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2025-03-21T14:48:29.943067
| 2020-02-27T12:05:56 |
353683
|
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|
Stack Exchange
|
Corepresentations on Hilbert modules
In the seminal paper "Unitaires multiplicatifs et dualité pour les produits croisés de $C^*$-algèbres" by Baaj and Skandalis, we find the following very general definition of what a corepresentation of a Hopf $C^*$-algebra (perhaps more properly called a "$C^*$-bialgebra") is:
There are no references given. I would justify the definition by using the isomorphism between $\mathcal L(H\otimes A)$ and $M(K(H)\otimes A)$, the latter algebra being where we usually take corepresentations to live.
I am interested in earlier, or contemporaneous, work which considers such a definition in more detail (or even just gives the definition).
That is, where does this definition come from? Is this the first paper to make it?
|
2025-03-21T14:48:29.943154
| 2020-02-27T14:09:26 |
353688
|
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"Peter Humphries",
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|
Stack Exchange
|
A line integral involving $\zeta(s)$
Let $x>1$ and $\zeta$ denote the Riemann zeta function. Is the equality
$$\int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\zeta(s)x^{s}}{s(s+1)(s+2)\cdots (s+k)} \mathrm{d}s = 2\pi i$$ possible for any integer $k>1$ where $\sigma_{0}>1$ ?
The left-hand side is equal to $\frac{2\pi i}{k!} \sum_{n \leq x} (x - n)^k$.
@PeterHumphries, any proof or reference ?
Chapter 5 of Montgomery-Vaughan - the bit on Cesaro sums with $a_n = 1$ for all $n$, so that $\alpha(s) = \zeta(s)$.
@Peter Humphreys, thanks, just checked MV. But it seems your summand should rather be $(1-n/x)^k$...
Whoops, yep, that's correct
|
2025-03-21T14:48:29.943354
| 2020-02-27T14:09:45 |
353689
|
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|
Stack Exchange
|
2D closed convex shape which minimizes average distance between points
For a 2D closed convex shape, with metric $d$ and fixed area $A$, we can calculate the average distance between random (interior) points. For different shapes, we will get different values for this average. We can ask, for which shape is this average a minimum (in terms of $A$). For the Euclidean metric, the answer is the circle. If we pick a different metric, what shape do we get?
A few comments about your question, concerning possibly implicit assumptions. First, when you say 2D you mean that the underlying set is $\mathbb{R}^2$? Or some other notion of dimension? Are you going to assume that the metric is translation invariant? or more general? Do you have a specific metric in mind? What is the measure you are going to use for defining "average distance"?
Yes, $\mathbb{R}^2$. General metric (not necessarily translation invariant). No specific metric in mind. The measure for distance is the metric.
How do you define area? Just the usual one for R2?
@MichaelMcGettrick: not the measure for distance. The measure for defining averages. (You are presumably integrating over an area and dividing by the total area for taking averages, what is the area measure?)
Ah, okay, I guess I have not thought about the metric for the area A, or for the averaging: But then one can still pose this question for discrete case, area corresponding to N points, divide by N for average (as replied by Joseph below).
Here is an answer for lattice points and using the $L_1$ (Manhattan) metric:
Demaine, Erik D., Sándor P. Fekete, Günter Rote, Nils Schweer, Daria Schymura, and Mariano Zelke. "Integer point sets minimizing average pairwise $L_1$ distance: What is the optimal shape of a town?." Computational Geometry 44, no. 2 (2011): 82-94.
PDF download.
The continuous version of the same problem is addressed in this paper:
Bender, Carl M., Michael A. Bender, Erik D. Demaine, and Sándor P. Fekete. "What is the optimal shape of a city?." Journal of Physics A: Mathematical and General 37, no. 1 (2003): 147. Journal link.
From the Abstract: "In this paper a nonlinear differential equation for the boundary curve of such a city is determined."
For an arbitrary metric and measure, this is going to be a very difficult question to answer. The shape of the minimizer will depend on the choice of metric and measure, if it exists at all. Joseph O'Rourke discussed the $L^1$ distance already, so I'm going to focus on distance functions which are either symmetric, or else induced by a smooth Riemannian metric. These sorts of questions are known as extremal problems [1] and are closely related to isoperimetric inequalities, which are an active area of research.
As noted in the comments, we need to specify which measure we use to compute averages. It appears that there are at least two natural choices of measure.
the Lebesgue measure on $\mathbb{R}^2$
the measure induced by a Riemannian metric
As I mentioned earlier, the particular shape of the minimizer will depend on the choice of distance function and measure. However, from a variational standpoint, there is a fairly general principle which is very helpful.
A convex region $S$ is a critical point for the average distance
between interior points iff the integral $\int_S d(p,x) \, d \mu(x)$ is constant for all $p \in \partial S$. Here, $\mu$ is the
measure used to determine averages.
In other words, a region $S$ is a possible minimizer for the average distance between points if the average distance between a point $p$ on the boundary of $S$ and interior points in $S$ does not depend on $p$. It's possible to prove this rigorously using variational techniques or using Crofton's differential equation (for a good exposition, see [2]). However, the basic intuition is that if this property does not hold, we can slightly change the shape of $S$ so as to reduce the total average distance.
From this, there are a few immediate consequences.
If the distance function is translation and rotationally symmetric and the measure we use is the Lebesgue measure on $\mathbb{R}^2$, the disk is a critical point. Oftentimes, this will be the unique critical point (and thus also the minimum), but any proof of this seems likely to depend on the particular details of the distance function.
If we consider a surface of constant curvature with its natural distance function and induced measure, the critical points for the average distance between points are geodesic balls. There is no guarantee that geodesic balls will be convex in the choice of coordinates you use, but they will be the minimizers nonetheless.
For more general (non-symmetric) Riemannian metrics (with their induced distance function and measure), this question is much more difficult, and the variational principles are less useful. However, with a lower bound on the sectional curvature, it is possible to get a lower bound on the areas of geodesic balls. Using this, it should be possible to establish uniform lower bound on the average distance between points in a region of area $A$. However, these sorts of estimates don't tell you what the minimizer looks like (or even if it exists at all).
It is possible to construct distance functions where no minimizer exists, by incorporating patches where the curvature is more and more negative.
[1] Bauer, Christina; Schneider, Rolf, Extremal problems for geometric probabilities involving convex bodies, Adv. Appl. Probab. 27, No. 1, 20-34 (1995). ZBL0827.52004.
[2] Eisenberg, Bennett; Sullivan, Rosemary, Crofton’s differential equation, Am. Math. Mon. 107, No. 2, 129-139 (2000). ZBL0986.60011.
|
2025-03-21T14:48:29.943757
| 2020-02-27T14:22:50 |
353690
|
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|
Stack Exchange
|
Disintegration associative
Is the disintegration of two borelian probabilities measures is associative ? It means if $\mu = \mu_{y}^{1} \oplus h_{\#}^{1}\mu$ and $ h_{\#}^{1}\mu = \mu_{y}^{2} \oplus h_{\#}^{2} h_{\#}^{1}\mu$. Then do we have
$$
\mu = \mu_{y}^{1} \oplus \mu_{y}^{2} \oplus (h^{2} \circ h^{1})_{\#} \mu
$$
Where $\mu$ is a borelian probability over $\mathbb{R}^{d}$ and I note $\mu = \mu_{y} \oplus h_{\#}\mu$ the disintegration of $\mu$ according to $h_{\#} \mu$ that gives the kernel $\mu_{y}$ with $y \in \mathbb{R}^{d}$.
EDIT : To be clearer, we have $\mu = \mu^{1,2} \oplus(h^{2} \circ h^{1})_{\#} \mu$. The question is
$$
\mu^{1,2}_{y}(A) =? \int \mu_{x}^{1}(A) d\mu_{y}^{2}(x)
$$
I have problems with your notation. What is $h_2 \circ h_1$? I think if you formulate (if possible) your question with stochastic kernels the answer is simply true. Have a look f.i. in the old book of Bertsekas/Shreve (1978), Stochastic Optimal Control: The discrete Time Case, in particular ch. 7.4.3 Stochastic Kernels.
I edit thank you. Here the things is we have to check if
$$
\mu_{y}^{1}\oplus \mu_{y}^{2} (\mathbb{R}^{n} - (h^{2}\circ h^{1})^{-1}({y}))=0
$$
And
$$
y \rightarrow \mu_{x}^{1} \oplus \mu_{y}^{2}
$$
Is measurable.
Obious edit :
$$
y \rightarrow \mu_{x}^{1} \oplus \mu_{y}^{2} (A)
$$
Is measurable for any borel set $A$.
Disintegration is associative, in the following sense: Suppose that we have a disintegration
$$\mu(dx)=\int_Y(\mu h_1^{-1})(dy)\mu_{h_1}(y,dx)$$
of a probability measure $\mu$ with a kernel $\mu_{h_1}$ -- meaning that $\int_X\mu(dx)f(x)=\int_Y(\mu h_1^{-1})(dy)\int_X\mu_{h_1}(y,dx)f(x)$ for all nonnegative measurable functions $f$ on $X$, and that we further have a disintegration
$$(\mu h_1^{-1})(dy)=\int_Z(\mu h_1^{-1}h_2^{-1})(dz)(\mu h_1^{-1})_{h_2}(z,dy) \\ =\int_Z(\mu(h_2\circ h_1)^{-1})(dz)(\mu h_1^{-1})_{h_2}(z,dy)$$
of the measure $\mu h_1^{-1}$.
Then we have the disintegration
$$\mu(dx)
=\int_Z(\mu(h_2\circ h_1)^{-1})(dz)(\mu_{h_1}*(\mu h_1^{-1})_{h_2})(z,dx),$$
where
$$(\mu_{h_1}*(\mu h_1^{-1})_{h_2})(z,dx):=\int_Y (\mu h_1^{-1})_{h_2}(z,dy)\; \mu_{h_1}(y,dx).$$
We have $\mu = \mu^{1,2} \oplus(h^{2} \circ h^{1}){#} \mu$. The question is
$$
\mu^{1,2}{y}(A) =? \int \mu_{x}^{1}(A) d\mu_{y}^{2}(x)
$$
@CechMS : I think you never said what you mean by $\oplus$, $\mu_x^1$, $\mu^{1,2}$, ... .
The identity $\mu = \mu_{y} \oplus h_{#}\mu$ means that ${ \mu_{y} ; y \in \mathbb{R}^{n} }$ is a probability kernel such that :
$$
\mu(A) = \int \mu_{y} (A) d h_{#} \mu
$$
And
$$
\mu_{y}(\mathbb{R}^{n} - h^{-1}({ y}) = 0
$$
We write $\mu = \mu_{y} \oplus \nu$ everytime you have $\nu = h_{#} \mu$ otherwise even over $\mathbb{R}^{n}$ it could not exist.
It is still unknown what you mean by $\mu_x^1$, $\mu^{1,2}$, ... . Also, the first displayed formula in your latest comment, with just one entry of $y$, is apparently written not the way it should be.
|
2025-03-21T14:48:29.943948
| 2020-02-27T14:33:38 |
353691
|
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|
Stack Exchange
|
Are simplicial abelian sheaves fibrant?
I want to show that simplicial abelian sheaves are fibrant. For this, I wonder whether a morphism between simplicial sheaves is a fibration iff it has RLP w.r.t. all morphisms like
$$\Lambda^n_k\times X\to\triangle^n\times X$$
where $X\in Sm/k$. Is this claim true?
Here weak equivalences are stalkwise weak equivalences and cofibrations are monomorphisms.
Fibrant in what model structure?
Simplicial abelian sheaves (and presheaves) are fibrant
in the projective model structure because
all simplicial abelian groups are fibrant.
Simplicial abelian sheaves are definitely
not fibrant in the local projective model structure
because sheaf cohomology groups can be nontrivial.
For a specific example, consider the simplicial abelian sheaf
Γ(Z[1]) on the site of smooth manifolds,
given by applying the Dold-Kan functor Γ
to the chain complex of sheaves Z[1] given by placing
the constant sheaf Z in chain degree 1.
If the simplicial abelian sheaf Γ(Z[1]) was fibrant in the local projective model structure,
than the sheaf cohomology of a smooth manifold M
in degree 1 with coefficients in Z
could be computed by evaluating Γ(Z[1]) on M
and then taking the homology group in degree 0.
However, the homology group of Γ(Z[1])(M) in degree 0 is 0 for any M,
a contradiction with the case M=S^1, for which H^1(S^1,Z)=Z.
I definitely agree with the first line. but I don't quite see the relation between being fibrant in the local projective model structure and having trivial sheaf cohomology.
@SimonHenry: I added a specific example.
I mean the injective model structure. That is why I want to find a system of generators.
@NanjunYang: The generators you wrote down are precisely the generating cofibrations for the projective and local projective model structures. The prevailing opinion with respect to the generating cofibrations for the injective model structure is that there is no explicit such a set of generators, excluding some special categories of presheaves (but Sm/k is not one of them).
So are the simplicial abelian sheaves still fibrant under injective model structure? I want to prove that the Eilenberg MacLane spectra are stably fibrant finally, thus I have to show that $K(A,n)$ is injectively fibrant. I've been using injective model structures.
Maybe $K(A,n)$ isn't injectively fibrant in general. I think I could avoid this statement.
@NanjunYang: In the motivic context, you would need K(A,n)=Γ(A[n]) to be locally injectively fibrant and not just injectively fibrant. The argument in my post shows that K(A,n)=Γ(A[n]) is not a local object, so cannot be locally injectively fibrant. Even ignoring locality, though, I am pretty sure that K(A,n)=Γ(A[n]) is not injectively fibrant. Why not use the projective local model structure instead, where the Eilenberg–MacLane spectra seem to be fibrant?
My final decision is to use the Brown-Gersten model structure. I think $K(A,n)$ is fibrant if $A$ is injective.
@NanjunYang: The Brown–Gersten model structure is the projective model structure on simplicial sheaves. Everything I said about the projective and local projective model structures applies also to the Brown–Gersten model structure.
|
2025-03-21T14:48:29.944169
| 2020-02-27T14:43:15 |
353693
|
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|
Stack Exchange
|
Slices vs harmonic functions — intuition/definitions question
One can identify functions on slices of the boolean hypercube $[n] \choose d$ with harmonic multilinear functions of degree at most $d$.
I don't really understand the intuition behind the harmonic condition—for all $x_i$ , $$ \frac{\partial P}{\partial x_i} = 0,$$ and I also have some trouble with the definitions. I am using the notation as in Filmus - Orthogonal basis for functions over a slice of the Boolean hypercube.
Let's focus on the $d=2$ case.
It seems to suggest that if $P=A_{ij} x_i x_j$ then $A$ is anti-symmetric. But thats means that $P=0$.
For example,
$$(a_{12}+a_{21}) x_2 =-(a_{13}+a_{31})x_3$$
but $x_2$ and $x_3$ are independent, so we can choose whatever we want.
It seems as though on the one hand $x_i$, determines if $i$ is in the slice.
So the functional $\langle e_{\{i,j\}} , \cdot \rangle$ translates to $x_i x_j$.
On the other hand, $x_i x_j$ is not in the space.
We do however have a natural transformation $B$ to $\chi_B$ (in the space) for every top set, but not for the set $\{1,2\}$.
Is it more suitable for SE mathematics? I wasn't sure
First, you copied incorrectly the harmonicity condition, which the paper writes as the sum $$ \sum_{i= 1}^n \frac{\partial P}{\partial x_i}= 0$$ not at all that it has vanishing gradient. Applying this to the polynomial of the form $P = A_{ij} x_i x_j$ implies that $$ \sum_{j,k} (A_{jk} + A_{kj}) x_j = 0 $$ which is not the same as require $A$ to be antisymmetric! You just need $\sum_k A_{jk} + A_{kj} = 0$ for all $j$ (which is satisfied when $A$ is antisymmetric, but not necessarily so)
|
2025-03-21T14:48:29.944324
| 2020-02-27T15:02:16 |
353694
|
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"Benjamin Steinberg",
"Mike Pierce",
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|
Stack Exchange
|
Meaning of an algebra having "sufficiently many primitive idempotents"?
This is a phrase Ringel uses a few times in his writing, and I'm not sure exactly what he means by it. The context is that we have a quiver $Q$ with path algebra $\mathbf{k}Q$. If $Q$ is not a finite quiver, then $\mathbf{k}Q$ will not have a unit, but it does have sufficiently many primitive idempotents. I'm not sure what constitutes "sufficiently many". In other writings he refers to a general $\mathbf{k}$-algebra having sufficiently many primitive idempotents, so maybe this just means the algebra is sufficiently like the path algebra of a quiver? I'm only familiar with this sort of a phrasing of a category having sufficiently many projectives/injectives.
The exact paper I'm looking at is Butler and Ringel's Auslander-Reiten Sequences with Few Middle Terms and Applications to String Algebras on page 157.
I don't know about the "primitive" part, but sometimes (in representation theory of p-adic groups, for example), sufficiently many idempotents means that for every finite collection of idempotents $e_1 , \ldots , e_n$ there exists an idempotent $e$ such that $e e_i = e_i e = e_i$ for all $1 \leq i \leq n$ (a typical thing to imagine is functions with compact support on a space w.r.t. pointwise multiplication - then characteristic functions of compacts are idempotents, and you can take unions of finite collections of compacts...)
To fill out @Sasha's remark, probably it should also be required that, for every element $r$ of the ring, there is an idempotent $e$ such that $er=re=r$. (I do not offhand know what "primitive" should mean here.)
For non unital rings this means $R\cong \bigoplus_{i\in I}Re_i$ with the $e_i$ orthogonal primitive idempotents. The index set can be infinite. This is clearly the case for the path algebra of a quiver.
The condition @PaulGattett mentions is usually called having local units
@paulgarrett Oh, of course. Thank you.
@paulgarrett In the context of path algebras, their is an idempotent element $e_i$ for each vertex $i$ of the quiver, and these idempotents are orthogonal. Since they're orthogonal any sum of them will be idempotent too, so primitive refers to these $e_i$, the idempotents that can't be written as a sum of two other non-trivial idempotents.
@MarkSapir Yeah if the quiver is finite, then everything is fine. It's the case where "if $Q$ is not a finite quiver ... $\mathbf{k}Q$ will have sufficiently many primitive idempotents." In this case there are infinitely many pairwise orthogonal primitive idempotents. One for each vertex. Is infinitely many sufficiently many? ;) Sasha's response makes sense in this case though, so I believe her right now.
Thanks for the info, @MikePierce!
Let me add that having the direct sum decomposition as a left module is equivalent to having it as a right module. And it implies Sasha and Paul's conditions which are much weaker.
I looked back at the references on rings with enough idempotents and I had misremembered things. One should assume that the ring decomposes as a direct sum of the e_iR as a right module too. You don't get it for free. I got confused because a line in a paper about Leavitt path algebras seemed to imply that.
|
2025-03-21T14:48:29.944562
| 2020-02-27T15:42:28 |
353697
|
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|
Stack Exchange
|
Connected components of a codimension one fiber for a finite morphism
Let $f:X \to Y$ be a finite surjective morphism from a $\mathbb{Q}$-factorial variety to a smooth variety. Let $D_Y$ be a prime divisor on $X$ and let $\bigcup D_i$ be the inverse image of $D_Y$. Do we know anything about the number of connected components of $\bigcup D_i$?
Does the number equal to the number of $D_i$'s? (i.e. $D_i$'s do not intersect with each other.)
Thanks!
It is bounded by the degree of the morphism.
But this is not sufficient... I expect this to be the number of $D_i$ or at least related to that...
No, the number of connected components does not equal the number of irreducible components. For example take $X \rightarrow \mathbf P^2$ a double cover branched over a smooth quartic curve $C$. For a line $L \subset \mathbf P^2$ that is bitangent to $C$ the inverse image of $L$ is a union of 2 intersecting copies of $\mathbf P^1$.
Yes, this is a good example.
If $f$ is a morphism of curves, there is no better bound than the degree of the morphism.
|
2025-03-21T14:48:29.944674
| 2020-02-27T16:18:22 |
353702
|
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"rpc"
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|
Stack Exchange
|
Divisibility properties of linear combinations of binomial coefficients
Let $p$ be a prime and $a_0,\ldots,a_n\geq 0$ be integers. Define
$$
S(a_0,\ldots,a_n)=\sum_{k=0}^n a_k\binom{n}{k}.
$$
I am trying to find out how much we know about
$$
v_p(S(a_0,\ldots,a_n)),
$$
where $v_p$ is the $p-$adic valuation, or identities involving $S(a_0,\ldots,a_n)$.
Here are a few classical ones, $S(1,x,\ldots,x^n)=(1+x)^n$ and $S(0,1,\ldots,n)=n2^{n-1}$.
$v_p$ is $p$-adic valuation?
Yes. I've added that to the original post. Thx!
It is known is that if $f(x)$ is the exponential generating function of the sequence $a_n$, then $f(x)\cdot e^x$ is the exponential generating function of $S(a_0,...,a_n)$. This observation brings a lot of classical cases.
An example with $n=pd$, $\nu_p(S) = 2d-1$, and $a_k$ depending only on $d$ but not $p$ is given in my paper: https://arxiv.org/abs/1602.02632
Needs details or clarity. Arbitrary number can be represented in this form $a_0=N$, $a_1=...=a_n=0.$
|
2025-03-21T14:48:29.944772
| 2020-02-27T17:34:09 |
353708
|
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}
|
Stack Exchange
|
Hardness results for approximating Hölder continuous functions
Let $f \in \mathrm{Lip}^{L,\alpha}[a,b]$, and let $f_{h} \in C^{L}$ be a spline which interpolates $f$ at $a + ih$. Then standard theorems (Daubechies & Lagarias, SIAM J. Math. Anal. 22 (1991) 1388-1410) show that
\begin{align*}
\left\| f - f_{h} \right\|_{\infty} \le Ch^{(\alpha+L)}
\end{align*}
I'm interested in if there are any theorems which bound the error from below, i.e., does there exist $C_1 > 0$ such that
\begin{align*}
\left\| f - f_{h} \right\|_{\infty} \ge C_1h^{(\alpha+L)}?
\end{align*}
Obviously, if $f$ is a polynomial of degree $L$, no such constant exists, so we first must assume $f \in \mathrm{Lip}^{L,\alpha}[a,b]$, but not in $f \in \mathrm{Lip}^{L,\alpha + \epsilon}[a,b]$. Second, the set of points at which $f$ is not infinitely differentiable must be infinite, or else we can just divide $[a,b]$ into subintervals where everything converges faster.
Essentially, I want to know if the Hölder regularity of $f$ somehow fundamentally limits how fast we can compute approximations to $f$. (For example, the same sort of convergence rates hold for trigonometric polynomial approximations as well; this is Jackson's theorem.)
If the set $K_f$ of points at which $f$ is not infinitely differentiable is infinite, at the limiting points of $K_f$ it seems unclear(less possible) if we can have such a lower bound.
|
2025-03-21T14:48:29.944881
| 2020-02-27T17:47:45 |
353710
|
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"url": "https://mathoverflow.net/questions/353710"
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|
Stack Exchange
|
Explicit formula for this distance between positive semi-definite matrices?
Let $A$ and $B$ in $\mathbb{R}^{d\times d}$ be positive semi-definite (psd) matrices and let $d\tau$ be the uniform probability distribution on the unit sphere $\mathbb{S}^{d-1}$ in $\mathbb{R}^d$. I am interested in the quantity
$$
S(A,B)^2 := \int_{\mathbb{S}^{d-1}} (\Vert \theta\Vert_A - \Vert \theta\Vert_B)^2 d\tau(\theta)
$$
where $\Vert \theta \Vert_A = \sqrt{\theta^\top A\theta}$ is the Mahalanobis (semi)-norm on $\mathbb{R}^d$. Is it possible to give a simpler expression for this quantity? Or at least non-trivial lower bounds?
Where this comes from: A well known distance between two psd matrices $A$ and $B$ is the Bures-Wasserstein distance given by
$$
D(A,B)^2 := \mathop{tr}{A} + \mathop{tr}{B} - 2 \mathop{tr}{(A^{1/2}BA^{1/2})^{1/2}}.
$$
It equals to the $2$-Wasserstein distance between the Gaussian distributions $\mathcal{N}(0,A)$ and $\mathcal{N}(0,B)$ (see, e.g.~Bhatia et al.). Here, I am interested by the "sliced" or "Radon transform" version of this distance, obtained by averaging the squared-distance over all $1$ dimensional projections of the Gaussians (which, for a projection direction $\theta \in \mathbb{S}^{d-1}$, are Gaussians of variance $\theta^\top A\theta$ and $\theta^\top B\theta$). The expression above indeed satisfies
$$
S(A,B)^2 = \int_{\mathbb{S}^{d-1}}D(\theta^\top A \theta, \theta^\top B \theta)^2 d\tau(\theta).
$$
Your problem can be reduced to that of computing the quantity $z(A,B) := \mathbb E_{\theta \sim \tau}[|\theta|_A|\theta|_B]$. This is because the other terms have trivial formulae, since they are expectations of quadratic forms of a random vector whose covariance matrix you know explicitly. I don't think you can compute $z(A,B)$ in closed-form for general $A$ and $B$. You can definitely obtain (possibly crude!) lower and upper bounds for $z(A,B)$ in terms of the eigenvalues of $A$ and $B$.
|
2025-03-21T14:48:29.945028
| 2020-02-27T18:05:52 |
353713
|
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"authors": [
"Klim Efremenko",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353713"
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|
Stack Exchange
|
Modular representations of GL(n,q)
I wonder what is a good source to read about Modular representations of GL(n,q).
The specific question I am interested in is $GL(n,q)$ acts on $X=F_q^n$ in a natural way. If say q is prime and $q>n$ then all subrepresentations of $F_q^X$ are polynomials of degree at most $d$.
We can consider the action of $GL(n,q)$ on $X\times X$ by $g\cdot(x,y)=(gx,gy)$.
Does it known or how one can calculate all subrepresentations of $F_q^{X\times X}$?
$F_q^X$ is just permutation representation
|
2025-03-21T14:48:29.945089
| 2020-02-27T18:42:51 |
353715
|
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|
Stack Exchange
|
An identity for polynomials over partitions
Given an integer partition $\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$ of $n$ where $\ell(\lambda)$ is the length of $\lambda$, associate its conjugate partition $\lambda'$. Denote by $\lambda''=\lambda',0$ found by appending one extra zero at the right end of $\lambda'$. Further, define the following two numerics $a(\lambda'')_j=\lambda_j''-\lambda_{j+1}''$ for $j=1,2,\dots,\ell(\lambda')$ and also that
$b(\lambda'')=\#\{j: a(\lambda'')_j>0\}$.
For example, if $\lambda=(4,2,1)$ then $\lambda'=(3,2,1,1)$ and $\lambda''=(3,2,1,1,0)$ and $a(\lambda'')=(1,1,0,1)$ and $b(\lambda'')=3$.
QUESTION. If $n=2^m$ then are these two polynomials equal?
$$\sum_{\lambda\vdash n}(q-1)^{2b(\lambda'')}q^{n-\ell(\lambda)}
\prod_{a(\lambda'')_j\geq1}\frac{q^{2a(\lambda'')_j}-1}{q^2-1}=(q-1)(q^{2n-1}-1).\tag1$$
Remark 1. To get some motivation, consider dividing the left-hand side of (1) by $(q-1)^2$, for any $n\in\mathbb{N}$. Taking the limit $q\rightarrow1$ in the resulting expression forces $b(\lambda'')=1$ which means the corresponding Young diagram of the partition $\lambda'$ (hence $\lambda$ itself) must be rectangular. Therefore, the final expression equals the sum of divisors (arithmetic) function
$$\sum_{d\,\vert\, n}d.$$
Remark 2. I also observe that if $q\rightarrow-1$ in (1), then the left-hand side counts the number of ways of writing $n\in\mathbb{N}$ as a sum of two squares, which is this sequence $r_2(n)$.
Some powers of $q-1$ cancel in the l.h.s. to become:
$$\sum_{\lambda\vdash n} q^{n-\ell(\lambda)}
\prod_{a(\lambda'')_j\geq1}\frac{(q^{2a(\lambda'')_j}-1)(q-1)}{q+1}$$
@MaxAlekseyev: that's nice.
Is this true for $n \neq 2^m$ too?
Equation (1) seems true for $n=2^m$ while Remark 1 and 2 should hold for any $n\in\mathbb{N}$. Does that answer your question?
Yes, your identity $(1)$ is true. We can give a proof as follows:
Let's denote the left hand side of your identity $(1)$ by $A_n(q)$. Starting with the identity
$$\prod_{i\geq 1}\left(1+\sum_{r\geq 1}a_r(x_1x_2\cdots x_i)^r\right)=\sum_{\lambda}\left(\prod_{j\geq 1}a_{\lambda_j-\lambda_{j+1}}\right)\left(\prod_{j\geq 1}x_j^{\lambda_j}\right)$$
where $a_0$ is taken to be $1$, and the $a_i, x_i$ are formal variables, we make the substitutions $a_r=(q-1)^2\frac{q^{2r}-1}{q^2-1}$ for $r\geq 1$, $x_1=t$ and $x_i=qt$ for $i\geq 2$. This turns the right hand side into a generating function for the $A_n(q)$ where $A_0(q)$ is taken to be $1$. More specifically it gives
$$\sum_{n\geq 0}A_n(q)t^n=\prod_{i\geq 1}\left(1+(q-1)^2\sum_{r\geq 1}\frac{q^{2r}-1}{q^2-1}(q^{i-1}t^i)^r\right)=\prod_{i\geq 1}\frac{(1-q^it^i)^2}{(1-q^{i-1}t^i)(1-q^{i+1}t^i)}$$
$$=(1-q)\frac{(qt;qt)^2_{\infty}}{(t;qt)_{\infty}(q;qt)_{\infty}}.$$
This final product has a Hecke-Rogers type expansion that was given by Andrews in "Hecke modular forms and the Kac-Peterson identities" (see Lemma 1).Using this expansion we get
$$\sum_{n\geq 0}A_n(q)t^n=(1-q)\sum_{N\in \mathbb Z, r\geq |N|}(-1)^{r+N}q^{-N}(qt)^{\frac{(r+N)(r-N+1)}{2}}$$
and if we focus on $n=2^m$ we notice that the only way we can have $(r+N)(r-N+1)=2^{m+1}$ is if $(r,N)=(2^m,2^m)$ or $(r,N)=(2^m, 1-2^m)$. This means that $$A_{2^m}(q)=(1-q)(1-q^{2^{m+1}-1})$$
as desired. The observations in the remark can also be deduced from this last summation.
Great job, thank you.
|
2025-03-21T14:48:29.945420
| 2020-02-27T19:22:38 |
353719
|
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"Dan Rust",
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|
Stack Exchange
|
Entropy-minimal subshifts
Consider a subshift $X \subset \left\{0, \ldots, M \right\}^{\mathbb{N}}$. $X$ is said to be entropy-minimal if every subshift $Y \subsetneq X$ satisfies that $$h_{\mathrm{top}}(Y) < h_{\mathrm{top}}(X).$$ Equivalently, $X$ is entropy-minimal if for every word $\omega \in \mathcal{L}(X)$ the subshift $$ X_{\omega} = \left\{x \in X : \omega \text{ is not a factor of } x \right\}$$ satisfies that $h_{\mathrm{top}}(X_{\omega}) < h_{\mathrm{top}}(X)$.
It is a "folklore" result that irreducible subshifts of finite type of are entropy-minimal as well as subshifts with the specification property. Weaker notions of the specification property and entropy-minimality have been studied recently by Climenhaga, García-Ramos and Pavlov.
What I am looking for is a concrete example of a (one-dimensional) subshift $X$ that is non-entropy-minimal, transitive, and has positive topological entropy. It will be interesting if such example is a binary subshift.
Could one not take something like a positive entropy Toeplitz shift, the Thue-Morse shift, and then an extra 'generating element' which cycles through the language of both as you shift in either direction (in order to achieve transitivity). Then, I'm pretty sure the only proper subspaces which are subshifts are the two minimal components, and adding the TM shift and the generating element won't generate enough new words to increase the entropy, so the Toeplitz shift has the same entropy as the whole shift.
Or in fact, far easier, forget the TM component. Just take a Toeplitz shift with entropy >0, then take any element of that shift and insert some new illegal word somewhere in the middle. This won't increase the entropy, by minimality of the Toeplitz shift, this new element will have a dense orbit.
Dan, I will check what are you saying. I'm not familiar with Toeplitz subshifts. Fortunately, I found your work, so I will read it. Thanks
Let $f$ be a sublinear function that tends to infinity, such as $f(n) = \sqrt{n}$. Define $X \subset \{0,1,2\}^{\mathbb{N}}$ by forbidding all long enough words $w$ with more than $f(|w|)$ occurrences of $2$. Then $X$ has entropy $\log 2$ and is mixing, and properly contains the binary full shift, which likewise has entropy $\log 2$.
If you need a binary example, just replace the alphabet with three suitable binary words, like $01$, $001$ and $0001$.
|
2025-03-21T14:48:29.945599
| 2020-02-27T19:50:11 |
353722
|
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"url": "https://mathoverflow.net/questions/353722"
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|
Stack Exchange
|
Proving a specific case of Robin's Inequality
Edit: It turns out that this is equivalent to the RH which gives the idea that this might a a little difficult to show. As such we could consider an even simpler case in which the number $n$ is squarefree (all values $k_j$ are equal to $1$. In previous papers it has been shown that squarefree numbers satisfy Robin's Inequality, but is this still the case for $2^kn$? If we make this loose condition we find our simpler form of
$$
\dfrac{2^{k+1}-1}{2^{k}}\prod_{j=1}^m\dfrac{p_j+1}{p_j} < e^\gamma\log\log\left(2^k\prod_{j=1}^m p_j\right)
$$
with a minima of $f(k)$ at
$$
k_{min} = -\dfrac{W_{-1}\left(-e^\gamma\prod_{j=1}^m \frac{1}{p_j+1}\right)+\log\left(\prod_{j=1}^m p_j\right)}{\log2}
$$
if we plug this in to our inequality we need to show that
$$
e^\gamma\log\left(-W\left(-e^\gamma\prod_{j=1}^m\frac{1}{p_j+1}\right)\right) > \dfrac{e^{-W\left(-e^\gamma\prod_{j=1}^m\frac{1}{p_j+1}\right)-\log\left(\prod_{j=1}^m p_j\right)+1}-1}{e^{-W\left(-e^\gamma\prod_{j=1}^m\frac{1}{p_j+1}\right)-\log\left(\prod_{j=1}^m p_j\right)}}\prod_{j=1}^m\dfrac{p_j+1}{p_j}
$$
which I will admit is disgustingly messy, but looks (at least naively to me) potentially tractable since prime product and inverse prime product series are well studied.
One reformulation of the Riemann Hypothesis is Robin's Inequality which states that for $n>5040$ the following holds iff the RH holds:
$$
\sigma(n) < e^\gamma n\log\log(n)
$$
where $\sigma$ is the sum of divisors function and $\gamma$ is the Euler Mascheroni Constant. Now for my specific case I want to show that given given some number $n=p_1^{k_1}p_2^{k_2}\ldots p_m^{m} > 5040$ where $p_j \neq 2$ is a prime number, if Robin's Inequality holds for $n$, then it must also hold for $2^k\cdot n$. Performing some algebra on the inequality we can see that this is the same as showing that if the following inequality holds
$$
\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)} < e^\gamma\log\log\left(\prod_{j=1}^m p_j^{k_j}\right)
$$
then this inequality must hold as well
$$
\dfrac{2^{k+1}-1}{2^{k}}\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)} < e^\gamma\log\log\left(2^k\prod_{j=1}^m p_j^{k_j}\right)
$$
I'll admit that I am not well versed in Analytic Number Theory, so this might be obvious and I have no idea, but so far I have only been able to show three fairly trivial things
According to numerical computations this seems to hold true. For large values of $n$ it appears that the R.H.S. is strictly larger that $2$ times the L.H.S. in the assumed inequality.
Since the left side is bounded with respect to $k$ and the right side is not, there must exist some $N$ for which if $k\geq N$ then the inequality holds. Therefore there are only finite cases for which this inequality may not hold.
Taking the difference of the left and right sides as this
$$
f(k) = e^\gamma\log\log\left(2^k\prod_{j=1}^m p_j^{k_j}\right) - \dfrac{2^{k+1}-1}{2^{k}}\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)}
$$
has a derivative where $f'(0) < 0$ and $f'(N) > 0$, as such there exists a local minima of $f(k)$ which we can find to be the following value
$$
k_{min} = -\dfrac{W_{-1}\left(-e^\gamma\prod_{j=1}^m \frac{p_j-1}{p_j^{k_j+1}-1}\right)+\log\left(\prod_{j=1}^m p_j^{k_j}\right)}{\log2}
$$
where $W_{-1}$ is the second, more negative, solution of the Lambert W function when the argument is between $0$ and $-\frac{1}{e}$. The derivative also appears to be strictly positive past $k_{min}$ as
$$
f'(k) = \dfrac{e^\gamma \log(2)}{k\log 2 + \log\left(\prod_{j=1}^m p_j^{k_j}\right)} - \dfrac{\log(2)}{2^k}\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)}
$$
as this will behave like $\frac{1}{k} - \frac{1}{2^k}$ where the negative part decreases significantly faster than the positive part.
If anyone could offer some potential insight that would be much appreciated!
By Theorem 1.2 in this paper, Robin's Inequality is true for every odd integer $n>10$. If we knew what the OP wants to prove, then we would also know Robin's Inequality for every integer $n$ whose odd part exceeds $5040$. In particular, we would know Robin's Inequality for every colossally abundant number exceeding $5040$, because each colossally abundant number divides the second next one (cf. Proposition 4 in this paper). So, by Proposition 1 in Section 3 of Robin's paper, we would know Robin's Inequality for every integer exceeding $5040$, which is equivalent to the Riemann Hypothesis.
In short, it is hopeless to prove what the OP wants to prove, because it implies the Riemann Hypothesis.
Well that certainly is a bummer, I thought it was a simpler form but it turned out to be identical haha!
|
2025-03-21T14:48:29.945859
| 2020-02-27T22:32:11 |
353730
|
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|
Stack Exchange
|
Eigenvalue distribution of a random matrix
Is there any closed form distribution formula for the distribution of the eigenvalues of $\mathbf{X}^\mathrm{H}\mathbf{X}$ where the entries of $\mathbf{X}$ are independent Gaussian random variables with different variances $\sigma_{ij}^2$. Moreover, in the diagonal factorization $\mathbf{X}^\mathrm{H}\mathbf{X}=\mathbf{U}\mathbf{\Lambda}\mathbf{U}^{\mathrm{H}}$, the unitary matrix $\mathbf{U}$ has Haar distribution? Is it independent of $\mathbf{\Lambda}$?
no, $U$ is not uniformly distributed in the unitary group (no Haar distribution) and eigenvectors and eigenvalues are not independent. There is not much more to say in this completely general case.
Is there any result for non-identical cases?
More of a comment to indicate how hopeless this is: Consider the edge case of $\sigma_{ij} = \lambda_i \delta_{ij}.$ The unitary matrix then is the identity matrix, so is quite far from Haar distributed. The eigenvalues are distributed normally with the covariance matrix $\mathbf{X},$ also quite far from the "standard" case. An interesting question is whether there is a less trivial case when there is a discrete set of $\mathbf{U}.$
Q: "Is there any result for non-identical cases?"
The eigenvalue distribution function is known if $\sigma^2_{ij}=w_i$ depends only on the row index, see for example arXiv:1310.2467
$$P(x_1,x_2,\ldots x_n)\propto\prod_{i<j}|x_i-x_j|^\beta\prod_k x_k^{1-2/\beta}\int dU \exp\left(-\tfrac{1}{2}\beta\,{\rm tr}\,UXX^H U^{\rm H}W^{-1}\right),$$
where $W={\rm diag}\,(w_1,w_2,\ldots w_n)$ and $\beta=1$ for real matrix elements, $\beta=2$ for complex matrix elements.
The measure $dU$ is the Haar measure in ${\rm O}(n)$ for $\beta=1$ and in ${\rm U}(n)$ for $\beta=2$. This is the socalled Harish-Chandra-Itzykson-Zuber integral, which has a closed-form expression for $\beta=2$ (it’s a ratio of determinants). Note that the integral depends only on the eigenvalues $x_i$ of $XX^H$. For $W$ proportional to the identity matrix we recover the usual Wishart distribution.
|
2025-03-21T14:48:29.946022
| 2020-02-27T22:35:06 |
353731
|
{
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"Michael Engelhardt",
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|
Stack Exchange
|
Is there any relation between moments of random matrix and its eigenvalue distribution?
Let $\mathbf{X}$ be a random matrix with independent Gaussian random variable entries with different variances $v_{ij}$. Also define $\mathbf{A}=\mathbf{X}^\mathrm{H}\mathbf{X}$. Is there any relation between the distribution of the eigenvalues of $\mathbf{A}$ and moments of $\mathbf{A}$, i.e.,
$$\mathbf{B}_m=\mathbb{E}[\mathbf{A}^m]$$ for integers $m\geq 1$?
I'm not familiar with the notion of a moment being matrix-valued. Are you sure you don't mean something like ${\rm Tr}((A^m)^*A^m)$?
Moreover, can you specify what class of random matrices you are considering? Square? Symmetric/hermitian? etc
If, as Yemon Choi suspects, your moments of $\mathbf{A} $ are traces of powers thereof, then you can just calculate them in the eigenbasis, where they are directly moments of the eigenvalue distribution.
What is the relation between $\mathrm{Tr}[A^m]$ and eigenvalue distribution? Could you please explain with more details?
the expectation of ${\rm tr}, A^m$ depends only on the marginal distribution of a single eigenvalue, so it contains no information on the correlations between the eigenvalues.
If you know the eigenvalue distribution $\rho_{A} (\lambda )$, then $\mbox{Tr} [A^m] = \int d\lambda \lambda^{m} \rho_{A} (\lambda )$. This is even true for a single matrix $A$, before averaging over an ensemble.
|
2025-03-21T14:48:29.946151
| 2020-02-27T23:14:31 |
353732
|
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|
Stack Exchange
|
Given an abelian galois map of curves, what are the principal divisors on the source fixed by the galois group?
Let $f:X\rightarrow Y$ be an abelian galois map (not necessarily unramified) of nonsingular complete curves over algebraically closed $k$, where the order of the galois group $A$ is coprime to the characteristic of $k$. We can view the function field $K(X)$ as a $K(Y)$ vector space, and by galois theory we know its structure as a $k[A]$ module is given by $K(Y)[A]$.
So given our assumptions, we see that $K(X)$ has a basis of eigenvectors for $A$, corresponding the the distinct one dimensional representations of $A$ over $k$. Explicitly, it has a basis of $f_\lambda\in K(X)$ such that $g.f_\lambda=\lambda(g).f$ for $\lambda(g)$ an element of $K(Y)$. So the divisors associated to these $f_\lambda$ are canonical up to principal divisors from $Y$, and we have one for each character of $A$.
My question is, can you describe these divisors more geometrically?
I am vaguely aware that all coverings of this form should come from isogenies of the jacobian, but my knowledge of abelian covers of curves is not great, so apologies if this question is something really simple.
Assume that $Gal(K(X)/K(Y))$ is cyclic of degree $n$. Then $K(X)=K(Y)[f^{1/n}]$ and $K(X)/K(Y)$ is unramified (so $X\to Y$ is a covering) iff every zeros and poles of $f$ are of order $ln$. Thus the unramified coverings of $Y$ correspond to the $D\in Div^0(Y)$ such that $nD\in Prin(Y)$ through $nD=Div(f)$ ie. to the points of order $n$ in $Pic^0(Y)$. The non-cyclic case is to generate an abelian extension from any finite subgroup of $Pic^0(Y)$.
Sorry, I'll edit the question, I used covering forgetting the technical meaning, I just mean a nonconstant map of curves, possibly with ramification. I'll think about your comment further, though I don't see how the noncyclic case works, but that's probably because I'm not comfortable with $Pic^0$.
For non-cyclic abelian unramified coverings let $K(X)=K(Y)[f_1^{1/n},\ldots,f_j^{1/n}]$ where $Div(f_j)$ correspond as above to cyclic coverings. I don't think the ramified cyclic extensions correspond to some points of the Jacobian (which is isomorphic to $Pic^0$) try with $K(Y)$ an elliptic curve, it is its own Jacobian and the finite subgroups correspond to isogenies which are unramified. It might work if you replace $Pic^0$ by the class group of $O_Y(Y-{p_1,\ldots,p_l})$ where $p_l$ contain the ramified points of your branched covering.
|
2025-03-21T14:48:29.946374
| 2020-02-28T01:40:48 |
353737
|
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"authors": [
"Eduardo Longa",
"Misha",
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|
Stack Exchange
|
Embeddedness and homology of a limit of minimal surfaces
Consider the following theorem, proved in
this paper:
Theorem (Theorem 6.1). Suppose we have a sequence $(\Sigma_j, \partial \Sigma_j) \subset (M, \partial M)$ of immersed free boundary minimal $k$-dimensional submanifolds, where $1 \leq k \leq n$, with uniformly bounded area and second fundamental form. Then, after passing to a subsequence, $(\Sigma_j, \partial \Sigma_j)$ converges smoothly and locally uniformly to $(\Sigma, \partial \Sigma) \subset (M, \partial M)$, which is a smooth immersed free boundary minimal $k$-dimensional submanifold.
My questions are:
Question 1: If we assume additionally that all the submanifolds in the sequence are embedded, is it true that the limit surface is also embedded?
Question 2: If $M$ has dimension $3$ and
$(\Sigma_j, \partial \Sigma_j)$ is a sequence of compact, connected, oriented and properly embedded free boundary minimal surfaces that converges as in the theorem to $(\Sigma, \partial \Sigma)$, is it true that there exists $N \geq 1$ such that $ [\Sigma_j] = [\Sigma] \in H_2(M, \partial M; \mathbb{Z})$ for all $j \geq N$?
Question 3: Let $M$ be compact, connected and oriented of dimension $3$ and $(\Sigma_j, \partial \Sigma_j)$ be a sequence of compact, connected, oriented and properly embedded free boundary minimal surfaces that converges as in the theorem to $(\Sigma, \partial \Sigma)$. If $[\Sigma_j] \neq 0 \in H_2(M, \partial M; \mathbb{Z})$ for every $j \geq 1$, is it true that $[\Sigma] \neq 0$?
What kind of limits do you mean? For instance, on a Moebius band with flat metric you can have a sequence of embeded closed geodesics converging (as maps) to a non-embedded one.
I admit this would be my 0th question. The authors just say what I transcribed.
(I think I am partly repeating what Misha said but:) The answer to Question 1. I think is trivially "no". These surfaces are arbitrary codimension so e.g. two disjoint lines can meet in the limit. For the second question I'm not sure. Does smooth convergence of the pair $(\Sigma_j, \partial \Sigma_j)$ mean that the boundaries converge smoothly in $\partial M$?
Yes, the boundaries converge smoothly.
|
2025-03-21T14:48:29.946558
| 2020-02-28T02:37:41 |
353738
|
{
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"Jeremy Rickard",
"Rahul Sarkar",
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|
Stack Exchange
|
Adding $n$-tuples over groups
Consider a finite abelian group $\mathcal{G}$. Let $S_0$ be a $n$-tuple of elements of $\mathcal{G}$, and let $S_i$ be the cyclically shifted version of $S_0$ by $i$ indices to the right. So for example if $\mathcal{G}$ = $\text{GF}(2)$, $n = 3$, and $S_0 = (1,0,0)$, then $S_2 = (0,0,1)$.
Now let $\mathcal{S} = \{S_i : 0 \leq i \leq n-1\}$. One can add two $n$-tuples in $\mathcal{S}$ by adding the elements over the group $\mathcal{G}$. So in the above example, $S_0 + S_2 = (1,0,1)$.
The basic question is this: Pick any $g \in \mathcal{G}$. I want to choose a subset $\mathcal{H} \subseteq \mathcal{S}$ so that the number of times the element $g$ appears in the tuple $\sum_{h \in \mathcal{H}} h$ (let's call this the $g$-count) is maximum. I'm not really interested right now in the subset itself, but really in the maximum $g$-count.
Does anybody know what kind of results are known about this problem in general? Or what kind of techniques do you suspect are needed to tackle this problem? I'd also be interested if any special cases are known.
Edit: In view of the answer below, I'd like to mention that a special case that I'm interested in is when $\mathcal{G} = \mathbb{F}_4$, with the group operation being addition in $\mathbb{F}_4$.
Is $\mathcal {G} $ abelian?
@JeremyRickard Yes, in the case I'm interested in $\mathcal{G}$ is abelian.
You should probably add that to the question, then. If $\mathcal{G} $ is not abelian, then it's not clear what $\sum_{h\in\mathcal{H}}h $ means.
@JeremyRickard edited the question. Thanks!
It might help to know more about which cases interest you.
Here is an NP complete problem:
given a set of integers which add to $2N$ , decide if there is a subset which adds to $N.$
This translates into deciding for your question if the $g=N$-count for a certain vector over $\mathbb{Z}_{2N}$ is positive or $0.$
Over $G=\mathbb{Z}_2$ the only cases are $g=0$ and $g=1.$ The length $n$ all $0$ vector arises from taking $\mathcal{S}=\emptyset.$ Adding all the vectors in $\mathcal{S}$ gives a constant vector. All $1$'s if $S_0$ has an odd number of $1$'s and all $0$ otherwise.
Just to make it more interesting one might require that $\sum_{h \in \mathcal{H}} h$ is not a constant vector.
The triple $(1,1,0)$ gives only even weight words so the length $24$ tuple $(1,1,0,1,1,0,\cdots$ only gives $4$ distinct sums $\sum_{h \in \mathcal{H}} h$ and the non-zero ones all have $16$ $1$'s.
Over $\mathbb{Z}_2$ the possible tuples $\sum_{h \in \mathcal{H}} h$ constitute a linear subspace of $\mathbb{Z_2}^n.$ This is a cylic binary code and the question of the weight enumerator polynomial is well studied. There is a binary Golay code where the tuples have length $24$ and each one which isn't constant has $8,12$ or $16.$ This is no better numerically than the relatively trivial construction above, but more impressive.
Over $\mathbb{Z}_p$ for a larger $p$ we would get a linear code by taking sums $\sum_{h \in \mathcal{H}} c_h h$ with coefficients $c_h \in \mathbb{Z}_p.$ but sticking to $\sum_{h \in \mathcal{H}} h$ gives a subset of this code. There is a ternary Golay code over $\mathbb{Z}_3$ with $n=12$ where no one of $0,1,2$ can show up more than $6$ times in a non-constant tuple.
In summary:
Whatever $S_0$ is, there is at least one $g'$ so that the $g'$-count is can be $n$. This is the sum of the entries of $S_0.$
Then the highest $g$ count is also the highest $g'-g$ count, use complementary sets $H$ and $\mathcal{S}-H.$
$g=0$ can always be obtained $n$ times by the empty sum.
We may assume that the entries of $S_0$ generate the entire group, since otherwise we can just use a smaller group.
It is worth looking at the theory of cyclic linear codes.
To see how small the number could be, consider well known codes.
In the particular case of $\mathbb{Z}_4$ this suggests:
Treat separately the nine cases of $g=0$, $g=2$ and $g=1,3$ and and $g'=0$,$g'=2$, $g'=1,3.$
Check Preperata, Kerdock and other interesting $\mathbb{Z}_4$ codes.
A special case that I'm interested in is when $\mathcal{G} = \mathbb{F}_4$, with the group operation being addition in $\mathbb{F}_4$. I have added a comment to the question, in case it helps.
|
2025-03-21T14:48:29.946877
| 2020-02-28T02:41:05 |
353740
|
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|
Stack Exchange
|
Singular schemes with a torus action and embedded points
I've got a couple rather geometric questions about the following setup.
Let $X$ be a scheme over an algebraically closed field ($\mathbb{C}$, say) with the action of a torus $T$, such that the action has finitely many fixed points. $X$ may have some mild singularities but is normal, Cohen-Macaulay, etc. Let $\mathscr{L}$ be a $T$-equivariant line bundle on $X$.
We can form the $T$-fixed subscheme $X^T$. In general, the $\mathbb{C}$ points of $X^T$ will be some discrete set, but $X^T$ need not be reduced, so we should think of $X^T$ as some nonreduced scheme supported at some finite set of points.
Furthermore, let me consider a rational curve $Y \hookrightarrow X$, stable under the $T$ action, joining two $T$-fixed points $p_1$ and $p_2$ in $X$. If I assume that $Z$, the component of $X^T$ supported at $p_2$, is nonreduced, I could consider the scheme $Y'=Y \cup_{p_2} Z$ (I am not sure exactly how to make this more precise: do I want a pushout of these two schemes?).
Question 1: In the above setup, am I justified in viewing this scheme $Y'$ as $Y$ along with some embedded points (corresponding to the non-reduced points over $p_2$)?
Question 2: Let $I$ be the ideal sheaf of $X^T$ in $X$. Assume that I know how $\mathscr{L}$ restricts to $Y$; $\mathscr{L}|_Y=\mathscr{O}_{\mathbb{P}^1}(n)$, say.
Now consider the restriction $(I \otimes \mathscr{L})|_{Y}$. The sections of this sheaf over $Y$ will be some subset of $\Gamma(Y,\mathscr{O}_Y(n))$, vanishing at the $T$-fixed points $p_1$ and $p_2$. Note moreover that $I \otimes \mathscr{L}|_Z=0$, since $I$ vanishes identically on the $T$-fixed locus. My question is, do I gain information about orders of vanishing of sections $s \in \Gamma(Y, (I \otimes \mathscr{L})|_Y)$ at $p_2$?
In other words, if a section vanishes at an embedded point of the curve, can I see that reflected in higher orders of vanishing of that section restricted to the closed points of the curve?
I'm happy to expand or clarify any bit of this (it's rather sketchy) and would be nearly as happy with a good reference as a direct answer.
|
2025-03-21T14:48:29.947412
| 2020-02-28T03:30:16 |
353741
|
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|
Stack Exchange
|
Commonly used metrics to compare two Young measures
Let $\Omega\subset \mathbb{R}^n$ be a bounded open set, $K\subset \mathbb{R}^d$ be a compact set, and $M_1(K)$ be the set of probability measures on $K$. Then a Young measure is defined as a $\textrm{weak}^*$ measurable map:
$$
\nu:x\in \Omega\mapsto \nu(x)\in M_1(K).
$$
Now I want to compare two Young measures $\nu$ and $\mu$. However, since there are many distances defined on $M_1(K)$, I am not sure what is the most commonly used metric for Young measures. Is the choice of metric problem dependent?
In particular, I am interested in the case where $K$ is not finite. Since if $K$ is a finite set with cardinality $N$, then I can identify $M_1(K)$ as the probability simplex in $\mathbb{R}^N$, and hence view $\nu$ as a function
$$
\nu:x\in \Omega\to \nu(x)\in \mathbb{R}^N,
$$
and use the metric $\|\nu-\mu\|=\sup_{x\in \Omega}\|\nu(x)-\mu(x)\|_{\mathbb{R}^N}$, where $\|\cdot\|_{\mathbb{R}^N}$ denotes the Euclidean norm on $\mathbb{R}^N$.
I found in the literature that one can use the norm
$\|\nu\|_{\textrm{TV}}=\sup_{x\in \Omega}|\nu(x)|_{\textrm{TV}}$, where $|\cdot|_{\textrm{TV}}$ denotes the total variation norm of a probability measure. However, the TV norm seems to be too strong for my application. In particular, $(M_1(K),|\cdot|_{\textrm{TV}})$ is not separable if $K$ is uncountable, so it is not easy to approximate $M_1(K)$ by a finite dimensional sub-manifold and obtain convergence in $\|\cdot\|_{\textrm{TV}}$-norm.
I may be biased here due to the fact that my privileged topics these days is optimal transport, but have you tried Monge-Kantorovich-Wasserstein distances? You can approximate any probability measure by empiric (point) measures, and in fact convergence in such distances is roughly equivalent to weak (AKA "narrow" or "weak-*") convergence of measures (plus convergence of some moments)
|
2025-03-21T14:48:29.947593
| 2020-02-28T03:47:05 |
353742
|
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"GH from MO",
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|
Stack Exchange
|
Funny questions about Moebius Function
I need to firstly claim that my research is not about number theory, however, I am pretty interested in it, especially funny questions in number theory, e.g. Kollatz Conjecture. Three years ago, I attended a lecture in which the lecturer discussed about entropy and natural numbers. Since then, I became very interested in randomness of Moebius function, I came up with three funny questions as the following:
(1) Given a finite $\{0, 1, -1\}$ sequence (e.g.$\{0,0,1,-1,0,-1,\cdots\,1\}$), can we find a consecutive natural number sequence such that values of Moebius function of this natural number sequence is the given $\{0, 1, -1\}$ sequence ? (existence)
If there exists such a natural number sequence, is it unique ? (uniqueness)
If we remove the restriction of consecutivity on natural number sequence, what about existence and uniqueness ?
(2) Given a finite $\{0, 1, -1\}$ $K$-sequence equipped with a spacing sequence (e.g. $\{n_{1},n_{2},\cdots,n_{K-1}\}$), can we find a natural number sequence such that values of Moebius function of this natural number sequence is the given $\{0, 1, -1\}$ sequence and this natural number sequence satisfies the given spacing sequence ?
(3) General Question: Is there some deterministic pattern or probabilistic properties for values of Moebius function of the whole natural numbers ?
I proposed these questions only out of curiosity!!!
Note that every fourth number has $\mu=0$.
Mobius function is completely deterministic, just like the sequence of primes is, but for purposes of heuristic it can be treated as a more or less random function. One interesting fact is that the summatory function of Mobius function, known as the Mertens function, satisfies the same bounds as a uniform random walk, $M(x)=O(x^{1/2+\varepsilon})$, iff the Riemann hypothesis holds.
It is trivial (by the pigeonhole principle) that there is a sequence of every finite length with entries from ${0,1,-1}$ that occurs as consecutive values of the Möbius function infinitely often. See also https://arxiv.org/abs/1904.05096 for state-of-the-art nontrivial results in the subject.
@Wojowu Thank you so much! I think I can propose a new hypothesis that Riemann hypothesis can appear everywhere in number theory. :)
@GHfromMO Thank you for your answer. I will check that paper later.
@cheng: It is a most natural thing that the Riemann Hypothesis is relevant for the Möbius function, since $\sum\mu(n)/n^s$ equals $1/\zeta(s)$.
@GHfromMO Yes. There are many formulas which link Riemann's $\zeta$ function with number theoretical functions such as Moebius function. What I feel surprised is there are so many equivalent versions of Riemann hypothesis. Maybe, from those formulas, it's a natural thing.
@cheng: Well, this is the power of complex analysis. For example, the poles of $1/\zeta(s)$ and $\zeta'(s)/\zeta(s)$ are the same as the zeros of $\zeta(s)$. If you are interested in the statistical behaviour of the Möbius function, then analytic number theory is your friend. See also this brand new preprint for a different, but related research direction: https://arxiv.org/abs/2002.12421
@Wojowu Yes. Mobius function is completely deterministic. However, it depends on prime decomposition of integer or identification of prime if the given integer is a prime. So, to some degree, Moebius function can characterize distribution of primes which is a central topic in number theory. I am interested in statistical behaviour of Moebius function, in particular, when the given integer is large enough, and prime decomposition is not tractable, if there exists an appropriate approximation to Moebius function $f : \mathbb{Z} \rightarrow {-1,0,1}$, that will be great.
Cont: Actually, if we can find such $f$ satisfying some accuracy requirement theoretically or practically, we find an appropriate approximation to distribution of primes. As inspired in comments by @Wojowu, Moebius function can be regarded as a discrete-time stochastic process $X_{t} (t \in \mathbb{Z}), X_{t} \in {-1,0,1}$, an interesting analogue is $X_{t}$ can be regarded as stock price trending symbol, when $X_{t} = 1$, it means stock price increases at time $t$; similarly, when $X_{t} = -1$, stock price decreases; $X_{t} = 0$ means no change or ignorable change in stock price.
Cont: Based on observations in the past, we build a model to make predictions in the future. Back to reality, there is infinite number of integers, however, computer can only handle with finite number of integers. So, analogue makes sense here. Also, from fundamental theorem of primes, "density" of primes is low, corresponding to the fact stock price is almost always changing, $X_{t} = 0$ is rare.
@GHfromMO Thank you! I am interested in number theory, very mysterious! However, my complex analysis is ...tears.. My major is statistics, so I will try to dive deeper from statistical or probabilistical perspective. Interesting findings are expected!
|
2025-03-21T14:48:29.948037
| 2020-02-28T04:36:17 |
353743
|
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|
Stack Exchange
|
Integrality of Atkin-Lehner operator for $\Gamma_1(N)$
A result due to B. Conrad (http://math.stanford.edu/~conrad/papers/prasanna-inv.pdf, Theorem A.1) states that the Atkin-Lehner operator $w_{Q,k}$ is $\mathbb{Z}[1/Q]$-integral on $M_k(\Gamma_0(N))$. In other words, if $f\in M_k(\Gamma_0(N))$ has coefficients in $\mathbb{Z}[1/Q]$ then $w_{Q,k}(f)$ has coefficients in $\mathbb{Z}[1/Q]$.
Does anyone know of a corresponding result over $\Gamma_1(N)$? My guess is that $w_{k,Q}$ is $\mathbb{Z}[1/Q][\zeta_Q]$-integral on $M_k(\Gamma_1(N))$ but I can't find a reference for this.
I am however, aware of a weaker result (Theorem 5.4 in https://arxiv.org/abs/1807.00391), which implies that $w_{k,Q}$ is $\mathbb{Q}(\zeta_Q)$-integral on $M_k(\Gamma_1(N))$.
In the preprint The Manin constant and the modular degree http://www.math.u-psud.fr/~cesnavicius/Manin-degree.pdf the authors have proved bounds on the denominators of modular forms at the cusps, using adelic techniques (see Section 4). Maybe you could email them and ask whether their results are sufficient to prove what you want.
It is quite straightforward to prove that $w_Q$ is $\mathbf{Z}[1/N, \zeta_Q]$-integral on forms of level $\Gamma_1(N)$; the hard work in Conrad's note is to deal with integrality at primes dividing $N$ but not $Q$. Would this weaker result be sufficient for your purposes?
@DavidLoeffler actually yes that result would be sufficient. Would you be able to provide an outline of why it's true?
Theorem. Let $\ell$ be prime, and $Q, R \ge 1$ such that $(\ell, Q, R)$ are pairwise coprime. Let $N = QR$ and for simplicity assume $N \ge 4$. Then $W_Q$ preserves $M_k(\Gamma_1(N), \mathbf{Z}[1/N, \zeta_Q])$.
Proof. Let $M_k^{\mathrm{wk}}(\Gamma_1(N), A)$ denote the space of weakly modular forms (possibly meromorphic at the cusps) with $q$-expansions in the ring $A$. If $A$ is a subring of $\mathbf{C}$ then
$$M_k(\Gamma_1(N), A) = M_k^{\mathrm{wk}}(\Gamma_1(N), A) \cap M_k(\Gamma_1(N), \mathbf{C})$$
so it suffces to show that $M_k^{\mathrm{wk}}(\Gamma_1(N), \mathbf{Z}[1/N, \zeta_Q])$ is stable under $W_Q$.
If $A$ is a $\mathbf{Z}[1/N]$-algebra, then elements of $M_k^{\mathrm{wk}}(\Gamma_1(N), A)$ can be interpreted as rules sending "test objects" $(E, P_Q, P_R, \omega) / B$ to elements of $B$. Here $B$ is a $A$-algebra, $E$ is an elliptic curve over $B$, $P_Q$ and $P_R$ are points of exact order $Q$, $R$ respectively, and $\omega$ is a global differential on $E$. These have to satisfy various conditions (the main ones are compatibility with base change and homogeneity in $\omega$ of weight $k$).
So it suffices to show that $W_Q$ makes sense on test objects if $A = \mathbf{Z}[1/N, \zeta_Q]$. The map will send $(E, P_Q, P_R, \omega) / B$ to $(E', P_Q', P_R', \omega')/B$, where all but one of these objects are simple to define:
$E' = E/\langle P_Q \rangle$
$P_R' = \pi(P_R)$, where $\pi : E \to E / \langle P_Q\rangle$ is the quotient map
$\omega' = \pi_*(\omega)$
The hard one is $P_Q'$: one checks that $E[Q] / \langle P_Q\rangle$ has a unique generator $P_Q'$ characterised by the Weil pairing $\langle P_Q, P_Q'\rangle = \zeta_Q$ (and since it is unique, its formation is compatible with base-change). $\square$
(Caveat: this construction gives an operator whose square is something like $\langle Q\rangle_R Q^k$ where $\langle -\rangle_R$ denotes the diamond operator for something that is 1 (mod Q) and Q (mod R). Some people prefer to normalise away the $Q^k$, but this may not be possible if $Q$ is odd without introducing an extraneous $\sqrt{Q}$ into your ring.)
Thank you very much, this is very useful! Do you know whether there is an even simpler integrality statement if we instead restrict to modular forms with character? That is, would the Atkin-Lehner operator $w_{Q,k}$ be $\mathbb{Z}[1/N,\zeta_d]$-integral for some $d$ related to the conductor of the character? If there isn't a clear answer to this I'm happy to make a new question.
If you're working with $\Gamma_0(N) \cap \Gamma_1(M)$ (i.e. level $N$ and characters of conductor dividing $M$) then the exact same argument shows $w_Q$ is defined over $\mathbf{Z}[1/N, \zeta_d]$ where $d = GCD(M, Q)$. (But this isn't really an "integrality" result as such, because the $1/N$ is still there, it's just controlling the roots of unity a little better.)
|
2025-03-21T14:48:29.948331
| 2020-02-28T04:58:46 |
353744
|
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|
Stack Exchange
|
Eight-dimensional hypersurface related to $SU(3)$
$\lambda_i$, $i=1,\ldots 8$ being Gell-Mann matrices, symmetric $d_{ijk}$ and antisymmetric $f_{ijk}$ $SU(3)$-tensors are defined through the multiplication law (see https://projecteuclid.org/euclid.cmp/1103841153 Einstein summation convention over repeated indexes is assumed)
$$\lambda_i\lambda_j=\frac{2}{3}\delta_{ij}+(d_{ijk}+if_{ijk})\lambda_k.$$
When studying the Three-Higgs-doublet models, the following hypersurface appears in the 9-dimmensional $(r_0,r_1,\ldots ,r_8)$ space (see, for example https://arxiv.org/abs/1901.11472 ):
$$d_{ijk}r_ir_jr_k+\frac{1}{2\sqrt{3}}r_0(r_0^2-3r_ir_i)=0.$$
Is anything known about such hypersurfaces in the mathematical literature?
|
2025-03-21T14:48:29.948404
| 2020-02-28T07:24:06 |
353748
|
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"Denis T",
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|
Stack Exchange
|
Controlling the intersection of two surfaces in $\mathbb{R}^3$
Let $F_1,F_2$ be two closed orientable surfaces embedded in $\mathbb{R}^3$ with genus $2g_1, 2g_2$, respectively (edit: with $g_1, g_2 \geq 1$). Is it possible to isotope around $F_1$ and $F_2$ so that $F_1 \cap F_2$ is a single curve that splits $F_1$ into two surfaces of genus $g_1$ and splits $F_2$ into two surfaces of genus $g_2$?
I would also be interested in the setting where we have any two closed $n$-manifolds in $\mathbb{R}^{n+1}$ and they intersect them in some $(n-1)$-manifold that splits both of the manifolds. What can be said about the pieces? Are there some obstructions to what can happen in this situation?
Certainly $F_i$ can be written as the gluing of two $g_i$ surfaces connected by a cylinder $C_i$ and it should not be difficult to make $C_1\cap C_2$ into a curve homeomorphic to a circle (two non crossing handles). Am I wrong here? In general I do not see how the topology of the intersection can have any effect on the global topology of the surface (which may be modified adding handles away from the intersection).
I think the answer is no. For example, if $g_2=0$ you would be saying that a surface of even genus has to be an embedded connect-sum of two surfaces of half the genus, but if your $2g_1$-genus surface is knotted, that usually would not be possible. I think you will have similar problems when $g_1>0$.
@RyanBudney Sorry I had in mind both genera nonzero - I also thought the answer should be no but I didn't know how to obstruct it.
I think that two Borromean surfaces of genus 3 are not disjointable in such way. Maybe someone more competent in 3 dim topology can actually prove it. Possibly, a genus 2 Hopf link surface is already a counterexample.
|
2025-03-21T14:48:29.948551
| 2020-02-28T08:32:29 |
353752
|
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"Fedor Petrov",
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|
Stack Exchange
|
Regarding upper semicontinuity of a function
Let $E$ be a linear subspace of $\mathbb{C}^{n\times n}$.
Define the function $\mu_E:\mathbb{C}^{n\times n}\longrightarrow \mathbb{R}_+$ as
$$
\mu_E(A)=\frac{1}{\inf\{\|X\|:X\in E\text{ and }\det(I_n-AX)=0\}},\quad A\in \mathbb{C}^{n\times n}.
$$ Where $\|.\|$ denotes the operator norm.
Can anyone help me in showing that $\mu_E$ is upper semicontinuous? I know that if $\|X\|<1$, then $I_n-X$ is invertible. But this involves invertibility of $I_n-AX$.
Also does $\mu_E$ satisfy the triangular inequality?
Note: I know that $\mu_E(A)=\|A\|$ if $E$ is the whole space $\mathbb{C}^{n\times n}$ and $\mu_E(A)$ is the spectral radius of $A$ if $E$ is the subspace of the scalar multiples of identity.
spectral radius does not satisfy triangle inequality
Yes. Correct. Noted.
Fix $A$ and take a sequence $A_k\to A$, denote $\mu_E(A_k)=a_k$. We have to prove that $\mu_E(A)\geqslant \limsup a_k$. Assume the contrary. Then
$a:=\limsup a_k>\mu_E(A)\geqslant 0$. Passing to a subsequence, we may suppose that $a_k>0$ for all $k$ and the limit $a=\lim a_k>\mu_E(A)$ exists (maybe infinite).
Choose $X_k\in E$ such that $\|X_k\|=a_k^{-1}$ and $\det(I_n-A_kX_k)=0$ (such minimizing matrices $X_k$ exist by compactness). Again by compactness, we may pass to a subsequence and suppose that the matrices $X_k$ converge (their norms $a_k^{-1}$ have finite limit $a^{-1}$) to certain matrix $X_0\in E$, $\|X_0\|=a^{-1}$. Then $\det(I_n-AX_0)=\lim \det(I_n-A_kX_k)=0$, therefore $\mu_E(A)\geqslant 1/\|X_0\|=a$, a contradiction.
I know that $\mu_E(A)=|A|$ if $E$ is the whole space $\mathbb{C}^{n\times n}$ and $\mu_E(A)$ is the spectral radius of $A$ if $E$ is the subspace of the scalar multiples of identity.
ah, it is chosen from $E$, sorry
Fedor's answer holds correct, so it should be accepted as such.
|
2025-03-21T14:48:29.948687
| 2020-02-28T08:36:26 |
353753
|
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|
Stack Exchange
|
A formula on a generically finite morphism
In Nakayama's book 2004, pg. 39, a formula is written:
Let $f:X \to Y$ be a generically finite and proper surjective morphism, $D$ be a Cartier divisor on $Y$. Then, $f_*f^*D=(\deg f) D$.
More precisely, we can compute in this way. Let $\sum_i D_i$ be a reduced divisor on $X$ of all prime divisors mapped onto $D$. Then, $f^*D= \sum_i r_iD_i$ where $r_i$ is the ramification index of $D_i$. Since $f_*D_i =(\deg f|_{D_i}) D$ by definition, we have $f_*f^*D= \sum_i r_i (\deg f|_{D_i}) D =(\deg f) D$.
Here is my question:
(1) why define $f_*D_i =(\deg f|_{D_i}) D$?
(2) why $\sum_i r_i(\deg f|_{D_i}) =\deg f$?
where can I find a detailed reference for these formulae? I try to search on Stacks but I cannot find a detailed answer.
Thank you!
Fulton, intersection theory.
|
2025-03-21T14:48:29.948774
| 2020-02-28T09:06:34 |
353754
|
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"Daniele Tampieri",
"Dave L Renfro",
"Igor Khavkine",
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|
Stack Exchange
|
Generalised limits via derivatives of integrals?
Assuming that $f$ is a continuous function, we have that
$$f(x) = \frac{d}{dx}\int f(t)\,dt.$$
Assuming instead that $f$ has a removable singularity at $x=a$, and is otherwise continuous, we have that
$$\lim_{x\to a}f(x) = \left.\frac{d}{dx} \int f(t)\,dt\right|_{x=a},$$
where the integral is either an improper Riemann integral, or a Lebesgue or Gauge integral.
But what if the singularity at $x=a$ is not removable? For instance, we have that
$$\left.\frac{d}{dx}\int \sin\Big(\frac{1}{t}\Big)\,dt\right|_{x=a}=0,$$
but the singularity at $x=0$ of $\sin(1/x)$ is not removable.
How does this relate to other notions of generalised limit, such as Cesaro's or Abel's?
[edit]
For clarification, the procedure above is, given a function $f$, and a real number $a$, to consider the following limit
$$\lim_{\epsilon \to 0} \frac{\int_a^{a+\epsilon}f(t)\,dt}{\epsilon}.$$
The result you get from $f(x) = \sin(1/x)$ and $a = 0$ comes from the fact that the integral of $f$ has a removable singularity at $x=0$, on which $\int f$ can still be differentiated, albeit the resulting derivative is discontinuous.
I don't know much about this, but the following survey paper might be worth looking at: Peter Bullen, Non-absolute integrals in the twentieth century, AMS Special Session on Nonabsolute Integration, 23-24 September 2000, 27 pages (has 195 references). Perhaps also the following paper, although it doesn't appear to be available online: Ralph Henstock, A short history of integration theory, Southeast Asian Bulletin of Mathematics 12 #2 (1988), 75-95 (has 262 references).
As far as I understand the question, this seems to be simply a rewording of Cesàro convergence, right?
An addition to the comment of @DaveLRenfro: the paper by Henstok is available as a compressed archive of .TIFF scans from the Southeast Asian Mathematical Society Bulletin web site, not from the Springer Verlag (the current distributor) web site.
@MateuszKwaśnicki Not sure, but the function $\frac{3 \cos(1/x^3)}{x^2}$ is unbounded, while the Cesaro limit only works for bounded functions. The generalised limit for this function at $x=0$ is $0$
@DaveLRenfro Thank you. This is an interesting reference
@DanieleTampieri Thank you for the reference
@jkabrg: I see, you're more interested in gauge integral.
Perhaps I'm missing something. But by the fundamental theorem of calculus, $\left.\frac{d}{dx}\right|_{x=a} \int^x \sin(1/t) dt = \sin(1/a)$. So I don't see how sending $a\to 0$ helps recover some kind of limit. Cesàro integration involves an extra averaging before taking the limit.
@IgorKhavkine Hopefully my edit will help clarify
I think that the concept of the limit of a distribution might be relevant. This can be defined elementarily as follows: a distribution $f$ on, say, $]0,1]$ has limit $a$ at zero if it has the form $D^p F$ (distributional derivative) near zero, where $$\lim_{x\to0}\frac{F(x)}{x^p}=\frac a{p!}.$$
Every continuous on ${]0,1}$ function is a distribution and it can happen that it has a limit in the distributional sense, but not in the classical one. Thus we have $\lim_{x\to \infty} \sin x=0$ in the distributional sense (forgive me for switching to limits at infinity for the purpose of this example).
The case where $p=1$ gives that the limit of a continuous $f$ is $a$ if $\lim_{x\to 0} \frac {F(x)}x = 0$ where $F$is a primitive (application: $f(x)= \sin{\frac 1 x}$).
Sorry, this should really be a comment.
|
2025-03-21T14:48:29.949040
| 2020-02-28T10:04:21 |
353758
|
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|
Stack Exchange
|
Gradient $L^p$ estimates for heat equation
I'm looking for a proof of the gradient estimate associated to the heat equation with Dirichlet boundary conditions, to see if I can express the constant $\color{red}{C}$.
$$\|e^{t\Delta_d}f\|_{W^{1,p}(\Omega)}\le \color{red}{ C}t^{-\frac{1}{2}}\|f\|_{L^p(\Omega)} $$
$\Omega$ is a bounded open set and $1\le p<\infty$.
Are you looking for an estimate where $C$ may depend on $\Omega$?
Let me comment on what I know in an open set $\Omega$, trying to control $C$. First of all, the heat semigroup can be expressed through a kernel $p$ which is pointwise dominated by the heat kernel in $\mathbb{R}^n$, written by Bazin, but this is not sufficient, because of $\nabla_x$. One can extend the Gaussian bound to complex times thus getting
$$|p(z,x,y)| \le C(\Re z)^{-d/2} \exp (-c\frac{|x-y|^2}{|z|})$$ if $z=x+iy$ and, say, $|y| \le x$, with constants $C,c$ depending only on the constants in the real estimate, hence independent of $\Omega$. See Chapter 6 of the book "Analysis of Heat Equations On Domains", by El Maati Ouhabaz. The constants can be made explicit by following the proof. Next, using Cauchy theorem for holomorphic functions one deduces the estimate $$|p_t(t,x,y)| \le \frac{C}{t}p(x,y,ct),$$ see again the same book. Since $p_t=\Delta_x p$, this gives $$\|\Delta e^{t\Delta}\|_p \le \frac{C}{t}$$ with $C$ still independent of $\Omega$ and computable. The last point needs the companion "elliptic" interpolation inequality for the gradient. In the whole space it is
$$\|\nabla u\|^2_p \le C \|u\|_p\|\Delta u\|_p$$ and is usually proved using the fundamental solution of the heat equation! When $p=2$ the above inequality holds in $\Omega$, with $C=1$, by integration by parts and, with some effort, can be extended to $1\le p<2$ without requiring any assumption on $\Omega$. However, if $p>2$ I do not see how to avoid some regularity of the domain. Note that the elliptic gradient estimate is equivalent to $$\|\nabla u\|_p \le \epsilon \|\Delta u\|_p + \frac{C}{\epsilon} \|u\|_p$$ for every $\epsilon >0$. This inequality holds in $\Omega$, if $\Omega$ has the extension property and the Laplacian is substituted by the full $W^{2,p}$-norm. Then one needs elliptic regularity up to the boundary and obtains the gradient estimate in the above form for small $\epsilon$. Large $\epsilon$ are then deduced by the semigroup law, since the heat semigroup decays exponentially in $\Omega$. It is also possible to prove pointwise bounds on $\nabla_x p$ (with $1/\sqrt t$ in front) by interpolating the estimates for $p$ and $\Delta p$ in small balls, but again one needs some regularity of the boundary to do it and the control on the constants is not very explicit.
Th fundamental solution of the heat equation in $\mathbb R^d$
is
$$
E_d(x,t)=H(t) (4πt)^{-d/2} e^{-\frac{\vert x\vert^2}{4t}},
$$
so that $\Vert E_d(\cdot,t)\Vert_{L^1(\mathbb R^d)}=1,$
$
\nabla_x E_d(\cdot,t)=-H(t) (4πt)^{-d/2} e^{-\frac{\vert x\vert^2}{4t}} \frac{x}{2t},
$
$$
\color{red} C=t^{1/2}\Vert \nabla_x E_d(\cdot,t)\Vert_{L^1(\mathbb R^d)}=\color{red}{\frac{\Gamma((1+d)/2)}{2^{d/2}\Gamma(d/2)}}.
$$
|
2025-03-21T14:48:29.949267
| 2020-02-28T10:15:59 |
353759
|
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|
Stack Exchange
|
Is there a name for a "convex hull with holes"?
If I have a (solid) 3d object, is there a name for the object created from it by taking the convex hull and subtracting from it all points that are on a straight line between any two points on the surface of the original object if that straight line is completely outside the original object (except those two points)?
Is this what you're looking for: http://www.kurims.kyoto-u.ac.jp/EMIS/journals/AG/4-1/4_61.pdf ? (my paper "An analogue of convexity for complements of amoebas of varieties of higher codimension, an answer to a question asked by B. Sturmfels")
Is the original object a polyhedron? Is it non-convex?
The original is an arbitrarily shaped 3d object, non-convex, with holes and/or non-contiguous.
So if you start with the surface of a smooth torus, you end up with a partially open torus, because you have removed some points of the surface surrounding the hole by lines slanting through the hole, tangent to two points? And the outer parts of the torus have no double-tangents and so remain? Whereas if it is polyhedral torus, the whole surface is removed, leaving an open polyhedral torus?
@JosephO'Rourke I always end up with an object that's larger than the one I started with, up to the full convex hull, if there are no "straight holes" in the object. E.g. a cube with a dent in one face will result in a cube (i.e. its convex hull), a ball with a straight round ("drilled") hole, will just stay the same, but a cube with a drilled hole with an internal cavity that's hit by the drill will become a cube with just the drilled hole (i.e. the cavity's filled except for the hole)
I'll try again. Take a cube. A line on the top face passes through two points on the surface. Is that line "completely outside the original object"? Your statement wording suggests Yes, in which case the surface is removed and we are left with a cube open at the boundary. If that line is not completely outside the original object, then no line passing through two points of the surface is ever completely outside, because it must touch at the two points.
@JosephO'Rourke I should add "except those two points", so the surface is kept.
|
2025-03-21T14:48:29.949442
| 2020-02-28T10:33:14 |
353760
|
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|
Stack Exchange
|
Size of minimum cut in random graph
Consider a uniform random tournament with $n$ vertices. (Between any two vertices $x,y$, with probability $0.5$ draw an edge from $x$ to $y$; otherwise draw an edge from $y$ to $x$.) The score of each vertex $x$ is the minimum number of edges that need to be deleted so that $x$ cannot reach some other vertex $y$. (So if $x$ already cannot reach some $y$, its score is $0$.)
Let $p(n)$ be the probability that the two highest scores are different. What is $\lim_{n\rightarrow\infty}p(n)$?
It is likely that all scores are positive (i.e. the graph is strongly connected), but I believe the two highest scores should be different (so the limit goes to $1$).
Yes, the limit goes to $1$.
An observation: by the Chernoff bound, with (very) high probability all vertices have degree between $0.49n$ and $0.51n$, so let's assume this holds in the following.
First, let's see that the number of edges needed to remove all directed paths from a given $x$ to a given $y$ is either the out-degree of $x$ or the in-degree of $y$, whichever is smaller.
To see this, suppose a smaller set $S$ of edges is removed. First, suppose $S$ leaves at least $\tfrac{n}{10}$ out-going edges from $x$ and at least $\tfrac{n}{10}$ edges going in to $y$. Pick $\tfrac{n}{20}$ out-neighbours of $x$ and a disjoint set of $\tfrac{n}{20}$ in-neighbours of $y$, and by the Chernoff bound with (very very) high probability there are more than $\tfrac{1}{1000}n^2$ edges from the first set to the second set in the tournament. These would have to all be in $S$, a contradiction.
We conclude that a linear sized set $S$ separating $x$ from $y$ has to contain all but at most $\tfrac{n}{10}$ edges either leaving $x$ or going into $y$. It can't be both, because then $S$ would have to be too big. Suppose it is the first; since $S$ is smaller than the out-neighbourhood of $x$, there is an out-neighbour $z$ of $x$ such that $(x,z)$ is not in $S$. Now $S$ can't contain many out-neighbours of $z$ (because it is small) so by the argument above there is a directed path from $z$ to $y$, so $S$ doesn't separate $x$ and $y$, a contradiction. The other case is symmetric.
OK, now the score of $x$ is whichever is smaller out of the out-degree of $x$ and the minimum in-degree of all $y\neq x$. Let $z$ be a vertex with minimum in-degree; then all $x\neq z$ have score at most the in-degree of $z$ (some may have smaller score; in fact there can only be very few such) and this score is slightly less than $\tfrac{n}{2}$. What is the score of $z$? This cannot be the outdegree of $z$, which is slightly bigger than $\tfrac{n}{2}$. So it's the in-degree of the vertex $z'$ of the tournament which has smallest in-degree and is not $z$.
That means that $z$ is a vertex with maximum score; it is the unique vertex with maximum score if it is the unique vertex with minimum in-degree, and otherwise it has the same score as most vertices in the graph (so in this case there are many vertices with maximum score).
In fact, the likely event is that $z$ is the unique vertex with maximum score. To see vaguely why this should be true, observe that the probability a given vertex has in-degree $t$ is $2^{1-n}\binom{n}{t}$, and the probability its in-degree is at most $t$ is $2^{1-n}\sum_{i\le t}\binom{n}{i}$. The in-degree of different vertices is close to independent, and so we expect the minimum to be around $\tfrac{n}{2}-\tilde{O}(\sqrt{n})$ (by calculation). For $t$ in this range, the probability of a given vertex having in-degree $t$ is something like $n^{-3/2}$, so the probability that two given vertices have degrees $t$ and $t'$ in this range is something like $n^{-3}$ (assuming, which is cheating, independence). That means that the expected number of pairs of vertices with degree $t$ and $t'$ is something like $n^{-1}$. Now consider pairs $(t,t')$ which are in the given range and differ by at most $10$. There are roughly $n^{1/2}$ such pairs (as opposed to about $n$ such pairs with no restriction on the difference) so the expected number of pairs of vertices with in-degree in the given range and differing by at most $10$ is something like $n^{-1/2}$. In particular, it's likely there is no such pair.
And in particular, it's likely the vertex with minimum in-degree is unique; the next smallest in-degree is likely more than $10$ larger.
For a proper proof of this last (not cheating on independence; the rest of the argument is rigorous), see Frieze and Karonski's book 'Random Graphs'; this is in the 'degrees of dense random graphs' section.
Maybe it's interesting to note that the answer is very different for the minimum score. With probability about $\tfrac12$ (and this is the limit) the minimum in-degree is smaller than any out-degree, and hence all but one vertex ($z$) have the same minimum score. When the minimum out-degree is smaller than the minimum in-degree, it's likely unique (as above) and hence the minimum score vertex is unique. It's unlikely that the minimum in-degree and minimum out-degree are the same.
Chernoff plus Union bound say that the statement holds for all pairs of disjoint subsets of this size, in particular for any choice of S you make.
No, but the proof goes through with tiny changes.
|
2025-03-21T14:48:29.949925
| 2020-02-28T11:03:35 |
353762
|
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|
Stack Exchange
|
Trivial homology with local system
Let $X$ be the classifying space of the Higman group $G$. It is well known that $G$ is an acyclic group
$$H_{\ast}(X;\mathbb{Z})=H_{\ast}(pt;\mathbb{Z}).$$
Now, suppose that $\mathcal{M}$ is a local system on the space $X$ such that
$$H_{i}(X;\mathcal{M})=0, \textrm{ for all $0\leq i$}.$$
Does such local system $\mathcal{M}$ on $X$ exist (other then $\mathcal{M}= 0$)
For $X = BG$ local systems on $X$ can be identified with $G$-modules, and homology with the derived tensor product $-\otimes^L_{\mathbb ZG}\mathbb Z$, i.e. $H_i(X;M) \cong \operatorname{Tor}^i_{\mathbb Z G}(M,\mathbb Z)$. One way to see this is to take the definition $H_i(X;M):= H_i(\mathcal S_*(\widetilde X)\otimes_{\mathbb Z\pi_1(X)} M)$, where $\pi_1(X)$ acts on (singular) chains on the universal cover $\widetilde X$ via deck transformations, and replace $\mathcal S_*(\widetilde X)$ with the cellular complex $C_*(\widetilde X)$ of the CW structure of the realization of the nerve of the groupoid $G//G$; then $C_*(\widetilde X)\otimes_{\mathbb Z G} M = \dots\to \mathbb Z[G^2]\otimes M\to \mathbb Z[G]\otimes M\to M$ is the bar complex computing group homology.
Let $G$ be an arbitrary group, and let $M = \operatorname{ker}(\mathbb Z G\to \mathbb Z)$ be the reduced group algebra, so that we have a short exact sequence
$$
0\to M\to \mathbb ZG\to\mathbb Z\to 0
$$
of $\mathbb ZG$-modules. This gives rise to a long exact sequence of Tor-groups, in particular a boundary operator $\partial:\operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z,\mathbb Z)\to \operatorname{Tor}^i_{\mathbb Z G}(M,\mathbb Z)$. Its kernel is the image of the map $0 = \operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z G,\mathbb Z)\to \operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z,\mathbb Z)$, so it is always injective; its cokernel is the kernel of the map $\operatorname{Tor}^{i}_{\mathbb Z G}(\mathbb Z G,\mathbb Z)\to \operatorname{Tor}^{i}_{\mathbb Z G}(\mathbb Z,\mathbb Z)$, which is an isomorphism for $i = 0$ and has the zero group as its codomain for $i > 0$, so it is always injective, so $\partial$ is always surjective and thus always an isomorphism.
In your example of the Higman group, we know that $H_i(X;\mathbb Z)$ is $\mathbb Z$ concentrated in degree $0$, so that $\partial$ is an isomorphism with the zero group, so that $H_i(X;M) = 0$ for all $i \ge 0$.
Thanks for all the details! Do you think that if we add the condition that $M$ is finitely generated as $\mathbb{Z}G$-module, then the conclusion will be that $M=0$ ? Since the kernel of $\mathbb{Z}G\rightarrow \mathbb{Z}$ is not finitely generated (hope I am not wrong)
@Bertram Arnold what is the map from singular chain complex to the cellular one (or the way around) that allows you to replace one by the other?
In general, there is of course no explicit direct map. In this case the CW structure is the canonical one on the geometric realization of a simplicial set, so that there is a canonical map from the cellular complex to the singular chains sending an n-cell to the non-degenerate simplex it corresponds to. This is a chain homotopy equivalence, and any homotopy inverse does the job.
|
2025-03-21T14:48:29.950145
| 2020-02-28T11:20:57 |
353764
|
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"Carlo Beenakker",
"Iosif Pinelis",
"Math_Y",
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"url": "https://mathoverflow.net/questions/353764"
}
|
Stack Exchange
|
Expectation of inverse of random matrices
Assume that $\mathbf{X}$ is a random positive-definite matrix. Then, is there any upper or lower bound on the expectation of the following expression
$$\mathbb{E}[\mathbf{X}^{-1}]-\alpha\mathbb{E}[\mathbf{X}^{-2}]$$
based on $\mathbb{E}[\mathbf{X}]$?
how might this work? the expectation of $X$ will not give you information on $X$ near zero, which you need to know the expectation of $X^{-1}$ and $X^{-2}$.
For example, something like Jensen's inequality. $\mathbb{E}[X^{-1}]\geq\mathbb{E}[X]^{-1}$.
Let us assume that $\alpha>0$. Then, by rescaling, without loss of generality $\alpha=1$. So, we have to provide an upper or lower bound on $Ef(X)$, where $X$ is a random $n\times n$ positive-definite matrix with a given mean $EX$ and
$$f(x):=\frac1x-\frac1{x^2}$$
for real $x>0$.
First of all, there is no finite lower bound here. Indeed, already for $n=1$, letting $P(X=t)=1/2=P(X=2-t)$ with $t\downarrow0$, we get $Ef(X)\to-\infty$.
However, we can get an upper bound on $Ef(X)$, which will be exact if for some $b\in(0,2]$ we have $P(X=bI)=1$, where $I$ is the $n\times n$ identity matrix. Indeed,
$$f'(x)=\frac{2-x}{x^3},\quad f''(x)=2\frac{x-3}{x^4},$$
so that $f$ is increasing on $(0,2]$, decreasing on $[2,\infty)$, and concave on $(0,2]$. So, for any $a\in(0,2]$ and all real $x>0$,
$$f(x)\le g_a(x):=f(a)+f'(a)(x-a).$$
So, by the spectral decomposition, $f(X)\le g_a(X)$, and hence
$$Ef(X)\le Eg_a(X)=B_a(EX):=(f(a)-f'(a)a)I+f'(a)EX.$$
The upper bound $B_a(EX)$ on $Ef(X)$ will be exact if $P(X=bI)=1$ for some $b\in(0,2]$ and $a=b$.
More generally, if e.g. $EX\le mI$ for some real $m>0$, then
$$Ef(X)\le\min_{a\in(0,2]}B_a(mI)
=\begin{cases}
\tfrac14\,I &\text{ if }m\ge2,\\
\tfrac{m-1}{m^2}\,I &\text{ if }0<m\le2.
\end{cases}$$
If now $\alpha\le0$, then the corresponding function
$$F(x):=\frac1x-\frac\alpha{x^2}$$
is convex, and hence $EF(X)\ge F(E(X)$ by Jensen's inequality. In this case, there is no upper bound, though -- as the above example with $n=1$, $P(X=t)=1/2=P(X=2-t)$, and $t\downarrow0$ shows.
this would be a lower bound for $\alpha<0$, right?
@CarloBeenakker : I have added the case $\alpha\le0$ as well.
|
2025-03-21T14:48:29.950309
| 2020-02-28T14:34:51 |
353770
|
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|
Stack Exchange
|
Topological weak mixing vs measure-theoretic weak mixing
Let $X$ be a compact metric space and $T$ a continuous map from $X$ to $X$. The system $(X,T)$ is called topologically weakly mixing if the product system $(X\times X,T\times T)$ is topologically transitive. My question is that: If $(X,T)$ is minimal and topologically weakly mixing, whether there exists an ergodic measure $\mu$ on $(X,T)$ such that $(X,T)$ is weakly mixing as a measure-preserving system, which means $(X\times X,T\times T,\mu\times \mu)$ is ergodic?
I guess there might be a counter-example but I do not have any such example in mind, nor can prove it. If this is not true, is it possible to add some proper condition to make this true but not trivially true?
I am sure the answer is no, but I can’t think of an immediate counterexample. Basically these topological notions are not able to detect whether there is a measure-theoretic rotation factor (which is what lack of weak mixing is).
Now I figure out that the answer of my question is false. In the paper ``Topological mixing and uniquely ergodic systems", Lehrer proved any measure-preserving system has a topologcally mixing strictly ergodic topological model, which implies that there exists a topologcally mixing strictly ergodic system $(X,T)$ such that with the unique invariant measure $\mu$, is measure-theoretically isomorphic to an irrational rotation on the circle, which is not measure weakly mixing. So my question is even false when topologically weakly mixing is replaced by topologcally mixing .
|
2025-03-21T14:48:29.950418
| 2020-02-28T14:48:56 |
353772
|
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|
Stack Exchange
|
Properties of solution to Burger's equation using Cole-Hopf transformation
I am currently looking at a $1$D-Burger's equation defined by
\begin{equation} \label{ex burgers}
\left\{
\begin{array}{ll}
{} & \frac{\partial V_m}{\partial t} (t,x) = \frac{\sigma^2}{2} \frac{\partial^2 V_m}{ \partial x^2} (t,x) - V_m(t,x) \frac{\partial V_m}{\partial x} (t,x), \quad \forall (t,x) \in (0,T] \times \mathbb{R},
\\
{} & {} \\
{} & V_m(0,x)= \int_{-\infty}^x \, m(dy), \quad \forall x \in \mathbb{R}, \end{array}
\right.
\end{equation}
where $m$ is a probability measure on $\mathbb{R}$. It is known that $V_m(t,\cdot)$ is a cumulative distribution function and satisfies the following closed form solution via Cole-Hopf transformation (https://projecteuclid.org/euclid.aoap/1034968229, page $822$)
$$ V_m(t,x) = \frac{V_1(t,x,m)}{V_2(t,x,m)}, $$
where
\begin{equation} V_1(t,x,m):= \int_{\mathbb{R}} \frac{x-y}{t} \exp \bigg\{ - \frac{1}{\sigma^2} \bigg[ \frac{(x-y)^2}{2t} + \int_{x}^y \int_{-\infty}^z m(d\theta) \,dz \bigg] \bigg\} \, dy \label{V1 } \end{equation}
and
\begin{equation} V_2(t,x,m):= \int_{\mathbb{R}} \exp \bigg\{ - \frac{1}{\sigma^2} \bigg[ \frac{(x-y)^2}{2t} + \int_{x}^y \int_{-\infty}^z m(d\theta) \,dz \bigg] \bigg\} \, dy. \label{V2 } \end{equation}
My question is how can we show the properties that $V_m(t,\cdot)$ is a cumulative distribution function, for any $t \in [0,T]$, i.e. that
$$ \lim_{x \to \infty} V_m(t,x) =1, \quad \quad \lim_{x \to -\infty} V_m(t,x) =0, \quad \quad V_m(t, \cdot) \text{ is increasing, } \quad \forall t,$$
based on the above representation formula of the solution. I believe that there are certain estimates that I am missing and I need those to prove properties of functions with a similar form.
|
2025-03-21T14:48:29.950537
| 2020-02-28T15:02:24 |
353773
|
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"gmvh",
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"url": "https://mathoverflow.net/questions/353773"
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|
Stack Exchange
|
Differential equation satisfied by linear combinations of eigenfunctions of linear differential operator
Let $D$ be a linear differential operator on $\mathcal{C}^\infty(\mathbb{R})$, and let $\mathcal{E}_\lambda=\{f\in\mathcal{C}^\infty(\mathbb{R})|Df=\lambda f\}$ be the space of eigenfunctions of $D$ to the eigenvalue $\lambda$. It is easy to see that $\bigcup_{\lambda}\mathcal{E}_\lambda$ can be characterized by the non-linear ODE $(Df)'f-f'Df=0$. Is there a similar non-linear ODE satisfied by linear combinations of the eigenfunctions, i.e. can $\mathrm{span}\bigcup_{\lambda}\mathcal{E}_\lambda$ be characterized by an ODE? (This is loosely related to this question.)
If $f$ is a linear combination of at most $N$ eigenfunctions, then $f$,$Df$,$D^2f$,...,$D^Nf$ are linearly dependent. Hence $W(f,Df,...,D^Nf)=0$, where $W$ denotes the Wronskian.
Many thanks for answering the question I should have asked! The solution is really rather obvious, I had been coming at this from a completely wrong angle.
The characterization of $\mathrm{span}\bigcup_\lambda\mathcal{E}_\lambda$ asked for in the question is generally not possible. Consider e.g. $D=\frac{\rm d^2}{{\rm d}x^2}$. Then already $\mathrm{span}\bigcup_{\lambda=-n^2,n\in\mathbb{Z}}\mathcal{E}_\lambda$ is dense in $L^2([-\pi;\pi])$, and there is no hope to find a continuous $F$ such that $F(f)=0$ for all $f\in \mathrm{span}\bigcup_{\lambda=-n^2,n\in\mathbb{Z}}\mathcal{E}_\lambda$, but $F(f)\not=0$ for other $f\in L^2([-\pi;\pi])$. Adding in the other eigenfunctions of $D$ can make the issue only worse, not better.
The question I should have asked is: Let $D$ and $\mathcal{E}_\lambda$ be as above, and let $N\in\mathbb{N}$. Is there some differential equation satisfied by all those $f\in\mathrm{span}\bigcup_\lambda\mathcal{E}_\lambda$ that can be written as linear combinations of elements of at most $N$ different $\mathcal{E}_\lambda$? That question doesn't look quite so hopeless.
|
2025-03-21T14:48:29.950672
| 2020-02-28T15:29:54 |
353775
|
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|
Stack Exchange
|
Wiman's method for bounding the rank of an elliptic curve
In 1945 Wiman [W] showed that certain elliptic curves $E$ over $\mathbf Q$ have rank* at least 4. (It seems this was the highest known rank of an elliptic curve over $\mathbf Q$ until 1974, when Penney--Pomerance found a curve of rank at least 6.)
The method of his proof appears to be rather elementary, defining a map from $E(Q)$ to a certain abelian 2-group $A$ by using the $p$-valuations of the $x$-coordinate of a point $(x,y) \in E(\mathbf Q)$ for various primes $p$.
However, due to some combination of my inadequate German and Wiman's somewhat archaic mathematical style, I cannot decipher the exact definition of the group $A$ and the map $E(\mathbf Q) \rightarrow A$. So I ask
Question: What is the precise definition, in modern terms, of Wiman's group $A$ and map $E(\mathbf Q) \rightarrow A$?
[W] Wiman, A., Über den Rang von Kurven
$y^2=x(x+a)(x+b)$, Acta Math. 76, 225-251 (1945). ZBL0061.07109.
*: A bit confusingly, "rank" is used in this paper to mean "minimal number of generators of $E(\mathbf Q)$" rather than "minimal number of generators of $E(\mathbf Q)/E(\mathbf Q)_{\operatorname{tors}}$", so to get the "correct" rank one has to subtract 2 from each of the values reported by Wiman.
"We need more Bort license plates in the Gift Shop. Repeat, we are sold out of Bort license plates."
I have not read all the details of the article, but most of what I see is just descent by the isogeny $[2]$. The map is the Kummer map $$E(\mathbb{Q})\to \mathbb{Q}^{\times}/\square \times \mathbb{Q}^{\times}/\square \times \mathbb{Q}^{\times}/\square$$ which sends $(x,y)$ to $(x-e_1,x-e_2,x-e_3)$. In modern terms this is the connecting homomorphism $E(\mathbb{Q})\to H^1\bigl(\mathbb{Q}, E[2]\bigr)$. The kernel of the map is $2E(\mathbb{Q})$ and the image in each of the component lies in the group generated by the prime divisors of $\Delta$. Looking at valuations at those primes corresponds to studying the image of the local map $E(\mathbb{Q}_p) \to \mathbb{Q}_p^{\times}/\square\times\dots$. So the article actually studies the $2$-Selmer group. Similar to Selmer's work around the same time using $3$-descent. The idea to define Selmer groups and relate it to Galois cohomology only appears (if I am not mistaken) later in the work of Cassels, Lang, Tate, ...
Very helpful, thank you.
|
2025-03-21T14:48:29.950842
| 2020-02-28T15:41:22 |
353776
|
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|
Stack Exchange
|
Sufficient and necessary condition for the continuity of an improper integral
Let $f(\cdot) \in \mathscr{C}\left( \mathcal{D}; \mathbb{R} \right)$ where $\mathcal{D} \subseteq \mathbb{R}$ is open with $0 \in \mathcal{D}$ and
$$ f(0) = 0, \quad \forall x \in \mathcal{D}\setminus \{ 0 \}, \; xf(x) < 0.$$ My question is that is there a sufficient and necessary condition on the structure of $f(\cdot)$ which guarantees that
\begin{equation}
\forall x \in \mathcal{D}, \; \int_{x}^{0} \dfrac{1}{f(z)} \mathsf{d}z < + \infty \quad \text{and} \quad \lim_{x \to 0} \int_{x}^{0} \dfrac{1}{f(z)} \mathsf{d}z = 0
\end{equation}
are true? Note that $\int_{x}^{0} 1/f(z) \mathsf{d}z$ is an improper integral since $f(0) = 0$.
An example satisfying the above two conditions is $f(x) = -\mathsf{sign} (x) | x |^a $ with $\mathcal{D} = \mathbb{R}$ and $0 < a < 1$, which leads to
\begin{equation}
\int_{x}^{0} \dfrac{1}{f(z)} \mathsf{d}z = -\int_{x}^{0} \dfrac{1}{\mathsf{sign} (z)| z |^a } \mathsf{d}z = | x |^{1-a} \quad \text{and} \quad \lim_{x \to 0} | x |^{1 - a} = 0.
\end{equation}
Having $\int_{-\epsilon}^{\epsilon} \frac{1}{|f(z)|},dz < \infty$ is sufficient by the dominated convergence theorem, and likely pretty close to necessary.
Thank you so much, it works.
|
2025-03-21T14:48:29.950946
| 2020-02-28T16:19:43 |
353778
|
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|
Stack Exchange
|
Galois action on morphism between $\overline{k}$ schemes
I have a question on a certain property of morphisms between schemes endowed with Galois action. The motivation arises from a comment by Phil Tosteson on this question.
Phil wrote: "If the map factors through the projection, it factors uniquely (the projection is dominant) . So the factorization is automatically galois invariant..."
I want to understand how the Galois action on the morphism is concretly described in the given setting in order to be able to talk about "Galois invariant morphisms".
What we know. Let $X,Y$ be $k$-varieties or more generally $k$-schemes. Let $\overline{k}$ be algebraic closure of $k$ and $Gal(\overline{k}/k)$ the Galois group. We consider the fiber products $X \times_{\operatorname{Spec} \ k} \operatorname{Spec} \ \overline{k}, Y \times_{\operatorname{Spec} \ k} \operatorname{Spec} \ \overline{k}$.
We introduce the abbreviation $\overline{X}:=X \times_{\operatorname{Spec} \ k} \operatorname{Spec} \ \overline{k}$.
Let $f: \overline{X} \to \overline{Y}$ a morphisms between the $\overline{k}$-schemes.
QUESTION: I order to decide if $f$ is "Galois invariant" we need to decede which Galois action on $f$ we consider. Do we have in this context "the action" (ie a canonical one)?
I know a couple possibilities to define an action of $f$ in different ways (see below) but I'm not sure which one is the standard one when the literature talks about "Galois invariant morphism".
Two ways I know to define an action on $f$:
1. Let $ \sigma \in Gal(\overline{k}/k)$. Then $\sigma$ acts on $\overline{X}$ via morphism $id_X \times (\operatorname{Spec} \ \sigma)$ with $\operatorname{Spec} \ \sigma: \operatorname{Spec} \ \overline{k} \to \operatorname{Spec} \ \overline{k}$. We use notation $\overline{\sigma}:=id_X \times (\operatorname{Spec} \ \sigma)$.
Then we can define on action on $f$ by $\sigma$ via "conjugation" $\sigma(f):= \overline{\sigma^{-1}} \circ f \circ\overline{\sigma}$
2. Consider the set of $\overline{k}$-valued points $\overline{X}(\overline{k}):=Hom(\operatorname{Spec} \ \overline{k}, \overline{X})$.
We can say two things:
-Galois group acts on $\overline{X}(\overline{k})$ via composition $\alpha \mapsto \alpha \circ (\operatorname{Spec} \ \sigma)$ for $\alpha \in \overline{X}(\overline{k})$ and $\sigma \in Gal(\overline{k}/k)$
-$f$ induces map $f(\bar{k}):\overline{X}(\overline{k}) \to \overline{Y}(\overline{k})$ by compostion $\alpha \mapsto f \circ \alpha$
Thus the Galois Group also acts on $f(\bar{k})$ via precomposition $f(\bar{k}) \mapsto f(\bar{k}) \circ (\operatorname{Spec} \ \sigma)$
The funny thing is that $\overline{X}(\overline{k})$ is dense in $\overline{X}$ and thus the Galois action from 2 on $f(\bar{k})$ induces by continuity & density a unique action on $f$.
That is we have (at least) two possibilities how Galois group could act on $Hom(\overline{X}, \overline{Y})$ and we can say that a $f \in Hom(\overline{X}, \overline{Y})$ is Gaois invariant if for every $\sigma(f)=f$ is fixed.
Back to my question: if we talk about "the" Galois action on $f$(or calling $f$ Galois invariant) which action is generally proposed if it is not explicitly explained as in my MO question: the 1 or the 2?
The first action is the usual one.
Actually, your second action is not well-defined. By definition, $\bar{X}(\bar{k})$ is the set of morphisms $\alpha:\operatorname{Spec}\bar{k}\to \bar{X}$ such that $\pi_2\circ\alpha$ is the identity on $\operatorname{Spec}\bar{k}$, where $\pi_2:\bar{X}\to\operatorname{Spec}\bar{k}$ is projection onto the second factor (that is, the structure map). If $\pi_2\circ\alpha$ is the identity, then $\pi_2\circ(\alpha\circ \operatorname{Spec}\sigma)=\operatorname{Spec}\sigma$ will not be the identity unless $\sigma=1$, so $\alpha\circ \operatorname{Spec}\sigma$ will typically not be in $\bar{X}(\bar{k})$. So your proposal 2 does not give a well-defined Galois action on $\bar{X}(\bar{k})$, nor on $\operatorname{Hom}(\bar{X},\bar{Y})$.
Yes I see. One remark on a slightly different situation (which motivated me to impose the wrong proposal 2): If we start instead of $\overline{X}$ with $k$-scheme $X$ and consider the points $X(\bar{k})$ then the Galois action on $X(\bar{k})$ by precomposition $\alpha \mapsto \alpha \circ (\operatorname{Spec} \ \sigma)$ seems in this case to work since in this case for $X$ we haven't more the projection $\pi_2$ forcing $\pi_2\circ\alpha$ beeing idenity. So if we have this Galois action on $X(\bar{k})$, can this action be extended to Galois action on $X$? That is if certain $\sigma$ acts as
described on $X(\bar{k})$, then can (possibly by a certain continuity argument) the action extended to such one $X$? The problem is the existence, since if such extension of the action exist then it must be unique by density of $X(\bar{k})$ as I remarked. But I'm not sure if we can expect that for arbitrary $k$-scheme such extension realizable. Do you know if it always work or one need to impose extra conditions of $X$ (e.g. certain finiteness conditions...)
|
2025-03-21T14:48:29.951251
| 2020-02-28T16:41:01 |
353781
|
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|
Stack Exchange
|
$Π_2$ strength of KP
I am looking for a characterization of the $Π_2$ statements provable in KP.
Here, KP (often denoted KPω) is the Kripke-Platek set theory, including infinity and full induction on ordinals. Here is my conjecture (which the answer(s) can tweak if appropriate).
Conjecture: A $Π_2$ statement is provable in KP iff it is provable in primitive recursive set theory plus (schema over $n$) $∀α∈\mathrm{Ord} \,\, Ω_{\text{BH},n}(α)∈\mathrm{Ord}$.
Here, $Ω_{\text{BH},n}(α)$ is the nth element of the standard fundamental sequence above $α$ for $Ω_\text{BH}(α)$ (without requiring for $Ω_\text{BH}(α)$ to be an ordinal), and $Ω_\text{BH}(α)$ is the analog of the Bachmann-Howard ordinal above $α$.
Primitive recursive set theory can be weakened to a weak base theory (extensionality, foundation (without full induction), infinity, closure under rudimentary functions) plus $∀x ∀y ∃z=L_{\mathrm{rank}(x)}(\mathrm{trcl}(y))$ where $\mathrm{trcl}$ is the transitive closure.
Related results
The Bachmann-Howard ordinal $Ω_\text{BH}$ is the $Π^1_1$ proof ordinal of KP. All $Π_2$ statements provable in KP hold in $L_{Ω_\text{BH}}$. (Note that $Π_2$ is very expressive in ZFC, but its expressiveness drops if all sets are countable, and especially if the universe is also not closed under hyperjumps.)
A somewhat related result is that all $Π_2^\mathcal{P}$ statements provable in $\text{KP}(\mathcal{P}) + AC$ hold in $V_{Ω_\text{BH}}$ (Relativized ordinal analysis: The case of Power Kripke-Platek set theory). This is despite $\text{KP}(\mathcal{P})$ including powerset and with collection expanded accordingly.
While proof ordinals are commonly used for $Π^1_1$ statements, their relevance is much broader. For example, not only is $Γ_0$ the proof ordinal of $\text{ATR}_0$, but over $\text{RCA}_0$,
$\text{ATR}_0 ⇔ ∀X \, (\mathrm{WO}(X) ⇔ \mathrm{WO}(φ_X(0)))$ (in $\text{RCA}_0$, $\mathrm{WO}$ means lack of infinite descending sequences; $φ$ is Veblen fix-point function), and also $∀X \, (\mathrm{WO}(X) ⇔ \mathrm{WO}(Γ(X)))$ iff every $X$ is inside an $ω\text{-model}$ of $\text{ATR}_0$ (Omega-models and well-ordering principles).
An application
KP is inconsistent with $V = L_{Ω_\text{BH}}$ since (provably in KP) it is not an admissible set/class. However, we can correct this by using (essentially) the well-founded part. Below, $\mathrm{Ord}=\mathrm{wfp}(Ω_\text{BH}(α))$ means that every ordinal is isomorphic to an initial segment of $Ω_\text{BH}(α)$, with $Ω_\text{BH}(α)$ being the linear order for the standard notation system for the analog of Bachmann-Howard ordinal above $α$.
Consequences of the conjecture:
* KP + $∃α∈\mathrm{Ord} \,\, \mathrm{Ord}=\mathrm{wfp}(Ω_\text{BH}(α))$ is $Π_2$ conservative over KP.
* For every statement $φ$, KP + $∃S (φ^S ∧ \mathrm{Ord}=\mathrm{wfp}(Ω_\text{BH}(\mathrm{rank}(S)))$ is $Σ_1$ conservative over KP + $∃S \, φ^S$
The proof uses that adding the same $Σ_2$ statement to both sides preserves $Π_2$-conservativity, as well as an easy conservation result for the weak base theory.
The significance of the consequences is the following. A canonical ordinal analysis of some natural set theory is expected to include all ordinals that have a canonical definition in the theory, and not just the recursive ones. Which ordinals have a canonical definition is vague, but we can make it more precise by requiring that, consistently with the theory, all ordinals are included.
And a positive answer to the question will imply that for example, KP + $L_α ⊨ \text{ZFC}$ is relatively consistent with the Bachmann-Howard notation system capturing all ordinals above $α$ in terms of ordinals below $α$. Now, a limitation of KP is that it does not prove existence of admissible ordinals above $α$, so the next step may be $Π^1_1-\text{CA}_0$, as described in the question Ordinal analysis and nonrecursive ordinals.
There is a paper "Classifying the provably total set functions of $\mathsf{KP}$ and $\mathsf{KP}(\mathcal{P})$" by J. Cook and M. Rathjen (see https://arxiv.org/pdf/1610.02194.pdf ). Their Theorem 6.2 isn't exactly what you need but fairly close. I think that with some additional efforts their technique should give the result that you want.
Also there is an unpublished result of mine about an analogue of Schmerl's formula for $\mathsf{KP}$ from which it is fairly easy to derive your conjectured result. See slides http://www.mathnet.ru:8080/PresentFiles/22127/kpomega_via_reflection.pdf . Also there are recordings of my talks (in Russian) on the subject http://www.mathnet.ru/php/seminars.phtml?option_lang=rus&presentid=22127 , http://www.mathnet.ru/php/seminars.phtml?option_lang=rus&presentid=22655
@FedorPakhomov Theorem 6.2 appears to work. If $\text{KP}⊢∀x∃y , φ(x,y)$ (with $Δ_0$ $φ$), we can consider $f$ with $f(x)$ being the least $L_α(x)⊨∃y , φ(x,y)$, so it suffices to consider functions. Also, the paper notes that for each individual proof, the result of Theorem 6.2 is provable in KP; and the theory in the conjecture should also work as it can set up enough of the notation system. (Also, your analysis of iterated reflection in subsystems of KP is interesting.) Would you like to expand all the comments into an answer?
|
2025-03-21T14:48:29.951603
| 2020-02-28T16:47:06 |
353782
|
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|
Stack Exchange
|
Partition of unity in $\mathbb{R}$ with additional conditions on the derivatives
Let $K\subseteq \mathbb{R}$ be locally compact without isolated points and $X$ an infinite dimentional Banach space. Then
$$C_{0}^{(1)}(K,X)=\{ f\in C_{0}(K,X): \text{$f$ is continuously differentiable and} \;f' \in C_{0}(K,X) \}$$
is a banach space considering the norm $\left \Vert f\right \Vert_M=\max \{\left \Vert f\right \Vert_{\infty}, \left \Vert f'\right \Vert_{\infty}\}.$
Here $C_{0}(K,X)$ stands for the banach space consisting of all functions that vanish at infinity.
Suppose $\phi\in C_{0}^{(1)}(K,X)$, $k \in K$ and consider the set
$$
(*) \quad \quad V_{k}=\left\{ s\in K:\left\Vert \phi(k)-\phi(s)\right\Vert <\varepsilon \; \text{and} \; \left\Vert \phi'(k)-\phi'(s)\right\Vert \; <\varepsilon \right\},
$$
then
$\{V_k:k\in K\} $ is an open covering of
$K$. It is a well known fact that partitions of unity subbordinate to
the cover exists, i.e. there exists smooth
functions $ \varphi_{k} : V_k \rightarrow [0,1] $ with compact
support such that
$$ \sum_{k=1}^{\infty} \varphi_k(x) = 1 \; \text{for all $x\in V_k$} \quad \text{and} \quad supp \; \varphi_{k} \subseteq V_k \; \text{for all $k \in K$}.$$
My question is the following: By assuming the conditions on $(*)$ is it possible to conclude that
$$ \sum_{k=1}^{\infty} \varphi'_k(x) = 1 \quad \text{and} \quad supp \; \varphi'_{k} \subseteq V_k \;?$$
where $\varphi'_{k}$ is the derivative of $\varphi_{k}$.
If is not possible ( I dont really know ), is it possible to make more assumptions on the sets $V_t$ (or perhaps $K$) so that can happen?
EDIT: As pointed out by the comments, $\sum_{k=1}^{\infty} \varphi'_k(x) = 0$
So I'll try to give context to what I wanna try to emulate:
Basically I want to check that the set $A=\left\{ \phi\left(\cdot\right)x:\phi\in C_{0}^{(1)}\left(K\right),x\in X\right\} $ is dense in $C_{0}^{(1)}(K,X)$.
We start again with $\{V_k:k\in K\}$ a open cover for $K$, since $K$ is locally compact then there exist $\left\{ W_{t}:t\in K\right\} $ open covering of $K$ such that every $ W_{t}$ is relatively compact. So that there exist functions (with compact support) $\left\{ \varphi_{t}:K\mapsto\mathbb{R}\right\} _{t\in K}$ such that $supp\;\varphi_{t}\subset W_{t}$ for all $t\in K$, then every $\varphi_{t}$ and it's derivative vanish at infinity, i.e $\varphi_{t}\in C_{0}^{(1)}(K,[0,1])$.
this is the important part: Let $t \in K $ and consider
$$
g(t)=\sum_{\alpha\in K}\varphi_{\alpha}(t)\phi(t_{\alpha})
$$
( the idea is to conclude $(1) \; \left\Vert \phi(t)-g(t)\right\Vert_{\infty}<\varepsilon $ and $\left\Vert \phi'(t)-g'(t)\right\Vert_{\infty}<\varepsilon$ so that $\left\Vert \phi(t)-g(t)\right\Vert_{M}<\varepsilon $.
This is the $(1)$ part: $ \left\Vert \phi(t)-g(t)\right\Vert_{\infty}<\varepsilon $
\begin{align}
\left\Vert \phi(t)-\sum_{\alpha\in K}\varphi_{\alpha}(t)\phi(t_{\alpha})\right\Vert _{\infty}&=\left\Vert \sum_{\alpha\in K}\left(\varphi_{\alpha}(t)\phi(t)-\varphi_{\alpha}(t)\phi(t_{\alpha})\right)\right\Vert _{\infty}\\&=\left\Vert \sum_{\alpha\in K}\varphi_{\alpha}(t)\left(\phi(t)-\phi(t_{\alpha})\right)\right\Vert _{\infty}\\&\leq\sum_{\alpha\in K}\varphi_{\alpha}(t)\left\Vert \phi(t)-\phi(t_{\alpha})\right\Vert_{\infty}
\end{align}
Where $\left\Vert \phi(t)-\phi(t_{\alpha})\right\Vert _{\infty}<\varepsilon$.
Now by the help from comments I dont thinkg the part $\left\Vert \phi'(t)-g'(t)\right\Vert_{\infty}<\varepsilon$ is plausible. Should I construct a new function $g$? Is there a easier way to check that $A$ (the closed linear subspace) is dense?
Don't you need some regularity assumption on $f$ to belong to your space $^{(1)}_0(,)$, like being $C^1$ or smooth ?
Yea. All $f$ in that space are $C^1$. I'll fix it.
(1) If the support of $\varphi_k$ is contained in $V_k$, then isn't the support of $\varphi_k'$ trivially contained in $V_k$ as well? (2) Intuitively, if $\sum \varphi_k = 1$, shouldn't we expect $\sum \varphi_k' = 0$ (e.g. if we could commute the derivative with the infinite sum) instead of 1?
You are right! I was trying to emulate a inequlity from the $\varphi_k$ to $\varphi'_k$ . I guess I have to work on it a little more.
|
2025-03-21T14:48:29.951974
| 2020-02-28T17:06:34 |
353784
|
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|
Stack Exchange
|
Locus of non-injective maps between vector bundles
Suppose $E,F$ are two vector bundles of ranks $e$ and $f$ with $e<f$ on a smooth projective surface $X$ over $\mathbb{C}$ and let $r:=f-e$. I am interested in estimating the (co)dimension of the locus of noninjective maps (as sheaf maps or vector bundle maps, does not matter for me too much) inside $Hom(E,F)$.
This has a clear answer when the bundle $\mathcal{Hom}(E,F)$ is globally generated - the codimension of maps which are not injective on at least one fiber $E|_x \to F|_x$ is greater or equal to $r-1$.
Is there anything that could be said in the case when $\mathcal{Hom}(E,F)$ fails to be globally generated? In the situation I have at hand, $r \geq 3$ and I have at least one injective vector bundle map $E \to F$ and I want to rule out the situation when the locus of non-injective maps is divisorial, i.e. codimension 1 inside $Hom(E,F)$.
Remark. Something that appears potentially useful is that in my situation $X=\mathbb{P}^1 \times \mathbb{P}^1$ and $E,F$ are homogenous with respect to the action of $Aut(X)$.
A priori no. Imagine that $F = E \oplus E'$ with $\dim Hom(E,E) = 1$ and $Hom(E,E') = 0$. Then $Hom(E,F)$ is a 1-dimensional space, and the locus of non-injective maps in it is the point 0, so it is divisorial.
@Sasha, thank you for your observation. Are you aware of any weaker conditions I should be looking for to rule out a divisorial locus? Bundles $E,F$ are composed of exceptional bundles in my case, but as far as I can tell, not much is known about global generation for $\mathcal{Hom}$-bundles for bundles from an arbitrary strong exceptional collection on $\mathbb{P}^1 \times \mathbb{P}^1$.
What do you mean by saying "composed of"?
If $E,F$ are direct sums of vector bundles from an exceptional collection then Sasha's comment applies directly to your situation. Indeed split $E = G_{i_1}\oplus ...\oplus G_{i_q}$ and $F = G_{j_1} \oplus ... \oplus G_{j_p}$. Either all maps from $Hom(E,F)$ are non injective or forall $i_k$ there exists $j_k$ such that $G_{i_k} = G_{j_k}$ then applies Sasha's comment to $Hom(G_{i_k}, G_{j_k})$.
@Sasha, I am working with resolutions of sheaves, arising from spectral sequences associated to an exceptional collection (and I pick a collection in such a way, that dual collection is strong).
@Libli, I am looking at a slightly different situation than what you are suggesting. A model case for me is resolutions of the form $0 \to \mathcal{O}(-1,-1)^a \to \mathcal{O}(-1,0)^b \oplus \mathcal{O}(0,-1)^c \oplus \mathcal{O}^d \to \mathcal{V} \to 0$. I want to replace the line bundles by objects from a different strong exceptional collection: $0 \to E_0^a \to E_1^b \oplus E_2^c \oplus E_3^d \to \mathcal{V} \to 0$.
|
2025-03-21T14:48:29.952164
| 2020-02-28T17:09:04 |
353785
|
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|
Stack Exchange
|
Reconstructing a sine wave using square waves and Möbius inversion: L² convergence?
Let $s$ be the (“square wave”) $1$-periodic real function such that $s(x) = 1$ if $0<x<\frac{1}{2}$ and $s(x) = -1$ if $\frac{1}{2}<x<1$ (and maybe $s(0)=s(\frac{1}{2})=0$ for the sake of definitiveness, but I'll only be using $s$ a.e.). In $L^2(\mathbb{R}/\mathbb{Z})$, we have
$$
s(x) = \frac{4}{\pi} \sum_{m=0}^{+\infty} \frac{\sin(2(2m+1)\pi x)}{2m+1}
\tag{*}
$$
by a straightforward computation of its Fourier coefficients. Now the Fourier coefficients in question are multiplicative, so I'd like to use the Möbius inversion formula to reconstruct the sine wave using the square waves $s(kx)$, namely, I would like to state:
$$
\frac{4}{\pi} \sin(2\pi x) = \sum_{m=0}^{+\infty} \frac{\mu(2m+1)\,s((2m+1)x)}{2m+1}
\tag{†}
$$
where $\mu$ is the Möbius function. Unless I am mistaken, (†) is clear if we just demand weak $L^2$ convergence of the RHS. But I don't know how to say more. Specifically:
Main question: Is there (norm) $L^2$ convergence in (†)?
Bonus points: What can be said about convergence almost everywhere? For even more credits, what can be said about convergence for a particular $x$ (maybe every $x\not\in\mathbb{Q}$)? For super extra credits, discuss uniform convergence (I know the Gibbs phenomenon prevents uniform convergence in (*), but I don't see why this would forbid it in (†)).
Here is a graph of the partial sum for $m\leq 12$ of the RHS of (†):
Remarks:
Graphs, as well as a rendering as a sound wave (which is interesting, if not very pleasant to listen to) of the first few partial sums of the RHS of (†) are given in this Twitter thread.
Lest anyone make the same mistake I made when thinking about this problem at first, the $s(kx)$ are not orthogonal in $L^2(\mathbb{R}/\mathbb{Z})$. (In fact, applying Gram-Schmidt to the $s(kx)$ produces a rather unexpected sequence of functions: see here for some graphs.)
I have a feeling this might be related to the question (which has independent interest) of finding an upper bound on $\sum_{k|n,\;k\leq B\;}\mu(k)$ as a function of $n$ and independently of $B$. Maybe I should ask this as a separate question?
The fact that the value at $0^+$ of the RHS of (†) converges to $0$ is equivalent to the prime number theorem, so the extra credits probably depend on some number theory, but I don't know what to think about mere $L^2$ convergence.
The RHS converges for every x, given that $\sum_{m=0}^M {\mu(2m+1) s((2m+1)x)}=o(M)$ for fixed x. The $o(M)$ result can be found in the footnote on p.3 of the paper Quadratic Uniformity of the Mobius Function, namely, "qualitative versions of these results
... hold for rougher classes such as the continuous class $C^0$, or even piecewise continuous classes, by standard limiting arguments."
|
2025-03-21T14:48:29.952383
| 2020-02-28T17:17:57 |
353786
|
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|
Stack Exchange
|
Problems in advanced calculus
I have been teaching Advanced Calculus at the University of Pittsburgh for many years. The course is intended both for advanced undergraduate students and the first year graduate students who have to pass the Preliminary Exam. This is a difficult course as you can see from the problems that we have on our exam:
http://www.mathematics.pitt.edu/graduate/graduate-handbook/sample-preliminary-exams
What bothers me quite a lot is the lack of a good collection of problems for functions of several variables. There are plenty of excellent collections for functions of one variable and for metric spaces, but there is almost nothing regarding good problems for functions of several variables. The only exception that I know is:
P. N. de Souza, J.-N. Silva,
Berkeley problems in mathematics.
Third edition. Problem Books in Mathematics. Springer-Verlag, New York, 2004.
This is an amazing collection of problems covering many areas of mathematics and what is important the problems have complete solutions.
Question. Do you know a good collection of problems for functions of several variables?
By this I mean a collection of problems that require deep understanding of the problem rather than a standard application of formulas and theorems. I believe that most of the problems in our Preliminary Exam in Analysis fell into this category. Ideally, I would prefer to have a collection with solutions or hints, as it would be very helpful for students (and for me as well).
There are many non published collections of problems available online. I am also interested in links to such collections.
While this might seem as a question that is not research level, I think otherwise. We teach Advanced Calculus to students and if we want them to be ready to do research in Analysis, we need to teach them with such problems.
Edit. I actually knew all the references listed in the answers. Clearly, the answers show that there is no good source of "ready to use" problems in Advanced Calculus of several variables.
I see a perfect fit with textbook-recommendation tag; why the close votes then?
@MateuszKwaśnicki Do you know any such collection of problems? I do not mind if it is written in Polish.
Fichtenholz wrote a three-volume textbook Differential and Integral Calculus, available in a number of languages, but not in English. This used to be "the textbook" for real analysis here in Wrocław, and quite likely not only here. It contains a lot of examples and problems, covering some more advanced topics and including a lot of really difficult exercises, but it is certainly more of a textbook than a collection of problems. And I am pretty much sure you know it already.
For what it's worth, while Volume I of Introduction to Calculus and Analysis by Richard Courant and Fritz John is an excellent source of "lower level" single variable calculus problems, the problems in Volume II (for which 119 pages are devoted to their solutions) are almost entirely straightforward and are primarily designed to reinforce mastery of the text material and techniques, and thus they are probably NOT the type of problems you're looking for.
Section 3.3 of Putnam and Beyond by Răzvan Gelca and Titu Andreescu (Springer, 2007) is entitled "Multivariable Differential and Integral Calculus" and has a number of interesting, non-routine problems.
You might want to have a look at "Multidimensional Real Analysis" parts I (differentiation) and II (integration). They have plenty of problems, and the books are aimed at mid/advanced undergraduates: the books were/are used at Utrecht University for a second-year math course, when students had already taken real (1D) analysis and vector calculus (without all the rigorous proofs).
By this assumptions, I think you can find many interesting problems in the real analysis books that have chapters about several variable calculus. For example, I suggest the book:
Problems and Solutions in Real Analysis, 2nd Edition, by Masayoshi Hata.
There are a lot of good problems with detailed solutions about several variable calculus in this book.
Also, you can see the section of problem and solution of the Mathematical Monthly journals and select the relevant problems.
Nice book, but there are only 5 problems for several variables :(
Yes! But there are many similar books. Also, these gives idea for designing problem by yourself.
Take a look at Schaum's Outline of Advanced Calculus. It has many solved problems, and many unsolved problems
"By this I mean a collection of problems that require deep understanding of the problem rather than a standard application of formulas and theorems." That's precisely what Schaum's Outline problems are not.
|
2025-03-21T14:48:29.952717
| 2020-02-28T17:38:39 |
353788
|
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|
Stack Exchange
|
Motivation for Karoubi envelope/ idempotent completion
This is the second part of my venture to become more comfortable with the concept of idempotent elements and idempotent splittings from category theoretical viewpoint. In the first part we considered the interpretation of idempotent elements & splitting from viewpoint of commutative algebra. The most fruitful analogy (at least for me) was that if we consider the category $\text{$R$-ModFree}$
of free $R$-modules, taking its completion means making it closed under taking direct summands. As the direct summands of free modules are exactly the projective modules completing means "to add some objects" which occur naturally as building blocks.
Now in case of commutative algebra projective objects allows to deal with projective resolutions and provide a framework for direct calculations of derived functors of right exact functors.
I read that there are a lot of constructions spreaded in a relatively wide areas of mathematics where one starts with a certain category $C$, construct from this one another say $F(C)$, and then pass to its idempotent completion $\widehat{F(C)}$.
Probably the most prominent example is the construction of pure motives where we start with category $(\operatorname{Sm}/k)$ of smooth varieties over a field $k$, then pass to category of correspondences $\operatorname{Cor}_k$, build its idempotent completion $\widehat{(\operatorname{Cor}_k)} $ and go ahead with the construction to build the category of Motives $\operatorname{Mot}_k$ and then, by trying to mimic the procedure of building the derived category, we arrive at the category of pure motives (of course that's just a very coarse overview).
The point of my interest is the necessity of taking idempotent completion in the intermediate step.
Of course, that's just an example, but similar strategies occur for example in $K$-theory when one study vector bundles or in constructions dealing with triangulated categories.
My Question: Can there be extracted a common motivation in these examples making the step that takes idempotent completion necessary or does it in every construction almost everywhere strongly depend on "what one wants"?
The only one "general mantra" that I found up to now having the $\text{$R$-Mod}$ example in mind was the necessity of projective objects in order to study right exact functors.
Question: Is this the only motivation or are there some other common deep reasons for the importance of taking idempotent completions?
The point of the idempotent completion is so that direct summands of objects of your category are now objects as well.
Idempotents appear very naturally when studying the cohomology of algebraic varieties. For example if some finite group acts on your variety then you can consider the piece of cohomology cut out by characters of this group (corresponding to idempotents in the group algebra).
@Sam Hopkins: that's true. The point is when we think about the couple of constructions I mentioned (e.g. the motivic case) where it become neccessary to deal with a category beeing closed under taking direct summands? What might fail if not not do it?
I think a very illuminating example is the construction of Deligne's category $\mathrm{Rep}(S_t)$ of "representations of the symmetric group $S_t$" where $t$ is a complex parameter. The idea is that for integer $n$, every representation of $S_n$ is a direct summand of a tensor power of the defining representation. So to mimic this for an arbitrary parameter $t$, first you create via diagrammatic rules a category whose objects correspond to tensor powers of the defining representation; then you take the Karoubian envelope to get all representations.
so if we recall that the motivation of study motives was to develop a "universal " cohomology theory that it seems that passing to idempotent closure allows some nice features. More simply question is if we don't pass to idempotent closure in the construction of pure motives, will we obtain still a theory that is just not so powerful or does the construction fails completly. I was reading Manin's paper on construction of motives but I nowhere found a step which would "totally" fail if we not make pass to the completion. So the question is if it is of "vital" or "enhencing" nature?
@SamHopkins: that's a really nice one! +1
Another example for motives: you need projectors in order to define even very basic objects. For example the Tate motive $\mathbb{Z}(1)$ (or rather its inverse, the Lefschetz motive), you need to decompose the cohomology of the projective line (or the multiplicative group). If you don't have the Tate motive then you can't define motivic cohomology for example.
@FrançoisBrunault: That was the devil in the detail I was looking for. Indeed $\mathbb{P}^1= [\mathbb{P}^1 \times e] =(\mathbb{P}^1, p) \oplus L$ decomposes by point & Lefschetz motive. This leads to Tate motive as $\mathbb{Z}(1)[2]=L$ are related by the shift. Thank you!
The "motivic motivation" is that by idempotent completing correspondences over a finite field one obtains a category of homological motives where Kunneth decompositions of diagonals are available. Moreover, over any field the category of numerical motives is abelian semi-simple.
The proof of the latter statement is relatively simple, and can probaly be generalized to other relevant settings. Yet I do not think that there exists any "deep" and general yoga that says that idempotent completions are crucially important (and that is really relevant for motives).
Another observation is that over a field of positive characteristic $p$ we don't know whether Voevodsky motives of arbitrary varieties belong to the (smallest strict) triangulated subcategory generated by motives of smooth projectives, but they belong to the subcategory generated by Chow motives (if the characteristic $p$ is invertible in the coefficient ring).
About the "deep yoga": According to the nLab, the karoubi envelope is a special case of the cauchy completion of a category. This completion has some characterisations that are not related specifically to idempotents. https://ncatlab.org/nlab/show/idempotent#the_universal_idempotentsplit_completion
I have extended my answer. I suspect that the "yoga things" mentioned in the comments is not really actual for motives.
One very general categorical observation is that the idempotent completion functor can be factorised by first passing from the given linear category $\mathcal{A}$ to the non-unital ring $\bigoplus_{X,Y \in \mathcal{A}}\mathcal{A}(X,Y) $, and then taking the linear category $\mathrm{Idem}(\bigoplus \mathcal{A})$ of idempotent elements. The functors $\bigoplus \dashv \mathrm{Idem}$ form an adjoint pair, and idempotent-complete linear categories are algebras for the resulting monad.
More generally, for a category $\mathcal{C}$ enriched in pointed sets, you first pass to the semigroup $\bigvee \mathcal{C}:= \coprod_{X,Y \in \mathcal{C}}\mathcal{C}(X,Y)$ in pointed sets, then take the category of idempotent elements.
On some level, this means that idempotent completion is the universal way to obtain invariants which are really non-unital in nature. You can recover things like idempotent-complete module categories of a unital ring $R$ without knowing its unit.
My understanding of the use of Karoubian completion for motives is that one would really like to have an abelian category of pure motives (modulo homological equivalence, say). However, we don't know how to adjoin all kernels and cokernels, and the Karoubian completion is the best we can do.
There is a hope for an abelian category of pure motives that has all the nice properties we want. There are many flavours of motives around (Chow, André, Nori, Voevodsky, ...), and each of them satisfies some but not all of the desired properties. You use whichever one is most convenient for your problem.
(As Mikhail Bondarko pointed out: Chow motives modulo numerical equivalence¹ are semisimple abelian, and this is basically the only way we know how to prove Chow motives form an abelian category. However, this result of Jannsen was only proven in 1992, so I don't think it was the original motivation.)
¹The problem with Chow motives modulo numerical equivalence is that it does not have a cohomological realisation, unless we prove standard conjecture D.
Yes, Jannsen result is not very old; that is why I did not put it on the first place in my answer.:) A terminological remark: I think that it is better to speak about numerical motives than about Chow motives modulo numerical equivalence. Moreover, Chow motives are certainly useful, and one does not realy want to replace them by an abelian category.
|
2025-03-21T14:48:29.953549
| 2020-02-28T18:46:15 |
353789
|
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|
Stack Exchange
|
Spacing of prime divisors
The following question naturally came up when dealing with 4-rank of certain class groups. In this case I want to inductively deal with some Legendre symbols, and to do so I want my squarefree integers to be "decently" spaced in the sense below.
Is there an absolute constant $C > 0$ such that for all functions $f$ going to infinity and almost all squarefree integers $n = p_1 \cdot \ldots \cdot p_r$ we have
$$
\prod_{i = 1}^k p_i < p_{k + 1}^C
$$
for all $f(n) \leq k < r$? What if we ask the question instead for almost all $k$?
What is $p_{i+1}$ in the RHS?
Should have been $p_{k+1}$!
And, I suppose, you need $k<r$ instead of $k\le r$? (What is $p_{k+1}$ if $k=r$?)
This is not true for the all $k<r$ problem. Consider random $n$ below $x$, and put $z=\log x$. How many prime factors would a random number have in $[z,z^e]$? This is approximately Poisson with parameter $\sum_{z <p \le z^e} 1/p \approx 1$. So with positive probability you would find numbers with as many prime factors from this interval as you care to specify, which means that no fixed value of $C$ would work.
Look in the work of Ford on the multiplication table problem. There are different ways of phrasing conditions of this type; for example recasting it in terms of the number of prime factors up to some point satisfying a ``barrier" condition. Perhaps one of those reformulations would be useful for you.
|
2025-03-21T14:48:29.953680
| 2020-02-28T19:00:06 |
353790
|
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|
Stack Exchange
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Homology of perfect complexes
I apologize in advance if this question is basic.
If $P_{\bullet}$ is a perfect complex over say a ring $R$ such that
$H_{i}(P_{\bullet})=0 $ if $i\neq n$
$H_{i}(P_{\bullet})=E$ if $i=n$
is $E$ a finitely generated $R$-module ?
What can we say about the homology of a generic perfect complex in general?
Yes. Let $... 0\to P_r \to ... \to P_0 \to 0 ...$ be a complex of projective modules of finite type and denote by $Z_*$ the cycles. If $n=0$ it is clear. If not, $0\to Z_1\to P_1\to P_0\to 0$ is exact and so $Z_1$ is projective and of finite type. Then if $n=1$, $H_1(P)$ is of finite type. If $n\neq 1$, $0\to Z_2\to P_2\to Z_1\to 0$ is exact. And so on.
So the "last" nonzero homology module is of finite type.
It is even finitely presented
See Lemma 14.1.27 of the book Derived Categories (also available at the arXiv at https://arxiv.org/abs/1610.09640).
Dear Amnon: the downvotes might have been confusing, but I suspect they are connected with an unwritten rule at MO to avoid the appearance of author self-promotion, particularly with regard to books for sale.
Thanks for the heads up! I'm glad the arxiv version saved the day.
|
2025-03-21T14:48:29.953791
| 2020-02-28T19:52:17 |
353795
|
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|
Stack Exchange
|
On properties on a certain functional
Consider the following function:
$$F(z) = \omega(z)\sin^2\left(\frac{c\Gamma(z)}{z}\right)$$
Here, $\omega(z)$ is a weight we have to construct and $c$ is a constant.
The following three conditions should meet for $\omega(z)$:
$$\omega(z)>\frac{1}{z},\ \forall z\in\mathbf{R}$$
( More generally this condition is added for divergence of $\int_c^\infty F(x)dx$ So , $\omega(z)$ can even be complex valued for real domain as long as the given integral is divergent )
$$\lim_{ y→∞}|F(x ± iy)|e^{−2πy }= 0$$
$$\int_0^\infty |F(x + iy) − F(x − iy)|e^{−2πy} dy<+\infty$$ for every $x≥1$ and tends to zero as $x\to\infty$.
Question : Can we Explicitly construct $\omega(z)$.
If anyone could omit the first condition and could find the weight satisfying condition 2 and 3 please mention.
(Is it even possible?)
See this MSE post ( see the part after UPDATE) and this MSE post for more details.
The question is inspired by following analysis:
\begin{align}f(x) = {} & \sin^2\left(\frac{π\Gamma(x)}{2x}\right)\\
\sum_{k=2}^p f(k)= {} & \frac{f(2) +f(p)}2 + \int_2^p f(x) \, dx \\ & {}+ i\int_0^∞\frac{f(2+iy) − f(2−iy)}{e^{2πy }− 1} \, dy +i \int_0^∞\frac{f(p-iy) − f(p+iy)}{e^{2πy }− 1} \, dy
\end{align}
As we can see the above summation is 'sort of' a prime counting function.
I'm trying to prove the infinitude of primes by trying to show that the above sum diverges as p tends to infinity.
In doing so I faced the following:
The complex integral in the RHS show oscillation of increasing amplitude. So to convert the summation into real integral I tried to attach weight such that the complex integral tends to zero as p tends to infinity and so we only remained with real integral and some constant terms.
Do you want $\omega(z)$ to be a holomorphic function? What domain?
|
2025-03-21T14:48:29.954068
| 2020-02-28T20:48:45 |
353801
|
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|
Stack Exchange
|
Detecting comprehension topologically
This question basically follows this earlier question of mine but shifting from standard systems of nonstandard models of $PA$ to $\omega$-models of $RCA_0$. For $X$ a Turing ideal we get the map $c_X$ on $2^\omega$ given by $c_X(x)=[b[x]]\cap X$ where $b$ is some computable bijection $\omega\cong 2^{<\omega}$; in the language of the linked question, this is a closed set pattern.
I'm trying to understand how much information this construction loses in the sense of second-order arithmetic. For $X$ a Turing ideal we also get an $\omega$-model of $RCA_0$ - which I'll freely conflate with $X$ itself - and I'm interested in what axioms of second-order arithmetic we can detect topologically.
(Below, $\sim$ denotes "difference by homeomorphism" - for $c,d$ closed set patterns on $\mathcal{X},\mathcal{Y}$ we write $c\sim d$ iff for some $H:\mathcal{X}\cong\mathcal{Y}$ we have $a\in c(b)\leftrightarrow H(a)\in d(H(b))$ for all $a,b\in\mathcal{X}$.)
Say that a theory of second-order arithmetic is topologically detectable if there is some $\sim$-respecting property of closed set patterns which holds of $c_X$ iff $X\models A$ for each Turing ideal $X$. For example:
$WKL_0$ is topologically detectable via "$c_X^{-1}(\emptyset)$ is open."
$ACA_0$ is topologically detectable in at least two ways (after adding $WKL_0$): via "$ran(c_X)$ is closed under (single) Cantor-Bendixson derivatives" and "for each $x\in X$ the set $\{y\in X: c(y)\supseteq c(x)\}$ is closed."
My question is essentially whether any other "weak combinatorial principles" are topologically detectable:
Is there some "reasonably natural" $A$ with $ACA_0\models A$ and $WKL_0\not\models_\omega A$ such that $A$ (or at least $WKL_0+A$) is topologically detectable?
(Here $\models_\omega$ is the restriction of $\models$ to $\omega$-models, which is needed to rule out e.g. $I\Sigma_{17}$. Note that if $A$ is topologically detectable then so is $WKL_0+A$, but the converse isn't obvious to me.)
The most tempting candidate is of course $RT^2_2$, but there are plenty of others.
Is there a good list of things that are know to follow from $ACA_0$ but not $WKL_0$? All of the examples I can find seem to be related to $RT_2^2$.
@JamesHanson Re: your first question, at a glance they don't need that but I was being lazy - we can always throw on "... and ${a: c_X(a)=\emptyset}$ is open" to whatever characterization we have, as long as we're shooting for a principle above $WKL_0$. Re: your second question, yes the paradigm is Ramsey theory - the reverse math zoo is a good source.
Statements about existence of $\omega$-models can be topologically detected.
Specifically, fix $X$ a Turing ideal. For $t\in X$ say that $t$ enumerates a family of sets if:
Exactly one $p\in c_X(t)\cap X$ has $c_X(p)=X$.
For every other $q\in c_X(t)$ we have $c_X(q)=\{a\}$ for some $a\in y$.
For each $a\in y$ there is exactly one $q\in c_X(t)$ with $c_X(q)=\{a\}$.
In such a case we say $t$ enumerates the family $$X_t:=\{q\in X: \exists a\in c_X(t)(c_X(a)=\{q\})\},$$ and we can talk about the induced closed set pattern coming from $X_t$. It's not hard to see$^*$ that every sequence of sets in $X$ (that is, the whole sequence is in $X$) corresponds to such an $X_t$, and this means:
If $A$ is a topologically detectable sentence, so is the statement $O_A$ = "Every real is contained in an $\omega$-model of $A$."
Taking $A=WKL_0$ then gives an affirmative answer to the question. Of course $RCA_0+O_{WKL_0}\vdash WKL_0$: if $X$ is an $\omega$-model of $RCA_0$ and $T$ is an infinite binary tree in $X$, then any $\omega$-model of $WKL_0$ containing $T$ also thinks $T$ is an infinite binary tree - and being a path through a tree is absolute between $\omega$-models. (More generally, we have $$RCA_0+ O_\varphi\vdash\varphi$$ for every $\varphi\in\Pi^1_2$.)
Meanwhile, $RCA_0+O_{RCA_0}\vdash WKL_0$, since from a coded $\omega$-model of $RCA_0$ we can whip up a $DNR_2$ function (and this relativizes). So this approach does not produce an example of a topologically detectable sentence incomparable with $WKL_0$.
$^*$Specifically, given a sequence of reals $F=(f_i)_{i\in\omega}$ let $s_i$ be the natural code for the tree $$\{\sigma\in 2^{<\omega}: \vert\sigma\vert<i\mbox{ or }\sigma\prec f\}.$$ The closure of $\{s_i: i\in\omega\}$ is the set of paths through a pruned tree $T$ - which is computable in $F$ - and $[T]$ has only one "extra" path, corresponding to $2^{<\omega}$.
Of course this isn't totally satisfying - something of Ramseyan flavor would be better, and I'll hold off on accepting this answer in the hopes of a better answer - but it does give something.
|
2025-03-21T14:48:29.954395
| 2020-02-28T21:10:38 |
353803
|
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|
Stack Exchange
|
Understanding the odd-dimensional index
Given a Dirac operator $D$ on a closed odd-dimensional manifold $M$, I've sometimes heard it said that the Fredholm index of $D$ vanishes because it is an ungraded self-adjoint operator, so that $\dim\ker D=\dim\ker D^*$.
However, it seems to me that this is not the correct way to think about things. For example, when one works with more general versions of the index, the odd-dimensional index does not necessarily vanish. Say $M$ is not necessarily compact but there is an action of a group $G$ on $M$ with compact quotient $M/G$. Then the index of $D$ lies in $K_1(C^*(G))$, where $C^*(G)$ is the group $C^*$-algebra of $G$. If one thinks of the index of $D$ as a difference between the kernel and the cokernel (in the sense of finitely generated projective $C^*(G)$-modules), then this would vanish also. But the index in this case should not always vanish.
The technical definition of the index in the odd-dimensional case is given in terms of the exponential map in $K$-theory. I would like to understand this more intuitively, much like how the boundary map in the even-dimensional case can be understood as giving the difference $\dim\ker D-\dim\ker D^*$.
It seems to me that the correct way to understand the odd-dimensional index should somehow involve suspensions and Toeplitz operators; but I cannot piece together exactly how the story should go. So, along these lines, I would like to ask a slightly vague question.
Question: How should one understand the index of Dirac operators on odd-dimensional manifolds?
To clarify, you are no longer asking about the Fredholm index of an elliptic (hence Fredholm) operator on a closed odd-dimensional manifold, but some other notion of an index.
Yes that is correct - I'm looking for an interpretation in the general case, including the closed case.
If $\dim M=8n+1$, the index of the real Dirac operator is given by the parity of $\dim\ker D$. Is that the kind of explanation you were hoping for?
The odd index also comes up naturally in families: if the odd index of a fibrewise operator on odd-dimensional fibres is nonzero, the kernel cannot form a vector bundle over the base. Asobserved by Johannes Ebert, this is an old, but not so well-known fact. It has been exploited by Anja Wittmann in arXiv:1503.02002, see also the references there.
@SebastianGoette Yes that is similar to the type of explanation I was looking for (although I'm not entirely sure yet what I should be looking for).
|
2025-03-21T14:48:29.954598
| 2020-02-28T21:22:15 |
353804
|
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|
Stack Exchange
|
Sum of squares and partitions
This is an off-shot from my previous post on MO.
Given an integer partition $\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$ of $n$, denote $\ell(\lambda)$ to be the length of $\lambda$.
Let $r_2(n)$ denote the number of ways of expressing $n$ as a sum of two squares of integers, look it up on OEIS.
Here is perhaps a "new formulation" of $r_2(n)$:
QUESTION. Is this true? If it is, please either provide a reference or a proof.
$$r_2(n)
=\sum_{\lambda\vdash n}(-1)^{n-\lambda_1}{\prod_{j=1}^{\ell(\lambda)}} \,4\,(\lambda_j-\lambda_{j+1})$$
where $\lambda_{\ell(\lambda)+1}=0$ and the product excludes $\lambda_j=\lambda_{j+1}$.
Example. Take $n=4$. The solutions to $4=x^2+y^2$ are $(\pm2,0), (0,\pm2)$ and hence $r_2(4)=4$. On the other hand, $\lambda=(4,0), (3,1,0), (2,2,0), (2,1,1,0), (1,1,1,1,0)\vdash 4$ so that
$$(-1)^{4-4}4\cdot4+(-1)^{4-3}4\cdot2\cdot4\cdot1+(-1)^{4-2}4\cdot2
+(-1)^{4-2}4\cdot1\cdot4\cdot1+(-1)^{4-1}4\cdot1=4.$$
How far did you test it?
At least as far as $n=50$, but I can bet it is true.
Start by checking that the following formal product can be expanded as a sum over partitions
$$\prod_{i\geq 1}\left(1+\sum_{r\geq 1}a_r(x_1x_2\cdots x_i)^r\right)=\sum_{\lambda}\left(\prod_{j\geq 1}a_{\lambda_j-\lambda_{j+1}}\right)\left(\prod_{j\geq 1}x_j^{\lambda_j}\right)$$
with the convention that $a_0=1$. The proof of this is really just a weighted version of the usual generating function for partitions.
If we set $x_1=t$ and $x_i=-t$ for $i\geq 2$, and all $a_r=4r$ for $r\geq 1$ then the right side becomes
$$\sum_{n\geq 0} t^n\sum_{\lambda\vdash n}(-1)^{n-\lambda_1}{\prod_{j=1}^{\ell(\lambda)}} \,4\,(\lambda_j-\lambda_{j+1})$$
and the identity says that this is equal to the product
$$\prod_{i\geq 1} \left(1+\frac{4(-1)^{i-1}t^i}{(1-(-1)^{i-1}t^i)^2}\right)=\frac{(1+t)^2}{(1-t)^2}\cdot \frac{(1-t^2)^2}{(1+t^2)^2}\cdot \frac{(1+t^3)^2}{(1-t^3)^2}\cdots$$
$$=\prod_{k\geq 1}\frac{(1-t^{2k})^{10}}{(1-t^k)^4(1-t^{4k})^4}.$$
This last final expression is a well known product formula for $r_2(n)$:
$$\sum_nr_2(n)t^n=\frac{\eta(t^2)^{10}}{\eta(t)^4\,\eta(t^4)^4}, \qquad
\text{where $\eta(t)=t^{\frac1{24}}\prod_{k=1}^{\infty}(1-t^k)$ is the Dedekind eta function}.$$
So your identity follows.
This is cool, thanks.
|
2025-03-21T14:48:29.954760
| 2020-02-28T21:43:01 |
353806
|
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Stack Exchange
|
Is there such a thing as a weighted Kan extension?
The title pretty much sums it up.
More in detail. Let $C$, $D$ and $E$ be categories, let $F:C\to D$ and $G:C\to E$ be functors, and let $P:C^{op}\to \mathrm{Set}$ be a presheaf. The colimit of $F$ in $D$ satisfies
$$
D(\mathrm{colim} \,F, d) \cong [C,D](F, d)
$$
for each object $d$ of $D$, where in the right-hand side $d$ denotes the constant functor, and $[C,D]$ the functor category.
This can be seen as a special case of a Kan extension, which satisfies
$$
[E,D](\mathrm{Lan}_G F, K) \cong [C,D](F-,K\circ G-)
$$
for each functor $K:E\to D$. Namely, by setting $E$ the terminal category we get exactly a colimit.
Just as well, a colimit is a special case of a weighted colimit, which satisfies
$$
D(\mathrm{colim}_W \,C, d) \cong [C^{op}, \mathrm{Set}](W-, D(F-, d))
$$
for each object $d$ of $D$. We get an ordinary colimit by setting $W$ to be the constant presheaf at the singleton.
Now, is there a common generalization?
Note that
In the Kan extension, the "dependent variable" of $F$ is paired to $K\circ G$, while in the weighted colimit, it is paired to $W$. So it's unclear how to fit both dependencies together.
One can express Kan extensions as particular weighted colimits - this is not what I'm asking.
(I could ask the same question for the enriched case.)
Any reference would also be welcome.
Have you seen Chapter 4 in Kelly's book on enriched categories?
Yes. Given $F:C\to D$ and a profunctor $H:E$ ⇸ $C$, i.e. a functor $H : C^{\rm op}\times E\to \rm Set$ (or to the enriching category $V$), the $H$-weighted colimit of $F$ is the functor $L : E \to D$ such that each value $L(e)$ is the $W(-,e)$-weighted colimit of $F$ (in a coherent way).
Of course, if $E$ is the unit category this reduces to an ordinary weighted colimit.
On the other hand, if $G:C\to E$ and $H(c,e) = E(G(c),e)$ is the corresponding representable profunctor, this reduces to a (pointwise) Kan extension.
There are real advantages of viewing weighted colimits and Kan extensions in this profunctory light. In particular, this is the natural definition of "weighted (co)limit" that makes sense in the abstract generality of a proarrow equipment or a Yoneda structure. In this paper I found it very useful to obtain a good notion of (co)limit in a new kind of category. It also has good formal properties for relating limits and colimits; see for instance Prop. 8.5 of ibid.
Thank you. Is the paper you mention a nice reference to learn about this approach in detail?
Well, yes and no -- it's mainly focused on the new kind of category, but it's probably as good as most anything else out there.
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2025-03-21T14:48:29.954972
| 2020-02-28T22:10:00 |
353807
|
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"Gil Kalai",
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}
|
Stack Exchange
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Examples for simplicial complexes in which every k-edge is contained in exactly $d$ $k+1$-edges
Are there any(other than the full complex/1-case)?
Is there a name for this ($k$-edge-regular I call it)?
Thanks.
There are many such examples. If d=2 (plus some connectivity) those are called pseudomanifolds, so there are many of those, and there are many examples for larger values of d. When every set of size k is a k-edge these are designs.
I don't think this answer is true. As t-design requires that every (t-1) subset of V is contained in exactly d t-edges. And there could be an hypergraph that is not a design with this property.
Yes, as I said in the answer those examples are designs when every set of $k$ is in the hypergraph. Examples like psudomanifolds, and buildings, and quotients of building are more general; they need not be designs.
Yes, sorry. I misunderstood this to suggest that the name is designs. I don't know why.
I have looked for such construction with 5 vertices(and less). There were none. But it is possible with 6. Here are possible lists of triplets, as returned by Wolfram Mathematica.
{{{1, 2, 3}, {1, 2, 4}, {1, 3, 5}, {1, 4, 6}, {1, 5, 6}, {2, 3, 6}
, {2, 4, 5}, {2, 5, 6}, {3, 4, 5}, {3, 4, 6}}, {{1, 2, 3}, {1, 2,4}
, {1, 3, 6}, {1, 4, 5}, {1, 5, 6}, {2, 3, 5}, {2, 4, 6}, {2, 5, 6}
, {3, 4, 5}, {3, 4, 6}}, {{1, 2, 3}, {1, 2, 5}, {1, 3, 4}, {1, 4,6}
, {1, 5, 6}, {2, 3, 6}, {2, 4, 5}, {2, 4, 6}, {3, 4, 5}, {3, 5, 6}
}, {{1, 2, 3}, {1, 2, 5}, {1, 3, 6}, {1, 4, 5}, {1, 4, 6}, {2, 3,4}
, {2, 4, 6}, {2, 5, 6}, {3, 4, 5}, {3, 5, 6}}, {{1, 2, 3}, {1, 2, 6}
, {1, 3, 4}, {1, 4, 5}, {1, 5, 6}, {2, 3, 5}, {2, 4, 5}, {2, 4, 6}
, {3, 4, 6}, {3, 5, 6}}, {{1, 2, 3}, {1, 2, 6}, {1, 3, 5}, {1,4, 5}
, {1, 4, 6}, {2, 3, 4}, {2, 4, 5}, {2, 5, 6}, {3, 4, 6}, {3, 5, 6}
}, {{1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 6}, {1, 5, 6}, {2,3, 5}
, {2, 3, 6}, {2, 4, 6}, {3, 4, 5}, {4, 5, 6}}, {{1, 2, 4}
, {1, 2, 5}, {1, 3, 5}, {1, 3, 6}, {1, 4, 6}, {2, 3, 4}, {2, 3, 6}
, {2, 5, 6}, {3, 4, 5}, {4, 5, 6}}, {{1, 2, 4}, {1, 2, 6}, {1, 3,4}
, {1, 3, 5}, {1, 5, 6}, {2, 3, 5}, {2, 3, 6}, {2, 4, 5}, {3, 4, 6}
, {4, 5, 6}}, {{1, 2, 4}, {1, 2, 6}, {1, 3, 5}, {1, 3, 6}, {1, 4,5}
, {2, 3, 4}, {2, 3, 5}, {2, 5, 6}, {3, 4, 6}, {4, 5, 6}}, {{1, 2, 5}
, {1, 2, 6}, {1, 3, 4}, {1, 3, 5}, {1, 4, 6}, {2, 3, 4}, {2, 3, 6}
, {2, 4, 5}, {3, 5, 6}, {4, 5, 6}}, {{1, 2, 5}, {1, 2, 6}, {1,3, 4}
, {1, 3, 6}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 6}, {3, 5, 6}
, {4, 5, 6}}, {{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 2, 6}, {1,3, 4}
, {1, 3, 5}, {1, 3, 6}, {1, 4, 5}, {1, 4, 6}, {1, 5, 6}, {2, 3, 4}
, {2, 3, 5}, {2, 3, 6}, {2, 4, 5}, {2, 4, 6}, {2, 5, 6}, {3, 4, 5}
, {3, 4, 6}, {3, 5, 6}, {4, 5, 6}}}
Also, Conlon's hypergraph construction satisfies it:
https://arxiv.org/abs/1709.10006
Some of them are probably isomorphic
These are called $d$-(upper) regular $k$-complexes as defined here
https://arxiv.org/abs/1607.07734
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2025-03-21T14:48:29.955151
| 2020-02-28T23:55:22 |
353812
|
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|
Stack Exchange
|
Paramodular forms with level and Iwahori subgroups?
Given an integer $N>0$, not necessarily prime, we have the paramodular group $K(N) \subset \text{Sp}_{4}(\mathbb{Q})$, which consists of matrices of the form
$$\begin{bmatrix} * & *N & * & *\\ * & * & * & \frac{*}{N} \\ * & *N & * & * \\ *N & *N & *N & * \end{bmatrix} \in \text{Sp}_{4}(\mathbb{Q}) \,\,\,\,\,\,\,\,\,\,\,\,\, * \in \mathbb{Z}$$
We define the group $K^{*}(N) = \langle K(N), V_{N} \rangle$ generated by the paramodular group, and the matrix
$$V_{N} = \frac{1}{\sqrt{N}}\begin{bmatrix} 0 & N & 0 & 0\\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -N & 0 \end{bmatrix}$$
You can equivalently generate $K^{*}(N)$ with the parabolic subgroup isomorphic to $\text{SL}_{2}(\mathbb{Z}) \rtimes \mathbb{Z}^{2}$ and $V_{N}$, i.e. $K^{*}(N) = \langle \text{SL}_{2}(\mathbb{Z}) \rtimes \mathbb{Z}^{2}, V_{N} \rangle$.
So I have a function which transforms correctly under the very similar group $\langle \Gamma_{0}(N) \rtimes \mathbb{Z}^{2}, V_{N} \rangle$. I'm wondering, is this a common group? And does it have a moduli interpretation in abelian surfaces? I thought that replacing $\text{SL}_{2}(\mathbb{Z})$ by $\Gamma_{0}(N)$ might correspond to abelian surfaces with partial level structure and polarization $(1,N)$?
I think what I'm after is close to the Iwahori subgroup (along with $V_{N}$). But not quite. The Iwahori subgroup is $B(N) = \Gamma_{0}^{(2)}(N) \cap K(N)$, where $\Gamma_{0}^{(2)}(N)$ is the obvious Siegel congruence subgroup. The group $\langle B(N), V_{N} \rangle$ is close, but we lose that one non-integral entry $\frac{*}{N}$ in the paramodular group. And that entry seems to be important for me.
@Kimball Right, it's in $\text{Sp}{4}(\mathbb{R})$. This $K^{*}(N)$ formed using $V{N}$ seems standard, for example see page 5 of (http://www.math.unt.edu/~schmidt/dimension_formulas/papers/2016_on_siegel_paramodular_forms_with_small_level.pdf)
Sorry, I misread your question.
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2025-03-21T14:48:29.955281
| 2020-02-29T00:10:09 |
353814
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353814"
}
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Stack Exchange
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Understanding fundamental group of Poincare homology sphere
I'm currently reading Knots, Links, Braids, and 3-Manifolds by V. V. Prasolov and A. B. Sossinsky. I have trouble understanding the following picture. The dashed line denotes a trefoil whose tubular neighborhood is to be cut out, and the thickened line denotes a longitude with frame number 1 around the dashed trefoil. Therefore, after pasting back the solid torus, the meridian would be pasted onto the thickened line therefore bounding a disk in the new 3-manifold. Hence we would have a nontrivial relation in the fundamental group. What I don't understand is the specific order of elements depicted in the circle on the right.
Any help is greatly appreciated. Thank you very much.
It turns out that the order is actually not that important: choose a vertex on the circle to start from and a direction to travel in, and then read letters. If the orientation of the edge disagrees with your direction of travel, invert the letter (so save yourself some trouble by traveling counter-clockwise).
If you and I happen to make identical choices except for a choice of starting vertex, then the words we write down will differ by conjugation (in the free group on the set of generators). If additionally we disagreed about the direction of travel, then one of us will have to invert our word. However, the relations we each come up with will be true in both presentations, because they only differ by conjugation and inversion in the free group.
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2025-03-21T14:48:29.955411
| 2020-02-29T02:51:28 |
353821
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353821"
}
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Stack Exchange
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Linear matrix recurrences with polynomial coefficients
I am interested in solving linear recurrences of the form
$$a_{n+1}=\sum_{i=0}^K n^i X_i + \sum_{i=0}^L n^i Y_i a_n \tag{1}$$
where the $Y_i$ are $N\times N$ matrices, and the $X_i$ and $a_n$ are $N\times 1$ column vectors. (I am specifically interested in the $K=1,L=2,N=3$ case).
In the scalar ($N=1$) case, there is literature on how to find a closed-form expression for $a_n$ in terms of hypergeometric functions (Petkovšek's algorithm), and indeed computer algebra software returns an explicit solution for (1) in the scalar $N=1,K=1,L=2$ case.
Are there known techniques for solving (1) in the matrix case?
|
2025-03-21T14:48:29.955480
| 2020-02-29T04:38:53 |
353826
|
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"Matthew Daws",
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"https://mathoverflow.net/users/4832"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353826"
}
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Stack Exchange
|
Limit of spectral projection of increasing sequence of positive operators
Let $\mathcal M$ be a von Neumann algebra. Suppose $(x_\alpha)\subset \mathcal M$ is bounded increasing net of positive operators converging to a positive operator $x\in\mathcal M.$ Is it true that $\chi_{[0,1]}(x_\alpha)$ is an increasing net of positive operators converging to $\chi_{[0,1]}(x)$?
I think this is isn't true even for a commutative Von Neumann algebra, where the spectral projections have a very concrete form.
If $x_1 = 1/2$ and $x_2 = 2$ then $\chi_{[0,1]}(x_i)$ is not increasing.
Another example to consider is $\mathcal{M} = L^\infty([0,1])$, $x_n(t) = (2 - \frac{1}{n})t$, $x(t)=2t$. We have $x_n$ increasing to $x$ (in norm, i.e. uniformly on $[0,1]$), but $\chi_{[0,1]}(x_n)$ do not converge in norm.
|
2025-03-21T14:48:29.955556
| 2020-02-29T05:48:10 |
353828
|
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"url": "https://mathoverflow.net/questions/353828"
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|
Stack Exchange
|
Normalizer of a Normalizer of a subgroup of a finite group with no elements of order $p^2$
Coming from a non-group theory background, I noticed that the finite groups I was dealing with seem to all have the following property. Let $G$ be a finite group, $H$ a subgroup. Then the normalizer $N_G(N_G(H))$ of the normalizer of $H$ is just $N_G(H)$. It seems to be an exercise in almost any group theory book that this is true if $H$ is a Sylow subgroup, and it is definitely not true in general. The easiest examples are the dihedral group $D_8$of order 8, and similar examples for groups of order $p^3$, for any odd prime $p$. But these groups have an element of order $p^2$. Is there an easy example where the $N_G(H) = N_G(N_G(H))$ does not hold for some subgroup $H$ of $G$, where $G$ has no elements of order $p^2$, for any prime $p$ (including 2)?
|
2025-03-21T14:48:29.955629
| 2020-02-29T06:18:04 |
353829
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353829"
}
|
Stack Exchange
|
Reference request- Shock development problem for the compressible Euler equation in 1D
I was wondering if there is any good reference discussing the shock development problem for Euler in 1D? Something in the spirit of Christodoulou's work on the same for higher dimension.
I am interested in the quantitative behavior of the solution to the Euler equation as the shock strength increases starting from when the characteristics first collide. Any help will be appreciated.
|
2025-03-21T14:48:29.955688
| 2020-02-29T08:50:38 |
353832
|
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"authors": [
"Deane Yang",
"Igor Belegradek",
"MathDG",
"https://mathoverflow.net/users/1573",
"https://mathoverflow.net/users/613",
"https://mathoverflow.net/users/90594"
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"url": "https://mathoverflow.net/questions/353832"
}
|
Stack Exchange
|
From Riemannian curvature to Ricci curvature in warped product manifold
Let $M=B \times_f F$ be a warped product of two pseudo-Riemannian manifolds. If $X, Y, Z \in L(B)$ and $U, V, W \in L(F)$, (with $L(B)$ and $ L(F)$ I mean the set of all horizontal and vertical lift to M), then we have:
(1) $R(V, W)U = {R(V, W)U}^F +
\frac{\langle \nabla f, \nabla f \rangle }{f^2}(\langle V, U \rangle W − \langle W, U \rangle V )$
(with ${R(V, W)U}^F$ is lift to $M$ to the curvature tensor of $F$).
and Ricci curvature:
(2) $\operatorname{Ric}(V, W)={\operatorname{Ric}(V,W)}^F-(\frac{\Delta f}{f}+(k-1)\frac{\langle \nabla f , \nabla f \rangle }{f^2})\langle V,W \rangle$.
(where $k$ is the dimension of $F$).
From (1) to (2), $k$ turns out (which is the dimension of the fiber manifold) and will be due to some trace, but I don't understand if it will be a trace where they get $k$ and then they find something like:
$k \frac{\langle \nabla f, \nabla f \rangle }{f^2}-\frac{\langle \nabla f, \nabla f \rangle }{f^2}$, then:
$(k-1)\frac{\langle \nabla f, \nabla f \rangle }{f^2}$.
Or a trace in which a dimension is excluded for some reason, therefore not get $k$, but $k-1$, then $(k-1)\frac{\langle \nabla f, \nabla f \rangle }{f^2}$.
How is that $(k-1)$ found?
I would try working out the case when both $B$ and $F$ have dimension $1$.
Write everything in terms of an orthonormal basis of $T_bB$ and another one of $T_fF$, and take the trace. A related observation is that the Ricci curvature of the $n$-sphere is $\mathrm{Rc}(V,W) = (n-1)g(V,W)$.
The n-sphere, $S^n$, with the round metric is Einstein with $k=n-1$, i.e. $Ric=kg$, then $Ric=(n-1)g$. Now if we take the trace we have:$R=n(n-1)$. Then for n-Sphere $(n-1)$ is the constant $k$, not the result of the trace.
The skew symmetry of the Riemann curvature tensor causes one of the terms in the trace to vanish. That's also why the Ricci curvature of the unit $k$-sphere is $k-1$ and not $k$.
This is all easier to see if you write the Riemann curvatures with respect to an orthonormal bases of the base, fiber, and product. The curvature components in the fiber directions contain a term like $$ \frac{\langle\nabla f,\nabla f\rangle}{f^2}(\delta_{ij}\delta_{ab}-\delta_{ia}\delta_{jb})$$
When you take the trace over the indices $a$ and $b$, you get the $k-1$.
Using (1), the relevant trace is the following, where $e_1, \dots, e_k$ is an orthonormal frame on $F$:
\begin{align*}
\mathrm{Ric}(V,W) &= \cdots - \sum_{i=1}^k \langle e_i, R(V,e_i)W\rangle\\
&=
\cdots - \sum_{i=1}^k\frac{\langle\nabla f,\nabla f\rangle}{f^2}(\langle e_i,e_i\rangle\langle V,W\rangle - \langle V, e_i\rangle\langle W,e_i\rangle)
\end{align*}
Dear prof. Yang, and how does the term $\frac{\Delta f}{f}$ come out?
In Besse "Einstein Manifold" in Corollary 9.105 you can find:
Here, you can guess how (9.105c), (9.105e) and (9.105d) can contribute to (9.106a) which is yours (2).
|
2025-03-21T14:48:29.955995
| 2020-02-29T09:54:40 |
353834
|
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|
Stack Exchange
|
Compare degrees of a finite extension of domains and quotient domains
Let $A \subset B$ be a finite (finite type + integral) extension of integral domains and let $\mathfrak{p} \subset A, \mathfrak{q} \subset B$ be prime ideals such that $\mathfrak{q} \bigcap A =\mathfrak{p}$. Let $A/\mathfrak{p} \subset B/\mathfrak{q}$ be the induced inclusion of domains. Suppose $A$ is integrally closed.
Here is my question:
Is $A/\mathfrak{p} \subset B/\mathfrak{q}$ always a finite extension? If so, do we have $[K(B):K(A)] \ge [K(B/\mathfrak{q}):K(A/\mathfrak{p})]$?
PS: I only know this result is trivial in algebraic geometry (for varieties over a field $k$, [Fulton]'s intersection theory has a very nice formula on comparing the degrees with the ramification indices. I believe this is also easy when $K(B/\mathfrak{q})/K(A/\mathfrak{p})$ is a simple extension. But I am curious how general can it be. )
PS2: I still don't know how to prove the general case. If $K(B/\mathfrak{q})/K(A/\mathfrak{p})$ is separable, then this is a result of [Atiyah,MacDonald, Proposition 5.15].
The answer first question is trivially positive. For the second, it is true if $A$ is noetherian and normal, but false in general (think of the normalization).
@Angelo Thank you! I try to write a proof myself. The condition A being normal is necessary, but I don't know why we need to assume noetherian. Would you please explain a little bit more? Thank you!
I still don't understand how to figure out this problem. Anyone knows?
|
2025-03-21T14:48:29.956110
| 2020-02-29T10:20:28 |
353835
|
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"url": "https://mathoverflow.net/questions/353835"
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|
Stack Exchange
|
Linear-time logspace encodable error correcting code with constant
Is there a binary code with (quasi)constant rate, constant relative distance, and an encoder that takes (quasi)linear time and logspace simultaneously? Note that there are no constraints on decodability - really, I just want a fast and logspace deterministic way to map input strings so that they far from each other. I'd also be okay with all the (quasi) parts being replaced with blowup that is less than $n^{\epsilon}$ for every $\epsilon< 0$.
Two codes I found that are close but not quite are:
Spielman's codes don't quite work out-of-the-box cause, while they nominally have logspace and linear time encoding and decoding with constant rate/distance, it seemed like they have superlinear construction time, and I think I need an algorithm that, given a message, outputs a codeword in linear time including any preprocessing.
I don't think the code described in Babai et al. works because I don't think encoding is naively in logspace.
Binary alphabet, or?
Yes, sorry - edited to make this clear
To the best of my understanding, polar codes have sublinear time complexity but memory requirement is much larger than logspace, probably also just sublinear but not better. so they would be ruled out
|
2025-03-21T14:48:29.956220
| 2020-02-29T10:24:13 |
353836
|
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|
Stack Exchange
|
A struggle with jets and Grothendieck vs Ehresmann connections
Let $X\to Y$ be a $C^\infty$ submersion. Consider the following two sheaves.
The sheaf on $Y$ comprised of jets of sections of $X\to Y$.
The sheaf on $X$ given by the quotient of $\Delta_{X/Y}^{-1}C^\infty_{X\times_YX}$ by the second power of the ideal of $C^\infty$ maps about the diagonal vanishing on the diagonal.
Consider the following three definitions of a connection on the submersion $X\to Y$.
Let $J^1_{X/Y}\to Y$ be the bundle of 1-jets of sections of $X\to Y$. Retaining the constant term of a Taylor expansion gives a bundle map $J^1_{X/Y}\to X$ over $Y$. A connection on $X\to Y$ is a section of this bundle map.
Write $\Delta^{(1)}_{Y}$ for the locally ringed space structure on the underlying topological space of $Y$ with structure sheaf given by the above quotient of $\Delta_{Y}^{-1}C^\infty_{Y\times Y}$. It comes with two canonical locally ringed space projections $p_1,p_2:\Delta_{Y}^{(1)}\rightrightarrows Y$. A Grothendieck connection on $X\to Y$ is an isomorphism between the pullbacks (in the category of locally ringed spaces) of $X\to Y$ along $p_1,p_2$ which restricts to the identity on the diagonal.
An Ehresmann connection on $f:X\to Y$ is a section of the differential $\mathrm df:\mathrm TX\to f^\ast \mathrm TY$ (in the category of $C^\infty$ vector bundles over $X$).
I understand the equivalence of the first and third definitions. For instance, starting with a connection in the sense of the first definition, an Ehresmann connection can be constructed using derivatives of local sections of $X\to Y$.
Question. How are Grothendieck connections related/equivalent to the first/third definition?
Added. For what it's worth, an explicit construction of pullbacks in the category of locally ringed spaces is sketched in this answer. In particular, it seems that on stalks, a Grothendieck connection would give an automorphism of the stalk $(C_{X,x}^\infty\otimes_{C^\infty_{Y,y}}\mathcal O_{\Delta^{(1)}_Y,y})_\mathfrak q$ where $fx=y$ and $\mathfrak q$ is a prime ideal which pulls back to the maximal ideals of $C_{X,x}^\infty,\mathcal O_{\Delta^{(1)}_Y,y}$. The structure sheaf of the first neighborhood of the diagonal of $Y$ consists of 1-jets of real $C^\infty$ map on $Y$, so in coordinates its stalk is isomorphic to the ring of dual numbers $\mathbb R[\varepsilon_1,\dots \varepsilon _{\dim Y}]$. Even on this level of stalks I don't know how to relate to a section of $\mathrm df:\mathrm TX\to X\times_Y\mathrm TY$, though the explicit construction of the pullback in $\mathsf{LRS}$ hints at viewing tangents as derivations.
I suppose $k=1$?
Yes. Removed that bit.
It might be useful to look at this question and its answers. https://mathoverflow.net/questions/68305/grothendieck-connections-and-jets?rq=1
Dear @TomGoodwillie I spent a while there but wasn't able to relate to Ehresmann connections.
|
2025-03-21T14:48:29.956398
| 2020-02-29T11:35:00 |
353839
|
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|
Stack Exchange
|
Proving that $(f,g)$ are Cauchy data for the Schrödinger equation iff $(f,g)$ satisfies an equation
I have to prove that if $f\in H^{1/2}(\partial\Omega)$ then $(f,g)$ are Cauchy
data for the Schrödinger equation if and only if
$$g=\gamma^{-1/2} \Lambda_{\gamma}(\gamma^{-1/2} f)+1/2 \gamma^{-1}\frac{\partial\gamma}{\partial\nu}f,$$
where $g$ satisfies $$g(\varphi)=\int_{\Omega}\nabla u\cdot\nabla\overline\phi-\int_{\Omega}\frac{-\Delta\gamma^{1/2}}{\gamma^{1/2}}u\overline\phi\,.$$
Then my try is considering a solution for the Schrodinger equation as $\omega=\gamma^{1/2}u$ and then using a Dirichlet condition $\gamma^{-1/2} f$. But then using the mapping $\Lambda_{\gamma}(\gamma^{-1/2} f)(\varphi)=\int_{\Omega}\gamma^{1/2}\nabla\omega\nabla\overline\phi$ the term of the $\gamma^{-1/2}$ cancelled and using the gradient of the product it is almost the left hand side, but the problem is with the term $1/2 \gamma^{-1}\frac{\partial\gamma}{\partial\nu}f$ because I do not know how to pass to an integral. I can not apply the Green formula because we do not have the integral of this term in order to obtain the laplacian.
Does someone know how to proceed from this step? Thanks in advance.
|
2025-03-21T14:48:29.956502
| 2020-02-29T12:41:52 |
353844
|
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|
Stack Exchange
|
Are there any techniques that can be used in the case when a Neumann series doesn't converge?
Suppose we have a bounded linear operator $A = A(\gamma):H_1\to H_2$ where $H_1$ and $H_2$ are Hilbert spaces and $\gamma>0$ is some parameter, and we are interested in the solution to
$$
(I-A)x = y.
$$
If $\|A\|<1$ we can use a Neumann series expansion and get a series representation:
\begin{align}
x & = (I-A)^{-1} y \\
& = \sum_{j=0}^\infty A^j y
\end{align}
Now, suppose that when $\gamma$ becomes small enough $\|A(\gamma)\|$ becomes greater than $1$ and the Neumann series won't converge; thus we can't get a series representation for $x$.
An example of this situation can be found in problems featuring scattering of waves among disjoint objects. If the solution is represented as a Neumann series, it can fail to converge if frequencies become high or if distances between objects become small.
Are any techniques that can used in such a situation to 'get around' the non-convergence of a Neumann series and obtain a series representation for $x$?
You would still win if the spectral radius of $A$ is less than $1$. If the spectral radius of $A$ is $\geq 1$ then $I$ might be in the spectrum of $A$ and then there is no hope of defining $(I-A)^{-1}$ as a bounded operator
BTW you have $I+A$ at one point and $I-A$ on the next line; I assume one of these is a typo?
For symmetric operators Hilbert Space and the Padé Approximant, non-paywalled Google Books preview starting p.197 . Not a "series representation".
@KeithMcClary I have seen Pade approximants mentioned in the context of accelerating the convergence a Neumann series but I didn't know they can also work when a Neumann series doesn't converge. So I'll take a look into this, although I am dealing with non-symmetric operators so maybe it can't be applied to my case.
If the spectrum of $A$ is contained in a disk $\{z: |z - a| \le r\}$ where $|1-a| > r$, then the series $\sum_{n=0}^\infty (1-a)^{-1-n} (A - a I)^n$ converges to $(I-A)^{-1}$.
Of course, one only has a chance if $1$ is not in the spectrum of $A$.
Robert Israel's answer gives a series that converges to the resolvent $(I-A)^{-1}$ if the spectrum of $A$ is, for instance, contained in a disk with radius larger then $1$, but centered sufficiently far in the left half plane.
Another method to obtain a series representation for $(I-A)^{-1}$ is based on the following fact:
Proposition. Let $\mu \in \mathbb{C}$ be in the resolvent set of $A$ and let $\lambda \in \mathbb{C}$ be a number such that $|\lambda - \mu| < \frac{1}{\|(\mu I - A)^{-1}\|}$. Then $\lambda$ is also in the resolvent set of $A$ and the resolvent of $A$ at $\lambda$ is given by
$$
(\lambda I - A)^{-1} = \sum_{k=0}^\infty (\mu - \lambda)^k (\mu I - A)^{-(k+1)}.
$$
Proof. This follows readily from the Neumann series expansion if we use that
$$
\lambda I - A = (\lambda - \mu) I + \mu I - A = (\mu I - A)^{-1}\Big( I - (\mu - \lambda)(\mu I - A) \Big).
$$
If one is interested in the case $\lambda = 1$, but the spectral radius of $A$ is $\ge 1$, one could for instance try the following procedure:
Choose a real number $r > r(A)$ (where $r(A)$ denotes the spectral radius) and use the Neumann series to compute $(rI - A)^{-1}$. In case that the interval $[1,r]$ is in the resolvent set of $A$, one can now move a bit left of $r$ and compute the resolvent at this new point by means of the above proposition. Then, again, one can move a bit more left, and iterate this procedure until one arrives at $1$. Thus, one obtains a "representation" of $(I - A)^{-1}$ by means of a finitely often iterated series expansion.
Whether this is useful or not depends of course on the application one has in mind. Sometimes this (or a related) technique can be quite useful for theoretical purposes; on the other hand, I would suspect that the procedure is completely unsuited for, say, numerical computations.
Nice! Does this approach still work if the spectrum is say the unit circle and you want a series for some of the resolvent which is inside the unit circle?
|
2025-03-21T14:48:29.956763
| 2020-02-29T12:53:19 |
353846
|
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"Joseph O'Rourke",
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|
Stack Exchange
|
Convex Triangulations II
While the original question Convex triangulations was aimed at the existence and calculation of convex triangulations of a given set of $n$ points in the Euclidean plane, I would like to ask the opposite
Question:
where do pointsets, that come close to the ideal of having a convex Delaunay triangulation, "naturally" occur?
It appears to me that the pointsets that are generated via Weighted Voronoi Stippling algorithms diagrams or the nuclei of 2D biological cell structures are almost free of non-convex maximal unions of Delaunay triangles with a common inner points as a vertex.
Perhaps the more fruitful direction is to try to expand @JanKyncl's comment at the originating posting that internal degrees need to be $\ge 5$, to understand the structure of these triangulations. Maybe only then will it be clearer where these "naturally occur."
|
2025-03-21T14:48:29.956848
| 2020-02-29T14:08:36 |
353849
|
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|
Stack Exchange
|
Positive-semidefiniteness of a Gram-like matrix
Let $x_0, x_1, \ldots x_{n-1}$ be arbitrary vectors in a complex Hilbert space. Define the $n \times n$ symmetric real matrix $M$ by $M_{ij} = \lvert \langle x_i, x_j \rangle \rvert^2$. Must $M$ be positive semidefinite?
Yes. The matrix $A$ with $a_{i,j} = \langle x_i,x_j \rangle$ is a Gram matrix and thus positive semidefinite, so $A^T = \overline{A}$ is positive semidefinite too. It then follows from the Schur product theorem that your matrix $M = A \circ \overline{A}$ (where $\circ$ denotes the entrywise/Hadamard product) is positive semidefinite too.
|
2025-03-21T14:48:29.956920
| 2020-02-29T14:48:44 |
353853
|
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"authors": [
"Jiří Rosický",
"Leonid Positselski",
"Martin Brandenburg",
"Tim Campion",
"https://mathoverflow.net/users/2106",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/2841",
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|
Stack Exchange
|
Coreflective subcategories in Grothendieck/locally presentable categories
This question is a reference request for the following result or two results, which I believe are rather easy to prove.
Lemma. Let $\mathcal K$ be a locally presentable category and $\mathcal A\subset\mathcal K$ be a coreflective full subcategory. Assume that the coreflector $C\colon\mathcal K\to \mathcal A$ is an accessible functor (e.g., when viewed as a functor $\mathcal K\to\mathcal K$; this means that there exists a cardinal $\lambda$ such that $C$ preserves $\lambda$-directed colimits). Then
The category $\mathcal A$ is locally presentable.
If $\mathcal K$ is a Grothendieck abelian category and $\mathcal A$ is closed under kernels in $\mathcal K$, then $\mathcal A$ is a Grothendieck abelian category, too.
Is there any relevant reference? I was only able to find Corollary 6.29 in the book of Adámek and Rosický "Locally presentable and accessible categories". This corollary claims, among other things, that any coreflective full subcategory $\mathcal A$ in a locally presentable category $\mathcal K$ is locally presentable, if one assumes Vopěnka's principle.
My lemma above does not depend on Vopěnka's principle or any other set-theoretical assumptions. Part 1. of it is an elementary version of this corollary from the book of Adámek and Rosický. Is there any other/better reference?
Some context: part 2. of the lemma is a generalization of Lemma 3.4 from my preprint S.Bazzoni, L.Positselski "Matlis category equivalences for a ring epimorphism", https://arxiv.org/abs/1907.04973 .
Since a coreflective full subcategory is the category of coalgebras for the induced idempotent comonad, 1. is answered in presentability rank of categories of coalgebras (the corresponding comonad is accessible).
I think the author asks for a reference.
I haven't looked, but I'm presuming that the linked paper shows that the coalgebras for an accessible comonad on an accessible category are accessible, which would be a perfectly adequate reference for (1). But it's equally the case that the coalgebras for an idempotent comonad are the inverter of the underlying copointed endofunctor. Inverters are PIE limits, so it follows from well known results which go back at least to Makkai and Pare that if the endofunctor is accessible, then the coreflective subcategory is accessible. Of course, it's also cocomplete if the original category is.
Actually, I think the canonical reference for the fact that accessible categories and accessible functors are closed under PIE limits might be Greg Bird's thesis, which should be available on Ross Street's website.
@TimCampion Thank you. It is becoming clearer now.
@JiříRosický Thank you. My understanding of it is much better now.
TimCampion Your answer is better and the reference to Bird's thesis is right. There is also F. Ulmer, Bialgebras in locally presentable categories, Univ. Wuppertal 1977 which was never published.
|
2025-03-21T14:48:29.957132
| 2020-02-29T15:40:43 |
353860
|
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"Anonymous amateur",
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|
Stack Exchange
|
Taylor expansion of determinant of Riemannian metric in normal coordinates up to higher order
Let $(M,g)$ be an $n$-dimensional Riemannian manifold. Let $p\in M$, and let $\{x^i\}_{i=1}^n$ be normal coordinates centered around $p$.
Using Jacobi field, one can show that the metric $g$ has the following Taylor expansion
\begin{align}
g_{ij}(x)&=\delta_{ij}-\frac{1}{3}R_{ipqj}(p)x^px^q-\frac{1}{6}\nabla_rR_{ipqj}(p)x^px^qx^r \\
&\qquad+\left(-\frac{1}{20}\nabla_r\nabla_rR_{ipqj}(p)+\frac{2}{45}g^{kl}R_{ipqk}R_{jrsl}(p)x^px^qx^rx^s\right)+O(|x|^5)
\end{align}
where $\nabla$ is the Levi-Civita connection of $(M,g)$, $R_{ijkl}$ are the (components of the) Riemann curvature tensor, while $x$ is a point near $p$ with coordinates $x^i$, and $|x|:=d(x,p)$, the radial distance from $p$.
Using this, together with the Jacobi formula for derivative of determinant function, one should be able to obtain the Taylor expansion of $\det(g_{ij})$. It is claimed (e.g. in Hamilton's Ricci flow page 59) that
\begin{align}
\det(g_{ij})(x)&=1-\frac{1}{3}R_{ij}(p)x^ix^j-\frac{1}{6}\nabla_kR_{ij}(p)x^ix^jx^k \\
&\quad-\left(\frac{1}{20}\nabla_l\nabla_kR_{ij}(p)+\frac{1}{90}g^{pq}g^{rs}R_{pijr}R_{qkls}(p)-\frac{1}{18}R_{ij}R_{kl}(p)\right)x^ix^jx^kx^l \\
&\quad+O(|x|^5)
\end{align}
where $R_{ij}$ are the (components of the) Ricci curvature tensor.
My question is that
How do we obtain the term
\begin{align}
\frac{1}{90}g^{pq}g^{rs}R_{pijr}R_{qkls}(p)-\frac{1}{18}R_{ij}R_{kl}(p)
\end{align}
I believe it should come from the term $\displaystyle\frac{2}{45}g^{kl}R_{ipqk}R_{jrsl}(p)$ in the expansion of $g_{ij}$. By using Jacobi's formula and evaluating at $p$ (since $p$ is the point where $x=0$, many terms will vanish), one should see that the coefficient of $x^ix^jx^kx^l$ is just
\begin{align}
\frac{1}{4!}g^{ab}\partial_i\partial_j\partial_k\partial_lg_{ab}(p)
\end{align}
where $\partial_i=\frac{\partial}{\partial x^i}$. In other word, I expect that we only need to take the trace (and take care of the counting of possibly repeated terms). But then tracing $$\displaystyle\frac{2}{45}g^{kl}R_{ipqk}R_{jrsl}(p)$$ doesn't seem to yield the desired result. In particular, I am not quite sure how $R_{ij}R_{kl}$ pops out.
Any comment, hint or answer are greatly appreciated.
After dropping the first order terms using the normal coordinate condition,
$$\partial^4_{ijkl} \det(g) = \partial^3_{ijk} (g^{-1} \partial_l g) = g^{-1} \partial^4_{ijkl} g + ( \partial^2_{ij} g^{-1} \partial^2_{kl} g + \partial^2_{ik} g^{-1} \partial^2_{jl} + \partial^2_{jk} g^{-1} \partial^2_{il} g)$$
Note that
$$ \partial^2_{ij} g^{-1} = -g^{-1} (\partial^2_{ij} g) g^{-1} $$
after dropping the first order terms. So you should get contributions both from the quartic part of the expansion of $g$ and the quadratic part of the expansion of $g$, the latter of which you don't seem to have accounted for.
Thanks so much. Yes my mistake was exactly the omission of the contributions of the quadratic part.
|
2025-03-21T14:48:29.957295
| 2020-02-29T16:31:45 |
353864
|
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|
Stack Exchange
|
Why is the exterior algebra symbol a Lambda?
My guess is that this is because the capital lambda looks like a wedge. In this case why is the symbol a wedge.
Does anyone have any sources for this?
It's not a $\Lambda$, it's a $\bigwedge$.
Probably this question belongs at https://hsm.stackexchange.com/
|
2025-03-21T14:48:29.957356
| 2020-02-29T17:29:46 |
353868
|
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|
Stack Exchange
|
Inductive definition of Bernstein polynomials
For $n\in \mathbb{N}$ let $B_n$ be the linear operator taking a function $f$ on the unit interval $I=[0,1]$ to its $n$-th Bernstein polynomial $B_nf$,
$$ B_nf(x):=\sum_{k=0}^n\binom{n}{k} f\Big(\frac{k}{n}\Big)x^k(1-x)^{n-k}\label{1}\tag{1}$$
The polynomial $B_nf(x)$ has a natural probabilistic interpretation, namely, it is the expected value of $f(\xi)$, where $\xi=\frac{1}{n}\sum_{j=1}^n \omega_j$ is the average of $n$ independent random variables with identical Bernoulli distribution of parameter $x$, that is, $\mathbb{P}(\omega_j=1)=x$. In fact, this is the starting point in the beautiful Bernstein's proof of the Weierstrass' density theorem via the WLLN. However, this question is about an alternative definition of the sequence $(B_n)_{n\ge0}$.
Let $D:C^1(I)\to C^0(I)$ be the derivative operator, and for all $n\ge1$, let $D_n:C^0(I)\to C^0(I)$ be the approximate discrete derivative given by the incremental ratio
$$D_nf(x):=\frac{f\big( \frac{n-1}{n} x+\frac{1}{n}\big)-f\big( \frac{n-1}{n} x\big)}{\frac{1}{n}}, $$
(which is well-defined for $f\in C^0(I)$ and $x\in I$).
It is easy to check that definition \eqref{1} implies
$$DB_n=B_{n-1}D_n\label{2}\tag{2}$$
together with:
$$B_0f(x)=B_nf(0)=f(0)\label{3}\tag{3}$$
Conversely these two imply formula \eqref{1}, as it follows immediately by induction, at least, if we already have it (quite a common situation of formulas proven by induction). Thus, since \eqref{2} and \eqref{3} characterize $(B_n)_n$, we may even take them as an inductive definition of $(B_n)_n$. Note that replacing $D_n$ with $D$ in \eqref{2} gives the analogous inductive definition for the Taylor polynomials in $0$. (Incidentally, formula \eqref{2} is relevant in the approximation theory, in that it implies that for $f\in C^k(I)$ one has $B_nf\to f$ in $C^k$: this by induction from the case $k=0$, since $D_n$ converges strongly to $D$. Also, it says that if some derivative $f^{(k)}$ is non-negative on $I$, so is $(B_nf)^{(k)}$.)
Question: How can we deduce naturally formula \eqref{1} (i.e., assuming we don't know it, and we do not have a crystal ball to guess it) from \eqref{2} and \eqref{3}?
$\newcommand{\De}{\Delta}$
Iterating your condition \eqref{2}, for $k=0,\dots,n$ we have
\begin{equation*}
D^kB_n=\frac{n!}{(n-k)!}\,B_{n-k}P_{n,k},\label{a}\tag{a}
\end{equation*}
where
\begin{equation*}
P_{n,k}:=\De_{n-k+1}\cdots\De_n,\quad \De_j:=\tfrac1j\,D_j.
\end{equation*}
By induction on $k=0,\dots,n$,
\begin{equation*}
(P_{n,k}f)(x)=\sum_{i=0}^k(-1)^{k-i}\binom ki f\Big(\frac{n-k}n\,x+\frac in\Big),\label{b}\tag{b}
\end{equation*}
whence, using \eqref{a} and taking your condition \eqref{3} into account, we have
\begin{equation*}
\frac{(n-k)!}{n!}\,(D^kB_n f)(0)=(B_{n-k}P_{n,k}f)(0)=(P_{n,k}f)(0)
=\sum_{i=0}^k(-1)^{k-i}\binom ki f\Big(\frac in\Big).
\end{equation*}
Also, using again \eqref{a} and \eqref{b}, and again taking your condition \eqref{3} into account, we have
\begin{equation*}
\frac1{n!}\,(D^nB_n f)(x)=(B_0P_{n,n} f)(x)=(P_{n,n} f)(0)
=\sum_{i=0}^n(-1)^{n-i}\binom ni f\Big(\frac in\Big),
\end{equation*}
a constant. So, $B_n f$ is a polynomial of degree $\le n$, and hence
\begin{align*}
(B_n f)(x)&=\sum_{k=0}^n \frac{(D^kB_n f)(0)}{k!}\,x^k \\
&=\sum_{k=0}^n\binom nk x^k
\sum_{i=0}^k(-1)^{k-i}\binom ki f\Big(\frac in\Big) \\
&=\sum_{i=0}^n f\Big(\frac in\Big)\sum_{k=i}^n (-1)^{k-i}\binom nk \binom ki x^k \\
&=\sum_{i=0}^n f\Big(\frac in\Big)\binom ni x^i(1-x)^{n-i},
\end{align*}
as desired.
thank you, very nice
A comment on Josif Pinelis' formula $(b)$ for $\Delta_{n-k+1} \dots\Delta_{n-1}\Delta_{n}$, which is a main point of the computation. Let $\{\tau_{a}\}_{a\in\mathbb{R}}$ and $\{\delta_{b}\}_{a\in\mathbb{R}_+}$ denote respectively the linear group of translations on functions (that we may think defined on the whole real line w.l.o.g.), $f(\cdot)\mapsto f(\cdot+a)$, and the linear group of dilations, $f(\cdot)\mapsto f(\cdot b)$. So $$\tau_{a+b}=\tau_a\tau_b,$$ $$\delta_{ab}=\delta_a\delta_b,$$ $$\tau_{ab}=\delta_b^{-1}\tau_a\delta_b$$
Since $\Delta_n:=\delta_{\frac{n-1}{n}}\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)$, moving all dilations on the left by the above relations imply nicely
$$\Delta_{n-k+1} \dots\Delta_{n-1}\Delta_{n}=\delta_{\frac{n-k}{n}}\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)^{k},$$
whence
$$\frac{1}{k!} D^kB_n=\frac{1}{k!}B_{n-k} D _{n-k+1} \dots D _{n-1} D _{n}=\Big({n\atop k}\Big)B_{n-k}\delta_{\frac{n-k}{n}}\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)^{k},$$
which we can expand to formula $(b)$.
edit. In fact we may skip the last expansion too, keeping all Josif's formulas on the level of operators. Since the $D_k$'s lower the degree of polynomials, $(2)$ and $(3)$ imply that $B_n$ takes values on polynomials of degree less than or equal to $n$, as said. So, for any $x$, denoting $e_x$ the evaluation form,
$$ e_xB_n=e_0\bigg[\sum_{k=0}^n \frac{x^k}{k!}D^kB_n\bigg]=e_0\bigg[\sum_{k=0}^n x^k \Big({n\atop k}\Big)B_{n-k}\delta_{\frac{n-k}{n}}\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)^{k}\bigg]=$$
$$=e_0\bigg[\sum_{k=0}^n \Big({n\atop k}\Big)x^k\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)^{k}\bigg]=e_0\bigg( \mathbb{1} + x \big(\tau_{\frac{1}{n}}-\mathbb{1}\big) \bigg)^n =e_0\bigg( x \tau_{\frac{1}{n}} + (1-x)\mathbb{1} \bigg)^n$$
$$=e_0\bigg(\sum_{k=0}^n \Big({n\atop k}\Big)x^k(1-x)^{n-k}\tau_{\frac{k}{n}} \bigg) $$
which indeed takes $f$ to the original $(B_nf)(x)$ given by $(1)$.
I had thought about trying something like this, but did not have enough of such clarity of thinking -- indeed, this is so simple and so nice! Just one thing here about former formula (11): it is now labeled (b), thanks to the nice editing by Daniele Tampieri.
|
2025-03-21T14:48:29.957676
| 2020-02-29T18:14:53 |
353871
|
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|
Stack Exchange
|
Borel rank collapse in Hilbert cube modulo $\sigma$-ideal generated by zero-dimensional sets
Both of the commonly studied $\sigma$-ideals (meager sets and null sets) in Polish spaces with a natural measure (i.e. $\mathbb{R}$, $[0,1]$, $[0,1]^\omega$, $2^{\omega}$, etc.) have the nice property that every Borel set is equivalent to a Borel set of uniformly bounded rank modulo the ideal, specifically for any Borel set $A$,
there is an open set $U$ such that $A \Delta U$ is meager, and
there is a $G_\delta$ set $B$ such that $A \Delta B$ is null,
where $\Delta$ is the symmetric difference.
A notable result in topological dimension theory is the fact that the Hilbert cube, $[0,1]^\omega$, is strongly infinite dimensional, meaning that it cannot be covered by a countable collection of zero-dimensional sets (a set is zero-dimensional if it has a basis of clopen sets in the induced subspace topology). Note that a union of zero-dimensional sets is not in general itself zero-dimensional. In fact, a metrizable space has topological dimension $n$ if and only if it can be written as a union of $n+1$ zero-dimensional subsets.
This implies that the collection of zero-dimensional subsets of $[0,1]^\omega$ generates a non-trivial $\sigma$-ideal, which doesn't seem to have a standard name. What is interesting about this $\sigma$-ideal is that it is orthogonal to both the meager set $\sigma$-ideal and the null set $\sigma$-ideal in the sense that it contains $([0,1] \setminus \mathbb{Q})^{\omega}$, which is both comeager and full measure.
Question: Let $\mathcal{Z}$ be the $\sigma$-ideal of subsets of $[0,1]^\omega$ generated by zero-dimensional subsets. Does there exist a countable ordinal $\alpha$ such that for any Borel set $A \subseteq [0,1]^\omega$ there exists a Borel set $B \subseteq [0,1]^\omega$ with Borel rank less than $\alpha$ such that $A\Delta B \in \mathcal{Z}$? If such an $\alpha$ exists, what is the optimal $\alpha$?
A possibly relevant result (EDIT: due to Smirnov) is that (with no set theoretic assumptions) every separable metric space is the union of $\aleph_1$ zero-dimensional subsets. I can't remember who showed this originally. I also can't remember if separability is necessary, but that doesn't matter for this question.
To get off the ground, can we find a closed set $E$ that is not of the form $E = U \Delta B$ for $U$ open and $B \in \mathcal{Z}$?
I think ${0} \times [0,1]^\omega$ should have that property?
The $\sigma$-ideal $\mathcal Z$ has Borel base consisting of $G_{\delta\sigma}$-sets. I do not know if this will help to answer the Question.
I believe that the answer to the Question should be negative: just take a set $H$ of high Borel class in $[0,1]$ and consider the set $A=H\times[0,1]^{\mathbb N}$ in $[0,1]^{\omega}$. How to find a Borel subset $B$ of low class with $A\Delta B\in\mathcal Z$?
The fact that each separable metrizable space is the union of $\aleph_1$ is an old result of Smirnov, cited as Problem 5.1.B in Engelking's "Theory of dimensions finite and infinite''. For more information on the $\sigma$-ideals on the Hilbert cube, see https://doi.org/10.1007/s11856-015-1235-z
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2025-03-21T14:48:29.958003
| 2020-02-29T18:51:07 |
353872
|
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Stack Exchange
|
The space $M_g$ with the complex structure induced from $T_g$ is a coarse moduli space for compact Riemann surfaces of genus $g$
Proposition: The space $M_g$ with the complex structure induced from $T_g$ is a coarse moduli space for compact Riemann surfaces of genus $g$.
In the proof of (1), I wonder why the holomorphic map $\tilde{f}: \tilde{B}\to T_g$ which is equivariant with respect to a homomorphism $\pi_1(B)\to \text{Mod}_g$?
Resourse:https://www.intlpress.com/site/pub/files/_fulltext/journals/iccm/2015/0003/0002/ICCM-2015-0003-0002-a005.pdf
The equivariance condition can be expressed in terms of a commutative diagram involving the map $\tilde{f}$. Using the universal property of the fine moduli space $\mathcal T_g$, you can express this condition as a condition involving the map $\tilde{\pi}$. What is this condition? Can you check it?
@WillSawin Can you be more precise? I can only see, by the universal property, there is a family of curves of genus $g$: $\phi:u\to T_g$, s.t. $\tilde{\pi}:X'\to \tilde{B}$ is the pullback of $\tilde{f}:\tilde{B}\to T_g$ along $\phi:u\to T_g$. I don't see the relation with $\pi_1(B)\to\text{Mod}_g$.
Suppose you have a homomorphism $\pi_1(B) \to \operatorname{Mod}_g$. What does it mean for $\tilde{f}$ to be equivariant with respect to this homomorphism?
@WillSawin My understanding is we have an induces holomorphic map of $\tilde{f}$, $B=\tilde{B}/\pi_1(B)\to T_g/\text{Mod}_g=M_g$. I still can't get your point...
By definition, a map is equivariant with respect to $h: \pi_1(B) \to \operatorname{Mod}_g$, if and only if, for any element $\sigma$ of $\pi_1(B)$, $\tilde{f} \circ \sigma = h(\sigma) \circ \tilde{f}$. Here we let $\sigma$ act on $\tilde{B}$ by the action of $\pi_1(B)$ on $\tilde{B}$ and similarly with $h(\sigma)$ acting on $ T_g$. Is that definition familiar to you?
This is an equality of two maps $\tilde{B} \to T_g$. By pullback, these maps define two families of Riemann surfaces with markings over $\tilde{B}$. What are those families? Because $T_g$ is a fine moduli space, the maps are equal if and only if the families are isomorphic by a map preserving these markings. Can you check this?
|
2025-03-21T14:48:29.958164
| 2020-02-29T19:36:42 |
353874
|
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Stack Exchange
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Example of closed 4 manifold with $\mathbb{S}^1$ action with 1 fixed point and free away from it
I am looking for a smooth closed 4-manifold $M$ with a distinguished point $x\in M$, endowed with an $\mathbb{S}^1$ action such that the stabilizer of $p\in M\setminus\{x\}$ is trivial and $x$ is fixed.
A naive attempt:
If we consider the action given by $\mathbb{S}^1$ on $\mathbb{S}^3\subset \mathbb{C}^2$ that gives the Hopf fibration we can extend this action to the 4-ball (it is an unitary action), and it will have the origin as single fixed point. Unfortunately this manifold is not closed.
Such a closed $4$-manifold does not exist, and this follows from:
Church, P., & Lamotke, K. (1974). Almost free actions on manifolds. Bulletin of the Australian Mathematical Society, 10(2), 177-196
Let me present the argument anyway. The answer breaks down into a local and global part. The local question is to understand what happens near the single fixed point of the $S^1$-action. The global question is about whether we can find, for each local model near the fixed point, a $4$-manifold which closes off the boundary of the model.
Claim 1: The only local model near the fixed point is the "naive attempt" from your question statement, i.e. the standard $S^1$-action on $\mathbb{C}^2$.
Proof of Claim 1: Up to coordinate change, we may assume the $S^1$-action acts by orthogonal transformations on $\mathbb{R}^4$. In particular, it preserves and is free on $S^3$, and the only free $S^1$-action on $S^3$ is the Hopf (or anti-Hopf if we invert one coordinate) fibration. So your naive attempt really is the only local model.
Claim 2: There is no closed $4$-manifold with the property you ask for.
Proof of Claim 2: Suppose there were such a closed $4$-manifold. Removing a ball around $p$ corresponding to the local model, $\widetilde{M} := M \setminus B_p$. Then we obtain a fibration
$$S^1 \rightarrow \widetilde{M} \rightarrow X,$$
such that on the boundary the fibration is just the Hopf fibration $\partial\widetilde{M} = S^3$ over $\partial{X} = S^2$. The fibration over $X$ comes with its classifying map $X \rightarrow BS^1$, and the composition
$$S^2 = \partial X \hookrightarrow X \rightarrow BS^1$$
classifies the Hopf fibration. But the image of $S^2$ under the classifying map for the Hopf fibration $S^2 \rightarrow BS^1$ (which in more down-to-earth language is just $\mathbb{C}\mathbb{P}^1 \hookrightarrow \mathbb{C}{P}^{\infty}$) represents a nontrivial homology class, while $X \rightarrow BS^1$ is a nullhomology of this class. So we arrive at a contradiction.
If you prefer characteristic classes, it suffices to consider the first Chern class in this argument. In addition, we see that if we have a $4$-manifold with an "almost free" $S^1$-action, then the number of fixed points is even, since each fixed point either adds or subtracts $1$ from the first Chern class, depending upon orientations. Conversely, this argument can be boosted to prove you can find a closed $4$-manifold with any even number of fixed points.
EDIT (responding to comment)
You can ask this question for other group actions, and again, there’s the local and global parts to consider. Let me do this for the case of $\mathbb{Z}_p$ (the cyclic group, not the $p$-adics for anybody who might be confused) which was asked in a comment. We claim that in this case, again, there is no $4$-manifold $M$ with a $\mathbb{Z}_p$ action which is free except for a single fixed point. (Hopefully I haven't made a mistake, which is entirely possible, so feel free to be skeptical. I'm sure an algebraic topologist on this site has a better argument for the last part.)
Again, we start with the local models, which arise from the representation theory of $\mathbb{Z}_p$. For any finite cyclic group, the irreducible real representations are either 1-dimensional (act by the +1 or -1, the latter if p is even) or 2-dimensional (a rotation of order $p$). Since you want every point to be free except the fixed point, you can't use the 1-dimensional representations, since they have order 1 and 2 respectively, so you have a direct sum of two rotations by some 2$\pi k_1/p$ and $2\pi k_2/p$ where $k_1$ and $k_2$ are coprime to $p$. Up to $\mathrm{Aut}(\mathbb{Z}_p)$, we may assume $k_1 = 1$, and we will simply write $k_2 = k$. At the boundary $S^3$ of this local model, we obtain the fibration
$$\mathbb{Z}_p \rightarrow S^3 \rightarrow L(p;k)$$
where $L(p;k)$ is a Lens space. Now for the global part. Assume there exists such a $4$-manifold $M$. Then we have the classifying map for the bundle $S^3 = \partial \widetilde{M} \rightarrow \partial X = L(p;k)$ factors through the classifying map for $X$ (which is now itself a $4$-manifold):
$$\phi \colon L(p;k) = \partial X \hookrightarrow X \rightarrow B\mathbb{Z}_p = K(\mathbb{Z}_p,1).$$
The cohomology ring $H^*(K(\mathbb{Z}_p,1);\mathbb{Z}_p)$ can be completely understood in terms of Steenrod squares and the Bockstein homomorphism, and in particular, we find that for $u \in H^1(K(\mathbb{Z}_p,1);\mathbb{Z}_p)$ the fundamental class, the class
$$v:= u \smile \beta(u) \in H^3(K(\mathbb{Z}_p;1)\mathbb{Z}_p)$$
is a nontrivial element (in fact a generator). Then one can check (e.g. from the cohomology ring structure of $L(p;k)$) that $\phi^*v \neq 0$ as well. Dually, the pushforward of the mod $p$ fundamental class represents a nontrivial element
$$\phi_*[L(p;k)] \neq 0 \in H_3(K(\mathbb{Z}_p,1);\mathbb{Z}_p),$$
and so $\phi$ cannot factor through a $4$-manifold $X$. So no such $4$-manifold $M$ exists.
Thanks for your reply, do you know if this holds also for $\mathbb{Z}/p\mathbb{Z}$ actions in place of $\mathbb{S}^1$? This should be harder as now $H^1(\partial X, \mathbb{Z}/p\mathbb{Z}) \to H^1(X, \mathbb{Z}/p\mathbb{Z})$ may not be null.
My edit should address your question - hopefully I haven't made a mistake. Essentially, the problem boils down to showing that the tautological inclusion map $L(p;k) \hookrightarrow B\mathbb{Z}_p$ (corresponding to the bundle $S^3 \rightarrow L(p;k)$) is homologically nontrivial.
Thank you for the addendum, I would accept your answer twice if possible.
|
2025-03-21T14:48:29.958553
| 2020-02-29T20:23:01 |
353876
|
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Stack Exchange
|
Cofinality for coends?
Recall that a functor $I \xrightarrow u J$ is cofinal if it has the property that for any functor $J \xrightarrow F C$, we have that $\varinjlim F \cong \varinjlim Fu$ via the canonical map, either side of the equation existing if the other does. This notion is very useful: it admits various reformulations which can be checked directly; there are numerous practical examples, and once a functor is known to be cofinal, the property which I've just treated as a definition becomes a great tool for computing colimits.
When passing from colimits to coends, it would be nice to have similarly powerful tools available. Coends can be reduced to colimits -- e.g. we have $\int^{j \in J} F(j,j) = \varinjlim(Tw(J) \to J^{op} \times J \xrightarrow F C)$, where $Tw(J)$ is the twisted arrow category. But from examples I've tried in the past, my impression is that it's relatively uncommon for a functor $I \xrightarrow u J$ to induce a cofinal functor $Tw(I) \xrightarrow{Tw(u)} Tw(J)$.
Question 1: What are some examples of functors $I \xrightarrow u J$ which induce cofinal functors $Tw(I) \xrightarrow {Tw(u)} Tw(J)$? Are they really quite rare?
Even if I'm right in thinking that such functors are rare, we shouldn't be deterred. After all, a coend is not the colimit of an arbitrary functor out of $Tw(J)$ -- but rather one which factors through $J^{op} \times J$. So there may still be functors $I \xrightarrow u J$ out there which always induce equivalences of coends, even if $Tw(I) \xrightarrow{Tw(u)} Tw(J)$ is not cofinal.
Question 2: What are some examples of functors $I \xrightarrow u J$ such that for any $J^{op} \times J \xrightarrow F C$, we have $\int^{j \in J} F(j,j) \cong \int^{i \in I} F(ui,ui)$ (via the canonical map), either side existing if the other does?
Finally, a more systematic question about these functors:
Question 3: Is it possible to give a direct combinatorial characterization of functors $u$ of the form described in Question 1 or Question 2 (analogous to the usual characterization of cofinal functors via connectedndess of slice categories)?
I'm also interested in versions of these questions other settings like enriched category theory or $\infty$-category theory.
Obviously, everything should have a dual story about limits and ends, too.
An answer would probably also help with https://mathoverflow.net/questions/352607/
It seems to me like the general finality criterion might be equivalent to the special case of the coend that computes Hom-sets in J, in case that's any easier... it seems hard at the moment to come up with examples though.
For question 1, examples are provided by functors which are universally cofinal (i.e. cofinal after any base change); this work for $\infty$-categories as well. For instance, any functor which is smooth or proper with weakly contractile fibers. For instance, if $X$ is a quasi-category, the canonical functor from the category of simplices of $X$ to $X$ has the property you want.
@Denis-CharlesCisinski Thanks! This would make a great answer, especially because I don't immediately see a proof -- e.g. $Tw(I) \to Tw(J)$ is not typically a base change of $I \to J$...
Another approach to cofinality for coends might be to consider them as left Kan extensions along the profunctor $H\colon J^\circ \times J \to 1$ given by the hom-sets of $J$. Precomposition with a double cell into $H$, in the double category of profunctors, preserves this left Kan extension whenever it satisfies some kind of Beck-Chevalley condition. The horizontal dual setting, preservation of right Kan extensions, is explained in Section 4 of link.
@RoaldKoudenburg Thanks! As mentioned in the CW answer below, Alex Campbell had the same idea, and it turned out to be very enlightening!
A sufficient condition for a functor $u:I\to J$ to induce a cofinal functor $Tw(I)\to Tw(J)$ is that $u$ is universally cofinal (i.e. any base change of $u$ is cofinal). Another sufficient condition is that $u$ is universally final. In fact, as may be seen in the proof below, we only need that the base change of $u$ (or of $u^{op}$, respectively) along any cartesian fibration is cofinal.
The key observation to understand this is the following. For any functor $u:I\to J$, there are two canonical functors factoring $Tw(u):Tw(I)\to Tw(J)$
$$
Tw(I)\to (J^{op}\times I)\times_{(J^{op}\times J)}Tw(J)
\quad \text{and} \quad
Tw(I)\to (I^{op}\times J)\times_{(J^{op}\times J)}Tw(J)
$$
which are both cofinal. [If we take the convention that $Tw(I)$ if a cartesian fibration over $I^{op}\times I$ this comes from Proposition 5.6.9 in this book on $\infty$-categories; note that the author of this book calls final what many other people call cofinal, as may be seen from Theorem 6.4.5 in loc. cit.]. Therefore, it is sufficient to prove that one of the projections
$$
(J^{op}\times I)\times_{(J^{op}\times J)}Tw(J)\to Tw(J)
\quad \text{or} \quad
(I^{op}\times J)\times_{(J^{op}\times J)}Tw(J)\to Tw(J)
$$
is cofinal. But the first (second) one is a pullback along the cartesian fibration $Tw(J)\to J$ (along the cartesian fibration $Tw(J)\to J^{op}$) of the functor $u$ (of the functor $u^{op}$, respectively).
A final remark on the proof: if we work in the model of quasi-categories say, then pullbacks along cartesian fibrations in the $1$-category of quasi-categories are homotopy pullbacks with respect to the Joyal model structure. In particular, in the proof above, it does not matter if we work with pullbacks in the $1$-categorical sense or in the $\infty$-categorical sense. That is also a way to see that the proof above is model free.
Finally, a sufficient condition for $u$ to satisfy the hypothesis above is that $u$ remains a weak homotopy equivalence after any base change. This condition is satisfied by any functor $u:I\to J$ which is smooth or proper with weakly contractible fibers (e.g. any cartesian or cocartesian fibration with weakly contractible fibers); this follows easily from Proposition 7.1.12 in loc. cit. An example which is not a (co)cartesian fibration is the functor $\Delta_{/ N(J)}\to J$ (sending a sequence of maps $j_0\to\cdots\to j_n$ to $j_n$) for any category $J$ (where $\Delta_{/ N(J)}$ is the category of simplices of the nerve of $J$); this belongs to a larger class of examples provided by Proposition 7.3.8 in loc. cit.
Offline, Alex Campbell independently suggested a similar approach to the one Roald mentions in the comments, and worked it out. Here are the results -- we work with ends rather than coends for simplicity:
We observe that if $I^{op} \times I \xrightarrow F C$ is a functor, then the end $\int_{i \in I} F(i,i)$ is precisely the limit $\varprojlim_{Hom_I} F$ of $F$ weighted by the Hom-functor $Hom_I: I^{op} \times I \to Set$. Thus, we can apply the weighted version of initiality (the "limit" version of cofinality -- the more modern thing seems to be to say "final" for what I called "cofinal" above), which says in general that
Initiality for Weighted Limits: Let $I \xrightarrow u J$ be a functor, let $\phi: I \to Set$ and $\psi: J \to Set$ be functors (which we regard as "weights" for weighted limits), and let $\eta: \phi u \Rightarrow \psi$ be a natural transformation. Then the following are equivalent:
For any $C$ and any functor $J \xrightarrow F C$, we have $\varprojlim_\psi F \cong \varprojlim_\phi F u$ via the canonical map induced by $\eta$, either side existing if the other does.
$\eta$ exhibits $\psi$ as the Left Kan extension $\psi = Lan_u \phi$ of $\phi$ along $u$.
In particular, we can apply this in the case where $\phi = Hom_I$, $\psi = Hom_J$, and $\eta$ is given by the action of the functor $u$. The left Kan extension can be computed explicitly via a coend formula, and the result is the following:
Proposition (Initiality for Ends): Let $I \xrightarrow u J$ be a functor. The following are equivalent:
For every functor $F: J^{\mathrm{op}} \times J \to C$, we have $\int_{j \in J} F(j,j) \cong \int_{i \in I} F(ui,ui)$ via the canonical map, either side existing if the other does.
For every $j,j' \in J$, the canonical map $\int^{i \in I}Hom_J(j,ui) \times Hom_J(ui,j') \to Hom_J(j,j')$ is an isomorphism.
There are various ways to reformulate (2). For instance,
The composite of profunctors $Hom_J(1,u) \circ_I Hom_J(u,1)$ is canonically isomorphic to $Hom_J$.
For every $j\xrightarrow \beta j' \in J$, the "category of $u$-factorizations" of $\beta$ -- whose objects consist of triples $i \in I, j \xrightarrow \alpha ui \xrightarrow {\alpha'} j'$ composing to $\beta$ (morphisms are the obvious thing) -- is connected.
[ABSV] For any $C$, the functor $Fun(u,C): Fun(J,C) \to Fun(I,C)$, given by precomposition with $u$, is fully faithful.
[ABSV again] The functor $u$ is absolutely dense, i.e. for any $j \in J$ we have $j = \varinjlim (u / j \to J)$ and the colimit is absolute.
In the ABSV paper linked to above, such functors are called "lax epimorphisms" in light of (5) above (the idea being that a "pseudo-epimorphism" is a functor $u$ such that $F(u,C)$ is always a pseudo-monomorphism, which has something to do with the core of the categories involved, but here we take into account non-invertible 2-cells of $Cat$). In light of (5) above, one might also say "co-fully-faithful" or something like that.
Any localization is an example of such a functor. So is any composite or transfinite composite of localizations. The transfinite composites of localizations form the left half of a factorization system on $Cat$ whose right half is the conservative functors, and it's not hard to see that if $u$ is co-fully-faithful, in the factorization $u = wv$ with $v$ being a transfinite composite of localizations and $w$ being conservative, both $v$ and $w$ are co-fully-faithful. Thus when we look beyond localizations, it seems the appropriate thing to ask is "which conservative functors are co-fully-faithful?". For example, I think the functor from a category to its idempotent completion is co-fully-faithful (while also being fully faithful and in particular conservative). I think that's about all there is to say about co-fully-faithful functors which are also fully-faithful -- any such functor induces an equivalence of idempotent completions (one way to see this is to use the absolute density condition above with the Yoneda embedding). But of course, there may be quite a lot of daylight between co-fully-faithful functors which are conservative and those which are fully faithful.
The co-fully-faithful functors also seem related to the "liberal" functors (functors $u$ such that $Fun(u,C)$ is always conservative) of CJSV: co-fully-faithful implies liberal but not conversely.
Of course, the enriched and $\infty$-categorical counterparts of all of this should be clear at a conceptual level, at any rate.
Note also that everything is self-dual: a functor $u$ is co-fully-faithful iff $u^{op}$ is, so the questions about ends and coends are actually equivalent (and not just dual).
Another way of thinking about these functors is given here: they are precisely those functors $u: I \to J$ such that the counit $u_! u^\ast \Rightarrow 1$ is an isomorphism, where we're talking about the adjunction $u_!: {Set^I}^\to_\leftarrow Set^J : u^\ast$ where $u^\ast$ is given by precomposition. Of course, this is just another way of saying that $u^\ast$ is fully faithful, but it shows that in (5) above it suffices to take $C = Set$.
Not an answer; too long for a comment.
The best I've encountered is the following: Proposition 5.1.7 of Kerler-Lyubashenko "Non-Semisimple TQFTs for 3-Manifolds with Corners" (I know...)
Definition. Assume $\mathcal C$ is a small category with finite coproducts and a small epi-generating set, i.e. a subset $\mathcal S \subseteq \mathcal C_0$ of objects such that
every $C\in\mathcal C$ admits an epimorphism
$$
h_X : \coprod_{i=1}^{n_X}U_i \to X
$$ for a finite subset $\{U_1,\dots,U_{n_X}\}\subset S$, and
every morphism $f : S \to X$ with domain in $\mathcal S$ admits a factorization $f = h_X \circ f' : S \to \coprod U_i \to X$.
If $T : \mathcal{C}^o \times \mathcal{C} \to \mathcal{D}$ is a functor that commutes with finite limits and colimits in both variables, and we consider its restriction $T|_{\mathcal{S}}$ to $\mathcal S$, then
$$
\int^\mathcal{C} T(C,C) \cong \int^{\cal S}T|_{\mathcal{S}}(S,S)
$$
(meaning that one coend exists iff so does the other, and the canonical map $\leftarrow$ is invertible)
In the book on TQFTs this is stated for additive categories; I didn't see how the argument properly adapted, can fail in a general setting. Anyway, it's a fairly artificial and rigid condition...
|
2025-03-21T14:48:29.959572
| 2020-02-29T20:25:27 |
353877
|
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|
Stack Exchange
|
Stacky proof of no elliptic curves over Z
It is a well known result that there are no Elliptic curves over the integers with every where good reduction. In fact this is even true for abelian varieties (and hence higher genus curves) but let me restrict my question to elliptic curves.
In terms of the moduli stack $M_{1,1}
$ of elliptic curves over the integers, this says that the stack has no integral sections.
Since $M_{1,1}$ is relatively uncomplicated as a stack (it is essentially the affine line with a few stacky points) , perhaps it is possible to find a direct proof of this result? Is such a proof known?
I expect the ramification arguments in the standard proof should correspond to some non trivial obstructions at the stacky points.
The stack $\mathcal M_{1,1}$ is not so simple at the (crucial) primes $2$ and $3$, so I'm not so sure a stacky proof would be nice.
I just came in to say what I see Will already said!
Ah, the word "essentially," bane of our lives.
Exercise: what are the Spec Z points of the weighted projective line P(4,6) with one point removed, which is what M_{1,1} would be if it weren't for those two little meddling primes?
@JSE by the universal property of weighted projective space, to get a integral point, I would have to give a line bundle on Z and 2 sections in its 4th and 6th power the don't simultaneously vanish at any prime. Since all line bundles are trivial, the powers don't do anything and this is the same as an integral point which is to say, 2 coprime integers. Do I have that right?
|
2025-03-21T14:48:29.959718
| 2020-02-29T20:47:25 |
353878
|
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|
Stack Exchange
|
Necessary conditions for the existence of solution of Sylvester equation AX=XB
Let's consider square matrices $A_{n \times n}$, $B_{n \times n}$ and $X_{n \times n}$ with elements from $\mathbb{R}$. Could you tell me please, what would be the necessary conditions for the existence of solution (may be not unique) of Sylvester equation:
$$
AX=XB.
$$
As I know, sufficient condition looks like (but probably it is a necessary and sufficient condition)
$$
\sigma_p(A) \cap \sigma_p(B) \neq \varnothing,
$$
here $\sigma_p(A)$ and $\sigma_p(B)$ are the spectra of matrices $A$ and $B$.
$X = 0$ works. What conditions on $X$ do you want?
This equation always has a solution: $X = O$. I'll assume throughout this answer that you're interested in a non-zero solution.
The equation $AX = XB$ is equivalent to $(A \otimes I - I \otimes B^T)\mathbf{x} = \mathbf{0}$, where $\otimes$ denotes the Kronecker product and $\mathbf{x}$ is the vectorization of $X$. Your question is thus equivalent to asking when the matrix $A \otimes I - I \otimes B^T$ is not invertible (i.e., when $0$ is not an eigenvalue of $A \otimes I - I \otimes B^T$).
Since the eigenvalues of $A \otimes I - I \otimes B^T$ are exactly the sums of the eigenvalues of $A$ and $-B$, the condition that you wrote ($\sigma_p(A) \cap \sigma_p(B) \neq \varnothing$) is in fact both necessary and sufficient.
Thank you for the answer.
@Randal'Thor - The letter $O$ seems to be a common notation for the zero matrix (to distinguish from the zero vector, for example).
|
2025-03-21T14:48:29.959850
| 2020-02-29T21:06:37 |
353879
|
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|
Stack Exchange
|
Interpolation for Sobolev spaces
How one can identify the following (complex) interpolation space
$$E_\theta :=[L^2(\Omega), H^2(\Omega)\cap H_0^1(\Omega)]_\theta,$$
where $\Omega$ is a regular domaine. After research, it seems that this depend on the position of $\theta \in (0,1)$:
for $0<\theta<1/4$, $E_\theta=H^{2\theta}(\Omega)$ and for $1/4 <\theta <1$ we have $E_\theta=H^{2\theta}_0(\Omega)$, while the case $\theta =1/4$ is critical.
Some inclusions are immediate while the others are not. Is there any elegant way to establish such identification?
Any reference would be helpful.
There is a answer to a related question with some info, maybe this is sufficiently elegant..
@Hannes thank you, but this gives just the link to domains of fractional powers and not the explicit characterization depending on $\theta$.
This is a result by R. Seeley: Interpolation in Lp with boundary conditions, Studia Mathematica, 1972.
The main ingredient of the proof is that step functions are pointwise multipliers in Hs for s<1/2.
Thank you, but it seems that the result was already proved by many others, e.g., Fujiwara, Grisvard... I'm looking for an elementary way to do it.
The book of Lions and Magenes is another reference for this kind of result. But what do you mean by "elementary"? Surely not using pre-calculus only?
@MichaelRenardy I meant not using heavy theories such as $H^\infty$ calculus, or a way that can be generalized to more complex spaces e.g., product spaces.
|
2025-03-21T14:48:29.960096
| 2020-02-29T22:00:15 |
353882
|
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|
Stack Exchange
|
Map from local systems to holomorphic line bundles on a curve
Let $X$ be a Riemann surface of genus $g > 0$. Let $S$ denote the set of local systems (locally constant sheaves) on $X$ with fiber $\mathbb{C}$. $S$ is in natural bijection with $H^1(X, \underline{\mathbb{C}^{\times}})) \cong (\mathbb{C}^{\times})^{2g}$, where $\underline{\mathbb{C}^{\times}}$ is the constant sheaf on $X$ with fiber $\mathbb{C}^{\times}$.
By the Riemann-Hilbert correspondence, each local system $\mathbb{L} \in S$ corresponds to a line bundle with a flat connection. This map is given by $\mathbb{L} \mapsto (\mathbb{L} \otimes_{\mathbb{C}} \mathcal{O}_X, \nabla = 1 \otimes d)$, where $d$ is the usual exterior derivative. Forgetting the connection gives a map from $S$ to the Picard group of $X$, $H^1(X, \mathcal{O}_X)$. This is the natural map $H^1(X, \underline{\mathbb{C}}^{\times}) \to H^1(X, \mathcal{O}_X^{*})$, so in particular it is a map of abelian groups.
What is the description of this map? In particular, I do not know how to show it has non-trivial image, i.e., that there is a non-trivial holomorphic line bundle that admits a flat connection.
It follows from Chern-Weil theory that the image is contained in the Jacobian (as the curvature of a connection is essentially the first Chern class). The identification of $S$ with $(\mathbb{C}^{\times})^{2g}$ is non-canonical, but the topology this gives to $S$ is canonical. I think one can show that this map is continuous by thinking about cocycles, which gives another proof that the image is in the Jacobian.
There is a typo., you want $Pic(X) = H^1(X,\mathcal{O}_X^)$. One can show, with a bit of Hodge theory, that $H^1(X,U(1))$ (resp. $H^1(X,\mathbb{C}^$)) maps bijectively (resp. surjectively) onto the Jacobian. I might add details later if I have time.
I think the following theorem answers your question.
Theorem: Let $X$ be a smooth, proper connected curve over $\mathbf C$ with a line bundle $\mathscr L$. Then $\mathscr L$ admits a flat connection $\nabla$ if and only if $c_1(\mathcal L)=0$.
Remark: A more general statement would be that a line bundle $\mathscr L$ on a compact Kahler manifold admits an integrable connection if and only if $c_1(\mathscr L)\otimes \mathbf Q=0$.
Before we go to the proof of this theorem, we recall that define the first Chern class as the connecting homomorphism
$$
c_1\colon \mathrm{Pic}_X=\mathrm{H}^1(X,\mathscr O_X^*) \to \mathrm{H}^1(X,\mathbf Z(1))
$$
that comes from the exponential short exact sequence
$$
0 \to \mathbf Z(1) \to \mathscr O_X \xrightarrow{exp(-)} \mathscr O_X^* \to 0 \ .
$$
Proof: The essential idea is to relate the exponential sequences for $\mathbf C$ and $\mathscr O_X$ with each other. More precisely, we have a morhpism of short exact sequences:
$$0 \to \mathbf Z(1) \to \mathscr O_X \to \mathscr O_X^* \to 0 \\
\downarrow \\
0 \to \mathbf Z(1) \to \underline{\mathbf C} \to \underline{\mathbf C}^* \to 0 \ .
$$
So we have a commutative diagram
$$
\mathrm{H}^1(X,\mathbf C) \to \mathrm{H}^1(X,\mathbf C^*) \xrightarrow{\delta} \mathrm{H}^2(X,\mathbf Z(1)) \\
\downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \downarrow{\mathrm{Id}} \\
\mathrm{H}^1(X,\mathscr O_X) \to \mathrm{H}^1(X,\mathscr O_X^*) \xrightarrow{c_1} \mathrm{H}^2(X,\mathbf Z(1) \ .
$$
Step 1. The map $\delta$ is the zero map.
Probably the easiest way is to say that $\mathrm{H}^1(X,\mathbf C^*)=(\mathbf C^*)^{2g}$ and $\mathrm{H}^2(X,\mathbf Z(1))=\mathbf Z$. But there are no non-trivial homomorphisms from $(\mathbf C^*)^{2g}$ to $\mathbf Z$.
Step 2. The map $c_1(\mathscr L)=0$ if $\mathscr L$ admits a flat connection.
Riemann-Hilbert correspondence says that $\mathscr L$ admits a flat connection if and only if it lies in the image of $\mathrm{H}^1(X,\mathbf C^*) \to \mathrm{H}^1(X,\mathscr O_X^*)$. Combining it with Step 1, we conclude the claim.
Step 3. If $c_1(\mathscr L)=0$ then $\mathscr L$ is in the image of $\mathrm{H}^1(X,\mathbf C^*) \to \mathrm{H}^1(X,\mathscr O_X^*)$.
This easily follows from the commutative diagram above and the fact that $\mathrm{H}^1(X,\mathbf C) \to \mathrm{H}^1(X,\mathscr O_X)$ is surjective. The latter fact is, in turn, a consequence of degeneration of the Hodge-to-de Rham spectral sequence
$$
\mathrm{E}^{p,q}_2=\mathrm{H}^q(X, \Omega^p_X) \Rightarrow \mathrm{H}^{p+q}(X, \mathbf C) \ .
$$
Step 4. If $c_1(\mathscr L)=0$ then $\mathscr L$ admits a flat connection.
Again, this is just a consequence of Step 3 and Riemann-Hilbert Correspondence.
|
2025-03-21T14:48:29.960393
| 2020-03-01T01:39:37 |
353888
|
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|
Stack Exchange
|
Connectedness, loops and formal moduli problems
Let $k$ be an algebraically closed field of characteristic zero. Formalizing a classical folk concept, Pridham and (in a different way,) Lurie defined a formal moduli problem (over $k$) to be a functor from local Artin CDGA's to homotopy types satisfying a certain sheaf condition. If the commutativity condition is weakened to an $E_n$ condition, any formal moduli problem is (uniquely) representable by an $E_n$ algebra; in the commutative case, a formal moduli problem is not necessarily representable by a CDGA, but rather by a (derived) Lie algebra.
There's an intuitive picture I like for this in a special case, and I want to understand how it fits into the general picture. Namely, say that $G$ is an affine algebraic Lie group. Let $BG$ be its classifying stack. Then the deformation problem of maps to $BG$ (relative to a choice of point $*\to BG$) is classified by the Lie group $\mathfrak{g}$.
Of course this picture involves a group rather than a "formal stack", which is my intuition for (the opposite category to) formal moduli problems, and I am curious how hard it is for a formal moduli problem to be representable by a group object.
From the point of view of spaces, going from groups to spaces "is not very hard": the category of groups is equivalent (via the delooping functor) to the category of connected pointed spaces.
On the other hand, for $E_n$ algebras, the category of cogroup objects is equivalent to the category of (suitably defined) $(n,1)$-commutative Hopf algebras, which is more closely related (via Koszul duality) to the category of $E_{n+1}$ algebras than to $E_n$ algebras; in particular, the delooping functor from group objects to $E_n$ algebras is far from fully faithful.
Now by formal nonsense, there is a loop-deloop pair of adjoint functors $$B:GFMP\leftrightarrows FMP:\Omega,$$ where $FMP$ is the category of ($E_\infty$) formal moduli problems and $GFMP$ is the category of cogroup objects in $FMP$. It seems natural to ask the following questions, to which I don't know the answer:
Is $B$ fully faithful, and if so what is its image?
Is there an analogue on the level of (co)group objects in Lie algebras to the Koszul duality functor from $(n,1)$ Hopf algebras to $E_{n+1}$ algebras?
A natural further question, which I suspect is harder, is to ask whether there is an "algebraic point of view" via Lie-like objects for $E_\infty$ objects in $FMP$.
How are you defining $B$? The nerve functor from groups to unpointed spaces doesn't preserve fibrations or homotopy fibre products. You need a smoothness condition on $G$ for $BG$ to be a "formal moduli problem". You might also find the original proof of the equivalence helpful; see Theorem 2.30 and Corollary 4.57 of http://arxiv.org/abs/0705.0344 .
and there's no need for your algebraically closed hypothesis.
@JonPridham Edited the question to incorporate your authorship. Let me know if there is a more context that I should be included (I don't know the timeline here). I was purposefully vague about the functors $\Omega$ and $B$, but are you saying that there are serious problems with defining a functor from group objects to objects which takes a geometric group $G$ to the moduli problem classified by $BG$ (principal bundles with trivialization at the marked point)?
Thanks (the timeline is 2007 vs 2010 for first appearance of the comparison, as it happens). There aren't serious problems defining the functor $B$ that you want; it's just the universal FMP under the presheaf given by applying $B$ objectwise. That tends to have non-trivial $\pi_0$, so is bigger than naive moduli of principal bundles over a fat point.
The presentation of the formal moduli problems story in Gaitsgory-Rosenblyum A Study in Derived Algebraic Geometry, Vol 2 may be what you are looking for. We review it here (in the case over $\mathrm{Spec}\, k$ for a field $k$ of characteristic zero, that the question concerns):
1. Looping/delooping equivalence in formal DAG
Just like the familiar adjoint equivalence in the homotopy theory of spaces
$$\mathrm B: \mathrm{Grp}_{\mathbb E_1}(\mathcal S)\simeq \mathcal S_*^{\ge 1}:\Omega,$$
there is an analogous adjoint equivalence in formal DAG
$$\mathrm B:\mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k) \simeq \mathrm{FMP}_k:\Omega,$$
where the loop space functor is in both cases given by $\Omega X = \mathrm{pt}\times_X \mathrm{pt}$, as per usual in homotopical/$\infty$-categorical things. Note that formal moduli problems are already inherently pointed, by the assumption that $X(k)$ is contractible.
2. Formal groups and Lie algebras
Now, $\mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k) = \mathrm{FGrp}_k$ is (an incarnation) of the $\infty$-category of (derived) formal groups over $k$. Thanks to the characteristic zero assumption, there is a further equivalence
$$
\mathrm{Lie}:\mathrm{FGrp}_k \simeq \mathrm{LieAlg}_k : \exp
$$
with derived Lie algebras (as modelled for instance by dg Lie algebras). Just like expected, the derived Lie algebra $\mathfrak g$ corresponding to the formal group $G$ is $\mathfrak g = T_{G, e}$, the tangent fiber at the unit.
3. Formal moduli problems and Lie algebras
The celebrated Lurie-Pridham identification between formal moduli problems and derived Lie algebras is precisely the composite of these two equivalences of $\infty$-categories
$$
\mathrm{FMP}_k\simeq \mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k)=\mathrm{FGrp}_k\simeq \mathrm{LieAlg}_k.
$$
That is
It sends a formal moduli problem $X$ to
$$
\mathrm{Lie}(\Omega X) = T_{\Omega X, e} = T_{X, x_0}[-1],
$$
where $x_0$ is the base-point of $X$, unique up to a contractible space of choices $X(\kappa)$. This shifted tangent fiber carries a canonical Lie algebra structure coming from (i.e. as the Lie algebra of) the $\mathbb E_1$-group structure of $\Omega X$.
The inverse equivalence $\Psi: \mathrm{LieAlg}_k \simeq \mathrm{FMP}_k$ is then given by $\Psi(\mathfrak g)= \mathrm B\exp(\mathfrak g)$.
4. Lurie's formula for $\Psi$
You may justifiably complain that this description of the inverse functor $\Psi$ does not look the same as the one in Lurie's writing. Let's see how to get it in that form.
Let's assume that the formal moduli problem $\Psi(\mathfrak g)$ is formally affine (true under some finiteness assumptions on $\mathfrak g$), in the sense that
$$
\mathrm B\exp(\mathfrak g)= (\mathrm{Spf}\,k)/\exp(\mathfrak g)\simeq \mathrm{Spf} \,k^{\mathfrak g}.
$$
Here of course the formal spectrum is the usual functor $\mathrm{Spf}:(\mathrm{CAlg}^\mathrm{aug}_k)^\mathrm{op}\to \mathrm{FMP}_k$. The derived Lie algebra invariants may be computed via the Chevalley-Eilenberg complex, thus $k^\mathfrak g\simeq {C}^*(\mathfrak g)$.
Then for any Artinian $k$-algebra $A$ we have
$$ (\Psi(\mathfrak g))(A) \simeq \mathrm{Map}_{\mathrm{CAlg}^\mathrm{aug}_k}({C}^*(\mathfrak g), A). \qquad \quad(1)
$$
This is why Lurie tells us to consider the "Koszul duality functor" $\mathfrak D: (\mathrm{CAlg}_k^\mathrm{aug})^\mathrm{op}\to \mathrm{LieAlg}_k,$ right-adjoint to the Chevalley-Eilenberg cochains functor. Indeed, (through a little use of the finiteness of $\mathfrak g$) we get
$$(\Psi(\mathfrak g))(A) \simeq \mathrm{Map}_{\mathrm{LieAlg}_k}( \mathfrak D(A), \mathfrak g),\qquad \qquad(2)
$$
which is how Lurie tells us to define $\Psi$.
Note that this is going the other way than the $C^*\dashv\mathfrak D$ adjunction. Indeed, without finiteness assumptions on $\mathfrak g$, the formal moduli problem $\Psi(\mathfrak g)$ will not necessarily be formally affine, and the formula (1) will not necessarily work. On the other hand, as Lurie teaches us, formula (2) will always work.
5. Loose ends
If I understand the original question correctly, this is presenting the formal moduli problem story precisely like the intuitive picture mentioned. Indeed: any formal moduli problem may be written as $X\simeq \mathrm B G$ for a derived formal group $G$, and the formal moduli problem (of mapping into) $\mathrm BG$ is classified by the Lie algebra $\mathfrak g$.
That said, the question makes analogy with the $\mathbb E_n$-algebra version of this story too. I am a little confused about the points raised - in particular, it seems like the following is asserted: a formal moduli problem on $\mathbb E_n$-algebras is represented by an $\mathbb E_n$-algebra. That is, so far as I understand, incorrect. Instead, any FMP on $\mathbb E_n$-algebras in classified by an $\mathbb E_n$-algebra, in a way somewhat analogous to the way that the functor $\mathfrak D$ presents formal moduli problems in the commutative case with Lie algebras.
In particular, there seems to be no need to think about cogroup objects, as the equivalence of $\infty$-categories $\mathrm{FMP}_k^{\mathbb E_n}\simeq \mathrm{Alg}^\mathrm{aug}_{\mathbb E_n}$ is covariant.
$\qquad$
Edit: A question was raised in the comments whether group objects in $\mathrm{FMP}_k^{\mathbb E_n}\simeq \mathrm{Alg}_{\mathbb E_n}^\mathrm{aug}$ correspond to $\mathbb E_{n+1}$-algebras. Unless I am misunderstanding the question, the answer is negative.
A monoid structure (of which a group structure is a particular example of) on an $\mathbb E_n$-algebra $A$ is given by a map $A\times A\to A$, plus coherence data. On the other hand, an additional $\mathbb E_1$-algebra structure on $A$ (which is equivalent to making it into an $\mathbb E_{n+1}$-algebra by Dunn Additivity) is given by a map $A\otimes_k A\to A$, plus coherence data. So a group object in $\mathbb E_n$-algebras, and an $\mathbb E_{n+1}$-algebra are different structures.
This might feel a little weird because we're used to tensor products corresponding to products of schemes. Alas, unlike the contravariant $\mathrm{Spec}$, the equivalence $\Psi:\mathrm{Alg}^{\mathrm{aug}}_{\mathbb E_n}\simeq \mathrm{FMP}_k^{\mathbb E_n}$ is covariant. (PS: another difficulty: the tensor product only becomes the coproduct on the level of $\mathbb E_\infty$-algebras, not $\mathbb E_n$-algebras, so even a contravariant equivalence wouldn't necessarily do the trick).
Rather, the situation is the same as in the $\mathbb E_\infty$-case above: there is an equivalence
$$\mathrm B:\mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k^{\mathbb E_n})\simeq \mathrm{FMP}_k^{\mathbb E_n}:\Omega.$$
This is a special case of the following general phenomenon: for any operad $\mathcal O$, the loops functor $\Omega: \mathrm{Alg}_{\mathcal O}(\mathrm{Mod}_k)\to \mathrm{Grp}_{\mathbb E_1}(\mathrm{Alg}_{\mathcal O}(\mathrm{Mod}_k))$ is an equivalence of $\infty$-categories (A Sudy in Derived Algebraic Geometry, Vol 2, Chapter 6, Proposition 1.6.4). The above claims are special cases for $\mathcal O$ the Lie operad and the $\mathbb E_n$-operad respectively.
Thanks! This is a funny property, that the delooping functor is an equivalence but the category is not stable. I wonder if there are other examples.
And thanks for the correction: $E_n$ moduli problems are represented by $E_n$ algebras via a duality functor, so presumably a group object (in a geometric sense) would be representable by an $E_{n+1}$ algebra.
Reiterating a comment I made on the question, BG is not good notation for an infinitesimal functor with many objects (essentially the Deligne groupoid in this case).
@JonPridham: Agreed! But I don't quite see why this objection applies to the case at hand. (Pardon if I am missing something!) One of the requirements in the definition of a formal moduli problem is that $X(k)$, which may be thought of as the space of $k$-valued points of $X$, must be contractible. So $X$ has only a single "object", at least up homotopy (which is the best we can ask for anyway). On the other hand, Gaitsgory-Roseblyum also have a version of the correspondence between Lie algebroids and some kind of multi-object FMPs, where the objection to using $B$ might well be applicable.
The point is that $X(k)$ is the only connected space; the spaces $X(A)$ for Artinian rings are not, unless the DGLA in concentrated in non-positive cochain degrees. Contrast this with the notation $BG$ for the stack of $G$-torsors, whose stalks are connected in the sense that $(BG)(A)= B(G(A))$ for any Henselian ring $A$.
|
2025-03-21T14:48:29.961151
| 2020-03-01T03:18:54 |
353891
|
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|
Stack Exchange
|
Atkin-Lehner operator on supercuspidals
Suppose $f$ is a normalized cuspidal eigenform of level $p^2N$ ($p\nmid N$) and trivial character, such that the corresponding representation at $p$ is supercuspidal. Now suppose $\chi$ is primitive Hecke character of conductor $p$. We can apply the usual twisting operator by $\chi$ or $\chi^{-1}$ to $f$ to obtain normalized eigenforms $f_\chi$ and $f_{\chi^{-1}}$ with character $\chi^2$ and $\chi^{-2}$ respectively. Now we consider the adelic Atkin-Lahner operator given by the matrix $\begin{pmatrix} 0 & 1 \\ p^2 & 0\end{pmatrix}$ at $p$. Then one sees that it maps $f_\chi$ to $f_{\chi^{-1}}\otimes(\chi^2\circ\det)$ with some scalar multiple. My question is what is this scalar multiple? In particular is it a $p$-adic unit? (edited based on comment)
What you are asking for is a formula for the local epsilon-factors $\varepsilon(\pi \otimes \chi)$ where $\pi = \pi_{f, p}$ is the local component of $f$ at $p$. This is a deep question: it has to be, in some sense, since you can recover $\pi$ uniquely if you know the epsilon-factors of all its twists (Jacquet's local converse theorem).
Anyway, since you are assuming that $f$ has level $Np^2$ and trivial character, the representation $\pi$ is not too nasty: it's a "depth 0 supercuspidal", arising from a character $\eta: \mathbf{F}_{p^2}^\times \to \mathbf{C}^\times$ trivial on $\mathbf{F}_p^\times$. There is a paper by Jared Weinstein and me which gives an algorithm for computing $\eta$. Once you have this, there is a formula for the epsilon-factors in terms of Gauss sums over $\mathbf{F}_{p^2}^\times$. Up to some normalisation factor, $\varepsilon(\pi \otimes \chi)$ will be something like $\sum_{x \in \mathbf{F}_{p^2}^\times} \eta(x) \chi(\mathrm{Nm}(x)) e^{2\pi i \mathrm{Tr}(x)/p}$.
|
2025-03-21T14:48:29.961320
| 2020-03-01T05:49:44 |
353898
|
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"Ali Enayat",
"Andrés E. Caicedo",
"Emil Jeřábek",
"François G. Dorais",
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|
Stack Exchange
|
Interpreting proper elementarily equivalent end extensions?
Is there a tuple of parameter-free formulas $\Phi$ and a nonstandard $M\models PA$ such that $\Phi^M\models PA$, the induced $M$-definable initial segment embedding $j_\Phi^M:M\rightarrow\Phi^M$ is non-surjective, and $M\equiv \Phi^M$?
(Here by "$\Phi^M\models PA$" I mean "$\Phi$ defines an interpretation of a structure in the language of arithmetic in $M$, and that structure satisfies $PA$." Per Joel's comment below, we may freely require additionally that $j^M_\Phi$ be elementary.)
If we allow parameters the answer is yes, but in both the arguments there parameters are absolutely essential. I vaguely recall$^1$ a not-too-hard negative proof via Kripke's notion of fulfillability (see Putnam or Quinsey) but I can't reconstruct it at the moment.
$^1$I also recalled seeing this about the version with parameters, which turned out to be bogus; I think the parameter-free version is what I was actually thinking of, but now I'm much less sure I'm not just making stuff up.
If there is an example, then won't there have to be one where $j$ is elementary, since we can go to the definable hull of $M$, thereby reducing to the case where $M$ is pointwise definable, which will imply that $j$ is elementary as a consequence of elementary equivalence $\equiv$.
@JoelDavidHamkins Derp, of course - fixed!
@NoahSchweber Back in the early 1960s Feferman proved that there is no nonstandard model of PA that is elementarily equivalent to the standard model of arithmetic, and whose addition and multiplication are arithmetically definable. This can be found in the book Computability and Logic (by Boolos et al). The proof is very similar to the proof given by Joel in his answer to your question.
@AliEnayat Oh man I love that book, but I haven't read it in forever and I didn't think to look there. Thanks!
@NoahSchweber PS to my earlier comment: As I have explained in my comment to Joel's answer, his argument can be made to work ONLY IF we assume that the standard model of arithmetic is an elementary submodel of $M$ (i.e., if we assume that $M$ satisfies "true arithmetic"). This was noted explicitly in Theorem 1.1 of the paper "Nonstandard models that are definable in models of Peano Arithmetic" (Ikeda + Tsuboi, MLQ, 2007). They pose your question as an open question in their paper in item 2 of Remark 3.6, and to my knowledge the problem is wide open.
This answer is an attempt at explaining my critical posted comments on Hamkins' proposed answer; it also expands my posted comments to the MO question. I will explain:
(a) the gap in Hamkins' answer, (b) how it can be fixed (at the cost of considerably strengthening the hypotheses of the question), and
(c) relevant history and literature.
(a) [the gap] Suppose $M$ is a model of $PA$ and $\Phi$ consists of pair of arithmetical ternary formulae $\phi_{\rm{add}}(x,y,z)$, $\phi_{\rm{mult}}(x,y,z)$ for how to re-interpret (the graphs of) addition and multiplication such that (1) through (3) below hold:
(1) The model $\Phi^M$ satisfies $PA$, i.e., ($M$, $\phi^{M}_{\rm{add}}(x,y,z)$, $\phi^{M}_{\rm{mult}}(x,y,z)) \models PA$.
(2) $M$ and $\Phi^M$ are elementarily equivalent.
(3) The $M$-induced initial segment embedding $j^M_{\Phi}:M \rightarrow \Phi^M$ is not surjective.
Then as observed by Hamkins in his answer, by replacing $M$ by its elementary submodel $M_0$ of $M$ consisting of definable elements of $M$, the above conditions remain true. So far, so good.
The gap in the proposed proof occurs at the end of its first paragraph, where it is asserted that $j^{M_0}_{\Phi}:M_0 \rightarrow \Phi^M_0$ is an elementary embedding. The reasoning given is: "Since $M_0$ is pointwise definable, however, elementary equivalence will imply full elementarity, since all parameters are definable". However, for $j^{M_0}_{\Phi}$ to be an elementary embedding of $M_0$ into $\Phi^{M_0}$ we need to know that if $a \in M_0$ is definable in $M$ (and also in $M_0$) as the unique $x$ satisfying $\psi(x)$, then $j(a)$ is the unique $x$ satisfying $\psi(x)$ in $\Phi^M_0$, which is not warranted by the mere assumption that $j$ is an initial embedding.
(b) [the fix]. The gap explained above can be readily seen to be fillable if WE ASSUME, FURTHERMORE, THAT $M$ DOES NOT CONTAIN ANY NONSTANDARD DEFINABLE ELEMENTS (equivalently: $M$ is elementarily equivalent to the standard model of arithmetic $\Bbb{N}$, which is often paraphrased as "$M$ is a model of true arithmetic"). The reasoning: initial embeddings are automatically $\Delta_0$-elementary, so this extra assumption guarantees that $ a\in M_0$ is definable in $M$ via "the least $x$ such that $\delta(x)$ holds" for some $\Delta_0$-formula $\delta(x)$, which in turn guarantees that $j(a)$ is the unique $x$ satisfying $\delta(x)$ in $\Phi^M_0$.
(c) [history and literature] A few years before the appearance of Tennenbaum's famous characterization of the standard model of arithmetic as the only model of $PA$ (up to isomorphism) whose addition and multiplication are recursive, Feferman published an abstract in 1958 (in the Notices of AMS) to announce his result that there is no nonstandard model of true arithmetic (i.e., a model elementarily equivalent to the standard model of arithmetic $\Bbb{N}$) whose addition and multiplication are arithmetically definable, i.e., he proved:
Theorem (Feferman, 1958) The standard model of arithmetic $\Bbb{N}$ cannot interpret a nonstandard model of true arithmetic.
An early exposition of the above theorem of Feferman can be found in this 1960 paper of Scott. The theorem is also exposited as Theorem 25.4a in this edition of the textbook Computability and Logic (by Boolos, Burgess, and Jeffrey), but mysteriously with no reference to Feferman.
Many years later, Feferman's theorem was rediscovered and fine-tuned in by K. Ikeda and A. Tsuboi, in their paper Nonstandard models that are definable in models of Peano Arithmetic, Math. Logic Quarterly, 2007. Theorem 1.1 of this paper says the following (in the language used here).
Theorem. (Ikeda and Tsuboi, 2007) Suppose $M$ is an elementary extension of $\Bbb{N}$ (i.e., $M$ is a model of true arithmetic), and suppose that $M$ interprets a model $\Phi^M$ such that $M$ and $\Phi^M$ are elementarily equivalent. Then the $M$-induced initial segment embedding $j^M_{\Phi}:M \rightarrow \Phi^M$ is an isomorphism.
The proof of Ikeda and Tsuboi for Theorem 1.1 is essentially the one obtained by gluing Hamkins' answer (to the MO question) to the fix (b) above.
Ikeda and Tsuboi also pose the MO question and answer discussed here as an open question in their paper in item 2 of Remark 3.6. To my knowledge the problem remains open.
Finally, let me emphasize that the above discussion all pertained to parameter-free interpretations, since it is well-known that if $M$ is a recursively saturated model of $PA$, then there is some $\Phi(x)$ and some parameter $m \in M$, such that $M$ and $\Phi(m)^{M}$ are elementarily equivalent, and the $M$-induced initial segment embedding $j^M_{\Phi(m)}:M \rightarrow \Phi(m)^{M}$ is a nonsurjective embedding. Here is the proof outline: by recursive saturation, Th($M$) is in the standard system of $M$, and coded by some $m \in M$ such that, in the eyes of $M$, $m$ codes a consistent theory. So by the arithmetized completeness theorem, $M$ can build a model of $m$ whose elementary diagram is definable in $M$; the induced initial embedding is not surjective by Tarski's undefinability of truth.
Dear Ali, you seem to be using the Markdown syntax for quotations for things that are not actual quotations. This is certainly confusing for some, so you may consider editing. On the other hand, the layout makes visual sense so you could just warn users of your nonstandard usage of blockquotes to avoid confusion.
@FrançoisG.Dorais Well, people use the Markdown quote syntax for all kinds of things, such as highlighting statements of theorems, and this does not seem to confuse anybody. There is no other option how to make a part of text visually stand out in Markdown (except for disasters like making it all bold).
@FrançoisG.Dorais Thanks for your comment. I have used blockquotes in many of my MO answers for highlight purposes, and this is the first time that I am hearing that this could be confused with actual quotations, so I am not adding any explanations since (a) I feel that the way the blockquotes are used make it very clear that they are used for highlighting, and (b) I would have to add similar explanations in my other answers.
@François Is this some new policy? I've been doing this for years here and in the other site. We are talking of hundreds of posts that would need changing.
@FrançoisG.Dorais The point is that some administrators have decided to completely change the formatting of this markdown option and destroyed a very useful tool instead of replacing it. It would be worth a meta discussion.
@AndrésE.Caicedo and YCor This is not a policy. There are no style policies on MO, which is one thing all admins agree on. My comment just reflects my own confusion after reading Ali's answer (well, just the second blockquote). Perhaps it was too late in the day to read MO, but I was indeed confused. (I'm still perplexed by the second blockquote but it's fine if Ali really wants it that way.)
There can be no such model.
The first observation is that if there is a model as you describe,
then I claim there will be an instance where $j$ is elementary. To
see this, suppose that $M$ is as in your question. Let $M_0$ be the
collection of definable elements of $M$. This is an elementary
substructure of $M$, and so it also has $\Phi^{M_0}$, which will be
a definable interpretation of an elementarily equivalent model. So
it will have its own $j_0:M_0\to \Phi^{M_0}$ mapping $M_0$ isomorphically to an initial segment of the interpreted model $\Phi^{M_0}$, and
$M_0\equiv\Phi^{M_0}$. Since $M_0$ is pointwise definable, however,
elementary equivalence will imply full elementarity, since all
parameters are definable.
Let us continue with the main argument, where we assume that
$M=M_0$ and $j$ is elementary. The definition $\Phi$ has some
definite complexity. Let $N=\Phi^M$ be the interpreted
model. Let $n$ be such that the interpretation of the decoding of finite sequences in $N$ has complexity $\Sigma_n$ in $M$. Let $c$ be a number in $N$ larger than every element of $M$
(that is, larger than $j(x)$ for every $x\in M$). Inside $N$, let
$t$ be the pseudo-finite enumeration of the $\Sigma_{n+1}$ diagram
of $N$ for parameters up to $c$ and formulas with code up to $c$.
Since $n$ is standard finite, this is possible to do inside $N$.
In $M$, now, we can tell whether a given $\Sigma_{n+1}$ formula
$\varphi$ is true at a particular $x$ by looking at whether
$(\varphi,j(x))$ is in $t$. By assumption, this can be done in a
$\Sigma_n$ manner. So $\Sigma_{n+1}$ truth is $\Sigma_n$ definable,
which is a contradiction, since the arithmetic hierarchy does not
collapse.
EDIT by NS:
In fact, what the argument above shows is that whenever $\Phi^M\models PA$ and $j_\Phi^M$ is non-surjective (that is, we drop the elementarity hypothesis) we get a "relative hierarchy collapse:"
There is some $i$ (depending only on the complexity of $\Phi$) such that for all $k$ the $\Sigma_k$-theory of $\Phi^M$ is $\Sigma_i$ over $M$.
Interestingly, this does not neccessarily appear uniform since the complexity of finding the relevant parameter in $\Phi^M$ seems to grow unboundedly with $k$. In both the answers to the earlier MSE question linked in the OP, we did have such uniformity (at least in some fixed parameter from $M$) but it's not clear that this always happens.
Yes, I think that is what I have proved. Feel free to edit!
I've added a bit - of course feel free to remove/edit my remark for any reason.
@JoelDavidHamkins There is a very subtle gap at the beginning of your proof, when you transition from the model $M$ to its elementary submodel $M_0$. The model $M_0$ can certainly be elementarily embedded into every model of Th($M$), but there is no guarantee that the embedding $j$, when acting on $M_0$ will carve out an elementary submodel of $\Phi^M$ UNLESS we assume, in addition, that $M_0$ consists of only standard natural numbers, i.e., if we assume that $M$ satisfies "true arithmetic". See also my second comment to Noah's question.
|
2025-03-21T14:48:29.962233
| 2020-03-01T07:25:49 |
353905
|
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|
Stack Exchange
|
Multiplying all the elements in a group
Let $G = \{ g_i | i = 1, ...,n \}$ be a finite group and denote by $G!$ the multiset consisting of all the products of all different elements of $G$ in any order, that is
$$ G! = [ \prod_i g_{\sigma(i)} | \sigma \in S_n] $$.
I'm interested in knowing how $G!$ behaves as a set, and also how often does every element appear (i.e., how it behaves as a multiset).
In the case of an abelian $G$, $G!$ is either 1 or (iff exists) the single element of order 2 in $G$.
We can use the abelian case to get a (seemingly tight) upper bound on $G!$ as a set, by considering modding by $[G,G]$: projecting every such product to the quotient, which is abelian, we get $a^{\#[G,G]} \in G/[G,G]$ where $a$ is the single element of order 2 in $G/[G,G]$ if it exists, and the identity otherwise. Thus, if either $a$ is the identity or $\#[G,G]$ is even, $G!$ is entirely contained in $[G,G]$, and otherwise contained in the coset corresponding to $a$.
The reason this bound seems tight is that it's generally easy to get elements in the commutator group (up to the element of order 2): if $a,b,a^{-1},b^{-1}$ are distinct, $[a,b] \in G!$ by putting every other element of $G$ right next to it's inverse except for $a^{\pm1}, b^{\pm1}$ and likewise for products of commutators and so on.
Is it always the case that $G!$ is a coset of the commutator group? How often does each element appear?
It might also be useful to look at the action of $Aut(G)$ on $G!$, but I'm not totally sure what can that tell us.
It would be useful to use distinct notation for the multiset and the underlying set.
Yes, your $G!$ (as a set) is always either $[G,G]$ (if the order of $G$ is odd, or its Sylow $2$-subgroup is non-cyclic) or $z[G,G]$ if $G$ has cyclic Sylow $2$-subgroup, where $z$ is the involution in the Sylow $2$-subgroup. This was apparently a conjecture of Golomb, proved by Denes and Hermann. (I think the multiplicity is just the order of the derived subgroup.)
@James Why not post that as a proper answer to the question?
Yes, your $G!$ (as a set) is always either $[G,G]$ (if the order of $G$ is odd, or its Sylow $2$-subgroup is non-cyclic) or $z[G,G]$ if $G$ has cyclic Sylow $2$-subgroup, where $z$ is the involution in the Sylow $2$-subgroup. This was apparently a question/conjecture of Golomb (see p. 973) and independently of Fuchs (in a 1964 seminar), proved by Dénes and Hermann.
(Addressing Mark Sapir's comment, I could not find a published reference for Fuchs, but did manage to track down this paper by Dénes and Keedwell which contains a discussion of some of the history of the question.)
|
2025-03-21T14:48:29.962453
| 2020-03-01T08:14:07 |
353906
|
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|
Stack Exchange
|
Reducedness of complete intersection
Let $X$ be a very general surface of degree $\ge 5$ in $\mathbb{P}^3$ and $ Y$ is arbitrary irreducible cubic hypersurface. Is $X \cap Y$ reduced ?
Yes (of course, $Y$ should be reduced). Since $X$ is very general we may assume $\operatorname{Pic}(X)=\mathbb{Z}\cdot [\mathscr{O}_X(1)]$. If the divisor $Y_{|X}$ on $X$ is not reduced, it is of the form $2H+H'$, where $H$ and $H'$ are hyperplane sections of $X$ (possibly equal). But since the restriction map $H^0(\mathbb{P}^3,\mathscr{O}_{\mathbb{P}^3}(3))\rightarrow H^0(X,\mathscr{O}_X(3))$ is injective, this implies that $Y$ is non reduced.
Do we need $Y$ to be reduced or irreducible is enough ?
If $Y$ is not reduced, there is no chance that $Y\cap X$ is reduced.
|
2025-03-21T14:48:29.962547
| 2020-03-01T11:59:43 |
353915
|
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|
Stack Exchange
|
Proof of: If $f(x)=p(x)+o(x^n)$ for $x \to 0$, then $b_{k}=\frac{f^{(k)}(0)}{k !} $ for $ k=0,1, \ldots, n$
Before the current problem I work on, I proved the following:
Let $q$ be a polynomial with $\deg(q) \le n$. If $q(x)=o(x^n)$ for $x \to 0$, then $q$ is the zero polynomial.
I have to use the above for the problem I'm working on currently now, which is
Let $f:I \to \mathbb{R}$ be a $C^n$ function, defined on a open interval $I$ and assume $0\in I$. Let $p(x)=\sum_{k=0}^{n} b_{k} x^{k}$ be a polynomial with $\deg(p) \le n$. Show if $f(x)=p(x)+o(x^n)$ for $x \to 0$, then $b_{k}=\frac{f^{(k)}(0)}{k !} \quad \text{for} \quad k=0,1, \ldots, n$.
My attempt
I know that $o(x^n)$ is 0 for $x\to 0$. So in this limit $f(x)=p(x)$. And (I assume) since $p$ is the Taylor polynomial it's equal to $f$ for $x\to 0$. So I only have to calculate the Taylor polynomial of order $n$ with center point $0$. That will end the proof, right? Also, which theorem am I applying? I can find anything in my textbook which looks like what I have done.
Also, there is another problem with my method. I assume that $f$ is $n$ times differential in the center point $x=0$ when I calculate the Taylor polynomial $p$ of order $n$. But it is only given that $p$ is $n$ times continuous. I'm not sure about what to do.
Why do you "have" to use your little lemma? Is this perhaps a homework problem?
After all all you have to do is bringing p from one side to the other
Let $T$ be the Taylor polynomial for $f$ of order $n$ at $0$, so that $f(x)=T(x)+o(x^n)$ (as $x\to0$). Comparing this with the condition $f(x)=p(x)+o(x^n)$, we see that $(p-T)(x)=o(x^n)$. Using now what you have proved, we see that $p=T$, and hence the polynomial $p$ has the same coefficients as $T$.
But I still have to assume that f is n times differentiable?
Also, I can see much difference between my way and your way
(i) As can be seen in the linked Wikipedia article, it is enough to assume that $f$ is $n$ times differentiable at $0$; your $C^n$ condition is a bit of an overkill. (ii) I don't understand what you mean by your way and my way, and whether the difference between them (whatever it may be) matters, as far as answering your question is concerned.
I’m not criticizing you in any way. I’m appreciate the help very much. I’m just wondering where my mistake occurred
I am unable to understand what you call "My attempt". In particular, I don't know what meaning I could attach even to the first sentence there: "I know that $o(x^n)$ is 0 for $x\to0$."
|
2025-03-21T14:48:29.962747
| 2020-03-01T12:10:37 |
353916
|
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|
Stack Exchange
|
(maximal) antichains with respect to two different partial orders on the same set
In my recent work I stumbled across a problem of this type:
G with two partial oders $\leq$ and $\preceq$ on every set, i.e. for every $n \in \mathbb{N}$ $A_n \subset A_{n+1}$ and $(A_n, \leq) $ and $(A_n, \preceq)$ are partially ordered sets (partial orders do not depend on $n$).
The question now is whether there exist for any $n \in \mathbb{N}$ non-trivial sets $B \in \mathcal{P}(A_n)$ which are maximal antichains with respect to $(A_n,\leq)$ and $(A_n,\preceq)$? If so, how many?
Thank you very much in advance!
Maximal antichains with respect to which order? What is $A$? It seems that the $A_n$'s play absolutely no role here.
Thank you for your comment, I edited the question to make things clearer and correct.
I honestly have no idea about why this sequence is even relevant to the question. You're now asking about a local statement that has nothing to do with other sets in your sequence.
Almost a trivial counterexample:
Let $A_n$ be $\{0,\dots,n+2\}$ with one order being the usual $\leq$, that is a linear order, and another being $=$, that is the discrete order.
Note that all the $A_n$ have at least two elements. But the only sets which are antichains in any $A_n$ are singletons, which are never maximal with respect to $=$.
|
2025-03-21T14:48:29.962907
| 2020-03-01T13:03:34 |
353918
|
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|
Stack Exchange
|
Do disjoint unions of stacks commute with finite fibre products?
Choose a big $\mathit{fppf}$-site $(\mathbf{Sch})_{\mathit{fppf}}$ and let $S$ be a scheme in that site.
Let $\{\mathcal{X}_i\mid i\in I\}$ be a family of stacks in groupoids over $S$ and let $\mathcal{Y}\to\mathcal{Z}$ be a morphism of stacks in groupoids over $S$.
Let $\mathcal{X}\colon\!\!=\coprod_{i\in I}\mathcal{X}_i$ be the disjoint union as described in [Champs algébriques, G.Laumon/ L.Moret-Bailly, (3.3)].
Let $-\times-$ denote the $2$-fibre product of stacks in groupoids over $S$.
Is there a canonical morphism of stacks
$$\coprod_{i\in I}(\mathcal{X}_i\times_{\mathcal{Z}}\mathcal{Y})\to\Big(\coprod_{i\in I}\mathcal{X}_i\Big)\times_{\mathcal{Z}}\mathcal{Y}\quad$$ an isomorphism?
Have you tried writing down suitable functors between the two sides?
Of course $\mathbb{Z}$ can be replaced by any index set.
@S.Carnahan I have tried to write down the morphism on $T$-valued points. Given $T$, on the left hand side we have $(T=\coprod_i T_i, (\xi_i){i\in I})$ where $\xi_i\in (\mathcal{X}i\times{\mathcal{Z}}\mathcal{Y})(T_i)$ i.e. $\xi_i=(x_i,y_i,\varphi_i)$ where $\varphi_i$ is an isomorphism $\rho{\mathcal{X}i}(x_i)\overset{\sim}{\to}\rho{\mathcal{Y}}(y_i)$ in $\mathcal{Z}{T_i}$ if $\rho{\mathcal{X}i},\rho{\mathcal{Y}}$ denote the given morphisms $\mathcal{X}_i\to \mathcal{Z}$ and $\mathcal{Y}\to \mathcal{Z}$ respectively. On the right we have $((T=\coprod_i T'_i, x'_i),y,\varphi)$.
However I do not see, how the tuple on the left is mapped to the one on the right. This is likely related to the fact, that I do not understand how disjoint unions work in the $2$-category of categories fibred in groupoids. For example, how does one define $\mathcal{X}_i\to \coprod_i \mathcal{X}_i$ ? My first guess would be to define $(\coprod_i \mathcal{X}_i)(T)\colon!!=\coprod_i( \mathcal{X}_i(T))$ for categories fibred in groupoids?
Let $\coprod_i -$ be the stack in groupoids associated to the coproduct in the $2$-category of categories fibred in groupoids. If I did understand the process of stackification correctly, then the map $\mathcal{X}i(T)\to (\coprod_i \mathcal{X}i)(T)$ is given by choosing $T=T_i$ and $T_j=\emptyset$ for $j\not= i$ and then mapping $x_i\in \mathcal{X}i(T)$ to the $T$-valued point $(T=\coprod_i T_i=T_i, (x_i\mid{T_i}){i\in I}={x_i\mid{T_i}})$!?
Your first guess is wrong even for schemes: Let each $X_i$ be a point, and let $T$ be 2 points. The description in the next comment is correct (modulo clumsy notation).
@S. Carnahan So the disjoint union of categories fibred in groupoids is defined not this way? I found this definition in the book [Champs algébriques, G.Laumon/ L.Moret-Bailly] in chapter two under (2.2.1). This is the only reasonable definition I can think of. I use this clumsy notation, because I want to understand how things work in very detail (I am an average level master student, not a working mathematician, and I deal with this the first time).
@Riza Hawkeye Thanks, this is nice to know, but I did not learn about higher topos yet and thats why I try to understand this first in a more elementary way.
@S.Carnahan The definition of the coproduct of categories fibred in groupoids that sdigr is indeed correct. The point is that the (2,1)-Yoneda embedding doesn't preserve colimits (nor does the (1,1)-Yoneda embedding!). The stackified (2,1)-Yoneda embedding, however, does commute with coproducts.
@RizaHawkeye The nontrivial part of this statement is not the universality of colimits in the 2,1-topos of groupoid stacks on Sch_fppf. It's that the inclusion of Artin stacks into this category preserves coproducts. This is an important property of the topology, and it is clearly false if one chooses the trivial topology (more or less Scott's comment).
@sdigr My mistake - your guess is correct.
Another way to think about it is how the extensive topology interacts with stackification for superextensive topologies. Daniel Schäppi showed something you might want to check out, see discussion at https://nforum.ncatlab.org/discussion/3907/stacks-on-superextensive-sites/
I would split this problem up into two parts (here, 'sheaf (of groupoids)' is used instead of stack in order to disambiguate between Algebraic stacks (geometric objects) and mere (pseudo-)functors satisfying descent). :
1.) Show that the inclusion of algebraic stacks into the category of fppf sheaves of groupoids on Sch preserves coproducts. This follows immediately from the fact that algebraic stacks are a full (2,1)-subcategory of fppf sheaves and that if $F,G$ are two algebraic stacks, their sheafy coproduct is representable by an algebraic stack. If D is a diagram landing in a full subcategory whose limit or colimit exists in the ambient category and is in the full subcategory, this is also a limit or colimit of the diagram landing in the full subcategory without reference to the ambient category. I think the proof here is immediate by taking a disjoint union of the atlases.
2.) Show that colimits are universal in (2,1)-stack topoi. This follows from the left-exactness of the stackification (2,1)-functor together with the altogether more obvious version of this fact for (2,1)-topoi of groupoid fibrations (also called (2,1)-presheaf topoi), where one can immediately reduce to proving the statement for groupoids pointwise.
An object over a scheme $T$ on the left is given by a decomposition of $T$ into a parametrized disjoint union $T_i$ of schemes, and a parametrized family of triples $(x_i, y_i, \phi_i)$, where $x_i$ is an object of $X_i$ over $T_i$, $y_i$ is an object of $Y$ over $T_i$, and $\phi_i$ is an isomorphism $\rho_{X_i}(x_i) \to \rho_Y(y_i)$ in $Z$ over $T_i$. A morphism over $id_T$ is a parametrized family of pairs of maps $(f_i: x_i \to x'_i, g_i: y_i \to y'_i)$ that satisfy suitable commutative square relations. In particular, if two objects come from unequal decompositions of $T$, then there are no morphisms between them. Let us omit discussion of other morphisms, and pretend the "fibered category" property takes care of them.
An object over a scheme $T$ on the right is given by a decomposition of $T$ into a parametrized disjoint union $T_i$ of schemes, and a tuple $((x_i), y, \phi)$, where $x_i$ is an object of $X_i$ over $T_i$, $y$ is an object of $Y$ over $T$, and $\phi$ is an isomorphism $\rho_{\coprod X_i}((x_i)) \to \rho_Y(y)$ in $Z$ over $T$. A morphism over $id_T$ is a pair $((f_i: x_i \to x'_i), g:y \to y')$ that satisfies conditions that I won't describe.
In order to match these data, we need to identify $y$ with the parametrized family $(y_i)$, and $\phi$ with $(\phi_i)$ for objects, and $g$ with $(g_i)$ for morphisms. This is just using the fibered category property: pulling back along the isomorphism $\coprod T_i \to T$ yields a decomposition of $y$ that is unique up to unique isomorphism. It might be helpful to check that the object $\rho_{\coprod X_i}((x_i))$ in $Z$ over $T$ is identified with the tuple $(\rho_{X_i}(x_i))$ of objects in $Z$ over $T_i$.
Thanks for the explanation. I understand, that one can pullback $y$ along $\coprod_i T_i\to T$ because of the fibre product property (after a choice of pullbacks was made in the very beginning). But given a family $(y_i)$ of $y_i\in \mathcal{Y}_{T_i}$ as on the left. How can one obtain $y$? I guess that more is needed than only the property that pullbacks exist uniquely. I guess that we have to keep track of the descent data coming from the stackification process aswell, do we not need this?
I don't understand the description of the $T$-valued points of a disjoint union $\coprod_{i\in I}\mathcal{X}i$ of stacks as the choice of a decomposition $T=\coprod{i\in I}T_i$ and $(x_i)_{i\in I}, x_i\in {\mathcal{X}i}{T_i}$, because if I take the construction of stackification as in Lemma 02ZN literally, there does not seem to be an obvious reason why one can choose the covering indexed by the same index set $I$ and it is also not clear to me, why the descent data is not mentioned in this description. I think this causes extra confusion to me.
@sdigr To get $y$ from $y_i$, you pull back along the isomorphism $T \to \coprod_i T_i$. As far as stackification is concerned, if you have a finer covering, then we are concerned with gluing objects over $T_i$ in $X_i$, and descent for these objects is handled by the stack property of $X_i$.
|
2025-03-21T14:48:29.963483
| 2020-03-01T13:34:57 |
353920
|
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|
Stack Exchange
|
Is there any general reference on matrix quadratic equations?
I am studying a problem where a quadratic matrix equation emerges. The equation is as follow (all capital letters are n by n matrices)
$(I-X^{\prime}L)X=D$
where $L$ and $D$ are both symmetric and positive definite. How much can I say about a solution $X$?
Assuming you're using $'$ for transpose, it's symmetric.
Yes, $\prime$ for transpose. Could there be any sufficient and necessary conditions that guarantee the existence of a solution?
I don't presume this would be easy but wonder if there could be any results about the existence of a solution. I am not familiar with matrix equations. In the case of 2 by 2, this would be a system of quadratic equations and may be reduced to a quartic equation.
You write in your answer that "in the original equation $X$ must be symmetric", but this is not stated in your question, and on the contrary you used $X'$ there which suggests that there is a difference between the two. Is $X$ symmetric or not? If the answer is yes, then this is equation is well studied (algebraic Riccati equation; see here).
Also, your title is a lot more general than your actual question.
Does this answer your question? How to solve a quadratic matrix equation with positive semidefinite constaint Not an exact duplicate (different definiteness assumptions) but things work in the same way and my solution applies.
@FedericoPoloni $X = D + X^T L X$ must be symmetric, so the original equation might as well assume it.
A good resource is : Abou-Kandil, Hisham, Gerhard Freiling, Vlad Ionescu, and Gerhard Jank. Matrix Riccati equations in control and systems theory. Birkhäuser, 2012.
I figure out something and would like to share and check whether it is correct.
It is perfectly analogous to real quadratic equations.
Since $L$ is symmetric and positive definite, they can be decomposed:
$L=U_L^{\prime} \Lambda_L U_L$ where $U_L$ is orthonormal, and $\Lambda_L$ is diagonal with positive entries.
Let $Q=U_L^{\prime} \Lambda_L^{-1/2} U_L$, which is the inverse of the square root of $L$, and it is symmetric.
Then the above equation is equivalent to
$\tilde{X}^{\prime} \tilde{X}-Q\tilde{X}=D$ where $\tilde{X}=Q^{-1}X$
Suppose, in addition, $D$ and $L$ are simultaneously diagonalizable, I conjecture that $\tilde{X}$ is symmetric. Therefore, the above is equivalent to
$(\tilde{X}-\frac{1}{2}Q)^{\prime}(\tilde{X}-\frac{1}{2}Q)=\frac{1}{4}L^{-1}-D$
Note that the right-hand side is symmetric; hence, there exists a real solution of $\tilde{X}$ if and only if the right-hand side is positive semi-definite.
And after a bit of algebra, if solutions exist, it would be
$X=\frac{1}{2}L^{-1}[I \pm (I-4LD)^{1/2}]$ where $(I-4LD)^{1/2}$ is the real p.s.d square root of $(I-4LD)$, which exists because $(I-4LD)^{1/2}$ is p.s.d. and symmetric.
Hence, there are at most two solutions.
Are all of these arguments correct?
I don't see why $X$ symmetric implies $\tilde{X}$ symmetric. A product of symmetric matrices is not symmetric.
You are right, I just notice that and add some assumption. @Federico Poloni
$X$ is symmetric because $X=D+X^{\prime}LX$ where the right-hand side is symmetric given that $L$ and $D$ is symmetric.
I see, thanks. Then I suggest you to take a look at the question I linked in the comment above. That should solve your problem.
Yes, I definitely will. Thank you. @Federico Poloni
|
2025-03-21T14:48:29.963772
| 2020-03-01T14:27:31 |
353922
|
{
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"Igor Khavkine",
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"Nate Eldredge",
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|
Stack Exchange
|
How to prove roots of several univariate polynomials is a solution of a multivariable polynomial
When I study the existence of certain spherical polygon, I have encountered the following problem.
Let $\gamma$ and $\delta$ be angles such that
$1216\cos^6\gamma-1056\cos^4\gamma+84\cos^2\gamma-1=0, \ \cos\gamma\approx -0.8839453420007384$.
$1216\sin^6\gamma-2592\sin^4\gamma+1620\sin^2\gamma-243=0, \
\sin\gamma\approx-0.4675902397989036$.
$1216\cos^6\delta-2496\cos^4\delta+1668\cos^2\delta-361=0,\
\cos\delta\approx 0.8469176972278768$.
$1216\sin^6\delta-1152\sin^4\delta+324\sin^2\delta-27=0,\
\sin\delta\approx
0.5317240018301108$.
When I put $\cos\gamma=-0.8839453420007384$, $\sin\gamma=-0.4675902397989036$, $\cos\delta= 0.8469176972278768$ and $\sin\delta=0.5317240018301108$ into
$(\dfrac{\sin\frac{2\pi}{9}}{6}-\dfrac{\cos\frac{2\pi}{9}}{6\sqrt{3}}-\dfrac{4}{3\sqrt{3}})\sin\delta\sin\gamma+(\dfrac{5\cos\frac{2\pi}{9}}{18}-\frac{\sin\frac{2\pi}{9}}{6\sqrt{3}}-\dfrac{4}{9})\cos\delta\sin\gamma$
$-(\dfrac{\sin\frac{2\pi}{9}}{2\sqrt{3}}+\dfrac{\cos\frac{2\pi}{9}}{2})\cos\gamma\sin\delta+(\dfrac{5\sin\frac{2\pi}{9}}{6}+\dfrac{\cos\frac{2\pi}{9}}{2\sqrt{3}})\cos\delta\cos\gamma$, I get $5.551115123125783\times 10^{-17}$.
How to prove/disprove
$(\dfrac{\sin\frac{2\pi}{9}}{6}-\dfrac{\cos\frac{2\pi}{9}}{6\sqrt{3}}-\dfrac{4}{3\sqrt{3}})\sin\delta\sin\gamma+(\dfrac{5\cos\frac{2\pi}{9}}{18}-\frac{\sin\frac{2\pi}{9}}{6\sqrt{3}}-\dfrac{4}{9})\cos\delta\sin\gamma$
$-(\dfrac{\sin\frac{2\pi}{9}}{2\sqrt{3}}+\dfrac{\cos\frac{2\pi}{9}}{2})\cos\gamma\sin\delta+(\dfrac{5\sin\frac{2\pi}{9}}{6}+\dfrac{\cos\frac{2\pi}{9}}{2\sqrt{3}})\cos\delta\cos\gamma=0$.
Remark: In order to make sure there is no typo, I include the picture of the expression and the computation result in wxmaxima:
where $cga=\cos\gamma$, $sga=\sin\gamma$, $cde=\cos\delta$ and $sde=\sin\delta$.
Note that $2 \ 3^{3/2}=2\times 3^{3/2}$.
Have you checked your conjecture with higher numerical precision?
Another remark is that in principle, you can compute $\sin \gamma$, $\sin \delta$, etc, in closed form, by solving cubics.
In principle, resultants should give a way to do this; however, with such a long expression, it would likely take a long time to write it out. I suspect there's some relatively simple trig identity behind this, if it's true.
@Nate Eldredge, I have tried to checked my higher numerical precision. However, wxmaxima seems to take around 16 decimal places in the process even when I put those cosine values and sine values with precision of 60 or more decimal places in the beginning. I think I may need to use another program to verify my conjecture. However, wxmaxima seems to take around 16 decimal places in the process even when I put those cosine values and sine values with precision of 60 or more decimal places in the beginning. I think I may use another program to verify my conjecture.
At first, I think I can prove/disprove it easily by writing out those sine/cosine values explicitly via solving cubic. Since those equations are cubic polynomials with 3 real roots in square of sine/cosine of $\gamma$ or $\delta$, even one can compute $sin\gamma$ in closed form, but it must be involved imaginary number i and look quite complication. That is why I takes it as the last resort. Thank you for your comments.
@user44191, Can you elaborate more about how to apply the resultants method to my problem? Do you have any good references about how to prove/disprove the above problem? Thank you.
Does your expression get any nicer if you rewrite everything in terms of $\sin(\gamma \pm \delta)$, $\cos(\gamma \pm\delta)$ using the sum and difference identities?
In Maxima, multiple precision floating point numbers should be entered with the notation #.##########b# (big floats) instead of #.####### or #.#########e# (regular floats). Use bfloat(...) to evaluate known constants and functions as big floats. The global variable fpprec controls the number of digits kept while working with big floats.
I think you can convert this to polynomial equations, that is, introduce sin(gamma), etc as quantities defined via equations, and then use a Groebner basis approach.
@JamesCheung The idea for resultants is just that everything involved is an algebraic number, so the left-hand side can in principle be expressed as the root of a polynomial using resultants (e.g. $\sqrt{2} + \sqrt{3}$ is a root of the polynomial $\text{Res}_y(y^2 - 2, (x - y)^2 - 3)$).
This is a standard problem of algebraic geometry, which admits a completely algorithmic solution. We note that $\cos\frac{2\pi}9$ is a root $x$ of the equation $1 - 6 x + 8 x^3 = 0$. The rest of the task a computer algebra program can do easily, as seen in the following image of a Mathematica notebook, confirming your conjecture by giving the exact (zero) value of your big expression:
This is a more hands-on answer than the previous one; it is based on resultants.
Here we again note that
$$x_*:=\cos\tfrac{2\pi}9$$
is a root of the equation
$$p_1(x):=1 - 6 x + 8 x^3 = 0.$$
Next, using the formula $\cos^2=\frac1{1+\tan^2}$, we see that
$$V_*:=\tan\gamma$$
is a root of the equation
$$P_2(V):=243 - 891 V^2 + 81 V^4 - V^6 = 0$$
and
$$T_*:=\tan\delta$$
is a root of the equation
$$P_3(T):=27 - 243 T^2 + 585 T^4 - 361 T^6 = 0.$$
Moreover, the pair $(x_*,y_*)$ with
$$y_*:=\sin\tfrac{2\pi}9$$
is a root of the equation
$$p_4(x,y):=x^2+y^2 - 1 = 0.$$
We have to show that the quadruple $(x_*,y_*,V_*,T_*)$ is a root $(x,y,V,T)$ of the equation
$$L:=L(x,y,V,T):=\left(\frac{x}{2 \sqrt{3}}+\frac{5 y}{6}\right)+V\left(\frac{5 x}{18}-\frac{y}{6
\sqrt{3}}-\frac{4}{9}\right)-T\left(\frac{x}{2}+\frac{y}{2 \sqrt{3}}\right)+T V \left(-\frac{x}{6
\sqrt{3}}+\frac{y}{6}-\frac{4}{3 \sqrt{3}}\right)=0;$$
this simplification of the big expression in question is crucial, as it allows to reduce the calculations to a feasible level.
Let us now quote the following property of the resultant:
"The resultant of two polynomials with coefficients in an integral domain is zero if and only if they have a common root in an algebraically closed field containing the coefficients."
Consider successively the resultants
$$r_1(c,x,V,T):=R(L(x,y,V,T)-c,p_4(x,y),y),$$
$$r_2(c,V,T):=R(r_1(c,x,V,T),p_1(x),x),$$
$$r_3(c,T):=R(r_2(c,V,T),P_2(V),V),$$
$$r_4(c):=R(r_3(c,T),P_3(T),T),$$
where $R(P,Q,w)$ denotes the resultant of polynomials $P$ and $Q$ with respect to a variable $w$. (It takes about 30 sec for Mathematica to compute each of the resultants $r_3(c,T)$ and $r_4(c)$.)
It follows from the quoted property of the resultant that the real number $L(x_*,y_*,V_*,T_*)$ must be in the set of all real roots of the polynomial $r_4(c)$. In fact, $r_4(c)$ has the root $0$ (of multiplicity $3$), whereas all the other $213$ real roots of $r_4(c)$ are $>1/2$ in absolute value. However, a direct numerical calculation shows that $|L(x_*,y_*,V_*,T_*)|\le1/2$.
Thus, $L(x_*,y_*,V_*,T_*)=0$, as desired.
|
2025-03-21T14:48:29.964304
| 2020-03-01T15:32:47 |
353923
|
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"S. Carnahan",
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|
Stack Exchange
|
Cartier duality and Frobenius on Witt vector schemes
Suppose for simplicity we are working over $\mathbb{F}_p$. Cartier duality is an antiequivalence between formal groups and affine group schemes over $Spec(\mathbb{F}_p)$. Let $\mathbb{W}_p(-)$ denote the Witt vector affine group scheme. It is well known that the Cartier dual $Map_{Grp}(\mathbb{W}, \mathbb{G}_m)$ to this is the Witt vector formal group $\widehat{\mathbb{W}_p}$. This can be defined by taking the colimit of all formal completions at the identity of the truncated Witt schemes and comes endowed with natural formal group structure.
I know that Cartier duality should be viewed as exchanging the Frobenius and Verschiebung maps. Does this mean that the Cartier dual of the natural Frobenius map $F: \mathbb{W}_p \to \mathbb{W}_p$ will be the Verschiebung map defined on the Witt vector formal scheme? This seems to be incompatible with the statement that the Cartier dual of the formal multiplicative group $\widehat{\mathbb{G}_m}$ is the subgroup scheme of "fixed points" of $F: \mathbb{W}_p(-) \to \mathbb{W}_p(-)$.
Can you say what duality does to F and V in the case of truncated Witt vectors in a bit more detail? Also, I thought the Cartier dual of $\widehat{\mathbb{G}_m}$ is $\mathbb{G}_a$. At least, that seems to be what you get from the topological dual Hopf algebra.
|
2025-03-21T14:48:29.964415
| 2020-03-01T16:16:06 |
353925
|
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"url": "https://mathoverflow.net/questions/353925"
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|
Stack Exchange
|
When is the symplectic reduction of an action reduced?
Let $X$ be a smooth affine variety with an action of a reductive linear algebraic group $G$ over the complex numbers. We have a moment map $\mu:T^*(X)\to \mathfrak{g}^*$ given by $\mu(x,\xi,Y)=\xi(\tilde{Y}_x)$ where $\tilde{Y}$ is the vector field on $X$ obtained by applying the differential of the action map $a:G\times X\to X$ to $Y\in T_xX\subseteq T_{(e,x)}G\times X$.
The (affine) symmplectic reduction of $X$ by the action of $G$ is given by the categorical quotient of the fiber of the moment map $\mu^{-1}(0)$ by the induced action of $G$..
Note that in this question I consider the schematic fiber of the moment map, which might in principle be non-reduced.
Question: Is there a classification the linear representations of $G$ for which the symplectic reduction of the action of $G$ on the representation space is reduced?
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2025-03-21T14:48:29.964511
| 2020-03-01T16:19:09 |
353926
|
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|
Stack Exchange
|
Why should it be hard to generalize Dvir's proof of the finite field Kakeya conjecture to the Euclidean case?
Let $q$ be prime and let
$q\delta \sim 1.$ Let $K$ be any set of $C_n\delta$-separated tubes in $B(0,2)$, where $C_n$ is some constant depending on $n$. Let us consider a grid of $q^n$ points scaled with $\delta^n$. We embed this evenly spaced orthogonal grid in $\mathbf{R}^n$ so that the origins of the two sets meet. Let $K' \subset K$ so that each tube in $K$ contains a line from $K'$. The lines $a + vt \in K'$ are such that if $v \in tu$ then $a + tu = a + vt$. So that we have $q^{n−1}$ lines in $K$ and the directions are $\delta$-separated (edit: not necessarily true if $K' \subset K$) . By Dvir's theorem we have $\lvert K'\rvert \ge \frac{q^n}{2^n}$. So we have at least that many points $x \in K' \subset K$ that are $\delta$-separated. Multiplying above by $\delta^n$ we have $\lvert K\rvert \ge \lvert K'\rvert\delta^n \ge q^n\delta^n \frac{1}{2^n} \gtrsim \frac{1}{2^n}$ , because the $C_n\delta$-tubes contain the $\delta$−tubes. Isn't this a straightforward generalization? (EDIT) As was pointed out, those "grid lines" are unlike euclidean lines. The grid lines can go to the end of the grid and then reappear somewhere else in the grid. If a grid line contains $m$ points, it seems to me that for those "grid lines" we may even need (in euclidean sense) $m$ parallel lines to the cover the grid line. This grows the number of (euclidean) lines and solves the mystery (for me). The solution: Let $K$ contain in addition to $\delta$-separated tubes essentially parallel tubes in order to achieve $K' \subset K.$ However, estimating how many parallel tubes you need may be difficult.
I don't entirely follow your proposed approach, but let me make a few remarks which might help you think about this:
(1) Clearly this can't work, since this argument purports to show that a Kakeya set must have positive Lebesgue measure which, in contrast to the finite field setting, we know is false in Euclidean space.
(2) Note that the discretization of a tube is not necessary an algebraic line or even approximately an algebraic line. Indeed consider R^2 and a tube oriented along the y axis. Now adjust the angle slightly in either direction. What does a discretization of this tube look like? Well it depends on the scales involved but, roughly in the case of a $\delta$ tube rotated by a $~\delta$ degree angle, the discretization will be the union of long segments of vertical lines (rectangles/blocks) stacked on top of each other slowly drifting in the direction of the rotation. This is not an algebraic line.
(3) In the discrete (finite field) setting two lines intersect at a single point. However, for reasons similar to the discussion in (2), two tubes can have substantial overlap. This is sometimes refereed to as "small angle issues" in the literature.
It does not imply that. It implies that there are infinitely many $\delta$ s.t the tube set is of positive measure. Has anyone showed that tube sets can be formed to make a small Kakeya set? IMO this implies that tubes dont cover Besicovitch set effectively. 2) The algeibrac lines belong to the tubes. There is that $C_n*\delta$ of room to play.3) Tube intersections dont matter. It matter that the tubes contain the points that are $\delta$-separated.
(1) Consider a $\delta$ neighborhood of a Besicovitch set, then this is a "tube set" with measure decreasing to 0 as $\delta \rightarrow 0$, contradicting the claim you want to prove. (2) Consider the example of a tube rotated by an angle of $\delta$ with the y axis in $R^2$. What algebraic line is contained within it? (2*) Also think about what a mod p line looks like when plotted under you mapping to $R^n$. It will look like (roughly) an arithmetic progression (with some wrap around) in each variable. Notably it will not be contained within a $\delta$ tube.
We know it will contain at least $\delta^{1-n}$ tubes and its measure tends to $0$. If the set were to contain more tubes, this would increase the measure not decrease it. 2) What does "$\delta *t (0,1)$ where $t \in F_q$" mean here; $\delta$ is a small real number, not an element of $F_q$?
Ok. I see , but that does not count as a proof. Please, provide a reference for your technique or make an answer. 2) $\delta*t(0,1)$ where $t \in F_q$. Sorry, I forgot to scale the lines. I was now talking about elements of $R$^n. .3) Why? It is a arithmetic progressin that is rescricted from the continuos line and belongs to a $\delta$-tube.
(2) the point is that your argument is conflating lines in $R^n$ with lines (arising via your mapping) from $F_p$, and the two are very different. If you don't see this, try to find a line in in $F_p$ contained within a neighborhood of $\delta * t(0,1)$. (3) I suggest you attempt to plot a (discritized) line from $F_p^2$ via your mapping into $R^2$ to see what it looks like.
Ok. I see what you mean. Why can`t I use $t$ from $F_p$ if the order of the field is $n$, where $p$ is a prime?
I'll point out Zhang (https://arxiv.org/abs/1403.1352) has developed a variant of Dvir's argument directly in $R^n$ that proves an analog of the Kakeya conjecture where tubes are replaced with Euclidean lines (and points). You can try to insert that into a discretization argument similar to what you are proposing here which will get you around some of the issues we just talked through. But then you start running into things like the "small angle issues" I alluded to with (3). Needless to say it doesn't prove the conjecture. However it might be instructive for you to think through.
I accept your answer and many thanks for you. Your method 1) seem to work for compact sets. My mistake ) was to estimate the number of lines wrongly. If $(x_1,..,x_n)$ is a point in the grid $[0,1]^n$ we can choose the points in about $q^{n-1}\Pi(q)$ different ways, where $\Pi(q)$ is the prime counting function. If one of the coordinates is a prime, $0$ or $1$ it generates its own line. So we have more than $\delta^{1-n}$ lines in our tube set, so it cant be $\delta$-separated (I considered only the points where one coordinate is $1$).
Moreover, most of the algebraic lines will be splitted in two because $a + tv$ will belong to $[0,2q]^n$. So we have essentially two sets. However, even smaller of those sets contain $\sim q^{n-1}\Pi(q)$ lines. If we can show that we have at least $\sim q^{n-1}\Pi(q)$ points we have the estimate with an logarithmic factor.
I don't know how to extend the proof. But those "small angle" issues can be solved by dividing the $\delta$-tubes to really small paraller $\delta'$-tubes, whit small overlap. Then all the angles will be very large compared to $\delta'$.
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2025-03-21T14:48:29.965244
| 2020-03-01T17:39:20 |
353931
|
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|
Stack Exchange
|
Eigenvalue and eigenfunction convergence
Consider a bounded Euclidean domain $\Omega \subset \mathbb{R}^n$ (for simplicity, let's say, $\Omega$ has smooth boundary and is simply connected). Let $p \in \Omega$ be a point, and call $\Omega_n = \Omega \setminus B(p, \frac{1}{n})$. Let $\Delta$ (respectively, $\Delta_n$) denote the Dirichlet Laplacian on $\Omega$ (respectively, $\Omega_n$). I am interested in when the eigenvalues and eigenfunctions of $-\Delta_n$ converge to the eigenvalues and eigenfunctions of $-\Delta$, and in what sense (norm etc.) the eigenfunction convergence happens. The same question can be asked about the Neumann Laplacian.
The chief reason for my interest is that I am studying problems in homogenization and inverse problems (for example, papers of Vogelius, Friedman, Ammari et al). In these problems the setting is typically more complex, and problems involve an increasing number of perforations with decreasing "size". I am trying to understand a "toy case" thoroughly.
Edit: The main difficulty in the question seems to be for the case of repeated eigenvalues.
The corresponding quadratic forms are all equal, but the domains increase. As long as the union of the Sobolev spaces $H_0^1(\Omega_n)$ is dense in $H_0^1(\Omega)$, the eigenvalues converge (by the min-max characterisation). Convergence of eigenfunctions is trickier if the eigenvalues are degenerate.
If I remember correctly, Jerison and Kenig developed "domain deformation" techniques in the 1990s. [D. Jerison and N. Nadirashvili, The “hot spots” conjecture for domains with two axes of symmetry,
J. Amer. Math. Soc. 13 (2000) 741–772] might be a good source of references. If I find time tomorrow, I will have a look at it.
@MateuszKwaśnicki By "degenerate" I guess you mean non-repeated eigenvalues, correct? If the eigenvalues are non-repeated, and the union of $H^1_0(\Omega_n)$ is dense in $H^1_0(\Omega)$, then the eigenfunctions attaining the min-max should also converge. Is that correct?
The only thing you need to show is that the convergence of the resolvent operator is sequentially compact. Then, general results about sequential compactness give you the result that the eigenvalues converges to the eigenvalues, in the sense that for any given open neighborhood of any eigenvalue in $\mathbb C$ of $T$ of multiplicity $d$, the limit compact operator, for $n$ large enough, $T_n$ (the resolvent compact operator) will have exactly $d$ eigenvalues in that neighbourhood.
Proving compactness just follows from Rellich-Kondrachov in this case.
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2025-03-21T14:48:29.965437
| 2020-03-01T19:49:47 |
353936
|
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|
Stack Exchange
|
Minimal radius of a ball admitting a trivialization of a vector bundle
Let $X$ be a compact Hausdorff space and $p : V \to X$ a complex vector bundle of rank $n$. For $r > 0$ let $B(r,x)$ denote the open ball of radius $r$ around $x$. Does there exist an $r$ such that, for any $r'$, $0< r' \leq r$ and any $x \in X$, we have $p^{-1}(B(x, r')) \cong B(x, r') \times \mathbb{C}^n$?
https://en.m.wikipedia.org/wiki/Lebesgue's_number_lemma
Looks good! Thanks!
|
2025-03-21T14:48:29.965519
| 2020-03-01T22:35:24 |
353943
|
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"Giorgio Metafune",
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Stack Exchange
|
Weak version of Karamata's Tauberian theorem
I first posted this on mathematics. However, I got no answer there and it seems adapted here too. Also, it seems to be harder than I first thought.
Karamata's Tauberian theorem states the following. Let $A(z)=\sum a_nz^n$ be a power series with non-negative coefficients $a_n$ and radius of convergence 1. Let $\beta>0$. Then, as the real variable $s\in [0,1]$ tends to 1, $\sum_{n\geq 0}a_ns^n\underset{s\to 1}{\sim} c/(1-s)^\beta$ if and only if $\sum_{k=0}^na_k\underset{n\to \infty}{\sim}c'n^{\beta}$, where $c$ and $c'$ are determine each other. Moreover, if $a_n$ is non-increasing, then one can replace $\sum_{k=0}^na_k$ with $a_n$, that is, we have $a_n\sim c''n^{\beta-1}$. See for example Corollary 1.7.3 in Bingham, Goldie and Teugel's book Regular variation.
I'm wondering if the following is true. For two functions $f$ and $g$, write $f\asymp g$ if there exists $C\geq 0$ such that $f\leq Cg$ and $g\leq Cf$. Let $A(z)=\sum a_nz^n$ be a power series with non-negative coefficients $a_n$ and radius of convergence 1. Then, $\sum_{n\geq 0}a_ns^n \asymp 1/(1-s)^\beta$ for $s\in [0,1]$ if and only if $\sum_{k=0}^na_k\asymp n^{\beta}$ if and only if $a_n\asymp n^{\beta-1}$. The implicit constants are asked not to depend on $s$ and $n$ respectively.
The asymptotics $a_n \approx n^{\beta}$ and $\sum_{k=0}^n a_k \approx n^{\beta}$ sound inconsistent; $a_n$ should behave like $n^{\beta-1}$. Also, if $(a_n)$ is decreasing, then $\beta$ can be at most 1.
I do not have access to Bingham–Goldie–Teugels book right now, but did you check in the chapter on "$O$-regular variation" (Chapter 2, I believe)? At least in the continuous case (the Laplace transform instead of the power series), both Karamata's Tauberian theorem and monotone density theorem have their couterparts for the "$O$-regular variation" (that is, roughly speaking, with "$\sim$" replaced by "$\asymp$").
@GiorgioMetafune Of course you're right, thank you very much. I'm particuarly interested in the case where $\beta <1$.
This seems to follow easily from de Haan–Stadtmüller Theorem; see Theorem 2.10.2 in the Bingham–Goldie-Teugels book:
Theorem: Let $U$ be non-decreasing, and vanish on $(-\infty, 0)$. The following are equivalent:
(i) $U \in OR$;
(ii) $\hat{U}(1/\cdot) \in OR$;
(iii) $\hat{U}(1/t) \asymp U(t)$ ($t \to \infty$).
Here $\hat{U}(\lambda) = \int_{[0, \infty)} e^{-\lambda x} U(dx)$ is the Laplace–Stieltjes transform of the measure $U(dx)$, and $OR$ stands for the class of $O$-regularly varying functions at infinity, that is, positive functions $f$ such that $$\limsup_{x \to \infty} \frac{f(\lambda x)}{f(x)} < \infty$$ for every $\lambda \in (0, \infty)$.
Now suppose that $$A(z) = \sum_{n = 0}^\infty a_n z^n $$ with $a_n \geqslant 0$. If we let $U(x) = \sum_{k = 0}^n a_k$ for $x \in [n, n+1)$, then
$$ \hat{U}(\lambda) = \sum_{n = 0}^\infty a_n e^{-\lambda n} = A(e^{-\lambda}) . $$
Thus the above theorem reads as follows:
Corollary: With the above notation, the following are equivalent:
(i) $\sum_{k = 0}^{\lfloor \cdot \rfloor} a_k \in OR$;
(ii) $A(e^{-1/\cdot}) \in OR$;
(iii) $A(e^{-1/t}) \asymp \sum_{k = 0}^{\lfloor t \rfloor} a_k$ ($t \to \infty$).
Clearly, if $$\sum_{k = 0}^n a_k \asymp n^\beta,$$ then (i) holds, and by (iii) we have $$A(e^{-1/t}) \asymp t^\beta.$$ On the other hand, if $$A(s) \asymp (1 - s)^{-\beta}$$ as $s \to 1^-$, then (ii) is satisfied, and by (iii) we have $$\sum_{k = 0}^n a_k \asymp (1 - e^{-1/n})^{-\beta} \asymp n^\beta.$$
Well thank you very much. I was not aware of this result and I should have looked more carefully at Bingham–Goldie-Teugels book first.
You are welcome. The Regular variation book is very useful, but I always have problems finding the result I need there.
Are you aware of any result in the regular variation book for sequences $a_n$ such that $U$ is not in OR ? I would be interested in particular in results for $a_n$ of the form $a_n=\mathrm{exp}(-n^c)$ with $0<c<1$.
Do you really mean $\exp(-n^c)$, and not $\exp(n^c)$? When $a_n$ is summable, then $\sum_{k = 1}^n a_k$ just converges to a constant as $n \to \infty$, and the theorem says nothing interesting. On the other hand, $a_n$ is decreasing, and so the generating function of $a_n$ should indeed be related to the decay rate of $a_n$. But I do not remember a result of that kind in BGT, and unfortunately I do not have enough time now to look it up.
Hello, thank you very much for your answer. No I really meant $\mathrm{exp}(-n^c)$, although $a_n=\mathrm{exp}(n^c)$ is also interesting. But in the convergent case, it is possible to replace the partial sum by the remainder, isn't it ? I mean, setting $U(x)=\sum_{n>x}a_n$, one still has $\hat{U}(\lambda)=A(\mathrm{e}^{-\lambda})$, or am I mistaken ?
Sorry, I meant $\hat{U}(\lambda)=A(1)-A(\mathrm{e}^{-\lambda})$. And so the asymptotics of the remainder are related to how $A((1-s))$ converges to $A(1)$ as $s$ tends to 0.
This is too long for a comment but I did not check all the details. I think that the solution can be found following the arguments in the book of Titchmarsh "The theory of functions" pp. 224 and following. First of all, the series expansion of $(1-x)^{-\beta}$ has positive coefficients $b_n \approx n^{\beta-1}$. Therefore, if $a_n \approx n^{\beta-1}$, then $F(x) \approx (1-x)^{-\beta}$. More generally, the same holds if $s_n:=\sum_{k=0}^n a_k \approx n^\beta$, using $F(x)=(1-x)\sum_{n=1}^\infty s_n x^{n-1}$. The converse, that is going from $F$ to $(s_n)$, should follow from the preceeding consideration by adapting the explanations in 7.52, where the author shows how to obatin weaker forms of Karamata's theorem with $\beta=1$, using more elementary tools .
Nice answer. Thank you very much.
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2025-03-21T14:48:29.965894
| 2020-03-01T22:40:24 |
353944
|
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|
Stack Exchange
|
Strong chains of uncountable functions and cardinal characteristics
A family of functions $\langle f_\alpha:\alpha<\kappa\rangle$ from $\omega_1$ to $\omega_1$ is called a strong chain if $\alpha<\beta<\kappa\Longrightarrow \{\xi<\omega_1: f_\beta(\xi)\leq f_\alpha(\xi)\}$ is finite. It is a result of Koszmider [1] that the existence of a strong chain of length $\omega_2$ is consistent. (Another presentation of this result is in [2], using a technique due to Neeman.)
The existence of a strong chain of length $\kappa$ implies that $\kappa\leq\frak{c}$ (as the restrictions of the functions to the first $\omega$ places are distinct), so in particular, if there is a strong chain of length $\omega_2$ then the Continuum Hypothesis fails.
Does the existence of a strong chain of length $\omega_2$ have any
influence on the configuration of standard cardinal characteristics of
the continuum?
[1] Koszmider, Piotr, On strong chains of uncountable functions, Isr. J. Math. 118, 289-315 (2000). ZBL0961.03039.
[2] Veličković, Boban; Venturi, Giorgio, Proper forcing remastered, Cummings, James (ed.) et al., Appalachian set theory 2006–2012. Based on the Appalachian set theory workshop series during the period 2006–2012. Cambridge: Cambridge University Press (ISBN 978-1-107-60850-4/pbk). London Mathematical Society Lecture Note Series 406, 331-362 (2013). ZBL1367.03094.
so Chang's Conjecture keeps this $\kappa$ small ($\aleph_1$) and CC is c.c.c indestructible so basically this $\kappa$ cannot bound any cardinal invariant. For the other direction, is it known if it is possible to have $\kappa\geq \omega_3$?
I don't the answer to the $\omega_3$ question. What I'm really curious about is if the existence of the strong $\omega_2$-chain implies that some cardinal characteristics must be greater than $\omega_1$ (we know that $\mathfrak{c}>\omega_1$, for example).
It seems that the forcing to add such an $\omega_2$ strong chain is strongly proper, so at least we can exclude invariants like $\mathfrak{b}$ (and many others) since no strongly proper forcing can add a dominating real (any real lives in a Cohen sub-extension but Cohen forcing can't add a dominating real).
Yes. And once you have a long strong chain, it will still exist in further ccc extensions, so you can achieve further configurations, say by adding a small dominating family.
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2025-03-21T14:48:29.966073
| 2020-03-02T01:19:51 |
353949
|
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Stack Exchange
|
Regular growth of ranks in Iwasawa tower
$\newcommand{\rank}{\operatorname{rank}}$Let $G=H \times K$ be a torsion free pro-$p$, $p$-adic Lie group. Let $H =\mathbb{Z}_p$, the ring of $p$-adic integers and $K$ is a non-commutative torsion free pro-$p$, $p$-adic Lie group. Let $H_n=p^n\mathbb{Z}_p$.
Let $M$ be a finitely general module over the Iwasawa Algebra $\Lambda(G)=\mathbb{Z}_p[[G]]$. Let $M_H=M/I(H)$, the co-invariance of $M$ with respect to the augmentation ideal of $I(H)$ of $H$.
Then I want to understand whether the following two statements are equivalent.
1) $M_H$ is a $\Lambda(K)$-module of rank $r$.
2) $M_{H_n}$ is a $\Lambda(K)$-module of rank $p^nr$ for all sufficiently large $n$.
Note that $M$ is a $\Lambda(G)$ module implies that $M$ is also a $\Lambda(G_n)$ module where $G_n: =H_n \times K$ and hence $G_n$ is of finite index in $G$. Therefore, $M_{H_n}$ is also a $\Lambda(K)$-module which is in (2) above.
If (1) and (2) above are not equivalent, which one implies the other? Any help will be greatly appreciated.
I know the following statement:
$\rank_{\Lambda(G)}M = [G:G_0]\rank_{\Lambda(G_0)}M$ if $G_0$ is an open normal subgroup of $G$ but this does not seem to help as, in (1) and (2), I am interested in the ranks over the same Iwasawa Algebra but I am taking the coinvariances differently.
|
2025-03-21T14:48:29.966188
| 2020-03-02T01:23:09 |
353950
|
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Stack Exchange
|
Aleksandrov's proof of the second order differentiability of convex functions
Aleksandrov [A], proved a remarkable property of convex functions.
Theorem. If $f:\mathbb{R}^n\to\mathbb{R}$ is convex, then for almost every $x\in\mathbb{R}^n$ there is $Df(x)\in\mathbb{R}^n$ and a symmetric $(n\times n)$ matrix $D^2f(x)$ such that
$$
\lim_{y\to x}
\frac{|f(y)-f(x)-Df(x)(y-x)-\frac{1}{2}(y-x)^TD^2f(x)(y-x)|}{|y-x|^2}=0.
$$
I know two proofs of this result. One based on the theory of maximal monotone functions and one based on the fact that the second order distributional derivatives of a convex function are Radon measure. Both proofs are mentioned in
Second order differentiability of convex functions. Since these proofs use relatively modern techniques not available during Aleksandrov's time, his argument must have been very different.
Question 1. Can you briefly explain what was the idea of the original proof due to Aleksandrov?
My guess would be that his proof was based on methods of differnetial geometry. What else could he use in those days?
Question 2. In there any textbook where I can find the original proof due to Aleksandrov?
[A] A. D. Alexandroff,
Almost everywhere existence of the second differential of a convex function and some properties of convex surfaces connected with it. (Russian)
Leningrad State Univ. Annals [Uchenye Zapiski] Math. Ser. 6, (1939), 3–35.
Sorry to bother. Do you have an electronic version of the reference [A]? I can't find it online. Thank you very much!
@foxell Unfortunately, I don't know where to find this paper.
@PiotrHajlasz Never mind--there is no English version online. https://mathoverflow.net/questions/353950/aleksandrovs-proof-of-the-second-order-differentiability-of-convex-functions?noredirect=1#comment1139339_353950
The paper On the second differentiability of convex surfaces by Bianchi, Colesanti, and Pucci (Geometriae Dedicata volume 60, pages 39–48 (1996)) concerns the proof of the Busemann-Feller-Alexandroff Theorem on the second order differentiability of convex functions. Its introduction gives brief synopses on several different methods of proof, including the original argument of Busemann-Feller and later Alexandroff, the two methods you mentioned in the other question (the monotone operator method of Rockafeller (using a result of Mignot); and the measure/distribution method of Reshetnyak), as well as a different one by Bangert (using almost purely differential geometric methods).
The paper gives also a new proof of the theorem, which is claimed to be in the same spirit of the original arguments of Busemann-Feller and Alexandroff. The authors considered the second order difference quotient of the convex function based at a point $x$, which they show has a limit a.e. as a convex function. This new convex function is related to the argument of Busemann-Feller in that the indicatrices constructed by Busemann-Feller are the 1-level-sets of this limited convex function.
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2025-03-21T14:48:29.966407
| 2020-03-02T01:56:02 |
353952
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{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Harry Gindi",
"https://mathoverflow.net/users/1353"
],
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"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:626776",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353952"
}
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Stack Exchange
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Which free strict $\omega$-categories are also free as weak $(\infty,\infty)$-categories?
There are a number of formalisms available for presenting free strict $\omega$-categories -- Street's parity complexes, Steiner's directed complexes, computads, polygraphs,... Typically one has a certain category $\mathcal C$ of combinatorial data, a straightforward notion of "map" from $C \in \mathcal C$ to a strict $\omega$-category $X$, and a free functor $F: \mathcal C \to \omega Cat$ with an explicit combinatorial description of $F$. The free functor will have the universal property that strict $\omega$-functors $F(C) \to X$ are in bijection with "maps" $C \to X$ whenever $X$ is a strict $\omega$-category.
Depending on one's choice of model, there may still be a clear notion of "map" from $C \in \mathcal C$ to a weak $(\infty,\infty)$-category $Y$. I'm interested to know conditions on $C \in \mathcal C$ guaranteeing that "maps" $C \to Y$ are in bijection with morphisms $i(F(C)) \to Y$, where $i: \omega Cat \to (\infty,\infty)Cat$ is the inclusion from strict $\omega$-categories to weak $(\infty,\infty)$-categories. Of course, this may be model-dependent. I find myself needing a result of this form in a particular setting, but I'd be interested seeing results of this kind for any choice of $\mathcal C$ and any model of weak $(\infty,\infty)$-categories -- I'd be particularly happy if the model of weak $(\infty,\infty)$-categories is nonalgebraic in nature.
Question: What is an example of
a category $\mathcal C$ of "presentations of certain strict $\omega$-categories",
a free functor $F: \mathcal C \to \omega Cat$ from $\mathcal C$ to strict $\omega$-categories with an explicit combinatorial description,
a 1-category $(\infty,\infty)Cat$ which "models" the $\infty$-category of weak $(\infty,\infty)$-categories (e.g. via a model structure or whatever) with inclusion functor $i: \omega Cat \to (\infty,\infty)Cat$,
a straightforward notion of "map" from objects $C \in \mathcal C$ to objects of $(\infty,\infty) Cat$,
and a (not completely vacuous) condition $\Phi$ on the objects of $\mathcal C$
such that
Objects $C \in \mathcal C$ satisfying $\Phi$ have the property that $Hom(iF(C),Y)$ is naturally isomorphic to the set of maps from $C$ to $Y$, for all (suitably fibrant, perhaps) $Y \in (\infty,\infty)Cat$?
I'm happy to see quite restrictive conditions $\Phi$; in fact in my case I don't need to understand much more than Gray tensor powers of the arrow category $\bullet \to \bullet$. I suspect that something along the lines of "$F(C)$ is gaunt" or "loop-free" or something may often do the trick, but I'd be happy with something more or less restrictive.
I suppose I'd also be happy to see examples with "$n$" replacing "$\omega$" and "$(\infty,n)$" replacing "$(\infty,\infty)$".
And let me stress that I'm not looking for some kind of fancy $\infty$-categorical freeness -- when I say $Hom(i(F(C)),Y)$ above, I mean $Hom$ in whatever 1-category is being used to model $(\infty,\infty)Cat$. Although if there are results showing something fancier, that would be interesting to hear about too.
I'm also interested in an answer to this question, but I think it's wide open. I tried working it out at one point, but I didn't get anything super nice. Free strict ω-categories on globular sets do work, but this is not enough to get loop-free Steiner complexes. I suspect that the answer might be free strict ω-categories on a polygraph (which does include Steiner complexes), but things get really hairy really fast. You would have to show levelwise that each free n-category on the n-1-polygraph can be constructed by gluing in higher dimensional analogues of horns.
The reason why this is highly problematic even for Steiner's diagrams is that cells are glued into cells freely generated in a lower dimension, then you have to generate more cells, then glue again, etc., and things get really crazy when you have to glue in cells along whiskerings.
One final comment, I think you could prove (with a lot of combinatorial work) the statement for 2-polygraphs using some of the recent work by Alex Campbell on (∞,2)-categories. I think (∞,n) is probably out of reach, however.
If you don't mind, I'll talk about strinct $\infty$-categories, but weak $(\infty,n)$-category to avoid discussing the 'problem' regarding the non uniqueness of the meaning of $(\infty,\infty)$-categories mentioned Here.
Also I don't think what follows completely answer the question, as Harry said in the comment, this is a quite open problem.
So, I think, a very simple, and 'essentially' unique, way to obtain such situation is to construct model categories for $(\infty,n)$-category, which are categories of either presheaf of set or presheaf of spaces over a small category $C$ of 'diagrams'. (You can generally jump from a description as presheaf of sets to a description as presheaf of spaces, using some variant of the simplicial completion techniques as illustrated for example in the case of the category $\Theta$ Here.)
Obviously, if you have such a model then the objects of your category $C$ have all the properties you expect. Also notes that generally presheaf over $C$, or at least some presheaves over $C$, still corresponds to some kind of diagrams and so you get this sort of construction not justs for objects of $C$ but also for more general diagrams build out of objects of $C$. (basically, the cofibrant object of your model structure).
(And of course model that are not exactly presheaves can also give some partial answer to your question as long as they contain some subcategories of cofibrant objects that can be thought of as diagrams.)
While producing a rigorous argument for the converse will be difficult, it definitely feels also true: as soon as you have a rich enough class of diagrams $C$ with the sort of properties you are requiring one should be able to prove some sort of Nerve theorem (an $\infty$-categorical version of monads with arities or Nervous monads) to show that $(\infty,n)$-categories can be represented as presehaves of spaces on $C$ satisfying some segal type conditions, for which you'll be able to build a projective/injective model structure on $[C^{op},sSet]$. This model structure will also often (especially if $C$ is rich enough) have a "simplicial decompletion" on the category of presheaf of sets of $C$.
So it remains to look at examples of such model for $(\infty,n)$-categories... But here much work is left to be done (and to be completely honest this is something that I'm very interested in and actively working on) One reason for this is that people have generally tried to come up with small and simple models, while here we ask about very big models.
The first that comes to mind is obviously the category $\Theta_n$ (see for example Dimitri's paper mentioned above) which obviously fits into this picture, and Verity complicial set models which does the job for Street Orientals, with the only problem that this model hasn't been compared to others ones.
But maybe the most interesting example already worked out regarding your question is the so-called "Colossal model" constructed in Barwick and Schommer-Pries' paper On the unicity of the theory of higher categories, which does this for a large class of gaunt categories. But I'll have to read again this paper before I can say something more precise here.
Finally my own work on polygraphs and the Simpson conjecture (here and enter link description here) is basicaly an attempt to prove this result for the very large class of all "non-unital polygraphs".
Here the hope is that $(\infty,n)$-category can be represented by a model structures on presehaves of sets and or space over the category of "plex" (the representable in the category of polygraphs) probably with some "stratification" in the spirit of the complicial model. So far I have been focusing mostly on modeling $\infty$-groupoids for simplicity, but I expect the extention to $\infty$-category will not be the hardest part.
Even in the groupoid case I'm stuck for technical reason for general polygraphs, and I can only make the theory works for "regular polygraphs", but this is still a fairly large class (containing $\Theta, \Delta$ and many other things and closed under the Gray tensor products) and I've shown that regular polygraphs form a presheaf category and carries a natural model structure that models all $\infty$-groupoids. I believe extending this to a model structure on "stratified regular polygraphs" modeling weak $(\infty,n)$-categories should be possible with a bit of work (I mean by that: it is probably a lot of work, but very feasible if someone want to spend some time on it) and this would constitute a very good answer to your question.
Also notes that in everything I have discussed above the functor "$F$" of your question is not really expecte to be fully faithful. I don't know if this is a requirement you have or if you are happy staying with "polygraphic morphisms".
So actually, this problem shows up when you try to use Barwick-Schommer-Pries to show that saturated weak complicial sets satisfy their axioms.
Important progress on this question has been made by Yuki Maehara, who shows in Orientals as free $\omega$-categories that the complicial nerve of the $n$th oriental is a complicial anodyne extension of the $n$-simplex (providing a fibrant replacement thereof in Verity's model structure on complicial sets). In other words, at least in the complicial set model, the orientals -- which a priori have a strict universal property -- have the expected weak universal property too.
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