added
stringdate 2025-03-12 15:57:16
2025-03-21 13:32:23
| created
timestamp[us]date 2008-09-06 22:17:14
2024-12-31 23:58:17
| id
stringlengths 1
7
| metadata
dict | source
stringclasses 1
value | text
stringlengths 59
10.4M
|
|---|---|---|---|---|---|
2025-03-21T14:48:31.611845
| 2020-07-26T13:49:52 |
366613
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Adam P. Goucher",
"Jianrong Li",
"https://mathoverflow.net/users/11877",
"https://mathoverflow.net/users/35959",
"https://mathoverflow.net/users/39521",
"user64494"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631478",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366613"
}
|
Stack Exchange
|
What is the name of the following root system?
The Dynkin diagram of the root system of affine $D_4$ is
$$
\circ \quad \circ \quad \circ \quad \circ \\
\circ
$$
where all of the four vertices in the first row connects to the vertex in the second row.
Is there are root system which has the following Dynkin diagram?
$$
\circ \quad \circ \quad \circ \quad \circ \quad \circ \\
\circ
$$
where all of the five vertices in the first row connects to the vertex in the second row.
If yes, what is the name of the root system? Thank you very much.
The corresponding Coxeter group is paracompact and hyperbolic. Apparently it's called $\overline{L_5}$ and it's mentioned here: https://en.wikipedia.org/wiki/Coxeter%E2%80%93Dynkin_diagram#Ranks_4%E2%80%9310
See $E5$ in https://en.wikipedia.org/wiki/Dynkin_diagram .
@user64494, I think it is $\overline{L_5}$ in Adam P. Goucher's comment.
|
2025-03-21T14:48:31.611927
| 2020-07-26T14:06:12 |
366614
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631479",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366614"
}
|
Stack Exchange
|
Holomorphic maps on moduli space and Deformation theory
Let $\mathcal{M},\mathcal{F}$ be the classifiying spaces (i.e. complex manifolds) of two (possibly) different moduli problem. To give a map
$$f:\mathcal{M}\rightarrow \mathcal{F}$$
means that for each object classified by $\mathcal{M}$, we must associate an object classified by $\mathcal{F}$. Now if I want to show that this map is holomorphic, then the way I've seen this often done is that you need to use some explicit local parametrization of both $\mathcal{M}$ and $\mathcal{F}$.
However, I wonder if this can not be done in a deformation theoretic way. Intuitively, infinitesimal deformations of an object $m\in\mathcal{M}$ corresponds to the tangent space of our manifold, and if we assume that our moduli space is smooth, it would thus also correspond to a "small" neighborhood of $m$. Since we only need to show that in sufficienttly small neighborhoods of point, the map is a power series, I feel like that the stalk $\mathcal{O}_{\mathcal{M},m}$ should "see" the holomorphicity.
As a concrete example, consider the moduli space of surfaces $\mathcal{M}_{g,n}$ and the flag variety $\mathcal{Fl}$. To each $[X,x_1,\ldots,x_n]\in \mathcal{M}_{g,n}$ we can associate the flag $F^p(H^k(X,\mathbb{C}))$. It is known by the work of Griffiths that this map is holomorphic. Can this be proven by studying how the period map behaves at the Teichmüller space $\mathcal{T}_g$?
|
2025-03-21T14:48:31.612054
| 2020-07-26T14:08:08 |
366615
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Duloren",
"Mikhail Borovoi",
"https://mathoverflow.net/users/131781",
"https://mathoverflow.net/users/161780",
"https://mathoverflow.net/users/4149",
"user131781"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631480",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366615"
}
|
Stack Exchange
|
Find tetrahedron vertex given 3 vertices of a face and the 3 opposite angles
I have the following tetrahedron:
which I know the coordinates of $P$, $Q$ and $R$ and the value of angles $\theta_0$, $\theta_1$ and $\theta_2$.
I need to find the coordinates of vertex $E$. Is that possible? If so, how can I find $E$?
So far, I have tried the following approaches:
Solving a system by using angles between vectors $\vec{ER}$, $\vec{EP}$ and $\vec{EQ}$:
$$\cos{\theta_0}=\frac{\vec{ER} \cdot \vec{EP}}{||\vec{ER}\||||\vec{EP}||}$$
$$\cos{\theta_1}=\frac{\vec{ER} \cdot \vec{EQ}}{||\vec{ER}\||||\vec{EQ}||}$$
$$\cos{\theta_2}=\frac{\vec{EP} \cdot \vec{EQ}}{||\vec{EP}\||||\vec{EQ}||}$$
By doing this I ended up with a system which I can't solve.
Solving system by using Law of Cosines between unknown edges $||\vec{ER}||$, $||\vec{EP}||$, $||\vec{EQ}||$ and known ones $||\vec{PQ}||$, $||\vec{QR}||$, $||\vec{RQ}||$:
$$||\vec{PQ}||^2=||\vec{EP}||^2 + ||\vec{EQ}||^2 - 2||\vec{EP}||||\vec{EQ}||\cos{\theta_0}$$
$$||\vec{QR}||^2=||\vec{EQ}||^2 + ||\vec{ER}||^2 - 2||\vec{EQ}||||\vec{ER}||\cos{\theta_1}$$
$$||\vec{RP}||^2=||\vec{ER}||^2 + ||\vec{EP}||^2 - 2||\vec{ER}||||\vec{EP}||\cos{\theta_2}$$
By doing this I ended up with a similar system as the previous one.
I also tried to use Sine Rule finding the circumradius for each face:
$$circumradius_{EPQ} = \frac{||\vec{PQ}||}{2sin(\theta_0)}$$
$$circumradius_{EQR} = \frac{||\vec{QR}||}{2sin(\theta_1)}$$
$$circumradius_{ERP} = \frac{||\vec{RP}||}{2sin(\theta_2)}$$
but it also actually didn't lead me to find a solution.
I'm not sure if there is a single solution for these settings or even if it is possible to find one. Any help is appreciated.
This nice question seems to be more suitable for Math StackExchange.com.
Hi @MikhailBorovoi thanks for the feedback, indeed I'm confused about the two sites. Could you clarify the difference briefly?
Humm ok I just found it https://math.meta.stackexchange.com/questions/41/differences-between-mathoverflow-and-math-stackexchange
Well you can suppose that your points have coordinates $(0,0,0),(1,0,0),(p,q,0),(r,s,t)$ where $p$ and $q$ are given. Your conditions then give three polynomial equations of fourth degree in the unknowns $r,s,t$. Not sure if this is of any use for your purposes.
Hi @user131781 thanks for the hint. This fourth degree 3-unknow system was what a have found. Honestly speaking I don't know if I have the required skill to solve it or if it doesn't have a solution. I m running it on Mathematica. 40 minutes and no response yet.
Not all polynomial equations (even in one unknown) have solutions in elementary functions of coefficients. However, for given numbers $p,q$ you can solve your system numerically.
@MikhailBorovoi I'm still on going but it seems that this problem has not a canonical solution. Wu-Ritt’s zero decomposition algorithm is my next station on this travel https://en.wikipedia.org/wiki/Wu%27s_method_of_characteristic_set
|
2025-03-21T14:48:31.612261
| 2020-07-26T14:47:21 |
366617
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Achim Krause",
"Andrej Bauer",
"Dan Ramras",
"Fernando Muro",
"LSpice",
"Matt Feller",
"R. Srivastava",
"https://mathoverflow.net/users/1176",
"https://mathoverflow.net/users/12166",
"https://mathoverflow.net/users/132451",
"https://mathoverflow.net/users/161330",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/39747",
"https://mathoverflow.net/users/4042",
"https://mathoverflow.net/users/41291",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631481",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366617"
}
|
Stack Exchange
|
Brouwer's fixed point theorem and the one-point topology
I posted this question last week on Math SE and got upvotes and helpful comments that allowed me to make the question more precise https://math.stackexchange.com/q/3765546/810513. As I did not get an answer to this question, please allow me to ask it on MO:
In the 2-dimensional case, Brouwer's fixed point theorem (BFPT) says that every continuous function $D^2\to D^2$ has a fixed point, where $D^2$ is the disk.
Now fix a particular topology: pick some point $x_0\in D^2$ and use it to define the one-point topology $\cal T_0$ on $D^2$: it includes all sets $A$ with $x_0\in A$, and the empty set. (This is indeed a topology, see for example https://en.wikipedia.org/wiki/Particular_point_topology).
With respect to $\cal T_0$, a self map $D^2\to D^2$ is continuous if and only if it is constant or has $x_0$ as a fixed point. So, for every self map on $D^2$, continuity with respect to $\cal T_0$ means that a fixed point exists. Hence the BFPT is trivially true, by definition of $\cal T_0$.
In conclusion, there are topologies where BFPT is a theorem that requires proof, and there is a topology $\cal T_0$ where BFPT is true just by definition.
This gives $\cal T_0$ a special place among all possible topologies on $D^2$: it is the topology that makes BFPT trivial. Does such a situation or property have a name? Does it have a category theory interpretation (maybe like "universal property")?
I feel there is a certain equivalence between BFPT and $\cal T_0$ here. They characterize each other in a certain way: $\cal T_0$ makes BFPT trivially true by definition, and BFPT links continuity and fixed points (like $\cal T_0$ does). Can this sense of equivalence be expressed rigorously?
I am struggling to express this property and my sense of "equivalence" more precisely and thus understand it better - I would be very grateful if someone could help me with this.
EDIT: Thank you for great comments! They are extremely helpful. I edited and corrected my question to reflect them.
"Continuity and having the fixed point $x_0$ are one and the same thing here.", that's not true, is it? For example if $x_1$ is any other point, the constant map to $x_1$ is also continuous with respect to your $x_0$-topology, since all preimages are either empty or everything. But this does not have $x_0$ as fixed point.
Maybe the problem is that literally as written, your topology is not a topology, since the empty set also needs to be open.
@AchimKrause thank vou very much for your remark about the empty set! I corrected it now in my question. The one-point topology which I am describing here should indeed be a proper topology, it is sometimes called the "particular point topology". Please see the related Wikipedia article.
It is still not true that continuity and having fixed point $x_0$ are one and the same thing for that topology. For any topology any point is the unique fixed point for some continuous map (e. g. the constant map with value this point).
What you really want to say is that a self map is continuous with respect to this topology if and only if it is either constant or has $x_0$ as a fixed point. Here is an argument that a continuous map which doesn't fix $x_0$ must be constant: if $f$ is continuous and sends $x_0$ to $x_1\neq x_0$, then for any $x_2\neq x_1$, the pre-image $f^{-1}({x_0, x_2})$ doesn't contain $x_0$, so must be empty. Thus $x_1$ is the only element of the image of $f$.
I don't sense any real equivalence here between continuity and having a fixed point, since there are many non-constant self maps which fix other points but don't fix $x_0$. I think the best we can say is "BFPT is easier to prove when our topology is a lot simpler." My guess is that there isn't anything deeper going on.
@MattFeller thanks a lot for your two comments! Also thanks to the others giving the same comment about the constant function - I will edit and correct. Regarding your second question, let me think about your statement „BFPT is easier to prove when our topology is a lot simpler“
Aside from the question for whether BFPT is true (trivially or not), I think your argument doesn't establish that it is. You say: every function with $x_0$ as fixed point is continuous in this topology (which is true); but you conclude: every function that is continuous in this topology has $x_0$ as fixed point. As others have pointed out, that's false, but, even if you revise to a true statement, it doesn't follow from your argument. (Could you also define mathematically "theorem" versus "trivially true"? It seems a judgement. Maybe Brouwer found it trivially true in usual topology!)
@R.Srivastava the constant function has a fixed point...
The first paragraph is perplexing to me. You write "This is valid for any choice of topology." But the BFPT is a fact about the standard Euclidean topology on $D^2$, and certainly fails for other topologies. To be explicit, consider the indiscrete topology on $D^2$, in which only the empty set and $D^2$ itself are open. Every function $D^2\rightarrow D^2$ is continuous in this topology, but there are certainly functions from $D^2$ to itself with no fixed point (e.g. a function that maps the origin to $(0,1)$ and rotates the rest of the disk by 90 degrees).
The example I gave above is explicit, but you can get lots of topologies on $D^2$ in which BFPT fails by simply looking for topological spaces of the same cardinality as $D^2$ for which BFPT fails, e.g. the real line. If you choose a bijection between $D^2$ and $\mathbb{R}$ (which presumably requires the Axiom of Choice), then there's a unique topology on $D^2$ making this function a homeomorphism, and the BFPT will fail (simply because on $\mathbb{R}$, the function $f(x) = x+1$ is continuous and has no fixed point).
(I now see that my comments are about a line added in a recent edit, which explains why the other commenters didn't complain about this.)
You say that every continuous $f : D^2 \to D^2$ has a fixed point, for any choice of topology on $D^2$. How about the discrete and the indiscrete topologies?
@DanRamras thank you very much for your comprehensive comment and your example. You are right! My statement was wrong and I have removed it now.
|
2025-03-21T14:48:31.612710
| 2020-07-26T14:55:56 |
366618
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"R. van Dobben de Bruyn",
"Stanley Yao Xiao",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/82179"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631482",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366618"
}
|
Stack Exchange
|
What aspects of Faltings' proof of Mordell's conjecture can be generalized to higher dimensional varieties?
Falting's proof of Mordell's conjecture is one of the greatest achievements in arithmetic geometry. Broadly speaking, it capitalizes on an earlier observation of Parshin, which reduces Mordell's conjecture to a conjecture of Shafarevich. The proof can be broken down as follows:
Given a curve $C/K$, $K$ a number field, Parshin's construction shows that for every $K$-point $x \in C$ there exists a curve $C^\prime(x)$ of fixed degree such that $C^\prime(x)$ admits a dominant map to $C$ branched only at $x$. Moreover, this dominant map is defined over a number field $L/K$ which is independent of $x$. Finally, for each $x \in C(K)$ the primes of bad reduction of the curve $C^\prime(x)/L$ lie above the corresponding set for $C/K$. This defines a finite-to-one map between the set of curves $\mathcal{C}/L$ having good reduction outside of a fixed set of primes and the set $C(K)$ and thus, reduces the proof of Mordell's conjecture to Shafarevich's conjecture.
Faltings broke down the proof of Shafarevich's conjecture into two components: the first is essentially Tate's conjecture for abelian varieties, which is the assertion that the $L$-function of an abelian variety exactly determines the isogeny class of the abelian variety.
The most novel part of Faltings' proof, arguably, is the statement that there are at most finitely many $K$-isomorphism classes of abelian varieties in a given isogeny class. This necessitated the notion of a new height which is now named after Faltings. In particular, he proved that Faltings' height is necessarily bounded within an isogeny class, and that non-isomorphic but isogenous abelian varieties have different enough Faltings' height. These facts imply that the number of isomorphism classes within an isogeny class is finite.
These statements together imply Mordell's conjecture, albeit in an fairly ineffective way (to be precise, it is known how to extract a numerical bound for the cardinality of $C(K)$ from this argument, but not the height of the points themselves).
What aspects of this argument can be generalized to varieties of general type of higher dimensions? I understand that Tate's conjecture outside of the abelian variety setting appears to be very very difficult. What about other aspects of this argument?
Note that it is not at all clear how the Tate conjecture would help in the absence of Parshin's trick (and possibly a Torelli theorem). But there are some analogues beyond the curve case of the finiteness of integral points in moduli, e.g. in the work of Lawrence and Venkatesh. As far as I know, these do not lead to new cases of the Bombieri–Lang–... conjectures.
Ah yes you are right. I forgot about this key aspect of Faltings' proof, which I believe has no analogue in the higher dimensional case.
|
2025-03-21T14:48:31.612954
| 2020-07-26T15:32:31 |
366622
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631483",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366622"
}
|
Stack Exchange
|
Schur Weyl duality for the supergroup $\text{GL}(m|n)$
Let $G$ be the supergroup $\text{GL}(m|n)$. It has a tautological representation $V= \mathbb{C}^{m|n}$.
For every natural number $d$ we have a natural map $$\Phi_d:\mathbb{C} S_d\to \text{End}_G(V^{\otimes d})$$
where $\sigma\in S_d$ is sent to the linear transformation given by tensor permuting $V^{\otimes d}$ according to $\sigma$. To what extent does Schur-Weyl duality generalise from the case $n=0$ to the case of super vector space? That is:
Question 1: Is $\Phi_d$ surjective?
Question 2: What is the kernel of $\Phi_d$ ? Can it also be described using some combinatorial condition similar to the case $n=0$?
Schur Weyl duality holds in the super case, as well. There is the double centralizer property, thus a positive answer to Q1, and also a characterization of the kernel as those ideals of $\mathbb C[S_d]$ which correspond to partitions that don't fit inside the (m,n)-hook.
See the paper "Hook Young diagrams with applications to combinatorics and to representations of Lie superalgebras" by Berele and Regev. For a more recent textbook treatment this is also done nicely in chapter 11 of Musson's book "Lie Superalgebras and Enveloping Algebras".
This is a result of Sergeev, but I can only find the article in Russian at the moment. If I remember right, the map is surjective, and the kernel can be worked out from the fact that a Schur functor applied to the standar representation of $\mathfrak{gl}(m|n)$ vanishes iff the Young diagram contains the box at position (m+1,n+1), i.e. if they do not fit in an "(m,n) hook." For example, see the discussion in these slides of Serganova. Note that the classical case the diagrams fitting in an (m,0)-hook are exactly the ones with m or fewer rows, while in the entirely odd case they need to have n or fewer columns.
|
2025-03-21T14:48:31.613135
| 2020-07-26T16:09:24 |
366623
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alkan",
"https://mathoverflow.net/users/122662",
"https://mathoverflow.net/users/161528",
"Đào Thanh Oai"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631484",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366623"
}
|
Stack Exchange
|
A closed chain of $2n+1$-gon around $2n+1$-points
I posed a generalization of Theorem 3.2 In my paper
Conjecture: Let $P_1, P_2,....,P_{2n+1}$ and $O$ be $2n+2$ points in plane. Construct a chain $2n+1$ regular ${2n+1}$-gons $A_{1\;1}A_{1\;2}...A_{1\;2n+1}$; ....;$A_{2n+1\;1}A_{2n+1\;2}...A_{2n+1\;2n+1}$ with center $A_1, A_2...., A_{2n+1}$ such that $A_{1\;1}=O$, $A_{1\;2}=P_1$, $A_{2\;1}=A_{1\;3}$, $A_{2\;2}=P_{2}$, $A_{i+1\;1}=A_{i\;3}$, $A_{i+1\;2}=P_{i+1}$ for $i=\overline{1\;2n}$
Continuing construct a chain $2n+1$ regular ${2n+1}$-gons $B_{11}B_{12}...B_{1\;2n+1}$; ....;$B_{2n+1\;1}B_{2n+1\;2}...B_{2n+1\;2n+1}$ with centers $B_1, B_2...., B_{2n+1}$, such that $B_{1\;1}=A_{2n+1\;3}$, $B_{1\;2}=P_1$, $B_{2\;1}=B_{1\;3}$, $B_{2\;2}=P_{2}$, $B_{i+1\;1}=B_{i\;3}$, $B_{i+1\;2}=P_{i+1}$ for $i=\overline{1\;2n}$.
Then $B_{2n+1\;3}=O$ and segments $A_1B_1=A_2B_2=...=A_{2n+1}B_{2n+1}$ and $\angle (A_iB_i, A_{i+1}B_{i+1})=\frac{(2n-1)\pi}{2n+1}$
Corollary: $B_{2n+1\;3}$ is fixed point when $P_1$, $P_2$, ....,$P_n$ be moved.
Question 1: Is the conjecture correct?
Question 2: Let $P_1$, $P_2$, ....,$P_n$ are fixed point in the plane, find position of $O$ such that $A_{2n+1\;3}=O$
See also:
Napoleon theorem
Van Aubel theorem
Petr–Douglas–Neumann theorem
PS: In the conjecture, all regular polygon is same direction.
Nice question !
Thanks you, I correction @FedorPetrov
You may easily calculate everything in complex numbers. Denote $m=2n+1$, $w=e^{2\pi i/n}$, $Q_i=A_{i,3}=A_{i+1,2}$ for $i=i,2,\ldots$. We may suppose that $P_{m+i}=P_i$ for $i=1,\ldots,m$ and we have one sequence of $2m$ polygons. Then we have to prove $Q_{2m}=O$ (let $O=0$ be the origin) and that $C_k:=A_{k+m}-A_k$ satisfy $C_{k+1}=-w C_k$. This follows from $Q_k-P_k=(P_k-Q_{k-1})w$, where $Q_0=0$. Dividing by $w^k$ this gives for the sequence $R_k:=(-1)^kQ_k/w^k$ the recurrence $R_k-R_{k-1}=(-1)^kP_k(1+w)/w^k=:x_k$. We have $x_{m+k}+x_k=0$, thus $R_{2m}=x_1+\ldots+x_{2m}=0$. Also $A_k=(Q_k-P_kw)/(1-w)$, thus $$C_k=A_{k+m}-A_k=\frac{1}{1-w}(Q_{k+m}-Q_k)=(-1)^{k}\frac{w^k}{1-w}(R_k+R_{m+k})=
(-1)^{k}\frac{w^k}{1-w}(x_1+\ldots+x_k+x_1+\ldots+x_{m+k})=
(-1)^{k}\frac{w^k}{1-w}(x_1+\ldots+x_m),$$
the result follows.
As for question 2, it asks when $x_1+\ldots+x_m=0$. Since $O$ is variable, we replace $P_i$ to $P_i-O$ and get the equation
$$
\sum_{k=1}^m (-1)^k (P_k-O)/w^k=0 \Leftrightarrow O=\frac{1+w}2\sum_{k=1}^m (-1)^{k-1}P_k w^{-k}.
$$
|
2025-03-21T14:48:31.613291
| 2020-07-26T16:39:50 |
366626
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Giorgio Metafune",
"Lin2568",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/151662"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631485",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366626"
}
|
Stack Exchange
|
With Khinchine's inequality, prove Fourier basis is unconditional in $L^{p}[0,1]$ only for $p=2$
I am trying to prove Problem 6.10 in "Classical and Multilinear Harmonic Analysis" by by Camil Muscalu and Wilhelm Schlag.
Problem
Problem 6.10. Let $1\le p < \infty$ and suppose that there exists a constant $C(p)$ such that
$$\sup_{\varepsilon_n=\pm1} \left\lVert \sum_{-N}^N \varepsilon_n \widehat f(n) e(n\theta) \right\rVert \le C(p) \|f\|_p \qquad \forall f\in L^p([0,1]), \forall N\ge1. \tag{6.33}$$
Show that necessarily $p=2$. Next show that (6.33) is equivalent to the property that $\sum_{-\infty}^\infty \varepsilon_n \widehat f(n) e(n\cdot)$ converges in $L^p([0,1])$ for each choice of signs $\varepsilon_n=\pm1$ and $f\in L^p([0,1])$. This latter property is called unconditional convergence, and this problem therefore amounts to proving that the exponential system is unconditional only for $p=2$. In Section 8.4 we shall see that Haar functions are unconditional on $L^p([0,1])$ for $1<p<\infty$. Hint: Use Khinchine's inequality
Here are some attempts,
\begin{align*}
\mathbb{E}\Big\|\sum_{n=-N}^N\epsilon_n\hat f(n)e(n\theta)\Big\|_p&=\mathbb{E}\Big(\int_0^1|\sum_{n=-N}^N\epsilon_n\hat f(n)e(n\theta)|^pd\theta\Big)^{1/p}\\
&\leq\Big(\mathbb{E}\int_0^1|\sum_{n=-N}^N\epsilon_n\hat f(n)e(n\theta)|^pd\theta\Big)^{1/p}\quad(\text{since $g(x)=x^{1/p}$ is concave for $p\geq1$})\\
&\leq C(\sum_{n=-N}^N|\hat f(n)|^2)^{1/2}\quad(\text{Khinchine's inequality})\\
&\leq C\|f\|_2
\end{align*}
But
$$
\mathbb{E}\Big\|\sum_{n=-N}^N\epsilon_n\hat f(n)e(n\theta)\Big\|_p\leq\sup_{\epsilon_n}\Big\|\sum_{n=-N}^N\epsilon_n\hat f(n)e(n\theta)\Big\|_p
$$
the inequality seems in the wrong direction. Any suggestion on how to do next? Thanks in advance!
You can find a proof in Chapter II.D of the book: Banach spaces for analysts, by P. Wojtaszczyk
Thanks for the reference!
|
2025-03-21T14:48:31.613539
| 2020-07-26T17:01:16 |
366627
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mark Schultz-Wu",
"Sourangshu Ghosh",
"https://mathoverflow.net/users/101207",
"https://mathoverflow.net/users/172447"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631486",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366627"
}
|
Stack Exchange
|
Functional equation for $\eta(s)$ following Riemann's $2^{nd}$ method
I'm crossposting.
Being
\begin{equation*}
\eta(s)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}=\frac{1}{1^{s}}-\frac{1}{2^{s}}+\frac{1}{3^{s}}-\frac{1}{4^{s}}+\cdots
\end{equation*}
and following Riemann's second method, (Edwards p.15), to obtain the functional equation for $\zeta(s)$ one
can think the same way and try the same aproach for $\eta(s)$.
Thus from
\begin{equation*}
\int_{0}^{\infty} \operatorname{exp}\left(-n^{2} \pi x\right) x^{s / 2-1} d x=\pi^{-s / 2} \Gamma\left(\frac{s}{2}\right)\frac{1}{n^{s}} \text { for } s>0
\end{equation*}
one can express $\eta(s)$ as
\begin{equation*}
\pi^{-s / 2} \Gamma \left(\frac{s}{2}\right)\underbrace{\left(1-\frac{1}{2^{s}}+\frac{1}{3^{s}}+\cdots\right)}_{\eta(s)} =\int_{0}^{\infty}\left(e^{-\pi 1^2 x}-e^{-\pi 2^2 x}+e^{-\pi 3^2 x}+\cdots\right)x^{s/2}\text{ }\frac{dx}{x}
\end{equation*}
How would one proceed from here to craft a functional equation for $\eta(s)$?
I'm interested in refferences and/or answers. Any of them will be very much apreciated.
Thanks.
A.Neves, Can I get your email id to contact you by email?
Ignoring technicalities of convergence, in Riemann's second proof, you start with the Poisson summation formula $\sum_{n\in\mathbb Z} f(n / x) = x \sum_{n\in\mathbb Z} \hat f (n x)$, take the Mellin transform of both sides, and use the self-dual function $f(x)=e^{-x^2}$.
To get the alternating sum you want, you could either change the function or change the summation formula. For the function, you could use something like $\sum_{n\in\mathbb Z} f(n / x) \exp(\pi i n)$, and do some computations. You could also take a twisted Poisson summation formula $\sum (-1)^n f(n) = \sum_{n \textrm{ odd}} \hat f(n/2)$, but the steps for proving that are identical to the manipulations done to derive the functional equation for $\eta(s)$ from the functional equation for $\zeta(s)$.
Furthermore, an inverse Mellin transform allows you to go in the converse direction: a functional equation of Dirichlet series gives a summation formula. If the gamma factor is different then it will not be a Fourier transform but a generalization. If the degree of the functional equation is $d$ then the sum will be over $d$-th roots of natural numbers instead of over natural numbers.
Isn't it $f(x) = \exp(-\pi x^2)$ which is self-dual? I think if one uses $\exp(-x^2)$ a slightly different form of Poisson summation is required, see here.
|
2025-03-21T14:48:31.613698
| 2020-07-26T17:10:22 |
366629
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mateusz Kwaśnicki",
"https://mathoverflow.net/users/108637"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631487",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366629"
}
|
Stack Exchange
|
Expansion of the Riemann-Liouville integral
Riemann-Liouville integral operator of the $\alpha$-th order ($\alpha > 0$) is defined as
$$
I^{\alpha} x(t) := \frac{1}{\Gamma(\alpha)}\int\limits_{0}^{t} (t-\tau)^{\alpha - 1} x(\tau)d\tau
$$
I would like to know are there some expansion formulas of this operator like:
$$
I^{\alpha} x(t) = f_0(\alpha,t) I^0 x(t) + f_1(\alpha,t) I^1 x(t) + \ldots
$$
where $I^0 x(t) = x(t)$, $I^1(t) = \int\limits_{0}^{t} x(\tau) d\tau$ and in general $I^{k} x(t) = I^1\left(I^{k-1}x(t)\right)$ for $k = 2,3,\ldots$. Coefficients $f_i(\alpha,t)$ are some known functions.
There seems to be no such expansion, as this would imply that $(t - \tau)^{\alpha-1}$ can be written as $\sum_k f_k(\alpha, t) (t - \tau)^{k-1}$, which is not the case. On the other hand, if $\alpha > 1$, one can get arbitrarily good approximations (say, with respect to the supremum norm on $[0, t]$) by using finitely many terms: simply approximate $(t - \tau)^{\alpha-1}$ on $[0, t]$ by a polynomial.
|
2025-03-21T14:48:31.613782
| 2020-07-26T17:35:11 |
366630
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"H A Helfgott",
"Ryan O'Donnell",
"https://mathoverflow.net/users/398",
"https://mathoverflow.net/users/658"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631488",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366630"
}
|
Stack Exchange
|
Closed paths, closed trails and traces
Let $A$ be the adjacency matrix of a (non-oriented) graph $\Gamma$. Then $\textrm{Tr} A^k$ equals both the sum $\sum_i \lambda_i^k$ of $k$th powers of eigenvalues of $A$, on the one hand, and the number of closed paths of length $k$ in $\Gamma$, on the other hand.
If we have a good bound on the latter, then we have a good bound on the former. For $k$ even, a good bound on the former quantity implies that there aren't too many large eigenvalues $\lambda_i$.
What happens if we have an even better bound on closed trails of length $k$, that is, closed paths where no edges are repeated?
Or what happens if we can give a very good bound on the number of closed paths that neither backtrack (i.e., an edge is never followed by its inverse) nor have long repeated sequences of edges (as in, an edge sequence $$(v_0,v_1), (v_1,v_2),\dotsc, (v_{l-1},v_l)$$ appearing at least twice, where $l$ is not that much smaller than $k$)? Can we then say something stronger about $A$?
Here, by "a very good bound", I mean a bound that is stronger ( = smaller) than would be possible for the number of closed paths of length $k$: there are always plenty of backtracking closed paths (at least $d^{k/2}$ starting from a vertex of degree $d$).
(I've asked related questions in the recent past: see Closed paths, traces and spectra and Ihara zeta function and closed paths and trails .)
Maybe you are familiar with all this, but you might like to check out (e.g.) Bordenave's recent simpler proof of Friedman's theorem, that in a random d-regular graph all nontrivial eigenvalues are at most 2sqrt(d-1)+o(1) with high probability. https://arxiv.org/abs/1502.04482 It contains a number of useful ideas, including bounding eigenvalues using Ihara--Bass and the non-backtracking operator, bounding only nontracking walks in which each edge is used at least twice, and getting improved bounds when the graph doesn't have two cycles in any small neighborhood.
@Ryan O'Donnell Actually, I hadn't read it - thanks. In the context I'm working on, I also manage to exclude nontracking walks in which an edge is used only once. I wonder whether the trick is the same - where does Bordenave does this?
He doesn't exactly, but what he does is morally strictly harder than handling random d-regular graphs with random edge signs. In the latter case, closed walks where every edge is used an even number of times naturally arise. See, e
Your comment was cut off?
e.g. my paper with Mohanty and Paredes on explicit Ramanujan graphs (linked on my homepage -- sorry, am typing this on my phone)
|
2025-03-21T14:48:31.613972
| 2020-07-26T18:12:57 |
366633
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aurel",
"Dianbin Bao",
"Kimball",
"OmniaOperator",
"Will Sawin",
"https://mathoverflow.net/users/160900",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/31134",
"https://mathoverflow.net/users/40821",
"https://mathoverflow.net/users/6518"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631489",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366633"
}
|
Stack Exchange
|
L functions of elliptic curves over quadratic fields
Let $E$ be an ellitpic curve over a quadratic field $K/\mathbb{Q}$. Then the L function of $E$ is defined as
$L(E_K,s)=\prod_{\mathfrak{p}\nmid \Delta}(1-a_{\mathfrak{p}}N(\mathfrak{p})^{-s}+N(\mathfrak{p})^{1-2s})\prod_{\mathfrak{p}|\Delta}(1-a_{\mathfrak{p}}N(\mathfrak{p})^{-s})$, where $a_{\mathfrak{p}}=N(\mathfrak{p})+1-\#E(\mathcal{O}_K/\mathfrak{p})$.
Given an elliptic curve explicitly, how to find effectively the Dirichlet series of $L(E_K,s)$ and its functional equation?
One can find the conductor with Tate's algorithm. Then look for Hilbert modular forms / Bianchi modular forms of the corresponding level. Or, if the conductor is small, look it up here https://www.lmfdb.org/EllipticCurve/
Many computer algebra systems can do this for you. For instance in pari/gp, use nfinit (for $K$) + ellinit (for $E$) + lfuncreate (for $L$) (look up the documentation to learn how to use those).
Can you be a little more precise about what you mean? How to find the first $n$ coefficients of the Dirichlet series from the $a_p$'s, and the analytic conductor and the root number? In what sense do you have the elliptic curve and what sense do you want the L-function?
$a_{\mathfrak{p}}$ can be computed by counting. To me it seems not effective to compute $a_n$ by the Euler product. I want the dirichelt series of the L function and its functional equation to compute numerically the critial values of the L functions. By the comment of Will above, it seems to corresponds to Hilbert modular forms or Bianchi modulars depending whether the quadratic field is real or complex?
In practice, computing critical values is more or less equivalent to computing the first $O(\sqrt N)$ L-series coefficients. I'm not sure it will ever be more efficient to compute critical values by a correspondence to a Hilbert/Bianchi form. (In the case over Q, this only occurs when there is an easy eta-product for the modular form, or possibly when there is CM (both are easy to compute in that case)). OTOH, some theoretical constructs (ie period relations) might be easier for the form rather than the curve.
|
2025-03-21T14:48:31.614141
| 2020-07-26T18:49:03 |
366636
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"AmorFati",
"Martin de Borbon",
"Robbixmaths",
"https://mathoverflow.net/users/105103",
"https://mathoverflow.net/users/127247",
"https://mathoverflow.net/users/161793"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631490",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366636"
}
|
Stack Exchange
|
Completion/Compactification of a Kähler metric on $\mathbb C^2$
Consider $\mathbb{C}^{2}$ equipped with the Kähler form
$$
\omega_{\mu}=\frac{i}{2} \partial \bar{\partial} \log \left(\left(1+|z|^{2}\right)^{\mu}+|w|^{2}\right),
$$
where $\mu$ is a positive real number.
Question. Does there exists a complete Kähler manifold $\left(M, \hat{\omega}_{\mu}\right)$ such that
$\mathbb{C}^{2} \subset M$,
$\mathbb{C}^{2}$ is dense in $M$,
$\hat{\omega}_{\left.\mu\right|_{\mathbb{C} 2}}=\omega_{\mu} ?$
Remark. The complex projective plane $M=\mathbb C P^{2}$ equipped with the Fubini-Study form is a positive answer to the question, when $\mu=1 .$ I suspect this is the only case but I am
not able to prove it.
This is an interesting problem, what is the motivation?
I think that if mu is not one, then the metric completion adds only two points, so it's not a manifold
@MartindeBorbon Thank you for the answer! But I did not understand your argument, could you elaborate it, please?
If you restrict to z=0 you add a point and get a sphere. If you restrict to w=0 you add another point and get a sphere (scaled by mu). I think, if mu is not 1, those are the only two points. Let me restrict to mu>1. Look at the distance from $(z_0, w_0)$ to $(z_0, 0)$ where $(z_0, w_0)$ belongs to the line $z= \lambda w$ for some fixed nonzero $\lambda$. Restrict to the line $z=z_0$, you get a sphere and the distance squared from $(z_0, w_0)$ to $(z_0, 0)$ is roughly $|z_0|^{1-\mu}$ so goes to zero as $z_0$ goes to infinity
Btw, your metric is toric, so one could look at the moment polytope and see it does not corresponds to a manifold
I realized that you add some other points when w grows like z to the power mu. So I don't know how to answer your question, sorry.
|
2025-03-21T14:48:31.614292
| 2020-07-26T19:10:52 |
366638
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"R. van Dobben de Bruyn",
"https://mathoverflow.net/users/108274",
"https://mathoverflow.net/users/82179",
"user267839"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631491",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366638"
}
|
Stack Exchange
|
Locally freeness of twisted direct images $\pi_* \mathcal{F}(i)$
I have a question about a step in the proof of the
Existence of Flattening Stratification I found in
Nitsure's paper here: https://arxiv.org/abs/math/0504590 This question is closely related to Local freeness of $\pi_*F(r)$ from flatness of $F$ where the proof of Lemma 3.2 was discussed
which we are probably going to apply here.
That's the converse implication of part (ii) (page 22):
If $\phi : T \to S$ is a morphism such that $\mathcal{F}_T$ is flat,
then
the Hilbert polynomial is locally constant over $T$. Let $T_f$
(this notation is introduced in part (i) of the proof)
be the open and closed
subscheme of $T$ where the Hilbert polynomial is $f$.
Clearly, the set map $|T_f | \to |S|$
factors via $|S_f |$.
Now the step I not understand:
But as the direct images $(\pi_T)_*\mathcal{F}_T(i)$ are locally
free of rank $f(i)$ on $T_t$, it follows
in fact that the schematic morphism $T_f \to S$ factors via $S_f$ ...
Question: Why flatness of $\mathcal{F}_T$ implies that all
twisted direct images $(\pi_T)_*\mathcal{F}_T(i)$ are locally
free?
Seemingly, Nitsure used implicitely here Lemma 3.2 + a certain exercise (p 15):
Lemma + Exercise imply that $\mathcal{F}$ is flat if and only if
there exist some integer $N$ such that for all $r ≥ N$ the direct image
$\pi_*\mathcal{F}(r)$ is locally free.
But, it says only that for flat $\mathcal{F}_T$ the
twisted direct images $(\pi_T)_*\mathcal{F}_T(r)$ are locally free
IF the $r$ twist is big enough! On the other hand in the proof is clamed
that $(\pi_T)_*\mathcal{F}_T(i)$ are locally free for ALL $i$'s.
This aspect confuses me. Does anybody know why assuming
$\mathcal{F}_T$ flat, implies that all twisted direct images
$(\pi_T)_*\mathcal{F}_T(i)$ are locally
free independend of $i$?
This is only true for $i \gg 0$, but that's also all you need. (The Hilbert polynomial is determined by its eventual values.)
...you mean in the sense if $f$ Hilbert has degree $n$ then it is uniquely determined by $f(N), f(1+N),... f(i+N),...,f(n-1+N)$ for arbitrary $N$, ie can be choosen big, that's the point, right? And in the quoted text the statement on locally freeness of $(\pi_T)_*\mathcal{F}_T(i)$ is just formulated awkwardly?
|
2025-03-21T14:48:31.614453
| 2020-07-26T19:48:36 |
366641
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631492",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366641"
}
|
Stack Exchange
|
Wasserstein space with strictly non-positive sectional curvature
Let $(X,d)$ be a (separable, complete) metric space with uniformly strictly non-positive curvature in the sense of Alexandrov, i.e. $(X,d)$ satisfies a $CAT(K)$ inequality for some $K<0$.
Does it hold that the 2-Wasserstein space on $(X,d)$ has strictly non-positive curvature?
I suspect this is false, but haven't found this result anywhere. Note that the analogous statement for $CAT(0)$ is definitely false, because the 2-Wasserstein space on $\mathbb{R}^d$ is positively curved (see for example Remark 2.10 in this paper by Bertrand and Kloeckner: https://arxiv.org/pdf/1010.0590.pdf).
Yes, it is false. Note that the space $W_2(X)$ contains the space $\tfrac1{\sqrt{n}}\cdot X^{\times n}/S_n$ where the group $S_n$ permutes the $X$-factors. While space $X^{\times n}$ is $\mathrm{CAT}(0)$, the quotient $X^{\times n}/S_n$ is not --- for example, take $X=$ Lobachevsky plane.
|
2025-03-21T14:48:31.614529
| 2020-07-26T20:06:49 |
366644
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631493",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366644"
}
|
Stack Exchange
|
Linearly dependent points and the uniform position theorem
One proof of the uniform position theorem (as stated in p. 109 or p. 113 in Section III.1 of "Geometry of Algebraic Curves") uses a monodromy argument. While this gives us something even more general than the usual statement of the uniform position theorem, it isn't something that is constructive.
What is currently known about the locus of hyperplanes whose intersection with a curve gives a linearly dependent collection of points? Is there an analogue involving higher dimensional varieties? The reason I'm not sure is that uniform position in higher dimensional varieties involves intersection with generic hyperplanes (Uniform position principle). As a start, there is generalization of the usual statement of the uniform position theorem involving divisors containing a given hyperplane section (p. 113 in ACGH above). However, it has been difficult to find concrete examples where this non-generic situation is considered.
|
2025-03-21T14:48:31.614621
| 2020-07-26T20:34:27 |
366646
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D.W.",
"Hans",
"RobPratt",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/141766",
"https://mathoverflow.net/users/32660",
"https://mathoverflow.net/users/37212"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631494",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366646"
}
|
Stack Exchange
|
Optimal rule for multiple stopping times for defect finding
Suppose a quality inspector is inspecting $b$ black amongst which $d_B$ are known to be defective and $w$ white gadgets amongst which $d_W$ are known to be defective. The gadgets come down along an assembly line one by one in uniformly distributed random order. As each gadget passes, the inspector observes its color, and he chooses to let the gadget pass or use a device to detect whether the gadget is defective. But he can only use the device a total of $n$ times. What is the optimal stopping rule to use the inspecting device to maximize the expected number of defective gadgets found.
Edit: As fedja points out below, there is a difference between detecting a defect by the device and deducing for sure by logic a defect. Both are legitimate objectives. The solution to the first is easier than that to the second. I for now choose the first definition, i.e., to define finding a defective gadget as only indicated by the device and use up one quota of using the device even if the quality inspector is sure by logic alone the gadget under inspection is defective.
Suppose at a pass the number of black gadgets already inspected by the device is $i_B$, amongst them $f_B$ are detected to be defective. Suppose the current passing gadget is black and yet to be inspected. Then the probability of this black gadget being defective is $p_B=\frac{d_B-f_B}{b-i_B}$. Symmetrical probability holds for if the current passing gadget is white.
I have a conjecture for the explicit solution, which is a greedy algorithm, as follows and am seeking a proof.
At a pass of the gadget, without loss of generality, suppose the current gadget which has not yet been inspected is black. Suppose there are $n_B$ black including the current one and $n_w$ white gadgets left. Suppose $i_B$ black and $i_W$ white gadgets have been inspected, amongst which $f_B$ black and $f_W$ have been found to be defective. If $p_B\ge p_W$ or $n_W=0$ the inspector inspect the current black gadget with the device. Otherwise, the inspector let the current black gadget pass without inspection.
I have set up the dynamic programming formulation but fail to see either the proof or a counterexample to my conjecture.
Cross-posted here
@RobPratt: It is just that so far it does not look like I can get an answer there.
Cross-posted: https://math.stackexchange.com/q/3769463/14578, https://mathoverflow.net/q/366646/37212, https://cstheory.stackexchange.com/q/47306/5038. Please do not post the same question on multiple sites. Please pick one. You can delete the question elsewhere if you decide you've posted it on the wrong site.
@D.W.: I noted the cross posting on the computer science site. That was how you came here. Also, 5 days have lapsed since I posted this question here before I post the question on the computer science site. Let me add more details to the question to answer your questions.
Your condition $p_b<p_w$ does not include $n_b$ and $n_w$ in any way. On the other hand, it is clear that if $n_w=0$, then just passing all remaining black gadgets is suboptimal. Do I misunderstand something about the proposed greedy algorithm?
@fedja: You are right. I have edited my conjectured algorithm. Would you like to have a crack at the problem? Thank you.
The current model seems too complicated to make a reasonable exact answer possible: all sorts of issues arise (like if $d_w=n_w$, you don't need a single test to reject all white gadgets). Writing the full set of dynamic programming equations is possible, of course, but the variables would be full distributions, not mere expectations. However, if it comes from a real life problem, we can try to make a shot at an approximate solution. Then it will be very helpful to know what range of parameters you are thinking of.
@fedja: First, do you agree with my dynamic programming solution in my answer below?
If the objective is to maximize the number of beeps on the detector (i.e., if you ignore special cases like the one I mentioned), then the recursion looks correct. One also needs to set the boundary values though to make it a full set of equations.
@fedja: The objective is quite clear from the question which is "to maximize the expected number of defective gadgets found" which is equivalent to what you say except it should be the expected number of beeps. I neglected to put the boundary condition as I thought it was quite obvious. I can put it up for completeness. As for the special case like the one you mentioned where say $n_W=0$, it should be a corollary derived from the recursion along with the boundary condition rather than an independent case.
"which is equivalent to what you say" But, as I said earlier, it is not: if we have, say, 4 white and 4 black gadgets remaining with 3 tests left, and we know that all whites are defective and 1 out of 4 blacks is defective, then we maximize the expected number of beeps by testing the white gadgets and maximize the expected number of "found" (in the sense "proved to be defective by direct measurement or logical derivation") gadgets by testing blacks. Am I making sense? The objective can be formally set to be either one, I just asked which one you had in mind.
@fedja: Yes, you are making perfect sense. Actually, now I remember this case did occur to me when I was thinking about this problem. But I forgot about it when I wrote the recursion. It is the expected number of beeps I am asking in the question. I will make the statement more rigorous later. However, the second question of "proving to be defective by direct measurement or logical derivation" is a more interesting objective. It seems only the case when, at least one color, is all defective would make a difference. Am I missing something again?
@fedja: In that case, the recursion has a "discontinuity" in treating the cases and needs a special clause.It is somewhat discomforting.
It seems only the case when, at least one color, is all defective would make a difference. In the recursive formula, yes, but it changes the answer even in the cases when we haven't arrived at that situation yet but are potentially close to it. The "beep objective" results in a cleaner recursion and simpler boundary conditions, so let's stick to it for now: I still do not see how to finish even this case :-)
I do not answer the ultimate question which is to prove or disprove the conjecture. But I hereby post the dynamic programming algorithm for solving any given set of parameters.
Let $\bigg(\begin{matrix}c_B&l_B&n_B \\c_W&l_W&n_W\end{matrix}\hspace{2ex} m\bigg)$ denote the maximal expected number of defects to be found with $n_B$ black and $n_W$ white gadgets to pass and $m$ inspections to use, where $c_B:=d_B-f_B$, $l_B:=b-i_B$, $c_W:=d_W-f_W$, $l_W:=b-i_W$. We obtain a recursion by conditioning on observing the color of the current passing gadget then taking the maximum of the expected maximum expectation of the defect detection based on maximizing the two choices of whether or not to use the device to detect the defect. At each pass of a gadget, the probability that the current gadget is black is $\frac{n_B}{n_B+n_W}$ while it is white is $\frac{n_W}{n_B+n_W}$. Conditioned on having observed the color of the passing gadget being black the probability of it being detected to be defective is $\frac{c_B}{l_B}$. Conditioned on having observed the color of the passing gadget being white the probability of it being detected to be defective is $\frac{c_W}{l_W}$. We have the recursion
\begin{align}
&\bigg(\begin{matrix}c_B&l_B&n_B \\c_W&l_W&n_W\end{matrix}\hspace{2ex} m\bigg) \nonumber \\[1ex]
=&\ \ \frac{n_B}{n_B+n_W} \max
\begin{cases}
\frac{c_B}{l_B}\bigg(1+\bigg(\begin{matrix}c_B-1&l_B-1&n_B-1 \\c_W&l_W&n_W\end{matrix}\hspace{2ex} m-1\bigg) \bigg) +\Big(1-\frac{c_B}{l_B}\Big)\times \\
\quad\quad\bigg(\begin{matrix}c_B&l_B-1&n_B-1 \\c_W&l_W&n_W\end{matrix}\hspace{2ex} m-1\bigg) \\[4.5ex]
\bigg(\begin{matrix}c_B&l_B&n_B-1 \\c_W&l_W&n_W\end{matrix}\hspace{2ex} m\bigg)
\end{cases} \nonumber \\[2ex]
&+ \frac{n_W}{n_B+n_W}\max
\begin{cases}
\frac{c_W}{l_W}\bigg(1+\bigg(\begin{matrix}c_B&l_B&n_B \\c_W-1&l_W-1&n_W-1\end{matrix}\hspace{2ex} m-1\bigg) \bigg) +\Big(1-\frac{c_W}{l_W}\Big)\times \\
\quad\quad\bigg(\begin{matrix}c_B&l_B&n_B \\c_W&l_W-1&n_W-1\end{matrix}\hspace{2ex} m-1\bigg) \\[4.5ex]
\bigg(\begin{matrix}c_B&l_B&n_B \\c_W&l_W&n_W-1\end{matrix}\hspace{2ex} m\bigg)
\end{cases} .
\end{align}
|
2025-03-21T14:48:31.615106
| 2020-07-26T21:55:58 |
366648
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631495",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366648"
}
|
Stack Exchange
|
Fourier mode decomposition and eigenvalues of Schroedinger operators with radial potential in N-dimensions
In the study the stability of minimal hipersurfaces $\Sigma \subset \mathbb{R}^{N+1}$ one is lead to study the Morse index of a Schroedinger operator $J := - \Delta_g + |A|^2$ (usually called Jacobi operator) with Dirichlet boundary conditions, where $g$ is the metric on $\Sigma$. If we know that $\Sigma$ is a rotationally symmetric, with polar coordiantes $(r,\theta)$ and $g = dr^2 + h(r) \, d\theta^2$, the potential $|A|^2$ is a radial function.
Are there standard references for studying the Fourier mode decomposition of the eigenvectors of $J$ in this $N$-dimensional but symmetric setting? In particular I am interested in the following questions:
$\bullet$ Is the $0$th-mode eigenspace unidimensional? Are the eigenspaces associated to radial eigenfunctions unidimensional for all modes?
$\bullet$ Is the eigenvalue of the radial eigenfunctions in the $k$-th mode always less than the one in the $(k+1)$-th mode?
$\bullet$ In the same mode, is the eigenvalue of a radial eigenvector less than the eigenvalue of the eigenvectors which include angular variables?
$\bullet$ Let us call radial index of $J$ the sum of the dimensions of all the eigenspaces of $J$ on $[c,d]\times \mathbb{S}^{N-1}$ associated to the negative eigenvalues and only radial eigenvectors. Can one easily compare the radial index of $J$ and the Morse index of the 1-dimensional operator $J_1 := - \partial_r^2 + |A(r)|^2$ on $[c,d]$?
I am new in this field. Those questions seem natural (or naive) to me and I think they could be easily answered (under the technical assumptions and usual hypotheses in the topic), if specific results on the stability of rotationally symmetric minimal hypersurfaces are known. I do not know where to study the subject, theory and examples of computations.
Finally, do the answers of the above questions change totally if one consider instead weighted minimal hypersurfaces with weight $e^{f(r)}$? In that case the functional spaces are weighted and the operator takes the form $J + b(r) \, \partial_r$, with $b$ a radial function but not necessarily non-negative (the function $b$ I have in mind is smooth, bounded and has two horizontal distinct asymptotes).
The general idea is to ask if the study of the radial equation and radial eigenvectors give bounds or relations for the computation of the index of instability of rotationally symmetry (possibly weighted, with radial weigth) minimal hipersurfaces.
|
2025-03-21T14:48:31.615400
| 2020-07-27T00:46:48 |
366654
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Will Sawin",
"https://mathoverflow.net/users/18060"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631496",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366654"
}
|
Stack Exchange
|
Tannakian group of Galois representations coming from geometry
Let $K$ be a number field. Let $G_K$ be its absolute Galois group.
Let $p$ be a rational prime.
Let $\mathcal{R}_{K,p}^g$ be the category of finite-dimensional continuous $p$-adic representations of $G_K$ that come from geometry. Thus, an object $V$ of $RG_{K}^p$ is isomorphic to a sub-quotient of $H^i_{ét}(X, \mathbb{Q}_p)(n)$ for some smooth projective variety $X$ and integers $i$ and $n$.
$\mathcal{R}_{K,p}^g$ should be a Tannakian category, and therefore, have a Tannakian fundamental group scheme, $\mathcal{G}_{K,p}^g$, associated to it.
How is $\mathcal{G}_{K,p}^g$ related to $G_K$?
Its component group is $G_K$ because all Galois representations with finite image come from geometry. I don't think there is any other useful relation.
|
2025-03-21T14:48:31.615818
| 2020-07-27T00:54:36 |
366655
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Konstantinos Kanakoglou",
"https://mathoverflow.net/users/85967"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631497",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366655"
}
|
Stack Exchange
|
Poincaré Recurrence Theorem for flows
Someone knows a book that have the proof of the Poincaré Recurrence Theorem for flows when the set have finite volume and when don't?
Apart from the initial papers of Poincare and the subsequent version of the proof by Caratheodory (you can find these references for example in: https://encyclopediaofmath.org/wiki/Poincar%C3%A9_return_theorem), i think you can find a proof for the ergodic "version" of the theorem at: P. Walters, An introduction to ergodic theory, Graduate Texts in Mathematics, Vol. 79. 1982.
|
2025-03-21T14:48:31.615922
| 2020-07-27T00:59:30 |
366656
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chen Wang",
"Gerry Myerson",
"LSpice",
"Tomita",
"Will Sawin",
"Zhi-Wei Sun",
"https://mathoverflow.net/users/124654",
"https://mathoverflow.net/users/150249",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/57993"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631498",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366656"
}
|
Stack Exchange
|
Is it true that $\{x^3-2x+y^3-2y+z^3-2z: x,y,z\in\mathbb Z\}=\mathbb Z$?
A well known conjecture states that
$$\{x^3+y^3+z^3:\ x,y,z\in\mathbb Z\}=\{m\in\mathbb Z:\ m\not\equiv\pm4\pmod 9\}.$$
For $m=33,\, 42$ an integer solution to the equation $x^3+y^3+z^3=m$ was only found last year.
In 2017, Tyrell asked whether
$$\left\{\frac{x(x+1)(x+2)}6+\frac{y(y+1)(y+2)}6+\frac{z(z+1)(z+2)}6:\ x,y,z\in\mathbb Z\right\}=\mathbb Z,$$ see the question with the website http://math.stackexchange.com/questions/2472205. Few weeks ago Alkan (cf. Numbers of the form $x^2(x-1) + y^2(y-1) + z^2(z-1)$ with $x,y,z\in\mathbb Z$) conjectured that
$$\left\{\frac{x^2(x-1)}2+\frac{y^2(y-1)}2+\frac{z^2(z-1)}2:\ x,y,z\in\mathbb Z\right\}=\mathbb Z.$$
I think it's interesting to find a cubic polynomial $P(x)$ with integer coefficients such that
$$\{P(x)+P(y)+P(z):\ x,y,z\in\mathbb Z\}=\mathbb Z.$$
This led me to pose the following conjecture.
Conjecture. Each $m\in\mathbb Z$ can be written as a sum of three numbers of the form $x^3-2x\ (x\in\mathbb Z)$. In other words, we have
$$\{x^3-2x+y^3-2y+z^3-2z: x,y,z\in\mathbb Z\}=\mathbb Z.$$
As $P(x)=x^3-2x$ is an odd function, the conjecture can be reduced to the case $m\in\mathbb N=\{0,1,2,\ldots\}$. Via computation I found that those natural numbers $n\le1000$ not in the set
$$\{x^3-2x+y^3-2y+z^3-2z:\ x,y,z\in\{-1000,\ldots,1000\}\}$$
are
\begin{gather}70,\ 75,\ 83,\ 86,\ 139,\ 185,\ 198,\ 237,\ 253,\ 262,\ 275,
\ 305,\ 338,\ 355,\ 362,
\\397, 414,\ 415,\ 422,\ 426,\ 457,\ 458,\ 509,\ 535,\ 558,\ 562,\ 564,\ 580,\ 583,
\\ 593, \ 613,\ 614,\ 635,\ 642,\ 673,\ 677,\ 684, \ 693,\ 697,\ 722,\ 735,\ 779,\ 782,
\\ 790,\ 791,\ 793,\ 807,\ 818,\ 850,\ 851,\ 870,\ 888,\ 898,\ 908,\ 943,\ 957.
\end{gather}
Let $S$ denote the set of these numbers.
QUESTION. Can we find an explicit solution of the equation
$$n=x^3-2x+y^3-2y+z^3-2z\ \ (x,y,z\in\mathbb Z)$$ for each $n\in S$?
Actually I tried $x^2(x-2)$ first, but I could not verify the representation for all $1\le n\le 40$. $P(x)=x^3-2x$ is an odd function, but $x^2(x-2)$ is not and hence less elegant.
After the computational reports from C. Wang, Deyi Chen and Tomita, now it remains to find a required representation for the following 10 numbers: $$185,\ 198,\ 338,\ 422,\ 426,\ 509,\ 558,\ 583,\ 614,\ 793.$$
@Zhi-Wei-Sun: Solutions for $n=185,198,614,793$ were found.
@Tomita Thank you very much for your computational results.
No solution for $338,422,426,509,558,583$ exists where one of $|x|,|y|,|z|$ does not exceed $6\times10^5$.
Solutions for n=422,426,509,583 were found.
$338= P(109043424)+ P(223729659)+ P(-232050701).$
Only $n=558$ remains.
@Tomita Great! Hope that you will find a solution for $n=558$ soon.
Unfortunately, solution for $n=558$ was not found with $|x|,|y|,|z|<10^{10}$.
I believe each integer can be written as $P(x)+P(y)+P(z)\ (x,y,z\in\mathbb Z)$ in infinitely many ways, where $P(x)=x^3-2x$. Solutions for $P(x)+P(y)+P(z)=0$ with $xyz\not=0$ are also interesting.
I think it strange that there is no solution for $n = 558$.
There should be a solution for $n=558$. Maybe the first solution is relatively large, but less than $10^{11}$ in my opinion. Note that the first solution of $33=x^3+y^3+z^3$ given by Booker is particularly large.
The way solutions appear sporadically, with no small solutions for particular values of $n$, suggests that there is no parametric solution as a function of $n$. I have no idea if it is possible to prove this. If there are no special solutions, then the number of solutions for a given $n$ up to a certain range could follow the "product of local densities" heuristic.
Comments from a message of Andrew Booker to me: "There is a standard method to search for solutions to $n=P(x)+P(y)+P(z)$ for $P(x)=x^3-2x$, based on the identity $n-P(z)=(x+y)(x^2-xy+y^2-2)$. Running through all candidate values for $z$ and all divisors of the LHS, we extract $x$ and $y$ from the RHS and test whether they are integers. Tomita is surely aware of this, and I assume that's what he is using to conduct his searches."
More comments from Andrew Booker: "The method as described uses factorization, which is quite slow; it can be made much faster by fixing the divisor $x+y$ and solving the polynomial congruence for $z$, as I did in the case of cubes."
I already searched the solution of $x^3-2x+y^3-2y+z^3-2z=558$ by solving $z^3-2z\equiv 558 \pmod {x+y}$ with $x+y<10^{10}$.
However my method is not complete.
I assume that $n=x+y=p \times q$ and $p$ is prime number,$q$ is composite number with $p<10^{8}$.
Hence my method does not work when $p>10^{8}$ or $n$ is prime number.
Since I used the two methods, LLL reduction and solving modulus equation, I'm sure there is no solution for n=558 with $|x|,|y|,|z|<10^{10}$.
I could not find a solution for $n=558$ using modulus method with $|x|,|y|,|z|<10^{11}$.
To verify this result, I'll use LLL method with PARI-GP, which takes about 20 days.
@Tomita Thank you for your great efforts. Hope that you could finally find a solution for $n=558$.
I finished searching the solution for $n=558$ using LLL method with $|x|,|y|,|z|<10^{11}$.
Unfortunately, the solution for $n=558$ was not found.
I hope someone will find it.
In a message to me on August 6, 2020, Prof. Andrew Sutherland wrote "If at some point in the future I return to this problem and your question about 558 remains open, I will give it a try (it should be feasible to check for solutions with $\min(|x|,|y|,|z|)$ up to $10^{16}$ without too much effort)."
0 = P(7) + P(10) + P(-11)
= P(3250) + P(2293) + P(-3593)
= P(6266) + P(13243) + P(-13695)
= P(11700) + P(13277) + P(-15797)
= P(37555) + P(131381) + P(-132396)
= P(747511) + P(1059490) + P(-1171307)
= P(5529835) + P(22681597) + P(-22790636)
= P(8042677) + P(13682243) + P(-14552100)
= P(14270088) + P(39054467) + P(-39679475)
= P(29292092) + P(81358953) + P(-82605425)
= P(42588445) + P(291524359) + P(-291827018)
= P(56973565) + P(71715599) + P(-82119294)
= P(35977605) + P(866776048) + P(-866796709)
= P(143141833) + P(102053460) + P(-158684449)
= P(784428376) + P(3091918585) + P(-3108657737)
= P(129810373) + P(136917575) + P(-168147294)
Though I expanded the search range to 10^10, no solution for n=558 was found.
On the other hand, there are many solutions for n=1 below.
1 = P(1439) + P(2554) + P(-2698)
= P(-1506) + P(-2432) + P(2611)
= P(-5214) + P(-11006) + P(11383)
= P(-8516) + P(-17400) + P(18055)
= P(13952) + P(70243) + P(-70426)
= P(18457) + P(10233) + P(-19451)
= P(18949) + P(56163) + P(-56873)
= P(21394) + P(107636) + P(-107917)
= P(21599) + P(61917) + P(-62781)
= P(75215) + P(256620) + P(-258756)
= P(132479) + P(517316) + P(-520196)
= P(525599) + P(2589115) + P(-2596315)
= P(697638) + P(803074) + P(-950033)
= P(-140064) + P(-198656) + P(219583)
= P(-198846) + P(-913333) + P(916464)
= P(-257810) + P(-1509380) + P(1511883)
= P(-617569) + P(-1930917) + P(1951749)
= P(-887510) + P(-1092290) + P(1260399)
= P(-931224) + P(-1288696) + P(1433823)
= P(1384739) + P(2458622) + P(-2597096)
= P(1602719) + P(9519294) + P(-9534414)
= P(4092479) + P(28437689) + P(-28465913)
= P(4875121) + P(2381859) + P(-5057717)
= P(9192959) + P(73135432) + P(-73183816)
= P(-1288696) + P(-931224) + P(1433823)
= P(-6063625) + P(-20241211) + P(20420995)
= P(-6919820) + P(-21816096) + P(22045735)
= P(-8121991) + P(-32025689) + P(32198879)
= P(18740159) + P(167927031) + P(-168004791)
= P(24544311) + P(124666228) + P(-124982552)
= P(62900639) + P(689911189) + P(-690085429)
= P(96931304) + P(198453683) + P(-205880474)
= P(-11745176) + P(-17900062) + P(19447931)
= P(-20241211) + P(-6063625) + P(20420995)
= P(-24301082) + P(-68349676) + P(69358667)
= P(-41154429) + P(-47640292) + P(56234034)
= P(-42083576) + P(-117387233) + P(119163144)
= P(-95843081) + P(-181052723) + P(189595899)
= P(106254719) + P(1271978124) + P(-1272225228)
= P(123437629) + P(177749151) + P(-195715037)
= P(-119444557) + P(-275690964) + P(282970698)
= P(-120282709) + P(-113449262) + P(147367664)
= P(-169017105) + P(-182314167) + P(221641415)
= P(-181052723) + P(-95843081) + P(189595899)
= P(-190571214) + P(-1169296181) + P(1170981088)
= P(-1129360025) + P(-3749403040) + P(3783251348)
Solution for n=338 was found using LLL algorithm for X^3+Y^3=1.
338= P(109043424)+ P(223729659)+ P(-232050701)
Only n=558 remains.
Added new solutions.
422= P(31441077)+ P(52488141)+ P(-56007428)
426= P(-11575473)+ P(-42374626)+ P(42660619)
509= P(4620839)+ P(7911642)+ P(-8405584)
583= P(-2697799)+ P(-3187069)+ P(3732685)
Added new solutions.
185=P(-14114372)+ P(-283189)+ P(14114410)
198=P(-142960)+ P(-613349)+ P(615927)
614=P(-412307)+ P(-16619)+ P(412316)
793=P(-296708)+ P(-387970)+ P(438851)
262 = P(10239) + P(5400) + P(-10717)
275 = P(38314) + P(4857) + P(-38340)
305 = P(8535) + P(5187) + P(-9131)
355 = P(2568) + P(982) + P(-2615)
362 = P(6547) + P(636) + P(-6549)
397 = P(-2029) + P(-973) + P(2101)
414 = P(1059) + P(576) + P(-1113)
457 = P(-7709) + P(-6134) + P(8832)
535 = P(-11999) + P(-2241) + P(12025)
562 = P(-3435) + P(-862) + P(3453)
564 = P(-848) + P(-751) + P(1011)
580 = P(-2295) + P(-825) + P(2330)
593 = P(1563) + P(458) + P(-1576)
613 = P(18873) + P(1623) + P(-18877)
635 = P(10566) + P(9745) + P(-12816)
642 = P(-5020) + P(-3871) + P(5693)
673 = P(4487) + P(566) + P(-4490)
677 = P(5967) + P(1087) + P(-5979)
684 = P(4316) + P(2750) + P(-4660)
693 = P(3575) + P(702) + P(-3584)
697 = P(-17181) + P(-2952) + P(17210)
722 = P(-1051) + P(-311) + P(1060)
735 = P(1934) + P(1460) + P(-2179)
779 = P(3781) + P(1593) + P(-3873)
790 = P(-152491) + P(-8563) + P(152500)
791 = P(11265) + P(8599) + P(-12735)
818 = P(2003) + P(874) + P(-2057)
850 = P(9047) + P(1510) + P(-9061)
851 = P(1105) + P(264) + P(-1110)
870 = P(6390) + P(1917) + P(-6447)
888 = P(3928) + P(1444) + P(-3992)
898 = P(1709) + P(929) + P(-1796)
908 = P(4950) + P(4172) + P(-5788)
943 = P(-5848) + P(-3743) + P(6320)
957 = P(-4297) + P(-3091) + P(4775)
Let $P(x):=x^3-2x$. Then
\begin{gather}
70=P(2714)+P(1367)+P(-2825),\\
75=P(16333)+P(14200)+P(-19328),\\
83=P(6714)+P(-6682)+P(-1627),\\
86=P(6413)+P(3721)+P(-6806).
\end{gather}
Dr. Deyi Chen has informed me that he has found $139=P(-105811)+P(105801)+P(6951)$.
More contributions from Dr. Deyi Chen: \begin{align}237=&P(-54523)+P(54267)+P(13147),\262=&P(-2719)+P(2712)+P(537),\275=&P(-38340)+P(38314)+P(4857),\305=&P(-9131)+P(8535)+P(5187),\355=&P(-2615)+P(2568)+P(982).\end{align}
Further contributions from Dr. Deyi Chen: \begin{align}253=&P(-63060)+P(61067)+P(28452),\415=&P(-80702)+P(79605)+P(27652),\458=&P(-32329)+P(-12710)+P(32971),\782=&P(-74024)+P(-56403)+P(83637),\807=&P(-68697)+P(62165)+P(43789).\end{align}
As Deyi Chen is not a user of MO, I just posted his contributions as comments (not an answer) after Wang's answer.
@MattF., since this one came before the other answer, I think it couldn't reasonably have been a comment on it …. (But why should it be?)
@Matt an upvote merely means "this answer is useful".
|
2025-03-21T14:48:31.616506
| 2020-07-27T02:46:20 |
366658
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"RyanChan",
"https://mathoverflow.net/users/149696",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631499",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366658"
}
|
Stack Exchange
|
Does the following function series converge?
Let
$$
f_n(x)=\frac{\frac{1}{(n-1)!}\sum_{k=0}^{\lfloor \alpha n-x\rfloor}C_{n-1}^{k}~(-1)^k(\alpha n-x-k)^{n-1}}{\frac{1}{n!}\sum_{k=0}^{\lfloor \alpha n\rfloor}C_{n}^{k}(-1)^k(\alpha n-k)^{n}},
$$ where
$x\in[0,1]$,
$C_{n}^{k}$ is the binominal coefficient,
and $\alpha$ is a constant such that $0 \le \alpha \le 1$.
Based on my intuition and numerical results, I guess the above series converges pointwise to a truncated exponential function $g(x)=A\exp(-\lambda x)$, where $A$ and $\lambda$ are parameters to be determined.
Question: How to prove or disprove this conjecture?
This question originates from my studies on the the marginal distribution of a uniform distribution defined over an $n$-dimensional simplex truncated by a unit cube, which can be defined as
$$
\mathscr{T}_n(t)=\bigg\{\vec{\mathbf{x}}:\sum_{i=1}^n x_i \le t, 0 \le x_i \le 1\bigg\}.
$$
Considering a uniform distribution over the domain $\mathscr{T}_n(\alpha n)$, I obtained the density function of the marginal distribution in any dimension as
$$
p(x)=f_n(x)=\frac{\text{vol}\left(\mathscr{T}_{n-1}\left(\alpha n-x\right)\right)}{\text{vol}\left(\mathscr{T}_{n}\left(\alpha n\right)\right)},$$
where $x\in[0,1]$. It is known that the marginal distribution of the joint random vector uniformly distributed over a simplex with a finite and nonzero mean value will converge to an exponential distribution. For this reason, I guess the considered series converges to the density function of a truncated exponential distribution.
In the denominator, do you mean $\sum_{k=0}^{\lfloor \alpha n\rfloor}$ or $\sum_{k=0}^{\lfloor \alpha n-x\rfloor}$?
@Iosif Pinelis The latter one is right.
$\newcommand{\si}{\sigma}$
By the Irwin--Hall formula, your first displayed ratio is
\begin{equation}
f_n(x)=\frac{P(S_{n-1}\le an-x)}{P(S_n\le an-x)}=\frac{P(S_{n-1}\le a(n-1)-(x-a))}{P(S_n\le an-x)},
\end{equation}
where $a:=\alpha\in[0,1]$, $x\ge0$, $S_n:=X_1+\dots+X_n$, and $X_1,\dots,X_n$ are iid random variables each uniformly distributed on $[0,1]$.
If $a=0$ then $P(S_n\le an-x)=0$ for $x\ge0$, so that $f_n(x)$ is undefined. If $a>1/2$ then, by the law of large numbers, $P(S_n\le an-y)\to1$ (as $n\to\infty$) for any fixed real $y$, so that $f_n(x)\to\frac11=1$. If $a=1/2$ then, by the central limit theorem, $P(S_n\le an-y)\to1/2$ for any fixed real $y$, so that $f_n(x)\to\frac{1/2}{1/2}=1$.
It remains to consider the nontrivial case when $a\in(0,1/2)$. Since $X_i$ equals $1-X_i$ in distribution, we have
\begin{equation}
f_n(x)=\frac{P(S_{n-1}\ge b(n-1)+(x-a))}{P(S_n\ge bn+x)},
\end{equation}
where
\begin{equation}
b:=1-a\in(1/2,1).
\end{equation}
By Theorem 1 by Petrov,
\begin{equation}
P(S_n\ge tn)\sim\frac{e^{nL_t(h_t)}}{h_t\si(h_t)\sqrt{2\pi n}} \tag{*}
\end{equation}
uniformly in $t$ in any closed subinterval of the interval $(1/2,1)$, where
\begin{equation}
L_t(h):=-ht+\ln R(h),\quad R(h):=Ee^{hX_1}=\frac{e^h-1}h,\quad\si(h):=m'(h),\quad m(h):=R'(h)/R(h)
\end{equation}
for real $h>0$, and $h_t\in(0,\infty)$ is the only root of the equation
\begin{equation}
m(h_t)=t.
\end{equation}
The functions $m$ and $\si$ (on $(0,\infty)$) are smooth, and $\si>0$. So, $m$ is a smooth increasing function, and hence the function $(1/2,1)\ni t\mapsto h_t$ is smooth. So, if $t\to t_0\in(1/2,1)$, then
\begin{equation}
h_t\si(h_t)\sim h_{t_0}\si(h_{t_0})
\end{equation}
and
\begin{equation}
\frac d{dt}L_t(h_t)=\frac{\partial L_t(h)}{\partial h}\Big|_{h=h_t}\;\frac{dh_t}{dt}-h_t
=(-t+m(h_t))\;\frac{dh_t}{dt}-h_t=-h_t\sim-h_{t_0},
\end{equation}
whence, by (*),
\begin{equation}
\frac{P(S_n\ge tn)}{P(S_n\ge t_0n)}\sim
\exp[-nh_{t_0}(t-t_0)(1+o(1))].
\end{equation}
Using this with $t_0=b$ and $t=b+x/n$, we get
\begin{equation}
\frac{P(S_n\ge bn+x)}{P(S_n\ge bn)}\sim
e^{-h_b x}
\end{equation}
for each real $x$. Hence,
\begin{align}
P(S_n\ge bn)&=\int_0^1 P(S_{n-1}\ge bn-z)\,dz \\
&=\int_0^1 P(S_{n-1}\ge b(n-1)+b-z)\,dz \\
&\sim P(S_{n-1}\ge b(n-1))\int_0^1 e^{-h_b(b-z)}\,dz \\
&=P(S_{n-1}\ge b(n-1))e^{-h_b b}R(h_b).
\end{align}
We conclude that
\begin{align}
f_n(x)&=\frac{P(S_{n-1}\ge b(n-1)+(x-a))}{P(S_n\ge bn+x)} \\
&=\frac{P(S_{n-1}\ge b(n-1)+(x-a))}{P(S_{n-1}\ge b(n-1))}
\frac{P(S_{n-1}\ge b(n-1))}{P(S_n\ge bn)}
\frac{P(S_n\ge bn)}{P(S_n\ge bn+x)} \\
&\sim e^{-h_b(x-a)}\frac{e^{h_b b}}{R(h_b)}\,e^{h_b x}
=\frac{e^{h_b}}{R(h_b)}
\end{align}
for each real $x$.
For an illustration, here are the graphs $\{(x,f_n(x)/\frac{e^{h_b}}{R(h_b)})\colon|x|<5\}$ with $a=0.25$ for $n=100$ (left) and $n=500$ (right):
Very helpful infinitesimal analysis based on the LDT. I have corrected a critical mistake in the previous post, but your method is also applicable for the edited question. Based on my own undestanding of your method, I have verified the edited function series indeed converges to the truncated exponential density function.
|
2025-03-21T14:48:31.616787
| 2020-07-27T04:28:39 |
366661
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Douglas Somerset",
"Jamie Gabe",
"Math Lover",
"Matthew Daws",
"Yemon Choi",
"https://mathoverflow.net/users/126109",
"https://mathoverflow.net/users/129638",
"https://mathoverflow.net/users/142780",
"https://mathoverflow.net/users/406",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631500",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366661"
}
|
Stack Exchange
|
Need reference for $\phi(Z)=Z'$ if and only if $\Phi: \operatorname{Prim}(Z')\to \operatorname{Prim}(Z)$ is injective
Let $A$ and $B$ be $C^{\ast}$-algebras with centers $Z$ and $Z'$ respectively. Let $\phi:A \to B$ be surjective $C^{\ast}$-morphism. Then
$\phi(Z)=Z'$ if and only if the map $\Phi: \operatorname{Prim}(Z') \to \operatorname{Prim}(Z)$ defined as $\Phi(J) = \phi^{-1}(J)$ is injective.
Can someone please give me reference for the above result?
The above result is mentioned without proof in the paper titled On the homomorphic image of Center of $C^{\ast}$-algebras by Vesterstrom.
Which parts of this have you been able to prove yourself? What have you tried so far?
Also, something looks wrong with your notation. What is $\tilde{\phi\vert_Z}$ supposed to be?
@YemonChoi: I got confused with the proof. Could not get much idea of the proof. I'm trying it. Regarding your second comment: fixed notation
The map $\tilde \phi$ isn't even well-defined. Sure, $\phi^{-1}(J)$ is a two-sided closed ideal, but will in general not be primitive (even when $J$ is). The primitive ideal space is not a functorial invariant for $C^\ast$-algebras, only in the commutative case.
@JamieGabe: I have fixed the notations.
The map $\Phi$ is not well-defined. Take The unital embedding $\mathbb C \oplus \mathbb C \to M_2(\mathbb C)$ as a counter example.
Argh. Sorry. I tried to correct the definition, and I'm not sure I've helped. I'll leave it up to @MathLover to try to be a little more precise about what exactly is meant.
@MatthewDaws: Probably I am missing something from the linked paper. The exact result is mentioned in Proposition $1$ of the linked paper.
The linked paper has an important condition which is missing from the question: $\phi$ is supposed to be a surjective $*$-homomorphism! This takes care of Jamie Gabe's objection (without $\phi$ being surjective, indeed $\Phi$ makes no sense). I am following Blackadar's book, II.6.5.4, if $J=\ker\phi$ then $B\cong A/J$ and $\newcommand{\prim}{\operatorname{Prim}}\prim(A/J) \cong {K\in\prim(A) : J\subseteq K }$. I think to understand Proposition 1 you'll need to know about the Dauns-Hofman Theorem.
@MatthewDaws: Sorry, my bad. Do you know about any reference where the proof is given?
The restated version of the question finally makes sense (it would have saved everyone time if a link to the relevant paper had been posted in the original question). Without claiming to immediately see a full proof, it should be noted that fo a commutative ${\rm C}^*$-algebra primitive ideals are the same as maximal ideals, and indeed the primitive ideal space corresponds naturally to the Gelfand spectrum. So the map $\Phi$ admits a very concrete description
Basically it is a theorem about commutative unital C$^*$-algebras (Vesterstrom also has a blanket assumption that $A$ and $B$ are unital).
We have a map $\phi: Z\to Z'$. So $\phi(Z)$ is a C$^*$-subalgebra of $Z'$, and $\phi(Z)=Z'$ if and only if $\phi(Z)$ separates the points of ${\rm Prim}(Z')$. For $J\in{\rm Prim}(Z')$, $\Phi(J)=\phi^{-1}(J)=\{ z\in Z: \phi(z)\in J\}$.
Hence for $J_1, J_2\in{\rm Prim}(Z')$, $\Phi(J_1)=\Phi(J_2)$ if and only if for all $z\in Z$, $\phi(z)\in J_1\Leftrightarrow \phi(z)\in J_2$, and this condition holds if and only if $\phi(Z)$ fails to separate $J_1$ and $J_2$. Thus $\phi$ is surjective if and only if $\Phi$ is injective.
Sorry where did you use the fact that $A$ and $B$ are unital $C^{\ast}-$ algebras?
I am not sure that I did, but Vesterstrom has it, so I felt safer with that hypothesis. In the non-unital case, the Stone-Weierstrass theorem requires "separates the points and vanishes nowhere", if I remember, so perhaps I did use it in line 2 of the second para.
|
2025-03-21T14:48:31.617023
| 2020-07-27T05:38:49 |
366662
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631501",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366662"
}
|
Stack Exchange
|
Transcendence detection from algebraic constructions
I have a simple image or intuition in my mind that I can't seem to shake off, so I thought I'd seek help here.
Suppose we don't know if $\alpha \in \mathbb{C} - \mathbb{Q}$ is algebraic or transcendental, and we'd dearly like to find out.
Now, if $\alpha$ is transcendental, the algebraic theory over (theory of polynomial equations with coefficients in) the field $\mathbb{Q}(\alpha)$ should be essentially a copy of the theory over $\mathbb{Q}$. On the other hand if $\alpha$ is algebraic but not rational, the theory should be over some non-trivial finite extension of $\mathbb{Q}$. Has this dichotomy been used in transcendence proofs (which I know nothing about, but I have a notion that they are quite analytic), or is this too simple-minded?
Another way to put it: suppose we take a system of polynomial equations $S$ over $\mathbb{Q}(t)$ where $t$ is an indeterminate, and then specialize whichever constructions we make from $S$ to $t$ equal to various elements of $\mathbb{C}$. The properties/invariants of these specializations should in general be different depending on whether $t$ is rational/transcendental or algebraic-but-not-rational. Is the difference detectable in some clever choice of S and one of its invariants which takes one set of values when $t$ is rational/transcendental and another disjoint set if $t$ is algebraic-but-not-rational and which can be decided to be in one set or the other without knowing a priori if the value of $t$ is rational, algebraic or transcendental? Or perhaps there is a "connection" on a bundle on the base Spec $\mathbb{Q}(t)$ constructed from $S$ whose singularities occur precisely when $t$ is an algebraic number? Some such thing ... may be it is hopelessly naïve and unworkable, or just plain wrong.
|
2025-03-21T14:48:31.617173
| 2020-07-27T06:24:11 |
366663
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"2734364041",
"Alapan Das",
"Ben Smith",
"Conor Grogan",
"Dave Benson",
"Fredrik Johansson",
"Gerry Myerson",
"Max Alekseyev",
"Robert Israel",
"Terry Tao",
"ho boon suan",
"https://mathoverflow.net/users/111215",
"https://mathoverflow.net/users/118745",
"https://mathoverflow.net/users/13650",
"https://mathoverflow.net/users/156029",
"https://mathoverflow.net/users/156215",
"https://mathoverflow.net/users/3684",
"https://mathoverflow.net/users/460592",
"https://mathoverflow.net/users/4854",
"https://mathoverflow.net/users/515074",
"https://mathoverflow.net/users/7076",
"https://mathoverflow.net/users/766"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631502",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366663"
}
|
Stack Exchange
|
Is $\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!}$ composite for $n\geq 4$?
Define $a_n$ as follows:
$$
a_1=1,\ \ a_{n+1}=na_n+1\
$$
At this time, the sequence $a_n$ is as follows:
$$
a_n=\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!}
$$
I made some discoveries about this sequence.
The first:$$a_k\equiv 0\pmod{m}\Rightarrow a_{k+Nm}\equiv 0\pmod{m}~~~~\forall k,m,N\in\mathbb{N}$$
The second:$$ n\geq 4\,\Rightarrow\,a_n ~\mathrm{is~composite} $$
I was able to prove the first, but not the second. My expectation is that the second is correct, but I'm not sure it can be proved. My friend used computer and check $a_n$ is composite for $4\leq n\leq 48$. After $a_{49}$, it is too large number to check on his computer. Please let me know if you come up with a proof method. Any help is welcome!
(I am a Japanese college student. I'm sorry for my poor English.)
This is http://oeis.org/A000522
$a_i$ is odd resp. even when $i$ is odd resp. even. So $a_i$ is certainly composite for all even $i$.
The series $a_{n+1}=\sum_{k=0}^{n} \frac{n!}{k!}$ can be written in another representation. That is $$a_{n+1}=2^n+\sum_{k=2}^{n} \binom{n}{k}2^{n-k}D_k$$. Where, $D_k$ is the number of derangements.
$a_n$ is composite for $4 \le n \le 2016$.
$a_{2017}$ appears to be prime (it passes a strong pseudoprime test). I have not tried to certify that it is prime (this would take a while as the number has 5789 digits).
A pity this wasn't discovered $3$ years ago.
Mathematica seems to think that $a_{2017}$ is prime. I used the FactorInteger command, and after about 10 seconds, it says that $a_{2017}$ is the only prime factor. Mathematica's factorization algorithm is usually not this efficient, which suggests that either (a) Mathematica failed fantastically, of (b) $a_{2017}$ is a prime of a special form. Maybe the algorithm works faster for numbers passing the pseudoprime test that you used.
Well, Magma didn't have such an easy time of it. It very quickly decided that $a_{2000}$ to $a_{2016}$ were composite, but at $a_{2017}$ my (quite powerful) laptop sat there for five minutes getting hotter and blowing out air before I aborted the computation. I suspect that Mathematica is lying to you about its confidence in its answer.
Yes, I believe Mathematica's FactorInt function with default settings is using the Lenstra elliptic-curve factorization method when run with input $a_{2017}$, which only shows that $a_{2017}$ has no small prime factors. To check if $a_{2017}$ is prime with certainty using Mathematica, one could try using the PrimalityProving package, though it would probably take a very, very, very long time to run.
This is broadly consistent with the (very crude) heuristic that each $a_n$ has a "probability" of $\approx \frac{1}{\log a_n} \approx \frac{1}{n \log n}$ of being prime; the sum $\sum_{n=2}^\infty \frac{1}{n \log n}$ diverges, but only double logarithmically, so one would tentatively expect an infinite number of counterexamples, but spread out extremely thinly, and it makes sense that the first one is only seen at $n \approx 2000$.
I just ran this independently for n=2017; confirmed that this number is in fact prime! So an is composite for 4≤n≤2016
I ran sympy.factorint on a i9-13900k for the 5789 digit number n=2017 and it only returned 1 and the number
That's does not bring much new as compared to the previous answer. Better be added as a comment, not an answer.
It would be nice to know what exactly you "ran".
I tried - wasn't able to add a comment due to my account being new.
I ran sympy.factorint on a i9-13900k for the 5789 digit number n=2017 and it only returned 1 and the number
As I mentioned in a comment to the accepted answer on this post, Lenstra ECM only finds relatively small factors and does not guarantee primality. See the sympy documentation for this function; you can verify this behavior by setting only the use_ecm flag to True for the factorint function.
That said, sympy.ntheory.primetest.isprime uses a few seconds to show that $a_{2017}$, if not a prime, is at least a strong Baillie–PSW pseudoprime. The BPSW test is known to be correct up to $n=2^{64}$, and we don't know if BPSW pseudoprimes (i.e., composite numbers that fool the BPSW test into thinking they are prime) exist. (Similar tests are used by say Mathematica's PrimeQ and Maple's isprime, so one has to be quite careful when using a computer to check if a large number is prime.)
|
2025-03-21T14:48:31.617612
| 2020-07-27T06:26:05 |
366664
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D. Hershko",
"Gabe Goldberg",
"Otto",
"https://mathoverflow.net/users/102684",
"https://mathoverflow.net/users/119731",
"https://mathoverflow.net/users/123559"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631503",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366664"
}
|
Stack Exchange
|
External ultrafilters definitions
I am reading a paper by Goldberg, in which he defines ultrafilter over a model of set theory (not transitive). These are the definitions:
I get the definition of an M-ultrafilter, it is a real subset of the boolean algebra $P^M(X)$ in which union and intersection are obviously defined in $M$.
What I don't get is the other definitions: in $\gamma$-completeness over $M$, why is it correct to look at $(\cap F)^M$ and ask if it is in $U$? I mean if $F$ is a real subset of $U$, that doesn't mean that all it's $M$-elements (the sets $M$ thinks that are elements of $F$) are in $U$.
Am I missing something? can someone elaborate on the definitions?
Thank you very much.
Where is the confusion? $F$ is a subset of $P^M(X)$ and $F$ is in $M$, so $M$ knows that $F$ consists of subsets of $X$. Hence $\bigcap F$ is a subset of $X$ in $M$, so we can ask if it is in U.
Does $M$ thinks that $F$ is a subset of $P^M(X)$? If yes than $M$ knows that $F$ consists of subsets of $X$. But this isn't implied by the fact that $F$ is an actual subset of $P^M(X)$
Why not? Any counterexample?
$P^M(X)$ is the collection of sets in $M$ that $M$ thinks are subsets of $X$. I couldn't see any problem
I guess there is something very basic I don't understand. Is $P^M(X)$ a set in $M$? is $P^M(X)$ really the collection you mentioned, or just $M$ thinks that?
I was sort of sloppy here. It is true that the definition is not literally correct. I wanted to identify elements of $M$ with their extensions in $V$ when it is unambiguous. So $F\subseteq U$ really means that $\text{ext}(F)\subseteq U$, where $\text{ext}(F) = {s : M\vDash s\in F}$. Feel free to email me if you have other questions like this.
When relativizing functions we need a formula defining the function (see the section on relativization in chapter IV of Kunen's Set Theory). For instance, $(\bigcap F)^M$ can be defined as the unique $x \in M$ such that
$$\forall z \in x \cap M \ \exists y \in F \cap M \ (z \in y) \, \wedge \, \forall w \in M \ [\forall z \in w \cap M \ \exists y \in F \cap M \ (z \in y) \, \rightarrow$$
$$\rightarrow \, \forall z \in w \cap M \ (z \in x).$$
This need not be the same as $\bigcap (F \cap M)$, because $M$ doesn't necessarily have witnesses for all elements of $\bigcap (F \cap M)$ being in $(\bigcap F)^M$.
The reason for asking that $(\bigcap F)^M$ be in $U$ is that $(\bigcap F)^M$ is the greatest lower bound of $F$ according to $M$.
|
2025-03-21T14:48:31.617801
| 2020-07-27T08:02:03 |
366668
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Manuel Lafond",
"Venugopal K",
"https://mathoverflow.net/users/161818",
"https://mathoverflow.net/users/56791"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631504",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366668"
}
|
Stack Exchange
|
Meaning of L-reduction from Dominating set problem
We are working in a variation of Locating dominating sets. Recently, we realized that the reduction from dominating set to our problem in proving its NP-completeness turns out to be also an L-reduction. This would mean that there is no PTAS for our problem since the dominating set problem does not have PTAS. Now we have two questions:
since the dominating set problem is W[2]-complete, does that follow to our problem because of the L-reduction?
we have proved that our problem is in XP (i.e, piecewise polynomial) and also in APX (there is a 2-approximation polynomial algorithm). But dominating set problem is known to be not in APX. Does this mean somewhere we might have gone wrong or is this perfectly alright?
(1) I'd say that it depends. L-reductions allow some leeway between the upper and lower bounds of solution sizes (the $\alpha$ and $\beta$ constants), whereas W[2]-hardness reductions use an exact value of $k$. If your proof show that G has a dom-set of size at most k iff your instance has solution size at most f(k) for some function f, then W[2]-hardness should hold. (2) APX membership is preserved between minimization problems, so if you have a minimization problem, something might be wrong. But if you have a maximization problem, you might still be OK.
@ManuelLafond Ours is a minimization problem and we have proved it to be APX complete by L-reduction from Dominating set-B (graphs bounded by vertex degree B, i.e, $\Delta(G)=B$ ) problem. In proving W-hardness, f(k) is generally like k or k+1 or k+2 and so on. Any idea how much can we stretch that f(k)? Can it be as big as something like $2k$ or say $\Deltak$? Thank you for the response!
DomSet-B is in APX, assuming B is a constant. So there seems to be no contradiction with the L-reduction. For W-hardness, you can stretch k to any function f(k) not depending on $n$. So $2k$ or even $2^{2^k}$ is fine but not $\Delta k$ unless $\Delta$ is bounded by $k$. Also, I don't know if DomSet-B is actually W[2]-hard - make sure this is the case if you reduce from that variant.
|
2025-03-21T14:48:31.617972
| 2020-07-27T08:04:00 |
366669
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Florian Lehner",
"Xin Zhang",
"https://mathoverflow.net/users/148974",
"https://mathoverflow.net/users/97426"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631505",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366669"
}
|
Stack Exchange
|
Looking for examples showing that the crossing number may not be realized by the drawings with local crossing number
The crossing number $cr(G)$ of a graph $G$ is the lowest number of edge crossings of a plane drawing of the graph $G$. The local crossing number of a drawing of a graph is the largest number of crossings on a single edge. The minimum local crossing in any drawing of a graph is the local crossing number for that graph.
I am looking for examples $G$ so that
$G$ has local crossing number $k$;
Any drawing of $G$ with local crossing number $k$ has more crossings then $cr(G)$,
especially for $k$ is small.
Can anyone find such examples? I believe that they exist.
Florian Lehner gave a nice example for $k=1$. Next, is there any example for $k=2$.
Couldn't you just replace every edge in my example by two parallel edges (or by two paths if you prefer simple graphs)?
I think the following graph works for $k = 1$:
It clearly has crossing number at most $2$ and local crossing number $1$.
In any drawing with $2$ or fewer crossings, the green cycles cannot cross (the spokes from the red vertices would create at least one additional crossing). Once we have embedded the green cycles and the black matching edges between them, each face meeting the "outside" of the green cycles is incident to at most two vertices of each cycle. Consequently, embedding a red vertex outside its cycle creates at least $3$ crossings (the spokes to the vertices not incident to the face where it was embedded). So the red vertices must be embedded inside their respective green cycles and thus the red edge crosses both green cycles.
I think this even shows that the drawing on the left is (up to isomorphism) the unique drawing with two crossings.
|
2025-03-21T14:48:31.618109
| 2020-07-27T09:02:24 |
366674
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anonymous",
"https://mathoverflow.net/users/89233"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631506",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366674"
}
|
Stack Exchange
|
Isomorphic hypergraphs of dense sets of Hausdorff spaces
If $H_i = (V_i, E_i)$ are hypergraphs for $i = 1,2$ , we say that they are isomorphic if there is a bijection $f:V_1 \to V_2$ such that for all $e\subseteq V_1$ we have $e\in E_1$ if and only if $f(e)\in E_2$.
If $(X,\tau)$ is a topological space, we let the dense set hypergraph ${\cal D}(X,\tau)$ be the collection of all dense subsets of $X$ with respect to $\tau$.
Note that for $X = \{0,1,2\}$ let $\tau_1 = \{\varnothing, \{1\}, X\}$ and $\tau_2 = \tau_1 \cup\big\{\{1,2\}\big\}$. We have $(X,\tau_1)\not\cong(X,\tau_2)$, but ${\cal D}(X,\tau_1)$ and ${\cal D}(X,\tau_2)$ are isomorphic (they are even equal).
Question. Let $(X_i, \tau_i)$ be Hausdorff spaces for $i=1,2$. If ${\cal D}(X_1,\tau_1)$ and ${\cal D}(X_1,\tau_2)$ are isomorphic, does this imply that $(X_1, \tau_1)\cong (X_2, \tau_2)$?
If $X_1$ and $X_2$ are non-homeomorphic countably infinite spaces each having exactly one non-isolated point, then they have the same dense subsets.
Counterexamples abound. Here are a few of them.
Theorem. If $\emptyset\ne X\subseteq\mathbb R$ and $X\subseteq\operatorname{cl}(\operatorname{int}(X))$, then $\mathcal D(X)\cong\mathcal D(\mathbb R)$.
Proof. Construct an infinite sequence of pairwise disjoint open intervals $I_n$ so that $\bigcup_{n=1}^\infty I_n$ is a dense subset of $X$, and $|X\setminus\bigcup_{n=1}^\infty I_n|=2^{\aleph_0}$. Then a set $D\subseteq X$ is dense in $X$ if and only if $D\cap I_n$ is dense in $I_n$ for each $n$.
Thus, if $X'$ is another set satisfying the same hyotheses, with an analogous sequence of intervals $I'_n$, then a bijection $f:X\to X'$ which maps each $I_n$ homeomorphically onto the corresponding $I'_n$ is an isomorphism from $\mathcal D(X)$ to $\mathcal D(X')$.
P.S. With a somewhat more complicated argument one can prove:
Theorem. If $X$ and $Y$ are nonempty Polish spaces with no isolated points, then $\mathcal D(X)\cong\mathcal D(Y)$.
The idea is to construct a family $(U_i:i\in I)$ of open subsets of $X$ and a family $(V_i:i\in I)$ of open subsets of $Y$ so that:
(1) a set $D\subseteq X$ is dense in $X$ if and only if $D\cap U_i\ne\emptyset$ for each $i\in I$;
(2) a set $D\subseteq Y$ is dense in $Y$ if and only if $D\cap V_i\ne\emptyset$ for each $i\in I$;
(3) there is a bijection $f:X\to Y$ such that $f[U_i]=V_i$ for each $i\in I$.
|
2025-03-21T14:48:31.618291
| 2020-07-27T09:10:49 |
366675
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Denis Nardin",
"Fernando Muro",
"Kevin Carlson",
"Lennart Meier",
"Mike Shulman",
"Mirco A. Mannucci",
"Urs Schreiber",
"https://mathoverflow.net/users/12166",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/2039",
"https://mathoverflow.net/users/381",
"https://mathoverflow.net/users/43000",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/49"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631507",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366675"
}
|
Stack Exchange
|
Grothendieck derivators vs $\infty$-categories
I have some questions on derivators and $(\infty,1)$-categories,
I would be grateful if someone could help me.
Is there some problems that $(\infty,1)$-categories/derivators can resolve but derivators/$(\infty,1)$-categories cannot resolve?
Why do so many people prefer $(\infty,1)$-categories than Grothendieck derivators?
Is there a good place to learn about $(\infty,1)$-categories than Grothendieck derivators but with a historical and comparing point of view?
Thank you in advance!!
Can one take a limit of a diagram of derivators?
The following paper shows interesting relations between both: Arlin, Kevin. 2020. “On the $\infty$-Categorical Whitehead Theorem and the Embedding of Quasicategories in Prederivators.” ArXiv:1612.06980 [Math], February. http://arxiv.org/abs/1612.06980.
Another version of Lennart's comment: can you talk about sheaves of derivators? One of the great strenghts of ∞-cats is that they work very well in families (so, for example, you can rephrase faithfully flat descent as "$\mathrm{QCoh}(-)$ is a sheaf")
I don't know that I have a really compelling answer here, but to Denis and Lennart's points the answer is: at best only in terms of homotopy limits in a model structure, which is a major advantage of $\infty$-categories. Derivators are better suited to working within a single homotopy theory at a time. Regarding your second question, well, since Lurie began writing there has been vastly more machinery developed for $\infty$-categories, and some things (see above) have been done only in that framework.
Deleted my answer. Now, Kevin, Harry and Mike, I would be very interested in reading your answers here.
Very roughly speaking, by [Ren09] the $2$-category of derivators is equivalent to:
Take the $(\infty,1)$-category of $(\infty,1)$-categories.
Truncate it to a $(2,1)$-category.
Perform a $2$-categorical localization inverting those $1$-morphisms which induce an equivalence on homotopy categories.
[Ren09] Renaudin, Olivier. 2009. “Plongement de Certaines Théories Homotopiques de Quillen Dans Les Dérivateurs.” Journal of Pure and Applied Algebra 213 (10): 1916–1935. https://doi.org/10.1016/j.jpaa.2009.02.014.
So, yes, you miss information but you do keep a lot. Also, very roughly speaking (and shifting dimensions by $-1$), derivators are to $\infty$-categories what homotopy theory is to topology. So, if you like homotopy theory, probably you should also like derivators as much.
@FernandoMuro Your statement is missing the very important adjective "locally presentable" in front of "derivators" and "$(\infty,1)$-categories". (I know you said "roughly speaking", but I think this is important not to omit. For one thing, a derivator is always complete and cocomplete, whereas an arbitrary $(\infty,1)$-category is not!)
It took me a while to figure out what you meant by "derivators are to ∞-categories what homotopy theory is to topology". Would a more precise version of the statement end with "...what -algebras are to spaces"?
@mikeshulman I actually said "very roughly" ;) thanks for stressing this part, it's difficult to sum up, better to look at the reference, which doesn't even talk about infinity categories!
My second comment is kind of infamous, one has to put a lot of will to understand it well, I'm tempted to erase it.
@FernandoMuro I think your pointer to Renaudin's result https://ncatlab.org/nlab/show/Ho(CombModCat) is the most pertinent reply here, since the only way to really know how derivators relate to ∞-categories is to compare the (large) categories (n-categories) which both form. The only trouble is that your comment takes for granted a fact whose proof seems to be sadly missing in the literature: That the 2-category of combinatorial model categories localized at Quillen equivalences is equivalent to the homotopy 2-category of presentable ∞-categories. This ought to be true, but is there a proof?
Excellent point @UrsSchreiber No, I don't know of any reference but it looks like something interesting to look at. I actually remember someone rising this question last time I talked about derivators a couple of years ago, in person.
The short answer is that $(\infty,1)$-categories are the "real" object of interest. Derivators are a tool for working with them that is sometimes (for some people) easier to use, but doesn't remember all the information and hence is not always applicable.
When you work with $(\infty,1)$-categories, you have to deal explicitly with higher-dimensional coherence all the time. Everything is determined only up to equivalence. This can be kind of a pain, so it's convenient to have 1-categorical structures that neverthless carry $\infty$-categorical information.
The classical kind of 1-categorical structure used for this is a Quillen model category. This carries "too much" information, in that two objects can be equivalent in an $(\infty,1)$-category but not isomorphic in the model category it presents, and similarly two model categories can present equivalent $(\infty,1)$-categories but not be equivalent categories. Thus a model category needs a notion of "weak equivalence" between its objects, and similarly we have a notion of "Quillen equivalence" between model categories. But a model category contains all the information of an $(\infty,1)$-category, and so we can work with all the higher coherences as necessary.
A derivator is sort of a "dual" to a model category: it carries "too little" information. An advantage is that it is not subject to the issues of weak equivalence: two objects of an $(\infty,1)$-category are equivalent if and only if they are isomorphic in the corresponding derivator, and similarly two (locally presentable) $(\infty,1)$-categories are equivalent if and only if their corresponding derivators are equivalent (in a 1-categorical sense: derivators are the objects of a 2-category just like 1-categories are, and here we mean equivalence internal to that 2-category).
But a derivator doesn't have enough information to do everything we might want to with higher coherences. It's essentially an enhancement of the homotopy 1-category (the quotient by homotopies) that remembers the notion of homotopy-coherent diagram, and therefore also of homotopy limit and colimit (and homotopy Kan extension). This is sufficient for a surprising amount of homotopy coherence, at least when working only within a single $(\infty,1)$-category. But there are some things it can't do (or not very well), notably those that involve diagrams of $(\infty,1)$-categories and things like functor $(\infty,1)$-categories.
So the answer to your first question is yes, there are things you can do with $(\infty,1)$-categories but not with derivators; but no, anything you can do with a derivator can also be done with an $(\infty,1)$-category. And I think this mostly answers your second question as well: given that derivators are not good enough for everything, even if they make certain things easier, it's understandable that many people prefer not to learn two different languages and stick to $(\infty,1)$-categories even if they happen to be doing something that might be a bit easier with derivators. (On the other hand, I personally feel that it is easier to be sloppy with $(\infty,1)$-categories --- partly because being precise is so much work --- and easier to be precise with derivators, which is one reason that I still use the latter sometimes.)
Historically, derivators were introduced (by Heller, Grothendieck, and Franke, fairly independently) before practical notions of $(\infty,1)$-category were available. So at the time they were the only way of getting rid of the "weak equivalences"; but even with that advantage they never really caught on. I'm not quite sure why not; perhaps the requisite 2-category theory was also offputting at the time?
As for references, I don't have any suggestions other than those on the nLab page. I particularly like the expository aspect of Moritz Groth's work. You may also be interested in this blog post of mine.
Mike, you got my like, for two reasons: 1) because you had the guts to post an answer (a very good one, but that is beside the point) rather than "dancing around" 2) the second one is that it helps me personally to understand the context. So: even though I used the inappropriate (but not totally inappropriate, I will explain later why) term "invariant", whereas I should have said "real", th bottom line is this: infinity cats are (at least at the current level of understanding) THE homotopy object, whereas both Quillen and associates and Derivators are TOOLS.
As for your reconstruction of the history and the role of weak equivalences, I have some reservations, on which I will comment over the weekend. Meanwhile I invite everyone to like it and the poster to accept it. BRAVO
My humble suggestion about why derivators never caught on is because working with them is really a lot harder than you seem to think, or at least this has been my experience. It might be that they are very natural for people well versed in 2-category theory though (which I am very much not).
@MircoA.Mannucci I don't object to the word "invariant". Regarding history, most of my answer is blatantly ahistorical. I included one paragraph about history because the OP mentioned wanting a historical point of view, but for the most part I didn't even try to be historically faithful.
@DenisNardin I've (co-)written four papers using derivators. My experience is that it takes a bit of getting used to to think of a "coherent diagram" as a primitive object rather than something "put together" out of objects, morphisms, and (perhaps) homotopies, but once that barrier is surmounted the technology is quite pleasant, especially in that one doesn't worry about things with names like left-quasi-marked-fibrations of simplicial sets but can just use standard 1-categorical ideas like adjunctions. Perhaps, as you say, some of that is due to my familiarity with 2-categories.
@MikeShulman thanks to your excellent post I felt motivated to revisit the definition of Derivator. Now I see why you folks objected to my supermarket definition: of course, the notion of pre-derivator is seemingly much broader than the one I implicitly assumed, ie the one which assumes a cat with weak equivalence, an builds the derivator from there. But now I have a question for you: is it true that all derivators with some extra conditions (TBD) are sub-derivators or quotients of the "representable derivator ", by this I mean the one coming from a weak equivalence?
in other words, how much broader is the standard definition? Of course I realize that I waved my hand with the TBD. There are likely plenty of derivators which are not related to the "representable", yet I am still curious to see a precise statement. Thanks in advance
@MircoA.Mannucci The only statement along those lines that I know of is the result of Renaudin, that the 2-categories of locally presentable derivators and of locally presentable $(\infty,1)$-categories (which he constructs using the 2-category of combinatorial model categories and Quillen adjunctions) are equivalent.
|
2025-03-21T14:48:31.619064
| 2020-07-27T09:43:35 |
366676
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631508",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366676"
}
|
Stack Exchange
|
$K$-group of category of bounded chain complexes of Projective modules with finite length homologies
For a Commutative Noetherian local ring $(R, \mathfrak m)$, let $K_0^{\mathfrak m}(R)$ denote the abelian Group generated by isomorphism classes of bounded chain complexes of finitely generated free modules with finite length homologies (i.e. the support of each homology is contained in $\{\mathfrak m\}$) subject to the following two relations: (1) $[P_{\bullet}]=0$ if $P_{\bullet}$ is exact. (2) $[P_{\bullet}]=[P'_{\bullet}]+[P''_{\bullet}]$ if there exists a short exact sequence of chain complexes (and chain maps) $0\to P'_{\bullet}\to P_{\bullet}\to P''_{\bullet}\to 0$ .
It can be easily shown that in $K_0^{\mathfrak m}(R)$, we have $[\sum P_{\bullet}]=-[P_{\bullet}]$ and $[P_{\bullet}]+[Q_{\bullet}]=[P_{\bullet}\oplus Q_{\bullet}]$ , so in particular, any element of $K_0^{\mathfrak m}(R)$ is of the form $[P_{\bullet}]$.
Now let $(R, \mathfrak m)$ be a Noetherian local domain of dimension $1$, with fraction field $F$.
So for any non-zero ideal $I$, the $R$-module $R/I$ has finite length, so in particular, for any $0\ne r\in R$, the bounded chain complex of f.g. free modules $...\to 0\to R \xrightarrow{.r} R\to 0\to ...$ has finite length homologies.
My claim is to show $K_0^{\mathfrak m}(R)\cong F^{\times}/R^{\times}$. So I define a map $F^{\times}\to K_0^{\mathfrak m}(R)$ by sending $a/b$ to
$[...\to 0\to R \xrightarrow{.a} R\to 0\to ...] - [...\to 0\to R \xrightarrow{.b} R\to 0\to ...]$ .
Now I can show that this map is well-defined , but I'm having trouble to show Surjectivity and that the kernel is $R^{\times}$ (the group of units of $R$).
Please help.
|
2025-03-21T14:48:31.619212
| 2020-07-27T11:33:13 |
366681
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ville Salo",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/17773",
"https://mathoverflow.net/users/52871",
"kodlu",
"lchen"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631509",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366681"
}
|
Stack Exchange
|
NP-hardness of a sequence problem
Given $n$ binary sequences $s_i$ ($1\le i\le n$) with common period $T$. Let $s_i^{t_i}$ denote the sequence obtained by cyclically shifting $s_i$ for $t_i$ bits. The $n$ sequences form a good system if under any combination of $\{t_i\}_{i=1}^n$, for each sequence $s_i$ there always exists $\tau_i$ such that $s_i^{t_i}(\tau_i)=1$ and $s_j^{t_j}(\tau_j)=0$ for $j\ne i$. For example, $s_1=1010$ and $s_2=1100$ is a good system, while $s_1=0001$ and $s_2=1000$ is not a good system.
Is the problem of deciding whether a system is good NP-hard?
The background of the problem is below. We want to design a code for each of the $n$ users. User $i$ with code $s_i$ transmits its packet in slot $t$ if $s_i(t)=1$. We want to check whether a set of codes can ensure that even the users are not time-synchronized, each of them can successfully transmit a packet under any clock drift among users. If two or more users transmit at the same slot, none of them succeeds.
you need to parse your sentences or define the condition for success properly, too many such that... such that, given that... Write it out in equations.
@kodlu I have reformulated the problem. Hope it is clear now.
@Ville Salo I have reformulated the problem. Hope it is clear now.
Why are 101 and 011 good? Shifting one gives 101, 101 so doesn't seem so.
And it seems picking the $\tau_i$ makes the shifting meaningless, so I still don't understand what you are after.
I have reformulated the problem
@VilleSalo $\tau_i$ allows locate the appropriate bit in the shifted sequence.
Ok so should it say there always exist $\tau_i$ such that $s_i^{t_i}(\tau_i) = 1$ and $s_j^{t_j}(\tau_i) = 0$ for $i \neq j$?
If I understand correctly that good means no sequence in the set is a union of shifts of others, I'm guessing you can ensure only one of the sequences has small enough support that it's possible to cover with the others, and then there should be some easy reduction. It's pretty much clear that goodness is (co-)NP-complete. (Even if not as I suggest here.)
I suppose I have at least a direct reduction from $3$-SAT, I can try to write it tomorrow if no one gives an answer before that. (Hopefully someone does, mine is not great.)
Isn't "We want to design a code for each of the $n$ users." + " We want to check whether a set of codes can ensure that..." equal to "For what $T$ and $n$ can we design a set of codes such that ... and how to do it?". If this equality holds, the question should, probably, be asked quite differently (indeed, why should you care whether some set of codes designed not by you is good or not if you can just set up the codes yourself?). And yes, one can easily reduce 3-SAT to your current problem, but Ville was the first to note that, so I'll leave the honor to him :-)
@fedja Thank you for your comments.
I agree with @fedja that if you want to find such a code, then you could've just asked for such a code. If you need many such codes, or are not sure yet what kind of additional properties you want for the codes, then it's useful to know the problem is in co-NP because computers are good at solving problems in co-NP. Knowing it's co-NP-complete is useful because it means you're not missing an obvious polynomial-time algorithm, unless we all are.
@Ichen Your are welcome, of course, but my main point was to ask if you are really interested in the efficient design of such codes (I.e., minimizing $T$, given $n$) rather than in the problem you posted. If you are, you are welcome to ask a separate question to that extent. My construction was pretty close to Ville's though I made some steps a bit differently (it is the same encoding with the same 2 extra precautions: bounding the shifts and making sure that only one user can be blocked).
FWIW, I'm pretty sure my $T = O(n^2)$ is not optimal (and I conjecture that @fedja has a $T = O(n)$ construction).
@VilleSalo I am inclined to say that $T=O(n^2)$ seems to be the lower-bound, $O(n)$ seems not achievable.
Well, maybe ask it separately, as has been suggested?
I assume you mean "there always exist $\tau_i$ such that $s_i^{t_i}(\tau_i) = 1$ and $s_j^{t_j}(\tau_i) = 0$ for $i \neq j$", i.e. you want that no matter how the sequences are shifted, each sequence has at least one bit which is zero in the other shifted sequences, and that's the slot when it manages to send its packet in your application.
(What you have written currently is "there always exist $\tau_i$ such that $s_i^{t_i}(\tau_i) = 1$ and $s_j^{t_j}(\tau_j) = 0$ for $i \neq j$". If they are chosen separately for each $i$ this just means each of the $s_i$ contain both $1$ and $0$. If they are chosen once and for all, this is impossible unless $n = 1$.)
Your problem as I interpret it is clearly in co-NP, as you check that all ($\forall$) shifts satisfy a (polynomial-time checkable) constraint, so it is probably not NP-hard, as that would collapse the polynomial hierarchy. I'll complement your problem and sketch a proof of NP-hardness of the resulting problem, meaning your problem is co-NP-complete.
Notation: On the set $X = \{0,1\}^{\mathbb{Z}_T}$ we have the shift action of $\mathbb{Z}_T = \mathbb{Z}/T\mathbb{Z}$ by $\sigma(s)_i = s_{i+1}, \sigma : X \to X$. For $s, s' \in X$ define $(s \cup s')_i = \max(s_i, s'_i)$. Write $s \leq s'$ for $\forall i: s_i \leq s'_i$.
The complemented problem: Consider a set of sequences $S = (s_i)_i$, $s_i \in X$. We say $i$ is a bad index for $S$ if $s_i \leq \bigcup_{j \neq i} \sigma^{t_j}(s_j)$ for some $t_j \in \mathbb{Z}_T$. We say $S$ is bad if there exists a bad index. Clearly $S$ is bad if and only if it is not good. The problem we prove NP-complete is identifying bad sets of sequences.
First, we will make sure $i = 1$ is the only possible bad index, i.e. $s_1$ is the only sequence that could possibly be the union of the others. For this, we will put an arithmetic progression $a_i$ in $s_i$, $i > 1$. This progression should be longer than $n$ and such that any other $s_j$ covers at most one element of it. I'll write some formulas for completeness.
Pick some $M$ (a parameter for future purposes). If $a_i$ is the sequence with support $\{kM(n^2+i) \;|\; k = 0,1,...,n+1\}$, then any shift of $a_i$ covers at most one position of any other $a_{i'}$: if $kM(n^2+i) = k'M(n^2+i')$, $k, k' \in \{1, ..., n+1\}$, $i, i' \in \{2, ..., n\}$ and $i' > i$, then $k/k' = (n^2+i')/(n^2+i) \in (1, \frac{n^2+n}{n^2+2}] \subset (1, \frac{n+1}{n})$, but clearly $k/k' > 1 \implies k/k' \geq (n+1)/n$. Now just include $a_i \leq s_i$ for each $i \geq 2$, and make sure that all other remaining things we include in the sequences $s_i$ fit within a single interval of length $Mn^2$ which is sufficiently far from $0$ (pick e.g. $T = 100 M n^3$ and there's plenty of space left, since the total length of $a_i$ is less than $2Mn^3$).
Now, consider a SAT instance with $n-1$ variables and clauses, $x_i, \phi_i, i \in \{2,...,n\}$. To reduce SAT, we want $\exists$ to have to make a binary choice for each $i > 1$, which will represent a choice between $x_i$ and $\neg x_i$. Pick arithmetic progressions $b_i$ similarly as we did with $a_i$ (but on a smaller scale; pick a suitable $M$ so we can do all that follows in an interval of length $Mn^2$ as we promised ourselves in the previous paragraph). The sequence $s_1$ contains one copy of $b_i$ while $s_i$ contains two copies of $b_i$ at distance $h$ from each other. If $\exists$ is to win, the copy of $b_i$ in $s_1$ has to be covered by one of the copies in $s_i$ (note that as long as $b_i$ fits in an interval of length $Mn^2$, the existing $a_j$-bits in the $s_j$ are not helpful for covering it).
Now, we can add for each clause of the SAT instance a single bit in $s_1$. These bits are in arithmetic progression with distance $2h$ between them. Depending on whether $x_j$ or $\neg x_j$ appears in the clause (or neither), we put a $1$ in the position in $s_j$ such that the correct clause bit is covered. (The bits coming from the choice $x_i = \top$ do not touch any clause bits if we choose the $x_i = \bot$ alignment for $s_i$, since this gives only a displacement of $h$; and vice versa for the $x_i = \top$ alignment.)
Thank you so much. Although I can see the intuition and logic of relating the problem to SAT, I fail to understand the intuition of proving $s_i$ as the only possible bad index, and the notion of $a_i$ (and the next paragraph).
If $s_1$ is the only one that could be covered, I just have to make sure, it's coverable iff the SAT instance is solvable. Otherwise I'd have to think about all $n$ sequnces.
The sequences $a_i$ are a simple set of sequences such that one is not covered by a union of others. I doubt the quadratic length is optimal, but the proof is very simple algebra for them (I wrote it).
Dunno what to clarify about the next paragraph (the one about $b_i$s I guess), except I'll fix a typo.
|
2025-03-21T14:48:31.619895
| 2020-07-27T11:33:24 |
366682
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/157982",
"vaoy"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631510",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366682"
}
|
Stack Exchange
|
When does the predictable $\sigma$-algebra $\mathcal{P}$ coincide with the optional $\sigma$-algebra $\mathcal{O}$?
The setup of my question is the following: Suppose that we have a measurable space $(\Omega,\mathcal{F})$ and a filtration $\mathbf{F} = (\mathcal{F}_t)_{t \geq 0}$ on it. Let $\mathcal{P}(\mathbf{F})$ be the predictable $\sigma$-algebra, that is, the $\sigma$-algebra generated by all real-valued left-continuous processes adapted to the filtration $\mathbf{F}$. Let $\mathcal{O}(\mathbf{F})$ be the optional $\sigma$-algebra, that is, the $\sigma$-algebra generated by all real-valued right-continuous process with left-limits adapted to the filtration $\mathbf{F}$.
I was trying to find conditions on the underlying measurable space $(\Omega,\mathcal{F})$ or the filtration $\mathbf{F}$, under which the predicatble $\sigma$-algebra $\mathcal{P}(\mathbf{F})$ coincides with the optional $\sigma$-algebra $\mathcal{O}(\mathbf{F})$. For example, if the filtered space $(\Omega,\mathcal{F})$ supports a process $X=(X_t)_{t \geq 0}$ with values in a metric space $(E,d)$ for example, such that the paths $\mathbb{R}_+ \ni t \mapsto X_t \in E$ are continuous, and the filtration $\mathbf{F} = (\mathcal{F}_t)_{t \geq 0}$ is generated by $X$, that is,
$$\mathcal{F}_t = \sigma(X_r : 0 \leq r \leq t),$$
does the equality
$$\mathcal{P}(\mathbf{F}) = \mathcal{O}(\mathbf{F})$$
hold?
If this condition is too weak, does there exist some other general condition under which the predictable $\sigma$-algebra coincides with the optional $\sigma$-algebra?
(Answering your comment)
Off the top of my head, I'd look in vol.2 of Probabilités et Potentiel (Dellacherie & Meyer) or in Limit Theorems for Stochastic Processes (Jacod & Shiryaev). Another convenient resource is the blog https://almostsure.wordpress.com of Geo. Lowther.
The key is that for a bounded rc martingale $M$, the predictable projection ${}^p\!M$ coincides with the left limit process $(M_{t-})$. Thus, if $\mathcal P =\mathcal O$ then each such $M$ coincides with its predictable projection and so is left continuous, hence continuous.
Conversely, if each bounded rc martingale is continuous, then each such martingale is predictable. In particular, for a bounded r.v. $Z$ the rc version of the martingale $t\mapsto E[Z|\mathcal F_t]$ must be predictable. Using this, Theorem IV-62 of Dellacheie & Meyer cited above yields that every stopping time is predictable, and this in turn implies that $\mathcal O =\mathcal P$.
A sufficient (and necessary) condition is that each bounded right-continuous martingale is continuous. This is true, for example, if the filtration is that of a Brownian motion.
The condition you suggest is too weak. Example: Let $U$ be uniformly distributed on $(0,1)$ and let $\xi$ be an independent random variable taking the two values $1$ and $-1$ with equal likelihood, both defined on some probability space $(\Omega,\mathcal F,\Bbb P)$. Take your filtration to be $\mathcal F_t:=\cap_{\epsilon>0}\mathcal F_{t+\epsilon}^o$, where $\mathcal F_t^o$ is defined by
$$
\mathcal F^o_t:=\sigma\{1_{\{U\le s\}}, 0\le s\le t; \xi1_{\{U\le t\}}),\qquad t\ge 0.
$$
Notice that $U$ is an $(\mathcal F_t)$-stopping time. Now define a continuous-path process $X$ by
$$
X_t=\cases{0,&$0\le t<U$,\cr \xi(t-U),&$U\le t$.\cr}
$$
This process generates $(\mathcal F_t)$ (modulo null sets) but the (right-continuous) martingale
$$
M_t:=\Bbb E[X_1|\mathcal F_t] = \cases{0,&$0\le t<U$,\cr X_1,&$U\le t$\cr}
$$
is not continuous.
can you recommend a reference for the sufficient and necessary condition that each bounded right-continuous martingale has to be continuous?
|
2025-03-21T14:48:31.620137
| 2020-07-27T11:47:34 |
366683
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Giorgio Metafune",
"Jochen Wengenroth",
"Narutaka OZAWA",
"erz",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/21051",
"https://mathoverflow.net/users/53155",
"https://mathoverflow.net/users/7591"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631511",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366683"
}
|
Stack Exchange
|
Supremum over which sets makes $H^{\infty}$ non-separable?
It is known that the space $H^{\infty}$ of bounded holomorphic functions on the unit disk $D$ is non-separable with respect to the supremum norm $\|\cdot\|_{\infty}^{D}$. Let $E\subset D$ be connected and not a singleton. Then, $\|\cdot\|_{\infty}^{E}$ (the supremum norm over $E$) is in fact a norm.
What are geometric conditions on $E$ equivalent to $H^{\infty}$ being non-separable with respect to $\|\cdot\|_{\infty}^{E}$?
If $\overline{E}\subset D$, then polynomials are dense with respect to $\|\cdot\|_{\infty}^{E}$, and so in this case $H^{\infty}$ is separable with respect to $\|\cdot\|_{\infty}^{E}$. Hence, we need $\overline{E}$ to intersect the unit circle $C$.
Carleson's interpolation theorem gives a sufficient geometric condition.
The proof I know to show the non separability of $H^\infty$ uses the functions $g_\xi (z)=\exp \left ((z+\xi)/(z-\xi) \right )$, $|\xi|=1$ . These are in $H^\infty$ and their mutual distance is bigger than 1; however this last is seen by taking radial (or non tangential limits). So we could consider the set $\Gamma$ of all points in $\partial D$ which are non tangential limits of points in $E$. If this set is uncountable, then $H^\infty$ is not separable under the norm $|\cdot |_\infty^E$.
@GiorgioMetafune I thought the same thing, but i don't have a good sense of the non-tangential limits. Also, perhaps this condition is too strong, it's not clear to me if $[0,1)$ works for my purpose..
You are right, the case of the interval is the first to be understood.
@JochenWengenroth thank you for suggestion! It was enough to completely answer my question
It also follows from Bourgain's theorem that $H^\infty$ is a Grothendieck space. MR0722264 (85j:46092) From this, non-separability follows if the restriction map from $H^\infty$ into $\ell_\infty(E)$ is not weakly compact, which can be checked when $\overline{E}$ intersects the unit circle.
Jochen Wengenroth suggested to look at Carleson's interpolation theorem, and it seems like it completely answers my question. Namely, the following is true.
Let $E$ be a subset of $D$. Then $H^\infty$ is non-separable with respect to $\|\cdot\|_\infty^E$ if and only if $\overline{E}$ intersects the unit circle.
Necessity follows from the fact that if $\overline{E}\subset D$, the polynomials are dense in $H^\infty$ with respect to $\|\cdot\|_\infty^E$, and so $H^\infty$ is separable in this case.
To show sufficiency let us recall the following corollary from the Carleson's interpolation theorem (combine theorems 9.1 and 9.2 from the book Duren - Theory of Hp Spaces): Let $\{z_n\}_n$ be a sequence in $D$ such that there is $0<c<1$ such that $1-|z_{n+1}|\le c(1-|z_{n}|)$, for every $n$. Then $\{z_n\}_n$ is an interpolation sequence, i.e. the operator $T:H^\infty\to l^\infty$ defined by $Tf=\{f(z_n)\}_n$ is a surjection.
Now fix some $z_1\in E$ and $0<c<1$. If $z_1,...,z_n$ are already chosen then $F=\{z\in D, ~|z|\ge 1-c(1-|z_n|)\}$ intersects with $E$ as $\overline{E}$ intersects the unit circle. Choose $z_{n+1}\in F\cap E$.
The sequence $B=\{z_n\}_n$ constructed this way satisfies the condition above, and so the operator $T$ is surjective. Moreover, $\|\cdot\|_\infty^B\le\|\cdot\|_\infty^E$, and so $T$ is a continuous map from $(H^\infty, \|\cdot\|_\infty^E)$ onto a non-separable Banach space. Thus, $(H^\infty, \|\cdot\|_\infty^E)$ is non-separable.
Remark. Using this fact one can now show that $H^\infty(U)$ is non-separable for every bounded open connected set $U$. Indeed, $(H^\infty(V),\|\cdot\|_\infty^U)$ embeds isometrically into $H^\infty(U)$, where $V$ is the complement to the unbounded component of $\mathbb{C}\backslash U$, and the former is non-countable as we have shown.
I missed the connectedness assumption in your question. The Carleson interpolation theorem gives a sufficient condition also for disconnected sets. But then necessity seems to be difficult.
I do not see the connectness assumption, too. If $E \subset D$ and its closure meets the boundary, you can always find a sequence $(z_n) \in E$ as above. Also, if $\bar E \subset D$, the density of polynomials follows without assuming connectness.
@JochenWengenroth you are right. Thank you!
@GiorgioMetafune you are also right! This is what happens when one has an extra nice condition - gets blinded somehow
|
2025-03-21T14:48:31.620434
| 2020-07-27T12:10:57 |
366685
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Moritz Firsching",
"Xie",
"https://mathoverflow.net/users/151117",
"https://mathoverflow.net/users/39495"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631512",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366685"
}
|
Stack Exchange
|
The best way for obtaining the canonical label of a matroid?
For a graph, we can canonical labelings with Nauty or other methods. But for a matroid, I do not find an easy way to labeling.
Dillon Mayhewa, Gordon F. Royle introduced the hyperplane graph of a matroid in their paper "Matroids with nine elements", which is the bipartite graph whose vertices are the elements and hyperplanes of the matroid and where a hyperplane-vertex is adjacent to an element-vertex if and only if the hyperplane contains the element. Then we can use Nauty to labelings the bipartite graph. But this method is a bit slow.
So I want to know is there any better method for canonical labeling of a matroid?
Depending on how your matroid is given, some other hash might be more suitable. How are your matroids given? Maybe relevant: http://www.cse.iitm.ac.in/~bvrr/jour/RS1.pdf
I find some sagemath codes in "https://oeis.org/A300985/a300985.sage.txt", where use canonical_label of an IncidenceStructure. It's enough for me.
|
2025-03-21T14:48:31.620553
| 2020-07-27T12:10:59 |
366686
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Kevin Carlson",
"Rune Haugseng",
"Simon Henry",
"Tim Campion",
"https://mathoverflow.net/users/1100",
"https://mathoverflow.net/users/22131",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/43000"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631513",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366686"
}
|
Stack Exchange
|
Enriched vs ordinary filtered colimits
Filtered categories can be defined as those categories $\mathbf{C}$ such that $\mathbf{C}$-indexed colimits in $\mathrm{Set}$ commute with finite limits.
Similarly, for categories enriched in $\mathbf{V}$ (where the appropriate notion of colimits is colimits weighted by enriched presheaves) one can define a presheaf $W \colon \mathbf{C}^{\mathrm{op}} \rightarrow \mathbf{V}$ to be ($\kappa$-)flat if $W$-weighted colimits in $\mathbf{V}$ commute with finite ($\kappa$-small) limits in $\mathbf{V}$ (for some regular cardinal $\kappa$). Borceux, Quinteiro, and Rosický take this as a starting point to develop a theory of accessible and presentable $\mathbf{V}$-categories in their paper "A theory of enriched sketches".
BQR show that in some ways flat weighted colimits are closely related to ordinary (conical) filtered colimits. For example, they show that if $\mathbf{C}$ has finite ($\kappa$-small) weighted limits, then a presheaf on $\mathbf{C}$ is ($\kappa$-)flat if and only if it is a ($\kappa$-)filtered ordinary colimit of representable presheaves. However, they give a counterexample that shows this need not be true for arbitrary $\mathbf{C}$ - but in this example it is still true that flat presheaves are filtered colimits of absolute colimits of representables.
Question 1: A $\kappa$-filtered ordinary colimit of absolute colimits of representables is always a $\kappa$-flat presheaf. Is anything further known (or expected) about the other direction, i.e. whether every $\kappa$-flat presheaf can be decomposed as such a colimit (or some variant involving two cardinals)?
Let me add a second closely related question that indicates why one might care about the first one. BQR prove that if $\mathbf{M}$ is a presentable $\mathbf{V}$-category then its underlying ordinary category is also presentable.
Question 2: Suppose $\mathbf{M}$ is a cocomplete $\mathbf{V}$-category whose underlying category is presentable. Does this imply that $\mathbf{M}$ is a presentable $\mathbf{V}$-category?
(This would be the case if the two classes of presheaves in the first question coincide.)
I doubt that Question 2 has an affirmative answer in general. In my head, the kingdom of Benabou cosmoi $(V, \otimes, I)$ is broken into four phyla, according to the answers to the following two questions: a.) Is $(V,\otimes, I)$ cartesian? and b.) Is $Hom(I,-): V \to Set$ conservative? If the answer to (b) is negative (e.g. when $V = (sSet, \times, \Delta[0])$ or $V = ([0,\infty], +, 0)$), then I think that the answer to your Question (2) is probably negative.
@TimCampion Is your second $V$ presentable?
@RuneHaugseng Any idempotent complete small category is accessible, and $[0,\infty]$ is cocomplete, so it's locally presentable.
@KevinArlin That doesn't sound quite right - an accessible category is required to have $\kappa$-filtered colimits for some regular cardinal $\kappa$. It should also be generated under $\kappa$-filtered colimits by a set of $\kappa$-compact objects, but in this example I guess every object is actually compact.
@RuneHaugseng No, it's really true. Try letting $\kappa$ be the cardinality of the set of morphisms.
Is the point that if you choose $\kappa$ big enough then any $\kappa$-filtered diagram is forced to eventually be constant?
It's something like that. If $J\to \mathcal C$ is $\kappa-$filtered where $|\mathcal C|<\kappa$ then there's a subgraph $J'\to J$ with $|J'|<\kappa$ such that the image of $J'$ coincides with the image of $J$, then a cocone over $J'$ is a cocone over $J$, and any diagram containing a cocone over itself has a retract of that cocone as a colimit.
For Q1: something related is dealt with in a context more general than the classical one by Adamek, Borceux, Lack and Rosicky in their paper "A classification of accessible categories". They replace finite or $\kappa$-small limits with an arbitrary class of limits $\mathbb{D}$, and consider a condition which they call soundness, one of whose consequences is a decomposition of every $\mathbb{D}$-flat weight as a suitably
"$\mathbb{D}$-filtered" colimit of representables.
This is all in the unenriched context, which is not what you want, but the point is that they make axiomatic assumptions which are more or less exactly what is needed to force the answer to your question 1 to be true. Make of that what you will, but it at least suggests that it's not automatic, and will probably require a bespoke argument in each situation.
For Q2: No. I guess the classical reference is Kelly's "Structures defined by finite limits in the enriched context." If $\mathcal V$ is a symmetric monoidal closed category which is locally $\kappa$-presentable as a closed category (i.e., it is locally $\kappa$-presentable and the $\kappa$-presentable objects are closed under the monoidal structure), then there is a good notion of locally $\lambda$-presentable $\mathcal V$-category: they are precisely the cocomplete $\mathcal V$-categories, whose underlying ordinary categories are locally $\lambda$-presentable, and whose $\lambda$-presentable objects are closed under tensors (=copowers) with $\lambda$-presentable objects of $\mathcal V$. Without this last condition, there is a gap through which to thread a negative answer to your question.
EDIT
I agree with Simon that if $\mathcal{C}$ is a cocomplete $\mathcal{V}$-category whose underlying category is locally presentable, then one can always find some $\kappa$ such that $\mathcal{C}$ is locally presentable as a $\mathcal{V}$-category, meaning that $\mathcal{C}_0$ is locally $\kappa$-presentable and the $\kappa$-presentable objects are closed under tensors by $\kappa$-presentable objects of $\mathcal{V}$.
Here, by saying that $X \in \mathcal C$ is $\kappa$-presentable, I just mean that $\mathcal C(X,\text{-}) \colon \mathcal{C} \rightarrow \mathcal V$ preserves conical filtered colimits (this is Kelly's definition). As Rune says, one could also talk of $X \in \mathcal C$ being $\kappa$-compact, meaning that $\mathcal C(X,\text{-})$ preserves $\kappa$-flat colimits. Since there is in no reason to believe that every $\kappa$-flat weight is a $\kappa$-filtered conical colimit of representables, these two notions will in general be distinct.
However, they coincide when $\mathcal C$ is locally $\kappa$-presentable as a $\mathcal V$-category: so all the $\kappa$-presentable objects are $\kappa$-compact in the enriched sense. This is actually in the BQR paper you cite (Lemma 6.5) and follows from the following fact. Let us write $\mathcal A$ for the essentially small full subcategory of $\kappa$-presentable objects. Clearly $\mathcal A$ has $\kappa$-small colimits, and $\mathcal C$ is the free completion $\kappa\text-\mathbf{Filt}(\mathcal A)$ of $\mathcal A$ under conical $\kappa$-filtered colimits. But in fact, $\mathcal C$ is also the free completion $\kappa\text-\mathbf{Flat}(\mathcal A)$ of $\mathcal A$ under $\kappa$-flat colimits. Given this, a functor out of $\mathcal C$ preserves conical $\kappa$-filtered colimits iff it is the left Kan extension of its own restriction to $\mathcal A$, iff it preserves $\kappa$-flat colimits: in particular, $\kappa$-presentability and $\kappa$-compactness in $\mathcal C$ will coincide.
That $\kappa\text-\mathbf{Flat}(\mathcal A) = \kappa\text-\mathbf{Filt}(\mathcal A)$ is proven in Theorem 6.11 of Kelly's "Structures defined by...", or equally by Prop 4.5 of BQR (as you mention in your question.)
Regarding to Q2, I'm not sure you are fully correct: My understanding is that while there is indeed a mismatch between locally $\lambda$-presentable in the enriched and unenriched case, it is still is the case that if $V$ itself is locally presentable presentable, then local presentability in the enriched case and unenriched case are equivalent, just for possibly different value of $\lambda$. The point is that if $M$ is $V$ enriched and locally presentable, then the tensor $V \otimes M \to M$ is accessible, hence preserve $\lambda$-presentable object for some $\lambda$.
I think @SimonHenry is right - you can always find some $\kappa$ such that the full subcategory of $M$ consisting of objects that are $\kappa$-compact in the underlying category of $M$ has $\kappa$-small $V$-colimits and generates $M$ under (conical) $\kappa$-filtered limits. However, it seems unclear whether you can always choose $\kappa$ such that these objects are also $\kappa$-compact in the enriched sense, i.e. mapping out commutes with $\kappa$-flat weighted limits, which would be required for $M$ to be $V$-presentable.
We deal with Question 1 in this paper, joint with Steve Lack.
In general it is not true that every $\kappa$-flat presheaf lies in the closure of the resperesntables under $\kappa$-filtered colimits and absolute colimits. That's the case for example when $\mathcal V=\mathbf{Set}^G$, for a non-trivial finite group $G$ (see Section 5 of the paper).
However, there are many bases of enrichments for which the equality holds. In fact, whenever $\mathcal V$ is locally $\kappa$-presentable as a closed category and $\mathcal V(I,-)$ is (weakly) cocontinuous and (weakly) strong monoidal, then every $\kappa$-flat presheaf is a $\kappa$-filtered colimit of representables. (In this case absolute colimits reduce to the usual splittings of idempotents.) Therefore the existence and preservation of $\kappa$-flat colimits is equivalent to that of $\kappa$-filtered colimits in any $\mathcal V$-category (for $\mathcal V$ as above). Examples of such bases are the cartesian closed categories $\mathbf{Set}$ of sets, $\mathbf{Cat}$ of small categories, $\mathbf{SSet}$ of simplicial sets, $\mathbf{2}$ of the free-living arrow, $\mathbf{Pos}$ of partially ordered set, etc.
We also investigate the case when $\kappa$-flat colimits are generated by $\kappa$-filtered colimits plus other absolute colimits (such as finite direct sums and copowers by dualizable objects). Examples of $\mathcal V$ for this setting include the monoidal categories $\mathbf{CMon}$ of commutative monoids, $\mathbf{Ab}$ of abelian groups, $R\textit{-}\mathbf{Mod}$ of $R$-modules for a commutativering $R$, and $\mathbf{GAb}$ of graded abelian groups.
|
2025-03-21T14:48:31.621446
| 2020-07-27T12:26:19 |
366687
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631514",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366687"
}
|
Stack Exchange
|
Conjecture about minimal number of edge crossings in complete bipartite graphs
I am interested in the status of the conjecture about the minimum number of edge crossings $cr(K_{m,n})$ in a drawing of the complete bipartite graph $K_{m,n}$.
The Wikipedia article https://en.wikipedia.org/wiki/Tur%C3%A1n%27s_brick_factory_problem led me to study the original papers of Zarankiewicz (On a problem of P. Turan concerning graphs) from 1954 and of Urbanik (Solution du problème posé par P. Turán) from 1955.
I wondered whether someone could tell me whether an asymptotic approach has been successfully attempted (letting $n\to\infty$). If so, I would be very interested in any references for that.
The Electronic Journal of Combinatorics has many Dynamic Surveys one of which is The Graph Crossing Number and its Variants: A Survey by Schaefer which first appeared in 2013 and has been updated as recently as Feb 14, 2020. From the bottom of page 40 onto page 41 you will find this conjecture for complete bipartite graphs discussed (with many references). As far as I can tell from the survey the conjecture is open (both for exact values and asymptotically).
One paper you might be interested in is Zarankiewiczʼs Conjecture is finite for each fixed $m$ by Christiana, Richter, and Salazar. This paper shows that for each $m$ if the conjecture holds up for all $n$ up to some very large $N(m)$ (which is an explicit value), then the conjecture is true for $K_{n,m}$ with any $n$.
This survey also references another survey Turán’s Brick Factory Problem: The Status of the Conjectures of Zarankiewicz and Hill by Székely. (I haven't be able to access this survey so I don't exactly what's inside.)
It is a fascinating conjecture. The following might be a good reference for you: In 1997, Richter & Thomassen showed that
$$\lim_{n\to\infty}cr(K_{n,n})\left(\begin{array}{c} n \\ 2 \end{array}\right)^{-2}$$
exists and is at most $1/4$. If the conjecture is true, the value of this limit is exactly $1/4$.
(R.B. Richter, C. Thomassen, "Relations between crossing numbers of complete and complete bipartite graphs" Amer. Math. Monthly , 104 (1997) pp. 131–137)
As of 2017, the exact value of this limit is still not known, see for example Gethner, Hogben, Lidick, Pfender, Ruiz, Young: „Crossing numbers of complete tripartite and balanced complete multipartite graphs“" Journal of Graph Theory 84, no. 4 (2017): 552-565. DOI: 10.1002/jgt.22041
|
2025-03-21T14:48:31.621646
| 2020-07-27T12:43:10 |
366689
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris Wuthrich",
"Yaakov Baruch",
"https://mathoverflow.net/users/2480",
"https://mathoverflow.net/users/5015"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631515",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366689"
}
|
Stack Exchange
|
Proof of a 'well-known' result of Shimura on periods of modular forms
It is often noted in the literature that there are certain complex periods that allow one to normalize the modular symbol associated to a modular form in such a way that its coefficients are algebraic. This process of normalization by complex periods is regularly attributed to Shimura, though I can't seem to find a concrete reference explaining this result.
More precisely, let $
\Gamma=\Gamma_0(N)$ and fix an eigenform $f\in S_k(\Gamma)$. The modular symbol $\xi_f\in \operatorname{Hom}_{\Gamma}(\operatorname{Div}^0(\mathbb{P}^1(\mathbb{Q})),V_{k-2}(\mathbb{C}))$, where $V_{k-2}(\mathbb{C})$ is the space of homogeneous polynomials with complex coefficients of degree $k-2$, attached to $f$ is defined by
$$
\xi_f(\{r\}-\{s\})=2\pi i \int_s^r f(z)(zX+Y)^{k-2}dz.
$$
One can expand this into a homogeneous polynomial $\sum_{j=0}^{k-2} c_jX^jY^{k-2-j}$ where $c_j=\binom{k-2}{j}2\pi i \int_s^rf(z)z^jdz$.
The matrix $\begin{pmatrix} -1 &0\\ 0&1\end{pmatrix}$ normalizes $\Gamma$,
so the modular symbols come equipped with an involution, and hence there is a unique eigenspace decomposition $\xi_f=\xi_f^++\xi_f^-$, with $\xi^\pm$ in the $\pm 1$-eigenspace.
The following theorem is stated in the literature (see, for example, [Greenberg-Stevens, 3.5.4], [Bertolini-Darmon,1.1], or [Pollack-Weston,page 7]).
Theorem. There exists complex numbers $\Omega_f^\pm$ such that $\xi_f^\pm/\Omega_f^\pm$ takes values in $V_{k-2}(K_f)$, where $K_f$ is the number field generated by the Fourier coefficients of $f$.
Greenberg-Stevens cite this 1977 paper of Shimura, Pollack-Weston cite Shimura's book on automorphic functions, and the Bertolini-Darmon does not give a reference. I could not find anything helpful in Shimura's automorphic function book, but I think theorem 1 from the 1977 paper is probably what we want. For simplicity, I state it below in the case where $f$ has rational coefficients.
Theorem. (Shimura, Theorem 1) Fix a primitive Dirichlet character $\chi$. There exist complex numbers $u_f^\pm$ such that
$$
\frac{L(f_\chi,j)}{u_f^\epsilon\tau(\chi)(2\pi i)^j}\in K_fK_\chi
$$
where $0< j< k$, $\epsilon$ is the sign of $\chi(-1)(-1)^j$, $\tau(\chi)$ is the classical Gauss sum, and $L(f_\chi,s)=\sum\chi(n)a_nn^{-s}$ is the $L$-function of $f$ twisted by $\chi$.
In fact, Shimura gives precise (though rather noncanonical) descriptions of these periods $u_f^\pm$: they are essentially the value of the $L$-function at $k-1$.
I would like to know how the first theorem stated above follows from this theorem 1 of Shimura.
It seems like a nontrivial exercise, or perhaps I am just having some trouble connecting the dots. I would also be content to see a reference which outlines a proof of the first theorem above.
My thoughts are roughly the following. With the notation as above, let $m$ be the conductor of $\chi$. I know that (see [Mazur-Tate-Teitelbaum, 8.6], for example) one has the following connection between coefficients of the modular symbols and special values of $L$-functions
$$
\frac{j!}{(-2\pi i)^{j+1}}\frac{m^{j+1}}{\tau(\bar \chi)}L(f_{\bar\chi},j+1) =\sum_{a\in (\mathbb{Z}/m\mathbb{Z})^\times}\chi(a)\int_{-a/m}^{i\infty}f(z)(mz+a)^j dz,
$$
for $0\leq j \leq k-2$. This tells us, for example, that certain weighted sums of the coefficients of $\xi_f(\{\infty\}-\{-a/m\})$ can be scaled to be algebraic. Even more, after writing down the symbols $\xi_f^\pm$, I can find periods $\Omega_f^\pm$ such that, roughly speaking, $$
\frac{1}{\Omega_f^\pm}\sum\chi(a)(\text{$j$th coefficient of $\xi_f^\pm(\{\infty\}-\{a/m\}$}) )
$$
is algebraic, but again, this only tells me that (a) certain weighted sums of the coefficients are algebraic, and (b) only gives information about the modular symbol evaluated at $\{\infty\}-\{a/m\}$, which as far as I can tell, is not the generality needed for the first theorem above.
(I posted this question on MSE a few days ago, but did not have much luck there. I hope re-posting it here is not too much of a faux pas.)
For (b): The path from $r$ to $s$ can be split up into the path from $r$ to $i\infty$ and the path from $i\infty$ to $s$.
For (a): If $\sum_a \chi(a) X_a = Y_\chi$ then $X_a = 1/d \sum_{\chi} \bar{\chi}(a) Y_{\chi}$. But maybe I misunderstand the actual question
Having ploughed through a paper of Shimura many years ago I don't know if MSE was really the right place for this question!
You can find a proof of this theorem (the first in the OP) in Section 5.3 of the following paper by Pasol and Popa: https://arxiv.org/abs/1202.5802
The idea is to use the action of Hecke operators. More precisely, the map $f \mapsto \xi_f^{\pm}$ is Hecke-equivariant, the Hecke operators preserve the rational structures of both sides, and the eigenspaces are 1-dimensional.
This theorem could also, in principle, be deduced from Shimura's theorem (Theorem 1 in the OP), but the proof I have in mind would be very technical. The idea is to start from the formula expressing the values $L(f,\chi,j+1)$ in terms of modular symbols and then take the inverse Fourier transform. But there are many technical problems due to the fact that the Dirichlet characters are not necessarily primitive, and Shimura's formula is a priori only for primitive characters. Nevertheless, in the case of weight 2, Merel has proved a completely general formula expressing modular symbols in terms of twisted $L$-values, see the article Symboles de Manin et valeurs de fonctions $L$.
|
2025-03-21T14:48:31.622139
| 2020-07-27T14:14:12 |
366700
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Benoît Kloeckner",
"https://mathoverflow.net/users/4961"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631516",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366700"
}
|
Stack Exchange
|
Characterization of metrics such that the volume of balls doesn't depend on their centers?
Given a finite metric space $(X,d)$,
when does it hold that for all $y\in X$ and $r>0$, $\#B(y,r)$ does not depend on $y$?
Here $ B(y,r):=\{x\in X: d(x,y)\le r\} $ denotes a ball of radius $r$ centered at $y$,
and $\#A$ denotes the cardinality of $A$.
An example that does not satisfy the above property is $Z$-distance on $\{0,1\}^n$.
It is defined as $ d_Z(x,y) := \max\{\#(\mathrm{supp}(x)\setminus\mathrm{supp}(y)),\#(\mathrm{supp}(y)\setminus\mathrm{supp}(x))\} $ for $x,y\in\{0,1\}^n$.
Here $ \mathrm{supp}(x):=\{i\in[n]:x_i\ne0\} $ denotes the support (set of nonzero locations) of a vector $x\in\{0,1\}^n$.
(The notation $Z$-distance is perhaps nonstandard. My motivation comes from coding theory, in particular error-correcting codes for $Z$-channels which only zero out bits but do not flip zeros to ones. One can check that $Z$-distance is indeed a metric.)
All normed spaces satisfy the above property.
It is easy to see that the above property is equivalent to the condition: for all $y,z\in X$ and $r>0$, $ \#(B(y,r)\setminus B(z,r)) = \#(B(z,r)\setminus B(y,r)) $.
However, I couldn't further simply it.
My question is: is there, or is it possible to have, a characterization of metrics satisfying the above property?
What kind of characterization are you looking for? All homogeneous space satisfy the condition, for example any group endowed with the length metric of a system of generators.
I do not know the answer in general, but in the case in which $X$ is a subset of $\mathbb{R}^n$ there are some known remarkable results:
If $X\subset\mathbb{R}^2$ is bounded, then $X$ consists of vertices of
a regular $n$-gon or $X$ is a union of the vertices of two regular
$n$-gons having having the same center and radius.
If $X\subset\mathbb{R}^n$ is bounded, then $X$ is contained in a sphere.
You can find proofs of these and other results in
B Kirchheim, D. Preiss,
Uniformly distributed measures in Euclidean spaces.
Math. Scand.
90 (2002), no. 1, 152–160.
The proofs are surprisingly difficult and deep. In view of these results I would expect that it is difficult to find a general and satisfactory answer.
|
2025-03-21T14:48:31.622319
| 2020-07-27T15:38:44 |
366707
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mikhail Borovoi",
"https://mathoverflow.net/users/4149"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631517",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366707"
}
|
Stack Exchange
|
Comparison between spinor representations in $\operatorname{SL}(2,\mathbb C)=\operatorname{Spin}(1,3)$ and $\operatorname{Spin}(4)$
$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SU{SU}$We know that
$$
\Spin(1,3)=\SL(2,\mathbb C)
$$
and
$$
\Spin(4)=\SU(2) \times \SU(2).
$$
The $\Spin(1,3)$ is the Lorentz version of Spin group, while
$\Spin(4)$ is the Euclidean version of Spin group.
The spinor representations of $\Spin(4)$ is essentially the spinor representation of $\Spin(4)=\SU(2) \times \SU(2)$ thus it is labeled by two components of spinors, each for the spinor representation of $\SU(2)$.
How do we understand the spinor representations of $\Spin(1,3)$? Note that $\Spin(1,3)$ is non-compact. How is this $\Spin(1,3)$ spinor related to the spinor of the compact group
$\Spin(4)$?
How do we understand the 2 dimensional spinor as complex Weyl and 4 dimensional spinor as complex Dirac but NO 4 dimensional real Majorana in $\Spin(4)$?
It is also said $\Spin(1,3)$ is the complexification of $\SU(2)$. How to see this from spinor representation living in the vector spaces?
How do we understand the 2 dimensional spinor as complex Weyl and 4 dimensional spinor as complex Dirac or real Majorana in $\Spin(3,1)$?
See https://en.wikipedia.org/wiki/Spinor#Spinors_in_representation_theory
Refs from google search:
http://www-personal.umich.edu/~williams/notes/spinor.pdf
http://scipp.ucsc.edu/~haber/ph251/Spinor_Shijun
https://en.wikipedia.org/wiki/Complexification_(Lie_group)
Of your three questions I understand only the first one. I understand the two "spinor" representations of the group $G={\rm SL}(2,{\Bbb C}) $ (regarded as a real algebraic group) as the two 2-dimensional complex representations $$\rho,\bar\rho,\colon G\to {\rm GL}(2,{\Bbb C})$$ given by the formulas $,\rho(g)=g,$ and $,\bar\rho(g)=\bar g$, where the bar over $g$ denotes the complex conjugation.
|
2025-03-21T14:48:31.622455
| 2020-07-27T15:41:28 |
366708
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Connor Malin",
"https://mathoverflow.net/users/134512"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631518",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366708"
}
|
Stack Exchange
|
Compactly supported chern character
It is a standard result that for a CW complex $X$, the chern character
$$\text{ch}: K^*(X)\otimes_{\mathbb{Z}} \mathbb{Q}\to H^*(X,\mathbb{Q})$$
induces an isomorphism. Suppose now that $X$ is an open manifold and consider the chern character with compact support
$$\text{ch}_{\text{cs}}: K^*_{\text{cs}}(X)\otimes_{\mathbb{Z}} \mathbb{Q}\to H^{*}_{\text{cs}}(X,\mathbb{Q})$$
Is it still true that it is an isomorphism? It seems to be so for the case of $\text{Tot}(E\to S)$ where $S$ is a compact space and $E$ is a vector bundle. I suppose this is well known, but I couldn't find a reference.
Compactly supported cohomology is the same as reduced cohomology of the one point compactification (for reasonable spaces at least). So the result for compactly supported is implied by the result for usual.
Yes, this is true. For any generalized cohomology theory $E$, the compactly supported $E$-cohomology of a space $X$ is
$$E_{\mathit{cs}}^*(X) := \varinjlim\limits_{K\subseteq X:\text{ $K$ compact}} E^*(X, X\setminus K).$$
The Chern character is a natural isomorphism of cohomology theories, so is compatible with the limit above, hence induces an isomorphism on compactly supported rational $K$-theory.
One reference for this definition of compactly supported generalized cohomology is Ranicki-Roe, “Surgery for Amateurs”, Remark 2.1.
|
2025-03-21T14:48:31.622583
| 2020-07-27T16:09:50 |
366711
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"https://mathoverflow.net/users/35959",
"https://mathoverflow.net/users/36721",
"user64494"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631519",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366711"
}
|
Stack Exchange
|
Uniform boundedness of integral?
I have perhaps a very simple question where I lack some inutition at the moment: Is the expression
$$\sup_{\alpha < 0, \lambda \in \mathbb N}\int_{-\infty}^{\alpha} e^{-\lambda t^4} \ dt \int_{\alpha}^0 e^{\lambda t^4} \ dt $$
finite or not?-My main concern is $\alpha$ close to zero since $t^4$ is not strongly convex (this is something that would've bailed me out I think, so for Gaussians I think I could show this boundedness).
Let $a:=\alpha$ and $x:=\lambda$. Using now the substitution $t=x^{-1/4}s$ and introducing $b:=x^{1/4}a$, we see that the problem is to show that $\sup_{b\le0,x\ge1}x^{-1/2}I(b)J(b)=\sup_{b\le0}I(b)J(b)<\infty$, where
$$I(b):=\int_b^0 e^{s^4}\,ds=\int_0^{-b}e^{s^4}\,ds,\quad J(b):=\int_{-\infty}^b e^{-s^4}\,ds$$
for $b\le0$.
By l'Hospital's rule,
$$I(b)\sim-b^{-3}e^{b^4}/4,\quad J(b)\sim -b^{-3}e^{-b^4}/4$$
as $b\to-\infty$. Thus, $I(b)J(b)\to0$ as $b\to-\infty$. Also, the functions $I$ and $J$ are continuous and hence locally bounded. Thus, indeed $\sup_{b\le0}I(b)J(b)<\infty$.
Do you take into account that $dt=\lambda^{1/4},ds$? I don't see $\lambda^{1/4}$ in the above.
Sorry, I don't see any denominator in the above. Could you elaborate your comment? TIA.
@user64494 : Oops! I have now inserted the missing factor $x^{-1/4}x^{-1/4}=x^{-1/2}$. Since this factor is bounded for $x\ge1$, it does not affect the conclusion.
Iosif Pinelis, can you do the right edit, in particular, defining $y$? I think, $y=b$ in your notation.
@user64494 : $y$ is a dummy variable, and thus can be replaced by any symbol. Later, we substitute $|b|$ for $y$. I only did not want to use $b$ in place of $y$, because $b<0$ and hence $b\ne|b|$.
@user64494 : I have removed $y$ and rewritten everything only in terms of $b$ (and $x$).
One more $y$ should be replaced in the above.
@user64494 : Thank you. Have done this too.
It should be $\le$ instead of $=$ in line 3.
@user64494 : The equality is correct, too, because of the $\sup$ on both sides of it.
Mathematica 12.0 does the job by
Integrate[Exp[\[Lambda]*t^4],{t, \[Alpha], 0}, Assumptions->\[Alpha]<0 && \[Lambda] >= 1]*
Integrate[Exp[-\[Lambda]*t^4],{t,-Infinity,\[Alpha]},Assumptions->\[Alpha]<0&&\[Lambda]>=1]
$$-\frac{(-1)^{3/4} \alpha E_{\frac{3}{4}}\left(\alpha ^4 \lambda \right) \left(\Gamma \left(\frac{1}{4},-\alpha ^4 \lambda \right)-\Gamma \left(\frac{1}{4}\right)\right)}{16 \sqrt[4]{\lambda }} $$
NMaximize[{%,\[Alpha]<0&&\[Lambda]>=1},\[Alpha],\[Lambda]},Method-> "DifferentialEvolution"]
$$\{0.209323,\{\alpha \to -0.476784,\lambda \to 1.\}\}$$
Addition. Maple confirms it by
DirectSearch:-Search((alpha, lambda) -> int(exp(-lambda*t^4), t = -infinity .. alpha, numeric)*int(exp(lambda*t^4), t = alpha .. 0, numeric), {-100 <= alpha, 1 <= lambda, alpha <= 0, lambda <= 100}, maximize);
$$[ 0.209323347704846, \left[ \begin {array}{c} - 0.476781454615864297
\\ 1.00000000002946488\end {array} \right] ,117]
$$
You know that NMaximize does not guarantee a true global maximum, right?
@Iosif Pinelis: Maple produces the same.
Does that Maple numerical routine guarantee a true global maximum? Does the coincidence (up to some, possibly high, accuracy) of the Mathematica and Maple results guarantee a true global maximum?
@Iosif Pinelis: Thank you for your interest to the answer. Of course, neither DirectSearch nor NMaximize guarantees a true global maximum for all cases. The default accuracy of the Search command is $10^{−6}$ both for maximum value and variables. The coincidence of the results produced by two independent CASes suggests that the upper bound under consideration does equal ≈0.2093, no more and no less
|
2025-03-21T14:48:31.622825
| 2020-07-27T16:12:46 |
366712
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerhard Paseman",
"https://mathoverflow.net/users/3402"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631520",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366712"
}
|
Stack Exchange
|
Ramsey's number R(4,4) with arithmetic progressions
Can 17 positive integers in arithmetic progression be found such that that no four of them have, pairwise, a common divisor greater than 1, but, likewise, no four of them are, pairwise, relatively prime?
Because R(4,4)=18, 18 such numbers are impossible.
At https://puzzling.stackexchange.com/questions/100391/seventeen-positive-integers/100477#100477 it has been shown that 17 numbers with the required property, but not necessarily in arithmetic progression, can be easily found based on Paley's graph of order 17. The question arises: could those 17 numbers be in arithmetic progression?
Do you have an example for five numbers for R(3,3)? Gerhard "Likes To Start Out Small" Paseman, 2020.07.27.
No, there cannot be 17 such numbers in arithmetic progression (and there cannot be 5 such numbers with the corresponding property for triples).
Suppose we have such an arithmetic progression of length $k$, say $x,x+d,\ldots,x+(k-1)d$. I claim that if a prime $p$ divides any two of them then either it divides all of them (which cannot be the case), or else $p<k$.
Indeed, if $p\mid x+id$ and $p\mid x+jd$ for $0\leq i<j<k$ then $p\mid \ell d$ for some $1\leq \ell<k$. If $p\mid d$ then $p\mid x$ and so all members of the AP are divisible by $p$. Therefore $p\mid \ell$ for some $1\leq \ell <k$, and hence $p<k$.
When $k=5$, the only possibilities are $p=2$ or $p=3$. Consider the graph on these 5 numbers. If there is no triangle/anti-triangle in this graph, every vertex has degree exactly two, and in particular must be divisible by either 2 or 3. By the pigeonhole principle at least 3 of the numbers are divisible by one of them, so there are three numbers with a common non-trivial divisor, contradiction.
Similarly, when $k=17$, the only primes are $2,3,5,7,11,13$. If we draw a graph as above, then to avoid a monochromatic 4-clique it must be the Paley graph of order 17, and in particular every vertex has degree at least one. Colour each vertex by the primes out of $2,3,5,7,11,13$ which divide that vertex, so this is a 6-colouring of the vertices (where a vertex may receive multiple colours). Clearly no colour can appear on more than 3 such vertices, and each vertex receives at least one colour.
It is easy to see that 5 of the colours must appear on exactly 3 of the vertices (since $2\times 2+4\times 3<17$), and the remaining colour appears on the remaining 2 vertices and at most one of the other 15 vertices. Choosing one vertex from each of the 4 colour classes of size 3 that don't receive this 6th colour produces 4 vertices that are pairwise jointly divisible by none of the primes $2,3,5,7,11,13$, and hence pairwise have no common divisor.
|
2025-03-21T14:48:31.623024
| 2020-07-27T16:15:36 |
366714
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Daniel Loughran",
"Dror Speiser",
"Jeremy Rouse",
"Stanley Yao Xiao",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/2024",
"https://mathoverflow.net/users/48142",
"https://mathoverflow.net/users/5101"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631521",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366714"
}
|
Stack Exchange
|
Quadratic non-residues in elliptic divisibility sequences
Let $E: y^2 = x^3 + ax + b$ be an elliptic curve over $\mathbb{Q}$ with $a,b \in \mathbb{Z}$. Recall that any rational point $P = (x,y)$ can be written uniquely as $P = (u/d^2, v/d^3)$ with $u,v,d \in \mathbb{Z}, d > 0, \gcd(u,d) = \gcd(v,d) = 1$. We write $d(P)$ for the number appearing in the denominator.
The sequence $d(nP)$ for $n \in \mathbb{N}$ is the elliptic divisibility sequence associated to $P$. My question is about existence of quadratic non-residues in these sequences modulo all but finitely many primes $p$.
Let $P$ be a point of infinite order. Does there exists a finite set of primes $S$ such that for all $p \notin S$, there is $n \in \mathbb{N}$ such that
$$\left( \frac{d(nP)}{p}\right) = -1 \quad ? \quad (*)$$
Remarks:
It is known that the sequence $d(nP)$ contains only finitely many squares, by Theorem 1.1 of
Everest, Reynolds, Stevens - On the denominators of rational points on elliptic curves.
In particular, for all but finitely
many $n$, one finds that $d(nP)$ is a quadratic non-residue modulo half of all primes.
I want to know that as one varies $n$ and takes all these primes together, one only
excludes finitely many primes.
There is a slightly different definition of elliptic divisibility sequences
in terms of division polynomials and recurrence sequences, which can differ
from the above sequence by a sign and some small primes (this is the original
definition of Ward). I'm also interested in the analogous problem
for such sequences.
Is this known for Lucas sequences?
Good question. I don't know.
Is it even known that the primes excluded have density zero?
@Stanley Yao Xiao up to some work reconciling the definition difference, I think so: combine the main result in Reductions of Points on Elliptic Curves by Akbary et al, with the main result in Character Sums with Division Polynomials by Shparlinski and Stange.
We already have some ideas how to prove this "density zero" result, so are really looking for a proof that there are only finitely many exceptions.
It appears that for a fixed $E$ and $P$, the primes for which there is no $n$ so that $\left(\frac{d(nP)}{p}\right) = -1$ can be reasonably large. I just looked at the example of $E : y^{2} = x^{3} - 2x$ and $P = (2,2)$, and found that there is no such $n$ for $p = 17$, $257$, $1009$, $1361$, $26881$, and $141041$.
Still, heuristically the number of such $p$ should be finite: each $p$ divides $d(n_pP)$ for some $n_p$, and then there's periodicity. $\left(\frac{d(nP)}{p}\right)=1$ for all $n<n_p$ occurs with chance $2^{n_p}$. So we can estimate an upper bound for the number of "bad" primes: $\sum_n \omega(d(nP))/2^n$, where $\omega(k)$ is the number of distinct primes dividing $k$. Since $d(nP)=O(e^{cn^2})$ for some $c$, and $\omega(k)=O(\log{k})$, this sum is finite.
|
2025-03-21T14:48:31.623222
| 2020-07-27T16:48:00 |
366715
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631522",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366715"
}
|
Stack Exchange
|
Deterministic Cipolla's algorithm
Cipolla's algorithm for computing square roots in a finite field
(Wikipedia)
requires finding $x$ such that $x^2 - a$ is not a square (for computing a square root of $a$). The usual way to do this is to try random values of $x$ until one verifies this condition.
What about trying $x = 0, 1, 2, ...$ (assuming the field is $GF(p)$ for prime $p$)? What is known about the smallest integer that verifies the condition (possibly assuming GRH)?
In Tonelli–Shanks' algorithm, we would instead be looking for $y$ that is not a square, and assuming GRH one can prove that the smallest such $y$ in $O(\log(p)^2)$ (even $\leq 2 \log(p)^2$ IIRC, this is due to Bach and can be found in
Prime Numbers A Computational Perspective
by Crandall and Pomerance). Without GRH, some much weaker bounds are available ($O(p^{1/2})$ is easy IIRC).
My question is whether a similar bound is known for the variant of Cipolla's algorithm that I described above.
Note that I'm not asking whether it is advisable to implement the algorithm like this. Selecting $x$ randomly is certainly easy enough and theoretically well grounded. However, this question seems natural to me.
I'm pretty sure that no good (meaning, polynomial in $\log(p)$) bound is known without GRH, since it would yield a simple, fully deterministic, polynomial time, algorithm for computing square roots in finite fields, and this would be widely known.
I tried reducing my question to Bach's result, but got stuck. If $z^2 = a$, then $x^2 - a$ is a square iff $x = z$ or $\frac{x + z}{x - z}$ is a square, and $f: x \mapsto \frac{x + z}{x - z}$ is a bijection on the projective line. This makes it easy to get a probability for a random $x$ to verify the condition, but $f(x)$ is not necessarily "small" when $x$ is, so I can't relate this to Bach's.
|
2025-03-21T14:48:31.623364
| 2020-07-27T17:30:26 |
366719
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerhard Paseman",
"Mark Lewko",
"Pietro Majer",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/630"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631523",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366719"
}
|
Stack Exchange
|
How many squares can be formed by $n$ points in general position in the plane?
[This is much in the spirit (but different from) the questions from different posters: How many squares can be formed by using n points? and How many squares can be formed by using n points: revisited?]
Let $A$ be a set of $n$ points in the plane in general position. By general position we mean that no $3$ points are co-linear. What is the maximum number of squares that can be formed with vertices in $A$?
I note that there are trivial upper and lower bounds for this problem:
[Trivial Upper Bound] Given $n$ arbitrary points in the plane, noting that any two points determine at most $3$ squares it follows that there are at most $O(n^2)$ squares with vertices in $A$.
[Trivial Lower Bound] Place four points at the corner of a square, and repeat taking care to avoid all lines generated by pairs of points already placed in the plane until we've placed $n$ points. This clearly gives a lower bound of $\Omega(n)$.
I can improve the implied constant in both the upper and lower bound by being a bit more clever. The problem, however, is to
Improve (asymptotically) on either the upper or lower bound just given.
Since there can't be three co-linear points, in fact two points determine at most 1 square (either an edge or a diagonal). Since a square has 6 pairs of vertices, this gives a trivial upper bound $n(n-1)/12$.
We can get a lower bound on the order of $n \log n$.
I'll describe how to arrange $4^n$ points in general position to get $n 4^{n-1}$ squares.
The arrangement is described recursively. For the base case $n=1$, we have $4^1 = 4$ points, and you can probably guess how we should arrange them to get $1 \cdot 4^{1-1} = 1$ squares. Now suppose we have an arrangement $A$ of $4^{n-1}$ points, in general position, giving us a total of $(n-1)4^{n-2}$ squares. Take $4$ copies of $A$ (a total of $4^n$ points). Place the $4$ copies of $A$ at the $4$ corners of a "large" square, and then rotate each copy of $A$ by a "random" angle $\theta$ (the same angle for each of our $4$ copies of $A$). This gives us our new arrangement of points.
If the square mentioned above is large enough, then no points from $3$ distinct copies of $A$ can lie on a line. And it is not hard to show that, with probability $1$, a randomly chosen $\theta$ will have the property that no two points from a given copy of $A$ will lie on a common line with a different copy of $A$. So for a "large" square and a "random" angle, we get a set of $4^n$ points in general position.
In each small copy of $A$, we get $(n-1)4^{n-2}$ small squares, for a total of $4(n-1)4^{n-2} = (n-1)4^{n-1}$ small squares in our new arrangement. In addition to these, we get $|A| = 4^{n-1}$ additional large squares, by connecting the $4$ corresponding points in each of our $4$ copies of $A$. This gives a total of $n4^{n-1}$ squares, as promised.
Because of rotational symmetry (around the center of the large square), you get at least twice that many large squares. Gerhard "Buy Four Get One Free" Paseman, 2020.07.27.
@Will: Very nice. This does formally satisfy my question of improving asymptotically on the trivial estimates in some direction. However I'm going to leave the question open in hopes that someone can further close the gulf between the upper and lower bounds.
|
2025-03-21T14:48:31.623596
| 2020-07-27T18:15:05 |
366726
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"HJRW",
"Neal",
"Paul Cusson",
"abx",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/143629",
"https://mathoverflow.net/users/1463",
"https://mathoverflow.net/users/20796",
"https://mathoverflow.net/users/40297",
"https://mathoverflow.net/users/40804",
"mme"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631524",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366726"
}
|
Stack Exchange
|
Learning from unsuccessful attempts at the Poincaré conjecture
This is reposted from MSE, but perhaps it is more appropriate to post it here. Let me know if I'm wrong.
Among the many unsuccessful attempts at solving the Poincaré conjecture, I'm wondering if there was an approach that went along the lines of showing that a closed simply connected $3$-manifold could be endowed with a Lie group structure. This might not be enough to solve it, since there are other compact $3$-dimensional Lie groups (if I'm not wrong), but it's tempting to think about since the only spheres which are Lie groups are of dimension $0, 1$ or $3$. This makes it feel like there is this special additional structure that can be used for the case $n=3$ as opposed to solving the generalized Poincaré conjecture, where no higher dimensional spheres are Lie groups.
So I'm wondering if someone tried this. If they did, what hurdle/subtlety did they find in the course of the attempt that prevented a solution? Did we learn something interesting from it?
for reference: https://math.stackexchange.com/questions/3764770/has-anyone-attempted-solving-the-poincaré-conjecture-using-lie-groups
There are no other compact simply connected Lie groups of dimension $3$.
I don't really know why you would imagine it would be any easier to construct a Lie group structure on a simply connected closed 3-manifold than finding a diffeomorphism to $S^3$ and pulling back the group structure by that diffeomorphism. How do you imagine simple connectedness help you?
@MikeMiller Personally I don't know either, my knowledge of the subjects here is sparse. Perhaps this could lead to another question, can one find examples where "unconventional" properties of a manifold make it a Lie group?
Tangentially you might find Stallings' paper how not to prove the Poincare conjecture a useful read
@Neal: I love that paper, but Stallings' approach is completely unrelated to the one proposed here (which, as Ben McKay points out, is morally very similar to Thurston's).
@HJRW That is why it is tangential to the conversation. :)
@Neal: I’d call it more antipodal than tangential. :)
Thurston approaches 3-manifolds by cutting them up along various surfaces (one first cuts along spheres [Kneser-Milnor] and then along tori [Jaco-Shalen-Johannson]) into pieces which each admit a locally homogeneous geometric structure, modelled on a homogeneous space with an invariant Riemannian metric. A compact, simply connected manifold with such a structure is homogeneous. But a Lie group is homogeneous under its own left translations (or right translations), with many invariant Riemannian metrics (given by left, or right, translation of any positive definite inner product on any one tangent space. So Thurston's programme contains your programme as a special case.
Thanks to Allen Hatcher for correcting my description of the decomposition.
|
2025-03-21T14:48:31.623929
| 2020-07-27T18:31:00 |
366729
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Beni Bogosel",
"Leonid Dworzanski",
"Robert Israel",
"Wojowu",
"https://mathoverflow.net/users/13093",
"https://mathoverflow.net/users/13650",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/3315"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631525",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366729"
}
|
Stack Exchange
|
What numbers (irrational) in radicals are expressible as trigonometric rational fraction with only rational multiplies of $\pi$?
What irrational expressions $A$ with radicals can be expressed as trigonometric rational fraction (not a series) with only rational multiplies of $\pi$.
Example:
$ \frac{1}{\sqrt5} = \frac{\sin\frac{\pi}{10}}{1- \sin\frac{\pi}{10}}$
Is there a general algorithm for such numbers to find such expressions?
UPD: Niven's theorem on Wiki [1] didn't help me to figure it out.
[1] https://en.wikipedia.org/wiki/Niven%27s_theorem
You realize that your question is too broad? What do you mean by finite trigonometric expressions? Which arguments are you allowed to use in the trigonometric functions? What coefficients can you put before the trigonometric functions? Etc.
Depending on how exactly you formalize this, a reasonably complete answer might be that those are precisely the numbers which are contained in some cyclotomic extension of $\mathbb Q$, or equivalently (by Kronecker-Weber theorem) in some abelian extension.
Presumably you don't want something like $x = \tan(\arctan(x))$?
@BeniBogosel, Yes I do. Thanks to your comment I've refined the question to "trigonometric rational fractions". I think that this is more precise - tell me if it is not.
@RobertIsrael, Yes, I do not.
So do you only allow trigonometric functions of rational multiples of $\pi$?
@RobertIsrael, Yes, only with rational multiples of $\pi$ - I had the same thought a few hours ago. Thanks - I've refined the question.
Then Wojowu's comment is a propos.
@Matt F: Since $\sqrt{17}$ is an algebraic integer with minimal polynomial $x^2-17$ having Galois group of order $2$, thus abelian, the Kronecker-Weber theorem says it is. In fact, if $w$ is a primitive $17$'th root of unity, $\sqrt{17} = \pm (1+2{w}^{3}+2{w}^{5}+2{w}^{6}+2{w}^{7}+2{w}^{10}+2{w}^{11}+2{w}^{12}+2{w}^{14})$.
|
2025-03-21T14:48:31.624075
| 2020-07-27T18:32:49 |
366730
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631526",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366730"
}
|
Stack Exchange
|
Some general questions about deformations
$\newcommand{\spec}[1]{\mathrm{spec}(#1)}$
$\newcommand{\ZZ}{\mathbb{Z}}$
$\newcommand{\QQ}{\mathbb{Q}}$
$\newcommand{\CC}{\mathbb{C}}$
These days I am reading in Kurke, Pfister, Roczen "Henselsche Ringe und algebraische Geometrie" the chapter 5. "Approximation theory and algebraization of Deformations" especially "5.7 Infinitesimal deformations of affine schemata" and "5.8 Infinitesimal deformations of proper schemata."
Thinking about deformations in general I came to some questions, which are so general, that I could not find a way to resolve them by googling key phrases (and only partially by looking into Hartshorne's Springer GTM book on Deformation theory).
If one starts with a variety (or an algebraic scheme = of finite type and separated over $k$) $X_0/k$ over an algebraically closed field $k$ and considers all mathematically interesting/mostly studied families of deformations $f:X \to S$, that is flat morphisms $f$ with
$f^{-1}(s_0) = X_0$ for a closed point $s_0 \in S$, what can one say about the $S$ chosen as base?
1a). Of course $S = \spec{A}$ with $A$ Artin ring over $k$, especially $A = k[t]/(t^n)$, also $A=k[[t]]$ and $A=k[[t_1,\ldots,t_r]]$ (with $r = \dim_k \mathrm{Ext}^1(\Omega_{X_0|k}, \mathcal{O}_{X_0})$ occur naturally, but what are typical non-local $S$?
1b). Is it more usual to have $S$ that are themselves nonsingular or are for global deformations singular $S$ sometimes "better" or "enforced"?
1c). If only $X_0/k$ is given initially, does there exist a "best" or "canonical" $S$, which is to $X_0$ like the domain of holomorphic continuation to the germ of a holomorphic function?
1d). Can such an $S$ or $f:X \to S$ be concretely computed, say from equations $f_i(x_1,\ldots,x_n)$ giving $X_0$ as an affine or projective scheme?
If $X_0$ is proper over $k$ and one considers deformations $f:X \to S$ of $X_0$ non-singular with $X$ non-singular over $k$, then $f_V:X - f^{-1}(V) \to S - V$ is smooth for a closed subset $V \subseteq S$. So if $k=\CC$ the map $f_V$ is a proper submersion and by Ehresmann's fibration theorem a fibration (of manifolds).
2a). Can/Should $S$ be chosen, starting from a given $X_0$, so that $V = \{ s_1,\ldots, s_r\}$ is finite set of points of $S$?
2b). Do the singular fibers $X_{s_i}$ determine $f:X \to S$ in this case totally?
2c). What is the relation between the homology groups
$H_p(X_s, \ZZ)$ and $H_p(X_{s_i},\ZZ)$ for $s \notin V$ (replace $\ZZ$ by $\QQ,\CC$)?
2d). The image of a map $\gamma:[0,1] \to S - V$ can be covered by patches $U_i \subseteq S$ with $f^{-1}(U_i) = U_i \times F$. The transition maps define a map $H_p(X_{\gamma(0)}, \ZZ) \to H_p(X_{\gamma(1)})$. How concretely can these maps be computed with a computer algebra system?
What if one asks, given a deformation $f:X \to S$ of $X_0/k$ whether an "object" associated with $X_0$ can be deformed along $f$ and in how many ways?
3a) Say $\mathcal{F}_0$ is a vector bundle on $X_0$. How can one compute the space of all vector bundles $\mathcal{F}$ of $X$, flat over $S$, that have $i_0^*(\mathcal{F}) = \mathcal{F}_0$, where $i_0:\spec{k} \to S$ and $X_0 = i_0^*(X)$?
Can one compute its dimension from some $\mathrm{Ext}$ or $H^p(X_0,-)$ groups, at least, when $X_0$ and $X$ are proper/projective over $S$?
3b) Another "object" would be a closed immersion $g_0:X_0 \to \mathbb{P}_k^N$. Can one compute whether there is an extension of $g_0$ to a closed immersion $g:X \to \mathbb{P}_S^N$? How to compute such an extension explicitly? How many extensions exist?
3c) An extension of 3b would be: Say we have to deformations $f:X \to S$ of $X_0$ and $g:Y \to S$ of $Y_0$ and a closed immersion $h_0:X_0 \to Y_0$. In how many ways can $h_0$ be extended to a closed immersion $h:X \to Y$ over $S$? Can one compute from explicit cohomology groups a criterion whether such an $h$ exists?
Please excuse that this question violates the norms for questions at mathoverflow probably in many ways, but I do not have the possibility to ask them to someone knowledgeable in a personal conversation, which would be surely more appropriate.
So I am happy with any partial answer and also, of course, with links to surveys (preferably accessible over the internet), which would give me a more advanced standpoint, from which I could put these questions into position and perhaps answer them myselves.
Thanks in advance!
|
2025-03-21T14:48:31.624369
| 2020-07-27T19:55:08 |
366733
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Fredrik Johansson",
"Sidharth Ghoshal",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/46536",
"https://mathoverflow.net/users/4854"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631527",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366733"
}
|
Stack Exchange
|
Does Rademacher's convergent series for p(n) define an analytic function?
Let $p(n)$ be the number of partitions of $n\geq 0$. We can let $n$ be
any complex number in Rademacher's convergent infinite series for
$p(n)$. (See e.g. equation (24) here.)
For what $n$ does it converge? Does it define an analytic function for
such $n$? If so, what are the properties of this analytic function
(singularities, branch points, domain of existence, etc.)? This question
asked for an analytic continuation of $p(n)$, so I am suggesting a
possible answer.
The attempt to define fractional derivatives in Rademacher's formula might suggest a connection to JM's family of solutions here: https://math.stackexchange.com/questions/34318/feeding-real-or-even-complex-numbers-to-the-integer-partition-function-pn in the case of $\rho \rightarrow 1$
Edit. We can write the series in the form
$$p(n)=\sum A_k(z)\frac{d}{dz}f(z/k^2),$$
where $|A_k(z)|\leq Ck^{1/2}e^{C_1(\Im z)^+},$ where $y^+=\max\{ y,0\},$ and $C_j$ are various positive absolute contants, $f(z)=(\sinh\sqrt{z})/\sqrt{z},\; z=C_2(n-1/24)$.
Notice that $f$ is an even entire function of order $1/2$, normal type so $f'(0)=0$. So
$$p(n)=\sum c_k(z)f'(z/k^2),$$
where $|c_k(z)|\leq C_3k^{-3/2}e^{C_1(\Im z)^+}$. Since $f'$ is an entire function of order $1/2$, normal type and $f'(0)=0$, we have an estimate of the form
$$|f'(z)|\leq C_2|z|e^{A\sqrt{|z|}},$$
and we see that the sum above must satisfy the estimate
$$|p(z)|\leq C_4e^{C_1(\Im z)^++A\sqrt{|z|}}.$$
Therefore $p(n)$ extends to an entire function in the plane of exponential, type. An interesting question about such a function is the location of its zeros,
and the related question, asymptotics for large $|z|$. One could try to plot $|p(z)|$ on a computer and look what to expect. Even a plot on the imaginary line may tell us something interesting.
Based on this argument, and pictures of Fredrik Johansson, I conjecture that zeros are asymptotic to some parabola $y=C\sqrt{x}$ in the right half-plane, and close to the real axis in the left half-plane (the pictures indicate infinitely many negative zeros).
To prove this one has to prove the asymptotics
$$p(z)=\exp\left(B(\Im z)^++A\Re\sqrt{z}+o(\sqrt{|z|})\right)$$
outside some small circles around zeros.
Remark. Unfortunately, this crude argument gives $C_1=B=2\pi$, and in such class of entire functions extension from integers to the complex plane is not unique.
Where did the dependence on $e^{\pi i n}$ in $a_k$ go in this analysis? You can clearly see the $e^{\pi |z|}$ growth in the imaginary direction in the plots I made. Can this be suppressed?
@Fredrik Johansson: Thanks. I did not notice $e^{2\pi i n}$ in the $A_k$. Now I corrected. As a result it has exponential type (not order 1/2, as I said before), and may grow in the upper half-plane.
Not a direct answer to the question, but a brief numerical exploration of this function.
First, a trivial observation: we can write either $e^{\pi i x}$ or $\cos(\pi x)$ in the formula for the exponential sum $A_k(n)$ in the Rademacher series. This makes no difference at integer $n$, but gives different generalizations for noninteger $n$. The cosine version has the nice property of being real-valued on the real line and conjugate symmetric.
Here is a plot of the cosine-extended $p(n)$ on the real line:
The exponential version (real and imaginary parts):
Either version of the function seems to have simple zeros at the negative integers $-1, -2, -3, ...$. (This is quite nice if correct, because it matches the obvious combinatorial interpretation $p(-n) = 0$.) The cosine version has additional zeros on the negative real line (the first near $-0.93$).
At half-integers, it appears that all terms in the cosine version of the Rademacher series except the first term vanish, and so one has a trivial closed-form evaluation of $p(k+\tfrac{1}{2})$, $k \in \mathbb{Z}$. This is not the case for the exponential version.
Taking this leading term as a cue for the asymptotics on the real line, the origin is a turning point between exponential growth to the right and $\text{oscillation} \cdot O(n^{-1})$ behavior to the left.
Viewed in the imaginary direction, the exponential version appears to grow exponentially as $n \to +i \infty$ but remains small as $n \to -i \infty$. Plot of the real and imaginary parts of $p(i x)$:
The cosine version looks like the upper-half-plane exponential version:
There are additional zeros in complex plane. Since the Rademacher series converges slowly when the imaginary part of $n$ is large, it's a bit difficult to explore these zeros numerically. Here is a low-resolution plot of the exponential version of $p(z)$ on $z \in [-4,4] + [-2,2] i$:
And the cosine version:
The slow convergence also makes it difficult to search numerically for other potential closed forms (it is expensive to get more than ~6 digits). A faster algorithm for computing $p(n)$ to high precision near the origin would be very exciting.
Unoptimized Python implementation that I used to create these plots: https://gist.github.com/fredrik-johansson/7c2711887811ef9f2d7038b8451a4e63
|
2025-03-21T14:48:31.624706
| 2020-07-27T19:56:54 |
366734
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Achim Krause",
"Moritz Firsching",
"Pietro Majer",
"https://mathoverflow.net/users/39495",
"https://mathoverflow.net/users/39747",
"https://mathoverflow.net/users/6101"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631528",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366734"
}
|
Stack Exchange
|
On a generalization of the Borsuk-Ulam theorem / Tucker's lemma for a map from simplex to its boundary
The Borsuk-Ulam theorem is equivalent to $S^{n-1}$ not being a retract of $B^n$. <totally wrong! or else 2+2=4 is equivalent to the Poincare conjecture/thm>
How shall i prove the following stronger version :
There is no continuous map $f:\Delta_n \to \partial \Delta_n$ such that every face is mapped to itself. <the Borsuk's homotopy extension thm since $f|S^{n-1}$ would be homotopic to the identity on $\ S^{n-1}$>
In this case, Tucker's lemma comes to mind, as it does say something similar.
Any thoughts?
If $f:\Delta_n\to \Delta_n$ maps every face $F$ to itself, one can consider the topological degree of $f_{|\partial F}:\partial F\to\partial F$ and prove by induction on dim(F) that it is $1$. So $f(F)=F$ for all faces, and in particular $f(\Delta_n)=\Delta_n$.
Does this answer your question? Map from simplex to itself that preserves sub-simplices
@Wlod AA, I'm pretty sure that way to use edits is against the rules here. If you have something to say, try comments or answers...
|
2025-03-21T14:48:31.624833
| 2020-07-27T20:57:36 |
366738
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Curaçao Hajek",
"Iosif Pinelis",
"LSpice",
"Mateusz Kwaśnicki",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/145572",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631529",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366738"
}
|
Stack Exchange
|
Can functions be differentiable on sets with empty interiors?
As a simple example, suppose we have a function $f: \mathbb{R}^3 \to \mathbb{R}$ defined on the set (and taking $+\infty$ everywhere else),
$$\{x \in \mathbb{R}^3| x_1 \in [-1, 1], x_2 \in [-1, 1], x_3 = 0\}$$
The set has no interior but a relative interior given by $(-1,1) \times (-1,1) \times \{0\}$.
Similarly, consider sets such as $\{x \in \mathbb{R}^3| \langle e, x\rangle = 1, x_i \geq 0\}$, where $e$ is the one-vector. Once again, it has no interior, but has a relative interior relative to the hyperplane $\langle e, x \rangle = 1$ given by $\{x \in \mathbb{R}^3| \langle e, x\rangle = 1, x_i > 0\}$,
Example functions could include:
$f(x) = \langle x, x \rangle$ for the first set
$f(x) = -\langle e, \ln(x) \rangle$ for the latter set
Are such function differentiable on such sets (i.e. the gradient exists)? If not, why? Can't seem to find any resource on this.
Edited per comment: Also, is it problematic if I were to pretend that the function was defined on the whole space, take the gradient there, and restrict it to the relative interior? For example, consider $f(x) = \langle x, x \rangle$ defined on the set $[-1, 1]^2 \times \{0\}$. What is wrong if I were to take the gradient as usual, $\nabla f(x) = 2 x$ and define it on the relative interior of the same set $(-1, 1)^2 \times \{0\}$?
It's not clear to me what you are asking. Are you asking for a function that is defined on all of $\mathbb R^3$ (as the notation $f : \mathbb R^3 \to \mathbb R$ indicates) but that is differentiable only on a set such as $[-1, 1]^2 \times {0}$, or are you looking for a definition of differentiability that works for functions defined only on $[-1, 1]^2 \times {0}$?
If the latter, it may be interesting to consider the notion of manifolds with boundary, of which both of your spaces are natural examples. Here is the first Google result for me for "smooth functions on manifolds with boundary": http://math.ucr.edu/~res/math260s10/manwithbdy.pdf .
In your second example $f(x) = -\langle e, \ln(x)\rangle$, do you mean to define $\ln(x_1, x_2, x_3) = (\ln(x_1), \ln(x_2), \ln(x_3))$? If so, then how do you define the function on the boundary?
@LSpice In the second example, this is the sum $-( \ln(x_1) + \ln(x_2) + \ln(x_3))$
@LSpice I guess a even lower example is whether a function, e.g., $f(x) = x_1^2 + x_2^2$ defined only for straight lines, e.g., it is only finite on the line $x_1 + x_2 = 1$ and $+infinity$ everywhere else. Is the function differentiable on that line in $\mathbb{R}^2$ (which also has no interior). Is the function differentiable. in the usual sense? And if there is a compatible definition in the case it is not.
Your $x \mapsto -(\ln(x_1) + \ln(x_2) + \ln(x_3))$ is still not defined (or at least not finite) on the boundary. Your second example definitely is not differentiable in the usual sense, because the usual definition of differentiability only handles finite functions (else how to define the difference quotient?). I really think you don't want to view your function as defined on a subset of $\mathbb R^2$ and extended by $\infty$ to all of $\mathbb R^2$; but, if you do, then you must call it $f : \mathbb R^2 \to \mathbb R \cup {+\infty}$, not $f : \mathbb R^2 \to \mathbb R$.
@LSpice Ok. I'm just going by the notation in https://web.stanford.edu/~boyd/cvxbook/. These sets are directly taken from the book in the section about relative interiors. I just don't know how differentiability is handled. For that particular case about the boundary, I guess we could set it to be $f: 0 \mapsto 0$. But this is not the only example, another example could be $\langle x, \ln(x) \rangle$, which has the boundary condition $\langle 0, \ln(0) \rangle = 0$
Ah, if you're going by a particular book, then you'll definitely want to edit that into the question, and, if possible, refer to any specific part of the book where this question arises. I also have no problem with the sets you mention, just with referring to functions on them as functions on the entire Euclidean space on which they live, or with functions extended by $\infty$ as valued in $\mathbb R$.
I am afraid the functions you list here are not differentiable in any usual sense. You seem to be concerned with the entropy function on a simplex, as in Norman's answer below. If yes, then you should describe your specific problem; right now this question makes little sense to me.
@MateuszKwaśnicki May I ask what doesn't make sense? As I described in the comment above a simple case is differentiability of $f(x) = x_1^2 + x_2^2$ on $x_1 + x_2 = 1$. This is sort of a yes or no answer. Yes, it is differentiable vs no it is not. LSpice has said they are not differentiable in the usual sense, and I agree with him. I don't think the answer below fully capture my concern.
@Cauchy'sCarrot: I did not mean this to be offensive. To me, a question "let $f$ be infinite except at a single point, is it differentiable" makes little sense if "differentiable" is understood in the usual sense, because the definition of differentiability simply makes no sense in this case. You can speak about directional derivatives, though.
I think what you are looking for is the standard definition of once-differentiable manifold with boundary.
In order to define derivative, you need a normed vector space. You need a vector space because differentiation is ${\Bbb R}$-linear. You want to preserve the linearity because in a sense, differentiation is linearization. You need the norm to take the limit.
The concept is generalized to manifolds by looking at the infinitesimal neighborhood of a point, the tangent space, which is a vector space.
As LSpice commented, your examples are manifolds with boundary. The boundary is itself of a manifold one dimension lower, so you can define derivative there.
You can also consider a boundary point as part of the whole space. There your "tangent space" is only half of a vector space. You can generalize linearity here also, if you like.
Finally, manifolds are define by charts and you want to make sure that your differentiation operator is defined consistently across the charts. This means that the transition maps should be differentiable.
You can parametrize such sets and then consider the differentiability with respect to the parameters. The differentiability property will be invariant with respect to diffeomorphisms: if two parametrizations are related by a diffeomorphism (that is, by a differentiable bijection whose inverse is also differentiable), then a function differentiable with respect to one of the two parametrizations will be differentiable with respect to the other parametrization. In general, a parametrization can be any bijection. Some parametrizations may be more useful/natural than others -- e.g., parametrizations that are homeomorphisms with respect to the natural topologies would usually be better than parametrizations that are not homeomorphisms.
E.g., you can parametrize the set $S:=\{x\in\mathbb R^3\colon x_1+x_2+x_3=1,x_i>0\ \forall i\}$ by the parametrization
$$S_1\ni(s,t)\mapsto\phi(s,t):=(s,t,1-s-t)\in S$$
or, e.g., the parametrization
$$S_2\ni(s,t)\mapsto\psi(s,t):=(1-s-t,s,t)\in S,$$
where $S_1:=S_2:=\{(s,t)\in\mathbb R^2\colon s>0,t>0,s+t<1\}$. These two parametrizations are equivalent, in the sense that they are related by a diffeomorphism -- here, specifically, by the diffeomorphism
$$S_2\ni(s,t)\mapsto g(s,t):=(1-s-t,s)\in S_1$$
in the sense that $\psi=\phi\circ g$ and hence $\phi=\psi\circ g^{-1}$.
$\big($In the above example, the domains $S_1$ and $S_2$ of the two different parametrizations $\phi$ and $\psi$ of the same set $S$ were the same.
In general, though, the domains of different parametrizations of the same set may be different. Even in the above example, another parametrization of $S$ is
$$S_3\ni(s,t)\mapsto\rho(s,t):=(s,t-s,1-t)\in S,$$
where $S_3:=\{(s,t)\in\mathbb R^2\colon0<s<t<1\}\ne S_1$. The parametrization $\rho$ is then equivalent to the parametrizations $\phi$ and $\psi$.$\big)$
A function $f\colon S\to\mathbb R$ may then be called differentiable if the function $f\circ\phi\colon S_1\to\mathbb R$ is differentiable or, equivalently, if the function $f\circ\theta$ is differentiable, where $\theta$ is any parametrization of $S$ equivalent to $\phi$. Then, by the chain rule, we also have
$$(f\circ\psi)'=(f\circ\phi\circ g)'=(f\circ\phi)'\circ g';$$
here, at each point of $S_2$, $g'$ is a linear operator from $\mathbb R^2$ to $\mathbb R^2$, and $(f\circ\phi)'$ is a linear operator from $\mathbb R^2$ to $\mathbb R$ (that is, a linear functional).
For instance, the function $S\ni x\mapsto f(x):=x_1^2+x_2x_3$ will be differentiable, because the function $S_1\ni(s,t)\mapsto (f\circ\phi)((s,t))=s^2+t(1-s-t)$ is differentiable or, equivalently, because the function $S_2\ni(s,t)\mapsto (f\circ\psi)((s,t))=(1-s-t)^2+st[=(f\circ\phi\circ g)((s,t))]$ is differentiable.
For further reading, see e.g. differentiation on manifolds.
I'm not sure this fully answers the question, because your example is of a manifold without boundary, whereas the question explicitly wants to consider differentiation on manifolds with boundary.
Is it problematic if I were to pretend that the function was defined on the entire set instead? For example, consider $f(x) = \langle x, x \rangle$ defined on the set $[-1, 1]^2 \times {0}$. What is wrong if I were to take the gradient as usual, $\nabla f(x) = 2 x$ and define it on the interior of the same set $(-1, 1)^2 \times {0}$?
With this choice, the functions $f : x \mapsto x_1^2 + x_2^2 + x_3^2$ and $g : x \mapsto x_1^2 + x_2^2$, which agree on your given domain, would be found to have different gradients there, so that the gradient could no longer be said to be a property of the restriction of the function to $[-1, 1]^2 \times {0}$.
@LSpice : Concerning your first comment, you are of course right. However, I thought the main idea should be first explained on the significantly easier case of a manifold without boundary.
@Cauchy'sCarrot : As noted in the above comment by user LSpice, the gradient will not in general be a property of the restriction of the function. However, the two functions in the example given in that comment have in fact the same gradient, $(2x_1,2x_2,0)$, on the set $T:=(-1,1)^2\times{0}$. However, the functions $x\mapsto x_1+x_2+x_3$ and $x\mapsto x_1+x_2$ will have the same restriction to the set $T$, but their gradients, $(1,1,1)$ and $(1,1,0)$, will be different (everywhere and hence) on $T$.
@LSpice : Concerning your second comment, please see my latter comment.
@IosifPinelis, right; thank you for replacing my wrong example with a correct one.
|
2025-03-21T14:48:31.625463
| 2020-07-27T21:24:38 |
366739
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Deane Yang",
"Dry Bones",
"Sebastian Goette",
"https://mathoverflow.net/users/108274",
"https://mathoverflow.net/users/613",
"https://mathoverflow.net/users/70808",
"https://mathoverflow.net/users/84756",
"user267839"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631530",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366739"
}
|
Stack Exchange
|
Fubini-Study metric induced by submersion
The Fubini-Study metric $g:=g_{FS}$ is the unique $U(n+1)$-invariant
Riemannian metric on the complex projective space $\mathbb{CP}^{n}$ the complex projective space
which by $U(n+1)$-invariance can be wlog definined on
tangent bundle over affine chart
$ U_0 :=\{Z \in \mathbb{CP}^n \ \vert
z_0 \neq 0 \} \subset \mathbb{CP}^n $ and then it extends uniquely by $U(n+1)$-action
to all affine pieces $U_i$.
So we can restict our considerations to tangent space $T_Z \mathbb{CP}^n$ over a point
$$Z =[1:z_1:....z_n] \in U_0 :=\{Z \in \mathbb{CP}^n \ \vert
z_0 \neq 0 \} \subset \mathbb{CP}^n$$
Let $\{\partial _{1},\ldots ,\partial _{n}\}$ the be frame of tangent space
$T_Z \mathbb{CP}^n$ at $Z$, ie the canonical $\mathbb{C}$-basis of
$T_Z \mathbb{CP}^n$. Then the Fubini Study metric is defined by
$$\tag{FS} g_{i{\bar {j}}}=h(\partial _{i},{\bar {\partial }}_{j})=
{\frac {(1+|\mathbf {z} |^{2})\delta _{i{\bar {j}}}-
{\bar {z}}_{i}z_{j}}{(1+|\mathbf {z} |^{2})^{2}}} $$
At first glance one can think that $g$ somehow "has hallen from heaven"
but it is well knows that it's not.
One way to derive it is using Kähler potentials and canonical
choice of a distinguished connection: Chern connection: See Here
the excellent answer by Arctic Char.
The concern of this question is to find out if it also possible to derive $g$ on
$\mathbb{CP}^{n}$ naturally by recognizing $(\mathbb{CP}^{n},g)$ as
Riemannian submersion of $(\mathbb{C}^{n+1} \backslash \{0\},h)$
where $h$ is to restriction of standard Hermitian metric on
$\mathbb{C}^{n+1}$ to $\mathbb{C}^{n+1} \backslash \{0\}$.
What do I mean by 'naturally"?
Well, $\mathbb{CP}^{n}$ can be considered as submersion of complex manifold
$\mathbb{C}^{n+1} \backslash \{0\}$ via canonical projection
$$\pi: \mathbb{C}^{n+1} \backslash \{0\} \to \mathbb{CP}^n$$
The complex manifold is moreover Hermitian since
$ \mathbb{C}^{n+1} \backslash \{0\}$ inherits naturally the standard
Hermitian metric $h$ on $ \mathbb{C}^{n+1} $.
Futhermore by $\pi$ the $\mathbb{CP}^{n}$ can also be considered as
quotient $ \mathbb{C}^{n+1} \backslash \{0\} / \mathbb{C}^*$.
By definition a Riemannian submersion $F: (M,g_M) \to (N, g_N)$
is a
submersion from one Riemannian manifold $M$ with Riemannian
metric $g^M$ to another Riemannian manifold $N$ with Riemannian
metric $g^N$
that respects the metrics, meaning that it is an
orthogonal projection on tangent spaces.
Namely that means that for every $x \in M$ and $y=F(X)$
the restriction of the differential map $dF_x$ to
$ker(dF_x)^{\perp} \subset T_x M $ gives
an isometry $d_x F: ker(dF_x)^{\perp} \to T_y N$
such that the metric $g^N$ is induced by $g^M$ is following sense:
Let $A, B \in T_y N$ and denote by $\overline{A}, \overline{B}
\in ker(dF_x)^{\perp}$ their unique preimages with respect the
isometry $d_x F: ker(dF_x)^{\perp} \to T_y N$.
Then
$$\tag{RieSub} g^N _y(A,B)= g^M _x(\overline{A}, \overline{B})$$
QUESTION:
How can be proved that the Fubini-Study metric $g$ arises exactly on this way? ie that
the canonical projection
$\pi: \mathbb{C}^{n+1} \backslash \{0\} \to \mathbb{CP}^n$
extends naturally to Riemannian submersion
$\pi: (\mathbb{C}^{n+1} \backslash \{0\},h) \to
(\mathbb{CP}^n, g)$
is the sense above, where $h$ is the restriction of the
stadard Hermitian metric on $\mathbb{C}^{n+1}$ and
$g$ is the Fubini- Study metric defined above in (FS).
In other words we need to verify that the metrics $h$ and $g$ are related
to each other as in (RieSub).
Take as $y= Z =[1:z_1:....z_n]$ and $x =(1,z_1,..., z_n) \in
\mathbb{C}^{n+1}$ and $\partial _{i}, \partial _{j} \in
T_y \mathbb{CP}^n$ element of the canonical basis of the tangent space
at $y$. Firstly we need an explicit identification between the tangent space
$T \mathbb{CP}^n$ and $ker(d \pi)^{\perp}$. We need this isometry explicitly
since in order to prove (FS) we have to work with unique preimages of elements
in $T_y \mathbb{CP}^n$ living in $ker(d \pi_x)^{\perp}$.
But I don't see how can I show that indeed
$$g(\partial _{i},{\bar {\partial }}_{j})=
{\frac {(1+|\mathbf {z} |^{2})\delta _{i{\bar {j}}}-
{\bar {z}}_{i}z_{j}}{(1+|\mathbf {z} |^{2})^{2}}} $$
holds. On the left side we have already
identified $\partial _{i}, \partial _{j} \in
T_y \mathbb{CP}^n$ with their unique preimages in
$ker(d_x \pi)^{\perp}$.
It can't be a Riemannian submersion. This is more easily seen in the case of the radial projection $\pi: \mathbb{R}^n\backslash{0} \rightarrow S^{n-1}$, which also is not a Riemannian submersion, because the restriction of $\pi$ to each sphere of radius $r$ is not an isometry. You have to put a scale invariant metric on $\mathbb{R}^n\backslash{0}$ or $\mathbb{C}^n\backslash{0}$ to make the projection map a Riemannian submersion.
Alternatively, you could restrict to the unit sphere, where $\pi$ becomes a Riemannian submersion. This may not look natural because the unit sphere is not a complex manifold and the standard Hermitian metric on $\mathbb C^n$ is somewhat arbitrary. But in a technical sense, the construction is still natural because the isometry group of $S^{n+1}$ maps surjectively to that of $\mathbb{CP}^n$.
@DeaneYang: Hi. What do you percisely mean by "scale invariant metric" on
a manifold $M$? Altough I think I intuitively understand what you mean, I nowhere
found literature defining it precisely.
Below I will sketch what I guess you mean, but
I'm not pretty sure if that's correct. The next important
question is if it possible to obtain a "scale invariant metric" $g_{si}$ on
$\mathbb{C}^n\backslash{0}$ by naive geometric modification of the stadard Hermitian metric $h$.
My guess : Do you mean that a metric $g$ on complex manifold
$M= \mathbb{C}^n\backslash{0}$
is a scale invariant metric if at every point $m \in M$,
arbitrary tangent vectors $X, Y \in T_m M$ at $m$ and
arbitrary scalars $a, b \in
\mathbb{C} \backslash{0}$
following holds:
Let decompose $X= X^{\perp}+ X^{\parallel}$ and
$Y= Y^{\perp}+ Y^{\parallel}$ with unique
$X^{\parallel}, Y^{\parallel} \in \mathbb{C} \cdot m$ and
$X^{\perp}, Y^{\perp} \in (\mathbb{C} \cdot m)^{\perp}$.
Then $g(X^{\perp}+ aX^{\parallel},Y^{\perp}+ bY^{\parallel})=
g(X^{\perp}+ X^{\parallel},Y^{\perp}+ Y^{\parallel})$
for arbitrary scalars $a, b \in
\mathbb{C} \backslash{0}$.
By linearity this implies that $g_m$ is constant on subpace
$\mathbb{C} \cdot m \subset T_m M$ for every $m \in M$.
Is this the definition of scale invariant metric you have in mind?
I apologize for not reading your latest comments carefully, but I've tried to elaborate on my comment in the answer below.
Let me start from scratch. Note that everything below uses only the definition of complex projective space and the natural Hermitian inner product on $\mathbb{C}^{n+1}$. Also, the construction is coordinate-independent in the sense that everything below can be done with an abstract complex vector space with a Hermitian inner product without using any basis.
Recall that canonical projection map is defined to be
\begin{align*}
\mathbb{C}^{n+1}\backslash\{0\} &\rightarrow \mathbb{C}P^n\\
z &\mapsto [z],
\end{align*}
where $[z] = \{ tz\ :\ t \in \mathbb{C} \}$. For each $z \ne 0$, the tangent space of $\mathbb{C}^{n+1}$ at $z$ splits naturally
$$
T_z\mathbb{C}^{n+1} = [z] \oplus z^\perp,
$$
where $z^\perp = \{ w \in \mathbb{C}^{n+1}\ :\ \langle w,z\rangle = 0 \}$. For each $z$, the pushforward map $\pi_*: T_z\mathbb{C}^{n+1} \rightarrow T_{[z]}\mathbb{C}P^n$ is an isomorphism if restricted to $z^\perp$.
We want to use the Hermitian inner product to define a Kähler metric $g$ on $\mathbb{C}P^n$. Observe that, given such a metric, $\pi^*g$ is an degenerate Hermitian $2$-tensor $h$ on $T_z\mathbb{C}$ that is scale invariant in the sense that if $R_t(z) = tz$, then $(R_t)^*h = h$. Conversely, a Hermitian metric on $\mathbb{C}^{n+1}$ defines one on $\mathbb{C}P^n$ only if its restriction to each $z^\perp$ is scale invariant.
If we want to define the metric using only the standard Hermitian inner product pointwise on $\mathbb{C}^{n+1}$, then the only possibility is one of the form
$$
g = f(|z|^2)|dz|^2
$$
restricted to $z^\perp$, for each $z \in \mathbb{C}^{n+1}\backslash\{0\}$. On the other hand, if you work out all the definitions carefully, you find that
$$
R_t^*g = f(|t|^2|z|^2)|t|^2|dz|^2,
$$
which equals $g$ if and only if there exists a real constant $c$ such that $f(|z|^2) = c|z|^{-2}$. By the observations above, this metric can be pushed down to a metric on $\mathbb{C}P^n$, and it can be verified that it is, up to a scale factor, the Fubini-Study metric.
Note that the explanation above is just the definition of a Riemannian submersion but for the specific situation here.
Thank you for your excellent answer! One detain I still not understand: the scale invariance is defined by $(R_t)^*h = h$. In your case this means that $f(|z|^2)|dz|^2 = R_t^*g = f(|t|^2|z|^2)|t|^2|dz|^2$, thus $ f(|z|^2)= f(|t|^2|z|^2)|t|^2$. Why this already imply $f(|z|^2) = |z|^{-2}$?
What I said in my answer was not exactly correct, I've revised it. As for your question, note that the identity for $f$ is equivalent to saying that $f(x) = f(ax)a^2$, for any positive $x, a$. Now set $y = ax$.
I'm not sure if this could be of any use here (maybe to simplify the notation?), but it's interesting to note that $T_{[z]}\Bbb CP^n$ is naturally identified with the vector space of linear maps ${\rm Hom}([z],z^{\perp})$. In general, $T_p{\rm Gr}(k,V)={\rm Hom}(p,V/p)$ even without the choice of a form in $V$, where ${\rm Gr}(k,V)$ denotes the Grassmannian of $k$-dimensional subspaces of a vector space $V$.
@DryBones, that's the right abstract definition of the tangent space, in the sense that it does not depend on coordinates or any inner product. So it's useful when you want to study properties of complex projective space that do not depend on the metric.
I still not understand how we obtain $f(|z|^2) = |z|^{-2}$.
In your answer you wrote that $g= R_t^*g$ if and only if there exist
a real constant $c$ such that $f(|z|^2)=c |z|^{-1}$. Clearly, one direction is trivial, since
$f(|z|^2)=c |z|^{-1}$ implies $g= R_t^*g$, but the other
direction I still not understand. As you said $g= R_t^*g$
is equivalent to $f(|z|^2)= f(|a|^2|z|^2)|a|^2$, but
why this already imply that $f$ has the shape
$f(|z|^2) = c|z|^{-2}$?
Choose $z$ so that $|z| = 1$.
I see, then $c= f(1)$, sorry for foolish question.
|
2025-03-21T14:48:31.626211
| 2020-07-27T21:53:26 |
366741
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Adittya Chaudhuri",
"David Roberts",
"LSpice",
"Sebastian Goette",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/70808",
"https://mathoverflow.net/users/86313"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631531",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366741"
}
|
Stack Exchange
|
What is the natural Lie groupoid structure on the Atiyah Lie groupoid of a principal $G$-bundle?
$\DeclareMathOperator\At{At}\DeclareMathOperator\Obj{Obj}\DeclareMathOperator\Mor{Mor}$According to https://ncatlab.org/nlab/show/Atiyah+Lie+groupoid#idea the Atiyah Lie groupoid $\At(P)$ of a principal $G$ bundle $\pi:P \rightarrow X$ is a category for which $$\Obj(\At(P))=\lbrace \pi^{-1}(x): x \in X \rbrace$$ and $$\Mor(\At(P))=\big\lbrace f:\pi^{-1}(x)\rightarrow \pi^{-1}(y): \text{$f$ is a $G$ equivariant morphism}\big\rbrace.$$ Structure maps of this category are easy to guess. Now it is easy to see that $\At(P)$ is indeed a groupoid.
Although it is mentioned in https://ncatlab.org/nlab/show/Atiyah+Lie+groupoid#idea that the Atiyah Lie groupoid is indeed a Lie groupoid, I am not able to guess appropriate smooth structures on $\Obj(\At(P))$ and $\Mor(\At(P))$ such that the source and the target maps are surjective submersions and other structure maps are smooth.
Is there any natural choice of such smooth structures on both $\Obj(\At(P))$ and $\Mor(\At(P))$ such that $At(P)$ is a Lie groupoid so that if someone talks about the Atiyah Lie groupoid of a principal $G$ bundle then he/she is precisely assuming those natural choice of smooth structures on $\Obj(\At(P))$ and $\Mor(\At(P))$?
I would also be very grateful if someone point me to any literature in this direction.
At least according to Wikipedia, the obvious place to start with the literature is Atiyah - Complex analytic connections in fibre bundles. Also, your definitions don't seem to match the nLab's, which declares $X$ the objects and $(P \times P)/G$ the morphisms, both with obvious smooth structure. (Iincidentally, note if desired you can replace \lbrace\rbrace by \{\}.)
@LSpice Sorry if I sound stupid but I asked about Atiyah Lie Groupoids not Atiyah algebroid. According to wikipedia reference Atiyah Lie Algeroid is the Lie Algebroid of a Gauge Groupoid of a Principal bundle. How from this I can guess the smooth structure on $Obj(At(P))$ and $MorAt(P)$? Can you please explain a bit in little detail?
@LSpice I was asking about the definition given in the section idea https://ncatlab.org/nlab/show/Atiyah+Lie+groupoid#idea not https://ncatlab.org/nlab/show/Atiyah+Lie+groupoid#over_the_pair_groupoid. So are you saying both are actually same notion and the definition given in the idea section is just an informal notion?
Yes, that is what I am saying.
(Also, you don't sound stupid; quite possibly my reference to Atiyah's paper is stupid. It's just that, if I'm looking for the literature on a concept with someone's name attached to it, I always start by looking at whether that person defined it and, if so, where. I haven't met this concept, so went to Wikipedia to see where Atiyah had defined it.)
@LSpice But even if those 2 notions are same I cannot immediately see how they are same. For the definition given in https://ncatlab.org/nlab/show/Atiyah+Lie+groupoid#over_the_pair_groupoid the smooth structure on Objects and morphisms are clear but from here how can I guess the smooth structures for the definition given in the "Idea" section https://ncatlab.org/nlab/show/Atiyah+Lie+groupoid#idea
There is no definition in the 'idea' section; it is just an idea, and, as not rigorously defined, cannot be checked against a rigorous definition.
@LSpice Ok I got your point. Thanks.
The objects as defined on the nLab and on Wikipedia give isomorphic sets. That the morphism sets are the same is less obvious, but it boils down to knowing that a map between principal homogeneous $G$-spaces is determined entirely by its value at a single point. It's worth thinking about the case of a trivialisable bundle first.
@DavidRoberts Thank you Sir.
Contrary to what is claimed in the comments, I would argue
that the definition given in nLab's Idea section is rigorous
enough to be an actual definition in a research-level paper,
possibly with an additional phrase thrown in like
“The sets of objects and morphisms are equipped with the obvious
smooth structures that turn this groupoid into a Lie groupoid.”
Let's see how these smooth structures are constructed.
Recall that the set of objects is $\{π^{−1}(x)\mid x∈X\}$,
i.e., the set of fibers of $P$.
Fibers are in a bijective correspondence with points in the base $X$,
and the latter is a smooth manifold.
The set of morphisms is $\{f\colon π^{−1}(x)→π^{−1}(y)\mid \text{$f$ is a $G$-equivariant morphism}\}$.
A morphism between two $G$-torsors $U→V$ is uniquely determined
by its value $v∈V$ at some point $u∈U$.
That is, for any pair $(u,v)∈U⨯V$ there is exactly one morphism
that sends $u↦v$.
The pair $(gu,gv)$ gives rise to the same morphism $U→V$ as $(u,v)$.
It is also easy to see that the converse is true:
$(u,v)$ and $(u',v')$ yield the same morphism if there is $g∈G$ such
that $(u',v')=(gu,gv)$.
Thus, the set of morphisms $U→V$ is $(U⨯V)/G$,
where $G$ acts on $U⨯V$ via $g(u,v)=(gu,gv)$.
The action of $G$ on $U⨯V$ is a smooth free proper action,
so the quotient $(U⨯V)/G$ is a smooth manifold
and the quotient map $U⨯V→(U⨯V)/G$ is a submersion.
From here, we see that the set of all morphisms
is $(P⨯P)/G$ and therefore possesses a canonical smooth structure.
The source and target maps are surjective submersions by the 2-out-of-3 property.
Thank you Sir very much for the answer!
Calling a fibre $X$ if the base is already called $X$ is a bit awkward. I suggest to pick different letters in the middle paragraph, maybe $F_x$ and $F_y$.
|
2025-03-21T14:48:31.626844
| 2020-07-27T22:22:22 |
366743
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631532",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366743"
}
|
Stack Exchange
|
Image restoration quality general lower bounds
A typical image restoration model posits that, starting from a true image $f = f(x,y)$, we observe
$$
\tilde f = f \star h + n
$$
where $\star$ is convolution, $h$ is the point spread function (caused, e.g., by diffraction in the optics, but hopefully close to a $\delta$ function), and $n$ is noise (typically assumed to be independent between pixels, and possibly i.i.d.).
Question. Given $h$ and a noise model, are there any usefully-computable lower bounds on reconstruction quality?
That is, if $\hat f$ is a reconstruction of the given image $f$, I'm looking for lower bounds on something like
$$
\sup_f \mathbb{E}\|\hat f - f\|_2,
$$
where the supremum is universal over the class of reconstruction algorithms. I'd expect the lower bound to mention things like $\|f\|_2$, $\|n\|_2$, $\|h\|_2$ (or possibly other moments, or even the exact shape of $h$). If limiting the class of $f$'s is helpful, I'm happy to do that.
The closest inequality I know is the Cramér-Rao lower bound, but that's for unbiased estimates; I don't necessarily care about whether the estimator is biased, although I'm potentially willing to constrain the estimator to e.g. be linear or smooth or local or some such.
|
2025-03-21T14:48:31.626972
| 2020-07-27T23:28:54 |
366746
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David E Speyer",
"Derek Holt",
"THC",
"https://mathoverflow.net/users/12884",
"https://mathoverflow.net/users/22377",
"https://mathoverflow.net/users/297",
"https://mathoverflow.net/users/35840",
"verret"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631533",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366746"
}
|
Stack Exchange
|
Presentations of $\mathbf{PGL}_3(\mathbb{F}_2)$ by three involutions
I am searching for (two) presentations of the group $\mathbf{PGL}_3(\mathbb{F}_2)$ for which the generators are involutions $a, b, c$, and such that the following relations are present [among extra relations, of course], in two separate cases:
REP. 1: $(ab)^4 = 1, (bc)^4 = 1, (ac)^2 = 1$;
REP. 2: $(ab)^3 = (bc)^3 = (ac)^3 = 1$.
Can anyone help me out here ? (Thanks !!)
I don't think these exist...
For research into problems like this, you really need to learn to use computer software, such as GAP or Magma. This is just a routine calculation.
This cannot be done. Let $G_1$ and $G_2$ be the groups
$$G_1 = \langle a,b,c | a^2 = b^2 = c^2 = (ab)^4 = (bc)^4 = (ac)^2 \rangle$$
$$G_2 = \langle a,b,c | a^2 = b^2 = c^2 = (ab)^3 = (bc)^3 = (ac)^3 \rangle.$$
You are looking for surjections from the $G_j$ onto $PGL_3(\mathbb{F}_2)$, which is the simple group of order $168$.
Now, $G_1$ and $G_2$ are Coxeter groups and, in fact, they are two of the affine Coxeter groups -- types $\tilde{B}_2$ and $\tilde{A}_2$ respectively. They are known to have the alternate descriptions $D_6 \ltimes \mathbb{Z}^2$ and $D_8 \ltimes \mathbb{Z}^2$, where $D_{2m}$ is the dihedral group of order $2m$.
In particular, these two groups are solvable. However, $PGL_3(\mathbb{F}_2)$ is not solvable, so it cannot be a quotient of a solvable group.
@DavidSpeyer : Great answer !! Would there be other groups generated by reflections around sides of Euclidean triangles which do map surjectively onto $\mathbf{PGL}_3(\mathbb{F}_2)$ ? Or does essentially the same problem occur in the other cases (if there are any, which I don't know actually) ? Thanks --
The group of symmetries of the Euclidean plane is solvable, so any subgroup of it is solvable. The hyperbolic plane would be more interesting here.
There is a pretty direct argument here. Note first that $G = {\rm PGL}(3,2) = {\rm SL}(3,2)$ is a finite simple group of order 168. Suppose it were so generated.
Note next that $a$ inverts both $ab$ and $ac$, so normalizes $\langle ab, ac \rangle$.
Now $bc = (ca)(ba)^{-1} \in \langle ab,ac \rangle$, so that $a$ normalizes $\langle ab,bc,ac \rangle.$ The reason I leave the redundant generator visible is that by symmetry, it follows that $b$ and $c$ both normalize $\langle ab,bc,ac \rangle.$
Thus $\langle ab,ac,bc \rangle$ is normal in $\langle a,b,c \rangle = G.$
Hence $\langle ab, bc, ac \rangle$ is either trivial or all of ${\rm PGL}(3,2)$, by simplicity.
Triviality of the subgroup $\langle ab,bc,ac \rangle$ yields $a = b = c$, so we have a contradiction.
Suppose then that $\langle ab,bc,ac \rangle = G.$ I give the complete argument in the case of REP2: let $x = ab, y = bc, z = ac.$ Then $x^{3} = y^{3} = z^{3}= 1$, by assumption, and $xy = z.$ The following argument was known to W. Burnside.
Now $\langle x^{-1}y, yx^{-1} \rangle$ is an Abelian normal subgroup of $G$, because $(xyx)(yxy) = 1$, as $xy$ has order $3$, yielding $yxy = (xyx)^{-1} = (x^{-1}y)(yx^{-1})$, while also $(yx^{-1})(x^{-1}y) = yxy$ because $x$ and $y$ have order 3.
Thus $x^{-1}y$ and $yx^{-1}$ commute. But $x(x^{-1}y)x^{-1} = yx^{-1},$ so that $x$ normalizes $\langle x^{-1}y, yx^{-1} \rangle$. Since $x^{-1}y$ certainly normalizes $\langle x^{-1}y, yx^{-1} \rangle,$ we see that $\langle x^{-1}y, yx^{-1} \rangle \lhd \langle x,y \rangle = G$, a contradiction.
There is a similar well-known argument to exclude the case of REP1, since then $\langle ab,ac,bc \rangle$ is a "$(2,4,4)$"-group
Later edit: Instead of using a generators and relations type argument, it is also possible to see via a character calculation that if $u,v,w$ are elements of order $4,4$ and $2$ respectively in $G$ with $uv = w$, then $\langle u,v \rangle$ is a cyclic $2$-group.
For $G$ has 6 complex irreducible characters of degrees $1,3,3,6,7,8$. Let $u$ be an element of order $4$ of $G$ and $w$ be an involution (there is a single class of elements of order $4$ and a single class of involutions in $G$). Only the irreducible characters $\chi$ of degrees $1,3,3$ and $7$ have $\chi(u)\chi(w) \neq 0$, and we find from the well-known character formula that the number of times $w$ is expressible as a product of two conjugates of $u$ in $G$ is $\frac{168}{16} [ 1 -\frac{1}{3}-\frac{1}{3} - \frac{1}{7}] = 2$. On the other hand, if we take $D = C_{G}(w)$ to be a Sylow $2$-subgroup in which $w$ is central, then $w$ is already expressible as $x^{2}$ and as $y^{2}$, where $x$ and $y$ are the two elements of order $4$ in $D$.
Hence whenever $w$ is expressible as a product of two elements of order $4$ in $G$, we see that these two elements of order $4$ are equal.
But rep 1 gives elements $ab, bc$ of order $4$ with $(ab)(bc) = ac$ of order $2$, and, as we have seen, we should have $\langle ab,bc \rangle = G$, a contradiction.
|
2025-03-21T14:48:31.627300
| 2020-07-28T02:22:14 |
366748
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Prof.Hijibiji",
"https://mathoverflow.net/users/161865"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631534",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366748"
}
|
Stack Exchange
|
Problem regarding Lebesgue measure in $\mathbb{R}^2$
Let $P=A_1\times A_2,$ where $A_1,A_2\subset \mathbb{R}$ are set of positive Lebesgue measure, and $Z\subset \mathbb{R}^2,$ be a set of zero Lebesgue measure. Can we always find positive Lebesgue measure sets $B_1,B_2\subset \mathbb{R}$ such that
$$B_1\times B_2 \subset \overline{P\setminus Z}?$$
What extra conditions ensure that the above is true?(I can show that the above is true if $P\setminus \overline{Z}$ is of positive measure then the above is true)
In this question
https://math.stackexchange.com/q/3767758/641816, it was shown that the result is true if $A_1=A_2=[0,1]$.
This is my attempt: Since $A_1,A_2$ are positive Lebesgue measure set we can find $a_1\in A_1, a_2\in A_2$ such that for any $r>0$ we have $B(a_1,r)\cap A_1, B(a_2,r)\cap A_2$ are sets of positive measure(in fact this phenomenon is true for almost every $a_1\in A_1,a_2\in A_2$).
Consider $$B_1^r=\overline{B(a_1,r)\cap A_1},\quad B_2^r=\overline{B(a_2,r)\cap A_2}$$
Then I think somehow one can show that there exits some $s,t>0$ such that $$B_1^s\times B_2^t\subset \overline{P\setminus Z}.$$
For $i=1,2$, let
$$Q:=C_1\times C_2,$$
where
$$C_i:=\{x\in A_i\colon\forall r>0\ |B(x,r)\cap A_i|>0\},$$
$B(x,r):=(x-r,x+r)$, and $|\cdot|$ denotes the Lebesgue measure in $\mathbb R^d$, for any $d\ge1$. Then $|C_i|=|A_i|>0$, by (say) the Lebesgue density theorem.
For all $(x_1,x_2)\in Q$, all real $r>0$, and all $i\in\{1,2\}$
$$|(B(x_1,r)\cap A_1)\times(B(x_2,r)\cap A_2)\setminus Z|
=|(B(x_1,r)\cap A_1)\times(B(x_2,r)\cap A_2)|
=|(B(x_1,r)\cap A_1)|\ |(B(x_2,r)\cap A_2)|>0.$$
So, for each $(x_1,x_2)\in Q$ and each $r>0$ there is some
$$(y_1,y_2)\in(B(x_1,r)\cap A_1)\times(B(x_2,r)\cap A_2)\setminus Z \\
=(B(x_1,r)\times B(x_2,r))\cap P\setminus Z.$$
Thus,
$$C_1\times C_2 =Q\subset \overline{P\setminus Z}$$
and $|C_i|=|A_i|>0$ for $i=1,2$, as desired.
Moreover you are saying that we won't lose any measure in obtaining a rectangle. That is great.
|
2025-03-21T14:48:31.627441
| 2020-07-28T03:30:43 |
366754
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Felipe Augusto de Figueiredo",
"Johannes Trost",
"Mateusz Kwaśnicki",
"https://mathoverflow.net/users/103291",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/37436"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631535",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366754"
}
|
Stack Exchange
|
Inaccurate results for the analytical expression of $\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right]$
I'm trying to plot a graph for the following expectation
$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right]=a 2^{-\frac{\kappa }{2}-1} b^{-\frac{\kappa }{2}} \theta ^{-\kappa } \left(\frac{\, _2F_2\left(\frac{\kappa }{2}+\frac{1}{2},\frac{\kappa }{2};\frac{1}{2},\frac{\kappa }{2}+1;\frac{1}{2 b \theta ^2}\right)}{\Gamma \left(\frac{\kappa }{2}+1\right)}-\frac{\kappa \, _2F_2\left(\frac{\kappa }{2}+\frac{1}{2},\frac{\kappa }{2}+1;\frac{3}{2},\frac{\kappa }{2}+\frac{3}{2};\frac{1}{2 b \theta ^2}\right)}{\sqrt{2} \sqrt{b} \theta \Gamma \left(\frac{\kappa +3}{2}\right)}\right)$$
where $a$ and $b$ are constant values, $\mathcal{Q}$ is the Gaussian Q-function, which is defined as $\mathcal{Q}(x) = \frac{1}{\sqrt{2 \pi}}\int_{x}^{\infty} e^{-u^2/2}du$ and $\gamma$ is a random variable with Gamma distribition, i.e., $f_{\gamma}(y) \sim \frac{1}{\Gamma(\kappa)\theta^{\kappa}} y^{\kappa-1} e^{-y/\theta} $ with $\kappa > 0$ and $\theta > 0$.
This equation was also found with Mathematica, so it seems to be correct.
Follows some examples, where I have checked the analytical results against the simulated ones.
When $\kappa = 12.85$, $\theta = 0.533397$, $a=3$ and $b = 1/5$ it returns the correct value $0.0218116$.
When $\kappa = 12.85$, $\theta = 0.475391$, $a=3$ and $b = 1/5$ it returns the correct value $0.0408816$.
When $\kappa = 12.85$, $\theta = 0.423692$, $a=3$ and $b = 1/5$ it returns the value $-1.49831$, which is negative. However, the correct result should be a value around $0.0585$.
When $\kappa = 12.85$, $\theta = 0.336551$, $a=3$ and $b = 1/5$ it returns the value $630902$. However, the correct result should be a value around $0.1277$.
Therefore, the issue happens as $\theta$ decreases. For values of $\theta > 0.423692$ the analytical matches the simulated results. The issue only happens when $\theta <= 0.423692$.
I'd like to know if that is an accuracy issue or if I'm missing something here and if there is a way to correctly plot a graph that matches the simulation.
This might be a bug in Mathematica kernel (I have already found one such issue with the $_2F_1$ function — Wolfram corrected it within a few months after I contacted them), but I suspect simple rounding error here: you seem to be subtracting huge values here, the hypergeometric function is of the order $10^11$ to $10^16$ for the given problematic parameters.
Maybe using a higher PrecisionGoal value helps. But this might also increase the calculation time considerably.
@MateuszKwaśnicki, I have also seen the same problem plotting with matlab... perhaps there is a way to simplify or derive the above expression another way with other functions....
@FelipeAugustodeFigueiredo: This is perhaps better suited to mathematica.se then. Anyway, have you tried working with higher precision? This can be done by writing SetPrecision[12.85, 50] etc. for all numeric constants.
@MateuszKwaśnicki, thanks for your suggestion, with that I was able to correctly get the expected results.
You just need to use a higher precision, using exact numbers when possible, as shown in the Mathematica work below (with $k:=\kappa$ and $t:=\theta$). (However, this question is indeed better suited for Mathematica SE.)
The accuracy issue with the evaluation of the hypergeometric function can be avoided for integer $\kappa$, since then the full expression reduces to an error function (see my answer to your previous question).
I tried this out for $\kappa=5$. Then
$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right]=\frac{a}{48 b^4 {\theta}^8} \left[\sqrt{\frac{2b}{\pi }} {\theta} \left(b {\theta}^2 \left(4 b {\theta}^2 \left(3 b {\theta}^2 \left(\sqrt{2 \pi b} {\theta}-2\right)+1\right)+1\right)+1\right)-e^{\frac{1}{2 b {\theta}^2}} \left(b {\theta}^2 \left(3 b {\theta}^2 \left(8 b^2 {\theta}^4-4 b {\theta}^2+1\right)+2\right)+1\right) \text{erfc}\left(\frac{1}{\sqrt{2b} {\theta}}\right)\right].$$
In the plot, for $a=3,b=1/5$ as a function of $\theta$, I compare the above expression with the error function (gold) with a numerical evaluation of the original expression with the hypergeometric function (blue). You see that the two evaluations agree for sufficiently large $\theta$, but then upon reducing $\theta$ the latter becomes numerically unstable while the former does not.
|
2025-03-21T14:48:31.627736
| 2020-07-28T05:37:56 |
366757
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Kyle Yip",
"https://mathoverflow.net/users/129185",
"https://mathoverflow.net/users/161854",
"mathworker21"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631536",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366757"
}
|
Stack Exchange
|
Equidistribution of $\{\sqrt{p}: p \text{ primes }\}$ modulo 1
I am trying to show $\{\sqrt{p}: p \text{ primes }\}$ is equidistributed modulo 1. Using Weyl's criterion, it is sufficient to show for each nonzero integer $k$,
\begin{equation}
\sum_{n \leq x}e(k\sqrt{n}) \Lambda(n) =o(x), \text{ as } x \to \infty.
\end{equation}
I found an exercise on chapter 13 "sum over primes" in Iwaniec H., Kowalski E. - Analytic number theory regarding this: it's the exercise 2 on page 348:
for any real number $\alpha \neq 0$,
$$
\sum_{n\le x} e(\alpha \sqrt{n}) \Lambda(n) \ll_\alpha x^{\frac{5}{6}} (\log x)^4.
$$
The idea is to use the following Vaughan's identity and estimate each sum separately: Let $y,z \geq 1$, then
$$
\Lambda(n)=\sum_{\substack{b \mid n\\ b \leq y}} \mu(b) \log \frac{n}{b}-\sum_{\substack{bc \mid n\\ b \leq y, c \leq z}} \mu(b) \Lambda(c)+\sum_{\substack{bc \mid n\\ b > y, c >z}} \mu(b) \Lambda(c)+R(n),
$$
where $R(n)=0$ when $n>z$, and $R(n)=\Lambda(n)$ when $n \leq z$.
I came up with the following lemmas as some basic preparation:
For any real number $\alpha> 0$,
$$
\sum_{n\le x} e(\alpha \sqrt{n}) \ll x^{\frac{1}{2}} \alpha^{-1}+\alpha^2,
$$
where the implicit constant does not depend on $\alpha$. (Using Van der Corput Method)
For any real number $\alpha > 0$, $N \leq x$, and any complex numbers $\beta_m,\gamma_n$ with $|\beta_m| \leq 1$ and $|\gamma_n| \leq 1$, we have
$$
A(M,N)=\sum_{M<m \leq 2M} \sum_{N<n \leq 2N} \beta_m \gamma_n e(\alpha \sqrt{mn}) \ll MN^{1/2} + M^{1/4} N^{3/4}\alpha^{-1/2}(\log N)^{1/2}+N^{5/4} \alpha.
$$
For any real number $\alpha > 0$, $N \leq x$, and any complex numbers $\beta_m$ with $|\beta_m| \leq 1$, we have
$$
B(x; N)=\sum_{N<n \leq 2N} |\sum_{mn \leq x} \beta_m e(\alpha \sqrt{mn})| \ll N^{1/2} x^{1/2}+x^{3/4} \alpha^{-1/2} (\log x)^{1/2}+x^{5/4} N^{-3/4} \alpha.
$$
(Use lemma 2 and lemma3, Estimate for bilinear form) For any real number $\alpha \neq 0$, and any complex numbers $\beta_m, \gamma_n$ with $|\beta_m| \leq 1, |\gamma_n| \leq 1$, $M,N<\sqrt{x}$, we have
$$
\sum_{\substack{mn \leq x\\ m>M,n>N}} \beta_m \gamma_n e(\alpha \sqrt{mn}) \ll x^{3/4}+x^{3/4} \alpha^{-1/2} (\log x)^{3/2}+x^{5/4} (N^{-3/4}+M^{-3/4})\alpha +x^{5/8}\alpha \log x +MN.
$$
By Vaughan's idenity, we have
\begin{align*}
\sum_{n\le x} e(\alpha \sqrt{n}) \Lambda(n)
&= \sum_{\substack{n\le x\\ b \mid n\\ b \leq y}} \mu(b) e(\alpha \sqrt{n}) \log \frac{n}{b}-\sum_{\substack{n\le x \\ bc \mid n\\ b \leq y, c \leq z}} \mu(b) \Lambda(c)e(\alpha \sqrt{n})\\
& \quad +\sum_{\substack{n\le x \\ bc \mid n\\ b > y, c >z}} \mu(b) \Lambda(c)e(\alpha \sqrt{n})+O(\sum_{n \leq z} \Lambda(n)).
\end{align*}
For the first sum, using lemma1 and partial summation, we get
\begin{align*}
\sum_{\substack{n\le x\\ b \mid n\\ b \leq y}} \mu(b) e(\alpha \sqrt{n}) \log \frac{n}{b}
&=\sum_{\substack{bl\le x\\ b \leq y}} \mu(b) e(\alpha \sqrt{bl}) \log l \\
&\ll \sum_{b \leq y} |\sum_{l \leq x/b} e(\alpha \sqrt{bl}) \log l |\\
&\ll_{\alpha} x^{1/2} \log x \log y+y^2 \log x.
\end{align*}
For the second sum, using lemma1 and partial summation, we get
\begin{align*}
\sum_{\substack{n\le x \\ bc \mid n\\ b \leq y, c \leq z}} \mu(b) \Lambda(c)e(\alpha \sqrt{n})
&=\sum_{\substack{bcl\le x \\ b \leq y, c \leq z}} \mu(b) \Lambda(c)e(\alpha \sqrt{bcl}) \\
&\ll \sum_{b \leq y, c \leq z} |\mu(b) \Lambda(c)| |\sum_{bcl \leq x} e(\alpha \sqrt{bcl})|\\
&\ll \sum_{b \leq y, c \leq z} |\sum_{bcl \leq x} e(\alpha \sqrt{bcl})| \Lambda(c)\\
&\ll_{\alpha} \sum_{b \leq y, c \leq z} (x^{1/2} (bc)^{-1}+bc) \Lambda(c)\\
&\ll x^{1/2} \log y\log z+y^2z^2.
\end{align*}
For the third sum, using lemma 4, it is easy to show that
\begin{align*}
\sum_{\substack{n\le x \\ bc \mid n\\ b > y, c >z}} \mu(b) \Lambda(c)e(\alpha \sqrt{n})
&=\log x \sum_{\substack{bk\le x \\ b > y, k >z}} \mu(b) \gamma_k e(\alpha \sqrt{bk}).
\ll_{\alpha}
x^{3/4} (\log x)^{5/2}+x^{5/4} (y^{-3/4}+z^{-3/4})\log x+yz\log x.
\end{align*}
However, it seems some of my estimates could be improved, because based on the above estimate, there is no way to choose y,z such that
$$
\sum_{n\le x} e(\alpha \sqrt{n}) \Lambda(n) =o(x).
$$
And the main issue here is that the $y^2z^2$ term from the second estimate seems to be too large, and this comes from the $\alpha^2$ term from lemma 1( the estimate for standard exponential sum). I had some ideas that could improve the $\alpha^2$ term to $\alpha^{3/2}$, but I guess $\alpha^{3/2}$ is still not good enough.
Any suggestions would be appreciated. Thank you very much！
Updated 1: I guess I figured out how to do this, for lemma 2 and lemma 3, we can take advantage that $n>N$ and use an improved version of lemma1.
Updated 2: Could someone point out what is the best known upper bound for
$$
\sum_{n\le x} e(\alpha \sqrt{n}) \Lambda(n)
$$
i don't see how $y^2z^2$ comes from the $\alpha^2$ term from lemma 1. specifically, I don't see how you got $x^{1/2} (bc)^{-1}+bc$.
For fixed $b,c$, $$|\sum_{bcl \leq x} e(\alpha \sqrt{bcl})|
\ll x^{1/2} (bc)^{-1} \alpha^{-1}+ \alpha^2 bc$$ follows easily from lemma 1. The sum is over $l \leq x/bc$, and the "$\alpha$" in lemma 1 will be replaced $\alpha \sqrt{bc}$.
I still don't see how you got the bound $x^{1/2}(bc)^{-1}+bc$. Should the $\ll$ be $\ll_\alpha$?
Oh yes, that's a typo. I have corrected that.
|
2025-03-21T14:48:31.628128
| 2020-07-28T05:48:37 |
366759
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jeremy Rickard",
"https://mathoverflow.net/users/22989"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631537",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366759"
}
|
Stack Exchange
|
Algebras with symmetric Cartan matrix
Let $A$ be a finite dimensional algebra with Cartan matrix $C_A$.$C_A$ being a symmetric matrix is equivalent to the Coxeter matrix being minus the identity matrix in case $A$ has finite global dimension and thus for finite global dimension having a symmetric Cartan matrix is a derived invariant.
Question 1: Is having a symmetric Cartan matrix a derived invariant for general algebras $A$?
edit: This is true by Jeremy Rickard's comment.
Question 2: Is it true for a general finite dimensional algebra with symmetric Cartan matrix that the dominant and Gorenstein dimensions are even in case they are finite?
Probably not, but maybe it is true when adding an extra condition. Is it true when the algebra additionally has finite global dimension?
(I edited question 2: Before it was about the finitistic dimension instead of the Gorenstein dimension, but there is an algebra with symmetric Cartan matrix and finitistic dimension 1).
Symmetry of the Cartan matrix is a derived invariant. If $C$ and $C'$ are the Cartan matrices of two derived equivalent finite dimensional algebras, then $C'=X^TCX$ for some invertible integer matrix $X$.
|
2025-03-21T14:48:31.628219
| 2020-07-28T06:30:10 |
366761
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jay Taylor",
"Mare",
"https://mathoverflow.net/users/22846",
"https://mathoverflow.net/users/61949"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631538",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366761"
}
|
Stack Exchange
|
Computation of the Lusztig a-function
See for example https://www.sciencedirect.com/science/article/pii/0021869387901542 for the definition of the Lusztig a-function.
Question 1: Is there a table for the values of Lusztig's a-function for a given Dynkin type?
Can one at least find those values when the corresponding simple Lie-algebra has rank <=4? I try to find a reference for the case $G_2$ especially.
Question 2: Is there a computer program to calculate those values for small Dynkin types?
If $W$ is a finite Weyl group then it is known that for $w \in W$ we have $\mathbf{a}(w) = \mathbf{a}{\lambda}$ for some complex irreducible character $\lambda \in \mathrm{Irr}(W)$, see Prop. 2.3.14 of the book by Geck–Jacon. The values $a{\lambda}$ have all been computed by Lusztig (naturally) and are contained in the CHEVIE software maintained by Jean Michel. See also Chapter 6 of the book by Geck–Pfeiffer. There combinatorial descriptions are given for $\mathbf{a}_{\lambda}$ when $W$ is classical.
@JayTaylor Thank you. That probably answers my questions, so you might want to turn this into an answer. Can you give an example, for example for G2, with which commands one can obtain those values using GAP?
In Addition to Geck-Pfeiffer: Small values of the a-function are also contained in Geck, Jacon - Representations of Hecke algebras at roots of unity. In particular, for $G_2$ it's in Table 1.3.; for $F_4$ it's table 1.2.; other values are available through combinatorially formulas (for example type $A$ is completely covered in section 2.8) and Remark 1.3.11 contains a fairly straight-forward algorithm to compute the a-function in general.
|
2025-03-21T14:48:31.628394
| 2020-07-28T07:51:28 |
366765
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abhiram Natarajan",
"Benedict W. J. Irwin",
"Carl-Fredrik Nyberg Brodda",
"Carlo Beenakker",
"Claus",
"Dmitry Kamenetsky",
"Hao S",
"Jan Kyncl",
"John Coleman",
"LSpice",
"Louis Thibault",
"Matt",
"Per Alexandersson",
"Terry Tao",
"Toffomat",
"Ville Salo",
"YCor",
"domotorp",
"efs",
"https://mathoverflow.net/users/103226",
"https://mathoverflow.net/users/1056",
"https://mathoverflow.net/users/106013",
"https://mathoverflow.net/users/109085",
"https://mathoverflow.net/users/110087",
"https://mathoverflow.net/users/111644",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/120914",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/134782",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/156936",
"https://mathoverflow.net/users/161329",
"https://mathoverflow.net/users/161819",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/24076",
"https://mathoverflow.net/users/56833",
"https://mathoverflow.net/users/766",
"https://mathoverflow.net/users/89084",
"https://mathoverflow.net/users/955",
"user161819"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631539",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366765"
}
|
Stack Exchange
|
Issue UPDATE: in graph theory, different definitions of edge crossing numbers - impact on applications?
QUICK FINAL UPDATE: Just wanted to thank you MO users for all your support. Special thanks for the fast answers, I've accepted first one, appreciated the clarity it gave me. I've updated my torus algorithm with ${\rm cr}(G)$. Works fine on my full test set, i.e. evidence for ${\rm cr}(G)={\rm pcr}(G)$ on torus. More on this later, will test sharper bound from last answer as well. I'm going to submit in time! Thanks again MO users for all your help!
Original post:
I apologize if „crisis“ is too strong a word, but I am in a mode of panic, if that's the right word: In two weeks, I should be submitting my Ph.D. Thesis, but I have just received bad news, or I should say information that makes me very concerned. It is really an emergency situation:
My thesis is in computer science, algorithms related to graph drawings on the sphere and the torus. One of the cornerstone mathematical results I am relying on is the graph edge crossing lemma (or edge crossing inequality). It gives a lower bound for the minimum number of edge crossings ${\rm cr}(G)$ for any drawing of the graph $G$ with $n$ vertices and $e$ edges
$${\rm cr}(G)\geq \frac{e^3}{64n^2}$$
for $e>4n$.
PROBLEM: I am reading in the article of Pach and Tóth that there is a possibility that mathematics papers on crossing numbers operate with different definitions. There is the crossing number ${\rm cr}(G)$ (minimum of edge crossings in a drawing of $G$), but also the pair crossing number
${\rm pcr}(G)$, the minimum number of edge pairs crossing in a drawing of $G$.
I double-checked my algorithms and, based on this definition, I clearly apply the pair crossing number ${\rm pcr}(G)$
CRITICAL QUESTION: Can you confirm to me that the edge crossing lemma remains valid on the sphere and the torus also for the pair crossing number ${\rm pcr}(G)$?
Reference: János Pach and Géza Tóth. Which crossing number is it anyway? J. Combin. Theory Ser. B, 80(2): 225–246, 2000.
And Wikipedia article as a starting point https://en.wikipedia.org/wiki/Crossing_number_inequality
I don't really know anything about crossing numbers, but I can appreciate how stressful this must be for you. I hope that you are able to patch things up in time!
@Carl-FredrikNybergBrodda You are right!! The situation is extreme, I just looked up the link from the answer below
What is the difference between pcr and cr? The names really do not show how they differ, at least not to me.
@PerAlexandersson --- as I understand it two edges may intersect multiple times; this multiplicity is counted in cr but not in pcr, hence pcr $\leq$ cr.
I suppose that the downvoter never felt any stress while doing his/her Ph.D Thesis...
Frege once had to write "A scientist can hardly meet with anything more undesirable than to have the foundations give way just as the work is finished. I was put in this position by a letter from Mr. Bertrand Russell when the work was nearly through the press. "...
If there is a wider question based on the 'submission of the thesis given a problem' aspect, it might be worth asking/checking on the Academia StackExchange. You've already done the work, so having to redo anything major is probably out of scope at this point (funding, time, ... etc.), and that shouldn't prevent you from submitting or getting the PhD. It's just a matter of clearly stating any assumptions made and the limitations surrounding the work, and preparing for the viva with this in mind. Don't fall into the trap of having the PhD drag on forever to make it 'perfect'.
Maybe you could accept the answer that is here, and then add your own answer after you successfully defend your dissertation. The vast majority of people who read this question are rooting for you.
I'd suggest (and would really appreciate) to shorten the title to its mathematical part: "In graph theory, different definitions of crossing numbers".
@YCor ok will make shorter
Look on the bright side - you've joined an awesome community and got many likes! :)
This awesome discussion was spotlighted on Hacker News: https://news.ycombinator.com/item?id=24049428 .
As a fellow Ph.D, albeit in a different field: (1) hang in there and (2) plan for a long, uneventful vacation after your defense. Right now you're neck deep in heroic self-sacrifice (as you should) in order to pull this off, but please hear me when I say "extraordinary effort requires extraordinary recovery". You'll need rest after the battle; please take care of yourself. Good hunting in the meantime. We're all rooting for you.
Very happy for you! Good luck.
A proof of the Strong Hanani-Tutte for the torus just appeared on arxiv: https://arxiv.org/abs/2009.01683
What is known about cr(G) vs pcr(G) do we even know they are in the same order? even for the planar case?
$\DeclareMathOperator\cr{cr}\DeclareMathOperator\pcr{pcr}$For the pair crossing number $\pcr(G)$, the short answer is yes the crossing lemma holds for drawings on the sphere, but it is not known whether it also holds on the torus.
The best and most current reference for you could be the survey article from Schaefer, updated in February 2020: “The Graph Crossing Number and its Variants: A Survey” from the Electronic Journal of Combinatorics
(https://doi.org/10.37236/2713).
The relevant pages for you are pages 5 and 6 with the following quote from Schaefer:
“Since the Hanani–Tutte theorem is not known to be true for the torus, this means that we do not currently have a proof of the crossing lemma for $\pcr$ or $\pcr_−$ on the torus.”
Generally, $\pcr(G)\leq \cr(G)$. It is still an open problem whether they are equal or not. The first proofs of the crossing lemma did not make the distinction. The first one to raise the ambiguity was Mohar (1995) in a conference talk.
The Pach and Tóth (2000) paper that you mention does make the distinction between $\pcr(G)$ and $\cr(G)$, and applies Hanani–Tutte in the proof of the crossing lemma, which ensures that it also holds for $\pcr(G)$.
The issue is that you can apply Hanani–Tutte for the sphere (and the projective plane), but you cannot apply it for the torus. For surfaces of genus $\geq4$ it is known to be false, see Fulek and Kynčl (2019). This means the torus is really “in-between”.
Edit: Adding the references
Bojan Mohar (1995): Problem mentioned at the special session on Topological Graph Theory, Mathfest, Burlington, Vermont. (cited from: L.A. Székely (2016): Turán’s Brick Factory Problem: The Status of the Conjectures of Zarankiewicz and Hill. In: R. Gera et al. (eds.)(2016): Graph Theory—favorite conjectures and open problems. 1.)
Hanani–Tutte Theorem https://en.wikipedia.org/wiki/Hanani%E2%80%93Tutte_theorem
Radoslav Fulek and Jan Kynčl (2019): Counterexample to an Extension of the Hanani–Tutte Theorem on the Surface of Genus 4. Combinatorica, 39(6):1267–1279
From OP's point of view this could be viewed as glass half-full rather than glass half-empty. Their dissertation results hold unequivocally on the sphere and might hold on the torus, though it is an open problem if they do. It is certainly legitimate to study what follows from a given conjecture being true. It could even be spun as a feature rather than a bug of the dissertation. If the results in fact fail on the torus then you know that the conjecture must be false. Potentially, it could open up a fruitful avenue of attack.
I think in algebraic topology literature, Hanani-Tutte is known as a version of the Flores-Van Kampen theorem, if that's a helpful link
Adding to my comment, here is a 2019 reference "Invariants of graph drawings in the plane" from A. Skopenkov https://arxiv.org/pdf/1805.10237.pdf
@ClausDollinger . The clarity is helpful. Thanks for your fast help. Much appreciated. I'm glad I can keep my sphere algorithm. I'm checking whether I can adapt my torus algorithm to ${\rm cr}(G)$ within the next 5 days
Assuming an unpublished Ramsey-type result by Robertson and Seymour about Kuratowski minors [FK18, Claim 5], which is now "folklore" in the graph-minor community,
an asymptotic variant of the crossing lemma, $\operatorname{cr}(G)\ge \Omega(e^3/n^2)$, is true even for the pair crossing number on a fixed surface, such as a torus.
With Radoslav Fulek [FK18, Corollary 9] we have shown that [FK18, Claim 5] implies an approximate version of the Hanani–Tutte theorem on orientable surfaces.
In particular, [FK18, Claim 5] implies that there is a constant $g$ such that for every graph $G$ that can be drawn on the torus with every pair of independent edges crossing an even number of times, $G$ can be drawn on the orientable surface of genus $g$ without crossings.
This gives an upper bound $3n + O(g)$ on the number of edges of every such graph $G$, and this can be used in the probabilistic proof of the crossing lemma, as described on p. 5-6 of Marcus Schaefer's survey [S20], mentioned in Claus Dollinger's answer. See also [SSSV96, Theorem 4.1].
References:
[FK18] https://dx.doi.org/10.4230/LIPIcs.SoCG.2018.40, https://arxiv.org/abs/1803.05085 - R. Fulek and J Kynčl, The $\mathbb Z_2$-genus of Kuratowski minors
[SSSV96] https://doi.org/10.1007/BF02086611 - F. Shahrokhi, L. A. Székely, O. Sýkora and I. Vrt'o, Drawings of graphs on surfaces with few crossings, Algorithmica 16, 118-131 (1996)
[S20] https://doi.org/10.37236/2713 - M. Schaefer, The Graph Crossing Number and its Variants: A Survey, The Electronic Journal of Combinatorics, DS21: Feb 14, 2020.
Edit:
"Strong Hanani-Tutte for the Torus" by Radoslav Fulek, Michael J. Pelsmajer and Marcus Schaefer has just appeared on arxiv: https://arxiv.org/abs/2009.01683
Is Schaefer's survey Hanani–Tutte and related results (MSN)?
I meant the survey about crossing numbers, mentioned in Claus Dollinger's answer. I will make an edit and add the reference to make it clear.
By the way, you know that diacrits like ˇ are allowed in your username if you want, right?
@JanKyncl Thank you for your help as well, much appreciated. My advisor says the faculty will not accept asymptotics. I'm checking whether I can put ${\rm cr}(G)$ into my torus algorithm
"the faculty will not accept asymptotics"?
@user161819 I wanted to make a comment but it got too long, so putting it as an answer. But please take it just as a comment for later, once everything is finished:
If I understand your comment to my answer correctly, you are aiming to change your algorithm for the torus so it works with ${\rm cr}(G)$. I think the whole MO community is keeping their fingers crossed, wishing you that you can successfully complete everything in time!
Looking at the far horizon, I wanted to make a suggestion to you. Once you have changed your torus algorithm and completed your thesis, you will have effectively two algorithms in your hands for the torus: The old one based on ${\rm pcr}(G)$ and the new one based on ${\rm cr}(G)$. I am saying the obvious here, keep both of them, they can really be fruitful for future research.
(1) Obviously, your two algorithms could support research on the big open question whether ${\rm pcr}(G)\stackrel{\rm ?}{=}{\rm cr}(G)$ or not. They
could produce experimental evidence, ideas, and insights for a
future proof of equality, or an actual counterexample. (Again, I am
saying the obvious here.)
(2) To really pressure-test ${\rm pcr}(G)\stackrel{\rm ?}{=}{\rm cr}(G)$ on the torus, it would be interesting to also try the
best known to date lower bound for ${\rm cr}(G)$
$$\frac{1}{29}\frac{e^3}{n^2}$$ for graphs with $e>7n$. This lower bound is from
Eyal Ackerman (2019): "On topological graphs with at most four
crossings per edge", Computational Geometry, 85: 101574, 31,
doi:10.1016/j.comgeo.2019.101574 (probably you are aware of it from
the Wikipedia article that you quoted).
I think your question and this whole topic are really important. László Székely calls it one of the "foundational problems" and devotes a whole section to it in his article Turán’s Brick Factory Problem: The Status of the Conjectures of Zarankiewicz and Hill. In: R. Gera et al. (eds.)(2016): Graph Theory—favorite conjectures and open problems. 1.)
For now, fingers crossed that you can complete your thesis in time!
Thanks for your comment!! Much appreciated. And thanks again for your help. I'm very interested in this Ackermann bound, will take a look at it
One good thing, my prototype for updated torus algorithm: tested on on first 2 graphs from my test set, and went ok
The term "Ackermann bound" generally refers to a bound of a type far, far worse than the one given here: https://en.wikipedia.org/wiki/Ackermann_function
@TerryTao Terry you are very right. Huge difference between the Wilhelm Ackermann bound that you mention and the Eyal Ackerman bound I am quoting here! Good to make the distinction.
|
2025-03-21T14:48:31.629229
| 2020-07-28T08:31:57 |
366768
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Houa",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/145357"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631540",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366768"
}
|
Stack Exchange
|
How to estimate the order of this integral with parameter
Some introduction:
Given a homogeneous structure called "dilation" in $R^n$: For $t\geq 0$
$$D_t: R^n\rightarrow R^n$$
$$D_t(x)=(t^{a_1}x_1,...,t^{a_n}x_n)$$
where $1=a_1\leq...\leq a_n$, and $a_i$ are all integers. And we call $Q=a_1+...+a_n$ the homogeneous dimension. In our problem, we only consider when $Q>n\geq 2$.
Now consider the integral:
$$J(r)=\int_{[0,1]^n}\frac{dx}{P(x,r)}=\int_{[0,1]^n}\frac{dx}{f_n(x)r^n+f_{n+1}(x)r^{n+1}+...+f_Q(x)r^Q}$$
where $f_k(x)$ satisfies:
(1) $f_k(D_t(x))=t^{Q-k}f_k(x)$ for all $x\in R^n$ and $t\geq0$
(2) $f_k(x)$ is the combination of some positive monomials. (Examples will be shown below)
(3) $f_Q(x)=Constant>0$. (This property follows from other theorems and propositions, but they are too many so I don't describe them here.)
Four examples are the followings:
(ex1) In $R^2$, $D_t(x)=(tx_1,t^2x_2)$, so $Q=3$. And Let $P(x,r)=x_1r^2+r^3$.
(ex2) In $R^3$, $D_t(x)=(tx_1,tx_2,t^2x)$, so $Q=4$. Let $P(x,r)=(x_1+x_2)r^3+r^4$
(ex3) In $R^3$, $D_t(x)=(t^{1}x_1,t^2x_2,t^{3}x_3)$, so $Q=6$. Let $P(x,r)= x_1^3r^3+(x_2+3x_1^2)r^4+5x_1r^5+3r^6$
(ex4) In $R^3$, $D_t(x)=(t^{1}x_1,t^2x_2,t^{3}x_3)$, so $Q=6$. Let $P(x,r)= x_1x_2r^3+(x_2+2x_1^2)r^4+3x_1r^5+r^6$
(You will find that $x_n$ doesn't make effort. In my work $x_n$ do make no sense in the integral but this follows from other theorems, and it doesn't matter here. )
Problem: Find the order of $J(r)$ when $r$ goes to $0^+$. Like the following description.
Attempt and information: I guess $J(r)=\frac{1}{r^\alpha}I(r)$, where the $\alpha$ is the "critical value", that is:
(i) $\liminf_\limits{r\rightarrow0^+}I(r)>0$.
(ii) for any $\epsilon>0$, $\lim_\limits{x\rightarrow0^+}r^\epsilon I(r)=0$.
I will give the reason why I guess so in the below. I can show that $g_p(r)=r^p J(r)$, then there exists $p_0$ s.t. when $a<p_0$, $\lim_\limits{r\rightarrow0^+}g_a(r)>0$ and when $a>p_0$, $\lim_\limits{r\rightarrow0^+}g_a(r)=0$. But I can't show $\lim_\limits{r\rightarrow0^+}g_{p_0}(r)>0$, that is, I can't show the (i) above. (see https://math.stackexchange.com/questions/3769564/how-to-find-the-critical-index-a-of-xafx) One gave a counterexample for the proposition in that link. But its counterexample will not appear in this problem. Because this is a rational fractional integral. The $I(r)$ I guess will be like the combination of $\log$ and $\arctan$.
The four example have the order estimates:
(ex1) We can calculate directly:
$$J(r)=\frac{1}{r^2}\ln(1+\frac{1}{r})=\frac{1}{r^2}I(r)$$
where $ I(r)$ satisfies (i)(ii) above.
(ex2)
$$J(r)=\frac{1}{r^3}I(r)$$
where $I(r)$ can be calculate or one can use Dominate convergence theorem to estimate that $I(r)$ satisfies (i)(ii)
(ex3)
$$J(r)=\frac{1}{r^{3+2/3}}I(r)$$
see https://math.stackexchange.com/questions/3718932/estimate-a-integral-with-parameter
(ex4)
$$J(r)=\frac{1}{r^{3}}I(r)$$
First
$$J(r)=\frac{1}{r^3}\int_{[0,1]^2}\frac{dxdy}{xy+(y+2x^2)r+3xr^2+r^3}=\frac{1}{r^3}I(r)$$
we can show $I(r)$ satisfies (i)(ii):
(i) change variables:
$$I(r)=\int_{0}^{1/r^2}\int_{0}^{1/r}\frac{dxdy}{xy+(y+2x^2)+3x+1}$$
and then obviously.
(ii) for $3>\epsilon>0$ (the part $\epsilon\geq 3$ follows from the part $3>\epsilon>0$),
$$r^\epsilon I(r)=\int_{[0,1]^2}\frac{r^\epsilon}{xy+(y+2x^2)r+3xr^2+r^3}dxdy=\int_{[0,1]^2}h_r(x,y)dxdy=\int_{(0,1)^2}h_r(x,y)dxdy$$
Pointwisely $\lim_\limits{r\rightarrow0^+}h_r(x,y)=0$ in $(0,1)^2$. Now look for a dominating function in $(0,1)^2$:
$$\frac{1}{h_r(x,y)}\geq \frac{xy}{r^\epsilon}+r^{3-\epsilon}\geq C(xy)^{1-\frac{\epsilon}{3}}$$
So $h_r(x,y)\leq \frac{C}{(xy)^{1-\frac{\epsilon}{3}}}$ in $(0,1)^2$, which is integrable. By DCT, we have $I(r)$ satisfying (i)(ii). But this method doesn't work in other examples like (ex3).
Based on the four examples, I guess $$J(r)=\frac{1}{r^\alpha}I(r).$$ But I can't show how to find the critical value $\alpha$ and even it's difficult to show the existence of critical value
It looks like you care only about the order of magnitude (i.e., an answer up to a constant factor), in which case it is fairly easy.
First, ignore all coefficients. Setting them to $1$ just changes the answer at most constant number of times. Now, suppose we have the denominator of the form $\sum_{(\alpha,\beta)} x^\alpha r^\beta$ where $\alpha$ is a multi-index with real entries and $\beta$ is a real number. The sum is assumed to be finite. Make the change of variable $x_j=e^{-y_j}$. Now, at each point, only the maximal term matters (up to a factor that is the total number of terms). In terms of $y$'s, the condition of maximality of $x^\alpha r^\beta$ is $y_j\ge 0$,
$$
\langle y,\alpha-\alpha'\rangle\le (\beta'-\beta)\log(1/r)
$$
for all $(\alpha',\beta')\ne(\alpha,\beta)$. This domain is just a fixed polyhedron $P_{\alpha,\beta}$ stretched $\log(1/r)$ times (we keep only those with non-empty interiors in what follows; also I call it a "polyhedron" though, technically, it can be unbounded). Thus,
$$
J(r)\asymp\sum_{(\alpha,\beta)}r^{-\beta}\int_{(log\frac 1r)P_{\alpha,\beta}}e^{\psi_{\alpha,\beta}(y)}\,dy
$$
where $\psi_{\alpha,\beta}(y)=\langle \alpha-e,y\rangle$, $e=(1,\dots,1)$.
Now the life becomes straightforward. All you need is to find the order of magnitude of each integral. I'll drop the indices $\alpha,\beta$ for brevity. Let $F$ be the face of $P$ on which $\psi$ attains its maximum $p$ and let $d$ be the dimension of $F$. If $\psi\equiv 0$ (i.e., $\alpha=e$), we just have $F=P$ and $\int_{(\log\frac 1r)P}e^{\psi}=V(P)\log^d(1/r)$. Consider now the non-trivial situation when $\psi$ is not $0$. Then we can rotate and shrink the coordinate system so that $-\psi(y)$ becomes a new variable $t$. Also we can shift $P$ along this coordinate so that the face $F$ lies on the corresponding coordinate hyperplane $\{t=0\}$. Then the integral in question is just
$$
e^{p\log(1/r)}(\log^{D-1}\frac 1r)\int_{0}^\infty e^{-t}S_P(\frac t{\log{1/r}})\,dt
$$
where $S_P(\tau)$ is the $D-1$-dimensional volume of the cross-section of $P$ by the hyperplane $\{t=\tau\}$. By the general convex geometry nonsense, for small $\tau$, $S_P(\tau)=v_d\tau^{D-1-d}+v_{d-1}\tau^{D-d}+\dots+v_0\tau^{D-1}$ where $v_d>0$ and then it becomes smaller (look up "mixed volumes" on Google if you are interested in the details), whence the leading term in the integral becomes $\log^d\frac 1r$ with some coefficient depending on $P$. Thus, the final answer for the integral we are interested in with the factor $r^{-\beta}$ is
$$
\asymp r^{-p_{\alpha,\beta}-\beta}\log^{d_{\alpha,\beta}}\frac 1r
$$
We have several competing terms like that, so the winning one is the one with largest $p+\beta$ and among those the one with the largest $d$.
In your last example $x_1x_2+x_1^2r+x_2r+x_1r^2+r^3$ (I ignore $r^3$ that can be carried out and all the coefficients), we have $5$ polyhedra and functionals (I drop the trivial restrictions $y_1,y_2\ge 0$):
$$
P_{1,1,0}=\{-y_1+y_2\le 1, y_1\le 1, y_2\le 2, y_1+y_2\le 3\},
\\
\psi_{1,1,0}(y)=0
\\
P_{2,0,1}=\{y_1-y_2\le -1, 2y_1-y_2\le 0, y_1\le 1,2y_1\le 2\},
\\
\psi_{2,0,1}(y)=y_1-y_2
\\
et\ cetera.
$$
Here $P_{1,1,0}$ dominates and yields $\log^2\frac 1r$ but it may be instructive to find the contribution of $P_{2,0,1}$. In this case (just draw the picture) $p=-1$, $\beta=1$, $d=1$, so we get $\log\frac 1r$.
Sorry I am difficult to understand some point could you give more details? (1) From "$x_j=e^{-y_j}$" to "where $\psi_{\alpha,\beta}(y)=<\alpha-e,y>$" . I am confused. For example (ex1) $\int_{0}^1\frac{dx}{r+x}$ (ignore $r^2$ that can be carried out). After $x=e^{-y}$, we have $\int_{0}^{+\infty}\frac{dy}{e^yr+1}$. By your fomula, It $\asymp \int_{\log(1/r)P_{0,1}}e^{-y}dy+\int_{\log(1/r)P_{1,0}}1dy$, so in this case what are $P_{0,1}$ and $P_{1,0}$? can they be written down? (question (2) in next comment)
And question (2): In your answer, for the last example (ex4), your answer show that the $I(r)\asymp r\log(\frac{1}{r})$, that is $J(r)\asymp\frac{1}{r^2}I(r)$. But in my description, (ex4) is $J(r)\asymp\frac{1}{r^3}I(r)$. Is there anything wrong in my description or something else?
@Houa Oops, I forgot about $r^\beta$ in my final answer. I edited. So the contribution of $P_2$ is actually $\log(1/r)$. However, it is still $P_1$ that dominates (contributing $\log^2(1/r)$, so $J(r)\asymp r^{-3}\log^2(1/r)$ in full agreement with what you wrote). As to question 1, by my (corrected) formula the integral is $\asymp r^{-1}\int_{\log(1/r)P_{0,1}}e^{-y}dy+\int_{\log(1/r)P_{1,0}}1dy$. From the definition with linear inequalities (just drop $\log(1/r)$ on the RHS), $P_{1,0}={0\le y\le 1}=[0,1]$ and $P_{0,1}={y>0,-y\le-1}=[1,+\infty]$ so you get your $\log(1/r)+1\asymp\log(1/r)$
@Houa Is it clearer now? If not, what should I clarify? (I addressed two explicit questions you asked but I suspect more clarification may be still needed, so don't hesitate to ask for it but try to explain what exactly you are confused about)
I've understood how to calculate $P_{\alpha,\beta}$ and use this method to calculate some examples. The answer is right. Your idea is wonderful! Now I want to know some details. Question (3): there will be some $P_{\alpha,\beta}$ with its n-dimension Lebesgue
measure zero. In this case, we can ignore that term, right? Because its integral is zero. Question (4): It seems that you didn't show the existence of maximum of $\psi_{\alpha,\beta}(y)$. It seems not obvious, for $P_{\alpha,\beta}$ may be unbounded.
Question (5): the changing of variables makes me confused. Do you rotate $P$ first then let $-\psi(y)=t$? For example , if $P={1<y_1<2, 1<y_2<3}$ and $\psi$ attains the max in line $l={y_2=3,1<y_1<2}$. Then we first rotate the rectangle and shift it s.t. $l$ lies in $y_1=0$? Then let $-\psi(y)=t$? In fact I don't know how you get the integral after changing the variable. Is there any formula? Question (6) : Is the expansion of $S_p$ one type of " Steiner formula" ? (In fact if I understand question (5) then I think I can almost understand the whole answer.)
(3) Yes, of course. As I said "ignore everything with empty interior". (4) It follows from the physical meaning of your particular problem: since you have a constant term, the integral should converge, and if $\psi$ is unbounded (or the face $F$ on which the maximum is attained is unbounded), you'll certainly just get $+\infty$ immediately. This can happen in the general setting I considered but your particular case is tame. (5) Yes, you can think of it as first rotating the rectangle and then shifting it or you can think of first shifting the origin and then rotating the coordinate frame.
@Houa (6) Yes, it is related to it (read the previous comment for (3-5); I forgot to address it to you). As to the integral, just integrate term by term. The claim is that if $f(t)$ is any function that equals $p(t)=\sum_k c_kt^k$ ($c_k>0$) near the origin and is always between $0$ and $p(t)$, then $\int_0^\infty f(\delta t)e^{-t},dt\approx \sum_k k!c_k\delta^k$ as $\delta\to 0$.
Thinks for answering patiently! To (5) yesterday, I try to do the changing of variable, could you point out something wrong for me? (For brevity let $b=\log(1/r)$). First $\int_{bP}e^{\psi(y)}dy=b^n\int_{P}e^{b\psi(y)}dy$, then let $-\psi(y)=t$, we have $b^n\int_{A(P)} e^{-bt} |det(A)|dtdt_2...dt_n=b^n|det(A)|\int_{-p}^\infty e^{-bt}S_P(t)dt$, where $A$ is a transform matrix (then |det(A)| will be ignored,) and since $-t\leq p$ now. Then let $t'=t+p$, we have $e^{bp}b^n\int_{0}^\infty e^{-bt'}S_P(t'-p)dt$. Lastly, let $bt'=u$ we have $e^{bp}b^{n-1}\int_{0}^\infty S_P(\frac{u}{b}-p)du$.
(going on with the last comment) in your answer is $S_P(t/b)$ but in my comment is $S_P(u/b-p)$, where $b=\log(1/r)$ and $p$ is the maximum. And others almost the same...(In last comment some typo mistake I can't edit…… The last integral is $e^{bp}b^{n-1}\int_{0}^\infty e^{-u}S_P(u/b-p)du$)
@Houa That's all right, but note that I also first shifted the origin to the point where the maximum is attained (so I should rather say "making $t=p-\psi(y)$ a new coordinate" if I had used the original $y$, not $y$ with respect to the shifted origin), in which case the shift in the argument of $S_P$ disappears and our formulae agree. I apologize for being somewhat confusing here
In fact neither $S_P(t/b)$ nor $S_P(t/b-p)$ changes the order of $J(r)$, so it doesn't matter :-). And some extra questions. (7) if, say $J(r)=\int_{(-1,1)^3}\frac{dxdydz}{|x-y|+|x^2-z|r+r^2}$, do you think the method you did in the answer still works here? Or we should find some new ways? (8) would you mind give your real name and institution to us? Since your idea helps us a lot with a lemma in our research, and we want to write an acknowledgement in paper (if we write this lemma in it). My email is in my userpage profile page
(7)(8) may be the last two questions and Thank you very much these days
@Houa (7). Your new $J(r)$ is more tricky. Part of what I said still applies, but now the maximality condition of each individual term is different because you can no longer say that there is no cancellation between the terms though most of the time it is still so, but you now encounter extra fancy regions where the largest terms may cancel. If you want to pursue this venue, just post a new question: this comment thread is already too long. (8) Just credit MO. They have guidelines for how to do it somewhere.
|
2025-03-21T14:48:31.630144
| 2020-07-28T08:53:52 |
366771
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Francesco Polizzi",
"Max Horn",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/7460",
"https://mathoverflow.net/users/8338"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631541",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366771"
}
|
Stack Exchange
|
Are these two kernels isomorphic groups?
We have a finitely presented, infinite group $\mathsf{B}$,
coming from a geometric topology problem (it is the quotient of a braid group for a genus 2 surface).
It is generated by elements
\begin{equation}
\{r_{11}, \, t_{11}, \, r_{12}, \, t_{12}, \, r_{21}, \, t_{21}, \,
r_{22}, \, t_{22}, \, z \},
\end{equation}
subject to the following relations (here $[a, \, b]=aba^{-1}b^{-1}$):
\begin{equation*}
\begin{aligned}
& z^2=1 \\
& [r_{12}^{-1}, \, t_{12}^{-1}] \, t_{12}^{-1} \,
[r_{11}^{-1}, \, t_{11}^{-1}] \, t_{11}^{-1}\, (t_{11} \, t_{12}) =
z & \\ & [r_{21}^{-1}, \, t_{21}] \; t_{21} \;
[r_{22}^{-1}, \, t_{22}] \, t_{22}\, (t_{22}^{-1} \, t_{21}^{-1})=
z^{-1} \\
& \\
& [r_{11}, \, r_{22}]=1 & &
[r_{12}, \, r_{22}]=1 \\
& [r_{11}, \, r_{21}]=1 &
& [r_{12}, \, r_{21}]=
z^{-1}\,r_{21}\,r_{22}^{-1}\,z\,r_{22}\,r_{21}^{-1} \\
& [r_{11}, \, t_{22}]=1 & &
[r_{12}, \, t_{22}]=z^{-1} \\
& [r_{11}, \, t_{21}]=z^{-1} & &
[r_{12}, \, t_{21}]=[z^{-1}, \, t_{21}] \\
& [r_{11}, \, z]=[r_{21}^{-1}, \, z] &
& [r_{12}, \, z]=[r_{22}^{-1}, \, z] \\
& \\
& [t_{11}, \, r_{22}]=1 & &
[t_{12}, \, r_{22}]= t_{22}^{-1}\, z \, t_{22} \\
& [t_{11}, \, r_{21}]=t_{21}^{-1}\, z \, t_{21}
& & [t_{12}, \, r_{21}]= [t_{22}^{-1}, \, z] \\
& [t_{11}, \, t_{22}]=1 & &
[t_{12}, \, t_{22}]=[t_{22}^{-1}, \, z] \\
& [t_{11}, \, t_{21}]=[t_{21}^{-1}, \, z] &
& [t_{12}, \, t_{21}]=
t_{22}^{-1}\, z \, t_{22} \, z^{-1} \, t_{21} \, z \, t_{22}^{-1}\,z^{-1}
\,t_{22}\,t_{21}^{-1} \\
& [t_{11}, \, z]=[t_{21}^{-1}, \, z] &
& [t_{12}, \, z]=[t_{22}^{-1}, \, z] \\
\end{aligned}
\end{equation*}
Next, we consider the two extra-special groups of order $32$, whose labels are $[32, \, 49]$ and $[32, \, 50]$ in GAP4 notation, and whose presentations are as follows:
\begin{equation}
\begin{split}
\mathsf{G}(32, \, 49) = \langle \, & \mathsf{r}_1, \, \mathsf{t}_1, \,
\mathsf{r}_2,\, \mathsf{t}_2, \, \mathsf{z} \; | \; \mathsf{r}_{j}^2 =
\mathsf{t}_{j}^2=\mathsf{z}^2=1, \\
& [\mathsf{r}_{j}, \, \mathsf{z}] = [\mathsf{t}_{j}, \, \mathsf{z}]= 1, \\
& [\mathsf{r}_j, \mathsf{r}_k]= [\mathsf{t}_j, \mathsf{t}_k] = 1, \\
& [\mathsf{r}_{j}, \,\mathsf{t}_{k}] =\mathsf{z}^{- \delta_{jk}} \, \rangle
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\mathsf{G}(32, \, 50) = \langle \, & \mathsf{r}_1, \, \mathsf{t}_1, \,
\mathsf{r}_2,\, \mathsf{t}_2, \, \mathsf{z} \; | \; \mathsf{r}_{1}^2 =
\mathsf{t}_{1}^2=\mathsf{z}^2=1, \; \mathsf{r}_{2}^2 =
\mathsf{t}_{2}^2=\mathsf{z}\\
& [\mathsf{r}_{j}, \, \mathsf{z}] = [\mathsf{t}_{j}, \, \mathsf{z}]= 1, \\
& [\mathsf{r}_j, \mathsf{r}_k]= [\mathsf{t}_j, \mathsf{t}_k] = 1, \\
& [\mathsf{r}_{j}, \,\mathsf{t}_{k}] =\mathsf{z}^{- \delta_{jk}} \, \rangle
\end{split}
\end{equation}
There are two surjective group homomorphisms $$\varphi_{49} \colon \mathsf{B} \to \mathsf{G}(32, \, 49), \quad \varphi_{50} \colon \mathsf{B} \to \mathsf{G}(32, \, 50),$$ expressed in both cases by
$$[r_{11}, \; t_{11}, \; r_{12}, \; t_{12}, \; r_{21}, \; t_{21}, \; r_{22}, \; t_{22}, \; z] \mapsto [\mathsf{t_2}, \; \mathsf{t}_1, \; \mathsf{t}_1 \mathsf{r}_2, \; \mathsf{r}_1, \; \mathsf{t}_2 \mathsf{r}_1, \; \mathsf{r}_2, \; \mathsf{t}_1, \; \mathsf{t}_2, \; \mathsf{z}].$$
Let us denote their kernels by $\mathsf{K}_{49}$ and $\mathsf{K}_{50}$, respectively.
Question. Are $\mathsf{K}_{49}$ and $\mathsf{K}_{50}$ isomorphic groups?
I have constructed $\mathsf{K}_{49}$ and $\mathsf{K}_{50}$ by using GAP4, and I tried to make some calculations on them. Unfortunately, GAP4 does not work with infinite groups so well as it works with finite groups, so the computational approach seems to be difficult (at least for me). The usual invariants that I was able to compute ($\mathtt{AbelianInvariants}$, $\mathtt{MaximalAbelianQuotient})$ were not useful in order to distinguish them.
Edit: Here is the GAP4 code I used in order to construct the group $\mathsf{B}$ and the two kernels.
ColorPrompt(true);
#redefine commutators
comm:=function(x, y) return x*y*x^-1*y^-1; end;
#Construction of the Group B
F:=FreeGroup("r11", "t11", "r12", "t12", "r21", "t21", "r22", "t22", "z");;
r11:=F.1;; t11:=F.2;; r12:=F.3;; t12:=F.4;;
r21:=F.5;; t21:=F.6;; r22:=F.7;; t22:=F.8;; z:=F.9;;
U:=z^2;;
S1:=comm(r12^(-1), t12^(-1))*t12^(-1)*comm(r11^(-1), t11^(-1))*t12*z^(-1);;
S2:=comm(r21^(-1), t21)*t21*comm(r22^(-1), t22)*t21^(-1)*z;;
R1:=comm(r11, r22);;
R2:=comm(r11, r21);;
R3:=comm(r11, t22);;
R4:=comm(r11, t21)*z;;
R5:=comm(r11, z)*comm(z, r21^(-1));;
R6:=comm(r12, r22);;
R7:=z^(-1)*r21*r22^(-1)*z*r22*r21^(-1)*comm(r21, r12);;
R8:=comm(r12, t22)*z;;
R9:=comm(r12, t21)*comm(t21, z^(-1));;
R10:=comm(r12, z)*comm(z, r22^(-1));;
T1:=comm(t11, r22);;
T2:=comm(t11, r21)*t21^(-1)*z^(-1)*t21;;
T3:=comm(t22, t11);;
T4:=comm(t11, t21)*comm(z, t21^(-1));;
T5:=comm(t11, z)*comm(z, t21^(-1));;
T6:=comm(t12, r22)*t22^(-1)*z^(-1)*t22;;
T7:=comm(t12, r21)*comm(z, t22^(-1));;
T8:=comm(t12, t22)*comm(z, t22^(-1));;
T9:=t22^(-1)*z*t22*z^(-1)*t21*z*t22^(-1)*z^(-1)*t22*t21^(-1)*comm(t21, t12);;
T10:=comm(t12, z)*comm(z, t22^(-1));;
B:=F/[U, S1, S2, R1, R2, R3, R4, R5, R6, R7, R8, R9, R10, T1, T2, T3, T4, T5, T6, T7, T8, T9, T10];;
r11:=B.1;; t11:=B.2;; r12:=B.3;; t12:=B.4;;
r21:=B.5;; t21:=B.6;; r22:=B.7;; t22:=B.8;; z:=B.9;;
#construction of G(32, 49)
f:=FreeGroup("r1", "t1", "r2", "t2", "s");
r1:=f.1;; t1:=f.2;; r2:=f.3;; t2:=f.4;; s:=f.5;;
g:=f/[r1^2, t1^2, r2^2, t2^2, s^2, comm(r1, s), comm(t1, s), comm(r2, s), comm(t2,s),
comm(r1, r2), comm(t1, t2), comm(r1, t1)*s, comm(r1, t2), comm(r2, t1), comm(r2, t2)*s];;
r1:=g.1;; t1:=g.2;; r2:=g.3;; t2:=g.4;; s:=g.5;;
IdSmallGroup(g); StructureDescription(g);
#construction of the group homomorphism B->G(32, 49)
hom49:=GroupHomomorphismByImages(B,g,[r11, t11, r12, t12, r21, t21, r22, t22, z],
[t2, t1, r2*t1, r1, r1*t2, r2, t1, t2, s]);
K49:=Kernel(hom49);
#construction of G(32, 50);
o:=FreeGroup("u1", "v1", "u2", "v2", "y");
u1:=o.1;; v1:=o.2;; u2:=o.3;; v2:=o.4;; y:=o.5;;
h:=o/[u1^2, v1^2, u2^2*(y^-1), v2^2*(y^-1), y^2, comm(u1, y), comm(v1, y), comm(u2, y), comm(v2,y), comm(u1, u2), comm(v1, v2), comm(u1, v1)*y, comm(u1, v2), comm(u2, v1), comm(u2, v2)*y];;
u1:=h.1;; v1:=h.2;; u2:=h.3;; v2:=h.4;; y:=h.5;;
IdSmallGroup(h); StructureDescription(h);
#construction of the group epimomorphism B->G(32, 50)
hom50:=GroupHomomorphismByImages(B,h,[r11, t11, r12, t12, r21, t21, r22, t22, z],
[v2, v1, u2*v1, u1, u1*v2, u2, v1, v2, y]);
K50:=Kernel(hom50);
You can ask GAP for a presentation of the kernels via IsomorphismFpGroup. However, since $B$ is infinite and the quotients are finite, the kernels will be infinite, too. So in general, I wouldn't expect a generic computation to be able to prove if they are isomorphic. If you are lucky, then indeed some invariant can distinguish them (you could try nilpotent quotients via the nq package). Your best hope likely is to try to exploit the structure of $B$ somehow. How exactly did you construct it?
Oh and for anybody willing to help you, it would tremendously help if you also provided the GAP code to enter $B$ and compute the two kernels. I certainly would play with it a bit if it was available, but I am not going to manually type in a presentation :-).
Regarding the group B, it comes from a geometric topology problem (it is the quotient of a braid group for a genus 2 surface).
@MaxHorn: You are absolutely right, I included the code. Thank you for the suggestion :-)
Even if the kernels have the same abelianization, you might say what this abelianization is?
@YCor: The abelianization in both cases is $\mathbb{Z}^8 \oplus (\mathbb{Z}_2)^4$.
OK. So in both case there's a characteristic subgroup of index $2^8$ in $K_i$ (mod out by derived subgroup of $K_i$, then by torsion, and then by squares, and keep the kernel). Maybe this is small enough to compute its abelianization? Also it's tempting to directly kill squares and keep the kernel, which now has index $2^{12}$ in $K_i$, it it's still amenable to computation.
|
2025-03-21T14:48:31.630518
| 2020-07-28T08:56:11 |
366772
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Seva",
"Wolfgang",
"https://mathoverflow.net/users/29783",
"https://mathoverflow.net/users/9924"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631542",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366772"
}
|
Stack Exchange
|
A question related to the Hofstadter–Conway \$10000 sequence
The Hofstadter–Conway \$10000 sequence is defined by the nested recurrence relation $$c(n) = c(c(n-1)) + c(n-c(n-1))$$ with $c(1) = c(2) = 1$. This sequence is A004001 and it is well-known that this sequence has many amazing properties which are investigated in a very interesting paper Kubo and Vakil - On Conway's recursive sequence.
A related curious sequence can be defined by similar recurrence relation as below $$c^*(n) = n - c^*(c^*(n-1)) - c^*(n-c^*(n-1))$$ with $c^*(1) = c^*(2) = 1$. I introduced this sequence in A287422 and recurrence is also investigated in terms of sensitivity of initial conditions selections
Alkan and Aybar - On Families of Solutions for Meta-Fibonacci Recursions Related to Hofstadter–Conway $10000 Sequence.
Conjecture. $c(n) - \frac{n}2$ $\ge$ $\left\lvert c^*(n) - \frac{n}2 \right\rvert$ for all $n \ge 1$. (It is checked up to $2^{32}$.)
Question. Can someone show that $c(n) - \frac{n}2$ $\ge$ $\left\lvert c^*(n) - \frac{n}2 \right\rvert$ for all $n \ge 1$ ?
Comments that are related to combinatorial characterization of $c^*(n)$ are also very welcome.
(I the share below graph in order to display behaviours of both sequences for $n \le 2^{10}$.
Red: $c(n) - \frac{n}2$, Black: $c^*(n) - \frac{n}2$.)
Note. I believe that it is also possible to see a way to connection of $c(n)$ and $c^*(n)$ thanks to certain auxiliary variants defined by A317754 and A317854, see the below graph (red: transformation of $c(n)$, black: transformation of $c^*(n)$). (These variants also make probably interesting sounds if one can think that recurrences that produce sequences with curious sounds are general of interest.)
Has anybody tried to plot the red arcs against the properly scaled Takagi function (better still, to plot the difference between the two)?
@Seva Sure enough, that readily comes to mind. On p.20-21 of the first quoted paper (excellent indeed!), a parametrization of the limit curve is developed, in terms of the erf function...! No reference to the Takagi/Blancmange curve though.
|
2025-03-21T14:48:31.630667
| 2020-07-28T09:07:55 |
366774
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Max",
"Yemon Choi",
"https://mathoverflow.net/users/161875",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631543",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366774"
}
|
Stack Exchange
|
How to introduce not-orthonormal base on Reproducing Kernel Hilbert Spaces?
I read some tutorial papers and slide，and find that the bases on Reproducing Kernel Hilbert Spaces always be orthonormal. For examples，you can refer to this link for the content about Reproducing Kernel Hilbert Spaces. What if the introduced base are not orthonormal? Any papers talk about this problems?
You need to be careful about what you mean by a "basis" when dealing with infinite-dimensional normed spaces: there are several different notions, and I think you should first look up which one fits your needs. Right now, since it is not clear if you know what a Schauder basis or an unconditional basis is, it is difficult to point you in an appropriate direction
In any case, if you are relatively inexperienced in mathematics, then perhaps you would do better to ask your question on http://math.stackexchange.com
@YemonChoi, thanks for your comments. You can check the link I refer in my questions. What makes I confuse is that if I introduce a non-orthonormal to replace the basis e_m in Eq.(1) mentioned in the link. Then, if the kernel still has the original property like: ()=⟨,⟩
|
2025-03-21T14:48:31.630782
| 2020-07-28T10:00:47 |
366777
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"KKD",
"LSpice",
"Vít Tuček",
"https://mathoverflow.net/users/135674",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/6818"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631544",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366777"
}
|
Stack Exchange
|
Morphism of Verma modules
$\DeclareMathOperator\Hom{Hom}$I'm trying to understand morphism of Verma modules and consider the following example.
PART 1:
Consider $\mathfrak{g}=\mathfrak{gl}_3$ over $\mathbb{C}$ with positive roots
\begin{equation*}\Phi_+=\{\alpha_1=(1,-1,0),\alpha_2=(1,0,-1),\alpha_3=(0,1,-1)\},\end{equation*}
which defines a cartan decomposition $\mathfrak{g}=\mathfrak{n}^- \oplus \mathfrak{h}\oplus \mathfrak{n}$. Then to the positive roots corresponding reflections are $s_{\alpha_1}=(1,2)$, $s_{\alpha_2}=(1,3)$ and $s_{\alpha_3}=(2,3)\in S_3$. Denoting by $\rho=\frac{1}{2} (\alpha_1 +\alpha_2 + \alpha_3)=(1,0,-1)$ half the sum of all positive roots, we have for weights $\lambda=(0,-1,1)$ and $\mu=(-1,-1,2)$, that
\begin{equation*}
\mu=s_{\alpha_2}((1,-1,0))-(1,0,-1)=s_{\alpha_2}(\lambda+\rho)-\rho=s_{\alpha_2}\cdot \lambda=\lambda -\alpha_2<\lambda.
\end{equation*}
Hence by a Theorem of Verma (Theorem 4.6 in [H]: Humphrey's "Representation of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$") there exist a morphism of Verma modules $\phi:M(\mu) \rightarrow M(\lambda)$, with respective maximal vectors $v_\mu$ and $v_\lambda$. The morphism $\phi$ is known to be injective (Theorem 4.2 in [H]) and we have $\phi(v_\mu)=u\cdot v_\lambda$ for a unique $u \in U(\mathfrak{n}^-)$, which also determines $\phi$. Furthermore $\dim(\Hom(M(\mu),M(\lambda))=1$, hence up to some scalar there is only one choice for $u$, which I'm trying to find.
My thoughts so far:
The Verma modules $M(\lambda)$ and $M(\mu)$ each have a unique simple submodule $L(\mu')$, which should be isomorphic/the same and is also a Verma module (Proposition 4.1 and Theorem 4.2 in [H]). By Theorem 4.8 in [H] $\mu'$ has to be antidominant. Hence $\mu'=(-2,0,2)$. According to the proof for $\dim(\Hom(M(\mu),M(\lambda))=1$ in [H], it is enough to understand how the simple module $L(\mu')$ is mapped to itself under $\phi$. As $\mu -\mu'=\alpha_1$ we have $\dim M(\mu)_{\mu'}=1$, hence the maximal vector of $L(\mu') \subset M(\mu)$ is $y_{\alpha_1}v_\mu$ with respect to $v_\mu$ and fixed choosen root vectors $y_{\alpha_i}$ of $\mathfrak{g}_{-\alpha_i} \subset \mathfrak{g}$. But then I struggle as we have for the equation $\lambda-\mu'=t_1\alpha_1+t_2\alpha_2+t_3\alpha_3$ with $t_i \geq 0$ two solutions, namely $(t_1,t_2,t_3) \in \{(2,0,1),(1,1,0)\}$. Hence $\dim M(\lambda)_{\mu'}=2$ and I don't know if $\phi(y_{\alpha_1}v_\mu)=c\cdot y_{a_1}^2y_{a_3}v_\lambda$ or $\phi(y_{\alpha_1}v_\mu)=c\cdot y_{a_1}y_{a_2}v_\lambda$ ($c$ some scalar). Or is this completely wrong?
SOLUTION PART 1: By the comments below it follows that $\phi(v_\mu)=c(y_{\alpha_1}y_{\alpha_3}+2y_{\alpha_2})v_\lambda$ for some $c \in \mathbb{C}$.
ADDENDUM PART 2: Consider then part of the (strong) BGG resolution (using the notation as in [H]) of the simple module $L((0,0,0))$
\begin{equation*}
C:M((-2,0,2)) \xrightarrow{\delta_3} M((-2,1,1)) \oplus M((-1,-1,2) \xrightarrow{\delta_{2_1}} M((0,-1,1))
\end{equation*}
with $\delta_{2_1}:M((-2,1,1)) \oplus M((-1,-1,2) \xrightarrow{\delta_{2}} M((0,-1,1)) \oplus M((-1,1,0)) \xrightarrow{\pi_1} M((0,-1,1))$.
Hence $\delta_{2_1}\circ \delta_3=0$. With the same arguments as in the comments, we have
\begin{align*}
\delta_3(v_{(-2,0,2)})&{}=(a_1y_{\alpha_3}v_{(-2,1,1)},a_2y_{\alpha_1}v_{(-1,-1,-2)}), \\
\delta_{2_1}(v_{(-2,1,1)},v_{(-1,-1,2)})&{}=(b_1y_{\alpha_1}^2+b_2(y_{\alpha_1}y_{\alpha_3}+2y_{\alpha_2}))v_{(0,-1,1)} \end{align*}
for some non-trivial scalars $a_i$, $b_i$.
So we would get
\begin{align*} 0&{}=\delta_{2_1}\circ \delta_3(v_{(-2,0,2)})=\delta_{2_1}(a_1y_{\alpha_3}v_{(-2,1,1)},a_2y_{\alpha_1}v_{(-1,-1,-2)})\\&{}=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_2b_2(y_{\alpha_1}^2y_{\alpha_3}+y_{\alpha_1}y_{\alpha_2}))v_{(0,-1,1)}\\
&{}=((a_1b_1+a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2}))v_{(0,-1,1)}.
\end{align*}
But why is the last term equal to zero for nontrivial $a_i$, $b_i$? I thought that $y_{\alpha_1}^2y_{\alpha_3}$ and $y_{\alpha_1}y_{\alpha_2}$ are linearly independent.
You referred to "the comments below", but there were (prior to this one) no comments. I assumed you meant @VitTuček's answer, so I edited in a link to it, as well as to Humphreys's book. (Also, a TeX comment: note that $\mathfrak g$=$\mathfrak{gl}_3$ $\mathfrak g$=$\mathfrak{gl}_3$ spaces poorly compared to $\mathfrak g=\mathfrak{gl}_3$ $\mathfrak g=\mathfrak{gl}_3$. I adjusted to the latter.)
PART 1:
The element $u$ must have weight $-\alpha_2$, since $\mu = \lambda - \alpha_2.$
In $U(\mathfrak{n^-})$ there are only two linearly independent elements that have such weight (assuming PBW basis with respect to fixed order of generators based on positive roots): $y_{\alpha_2}$ and $y_{\alpha_1}y_{\alpha_3}.$ Hence the sought element $u$ is a linear combination of such
$$
u = a y_{\alpha_2} + b y_{\alpha_1}y_{\alpha_3}.
$$
Since this has to be the image of the higest weight vector of $M(\mu)$ we must have $x_{\alpha_1} u = 0$ and $x_{\alpha_3} u = 0.$ Writing it out and using relations defining the Verma module and the Lie algebra, we end up with system of 2 linear equations for 2 unknowns. E.g. we have
$$
x_{\alpha_1} (ay_{\alpha_2}v_\lambda) = (a[x_{\alpha_1}, y_{\alpha_2}] + ay_{\alpha_2} x_{\alpha_1})v_\lambda
$$
where the first term on the right hand side is either zero, or some element of Cartan subalgebra acting on $v_\lambda$, and the second term is zero from the definition of the Verma module.
PART 2:
I think you made a mistake in your calculations. For any $U(\mathfrak{g})$-homomorphism $\varphi$ we have $\varphi(u v) = u \varphi(v)$. Hence the composition going through $M(-2, 1, 1)$ is equal to $$
a_1y_{\alpha_3} \delta_{2_1}(v_{(-2, 1, 1)}) = a_1b_1 y_{\alpha_3} y_{\alpha_1}^2 v_{(0, -1, 1)}.
$$
The elements $y_{\alpha_1}$ and $y_{\alpha_3}$ do not commute, in fact $[y_{\alpha_1}, y_{\alpha_3}]$ should be a multiple of $y_{\alpha_2}.$
Similarly, the composition going through $M(-1,-1,2)$ equals
$$
a_2y_{\alpha_1} \delta_{2_1}(v_{(-1,-1,2)}) = a_2 b_2 y_{\alpha_1}(2y_{\alpha_1} y_{\alpha_3} + y_{\alpha_2}) v_{(0, -1, 1)}.
$$
Maybe this questions is dumb, but why you are no considering "$x_{\alpha_2}u=0$" too?
@CJS Because $x_{\alpha_2} = [x_{\alpha_1}, x_{\alpha_3}].$ In general, it is sufficient to consider only action by elements corresponding to simple roots. Alas, this straightforward strategy of finding $u$ gets unfeasible pretty quickly because dimension of graded-homogeneous components of $U(\mathfrak{n}^-)$ grows very fast.
Thanks for all these answers. At least I understood what's going on.
So after doing the computation it turns out, that $a=2$ and $b=1$ is a solution (and any multiple of (2,1))
I added some part to my original question.
@CJS This is generally discouraged on this site as it can lead to Questions and Answers which are hard to parse for people coming late to the party. I've amended my answe but since it's midnight for me, I will leave the rest of the calculations to you. If you do them, please share the result with us. Also, I vaguely remember that the choice of constants in these "diamonds" is dealt with in the original BGG article.
For part 2:
With
\begin{equation*}
y_{\alpha_1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},y_{\alpha_2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},y_{\alpha_3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}
\end{equation*}
we have \begin{align}
[y_{\alpha_1},y_{\alpha_2}]&= 0, & (1)\\
[y_{\alpha_1},y_{\alpha_3}]&= -y_{\alpha_2}. & (2)
\end{align}
Then
\begin{align}
0 &=\delta_{21}\circ \delta_3(v_{(−2,0,2)})=\delta_{21}(a_1y_{\alpha_3}v_{(−2,1,1)},a_2y_{\alpha_1}v_{(−1,−1,−2)})\\
&=(a_1y_{\alpha_3}b_1y_{\alpha_1}^2+a_2y_{\alpha_1}b_2(y_{\alpha_1}y_{\alpha_3}+2y_{\alpha_2}))v_{(0,-1,1)} \\
&=(a_1b_1y_{\alpha_3}y_{\alpha_1}^2+ a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}\\
\end{align}
with $(2)$ follows
\begin{align}
&=(a_1b_1(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})y_{\alpha_1}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\
&=(a_1b_1y_{\alpha_1}y_{\alpha_3}y_{\alpha_1}+a_1b_1y_{\alpha_2}y_{\alpha_1}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}
\end{align}
Applying $(2)$ again and additionally $(1)$ we get
\begin{align*}
&=(a_1b_1y_{\alpha_1}(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\
&=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\
&=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+2a_1b_1y_{\alpha_1}y_{\alpha_2}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\
&=((a_1b_1+a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+(2a_1b_1+2a_2b_2)y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}.
\end{align*}
and $a_1b_1=-a_2b_2$ is enough. As in [H| mentioned it is possible to choose $a_i,b_i \in \{-1,1\}$.
|
2025-03-21T14:48:31.631230
| 2020-07-28T11:18:19 |
366781
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dieter Kadelka",
"Giuliosky",
"Piotr Hajlasz",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/121665",
"https://mathoverflow.net/users/161651"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631545",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366781"
}
|
Stack Exchange
|
Measurability of infimum function
In Theorem 14.37 of Variational Analysis by Rockafellar and Wets, it shows that for any normal integrand $f:T\times \mathbb{R}\to\mathbb{R}$, the function $p:T\to \mathbb{R}$ given by
$p(t):=\inf f(t,⋅)$
is measurable. I am struggling to understand a step of the proof. In particular, it says that, since the closure correspondence:
$$
\text{Cl}R:t\mapsto \left\lbrace \alpha\in \mathbb{R}:p(t)\leq \alpha\right\rbrace
$$
is closed-valued and measurable, then $p$ is measurable.
I don't have the book. What is $T$ and what is a normal integrand?
Oh sorry, I should have made a more precise answer. $(T,\mathcal{F})$ is an arbitrary measurable space, while a normal integrand is a function whose epigraphical correspondence is closed-valued and measurable. In particular, $f:T\times \mathbb{R}\to\mathbb{R}$ is a normal integrand, if the correspondence: $\Gamma_f:t\mapsto \left\lbrace (x,\alpha)\in \mathbb{R}^2:f(t,x)\leq \alpha\right\rbrace$ is closed-valued and measurable.
@Giuliosky Just rewrite the question making all definitions and terms used clear. Adding explanations in comments is not a good idea since the question itself should be self-contained.
|
2025-03-21T14:48:31.631328
| 2020-07-28T11:36:14 |
366783
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David White",
"https://mathoverflow.net/users/11540"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631546",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366783"
}
|
Stack Exchange
|
Coherence for pseudomonads and their pseudoalgebras
Let $\mathcal K$ be a bicategory. For every pseudomonad $T : \mathcal K \to \mathcal K$, does there exist a 2-monad $S : \mathcal C \to \mathcal C$, where $\mathcal C$ is a 2-category biequivalent to $\mathcal K$, such that the bicategory of pseudoalgebras $T\text{-}\mathrm{Alg}$ is biequivalent to the 2-category of pseudoalgebras $S\text{-}\mathrm{Alg}$?
That is, can we always choose to work with 2-monads rather than pseudomonads, so long as we consider their pseudoalgebras, rather than their strict algebras?
If this is true, does the statement continue to hold for the 2-category of strict algebras $S\text{-}\mathrm{Alg}_s$? (Not every pseudoalgebra is equivalent to a strict algebra, but this fact does not take into account changing the 2-monad: I imagine this is false, but it would be good to have a reference.)
A partial answer appears in Theorem 0.1 of https://www.math.uchicago.edu/~may/PAPERS/AddCat1.pdf With conditions on $K$ and $T$.
|
2025-03-21T14:48:31.631410
| 2020-07-28T13:02:20 |
366787
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dianbin Bao",
"OmniaOperator",
"https://mathoverflow.net/users/160900",
"https://mathoverflow.net/users/31134"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631547",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366787"
}
|
Stack Exchange
|
Rankin-Selberg method and Symmetric power of elliptic curves
Let $E$ be an elliptic curve over $\mathbb{Q}$ with conductor $N$. Let $f=\sum a_n q^n$ be the weight 2 modular form corresponding to $E$.
Define $L_2(f,s)=\zeta(s-1)L(Sym^2(E),s)$.
The following identity is mentioned in Zagier's paper "Classical and alliptic polylogarithms and special values of L-series":
$L_2(f,s)=(1+N^{1-s})\zeta(2s-2)\sum_{n=1}^{\infty}\frac{a_n^2}{n^s}$.
I would like to find a proof of this identity. Thanks
Finally found a reference. See page 43 of Number theory 3 by Kurokawa,Kurihara and Saito (Translations of mathematical monographs Volume 242). The factor $(1+N^{1-s})$ comes from the primes of bad reduction and works for the case $N=37$. For different conductor, the formula may look different at the bad primes.
In (0.2) of "On the Holomorphy of Certain Dirichlet Series", Shimura says that "it can easily be seen that"... (though he does give the relevant Euler factor for the proof).
|
2025-03-21T14:48:31.631753
| 2020-07-28T13:07:56 |
366788
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andi Bauer",
"https://mathoverflow.net/users/115363"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631548",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366788"
}
|
Stack Exchange
|
Symmetric group in terms of block permutations
For $i+j+k=N$, consider the permutation $\Pi_{i,j,k}\in S_N$, which keeps the numbers $0,\ldots,i-1$ fixed, and exchanges the numbers $i,\ldots,i+j-1$ with the numbers $i+j,\ldots,i+j+k-1$.
$$\Pi_{i,j,k}(x)=
\begin{cases}
x & \text{if}\quad 0\leq x<i\\
x+k&\text{if}\quad i\leq x<i+j\\
x-j&\text{if}\quad i+j\leq x<i+j+k
\end{cases}
$$
$i=0$ is allowed. Rather obviously, $\Pi_{i,j,k}$ for different $i,j,k$ generate the whole symmetric group $S_N$. Is there simple set of relations for those generators, such that $S_N$ is the free group over the $\Pi_{i,j,k}$ modulo those relations?
I'm talking about relations like $\Pi_{0,i,j+k+l}\Pi_{j,k,i+l}\Pi_{0,k+j+l,i}\Pi_{i+k,j,l}=\Pi_{i,j+k,l}$ (for all $i$, $j$, $k$, $l$).
|
2025-03-21T14:48:31.631964
| 2020-07-28T13:18:19 |
366789
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Deane Yang",
"Igor Khavkine",
"Michele",
"https://mathoverflow.net/users/2622",
"https://mathoverflow.net/users/613",
"https://mathoverflow.net/users/61364"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631549",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366789"
}
|
Stack Exchange
|
Local solvability and Cauchy-Kovalevskaya theorem for PDEs
I am trying to understand the exact implications between local solvability and a general version of the Cauchy-Kovalevskaya (CK) theorem, in the context of PDEs.
Let $\Delta(x,u^{(n)})=0$ be a system of PDEs of order $n$. Here $x$ is the vector of independent variables, and $u^{(n)}$ is the vector of dependent variables and all their derivatives up to order $n$. According to Olver ("Applications of Lie groups to Differential Equations", 2/e, Ch. 2, Def. 2.70), $\Delta$ is locally solvable if the variety it induces on $\mathbb{R}^{|x|+|u^{(n)}|}$
$V(\Delta):=\{(x_0,u_0^{(n)}):\Delta(x_0,u_0^{(n)})=0\}$
coincides with its solution variety
$S(\Delta):={\small \{(x_0,u_0^{(n)})\in V(\Delta) : \text{ $\exists$ an analytic solution $U$ of $\Delta$ in a neighborhood of $x_0$
s.t. } U^{(n)}(x_0)=u_0^{(n)} \}}$
Olver shows that if $\Delta$ is in Kovalevskaya form then it is locally solvable (Corollary 2.74, p. 163; this is indeed an easy consequence of CK theorem). He then states that the same result still holds when $\Delta$ is in general Kovalevskaya form:
I am struggling to convince myself of the validity of this statement. For instance, consider the 2nd order system $\Delta$ in the independent variables $t,x$ and the dependent variables $u,v$:
\begin{align*}
u_t & = v\\
v_{tt} &= u_x\,.
\end{align*}
From what I gather, this system is not locally solvable. Indeed, there are differential consequencences that are not captured algebraically by the two equations above, such as $u_{tx}=v_x$. So there are points $(x_0,u_0^{(2)})\in V(\Delta)$ s.t. (with obvious notation) $u_{0,tx}\neq v_{0,x}$, hence not in $S(\Delta)$.
Yet, $\Delta$ is in general Kovalevskaya form, is it?
Edit: for reference, I paste Olver's original definition. Note that $pr^{(n)}f$ denotes the $n$-th prolongation of $f$.
Couldn't you find all solutions by solving the first order system \begin{align*}
u_t &= v\
v_t &= w\
w_t &= u_x
\end{align*}
with initial data $(u,v,w)$ along $t = 0$?
Your system is "equivalent" to the old one in terms of solutions. However the point here is not finding the solutions, but solving the apparent contradiction I have pointed out. Cheers
Why is this system not locally solvable? Doesn’t $u_{0,tx} = v_{0,x}$ follow from the assumption that the equations hold at $(x_0,t_0)$ up to second order.
This is in the definitiion of $S(\Delta)$, but not in that of $V(\Delta)$. That is, there are 2nd order differential consequences (e.g. $u_{tx}=v_x$) that are not algebraic consequences.
Here, it appears that local solvability means that any second order jet that solves the system can be extended to a local smooth solution (which is not the standard definition of local solvability). To be a second order solution, the three equations and the first partial derivatives with respect to both $x$ and $t$ of the first two equations must hold. In particular, if a 2-jet is a solution at $(x_0,t_0)$, $u_{tx} = v_x$ then holds at $(x_0,t_0)$. So in fact this equation is a consequence of the jet being in $S(\Delta)$.
It may not be the 'standard definition' of local solvability, but it is the way it is defined Olver's textbook. I copy it below:
I have pasted the original definition into my question. In short, for l.s. we require $V(\Delta)=S(\Delta)$ on the 2nd order jet, which does not seem to be the case for my 2-equations system. Of course it is the case for your 3-equations system, but this does not solve my doubt about Olver's statement.
I would second Deane's comment that Olver's defnition is not the most common/reasonable one. In fact, it reads a little "lazy," which surprises me. Here's a better definition. Locally solvable: every formal power series solution (aka an $\infty$-jet at a point) is the Taylor series of a smooth solution (locally extends to a smooth solution). Dropping this definition into your example removes any contradictions. Note that the extended Kovalevskaya form precisely guarantees a unique formal solution for any free initial data (of appropriate order for $u$ and $v$).
@Igor. Ok, so to be specific, you propose to replace the definition of $S(\Delta)$ given in my question with $S(\Delta):=...$? As an aside, concerning the extended Kovalevskaya form, my understanding is that it additionally guarantees analiticity (given analytic initial data). If one is interested in formal power series solutions, disregarding analyticity, there are more generous formats. See e.g. here link.
@Michele To be specific, replace $V(\Delta)$ by the locus of $\infty$-jets cut out by $\Delta=0$ and all of its differential consequences, at the same time replace $S(\Delta)$ by the locus of $\infty$-jets of locally defined smooth/analytic (as you prefer) solutions. Then $S(\Delta) = V(\Delta)$ is a good definition of local solvability. Sure the ext-K form gives you lots of information, but also including the ability to specify formal solutions by finite amounts of data (the free initial data), as I wrote earlier. The hypotheses of the Cartan-Khäler theorem do that in even more generality.
@Igor. This fills the gap between $S(\Delta)$ and $V(\Delta)$, at the cost of forcing infinite jets in the definition of $V(\Delta)$, thus spoiling its finitary, 'algebraic' nature. For instance, suppose $\Delta$ is polynomial, then it may be useful to reason on $V(\Delta)$ in algebraic-geometric terms, that is as the affine variety induced by $\Delta$ on $n$-th order jets. Note that local solvability (Olver style) is central also in differential-completion algorithms for PDEs, like Cartan-Kuranishi: see e.g. p. 15 [link] (https://www.maths.dundee.ac.uk/plin/SiAM_ReidLinWittkopf_01.pdf ).
@Michele There is no argument against both subjective and temporal fluctuations in what is or is not a good definition. There are many ways in which a definition may be modified to suit one's taste. Is that the philosophical direction you intended your question to go in? The gap between $\infty$-jets and finite order data is closed by the notion of an involutive form of a PDE, about which many volumes have been written. Your last reference is an example. Another notable reference is Involution by W. Seiler (Springer, 2010).
@Igor. Just to clarify, my direction is merely pragmatical. In particular, again restricting to the polynomial case, being able to use certain well-known algebraic-geometric notions (notably, Groebner bases) for algorithmic purposes. Just to make a simple example, under suitable conditions, the set of invariant polynomials of $\Delta$ --- those identically 0 on solutions of $\Delta$ --- coincide with the ideal generated by $\Delta$, on $n$-order jets. This in turn may be useful in direct methods to search for conservation laws.
@Michele There is no reason to be hypothetical. All of the tools that you have mentioned are in fact already being used in the relevant literature, without any compunctions about using $\infty$-jets when convenient. This discussion seems to be veering more into philosophy. I'm not sure what other practical information I could add, besides what I already wrote.
The system you wrote down
First, let's assume everything is smooth.
\begin{align*}
u_t &= v\\
v_{tt} &= u_x\\
\end{align*}
is equivalent to the first order system
\begin{align*}
u_t &= v\\
v_t &= w\\
w_t &= u_x
\end{align*}
in the sense that $(u,v)$ is a solution to the first system if and only if $(u,v,w)$ is a solution to the second system, where we set $w = v_t$.
By Cauchy-Kovalevski, given any analytic functions $u_0(x), v_0(x), w_0(x)$, there exists a unique analytic solution $(u,v,w)$ to the second system such that $u(x,0) = u_0(x), v(x,0) = v_0(x), w(x,0) = w_0(x)$. This is equivalent to saying that given any analytic functions $u_0, v_0, w_0$, there exists a unique analytic solution $(u,v)$ to the first system such that $u(x,0) = u_0(x)$, $v(x,0) = v_0(x)$, and $v_t(x,0) = w(x)$, which is what Olver asserts. The fact that $u_{tx} = v_x$ is a consequence of the equations and need not be specified in the initial data.
As for Olver's definition of local solvability, an element of $V(\Delta)$ for the original system consists of a $2$-jet $(x_0,t_0,u(x_0,t_0), u_x(x_0,t_0), u_t(x_0,t_0), v(x_0,t_0), v_x(x_0,t_0), v_t(x_0,t_0))$ that satisfies the system up to second order. On other words, at $(x_0,t_0)$,
\begin{align*}
u_t &= v\\
u_{tx} &= v_x\\
u_{tt} &= v_t\\
v_{tt} &= u_x
\end{align*}
In particular, if a $2$-jet lies in $V(\Delta)$, then $u_{tx} = v_x$ does hold for that $2$-jet at $(x_0,t_0)$. Given such a $2$-jet, you can extend $u$, $v$, and $v_t$ arbitrarily to initial data along $t = 0$ and solve the system as described above. In particular, the initial data is assumed to satisfy $u_{tx} = v_x$ at $(x_0,t_0)$.
|
2025-03-21T14:48:31.632439
| 2020-07-28T14:18:10 |
366792
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Javi",
"https://mathoverflow.net/users/134242",
"https://mathoverflow.net/users/144957",
"pupshaw"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631550",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366792"
}
|
Stack Exchange
|
Detailed proof of $\mathfrak{s}^{-1}\mathrm{End}_V\cong \mathrm{End}_{\Sigma V}$
I asked this question on MSE but I want to ask it again here with some more context sine it received no answers. In Chapter 3 (Algebra) of the book Operads in Algebra, Topology and Physics by Markl, Shnider and Staffesh there is the Lemma 3.16 where the authors state the isomorphism of operads $\mathfrak{s}^{-1}\mathrm{End}_V\cong \mathrm{End}_{\Sigma V}$. For the definition of the operadic suspension $\mathfrak{s}$ and other notation please see my previous question.
But the proof they give is not very explicit, since they give (non-explicit) isomorphisms of graded modules and don't really justify that they are morphisms of operads. I tried to show that and I came across a problem trying to show that the isomorphism commutes with the action of the symmetric group. This is where my MSE questions begins, so I am going to copy it here.
Let $V$ be a graded vector space and $\text{End}_V(n)=\hom(V^{\otimes n},V)$. There is a natural action of the symmetric group $S_n$ on $\text{End}_V(n)$ by permuting the arguments, i.e. if $f\in \text{End}_V(n)$ and $\sigma\in S_n$, $(f\sigma)(v_1\otimes\cdots\otimes v_n)=\varepsilon(\sigma)f(v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)})$, where $\varepsilon(\sigma)$ is the Koszul sign produced by permuting $v_1,\dots,v_n$ via $\sigma$. We can twist this action by the sign of $\sigma$, i.e. we consider the action
$$(f\sigma)(v_1\otimes\cdots\otimes v_n)=(-1)^{\sigma}\varepsilon(\sigma)f(v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)})$$
If $\Sigma V$ is the suspension of $V$, we consider the natural action of the symmetric group on $\text{End}_{\Sigma V}(n)$ (the first I defined, without the twist).
There is a map $\phi:\text{End}_{\Sigma V}(n)\to \text{End}_V(n)$ given by $f\mapsto \Sigma^{-1}\circ f\circ\Sigma^{\otimes n}$ which is indeed an isomorphism of graded modules.
I need to show that $\phi$ commutes wit the action of the symmetric group, where we have the natural action on the domain and the twisted action on the codomain.
I can show this for transpositions of the form $\sigma=(i\ i+1)$. On the one hand,
$$\phi(f\sigma)(v_1\otimes\cdots\otimes v_n)=(-1)^{\sum_{j=1}^n (n-j)v_j}\Sigma^{-1}\circ (f\sigma)(\Sigma v_1\otimes\cdots\otimes \Sigma v_n)=$$
$$(-1)^{\sum_{j=1}^n (n-j)v_j+(v_i-1)(v_{i+1}-1)}\Sigma^{-1}\circ f(\Sigma v_1\otimes\cdots\otimes\Sigma v_{i+1}\otimes\Sigma v_i\otimes\cdots\otimes \Sigma v_n).$$
On the other hand
$$(\phi(f)\sigma) (v_1\otimes\cdots\otimes v_n)=(-1)^{v_iv_{i+1}-1}\Sigma^{-1}\circ f\circ \Sigma^{\otimes n}(v_1\otimes\cdots\otimes v_{i+1}\otimes v_i\otimes\cdots\otimes v_n)=$$
$$(-1)^{v_iv_{i+1}-1+\sum_{j\neq i,i+1}(n-j)v_j +(n-i-1)v_i+(n-i)v_{i+1}}\Sigma^{-1}\circ f(\Sigma v_1\otimes\cdots\otimes \Sigma v_{i+1}\otimes \Sigma v_i\otimes\cdots\otimes \Sigma v_n).$$
Now I just have to check that the signs are the same. Modulo $2$, the sign of the first map is
$$v_iv_{i+1}+v_i+v_{i+1}-1+\sum_{j=1}^n(n-j)v_j=$$
$$v_iv_{i+1}-1+\sum_{j\neq i,i+1}^n(n-j)v_j+(n-i-1)v_i+(n-i)v_{i+1},$$
which indeed coincide with the sign on the second map.
Question: Since these transpositions generate the symmetric group I feel that I should be able to conclude that the action commutes with $\phi$, but I don't know how to do it.
I would recommend establishing this at the level of universal properties. Writing (SSeq, #) for the monoidal category of symmetric sequences, and embedding S and V as pure unary sequences, we can observe that naturally in X, SSeq(X, End(S # V)) = SSeq( X # S # V, S # V) = SSeq(S^{-1} # X # S # V, V) = SSeq(S^{-1} # X # S, End(V)) = SSeq(X, S # End(V) # S^{-1}) which is indeed the desired desuspension. one only needs the associativity of composition and the universal property of endomorphisms.
Your actual Question has nothing to do with operads. Perhaps it is clarifying to consider the following more general setting: let $G$ be a group, $X$ and $Y$ be right $G$-sets, and $f : X \to Y$ be a function. If $g, h \in G$ and $f$ commutes with the actions of $g$ and of $h$ then it commutes with the action of $gh$:
$$f(x) \cdot (gh) = (f(x) \cdot g) \cdot h = f(x \cdot g) \cdot h = f((x \cdot g) \cdot h) = f(x \cdot (gh)).$$
So if $f$ commutes with a set of elements which generate $G$ then it commutes with all elements of $G$.
You were right, seeing the problem in a more general setting was clarifying. Thank you.
|
2025-03-21T14:48:31.632698
| 2020-07-28T14:27:15 |
366793
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Connor Malin",
"Michael Albanese",
"Nanjun Yang",
"https://mathoverflow.net/users/134512",
"https://mathoverflow.net/users/149491",
"https://mathoverflow.net/users/21564"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631551",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366793"
}
|
Stack Exchange
|
Cohomological behavior of the embedding $Gr(3,5)\to Gr(3,6)$
The following question is particularly interesting for me:
Does the natural map $Gr(3,5)\to Gr(3,6)$ induce a surjection
$$H^4(Gr(3,6),\mathbb{Z})\to H^4(Gr(3,5),\mathbb{Z})?$$
Here $Gr(k,n)$ means the real grassmannian of rank $k$.
Cells of the grassmannian correspond to restricted partitions, is there a combinatorial interpretation of the cellular boundary map? It appears that even in the $\mathbb{Z}/2$ case it is not just summing over all partitions obtained by subtracting one from a term of the partition.
@ConnorMalin The statement is true if you replace singular cohomologies by Chow groups since the morphism of Chow motives splits.
I think $H^i(Gr(k,n),\mathbb{Z})\to H^i(Gr(k,n-1),\mathbb{Z})$ is surjective if $n-k$ is odd and $H^i(Gr(k,n),\mathbb{Z})\to H^i(Gr(k-1,n-1),\mathbb{Z})$ is surjective if $n-k$ is even.
Would you mind providing some explanation as to why you think this is the case?
@MichaelAlbanese You could use M. Wendt's results on Chow-Witt rings of Grassmannians. The statements above haven't been published yet:)
Let me give more details on Nanjun Yang's answer (which I think is correct). There is no need for any results concerning Chow-Witt rings of Grassmannians, the surjectivity can be deduced from the results of Casian and Kodama linked in Danny Ruberman's answer to this MO-question (or other sources of specific knowledge about the structure of the cohomology of the real Grassmannians).
Here's the relevant facts on the cohomology of the real Grassmannians ${\rm Gr}_k(\mathbb{R}^n)$. The non-torsion classes are generated by Pontryagin classes of the tautological sub- and quotient bundle, plus an additional class in degree $n-1$ in the case both $k$ and $n-k$ are odd (not a characteristic class but related to an Euler class on some other Grassmannian). All torsion is 2-torsion, and it's given by the image of the integral Bockstein from the mod 2 cohomology (i.e. generated by Bocksteins of Stiefel-Whitney classes). Relations between these generators all come in some way from the Whitney sum formula.
In the specific case we have ${\rm H}^4({\rm Gr}(3,5),\mathbb{Z})=\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$. A generator of the $\mathbb{Z}$-summand is the first Pontryagin class of either of the tautological bundles. A generator for the 2-torsion summand is the Bockstein of ${\rm w}_1^3$ (with ${\rm w}_1$ the first Stiefel-Whitney class of either of the tautological bundles). In Young diagram terms, this corresponds to the partition $(2,1,1)$. This can be explicitly found (for ${\rm Gr}(2,5)={\rm Gr}(3,5)$ in the paper of Casian and Kodama.
On the other hand we have ${\rm H}^4({\rm Gr}(3,6),\mathbb{Z})=\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}^{\oplus 2}$, generated again by the Pontryagin class and classes corresponding to the partitions $(3,1)$ and $(2,1,1)$. The restriction morphism induced from the inclusion ${\rm Gr}(3,5)\hookrightarrow{\rm Gr}(3,6)$ maps characteristic classes of the tautological bundles on ${\rm Gr}(3,6)$ to the corresponding characteristic classes of the tautological bundles on ${\rm Gr}(3,5)$. Because Pontryagin and Stiefel-Whitney classes are stable, we get the required surjectivity.
The general moral here is that we get surjectivity whenever the cohomology group of the smaller Grassmannian is generated by stable characteristic classes. The only problematic class is the non-characteristic class in degree $n-1$ (corresponding to the maximal hook partition). That only exists on ${\rm Gr}(k,n)$ whenever $k$ and $n-k$ are both odd. That is exactly precluded by the conditions in Nanjun Yang's answer.
|
2025-03-21T14:48:31.632936
| 2020-07-28T14:52:45 |
366796
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631552",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366796"
}
|
Stack Exchange
|
Proving Hall's marriage theorem using Sperner's lemma
In the paper Hall's theorem for hypergraphs (Aharoni and Haxell, 2000),
the authors prove a theorem on the existence of perfect matchings in bipartite hypergraphs, using Sperner's lemma. At the last page (6), they say that "we have here a topological proof of Hall's theorem" (for bipartite graphs). I thought it should be easy to write this proof explicitly, since a simple bipartite graph is just a bipartite hypergraph in which each hyperedge is of size 2. But there is a problem: during the proof, the authors assume that the sets of neighbors (i.e., the sets $N(x)$ for each vertex $x\in X$, where $X$ is one part of the hypergraph) are pairwise-disjoint. For a hypergraph, this assumption is without loss of generality, since we can add dummy vertices to edges, and this does not affect the theorem conditions or concludion. But in a graph, we cannot add vertices to edges.
So my question is: is there an explicitly-written proof of Hall's marriage theorem, using Sperner's lemma (or a similar topological theorem)?
Penny Haxell's 2011 paper On Forming Committees in the American Mathematical Monthly explicitly uses Sperner's lemma to prove Hall's theorem for bipartite graphs (see theorem 4.1 and 4.2).
In addition to Carlo Beenakker's answer that gives Hall via Sperner directly, I think you can also get it by applying Hall's Theorem for Hypergraphs as follows. Let $G$ be a bipartite graph with partite sets $X, Y$, and write $V(X) = \{x_1, \ldots, x_n\}$. For each $i$, define a $1$-uniform hypergraph $H_i$ with vertex set $N(x_i)$ and edge set $\{ \{y\} : y \in N(x_i) \}$.
Letting $\mathcal{A} = \{H_1, \ldots, H_n\}$, we see that $\mathcal{A}$ has a disjoint set of representatives if and only if $G$ has a matching that saturates $X$. Now the condition of Aharoni and Haxell's Theorem 1.1 applied to the family $\mathcal{A}$ is clearly equivalent to Hall's Condition on $G$. Since classical Hall's Theorem falls out of the hypergraph version so quickly, I think it's fair to think of the topological proof of the hypergraph version as also being a topological proof of Hall's theorem.
|
2025-03-21T14:48:31.633109
| 2020-07-28T14:55:14 |
366797
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Andrej Bauer",
"Franka Waaldijk",
"Nik Weaver",
"fedja",
"https://mathoverflow.net/users/101577",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/1176",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/36886"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631553",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366797"
}
|
Stack Exchange
|
Algorithm/iterative procedure for constructing hypercyclic vectors?
Let $B$ be a separable Banach space and let $L:B\rightarrow B$ be a hypercyclic operator; here I use the definition of hypercyclicity given implicitly by Birkhoff's Transitivity Theorem: continuous linear operator for which every two distinct non-empty open subsets $U$ and $V$ there is some $N\in \mathbb{N}$ such that $L^N(U)\cap V \neq \emptyset$.
Also, let $b \in B$ be given and be such that $L^n(b)\neq L^m(b)$ for any $n,m \in \mathbb{N}$ with $n\neq m$.
Is there an iterative procedure/algorithm for constructing a sequence $\{b_n\}_{n \in \mathbb{N}}$, using only $b$, $L$, and the vector space structure on $B$ such that $\lim\limits_{n\to\infty} b_n$ converges in $B$ to a hypercyclic vector? I'm looking especially for something constructive.
Some special cases of such constructions are:
Chaos on function spaces
Explicit constructions of dense common hypercyclic subspaces
Are there other known procedures or broader (constructive/algorithmic) methods?
Definitely some of them but usually not all of them. In what particular terms do you want us to describe them? I mean, take the usual example of twice the backward shift in $\ell^2$. What answer do you expect here?
Another "what can be said" question ... these are not very satisfying.
@NikWeaver I made the question much more precise now. I'm looking for an algorithm to construct a hypercylic vector.
@fedja I was mostly looking for a way to construct them, beginning from a reasonable starting point in $B$.
You probably should make your question more precise in a constructive sense. In constructive math, the statement "$L$ is a hypercyclic operator" should be equivalent to having an explicit construction of a hypercyclic vector, because existence implies explicit constructibility. This renders your question difficult to answer in a meaningful way. For example, a more meaningful constructive question would be: `Is there a constructive proof of Kitai's Criterion?'.
@FrankaWaaldijk But is Kitai's Criterion necessary and sufficient, I think there are examples of hypercyclic operators which do not satify it?
Sure, but one step at a time is already quite a constructive pace in this area! The second cited paper above actually shows (I believe...) how to prove constructively that an infinite dimensional Banach space with basis admitting a weighted backward shift, is hypercyclic. Any other constructive criterions for hypercyclicity would be most welcome, even if they are not exhaustive. It does not seem likely that you will find an exhaustive criterion, as long as there is no such thing even in classical mathematics...
Fair, I guess what I'm trying to ask is if there is a constructive version of the second paper's result when the space doesn't necessarily admit a generalized weighted-backshift operator
OK that's maybe helpful... but do you know the answer in the classical case? Is it true that any continuous linear $L$ on a Banach space with a Schauder basis is hypercyclic classically, when given a ${b}$ such as you describe? If that is not the case, or if we don't know, then one needs to know some explicit criterion beforehand (different from "$L$ is hypercyclic" since that implies that you already have your hypercyclic vector).
Actually, we don't need to have it explicitly before hand since by 'L is hypercylic' I mean L is continuous, linear, and topologically transitive (then we know such a vector exists by Birkhoff but it doesn't mean I have it). (This is in the classical case since Birkhoff uses Baire so it is non-constructive).
Then please define 'topologically transitive' constructively in your question, because the most common definition I see is again one where an explicit existence of a dense trajectory is used... It may seem a bit like nit-picking, but really it isn't, because in order to have an algorithm, one must know the given data quite precisely.
@FrankaWaaldijk Yes you're right, I make the updated.
Let's constructivize a bit more. In the first paragraph, you should say "inhabited" instead of "non-empty" and "$L^N(U) \cap V$ is inhabited" instead of $L^N(U) \cap V \neq \emptyset$." I also worry about "distinct $U$ and $V$", should that be just "$U \cap V = \emptyset$" or something stronger, such as $\forall x \in U, y \in V, |x - y| > 0$?
Similarly, we can consider strenghening $L^n(B) \neq L^m(b)$ to $| L^n(b) - L^m(b) | > 0$.
@AndrejBauer I'm not sure I see why we should use the condition $|x-y|>0$, what is the intuition/motivation?
In construcive mathematics it often makes sense to use an apartness relation instead of inequality. In a normed space that would be $| x - y | > 0$. For instance, if you know $x \neq 0$ you cannot yet normalize $x$ to $x / | x |$, but if you know $| x | > 0$ then you can.
Ah, I didn't know this. I'll make the update shortly. Thanks for the insight!
|
2025-03-21T14:48:31.633461
| 2020-07-28T15:23:42 |
366799
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Benoît Kloeckner",
"Emil Jeřábek",
"Gerhard Paseman",
"Iosif Pinelis",
"Mateusz Kwaśnicki",
"fedja",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/155651",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/4961",
"user655870"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631554",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366799"
}
|
Stack Exchange
|
What is the probability that a random chord in a sphere touches opposite hemispheres?
(edited) Consider the unit sphere $\mathbb{S}^2\subset \mathbb{R}^3$, and its upper $(z>0)$ and lower $(z<0)$ hemispheres.
Draw two independent, uniformly distributed points $X,Y$ on $\mathbb{S}^2$.
Given $\theta\in[0,\pi/2]$, what is the probability that $X$ and $Y$ belong to different hemispheres among the above two, conditioning by the event that the chord $[X,Y]$ makes an angle $\theta$ with the $z$-axis?
Numerically I find that this probability is $\cos{\theta}$.
Could anyone help me justify this $\cos{\theta}$?
"consider a random chord which makes an angle $\theta$ with the $z$ axis". This definition of the probability space is too vague to allow for any more meaningful answer than "something between $0$ and $1$"). Also, I suspect that MSE is a more appropriate place for this question unless your probability distribution is very fancy.
Can you describe your numerical work in detail?
I don’t understand the question. As written, the distribution does not depend on $\theta$.
@EmilJeřábek : As I understand it, the question is whether the conditional probability that the random chord touches both hemispheres given that the chord makes an angle $\theta$ with the $z$ axis is $\cos\theta$.
@Iosif: I generate random gaussian 3D vectors, which after normalisation are uniformly distributed on a sphere. Then I randomly pick two of them, and from their coordinates I check if they belong to opposite or same hemispheres, and what angle the corresponding chord makes with the z axis. I repeat the operation and do statistics on the results, and that's what I find.
@Emil: The probability does depend on $\theta$, imagine two extreme cases: If $\theta$ is zero (the chord is vertical) it necessarily has to touch both hemispheres. If $\theta$ is $\pi/2$ (the chord is horizontal) it can only start and end on one same hemisphere.
@user655870 : By "generate", do you mean "simulate", i.e., "get corresponding pseudo-random numbers"?
@Iosif: Yes indeed...
@user655870 : You estimate the conditional probability (given an event of probability $0$) by the ratio $r=num/den$ of two small probabilities, $num$ and $den$. Each of these two probabilities is an integral over a comparatively small region in the huge 6D space. Therefore, I think it is very hard to estimate $num$ and $den$ accurately enough by simulation. So, I doubt that such simulation can be reliable enough.
I took the liberty to integrate the edit to the question itself. It should be clearer this way, but I might have been heavy-handed so let me know if you prefer me to reverse part or all of the edit.
@Benoît It's very good this way thank you.
What do you do if the chord is skew to the z-axis? Gerhard "Not Sure Of The Condition" Paseman, 2020.07.28.
With the new formulation it is, indeed, true. The straightforward computation results in the integral $\frac 2\pi\int_0^{\frac\pi 2}\frac {d\alpha}{1+\tan^2\theta\cos^2\alpha}$, which is elementary but somewhat unpleasant to compute (the usual trig substitution $z=\tan\alpha$). However such a simple answer should have an equally simple explanation, so I'd rather wait for someone to find a "no pen or paper solution"
Could you explain how you obtain this integral?
This is not a true "no pen or paper" solution requested by fedja, but at least it avoids integrals. :-)
Let $X$ and $Y$ be independent random vectors on the unit sphere. Write $E = (X - Y) / |X - Y|$ for the unit vector parallel to the chord $XY$, and $Z = \tfrac{1}{2} (X + Y)$.
Claim: Conditionally on $E = e$, the projection $Z = \tfrac{1}{2}(X + Y)$ of the chord $XY$ onto its perpendicular bisector plane $\pi_e = \{v : v \perp e\}$ (the entire chord projects onto a single point) is uniformly distributed over the unit disk in $\pi_e$.
Given the above claim, the proof is straightforward. Indeed: given any unit vector $e$ such that $\theta = \arcsin |e \cdot (0,0,1)|$ as in the question, and conditionally on $E = e$, $X$ and $Y$ belong to different hemispheres if and only if $Z$ belongs to an ellipse, which is the projection of the equator of the unit sphere onto $\pi_e$. This ellipse has semi-axes $1$ and $\cos \theta$ (this becomes pretty clear if one draws a picture). The area of this ellipse is equal to $\cos \theta$ times the area of the unit disk, and the desired result follows. Thus, it remains to prove the claim.
Proof of the claim: Since the random variable $X \cdot Y$ is uniformly distributed over $[-1, 1]$ (Archimedes's theorem!), the random variable
$$ \|Z\|^2 = \|\tfrac{1}{2} (X + Y)\|^2 = \tfrac{1}{2} (1 - X \cdot Y) $$
is uniformly distributed over $[0, 1]$. By rotational symmetry, $\|Z\|^2$ and $E$ are independent. It follows that conditionally on $E = e$, $\|Z\|^2$ is uniformly distributed over $[0, 1]$. Again by symmetry, the conditional distribution of $Z$ (given $E = e$) is invariant under rotations of $\pi_e$, and so it follows that this conditional distribution is uniform over the unit disk on $\pi_e$, as desired.
Thanks Mateusz, there are many things I don't get:
What do you exactly mean by "the projection Z of the chord XY"? I get what projecting a point or a vector means, but a chord? If it's the same Z as defined before, then I would say that is the middle of the chord, is it not?
I'm not sure what plane $\pi_e$ you're talking about, there's an infinity perpendicular to e, do you implicitly mean the one passing by the origin?
Shouldn't it be arccos and not arcsin?
@user655870: Since $e$ is parallel to the chord $XY$, the plane $\pi_e$ is perpendicular to it, and hence the entire chord projects onto a single point $Z$. And indeed $Z = \tfrac{1}{2} (X + Y)$: since $|X| = |Y| = 1$, the line passing through the origin and the midpoint $\tfrac{1}{2} (X + Y)$ of $XY$ is perpendicular to $XY$, and hence it lies in the plane $\pi_e$. I'll try to add some pictures to my answer in a few minutes. And please do ask if anything is unclear.
Thanks for the clarifications and pictures! Now I see. I'm still puzzled about how you define your $\theta$, in my case it's the one formed with the z axis (perp to the equatorial plane), did I miss something basic again?
I remember going in the direction of your claim in the following way: Imagine you distribute random points on the surface of the sphere with uniform density, then you link them by pairs in a given direction so as to make all possible parallel chords in that given direction. Then I thought the $\cos{\theta}$ would be obvious if the density of chords crossing $\pi_e$ was uniform across $\pi_e$, but I thought it couldn't be uniform, because on the edge for example, isn't it infinite?
I'm trying to reconcile your claim with my intuition, when I look at a (transparent) spherical shell, the density of protected matter is infinite on the edge. So there should be an infinite density of chords there, no? I can't picture how it can be uniform.
@user655870: This is the fun part! If one projects $X$ onto a fixed plane, then indeed the distribution concentrates near the boundary. More precisely, the density function is proportional to $1/sqrt{1-x^2-y^2}$. However, here we project on a random hyperplane $\pi_E$, and condition on the event $E = e$. This conditioning changes the distribution: given that $E = e$, the random variable $X$ is no longer uniformly distributed over the unit sphere! The "algebraic" argument used in the "proof of the claim" is not very intuitive, so here is a way to imagine that. (1/2)
Suppose that we condition on $E$ being very close to $e = (0,0,1)$ (rather than exactly equal to $e$). If $X$ is fixed, then $Y$ is necessarily in a very narrow, vertically aligned cone with an apex at $X$. The probability of this event is the surface area of the intersection of the sphere and the cone, and it depends on $X$: if $X$ is close to the pole, the intersection will be large, while if $X$ is near the equator, then the intersection can get very small. (2/2)
|
2025-03-21T14:48:31.634143
| 2020-07-28T16:35:42 |
366807
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"curious math guy",
"https://mathoverflow.net/users/152554"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631555",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366807"
}
|
Stack Exchange
|
Complex Geometric Interpretation of Mordell conjecture
The Mordell conjecture/Falting's Theorem says that any smooth projective curve $X$ of genus $g\geq 2$ over $\mathbb{Q}$ has finitely many integer points (using the valuatlive criterion).
We can of course apply the analytification functor then the sturcture morphism of $X$ becomes
$$X(\mathbb{C})^{an}\rightarrow \text{Spec}(\mathbb{Z})(\mathbb{C})^{an}$$
and an integer point corresponds to a section of this morphism. However $\text{Spec}(\mathbb{Z})(\mathbb{C})^{an}$ is just a point, so this is not an interesting geometric object. However, on an intuitive level, the specta of Dedekind domains correspond to $1$-dimensional curves.
My question therefore is if there exists a (complex) geometric interpretation of the Mordell Conjecture?
The most naive interpretation I have and can thus hope for this would be that there exist only finitely many maps
$$\mathbb{C}\rightarrow X(\mathbb{C})^{an}.$$
Is there anything in that direction?
@crispr Yes, I'm aware of the function field analogy and the Mordell conjecture in that setting.
I guess my question could be clarified by asking "What is the complex analogue of a integer point?"
|
2025-03-21T14:48:31.634252
| 2020-07-28T17:48:51 |
366810
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Sebastien Palcoux",
"https://mathoverflow.net/users/34538"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631556",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366810"
}
|
Stack Exchange
|
Explicit tensor product decomposition for the representations of PSL(2,q)
$\DeclareMathOperator\PSL{PSL}$Let the type of the character theory of a finite group $G$ be the list $[[d_1,n_1], \dotsc, [d_k,n_k]]$ with $1=d_1 < \dotsb < d_k$ and $n_i$ the number of irreducible characters of degree $d_i$.
There is a global understanding of the character theory of the finite groups $\PSL(2,q)$ with $q$ a prime power,
if $q \equiv 0 \pmod2$, it is of type $[[1,1],[q-1,\frac{q}{2}],[q,1],[q+1,\frac{q-2}{2}]]$,
if $q \equiv 1 \pmod4$, it is of type $[[1,1],[\frac{q+1}{2},2],[q-1,\frac{q-1}{4}],[q,1],[q+1,\frac{q-5}{4}]]$,
if $q \equiv 3 \pmod4$, it is of type $[[1,1],[\frac{q-1}{2},2],[q-1,\frac{q-3}{4}],[q,1],[q+1,\frac{q-3}{4}]]$,
and according to these three cases, there is a global way to write the character table of $\PSL(2,q)$. It was first computed by H. Jordan (Group-Characters of Various Types of Linear Groups, 1907). The table is directly available in Character table of $\PSL(2, \mathbb F_q)$ of J. Adams (see also his detailed note Character tables for $\operatorname{GL}(2)$, $\operatorname{SL}(2)$, $\operatorname{PGL}(2)$, and $\PSL(2)$ over a finite field for the proof).
The character table of a finite group $G$ and the Schur orthogonality relations provide the Grothendieck ring (where $n_{i,j}^k \in \mathbb{N}$): $$ \chi_i \chi_j = \sum_k n_{i,j}^k \chi_k. $$
Let $U_i$ be an irreducible unitary representation related to the character $\chi_i$. Then there are $G$-isomorphisms $\Phi_{i,j}$:
$$ \Phi_{i,j} : U_i \otimes U_j \to \bigoplus_k M_{i,j}^k \otimes U_k $$ with $M_{i,j}^k$ a multiplicity space of dimension $n_{i,j}^k$.
Question: What are explicit maps $\Phi_{i,j}$ for $G = \PSL(2,q)$, with a global understanding (as for the character
table)?
There are two typos in the second table that Jeffrey Adams put here (I just pointed out to him): first line: (q-7)/4 should be (q-3)/4; last column, middle line: 1 should be -1.
|
2025-03-21T14:48:31.634415
| 2020-07-28T18:21:58 |
366812
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Deane Yang",
"Giorgio Metafune",
"Guy Fsone",
"Mateusz Kwaśnicki",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/112207",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/33741",
"https://mathoverflow.net/users/613",
"leo monsaingeon"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631557",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366812"
}
|
Stack Exchange
|
Optimizing the gradient norm on the unit sphere
Let $ \Bbb S^{d-1}=\{(x_1,\cdots ,x_d): x_1^2+ \cdots +x_d^2=1\}\subset \Bbb R^d$ be the unit
sphere. Let $\nabla u= (\partial_{x_1}u,\cdots, \partial_{x_d}u)$ be the gradient of a function $u\in C_c^\infty(\Omega)$ with $\Omega \subset \Bbb R^d$ open. For $e\in \Bbb S^{d-1}$, we write $$\nabla u(x)\cdot e = \partial_{x_1}u(x) e_1+\cdots+\partial_{x_d}u(x) e_d.$$
Then, for $p\geq 1$, what is the following quantity?
\begin{split}
I(u) =\sup_{e\in \mathbb{S}^{d-1}}\int_\Omega |\nabla u(x)\cdot e|^pdx
\end{split}
I have tried to use the fact that for all $z\in \Bbb R^d$ we have, $$|z|=\sup_{e\in \mathbb{S}^{d-1}}|z\cdot e|$$ And only got the upper estimate $$I(u)\leq \int_\Omega |\nabla u(x)|^pdx$$
In fact I am expecting an equality here.
There is no hope for equality here: unless almost all the vectors $\nabla u(x)$ are parallel to each other, for every $e$ the inequality $|\nabla u(x) \cdot e| \leqslant |\nabla u(x)|$ is strict on a set of positive measure, and hence the inequality $I(u) \leqslant \int |\nabla u|^p$ is in fact strict.
it seems plausible that the direction $e\in \mathbb S^{d-1}$ achieving the sup should be the average gradient correctly notmalized, i-e $e=\frac{\int \nabla u}{\left|\int \nabla u\right|}$ (provided the average is not zero, of course). Has anyone thought of that?
huuuum, nope, this is only true for $p=1$, for $p>1$ some nonlinear effects come into play and a maximizer $e\in\mathbb S^{d-1}$ must satisfy the nonlinear integral condition that it must be colinear to $e\propto \int_\Omega p|\nabla u(x)\cdot e|^{p-1} \nabla u(x),dx$
@leo monsaingeon True, and also for $p=1$ this is true without the absolute value.
@GiorgioMetafune : yes, for $p=1$ my guess that $e=\pm \frac{\int \nabla u}{\left|\int\nabla u\right|}$ is correct, which is consistent with taking $p\to 1$ in the above integral condition.
@Guy Fsone: out of curiosity, would you mind telling us in which context you encountered this functional?
SHORT COMMENT: If you average instead of taking the sup over $e \in S^{d-1}$, then you do get what you want. That's because $$\int_{S^{d-1}}\int_{\mathbb{R}^d} |e\cdot\nabla f|^p,dx,de = \int_{\mathbb{R}^d}\int_{S^{d-1}} |e\cdot \nabla f|^p,de,dx $$ and, since $$\int_{S^{d-1}} |e\cdot v|^p,de$$ is a rotationally invariant function of $v$, it has to be a dimensional constant times $|v|^p$.
I do not believe there is a simple formula to express $I(u)$, but for sure for most of the functions the inequality
$$
I(u)<\int_\Omega |\nabla u|^p\, dx
$$
is sharp. For example if $\Omega=B$ is a ball and $u(x)=f(|x|)$ is a radial function, then $\nabla u$ is a vector field orthogonal to the sphere $\mathbb{S}^{d-1}$. Then $|\nabla u(x)|=|\nabla u(x)\cdot e|$ only for points $x$ on the line passing through $0$ and parallel to $e$ for all other points the inequality is sharp.
To have equality you would have to have supremum under the sign of the integral, but that would be a different expression.
Of course, with the supremum inside the problem becomes blatantly obvious.
Actually for radial functions, the integral defining $I(u)$ is Independent of $e$ and can be computed in spherical coordinates.
@Piotr Hajlasz Please do you have any suggestion concerning this question? https://mathoverflow.net/q/368396/112207. I have seen a paper of yours on a related topic for Sobolev spaces.
A bit of a longer comment and expansion on Piotr's answer: if you restrict to the case $p$ is even, you can write your integral as
$$ \int |\nabla u \cdot e|^p = \int (\nabla u \otimes \cdots \otimes \nabla u) \cdot (e \otimes \cdots \otimes e) $$
where the tensor product is taken over $p$ times. Since the arguments are manifestly symmetric, you have
$$ I(u) \leq \sup_{T \in \mathcal{S}^P\mathbb{R}^d, |T| = 1} \int (\otimes^p \nabla u) \cdot T =: J(u) $$
Here $\mathcal{S}^p\mathbb{R}^d$ is the set of symmetric rank $p$ tensors over $\mathbb{R}^d$, with induced inner product. The optimization of $J(u)$ is a linear problem and can be easily solved with
$$ T = \frac{\int \otimes^p \nabla u}{| \int \otimes^p \nabla u|}$$
and $J(u) = | \int \otimes^p \nabla u |$.
If you are lucky enough that this $T$ is a pure tensor (of the form $e \otimes e \otimes \cdots \otimes e$) then $I(u) = J(u) = | \int \otimes^p \nabla u |$. However, for generic $u$ this cannot be expected, and you get $I(u) < J(u)$ in this case.
As a final note, you have that
$$ \left| \int \otimes^p \nabla u \right| \leq \int |\otimes^p \nabla u| = \int |\nabla u|^p $$
The inequality step is strict as long as $\nabla u$ is not almost everywhere parallel to a fixed vector. So this method, while it doesn't give you the exact value of $I(u)$, can at times give you a slightly improved upper bound.
Thanks for this extended idea.
More of a comment than an answer, really, but too long for the comment box: For a fixed smooth function $u$ the map
$$
e\mapsto U(e):=\int_\Omega |\nabla u(x)\cdot e|^p dx
$$
is differentiable as a function of $e\in\mathbb R^{d}$, and its differential in the direction $h$ is simply
$$
DU(e).h
=\int_\Omega p|\nabla u(x)\cdot e|^{p-2} \nabla u(x)\cdot h\, dx
=\left(\int_\Omega p|\nabla u(x)\cdot e|^{p-2} \nabla u(x)\,dx\right)\cdot h
$$
A maximizer $e$ on the unit sphere must then satisfy the first-order optimality condition $ DU(e)\cdot h=0$ for all tangent directions $h\in T_e\mathbb S^{d-1}\Leftrightarrow h\cdot e=0$, which means here that $DU(e)$ must be colinear to $e$. In other words,
$$
e=C \int_\Omega p|\nabla u(x)\cdot e|^{p-2} \nabla u(x)\,dx
$$
for some normalization constant $C=\frac{\pm1}{\left|\int_\Omega p|\nabla u(x)\cdot e|^{p-2} \nabla u(x)\,dx\right|}$ (provided the denominator does not vanish).
Of course we have a $\pm$ degree of freedom due to the invariance $U(e)=U(-e)$.
I don't know how much one can exatract from this integral condition, but at least it is clear that the reasonable guess $e=C\int |\nabla u|^{p-2}\nabla u$ is too naive and does not work (since it does not satisfy a priori this integral condition).
Note that for $p=2$ the solution is obviously given by $e= C\int \nabla u$, the average gradient (provided it is not zero, of course), so the functional is somehow the "directional $TV$ norm" $I(u)=\int |\partial_e u|$ in the average (most varying) direction $e=C\int \nabla u$.
Interesting functional!
In the formula for the differential there should be $|\nabla u \cdot e|^{p-2} (\nabla u \cdot e)$
LONG COMMENT: My collaborators and I have in fact studied the function
$$ x\in \mathbb{R}^d \mapsto \left(\int_{\mathbb{R}^d} |x\cdot\nabla f|^p\,dx\right)^{1/p} $$
quite extensively, for example in this paper. Here are a few highlights:
It defines a norm on $\mathbb{R}^d$. The unit ball of this norm is called the $L^p$ polar projection body of the function $f$. The unit ball of the dual norm is called the $L^p$ projection body. This generalizes the definition of the $L^p$ projection and polar projection bodies of a convex body, which were studied here.
If we denote the norm by $\|\cdot\|_{f,p}$, then the sup of $\|e\|_{f,p}$ over all unit vectors $e$ is half the diameter of the polar projection body, where the diameter is defined to be the maximum distance between two parallel supporting hyperplanes.
Also, note that the quantity
$$ \int_{S^{d-1}} \|e\|_{f,p}^{-n}\,de = \int_{S^d}\left(\int_{\mathbb{R}^d} |e\cdot\nabla f|^p\,dx\right)^{-n/p}\,de $$
is the volume of the polar projection body. Although this does not define a norm on any function space, we proved that it satisfies a sharp affine Sobolev inequality. We call the inequality affine, because this quantity is invariant under the action of $SL(d)$ acting on $\mathbb{R}^d$. It is also not hard to show that this inequality implies the standard sharp Sobolev inequality on $\mathbb{R}^d$ that uses the standard Euclidean norm.
A paper that discuss this function, when $f$ is a probability density function can be found here:
E. Lutwak, D. Yang, G. Zhang. Moment-entropy inequalities, Annals of Probability 32 (2004) 757-774.
|
2025-03-21T14:48:31.634959
| 2020-07-28T18:41:13 |
366813
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Karagila",
"Mark Wildon",
"https://mathoverflow.net/users/7206",
"https://mathoverflow.net/users/7709"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631558",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366813"
}
|
Stack Exchange
|
Partition $\mathfrak c$-dense set to $\mathfrak c$- many dense set
A subset of $A\subset\mathbb R$ is called $\mathfrak c$-dense if $ |A\cap I|=\mathfrak c$ for any none open interval $I\subset\mathbb R.$ Then, there is a partition for $A$ to continuum many dense set. Here is my idea: Let $\{I_{\xi}\colon \ \xi<\mathfrak c\}$ be enumeration of all open interval subset $R$ and since $A$ is $\mathfrak c$-dense, so we have $A_{\xi}= A\cap I_{\xi}$ and $|A_{\xi}|=\mathfrak c.$ Put $\mathcal A=\{A_\xi\colon \ \xi<\mathfrak c\}.$ Let $\{\langle r_\xi, A_\xi\rangle\colon \ \xi<\mathfrak c\}$ be an enumeration of $A\times \mathcal A$ with $\mathfrak c$-many repetitions, since $|A\times \mathcal A|=\mathfrak c$. By induction on $\xi<\mathfrak c$, choose a sequence $$ x_\xi\in A_\xi\setminus \{x_\zeta\colon \ \zeta<\xi\}$$Now, for $r\in A$, let $D_r=\{x_\xi\colon r_\xi=r\}$. Notice that the points $x_\xi$ are all distinct. so the sets $D_r$'s are pairwise disjoint and it is to check they are dense. Now, if $$A\setminus\bigcup D_r=B.$$ So we can add the set $B$ to one of the sets $D_r$ and this finishes the proof. Please let me know if there is some thing wrong. Thank you in advance.
https://math.stackexchange.com/questions/3772354/partition-mathfrak-c-dense-set-to-mathfrak-c-many-dense-set
I’m voting to close this question because it has been posted on MSE where it seems better suited.
|
2025-03-21T14:48:31.635096
| 2020-07-28T19:10:38 |
366816
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gro-Tsen",
"Johannes Trost",
"OmniaOperator",
"Sara",
"Tanya Vladi",
"https://mathoverflow.net/users/158421",
"https://mathoverflow.net/users/160900",
"https://mathoverflow.net/users/161896",
"https://mathoverflow.net/users/17064",
"https://mathoverflow.net/users/37436"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631559",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366816"
}
|
Stack Exchange
|
What is the asymptotics of the Fourier transform of $\exp(-x^4)$ for large wave numbers?
The Fourier transform of $\exp(-x^4)$ has an analytical expression, it's the difference of two generalized hypergeometric functions:
$\int d x \ e^{-x^4} e^{ikx} = 2 \ \Gamma(\frac{5}{4}) \ _0F_2(;\frac{1}{2},\frac{3}{4};\frac{k^4}{256}) - \frac{k^2}{4} \Gamma(\frac{3}{4}) _0F_2(;\frac{5}{4},\frac{3}{2};\frac{k^4}{256})$
The asymptotics of the 0F2 is known, but the two series for large k just cancel, so how can one get the asymptotics of the Fourier transform of $\exp(-x^4)$ as $k \rightarrow \infty$?
You can find relevant expressions here https://arxiv.org/pdf/0911.4796.pdf
Thank you. However the expressions in the paper you linked considers the Fourier transform of $exp(-x^{\beta})$ where $\beta < 2$, they seem not to apply to $\beta = 4$.
What do you mean that the "two series for large k just cancel"? Are you saying that they have the same asymptotic expansion about $\infty$, or just the same leading term?
This https://www.sciencedirect.com/science/article/abs/pii/S0096300314006572 is behind a paywall. However the abstract is accessible and gives the asymptotic result.
Perfect, thank you!
For completeness of MathOverflow, the Fourier transform $\Phi$ in question, according to the paper cited by Johannes Trost, satisfies $\Phi(k) \sim 2^{7/6} , \sqrt{\pi/3}, k^{-1/3} \exp(-(2^{1/3},3/16),k^{4/3}) , \cos((3^{3/2},2^{1/3}/16),k^{4/3}-\pi/6)$.
|
2025-03-21T14:48:31.635222
| 2020-07-28T19:59:39 |
366817
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631560",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366817"
}
|
Stack Exchange
|
Convergence of iteration of a convex program
Let $\mathbf{V} \in \mathbb{R}_{+}^{n \times m}, \ \ \mathbf{E} \in \mathbb{R}_{+}^{n \times m}$, with $\mathbf{V} \mathbf{1}_{m} = \mathbf{1}_{n}$ and $\mathbf{E}^{T} \mathbf{1}_{n} = \mathbf{1}_{m}$ where $\mathbf{1}_{m}$ is the vector of ones of size $m$.
Define the function $f: \mathbb{R}^m \to \mathbb{R}^m$:
\begin{align}
f(\mathbf{q}) = \arg \min_{\mathbf{p}} \left\{ \sum_{j = 1}^m p_j - \sum_{i = 1}^n
\left(\sum_{j = 1}^m E_{ij} q_j\right) \log{\left(\min_j \{ \frac{p_j}{V_{ij}} \} \right)} \right\}
\end{align}
Define the following iterative process:
\begin{align}
\mathbf{p}(0) = (\frac{1}{m}, \frac{1}{m}, \dots, \frac{1}{m})^T \\
\mathbf{p}(t) = f(\mathbf{p}(t-1))
\end{align}
My goal is to understand the convergence properties of the above iterative process. That is, I want to understand for which $\mathbf{V}$ and $\mathbf{E}$ this iterative process converges to the fixed point of $f$. I have around 100,000 experiments where I initialize $\mathbf{E}$ and $\mathbf{V}$ randomly and the process always converges. However, I also have the following "hand-crafted" $\mathbf{V}$ and $\mathbf{E}$ for which the process does not converge:
\begin{align}
\mathbf{V} = \begin{bmatrix}
1 & 0\\
0 & 1\\
\end{bmatrix}\\
\mathbf{E} = \begin{bmatrix}
0 & 1\\
1 & 0\\
\end{bmatrix}
\end{align}
These experiments seems to suggest to me that for randomly initialized $\mathbf{V}$ and $\mathbf{E}$ the probability of convergence of the iterative process is 1. That is, the probability of getting $\mathbf{V}$ and $\mathbf{E}$ such that the process does not converge is 0.
I want to answer the following questions: Which conditions must $\mathbf{V}$ and $\mathbf{E}$ satisfy for the process to converge.
The most recent approach I took was to use this paper to find the jacobian of $f$ to see for which values of $\mathbf{V}$ and $\mathbf{E}$ the jacobian of the function is less than 1 but the expression I get is too complicated to understand the values of $\mathbf{V}$ and $\mathbf{E}$ for which the process. I am not even sure how to approach the problem and appreciate any pointers.
|
2025-03-21T14:48:31.635384
| 2020-07-28T20:04:05 |
366818
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris Wuthrich",
"Somos",
"https://mathoverflow.net/users/113409",
"https://mathoverflow.net/users/5015"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631561",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366818"
}
|
Stack Exchange
|
Reference for the $3$-series of an elliptic formal group law
The $3$-series of the formal group law of the Weierstrass curve $y^2 = x^3 + a_2 x^2 + a_4 x$ begins
$$
[3](z) = 3 z - 8 a_2 z^3 + (24 a_2^2 - 96 a_4) z^5 - (72 a_2^3 - 288 a_2 a_4) z^7 + (216 a_2^4 - 1472 a_2^2 a_4 + 2432 a_4^2) z^9 + \dots
$$
(assuming I did not make a mistake). Is there a published reference that contains this information, including the coefficient of $z^9$?
sage: R.<a2,a4> = QQ[]; E = EllipticCurve([0,a2,0,a4,0]); E.formal_group().mult_by_n(3)
In PARI/GP it is $\texttt{E=ellinit([0,a2,0,a4,0]); m=9; subst(ellformalexp(E,m),x,3*ellformallog(E,m))}$.
|
2025-03-21T14:48:31.635458
| 2020-07-28T20:05:44 |
366819
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Handelman",
"LSpice",
"Nathaniel Johnston",
"Zach Teitler",
"darij grinberg",
"https://mathoverflow.net/users/11236",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/42278",
"https://mathoverflow.net/users/88133"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631562",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366819"
}
|
Stack Exchange
|
Eigenvectors of a matrix
Let $M$ be a square matrix of order $n\times d$. Let $\xi_{1},\dots,\xi_{n\times d}$ be an orthonormal basis of $\mathbb{R}^{n\times d}$ formed of eigenvectors of $M$. We have
$$\xi_{i}=(\lambda_1, 0, \dots, 0) + \cdots + (0, \dots,0,\lambda_n)$$
where $0, \lambda_j\in \mathbb{R}^{d}$. We can write $\xi^{i}$ as
$$\xi_{i}=\alpha_{i,1}(\tilde{\lambda}_1,0,\,.\,.\,.\,,0)+\alpha_{i,2}(0,\tilde{\lambda}_2,0,\,.\,.\,.\,,0)+.\,.\,.\,+\sqrt{1-\alpha_{i,1}^2-\,.\,.\,.\,-\alpha_{i,n-1}^2}(0,\,.\,.\,.\,,0,\tilde{\lambda}_n).$$
We want to prove that there exist $\alpha_0>0$ small enough and $i_1,\,.\,.\,.\,,i_{(n-1)\times d}$ such that
$$\alpha_{i_k,l}>\alpha_0\qquad\text{for}\;k=1,\,.\,.\,.\,,(n-1)\times d\;\text{and}\; l=1,\,.\,.\,.\,n-1.$$
What is $\alpha_0$? As it’s currently written you can just choose any positive number smaller than the other $\alpha$s, as long as they are nonzero.
What is a square $n \times d$ matrix? This means nothing to me.
@David Handelman - They mean an $nd \times nd$ matrix. They're using $\times$ to mean multiplication, not separation of matrix dimensions.
(To be fair, there was an explanation that the confusing terminology "square $n \times d$ matrix" meant $n \times d$ rows and $n \times d$ columns, but, for some reason, @vidyarthi edited it out.)
The $+$ signs mean $\oplus$, right? (Else the lengths of the vectors don't match.) Also, what does "small enough" mean in an existence statement?
@darijgrinberg - My understanding of the question is that the "zeros" are d-dimensional zero vectors (and the $\lambda$s are similarly d-dimensional), so the $+$ signs really are addition (of vectors living in $\mathbb{R}^{nd}$).
|
2025-03-21T14:48:31.635609
| 2020-07-28T20:32:21 |
366820
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"AmorFati",
"Asvin",
"Donu Arapura",
"Joe Silverman",
"Sebastian",
"Tabes Bridges",
"https://mathoverflow.net/users/105103",
"https://mathoverflow.net/users/11926",
"https://mathoverflow.net/users/27219",
"https://mathoverflow.net/users/4144",
"https://mathoverflow.net/users/4572",
"https://mathoverflow.net/users/58001"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631563",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366820"
}
|
Stack Exchange
|
Concerning the omnipresence of hyper-elliptic curves in the construction of examples
Vague rambling: I hate asking these types of questions, but I feel that I would benefit immensely from hearing some discussion of the use of hyperelliptic curves in constructing certain examples. What do I mean? Of course, as mathematicians, and especially as (aspiring) algebraic geometers, either while reading, examples are essential. Often times, however, hyperelliptic curves are mentioned. They are mentioned so often (from what I see) that they must have some property that is very useful when trying to construct examples.
Let us consider an actual instance of this: Following Campana's survey on special manifolds, define a Bogomolov sheaf to be a saturated, coherent subsheaf $\mathcal{L} \subset \Omega_X^p$ of rank one over a compact Kähler manifold $X$ such that the Kodaira dimension $\kappa(X, \mathcal{L})=p>0$. The manifold $X$ is said to be special if $X$ admits no Bogomolov sheaves.
We'll construct an example of a Bogomolov sheaf as follows: Let $E$ be an elliptic curve and $\tau$ a translation of order $2$ on $E$. Let $C$ be a hyperelliptic curve with an involution $\eta$. Set $X : = (E \times C)/\langle \tau, \eta \rangle$ and let $f : X \to B$ be the Moishezon--Iitaka map onto some curve $B$. The saturation of $f^{\ast}K_B$ in $\Omega_X^1$ has Kodaira dimension $\kappa(X, f^{\ast}K_B)=1$ and is therefore a Bogomolov sheaf. In fact, it is the only one on $X$.
Question: What properties do hyperelliptic curves have that make them useful in the construction of examples?
Feel free to downvote my question if this question is better suited for math.stackexchange. The reason for posting the problem to overflow is that the motivation is clearly at the level of research mathematics. And the question is, fundamentally, one of motivation.
Properties besides being very explicit?.
@DonuArapura From a geometric perspective, I find elliptic curves much more explicit: they are flat and have genus $1$!
@DonuArapura I will try to be more precise in my question. Again, I apologize for the vague nature of the question. It clearly reflects my lack of understanding of hyperelliptic curves.
If you want to get your hands on a higher genus curve for whatever reason, then a hyperelliptic curve is a good thing to look at: the $1$-forms on it is easy to describe, Mumford gave an explicit construction of its Jacobian...
There's also the fact that every genus 2 curve is hyperelliptic, so you're getting another whole genus. OTOH, there are certainly theorems along the lines of "If C is hyperelliptic, then X is true, otherwise Y is true." A larger class of curves that is easier to deal with are cyclic covers of $\mathbb P^1$ (often called "super-elliptic curves"). Also, by analogy if you've studied number theory, after $\mathbb Q$, it's very instructive to study quadratic extensions of $\mathbb Q$; replace $\mathbb Q$ with $k(T)$, voila, you have hyperelliptic function fields.
Also, hyperelliptic (and superelliptic) curves have (lots of) automorphisms. Automorphisms can be really useful!
@DonuArapura Thank you
@JoeSilverman Thank you
@Asvin Thank you! Do you have an illustrative example?
Superelliptic curves (in characteristic not dividing the degree of the cover) have the form $$Y^n=f(X).$$ They thus admit automorphisms $(X,Y)\to(X,\zeta Y)$, where $\zeta$ is an $n$th root of unity. More intrinsically, since there is a map $C\to\mathbb P^1$ that is Galois with group $G\cong\mathbb Z/n\mathbb Z$, the elements of $G$ are automorphisms of $C$. And depending on the shape of $f(X)$, it's possible for $\operatorname{Aut}(C)$ to be even larger.
Besides the fact that one can gets his/her hands on hyper-elliptic curves, it is worth noticing that they are general enough to be interesting for various reasons and in many different situations.
Hyperelliptic curves are unique among curves of general type in that their canonical bundle is not very ample; this property has always seemed particularly relevant to me. It might also be relevant that hyperelliptic curves are not complete intersections, so they are an explicit way of getting around a property that is "hard to avoid" most of the time.
|
2025-03-21T14:48:31.636066
| 2020-07-28T21:16:46 |
366821
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gabriel Soranzo",
"Will Sawin",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/34066"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631564",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366821"
}
|
Stack Exchange
|
Ramification and reduction
Let $K$ a local field ($K$ finit extension of $\mathbb{Q}_p$), $\mathcal{O}_K$ the integer of $K$ and $k$ the residue field of $\mathcal{O}_K$.
Let $\psi:\mathbb{P}^1_K\to\mathbb{P}^1_K$ a finit separable morphism, $\widetilde{\psi}=\Psi:\mathbb{P}^1_{\mathcal{O}_K}\to\mathbb{P}^1_{\mathcal{O}_K}$ a model of $\psi$ that is $\Psi$ is the extension of scalar of $\psi$ ie $\Psi=\psi\times_{\mathcal{O}_K}\text{Id}_K$.
$$ \require{AMScd}
\begin{CD}
\mathbb{P}^1_K @>{\psi}>> \mathbb{P}^1_K\\
@VV{\alpha}V @VVV \\
\mathbb{P}^1_{\mathcal{O}_K} @>{\Psi}>> \mathbb{P}^1_{\mathcal{O}_K}
\end{CD} $$
Let $\overline{\Psi}=\Psi\times_{\mathcal{O}_K}\text{Id}_k$ the reduction of $\Psi$.
$$ \require{AMScd}
\begin{CD}
\mathbb{P}^1_k @>{\overline{\Psi}}>> \mathbb{P}^1_k\\
@VV{i}V @VVV \\
\mathbb{P}^1_{\mathcal{O}_K} @>{\Psi}>> \mathbb{P}^1_{\mathcal{O}_K}
\end{CD} $$
If the branching points (ie ramification points) $P_1,\ldots,P_n$ of $\psi$ are $K$-rationnals, as $\mathbb{P}^1_{\mathcal{O}_K}(\mathcal{O}_K)=\mathbb{P}^1_K(K)$ (by mutliplication of denominators) one can take their reductions $\overline{P_1},\ldots,\overline{P_n}\in\mathbb{P}^1_k(k)$.
Question: I'd like to prove that if the ramification indices of the $P_i$ are resp. $e_i$, they are the same for the $\overline{P_i}$ and if there are ``coalescence'' then the ramification indices of the resulting ramification point $\overline{Q}$ is the sum of the indices $e_i$ for which $\overline{P_i}=\overline{Q}$. I don't have the beginning of an explanation of that, if it's true...
I guess that we shouldn't have wild ramification so the sums of $e_i$ of point that collapse in the same point shouldn't be nul in $k$.
I guess that a general reference for that is SGA1 (Exposé X) but for the moment it's to difficult for me... If someone has a simpler reference for my specific case I'l take it! Thanks!
If you find this question to easy for mathoverflow feel free to answer here in mathstackexchange and tell me in a comment.
In your setting, you can just do everything concretely using the derivative.
The correct statement is for $\overline{Q}$ in $\mathbb P^1_k$,
$$e(\overline{Q}) + \operatorname{swan}(\overline{Q}) = 1 + \sum_{\substack{ i \in \{1,\dots n \} \\ \overline{P}_i = \overline{Q} }} (e_i - 1).$$
This is under your assumptions except that we need to assume $\overline{\Psi}$ is separable.
To prove this, first we can assume by a change of variables that $\overline{\Psi}(\overline{Q}) \neq \infty$. Then express $\widetilde{\psi}$ as a rational function $f$ in $\mathbb Z_p[X]$, without a pole at $\overline{Q}$, and thus without a pole at any of the $P_i$ that reduce to $Q$. Now consider its derivative $\frac{df}{dx}$.
In characteristic zero, this function vanishes exactly at the ramification points $P_1,\dots, P_n$, and its order of vanishing at $P_i$ is $e_i-1$.
In characteristic $p$, its order of vanishing at a point is $e$ plus the Swan conductor minus $1$.
Now we just need to know that the order of vanishing of $\frac{df}{dx}$ at $\overline{Q}$ is the sum of its orders of vanishings at $P_i$ for all the $P_i$ that reduce to $Q$. This follows from factoring the numerator of $\frac{df}{dx}$ into linear factors, and noting that the order of vanishing is the number of linear factors that vanish at a point.
We can't rule out wild ramification here, as the example $x^p-x$ (for $K = \mathbb Q ( p^{1/(p-1)})$) shows. In that case we have one point $\infty$ with $e=p$ and $p-1$ points (the $p-1$st roots of $p^{-1}$) with $e=2$, that all reduce to $\infty$, and in the reduction, $\infty$ has $e=p$ and $\operatorname{swan}=p-1$.
Thanks! Could you tell me in short words what is the Swan conductor or give me a reference?
@Macadam In this case we can just define the Swan conductor as the order of vanishing of the derivative minus $e-1$, which is $0$ in the case of tame ramification and $>0$ in the case of wild ramification.
Any reference on Riemann-Hurwitz in characteristic $p$ should explain what you need here, because it's mostly the same ideas.
Excuse my stupidity, but In a simple exemple, $P=x^2(x-3)^2$ in $\mathbb{Q}_3$, one hase $P_1=0$ and $P_2=3$ with $\overline{P_1}=\overline{P_2}=\overline{0}$ in $\mathbb{F}_3$, $e_1=2$, $e_2=2$ and the resulting ramification is $4$ because $\overline{P}=x^4$, correct? But I can't see it with your formula?
@Macadam The derivative of $x^2 (x-3)^2 = 2 x (x-3)^2 + 2 x^2 (x-3) = 2x (x-3) (2x-3)$. There is a third ramification point $3/2$ which also has $e=2$.
Ok and $3/2$ is $[2:3]$ so reduce to $[2:0]=[1:0]$ in $\mathbb{P}^1_{\mathbb{F}_3}$. So we have 3 points $0$, $3$ and $3/2$ which goes to $\overline{0}$ all with $e=2$ and $e(\overline{0})=4$ because the function becomes $z^4$ which is coherent with $1+(2-1)+(2-1)+(2-1)$.
I need some details with the reasonning with the rationnal function $f$: I'm agree that we can write $f=\pi^k P/Q$ with $\pi$ an uniformizer of $\mathcal{O}_K$, $q\in\mathbb{Z}$, $P,Q\in\mathcal{O}_K[X]$ primives and prime between them so $\widetilde{\psi}$ has a sens (not as rationnal function but with $(\pi^k P,Q)$ or $(P,\pi^kQ)$ global section generating the invertible sheaf $\mathcal{O}(d)$). So it has a sens to reduce $f$ in the residue field $k$.
But I see two problems with that: if $q\neq 0$ the resulting reduced morphism is $(0,\overline{Q})$ or $(\overline{P},0)$ which is constant (should I add the hypothesis $\overline{\psi}$ not constant?). Second problem: what if the points are not in $\mathcal{O}K$: they are in $\mathbb{P}^1{\mathcal{O}_K}$ and not $\infty$ but they should be for example $[1:u/\pi]$ and so we can not reduce the linear factor $(X-u/\pi)$ in $f'$.
@Macadam If $\Psi$ is a well-defined morphism on $\mathbb P^1_{\mathcal O_K}$ then $\overline{\psi}$ cannot be constant (using that $\psi$ is finite, thus surjective.) For instance because $\overline{\psi}$ is a map from a projective variety, hence proper, so its image is closed. If the points are not in $\mathcal O_K$ (i.e. reduce mod $\pi$ to $\infty$) you get factors like $(cX -1)$ for $c \in \mathcal O_K$ and the same argument works. Alternately, you can extend to a larger field, which gives you enough freedom to change the variables so that none of the points reduce to $\infty$.
For the 1st pb: the fact that the image is closed is not contradictory with the constancy of the map, or I missed something? For the second pb: factors like $(cX-1)$ can reduced to $-1$ if $c\notin\mathcal{O}_K^*$ so disapear in $f'$. Maybe all this stuff work if the reduction is good? If we enlarge the field the $u/\pi$ becomes $u/\eta^q$ and the problem stay. Excuse my naive questions...
@Macadam It is contradictory with the map being surjective in characteristic $0$ and constant in characteristic $p$ (then, what is the image? Is it closed?) The factors disappearing is OK, it just means the ramification points are going off to infinity. The field enlargement is to make a sufficient linear change of variables (rational linear transformation in $X$) that the points don't go off to $\infty$.
1st pb: I'm agree if you mean that the image of $\Psi$ is closed: the points of $\mathbb{P}^1_\mathbb{K}$ are not closed in $\mathbb{P}^1_{\mathcal{O}_K}$ so the image of $\overline{\varphi}$ contains points of the closure of points of $\mathbb{P}^1_K$ so $\overline{\varphi}$ is not closed. For the 2nd: Ok! I'd so like to have better algebraic tools in disposal so I don't have to use equations for the morphism $f$...
@Macadam If you look at proofs of Riemann-Hurwitz you will see the more abstract tools that are needed.
A last think (I promise) it's important for my (master) thesis: for me there was always unique a morphism $\Psi:\mathbb{P}^1_{\mathcal{O}K}\to\mathbb{P}^1{\mathcal{O}K}$ such that $\psi$ was its extension of scalar with the trick $f=\pi P/Q$ as I said upper. But here you telled me that $\Psi$ may be not well defined: I can't see why because in all cases $\pi^k P$ and $Q$ are coprimes ($P$ and $Q$ are coprime and primitives so $\pi^k P$ ans $Q$ stay coprimes...) What's the problem? There are global sections generating the invertible sheaf $\mathcal{O}{\mathbb{P}^1_{\mathcal{O}_K}}(d)$?
@Macadam The issue is the morphism will not be well-defined at the vanishing points of $Q$ mod $\pi$ since $\pi P$ and $Q$ both vanish there. In general, for a morphism $a/b$ to be well-defined we want the ideal generated by $a$ and $b$ to be the unit ideal.
|
2025-03-21T14:48:31.636620
| 2020-07-28T22:13:04 |
366823
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gilles Mordant",
"Iosif Pinelis",
"https://mathoverflow.net/users/125260",
"https://mathoverflow.net/users/153595",
"https://mathoverflow.net/users/36721",
"香结丁"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631565",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366823"
}
|
Stack Exchange
|
The power of chi-square test
Under the null hypothesis, if we have
$$\sqrt{n} \vec{x} \, \rightarrow_d \, N(0, I_p),$$
the test statistic can be construct as:
$$\hat{\Psi} = n \vec{x}^{\top} \vec{x} \, \rightarrow_d \,\chi^2_p.$$
And we reject the null hypothesis if $\hat{\Psi} > \chi^2_{p, 1 - \alpha}$ under level $\alpha$.
Now, if under the alternative hypothesis $H_1$,
$$\sqrt{n} \left( \vec{x} - \vec{\mu} \right)\, \rightarrow_d \, N(0, \Sigma),$$
I want to know the power of using test statistic $\hat{\Psi}$, i.e.
$$\mathrm{P} \left( \left. \hat{\Psi} > \chi^2_{p, 1 - \alpha} \right| H_1 \right) = ?$$
I know under $H_1$, $\hat{\Psi}$ can be decomposed as:
$$\hat{\Psi} \, \rightarrow_d \, \sum_{j = 1}^p \xi_j^2, \quad \left( \xi_1, \cdots, \xi_p \right)^{\top} \, \sim \, N(\vec{\mu}, \Sigma),$$
but I don't know how to deal with this square sum. Can anyone help me? Thanks a lot!!
It seems we cannot use the above method to calculate the power of this test. Since $n^{-1} \hat{\Psi} , \rightarrow_p , \vec{\mu}^{\top}\vec{\mu}$, I guess, under $H_1$, $n^{-1/2} \left( \hat{\Psi} - \vec{\mu}^{\top}\vec{\mu} \right)$ is asymptotically normal? So the power goes to 1 when $n \rightarrow \infty$, it that right?
If $\Sigma=I_p$, then the distribution of
$\sum_{j=1}^p\xi_j^2$ for $(\xi_1,\cdots,\xi_p)^\top\sim N(\vec{\mu},\Sigma)$ is the non-central chi-square distribution with $p$ degrees of freedom and non-centrality parameter $\vec{\mu}^\top\vec{\mu}$.
If $\Sigma\ne I_p$, then the distribution of
$\sum_{j=1}^p\xi_j^2$ for $(\xi_1,\cdots,\xi_p)^\top\sim N(\vec{\mu},\Sigma)$ has no name or closed-form expression of its pdf or cdf -- even when $\vec{\mu}=\vec0$.
Responding to a comment by the OP: Let $x_n:=\vec x$ and $\mu:=\vec\mu$. If $$Z_n:=\sqrt n(x_n-\mu) \to_d Z\sim N(0,\Sigma), \tag{1}$$
then, for any fixed $\mu\ne0:=\vec0$,
$$nx_n^\top x_n=(\sqrt n\mu+Z_n)^\top(\sqrt n\mu+Z_n)
=n\mu_n^\top\mu_n+O_P(\sqrt n),$$
so that $nx_n^\top x_n$ converges in probability to $\infty$, rather than to a finite random variable.
To get a finite random variable in the limit, you need to consider alternative hypotheses close enough to the null one; that is, in this case, you need to consider nonzero alternative asymptotic mean vectors close enough to $0$. In particular, it will make sense to fix some $\mu\ne0$ and consider the alternative values $\mu/\sqrt n$ of the asymptotic mean.
So, now instead of (1) we are assuming that
$$Z_n:=\sqrt n x_n-\mu=\sqrt n(x_n-\mu/\sqrt n) \to_d Z\sim N(0,\Sigma).$$
Then
$$nx_n^\top x_n=(\mu+Z_n)^\top(\mu+Z_n)\to_d (\mu+Z)^\top(\mu+Z),$$
and $\mu+Z\sim N(\mu,\Sigma)$, so that the limit random variable
$(\mu+Z)^\top(\mu+Z)$ will have the non-central chi-square distribution with $p$ degrees of freedom and non-centrality parameter $\mu^\top\mu$.
Thanks so much. But is there a feasible method to tackle $$\mathrm{P} \left( \sum_{j
= 1}^p \xi_j^2 > \chi_{p, 1 - \alpha}^2\right)$$ when $\Sigma \neq I_p$?
@香结丁 : As I said, if $\Sigma\ne I$, then little can be said even for $\vec\mu=\vec0$; see e.g. https://link.springer.com/article/10.1007%2Fs00440-003-0262-6 for some inequalities.
@ Iosif Appreciate your helping! It seems we cannot use this method to calculate the power. Since $n^{-1} \hat{\Psi} , \rightarrow_p , \vec{\mu}^{\top}\vec{\mu}$, I guess, under $H_1$, $n^{-1/2} \left( \hat{\Psi} - \vec{\mu}^{\top}\vec{\mu} \right)$ is asymptotically normal? It that right?
Thanks for your generous help! Now I know what I should do!
@香结丁 : As a relatively new user, you may want to look at these guidelines
Iosif Pinelis' answer says it all. Still, if you are more interested in numerical computations, the following article may help you. https://www.jstor.org/stable/2332763?seq=1#metadata_info_tab_contents
There even is an R package that could serve your needs : https://cran.r-project.org/web/packages/CompQuadForm/index.html.
@IosifPinelis I have accepted your answer as well as who solved my previous questions. Thanks so much for your reminder.
@GillesMordant In my study case, numerical computation is indispensable. Thanks for your prompt guide.
|
2025-03-21T14:48:31.636902
| 2020-07-28T23:04:47 |
366824
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben MacAdam",
"Simon Henry",
"https://mathoverflow.net/users/22131",
"https://mathoverflow.net/users/75783"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631566",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366824"
}
|
Stack Exchange
|
Groupoids as models of symmetric simplicial sets
In the Elephant, Peter Johnston remarks that internal categories may be regarded as simplicial objects that “preserve all limits that happen to exist in $\Delta^{op}$“ (I guess you might call this a flat functor). This is because the join in $\Delta$ is a limit.
Does a similar statement exist for the symmetric simplicial set/groupoid correspondence? It’s clear that the nerve of a groupoid yields a symmetric simplicial set, the question is then whether or not they correspond to flat functors on the site for symmetric simplicial sets.
Note that there is already a notion of "flat functor" which is indeed quite close to what you are talking about (https://ncatlab.org/nlab/show/flat+functor), but these are not what you are talking about. Informally, a flat functor preserve all limits, "even these that do not exist". Here you just have a limit preserving functor.
You can definitely characterize groupoids as presheaves on $Fin_+$ preserving some colimtis (i.e. sending some colimits in $Fin_+$ to limits in Set). In fact Groupoids are the presheaf on $Fin_+$ that preserve the colimits comming from $\Delta$.
However, the Category $Fin_+$ has much more colimits than $\Delta$, in fact it all non-empty finite colimits. Groupoids do not preserves most of these colimits.
For example, in $Fin_+$ you have $\{1\} \coprod \{1\} \simeq \{1,2\}$ , but given a a groupoid $G$, $G(\{1\})$ is the set of objects, and $G(\{1,2\})$ is the set of arrow, so the isomorphisms $G(\{1,2\}) \simeq G(\{1\}) \times G(\{1\})$ means that your groupoid has a unique arrow between any two objects.
In fact, I claim that the colimits in $Fin_+$ that exists are exactly the finite non-empty colimits, and the category of presheaf preserving them is equivalent to the category of sets. In terms of the usual description of groupoids as presheaf on $Fin_+$, these corresponds to anti-discrete groupoids.
Of course, so if you preserve products then you get the pair groupoid, whereas if you only preserve the necessary pullbacks your models are groupoids.
|
2025-03-21T14:48:31.637074
| 2020-07-29T00:09:45 |
366828
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631567",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366828"
}
|
Stack Exchange
|
What are the hypotheses we should add for the generalizations of Furstenberg recurrence theorem?
In my question here I suggest a possibility for generalization of Furstenberg recurrence theorem needing some hypothesis for that generalization to be hold in the side of convergence of the below average (Multiple recurrence theorem) and existence of positive integer $n$ such that we have positive measure (Furstenberg recurrence theorem)
Furstenberg recurrence theorem: Let $E$ be a subset of a probability space $(X,\mu)$ of positive measure, and let $T: X \to X$ be an invertible measure-preserving shift. Then for any $k \geq 1$ there exists a positive integer $n$ such that $E \cap T^n E \cap T^{2n} E \cap \dots \cap T^{(k-1) n} E$ has positive measure.
Assume we try to generalise this recurrence theorem by the substitution
of linear terms $n,2n,\cdots (k-1)n $ by integer-valued polynomials $p_1(n),p_2(n),\cdots p_i(n),1\leq i \leq n$ of degree ($d \geq 1$), My question here is:
Question: What are the hypothesis we should add to that generalisation to have existence of positive integer $n$ with $E \cap T^{p_1(n)} E \cap T^{p_2(n)} E \cap \dots \cap T^{(k-1) p_i(n)} E$ has a positive measure ?
Probably this question have the same meaning with asking about necessary hypothesis that we should add for the convergence of the lower limit (average) to be hold in $L^2(\mu)$ which is defined by :
$\lim_{N\to \infty} \inf\frac{1}{N} (\sum_{}^{}E \cap T^{p_1(n)} E \cap T^{p_2(n)} E \cap \dots \cap T^{(k-1) p_i(n)} E)>0$ ?
The extension you want was proved in the 1990s by Bergelson and Leibman. See [1] and also further developments in [2].
[1] Bergelson, Vitaly, and Alexander Leibman. "Polynomial extensions of van der Waerden’s and Szemerédi’s theorems." Journal of the American Mathematical Society 9, no. 3 (1996): 725-753.
[2] Bergelson, Vitaly, Alexander Leibman, and Emmanuel Lesigne. "Intersective polynomials and the polynomial Szemerédi theorem." Advances in Mathematics 219, no. 1 (2008): 369-388.
|
2025-03-21T14:48:31.637218
| 2020-07-29T00:27:02 |
366830
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Willie Wong",
"https://mathoverflow.net/users/3948"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631568",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366830"
}
|
Stack Exchange
|
Angular excitations and Schrodinger operators with radial potential in N-dimensions
Can someone please explain the following in mathematical language?
"First of all, angular excitations only push the energy up, never down, so it is enough to analyze spherically symmetric s-waves...", c.f. first answer in
https://physics.stackexchange.com/questions/370901/solution-for-inverse-square-potential-in-d-3-spatial-dimensions-in-quantum-mec
Is that assertion also valid in $N$ dimensions? And more generally for a Schrodinger operator in a rotationally symmetric riemannian manifold with radial potential? In that case, how to prove it? (references are welcome).
In fact, I would like to apply that kind of results in the study of stability of minimal surfaces. More details and context in this related question:
Fourier mode decomposition and eigenvalues of Schroedinger operators with radial potential in N-dimensions
Yes. If you have a rotationally symmetric Riemannian manifold with a radial potential, so you have a Schrodinger operator $H = -\Delta + V$, then $\langle f, Hf\rangle \geq \langle \bar{f}, H\bar{f}\rangle$ where $\bar{f}$ is the spherical average of $f$.
|
2025-03-21T14:48:31.637674
| 2020-07-29T02:06:08 |
366834
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631569",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366834"
}
|
Stack Exchange
|
How did Jouanolou define the cup product with no finiteness hypotheses in SGA 5?
In SGA 5 Exposé VII, at the beginning of §2, Jouanolou lets $X$ and $Y$ denote two schemes, $f:X\rightarrow Y$ a morphism, and $A$ the ring $\mathbf{Z}/\nu\mathbf{Z}$ where $\nu$ is an integer prime to the residual characteristics of $X$ and $Y$. He defines an arrow
$$Rf_*(A_X)\otimes^L Rf_*(A_X)\to Rf_*(A_X)\tag{*}$$
by first defining an arrow
$$f_*C^\cdot(X,A_X)\otimes f_*C^\cdot(X,A_X)\to f_*C^\cdot(X,A_X)$$
where $C^\cdot(X,A_X)$ is the Godement resolution of the sheaf $A_X$ so that $f_*C^\cdot(X,A_X)$ computes $Rf_*(A_X)$.
He then writes that one can obtain (*) using flat resolutions of the components on the left. Under some finiteness hypotheses that assure that $Rf_*(A_X)$ is in $D^-(X)$, I see this no problem. Actually, using the result of Spaltenstein published in ’88, one can use a K-flat resolution to define $\otimes^L$ on all of $D(X)\times D(X)$, but this definition hadn’t been made by the time Jouanolou’s exposé was published. So, what did Jouanolou have in mind here? Thanks for your help!
|
2025-03-21T14:48:31.637780
| 2020-07-29T03:15:32 |
366837
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ian Agol",
"https://mathoverflow.net/users/1345"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631570",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366837"
}
|
Stack Exchange
|
3-manifold foliated by circles is Seifert fibered
Let $M$ be a compact 3-manifold with boundary equipped with a 1-dimensional foliation all of whose leaves are circles. An old theorem of Epstein says that $M$ is a Seifert fibered space.
The proof of Epstein’s theorem is quite complicated, and it predates the modern theory of 3-manifolds centered around Thurston’s work and geometrization.
Question: Are there any alternate proofs now of Epstein’s theorem?
I'm not sure that newer developments, such as the geometrization theorem, might help with this problem. The issue seems to be the boundedness of the lengths of the leaves, which fails for foliations by circles of a compact 5-manifold by a result of Sullivan. On the other hand, Edwards-Millet-Sullivan showed more generally that codimension-2 foliations of manifolds with compact leaves have leaves of bounded volume.
https://doi.org/10.1016/0040-9383(77)90028-3
Note that they incorporate a simplification to part of Epstein's proof due to Norris Weaver. https://www.jstor.org/stable/1970855
|
2025-03-21T14:48:31.637881
| 2020-07-29T03:37:09 |
366838
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Hansen",
"Stefan Kohl",
"abx",
"curious math guy",
"https://mathoverflow.net/users/1464",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/28104",
"https://mathoverflow.net/users/40297"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631571",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366838"
}
|
Stack Exchange
|
Boundary of Siegel modular variety
The moduli space of curves has a compactification whose boundary can be understood as the product of moduli spaces of curves of lower genus. Therefore (perhaps naively) one might hope that there exists a compactification of $A_g$ whose boundary can be understood as in terms of the moduli of abelian varieties of lower dimension. Is there any such compactification?
There are many! This is a long chapter of algebraic geometry. You might have a look at this paper for instance to get some idea.
The minimal ("Baily-Borel-Satake") compactification of $\mathcal{A}g$ has boundary $\coprod{i<g} \mathcal{A}_i$ (with a certain topology). Maybe that's the sort of thing you want?
Thank you very much to the both of you! This is exactly what I wanted!
I unprotected this question because it seems the only reason it got protected was that it happened to be hit by two spam answers by one user, and because I don't see a reason that this question is more likely to receive further spam answers than just any other question.
|
2025-03-21T14:48:31.637999
| 2020-07-29T04:46:22 |
366839
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Min Wu",
"https://mathoverflow.net/users/123506",
"https://mathoverflow.net/users/970",
"j.p."
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631572",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366839"
}
|
Stack Exchange
|
Not sure whether I find a counterexample to poset fiber theorem
I am trying to implement Theorem 1.1 in the paper "Poset Fiber Theorems" by Bjorner, Wachs, and Welker. https://www.researchgate.net/publication/228782786_Poset_fiber_theorems
I am pretty excited to learn about the theorem, which is a generalization of the well-known Quillen fiber lemma and should be very useful for computation. But when I implement the theorem to the following example, the result seems inconsistent with my knowledge:
The posets are as shown in their Hasse diagrams and how the map is defined is: fixing $3,6,9,10$, mapping $11$ to a new minimum point $0$, and mapping other points downward by 1 level to the unique choice in $\{3,6,9,10\}$. From the construction, it is order-preserving and hence a poset map.
The conditions of Theorem 1.1 are satisfied as shown in the table on the right. Since $Q$ is the cone over the order complex $\Delta(\{3,6,9,10\})$, it is contractible. Thus, applying Theorem 1.1 in the paper, $\Delta(P)$, the order complex of $P$, is homotopic equivalent to a wedge of $\Delta(f^{-1}(Q_{\leq q}))\star\Delta(Q_{>q})$, where $\star$ denotes the join operation.
I am only interested in the homology group over fields, so I will use Betti numbers for the rest of the discussion. I also use Kunneth's formula for the join of simplicial complexes (see, e.g., the very beginning of https://arxiv.org/abs/math/0412552) in the case of field coefficients:
For $q = 0$, the reduced Betti numbers for $\Delta(f^{-1}(Q_{\leq q}))\star\Delta(Q_{>q})$ are all zeros.
For $q = 9$ or $10$, the reduced Betti numbers are $0,0,1$ (i.e. $H_k$ for $k = 0, 1, 2$).
For $q = 3$ or $6$, $f^{-1}(Q_{\leq q})$ is isomorphic to the face poset of a $2$-simplex. Hence, the reduced Betti numbers of $\Delta(f^{-1}(Q_{\leq q}))\star\Delta(Q_{>q})$ are all zeros.
Thus, by Theorem 1.1 in the paper, the reduced Betti numbers of $\Delta(P)$ are 0, 0, 2 (i.e. $H_k$ for $k = 0, 1, 2$) and $0$ for other dimensions.
However, it can be noted that $P$ is isomorphic to the face poset of the CW complex obtained by taking two copies of a $2$-simplex and identifying only their corresponding vertices. (You may visualize it as a pair of panties.) Thus, the reduced Betti numbers are 0, 2, 0 (i.e. $H_k$ for $k = 0, 1, 2$) and $0$ for other dimensions. I also verify the result by computer.
I have been checking logical bugs the whole day and cannot find a mistake. I am relatively new to these poset topology things and afraid that I have made some naive/stupid mistakes. Any comments, questions, suspicion about my computations, or pointing out my mistakes are welcome. Thank you.
The preimage of $Q_{\le 3}$ should contain everything NOT mapped to $6$. Aren't you missing $7$ and $8$? (same for $6$ instead of $3$)
I think you are right. LOL Thanks for pointing out. This saves me a lot of time.
I corrected my computation and it turns out the assumptions required by Theorem 1.1 in the paper are not satisfied. Many thanks to everyone, who spent time reading my long writing. Cheers.
|
2025-03-21T14:48:31.638237
| 2020-07-29T08:05:42 |
366842
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631573",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366842"
}
|
Stack Exchange
|
Extending a continuous complex valuation to a complex Borel measure
Let $X$ be a locally compact Polish space and let $\mathcal{O}(X)$ be the set of open subsets of $X$. A complex valuation on $X$ is a function $v: \mathcal{O}(X) \to \mathbb{C}$ such that $v(\varnothing)=0$ and $v(U \cup V)+v(U\cap V) = v(U)+v(V)$. Further, $v$ is called continuous if for a sequence $U_0 \subseteq U_1 \subseteq U_2 \subseteq \dots$ of open sets we have that $v(U_n)$ converges to $v(\bigcup_i U_i)$ for $n \to \infty$.
Can a continuous valuation be extended to a complex Borel measure on $X$?
For valuations with values in $[0,+\infty)$ that are monotonous, I believe this follows from Theorem 4.4 in Extension of valuations on locally compact sober spaces by Alvarez-Manilla.
In the special case $X = \mathbb{R}$, I believe the answer is yes. I'm still interested in more general cases as well.
The proof for $X = \mathbb{R}$ goes as follows. Comments, shorter proofs or references are welcome.
Suppose that $v : \mathcal{O}(X) \to \mathbb{C}$ is a continuous complex valuation. Then we can define its real part $\mathrm{Re}(v)$ and imaginary part $\mathrm{Im}(v)$ via $\mathrm{Re}(v)(U) = \mathrm{Re}(v(U))$ and $\mathrm{Im}(v)(U) = \mathrm{Im}(v(U))$. This gives two valuations with values in $\mathbb{R}$, and we can write $v = \mathrm{Re}(v) + i\, \mathrm{Im}(v)$. So if we want to show that $v$ can be extended to a complex Borel measure, we can assume without loss of generality that $v$ takes values in $\mathbb{R}$.
We now show that there are only countably many points that have a "point mass" associated to it. Let $v : \mathcal{O}(X) \to \mathbb{R}$ be a continuous valuation, and define $w(x) = v(\mathbb{R})-v(\mathbb{R}-\{x\})$ for every real number $x$. Further, we write $W = \{ x \in \mathbb{R} : w(x) \neq 0\}$. We show that $W$ is countable. If it was not, then there would be an $N \in \mathbb{N}$ such that $W’ = \{ x \in \mathbb{R} : w(x) > 1/N \}$ is still uncountable. Uncountable subsets of $\mathbb{R}$ have a limit point, so we take a limit point $\tilde{x} \in W'$ and a sequence $(x_n)_n$ in $W’$ such that $x_n$ converge to $\tilde{x}$ and such that $W’’ = \{ x_1, x_2, x_3, \dots \} \cup \{ \tilde{x} \}$ is homeomorphic to $\{0,1,1/2,1/3,\dots\} \cup \{ 0 \}$. Now set $U_0 = \mathbb{R} - W’’$ and $U_n = U_0 \cup \{x_1, \dots, x_n\}$ for each $n>0$. Then $$U_n \cap (\mathbb{R} - \{x_n\}) = U_{n-1}$$ and $U_n \cup (\mathbb{R}-\{x_n\}) = \mathbb{R}$, so we see that $v(U_{n-1}) + v(\mathbb{R}) = v(U_n) + v(\mathbb{R}-\{x_n\})$, and as a result $v(U_n) = v(U_{n-1}) + w(x_n) \geq v(U_{n-1}) + 1/N$. But because $v$ is continuous and $U_n$ is an increasing union of open subsets, $v(U_n)$ should converge, a contradiction. So $W$ is countable.
We now show that $v$ extends to a (signed) Borel measure on $\mathbb{R}$. Define the cumulative distribution function $F(t) = v(\mathbb{R})-v((t,+\infty))$. Continuity of $v$ shows that $F$ is right continuous. We want to construct a (signed) Borel measure that has $F$ as cumulative distribution function. Since $F$ is right continuous, a Borel measure like this exists if and only if $F$ is of bounded variation (this is explained well here). Suppose that $F$ is not of bounded variation. Then we can find $a_0 < a_1 < \dots < a_n$ such that $\sum_{i = 0}^{n-1} | F(a_{i+1}) - F(a_i) | > 1$. We claim that we can assume $a_0,\dots,a_n \notin W$, for $W$ as defined above. Then because $a_i \notin W$ for each $a_i$, we find that $F(a_{i+1}) - F(a_i) = v((a_i,a_{i+1}))$ for each $i$. To make sure that $a_i \notin W$, we can replace $a_i$ by $a_i + \epsilon_i \notin W$ for $\epsilon>0$, with $\epsilon$ small enough such that still $\sum_{i = 0}^{n-1} |v((a_i+\epsilon_i,a_{i+1}+\epsilon_{i+1}))| = \sum_{i = 0}^{n-1} | F(a_{i+1}+\epsilon_{i+1}) - F(a_i+\epsilon_i) | > 1$. This is possible because $W$ is countable and $F$ is right continuous. If we now consider the disjoint open intervals $I_1^{(0)},\dots,I_{n_0}^{(0)}$ given by $(-\infty, a_0+\epsilon_0)$, $(a_0+\epsilon_0, a_1+\epsilon_1)$, …, $(a_{n-1}+\epsilon_{n-1}, a_n + \epsilon_n)$, $(a_n+\epsilon_n,+\infty)$, then we still have $\sum_{j=1}^{n_0} |v(I_j)| > 1$.
Now $F$ again fails to be of unbounded variation over at least one of the intervals $I^{(0)}_j$. We choose precisely one such interval, and we subdivide this particular interval further in the way described above, such that we end up in total with a new set of open intervals $I_1^{(1)}, \dots, I_{n_1}^{(1)}$ that again partition $\mathbb{R}$ (up to finitely many points) and such that $\sum_{j =1}^{n_1} |v(I_j)| > 2$. By repeating this process, we make in step $k$ an open partition $I_1^{(k)}, \dots, I_{n_k}^{(k)}$ of $\mathbb{R}$ (up to finitely many points) with $A(k) = \sum_{j=1}^{n_k} |v(I_j^{(k)})| > k$.
After renumbering, we can assume that the interval in step $k$ that gets subdivided in step $k+1$, is precisely the interval $I_1^{(k)}$. Further, by choosing the intervals small enough in each step, we can assume that $v(I_1^{(k)})$ converges to a constant $C \in \mathbb{R}$ as $k$ goes to infinity. Now let $\Omega$ be the set of intervals that do not have a further subdivision, i.e. the intervals of the form $I_j^{(k)}$ for some $k$ and $j\geq 2$. Then $\sum_{I \in \Omega} v(I)$ converges unconditionally to $v(U)$ with $U = \bigsqcup_{I \in \Omega} I$. But then the series is also absolutely convergent by Riemann's Rearrangement Theorem, so $A = \sum_{I \in \Omega} |v(I)| < +\infty$. We now see that $k < A(k) = \sum_{j=1}^{n_k} |v(I_j^{(k)})| \leq A + |v(I_1^{(k)})|$. Because $|v(I_1^{(k)})|$ converges to $C$ for $k \to +\infty$, this gives a contradiction.
Because $F$ is right continuous and of bounded variation, there is a continuous Borel measure $\mu$ such that $F(t) = \mu((-\infty, t])$. It remains to show that $\mu$ restricts to $v$ on open subsets. First one can show that $\mu(\mathbb{R}) = v(\mathbb{R}) = \lim_{x\to+\infty} F(x)$. Then from the definition of $F$ it follows that $\mu((a,+\infty))=v((a,+\infty))$ for all $a \in \mathbb{R}$. Further, $$\begin{align*}\mu((-\infty,b)) &= \lim_{\epsilon \to 0} \mu((-\infty,b-\epsilon]) = \lim_{\epsilon \to 0} F(b-\epsilon) = \lim_{\epsilon \to 0} v(\mathbb{R}) - v((b-\epsilon,+\infty)) \\ &= \lim_{\epsilon \to 0} v((-\infty,b-\epsilon)) = v((-\infty,b))\end{align*},$$ where in the second last equality we use that we can take the limit over those $\epsilon > 0$ with $b-\epsilon \notin W$. Since $\mu$ and $v$ agree on the intervals $(a,+\infty)$ and $(-\infty,b)$, they agree on all open sets.
|
2025-03-21T14:48:31.638750
| 2020-07-29T10:08:17 |
366843
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"GH from MO",
"Gerry Myerson",
"Ofir Gorodetsky",
"Sylvain JULIEN",
"Terry Tao",
"Tom Copeland",
"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/12178",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/31469",
"https://mathoverflow.net/users/51189",
"https://mathoverflow.net/users/766",
"https://mathoverflow.net/users/84768",
"reuns",
"zeraoulia rafik"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631574",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366843"
}
|
Stack Exchange
|
Reference request for the explicit formula for $\sum_{n\leq x} \Lambda(n)n^{-s}$
Denote by $\Lambda(n)$ the von Mangoldt function, which is equal to $\log p$ if $p\geq 2$ is a prime, and $0$ otherwise. Let $\rho$ denote a complex zero of the Riemann $\zeta$-function. If I recall well, I once heard sometime ago that
$$\sum_{n\leq x} \Lambda(n)n^{-s} = -\frac{\zeta'}{\zeta}(s) + \frac{x^{1-s}}{1-s} - \sum_{|\mathrm{Im} (\rho)| \leq x} \frac{x^{\rho-s}}{\rho-s} + O(\log^{2}x)$$ for $s\neq 1, s\neq \rho$ and $s\neq -2k$ ($k\in \mathbb{N}$).
Does anyone have a reference for this result?
Riemann's habilitation memoir from 1859?
@Sylvain von Mangoldt's work came 30 years after Riemann's.
Have you checked (Hardy and Wright 1979, p. 340; Edwards 2001, p. 51) ?
As stated the asymptotic is only valid in the regime where $s$ is fixed and $x$ is sent to infinity (or to put it another way, the implied constant in $O(\log^2 x )$ will depend on $s$). At this level of non-uniformity the $-\frac{\zeta'}{\zeta}(s)$ term can be absorbed into the error. There are more uniform versions of this formula (see e.g., Exercise 27 of https://terrytao.wordpress.com/2014/12/09/254a-notes-2-complex-analytic-multiplicative-number-theory/ ) but the restriction $|\mathrm{Im} \rho| \leq x$ should be replaced by $|\mathrm{Im} (\rho-s)| \leq x$.
In my notes I referred to this sort of formula as a "truncated Landau explicit formula" but after looking through the 1911 and 1912 papers of Landau on the subject it does not appear that this formula appears in this exact form, though many related formulae of this type do.
@TerryTao: Thanks for clarifying this. In my response, I indicated the obvious approach, but I had no patience/time to work out the details (and I was honest about this).
@TerryTao The non-truncated version is indeed due to Landau. I needed this formula recently and a colleague (J. Merikoski) pointed out to me that it is used in H. L. Montgomery's original paper on pair correlation in number theory, "The Pair correlation of zeros of the zeta function" (Analytic Number Theory, Proc. Sympos. Pure Math. 24, St. Louis Univ. Missouri 1972, 181-193, 1973). There, a precise reference is given to a book of Landau: page 353 of "Handbuch der Lehre von der Verteilung der Primzahlen", Teubner, Berlin, 1909.
@OfirGorodetsky Thanks for clearing up the reference! I don't recall where I first found this formula attributed to Landau but it is good to actually have a concrete citation to back it up.
A brief historical overview of related work with standard refs, including those of Landau and Montgomery that OG points out, is "A computational history of prime numbers and Riemann zeros" by Pieter Moree, Izabela Petrykiewicz, and Alisa Sedunova (https://arxiv.org/abs/1810.05244).
I don't know a reference off-hand, but here is a sketch of the proof (the details need to be checked carefully, and I have not done it). One can start from
$$\sum_{n\leq x}\frac{\Lambda(n)}{n^s}=\frac{1}{2\pi i}\int_{(\sigma)}-\frac{\zeta'(z)}{\zeta(z)}\cdot\frac{x^{z-s}}{z-s}\,dz,\qquad\sigma>\max(1,\mathrm{Re}\,s),$$
which is a variant of Theorem 5.1 in Montgomery-Vaughan: Multiplicative number theory I, and can be proved in the same way. As in the theorem, the RHS is understood as a Cauchy principal value, while in case of $x\in\mathbb{N}$ the term corresponding to $n=x$ in the LHS is counted with weight $1/2$.
The integrand is meromorphic with simple poles at $z=s$, $z=1$, and $z=\rho$. The corresponding residues are $-\zeta'(s)/\zeta(s)$, $x^{1-s}/(1-s)$, and $-m_\rho\cdot x^{\rho-s}/(\rho-s)$, where $m_\rho$ is the multiplicity of $\rho$. So one can derive the OP's display by performing the following steps:
Truncate the integral on the RHS to $|\mathrm{Im}\,z|\leq x$ and estimate the error introduced. Perturb $x$ slightly if it is very close to some $\mathrm{Im}\,\rho$.
Extend the truncated contour (which is a vertical line segment) to a rectangle containing the points $s=1$ and $s=0$, hence all the $\rho$'s with $|\mathrm{Im}\,\rho|\leq x$. By the Residue Theorem, the integral weighted by $1/(2\pi i)$ equals the sum of corresponding residues listed above.
Estimate the contribution of the horizontal line segments of the rectangular contour, as well as of the vertical line segment to the left of $s=0$.
The LHS equals the sum of residues listed in item 2, up to the error terms listed in items 1 and 3.
P.S. See also Terry Tao's valuable remarks below the original post.
The non-truncated version of this estimate is due to E. Landau, see page 353 of his book "Handbuch der Lehre von der Verteilung der Primzahlen", Teubner, Berlin, 1909. It states that for $s \neq 1$ for which $\zeta(s) \neq 0$ one has
$${\sum_{n \le x}}'\frac{\Lambda(n)}{n^s} = \frac{x^{1-s}}{1-s}-\frac{\zeta'}{\zeta}(s)-\sum_{\rho}\frac{x^{\rho-s}}{\rho-s} + \sum_{k=1}^{\infty} \frac{x^{-2k-s}}{2k+s},\tag{$\star$}$$
where the $\prime$ in the sum indicates that the last term is counted with weight $1/2$ if $x$ is a positive integer. Here the sum is over non-trivial zeros of $\zeta$, as usual.
This is also given as Exercise 4 in Chapter 12, section 1 of "Multiplicative Number Theory I" by Montgomery--Vaughan.
One should also say that $(\star)$ holds for $s=1$ as well, if one interprets $x^{1-s}/(1-s) - \zeta'(s)/\zeta(s)$ as its limit at $s=1$, which is $\log x -\gamma$.
A truncated version has been used many times in the literature (e.g. Lemma 6 of the paper "On integers free of large prime factors" by A. Hildebrand and G. Tenenbaum, Trans. Am. Math. Soc. 296, 265-290, 1986) and is known to experts. Apart from the blog post Terry Tao mentioned in the comments, and the proof sketch given in GH from MO's answer, a similar proof can also be found in the appendix of the arXiv preprint 2211.08973. It states that for $x \ge 4$ and $T \ge 2 + |\Im s|$ we have
$${\sum_{n \le x}}'\frac{\Lambda(n)}{n^s} = \frac{x^{1-s}}{1-s}-\frac{\zeta'}{\zeta}(s)-\sum_{|\Im(\rho-s)|\le T}\frac{x^{\rho-s}}{\rho-s} + \sum_{k=1}^{\infty} \frac{x^{-2k-s}}{2k+s} + R(x,T)$$
if $\zeta(s)\neq 0$, where $R$ satisfies the bound
$$ R(x,T) \ll (\log x) x^{\prime -\Re s}\min\left\{ 1, \frac{x}{T\langle x \rangle}\right\} + \frac{\log^2 (xT)}{T}\left( 2^{|\Re s| } x^{1-\Re s}+\frac{2^{-\Re s}}{\log x}\right)$$
with an absolute implied constant, where $x'$ is the prime power closest to $x$ not equal to $x$. For $s=0$ this recovers the usual explicit formula (Theorem 12.5 from Montgomery--Vaughan), and its proof is the same. Taking $T \to \infty$ recovers $(\star)$. Here $\langle x \rangle$ is as in the statement of Theorem 12.5: it's the distance of $x$ to the nearest prime power not equal to $x$.
I'm curious why you use both $\rho+s$ and $\rho-s$ this formula; it would have been more aesthetic to me to just use $\rho-s$ (though in the regime you indicate where $\Im s$ is small compared to $T$, it makes little difference).
@TerryTao Oh, that's an embarrassing mathematical sign mistake on my part that I didn't catch until now, thanks for pointing it out. I'll update the sign in the next revision of the preprint. (Fortunately, as you say, this sign mistake is inconsequential -- in the considered range, the error incurred by replacing a sum over $|\Im(\rho+s)|<T$ with a sum over $|\Im(\rho-s)|<T$ can be absorbed in the bound for $R(x,T)$.)
Need to add $1_{x >1}$ in the explicit formulas. The inverse Mellin transform of $1/(z-\rho)$ on $\Re(z)=\sigma > \Re(\rho)$ is $1_{x >1} x^\rho$.
@reuns Note that I already assume $x \ge 4$. (This could be relaxed to $x>c$ for fixed $c>1$, as in the statement of Theorem 12.5 in Montgomery--Vaughan.)
|
2025-03-21T14:48:31.639272
| 2020-07-29T10:25:47 |
366846
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Brendan McKay",
"Florian Lehner",
"https://mathoverflow.net/users/100231",
"https://mathoverflow.net/users/9025",
"https://mathoverflow.net/users/97426",
"vidyarthi"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631575",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366846"
}
|
Stack Exchange
|
Decomposition of regular graphs
Let $G$ be a regular simple graph with degree $\Delta=n-k-1$ and order $m$. Let $C_k$ be the regular graph which is formed by removing a $k$-factor from the complete graph $K_{n}$. I think we could always find a proper induced subgraph of $C_k$ with maximum degree at least $\ge\frac{\Delta}{2}$ as a subgraph of the graph $G$. Is this true?
If this be true, then I think to find the invariants of $G$, it suffices to find the invariants of $C_k$. Then, the invariants of $G$ would be related linearly with that of $C_k$. For example, the chromatic number/ index seems to be closely related to the number of disjoint copies of $C_k$ which occurs as an induced graph and $m$. In a way, I think this could be related to the tree-decomposition of the graph $G$. Any light on this? Thanks beforehand.
There is more than one graph that can be obtained from $K_n$ by removing a $k$-factor (for $k = 2$, e.g. removing a Hamilton cycle vs removing two disjoint cycles covering all vertices gives different graphs). Neither of these graphs is a subgraph of the other.
@FlorianLehner thanks! so for regular graphs which contain certain copies of $C_k$ with a removal of a fixed $k$ factor, can we relate the graph invariant of $G$ with that of $C_k$?
I don't get it. $G$ and $C_k$ are both regular graphs of degree $\Delta$, so one can't be an induced subgraph of the other unless they are equal. I suspect you are not asking the question you intend to ask.
@BrendanMcKay thanks! edited the post. The comment by Florian answers my main question anyways.
Sorry, but removing "induced" doesn't help.
@BrendanMcKay again edited. I think it makes sense now.
It makes sense, but the answer is "no". The simplest example is if $\Delta=k$ then $C_k$ could be the complement of $G$ and so have no edges in common with $G$. If $\Delta$ is not far from $k$, it can still be true that $C_k$ and $G$ have too few common edges.
The answer is no.
Let $G$ be the Hoffman-Singleton graph (hence n=50, k=42). Let $C_k$ be the disjoint union of 5 $K_8$s and a $K_{10}-C$ ($K_{10}$ with a 10-cycle removed). Any proper induced subgraph of $C_k$ with maximum degree at least 4 will contain a $C_3$ or $C_4$, which $G$ does not contain.
|
2025-03-21T14:48:31.639464
| 2020-07-29T10:58:38 |
366847
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Matthew Daws",
"Michael Greinecker",
"https://mathoverflow.net/users/35357",
"https://mathoverflow.net/users/406"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631576",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366847"
}
|
Stack Exchange
|
Openness of the set of injective functions in $C(\mathbb{R})$?
Let $C(\mathbb{R})$ be equipped with the topology of compact convergence (or equivalently the compact-open topology). Then, is the subset $\left\{f\in C(\mathbb{R}):
\text{$f$ injective}
\right\}$ an open subset therein?
You can easily find a sequence of continuous functions $f_n$ each $[0,1]\rightarrow [0,1]$ with $f_n(0)=0, f_n(1)=1$, each non-injective, and with $f_n(x)\rightarrow x$ uniformly. Now extend each function to $\mathbb R$ by defining it as the identity outside $[0,1]$. This gives a sequence of non-injectives which converges uniformly on compacta to an injective. So the complement of your set is not closed.
A continuous function on the reals is injective iff it is either strictly decreasing or strictly increasing.
Based on Matthew's post here we go:
Let $f_n(x)\triangleq \left|\frac{x}{n+1}\cos(\frac{x-1}{n})\right| + \left(1-\frac1{n+1}\right) x$ and $f(x)=x$ and $\sup_{x \in [0,1] }\|f_n(x)-f(x)\| \in \mathscr{O}(n^{-1})$.
This provides a counter example on $C((0,1))$ and then just use the homeomorphism:
$$
\begin{aligned}
C(0,1) &\rightarrow C(\mathbb{R})\\
f &\mapsto f \circ \frac{1}{1+ \exp(-x)},
\end{aligned}
$$
to get the conclusion. (Note, that is preserves the class of injective maps).
A simpler example is $f_\epsilon(x)=x^3+\epsilon x$ which is injective iff $\epsilon\geq 0$.
I think that this is a kind of question (which is not research level and) which should not be answered by a counterexample but by an explanation: What would it mean that, e.g., the identity map is an interior point of the set of injective continuous functions? There should be a compact interval $[-n,n]$ and $\varepsilon>0$ such that any function $f\in C(\mathbb R)$ with $|f(x)-x|<\varepsilon$ for $x\in [-n,n]$ is injective. This is absurd because you have no condition at all outside $[-n,n]$. But also inside the interval one can easily modify the graph of the identity to get a non-injective function which $\varepsilon$-close to the identity.
|
2025-03-21T14:48:31.639628
| 2020-07-29T11:03:52 |
366848
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"HenrikRüping",
"https://mathoverflow.net/users/3969"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631577",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366848"
}
|
Stack Exchange
|
Rank of sum of two matrices
Given matrices $A, B \in \Bbb R^{3 \times 3}$ whose ranks satisfy $\mbox{rank} (A), \mbox{rank} (B) \geq 2$, I would like to prove that for large (or small) enough scalar $\alpha \in \mathbb{R} \setminus \{0\}$ the following does hold.
$$\mbox{rank} (A+\alpha B) \geq 2$$
This seems to be true by hand waving argument, but I would like to find some short proof or reference that formally proves it. Thanks.
Since rank$(A)\geq 2$, the matrix $A$ has a $2\times 2$ submatrix $a$ with nonzero determinant; the determinant is a continuous function of the matrix elements, so adding a sufficiently small perturbation $\alpha B$ to $A$ will leave $\det a\neq 0$ and hence the rank of $A+\alpha B$ remains $\geq 2$.
If instead of small $\alpha$ you wish to take large $\alpha$, define $B'=\alpha B$, $A'=\alpha A$ and work with $B'+(1/\alpha)A'$, where $B'$ has rank $\geq 2$.
By the way, this argument shows that the rank function is lower-semicontinuous.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.