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2025-03-21T14:48:29.967126
| 2020-03-02T02:09:48 |
353953
|
{
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|
Stack Exchange
|
non-singular divisors of the jacobian variety
Let $X$ be a smooth, projective curve of genus at least $4$. The well-known divisor $\theta$ of the associated Jacobian variety is $\mathrm{Jac}(X)$ is singular and also ample. The $\theta$ divisor also has a very explicit desciption in terms of the Abel-Jacobi map. Of course, as $\theta$ is ample, a multiple of it is linearly equivalent to something smooth. I am looking for a non-singular divisor of $\mathrm{Jac}(X)$ with an "explicit" desciption. My main motivation is to find a non-singular divisor, so that I can compute the Hodge numbers of the divisor. I keep the question a little vague to encourage different answers. Any reference/hints will be most welcome.
For a complex curve, do you know how to construct immersion or embedding into $\Bbb{P}^n$ given by homogeneous polynomials in theta functions of the lattice ? (I don't) The zero of one of them will be ample or very ample
A general element of $|2\theta|$ is smooth, and it is fairly easy to compute its Hodge numbers via the standard exact sequence. I don't understand what else you might want.
@abx Could you suggest some reference or hints on the exact sequence you mention.
For $D$ a smooth divisor in a variety $X$, use the exact sequence $0\rightarrow \mathscr{O}D(-D)\rightarrow \Omega ^1{X|D}\rightarrow \Omega ^1_D\rightarrow 0$.
@abx I am not an expert on abelian varieties. Could you please suggest some reference for the proof that $2\theta$ is very ample (or a general member of the associated linear system is smooth). I would like to understand the general member of the linear system, but I cannot find a good reference.
$|2\theta|$ is not very ample, but base-point free. You'll find this in any book on abelian varieties, e.g. Mumford or Birkenhake-Lange.
|
2025-03-21T14:48:29.967278
| 2020-03-02T05:26:39 |
353959
|
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|
Stack Exchange
|
Show the coordinate distribution has a very large sub-gaussian norm
Consider a random vector X with the coordinate distribution is uniformly distributed in the set $\{\sqrt{n}e_i : i = 1,..., n\}$, where $e_i$ denotes the n-element set of the canonical basis vectors in $R^n$. Show that $ \parallel X \parallel_{ \psi 2}\asymp \sqrt{\frac{n}{{ log n}}}.$
By the definition of the sub-gaussian norm of a random vector, $\parallel X\parallel_{ \psi_2}=\sup_{x∈S^{n−1}}\parallel <X, x>\parallel_{\psi 2} $, for all $x ∈ R^n$.
I tried to consider $\sup_{x∈S^{n−1}}\parallel <X, x>\parallel_{\psi 2}=\sup_{x∈S^{n−1}} \parallel \sum_{i=1}^n x_iX_i\parallel_{\psi_2}$, while I am not sure how to precede next? Thank you!
This is a result I read from section 3.4.2 of book "high dimensional probability" by Vershynin. I am trying to prove this argument.
The subgaussian norm of a real-valued random variable $Y$ is
$$\|Y\|:=\|Y\|_{\psi_2}:=\inf\{t>0\colon Ee^{Y^2/t^2}\le2\}.$$
If $Y$ is such that $P(Y=0)<1$ and $Ee^{Y^2/t^2}<\infty$ for all real $t>0$, then
$Ee^{Y^2/t^2}$ continuously decreases in real $t>0$ from $\infty$ to $0$, so that $\|Y\|$ is the unique positive root of the equation
$$Ee^{Y^2/\|Y\|^2}=2.$$
So, for $x=(x_1,\dots,x_n)\in S^{n-1}$ and $\nu_x:=\|X\cdot x\|$, where $X\cdot x:=\langle X,x\rangle$, we have
$$2=E\exp\frac{(X\cdot x)^2}{\|X\cdot x\|^2}
=\frac1n\,\sum_{i=1}^n\exp\frac{nx_i^2}{\nu_x^2}
\le\frac1n\,\Big(n-1+\exp\frac n{\nu_x^2}\Big), \tag{1}$$
because $\sum_{i=1}^n\exp\frac{nu_i}{\nu_x^2}$ is convex in $u=(u_1,\dots,u_n)$ in the simplex that is the image of $S^{n-1}$ under the map $(x_1,\dots,x_n)\mapsto(x_1^2,\dots,x_n^2)$, and hence $\sum_{i=1}^n\exp\frac{nu_i}{\nu_x^2}$ attains its maximum on this simplex at one of the vertices of this simplex, which latter are the standard basis vectors $e_1,\dots,e_n$ in $\mathbb R^n$.
It follows from (1) that
$$\nu_x\le\sqrt{\frac n{\ln(n+1)}}$$
for all $x\in S^{n-1}$, and this upper bound on $\nu_x$ is attained if $x$ is one of the $e_i$'s. Thus,
$$\|X\|=\max_{x\in S^{n-1}}\nu_x=\sqrt{\frac n{\ln(n+1)}}\sim \sqrt{\frac n{\ln n}}$$
as $n\to\infty$, as desired.
Thank you so much!
|
2025-03-21T14:48:29.967435
| 2020-03-02T06:13:07 |
353960
|
{
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"authors": [
"Dan Turetsky",
"Noah Schweber",
"SSequence",
"Wojowu",
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"url": "https://mathoverflow.net/questions/353960"
}
|
Stack Exchange
|
How far does this restricted definition on $\mathcal{O}$ goes?
$\mathcal{O}$ notation describes an onto function $f:\mathcal{O} \rightarrow \omega_{CK}$. In calculating all values $n \in \mathbb{N}$ such that $f(n)=\alpha$, when $\alpha$ is a limit, all indexes $e$ of ordinary programs are considered such that $\phi_e(i)=n_i$ (with $i \in \mathbb{N}$). The values $n_i$ must satisfy the condition that $f(n_i)=\alpha_i$ with the $\alpha_i$'s forming a (fundamental) sequence for $\alpha$.
I am just interested in the variation where we don't consider indexes of normal programs but instead consider indexes of primitive recursive functions/programs (given a suitable indexing for them).
(Q1) How far would such a variation go? That is, what is the smallest ordinal it can't represent.
(Q2) Consider the notation system (described by a 1-1 function $g:\beta \rightarrow \mathbb{N}$) assigning a unique number to each ordinal in which on limit values $\alpha<\beta$ we seek the p.r. function with smallest index $e$ satisfying $\phi_e(i)=g(\alpha_i)$ where the $\alpha_i$'s must form a (fundamental) sequence for $\alpha$.
At what value $\beta$ would such a system stop?
I don't have proficiency with the underlying theory so if the question seems posed in a strange manner then that might be the reason.
If I recall correctly, every computable ordinal is an order type of a primitive recursive well-order. So the primitive recursive variant also goes all the way up to $\omega_1^{CK}$.
Indeed, since we can always kill time by enumerating successors while we wait for a given recursive function to converge.
@DanTuretsky I have a small additional question. Would the system such as one in (Q2) also go to $\omega_{CK}$ [both for p.r. and ordinary programs]?
@Wojowu In fact, we can bring this down to polynomial-time computability (or indeed even less!) - $\omega_1^{CK}$ is unspeakably robust.
For Q2, the answer is $\omega^2$, for both recursive and primitive recursive notations. It's not hard to see that every ordinal below $\omega^2$ can be reached.
To show that $\omega^2$ cannot be reached, the argument is the same for both primitive recursive and full recursive. For each $a$, we construct a $b$ ensuring that $a$ is not a notation of this sort for $\omega^2$. We will construct $b$ knowing its own index. For recursive notations, this is an application of the recursion theorem. For p.r. notations, even though we don't have the recursion theorem, this is still possible so long as we're willing to specify an a priori p.r. bound on the runtime. Since we intend to kill time by enumerating successors, an exponential bound will be fine.
Our first goal is to arrange that $b$ codes $\gamma+\omega$, where $\gamma$ is greatest such that some $c < b$ with $c <_\mathcal{O} a$ codes $\gamma$. So we unwrap the notation $a$ while enumerating successors. As soon as we see some $c < b$ with $c <_{\mathcal O} a$ and $|c|$ larger than any of the others yet seen, a c.e. event, we include $c$ in the sequence we're enumerating.
As soon as we see $b <_\mathcal{O} a$, we do something to ensure that $b$ is not a valid notation, e.g. break the monotonicity of our sequence.
If $a$ is a notation for $\omega^2$ with $b \not <_\mathcal{O} a$, then $\gamma+\omega < |a|$, and $b$ is the least notation for $\gamma+\omega$ (for any smaller notation would be included in the definition of $\gamma$, contradicting our choice of $\gamma$). So if $a$ is a notation of the desired sort for $\omega^2$, then $b <_\mathcal{O} a$. But then $b$ is not a notation, by construction, so $a$ is not a notation.
Very nice answer! I have to admit though that I am finding it quite difficult to understand the details in your answer due to my lack of understanding. But anyway, the answer to question is still quite useful/informative regardless. [and sry if you have flood of notifications ..... the 5 min. edit time-limit is difficult for me to manage]
|
2025-03-21T14:48:29.967724
| 2020-03-02T08:10:03 |
353962
|
{
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"Brendan McKay",
"Gerry Myerson",
"https://mathoverflow.net/users/3684",
"https://mathoverflow.net/users/9025"
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"url": "https://mathoverflow.net/questions/353962"
}
|
Stack Exchange
|
Unique maximum degree in tournament
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 $x$.) Let $p(n)$ denote the probability that there is only one vertex with the maximum out-degree. What is $\lim_{n\rightarrow\infty}p(n)$?
I would be surprised if this statement hasn't been written and proved anywhere, but I can't find it. Does anyone know of a reference, or a result that implies it?
The number of tournaments with a unique winner is tabulated at http://oeis.org/A013976 – the total number of tournaments on $n$ vertices is $2^{n(n-1)/2}$, so you can do some calculations to see whether it looks like the probability is going to one.
I get $p(16)$ to be about $0.64$. Why are you sure $p(n)\to1$?
Not only the maximum outdegree but quite a few of the largest outdegrees are unique. This is known and I'm looking for a place it is written down. It can be proved by fairly elementary means.
Hmmm, not in the places I looked. The proof is almost the same as proving that the maximum degree vertex of a random undirected graph is unique, in Bollobas' book "Random Graphs". Convergence is slow: for tournaments it passes 90% at around 900 vertices.
I can't find a published proof of this known result, but here is a close miss.
In this paper, page 256, is a short proof that a random undirected graph has a unique vertex of maximum degree almost surely. If you replace "graph" by "tournament" and "degree" by "out-degree", the exact same proof works for tournaments.
The reason this works is that the degree of a given vertex in a random graph, or out-degree in a random tournament, have the same binomial distribution Bin$(n-1,\frac12)$. Moreover, the degrees/out-degrees of different vertices are almost independent.
|
2025-03-21T14:48:29.967879
| 2020-03-02T09:50:41 |
353965
|
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|
Stack Exchange
|
Is this ordering on the set of all covers of $\omega$ a (complete) lattice?
Let ${\frak C} \subseteq {\cal P}({\cal P}(\omega))$ be the collection of all covers of $\omega$ (that is, ${\cal C} \in {\frak C}$ iff $\bigcup {\cal C} = \omega$.)
We define the following binary relation on ${\frak C} = $: For ${\cal A}, {\cal B} \in {\frak R}$ we say ${\cal A} \leq_\text{r} {\cal B}$ if ${\cal A}$ is a refinement of ${\cal B}$, that is for all $A\in {\cal A}$ there is $B\in {\cal B}$ with $A\subseteq B$. The elements $\{\omega\}, \{\omega, \{0\}\}\in{\frak C}$ show that the relation $\leq_\text{r}$ is not anti-symmetric, so we set ${\cal A}\simeq_\text{r} {\cal B}$ if ${\cal A} \leq_\text{r}{\cal B}$ and ${\cal B} \leq_\text{r}{\cal A}$. So we get a poset $$({\frak C}/\simeq_\text{r},\leq_\text{r}),$$ where $\leq_\text{r}$ applies to equivalence classes in the usual way.
Question. Is the poset $({\frak C}/\simeq_\text{r},\leq_\text{r})$ a (complete) lattice?
Yes. The l.u.b. of $\mathcal A$ and $\mathcal B$ is $\mathcal A \cup \mathcal B$. The g.l.b. of $\mathcal A$ and $\mathcal B$ is $\{A\cap B: A\in\mathcal A , B\in\mathcal B\}$.
We can even generalize by replacing $\subseteq$ by an arbitrary meet-semilattice preordering, in which case
The l.u.b. of $\mathcal A$ and $\mathcal B$ is still $\mathcal A \cup \mathcal B$.
The g.l.b. of $\mathcal A$ and $\mathcal B$ is now the equivalence closure of $\{A\wedge B: A\in\mathcal A , B\in\mathcal B\}$ where $A\wedge B$ is a representative of the meet of $A$ and $B$.
One example of this is to replace $\subseteq$ by Turing reducibility. Then $A\wedge B$ can be a set encoding both the information in $A$ and $B$. The resulting structure is then called the Muchnik degrees (if we ignore the covering $\omega$ condition I guess).
A cover of $\omega$ is a subset $C\subseteq {\mathcal P}(\omega)$ satisfying $\bigcup C=\omega$. For any such $C$, let $\widehat{C}$ be the order ideal (=lower set) that it generates. $A\leq_r B$ iff $\widehat{A}\subseteq \widehat{B}$. Hence the map $C\mapsto \widehat{C}$ is an order preserving and reflecting map of the poset $\frak{C}$ to the lattice of order ideals of ${\mathcal P}(\omega)$. The kernel of the map is the relation $\simeq_r$.
There is a least element in the image of this map, which is the order ideal of ${\mathcal P}(\omega)$ generated by the singletons. Also, the image is closed upward in the lattice of order ideals of ${\mathcal P}(\omega)$. This shows that $\frak{C}$ is isomorphic to a principal filter of the lattice of order ideals of ${\mathcal P}(\omega)$.
The advantage from passing from covers to the order ideals they generate is that the order becomes containment and the lattice operations become union and intersection. This makes it clear that the resulting lattice is a complete (in fact algebraic) distributive lattice.
|
2025-03-21T14:48:29.968092
| 2020-02-23T23:45:07 |
353414
|
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|
Stack Exchange
|
Is $\mathbb{C}^n$ rigid?
Let $\pi:X\to S$ be a smooth family of complex manifolds (in the sense of deformation theory) such that $\pi^{-1}(0)\cong\mathbb{C}^n$ and $S\subset \mathbb{C}$ is a neighborhood of $0$. Is $\pi$ trivial? That is, is $X\cong \mathbb{C}^n\times S$ possibly after shrinking $S$?
I know that a smooth family of compact complex manifolds with $H^1(M,\mathcal{T}_M)=0$ is trivial (where $M=\pi^{-1}(0)$) but I'm not sure whether this extends to this non-compact situation.
Example. $\pi: \{(z,w)\in {\mathbb C}^2: |zw|<1\}\to {\mathbb C}$, $\pi(z,w)=z$.
Edit. Similarly, to get a nontrivial deformation of ${\mathbb C}^n$, consider
$$
X=\{(z_0, z_1,...,z_n)\in {\mathbb C}^{n+1}: |z_0 z_1|<1\}
$$
and let $\pi$ be the projection of $X$ to ${\mathbb C}$ which the 1-st coordinate line in ${\mathbb C}^{n+1}$. Then $\pi^{-1}(t)$ (for $t\ne 0$) will be biholomorphic to $B \times {\mathbb C}^{n-1}$, where $B\subset {\mathbb C}$ is an open disk. Hence, these fibers are not biholomorphic to $\pi^{-1}(0)={\mathbb C}^{n}$.
A better question, I think, is if every Stein manifold of positive dimension admits a nontrivial deformation. I suspect that this is known.
|
2025-03-21T14:48:29.968196
| 2020-02-24T00:16:55 |
353415
|
{
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"Vít Tuček",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/42804",
"https://mathoverflow.net/users/6818",
"user42804",
"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
A conjecture about the barycenter of a polytope
Could someone help me with the following conjecture? Thanks a lot!
Suppose I have a polytope $\Delta$ in $\mathbb R^n (n\geq 2)$ with coordinates $(x_1,x_2,\cdots,x_n)$ defined by linear inequalities as follows.
\begin{equation}
\begin{split}
\Delta=\{(x_1,\cdots,&x_n)|0\leq x_1\leq n+1,\,0\leq x_2\leq 2(n-1)+1,\,\cdots,\\
&0\leq x_i\leq i(n+1-i)+1,\,\cdots,0\leq x_n\leq n+1,\\
&f_1\geq 0,\,f_2\geq 0,\,\cdots,f_n\geq 0\},
\end{split}
\end{equation}
where, under the convention $x_0=x_{n+1}=0$,
\begin{equation}
\begin{split}
&f_1(x_1,\cdots,x_n)=2x_1-x_2-x_0=2x_1-x_2,\\
&f_2(x_1,\cdots,x_n)=2x_2-x_3-x_1,\\
&\cdots\\
&f_j(x_1,\cdots,x_n)=2x_j-x_{j+1}-x_{j-1},\\
&\cdots\\
&f_{n-1}(x_1,\cdots,x_n)=2x_{n-1}-x_n-x_{n-2},\\
&f_n(x_1,\cdots,x_n)=2x_n-x_{n+1}-x_{n-1}=2x_n-x_{n-1}.\\
\end{split}
\end{equation}
Let $\rho(x_1,\cdots,x_n)$ be the density function over $\Delta$ defined as follows,
$$\rho(x_1,\cdots,x_n):=\prod_{1\leq i<j\leq n+1}((x_i-x_{i-1})-(x_j-x_{j-1}))^2.$$
Here we also use the the convention that $x_0=x_{n+1}=0$.
The barycenter $\overline X=(\overline x_1,\cdots,\overline x_n)$ of $\Delta$ with respect to $\rho$ is given by the following formula
$$\overline x_i:=\frac{\int_{\Delta}x_i\cdot\rho(x_1,\cdots,x_n)\, dx_1dx_2\cdots dx_n}{\int_{\Delta}\rho(x_1,\cdots,x_n)\, dx_1dx_2\cdots dx_n}$$
where $i=1,\cdots,n$.
Then I would like to know if the following conjecture is true. (I have check the case $n=2,3,4$ by mathematica but do not have any clue how to prove it or disprove it.)
$${\bf Conjecture}:\overline x_i>i(n+1-i)\,\,{\rm for\,}i=1,\cdots,n.$$
How does this relate to Lie algebras?
The density function is the square of the product of the inner products with all the positive roots.
@user42804 Positive roots (assuming we are in the $A_{n-1}$ case are $\epsilon_i - \epsilon_j$ but your product involves also coordinates $i-1$ and $j-1.$ Anyway, tag root-systems is more appropriate unless you can present the connection to Lie algebras.
|
2025-03-21T14:48:29.968341
| 2020-02-24T01:38:46 |
353421
|
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"Matthew Daws",
"NewB",
"Yemon Choi",
"https://mathoverflow.net/users/145729",
"https://mathoverflow.net/users/406",
"https://mathoverflow.net/users/763"
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|
Stack Exchange
|
Sequence of Hilbert Schmidt operators
Consider the Banach space $\mathcal K=S_2(H)$ of Hilbert Schmidt operators on a Hilbert space $H$. I am looking for an example of two pairs of sequences $\{T^{(i)}\},\{\tilde T^{(j)}\}$ and $\{S^{(i)}\},\{\tilde S^{(j)}\}$ in the unit ball of $\mathcal K$ and an anti-linear operator $\phi:\mathcal K\to \mathcal K$ such that the both iterated limits exists but $$\lim_i\lim_j\sum_{r,s}T^{(i)}_{rs}\tilde T^{(j)}_{rs}\overline{\phi(S^{(i)}\star \tilde S^{(j)})_{rs}}\neq \lim_j\lim_i\sum_{r,s}T^{(i)}_{rs}\tilde T^{(j)}_{rs}\overline{\phi(S^{(i)}\star \tilde S^{(j)})_{rs}}$$
Where $T_{rs}$ denote the $r\times s$ entry in the matrix of $T$ and "$\star$" denotes the Schur product of operators(entrywise product of matrices).
(Or otherwise, prove that these limits are always equal irrespective of the choice of sequences in unit ball and $\phi$).
I guess $H$ comes with a fixed orthonormal basis, which allows us to think of elements of $\mathcal K$ as matrices. I think the "anti-linear operator" a bit unnatural and unmotivated. Given $\phi$ cannot I define $\psi:\mathcal K\rightarrow\mathcal K$ by $\psi(T){rs} = \overline{\phi(T){rs}}$. Then $\psi$ is bounded exactly when $\phi$ is bounded (I guess $\phi$ is bounded?) and is linear. Why couldn't we work with $\psi$ to start with?
Actually, I am dealing with the bilinear forms of type $m:\mathcal K\times\mathcal K\to \mathbb C$ such that $m(S,T)=\left<T,\phi(S)\right>=\sum_{r,s}T_{rs}\overline{\phi(S)_{rs}}$, hence I need it to be anti-linear bounded map.
But I guess what you said is true. A linear and bounded operator would suffice for a counterexample.
Is this the paper you refer to when you say (in the bounty) 'Need this for my paper"? https://arxiv.org/abs/2001.00830
Yes it is @yemon
The limits are always the same.
As $\mathcal K = S_2(H)$ is a Hilbert space, and as only the Schur product is involved anywhere, we can infact identify $\mathcal K$ with $\ell^2 = \ell^2(\mathbb N)$ and consider the pointwise product of vectors in $\ell^2$.
Let $x=(x_r)\in\ell^2$ and let $(y^{(i)})$ be a bounded sequence in $\ell^2$. Thus, for each $r$ the scalar sequence $(y^{(i)}_r)$ is bounded, so by moving to subsequences and using a diagonal argument, we may suppose that $y^{(i)}_r\rightarrow y_r$, say, as $i\rightarrow\infty$. Set $y=(y_r)$ and observe that for any $R$,
$$ \sum_{r=1}^R |y_r|^2 = \sum_{r=1}^R \lim_i |y^{(i)}_r|^2
= \lim_i \sum_{r=1}^R |y^{(i)}_r|^2
\leq \lim_i \|y^{(i)}\|_2^2 < \infty. $$
As $R$ was arbitrary, we conclude that $y\in\ell^2$.
Then, for $R>0$,
\begin{align*} \lim_i \|xy^{(i)} - xy\|_2^2
&= \lim_i \sum_r |x_r y^{(i)}_r - x_r y_r|^2 \\\\
&= \sum_{r=1}^R \lim_i |x_r y^{(i)}_r - x_r y_r|^2
+ \lim_i \sum_{r>R} |x_r y^{(i)}_r - x_r y_r|^2 \\\\
&= \lim_i \sum_{r>R} |x_r y^{(i)}_r - x_r y_r|^2 \\\\
&\leq \lim_i \Big(\sum_{r>R} |y^{(i)}_r - y_r|^2\Big) \Big( \sup_{r>R} |x_r| \Big).
\end{align*}
We can make this arbitrarily small by choosing $R$ large. We conclude that $xy^{(i)} \rightarrow xy$ in norm.
Given bounded sequences $(x^{(i)}), (y^{(i)})$ in $\ell^2$, let $x,y$ be the pointwise limits, as in the previous paragraph. Let $z\in\ell^2$, and consider
\begin{align*} \lim_i \sum_r x^{(i)}_r z_r y^{(i)}_r. \end{align*}
As $x^{(i)} y^{(i)}$ is in $\ell^1$ and $(z_r)$ is bounded, this sum make sense, and we can copy the argument to show that
\begin{align*} \lim_i \sum_r x^{(i)}_r z_r y^{(i)}_r
= \sum_r x_r z_r y_r
. \end{align*}
Thus, given bounded sequences $(x^{(i)}), (y^{(i)}), (a^{(i)}), (b^{(i)})$ in $ell^2$, let $x,y,a,b$ be the pointwise limits, as in the previous paragraph. Given a bounded linear map $\phi$ on $\ell^2$, we see that as $a^{(i)} b^{(j)} \rightarrow a b^{(j)}$ in norm, as $i\rightarrow\infty$, also $\phi(a^{(i)} b^{(j)}) \rightarrow \phi(a b^{(j)})$ in norm. Hence, by the same argument,
\begin{align*}
\lim_j \lim_i \sum_r x^{(i)}_r y^{(j)}_r \phi(a^{(i)} b^{(j)})_r
= \lim_j \sum_r x_r y^{(j)}_r \phi(a b^{(j)})_r
= \sum_r x_r y_r \phi(a b)_r.
\end{align*}
By symmetry, we get the same answer with the limits taken in the other order.
|
2025-03-21T14:48:29.968709
| 2020-02-24T02:45:16 |
353424
|
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|
Stack Exchange
|
Probability distribution of random products of elements of a generating set of a finite non-abelian group
Let $G$ be a finite non-abelian group, and consider a choice of $N$ distinct elements $g_{0},g_{1},\ldots,g_{N-1}\in G$ that generate $G$. Now, let $t$ be an arbitrary positive integer, and let $d_{1},\ldots,d_{K}\in\left\{ 0,\ldots,N-1\right\}$ be the $N$-ary digits of $t$, so that:$$t=\sum_{k=1}^{K}d_{k}N^{k-1}$$ is then the $N$-ary/$N$-adic representation of $t$. Next, letting $\mathbb{N}_{1}$ denote the positive integers, define the map $\chi:\mathbb{N}_{1}\rightarrow G$ by: $$\chi\left(\sum_{k=1}^{K}d_{k}N^{k-1}\right)=\prod_{k=1}^{K}g_{d_{k}}=g_{d_{1}}g_{d_{2}}\cdots g_{d_{K}}$$ Finally, make $\mathbb{N}_{1}$ into a probability space by equipping it with the probability measure defined by the upper density of a set of integers: $$\textrm{P}\left(A\right)=\limsup_{n\rightarrow\infty}\frac{\left|A\cap\left\{ 1,2,\ldots,n\right\} \right|}{n},\textrm{ }\forall A\subseteq\mathbb{N}_{1}$$
With this set-up, we can then think of $\chi$ as a random variable on $\mathbb{N}_{1}$. That being so, is it then true that $\chi$ will be uniformly distributed?—i.e.: $$\textrm{P}\left(\chi=g\right)\overset{?}{=}\frac{1}{\left|G\right|},\textrm{ }\forall g\in G$$
My intuition has convinced me that this must be the case whenever $G$ is abelian, however, I cannot help but feel that in the non-abelian case, the probability distribution of $\chi$ somehow depends on the particular choice of the generating set $\left\{ g_{0},\ldots,g_{N-1}\right\}$.
To begin with, endowing the set of integers with the upper density is quite far from making it a probability space. Nonetheless, the question you ask still makes sense. Namely, you consider the $G$-valued sequence defined by your function $\chi$ and just ask whether the empirical frequencies of its values converge to the uniform distribution.
The answer is indeed "yes". The proof consists of 3 steps.
(1) Take $n=N^d$ to be the $d$-th power of $N$. Then the empirical distribution at time $n$ is just the $d$-fold convolution of the uniform measure on your generating set, which is well-known to converge to the uniform distribution on $G$.
(2) The uniform distribution on $G$ is preserved under the multiplication by any group element, which implies that for any $N$-rational numbers $0\le \alpha<\beta\le 1$ the empirical frequencies over the sequence of intervals
$[N^d\alpha, N^d\beta]$ converge to the uniform distribution as well.
(3) It remains to deduce the claim from (2).
"To begin with, endowing the set of integers with the upper density is quite far from making it a probability space" — As I'm not quite good in probability, would you mind explaining why? Since the upper density is defined for all infinite sets of positive integers, doesn't that make the set of "measurable" sets into a sigma algebra?
The point here is $\sigma$-additivity (rather than just the presence of a $\sigma$-algebra). The "measure" you are defining is not only not $\sigma$-additive (look at the partition of the integers into one-point sets), but is not even finitely additive (it is possible that both a subset of integers and its complement simultaneously have upper density 1).
What about if we instead look at natural density (where both upper and lower density exist and are the same)?
In this case you obtain a finitely additive measure on the algebra (not $\sigma$-algebra!) of subsets which have a density. By using the Hahn-Banach theorem you can extend it to a finitely additive measure on all subsets of $\mathbb N$. However, this is not of much use as finitely additive measures are quite different from the "usual" $\sigma$-additive ones and are not really used in the probability theory.
|
2025-03-21T14:48:29.968960
| 2020-02-24T03:30:19 |
353425
|
{
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"Anton Petrunin",
"Robert Israel",
"Wlod AA",
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"url": "https://mathoverflow.net/questions/353425"
}
|
Stack Exchange
|
$\epsilon$-net under Hausdorff distance
Consider linear subspaces of $\mathbb{R}^n$. For two subspaces $X$ and $Y$, we define their Hausdorff distance as
$$
{\displaystyle d_{\mathrm {H} }(X,Y)=\max \left\{\,\sup _{x\in X, |x|_2=1}\inf _{y\in Y,|y|_2=1}d(x,y),\,\sup _{y\in Y,|y|_2=1}\inf _{x\in X, |x|_2=1}d(x,y)\,\right\}.\!}
$$
For $\epsilon>0$, the set of subspaces $X_1,\dots,X_m$ is called an $\epsilon$-net under Hausdorff distance if for any subspace $Y$, there is some $X_i$ such that
$$
d_{\mathrm {H} }(X_i,Y)<\epsilon.
$$
The questions is to provide upper and lower bound of $m$.
It is direct to observe that if $X$ and $Y$ have different dimensions, their distance is always greater than 1.
Therefore, we only need to consider $m=\sum_{k=1}^n m_k$, where $m_k$ denotes the $\epsilon$-net of fixed ($k$) dimensional subspaces.
Also, it would be interesting to consider another definition of Hausdorff distance as
$$
{\displaystyle d_{\mathrm {H} }(X,Y)=\max \left\{\,\sup _{x\in X, |x|_2\leq 1}\inf _{y\in Y,|y|_2\leq 1}d(x,y),\,\sup _{y\in Y,|y|_2\leq1}\inf _{x\in X, |x|_2\leq 1}d(x,y)\,\right\}.\!}
$$
Would the difference between the distance induces a significant difference between the bound of the $\epsilon$-net?
I guess you assume that $X$ and $Y$ are bounded (and maybe closed and nonempty). Still the space of such sets is not compact, so it has no nets. If you restrict to subsets in a bounded set $B$, with an $\varepsilon$-net $N$, then the set of all nonempty subsets in $N$ is an $\varepsilon$-net in the space of subsets in $B$. So, if $n=|N|$, then $m=2^n-1$.
Perhaps it'd be simpler to restrict the linear subspaces to dimension $\ >0\ $ and $\ <n.$ Furthermore, one could concentrate on the subspaces of a fixed dimension $\ k,\ $ thus looking at $\ m_1\ldots m_{n-1}.\ $ Then, it'd be rather straightforward to pass to the overall $\ m.$
@AntonPetrunin, everything seems to be fine (but for taking certain terminology liberty). Of course, we could more properly talk about the Hausdorff distances between the respective unit spheres.
By "upper bound of $m$", I think you mean "upper bound on the least $m$ for which such an $\epsilon$-net exists". Otherwise, given any $\epsilon$-net, you can always enlarge it as much as you wish by throwing in more subspaces.
|
2025-03-21T14:48:29.969132
| 2020-02-24T09:18:07 |
353428
|
{
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"Noah Schweber",
"Zuhair Al-Johar",
"https://mathoverflow.net/users/8133",
"https://mathoverflow.net/users/95347"
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"url": "https://mathoverflow.net/questions/353428"
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|
Stack Exchange
|
Can we have the well founded world of NF obeying ZF?
The following question is about the possibility of having a world of sets obeying new foundations "NF" with their well founded sets obeying rules of ZF. It uses the revised version of Quines $``ML"$ (Mathematical Logic (chapter 4)) system, in order to define well foundedness in a faithful manner, and then adds axioms of size and infinity over the well founded sector.
FORMAL EXPOSITION:
Language: First order logic with equality and membership with extra-logical axioms of:
Extensionality: $\forall z (z \in x \leftrightarrow z \in y) \to x=y$
Classes: if $\phi$ is a formula in which $x$ doesn't occur free, then $(\exists x \forall y (y \in x \leftrightarrow set(y) \land \phi))$
Where: $set(y) \iff \exists z(y \in z)$
Define: $x=V \iff \forall y \ (set(y) \to y \in x)$
Define: $x=\{y \in V| \phi\} \iff \forall y (y \in x \leftrightarrow y \in V \land \phi)$
Stratification: if $\phi(y,x_1,..,x_n)$ is a stratified formula in which all quantifiers are bounded by $V$, and all free variables of it are among symbols $``y,x_1,..,x_n"$, then:$$\forall x_1 \in V,...,\forall x_n \in V (\{y\in V| \phi\} \in V)$$
Size: $x,y\text{ are well founded} \land |x|=|y| \land x \in V \to y \in V$
Where: $\text{well founded} (x) \iff \\\not \exists d (x \cap d \neq \emptyset \land \forall m \in d \exists n \in d (n \in m))$
Infinity: $\omega \in V$
Where $\omega$ is the set of all finite Von Neumann ordinals.
Question 1: Is this theory consistent relative to consistency of $NF$ and $ZF$?
Question 2: if we weaken Extensionality to weak Extensionality of $NFU$, as to allow Ur-elements. Would that be consistent relative to $NFU$ and $ZFU$?
$V$ in ML has well defined, it is the SET of all sets. I'll add it
@NoahSchweber. Yes! I've edited my posting to clarify this issue.
This is an interesting question - I think it might be worth allowing urelements, though, given that NF is comparatively ill-understood.
@NoahSchweber, Yes. correct. NFU is more understood than NF.
@NoahSchweber, the other direction in the definition of $V$ is trivial!
Derp, not my finest moment!
@MattF. is there a limit to the total number of times of editing a question on MathOverflow.
|
2025-03-21T14:48:29.969311
| 2020-02-24T09:51:04 |
353429
|
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"Yoav Kallus",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353429"
}
|
Stack Exchange
|
Minimizing expected mutual distances in spherical regions
Suppose I take the unit sphere in $d$ dimensions, and I take some subset $A$ of the sphere of fixed relative volume $V$. Now from this set $A$ I draw two vectors, uniformly at random, and I look at their mutual distance (or angle). Depending on the shape of $A$, the pairwise distances obtained from this process may be very large or very small.
My intuition says that if we want to minimize the expected value of the pairwise distances, or if we want to maximize the probability of getting a distance below some threshold, then the best shape for $A$ is always a spherical cap, i.e. all vectors on the sphere with first coordinate larger than some parameter $\alpha$.
Can this be proven? Has something similar been shown before? Or can anyone provide references or appropriate keywords to search for, to find relevant literature on this topic?
You are looking for optima of $\int V(|x-y|)d\mu(x)d\mu(y)$ over (in your case, uniform) measures $\mu$. I found this kind of question is usually the domain of something called "potential theory" by practitioners. Hopefully this is a useful keyword.
I think all you need is a spherical version of Riesz rearrangement inequality (https://en.wikipedia.org/wiki/Riesz_rearrangement_inequality), like theorem 1.1 of https://arxiv.org/abs/1810.06813v1
|
2025-03-21T14:48:29.969442
| 2020-02-24T10:06:04 |
353430
|
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"B K",
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"url": "https://mathoverflow.net/questions/353430"
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|
Stack Exchange
|
Asymptotic decay rate of an oscillatory integral
Consider the following oscillatory integral
$$
I(n):=\int_{-\pi}^\pi\int_{-\pi}^\pi e^{i n(x+y)}\frac
{(1 - \cos(2x)) (1 - \cos(2y))}
{2k - (\cos x + \cos y)}\ \mathrm{d}x\,\mathrm{d}y.
$$
where $n\in\mathbb{N}$, $n>2$, $k\in\mathbb{R}$, $k>1$, and $i$ is the imaginary unit.
My question. Is it possible to characterize the asymptotic decay rate of $I(n)$ as a function of the integer parameter $n$?
The function $(1-\cos(2x))(1-\cos(2y)) / (2k - \cos x - \cos y)$, by your assumption, is a smooth function on the torus. So its Fourier transform decays faster than any polynomial, no?
@WillieWong Yes, correct. However, I would like to explicitly characterize the decay rate of $I(n)$ as a function of parameter $n$.
@WillieWong not sure I understand your point. To clarify, I would like to find an asymptotic estimate of the form $I(n)\sim f(n)$, where $f(n)$ decays exponentially. Numerics suggests that $f(n)=c^{2n}/\sqrt{n}$, where $c=k-\sqrt{k^2-1}$ but I'm looking for a formal proof.
As often with parameter-dependent integrals, Feyman's trick (taking derivatives with respect to the parameter $k$) might be useful here.
These are things that you could've included in your question to start with. Frequently when people talk about asymptotics of oscillatory integral they are talking about stationary phase issues. In your case you might as well just titled the question about asymptotic decay rates of the Fourier series.
@WillieWong Yes, you are right. I modified the title.
The asymptotics for large $n$ is
$$ J(n,\kappa) := \Big(\frac{2}{\pi}\Big)^2 \int_{-\pi}^\pi \int_{-\pi}^\pi
\exp{(i\,n(x+y))}\frac{\sin^2x\,\sin^2y} {2\kappa - (\cos{x}+\cos{y}) }\, dx \,dy \sim$$
$$ \sim \frac{8}{\sqrt{\pi \kappa n}}(\kappa^2-1)^{7/4} (\kappa - \sqrt{\kappa^2-1})^{2n}\quad, \quad (\kappa>1)$$
The proof consists of 5 parts. The first part is to write the 2-dimensional fourier transform in a 1-dimensional form, albeit with special functions as part of the integrand. In Parts 2-5 asymptotic analysis will be undertaken. The proof is already long, and I'm not going to justify every switch of $\sum$ and $\int$, nor do some of the details that a rigorous saddle point analysis requires, like proving the tails decay sufficiently fast. I have checked the result numerically, and for $\kappa$ = 2.2, I get the following errors between the true value and the asymptotic approximation: n=50, 2.46%; n=100, 1.23%, n=200, 0.61%
Part 1: Show
$$ J(n,\kappa) = \int_0^\infty \exp{(-2\kappa t)}\Big(2I_n(t) - I_{n-2}(t) - I_{n+2}(t) \Big)^2 dt $$
where $I_n(t)$ is the modified Bessel function.
With $r=1/(2\kappa)$,expand the integrand in a power series in $r$ to get
$$ J(n,\kappa) =r \sum_{m=0}^\infty r^m \Big(\frac{2}{\pi}\Big)^2 \int_{-\pi}^\pi \int_{-\pi}^\pi \big(\cos{x}+\cos{y})^m \sin^2{x} \sin^2{y} \,e^{i\,n(x+y)} dx \, dy =$$
$$ = r \sum_{m=0}^\infty r^m m! \sum_{k=0}^m b_{m-k} \, b_k \, \text{ where }b_k=\frac{2}{k!\,\pi}\int_{-\pi}^\pi \cos^k y \sin^2 y \, e^{i\,n y} dy$$
By Cauchy product of power series, where $B(u)=\sum_k u^k\,b_k,$
$$J(n,\kappa) =r \sum_{m=0}^\infty r^m m! [u^m] B(u)^2 $$
where the 'coefficient of' operator has been used. By the Borel transform we get
$$J(n,\kappa) = \int_0^\infty dt \exp{(-2\kappa t)} \sum_{m=0}^\infty t^m [u^m] B(u)^2 = \int_0^\infty \exp{(-2\kappa t)}B(t)^2 dt$$
We work on $B(t)$ now.
$$ B(t) = \frac{2}{\pi} \int_{-\pi}^\pi dy \, e^{i\,n\,y} \sin^2 y \sum_{k=0}^\infty\frac{(t\, \cos{y})^k}{k!} $$
The interior summation is $\exp{(t \cos{y} )}$ which also has an expansion involving modified Bessel functions
$$ \exp{(t \cos{y} )} = I_0(t)+2\sum_{j=1}^\infty I_j(t) \cos{j\,y} $$
We thus must consider the integral
$$\frac{2}{\pi} \int_{-\pi}^\pi dy \, e^{i n \, y} \sin^2{y} \cos{j\,y} $$
Mathematica knows that $$\frac{2}{\pi}\int_{-\pi}^\pi e^{i n \, y} \sin^2{y} \, dy = \frac{-2}{\pi} \frac{4 \sin{(\pi z)}}{z(z+2)(z-2)} $$
which, for integer $z,$ is non-zero only if $z=0$ or $\pm 2.$ The penultimate formula says we have $z=n \pm j,$ and the formula under the 'Part 1' heading follows by keeping only the non-zero terms in the infinite sum over Bessel functions.
Part 2: a useful simplification
$$2I_n(u)-I_{n-2}(u)-I_{n+2}(u)=\frac{2}{u}\Big( (n+1)I_{n+1}(u)-(n-1)I_{n-1}(u) \Big) $$
This follows from applying the recursion
$I_{n-1}(u)-I_{n+1}(u)=\frac{2n}{u}I_n(u)$ twice. The equation of Part 1 will thus consist of two BesselI's squared, and a cross term.
Part 3: Show
$$\int_0^\infty e^{-2 \kappa\,u}\,I_n(u)^2 \frac{du}{u^2}= \frac{1}{n\,\pi(4n^2-1)}
\int_0^1 \frac{x^{-n}}{\sqrt{x(1-x)}}\big(\kappa-\sqrt{\kappa^2-x}\big)^{2n}
\big(\kappa + 2n\sqrt{\kappa^2 - x} \big)\,dx$$
Use Gradshteyn 6.592.6 and 6.611.4, respectively:
$$\frac{1}{\pi} \int_0^1 \frac{dx}{\sqrt{x(1-x)}}I_{2n}(2u\sqrt{x}) = I_n(u)^2 \, , $$
$$ \int_0^\infty e^{-a\,x} I_{\nu}(b\,x)\,dx = b^{-\nu}\frac{\big(a-\sqrt{a^2-b^2}\big)^{\nu}}{\sqrt{a^2-b^2}} .$$
Then integrate twice with respect to the argument of the exponential ($a,$ to begin with).
Part 4: Asymptotics of 'cross term'
$$M_n:=\int_0^\infty e^{-2 \kappa\,u}\,I_{n-1}(u)\,I_{n+1}(u) \frac{du}{u^2}=
\int_0^\infty e^{-2 \kappa\,u}\,I_{n}(u)^2 \frac{du}{u^2} \big(1+\cal{O}(1/n)\big) $$
Use the series form for product of Bessel functions, Gradsteyn 8.442.1. Then
$$M_n=\int_0^\infty \frac{du}{u^2}e^{-2 \kappa\,u} \sum_{m=0}^\infty \frac{(u/2)^{2n+2m}}{m!} \frac{(2n+m+1)_m}{(n+m+1)!(n+m-1)!} =$$
$$=\int_0^\infty \frac{du}{u^2}e^{-2 \kappa\,u} \sum_{m=0}^\infty \frac{(u/2)^{2n+2m}}{m!} \frac{(2n+m+1)_m}{(n+m)!(n+m)!}\Big(1 - \frac{1}{n+m+1} \Big)$$
$$=\int_0^\infty \frac{du}{u^2}e^{-2 \kappa\,u} I_n(u)^2\big(1+\cal{O}(1/n)\big) $$
Part 5: Put pieces together
$$ J(n,\kappa) \sim 4\int_0^\infty \frac{du}{u^2}e^{-2 \kappa u}\Big( \big((n+1)I_{n+1}(u)\big)^2 + \big((n-1)I_{n-1}(u)\big)^2 -2(n+1)(n-1)I_{n}(u)^2\Big)$$ and using the result of Part 3,
$$\frac{\pi}{4}J(n,\kappa)\sim \int_0^1 \frac{dx\,x^{-n}}{\sqrt{x(1-x)}}\big( \kappa - \sqrt{\kappa^2-x}\big)^{2n} \cdot$$
$$\cdot \Big\{ \frac{n+1}{4(n+1)^2-1} \frac{\big( \kappa - \sqrt{\kappa^2-x}\big)^2}{x}\big(\kappa+2(n+1)\sqrt{\kappa^2-x}\big) - $$
$$ \frac{n-1}{4(n-1)^2-1} \frac{x}{\big( \kappa - \sqrt{\kappa^2-x}\big)^2}\big(\kappa+2(n-1)\sqrt{\kappa^2-x}\big) + $$
$$ \frac{2(n+1)(n-1)}{n(4n^2-1)} \big(\kappa+2n\sqrt{\kappa^2-x}\big) \Big\} $$
Expand curly brackets as $n \to \infty.$
$$ \Big\{...\Big\}=\frac{2}{x}\big(\kappa^2-x\big)^{3/2}+\kappa(\kappa^2/x-1)\frac{1}{n}+...$$ Since earlier we approximated the cross-term to order 1/n, we'll keep only the first term in this expansion. By letting $x \to u^2$ then
$$ \frac{\pi}{16} J(n,\kappa) \sim \int_0^1 \frac{du \, u}{\sqrt{(1-u)(1+u)}}\big(\kappa/{u}-\sqrt{ (\kappa/u)^2-1}\big)^{2n}\big((\kappa/u)^2-1)^{3/2} $$
Use the following transformation, good for positive functions $g,$
$$\int_0^1g(t)/\sqrt{1-t}=2\int_0^1 g(4u(1-u))\,du$$
This has the effect of putting the strong growth behavior at the endpoint of the domain as $u \to 1$ to the region near $u=1/2.$ In fact, the following integral is well-setup for a classic saddle point analysis:
$$ \frac{\pi}{128}J(n,\kappa) \sim \int_0^1 \frac{du \, u(1-u)}{\sqrt{1+4u(1-u)}} \Big(\big(\frac{\kappa}{4u(1-u)}\big)^2-1\Big)^{3/2}\cdot $$
$$\Big(\frac{\kappa}{4u(1-u)}-\sqrt{ \big(\frac{\kappa}{4u(1-u)}\big)^2-1}\Big)^{2n}$$
Define
$$h(u)=-\log\Big(\frac{\kappa}{4u(1-u)}-\sqrt{ \big(\frac{\kappa}{4u(1-u)}\big)^2-1}\,\Big) $$
We find $h'(u)=0 \implies u=1/2.$ A Taylor expansion about u=1/2 leads to
$$ h(u)=-\log\Big(\kappa-\sqrt{\kappa^2-1}\Big) +\frac{4\kappa}{\sqrt{\kappa^2-1}}(u-1/2)^2+\cal{O}((u-1/2)^4) $$
Define $g(u)$ as below and keep its first term in a Taylor expansion about u=1/2
$$ g(u) = \frac{ u(1-u)}{\sqrt{1+4u(1-u)}} \Big(\big(\frac{\kappa}{4u(1-u)}\big)^2-1\Big)^{3/2}=\frac{\big(\kappa^2-1)^{3/2}}{4\sqrt{2}}+\cal{O}((u-1/2)^2)$$
Then
$$ \frac{\pi}{128}J(n,\kappa) \sim \Big(\kappa-\sqrt{\kappa^2-1}\Big)^{2n}
\frac{\big(\kappa^2-1)^{3/2}}{4\sqrt{2}} \int_{-\infty}^\infty
\exp{\Big(\frac{-8\kappa \,n}{\sqrt{\kappa^2-1}}(u-1/2)^2 \Big)} du $$
Use of the Gaussian integral and algebra completes the proof.
The leading order exponential can be (almost) found by Paley-Wiener (which also applies to Fourier series).
Your integrand extends to a complex analytic function on the strip $$|\Im x|, |\Im y| < a = \cosh^{-1}(k) = \ln(k + \sqrt{k^2 - 1})$$
So the Payley-Wiener theorem says that (after performing the Fourier transform separately in $x$ and $y$) for such functions the Fourier series has the decay
$$ I(n) \lesssim_{\epsilon} e^{-2an + n\epsilon} $$
for any $\epsilon > 0$. (The factor of 2 coming from the fact that you are looking at the Fourier series in $x$ times the Fourier series in $y$.) Plugging in you get
$$ I(n) \lesssim_\epsilon \left(\frac{1}{k + \sqrt{k^2 - 1}} \right)^{2n} \cdot e^{n\epsilon} $$
(Note that $(k + \sqrt{k^2 - 1})^{-1} = k - \sqrt{k^2 - 1}$.)
For the precise leading order you will probably need to do some contour integrals by hand.
Thanks! One quick question: what does $\lesssim_\varepsilon$ mean?
bounded above by, up to a constant depending on $\epsilon$. (So I guess I could've written: for every $\epsilon > 0$ there exists a $C(\epsilon)$ (independent of $n$) such that $I(n) \leq C(\epsilon) e^{-2an + n\epsilon}$.) In general the constant $C(\epsilon)$ may blow-up as $\epsilon \searrow 0$; in your case since you have a very explicit integrand you may be able to show by doing the contour integral that there is a uniform bound on $C(\epsilon)$ in which case you can get rid of the $e^{n\epsilon}$ loss. I am not sure how to get the $\sqrt{n}$ factor though.
I see, thanks. So, assuming that I could get rid of the $e^{n\epsilon}$ term, what I get is the upper bound $I(n)\le (k-\sqrt{k^2-1})^{2n}$. Is it possible to show that such a bound is tight? (btw, I'm not completely sure about the $\sqrt{n}$ factor: it's just a numerical guess)
In terms of rates: yes. Paley-Wiener is if-and-only-if: is if you have a strictly faster exponential decay rate, it will contradict the fact that your integrand has poles when \cos(x) = \cos(y) = k.
Since I'm not familiar with Paley-Wiener theorem, could you suggest a good reference on this? Thanks.
http://www.math.lsa.umich.edu/~rauch/555/fouriercomplex.pdf gets to it very quickly (as an exercise with hints).
|
2025-03-21T14:48:29.970014
| 2020-02-24T10:32:44 |
353433
|
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|
Stack Exchange
|
Bound for $\left|\sum_{\substack{1\leq d\leq x\\(d,r)=1}}\frac{G_d}{d^{1+\varepsilon}}\right|$, where $G_d$ denote the Gregory coefficients
In this post we denote the Gregory coefficients, or reciprocal logarithmic numbers, this Wikipedia Gregory coefficients as $G_k$, for integers $k\geq 1$. I would like to know if it is possible to get upper bounds for next two sequences.
Question. We denote the greatest common factor of two positive integers $a\geq 1$ and $b\geq 1$ as $(a,b)$. Provide the calculations (or hints for one of these similar problems) to get $$\left|\sum_{\substack{1\leq d\leq x\\(d,r)=1}}\frac{G_d}{d^{1+\varepsilon}}\right|\leq \text{upper bound}=\text{upper bound}(\varepsilon,x,r)\tag{1}$$ or for the sequence
$$\left|\sum_{\substack{1\leq d\leq x\\(d,r)=1}}\frac{G_d}{d^{1+\varepsilon}}\log\frac{x}{d}\right|\leq \text{upper bound}=\text{upper bound}(\varepsilon,x,r)\tag{2}$$
for reals $\varepsilon>0$ and $x\geq 1$, and integers $r\geq 1$. Many thanks.
Here thus $\text{upper bound}(\varepsilon,x,r)$ denotes a suitable function of $\epsilon$, $x$ and $r$, and I'm asking it (if possible/feasible) in the spirit of Theorem 1.1 and Theorem 1.2 from [1].
As was said if the calculations are similar, provide the reasonings for one of these problems and if you want hints for the other. Any case, I'm asking about what work can be done for some of these problems.
References:
[1] Olivier Ramaré, Some elementary explicit bounds for two mollifications of the Moebius function, Funct. Approx. Comment. Math. Volume 49, Number 2 (2013), pp 229-240.
I know the reference from Project Euclid.
I hope that my Question is clear, interesting and at research level, please feel free to add your feedback in comments.
|
2025-03-21T14:48:29.970142
| 2020-02-24T10:45:56 |
353434
|
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|
Stack Exchange
|
Borderline Collatz-like problems
The usual Collatz map is $C:n \mapsto n/2$ if $n$ even, $(3n+1)/2$ if $n$ odd. Let $f^{\circ (r+1)}:=f \circ f^{\circ r}$.
We suspect that for every fixed $n>0$, the sequence $C^{\circ r}(n)$ never diverges to infinity (and more strongly, always converges to $1$), because (together with experimental checking) heuristically the application of $C$ to a random number multiplies it in geometric average by $(\frac{1}{2} \cdot \frac{3}{2})^{1/2} = \frac{\sqrt{3}}{2} < 1$.
The (borderline) Collatz-like problems:
A map $f: \mathbb{N} \to \mathbb{N}$ will be called a Collatz-like map if $$ 0 \neq \lim_{n \to \infty} \left( \prod_{r=1}^n \frac{f(r)}{r} \right)^{1/n} \le 1 \ \ \ \ \ (*) $$ If the inequality $(*)$ is an equality then the map $f$ will be called a borderline Collatz-like map.
For each (borderline) Collatz-like map $f$, we have the (borderline) Collatz-like problem asking whether its iterations diverges nowhere to infinity, i.e. $$\forall n>0, \ \exists m,r>0 \text{ with } f^{\circ (m+r)}(n) = f^{\circ m}(n).$$
If the answer is yes, then let us call $f$ an acceptable (borderline) Collatz-like map.
This post focus on a specific family of borderline Collatz-like problems:
For any given $\alpha >0$, let us consider the following map $$
f_{\alpha}: n \mapsto \left\{
\begin{array}{ll}
\left \lfloor{n\alpha} \right \rfloor & \text{ if } n \text{ even,} \\
\left \lfloor{n/\alpha} \right \rfloor & \text{ if } n \text{ odd.}
\end{array}
\right.
$$
The map $f_{\alpha}$ is borderline Collatz-like. Let $S$ be the set of $\alpha>0$ for which $f_{\alpha}$ is acceptable.
Question: What is the set $S$, explicitly?
It is obvious to find some $\alpha$ in $S$, and some other out, for example $S \cap \mathbb{Z}_{>0} = \{ 1 \}$.
Proposition: $3/2 \in S$.
Proof: If $n$ is even then $n=2^ra$ with $a$ odd and $r>0$. Then $f_{3/2}(n)=2^{r-1}3a$, $f^{\circ r}_{3/2}(n)=3^ra$ and $f^{\circ(r+1)}_{3/2}(n)=3^{r-1}2a = f^{\circ(r-1)}_{3/2}(n)$. Next, if $n$ is odd, then $n=6k+i$ with $i \in \{1,3,5\}$, and $\left \lfloor{2n/3} \right \rfloor = \left \lfloor{2(6k+i)/3} \right \rfloor = 6k + \left \lfloor{2i/3} \right \rfloor$, but $\left \lfloor{2i/3} \right \rfloor = 0,2,3$ for $i=1,3,5$. So the cases $i=1,3$ reduce to the even case, next, if $i=5$ then $\left \lfloor{2n/3} \right \rfloor = 6k+3$. $\square$
Remark: See this comment of user35593 which proves that $\alpha = 2/3$ is also in $S$.
Now what about $\alpha$ irractional? This post (and its comments) provides a family of quadratic integers not in $S$ (one of which being the golden ratio $\phi$).
Below are the pictures for $\alpha = \pi, 1/\pi$ and $\sqrt{2}$, of the plot of the map which for every fixed $n$ gives the minimal $m$ such that $f_{\alpha}^{\circ (m+r)}(n) = f_{\alpha}^{\circ m}(n)$ for some $r$.
Subquestion 1: Is it true that $\{\pi, 1/\pi,\sqrt{2} \} \subset S$?
Then, we could expect that $\alpha = 1/\sqrt{2}$ is also in $S$, but in fact it seems not! The following picture shows the value of $\log_{10}(f_{\alpha}^{\circ r}(n))$ for $\alpha = 2^{-1/2}$, $r=200$ and $n<20000$.
Subquestion 2: Is it true that $2^{-1/2} \not \in S$?
[for a focus on this specific problem, see this post]
With $n=15$ we get the following loop of length $33$
$$15,
21,
29,
41,
57,
80,
56,
39,
55,
77,
108,
76,
53,
74,
52,
36,
25,
35,
49,
69,
97,
137,
193,
272,
192,
135,
190,
134,
94,
66,
46,
32,
22,
15, \dots
$$The smallest $n$ for which the iterations of $f_{2^{-1/2}}$ seems to diverge to infinity is $73$: $$73, 103, 145, 205, 289, 408, 288, 203, 287, \dots ,<PHONE_NUMBER>71,<PHONE_NUMBER>03, \dots$$
The difference of shape of the pictures for $\pi$ and $\sqrt{2}$ is surprising. Below is the picture for $\alpha = \sqrt{2}^{\sqrt{2}}$, it is also unsimilar to those above:
Moreover, the fact that $\sqrt{2}, \pi$ seem to be in $S$ and $\sqrt{2}/2, \phi$ not, is also surprising because then the answer would be very irregular on the irrational numbers.
All these surprises lead to feel that the answer to the main question could be reachable.
Extended family of (borderline) Collatz-like maps
We can extend the above family as follows: let the tuple $A=(\alpha_1, \alpha_2, \dots , \alpha_r)$ such that $\alpha_i>0$ for all $i$ and $\prod_i \alpha_i \le 1 $. Consider the map $f_A: n \mapsto \left \lfloor{n\alpha_i} \right \rfloor \text{ if } n \equiv i \mod (r)$.
Problem: What are the tuples $A$ for which the map $f_A$ is acceptable?
Other borderline Collatz-like maps
Here are other examples of boderline Collatz-like maps, provided by Tom Crawford in this Numberphile video (which inspired this post). Let $\alpha$ be a normal number in base $10$ (you can think to $\pi$, which is strongly suspected to be normal). Consider the function $g_{\alpha}(n)$ equals to the position of the first occurence of $n$ (written in base $10$) in the decimal digits of $\alpha$. Because $\alpha$ is normal, $g_{\alpha}$ should satisfy $(*)$. Note that $g_{\pi}$ is available in OEIS A014777.
Here is an other Collatz-like map involving the set of prime numbers $\mathbb{P}$:
First consider the usual bijection $b: \mathbb{P} \to \mathbb{N}$, next the map $g: \mathbb{P} \to \mathbb{P}$ with $$g(p):= \text{ the greatest prime factor of } 2p+1.$$ Then $f:= b \circ g \circ b^{-1}$ should also be a (borderline?) Collatz-like map.
For more details on this map and its generailzations, see this post.
For $\alpha=\frac{2}{3}$ if you start with an even number you get a smaller number in one step. If we start with an odd number >1 we can write it as $u2^k+1$ where u is odd and $k>0$. Then after $k$ steps we get the even number $u3^k+1$, after $k+1$ steps $2u3^{k-1}$ and after $2k$ steps $u2^k$ so 1 smaller than we started with. Hence we arrived at a smaller number and the sequence will always end at $1$.
@user35593: Right! The irregularity of the picture for $2/3$ was a bad indicator on the complexity of this case.
Tenth version of the question in a day.
@GerryMyerson https://meta.mathoverflow.net/q/192/34538
What's your point, Sebastien?
@GerryMyerson: here is not the right place to discuss that, but as explained by Scott in this meta-post, the installation of a "minor edit" check box should dissolve all these tensions due to multiple editing.
Maybe so, but in the meantime....anyway, most of the edits here have not been minor.
|
2025-03-21T14:48:29.971029
| 2020-02-24T10:50:02 |
353435
|
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|
Stack Exchange
|
Every complex number has a square root via LLPO without weak countable choice
Is it possible to prove that every complex number has a square root using analytic LLPO, but avoiding Weak Countable Choice or Excluded Middle? Unique Choice is allowed.
(Analytic LLPO is the statement that given any pair of real numbers $x$ and $y$, either $x \leq y$ or $x \geq y$. This statement is non-constructive, but still weaker than other statements like LPO or Excluded Middle.)
It is true in Johnstone's Topological Topos. This is because the Fundamental Theorem of Algebra is true for Cauchy Real numbers, and Cauchy reals are isomorphic to Dedekind reals in the Topological Topos. But this reasoning doesn't seem to work unless Cauchy is iso to Dedekind.
Yes it is but there is no extensional square root function unless we also have LPO.
Note that the squaring function is a bijection from $Q_{+} = \{x + iy \mid x \geq 0, y \geq 0\}$ onto $H_{+} = \{x + iy \mid y \geq 0\}$. Similarly it is a bijection from $Q_{-} = \{x + iy \mid x \geq 0, y \leq 0\}$ onto $H_{-} = \{x + iy \mid y \leq 0\}$. Given LLPO, $H_{+} \cup H_{-} = \mathbb{C}$ and that is enough to prove the existence of square roots. The confusing part is that for a negative real number $x$, we obtain the square root $+ i\sqrt{-x}$ or $-i\sqrt{-x}$ depending on whether LLPO gives $x \in H_{+}$ or $x \in H_{-}$. So this does not give an extensional square root function.
Interestingly, this argument gives the stronger conclusion that every complex number has a square root in $$Q_{+} \cup Q_{-} =\{ x + iy \mid x \geq 0\}. $$ This stronger statement turns out to be equivalent to LLPO. Indeed, given a small enough real number $x$, if the square root of $-1+ix$ in $Q_{+} \cup Q_{-}$ is close to $i$ then $x \geq 0$ and similarly if that square root is close to $-i$ then $x \leq 0$.
Do you have a reference to fill in the gap of that last sentence?
@cody The gap? What am I missing?
The fact that you need LPO to build an extensional square root function doesn't seem to obviously follow from this argument.
@cody Ah, the other last sentence! That's basically from Grilliot's trick: an extensional square root function has to be discontinuous somewhere.
|
2025-03-21T14:48:29.971198
| 2020-02-24T10:55:01 |
353437
|
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"Gilles Mordant",
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|
Stack Exchange
|
A new "adversarial" Wasserstein distance?
Let us consider $\mu_1, \mu_2$ and $\mu_3$ three probability measures living on $[0,1]^{k_1}, [0,1]^{k_2}$ and $[0,1]^k$respectively, with $k_1 +k_2=k$. Let us denote by $\Gamma(\mu,\nu)$ the set of measures with prescribed marginals $\mu$ and $\nu$.
What is known about
$$\sup_{\pi \in \Gamma(\mu_1,\mu_2)}\inf_{\nu \in \Gamma(\pi,\mu_3)} \int_{[0,1]^k\times[0,1]^k} \lvert x-y\rvert² d\nu(x,y) \ ?$$
Further assume that one discretises such measures (by $n$ points, say).
Is there any numerical scheme to solve this discretised problem, alike the Hungarian algorithm in the classical case ?
I can figure out that one will be looking for a $n\times n \times n $ object containing only zeros and ones, with sum conditions but I am starting to be stuck at this point.
My browsing of the literature did not yield much so far.
You are looking for a coupling $\pi$ which maximizes the Wasserstein distance to a given measure $\mu_3$. A special case of that is the question https://mathoverflow.net/q/264231/106046 If you are only interested in numerics, I think by discretising the problem and using the dual formulation for the inner problem, you end up with a linear program.
thank you for the fast reply. I am digging into this
|
2025-03-21T14:48:29.971314
| 2020-02-24T11:26:44 |
353438
|
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|
Stack Exchange
|
Extended Kalman Filter and its State Transition Matrix
Sorry for what might be a long post, I want to give background.
Initially I had regular Kalman filter, and the state model was defined by Newtonian kinematics, with initial position 0 and speed of 2. I was tracking position (x) and velocity (v), i.e. my state vector is $\begin{bmatrix}
x & \dot x \\
\end{bmatrix}^T$:
$$x = x_0 + v_0t$$
$$v = v_0$$
This resulted in a State Transition Matrix:
$$ \begin{bmatrix}
1 & \Delta t\\
0 & 1\\
\end{bmatrix}$$
Now I am trying to implement Extended Kalman Filter. I have given system Acceleration of 2, so that equations go like this together with plugging in initial speed and acceleration:
$$x = x_0 + v_0t + \frac12at^2 \rightarrow x = 2t + t^2$$
$$v = v_0 + at \rightarrow v = 2 + 2t$$
Now I need to find Jacobian Matrix with respect to my state vector and I understand what it is, however, I do not understand, how do I find my State Transition Matrix, if equations that I have are expressed in terms of time and not in terms of the state variables. I assume that first line in State Transition Matrix remains the same, since position changes in same way, i.e:
$$position = previous\,position + \Delta time * speed\, over\, that \, time\, period$$
It is the speed that is changing. But I don't know how to define it in my State Transition Matrix. From what I understand, my STM will be different every epoch, am I right? As I said, I know that I need to find Jacobian and know what it is, but I don't know how to find it in this particular case.
Thank you.
|
2025-03-21T14:48:29.971429
| 2020-02-24T11:50:04 |
353439
|
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"Alexey Ustinov",
"Bazin",
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"sort": "votes",
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|
Stack Exchange
|
Stationary phase method for $\varphi''(x_0)= 0$
Stationary phase method (in the usual setup) gives asymptotic for
$$
I(\lambda)=\int_{a}^{b} f(t) e^{i \lambda \varphi(t)} d t,
$$
when at any stationary point $x_0$ ($\varphi'(x_0)=0$) second derivative
does not vanishe ($\varphi''(x_0)\ne 0$). Is it possible to find in the literature asymptotic formula for $I(\lambda)$ with explicit error term (not like $1+o(1)$) in the case $\varphi'(x_0)=\varphi''(x_0)= 0$, $\varphi'''(x_0)\ne 0$?
Let me assume that $a=-\infty, b=+\infty, x_0=0$ and $f$ smooth and compactly supported near 0. Then after a suitable change of variable, you get that
$
I(\lambda)=\int g(t) e^{i\lambda t^3/3} dt,
$
with $g$ smooth and compactly supported
and applying Plancherel formula you get
$$
I(\lambda)=\int \hat g(\tau) A(\lambda ^{-1/3}\tau) d\tau \lambda ^{-1/3}=\lambda ^{-1/3}\psi(\lambda ^{-1/3}),
$$
where $A$ is the Airy function (the inverse Fourier transform of $t\mapsto e^{it^3/3}$). Then you may apply what is known on the Airy function on the real line or simply use that
$$
I(\lambda)=\lambda ^{-1/3} \psi(0)+O(\lambda^{-2/3}), \quad \psi (0)= g(0) A(0),\quad
A(0)=3^{-1/6}\Gamma(1/3)/(2π).
$$
To get an explicit error term, you may use the explicit expansion of the Airy function.
Your argument is very nice, but I'm looking for the formula for finite $a$ and $b$. It should depend on the values of $\varphi'(t)$ and $\varphi''(t)$ at boundary points.
@AlexeyUstinov Plancherel formula can be used as well when there are boundary points by calculating the Fourier transform of $f\mathbf 1_{[a,b]}$. Assuming for instance that $x_0$ is in the interior of $(a,b)$, you may put a cut-off function and deal with a more standard stationary phase argument at the boundary.
|
2025-03-21T14:48:29.971560
| 2020-02-24T12:10:40 |
353440
|
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|
Stack Exchange
|
Is there a strictly increasing differentiable function maps positively measurable set to zero measure set?
Let $g(t)$ be a strictly increasing differentiable function. Can it map positively measurable set to zero measurable set?
It's obviously that $\{g'>0\}$ is dense. If I can prove that the Lebesgue measure $m(\{g'=0\}) = 0$, then for every set with positive measure, there will be a positively measurable subset with $g'>\epsilon$ on it, and consequently maps the set to nonzero measure set(It's a theorem and I forget it's name).
The question derives from my textbook, which says if $g(t)$ is a strictly increasing differentiable function and Riemann integrable, and $f(x)$ is Riemann integrable, then
$$\int f(x) = \int f(g(t))g'(t)$$
All functions defined on suitable closed set.
It seems that $f(g(t))$ may even be not integrable and that is totally a typo. But with my intuition of measure theory, this might be true since $g$ is differentiable.
@GeraldEdgar It's not the same thing. I am saying mapping sets of positive measure to sets of measure zero instead of mapping sets of measure zero to sets of positive measure
There are strictly increasing $C^1$ functions that map sets of positive measure to sets of measure zero. Here is a construction:
Let $C\subset [0,1]$ be a Cantor set of positive measure. For a construction, see https://en.wikipedia.org/wiki/Smith-Volterra-Cantor_set. Let $g(x)=\operatorname{dist}(x,C)$. The function $g$ is clearly continuous and equal zero on $C$. In fact $g$ is a $1$-Lipschitz function. Let
$$
f(x)=\int_0^x g(t)\, dt.
$$
The function $f$ is $C^1$ and it is strictly increasing. Indeed, if $y>x$, then
$$
f(y)-f(x)=\int_x^y g(t)\, dx>0
$$
because the interval $[x,y]$ is not contained in the Cantor set $C$ and therefore it contains an interval where $g$ is positive.
On the other hand $f'=g=0$ on $C$ which has positive measure and $f(C)$ has measure zero since $m(f(C))=\int_C f'(t)\, dt=\int_C g(t)\, dt=0$.
As was pointed out by Mateusz Kwaśnicki in his comment, this construction gives the following result:
Theorem. Let $f$ be as above. Then there is a Riemann integrable function $h$ such that $h\circ f$ is not Riemann integrable.
Proof. The set $f(C)$ is homeomorphic to the Cantor set ($f$ is strictly increasing so it is a homeomorphism) and has measure zero as explained above. Let
$$
h(x)=\begin{cases}
1 & \text{if $x\in f(C)$}\\
0 & \text{if $x\not\in f(C)$.}
\end{cases}
$$
The function $h$ is Riemann integrable with the integral equal zero since it is bounded and continuous outside the set $f(C)$ of measure zero (because $\mathbb{R}\setminus f(C)$ is open and $h=0$ there). However,
$$
(h\circ f)(x)=\begin{cases}
1 & \text{if $x\in C$}\\
0 & \text{if $x\not\in C$.}
\end{cases}
$$
is not Riemann integrable since it is discontinuous on a set $C$ of positive measure. $\Box$
In case one needs a paper reference, virtually the same construction is carried out in Real Analysis Exchange 22(1) (1996-97): 404–405 by Javier Fernández de Bobadilla de Olazabal.
Furthermore, if you modify your construction just a little bit to make $g$ be a positive $C^\infty$ “bump” on each interval (connected component) of the complement of $C$, with all derivatives tending to $0$ at the edges of the interval, and with the height of the bumps tending fast enough to $0$, you can get $g$ hence $f$ to be both $C^\infty$, still with the same property.
|
2025-03-21T14:48:29.971772
| 2020-02-24T12:37:13 |
353441
|
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|
Stack Exchange
|
Density of bipartite $d$-degenerate graph
A graph $G$ is $d$-degenerate if every subgraph of $G$ contains a vertex of degree at most $d$. It is known that an $n$-vertex $d$-degenerate graph has at most $d(n-1)$ edges. However, if we are given a bipartite $d$-degenerate graph $(H:m,n)$, what is its maximum number of edges?
Theorem. A $d$-degenerate $n$-vertex bipartite graph has at most $\lceil \frac{n}{2} \rceil \lfloor \frac{n}{2} \rfloor$ edges if $n < 2d$ and at most $d(n-d)$ edges if $n \geq 2d$. Moreover, both these bounds are tight.
Proof. Suppose $G$ is a $d$-degenerate $n$-vertex bipartite graph. Every $n$-vertex bipartite graph contains at most $\lceil \frac{n}{2} \rceil \lfloor \frac{n}{2} \rfloor$ edges. If $n \leq 2d$, then $K_{\lceil \frac{n}{2} \rceil, \lfloor \frac{n}{2} \rfloor}$ is $d$-degenerate, so this bound is tight. If $n \geq 2d$, we proceed by induction on $n$. The base case of $n=2d$ has already been established by the previous argument. Thus, we may assume that $n > 2d$. Let $x$ be a vertex of degree at most $d$ in $G$. By induction, $G-x$ contains at most $d(n-1-d)$ edges, so $G$ has at most $d(n-d)$ edges. This bound is tight as demonstrated by $K_{d, n-d}$. $\square$
Note that if $d$ is a constant, then the bound is not much better than for general graphs. A $d$-degenerate graph contains at most $\binom{n}{2}$ edges if $n \leq d$ and at most $d(n-d)+\binom{d}{2}$ edges if $n > d$. These bounds are tight, since $K_n$ is $d$-degenerate if $n \leq d$ and the join of $K_d$ and a stable set of size $n-d$ is $d$-degenerate if $n > d$.
|
2025-03-21T14:48:29.971880
| 2020-02-24T15:26:33 |
353446
|
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"Dmytro Taranovsky",
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|
Stack Exchange
|
$\Sigma^2_2$ absoluteness and $\diamondsuit$
This question touches on recent questions concerning the role of $\diamondsuit$ and the Continuum Hypothesis. In particular, I would like to know more information about a conjecture of Woodin on the relationship between $\Sigma^2_2$-absoluteness and $\diamondsuit$ in the presence of large cardinals.
Quoting Andres Caicedo (in an answer to this question):
Beyond $\Sigma^2_1$, there is much speculation. It is expected some strengthening of diamond will be maximal for $\Sigma^2_2$, and we will get a similar theorem, but beyond $\Sigma^2_2$ this starts to conflict with other conjectures.
What is the current state of affairs? Is the answer known for $\diamondsuit$ itself?
The conditional absoluteness is still conjectural. See my new question for a partial explanation of the current state.
Ah, it is good to know it is still not settled. I saw an abstract for a talk in Toronto last year that claimed a counterexample existed, but couldn't track down any more information.
|
2025-03-21T14:48:29.971975
| 2020-02-24T15:28:26 |
353447
|
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"Fabrice Rouillier",
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"url": "https://mathoverflow.net/questions/353447"
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|
Stack Exchange
|
Using rational univariate representation to obtain approximated solution of the system of multivariable polynomials with error $<10^{-16}$
This post is a follow up question of my previous post (Accuracy of solutions of system of multivariable polynomials).
According to wikipedia about Rational univariate representation (RUR):
Consider a zero-dimensional system of polynomials in variable $x_i$ for $1\le i\le n$.
Then we can express $x_i$ in terms of rational function of a root of some
univariate polynomial. More precisely:
A RUR of a zero-dimensional system consists in a linear combination $x_0$ of the variables, called separating variable, and a system of equations
\begin{cases}h(x_{0})=0\\x_{1}=g_{1}(x_{0})/g_{0}(x_{0})\\\quad \vdots \\x_{n}=g_{n}(x_{0})/g_{0}(x_{0}),\end{cases}
where $h$ is a univariate polynomial of degree $D$ and $g_0, \cdots, g_n$ are univariate polynomials of degree less than $D$.
For zero-dimensional systems, the RUR allows retrieval of the numeric values of the solutions by solving a single univariate polynomial and substituting them in rational functions. This allows production of certified approximations of the solutions to any given precision.
For more details about RUR, see this paper: Solving Zero-dimensional Polynomial Systems through the Rational Univariate Representation---Fabrice Rouillier.
Moreover, MPSolve can compute the roots of univariate polynomials with guaranteed precision (Thanks to comment of François Brunault). So, one can approximate $x_0$ to any precision.
However, I still do not know how to approximate $x_i$ to any precision.
Then I find out a paper
A NEW ALGORITHM FOR REAL ROOTS OF A ZERO-DIMENSIONAL SYSTEM BY A LINEAR SEPARATING MAP---YUJI KONDOH, TOMOKATSU SAITO and TAKU TAKESHIMA
about how to approximate $x_i$ to any precision. But it does not explicitly mention how.
According to the paper, it mentioned that
What is the necessary procedure that is equivalent to determine the sign of a algebraic number mentioned in the last sentence, in order to make sure one can get all the approximated solutions with error $<10^{-16}$?
Are there any shortcut to ensure the errors involved in the approximated solution of the original zero-dimensional system are $<10^{-16}$?
Using interval arithmetic will solve your problem.
Have a look at the documentation here.
|
2025-03-21T14:48:29.972136
| 2020-02-24T17:13:45 |
353452
|
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|
Stack Exchange
|
Very weak Riemann-Roch on curves (by J. Kollar)
I have a question on an unequality used in the proof of the Very weak Riemann-Roch on curves in Janos Kollar's Lecture on Resolution of Singularities (page 14):
1.13 (Very weak Riemann-Roch on curves). Let $C$ be an irrecucible, reduced, projective curve over an alg closed field $k$. We claim that for any ample line bundle $L$,
$$h^0(C,L^m) \ge m \cdot deg \ L +1 -\binom {\operatorname{deg} \ L-1}{2}$$
for $ m \ge \operatorname{deg} \ L$.
(proof) Indeed, embed $C$ into $\mathbb{P}^n$ by $L$ (since $L$ ample that works), and then project it generically to a plane curve of degree $\operatorname{deg} \ L$, $\pi: C \to C' \subset \mathbb{P}^2$
few explanations: i.e. $\pi$ is the composition of the embedding $i:C \to \mathbb{P}^n$ and and projection of $\mathbb{P}^n$ to a plane $\cong \mathbb{P}^2$; it's only a rational map but if we restrict it to $C$ and chose the projection planes in clever way (see my comments below) then the projection $C \dashrightarrow \mathbb{P}^2$ becomes a morphism. $C'$ is the image of $C$ and $C'$ is birational to $C$. we continue the proof...
Now, for $m \ge \operatorname{deg} \ L$,
$$h^0(C,L^m) \ge h^0(C', \mathcal{O}_{\mathbb{P}^2}(m) \vert _{C'}) (???) \\ \ge h^0(\mathbb{P}^2, \mathcal{O}_{\mathbb{P}^2}(m)) -h^0(\mathbb{P}^2, \mathcal{O}_{\mathbb{P}^2}(m- \operatorname{deg} \ L)) \\ \ge m \cdot \operatorname{deg} \ L +1 - \binom {\operatorname{deg} \ L-1}{2}.
$$
[...]
Question: I not understand the first unequality
$$h^0(C,L^m) \ge h^0(C', \mathcal{O}_{\mathbb{P}^2}(m) \vert _{C'}).$$
more concretly, I don't see how $\pi$ induces an appropriate exact sequence which allows to relate dimensions of sheaf cohomology groups $H^0(C,L^m)$ and $ H^0(C', \mathcal{O}_{\mathbb{P}^2}(m) \vert _{C'})$
In contrast, the second equality is clearly induced by $0 \to O(m-\operatorname{deg} \ L) \to O(m) \to O(m) \vert _{C'} \to 0$. Ok, on first one I have no idea.
A couple words about the projection $\mathbb{P}^n \dashrightarrow \mathbb{P}^2$: that's nothing but the composition of the projections explaned in example 1.9 page 12: Let $p \in \mathbb{P}^n$ and $\mathbb{P}^{n-1} \cong H \subset \mathbb{P}^n$ a hyperplane not containing $p$. This induces the projection $\pi_{p,H}: \mathbb{P}^n \dashrightarrow \mathbb{P}^{n-1}$ (in the book the construction is described concretly). Moreover the book tells that sophisticated choises of points and hyperplanes allows to project down a curve $C$ birationally to $C'$ by iterating the process above & restricting to $C$.
The projection $\pi \colon C \to C'$ is induced by a $2$-dimensional sub-linear system of the complete linear system $|L|$, so $C'$ is not linearly normal (unless $H^0(C, \, L)=3$) and $H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(1)|_{C'})$ has the same dimension as $H^0(C, \, L)$.
It follows that every global section $\sigma' \in H^0(C', \, \mathcal{O}_{\mathbb{P}^2}(m)|_{C'})$ gives a section $\sigma \in H^0(C, \, L^m)$ such that $\pi(\mathrm{div}(\sigma)) = \mathrm{div} (\sigma')$. This yields the desired inequality.
When you talking about "dimension" of sub-linear system of cls $\vert L \vert$ you mean the "projective" dimension, in sense of "dimension" of projective subspace of $H^0(C,L) - {0}/k^*$, not as $k$-subspace of $H^0(C,L)$, right? Question: Why the projection $\pi: C \to C'$ is induced by a $2$-dimensional sub-linear system? I assumed that $\pi$ arises from the composition $ C \subset \mathbb{P}^n \dashrightarrow \mathbb{P}^2$, right? And the space defining $\pi$ we obtain by taking the linear $k$-subspace $<\pi^*X_i \vert i=0,1,2>_k \subset H^0(C, L)$ and then projectivize it.
Why it is $2$-dimensional? Indeed, it is generated by pullbacks of the $X_i \in H^0(\mathbb{P}^2, \mathcal{O}(1)_{\mathbb{P}^2})=kX_0 \oplus kX_1 \oplus kX_2$. After prozectivizing $<\pi^X_i \vert i=0,1,2>_k-{0}/k^$ is at most $3-1=2$ dimensional. So why for example it cannot be $1$ dimensional?
Second question: Why assumption $\dim_k H^0(C, , L)=3$ would imply $C \cong C'$?
The projection is given by $x \mapsto [\sigma_0(x):\sigma_1(x) : \sigma_2(x)]$, where $\langle \sigma_0,, \sigma_1, , \sigma_2 \rangle$ span a $3$-dimensional subspace of $H^0(C, , L)$, so a $2$-dimensional sub-system of $|L|$. If the image is $1$-dimensional, you are actually projecting on a line, not on the plane: this is a special situation, that does not happen for the general projection.
Regarding the second question, you are right: also in this case, $C'$ is in general only birational to $C$. I edited the post.
So we can conclude from this that asigning $X_i \to \pi^*X_i = \sigma_i$ gives an injection $kX_0 \oplus kX_1 \oplus kX_2=H^0(\mathbb{P}^2, \mathcal{O}(1){\mathbb{P}^2}) \to H^0(C, L)$ since $\dim_k(H^0(\mathbb{P}^2, \mathcal{O}(1){\mathbb{P}^2}))=3$. By construction it factorizes through $H^0(C', , \mathcal{O}{\mathbb{P}^2}(1)|{C'})$. What I not understand is why this already imply that the induced map $H^0(C', , \mathcal{O}{\mathbb{P}^2}(1)|{C'}) \to H^0(C, L)$ is also injective.
And when we know it for $m=1$, how to see that this injection also holds for every twisted version $H^0(C', , \mathcal{O}{\mathbb{P}^2}(m)|{C'}) \to H^0(C, L^m)$?
Hi, meanwhile I found out that that one can also argue as follows: I think that the image $C'$ is implicitly endowed with unique reduced structure and since $C$ is irreducible, $C'$ is integral and the structure morphism $O_{C'} \to \pi_*O_C$ has a kernel is nilpotent since $\pi$ is dominant. Thus $O_{C'} \to \pi_*O_C$ is injective. And obviously the injectivity is preserved under twists by $O_{C'}(m)$ and we are done.
Nevertheless in still not understand the argument from your answer and I'm really curious to find out how it works. You say that essentially the injectivity of $H^0(C', , \mathcal{O}{\mathbb{P}^2}(1)|{C'}) \to H^0(C,L)$ follows from that $C'$ is not linearly normal. Why?
What do we know? Well, we assumed that $x \mapsto [\sigma_0(x):\sigma_1(x) : \sigma_2(x)]$ not factors through a line, i.e. $C'$ is not degenerated and $H^0(\mathbb{P}^2, \mathcal{O}(1){\mathbb{P}^2}) \to H^0(C, L)$ is injective and trivially $H^0(\mathbb{P}^2, \mathcal{O}(1){\mathbb{P}^2}) \to H^0(C', , \mathcal{O}{\mathbb{P}^2}(1)|{C'})$ is injective too by the same argument with non degeneracy. But $C'$ is not linearly normal is equivalent to $H^0(\mathbb{P}^2, \mathcal{O}(1){\mathbb{P}^2}) \to H^0(C', , \mathcal{O}{\mathbb{P}^2}(1)|_{C'})$ not surjective, thus we
not hit all sections in $H^0(C', , \mathcal{O}{\mathbb{P}^2}(1)|{C'})$. So it seems that we don't have any "control" over what happens with the images of these sections under the map $H^0(C', , \mathcal{O}{\mathbb{P}^2}(1)|{C'}) \to H^0(C,L)$. That is I not understand why $C′$ is not linearly normal implies the injectivity of this map. On the other hand if $C'$ would be linearly normal, then $H^0(\mathbb{P}^2, \mathcal{O}(1){\mathbb{P}^2}) \cong H^0(C', , \mathcal{O}{\mathbb{P}^2}(1)|_{C'})$ and the desired map would be injective, but
the way you conclude the injectivity as consequence from assumption that $C'$ is not lineary normal is not clear to me. Could you explain it more detailed? Sorry, for frequently annouying but I absolutely wants to figure out how your argument works.
|
2025-03-21T14:48:29.972551
| 2020-02-24T18:17:37 |
353459
|
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"Anton Petrunin",
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|
Stack Exchange
|
This almost periodic condition implies equicontinuity?
Let $X$ be a metrizable compact space and $T\colon X\to X$ a minimal homeomorphism, i.e.
$$ \mathrm{orb}(x) := \{T^kx:k\in\mathbb{Z}\}$$
is dense in $X$ for every $x \in X$. Assume that the following condition is met:
There exist $\varepsilon_n \to 0$ and $s_n \in \mathbb{N}$ such that $d(T^{s_n}x,x) < \varepsilon_n$, for every $n \in \mathbb{N}$ and $x \in X$.
This implies that $X$ is equicontinuous? When $X$ is a subshift it is easy to see that this is true (beacuse, then, every point is periodic), but for a general system I couldn't prove it.
Hi, Ia writing just to signal a little typo: perhaps it is minimal, not minimial.
Here is a relevant statement: If $T$ is an isometry and $X$ is proper, then $X$ is compact --- it is simple, but not trivial; see “On conditions under which...” by Aleksander Całka.
@AntonPetrunin you are right! In fact in every equicontinuous system this can happen. What about if this condition implies equicontinuity? I will edit the question.
This is called uniform topological rigidity if I recall correctly, and https://arxiv.org/abs/1508.03366 should contain some info.
I think your specific question is answered by their statement that rigidity of minimal systems does not imply uniform rigidity, but I can't check now.
@VilleSalo My condition is equivalent to uniform rigidity, and in the arxiv article they say that there are uniformly rigid and weakly mixing actions. So, the answer to my question is negative. Thanks!
I checked now, and I think I was wrong: the statement that "rigidity of minimal systems does not imply uniform rigidity" does not solve your problem, exactly because you're already looking for uniform rigidity + extra properties. The main result of the paper also does not solve the problem, I think, because they don't construct minimal models (I haven't read the construction, but they don't seem to state it is minimal, and someone once asked me if it can be made minimal (I didn't know)).
In general, such a homeomorphism is not necessary equicontinuous.
The existence of such examples on $X=\mathbb{T}^2$, i.e. the $2$-torus, can be shown as follows: let $\mathcal{O}$ be the $C^\infty$ closure of the set $\{h\circ R_\alpha\circ h^{-1} : h\in\mathrm{Diff}^\infty(\mathbb{T^2}),\ \alpha\in\mathbb{T}^2\}$, where $\mathbb{T}^2$ denotes the $2$-torus and $R_\alpha : \mathbb{T}^2\ni x\mapsto x+\alpha$.
Fathi and Herman showed in
Fathi, Albert; Herman, Michael R., Existence de difféomorphismes minimaux, Astérisque 49(1977), 37-59 (1978). ZBL0374.58010.
that there is residual set $C_0\subset \mathcal{O}$ such that every diffeomorphism of $C_0$ is minimal.
On the other hand, in
Kocsard, Alejandro; Koropecki, Andrés, A mixing-like property and inexistence of invariant foliations for minimal diffeomorphisms of the 2-torus, Proc. Am. Math. Soc. 137, No. 10, 3379-3386 (2009). ZBL1179.37063.
we proved the existence of a residual set $C_1\subset\mathcal{O}$ such that any $f\in C_1$ is weak-spreading, i.e. if $\tilde f\colon\mathbb{R}^2\to\mathbb{R}^2$ is a lift of $f$, then for every non-empty open set $U\subset\mathbb{R}^2$, every $\epsilon>0$ and any $R>0$, there exist $n>0$ and a ball $B_R\subset\mathbb{R}^2$ of radius $R$ such that $\tilde f^n(U)$ is $\epsilon$-dense in $B_R$. It is clear that every weak spreading homeomorphism is not equicontinuous.
Finally, one can show that rigid diffeomorphisms are generic in $\mathcal{O}$, i.e. there is residual set $C_2\subset\mathcal{O}$ such that for every $f\in C_2$ there is a sequence of natural numbers $n_j\to \infty$ so that $f^{n_j}\to id$ in the $C^0$ uniformly, when $j\to \infty$.
So, any diffeomorphism in $C_0\cap C_1\cap C_2$ is minimal and rigid, but not equicontinuous.
Thank you very much!
|
2025-03-21T14:48:29.972927
| 2020-02-24T18:21:10 |
353460
|
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|
Stack Exchange
|
A name for this kind of lax 2-limit
Consider the following statement of a universal property in a 2-category:
Consider the situation of lax squares:
then what is the name for a universal object $\ell$ equipped with a lax square over the cospan $X\to S \leftarrow S'$ such that in the lax square above, there exists a unique dotted map $X\to \ell$, and a unique 2-cell $(X'\to \ell \to X) \Rightarrow (X'\to X)$ as in the diagram
such that the pasting composite of the 2-cells is equal to the 2-cell we started with?
(sorry for the images, Mathjax had some trouble making this diagram reasonable)
Does it have a name? It's describing something like a 'lax base change', but it's not really a comma object.
Bonus question: Is there a nice and simple construction of this thing in Cat?
In the 2nd diagram, it should be $X'$ in the top left corner.
It seems to me that such constructions need not exist. In Cat, let $S$ be terminal. Then any choice of the 2-morphism with 0-domain $X’$, which you assume to exist uniquely, is as good as any other, the pasting always giving the unique 2-morphism with 0-codomain $S=*$.
To explain this, it looks to me as if you are asking precisely that the lax limit of the arrow $\ell\to X$ should be the comma object of the original span. But not every projection out of a comma object factors through a projection out of the limit. Indeed, thinking again of $S=*$, the comma object is $X\times S’$and the projection $X\times S’\to X$ need not factor this way, unless I’m missing something-for instance try $X$ an arrow and $S’$ discrete.
I figured it might exist because Cat is 2-complete and you can see this thing as some kind of 2-Kan extension of a certain shape of 2-diagram, but I guess that doesn't work the same way in 2-category land.
|
2025-03-21T14:48:29.973093
| 2020-02-24T18:22:51 |
353461
|
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|
Stack Exchange
|
On partial sum estimate on the series $S(p,q;s)=\sum_1^q\frac{\sin^2(\frac{p\Gamma(n)}{n})}{n^s}$ and other Generalizations
Consider the following sum :
$$S(p,q;s)=\sum_{n=1}^q\frac{\sin^2(\frac{p\Gamma(n)}{n})}{n^s}$$
Here , $p$ is a variable w.r.t which we are going to analyse the sum.
$s$ is another parameter with domain $s\in(0,1]$.
I tried to use Abel - Plana summation formula (ABSF) for partial sum :
we can get some crude estimates on integral :
$$I_1(q)= \int_1^q\frac{\sin^2(\frac{p\Gamma(t)}{t})}{t^s}dt $$
as $q\rightarrow \infty $
( Integrand oscillates very wildly in right half plane )
But it seems that second integral in ABSF is impregnable .
Second integral in ABSF:
$$I_2(x)=\int_0^\infty \frac{F(x + iy,s) − F(x − iy,s)}{e^{2πy}-1}dy$$
Where , $F(z)=\frac{\sin^2(\frac{p\Gamma(z)}{z})}{z^s} $
I tried to get estimate on this as $x\rightarrow \infty $ but in vain .
The reason I chose ABSF is that $\Gamma$ is a 'nice' function in terms of complex variables and due to this I'm optimistic about the $I_2$
The importance of this function lies in the fact that , See for ex . for $p=π/2$ the $\sin^2$ term is finite for primes and zero for non primes . So I mentioned 'sharp' for this purpose . I need critical details of $I_2$.
( I'm just following the advice of F.R.Villegas to generalize the series with such parametrization.)
Question : Can we get an 'Sharp' estimates on the sum w.r.t parameters $p$ and $s$?
Can we prove the divergence of series (as a whole) for specified conditions on $p$?
(Also , I calculated various values of $I_2(x)$ for various $x$'s and test parameters .)
Edit:
More generally we can consider the following:
$$F(z) = \omega(z)\sin^2\left(\frac{π\Gamma(z)}{2z}\right)$$
Here, $\omega(z)$ is a weight we have to construct.
I try to construct the $\omega(z)$ s.t.
$$\sum_2^n F(n)= \int_2^n F(x)dx + A$$
Here A is constant.
For this to be true the following three conditions should meet for $\omega(z)$ in context of APSF:
$$\omega(z)>\frac{1}{z},\ \forall z\in\mathbf{R}$$
( 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 $x\to\infty$.
Can we Explicitly construct $\omega(z)$?
Even if one can omit 1st condition and able to construct the weight s.t. it follows condition 2,3 please mention. ( i.e in this case the integral $\int_2^\infty F(x)dx$ is convergent.
Also one can generalize further:
$$F(z) = {\phi(\sin^2[π\Gamma(z)/(2z)])}$$
S.t.
(1)$ ϕ(x)=0$ if x is zero ; and 'suitably' finite otherwise
(Here , 'suitably' means a value which guarantees the expected divergence of sum (very close to 1 or greater than or equal to 1) )
(2) condition (3) holds for such function.
Could we make above analysis workable?
Also, I think the condition (3) is very hard to achieve because of the complex roots of equations
$\Gamma(z)/(z)$=even integer
Also if condition (3) is not possible in any way, can we get estimate on analogues $I_2(x)$ with use of suitable weight which makes things easier?
If this could be achieved then we are able to get the estimates on primes using purely analytic information ( no number theoretic information like Euler product ) . And this seems (although extremely hard but,) possible as due to relatively elementary nature of summand and integrals .
Note: I know this question received negative reviews (due to both my behavior and insufficient information ) but please consider the importance of question.
Also see: https://math.stackexchange.com/q/3570663/789323
I would think this is divergent for any $s\in(0,1]$.
Thank you everyone who upvoted! At least question now maintain a zero score.
I would just approximate $\sin^2$ by 1/2, it oscillates rapidly and the average over one period should be a sensible estimate, then
$$S(p,q,s)\approx \tfrac{1}{2}H_q^{(s)},$$
the Harmonic number. For large $q$ and $s<1$ this gives
$$S(p,q,s)\rightarrow q^{-s} \left(\frac{q}{2 (1-s)}+\tfrac{1}{4}+{\cal O}(q^{-2})\right)+\tfrac{1}{2}\zeta (s),$$
while for $s=1$ one has
$$S(p,q,1)\rightarrow=\tfrac{1}{2} \ln q+\tfrac{1}{2}\gamma_{\rm Euler}+\frac{1}{4 q}+{\cal O}(q^{-2}).$$
Here is a comparison for $p=1$, $s=1/2$ (blue the exact sum, orange the large $q$ asymptotics).
thank you for the answer. But, the main purpose of the question is very different . See for ex . for $p=\pi/2$ the $\sin²$ term is finite for primes and zero for non prime . So I mentioned 'sharp' for this purpose . I need critical details of $I_2$. (The sum is very delicate for such critical values )I'm just following the advice of F.R.Villegas to generalize the series with such parametrization
OK, this was not clear at all to me from the question, apologies, the estimate I gave in this answer applies to generic values of $p$; so you want to be able to estimate something like $\sum_{k=\text{prime}}k^{-s}\cos^2(\pi/2k)$.
|
2025-03-21T14:48:29.973431
| 2020-02-24T18:53:52 |
353464
|
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|
Stack Exchange
|
When are projective modules closed under highly-filtered colimits?
Let $R$ be a ring. Let $Mod(R)$ be the category of left $R$-modules, and let $Proj(R) \subseteq Mod(R)$ be the full subcategory of projective $R$-modules. Let's say that $R$-projectives are closed under highly-filtered colimits if there exists a cardinal $\kappa$ such that $Proj(R)$ is closed under $\kappa$-filtered colimits in $Mod(R)$. For example,
When $R$ is a field (or division ring), we have $Proj(R) = Mod(R)$, and so vacuously we have that $R$-projectives are closed under highly-filtered colimits.
When $R = \mathbb Z$, the question (perhaps surprisingly) depends on set theory
Under the anti-large cardinal hypothesis $V=L$, Eklof and Mekler showed that there exist aribtrarily large non-free abelian groups all of whose smaller subgroups are free and it follows that $\mathbb Z$-projectives are not closed under highly-filtered colimits.
Whereas if there exists a strongly compact cardinal, then it follows from the powerful image theorem of Makkai and Pare that free abelian groups are an accessible, accessibly embedded subcategory of all abelian groups, and in particular $\mathbb Z$-projectives are closed under highly-filtered colimits.
The positive result of Makkai and Pare generalizes immediately to show that if $R$ is a PID (in the noncommutative case I'm not sure of the terminology, but the hypothesis is that every submodule of a projective $R$-module is free), then assuming the existence of a sufficiently-large strongly compact cardinal (I think the cardinal just has to be bigger than the cardinality of $R$) we have that $R$-projectives are closed under highly-filtered colimits.
In fact, Rosicky and Brooke-Taylor's "$\lambda$-pure" version of the powerful image theorem shows that if there is a sufficiently large strongly compact cardinal, then projective $R$-modules are closed under highly-filtered colimits provided that $R$ is of global projective dimension $\leq 1$ (every submodule of a projective module is projective rather than free), so this result holds even for some $R$ which are not domains.
Question:
What are some other rings $R$ for which we can say (perhaps under set-theoretical hypotheses) whether $R$-projectives are closed under highly-filtered colimits?
Are there any rings $R$ other than division rings for which the question "are $R$-projectives closed under highly-filtered colimits?" is decidable in ZFC?
Is the precise large-cardinal strength of "$\mathbb Z$-projectives are closed under highly-filtered colimits" -- or (equivalently, I think) "any sufficiently-large abelian group all of whose smaller subgroups are free, is free" -- known?
Let $R$ be a ring and $\kappa$ be a strongly compact cardinal such that $|R|<\kappa$. Then the class of all projective $R$-modules is closed under $\kappa$-filtered colimits.
This is Theorem 3.3 in the recent preprint of J. Šaroch and J. Trlifaj "Test sets for factorization properties of modules", https://arxiv.org/abs/1912.03749
For any left perfect ring $R$ (which includes both all left Artinian and all right Artinian rings), all flat left $R$-modules are projective, so the class of projective left $R$-modules is closed under $\aleph_0$-filtered colimits. For example, this applies to all the rings $\mathbb Z/n\mathbb Z$, $n\ge2$ (many of which are not division rings).
Thanks, this is fantastic! I will hold off for a bit on accepting in hopes of hearing about more ZFC examples and about negative results under anti-large-cardinal hypotheses, and about more precise consistency-strength calibrations, but I think this is primarily what I was looking for.
Actually, your (1) is stronger than I expected might be known. Do you know whether large cardinal hypotheses imply more generally that every small-projectivity class in a locally presentable category is accessibly embedded -- or perhaps even that the left half of an accessible wfs is accessibly embedded in the arrow category?
No, concerning (1), this is just a new (and very interesting) result which I happen to have heard about. I am not an expert on such things, have not studied the details, and do not know what the potential for generalizations of this theorem may be.
@TimCampion The assertion that "every small projectivity class in an accessible category is accessibly embedded" follows from Vopěnka's Principle (and is independent of ZFC). See our new preprint: https://arxiv.org/abs/2201.06782
@SeanCox Oh wow, that's really cool! Thanks for sharing! This is relevant to another MO question I once asked too.
|
2025-03-21T14:48:29.973725
| 2020-02-24T20:08:58 |
353470
|
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|
Stack Exchange
|
Can the immediate basin of attraction of super-attracting fixed point at 0 of a polynomial contain non-zero roots?
Let $f$ be a polynomial with a super attracting fixed point at $x=0$. Can the immediate basin of attraction of the fixed point contain other roots? If so, please provide a specific example with the immediate basin of attraction. If not, why?
How about $z^3-z^2$?
@Mark is zero an attracting fixed point? Is, say, $z=1/2$ attracted to zero?
@Mark is 1 in the immediate basin of attraction of 0? It is not obvious
what does super attracting mean ?
@Piyush A fixed point is super-attracting if the derivative is 0. It means the points in the basin of attraction approach specifically at an exponential rate. In this case, for $z$ in the basin of attraction,$ |f(z)-0|\approx b|z-0|^p$ for some constant p.
Let $f(z)=az^2(z-1)$. Zero is superattracting. Now choose $a$ so small that
$|f(z)|<|z|/2$ for $|z|<2$. Then the root $z_0=1$ is in the immediate domain of
attraction.
|
2025-03-21T14:48:29.973825
| 2020-02-24T22:57:28 |
353481
|
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"T. Amdeberhan",
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|
Stack Exchange
|
Asymptotics for certain integrals
I stumbled on the following problem, if you can see a way through it.
Let $x$ be a real variable and fix a real value $\frac14\leq\nu\leq\frac34$.
QUESTION. For $x\rightarrow0$, does there exist a constant $C>0$ (independent of $x$ that is) with the below property?
$$\int_{\mathbb{R}}\left\{\int_0^{\infty}
\left(e^{-\frac12\left[\frac{x^2(1+y)^2}{z^2}+z\right]}
-e^{-\frac12\left[\frac{x^2y^2}{z^2}+z\right]}
\right)z^{-\nu}dz\right\}^2dy \,\,\,\,\sim \,\,\,C\,\vert x\vert.$$
Replacing $y$ with $y-\frac 12$, we get the integral
$$
2\int_0^\infty\left[\int_0^\infty z^{-\nu}e^{-\frac z2}e^{-x^2\frac{y^2+0.25}{2z^2}}
\left(e^{\frac{x^2y}{2z^2}}-e^{-\frac{x^2y}{2z^2}}\right)dz\right]^2\,dy
$$
Now, for small $x>0$, reduce the integration to $x\le z\le 2x$, $1\le y\le 2$. Then the inner integral is about $x^{1-\nu}$ and the final answer is at least $x^{2-2\nu}$, which is much larger than $x$ for $\nu>\frac 12$. Off hand, it looks like the region I described gives the main contribution in all cases but I haven't checked it.
Thank you much!
|
2025-03-21T14:48:29.973929
| 2020-02-24T23:42:46 |
353482
|
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"Anthony Quas",
"Austin Alderete",
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|
Stack Exchange
|
Upper-bounding $\dim \text{span}\{v_1,\dots,v_n\}$ in terms of $\dim \text{span}$ of subsets
I asked this question on Stack Exchange two weeks ago, and didn't get any answers, so I'm shamelessly reposting it here.
Let $S=\{v_1,\dots, v_n\} \subset V$ be a set of nonzero vectors in a vector space (we can take the field to be algebraically closed). It is easy to prove that if there exists a sequence $S_1,\dots, S_r \subseteq S$ such that
$\dim\text{span}S_j=1$ for all $j=1,\dots, r$.
$S_j \cap S_{j+1} \neq \emptyset$ for all $j=1,\dots, r-1$.
$\bigcup_{j=1}^r S_j=S$.
then $\dim\text{span}S=1$.
Essentially, this gives a way to upper bound $\dim\text{span}S$ in terms of $\dim\text{span}$ of subsets obeying certain properties. My open-ended question is whether there are "interesting" generalizations of this to the case in which we want to upper bound $\dim\text{span}S\leq d$ for some $d\geq 1$, in terms of subsets obeying certain properties.
Of course, there are trivial results in this vein: If every subset of size $\leq n-1$ has dimension $\leq d$ for some $d<n-1$, then $\dim\text{span}S\leq d$. I am looking for something more useful, that hopefully looks something like the above result for $d=1$.
I've included a link to avoid possible duplication of effort. It would be good to do the same thing in the other direction.
Done. Thank you
If condition 1 is replace by dim span $S_j = 2$ for $j=1,\ldots,r$, then dim span $S$ can be anything from 2 to $r$.
That’s a nice observation, thanks!
Let $S = \{v_1,...,v_n \} \subseteq V$ be a set of nonzero vectors in a vector space over an algebraically closed field and suppose that we have $S_1,...,S_r \subseteq S$ such that $S = \bigcup_{i=1}^r S_i$.
For each $S_i$, set $d_i = \mathrm{dim} \mathrm{span} (S_i)$ and $c_i = \dim \mathrm{span} (S_i \cap S_{i+1})$, (we omit $c_r$). Then we have
$$
\mathrm{dim} \mathrm{span}(S) = \mathrm{dim} \mathrm{span} \left( \bigcup_{i=1}^r S_i \right) = \sum_{i=1}^r \mathrm{dim} \mathrm{span}(S_i) - \sum_{1 \leq i <j \leq r} \mathrm{dim} \mathrm{span} (S_i \cap S_j)
\leq \sum_{i=1}^r \mathrm{dim} \mathrm{span}(S_i) - \sum_{i = 1}^{r-1} \mathrm{dim} \mathrm{span} (S_i \cap S_{i+1})
=
\sum_{i=1}^r d_i - \sum_{i=1}^{r-1} c_i
$$
Therefore,
$$
\dim \mathrm{span}(S) \leq \sum_{i=1}^r d_i - \sum_{i=1}^{r-1} c_i.
$$
At this point, we note that $d_i \geq c_i$ and $|S_i \cap S_{i+1}| \geq c_i$. The above gives us an upper-bound for the dimension of $S$, even if it's not very good. The special case is your original theorem: let $d_i = 1$ for all $i$ and let $S_i \cap S_{i+1} \neq \emptyset$. By the assumption that our vectors are non-zero, $\dim \mathrm{span}(S_i \cap S_{i+1}) \geq 1$. So $c_i \geq 1$ and $c_i \leq d_i$, so $c_i = 1$. Then $\dim \mathrm{span}(S) = 1$ immediately follows.
In the comment left by Brendan McKay, we have $d_i = 2$ and $1 \leq c_i \leq 2$, so $\dim \mathrm{span}(S)$ is between $r$ and $2$ inclusive.
Hopefully, this is a generalization you're satisfied with even if in general, it gives a huge range.
Thanks. Your bound is correct but your proof is not. The set S=S_1=S_2=S_3={e_1} contradicts your second equality.
Ah, I see the error now. So right now this only works if every triple intersection is the empty set. When I'm at my computer I'll try and fix the proof, thanks!
|
2025-03-21T14:48:29.974166
| 2020-02-25T00:13:48 |
353484
|
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"Anton Petrunin",
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|
Stack Exchange
|
On some characteristics of continuous maps $S^n \to \mathbb{R}^n$
I've asked this question about two month ago in math exchange but there were no answer to it.
Any information or paper relating to this question is appreciated.
By the Borsuk-Ulam theorem we know that every continuous map $M:S^n\to \mathbb{R}^n$ has a point in its range that is the image of two points in the domain. By considering this, I am curious to see what amount of this double points would exist for such maps and what we can say about the set of points in the range which are the image of one or odd number of points in the domain for all maps.
It is known that for every continuous non-one element range map $f_1:S^1\to \mathbb{R}$ there are properties which state that for an even $e\in \mathbb{N}$ the set of points in its range which their preimage have exactly $e$ points is uncountable and for evey odd $o\in \mathbb{N}$ the $o$-point set in the codomain $\mathbb{R}$ has countable elements.
Also intuition maybe says something similar but more complex about two dimensional continuous maps $f_2:S^2\to \mathbb{R}^2$, where the dimension of $f_2(S^2)$ is $2$ (the same as codomain $\mathbb{R}^2$ dimension).
It states that the $o$-point sets in $\mathbb{R}^2$ where $o\in \mathbb{N}$ is odd have one dimensional loop structure and for at least one even $e\in \mathbb{N}$ the $e$-point set has a two dimensional disc structure for each of its connected components.
Questions:
First: can anyone present more rigorous statements about the latter case above? (with proof)
Second: In general does there any invariant of the continuous maps relate to aggregation of characteristics info (structure, size, dimension ,...) of $n$-point ($\forall n\in \mathbb{N}$) and uncount-point sets of the map?
Note: we define $n$-point set for a map $f:X\to Y$ is the set of all points such $p\in Y$ that the preimage $f^{-1}(\{p\})$ consists of exactly $n$ points in $X$, if the preimage $f^{-1}(\{p\})$ consist of uncountable points in the domain then we call it a uncount-point.
For a typical continuous map $f\colon\mathbb{S}^1\to \mathbb{R}$, the preimage of $f^{-1}{x}$ typical point $x$ is either empty or a Cantor set. Doesn't it contradict your statement? (I was not able to see its precise meaning)
@AntonPetrunin sorry I can't understand of what you call "typical".
@AntonPetrunin I see it now ,good question, I think what you mentioned may add additional info about uncount-point set for a typical continuous function that should be considered from the point of question 2.
|
2025-03-21T14:48:29.974363
| 2020-02-25T01:17:08 |
353486
|
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"David Roberts",
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|
Stack Exchange
|
What other lattices are obtainable from this noncommutative ring?
Here I will regard $SU(2)$ as the multiplicative group of unit quaternions.
There are just three conjugacy classes of finite subgroups $G < SU(2)$ where $[G:C] > 2$ for all cyclic subgroups $C < G$:
(i) If $|G| = 24$, $G \cong SL_{2}(3)$. This $G$ has a normal subgroup of order 8 which can be taken to be $Q_{8} = \{ \pm{1}, \pm{i}, \pm{j}, \pm{k} \}$. The elements of $G$ outside this subgroup of index 3 can be taken to be $\{ \pm{\frac{1}{2}}\pm{\frac{i}{2}}\pm{\frac{j}{2}}\pm{\frac{k}{2}} \}$, where the signs are taken every possible way.
(ii) If $|G| = 120$, $G \cong SL_{2}(5)$. The elements of this group can be described in terms of $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$, but I won't go into that here.
(iii) If $|G| = 48$, $G$ is a double cover of $S_{4}$ sometimes called the binary octahedral group. This $G$ has a subgroup of order 24 which can be taken to consist of the elements described in (i). The extra elements of $G$ can be taken to be $\frac{u+v}{\sqrt{2}}$, where $u$ and $v$ are non-proportional elements of $Q_{8}$.
The Hurwitz ring of integral quaternions is the ring consisting of finite sums of elements of the group described in (i). Wilson, in The Finite Simple Groups, calls this ring the tetrian ring.
The ring consisting of finite sums of elements of (ii) is sometimes called the icosian ring. Wilson describes how to describe the Leech lattice as a 3-dimensional lattice over this ring, and SPLAG describes how to identify this ring with the $E_{8}$ lattice.
The ring I want to ask about here is the ring consisting of finite sums of elements of the group in (iii), which I've called the octian ring. Here, I'll use $\mathbb{O}$ to denote the octian ring (also used elsewhere for the octonions, but all rings discussed here are associative, so there's no danger of confusion). If I have computed correctly, it can be identified with the $E_{8}$ lattice as follows:
For the purpose of this construction, we define the Euclidean norm of an element of $\mathbb{Q}(\sqrt{2})$ by $EN(u+v\sqrt{2}) = u+v$, where $u,v \in \mathbb{Q}$. The quaternion norm is the usual $QN(a + bi + cj + dk) = a^{2}+b^{2}+c^{2}+d^{2}$, where $a,b,c,d \in \mathbb{R}$. In fact, the quaternion norm of an element of $\mathbb{O}$ is always in $\mathbb{Z}[\sqrt{2}]$, but establishing that takes a little more work.
For $x,y \in \mathbb{O}$, the inner product that identifies $\mathbb{O}$ with the $E_{8}$ lattice is
$$\langle x,y\rangle = EN(QN(x+y)) - EN(QN(x)) - EN(QN(y)).$$
An integral basis for $\mathbb{O}$ can be given by:
$$\begin{array}{ccl}
v_{1} &=& \frac{1}{2}+\frac{i}{2}+\frac{j}{2}+\frac{k}{2} \\
v_{2} &=& \frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} \\
v_{3} &=&i \\
v_{4} &=&\frac{i}{\sqrt{2}}+\frac{j}{\sqrt{2}} \\
v_{5} &=&j \\
v_{6} &=&\frac{j}{\sqrt{2}}+\frac{k}{\sqrt{2}} \\
v_{7} &=&k \\
v_{8} &=&k\sqrt{2}.
\end{array}$$
With respect to this basis, the Gram matrix of the lattice with which $\mathbb{O}$ becomes identified is (if I computed correctly)
$$\begin{pmatrix} 2 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 2 & 1 & 1 & 0 & 0 & 0 & 0 \\
1 & 1 & 2 & 1 & 0 & 0 & 0 & 0 \\
1 & 1 & 1 & 2 & 1 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 2 & 1 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 2 & 1 & 2 \\
1 & 0 & 0 & 0 & 0 & 1 & 2 & 2 \\
1 & 0 & 0 & 0 & 0 & 2 & 2 & 4 \end{pmatrix}$$
To verify that this identifies $\mathbb{O}$ with the $E_{8}$ lattice, it suffices to verify that the lattice thus obtained is positive definite, even, and unimodular.
Positive definiteness: Any element of $\mathbb{O}$ has a quaternion norm which is a sum of 4 squares of elements of $\mathbb{Q}(\sqrt{2})$. The Euclidean norm of $(s+t\sqrt{2})^{2}$ is $s^{2}+2st+2t^{2}$, which is positive for $s,t \in \mathbb{Q}$ except when $s=t=0$. Therefore any nonzero element of $\mathbb{O}$ has positive inner product with itself, as claimed.
Evenness: All of the entries of the Gram matrix are integers, and all of the diagonal entries are even. This is sufficient to verify evenness.
A proof which does not depend on the computation of the Gram matrix notes that $<x,x> = EN(QN(2x)) - 2EN(QN(x)) = 2EN(QN(x))$. Since $QN(x) \in \mathbb{Z}[\sqrt{2}]$ for all $x \in \mathbb{O}$, $EN(QN(x)) \in \mathbb{Z}$ and evenness follows.
Unimodularity: If I entered this correctly into a determinant calculator, the determinant of this Gram matrix is 1.
A proof which doesn't depend on calculation of an 8x8 Gram matrix determinant can be given as follows:
First, consider the sublattice spanned by $\{ 1, \sqrt{2}, i, i\sqrt{2}, j, j\sqrt{2}, k, k\sqrt{2} \}$.
The planes spanned by $\{ 1, \sqrt{2} \}$, $\{ i, i\sqrt{2} \}$, $\{ j, j\sqrt{2} \}$, and $\{ k, k\sqrt{2} \}$ are mutually perpendicular, so the volume of a fundamental region of the sublattice we are considering is the product of the areas of the fundamental regions of the indicated 2-dimensional sublattices within those planes.
It is an easier computation that the Gram matrix of any of those 2-dimensional sublattices is
$$\begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix}$$
The determinant of that Gram matrix is 4, which means the area of a fundamental region of any of those 2-dimensional sublattices is 2. Then the volume of a fundamental region of that 8-dimensional sublattice is $2^{4}=16$.
$\mathbb{O}$ is obtained by extending that sublattice by sums of evenly many of $\{ \frac{1}{\sqrt{2}}, \frac{i}{\sqrt{2}}, \frac{j}{\sqrt{2}}, \frac{k}{\sqrt{2}} \}$, and also by $\frac{1}{2} + \frac{i}{2} + \frac{j}{2} + \frac{k}{2} $. These reduce the volume of a fundamental region by a factor of $2^{3} \cdot 2 = 16$, so the final volume of a fundamental region is 1.
I hope the computation I detailed above isn't erroneous in a serious way, but I also want to ask:
What other discrete additive subgroups of $\mathbb{R}^{8n}$ can be identified (up to congruence) with free modules over $\mathbb{O}$?
Addendum: The minimal vectors of $E_{8}$
I want to describe how the minimal vectors of $E_{8}$ can be seen in this
construction.
Recall that an algebraic number field is totally real if every embedding of it into $\mathbb{C}$ lands in $\mathbb{R}$. Within a totally real field, an element may be called totally positive if it and all its Galois conjugates are positive real numbers. Any nonzero square, or nonzero sum of squares, of elements of a totally real field will be totally positive.
$\mathbb{Q}(\sqrt{2})$ is a totally real field. Therefore quaternion norms of nonzero elements of $\mathbb{O}$ are totally positive elements of $\mathbb{Z}[\sqrt{2}]$. The minimal vectors of the lattice obtained this way from $\mathbb{O}$ are the $x$ such that $2 = \langle x,x\rangle = EN(QN(2x)) - 2EN(QN(x)) = 2EN(QN(x))$, so they come from $\alpha \in \mathbb{Z}[\sqrt{2}]$ such that $\alpha$ is totally positive, and $EN(\alpha) = 1$. There are just three such $\alpha$: $1$, $2-\sqrt{2}$, and $3-2\sqrt{2}$.
Quaternion norm $1$:
These are just the unit quaternions in $\mathbb{O}$, which form the binary octahedral group. There are 48 of these.
Quaternion norm $2-\sqrt{2}$:
Some of these have the form $(1-\frac{1}{\sqrt{2}})u + \frac{v}{\sqrt{2}}$, where $u,v \in Q_{8}$ are non-proportional elements as before. There are $4 \cdot 3 \cdot 2^{2} = 48$ of these. The others are the images of $1-\frac{1}{\sqrt{2}} +(1-\frac{1}{\sqrt{2}})i + \frac{j}{2} + \frac{k}{2}$ under the signed permutations of $\{ 1,i,j,k \}$. There are $\frac{4! \cdot 2^{4}}{2^{2}} = 96$ of these.
Quaternion norm $3-2\sqrt{2}$:
Some of these are of the form $(1-\frac{1}{\sqrt{2}})u + (1-\frac{1}{\sqrt{2}})v$, where again $u$ and $v$ are non-proportional elements of $Q_{8}$. There are $\binom{4}{2}\cdot 2^{2} = 24$ of these. Others are of the form $\pm (1-\frac{1}{\sqrt{2}}) \pm (1-\frac{1}{\sqrt{2}})i \pm (1-\frac{1}{\sqrt{2}})j \pm (1-\frac{1}{\sqrt{2}})k$. Since the ambiguous signs can be chosen every possible way, there are $2^{4} = 16$ of these. The last ones are the set $(\sqrt{2}-1)Q_{8}$. There are $8$ of these.
(More generally, all of the minimal vectors whose quaternion norm is $3-2\sqrt{2}$ are elements of the binary octahedral group, scaled by a factor of $\sqrt{2}-1$. Possibly this should replace the three cases above.)
Taken together, these account for $48+48+96+24+16+8=240$ vectors, which is the full count of minimal vectors of the $E_{8}$ lattice.
Should the Euclidean norm of an element of $\mathbb{Q}(\sqrt{2})$ be $|u|+|v|$, or $u^2 + v^2$? I only ask, since otherwise it's not non-negative or non-degenerate.
I don't know what kind of answer you're looking for, since you have characterized the desired class of lattices $L$ pretty well already as those containing the designated quaternion order inside their endomorphism ring $\mathrm{End}(L)$.
To recover many of our favorite lattices over (possibly noncommutative) rings along the lines that you have suggested for $E_8$, see Martinet, Perfect lattices in Euclidean spaces; chapter 8 considers "Hermitian structures" on lattices (including the noncommutative case), and section 8.2 recovers $E_8$.
Perhaps another way to think about this is to think about things just in terms of automorphisms? Are you looking for a characterization in terms of the integral representation theory of the binary octahedral group $G=2O$? In 11.5.4 and Proposition 32.7.1(b) of my book (http://quatalg.org), what you are asking for is pretty close to asking for $L$ to be a $\mathbb{Z}[\sqrt{2}][G]$-module in a specified way?
I get a Bad Gateway from that link. This one works for me https://math.dartmouth.edu/~jvoight/quat.html
I wrote an expository post asking people to check this construction here:
From the octahedron to E8.
So far one person has checked it with the help of a computer.
In your blog post, you asked if this had been known/done before. This paper from 1999 uses the terms "octian" and "octian ring" (so the naming is consistent):
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjG17v30pHrAhUERK0KHdejCsEQFjACegQIARAB&url=https%3A%2F%2Fwww.sciencedirect.com%2Fscience%2Farticle%2Fpii%2FS002437959900107X%2Fpdf%3Fmd5%3Da4eff433e5201cc74d335f4ff4b2ca92%26pid%3D1-s2.0-S002437959900107X-main.pdf%26_valck%3D1&usg=AOvVaw2WzOuVuAFy4s_tom9voi_5
But these authors' main concern is hyperbolic 4- and 5-space, not Euclidean 8-space.
A post in which John Baez cites Greg Egan
describes how to obtain the Leech lattice from the $E_{8}$ lattice using an orientation-preserving rigid motion $R: \mathbb{R}^{8} \to \mathbb{R}^{8}$ such that:
$< r_{i} , \sqrt{2}R r_{i} > = < r_{i} , \sqrt{2}R^{-1} r_{i} > = 1$
for all simple roots $r_{i}$ of the $E_{8}$ lattice.
(This depends on the choice of a set of simple roots among the minimal vectors of the lattice, but the reasoning that follows, which applies to all minimal vectors, obviates the need for such a choice.)
If we wish to describe such a motion in octian terms, then it's good to note the following:
Lemma. Let $\beta \in \mathbb{Q}\otimes\mathbb{O}$ be such that $QN(\beta)=1$. Then the map $x \to \beta \cdot x$ is a rigid motion from $\mathbb{Q}\otimes\mathbb{O}$ to itself, in the sense defined by the octian inner product.
Proof. $< \beta\cdot u, \beta\cdot v> = EN(QN(\beta\cdot u + \beta\cdot v)) - EN(QN(\beta \cdot u)) - EN(QN(\beta \cdot v))=\
EN( QN(\beta)QN(u+v) ) - EN(QN(\beta)QN(u)) - EN(QN(\beta)QN(v)) =\ EN(QN(u+v)) - EN(QN(u)) - EN(QN(v)) = <u,v>$. (The same argument applies to right-multiplications by a fixed unit quaternion in the skew-field, so it's omitted.)
One possible $\beta$ that allows the Baez-Egan construction to be applied here is
$\beta = \frac{1}{\sqrt{8}} + i(\frac{1}{2} + \frac{1}{\sqrt{8}}) + j(\frac{1}{2} - \frac{1}{\sqrt{8}}) + \frac{k}{\sqrt{8}}$, for which we have
$\beta^{-1} = \frac{1}{\sqrt{8}} - i(\frac{1}{2} + \frac{1}{\sqrt{8}}) - j(\frac{1}{2} - \frac{1}{\sqrt{8}}) - \frac{k}{\sqrt{8}}$.
Before proceeding further, note that $\sqrt{2}\beta \in \mathbb{O}$, so $x \to \sqrt{2}\beta \cdot x$ carries $\mathbb{O}$ into itself.
As for showing that the inner products work out as required by Baez and Egan, we show that $< \sqrt{2}\beta u, u > = 1$ for any $u \in \mathbb{O}$ such that $< u,u > = 2$ (or, equivalently, $EN(QN(u)) = 1$):
$<\sqrt{2}\beta u, u> = -< \sqrt{2}\beta u, -u> = -[ EN(QN( \sqrt{2}\beta u- u )) - EN(QN( \sqrt{2}\beta u )) - EN(QN( -u )) ] =\
-EN(QN( \sqrt{2}\beta u - u )) + EN(QN( \sqrt{2}\beta u )) + EN(QN(u)) =\ -EN(QN( \sqrt{2}\beta - 1)QN(u)) + EN(QN( \sqrt{2}\beta )QN(u)) + EN(QN(u))$.
Note that $Re(\sqrt{2}\beta) = \frac{1}{2}$, so it's easy to see that $QN( \sqrt{2}\beta ) = QN( \sqrt{2}\beta - 1 )$.
Then the preceding expression for $< \sqrt{2}\beta u, u >$ simplifies to
$EN(QN(u)) = 1$, and we are done.
It remains to show that $x \to \beta \cdot x$ is orientation-preserving, as a rigid motion of $\mathbb{Q}\otimes\mathbb{O}$.
An easy way to see it is this:
If $\beta$ were orientation-reversing, its $-1$-eigenspace would be odd-dimensional and therefore remain nonempty when the zero vector is removed. A nonzero unit vector in this eigenspace could be approximated arbitrarily well by a unit (in the $QN$ sense) element of $\mathbb{Q}\otimes\mathbb{O}$. Then we would have $< \beta u, u >$ being negative for some $u \in \mathbb{Q}\otimes\mathbb{O}$ with $< u, u> = 1$.
But actually $< \beta u, u > = < \beta u u^{-1}, u u^{-1} > = < \beta, 1 > = 2 Re(\beta) = \frac{1}{\sqrt{2}} > 0$, for a contradiction.
|
2025-03-21T14:48:29.975209
| 2020-02-25T06:16:49 |
353492
|
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|
Stack Exchange
|
The status of the journal “Forum Geometricorum”
The online journal Forum Geometricorum is a sort of central organ of elementary geometry (mainly triangle geometry and related topics). It has been published regularly since 2000 but seems to have trailed off about a year ago, without any announcement about its future. Can anybody on this forum provide relevant information?
(I am not sure whether this is an appropriate query for MO. If not, feel free to delete. I tried mailing the editors but never received a reply).
FWIW, it seems that Paul Yiu has retired from FAU.
Thanks You. I like the journal, but maybe Professor Paul Yiu's health is not good
My guess would be that at least one of these people would be able to give a better answer: http://forumgeom.fau.edu/editors.html
I guessed this too and acted on it without receiving a response. I had hoped to preempt this kind of reaction by mentioning this fact in my post.
The wrecking crew has been at work again— the added tag „metric geometry“ has less than nothing to do with my post.
@user131781 As you can see from the tag description, mg.metric-geometry is the top-level tag which covers Euclidean geometry.
"removed body since it has been interfered with" Huh?
rolled back --- please don't vandalize a post.
@user131781 If even the editors of the journal are not able to give you an answer, what makes you think anyone else will?
|
2025-03-21T14:48:29.975403
| 2020-02-25T06:55:25 |
353493
|
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Unexpected behavior involving √2 and parity
This post makes a focus on a very specific part of that long post. Consider the following map:
$$f: n \mapsto \left\{
\begin{array}{ll}
\left \lfloor{n/\sqrt{2}} \right \rfloor & \text{ if } n \text{ even,} \\
\left \lfloor{n\sqrt{2}} \right \rfloor & \text{ if } n \text{ odd.}
\end{array}
\right.$$
Let $f^{\circ (r+1)}:=f \circ f^{\circ r}$, consider the orbit of $n=73$ for iterations of $f$, i.e. the sequence $f^{\circ r}(73)$: $$73, 103, 145, 205, 289, 408, 288, 203, 287, 405, 572, 404, 285, 403, 569, 804, 568, 401, \dots$$
It seems that this sequence diverges to infinity exponentially, and in particular, never reaches a cycle. Let illustrate that with the following picture of $(f^{\circ r}(73))^{1/r}$, with $200<r<20000$:
According to the above picture, it seems that $f^{\circ r}(73) \sim \delta^r$ with $\delta \sim 1.02$.
Now consider the probability of the $m$ first terms of the sequence $f^{\circ r}(73)$ to be even: $$p_{0}(m):= \frac{|\{ r<m \ | \ f^{\circ r}(73) \text{ is even}\}|}{m}.$$
Then $p_1(m):=1-p_0(m)$ is the probability of the $m$ first terms of $f^{\circ r}(73)$ to be odd.
If we compute the values of $p_i(m)$ for $m=10^{\ell}$, $\ell=1,\dots, 5$, we get something unexpected:
$$\scriptsize{ \begin{array}{c|c}
\ell & p_0(10^{\ell}) &p_1(10^{\ell}) \newline \hline
1 &0.2&0.8 \newline \hline
2 &0.45&0.55 \newline \hline
3 &0.467&0.533 \newline \hline
4 &0.4700&0.5300 \newline \hline
5 &0.46410&0.53590 \newline \hline
6 & 0.465476& 0.534524
\end{array} }$$ (the line for $\ell = 6$ was computed by Gottfried Helms, see the comments)
It is unexpected because it seems that $p_0(m)$ does not converge to $1/2$, but to $\alpha \sim 0.465$.
It matches with the above observation because $$ \delta \sim 1.02 \sim \sqrt{2}^{(0.535-0.465)} = \sqrt{2}^{(1-2 \times 0.465)} \sim \sqrt{2}^{(1-2\alpha)}.$$
Question: Is it true that $f^{\circ r}(73)$ never reach a cycle, that $(f^{\circ r}(73))^{1/r}$ converges to $\delta \sim 1.02$, that $p_0(m)$ converges to $\alpha \sim 0.465$, and that $\delta^24^{\alpha} = 2$? What are the exact values of $\delta$ and $\alpha$? (or better approximations?)
The following picture provides the values of $p_0(m)$ for $100 < m < 20000$:
Note that this phenomenon is not specific to $n=73$, but seems to happen as frequently as $n$ is big, and then, the analogous probability seems to converge to the same $\alpha$. If $n <100$, then it happens for $n=73$ only, but for $n<200$, it happens for $n=73, 103, 104, 105, 107, 141, 145, 146, 147, $ $ 148, 149, 151, 152, 153, 155, 161, 175, 199$; and for $10000 \le n < 11000$, to exactly $954$ ones.
Below is the picture as above but for $n=123456789$:
Alternative question: Is it true that the set of $n$ for which the above phenomenon happens has natural density one? Is it cofinite? When it happens, does it involves the same constant $\alpha$?
There are exactly $1535$ numbers $n<10000$ for which the above phenomenon does not happen. The next picture displays for such $n$ the minimal $m$ (in blue) such that $f^{\circ m}(n) = f^{\circ (m+r)}(n)$ for some $r>0$, together with the miniman such $r$ (in red):
In fact all these numbers (as first terms) reach the following cycle of length $33$:
$$(15,21,29,41,57,80,56,39,55,77,108,76,53,74,52,36,25,35,49,69,97,137,193,272,192,135,190,134,94,66,46,32,22)$$
except the following ones: $$7, 8, 9, 10, 12, 13, 14, 18, 19, 20, 26, 27, 28, 38, 40, 54,$$ which reach $(5,7,9,12,8)$, and that ones $1, 2, 3, 4, 6$ which reach $(1)$, and $f(0)=0$.
If the pattern continues like above up to infinity, they must have infinity many such $n$.
Bonus question: Are there infinitely many $n$ reaching a cycle? Do they all reach the above cycle of length $33$ (except the
few ones mentioned above)? What is the formula of these numbers $n$?
Below is their counting function (it looks logarithmic):
Fascinating! The pattern for the $n$ with periodicity is of course close to self-similar with a ratio of $\sqrt2$, as might be expected. Just out of curiosity: could you add a graphic which displays for each of those $n$ the minimal period $r$?
Out of curiosity: the mapping $f$ looks like it would be expected 'on average' to be slightly contracting, since there are floor operations in the definitions of both the even and odd branches, and without those floor operations one would expect $f()$ to behave like a random process with (geometric) mean equal to its initial value. Have you looked for cycles in the iterates of $f()$ starting from 73, and is it possible that most of the numbers are falling into the same cycle? That would explain the commonality in the limit...
@StevenStadnicki No cycle found starting from $73$. There is a cycle of length $33$ starting from $15$. There are additional details in the long post cited at the beginning. You are right in the sense that it would be more precise to first ask whether the process never reach a cycle from $73$. Next for those never reaching a cycle, whether that probability converges to that common $\alpha$. What I have in mind when I write ‘phenomenon’ is precisely ‘never reaching a cycle’. If it reaches a cycle then the limit can also be not $1/2$ (and should) but must be closer to $1/2$ than when no cycle.
I see; thank you for the pointer. I hadn't looked at the previous post, apologies. You don't really talk about process in either post, though — do you have your source available for these explorations? How many iterations did you go on $n=73$, and what was the maximum magnitude of $f^{\circ r}(n)$ you found over those iterations? Do you have any plots of, e.g. $\log(f^{\circ r}(73))$ for $r=1..10^k$ for some $k$? It would be good to see what the iterate trajectory looks like.
Perhaps I made a mistake, but I get a different value of $p_0(10^4)$ (0.4912 instead of 0.4700). For larger $m$ the value seems even closer to 1/2. As a control the computation for $m=10^4$ has a maximum of $f^{\circ k}(73)$ equal to $1341801280048839911857201274496$ (for $k=2584$), so roundoff errors when computing $n\cdot \sqrt{2}^{\pm 1}$ could have occured in your computation. Could some confirm the above?
@Kasper Andersen The problem seems to be the floating point arithmetic you use. I calculated (with python and the module mpmath) $p_0(10^4)$ and $p_0(10^5)$ with a precision of 100,1000,10000 decimal places and got for 10000 decimal places the same values as in the table. But for fewer decimal places the values were different, f.i. with double precision the values seemed to converge to $0.5$.
@Sebastien Palcoux Can you tell which arithmetic you use for your calculations? I suspect that your results are not artefacts resulting from imprecise arithmetic, the results are "too good".
@KasperAndersen For f[n_]:=Floor[2^Mod[n,2] n/Sqrt[2]] on Mathematica I get for $f^{\circ2584}(73)$ the value$$670900640024419955928600637257.$$This is almost half of your value but not quite:$$1341801280048839911857201274496/2=670900640024419955928600637248$$
@მამუკა ჯიბლაძე : Sorry for the confusion. I though I was using Magma for the computation with quite large precision, unfortunately the way to compute $\sqrt{2}$ is not RealField(d)!Sqrt(2) but Sqrt(RealField(d)!2) (d is the precision). So now my results are in happy agreement with the table and also with your result for $f^{\circ 2584}(73)$.
@Wolfgang: Done!
Can one somehow count to how many orbits do numbers up to $n$ belong? For example, you say that you only found four orbits that end in a cycle (or loop).
Also $73,103,104,105,141,145,146,147,148,149,199$ are all on the same orbit; and $107,151,152,153,155$ are on the same orbit too. Is it clear that these are not all on a single orbit? Is it even clear whether there are infinitely many orbits?
@მამუკაჯიბლაძე: I did not say that I found only four orbits reaching a cycle, I said that I found four cycles, and that there may have no other. Now this depends on what you mean by orbit... anyway, it seems relevant to ask whether the partition associated to the equivalence relation generated by $f(n)=m$, admits finitely many components.
Yes thanks that's what I mean. Clearly any distinct cycles belong to separate equivalence classes in your sense; thus so far you established that there are at least four distinct equivalence classes. But if I understand correctly, so far it is not even clear whether there are any other equivalence classes?
@მამუკაჯიბლაძე: Four distinct equivalence classes reaching cycles, and at least one which does no. The question is : are all the numbers non-reaching a cycle, in the same class? Very interesting!
Thanks. The "miniman" 5-cycle should probably be 12,8,5,7,9.
@Wolfgang Yes, fixed!
Actually $73$ and $107$ are in the same equivalence class: $f^{\circ57}(107)=f(f(73))$
@მამუკაჯიბლაძე: I tried $73$ and $123456789$, but up to the 10000th iteration they share no point. If we assume that for a subset of numbers $n$ of natural density one, the sequence $f^{\circ r}(n)$ has exponential grows (expected $\sim 1.02^r$), then I think that an argument of density can prove that there must have infinitely many equivalence classes.
Going in a different direction: If you take disjoint intervals, say $I_r:=[5\sqrt2^{,r},5\sqrt2^{,r+1}]$, then the "point clouds" in your diagrams of the $n$'s reaching a cycle should be entirely inside each $I_r$. If you count those points in each $I_r$, how do their counts grow? I guess like some $\alpha^r$, but with $\alpha$ much smaller than $\sqrt2$.
@Wolfgang: This $5$ comes from where? Should it be exact or an approximation? Here is the counting your asked: [6, 1, 2, 3, 5, 7, 11, 15, 22, 28, 36, 42, 50, 61, 74, 90, 105, 122, 137, 153, 168, 182, 193] together with the corresponding approximation of your $\alpha$ (different from what is called $\alpha$ in the post):[6.00, 1.00, 1.26, 1.32, 1.38, 1.38, 1.41, 1.40, 1.41, 1.40, 1.39, 1.37, 1.35, 1.34, 1.33, 1.32, 1.31, 1.31, 1.30, 1.29, 1.28, 1.27, 1.26].
Thank you! $5$ is just an approximation. Oh yes, "$\alpha$" was a bit unfortunate. Say $\beta$ then, and rather $\beta^{r+3}$ than $\beta^{r}$, as $5\approx\sqrt2^{,3}$ (always expecting such things to start "at the origin"...). But that should make hardly a difference, of course. The choice of my intervals (with "$5$") was just to exclude "boundary effects", as those points come already in clouds, so why not try to capture those clouds inside the intervals.
Now I see that this $\beta$ is after all not that much smaller than $\sqrt2$, but it does seem to slowly decrease. Till where??
BTW I guess I thought of those intervals because of the
"self similar sequences" that came to my mind when seeing your images.
Using Pari/GP and setting internal precision to 15000 decimal digits I could get $p_1(10^6) = 0.534524$ The value of $f°^{1e6}(73)$ is about 3.89439e10394, (thus one needs such a big precision) and the $\log_{73}()$ of it is 5578.52.
@GottfriedHelms: Thanks! My laptop was not able to get $ℓ=6$. The computation for general $\ell$ should require a precision of about $1.5 \times 10^{\ell-2}$ decimal digits.
Are there cases, which have no precedessor? For instance, $a_1=300$ traverses to $73$ by $6$ steps - but I didn't find a number $a_0$ which goes to $a_1=300$ . If there is indeed no such number - can we characterize the type of numbers which have no precedessor?
@მამუკაჯიბლაძე: you may be interested in the answer I just posted.
@GottfriedHelms: you may be interested in the answer I just posted.
Here is a heuristic answer inspired by this comment of Lucia.
First, let assume that the probabilty for an integer $n$ to be odd is $\frac{1}{2}$, and that the probabilty for $f(n)$ to be odd when $n$ is even (resp. odd) is also $\frac{1}{2}$. We will observe that (surprisingly) it is no more $\frac{1}{2}$ for $f^{\circ r}(n)$ when $r \ge 2$ (in some sense, the probability does not commute with the composition of $f$ with itself).
if $n$ and $m=f(n)$ are even: note that $\frac{n}{\sqrt{2}} = m+\theta$ (with $0 < \theta < 1$) so that $m=\frac{n}{\sqrt{2}}- \theta$, then $$f^{\circ 2}(n) = f(m) = \left \lfloor{\frac{m}{\sqrt{2}}} \right \rfloor = \left \lfloor{\frac{\frac{n}{\sqrt{2}}- \theta}{\sqrt{2}}} \right \rfloor = \left \lfloor \frac{n}{2} - \frac{\theta}{\sqrt{2}}\right \rfloor$$ but $\frac{n}{2}$ is even with probability $\frac{1}{2}$, so in this case, $f^{\circ 2}(n)$ is odd with probability $\frac{1}{2}$.
if $n$ is even and $m=f(n)$ is odd: $$f^{\circ 2}(n) = f(m) = \left \lfloor\sqrt{2}m \right \rfloor = \left \lfloor \sqrt{2}(\frac{n}{\sqrt{2}} - \theta) \right \rfloor = \left \lfloor n - \sqrt{2} \theta) \right \rfloor$$ but $n$ is even and the probability for $0<\sqrt{2} \theta<1$ is $\frac{\sqrt{2}}{2}$ (because $\theta$ is assumed statistically equidistributed on the open interval $(0,1)$), so $f^{\circ 2}(n)$ is odd with probability
$\frac{\sqrt{2}}{2}$.
if $n$ is odd and $m=f(n)$ is even:
$$f^{\circ 2}(n) = f(m) = \left \lfloor{\frac{m}{\sqrt{2}}} \right \rfloor = \left \lfloor{\frac{\sqrt{2}n-\theta}{\sqrt{2}}} \right \rfloor = \left \lfloor n - \frac{\theta}{\sqrt{2}} \right \rfloor $$
but $n$ is odd and $0 < \frac{\theta}{\sqrt{2}}<1$, so $f^{\circ 2}(n)$ is even.
if $n$ is odd and $m=f(n)$ is odd:
$$f^{\circ 2}(n) = f(m) = \left \lfloor \sqrt{2} m \right \rfloor = \left \lfloor \sqrt{2} (\sqrt{2}n-\theta) \right \rfloor = \left \lfloor 2n - \sqrt{2} \theta \right \rfloor $$
but $2n$ is even and the probability for $0<\sqrt{2} \theta<1$ is $\frac{\sqrt{2}}{2}$, so $f^{\circ 2}(n)$ is odd with probability $\frac{\sqrt{2}}{2}$.
By combining these four cases together, we deduce that the probability for $f^{\circ 2}(n)$ to be odd is $$\frac{1}{2} \times \frac{1}{2} \times (\frac{1}{2} + \frac{\sqrt{2}}{2} + 0 + \frac{\sqrt{2}}{2}) = \frac{2\sqrt{2}+1}{8}$$
By continuing in the same way, we get that the probability for $f^{\circ 3}(n)$ to be odd is:
$$ \frac{1}{4} (\frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2} + 1\frac{1}{2} + \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}+7}{16}$$
For $2 \le r \le 24$, we computed the probability $p_r$ for $f^{\circ r}(n)$ to be odd (see Appendix). It seems (experimentally) that $p_r$ converges to a number $\simeq 0.532288725 \simeq \frac{8+3\sqrt{2}}{23}$ by Inverse Symbolic Calculator. This leads to the following question/conjecture:
$$\lim_{r \to \infty}p_r = \frac{8+3\sqrt{2}}{23} \ \ ?$$
If so, consider the number $\alpha$ mentioned in the main post, then $$\alpha = 1-\frac{8+3\sqrt{2}}{23} = \frac{15-3\sqrt{2}}{23} \simeq 0.467711,$$ which matches with the computation in the main post. And next, we would have:
$$ \delta = \frac{\sqrt{2}}{2^{\alpha}}= 2^{\frac{1}{2}-\alpha} = 2^{\frac{6\sqrt{2}-7}{46}} \simeq 1.022633$$
Appendix
Computation
sage: for i in range(3,26):
....: print(sq2(i))
....:
[1/4*sqrt(2) + 1/8, 0.478553390593274]
[1/16*sqrt(2) + 7/16, 0.525888347648318]
[3/32*sqrt(2) + 13/32, 0.538832521472478]
[15/64*sqrt(2) + 13/64, 0.534581303681194]
[5/128*sqrt(2) + 61/128, 0.531805217280199]
[39/256*sqrt(2) + 81/256, 0.531852847392776]
[93/512*sqrt(2) + 141/512, 0.532269260352925]
[51/1024*sqrt(2) + 473/1024, 0.532348527032254]
[377/2048*sqrt(2) + 557/2048, 0.532303961432938]
[551/4096*sqrt(2) + 1401/4096, 0.532283123258685]
[653/8192*sqrt(2) + 3437/8192, 0.532285334012406]
[3083/16384*sqrt(2) + 4361/16384, 0.532288843554459]
[3409/32768*sqrt(2) + 12621/32768, 0.532289246647030]
[7407/65536*sqrt(2) + 24409/65536, 0.532288816169701]
[22805/131072*sqrt(2) + 37517/131072, 0.532288667983386]
[24307/262144*sqrt(2) + 105161/262144, 0.532288700334941]
[72761/524288*sqrt(2) + 176173/524288, 0.532288728736551]
[159959/1048576*sqrt(2) + 331929/1048576, 0.532288729880941]
[202621/2097152*sqrt(2) + 829741/2097152, 0.532288725958633]
[639131/4194304*sqrt(2) + 1328713/4194304, 0.532288724978704]
[1114081/8388608*sqrt(2) + 2889613/8388608, 0.532288725350163]
[1825983/16777216*sqrt(2) + 6347993/16777216, 0.532288725570602]
[5183461/33554432*sqrt(2) + 10530125/33554432, 0.532288725561857]
Code
def sq2(n):
c=0
for i in range(2^n):
l=list(Integer(i).digits(base=2,padto=n))
if l[-1]==1:
cc=1/4
for j in range(n-2):
ll=[l[j],l[j+1],l[j+2]]
if ll==[0,0,0]:
cc*=1/2
if ll==[0,0,1]:
cc*=1/2
if ll==[0,1,0]:
cc*=(1-sqrt(2)/2)
if ll==[0,1,1]:
cc*=sqrt(2)/2
if ll==[1,0,0]:
cc*=1
if ll==[1,0,1]:
cc=0
break
if ll==[1,1,0]:
cc*=(1-sqrt(2)/2)
if ll==[1,1,1]:
cc*=sqrt(2)/2
c+=cc
return [c.expand(),c.n()]
You've had a really good idea! It seems, the problem is resolved? I like the days when I've shared the intense explorations here. Thank you for the "ping" :-)
@GottfriedHelms: it is resolved heuristically only.
A list of predecessors as mentioned in my comment.
I document pairs of $(m,n)$ for consecutive $m$ and their 1-step predecessors $n$ such that $f(n)=m$. The value $n=0$ indicates, that $m$ has no predecessor. I didn't reflect, that one $m$ can have two predecessors, but if $n/2$ is odd, then $n/2$ is a second predecessor.(This makes the table more interesting, because all odd predecessors $n$ are overwritten by the even predecessors $2n$...
Moreover, a nearly periodic structure occurs. I tried to resemble this by the arrangement of three or four columns of $(m,n)$ such that the first column contains all $m$ which have no predecessor. The basic pattern is not really periodic, but has super-patterns which again seem to be periodic but actually aren't. This pattern-superpattern-structure is also recursive. It reminds me of a similar structure when I looked at $\beta=\log_2(3)$ and found a similar style of pattern-superpattern-supersuperpattern-... and is there related to the continued fraction of $\beta$.
So I think we'll get no nice description for the cases $m$ which have no predecessor...
m n m n m n m n
-------------------------------------------------------------
1 2 2 4
3 0 4 6 5 8
6 0 7 10 8 12 9 14
10 0 11 16 12 18
13 0 14 20 15 22 16 24
17 0 18 26 19 28
20 0 21 30 22 32
23 0 24 34 25 36 26 38
27 0 28 40 29 42
30 0 31 44 32 46 33 48
34 0 35 50 36 52
37 0 38 54 39 56
40 0 41 58 42 60 43 62
44 0 45 64 46 66
47 0 48 68 49 70 50 72
51 0 52 74 53 76
54 0 55 78 56 80 57 82
58 0 59 84 60 86
61 0 62 88 63 90
64 0 65 92 66 94 67 96
Update Some more explanation on the idea of "recursive aperiodic pattern".
If we list the values $m$ which have no predecessor, we get
m_k: 3, 6,10,13, 17,20,23,27,30,...
Writing the differences (I have prepended a zero-value to the above list of $m_k$)
,3,3,4 ,3,4 ,3,3,4 ,3,4 ,3,3,4 ,3,4 ,3,4 ,3,3,4 , ...
We note, that we have a pattern of two different words: 3,3,4 and 3,4 repeating, but aperiodical. Let's denote the longer one with the capital A and the shorter one with the small a (and A means a difference of 10 and a of 7).
We get
Aa Aa Aaa
Aa Aaa
Aa Aa Aaa
Aa Aaa
Aa Aa Aaa
Aa Aaa
Aa ...
Again we find only two kind of "words". Let's them shorten by Aaa=B and Aa=b. B means now a difference of 24, b of 17.
Then we get
bbB bB
bbB bB
bbB bB bB
bbB bB
bbB bB bB
bbB bB
bbB bB
bbB bB bB
...
Next obvious step gives
Cc Cc Ccc
Cc Ccc
Cc Cc Ccc
Cc Ccc
Cc Cc Ccc
Cc Ccc
...
with c representing a difference of 17+24=41 and C of 17+17+24=58.
And so on.
If I recall correctly, then with the mentioned case of working with $\beta = \log_2(3)$ the same style of recursive pattern reflected the convergents of the continued fractions of $\beta$.
The first few differences here match the convergents of the continued fraction of $\sqrt2$ so far:
a b c short patterns
-------------------------------------
[1 1 3 7 17 41 99 239 577 ... ] convergents of contfrac(sqrt(2))
[0 1 2 5 12 29 70 169 408 ... ]
-------------------------------------...
A/2 B/2 C/2 long patterns
Update 2 The above can be explained by the following:
a number of the form $\lfloor2k\sqrt2\rfloor$ has exactly one predecessor $4k$;
a number of the form $\lfloor(2k-1)\sqrt2\rfloor$ has exactly two predecessors $2k-1$ and $4k-2$;
a number has no predecessors iff it has form $\lfloor n(2+\sqrt2)\rfloor$.
The first two statements are easily checked, while the third follows from the Beatty theorem, as explained in another answer by @Dattier
Update 3 Using a back-step algorithm (recursive) it seems I've got the predecessing tree of $m=73$. If no bugs, then this tree would also be complete. (But my routine may still be buggy, please check the results!)
The back-steps go from top-right south-west (antidiagonal) downwards. When there are two possible predecessors, they occur in the same column, but on separate rows.
If there is a predecessor without further predecessor, a short line (---) is printed.
73 <--- start
104
148
105 ---
210
149
212
300 ---
298
211 ---
422
299
424
600 ---
598
423 ---
846 ---
---------------------------- tree seems to be complete (please check for errors!)
Interesting! A number $m$ admits no predecessor iff the interval $[m\sqrt{2},(m+1)\sqrt{2}]$ admits no even number and the interval $[m/\sqrt{2},(m+1)/\sqrt{2}]$ admits no odd number. There are exactly $r_{\ell}$ such numbers $m<10^{\ell}$ with $\ell=1,2,3,4,5,6$ and $r_{\ell}=2,29,292,2928,29289,292893$. Strangely, for $\ell \le 6$ we observe that $r_{\ell-1} = ⌊ r_{\ell}/10⌋$, it should be true in general, I don't know why but it should be due to some pattern.
Numbers without predecessors are described in the answer by @Dattier as those of the form $\lfloor n(2+\sqrt2)\rfloor$
Moreover it seems that numbers with one predecessor are those of the form $\lfloor2k\sqrt2\rfloor$ and numbers with two predecessors those of the form $\lfloor(2k-1)\sqrt2\rfloor$
In fact, predecessor of $\lfloor2k\sqrt2\rfloor$ is $4k$ while predecessors of $\lfloor(2k-1)\sqrt2\rfloor$ are $2k-1$ and $4k-2$.
@მამუკაჯიბლაძე - very nice observations, indeed! For ease-of-reading: should I incorporate your results in my answer or even better won't you like to compose an own answer of yours?
I believe it is better to extend your answer, not to have too many partial answers. Note that relevant keywords here are Sturmian sequences and Beatty sequences (the latter mentioned in another answer)
@მამუკაჯიბლაძე - you might even incorporate this in my answer yourself. (I'm already a bit tired due to eyes-weakness)
Well I added it to the end although I am not sure how to do it better
@მამუკაჯიბლაძე - all well, thanks!
@მამუკაჯიბლაძე: the formula for the numbers without predecessor you pointed out from Dattier answer explains the cardinal I computed in the first comment as $1/(2+\sqrt{2}) = 1-\sqrt{2}/2 = 0.2928932188\cdots$
Do you expect that these investigations will lead to an answer of the question?
@Sebastien - primarily I try to get familiar with the whole system. My backtracer showed (immediately) that the predecessor-tree of the small cycle (at $5$) is finite - which is different compaed to the Collatz-type tree-structures. For the 3x-1 and the 5x+1 problem the predecessortrees of all cycles are always infinite. Q: has also the 33-step tree here only a finite precedessor-tree? The backtracing-algorithm might give numerical evidence if it is so. If all cycles must have finite predecessor-trees, are there more cycles? And so on... It's just exploratory so far.
... but, well, of course I don't want to spoil your track so far!
No problem, I appreciate your investigation, I just wanted to know whether you have expectation.
All branches going to the cycle $15,21,...,22,15$ are finite except, possibly, one, joining the cycle at $80=f(114)$. Successive $f$-inverse images of ${114}$ go like$${81,162},{115,230},{163,164,326},{231,232,462},{327,328,654},{463,464,926},{655,656,1310},{927,928,1854},{1311,1312,2622},{1856},{1313,2626},{929,1857,1858,3714},...$$
I confirm the calculation with the backtracking tree of $73$. The backtracking tree for $f(73)=103$ is finite too. However that for $f(f(73))=145$ seems to be infinite
Just in case - a quick algorithm for $f^{-1}(n)$: let $m=\lceil\frac n{\sqrt2}\rceil$. Then $f^{-1}(n)=\varnothing$ if $m\sqrt2\geqslant n+1$; otherwise it is ${2m}$ if $m$ is even and ${m,2m}$ if $m$ is odd.
80is the only entry of the 33-step cycle which has not short finiteteness in backtracing. Document: 80 (part of 33-step-cycle)
...<-- 114 <-- 162 <-- 230 <-- 326 <-- 462 <-- 654 <-- 926 <-- 1310 <-- 927 <-- 1312 <-- 1856 <-- 1313 <-- 929 <-- 657 : split into two branches of each infinite?
Branch 1 (657)<-- 465 <-- 329 <-- 233 <-- 165 <-- 234 <-- 332 <-- 470 <-- 333 <-- 472 <-- 668 <-- infinite ???
Branch 2 (657)<-- 930 <-- 1316 <-- 1862 <-- 2634 <-- 3726 <-- 5270 <-- 7454 <-- 10542 <-- 7455 <-- 10544 <-- infinite ???
Seems like the inverse tree for 114 has infinitely many infinite branches, as well as infinitely many finite ones
Hmm having calculated more now I am not so sure about it. Actually your second branch dies out in 52 steps after 10544. However the first branch further branches at 1449 into 1025 <-- 725 <-- 513 <-- 363 <-- 514 <--... and 2050 <-- 2900 <-- 4102 <-- 5802 <-- 8206 <--...; whether either of them is infinite, I don't know
@მამუკაჯიბლაძე - I began to guess, that there will be exactly 1 infinite backwards branch begining at 80 - just because I looked at the patterns of the died/finite branches. Because the steps increase by a relatively small amount the elements of a branch lie near together and this might be too dense for the existence of multiple infinite branches. But I've not yet an estimate for the according densities/distributions. The reverse Collatz-tree has a much wider stepping and so any subtree can have/has infinite subtrees.
Difficult to say. The branch at 1025 branches still further at 195185 while the branch at 2050 branches at 16417, etc. I don't see any method to guess whether any of these branches goes on infinitely
You can say with Beatty theorem : $A=\{E(n(\sqrt{2}+2)) \text{ ; } n\in\mathbb N^*\}$ and $B=\{E(n \sqrt{2});n\in\mathbb N^*\}$ is a partition of $\mathbb N^*$
And we have $E(n(\sqrt{2}+2))=2n+E(n\sqrt{2})$
with $E$ is the function integer part
What is $E(x)$? How does this answer the question?
|
2025-03-21T14:48:29.977503
| 2020-02-25T08:22:37 |
353495
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353495"
}
|
Stack Exchange
|
Associated barrelled topology of norm topology on $C_c(X)$
Let $X$ be a locally compact Hausdorff space, $C(X; K)$ the Banach space of continuous functions on $X$ with support in $K$, for compact $K \subseteq X$, and $C_c(X) = \lim_K C(X; K)$ the locally convex inductive limit with inclusion linking mappings $C(X; K) \to C_c(X)$. Denote the locally convex inductive limit topology on $C_c(X)$ by $\tau$, so that $\tau$ is the finest locally convex topology on $C_c(X)$ such that $C(X; K) \to C_c(X)$ is continuous. Denote also the norm topology by $\eta$. Then $\eta \subseteq \tau$. Moreover, $\tau$ is ultrabornological (therefore also barrelled and bornological), but $\eta$ is generally not barrelled.
Define $\eta^\beta$ as the strong topology $\beta(C_c, (C_c, \eta)')$ and $\eta^b$ as the associated barrelled topology (the coarsest barrelled topology finer than $\eta$; note that the finest locally convex topology is barrelled) and similarly $\eta^{ub}$ as the associated ultrabornological topology. Then
$$\eta \subseteq \eta^\beta \subseteq \eta^b \subseteq \eta^{ub} \subseteq \tau.$$
I would like to know whether $\eta^\beta = \tau$ or maybe just $\eta^b = \tau$ or at least $\eta^{ub} = \tau$. What if $X$ is assumed to be paracompact?
For $\sigma$-compact $X$ it holds $\eta^\beta = \tau$, see here (the proof uses a weighted seminorms approach).
Note also that for $X = [0, \omega_1)$ (the first uncountable ordinal), it holds $C_c(X) = C_0(X)$ (because continuous functions on $X$ are eventually constant) and therefore the norm topology $\eta$ is a Banach space topology. $X$ is not paracompact.
In a question like this it is often a good strategy to start by looking at the simplest non trivial case. Here this would be a discrete space, say the positive integers. Then the situation is quite transparent and might provide hints towards the general solution.
@user131781 I must regret, that I am a novice to locally convex spaces, but I think that the question fits MO better than MSE. For countable discrete spaces (they are $\sigma$-compact), this follows from the link above: $\eta^\beta = \tau$. For a general discrete space $X$ with cardinality $d$ (so that $C_c = \varphi_d$ and $(C_c, \tau)' = \omega_d$, $\tau$ is the finest lc top.) one can find in [Köthe, "Topological Vector Spaces II", §34.10 Remark] that $\sigma(\varphi_d, \varphi_d)^b = \beta(\varphi_d, \omega_d) = \tau$, hence $\eta^b = \tau$. (I don't know whether $\eta^\beta = \tau$).
If I am not mistaken, for a discrete $X$ it also holds $\eta^\beta = \tau$ because $\beta(\varphi_d, \varphi_d) = \beta(\varphi_d, \omega_d)$: every bounded set $B \subseteq \omega_d$ is contained in the closure of a bounded set $A \subseteq \varphi_d$: set $A := { (a_x){x \in X} \in \varphi_d \mid \exists (b_x){x \in X} \in B, , F \subseteq X \textrm{ finite}: a_x = b_x \textrm{ for } x \in F, a_x = 0 \textrm{ else} }$.
The reason I mentioned the discrete case is that it suggests the following path to a solution for the paracompact one. Consider next a direct sum of compacta where similar methods apply. Finally, reduce the general (paracompact) case to the latter by using the technique of partitions of unity (for inductive and projective limits of locally convex spaces—de Wilde). Since I haven’t sat down to write this up, I will leave it as a comment. I have no ideas about the non-paracompact case.
|
2025-03-21T14:48:29.977749
| 2020-02-25T08:48:07 |
353500
|
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"authors": [
"Adittya Chaudhuri",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353500"
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|
Stack Exchange
|
Weak 2-groups and non-abelian gerbe over a manifold
In literature, a strict 2-group is defined as a group object in the category of categories or groupoids. It has many well known equivalent descriptions viz:
1. A strict monoidal category in which all objects and morphisms are invertible.
2 A crossed module.
3 A strict 2-category with one object in which all 1-morphisms and 2-morphisms
are invertible.
4 A category object in Grp
I am here interested in especially 2 .
Now from Brown-Spencer result, we know that that the category of crossed module is equivalent to the category of strict 2- groups.
Now there is a notion of cocycle description of non-abelian gerbe over a manifold $M$ given by the following data:
Let $\lbrace U_{\alpha}\rbrace$ ($\alpha \in I$) be an open cover of $M$.Let $W$ be a Strict 2-group and let $(G, H,\tau,\alpha)$ be its corresponding crossed module of groups where $\tau:H \rightarrow G$ and $\alpha:G \rightarrow \mathrm{Aut}(H)$. Let $g_{ij}: U_i \cap U_j \rightarrow G$ and $h_{ijk}:U_i\cap U_j \cap U_k\rightarrow H$ (where $i,j,k \in I$) be a collection of functions such that $g_{ij}.g_{jk}=\tau(h_{ijk}).g_{ik}$ on $U_i\cap U_j \cap U_k$.
Recently in higher gauge theory, it had been found that there are some relations between the above cocycle description of non-abelian gerbes over a manifold and local description of principal 2-bundles over a manifold or loop space of a manifold (where the structure group is the strict 2-group corresponding to the crossed module defined in gerbe.)
Now there is a notion of weak 2-group which is defined by weakening the description (1) i.e a weak 2- group is a weak monoidal category $(M,\times,1)$ in which all morphisms are invertible and for each object $x \in M$ there exist an object $y$ such that $x \times y$ is isomorphic to 1 and $y \times x$ is isomorphic to 1. Now if we have a coherent choice of inverses satisfying some coherent diagrams then we call a weak 2-group a coherent 2-group.
My questions are the following:
A. Is there any known equivalent description of weak or coherent 2-groups analogous to crossed module description of strict 2-Groups?
B Is there any weaker notion of a cocycle description of non-abelian gerbe over a manifold where the crossed module is replaced by a suitable description of weak or coherent 2-group?
C Is there any relation between the local description of a principal 2-bundle (where the structure group is a weak or coherent 2-Group) over a manifold and the notion of gerbe (suitably weakened) over a manifold?
For the definition of principal 2-bundle , I have followed the following references:
https://arxiv.org/pdf/0803.3692.pdf (by Wockel)
http://math.ucr.edu/home/baez/2conn.pdf ( Baez and Schrieber)
For the definition of cocycle description of non-abelian gerbe over a manifold I have followed:
http://math.ucr.edu/home/baez/2conn.pdf (Baez and Schrieber)
For the definition of weak 2-groups and strict 2-Groups, I have followed
https://arxiv.org/pdf/math/0307200.pdf (Baez and Lauda).
Also in the paper 'Higher gauge theory: 2 connection on 2 Bundles' https://arxiv.org/abs/hep-th/0412325 (Baez and Scrieber) in the section 'Discussion and open problems' a question had been proposed that
"How does the discussion in this paper generalize when the standard fiber F of the
2-bundle is allowed to be something which is not a strict 2-group?"
It would be really helpful to know about the current status of the above question.
Any clarification regarding my question is highly appreciated.
Thanks in advance.
You can remove “ In literature, a strict 2-group is defined as a group object in the category of categories or groupoids. It has many well known equivalent descriptions viz: 1. A strict monoidal category in which all objects and morphisms are invertible. 2 A crossed module. 3 A strict 2-category with one object in which all 1-morphisms and 2-morphisms are invertible. 4 A category object in Grp I am here interested in especially 2. Now from Brown-Spencer result, we know that that the category of crossed module is equivalent to the category of strict 2- groups.”
That does not add any extra clarity for the question.. removing that part reduce the length making it easier to read
@PraphullaKoushik Thanks for your kind suggestion. But I feel it is necessary. The only definition of weak 2 group I know is as a weak monoidal category whereas in the proof(that I know) of Brown-Spencer result they have treated Strict 2 group as a group object in Cat . So to get the answer to (A) we need to mention that strict 2 group is same as a Strict Monoidal category with all objects and morphisms invertible. The things that I feel unnecessary to this question is (3) and (4). Still I feel its better to mention all the descriptions that I know of Strict 2 group.
:) as you wish.
I'm also interested in the analogue of a weak crossed module.
@Student To "some extent" it is there is the section 8.3 of https://arxiv.org/pdf/math/0307200.pdf . Especially please check Theorem 43 and Corollary 44
The paper https://arxiv.org/abs/1403.7185 by Jurco, Saemann, and Wolf seems to answer at least your questions A and B, for what they call "semi-strict 2-groups", a concept that for me seems to be the same as "coherent 2-groups".
@KonradWaldorf Thank you Sir.
|
2025-03-21T14:48:29.978042
| 2020-02-25T08:48:40 |
353501
|
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],
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"BCLC",
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|
Stack Exchange
|
Are anti-linear maps/semi-linear, such as conjugations, linear in other almost complex structures?
I have asked this on mse, but I did not get any responses even after a bounty.
I started studying the book of Daniel Huybrechts, Complex Geometry An Introduction. I tried studying backwards as much as possible, but I have been stuck on the concepts of almost complex structures and complexification. I have studied several books and articles on the matter including ones by Keith Conrad, Jordan Bell, Gregory W. Moore, Steven Roman, Suetin, Kostrikin and Mainin, Gauthier
I have several questions on the concepts of almost complex structures and complexification. Here are some:
Assumptions and notations: Let $V$ be a $\mathbb C$-vector space. Let $V_{\mathbb R}$ be the realification of $V$. For any almost complex structure $I$ on $V_{\mathbb R}$, denote by $(V_{\mathbb R},I)$ as the unique $\mathbb C$-vector space whose complex structure is given $(a+bi) \cdot v := av + bI(v)$. Let $i^{\sharp}$ be the unique almost complex structure on $V_{\mathbb R}$ such that $V=(V_{\mathbb R},i^{\sharp})$.
Let $W$ be an $\mathbb R$-vector space. Let $W^{\mathbb C}$ denote the complexification of $W$ given by $W^{\mathbb C} := (W^2,J)$, where $J$ is the canonical almost complex structure on $W^2$ given by $J(v,w):=(-w,v)$. Let $\chi: W^2 \to W^2$, $\chi(v,w):=(v,-w)$
For any map $f: V_{\mathbb R} \to V_{\mathbb R}$ and for any almost complex structure $I$ on $V_{\mathbb R}$, denote by $f^I$ as the unique map $f^I: (V_{\mathbb R}, I) \to (V_{\mathbb R}, I)$ such that $(f^I)_{\mathbb R} = f$. With this notation, the conditions '$f$ is $\mathbb C$-linear with respect to $I$' and '$f$ is $\mathbb C$-anti-linear with respect to $I$' are shortened to, respectively, '$f^I$ is $\mathbb C$-linear' and '$f^I$ is $\mathbb C$-anti-linear'.
The complexification, under $J$, of any $g \in End_{\mathbb R}W$ is $g^{\mathbb C} := (g \oplus g)^J$, i.e. the unique $\mathbb C$-linear map on $W^{\mathbb C}$ such that $(g^{\mathbb C})_{\mathbb R} = g \oplus g$
Let $\sigma: V_{\mathbb R}^2 \to V_{\mathbb R}^2$, $\gamma: W^2 \to W^2$ and $\eta: V_{\mathbb R} \to V_{\mathbb R}$ be any maps such that $\sigma^J$, $\gamma^J$ and $\eta^{i^{\sharp}}$ are conjugations. (The $J$'s are of course different, but they have the same formula.)
Questions:
For $\sigma$, does there exist an almost complex structure $I$ on $V_{\mathbb R}^2$ such that $\sigma^I$ is $\mathbb C$-linear, and why/why not?
Whenever we have such an $I$, is $I$ necessarily $I=k \oplus h$ for some almost complex structures $k$ and $h$?
For $\gamma$, does there exist an almost complex structure $K$ on $W^2$ such that $\gamma^K$ is $\mathbb C$-linear, and why/why not?
Note: I think the answer to Question 3 is no if the answer to Question 1 is no. However, I think Question 3 is answered affirmatively and with explanation if the answer to Question is 1 is yes and the answer to Questions 2 is no.
For $\eta$, does there exist an almost complex structure $H$ on $V_{\mathbb R}$ such that $\gamma^K$ is $\mathbb C$-linear, and why/why not?
Note: I think the answer to Question 4 is no if the answer to Question 3 is no.
Observations that led to above questions:
$\chi^J$ is a conjugation, on $(V_{\mathbb R})^{\mathbb C}$, called the standard conjugation on $(V_{\mathbb R})^{\mathbb C}$.
Let $\hat i: V_{\mathbb R}^2 \to V_{\mathbb R}^2$, $\hat i := i^{\sharp} \oplus i^{\sharp}$. $\hat i$ is an almost complex structure on $V_{\mathbb R}^2$.
While $\chi^J$ and $\chi^{-J}$ are $\mathbb C$-anti-linear, we have that $\chi^{\hat i}$ is $\mathbb C$-linear.
$k$ and $h$ are almost complex structures on $V_{\mathbb R}$ if and only if $k \oplus h$ is an almost complex structure on $V_{\mathbb R}^2$
Actually, I think $\chi^{k \oplus h}$ is $\mathbb C$-linear, for any almost complex structures $k$ and $h$ on $V_{\mathbb R}$, not just $k=h=i^{\sharp}$.
You change complex linearity to conjugate linearity, and vice versa, by replacing $I$ by $-I$, but only on the domain or the range independently. If you want to change them both, as the same vector space with the same complex structure, it is trickier.
For question 3, a complex linear map, when realified, can only have an even number of -1 eigenvalues, so a conjugation can't be complex linear on $\mathbb{R}^2$, for example, for any complex structure. On the other hand, in any real dimension which is a multiple of 4 you can clearly have such a complex structure, and there are many.
If you have a real linear map with simple eigenvalue, to become complex linear, the necessary and sufficient condition is that the real eigenvalues have even multiplicities. You can pick any complex structure on each (even dimensional) real eigenspace, and then pick out any complex eigenvalues in conjugate pairs, making one of them into an $\sqrt{-1}$ eigenspace, and the other into a $-\sqrt{-1}$ eigenspace. For generalized eigenvalues, it is more complicated.
It is easier to work in a complex linear coframing; see my lecture notes (https://arxiv.org/abs/1706.09697) where I compute some examples. You avoid this $J$ thing almost entirely.
Thanks Ben McKay. I have yet to analyse your answer.
|
2025-03-21T14:48:29.978326
| 2020-02-25T09:44:34 |
353503
|
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|
Stack Exchange
|
How to solve numerically a system of 3 interdependent non-linear ordinary differential equations?
As per title, I need to solve this:
$$
\begin{cases}
\frac{d^2V}{dx^2} = -\frac{q}{\epsilon}\left[p - n + \frac{N_0}{1+c_pp+c_nn}\right] \\\\
\frac{d}{dx}\left[\mu_nn\frac{dV}{dx} + D_n\frac{dn}{dx}\right] - \left[\frac{n-n_0}{\tau_n} - G_{op}\right] = 0 \\\\
\frac{d}{dx}\left[\mu_pp\frac{dV}{dx} - D_p\frac{dp}{dx}\right] + \left[\frac{p-p_0}{\tau_p} - G_{op}\right] = 0
\end{cases}
$$
where V (electrostatic potential), n (total free electrons density) and p (total free holes density) are the unknown functions.
Being in 1D and in static conditions, all the functions depend on just the spatial variable "x" (the domain goes from "0" to "L").
Finally, the boundary conditions are:
$$
\begin{cases}
V(x=0) = V_{left} (known) \\\\
V(x=L) = V_{right} (known) \\\\
\frac{dp}{dx}(x=0) = 0 \\\\
\frac{dn}{dx}(x=0) = 0 \\\\
\frac{dp}{dx}(x=L) = 0 \\\\
\frac{dn}{dx}(x=L) = 0
\end{cases}
$$
It is basically a Poisson equation coupled with drift-diffusion + continuity equations for electrons and holes in a crystalline semiconductor.
Device Modeling? There are several monographs published by the Wien branch of Springer Verlag that describe comprehensively the theory and the finite difference schemes used to numerically solve the drift-diffusion equations coupled. with the nonlinear Poisson equation, for example "Semiconductor equations" authored by Markowich, Ringhofer and Schmeiser (1990)
Matlab bvp4c can solve such equation
Are all the coefficients $\mu_n,D_n,\mu_p,D_p,\tau_n,\tau_p$ constants, or functions of $x$?
I tried an iterative approach for which I set an initial guess for "n" and "p" (which are the functions I have at equilibrium), solve Poisson with boundary conditions slightly changed from equilibrium conditions, put the resulting potential profile in the 2 drift-diffusion + continuity equations and derive new guesses for holes and electrons densities to put again in the Poisson equation: iterate until it converges.
Then change again the potential boundary conditions and do it all again using as initial guess for "n" and "p" the results of the previous iterative procedure.
Finally, the potential boundary conditions are slowly brought to the desired value.
Practically, I solve the system at conditions slowly varying from equilibrium in an iterative way.
It converges but gives crazy (and wrong) results. I can provide the Matlab file!
As has already been pointed out, there are entire books about this type of problem. If your difficulties result from small diffusion coefficients or such, these resources certainly have hints how to deal with that. If you simply have a bug in your code, this is not the right place to get it debugged.
|
2025-03-21T14:48:29.978534
| 2020-02-25T10:27:31 |
353508
|
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|
Stack Exchange
|
Evaluation of the quality of research articles submitted in mathematical journals: how do they do that?
I would like to know as curiosity how the editorial board or editors* of a mathematical journal evaluate the quality, let's say in colloquial words the importance, of papers or articles.
Question. I would like to know how is evaluate the quality of an article submitted in a journal. Are there criteria to evaluate it? Many thanks.
I think that it must be a difficult task to evaluate the quality of a mathematical paper due how is abstract (the high level of abstraction of research in mathematics) the work of professional mathematicians. Are there criteria to evaluate it, or is it just the experience, knowledges and good work of the people working as publishers?
I'm curious about it but I think that this is an interesting and potentially useful post for other. As soon as I can I should accept an answer.
If there is suitable references in the literature about how they do this work, feel free to refer the literature, answering this question as a reference request, and I try to search and read it from the literature.
*I don't know what is the role of each person working in the edition of a mathematical journal.
Thanks in advance to all people adding contributions to this post, I hope that my post is suitable for MathOverflow.
I think this works similarly to how it works in other disciplines. "Interesting" roughly means "useful for what active communities of mathematicians are working on", and is thus not a static or timeless thing. This is a necessity of the economy of ideas. It leads to some obvious political problems, but otherwise imagine the difficulty of identifying novelty in a reasonable time...
@JonBannon I agree, it's very subjective and dependent on the surrounding mathematical developments, but I think one can be a slightly more specific: (1) To what extent does it address a problem/methods that others are interested in? (2) What potential does it have to inspire future research?
Aside from literally having the word "mathematics" in it, how is this about research-level mathematics? It seems that the same question could be asked almost literatim about pretty much any discipline.
I think that this post @LSpice is potentially useful or interesting for many mathematicians, in particular for those who start to research in mathematics.
Assume we are talking about a good journal with a large editorial board representing a wide scope of mathematical interests. I will describe both the role of the editors and the role of the referees. This is my personal viewpoint and others might have different opinion/experience.
The role of the editors.
Good journals can accept only about 20% of submitted papers. This is not easy to reject 80% of papers and it often results in rejecting really good papers. Everyone understands that. The procedure of evaluating the papers by the editors is more or less as follows:
The editors have many years of research experience and (hopefully) developed a good mathematical taste. If the paper is close to the research interest of the editor then he or she can relatively easily identify the papers that are not particularly interesting. The reason for not being interesting can be based (for example) on the following criterion:
The result is not well motivated. It is very technical and follows more or less standard arguments. The authors simply take a known result, and prove a new result by slightly modifying the given assumptions. Usually it means that they make the statement more complicated and in a sense more general. Often, they neither have interesting examples supporting such generalizations nor indication of possible applications.
Unfortunately, most of the papers fell into this category. If the editor is sure that this is the case, then he or she rejects the paper without sending it to a referee. Then the authors usually get a rejection notice similar to this one:
We regret that we cannot consider it, in part because at present
we have a large backlog of excellent articles awaiting publication.
We are thus forced to return articles that might otherwise be considered.
If the editor is not sure about the quality of the paper, then they ask an expert (or several experts) for a quick opinion:
I wonder if you could make a quick, informal assessment of it. Are the results strong enough to warrant sending the article to a referee? Because of our backlog, we like to send to referees only articles that appear to be of very good to outstanding quality.
In that case the expert evaluating the paper is not asked to check all the proofs but to make a judgement based on the criterion explained above. This is an easy task for an expert. If the expert writes a negative opinion, then the authors receive a rejection notice often phrased the way as the rejection notice listed above.
For top journals all experts have to write a positive opinion before the paper is sent to referees.
If the experts' opinion is positive, then the paper is sent to a referee or to many referees. The most extreme case that I know of was a panel of 12 referees who took several years to evaluate the paper (this was when Thomas Hales proved the famous Kepler conjecture). For top quality journals all referees must write positive reports before the paper is accepted (once I received 6 reports, 4 positive and 2 not so positive and the paper was rejected).
Let me also add that the editorial boards are structured basically in two different ways. (1) The authors are asked to choose an editor from the list of editors and submit the paper directly to them. Then the editor who receives the paper handles the submission process according to the rules explained above. (2) The authors submit the paper to the main editor or just to the journal and then the main editor either rejects the paper by themselves or he/she sends it to one of the editors from the editorial board and that editor applies the rules listed above.
Of course some of the journals might have a slightly different approach than the one explained here. There is no a canonical solution and what I wrote is a somewhat a simplified version of the process that is applied in reality.
The role of referees. A paper passed through an initial screening and it was sent to a referee. This is the most unpleasant part of the process. A referee spends a lot of time to read the paper, they are not paid for this job and since their work is anonymous, they do not get any recognition for what they do.
What is the referee required to do? First of all, the referee has to assess originality of the results and whether the results are interesting enough. This part is the same as the one in the initial screening when the paper is sent to an expert for a quick opinion. Secondly, the referee is required to read the paper and check details. Let's be clear about that. Unless the paper is directly related to the research of the referee and he or she really wants to understand the details, there is no way the referee can check all details. Since I cannot speak for other people, let me say what I do in this situation.
My answer will only be a simplified version of the real process of the refereeing a paper, just a main idea of what I do.
I go through the whole paper (or most of the paper) to have a good idea of what it is all about, to see a big picture not only of the meaning of the theorems, but also a big picture of the techniques used in the proof. Then I check carefully details of many/some arguments while for other arguments I briefly skim over. If the argument seems reasonable and believable to me I do not bother checking it very carefully. If all details that I check are correct and if all other arguments seem reasonable I am content. In this case, if I like the statement of the main result, I accept the paper. If however, some arguments seem fishy to me, then I check them carefully. This is a point where often I ask the authors for further clarifications. If I really cannot pass through the paper, because I think it has mistakes or if it is written in an unreadable way, I often reject the paper.
The biggest problem is when I am convinced that the result proved in a paper is of an outstanding quality, but the paper is very difficult and for that reason not very easy to read. Then, hmm... Then, it is not easy and I often struggle with making a right decision.
This answer is informative, but I feel like it doesn't clearly express the referees' role in evaluating the quality.
@Kimball Good point. The question was mostly about the role of the editorial board, but you are right I should also say about the role of referees. I will try to do it later when I will have time.
If you move $\epsilon \geq 0$ forward in the working area of some influential editors following his/her ideas and citing a lot of his/her papers, that's a fantastic paper! If you stride in an area that is not well-perceived (or well-known) by the editors, that's a minor and uninteresting paper.
Otherwise, you can also solve a one-hundred years open question on which hundreds of mathematicians have worked on without success. Everyone would agree that it makes a great paper. But it seems these days that papers of very high quality are published much more often than interesting questions do appear...
What do you mean by your last sentence?
@user2520938 : I think Libli is writing in a tongue-in-cheek style. "Papers of very high quality" is a sarcastic phrase, referring to epsilonic improvements in areas of interest to the powers that be, while "interesting questions" are major conjectures on the level of the Riemann hypothesis.
+1 for using $\epsilon \geq 0$ rather than $\epsilon > 0.$ Actually, I think in some (nontrivially many) cases $\epsilon < 0$ applies, especially if one factors in criteria such as @Piotr Hajlasz's observation: "Usually it means that they make the statement more complicated and in a sense more general $[\cdots]$ they neither have interesting examples supporting such generalizations nor indication of possible applications".
@TimothyChow that's what I thought, but I wasn't sure.
@DaveLRenfro : Thanks! That's somehow the crux of my "proof" :p
(i) I think almost every journal has a declared "Aims and scope" policy, and how well the submitted paper fits that policy is perhaps the important criterion in the editors' judgment of how good the paper would be for the readers of the journal. (ii) Obviously, for any reputable mathematical journal, the mathematics in the paper must be correct and nontrivial.
Nowadays, in the era of citation indexes, the main criterion, after the criteria (i) and (ii) above, seems to be how "topical" the paper is, that is, how high is the level of current interest the paper may attract.
So you’re suggesting that it helps for a paper to be sufficiently modish?
@Lubin : I am not suggesting that one should write modish papers. However, if a paper is "topical", it does seem to significantly increase its chances to be accepted, in most journals.
|
2025-03-21T14:48:29.979489
| 2020-02-25T11:18:56 |
353514
|
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|
Stack Exchange
|
What's the smallest $\lambda$-calculus term not known to have a normal form?
For Turing Machines, the question of halting behavior of small TMs has been well studied in the context of the Busy Beaver function, which maps n to the longest output or running time of any halting n state TM. The smallest TM with unknown halting behavior has 5 states. It takes $5\cdot2\cdot\log_2(5\cdot4 + 4)$ or roughly 46 bits to describe, 50 bits if we assume a straightforward encoding.
In comparison, $\lambda$-calculus terms have a simple binary encoding: $00$ for lambda, $01$ for application, and $1^n0$ for variable with de Bruijn index $n$. It's natural to define a $\lambda$-calculus analog of the busy beaver function as the maximum normal form size of any size $n$ closed lambda term.
As the smallest closed lambda term is $\lambda\,1$, with encoding $0010$, we determine
$$
BB_{\lambda}(4) = 4
$$
The next smallest ones, $\lambda\,\lambda\,1$ and $\lambda\,\lambda\,2$ are similarly already in normal form, and give
$$
BB_{\lambda}(6) = 6,\qquad BB_{\lambda}(7) = 7
$$
$BB_{\lambda}(n)$ will have to remain undefined for $n < 4$ or $n = 5$.
The first enlarged normal form shows up at $\lambda\,(\lambda\,1\,1)\,(1\,(\lambda\,2))$ which gives
$$
BB_{\lambda}(21) = 22
$$
$BB_{\lambda}$ starts to grow rapidly at $n \geq 30$, since tripling Church numeral two, $(\lambda\,1\,1\,1)\,(\lambda\,\lambda\,2\,(2\,1))$ with normal form Church numeral $2^{2^2}= 16$, gives
$$
BB_{\lambda}(30) \geq 5 \cdot 16 + 6 = 86
$$
and quadrupling/quintupling give
$$
BB_{\lambda}(34) \geq 5 \cdot 2^{16} + 6
$$
$$
BB_{\lambda}(38) \geq 5 \cdot 2^{2^{16}} + 6
$$
which exceed the TM Busy Beavers for 4 and 5 states.
An Ackermann-like function takes a mere 29 bits.
A twisted application to Church numeral $2$ yields a $BB_{\lambda}(51)$ exceeding 2↑↑↑↑5.
Graham's number is exceeded in at most 49 bits, giving
$$
BB_{\lambda}(49) \geq 5 \cdot G + 6
$$
(compared with a 16 state TM that needs over 192 bits to describe).
What's the smallest n for which $BB_{\lambda}(n)$ is unknown in ZFC?
One upper bound is 213 bits.
Let's try to narrow it down some more.
Function $BB_{\lambda}$ has been added to the Online Encyclopedia for Integer Sequences.
Are all those slashes supposed to represent lambda?
Yes, they were. I replaced them by actual lambdas now.
Shouldn't $\lambda 1$ have encoding 00110 (of length five), or am I missing something in the notation here?
When i changed from 0-based to 1-based variables, I forgot to update the encoding. It's fixed now.
It seems that application is also the other way round... quoting yourself :-)
oops; good catch! will fix right away.
Curiously, triple Church_2 is not the best 30 bit beaver. A very minor tweak gives a 160 bit normal form!
For an upper bound, do you know what a lambda calculus formulation of twin primes / Goldbach / similar number-theoretic statements might look like?
Yes; Goldbach is 267 bits, as can be seen at https://github.com/tromp/AIT/blob/master/goldbach.gif which is compiled from goldbach.lam; larger than the 213 bit Laver statement above.
I also have an upper bound of 247 bits for finding an odd perfect number at https://github.com/tromp/AIT/blob/master/oddperfect.lam
By “not known to halt”, do you mean “unknown halting behavior” or “known to not halt or unknown halting behavior”?
I mean "unknown halting behaviour". We know that (\ 1 1)(\ 1 1) is the smallest nonhalting term, of size 18.
Does "unknown" mean "unknown in ZFC"? I infer that this is the case from the upper bound being a reference to a program which is known to halt under the assumption of a rank-into-rank cardinal.
Yes, I meant unknown in ZFC. I'll make that explicit.
|
2025-03-21T14:48:29.979764
| 2020-02-25T11:55:50 |
353517
|
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|
Stack Exchange
|
Determinant of a block matrix with dissimilar elements
I am looking for an expression that convert calculation of given quadratic form into determinant of some block matrix. I see this in that form (that may be incorrect):
$x^T A x = \begin{vmatrix} x^T & 0 & 0 \\ 0 & A & 0 \\ 0 & 0 & x \end{vmatrix}$
To clarify. The above matrix is a block matrix with “additional” block entries on the diagonal of the matrix. There are two 3-dimensional vectors $x$ and a matrix $A$ of size $3\times3$.
My first port of call was the application of the determinant formula for block matrices, given in the next link:
https://en.wikipedia.org/wiki/Determinant#Block_matrices
Firstly, I did not find the application of the above formulas to the structure with which I am dealing, and secondly, the standard det commands in Mathkad and Matlab do not work with such structures. I was wondering if there was an article among mathematicians or a well-known formula that would facilitate this task.
The task for me now is to somehow find an expression for the determinant, so that the solution to the problem consists of at least a combination of “det” commands Mathkad and Matlab.
I don't get at all that formula. $x^T A x$ surely isn't equal to $\begin{vmatrix} x^T & 0 & 0 \ 0 & A & 0 \ 0 & 0 & x \end{vmatrix}$. Actually the RHS is zero, because the matrix is upper triangular with zeros on its diagonal.
This is a conditional entry form. Made for illustration purposes only. If we assume that $a = x$, $b = A$, $c = x^T$, then the determinant is equal to $a b c$. I want to understand if it is possible to turn the calculation of the quadratic form into the calculation of the determinant of some block matrix?
It's hard for me to understand what you are asking, but maybe this answers your question. One has
$$
-x^TAx = M / A^{-1},
$$
where
$$
M = \begin{bmatrix}A^{-1} & x\\ x^T & 0 \end{bmatrix}
$$
and the symbol $/$ denotes the Schur complement.
One has
$$
\det(M / A^{-1}) = \det(M) / \det(A^{-1}) = \det(M)\det(A),
$$
so you get
$$
x^TAx = -\det(M)\det(A).
$$
Side note: if your goal is actually computing things numerically, then using determinants is usually considered a bad idea. The good thing about determinants is that they give you closed forms of stuff, but when it comes to scientific computing you are almost always better off without using them.
Yes, I was looking for something like that. I want to clarify - how is the determinant of the $M$ calculated, how is the determinant of a block matrix?
@dtn A block matrix is just a matrix that is obtained by stacking the corresponding blocks. Its determinant does not have a particular definition, it is just the 'usual' determinant of a matrix.
I already checked this with calculations. It really is. No special formulas are needed to calculate the determinants. In general, you helped me by suggesting what the Shur complement is. Thank you very much!
Always happy to help!
|
2025-03-21T14:48:29.980001
| 2020-02-25T12:05:27 |
353519
|
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|
Stack Exchange
|
Uniqueness constraints for Delaunay triangulation
Commonly the assumption that is made on point sets that shall be Delaunay-triangulated is that no three are collinear and no four are cocircular.
Those assumptions are however too restrictive: if the pointset consists of the corners of a regular n-gon plus the center of the circle on which those corners lie, then the Delaunay triangulation is unambiguous and not degenerate in any sense.
The correct condition would rather be that no four points are cocircular on a circle with empty interior.
Questions:
have the original conditions regarding cocircularity been formulated correctly and only later has the emptyness condition been brushed under the carpet?
What are examples of publications that make the correct assumptions when discussing the Delaunay triangulation?
I think it is well-understood that "no four points cocircular" avoids Delaunay triangulation degeneracies. It is also well-understood that one can have four cocircular points that don't cause any degeneracies, as in your example.
Perhaps this paper addresses your concern.
The authors explicitly define the tolerance of the
Delaunay triangulation of a point set as the smallest perturbation of
the points that causes a diagonal flip (and they justify this definition).
This naturally leads to studying the annulus illustrated below.
Abellanas, Manuel, Ferran Hurtado, and Pedro A. Ramos. "Structural tolerance and Delaunay triangulation." Information Processing Letters 71, no. 5-6 (1999): 221-227.
PDF download.
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2025-03-21T14:48:29.980143
| 2020-02-25T12:20:48 |
353520
|
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|
Stack Exchange
|
Kodaira dimensions of push-forward via finite map
Let $f:X \to Y$ be a finite map from a normal projective variety to a smooth projective variety, $D$ be a Cartier divisor on $X$. Do we have any relation between $\kappa(X,D)$ and $\kappa(Y,f_*D)$?
There is an obvious relation: the pushforward map $f_*:|mD|\rightarrow |f_*(mD)|=|mf_*D|$ is injective, hence $\kappa (X,D)\leq \kappa (X,f_*D)$. It is easy to see that you cannot get more: for instance, take for $f$ a general projection from a cubic surface $X\subset \mathbb{P}^3$ to $\mathbb{P}^2$, and for $D$ a line in $S$. Then $\kappa (S,D)=0$, but $f_*D$ is a line in $\mathbb{P}^2$, so $\kappa (\mathbb{P}^2,f_*D)= 2$.
why $f_*$ is injective? Is there a reference?
I can imagine that $|mD| \to |mf^*f_D|$ is injective but I am not sure if $f_$ is injective.
@Hu Zhengyu: It is finite because $f$ is finite, on the other hand it is linear.
I think $f_$ is not injective. If you take $D=0$, then $f_$ would map some nonzero rational function of $X$ to zero since $K(X)$ is obviously bigger than $K(Y)$. You just said $f$ is finite? What is the definition of finite for a linear map?
|
2025-03-21T14:48:29.980256
| 2020-02-25T13:38:18 |
353524
|
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"Lucia",
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|
Stack Exchange
|
The irrational numbers α such that n odd and m=⌊nα⌋ odd implies ⌊mα⌋ odd
This post is the analogous of that one (about $\sqrt{2}$) but with a much stronger expectation here.
We observed, and then this comment of Lucia proved, that for $\phi$ the golden ratio, if $n$ and $m=\lfloor n\phi\rfloor$ are odd then so is $\lfloor m\phi\rfloor$. Next EmilJeřábek3.0 provided a family of quadratic algebraic integers satisfying this property (also in comments, by extended Lucia's proof).
Let $L$ be the set of irrational positive numbers $\alpha$ satisfying the property that if $n$ and $m=⌊nα⌋$ are odd then so is $⌊mα⌋$.
Question: What is the set $L$, explicitly?
Remark: Then $L \cap S = \emptyset$, with $S$ defined in this post on the borderline Collatz-like problems.
Yes, by some easy algebra. Write $n\phi = m+\theta$, and then $m\phi = n\phi^2 -\theta\phi = n\phi + n - \theta \phi = m+n +\theta -\theta \phi$. This is a little less than the even number $m+n$.
@EmilJeřábek3.0: Correct! It should be asked whether it is unique up to the addition of an even integer.
It is far from unique. If I didn’t make a numerical mistake (which should only affect the side conditions anyway), Lucia’s argument applies to all numbers of the form $\frac12\bigl(b+\sqrt{b^2+4c}\bigr)$ where $b,c$ are integers such that $b>0$, $b^2+4c$ is a positive nonsquare, and either $b\ge c>0$ and $b\equiv c\pmod2$, or $c<0$ and $b\not\equiv c\pmod2$.
It appears I did make a mistake: the second disjunct should read $1-b<c<0$ and $b\not\equiv c\pmod2$. (This implies $b^2+4c>0$, so this condition is redundant.)
@EmilJeřábek3.0: Ok! Do you expect the existence of something else? Do you expect that the full classification is reachable?
I don’t know. I just tried to push Lucia’s argument to its limit. The argument in fact gives that if $\alpha^2=b\alpha+c$, then $m\alpha=bm+cn+(b-\alpha)\theta$. The conditions I gave ensure that $|b-\alpha|<1$ and that it has the right sign w.r.t the parity of $bm+cn$ (i.e., $b+c$); thus, if correct, I think that this might give a full classification for the limited case when $\alpha$ is a quadratic algebraic integer. Whether there are examples of a different nature, I don’t know.
Sorry, the closure under addition of even integers is not actually true.
@EmilJeřábek3.0 I checked that, it is actually true. What is your counter-example?
Actually, pretty much all such numbers are conterexamples. But to be completely explicit, just take $\alpha=\phi+2$, $n=1$, $m=\lfloor \alpha\rfloor=3$, then $\lfloor m\alpha\rfloor=10$. Here is what goes wrong: assume $\alpha'$ satisfies the property and $\alpha=\alpha'+2k$. Let $n$ and $m=\lfloor n\alpha\rfloor$ be odd. Then $m'=\lfloor n\alpha'\rfloor=m-2kn$ is odd, hence $\lfloor m'\alpha'\rfloor$ is odd (by assumption), hence $\lfloor m'\alpha\rfloor=\lfloor m'\alpha'\rfloor+2km'$ is odd, but there is no reason for $\lfloor m\alpha\rfloor=\lfloor m'\alpha+2k\alpha\rfloor$ to be odd.
@EmilJeřábek3.0: Got it! Funny trap!
@Lucia: you may be interested in this answer inspired by your comment above.
|
2025-03-21T14:48:29.980497
| 2020-02-25T14:48:33 |
353530
|
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|
Stack Exchange
|
What is the closed form of this function?
It is well-known that the binomial coefficient has some monotonicity,
and that can be used to find the maximum and minimum of binomial coefficients.
Similarly, now let $\left(\delta_{i,j}\right)_{6\times6}$ be a binary
matrix with $\delta_{i,i}=1$ and $\delta_{i,j}=\delta_{j,i}=1$ or
$0$ when $i\ne j$ and $\delta_{i}=\sum_{j=1}^{6}\delta_{i,j}$.
Let's define
$$
f(k):=\min_{\delta_{1}+\delta_{2}+\delta_{3}+\delta_{4}+\delta_{5}+\delta_{6}=2k}\delta_{1}\delta_{2}\delta_{3}\delta_{4}\delta_{5}\delta_{6}
$$
for integer $k$ , $3\le k\le18$, then what is the closed form or
combinatorial expression of $f(k)$ as a function of $k$?
Thanks.
I deleted my comment because there's something still unclear in your problem statement. For odd $k$, there is no matrix of the type you describe, because of the constraint $\delta_{i,j} = \delta_{j,i} $, if indeed that's what you want.
@MichaelEngelhardt Thank you for your comment, and it has now been revised. Thanks.
Just so I understand the idea here — $\delta_{i,j}$ is just a symmetric (0,1) matrix with all diagonal entries 1? Or are there any other constraints?
related https://math.stackexchange.com/q/3559389/87355
The minimum is achieved by filling as large an upper left square block as possible with entries 1, and partially filling the next row below that block (and, symmetrically, the next column to the right of that block) with the remaining available entries 1. The value of $f(k)$ achieved in this way is
$$
f(k)=c(k)^{c(k)} \cdot [(c(k)+1)/c(k)]^{r(k)} \cdot [r(k)+1]
$$
where the aforementioned upper left square block has size $c(k) \times c(k)$, and $2r(k)$ is the number of residual entries 1 available to partially fill the next row and column, as described. It remains to give explicit expressions for $c(k)$ and $r(k)$. There are probably prettier and more succinct forms, but quickly one could just specify
$$
c(k)=1+\frac{k}{4}+\frac{k}{6}+\frac{k}{9}+\frac{k}{13}-\frac{k}{8}-2\left( \frac{k}{12} \right) -\frac{k}{18}
$$
where division is to be interpreted as integer division, and
$$
r(k)=k-3+(1-c(k))c(k)/2
$$
I have verified that these expressions give the correct minimal values by explicit computation of all possibilities.
An alternative expression for $c(k)$ is
$$
c(k) = \mbox{int} \left[ \frac{1}{2} + \sqrt{\frac{1}{4} + 2k-6} \right]
$$
Thank you for your answer, but is the minimum "achieved by filling as large an upper left square block as possible with entries 1" for $k=6$? Thanks.
@EmmaT - Why are you asking these questions? My answer is valid for all $k$. For $k=6$, you have a $3\times 3$ block, and you can easily convince yourself that any relocation of entries will increase the value of the product.
Sorry that I confused the minimal configuration with another configuration for k=6, but how about $k=7$ (do you get 108 by "filling as large an upper left square block as possible with entries 1", and partially filling the next row below that block (and, symmetrically, the next column to the right of that block) with the remaining available entries 1")? Thanks.
Could you possibly express $f$ by combinatorial numbers or perhaps using multinomial coefficients? Thanks a lot.
@EmmaT - For $k=7$, the minimal configuration yields 72, not 108. You can do it in your head - the $\delta_{i} $ are 4,3,3,2,1,1. I don't understand what you're aiming at regarding combinatorial numbers. I've given the result as a function of $k$. If you wish, replace $k$ by the binomial coefficient $\left( k \atop 1 \right)$. One can surely obfuscate it some more such as to seem less facetious.
Now I see what you mean by "partially filling the next row below that block (and, symmetrically, the next column to the right of that block) with the remaining available entries 1 ". As I checked for many cases the statement holds now, but what is the mathematical or rigorous reason why any other configuration will yield a larger product? Thanks a lot.
@EmmaT - The reason is essentially that, for $a\geq b$, $(a+1)(b-1)<ab$, so, roughly speaking, you want to partition the $\delta_{i} $ into two sets which are maximal and minimal, respectively. The symmetry requirement correlates entries a little bit, leading to a preference for keeping as many $\delta_{i} =1$ as possible even if it means keeping the large $\delta_{i} $ smaller than 6. Note of course that the problem has symmetries w.r.t. permuting rows and columns, but the arrangement I chose is probably the simplest if you want to do inductive proofs, say.
@EmmaT - I've added an alternative expression for $c(k)$ to the answer. Shorter, but requires more than the four elementary operations of arithmetic.
Thanks, but I had obtained the alternative expression by the time when the question in my earlier comment was deleted. However, how would you use your reason $(a+1)(b−1)<ab$ to explain the comparison on the configurations (5,2,2,2,2,1) and (4,3,3,2,1,1)? Thanks a lot.
@EmmaT - This is what I mean by the symmetry requirement correlating entries; in general, you don't get to do $(a+1)(b-1)<ab$ considerations in isolation, but you have to consider the relative merits of two such tradeoffs. In your example, you move an entry from column 5 to 3, gaining you a factor 4/3, but you simultaneously have to move an entry from column 1 to 2, losing you a factor 5/6. The merit of changing the value of a $\delta_{i} $ from 2 to 1 is always the greatest, hence my statement "keeping as many $\delta_{i} =1$ as possible".
|
2025-03-21T14:48:29.980868
| 2020-02-25T15:06:06 |
353531
|
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|
Stack Exchange
|
Point wise convergence of Laplace transform and convergence of functions
Assume that functions $f_n(t), f(t)\in C_b(R_+)$. For every $\lambda >0$, we have
$$
\bigg|\int_0^\infty e^{-\lambda t}f_n(t)d t-\int_0^\infty e^{-\lambda t}f(t)d t\bigg|\leq C_\lambda n^{-1},
$$
where $C_\lambda>0$ is a constant. Can we get that for every $t>0$,
$f_n(t)$ converges to $f(t)$? Is it possible to get the order of convergence?
If $f_n(t) = \sin(n t)$ and $f(t) = 0$, then the left-hand side is $n/(\lambda^2 + n^2)$, which is $O(1/n)$ despite the fact that $f_n(t)$ does not converge to $f(t)$ pointwise. Or am I missing something?
|
2025-03-21T14:48:29.981048
| 2020-02-25T15:46:17 |
353534
|
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|
Stack Exchange
|
roots and embeddings
Let $G$ be a connected reductive group over an algebraically closed field, can we always find an embedding let $\rho:G\rightarrow GL_n$, that sends a Borel pair $(B_G,T_G)$ to $(B,T)$ and center to center,
such that the image in $X^{*}(T_G)$ of the simple roots of GL_n is included inside the positive span of simple roots of $G$?
This question is in fact related to functoriality for the wonderful compactification (see Thm. 4.7):
https://www.ias.ac.in/public/Volumes/pmsc/109/03/0241-0256.pdf
If the center of $G$ is not cyclic, clearly you can not do that.
I would already be satisfied with G adjoint. In fact, a case by case check shows that it is ok for classical groups, so I would like to know what happens for exceptional types.
|
2025-03-21T14:48:29.981134
| 2020-02-25T16:52:36 |
353535
|
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"Daniele Tampieri",
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"url": "https://mathoverflow.net/questions/353535"
}
|
Stack Exchange
|
Very weak solution to parabolic PDE (pointwise a.e. in time with time derivative on test function)
Consider the parabolic PDE
$$u' + Au = 0$$
as an equality in $L^2(0,T;V^*)$ for some Hilbert space $V$ with $A\colon L^2(0,T;V) \to L^2(0,T;V^*)$ a coercive, bounded linear operator. Here $u'$ is the weak time derivative in the usual sense.
We can write a very weak formulation of the equation as
$$-\int_0^T \langle v'(t), u(t) \rangle + \int_0^T \langle Au(t), v(t) \rangle = 0\tag{1}$$
for every $v$ sufficiently smooth with $v(T)=v(0)=0$.
Consider the pointwise a.e. in time equality
$$-\langle v'(t), u(t) \rangle + \langle Au(t), v(t) \rangle = 0.$$
Suppose that solutions and data are smooth.
Is there any theory on such weak formulations and the relation to (1)? Clearly, it implies (1), but there doesn't seem any way to it from (1) (or some form of the original PDE). Is it a useful concept in any way (even for theoretical reasons)?
I'm not an expert but, since (1) is a linear evolution equation, I think most of the studies on its generalized solutions adopt the concept of mild solution, i.e. interpret it as a solution of the following vector-valued integral equation $$ u(T)-u_0=\int\limits_0^T!! Au(t),\mathrm{d} t.$$ However, it seems to me that the concept of very weak solution is currently adopted in several recent studies on the Navier-Stokes equations (which is however non-linear), as you can see by googling the relevant words.
|
2025-03-21T14:48:29.981253
| 2020-02-25T17:19:26 |
353538
|
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|
Stack Exchange
|
Generalized "Homology Whitehead" -- How much does stabilization remember?
Classically, the (non-local-coefficients) homology Whitehead theorem says that if $X \xrightarrow f Y$ is a map of simple spaces, and if the induced map $H_\ast(X;\mathbb Z) \to H_\ast(Y;\mathbb Z)$ is an isomorphism, then $f$ is a weak homotopy equivalence.
Conceptually for me, the essence of this theorem is that we have a stable invariant ($H_\ast(-;\mathbb Z)$), and we identify a (reasonably large) class of spaces (the simple spaces) such that our invariant detects equivalences when restricted to this class. I'm wondering how generally a statement of this form holds in a fairly general $\infty$-category $\mathcal C$ in place of $Spaces$.
For my purposes, I'm not particularly concerned with which stable invariant we use, so we might as well restrict attention to the universal case. Moreover, there are two general forms of "stabilization" which come to mind -- the category of spectrum objects $Sp(\mathcal C) = \varprojlim (\cdots \xrightarrow \Omega \mathcal C_\ast \xrightarrow \Omega \mathcal C_\ast)$, and the Spanier-Whitehead category $SW(\mathcal C) = \varinjlim (\mathcal C_\ast \xrightarrow \Sigma \mathcal C_\ast \xrightarrow \Sigma \cdots)$ (I use $\mathcal C_\ast$ to denote the $\infty$-category of pointed objects in $\mathcal C$). But we're eventually passing to some subcategory anyway, so we can reduce the $SW$ notion to the $Sp$ notion if we start out by replacing $\mathcal C$ with $Ind(\mathcal C)$ via the equation $Ind(SW(\mathcal C)) = Sp(Ind(\mathcal C))$.
Thus we are led to the following formulation:
Question: Let $\mathcal C$ be a presentable $\infty$-category. Can we identify a (reasonably large) full subcategory $\mathcal D \subseteq \mathcal C$ such that the composite functor $\mathcal D \to \mathcal C \xrightarrow {\Sigma^\infty_+} Sp(\mathcal C)$ is conservative? In particular, is this the case for $\mathcal D$ being one of the following?
The 1-fold suspension objects?
The 1-fold loop objects?
The 1-connected objects?
Here, a 1-fold suspension object is simply an object of the form $X = \Sigma Y$ for some $Y \in \mathcal C$; a 1-fold loop object is an object of the form $X = \Omega Y$ where $Y \in \mathcal C_\ast$ is a pointed object of $\mathcal C$. A 1-truncated morphism $W \to Z$ is a morphism such that for every $C \in \mathcal C$, the map $\mathcal C(C,W) \to \mathcal C(C,Z)$ has 1-truncated fibers, a morphism is 1-connected if it is left orthgonal to the 1-truncated morphisms, and an object $X$ is 1-connected if the map $X \to 1$ is 1-connected, where $1$ is the terminal object.
As a sanity check, I think each of my candidates for $\mathcal D$ are trivial when $\mathcal C$ has discrete hom-spaces, which is a good thing because in this case $Sp(\mathcal C)$ is also trivial.
Is it important to you that $\mathcal{C}$ be an arbitrary locally presentable $(\infty,1)$-category, or would it be sufficient to consider the case when it is an $(\infty,1)$-topos? In the latter case, the result would follow from a proof of the homology Whitehead theorem in homotopy type theory; I don't know if that exists in the literature yet, but it's at least not too far off.
Taking $\mathcal{C}$ to be $n$-groupoids (aka $n$-truncated spaces), it seems to me that all of your criteria fail (because the stabilization is trivial).
@MikeShulman It would be nice to understand the $\infty$-topos case, although I am interested in going beyond this.
@LennartMeier That's a very good point. Offhand, I can't think of a natural restriction to put on $\mathcal C$ to rule this out, other than simply asking $\mathcal C$ to be an $\infty$-topos...
Another question is whether hypercompleteness will matter. I looked back over my notes and I believe I know how to prove a "homology Whitehead theorem" in HoTT saying that if a map of 1-connected types induces an isomorphism on integral homology, or on cohomology with coefficients in all abelian groups, then the map is $\infty$-connected (i.e. induces an isomorphism on all homotopy groups). If this works, then it would imply a positive answer in the case of $\infty$-toposes where $\mathcal{D}$ is the hypercomplete 1-connected objects. But I don't know how to do better right now.
The cohomology version of @MikeShulman's argument was written up by my student Luis Scoccola on pages 20 and 21 of https://arxiv.org/abs/1903.03245 . He cites Mike's blog post, where these ideas originated.
Putting together my and Dan's comments deserves to be called an answer. Namely:
If $\mathcal{C}$ is an $(\infty,1)$-topos, then the statement is true when $\mathcal{D}$ is the class of hypercomplete, pointed, nilpotent objects. Hypercompleteness is the usual $(\infty,1)$-categorical notion, pointed is obvious, while "nilpotent" here means that the internal group object $\pi_1(X)$ is nilpotent and acts nilpotently on each internal abelian group object $\pi_n(X)$, in an appropriate internal sense.
For a proof, see section 3 of Nilpotent Types and Fracture Squares in Homotopy Type Theory by Luis Scoccola, which proves in homotopy type theory that any cohomology isomorphism between pointed nilpotent types is $\infty$-connected — hence an equivalence if the types are hypercomplete. Then we get the result for $(\infty,1)$-toposes by interpreting homotopy type theory internally therein, as shown here for universes and here for higher inductive types. (Those papers don't yet quite complete the interpretation by showing that the universe is closed under HITs, but I doubt that this proof depends crucially on that.)
Note that the assumption used here is weaker than yours, namely that an isomorphism is induced only on internal cohomology group objects with coefficients in all internal abelian group objects. It seems likely that with your stronger assumption hypercompleteness could be removed from the result, but probably a different method would be required.
|
2025-03-21T14:48:29.981898
| 2020-02-25T17:19:43 |
353539
|
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|
Stack Exchange
|
About generalized binomial theorem and Grünwald-Letnikov fractional derivative
I have run into a problem while computing the fractional derivatives of order $\alpha$ for the Riemann zeta function. My Theorem states
Let $s\in\mathbb{C}$, $\mathfrak{Re}(s)>1$, then the fractional derivative of order $\alpha$ of the Riemann zeta function is given by
$$\zeta^{(\alpha)}(s)\equiv\sum_{k=2}^{\infty}\frac{e^{i \pi \alpha}\log^{\alpha}(k)}{k^s}\text{ .} $$
Research by Guariglia showed that this is indeed the desired expression for the frac. der. of zeta. I tried to prove it the following way:
Proof. We proceed by employing the fact that $| \hspace{0.5mm}.\hspace{0.5mm} |$ is a metric in $\mathbb{R}$. Let $\zeta_N(s)=\sum\limits_{k=1}^{N}\dfrac{1}{k^s}$, where $\lim\limits_{N\rightarrow\infty}\zeta_N(s)=\zeta(s)$, then
\begin{align*}
|\zeta^{(\alpha)}(s)-\mathfrak{D}_{s}^{\alpha}\zeta_N(s)|&=\Big| \zeta^{(\alpha)}(s) - \lim _{h \rightarrow 0^+} \frac{1}{h^\alpha}\sum_{m = 0}^{\infty}(-1)^{m}{\alpha\choose m} \zeta_N(s-m h) \Big|\\
&=\Big| \zeta^{(\alpha)}(s) - \lim _{h \rightarrow 0^+} \frac{1}{h^\alpha}\sum_{m = 0}^{\infty}(-1)^{m}{\alpha\choose m} \sum_{k=1}^{N}\frac{1}{k^{s-m h}} \Big|\\
&=\Big| \zeta^{(\alpha)}(s) -\sum_{k=1}^{N}\frac{1}{k^{s}}\left[\lim _{h \rightarrow 0^+}\frac{1}{h^\alpha}\sum_{m = 0}^{\infty}(-1)^{m}{\alpha\choose m}k^{mh} \right]\Big|\\
&=\Big| \zeta^{(\alpha)}(s) - \sum_{k=1}^{N}\frac{1}{k^{s}}\left[\lim _{h \rightarrow 0^+}\frac{1}{h^\alpha}\sum_{m = 0}^{\infty}{\alpha\choose m}(-k^{h})^m \right] \Big|\\
\end{align*}
Comparing the last series in the last line with the generalized binomial theorem one can notice, that it goes to $(1-k^h)^{\alpha}$. Overall $\zeta_N^{(\alpha)}(s)$ now equates to
$$\zeta^{(\alpha)}_N(s)=\sum_{k=1}^{N}\frac{1}{k^{s}}\lim _{h \rightarrow 0^+}\left(\frac{1-k^h}{h}\right)^\alpha = \sum_{k=1}^{N}(-1)^\alpha \frac{1}{k^{s}}\left(\lim _{h \rightarrow 0^+}\frac{k^h-1}{h}\right)^\alpha\text{ .}$$
The next step is to realize, that the expression $\lim _{h \rightarrow 0^+}\dfrac{k^h-1}{h}$ is simply the limit definition of the natural logarithm. Thus we overall arrive at
\begin{align*}
|\zeta^{(\alpha)}(s)-\mathfrak{D}_{s}^{\alpha}\zeta_N(s)|&=\Big|\sum_{k=1}^{\infty}e^{i\pi\alpha} \frac{\log^\alpha(k)}{k^{s}}-\sum_{k=1}^{N}(-1)^{\alpha} \frac{\log^\alpha(k)}{k^{s}}\Big|\\
&=\Big|\sum_{k=N+1}^{\infty}e^{i\pi\alpha} \frac{\log^\alpha(k)}{k^{s}}\Big|\\
&\le\sum_{k=N+1}^{\infty} \Big|e^{i\pi\alpha} \frac{\log^\alpha(k)}{k^{s}}\Big|\\
&=\sum_{k=N+1}^{\infty} \frac{\log^\alpha(k)}{k^{\mathfrak{Re}(s)}}\\
&<\sum_{k=N+1}^{\infty} \frac{k^\alpha}{k^{\mathfrak{Re}(s)}}\\
&=\zeta(\mathfrak{Re}(s)-\alpha)-\zeta_N(\mathfrak{Re}(s)-\alpha)\text{ ,}
\end{align*}
where the last inequality followed from the fact that $k>\log(k)$ for all $k\in\mathbb{N}$. Notice that when we let $N\rightarrow\infty$ our upper estimate also goes to zero, meaning\
$$\lim\limits_{N\rightarrow\infty}|\zeta^{(\alpha)}(s)-\mathfrak{D}_{s}^{\alpha}\zeta_N(s)|<\lim\limits_{N\rightarrow\infty}\zeta(\mathfrak{Re}(s)-\alpha)-\zeta_N(\mathfrak{Re}(s)-\alpha)=0\text{ .}$$
By the properties of a metric this implies that
$$\lim\limits_{N\rightarrow\infty}\mathfrak{D}_{s}^{\alpha}\zeta_N(s)=\lim\limits_{N\rightarrow\infty}\zeta^{(\alpha)}(s)=\zeta^{(\alpha)}(s)$$
and hence the claim follows.
Now here's the problem: The generalized binomial theorem only converges to $(1-k^h)^\alpha$ for $|-k^h|<1$ which is clearly not the case. But for some reason the limit of h and the corresponding 1/h keep the whole expression regularized and make it go to $(-1)^\alpha\log^\alpha(k)$.
Is there some rigorous way to justify what is happening here? I am trying to resolve this issue for several days now but nothing seems to be working. Help is highly appreciated, thank you!
Best regards,
--Jens
Isn't it known that $\mathfrak{D}_{s}^{\alpha}k^{-s}=e^{i\pi\alpha}\log^{\alpha}(k)k^{-s}$? Using this directly along with absolute convergence for $\Re(s)>1$ seems like it would do the trick...
One more thing: What is the "$\equiv$" meant to symbolize in your first equation? I am most used to that meaning that the two concepts are defined to be equal which does not seem like the case here
|
2025-03-21T14:48:29.982130
| 2020-02-25T17:24:41 |
353541
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353541"
}
|
Stack Exchange
|
Zeroth order method with near-optimal rate that works in practice?
I want to find a ZO (zeroth-order, i.e. no access to gradient) algorithm to minimize a strongly-convex deterministic objective (say, as a sum of smooth and nonsmooth proximable functions). I want such algorithm to have at most $O(n)$ worse theoretical convergence rate than the optimal first-order algorithm which has $O(\sqrt{\frac{L}{\mu}} \log\frac{1}{\epsilon})$ rate, where $n$ is variable dimension, and $L$, $\mu$ are smoothness and strongly-convex constants. I also want this algorithm to work well in practice.
In this paper Random gradient-free minimization of convex functions, Nesterov and Spokoiny gave an algorithm (see algorithm and theorem 9 in: section 6, accelerated random search) that has $O(n\sqrt{\frac{L}{\mu}} \log\frac{1}{\epsilon})$ rate I want, but in practice, even for a toy problem, it runs notoriously slowly and never converges to the $10^{-3}$ accuracy I want. I tried many other ZO methods, they all have very slow practical convergence.
Does anyone know any ZO method (for strongly-convex minimization) that actually works in practice? Please let me know, thank you so much : )
|
2025-03-21T14:48:29.982235
| 2020-02-25T17:27:08 |
353542
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353542"
}
|
Stack Exchange
|
Steering an ODE out of a ball
Suppose that $b:\mathbb{R}\to\mathbb{R}$ is locally Lipschitz and of polynomial growth. Suppose further that there are constants $C_1,C_2>0$ such that $(x-y)(b(x)-b(y))\leq C_1-C_2(x-y)^2$ for all $x\in\mathbb{R}$. The ODE
$$
\dot{x}(t)=b(x(t))+d(t)+u(t),\quad x(0)=x_0,\qquad t\in[0,1],
$$
is then seen to have a unique solution for any integrable functions $d,u:[0,1]\to\mathbb{R}$.
Let us denote $\mathcal{C}_0^\infty([0,1])=\{f\in \mathcal{C}^\infty([0,1]):\, f(0)=0\}$.
Is under this set of conditions the following true?
Claim: Let $R>0$. Then there are $M,\delta>0$ such that given any initial condition $x_0\in\mathbb{R}$ and $d\in\mathcal{C}_0^\infty([0,1])$, we can find a function $u:[0,1]\to\mathbb{R}$ with $\|u\|_\infty\leq M$ and times $0\leq t_1<\cdots<t_n\leq 1$ such that
$$
\sum_{i=1}^{n-1}t_{i+1}-t_i\geq\delta
$$
and $|x(t)|\geq R$ for all $t\in [t_1,t_2]\cup\cdots\cup[t_{n-1},t_n]$.
I managed to prove this for $b$ globally Lipschitz, which is essentially enough to get it under the additional assumption $\inf_x b^\prime(x)$ (since then, in combination with the one-sided Lipschitz condition, the derivative is bounded).
The bounty is for proving that if $\inf_x b^\prime(x)=-\infty$, then it is not possible to find uniform $M,\delta$. For this, I'd content myself with an example, e.g. $b(x)=x-x^3$.
The property in question holds if $\inf_{\mathbb R}b′>−\infty$ and does not hold if, say, $\lim_{x\to\infty}b'(x)=-\infty$ (so $−x$ is OK, but $−x−x^3$ is not)
Thanks for the answer. Do you have an easy argument which shows that the property fails for $-x-x^3$?
Do you have an easy argument...
Sorry for the late reply, but it is, actually, a rather simple story. Let $R=1$ and suppose that we have declared some $M$ and $\delta$. Then the adversary starts at some $x_0\in(-\frac 12,\frac 12)$ and he has some guaranteed time $T=T(b,M)>0$ during which the solution stays in $(-1,1)$ if he keeps $d=0$ on that interval of time. What he does after that is to choose some $L>0$ and make $d$ mimic a big multiple of the $\delta$-measure to achieve a nearly instantaneous shift of the solution by a fixed large constant, so that we find ourselves in, say, $2$-neighborhood of some point $x_1$ such that $b'(x)<-L$ whenever $|x-x_1|<2$. He then switches $d$ to $-b(x_1)$ so that $x_1$ becomes an equilibrium point without control and the convergence to that equilibrium from its $2$-neighborhood is exponential with factor $L$ in the exponent, so in time $C/L$ we reach the $1/4$-neighborhood of $x_1$. If $L$ is large enough (in terms of $M$), our control is next to useless in this area and we still find ourselves in the $1/3$-neighborhood of $x_1$ in time $C/L$ with fixed known $C>0$. The adversary then uses $d$ once more to inflict an almost instantaneous shift by almost $-x_1$ and the solution returns to the $1/2$-neighborhood of the origin, so the cycle can be repeated. All he needs to ensure is that the stabilization time $C/L$ is less than $\frac\delta 2T$, say (which can be done by choosing $L$ large enough) and that the "almost instantaneous shifts" are fast and precise enough.
I love the way you write your answers.
|
2025-03-21T14:48:29.982469
| 2020-02-25T18:28:42 |
353546
|
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"Praphulla Koushik",
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"url": "https://mathoverflow.net/questions/353546"
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|
Stack Exchange
|
Lie group (topological group) action on differentiable stack (topological stack)
Let $G$ be a Lie group and $\mathcal{D}$ be a differentiable stack (I am also ok to start with a topological group and topological stack).
I have seen someone mentioning somewhere that the notion of group action on stacks appeared first in Group Actions on Stacks and Applications by M. Romagny. The below definition of group action on a differentiable stack is from Group actions on stacks and applications to equivariant string topology for stacks by Gregory Ginot and Behrang Noohi.
A Lie group action on a differentiable stack is given by a morphism stacks $\alpha: G\times \mathcal{D}\rightarrow \mathcal{D}$ satisfying some conditions. Though they did not specify, I am believe that by $G$ they mean the stack $[*/G]$, so an action of a Lie group $G$ on a differentiable stack $\mathcal{D}$ is a morphism of stacks $\alpha: [*/G]\times \mathcal{D}\rightarrow \mathcal{D}$ satisfying some conditions (correct me if I am wrong).
Questions :
Is there any notion of Lie group $G$ action on a Lie groupoid $[\mathcal{G}_1\rightrightarrows \mathcal{G}_0]$? Would a pair of maps $(G\times \mathcal{G}_1\rightarrow \mathcal{G}_1, G\times \mathcal{G}_0\rightarrow \mathcal{G}_0)$ giving an action of Lie group on the manifolds $\mathcal{G}_1,\mathcal{G}_0$ compatible with source, target etc maps of Lie groupoid, a good notion of Lie group action on a manifold?
Is the notion of Lie group action on a differentiable stack mentioned above deduced/inspired from some notion of Lie group action on a Lie groupoid, in the sense that this notion of Lie group action on Lie groupoid is Moria invariant giving an action of Lie group on a differentiable stack?
Is this definition of Lie group action on a differentiable stack directly/indirectly related to the notion of action of a group object on an object of a category as mentioned in Definition $2.15$ of Notes on Grothendieck topologies, fibered categories
and descent theory?
Thanks @YCor for the edit :)
I am not sure that $[/G]$ is the correct object to consider. If you want to recover the action of a group on a manifold in the case that $\mathcal D$ is just a manifold, you really need $G$ to be a Lie group, that is a manifold with extra structure maps (like $G\times G\to G$). I don't think $[/G]$ would do the job.
@SebastianGoette That seem to be correct.. :) :) Thank you.. How should I interpret the map $G\times \mathcal{D}\rightarrow \mathcal{D}$ as?
I was hoping to see a good answer to your question by somebody else ...
@SebastianGoette Oh. Please let me know if you have any favorite reference for this set up?
I think that the $G$-action on $\mathcal D$ should be captured by a map $[\mathcal D/G]\to[*/G]$, not the one that you wrote.
@Z.M can you please explain in some more words what you mean..
$G$-torsors on $X$ 1-1 correspond to maps $X\to[/G]$ via pulling back the "universal $G$-bundle" $\to[*/G]$, and the pair of $Y$ along with a $G$-action 1-1 corresponds to the pair of $X$ along with a $G$-torsor on $X$ in the following way: $Y=[X/G]$ and $X$ defines a $G$-torsor on $Y$; and a $G$-torsor on $Y$ defines a $G$-bundle $X\to Y$ and thus $X$ carries a $G$-action. Thia holds in any topos, and in particular, in the topos of differential manifolds.
@Z.M at the level of smooth manifolds, it is clear.. the question was how to understand this at the level of differentiable stack... For a differentiable stack D, a Lie group G, how to think about action of G on D.. one way is to see G as a differentiable stack BG and consider a morphism of stacks $BG\times D\rightarrow D$ (along with some other data)... Do you have any suggestions regarding this idea..
As I said, it should be $G\times D\to D$ (with data of coherent homotopy), not $BG$ ($G$ is the underlying manifold). A general formalism is given in Lurie's Higher Algebra §4.2 (where we see $D$ as a left $G$-module in the Cartesian monoidal structure).
|
2025-03-21T14:48:29.982715
| 2020-02-25T19:21:34 |
353550
|
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"authors": [
"Karl Schwede",
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"url": "https://mathoverflow.net/questions/353550"
}
|
Stack Exchange
|
Existence of ample divisor with smooth image
Let $f:X \rightarrow Y$ be a proper, generically finite morphism between smooth, projective varieties.
Is there an irreducible ample divisor $D$ on $X$ such that $f(D)$ (with the induced reduced scheme structure) is a smooth divisor?
(Note: originally posted here)
Do you mean f(D) set theoretically?
@KarlSchwede I think that is what I mean. Is that not equivalent to the current form of the question?
Sometimes people just mean $f(D)$ as a divisor (in other words throw out any irreducible components of $f(D)$ that are not codimension 1).
@KarlSchwede I'm going to add the assumption that $D$ is irreducible to the question.
Do you have any control at all of what you blew up? Can you assume that $f$ is an isomorphism outside of some non-singular subvariety for instance?
@KarlSchwede My particular situation is as general as the one in the question. But I'd be interested in any solutions in more special cases (like the one you mention).
Ok, the reason I ask is that I think if you have $Z$ a smooth subvariety of $Y$, then the general member of the linear system $H^0(X, O(nA) \otimes I_Z) \subseteq H^0(X, O(nA))$ is still nonsingular for $n \gg 0$ assuming $A$ is ample. This looks pretty close to the linear system you'll get via the pushdown, at least in the birational case.
@KarlSchwede That is helpful, thank you.
|
2025-03-21T14:48:29.982839
| 2020-02-25T20:19:20 |
353553
|
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"Dan Rust",
"KhashF",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353553"
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|
Stack Exchange
|
Examples of minimal topological systems which are not intrinsically ergodic
Consider dynamical systems $(X,T)$ where $X$ is a compact metric space, $T:X\rightarrow X$ is continuous, the system is minimal and finally, $0<h_{\rm{top}}(X)<\infty$. I am looking for examples of such systems that do not admit a measure of maximal entropy (mme). Non-minimal topological systems without mme are easy to construct, so I am asking for examples of minimal systems without mme.
It looks like the answer is given here https://mathoverflow.net/questions/43564/transitive-shifts-with-multiple-fully-supported-mmes
@DanRust Thanks for the link. But doesn't that post discuss systems with multiple MME's? I am looking for a system without any MME.
Apologies, I misread.
The question
Transitive shifts with multiple fully supported MMEs
that @DanRust mentioned above also discusses MME's, but it is concerned with a different way that a system may fail to be intrinsically ergodic, i.e. having multiple MME's. Grillenberger's name comes up in that post. After some search I found this article of his where he constructs examples I was looking for as certain shift systems.
|
2025-03-21T14:48:29.982931
| 2020-02-25T20:34:28 |
353554
|
{
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"authors": [
"Aster Phoenix",
"Chris Wuthrich",
"Jef",
"https://mathoverflow.net/users/110362",
"https://mathoverflow.net/users/143969",
"https://mathoverflow.net/users/5015",
"https://mathoverflow.net/users/84768",
"reuns"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353554"
}
|
Stack Exchange
|
The minimal equation of the Frey curve
In the paper of G.Frey there is a link between stable elliptic curves and certain Diophantine equations. The Frey curve of the equation $A-B=C$ is
$$E :\;y^2=x(x-A)(x-B)$$
where $A=a^p$, $B=b^p$, $C=c^p$.
And he define also the minimal equation of $E$ by the the change of variable $x=4X$ and $y=4X+8Y$ and the the equation of $E$ becomes
$$ Y^2+XY=X^3+ \frac{A+B-1}{4}X^2+\frac{AB}{16}X$$
And the discriminant is $\Delta= \frac{(ABC)^2 }{2^8}$.
My question is why Frey make this change of variable and define two equations of the curve? and why he takes specifically this change $x=4X$, $y=4X+8Y$ what is the aim of all this ?
A naive guess would be that the second equation is more suited for studying its reduction mod $2$.
The paper is there, he is proving that the curve has stable reduction (ie. good or multiplicative) at every prime which is why he needs a model at $2$. Then he is looking at the properties of the morphism $X_0(N)\to E$ assuming the curve is modular.
If you want to understand in general how one obtains the "minimal Weierstrass equation" you should look up Tate's algorithm (eg in Silverman 2, chapter IV.9).
@reuns .@chris withrich so i undestand that when the curve has bad reduction at some point in this we should made miminal model
|
2025-03-21T14:48:29.983065
| 2020-03-02T11:45:07 |
353973
|
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"Fedor Petrov",
"Florian Felix",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353973"
}
|
Stack Exchange
|
Sequences of positive integers $(a_{k})_{k \in \omega}$ that only give finitely many zeros modulo $p_{k}$ in total for all polynomials
Let $(a_{k})_{k \in \omega}$ be a sequence of positive integers such that $a_{k} < p_{k}$, $a_{k} \leq a_{k+1}$ and $\lim_{k \rightarrow \infty} a_{k}=\infty$ where $p_{k}$ is the k-th prime number. Let $f(x)= \sum_{i=0}^{n}b_{i}X^{i}$ be any non-zero polynomial with $b_{i} \in \mathbb{Z}$.
Can we give a nice criteria that the sequence $(a_{k})$ is so that $f(a_{k})\equiv 0 \ (\text{mod} p_{k})$ for infinitely many primes $p_{k}$? (Editnote: Originally I asked if there is an example of such a sequence, but this was not really the intended goal.)
If $(a_{k})$ grows really slow, for example if we start with $1$ and just add one after exponentially increasing intervals, this can not be because the prime numbers outgrow $f(a_{k})$ in any case, so we don't reduce anything at a certain point and because a polynomial has only finitely many integer zeros we are done. But what if we choose for example $a_{k}=k$?
what if $f(x)=x$?
I misformulated the question, thank you! I meant if there can be a sequence such that a polynomial exists such that.... I edited it.
I don't know how much freedom you have in the choice of the sequence and/or the polynomial but I think that you can find examples.
The simplest one I can think of is the sequence with $a_1 = 1$ and $a_k = \frac{p_k+1}{2}$, together with the polynomial $f(x) = 4x - 2$. Then you have that for all $k$, $f(a_k)$ is divisible by $p_k$.
Ah yes, of course. This works, but I think I would be interested in general criteria when this can not happen that are a bit stronger than "when it grows very very slow", thank you anyway!
Ok, I thought it was probably not the answer you expected, I see you have now updated your question. According to your comment shouldn't you phrase your question as: is it possible that for any polynomial $f$, $p_k$ divides $f(a_k)$ for only finitely many $p_k$?
I already know that this is possible by choosing $a_{k}$ to grow really really slow (so slow that every polynomial will have good reduction from some point on), the question is rather "Can one do better? And is there a general criteria?" For example, as stated, I would be interested if $a_{k}=k$ works. Even if I assume that the answer to the question for general criteria will be "No."
|
2025-03-21T14:48:29.983348
| 2020-03-02T13:14:59 |
353977
|
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|
Stack Exchange
|
Connections between eigenvalues of $B$ and $A+iB$
Consider two symmetric and real matrices $A,B\in\mathbb{R}^n$ and definie $A+iB$. Note that $A+iB$ is not hermitian in this case. There are many results based on Brendixson and Courant-Fischer, saying, that for every eigenvalue $\lambda+i\mu$ of $A+iB$ and $\beta_{min},\beta_{max}$ the minimal and maximal eigenvalue of $B$, we get that
$\beta_{min}\leq \mu\leq \beta_{max}$.
My question now is: Are there any reversal inequalities? What can we say about the eigenvalues of $B$ knowing those of $A+iB$?. The case in which I am interested the most is the one with $\mu>0$ for all eigenvalues $\lambda+i\mu$ of $A+iB$.
Do you have any idea how to do this?
Thanks for your help!
|
2025-03-21T14:48:29.983427
| 2020-03-02T13:31:28 |
353978
|
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"Daniele Tampieri",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353978"
}
|
Stack Exchange
|
Fixing constants of a series solution of a fourth-order PDE
The following is the PDE I want to solve,
$$\left(1+x^{2}\right)^{2}y_{xxxx}+8x\left(1+x^{2}\right)y_{xxx} + 4\left(1+3x^{2}\right)y_{xx} + K\left[2x yy_{xx}+\left(1+x^{2}\right)\left(yy_{xxx} + y_{x}y_{xx}\right) - 4(1+3x^{2})y_{xxK} - 4x(1+x^{2})y_{xxxK}\right]=0,\tag{1}$$
This is a fourth-order nonlinear PDE whose boundary conditions read
$$x=0: y_{x} = 0, y+Ky_{K}=0;\ x\to\infty: y_{x}\to1,y+Ky_{K}\to x $$
The following Ansatz satisfies the boundary conditions at $x=0$
$$y(x) = \sum_{m,n=0}^{\infty}a_{n}(K)b_{m}(x)\tag{2},$$
where
$$ a_{n}(K) = \sum_{i=0}^{\infty}\zeta_{n,i}K^{i},\\
b_{n}(x) = \sum_{j=0}^{\infty}\lambda_{m,j}\left(x\ \tan^{-1}\left( x^{j}\right) - \frac{jx^{j+1}}{j+1}\ _{2}F_{1}\left[\frac{j+1}{2j},1;\frac{1}{2}\left(3+\frac{1}{j} \right);-x^{2j}\right] \right).$$
Here, $\zeta_{n,i}$ and $\lambda_{m,j}$ are constants. with $x\to\infty$ the ansatz yields the following relations between the series constants,
$$ \sum_{m,n=0}^{\infty}\sum_{i=0}^{\infty}\zeta_{n,i}K^{i}\left(\frac{\lambda_{m,0}}{2} + \sum_{j=1}^{\infty}\lambda_{m,j} \right) = \frac{2}{\pi}\tag{3}, $$
$$ \frac{1}{2}\sum_{m,n=0}^{\infty}\sum_{i=0}^{\infty}\zeta_{n,i}K^{i}\lambda_{m,0} + \frac{\pi}{2}\sum_{m,n =0}^{\infty}\sum_{i=0}^{\infty}i\zeta_{n,i}K^{i}\left(\frac{\lambda_{m,0}}{2} + \sum_{j=1}^{\infty}\lambda_{m,j} \right) = x\tag{4}.$$
How do I determine the series constants (up to a summation index of $2$)? I couldn't do the same via Mathematica and hand calculations by plugging the ansatz into the PDE seem too messy (but I believe there must be a simpler way which I can't seem to figure out). Since I am interested in solutions only up to a summation index of $2$, I can rely on the equations $(3)$ and $(4)$ to help me find relations between the $0,1,2$ index constants but I can't seem to go further.
I don't understand why this is a PDE: I see only derivations respect to the $x$ variable, therefore I'd expect it to be a ODE and your problem to be a boundary value problem for an higher order ODE. Could you clarify this point?
@DanieleTampieri Just saw that I didn't copy the PDE correctly from my LaTeX script. It is a PDE and I have appended the terms dependent on "K". Sorry about that.
|
2025-03-21T14:48:29.983570
| 2020-03-02T14:57:56 |
353982
|
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"url": "https://mathoverflow.net/questions/353982"
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|
Stack Exchange
|
Vector bundles on $K3$ surface
$\underline {Background}$ : Let $$ be a vector bundle on a $3$ surface which is globally generated away from a finite set of points. We assume that $^0(^∗)=0$. Then from the exact sequence
$0 \to \to ^0()\otimes \mathcal O_ \to \to _→0$
(where $,_$ are respectively kernel and cokernel of the evaluation map) we have another exact sequence
$0 \to ^∗ \to ^0()^∗ \otimes \mathcal O_ \to ^∗ \to \mathcal E ^2(_,\mathcal O_) \to 0$.
Let's denote $_:=\mathcal E^2(_,\mathcal O_)$.
Then we essentially end up having following two S.E.S of vector spaces:
$0 \to (^0())^∗ \to ^0() \to ^1(^∗)\to 0.....(1)$
and
$ 0\to ^0() \to ^0(^∗) \to ^0(_)→0....(2)$
where $=(^0()^* \otimes \mathcal O_)/^∗$.
At this point it's true that " the vector space $=^0()^∗\oplus ^0(_)$ generates $^∗$" using the S.E.S $(1)$ and $(2)$
and the previous discussion.
$\underline {Question(1)}$: Is the following argument for the above fact correct?
$\underline {Attempt}$: We have the exact sequence
$0 \to K \to ((H^0(E))^* \oplus H^0(R_E)) \otimes \mathcal O_X \to V^* \to cokerf \to 0$(where $f$ is the the evaluation map at the middle).
If we assume that $\mathcal Hom(cokerf, \mathcal O_X)$ and $\mathcal Ext^1(cokerf, \mathcal O_X) = 0$(being supported on a finite set(why?)). Then one can show that $V^{**} \cong \mathcal Hom(Imf , \mathcal O_X)$. From there can one conclude $Imf =V^*$?
$\underline {Question(2)}$: How from the previous discussion we end up having the exact sequence : $0→^∗→G^∗→^0(_)\otimes \mathcal O_X→_→0$?
(where $G$ is the dual of the kernel of the surjective evaluation map for $V^*$)
$\underline {Attempt}$: We have the exact sequence
$0 \to K^{'} \to H^0(R_E)) \otimes \mathcal O_X \to R_E \to cokerg \to 0$(where $g$ is the the evaluation map at the middle).
If we assume that $\mathcal Hom(cokerg, \mathcal O_X)$ and $\mathcal Ext^1(cokerg, \mathcal O_X) = 0$(being supported on a finite set(why?)). Then one can show that $R_E^{*} \cong \mathcal Hom(Im(g) , \mathcal O_X)$. From there can one conclude $R_E = Im(g)$?. If we assume that this gives the surjection of $g$, then we already have the right end of the desired exact sequence. Also we can give a map $G^* \to H^0(R_E) \otimes \mathcal O_X$. But I don,t see how to show exactness at that point and also how to complete the sequence at the left.
Any help from anyone is welcome.
|
2025-03-21T14:48:29.983717
| 2020-03-02T15:38:42 |
353986
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353986"
}
|
Stack Exchange
|
Is the switch automorphism inner for continuous-trace $C^*$-algebras?
If $R$ is a commutative ring, and $A$ is an Azumaya algebra over $R$, then the switch (or flip, or exchange, etc.) automorphism of $A\otimes_R A$, given by $a\otimes b\mapsto b\otimes a$, is inner: it is the conjugation by the so-called Goldman element $g\in (A\otimes_R A)^\times$. This element is an important feature of Azumaya algebras, and can be defined as the element corresponding to the reduced trace of $A$ under the natural $R$-module identification $A\otimes_R A\simeq \operatorname{End}_R(A)$ (where the reduced trace is seen as a $R$-linear map $A\to R\subset A$).
Now in the context of $C^*$-algebras, if $R$ is a commutative $C^*$-algebra (corresponding to some $C_0(T)$), my understanding is that the equivalent of Azumaya algebras over $R$ are the continuous-trace algebras $A$ with spectrum $T$. For instance, $A$ is locally Morita-equivalent to $R$; it also has a Morita-inverse: $A\otimes_T \bar{A}$ is Morita-equivalent to $R$, where $\otimes_T$ is the balanced tensor product relative to the spectrum $T$.
The theory of continuous-trace $C^*$ algebras has many similarities to the theory of Azumaya algebras, but I haven't been able to determine if they share this property:
Is there a "Goldman element" $g\in (A\otimes_T A)^\times$ such that the conjugation by $g$ is the switch automorphism? If so, can this element be chosen unitary?
I would be fine with restricting to the case where $T$ is compact (so $R$ has a unit).
The algebraic construction of the element can almost, but not quite exactly, be imitated, since $A\otimes_T A$ can identified as a $C_0(T)$-module with $\mathcal{K}(Y_A)$ where
$$Y_A = \{a\in A\,|\, (t\mapsto tr(aa^*)(t))\in C_0(T)\}$$
is a right Hilbert $C_0(T)$-module. This is similar to $A\otimes_R A\simeq \operatorname{End}_R(A)$ in the algebraic case, but there is no obvious "trace" element in $\mathcal{K}(Y_A)$.
|
2025-03-21T14:48:29.983861
| 2020-03-02T15:59:29 |
353987
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"François Brunault",
"Maksym Voznyy",
"Noam D. Elkies",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/14830",
"https://mathoverflow.net/users/6506",
"https://mathoverflow.net/users/95511"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/353987"
}
|
Stack Exchange
|
A generator needed for a Z/6 elliptic curve
We are searching for rank $8$ elliptic curves with the torsion subgroup $\mathbb{Z}/6$ using newly discovered families similar to Kihara's as described in
A. Dujella, J.C. Peral, P. Tadić, Elliptic curves with torsion group $\mathbb{Z}/6\mathbb{Z}$, Glas. Mat. Ser. III 51 (2016), 321-333 doi:10.3336/gm.51.2.03, 1503.03667
and came across a curve
[0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]
Both Magma Calculator and mwrank (with $-b14$) return $7$ generators for this curve:
SetClassGroupBounds("GRH");
E:=EllipticCurve([0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]);
MordellWeilShaInformation(E);
Using model [ 0, 1, 0, -60313024735007362096072931339173916555726439220,<PHONE_NUMBER>437764940265112114690488828303891527290367159038723810320068 ]
Torsion Subgroup = Z/6
The 2-Selmer group has rank 9
New point of infinite order (x =<PHONE_NUMBER>960497219694209104164/9078169)
New point of infinite order (x = -146551684206472947976069)
New point of infinite order (x = -151681843950496144133344)
New point of infinite order (x =<PHONE_NUMBER>71733499670296)
New point of infinite order (x =<PHONE_NUMBER>343712794396956)
New point of infinite order (x =<PHONE_NUMBER>3919437193768749863407/42159049)
New point of infinite order (x =<PHONE_NUMBER>402839599279348827593704/4311629569)
After 2-descent:
7 <= Rank(E) <= 8
Sha(E)[2] <= (Z/2)^1
(Searched up to height 10000 on the 2-coverings.)
Both Magma and mwrank return $8$ for the upper bound on rank:
E:=EllipticCurve([0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]);
TwoPowerIsogenyDescentRankBound(E);
8 [ 3, 3, 3, 3, 3 ]
[ 7, 7, 7, 7, 7 ]
mwrank -v0 -p200 -s
[0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]
Version compiled on Oct 29 2018 at 22:35:09 by GCC 7.3.0
using NTL bigints and NTL real and complex multiprecision floating point
Enter curve: [0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068]
Curve [0,1,0,-60313024735007362096072931339173916555726439220,3303762732437764940265112114690488828303891527290367159038723810320068] : selmer-rank = 9
upper bound on rank = 8
Considering parity, there should be one more generator on the curve.
Is there a way to find it?
We would greatly appreciate any hint leading to the discovery of the extra generator.
A bounty of $100$ will be offered for obtaining it.
Also, if you can compute an extra generator, your name will be published at the bottom of the page here: https://web.math.pmf.unizg.hr/~duje/tors/z6.html.
So far, we are unsuccsessful applying Zev Klagsbrun's approach.
I'd appreciate the title not to contain the OP's opinion of the difficulty of the question (and other meta information).
Thank you for the comment. The title has been edited.
Did you try applying either method to the 3-isogenous curve E' ? It's possible that it will turn up one or more points whose image(s) in E aren't all in the rank-7 subgroup you've already found.
@Noam D. Elkies: Do you mean 2-isogenous curve E'? If you mean 3-isogenous, how would I produce it?
@Noam D. Elkies: We tried both Magma Calculator and mwrank (with -b14) on the 2-isogenous curve E'=[0,1,0,-52962146173716537846156113928697292067549954365,4689992072216871348070877367918171236315109027756733265235941818761900]. Only 7 generators were produced.
No, I mean 3-isogenous. Since there's a 3-torsion point there's a 3-isogenous curve. Using the 2-isogenous curve will produce nothing new because mwrank already works with a 2-isogenous pair when there's a rational 2-torsion point. But the subgroup that pulls back to the 3-isogenous curve E' is easier to compute on E', and may include something new.
@Noam D. Elkies: Thank you for the clarification. How would we get the 3-isogenous curve E' explicitly in SageMath or Magma?
@Noam D. Elkies: Also, is (-269400128398817960910169,0) a 3-torsion point you are referring to?
@Noam D. Elkies: For the suggested 3-torsion point above, E' must be [0,1,0,-2158797086129675572238568551455780340059707846220,-1220725611278427420118075288418495156876490140275772729026826816863424332]. Am I right? Magma Calculator's answer is still 7 <= Rank(E) <= 8.
Is it "the same" rank-7 subgroup, though? One easy test is to compare the regulators. If their ratio is a small power of 3 then it must be the same group. If it looks like a random real number then mapping back to E should give something independent of your known rank-7 group on E.
@Noam D. Elkies: Sorry, so far I am not sure what I am doing at all. Before checking that the rank-7 subgroup is "the same", I would really appreciate if somebody checked the 3-torsion point and the 3-isogenous curve E' for me please. The above point is probably wrong and the curve could also be [0,1,0,-146514059534015623106469373299474230011420497965,-15603537836846114118299910584749213985141648571195287986036304039959680]. I am not sure.
Your computation seems correct. Here is the Magma code to compute isogenies using Vélu's formulas: F3 := DivisionPolynomial(E,3);
F := Factorization(F3)[1][1];
E3, phi := IsogenyFromKernel(E,F);
@François Brunault: Thank you so much! Your code really helps on the Magma side. I did my calculations in SageMath, now I know how to check them.
The eights generator was successfully computed by Zev Klagsbrun! It is (-825240948184709245504424305468714422091334601593572177730440863632644222609685944791784/3476235906470724630424489414741262801265768781870915852371028441:13350668242169750290895245364589704907786881444602849078579635591912831036781435950810619013243973916848136198892088368285547068910/204957522013550460549477456089656119391219129674638303351158680324230445521351939753256298608739:1) and has canonical height 169.76. I am ready to award the 100 bounty as promised when the answer is posted.
In this case, it pays to work on the curve $E'$ that is 2-isogenous to $E$, which is given by the equation
$$
y^2 = x^3 +<PHONE_NUMBER>98226941365253x^2+<PHONE_NUMBER>258164849983363482095324897635296971x.
$$
It is relatively easy to find the 7 independent points
(110776963853866550724016 : -80505869468630089210131377497504980 : 1),
(32748343004768539696321 :<PHONE_NUMBER>9610504103217516288626615 : 1),
(33849219894746702769856 : -23485175376256902691040244680173080 : 1),
(83632124830832490757371607899/221801449 :<PHONE_NUMBER>678270035941408483841167567865756495/3303288979957 : 1),
(708467443931328192488072059/9 :<PHONE_NUMBER>2867793914636996097237707997855/27 : 1),
(82669556642133426513025 : -58721781791249872164660769545564855 : 1),
(15653787556726119039025946369377024/458311398169 :<PHONE_NUMBER>260403572244890413120572309377214021131920/310270858512236803 : 1)
on $E'$.
To find the final generator for $E'$, we need to apply 4-descent. We can find the final point by applying 4-descent to any number of 2-covers of $E'$. I got a hit on the second one I tried, which was given by
$$
C: y^2 =<PHONE_NUMBER>5797531781124x^4 +<PHONE_NUMBER>16514838017136x^3 -<PHONE_NUMBER>39703266788148x^2 -<PHONE_NUMBER>233000587306308x +<PHONE_NUMBER>121696083297961.
$$
One 4-covering of $C$ is given by the intersection of the quadrics
$$
26478x^2 + 149391xy + 147873xz + 592534xw - 114021y^2 - 58434yz + 336829yw + 118629z^2 + 510074zw + 438488w^2
$$
and
$$
148610x^2 + 19042xy + 1022361xz - 1631065xw - 112833y^2 + 39500yz + 44513yw - 182441z^2 + 822710zw + 972880w^2
$$
in $\mathbb{P}^3$. This has a rational point at (9608:-18440:7168:6485).
The corresponding rational point $Q$ on $E'$ is given by
(-1117913472469704108682566452343804675555656978809017451310789/5063218268474760928272238797769806961 :<PHONE_NUMBER>298703272836151916049253048501046484070626117498677044566639368499019825133763515/11393049242085218620499121263510902732524161971013459959 : 1),
which has canonical height $106.35$. Using the other generators, we can find the somewhat smaller point $Q'$
(-101978171213582065261997068234175010796855613305889831/9689020674223383052192046420224 :<PHONE_NUMBER>70312399327355677945959801245147861170127803803586963373362965000839965/30159200458008252634403235853082432918163525632 : 1)
having canonical height $91.18$ that is independent of the first seven generators.
The discriminant of $E'$ is somewhat smaller than that of $E$, so arguably $E'$ is the model that you'd want to present. However, if you want to work on $E$, then we can map the generators of $E'$ to $E$. A "nice" choice of points that generator $E(\mathbb{Q})$ modulo torsion is:
(225814423074482435222996 : -34626227198383025293628840137707870 : 1),
(-151681843950496144133344 :<PHONE_NUMBER>0541536347147233612637650 : 1),
(-209328450874403020690564 :<PHONE_NUMBER>4719522705484038947202890 : 1),
(30089244387022819458738989/4 : -164963185085547559044466862956833045285/8 : 1),
(1521645572343712794396956 : -1853309164504557572407788096216777750 : 1),
(-108427296385410370841188 :<PHONE_NUMBER>0027620341222482486229910 : 1),
(50259473549510044315364816963047/529 : -356309090633909063646983926957844863796356762164/12167 : 1),
(-825240948184709245504424305468714422091334601593572177730440863632644222609685944791784/3476235906470724630424489414741262801265768781870915852371028441 :<PHONE_NUMBER>2169750290895245364589704907786881444602849078579635591912831036781435950810619013243973916848136198892088368285547068910/204957522013550460549477456089656119391219129674638303351158680324230445521351939753256298608739 : 1).
The final generator has canonical height $169.76$, which means that it would be more difficult to find by searching directly on 4-coverings of $E$.
|
2025-03-21T14:48:29.984308
| 2020-03-02T16:34:37 |
353990
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/353990"
}
|
Stack Exchange
|
A question on p-rationality of number fields
Let $p$ be an odd regular prime and $F$ be a $p$-rational number field containing $\mu_p$. Equivalently, there is a unique prime $\mathfrak{p}$ above $p$ in $F$ and the $p$-class group is generated by $\mathfrak{p}$ [Gras' book on Class Field Theory IV.3.5]. Let $F_\mathfrak{p}$ be the maximal pro-$p$ unramified outside $\mathfrak{p}$ extension of $F$. Is it guaranteed that the inertia group of $\mathfrak{p}$ has finite index in $Gal(F_\mathfrak{p}/F)$? Somehow is it true that in a cyclic $p$-extension of a $p$-rational field, the unique prime above $p$ must ramify and can not remain inert?
I asked your question to my advisor Professor Movahhedi who introduced and studied the notion of $p$-rationality in his thesis.
Here I share his answer with you.
The two questions are somehow of different nature.
Denote by $F_{S_p}$ the maximal $p$-extension of the $p$-rational filed $F$ unramified outside the prime(s) above $p$. Suppose that $F_{S_p}$ has only one prime above $p$; for instance, when our $p$-rational field $F$ contains the $p^{th}$-roots of unity $\mu_p$ as is assumed. Denote by ${\mathfrak p}$ and ${\cal P}$ the primes above $p$ in $F$ and $F_{S_p}$ respectively.
Then the corresponding decomposition group $D({\cal P}/{\mathfrak p})$ is the whole Galois group $G_{S_p}:=Gal(F_{S_p}/F)$ which can also be identified to the local Galois group after completion at the primes above $p$.
The corresponding inertia group $I({\cal P}/{\mathfrak p})$ is normal in the decomposition group $D({\cal P}/{\mathfrak p})$ and we have a short exact sequence
$$0 \rightarrow I({\cal P}/{\mathfrak p}) \rightarrow G_{S_p}=D({\cal P}/{\mathfrak p}) \rightarrow Gal(E/F) \rightarrow 0, $$
where the inertia field $E$ is the maximal unramified extension of $F$ contained in $F_{S_p}$ since ${\cal P}$ is the only ramified prime in $F_S/F$.
Therefore $E/F$ is a finite (cyclic) extension as was to be shown: the inertia group $I({\cal P}/{\mathfrak p})$ is a normal subgroup of finite index in $G_{S_p}=D({\cal P}/{\mathfrak p})$. Moreover the quotient is cyclic.
We also notice that, by maximality of $F_{S_p}$, it is clear that $E$ is in fact the maximal unramified $p$-extension of $F$.
As far as the second question is concerned, the answer is negative. For this, you first need to be familiar with the notion of primitive sets of primes which play an important role in the going-up properties :
Definition (Movahhedi's thesis, Definition 1, page 42).
A set $S$ of primes of $F$ containing $S_p$ is called primitive for ($F, \; p$) if
the Frobenius $\sigma_v({{\tilde{F}}}_1/F)$ "attached" to the primes $v$ in $S-S_p$ generate an ${\bf F}_p$-subspace of ${\rm Gal}({{\tilde{F}}}_1/F)$ of dimension the
cardinality of $S-S_p$, where ${{\tilde{F}}}_1$ is the compositum of
the first layers of the ${\bf Z}_p$-extensions of $F$.
Now, for each odd regular prime $p$, there exist infinitely many $p$-rational fields $F \supset \mu_p$ admitting a cyclic extension of degree $p$ (contained in $F_{S_p}$) in which the $p$-adic prime of $F$ is inert. Namely, start with the cyclotomic field $k:=Q(\mu_p)$ which is $p$-rational. Consider a maximal primitive set $T:=\{{\mathfrak p, \mathfrak L_1, \mathfrak L_2, \cdots, \mathfrak L_r}\}$ for $(k,p)$, where $r:=(p+1)/2$ is the number of independent $Z_p$-extensions of $k$ and ${\mathfrak p}=(\zeta_p-1)$ is the $p$-adic prime of $k$. By the Čebotarev density theorem, infinitely many such primitive sets exist. Since $k$ is $p$-principal (i.e. $p$ does not divide the class number of $k$), for each $i= 1, 2, \cdots, r$, there exists an integer $\alpha_i$ in $k$ such that ${\mathfrak L_i^{t_i} = (\alpha_i)}$ for some prime-to-$p$ integer $t_i$. Now consider the field $F$ obtained by adding to $k$ a $p$-th root of $a:=(\zeta_p-1)\alpha_1 \alpha_2 \cdots \alpha_r$, where $\zeta_p \neq 1$ is a $p$-th root of unity : $F=Q(\zeta_p, \sqrt[p]{a}).$
Then the cyclic Kummer extension $F/k$ is ramified exactly at the primes in the primitive set $T$ (notice that the binomial $X^p-a$ is of Eisenstein type with respect to each prime in $T$) and unramified elsewhere.
Therefore, $F$ is $p$-rational according to the following theorem giving the necessary and sufficient condition for the $p$-rationality to go up along a finite $p$-extension.
Theorem (Movahhedi's thesis, Theorem 2, page 50).
Let $F/k$ be a finite (Galois) $p$-extension of number fields.
Then $F$ is $p$-rational precisely when $k$ is $p$-rational and the set $S_k$ of those primes in $k$ lying above $p$ or ramified in $F$ is primitive for ($k,p$).
Moreover, if $F$ is $p$-rational, then the extension $S_F$ of $S_k$ to $F$ remains primitive for ($F,p$).
Besides, since $p$ does not divide the class number of $k$, applying Chevalley's genus formula to the cyclic extension $F/k$, we have the following equality for the order of the class number of $F$ fixed by the Galois group $G:=Gal(F/k)$
$$|Cl(F)^G|=\frac{\prod_{v}e_v(F/k)}{\textstyle [F:k][U_k:U_k\cap N_{F/k}(F^*)]} $$
with the usual notations :
$e_v(F/k)$ is the ramification index of $v$, $U_k$ is the unit group and $N_{F/k}$ is the norm map corresponding to the extension $F/k$.
An easy calculation (maximality of the chosen primitve set $T$ and Dirichlet's unit theorem) then shows that, in our situation, $p$ divides the class number of $F$. Therefore, there exists a non-trivial cyclic $p$-extension of $F$ (namely the $p$-Hilbert class field of $F$) contained in $F_{S_p}$ in which the $p$-adic prime ${\mathfrak p}$ is inert.
Notice that smaller primitive sets may produce a field $F$ with the same properties if the norm index $[U_k:U_k\cap N_{F/k}(F^*)]$ is small.
Numerically, one can take $p=3$, $k:=Q(\zeta_3)$ and let ${\mathfrak L_7}$ and $\mathfrak L_ {19}$ each be one of the primes above $7$ and above $19$, respectively. Then the set $T=\{{\mathfrak p, \mathfrak L_7, \mathfrak L_ {19}}\}$ is a maximal primitive set for $(k,3)$
[see Movahhedi's thesis, bottom of page 49].
It is not hard to see that we can take ${\mathfrak L_7}=(\zeta_3-2)$ and ${\mathfrak L_{19}}=(\zeta_3-7)$, so that
$$a:=(\zeta_3-1) (\zeta_3-2) (\zeta_3-7)= 33\zeta_3-3.$$
Hence, one corresponding such field is $F=Q(\zeta_3, \sqrt[3]{33\zeta_3-3})$, which is both $3$-rational and admits an unramified cyclic extension $E$ of degree $3$ (inside $F_{S_p}$) in which $3$ remains inert. This field $F$ is obtained by adding to $Q$ a root of $X^6 + 39X^3 + 1197$ which is the minimal polynomial of $\zeta_3 \sqrt[3]{33\zeta_3-3}$.
One could also take for $F$ the "smaller" field $F:=Q(\zeta_3, \sqrt[3]{\zeta_3-7})$, in which only the primes ${\mathfrak p}=(\zeta_3-1)$ and ${\mathfrak L_{19}}=(\zeta_3-7)$ are ramified and have the same properties :
$F$ is both $3$-rational and admits an unramified cyclic extension $E$ of degree $3$ (inside $F_{S_p}$) in which $3$ remains inert. Indeed, $F$ is $3$-rational according to the above theorem since the set $\{{\mathfrak p, \mathfrak L_ {19}}\}$ is , a fortiori, primitive for ($k,3$). On the other hand, since $N_{F/k}(2+\sqrt[3]{\zeta_3-7})=1+\zeta_3$, then $N_{F/k}((2+\sqrt[3]{\zeta_3-7})^2)=\zeta_3$ and the norm index in the above genus formula is therefore trivial. Accordingly, the class number of $F$ is divisible by $3$. Notice also that this $3$-rational field $F$ is obtained by adding to $Q$ a root of $X^6 +15X^3 + 57$ to $Q$ which is the minimal polynomial of $\zeta_3 \sqrt[3]{\zeta_3-7}$.
|
2025-03-21T14:48:29.984722
| 2020-03-02T17:29:00 |
353993
|
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|
Stack Exchange
|
Holomorphic local trivialization of a principal toric bundle
Let $G$ be an even-dimensional compact Lie group with Lie algebra $\mathfrak{g}$ and let $T \subset G$ be a maximal torus with Lie algebra $\mathfrak{t}$.
We can construct a left-invariant complex structure over $G$ in the following way. Let $\Sigma$ be the set of roots associated to the Cartan algebra $\mathbb C \mathfrak{t}$ (the complexification of $\mathfrak{t}$) and $\mathfrak g_\alpha$ the eigenspaces associated to the root $\alpha \in \Sigma$. We denote by $\Sigma_+$ a choice of positive roots of $\Sigma$.
Now, let $\mathfrak{m} \subset \mathbb C \mathfrak{t}$ be any subalgebra such that $\mathfrak{m} \oplus \overline{\mathfrak{m}} = \mathbb C \mathfrak{t}$, we then define $\mathfrak h = \mathfrak{m} \oplus \bigoplus_{\alpha \in \Sigma_+} \mathfrak g_\alpha$. The Lie algebra $\mathfrak h$ defines a complex structure on $G$ such that every left-translation of $G$ is holomorphic. And since $T$ is abelian, the structure $\mathfrak m$ is a bi-invariant complex structure of $T$ and thus $T$ is a complex Lie group.
In [1], it is said that $T \to G \to G / T$ is ``obviously'' a holomorphic principal bundle. I tried to verify this claim, I can prove that $G$ can be regarded as a smooth principal bundle with structure $T$, but I could not prove that this bundles is holomorphically locally trivializable.
How can I prove that such bundle is holomorphically locally trivializable? Are there some references I could follow?
I do not know if you could call this bundle a holomorphic principal bundle without this property, the the authors of [1] need to apply the Borel spectral sequence, and in the construction of such spectral sequence, it is necessary to use a local trivialization compatible with the complex structure of the fiber and of the base space.
[1]: Vanishing theorems on Hermitian manifolds - Bogdan Alexandrov and Stefan Ivanov (https://www.sciencedirect.com/science/article/pii/S0926224501000444)
Proposition 4.26 in Félix, Oprea, Tanré, Algebraic Models in Geometry applied to $T \to G \to G/T$ shows what you want since $G/T$ is Kähler manifold. (See also Example 4.32.)
Thank you very much for your answer. Unfortunately, the mentioned proposition does not answer the question. The proposition just explain how to construct a complex structure on G, which the authors claim that could be chosen to be either left or right-invariant. Still, it is not clear why this bundle would by holomorphically locally trivializable. Also, it is not clear that (in the case of G having a left-invariant structure) why the right action by T would be holomorphic (also necessary for it to be a holomorphic bundle).
|
2025-03-21T14:48:29.985007
| 2020-03-02T17:30:14 |
353994
|
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|
Stack Exchange
|
sequential learning reference request
I would like to find a book for master math student about the following topics.
I don't know the field and I don't want to be lost in details so if it could be an straight forward please.
I'm a math studen but I would like to apply myselft to artificial intelligence so, I would like the books is applied and contains images. Illustrations (graph of function, schema) are important for me as exemples and remarks.
1) Sequential learning problems, Stochastic Gradient Optimisation.
2) Learning with expert advices, mirror
descent.
3) Learning in an unknown environnement. Exploration/exploitation
trade-off, Information-Theoretic lower bounds.
Thank you for your help.
|
2025-03-21T14:48:29.985088
| 2020-03-02T18:31:57 |
353998
|
{
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"authors": [
"Andreas Blass",
"Andrés E. Caicedo",
"Goldstern",
"Hannes Jakob",
"Noah Schweber",
"Todd Eisworth",
"https://mathoverflow.net/users/138274",
"https://mathoverflow.net/users/14915",
"https://mathoverflow.net/users/18128",
"https://mathoverflow.net/users/6085",
"https://mathoverflow.net/users/6794",
"https://mathoverflow.net/users/8133"
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"url": "https://mathoverflow.net/questions/353998"
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|
Stack Exchange
|
Are there forcing notions adding $\kappa$ random, sacks, prikry, or Mathias reals?
This is my first question here, so if I am doing things incorrectly, please let me know. Now on to the question:
The forcing notion $Fn(\kappa,2)$, which constists of partial functions from $\kappa$ to $2$ with finite support, ordered by reverse inclusion, is known to add $\kappa$ new cohen reals.
Are there similare forcing notions adding $\kappa$ new random, sacks, prikry, or Mathias reals?
Misread this at first as to asking about such forcings for cardinal $\kappa$, instead of adding $\kappa$ many reals.
"Notion", rather than "notation".
The reason $Fn(\kappa,2)$ works for adding $\kappa$ new cohen reals is, because there exists a bijection between $\kappa$ and $\kappa\times\omega$, therefore the forcing $Fn(\kappa,2)$ is isomorphic (as a partial order) to $Fn(\kappa\times\omega,2)$ which itself is isomorphic to a Finite Support iteration of Length $\kappa$ of Cohen Forcing.
Similarly, for any forcing of your desired type (be it random, sacks, ... forcing), you could use a Finite Support iteration of length $\kappa$ to obtain your desired forcing. If you need an introduction to Forcing Iterations, Jechs book on Set Theory is a good start.
Why do we need to use iterations instead of products here?
A finite support iteration of non-ccc forcings, such as Mathias forcing or Sacks forcing, will (after $\omega$ many steps) always collapse $\omega_1$. This is usually an undesirable feature, and this is a reason why countable support iteration was invented. (But CS iterations has its own problems...)
@NoahSchweber For example, if you take the product of two Mathias forcings, it will lose some of the interesting properties that a single Mathias forcing or a CS iteration of Mathias forcing has, such as the Laver property. (The product of two dominating forcing notions, or of three unbounded forcing notions, will add a Cohen real.) - Also, both FS and CS products of infinitely many Mathias forcings will collapse $\aleph_1$. - Of course there are other produts (e.g. mixed support)
@Goldstern Sure, it's better to do so, but with the product you still add $\kappa$-many reals which are whatever-generic over $V$.
In this case using FS Iterations also highlights the similarity, as $Fn(\kappa\times\omega,2)$ is simply an FS Iteration of $Fn(\omega,2)$ of Length $\kappa$.
In addition to products and the iterations already mentioned, random reals admit the option of forcing with the measure algebra of $2^\kappa$.
Could you elaborate on the existence of a measure algebra on $2^{\kappa}$? As far as i know, there is currently no known way of constructing a (non-degenerate) measure on $2^{\kappa}$.
@HannesJakob Take the product measure induced from the uniform measure on $2$.
Of course, im sorry, i confounded $2^{\kappa}$ and $\kappa^{\kappa}$.
|
2025-03-21T14:48:29.985300
| 2020-03-02T18:39:14 |
353999
|
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|
Stack Exchange
|
Some questions regarding computation of the Mordell-Weil group
I was reading the theory relevant to Selmer and Shafarevich-Tate groups from Silverman's AEC. And I have a lot of doubts related to these topics:
First, I don't understand the reasoning behind the name 'homogeneous spaces'? Does it have something to do with homogeneous spaces from Lie theory?
Let's talk by 'descent via isogeny' method now. Some of the ideas are still hazy to me and I am having difficulty in tying together all the things discussed in Chapter X of the book.
Here is what I understood: We want to compute $E(K)$ for an arbitrary number field, $K$, but the whole argument in the Mordell-Weil theorem tells us that it is sufficient to compute the weak Mordell-Weil group, $E(K)/ mE(K)$ ($m$ is as given by the descent theorem and is not arbitrary). Now, using the following exact sequence:
$$ 0 \longrightarrow \frac{E'(K)[\hat{\phi}]}{\phi(E(K)[m])} \longrightarrow \frac{E'(K)}{\phi(E(K))} \longrightarrow \frac{E(K)}{mE(K)} \longrightarrow \frac{E(K)}{\hat{\phi}(E'(K)} \longrightarrow 0 $$
w see that to compute $\frac{E(K)}{mE(K)}$, it is sufficient to compute $\frac{E'(K)}{\phi(E(K))}$ and $\frac{E(K)}{\hat{\phi}(E'(K)}$. Further, using the exact sequence:
$$ 0 \longrightarrow \frac{E'(K)}{\phi(E(K))} \longrightarrow S^{(\phi)}(E/K) \longrightarrow \Sha(E/K)[\phi] $$
in certain cases, like when Sha group is trivial we are able to compute to our group $\frac{E'(K)}{\phi(E(K))}$ (and similarly the other one and hence the group $E(K)/mE(K)$. Now my questions are:
Why do we use an isogeny $\phi$ with degree $m$ instead of directly using the isogeny 'multiplication-by-$m$'?
In general, it is not that easy to compute E(K) as the example 4.10 in Chapter X of the book shows, right? I mean in that particular example, we didn't have to go through the whole descent procedure to compute E(K) from E(K)/2E(K) but that need not happen in general, right? Also, in the same example, we picked a 2-isogeny but $m$ need not be $2$ here, right?
Also is it correct that 'descent via isogeny' method, that is computing the abovementioned two groups work when there is at least one $\phi$-torsion point that is also $K$-rational? (This doesn't really seem to be a very effective way to compute the weak Mordell-Weil group to me though.)
And as for its name (2-descent via isogeny), is it because we are reducing the problem of finding $E(K)/mE(K)$ to first finding two Selmer groups and then their corresponding quotient groups?
It would be great if someone here could help confirm or deny things I have said above. They will help a great deal in understanding this whole discussion of Mordell-Weil group computation.
And as always, thank you all for reading and responding to my queries!
Edit: I forgot that I was on mathoverflow and not math StackExchange. While I repost this to the math StackExchange, I think it would be okay to let it remain here as well?
To the question at the top: the definition is exactly the same as that in Lie theory, except we require the maps to be algebraic instead of just continuous/smooth - it's a space with a free and transitive action of our elliptic curve.
The questions below: 1. The smaller the degree of the isogeny the easier it is to compute. We always have $[m]$ but if $m>10$ there is little hope to be able to compute it directly. So if there is an isogeny of prime degree (e.g. $2$ or $3$) that is defined over $\mathbb{Q}$ then use it. --- 2. Right. --- 3. Descent by isogeny can be done if the isogeny is defined over $K$ that is a weaker assumption than saying that the kernel contains a $k$-rational point. -- 4. Well, yes.
For answer 1. I don't see why direct computation of $E(K)/mE(K)$ would be difficult in case m>10, is it because then it has more elements ($m^2$) than the group $E'(K)/\phi(E(K))$ (and also the other one)does ($m$) and thus overall it's difficult to compute the bigger group?
|
2025-03-21T14:48:29.985559
| 2020-03-02T18:50:37 |
354000
|
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|
Stack Exchange
|
A question about Stroock's notes on the Weyl lemma
On p.4 of these notes, D. Stroock gives a quick and efficient construction of the Markov transition functions of a certain diffusion. The idea of his construction (on page 4) is to 'freeze' the coefficients of the Kolmogorov forward equation during "inter-dyadic" time-intervals $[m2^{-n},(m+1)2^{-n})$ whereby he obtains an approximate transition functions $P_n(x_2|x_1,t)$ simply by an extended convolution of Gaussians. Towards the bottom third of p.4 he says:
Using elementary facts about weak convergence of probability measures, one can
show that there is a continuous map $(t, x)|\mapsto P(.|x_0,t)$ such that $P_n(.|x_0,t) → P(.|x_0,t)$ uniformly on compacts.
I think I can follow him in that rush of thought insofar that I believe the family $\{P_n(.|x_0,t)\}_n$ is easily shown to be tight. Prokhorov's theorem can subsequently get some things done: there exists a weakly converging subsequence of the measures $P_n(x|x_0,t)dx$. Subsequently, I see a way to show that in fact the sequence must converge in its entirety. I however don't see how he finds that the limit has a continuous density $P(.|x_0,t)$ and how $P_n(.|x_0,t) \to P(.|x_0,t)$ uniformly on compacts. I know that it could be accomplished if one uses Arzelà-Ascoli, which requires as a pre-condition the equicontinuity and boundedness of the functions $\{P_n(.|x_0,t)\}_n$, which I don't know how to prove and which don't seem to fit well under the nomer "elementary facts about weak convergence" so that I think that I might be on the wrong track.
I don't think that he ever claims that the $P(t,x)$ have a continuous density, but that the continuity he's referring to is the continuity $(t,x) \mapsto P(t,x)$ from $\mathbb{R}^{N+1}$ into the space of probability measures (with the topology of weak convergence).
I see, but why is the continuity of that dependence on $(t,x)$ and its uniform convergence on compact $x$-regions relevant in the immediate sequel of the discussion? I.e. he writes "
Moreover, because, uniformly on compacts, $\langle L_y\phi, \Gamma({\tau}_n,y) \rangle \to L\phi$". I don't see why you have to bother about the $x$-dependence to establish that fact (rather the increasing concentration of $\Gamma({\tau}_n,y)$ seems what matters).
I may again misunderstand his purposes (or just be mistaken). But I see no other occurrence in the subsequent discussion where facts about the dependence on $(t,x)$ seem to matter.
|
2025-03-21T14:48:29.985746
| 2020-03-02T19:54:53 |
354005
|
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|
Stack Exchange
|
Fixed part of a line bundle on a K3 surface
This question comes from Huybrechts' lecture notes on K3 surfaces, more specifically, chapter 2.
Let $ X $ be a K3 surface (over an algebraically closed field $ k $) and $ L $ a line bundle on $ X $. The base locus of the linear system $ |L| $ is defined as a closed subscheme of $ X $ by $$ \text{Bs} (L) := \cap_{s \in H^0(X,L)} Z(s) $$ where $ Z(s) $ is the zero locus of the section $ s $.
On a surface $ X $, the base locus may have components of dimension zero and one. Let $ F $ be the one-dimensional part, called the fixed part of $ L $.
(1) Why is $ F $ a divisor on $ X $?
I fail to see why $ F $ should necessarily be a divisor, the one dimensional component may have something bad like embedded points. But provided that it is one,
(2) Why is $ h^0(X,F) = 1 $?
Huybrechts' doesn't really give any explanation and uses this 'fact' later to prove that $ F $ is a sum of smooth rational curves. I would really appreciate any help.
(1) I guess, as a scheme, the base locus might have embedded points. But the fixed part is defined as the pure 1-dimensional part of the base locus scheme.
(2) If $F$ is the fixed part, it means that every divisor in the linear system can be written as
$$
D = D' + F.
$$
One can also assume that $F$ has no common components with $D'$.
If $h^0(X,F) > 1$ then $F$ is linearly equivalent to some $F'$, and then the original linear system also contains the divisor
$$
D' + F'.
$$
But $D' + F'$ does not contain $F$, hence the fixed part of the original linear system is strictly smaller than $F$.
Thank you for the answer. I had a question about the assumption of $ D' $ and $ F $ not containing a common component. This seems to rely on the following fact: if the field is infinite, then a finite dimensional vector space is never a finite union of proper subspaces. Or is there some other way to see this?
|
2025-03-21T14:48:29.985897
| 2020-03-02T21:21:36 |
354011
|
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|
Stack Exchange
|
Equivalence of the union-closed sets conjecture that is locally weaker of any use?
Let $F$ be a union-closed family. We call $F$ minimal if for every $x\in \cup(F)$ we find $S\in F$ such that $S\backslash \{x\} \in F$. It is sufficient to proof the union-closed sets conjecture for minimal families as explained here. In the following we will always assume $F$ to be minimal.
Given $X\subsetneq \cup (F)$ and $S\in F$ let $S_X$ be the maximal element in $F$ with $S_X \subseteq S\cup X$. We now define $F\backslash X:= \{S\backslash X \mid S\in F\}$, $F_X^{\wedge}:=\{S_X \mid S\in F \}$, $F_X^{\vee}:=F\backslash F_X^{\wedge}$ and $(F_X^{\vee})_x := \{ S\in F_X^{\vee} \mid x\in S\}$ for $x\in X^{c}$ (complement in $\cup(F)$). One can make the following two observations:
(1) We have $|F\backslash X| = |F|-|F_X^{\vee}|$ and $|(F\backslash X)_x| = |F_x| - |(F_X^{\vee})_x|$.
By that we will proof that the union-closed sets conjecture is equivalent to the following statement. For all union-closed families $F$ we find $X\subsetneq F$ such that for all $x\in X^{c}$ we have $|(F_X^{\vee})_x| \geq \frac{|F_X^{\vee}|}{2}$.
Proof. $"\Rightarrow":$ Let $F$ be a union-closed family that satisfies the union-closed sets conjecture for the element $x\in \cup(F)$. By (1) it follows that $|F_{\{x\}^c}^{\vee}| = |F|-2$ and $|(F_{\{x\}^c}^{\vee})_x| = |F_x|-1$. We thus have $|(F_{\{x\}^c}^{\vee})_x| = |F_x|-1 \geq \frac{|F|-2}{2} = \frac{|F_{\{x\}^c}^{\vee}|}{2}$.
$"\Leftarrow":$ Since $F$ is minimal we have $|F\backslash X| < |F|$ so we can inductively assume that $F\backslash X$ satisfies the union-closed set conjecture for say $x\in X^{c}$. By (1) it follows that $|F_x|-|(F_X^{\vee})_x|=|(F\backslash X)_x | \geq \frac{|F\backslash X|}{2} = \frac{|F|-|F_X^{\vee}|}{2}$ but we also have $|(F_X^{\vee})_x| \geq \frac{|F_X^{\vee}|}{2}$ and thus $|F_x| \geq \frac{|F|}{2}$.
Since the proof for $"\Leftarrow"$ used induction it would not work if we fix a family $F$ however the proof for $"\Rightarrow"$ would. So in a sense the newly introduced statement is weaker locally and thus (maybe?) easier to proof. Could it be of any use in proving the union-closed sets conjecture?
It does not follow that the empty set is in F. Take F to be all subsets of the union having at least one element less than the union. The empty set is in F in the case that union is a singleton, and not for any larger set. Gerhard "Is Not Following The Link" Paseman, 2020.03.02.
Thanks for the correction.
|
2025-03-21T14:48:29.986063
| 2020-03-02T21:25:12 |
354012
|
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|
Stack Exchange
|
Invariant complex structures for simple Lie groups
For which simple Lie groups there exists a left-invariant complex structure?
Clearly for the complex ones, so the question is probably whether these are the only ones? (could naturally be asked for semisimple groups too)
This MathSE answer can be relevant.
|
2025-03-21T14:48:29.986117
| 2020-03-02T23:13:20 |
354015
|
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}
|
Stack Exchange
|
Rigorous error estimate for semi-discrete heat equation in bounded domain
Let $\Omega$ be a bounded Lipschitz domain in $\mathbb R^N$ and $u_h$ be a solution of
$$
\begin{cases}
\partial_t u_h -\Delta_h u_h = f(x) & \text{ in } \Omega_h\\
u_h=0 &\text{ in } \partial \Omega_h \setminus \Omega
\end{cases}$$ where $\Delta_h$ is the finite difference approximation of the Laplace operator, $\Omega_h$ is an extension of $\Omega$:
$$\Omega_h=\{x \in \mathbb R^N: dist(x,\Omega) < h \}.$$
How can we estimate the error $\Vert u_h - u\Vert_{L^\infty(\bar \Omega)}$, where $u$ is the solution of $$
\begin{cases}
\partial_t u -\Delta u = f & \text{ in } \Omega\\
u=0 &\text{ in } \partial \Omega
\end{cases}$$
(i.e. the corresponding continuous problem)?
There are entire books on finite difference methods for PDEs. Also, it is by no means clear what "the" finite difference approximation of the Laplace operator is on a domain that is not a rectangle.
@MichaelRenardy Yes, there are many books on the topic, but I cannot find a rigorous error estimate anywhere. For the definition, we extend the domain and set $u_h = 0$ in $\Omega_h \setminus \Omega$.
|
2025-03-21T14:48:29.986225
| 2020-03-03T02:21:16 |
354017
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354017"
}
|
Stack Exchange
|
A possible generalization of Gauss Lucas theorem to higher dimension
A real half space in $\mathbb{C}^2$ is $$\{(z,w)\in \mathbb{C}^2\mid \phi(z,w) > \lambda\}$$ where $\lambda$ is a real number and $\phi$ is a $\mathbb{R}$- linear functional from $\mathbb{C}^2$ to $\mathbb{R}$.
Assume that $p(x,y)\in \mathbb{C}[x,y]$ has all its roots in a real half space $K$. Is it true that all critical points of $p$ must necessarily lie in $K$?
Can you provide an example?
@geocalc33 As trivial example $z=0$.There is no critical point.
The real half space is $Re(z)>-1$
You might be able to use a convex space argument here because the real half space is a convex set. Given a set $X,$ a convexity over $X$ is a collection $$ of subsets of X satisfying the $(X, ).$ It's hard to refine the convex sets in question in higher dimensions. I would suggest reading this paper, which delves into generalisations of Gauss Lucas: researchgate.net/publication/221966131_Multivariate_Gauss-Lucas_Theorems
and this: [4] A. W. Goodman: Remarks on the Gauss-Lucas theorem in higher dimensional space,Proc. Amer. Math. Soc. 55 (1976), 97–12
There is a very pretty multivariate extension of Gauss-Lucas proved by Marek Kanter here As far as I can tell, the paper has not been published, but the proof of the main theorem is half a page, so...
|
2025-03-21T14:48:29.986346
| 2020-03-03T02:24:02 |
354018
|
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|
Stack Exchange
|
Elliptic curves and its Neron model
Let $E$ be an elliptic curve over $\mathbb{Q}$. For a prime $p$, let $\mathcal{E}_p$ denote its Neron model over $\mathbb{Z}_p$. Also, let $\Phi_p(E)$ denote the component group of $\mathcal{E}_p$.
The structure of $\Phi_p(E)$ is well-known, and I want to study it when $E$ has multiplicative reduction at $p$. First, if $E$ has split multiplicative reduction at $p$, then it is known that $\Phi_p(E) \simeq \mathbb{Z}/{n\mathbb{Z}}$, where $n$ is the (normalized) $p$-adic valuation of the discriminant of $E$. (I will admit this fact.) Next, if $E$ has non-split multiplicative reduction at $p$, then $\Phi_p(E) \simeq \mathbb{Z}/{m\mathbb{Z}}$, where $m=1$ or $2$ such that $m \equiv n \pmod 2$.
Here is my question: Suppose that $E$ has non-split multiplicative reduction at $p$. As above, we use the same notation ($m$, $n$, $\Phi_p(E)$ etc).
We know the following.
There is a unramified quadratic extension $L/{\mathbb{Q}_p}$ such that $E/{\mathcal{O}_L}$ has split multiplicative reduction.
Neron model does not change under etale base change.
If the above two are correct, then the component group $\Phi_p(E)$ can be computed using the Neron model of $E/L$, where $E$ has split multiplicative reduction. Since the $p$-adic valuation does not change under the unramified extension, the component group of the Neron model of $E/L$ is isomorphic to $\mathbb{Z}/{n\mathbb{Z}}$. Thus, $\Phi_p(E)$ is also isomorphic to $\mathbb{Z}/{n\mathbb{Z}}$, which is not true if $n>2$.
I don't know where my argument fails. Please correct me!
The Galois group of $L/{\mathbb Q_p}$ acts on the component group over $L$ and the component group over $\mathbb Q_p$ should be the subgroup of elements fixed by the Galois group.
|
2025-03-21T14:48:29.986485
| 2020-03-03T05:47:06 |
354024
|
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|
Stack Exchange
|
Asymptotic number of $3$-AP's in a set $A\subseteq\mathbb{F}_{p}^{n}$ of density $\epsilon$
Problem: Let $p$ be an odd prime number and consider the $n$-dimensional vector space over the field with $p$ elements. I want to prove that the number of $3$-term arithmetic progressions in a subset
$A\subseteq\mathbb{F}_{p}^{n}$ is $cp^{2n}$ for every $n\geq N_{0}$, for sufficiently large $N_{0}$ and a constant $c>0$.
Fact 1: I shall use Meshulam's theorem which states that if a subset $B\subseteq\mathbb{F}_{p}^{n}$ has density bigger than $2/n$, i.e. $|B|> \frac{2p^{n}}{n}$, then $Β$ contains a $3$-term arithmetic progression.
Fact 2: I also may use the following fact: First observe that if $U$ is a subspace of $\mathbb{F}_{p}^{n}$, then there are $p^{n-k}$ distinct cosets of $U$.
Now, let $A\subseteq\mathbb{F}_{p}^{n}$ be a subset of density $\epsilon>0$.
Then, there are at least $\frac{\epsilon}{2}p^{n-k}$ cosets $V$ of $U$ such that $|A\cap V|\geq \frac{\epsilon}{2}p^{k}$.
As an affine subspace of $\mathbb{F}_{p}^{n}$ is just a coset of a subspace of $\mathbb{F}_{p}^{n}$, we have that the above implies that there exist at least $\frac{\epsilon}{2}f(k,n)$ $k$-dimensinal affine subspaces $V$ of $\mathbb{F}_{p}^{n}$ such that $|A\cap V|\geq \frac{\epsilon}{2}p^{k}$, where $f(k,n)$ is the total number of the $k$-dimensional affine subspaces of $\mathbb{F}_{p}^{n}$. This number equals to
\begin{equation}
f(k,n)=p^{n-k}\binom{n}{k}_{p}
\end{equation}
where $\binom{n}{k}_{p}$ is the Gaussian coefficient and id defined as
\begin{equation}
\binom{n}{k}_{p}=\frac{(p^{n}-1)(p^{n}-p)\cdots(p^{n}-p^{k-1})}{(p^{k}-1)(p^{k}-p)\cdots (p^{k}-p^{k-1})}.
\end{equation}
My attempt goes as follows: Let $n$ be large enough so that there exists $k<n$ such that $\frac{\epsilon}{2}>\frac{2}{k}$. Then, we consider $p^{n-k}$
$k$-dimensional affine subspaces $V$ of $\mathbb{F}_{p}^{n}$ pairwise disjoint (we can take such by taking the $p^{n-k}$ of a $k$-dimensional subspace). We know that there $\frac{\epsilon}{2}p^{n-k}$ of them such that $|A\cap V|\geq \frac{\epsilon}{2}p^{k}>\frac{2p^{k}}{k}$. Then, considering each $A\cap V$ as a subspace of $V\cong\mathbb{F}_{p}^{n}$, Meshulam's theorem implies that each one of them contains a $3$-AP and since they are pairwise disjoint there are $\frac{\epsilon}{2}p^{n-k}$ $3$-AP's in $A$.
That is as far as I get. My idea is to use the more powerful statement of Fact 2, where by finding $cp^{2n}$ disjoint $k$-dimensional affine subspaces of $\mathbb{F}_{p}^{n}$ it implies the desired. But I have no clue how to find them. Any ideas?
Are you aware of Varnavides' paper https://academic.oup.com/jlms/article-abstract/s1-34/3/358/847953?redirectedFrom=fulltext , or, say, https://arxiv.org/pdf/1203.2383.pdf by Serra and Vena?
No, these two papers do not refer to $\mathbb{F}_{p}^{n}$.
Sure, they do not solve the specific problem you posed. But they suggest the directions and the circle of ideas.
I found the wording of the question a little confusing. But if I have understood it correctly, you want some result saying that large subsets contain many progression? Perhaps Corollary 3.2 in this paper answers your question...https://arxiv.org/pdf/1905.08457.pdf
Yeah, Corollary 3.2 answers my question, though it is based upon the work of E. Croot, V. Lev, P. P. Pach and J. Ellenberg and D. Gijswijt, rather thatn Meshulam's as I wanted to. Thank you. Yet, I read again my question and I do not seem to understand what do you mean by "confusing wording".
The recent advances of Croot-Lev-Pach and Ellenberg-Gijswijt aren't needed here, the Meshulam bound is enough, and the ideas you mention in the question. The weakness in your approach is trying to take the affine subspaces pairwise disjoint. You can improve this to get the result you want by selecting affine subspaces uniformly at random instead, and using the first moment method.
I was going to comment with a link to where this Varnavides idea is written up, but to my surprise I couldn't find one simply done in the case of $\mathbb{F}_p^n$, so I thought I'd sketch the idea here. (Of course none of this is original to me, it's one of those proofs that is well-known in the field, and is a routine generalisation of the Varnavides proof.)
Let $A\subset \mathbb{F}_p^n$ be a set with density $\lvert A\rvert/p^n=\epsilon$. Let $k$ be some integer to be chosen later, and let $T$ be the number of three-term arithmetic progressions in $A$ (where here I'm only talking about genuine 3APs, i.e. $x,x+d,x+2d$ with $d\neq 0$).
Let $U$ be an affine subspace of $\mathbb{F}_p^n$ of dimension $k<n$ chosen uniformly at random. We compute the expected number of 3APs in $A\cap U$ in two different ways.
Let $q$ be the probability that a fixed 3AP is in $U$ (this is clearly independent of which 3AP we're talking about). Then by linearity of expectation the expected number of 3APs in $U\cap A$ is just $qT$.
On the other hand, the expected density of $A\cap U$ in $U$ is $\epsilon$. We convert this into a lower bound for the expected number of 3APs as follows. Let $L$ be the total number of affine subspaces we're choosing from, and let $L'$ be the number of such subspaces $U$ where $\lvert A\cap U\rvert \geq \frac{2}{k}\lvert U\rvert$ (such subspaces are 'good'). In particular, by Meshulam's bound, any such $U$ has the property that $A\cap U$ contains at least one non-trivial three-term arithmetic progression
We know that $\sum_{U} \lvert A\cap U\rvert \geq \epsilon p^k L$. The contribution from non-good $U$ is at most $\frac{2}{k}(L-L')p^k$. The contribution from good $U$ is trivially at most $L'p^k$. Therefore,
$$ L'+\frac{2}{k}(L-L')\geq \epsilon L,$$
and hence after rearranging, assuming $\epsilon\geq 4/k$ and $k\geq 4$, say, $L'\geq \frac{\epsilon}{4}L$, and hence the probability that $U$ is good is $\geq \epsilon/4$.
Since any good affine subspace contains at least one 3AP, the expected number of 3APs in $U\cap A$ is $\geq \frac{\epsilon}{4}$. Comparing this to the other calculation, we see that $T\geq \epsilon/4q$. We can calculate $q$ as follows.
We know that the total number of 3APs in $\mathbb{F}_p^n$ is exactly $p^n(p^n-1)$. Similarly the number in any fixed affine subspace of dimension $k$ is exactly $p^{k}(p^{k}-1)$. Therefore
$$ q p^n(p^n-1)= p^{k}(p^{k}-1),$$
and so
$$ q = \frac{p^{k}-1}{p^{n-k}(p^n-1)}\ll p^{-2n+2k}.$$
Therefore $T \gg p^{2n-2k}$. Our requirement on $k$ was that $\epsilon \geq 4/k$, and so we can select some $k=O(\epsilon^{-1})$, and hence
$$ T \gg \epsilon p^{O(\epsilon^{-1})}p^{2n} $$
as required.
Notice that the type of dependence on $\epsilon$ was dependent on Meshulam's bound, but for a qualitative bound (so just $T\geq c(\epsilon)p^{2n}$ for some $c_\epsilon$ depending only on $\epsilon$) any density result will do. Similarly, the more powerful result of Ellenberg-Gijswijt yield a correspondingly better dependence on $\epsilon$.
Hi Thomas forgive me if I am misunderstanding your argument but I am not able to see why it follows from Meshulam's theorem that if the expected density of $A \cap U$ in $U$ is high enough to get a single 3-AP then the expected number of 3-APs in $A \cap U$ must be at least 1. Isn't it possible for $A$ to be concentrated in a small number of subspaces in which case Meshulam's criteria would not be triggered very often which would lead to a lower average?
Yes, you're absolutely right - one needs to use some kind of 'popularity principle' to show that there is in fact a high probability that $A\cap U$ has large enough density, which would then give the required lower bound for the expected count of 3APs. I've corrected this in the answer.
|
2025-03-21T14:48:29.987298
| 2020-03-03T06:02:16 |
354025
|
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|
Stack Exchange
|
General linear group action on extensions of finite fields
Let $q$ be a prime power. Let $\mathbb{F}_q$ be the finite field with $q$ elements. Then $\mathbb{F}_{q^n}$ is a field extension of $\mathbb{F}_q$ of degree $n$ and can be considered as an $n$-dimensional vector space $V$ over $\mathbb{F}_q$. Now consider the action of $GL(V)$ on $V$. Any element $x$ in $\mathbb{F}_{q^n}$ acts on itself by multiplication and hence defines an element in $GL(V)$; we denote this element as $g_x$. Suppose now an element $h\in GL(V)$ commutes with $g_x$ for some $x\neq0,1\in \mathbb{F}_{q^n}$. Does it imply that $h=g_y$ for some $y\in \mathbb{F}_{q^n}$ as well? Thanks.
If $\Bbb{F}{q}(x)$ is smaller than $\Bbb{F}{q^n}$ then not necessarily (try with $x=1$).
If $\Bbb{F}{q}(x)=\Bbb{F}{q^n}$ then yes if the minimal polynomial of $h$ is irreducible ?
Sorry I should have said x not equal to 1 or 0.
I see. The answer should be false in general. But when would it be true?
Let $F$ be any field and $F<E$ a finite field extension.
Fix $x\in E^*$ and consider the multiplication operator $g_x\in \text{GL}_F(E)$,
the group of invertible $F$-linear transformations of $E$.
The centralizer of $g_x$ could be naturally identified with the subgroup $\text{GL}_{F[x]}(E)<\text{GL}_F(E)$, where $F[x]<E$ is the subfield of $E$ generated by $x$ over $F$. We thus have
$$ E^* \simeq \text{GL}_E(E)<\text{GL}_{F[x]}(E)<\text{GL}_F(E).$$
The inclusion $\text{GL}_E(E)<\text{GL}_{F[x]}(E)$ is an equation iff $F[x]=E$.
To see the "only if" part it is enough to recall that $\text{GL}_n$ is never commutative for $n\geq 2$.
Thus every element in the centralizer of $g_x$ is of the form $g_y$ for $y\in E^*$ iff $E=F[x]$, that is $x$ generates $E$ over $F$.
Specializing to $F=\mathbb{F}_q$ and $E=\mathbb{F}_{q^n}$,
we get that every element in the centralizer of $g_x$ is of the form $g_y$ for $y\in \mathbb{F}_{q^n}^*$ iff $x$ generates $\mathbb{F}_{q^n}$ over $\mathbb{F}_q$.
Uri is right the result is immediate because "$h\in End_{\Bbb{F}q}(\Bbb{F}{q^n})$ commutes with $g_x$" is the definition of "$h$ is $\Bbb{F}_q[x]$ linear" thus it is defined by its action on a $\Bbb{F}q[x]$ basis of $\Bbb{F}{q^n}$.
Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$.
If $k(x)$ is smaller than $K$ then every $k(x)$-linear endomorphism of $K$ commutes with $x$, most of them are not of the form $g_x$.
Assume that $K=k(x)$
$g_x$ has an eigenvector $v\in K^n$ with eigenvalue $x$. Since $g_x\in M_n(k)$ then $$g_x(v^{q^l}) =(g_x(v))^{q^l}=(xv)^{q^l}=x^{q^l}v^{q^l}$$
Since the $x^{q^l}$ are distinct the $v^{q^l}$ must be distinct and we have diagonalized $g_x$ $$g_x(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l x^{q^l} v^{q^l}$$
If $h\in M_n(k)$ commutes with $g_x$ then $h$ has an eigenvector which is an eigenvector of $g_x$, wlog we can assume it is $v$ so that $h(v)=av$ for some $a \in F$ where $F$ is the splitting field of $h$'s minimal polynomial. As before we obtain the diagonalization from $$h(v)=av \implies h(v^{q^l}) = a^{q^l} v^{q^l}, \qquad h(\sum_{l=1}^n c_l v^{q^l})=\sum_{l=1}^n c_l a^{q^l} v^{q^l}$$
$v^{q^n}= v$ implies that $a^{q^n}=a$ thus $a\in K$. Whence $a = f(x)$ for some $f\in k[T]$ and $$h = f(g_x)=g_a$$
|
2025-03-21T14:48:29.987516
| 2020-03-03T06:22:03 |
354027
|
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|
Stack Exchange
|
Do Neron-Severi groups of smooth projective unirational varieties contain $\ell$-torsion?
Let $X$ be a smooth projective unirational variety over an algebraically closed field of characteristic $p>0$, and $\ell\neq p$ a prime. My question: can the Neron-Severi group of $X$ contain (non-zero) $\ell$-torsion? This appears to be closely related to the presense of torsion in the etale cohomology group $H^2_{et}(X,\mathbb{Z}_l)$.
It seems that existence of $\ell$-torsion is possible. Here's one example.
Let $k$ be an algebraically closed field of characteristic $p$. In the paper "An Example of Unirational Surfaces in Characteristic p." (https://eudml.org/doc/162649) Shioda proves that a hypersurface $X_n = \{x_1^n + x_2^n + x_3^n + x_4^n = 0\} \subset \mathbb{P}^3_k$ is unirational if $p+1$ is divisible by $n$. For $n = \ell$, where $\ell \neq p$ is a prime number, such unirational hypersurface admits a free action of the group $\mathbb{Z}/\ell$ given by the formula:
$$
k\in \mathbb{Z}/\ell \quad k \cdot (x_1,\ldots,x_4) = (\xi^{ka_1} x_1, \xi^{ka_2} x_2, \xi^{ka_3} x_3, \xi^{ka_4} x_4),
$$
for $\xi$ a primitive $\ell$-th root of unity and $\{a_i\}$ pairwise distinct integers not divisible by $\ell$.
The quotient $Y_\ell = X_\ell / \mathbb{Z}/\ell$ then has non-trivial $H^1_{et}(Y_\ell,\mathbb{Z}/\ell)$ and hence non-trivial $\ell$-torsion in the Picard group. Since $H^1(X_\ell,\mathcal{O}_{X_\ell}) = 0$ and $p$ is coprime to $\ell$, we see that $H^1(Y_\ell,\mathcal{O}_{Y_\ell}) = 0$ and hence $Pic^0(Y_\ell) = 0$. Consequently our $\ell$-torsion in $Pic(Y_\ell)$ yields $\ell$-torsion in $NS(Y_\ell) = Pic(Y_\ell)/Pic^0(Y_\ell)$.
|
2025-03-21T14:48:29.987639
| 2020-03-03T07:33:59 |
354030
|
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|
Stack Exchange
|
Degree of normal closure of number fields
Let $K\subset E\subset F$ be a tower of number fields (i.e. finite extensions of $\mathbb Q$). Assume $E/K$ and $F/E$ are non-trivial normal extensions, and let $F^*/K$ be the normal closure of $F/K$.
What are the (optimal) bounds for $[F^*:K]$ in terms of $[F:E]$ and $[E:K]$?
From the fundamental theorem of Galois theory, it follows that $\mathrm{Gal}(F^*/K)$ is a subgroup of the wreath product $\mathrm{Gal}(F/E) \wr \mathrm{Gal}(E/K)$ and moreover this will be the generic situation, giving the optimal bound $[E:K] \cdot [F:E]^{[E:K]}$.
Great! Thanks a lot for your very good answer. Two follow up questions: What happens with the bounds when we remove the assumption that E/K is Galois? What do you mean by generic?
If you go through the exercise of actually understanding my comment, the answers to your followup should also be clear.
Ok, I will try my best. Thanks.
|
2025-03-21T14:48:29.987736
| 2020-03-03T08:20:22 |
354033
|
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"url": "https://mathoverflow.net/questions/354033"
}
|
Stack Exchange
|
What is the motivation to define measure valued solutions to a PDE model?
Consider the model
$$\partial _{t}\mu + \partial_{x}(b(t,\mu)\mu)+c(t,\mu) \mu=0,~~~\mu \in \mathcal{M}^{+}(\mathbb{R}^{+}), t \in [0,T], x \in \mathbb{R}^{+} $$
$$ \mu(0)=\mu_{0} $$
where $ \mu (t)$ is the measure determining the distribution of the population with respect to the structural variable $x$ and $b,c$ are vital rates.
I want to know what it means for a solution to be measure-valued i.e. solution is a measure?
If the solution (in weak sense) is measurable function then it makes sense to me but solution as a measure is really confusing me, because measure is defined on a set and solution we usually need at every point of domain.
Perhaps you are just confused by the terminology? The domain of your solution is a set, $\mathbb{R}^+$ in this case. While pointwise values $\mu(t,x)$ do not make sense for a measure, what does make sense is any integral of the form $\int_a^b w(x)\mu(t,x)$, giving you a weighted average of $w(x)$ over the interval $x\in[a,b]$ with respect to $\mu(t)$, a "population distribution" in your case. Perhaps such weighted averages are all that you need from your solution.
@IgorKhavkine yes, that is what I was missing, because solution here in my case is a solution of the population model.
I don't know this concrete equation but I know many similar. The thing that you should remember (and @IgorKhavkine explained it well) is that if you work on PDE problems of this kind, you can't only have pointwise solutions (that is kinda rare). You'll usually have some weak solutions such as measure-valued solutions, solutions in the distribution sense, etc.
|
2025-03-21T14:48:29.987892
| 2020-03-03T08:34:16 |
354034
|
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|
Stack Exchange
|
On what proper Gromov hyperbolic space does a free product act?
Per Bowditch, a group is relatively hyperbolic if it acts geometrically finitely on a proper geodesic Gromov hyperbolic space.
A free product of two (or finitely many) finitely generated groups is well known to be relatively hyperbolic. It also acts on an associated Bass-Serre tree which is however locally infinite.
My question: Is there a nice explicit description of a proper geodesic Gromov hyperbolic space on which the free product acts geometrically finitely?
So, how does the proof of relative hyperbolicity for free products work, if not by exhibiting an action?
In his paper "Relatively Hyperbolic Groups" Bowditch proved that relative hyperbolicity (with the above definition) is equivalent to admitting an action on a (locally infinite) hyperbolic graph K such that the following condition hold.
All edge stabilizers are finite.
The number of orbits of edges is finite.
The graph K is fine, that is, for every n ∈ N, any edge of K is contained
in finitely many circuits of length n. (Here circuit means a cycle without self–
intersections).
But I wanted a very explicit description of the action on a proper hyperbolic metric space. It feels like there should be some action on a finite valence tree lurking...
@ThiKu probably because in the definition of relative hyperbolicity, no properness (of the space) is required.
@ThiKu Also because it immediately follows from Farb's definition of relative hyperbolicity, which is known to be equivalent to Bowditch definition (which is approximately what Yellow Pig answered in his/her comment above).
@YellowPig, I think your intuition isn't right here. A proper action on a finite-valence tree would imply that your group is virtually free.
@YCor What do you mean by 'no properness" is required ? In Farb or Osin's formulation, there is no properness indeed, you just ask that the coned-off graph is hyperbolic and that the BCP property is satisfied. However, in Bowditch definition, you need a geometricaly finite action on a proper, geodesic, hyperbolic metric space.
@HJRW But the action is not required to be proper, right ? It's required to be geometrically finite. Am I missing something ?
@M.Dus -- I thought that a geometrically finite action has to be proper; is that not the case? Anyway, I'm pretty sceptical that there can be a geometrically finite action of a finitely generated group on a finite-valence tree unless the group is virtually free.
@HJRW Hum you're right I got confused. My bad. Also, maybe the OP might clarify what should be "some action on a finite tree lurking". Is it an actual action on a tree ? Is it an action somehow involving a tree ? For instance, is AGenevois answering your question ? I'm pretty sure the OP was aware of Groves and Manning's construction...
@M.Dus -- what makes you think that? I see nothing in their profile history which suggests they had already seen the Groves--Manning construction.
I think that the characterisation of relatively hyperbolic groups given in Groves and Manning's article Dehn filling in relatively hyperbolic groups answers your question. I give a few details:
Let $G:=\underset{1 \leq i \leq n}{\ast} A_i$ be a free product of $n$ finitely generated groups. For every $1 \leq i \leq n$, fix a finite generating set $S_i$ of $A_i$. Clearly, $S:= \bigcup\limits_{i=1}^n S_i$ is a finite generating set of $G$. The Cayley graph $\mathrm{Cayl}(G,S)$ is naturally a tree of spaces, the vertex-spaces being Cayley graphs of $A_i$'s. Now, the idea is to glue "horoballs" on the vertex-spaces. More precisely, consider
$$X:= \left( \mathrm{Cayl}(G,S) \cup \bigcup_\limits{g \in G} \bigcup\limits_{i=1}^n \mathcal{H}(gA_i) \right) / \sim,$$
where the combinatorial horoball $\mathcal{H}(gA_i)$ over $\mathrm{Cayl}(A_i,S_i)$ is glued on $gA_i$.
The combinatorial horoball $\mathcal{H}(Y)$ over a graph $Y$ is defined as follows. The vertex-set of $\mathcal{H}(Y)$ is $Y \times \mathbb{N}$. If $u$ and $v$ are two adjacent vertices of $Y$, connect $(u,0)$ and $(v,0)$ with an edge. Also, for every $k \geq 0$ and for every vertex $u \in Y$, connect $(u,k)$ and $(u,k+1)$ with an edge. Finally, for every $k \geq 0$, if $u,v \in Y$ are two vertices satisfying $d_Y(u,v) \leq 2^k$, connect $(u,k)$ and $(v,k)$ with an edge.
It turns out that $X$ is a proper hyperbolic space, and $G$ naturally acts on it.
This is a nice answer, but I think the OP was hoping for another hyperbolic space whose description is well-known. Note that you can also glue ideal triangles rather than combinatorial horoballs, using Bowditch's construction.
Gluing horoballs is a pretty well known operation, going back pretty far into the history of Gromov hyperbolic spaces.
Too long for a comment. Here is another approach when the free factors are torsion-free abelian groups. Let $G=\mathbb{Z}^{d_1}*\mathbb{Z}^{d_2}...*\mathbb{Z}^{d_n}$. You can realize $G$ as a generalized Schottky group, acting on the real hyperbolic space $H^n$ for some $n$ via a geometrically finite action. You first take a geometrically finite kleinian group $G_0$ whose parabolic subgroups are exactly $\mathbb{Z}^{d_1}$,...,$\mathbb{Z}^{d_n}$. Now for each subgroup $\mathbb{Z}^{d_i}$, you choose some large $k_i$ and consider powers $e_1^{k_i},...,e_{d_i}^{k_i}$ of the standard generators $e_1$,...,$e_{d_i}$. If those $k_i$ are chosen sufficiently large, then an easy application of the ping-pong lemma shows that the subgroup $G_1$ generated by these elements is a free product.
Since the subgroup of $\mathbb{Z}^{d_i}$ generated by $e_1^{k_i}$,...,$e_{d_i}^{k_i}$ is still isomorphic to $\mathbb{Z}^{d_i}$, $G_1$ is isomorphic to $G$.
Note that you can play the same game with torsion-free nilpotent groups, replacing the real hyperbolic plane by some simply connected Riemannian manifold of pinch negative curvature, but there are some differences. To simplify, assume that $G$ is of the form $G=G_1*\mathbb{Z}$, where $G_1$ is torsion-free nilpotent.
Then, $G_1$ can be realized as the stabilizer of a cusp in a finite volume manifold of pinched negative curvature. This is a consequence of several results, including those of P. Ontaneda's spectacular paper Pinched smooth hyperbolization, see this question on mathoverflow for more details. You then choose a loxodromic element $g$ and applies the ping-pong lemma to $g$ and a subgroup of $G_1$ generated by elements that are far from the identity. You thus get a free product of the form $\mathbb{Z}*G_1'$. However, the main difference is that it is not clear to me if you can always have $G_1'$ isomorphic to $G_1$. To my knowledge, different lattices in the same nilpotent Lie group can be non-isomorphic, see for instance there.
|
2025-03-21T14:48:29.988323
| 2020-03-03T09:45:18 |
354040
|
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|
Stack Exchange
|
Given $n, c$ find $a>1,b$ such that $b ^ a \equiv c \pmod n$
Given a natural number $n$ (of unknown factorization) and an arbitrary number $c \in \mathbb{Z}^*_n$ (the set of natural numbers smaller than $n$ and coprime to it), is there an efficient algorithm which outputs numbers $a \in \{2,\ldots,n-1\}$ and $b \in \mathbb{Z}^*_n$ such that:
$$b ^ a \equiv c \pmod n$$
Note 1: In my problem, $n$ is the product of two safe primes.
Example: Let $n=77$ and $c=2$. The set of all admissible pairs $(b,a)$ is:
$$\{ (18, 23), (30, 7), (39, 29), (46, 11), (51, 13), (72, 19), (74, 17) \} \,.$$
For the same $n$ and $c=15$, the set of all admissible pairs $(b,a)$ is:
$$\{ (2, 12), (3, 24), (4, 6), (5, 18), (6, 8), (8, 4), (9, 12), (13, 2), (16, 3), (17, 18), (18, 6), (19, 24), (20, 2), (24, 12), (25, 9), (26, 6), (27, 8), (29, 6), (30, 24), (31, 12), (36, 4), (37, 6), (38, 18), (39, 18), (40, 6), (41, 4), (46, 12), (47, 24), (48, 6), (50, 8), (51, 6), (52, 24), (53, 12), (57, 2), (58, 9), (59, 6), (60, 3), (61, 18), (62, 6), (64, 2), (68, 12), (69, 4), (71, 3), (72, 18), (73, 6), (74, 24), (75, 12) \} \,.$$
Note 2: One can assume that $c$ is chosen randomly, and the algorithm is probabilistic and may output the right answer with good probability.
$~$
Note 3: If $b$ is fixed, the problem is the discrete log, and is known to be hard. If $a$ is fixed, the problem might be a variant of RSA or Rabin cryptosystem, both of which are considered hard. I hope that by giving the algorithm the freedom of picking both $a,b$, it is solvable efficiently.
I presume that choosing $b=c$ and $a=(n-1)!+1$ is cheating?
From your examples, it looks like you are insisting on $a<n$ and $b<n$. Perhaps you should state your restrictions explicitly.
$b<n$ is without loss of generality. And an exponentially large $a$ will require an exponential-time algorithm, which does not count as efficient. If there is a polynomial-time algorithm (possibly randomized), that by itself imposes a bound $a\le2^{(\log n)^k}$ for some constant $k$.
@JeremyRickard: Not cheating, but it's not admissible as no efficient algorithm can output such a large number. See Emil's comment for further description.
@GerryMyerson: As per Emil's comment, it might be unnecessary, though I edited the question to reflect the conditions.
After much struggling, I found out that this is currently deemed a hard problem; so there is no known efficient algorithm to solve it.
From the Encyclopedia of Cryptography and Security (2011):
Strong RSA Assumption
The Strong RSA Assumption was introduced by Baric
and Pfitzmann [3] and by Fujisaki and Okamoto [18] (see
also [13]).
This assumption differs from the RSA Assumption in
that the adversary can select the public exponent e. The
adversary’s task is to compute, given a modulus n and a
ciphertext C, any plaintext M and (odd) public exponent
e ≥ 3 such that $C = M^e \pmod n$. This may well be easier
than solving the RSA Problem, so the assumption that it is
hard is a stronger assumption than the RSA Assumption.
The Strong RSA Assumption is the basis for a variety of
cryptographic constructions.
with references as follows:
Barić N, Pfitzmann B (1997) Collision-free accumulators and
fail-stop signature schemes without trees. In: Fumy W (ed)
Advances in cryptology – EUROCRYPT’97. Lecture notes in
computer science, vol 1233. Springer, Berlin, pp 480–494.
$~$
Cramer R, Shoup V (2000) Signature schemes based on the
strong RSA assumption. ACM Trans Inform Syst Sec 3(3):
161–185.
$~$
Fujisaki E, Okamoto T (1997) Statistical zero knowledge protocols to prove modular polynomial relations. In: Kaliski BS
Jr (ed) Advances in cryptology – CRYPTO’97. Lecture notes in
computer science, vol 1294. Springer, Berlin, pp 16–30.
|
2025-03-21T14:48:29.988564
| 2020-03-03T10:25:22 |
354041
|
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|
Stack Exchange
|
Contraction criticality and edge-adding criticality for Hadwiger number
Let $G=(V,E)$ be a connected, simple, finite, undirected graph. The Hadwiger number $\eta(G)$ is the maximum integer $n\in \mathbb{N}$ such that $K_n$ is a minor of $G$.
We say that $G$ is contraction-Hadwiger-critical (cHc) if contracting any pair of non-adjacent vertices increases the Hadwiger number. (The icosahedron graph is an example of a contraction-critical graph.)
Morever, we call $G$ edge-Hadwiger-critical (eHc) if adding an edge between any pair of non-adjacent vertices increases the Hadwiger number.
It is not hard to show that any cHc graph is eHc.
Does the converse hold?
No, $K_n$ minus an edge is eHc but not cHc.
Thanks - will delete the question
|
2025-03-21T14:48:29.988642
| 2020-03-03T10:42:16 |
354043
|
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|
Stack Exchange
|
anticanonical divisors
Let $k=\mathbb{C}$. Call a smooth projective surface $X$ an anti-canonical divisor if there is a smooth projective $3$-fold $Y$ with a section $s \in H^{0}(Y,-K_{Y})$ such that $\{s=0\} \cong X$.
Question: Are anti-canonical divisors classified?
I asked this on math stack exchange 7 months ago with no answer (https://math.stackexchange.com/questions/3311295/smooth-surfaces-appearing-as-an-anticanonical-section) so I ask it here.
What do you mean by "classified"? By the adjunction formula your surface has trivial canonical bundle, hence is a K3 or an abelian surface, and both cases can occur. What else?
I mean classification up to isomorphism.
Is "K3 or abelian" a satisfactory answer? If not, what kind of description do you expect?
It seems clear that every abelian surface $X$ can be realized as an anti-canonical divisor since we can view $X$ as an etale double cover of another abelian surface $Y$ and so $X$ embeds as an anti-canonical divisor in the threefold $\mathbb{P}(\mathcal{O}_{Y}\oplus L)$ where $L \to Y$ is the two torsion line bundle corresponding to the cover $X \to Y$. So the OP seems to be asking - which smooth K3 surfaces can be realized as anti-canonical divisors of smooth threefolds. I am not sure where this is going since obviously every K3 is a connected component of an anti-canonical divisor.
If you also require that $-K_Y$ is ample, so that $Y$ is a Fano threefold, then say in Picard rank one, K3 surface $X$ will be of degree 2, 4, 6, 8, 10, 12, 14, 16, 18, 22, which is part of the classification of Fano threefolds. Beauville generalizes this for arbitrary Picard rank (keeping the Fano assumption): https://arxiv.org/pdf/math/0211313.pdf
|
2025-03-21T14:48:29.988778
| 2020-03-03T11:18:07 |
354044
|
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|
Stack Exchange
|
Finite dimensional algebras over $\mathbb{Q}$
It is known that a finite dimensional basic algebra over an algebraically closed field is isomorphic to the path algebra of a finite quiver modulo an admissible ideal.
Question 1: Is the same true for algebras over $\mathbb{Q}$? If not, are there suitable further assumptions that would guarantee this?
Question 2: Are there some general useful tools to compute the dimension of the algebra, given the quiver and the ideal? (In concrete, easy cases it might be doable by hand, but I am looking for methods that are more broadly applicable)
Question 1: No, take any finite field extension of $\mathbb{Q}$. It is basic but has a simple module that is not 1-dimensional and thus it is not of the form $KQ/I$ for $I$ admissible (since all simple modules are 1-dimensional for algebras of the form $KQ/I$).
For any field $K$, a basic algebra is isomorphic to a quiver algebra with admissible ideal if and only if the algebra $A$ is split (also called elementary sometimes), meaning that $A/J$ is isomorphic to matrix algebras over $K$. You can find this result in most representation theory books (of finite dimensional algebras) such as in the book of Auslander,Reiten and Smalo or the book by Kirichenko and Drozd.
Question 2: The best (computer) tool is the GAP-package qpa: https://folk.ntnu.no/oyvinso/QPA/ . Given a quiver and admissible ideal, you can calculate many information for the algebra, including the dimension.
Thanks a lot for your answer. With the sentence "$A/J$ is isomorphic to matrix algebras over $K$" do you mean that the quotient of $A$ by its Jacobson radical is isomorphic to some subalgebra of $M^{n \times n}(K)$ for some $n$?
He meant "a direct product of matrix algebras", i.e. the skew fields $D_i$ in the Wedderburn decomposition $A/J \cong \prod_{i=1}^n D_i^{k_i\times k_i}$ are all equal to $K$ itself. This is always the case for algebraically closed $K$ (because the $D_i$ are of finite dimension over $K$) but not in general.
|
2025-03-21T14:48:29.988928
| 2020-03-03T11:45:15 |
354045
|
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|
Stack Exchange
|
Is keeping the kernel fixed an open condition for maps of vector bundles?
More precisely, let $M$ be a smooth manifold, $E_1$, $E_2$ vector bundles over $M$, and consider a $C^\infty(M)$-linear map $A:\Gamma(E_1) \to \Gamma(E_2)$ of vector bundles.
Now consider the subspace consisting of $C^\infty(M)$-linear maps $B:\Gamma(E_1) \to \Gamma(E_2)$ satisfying $\ker(A) \subseteq \ker(B)$.
Is the subset of maps $B$ as above satisfying $\ker(A) = \ker(B)$ open (in some $C^p$ topology) inside the set of maps such that $\ker(A) \subseteq \ker(B)$?
I believe this should be true as reducing modulo $\ker(A)$, the question translates into a map nearby an injective map (the map induced by $A$) should still be injective, but I am not entirely sure if this idea still holds in the infinite-dimensional setting.
Thanks in advance!
|
2025-03-21T14:48:29.989130
| 2020-03-03T12:02:01 |
354048
|
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|
Stack Exchange
|
Explanation of $\sigma$-weak topology von a von Neumann algebra
Let $A$ be a von Neumann algebra. I want to understand the precise meaning of the $\sigma$-weak topology on $A$. What I understand so far is the following: The $\sigma$-weak topology, which we will denote by $\tau$, is the weak$^*$-topology on $A$ (which can be defined since every von Neumann algebra has a (unique) predual $A_*$). It seems that $A_*$ is the Banach space of ultraweakly continuous linear functionals on $A$. But how to understand the ultraweak topology on $A$? Do one use the fact that $A$ is continuously embedded into some $B(H)$?
In addition I have also some other questions regarding the structure of the $\sigma$-weak topology:
Is the $\sigma$-weak topology a Hausdorff topology on $A$ and possibly sequentially complete on norm-bounded sets? Is this topology even norming on $A$, i.e., for all $x\in A$ one has that
$$
||x||=\sup_{\substack{\varphi\in(X,\tau)'\\||\varphi||\leq1}}{|\varphi(x)|}.
$$
Probably these questions are all trivial to answer, however it would be great to have some references. Could you help me with this? Thank you very much in advance.
This isn't research level, but yes, every von Neumann algebra can be realized as a weak* closed subalgebra of some $B(H)$, and its weak* topology is the inherited weak* topology. That should answer your other questions too,
Regarding references: my guess is that Kadison+Ringrose's 2-volume book would explain the relationship between these topologies, but I don't have a copy at hand to check. There might also be something in Murphy's book on Cstar algebras
This is not really research level, and is probably better suited to math.stackexchange, but here's an answer anyway. I will take as given that you know the notation $\sigma(E^*,E)$ for the weak-* topology on $E^*$, or more generally $\sigma(F,E)$ on a Banach space $F$ isomorphic to the dual space of a Banach space $E$.
For any Hilbert space $\newcommand{\Hil}{\mathcal{H}}\Hil$, the space of bounded operators $B(\Hil)$ is isomorphic to the dual space of the space of trace-class operators $B_*(\Hil)$ by the pairing $\langle T, \rho \rangle = \mathrm{tr}(T\rho)$, for $T \in B(\Hil)$ and $\rho \in B_*(\Hil)$. The weak-* topology $\sigma(B(\Hil),B_*(\Hil))$ on $B(\Hil)$ can be called the $\sigma$-weak or ultraweak topology, according to the author's preference, i.e. these are two names for the same thing.
When we have a C$^*$-algebra $A$ that is isometrically isomorphic to the dual space of a Banach space $A_*$, then we say it is a W$^*$-algebra, and the weak-* topology $\sigma(A,A_*)$ is also called the $\sigma$-weak or ultraweak topology.
If $A \subseteq B(\Hil)$ is a von Neumann algebra, then we can consider the restriction of $\sigma(B(\Hil),B_*(\Hil))$ to $A$. We can then define the space $A_*$ to be the set of linear functionals continuous in this topology. It is then a (nontrivial) theorem, that $A_*$ is a closed subspace of $A^*$, and $A$ is isomorphic to the dual space of $A_*$ under the restriction of the double dual embedding, and the topologies $\sigma(B(\Hil),B_*(\Hil))$ and $\sigma(A,A_*)$ agree on $A$, justifying why they're both called the $\sigma$-weak/ultraweak topology.
The questions in your second paragraph are all consequences of being a weak-* topology (the Banach-Alaoglu theorem implies that weak-* closed norm-bounded sets are complete, because they're compact).
I hope this helps you to understand standard references like Takesaki's Theory of Operator Algebras, Dixmier's Von Neumann Algebras, Kadison and Ringrose's Fundamentals of the Theory of Operator Algebras.
|
2025-03-21T14:48:29.989393
| 2020-03-03T12:13:29 |
354049
|
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"Adittya Chaudhuri",
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"Praphulla Koushik",
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}
|
Stack Exchange
|
Why the third stage of Cech nerve a Lie 2-groupoid?
In the page https://ncatlab.org/nlab/show/Lie+2-groupoid the Lie 2-groupoid is defined as the 2 truncated $\infty$-Lie groupoid.
I am not much comfortable with the language of higher category theory yet. But after reading some links in ncatlab I felt Lie 2-Groupoid is the same as a 2-groupoid( a 2 category whose both 1 morphisms and 2 morphisms are invertible) internal to the category of smooth manifolds.
Am I right??
Now on the page https://ncatlab.org/nlab/show/infinity-Chern-Weil+theory+introduction#Cech2Cocycles it is mentioned that the Cech groupoid of a manifold $X$ with a cover $U_{\alpha}$ can be thought as a Lie 2 -groupoid by considering the third stage of the full Cech Nerve.
I can understand that the Cech Groupoid $(\sqcup{U_{i}\cap U_{j}}\rightrightarrows \sqcup U_k)$ is a Lie groupoid. But I am not able to understand how $(\sqcup{U_{i}\cap U_{j}\cap U_{k}} \substack{\textstyle\rightarrow\\[-0.6ex]
\textstyle\rightarrow \\[-0.6ex]
\textstyle\rightarrow}
\sqcup{U_{i}\cap U_{j}}\rightrightarrows \sqcup U_k)$ is a Lie 2-groupoid (in the sense I have understod the definition of Lie 2-groupoid). Infact I am not able to guess what can be the 2-morphisms in $(\sqcup{U_{i}\cap U_{j}\cap U_{k}} \substack{\textstyle\rightarrow\\[-0.6ex]
\textstyle\rightarrow \\[-0.6ex]
\textstyle\rightarrow}
\sqcup{U_{i}\cap U_{j}}\rightrightarrows \sqcup U_k)$ (when described as a Lie 2-groupoid).
So is my understanding of Lie 2-groupoid wrong? If not then what is the 2-categorical description of $(\sqcup{U_{i}\cap U_{j}\cap U_{k}} \substack{\textstyle\rightarrow\\[-0.6ex]
\textstyle\rightarrow \\[-0.6ex]
\textstyle\rightarrow}
\sqcup{U_{i}\cap U_{j}}\rightrightarrows \sqcup U_k)$?
Thanks in Advance.
Check last 1/4 th of page 9 and first 1/4 th of page 10 in https://arxiv.org/abs/1706.07152
what is the 1 morphism in Cech groupoid?
The space of all 2-morphisms in the Lie 2-groupoid under consideration is the disjoint union of U_i ∩ U_j ∩ U_k. The correspondence between various flavors of 2-categories and simplicial objects is explained in https://ncatlab.org/nlab/show/geometric+nerve+of+a+bicategory
@PraphullaKoushik Thanks for the link. From the link, it seems what I guessed about the definition of Lie 2-Groupoid is right.
@DmitriPavlov Thanks for the comment and the link. I also guessed that the space of all 2-morphisms is $(\sqcup{U_{i}\cap U_{j}\cap U_{k}})$. But I am not able to figure out that for a general 2 morphism $(x,i,j,k)$ what will be its source(1-morphism) and target(1-morphism)? I hope the space of 1-morphisms will be $(\sqcup{U_{i}\cap U_{j}})$ where the source and target of a general 1-morphism $(x,i,j)$ are $(x,i)$ and $(x,j)$ respectively.
@PraphullaKoushik Please see the comment above.
@AdittyaChaudhuri: No, it doesn't work this way. The 2-categorical model used here is simplicial, so a 2-morphism is an arrow fh→g, where f, g, h are 1-morphisms given by the three simplicial faces of this 2-simplex. In the case under consideration, these three faces are given by mapping U_i∩U_j∩U_k into U_j∩U_k, U_i∩U_k, and U_i∩U_j respectively.
@DmitriPavlov Thanks for the comment. I got where I was making the mistake. I am trying to understand the 2 categorical simplicial model(as you have mentioned in the comment. )
@DmitriPavlov Sir, I wrote an answer(with some doubts stated) in response to your comment above. Please can you check once in your free time and let me know whether it makes sense or not? ... Thanks in advance.
This is not an answer, just too long for a comment (could be slightly misleading). This is precisely what Dimitri Pavlov has mentioned in his comment.
In general, the description of $2$-category $\mathcal{C}$ goes as follows.
a collection of objects.. Let $A$ be an object of $\mathcal{C}$...
a $1$-morphism is between "two" objects $A\rightarrow B$...
a $2$-morphism is between "two" $1$-morphisms as in the following diagram
In simplicial model, $1$-morphism is between two objects. But, $2$-morphism is not betwen two $1$-morphisms but between three $1$-morphisms as in the following diagram:
By definition, Cech nerve is a simplicial object. So, $2$-morphism is between three $1$-morphisms.
Yes, I can understand! But I am trying to now understand how we are seeing the 3rd stage of simplical object Cech Nerve as a 2 categorical model.
I don't know if I understand your question correctly.. You do not see third stage as a 2 category.. you see first second third stage combined together as a 2 category...
I mean to say that only.
But I have said that in the above answer :O
yes that is the same thing Dimitri Pavlov mentioned before in the comment. But I am trying to understand how it is following from the geometric nerve construction of bicategories. I almost got it. (:
I hope some one more experienced than me can clear your doubt... ask Dimitri pavlov if he can make it as an answer adding some details..
Since its too long for a comment, I had to post it as an answer in response to the comment made by Sir Dmitri Pavlov.
I got what you mentioned. You mean to say that we can treat a $(\sqcup{U_{i}\cap U_{j}\cap U_{k}} \substack{\textstyle\rightarrow\\[-0.6ex]
\textstyle\rightarrow \\[-0.6ex]
\textstyle\rightarrow}
\sqcup{U_{i}\cap U_{j}}\rightrightarrows \sqcup U_k)$ as a part of simplicial set data or in other words the whole Cech nerve can be described as a simplicial set in the following way:
$[o]\mapsto \sqcup{U_{i}}$,
$[1] \mapsto \sqcup{U_{i}\cap U_{j}}$
$[2]\mapsto \sqcup{U_{i}\cap U_{j}\cap U_{k}}$
and so on... (where $[0],[1],[2],..$ are objects of the simplex category).
where we treat elements of $\sqcup{U_{i}}$ as our objects, elements of $\sqcup{U_{i}\cap U_{j}}$ as our 1-morphisms(two face maps $d_{0}$ and $d_{1}$ are source and targets), elements of $ \sqcup{U_{i}\cap U_{j}\cap U_{k}}$ as 2 morphisms (here we consider 3 face maps $d_{0}$,$d_{1}$, and $d_{2}$).
Now I have the following doubt:
We know that there exist full, faithful Nerve functors $N_1:cat\rightarrow Ssets$, $N_{2}:Bicat\rightarrow Ssets$,.... and so on....(I am sure about $N_{1}$ and $N_{2}$ but not about higher values).
Does there exist a $c\in Bicat$ such that $N_{2}(c)$ is isomorphic to the Cech Nerve? Then we can think that $c$ as my Lie 2-Groupoid (with appropriate internalization) which represents the Cech Nerve.
Nerves of bicategories were characterized by Duskin in Theorem 8.6 of his paper Simplicial matrices and the nerves of weak n-categories I: nerves of bicategories. However, there is no reason to prioritize the bicategorical model over the simplicial model; in fact, the latter is what typically is used in the realm of higher Lie theory and stacks.
@DmitriPavlov Thanks. So that means in the construction of Principal 2-bundle as mentioned in https://ncatlab.org/nlab/show/infinity-Chern-Weil+theory+introduction#Cech2Cocycles the morphism $g:C(U)\rightarrow B^{2}U(1)$ is actually a morphism of simplicial sets?(where C(U) is the Cech Nerve and $B^{2}U(1)$ is the one object 2 groupoid)...Then I guess they actually treated $B^{2}U(1)$ as the nerve of Lie 2-groupoid $B^{2}U(1)$?? But then it means they actually consider the whole Cech nerve ( not only upto its 3rd stage) as the former is a simplicial set and the later is not.
Not simplicial sets, but rather simplicial presheaves on the site of smooth manifolds. Both Č(U) and B^2 U(1) are such simplicial presheaves. There is no need to consider the whole Čech nerve because the target B^2 U(1) is 2-truncated, so only simplicial levels 0, 1, 2, and 3 of Č(U) matter.
@DmitriPavlov Thanks. (As I already mentioned that I am not well versed with the language of higher category yet.). So I am trying to understand your comment. I can understand that $B^{2}U(1)$ is 2-truncated. But my question is "when you dont consider the whole Cech nerve" then are you assuming that all the higher morphisms(greater than 2) in Cech nerve are trivial?? (I apologize priorly if I sound stupid. I started learning higher category very recently. )
All simplicial levels of the Čech nerve are nontrivial. However, if you map into a 2-truncated object, then all simplicial levels of the source above level 3 get thrown away, whereas the 3rd level gets collapsed onto the 2nd (i.e., two 2-simplices get identified if they are connected by a homotopy presented by a 3-simplex).
@DmitriPavlov Thank you Sir. If possible can you please fill little detail about how "all simplicial levels of the source above level 3 get thrown away, whereas the 3rd level gets collapsed onto the 2nd"?.. Also since I am a beginner in this area it would be of immense help if you can suggest some literature which deals with this kind of question(setup).
A high-level discussion can be found in https://ncatlab.org/nlab/show/n-truncated+object+of+an+%28infinity%2C1%29-category#truncation. The article by Rezk http://www.math.uiuc.edu/~rezk/homotopy-topos-sketch.pdf contains details in Section 7.
@DmitriPavlov Thank you Sir.
First, let me elaborate on how we obtain a groupoid by the 2-truncation of Čech nerve. Then in a similar manner, you can look at 3-truncation, 4-truncation and so on which in return produces 2-groupoid and its higher-order counterparts. In fact, the full Čech nerve is an example of an infinity groupoid (a simplicial set).
Given a manifold $M$ and a good open cover $\{U_i\}$ on it, the $2$-coskeleton $\text{Č}\left(\{U_i\}\right)_{\le2}$ of the full Čech nerve of the covering form a groupoid usually called the Čech groupoid or covering groupoid.
Its set of objects is the disjoint union $\bigsqcup_{i}U_i$ of the covering patches (open sets), and the set of morphisms is the disjoint union of the intersections $\bigsqcup_{i,j}U_i\cap U_j$ of these patches, i.e., an object is a pair $(x,i)=x_i$ where $x\in U_i$ and there is a unique morphism $(x,i,j)=x_{ij}:x_j\to x_i$ for all pairs of objects with $x\in U_i\cap U_j=U_i\underset{M}{\times}U_j.$ The composition (multiplication) of morphisms is a commutative triangle
satisfying $$x_{ij}x_{jk}=x_{ik}$$ for all $x\in U_i\cap U_j\cap U_k.$ In other words $m(x_{jk},x_{ij})=x_{ik}$ The source and target maps are given by $$s(x_{ij})=x_j,\, t(x_{ij})=x_i$$ for all $x\in U_i\cap U_j.$ Finally, in order to be a Lie groupoid you need to show that these structure maps are contunuous and it is enough to verify this for the source map.
How does this answer the question? The OP already knows how to interpret the 2-coskeleton as a Lie groupoid, the question was how to generalize this to the 3-coskeleton. This is precisely what you omitted when you said "Then in a similar manner, you can look at…".
|
2025-03-21T14:48:29.990020
| 2020-03-03T13:25:06 |
354059
|
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"authors": [
"Asaf Karagila",
"https://mathoverflow.net/users/14340",
"https://mathoverflow.net/users/7206",
"喻 良"
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|
Stack Exchange
|
The measure of ideals generated by random reals
We assume that for every real $x$, $L[x]$ only contains countably many reals.
Given a set $X$ of reals, then $L$-ideal generated by $X$ is the smallest set $I$ of reals so that
For any reals $x\in I$ and $y$, $y\in L[x]$ implies $y\in I$; and
For any finite $F\subseteq X$, there is a real $z\in I$ so that $F\subseteq L[z]$.
The question is
Question: Given a null set $X$ only containing $L$-random reals, must the $L$-ideal $I$ generated by $X$ be null?
Note that, given the set $X$ as in the question, the $L$-upward closure $U_X=\{y\mid \exists x\in X(x\in L[y])\}$ must be null.
In other words, $\omega_1$ is inaccessible to reals, yes?
@AsafKaragila yes
OK, I believe the question has a negative answer.
The question has a negative answer. The technique is essentially due to Jockusch and Posner.
Proof: Let $x$ be a real in which every constructible real is recursive. Now $$A=\{r\mid r\mbox{ is Martin-L\" of random relative to }x \wedge x\oplus r\geq_T x',\mbox{ the Turing jump of }x.\}$$ Then $A$ is null, $r\oplus x$ ranges over the upper cone $\geq_T x'$, and only contains $L$-random reals. Now for any $z\geq_T x'$ and $r \in A$, we have that
(1). $r\triangle x \in A$; and
(2). $(r\triangle x) \oplus r\equiv_T x\oplus r\geq_T x'$.
So the $L$-ideal generated by $A$ is $\mathbb{R}$.
|
2025-03-21T14:48:29.990148
| 2020-03-03T15:07:23 |
354065
|
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"Emil Jeřábek",
"ervx",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/354065"
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|
Stack Exchange
|
Is a finite increasing chain of closed sets the closure of the union of the interiors of the relative complements?
Let $X$ be a topological space. Suppose there are closed subsets $X=:F_{k}\supseteq F_{k-1}\supseteq\cdots\supseteq F_{1}\supseteq F_{0}:=\emptyset$. Is it true that
$\overline{\bigcup_{j=1}^{k}\operatorname{Int}(F_{j}\setminus F_{j-1})}=X$?
If $k=1$, the result is trivial. For $k>1$, we have
$$
\overline{\bigcup_{j=1}^{k}\operatorname{Int}(F_{j}\setminus F_{j-1})}=\overline{\operatorname{Int}(X\setminus F_{k-1})}\cup\overline{\bigcup_{j=1}^{k-1}\operatorname{Int}(F_{j}\setminus F_{j-1})}=\overline{X\setminus F_{k-1}}\cup\overline{\bigcup_{j=1}^{k-1}\operatorname{Int}(F_{j}\setminus F_{j-1})}.
$$
When $k=2$, the RHS simplifies to
$$
\overline{X\setminus F_{1}}\cup\overline{\operatorname{Int}(F_{1})}=\overline{X\setminus\operatorname{bd}(F_{1})}=X.
$$
I'm not sure if it's possible to carry out a similar reduction when $k>2$; or, if this result is indeed true. Any help is appreciated.
Thank you.
This is true, even if the sets do not form a chain. It follows by induction on $k$ using the observation that $\mathrm{int}(A\setminus B)\subseteq\overline{\mathrm{int}(A\setminus C)\cup\mathrm{int}(C\setminus B)}$ for closed $C$.
I see. Thank you, this was very helpful!
|
2025-03-21T14:48:29.990240
| 2020-03-03T15:17:29 |
354067
|
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"url": "https://mathoverflow.net/questions/354067"
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|
Stack Exchange
|
Spectral theorems for generalized Hermitian matrices
Let $k$ be a field, and let $\sigma$ be a nontrivial involutory automorphism of $k$. Let $A$ be a square matrix with entries in $k$, such that $(A^{\sigma})^T = A$; here $A^\sigma$ means the matrix $(a_{ij}^\sigma)$, where $A = (a_{ij})$, and the $T$-exponent means "transposed of."
Example: $k = \mathbb{C}$ and $\sigma$ is complex conjugation. In that case, $A$ is called a Hermitian matrix.
Question: Is there a spectral theorem known which naturally generalizes the spectral theorem for complex Hermitian matrices?
What about if $k$ is algebraically closed, or real-closed?
|
2025-03-21T14:48:29.990313
| 2020-03-03T15:44:10 |
354072
|
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"Martin Brandenburg",
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"url": "https://mathoverflow.net/questions/354072"
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|
Stack Exchange
|
An example of a commutative ring with an annihilator condition
I am looking for an example of a non-zero commutative ring $R $ with identity such that $R $ has no idempotent elements, and there exists a non-nilpotent element $s\in R $ such that for each $n\in\mathbb{ N }$ there exists $r\in R$ with $r (r-1)s=0$, $rs^n\not=0 $ and $(r-1)s^n\not=0$.
Please provide more context. Why are you interested in such an example? And what have you tried so far? Recently there was a similar question here, was this also asked by you under a different account?
For this type of problem, I find that free algebras modulo the relations you want often suffice. That is the case here. Take $R=\mathbb{Z}\langle r,s : sr=rs,r^2s=rs\rangle$. We check that
$$
r^2(sr)=r^3s=r^2s=rs \text{ and } (r^2s)r=rsr=r^2s=rs
$$
as well as
$$
s(r^2s)=srs=rs^2 \text{ and } (sr)rs=rsrs=r^2s^2=rs^2.
$$
So the two overlaps between the two relations both resolve to the same expression, and so Bergman's Diamond Lemma tells us that we can write elements of $R$ in a unique reduced form as
$$
t=f(r) + sg(s) + rsh(s)
$$
where $f(x),g(x),h(x)\in \mathbb{Z}[x]$ are polynomials. We achieve this normal form by repeatedly applying the two rules $sr\mapsto rs$ and $r^2s\mapsto rs$ to simplify.
Let us now show that there are no nontrivial idempotents. If we were to have $t^2=t$, then $f(r)^2=f(r)$ (work modulo the ideal generated by $s$), and so $f(r)\in \{0,1\}$ (since those are the only idempotents in $\mathbb{Z}[r]$). After replacing $t$ by $1-t$ as necessary, we may assume $f(r)=0$. Now, $t^2=s^2(g(s)+rh(s))^2$ has degree (in $s$) larger than $t$, unless $g(s)=h(s)=0$. Thus the only idempotents are $0,1$.
It is also clear that $rs^n\neq 0$ and $rs^n\neq s^n$, from the normal form, as desired.
|
2025-03-21T14:48:29.990443
| 2020-03-03T17:02:36 |
354074
|
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"Alex Townsend-Teague",
"Autumn Kent",
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|
Stack Exchange
|
A version of Hurwitz' theorem in terms of Euler characteristic
Page 203 of Farb and Margalit's Primer on Mapping Class Groups contains the result:
Let $g ≥ 2$. The order of any finite subgroup of $MCG(S_g)$ is at most $84(g − 1)$.
I've been told by my supervisor that a version exists in terms of the Euler characteristic of a surface $S$; that is, the following result is true:
Let $\chi(S) < 0$. The order of any finite subgroup of $MCG(S)$ is at most $42\left|\chi(S)\right|$.
However, I can't find this anywhere in the literature! Does anyone know of a reference for this, or is anyone able to nudge me towards a proof (or counterexample)? Many thanks.
@PhilTosteson They mean the case when the surface isn't closed.
The proof using the Gauss Bonnet theorem will give this to you in the finite type case. (https://en.wikipedia.org/wiki/Hurwitz%27s_automorphisms_theorem)
@Autumn Kent: Ahhhhh I see, perfect - thanks!
|
2025-03-21T14:48:29.990558
| 2020-03-03T17:33:28 |
354076
|
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|
Stack Exchange
|
Reason to apply the Koszul sign rule everywhere in graded contexts
The Koszul sign rule is a sign rule that arises from graded-commutative algebras. For instance, let $\bigwedge(x_1,\dots, x_n)$ be the free graded-commutative algebra generated by $n$ elements of respective degrees $\lvert x_i\rvert$. Then, the sign $\varepsilon(\sigma)$ of a permutation $\sigma$ on $(x_1,\dotsc, x_n)$ is given by
$$x_1\wedge\dotsb\wedge x_n=\varepsilon(\sigma)x_{\sigma(1)}\wedge\dotsb\wedge x_{\sigma(n)},$$
which comes from the fact that in a graded-commutative algebra one has by definition $a\wedge b = (-1)^{\lvert a\rvert\lvert b\rvert}b\wedge a$.
There is also an antisymmetric Koszul sign rule which arises from graded-anticommutative algebras and it's just the previous sign times the sign of the permutation. Both signs are used for instance in Lada and Markl - Symmetric brace algebras.
However, I've been seeing the Koszul sign rule used in any graded context and even for operations that are not products in some algebra. For example, from Roitzheim and Whitehouse - Uniqueness of $A_\infty$-structures and Hochschild cohomology, given graded maps of graded algebras $f,g:A\to B$, if we want to evaluate $f\otimes g$ in an element $x\otimes y$, apparently we need to apply the sign rule to get
$$(f\otimes g)(x\otimes y)=(-1)^{\lvert x\rvert\lvert g\rvert}f(x)\otimes g(y),$$
but I see no mathematical reason to do that, it just seems to be a convention.
A more complex example of application of the Koszul sign rule is in the definition of brace algebra (also in the Lada and Markl paper).
I could give many more examples. In some of them I can understand the reason. For instance, the differential of a tensor product of complexes $C$ and $D$ cannot simply be $d_C\otimes 1_D+ 1_C\otimes d_D$ (it can be defined this way if we use the sign rule when we apply it to elements, but in any case it needs the sign). But maps in general need not be differentials. In other cases, the signs appear in nature and one use this sign rule to justify them, as in $A_{\infty}$-algebras, but this feels too artificial for me and doesn't really explain why we should use that sign rule.
So, in the end, every time there is a sequence $(x_1,\dotsc, x_n)$ of graded objects of any kind and not necessarily all of them of the same kind (elements, maps, operations, …), and related in any way (they can be multiplied, or applied, or whatever), we use the Koszul sign rule to permute the sequence.
To me all of this seems more philosophical than mathematical, and as I said it feels to be just a convention. But, is there some general mathematical reason to use the sign rule in such an extensive way? And if it's just a convention, why should we use it? From my experience, it gets very messy when it comes to applying the sign rule to larger formulas, and in the end everything is just a $+$ or $-$ sign, so I see no advantage.
@NajibIdrissi The signs come from the sign rule applied to shifted maps, so if I understand correctly, without the sign rule there would be no signs at all.
A precise statement of the conventions (which Jesse is referring to) is that the authors are using the symmetric monoidal structure on graded vector spaces, where the braiding map,, $\tau: V \otimes W \to W \otimes V$, is given by $$v \otimes w \mapsto (-1)^{{\rm deg} ~w ~{\rm deg}~ v} w \otimes v.$$
Roughly, what it means to use this symmetric monoidal structure is that you have to make all of your definitions diagramatically, using only $\tau$ to exchange symbols.
For instance suppose we have two algebras $A, B$ and an $A$ module $M$ and a $B$ module $N$. Then if $A,B,M,N$ were ordinary vector spaces, we are used to the fact that $M \otimes N$ is an $A \otimes B$ module. In the graded context, under the Koszul conventions, we define the action $$A \otimes B \otimes M \otimes N \to A \otimes M \otimes B \otimes N \to M \otimes N,$$ where in the first step we have used $1 \otimes \tau \otimes 1.$ Something quite similar is happening in the example your mention.
So far, this is more of a unified answer to how rather than why people use this convention.
For the question of why, the main reason the Koszul convention is useful in homological algebra has to do with the origin of homological algebra---topology.
Consider $\mathbb R^{p +q}$ with its standard orientation. Then the switching map $$\tau: \mathbb R^p \times \mathbb R^q \to \mathbb R^{q} \times \mathbb R^p$$ multiplies this orientation by $(-1)^{p q}$. This fundamental fact manifests itself in several ways.
One is that homology functor $H_*(-, k)$ from topological spaces to graded vector spaces is symmetric monoidal, but only with respect to the Koszul sign rule. This means that if one has an algebraic structure on a topological space $X$, then $H_*(X)$ naturally carries the same algebraic structure, but with respect to the Koszul sign rule. For instance, $X$ is always a co-commutative coalgebra, so $H_*(X)$ becomes a graded co-commutative coalgebra (with sign conventions according to the Koszul rule).
Something similar happens with the $A_\infty$ operad. Namely, the $A_\infty$ operad is the $dg$ operad obtained by taking the cellular homology of a (cellular) operad in topological spaces. The orientations of the cells of this operad explain the signs which arise.
There is also the monoidal Dold Kan correspondence, which you can read about on the nLab.
At the end of the day, it is just a convention (and not always the right one) but the relation with topology explains why people like to use it systematically.
This was the kind of answer I was hoping for, thank you very much!
This is not a complete answer (none will be, really), but there is a definite reason for applying the specific sign convention you've described just when considering graded maps of graded vector spaces with the signed braiding $\tau_{A,B}:A\otimes B\to B\otimes A$.
With homogeneous maps $f : A_\bullet \to B_{\bullet-n} $ and $g : X_\bullet \to Y_{\bullet+m}$, there are two competing ways to try interchanging the roles of $f$ and $g$: one can consider together with $\tau_{A,X}(a\otimes x) = (-1)^{|a||x|}(x\otimes a)$ Alternatively, there's an evaluate-first ask-questions-later approach, $\tau_{B,Y} (f a)\otimes (g x) = (-1)^{(|f|+|a|)(|g|+|x|)} (g x)\otimes(f a) $.
Now, conceivably, one could push all the extra signums, the $|f||g|+|f||x|+|g||a|$ into just the interchange $f\otimes g \;\mathrm{vs}\; g\otimes f$, but on the whole it seems to be cleaner to remark that evaluating $(f\otimes g)(a\otimes x) = \mathrm{ev} (f\otimes g\otimes a\otimes x)$ already involves interchanging $g$ and $a$, and similarly evaluating $(g\otimes f)(x\otimes a) = \mathrm{ev} (g\otimes f \otimes x\otimes a) $ involves interchanging $f$ and $x$.
In your second paragraph, should "there are two competing ways …: one can consider … Alternatively …" be something like "there are two competing ways …: one can consider …; alternatively, …"? That is, is that all one sentence?
|
2025-03-21T14:48:29.991104
| 2020-03-03T17:35:56 |
354077
|
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|
Stack Exchange
|
Shrinking and stretching of vector bundles
Let $M$ be a manifold, $p:E\to M$ a rank $d$ vector bundle. Suppose that $U \subset E$ is an open subset such that $U \cap p^{-1}(x)$ is nonempty and convex for all $x \in M$. Is it true that $U \to M$ is a fiber bundle with fiber $\mathbb R^d$? And that $U \cong E$ as fiber bundles? We may assume with no loss of generality that $U$ contains the zero section.
This seems like a statement that could be a lemma in any number of textbooks (if true), e.g. in connection with the tubular neighborhood theorem, but I haven't seen it anywhere. Lang proves in his differential geometry book that any vector bundle over a manifold is what he calls compressible, meaning that any open neighborhood of the zero section of $E$ can be shrunk to a smaller open neighborhood which is diffeomorphic to $E$ as a bundle over $M$.
Kosinski has a theorem like this in his differentiable manifolds textbook. But rather than your fibrewise condition he talks about $M$ being a deformation-retract of $E$. I think he uses $h$-cobordism or minimal handle decompositions, though, so it is a little different than your context.
One can try to prove this by constructing an exhaustion of $U$ by a sequence of $V_i$, such that 0) $V_0$ is a smooth section of $E$ lying in $U$ 1) For $i>0$ each $V_i$ is a smooth closed submanifold of $U$ with boundary. 2) $V_i$ lies in the interior of $V_{i+1}$ 3) The intersection of $V_i$ with each fiber is a compact convex subset 4) $\cup_i V_i=U$. Note that it is easy to find a smooth section, using the partition of unity (+ convexity). As for constructing these $V_i$, this also looks doable in the same way replacing the sum by the Minkowsky sum. I can try to write this down.
Dear Dmitri, that's a very nice idea. Just so I understand, the idea is then that each $V_i$ is diffeomorphic to a disk bundle of radius $i$ around the zero section for some Riemannian metric on $M$? Is it clear that this is going to be the case?
Dear Dan, yes the radius $i$ ball sub-bundle for some Euclidean metric on $E$. I think this is true, if you have a smooth compact convex set $B\subset \mathbb R^n$ (with non-zero interior), containing $0$, then you can construct a smooth map from $B$ to the unit ball, that will be radial, i.e., it will send any straight segment through $0$ to a radius of the unit ball. One can take it isometric close to $0$ and then adjust smoothly, so $\partial B$ goes to $\mathbb S^{n-1}$. This can be also done smoothly in family.
Lovely - yes, that makes sense!
Since there are no references so far, let me give a sketch proof along the lines of my comment. I'll assume that $M$ is compact.
Let's show first that there is a smooth section of $E$ lying in $U$. Indeed, for any point $x\in M$ there is a neighbourhood $U_x$ with a section $s_x$. Take a finite cover $U_i$ of $M$ by such neighbourhoods and take the corresponding partition $1=\sum f_i$ of unity. Then by convexity $\sum s_i f_i$ is a smooth section lying in $U$.
Clearly we can assume that $s$ is the zero section (by taking an appropriate fiberwise diffeo), we will assume this from now on.
Now we will construct an exhaustion of $U$ by an increasing sequence of fiber-wise compact convex subsets $0\subset {\cal B_1}\subset ... \subset {\cal B_i}\subset ...$ so that $U=\cup_i {\cal B_i}$.
Let me show first how to construct one such subset ${\cal B_1}\subset U$.
For every point $p\in M$ let us choose some covex compact subset $B_p$ with smooth boundary in the fiber $U_p$. Then, since $U$ is open, there is an open neighbourhood $V_p$ of $p$ in $M$ such that over this neighbourhood there is a smoothly varying family of $B_x$ ($x\in V_p$), such that $B_x\subset U_x$. Take a finite cover of $M$ by such $V_i's$, let $\phi_i$ be the partition of unity. Then the sum
$${\cal B_1}=\sum_i \phi_i B_i (x)$$
is the desired subset $B\subset U$. Here by sum I mean the Minkowski sum.
It is clear that the interior of $\cal B_1$ is diffeomorphic to the bundle of vectors of length less than $1$ in $E$ (for some fiber-wise Euclidean metric). So the only need to construct a family of $\cal B_i$ that will exhaust $U$. This can be done as in 1).
Thanks! For completeness, I guess a natural way to ensure that this construction exhausts $U$ is to put a Riemannian metric on the total space of $E$ and then in step 1) note that we can make sure that each $B_x$ contains all vectors whose distance to the complement of $U$ is at least $1/n$.
Dan, yes, for example, a good metric would be such that each fiber $\mathbb R^n$ is isometric to an open half of a unite sphere (it is identified with $\mathbb R^n$ by the projection from its centre). Such a metric is good because convex subsets of $\mathbb R^n$ correspond to convex subsets of the half-sphere.
|
2025-03-21T14:48:29.991463
| 2020-03-03T17:44:51 |
354078
|
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|
Stack Exchange
|
Riemannian submanifolds of Euclidean space admitting Lipschitz extension of Lipschitz functions, and converse statement
Let $M \subset \mathbb{R}^p$ be a Riemannian submanifold. In what follows, when we talk about a Lipschitz function $f$ on $M$, namely $f: M \to \mathbb{R}$, we will assume there is a $L > 0$ so that:
$$ \lvert f(x) - f(y)\rvert \leq Ld_M(x,y)\quad \forall x, y \in M. $$
My question is: What is either a necessary and sufficient condition (preferred) OR if not possible, then at least a sufficient condition that every Lipschitz function on $M$ can be extended to a Lipschitz function on $\mathbb{R}^p$? So to write mathematically, what is or are sufficient condition(s) on $M$ so that for every Lipschitz $f: M \to \mathbb{R}$, there exists an extension $\tilde{f}$ of $f$ so that $\tilde{f}$ is also Lipschitz (possibly with a bigger Lipschitz constant)?
On the other side, what is a necessary and sufficient condition on $M$ so that every Lipschitz function on $\mathbb{R}^p$ restricts to a Lipschitz function on $M$? I feel like for a compact manifold, this will be true, as there we will have:
$$C_1\lVert x- y\rVert \leq d_M(x,y) \leq C_2\lVert x-y\rVert,$$
because the exponential map is smooth, so the norm of its gradient will have an upper and a lower bound on compact $M$, resulting in the above inequality. Is it true though?
For future reference, it's "Lipschitz"; I have edited accordingly. I also changed "sufficient or necessary and sufficient" to just "sufficient".
No problme with Lipscitz, thanks. May I ask why are you editing the necessary and sufficient part? It'd be nice to have an equivalent condition, no?
I agree, but asking for a condition that is sufficient or necessary and sufficient is the same as asking for a condition that is sufficient (and may also be necessary). If you think it makes it clearer, then please feel free to put it back in.
@LSpice OKay. Sorry but I'm most happy with necessary and sufficient, slighly less happy with sufficient. So I'm putting it back in.
Say $M$ admits universal Lipschitz extension if, for any Lipschitz $f : M \to \mathbb R$, there exists Lipschitz$F : \mathbb R^n \to \mathbb R$ such that $F|_M = f$. $M$ admits universal Lipschitz extension if and only if there exists $C$ such that $d_M(x,y) \leq C \|x-y\|$ for all $x,y \in M$ (note that it always holds that $\|x-y\| \leq d_M(x,y)$).
In the forward direction, suppose $d_M(x,y) \leq C \|x-y\|$ for all $x,y \in M$. Let $f : M \to \mathbb R$ be $L$-Lipschitz with respect to $d_M$. Then $f$ is $CL$-Lipschitz with respect to the Euclidean metric on $M$, and therefore by McShane's theorem, the function
$$F(x) = \sup\{f(y) + CL \|y-x\| : y \in M\}$$
is $CL$ Lipschitz and extends $f$.
For the other direction, suppose without loss of generality that $0 \in M$. Let $\mathcal L_M$ be the Banach space of Lipschitz functions $f$ on $M$ (Lipschitz with respect to $d_M$) such that $f(0) = 0$, equipped with the norm $\|f\|_{\mathcal L_M} = \text{Lip}(f)$. Let $\mathcal L_{\mathbb R^n}$ be the Banach space o\f Lipschitz functions on $\mathbb R^n$ equipped with the Lipschitz norm $\|f\|_{\mathcal L_{\mathbb R^n}}$. The restriction map $R : \mathcal L_{\mathbb R^n} \to \mathcal L_M$, where $R(F) = F|_M$, is $1$-Lipschitz. Provided that $M$ admits universal Lipschitz extension, $R$ is surjective, and so by the open mapping theorem there exists $C \in (0,\infty)$ such that for any $f \in \mathcal L_M$, there exists $F \in \mathcal L_{\mathbb R^n}$ such that $RF = f$ and $\|F\|_{\mathcal L_{\mathbb R^n}} \leq C \|f\|_{\mathcal L_M}$. Now, suppose for contradiction that there exist $x_n,y_n \in M$ such that $d_M(x_n,y_n) \geq n \|x_n - y_n\|$. The function $f_n(y) = d_M(x_n,y) - d(x_n,0)$ is in $\mathcal L_M$ with $\|f_n\|_{\mathcal L_m} = 1$, but for any $F \in \mathcal L_{\mathbb R^n}$ with $R F = f_n$, $$\|F\|_{\mathcal L_{\mathbb R^n}} \geq \frac{|f_n(y_n) - f_n(x_n)|}{\|x_n - y_n\|} = \frac{d_M(x_n,y_n)}{\|x_n - y_n\|}\geq n.$$
Open mapping theorem guarantees boundedness from below only for injections. I think you should instead of $\mathcal{L}{\mathbf{R}^n}$ consider $\mathcal{L}{(M,|\cdot|)}$.
Thanks, you are right of course. Edited.
@Justthisguy Thank you for the answer! Quick question tio brush up my geometry: in the last line of the first paragraph of your answer, you wrote: "note that it always holds that $∥x−y∥≤d_M(x,y)$". Why is it true?
The distance on $M$ is the inf of the lengths of paths in $M$ connecting $x$ and $y$. The distance in $\mathbb R^n$ between $x$ and $y$ is the inf over the lengths of paths in $\mathbb R^n$ connecting $x$ and $y$, so it is in the inf over a larger set and therefore bounded above by $d_M(x,y)$.
@Justthisguy Thanks, but these two lengths of paths are w.r.t. different Rielannian metrics, namely induced Riemannian metric on the manifold $M$ versus the canonical Euclidean matric in $\mathbb{R}^n$, so, sorry but I'm not sure how the larger set arguments go through? For a positively curved part of the manifold, path lengths or distances will increase from the Euclidean lengths/distances, but for the part that's negatively curved, the distance will decrease from the Euclidean distance, no?
The Riemannian metric on $M$ is the restriction of the Euclidean metric to the tangent space of $M$. Infinitesimally, the distances are equal, and so they give the same answer for the lengths of paths. That is, for any $\gamma : [0,1] \to M$, $|\gamma'|$ is the same whether you view $\gamma$ as taking values in $M$ or taking values in $\mathbb R^n$, and the length of $\gamma$ is just $\int\limits_0^1 |\gamma'(t)| dt$.
@Justthisguy Thanks, but then by what you wrote (" |γ′| is the same whether you view γ as taking values in $M= M^d$ or taking values in $\mathbb{R}^n$"), for every $v \in T_p{M^d}$ with co-ordinate vector w.r.t. a basis again dented by $v$, we'll have: $||v||^2_{\mathbb{R}^d} = v^{T}Av $ where $A$ is a symetric, positive definite matrix that represent the symmetric positive definite bilinear form on $T_p{M}$ that's the restriction of the innner product on $T_p{M}$. Sorry but I don't see why that's true? Why would the Euclidean norm of a vector and the inner product norm of a vector coincide?
Let us continue this discussion in chat.
@Justthisguy sure sent you a message in private chat if you so prefer, but I don't see the LaTeX symbols compiling there.
|
2025-03-21T14:48:29.992058
| 2020-03-03T18:38:16 |
354080
|
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|
Stack Exchange
|
Concerning $k \subset L \subset k(x,y)$
The following is a known result in algebraic geometry:
Let $k$ be an algebraically closed field of characteristic zero (for example, $k=\mathbb{C}$).
Let $L$ be a field such that $k \subset L \subset k(x,y)$
and $L$ is of transcendence degree two over $k$.
Then there exist $h_1,h_2 \in k(x,y)$ such that $L=k(h_1,h_2)$.
Is it possible to find $g_1,g_2 \in k[x,y]$ such that $L=k(g_1,g_2)$?
The motivation is the following result: If $k \subset L \subset k(x,y)$ is of transcendence
degree one over $k$, then $L=k(h)$, where $h \in k[x,y]$; see this answer. Perhaps the arguments in that answer are also applicable here?
I have asked the above question in MSE, with no comments. See also this question.
Thank you very much!
@YCor, thank you very much for adding the highly relevant tags.
No. $M\mathrel{:=}k(x,y)$ has a $k$-automorphism $\sigma:x\mapsto 1/x,\,y\mapsto 1/y$, of order 2. Let $G\mathrel{:=}\langle\sigma\rangle$, and put $L\mathrel{:=}M^{G}$, the fixed field. The elements $x+1/x$ and $y+1/y$ of $L$ are algebraically independent over $k$, hence $L$ has transcendency degree 2. However, no non-constant polynomial $g\in k[x,y]$ can be in $L$ (that is, be invariant under $\sigma$).
By the same token, the automorphism $\tau:x\mapsto 1/y,\,y\mapsto 1/x$, fixes the field $k(x/y)$, which has transcendency degree 1 over $k$, but it cannot fix any non-constant polynomial. So $k(x/y)\neq k(h)$ for every $h\in k[x,y]$. (Here, $M$ is again quadratic over the fixed field of $\tau$, so that the latter is of tr. deg. 2 over $k$ again. A transcendental element independent of $x/y$ is $x+1/y$, for instance).
Thank you for the nice answer!
|
2025-03-21T14:48:29.992195
| 2020-03-03T18:39:27 |
354081
|
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|
Stack Exchange
|
About the number of primes which are the sum of 3 consecutive primes (OEIS A034962)
I made some numerical simulations about the number of primes which are the sum of 3 consecutive primes (OEIS A034962), that is for instance:
$$5+7+11=23$$
$$7+11+13=31$$
$$11+13+17=41$$
$$17+19+23=59$$
$$19+23+29=71$$
$$23+29+31=83$$
$$29+31+37=97$$
$$...$$
The number of such triplets, till a certain integer n, seems to be well approximated by the following function:
$$\frac{e\cdot\pi(n)}{\log(n)}$$
If the previous was true, the density of such primes in the whole set
of prime numbers would be comparable to that of primes inside the set
of naturals.
I ask if this estimate is correct and some references about this matter.
Thanks.
is your use of $\pi$ supposed to be a capital $\prod$ to indicate a product?
Not at all. It indicates the primes counting function. See also Lucia's answer.
Thanks - new notation for me!
The question asks how many primes $p_n \le x$ are there such that $p_n + p_{n+1}+p_{n+2}$ is also prime. This is beyond our reach to answer, but one can use Hardy-Littlewood type heuristics to attack this. Since $p_n + p_{n+1}+ p_{n+2}$ is roughly of size $x$, it has about $1/\log x$ chance of being prime, and so one should expect the answer to be on the scale $\pi(x)/\log x \approx x/(\log x)^2$. But the asymptotic will be a little different, since $p_n + p_{n+1} +p_{n+2}$ is not quite random -- for example it will always be odd (omitting $2+3+5$).
Let's flesh out the usual heuristic. To be prime, a number must be coprime to each prime number $\ell$. A random integer has a chance $(1-1/\ell)$ of being coprime to $\ell$. What is the chance that $p_n+ p_{n+1} + p_{n+2}$ is coprime to $\ell$? Each of the primes $p_n$, $p_{n+1}$, $p_{n+2}$ itself lies in some reduced residue class $\mod \ell$. Assuming these possibilities are equally distributed, one should get the chance
$$
\frac{1}{(\ell-1)^3} \sum_{\substack{a, b, c=1 \\ (a+b+c,\ell)=1}}^{\ell-1} 1.
$$
If $a$ and $b$ are given with $a\not\equiv -b \mod \ell$, then $c$ has $\ell-2$ possible residue classes, and if $a\equiv -b \mod \ell$ then $c$ has $\ell-1$ possible residue classes. So the answer is
$$
\frac{1}{(\ell-1)^3} \Big( (\ell-1)(\ell-2)(\ell-2) + (\ell-1)(\ell-1)\Big)=
\frac{\ell^2-3\ell+3}{(\ell-1)^2}.
$$
So we adjust the random probability at $\ell$ by the factor
$$
\Big(1-\frac {1}{\ell}\Big)^{-1} \frac{(\ell^2-3\ell+3)}{(\ell-1)^2}= 1 +\frac{1}{(\ell-1)^3}.
$$
For example, when $\ell=2$, this adjustment factor is $2$ reflecting the fact that $p_n+p_{n+1}+p_{n+2}$ is guaranteed to be odd.
Then the analog of the Hardy--Littlewood conjecture will be
$$
\sim \prod_{\ell} \Big(1+\frac{1}{(\ell-1)^3} \Big) \frac{x}{(\log x)^2}.
$$
The constant in the product is approximately $2.3$; definitely not $e$.
There are interesting biases concerning why this conjecture would be an underestimate. Work of Lemke Oliver and Soundararajan makes precise conjectures on the distribution of consecutive primes in arithmetic progressions. For example, if we look at $\ell=3$, the sum $p_n + p_{n+1} +p_{n+2}$ can be a multiple of $3$ only if all three primes are congruent to each other $\mod 3$. This configuration is not preferred, and there are significant biases against it for small numbers! Similarly for other $\ell$ also, there are biases that indicate that the actual probability would be a tiny bit larger (with the effect wearing off slowly for large $x$). In other words, one might be able to identify lower order terms on the scale of $x(\log \log x)/(\log x)^3$, which might explain why the numerics suggest a larger value for the constant.
|
2025-03-21T14:48:29.992506
| 2020-03-03T19:04:57 |
354082
|
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|
Stack Exchange
|
Relevant mathematics to the recent coronavirus outbreak
I would like to ask about (old* and new) reliable mathematical literature relevant to various mathematical aspects of the recent coronavirus outbreak: In particular, standard statistical/mathematical models that are used to predict the spread, mathematical studies of effectiveness of various strategies, etc.
*(Added) By old I also mean well-established models.
Out of respect though , if this question was asked by someone anonymous it will be closed in seconds .
Mathematically, I doubt that there's anything particularly new about this coronavirus. Mathematical models of epidemics are well-established.
Of course we'd like to know the parameters (and to what extent something can be done about them). See Wikipedia
Random walk on graphs?
There are many. Here is a recent article of my friend: https://bmcinfectdis.biomedcentral.com/articles/10.1186/1471-2334-3-19
@MarkSapir I have to disagree - I think this question is simply far too broad for MO. (I won't vote to close because I have absolutely no background in the relevant topic, however.)
Regarding new research motivated by the outbreak, I found 3 papers on the arxive https://arxiv.org/search/advanced?advanced=&terms-0-operator=AND&terms-0-term=coronavirus&terms-0-field=abstract&classification-mathematics=y&classification-physics_archives=all&classification-include_cross_list=include&date-filter_by=all_dates&date-year=&date-from_date=&date-to_date=&date-date_type=submitted_date&abstracts=show&size=50&order=-announced_date_first
I liked this question, I wish it wasn't closed. Not everything that can be known about the mathematics of pandemics has been discovered and the novel coronavirus is as good an opportunity as any to motivate new results.
PLOS started the "Mathematical Modelling of Infectious Disease Dynamics Collection", about the spreading some literature can be found here, values for R0[4,5],CFR could be interesting too, google has lots of stuff but more qualitatively could be useful this course by CDC
@Bambi , In retrospect, the question was closed and also the users' votes were not favorable. So I don't think this is a case where anonymous user would have great disadvantage. Probably I could have done a better job in writing the question (so I did not vote to reopen myself) and I hope (as a non expert) that at some time better questions will be asked both regarding well-established models of epidemics, and about specific mathematical issues related to the current outbreak.
@Gil Kalai First of all I have to tell you this , I've enormous respect for you and I'm follower of your work on quantum computing . I didn't expect that this comment will get such attention. I just mentioned this because I've seen various questions of above form getting closed immediately on MO . Again apologies if I'm being rude .( After this comment three of my own MO questions got downvotes )( I didn't even vote to close it )
@Bambi, no problems...
@GilKalai if you expand to include q-bio you go from 3 hits to 24, many of which are at least math-adjacent if not outright actual math.
Not epidemics and more cs than math but could be interesting anyway, some data on the protein structure is being released now and this is done by alphafold + which changed protein prediction in 2018 just as imagenet changed visual recognition back in 2012
I was thinking to ask such question, but I afraid of that it is not suitable for here. Anyway, there are many good epidemics modeling by the NetLogo software for HIV, Ebola, flue and many others.
I think by redefining and tuning the existing model, we can find some good new models for COVID-19.
Possibly unrelated (I'm not a professional mathematician and I don't know the following resources) to your specific question about the coronavirus outbreak in this site is that the official channel Centre International de Rencontres Mathématiques is editing from its YouTube channel recent videos that maybe are related (in my view) to models of epidemics, for example by Malwina Luczak and other by Simon Frost. I have not looked for more references as video lectures in other math channels that I know. Isn't required a response of this message, and I hope don't disturb.
An interesting MSRI (online) lecture by Nicholas Jewell https://youtu.be/MZ957qhzcjI
I tried to answer you in
https://mathoverflow.net/questions/355489/suggestings-for-reducing-the-transmission-rate
On Terry Tao's blog: Polymath proposal: clearinghouse for crowdsourcing COVID-19 data and data cleaning requests. https://terrytao.wordpress.com/2020/03/25/polymath-proposal-clearinghouse-for-crowdsourcing-covid-19-data-and-data-cleaning-requests/
IEEE is giving free access to all COVID-19-related papers. https://ieeexplore.ieee.org/search/searchresult.jsp?newsearch=true&queryText=%22Free%20Promotions%22:COVID-19
The Centre for the Mathematical Modelling of Infectious Diseases recently made a study on the under-reporting of cases by country (here is a article but some of what is written may differ on more recent data), this is not math focused but hopefully still useful
Maybe relevant:
Yu Chen, Jin Cheng, Yu Jiang, Keji Liu, A Time Delay Dynamical Model for Outbreak of 2019-nCoV and the Parameter Identification, https://arxiv.org/abs/2002.00418
Thanks for the answer, Martin!
The following paper is a little strange, since it dates back to 2015, but has some valuable data:
A SARS-like cluster of circulating bat coronaviruses shows potential for human emergence, Nature Medicine, 2015.
Here is another case from february 2019 "Bat Coronaviruses in China" from the abstract "Thus, it is highly likely that future SARS- or MERS-like coronavirus outbreaks will originate from bats, and there is an increased probability that this will occur in China."
@Daniel D. And it is more strange why we do not have any kind of vaccination by these predictions!?
There exists the "Coalition for Epidemic Preparedness Innovations (CEPI)" which was created after the Ebola outbreak and is leading the efforts in the creation of a vaccine but is not the only one so there are a lot of news and I would say is not clear were this race is going yet, on the other hand for MERS there was no vaccine either so it seems to me what was done first was to test if the treatments for MERS were also effective against COVID-19 in particular Remdesivir and chloroquine
For completeness sake let me add a another paper that I have read on the news more or less had foreseen the situation:Severe Acute Respiratory Syndrome Coronavirus as an Agent of Emerging and Reemerging Infection
The book by Gábor Csárdi, Tamás Nepusz, Edoardo Airoldi,
Statistical Network Analysis with igraph
Based around popular software library igraph, Wikipedia link
contains whole chapter with source codes (in R) on Epidemics on networks in particular 6.5 Vaccination strategies
Let me quote the content of the chapter:
6 Epidemics on networks
6.2 Branching processes
6.3 Compartmental models on homogeneous populations
6.3.1 The susceptible-infected-recovered (SIR) model
6.3.2 The susceptible-infected-susceptible (SIS) model
6.3.3 The susceptible-infected-recovered-susceptible (SIRS)model
6.4 Compartmental models on networks
6.4.1 A general framework for compartmental models onnetworks
6.4.2 Epidemics on regular and geometric networks . . . . . . . . 180
6.4.3 Epidemics on scale-free networks . . . . . . . . . . . . . . . . . . . 187
6.5 Vaccination strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Thanks for the answer, Alexander!
Here is a recent paper written by mathematicians: Risk Assessment of Novel Coronavirus COVID-19 Outbreaks Outside China.
Thanks for the answer, Matt!
@GilKalai: My name is not Matt (and I wrote this answer), but thanks nevertheless.
Thanks for the answer, GH from MO!
Recently found this :
https://staff.math.su.se/tom.britton/
Maybe relevant.
Here is a related videotaped lecture by Britton, https://youtu.be/gSqIwXl6IjQ
@GilKalai Wow , thanks , I didn't know about video
The following paper is extremely important because it has informed the decisions of the UK government that realised (announced) on Monday 16/03/2020 that it can not afford "Herd immunity". The paper only shows the outcomes of the model and speaks about its parameters. It would of course be extremely interesting to know what exactly is the mathematics behind it. Mathematicians should try to read it.
https://www.imperial.ac.uk/media/imperial-college/medicine/sph/ide/gida-fellowships/Imperial-College-COVID19-NPI-modelling-16-03-2020.pdf?fbclid=IwAR2Ca5Ki23DWn-EGWeB3yaNE4f9GmnUcEWU_S60lsDC230AKUg4v_w82qeE
Thanks for the answer, aglearner!
Here is a related mathematical discussion on Gowers' blog: https://gowers.wordpress.com/2020/03/28/how-long-should-a-lockdown-relaxation-cycle-last/
Volume 12 of What's Happening in the Mathematical Sciences by Dana Mackenzie has three chapters devoted to mathematical modeling related to the coronavirus pandemic.
Fifty Ways to Beat a Virus (Part 1)
In 2020, the world for the first time in a century confronted
a global pandemic that would claim millions of lives. While
students were sent home and campuses closed, many mathematicians found an opportunity to join the fight against
COVID-19. Even simple differential equation models can teach
us important lessons about the exponential growth of a new
epidemic and the importance of threshold behavior. Using
more elaborate (and realistic) models, two mathematical
modeling groups, in Texas and Illinois, had a profound and
positive effect on the management of the epidemic by local and
state authorities during the first waves.
Fifty Ways to Beat a Virus (Part 2)
Continuing the previous chapter, Part 2 discusses the problems
confronted by mathematicians and epidemiologists in the later
part of 2020 and in early 2021. How could universities re-open
safely? How does the uncontrolled spread of an epidemic in
prisons affect the surrounding community, and what can be
done about it? And the biggest question: could vaccination
bring the epidemic under control? Even though the coronavirus
kept throwing surprises at us, mathematicians did a surprisingly good job of developing strategies, giving realistic answers
and highlighting the main reasons for uncertainty.
Fifty Ways to Beat a Virus (Part 3)
Another important front in the battle against COVID-19 was
to understand how the infection progresses within the human
body. Why do some people have life-threatening symptoms,
while others have none at all? An online group called the
“Immune Gals” highlighted the delayed release of interferon
as a characteristic marker of severe cases. Another mathematician used the tools of graph theory to identify parts of the
viral RNA that are especially vulnerable to attack by drugs or
gene therapy. And a third group adapted a machine-learning
language model to detect escape variants of coronavirus. In
effect, they taught the computer to “speak virus.”
|
2025-03-21T14:48:29.993743
| 2020-03-03T19:49:43 |
354085
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Capublanca",
"Piero D'Ancona",
"Willie Wong",
"https://mathoverflow.net/users/3948",
"https://mathoverflow.net/users/54552",
"https://mathoverflow.net/users/7294"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:626876",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/354085"
}
|
Stack Exchange
|
Smoothing-Strichartz estimates for the heat-Schrodinger evolution
Consider the heat-Schrodinger evolution $e^{(1+i)t\Delta}$, $t\geq 0$. For simplicity, suppose to work in dimension three. Due to the Strichartz estimates for the Schrodinger equation, and the smoothing effect of the heat flow, I expect that the following estimate holds true:
\begin{equation}\label{tre}
(1)\qquad \left\|\int_0^te^{(1+i)(t-s)\Delta}F(s)ds\right\|_{L^2((0,t),W^{1,6}(\mathbb{R}^3))}\lesssim \|F\|_{L^2((0,t),L^2(\mathbb{R}^3))}
\end{equation}
Nevertheless, I have not been able to prove it.
Is estimate (1) actually true? In case, is it already known in the literature?
Thank you for your suggestions.
Sure, thank you. I edited
I think the estimate you want does not really require Strichartz.
First, your estimate is equivalent to the following, which is a bit easier for me to think about: let $u$ be the solution to the equation
$$ \partial_t u - (1 + i) \triangle u = F \tag{*}$$
with initial data
$$ u(0,x) \equiv 0 $$
Then the desired estimate is
$$ \| u \|_{L^2_t W^{1,6}_x} \leq \|F\|_{L^2_t L^2_x}$$
We can derive this estimate by multiplying the equation (*) by $\triangle \bar{u}$ and taking the real part, which gives
$$ \partial_t u \triangle \bar{u} + \partial_t \bar{u} \triangle u - (1+i) \triangle u \triangle \bar{u} - (1-i) \triangle u \triangle\bar{u} = F \triangle \bar{u} + \bar{F} \triangle u $$
Integrate by parts in space we get
$$ \partial_t \| \nabla u\|^2_{L^2_x} + 2 \|\triangle u\|_{L^2_x}^2 = - \int_{\mathbb{R}^3} F\triangle \bar{u} + \bar{F} \triangle u ~dx$$
Integrate between time 0 and time T you get (using the triviality of the data)
$$ \|\nabla u(T)\|_{L^2_x}^2 + 2 \int_0^T \| \triangle u\|^2_{L^2_x} ~dt \leq \int_{[0,T]\times \mathbb{R}^3} 2 |F|\cdot |\triangle u| ~dx~dt (**)$$
Applying Young's inequality on the right, you can pull out $\epsilon \|\triangle u\|_{L^2_t L^2_x}$ which can be absorbed on the left hand side, and this reduces to
$$ \| \triangle u\|_{L^2_t L^2_x} \lesssim \| F\|_{L^2_t L^2_x} $$
Finally use elliptic estimates to bound $\|\nabla^2 u\|$ by $\|\triangle u\|$, and then Sobolev inequality in space gets you
$$ \| \nabla u\|_{L^2_t L^6_x} \lesssim \|F\|_{L^2_t L^2_x} $$
This takes care the homogeneous part of the Sobolev norm.
I don't think the inhomogeneous part holds. It doesn't have the correct scaling.
If you rescale $u(t,x) \mapsto u(\lambda^2 t, \lambda x)$ by the parabolic scaling, then $F(t,x) \mapsto \lambda^2 F(\lambda^2 t, \lambda x)$. The $L^2_t L^2_x$ norm of $F$ scales like $\lambda^{-1/2}$, but the $L^2_t L^6_x$ norm of $u$ scales like $\lambda^{-3/2}$. The term with the correct scaling should be $\|F\|_{L^1_t L^2_x}$. And the estimate
$$ \| u\|_{L^2_t L^6_x} \lesssim \|\nabla u\|_{L^2_t L^2_x} \lesssim \|F\|_{L^1_t L^2_x} $$
can be proven by an analogous manner as above but instead of multiplying by $\triangle u$, multiply by $u$.
Incidentally: both estimates also hold for the heat equation, without the Schroedinger part.
Addendum: I just noticed that your inequality is only stated for $L^2((0,T),X)$ for some Banach space $X$. Do you only care about the local estimate where your constant can depend on the length of the interval? Or do you actually want uniform estimates for all $T$? I ask because quite obviously on a fixed time interval you have that $\|F\|_{L^1_t L^2_x} \lesssim \|F\|_{L^2_t L^2_x}$, and in fact you can get the same result by noting that (**) also implies $$\|u\|_{L^\infty_t L^6_x} \lesssim \|F\|_{L^2_t L^2_x}$$ and can be easily localized to intervals.
Nice. I think for the lower order term you might need some weight at the RHS
@PieroD'Ancona: I don't quite understand what you mean. Can you say a couple sentences more? What I imagine: multiplying by $u$ and IBP gives us that $$|u(T)|{L^2}^2 + | \nabla u|{L^2_{t,x}}^2 \leq | F u |{L^1{t,x}} $$ the right hand side bound is bounded by $| F|{L^1_t L^2_x} |u|{L^\infty_t L^2_x}$ which implies $$ |u|{L^\infty_t L^2_x} \leq |F|{L^1_t L^2_x}$$ which gives the desired result. I don't think weights are necessary. But possibly I misunderstand your comment.
What I mean is that another way to achieve the correct scaling is, say, using a weighted norm such as $||x|F|_{L^2L^2}$. I think the estimate for the lower order term might be true also with this right hand side
@PieroD'Ancona: ah, I see. For the pure Schrodinger this would be an end-point case if I am not mistaken. (It almost follows from my paper https://arxiv.org/abs/1701.01460v4 but it would be the endpoint case which I didn't do).
Nice work, thank you. I was indeed interested only in the homogeneous estimates, but also the suggestion by Piero for the lower order term is very interesting. I think that, if one considers the evolution $e^{(i+\varepsilon)t\Delta}$, where $\varepsilon>0$ is a small parameter, your approach produces a divergence rate as $\varepsilon\to 0$ which can be improved by using the Strichartz estimates for the Schrodinger equation. Maybe it could be interesting (and not so trivial) to establish the optimal divergence rate as $\varepsilon\to 0$.
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