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2025-03-21T14:48:31.976452
| 2020-09-07T08:55:19 |
371072
|
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"Dieter Kadelka",
"Erel Segal-Halevi",
"Fedor Petrov",
"Iosif Pinelis",
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|
Stack Exchange
|
Can you solve this problem using a finite number of queries?
Let $g:[0,1]\to[0,1]$ be a continuous monotonically-increasing function. You can access $g$ using queries of two kinds:
Given $x\in[0,1]$, return $g(x)$.
Given $y\in[0,1]$, return $g^{-1}(y)$.
Given fixed parameters $s,t\in (0,1)$, can you find, using finitely many queries, a point $x$ for which
$$
g(x+s) - g(x) < t
$$
(if such $x$ exists)?
Example: if $g$ is the function below, $s=0.3$ and $t>0.1$, then $x=0.4$ is a solution as $g(x+s)-g(x)=0.1$. If $t\leq 0.1$ then there is no solution.
I guess that you would not like to add continuity as an assumption?
@JochenWengenroth good question. I added a continuity assumption, but not sure it matters.
if it is continuous, then such $x$ form an open set, how can it be a single point?
For continuous $g$ I believe that a quite simple ''bisection algorithm'' does the job.
I think that $s,t \in (0,1]$? Does $s = 0$ or $t = 0$ make any sense?
@FedorPetrov you are right, I had in mind a different question. I removed the incorrect claim.
The question is stated informally, using the terms "queries" and "access".
Here is how I formally interpret it:
Take any $s$ and $t$ in $(0,1)$. Let $G_{s,t}$ be the set of all continuous strictly increasing functions $g\colon[0,1]\to[0,1]$ such that the set
$$E:=E_{s,t}(g):=\{x\in[0,1-s]\colon g(x+s)-g(x)<t\}$$
is nonempty. Do there exist sequences $(x_j)_{j=1}^\infty$ and $(y_j)_{j=1}^\infty$ in $[0,1]$ such that for any $g\in G_{s,t}$ there is a natural $n$ such that the following implication holds: If a function $h\colon[0,1]\to[0,1]$ is continuous and strictly increasing and for all $j\in[n]:=\{1,\dots,n\}$ we have $h(x_j)=g(x_j)$ and $h^{-1}(y_j)=g^{-1}(y_j)$, then ($h\in G_{s,t}$ and) for some $k\in[n]$ we have $x_k\in E_{s,t}(h)$?
Then the answer is yes.
Indeed, informally, let the $j$th query give us the values $g(q_j)$ and $g(q_j+s)$, where $(q_j)_{j=1}^\infty$ is an enumeration of the set of all rational numbers in the interval $[0,1-s]$. Take any $g\in G_{s,t}$, so that $E_{s,t}(g)\ne\emptyset$. Since $g$ is continuous, the set $E_{s,t}(g)\subseteq[0,1-s]$ is open in $[0,1-s]$, and hence $q_n\in E_{s,t}(g)$ for some natural $n$. So, we will find the point $q_n\in E_{s,t}(g)$ on our $n$th query.
Formally, let $(y_j)_{j=1}^\infty$ be any sequence in $[0,1]$ (it will be of no use to us). For each natural $j$, let $x_{2j-1}:=q_j$ and $x_{2j}:=q_j+s$.
Take any $g\in G_{s,t}$. Then, as noted above, $q_n\in E_{s,t}(g)$ for some natural $n$.
Now, if a function $h\colon[0,1]\to[0,1]$ is continuous and strictly increasing and for all $j\in[2n]$ we have $h(x_j)=g(x_j)$ and $h^{-1}(y_j)=g^{-1}(y_j)$, then for $k=2n-1(\in[2n])$ we have $x_k=x_{2n-1}=q_n\in[0,1-s]$ and $h(x_k+s)-h(x_k)=h(x_{2n})-h(x_{2n-1})=g(x_{2n})-g(x_{2n-1})=g(q_n+s)-g(q_n)<t$ (because $q_n\in E_{s,t}(g)$), so that $x_k\in E_{s,t}(h)$ (and $h\in G_{s,t}$). Thus, the implication in question holds.
The monotonicity condition on $g$ or the knowledge of values of $g^{-1}$ was not actually needed in this proof.
Thanks! But this solution assumes that such a point $x$ exists. What if we do not make this assumption, and need to also decide whether or not it exists?
@ErelSegal-Halevi : You did say "if such x exists". The bulk of the work here was to formalize the problem. Anyway, I think that without the existence assumption and with a reasonable formal interpretation, the answer will change to "no". However, I'd suggest that you post, separately, a quite formally presented question without the existence assumption, avoiding such terms as "query", "access", etc. or perhaps using them only for an accompanying informal version of the question.
@ErelSegal-Halevi : A negative answer to the question without the existence assumption is now posted at https://mathoverflow.net/questions/371123/detecting-slow-growth-in-a-finite-number-of-queries/371124#371124
Wonderful, thanks so much
Iosif Pinelis proved that, when a solution is guaranteed to exist, it can be found using finitely many queries.
When a solution is not guaranteed to exist, then it may be impossible to decide whether or not it exists with finitely many queries. I could prove it for the special case $t = s$. Suppose that, after some $n$ queries, for every $j\in [n]$, the answer for query $x_j$ is $g(x_j)=x_j$ and the answer for query $y_j$ is $g^{-1}(y_j)=y_j$. Then, it is possible that $g(x)\equiv x$, in which case no solution exists. However, it is also possible that $g(x)$ is slightly different than $x$ in some open interval that does not contain any $x_j$ or $y_j$. In this case a solution exists.
When $t<s$ and a solution is not guaranteed to exist, Iosif Pinelis proved that the problem may not be decided using a finite number of non-adaptive queries (queries that must be determined in advance, and may not depend on answers to previous queries). The idea is that, for every finite number $n$ of queries, there is a piecewise-linear function $g$ for which no solution exists, and a slight modification of it - that does not change the answer to any of the $n$ queries - yields a function $h$ for which a solution exists.
A remaining open case is that of adaptive queries, in which each query may depend on answers to previous queries. When $t<s$, I do not know if it is possible to decide with finitely-many queries.
|
2025-03-21T14:48:31.976783
| 2020-09-07T10:11:38 |
371075
|
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"Enrico",
"Federico Fallucca",
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|
Stack Exchange
|
Minimal model vs canonical model of a surface
When I have a projective surface $X$, for simplicity smooth, I can find a simpler smooth surface on its binational class. In this way we find in a finite number of steps the simplest surface $Y$, i.e. a smooth surface birational equivalent to $X$ with no exceptional curves in the negative part of the Mori cone with respect the canonical divisor $K_Y$. This surface $Y$ is a ruled surface or its canonical divisor $K_Y$ is nef.
In the last case we define $Y$ as the minimal model of $X$ and one can prove this is the unique smooth surface on the birational class of $X$ with canonic divisor nef, up to isomorphism of surfaces.
By other side one can consider the canonical model of $X$, i.e
$$\operatorname{Proj}( \oplus_{n}H^0(X, K_X^n)) $$
and we define the canonical dimension, or Kodaira dimension of $X$ as the dimension of its canonical model.
If I understood in the right way, the canonical dimension of $X$ is always greater or equal than the canonical dimension. When the two dimensions are the same, i.e. the canonical dimension is equal to $2$, then $X$ is called surface of general type, right?
In this case I have two new surfaces, the minimal model $Y$ of $X$ and the canonical model. Is there a relationship between them?
Yes. The canonical model, say $\mathscr{X}$, is the same for all surfaces in the birational class of $X$, in particular for $Y$. Therefore there is a morphism $Y\rightarrow \mathscr{X}$; it is birational and contracts the $(-2)$-curves.
But $Y$ is the minimal model and $\mathscr{X}$ is smooth, so this would be $Y\cong \mathscr{X}$, right?
No, $\mathscr{X}$ is not smooth. It has rational singularities, which are resolved in $Y$.
And what is the reasoning for which one would consider a singular surface rather than the beautiful minimal smooth model $Y$?
Several possible reasons:
Rational singularities are not at all "bad" (by at least a great number of definitions of "bad")
The canonical ring is (at least sometimes) easier to compute. Useful if you want to have an explicit model for your surface.
Beauty is subjective :)
(and so on...)
Thank you very much, now I have clearer ideas
|
2025-03-21T14:48:31.976964
| 2020-09-07T10:21:16 |
371077
|
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"authors": [
"Alexandre Eremenko",
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|
Stack Exchange
|
On the upper bound for the real part of the logarithmic derivative of a polynomial
Let $P(z)=a_0+a_1z+\cdots+a_nz^n, a_n\neq 0$ with $\max_{|z|=1}|P(z)|=1.$ May it be true that for all $z$ on $|z|=1,$ for which $P(z)\neq 0,$
$$\Re{\frac{zP'(z)}{P(z)}}\leq n-\frac{1+|a_0|-|a_n|}{1+|a_0|+|a_n|}?$$ or else any counter example?
No. If this were true for all polynomials with $P(z)\neq 0$, then the same would be true for all polynomials, by continuity. But it is evidently wrong for some polynomials with $P(z)=0$, for example for $P(z)=z-1$, and $z=1$.
@AlexandreEremenko $z-1$ does not satisfy the last assumption in the first line of the post, does it?
@AlexandreEremenko I did. The real part is exactly at $1/2$ (the circle goes exactly to the critical line).
@fedja That is right.
@fedja,Any polynomial P(z) having all its zeros outside $|z|<1$ satisfies $\Re(zP'(z)/P(z))\leq n/2$ for all z on $|z|=1$ for which $P(z)\neq 0.$
Polynomial $P(z)=(2/3)z-1/3$ satisfies all conditions. Your inequality is violated for $z=1$.
|
2025-03-21T14:48:31.977065
| 2020-09-07T10:34:22 |
371078
|
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"Doriano Brogioli",
"Robert Bryant",
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|
Stack Exchange
|
Relation between separation of variables and Hessian properties
I have a function $f(x,y)$, where both $x$ and $y$ are $n$-dimensional vectors, $n\ge 2$. I know that this function
has the following property:
$$
\frac{\partial}{\partial x_j} \frac{\partial}{\partial y_k} f =
a_j(x,y) b_k(x,y)
$$
This can be expressed saying that the $n\times n$ block of the Hessian, out of diagonal, is a diadyc matrix. Examples of such a $f$ are:
$$
f(x,y) = A(x)B(y) + C(x) + D(y)
$$
Actually, this example is similar to the form used in the "separation of variables" method for solving partial differential equations.
The second equation implies the first. My question is: is it also true that the first equation implies the second? Or a counter-example can be found?
The answer is 'no, the first equation does not imply the second when $n=2$'.
The reason is that when $n=2$, the equation is essentially equivalent to requiring that the off-diagonal $n$-by-$n$ block of the Hessian of $f$ have determinant equal to zero.
This is one (non-linear) second-order equation for $f$ as a function on $\mathbb{R}^4$, and it is easy to show that the local solutions depend on 3 functions of 4 variables in Cartan's sense. (Basically, one can prescribe $f$ and its first normal derivative along a generic hypersurface in $\mathbb{R}^4$ subject to some generic conditions.)
However, the second form depends only on 4 functions of 2 variables, so not every local solution can be put in the second form, even locally.
In higher dimensions, the condition is equivalent to requiring that that $n$-by-$n$ matrix have rank at most equal to $1$. This is $(n{-}1)^2$ second order equations for $f$, so it's overdetermined when $n>2$. One would need to do the Cartan-Kähler analysis of this system to determine the generality of its space of solutions, but, probably, the answer is still 'no'.
Could you please write an explicit example of $f$ which meets the first equation but not the second, for the 2D case?
@DorianoBrogioli: Yes, for example, let $f = x_1^2y_1+x_1y_1^2$.
It seems that this very simple example works for every $n$.
@DorianoBrogioli: Yes, that's true. In fact, any $f(x_1,y_1)$ satisfies the equation, but the generic one is not of the desired form.
|
2025-03-21T14:48:31.977237
| 2020-09-07T11:13:16 |
371081
|
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"Peter Humphries",
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|
Stack Exchange
|
Decomposition of the Schwartz space as a representation for the orthogonal group
The Schwartz space $\mathcal{S}(\mathbb{R}^n)$ is naturally a $O_n(\mathbb{R})$-representation. I'm assuming that this is a relatively well-behaved representation among the infinite-dimensional ones and that someone has already thought through the decomposition of this representation into irreducibles in a suitable, infinite-dimensional sense.
My naive conjecture would be that $\mathcal{S}(\mathbb{R}^n)$ is a specific completion of the sum over most/all irreducibles of $O_n$; probably a subspace like $\mathbb{C}[x_1,\ldots,x_n]\cdot e^{-\|x\|^2}$ gets completed and probably the completion step works something like the rapidly decreasing sequences space $\mathcal{s}(\mathbb{N}^n)$.
Is this true? Can you point me to a citable resource for this?
Additionally I'm looking for a description of how the decomposition into irreducibles plays with the Fourier transform and the multiplication on $\mathcal{S}(\mathbb{R}^n)$?
Well, the decomposition of $L^2(S^{n - 1})$ as a $\mathrm{O}(n)$-module is simply the theory of spherical harmonics. You can extend these radially via a Gaussian. With regards to your last question, see Hecke's identity for homogeneous harmonic polynomials; it's somewhere in Stein and Weiss.
|
2025-03-21T14:48:31.977348
| 2020-09-07T11:17:59 |
371082
|
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|
Stack Exchange
|
Perimeter decreases under intersection with half spaces
The main thing i need to prove is the following assertion:
Let $E\subset R^N$ be a set of finite perimeter and $H=\{x\in R^N : x\cdot e < t \}$ for $t\in R$ and $e\in S^{N-1}$.
Then prove that $$ \text{Per}(E\cap H)\leq\text{Per}(E), $$
by using the following decomposition of the Gauss Green measure of $E\cap H$
$$ \mu_{E\cap H}= (\mu_{E})_{|_{H}} + e H^{N-1}_{|_{E\cap\partial H}}. $$
For the convenience of the reader i recall that E is of locally finite perimeter if there exists a vector-valued Radon measure $\mu_E$ such that $\int_{E} \text{div}Tdx=\int_{R^N} T\cdot d\mu_E$ for every smooth compactly supported vector field $T$, in this case $\text{Per}(E;A)=|\mu_E|(A)$.
The hint of the book is the following: Use the recalled decomposition with the vector field $T(x)=e$ for each $x$ (clearly a truncation argument is required).
My attempt is like the following, i consider $T_M(x)= \phi(|x|)e$ ($M>0$) where $\phi\in C^{\infty}(R)$ is such that $\phi=1$ on $(-\infty, M]$ and $\phi=0$ on $[M+1,\infty)$. Using this vector field in the decomposition and passing to the limit for $M\to\infty$ i can get easily
$$H^{N-1}(E\cap \partial H)\leq \text{Per}(E;H). $$
I can't fine out how to prove the derired inequality
|
2025-03-21T14:48:31.977466
| 2020-09-07T11:43:27 |
371083
|
{
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"Jérémy Blanc",
"Pop",
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|
Stack Exchange
|
(stable)-base locus on fibres
Let $X\to Y$ be a morphism between projective varieties, with general fibre being smooth and $Y$ being a smooth curve. Let $D$ be a divisor on $X$. Is is true that for a general fibre $F$ and $m\ge 1$ big enough, the restriction of the base-locus of $mD$ to $F$ is equal to the base-locus of $m\cdot D|_F$ ? Or equivalently, is the restriction of the (stable) base-locus equal to the (stable) base-locus of the restriction.
PS: If the answer is no, is it OK when assume that $D$ is big?
First, there is a small inclarity in the question with the meaning of "restriction of the base-locus". Since the base locus is in general just a closed set, of possibly arbitrary codimension, it is not clear to me whether "restriction" is just meant to mean set-theoretic intersection, or something more sophisticated.
In any case, even in the simplest situation where the base locus is a codimension 1, the answer to both questions is no.
For the first version, let $X$ be $\mathbf P^2$ blown up in 9 points which are the intersection of two smooth cubics. Let $X \rightarrow \mathbf P^1$ be the elliptic fibration, and let $D$ be one of the exceptional divisors of $X \rightarrow \mathbf P^2$. Then $mD$ is fixed for all $m \geq 0$, but for a smooth fibre $F$ the restriction $D_{|F}$ is a point on an elliptic curve, so $2D_{|F}$ is basepoint free.
For the second version, fix an ample divisor $A$ on $X$ as above. For $D$ as before and for any $n \geq 1$, the divisor $A+nD$ is big. For $n$ sufficiently large, we have $(A + nD) \cdot D = A \cdot D -n <0$, so for such $n$ and sufficiently large $m$, the base locus of $m(A+nD)$ is again $D$. On the other hand, for any $n \geq 1$ the restriction of $A+nD$ to $F$ is basepoint free.
Update: Jérémy asks further for an example where the stable base locus of $D$ is codimension 2. This is a little trickier, but I think one can cook it up starting from the example constructed by Totaro in Theorem 6.1 of the paper:
Totaro, Burt, Hilbert’s 14th problem over finite fields and a conjecture on the cone of curves, Compos. Math. 144, No. 5, 1176-1198 (2008). ZBL1157.13006.
In this example we have a fibration $W \rightarrow \mathbf P^3$ whose general fibre $F$ is an abelian surface. The key point is that there is a pseudo-isomorphism $X \dashrightarrow \mathbf W$ given by the inverse flip of 45 disjoint curves on another variety $X$, which is the blowup of $\mathbf P^5$ in a certain arrangement of 9 points.
So start with a very ample divisor $H$ on $X$ say. Then its proper transform $\widetilde{H}$ on $W$ will be movable, with stable base locus $Bs(\widetilde{H})$ consisting of exactly the indeterminacy locus of $W \dashrightarrow X$. This is a union of 45 copies of $\mathbf P^3$ on $W$, each of them a section of $W \rightarrow \mathbf P^3$. So the intersection of $Bs(\widetilde{H})$ with a general fibre $F$ consists of 45 points on $F$.
On the the other hand the restriction $\widetilde{H}_{|F}$ is an effective divisor on the abelian surface $F$, so it is semi-ample.
Now I missed one condition of the original question, namely that the base $Y$ should be a curve. So take $Y \subset \mathbf P^3$ to be a general line, and restrict the fibration $W \rightarrow \mathbf P^3$ to $Y$, to get a fibration $V \rightarrow Y$ say. Then taking the divisor $\widetilde{H}_{|V}$ the same argument as before will work.
Addendum: The examples above are (I think) correct, but I am kind of embarrasssed by how unnecessarily complicated they are. My compulsion to put the blowup of $\mathbf P^2$ in 9 points in every answer I write led me to overlook much simpler examples that fit the bill just as well. So let me add those in now. (And then I will leave this answer alone.)
For the first version, now let $X$ just be $\mathbf P^2$ blown up in a single point, and $f:X \rightarrow \mathbf P^1$ the projective bundle map. Again let $D$ be the exceptional divisor of $X \rightarrow \mathbf P^2$. Then everything in the previous answer works the same way. In this case the fibre $F$ is $\mathbf P^1$, so $D_{|F}$ is already basepoint-free, even very ample.
For the second version, where the stable base locus of $D$ is meant to have codimension 2, we can take $X$ to be the projective bundle
$$X = \mathbf P_{\mathbf P^1} \left( O \oplus O(-1) \oplus O(-1) \right) $$
and our morphism is again the bundle map $X \rightarrow \mathbf P^1$.
The point is that $X$ is a compactification of the total space of the bundle $O(-1)\oplus O(-1)$ on $\mathbf P^1$. So it contains a curve $C$ which is a copy of $\mathbf P^1$ with normal bundle $O(-1) \oplus O(-1)$. I claim this curve can be flopped to give a pseudoisomorphism $X \dashrightarrow W$, and taking a very ample divisor on $W$ and its proper transform on $X$ we again get a divisor class $D$ on $X$ whose base locus is precisely $C$, but whose restriction to any fibre $F \cong \mathbf P^2$ is basepoint-free, even very ample.
These examples have the virtue of being much simpler than the first set, but also showing that this behaviour is probably very common --- you don't need special conditions like having a fibration with fibres which are abelian varieties, for example.
Thanks for the nice answer. Do you have also examples where the base-locus consists of codimension 2 subsets (no fixed component) ?
@JérémyBlanc: see the update. There might be simpler examples.
Thanks a lot. This is quite nice!
@JérémyBlanc: I added a further (final!) update with some simpler examples.
|
2025-03-21T14:48:31.977834
| 2020-09-07T11:56:53 |
371085
|
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|
Stack Exchange
|
Isomorphism of algebras depending on the field
Let $X$ be a finite set of primes (where 0 is also a prime, but 1 is not!).
Are there two quiver algebra $A_1=KQ/I_1$ and $A_2=KQ/I_2$ such that the coefficeints of the generators of the ideals $I_1$ and $I_2$ contain only field elements 1 and -1 and are admissible for any field $K$ and we have $KQ/I_1 \cong KQ/I_2$ iff $K$ has characteristic $p \notin X$?
I am only aware of such an example for $X= \{2 \}$, namely the penny farthing algebras. Are there any other examples?
Is it true that in case $A_1$ and $A_2$ are representation-finite for any $K$, such a thing can only happen for $X= \{ 2\}$?
Here is the example of the penny-farthing algebras:
Let $Q$ be the quiver with 2 vertices 1 and 2 and arrows $a$ from 2 to 1 and $b$ from 1 to 2 and $y$ from 2 to 2. Let $I_1=\langle ab-y^2,ba\rangle$ and $I_2=\langle ab-y^2,y^4,ba-bya\rangle$.
Question: What is the best way to show that $KQ/I_2$ is isomorphic to $KQ/I_2$ iff $K$ has characteristic different from 2?
Of course it is possible with brute froce, but maybe there is an elementary nice way (I am not aware of an elementary invariant to distinguish those two algebras in characteristic 2).
This is exercise 8 in chapter III the Representation theory of Artin algebras book by Auslander, Reiten and Smalo.
Something's wrong here. The only ideal containing $\pm 1$ is $A$ itself. Do you mean "$I_i$ is generated by polynomials whose coefficients are either $\pm 1$ or $0$" or something along those lines?
@JohannesHahn Yes, thanks.
|
2025-03-21T14:48:31.978092
| 2020-09-07T12:46:31 |
371091
|
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"Nulhomologous",
"Will Sawin",
"abx",
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"url": "https://mathoverflow.net/questions/371091"
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|
Stack Exchange
|
Does the genus of a variety depend on the base field?
"Does the (arithmetic) genus of a variety depend on the base field?"
So the question arises from a definition of the elliptic curve, the Hartshorne's book just says that the elliptic curve is a curve of genus 1. (316p) Indeed, this definition is probably for the elliptic curve over C. Then I found the wikipedia says so for general fields. (https://en.wikipedia.org/wiki/Elliptic_curve)
Formally, an elliptic curve is a smooth, projective, algebraic curve of genus one
For this definition to make sense, the (arithmetic) genus should be invariant over base fields since we call $y^2z=x^3+axz^2+bz^3$ as an 'elliptic curve', no matter what the base field is. Then I got to wonder if the genus is invariant over base field in general.
What definition of genus are you using? This is not hard to see for any of them, but the exact proof varies depending on the definition.
@WillSawin I'm talking about the arithmetic genus (https://en.wikipedia.org/wiki/Arithmetic_genus). I found the definition at the problem 5.3 (230p) of the Algebraic Geometry(Hartshorne).
Use Proposition III.9.3 in Hartshorne: "cohomology commutes with flat base extension".
On the other hand, if you use "genus" for the relative genus, so the number it appears in the Riemann-Roch formula $\ell(D)=deg(D)+1-g$ for divisors $D$ defined over the base field, with degree suffiently large and $\ell(D)$ the dimension as $K$-vector spaces, of the space of functions on the curve, defined over $K$ and with polar divisor bounded by $D$, then this genus can change under finite inseparable extensions.
I have a feeling that the arithmetic genus does depend on the base field. For instance if you look over $\overline{\mathbb{F}_3}(t)$ the curve given by $y^2=x^3+t$ should have genus one. However, after inseparable base change such as $\overline{\mathbb{F}_3}(s)$ with $t=s^3$ it becomes genus zero: $y^2=x^3+s^3=(x+s)^3$. This agrees with @Nulhomologous comment.
The curve has already genus zero (and is not smooth) over $\overline{\mathbb{F}_3}(t)$.
@abx I agree that the curve is not smooth, but "Riemann-Roch Theory" should work for normal curves, the point being that the two notions are not the same for curves defined over an imperfect field such as $K=\overline{\mathbb{F}_3}(t)$. So $C/K$ is normal with genus one but "$C\otimes K^{1/3}$" is genus zero. This is an example of a Non-Conservative curve. The terminology is due to John Tate, as far as I know.
I will try to answer the question of the title (so, if genus changes under base extension), but just for curves.
Answer 1: If the curve is smooth, projective and geometrically irreducible over a field, the genus does not change under a base extension.
For a reference, you can see Lemma 53.8.2. plus Lemma 33.26.2., which shows that such curves verify the hypothesis of the lemma 53.8.2.
Since this is what you asked, I could stop here. But I want to explain why there are a lot of results about genus changing under base extension. And the explanation is:
Answer 2: If the curve is regular and projective over a nonperfect field, then the genus can change under an inseparable extension. Such curves that change the genus are called "non-conservative".
Notice that over a nonperfect ground field, one needs to distinguish between regular (all the local rings are regular) and smooth (Jacobian condition or, equivalently, the base change to the algebraic closure is regular).
Why does one care about regular but non-smooth curves? Because of the famous equivalence between "extension $K/k$ of transcendence degree 1" and "regular projective curves over $k$", or, if you like, because normal projective curves are regular, but not necessarily smooth (see the stacks project: 53.2 Curves and function fields).
The curve given by @amateur in the comments, so the projective regular curve determined by the equation $y^2=x^3+t$ over $\mathbb{F}_3(t)$, is an example: it is regular, but not smooth. It has arithmetic genus 1, but when base changed to $\mathbb{F}_3(t^{\frac 13})$ (and desingularized) it has arithmetic genus 0. So, using the definition "an elliptic curve is a smooth, projective, algebraic curve of genus one with a fixed point", such a curve is not an elliptic curve, although it has a nice looking Weierstrass equation.
|
2025-03-21T14:48:31.978393
| 2020-09-07T12:48:21 |
371092
|
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"url": "https://mathoverflow.net/questions/371092"
}
|
Stack Exchange
|
Intuition for conformal nets
I was planning on reading the work of Arthur Bartels, Christopher L. Douglas and André Henriques on the 3-category of conformal nets as discussed in these papers: Coordinate-free nets, Conformal blocks, Fusion of defects, The 3-category and Dualizability.
I wanted to know how the structure of a conformal net arises from the physical notion of a conformal field theory.
What physical notion of CFT is your starting point?
So I'm familiar with the physicists approach and the definition as cobordisms of 1 dimensional complex manifolds (Segal's definition). Either of these would be good starting points for me.
I'm very very far from an expert on the subject but this might be helpful
You might find André Henriques' lecture notes https://arxiv.org/abs/1304.7328 useful
|
2025-03-21T14:48:31.978492
| 2020-09-07T13:20:00 |
371093
|
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|
Stack Exchange
|
Function series of normal lower semi-continuous functions
For a real-valued $f$ on a topological space $X$, the upper limit of $f$ at $x\in X$ is
defined as follows:
$
f^{\ast }\left( x\right) =\inf \left\{ \sup \left\{ f\left( y\right) :y\in
U\right\} :U\in \mathcal{N}\left( x\right) \right\}
$,
where $\mathcal{N}\left( x\right) $ is the neighborhood system at $x$. The
lower limit of $f$ at $x$ is defined dually and denoted by $f_{\ast }$. The $%
f^{\ast }$ and $f_{\ast }$ are extended real-valued functions on $X$ are
respectively upper semi-continuous and lower semi-continuous.
A real-valued function is called normal lower semi-continuous if $\left(
f^{\ast }\right) _{\ast }=f$ at each point of $X$.
We know the next theorem
Theorem: An lower semi-contionuous function $f$ on $X$ is normal iff for
each real number $\lambda $, $\left\{ x\in X:f\left( x\right) <\lambda
\right\} $ is a union of regular closed sets. (A set equals the closure of
its interior is called regular closed)
It well known that the characteristic function of a set is lower
semi-continuous if and only if the set is open.
Suppose that there exists a decreasing sequence $\left\{
U_{n}\right\} $ of open sets. Define
$f_{n}\left( x\right) =\left\{
\begin{array}{c}
1, \\
0,
\end{array}%
\right.
\begin{array}{c}
x\in X\backslash cl\left( U_{n}\right) \\
x\in cl\left( U_{n}\right)
\end{array}%
$
Then each $f_{n}$ is normal lower semi-continuous function. Set $f\left(
x\right) =\sum\limits_{n\geq 1}2^{-n}f_{n}\left( x\right) $. Then $f$ is
also normal semi-contionus function.
Being semi-continuous function of $f$ is clear. But I can't prove to be
normal of $f$.
It is clear that $\left\{ x\in X:f\left( x\right) >\lambda \right\}
=\bigcup\limits_{n\geq 1}\left\{ x\in X:g_{n}\left( x\right) >\lambda
\right\} $, where $g_{n}\left( x\right) =2^{-1}f_{1}\left( x\right)
+2^{-2}f_{2}\left( x\right) +...+2^{-n}f_{n}\left( x\right) $.
Moreover, we write $$\left\{ x\in X:f\left( x\right) <\lambda \right\} =\bigcup _{\lambda ^{\prime
}<\lambda }\left\{ x\in X:f\left( x\right) \leq \lambda ^{\prime }\right\} $$
Since $$\left\{ x\in X:f\left( x\right) \leq \lambda ^{\prime }\right\}
=\bigcap _{n\geq 1}\left\{ x\in X:g_{n}\left( x\right) \leq \lambda ^{\prime
}\right\} $$, we have
$$
\left\{ x\in X:f\left( x\right) <\lambda \right\} =\bigcup _{\lambda ^{\prime
}<\lambda }\bigcap _{n\geq 1}\left\{ x\in X:g_{n}\left( x\right) \leq \lambda
^{\prime }\right\}
$$
Therefore, I need to show that $\bigcap _{n\geq 1}\left\{ x\in X:g_{n}\left(
x\right) \leq \lambda ^{\prime }\right\} $ is regular closed set or a union
of regular closed sets.
How can I prove this?
Since $\left\{ x\in X:f\left( x\right) <\lambda \right\} =\left\{
\begin{array}{cl}
\emptyset, & \lambda\le 0, \\
X, & \lambda>1 \\
\overline{U}_k, & \frac 1{2^k}<\lambda\le \frac1{2^{k-1}}
\end{array}
\right.
$
we have $f$ is a normal function.
|
2025-03-21T14:48:31.978664
| 2020-09-07T14:08:47 |
371099
|
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"Andreas Blass",
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"Sergey Grigoryants",
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|
Stack Exchange
|
Unbounded $\omega_1$-sequence in $^*\mathbb{N}$
Let $\mathcal{F}$ be a non-principal ultrafilter on $\omega$. Let $^*\mathbb{N}$ = $\mathbb{N}^\omega/\mathcal{F}$ be an ultrapower. Let $\{n_\alpha\}_{\alpha\in\omega_1}$ be a strictly increasing sequence in $^*\mathbb{N}$.
Assuming ZFC, would this sequence be unbounded in $^*\mathbb{N}$, i.e. $\forall n \in {^*\mathbb{N}}\ \exists \alpha \in \omega_1\ n_\alpha > n $? Does it depend on CH?
I don't think (2) will play any role (if $u$ satisfies (1), it's easy to produce $u'$ satisfying (1) and (2), and such that $u$ is bounded iff $u'$ is bounded).
The set of $n\in{^*\mathbb{N}}$ such that $\lim_{m\to\omega} n(m)/m=0$ is convex but has uncountable cofinality. So there exists $(n_\alpha)$ satisfying (1) such that each $n_\alpha$ is less than the sequence $m\mapsto m$.
The question seems to be whether every sequence is unbounded, which is purportedly negatively answered here. Isn't the question rather whether some sequence is unbounded (which would better fit the title)?
@Ycor thank you for your first note. I removed the redundant condition. But how uncountable cofinality implies the existence of strictly increasing $\omega_1$ sequence?
If $X$ is a totally ordered set of uncountable cofinality, define (with choice) by transfinite induction $x_\alpha$ as some element $>x_\beta$ for all $\beta<\alpha$, for $\alpha\in\omega_1$.
Sorry, maybe I don't understand, but this claim is not true for arbitrary totally ordered uncountable set. For example, there is no strictly increasing $\omega_1$ sequence of reals.
You seem to be misreading cofinality as cardinality.
By "$X$ has uncountable cofinality" I mean "for every countable subset $Y\subset X$ there exists $x\in X$ such that $\forall y\in Y$, $y<x$".
@Ycor thank you now I see the whole point. But how to prove, that set of $n \in *\mathbb{N}$ such that $\frac{n(m)}{m} \to 0$ has uncountable cofinality?
Let $(u_n)$ be a sequence of such sequences. I guess something like $u(m)=\max_{k\le m}\lceil 2^{-k}u_k(m)\rceil$ should work.
Let $(u_n)$ be a sequence of such sequences. I guess: set $u(m)=1+\max_{k\le m}\lceil 2^{-k}u_k(m)\rceil$ satisfies $u\ge 2^{-k}u_k$ for all $k$ and $u(m)=o(m)$ when $m\to\omega$, so if $v(m)=m/\sqrt{m/v(m)}$, then it should work.
@YCor, Sorry, maybe I don't understand, but I don't think this construction is working. For example, let's take $$u_n(k) =
\begin{cases}
2^{2k},\quad k = n \
\floor{\sqrt{k}},\quad else
\end{cases}$$
In this case, $u(m) = 1 + 2^m$, hence $u(m) \neq o(m)$.
I'm maybe too hasty. Possibly one should first replace $m\mapsto u_n(m)$ with $\min(u_n(m),m2^{-n})$.
If HC (continuum hypothesis in French) holds, then some of those sequences are cofinal whereas some are not.
Indeed, HC implies that the corresponding ultrapower$\ ^*\mathbb{R}$ of $\mathbb{R}$ is a saturated ordered field with cardinal $\omega_1$. This is the same for the field $\mathbf{No}(\omega_1)$ of surreal numbers with countable birthday. So they are isomorphic.
In $\mathbf{No}(\omega_1)$, there is a strictly increasing and cofinal embedding $x \mapsto \omega^x: \mathbf{No}(\omega_1) \rightarrow \mathbf{No}(\omega_1)^{>0}$ which satisfies in particular $\forall x,y(0\leq x<y\Longrightarrow \omega^x+1<\omega^y)$. So taking integer parts in $^*\mathbb{N}$, we obtain a cofinal order embedding $ x \mapsto \left\lfloor \omega^x \right\rfloor: \mathbf{No}(\omega_1)^{\geq 0} \longrightarrow\ ^*\mathbb{N}$.
Since there are copies of $\omega_1$ in $\mathbf{No}(\omega_1)^{\geq 0}$ which are cofinal, and others which are bounded, this yields cofinal and bounded $\omega_1$-sequences in$\ ^*\mathbb{N}$.
In ZFC, I think (but I am not sure) that it is consistent that the cofinality of$\ ^*\mathbb{N}$ be $\omega_2$, meaning that each $\omega_1$-sequence would be bounded.
What does it mean to say HC holds?
@AlexKruckman HC means CH (as in "hypothèse du continu" and probably similar things in other languages).
@AlexKruckman I didn't even realize I wrote HC, which is indeed the french acronym for the continuum hypothesis.
"each $\omega_1$-sequence is bounded" literally means that the cofinality of $^*\mathbf{N}$ is $\ge\omega_2$.
You might get information about cofinality of $^*\mathbf{N}$ from https://projecteuclid.org/download/pdf_1/euclid.ndjfl/1093635237
@YCor Thanks, I'll try to dinf out tomorrow if the cardinal $d$ they are mentionning can indeed take any value between $\aleph_1$ and $2^{\aleph_0}$.
|
2025-03-21T14:48:31.978960
| 2020-09-07T15:03:22 |
371101
|
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"Nulhomologous",
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|
Stack Exchange
|
Finite flat group schemes over $\mathbb{Z}$ that are extensions of $\mu_p$ by $\mathbb{Z}/p\mathbb{Z}$
Suppose $X$ is a finite flat group scheme over $\mathbb Z$, killed by a prime number $p$ and such that there exists an extension as finite flat group schemes defined over $\mathbb Z$:
$$0\to \mathbb{Z}/p\mathbb{Z}\to X \to \mu_p \to 1.$$
Question: Can we conclude that $X\cong \mathbb{Z}/p\mathbb{Z}\times \mu_p$ over $\mathbb{Z}$?
I know that the answer to this question is negative in general if you consider it over $\mathbb Q$, since you can take $X=E[7]$, the group scheme of $p$-torsion points of an elliptic curve with a $7$-torsion point defined over $\mathbb Q$, since we have such elliptic curves but no such curve with $E[7]\cong \mathbb{Z}/7\mathbb{Z}\times \mu_7$. Of course you can find examples easily for $p=2,3,5$, and probably for infinitely many prime numbers.
On the other hand, over the finite field $\mathbb{F}_p$ the answer is positive, since $\mu_p$ is connected and $\mathbb{Z}/p\mathbb{Z}$ étale, and one could use the connected-étale exact sequence of $X$ to get an splitting of the exact sequence above.
If the answer to the question is affirmative, I will be also interested for what other ring of integers the result is true. I suspect it should be related to the fact that $\mathbb{Q}$ has no unramified extensions.
It is proved in step 3 and 4 of section 3.4.3 in J-M. Fontaine. Il n’y a pas de variété abélienne sur Z. Invent. Math., 81(3):515–538, 198 (using the ramification bound in that paper) that:
For $E=\mathbb Q$ and $\mathbb Q(\sqrt{-1})$, $\mathbb Q(\sqrt{-3})$, in the category of finite flat group schemes over $O_E$ killed by $p$, there is no non-trivial extension of $\mu_p$ by $\mathbb Z / p \mathbb Z$ for $p=3, 5, 7, 11, 13, 17$.
Many thanks for the reference. After reading it, it seems to me that Fontaine proved it for $E=\mathbb{Q}, \mathbb{Q}(\sqrt{-1}), \mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{5})$ (it needs a more delicate argument for $p=3$ and this last field), not for a general number field... In the result it is under the hypothesis of his theorem 4.
@Nulhomologous You're right, I forgot to put the "small" restriction. But his method can be improved to give more results, see Abrashkin‘s works.
I am sorry to say that after reading in detail the proof by Fontaine, I must say that he only proves the result for $K=\mathbb{Q}$ and $p=3, 5, 7, 11, 13, 17$ (and for the other fields, for a more restrictive list of primes). This is because he needs his lemma 3.4.2., which in turn uses the tables by Diaz y Diaz only for that primes. I am not sure Abrashkin says something about this problem on some of his papers...
@Nulhomologous See Serre's Duke87 https://aghitza.org/pdf/translation-serre-duke.pdf 4.5 Group schemes of type (p, p) over Z, where he uses Fontaine's proof to prove the result for any p>=3.
I am surely confused, but Serre's proof of the result (lemma 3 in section 4.5) only says that Fontaine proved it and no other argument...
I downvoted since it seems the argument is incomplete.
The connected–étale sequence holds over $\mathbb{Z}_p$ (see for example Tate's "Finite flat group schemes" article), thus $X$ splits over $\mathbb{Z}_p$.
Combining the above with the fact $X$ is étale over $\mathbb{Z}[\frac{1}{p}]$, we find $\mathbb{Q}(X) = \mathbb{Q}(\mu_p)$.
As $X$ is killed by $p$ over $\mathbb{Z}_p$, it follows by flatness that $X$ is killed by $p$ over $\mathbb{Z}$, and in particular over $\mathbb{Z}[\frac{1}{p}]$.
These two statements give us that $X$ is split over $\mathbb{Z}[\frac{1}{p}]$ (see Section 3.6 in Tate's article).
We now need some gluing argument to conclude $X$ is split over $\mathbb{Z}$. This can be done using Corollary 2.4 in Schoof's "Abelian varieties over cyclotomic fields with good reduction everywhere" paper.
|
2025-03-21T14:48:31.979221
| 2020-09-07T15:03:39 |
371102
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371102"
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|
Stack Exchange
|
Difficulty of homeomorphism of effective Cantor dynamics
Let $X = \{0,1\}^{\mathbb{N}}$ with the product topology. Given a Turing machine $M$ and $x \in X$, define $M(x) \in \{0,1\}^* \cup X$ as the sequence of bits output by $M$ when given an oracle for $x$.
Given two Turing machines $M_1, M_2$, how difficult is it to determine whether they define homeomorphisms $f_1, f_2 : X \to X$ such that the topological dynamical systems $(f_1, X)$ and $(f_2, X)$ are isomorphic?
By $M_i$ defines a homeomorphism I mean $M_i(x) \in X$ for all $x \in X$ and that this map is a homeomorphism on $X$ (equivalently bijective). By isomorphism I mean a homeomorphism $\phi : X \to X$ that intertwines the actions. By how difficult I mean the smallest level of the arithmetical/hyperarithmetical/analytical hierarchy (of sets of natural numbers) that contains the set of Gödel numbers of such pairs of machines $(M_1, M_2)$.
The set of pairs such that both $M_i$ are homeomorphisms is low in the arithmetical hierarchy: having image in $X$, being injective and being surjective are all $\Pi^0_2$ properties, so defining a homeomorphism is too. I am mainly interested in how difficult isomorphism is. Clearly it is $\Sigma^1_1$, is it $\Sigma^1_1$-complete?
|
2025-03-21T14:48:31.979323
| 2020-09-07T15:28:16 |
371104
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371104"
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|
Stack Exchange
|
Weak relaxation of a strongly lower semi-continuous functional
Let $F$ be a lower semicontinuous functional on a Banach space $X$, wrt its strong topology. Is there a known form for the relaxation (lower semicontinuous envelope) of $F$ with respect to the weak topology on $X$?
|
2025-03-21T14:48:31.979369
| 2020-09-07T16:29:50 |
371110
|
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"url": "https://mathoverflow.net/questions/371110"
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|
Stack Exchange
|
Reference for representations of $\text{GL}_n(\mathbb{Z}/p^k\mathbb{Z})$ over $\mathbb{F}_p$
I was looking for a reference on representations of $GL_n(\mathbb{Z}/p^k\mathbb{Z})$ over $\mathbb{F}_p$. In particular what the irreducible and indecomposable representations look like.
Over $\mathbb{C}$, this problem is open for $n \geq 3$ and $k$ arbitrary.
For irreducible representations, it reduces to the case k=1 (where a lot of things are known thank to the theory of polynomial representations, for example, even if the dimensions of the irreducible representations is generally unknown) because the kernel of the canonical surjective group morphism onto $GL_n(\mathbb{Z}/p\mathbb{Z})$ is a $p$-group.
Ahh didn't know the problem was still open. Thanks for the replies.
|
2025-03-21T14:48:31.979447
| 2020-09-07T17:17:49 |
371116
|
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|
Stack Exchange
|
Why do theorems involving $V(p)$ always come in the form $\sum_{p<Q}pV(p)<B$
The quantity
$$V(p)=\sum_{r=0}^{p-1}\left|\sum_{\substack{n=1 \\n \equiv r\mathrm{mod}(p)}}^N a_n-p^{-1}\sum_{n=1}^N a_n
\right|^2$$
has been well studied by many number theorists (see this survey for example), and all of the inequalities I have seen involving it have come in the form
\begin{equation}
\sum_{p<Q}pV(p)<B\tag{1}
\end{equation}
Where $Q=Q(N)$ limits the growth on $p$ and $B=B(N)$ is the actual bound. Examining $\frac{p^2V(p)}{N^2}$ we see that
$$\frac{p^2V(p)}{N^2}=\sum_{r=0}^{p-1}\left|\frac{p}{N}\sum_{\substack{n=1 \\n \equiv r\mathrm{mod}(p)}}^N a_n-\frac{1}{N}\sum_{n=1}^N a_n
\right|^2$$
and if we assume that the quantities
\begin{equation}
\lim_{N\to\infty}\frac{p}{N}\sum_{\substack{n=1 \\n \equiv r\mathrm{mod}(p)}}^N a_n,\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N a_n\tag{2}
\end{equation}
exists then we have that $V'(p)=\lim_{N\to\infty}\frac{p^2V(p)}{N^2}$ exists. This quantity is easier to study since it is generally bounded/small as $N\to\infty$ which can be seen from the fact that
$$\sum_p \frac{V'(p)}{p}$$
converges if $a_n$ takes on the values $0,1$ as can be seen from the inequality
$$\sum_{p\leq Q}pV(p)\leq (N+Q^2-1)\sum_{n=1}^N|a_n|^2$$
To have a more accurate study of $V(p)$ on its own, I would like to study $V'(p)$ on its own. More specifically, since $\sum_p \frac{V'(p)}{p}$ converges we have that
\begin{equation}
\lim_{Q\to\infty}\frac{1}{\pi(Q)}\sum_{p<Q}V'(p)=0 \tag{4}
\end{equation}
with an unknown rate of convergence. Here is my question
Is there are function $\epsilon(Q)\to0$ for which
\begin{equation}
\lim_{Q\to\infty}\frac{1}{\pi(Q)}\sum_{p<Q}V'(p)<\epsilon(Q)
\end{equation}
With the conditions that $a_n$ takes on the values $0,1$ and the sums in (2) converge? In other words, does (3) decay at a global rate?
Does $p$ denote a prime variable in your question, and is $\pi(\cdot)$ the prime counting function? What properties does the sequence ${a_n}$ have?
@StanleyYaoXiao Yes, $p$ denotes a prime and $\pi(\cdot)$ is the prime counting function. Other than the convergence of the sums in (2) and that {a_n} takes on only the values $0,1$, there are no conditions on $a_n$.
|
2025-03-21T14:48:31.979597
| 2020-09-07T17:33:26 |
371118
|
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|
Stack Exchange
|
Computing adjacency matrix eigenvalues by counting closed walks
Let $G$ be a finite undirected graph. A closed walk in $G$ is a walk from any vertex of $G$ to itself. It is relatively straightforward to show that the total number of closed walks of length $k$ in $G$ is:
$$ \lambda_1^k + \lambda_2^k + \cdots + \lambda_n^k,$$
where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of the adjaceny matrix of $G$, with mutliplicity.
Often this is thought of as a way to count walks using linear algebra, but in principle it is possible to apply this result "in the other direction" and compute the eigenvalues of the adjacency matrix of $G$ by counting closed walks via some other (e.g., direct combinatorial) argument.
For example, it is possible to find the eigenvalues of the adjacency matrix of the hypercube graph this way (see Stanley, "Enumerative Combinatorics," Vol. 1, 2nd Edition, Chapter 4 Exercise 68; see also Number of closed walks on an $n$-cube). Even simpler would be the case of the complete bipartite graph (ibid. Exercise 69). And for the complete graph this is almost trivial.
Question: Are there some other nice examples of (families of) graphs whose adjacency matrix eigenvalues can be computed via combinatorial walk counting? Especially, are there any graphs where this is the only known way to find a formula for the adjacency matrix eigenvalues?
I would imagine that you can count the number of closed walks on a k-partite graph too, similar to a bipartite graph. Is there a version for infinite graphs - if so, the Rado graph seems doable too.
@Asvin: Perhaps you can make sense of closed walk counting for infinite graphs using some kind of normalization procedure, but naively even for $k=0$ we get $\infty$-many walks.
Alternatively, you could make asymptotic statements: As the number of vertices goes to infinity, the number of closed walks of a random graph approaches something with probability one and so the eigenvalues also approach something with probability one... Of course, I think it isn't too hard to calculate the eigenvalues directly in either of these examples.
Exercises 1.6, 1.7, 1.8, and 1.10 of my book Algebraic Combinatorics, second ed., can be done by combinatorial reasoning, though there are also linear algebraic proofs.
|
2025-03-21T14:48:31.979771
| 2020-09-07T18:50:30 |
371123
|
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"Erel Segal-Halevi",
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"Rodrigo de Azevedo",
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|
Stack Exchange
|
Detecting slow growth in a finite number of queries
The following question was asked at Can you solve this problem using a finite number of queries?
:
Let $g:[0,1]\to[0,1]$ be a continuous monotonically-increasing function. You can access $g$ using queries of two kinds:
Given $x\in[0,1]$, return $g(x)$.
Given $y\in[0,1]$, return $g^{-1}(y)$.
Given fixed parameters $s,t\in (0,1)$, can you find, using finitely many queries, a point $x$ for which
$$
g(x+s) - g(x) < t
$$
(if such $x$ exists)?
On the same page, this question was answered, affirmatively.
In a comment, the OP then asked what will happen without the assumption that "such $x$ exists".
It will be shown here that, with a reasonable formal interpretation, the answer will change to "no".
How is this related to approximation algorithms?
@RodrigodeAzevedo : Here I used the tags used for the linked question. I think the approximation algorithm here is given by the choice of the query points, to approximate the functions $g$ and $g^{-1}$ by their restrictions to the corresponding sets of the query points, in order to detect slow growth.
I don't want to be too annoying and split hairs that no one cares about, but both questions have a computer science "feel" combined with real analysis. However, an algorithm that approximates something is not an approximation algorithm as used by computer scientists — the tag description has a CS bent. The semidefinite programming approach to MAX-CUT is an approximation algorithm, however.
@RodrigodeAzevedo : I have had very little experience in computer science. However, it appears that the author of the linked question (whose tags I borrowed here) is a computer scientist. So, it appears to me that for some computer scientists the approximation-algorithms tag seems appropriate.
First of all, let us formally interpret the question, as follows:
Take any $s$ and $t$ in $(0,1)$. Let $CI_{s,t}$ be the set of all continuous strictly increasing functions $g\colon[0,1]\to[0,1]$. Let $G_{s,t}$ be the set of all functions $g\in CI_{s,t}$
such that the set
$$E_{s,t}(g):=\{x\in[0,1-s]\colon g(x+s)-g(x)<t\}$$
is nonempty. Do there exist sequences $(x_j)_{j=1}^\infty$ and $(y_j)_{j=1}^\infty$ in $[0,1]$ such that for any $g\in CI_{s,t}$ there is a natural $n$ such that the following implication holds: If for some function $h\in CI_{s,t}$ and for all $j\in[n]:=\{1,\dots,n\}$ we have $h(x_j)=g(x_j)$ and $h^{-1}(y_j)=g^{-1}(y_j)$, then
(i) if $g\in G_{s,t}$ then ($h\in G_{s,t}$ and) for some $k\in[n]$ we
have $x_k\in E_{s,t}(h)$;
(ii) if $g\notin G_{s,t}$ then $h\notin G_{s,t}$.
The answer is now no, in general.
Indeed, take any $s,t$ such that $0<t<s<1$. Take any sequences $(x_j)_{j=1}^\infty$ and $(y_j)_{j=1}^\infty$ in $[0,1]$. Take any natural $n$.
Consider the set $P_{s,t}$ of all pairs $(a,b)$ such that
$$0<a<a+s<1\ \&\ 0<b<b+t<1\ \&\ \min\Big(\frac{b}{a},\frac{1-b-t}{1-a-s}\Big)>\frac{t}{s}.$$
The set $P_{s,t}$ is nonempty and open; in fact,
$$(a,b)\in P_{s,t}\iff \Big(0<a<1-s\ \&\ \frac{a t}{s}<b<\frac{a t+s-t}{s}\Big).$$
Take now any pair $(a,b)\in P_{s,t}$ such that $a\notin\big\{x_j\colon j\in[n]:=\{1,\dots,n\}\big\}$ and $b\notin\{y_j\colon j\in[n]\}$; such a pair $(a,b)$ exists, since $P_{s,t}$ is nonempty and open.
Next, let $g=g_{a,b}=g_{s,t,a,b}$ be the function whose graph is the union of the straight line segments successively connecting the points $(0,0),(a,b),(a+s,b+t),(1,1)$. Then $g\in CI_{s,t}\setminus G_{s,t}$.
Let
$$x_{n,a}:=\min\{x_j\colon j\in[n],x_j>a\},\quad x_{n,a}^-:=\max\{x_j\colon j\in[n],x_j<a\},\quad y_{n,b}:=\min\{y_j\colon j\in[n],y_j>b\}.$$
Then $x_{n,a}^-<a<x_{n,a}$ and $y_{n,b}>b$. Since $g$ is strictly increasing, there is some $c$ such that
$$b=g(a)<c<\min[g(x_{n,a}),y_{n,a}].$$
For such $c$ and all $x\in[0,1]$, let
$h$ be the function whose graph is the union of the straight line segments successively connecting the points $(0,0),(x_{n,a}^-,g(x_{n,a}^-)),(a,c),(x_{n,a},g(x_{n,a})),(a+s,b+t),(1,1)$. Then $h(x_j)=g(x_j)$ and $h^{-1}(y_j)=g^{-1}(y_j)$ for all $j\in[n]$. However, $h(a+s)-h(a)=g(a+s)-c<g(a+s)-g(a)=t$, so that $h\in G_{s,t}$, whereas $g\notin G_{s,t}$. Thus, conclusion (ii) of the implication in the highlighted formalization of the question fails to hold. $\Box$
The graphs of $g$ (blue) and $h$ (gold) for $s=4/10,t=2/10,a=3/10,b=5/10,x_{n,a}^-=2/10,x_{n,a}=4/10,y_{n,a}>55/100$ are shown below.
Very interesting! In the language of queries, it seems that your formalization corresponds to "non-adaptive queries". This means that the algorithm has to decide in advance what queries to ask, i.e., the sequences $x_n$ and $y_n$ are pre-determined. There is a different model called "adaptive queries", in which the algorithm may decide what queries to ask based on the replies to previous queries, i.e., $x_j$ and $y_j$ may depend on $g(x_1),\ldots,g(x_{j-1})$ and $g(y_1),\ldots,g(y_{j-1})$. When $s=t$, there is no finite solution even in the adaptive query model (see the original question).
@ErelSegal-Halevi : I'd guess that even for general $s,t$ there is no finite solution even in the adaptive query model.
I agree, and added a proof. Sorry for using the language of queries - this is the language I am used to. It comes from papers such as this: https://www.combinatorics.org/ojs/index.php/eljc/article/view/735
Here is a proof that, even with "adaptive queries" (queries that may depend on answers to previous queries, rather than be set in advance), a finite algorithm may not exist.
Pick some $s'\in(s,1)$, and define the following piecewise-linear function:
$$
g_0(x) :=
\begin{cases}
(t/s)\cdot x & x \leq s'
\\
(s' t / s) + \frac{1-(s' t / s)}{1-s'} \cdot (x-s') & x\geq s'
\end{cases}
$$
Note that $g_0(0)=0, g_0(1)=1$, there are uncountably many $x$ for which $g_0(x+s)-g_0(x) = t$, but no $x$ for which $g_0(x+s)-g_0(x) < t$.
Suppose that the answers to all queries are as if $g\equiv g_0$. After finitely many queries, it is possible that indeed $g = g_0$, in which case there is no solution. However, after finitely many queries, there are uncountably many points $x\in [0,s'-s]$ that did not participate in any query. By slightly increasing the value of $g_0(x)$ while keeping the function continuous, as in the figure in Iosif's answer, we get another function $g_1$, for which $g_1(x+s)-g_1(x)<t$.
While the question has been answered, it is interesting to check what happens if we slightly change the condition, from $g(x+s)-g(x)<t$ to $g(x+s)-g(x)\leq t$. The above proof does not work.
However, I still think that it is impossible to decide if such $x$ exists with finitely-many queries. Fixing $s$ and $t$, for every $z\in[0,1-s]$, let $G_z$ be the set of continuous functions $g_z$ for which:
$$
g_z(x+s) - g_z(x) > t ~~~~ x\neq z
\\
g_z(x+s) - g_z(x) = t ~~~~ x = z
$$
(it should be possible to construct such continuous functions; I do not have the exact construction now).
To prove impossibility, we can use an adversary argument: we show that, for any algorithm for asking adaptive queries, an adversary can answer the queries in such a way that the algorithm will never know if a solution exists or not.
The adversary works as follows: he picks an arbitrary $z\in[0,1-s]$, and an arbitrary $g_z\in G_z$, and answers all queries as if $g \equiv g_z$, as long as the queries do not involve the point $z$ itself. In case a query does involve the point $z$, the adversary picks a nearby point $z'$, that is not equal to any recorded point (any point that appeared in a previous query). He constructs a new function $g_{z'}\in G_{z'}$, that coincides with $g_z$ in all recorded points (there are finitely many such points, so it should be possible to construct such a continuous function). The adversary can keep switching functions forever, and the algorithm will never know the actual $z$, and thus will never know if a solution exists.
|
2025-03-21T14:48:31.980335
| 2020-09-07T19:10:12 |
371126
|
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|
Stack Exchange
|
On dissecting rectangles into rectangles
It is well known that, for any two rectangles of the same area, the first can be cut into a finite number of polygonal pieces and reassembled into the other (for example, by Montucla's dissection).
Questions: Can we do the same if only rectilinear polygons can be used as intermediate pieces? Or if only rectangles can be used as intermediate pieces?
|
2025-03-21T14:48:31.980393
| 2020-09-07T19:38:55 |
371128
|
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"authors": [
"Ivo Terek",
"Qiaochu Yuan",
"Tim Phalange",
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|
Stack Exchange
|
What is the relationship between Riemannian and sympletic musical isomorphisms on the cotangent bundle?
Let $M$ be a smooth manifold. Its cotangent bundle naturally has a symplectic structure, and this gives rise to musical isomorphisms. These musical isomorphisms are the ones from physics that relate the Hamiltonian function and the Hamiltonian vector field.
What I am confused about is that if $M$ is a Riemannian manifold, then its cotangent bundle has an induced Riemannian metric. This means that in addition to the symplectic musical isomorphisms on the cotangent bundle of $M$, you have Riemannian musical isomorphisms. Since the symplectic musical isomorphisms do not depend on the Riemannian metric, there's no reason to think that they would be the same. But still, I wonder, is there any significance to these Riemannian musical isomorphisms? Do they have any physical significance for example? Is there an obvious relationship between them and the symplectic musical isomorphisms?
Consider the local case of a single finite-dimensional vector space. A symplectic form gives some identification with the dual. An unrelated inner product gives another. Without positing a relationship between the two (e.g. that they arise from a complex inner product) what else is there to say?
Let me ask more pointedly, then: why are the musical isomorphisms coming from the symplectic structure the ones that arise in Hamiltonian mechanics? What makes them more natural in this setting to the point of us ignoring the Riemannian musical isomorphisms? Is this just a cosmetic choice, or is there something more meaningful here?
The symplectic one is the one coming from the Poisson bracket, and the Poisson bracket’s alternating property is what makes Noether’s theorem true. In the generality of a Poisson manifold or Poisson algebra we can still do Hamiltonian mechanics without a symplectic musical isomorphism; the map only needs to go from the differential of the Haniltonian to the Hamiltonian vector field, not the other way.
I'm confused about your comment. The Poisson bracket is simply the Lie bracket by another name under the symplectic musical isomorphism, no? You can just as easily look at the Lie bracket under a Riemannian musical isomorphism... You'll still get anti-commutativity because that's a property of the Lie bracket...
I'd expect some concrete relation if, e.g., $M$ is Kähler. A priori the metric and the symplectic form can be completely unrelated (this is what Qiaochu is saying). Also I'd be careful talking about a metric on the manifold $T^*M$ (I'm sure this is not what you meant -- the Riemann extension of the Levi-Civita connection of the original metric in $M$ is a neutral pseudo-Riemannian metric on the manifold $T^*M$). Lastly, the Poisson bracket is a particular type of Lie bracket only, in the algebra of smooth functions, not directly on arbitrary vector fields.
|
2025-03-21T14:48:31.980635
| 2020-09-07T19:39:58 |
371129
|
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"Fedor Pakhomov",
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|
Stack Exchange
|
Where can I read about Veblen functions / klammersymbols beyond the large Veblen ordinal?
So, I'm not sure to what extent this is a thing. John Baez mentions in this blog post that common large countable ordinals beyond the large Veblen ordinal can also "be defined as fixed points". He doesn't expand on that but I take it that means fixed points of a further Veblen-like / klammersymbol-like construction.
But, I haven't been able to find any good account of this. Is this a known/standard thing? I want to know just how far the Veblen construction can be pushed -- like, beyond the large Veblen ordinal, is there an "ultimate Veblen ordinal" somewhere; and might it be equal to an already-named / well-known one, such as Bachmann-Howard? (I suppose a negative answer here would be if arbitrarily large computable ordinals could be obtained this way, making the "ultimate Veblen ordinal" just $\omega_1^{\mathrm{CK}}$.)
Now I can certainly think of ways of continuing beyond the large Veblen ordinal myself... but I don't really want to reinvent the wheel here when I expect others have likely already done it better. So, is there a good account of this somewhere?
Thank you all!
If one takes Veblen-like construction as an informal concept, then it would be hard to justify some $\alpha<\omega_1^{\mathsf{CK}}$ to be the ultimate Veblen ordinal (the suprema of all ordinals obtained by Veblen-style constructions). The reason for this is that it wouldn't be clear why one shouldn't be allowed to make the next step of diagonalization and overcome this particular ordinal.
However, I think that it is possible to give Bachmann-Howard ordinal as a (non-least) upper bound for reasonable constructions of this kind. The motivation here is that it is reasonable to expect that this constructions should be formalizable in Kripke-Platek set theory with infinity $\mathsf{KP}\omega$ and the proof-theoretic ordinal of $\mathsf{KP}\omega$ is Bachmann-Howard ordinal.
I think a link to original paper (of howard I think?) was posted on fom (if I am re-calling correctly). I don't think it was that long ago (maybe around an year or bit more). I will try to find and post the link (in few days perhaps). Other than that, one could probably find a number of relevant topics by using a search or following the links in some of the topics under "Related".
Here is the link I was referring to in comment above: https://arxiv.org/abs/1903.04609
@FedorPakhomov: Yes, that is the obvious problem; but I thought maybe if at some point there were some component to the construction which there were truly no way to iterate more than finitely many times, then you could take the limit of those and then be unable to continue further, as you simply wouldn't have enough to find a fixed-point. But I don't have a good idea if that would happen or not.
@FedorPakhomov Also like -- it has to give out somewhere on account of it certainly can't get you past $\omega_1^{CK}$! But if it can't, then there must be some concrete way in which the construction fails at that point; and since the construction of taking derivatives of appropriate sorts doesn't seem obviously related to computability, this raises the possibility that maybe that failure happens somewhere earlier?
@HarryAltman I think that it is very unlikely that there is some concrete point of failure. But in my opinion the reasoning that either there is a concrete point of failure or we could reach unlimitely large ordinals isn't valid. This is because we are dealing with an informal notion here. Furthermore, it is rather unlikely that any particular formalization would be completely satisfactory (precisely because then it would lead to a concrete point of failure, which we would be able to diagonalize against).
@HarryAltman However, this doesn't mean that we couldn't give some upper bound. We just need to have some necessary but not sufficient conditions for some construction to count as Veblen-like. I think that formalizability in $\mathsf{KP}\omega$ probably should be a necessary condition (which gives B-H ordinal as an upper bound). Basically, in order to show this (in an informal sense) we need to demonstrate that all the types of constructions and manipulations with them that could be building blocks of a Veblen-like construction are formalizable in $\mathsf{KP}\omega$.
@HarryAltman Essentially here I am using the same kind of argument as M. Rathjen have used in his paper on limits of Martin-Löf type theory http://www1.maths.leeds.ac.uk/~rathjen/EHPanthology.pdf , where he argued that the strengths of this kinds of theories are bounded by the strength of the theory $\mathsf{KPM}$.
OK I'm confused by the Howard paper. He constructs $\Gamma_0$ in section 3, calling it $\lambda$, but I don't really follow the rest. Why does section 4 involve uncountables (and even $\omega_2$?!)? And the subscripts here are individual ordinals, rather than there being multiple subscripts? I'm confused; I don't really understand how this fits with what I know.
Oh, I think I see -- if you think in terms of sets and intersections rather than functions and fixed points and limits, the suggestion I gave above about how it could possibly terminate makes no sense. So I guess it has to go to ω_1^{CK}$ after all, after that many steps (using some notional setup where we've, like, linearized the sequence of steps)...
Here is some relevant information by a grandmaster on the subject:
http://www.mathematik.uni-muenchen.de/~buchholz/articles/jaegerfestschr_buchholz3.pdf
OK, taking a look at this I'm afraid I don't understand this. He explains how to construct $\Gamma_\alpha$ in the preliminaries, but I'm having trouble following it beyond that. His $\phi$ only has one subscript, but (going by the equation before Lemma 1.6) it acts like multiple (i.e., acts like Klammersymbols) if you go into uncountables? I don't really understand how to get beyond that though, especially if merely getting beyond (the equivalent of) a single subscript requires getting to $\Omega$! I'd hope for each step to be somehow expressible in terms of things constructed so far...
|
2025-03-21T14:48:31.981081
| 2020-09-07T21:16:59 |
371136
|
{
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|
Stack Exchange
|
History of the notation for scalar product
What's the history of the development of the notation for (real or hermitian) scalar product? In particular,
Did "bra-ket" notations, such as $\langle u\mid v\rangle$ or $(u\mid v)$, first arise with Paul Dirac in the context of quantum mechanics?
Did the $\langle u, v\rangle$ notation appear later as a modification of the one with the vertical bar or vice versa?
Was the "dot product" notation $\overrightarrow{u}\bullet\overrightarrow{u}$ introduced before the "bra-ket" ones?
Yes, the bra-ket was introduced by Dirac and made standard in his celebrated book on QM . However, looks like Grassman had a similar notation long before that ( [ v, w] rather than <v, w>
Except that the purpose of Dirac's notation was to separate the two vectors in the notation. Apparently he went round saying "I invented the bra" (see the depressing biography "The Strangest Man" by Graham Farmelo).
Cajori, A History of Mathematical Notation § 506 (vol 2) attributes to Grassmann the notations $a \times b$ (1848) and $[a|b]$ (1862) for the scalar product, to Heaviside and others the $a|b$ in the 1890s, to Lorentz $(a,b)$ in the early 1900s.
Isn't the historical spelling "Grassmann"?
Yes. I am guilty of the same typo: Grassmann, ie Graßmann
Of course, thank you
|
2025-03-21T14:48:31.981229
| 2020-09-07T21:50:05 |
371139
|
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|
Stack Exchange
|
Proving an identity used in general relativity
I need to prove the following identity for scalar field ($\phi:M\rightarrow R$) in curved spacetime without torsion called $M$
$\nabla_{\mu}[\Box \phi \nabla^{\mu}\phi-\frac{1}{2}\nabla^{\mu}(\nabla \phi)^{2}]=(\Box \phi)^{2}-
R_{\mu\nu}\nabla^{\mu}\phi\nabla^{\nu}\phi- \nabla^{\mu}\nabla^{\nu}\phi \nabla_{\mu}\nabla_{\nu}\phi$.
I opened up the total derivative and the term $(\Box \phi)^{2}$ appears but the others I cannot get the right combination. I tried to use the following identity to make appears the terms $R_{\mu\nu}\nabla^{\mu}\phi\nabla^{\nu}\phi$, namely, $[\nabla_\mu, \nabla_\nu]X_{\alpha}=-R^{\kappa}_{\alpha\mu\nu}X_{\kappa}$, but I got stuck at some point, and the third term does not appear at all.
Any help would be much appreciated.
Is your left-most term missing a $\phi $?
Would the Bianchi identity be of some help?
Dear Michael, yes there was a typo.
Dear Sylvain, I tried to use the Bianchi identity but I could not get the right answer.
This kind of trick appears in Hordensky theory.
This is the differential form of the Reilly formula. It holds for a function on any pseudo-Riemannian manifold. (Robert C. Reilly. Applications of the Hessian operator in a Riemannian manifold, Indiana Univ. Math. J. 26 (1977), no. 3, 459–472, doi:10.1512/iumj.1977.26.26036)
Use the product rule to say
$$(\Delta f)^2=\operatorname{div}(\Delta f\cdot\nabla f)-\langle\nabla f,\nabla\Delta f\rangle.$$
Use the commutation formula for covariant derivatives to replace the last term by
$$\langle\nabla f,\nabla\Delta f\rangle=\langle\nabla f,\Delta\nabla f\rangle-\operatorname{Ric}(\nabla f,\nabla f).$$
Use the product rule to replace the second to last term by
$$\langle\nabla f,\Delta\nabla f\rangle=\operatorname{div}\big(\nabla^2f(\nabla f,\cdot)\big)-|\nabla\nabla f|^2.$$
Finally $\nabla^2f(\nabla f,\cdot)=\frac{1}{2}\nabla|\nabla f|^2$. This gives your formula.
Edit. As pointed out by Jeffrey Case below, this also follows from the Bochner formula
$$\frac{1}{2}\Delta|\nabla f|^2=|\nabla\nabla f|^2+\langle\nabla\Delta f,\nabla f\rangle+\operatorname{Ric}(\nabla f,\nabla f),$$
where you just need to use the very first line above to replace the middle term on the RHS. The proof of the Bochner formula is by the other lines above
I think it is better to call this (a variant of) the Bochner formula. The formula in the original post will (essentially) appear in any textbook under this name. "Reilly formula" usually refers to the result after integrating on a manifold with boundary and performing further algebraic manipulations on the boundary.
That's fair. When I called it the Reilly formula I was thinking of the case of a closed manifold
Dear Quarto Bendir, thank you so much for your help and specially for proving the mathematical reference.
|
2025-03-21T14:48:31.981456
| 2020-09-07T22:16:31 |
371140
|
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|
Stack Exchange
|
Can every Hodge structure be polarized?
I suspect this is very elementary, but it is not stated anywhere. A Hodge structure of weight $k$ consists of a finite rank lattice $H_{\mathbb{Z}}$ together with a decomposition of its complexification $H : = H_{\mathbb{Z}} \otimes \mathbb{C}$, $$H = \bigoplus_{p+q=k} H^{p,q},$$ with $H^{p,q} = \overline{H^{p,q}}$. Set $F^p : = \bigoplus_{r \geq p} H^{r,k-r}$.
A blinear form $Q$ on $H$ polarizes $H$ if
$Q(u,v) = (-1)^k Q(v,u)$,
$Q(F^p, F^{k-p+1})=0$,
The Hermitian form $Q((\sqrt{-1})^{p-q} \cdot, \overline{\cdot})$ is positive definite.
Question: Can every Hodge structure be polarized?
I guess you may be right it’s hard to find, but the answer is no. Hodge structures of type (1,0), (0,1) are equivalent to complex tori, and the polarizable ones correspond to abelian varieties.
However the OP does not mention that $Q$ takes integral values on $H$; if you don't put that condition, finding such a $Q$ is just an exercise in linear algebra.
Let me summarize comments.
abx points out that $Q$ should be integer valued, otherwise it's not an interesting notion.
(Assuming integrality) the answer is no because the categories of Hodge structures of type $\{(1,0), (0,1)\}$ and complex tori are equivalent. The polarizable ones correspond to abelian varieties, which is a proper subset.
Thanks for your answer. The equivalence between Hodge structures of type ${ (1,0), (0,1) }$ and complex tori is given as follows? Let $H_{\mathbb{Z}}$ be a lattice of rank $2n$ with $\phi : H_{\mathbb{Z}} \to H_{\mathbb{Z}}$ an endomorphism. Take the eigenvalues $\lambda_1, ..., \lambda_{2n}$ of $\phi$ to be distinct with non-zero imaginary part. Let $H^{1,0}$ be the eigenspace of $\lambda_1, ..., \lambda_n$, such that $\lambda_i \neq \overline{\lambda_j}$ for any $1 \leq j \leq n$. The torus is then $T = H/(H^{1,0} \oplus H_{\mathbb{Z}})$?
|
2025-03-21T14:48:31.981617
| 2020-09-07T22:21:45 |
371141
|
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|
Stack Exchange
|
Combinatorics and geometry underlying a refined Pascal matrix/Newton identities
The partition polynomials of OEIS A263633 give the coefficients of the power series/o.g.f of the multiplicative inverse (reciprocal) of a power series/o.g.f. and so give the Newton identities for transforming between complete homogeneous symmetric polynomials/functions and elementary symmetric polynomials/functions. Certain Koszul duals are related to this.
After reading Stanley's answer, I guess I need to rephrase the question to lean away from algebraic combinatorics towards geometric combinatorics. For example, the square of the lower triangular Pascal matrix is a convolution of binomial coefficients and also the face vectors of the hypersquares, or hypercubes. Another example, the refined associahedra partition face polynomials are combinations of averaged factorials that give a prescription for compositional inversion of o.g.f.s--it's the connection to the combinatorics of the faces--the geometry--that I would like emphasized.
On my blog "Shadows of Simplicity" I posted "Squaring Triangles" to explain to motivated high school students the relation between the algebraic and geometric combinatorics. Ideally, that is what I'm looking for here.
The algebraic combinatorics of the complementary reciprocal of a Taylor series/e.g.f. is governed by the antipode/refined Euler characteristic classes of the permutahedra or, equivalently, by surjective mappings, so I have an indirect geometric combinatorial interpretation of 'scaled' versions of the Newton identities, but I'm looking for more direct interpretations.
What geometric structures are enumerated by the integer coefficients of these partition polynomials for conversion of an o.g.f. into a reciprocal o.g.f.?
One answer from A133314 is colored surjections, where the arrows mapping into an element of the image induce a linear order by color that are then permuted. The resulting partition polynomials of A263633 are then the coefficients of the e.g.f. enumerating these mappings, i.e. the scaled, signed face partition polynomials of the permutahedra divided by n!.
in the links, I show that the Pascal lower triangular matrix itself, ignoring the initial column of ones, contains the face vectors of the hypertriangles. E.g., (4,6,4,1) is the f-vector for the tetrahedron, encompassing 4 vertices, 6 edges, 4 triangles as facets, and 1 tetrahedron.
Related: https://mathoverflow.net/questions/243185/combinatorial-interpretation-for-coefficients-of-reciprocal-of-power-series?rq=1
This is the first I've heard of that particular blog, and I notice that you've got lines that say things like $ n!\sum_{k=0}^n \binom{n-n}{n-k} \frac{(-x)^k}{k!} $ instead of $ \displaystyle n!\sum_{k=0}^n \binom{n-n}{n-k} \frac{(-x)^k}{k!}. $ Note not only the difference in sizes of some things, but the difference in formatting of the subscripts and superscripts on the summation sign, thus $\sum_{k=0}^n,$ with the subscript and superscript to the right of $\sum,$ versus $\displaystyle \sum_{k=0}^n.$ Why put them in inline style rather than display style when they're on a line by themselves?
@MichaelHardy, limited personal time and availability of free, nice, efficient apps for publishing large pdf files in LaTex. I've got tons of written material that I haven't published yet. Any recommendations for stand-alone apps or for add-ons for Google docs? Of course, I use cut and paste for related entries / use (MO, MSE, OEIS, pdf, WordPress, Desmos, Wolfram Alpha) and don't regard it a necessity to perfect the formatting for each venue for aesthetic tastes.
(cont.) In fact, some users have edited some of my posts beautifully but making the lines then extremely difficult to cut and paste for use in other venues or even for calculations via substitutions in specific applications. // Some twenty years ago I used a free app that made it very easy to produce reasonably-sized pdfs with nicely formatted multi-colored equations, but it disappeared a couple of years later and I haven't found a comparable one since.
I don't know any such apps. I just use LaTeX and MathJax.
Since
$$ \frac{1}{1+a_1x+a_2x^2+\cdots} = \sum_{k\geq 0}(-1)^k(a_1x+a_2x^2+\cdots)^k, $$
the coefficient of $a_1^{c_1}a_2^{c_2}\cdots x^n$, where $n=\sum ic_i$, is just $(-1)^k$ times the multinomial coefficient ${k\choose c_1,c_2,\dots}$, where $k=\sum c_i$.
Thanks. Yes, that is the algebraic combinatorics. I refined my question to be clearer about what I'm looking for.
|
2025-03-21T14:48:31.982210
| 2020-09-07T23:57:06 |
371145
|
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|
Stack Exchange
|
Simple closed forms for sums such as $\sum_{k=1}^\infty \frac{(-1)^{k+1}}{qk - p}$ and related integrals
My goal here is to get a simple expression for $\zeta(3)$. This is a follow up to my previous question posted here. Any Taylor-like expansion from everything I tried won't make it. So this is my last trick hoping I get something interesting.
By simple, I mean a finite sum involving only logarithms and trigonometric functions. Here $p, q$ are integers with $q>0$ and $p<q$. Let us define
$$G_1(p,q)=\sum_{k=1}^\infty \frac{(-1)^{k+1} q}{qk-p}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k-\frac{p}{q}}$$
$$G_2(p,q)=\sum_{k=1}^\infty \frac{(-1)^{k+1} q^2k}{(qk)^2-p^2}
=\sum_{k=1}^\infty \frac{(-1)^{k+1} k}{k^2-\Big(\frac{p}{q}\Big)^2}$$
$$G_3(p,q)= \frac{G_2(p,q)-\log 2}{p^2/q^2} =
\sum_{k=1}^\infty \frac{(-1)^{k+1} }{k\Big[k^2-\Big(\frac{p}{q}\Big)^2\Big]}$$
It is easy to establish the following:
$$G_1(p,q) = \int_0^\infty \frac{\exp{\Big(\frac{p}{q}\cdot x\Big)}}{1+\exp x}dx$$
$$G_2(p,q)=\int_0^\infty \frac{\cosh{\Big(\frac{p}{q}\cdot x\Big)}}{1+\exp x}dx.$$
$$\lim_{q\rightarrow\infty} G_3(1,q) =\frac{3\zeta(3)}{4}$$
Interestingly (see Mathematica computation here), we have:
$$\sum_{k=1}^\infty \frac{(-1)^k }{k^2-\Big(\frac{p}{q}\Big)^2} =\frac{3}{2}\Big(\frac{p}{q}\Big)^2 +\frac{\pi}{2}\Big(\frac{p}{q}\Big)^{-1}\csc\Big(\pi\cdot \frac{p}{q}\Big).$$
Unfortunately, I could not find such formulas for $G_1$ or $G_2$. The last formula is especially appealing in the following sense. Assume $p,q\rightarrow\infty$ in such a way that $\frac{p}{q}\rightarrow\alpha$ where $\alpha$ is an irrational number. Then we have a simple closed form for the sum even if $\frac{p}{q}$ is replaced by an irrational number.
My question
Is it possible to obtain such a simple expressions for $G_1$ and $G_2$, maybe a sum involving $q+1$ terms? Both integrals can be computed in closed form, even the indefinite integrals, when $p,q$ are integers with $p<q$ and $q>0$. I obtained a closed form for $G_2$ (see next section) and my guess is that $G_1$, even though a bit trickier, has also a simple closed form, see here or picture below for the case $q=16, p=1$. The next section provides hints about how to solve this problem.
The last section is about my second question: there is something that looks very mysterious to me, and maybe someone can provide some insights about that mystery.
Towards a solution
For $G_1$, we have, using integral-calculator.com (based on the Maxima symbolic math solver):
$$\int \frac{\exp(px/q)}{1+e^x}dx=\sum_{\left\{w:\>w^q+1=0\right\}} w^{p-q}
\cdot\log\Big(\Big|e^{x/q}-w\Big|\Big)+C.$$
The sum is over all $q$ (mostly complex) roots of $w^q+1=0$. I don't know what the symbol $|\cdot|$ stands for in the complex logarithm function. A similar formula, albeit more intricate, also exists for $G_2$, and eventually (after considerable cleaning and assuming $|\cdot|$ is the absolute value) it leads to:
$$G_2(p,q)=\frac{q}{2p}+\frac{1}{2}\sum_{j=0}^{q-1}\cos\Big[(2j+1)\frac{p\pi}{q}\Big]\log\Big(1-\cos\Big[(2j+1)\frac{\pi}{q}\Big] \Big).$$
Below is a bar chart showing the values of the $q$ terms in the summation, from $j=0$ on the far left to $j=q-1$ on the far right on the X-axis. Here $q=1103$ and $p=799$.
Note that I did not really prove the result. All I did was to use a symbolic math calculator for the indefinite integrals and for the few definite integrals that it was able to solve exactly (small values of $p$ and $q$). I found a pattern in the indefinite integrals when $q$ is a power of 2, turned to the definite integrals, plugged in the unitary complex roots in the formula, did a lot of cleaning, and tested my formula for various $p,q$ against values that were obtained numerically. It worked, and it even worked when $q$ is not a power of 2. Below is the code to compute $G_2(p,q)$. It is trivial, and the only reason I provide it is in case my formula has a typo: the code below is correct for sure.
$pi=3.141592653589793238462643383279;
$q=11;
$p=7;
$sum=0;
for ($j=0; $j<$q; $j++) {
$theta=((2*$j+1)*$pi)*($p/$q);
$theta2=((2*$j+1)*$pi)/$q;
$sum+=(cos($theta)*log(1-cos($theta2)))/2;
}
$sum+=($q/(2*$p));
print "($p/$q): $sum\n";
My second question
If you look at my formula for $G_2(p,q)$ in the previous section, it does not seem to be a function of $\frac{p}{q}$. Yet I know it must be one. How can I write $G_2(p,q)$ explicitly as a function of $\frac{p}{q}$ only, say $G_2(\frac{p}{q})$? I am also interested in some expansion of $G_2$ when $p,q\rightarrow\infty$ and $\frac{p}{q}\rightarrow\alpha$.
Note that
$$2\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k-a}
=2\sum_{j=1}^n\Big(\frac1{2j-1-a}-\frac1{2j-a}\Big)
=\sum_{j=1}^n\Big(\frac1{j-(1+a)/2}-\frac1{j-a/2}\Big).$$
Also,
$$\sum_{j=1}^n\frac1{j+b}=\ln n-\psi(1+b)+o(1)$$
(as $n\to\infty$), where $\psi$ is the digamma function. So, your $G_1$ is
$$g_1(a):=\tfrac12\,[\psi(1-a/2)-\psi(1/2-a/2)],$$
where $a:=p/q$.
$G_2$ can be handled similarly, by first using the partial fraction decomposition
$$2\frac k{k^2-a^2}=\frac1{k-a}+\frac1{k+a}.$$
So, your $G_2$ is
$$\tfrac12\,[g_1(a)+g_1(-a)]=\tfrac14\,[\psi(1-a/2)-\psi(1/2-a/2)+\psi(1+a/2)-\psi(1/2+a/2)],$$
where again $a=p/q$.
Now, to rewrite these expressions for $G_1$ and $G_2$ in terms of the logarithmic and trigonometric functions, use the trivial identity $\psi(z+1)=\psi(z)+1/z$ and the Gauss digamma theorem.
In particular, we get
$$G_1=\frac12\,\ln\frac{q-p}{2q-p} \\
+\frac\pi4\,\cot\frac{(q-p)\pi}{2q}-\frac\pi4\,\cot\frac{(2q-p)\pi}{2q}\\
+\sum_{n=1}^{q-1}\Big(\cos\frac{(2q-p)\pi n}q-\cos\frac{(q-p)\pi n}q\Big)
\ln\sin\frac{\pi n}{2q} \\
=\frac12\,\ln\frac{q-p}{2q-p} \\
+\frac\pi4\,\tan\frac{p\pi}{2q}+\frac\pi4\,\cot\frac{p\pi}{2q} \\
+\sum_{n=1}^{q-1}[1-(-1)^n]\cos\frac{p\pi n}q\,
\ln\sin\frac{\pi n}{2q},$$
as desired. Similarly, for $G_2$, using the identity $\psi(1+a/2)=\psi(a/2)+2/a$.
@ Iosif: thanks for your insights. I know these sums can be expressed in terms of digamma functions or their derivatives. But I want a formula based on simpler functions - log or trigonometric functions. Essentially, my goal is to get an exact original formula for $\zeta(3)$, as per my last post (see https://dsc.news/3i9P1kh).
Thanks for the link to the Gauss digamma theorem.
There are numerical algorithms for determining whether given constants, like $\zeta(3)$, have expressions in terms of simple functions. Might be worth looking at the work of David Bailey, Jonathan Borwein and the like.
@ Gerry: Thank you. I know you can test trillions of combinations of "simple" numbers to see if one matches $\zeta(3)$ (or any other constant). I reached out to David a while back about a different issue, he is a very nice guy. Same with Andrew Granville. At the end, this is not really a calculus problem (however advanced it might look) but a pure number theory problem.
@VincentGranville : I have now written out the expression of for $G_1$ in terms of the logarithmic and trigonometric functions, as you apparently desired. The case of $G_2$ is quite similar, since $\psi(1+a/2)=\psi(a/2)+2/a$.
@GerryMyerson : Thank you for your comment.
@ Iosif: wonderful, thank you! You can simplify the $\cos A - \cos B$ expression in parenthesis: it becomes a product of two sin functions. My guess is that one factor becomes +1 or -1.
@VincentGranville : I have now simplified the expression for $G_1$, using expressions for $f(t+k\pi/2)$, where $f$ is a trigonometric function function and $k$ is an integer.
|
2025-03-21T14:48:31.982774
| 2020-09-07T23:58:49 |
371146
|
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|
Stack Exchange
|
How many Lie and associative algebras over a finite field are there?
This question is related to the following general question:
Given a variety of (non-associative) algebras $\mathcal V$, a finite field $\mathbb{F}_q$, with $q$ elements, and a positive integer $n$, how many $n$-dimensional $\mathbb F_q$-algebras in $\mathcal V$ are there?
It is well known that if $A$ is an $n$-dimensional algebra over a field $F$, with basis $\{e_1, \dots, e_n\}$ then its algebra structure is uniquely determined by the $n^3$-tuple $(\alpha_{ij}^{(k)})\in F^{n^3}$, defined by $e_i e_j=\sum_{k=1}^n \alpha_{ij}^{(k)} e_k$.
So the general question above can be reformulated as:
How many of these $n^3$-tuples of elements of $\mathbb F_q$ define algebras in $\mathcal V$?
Or in another language: given an arbitrary $n$-dimensional algebra, what is the probability that it lies in $\mathcal V$?
Let us denote such number by $N_{q,n}(\mathcal V)$.
Some examples are simple to compute.
For example, if $q$ is odd, one can easily show that if $\mathcal V$ is the variety of anticommutative algebras (i.e., the class of all algebras satisfying the identity $xy+yx=0$), then $N_{q,n}(\mathcal V)=q^{n^2(n-1)/2}$ and if $\mathcal C$ is the variety of commutative algebras, then $N_{q,n}(\mathcal C)=q^{n^2(n+1)/2}$.
But other examples seem to be much more difficult, for example for the varieties of Lie and associative algebras.
So my main questions (for now) are the following:
How many Lie algebras of dimension $n$ over a field with $q$ elements are there?
How many associative algebras of dimension $n$ over a field with $q$ elements are there?
I'd like to stress that I am not interested in isomorphism classes, but in the number of such algebras only (that is to say this is a problem of combinatorics and not of algebra).
Finally, I would like to remark that I have considered the possibility to write a computer program to compute some cases (example for $q=3$ and $n \leq 6$), so I could have a guess of the general answer, but in a first look I realized that this is would take too much time.
EDIT:
The answer I am expecting is an explicit formula for $N_{q,n}(\mathcal V)$ when $\mathcal V$ is the variety of associative or Lie algebras.
There's almost certainly no chance of an exact answer here... so you might want to be more specific about what kind of answer you'd like
An algebra is determined by its $n^3$-tuple, but not every $n^3$-tuple determines an algebra, right? Unless I'm off on that, maybe the first question is how many $n$-dimensional algebras there are, full stop. And then, of course, among the ones that are algebras, there will be isomorphisms. Do you want to distinguish isomorphic algebras?
Oops, it gradually sunk in that you mean "algebra" in the "universal algebra" sense, so that you genuinely meant all of the $n^3$ tuples to define an algebra. (At first I didn't register the word 'Lie' as an option indicating you didn't just want associative algebras ….) But then there's still the issue of whether to distinguish isomorphic algebras.
@LSpice: they meant an algebra over a field (https://en.wikipedia.org/wiki/Algebra_over_a_field), not necessarily associative. I'm not sure how that's related to "universal algebra"?
@SamHopkins, I shouldn't go on too long since I got it wrong, and now understand, but, to me, 'associative' is always part of 'algebra' (unless one explicitly says 'non-associative'), and 'unital' often is, and of course neither of those properties is satisfied for a random $n^3$-tuple. Since the question mentioned varieties of algebras, I figured it came from the point of view of universal algebra, which is the only place I have encountered this much more general definition. Anyway, it was my mistake, and I understand now.
Yes, I mean algebra over a field - i.e., just a vector space with a bilinear map.
As regards your edit (you want an explicit formula, not just estimates): it's clearly out of reach at the moment.
An explicit formula seems quite hopeless except maybe for very small values of $n$, say $n \le 3$?
Ok, I agree that for now explicit formulas are just a dream.
Bjorn Poonen addresses this question for commutative (associative, unital) algebras in The moduli space of commutative algebras of finite rank; asymptotically we have
$$q^{\frac{2}{27} n^3 + O(n^{8/3})}$$
such algebras (Theorem 10.9). Bjorn also gives a more precise lower bound on the dimension of the corresponding affine scheme in Theorem 9.2 which is a collection of three polynomials with leading term $\frac{2}{27} n^3$ depending on the value of $n \bmod 3$. The $\frac{2}{27}$ may seem familiar from a corresponding count of the number of finite $p$-groups and it happens for very similar reasons as he discusses in Section 10:
The approach towards both those results is to adapt the proof (begun in [Hig60] and completed in [Sim65]) that the number of $p$-groups of order $p^n$ is $p^{ \frac{2}{27} n^3 + O(n^{8/3})}$. As suggested to us by Hendrik Lenstra, there is an analogy between the powers of the maximal ideal of a local finite-rank $k$-algebra and the descending $p$-central series of a $p$-group. Although there seems to be no direct connection between finite-rank $k$-algebras and finite $p$-groups, the combinatorial structure in the two enumeration proofs are nearly identical.
He also cites An estimate of the number of parameters defining an $e$-dimensional algebra by Yuri Neretin (which is in Russian, sadly for me) as addressing the Lie and associative cases; I'm not sure if the estimates immediately carry through to a finite field but if they do the answer is the same for Lie algebras and for associative algebras it is
$$q^{ \frac{4}{27} n^3 + O(n^{8/3}) }.$$
Presumably the analogous structure for Lie algebras making the answer similar is the descending central series for a nilpotent Lie algebra. For the associative case maybe it's something like powers of the Jacobson radical?
Note also that because $\frac{8}{3} > 2$ the error term in the exponent absorbs multiplicative factors as large as $q^{O(n^2)}$ so these asymptotics hold regardless of whether or not we quotient by the action of $GL_n(\mathbb{F}_q)$ (which is equivalent to asking for the isomorphism classification), which you might see as unsatisfactorily lenient but I think these are state of the art.
Edit: The lower bound for Lie algebras is easy enough to give here; it's very similar to the argument for finite $p$-groups and for commutative algebras but, I think, simpler. We consider only 2-step nilpotent Lie algebras $L$ of some dimension $n$, which arise as a central extension
$$0 \to [L, L] \to L \to A \to 0$$
of an abelian Lie algebra $A$ (the abelianization) by another abelian Lie algebra $[L, L]$ (the commutator; I am not using fraktur here to save typing). Explicitly, the Lie bracket $[-, -]$ factors through $A$ and lands in $[L, L]$, and so the only constraint on it is that it's a surjective alternating map $\wedge^2(A) \to [L, L]$; given any such map we can construct a Lie bracket which trivially satisfies the Jacobi identity because all triple commutators vanish by 2-step nilpotence. This is a mild generalization of the construction of the Heisenberg algebra where $\dim [L, L] = 1$.
So, fixing the vector space $L$, we put a 2-step nilpotent Lie algebra structure on $L$ by first choosing a subspace $[L, L]$ we want to be the commutator and then choosing a surjection $\wedge^2(L/[L, L]) \to [L, L]$. In general the space of surjections from a f.d. vector space $V$ to a f.d. vector space $W$ admits a free action by $GL(W)$ and the quotient by this action is the Grassmannian of codimension $\dim W$ subspaces of $V$. So, setting $b = \dim [L, L]$, the space of choices we have available is the triple of choices of
a $b$-dimensional subspace $[L, L]$ of $L$,
a $b$-codimensional subspace of $\wedge^2(L/[L, L])$, and
an isomorphism between the first choice and the quotient by the second choice.
Write $a = n - b = \dim L/[L, L] = \dim A$, so that $a + b = n$. Over $\mathbb{F}_q$ there are exactly
$${n \choose b}_q { {a \choose 2} \choose b}_q |GL_b(\mathbb{F}_q)|$$
ways to make the above choices. Now our job is to find $a, b$ which maximizes this, or at least which makes it quite big since we're aiming for a lower bound. The leading term in $q$ is $q$ to the power of
$$ab + \left( {a \choose 2} - b \right) b + b^2 = \frac{a(a+1)b}{2}.$$
Subject to the constraint that $a + b = n$ this is maximized when $a \approx \frac{2n}{3}, b \approx \frac{n}{3}$, and we could be more careful depending on the value of $n \bmod 3$ if desired. Let's instead restrict to the case that $3 \mid n$ so that we can divide by $3$ exactly, and also take the liberty of dividing by $(q - 1)^b$ so that what remains is a polynomial in $q$ with nonnegative coefficients and so the leading term is a true lower bound. We get that there are at least
$$q^{ \frac{2}{27} n^3 + \frac{n^2}{9} - \frac{n}{3}}$$
2-step nilpotent Lie brackets on $\mathbb{F}_q^n$ when $3 \mid n$.
To get a lower bound on the number of isomorphism classes we quotient badly by the action of $GL_n(\mathbb{F}_q)$. At this point we can actually restore the factor of $(q - 1)^b$ we lost above (although it doesn't matter too much either way); it's not hard to show that $\frac{|GL_n(\mathbb{F}_q)|}{|GL_b(\mathbb{F}_q)|} \le q^{n^2 - b^2}$, so we can then divide by $|GL_b(\mathbb{F}_q)|$ and then by $q^{n^2 - b^2}$ to get a lower bound, whereupon what remains is a polynomial in $q$ with non-negative coefficients which can be bounded from below by its leading term again. We get that there are at least
$$q^{ \frac{2}{27} n^3 - \frac{8n^2}{9}}$$
isomorphism classes of 2-step nilpotent Lie algebras of dimension $3 \mid n$ over $\mathbb{F}_q$. It is maybe surprising that it's possible to prove a matching upper bound, at least up to leading order in the exponent; I don't know what that argument looks like in detail.
For small values of $n$ it would be feasible to not only maximize but sum over all $a + b = n$ above and so compute the exact number of 2-step nilpotent Lie brackets. The truly brave who wanted to compute the exact number of isomorphism classes could hope to apply Burnside's lemma...
The lower bound for finite groups is very similar, using 2-step nilpotent $p$-groups arising as the central extension of a f.d. $\mathbb{F}_p$-vector space by another one, etc.
In the Lie case, one chooses for $m<n$ the $(n-m)$-Grassmanian in $\Lambda^2(K^n)$. This gives as many 2-step $n$-dim nilpotent Lie algebras. It seems to be maximal around $m=2n/3$ (whence the mod 3 discussion), in which case this Grassmannian has dimension $(2/27)n^3+O(n^2)$. This also gives the lower bounds for $p$-groups (at least for odd $p$), yielding $p^{(2/27)n^3+O(n^2)}$ such groups (2-nilpotent of exponent $p$). The estimate in $S^2(K^n)$ instead of $\Lambda^2(K^n)$ seems roughly the same, yielding as many commutative associative nilpotent algebras (or local, if one requires a unit).
I'm more surprised by Neretin's lower bound $4/27$: it should come either from non-nilpotent Lie algebras, or nilpotent of unbounded nilpotency length (as otherwise it would provide for large $p$, as many $p$-groups of order $p^n$)
The $\frac{4}{27}$ confuses me also but I am very unfamiliar with the structure theory of finite-dimensional algebras. I guess such an algebra $A$ over $\mathbb{F}q$ looks like a finite product $A/J(A)$ of matrix rings $M_n(\mathbb{F}{q^i})$ extended by the Jacobson radical $J(A)$ which should in particular be nilpotent... I guess after two square-zero extensions $A/J^3(A) \to A/J^2(A) \to A/J(A)$ complicated stuff is happening?
By the way, the existence of a component of dimension $\ge 2n^3/27$ in the variety of Lie algebra laws in dimension $n$ was observed in: M. Vergne, Réductibilité de la variété des algèbres de Lie nilpotentes. (French)
C. R. Acad. Sci. Paris Sér. A-B 263 (1966), A4–A6.
@Qiaochu Yuan Thank you for your answer and for the references. I'll try to read it carefully later.
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2025-03-21T14:48:31.983504
| 2020-09-08T01:31:54 |
371151
|
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|
Stack Exchange
|
Boundedness of Calderon-Zygmund type operator
I am trying to prove the following fact.
Suppose $\varphi\in C_c^\infty(\mathbb{R}^1)$. Define
$$T(u)(x):=P.V.\int_{\mathbb{R}^1}\frac{\varphi(x)-\varphi(y)}{|x-y|^{\frac32}}u(y)dy$$
where P.V. means principle value. Then (I guess)
$$||T(u)||_{L^2{(\mathbb{R}^1)}}\leq C||u||_{L^2(\mathbb{R}^1)}.$$
My attempt: Notice $|T(u)|\leq C\int_{\mathbb{R}^1}\frac{|u(y)|}{|x-y|^{1/2}}dy=XI_{\frac12}(|u|)$, here $I_{\alpha}$ is the Riesz potential. Using the boundedness of Riesz potential $I_{1/2}$ from $\mathcal{H}^1(\mathbb{R}^1)$ to $L^2(\mathbb{R}^1)$, one gets
$$||T(u)||_{L^2(\mathbb{R}^1)}\leq C||u||_{\mathcal{H}^1(\mathbb{R}^1)}.$$
My approach does not give the bound of in $L^2\to L^2$. I think the reason is that I did not use the cancellation of the kernel. I am hoping to get some analogous results to the Calderon-Zygmund operator. However, my knowledge about that is limited to the classical theory of Calderon-Zygmund operator.
Does anyone know the references study this type of operator? Or any idea how to prove the fact.
Or my guess is wrong? Any comment is welcome.
You are not using that $\varphi$ is compactly supported. Apply the Schur test. ;-)
@fedja Got you. Thank you!!
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2025-03-21T14:48:31.983602
| 2020-09-08T02:09:46 |
371152
|
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|
Stack Exchange
|
Commutator estimates regarding pseudo-differential operators
Forgive me if what I'm asking is too naive for Mathoverflow.
Given a compact Riemannian manifold $(M,g)$ with Hodge Laplacian $\Delta$. Recall that the Riesz potential $(-\Delta)^{-\frac{1}{2}}$ defined by
\begin{equation}
(-\Delta)^{-\frac{1}{2}} = \int_{0}^{\infty}e^{-t\Delta}\frac{dt}{\sqrt{t}}
\end{equation}
is a pseudo-differential operator on $M$ of order one. I'm wondering whether there exists a commutator identity/estimates regarding the Riesz potential and the Riemannian gradient $\nabla$. In other words, let $r$ be an $(s,t)$-tensor, then what can we say about $[\nabla, (-\Delta)^{-\frac{1}{2}}]r$ in terms of the norm of an $(s+1,t)$-tensor? When commuting two full covariant derivatives, the commutator is usually controlled by the curvature term. But I'm not familiar with the pseudo-differential version of the story. Any insight or help would be appreciated.
|
2025-03-21T14:48:31.983692
| 2020-09-08T02:37:37 |
371153
|
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"Mark Schultz-Wu",
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|
Stack Exchange
|
What is the term for convoluting but scaling the time domain instead of shifting?
Given that the convolution definition as far as I am aware is:
$(f*g)(t) = \int_{-\infty}^\infty f(\tau)g(t-\tau)d\tau$
Here I see that the functions f and g are being shifted across the time domain. I wonder if there is an equivalent for the scaling the time domain such at:
$(f?g)(t) = \int_{-\infty}^\infty f(\tau)g(t\tau)d\tau$
In my mind this would be similar to how a Fourier transform works (e.g. $g(t\tau)$ would be $cis(t\tau)$. I am not sure what kind of applications this could have in non-periodic data but I am still curious nontheless if there is a proper term for it.
The "multiplicative" version of the Fourier transform is the Mellin transform. This has a corresponding convolution theorem (ctrl+F "mellin convolution"), which obeys $\mathcal{M}[f]\mathcal{M}[g] = \mathcal{M}[f\ast_{\mathrm{Mellin}}g]$. In generally one can define an analogue of the Fourier transform (which leads to a convolution theorem) for examining symmetries other than "Shifting" or "Scaling". This is done in Harmonic Analysis.
@Mark Is there something analogous for two separate functions? After looking at the Mellin transform it looks like this is for a single function much like a fourier transform or laplace transform.
The convolution can be defined for functions on groups, typically abelian locally compact groups with Haar measure. The cases you mention are those of the real line, with addition, and the positive reals with multiplication. The general situation is covered in any text on abstract harmonic analysis, as mentioned above.
@Saxpy The Mellin transform, like the Fourier transform, is defined for a single function. Mellin convolution, like Fourier convolution, is defined for pairs of functions. The particular convolution is $(f\ast_{\mathrm{Mellin}}g)(y) = \int_{0}^\infty f(x) g(y/x) \frac{\mathrm{d} x}{x}$, which can be found on the page I linked (or here). As mentioned, you can preform this analogous setup over other locally compact abelian groups (other than $(\mathbb{R}, +)$ or $(\mathbb{R}^\times, \times)$).
@Mark I see. I need to do a bit more studying as I am not familiar with the term locally compact abelian group. That being said, Mellin convolution was in fact the term I was looking for, thank you. I will look into this topic further.
|
2025-03-21T14:48:31.983877
| 2020-09-08T04:03:43 |
371155
|
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|
Stack Exchange
|
Collatz conjecture in all its variants
There are all kinds of execution variants to the collatz conjecture for when hitting an odd number:
$3n+1$ or $3n+3^a$ or $1.5n + 0.5$ or $1.5n + 1.5$... . The assumption is: proving any of them will prove all of them.
I have been experimenting with different executions and stumbled upon the following: if divisible by $3$: divide by $3$, else if divisible by $2$: divide by $2$, else: execute $5n +1$. see results here
It seems to follow the same usual collatz patterns of hitting $1$ eventually, whereas any other variant such as $7n+1$, & $11n +1$... seem to have infinite sequences (never hitting $1$).
My question is:
If the original conjecture ($3n+1$) is proven, will it also prove what I have stumbled upon or will it be a totally different conjecture?
Update: I am well aware of the Collatz patterns for when there are only 2 rules, however in my experiment there are 3 rules and if the third rule is: execute $5n +1$, all sequences seem to collapse all the way to $1$.
Hard to say without seeing the proof. The basic questions are which are the key things that make the proof work? Do those things apply to your modified Collatz sequence? As you can imagine, this is not something that can be answered without having a proof of Collatz in front of you.
I have attached a link for the first 200 integers, however it seems consistent for the first 1000 integers (I can't submit it because the script keeps braking)
For any piecewise linear Collatzish problem, there's a simple heuristic which suggests whether sequence goes off to infinity or collapses down to small cycles. There is plenty of literature to consult on this.
I understand that, but this heuristic seems to suggest a collapse all the way to $1$ and the set of instructions are slightly different from a "regular" Collatzish problem, In a "regular" Collatzish problem, you have 2 instructions rules: when odd and when even. Here there are 3 instructions rules.
It's very difficult to say unless two problems are provably equivalent. For instance it was widely expected that the twin prime problem and the binary Goldbach problem are basically equivalent (influenced by J.R. Chen's monumental work which used the same method to obtain roughly the same almost-result for both), but perhaps this has changed since we now know bounded gaps between primes while similar progress towards Goldbach has not been made.
Collatz-like problems with three or four or any finite number of rules, have been studied, heuristics formulated, etc. The literature is enormous, please have a look at it. You can start with Jeff Lagarias' book, it will point you toward other resources. Also, if you want to be sure I see a comment intended for me, you have to put @Gerry in it.
|
2025-03-21T14:48:31.984084
| 2020-09-08T04:22:18 |
371157
|
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|
Stack Exchange
|
A primality criterion for specific class of $N=4kp^n+1$
Can you provide a proof for the following claim:
Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4}\right)^m\right)$ .
Let $N= 4kp^n+1 $ such that $p$ is a prime number greater than $3$ , $k$ is an odd natural number , $3 \nmid k$ , $n\ge1$ and $4k <p^n$. Let $S_i=P_p(S_{i-1})$ with $S_0=P_{k}(10)$, then $N$ is a prime iff $S_{n} \equiv 0 \pmod{N}$ .
You can run this test here. I have verified this claim for $k \in [1,300]$ with $p \in [5,109]$ and $n \in [1,50]$ .
How did you come up with this conjecture?
@MiloMoses This was inspired by Lucas-Lehmer-Riesel test.
It is rather standard, for such tests, to prove that if $N$ is prime then it divides $S_n$, but the opposite direction looks difficult, and possibly even not true. Say, if $N$ is Carmichael number plus some other restrictions, it is a counterexample.
|
2025-03-21T14:48:31.984181
| 2020-09-08T04:35:35 |
371158
|
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|
Stack Exchange
|
Neighbor count in sphere packing in N dimensions
So I'm really interested in building a mathematical model for how powerful computer chips could be given extra spatial dimensions. Obviously this is a squishy problem, since "computer chips" is not a rigorously defined entity. But I want to at least characterize the shape of the curve - i.e. linear, polynomial, exponential, super-exponential.
I think the starting place to start thinking about this is how many neighbors a sphere has in an optimal packing in N dimensions (because chips want to pack logic circuits in this way). To my surprise, there's not a closed form solution to this problem, and the optimal packing in higher dimensions in general is not even known.
That's crazy and surprising! Why is this the case? Looking for some intuition here. I'm not a real mathematician, please add color. Is there a connection with this packing and the sporadic groups? I browsed a couple of papers and some common characters appear, like the Leech Lattice.
Why is this the case? Well, for context, we didn't know(1) until the late 90s the optimal packing in 3d. That we now know it in 8 and 24 dimensions is mildly astounding, and due to properties extra special to those dimensions. [(1) depending if you believe that Hales' original proof was enough or not; the referees of his paper couldn't certify his code, hence the more recent formalised proof]
I don't know what you mean by "this packing" in the last paragraph, but I guess you mean that we know the densest packing in 8 and 24 dimensions, as I said. The connections to sporadic (I guess you mean finite simple) groups are not obvious. The symmetries of the Leech lattice form a finite group that maps two-to-one onto the sporadic simple group $Co_1$. It also connects with the other sporadics in the "2nd generation of the so-called Happy Family".
If you can get a copy of Sphere packing, lattices and groups (https://doi.org/10.1007/978-1-4757-6568-7) it has a lot of info, not at the right level, perhaps, but with lots of references. Cohn has a nice 2014 survey https://arxiv.org/abs/1603.05202, predating the work Viazovska (+et al) establishing the d=8,24 cases. See Fig 1 on p53
|
2025-03-21T14:48:31.984485
| 2020-09-08T04:57:20 |
371159
|
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|
Stack Exchange
|
Value of an integral
I need to verify the value of the following integral
$$ 4n(n-1)\int_0^1 \frac{1}{8t^3}\left[\frac{(2t-t^2)^{n+1}}{(n+1)}-\frac{t^{2n+2}}{n+1}-t^4\{\frac{(2t-t^2)^{n-1}}{n-1}-\frac{t^{2n-2}}{n-1} \} \right] dt.$$
The integrand (factor of $4n(n-1)$) included) is the pdf of certain random variable for $n\geq 3$ and hence I expect it be 1. If somebody can kindly put it into some computer algebra system like MATHEMATICA, I would be most obliged. I do not have access to any CAS software.
PS:-I do not know of any free CAS software. If there is any somebody may please share
Your integral equals $1$. See the proof below.
I suppose that the round bracket in $t^4{\frac{(2t-t^2)^{n-1}}{n-1})$ is redundant (the one on the right). But I did not want to edit this, since it is better if you clarify what is the intended function. I am referring to the current revision.
@MartinSleziak: I omitted the closing round bracket (which has no opening counterpart), and proved that the resulting expression equals $1$. See below.
The question was posted also on [math.se]: Value of an integral in the interval (0,1). I will add a pointer to this post on meta: Cross posts to Math SE.
If it is a pdf, this itself proves that the integral equals 1, "expect" is underestimating.
@FedorPetrov: I thought the same. Then I thought maybe the OP just wanted to double check, so I went ahead and proved it. It was a fun exercise.
@FedorPetrov what pdf is it? (any name)
@vidyarthi I do not know, that's the OP claim.
@FedorPetrov oh! I didnt see it. Then wasnt it too trivial?
@vidyarthi: Probability density functions have integral $1$ on $\mathbb{R}$. However, the OP did not explain why the integrand is a probability density function.
Thank u all of you for your valuable comments and wonderful responses.And also my apologies for cross posting.Actually I do not know how to add a link from one SE to another.Next time i post i will bear the rules in mind. @vidyarthi I know the inegral has to be 1,I only wanted to double check to see whether my calculations are correct.This is the pdf of the range of a certain sequence of random variables. I am especially thankful to @{GH from MO} for making it elegantly simple
@sajjadveeri you can create a free basic account through Wolfram's cloud portal https://www.wolframcloud.com/. It lets you run your own Mathematica notebooks.
The integral can be rewritten as
\begin{align*}
I&=\frac{n(n-1)}{2}\int_0^1\frac{t^{n-2}(2-t)^{n+1}-t^{2n-1}}{n+1}-\frac{t^n(2-t)^{n-1}-t^{2n-1}}{n-1}\,dt\\[6pt]
&=\frac{1}{2n+2}+\frac{n(n-1)}{2}\int_0^1\frac{t^{n-2}(2-t)^{n+1}}{n+1}-\frac{t^n(2-t)^{n-1}}{n-1}\,dt.
\end{align*}
Integrating by parts, we obtain
$$\int_0^1\frac{t^{n-2}(2-t)^{n+1}}{n+1}\,dt=\frac{1}{n^2-1}+\int_0^1\frac{t^{n-1}(2-t)^n}{n-1}\,dt.$$
Therefore,
\begin{align*}
I&=\frac{1}{2}+\frac{n}{2}\int_0^1t^{n-1}(2-t)^n-t^n(2-t)^{n-1}\,dt\\[6pt]
&=\frac{1}{2}+\frac{1}{2}\int_0^1(t^n(2-t)^n)'\,dt=\frac{1}{2}+\frac{1}{2}=1.
\end{align*}
P.S. You can use SageMath and WolframAlpha for symbolic calculations. Both are free.
It seems that yesterday I inadvertently put a -1 to your answer. the only way to correct is making a trivial edit. sorry!
@PietroMajer: This is very kind of you, thanks!
It seems your conjecture is true. Mathematica gives the result
$$
(1 + 4^n (-1 + n) n \mbox{Beta} [1/2, -1 + n, 2 + n] -
4^n n (1 + n) \mbox{Beta} [1/2, 1 + n, n])/(2 (1 + n))
$$
in terms of the incomplete Beta function, and putting in random integers $\geq 3$ always yields 1 (I haven't managed to get Mathematica to spit that out as a general result for arbitrary $n$).
I proved the conjecture in my response.
@GHfromMO - yes, very nice, you have my upvote. We posted simultaneously. I considered deleting my answer, but decided against it, since at least I answered the OP's literal request :-D
You can use CoCalc.
For instance, type
integral(x^2,x)
and get
1/3*x^3
It also permits symbolic parameters.
Input:
f(x,n)=x^2+n
integral(f(x,n),x)
Output:
1/3*x^3+n*x
|
2025-03-21T14:48:31.984795
| 2020-09-08T05:06:53 |
371160
|
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}
|
Stack Exchange
|
Finding super(sub)-harmonic functions for an elliptic operator
I am looking for a super(sub) harmonic function for an elliptic operator.
Let $n$ be a positive integer. We denote by $(\cdot,\cdot)$ and $|\cdot|$ the standard inner product and norm on $\mathbb{R}^n$, respectively. We denote by $U \subset \mathbb{R}^n$ the open unit ball centered at the origin. The elliptic operator $\mathcal{L}$ is defined as follows:
\begin{align*}
\mathcal{L}f(x)=(1-|x|^2)\Delta f-c((x-\theta),\nabla f),\quad x \in U,
\end{align*}
where $\theta \in U$ and $c$ is a positive constant.
My question. Can we find a smooth and nonnegative function $f\colon U \to \mathbb{R}$ and $\varepsilon>0$ such that $\mathcal{L}f \ge \varepsilon$ on $U $ ?
Needless to say, the function $f$ may depend on $\theta$. If necessarily, the ranges of $c$ and $|\theta|$ may be limited. If we find such a function, in a sense, we can understand the boundary behavior of the diffusion process associated with $\mathcal{L}$.
If $\theta=0$, we can find such a function. Indeed, if we set $f=\alpha^{-1}\{1-(1-|x|^2)^{\alpha}\}$, $\alpha \in (0,1)$ (there may be something simpler than this), we obtain that
\begin{equation}
\mathcal{L}f=4\{(1-c/2)-\alpha \}|x|^2(1-|x|^2)^{\alpha-1}+2n(1-|x|^2)^{\alpha}.
\end{equation}
Therefore, if $c<2$ and $\alpha \in (0,1-c/2)$, we find that $f$ possesses the desired property (in fact, $c=2$ is a border in a sense).
If $\theta \neq 0$, however, I could not find a function satisfying the above conditions.
If you find one, please let me know.
Do you know what happens if $\theta=0$ and $c\geq 2$?
@GiorgioMetafune Thank you for your comment. There will be no function satisfying the conditions if $c \ge 2$ and $\theta=0$. This is based on probabilistic considerations.
It seems that your barrier works if $c(1+|\theta|) <2$; do you find the same?
@GiorgioMetafune I don't find it. Could you tell me the reason?
@GiorgioMetafune Your result is consistent with the case of $\theta=0$.
I find $Lf=\left(4(1-c/2-\alpha)r^2+2cx\cdot \theta +2n (1-r^2)\right )(1-r^2)^{\alpha-1}$. Then $(4-4\alpha -2c)r^2+2cx\cdot \theta +2n (1-r^2) \ge (4-4\alpha -2c)r^2-2rc|\theta|+2n (1-r^2)$ and this last expressione is nonegative if $c(1+|\theta|) <2$, choosing $\alpha$ small as you did.
@GiorgioMetafune Thank you for letting me know. $f=\alpha^{-1}(1-|x|^2)^{\alpha}$ ? This seems to be the simplest.
Yes, exactly. The same you wrote
If you have a second order elliptic operator L on a smooth noncompact connected manifold then you can always find a smooth function f>0 such that Lf > 0 . See the paper by Napier and myself in L'Enseignment Mathematique vol 50 2004 pages 367-390 .
Here the problem is the degeneracy of the elliptic part at the boundary. I guess that your result holds for uniformly elliptic (or strictly elliptic) operators, that is when the ellipticity constant is bounded above and below from 0.
The result of Napier and Myself does not assume strict or uniform ellipticity. It uses a Runge style argument.
I see, but then I do not understand the answer of @sharpe to a comment above, saying that such a function does not exist if $\theta=0, c \ge 2$. Unless he assumes boundedness and your construction yields (possibly) unbounded functions.
Yes the function will be unbounded. In fact you can make it so that sublevel sets are relatively compact.
@MohanRamachandran Thank you for your comment. If $\theta=0$ and $c\ge 2$, I thought that functions with the above conditions do not exit. But it was different.
|
2025-03-21T14:48:31.985040
| 2020-09-08T08:13:06 |
371166
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371166"
}
|
Stack Exchange
|
Computing Hodge numbers by point counting
In the lecture note of Bhatt from Arizona winter school 2017, there is an exercise which claims if X is a proper smooth scheme defined over $\mathbb{Z}[1/N]$ and if there is a polynomial $P$ such that for every prime $p$ coprime to $N$ we have $X(F_p)=P(p)$ then the Hodge numbers $h^{i,j}=0,i\not =j $
I do not know how to attack this problem because if you want to use zeta functions and Weil conjectures you need the number of points of X over all finite fields. But I do not have any counter example. So is there a typo in this exercise or can someone hint how to prove the claim ?
Let $H = \bigoplus_i H^i(X_{\overline{\mathbf{Q}}}, \mathbf{Q}_\ell)$, a graded Galois rep. The assumption on $X(\mathbf{F}_p)$ tells you the (graded) trace of ${\rm Frob}_p$ on $H$. By Chebotarev density, the conj classes ${\rm Frob}p$ ($p>N$) are dense in the Galois group of $\mathbf{Q}$. Now using the polynomial $P$ you can build a graded Galois rep $H'$ which is a direct sum of $\mathbf{Q}\ell(i)$'s which has the same Frobenius traces for $p>N$. Therefore $H$ and $H'$ have the same semisimplification. Finally, $p$-adic (for $p=\ell$ large) Hodge theory gives you the Hodge numbers.
@PitorAchinger I'm not familiar with graded representations. Is the equality of graded trace is enough to identify two such representations?if so why both of the notes and in the original argument of katz use weil conjectures
yes, the Weil conjectures are needed to take care of the grading. Sorry for the sloppiness.
|
2025-03-21T14:48:31.985177
| 2020-09-08T08:35:20 |
371167
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371167"
}
|
Stack Exchange
|
Berkovich Integration on algebraic curves
Berkovich developed a theory of integrating one-forms on his analytic spaces in his book "Integration of One-forms on $P$-adic analytic spaces". As this book is difficult to digest for me, I am wondering how this theory breaks down, if one just considers analytifications of smooth projective curves. More specifically, I am wondering if his theory can be used to translate some of the result over $\mathbb C$ (which I will recall below) to the $p$-adic world.
In the complex world the moduli space $\Omega M_g$ of pairs $(X, \omega)$, where $X$ is a smooth projective curve of genus $g$ and $0 \neq \omega\in H^0(X, \Omega_X)$ is a holomorphic differential on $X$, is a well studied object. The space $\Omega M_g$ is a complex orbifold and the points are called translation surfaces. One first result is that every translation surface can be represented as a finite union of polygons in the complex plane with edge identifications. One gets this equivalence by integrating (using $\omega$) along paths between the zeros of $\omega$.
$\Omega M_g$ comes equipped with a natural stratification: Let $\kappa$ be a partition of $2g-2$ (the number of zeros of $\omega$ counting multiplicity). Then $\mathcal H(\kappa)$ is the subset of $\Omega M_g$ containing the points $(X,\omega)$ such that the order of zeros of $\omega$ corresponds to the partition $\kappa$. This subset $\mathcal H (\kappa)$ is itself a complex orbifold. Roughly speaking, charts are given by integrating the same paths with respect to different differentials.
If you want more details on this topic, I would suggest having a look at a nice overview paper by: Alex Wright
I would like to bring those two results over to the $p$-adic world ($\mathbb C_p$), so let me restate my question:
Using Berkovich integration on the analytification of a projective smooth curve $X$, is there a nice geometric description of the pair $(X^{an},\omega)$ (where $\omega$ is a global section of the differentials on $X^{an}$)?
On the strata of $\Omega M_g$ (which exists algebraically) can we find some kind of "coordinates" by integrating using the differentials?
I would appreciate any kind of feedback, whether those results are clearly unobtainable or might very well be possible.
Just a comment on your first question. Which kind of ``nice geometric description'' do you have in mind? I am not sure what the analogue of a polygon should be in Berkovich geometry and gluing edges can certainly be problematic. Maybe gluing discs along their boundaries? Do you have any examples where you see how things work?
@JérômePoineau Unfortunately I don't have any examples. One could certainly make something similar as the second definition of translation surface in the paper of Wright: By locally integrating using $\omega$ (as one would do over $\mathbb C$) one can get a $p$-adic manifold, i.e. charts to $\mathbb C_p$. But since $\mathbb C_p$ is totally disconnected this is not really usefull I think...
|
2025-03-21T14:48:31.985386
| 2020-09-08T09:32:04 |
371171
|
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"url": "https://mathoverflow.net/questions/371171"
}
|
Stack Exchange
|
Euler systems over abelian number fields
Im confused with the following statement:
Coleman’s conjecture concerning circular distribution imply that Euler systems over abelian number fields arise in “an elementary” way from the theory of cyclotomic units.
What is that “elementary way”?
I would appreciate if someone could help.
|
2025-03-21T14:48:31.985441
| 2020-09-08T10:34:37 |
371172
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371172"
}
|
Stack Exchange
|
Growth of (integral of) Laplace transform of a function of compact support as $Re \to -\infty$
Let $f:[0,\infty)\to \mathbb{R}$ be supported on $[0,1]$, with $\int_0^1 f(x) dx = 1$. Let $\mathcal{L} f$ be its Laplace transform. How slowly may
$$\int_{-\infty}^\infty |\mathcal{L} f(\sigma+i t)| dt$$
grow as $\sigma\to -\infty$? It is clear it could be $\ll e^{\epsilon |\sigma|}$ (just let $f$ be supported on $[0,\epsilon]$). Could it grow polynomially on $|\sigma|$? Linearly on $|\sigma|$?
It has exponential growth: if your integral is $O(e^{-\sigma\epsilon}), \sigma\to -\infty,$ then the
support of your function $f$ is contained in $[0,\epsilon]$. This follows
from the inversion formula for the Laplace transform:
$$f(x)=\lim_{r\to\infty}\frac{1}{2\pi i}\int_{b-ir}^{b+ir}Lf(s)e^{sx}ds,$$
where the integration is on any vertical line $\Re s=b$.
Under your assumptions,
$$|f(x)\leq\frac{1}{2\pi} e^{-b\epsilon}e^{bx},$$
for every $b<0$, therefore, by letting $b\to-\infty$ we obtain $f(x)=0$ when $x>\epsilon$.
The Fourier inversion formula says that $$f(x)e^{-\sigma x} = (2\pi)^{-1} \int \mathcal{L}f(\sigma+it) e^{itx} dt$$
so if your integral is $I(\sigma)$, one sees that
$$|f(x)| \le (2\pi)^{-1} e^{\sigma x} I(\sigma)$$
It follows that if $I(\sigma)$ is $O(e^{\varepsilon\sigma})$, then f is supported on $[0,\varepsilon]$. So that can't hold for ALL $\varepsilon$. Thus the behavior that you mention is best possible.
|
2025-03-21T14:48:31.985547
| 2020-09-08T10:42:48 |
371173
|
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"Carlo Beenakker",
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"Richard Stanley",
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|
Stack Exchange
|
Evans conjecture for symmetric latin squares
The Evans conjecture ( which was proved later by Smetaniuk) states that for any $n$, if at most $n-1$ entries of a partial $n\times n$ latin square are filled, it can be completed to the full latin square.
My question pertains to whether this is applicable to symmetric (or commutative) latin square? That is, given $n\times n$ latin square, with at most $n-1$ prefilled entries ( which are symmetric with respect to diagonal), can it be completed to a symmetric latin square? Note that this corresponds to pre- total coloring of a complete graph of order $n$ with at most $n-1$ previous filled colors. Any hints? Thanks beforehand.
No, when it comes to symmetric latin squares it is no longer true that as many as $n-1$ cells can be prescribed unconditionally. This is explained in the Ph.D. thesis of Matthew Henderson.
The key point here is that in a symmetric latin square, precisely
because of the symmetry, every symbol $\sigma$ occurs an even number
of times in cells outside of the main diagonal. Therefore, as every
symbol $\sigma$ occurs $n$ times in total it follows that the number
of cells of the main diagonal containing symbol $\sigma$ is congruent
to $n$ modulo 2. A partial latin square can be incompletable because
there are more symbols which occur on the main diagonal a number of
times (zero included) incongruent to $n$ modulo 2 than there are
empty cells on the main diagonal.
If this obstruction does not occur the diagonal is called "admissible". For $n$ odd a diagonal is admissible if and only if no symbol occurs more than once on it.
The generalization of Evans theorem to symmetric latin squares by Andersen and Hilton is that the symmetric latin square is completable if the diagonal is admissible and at most $n-1$ cells in total are prescribed.
In response to the comment by Richard Stanley: it is not sufficient to only count cells on or above the main diagonal, as this counter example shows. The diagonal is admissible ($n=5$ is odd and no symbol appears more than once on the diagonal), and the number of prescribed cells on or above the main diagonal is $n-1=4$, and yet the latin square is not completable.
Is this the only obstruction to completing a partial symmetric latin square with $n-1$ symbols? This might be the correct analogue of Evans conjecture for symmetric latin squares. Even more strongly, is it possible that you can specify $n-1$ symbols on or above the main diagonal, as long as you satisfy: (1) when you reflect the symbols about the main diagonal, we have a partial latin square, and (2) Carlo's condition?
The problem mentions "Evans conjecture", this answer mentions "Evans theorem". Please, @RichardStanley or anyone else, clarify for me the relation (if any) between the two.
@GerryMyerson --- the completability condition for nonsymmetric Latin squares was conjectured by Evans in 1960 and proven by Smetaniuk in 1981.
OK, so, there isn't an Evans theorem?
I'm not sure what the common practice is here, if a conjecture is proven can we then change its name into a theorem? Fermat's last theorem was called a conjecture in older books, but I guess it would be strange to keep calling it a conjecture now. There could be a confusion with other theorems of Evans, but the only one I found was the Evans-Krylov theorem.
Surely it makes more sense to call it Smetaniuk's Theorem?
|
2025-03-21T14:48:31.985798
| 2020-09-08T11:07:08 |
371175
|
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"url": "https://mathoverflow.net/questions/371175"
}
|
Stack Exchange
|
Factorization of a bilinear matrix-valued function
Suppose that $F(u, v) = \sum_{i}\sum_j u_i v_i C_{ij}$ is a bilinear matrix-valued function,
where $C_{ij}$ are known matrices.
Is there a relatively easy way to factorize $F$ so that
the $u$ and $v$ variables are separated? That is, to find matrices $A_i$ and $B_j$ such that
$F(u,v) = (\sum_i u_i A_i)(\sum_j v_j B_j)$.
|
2025-03-21T14:48:31.985853
| 2020-09-08T11:55:58 |
371181
|
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"Francesco Polizzi",
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"YCor",
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|
Stack Exchange
|
CCT groups of order $\leq 32$
A finite, non-abelian group $G$ is said to be a center
commutative-transitive group
$($or a CCT-group, for short$)$ if commutativity is a transitive
relation on the set on non-central elements. In other words, if $x, \, y,
\, z \in G-Z(G)$ and $[x, \, y]=[y, \, z]=1$, then $[x, \, z]=1$.
A quick search with GAP4 allows to prove the following
Proposition. Let $G$ be a non-abelian finite group with $|G| <32$; then $G$ is a CCT-group, unless it is isomorphic to $S_4$. In the case $|G|=32$, there are
precisely seven groups that are not CCT: the two extra-special groups (whose nilpotency class is $2$), and five further groups having nilpotency class $3$.
I tried to give a computer-independent proof, but it turned very soon into a lenghty and messy case-by case analysis. So let me ask the following
Question.
Is there any short conceptual proof of the proposition above?
Does the result in the proposition above appear somewhere in the literature?
It's quite easy to check for groups whose order is a product of at most 3 primes (equal or not). Below $32$ this only excludes $16$ and $24$.
May I ask you your argument?
@YCor: oh, right, thanks. In fact, a characterization of CCT-groups is that the centralizer of any non-central element is abelian, and this is true (by your argument) for groups whose order is the product of at most three primes.
Eventually I posted an answer, also covering order $p^4$.
The paper D. M. Rocke, p-groups with abelian centralizers, Proc. London Math. Soc. 30 (1975), 55-75, describes, to a certain extent, finite $p$-groups with your property. Abdollahi et al. (Non-commuting graph of a group, J. Algebra 298 (2006), 468-492) call these groups AC-groups, and give some information about them. Otherwise, CCT-groups are precisely groups with the property that all centralizers of non-central elements are abelian. This can be used together with GAP to efficiently scan through the database of small groups and look for the CCT ones.
Proposition.
(a) Let $G$ be a group of order a product of $\le 3$ primes (possibly equal). Then $G$ is CCT (the smallest numbers not of this form are $16$, $24$, $32$, $36$, $48$, $54$, $60$).
(b) Let $G$ be a group of order $p^4$, $p$ prime. Then $G$ is CCT.
Proof. (a) Suppose by contradiction that $G$, of order a product of $\le 3$ (possibly equal) primes is not CCT. So there are $x,y,z$ with $y$ non-central, such that $x,z$ don't commute and both commute with $y$. Let $C$ be the centralizer of $y$: its order is a product of at most two primes, and it has a nontrivial center (as $1\neq y\in C$), so $C$ is abelian. Contradiction as $x,z\in C$.
(b) Let $x,y,z$ be as in (a). Then they generate a non-abelian subgroup $N$, which is not all of $G$, since $y$ is not central in $G$. Hence $N$ has order $p^3$, and hence its center $Z$ has order $p$. As a subgroup of index $p$ in a finite $p$-group, it is normal. The $G$-action by conjugation on $N$ preserves $Z$, whose automorphism group has order $p-1$. Hence the $G$-action on $Z$ is trivial, so $y$ is central, contradiction.
Corollary. Every group of order $<32$ and $\neq 24$ is CCT.
Well, you are looking at small groups, so you might expect to have to do some case-by case analysis. But here are a few general remarks. A direct product $A \times B$ is a CCT group if and only either if $A,B$ both are, or if one of $A,B$ is Abelian and the other is CCT. So, a non-Abelian nilpotent group is a CCT group if each if its Sylow subgroups is CCT or Abelian (but not all are Abelian).
A dihedral group $D$ is a CCT group. This is clear if $|D| = 2n$ with $n$ odd. If $D = \langle t \rangle N$ with $[D:N] = 2$ and $t$ inverting each element of $N$, then $Z(D)$ has order $2$, and $Z(D) \leq N$. Each element $x \in N \setminus Z(D)$ has centralizer $N$, while each element $y \in D \setminus N$ has centralizer $\langle y \rangle Z(D)$.
Any Frobenius group with Abelian kernel and Abelian complement is a CCT group (recall that a Frobenius group is a group of the form $G= KH$ with $K \lhd G$ and $H \cap H^k = 1$ for each $k \in K \setminus \{1\}$.
Any non-Abelian $p$-group whose maximal subgroups are all Abelian is a CCT group.
|
2025-03-21T14:48:31.986158
| 2020-09-08T12:23:15 |
371184
|
{
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|
Stack Exchange
|
Zariski closure and complex curve selection
Let $X\subset \mathbb{C}^n$ be a quasi affine variety, and let $x\in \overline{X}\setminus X$. By the real analytic curve selection lemma, there exists a real analytic curve $\gamma:[0,\epsilon)\to \bar{X}$ with $\gamma(0)=x$ and $\gamma((0,\epsilon))\subset X$.
My question is if is also true that there exists a map $Spec(\mathbb{C}[[t]])\to \overline{X}$ satisfying the analogous properties.
If I've got my terminology right, a very, very special case of this is Lemma 3.1 and Theorem 3.4 of Kempf - Instability in invariant theory.
What am I missing? I would think that there has to be a complex algebraic curve contained in $\bar X$, not contained in $X$, containing the point $x$; and that normalizing this would provide not just a formal curve but a complex analytic curve with the analogous property..
@TomGoodwillie You are not missing anything, this is I think the most concise argument to show the claim.
Depending possibly on exactly how you define everything I think the answer is yes in a more general situation:
$\textbf{Proposition:}$ Let $\overline{X}$ be a finite type scheme over $\mathbb{C}$, and $X \subset \overline{X}$ a dense open, and $x \in \overline{X} \setminus X$ a closed point. Then there exists a map $\mathrm{Spec}(\mathbb{C}[[t]]) \to \overline{X}$ such that $(t) \mapsto x$ and $(0) \mapsto y$ for some $y \in X$ which will be the generic point of a curve passing through $x$.
Notice that any map $\mathrm{Spec}(\mathbb{C}[[t]]) \to \overline{X}$ factors through $\mathrm{Spec}(\mathbb{C}[[t]]) \to \mathrm{Spec}(\mathcal{O}_{X,x}) \to \overline{X}$ where $\mathcal{O}_{X,x} \to \mathbb{C}[[t]]$ is a local map (such that $(t) \mapsto x$). Let $A = \mathcal{O}_{X,x}$.
There is a nonempty principal open $D(f) \subset X \cap \mathrm{Spec}(\mathcal{O}_{X,x})$ for some $f \in A$ with $f \notin \mathrm{nilrad}(A)$ but $x \notin D(f)$ since $x \in \overline{X} \setminus X$. Now we apply the following lemma:
$\textbf{Lemma:}$ Let $(A, \mathfrak{m})$ be a Noetherian local ring with finite dimension $\mathrm{dim}(A) \ge 1$ and let $f \in \mathfrak{m}$ with $f \notin \mathrm{nilrad}(A)$. Then there exists a prime ideal $\mathfrak{p} \subset A$ such that $f \notin \mathfrak{p}$ and $\mathrm{dim}(A / \mathfrak{p}) = 1$.
Proof: there is a prime $\mathfrak{p}_0 \subset A$ with $f \notin \mathfrak{p}_0$ because $f \notin \mathrm{nilrad}(A)$. Therefore replacing $A$ by $A / \mathfrak{p}_0$ we can assume $A$ is a domain. Now we proceed by induction on $\mathrm{dim}(A) = n$. If $\mathrm{dim}(A) = 1$ then we have prime ideals $(0) \subsetneq \mathfrak{m}$ so taking $\mathfrak{p} = (0)$ the result follows. Otherwise, choose a maximal length chain of primes,
$$ (0) \subsetneq \mathfrak{p}_1 \subsetneq \mathfrak{p}_2 \subsetneq \cdots \subsetneq \mathfrak{p}_n = \mathfrak{m} $$
Then there are infinitely many prime ideals $\mathfrak{p}_1'$ strictly between $(0)$ and $\mathfrak{p}_2$ which all must have height one since this chain is maximal length (Kaplansky, Commutative Rings, Thm. 144). And thus some $\mathfrak{p}_1'$ does not contain $f$ (Kaplansky, Commutative Rings). Therefore, $A' = A / \mathfrak{p}_1'$ satisfies the hypotheses and $\mathrm{dim}(A') = n - 1$ so we conclude.
Returning to the propositon, we get a prime $\mathfrak{p} \in D(f) \subset X$ with $B = A / \mathfrak{p}$ a Noetherian local domain of dimension one. Let $C$ be the integral closure of $B$ in $\mathrm{Frac}(B)$ so $B \subset C$ is an integral extension of Noetherian domains and thus $\mathrm{dim}(C) = 1$ and $\mathrm{Spec}(C) \to \mathrm{Spec}(B)$ is surjective so there is a maximal ideal $\mathfrak{m}' \subset C$ above $\mathfrak{m}$ and thus $B \to C_{\mathfrak{m}'}$ is a local inclusion of Noertherian domains where $ C_{\mathfrak{m}'}$ is a DVR since it is integrally closed. Then $\widehat{C_{\mathfrak{m}'}} \cong \mathbb{C}[[t]]$ by the Cohen structure theorem. So we get a local injection $B \to C \to C_{\mathfrak{m}'} \to \mathbb{C}[[t]]$. Thus the map,
$$ \mathrm{Spec}(\mathbb{C}[[t]]) \to \mathrm{Spec}(\mathcal{O}_{X,x}/\mathfrak{p}) \to \mathrm{Spec}(\mathcal{O}_{X,x}) \to \overline{X} $$
sends $(t) \mapsto x$ and $(0) \mapsto \mathfrak{p} \in D(f)$.
|
2025-03-21T14:48:31.986540
| 2020-09-08T13:44:05 |
371190
|
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|
Stack Exchange
|
Independence number of $C_4$-free graphs
It's well known that a $C_4$-free graph of order $n$ has average degree $O(\sqrt{n})$, and it follows that the independence number is $\Omega(\sqrt{n})$.
This bound cannot be improved over $\Theta(n^{\frac34})$: A polarity graph of a projective plane of order $q$ has order $q^2+q+1$,degree $q+1$ and minimum eigenvalue $-\sqrt q$, so by the Hoffman bound, the independence number is at most $O(q^\frac32)=O(n^\frac34)$.
Question: Is it possible to get better bounds like $\alpha(G)=\Omega(|G|^a)$ with $a>\frac12$?
Any answer giving proof or disproof about whether it is possible to attain $a=\frac58$ would be accepted.
Maybe it's a typo, but $O(n^{3/4})$ is consistent with $\Theta(n^{3/4})$, so why should the polarity graph exclude that $a = 3/4$? (I can see why it excludes $a>3/4$)
Thanks for your correction. The post is now edited.
If we denote $m=\alpha(G)+1$, then our graph does not contain $C_4$ and its complement does not contain $K_m$, thus $n<R(C_4,K_m)$ (and viceversa, if $n<R(C_4,K_m)$, there exists a graph on $n$ vertices without $C_4$ such that $\alpha(G)\leqslant m-1$). So this question is about $C_4$ and $K_m$ Ramsey number. J. Spencer (Asymptotic Lower Bounds for Ramsey Functions. Discrete Mathematics. 20 (1977), 69-76) proved that $R(C_4,K_m)\geqslant m^{3/2-o(1)}$, providing an example of the graph with $n$ vertices without $C_4$ and with $\alpha(G)\leqslant n^{2/3+o(1)}$. The currently best lower bound is, as far as I know, $R(C_4,K_m)\leqslant m^{2-o(1)}$ (see Y. Caro, Y. Li, C. C. Rousseau, and Y. Zhang. Asymptotic bounds for some bipartite graph - complete graph Ramsey numbers. Discrete Mathematics. 220 (2000), 51-56) that results in $\alpha(G)\geqslant n^{1/2+o(1)}$, there are some log improvements with respect to $n^{1/2}$ bound from OP). Since $5/8$ lies between $2/3$ and $1/2$, your question seems to be open.
|
2025-03-21T14:48:31.986684
| 2020-09-08T13:47:14 |
371191
|
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|
Stack Exchange
|
A unique equilibrium state which does not have Gibbs property
Let $T:\Sigma \rightarrow \Sigma$ be a topologically mixing subshift of finite type and let $f:\Sigma \rightarrow \mathbb{R}$ be a continuous functions over $(T, \Sigma)$. Assume that there is a unique equilibrium measure $\mu$ for $f$ because of some reason.
$\textit{Question}:$ Does $\mu$ necessarily have Gibbs property?
I guess the answer is no, but I can't find a reference.
The measure $\mu$ does not necessarily have the Gibbs property. In fact, it has the Gibbs property if and only if $f$ has the Bowen property: $\sup_n \sup \{ |S_n f(x) - S_n f(y)| : x_1 \dots x_n = y_1 \dots y_n \} < \infty$. Every such $f$ has a unique equilibrium measure, but there are some potentials without the Bowen property that still have unique equilibrium measures.
$\mu$ Gibbs iff $f$ Bowen. The Gibbs property requires that there be $K>0$ such that for every $x\in \Sigma$ we have
$$
K^{-1}\leq \frac{\mu[x_1\dots x_n]}{e^{-nP(f) + S_nf(x)}} \leq K.
$$
Given $x,y \in \Sigma$ with $x_1\dots x_n = y_1 \dots y_n$, the only quantity in the corresponding inequalities that can vary is $S_n f$, and comparing them gives
$$
K^{-2} \leq e^{S_n f(x) - S_n f(y)} \leq K^2.
$$
Thus $|S_n f(x) - S_n f(y)| \leq 2\log K$, which proves the Bowen property. The other direction is classical; see
Bowen, Rufus, Some systems with unique equilibrium states, Math. Syst. Theory 8(1974), 193-202 (1975). ZBL0299.54031.
which gives a more general result (expansive systems with specification, which includes mixing SFTs).
An example of a non-Bowen potential that has a unique equilibrium state.
Hofbauer, Franz, Examples for the nonuniqueness of the equilibrium state, Trans. Am. Math. Soc. 228, 223-241 (1977). ZBL0355.28010.
The example there is the full shift on two symbols 0,1, and the potential is $f(x) = a_k$ whenever $x = 1^k 0\dots$, where $a_k$ is a sequence of real numbers converging to $0$. (Also $f(1^\infty) = 0$.) Writing $s_k = a_0 + \cdots + a_k$, the table on page 239 of that paper is useful. The potential $f$ has the Bowen property iff $\sum a_k$ converges, but there are examples where $\sum a_k$ diverges and $f$ still has a unique equilibrium measure.
It is often the case that unique equilibrium measures, including the ones in Hofbauer's paper, satisfy a "non-uniform" Gibbs property: see
Climenhaga, Vaughn; Thompson, Daniel J., Equilibrium states beyond specification and the Bowen property, J. Lond. Math. Soc., II. Ser. 87, No. 2, 401-427 (2013). ZBL1276.37023.
Thank you very much for your great answer.
|
2025-03-21T14:48:31.986867
| 2020-09-08T13:57:55 |
371192
|
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|
Stack Exchange
|
Sufficient condition for asymptotic-$\ell_{p}$ in terms of spreading models?
Let $(X,\|\cdot\|)$ be a Banach space with a Schauder basis and fix $p\in[1,\infty]$. Suppose that $X$ is asymptotic-$\ell_{p}$ with respect to this basis. It is known that the closed linear span of every (nontrivial) spreading model of $X$ is isomorphic to $\ell_{p}$ if $X$ is reflexive and at least contains an isomorphic copy of $\ell_{p}$ in general (replace $\ell_{p}$ by $c_{0}$ if $p=\infty$). In other words, the global asymptotic geometry of $X$ gives some information about the local asymptotic geometry.
Do there exist any known converse results? For example, are there general hypotheses that, in combination with the closed linear span of every spreading model containing an isomorphic copy of $\ell_{p}$, ensure that $X$ itself will be asymptotic-$\ell_{p}$?
Note that this question is cross-listed here on MSE.
It's good to see questions about asymptotic $\ell_p$ spaces. There is a lot of interesting literature that might help you. You should start by looking at the introduction to this paper: https://arxiv.org/abs/1902.10092 and references. I don't completely understand you statement: "It is known that the closed linear span of every (nontrivial) spreading model of is isomorphic to ℓ if is reflexive and at least contains an isomorphic copy of ℓ in general." Can you clarify?
Thanks for your comment and reference suggestion! To clarify my statement, it is Corollary 3.4.6 and the following paragraph of this thesis.
The answer to the question you formulated is no in a very strong sense. For all $1<p<\infty$ there exists a reflexive space $X$ with an unconditional
basis so that $X$ for all $\varepsilon>0$ every normalized weakly null sequence in $X$ admits a subsequence $1+\varepsilon$-equivalent to the unit vector basis of $\ell_p$ (so all spreading models are isomorphic to $\ell_p$) and yet its asymptotic structure contains $\ell_q^n$'s for some $p\neq q$. See Example 4.2 of this paper of Odell and Schlumprecht
There are many variation of the problem relating different asymptotic structures, and often the right question is to ask for a subspace with a better asymptotic structure under the assumption of, say, all spreading models being isomorphic to some $\ell_p$. The remarkable paper of Argyros and Motakis (that Kevin already referred to) gives some definite answers to some of these difficult questions. See the references therein to discover older results.
Thanks for your helpful answer! As it happens, I am interested in Banach spaces that are known to admit $\ell_{1}$ as a unique spreading model and that (I believe) cannot contain $\ell_{p}^{n}$'s.
You need to be more specific. It is trivial to give counterexample to this. Take $\ell_1$-sum of $\ell^n_p$'s, $\left(\sum_n \ell^n_p\right)_{\ell_1}$
You also want to ask copies of $\ell^n_p$'s as blocks in the space, otherwise you can always find them in $\ell_1$ for $p\le 2$.
|
2025-03-21T14:48:31.987095
| 2020-09-08T14:31:56 |
371193
|
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|
Stack Exchange
|
If the universal cover of a manifold is spin, must it admit a finite cover which is spin?
If $M$ is non-orientable, then it has a finite cover which is orientable (in particular, the orientable double cover).
If $M$ is non-spin, then it does not necessarily have a finite cover which is spin, e.g. $M = \mathbb{CP}^2$. As a cover of a spin manifold is spin, a necessary condition for $M$ to admit such a finite cover is that its universal cover is spin (which is not the case in the previous example). In analogy with the first sentence, note that the universal cover is always orientable.
Let $M$ be a closed smooth manifold whose universal cover is spin. Is there a finite cover of $M$ which is spin?
This question is partially motivated by the study of positive scalar curvature. The Dirac operator on a spin manifold can be used to obtain obstructions to positive scalar curvature à la Lichnerowicz, Hitchin, Gromov & Lawson, Rosenberg, etc. More generally, these techniques can be applied to manifolds which admit a spin cover. Things are generally more difficult in the non-compact case than in the compact case, so if the answer to my question were 'yes', we could just pass to a compact cover which is spin and apply the techniques there (as opposed to passing to the potentially non-compact universal cover).
You can even get an example with contractible universal covering. I will write details later when I have more time, or someone else can do this using Davis' trick.
@MoisheKohan: I would be very interested to see such examples.
No, this is not true: for each dimension $d \geq 4$, there is a closed, oriented $d$-manifold which is not spin, whose universal cover is spin, but which does not have a finite cover that is spin.
The reason is simply that there are finitely presented groups which have no nontrivial finite quotient.
One example is Higman's group $H$, see
https://en.wikipedia.org/wiki/Higman_group.
The key features of $H$ are:
$H$ is infinite,
$H$ does not admit a nontrivial finite quotient,
$H$ is acyclic,
$H$ has a classifying space $BH$ which is a finite $2$-dimensional CW-complex.
The proof of 1,2 can be found in Tao's blog https://terrytao.wordpress.com/2008/10/06/finite-subsets-of-groups-with-no-finite-models/,
and the proof of 3,4 in ''The topology of discrete groups'' by Baumslag, Dyer, Heller).
Now pick an element $1 \neq x \in H$, which induces an injective homomorphism $\mathbb{Z} \to H$ and form
the amalgamated product $G=H \ast_{\mathbb{Z}} H$. The group $G$ is infinite and has no nontrivial homomorphism to a finite group $F$, since any homomorphism
$G \to F$ must vanish on the two copies of $H$.
The pushout $BH \cup_{S^1} BH$ is aspherical by Whiteheads asphericity theorem and hence a model for $BG$.
I have designed things so that $H_2(BG) \cong \mathbb{Z}$ and all other homology groups are trivial.
In particular, $G$ is perfect, and the Quillen plus construction $BG^+$ must be homotopy equivalent to $S^2$, so that there is a homology equivalence
$f:BG \to S^2$. Now let $V \to S^2$ be the nontrivial oriented vector bundle of rank $d$, which has $w_2 (V) \neq 0$. It follows that the vector bundle
$f^\ast V \to BG$ is not spin. $BG$ has no nontrivial cover, and $BG$ is aspherical, so the pullback of $f^\ast V$ to the universal cover is trivial.
Now there exists, when $d \geq 4$, a closed $d$-manifold $M$ with a $2$-connected map $\ell: M \to BG$ and a bundle isomorphism $TM\oplus \mathbb{R}\cong \ell^\ast f^\ast V \oplus \mathbb{R}$.
This is achieved by surgery below the middle dimension. In particular, $\pi_1 (M)\cong G$. Hence $\pi_1(M)$ has no nontrivial finite index normal subgroup, and therefore no nontrivial finite index subgroup at all. It follows that $M$ does not have a nontrivial finite cover.
By construction, $w_2 (TM) \neq 0$, but the universal cover of $M$ is stably parallelizable.
Thanks for your answer. I think I follow the construction up until the penultimate paragraph. I am trying to understand the construction of $M$ via Theorem 1.2 on page 22 here. Is there a more direct result I can apply to deduce the existence of $M$?
I don't think you can get it in a more direct way, because you also want to control the tangent/normal bundle along the process. An easier strategy to construct a manifold with fundamental group $G$ which is often mentioned is to embed the 2-skeleton of BG into R^5 (or some larger R^n), to take a regular neighborhood of the image and to take its boundary. The result would, however, be stably parallelizable.
I know this is technique is relatively standard, but it's my first time seeing it. I don't see how the aforementioned theorem allows us to deduce that there is a $2$-connected map $\ell : M \to BG$ with $TM\cong\ell^*f^*V$. Rather, from the definition of normal map on page 19, it seems to me that one obtains a $2$-connected map $\ell : M \to BG$ with $TM\oplus\ell^*f^*V$ stably trivial. What am I missing?
One only gets an isomorphism $TM \oplus \mathbb{R}\cong \ell^\ast f^\ast V\oplus \mathbb{R}$; I have corrected the statement. To deduce my claim from the result in Wall's book, you pick a complementary bundle $V^\bot$ of large dimension, apply Wall's result to get $TM \oplus \ell^\ast f^\ast V^\bot$ stably trivial, add further trivial bundles to $V^\bot$ to make the sum actually trivial. Add a copy of $\ell^\ast f^\ast V$ to both sides of the equation and get that $TM \oplus \mathbb{R}^n \cong \ell^\ast f^\ast V \oplus \mathbb{R}^n$ for some large $n$. (ctd.)
Then by obstruction theory (compute the connectivity of the map $BO(d+n-1)\to BO(d+n)$ and recall that $M$ is $d$-dimensional, you can cancel all of the extra summands except for the last one. Without adding the extra copy, the statement becomes false, as you can see when $d=4$ and you apply this surgery below the middle dimension to the trivial bundle $\mathbb{R}^4 \to \ast$: there does not exist a simply connected $4$-manifold which is parallelizable, by an Euler number argument.
Thanks for fleshing out the details, I can now see why $w_2(TM) \neq 0$. Is the universal cover of $M$ parallelisable or only stably parallelisable? I can show the latter using the isomorphism $TM\oplus\mathbb{R} \cong \ell^*f^*V\oplus\mathbb{R}$.
Of course the universal cover is only stably parallelizable.
What about the other way? If a universal cover of M is not spin does that imply M is not spin?
@R.Rankin: The projection $\pi:\widetilde{M}\rightarrow M$ from the universal cover $\widetilde{M}$ of $M$ to $M$ is a local diffeomorphism, so that, $\pi^\ast(TM) = T\widetilde{M}$. Now compute $w_2$ of both sides and use naturality.
As promised, here is my solution based on the Davis trick. First, there is a very general construction of PL aspherical 4-manifolds (it also works in higher dimensions). Start with a finite aspherical 2-dimensional CW complex $W$. Up to homotopy, $W$ always embeds in the Euclidean 4-space $E^4$ (I think, this is due to Stallings). Take such an embedding and let $N=N(W)$ denote a regular neighborhood of $W$ in $E^4$. Now, apply "Davis trick" to $N$: Introduce a reflection orbifold structure on the boundary of $N$ such that the corresponding stratification of the boundary is dual to a triangulation of $\partial N$. The resulting orbifold ${\mathcal O}$ is very good (admits a finite orientable manifold-covering $M\to {\mathcal O}$) and its universal covering (same for $M$ and for ${\mathcal O}$) is contractible. As a bonus, $\pi_1(W)$ embeds in $\pi_1(M)< \pi_1({\mathcal O})$. For details see
Mess, Geoffrey, Examples of Poincaré duality groups, Proc. Am. Math. Soc. 110, No. 4, 1145-1146 (1990). ZBL0709.57025.
and, of course, the original paper by Mike Davis from 1983. (Actually, it was Bill Thurston who came up with this trick in the context of 3-manifolds: He used it for the proof of his hyperbolization theorem.) This construction allows one to embed 2-dimensional finitely presented groups with "exotic properties" in fundamental groups of closed aspherical PL manifolds.
I will use a relative version of this construction. Start with a closed connected oriented surface of genus $\ge 1$; I'll take the torus $T^2$. Let $E\to T^2$ be the 2-disk bundle over $T^2$ with the Euler number $\pm 1$. The boundary of the 4-manifold $E$ is a 3-dimensional Nil-manifold: The total space of a nontrivial circle bundle over the torus. The group $\pi_1(\partial E)$ has two generators $a, b$, and $\pi_1(\partial E)$ has the presentation
$$
\langle a, b| [a,b]=t, [a,t]=1, [b,t]=1\rangle.
$$
Represent $a, b$ by simple disjoint loops $\alpha, \beta$ in $\partial E$. Now, take your favorite finite 2-dimensional aspherical complex $W$ whose fundamental group is nontrivial and has no proper finite index subgroups (I care only about the homotopy type of $W$). The standard example is the presentation complex of Higman group. But there are many other examples. As before, embed $W$ in $E^4$, take a regular neighborhood $N$ of $W$ in $E^4$. Then $\pi_1(\partial N)$ maps nontrivially to $\pi_1(W)$. Pick two simple loops $\alpha', \beta'\subset \partial N$ which map nontrivially to $\pi_1(W)$ (you can take the same loop).
Now, take two copies $N_a, N_b$ of $N$ and attach them to $E$ by identifying a regular neighborhood of $\alpha'$ to that of $\alpha$ for $N_a$ and identifying a regular neighborhood of $\beta'$ to that of $\beta$ for $N_b$. The result is a compact PL aspherical 4-manifold with boundary $Z$. The $\pi_1(Z)$ is an amalgam of $\pi_1(E)\cong {\mathbb Z}^2$ with two copies of $\pi_1(N)$ (along infinite cyclic subgroups).
For each homomorphism to a finite group
$$
\phi: \pi_1(Z)\to \Phi
$$
the subgroups $\pi_1(N_a), \pi_1(N_b)$ will have to map trivially. Hence, $a$ and $b$ will have to map trivially as well. Since $a, b$ generate $\pi_1(E)$, $\pi_1(Z)$ has no nontrivial homomorphisms to finite groups. Now, apply Davis trick to $Z$. The result is an orbifold ${\mathcal O}$. Since $Z$ was aspherical, so is ${\mathcal O}$ (i.e. it has contractible universal covering space).
Take a finite orientable manifold-covering $M\to {\mathcal O}$. Then $M$, of course, has contractible (hence, spin) universal covering. I claim that $M$ has no finite spin-covering spaces. Indeed, for each finite-sheeted covering $p: M'\to M$, the manifold $int(Z)\subset M$ has to lift trivially; more precisely, $p$ restricts to a trivial covering
$$
p^{-1}(int Z)\to int Z.$$
This is because $\pi_1(Z)$ has no nontrivial homomorphisms to finite groups. Thus, $M'$ contains a copy of $E$. In particular, $M'$ contains a 2-torus with odd self-intersection, i.e. the intersection form of $M'$ is not even, i.e. $M'$ is not spin.
I was working in the PL category but in dimension 4, PL is the same DIFF, so you get a smooth example as well.
Edit. Lemma. Let $M$ be a triangulated manifold, $W\subset M$ is a subcomplex and $N=N(W)$ is the regular neighborhood of $W$ in $M$. Then the inclusion map $W\to N$ is a homotopy-equivalence; if $W$ is connected and has codimension $\ge 2$ in $M$ then $\partial N$ is connected and the induced map $\pi_1(\partial N)\to \pi_1(W)$ is surjective.
Proof. The homotopy-equivalence part is standard and holds for general simplicial complexes $M$, not just for manifolds. Moreover, the inclusion map $\partial N\to (N \setminus W)$ is also a homotopy-equivalence. (Both are proven using "straight-line homotopy.")
I will prove the second part. Take an arc $\alpha$ in $N$ connecting two points $x, y\in \partial N$. Since $W$ has codimension $\ge 2$, taking $\alpha$ in general position, we see that it will be disjoint from $W$, hence, is homotopic relative to $\{x, y\}$ to an arc in $\partial N$.
(I am using here and below the h.e. $\partial N\to N-W$.) Thus, $\partial N$ is connected. Next, let $\alpha$ be a loop in $N$ based at $x\in \partial N$. By the same argument, $\alpha$ is homotopic to a loop based at $x$ and contained in $N-W$, hence, to a loop in $\partial N$.
I think you mean $\pi_1(\partial E)$ has three generators $a$, $b$, and $t$. In the following paragraph, how do you know that $\pi_1(\partial N)$ maps non-trivially to $\pi_1(W)$? Is the inclusion map $\partial N \hookrightarrow N$ followed by the deformation retraction $N \to W$ non-trivial on $\pi_1$?
@MichaelAlbanese If you prefer three, but two suffice, since [,]=.
[Feel free to ignore this question] If you don't include $t$ as a generator, what does $[a, b] = t$ mean ($t$ is not a word in $a$ and $b$)?
Am I correct in saying that the reason you apply the Davis trick to $Z$ is to obtain a closed aspherical manifold $M$ (as opposed to $Z$ which is an aspherical manifold with boundary)?
@MichaelAlbanese: $t$ is a letter in a presentation: You can eliminate it if you like by writing down relators of the form $[a,[a,b]]=1$. The bottom line is that you need just two generators.
@MichaelAlbanese: Right.
@MichaelAlbanese: One common way to eliminate the boundary is to double the manifold along the boundary. However, you typically loose asphericity. Davis trick is a form of doubling which preserves asphericity.
I see, $\langle a, b \mid [a, [a, b]] = 1, [b, [a, b]] = 1\rangle$ is a valid presentation. Thanks yet again for your help. I will try to remember this trick.
Another reference for the reflection trick is Chapter 11 of Davis' The Geometry and Topology of Coxeter Groups.
|
2025-03-21T14:48:31.988242
| 2020-09-08T14:40:17 |
371194
|
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|
Stack Exchange
|
Classification of similarity transformations of Minkowski space
Consider $\mathbb{R}^{n+1}$ equipped with the Minkowski (sign indefinite) metric:
$$g=(x^0)^2-(x^1)^2-\dots -(x^n)^2.$$
Is there a classification of diffeomorphisms $F\colon \mathbb{R}^{n+1}\tilde\to \mathbb{R}^{n+1}$ with the property $F^*g=a\cdot g$, where $a$ is a constant?
If $a>0$, since $a$ is constant, you can just compose $F$ with a suitable rescaling to get $a=1$, and then $F$ is an isometry. So isometries composed with dilations. If $a<0$, not possible for $n>1$ because it changes signature.
Thanks. Is there an explicit classification of isometries? Are they necessarily affine?
Isometries preserve the Levi-Civita connection, which is the standard flaf connection of Euclidean space, and preserving that is precisely being affine. So then you see that the isometries are Poincare transformations, i.e. translations and Lorentz transformations.
|
2025-03-21T14:48:31.988350
| 2020-09-08T14:54:05 |
371196
|
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|
Stack Exchange
|
Flat with geometrically connected and proper fibers is proper
Let $S$ be a locally Noetherian scheme.
Lemma 1. If $f: X \to S$ a proper flat morphism, then it has proper fibers, and if furthermore the generic fibers are geometrically connected all its fibers are (by semi-continuity, EGA IV.15.5.4).
There is a converse:
Lemma 2. if $f$ is flat (or even just univ. submersive), finite type separated, has proper and geometrically connected fibers then it is proper (EGA IV.15.7.10).
If $S$ is a dvr, there is an intermediate result:
Lemma 3. Let $f:X \to S$ be flat separated of finite type, S a dvr. If the special fiber is proper and the generic fiber is geometrically connected then $f$ is proper. Hence the special fiber is geometrically connected by the result above.
Proof:
Indeed the proof is in Serre-Tate, Good reduction of abelian varieties, Lemma 3. By fpqc descent they reduce to $S=\mathop{Spec} \hat{A}$ the spectrum of a complete noetherian ring, and use lifting of proper components of the special fiber. (The Lemma assumes that $f$ is smooth but I think they only use that it is univ open).
So the following question is natural:
Question. if $f:X \to S$ is flat separated, of finite type with geometrically connected generic fibers. If the fiber at $s$ is proper, is $f$ proper at $s$ (that is proper in a neighborhood of $s$)?
Essentially I'd like to combine the local proper valuative criterion (of EGA IV.15.7.5) with the version https://stacks.math.columbia.edu/tag/0894 where we can restrict to a dense open to reduce to the previous case. Does this work?
I think that if the fiber is proper at $s$ and all fibers of generisations of $s$ are geometrically connected, then $f$ is indeed proper at $s$ (so fibers of generisations of $s$ are proper). Indeed using the local valuative criterion we reduce exactly to Lemma 3.
Conversely if all fibers of generisations of $s$ are propers and the generic fibers (we only need this for generic $\eta$ specializing to $s$) are geometrically connected then $f$ is proper too. Indeed we take a chain $s_0 \to s_1 \to \dots s_n=s$ and we Lemma 3 to conclude that $f$ is proper at $s_1$, so its fiber is geometrically connected, so $f$ is proper at $s_2$, and so on.
EGA IV.15.7.8 applies only when $f$ is quasi-finite.
Sorry I meant EGA IV.15.5.4 instead of EGA IV.15.5.9, I'll edit
The problem is local on $S$, so assume that $S$ is an affine scheme. Now factor $f$ as the composition of a dense open immersion and a proper morphism via Nagata compactification. Finally, consider the closed image in $S$ of the closed complement of the open immersion.
@Jason: sorry I was not able to complete your proof. But I think I found another one.
So I think I have the following proof:
since the condition is topological, we can assume that everything is reduced. Since $f$ is separated, it suffices to prove that $f$ is proper at $y$ at each irreducible component of $X$. So we reduce to $S$ and $X$ integral and $f$ dominant. But in this case we can use EGA IV.15.7.1, which states that it suffices to check the valuative criterion for $S'$ a dvr when the generic point $\eta$ of $S'$ maps to the generic point of $X$ hence also to the generic point of $Y$, and the closed point $s$ of $S'$ maps to $Y$. But then we are reduced to Lemma 3.
|
2025-03-21T14:48:31.988601
| 2020-09-08T15:24:02 |
371201
|
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|
Stack Exchange
|
Reference on two numbers associated to a module of finite homological dimension
Let $A$ be a finite dimensional algebra over a field $K$ with a module $M$ which has finite projective dimension and finite injective dimension.
Let $n \geq 1$.
Let $(P_i)$ be a minimal projective resolution of $M$ and $(I_i)$ a minimal injective coresolution of $M$.
Let $u_M^n:=dim(P_0)-dim(P_1)+dim(P_2)-dim(P_3)+....+(-1)^n dim(P_n)$ and $t_M^n:=dim(I_0)-dim(I_1)+dim(I_2)-dim(I_3)+...+(-1)^ndim(I_n)$ ($n$ might not be the projective/injective dimension of $M$ but smaller).
Question: Is there a name for the numbers $u_M^n$ and $t_M^n$ associated to $M$? Have they been studied in the literature?
Aren't they just equal to $dim(M)$? I must be missing something.
@JohnPalmieri I made the question more precise by adding an $n$ (which might be smaller than the projective dimension).
|
2025-03-21T14:48:31.988806
| 2020-09-08T15:25:58 |
371202
|
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"Gerry Myerson",
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|
Stack Exchange
|
Vertex configuration to tile repeat unit
I am working with elongated triangular tiling which has a vertex configuration of <IP_ADDRESS>.4 and noticed that the representative symmetry is not the repeat unit. Is there a general formula to convert symmetry to repeat tile (or specific to for semiregular tiling)?
"elongated triangular tiling", "representative symmetry", "repeat unit". Are these standard terms in the study of tilings, or are they terms you've come up with that other people might not understand?
Good question Gerry, I got them from Wiki. https://en.wikipedia.org/wiki/List_of_Euclidean_uniform_tilings
Since posting, I still do not know the answers, but I did find http://gruze.org/tilings/semiregular by Kevin Jardine where he lists the repeat unit as translational units.
"translational unit", another term I think I've never seen.
|
2025-03-21T14:48:31.988919
| 2020-09-08T16:08:54 |
371205
|
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|
Stack Exchange
|
Is there a concept of limit of formulas
I wonder if there is a notion like the limit of formulas (and structures) because I believe it is important in describing countable structures (from finite structures). (For more detail, see this paper.) Now I give an excellent example illustrating it. The example is the so-called Quine atom. Let
$$
I_n=\{I_{n-1}\},\quad\text{for } n\geqslant1, \:\text{and } I_0=G_0.
$$
Where $G_0\neq\{G_0\}$. By repeatedly applying the axiom of extensionality, we get
\begin{align}
\phi_n&\iff\exists !\:y_n\in I_n
\\
&\iff \exists !y_n(y_n=I_{n-1})
\\
&\iff\exists !y_n((\exists !y_{n-1}\in y_n)(y_{n-1}\in I_{n-1})\land (\exists !y_{n-1}\in I_{n-1})(y_{n-1}\in y_n))
\\
&\iff\exists !y_n \exists !y_{n-1}(y_{n-1}\in y_n\land y_{n-1}\in I_{n-1})
\\
&\iff\exists !y_n \exists !y_{n-1}(y_{n-1}\in y_n)\land \exists !y_{n-1}(y_{n-1}=I_{n-2})
\\
&\iff\exists !y_n \exists !y_{n-1}(y_{n-1}\in y_n)\land\exists !y_{n-1}((\exists !y_{n-2}\in y_{n-1})(y_{n-2}\in I_{n-2})\land(\exists !y_{n-2}\in I_{n-2})(y_{n-2}\in y_{n-1}))
\\
&\iff\exists !y_n \exists !y_{n-1}(y_{n-1}\in y_n)\land \exists !y_{n-1}\exists !y_{n-2}(y_{n-2}\in y_{n-1}\land y_{n-2}\in I_{n-2})
\\
&\quad\:\vdots
\\
&\iff \exists !y_n \exists !y_{n-1}(y_{n-1}\in y_n)\land \bigwedge_{1\leqslant j\leqslant n-1}\exists !y_j\exists !y_{j-1}(y_{j-1}\in y_{j})\quad\quad\quad\quad\text{(1)}
\end{align}
Clearly, for any $n<\omega$, $I_{n}\models \phi_n$. This suggests that we can define the limit of formulas as follows.
Edit: As pointed out in @Goldstern's example, formulas for limit can not be too general. So we limit them to a $\aleph_0$-categorical theory.
Definition 1: Suppose for any $n<\omega$, $\varphi_{n}$ is in a $\aleph_0$-categorical theory $T$ in $L$ (an infinitary language of $L_{\omega_1, \omega}$) and $M_{n}\models \varphi_{n}$. If for any $n$, there is a $N_n$ that for any $k>N_n,\:M_{k}\models \varphi_{n}$, then $\lim_\limits{n\to\omega}\varphi_{n}$ is a (unique) formula (up to equivalence) in $L$, and $\lim_\limits{n\to\omega}M_{n}$ is a (unique) $L$-structure. Let $\lim_\limits{n\to\omega}\varphi_{n}=\varphi_{\omega}$ and $\lim_\limits{n\to\omega}M_{n}=I_{\omega}$. Then $M_{\omega}\models \varphi_{\omega}$.
Also the follow axioms hold for the limit of formulas.
$\lim_\limits{n\to\omega}(\phi_{n}\land\varphi_n)=\lim_\limits{n\to\omega}\phi_{n}\land\lim_\limits{n\to\omega}\varphi_n$
$\lim_\limits{n\to\omega}\neg\phi_{n}=\neg\lim_\limits{n\to\omega}\phi_{n}$
$\lim_\limits{n\to\omega}\exists x\phi_{n}=\exists x\lim_\limits{n\to\omega}\phi_{n}$
$\lim_\limits{n\to\omega}\bigwedge_\limits{1\leqslant j\leqslant n}\phi_j=\bigwedge_\limits{ n<\omega}\phi_n$
We can prove from the above axioms that
$\lim_\limits{n\to\omega}(M_{n-1}\in M_n)=\lim_\limits{n\to\omega}M_{n-1}\in \lim_\limits{n\to\omega}M_{n}$
$\lim_\limits{n\to\omega}\exists x_n(M_{n-1}\in M_n)=\exists\lim_\limits{n\to\omega} x_n(\lim_\limits{n\to\omega}M_{n-1}\in \lim_\limits{n\to\omega}M_{n})$
(For detailed proofs, see this paper.)
$\operatorname{Th}(I_n)$ is $\aleph_0$-categorical because any $I_i$ can be mapped one-on-one to $I_j$, and so is homogeneous. Since the language of set theory is finite relational, $\operatorname{Th}(I_n)$ is $\aleph_0$-categorical. By definition 1 and (1), we can see that $\lim_\limits{n\to\omega}\phi_{n}$ and $\lim_\limits{n\to\omega}I_{n}$ both are unique. Let $\lim_\limits{n\to\omega}\phi_{n}=\phi_{\omega}$ and $\lim_\limits{n\to\omega}I_{n}=I_{\omega}$. Then $I_{\omega}\models \varphi_{\omega}$. Furthermore
\begin{align}
\phi_{\omega}&=\lim_\limits{n\to\omega}\phi_{n}
\\
&=\lim_\limits{n\to\omega}\exists !y_n \exists !y_{n-1}(y_{n-1}\in y_n)\land \lim_\limits{n\to\omega}\bigwedge_{1\leqslant j\leqslant n-1}\exists !y_j\exists !y_{j-1}(y_{j-1}\in y_{j})
\\
&=\exists !I_{\omega}(I_{\omega}\in I_{\omega})\land \bigwedge_{n<\omega}\exists !y_n\exists !y_{n-1}(y_{n-1}\in y_{n})\quad\quad\quad(\lim_\limits{n\to\omega}y_n=I_{\omega})
\end{align}
Thus $I_{\omega}=\{I_{\omega}\}$, i.e. $I_{\omega}$ is a Quine atom.
Since the limit of formulas (for finite structures) can completely describe the Quine atom, I believe that it (will) play a significant role in the investigation of countable structures. I'd like to confirm that the above notion of limit of formulas and reasoning are not available in current model theory and so belong to a new field of research.
Edit: Next I will give two examples on how to apply the limit of formulas to studying some known results in model theory.
Example 1: There is arbitrary large number in nonstandard number theory.
Let $\phi_n=\exists x\bigwedge_\limits{m\leqslant n}(x>m)$ and $M_n\models \phi_n$. Since for any $k>n, \:x>k\to x>n, \:M_k\models \phi_n$. So the limit of $\phi_n$ exists and
$$
\lim_\limits{n\to\omega}\exists x\bigwedge_\limits{m\leqslant n}(x>m)=\exists x\bigwedge_\limits{n<\omega}(x>n)
$$
And there is a $M\models \exists x\bigwedge_\limits{n<\omega}(x>n)$.
The second example shows that theory of DLO without endpoints is the limit of union of integers shifts.
Example 2: Suppose $T$ is the theory of DLO without endpoints and $M_n=\bigcup_\limits{1\leqslant j<n}(\Bbb{Z}+j/n)$. Then $\lim_\limits{n\to\omega}\bigcup_\limits{n<\omega}M_n=\Bbb{Q}$ and $T = Th(\Bbb{Q})$.
Suppose $\varphi_n,\phi_n,\delta_n$ are sentences specifying the properties of linear ordering, a dense subset and set without endpoints for $M_n$. Then
\begin{align}
\varphi_n &= (∀x,y,z∈M_n)(x≤x∧(x≤y∧y≤x⇒x=y)∧(x≤y∧y≤z⇒x≤z))
\\
\phi_n &= (∀x,y∈M_n)(x<y⇒(∃z∈M_{l})(l>n∧x<z<y))
\\
\delta_n &= (∀x∈M_n)((∃y∈M_n)(y<x)∧(∃y∈M_{n})(x<y))
\end{align}
For any $x,y∈M_{n}\:(x<y)$, set $N_{n}=2n$. Then $∀k>N_{n},\:∃z∈M_{k}$ that $x<z<y$, i.e. $M_{k}\models φ_{n}$. Since $ℤ\modelsϕ_{n}∧δ_{n},\:M_{k}\models ϕ_{n}∧φ_{n}∧δ_{n}$ and $T$ is $ℵ_0$-categorical, $\lim_\limits{n\to\omega}M_n$ is unique. And
\begin{align}
\lim_\limits{n\to\omega}ϕ_{n} &=(∀x,y,z∈\lim_\limits{n\to\omega}M_n)(x≤x∧(x≤y∧y≤x⇒x=y)∧(x≤y∧y≤z⇒x≤z))
\\
\lim_\limits{n\to\omega}\phi_{n} &=(∀x,y∈\lim_\limits{n\to\omega}M_n)(x<y⇒(∃z∈\lim_\limits{n\to\omega}M_n)(x<z<y))
\\
\lim_\limits{n\to\omega}δ_{n} &=(∀x∈\lim_\limits{n\to\omega}M_n)((∃y∈\lim_\limits{n\to\omega}M_n)(y<x)∧(∃y∈\lim_\limits{n\to\omega}M_n)(x<y))
\end{align}
Since $\lim_\limits{n\to\omega}ϕ_{n},\lim_\limits{n\to\omega}φ_{n}, \lim_\limits{n\to\omega}δ_{n}$ are axioms of $Th(\Bbb{Q}),\: \lim_\limits{n\to\omega}M_n=\Bbb{Q}$ and $T = Th(\Bbb{Q})$.
Your "Definition 1" does not define what "$\lim_n \varphi_n$" is, it just claims that this limit has a certain property. Please clarify.
Instead of taking the limit of formulas, would you be interested in taking a limit (indeed, liminf) of the theories of the structures as $\displaystyle \bigcup_{k \geq 1 } \bigcap_{n \geq k} Th(\mathcal{M}_n)$? This seems to be close to what you want.
What is the motivation for axiom 3: $\lim\limits_{n\to\omega} \exists x\phi_n(x) = \exists x \lim\limits_{n\to\omega} \phi_n(x)$?
@François G. Dorais, it is used to prove $\lim_\limits{n\to\omega}\exists !y_n \exists !y_{n-1}(y_{n-1}\in y_n)=\exists !I_{\omega}(I_{\omega}\in I_{\omega})$
@hermes I don't follow why you think it is virtually impossible or too complicated: If your "limit" existence condition is satisfied for a sequence $(\varphi_i: i \in \omega)$, then ${\varphi_i: i \in \omega} \subseteq \bigcup_{k \geq 1} \bigcap_{n \geq k} Th(M_n)$. Conversely, any sequence obtained from formulas in $\bigcup_{k \geq 1} \bigcap_{n \geq k} Th(M_n)$ will automatically satisfy your limit existence condition.
@Burak, yes, it is working as an alternative. So this shows that the limit of formulas is analogous to the limit of a sequence in analysis.
I am not sure what you mean by $G_0$. If you mean an element $x$ satisfying $x={x}$, then in your example you have $I_0 = {I_0} = I_1 = I_2 = \cdots$. But then you are looking at the limit of a constant sequence, which really gives no insight into what you mean by "limit".
@hermes If you don't give a definition of "limit of a sequence of formulas", your question might be closed as "unclear what you are asking".
@Goldstern, limit of a sequence of formulas is a (unique) formula. I modified definition 1. Also $G_0≠{G_0}$
What you wrote makes less sense now. What does it mean to be a formula of a theory? To be a consequence of the theory, or to be a formula in the language of the theory? Since you want to have models of the formulas $\varphi_n$, it seems you mean the former. But then you have problems with the definition of limits involving negation.
It should be a formula of a $\aleph_0$-categorical theory which is $L_{\omega_1, \omega}$ because it has a unique countable model. The definition of negation is consistent with that of general model theory, i.e. $M\models \neg \phi$ iff $M\nvDash \phi$. The limit of formulas is natural because an infinitary formula can be form from a sequence of finite formulas through the limit operation.
I add two examples to show how to apply the limit of formulas to studying some current results in model theory. This method needs be tested on more works to demonstrate its validity and power.
Your use of standard notation is so peculiar that it is almost impossible to tell what you are saying.
Here I assume that an infinitary formula can be formed from a sequence of finitary formulas through the limit operation. It could be a new feature.
@hermes Saying that your paper is “available upon request” is a bit odd, like you want someone to beg you for it — if the paper is relevant to the question, it’s customary to simply post the arxiv link somewhere in the body. Then inquiring minds can simply click the link to check out your paper if they desire, rather than having to waste time asking for a copy and waiting for a reply. (not a downvoter or upvoter yet here, just mildly interested and following along. if experts in the field agree that it’s a well-posed question i’ll upvote, and they could decide more quickly with a link)
@Alec Rhea, the file is available now, just click the link in the post.
One could ask if there are natural interesting compactifications of the set of formulas. For instance, given a structure $M$ one can map a formula $f(x)$ to the set of $t\in M$ such $f(t)$ is true. This gives a compactification (for each $M$). Also given a family $(M_i)$ of structures, one can map a sentence $F$ to the set of $i$ such that $M_i$ satisfies $F$.
@There is a natural compactification of the set of formulas (over some fixed set of variables) in first-order logic. It's the collection of partial types (i.e. closed subsets of type space) with the Vietoris topology.
I am not sure which set-theoretic axioms you want to use. Certainly not foundation, but I guess that at least the singleton axiom is allowed.
Consider $M_0:= \{x\}$, where $x$ is any element satisfying $x\not=\{x\}$. $M_{n+1}:=\{M_n\}$.
Let $\varphi_n$ be $\phi_n \wedge \psi$, where $\phi_n$ is your formula, and
$\psi$ says the unique element is not an element of itself. Then by your arguments the limit $M_\omega$ must satisfy $M_\omega = \{M_\omega\}$, but it must also satisfy $\psi$, which is a contradiction.
So your axioms and rules lead to a proof of $x=\{x\}$ for all $x$.
It may be that I missed something. As a matter of fact, I am only guessing at what formal language you are using.
@ Goldstern, thanks. This example is right which means that formulas for limit can not be too general. So I modified definition 1 to limit them to a $\aleph_0$-categorical theory, which is atomic and exists a complete formula. Since $\psi$ in your example does not follow from the complete formula, the problem is avoided.
The structures $M_i$ in my example all have exactly one element, so their theory is $\aleph_0$-categorical (in the first order sense). The limit of my formulas is (following your rules) $\phi_\omega\wedge \psi$.
Your structure is the same as $I_n$. Since $Th(I_n)$ is $ℵ_0$-categorical, it is atomic and has a complete formula $ϕ_ω$. If $ψ=∃x(x≠{x})$, then $ϕ_ω∧ψ$ has no contradiction. However, if $ψ=∀x(x≠{x})$, then $ψ∉Th(I_n)$ because $ϕ_ω→¬ψ$.
The formal language used here is the infinitary language of set theory ($L'{\omega_1,\omega}$). However, it can be extended to a general language of $L{\omega_1,\omega}$.
Also the axioms of this set theory is basically ZF minus the axiom of regularity, where the axiom of extensionality in ZF needs be modified to accommodate NWF sets like Quine atoms. The axiom of union needs change too. The axiom of regularity is dropped and replaced by a new axiom stating that there are NWF sets. The rest axioms (pairing, power set, infinity, replacement, separation) remain the same as those of ZF. See the file listed in the post for more details.
I have prepared for counter-examples for the limit of formulas, but I can not find them. This allows me to improve the limit of formulas. So I accept this answer.
|
2025-03-21T14:48:31.989675
| 2020-09-08T18:01:37 |
371213
|
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"Iosif Pinelis",
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|
Stack Exchange
|
Extension of Bernstein’s Inequality when the random variable is bounded with large probability
Bernstein’s Inequality can be stated as follows : Let $x_1, x_2, \dots, x_n$ be independent bounded random variables such that $\mathbb{E}[x_i] = 0$ and $|x_i| \leq \zeta$ with probability $1$ and let $\sigma^2 = \tfrac{1}{n}\sum_{1}^{n} Var\{x_i\}$. Then for any $\epsilon > 0$, we have
$$
\mathbb{P} \left[ \frac{1}{n} \sum_{i=1}^{n} x_i \geq \epsilon \right] \leq \exp{\left\{ \frac{-n \epsilon^2}{ 2 \sigma^2 + 2\zeta \epsilon/3} \right\}}
$$
If instead of $|x_i| \leq \zeta$ with probability $1$, it is the case that $ \sum_{i=1}^{n} \mathbb{P}\left\{ |x_i| > \zeta \right\} \leq \delta$, then is the following applicable?
$$
\mathbb{P} \left[ \frac{1}{n} \sum_{i=1}^{n} x_i \geq \epsilon \right] \leq \exp{\left\{ \frac{-n \epsilon^2}{ 2 \sigma^2 + 2\zeta \epsilon/3} \right\}} + \delta \qquad \qquad \qquad \qquad (1)
$$
I think, the above extension is similar to the extension of the Azuma-Hoeffding inequality proved in Theorem 32 of Chung and Lu(2006). The question concerning the extension of the Azuma-Hoeffding inequality was also asked here.
Is it possible to extend Bernstein’s Inequality as eq. $(1)$ following the lines of the proof of Theorem 32 in Chung and Lu(2006)?
Addendum : The Freedman inequality for martingales is a result like the Bernstein inequality but where the random variables can be dependent. Is an extension like the above possible for Freedman inequality?
For reference Freedman inequality (Theorem 1.6 in Freedman (1975)] ) can be stated as follows: let $x_1, x_2, \dots $ be a martingale difference sequence and $|x_i| \leq \zeta$ for all $i$. Then
$$
\mathbb{P} \left[ \sum_{i=1}^{n} x_i \geq \epsilon, \sum_{i=1}^{n} var(x_i | \mathcal{F}_{i-1}) \leq \eta \right] \leq \exp{\left\{ \frac{-\epsilon^2}{2\eta + 2 \zeta \epsilon/3 } \right\}}.
$$
If instead of $|x_i| \leq \zeta$ with probability $1$, it is the case that $ \sum_{i=1}^{n} \mathbb{P}\left\{ |x_i| > \zeta \right\} \leq \delta$, then is the following applicable?
$$
\mathbb{P} \left[ \sum_{i=1}^{n} x_i \geq \epsilon, \sum_{i=1}^{n} var(x_i | \mathcal{F}_{i-1}) \leq \eta \right] \leq \exp{\left\{ \frac{-\epsilon^2}{2\eta + 2 \zeta \epsilon/3 } \right\}} + \delta \qquad \qquad (2)
$$
Thank you.
For $\delta=1$, it looks like your eq. (1) doesn't recover Bernstein's inequality -- is there maybe a typo somewhere?
@gmvh It, obviously, should be $1-\delta$ in the assumption but I agree that it would be nice to fix it.
By "$|x_i| \leq \zeta$ with probability $\delta$", do you mean "$P(\max_i|x_i| \leq \zeta)=1-\delta$ or "$P(|x_i| \leq \zeta)=1-\delta$ for all $i$" or ...?
Thanks for pointing out the typo @fedja.
@IosifPinelis, thanks. I have clarified it now.
I don't see the ambiguity resolved. Which one of the first two versions I listed do you have in mind?
@IosifPinelis, I have made it more explicit. Please let me know if it's unambiguous now. I was also wondering if Bernstein inequality could work with different upper bounds on $|x_i|$ i.e $|x_i| \leq \zeta_i $ instead of $|x_i| \leq \zeta $. But I see you have already answered that it's not possible here.
@Komkom : Also, you are citing Freedman's Theorem 1.6 incorrectly: instead of $x_i^2$, you need $Var(x_i|\mathcal F_{i-1})$.
Doesn't your (1) follow from using Freedman (the correct version, as Iosif Pinelis points out) instead of Azuma in the Chung-Lu argument, and noting $\delta$ is an upper bound (union) for $Pr(B)$ in their paper?
As a cheap justification, note that you can couple $x_i$ to $y_i$ defined to be equal to $x_i$ if $|x_i|\le\zeta$ and $0$ otherwise. If the $y_i$ have mean zero, then they are by Bernstein concentrated as you desire, and the failure probability of the coupling is at most $\delta$, so you get (1).
Of course, in general the $y_i$ will not have mean zero, so they concentrate to whatever the sum of their expectations actually is. This can only be a problem if the sum of expectations is far from zero. But for that to happen, the $x_i$ have to have a significant contribution to expectation from when they come out bigger than $\zeta$ (or smaller, in the other direction). That blows up the variance of the $x_i$ (compared to that of the $y_i$); the $y_i$ will be more concentrated when you use the correct variance. So... if you want to know if (1) will go wrong,
then the following random variables are what you should look at. Take $\zeta=1$ (this just sets the scale), and each $x_i$ to be $-t$ with probability $1/(t+1)$, and otherwise $1$. And suppose you have $n$ variables. Then we have to take $\delta=n/(t+1)$, so for there to be any way for (1) to fail, we need $n<t+1$. Since the sum of the $x_i$ is likely to be about $1$, we should take $\varepsilon<1$, and (it's easy to check) say $0.99$ is about as good as we can do.
Now we can write out the LHS of (1) explicitly - the probability is the same as that some $x_i$ comes out to $-t$
and the right hand side is easy enough to figure out. Using Taylor to expand the exponential, you (just!) get enough cancellation between $\delta$ and the negative exponential that (1) is true for these particular random variables.
@user36212, Thanks for your comments. Using the same argument as in the Chung-Lu paper (pr(B) for union probability of bad event), could (2) be proved as well? For my purpose (2) is more relevant as the random variables (r.v.'s) are dependent rather than independent.
Wrt my previous previous comment, $B$ is the bad set associated with the following : $|x_i| \leq \zeta$ and $var(x_i | \mathcal{F_{i-1}}) \leq \sigma_i^2$. And in (2), $\eta = \sum_{i=1}^{n} \sigma^2$.
$\newcommand{\de}{\delta}$Your inequality (2) does hold. Actually, a better and more general bound holds. First here, let us standardize and simplify notations. Let us use $X_i$ instead of $x_i$, $x$ instead of $\epsilon$, $y>0$ instead of $\zeta$, $B^2>0$ instead of $\eta$, $Var_{i-1}\,\cdot$ instead of $var(\cdot|\mathcal{F}_{i-1})$, and $E_{i-1}\,\cdot$ instead of $E(\cdot|\mathcal{F}_{i-1})$.
Instead of the conditions that the $x_i$'s are martingale differences and $\sum_{i=1}^n P(|x_i|>\zeta)\le\de$, let us use the more general conditions that the $X_i$'s are supermartingale differences and
\begin{equation}
P(\max_{i=1}^n X_i>y)\le\de.\tag{1}
\end{equation}
Let also $Y_i:=X_i\,1(X_i\le y)$ and $Z_i:=Y_i\,1(V_i\le B^2)=X_i\,1(X_i\le y,V_i\le B^2)$, where
$$V_i:=\sum_{j=1}^i E_{j-1}\,Y_j^2.$$
Note that $V_i$ is no greater than $\sum_{j=1}^i E_{j-1}\,X_j^2$, which latter coincides with $\sum_{j=1}^i Var_{j-1}\,X_j$ in the special case when the $X_i$'s are martingale differences.
By (1),
$$P\Big(\sum_{i=1}^n X_i\ge x,V_n\le B^2\Big)\le P\Big(\sum_{i=1}^n Y_i\ge x,V_n\le B^2\Big)+\de.$$
Obviously, $V_i\le V_n$ for $i\le n$. So,
$$P\Big(\sum_{i=1}^n Y_i\ge x,V_n\le B^2\Big)\le P\Big(\sum_{i=1}^n Z_i\ge x\Big).$$
Next, $E_{i-1}Z_i=1(V_i\le B^2)E_{i-1}Y_i\le 1(V_i\le B^2)E_{i-1}X_i\le0$, so that the $Z_i$'s are supermartingale differences. Also, $E_{i-1}Z_i^2\le E_{i-1}X_i^2$. So, by Theorem 8.2 on page 1702, we have the Hoeffding-type inequality
\begin{equation*}
P\Big(\sum_{i=1}^n Z_i\ge x\Big)\le\exp\Big\{\frac{B^2}{y^2}\psi\Big(\frac{xy}{B^2}\Big)\Big\},
\end{equation*}
where $\psi(u):=u-(1+u)\ln(1+u)$. Collecting the pieces, we get
\begin{equation*}
P\Big(\sum_{i=1}^n X_i\ge x,V_n\le B^2\Big)\le\exp\Big\{\frac{B^2}{y^2}\psi\Big(\frac{xy}{B^2}\Big)\Big\}+\de. \tag{2}
\end{equation*}
The latter bound is better than the Bernstein-type bound
\begin{equation*}
\exp\Big\{-\frac{x^2}{2B^2+2xy/3}\Big\}+\de, \tag{3}
\end{equation*}
because, as shown in Theorem 3, the Hoeffding-type bound is the best exponential bound in its terms. Another, direct way to see that the bound in (2) is better than (3) is to use the inequality $\psi(u)\le-u^2/(2+2u/3)$ for real $u\ge0$.
Thus, the bound in (2) is better and more general than what you wanted.
Thank you very much for the detailed answer. I read Theorem 3 but I don't understand how it shows that the obtained bound is the best exponential bound. Could you please elaborate?
@Komkom : The inequality sign in Theorem 3 (and in other results in that paper) is, not $\leqq$, but $\overset{(T)}\leqq$, which means that the bounds are exact (best possible), in the sense defined on the first page of that paper. The letter T is the first letter of the Russian word "точный", meaning "exact".
|
2025-03-21T14:48:31.990182
| 2020-09-08T18:29:06 |
371215
|
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|
Stack Exchange
|
Self-map of short exact sequences
Consider the commutative diagram of finite abelian groups
$\require{AMScd}$
\begin{CD}
0@>>> A @>i>> B@>\pi>> C@>>> 0\\
\ @VV 0 V@VVfV@VV 0 V\\
0@>>>A @>>i> B@>>\pi> C@>>> 0
\end{CD}
where all maps are homomorphisms, the rows are exact, and the leftmost and the rightmost vertical map are zero? Is the middle map $f$ also zero?
It is clear that $f\circ f=0$ but this is all that seems to follow from lazy diagram chasing, hence I suspect $f$ need not be zero, in general. If so, what is a counterexample?
Take the sequence $0\to Z/2Z\to Z/4Z\to Z/2Z\to 0$ and the vertical map multiplication by 2.
Oh, thanks, I should have seen this myself.
Or take $0\to A\to A\oplus C\to C\to0$, then $f(a,c)=(h(c),0)$ for any homomorphism $h:C\to A$
|
2025-03-21T14:48:31.990275
| 2020-09-08T18:42:31 |
371217
|
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"url": "https://mathoverflow.net/questions/371217"
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|
Stack Exchange
|
Classification of conformal diffeomorphisms of Minkowski space, part 2
This is a continuation of Classification of conformal diffeomorphisms of Minkowski space
Consider $\mathbb{R}^{n+1}$ equipped with the Minkowski (sign indefinite) metric:
$$g=(x^0)^2-(x^1)^2-\dots -(x^n)^2.$$
Is there a classification of diffeomorphisms $F\colon \mathbb{R}^{n+1}\tilde\to \mathbb{R}^{n+1}$ with the property $F^*g=a\cdot g$, where $a$ is a function?
If $n=1$, let $u=x^0+x^1$, $v=x^0-x^1$. Then the maps $(U,V)=(U(u),V(v))$, for any diffeomorphisms $U(u)$ and $V(v)$ of the real number line, are conformal, an infinite dimensional group. For $n>1$, the group is finite dimensional, as this is a Klein geometry, and consists of those conformal maps of the conformal compactification which preserve this affine open set. But I will have to think of a good reference.
@BenMcKay Maybe something is in the book on Cartan geometries by Sharpe?
My impression is that the answer is that, for dimensions $p+q\geq 3$, strictly speaking every global conformal transformation of $\mathbb R^{p,q}$ is a similarity transformation -- you need to compactify for the "interesting" conformal transformations to be globally defined. A reference I was told has this information is Schottenloher, "A mathematical introduction to conformal field theory". Once you compactify, the global conformal group for $\mathbb R^{p,q}$ is supposed to be $SO(p+1,q+1)$.
For every $n\ge 2$, each conformal transformation of ${\mathbb R}^{1,n}$ has constant conformal factor; in other words, it is an affine transformation of the form ${\mathbf x}\mapsto aU{\mathbf x} +{\mathbf b}$, where $a>0$, $U\in O(1,n)$, ${\mathbf b}\in {\mathbb R}^{n+1}$.
When $n=1$, there are more conformal transformations; after your rotate the coordinates so that the invariant bilinear form becomes $xy$, the group $Conf_+({\mathbb R}^{1,1})$ (of orientation-preserving conformal diffeomorphisms) consists exactly of diffeomorphisms of the form $(x,y)\mapsto (f(x),g(y))$, where $f, g$ are diffeomorphisms ${\mathbb R}\to {\mathbb R}$ and they either both preserve or both reverse orientation.
The case $n=0$, I hope, is clear.
For details, see:
Martin Schottenloher, The conformal group, chapter 2 of "A mathematical introduction to conformal field theory," 2008.
See my answer to the conformal group of $S^n$ where you can find explanation for the the positive definite case and a link to a paper that actually works in arbitrary signature.
Here is a short summary of the argument: Rewrite the problem using definition of the pullback as a system of PDEs for $F$. Differentiate these equations and take clever sums. Use the original system to keep only derivatives which are not determined by the original equations. Repeat. Lo and behold! You obtained no more "undetermined" derivatives! It then follows that the solution space is finite-dimensional and you can write it's basis explicitly.
It's roughly a one page calculation plus half a page of text. For details, see the article by Slovák that I link to in my previous answer.
Do you have a more direct and explicit answer? Thank you.
@makt This argument is as direct as it gets. Expand the definitions. Calculate. Calculate some more. Write down all the solutions. The end. It's all pretty explicit in the article by Slovák.
The Liouville theorem 5.4 in Slovak's article gives the conformal group by giving a list of generators. I believe makt would like a parametrization of the group, e.g. the isometry group is more precisely described as mappings "Ax+b" rather than "the group generated by rotations and translations"
|
2025-03-21T14:48:31.990642
| 2020-09-08T18:46:51 |
371218
|
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|
Stack Exchange
|
Forcing zero subset sums in zero characteristic
$\newcommand{\ve}{\varepsilon}$
A problem from the year 2003 Miklos Schweitzer exam (see also this MO post) goes, essentially, as follows:
If $b_1,\dotsc,b_k\in\mathbb F_p^n$ are vectors with $b_1^\perp\cup\dotsb\cup b_k^\perp=\mathbb F_p^n$ (meaning that every vector from $\mathbb F_p^n$ is orthogonal to at least one of the vectors $b_i$), then there exist $\ve_1,\dotsc,\ve_k\in\{0,1\}$ not all equal to $0$ such that $\ve_1b_1+\dotsb+\ve_kb_k=0$.
What is the zero-characteristic analogue of this fact? Are there any reasonable assumptions that guarantee the existence of not-all-zero $\ve_1,\dotsc,\ve_k\in\{0,1\}$ with $\ve_1b_1+\dotsb+\ve_kb_k=0$, say, over the reals?
Notice that for a vector space $V$ over a zero-characteristic field, the assumption $b_1^\perp\cup\dotsc\cup b_k^\perp=V$ becomes totally unreasonable (simply meaning that one of the vectors $b_i$ is $0$).
|
2025-03-21T14:48:31.990729
| 2020-09-08T18:56:57 |
371219
|
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"Rohil Prasad",
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|
Stack Exchange
|
Gromov's compactness theorem via Sacks-Uhlenbeck and Schoen-Uhlenbeck
Gromov's compactness theorem for pseudo-holomorphic curves, in section 1.5 of "Pseudoholomorphic curves in symplectic manifolds," is very well known. I'm aware of the following two proofs:
J.G. Wolfson. Gromov's compactness of pseudo-holomorphic curves and symplectic geometry. J. Differential Geom. Volume 28, Number 3 (1988), 383-405.
Rugang Ye. Gromov's compactness theorem for pseudo holomorphic curves. Trans. Amer. Math. Soc. 342 (1994), 671-694.
Wolfson's proof applies the methods of
J. Sacks and K. Uhlenbeck. The existence of minimal immersions of 2-spheres. Ann. of Math. (2) 113 (1981), no. 1, 1–24.
Richard Schoen and Karen Uhlenbeck. A regularity theory for harmonic maps. J. Differential Geometry 17 (1982), no. 2, 307–335.
which seems quite natural as Gromov's theorem is an adaptation of the Sacks-Uhlenbeck phenomena for harmonic maps to pseudoholomorphic curves. Ye explains that Gromov's proof in the case of manifolds with boundary (section 1.5D2 of his paper) does not work in general. He also explains that Wolfson's result is weaker than Gromov's, since it does not recover area convergence; and that Wolfson's methods do not immediately extend to the case of manifolds with boundary. Ye gives a proof of Gromov's theorem for manifolds with and without boundary, and with area convergence. (To my knowledge it is the first to do so, if Gromov's only works for closed manifolds.)
My questions are:
Is it possible to extend Wolfson's methods to manifolds with boundary, and has anyone done so? I am thinking of the following article as inspiration: Richard Schoen and Karen Uhlenbeck. Boundary regularity and the Dirichlet problem for harmonic maps. J. Differential Geom. 18 (1983), no. 2, 253–268.
Is it possible to extend Wolfson's methods to recover area convergence, and has anyone done so? The issue seems conceptually parallel to the problem of energy quantization for harmonic maps, which has been widely studied. The following article, for instance, applies the Sacks-Uhlenbeck theory to get such a result.Yum Tong Siu and Shing Tung Yau. Compact Kähler manifolds of positive bisectional curvature. Invent. Math. 59 (1980), no. 2, 189–204.
Are there other proofs (with details) that give Ye's result?
It's not exactly what you're looking for, but Joel Fish's "Target-local Gromov compactness theorem" (https://arxiv.org/abs/0912.4435) gives a Gromov compactness theorem for $J$-holomorphic curves with boundary that live inside a manifold with boundary. It relies on the interpretation of $J$-holomorphic curves as surfaces satisfying an inhomogeneous mean curvature equation.
|
2025-03-21T14:48:31.990904
| 2020-09-08T20:32:12 |
371225
|
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|
Stack Exchange
|
Green's function for Robin boundary condition
Let $\Omega$ be a bounded subset in $\mathbb{R}^n$, $n \ge 3$, with smooth boundary $\partial \Omega$. Assume given $a^{ij} \in W^{2, p}(\Omega)$ with $a^{ij} = a^{ji}$, $f \in L^p(\Omega)$ and $g \in L^q(\partial \Omega)$ with $p > n/2$ and $q > n-1$. Assume further that there exist constants $\lambda$ and $\Lambda$ such that, for all $x \in \Omega$ and $v \in \mathbb{R}^n$, we have
$$
\lambda |v|^2 \le a^{ij} v_i v_j \le \Lambda |v|^2.
$$
We consider the elliptic PDE
$$
\begin{aligned}
- \frac{\partial}{\partial x^j} \left( a^{ij} \frac{\partial u}{\partial x^i} \right) + f u &= h,\\
a^{ij} \nu_j \frac{\partial u}{\partial x^i} + g u &= 0,
\end{aligned}
$$
where $\nu$ is the unit outgoing normal to $\partial \Omega$ and $h \in L^r(\Omega)$ is given, $r \in (1, \infty)$.
I assume that the quadratic form
$$
u \mapsto \frac{1}{2} \int_{\Omega} \left[a^{ij} \frac{\partial u}{\partial x^i} \frac{\partial u}{\partial x^j} + f u^2 \right] dx + \frac{1}{2}\int_{\partial \Omega} g u^2 dx
$$
(naturally associated to this problem) is coercive on $H^1(\Omega)$. So, in particular, the homogeneous system (i.e. $h \equiv 0$) only admits the trivial solution $u \equiv 0$.
As is "well known" (any reference for this fact is welcome), we can write
$$
u(x) = \int_{\Omega} G(x, y) h(y) dy
$$
where $G$ is the associated Green function.
How can we prove, in such a low regularity context, that $G$ is bounded from below by a positive constant?
(Regularity for $g$ can be improved to e.g. $g \in W^{1-1/p, p}(\partial \Omega)$ if needed).
Any reply will be appreciated!
It is not clear to me why you assume such a regularity on $f,g$ but smoothness on $(a_{ij})$. Does the problem concern with regularity or the type of boundary conditions? Moreover; do you look for a bound from below up to the boudary or only in the interior? It would be nice if you say what is known in special situations.
The origin of the problem is related the conformal Laplacian on a manifold. If the metric belongs to $W^{2, p}$, then the scalar curvature is in $L^p$ and the mean curvature of the boundary is in $W^{1 - 1/p, p}$.
I am looking for a lower bound of the Green function up to the boundary. Otherwise this is just some sort of Harnack inequality.
Thank you very much for youre answer. What is known if also $f,g$ are more regular?
I would say that this is certainly settled for higher regularly but I would like to have a good reference for it anyway.
|
2025-03-21T14:48:31.991092
| 2020-09-08T20:44:08 |
371226
|
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|
Stack Exchange
|
Smoothness and decay correspondence for Laplace transform
For the Fourier transform, there are various theorems formalizing a correspondence between the smoothness of a function and the rate of decay of its Fourier transform. For example, if a function is $n$ times continuously differentiable, then its Fourier transform asymptotically decays like a reciprocal $n$th power, something like this.
For the unilateral Laplace transform, are there corresponding results? In particular, are there natural conditions on a function which entail any desired particular decay rate on its unilateral Laplace transform?
The decay rate of Laplace transform is related to the behavior of function near 0, but not to smoothness at other points. For example, vanishing of $f$ to order $a$ at $0$ implies $|Lf(s)|<Cs^{-a-1},; s\to+\infty$. Vanishing of $f$ on an interval near zero is equivalent to exponential decay of the Laplace transform.
Oh, that's a great criterion, thanks! Where can I find a reference with more details on this?
Any book which treats Laplace transform. For example Folland, Fourier Analysis.
Thanks! I've found the result there. I realize this was a very basic question to ask on MathOverflow, but I didn't know the answer, and now I do!
|
2025-03-21T14:48:31.991206
| 2020-09-08T21:39:36 |
371228
|
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|
Stack Exchange
|
Does any group action of $\mathbb{F}_p^\mathbb{N}$ on $\mathbb{F}_p^\mathbb{N}$ have non-trivial fixed points?
It is well-known that if $G$ is a finite $p$-group acting on a non-zero $\mathbb{F}_p$-vector space $V$, then $V^G \neq \{0\}$.
My question is about a generalization of this result when $G = V = \mathbb{F}_p^\mathbb{N}$ (no topology involved).
(A counter-example with $G = \mathbb{F}_p^{(\mathbb{N})}$ would provide a negative answer since the latter is a direct factor of $\mathbb{F}_p^\mathbb{N}$.)
For every group $G$, the left action $g\cdot f(x)=f(g^{-1}x)$ yields an action on the space $\mathbf{F}_p^{(G)}$ of finitely supported functions $G\to \mathbf{F}_p$. If $G$ is infinite, this action has no nonzero fixed point.
If $G=\mathbf{F}_p^{(\alpha)}$ with $\alpha$ infinite, $G$ itself has cardinal $\alpha$ and hence $\mathbf{F}_p^{(G)}$ is isomorphic to $G$. Hence $G$ has an action on $G$ by group automorphisms, with no nonzero fixed point. This applies to your particular case $\alpha=2^{\aleph_0}$.
Added: in a less specific context: the argument shows that:
For every infinite group $G$, for every field $K$, and cardinal $\alpha\ge |G|$, there exists a representation of $G$ in $K^{(\alpha)}$ with no nonzero finite-dimensional subrepresentation.
The above shows it for $\alpha=|G|$, and the general case follows by taking direct sums of copies of such a representation.
|
2025-03-21T14:48:31.991316
| 2020-09-08T22:16:13 |
371232
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371232"
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|
Stack Exchange
|
Morphism of pseudomonads induces pullback functors between pseudoalgebras
If $S$ and $T$ are monads on a category $C$, and $\lambda:S\to T$ is a morphism of monads, it is well-known that there is a functor $\lambda^*:C^T\to C^S$ which assigns to the $T$-algebra $(A,a:TA\to A)$ the $S$-algebra $(A,a\circ\lambda:SA\to A)$.
Does a similar phenomenon happen for pseudomonads and their pseudoalgebras (as defined for example here)?
A reference would be welcome too.
Yes, Theorem 3.4 of Gambino–Lobbia's On the formal theory of pseudomonads and pseudodistributive laws establishes that pseudomonad morphisms are in correspondence with liftings to pseudoalgebras. They work more generally in the setting of pseudomonads in a Gray-category, and with pseudomonads on different objects, rather than pseudomonads on a fixed object, but the theorem specialises to the result you are looking for.
|
2025-03-21T14:48:31.991400
| 2020-09-08T22:16:45 |
371233
|
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"url": "https://mathoverflow.net/questions/371233"
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|
Stack Exchange
|
Littlewood Offord for odd number of real numbers
One version of the Littlewood-Offord problem states that if $a_1, \ldots, a_n$ are real numbers with absolute value at least $1$, then the number of sequences $\varepsilon_1, \ldots, \varepsilon_n$ with $\varepsilon_i = \pm 1$ such that
$$
-1 < \sum_{i=1}^{n} \varepsilon_i a_i < 1
$$
is at most $\binom{n}{\lfloor n/2 \rfloor}$.
This bound is sharp for even $n$, obtained by taking $a_i = 1$ for all $1 \le i \le n$. My question is about the sharpness for odd $n$. By taking $a_1 = 2$ and $a_i = 1$ for $2 \le i \le n$, we obtain that the number of sequences in such case is $2 \binom{n-1}{\lfloor (n+1)/2 \rfloor}$. However I couldn't find a better example for the odd $n$ case and so my question is: is this example for odd $n$ sharp?
Thanks in advance!
|
2025-03-21T14:48:31.991480
| 2020-09-08T22:45:42 |
371235
|
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|
Stack Exchange
|
Identifying Lie groupoids among smooth groupoids
I have been approaching groupoids in the category of smooth manifolds using methods from essentially algebraic theories/limit sketches. Are there any results that identify Lie groupoids amongst internal groupoids in the category of smooth manifolds (without the advance knowledge that $s,t: G \to M$ are submersions).
Are you defining, as part of the data, additional to $X_1$ and $X_0$ and source/target etc, a manifold $X_2$ and three maps $X_2 \to X_1$ (two "projections" and "composition") so that $X_2 \simeq X_1\times_{X_0} X_1$? It's not clear to me how one extracts from the existence of a pullback square of manifolds that both of the maps in the cospan part are submersions.
In practice, I have a (symmetric) simplicial manifold that satisfies some version of the Segal conditions. I should edit my question - I guess where I'm confused is whether the conditions for "geometric groupoids" in papers by Henriques or Getzler and Behrend will force a simplicial manifold to be a Lie groupoid.
Oh, ok. I was assuming you were working from the older Eheresmann-style sketch, where ambient cats without finite limits were the norm, so the required limits and their cones were built in. Since you can say an internal groupoid is an appropriately coskeletal simplicial object, you can truncate just above that, and only deal with a finite diagram (I almost wrote "Segal condition" but decided to write out the only pertinent iso!). But saying what "coskeletal" means might require saying something is a surjective submersion, I can't recall right now if it's that, or if it's just an iso.
Well I'm using a Kelly-style enriched sketch so I'm able to cheat a bit and force the nerve of the groupoid to exist, and further require that all of the pullbacks be preserved by the tangent functor.
Well this is embarrassing, I asked this question after spending a weekend thinking about it staring at the definition all day and two hours later I have a partial answer - it seems that this is equivalent to requiring the tangent projection $p: T\mathcal{G} \to \mathcal{G}$ be a bifibration.
Recall from the theory of Grothendieck fibrations and opfibrations, for any functor $q: \mathbb{C} \to \mathbb{D}$, if $u:q^{-1}(A) \to q^{-1}(B)$ over $f: A \to B$ is an isomorphism, then $u$ is both cartesian and cocartesian. For a functor between groupoids, any map is (co)cartesian above its image.
A morphism $f: M \to N$ of smooth manifolds is a submersion if the naturality square with the tangent projection $p: T \Rightarrow id$ is a weak pullback. So for any $(a,b):X \to M \times TN$ where $f(a) = p(b)$, there exists a map $c: X \to TM$ so that $p(c) = a, Tf (c) = b$.
Now, recall that our smooth groupoid $\mathcal{G}$ has all of its pullbacks preserved by the tangent functor. This means that we have a functor $p: T \mathcal{G} \to \mathcal{G}$. The condition that target be a submersion is equivalent to asking for any map $a \xrightarrow{w} b$ and tangent vector $\gamma:TM$ so that $p(\gamma) = b$, there is a map $\gamma' \xrightarrow{\omega} \gamma$ above $w$. Because $\omega$ is an isomorphism, it will be a cartesian map, so this is equivalent to requiring that $p$ be a fibration. A similar line of reasoning for the source map leads to the conclusion that $p$ is also an opfibration. If $s,t$ are both submersions, then $p$ is a bifibration.
|
2025-03-21T14:48:31.991707
| 2020-09-08T23:49:07 |
371236
|
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|
Stack Exchange
|
Combinatorial Summation $\frac{1}{n} \sum_{k=n+1}^{2n} (k-n)\binom{2n}{k}$
For a particular problem, I reached until this point where eventually I have to prove this summation
$$
\frac{1}{n} \left ( \binom{2n}{n+1} + 2\binom{2n}{n+2} + 3\binom{2n}{n+3} + \dots + n\binom{2n}{2n} \right ) = \frac{1}{2}\binom{2n}{n}
$$
I've tried to form a differentiating function that would result in this but no luck with that.
Maybe it can be proved by induction on $n$? Iterating the recursion
${x \choose y} = {x-1 \choose y-1} + {x-1 \choose y}$ we get
${x \choose y} = {x-2 \choose y-2} + 2 {x-2 \choose y-1} + {x-2 \choose y}$.
Apply this to each term $(k-n) {2n \choose k}$ to get something very close to
the sum for $n-1$.
This smells like a homework question, which is not what this site is for. You could try asking at math.stackexchange.com, but in order for that to be well received, you will show that you have put in some serious effort first (describe what you have tried in detail etc.)
The left hand side can be rewritten as
\begin{align*}
\sum_{j=0}^n\left(1-\frac{j}{n}\right)\binom{2n}{j}
&=\sum_{j=0}^n\binom{2n}{j}-\sum_{j=0}^n\frac{j}{n}\binom{2n}{j}\\[6pt]
&=\sum_{j=0}^n\binom{2n}{j}-2\sum_{j=1}^n\binom{2n-1}{j-1}\\[6pt]
&=\frac{2^{2n}+\binom{2n}{n}}{2}-2^{2n-1}\\[6pt]
&=\frac{1}{2}\binom{2n}{n}.
\end{align*}
Done.
|
2025-03-21T14:48:31.991816
| 2020-09-09T00:08:55 |
371237
|
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|
Stack Exchange
|
Poincare inequality for martingales
This is a vague question but here we go: Is there a form of the Poincaré inequality that is better suited for martingales?
For example, the Poincaré inequality for the boolean cube states that for any function $f:\{-1,1\}^n \rightarrow \mathbb{R}$ we have $$Var(f(\epsilon)) \le \mathbb{E} \|Df(\epsilon)\|_2^2,$$ where we use the uniform measure in the cube and define the partial derivative vector $D f(\epsilon)$ via $$D_i f(\epsilon) = \frac{f(\epsilon_1, \ldots, \epsilon_i,\ldots, \epsilon_n) - f(\epsilon_1, \ldots, -\epsilon_i,\ldots, \epsilon_n)}{2},$$ i.e., flipping just the $i$th bit.
Now suppose that $(M_i)_i$ is a martingale adapted to the sequence $\epsilon_1,\ldots, \epsilon_n$ (uniform in the cube) and we are interested in the variance of $g(M_n)$. While this is a function of $\epsilon$ and we can apply the Poincaré inequality, the discrete derivative above does not seem to be appropriate quantity to bound the variance.
Concretely, the Poincaré inequality gives the right bound for linear functions: $$Var\Big(\sum_i a_i \epsilon_i\Big) \le \|a\|_2^2.$$ But consider a Paley-Walsh martingale $$M_n = \sum_{i \le n} V_i \epsilon_i,$$ where $V_i$ is a function of $\epsilon_1, \ldots, \epsilon_{i-1}$. While by orthogonality we have a similar expression as above $$Var(M_n) = \sum_i \mathbb{E}(V^2_i),$$ this does not seem to be captured by applying the above Poincaré inequality with $f = M_n$ (notice $D_i M_n$ is not $V_i \epsilon_i$ because flipping $\epsilon_i$ may change $V_{i+1}, \ldots$).
The problem seems to be that in the martingale case changing one bit changes the whole future of the martingale difference sequence; this makes the discrete gradient $D M_n$ too big compared to the actual size of the increments of the martingale.
Any pointers on what is the right way of looking at this are appreciated!
My intuition is that the Poincaré inequality is just not very sharp; it's based on the crude idea of bounding a matrix below by its (second) smallest eigenvalue. The right analogy may be an inequality of a different type, though I don't see off the top of my head what the right one is.
Thanks for your comment Nate. You are totally right that such eigenvalue bound for the generator of this process seems inefficient for martingale-looking functions, but maybe there is another diffusion process that is more aware of the filtration (e.g. first continuously change $\epsilon_n$ to a fresh $\epsilon'n$, and then move to $\epsilon{n-1}$, etc). That would be a nice mix of the Lindeberg- and Markov semigroup-type interpolations.
I think the knowledge that $(M_i)$ is a martingale is not particularly helpful as far as the variance of an arbitrary nonlinear function of $M_n$ is a concerned.
|
2025-03-21T14:48:31.992021
| 2020-09-09T00:43:08 |
371239
|
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|
Stack Exchange
|
Sierpinski's characterization of $F_{\sigma\delta}$ spaces
According to [2]: Let $X$ be a space. We call a system $(X_s)_{s\in T}$ a Sierpinski
stratification of $X$ if $T$ is a nonempty tree over a countable alphabet and $X_s$ is a closed subset of $X$ for each $s \in T$ such that:
(i) $X_{\varnothing} = X$ and $X_s =\bigcup \{X_t : t \in \text{succ}(s)\}$ for all $s \in T$, and
(ii) if $\sigma\in [T]$ then the sequence $X_{σ\restriction 0},X_{σ\restriction 1},...$ converges to a point $x_{\sigma} \in X$.
Here $[T]$ is the set of all infinite branches of $T$.
The authors of [2] state that in [1] Sierpinski proved: A (separable metrizable) space $X$ is absolute $F_{\sigma\delta}$ if and only if $X$ has a Sierpinski stratification.
My question concerns replacing condition (ii) with:
(ii') if $\sigma\in [T]$ then $\bigcap_{n<\omega} X_{σ\restriction n}$ is a singleton.
Apparently, if $(X_s)_{s\in T}$ is a Sierpinski stratification of $X$ then (ii') holds. My main question is whether the definition above is equivalent to the definition with (ii') in the place of (ii). Can this be proved simply by refining the stratification so that at level $n$ of the tree, we intersect each $X_s$ with sets from a countable cover of $X$, such that each set in the cover is a closed neighborhood of diameter $<1/n$?
Also I noticed that neither of these definitions will be identical to what Sierpinski describes in [1]. Is there a more recent reference where Sierpinski's theorem and its variants are proved? Surely there have been more proofs since 1924!
[1] Sierpiński, W., Sur une définition topologique des ensembles (F_{\sigma \delta})., Fund. math. 6, 24-29 (1924). ZBL50.0141.03.
[2] Dijkstra, Jan J.; van Mill, Jan, Erdős space and homeomorphism groups of manifolds, Mem. Am. Math. Soc. 979, v, 62 p. (2010). ZBL1204.57041.
What do you mean that the sequence of sets converges to a point? I would interpret that as meaning that their intersection is the singleton. So then (ii) and (ii)’ are reformulations.
Suerpiński himself states it as: if we pick a point $p_n$ from each $X_{\sigma\restriction n}$ then $(p_n)$ converges to some point $x$ in the space.
@HennoBrandsma I think that converging to a point means that every neighborhood of the point contains all but finitely many of the sets $X_{\sigma\restriction n}$. So this condition is stronger than Sierpinski's. I think they are likely equivalent, whereas the singleton intersection property might be too weak.
The answer seems to be no, even assuming each $X_s$ is non-empty. Take any $F_{\sigma\delta}$-space $X$ and any non-$F_{\sigma\delta}$-space $Y$. Then $X\times Y$ has a stratification (of non-empty sets) satisfying (ii'), but $X\times Y$ is not $F_{\sigma\delta}$ and therefore does not have a Sierpinski stratification.
|
2025-03-21T14:48:31.992222
| 2020-09-09T01:30:11 |
371240
|
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|
Stack Exchange
|
Which sums-of-two-squares are totients?
Consider the two subsets of the natural numbers $A$, $B$ given by
$$A = \{n \in \mathbb{N} \mathrel: \text{$\exists x,y \in \mathbb{Z}$ s.t. $n = x^2 + y^2$}\}$$
and
$$B = \{n \in \mathbb{N} : \exists m \in \mathbb{N} \text{ s.t. } n = \phi(m)\}$$
where $\phi$ is the Euler totient function.
We know that the intersection of the two sets $A$, $B$ contains infinitely many elements. Indeed, for $C \subset \mathbb{N}$, put $C(X)$ for the number of elements in $C$ satisfying $n \leq X$. Then
$$(A \cap B)(X) \gg X (\log X)^{-3/2},$$
since for a prime $p$ of the form $p = x^2 + y^2 + 1$, we have $p-1 \in A \cap B$, since $\phi(p) = p-1$. The density comes from a theorem of H. Iwaniec from 1972; see also the update On a theorem of Bredihin and Linnik by Friedlander and Iwaniec.
By multiplicativity, if $p_1, \cdots, p_k$ are primes of the above form then
$$\displaystyle \phi(p_1 \cdots p_k) \in A \cap B.$$
Is there a nice way to characterize the set $A \cap B$? Failing that, what is a better estimate for the density of $A \cap B$?
I did some very iffy heuristics (based on a number in $[1,X]$ has $\log\log X$ prime factors, which I assumed were like one in each $[e^{e^i},e^{e^{i+1}}]$ for $i=1,\ldots,\log\log X$) and ended up with an estimate for $(A\cap B)(X)$ which is smaller than your lower bound(!) My tentative conclusion from this is that the dominant term in $(A\cap B)(X)$ actually comes from the $\phi(p)$ that you're already looking at.
@AnthonyQuas the example extends to numbers of the form $n = p_1 \cdots p_k$, where $p_j = x_j^2 + y_j^2 + 1$, by multiplicativity
Banks, Luca, Saidak, and Shparlinski showed that the number of $m\le X$ with $\phi(m)$ a sum of two squares is $O(X/(\log X)^{3/2})$. Since $\phi(m) \le x$ implies that $m \ll x\log\log{x}$, this gives an upper bound of the shape $x/(\log{x})^{3/2+o(1)}$ for the count you are after. Probably one can remove the "$o(1)$", and get $\ll x/(\log{x})^{3/2}$, by fine tuning the BLSS arguments. (This is suggested by the very rapid decay of the distribution function of $m/\phi(m)$.)
|
2025-03-21T14:48:31.992391
| 2020-09-09T02:23:44 |
371244
|
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|
Stack Exchange
|
Hodge structure and rational coefficients
Suppose $X$ is a complex projective variety with a model $X_\mathbb{Q}$ defined over the rational numbers. Then there is a rational de Rham lattice $H^k_{dR}(X_\mathbb{Q}, \mathbb{Q})\subset H^k(X, \mathbb{C})$ coming from the hypercohomology of the de Rham complex with rational coefficients. Hodge theory implies a formality quasiisomorphism $H^*(X, \mathbb{C})\to C^*(X, \mathbb{C})$ which is compatible with multiplication, and even (see this paper of Guillen, Navarro, Pascual and Roig) functorial in a dg sense with respect to maps between projective varieties. Now I want to understand the map from the sub-lattice $H^*_{dR}(X_{\mathbb{Q}}, \mathbb{Q})\to C^*(X, \mathbb{C}).$ I have two questions.
Is there a way to characterize the forms in the image of $H^*(X_\mathbb{Q}, \mathbb{Q})$ in terms of Dolbeault forms? For example are they in some sense local linear combinations of rational holomorphic forms wedged with rational antiholomorphic forms?
Is there a canonical way to relate $H^*_{dR}(X_\mathbb{Q}, \mathbb{Q})$ with the hypercohomology complex $C^*_{dR}(X_{\mathbb{Q}}. \mathbb{Q})$ via a canonical chain of quasiisomorphisms? By canonical I mean at least compatible with rational closed immersions, but hopefully more generally with all maps of rational projective varieties?
is the map $H^(X, \mathbb{C})\to C^(X, \mathbb{C})$ given by taking harmonic representatives?
@ranicky it is.
Sullivan constructed a rational DGA via polynomial forms with respect to a triangulation (c.f. Griffiths, Morgan, Rational homotopy theory and differential forms). Would that work for you?
|
2025-03-21T14:48:31.992632
| 2020-09-09T04:01:35 |
371247
|
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|
Stack Exchange
|
A Green's function for the Laplacian on k-forms
Let $X$ be a compact, oriented, Riemannian $n$-fold. Then we have a Laplacian operator $\Delta = d d^{\ast} + d^{\ast} d$ from $\Omega^k(X)$ to itself. We have the Hodge decomposition $\Omega^k(X) = \mathrm{Ker}(\Delta) \oplus d \Omega^{k-1}(X) \oplus d^{\ast} \Omega^{k+1}(X)$; these summands are called harmonic forms, exact forms and co-exact forms. The map $\Delta$ takes each summand to itself, acting by $0$ on the harmonic forms and bijectively on the other two summands. Let $\Gamma : \Omega^k(X) \to \Omega^k(X)$ be the operator which is $0$ on the harmonic forms, and which is inverse to $\Delta$ on the exact and the co-exact forms.
When $k=0$, the operator $\Gamma$ can be described explicitly using the Green's function. This is a function $G(x,y)$ on $(X \times X) \setminus (\mbox{diagonal})$ such that $\Gamma(f)(y) = \int_{x \in X} G(x,y) f(x) \ \mathrm{Vol}$, where $\mathrm{Vol}$ is the $n$-form coming from the Riemannian metric and the orientation of $X$.
I'd like to know the analogous statement for other values of $k$.
Here is what I suspect is the answer. Recall that, if $\pi : Y \to Z$ is a proper submersion from
a $(d+e)$-fold to a $d$-fold, with oriented fibers, then we get a map $\pi_{\ast} : \Omega^{k+e}(Y) \to \Omega^k(Z)$ such that, for any $k$-dimensional cell $U$ sitting in $Z$, we have $\int_U \pi_{\ast}(\omega) = \int_{\pi^{-1}(U)} \omega$. Let $\pi_1$ and $\pi_2$ be the coordinate projections from $X \times X$ to $X$. Then I suspect there is some $n$-form $\gamma$ on $(X \times X) \setminus (\mbox{diagonal})$ such that $\Gamma(\omega) = (\pi_2)_{\ast} (\gamma \wedge \pi_1^{\ast} (\omega))$. If we have orthonormal eigenbases $\omega_{k,j}$ of $\Delta$ on $\Omega^k(X)$, with eigenvalues $\lambda_{k,j}$, then I expect
$$\gamma = \sum \frac{1}{\lambda_{k,j}} (\pi_1^{\ast} \omega_{k,j}) \wedge (\pi_2^{\ast} (\omega_{k,j}^{\ast})). \qquad (1)$$
Here $\omega_{k,j}^{\ast}$ is Hodge star. The letter $\gamma$ is meant to suggest that this is a generalization of the Green's function.
So first question
Is this right? For example, does the sum in (1) converge, and in what sense?
Then the second question:
How do I get a hold of this $n$-form $\gamma$ explicitly? Does it obey some differential equation? For example, what is it for the $n$-sphere?
I started wondering about this stuff because I wrote an answer over on math.stackexchange about how to compute Hodge decomposition on a torus, and then I realized that I wouldn't know how to do it for a sphere. I do know how to compute the classical Green's function for a sphere; see here.
There is a paper by Stern, Fixed Point Theorems from a de Rham Perspective, that uses "Schwarz forms" for the heat kernel (replacing your $1/\lambda$ by $e^{-t \lambda}$). Your sum seems formally correct, but I'm not sure about convergence. It also seems to me there must be some relevant abstract theorems in functional analysis about integral kernels, but I'm rusty on this. The inverse of the Laplacian is a pseudodifferential operator of order -2, of course.
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2025-03-21T14:48:31.992841
| 2020-09-09T05:59:54 |
371249
|
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|
Stack Exchange
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What do the eigenvalues of a random element of $\mathbb Z_\ell[\Gamma]$ look like?
Let $\Gamma = \varprojlim \Gamma_n$ be a profinite group with $\Gamma_n$ finite quotients. For concreteness, let us fix $\Gamma_n = \operatorname{PGL}_2(\mathbb Z/\ell^n)$ so $\Gamma = \operatorname{PGL}_2(\mathbb Z_\ell)$.
Let $R = \mathbb Z_\ell[[\Gamma]]$ be the completed group algebra and consider its action on itself. If we pick a random element $\gamma \in R$, what does the distribution of its eigenvalues look like?
More precisely, call the image of $\gamma$ in $R_n = \mathbb Z_\ell[\Gamma_n]$ as $\gamma_n$. Now $\gamma_n$ acts linearly on the finite free $\mathbb Z_\ell$-module $R_n$ and let its set of eigenvalues be $S_n$. Can we say anything about the distribution of $S_n$ as $n\to \infty$?
For a simpler situation, we can consider $\Gamma = \mathbb Z_\ell$ and then $R \cong \mathbb Z_\ell[[t]]$ and if $\gamma = f(t)$, then $\gamma_n$ will have eigenvalues $f(\zeta_{\ell^n}^i-1)$ for $i=1,\dots, n$ (unless I made a mistake in the computation).
I am not even sure what we can say about this set as $n\to \infty$.
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2025-03-21T14:48:31.992928
| 2020-09-09T08:14:37 |
371252
|
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|
Stack Exchange
|
Control and observability of Clifford algebra and quaternion valued systems?
Good evening everybody, I'm asking you if there is a mathematical theory about control and observability of differential systems within the quaternionic and Clifford (geometric) algebras. And if so, please provide me some literature. Thank you.
H. Banouh.
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2025-03-21T14:48:31.992978
| 2020-09-09T09:10:57 |
371254
|
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|
Stack Exchange
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On a limit for the resolvent norm
Let $H$ be a complex, infinite dimensional, separable Hilbert space. Fix any two nonzero operators $A,B \in B(H)$ such that $B$ is not a scalar multiple of $A$. It is well known that:
$$ \| R_A (z) \| \rightarrow 0 \quad \text{when}\, |z| \rightarrow +\infty $$
This easily follows from:
$$\|R_A(z) \| \leq \frac{1}{|z| - \| A \|} \quad \forall z \in \mathbb{C} : |z| > \| A \|$$
Now consider the operator $A+ cB$ where $c$ is a positive real number, and fix some complex number $z$. Can we say that, for all $\epsilon >0$, there exists $M>0$ such that:
$$ c > M \, \, \text{such that} \, \, z \not \in \sigma(A+cB) \Rightarrow \|R_{A+cB} (z)\| < \epsilon$$
Intuitively, for $c \rightarrow +\infty$, we would have:
$$ R_{A+cB}(z)=(z-A-cB)^{-1}= c^{-1}(z/c - A/c -B)^{-1} "\rightarrow" 0$$
because $|z/c| \rightarrow 0$, $\|A\| / c \rightarrow 0$, $1/c \rightarrow 0$ and $B$ is fixed. However, I'm not aware of any reference for a result of this kind. Does anybody have a reference or a proof of this?
You need B to be invertible for this to work.
As Michael already observed: It is simple to construct counterexamples if $B$ has a nontrivial kernel $N(B)$ and $A$ maps $N(B)$ into itself, since on $N(B)$ the size of $c$ plays no role.
But even if $B$ is an isomorphism, even the identity $B=I$, the conjecture is false: $z-A-cB=(z-c)-A$. The maximum of the norm of the inverse (over all $z$ from the corresponding resolvent set) is clearly independent of $c$.
Anyway, if $B$ is an isomorphism, there are some estimates:
There holds $(z-A-cB)^{-1}=B^{-1}((z-A)B^{-1}-c)^{-1}$, and one can use the resolvent estimate mentioned in the question to obtain
$$\|R_{A+cB}(z)\|\le\|B^{-1}\|\frac1{|c|-\|(z-A)B^{-1}\|}.$$
This gives an estimate if $|c|$ is large compared to $|z|\|B^{-1}\|$.
If $|z|$ is large compared to $|c|\|B\|$, the estimate from the question can be used.
If neither is the case, there need not be a good estimate as mentioned above ($B=I$).
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2025-03-21T14:48:31.993124
| 2020-09-09T09:20:34 |
371255
|
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|
Stack Exchange
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Infinite dimensional irreducible representations of a tensor product
The second part of Theorem 3.10.2 of "Introduction to representation
theory" by Etingof, Golberg, Hensel, Liu, Schwender, Vaintrob and Yudovina states that
if $A$ and $B$ are $k$-algebras ($k$ an algebraically closed field) and $M$ is an irreducible finite dimensional representation of $A\otimes_k B,$ then $M\cong V\otimes_k W$ where $V$ and $W$ are finite dimensional irreducible representations of $A$ and $B$ respectively.
My question is about the first part of the remark following this theorem. This remark states that the previous proposition fails for infinite dimensional representations, "e.g. it fails when A is the Weyl algebra
in characteristic zero." I don't see how to construct an irreducible infinite dimensional representation $M$ of $A\otimes B,$
where $A$ is the Weyl algebra, such that $M\ncong V\otimes_k W$.
(I asked the same question on Math.SE more than one year ago without receiving answers, also after starting bounties)
Welcome. You should have asked it here sooner! It seems kind of silly to wait a whole year. Also, we don't use 'et al', so I fixed that for you =].
Over on math.SE Nate suggests the module over the Weyl algebra $A_2$ generated by $e^{xy}$ which seems like a sensible guess to me but I'm not sure how to prove that it's irreducible: https://math.stackexchange.com/questions/3349346/infinite-dimensional-irreducible-representations-of-a-tensor-product
Nate's suggestion on math.SE works. We'll show that if $A = k[x, \partial_x]$ and $B = k[y, \partial_y]$ are both taken to be the Weyl algebra, then the module over $A_2 = A \otimes B \cong k[x, \partial_x, y, \partial_y]$ generated by $e^{xy}$ is 1) simple and 2) not a tensor product of simple modules of $A$ and $B$.
Explicitly this module $M$ consists of elements of the form $f(x, y) e^{xy}$ where $f$ is a polynomial, with the obvious action by multiplication and differentiation. Abstractly it is the quotient of $k[x, \partial_x, y, \partial_y]$ by the left ideal $(x - \partial_y, y - \partial_x)$. We can show very straightforwardly that every nonzero element of $M$ generates it, by computing that
$$(\partial_x - y) x^i y^j e^{xy} = ix^{i-1} y e^{xy}$$
and similarly that
$$(\partial_y - x) x^i y^j e^{xy} = j x^i y^{j-1} e^{xy}.$$
In other words, this module is the pullback of the usual module $k[x, y]$ under the (edit: inverse of the) automorphism $A_2 \to A_2$ sending $x \mapsto x, y \mapsto y, \partial_x \mapsto \partial_x - y, \partial_y \mapsto \partial_y - x$. Now $k[x, y]$ is irreducible which means so is this module (explicitly, every nonzero element is a generator because we can repeatedly differentiate resp. apply the above maps to get to $1$ resp. $e^{xy}$), and $k[x, y]$ is a tensor product $k[x] \otimes k[y]$.
But we can show that $M$ is not such a tensor product (this property is not invariant under twisting by automorphisms). If it were such a tensor product $V \otimes W$, then a pure vector $v \otimes w$ would have the property that there is some differential operator $a(x, \partial_x) \in k[x, \partial_x]$ such that $(a \otimes 1)(v \otimes w) = av \otimes w = 0$; in words, it would satisfy a polynomial differential equation involving $x$ only.
No nonzero element of $M$ has this property. The key point is that every nonzero element of $k[x, \partial_x]$ makes elements bigger in the lex order: we have
$$x \left( x^i y^j e^{xy} \right) = x^{i+1} y^j e^{xy}$$
$$\partial_x \left( x^i y^j e^{xy} \right) = ix^{i-1} y^j e^{xy} + x^i y^{j+1} e^{xy}$$
and so formally, writing an arbitrary element $a \in k[x, \partial_x]$ as a sum $\sum a_{k, \ell} \partial_x^k x^{\ell}$, we see that if $\partial_x^k x^{\ell}$ is the largest monomial in $a$ in the lex order where $x > \partial_x$, then $a(x^i y^j e^{xy})$ has largest monomial $x^{i + \ell} y^{j + k}$, and in particular it does not vanish, so the same is true if $x^i y^j e^{xy}$ is replaced by any other element with the same largest monomial.
Nice. Yea this is exactly what I was thinking. I had meant to come back and flesh this out earlier, but I forgot until now.
What if $V$ is a faithful module of $k[x, \partial_x]$?
@Vit Well, the Weyl algebra is a simple algebra, so all nonzero modules are faithful. But being faithful isn't a problem. Having an element with zero annihilator would be a problem, but that's not possible for a simple module.
@JeremyRickard Yeah, faithfulness is not the property I want. What I should've really asked is why is there always $a$ that annihilates $v$. I still don't see it.
@Vit Otherwise the (left) submodule generated by $v$ is isomorphic to the free module $k[x,\partial_x]$, which is not irreducible since $k[x]$ is a proper submodule.
@JeremyRickard Thanks!
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2025-03-21T14:48:31.993682
| 2020-09-09T11:33:24 |
371258
|
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|
Stack Exchange
|
A class in the motivic cohomology group $H^{0,1}(\operatorname{Spec}k;\mathbb{Z}/p)$
In the following paper
N. Yagita, Examples for the Mod p Motivic Cohomology of Classifying Spaces,
on the first page, below display (1.1), it says "It is known that there is an element $\tau\in H^{0,1}(\operatorname{Spec}k;\mathbb{Z}/p)$...".
So, where and how exactly is this class $\tau$ defined?
Since there is no motivic stable cohomology operation of bidegree $(0,1)$, is it true that this class $\tau$ vanishes after stabilization?
The definition of the class is actually given in the cited sentence. For the relevant motivic cohomology group we have
$${\rm H}^{0,1}({\rm Spec} k,\mathbb{Z}/p\mathbb{Z})\cong {\rm H}^0_{\rm ét}({\rm Spec } k,\mu_p)\cong \mu_p(k)$$
The complex realization (denoted $t_{\mathbb{C}}^{\ast,\ast}$ in Yagita's paper) in this case boils down to an identification $\mu_p(k)\cong\mathbb{Z}/p\mathbb{Z}$, hence the class $\tau$ is basically a choice of primitive $p$-th root of unity in $k$.
The notation is pretty standard in the literature, it already appears in Voevodsky's "Motivic cohomology with $\mathbb{Z}/2$-coefficients" (but of course, in the $p=2$ situation, there is only one choice of primitive 2nd root of unity).
Not sure what is meant with the question on stabilization. Stability in the context of cohomology operations refers to compatibility with the suspension isomorphisms. Of course, as a class in the motivic cohomology of the base field, the class $\tau$ doesn't die under the suspension isomorphisms. I think there's some misconception concerning the motivic cohomology operations in the question. The motivic Steenrod algebra is an algebra over the motivic cohomology of the base field, hence "multiplication by $\tau$" actually is a bistable motivic cohomology operation of bidegree $(0,1)$.
Thank you for your answer! I do apologize for the second question, which, as you pointed out, involves some misconception. What I meant to ask is answered by the sentence "Of course, as a class...", which, as I realize now, is tautology.
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2025-03-21T14:48:31.993836
| 2020-09-09T13:22:44 |
371260
|
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|
Stack Exchange
|
Separating a lattice simplex from a lattice polytope
Let $P\subset\mathbb{R}^n$ be a convex lattice polytope.
Do there always exist a lattice simplex $\Delta\subset P$ and an affine hyperplane $H\subset\mathbb{R}^n$ separating $\Delta$ from the convex hull of the integer points of $P\setminus \Delta$?
This is equivalent to say that there exist a degree one polynomial $h:\mathbb{R}^n\rightarrow\mathbb{R}$ that is positive on all the integer points of $\Delta$ and negative on all the integer points of $P\setminus \Delta$.
Yes, it is convex.
Just to double check, a "lattice simplex" means a simplex with integer vertices, that doesn't contain integer points apart from the vertices? I.e., you don't assume that its volume is $\frac{1}{n!}$ (otherwise the statement would not hold).
@DmitriPanov: Usually a lattice simplex just means a simplex whose vertices are integral (lattice points). They might be empty or not.
I'm not sure if there's a standard term for simplices unimodularly equivalent to the standard simplex (these would be the ones with volume $1/n!$).
Oh, maybe "unit simplex" is sometimes used for those (smallest lattice simplices).
Yes, "lattice simplex" means a simplex with integer vertices, that does not contain integer points apart from the vertices.
@F_L: That would usually be called an "empty lattice simplex"; see, e.g., https://arxiv.org/abs/1908.08933.
This is possible and here is how to do this. We will use an inductive argument, assume that the statement holds for polytops of dimension $<n$ and prove it for dimension $n$.
Take any vertex $v$ of the $n$ dimensional polytop $P$ and denote by $v_1,\ldots, v_m$ all the end-points of all the edges of $P$ starting at $v$. Let $P'$ be the convex hull of $v,v_1,\ldots, v_m$. Let $P''$ be the convex hull of all the integer points in $P'$ except $v$. Clearly, $P''$ doesn't contain $v$.
Now, take any face of $P''$ that is "visible from $v$", i.e. you can connect it with $v$ by a straight segment that doesn't intersect $P''$ in its interior point (if $P''$ is degenerate and has dimension $n-1$, we take the whole $P''$ as such a face). Denote by $H$ the hyperplane that contains this face. It follows from the construction, that $H$ intersects only whose edges of $P$ that are adjacent to $v$.
Now let's cut $P$ along $H$ and take the part that contains $v$, and call it $Q$. Denote by $F$ the face of $Q$ that lies in $H$. All its vertices lie on the edges of $P$ adjacent to $v$. It is easy to see that $Q$ has a structure of a cone over $F$ with vertex $v$. By construction, integer points in $Q$ is the union of those in $F$ with $v$. Next, apply to $F$ the inductive step and cut a simplex out of it by a certain hyperplane $H'$ (of dimension $n-2$) contained in $H$. To finish, rotate a tiny bit $H$ around $H'$. This is the hyperplane we were looking for.
ah, sorry, I misread the definition of $P''$
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2025-03-21T14:48:31.994049
| 2020-09-09T13:23:18 |
371261
|
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|
Stack Exchange
|
Cohomogeneity one action on $S^7$-bundles over $S^8$
Is it known if the total space of an $S^7$-bundles over $S^8$ with structure group $SO(8)$ admits a cohomogeneity one action?
I don't think the argument is written down anywhere, but not all $S^7$-bundles over $S^8$ admit cohomogeneity one actions.
Focusing on those with Euler class $\pm 1$, the resulting total space $E$ is a homotopy sphere. Homotopy spheres admitting cohomogneity one actions have been classified by Straume in
Compact connected Lie transformation groups on spheres with low cohomogeneity I. Mem. Amer. Math. Soc., 119(569), 1996.
In particular, Straume shows that in dimensions not of the form $4k+1\geq 5$, that a homotopy sphere admitting a cohomogeneity one action is diffeomorphic to the standard sphere. (In dimension $4k+1$, one can also get Brieskorn varieties.)
So, in dimension $15$, the only homotopy sphere which admits a cohomogeneity one action is the standard sphere. Since there are exotic spheres which are the total space of an $S^7$-bundle over $S^8$ (as shown by Shimada in
Differentiable structures on the 15-sphere and Pontrjagin classes of certain manifolds, Nagoya Math. J. 12 (1957), 59–69. MR0096223 (20 #2715) Zbl 0145.20303,
the result follows.
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2025-03-21T14:48:31.994148
| 2020-09-09T13:44:09 |
371263
|
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"prochet"
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|
Stack Exchange
|
formal smoothness and cotangent complex
If $k$ is a field and $A$ is a formally smooth $k$-algebra, then we know that $\Omega^{1}_{A/k}$ is projective. What about its cotangent complex $L_{A/k}$? When is it quasi-isomorphic to $\Omega^{1}_{A/k}$ put in degree zero? If $A$ is noetherian, it is always true, by Popescu's theorem and the reduction to the finite type case.
In general, I would expect to be wrong but is it true in the intermediate case where $A=colim B_i$ where the $B_i$'s are $k$-algebra of finite type with smooth transition maps?
The cotangent functor is left Quillen, so $\mathbb{L}{\varinjlim B_i} \simeq \varinjlim (\mathbb{L}{B_i}\otimes_{B_i}^{\mathbf{L}}A)$. Taking all the transition maps to be smooth, that makes the ind-smooth case OK.
transition maps are smooth but one $B_i$ could be singular, so that every other $B_j$ has the same singularities as $B_i$.
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2025-03-21T14:48:31.994237
| 2020-09-09T14:20:14 |
371264
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|
Stack Exchange
|
Bayesian inference of stochastically evolving model parameters
I have a question related to self-calibration in radio interferometry, but I will try to phrase it as generic as possible. I have a set of data points, $D = \{ d_{0, t_0}, d_{1, t_0}, ..., d_{M, t_0}, d_{0, t_1}, d_{1, t_1}, ..., d_{M, t_1}, ..., d_{0, t_n}, d_{1, t_n}, ... d_{M, t_n} \}$, taken at different time intervals, $t = 0, 1, ..., n$ to which I want to fit a model. For each time interval I have $N$ model parameters $\theta = \{\theta_{0, t_0}, \theta_{1, t_0}, ..., \theta_{N, t_0} \}$ and $M = N ( N - 1 ) / 2$ data points.
I have generated some noisy simulated data from my model so that the model parameters evolve with time in a stochastic manner (which is close to real observations; see Figure 1 for the evolution of my simulated model parameters) and my aim is to infer those parameters. Due to SNR restrictions, I can not infer the model parameters in every single time interval, so instead what I am doing is to group together data points taken e.g. at 5 consecutive time intervals and minimise the likelihood of my model given the data to find the average $\bar{\theta}_{k, t^{'}_0}$, $(k = 0, 1, ..., N)$ for $t_0< t < t_5$ and so on. In this way I infer the average set of $N$ model parameters for each binned up subset of my data, independently (seems to be working fine).
However, my approach does not take advantage of the fact that the evolution of the model parameters is a stochastic process, which means $\theta_{k, t_{n+1}}$ is not independent from $\theta_{k, t_{n}}$. I am wondering if I can solve for all the model parameters simultaneously and introduce Gaussian priors so that the prior for the k'th model parameter at $t_n$ is centered at $\theta_{k, t_{n - 1}}$. In this case would the expression to minimize look something like this,
\begin{equation} \label{eq:self_calibration}
P({\theta} | d) = \prod_{n=0} \left( L(d_n | {\theta_n}) \prod_{k=1} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{\frac{(\theta_{k, t_n} - \theta_{k, t_{n-1}})^2}{\sigma^2}} \right)
\end{equation}
where $d_n = \{ d_{0, t_n}, d_{1, t_n}, ... d_{M, t_n} \}$, $\theta_n = \{\theta_{0, t_n}, \theta_{1, t_n}, ..., \theta_{N, t_n} \}$ and $\sigma$ will be fixed to some value. Does my problem fall into a particular category of bayesian statistics problems, so I can read up more about this?
Have you tried asking at CrossValidated [https://stats.stackexchange.com/]?
|
2025-03-21T14:48:31.994410
| 2020-09-09T14:40:17 |
371266
|
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|
Stack Exchange
|
Extending a monoid object in a category
A monoid object in a pointed category $\mathcal{C}$ is an object $M$ equipped
with a multiplication morphism $\mu: M\times M\to M$ that is associative and unital, meaning that the diagrams that express those properties commute. A (two-sided) $M$ "module" also can be formulated in terms of arrows: we need action map $\alpha_R:X\times M \to X$ and $\alpha_L:M\times X\to X$ that are associative and unital. The arrows $M \to M\times X\to X$ and $M\to X\times M \to X$ should be equal; let's call it $t:M\to X$.
Now let $t: M\to X$ be the morphism from the monoid $M$ to its two-sided module $X$. I'd like to find an "extension" $e:M\to N$ of $M$ using $t$. The properties that I want for the extension are
$N$ should be a monoid object and $e$ should be a homomorphism
$e$ should factor $M\xrightarrow{t} T\xrightarrow{f}N$
if $h:M\to Q$ is a monoid homomorphism that factors $M\xrightarrow{t}T\xrightarrow{k} Q$ for some morphism $k:T\to Q$, then there is a unique homomorphism $g_k: N\to Q$ such that $k =g_k\circ f$.
I have a plan for how to make this construction. Set $N(0) = M$ and inductively define $N(k)$ by forming the pushouts of the diagrams
$$
N(k)
\longleftarrow
(N(k)\times_{M} N(0))
\cup
(N(k-1) \times_{M} N(1))
\longrightarrow
N(k)\times_{M} N(1)
$$
Then $N = \mathrm{colim}\, N(k)$ should do the job.
Some notes:
The notation $A \times_M B$ indicates a "tensor product of modules" defined to be the pushout of $A\times B \gets A\times M \times B\to A\times B$, using the action map of $A$ on the left and $B$ on the right.
I am using union as shorthand for a pushout.
Partial multiplications $N(k)\times_M N(\ell) \to N(k+\ell)$ would have to be defined along the construction.
I don't have any serious fears about this construction; but rather than work it all out and write it all down, I'd prefer a good reference.
Question: Is there a good reference for monoid extensions in this sort of categorical generality?
It sounds like you want a left adjoint for some 'forgetful'-looking functor (maybe from monoids under M to M-bimodules under M?). This seems to preserve all limits and all filtered colimits (I think?). So if we assume C is presentable, does that do it?
I thought about (I think) the exact same question just a few days ago. It works in a monoidal closed category, and the nice thing (which I assume is what you are interested in ?) is that the tranisiton map N(k) -> N(k+1) is a pushout of the k-fold "corner-product/Leibniz-product" of N-> X with itself. So that for e.g., in a monoidal model category, if N-> X is a (trivial) cofibration, then all the N(k)->N(k+1) (and their composite X-> Q) are as well. But unfortunately I havn't found a reference for this so far.
... Of course you need some assumptions about the product commuting to (some) colimits for this to work. But assuming monoidal closed is enough.
@SimonHenry Yes, I'm happy to load $\mathcal{C}$ with properties as need arises.
I don't think the question as you asked with the construction you are describing as been explicitely treated in the literature (though it very well could be).
What has been discused a lot in the litterature is special case where $M$ is the trivial monoid (the terminal object, or more generally the unit for the product). This is the construction of the "Free monoid on a pointed object". See for example this paper of S.Lack and its references.
The construction in the paper of Lack do not seem to be quite the same as the one you are after though.
I suspect the construction you described can be extracted from Kelly's transfinite paper.
Specifically, he shows in section 23 how to construct this free monoid using the construction of the free algebra on a pointed endofunctor, which I think gives exactly the construction you are describing, but it is not said so explicitely.
Now, the more general case you are talking about (with M non trivial) can be recovered from this special case as follows: The construction I described above do not assume that we work with the cartesian product: any monoidal structure (with some colimit preservation properties) will suffice.
Because of this, you can work in the category of $M$-bimodule, with the tensor product $A \otimes_M B$ (which use the right $M$-module structure on $A$ and the left $M$-module structure on $B$ to define the tensor product, and the other two to endow the tensor product with a bimodule structure).
The unit object of $M$ with its obvious bimodule structure.
A "pointed object" here is exactly a bimodule $X$ with a bimodule morphism $M \to X$ as you describe.
A monoid in this monoidal category is the same as a monoid $N$ with a morphism of monoid $M \to N$. So what you are after is the construction of the free monoid on a pointed object in this category of bimodules, which can be constructed using Kelly's or Lack's paper (and probably other references).
In terms of assumption, you need some assumption of existence of colimits (to define the tensor product of bimodule and to perform your construction) and of preservation of colimits under products for the construction itself to work (and maybe also for the tensor product of $M$-bimodule to be a monoidal structure ?). The precise assumption you need depends on which construction exactly you are using, but the safe thing to do is to assume that you work in a cartesian (or monoidal) closed category.
|
2025-03-21T14:48:31.994875
| 2020-09-09T14:59:55 |
371267
|
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|
Stack Exchange
|
Cusps of hyperbolic surfaces under finite covers
The following statement seems true, but I don't know a proof or a reference for it (and I would like one).
Let $\Gamma< \operatorname{PSL}(2,\mathbb R)$ be a nonuniform lattice with one cusp. We may conjugate $\Gamma$ so that an element
$
\begin{pmatrix}
1 & s\\
0 & 1
\end{pmatrix}
$ generates the cusp of $\Gamma$. Let $\Gamma'$ be such that $\Gamma<\Gamma'$ be an index $2$ subgroup. Then, the element
$
\begin{pmatrix}
1 & s/2\\
0 & 1
\end{pmatrix}
$ generates the cusp of $\Gamma'$.
I imagine a similar statement would be true if both $\Gamma$ and $\Gamma'$ had two cusps.
I think you should have -1 on the diagonal.
Sorry I meant that $\Gamma< \operatorname{PSL}(2,\mathbb{R})$, and I edited the question to reflect this. So I don't think such things matter.
Assume just that $\Gamma$ has index $k$ in $\Gamma'$. Let $C \subset \mathbb R \cup \{\infty\}$ be the set of parabolic points for the action of $\Gamma$. Then $C$ is also the set of parabolic points for the action of $\Gamma'$, because if $\gamma \in \Gamma'$ is parabolic with fixed point $x$ then for some integer $i \ge 1$, $\gamma^i \in \Gamma$ is parabolic with the same fixed point $x$.
In general every $\Gamma'$ orbit of $C$ decomposes as a union of $\Gamma$ orbits, so
$$\#\text{cusps}(\Gamma') \le \#\text{cusps}(\Gamma)
$$
with equality if and only if the every $\Gamma'$ orbit of $C$ is just a single $\Gamma$-orbit. So assuming that
$$\#\text{cusps}(\Gamma') = \#\text{cusps}(\Gamma)
$$
it follows for each $x \in C$ the group $\text{Stab}(x;\Gamma)$ has index $k$ in $\text{Stab}(x;\Gamma')$.
For the case $x=\infty \in C$ it follows that if $\begin{pmatrix}
1 & s\\
0 & 1
\end{pmatrix}$ generates $\text{Stab}(\infty;\Gamma)$ then $\begin{pmatrix}
1 & s/k\\
0 & 1
\end{pmatrix}$ generates $\text{Stab}(\infty;\Gamma')$.
|
2025-03-21T14:48:31.995009
| 2020-09-09T15:26:51 |
371269
|
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|
Stack Exchange
|
A composition of a simple extension and a separable extension is simple
Let $K/L/M$ be a tower of finite field extensions with $K/L$ separable and $L/M$ simple (in the sense of being generated by a single element). How does one show that $K/M$ is also simple?
I know that this is standard material, but I am trying to understand the argument in the context of residue fields of smooth curves over a field.
Here is what I know. One should probably use Artin's primitive element theorem and so the point is to prove that $K/M$ has only finitely many field subextensions $K'/M$. One may replace $M$ by $K' \cap L$ to reduce to looking for only those subextensions that are disjoint from $L$. How does one see that there are only finitely many such $K'$? (I think I know how to conclude by using flat descent for étale algebras, but it would be ridiculous to argue like this: what is the correct argument?)
I would strongly encourage removing the argumentation in the second paragraph. I think that "I know that this is standard material, but I am trying to understand the argument in the context of residue fields of …" is much less confrontational and says the same, and anyone who's not swayed by it also won't be swayed by the longer current second paragraph.
The claim would follow from the following form of the primitive element theorem: If $L=K(a_1,\dots,a_n)$ with $a_1,\dots,a_{n-1}$ separable algebraic over $K$ and $a_n$ algebraic over $K$, then $L=K(a)$ for some $a$. Would that answer your question?
In the form suggested by @ArnoFehm (Theorem 5.1 of Fields and Galois theory (4.61)), Milne gives an elegant proof (probably the standard one, but I don't have any other reference to hand to check): the finitude of intermediate fields is obvious if the ground field is finite; otherwise, a generic linear combination of the $a_i$'s is a primitive element.
Thank you very much for your help.
|
2025-03-21T14:48:31.995171
| 2020-09-09T16:14:32 |
371272
|
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|
Stack Exchange
|
What about zeros of higher dimensional power series?
Let $K$ be a p-adic field, $O$ be the ring of integers and let $m$ be its maximal ideal. Let $f(X)=(f_1(X),~f_2(X), \cdots, f_n(X)) \in O[[X]]^n,\ X=(x_1,~x_2, \cdots, x_n)$ is a power series.
I am interested in the zeros of $f(X)$.
Now, $f(X)=0 \Leftrightarrow f_i(X)=0 , \ i=1,2,....,n.$
Now suppose I know the zeros of all component power series $f_i(X)=f_i(x_1,~x_2, \cdots, x_n)$ in $m$.
Does this constitute all zeros of $f(X)$ in $m$?
Isn't there matter of common zeros which are common to all $f_i(X)$ in $m$ ?
Suppose I have $n_i$ zeros of each $f_i(X)$ in maximal ideal $m$,
i.e., $f_1(X)$ has $n_1$ zeros in $m$, $f_2(X)$ has $n_2$ zeros in $m$ and so on.
Question:
Is it enough information to know about roots of $f(X)$ from these information ?
How many roots of $f(X)$ are there in maximal ideal $m$ ?
Edit: I am considering only power series not formal power series.
In general, the notion of zero of a formal power series is meaningless.
@AbdelmalekAbdesselam, probably Masmath means to speak only of substituting $x$-values in $m$.
@AbdelmalekAbdesselam, assuming they converge on the maximal ideal of p-adic field.
You have essentially answered your own question. For $n = 1$, for example, $f_i(X) = (X - x_{i 1})\dotsb(X - x_{i n_i})$ satisfies your conditions no matter what $x_{i 1}, \dotsc, x_{i n_i}$ we choose, and the choices can be made so as to have as few or (within obvious sanity checks) as many common zeroes as desired.
@LSpice, for $n>1$, we have to select the common zeros of each component power series. Am I right ?
Yes; there is no difference in this respect between $n = 1$ and $n > 1$. This is just definitions.
@LSpice, You said no difference. But for $n=1$, we have just $f(X)=f_1(x_1,x_2, \cdots, x_n)$, hence all roots of $f_1$ are roots of $f$. But for $n \geq 2$, we are taking the common roots of component power series. Isn't the difference? Can you please clear it ?
Oh, I missed that you were insisting that the number of polynomials be the same as the number of variables. (So, maybe, did you, since you wrote $f_1(x_1, x_2, \dotsc, x_n)$ instead of $f_1(x_1)$?) Yes, I agree that for $n > 1$ it is about common $0$s.
@LSpice, aah, I understand what you were saying. For $n=1$, you chosen $f(x_1)=(f_1(x_1),~f_2(x_2), \cdots, f_n(x_n)) \in O_K[[x_1,x_2, \cdots, x_n]]^1$.. Am I right, you were considering this one ? or Am I missing your above comment $f_i(X)=(X-x_{i1}) \cdots (X-x_{in})$ ?
I did not realise that you wanted the number of polynomials to be the same as the number of variables, so I was choosing $(f_1(x_1), \dotsc, f_m(x_1)) \in O[[x_1]]^m$ for arbitrary $m$. (Note my $f_i$ have degree $n_i$, not necessarily $n$.) Anyway, it doesn't much matter; the point is that you can manufacture polynomials with essentially arbitrary (and arbitrarily intersecting $0$ sets), so knowing the numbers of individual $0$s of several polynomials tells you essentially nothing about the number of their common $0$s (except crude upper bounds).
@LSpice, ok thanks. Little correction in your comment , it is not polynomial but power series
Yes, exactly. If we can already find polynomials exhibiting arbitrarily complicated behaviour, then there's no hope to be able to say anything better for power series.
|
2025-03-21T14:48:31.995397
| 2020-09-09T16:33:35 |
371274
|
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|
Stack Exchange
|
Projection onto column space perturbed by Gaussian noise
Suppose we have a matrix $X\in\mathbb{R}^{m\times n}$ (with $n \le m$) with iid standard Gaussian entries, and suppose we have noise matrix $W\in\mathbb{R}^{m\times n}$ with iid Gaussian entries, but with some small variance $\sigma_W < 1$. We know that the columns of $X$ and $X+W$ are linearly independent almost surely, so they form a basis. I am interested in knowing how the noise $W$ changes the projection matrix $X(X^TX)^{-1}X^T$ of $X$. For instance, denoting $X_W = X+W$ and $x_j$ as the jth column of $X$, can we say anything about $X_W(X_W^TX_W)^{-1}X_W^Tx_j$? I am especially interested in references that discuss this kind of problem. I know standard matrix perturbation theory results could be applied on this, but I am in particular interested in the high-dimension regime, that is, when $n$ and/or $m$ are large, are there concentration results available?
If you can reframe your question to be about the singular vectors of $X$ and $X_W$, there is some nice work by Vu that tells you how singluar vectors behave under random perturbation. In particular, the bounds are better than 'worse case' bounds from Wey's inequality for instance. The paper is here: https://arxiv.org/pdf/1004.2000.pdf
@SandeepSilwal This is a very interesting work, thank you for the reference!
|
2025-03-21T14:48:31.995525
| 2020-09-09T16:41:12 |
371275
|
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"user1504",
"zeraoulia rafik"
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|
Stack Exchange
|
What do you like in the mathematics of Vaughan Jones? And how Vaughan Jones liked mathematics to be?
Edit: Directed to mathematicians ,I think this is a suitable time to open this question to know more about the creative mathematician Vaughan Jones Probably his students are ready for getting answers from them right now .
From Terry Tao's Blog, Vaughan Jones, who made fundamental contributions in operator algebras and knot theory (in particular developing a surprising connection between the two), died this week, aged 67.
Vaughan Jones was a very creative mathematician and he was arguably an extreme point in the convex hull of all mathematicians. He will very much be missed. And as a scientific family from MO we extend our sincere condolences to his family, friends and students.
And here is my big list/simple question to know more about this creative mathematician is:
Question: What do you like in the mathematics of Vaughan Jones? and how Vaughan Jones liked mathematics to be ?
Addendum: This part is directed to his students and his colleagues. As Jon Bannon suggested and pointed out in the comments, since Vaughan Jones was a very creative mathematician we may get from Vaughan Jones' students or by his colleagues the state of the art of his vision? And how Vaughan Jones saw what the applications of planar algebra in Quantum theory should be? The best example is the reformulation of Riemann hypothesis using planar algebra. More than that, how Vaughan Jones saw the connection between planar algebra and number theory should be used to solve biggest open problems in pure mathematics like Riemann hypothesis and irrationality measure of pi? Also from the point of view of Vaughan Jones does the expedition growth of mathematics need more development of knot theory and its connection with other area of mathematics and physics?
His known results are pretty famous. In terms of 'unpublished results', I'm not sure, but Tao already mentions a good one in the blog post you have linked to.
@Tao said in the linked post " I have not had much occasion to directly use much of Vaughan’s mathematical contributions"
Dear Zeraoulia: Just a suggestion. Vaughan was a very creative mathematician. I would love to have a question asking about the state of the art of his vision/ the things he would like to have seen accomplished. Maybe some of his students or close colleagues might be able to provide some of these things on MO for future activity. For example, his theory of planar algebras may point to things many of us are unaware of...in any case, Jones will be sorely missed.
In my last year of high school (12th grade), I took a couple of classes at UC Berkeley, including discrete math and proofs (mostly for computer science students) taught by Vaughan Jones. That was a great class!
@John Bannon , I already what you have suggested to me , I wish that I covred your needs
A meta discussion about memorial questions is here: https://meta.mathoverflow.net/questions/4619/ron-graham-1935-2020-and-mo-memorial-questions
I agree with Matt F. "Lesser-known results" worked very well for John H. Conway but I don't think it works equally well for all mathematicians, and I don't think that every time a famous mathematician dies, there should automatically be an MO question asking for lesser-known results.
@TimothyChow, An MO question about every mathematician dies offer students and paople aknowldgemnts and checking his area of reasearch encourage students for research , The nice thing I saw in the blog of Terry Tao always share contributions and information about new died mathematician
@zeraouliarafik : Even if I accept everything you say in this last comment, it does not justify the title "Lesser-known results."
@TimothyChow, I agree with you I have edited the question turning it big list simple question to avoid iteration of this question in MO
It would be nice of people were not doing these knee-jerk reflex "close" votes on such questions; we have nothing to lose by learning about a mathematician's works and if someone in the community decides to honor them like this, then why not? why be so uptight and caught up in own self-importance to strike with your "close" hammers, being a bit more gracious for such instances only makes you more human.
This question feels...premature. There will come a day when it's time to write about Jones' mathematical contributions, but for most of his students and colleagues, he was a friend and news of his death was a shock. (It's possible that some of them learned about his death from this MO post.) Most of the people who could answer this question well are busy processing grief right now, and it feels a bit ghoulish to request that they amuse us by filling in answers to a big list question. Right now, I think this question falls afoul of Paul Siegel's 'tasteful' criterion.
@user1504 Really am confused and am sorry in the same time for mathematicians who they are voted to close this question, I think it is very appropriate to MO website to ask about advices or recomondation of mathematician who spent his life to serve scientific life and particulary mathematics .
I have to agree in a sense that this is a bit jarring to see such a question so quickly. I'm even still mourning another mathematician's passing and never found it fitting to compose a memorial question for him. I tried to write a reflective piece to deal with it, but stalled and hope one day to do so well. There is, though, a desire/impulse to immediately pay one's respects or deal with the grief by posting or writing something...even right away. It may feel as if one can hold onto the experience of hearing a mathematician's ideas one final time as a farewell.
@JonBannon, I think it is the suitable time for this question to be open because opening this question means encourage students and researchers to get benefits and acknowledgement about knot theory and planar algebera .
@zeraoulia rafik: The idea of encouraging students and researchers getting benefits and acknowledgements from the work of a recently deceased mathematician makes such students and researchers somewhat like a pack of hyenas. I think this is precisely the point. A good question here should celebrate the life and work of the mathematician, and early on the loss is a bit fresh to do so without it seeming a bit grotesque.
You used to bump the question by repeated edits. I hope you wont do again such a series of repeated edits. In addition I believe that the choice of the questions in the title is particularly poor and awkward.
Thanks Ycor , I think I will ask another with other reformulation
Your previous edits already include several attempts... You might also consider waiting until somebody else asks such a question.
|
2025-03-21T14:48:31.996028
| 2020-09-09T17:12:50 |
371278
|
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"Daniele Tampieri",
"GH from MO",
"Stanley Yao Xiao",
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|
Stack Exchange
|
The best estimation of the function $\vartheta(x;q,a)$
Consider the function $$\vartheta(x;q,a)=\sum_{p \leq x ,q|(p-a)}\log p=\frac{x}{\phi(q)}+O(\frac{x}{(\log x)^C}).$$
If the Riemann Hypothesis is ture, in the case $q=a=1$ we have $$|\psi(x)-x|<\frac{\sqrt{x}(\log x)^2}{8\pi}.$$
Now in the general case how well can the estimation be? (Can assume some further hypothesis like the GRH in the comment)
You will need GRH for Dirichlet $L$-functions, not just RH.
Hi ZZP, and welcome to the MathOverflow. This comment is just to say that there's a typo in the first text line below the first formula (true, not ture) and if you what to scale properly the round brackets in the big-$O$ term in said formula, you could use the TeX commands \left and \right placing them respectively before the opening bracket and the closing bracket.
It depends on what you want. On the one hand Montgomery conjectures that the estimate
$$\displaystyle \left \lvert \vartheta(x; q, a) - x \right \rvert = O_\epsilon \left(q^{-1/2} x^{1/2 + \epsilon} \right),$$
holds under GRH for Dirichlet $L$-functions for $q \ll_\epsilon x^{1 - \epsilon}$, and this error term is best possible for $q$ close to $x$ in size. On the other hand Iwaniec and Kowalski's book contains the following estimate from the same hypothesis:
$$\displaystyle \left \lvert \vartheta(x; q, a) - \frac{x}{\phi(q)} \right \rvert = O \left(x^{1/2} (\log qx) \right).$$
I am sure if one is careful one can track down an explicit implied constant, but I am not aware of this being done. Since most likely one cannot get away from dependence either on $q$ or some $\epsilon$, this may not be an interesting problem.
Thank you, it's helpful.
The first display does not seem right. For $x=1$ the LHS is $1$, while the RHS is $O(q^{-1/2})$, a contradiction.
@GHfromMO Perhaps I missed an assumption... looking at the statement it seems one should have to assume $q = O_\epsilon (x^{1- \epsilon})$. Indeed, one would not expect uniformity beyond that, since even Elliott-Halberstam only asserts moduli up to that range.
|
2025-03-21T14:48:31.996216
| 2020-09-09T17:36:36 |
371281
|
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|
Stack Exchange
|
Does there exist a complete metric space which is Rothberger (or Menger) but not Hurewicz?
A topological space $X$ is said to be a
Menger space if for each sequence $(\mathcal{U}_n)$ of open covers of $X$ there is a sequence $(\mathcal{V}_n)$ such that for each $n$ $\mathcal{V}_n$ is a finite subset of $\mathcal{U}_n$ and $\cup_{n\in\mathbb{N}}\mathcal{V}_n$ is an open cover of $X$.
Hurewicz space if for each sequence $(\mathcal{U}_n)$ of open covers of $X$ there is a sequence $(\mathcal{V}_n)$ such that for each $n$ $\mathcal{V}_n$ is a finite subset of $\mathcal{U}_n$ and each $x\in X$ belongs to $\cup\mathcal{V}_n$ for all but finitely many $n$.
Rothberger space if for each sequence $(\mathcal{U}_n)$ of open covers of $X$ there is a sequence $(U_n)$ such that for each $n$ $U_n\in\mathcal{U}_n$ and $\cup_{n\in\mathbb{N}}U_n=X$.
I did not find an example of a complete metric space which is Rothberger (or Menger) but not Hurewicz.
I have already posted this question on MSE but did not get answer. https://math.stackexchange.com/questions/3817583/does-there-exist-a-complete-metric-space-which-is-rothberger-but-not-hurewicz https://math.stackexchange.com/questions/3817530/does-there-exist-a-complete-metric-space-which-is-menger-but-not-hurewicz
If I remember well, a complete metric space is Menger iff it is $\sigma$-compact (shown by Hurewicz himself). Hence a complete metric Menger space is Hurewicz.
@Mathieu This should be the paper by Hurewicz in which the implication you mentioned is proved, but my German is a bit rusty. In any case more recently the result was extended to all Cech-complete spaces by Tall and Tokgöz
If a metric space is a Rothberger space, then it is separable, and it contains no subspace homeomorphic to the Cantor set. Hence a complete metric space is a Rothberger space if and only if it is countable and therefore a Hurewicz space. (I posted this as an answer which I deleted it because I saw it was already answered in comments.}
Definition 2.Hurewicz space, "but finitely many " -- is there a typo?
|
2025-03-21T14:48:31.996360
| 2020-09-09T19:01:18 |
371286
|
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|
Stack Exchange
|
What is a Shimura variety?
It so look like that Shimura variety is a modular space that contains classes of zeros of modular functions that have certain symmetry that have structure of Hodge type.
Is this true?
https://www.jmilne.org/math/xnotes/svi.pdf
In my opinion, to have a reasonable answer, this question needs to include more information about what you already know about Shimura varieties: What examples are you familiar with? Do you understand the pieces of the definition on a formal level and are looking for intuition about how they fit together, or is there some part of the definition you already don't understand? It would also be helpful to expand on terms in your question like "modular space", "contains classes of zeroes", "have certain symmetry".
I understood the definition of modular functions well,I also understood the idea of classes of varieties as quotient spaces,I understood the idea of a Hodge structure but I don't see clearly the intuition of Shimura variety as it fit into all these ideas...
Do you understand Hilbert modular surfaces and moduli spaces of principally polarized abelian varieties?
Know some Hilbert modular surface but not polarized abellian variety
Shimura variety.
As well as Milne's lecture notes, you could also try this two-page overview article in the AMS Notices, https://www.ams.org/notices/201211/rtx121101560p.pdf, whose title is precisely "What is ... a Shimura variety?"
@DavidLoeffler.thank you it clear some confusion
|
2025-03-21T14:48:31.996496
| 2020-09-09T20:21:50 |
371292
|
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"Mirco A. Mannucci",
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|
Stack Exchange
|
Subobject classifiers with a quantale structure
Quantales were kind of popular a few decades ago, as some sort of "quantum" version of the much more widespread concept of locale (Mulvey introduced them to study $C^*$ algebras). I was under the impression that they more or less faded away, but, much to my delight, it looks like they are back (see for instance the interesting monograph by Pedro Resende here).
Somehow, since I found out about their existence, I had the idea that, just like in Topos Theory the structure of sub-objects (and thus also of their classifier) is a Heyting Algebra, some other sufficiently rich categories, for instance some monoidal closed categories, should have a "sub-object classier" whose structure is that of a quantale.
Here is the question:
Is there any category $C$ such that it admits a contra-variant $SUB(-)$ functor
factoring through the category of quantales, and moreover such that $SUB(-)$ is representable by "characteristic maps" into a distinguished object $\Omega$ ?
PRE-EMPTIVE CLARIFICATION: I am not interested in topoi with additional structure, so that the quantale comes up on top of the underlying heyting algebra. Rather, I would like the sub-object functor to have a native quantale structure. For fans of Linear Logic, I could say that I am interested in a category whose internal logic is linear in the full sense (any type, ie any object, has a quantale of predicates associated to it. Representability would extend the power to higher order linear logic)
ADDENDUM
Waiting for an answer, I have stumbled upon an old article by Borceux and others
here, which describes Quantale Sets. The idea is to build quantale sets just like one builds localic topoi. What is interesting is that this cat is kind of a quantale version of a topos, and does indeed have an internal quantale which acts as a sub-object classifier. Problem solved? Not quite. I still would like to have a general way to describe "Qopoi" (this is my own pet name for quantale version of a topos)
A tautological note: quantales are described by an infinitary Lawvere theory -- i.e. a quantale is just a set $X$ with certain infinitary operation $X^\kappa \to X$ satisfying certain equations. So if $\mathcal C$ is a category with all products, we can define the notion of an internal quantale object $\Omega$ in $\mathcal C$. Moreover, the functor $Hom_{\mathcal C}(-,\Omega): \mathcal C^{op} \to Set$ lifts to a quantale-valued functor just when $\Omega$ is an internal quantale object.
So unless you can formulate some more precise conditions on the functor $SUB$, the answer to the question is just "for any $\mathcal C$ with an internal quantale object $\Omega$, we have exactly this situation by defining $SUB = Hom_{\mathcal C}(-,\Omega)$." For instance, we could take $\mathcal C = Set$ and $\Omega = ([0,\infty], \geq, +)$, and then $SUB = Map(-,[0,\infty])$ is some kind of "generalized subobject" functor. Perhaps you want $SUB$ to literally be the "subobject poset" functor? In that case, things seem rather restrictive...
Good try Tim! But alas (or perhaps luckily) the story is a bit more involved. I want a cat $C$ where the REAL contra-variant Sub(-) functor does exist, which in particular entails that C has some degree of regularity. Now, that Sub(-) functor does not simply land in Sets, but I want it to have a full quantale structure. Then, assuming that such a category exists (to be honest I have a few candidates, but I am waiting to see what folks say), I want more: I want that this "quantale-enriched " functors be repesentable, in other words that
there exist a quantale object in C doing what is supposed to do with respect to SUB(-). To be a bit more clear, what I am actually after is a "universe of sets" much like a topos, except its internal logic is quantale based, rather than heyting. Please take a look at the 1989 article by Borceux for some idea (take a look at my comments in the addendum)
PS why too restrictive? Please elaborate on that. To me it does not look restrictive at all, actually the candidate category is a generalization of the topos itself (locales are just special varieties in the category of quantales). The topos is basically the classical-intuitionistic version of what I am after
Maybe not "too" restrictive, but it might be nice to have a bit more wiggle room. E.g. the poset of regular subobjects might be smaller. As I think about it, I'm surprised how difficult it is to even come up with examples of categories where the subobject lattices have any kind of multiplication on them not derived from the lattice structure. The closest thing I can think of is to let $\mathcal C = Monoids$ and let "$SUB$" be the collection of all subsets (not required to be submonoids) -- this has a quantale structure but is not representable.
Ok Tim, I think we are finally playing. Finally someone who is not scared to play! Very good. You are right, if C has only lims, colims, all that jazz, no way to get the TENSOR on subobjects. But how about starting from some C who has a tensor product already, say R-modules, and begin from there? R-mods does to cut it yet, but it will give you some intuition. Last note: you need to TENSOR sub-objects and get ANOTHER subobject.
Hmmm... If $SUB$ is really the subobject functor, then $\Omega$ must be a subobject classifier in the usual sense. I don't know many examples of categories with subobject classifiers which are not actually toposes. I think I can show under a local presentability hypothesis that any category with a subobject classifier has disjoint coproducts and a strict initial object, which is a modest step towards actually being a topos. So most likely any example will actually be a topos where the subobject classifier happens to have an exotic multiplication for some reason.
Ok, whereas I have no doubts that I can have examples of cats where SUB(-) is what I want it to be, I have my own doubts about $\Omega$. If you come up with a proof of an obstruction, please post it as an answer. PS In the article above, though, looks like Borceux has a quantalic $\Omega$, so ....
@TimCampion so, looks like after your last MO question and subsequent (slick) answer there is absolutely no doubt: there ARE cats with a sub classifier and with some limits which are neither topoi, nor are embeddable in them. The key point is that one has to drop the cartesian closedness.
|
2025-03-21T14:48:31.997040
| 2020-09-09T21:39:32 |
371296
|
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|
Stack Exchange
|
Does cohomology and base change hold if supported at a point?
I have a flat, quasicompact, and separated map $p : X \to \mathbb{A}^1$ and I know that $R^i p_* \mathcal{O}_X$ vanishes everywhere except possibly $0 \in \mathbb{A}^1$.
Q1: Does "cohomology and base change" automatically hold here? I.e., is the natural map
$(R^i p_* \mathcal{O}_X)_s \otimes_{\mathcal{O}_{\mathbb{A}^1, s}} k(s) \to H^i(X_s, \mathcal{O}_{X, s})$
an isomorphism for all $s \in \mathbb{A}^1$? This is clear for all nonzero $s$ and I believe it should follow for $0 \in \mathbb{A}^1$ automatically.
Q2: Does it then follow that $R^i p_* \mathcal{O}_X = 0$? Theorem 1.1 in Conrad's notes would then imply that $R^i p_* \mathcal{O}_X$ was locally free, hence constant rank 0, hence 0.
Q3: If $X \overset{v}{\to} Y \to \mathbb{A}^1$ are proper maps and both $X$ and $Y$ are flat over $\mathbb{A}^1$ ($v$ is proper birational, n.n. flat), then does cohomology of $\mathcal{O}_X$ along $v$ commute with base change in $\mathbb{A}^1$? The versions of Cohomology and Base Change I've seen have one map which must be both proper and flat, as opposed to a composite being flat and the first map proper. This is my original situation, and I got to the above by localizing in $Y$.
I've been trying to do this descending induction, where $R^N p_* \mathcal{O}_X$ is zero for $N \gg 0$ for dimensional reasons, so $R^{N-1} p_* \mathcal{O}_X$ commutes with base change, but I also need $R^{N-2} p_* \mathcal{O}_X$ to commute with base change to conclude $R^{N-1}p_* \mathcal{O}_X$ was locally free, hence zero, and continue.
It seems to me that you are assuming your hypothesis also for $i=0$, which is impossible — $p_*\mathscr{O}X$ contains $\mathscr{O}{\mathbb{A}^1}$.
What happens for the projection from $\mathbb{A}^3\setminus{0}$ to $\mathbb{A}^1$?
@abx I understand, but I'd like to at least get it for $i \geq 2$ if possible -- I'll take what I can get.
@PiotrAchinger Thank you for bringing up this example. I thought I checked it but I made a mistake. I don't think that arises in my situation because I got there by (Zariski) localizing a genuinely proper map $X \to Y$ in $Y$ and then I have an affine $Y$ over $\mathbb{A}^1$ (as in Q3). I don't believe $\mathbb{A}^3 \setminus 0$ can factor as a projective map to an affine over $\mathbb{A}^1$ because the map to an affine would have to factor through $\mathbb{A}^3$ and then the inclusion $\mathbb{A}^3 \setminus 0 \subseteq \mathbb{A}^3$ would have to be closed. I'll rewrite the question better.
|
2025-03-21T14:48:31.997223
| 2020-09-09T22:02:17 |
371297
|
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|
Stack Exchange
|
Integers $n$ such that $n^d + (n+1)^d$ is never prime
Let us call an integer $n>0$ pure if for all integers $d>0$ we have that $n^d + (n+1)^d$ is not prime. Is the set of pure integers non-empty? Is it computable?
$n^d+(n+1)^d$ is surely composite if $d$ is not a power of 2. So, one can restrict attention to the case $d=2^k$. That is, $n$ is pure iff the corresponding generalized Fermat numbers are all composite.
You may also check the factorisation of fermat numbers
The OP is allowing d to be 1, so the smallest candidate for purity, according to OEIS, is 28
@paulMonsky: Good point! Then a bettert OEIS reference is A057856, which by the way states a conjecture implying that there are no pure numbers.
|
2025-03-21T14:48:31.997318
| 2020-09-09T23:25:34 |
371301
|
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|
Stack Exchange
|
Find $a$ satisfying $x \cup_1 y = \delta a$ when $x,y \in Z^2(G,\mathbb{Z}_2)$
Let $G$ be a finite group. Let $x,y \in Z^2(G,\mathbb{Z}_2)$ be 2-cocycles. Find $a \in C^2(G,\mathbb{Z}_2)$ such that
\begin{align}
x \cup_1 y = \delta a.
\end{align}
Is there a general solution? Is it possible to know when a solution exists?
Where:
\begin{align}
[x \cup_1 y](g,h,k) &= x(gh,k)y(g,h) + x(g,hk)y(h,k)\\
\delta a &= a(g,h)+a(gh,k)-a(g,hk)-a(h,k)
\end{align}
What is $\cup_1$ here? Is it the cup product? Isn't the cup product of two $2$-cocycles a $4$-cocycle?
@QiaochuYuan I added an explicit expression. You are right, the cup product of two 2-cocycles is a 4-cocycle. The $\cup_1$ product of two 2-cocycles is a 3-cochain.
The $\cup_1$-product is not always a coboundary. For example, for a cocycle $x\in Z^2$, $x\cup_1 x\in Z^3$ is a representative of the class $\operatorname{Sq}^1 x$. This can be non-trivial already for $G=C_2 \times C_2$.
In general, $x\cup_1 y$ (for different cocycles $x, y$) is not even a cocycle, so in that generality there is no such $a$ for trivial reasons.
Good point! Thanks.
|
2025-03-21T14:48:31.997420
| 2020-09-09T23:47:46 |
371302
|
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|
Stack Exchange
|
Counting the number of linearly independent primitive elements of a field extension
Let $L/K$ be a field extension with extension degree $n>1$. We say $L/K$ is simple if $L=K(a)$ for some a in $L$. In this case, $a$ is called a primitive element. My question is now about the dimension of the largest K-subspace of L whose nonzero elements are primitive, in the context of a finite extension of a finite field.
An element is not simple if and only if it lies in a proper sub-extension. The sub-extensions are parameterised by the divisors of the degree of the extension. Now do inclusion–exclusion.
It is true, but then how do you compute the dimension of the subspace generated by simple elements? In case $K$ is infinite, things work as one can use a well-known linear algebra theorem that asserts a vector space over an infinite field cannot be written as a union of its subspaces. We don't have such result for finite finite base field. See my manuscript about this: https://arxiv.org/pdf/1912.05411.pdf
Ah, got it. I was reading the question in the title, rather than the one in the body. I agree that this is less obvious.
A related question: https://mathoverflow.net/questions/333715/enumeration-of-field-generators-of-a-finite-field-over-mathbbf-q-which-a
Every finite extension of a finite field is simple, isn't it? If the degree $n$ is prime, then every element of $L$ that's not in $K$ is a simple element. It should be possible to write $L$ as a direct sum of $K$ and a vector space $V$ of dimension $n-1$ in this case, and all nonzero elements of $V$ will be simple.
Sure, every finite extension of a finite field is simple. The case that extension degree is prime is very special.
The standard terminology is that $a$ is then a primitive element.
Thanks, will edit it.
There still seem to be two different questions being asked— do you want $V\subseteq L$ such that every nonzero $a\in V$ is primitive, or just that every element of some basis for $V$ is primitive?
@Tim Campion I understand that these two problems are different but they are pretty related. What I want is to find the largest subspace $V$ whose every nonzero element is primitive.
Very nice question, Are you assuming the base field is finite ?
|
2025-03-21T14:48:31.997614
| 2020-09-10T04:00:24 |
371310
|
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|
Stack Exchange
|
Are there axioms satisfied in commutative rings and distributive lattices but not satisfied in commutative semirings?
Consider the language of rigs (also called semirings): it has constants $0$ and $1$ and binary operations $+$ and $\times$. The theory of commutative rigs is generated by the usual axioms: $+$ is associative, commutative, and has unit $0$; $\times$ is associative, commutative, and has unit $1$; $\times$ distributes over $+$; and $0$ is absorbing for $\times$.
Every commutative ring is a commutative rig (of course), and every distributive lattice as well (interpreting $\bot$ as $0$, $\top$ as $1$, $\vee$ as $+$, and $\wedge$ as $\times$). In fact, the category of commutative rings is a full reflective subcategory of the category of commutative rigs, as is the category of distributive lattices. The intersection of the two is trivial, in the sense that only the trivial algebra is both a ring and a lattice. (In a lattice, $\top \vee \top = \top$; but in a ring $1 + 1 = 1$ implies $0 = 1$.) What I am wondering is how close do these two subcategories come to capturing "all" the possible behaviour of commutative rigs. More precisely:
Question 1. Is there a Horn clause in the language of rigs that is true in every commutative ring and every distributive lattice but false in some commutative rig?
Since commutative rings are not axiomatisable in the language of rigs using only Horn clauses, I would also be interested to hear about, say, cartesian sequents instead of Horn clauses. This can be phrased category theoretically:
Question 2. Is there a full reflective subcategory $\mathcal{C}$ of the category of commutative rigs that is closed under filtered colimits and contains the subcategories of commutative rings and distributive lattices but is not the whole category? (Furthermore, can we choose such a $\mathcal{C}$ so that the reflection of $\mathbb{N} [x]$ (= the free commutative rig on one generator) represents a monadic functor $\mathcal{C} \to \textbf{Set}$?)
I don't want to be too permissive, however – since commutative rings and distributive lattices can both be axiomatised by a single first order sentence in the language of rigs, taking their disjunction yields a sentence that is true in only commutative rings and distributive lattices but false in general commutative rigs.
Here is an example of a first-order axiom that is true in commutative rings and distributive lattices that is false in some commutative rig:
For all $a$ and $b$, there exist $c$ and $d$ such that $(c + d) a + d b = b$.
This axiom amounts to saying that every ideal is subtractive (which is a second-order axiom prima facie); it is the case that every ideal in a commutative ring or distributive lattice is subtractive. The way I prefer to think about it is that in a commutative ring, ideals are automatically subtractive because $-1$ exists, and in a distributive lattice, ideals are automatically subtractive because they are downward closed. As it turns out, this can be expressed as a first-order sentence, albeit not a cartesian sequent.
How about $x+z=y+z, x\times z = y \times z \vdash x = y$? There should be a rig where that doesn't hold, but it is true in every commutative ring and every distributive lattice.
Very nice! That axiom says that for a ring or a distributive lattice $(-) + z$ and $(-) \times z$ are jointly injective, which seems to be worth thinking about. It fails in $\mathbb{N} \cup { +\infty }$ (which is a semiring if we define $0 \times {+ \infty} = 0$). I will accept that as answer.
Following François's suggestion, I ran alg to find a unital commutative semiring which fails to satisfy
$$
\forall x\, y\, z,\; x + z = y + z \land x \times z = y \times z \Rightarrow x = y.
\tag{1}
$$
The smallest one has size 3. Here is the output of the program, cut off after the first example.
Theory unital_commutative_semiring.
Constant 0 1.
Binary + *.
Axiom: 0 + x = x.
Axiom: x + 0 = x.
Axiom: x + (y + z) = (x + y) + z.
Axiom: x + y = y + x.
Axiom: 1 * x = x.
Axiom: x * 1 = x.
Axiom: x * (y * z) = (x * y) * z.
Axiom: x * y = y * x.
Axiom: (x + y) * z = x * z + y * z.
Axiom: x * (y + z) = x * y + x * z.
Axiom: 0 * x = 0.
Axiom: x * 0 = 0.
# Extra command-line axioms
Axiom: not forall x y z, x + z = y + z /\ x * z = y * z => x = y.
# Size 3
### unital_commutative_semiring_with_extras_3_1
+ | 0 1 a
--+---------
0 | 0 1 a
1 | 1 1 1
a | a 1 a
* | 0 1 a
--+---------
0 | 0 0 0
1 | 0 1 a
a | 0 a 0
We can also count the structures. For commutative unital semirings the counts are
size | count
-----|------
2 | 2
3 | 6
4 | 36
5 | 228
6 | 2075
The counts for commutative unital semirings that also satisfy (1):
size | count
-----|------
2 | 2
3 | 3
4 | 9
5 | 12
6 | 23
So clearly there will be some, but it looks like not too many for small sizes.
It is a bit weird that $2, 6, 36, 228, 2075$ is not in OEIS.
The example you give is even an affine quantale! Interesting...
The example could perhaps be concisely described as “the three-element semiring {0,1,infinitesimal}”.
A square-nilpotent infinitesimal no less.
Andrej, thank you for alg! It's a really wonderful tool, I wish I had remembered it when I made my suggestion in the comments since I did look at it a while ago but I forgot about it... Thank you for working it out for us!
There's an easy way to remedy the absence from OEIS...
|
2025-03-21T14:48:31.997974
| 2020-09-10T04:11:24 |
371311
|
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"Federico Poloni",
"Nathaniel Johnston",
"ZHz",
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"https://mathoverflow.net/users/164296",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371311"
}
|
Stack Exchange
|
How to check if there exists a non-negative element in an affine space?
Suppose that a $n$-dimensional affine space is defined with vectors $e_i=(0,\dots,1,\dots,0), i=1,\dots,n$ (the $1$ stands in the $i$-th place), i.e.
$$\mathcal{A}=\left\{\alpha_1e_1+\alpha_2e_2+\dots+\alpha_ne_n\mid\sum_{k=1}^n\alpha_k=1\right\}$$
and there is an affine subspace, e.g. $Ax+c$, in $\mathcal{A}$. The affine subspace is treated as a linear space $Ax$ shifted by vector $c$. In such case, linear space $Ax$ should be parallel to $\mathcal{A}$. The problem is that if there exist a non-negative element $x'$ in the affine subspace, i.e. $Ax'+c\ge0$. If not, it's desired to find out the closest point to achieve that goal.
The non-negativity requirement in the affine space $\mathcal{A}$ is equivelent to the definition of a standard $n$-dimensional simplex $$\Delta^n=\left\{\alpha_1e_1+\alpha_2e_2+\dots+\alpha_ne_n\mid\sum_{k=1}^n\alpha_k=1,\; \forall \alpha_k\ge0 \right\}$$
From the geometrical point of view, the problem is equivelent to check if a line or hyperplan, etc, intersects the simplex in the same affine space $\mathcal{A}$? if not, we want to find out the closest point to the simplex. Illustration
Is there some simple and efficient way to do so?
A quantifier is missing in $Ax+c \geq 0$. Do you want to check if there exists a non-negative element in an affine space? What is "a basis parallel to $\mathcal{A}$"?
@FedericoPoloni Yes, i want to check if there exists a non-negative element in the affine subspace in $\mathcal{A}$. sorry for that incorrect expressions. I treat the affine subspace as a linear space $Ax$ shifted by vector $c$. in such case, linear space should be parallel to $\mathcal{A}$.
Please try and edit the question to make it clearer and specify these details directly in the text.
This problem is a linear program.
@NathanielJohnston Could you give me some hints on how to define the objective and constraints? Thank you in advance.
|
2025-03-21T14:48:31.998130
| 2020-09-10T07:46:41 |
371317
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Dominic van der Zypen",
"bof",
"https://mathoverflow.net/users/43266",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371317"
}
|
Stack Exchange
|
Singular cardinal $\kappa$ with projective plane such that all edges have cardinality $<\kappa$
Is there an infinite singular cardinal $\kappa$ such that there is a set $E\subseteq{\cal P}(\kappa)$ with the following properties?
$|e| < \kappa$ for all $e\in E$,
whenever $\alpha\neq\beta\in \kappa$ there is $e\in E$ with $\{\alpha,\beta\} \subseteq e$, and
if $e_1\neq e_2\in E$ then $|e_1\cap e_2| = 1$.
(There can be no infinite regular cardinal with this property.)
Right :) Will amend
I thought this question sounded familiar, but I couldn't place it until now. The answer to this question follows trivially from the answer to this old question of yours: https://mathoverflow.net/questions/361955/subset-of-kappa-kappa-with-linear-intersection Namely, it follows from (1) and (2) that each point is on at least two lines.
Does this answer your question? Subset of $[\kappa]^{<\kappa}$ with linear intersection
Why has nobody else voted to close this question as a duplicate? Is it not a duplicate? What am I missing?
The answer is no. Let $\kappa$ be any infinite cardinal, regular or singular, and assume for a contradiction that there is a set $E\subseteq\mathcal P(\kappa)$ satisfying your conditions. I will call the elements of $\kappa$ points and the elements of $E$ lines.
The lines do not all go through one point: Given a point $\alpha$, choose a point $\beta\ne\alpha$ and a point $\gamma$ not on the line through $\alpha$ and $\beta$; the line through $\beta$ and $\gamma$ does not go through $\alpha$.
There are $\lt\kappa$ lines through any point: Consider any point $\alpha$ and let $\lambda$ be the number of lines through $\alpha$. Choose a line $e$ which does not go through $\alpha$. Since each line through $\alpha$ meets $e$ in a different point, $\lambda\le|e|\lt\kappa$.
Now choose two distinct points $\alpha$ and $\beta$. Say there are $\lambda$ lines through $\alpha$ and $\mu$ lines through $\beta$. Let $e$ be the line through $\alpha$ and $\beta$. Now every point which is not on the line $e$ is the point of intersection of a line through $\alpha$ and a line through $\beta$. Hence $\kappa\le|e|+\lambda\mu\lt\kappa$ which is absurd.
This argument is adapted from the proof that a finite projective plane of order $n$ has $n^2+n+1$ points. In that case we have $|e|=\lambda=\mu=n+1$ and the number of points is exactly $|e|+(\lambda-1)(\mu-1)=n^2+n+1$.
P.S. The answer is still no if condition (2) is weakened to "for every $\alpha\in\kappa$ we have $|\{e\in E:\alpha\in e\}|\gt1$". This more general result was proved in my answer to this old question.
|
2025-03-21T14:48:31.998304
| 2020-09-10T07:58:42 |
371319
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371319"
}
|
Stack Exchange
|
Doubt when calculating the S-transform of Hida differential operator
Assume we have a Hida test function $\varphi\in (\mathcal S)$, and $y\in \mathcal S'(\mathbb R)$. Define the Gateaux directional derivative of $\varphi$ (in the direction of $y$) by:
$$D_y\varphi(x):=\sum_{n=1}^{\infty} n \langle :x ^{\otimes (n-1)}:,\langle y, f_n\rangle\rangle,$$
where the sequence of (symmetric) $f_n$'s is the element in the Fock space that corresponds to $\varphi$ (via the Ito-Segal isomorphism).
At this point I need to calculate the $S$-transform of this element, but I am having some problems (probably trivial stuff) with the calculations:
I have that by definition of the $S$-transform we have
$$S(D_y\varphi(x))(\xi)=\sum_{n=1}^{\infty} n\big\langle \langle y,f_n\rangle, \xi^{\otimes (n-1)}\big\rangle.$$
From this I should be able to obtain
$$\sum_{n=1}^{\infty} n\big\langle y\hat{\otimes} \xi^{\otimes (n-1)},f_n \big\rangle,$$
but this very last step is not clear to me and I haven't been able to find an explanation.
N.Obata justifies this step by using contraction tensors but prefer to avoid that formulation.
Could you please explain me this? I have the feeling that it's something straightforward but I haven't been able to get it.
Thanks in advance!
EDIT:
A colleague told me that this could be a consequence of the Kernel Theorem , but honestly I don't see how.
EDIT 2:
So this is very informal but if one thinks of the pairing $\langle \cdot,\cdot\rangle$ as the inner product in $(L^2)$ (which is not entirely correct since one of the element could be outside $(L^2)$ i.e. if $y\in S'(\mathbb R)\backslash L^2(\mathbb R)$) we can write that:
\begin{align}
&\big\langle \langle y,f_n\rangle, \xi^{\otimes (n-1)}\big\rangle\\
=&\int_{\mathbb R^{n-1}} \int_{\mathbb R} f_n(t_1,\cdots,t_n)y(t_1)\xi^{\otimes (n-1)}(t_2,\cdots,t_n) dt_1d(t_2,\cdots,t_n)\\
=&\int_{\mathbb R^{n}}f_n(t_1,\cdots,t_n)y(t_1)\xi^{\otimes (n-1)}(t_2,\cdots,t_n) d(t_1,t_2,\cdots,t_n)\\
=&\langle y\otimes \xi^{\otimes (n-1)},f_n\rangle,
\end{align}
But again this is informal and I don't know if I can justify all steps, what do you think?
|
2025-03-21T14:48:31.998450
| 2020-09-10T11:29:29 |
371326
|
{
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"authors": [
"Peter Humphries",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371326"
}
|
Stack Exchange
|
Generalization of the The Liouville Lambda function
Let $n=p^{\alpha_1}_1 \cdots p^{\alpha_m}_m,$ and define
$$\lambda_k(n)= (-1)^{ [\frac{\Omega(n)}{k} ]},$$
where $\Omega(n)= \alpha_1 + \cdots + \alpha_k,$ and $[\cdot]$ is the floor function.
For $k=1$, $\lambda_1$ is the Liouvilles Lambda function. For $k=2$:
$\lambda_2(1)=1, \hspace{2 mm} \lambda_2(p_1)=1, \hspace{2 mm} \lambda_2(p_1p_2)=-1, \hspace{2 mm} \lambda_2(p_1p_2p_3)=-1 \hspace{2 mm} \text{ and so on...}.$
-Is there anythig known about this function?
-If for $\Re(s)>1$ we define $$L(s, \lambda_2):= \sum_{n=1}^{\infty} \frac{\lambda_2(n)}{n^s},$$
Is there any connection between $L_{\lambda_2}(s)$ and the zeros of the Riemann zeta function?
-I guess (without knowing how to prove) that $$\sum_{n<x} \lambda_2(n) = o(x),$$
smilar to $\lambda_1$ should we expect square root cancelation on sum of $\lambda_2$ as well?
Let's just consider the case $k=2$; you can try to generalize this argument for larger $k$. For $k=2$,
$$
\sum_{n\le x} \lambda_2(n) = \sum_{\substack{ n\le x \\ \Omega(n) = 0,1 \mod 4}} 1 - \sum_{\substack{ n\le x \\ \Omega(n) = 2,3 \mod 4}} 1.
$$
This can be expressed as
$$
\text{Re} \sum_{n\le x} i^{\Omega(n)} + \text{Im} \sum_{n\le x}i^{\Omega(n)},
$$
and now one can use Selberg--Delange. This will show an asymptotic expansion for $\sum_{n\le x} i^{\Omega(n)}$, and the leading term in the asymptotic is
$$
C x (\log x)^{i-1},
$$
for a suitable constant $C$. Thus you can find an asymptotic for your sum, which will be of the form
$$
A \cos(\log \log x) \frac{x}{\log x} + B \sin(\log \log x) \frac{x}{\log x},
$$
for suitable constants $A$ and $B$. Weird! But then you're also looking at a weird object!
Indeed, it's closely related to what Tian An Wong, Snehal Shekatkar, and I studied in this paper (again using Selberg-Delange): https://doi.org/10.5802/jtnb.1066
|
2025-03-21T14:48:31.998581
| 2020-09-10T12:02:57 |
371330
|
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"authors": [
"Dieter Kadelka",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/153595",
"香结丁"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371330"
}
|
Stack Exchange
|
How to use the mixed normal distribution to construct a proper statistics?
For a random vector $\xi_n \in \mathbb{R}^p$, if $\xi_n \rightarrow_d N(\mu, \Sigma)$, we can construct
\begin{equation*}
\Psi := \xi_n^{\top} \widehat{\Sigma}^{-1} \xi_n
\end{equation*}
for testing whether $\mu = 0$, where $\widehat{\Sigma}$ is an reasonable estimate of $\Sigma$.
Now assume $\xi_n$ converge to a mixed normal distribution, i.e.
\begin{equation*}
\xi_n \rightarrow_d \sum_{k = 1}^K w_k Z_k
\end{equation*}
where $w_1 +\cdots + w_K = 1$ and $Z_1, \cdots, Z_K$ are independently distributed as $N(0, \Sigma_k)$. Since the elements in one $\Sigma_k$ are quite different from those in the other, we cannot write this mixed normal distribution as a single normal distribution. In this case, I try to construct
$$\tilde{\Psi} = \sum_{k = 1}^K w_k \xi_n^{\top} \widehat{\Sigma}_k^{-1} \xi_n$$
as the statistic. However, it seems to be difficult to analyze the properties of this statistic since the components on the right side are NOT independent from each other.
Do anyone know how to construct a proper statistic in this case. Thanks so much!
I don't understand. $\sum_{k=1}^K w_kZ_k$ is again $N(0,\Sigma)$-distributed with some appropriate $\Sigma$, if $w_i$ are constant. So what is your assumption about $w_i$?
@DieterKadelka Thank you! $w_k$ indeed are constant. Do you mean the weighted sum is again normal distribution, even the elements in each $\Sigma_k$ are quite different from those in another one? If so, is there a possible way to calculate the elements in $\Sigma$ exactly?
The distribution is $N(0,\Sigma)$ with $\Sigma = \sum_{k=1}^K w_k^2 \Sigma_k$. This holds if the $Z_k$ are independent.
@DieterKadelka I see, I deemed only one-dimensional normal distributions have this property before. Thanks so much!
|
2025-03-21T14:48:31.998835
| 2020-09-10T12:10:00 |
371331
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371331"
}
|
Stack Exchange
|
Ideas on how to prove Pythagorean identity involving Wasserstein distances?
I conjectured earlier that if $P$ and $Q$ were two probability measures, then we could show
$$W^2(P,Q) = \min_{T} [d^2(P,T_{\#}P) + W^2(T_{\#}P,Q)]$$ where $W^2(P,Q)$ denotes the squared Wasserstein-2 distance between $P$ and $Q$. Furthermore, $d^2(P,T_{\#}P) = E_{x\sim P} [ \left\| Tx - x \right\|^2 ]$.
I originally thought I could use some results involving the Pythagorean theorem in an information space with Bregman divergences as seen here for KL-divergence and here, but the Wasserstein distance isn't a type of Bregman divergence.
Any thoughts on how I could try to prove the above?
$\newcommand\R{\mathbb R}\newcommand\B{\mathcal B}\newcommand\Si{\Sigma}\newcommand\ga{\gamma}$
Your conjecture is false in general.
E.g., suppose that the underlying measurable space
$(S,\Si)$ on which $P$ and $Q$ are defined is $(\R,\B(\R))$, where $\B(\R)$ is the Borel $\sigma$-algebra over $\R$. Let $P$ be the uniform distribution on $(0,1)$. Then any distribution $R$ on $(\R,\B(\R))$ is of the form $T_{\#}P$ for some $T$; namely, $T$ is the quantile transformation given by the formula
$$T(u)=H^{-1}(u):=\inf\{x\in\R\colon H(x)\ge u\}$$
for $u\in(0,1)$, where $H$ is the cdf of the distribution $R$. Let now $Q$ be any distribution on $(\R,\B(\R))$ other than $P$. There are real-valued random variables (r.v.'s) $X$ and $Y$ with respective distributions $P$ and $Q$ such that $W^2(P,Q)=E(X-Y)^2$; for instance, take any r.v. $X\sim P$ and let then $Y:=G^{-1}(X)$, where $G$ is the cdf of $Q$. Let now $Z:=(X+Y)/2$ and let $R$ be the distribution of $Z$. Then the right-hand side of your conjectured identity is
$$\le d^2(P,R) + W^2(R,Q)\le E(X-Z)^2+E(Z-Y)^2=\frac12\,E(X-Y)^2=\frac12\,W^2(P,Q)<W^2(P,Q);$$
the latter inequality holding because $Q\ne P$. So, the right-hand side of your conjectured identity is strictly less than its left-hand side.
A correct version of your conjectured identity is a direct application of the triangle inequality in $L^2$, as follows:
$$W(P,Q)=\inf_T[d(P,T_{\#}P)+W(T_{\#}P,Q)].\tag{1}$$
Indeed,
$$d(P,T_{\#}P)+W(T_{\#}P,Q)=\inf_\ga\Big[\sqrt{\int_{S\times S}(x-T(x))^2\ga(dx,dy)}+\sqrt{\int_{S\times S}(T(x)-y)^2\ga(dx,dy)}\Big]
\ge\inf_\ga\sqrt{\int_{S\times S}(x-y)^2\ga(dx,dy)}=W(P,Q),$$
where $\inf\limits_\ga$ is taken over all couplings $\ga$ of $P$ and $Q$. So, the right-hand side of (1) is no less than its left-hand side. On the other hand, taking $T$ to be the identity map of $S$, we see that the right-hand side of (1) is no greater than its left-hand side. Thus, (1) is proved.
|
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