
|
2025-03-21T14:48:31.388940
| 2020-06-28T18:37:34 |
364367
|
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"authors": [
"Federico Poloni",
"Iosif Pinelis",
"Michael Renardy",
"Robert Israel",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364367"
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|
Stack Exchange
|
Trace inequality under consideration of definiteness
Let $G \in \mathbb{R}^{3 \times 3}$ a symmetric, but indefinite matrix and $U \in \mathbb{R}^{3\times 3}$ a symmetric and positive definite matrix. I would like to prove the inequality
$$ \text{Tr} \left( G^2 \right) \leq \text{Tr} \left( GUGU^{-1} \right). $$
If $U$ and $G$ commute, both sides of the inequality are obviously equal. However for more general cases,
I have tried to rearrange the inequality to
$$ \text{Tr}(\underbrace{[UG-GU]}_{\text{skew-symmetric}}\ GU^{-1}) \leq 0 $$
and then using the Cauchy-Schwarz inequality. Unfortunately, I have not found a solution yet.
You may assume U is diagonal. Then write everything out in components.
Do you have ample computational evidence that the inequality holds? For instance, have you tried 1000 random-generated examples on a computer?
I have tried it with matlab and and a lot of variations, all of them were true. I have forgotten to say that both sides are of course larger than zero
Yes I Have tried it with random matrices which fulfill the restrictions stated above. All of them were fulfilled.
Write
$$G=\left(
\begin{array}{ccc}
a & b & c \\
b & d & e \\
c & e & f \\
\end{array}
\right)
$$
and, without loss of generality,
$$
U=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & u & 0 \\
0 & 0 & v \\
\end{array}
\right),$$
where $u,v>0$. Then
$$\text{Tr}(GUGU^{-1})-\text{Tr}(G^2)
=\frac{b^2 (u-1)^2 v+c^2 u (v-1)^2+e^2 (u-v)^2}{u v},$$
which is manifestly $\ge0$, as desired.
Moreover, this works for $n \times n$ matrices, not just $3\times 3$. You get
$$ \text{Tr}(GUGU^{-1} - G^2) = \sum_{i<j} \frac{(u_i - u_j)^2 g_{ij}^2}{u_i u_j}$$
@RobertIsrael : Good point!
|
2025-03-21T14:48:31.389086
| 2020-06-28T19:35:38 |
364368
|
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"authors": [
"Gerry Myerson",
"R. van Dobben de Bruyn",
"Sebastien Palcoux",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364368"
}
|
Stack Exchange
|
On the sum of the subgroup orders of a finite group
Let $G$ be a finite group. Consider the function providing the sum of the subgroups orders
$$\sigma(G) = \sum_{H \le G} |H|.$$
Note that if $C_n$ is cyclic of order $n$ then $\sigma(C_n) = \sigma(n)$, with $\sigma$ the usual divisor function. Consider the functions $$\sigma_{-}(n)= \min_{|G|=n} \sigma(G), \ \ \ \ \ \ \sigma_{+}(n)= \max_{|G|=n} \sigma(G). $$
This post is about a characterization of the extremizers, i.e. the finite groups $G$ such that $\sigma(G) = \sigma_{\pm}(|G|)$.
$$\begin{array}{c|c} n&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15 \newline \hline
\sigma_{-}(n)&1&3&4&7&6&12&8&15&13&18&12&28&14&24&24\newline \hline
\sigma(n)&1&3&4&7&6&12&8&15&13&18&12&28&14&24&24 \newline \hline
\sigma_{+}(n)&1&3&4&11&6&16&8&51&22&26&12&60&14&36&24 \end{array}$$
We can observe in above table that $\sigma_{-}(n) = \sigma(n)$, and it holds for all $n < 256=2^8$ (by checking on GAP).
Question 1: What are the finite groups $G$ such that $\sigma(G) = \sigma_{-}(|G|)$? Exactly the cyclic groups?
Next, consider the prime factorization of $n$ $$n=\prod_{i=1}^r p_i^{n_i},$$ then, candidates which come in mind for $\sigma(G) = \sigma_{+}(|G|)$ are the product of prime order cyclic groups: $$G = \prod_{i=1}^r C_{p_i}^{n_i}.$$
It works often but not always, as $\sigma(S_3) = \sigma_{+}(|S_3|) = \sigma_{+}(6) = 16$ whereas $\sigma(C_2 \times C_3) = 12$; but $S_3 = C_3⋊C_2$, moreover, for $n \le 60$, all the models I found are semi-direct product of prime order cyclic groups.
Question 2: What are the finite groups $G$ such that $\sigma(G) = \sigma_{+}(|G|)$?
Are there semi-direct product of prime order cyclic groups? Or at least supersolvable?
Sebastien has put $\sigma_+$ up on the OEIS, http://oeis.org/A335887
Partial answer:
The answer to Q1 is "yes" if one restricts $G$ to supersolvable groups or more generally groups that satisfy the converse of Lagrange's theorem.
If $G$ has the property that for every divisor of the group order there is at least one subgroup of that order, then in particular $\sigma(G)\geq\sigma(|G|)$. Furthermore, if $\sigma(G)=\sigma(|G|)=\sigma(\mathbb{Z}/|G|)$ holds, then there is exactly one subgroup of every order. In particular, all sylow groups are normal so that the group is the direct product of its sylows and inside every sylow subgroup, there is exactly one maximal subgroup so that it is cyclic.
Another observation: $\sigma$ is "multiplicative" for groups similar to how $\sigma$ is multiplicative for groups, i.e.
$gcd(|G|,|H|)=1 \implies \sigma(G\times H)=\sigma(G)\cdot\sigma(H)$
This follows from the fact that in these situations, every subgroup of $G\times H$ is uniquely decomposable as $G_0\times H_0$ with $G_0\leq G$ and $H_0\leq H$.
More generally, one can observe that $\sigma(G\rtimes H) = \sum_{G_0\leq G} |G_0| \cdot\sigma(N_H(G_0))$ holds for $gcd(|G|,|H|)=1$, because in this case every subgroup uniquely decomposes as $G_0\rtimes H_0$ by the Schur-Zassenhaus theorem. The right hand side is less than or equal to $\sigma(G)\cdot\sigma(H)$ with equality iff all subgroups of $G$ are $H$-invariant.
Here is a small remark: let $n$ be a number which is not abundant (that is, the sum of the proper divisors of $n$ is at most $n$). Suppose further that there is a non-cyclic finite group $G$ of order $n$. I claim that $\sigma(G) > \sigma(n)$.
Since $G$ is not cyclic, $\langle x \rangle$ is a proper subgroup of $G$ for each non-identity element $x \in G$. Hence each $x \in G$ is contained in some maximal (proper) subgroup of $G.$
Note also that $G$ has more than one maximal subgroup ( for if there is only one maximal subgroup $M$ of $G$, then as noted above, $M$ must contain each non-identity element of $G$, and it certainly contains the identity. This contradicts the fact that $M$ is proper).
Hence if we add the order of all maximal subgroups of $G$, we obtain an integer greater than $|G|$, since we count each non-identity element of $G$ at least once, and we count the identity element more than once. Hence we have $\sigma(G) > 2n \geq \sigma(n)$.
In particular, if $n$ is an integer greater than one which is not abundant, then we do have $\sigma_{-}(n) = \sigma(n)$, and only for cyclic groups $G$ of order $n$ do we attain $\sigma(G) = \sigma(|G|).$ This answers question 1 for groups $G$ whose orders are not abundant numbers. Furthermore, it shows that $\sigma_{-}(n) = \sigma(n)$ for any integer $n >1$ which is not abundant.
Good! Your redaction can be shorten by writting: $G$ non-cyclic iff $G$ equals the union of its maximal subgroups (which share the trivial element). Then for $G$ non-cyclic, the sum of the order of its maximal subgroups must be greater than $|G|$, so that $\sigma(G) > 2|G|$ .
Another way to get the same bound: every element is contained in the subgroup it generates and in the whole group (and these are different if $G$ is not cyclic). (By double counting, $\sigma(G)/|G|$ is the average number of groups containing a given element. I haven't been able to put this method to any further use.)
|
2025-03-21T14:48:31.389428
| 2020-06-28T23:00:12 |
364373
|
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"authors": [
"Anton Petrunin",
"Brendan McKay",
"M. Winter",
"bof",
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|
Stack Exchange
|
Notation for induced subgraphs
For a graph $G=(V,E)$, is there a standard notation for the induced subgraph on $V \setminus \{v,w\}$ where $v,w$ are the endpoints of some edge $e$? I know $G[V \setminus \{v,w\}]$ is an option, but I’m looking for something more compact. $G \setminus e$ is used for deletion and $G/e$ is used for contraction, so I’m inclined to use $G-e$, but perhaps that already has another meaning.
Isn't $G-e$ used for the spanning subgraph where only the edge is deleted and the vertices remain? How about $G-u-v$?
Once I used $G-[e]$; page 38 here: https://arxiv.org/abs/1812.06627
As bof said, definitely do not used $G-e$ as that is extremely widespread as meaning that only the edge is removed and not its end-vertices.
I myself would probably use $G-{v,w}$ as I always avoid revealing the implementation of my edges as sets, that is, I never state that $e={v,w}$. And then $G-S$ is equally well defined for any set $S$ of vertices.
|
2025-03-21T14:48:31.389539
| 2020-06-28T23:42:58 |
364374
|
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"Francois Ziegler",
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|
Stack Exchange
|
Jordan normal form in a reductive group
Let $G$ be a connected complex reductive group. Given an element $g \in G$, there is a canonical decomposition $g = s u$ such that $s$ is semisimple, $u$ is unipotent and $s$ and $u$ commute. Similarly, for $x \in \frak{g}$ in the Lie algebra, there is a canonical decomposition $x = s + n$, where $s$ is semisimple, $n$ is nilpotent and $[s,n] = 0$.
This is a generalisation of the Jordan normal form for matrices. However, in the case of matrices, the nilpotent component $x = s + n$ in the Jordan normal form has a special form, namely $n$ is block upper triangular with $1$'s on the upper diagonal.
My question is to what extent this more refined normal form generalises to reductive groups?
For example, let $n \in \frak{g}$ be a nilpotent element. Is it possible to find a maximal torus $T \subset G$, and a set of simple roots $\alpha_{1}, ..., \alpha_{k}$, such that
$n = \sum_{i} \epsilon_{i} \alpha_{i}$,
where each $\epsilon_{i} = 0$ or $1$? Maybe more precisely, whats the minimal number of simple roots required in such a decomposition (where we allow the choice of maximal torus and simple roots to vary as needed).
In fact, I this is enough to imply the general case. This can be seen as follows. Let $x = s + n \in \frak{g}$ be the canonical decomposition. The centralizer $H = C_{G}(s)$ is a connected and reductive subgroup of $G$. It also has maximal rank, because it contains any maximal torus containing $s$ (i.e. its Lie algebra contains $s$). But now $n \in \frak{h}$, and the roots of $H$ are a subset of the roots of $G$. So now we can apply any result about decomposing nilpotent elements inside of $H$, and this will be compatible with $s$.
Note: I also asked this question on math stack exchange.
This sounds like you are after the classification theory of (mainly nilpotent, indeed) orbits — theories of Jacobson, Morozov, Dynkin, Kostant, Bala, Carter,... as exposed in e.g. Collingwood-McGovern (1993). N.B. Cross-posting after < 2 hours is too early.
|
2025-03-21T14:48:31.389803
| 2020-06-28T23:56:43 |
364375
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364375"
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|
Stack Exchange
|
Bruhat-Tits theory: how does the normalizer act on an apartment?
Let $G$ be the points of a split, simply connected, semisimple algebraic group over a $p$-adic field $k$. Let $T$ be a maximal torus of $G$, $T_c$ its unique maximal compact subgroup, and $N$ the normalizer of $T$ in $G$. The canonical apartment of $G$ should be the vector space $V = X_{\ast}(T) \otimes \mathbb R$, where $X_{\ast}(T)$ is the lattice of cocharacters of $T$.
There is a natural action of $N$ as a group of vector space automorphisms of $V$, induced from the action of $N$ on cocharacters of $T$:
$$ n. \gamma(x) = n\gamma(x)n^{-1} \tag{$n \in N, x \in k^{\ast}, \gamma \in X_{\ast}(T)$}$$
However, in Bruhat-Tits theory there should be a more general action of $N$ as a group of affine space automorphisms of $V$, satisfying the following properties:
The translations on $V$ should come from $T$, and the kernel of the action on $N$ should be $T_c$.
Since every affine space automorphism of $V$ is a linear transformation (the "linear part") followed by a translation, the linear part of the automorphism induced by any $n \in N$ should be the one I just described above (coming from the action $n.\gamma(x) = n\gamma(x)n^{-1}$).
The action should depend on some sort of choice of root vectors $x_{\alpha}: k \xrightarrow{\cong} U_{\alpha}$, where $\alpha$ is a root of $T$ in $G$, and $U_{\alpha}$ is the root subgroup of $\alpha$. The action should possibly depend on a choice of Borel subgroup, I'm not sure.
What should the action of $N$ on $V$ as a group of affine space automorphisms be? What is the cleanest way of defining it?
The choice of a Borel subgroup (that is, a set of simple roots $\Delta$) gives an action of $T$ on $V$ by translations. If $\varpi_{\alpha}^{\vee}$ is the coweight corresponding to a simple root $\alpha$, then to $t \in T$ we can associate the translation
$$v \mapsto v + \sum\limits_{\alpha \in \Delta} \operatorname{ord}_k(\alpha(t)) \varpi_{\alpha}^{\vee}. \tag{$v \in V$}$$
If the normalizer $N$ split as a direct product $N/T \times T$, there is an easy way of defining the action of $N$, since the Weyl group $N/T$ already acts linearly on $V$. However, $N$ does not typically split in this way. I have some idea of how to define the action in general, although it depends on a choice of Borel subgroup $B = TU$ as well as a choice of "canonical Weyl group representatives" depending on a "splitting," i.e. choice of simple root vectors.
|
2025-03-21T14:48:31.390000
| 2020-06-29T00:25:18 |
364377
|
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|
Stack Exchange
|
Can an orthogonal matrix move monotonically toward a signed permutation matrix?
The question is motivated by this question on Mathematics SE.
Let $A \in O(n)$ be an orthogonal matrix that is not a signed permutation matrix, and let $P$ be the nearest signed permutation matrix to $A$ (for 'nearest', use
the distance induced by the Frobenius norm, i.e., $d(A,P)=||A-P||_F$). Then can $A$ move 'monotonically' to $P$? I.e., in every neighborhood of $A$, does there exist $B \in O(n)$ such that：
(1) $|b_{ij}| \geq |a_{ij}|$ at every non-zero entry $(i,j)$ of $P$ (and at least one inequality is strict), and
(2) $|b_{ij}| \leq |a_{ij}|$ at every other entry
Note that if we remove the condition that $P$ is the nearest signed permutation matrix to $A$ , then the claim is not true and a counterexample is given in the original question. Also note that the claim is true if $A$ is sufficiently close to $P$, as we can form a path from $A$ to $P$ by using exponential maps, say a path $B_t$ where $B_0 = A$ and $B_1 = P$. Since the entries of $B_t$ is analytic in $t$, in a small enough neighborhood the entries of $B_t$ would be monotonic in $t$, which shows every matrix in the path satisfies our property.
I am leaning toward that the claim is correct, but I am not sure. Any thoughts?
Edit:
Here is a weaker problem: for every $A\in O(n)$, $\textit{does there always exist}$ a signed permutation matrix $P$ where $A$ could move 'monotonically' to $P$ in the sense described above?
A possible approach is that, let $$C_P=\{B\in M_{n\times n}\mid B \text{ has the same sign as }A\text{ at the non-zero entries of }P, B \text{ has different sign from }A\text{ at the zero entries of }P\}.$$
Because the tangent space of $A$ is $n(n-1)/2$ dimensional, we would be done if we can prove that every $n(n-1)/2$ dimensional subspace of $M_{n\times n}$ intersects $\bigcup_{P\text{ is a permutation matrix}}C_P$ untrivially.
Edit 2:
Maybe the original question is still too strong, so I would like to weaken (2) to be:
(2) $|b_{ij}| \leq |a_{ij}|+\epsilon$ at every other entry
Then for every $\epsilon>0$, does such $B$ exists? Since we have some freedom here, maybe Gram-Schmidt would work?
When you say "any distance induced by a matrix norm" am I allowed to use the operator norm, i.e. the one for which every orthogonal matrix has norm 1?
Question also posted several hours ago to https://math.stackexchange.com/questions/3737954/can-a-orthogonal-matrix-move-monotonically-toward-a-signed-permutation-matrix
@YemonChoi Maybe it's better to fix one. I editted the question.
|
2025-03-21T14:48:31.390191
| 2020-06-29T00:25:25 |
364378
|
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|
Stack Exchange
|
Derivative of matrix argument function with respect to eigenvalues of argument
Let $\mathsf{SPD}_n$ denote the set of all real symmetric and positive definite $n\times n$ matrices. This set is convex so for every $A,B\in\mathsf{SPD}_n$ there exists a smooth path $\varphi:[0,1]\rightarrow\mathsf{SPD}_n$, i.e. $\varphi$ is a smooth function with $\varphi(0) = A$ and $\varphi(1) = B$.
Let $f:\mathsf{SPD}_n\rightarrow \mathbb R$. I am interested in the derivative: $\frac{\partial}{\partial t}f(\varphi(t))$. By the usual chain rule, this is $$\mathfrak Df(\varphi(t))\cdot\dot\varphi(t),$$
where $\mathfrak Df$ is the derivative of $f$ with respect to its argument (i.e. a matrix) and $\dot\varphi$ the derivative along the path. Note that $\cdot$ denotes the Frobenius inner product. This is my setting.
Since $\varphi(t)$ is a symmetric matrix for every $t$, there should exist orthonormal matrices $V(t)$ and diagonal matrices $\Lambda(t)$ such that $\varphi(t) = V(t)\Lambda(t)V(t)'$. Is there any chance to restate my setting (perhaps under additional assumptions) such that $$\frac{\partial}{\partial t}f(\varphi(t)) = \mathfrak Df(\varphi(t))\cdot\dot\psi(t),$$
where $\dot\psi$ is a vector valued function and $$
\mathfrak Df(M) = \left[\begin{array}{c} \frac{\partial}{\partial\lambda_1}f(M) \\ \frac{\partial}{\partial\lambda_2}f(M) \\\vdots \\ \frac{\partial}{\partial\lambda_n}f(M)\end{array}\right]
$$ now is the vector of derivatives wrt to the eigenvalues of the argument of $f$? Consequently, $\cdot$ denotes now the usual Euclidean inner product.
In words: can $\frac{\partial}{\partial t}f(\varphi(t))$ can be represented by $$\text{derivative of $f$ wrt eigenvalues of argument} \times \text{something} $$
?
Background: The above is trivially true if I consider the set of diagonal matrices (because eigenvalues and diagonal elements elements coincide). Now I want to adapt the setting for spd matrices. In my application, $A$ and $B$ are covariance matrices. $A$ is the sample covariance and $B$ is a "target" covariance. On the way (i.e. along $\varphi$) from $A$ to $B$ dimension reduction should be performed. The dimension reduction works similarily to PCA which is why I need to take the derivative wrt eigenvalues: if $\left\vert\frac{\partial}{\partial\lambda_j(t)}f(\varphi(t)\right\vert<\xi$ for some threshold $\xi>0$, "dimension" $j$ is ignored.
You can use the multidimensional chain rule and then product rule (since you have written $\varphi(t)$ as a product of three matrices). Product rule for matrices is $\frac{d}{dt}(AB)=\frac{dA}{dt}B+A\frac{dB}{dt}$
computing the ordinary derivative is no problem. My problem is that I want to express the derivative in terms of the derivative with respect to the eigenvalues of the argument.
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2025-03-21T14:48:31.390382
| 2020-06-29T01:07:49 |
364381
|
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|
Stack Exchange
|
When did the main conjecture in Vinogradov's mean value theorem first appear in literature?
Recently I was asked about the history of Vinogradov's mean value theorem that I was hoping someone here could clarify. Let me first start with some terminology. Let $J_{s, k}(X)$ be the number of $2s$-tuples $(x_1, \ldots, x_s, y_1, \ldots, y_s)$ such that
\begin{align*}
x_1 + \cdots + x_{s} &= y_{1} + \cdots + y_{s}\\
x_{1}^{2} + \cdots + x_{s}^{2} &= y_{1}^{2} + \cdots + y_{s}^{2}\\
&\vdots\\
x_{1}^{k} + \cdots + x_{s}^{k} &= y_{1}^{k} + \cdots + y_{s}^{k}
\end{align*}
for $1 \leq x_{i}, y_{i} \leq X$. It is not hard to see that $$J_{s, k}(X) \gtrsim_{s, k} X^{s} + X^{2s - \frac{1}{2}k(k + 1)}.$$
The now proven Main Conjecture in Vinogradov's Mean Value Theorem is that this lower bound is essentially an upper bound. More precisely, the conjecture was:
Conjecture: For every $\varepsilon > 0$, $$J_{s, k}(X) \lesssim_{\varepsilon, s, k} X^{\varepsilon}(X^{s} + X^{2s - \frac{1}{2}k(k + 1)}).$$
This conjecture follows from classical methods for $k = 2$, first proven by Wooley for $k = 3$ using efficient congruencing in 2014 and then proven by Bourgain, Demeter, and Guth for $k \geq 4$ using decoupling methods in 2015.
My question is: when did this conjecture as stated above first appear in the literature?
Looking through Vinogradov's 1935 paper "New estimates for Weyl sums", it seems that this conjecture is not stated. The term "Vinogradov's Mean Value Theorem" referring to any bound of the form $J_{s, k}(X)\lesssim X^{2s - \frac{1}{2}k(k + 1) + \Delta_{s, k}}$ for some $\Delta_{s, k}$ positive and $s \gtrsim k^{2}\log k$ seems to appear in print as early as 1947 or 1948 in these two works by Hua:
Page 49 of the Russian version of his Additive theory of prime numbers (http://mi.mathnet.ru/eng/tm1019)
In Hua's paper "An improvement of Vinogradov's mean-value theorem and several applications" (https://doi.org/10.1093/qmath/os-20.1.48)
Though in both, it seems to imply that this term was in use as early as 1940. However neither also state the conjecture as mentioned above.
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2025-03-21T14:48:31.390544
| 2020-06-29T04:01:34 |
364383
|
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|
Stack Exchange
|
Natural candidates for sub-half-exponential which limit to half-exponential function from below
There are no closed form candidates for half-exponential functions "Closed-form" functions with half-exponential growth.
However sub-half-exponentials (functions whose composition grows slower than exponential) have lot of candidates.
Eg: $n$ itself.
A little thought gives $f(0,a,n)={n^a}$, $f(1,a,n)={2^{(\log n)^a}}$, $f(2,a,n)={2^{2^{(\log\log n)^a}}}$, $\dots$, $f(k,a,n)={2^{^{\dots}{^{{2^{(\underbrace{\log\dots\log}_{\text{k }} n)^a}}}}}}$ etc. at fixed $a\in(1,\infty)$ and fixed $k\in\mathbb Z\cap[1,\infty]$.
The functions grow faster as $k$ increases.
Define $g_{lower}(a,n)=\lim_{k\rightarrow\infty}f(k,a,n)$.
Is $g_{lower}(a,n)$ almost the half-exponential function? That is, does $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n^{1-\epsilon})}$ hold at every $a\in(0,1)$ and at every $\epsilon>0$?
I do not think inductive arguments depending on finite $k$ work when $k\rightarrow\infty$ without knowing speed of $f(k,a,n)$'s increase as $k$ increases. It is possible $g_{lower}(a,n)$ is the half-exponential function.
It is also possible $g_{upper}(a,n)$ and $g_{lower}(a,n)$ approach each other where $g_{upper}(a,n)$ is defined in Natural candidates for super-half-exponential which limit to half-exponential function from above.
That is $g_{lower}(a,g_{lower}(a,n))=2^{\Omega(n)}$ holds which means $g_{lower}(a,n)$ is the half-exponential function.
Are there other natural sequence of function candidates $h(k,n)$ (not of form form $f(k,a',n)$ where $a'\in(0,a)$) with
$$f(k,a,n)\ll h(k,n)\ll f(k+1,a,n)$$
$$h(k,h(k,n))=2^{\Omega(n)}$$
at every fixed $k\in\mathbb Z\cap[1,\infty]$?
At a superficial glance, this looks identical to https://mathoverflow.net/questions/364354/natural-candidates-for-super-half-exponential-which-limit-to-half-exponential-fu . Did you double-post?
This is lower bound and other is upper bound for half-exponential.
Not an answer Merely remarks.
Let me use superscript $[k]$ for $k$-fold composition. $\log^{[3]} n$ means $\log\log\log n$.
As I remarked on the other question, for fixed $a$ and $n$, the value $f(k,a,n)$ becomes complex for large $k$. Sequence $\log^{[k]}(n)$ decreases as $k$ increases, until it becomes negative, and then complex.
I have worked with so-called transseries. In that formalism, we are interested in the rate of growth, not the initial part of the function.
Edgar, G. A., Transseries for beginners, Real Anal. Exch. 35(2009-2010), No. 2, 253-310 (2010). ZBL1218.41019.
My work on fractional iteration applies only to "exponentiality zero", so does not include $\exp^{[1/2]}$.
Edgar, G. A., Fractional iteration of series and transseries, Trans. Am. Math. Soc. 365, No. 11, 5805-5832 (2013). ZBL1283.30001.
Some remarks on the question here. Is there some reason to use exponentiation base $2$ and not $e$? I will use exponential and logarithm base $e$ instead of base $2$. Let me shift your index $k$ by $1$. Then define
\begin{align}
f(0,a,n) &:= an,
\\
f(1,a,n) &:= \exp f(0,a,\log n) = \exp(a\log n) = n^a
\\
f(2,a,n) &:= \exp f(1,a,\log n) = \exp(\exp(a \log(\log n)))
=\exp((\log n)^a)
\\ &\qquad \dots
\\
f(k,a,n) &:= \exp^{[k]}(a\log^{[k]}(n)) = \exp^{[k-1]}\Big(\big(\log^{[k-1]}(n)\big)^a\Big)
\end{align}
Write $M_a$ for the multiplication function: $M_a(n) = an$.
We are interested in a "limit" in some sense as $k \to \infty$.
I doubt that it converges according to the natural topologies
for transseries.
Note that
\begin{align}
f(k,a,n) &= \exp^{[k]}\circ M_a\circ \log^{[k]} (n)
\\
f(k,a,f(k,a,n)) &=\exp^{[k]}\circ M_a\circ \log^{[k]}\circ \exp^{[k]}\circ M_a\circ \log^{[k]} (n)
\\ &= \exp^{[k]}\circ M_a\circ M_a\circ \log^{[k]} (n)
\\ &= \exp^{[k]}\circ M_{a^2}\circ \log^{[k]} (n)
\\ &= f(k,a^2,n)
\end{align}
So if $g(a,n) = \lim_{k \to \infty} f(k,a,n)$ in some sense, then
we might expect that
$$
g(a,g(a,n)) = g(a^2,n) .
$$
So: prove by induction on $k$ that $f(k,a,n) < e^n$. Then conclude that $g(a,n) < e^n$, and $g(a,g(a,n)) = g(a^2,n) < e^n$ (good for what we want).
But also
$$
g(a,g(a,g(a,n))) = g(a^3,n) < e^n
\\
g(a,g(a,g(a,g(a,n)))) = g(a^4,n) < e^n
$$
So that second iterate $g(a,g(a,n))$ is much, much slower than $e^n$.
So $f(k,a,n)$ is not a sub half exponential function that approaches the half exponentials in the limit from below in limit $k\rightarrow\infty$? Instead of $2$ anything larger than $1$ is fine. I meant $n$ grows and so being complex is not considered.
We use $a>1$ so $an>n$. So $an < n^a < \exp((\log n)^a) < \exp(\exp((\log\log n)^a))$ and so on (for large $n$).
Ok so at $k\rightarrow\infty$ we will get half-exponential?
My argument suggests that the limit when $k \to \infty$ is still far short of half-exponential.
It is not clear whether 'suggests' is a formal proof since induction works for every finite $k<\infty$. What we have is a limit and I think we need to describe how fast we approach the goal to be formal. Is what you have a formal proof or just an intuition that is intended to be extrapolated somehow?
I think your argument works only for finite $k$.
|
2025-03-21T14:48:31.390860
| 2020-06-29T05:57:13 |
364385
|
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|
Stack Exchange
|
When is the inside of a Jordan curve open?
I'm working purely on intuition here. The Jordan curve theorem states that a Jordan curve separates the plane into a bounded component and an infinite component. For toy curves, it seems like this bounded component is always open. But in pathological cases, like an Osgood curve which has positive measure, clearly the inside cannot be open since it does not contain an open ball (I think).
Are there examples of Jordan curves with measure $0$ that don't have an open inside? Do Jordan curves with positive measure never have an open inside? More importantly, if the inside is open, does it guarantee that the curve is "non-pathological"?
EDIT: Perhaps my intuition was wrong. According to MO user Timothy Chow in this post, "The Jordan curve theorem was strengthened by Schoenflies to the statement that the two components are homeomorphic to the inside and outside of a circle." By Brouwer's invariance of domain theorem, this implies that the inside component of a Jordan curve is open, if I understand everything correctly.
the complement to a jordan curve is open. every component of an open set in the plane is open
The circle $\mathbb{S}^1$ is compact, and so is any continuous image of it in the plane, hence closed. So the complement of the image is open. Connected components of an open subset of a locally connected space are open (voting to close)
Of course, I wasn't thinking. Thanks for clarifying this anyway.
If we take the standard definition of a Jordan curve, that is a homeomorphic image of the unit sphere, then every Jordan curve in the Euclidean plane is compact. Given that every compact subset of a Hausdorff space is closed, every Jordan curve (in Euclidean space) is also closed, and so the complement is open. It follows that both the inner and outer components of the complement are open.
In short, the inner component of a Jordan curve in the plane is always open.
Complement.${}$
Oops, same mistake twice. Thanks
In the linked example, the Osgood curve shown is actually an "Osgood arc". You should take four copies (say, rotate the given arc through multiples of $\pi/2$) and glue them end to end. Then you will have a Jordan curve with an inside and an outside. Both will be open.
|
2025-03-21T14:48:31.391069
| 2020-06-29T07:03:59 |
364390
|
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"Bill Johnson",
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|
Stack Exchange
|
Renorming of $C[0,1]$ for a strictly convex dual
Let $C[0,1]$ be the space of all Real valued continuous functions on $[0,1]$ with the usual supremum norm. Does there exist an equivalent renorming on $C[0,1]$ such that the corresponding dual norm is strictly convex?
The dual of every separabel Banach space has an eqivalent strictly convex norm, but I understand that this is not you question.
Thank you Jochen for pointing out. I was a bit confused about Hahn-Banach smoothness and smoothness. A Banach space $X$ is said to be Hahn-Banach smooth if every $x^\in X^$ has unique norm preserving extension to $X^{**}$. Similarly a weaker version of this property can also be defined for the norm attaining functionals. If $X^*$ is strictly convex then any subspace of $X$ is Hahn-Banach smooth BUT this property does not imply (weakly) Hahn-Banach smoothness of $X$.
My earlier question can be reformulated as follows. Does there exist an equivalent renorming on $C[0,1]$ which makes it (weakly) Hahn-Banach smooth? Probably the answer to this problem is 'No' because the dual of any weakly Hahn-Banach smooth space has RNP.
One typically equivalently renorms a Banach space $Y$ to be strictly convex by finding an injective operator $S$ from $Y$ into some strictly convex space $Z$ and defining the new norm on $Y$ by $\|y\| +\|Sy\|$. When $Y$ and $Z$ are dual spaces and $S$ is weak$^*$ to weak$^*$ continuous, the new norm is a dual norm (the new unit ball of $X^*$ is weak$^*$ closed because $T^*$ is weak$^*$ continuous).
So let $X$ be any separable space and take an operator $T:\ell^2 \to X$ that has dense range. Renorm $X^*$ by
$!F! := \| F \|_{X^*} + \|T^*F\|_2$.
Bill, your exclamation marks would have surely won a contest for the funniest norm symbol, if only there were one! :-)
@TomaszKania. I think I first saw $! \cdot !$
in my undergraduate days when I took a course in functional analysis from Wilansky's book.
|
2025-03-21T14:48:31.391250
| 2020-06-29T08:20:02 |
364393
|
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"authors": [
"Aryeh Kontorovich",
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|
Stack Exchange
|
Measurable total order
Under what conditions on a metric space $X$, equipped with the Borel $\sigma$-algebra, does there exist a measurable total ordering of the elements of $X$?
By "measurable total ordering" we mean that any initial segment $I_y:=\{x: x<y\}$ is Borel-measurable.
Edit: We know that separability is sufficient for a measurable total order to exist.
Edit II: Vladimir Pestov (private communication) has shown that a measurable total order always exists; will post answer soon with link to full paper.
What kind of result are you looking for? Separability allows of course for a measurable embedding in $\mathbb{R}$, but there is no canonical non-separable totally ordered set.
Maybe an instructive counter-example (i.e., one where there is no measurable total order under the stated conditions)?
If $\dim X$ is finite then $X$ is the union of finitely many strongly zero-dimensional subspaces, all but one $F_\sigma$, the other $G_\delta$. Each is orderable so lay them end to end.
@KPHart see edited question
As mentioned in the OP, Vladimir Pestov has answered the question affirmatively. See Appendix D here:
https://arxiv.org/pdf/1906.09855.pdf
|
2025-03-21T14:48:31.391367
| 2020-06-29T08:28:10 |
364394
|
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|
Stack Exchange
|
Can we show that this transition semigroup preserves a certain Wasserstein space?
Let $E$ be a separable $\mathbb R$-Banach space, $v:E\to[1,\infty)$ be continuous, $$\rho(x,y):=\inf_{\substack{\gamma\:\in\:C^1([0,\:1],\:E)\\ \gamma(0)\:=\:x\\ \gamma(1)\:=\:y}}\int_0^1v\left(\gamma(t)\right)\left\|\gamma'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E,$$ $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space, $X:\Omega\times[0,\infty)\times E\to E$ be a stochastic flow, $$X^x_t:=X(\;\cdot\;,t,x)\;\;\;\text{for }(t,x)\in[0,\infty)\times E$$ and $$\kappa_t(x,B):=\operatorname P\left[X^x_t\in B\right]\;\;\;\text{for }(x,B)\in E\times\mathcal B(E)\text{ and }t\ge0.$$
Assume $$\operatorname E[v(X^x_t)]\le cv^{\lambda(t)}(x)\;\;\;\text{for all }(t,x)\in[0,\infty)\times E\tag1$$ for some $c>0$ and decreasing $\lambda:[0,\infty)\to[0,1]$. By $(1)$, $$\operatorname E[\rho(X^x_t,X^y_t)]\le c\rho(x,y)\tag2$$ for all $x,y\in E$ and $t\in[0,1]$.
Let, $\mathcal M_1$ denote the set of probability measures on $(E,\mathcal B(E))$, $\operatorname W_\rho$ denote the Wasserstein metric associated with $\rho$ and $$\mathcal S^1:=\{\mu\in\mathcal M_1\mid\exists y\in E:(\mu\otimes\delta_y)\rho<\infty\}.$$ By $(2)$, $$\operatorname W_\rho(\delta_x\kappa_t,\delta_y\kappa_t)\le c\operatorname W_\rho(\delta_x,\delta_y)\tag3$$ for all $x,y\in E$ and $t\in[0,1]$.
Let $t\ge0$. Can we show that $\kappa_t^\ast$ is $\mathcal S^1$-preserving? Or even that $\kappa_t^\ast\mathcal M_1\subseteq\mathcal S^1$?
I'm quite sure that at least the $\mathcal S^1$-preserving claim is true. If $\mu\in\mathcal M_1$, then we need to show that there is a $y\in E$ with $(\mu\kappa_t\otimes\delta_y)\rho_r<\infty$. Maybe we can pick $y=0$.
EDIT 1: Assume $\delta_x\kappa_t\in S^1$ for all $x\in E$ and $t\ge0$.
EDIT 2: Assume there are nondecreasing $v_i:[0,\infty)\to(1,\infty)$ with $v_1(\left\|x\right\|_E)\le v(x)\le v_2(\left\|x\right\|_E)$ for all $x\in E$ and $rv_2(r)\le \alpha v_1^\beta(r)$ for all $r>0$ for some $\alpha\ge0$ and $\beta\ge1$. Assume further that $\operatorname E[V^\theta(X^x_t)]\le\eta v^{\beta\lambda(t)}(x)$ for all $x\in E$ and $t\ge0$.
Then we easily see $\rho(0,x)\le\alpha v^\beta(x)$ for all $x\in E$. Now, since $\lambda$ is decreasing, it must hold $\lambda(t)\to0$ as $t\to\infty$ and hence $$\operatorname W_\rho(\mu\kappa_t,\delta_0)=\int\mu({\rm d}x)\operatorname E[\rho(0,X^x_t)]\le\alpha\eta\int\mu({\rm d}x)v^{\beta\lambda(t)}(x)\xrightarrow{t\to\infty}1\tag4$$ by monotone convergence for all $\mu\in\mathcal M_1$ and $t\ge0$. This should yield that $\kappa_t^\ast$ maps $\mathcal M_1$ to $\mathcal S^1$ for all $t\ge0$.
In the equation between (2) and (3), I guess that $\rho_r$ is simply $\rho$ (i.e. $\mathcal{S}^1$ is the set of measures with finite first moment).
I do not see how you get (2) from (1). For example, if $\lambda\equiv 0$ and $v\equiv 1$, (1) is satisfied but (2) can be false.
@BenoîtKloeckner Regarding your first comment: Yes, it should be $\rho$ in the definition of $\mathcal S^1$. Regarding your second comment: $\lambda$ should be decreasing (not nonincreasing). So, $\lambda\equiv0$ is not a valid choice.
That $\lambda$ is decreasing actually changes nothing when $v\equiv 1$. I see no way (2) could follow from (1), and I cannot guess a variation of the hypotheses that would change this. (1) is about the local geometry at $x$ and $X_x^t$, while (2) needs control over a whole curve.
There are some issues that I point out in comments, but assuming (3) you would get $\mathcal{S}^1$-preservation easily by convexity of Wasserstein distance, assuming that for at least one $x\in E$ you have $\delta_x\kappa_t\in\mathcal{S}^1$.
1. Convexity of $\mathrm{W}_\rho$ enables us to turn (3) into
$$\mathrm{W}_\rho(\mu\kappa_t,\nu\kappa_t) \le c\mathrm{W}_\rho(\mu,\nu).$$
(Let indeed $\mu,\nu\in\mathcal{S}^1$, and for each $t$ and each $(x,y)$ choose (measurably) optimal transport plans $\eta_{x,y}^t$ between $\delta_x\kappa_t$ and $\delta_y\kappa_t$. Let $\zeta$ be an optimal transport plan from $\mu$ to $\nu$; Then $\int \eta_{x,y}^t d\zeta(x,y)$ is a transport plan from $\mu\kappa_t$ to $\nu\kappa_t$, so that
\begin{align*}
\mathrm{W}_\rho(\mu\kappa_t,\nu\kappa_t) &\le \iint \rho(x',y') d\eta_{x,y}^t(x',y') d\zeta(x,y) \\
&\le \int \mathrm{W}_\rho(\delta_x\kappa_t,\delta_y\kappa_t) d\zeta(x,y) \\
&\le \int c\mathrm{W}_\rho(\delta_x,\delta_y) d\zeta(x,y) = c\int \rho(x,y) d\zeta(x,y) = c\mathrm{W}_\rho(\mu,\nu)
\end{align*}
as claimed.)
2. Then if for some $x\in E$, $\delta_x\kappa_t\in \mathcal{S}^1$, for all $\mu\in\mathcal{S}^1$ we have
$$ \mathrm{W}_\rho(\mu\kappa_t,\delta_x\kappa_t)\le c\mathrm{W}_\rho(\mu,\delta_x) <\infty$$
and thus $\mu\kappa_t\in\mathcal{S}^1$, as you wished.
3. You do need the additional assumption on $\delta_x\kappa_t$, (1) or (2) are not enough. Take $v\equiv 1$ and let $\kappa_t$ be a Markov kernel sending $\delta_x$ to some distribution with infinite first moment, translated by $x$. Then you have obviously (1) and (2), but you do not have $\mathcal{S}^1$ preservation.
4. You cannot expect to have $\mathcal{M}_1$ sent into $\mathcal{S}^1$ without additional assumption: the trivial dynamics $\delta_x\kappa_t=\delta_x$ satisfies your assumption.
Regarding 2.: I guess you conclude by the triangle inequality together with the assumption $\delta_x\kappa_t\in S^1$ which implies that $\operatorname W_\rho(\delta_x,\delta_x\kappa_t)<\infty$ and hence $\operatorname W_\rho(\mu\kappa_t,\delta_x)\le\operatorname W_\rho(\mu\kappa_t,\delta_x\kappa_t)+\operatorname W_\rho(\delta_x\kappa_t,\delta_x)<\infty$.
Please take note of my second edit. Under the additional assumptions, $\kappa_t^\ast$ should actually map $\mathcal M_1$ to $\mathcal S^1$.
I would like to know if you agree to my second edit. Am I missing something?
@0xbadf00d I do not know what $V^\theta$ is, and you seem to have produced your own answer about that. MO is not for others to check your details.
|
2025-03-21T14:48:31.391802
| 2020-06-29T08:57:02 |
364395
|
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|
Stack Exchange
|
Terminology: Existence + Representation
I'm looking to describe a result in a recent paper of mine, but I don't know if there is a term used for a result which is both an existence theorem and a representation theorem.
Specifically, the result is of the form:
Given a class of objects $\mathcal{C}$ there exists an object $X \in \mathcal{C}$ with property $\mathcal{P}$. Moreover, $X$ can has the following representation...
Is there a word which summarizes this existence + guarantee of a specific representation...maybe something like "descriptive existence" theorem?
What About Construction?
Ya but my issue with construction is that there should be a concrete structure given by the theorem; whereas in my case I have existence + know its form but the proof is non-construtive (so at-the end the reader would not be able to concretly write everything about the object down).
consider realization, substantiation , or demonstration
|
2025-03-21T14:48:31.391907
| 2020-06-29T09:30:49 |
364398
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364398"
}
|
Stack Exchange
|
The Thom map for the Brown-Peterson cohomology
For a prime number $p$ and the Brown-Peterson spectrum $BP$, let $T:BP\to H\mathbb{Z}_{(p)}$ be the Thom map, and $T':BP\to H\mathbb{Z}_p$ be the mop $p$ reduction of $T$. Tamanoi (1) determined the image of
$$T'_*:BP^*(K(\mathbb{Z}_{(p)},n+2))\to H\mathbb{Z}_p^*(K(\mathbb{Z}_{(p)},n+2))$$
for $n\geq 1$.
My question is, whether the same has been considered, or follows easily from the above, for
$$T_*:BP^*(K(\mathbb{Z}_{(p)},n+2))\to H\mathbb{Z}_{(p)}^*(K(\mathbb{Z}_{(p)},n+2)).$$
In particular, I would like to know if there is any nontrivial class of dimension $n+2$ in the image.
Some computation seems to suggest that for n=1, we have $p\iota_{n+2}\in \operatorname{Im}T_*$ where $\iota_{n+2}\in H\mathbb{Z}_{(p)}^*(K(\mathbb{Z}_{(p)},n+2))$ is the fundamental class, but of course I could be missing something.
Ravenel-Wilson-Yagita proved (maybe it is already in Tamanoi) that the $BP$ cohomology of Eilenberg-Maclane spaces is generated by the elements that get detected in mod $p$ cohomology. So $p\iota _{n+2}$ isn't in the image unless $\iota _{(n+2)}$ is, which only happens for $n=0$.
I guess the above arguments can be extended to determine the image of $T_*$.
Here is an "answer" which may be or not be good enough for your purpose, but which is easy to prove.
Let's start with Ravenel-Wilson-Yagita Theorem 1.20. Applied to Eilenberg-Maclane spaces, it implies that their $BP$ cohomology is generated by the elements detected by mod $p$ cohomology, in fact the elements maps to the set of elements described by Tamanoi.
The information above suffices to conclude that as a graded abelian groups, the image of $BP^*(K(Z_{(p)},n)\to H^*(K(Z_{(p)},n),Z_{(p)})$ is generated by some lift of Tamanoi's generator. In particular we can already conclude that $p\iota _{n+2}$ is not in the image if $n>0$.
I am not completely sure but probably these elements are torsion-free, which should give the additive structure of the image. If not, just study the the Bockstein spectral sequence.
If you need to determine concretely which elements of $H^*(K(Z_{(p)},n),Z_{(p)})$
are actually in the image, basically you will need to follow what Tamanoi did (fortunately not all). Step 1) find a description of what he calls $BP$-fundamental class in $H^*(K(Z_{(p)},n),Z_{(p)})$. Step 2) Find out which cohomology operations in $HZ_{(p)}^*HZ_{(p)}$ are covered by Landweber-Novikov operations. Step 3) find what the operations in the step 2 do to the element in the step 1.
Thank you for your answer! I still need some time to entirely convince myself of your argument though.
|
2025-03-21T14:48:31.392077
| 2020-06-29T09:37:03 |
364400
|
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"sort": "votes",
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|
Stack Exchange
|
Does this hereditary size based definition of cardinality work under grounds weaker than regularity and choice?
To $\sf ZF - Regularity$ add the following axiom:
Hereditary size: $\forall x \ \exists H_x \ \exists f (f: x \rightarrowtail H_x)$
Where: $H_x= \{y: \forall z \in TC(\{y\})\exists f (f: z \rightarrowtail x)\}$
Where $TC$ is the transitive closure operator on sets defined in the usual manner.
Is this axiom equivalent to $\sf AC$ over the rest of axioms of $\sf ZF - Regularity$?
The reason for asking this question, is that if the above proves to be weaker than $\sf AC$ then we have a sort of a definition for cardinality in absence of regularity and choice but with the above axiom holding. That definition would be:
The cardinality of a set x is the set of all sets equi-numeorus (i.e. bijective) to x that are hereditarily subnumerous (injective) to x.
|
2025-03-21T14:48:31.392160
| 2020-06-29T09:54:54 |
364401
|
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"Carlo Beenakker",
"Michael Renardy",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364401"
}
|
Stack Exchange
|
Linearization Navier-Stokes
I am considering the classical form of the stationary Navier-Stokes equation given by
$$\frac{1}{Re} (\nabla v, \nabla \phi) + ((v \cdot \nabla) v, \phi) - (p, \nabla \cdot \phi) = (f,\phi);\qquad (\nabla \cdot v, \xi) = 0.$$
Now, for suitable Reynolds numbers, one can treat the nonlinearity by linearization, e.g. Stokes linearization $(v^{k-1} \cdot \nabla) v^{k-1}$ or Oseen linearization $(v^{k-1} \cdot \nabla) v^k$.
My question: is it also possible to consider a linearization of the form: $(v^k \cdot \nabla) v^{k-1}$ ?
I thought the Oseen approximation substitutes $v=U+\delta v$ and linearizes in $\delta v$ assuming $\nabla U=0$, so $(v\cdot\nabla)v\rightarrow (U\cdot\nabla)\delta v$ without a term $(\delta v\cdot\nabla)U$.
Of course it is possible, but what would be the point?
|
2025-03-21T14:48:31.392241
| 2020-06-30T23:33:48 |
364547
|
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"Beginner Samya",
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"mathbeginner"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364547"
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|
Stack Exchange
|
Action of a finite group on a finite factor
Question: Let $G$ be a finite group and let $P$ be a $\rm II_1$ factor. Assume that $G$ acts on $P$ in a trace-preserving manner, such that the crossed product algebra $P \rtimes G$ is a factor. Is $G \curvearrowright^{\sigma} P$ outer?
Motivation: It's mentioned in Example 2.3.3(b) in Jones-Sunder's book that the above result is true. However, the proof is not given. I tried to prove it myself, but got stuck. Below is my attempt.
Attempt: Suppose $G \curvearrowright^{\sigma} P$ is not outer. Then there exists $g \in G$, $g \neq e$ such that $\sigma_g$ is inner. Assume that $\sigma_g =id$. Let $v =\sum_{h \in G} u_h u_g u_{h^{-1}}$. Then, it's easy to verify that $v $ lies in the center of $P \rtimes G$, and hence is a scalar. Now, let $H=C_G(g)$. Then, $v= \sum_{h \in H} u_g + \sum_{h \in G \setminus H} u_{hgh^{-1}}=c \in \mathbb C$. Therefore, $|H| + \sum_{h \in G \setminus H} u_{hgh^{-1}g^{-1}}=cu_{g^{-1}} $. Taking traces on both sides, we get that $|H|=0$, which is a contradiction.
I tried a similar trick when $\sigma_g=ad(u)$ for some unitary $u \in P$. I took $v= \sum_{h \in G} u_h u^{\ast}u_{gh^{-1}}$. I can show that $v $ lies in the center of $P \rtimes G$, and hence is a scalar. However, I got stuck after that.
It was Keshab
(Chandra Bakshi) who
found this problem.
We were talking about
it a few days ago, he
had a doubt about
this assertion in the
book of Jones and
Sunder.
I think that the claim in the question is false. One can construct a counterexample as follows. First assume in general that $G$ is a finite abelian group of order $n$ and that $\Omega : G \times G \to S^1$ is a bicharacter (i.e. a map that is multiplicative in both variables). Define the projective representation $U : G \to U(\ell^2(G))$ by
$$U(g)e_k = \overline{\Omega(g,k)} e_{gk}$$
where $(e_k)_{k \in G}$ is the standard orthonormal basis of $G$. One checks that
$$U(g) U(h) = \overline{\Omega(g,h)} U(gh)$$
for all $g,h \in G$. Identify $\ell^2(G) \cong \mathbb{C}^n$ and let $P_0$ be any $II_1$ factor. Define $P = M_n(\mathbb{C}) \otimes P_0$ and the action $G \curvearrowright^\sigma P$ by $\sigma_g = \operatorname{Ad}(U(g) \otimes 1)$. Then every automorphism $\sigma_g$ is inner by construction.
Consider the twisted group von Neumann algebra $L_\Omega(G)$ generated by canonical unitary operators $(v_g)_{g \in G}$ satisfying
$$v_g v_h = \Omega(g,h) v_{gh} \; .$$
There then is a $*$-isomorphism
$$\pi : P \rtimes_\sigma G \to P \otimes L_\Omega(G) : \pi(a u_g) = a(U(g) \otimes 1) \otimes v_g$$
for all $a \in P$ and $g \in G$.
To produce a counterexample, it now suffices to give an example where $L_\Omega(G) \cong M_n(\mathbb{C})$. This happens for instance when $\Gamma$ is a finite abelian group and $G = \Gamma \times \widehat{\Gamma}$ with
$$\Omega : G \times G \to S^1 : \Omega((g,\omega),(h,\eta)) = \omega(h) \; .$$
Defining the unitary operators $W(g,\omega)$ on $\ell^2(\Gamma)$ by
$$W(g,\omega)e_h = \omega(h) e_{gh} \; ,$$
it follows that
$$\theta : L_\Omega(G) \to B(\ell^2(\Gamma)) \cong M_n(\mathbb{C}) : \theta(v_{(g,\omega)}) = W(g,\omega)$$
is a $*$-isomorphism.
So for instance the group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ admits an inner action on a $II_1$ factor such that the crossed product is a factor.
Thank you for the answer sir.
Is the definition of bicharacter as following: $\Omega((a, b)(c, d))=\Omega(a,b)\Omega(c, d)$?
|
2025-03-21T14:48:31.392461
| 2020-07-01T00:08:13 |
364548
|
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|
Stack Exchange
|
Atiyah-Singer theorem in heat kernels and Dirac operators
I'm reading "Heat kernels and Dirac operators" by Berline, Getzler and Vergne.
I have some trouble to understand a identity on the bottom of page 146 which is essential for the proof of the Atiyah-Singer theorem:
If $a \in \Gamma(M, C(M))$ and $b \in \Gamma(M, \operatorname{End}_{C(M)}(\mathcal{E}))$,
then the point-wise supertrace of the section
$a \otimes b \in \Gamma(M, C(M) \otimes \operatorname{End}_{C(M)}(\mathcal{E}))
\cong \Gamma(M, \operatorname{End}(\mathcal{E}))$ was
shown in (3.21) to equal the Berezin integral
$$\operatorname{Str}_{\mathcal{E}}(a(x) \otimes b(x))=
(-2i)^{n/2} \sigma_n(a(x)) \operatorname{Str}_{\mathcal{E}/S}(b(x))$$
I would like try to lose shortly some words on some used notations (not all eg super trace and construction of spinor bundles are well explaned in the book):
Let $M$ be a compact oriented Riemannian manifold of even dimension $n$.
The Clifford bundle $C(M)$ of $M$ is the bundle of Clifford algebras over $M$
whose fibre at $x \in M$ is the Clifford algebra $C(T^*_xM)$
of the Euclidean spaces $T^*_x M$. The Clifford bundle $C(M)$ is an associated
bundle to the orthonormal frame bundle,
$$C(M)=O(M) \times_{O(n)} C(\mathbb{R}^n).$$
Let $S$ be the spin group accociated to Clifford algebra $C(\mathbb{R}^n)$. Assume,
that associated spinor bundle
$$\mathcal{S}= \operatorname{Spin}(M) \times_{\operatorname{Spin}(n)} S $$
on $M$ exists globally and let $\mathcal{W}$ another
arbitrary finite dimensional vector bundle. Clearly $\mathcal{S}$
is a Clifford module, since the action of $C(\mathbb{R}^n)$ on $S$
leads to an action of the associated bundle $C(M)$.
Let $\mathcal{W}$ another
arbitrary finite dimensional vector bundle on $M$ and define
$\mathcal{E}= \mathcal{W} \otimes \mathcal{S}$.
Observe that since $\operatorname{End}(\mathcal{S})= C(M)$ we obtain
$\operatorname{End}_{C(M)}(\mathcal{E})= \operatorname{End}(\mathcal{E})$.
If we now come back to indenity
$$\operatorname{Str}_{\mathcal{E}}(a(x) \otimes b(x))=
(-2i)^{n/2} \sigma_n(a(x)) \operatorname{Str}_{\mathcal{E}/S}(b(x))$$
I not understand, let take a look into proposition 3.21 which is proposed to
imply it:
Proposition 3.21. let $V$ is an even-dimensional oriented Euclidean vector and
$Q$ the induced bilinear form on $V \otimes \mathbb{C}$. Then if
the quadratic form $Q$ is non-degenerate, then there is,
up to a constant factor, a unique supertrace on $C(V)$, equal to $T \circ \sigma$. The
supertrace $\operatorname{Str}(a)$ defined in (3.8) equals
$$\operatorname{Str}(a)=(-2i)^{n/2} T \circ \sigma(a).$$
Here $\sigma$ is the symbol map $\sigma: C(V) \otimes \mathbb{C}
\to \wedge(V \otimes \mathbb{C})$ (page 104) ater complexification of $V$ and $T$ Berezin integral (page 42).
Now back to
$\operatorname{Str}_{\mathcal{E}}(a(x) \otimes b(x))=
(-2i)^{n/2} \sigma_n(a(x)) \operatorname{Str}_{\mathcal{E}/S}(b(x))$. What
happens with Berezin on the right side?
|
2025-03-21T14:48:31.392646
| 2020-07-01T01:12:47 |
364549
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364549"
}
|
Stack Exchange
|
Is linear combination of dependent exponential family random variables(multinomial, normal) in exponential family?
As well known to us, multinomial distribution and normal distribution belong to exponential family. Define
$$Z = \sum_{i=1}^{n}a_{i}(X_{i} - b_{i}) + \sum_{j=1}^{m}c_{j}Y_{j},$$
where $X_{i}$ is multinomial random variable, $Y_{j}$ is normal random variable, $a_{i}, b_{i}, c_{j}$ are constants. Additionally, $X_{i}, Y_{j} (i = 1,2,\cdots,n, j = 1,2,\cdots,m)$ are dependent.
Question: does the distribution of random variable $Z$ belong to exponential family?
Furthermore, in general, does the distribution of linear combination of exponential family random variables belong to exponential family?
If yes, can you provide a rigorous proof or a reference? Otherwise, can you provide a counterexample?
Thank you in advance!
|
2025-03-21T14:48:31.392710
| 2020-07-01T01:49:52 |
364550
|
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"Daniel Pomerleano",
"Hacon",
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"https://mathoverflow.net/users/6986"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364550"
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|
Stack Exchange
|
Log Calabi-Yau variety diffeomorphic to an algebraic torus
Let $U$ be a complex affine log Calabi-Yau variety, which I take to mean a smooth affine variety which admits a compactification in a smooth projective variety $X$ with an snc anti-canonical compactifying divisor $D$ (this seems to be a bit more stringent than what is sometimes meant by log CY).
Suppose $U$ is diffeomorphic to $(\mathbb{C}^*)^n$. Is it known that $U$ is algebraically an affine torus? I'm also curious about the same question with "diffeomorphic" replaced by "homotopy equivalent."
I should add that if $\operatorname{dim}(U)=2$, I'm pretty sure one could prove the answer is ``yes" just by classification.
Maybe http://www.numdam.org/article/CM_1981__43_2_253_0.pdf answers your question? It seems to me that we have a quasi-Albanese morphism $a:U\to A$ where $\dim A=\bar q(U)=\dim U$. By Theorem 28, $a$ is an open algebraic fiber space and by Corollary 29 it is birational
@Hacon Sorry for being dense. I can see that this is suggestive, but how does one actually conclude the result --- how do we get that $a$ is actually an isomorphism (as opposed to just birational)?
Oh, I think this follows by the properties of the quasi-Albanese mmorphism. Since $q(X)=0$, then $A=(\mathbb C^*)^n$. Since $a:U\to A$ is the quasi-Albanese morphism, by the universal property of the quasi-Albenese ((1) of page 271 of Kawamata's paper applied to $\beta = id _U$) there is a morphism $f:A\to U$ such that $id_U=f\circ a$, hence $a$ is an isomorphism.
|
2025-03-21T14:48:31.392824
| 2020-07-01T01:55:01 |
364551
|
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"authors": [
"Aurelio",
"Francesco Polizzi",
"https://mathoverflow.net/users/126773",
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"sort": "votes",
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|
Stack Exchange
|
Question about automorphism functor in Sernesi's "Deformations of algebraic schemes"
Let $X$ denote an algebraic scheme over $\operatorname{Spec} k$ such that its deformation functor $\operatorname{Def}_X$ has a semi-universal couple $(R,u)$, where $R$ is an Artinian $k$-algebra and $u \in \operatorname{Def}_X(R)$.
On pg. 91 of the book "Deformations of algebraic schemes" by E. Sernesi the following automorphism functor is introduced
$$ \operatorname{Aut}_u: \mathcal{A}_R \to \operatorname{Set} $$
$$ \operatorname{Aut}_u(A) = \text{the group of automorphisms of the deformations} \ \mathcal{X}_A$$
My question concerns the following proposition:
$\textbf{Proposition 2.6.2} $ If $X$ is projective, then
$\operatorname{Aut}_u$ has $H^0(X,T_X)$ as tangent space.
The proof of the proposition concludes (Eqn 2.29, pg. 92) with the claim that $$Aut_u(k[\epsilon]) \cong H^0(X, T_X).$$
However, $k[\epsilon]$ does not have a natural $R$-algebra structure so writing $Aut_u(k[\epsilon])$ doesn't really make sense.
What did Sernesi mean when he wrote this?
It seems to me that a $k$-algebra homomorphism $R \longrightarrow k[\varepsilon]$ exist by semi-universality, or am I missing something?
It appear that the same problem would occur at the beginning of the proof, when Sernesi shows that $\mathrm{Aut}_u$ satisfies condition $H_0$ (as in pag45), that is $\mathrm{Aut}_u(k)$ is the identity.
In my mind, geometrically, in the case of complex projective varieties this corresponds to a morphism of schemes $\mathrm{Spec}(k[\varepsilon]) \longrightarrow \mathrm{Def}(X)$, where $\mathrm{Def}(X)$ is (the germ of) the Kuranishi family of $X$.
In fact, there is a natural such structure, namely
$$ R \to k \to k[\varepsilon], $$
corresponding to the constant deformation $X\otimes k[\varepsilon]$, and indeed the automorphisms of $X\otimes k[\varepsilon]$ over $k[\varepsilon]$ restricting to the identity on $X$ correspond to derivations $\mathcal{O}_X \to \mathcal{O}_X$.
|
2025-03-21T14:48:31.392964
| 2020-07-01T04:34:58 |
364554
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364554"
}
|
Stack Exchange
|
Does a compact ANR have a local equiconnecting function which connects distinct points by simple paths?
It is known that if $X$ is a (metric) ANR, then $X$ is locally equiconnected, that is, there is a neighborhood $V$ of the diagonal $\Delta X \subseteq X \times X$ and a continuous function $$f \colon V \times [0,1] \rightarrow X$$
such that
For every $(x,y) \in V$, the path $f(x,y,-) \colon [0,1] \rightarrow X$ starts at $x$ and ends at $y$.
For every $x \in X$, the path $f(x,x,-) \colon [0,1] \rightarrow X$ is the constant path at $x$.
[Side note: Local equiconnectivity is equivalent to the diagonal map $\Delta \colon X \rightarrow X \times X$ being a Hurewicz cofibration.]
Let us also assume that $X$ is compact. My question is: Can we choose the $V$ and $f$ such that when $x \neq y$ in the 1st condition, the path connecting them is a simple path?
Remark: It follows from Lemma 2.1 of the paper "A remark on simple path fields in polyhedra of characteristic zero" by Fadell that the answer is yes when $X$ is a finite simplicial complex. I am interested in a (strict) generalization of this result.
|
2025-03-21T14:48:31.393064
| 2020-07-01T05:34:54 |
364555
|
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|
Stack Exchange
|
Variation of Euler characteristic when the sheaf is not flat
Let $f:X \to Y$ be a flat, projective morphism with $Y$ integral and every fiber of $f$ normal and integral. Let $F$ be a torsion-free, coherent sheaf on $X$ (not necessarily flat over $Y$). Then, is the function $y \mapsto \chi(F|_{X_y})$, upper semi-continuous, where $X_y:=f^{-1}(y)$?
Also, given any discrete valuation ring $R$ and a morphism $g:\mathrm{Spec}(R) \to Y$ with the
generic point of $\mathrm{Spec}(R)$ mapping to the generic point of $Y$. Denote by $F_R$ the pull-back of the $F$ to $X_R$ via the morphism $g$. We know that if $F_R$ is torsion-free, then $F_R$ is flat over $\mathrm{Spec}(R)$, hence the Euler characteristic of $F_K$ is the same as that of $F_k$ (here $K$ and $k$ are the fraction field and residue field of $R$, respectively).
My question is: if $F_R$ is not torsion-free, then is the support of the
torsion sub-sheaf of $F_R$ contained in the closed fiber $X_k$? If necessary, assume that the restriction of $F$ to the generic fiber $X_\eta$ of $f$ is torsion-free.
Not an answer, but the flattening stratification (https://stacks.math.columbia.edu/tag/0521, https://stacks.math.columbia.edu/tag/052F) might be useful. I think it at least shows that the function $\chi(F|_{X_y})$ is constructible?
Kollar in his article mentions that in general Euler characteristic is lower semi-continuous without the assumption of flatness. However, there is no proof of this statement in the article.
|
2025-03-21T14:48:31.393184
| 2020-07-01T06:21:51 |
364559
|
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|
Stack Exchange
|
General asymptotic result in additive combinatorics (sums of sets)
Let $S_1,\cdots,S_k$ be $k$ infinite sets of positive integers. Let $N_i(z)$ be the numbers of elements in $S_i$ that are less or equal to $z$. Let us further assume that
$$N_i(S) \sim \frac{a_i z^{b_i}}{(\log z)^{c_i}} \mbox{ as } z\rightarrow \infty,$$
where $a_i>0, 0 < b_i \leq 1, c_i \geq 0 (i=1,\cdots k)$ are constants. Also, let
$r(z)$ be the number of solutions to $x_1+\cdots + x_k\leq z$, with
$x_i \in S_i$ (here $z$ is an integer).
$t(z)$ be the number of solutions to $x_1+\cdots + x_k = z$, with
$x_i \in S_i$ (here $z$ is an integer).
Note that $t(z) = r(z) - r(z-1) \sim r'(z)$ asymptotically on average, where $r'$ denotes the derivative.
Question: Is the following result correct?
$$r(z) \sim \frac{\prod_{i=1}^k a_i \Gamma(b_i+1)}{\Gamma(1+\sum_{i=1}^k b_i)} \cdot \frac{z^{b_1 + \cdots + b_k}}{(\log z)^{c_1 +\cdots + c_k}} \mbox{ as } z \rightarrow \infty. $$
This covers many cases such as
Sums of squares
Sums of primes
Sums of powers
Sums of a square and a prime
Sums of superprimes
However it assumes that there are no congruence restrictions (for instance sums of odd integers can not be an odd integer, thus it does not work for sums of two odd primes, but it does for sums of two pseudo-primes depending on how you define pseudo-primes). See a proof for the case $k=2$, here. Also, is it true that if $r'(z) \rightarrow\infty$ as $z\rightarrow\infty$, then almost all integers can be written as
$$z = x_1+\cdots + x_k, \mbox{ with } x_i \in S_i.$$
By almost, I mean all but a finite number, or maybe all but a small infinite subset of positive integers, of density zero. If this was true, we would have the following result, maybe not that hard to prove:
Almost all positive integers can be written as $x_1^{d_1} + \cdots + x_k^{d_k}$ if $\frac{1}{d_1} + \cdots + \frac{1}{d_k}> 1$.
Here the $d_i$'s and $x_i$'s are positive integers. The constants $d_1,\cdots,d_k$ are fixed. This is compatible with a conjecture associated with the Waring problem, in the absence of congruence restrictions (see Wikipedia entry, check out the section Lower Bounds for $G(k)$). Namely, the fact that almost all positive integers can be written as $x_1^k + x_2^k +\cdots + x_{k+1}^k$ in the absence of congruence restrictions (e.g. not true for $k=2, 4$ due to congruence restrictions, but potentially true for $k=3,5,7, 11, 13$ etc.)
Update on 7/9/2022:
The formula for $r(z)$ provides an asymptotic upper bound to the number of elements in $S=S_1+\cdots +S_k$ that are smaller than $z$ (here $k$ is fixed). The actual value can be lower by an order of magnitude: this is the case for sums of squares. See also the multivariate Euler-Maclaurin summation formula, here.
I'm slightly confused because you answer your own question in the negative (is this asymptotic true - not necessarily, because there may be congruence restrictions). Do you mean further assume some kind equidistribution amongst different congruence classes, to e.g. rule out primes?
Yes you must assume some kind of equidistribution and define that concept.
Your guess for $r(z)$ is conjecturally correct, once corrected for congruence class restrictions (the way in which one does so is not obvious, and is a story in itself). This is a form of the heuristic which has been used in additive number theory, going back to Hardy and Littlewood, to make various conjectures about the count of solutions in Goldbach's conjecture and similar problems.
Of course, a conjecture and the heuristics leading to it are a long way from a theorem and a proof! Indeed, the last 100 years of the circle method have, in some sense, been an attempt to prove such conjectures, with varying degrees of success.
Could you share a reference about the definition of "no congruence class restrictions" assuming there is one generally agreed upon? Thank you.
|
2025-03-21T14:48:31.393429
| 2020-07-01T08:46:44 |
364565
|
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|
Stack Exchange
|
Off-diagonalize a matrix
Consider a self-adjoint matrix $M$ that has block form
$$M = \begin{pmatrix} M_{11} & M_{12} \\ M_{12}^* & M_{11} \end{pmatrix}.$$
I am wondering if there exists any criterion to decide if this matrix can be transformed by some invertible matrix $T$
such that $$TMT^{-1} = \begin{pmatrix}0 & C \\ C^* & 0 \end{pmatrix}$$
for some suitable matrix $C?$
Notice that one restriction that $\begin{pmatrix}0 & C \\ C^* & 0 \end{pmatrix}$ already puts is that the spectrum of $M$ has to be symmetric with respect to zero as conjugation by $$\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$$ shows.
As a first step, one might ask when we can achieve a form
$$TMT^{-1} = \begin{pmatrix}0 & C \\ D & 0 \end{pmatrix}$$
where $C$ and $D$ are arbitrary matrices?
Another common term for this form (block) anti-diagonal.
Looks like you're working in quantum mechanics?
This is a so-called chiral symmetry. The restriction on the symmetry of the spectrum of $M$ is the only restriction you need, you can then bring $M$ to the desired off-diagonal form by a unitary transformation:
$$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$
for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$.
Here $U$ is the unitary matrix of eigenvectors of $M$; the eigenvalues are contained in the diagonal matrix $\Lambda$.
|
2025-03-21T14:48:31.393918
| 2020-07-01T09:15:41 |
364567
|
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|
Stack Exchange
|
A Fredholm equation with non-separable kernel
I'm trying to solve this form of Fredholm equation:
$$
g(v)=f_1(v)+\int\limits_{0}^{v_\mathrm{th}} g(v_s)\frac{e^{-\tfrac{\big[(v-v_\mathrm{init})-(1-v_\mathrm{leak})(v_s-v_\mathrm{init})\big]^2}{2v_\mathrm{step}^2}}+e^{-\tfrac{\big[(v+v_\mathrm{init})+(1-v_\mathrm{leak})(v_s-v_\mathrm{init})\big]^2}{2v_\mathrm{step}^2}}}{\sqrt{2\pi}v_\mathrm{step}}\mathrm{d} v_s,
$$
where,
$v_\mathrm{init}$, $v_\mathit{th}$, $v_\mathrm{step}$, $v_\mathrm{leak}$ are constant and
$f_1$ has the following form:
$$
f_1(v)=\frac{e^{-\tfrac{(v-v_\mathrm{init})^2}{2v_\mathrm{step}^2}}+e^{-\tfrac{(v+v_\mathrm{init})^2}{2v_\mathrm{step}^2}}}{\sqrt{2\pi}v_\mathrm{step}}
$$
Are there any methods or theorems that can solve this problem?
Since I'm freshman in university, it's very challenging for me. But I really want to untangle this problem.
Can you help me to solve this kind of equation or let me know what theory I should study?
I would be grateful if you could just give me directions, so I can know where to start studying. Sorry for my bad English.
Where is the equation?
I signed up yesterday, and I think I was restricted to upload figures. Thank you for editing my question.
Are you sure the second exponential term is correct in the kernel? Looks there is no connection between kernel and your f1 function
Oh, I think the editor made a mistake by changing the picture to an expression. I edited second term in kernel, so that K(v,vinit)=f1(v).
|
2025-03-21T14:48:31.394039
| 2020-07-01T11:36:11 |
364575
|
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"Eigil Fjeldgren Rischel",
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|
Stack Exchange
|
Left Kan extensions of "strong" monoidal functors
Consider the 2-category $\mathsf{MonCat}$ where objects are monoidal categories,
1-cells are strong monoidal functors, and 2-cells are monoidal natural transformations.
Given arrows $f: \mathsf{C} \to \mathsf{D}, g: \mathsf{C} \to \mathsf{E}$ in this category,
we can obviously ask for a left Kan extension $Lan_fg : \mathsf{D} \to \mathsf{E}$.
If we instead work in $\mathsf{MonCat}^\mathrm{lax}$, the 2-category of monoidal categories, lax monoidal functors and monoidal natural transformations, this left Kan extension is known to exist under some reasonable assumptions.
Questions:
Under reasonable assumptions on the categories involved, does the "strong monoidal Kan extension" exist?
Does the "lax monoidal Kan extension" recover the strong version in the case where the two input functors are strong monoidal? In other words, are the subcategories of strong monoidal functors stable under left Kan extension along strong monoidal functors?
At the highest level of generality, we can consider lax or strong $\mathcal{O}$-monoidal functors between $\mathcal{O}$-monoidal $\infty$-categories, in the sense of Lurie.
In that context, the lax left kan extensions seem to correspond to Lurie's "operadic Kan extensions", but again there does not seem to be a strong version of this in the literature.
Related, but not related: on the problem of when a product preserving functor is still product preserving when left Kan-extended, and on the fact that the previos fact is a "miracle"
Also, see here! https://arxiv.org/pdf/1809.10481.pdf I wasn't aware of the existence of this note.
@fosco: Yes, I saw that note - it seems to say something about when the ordinary kan extension is also a "lax monoidal kan extension" - but I'm not having much luck figuring out if it's also strong when you extend a strong functor
This paper studies the lifting of Kan extensions that are "algebraic" over a monad on double categories; section 2 of this paper by Weber has an analogous 2-categorical approach. Roughly, to lift lax monoidal structure it suffices that the tensor of $E$ preserves the left Kan cell in both variables; to lift the colax structure it suffices that the left Kan cell is preserved by composition with the colax structure cell of $f$. In the pseudo case having both assumptions is sufficient.
|
2025-03-21T14:48:31.394210
| 2020-07-01T12:03:57 |
364576
|
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"Jon Bannon",
"MaoWao",
"Narutaka OZAWA",
"YCor",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364576"
}
|
Stack Exchange
|
Are groups with the Haagerup property hyperlinear?
In his 2008 paper Hyperlinear and Sofic Groups: A Brief Guide, Pestov asked (Open Question 9.5) whether every group with the Haagerup property is hyperlinear (or sofic). Has this question been answered in the meanwhile?
A short recap of the relevant definitions: A group is hyperlinear (sofic) if it embeds into the metric ultraproduct of unitary groups equipped with the normalized Hilbert-Schmidt distance (symmetric groups with the normalized Hamming distance).
A group has the Haagerup property if there is a sequence of positive definite functions that vanish at infinity and converge pointwise to the constant function $1$.
I believe this is still open.
For an irreducible lattice in a nonlinear Lie group such as a finite cover of $\mathrm{SU}(2,1)^k$ I guess hyperlinearity is unknown. (I'm not sure if non-residually finite lattices are known in such groups, but it sounds plausible).
It's not known if Thompson's group F is hyperlinear.
Thank you all for your comments. Especially @NarutakaOZAWA 's comment seems as good an answer I can expect. If you want to post it as an answer, I would be glad to accept it.
Thompson's group $F$ has the Haagerup property [1], but it is not known if it is hyperlinear according to Narutaka Ozawa's comment.
[1] Farley. Finiteness and CAT(0) properties of diagram groups. Topology, 2003.
|
2025-03-21T14:48:31.394333
| 2020-07-01T13:46:00 |
364581
|
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"Γιώργος Πλούσος"
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"url": "https://mathoverflow.net/questions/364581"
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|
Stack Exchange
|
Is this model of converting integers to Gray code correct?
The model shown in the figure converts all numbers that have k digits in the binary system to Gray code without any calculation, but I have no proof that guarantees this claim.
Here is some information on how to use it.
Conversion model of all integers that have k digits in the binary system in Gray code.
Rules
You will need k columns of numbers. The numbers to be encoded are arranged in the last column.
The conversion is done gradually by transferring groups of numbers to the adjacent columns in the way the arrows show. There are two types of paired arrows, parallel (=) and intersecting (×). Symbolically, the (=) and (×) are considered inverse of each other.
Each column with arrows is formed by an exact copy of the previous column and by a copy of the previous column that has inverted pairs of arrows, which is placed below the exact copy.
Observation
If we set "=" = 0 and "×" = 1, then the successive columns containing arrows form the Thue Morse sequence which essentially forms the rules for converting integers to Gray code.
I use the following mnemonic trick/inductive definition whenever I encounter gray codes: To get the Gray codes for the numbers $2^n,\ldots,2^{n+1}-1$, take the Gray codes of the numbers $1,..2^n-1$ reverse that list and change the new first bit to 1. Maybe you can prove things inductively using that description.
Yes, the proposed scheme always produces a Gray code. This can be proved by induction on $k$ as follows.
Proof.
The base cases of $k=1,2$ are trivial. Assume that a Gray code is produced for integers with $k=s\geq 2$ bits, and let's prove the same for $k=s+1$.
Let $c_1, \dots, c_{s+1}$ denote the columns. We view the column $c_s$ as formed by blocks of size 2, where the $i$-th block consists of numbers $2^s+2i-2$ and $2^s+2i-1$ is some order ($i=1,2,\dots,2^{s-1}$). It is easy to see that each column $c_j$ for $j<s$ represents a permutation of the same blocks.
Replacing $i$-th block in $c_s$ with a single number $2^{s-1}+i-1$ and using the induction assumption for $k=s$, we conclude that $c_1$ forms a Gray code unless we have a change in the numbers parity (i.e., unequal least significant bits) when we go from some block to the next one. It remains to show that such change in parity never happens.
Call a block positive is the numbers in it appears in their natural order, and call it negative otherwise. Absence of the parity change between adjacent blocks in $c_1$ means that positive and negative blocks in it alternate. Below we show that this property holds already for $c_{s-1}$, and then it trivially propagates to all $c_j$ with $j<s-1$ (including $c_1$).
Let $(\tau_i)_{i\geq 0}$ denotes Thue-Morse sequence. By construction, the $i$-th block in $c_s$ is positive iff $\tau_{i-1}=1$. Furthermore, the $i$-th block in $c_s$ retains its position in $c_{s-1}$ iff $\tau_{\lfloor(i-1)/2\rfloor}=1$, otherwise its position shifts by 1 (down or up). Since $\tau_{i-1} + \tau_{\lfloor(i-1)/2\rfloor} \equiv i-1\pmod{2}$, we get that in $c_{s-1}$ positive blocks appear at odd positions and negative blocks appear at even positions, i.e. positive and negative blocks alternate. QED
The proof inspires self confidence. I can't follow the subtle arguments but it shows great job. Thank you, as well as everyone who contributed, in any way, to this thread.
@ΓιώργοςΠλούσος: The proof would be easier to follow if you execute it on one of your examples for $k=4$ or $k=5$. Let me know if any particular argument is unclear.
Your proof is perfect, dear Max, written clearly and with emphasis on detail, I just meant that I have shortcomings in mathematics. I'm more of a visual guy and I have to get used to this notation.
|
2025-03-21T14:48:31.394618
| 2020-07-01T15:27:59 |
364585
|
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|
Stack Exchange
|
Continuous version of the fundamental theorem of invariant theory for the orthogonal group
A standard result in the invariant theory of the orthogonal group states the following.
Theorem
Let $(E, \langle .,. \rangle)$ be an n-dimensional euclidean vector space,
let $f : E^m \rightarrow {\bf R}$
a polynomial function satisfying
$f(g(v_1), ... g(v_m)) = f(v_1,...,v_m)$
for all isometries $g$ of $E$ and $v_1$,..., $v_m \in E$.
Then such a function is a polynomial function
in the quantities $\{\langle{v_i}{v_j}\rangle\}_{i,j = 1...m}$.
Does the theorem holds in the topological setting,
namely when polynomial is replaced by continuous ?
My guess is that it should be true and the proof should be simpler than its algebraic counterpart, maybe a short computation using SVD.
All references I know present the algebraic proof though. Same question in the differential setting.
Yes. It suffices to show that if one has a sequence $\vec v^{(n)} = (v^{(n)}_1,\dots,v^{(n)}_m) \in E^m$ whose Gram matrix $(\langle v^{(n)}_i, v^{(n)}_j \rangle)_{i,j=1,\dots,m}$ converges to a Gram matrix $(\langle v_i, v_j \rangle)_{i,j=1,\dots,m}$ of a tuple $\vec v = (v_1,\dots,v_m) \in E^m$, then after applying linear isometries to each of the $\vec v^{(n)}$, that $\vec v^{(n)}$ converges to $\vec v$.
By permuting indices we may assume that $v_1,\dots,v_k$ are linearly independent, and $v_{k+1},\dots,v_m$ are in the span of $v_1,\dots,v_k$, for some $0 \leq k \leq m$. Applying the Gram-Schmidt process to $v_1,\dots,v_k$ and transforming $v^{(n)}_1,\dots,v^{(n)}_k$ appropriately we may assume that $v_1,\dots,v_k$ are orthonormal. It is not difficult to inductively apply isometries to the $\vec v^{(n)}$ so that $\vec v^{(n)}_i$ converges to $v_i$ for $i=1,\dots,k$. If we write each $v_j, j=k+1,\dots,m$ as a linear combination $v_j = a_{j1} v_1 + \dots + a_{jk} v_k$ of $v_1,\dots,v_k$, then from the Gram matrix convergence we see that $\| v_j^{(n)} - (a_{j1} v^{(n)}_1 + \dots + a_{jk} v^{(n)}_k) \|^2$ converges to zero, hence $v_j^{(n)}$ converges to $v_j$ by the triangle inequality, and the claim follows.
Reflecting on the answer of Terence Tao, I guess it boils down to
the fact that an injective proper map between locally compact spaces
is a homeomorphism onto its image. Since we are working with
${\bf R}^n$ here, there is a simple characterisation of proper maps
that leads to the following statement.
Let $\Phi : {\bf R}^n \longrightarrow {\bf R}^k$ be a continuous
map satisfying
$$
\|\Phi(x)\| \longrightarrow \infty \quad when \quad {\|x\| \rightarrow \infty}.
$$
Let us define the fiber relation on ${\bf R}^n$ by
$x \sim x' \iff \Phi(x) = \Phi(x').$
Then $({\bf R}^n/\sim)$ is a locally compact metric space and
$\bar{\Phi} : ({\bf R}^n/\sim) \longrightarrow \Phi({\bf R}^n)$
is a homeomorphism.
The condition on the norm is there to ensure that
for all compact set $K \subset {\bf R}^k$, $\Phi^{-1}(K)$ is
closed and bounded (hence compact). In particular, the fibers
$\Phi^{-1}(\{y\})$ are compact and thus we can define
a distance on the quotient as follows:
$$
d(\bar{x}, \bar{x}') = d(\Phi^{-1}(\{\bar{\Phi}(\bar{x})\}), \Phi^{-1}(\{\bar{\Phi}(\bar{x}')\})).
$$
For the problem at hand, we take
$\Phi(v_1,...,v_l) = (\langle v_i, v_j \rangle)$
and note that the fibers of $\Phi$ are the orbits of the elements
of ${\bf R}^n$ under the action of the orthogonal group.
Then, for any invariant $f$, we have
$$
f(v_1,...,v_l) = \bar{f}(\bar{\Phi}^{-1}(\bar{\Phi}(\pi(v_1),..., \pi(v_l))))
= \bar{f} \circ \bar{\Phi}^{-1}(\langle v_i, v_j\rangle).
$$
The norm condition also ensures that $\Phi({\bf R}^n)$ is closed,
so $\bar{f} \circ \bar{\Phi}^{-1}$ can be extended to all ${\bf R}^k$ if needed.
|
2025-03-21T14:48:31.394855
| 2020-07-01T15:31:14 |
364586
|
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|
Stack Exchange
|
Permanent of a Kronecker product of matrices
It is well known that $\det(A \otimes B) = \det(A)^m \det(B)^n$ when $A$ and $B$ are square matrices of size $n$ and $m$ where $\otimes$ denotes the Kronecker product.
Question: Is there a similar formula for the permanent $per(A \otimes B) $ (instead of the determinant)?
One motivation is to calculate the permanent of the character table of an abelian group since the permanent for cyclic groups is known and the character table of a direct product $G_1 \times G_2$ of groups is given as the Kronecker product of the character table of $G_1$ and $G_2$.
An equality of polynomial nature looks tricky in general: If $a$ and $b$ are integers greater than $1$, and $J_{a} $ and $J_{b}$ are the square all $1$'s matrices of respective sizes $a$ and $b$, then $J_{a},J_{b}$ and $J_{a} \otimes J_{b}$ have respective permanents $a!,b!$ and $(ab)!$. Note that there is always a prime greater then either of $a$ and $b$, but less than $ab$.
In this paper:
R. A. Brualdi, Permanent of the direct product of matrices, Pacific J. Math. 16 (1966), 471482
(the Kronecker product is called here 'direct product') it is shown that, if $A$, $B$ are nonnegative matrices with order $m$, $n$, respectively, it holds true that:
$$ \operatorname{per}(A \otimes B) \geq \operatorname{per}(A)^n \operatorname{per}(B)^m $$
where the equality holds iff $A$ or $B$ has at most one nonzero term in its permanent expression. Moreover, there exists a minimal number, denoted in the paper by $K_{m,n}$, such that:
$$ \operatorname{per}(A \otimes B) \leq K_{m,n} \operatorname{per}(A)^n \operatorname{per}(B)^m $$
These numbers satisfy the inequality:
$$ K_{m,n} \geq \frac{(mn)!}{(m!)^n (n!)^m} $$
and it is conjectured that we actually have an equality.
Furthermore, in the paper:
Marvin Marcus, Permanents ot direct products, Proc. Amer. Math. Soc. 17:226-231
(1966)
it is proven that if $A$, $B$ are positive semidefinite Hermitian square matrices of order $m$, $n$, respectively, then:
$$ \operatorname{per}(A \otimes B) \geq \left (\frac{1}{n!} \right )^m \left (\frac{1}{m!} \right)^n \operatorname{per}(A)^n \operatorname{per}(B)^m $$
Equality holds iff at least one of $A$, $B$ has a zero row.
|
2025-03-21T14:48:31.395003
| 2020-07-01T15:34:26 |
364587
|
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|
Stack Exchange
|
Flat function with a spectral gap
I am looking for a sequence of functions $f_n,n\geq 1$ in $L^2(\mathbb R)$ such that $f_n$ is equal to $1$ on $[-n,n]$ and $\hat{f_n}$ vanishes on $[-1,1]$.
Actually, I would also like $f_n$ to be $C_c^2$ but I can relax the spectral gap assumption to $\int_{-1}^1 \hat{f_n}^2$ can be arbitrarily small (i.e. smaller than $a_n$ for any sequence of positive numbers)
EDIT: I am also interested in a result saying that this is impossible
Are you imposing any conditions on how much $f_n$ may grow at infinity as $n\to\infty$? If you impose such conditions, then $f_n$ converges to 1 in ${\cal S}'$. This would imply that $\hat f_n$ converges to a delta function, which is not consistent with vanishing on $[-1,1]$.
Maybe this is helpful: https://math.stackexchange.com/q/2332292/99220 https://mathoverflow.net/q/275072/107094
I am not imposing any condition, but as I indicated in my edit I am also interested in a negative result
|
2025-03-21T14:48:31.395095
| 2020-07-01T16:07:37 |
364590
|
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"Padraig Ó Catháin",
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|
Stack Exchange
|
Determining the irreducible invariant subspaces of a permutation action by computing eigenspaces of a matrix
Let $\Sigma\subseteq\mathrm{Sym}(n)$ be a permutation group on $N:=\{1,...,n\}$.
My goal is to determine the irreducible invariant subspaces of the permutation action of $\Sigma$ on $\Bbb R^n$, and I wonder whether the following technique always works.
Consider the element-wise action of $\Sigma$ on the subsets $\{i,j\}\subseteq N$ for $i,j\in N$.
Let $o(i,j)$ denote the orbit of $\{i,j\}$ w.r.t. this action, and let $\mathcal O(\Sigma)$ denote the set of all these orbits.
Now let $R:= \Bbb R[\,x_o\!\mid\! o\in\mathcal O(\Sigma)\,]$ be the polynomial ring over $\Bbb R$ with one variable $x_o$ for each orbit, and consider the matrix $M\in R^{n\times n}$ with components $ M_{ij} = x_{o(i,j)}$.
Question: Is is true that that the eigenspaces of this matrix $M$ are exactly the irreducible invariant subspaces of the permutation action of $\Sigma$ on $\Bbb R^n$?
The matrix is symmetric, and thus the eigenspaces are real and mutually orthogonal.
Each eigenspace is indeed $\Sigma$-invariant, but is it also irreducible?
I have applied this idea computationally and it never failed me.
In practice, I do not use variables $x_o$, but some algebraically independent number (or just some random numbers). But does it work always?
The orbits of $G$ on pairs ${x,y}$ of points in the domain are called orbitals. The matrices associated with the orbitals span the centralizer algebra. The matrix $M$ in your question is somehow a generic element of the centraliser algebra. Thinking about the permutation representation with respect to a basis in which it is a direct sum of irreducible representations, you can work out an explicit description of the centraliser algebra - the books of Wielandt or Cameron, both called "Permutation groups", should have details.
If the answer to your precise question is not 'yes' then some minor modification should be true, though note the orbital matrices are not always symmetric.
|
2025-03-21T14:48:31.395250
| 2020-07-01T16:15:05 |
364591
|
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|
Stack Exchange
|
how to prove the binomial equation below
I tried to open up all binomial expressions but things got more complicated. I could not find an appropriate solution.I'm just stuck and trying to find a solution for like 2 hours.I would be very happy if you come up with an ending solution I don't think i can improve my progress any longer
n is a positive integer
${n\choose 1}-2{n\choose2}+3{n\choose 3}+... (-1)^nn{n\choose n}=0$
${n-1\choose 0 }+ {n-1\choose 1} -2{n-1\choose 1} -2 { n-1\choose 2}+ 3{n-1\choose 2} +3{n-1\choose 3 }..$
I've just separated the expressions like this but still could not manage to find a solution.
A quick proof is to differentiate $(1+x)^n=\sum\binom{n}{k}x^{k}$ and then substitute $x=-1$. But this problem is not appropriate for this site, which is for research level math. It would be appropriate on MathStackExchange.
Note that $j{n\choose j}=n{n-1\choose j-1}$, hence
$$\sum_{j=1}^n (-1)^{j+1}j{n\choose j}=n\sum_{j=1}^n (-1)^{j+1}{n-1\choose j-1}=n\sum_{j=0}^{n-1} (-1)^{j}{n-1\choose j}=n(1-1)^{n-1}=0,$$
for $n\geq 2$. (The identity does not hold for $n=1$.)
|
2025-03-21T14:48:31.395343
| 2020-07-01T16:18:58 |
364592
|
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|
Stack Exchange
|
Injectivity of analytic functions
Suppose $f : \mathbb{R} \rightarrow \mathbb{R}^n$ is a real analytic function on $(a, \infty)$. I have two questions:
Suppose $||f(x)|| \rightarrow \infty$ as $x \rightarrow \infty$. I know without further conditions $f$ can be not injective (for e.g. take $f(x) = x + 2 \sin(x)$ for $n=1$). I want to know if any generic conditions are known such that $f$ is eventually injective (meaning that there exists $b > a$ s.t $f$ is injective on $(b, \infty)$.
Suppose $f(x)$ converges as $x \rightarrow \infty$, and $(\partial^m f / \partial x^m)(x) \rightarrow 0$ as $x \rightarrow \infty$, for all $m \geq 1$. Again without further conditions $f$ is not eventually injective (for e.g. $f(x) = e^{-x} \sin (x)$ shows it is not true generically). So again my question is whether there are some simple conditions known that makes $f$ eventually injective.
In 1, with "cannot be injective" you presumably mean "can be not injective". Your counterexample also doesn't work since $|x\sin(x)|$ doesn't tend to infinity. However, $f(x)=x+2\sin x$ will work for that.
Thanks @Wojowu. Updated the question.
If $n>1$, injectivity would be generic. For n=1, monotonicity seems to be what you are looking for.
@MichaelRenardy I myself would expect injectivity to be generic for $n>2$, but not for $n=2$, by analogy with random paths. I don't know of a way to formalize this statement though, especially if we condition that the functions tend to infinity. Do you have any reason (be it informal) to expect injectivity for $n=2$?
It can't be generically injective for $n=2$, e.g. we can force $f$ to be non-injective by requiring $|f(x) - g(x)| < 1/3$ for $-\pi \le x \le \pi$, where $g(x) = [\sin(x),\sin(2x)]$.
Yes, injectivity would only be generic for $n>2$. The condition $|f|\to\infty$ does not help. We can still have a path go to infinity and cross itself infinitely often.
For $n=1$, such simple (non-tautological) conditions do not exist, because real-analytic functions can mimic any $C^1$ functions in terms of their monotonicity patterns and the limit value at $\infty-$, simultaneously. So, the real-analyticity condition does not help at all; it is far from any kind of a rigidity condition.
Indeed, for any $a\in\mathbb R$, take any $C^1$ function $g\colon[a,\infty)\to\mathbb R$ such that $\exists\ g(\infty-)\in[-\infty,\infty]$ and for some increasing to $\infty$ sequence $(x_j)$ in $[a,\infty)$ we have $(-1)^j g'(x_j)>0$ for all $j$. Extend $g$ to a $C^1$ function on $\mathbb R$ and denote the extension still by $g$.
Take now any continuous function $h\colon\mathbb R\to(0,\infty)$ such that $h(x_j)<|g'(x_j)|$ for all $j$ and $h(\infty-)=0$. Then, according to a theorem by Whitney (see e.g. the first paragraph of this paper), there is a real-analytic function $f\colon\mathbb R\to\mathbb R$ such that $|f-g|+|f'-g'|\le h$ and hence $f(\infty-)=g(\infty-)$ and $(-1)^j f'(x_j)>0$ for all $j$ -- so that the real-analytic function $f$ indeed mimics the monotonicity pattern of the $C^1$ function $g$ and has the same limit value at $\infty-$.
|
2025-03-21T14:48:31.395543
| 2020-07-01T17:39:01 |
364596
|
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"Carlo Beenakker",
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|
Stack Exchange
|
Recommendations for mathematical essayists
I was wondering if people had recommendations for mathematical essays (by this I mean essays on a mathematical topic, not necessarily essays written by mathematicians).
A person who I used to find entertaining to read at high school was Isaac Asimov, although for some reason he did not write many essays on pure mathematics. In particular I remember the essay collection X Stands for Unknown, which had several entertaining mathematics essays, most of them with a combinatorial flavour if I remember rightly.
there is an extensive list at https://mathoverflow.net/q/220052/11260
Martin Gardner, maybe? But does this question belong to Mathoverflow?
Gian-Carlo Rota wrote beautifully and widely, my personal favourite.
Ed Sandifer and his column in AMM "How Euler Did It" ("HEDI").
|
2025-03-21T14:48:31.395652
| 2020-07-01T18:00:02 |
364600
|
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|
Stack Exchange
|
Homotopy group action and equivariant cohomology theories
Many of the introductory notes on generalized equivariant cohomology theories assume that one is working over the category of $G$-spaces or $G$-spectra. However, one thing that concerns me is that the action of $G$ is always strict. A $G$-space $X$ is given by a group homomorphism $G\to \text{Aut}(X)$, where $\text{Aut}(-)$ denotes the group of continuous automorphisms.
If instead I want to allow $\sigma:G\times X\to X$ to solve $\sigma(e)\sim \text{id}_X$ and
$$\sigma\circ(\text{id}_G\times \sigma)\sim \sigma\circ (\mu_G\times \text{id}_X)\,, $$
only up to homotopies ($\mu_G$ here is the multiplication on $G$) and possibly have higher homotopies I need to think about $\infty$-groupoids, as this paper shows that there is an obstruction to strictifying homotopy group actions.
Is there a well defined notion of equivariant cohomology theories in this setting?
For an $\infty$-groupoid, one can take its homotopy quotient (the colimit). Can one define the equivariant cohomology as the cohomology of this quotient?
A homotopy coherent group action is only enough to reconstruct Borel cohomology theories, because it is not enough to describe how the "fixed point spaces" should interact for various subgroups $H \subset G$. A homotopy coherent group action may be strictified in the sense that there will be a space $EG \times_G X$ with strict $G$-action and a continuous (coherently equivariant) map $EG \times_G X \to X$ which is a nonequivariant homotopy equivalence.
The right notion of homotopy coherent $G$-space to recover Bredon equivariant cohomology is not obvious to me, but it's probably encoded as some fibering over the orbit category. I'm sure an expert will arrive shortly to explain the right notion.
What does $EG\times_G X$ mean in this case? The diagonal "action" won't give me a relations as it isn't really an action. Is it just notation?
It's just me being lazy in notation. It's a bar construction $B(G,G,X)$, which makes sense in the coherent setting.
As shown by Dwyer and Kan in 1980s, any homotopy coherent ∞-group action can be strictified to a strict action of a simplicial group on a simplicial set, or a topological group on a topological space. So passing to the homotopy coherent setting yields nothing new. As already observed in other comments, for Bredon equivariance one really needs presheaves on the orbit category.
@DmitriPavlov which paper of Dwyer and Kan is this? Unfortunately, the 80s does not narrow it down too much for them two!
@JackDavidson: Dwyer–Kan, Equivalences between homotopy theories of diagrams, Theorem 2.4.
From modern perspective this is much more straightforward than the "genuine" version you described above the question. Naive $G$-spaces are just functors $BG\to \cal{S}$ among infinity categories. $G$-spectra are just functors $BG\to \mathrm{Sp}$. You can think of a $G$-spectrum as a functor on $G$-spaces by
$E \mapsto (X\mapsto \mathrm{Map}_{\mathrm{Sp}^{BG}}(\mathbb{S}[X],E))$ where $\mathbb{S}[-]= \Sigma^{\infty}$ is the stabilization functor, applied pointwise to functors from $BG$. Hence after accepting some notions like functors and stabilization in infinity category theory you immediately get a theory of equivariant stuff of this "up to homotopy" flavour. In particular, if $E$ has trivial $G$-action then by trivial-colimit adjunction and colimits preservation of the stabilization we get $$\mathrm{Map}_{\mathrm{Sp}^{BG}}(\mathbb{S}[X],E)\simeq \mathrm{Map}_{\mathrm{Sp}}(\mathbb{S}[X]_{hG},E)\simeq \mathrm{Map}_{\mathrm{Sp}}(\mathbb{S}[X_{hG}],E)$$ and you indeed get the cohomology of the homotopy quotient.
In some sense, the surprising thing from this modern perspective is the existence of the "strict" version, which is slightly harder to define internally to modern homotopy theory, even though it is doable.
Thank you for the quick answer. I am just unsure about what the notation $(-)_{hG}$ means. Could you also give me a reference where this modern perspective is discussed?
Unfortunately im horrific with references. It is centainly extractable from Lurie's HTT and HA, but I don't know of a good concise reference. Hopefuly someone who know the literature better will do a better job here! In any case, this notation is for the homotopy quotient, i.e. for example one way to compute it is $(X\times EG / G)$ where $EG$ is a contractible free $G$-space.
Much has already been said in the other answers and comments, but let me summarize a few points.
One way to obtain from a category a 'homotopy theory' (aka an $\infty$-category) is to specify a notion of weak equivalence. On the category of $G$-spaces (i.e. topological spaces with strict $G$-action), two of the major notions of weak equivalences are the following:
A map $X \to Y$ of $G$-spaces is a weak equivalence if the underlying map of spaces is a weak homotopy equivalence, or
a map $X \to Y$ of $G$-spaces is a weak equivalence if the maps $X^H \to Y^H$ are weak homotopy equivalences for all subgroups $H\subset G$.
More generally, you could specify a family $\mathcal{F}$ of subgroups of $G$ and you demand that you have a weak equivalence on $H$-fixed points for all $H\in \mathcal{F}$, but let's focus on the two cases above and call them underlying and genuine.
(Edit: Reacting to Denis's comment a clarification: Why should we consider these two kinds of equivalences? Geometrically, $G$-homotopy equivalences (i.e. we have an equivariant homotopy inverse and the homotopies are also equivariant) are maybe the most relevant notion. As in non-equivariant topology, there is a Whitehead theorem showing that genuine weak equivalences between $G$-CW complexes are $G$-homotopy equivalences. Illman's theorem shows that every compact $G$-manifold has the structure of a $G$-CW complex, so one can say that most nice $G$-spaces have the structure of a $G$-CW complex. If we want a Whitehead theorem for underlying equivalences instead, we must demand that the $G$-action is free though. Sometimes we are happy to do this, but often this is too restrictive. The different families $\mathcal{F}$ correspond to allowing different families of isotropy. )
It is the $\infty$-category associated with the underlying equivalences that can be modelled by homotopy coherent actions. Taking the coherent nerve of the simplicial category of spaces $\mathcal{S}$, we obtain the $\infty$-category of spaces and the $\infty$-category of spaces with homotopy coherent $G$-action is then modelled/defined as simplicial set maps (aka functors) from $BG$ into this coherent nerve. (If we fix $X$, this is the same as simplicial set maps from $BG$ into $B$ of the homotopy automorphisms of $X$.) This $\infty$-category is equivalent to that associated with $G$-spaces and underlying equivalences. (It is nothing special about starting with a group here. We can instead take functors from an arbitrary small category $\mathcal{C}$ into topological spaces and have a similar story using $B\mathcal{C}$. See e.g. Proposition <IP_ADDRESS> of Higher Topos Theory.)
We cannot, however, recover from the homotopy coherent action the data of the fixed points $X^H$. If we want to model this homotopy-coherently, we need not only $X$ with a homotopy coherent $G$-action, but we also need all spaces of fixed points $X^H$ with their residual actions and all the restriction maps between them. This can be modelled as a functor from the (nerve of the) orbit category $\mathrm{Orb}_G$ of $G$ into $\mathcal{S}$. In the background is Elmendorf's theorem that shows that there is a Quillen equivalence between $G$-spaces with genuine equivalences and functors from $\mathrm{Orb}_G$ to $\mathrm{Top}$ with underlying equivalences (the Quillen equivalence being given by associating to $G/H$ the fixed points $X^H$). Then one can apply e.g. Proposition <IP_ADDRESS> of HTT again.
As already remarked by others, some equivariant cohomology theories are only sensitive to underlying equivalences (Borel theories), while others are only invariant under genuine equivalences. The latter are actually more frequent (Bredon cohomology, equivariant K-theory, equivariant bordism...).
The story for spectra is a bit more complicated because there are even more types of weak equivalences one can put on, say, orthogonal spectra with a $G$-action. In Shachar's answer, he describes the case corresponding to underlying equivalences. Genuine equivalences (with respect to a complete universe) require more work. For finite groups, one can consider functors from the Burnside category -- this is the perspective of viewing $G$-spectra as spectral Mackey functors. But this is maybe leading too far here.
It might be helpful to remark that genuine weak equivalences of $G$-spaces (those that are weak equivalences on all fixed points) are precisely those that have an inverse up to equivariant homotopy for "nice" spaces (i.e. $G$-CW-complexes).
Since the OP asked for references, and about $(-)_{hG}$ in particular, I'll mention a few.
Schwede has very clear lecture notes about the basics of equivariant (stable) homotopy theory.
Many more references can be found in this syllabus, including specific references to Lurie's work.
Lastly, Paul VanKoughnett created a wonderful series of lectures, and this one carefully goes through homotopy fixed points and homotopy orbits, i.e. $(-)^{hG}$ and $(-)_{hG}$.
|
2025-03-21T14:48:31.396330
| 2020-07-01T18:43:13 |
364603
|
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|
Stack Exchange
|
The largest group acting on a non-orientable surface of genus 5
Let $N_5$ denote the non-orientable surface of genus 5.
In Conder's database https://www.math.auckland.ac.nz/~conder/BigSurfaceActions-Genus2to101-ByGenus.txt we can see that the biggest finite group $F$ acting on $N_5$ has order 120. Moreover, the quotient has signature $(0; +; [-]; \{(2,4,5)\})$.
Is there a very concrete description of this group $F$?
To be even more concrete. I would like to the length $n$ of the largest chain of subgroups $1=F_0<F_1<\cdots< F_n=F$ of $F$. Since $120=(2^3)(3)(5)$ then $n\leq 5$. Is it $n=5$? Is $n$ strictly less than 5?
Thank you!
I don't understand the notation well enough to answer your first question. As to the second, it seems to me that, no matter which group on the genus 5 list one considers, the only possible composition factors are groups of prime order and the alternating group $A_5$. As $A_5$ has a chain $1<Z_2<Z_2 \times Z_2<A_4<A_5$, it follows that every one of the groups in question has a chain of subgroups that is as long as Lagrange's Theorem will allow.
Eventually it's just a basic question about a certain group of order 120, which you should make more explicit, but actually from John Shareshian's answer, it follows that every group of order 120 has a chain of the largest possible size (as it's always true for solvable groups, and for $A_5$).
The group $F$ is isomorphic to the symmetric group $S_5$.
In fact, since $N_5$ is non-orientable of genus $5$, both $F$ and the extended group $F^*$ (of order twice the order of $F$) act on its orientable double cover, that has genus $4$. In Conder's database, this is expressed by saying that the action of $F$ in genus $4$ is reflexible and that there is a non-orientable quotient.
Looking at the actions of groups of order $120$ on orientable surfaces of genus $4$ that have these properties, we can find the presentation of $F$; in fact, there is a unique such a group:
Now a simple computation with GAP4 does the job.
Thank you Francesco. Your answer is exactly the kind of answer I was expecting.
|
2025-03-21T14:48:31.396496
| 2020-07-01T18:43:21 |
364604
|
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|
Stack Exchange
|
On isoclinism classes of finite p-groups
With reference to
James, Rodney, The groups of order (p^6) ((p) an odd prime)., Math. Comput. 34, 613-637 (1980). ZBL0428.20013., My question is can we get isoclinism class $\phi_2$ for a finite p-group of order $p^n$ ? Here, notation $ \phi_s(m_1,m_2...m_r)x_t $ denotes isoclinic class $\phi_2$ of type $(m_1,m_2...m_r)$ and $x_t$ denotes the genus of the group (defined in the article I have mentioned).
|
2025-03-21T14:48:31.396561
| 2020-07-01T19:27:14 |
364607
|
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"Ian Agol",
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|
Stack Exchange
|
Is there an orientable prime manifold covered by a non-prime manifold?
A manifold is called prime if whenever it is homeomorphic to a connected sum, one of the two factors is homeomorphic to a sphere.
Is there an example of a finite covering $\pi : N \to M$ of closed orientable manifolds where $M$ is prime and $N$ is not?
There are no examples in dimensions two or three. If one is willing to forgo the orientability requirement, then there are examples in dimension three. In this paper, Row constructs infinitely many topologically distinct, irreducible (and hence prime), closed 3-manifolds with the property that none of their orientable covering spaces are prime.
There are examples where $N$ is prime and $M$ is not, such as the double covering $\pi : S^1\times S^2 \to \mathbb{RP}^3\#\mathbb{RP}^3$.
There are examples analogous to Row's in dimensions $n>2$ which are orientable when $n$ is even. I'll give a bit of motivation for the example at the end.
Consider the action of the group $G= \mathbb{Z}^n\rtimes \{\pm I\}=\{ x \mapsto \pm x+ m, m\in \mathbb{Z}^n\}$ on $\mathbb{R}^n$. The subgroup $G_{m/2}=\{x,-x+m\}, m\in \mathbb{Z}^n $ is the stabilizer of $m/2\in \frac12\mathbb{Z}^n$. Remove open balls of radius $r<1/4$ about the lattice points $\frac12\mathbb{Z}^n$ to get the simply-connected manifold $V= \mathbb{R}^n -\mathcal{N}_r(\frac12\mathbb{Z}^n)$. When $n$ is even, $V$ admits an orientation which is $G$-invariant. This induces an orientation on $\partial V$. Since $G_{m/2}$ acts as the antipodal map on the sphere of radius $r$ about $m/2$, the quotient $W'=V/G$ will be a manifold with $2^n$ boundary components (corresponding to $\frac12\mathbb{Z}^n/\mathbb{Z}^n \cong (\mathbb{Z}/2\mathbb{Z})^n$) each of which is homeomorphic to $\mathbb{RP}^{n-1}$. The fundamental group of each boundary component will correspond to some $G_{m/2}$ up to conjugacy. $W'$ has a 2-fold cover $V/\mathbb{Z}^n$ which is homeomorphic to $T^n$ punctured at $2^n$ balls.
Take the $2^n$ boundary components of $W'$, and glue them together in pairs, so that when $n$ is even, the induced orientations get reversed, to get a manifold $W$. For concreteness, let's say that we identify the boundary components corresponding to $m/2+\mathbb{Z}^n$ and $m/2+\frac12^n +\mathbb{Z}^n$, inducing a homomorphism $\alpha_m:G_{m/2}\to G_{m/2+\frac12^n}$. In even dimensions, $W$ will be orientable. Since $\pi_1W'= G$, and the subgroup of a boundary component corresponding to the coset $m/2+\mathbb{Z}^n$ will be conjugate to $G_{m/2}$, we see that $\pi_1 W = G \ast_{m\in 0\times\{0,1\}^{n-1}} \alpha_m$ is a multiple HNN extension by the isomorphisms pairing the subgroups. Each HNN extension will introduce a new group element $t_m$ along with a relation of the form $t_mG_{m/2}t_m^{-1}=G_{m/2+\frac12^n}$. So we can give a relative presentation for the fundamental group as
$$\pi_1 W \cong \langle G, t_m | t_mG_{m/2}t_m^{-1}=G_{m/2+\frac12^n}, m\in 0\times\{0,1\}^{n-1}\rangle.$$ Note that there is some choice here of subgroup representative up to conjugacy which does not affect the overall group isomorphism type.
The claim is that $\pi_1 W$ does not split as a free product. This follows from the Kurosh subgroup theorem, and will be proved below.
Now suppose that $W$ is a non-trivial connect sum $W= W_1 \# W_2$. Then $\pi_1(W)=\pi_1(W_1)\ast \pi_1(W_2)$ by the Seifert-van Kampen theorem. Since $\pi_1(W)$ is not a non-trivial free product, that means that $\pi_1(W_1)=1$ (possibly after reindexing).
We need to show that $W_1'=W_1\backslash D^n$ is homeomorphic to the $n$-ball, and hence $W_1$ is the $n$-sphere. $W_1'$ lifts to the double cover of $W$ coming from the homomorphism $\pi_1(W)\to \mathbb{Z}/2\mathbb{Z}$, which is homeomorphic to $T^n \#( S^{n-1}\times S^1)^{\# 2^{n-1}}$ (a non-prime manifold). In turn, $W_1'$ lifts to the universal cover of this manifold which is a submanifold of $\mathbb{R}^n$ (because it is an infinite connect sum of $\mathbb{R}^n$s). Hence $W_1'$ is an $n$-ball by the Schoenflies Theorem, and we see that $W$ is irreducible and orientable when $n>2$ is even.
Now let's see why $\pi_1 W$ is freely indecomposable. Suppose that $\pi_1 W=A\ast B$. Since $ G < \pi_1 W$ is freely indecomposable, by the Kurosh subgroup theorem $G$ is conjugate to a subgroup of $A$ or $B$, let's say $A$. Moreover, the group $H=\pi_1 W/ \ll \mathbb{Z}^n \gg$ obtained by killing $\mathbb{Z}^n$ will be isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}^{\ast 2^{n-1}}$ since $G/\mathbb{Z}^n\cong \mathbb{Z}/2\mathbb{Z}$. Thus we see that the image $\overline{A}$ of $A$ in $H$ will contain $\mathbb{Z}/2\mathbb{Z}$, and hence will be non-trivial. Moreover, the quotient $H$ will split as a free product $\overline{A}\ast B$. However, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}^{\ast 2^{n-1}}$ is not a free product, since it has a non-trivial center, a contradiction.
Motivation
If a finitely generated group $G$ splits as a free product, then any Cayley graph for $G$ (associated to a finite generating set) has more than one end. If $G< G'$ is finite index, then the Cayley graphs of $G$ and $G'$ are almost equivalent (quasi-isometric), in fact a Cayley graph for $G$ can be obtained from one for $G'$ by collapsing some finite trees equivariantly (this is basically the Reidemeister-Schreier method).
Hence if $G$ has more than one end, so does $G'$.
Now a theorem of Stallings implies that if a group $G'$ has more than one end, then $G'$ is a graph of groups with finite edge groups. Thus, in this example, we found a manifold whose fundamental group is an HNN extension over $\mathbb{Z}/2\mathbb{Z}$ subgroups, but itself is not a free product. But it has an index 2 subgroup that does split as a free product.
If an $n$-manifold $M$ is a connect sum, and $\pi_k(M)=0$ for $k < n-1$, then a similar argument shows that $\pi_1(M)=A\ast B$ is a non-trivial free product. So any manifold finitely covered by such an $M$ will have fundamental group splitting over a finite group. One can probably find many more examples with such properties. I don't know how to find an example which is a connect sum with simply-connected summands, but finitely covers a prime manifold.
Thanks for your answer, it will take me some time to fully digest it. First of all, is $\pi_1(\mathcal{O})$ the orbifold fundamental group or the regular fundamental group? Second, I am not that familiar with HNN extensions, so I apologise if this is trivial, but what is the normal subgroup $\mathbb{Z}/2\mathbb{Z}$? Does the subgroup $\mathbb{Z}/2\mathbb{Z}$ in $\pi_1(\mathcal{O})$ become normal in the iterated HNN extension?
@MichaelAlbanese: Your first question is correct about the orbifold fundamental group - I got rid of the intermediate orbifold which is a useful trick to come up with such examples, but does not play a role in the end. Also, the subgroup is not normal: the group is a semidirect product, but there are many lifts of the {±} group. So I've modified the argument to first kill $\mathbb{Z}^n$ before obtaining a similar contradiction. Let me know if you need further clarification or see other mistakes.
Thanks a lot for taking the time to expand on your answer, I appreciate it. It's not clear to me why the universal cover of $T^n#(S^{n−1}\times S^1)^{# 2^{n−1}}$ is a submanifold of $\mathbb{R}^n$. Even the description as an infinite connected sum of $\mathbb{R}^n$'s is mysterious to me (in part because the universal cover of $S^{n-1}\times S^1$ is $\mathbb{R}^n\setminus{0}$).
@MichaelAlbanese $R^n-0 = R^n# R^n$. So this holds for $S^{n-1}\times S^1$. Actually, any compact submanifold in the universal cover will live inside a connect sum of finitely many $R^n$s which is just $R^n$ punctured at some points.
Please let me know if my interpretation of your use of the Schoenflies Theorem is correct. As $W_1$ is simply connected, the manifold $W_1\setminus D^n$ is a submanifold of the universal cover (here $D^n$ denotes the open $n$-ball). As you mentioned above, it follows that $W_1\setminus D^n$ embeds in $\mathbb{R}^n$, and hence there is an embedding $S^{n-1} = \partial(W_1\setminus D^n) \to \mathbb{R}^n$. By the Schoenflies Theorem, $S^{n-1}$ bounds a ball, so $W_1\setminus D^n$ is a closed ball and $W_1 = S^n$. Is that what you had in mind?
@MichaelAlbanese: Yes, thanks for the correction, I was confusing $W_1$ and $W_1\backslash D^n$. I'll fix that.
Thanks for all your help explaining the details to me, I really appreciate it.
In the final paragraph, what do you mean by "splitting over a finite group"? Finding an example where the connected sum is simply connected sounds like an interesting problem, although it is probably much more difficult.
Splitting over a finite group means as an amalgamated free product or HNN extension over a finite subgroup. See the discussion here: https://en.wikipedia.org/wiki/Stallings_theorem_about_ends_of_groups#Cuts_and_splittings_over_finite_groups
|
2025-03-21T14:48:31.397083
| 2020-07-01T20:32:32 |
364611
|
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|
Stack Exchange
|
Frechet-Urysohn quotient of second countable locally compact Hausdorff space
In this paper from 2010 https://cmuc.karlin.mff.cuni.cz/pdf/cmuc1001/arhangav.pdf Arhangelskii asks if there is a quotient of a second countable locally compact Hausdorff space which is Frechet-Urysohn and Tychonoff but not first countable at any point (Problems 5.14 and 5.15).
Is this problem still open?
|
2025-03-21T14:48:31.397147
| 2020-07-01T20:49:03 |
364613
|
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|
Stack Exchange
|
A generalization of strong primes
In this post we denote the sequence of prime numbers as $p_k$ for integers $k\geq 1$. I don't know if the following definition is in the literature.
Definition. We define the $\theta$-strong primes, or strong primes at level $\theta$, as the sequence of those prime numbers $p_n$ that satisfy the inequality $$p_n>\theta\, p_{n-1}+(1-\theta)p_{n+1}\tag{1}$$
for some fixed real number $0<\theta<1$.
Remark. My definition arises as a generalization of the known as strong primes. See the definition in number theory of strong primes from the Wikipedia Strong prime.
Question. I would like to know, if my question is interesting, if it is possible to prove that there exists some $\hat\theta$ for which the corresponding sequence of prime numbers has finitely/infinitely many terms. Many thanks.
Here I emphasize that $0<\theta<1$.
I have no intuition if my question is interesting, I don't know what can be a more interesting question at research level about the study of these sequences $(1)$ and their corresponding $\theta$. I did some calculations for $\theta\neq\frac{1}{2}$ a rational number.
Please if the question does not have the best mathematical content feel free to comment it. After I wrote the definition I thought it might be interesting to ask about this definition. I don't know what can be a more interesting question at research level about the study of these sequences related to their corresponding $\theta$.
since we know there are infinitely many consecutive primes with finite gap $C$ (I think smallest $C$ is $246$ now but not sure), it is trivial that if $2\theta > (1-\theta)C$ the $p_n$'s for which $p_{n+1}-p_n \le C$ work since $p_n-p_{n-1} \ge 2$ so $\theta(p_n-p_{n-1}) >C(1-\theta) \ge (1-\theta)(p_{n+1}-p_n)$; for $246$ we get $1 > \theta > 123/124$ satisfies the requirement with infinite $p_n$'s
Many thanks for your excellent contribution, feel free to made your comment as an answer @Conrad
Regarding any $\hat\theta$ for which the prime numbers sequence has finitely/infinitely many terms, consider one which has only finitely many terms. There would then exist a prime index $m$ for which all $n \gt m$ gives
$$p_n \le \hat\theta\, p_{n-1} + (1 - \hat\theta)p_{n+1} \tag{1}$$
Using the standard definition of prime gaps of
$$g_n = p_{n+1} - p_{n} \tag{2}\label{eq2A}$$
we then have
$$\begin{equation}\begin{aligned}
p_n & \le \hat\theta(p_{n} - g_{n-1}) + (1 - \hat\theta)(p_{n} + g_{n}) \\
p_n & \le \hat\theta p_{n} - \hat\theta g_{n-1} + (1 - \hat\theta)p_{n} + (1 - \hat\theta)g_{n} \\
\hat\theta g_{n-1} & \le (1 - \hat\theta)g_{n} \\
g_n & \ge \left(\frac{\hat\theta}{1 - \hat\theta}\right)g_{n-1}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
If $\hat\theta \gt 0.5$, then $\frac{\hat\theta}{1 - \hat\theta} \gt 1$, so \eqref{eq3A} shows the prime gaps are strictly increasing for all $n \gt m$. However, this contradicts that there are infinitely many prime gaps of at most $246$ (e.g., see Bounded gaps between primes). If $\hat\theta = 0.5$ instead, then $\frac{\hat\theta}{1 - \hat\theta} = 1$, so the prime gaps are non-decreasing, but this is also not possible since prime gaps can become arbitrarily large and, thus, must decrease later to become the smaller prime gaps, e.g., those at most $246$.
This means the original assumption of there being only a finite number of primes in the sequence must be incorrect, i.e., there are infinitely many $\theta$-strong primes for any $\hat\theta \ge 0.5$.
Update: The PDF version of the arXiv article On the ratio of consecutive gaps between primes, near the bottom of page $8$, states about Erdős
He mentioned 60 years ago [Erd5]: "One would of course conjecture that
$$\underset{n \to \infty}{\lim \inf}\frac{d_{n+1}}{d_n} = 0 \; \text{ and } \; \underset{n \to \infty}{\lim \sup}\frac{d_{n+1}}{d_n} = \infty \tag{2.2}$$
but these conjectures seem very difficult to prove." Based on a generalization
of the method of Zhang [Zha] the author proved (2.2) in [Pin2]
where [Pin2] is
J. Pintz, Polignac numbers, conjectures of Erdős on gaps between primes and the bounded gap conjecture. arXiv: 1305.6289 [math.NT] 27 May 2013.
i.e., here.
Note their $d_n$ is the same as $g_n$ in \eqref{eq2A}. If the first part of (2.2) is true, then no matter how close $\frac{\hat\theta}{1 - \hat\theta}$ is to $0$, \eqref{eq3A} cannot always being true for all $n \gt m$ for any integer $m$. This means there are infinitely many $\theta$-strong primes for all $0 \lt \hat\theta \lt 1$.
I haven't read the article or references to try to verify the validity of the claim. However, note the article seems to be basically very similar to at least part of the Springer Link book On the Ratio of Consecutive Gaps Between Primes, but I haven't paid to get a chapter or the entire e-book, or a physical copy of the book, to check on this.
Many thanks for your excellent answer, in a similar way than previous answer I'm understanding all details. Many thanks, are very good answers.
@user142929 You're welcome. Note I've updated my answer with a reference to an article which contains a result that shows all $\theta$-strong prime sequences have an infinite # of primes.
As requested I put my comment as a (partial) answer:
Since we know there are infinitely many consecutive primes with finite gap $C$ (smallest proved $C$ as of now seems to be $246$), any $\theta$ for which $2\theta >(1-\theta)C$ works for the $p_n$'s precisely the primes for which $p_{n+1}-p_n \le C, n \ge 2$ since then $p_n-p_{n-1} \ge 2$ so:
$\theta(p_n-p_{n-1}) \ge 2\theta > C(1-\theta) \ge (1-\theta)(p_{n+1}-p_n)$
For $C=246$ we get that $123/124 < \theta <1$ satisfies the requirement of the OP with infinitely many $p_n$'s
Many thanks for your excellent answer. I'm going to study the details.
|
2025-03-21T14:48:31.397511
| 2020-07-01T21:27:54 |
364614
|
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|
Stack Exchange
|
Chromatic number of square of a tree
What is an upper bound on the chromatic number of the square of a tree on $n$ vertices? Note that the power of the graph is considered in this sense.
If the tree were a path, then it is easy to see that the chromatic number is $3$ if the order is a multiple of $3$. This is because a path of order a multiple of $3$ has a triangle, therefore should require at least $3$ colors. Next, the square of a cycle on $n$ vertices where $n$ is divisible by $3$, has chromatic number $3$. In other cases, I think it equals $\Delta+1$, where $\Delta$ be the maximum degree of the tree. This is because, each star of order $\Delta$ in the tree induces a clique of order $\Delta+1$ in the square graph. But, can it be more than $\Delta+1$. Specifically, the maximum degree of the square graph is $2\Delta$ where $\Delta $ be the maximum degree of the tree. Any hints? Thanks beforehand.
Have you tried induction? It seems just removing a leaf should give you a matching upper bound..
@JoshuaErde you mean the upper bound of $\Delta+1$ is right?
The particular case of the square of a tree is easy to handle by producing a greedy $(\Delta+1)$-coloring starting from a root vertex and extending. However, much stronger results are known:
The $k$-th power of a tree was shown to be chordal in
Y.-L. Lin, S. Skiena, "Algorithms for Square Roots of Graphs", SIAM Journal of Discrete Mathematics, 8(1), 99-118, 1995
and even strongly chordal in
D. G. Corneil, P. E. Kearney, "Tree Powers", Journal of Algorithms,
29,111-131, 1998
Chordal graphs are perfect, so their chormatic number is the same as the size of the largest clique. The largest clique in the k-th power of a tree $T$ is the largest $k$-ball in $T$, where a $k$-ball centered at a vertex $v$ is the set of all vertices of $T$ which are at a distance $\le k$ from $v$.
great! Is there any such result for bipartite graphs?
The girth of the $k$-th power of a large cycle is large, and it particular it contains a large chordless cycle (so it is not chordal).
Moreover, the square of a 10-cycle contains chordless 5-cycles and hence is not perfect.
|
2025-03-21T14:48:31.397819
| 2020-07-01T22:18:45 |
364616
|
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|
Stack Exchange
|
Measure theory on abstract Boolean ring
Since a σ-algebra in measure theory is indeed an algebra over $\mathbb{Z}_2$ with addition given by symmetric difference and multiplication given by intersection, does it mean we can put measure on any $\mathbb{Z}_2$ algebra (aka Boolean ring)? In particular, it would mean that we can define integration on any idempotent ring with all results from measure theory holding. Am I missing anything, and has this been explored?
What would you be trying to integrate? One integrates functions (which are not elements of the $\sigma$-algebra), not subsets. You could maybe build some formal theory (consider the completion of the group algebra of your ring in some appropriate topology …), but the Stone representation theorem says that any algebra is an algebra of subsets anyway, so it seems like you'd be quite close to doing measure theory while carefully avoiding naming the set.
Do you want your measure theory to be $\sigma$-additive?
You wrote "$\mathbf{Z}_2$-algebra (aka Boolean ring)" but obviously not every $\mathbf{Z}_2$-algebra is a Boolean ring.
@LSpice You can identify a measurable function with the function that maps Borel sets to their preimages. This gives you a ring homomorphism from the Borel sets to the ring of measurable sets and you can build a theory of integration for such homomorphisms.
@LSpice, yes, I did not think about what I am trying to integrate. Indeed, we need a function not on sets but on actual elements. YCor, I got confused in definitions. A sigma-algebra is a Z_2-algebra, which is also a Boolean ring, i.e. 2x = 0 and x^2 = x for every element x. Also, the point about countable additivity is good.
You want to look at Fred Lintons thesis and his paper "Functorial measure theory" in
the Irvine Proceedings, Thompson , Washington D.C., !966.
See John M. H. Olmsted, Lebesgue theory on a Boolean algebra, Transactions of the American Mathematical Society 51 #1 (January 1942), pp.164-193 AND
Roman Sikorski, The integral in a Boolean algebra, Colloquium Mathematicum 2 #1 (1949), pp. 20-26.
According to Proposition 416Q(b) in Fremlin's Measure Theory,
finitely additive functionals A→[0,∞) are in a canonical bijective correspondence with finite Radon measures on the Stone space Spec(A) of A,
which is a compact Hausdorff totally disconnected topological space.
This means that we can integrate any continuous function Spec(A)→R with respect to a finitely additive functional on A.
This type of integrand is very general, and, in particular, includes all continuous homomorphisms from the Boolean algebra of Borel subsets of R to the Boolean algebra A,
as alluded to in the comments.
Fremlin's book can be found here: https://www1.essex.ac.uk/maths/people/fremlin/mt.htm
|
2025-03-21T14:48:31.398049
| 2020-07-01T23:09:12 |
364618
|
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|
Stack Exchange
|
Full measure properties for Zariski open subsets in $p$-adic situation
Let $F$ be a $p$-adic field and let $X$ be a smooth integral variety over $F$ (I am chiefly interested in the case when $X$ is a connected reductive group over $F$). Let $U$ be a non-empty open subset of $X$ with complement $Z$.
We can endow $X(F)$ with the Serre–Oesterlé measure (e.g., as in [1, Section 2.2] or [2, Section 7.4])—this is just the standard measure coming from a top form of $X$).
My question is then whether one knows a simple proof/reference for the following:
Claim: The subset $Z(F)$ of $X(F)$ has measure zero.
I think this is proven in [1, Lemma 2.14]—but this is concerned with a more specific context which makes it non-ideal as a reference.
Any help is appreciated!
[1] Magni - $p$-adic integration and birational Calabi–Yau varieties.
[2] Igusa, J.I., 2007. An introduction to the theory of local zeta functions (Vol. 14). American Mathematical Soc.
The simplest way I know of is to use charts to reduce to the case when $X$ is an affine space, then use induction. But perhaps there is a more canonical proof.
@WSL I guess this is essentially the proof in the linked article [1] but, again, it's obfuscated by the particular context it's discussing the result in. Do you know a reference for this proof?
A terrible way to argue in the specific case when $X$ is a group is to take $d$ to be the minimal dimension of a Zariski-closed subset with non-zero measure and for an irreducible $Z$ of such dimension consider the union of $>\mu(X(F))/\mu(Z(F))$ pairwise distinct translates of $Z$. By the definition of $d$ all intersections between different translates have measure zero, so this union ought to have measure $>\mu(X(F))$ by the inclusion-exclusion formula, which is absurd.
@SashaP, why can you assume $\mu(X(F))<\infty$? Alex Youcis, i don’t know a reference where this is spelled out but I’ll try to write it up unless someone beats me to it.
@WSL Yes, I wasn't completely careful, to account for that issue let's instead run the above argument with $g^{-1}(Z(F)\cap gU)$ and $U$ in place of $Z(F)$ and $X(F)$ where $U$ is a compact open subgroup of $X(F)$ and $g$ is an arbitrary element. The result for $Z(F)$ will follow as this set is the union of $g^{-1}(Z(F)\cap gU)$ as $g$ runs over coset representatives of $U$ in $X(F)$.
If $X$ is a $\sigma$-compact $p$-adic analytic manifold, there is a well-defined notion of measure zero subsets (as the measure inherited from charts is unique up to equivalence), and if $Y$ is the zero set of an analytic map (= locally sum of its Taylor expansion) and $Y$ has empty interior, I guess it should have measure zero.
@SashaP Thanks for your suggestion. For my own sake I'd prefer to have a proof that works for general $X$, just because I do occasionally also use it in that case (even though, as I said, my main interest is groups). Still, if you want to write this as an answer, it might be useful to some readers and I'll accept it if someone else doesn't provide the full answer.
@WSL Yeah, if you'd be willing to write a fairly precise answer that would be much appreciated. Of course, I'd be very happy to save you the hassle if you just knew a reference. Thanks!
I remember working on this for my PhD thesis. I regret to say that I managed to write the proof up in such a way that the main technical difficulty was hidden in a plausible-looking sentence. But the main idea I had was using the implicit function theorem for smooth morphisms (such as $X \rightarrow \operatorname{Spec}(k)$ where $k$ is a local field): for smooth varieties $X$, for every point $x \in X(k)$ there are functions $g_1,\ldots,g_d \in k(X)$ with $d = \dim X$ such that the map $X(k) \rightarrow k^d$ given by $x \mapsto (g_1(x),\ldots,g_d(x))$ is a local homeomorphism.
What I call the implicit function theorem here can be seen as the conjunction of a structure theorem for smooth morphisms (such as described here: https://stacks.math.columbia.edu/tag/039P) with the fact that etale morphisms are local homeomorphisms when evaluated on points over local fields. Perhaps this idea allows you to reduce to the affine case, as @WSL suggested, because under the identification mentioned the subvariety $Z$ is identified with an algebraic subvariety of $k^d$.
@RP_ Thanks for your input. This is essentially the idea I think in the above linked article--using the definition of smoothness you mentioned is how you get 'charts' on the $p$-adic manifold $X(F)$, but I just don't want to include this in the article I'm writing so just want a reference and don't want to have to reproduce a proof. Thanks again!
I'm really sorry to ask a foolish question. Why doesn't your $Z$ directly satisfy the conditions of [1, Lemma 2.14], so that you can simply cite it directly?
@LSpice Hey Loren. It's because he's starting with a smooth model over $\mathcal{O}_K$ and that $Z$ also has a smooth model over $\mathcal{O}_K$. If you look at his proof I don't really think this gets used, and it does show my desired claim. But, because he technically starts with those assumptions it's a pretty non-canonical reference for a beginner. I was hopting there was just some fairly basic reference to this result (e.g. it's sketched in a paper of Harish-Chandra) or that someone could give a comprehensive proof here that I could reference/copy. This is for a set of notes I'm writing.
@AlexYoucis, thanks! Have you looked at any of the work of Denef and Loeser? I went hunting and couldn't find it, but it seems like they must need exactly this sort of result for their theory of motivic integration.
@LSpice Yeah, I spent hours trawling though Denef, Loeser, Clucker, etc. I know it's buried in there morally but I can't find an exact statement. I need it in a more general context, but certainly Harish-Chandra et. al need it for the full-measure of G^reg(F), so I imagine they must also write it down somehwere. Thanks again!
|
2025-03-21T14:48:31.398474
| 2020-07-02T02:33:46 |
364624
|
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|
Stack Exchange
|
Behavior of invariants under reduction mod p
Let $R$ be a finitely generated $\mathbb{Z}$-algebra with an [edit: linear algebraic] action of $G(\mathbb{Z})$ where $G$ is a split simply-connected semisimple group.
Then for any prime $p$ we have a map $R^{G(\mathbb{Z})} \otimes \mathbb{F}_p \rightarrow (R \otimes \mathbb{F}_p)^{G(\mathbb{F}_p)}$. Is this map necessarily surjective for sufficiently large $p$?
Comments:
(1) The simply-connectedness assumption may seem weird; it is made to ensure that $G(\mathbb{Z}) \rightarrow G(\mathbb{F}_p)$ is surjective so that there is a map at all.
(2) If $G$ is a finite group, then the answer is yes by an averaging argument.
(3) If $G$ is unipotent, then the answer is no. For example, take $x \mapsto x+1$ acting on $k[x]$; there are many invariants in positive characteristic (Artin-Schreier covers!).
Do you mean just an abstract action of $G(\mathbb Z)$, or an algebraic action of $G$?
Also, how does the averaging in (2) work for orbits of size divisible by $p$?
Algebraic action. Regarding (2), I'm still referring to the statement for "sufficiently large $p$".
Compare https://mathoverflow.net/questions/350590/lifting-g-invariants-from-characteristic-p-gg-0-to-characteristic-0-for-a-re/350791#350791.
If a reductive group is split over $\mathbb{F}_p$ and acts as such on a module, then a root subgroup fixes a weight vector if and only if the $\mathbb{F}_p$ points of the root subgroup fix the vector.
Thanks very much for the link!
No.
Let $G=SL_n$, acting on its defining representation $V$, with $n\geq2$.
Let $R=\mathbb{Z}[X_1,\dots,X_n]$ be the obvious $\mathbb{Z}$-form of the ring
of polynomial functions on $V$. Let $p$ be a prime. For any $f\in R/pR$ the
product over all $g\in SL_n(\mathbb{F}_p)$ of $f\circ g$ is invariant under
$SL_n(\mathbb{F}_p)$. But there are no nontrivial $G(\mathbb{Z})$-invariants in $R$
because $G(\mathbb{Z})$ has a Zariski dense orbit in $V$.
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2025-03-21T14:48:31.398634
| 2020-07-02T02:46:43 |
364627
|
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|
Stack Exchange
|
Computation on homotopy colimit cocomplete triangulated categories
I have a couple of questions about dealing with homotopy (co)limits cocomplete triangulated categories.
Question I:The first one concerns a comment by Peter Arndt in this discussion about derived categories regarding posibibility to calculate the homotopy colimit when working with nice enough category. Peter wrote:
I also find this a very enlightening view point, but just for the record: Ho(co)lims in cocomplete triangulated categories are MUCH easier to compute by completing the right map to an exact triangle than by going via a simplicial (or any other) enrichment...
Where I can look up the theoretical background explaining that applying successively these steps we indeed obtain an object homotopic to homotopical (co)limit. In other words why this cooking recipe work?
Question 2: searching for an answer for my first question I found in this paper on Homotopy limits in triangulated categories by Bökstedt & Neeman an approach by so called 'Totalization of a complex'.
The steps in the construction look quite similar to the step Peter described and the constructed object is also described as homotopical colimit.
Question: How close the construction in the paper is to that one in first question. The principal aspect that confuses me is that the construction in the paper (as well the paper) not explicitly working with simplicial enrichments of homs.
Is using simplicial enrichment a more 'modern' approach to obtain the same object? And how would it flow into the construction?
Where I can look up the theoretical background explaining that applying successively these steps we indeed obtain an object homotopic to homotopical (co)limit. In other words why this cooking recipe work?
The recipe under discussion computes the homotopy colimit
of a sequence $X_0→X_1→X_2→⋯$ as the homotopy cofiber
of the shift map $⨁_{i≥0}X_i→⨁_{i≥0}$.
The shift map is the difference of the identity map and the map induced by transition maps to the next degree.
The homotopy cofiber of this difference can be computed
as the homotopy coequalizer of the two maps under consideration.
The latter homotopy coequalizer of two maps between direct sums (i.e., homotopy coproducts) can be rewritten as the homotopy colimit of a single diagram
indexed by a category $I$.
The latter category $I$ has a canonical functor $I→\{0→1→2→⋯\}$,
which is a homotopy final functor (the comma categories can
be easily checked to be contractible), hence the induced map
on homotopy colimits is a weak equivalence.
The criterion for homotopy finality can be found,
e.g., in Lurie's Higher Topos Theory (Proposition <IP_ADDRESS>), in Cisinski's book, and in many other places.
How close the construction in the paper is to that one in first question. The principal aspect that confuses me is that the construction in the paper (as well the paper) not explicitly working with simplicial enrichments of homs.
Is using simplicial enrichment a more 'modern' approach to obtain the same object? And how would it flow into the construction?
We don't see enrichments because the diagrams involved
are extremely special: they are sequences $X_0→X_1→X_2→⋯$
in which there are no nontrivial (homotopy) commutativity (or coherence) relations.
In this (very special) case one can show that a sequential
diagram $X_0→X_1→X_2→⋯$ in a triangulated category is the same
data as a weak equivalence class of sequential diagrams
in a stable model category that models the triangulated category.
This is part of the reason why one can compute the homotopy
colimit inside the triangulated category in this (very special) case.
Any time there is a nontrivial commutativity (coherence) involved
(e.g., when computing the homotopy colimit of a simplicial diagram),
the whole machinery of triangulated categories breaks down.
Indeed, it is not even possible to say what a (homotopy coherent) simplicial object
in a triangulated category is, since the necessary data of coherences
is simply not present in a strict functor from Δ^op to the triangulated category,
and the relevant information is altogether missing from the triangulated
category.
This is part of the reason why constructing enhancements of triangulated categories is often necessary.
But then again, one might as well work with the original stable model (or relative) category.
For more information, see the homotopy theory FAQ.
Thank you a lot for the answer. One remark on your last paragraph: Do you mean that the message is that if proper coherence (=non triv commut) occures, the data itself not contains not enough intrinsical information to allow to form a homotopy colimit in the way Peter described it? ie that to compute the homotopy colimit as Peter described without enrichment the described object might be not exist/not well defined? That is to make it possible to do it nevertheless, there is a machinery, the simplicial enrichment https://ncatlab.org/nlab/show/simplicial+object#simplicial_enrichment that
"artificially" endows the involved object with "enough" simplicial structure that then allows to build the homotopy colimit in the described way. Is this the point or did I misunderstood the issue?
In simpler analogy: If I want to associate to a certain mathematical object $M$ a group theoretical invariant, then firstly it's meaningful to develop machinery to associate a group $G(M)$ to $M$ in well defined way and then consider the invariants of $G(M)$. Is this the message behind the simplicial enrichment above?
@user7391733: Yes, the data itself is insufficient. You cannot define homotopy coherent commutative squares in a triangulated category, for example, because such a square is a quadruple of morphisms f,g,h,k together with a homotopy fg→hk, and there is no way to say what this homotopy is. The point of various enhancements and enrichments is that you move away from bare triangulated categories and add more data to obtain homotopy coherent commutative diagrams.
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2025-03-21T14:48:31.399031
| 2020-07-02T04:54:39 |
364633
|
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|
Stack Exchange
|
A question regarding an analog of Young symmetrizer: the product row and column preserving subgroups without sign representation
Consider a rectangular Young diagram $\lambda$ with $n = pq$ boxes, with $p$ rows and $q$ columns. If $C$ is the column preserving subgroup of $\lambda$ and $R$ is the row preserving subgroup, then we normaly construct the Young symmetrizer by defining
$$b = \sum_{g \in C} \text{sign}(g) g,$$ and, $$a = \sum_{g \in R} g,$$ and the Young symmetrizer is $$c = ba.$$
I am studying a slightly different problem, where one defines $b$ without $\text{sign}(g)$:
$$b' = \sum_{g \in C} g,$$ and then the analog of the Young symmetrizer is $c' =b' a$, the product of row and column preserving subgroups.
It is easy to see that unlike $c$, $c'$ is not an idempotent, i.e., $c' c' \neq \alpha c'$ for any $\alpha$. Nevertheless, I would like to compute $\frac{1}{n!}\text{tr} (c' c')$ in the regular representation, i.e. the number of $(r_1,c_1,r_2,c_2)$ with $r_1,r_2 \in R$ and $c_1,c_2 \in C$ where $r_1 c_1 r_2 c_2 = id$.
Indeed, one simple lower bound for $\frac{1}{n!}\text{tr} (c' c')$ is $\frac{1}{n!}\text{tr} (c c) = \text{hook length}(\lambda) \times \frac{1}{n!}\text{tr}(c) =\text{hook length}(\lambda).$
Is there anything known in the literature about the unsigned row and column product $c'$? or the particular trace that I wish to compute?
|
2025-03-21T14:48:31.399159
| 2020-07-02T05:28:57 |
364635
|
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|
Stack Exchange
|
Integer partitions into restricted parts
Given a linear diophantine equation $$x_1+\dots+x_n=m\leq nn'$$ how many solutions does it have with each $x_i\in[0,n']\cap\mathbb Z$? Looking for asymptotics that parametrizes well with both $n$ and $n'$ over different ranges for both situations
$x_1\leq\dots\leq x_n$ and
unordered.
Is $m$ a parameter of the problem, or can it take any value within the constraints?
Supposing they satisfy the conditions, are (2,1,1) and (1,2,1) both counted as solutions?
If so, this is a question about integer compositions rather than partitions.
@BrendanMcKay $m$ is any value within the constraints.
@BrianHopkins Yes it would be nice to know for both scenarios. When $x_i\leq x_{i+1}$ is forced (partitions) and not forced (compositions).
Since $m$ is a dummy variable (i.e. a bound variable) and $n,n'$ are "real" variables (i.e. they are free) perhaps we should rewrite the problem accordingly as
$``$compute the following
$$ f(y,z) = \#\left\lbrace (x_1,... , x_y )\mid x_1 + ... + x_y = m,\ x_i \in \mathbb{N},\ m \leq yz ,\ i < j \implies x_i \leq x_j \right\rbrace."$$
We can let $f'$ be the number if we omit the condition $i < j \implies x_i \leq x_j$ (i.e. $f$ corresponds to partitions and $f'$ corresponds to compositions). Intuition tells us that $f$ will likely have a closed form solution and $f'$ will probably only have solution expressible by generating functions (which are perfect for asymptotics). Let us confirm our intuitions:
Compositions (i.e. $f'$)
It is straightforward to see that if we let
$$g(y,m) = \#\left\lbrace (x_1,..., x_y ) \mid x_1 + ... + x_y = m, \ x_i \in \mathbb{N}\right\rbrace,$$
then we have the identity
$$f'(y,z) = \sum_{m=0}^{yz}g(y,m).$$
It is well known, see the famous stars and bars method in feller's introduction to probability theory, that
$$g(y,m)= \binom{y+m-1}{m},$$
so that $$f'(y,z) = \sum_{m=0}^{yz}\binom{y+m-1}{m}=
\binom{yz+y}{yz},$$
(or, using your notation, $f'(n,n')=\binom{nn'+n}{nn'}$)
where the last identity can be derived as a consequence of Chu Shih-Chieh's identity, see example 2.5.1 of Chuan-Chong & Khee-Meng's text on combinatorics. I also strongly recommend taking a peek at Flajolet & Sedgewick's text on analytic combinatorics for asymptotics and the more abstract symbolic method/species style which is necessary for the more difficult analysis of $f$.
Partitions (i.e. $f$)
It is straightforward to see that if we let
$$g(y,m) = \#\left\lbrace (x_1,... , x_y ) \mid x_1 + ... + x_y = m, \ x_i \in \mathbb{N}, \ i < j \implies x_i \leq x_j\right\rbrace,$$
then we have the identity
$$f(y,z) = \sum_{m=0}^{yz}g(y,m).$$
It is well known that if we define the generating function $\mathcal{G}_y$ as
$$\mathcal{G}_y(x) = \sum_{m \in \mathbb{N}} g(y,m) x^m,$$
then we have that
$$\mathcal{G}_y(x) = \prod_{k=1}^{y}\frac{1}{1-x^k},$$
see Flajolet & Sedgewick's text on analytic combinatorics or Andrews elementary text on partitions. One way to see this is to notice the following famous theorem attributed to Euler
The number of partitions of a number $n$ into at most $l$ parts is equal to the number of partitions of a number $n$ into parts all bounded by $l$
and the result follows by elementary generating function magic. Finally, theorem 5.1.1 of Chuan-Chong & Khee-Meng's text on combinatorics states that
$$\mathcal{A}(x) = \sum_{n \in \mathbb{N}} a_n x^n \implies \frac{1}{1-x}\mathcal{A}(x) = \sum_{n \in \mathbb{N}} \left(\sum_{k \leq n} a_k \right) x^n;
$$
therefore, if we define the generating function $\mathcal{F}_y$ as
$$\mathcal{F}_{y}(x) = \sum_{n \in \mathbb{N}} f(y,n) x^n,$$
then we have that $$\mathcal{F}_{y}(x) = \frac{1}{1-x}\mathcal{G}_y(x)
= \frac{1}{1-x}\prod_{k=1}^{y}\frac{1}{1-x^k}. $$
More explicitly we have that
$$f(y,z) = [x^{yz}] \mathcal{F}_{y}(x) =[x^{yz}] \left(\frac{1}{1-x}\prod_{k=1}^{y}\frac{1}{1-x^k}\right)$$
where the operator $[x^{k}] $ is defined as follows:
$$\mathcal{A}(x) = a_0+a_1x+ ... +a_nx^n+ ... \implies [x^{k}]\mathcal{A}(x) = a_k.$$ For the asymptotics please consult Flajolet & Sedgewick's text on analytic combinatorics where you will find a wealth of information and techniques for extracting the asymptoics of the coefficients of $\mathcal{F}_{y}(x) $.
Is there a clean asymptotic expression for $f$ just as $f'$ does?
So $f′(n,n′)$ is only exponential in $n$ because $\binom{nn′+n}{nn′}\approx(1+\frac1{n′})^{nn′}\rightarrow e^n$? That is surprising isn't it?
I think it is $e^{n+nn'}$. I think I made a mistake.
I doubt there is a nice expression for $f$ since partitions tend to not have nice expressions in general but there is a lot of literature written about them. The most famous asymptotic formula known for the partition function is $p(n) \sim \frac{1}{4n\sqrt{3}}\exp\left(\pi \sqrt{\frac{2n}{3}}\right) $ which is due to Hardy and Ramanujan, see https://en.wikipedia.org/wiki/Partition_function_(number_theory). Chapter 7 of https://g.co/kgs/pgs3uM also has a lot of material on partitions, mostly algebraic. The definitive text is probably Andrews "the theory of partitions" https://g.co/kgs/HPVXS4
By nice expression I was thinking a first order asymptotic.
Also I am wondering what happens if we constraint the difference between maximum and minimum terms? https://mathoverflow.net/questions/365176/integer-partitions-and-compositions-into-restricted-parts-with-a-difference-cond
I believe this is discussed in I.16 of ch. 1 sec. 3 of analytic combinatorics by Flajolet and Sedgewick which you can download for free, legally, off of one of the authors' websites here https://ac.cs.princeton.edu/home/ (sadly the other author passed away recently). Of course, you have to perform some simple manipulations to make it work; perhaps subtracting the minimum from all of the parts to transform it into the case $\min_i x_i' =0$.
|
2025-03-21T14:48:31.399790
| 2020-07-02T06:52:28 |
364636
|
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|
Stack Exchange
|
Depth of modules and regular sequences of endomorphisms
Let $(R, \mathfrak{m})$ be a Noetherian local ring and $M$ is a finitely generated $R$-module of depth $t$. It is well known that every maximal regular sequence of $M$ has length $t$. Recall that $x_1, \dotsc, x_t \in \mathfrak{m}$ is an $M$-regular sequence if $x_i$ is a non-zero divisor of $M/(x_1, \dotsc, x_{i-1})M$ for all $i = 1, \dotsc, t$, i.e. the multiplicative map $x_i: M/(x_1, \dotsc, x_{i-1})M \to M/(x_1, \dotsc, x_{i-1})M$ is injective.
Now we consider a sequence of endomorphisms instead of multiplications.
Definition. A sequence of endomorphism $\varphi_1, \dotsc, \varphi_t \in \operatorname{End}(M)$ is called a $M$-regular sequence if
(1) For all $i = 1, \dotsc, t$, $\operatorname{Im}(\varphi_i) \subseteq \mathfrak{m}M$.
(2) For all $i =1, \dotsc, t$, $\varphi_i$ induces an injective endomorphism on $M/(\operatorname{Im}(\varphi_1), \ldots, \operatorname{Im}(\varphi_{i-1}))$.
Question 1. Let $(R, \mathfrak{m})$ be a Noetherian local ring and $M$ be a finitely generated $R$-module of depth $t$. Does every maximal $M$-regular sequence of endomorphims of $M$ have length $t$?
Update: Based on Mohan's answer we will assume the our endomorphisms commute. It is natural to ask the following question.
Question 2. Suppose $\varphi_1, \ldots, \varphi_t$ is an $M$-regular sequence of endomorphisms. Is every permutation of $\varphi_1, \ldots, \varphi_t$ an $M$-regular sequence of endomorphisms?
If $\varphi_i\varphi_j=\varphi_j\varphi_i$ for all $i,j$, this seems correct.
Could you provide your proof, thanks.
This is not an answer to your question, but it may be of interest:
https://doi.org/10.1016/S0764-4442(00)87486-8
Let me give one proof of what I said in the comment. Proof is by induction on the depth. Endomorphism of a module, to avoid repetition, will mean an injective map with image
contained in maximal ideal times the module.
First we deal with depth zero. Then I claim there are no such endomorphisms. If $\phi:M\to M$ is any endomorphism, and $N\subset M$ is the maximal finite length submodule with $M/N$ having positive depth, it is immediate that $\phi(N)\subset N$. But $\phi$ is injective implies, by length considerations, $\phi(N)=N$. So, we get $\phi^r(N)=N$ for all $r$. Since $\phi(M)\subset\mathfrak{m}M$, we see that $N\subset \mathfrak{m}^rM$ for all $r$. But this implies $N=0$ and contradicting our assumption on $M$.
So, assume by induction we have proved this for all smaller depth and now let $M$ have depth $t>0$. Let $\phi_i, 1\leq i\leq r$ be a maximal sequence as you have. If $r<t$, you can easily check that $M/(\phi_1(M),\ldots,\phi_r(M))$ has depth $t-r>0$ since $\phi_i\phi_j=\phi_j\phi_i$. Then picking a non-zero divisor $x\in \mathfrak{m}$ for this module, we see that we can take $\phi_{r+1}$ to be multiplication by $x$. So, we may assume $r\geq t$. But, same argument says, $M/(\phi_1(M),\ldots, \phi_t(M))$ has depth zero and so by the first argument, $r=t$.
Can you say a bit more on why depth of $M/\phi_1(M)$ drops by 1?
@HailongDao One has an exact sequence $0\to M\stackrel{\phi}{\to} M\to M/\phi(M)\to 0$. Apply $\operatorname{Ext}(R/\mathfrak{m}, *)$.
But then 1) you don't really use $\phi_i\phi_j=\phi_j\phi_i$? and 2) one still needs to show that the induced map $Ext^n(k,M) \to Ext^n(k,M)$ has non-trivial kernel. I don't see how to prove it quickly (it is tempting to say that since $\phi(M)\subset m\phi(M)$, this map is zero like the case of multipliciation by $x$, but I am not sure how to prove it.
@HailongDao No, I use $\phi_i\phi_j=\phi_j\phi_i$ for induction. Once you have it, you can think of $\phi_i, i>1$ as endomorphisms of $M/\phi_1(M)$ and then use induction. The ext argument says $M/\phi_1(M)$ has depth at least $t-1$. If depth is greater, then, take general $x_i\in\mathfrak{m}$ so that we have an exact sequence $0\to M/(x_1,\ldots,x_t)M\to M/(x_1,\ldots,x_t)M\to M/(x_1,\ldots, x_t)M+\phi_1(M)\to 0$ This is impossible, since depth $M/(x_1,\ldots,x_t)M$ has depth zero, by the first step of the argument.
Nice argument! I added one more question that inspired by regular sequences.
The condition that $\phi_{i+1}$ is an endormorphism of $M/(\phi_1(M),...,\phi_i(M))$ was already assumed. I see the rest of the argument now, nice!
To prove the full statement, the key point is:
Lemma: Let $\phi: M\to M$ be a map such that $\phi(M)\subset mM$. Then the induced map $\phi_i: H^i_m(M) \to H^i_m(M)$ on each local cohomology module satisfies: $\ker(\phi_i)$ is nonzero if $H_m^i(M)\neq 0$.
Proof: Let $N=H^i_m(M)$. Let $z\in N$ be non-zero. By definition of local cohomology, one can check that $\phi_i^r(z)=0$ for $r\gg0$ (write $z=(\frac{a_1}{x_1},...\frac{a_n}{x_n})$ with $a_j\in M$ and $x_j\in m$, then $\phi_i^r(z)=(\frac{\phi_i^r(a_1)}{x_1},...\frac{\phi_i^r(a_n)}{x_n})$ . Let $L$ be the $N$-submodule $N\cap(\oplus \frac{M}{x_j})$, namely the collection of elements $(\frac{b_1}{x_1},...\frac{b_n}{x_n})$ in $N$ with $b_j\in M$. As the $x_j$ are fixed, $L$ is a finitely generated submodule of $H^i_m(M)$ and therefore has finite length. Then $\phi_i^r(z)\in m^rL=0$ for $r\gg0$). Choose $r$ smallest, then $\phi_i^{r-1}(z)\in \ker(\phi_i)$.
Now, if you have an injective map $\phi$, then the Lemma implies that $M$ has positive depth since we can apply it with $i=0$ to get that $H^0_m(M)=0$ (this is similar to that part in Mohan's proof). The long exact sequence of local cohomology coming from $0 \to M \to M \to M/\phi(M)\to 0$ and the Lemma again tells us that $depth(M/\phi(M))= depth(M)-1$. Induction finishes the statement.
Nice! but we only have $z = \frac{a}{x_1^{t_1}... x_n^{t_n}}$ for top local cohomology. I think we need filter regular sequence and Nagel-Schenzel's isomorphism here to prove $H^i_{\mathfrak{m}}(R)$ is $\varphi$-torsion.
@PhamHungQuy: Oh, yes, I wrote that with top one in mind. But I don't think you need NS. I edited the proof, better now?
Uhm, I not sure since $H^i_{\mathfrak{m}}(R)$ is the cohomology and $\oplus M_{x_j}$ is a term in the complex. It should need more details.
It's $\frac{M}{x}$ is just ${\frac{a}{x}}, a\in M$, not the localization. Just look at all the elements in $H_i^m(M)$ of that form.
Here I write a sketch of proof which answers both the questions. Consider the $R$ sub-algebra $S$ of $\operatorname{End} M$ generated by the $\phi_i$s. Then, by our assumption, $S$ is commutative, it is a finite type $R$-module and using the assumption $\phi_i(M)\subset\mathfrak{m}M$, it is also local. $M$ is naturally an $S$-module. Under these hypotheses, it is easy to check that $\operatorname{depth}_R M=\operatorname{depth}_S M$ and that you can easily see answers both the questions.
Great! thank you for your insight. In this point of view, all questions are clear. For example Hailong's remark on local cohomology. Another one is the Krull principal theorem: $\dim M/\varphi(M) \le \dim M - 1$.
Sorry, it should be $\dim M/\varphi(M) \ge \dim M - 1$.
Ok, let me answer my second question. What we discuss here says that we can generalize several concepts of sequence of elements for endomorphisms.
For my second question, it is enough to consider two endomorphisms $\varphi_1, \varphi_2$. Suppose $\varphi_1$ is not injective on $M/\varphi_2(M)$. Then we have $x \in M$ such that $x \notin \varphi_2(M)$ but $\varphi_1(x) = \varphi_2(y)$ for some $y$. Since $\varphi_1, \varphi_2$ is an $M$-regular sequence we have $y = \varphi_1(z)$ for some $x$. Thus $\varphi_1(x) = \varphi_2(y) = \varphi_2(\varphi_1(z)) = \varphi_1(\varphi_2(z))$ by commutativity. Thus $x = \varphi_2(z)$ by the injectivity of $\varphi_1$, a contradiction.
We are going to prove that $\varphi_2$ is injective on $M$. Consider the short exact sequence
$$0 \to M \overset{\varphi_1}{\to} M \to M/\varphi_1(M) \to 0.$$
Since $\varphi_2$ acts injective on $M/\varphi_1(M)$ we have an isophophism $\mathrm{ker}(\varphi_2) \overset{\varphi_1}{\to} \mathrm{ker}(\varphi_2)$. Repeating Mohan's argument for the $depth = 0$ case we have $\mathrm{ker}(\varphi_2) = 0$. The proof is complete.
|
2025-03-21T14:48:31.400383
| 2020-07-02T07:30:11 |
364638
|
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"Arie Rokach",
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|
Stack Exchange
|
ratio between a polygon bounded in another polygon
Let A be a convex polygon with area SA. Construct a new polygon B by orderly connecting the midpoints of the segments of A. Denote the area of B by SB. Claim : the ratio SB/SA is constant for all polygons of the same number of vertices. This ratio is easy to calculate for regular polygons. According to the avove claim , it is identical for all polygons with the same number of vertices. For example , for Hexagons , this ratio is 0.75. This claim can be observed empirically by using GEOGEBRA.
All you need to do is note that if you regard the set of vertices as an element of $2n$-dimensional space in the usual way, then the vertex transformation is represented by a simple matrix. Then use the formula $$A=1/2 \sum (x_iy_{i+1}-x_{i+1}y_i)$$ for the area of the polyhedron.
Added: as mentioned below, the statement of the OP is incorrect, but the method suggested here does provide a formula for the required area. It is of the form $A/2$ plus a correcting term which can be interpreted as a sum of areas of associated triangles. Not sure if it is much use though.
This is true only for triangles and quadrilaterals. These shapes have area ratios of 1/4 and 1/2 respectively. It does not hold for polygons with more than 4 sides.
To prove this consider the pentagon consisting of a unit square with a triangle of height $h$ placed on top:
Then the area of the midpoint polygon is $5/8+3h/8$ and the original polygon, $1+h/2$. Therefore the ratio is $\frac{3h+5}{4h+8}$ which is clearly not constant as we vary $h$. Similar arguments show that the ratio is non-constant for more vertices.
For example by adding further vertices onto the bottom side of the pentagon, keeping it a single line, we keep the area of the polygon the same and increase the midpoint area by a fixed amount independent of $h$. This produces a ratio of the form $\frac{3h+a}{4h+b}$ for some $a$ and $b$ which again is non-constant for varying $h$.
correct! you are right. i try to change the claim. Thanks a lot!
|
2025-03-21T14:48:31.400567
| 2020-07-02T08:41:07 |
364641
|
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|
Stack Exchange
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Consequences of Gromov's Conjecture
In Peter Petersen words, Gromov Betti number estimate is considered one of the deepest and most beautiful results in Riemannian geometry; which asserts that
Theorem (Gromov 1981). There is a constant $C(n)$ such that
any complete manifold $(M, g)$ with $\sec\geq 0$ and for any field $\Bbb F$ of coefficients satisfies
$$\sum_{i=0}^n b_i(M,\Bbb F)\leq C(n).$$
and conjectured that the best upper bound is $C(n)=2^n$.
I want to know
What would be the consequences if Gromov upper bound Conjecture be true? Will the new results be different from the non exact upper bound $C(n)$?
The $2^n$ bound would show that many known simply-connected manifolds admit no metric of nonnegative curvature. Gromov's constant $C(n)$ is explicit and huge (if memory serves, it is double exponential in $n$). There are many manifolds that don't violate Gromov's bound but would violate the $2^n$ bound. Mayer-Vietoris relates Betti numbers of connected sum to Betti numbers of the summand, and roughly they add up (one has to subtract/add $\le 2$ in some cases). Start taking e.g. iterated commected sum of $CP^n$ and see how quickly it fails the $2^n$ bound.
I guess $n$ is the dimension of $M$, right?
@EduardoLonga: yes, $n=dim(M)$ so I should be talking about $CP^{n/2}$. Gromov's theorem is about closed manifolds. Any open complete nonnegatively curved manifold deformation retracts onto a compact boundaryless nonnegatively curved submanifold, so $n$ is the dimension of that submanifold.
As Igor mentioned knowing the optimal bound is always better than knowing a non-optimal one such as the bound provided by Gromov's proof. It rules out a lot more examples. A proof of the sharp bound would also likely imply a rigidity result that if the sum of the Betti numbers is exactly $2^n$ then $M$ is a torus. This is quite out of reach with the currently known bound.
Conceptually more interesting is the relation of Gromov's conjecture to several other conjectures.
The strongest of these is a conjecture of Bott that a simply connected nonnegatively curved manifold is rationally elliptic. This means that that the total sum of ranks of rational homotopy groups is finite. This holds for homogeneous spaces for example. Rational ellipticity is a very strong condition.
It's been extensively studied in rational homotopy theory.
It in particular implies that the sum of the rational Betti numbers is at most $2^n$. So Bott's conjecture would imply Gromov's conjecture over $\mathbb Q$.
Ellipticity of $M$ also implies that $\chi(M)\ge 0$. So Bott's conjecture also implies Chern's conjecture that a nonnegatively curved manifold must have nonnegative Euler characteristic. Moreover it's known that a rationally elliptic space $M$ has $\chi(M)> 0$ iff all odd Betti numbers of $M$ vanish. This is also conjectured but not known for manifolds of nonnegative curvature.
Lastly, let me mention that Bott's conjecture is much stronger than Gromov's conjecture. For example it's easy to see that connected sum of at least 3 copies of $\mathbb{CP}^n$ is rationally hyperbolic (i.e. not elliptic) so it should not admit nonnegative sectional curvature according to Bott's conjecture. On the other hand Gromov's conjecture does not rule out the connected sum of $k$ copies of $\mathbb{CP}^n$ unless $k> \frac{2^{2n}-2}{n-1}$.
|
2025-03-21T14:48:31.400820
| 2020-07-02T09:23:53 |
364643
|
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"authors": [
"Alex Gavrilov",
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|
Stack Exchange
|
Isometric embedding of the modular surface
Is there an isometric embedding of the modular surface $X(1)=PSL(2,\mathbb{Z})\backslash \,\mathbb{H}$ into the Euclidean 3-space? For all I know this may be an open problem but I am also curious if anyone studied
it numerically or maybe even made a physical model of it. (Which would probably
look a little scary, with two horns (conical points) and a tail (cusp).)
P.S. To make it clear, I mean a $C^\infty$ embedding outside the conical points.
Sounds like an interesting question to investigate.
Nitpicking : $X$'s usually denote the compact modular curves, and $Y$'s the ones with cusps, so $Y(1)$ would be the currently standard notation.
Anyway, +1 for the very natural and interesting question.
There is no isometric immersion, let alone embedding, of $X(1)$ into Euclidean $3$-space. Here is a sketch of an argument:
First, let $\mathbb{H}\subset\mathbb{C}$ be the upper half plane endowed with the standard metric $(\mathrm{d}x^2+\mathrm{d}y^2)/y^2$ where $z = x+ i\,y$ with $y>0$. A fundamental domain for the action of $\mathrm{PSL}(2,\mathbb{Z})$ on $\mathbb{H}$ is then defined by the inequalities $|z|\ge 1$ and $|x|\le \tfrac12$. Then one identifies $\tfrac12+i\,y$ with $-\tfrac12+i\,y$ and $\cos\theta + i\,\sin\theta$ with $-\cos\theta + i\,\sin\theta$. The 'conical points' are $z_2 \equiv i$ (of order $2$) and $z_3 \equiv \tfrac12 + i\tfrac{\sqrt3}2$ of order $3$, and the 'cusp' point is $z_1 \equiv +i\,\infty$.
Now suppose that a smooth isometric immersion $f:X(1)\setminus\{z_1,z_2,z_3\}\to\mathbb{E}^3$ exists. Fix a point $z\in X(1)$ distinct from the three $z_i$. There will be a hyperbolic disk $D_r(z)$ of some radius $r>0$ about $z$ that does not contain any of the $z_i$. Because the Gauss curvature of $X(1)$ is $K=-1$, the convex hull of the $f$-image of $D_r(z)$ will contain an Euclidean ball of some positive radius $R>0$.
Meanwhile, let $\epsilon>0$ be a very small positive number and consider the subset $M_\epsilon\subset X(1)$ that consists of the $z = x+i\,y$ that satisfy $y\le 1/\epsilon$ and $d(z,z_2)\ge \epsilon$ and $d(z,z_3)\ge \epsilon$, where $d(z,w)$ is the hyperbolic distance between $z$ and $w$. This $M_\epsilon$ is a compact smooth surface whose boundary $\partial M_\epsilon$ consists of three disjoint circles:
$C_1$ (the points of the form $z = x+i/\epsilon$), which has length $\epsilon$;
$C_2$ (the points where $d(z,z_2)= \epsilon$), which has length $\pi\sinh\epsilon$, and
$C_3$ (the points where $d(z,z_3) = \epsilon$), which has length $\tfrac23\pi\sinh\epsilon$.
In particular, when $\epsilon>0$ is taken to be sufficiently small, each of these curves has total length less than $4\epsilon$.
Thus, each $f(C_i)$ must therefore lie in an Euclidean ball $B_i$ of radius at most $4\epsilon$. Hence the $f$-image of the boundary $\partial M_\epsilon$ must lie in an infinite 'slab' of thickness at most $4\epsilon$. (Just take a plane that passes through the centers of the three balls $B_i$ of radius $4\epsilon$ and look at the $4\epsilon$-neighborhood of that plane.)
Now, because the Gauss curvature of $M_\epsilon$ is strictly negative, the $f$-image of $M_\epsilon$ must lie within the convex hull of the image of $\partial M_\epsilon$. In particular, it must lie in the infinite slab of thickness at most $4\epsilon$.
However, if we take $\epsilon<R/4$ sufficiently small, the disk $D_r(z)$ will lie entirely within $M_\epsilon$ and hence the $f$-image of $D_r(z)$, whose convex hull contains an Euclidean ball of radius $R$, must lie in an infinite slab of thickness at most $4\epsilon<R$, which is obviously impossible.
Thus, such an $f$ cannot exist.
About the convex hull, could you give a reference?
@AlexGavrilov: Are you asking about "For an isometric immersion of a compact surface with negative Gauss curvature into $\mathbb{E}^3$, the image lies in the convex hull of the image of the boundary"? This is a common exercise: If $f:M\to\mathbb{E}^3$ is an isometric immersion and $h:\mathbb{E}^3\to\mathbb{R}$ is a linear function such that $h\circ f$ is not positive on $\partial M$, then $h\circ f$ is not positive on $M$ since, otherwise, if $h\circ f$ has a positive maximum at $p\in M$, then $f(M)$ will lie on one side of the image tangent plane at $f(p)$, which is impossible since $K<0$.
What about $Y(3)$ ? Conformally, it is a regular tetrahedron without its vertices, quotient of the upper half plane by the torsion-free group $\Gamma(3)$, kernel of the surjective morphism $PSL_2(\mathbb{Z})\to PSL_2(\mathbb{Z}/3)\simeq A_4$.
That would be a great example !
@BS.: Actually, I just realized (silly me) that an isometric immersion of $Y(3)$ is ruled out by Hilbert's Theorem. The point is that the simply-connected cover of $Y(3)$ is the entire Poincaré disk, and Hilbert's Theorem (1901) is that there is no isometric immersion of the entire Poincaré disk into $\mathbb{E}^3$. (Hilbert's Theorem didn't apply to $X(1)$ because of the two conical points. In particular, the composition $\mathbb{H}\to X(1)\to \mathbb{E}^3$ is not an immersion.)
Correction: The following idea doesn't work as stated, because (as Robert points out) the cusp and cones allow points where mean curvature is not defined.
If you could isometrically and smoothly embed the modular surface, you would smoothly and locally isometrically immerse the hyperbolic plane, which is impossible by a theorem of Hilbert.
Actually, Hilbert's theorem only applies to immersions of the entire hyperbolic plane. I don't see how Hilbert's proof would generalize to cover the case of "immersions" that have conical points or 'spikes' (i.e., cusps). For example, the mean curvature need not be smooth or even continuous at the conical points. For comparison, consider that the flat plane modulo the involution $v\mapsto -v$ can be isometrically embedded as a cone in $\mathbb{R}^3$, but the mean curvature blows up at the vertex of the cone even though $K\equiv0$.
@RobertBryant Ok, I will leave this up, but it is wrong for the reasons you point out. I think that Hilbert's result can be strengthened to prevent isometric immersions of sufficiently wide strips of the hyperbolic plane, but I am not sure if that is relevant.
|
2025-03-21T14:48:31.401222
| 2020-07-02T09:34:32 |
364644
|
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|
Stack Exchange
|
Fourier transform on finite groups in characteristic $p>0$
Is there a Fourier theory for finite groups in characteristic $p>0$? Assume that $p$ divides the order $|G|$ of finite groups (or just work with $p$-groups), i.e., in a modular representation-theoretic setting.
If no such full theory, maybe just for abelian case? Thank you for any relevant references.
It might help if you described what you might expect or hope for from such a theory.. There are orthogonality relations due to Brauer for modular characters.
Using the Brauer character table one can express any class function defined just on the $p'$-classes (i.e. the conjugacy classes of elements of order not divisible by $p$) as a linear combination of irreducible characters. In the abelian case this isn't much more than the complex character theory applied to the subgroup of $p'$-elements. For general class functions there just aren't enough irreducible characters to go around. Geoff Robinson has mentioned orthogonality relations: these involve projective characters and might serve as a partial substitute.
@GeoffRobinson In fact I am looking for some kind of fast Fourier transform with coefficients in characteristic $p$, but I don't know how exactly Fourier transform is formulated in char $p$. Since it involves modular representations, I would like to know what the dual side should be, which is no longer the same as in ordinary case.
@MarkWildon Exactly, in general there will be not enough irreducible characters. I'm not familiar with modular representation theory, my naive guess is that the dual side of Fourier transform should include irreducible and indecomposable representations, but I don't know if that makes sense...
I don't know exactly what you mean by Fourier theory, but the analogue of Pontyagin duality for finite commutative group schemes is Cartier duality: $G^\vee = \mathbf{Hom}(G,\mathbf G_{\operatorname{m}})$. The dual of the discrete group $\mathbf Z/p$ is the nonreduced local group scheme $\pmb\mu_p = \operatorname{Spec} k[x]/(x^p-1)$, so you'll probably have to leave the world of abstract groups.
@R.vanDobbendeBruyn Yes I know Cartier duality. I expect Fourier transform in char $p$ to envolve finite group schemes, though I still don't know what it is like, even in abelian case.
Here is a brief outline of the modular theory developed by R. Brauer : First of all, is necessary to take complex valued class functions. Then (after certain choices of prime ideal containing the rational prime $p$, etc. ), there are $m$ irreducible Brauer characters $\{\phi_{i}: 1 \leq i \leq m \}$, where $G$ has $m$ conjugacy classes of $p$-regular elements. There is a "dual basis" of Brauer characters of projective indecomposable modules $\{ \Phi_i : 1 \leq i \leq m \}$. Any complex valued class-function $\psi$ defined only on $p$-regular elements is uniquely expressible in the form $\sum_{i = 1}^{m} \langle \psi, \Phi_{i} \rangle \phi_{i}$, where for two complex valued class functions $\alpha, \beta$ defined only on $p$-regular elements, we define
$\langle \alpha, \beta \rangle $ to be the complex number $|G|^{-1}\sum_{g \in G_{p^{\prime}}} \alpha(g) \beta(g^{-1})$, where $G_{p^{\prime}}$ is the set of $p$-regular $g \in G$.
Each $\phi_{i}$ is obtained from an irreducible representation $\sigma_{i}$ of $G$ over an algebraically closed field $F$ of characteristic $p$. For $g \in G$ of order prime to $p$, $\phi_{i}(g)$ is obtained by lifting the eigenvalues to $g$ in that representation to $p^{\prime}$-roots of unity in $\mathbb{C}$ in a consistent fashion and adding them.
Each $\Phi_{i}$ is defined in an analogous manner, except that we must use the projective cover (as $FG$-module) of the irreducible module associated to $\sigma_{i}$. It is the case that we always have $\langle \phi_{i}, \Phi_{j} \rangle = \delta_{i,j}$ for $1 \leq i,j \leq m$.
It is worth remarking that if we try to do this for a prime $p$ which does not divide $|G|$, then we find that $\phi_{i} = \Phi_{i}$ for each $i$, and that the $\phi_{i}$ obtained are just the complex irreducible characters of $G$ (and all elements of $G$ are $p$-regular in this case). This is essentially because all $FG$-modules are projective when $p$ does not divide $|G|$.
Thank you for the outline! It seems Brauer's theory won't say much if there are no $p$-regular elements, e.g. for $p$-groups, is that right?
Yes, that's right, more or less, except that the identity is always a $p$-regular element. You might want to look at things like Burnside Rings instead of character rings, depending on what you are looking for.
|
2025-03-21T14:48:31.401570
| 2020-07-02T10:29:16 |
364646
|
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|
Stack Exchange
|
What is the appropriate notion of weakly equivalent or Morita equivalent categories internal to a category of generalized smooth spaces?
Let $G$ and $H$ be Lie groupoids. We know that there are two notions of equivalences of Lie groupoids:
Strongly equivalent Lie groupoids: (My terminology)
A homomorphism $\phi:G \rightarrow H$ of Lie groupoids is called a strong equivalence if there is a Lie groupoid homomorphism $\psi:H \rightarrow G$ and natural transformation of Lie groupoid homomorphism $T: \phi \circ \psi \Rightarrow \mathrm{id}_H$ and $S: \psi \circ \phi \rightarrow \mathrm{id}_G$. In this case $G$ and $H$ is said to be strongly equivalent Lie groupoids.
Weakly Equivalent or Morita Equivalent Lie groupoids :
A homomorphism $\phi:G \rightarrow H$ of Lie groupoids is called a weak equivalence if it satisfies the following two conditions
where $H_0$, $H_1$ are object set and morphism set of Lie groupoid H respectively. Similar meaning holds for symbols $G_0$ and $G_1$. Here symbols $s$ and $t$ are source and target maps respectively. The notation $pr_1$ is the projection to the first factor from the fibre product. from t. Here the condition (ES) says about essential surjectivity and the condition (FF) says about full faithfulness.
One says that two Lie Groupoids $G$ and $H$ are weakly equivalent or Morita equivalent if there exist weak equivalences $\phi:P \rightarrow G$ and $\phi':P \rightarrow H$ for a third Lie groupoid $P$.
(According to https://ncatlab.org/nlab/show/Lie+groupoid#2CatOfGrpds one motivation for introducing Morita equivalence is the failure of the axiom of choice in the category of smooth manifolds )
What I am looking for:
Now let we replace $G$ and $H$ by categories $G'$ and $H'$ which are categories internal to a category of generalized smooth spaces (For example, category of Chen spaces or category of diffeological spaces... etc). For instance, our categories $G'$ , $H'$ can be path groupoids.
Analogous to the case of Lie groupoids I can easily define the notion of Strongly equivalent categories internal to a category of generalized smooth spaces.
Now if I assume that the axiom of choice fails also in the category of generalized smooth spaces then it seems reasonable to introduce a notion of weakly equivalent or some sort of Morita equivalent categories internal to a category of generalized smooth spaces.
But it seems that we cannot directly define the notion of weakly equivalent or Morita equivalent categories internal to a category of Generalized Smooth Spaces in an analogous way as we have done for Lie Groupoids. Precisely in the condition of essential surjectivity (ES) we need a notion of surjective submersion but I don't know the analogue of surjective submersion for generalized smooth spaces
I heard that Morita equivalence of Lie groupoids are actually something called "Anaequivalences" between Lie groupoids.(Though I don't have much idea about anafunctors and anaequivalences).
So my guess is that the appropriate notion of weakly equivalent or Morita equivalent categories internal to a category of generalized smooth spaces has something to do with anaequivalence between categories internal to a category of generalized smooth spaces. Is it correct?
My Question is the following:
What is the appropriate notion of weakly equivalent or Morita equivalent categories internal to a category of generalized smooth spaces?
EDIT:
In the comments section after the answer by David Roberts we also had a discussion on the following two questions:
Let $F: G \rightarrow H$ be a Lie groupoid Homomorphism such that $F$ is fully faithful and essentially surjective as a functor between the underlying categories. Let us also assume the $G$ and $H$ are not Morita Equivalent. Then what are the properties that Lie groupoids $G$ and $H$ has in common apart from the trivial fact that they have equivalent underlying categories?
In papers on Higher gauge theory like Principal 2 bundles and their Gauge 2 groups by Christoph Wockel https://arxiv.org/abs/0803.3692 and the paper Higher Gauge theory: 2-connections by Baez and Schreiber https://arxiv.org/abs/hep-th/0412325 why strong equivalence is preferred over weak equivalence in the notion of Local triviality for Principal-2 bundles over a manifold? (Here equivalence means equivalence between categories internal to a category of generalized smooth spaces)
My deep apology for asking two sufficiently different (from the original) question in the comments section.
Thank you.
"Now if I assume that the Axiom of choice fails also in the category of generalized smooth spaces" <-- it definitely fails, no need to assume it.
@DavidRoberts Thanks.. I just said to be safe!!
@DavidRoberts: Why does it fails? Could you explain it bit more.
@Bumblebee the category of manifolds is a full subcategory of that of generalized smooth spaces, so it suffices to give a surjective submersion between manifolds with no smooth section. Take the squaring map $U(1) \to U(1)$.
In place of a detailed answer, let me point to Internal categories, anafunctors and localisations, but more specific to your case is the diffeological groupoids in Smooth loop stacks of differentiable stacks and gerbes.
To answer a more specific question here:
Precisely in the condition of essential surjectivity (ES) we need a notion of surjective submersion but I don't know the analogue of surjective submersion for generalized smooth spaces
For diffeological spaces, and I would imagine any generalised smooth spaces that can be considered as perhaps special sheaves on the category of manifolds, the type of map you want is subduction. I don't have a good canonical (nLab!) reference, but there some discussion in this answer, and such maps appear in Konrad Waldorf's work on gerbes. Subductions is also discussed (briefly) in the second linked paper above.
I guess I should add answers to the additional questions here, rather than leaving hints languishing in the comments.
Since you have group homomorphisms $G(x,x)\to H(Fx,Fx)$ that are smooth and bijective, and assuming you are dealing with finite-dimensional Lie groupoids, these are isomorphisms on the associated Lie algebras, hence a bijective local diffeomorphism, hence a diffeomorphism. It follows that since $G(x,y)$ is a principal homogeneous $G(x,x)$-space, and $H(Fx,Fy)$ is a principal homogeneous $H(Fx,Fx)$-space, and $G(x,y)\to H(Fx,Fy)$ is smooth and equivariant with respect to the groups and the isomorphism between them, then $G(x,y)\to H(Fx,Fy)$ is a diffeomorphism (essentially because it factors through a sequence of three diffeomorphisms involving all the data so far). Picking single isomorphisms $Fx\to x'$ and $Fy\to y'$ in $H$, for arbitrary objects $x',y'$ in $H$, we get a diffeomorphism $H(x',y') \simeq H(Fx,Fy)$, and hence every hom-manifold of $H$ is diffeomorphic to one in $G$. However, the diffeomorphisms don't necessarily assemble into a smooth family of diffeomorphisms indexed by the square of the appropriate object manifold.
On the other hand, while you should get a bijective map between the topological quotient spaces of orbits, it doesn't necessarily follow it's a homeomorphism, since the construction of the inverse uses continuous local sections arising from the Morita equivalence, not available here.
I don't know why the authors used strong equivalence, rather than weak equivalence, but it is the wrong notion. The weak version is correct, there are specific examples where it is clear that the weak version is in fact necessary. Wockel's paper detailing a principal 2-bundle over $\Omega G$, for $G$ a compact simple simply-connected Lie group, for example, states that it uses the strict notion of local triviality, but in fact it proves the weak version of local triviality, at the cost of some extra gymnastics of passing to a weakly equivalent Lie groupoid at one point.
Thank you very much for the references. I was looking in the papers Principal 2 bundles and their Gauge 2 groups by Christoph Wockel https://arxiv.org/pdf/0803.3692.pdf and the paper Higher Gauge theory 2-connections by Baez and Schreiber...In these papers they used the Equivalence of categories(int to generalized smooth spaces) for the notion of local triviality in principal 2 bundles over manifold(Considered as a discrete smooth category)..My question is why did not they go for more appropriate notion of equivalences(Like weak equivalences.. if they really makes sense in this setup)?
Link for the paper Higher gauge theory 2 connection I mentioned in the previous comment http://math.ucr.edu/home/baez/2conn.pdf
In Principal 2 bundles and their Gauge 2 groups https://arxiv.org/pdf/0803.3692.pdf local triviality issue is discussed in pg 6, Definition 1.8 and in Higher Gauge theory 2-connections http://math.ucr.edu/home/baez/2conn.pdf the notion of local triviality is mentioned in pg-31 Definition 30
So my question is the following: In the context of Higher Gauge theory is Strong Equivalence preferred over weak equivalence? If not (I expect so) then is it due to technical difficulties or some other reasons people don't use weak equivalences for the notion of local triviality in Principal 2- bundles?
I lost my comment I made. Basically, weak equivalence is the right one, strong equivalence is too strong. I don't know why the papers you give use strong equivalence, apart from they fact they are reasonably early in the study of higher geometry. Toby Bartels, in his thesis with John Baez on 2-bundles, uses the weak notion. This is roughly contemporary, but separate from, the technical development in the Baez–Schreiber paper.
There are examples of 2-bundles that need weak equivalence to get local triviality, for instance in A topological fibrewise fundamental groupoid I show how one can take a homotopy local trivial fibration and get a topological 2-bundle whose fibres are the fundamental groupoids of the original fibration. This would fail if strong equivalence was used.
Thank you very much for the comments about the local triviality of 2 bundles and for the added reference in the answer. I am now reading the references you provided.
I am just curious to know about the following: Let $F:G \rightarrow H$ be a Lie Groupoid homomorphism such that $F$ is essentially surjective and Fully Faithful. Since one of the reason to introduce the notion of Morita equivalence is the failure of Axiom of choice in the MAN. I want to know "How much close is $F$ to say that Lie groupoids "$G$" and "$H$" are actually same? It may sound stupid!!.. But what I want to know is precisely.... what are the properties that Morita equivalent pair of Lie groupoids $G$ and $H$(if exist) will share but "$F$" (ES, FF ) will fail to preserve those?
In the previous comment the notation $"F"$ (ES,FF) means that $F$ is a Lie groupoid homomorphism and is fully faithful and essentially surjective as a functor between the underlying categories.
And lastly.. Let $G$ and $H$ be a pair of Lie groupoids which are not Morita Equivalent. But let we know that there exists a Lie Groupoid homomorphism $F:G \rightarrow H$ such that $F$ is fully faithful and essentially surjective as a functor between the underlying categories.. What properties of Lie groupoids we can expect $G$ and $H$ to share apart from the trivial fact that the underlying categories of $G$ and $H$ are equivalent?
For your first question, they have homeomorphic topological spaces of orbits (and this is induced by the given functor), and diffeomorphic hom-manifolds $G(x,y) \xrightarrow{\simeq} H(Fx,Fy)$. In particular the automorphism groups of $x$ and $Fx$ are isomorphic as Lie groups. But it doesn't make sense to ask what $G$ and $H$ can have in common by virtue of the fact they are Morita equivalent _via $F$, but $F$ doesn't preserve it. For your second question: nothing, basically.
Actually, I misread your second question. Since you have group homomorphisms $G(x,x) \to H(Fx,Fx)$ that are smooth and bijective, and assuming you are dealing with finite-dimensional Lie groupoids, these are isomorphisms on the associated Lie algebras, hence a bijective local diffeomorphism, hence a diffeomorphism. But while you should get a bijective map between the topological quotient spaces of orbits, it doesn't necessarily follow it's a homeomorphism, since the construction of the inverse uses continuous local sections, not available here.
But new/additional questions you should not leave in the comments. In future, it's worth editing your original question, or asking a new one, if it's sufficiently different.
My apologies for asking about issues which are too different from what I asked in the original question. I understand, it can create confusion for the future readers. In future I will not do it.
Just to confirm what you said in the previous comment about my 2nd question: Let $F:G \rightarrow H$ be a Lie groupoid Homomorphism which is essentially surjective and fully faithful as a functor but let us also assume that $G$ and $H$ are not Morita equivalent. Then we have a diffeomorphism $G(x,y) \rightarrow H(Fx,Fy)$ and also we can say $Aut(X)$ and $Aut(Fx)$ are isomorphic as Lie groups. But though we get a bijective continuous map between topological quotient spaces of orbits it does not necessarily follows that it is a homeomorphism. Am I correct?
One confusion: Even if Lie Groupoids $G$ and $H$ are finite dimensional Lie groupoids. does it imply that $G(x,y)$ and $H(Fx,Fy)$ are finite dimensional manifolds? Also are $G(x,x)$ and $H(Fx,Fx)$ finite dimensional Lie groups? (Or you have assumed they are finite dimensional in your comments... Is it?)
According to what you suggested I edited my question.
There are many bijective continuous maps that are not homeomorphisms. I haven't said anything about the hom-manifolds in general, just the automorphism groups. Submanifolds of finite dimensional manifolds are finite dimensional. The automorphism groups are Lie groups (hence manifolds) by results about Lie groupoids. The hom manifolds, topological spaces to start with, are principal homogeneous spaces for Lie groups, hence homeomorphic (as spaces) to the Lie groups, hence manifolds.
Thank you very much!
Apologies for the late answer, I wish I'd found this earlier!
My MSc thesis was actually mainly devoted to developing a notion of Morita equivalence for diffeological groupoids! I don't think I'll have much to add to the answers by David Roberts and Joel Villatoro, and diffeological groupoids are clearly more specific than what you're looking for, but for what it's worth I'd like to contribute my two cents. You can find my thesis here:
Diffeology, Groupoids & Morita Equivalence. I also wrote a paper about the main result (a "Morita Theorem"), which is available on the arXiv here: arXiv:2007.09901 (and which has been accepted for publication in the Cahiers)!
The approach I focussed on in my thesis is that of biprincipal bibundles. This is a slightly different point of view than your question, but it turns out that this gives a notion of Morita equivalence that is equivalent (no pun intended) to the definition using weak equivalences (see Section 5.1.3 of the thesis). The paper by Meyer and Zhu, mentioned in one of David Roberts' comments, also uses this point of view. I don't know to what extent their theory can be used for diffeology. I mainly wanted to focus on this part of your question:
"...I don't know the analogue of surjective submersion for generalized smooth spaces."
As David Roberts mentions, one sensible option is to replace surjective submersions by subductions. The entire point of my thesis is essentially to show that the Lie groupoid theory still works if you do this. (If you're looking for a reference on subductions, Section 2.6 of my thesis discusses them in some detail.) This also leads us to an important point that Joel Villatoro raises:
"By the way, for my part I would recommend taking surjective local subductions as your submersion for the diffeological category."
The motivation for this could be that the local subductions between smooth manifolds are exactly the surjective submersions (this is proved in the Diffeology textbook by Iglesias-Zemmour), hence directly generalising surjective submersions to the diffeological setting (which is what we were after!). However, as just mentioned, choosing subductions still makes everything work. Besides this, here are two more reasons to consider choosing subductions over local subductions:
There are naturally occurring bundle-like objects in diffeology that are subductions, but not local subductions. The main examples I have in mind are the internal tangent bundles. For example, if you consider diffeological space that is the union of the two coordinate axes in $\mathbb{R}^2$, its internal tangent bundle is 2-dimensional at the origin, but 1-dimensional everywhere else (see Example 3.17 in arXiv:1411.5425 by Christensen and Wu). The internal tangent bundle of this space is then a subduction, but not a local subduction (thanks to an argument by Christensen). If we want to study such objects to occur in our theory of diffeological groupoid bundles and Morita equivalence, we need to allow subductions and not just local subductions.
The main reason that we assume the source and target maps of a Lie groupoid $G\rightrightarrows G_0$ are surjective submersions is to ensure that the fibred product $G\times_{G_0}G$ of composable arrows is again a smooth manifold. Since the category $\mathbf{Diffeol}$ of diffeological spaces is (co)complete, this assumption becomes redundant. However, the source and target maps of a diffeological groupoid are always subductions!
I elaborate on the choice of subductions over local ones in Sections 4.2 and 4.4.3 in my thesis. My answer to your main question in the setting of diffeology is therefore the same as that of David Roberts: an appropriate notion of Morita equivalence for diffeological groupoids is exactly as for Lie groupoids, but with surjective submersion replaced by subduction.
As to a more structured approach to generalising surjective submersions to the setting of generalised smooth spaces (your actual main question): I believe that the subductions in the category $\mathbf{Diffeol}$ of diffeological spaces are exactly the strong epimorphisms, cf. Proposition 37 in arXiv:0807.1704 by Baez and Hoffnung. In a more abstract setting of generalised smooth spaces you could therefore consider trying to use the strong epimorphisms to replace the surjective submersions!
Thanks a lot for the answer and the references!
I hope it was of some use!
I know this is a little late but I discuss this in the first two chapters of my thesis here:
https://arxiv.org/abs/1806.01939
Basically, as you mentioned, what you need is a notion of surjective submersion which generalizes surjective submersions of smooth manifolds. Once you have that, the definition falls out of it by the usual theory. In my thesis, I talk about the case where we are given a site, equipped with a distinguished set of morphisms which are the 'submersions'. That distinguished set of morphisms has to have a few properties which you can find in the definition of good site in the first chapter of my thesis.
The short version is that your category needs to be reasonably compatible with the grothendiek topology (i.e. morphisms are characterized by locally) and your notion of surjective submersions should generate the Grothendiek topology.
The other main property is that if you have a bunch of submersions $s_i \colon P_i \to B$ with images covering $B$ and some coherent transition maps, you should be able to glue the $P_i$ into a single submersion $P \to B$. Lastly, you need that if $f \circ g $ is a submersion then $f$ is a submersion.
The main difference between my thesis and the paper of Roberts and Vozzo is that they focus on when the category can be localized by a category of fractions method. My thesis is mainly concerned with constructing a 2-categorical equivalence between bibundles of internal groupoids and presentable sheaves of groupoids.
By the way, for my part I would recommend taking surjective local subductions as your submersion for the diffeological category. That's my two-cents anyway.
Thanks for linking to your thesis! Meyer and Zhu did a lot of this type of thing in http://www.tac.mta.ca/tac/volumes/30/55/30-55abs.html, if that useful for you. Might I point out Definition 1.3.2 (GS5) goes back to a paper of Bénabou from 1975? your category needs to be reasonably compatible with the grothendiek topology (i.e. morphisms are characterized by locally) <-- looks like asking for the (pre)topology to be subcanonical, to me.
The other main property is that if you have a bunch of submersions ... <- I can't find this in your thesis. It seems like asking for a superextensive site (see eg the appendix of https://arxiv.org/abs/1101.2363)
You're right about the sub-canonical thing. At the time I wrote the thesis I had misunderstood the definition of sub-canonical and thought that it implied that arbitrary gluing of objects was possible, which seemed to exclude the category of Hausdorff manifolds.
The second property is encoded in the condition that the functor F:Sub to C is a stack. I dont think it's the same as being superextensive.
ok, thanks, I'll think about that.
@JoelVillatoro Thanks for the answer and the link to your thesis!
@JoelVillatoro Thank you very much to both Roberts and Villatoro for the helpful and illuminating discussion in the comments section.
@DavidRoberts Thank you very much to both Roberts and Villatoro for the helpful and illuminating discussion in the comments section.
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2025-03-21T14:48:31.402898
| 2020-07-02T11:53:24 |
364650
|
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Stack Exchange
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Is $\beta^{*}(w_{2k-2}) = 0$ for an open orientable $2k$-manifold?
This question is motivated by the vector field question I asked recently. Panagiotis Konstantis answered this question for odd manifolds and I am trying to figure out the even case.
Let $M$ be a smooth oriented manifold (without boundary) of even dimension $2k$ with $k \geq 2$. Steenrod showed that the primary obstruction for lifting the tangent bundle $\tau\colon M \rightarrow BO(2k)$ along the fibration $$V_2(\mathbb{R}^{2k}) \rightarrow BO(2k-2) \rightarrow BO(2k)$$
is $$ \beta^*(w_{2k-2}) \in H^{2k-1}(M; \pi_{2k-2}(V_2(\mathbb{R}^{2k})) = H^{2k-1}(M; \mathbb{Z})\,,$$
where $\beta^*$ is the Bockstein operator and $w_{2k-2}$ is the $(2k-2)$th Stiefel-Whitney class of $M$.
Now Theorem 2 of Massey's "On the Stiefel-Whitney classes of a manifold II" paper says that this class vanishes when $M$ is closed. Can we say the same for open $M$? If the cohomology class had field coefficients, we could argue the vanishing as in this answer and the comments under it. But here the class is integral. Can we salvage the situation using other properties the problem has (such as $\beta^*(w_{2k-2})$ being 2-torsion)?
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2025-03-21T14:48:31.403024
| 2020-07-02T12:21:53 |
364652
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Stack Exchange
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Mean square estimate for the Kloosterman sums
For $m,n\in \mathbb{N}$, denote the Kloosterman sum
$$S(m,n;c)=\sum_{a\bmod c}e\left( \frac{ma+n \overline{a}}{c}\right),$$where $\overline{a}$
denotes the multiplicative inverse of $a\bmod c$.
Does any expert here know something upon the non-trivial power-saving bound for the mean square estimate of the Kloosterman sum, that is, for $m,n$ fixed, whether or not one has an estimate
$$\sum_{c\le x}\frac{|S(m,n;c)|^2}{c}\ll_{m,n}\,\,x^{\theta}\tag{$\ast$}$$
for some $\theta<1$?
As far as we know, the non-trivial bound for the first moment is due to Kuznetzov who showed $$\sum_{c\le x}\frac{S(m,n;c)}{c}\ll_{m,n}\,\,x^{\frac{1}{6}}\,\log ^{\frac{1}{3}}x.$$ The Linnik’s conjecture asserts that $$\sum_{c\le x}\frac{S(m,n;c)}{c}\ll_{m,n}\,\,x^{\varepsilon}$$ for any $\varepsilon>0$. For detailed description on the first moment estimate, one may see Sarnak and Tsimerman's paper: https://www.researchgate.net/publication/225888754_On_Linnik_and_Selberg%27s_Conjecture_About_Sums_of_Kloosterman_Sums
Recently I encounter this tricky sum as in ( $\ast$), for which one need to get a saving, compared with the trivial bound $x^{1+\varepsilon}$. There does not seem to be a reference available in the literature. If any expert here have some strategies or references, please give a guide. Many thanks.
Your any opinions are highly appreciated. Thanks in advance.
You cannot have estimates like (*) for any $\theta<1$. Fouvry and Michel showed that (see Theorem 1.2 there)
$$
\sum_{c\le x} |S(m,n;c)|/\sqrt{c} \gg_k \frac{x}{\log x} (\log \log x)^k,
$$
for any natural number $k$. By Cauchy-Schwarz one also gets a lower bound for the second moment.
They also note that one can get an upper bound for this quantity, saving a power of log. I would guess that
$$
\sum_{c\le x} |S(m,n;c)|^2/c
$$
is of the order $x$ (assume $mn \neq 0$).
Dear Lucia, thanks for timely explanation, much obliged.
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2025-03-21T14:48:31.403197
| 2020-07-02T12:26:40 |
364653
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"Bjørn Kjos-Hanssen",
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Stack Exchange
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"Arithmetically diverse" infinite binary string
For $a,b \in \omega$ with $a > 0$, let $f_{a,b}: \omega\to\omega$ be defined by $n \mapsto an+b$. What is an example of an infinite binary string $s:\omega\to\{0,1\}$ with the following property?
Whenever $(a,b), (a_1,b_1)\in (\omega\setminus\{0\})\times \omega$ with $(a,b)\neq (a_1,b_1)$, then $s\circ f_{a,b} \neq s\circ f_{a_1,b_1}$.
That's right @bof - sorry for the bad variable choice, have changed the string variable to $s$.
Let $s$ consist of $2^{2^k}$ zeros, followed by the same number of ones, for increasing $k$:
$$0^21^2\, 0^{16}1^{16}\, 0^{256}1^{256}\,0^{65536}1^{65536}\dots$$
Observe that all but finitely many blocks of 1s of $s\circ f_{a,b}$ have size of the form $\lfloor2^{2^{k}}/a\rfloor$ or $\lceil 2^{2^k}/a\rceil$.
We claim that we cannot keep having $k>j$ and $2^{2^k}/{a_1}\le 2^{2^j}/{a_2}+1$ (and therefore all $s\circ f_{a,b}$ are distinct).
Indeed, when this happens then $2^{2^k-2^j}\le a_1/{a_2}+a_1 2^{-2^j}$ is bounded. But as $k>j\to\infty$, $2^{2^k-2^j}$ is unbounded.
(Note that $2^k/a_1=2^{k-1}/a_2$ with $a_1=4$ and $a_2=2$, so a single-exponential $2^k$ in place of a double-exponential $2^{2^k}$ is not enough for this construction.)
Very neat construction - thank you very much!
@DominicvanderZypen thanks for another interesting question.
2 approaches:
Generate $b$ randomly, having $b(n)$ be $0$ with probability $1/2$ and having each $b(n)$ be independent. Then for each $(a,b)$ and $(a_1,b_1)$, the probability that
$$
b\circ f_{a,b}=b\circ f_{a_1,b_1}
$$
is $0$. Summing over all countably many such pairs, we still get a probability of $0$, so with probability $1$, our string works.
Generate $b$ greedily. Go through pairs $(a,b),(a_1,b_1)$ in some order and for each, choose two entries in our string very far out to force disagreement.
So what does $s$ look like concretely? Given $n\in \omega$ what is $s(n)$?
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2025-03-21T14:48:31.403347
| 2020-07-02T12:52:20 |
364657
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Stack Exchange
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Self avoiding walks and context free languages
Let $G$ be an infinite, locally finite, connected graph whose arcs (oriented edges) are labelled by letters in a finite alphabet $\Sigma$ such that arcs starting in the same vertex are labelled by different letters. Then every self avoiding walk starting at a given vertex $o$ can be encoded as a string over the alphabet $\Sigma$ (by reading the labels of arcs along the walk), denote the language of all such strings by $L_{\mathrm{SAW(o)}}$.
Assume that the labelled graph $G$ is quasi-transitive, that is, the group of label-preserving automorphisms acts with finitely many orbits on the set of vertices. It is known that if $L_{\mathrm{SAW(o)}}$ is a context free language, then so is $L_{\mathrm{SAW(o')}}$ for any other vertex $o'$ of $G$.
Question: Is there an elementary/direct proof of this fact? Perhaps even one that does not require quasi-transitivity? Clearly, it is enough to prove it for the case when $o'$ is a neighbour of $o$.
The only proof I am aware of is based on a characterisation of graphs with context free $L_{\mathrm{SAW(o)}}$ from the paper The language of self avoiding walks by C. Lindorfer and W. Woess. They show that $L_{\mathrm{SAW(o)}}$ is context free if and only if all ends of $G$ have size at most two (and the latter condition does not depend on $o$).
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2025-03-21T14:48:31.403487
| 2020-07-02T13:47:21 |
364660
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|
Stack Exchange
|
Cyclic codes: sparse codewords not orthogonal to the all-ones vector
Is it true that for any sufficiently large prime $p$, there exists a prime $q\ne p$ and a cyclic code of length $p$ over $\mathbb{F}_q$ that contains a codeword of Hamming weight at most ord$_p(q)$ and which is not orthogonal to the all-ones vector?
Viewed as an element of $\mathbb{F}_q[x]/(x^p-1)$, the second condition asks that the codeword is not a multiple of $(x-1)$. Here ord$_p(q)$ is the multiplicative order of $q$ modulo $p$.
Numerical experiments give the impression that the above is true for a codeword corresponding to an irreducible factor of $(x^p - 1)/(x-1) \in \mathbb{F}_q[x]$ (for some prime $q$).
As motivation for the condition on $\mathrm{ord}_p(q)$, observe that if $q$ has order $s$ mod $p$ so $p$ divides $q^s-1$ then $x^p-1$ has a factor, say $g(x)$, of degree $s$ in $\mathbb{F}_q[x]$. Now if the Hamming weight of $g(x)$ is $s$ or less, we're done. But in the worst case $g(x)$ has Hamming weight $s+1$, so this method fails. So it seems to me the real question is: when is there a prime $q$ such that one of the $(p-1)/\mathrm{ord}_p(q)$ irreducible factors of $1+x+\cdots+x^{p-1} = (x^p-1)/(x-1)$ in $\mathbb{F}_q$ has Hamming weight at most its degree, i.e. at most $\mathrm{ord}_p(q)$?
Thanks for the quick answer! I agree with the observation. The numerics seem to suggest that your reformulation has a positive answer. But the type of codeword I’m after is allowed to be a multiple of an irreducible factor (provided 1 is not a root). That might make the problem easier.
Instead of positive answer I meant to say that the answer might be “when p is big enough”.
One way this could work: choose $q$ so that $p$ divides $q^2+q+1$ but not $q-1$. Then $(x^p-1)/(x-1)$ splits into cubic factors in $\mathbb{F}_q$. Each factor splits as $(x-\zeta)(x-\zeta^q)(x-\zeta^{q^2})$ in a cubic extension field $\mathbb{F}_q(\zeta)$, so has norm $(-1)^3 \zeta^{1+q+q^2}$; since $p$ divides $1+q+q^2$ this is $-1$, and the cubic factor in $\mathbb{F}_q[x]$ is $x^3 + ax^2 + bx - 1$. Now if we could choose the cubic factor to have $0$ trace, we'd get a generating polynomial of Hamming weight $\le 3$, as required.
I am not completely sure if I understand what you mean by that the cubic factor has zero trace. This sounds a bit related to showing that if p does not divide $t = $ord$p(q)$, then the trace Tr${\mathbb{F}_{q^t}/\mathbb{F}_q}$ has a primitive $p$'th root of unity as a root. If there is a $q$ for which this is true, then that would work since the trace has "Hamming weight" $t$ and is not zero mod $p$ at 1. Is this indeed related to what you suggest?
I think I used 'trace' in a rather confusing way. All I meant was that if we could pick a cubic factor with $a=0$, then that would work.
Stupid question is $q<p,$ or $p>q,$ or either is possible?
The examples I've seen all have $q<p$. But I don't know if $p>q$ is impossible.
@MarkWildon: I agree that that would work. I guess more generally you maybe meant the trace of a primitive $p$'th root of unity $\alpha$? If ord$_p(q) = t$, then if I'm not mistaken, the coefficient of $x^{t-1}$ in the minimal polynomial of $\alpha$ will be zero, and so this minimal polynomial will work.
I think you are mistaken, or I am misunderstanding. For a small example, take $p=7$ and $q=2$, so $\mathrm{ord}7(2) = 3$ and $x^7-1 = (x+1)(x^3+x+1)(x^3+x^2+1)$. Each cubic factor has as its roots $\zeta$, $\zeta^2$, $\zeta^4$ for some primitive $7$th root of unity (since $p$ is prime, any $p$th root of unity except $1$ is primitive). Moreover both factors are primitive in the sense their roots generate the extension field $\mathbb{F}{2^3}$. But $x^3+x^2+1$ has a non-zero coefficient of $x^{3-1}$.
Yes you’re completely right, sorry. I should have said that the coefficient of $x^{t-1}$ is the trace of $\alpha$.
Incidentally I just found a paper of Cohen (https://www.sciencedirect.com/science/article/pii/0012365X90902154) on primitive elements with arbitrary trace. (There are a number of later works showing more general results.) I wonder if this can be adapted to elements of order $p$.
Are only interested in the minimal cyclic codes? For if we can use large cyclic codes the question becomes trivial by the cheating example of the universal code. That is cyclic and contains the word $x$ of Hamming weight one :-)
This doesn't sound like what I'm after. I am not sure what minimal and universal cyclic codes are to be honest. Is my question perhaps ill stated? I do want the codeword (polynomial) to have a $p$th root of unity as a root.
|
2025-03-21T14:48:31.403842
| 2020-07-02T13:48:47 |
364661
|
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"Michael Albanese",
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|
Stack Exchange
|
Chern number on non-spin manifold
Let $M^4$ be an orientable closed 4-manifold and $c_1$ be the first Chern class of a complex line bundle on $M^4$. Let $b$ be the mod 2 reduction of $c_1$, ie $b=c_1$ mod 2.
We have a relation $w_2 b = b^2$, where $w_n$ is the $n^\text{th}$ Stiefel-Whitney class of the tangent bundle of $M^4$. This implies that if $M^4$ is spin, then the Chern number
on $M^4$ must be even, ie $\int_{M^4} c_1^2 =0$ mod 2.
My question is that for any $M^4$ that is not spin, can we always find a complex line bundle on $M^4$, such that the Chern number on $M^4$ is odd, ie $\int_{M^4} c_1^2 =1$ mod 2.
The Enriques algebraic surface has even intersection form (i.e. for any class $\beta \in H^{2}(M,\mathbb{Z})$, $\int_{M^{4}} \beta^2$ is even) but is not spin by Rokhlin's theorem since the signature of the intersection form is $8$.
A simply connected $4$-manifold is spin $\iff$ the intersection form is even (which doesn't apply to the Enriques surface which has $\pi_{1} = \mathbb{Z}_{2}$).
EDIT: This does not work, in general, as explained by Michael Albanese's comment. Thanks!
If $M$ is not spin, then $w_2(M) \neq 0$. But $w_2$ agrees with $v_2$, the second Wu class, which measures whether the intersection form of $M$ is even or odd. Thus, we can find an element $\alpha \in H^2(M;\mathbb Z)$ such that $\alpha^2$ is an odd number times the cohomological fundamental class of $M$. Now represent $\alpha$ by a map $M \to K(\mathbb Z;2) = BU(1)$, i.e., a complex line bundle $E$ on $M$, then $c_1(E) = \alpha$ is as desired.
If $H_1(M; \mathbb{Z})$ has no $2$-torsion, then $w_2(M) \neq 0$ implies $M$ has odd intersection form. However, as the example in Nick L's answer demonstrates, it is not true in general. The issue is that if $w_2(M) \neq 0$, then by Poincaré duality there is $x \in H^2(M; \mathbb{Z}_2)$ with $w_2(M) \cup x \neq 0$. Now $x$ lifts to an integral cohomology class if and only if $\beta(x) \in H^3(M; \mathbb{Z})$ is zero where $\beta$ denotes the Bockstein. As $\beta(x)$ is $2$-torsion, this is automatic if $H_1(M; \mathbb{Z})$ has no $2$-torsion as $H^3(M; \mathbb{Z}) \cong H_1(M; \mathbb{Z})$.
Actually, this is the answer I want. It shows that we can find a complex line bundle on $M^4$ such that $\int_{M^4} c_1^2 = $ intersection number. Thanks!
|
2025-03-21T14:48:31.404022
| 2020-07-02T14:09:50 |
364664
|
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"Fedor Petrov",
"H A Helfgott",
"Sam Hopkins",
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|
Stack Exchange
|
Alternating sum over collections closed under containment
Let $\mathscr{C}$ be a collection of subsets of a finite set $P$. Assume $\mathscr{C}$ is closed under containment: if $S\subset P$ is in $\mathscr{C}$, then every set $S'\subset P$ containing $S$ is in $\mathscr{C}$.
What can we say about
$$\sum_{S\in \mathscr{C}} (-1)^{|S|},$$ where $|S|$ is the number of elements of $S$? In particular, is the absolute value of this sum bounded
by the number of minimal elements of $\mathscr{C}$, i.e.,
$$\left|\sum_{S\in \mathscr{C}} (-1)^{|S|}\right| \leq |\{S\in \mathscr{C}:\not \exists S'\subsetneq S \;\text{s.t.}\; S'\in \mathscr{C}\}|?$$
$P = {1,2,3,4,5,6,7,8}, \mathscr{C} = {\emptyset,{1,2},{1,2,3,4},{1,2,3,4,5,6},{1,2,3,4,5,6,7,8}}$ gives $|\sum_{S \in \mathscr{C}} (-1)^{|S|}| = 5$, but the number of minimal elements of $\mathscr{C}$ is $1$. Did I make a mistake?
No, I did. Let me add a condition.
Keywords: "order filter (dually, order ideal) in the Boolean lattice."
It's likely your question has a topological interpretation, via the Mobius function. I think you are essentially asking for the Mobius function of $\hat{0}\cup I \cup \hat{1}$ where $I$ is an order ideal of the Boolean lattice, and asking if it is bounded by the number of maximal elements of $I$. Probably some nice topological property of these posets (e.g. 'shellability') can help you.
See https://www.mi.fu-berlin.de/math/groups/discgeom/ziegler/Preprintfiles/018PREPRINT.pdf
Why would an order ideal be shellable? The minimal elements of $\mathscr{C}$ don't necessarily have the same number of elements. Also thanks for Sagan-Yeh-Ziegler - but that seems to give a bound that grows (exponentially) with $|P|$ and does not depend on the number of minimal elements of $\mathscr{C}$.
@HAHelfgott: yes, they give a bound on this value over all possible choices of $\mathcal{C}$, but maybe their techniques can be used to give what you want too. And yes, sorry, I meant somehow use shellability of the Boolean lattices (SYZ do seem to use shellability at one point in their argument).
There is an interpretation of your inequality using simplicial complex and $h$-vector. Namely, let $\Delta$ be the set of all $P-S$ with $S$ in your set. Then $\Delta$ is a collection closed under inclusion, hence a simplicial complex.
The invariant you are interested in is, up to sign, the alternating sum of $f_i$: the number of $i$-dim faces of $\Delta$. Then, by a well-known formula, it is equal to $|h_d|$, where $h_i$ forms the $h$-vector of $\Delta$, and $d-1$ is the dimension of $\Delta$.
The right hand side of the inequality you are interested in is the number of facets (maximal elements) of $\Delta$. As people have pointed out, in general the inequality you want does not hold.
However, I would like to point out that it is likely to hold under certain extra topological/homological assumptions. For instance, if $\Delta$ is Cohen-Macaulay, then all the $h_i$ are non-negative, and the number of facets is $f_{d-1}=\sum_{i\geq 0} h_i\geq h_d$, which is what you need. In fact, as $h_0=1$ and $h_1=n-d$ where $n=|P|$, you get something a little stronger.
One can prove other inequalities for $h_i$ under weaker conditions. For example, if $\Delta$ satisfies Serre's conditions $(S_{r})$, one still have non-negativity of $h_{\leq r}$, a result first proved by Murai-Terai. I discussed some of them in a recent talk (but it is perhaps a bit algebraic).
I am afraid that this inequality does not hold. Let $S_1,\ldots,S_m$ be the minimal elements of $\mathscr{C}=\{S:\exists i\in \{1,\ldots,m\} \, \text{such that}\, S_i\subset S\}$. We have by inclusion-exclusion
$$
\sum_{S\in \mathscr{C}} (-1)^S=\sum_{I\subset \{1,\ldots,m\}, I\ne \emptyset} (-1)^{|I|-1}\sum_{T:\cup_{i\in I} S_i\subset T} (-1)^{|T|}.
$$
The inner sum equals 0 unless $\cup_{i\in I} S_i=P$, otherwise it equals $(-1)^{|P|}$. Therefore up to sign your sum equals
$$
\sum_{I\subset \{1,\ldots,m\}, \cup_{i\in I} S_i=P} (-1)^{|I|-1}.
$$
Imagine that the union of any 17 sets from the collection $\{S_1,\ldots,S_m\}$ equals $P$, but the union of no 16 sets $S_i$ equals $P$. This may be achieved by choosing for each 16 sets $S_i$'s a special element which they do not contain, and letting $P$ to be equal to the set of all these elements. Then your sum is some polynomial of degree 16 in $m$.
It would be interesting to compare this to the bound over all $\mathscr{C}$ from the Sagan-Yeh-Ziegler paper.
it looks that this construction gives an example for about ${m-1\choose \lceil (m-1)/2\rceil}$, do they claim that this is an upper bound also?
Yes, I believe so.
|
2025-03-21T14:48:31.404452
| 2020-07-02T14:19:54 |
364665
|
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"JoshuaZ",
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|
Stack Exchange
|
Is there any good reference on the Bayesian view that can be helpful for reading papers on the number theory using heuristic arguments?
Nowadays there are many papers on the number theory using heuristics.
I have read some of them.
But I have no clear understanding of the Bayesian Probability(subjective probability).
The concept of using probabilistic method for a deterministic object is not clear at least to me.
Is there any good exposition on the Bayesian Probability especially for number theorists?
Thank you.
Possibly relevant paper https://arxiv.org/abs/1707.08747 .
Surjective or subjective?
@SylvainJULIEN I meant subjective, I've edited it. thank you.
By the way, it would be interesting to establish some kind of "extremal probability entropy principle", which would allow to determine which probability distribution is best suited to apprehend a deterministic mathematical phenomenon heuristically.
@SylvainJULIEN I agree with that. And I hope those discussions could become easily accessible to number theorists. Is the 'extremal probability entropy principle' a well established concept? If so can you suggest a good reference please?
I don't know actually. I was just trying to draw a parallel with both the maximum entropy principle and the extremal action principle in physics (I studied physics). A fun fact is that these two principles can be expressed as $\delta S=0$ with of course different meanings for the symbol $S$! But the Shannon entropy of a probability distribution is well defined, so you should be able to find relevant references about it.
|
2025-03-21T14:48:31.404579
| 2020-07-02T14:38:23 |
364666
|
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|
Stack Exchange
|
A question on cokernel
I was reading the proof of Theorem 7.12.11 in the book "Tensor categories" by Etingof et al. Let $\mathcal{C}$ be a finite multi-tensor category, and $A$ be an algebra in $\mathcal{C}$. Let $Bimod(A)$ be the category of $A$-$A$-bimodules in $\mathcal{C}$ and $Bimod(A^R \otimes A)$ be the category of $A^R \otimes A$-bimodules in $\mathcal{C}$. Here $A^R$ is the right adjoint of $A$ and the multiplication of the algebra $A^R \otimes A$ is induced by the evaluation $ev: A \otimes A^R \to 1$, $1$ is the tensor unit of $\mathcal{C}$.
It seems that the proof of Theorem 7.12.11 implies that the functor
$$
\mathcal{C} \xrightarrow{\simeq} Bimod(A^R \otimes A)\ ,\qquad x \mapsto A^R \otimes x \otimes A
$$
is a tensor equivalence of the categories. I know that the functor is fully faithful and essentially surjective. But I do not know how to show that the functor is a monoidal functor. More explicitly, I would like to know how to show that
$$
(A^R \otimes x \otimes A) \otimes_{A^R \otimes A} (A^R \otimes y \otimes A) \simeq A^R \otimes x \otimes y \otimes A,
$$
where $(A^R \otimes x \otimes A) \otimes_{A^R \otimes A} (A^R \otimes y \otimes A)$ is the coequaliser of the morphisms:
$$
1_{A^R \otimes x} \otimes ev \otimes 1_{A \otimes A^R \otimes y \otimes A}
$$
and
$$
1_{A^R \otimes x \otimes A \otimes A^R} \otimes ev \otimes 1_{y \otimes A}
$$
i guess that "essentially subjective" is "essentially surjective" ?
|
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